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Higher Engineering Mathematics
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In memory of Elizabeth
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Newnes is an imprint of Elsevier, The Boulevard, Langford Lane, Kidlington, Oxford OX5 1GB, UK, 30 Corporate Drive, Suite 400, Burlington, MA 01803, USA, First edition 2010, Copyright © 2010, John Bird, Published by Elsevier Ltd. All rights reserved., The right of John Bird to be identified as the author of this work has been asserted in accordance with, the Copyright, Designs and Patents Act 1988., No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form, or by any means electronic, mechanical, photocopying, recording or otherwise without the prior written, permission of the publisher., Permissions may be sought directly from Elsevier’s Science & Technology Rights Department in Oxford,, UK: phone (+44) (0) 1865 843830; fax (+44) (0) 1865 853333; email: permissions@elsevier.com., Alternatively you can submit your request online by visiting the Elsevier web site at, http://elsevier.com/locate/permissions, and selecting Obtaining permission to use Elsevier material., Notice, No responsibility is assumed by the publisher for any injury and/or damage to persons or property as a matter, of products liability, negligence or otherwise, or from any use or operation of any methods, products,, instructions or ideas contained in the material herein. Because of rapid advances in the medical sciences,, in particular, independent verification of diagnoses and drug dosages should be made., British Library Cataloguing-in-Publication Data, A catalogue record for this book is available from the British Library., Library of Congress Cataloging-in-Publication Data, A catalogue record for this book is available from the Library of Congress., ISBN: 978-1-85-617767-2, , For information on all Newnes publications, visit our Web site at www.elsevierdirect.com, , Typeset by: diacriTech, India, Printed and bound in China, 10 11 12 13 14 15 10 9 8 7 6 5 4 3 2 1
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Contents, Preface, , xiii, , Syllabus guidance, , xv, , 1, , 1, 1, 1, 3, 6, 8, 10, , 2, , 3, , 4, , Algebra, 1.1 Introduction, 1.2 Revision of basic laws, 1.3 Revision of equations, 1.4 Polynomial division, 1.5 The factor theorem, 1.6 The remainder theorem, Partial fractions, 2.1 Introduction to partial fractions, 2.2 Worked problems on partial fractions with, linear factors, 2.3 Worked problems on partial fractions with, repeated linear factors, 2.4 Worked problems on partial fractions with, quadratic factors, , 17, , Logarithms, 3.1 Introduction to logarithms, 3.2 Laws of logarithms, 3.3 Indicial equations, 3.4 Graphs of logarithmic functions, , 20, 20, 22, 24, 25, , Exponential functions, 4.1 Introduction to exponential functions, 4.2 The power series for e x, 4.3 Graphs of exponential functions, 4.4 Napierian logarithms, 4.5 Laws of growth and decay, 4.6 Reduction of exponential laws to, linear form, , 27, 27, 28, 29, 31, 34, , Revision Test 1, , 5, , Hyperbolic functions, 5.1 Introduction to hyperbolic functions, 5.2 Graphs of hyperbolic functions, 5.3 Hyperbolic identities, 5.4 Solving equations involving hyperbolic, functions, 5.5 Series expansions for cosh x and sinh x, , 13, 13, , 6, , 7, , 13, 16, , 37, 40, , 41, 41, 43, 45, 47, 49, , Arithmetic and geometric progressions, 6.1 Arithmetic progressions, 6.2 Worked problems on arithmetic, progressions, 6.3 Further worked problems on arithmetic, progressions, 6.4 Geometric progressions, 6.5 Worked problems on geometric, progressions, 6.6 Further worked problems on geometric, progressions, The binomial series, 7.1 Pascal’s triangle, 7.2 The binomial series, 7.3 Worked problems on the binomial series, 7.4 Further worked problems on the binomial, series, 7.5 Practical problems involving the binomial, theorem, Revision Test 2, , 8, , Maclaurin’s series, 8.1 Introduction, 8.2 Derivation of Maclaurin’s theorem, 8.3 Conditions of Maclaurin’s series, 8.4 Worked problems on Maclaurin’s series, 8.5 Numerical integration using Maclaurin’s, series, 8.6 Limiting values, 9 Solving equations by iterative methods, 9.1 Introduction to iterative methods, 9.2 The bisection method, 9.3 An algebraic method of successive, approximations, 9.4 The Newton-Raphson method, , 10 Binary, octal and hexadecimal, 10.1 Introduction, 10.2 Binary numbers, 10.3 Octal numbers, 10.4 Hexadecimal numbers, Revision Test 3, , 51, 51, 51, 52, 54, 55, 56, 58, 58, 59, 59, 62, 64, 67, 68, 68, 68, 69, 69, 73, 74, 77, 77, 77, 81, 84, 87, 87, 87, 90, 92, 96
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vi Contents, 11 Introduction to trigonometry, 11.1 Trigonometry, 11.2 The theorem of Pythagoras, 11.3 Trigonometric ratios of acute angles, 11.4 Evaluating trigonometric ratios, 11.5 Solution of right-angled triangles, 11.6 Angles of elevation and depression, 11.7 Sine and cosine rules, 11.8 Area of any triangle, 11.9 Worked problems on the solution of, triangles and finding their areas, 11.10 Further worked problems on solving, triangles and finding their areas, 11.11 Practical situations involving, trigonometry, 11.12 Further practical situations involving, trigonometry, , 97, 97, 97, 98, 100, 105, 106, 108, 108, , 12 Cartesian and polar co-ordinates, 12.1 Introduction, 12.2 Changing from Cartesian into polar, co-ordinates, 12.3 Changing from polar into Cartesian, co-ordinates, 12.4 Use of Pol/Rec functions on calculators, , 117, 117, , 119, 120, , 13 The circle and its properties, 13.1 Introduction, 13.2 Properties of circles, 13.3 Radians and degrees, 13.4 Arc length and area of circles and sectors, 13.5 The equation of a circle, 13.6 Linear and angular velocity, 13.7 Centripetal force, , 122, 122, 122, 123, 124, 127, 129, 130, , Revision Test 4, , 109, , 15.5 Worked problems (ii) on trigonometric, equations, 15.6 Worked problems (iii) on trigonometric, equations, 15.7 Worked problems (iv) on trigonometric, equations, 16 The relationship between trigonometric and, hyperbolic functions, 16.1 The relationship between trigonometric, and hyperbolic functions, 16.2 Hyperbolic identities, , 156, 157, 157, , 159, 159, 160, , 110, 111, 113, , 117, , 133, , 14 Trigonometric waveforms, 14.1 Graphs of trigonometric functions, 14.2 Angles of any magnitude, 14.3 The production of a sine and cosine wave, 14.4 Sine and cosine curves, 14.5 Sinusoidal form A sin(ωt ± α), 14.6 Harmonic synthesis with complex, waveforms, , 134, 134, 135, 137, 138, 143, , 15 Trigonometric identities and equations, 15.1 Trigonometric identities, 15.2 Worked problems on trigonometric, identities, 15.3 Trigonometric equations, 15.4 Worked problems (i) on trigonometric, equations, , 152, 152, , 146, , 152, 154, 154, , 17 Compound angles, 17.1 Compound angle formulae, 17.2 Conversion of a sinωt + b cosωt into, R sin(ωt + α), 17.3 Double angles, 17.4 Changing products of sines and cosines, into sums or differences, 17.5 Changing sums or differences of sines and, cosines into products, 17.6 Power waveforms in a.c. circuits, Revision Test 5, , 163, 163, 165, 169, 170, 171, 173, 177, , 18 Functions and their curves, 18.1 Standard curves, 18.2 Simple transformations, 18.3 Periodic functions, 18.4 Continuous and discontinuous functions, 18.5 Even and odd functions, 18.6 Inverse functions, 18.7 Asymptotes, 18.8 Brief guide to curve sketching, 18.9 Worked problems on curve sketching, , 178, 178, 181, 186, 186, 186, 188, 190, 196, 197, , 19 Irregular areas, volumes and mean values of, waveforms, 19.1 Areas of irregular figures, 19.2 Volumes of irregular solids, 19.3 The mean or average value of a waveform, , 203, 203, 205, 206, , Revision Test 6, , 20 Complex numbers, 20.1 Cartesian complex numbers, 20.2 The Argand diagram, 20.3 Addition and subtraction of complex, numbers, 20.4 Multiplication and division of complex, numbers, , 212, , 213, 213, 214, 214, 216
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vii, , Contents, 20.5, 20.6, 20.7, 20.8, , Complex equations, The polar form of a complex number, Multiplication and division in polar form, Applications of complex numbers, , 217, 218, 220, 221, , 21 De Moivre’s theorem, 21.1 Introduction, 21.2 Powers of complex numbers, 21.3 Roots of complex numbers, 21.4 The exponential form of a complex, number, , 225, 225, 225, 226, , 22 The theory of matrices and determinants, 22.1 Matrix notation, 22.2 Addition, subtraction and multiplication, of matrices, 22.3 The unit matrix, 22.4 The determinant of a 2 by 2 matrix, 22.5 The inverse or reciprocal of a 2 by 2 matrix, 22.6 The determinant of a 3 by 3 matrix, 22.7 The inverse or reciprocal of a 3 by 3 matrix, , 231, 231, , 23 The solution of simultaneous equations by, matrices and determinants, 23.1 Solution of simultaneous equations by, matrices, 23.2 Solution of simultaneous equations by, determinants, 23.3 Solution of simultaneous equations using, Cramers rule, 23.4 Solution of simultaneous equations using, the Gaussian elimination method, Revision Test 7, , 24 Vectors, 24.1, 24.2, 24.3, 24.4, 24.5, 24.6, 24.7, 24.8, 24.9, , Introduction, Scalars and vectors, Drawing a vector, Addition of vectors by drawing, Resolving vectors into horizontal and, vertical components, Addition of vectors by calculation, Vector subtraction, Relative velocity, i, j and k notation, , 25 Methods of adding alternating waveforms, 25.1 Combination of two periodic functions, 25.2 Plotting periodic functions, 25.3 Determining resultant phasors by drawing, , 228, , 231, 235, 235, 236, 237, 239, , 25.4 Determining resultant phasors by the sine, and cosine rules, 268, 25.5 Determining resultant phasors by, horizontal and vertical components, 270, 25.6 Determining resultant phasors by complex, numbers, 272, 26 Scalar and vector products, 26.1 The unit triad, 26.2 The scalar product of two vectors, 26.3 Vector products, 26.4 Vector equation of a line, Revision Test 8, , 275, 275, 276, 280, 283, 286, , 27 Methods of differentiation, 27.1 Introduction to calculus, 27.2 The gradient of a curve, 27.3 Differentiation from first principles, 27.4 Differentiation of common functions, 27.5 Differentiation of a product, 27.6 Differentiation of a quotient, 27.7 Function of a function, 27.8 Successive differentiation, , 287, 287, 287, 288, 289, 292, 293, 295, 296, , 28 Some applications of differentiation, 28.1 Rates of change, 28.2 Velocity and acceleration, 28.3 Turning points, 28.4 Practical problems involving maximum, and minimum values, 28.5 Tangents and normals, 28.6 Small changes, , 299, 299, 300, 303, , 29 Differentiation of parametric equations, 29.1 Introduction to parametric equations, 29.2 Some common parametric equations, 29.3 Differentiation in parameters, 29.4 Further worked problems on, differentiation of parametric equations, , 315, 315, 315, 315, , 254, 255, 260, 262, 263, , 30 Differentiation of implicit functions, 30.1 Implicit functions, 30.2 Differentiating implicit functions, 30.3 Differentiating implicit functions, containing products and quotients, 30.4 Further implicit differentiation, , 320, 320, 320, 321, 322, , 265, 265, 265, 267, , 31 Logarithmic differentiation, 31.1 Introduction to logarithmic differentiation, 31.2 Laws of logarithms, 31.3 Differentiation of logarithmic functions, , 325, 325, 325, 325, , 241, 241, 243, 247, 248, , 307, 311, 312, , 250, , 251, 251, 251, 251, 252, , 318
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viii Contents, 31.4 Differentiation of further logarithmic, functions, 31.5 Differentiation of [ f (x)]x, Revision Test 9, 32 Differentiation of hyperbolic functions, 32.1 Standard differential coefficients of, hyperbolic functions, 32.2 Further worked problems on, differentiation of hyperbolic functions, 33 Differentiation of inverse trigonometric and, hyperbolic functions, 33.1 Inverse functions, 33.2 Differentiation of inverse trigonometric, functions, 33.3 Logarithmic forms of the inverse, hyperbolic functions, 33.4 Differentiation of inverse hyperbolic, functions, 34 Partial differentiation, 34.1 Introduction to partial derivatives, 34.2 First order partial derivatives, 34.3 Second order partial derivatives, 35 Total differential, rates of change and small, changes, 35.1 Total differential, 35.2 Rates of change, 35.3 Small changes, 36 Maxima, minima and saddle points for functions, of two variables, 36.1 Functions of two independent variables, 36.2 Maxima, minima and saddle points, 36.3 Procedure to determine maxima, minima, and saddle points for functions of two, variables, 36.4 Worked problems on maxima, minima, and saddle points for functions of two, variables, 36.5 Further worked problems on maxima,, minima and saddle points for functions of, two variables, Revision Test 10, 37 Standard integration, 37.1 The process of integration, 37.2 The general solution of integrals of the, form ax n, 37.3 Standard integrals, 37.4 Definite integrals, , 326, 328, 330, 331, 331, 332, 334, 334, 334, , 38 Some applications of integration, 38.1 Introduction, 38.2 Areas under and between curves, 38.3 Mean and r.m.s. values, 38.4 Volumes of solids of revolution, 38.5 Centroids, 38.6 Theorem of Pappus, 38.7 Second moments of area of regular, sections, , 375, 375, 375, 377, 378, 380, 381, , 39 Integration using algebraic substitutions, 39.1 Introduction, 39.2 Algebraic substitutions, 39.3 Worked problems on integration using, algebraic substitutions, 39.4 Further worked problems on integration, using algebraic substitutions, 39.5 Change of limits, , 392, 392, 392, , 383, , 392, 394, 395, , 339, 341, 345, 345, 345, 348, 351, 351, 352, 354, 357, 357, 358, , 359, , 359, , 361, 367, 368, 368, 368, 369, 372, , Revision Test 11, 40 Integration using trigonometric and hyperbolic, substitutions, 40.1 Introduction, 40.2 Worked problems on integration of sin2 x,, cos2 x, tan2 x and cot2 x, 40.3 Worked problems on powers of sines and, cosines, 40.4 Worked problems on integration of, products of sines and cosines, 40.5 Worked problems on integration using the, sin θ substitution, 40.6 Worked problems on integration using, tan θ substitution, 40.7 Worked problems on integration using the, sinh θ substitution, 40.8 Worked problems on integration using the, cosh θ substitution, , 397, , 398, 398, 398, 400, 401, 402, 404, 404, 406, , 41 Integration using partial fractions, 41.1 Introduction, 41.2 Worked problems on integration using, partial fractions with linear factors, 41.3 Worked problems on integration using, partial fractions with repeated linear, factors, 41.4 Worked problems on integration using, partial fractions with quadratic factors, , 412, , θ, 42 The t = tan substitution, 2, 42.1 Introduction, , 414, 414, , θ, 42.2 Worked problems on the t = tan, 2, substitution, , 409, 409, 409, , 411, , 415
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Contents, θ, 42.3 Further worked problems on the t = tan, 2, substitution, Revision Test 12, , 416, 419, , 43 Integration by parts, 43.1 Introduction, 43.2 Worked problems on integration by parts, 43.3 Further worked problems on integration, by parts, , 420, 420, 420, , 44 Reduction formulae, 44.1 Introduction, 44.2 Using reduction, formulae for integrals of, , the form x n e x dx, 44.3 Using reduction, formulae for integrals of, , the form x n cos x dx and x n sin x dx, 44.4 Using reduction, formulae for integrals of, , the form sinn x dx and cosn x dx, 44.5 Further reduction formulae, , 426, 426, , 429, 432, , 45 Numerical integration, 45.1 Introduction, 45.2 The trapezoidal rule, 45.3 The mid-ordinate rule, 45.4 Simpson’s rule, , 435, 435, 435, 437, 439, , Revision Test 13, , 46 Solution of first order differential equations by, separation of variables, 46.1 Family of curves, 46.2 Differential equations, 46.3 The solution of equations of the form, dy, = f (x), dx, 46.4 The solution of equations of the form, dy, = f (y), dx, 46.5 The solution of equations of the form, dy, = f (x) · f (y), dx, 47 Homogeneous first order differential equations, 47.1 Introduction, 47.2 Procedure to solve differential equations, dy, =Q, of the form P, dx, 47.3 Worked problems on homogeneous first, order differential equations, 47.4 Further worked problems on homogeneous, first order differential equations, , 422, , 426, 427, , 443, , 444, 444, 445, 445, 447, 449, 452, 452, 452, 452, 454, , 48 Linear first order differential equations, 48.1 Introduction, 48.2 Procedure to solve differential equations, dy, + Py = Q, of the form, dx, 48.3 Worked problems on linear first order, differential equations, 48.4 Further worked problems on linear first, order differential equations, 49 Numerical methods for first order differential, equations, 49.1 Introduction, 49.2 Euler’s method, 49.3 Worked problems on Euler’s method, 49.4 An improved Euler method, 49.5 The Runge-Kutta method, Revision Test 14, 50 Second order differential equations of the form, dy, d2 y, a 2 + b + cy= 0, dx, dx, 50.1 Introduction, 50.2 Procedure to solve differential equations, dy, d2 y, of the form a 2 + b + cy = 0, dx, dx, 50.3 Worked problems on differential equations, dy, d2 y, of the form a 2 + b + cy = 0, dx, dx, 50.4 Further worked problems on practical, differential equations of the form, dy, d2 y, a 2 + b + cy = 0, dx, dx, 51 Second order differential equations of the form, dy, d2 y, a 2 + b + cy= f (x), dx, dx, 51.1 Complementary function and particular, integral, 51.2 Procedure to solve differential equations, d2 y, dy, of the form a 2 + b + cy = f (x), dx, dx, 51.3 Worked problems on differential equations, dy, d2 y, of the form a 2 + b + cy = f (x), dx, dx, where f (x) is a constant or polynomial, 51.4 Worked problems on differential equations, dy, d2 y, of the form a 2 + b + cy = f (x), dx, dx, where f (x) is an exponential function, 51.5 Worked problems on differential equations, dy, d2 y, of the form a 2 + b + cy = f (x), dx, dx, where f (x) is a sine or cosine function, , 456, 456, 457, 457, 458, , 461, 461, 461, 462, 466, 471, 476, , 477, 477, 478, , 478, , 480, , 483, , 483, 483, , 484, , 486, , 488, , ix
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x Contents, 51.6 Worked problems on differential equations, dy, d2 y, of the form a 2 + b + cy = f (x), dx, dx, where f (x) is a sum or a product, 490, 52 Power series methods of solving ordinary, differential equations, 52.1 Introduction, 52.2 Higher order differential coefficients as, series, 52.3 Leibniz’s theorem, 52.4 Power series solution by the, Leibniz–Maclaurin method, 52.5 Power series solution by the Frobenius, method, 52.6 Bessel’s equation and Bessel’s functions, 52.7 Legendre’s equation and Legendre, polynomials, 53 An introduction to partial differential equations, 53.1 Introduction, 53.2 Partial integration, 53.3 Solution of partial differential equations, by direct partial integration, 53.4 Some important engineering partial, differential equations, 53.5 Separating the variables, 53.6 The wave equation, 53.7 The heat conduction equation, 53.8 Laplace’s equation, Revision Test 15, , 493, 493, 493, 495, 497, 500, 506, 511, 515, 515, 515, , 556, 556, 559, , 58 The normal distribution, 58.1 Introduction to the normal distribution, 58.2 Testing for a normal distribution, , 562, 562, 566, , 59 Linear correlation, 59.1 Introduction to linear correlation, 59.2 The product-moment formula for, determining the linear correlation, coefficient, 59.3 The significance of a coefficient of, correlation, 59.4 Worked problems on linear correlation, , 570, 570, , 60 Linear regression, 60.1 Introduction to linear regression, 60.2 The least-squares regression lines, 60.3 Worked problems on linear regression, , 575, 575, 575, 576, , Revision Test 17, , 570, 571, 571, , 581, , 516, 518, 518, 519, 523, 525, 528, , 54 Presentation of statistical data, 54.1 Some statistical terminology, 54.2 Presentation of ungrouped data, 54.3 Presentation of grouped data, , 529, 529, 530, 534, , 55 Measures of central tendency and dispersion, 55.1 Measures of central tendency, 55.2 Mean, median and mode for discrete data, 55.3 Mean, median and mode for grouped data, 55.4 Standard deviation, 55.5 Quartiles, deciles and percentiles, , 541, 541, 541, 542, 544, 546, , 56 Probability, 56.1 Introduction to probability, 56.2 Laws of probability, 56.3 Worked problems on probability, 56.4 Further worked problems on probability, , 548, 548, 549, 549, 551, , Revision Test 16, , 57 The binomial and Poisson distributions, 57.1 The binomial distribution, 57.2 The Poisson distribution, , 554, , 61 Introduction to Laplace transforms, 61.1 Introduction, 61.2 Definition of a Laplace transform, 61.3 Linearity property of the Laplace, transform, 61.4 Laplace transforms of elementary, functions, 61.5 Worked problems on standard Laplace, transforms, , 582, 582, 582, , 62 Properties of Laplace transforms, 62.1 The Laplace transform of eat f (t), 62.2 Laplace transforms of the form eat f (t), 62.3 The Laplace transforms of derivatives, 62.4 The initial and final value theorems, , 587, 587, 587, 589, 591, , 63 Inverse Laplace transforms, 63.1 Definition of the inverse Laplace transform, 63.2 Inverse Laplace transforms of simple, functions, 63.3 Inverse Laplace transforms using partial, fractions, 63.4 Poles and zeros, , 593, 593, , 64 The solution of differential equations using, Laplace transforms, 64.1 Introduction, 64.2 Procedure to solve differential equations, by using Laplace transforms, 64.3 Worked problems on solving differential, equations using Laplace transforms, , 582, 582, 583, , 593, 596, 598, 600, 600, 600, 600
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Contents, 65 The solution of simultaneous differential, equations using Laplace transforms, 65.1 Introduction, 65.2 Procedure to solve simultaneous, differential equations using Laplace, transforms, 65.3 Worked problems on solving simultaneous, differential equations by using Laplace, transforms, Revision Test 18, 66 Fourier series for periodic functions of, period 2π, 66.1 Introduction, 66.2 Periodic functions, 66.3 Fourier series, 66.4 Worked problems on Fourier series of, periodic functions of period 2π, 67 Fourier series for a non-periodic function over, range 2π, 67.1 Expansion of non-periodic functions, 67.2 Worked problems on Fourier series of, non-periodic functions over a range of 2π, 68 Even and odd functions and half-range, Fourier series, 68.1 Even and odd functions, , 68.2 Fourier cosine and Fourier sine series, 68.3 Half-range Fourier series, , 605, 605, , 605, , 605, 610, , 611, 611, 611, 611, 612, 617, 617, 617, 623, 623, , 623, 626, , 69 Fourier series over any range, 69.1 Expansion of a periodic function of, period L, 69.2 Half-range Fourier series for functions, defined over range L, , 630, , 70 A numerical method of harmonic analysis, 70.1 Introduction, 70.2 Harmonic analysis on data given in tabular, or graphical form, 70.3 Complex waveform considerations, , 637, 637, 637, 641, , 71 The complex or exponential form of a, Fourier series, 71.1 Introduction, 71.2 Exponential or complex notation, 71.3 The complex coefficients, 71.4 Symmetry relationships, 71.5 The frequency spectrum, 71.6 Phasors, , 644, 644, 644, 645, 649, 652, 653, , Revision Test 19, , 630, 634, , 658, , Essential formulae, , 659, , Index, , 675, , xi
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xii Contents, , Website Chapters, , 72 Inequalities, 72.1 Introduction to inequalities, 72.2 Simple inequalities, 72.3 Inequalities involving a modulus, 72.4 Inequalities involving quotients, 72.5 Inequalities involving square functions, 72.6 Quadratic inequalities, , 1, 1, 1, 2, 3, 4, 5, , 73 Boolean algebra and logic circuits, 73.1 Boolean algebra and switching circuits, 73.2 Simplifying Boolean expressions, 73.3 Laws and rules of Boolean algebra, 73.4 De Morgan’s laws, 73.5 Karnaugh maps, 73.6 Logic circuits, 73.7 Universal logic gates, , 7, 7, 12, 12, 14, 15, 19, 23, , Revision Test 20, 74 Sampling and estimation theories, 74.1 Introduction, 74.2 Sampling distributions, , 28, 29, 29, 29, , 74.3 The sampling distribution of the means, 74.4 The estimation of population parameters, based on a large sample size, 74.5 Estimating the mean of a population based, on a small sample size, , 29, 33, 38, , 75 Significance testing, 75.1 Hypotheses, 75.2 Type I and Type II errors, 75.3 Significance tests for population means, 75.4 Comparing two sample means, , 42, 42, 42, 49, 54, , 76 Chi-square and distribution-free tests, 76.1 Chi-square values, 76.2 Fitting data to theoretical distributions, 76.3 Introduction to distribution-free tests, 76.4 The sign test, 76.5 Wilcoxon signed-rank test, 76.6 The Mann-Whitney test, , 59, 59, 60, 67, 68, 71, 75, , Revision Test 21, , 82
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Preface, This sixth edition of ‘Higher Engineering Mathematics’ covers essential mathematical material suitable, for students studying Degrees, Foundation Degrees,, Higher National Certificate and Diploma courses in, Engineering disciplines., In this edition the material has been ordered into the, following twelve convenient categories: number and, algebra, geometry and trigonometry, graphs, complex, numbers, matrices and determinants, vector geometry,, differential calculus, integral calculus, differential equations, statistics and probability, Laplace transforms and, Fourier series. New material has been added on logarithms and exponential functions, binary, octal and, hexadecimal, vectors and methods of adding alternating waveforms. Another feature is that a free Internet, download is available of a sample (over 1100) of the, further problems contained in the book., The primary aim of the material in this text is to, provide the fundamental analytical and underpinning, knowledge and techniques needed to successfully complete scientific and engineering principles modules of, Degree, Foundation Degree and Higher National Engineering programmes. The material has been designed, to enable students to use techniques learned for the, analysis, modelling and solution of realistic engineering, problems at Degree and Higher National level. It also, aims to provide some of the more advanced knowledge, required for those wishing to pursue careers in mechanical engineering, aeronautical engineering, electronics,, communications engineering, systems engineering and, all variants of control engineering., In Higher Engineering Mathematics 6th Edition, theory is introduced in each chapter by a full outline of, essential definitions, formulae, laws, procedures etc., The theory is kept to a minimum, for problem solving is, extensively used to establish and exemplify the theory., It is intended that readers will gain real understanding through seeing problems solved and then through, solving similar problems themselves., Access to software packages such as Maple, Mathematica and Derive, or a graphics calculator, will enhance, understanding of some of the topics in this text., , Each topic considered in the text is presented in a way, that assumes in the reader only knowledge attained in, BTEC National Certificate/Diploma, or similar, in an, Engineering discipline., ‘Higher Engineering Mathematics 6th Edition’ provides a follow-up to ‘Engineering Mathematics 6th, Edition’., This textbook contains some 900 worked problems, followed by over 1760 further problems (with, answers), arranged within 238 Exercises. Some 432, line diagrams further enhance understanding., A sample of worked solutions to over 1100 of the further problems has been prepared and can be accessed, free via the Internet (see next page)., At the end of the text, a list of Essential Formulae is, included for convenience of reference., At intervals throughout the text are some 19 Revision, Tests (plus two more in the website chapters) to check, understanding. For example, Revision Test 1 covers, the material in Chapters 1 to 4, Revision Test 2 covers the material in Chapters 5 to 7, Revision Test 3, covers the material in Chapters 8 to 10, and so on. An, Instructor’s Manual, containing full solutions to the, Revision Tests, is available free to lecturers adopting, this text (see next page)., Due to restriction of extent, five chapters that appeared, in the fifth edition have been removed from the text, and placed on the website. For chapters on Inequalities, Boolean algebra and logic circuits, Sampling and, estimation theories, Significance testing and Chi-square, and distribution-free tests (see next page)., ‘Learning by example’ is at the heart of ‘Higher, Engineering Mathematics 6th Edition’., , JOHN BIRD, Royal Naval School of Marine Engineering,, HMS Sultan,, formerly University of Portsmouth, and Highbury College, Portsmouth
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xiv Preface, Free web downloads, Extra material available on the Internet at:, www.booksite.elsevier.com/newnes/bird., It is recognised that the level of understanding, of algebra on entry to higher courses is often, inadequate. Since algebra provides the basis of so, much of higher engineering studies, it is a situation, that often needs urgent attention. Lack of space, has prevented the inclusion of more basic algebra, topics in this textbook; it is for this reason that, some algebra topics – solution of simple, simultaneous and quadratic equations and transposition, of formulae – have been made available to all via, the Internet. Also included is a Remedial Algebra, Revision Test to test understanding. To access the, Algebra material visit the website., Five extra chapters, Chapters on Inequalities, Boolean Algebra and, logic circuits, Sampling and Estimation theories, Significance testing, and Chi-square and, distribution-free tests are available to download at, the website., , Sample of worked Solutions to Exercises, Within the text (plus the website chapters) are, some 1900 further problems arranged within, 260 Exercises. A sample of over 1100 worked, solutions has been prepared and can be accessed, free via the Internet. To access these worked, solutions visit the website., Instructor’s manual, This provides fully worked solutions and mark, scheme for all the Revision Tests in this book, (plus 2 from the website chapters), together with, solutions to the Remedial Algebra Revision Test, mentioned above. The material is available to lecturers only. To obtain a password please visit the, website with the following details: course title,, number of students, your job title and work postal, address., To download the Instructor’s Manual visit the, website and enter the book title in the search box.
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Syllabus Guidance, This textbook is written for undergraduate engineering degree and foundation degree courses;, however, it is also most appropriate for HNC/D studies and three syllabuses are covered., , The appropriate chapters for these three syllabuses are shown in the table below., Chapter, , Analytical, Methods, for, Engineers, , Further, Analytical, Methods for, Engineers, , 1., , Algebra, , ×, , 2., , Partial fractions, , ×, , 3., , Logarithms, , ×, , 4., , Exponential functions, , ×, , 5., , Hyperbolic functions, , ×, , 6., , Arithmetic and geometric progressions, , ×, , 7., , The binomial series, , ×, , 8., , Maclaurin’s series, , ×, , 9., , Solving equations by iterative methods, , ×, , 10., , Binary, octal and hexadecimal, , ×, , 11., , Introduction to trigonometry, , ×, , 12., , Cartesian and polar co-ordinates, , ×, , 13., , The circle and its properties, , ×, , 14., , Trigonometric waveforms, , ×, , 15., , Trigonometric identities and equations, , ×, , 16., , The relationship between trigonometric and hyperbolic, functions, , ×, , 17., , Compound angles, , ×, , 18., , Functions and their curves, , ×, , 19., , Irregular areas, volumes and mean values of waveforms, , ×, , 20., , Complex numbers, , ×, , 21., , De Moivre’s theorem, , ×, , 22., , The theory of matrices and determinants, , ×, , 23., , The solution of simultaneous equations by matrices and, determinants, , ×, , 24., , Vectors, , ×, , 25., , Methods of adding alternating waveforms, , ×, , Engineering, Mathematics, , (Continued )
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xvi Syllabus Guidance, Chapter, , Analytical, Methods, for, Engineers, , Further, Analytical, Methods for, Engineers, , Engineering, Mathematics, , ×, , 26., , Scalar and vector products, , 27., , Methods of differentiation, , ×, , 28., , Some applications of differentiation, , ×, , 29., , Differentiation of parametric equations, , 30., , Differentiation of implicit functions, , ×, , 31., , Logarithmic differentiation, , ×, , 32., , Differentiation of hyperbolic functions, , ×, , 33., , Differentiation of inverse trigonometric and hyperbolic, functions, , ×, , 34., , Partial differentiation, , ×, , 35., , Total differential, rates of change and small changes, , ×, , 36., , Maxima, minima and saddle points for functions of two, variables, , ×, , 37., , Standard integration, , ×, , 38., , Some applications of integration, , ×, , 39., , Integration using algebraic substitutions, , ×, , 40., , Integration using trigonometric and hyperbolic, substitutions, , ×, , 41., , Integration using partial fractions, , ×, , 42., , The t = tan θ/2 substitution, , 43., , Integration by parts, , ×, , 44., , Reduction formulae, , ×, , 45., , Numerical integration, , ×, , 46., , Solution of first order differential equations by separation of, variables, , ×, , 47., , Homogeneous first order differential equations, , 48., , Linear first order differential equations, , ×, , 49., , Numerical methods for first order differential equations, , ×, , 50., , Second order differential equations of the form, d2 y, dy, + cy = 0, a 2 +b, dx, dx, , ×, , 51., , Second order differential equations of the form, d2 y, dy, a 2 +b, + cy = f (x), dx, dx, , ×, , 52., , Power series methods of solving ordinary differential equations, , ×, , 53., , An introduction to partial differential equations, , ×, , 54., , Presentation of statistical data, , ×, , ×, (Continued )
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Syllabus Guidance, Chapter, , Analytical, Methods, for, Engineers, , Further, Analytical, Methods for, Engineers, , Engineering, Mathematics, , 55., , Measures of central tendency and dispersion, , ×, , 56., , Probability, , ×, , 57., , The binomial and Poisson distributions, , ×, , 58., , The normal distribution, , ×, , 59., , Linear correlation, , ×, , 60., , Linear regression, , ×, , 61., , Introduction to Laplace transforms, , ×, , 62., , Properties of Laplace transforms, , ×, , 63., , Inverse Laplace transforms, , ×, , 64., , Solution of differential equations using Laplace transforms, , ×, , 65., , The solution of simultaneous differential equations using, Laplace transforms, , ×, , 66., , Fourier series for periodic functions of period 2π, , ×, , 67., , Fourier series for non-periodic functions over range 2π, , ×, , 68., , Even and odd functions and half-range Fourier series, , ×, , 69., , Fourier series over any range, , ×, , 70., , A numerical method of harmonic analysis, , ×, , 71., , The complex or exponential form of a Fourier series, , ×, , Website Chapters, 72., , Inequalities, , 73., , Boolean algebra and logic circuits, , 74., , Sampling and estimation theories, , ×, , 75., , Significance testing, , ×, , 76., , Chi-square and distribution-free tests, , ×, , ×, , xvii
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Chapter 1, , Algebra, 1.1, , 3x + 2y, x−y, , Introduction, , In this chapter, polynomial division and the factor, and remainder theorems are explained (in Sections 1.4, to 1.6). However, before this, some essential algebra, revision on basic laws and equations is included., For further Algebra revision, go to website:, http://books.elsevier.com/companions/0750681527, , 1.2, , Multiply by x → 3x 2 + 2x y, Multiply by −y →, , 3x 2 − xy − 2y 2, , Adding gives:, Alternatively,, , (3x + 2y)(x − y) = 3x 2 − 3x y + 2x y − 2y 2, , Revision of basic laws, , = 3x 2 − xy − 2y 2, , (a) Basic operations and laws of indices, The laws of indices are:, (i) a m × a n = a m+n, (iii), , (a m )n, , (v), , a −n, , =, , a m×n, , 1, = n, a, , am, (ii), = a m−n, an, √, m, (iv) a n = n a m, (vi), , a0, , Problem 3. Simplify, a = 3, b =, , 1, 8, , and c = 2., , When a = 3, b =, Problem 1. Evaluate, b = 12 and c = 1 12, , when a = 2,, , , 3, 3, 3, 1, 4a bc − 2ac = 4(2), − 2(2), 2, 2, 2, 2, , 3, , 2, , =, , a 3 b 2 c4, and evaluate when, abc−2, , a 3 b 2 c4, = a 3−1b2−1c4−(−2) = a 2 bc6, abc−2, , =1, , 4a 2 bc3−2ac, , − 3x y − 2y 2, , 4 × 2 × 2 × 3 × 3 × 3 12, −, 2×2×2×2, 2, , = 27 − 6 = 21, Problem 2. Multiply 3x + 2y by x − y., , and c = 2,, , , a 2 bc6 = (3)2 18 (2)6 = (9) 18 (64) = 72, 1, 8, , Problem 4. Simplify, , x 2 y3 + x y2, xy, , x 2 y3 + x y2, x 2 y3 x y2, =, +, xy, xy, xy, = x 2−1 y 3−1 + x 1−1 y 2−1, = xy 2 + y or y(xy + 1)
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4 Higher Engineering Mathematics, , i.e., , √, , √, t +3, √, =2 t, t, √, √, t +3= 2 t, √ √, 3= 2 t − t, √, 3= t, , and, , 9= t, , √, t, i.e., and, , Rearranging gives: d 2 p + D 2 p = D 2 f − d2 f, Factorizing gives:, , Exercise 3 Further problems on simple, equations and transposition of formulae, , Transpose the formula v = u +, , ft, m, , In problems 1 to 4 solve the equations, 1. 3x − 2 − 5x = 2x − 4., , ft, ft, = v from which,, = v−u, m, m, , ft, = m(v − u), and, m, m, , u+, , i.e., , f t = m(v − u), , and, , m, f = (v − u), t, , X 2 = Z 2 − R 2 and reactance X =, D, =, d, express p in terms of D, d and f ., Given that, , , , 3., 4., , Rearranging gives:, Squaring both sides gives:, , , , Z2 − R2, , , f +p, ,, f −p, , , , , , [−3], , R 2 + X 2 = Z and squaring both sides gives, R 2 + X 2 = Z 2 , from which,, , 1, 2, , 2. 8 + 4(x − 1) − 5(x − 3) = 2(5 − 2x)., , Problem 16. √The impedance of an a.c. circuit is, given by Z = R 2 + X 2 . Make the reactance X the, subject., , Problem 17., , f (D 2 − d2 ), (d2 + D2 ), , Now try the following exercise, , to make f the subject., , , , p=, , and, , (b) Transposition of formulae, Problem 15., , p(d 2 + D 2 ) = f (D 2 − d2 ), , , D, f +p, =, f −p, d, f +p, D2, = 2, f −p, d, , ‘Cross-multiplying’ gives:, d2 ( f + p) = D 2 ( f − p), Removing brackets gives:, d2 f + d2 p = D 2 f − D 2 p, , 1, 1, +, = 0., 3a − 2 5a + 3, √, 3 t, √ = −6., 1− t, , − 18, , , , [4], , 3(F − f ), . for f ., 5. Transpose y =, L, , yL, 3F − yL, or f = F −, f =, 3, 3, , 1, 6. Make l the subject of t = 2π, ., g , t 2g, l=, 4π 2, μL, for L., 7. Transpose m =, L + rC R, , mrC R, L=, μ−m, 8. Make r the subject of the formula, , , x−y, x, 1 + r2, ., r, =, =, y, 1 − r2, x+y, , (c) Simultaneous equations, Problem 18., , Solve the simultaneous equations:, 7x − 2y = 26, , (1), , 6x + 5y = 29., , (2)
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Algebra, 5 × equation (1) gives:, 35x − 10y = 130, , The factors of −4 are +1 and −4 or −1 and, +4, or −2 and +2. Remembering that the product of the two inner terms added to the product, of the two outer terms must equal −11x, the only, combination to give this is +1 and −4, i.e.,, , (3), , 2 × equation (2) gives:, 12x + 10y = 58, , (4), , equation (3) +equation (4) gives:, , 3x 2 − 11x − 4 = (3x + 1)(x − 4), , 47x + 0 = 188, 188, from which,, x=, =4, 47, Substituting x = 4 in equation (1) gives:, , (3x + 1)(x − 4) = 0 hence, , Thus, either, , (x − 4) = 0 i.e. x = 4, , or, , 28 − 2y = 26, , (b) 4x 2 + 8x + 3 = (2x + 3)(2x + 1), , from which, 28 − 26 = 2y and y = 1, , (2x + 3)(2x + 1) = 0 hence, , Thus, , Problem 19. Solve, x 5, + =y, 8 2, y, 11 + = 3x., 3, , (3x + 1) = 0 i.e. x = − 13, , (1), (2), , either, , (2x + 3) = 0 i.e. x = − 32, , or, , (2x + 1) = 0 i.e. x = − 12, , Problem 21. The roots of a quadratic equation, are 13 and −2. Determine the equation in x., , 8 × equation (1) gives:, , x + 20 = 8y, , (3), , 3 × equation (2) gives:, , 33 + y = 9x, , (4), , i.e., , x − 8y = −20, , (5), , and, , 9x − y = 33, , (6), , i.e. x 2 + 2x − 13 x − 23 = 0, , (7), , i.e., , x 2 + 53 x − 23 = 0, , or, , 3x2 + 5x −2 = 0, , 8 × equation (6) gives: 72x − 8y = 264, Equation (7) − equation (5) gives:, 71x = 284, 284, =4, 71, Substituting x = 4 in equation (5) gives:, x=, , from which,, , 4 − 8y = −20, 4 + 20 = 8y and y = 3, , from which,, , (d) Quadratic equations, Problem 20. Solve the following equations by, factorization:, (a) 3x 2 − 11x − 4 = 0, (b) 4x 2 + 8x + 3 = 0., (a), , The factors of 3x 2 are 3x and x and these are placed, in brackets thus:, (3x, , )(x, , ), , If, , 1, 3, , and −2 are the roots of a quadratic equation then,, (x − 13 )(x + 2) = 0, , Problem 22. Solve 4x 2 + 7x + 2 = 0 giving the, answer correct to 2 decimal places., From the quadratic formula if ax 2 + bx + c = 0 then,, √, −b ± b2 − 4ac, x=, 2a, Hence if 4x 2 + 7x + 2 = 0, , −7 ± 72 − 4(4)(2), then x =, 2(4), √, −7 ± 17, =, 8, −7 ± 4.123, =, 8, −7 + 4.123, −7 − 4.123, =, or, 8, 8, i.e. x = −0.36 or −1.39, , 5
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6 Higher Engineering Mathematics, Now try the following exercise, , For example,, , Exercise 4 Further problems on, simultaneous and quadratic equations, In problems 1 to 3, solve the simultaneous equations, , 13, ——–, 16 208, 16, 48, 48, —, ··, —, , 1. 8x − 3y = 51, 3x + 4y = 14., , 208, is achieved as follows:, 16, , [x = 6, y = −1], , (1) 16 divided into 2 won’t go, 2. 5a = 1 − 3b, 2b + a + 4 = 0., 3., , [a = 2, b = −3], , x 2y, 49, +, =, 5, 3, 15, , (2) 16 divided into 20 goes 1, (3) Put 1 above the zero, (4) Multiply 16 by 1 giving 16, (5) Subtract 16 from 20 giving 4, , 3x, y 5, − + = 0., 7, 2 7, , [x = 3, y = 4], , (6) Bring down the 8, (7) 16 divided into 48 goes 3 times, , 4. Solve the following quadratic equations by, factorization:, (a) x 2 + 4x − 32 = 0, , [(a) 4, −8 (b) 54 , − 32 ], 5. Determine the quadratic equation in x whose, roots are 2 and −5., [x 2 + 3x − 10 = 0], 6. Solve the following quadratic equations, correct to 3 decimal places:, (a), , −4 = 0, , (b) 4t 2 − 11t + 3 = 0., (a) 0.637, −3.137, (b) 2.443, 0.307, , 1.4, , (9) 3 × 16 = 48, (10) 48 − 48 = 0, , (b) 8x 2 + 2x − 15 = 0., , 2x 2 + 5x, , (8) Put the 3 above the 8, , Hence, Similarly,, , 208, = 13 exactly, 16, , 172, is laid out as follows:, 15, , 11, ——–, 15 172, 15, 22, 15, —, 7, —, 7, 7, 172, = 11 remainder 7 or 11 +, = 11, Hence, 15, 15, 15, Below are some examples of division in algebra, which, in some respects, is similar to long division with, numbers., (Note that a polynomial is an expression of the, form, , Polynomial division, f (x) = a + bx + cx 2 + d x 3 + · · ·, , Before looking at long division in algebra let us revise, long division with numbers (we may have forgotten,, since calculators do the job for us!), , and polynomial division is sometimes required when, resolving into partial fractions—see Chapter 2.)
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Algebra, Problem 23. Divide 2x 2 + x − 3 by x − 1., , (3) Subtract, (4), , 2x 2 + x − 3 is called the dividend and x − 1 the divisor. The usual layout is shown below with the dividend, and divisor both arranged in descending powers of the, symbols., 2x + 3, , ——————–, x − 1 2x 2 + x − 3, 2x 2 − 2x, 3x − 3, 3x − 3, ———, · ·, ———, Dividing the first term of the dividend by the first term, 2x 2, gives 2x, which is put above, of the divisor, i.e., x, the first term of the dividend as shown. The divisor, is then multiplied by 2x, i.e. 2x(x − 1) = 2x 2 − 2x,, which is placed under the dividend as shown. Subtracting gives 3x − 3. The process is then repeated, i.e. the, first term of the divisor, x, is divided into 3x, giving, +3, which is placed above the dividend as shown. Then, 3(x − 1) = 3x − 3 which is placed under the 3x − 3. The, remainder, on subtraction, is zero, which completes the, process., , x into −2x 2 goes −2x. Put −2x above the, dividend, , (5) −2x(x + 1) = −2x 2 − 2x, (6) Subtract, (7), , x into 5x goes 5. Put 5 above the dividend, , (8) 5(x + 1) = 5x + 5, (9) Subtract, Thus, , 3x 3 + x 2 + 3x + 5, = 3x 2 − 2x + 5, x +1, , Problem 25. Simplify, , (1) (4) (7), x 2 − x y + y2, , —————————–, x + y x 3 + 0 + 0 + y3, x3 + x2 y, − x2 y, + y3, − x 2 y − x y2, ———————, x y2 + y3, x y2 + y3, ———–, · ·, ———–, , Thus (2x 2 + x − 3) ÷ (x − 1) = (2x + 3), [A check can be made on this answer by multiplying, (2x + 3) by (x − 1) which equals 2x 2 + x − 3], Problem 24. Divide 3x 3 + x 2 + 3x + 5 by x + 1., (1) (4) (7), 3x 2 − 2x + 5, , —————————, x + 1 3x 3 + x 2 + 3x + 5, 3x 3 + 3x 2, − 2x 2 + 3x + 5, − 2x 2 − 2x, ————–, 5x + 5, 5x + 5, ———, · ·, ———, (1), , x into 3x 3 goes 3x 2 . Put 3x 2 above 3x 3, , (2) 3x 2 (x + 1) = 3x 3 + 3x 2, , x 3 + y3, ., x+y, , (1), , x into x 3 goes x 2 . Put x 2 above x 3 of dividend, , (2), , x 2 (x + y) = x 3 + x 2 y, , (3) Subtract, (4), , x into −x 2 y goes −x y. Put −x y above dividend, , (5) −x y(x + y) = −x 2 y − x y 2, (6) Subtract, (7), , x into x y 2 goes y 2 . Put y 2 above dividend, , (8), , y 2 (x + y) = x y 2 + y 3, , (9) Subtract, Thus, x 3 + y3, = x 2 − xy + y 2, x+y, , 7
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8 Higher Engineering Mathematics, The zero’s shown in the dividend are not normally, shown, but are included to clarify the subtraction process, and to keep similar terms in their respective columns., Problem 26., , Divide (x 2 + 3x − 2) by (x − 2)., , x +5, , ——————–, x − 2 x 2 + 3x − 2, x 2 − 2x, , 14x 2 − 19x − 3, ., 2x − 3, , [7x + 1], , 6. Find (5x 2 − x + 4) ÷ (x − 1)., , 5x + 4 +, , 8, x −1, , 7. Divide (3x 3 + 2x 2 − 5x + 4) by (x + 2)., , 2, 3x 2 − 4x + 3 −, x +2, , Hence, 8, x 2 + 3x − 2, =x +5+, x −2, x−2, Divide 4a 3 − 6a 2 b + 5b 3 by, , 2a − 2ab − b, , ———————————, 2a − b 4a 3 − 6a 2 b, + 5b 3, 3, 2, 4a − 2a b, 2, , 4. Find, , 5. Divide (x 3 + 3x 2 y + 3x y 2 + y 3 ) by (x + y)., [x 2 + 2x y + y 2 ], , 5x − 2, 5x − 10, ———, 8, ———, , Problem 27., 2a − b., , 3. Determine (10x 2 + 11x − 6) ÷ (2x + 3)., [5x − 2], , 8. Determine (5x 4 + 3x 3 − 2x + 1)/(x − 3)., , 481, 5x 3 + 18x 2 + 54x + 160 +, x −3, , 2, , 1.5, , −4a 2 b, + 5b3, 2, 2, −4a b + 2ab, ———— 2, −2ab + 5b 3, −2ab2 + b 3, —————–, 4b 3, —————–, , There is a simple relationship between the factors of, a quadratic expression and the roots of the equation, obtained by equating the expression to zero., For example, consider the quadratic equation, x 2 + 2x − 8 = 0., To solve this we may factorize the quadratic expression, x 2 + 2x − 8 giving (x − 2)(x + 4)., Hence (x − 2)(x + 4) = 0., Then, if the product of two numbers is zero, one or both, of those numbers must equal zero. Therefore,, , Thus, 4a 3 − 6a 2 b + 5b 3, 2a − b, = 2a 2 − 2ab − b2 +, , either (x − 2) = 0, from which, x = 2, or, (x + 4) = 0, from which, x = −4, , 4b3, 2a − b, , It is clear then that a factor of (x − 2) indicates a root, of +2, while a factor of (x + 4) indicates a root of −4., In general, we can therefore say that:, , Now try the following exercise, Exercise 5 Further problems on polynomial, division, 1. Divide (2x 2 + x y − y 2 ) by (x + y)., 2. Divide (3x 2 + 5x − 2) by (x + 2)., , The factor theorem, , [2x − y], [3x − 1], , a factor of (x − a) corresponds to a, root of x = a, In practice, we always deduce the roots of a simple, quadratic equation from the factors of the quadratic, expression, as in the above example. However, we could, reverse this process. If, by trial and error, we could determine that x = 2 is a root of the equation x 2 + 2x − 8 = 0, we could deduce at once that (x − 2) is a factor of the
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Algebra, expression x 2 + 2x − 8. We wouldn’t normally solve, quadratic equations this way — but suppose we have, to factorize a cubic expression (i.e. one in which the, highest power of the variable is 3). A cubic equation, might have three simple linear factors and the difficulty, of discovering all these factors by trial and error would, be considerable. It is to deal with this kind of case that, we use the factor theorem. This is just a generalized, version of what we established above for the quadratic, expression. The factor theorem provides a method of, factorizing any polynomial, f (x), which has simple, factors., A statement of the factor theorem says:, ‘if x = a is a root of the equation, f (x) = 0, then (x − a) is a factor of f (x)’, The following worked problems show the use of the, factor theorem., Problem 28. Factorize x 3 − 7x − 6 and use it to, solve the cubic equation x 3 − 7x − 6 = 0., Let, , f (x) = x 3 − 7x − 6, , If x = 1, then f (1) = 13 − 7(1) − 6 = −12, If x = 2, then f (2) = 23 − 7(2) − 6 = −12, If x = 3, then f (3), , = 33 − 7(3) − 6, , =0, , If f (3) = 0, then (x − 3) is a factor — from the factor, theorem., We have a choice now. We can divide x 3 − 7x − 6 by, (x − 3) or we could continue our ‘trial and error’ by substituting further values for x in the given expression —, and hope to arrive at f (x) = 0., Let us do both ways. Firstly, dividing out gives:, x + 3x + 2, , —————————, x − 3 x 3 − 0 − 7x − 6, x 3 − 3x 2, 2, , 3x 2 − 7x − 6, 3x 2 − 9x, ————, 2x − 6, 2x − 6, ———, · ·, ———, x 3 − 7x − 6, = x 2 + 3x + 2, Hence, x −3, i.e., , x 3 − 7x − 6 = (x − 3)(x 2 + 3x + 2), , x 2 + 3x + 2 factorizes ‘on sight’ as (x + 1)(x + 2)., Therefore, x 3 − 7x − 6 = (x − 3)(x + 1)(x + 2), A second method is to continue to substitute values of, x into f (x)., Our expression for f (3) was 33 − 7(3) − 6. We can, see that if we continue with positive values of x the, first term will predominate such that f (x) will not, be zero., Therefore let us try some negative values for x., Therefore f (−1) = (−1)3 − 7(−1) − 6 = 0; hence, (x + 1) is a factor (as shown above). Also, f (−2) = (−2)3 − 7(−2) − 6 = 0; hence (x + 2) is, a factor (also as shown above)., To solve x 3 − 7x − 6 = 0, we substitute the factors, i.e.,, (x − 3)(x + 1)(x + 2) = 0, from which, x = 3, x = −1 and x = −2., Note that the values of x, i.e. 3, −1 and −2, are, all factors of the constant term, i.e. the 6. This can, give us a clue as to what values of x we should, consider., Problem 29. Solve the cubic equation, x 3 − 2x 2 − 5x + 6 = 0 by using the factor theorem., Let f (x) = x 3 − 2x 2 − 5x + 6 and let us substitute, simple values of x like 1, 2, 3, −1, −2, and so on., f (1) = 13 − 2(1)2 − 5(1) + 6 = 0,, hence (x − 1) is a factor, f (2) = 23 − 2(2)2 − 5(2) + 6 = 0, f (3) = 33 − 2(3)2 − 5(3) + 6 = 0,, hence (x − 3) is a factor, f (−1) = (−1)3 − 2(−1)2 − 5(−1) + 6 = 0, f (−2) = (−2)3 − 2(−2)2 − 5(−2) + 6 = 0,, hence (x + 2) is a factor, x 3 − 2x 2, , − 5x + 6 = (x − 1)(x − 3)(x + 2), Hence, Therefore if x 3 − 2x 2 − 5x + 6 = 0, then, (x − 1)(x − 3)(x + 2) = 0, from which, x = 1, x = 3 and x = −2, Alternatively, having obtained one factor, i.e., (x − 1) we could divide this into (x 3 − 2x 2 − 5x + 6), as follows:, , 9
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10 Higher Engineering Mathematics, x − x −6, ————————–, x − 1 x 3 − 2x 2 − 5x + 6, x3 − x2, 2, , 1.6, , Dividing a general quadratic expression, (ax 2 + bx + c) by (x − p), where p is any whole, number, by long division (see section 1.3) gives:, ax + (b + ap), ————————————–, x − p ax 2 + bx, +c, ax 2 − apx, , − x 2 − 5x + 6, − x2 + x, ————–, − 6x + 6, − 6x + 6, ———–, · ·, ———–, Hence x 3 − 2x 2 − 5x + 6, = (x − 1)(x 2 − x − 6), = (x − 1)(x − 3)(x + 2), Summarizing, the factor theorem provides us with a, method of factorizing simple expressions, and an alternative, in certain circumstances, to polynomial division., , Now try the following exercise, , Exercise 6 Further problems on the factor, theorem, Use the factor theorem to factorize the expressions, given in problems 1 to 4., 1., , x 2 + 2x − 3, , 2., , x 3 + x 2 − 4x − 4, , [(x − 1)(x + 3)], , 3. 2x 3 + 5x 2 − 4x − 7, , The remainder theorem, , [(x + 1)(x + 2)(x − 2)], [(x + 1)(2x 2 + 3x − 7)], , 4. 2x 3 − x 2 − 16x + 15, [(x − 1)(x + 3)(2x − 5)], 5. Use the factor theorem to factorize, x 3 + 4x 2 + x − 6 and hence solve the cubic, equation x 3 + 4x 2 + x − 6 = 0., ⎤, ⎡ 3, x + 4x 2 + x − 6, ⎥, ⎢, = (x − 1)(x + 3)(x + 2) ⎦, ⎣, x = 1, x = −3 and x = −2, 6. Solve the equation x 3 − 2x 2 − x + 2 = 0., [x = 1, x = 2 and x = −1], , (b + ap)x + c, (b + ap)x − (b + ap) p, —————————–, c + (b + ap) p, —————————–, The remainder, c + (b + ap) p = c + bp + ap 2 or, ap2 + bp + c. This is, in fact, what the remainder, theorem states, i.e.,, ‘if (ax 2 + bx + c) is divided by (x − p),, the remainder will be ap 2 + bp + c’, If, in the dividend (ax 2 + bx + c), we substitute p for, x we get the remainder ap2 + bp + c., For example, when (3x 2 − 4x + 5) is divided by, (x − 2) the remainder is ap2 + bp + c (where a = 3,, b = −4, c = 5 and p = 2),, i.e. the remainder is, 3(2)2 + (−4)(2) + 5 = 12 − 8 + 5 = 9, We can check this by dividing (3x 2 − 4x + 5) by, (x − 2) by long division:, 3x + 2, , ——————–, x − 2 3x 2 − 4x + 5, 3x 2 − 6x, 2x + 5, 2x − 4, ———, 9, ———, Similarly, when (4x 2 − 7x + 9) is divided by (x + 3),, the remainder is ap 2 + bp + c, (where a = 4, b = −7,, c = 9 and p = −3) i.e. the remainder is, 4(−3)2 + (−7)(−3) + 9 = 36 + 21 + 9 = 66., Also, when (x 2 + 3x − 2) is divided by (x − 1), the, remainder is 1(1)2 + 3(1) − 2 = 2., It is not particularly useful, on its own, to know, the remainder of an algebraic division. However, if the, remainder should be zero then (x − p) is a factor. This, is very useful therefore when factorizing expressions., For example, when (2x 2 + x − 3) is divided by, (x − 1), the remainder is 2(1)2 + 1(1) − 3 = 0, which, means that (x − 1) is a factor of (2x 2 + x − 3).
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Algebra, In this case the other factor is (2x + 3), i.e.,, , i.e. the remainder = (1)(1)3 + (−2)(1)2, + (−5)(1) + 6, , (2x 2 + x − 3) = (x − 1)(2x − 3), The remainder theorem may also be stated for a cubic, equation as:, ‘if (ax 3 + bx 2 + cx + d) is divided by, (x − p), the remainder will be, ap 3 + bp 2 + cp + d’, As before, the remainder may be obtained by substituting p for x in the dividend., For example, when (3x 3 + 2x 2 − x + 4) is divided, by (x − 1), the remainder is ap 3 + bp2 + cp + d, (where a = 3, b = 2, c = −1, d = 4 and p = 1),, i.e. the remainder is 3(1)3 + 2(1)2 + (−1)(1) + 4 =, 3 + 2 − 1 + 4 = 8., Similarly, when (x 3 − 7x − 6) is divided by (x − 3),, the remainder is 1(3)3 + 0(3)2 − 7(3) − 6 = 0, which, means that (x − 3) is a factor of (x 3 − 7x − 6)., Here are some more examples on the remainder, theorem., Problem 30. Without dividing out, find the, remainder when 2x 2 − 3x + 4 is divided by (x − 2)., By the remainder theorem, the remainder is given by, ap 2 + bp + c, where a = 2, b = −3, c = 4 and p = 2., Hence the remainder is:, 2(2)2 + (−3)(2) + 4 = 8 − 6 + 4 = 6, Problem 31. Use the remainder theorem to, determine the remainder when, (3x 3 − 2x 2 + x − 5) is divided by (x + 2)., By the remainder theorem, the remainder is given by, ap 3 + bp2 + cp + d, where a = 3, b = −2, c = 1, d =, −5 and p = −2., Hence the remainder is:, 3(−2)3 + (−2)(−2)2 + (1)(−2) + (−5), = −24 − 8 − 2 − 5, = −39, Problem 32. Determine the remainder when, (x 3 − 2x 2 − 5x + 6) is divided by (a) (x − 1) and, (b) (x + 2). Hence factorize the cubic expression., (a), , When (x 3 − 2x 2 − 5x + 6) is divided by (x − 1),, the remainder is given by ap 3 + bp2 + cp + d,, where a = 1, b = −2, c = −5, d = 6 and p = 1,, , 11, , = 1−2−5+6 = 0, Hence (x − 1) is a factor of (x 3 − 2x 2 − 5x + 6)., (b) When (x 3 − 2x 2 − 5x + 6) is divided by (x + 2),, the remainder is given by, (1)(−2)3 + (−2)(−2)2 + (−5)(−2) + 6, = −8 − 8 + 10 + 6 = 0, Hence (x + 2) is also a factor of (x 3 − 2x 2 −, 5x + 6). Therefore (x − 1)(x + 2)(x ) = x 3 −, 2x 2 − 5x + 6. To determine the third factor (shown, blank) we could, (i) divide (x 3 − 2x 2 − 5x + 6) by, (x − 1)(x + 2)., or (ii) use the factor theorem where f (x) =, x 3 − 2x 2 − 5x + 6 and hoping to choose, a value of x which makes f (x) = 0., or (iii) use the remainder theorem, again hoping, to choose a factor (x − p) which makes, the remainder zero., (i) Dividing (x 3 − 2x 2 − 5x + 6) by, (x 2 + x − 2) gives:, x −3, , ————————–, x 2 + x − 2 x 3 − 2x 2 − 5x + 6, x 3 + x 2 − 2x, ——————, −3x 2 − 3x + 6, −3x 2 − 3x + 6, ——————–, ·, ·, ·, ——————–, Thus (x 3 − 2x 2 − 5x + 6), = (x − 1)(x + 2)(x − 3), (ii) Using the factor theorem, we let, f (x) = x 3 − 2x 2 − 5x + 6, Then f (3) = 33 − 2(3)2 − 5(3) + 6, = 27 − 18 − 15 + 6 = 0, Hence (x − 3) is a factor., (iii) Using the remainder theorem, when, (x 3 − 2x 2 − 5x + 6) is divided by, (x − 3), the remainder is given by
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12 Higher Engineering Mathematics, ap3 + bp2 + cp + d, where a = 1,, b = −2, c = −5, d = 6 and p = 3., Hence the remainder is:, 1(3)3 + (−2)(3)2 + (−5)(3) + 6, = 27 − 18 − 15 + 6 = 0, Hence (x − 3) is a factor., Thus (x 3 − 2x 2 − 5x + 6), = (x − 1)(x + 2)(x − 3), , Now try the following exercise, Exercise 7 Further problems on the, remainder theorem, 1. Find the remainder when 3x 2 − 4x + 2 is, divided by, (a) (x − 2) (b) (x + 1)., , [(a) 6 (b) 9], , 2. Determine the remainder when, x 3 − 6x 2 + x − 5 is divided by, (a) (x + 2) (b) (x − 3)., , [(a) −39 (b) −29], , 3. Use the remainder theorem to find the factors, of x 3 − 6x 2 + 11x − 6., [(x − 1)(x − 2)(x − 3)], 4. Determine the factors of x 3 + 7x 2 + 14x + 8, and hence solve the cubic equation, x 3 + 7x 2 + 14x + 8 = 0., [x = −1, x = −2 and x = −4], 5. Determine the value of ‘a’ if (x + 2) is a, factor of (x 3 − ax 2 + 7x + 10)., [a = −3], 6. Using the remainder theorem, solve the, equation 2x 3 − x 2 − 7x + 6 = 0., [x = 1, x = −2 and x = 1.5]
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Chapter 2, , Partial fractions, 2.1, , When the degree of the numerator is equal to or higher, than the degree of the denominator, the numerator must, be divided by the denominator until the remainder is of, less degree than the denominator (see Problems 3 and 4)., There are basically three types of partial fraction, and the form of partial fraction used is summarized in, Table 2.1, where f (x) is assumed to be of less degree, than the relevant denominator and A, B and C are, constants to be determined., (In the latter type in Table 2.1, ax 2 + bx + c is a, quadratic expression which does not factorize without, containing surds or imaginary terms.), Resolving an algebraic expression into partial fractions is used as a preliminary to integrating certain, functions (see Chapter 41) and in determining inverse, Laplace transforms (see Chapter 63)., , Introduction to partial fractions, , By algebraic addition,, 1, 3, (x + 1) + 3(x − 2), +, =, x −2 x +1, (x − 2)(x + 1), =, , 4x − 5, x2 − x − 2, , The reverse process of moving from, , 4x − 5, −2, , x2 − x, , 1, 3, +, is called resolving into partial, x −2 x +1, fractions., In order to resolve an algebraic expression into partial, fractions:, to, , (i) the denominator must factorize (in the above, example, x 2 − x − 2 factorizes as (x − 2) (x + 1)),, and, , 2.2 Worked problems on partial, fractions with linear factors, , (ii) the numerator must be at least one degree less than, the denominator (in the above example (4x − 5) is, of degree 1 since the highest powered x term is x 1, and (x 2 − x − 2) is of degree 2)., , Problem 1. Resolve, fractions., , 11 − 3x, into partial, x 2 + 2x − 3, , Table 2.1, Type, , Denominator containing, , 1, , Linear factors, (see Problems 1 to 4), , 2, , Repeated linear factors, (see Problems 5 to 7), , 3, , Quadratic factors, (see Problems 8 and 9), , Expression, , Form of partial fraction, , f (x), (x + a)(x − b)(x + c), , A, B, C, +, +, (x + a) (x − b) (x + c), , f (x), (x + a)3, , A, C, B, +, +, 2, (x + a) (x + a), (x + a)3, , f (x), + c)(x + d), , (ax 2 + bx, , Ax + B, C, +, + bx + c) (x + d), , (ax 2
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14 Higher Engineering Mathematics, The denominator factorizes as (x − 1) (x + 3) and the, numerator is of less degree than the denominator. Thus, 11 − 3x, may be resolved into partial fractions., x 2 + 2x − 3, Let, , Let, , 2x 2 − 9x − 35, (x + 1)(x − 2)(x + 3), ≡, , 11 − 3x, 11 − 3x, ≡, − 3 (x − 1)(x + 3), , , , x 2 + 2x, , ≡, , A, B, +, (x − 1) (x + 3), , where A and B are constants to be determined,, 11 − 3x, A(x + 3) + B(x − 1), i.e., ≡, ,, (x − 1)(x + 3), (x − 1)(x + 3), by algebraic addition., Since the denominators are the same on each side, of the identity then the numerators are equal to each, other., , ≡, , When x = 1, then, 11 −3(1) ≡ A(1 + 3) + B(0), 8 = 4A, A =2, , i.e., i.e., , 2x 2 − 9x − 35 ≡ A(x − 2)(x + 3), + B(x + 1)(x + 3) + C(x + 1)(x − 2), Let x = − 1. Then, , i.e., , B = −5, , Thus, , 2(−1)2 − 9(−1) − 35 ≡ A(−3)(2), + B(0)(2) +C(0)(−3), , B=, , −45, = −3, 15, , Let x = − 3. Then, , 2, 5, 2(x + 3) − 5(x − 1), −, =, (x − 1) (x + 3), (x − 1)(x + 3), =, , 11 − 3x, x 2 + 2x − 3, , 2x 2 − 9x − 35, into, (x + 1)(x − 2)(x + 3), the sum of three partial fractions., Convert, , −24, =4, −6, , −45 = 15B, , i.e., i.e., , 2, 5, ≡, −, (x − 1) (x + 3), , Problem 2., , A=, , i.e., , 11 − 3x, −5, 2, +, ≡, x2 + 2x − 3 (x − 1) (x + 3), , , Check:, , −24 = −6 A, , i.e., , 2(2)2 − 9(2) − 35 ≡ A(0)(5) + B(3)(5) + C(3)(0), , 11 −3(−3) ≡ A(0) + B(−3 −1), 20 = −4B, , , , Let x = 2. Then, , When x = −3, then, , i.e., , A(x − 2)(x + 3) + B(x + 1)(x + 3), + C(x + 1)(x − 2), (x + 1)(x − 2)(x + 3), , by algebraic addition., Equating the numerators gives:, , Thus, 11 −3x ≡ A(x + 3) + B(x − 1), To determine constants A and B, values of x are chosen, to make the term in A or B equal to zero., , A, B, C, +, +, (x + 1) (x − 2) (x + 3), , 2(−3)2 − 9(−3) − 35 ≡ A(−5)(0) + B(−2)(0), + C(−2)(−5), i.e., , 10 = 10C, , i.e., , C =1, , Thus, , 2x 2 − 9x − 35, (x + 1)(x − 2)(x + 3), ≡, , 3, 1, 4, −, +, (x + 1) (x − 2), (x + 3), , Problem 3., fractions., , Resolve, , x2, , x2 + 1, into partial, − 3x + 2
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Partial fractions, The denominator is of the same degree as the numerator., Thus dividing out gives:, x 2 − 3x + 2, , 1, +1, x2, x 2 − 3x + 2, —————, 3x − 1, ———, , Thus, , x − 10, x 3 − 2x 2 − 4x − 4, ≡ x −3+ 2, 2, x +x −2, x +x −2, ≡ x −3+, , Let, , x − 10, A, B, ≡, +, (x + 2)(x − 1) (x + 2) (x − 1), ≡, , For more on polynomial division, see Section 1.4,, page 6., Hence, , 3x − 1, x2 + 1, ≡1 + 2, 2, x − 3x + 2, x − 3x + 2, 3x − 1, ≡1 +, (x − 1)(x − 2), , A, B, 3x − 1, ≡, +, Let, (x − 1)(x − 2) (x − 1) (x − 2), ≡, , A(x − 2) + B(x − 1), (x − 1)(x − 2), , Equating numerators gives:, , x − 10 ≡ A(x − 1) + B(x + 2), Let x = −2. Then, , −12 = −3 A, A= 4, , i.e., Let x = 1. Then, , −9 = 3B, B = −3, , i.e., Hence, , x − 10, 4, 3, ≡, −, (x + 2)(x − 1) (x + 2) (x − 1), , Thus, , x3 − 2 x2 − 4x − 4, x2 + x − 2, ≡x−3+, , Let x = 1. Then 2 = −A, A = −2, , A(x − 1) + B(x + 2), (x + 2)(x − 1), , Equating the numerators gives:, , 3x − 1 ≡ A(x − 2) + B(x − 1), , i.e., , x − 10, (x + 2)(x − 1), , 4, 3, −, (x + 2) (x − 1), , Now try the following exercise, , Let x = 2. Then 5 = B, −2, 5, 3x − 1, ≡, +, Hence, (x − 1)(x − 2) (x − 1) (x − 2), Thus, , 2, 5, x2 + 1, ≡ 1−, +, 2, x − 3x + 2, (x−1) (x−2), , Problem 4. Express, fractions., , x 3 − 2x 2 − 4x − 4, in partial, x2 + x − 2, , The numerator is of higher degree than the denominator., Thus dividing out gives:, x −3, x 2 + x − 2 x 3 − 2x 2 − 4x − 4, x 3 + x 2 − 2x, ——————, − 3x 2 − 2x − 4, − 3x 2 − 3x + 6, ———————, x − 10, , Exercise 8 Further problems on partial, fractions with linear factors, Resolve the following into partial fractions., , 2, 2, 12, −, 1., 2, x −9, (x − 3) (x + 3), , , 2., , 4(x − 4), 2, x − 2x − 3, , 3., , x 2 − 3x + 6, x(x − 2)(x − 1), , 4., , 3(2x 2 − 8x − 1), (x + 4)(x + 1)(2x − 1), , , , , 5, 1, −, (x + 1) (x − 3), , 3, 2, 4, +, −, x (x − 2) (x − 1), , 7, 3, 2, −, −, (x + 4) (x + 1) (2x − 1), , 15
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16 Higher Engineering Mathematics, , 5., , x 2 + 9x + 8, x2 + x − 6, , 6., , x 2 − x − 14, x 2 − 2x − 3, , 7., , , , , , When A = 2 and B = 7,, 2, 6, 1+, +, (x + 3) (x − 2), 2, 3, 1−, +, (x − 3) (x + 1), , 3x 3 − 2x 2 − 16x + 20, (x − 2)(x + 2), , 3x − 2 +, , R.H.S. = −2(2) + 7 = 3 = L.H.S.], Hence, , 5x 2 − 2x − 19, as the sum, (x + 3)(x − 1)2, of three partial fractions., , Problem 6., 5, 1, −, (x − 2) (x + 2), , Worked problems on partial, fractions with repeated linear, factors, , Problem 5., , Resolve, , fractions., , 2x + 3, into partial, (x − 2)2, , The denominator contains a repeated linear factor,, (x − 2)2 ., A, 2x + 3, B, ≡, Let, +, (x − 2)2 (x − 2) (x − 2)2, A(x − 2) + B, (x − 2)2, , ≡, , Equating the numerators gives:, 2x + 3 ≡ A(x − 2) + B, Let x = 2. Then, , 7 = A(0) + B, , i.e., , B =7, , 2x + 3 ≡ A(x − 2) + B ≡ Ax − 2 A + B, Since an identity is true for all values of the, unknown, the coefficients of similar terms may be, equated., Hence, equating the coefficients of x gives: 2 = A., [Also, as a check, equating the constant terms gives:, , Express, , The denominator is a combination of a linear factor and, a repeated linear factor., Let, , 2.3, , 2, 7, 2x + 3, ≡, +, (x − 2)2 (x − 2) (x − 2)2, , 5x 2 − 2x − 19, (x + 3)(x − 1)2, ≡, , A, B, C, +, +, (x + 3) (x − 1) (x − 1)2, , ≡, , A(x − 1)2 + B(x + 3)(x − 1) + C(x + 3), (x + 3)(x − 1)2, , by algebraic addition., Equating the numerators gives:, 5x 2 − 2x − 19 ≡ A(x − 1)2 + B(x + 3)(x − 1), + C(x + 3), Let x = −3. Then, 5(−3)2 − 2(−3) − 19 ≡ A(−4)2 + B(0)(−4), + C(0), i.e., 32 = 16 A, i.e., A= 2, Let x = 1. Then, 5(1)2 − 2(1) − 19 ≡ A(0)2 + B(4)(0) + C(4), i.e., −16 = 4C, i.e., C = −4, Without expanding the RHS of equation (1) it can, be seen that equating the coefficients of x 2 gives:, 5 = A + B, and since A = 2, B = 3., [Check: Identity (1) may be expressed as:, 5x 2 − 2x − 19 ≡ A(x 2 − 2x + 1), + B(x 2 + 2x − 3) + C(x + 3), i.e. 5x 2 − 2x − 19 ≡ Ax 2 − 2 Ax + A + Bx 2 + 2Bx, , 3 = −2 A + B, , (1), , − 3B + Cx + 3C
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Partial fractions, Equating the x term coefficients gives:, , Equating the coefficients of x terms gives:, 16 = 6 A + B, , −2 ≡ −2 A + 2B + C, , Since A = 3, B = −2, , When A = 2, B = 3 and C = −4 then, , [Check: equating the constant terms gives:, , −2 A + 2B + C = −2(2) + 2(3) − 4, , 15 = 9 A + 3B + C, , = −2 = LHS, , When A = 3, B = −2 and C = −6,, , Equating the constant term gives:, , 9 A + 3B + C = 9(3) + 3(−2) + (−6), , −19 ≡ A − 3B + 3C, , = 27 − 6 − 6 = 15 = LHS], , RHS = 2 − 3(3) + 3(−4) = 2 − 9 − 12, = −19 = LHS], , Hence, , Thus, , 5x2 − 2x − 19, (x + 3)(x − 1)2, ≡, , 2, 3, 4, +, −, (x + 3) (x − 1) (x − 1)2, , Now try the following exercise, , 3x 2 + 16x + 15, Problem 7. Resolve, into partial, (x + 3)3, fractions., , Let, , 3x 2 + 16x + 15, (x + 3)3, , Exercise 9 Further problems on partial, fractions with linear factors, , 4, 4x − 3, 7, 1., −, 2, (x + 1), (x + 1) (x + 1)2, 2., , ≡, , A, C, B, +, +, (x + 3) (x + 3)2 (x + 3)3, , ≡, , A(x + 3)2 + B(x + 3) + C, (x + 3)3, , 3., , Equating the numerators gives:, 3x 2 + 16x + 15 ≡ A(x + 3)2 + B(x + 3) + C, Let x = −3. Then, , 3x2 + 16x + 15, (x + 3)3, 3, 6, 2, ≡, −, −, 2, (x + 3) (x + 3), (x + 3)3, , (1), , 4., , , , x 2 + 7x + 3, x 2 (x + 3), , 1, 2, 1, + −, x 2 x (x + 3), , 5x 2 − 30x + 44, (x − 2)3, , 5, 4, 10, +, −, (x − 2) (x − 2)2 (x − 2)3, 18 + 21x − x 2, (x − 5)(x + 2)2, , , 2, 3, 4, −, +, (x − 5) (x + 2) (x + 2)2, , 3(−3)2 + 16(−3) + 15 ≡ A(0)2 + B(0) + C, i.e., −6 = C, Identity (1) may be expanded as:, 3x 2 + 16x + 15 ≡ A(x 2 + 6x + 9), + B(x + 3) + C, , 2.4 Worked problems on partial, fractions with quadratic factors, , i.e. 3x 2 + 16x + 15 ≡ Ax 2 + 6 Ax + 9 A, + Bx + 3B + C, Equating the coefficients of x 2 terms gives: 3 = A, , Problem 8. Express, fractions., , 7x 2 + 5x + 13, in partial, (x 2 + 2)(x + 1), , 17
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18 Higher Engineering Mathematics, The denominator is a combination of a quadratic factor,, (x 2 + 2), which does not factorize without introducing imaginary surd terms, and a linear factor, (x + 1)., Let,, , Equating the numerators gives:, 3 + 6x + 4x 2 − 2x 3 ≡ Ax(x 2 + 3) + B(x 2 + 3), + (Cx + D)x 2, , 7x 2 + 5x + 13, Ax + B, C, ≡ 2, +, 2, (x + 2)(x + 1) (x + 2) (x + 1), ≡, , (Ax + B)(x + 1) + C(x 2 + 2), (x 2 + 2)(x + 1), , ≡ Ax 3 + 3 Ax + Bx 2 + 3B, + Cx 3 + Dx 2, Let x = 0. Then 3 = 3B, i.e., , Equating numerators gives:, 7x 2 + 5x + 13 ≡ (Ax + B)(x + 1) + C(x 2 + 2) (1), , Equating the coefficients of x 3 terms gives:, , Let x = −1. Then, , −2 = A + C, , 7(−1)2 + 5(−1) + 13 ≡ (Ax, , + B)(0) + C(1 + 2), , 15 = 3C, C= 5, , i.e., i.e., , B=1, , Equating the coefficients of x 2 terms gives:, 4= B+D, Since B = 1, D = 3, , Identity (1) may be expanded as:, 7x 2 + 5x + 13 ≡ Ax 2 + Ax + Bx + B + Cx 2 + 2C, , Equating the coefficients of x terms gives:, , Equating the coefficients of x 2 terms gives:, 7 = A + C, and since C = 5, A = 2, Equating the coefficients of x terms gives:, 5 = A + B, and since A = 2, B = 3, , 6 = 3A, A=2, , i.e., , From equation (1), since A = 2, C = −4, Hence, , [Check: equating the constant terms gives:, , 3 + 6 x + 4x2 − 2 x3, −4x + 3, 2, 1, ≡ + 2+ 2, x2 (x2 + 3), x x, x +3, , 13 = B + 2C, , ≡, , When B = 3 and C = 5,, B + 2C = 3 + 10 = 13 = LHS], Hence, , 7x2 + 5x + 13, (x2 + 2)(x + 1), , Problem 9., , ≡, , Resolve, , partial fractions., , 2x + 3, 5, +, ( x2 + 2) (x + 1), , 3 + 6x + 4x 2 − 2x 3, into, x 2 (x 2 + 3), , Terms such as x 2 may be treated as (x + 0)2 , i.e. they, are repeated linear factors., Let, , Now try the following exercise, Exercise 10 Further problems on partial, fractions with quadratic factors, , 2x + 3, 1, x 2 − x − 13, −, 1., (x 2 + 7)(x − 2), (x 2 + 7) (x − 2), , 2., , 6x − 5, (x − 4)(x 2 + 3), , 3., , 15 + 5x + 5x 2 − 4x 3, x 2 (x 2 + 5), , Cx + D, A, B, 3 + 6x + 4x 2 − 2x 3, ≡ + 2+ 2, 2, 2, x (x + 3), x, x, (x + 3), Ax(x 2 + 3) + B(x 2 + 3) + (Cx + D)x 2, ≡, x 2 (x 2 + 3), , 2, 3 − 4x, 1, + 2+ 2, x x, x +3, , , , , , 1, 2−x, +, (x − 4) (x 2 + 3), 1, 2 − 5x, 3, +, +, x x 2 (x 2 + 5), , (1)
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Partial fractions, , 4., , x 3 + 4x 2 + 20x − 7, (x − 1)2 (x 2 + 8), , , 3, 1 − 2x, 2, +, +, (x − 1) (x − 1)2 (x 2 + 8), , 5. When solving the differential equation, d2θ, dθ, − 6 − 10θ = 20 − e2t by Laplace, dt 2, dt, transforms, for given boundary conditions, the, , following expression for L{θ} results:, 39 2, s + 42s − 40, 2, L{θ} =, s(s − 2)(s 2 − 6s + 10), 4s 3 −, , Show that the expression can be resolved into, partial fractions to give:, L{θ} =, , 1, 5s − 3, 2, −, +, 2, s 2(s − 2) 2(s − 6s + 10), , 19
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Chapter 3, , Logarithms, 3.1, , In another example, if we write down that 64 = 82, then the equivalent statement using logarithms is:, , Introduction to logarithms, , With the use of calculators firmly established, logarithmic tables are now rarely used for calculation. However,, the theory of logarithms is important, for there are several scientific and engineering laws that involve the rules, of logarithms., From the laws of indices:, , 16 = 2, , 4, , The number 4 is called the power or the exponent or, the index. In the expression 24 , the number 2 is called, the base., In another example:, , 64 = 82, , In this example, 2 is the power, or exponent, or index., The number 8 is the base., What is a logarithm?, Consider the expression 16 = 24., An alternative, yet equivalent, way of writing this, expression is: log2 16 = 4., This is stated as ‘log to the base 2 of 16 equals 4’., We see that the logarithm is the same as the power, or index in the original expression. It is the base in, the original expression which becomes the base of the, logarithm., The two statements: 16 = 24 and log2 16 = 4 are, equivalent., If we write either of them, we are automatically implying the other., In general, if a number y can be written in the form, a x , then the index ‘x’ is called the ‘logarithm of y to the, base of a’,, i.e., , if y = a x then x = loga y, , log8 64 = 2, In another example, if we write down that: log3 81 =4, then the equivalent statement using powers is:, 34 = 81, So the two sets of statements, one involving powers, and one involving logarithms, are equivalent., Common logarithms, From above, if we write down that: 1000 = 103 , then, 3 = log10 1000, This may be checked using the ‘log’ button on your, calculator., Logarithms having a base of 10 are called common, logarithms and log10 is often abbreviated to lg., The following values may be checked by using a, calculator:, lg 27.5 = 1.4393 . . ., lg 378.1 = 2.5776 . . ., and lg 0.0204 = −1.6903 . . ., Napierian logarithms, Logarithms having a base of e (where ‘e’ is a mathematical constant approximately equal to 2.7183) are, called hyperbolic, Napierian or natural logarithms,, and loge is usually abbreviated to ln., The following values may be checked by using a, calculator:, ln 3.65 = 1.2947 . . ., ln 417.3 = 6.0338 . . ., and ln 0.182 = −1.7037 . . ., More on Napierian logarithms is explained in Chapter 4, following., Here are some worked problems to help understanding of logarithms.
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Logarithms, Problem 1. Evaluate log3 9., , Problem 6. Evaluate log3, , Let x = log3 9 then 3 x = 9, , from the definition, of a logarithm,, , 3 x = 32, , i.e., , 1, ., 81, , Let x = log3, , from which, x = 2, , 1, 1, 1, then 3 x =, = 4 = 3−4, 81, 81 3, from which, x = −4, , log3 9 = 2, , Hence,, , Hence,, , log3, , 1, = −4, 81, , Problem 2. Evaluate log10 10., Problem 7. Solve the equation: lg x = 3., Let x = log10 10 then 10 x = 10, , from the, , definition of a logarithm,, 10 = 10, x, , i.e., Hence,, , from which, x = 1, , 1, , log10 10 = 1, , (which may be checked, by a calculator), , Problem 3. Evaluate log16 8., , If lg x = 3 then log10 x = 3, and, , x = 103, , i.e. x = 1000, , Problem 8. Solve the equation: log2 x = 5., If log2 x = 5 then x = 25 = 32, , Let x = log16 8 then 16 x = 8, , from the definition, , Problem 9. Solve the equation: log5 x = −2., , of a logarithm,, i.e. (24 )x = 23 i.e. 24x = 23 from the laws of indices,, from which,, Hence,, , 4x = 3 and x =, log16 8 =, , If log5 x = −2 then x = 5−2 =, , 3, 4, , 3, 4, , 1, 1, =, 52 25, , Now try the following exercise, , Problem 4. Evaluate lg 0.001., then 10x = 0.001, , Let x = lg 0.001 = log10 0.001, i.e., Hence,, , 10 x = 10−3, , from which, x = −3, , lg 0.001 = −3 (which may be checked, , Exercise 11, logarithms, , Further problems on laws of, , In Problems 1 to 11, evaluate the given, expressions:, 1. log10 10000, , [4], , 2. log2 16, , 3. log5 125, , [3], , 1, 3, , 4. log2 18, , by a calculator), Problem 5. Evaluate ln e., 5. log8 2, Let x = ln e = loge e then ex = e, i.e., Hence,, , 7. lg 100, , ex = e1, from which, x = 1, ln e = 1 (which may be checked, by a calculator), , 9. log4 8, 11. ln e2, , [2], , 1, 1, 2, [2], , [4], [−3], , 6. log7 343, 8. lg 0.01, , [3], [−2], , , 10. log27 3, , 1, 3, , 21
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22 Higher Engineering Mathematics, The following may be checked using a calculator:, In Problems 12 to 18 solve the equations:, 12., , log10 x = 4, , 13., , lg x = 5, , 14., , log3 x = 2, , 15., , 1, log4 x = −2, 2, , 16., , lg x = −2, , 17., , log8 x = −, , 18., , ln x = 3, , lg 52 = lg 25 = 1.39794. . ., , [10000], [100000], [9], , , 1, 32, , [0.01], , 1, 16, , 4, 3, , [e3 ], , Also, 2 lg 5 = 2 × 0.69897. . . = 1.39794. . ., lg 52 = 2 lg 5, , Hence,, , Here are some worked problems to help understanding of the laws of logarithms., Problem 10. Write log 4 + log 7 as the logarithm, of a single number., log 4 + log 7 = log (7 × 4), by the first law of logarithms, = log 28, , 3.2, , Laws of logarithms, , There are three laws of logarithms, which apply to any, base:, (i) To multiply two numbers:, , Problem 11. Write log 16 − log 2 as the logarithm of a single number., , , 16, log 16 − log 2 = log, 2, , by the second law of logarithms, , log (A × B) = log A + log B, The following may be checked by using a calculator:, lg 10 = 1, Also, lg 5 + lg 2 = 0.69897. . ., + 0.301029. . . = 1, Hence,, lg (5 × 2) = lg 10 = lg 5 + lg 2, (ii) To divide two numbers:, , A, = log A − log B, log, B, The following may be checked using a calculator:, , 5, = ln 2.5 = 0.91629. . ., ln, 2, Also,, Hence,, , ln 5 − ln 2 = 1.60943. . . − 0.69314. . ., = 0.91629. . ., , 5, = ln 5 − ln 2, ln, 2, , (iii) To raise a number to a power:, log An = n log A, , , , = log 8, Problem 12. Write 2 log 3 as the logarithm of a, single number., 2 log 3 = log 32, , by the third law of logarithms, , = log 9, 1, Problem 13. Write log 25 as the logarithm of a, 2, single number., 1, 1, log 25 = log 25 2 by the third law of logarithms, 2, √, = log 25 = log 5, , Problem 14., , Simplify: log 64 − log 128 + log32., , 64 = 26, 128 = 27 and 32 = 25, Hence, log 64 − log 128 + log32, = log 26 − log 27 + log 25
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Logarithms, = 6 log2 − 7 log 2 + 5 log2, by the third law of logarithms, = 4 log 2, 1, 1, Problem 15. Write log16 + log27 − 2 log5, 2, 3, as the logarithm of a single number., 1, 1, log 16 + log 27 − 2 log5, 2, 3, 1, , 1, , = log 23 + log 5 4 − log 34, by the laws of indices, , , √, 4, 1, 8× 5, = 3 log 2 + log 5 − 4 log 3, i.e. log, 81, 4, by the third law of logarithms, Problem 18. Evaluate:, log 25 − log125 + 12 log 625, ., 3 log5, , 1, , = log 16 2 + log 27 3 − log 52, by the third law of logarithms, √, √, 3, = log 16 + log 27 − log 25, by the laws of indices, , log 25 − log125 + 21 log 625, 3 log5, , = log4 + log 3 − log 25, , , 4×3, = log, 25, by the first and second laws of logarithms, , 12, = log, = log 0.48, 25, , =, , log 52 − log 53 + 21 log 54, 3 log5, , =, , 2 log5 − 3 log 5 + 42 log 5 1 log5 1, =, =, 3 log5, 3 log5 3, , Problem 19. Solve the equation:, log(x − 1) + log(x + 8) = 2 log(x + 2)., LHS = log (x − 1) + log(x + 8), , Problem 16. Write (a) log30 (b) log 450 in terms, of log 2, log3 and log 5 to any base., , = log (x − 1)(x + 8), from the first law of logarithms, , (a) log 30 = log(2 × 15) = log(2 × 3 × 5), , = log (x 2 + 7x − 8), , = log 2 + log 3 + log 5, by the first law of logarithms, , RHS = 2 log(x + 2) = log (x + 2)2, , (b) log 450 = log(2 × 225) = log(2 × 3 × 75), , from the third law of logarithms, , = log(2 × 3 × 3 × 25), , = log(x 2 + 4x + 4), , = log(2 × 32 × 52), = log2 + log 32 + log 52, by the first law of logarithms, , Hence,, , log(x 2 + 7x − 8) = log (x 2 + 4x + 4), x 2 + 7x − 8 = x 2 + 4x + 4, , i.e. log 450 = log 2 + 2 log 3 + 2 log 5, by the third law of logarithms, , from which,, i.e., , 7x − 8 = 4x + 4, , , √, 4, 8× 5, in terms of, Problem 17. Write log, 81, log 2, log3 and log 5 to any base., , i.e., , 3x = 12, , and, , x=4, , , , , √, 4, √, 8× 5, 4, = log 8 + log 5 − log 81, log, 81, by the first and second, laws of logarithms, , , Problem 20. Solve the equation:, , 1, log 4 = log x., 2, , 1, 1, log 4 = log4 2 from the third law of logarithms, √, 2, = log 4 from the laws of indices, , 23
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24 Higher Engineering Mathematics, 1, log4 = log x, 2, √, log 4 = log x, , Hence,, becomes, , 1, 1, log 8 − log81 + log 27, 3, 2, 1, log 4 − 2 log 3 + log45, 9., 2, 1, 10., log 16 + 2 log3 − log 18, 4, 11. 2 log2 + log 5 − log 10, 8., , log2 = log x, , i.e., , 2=x, , from which,, , i.e. the solution of the equation is: x = 2, Problem, 21., Solve the equation:, , log x 2 − 3 − log x = log2., , x2 − 3, =2, x, , Rearranging gives:, , x 2 − 3 = 2x, , 13., , log 64 + log 32 − log 128, [log16 or log24 or 4 log2], , 14., , log 8 − log4 + log 32, [log64 or log 26 or 6 log2], , 16., , x = −1 is not a valid solution since the logarithm of a, negative number has no real root., Hence, the solution of the equation is: x = 3, Now try the following exercise, Exercise 12, logarithms, , log 27 − log9 + log 81, [log 243 or log 35 or 5 log3], , 15., , x = 3 or x = −1, , from which,, , [log 2], , 12., , (x − 3)(x + 1) = 0, , Factorizing gives:, , [log 1 = 0], , Evaluate the expressions given in Problems 15, and 16:, , x 2 − 2x − 3 = 0, , and, , [log 10], , Simplify the expressions given in Problems 12, to 14:, , 2, , , , x −3, log x 2 − 3 − log x = log, x, from the second law of logarithms, 2, , x −3, Hence,, = log 2, log, x, from which,, , [log 6], , Further problems on laws of, , In Problems 1 to 11, write as the logarithm of a, single number:, , 1, 1, 2 log 16 − 3 log 8, , log 4, log 9 − log3 + 12 log 81, 2 log3, , [0.5], [1.5], , Solve the equations given in Problems 17 to 22:, 17., , log x 4 − log x 3 = log5x − log 2x, , 18., , log 2t 3 − log t = log 16 + logt, , 19., , 2 logb 2 − 3 logb, , 20., , log (x + 1) + log(x − 1) = log 3, , 21., 22., , = log8b − log 4b, , 1, log 27 = log(0.5a), 3 , , log x 2 − 5 − log x = log 4, , [x = 2.5], [t = 8], [b = 2], [x = 2], [a = 6], [x = 5], , 1., , log 2 + log 3, , [log 6], , 2., , log 3 + log 5, , [log 15], , 3., , log 3 + log 4 − log 6, , [log 2], , 4., , log 7 + log 21 − log49, , [log 3], , 5., , 2 log 2 + log 3, , 6., , 2 log 2 + 3 log5, , [log 500], , The laws of logarithms may be used to solve certain, equations involving powers—called indicial equations. For example, to solve, say, 3 x = 27, logarithms to a base of 10 are taken of both sides,, , 7., , 1, 2 log 5 − log 81 + log 36, 2, , [log 100], , i.e. log10 3x = log10 27, , [log 12], , 3.3, , Indicial equations, , and x log10 3 = log10 27, by the third law of logarithms
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Logarithms, Rearranging gives, x=, , log10 27 1.43136 . . ., =, =3, log10 3, 0.4771 . . ., , which may be readily checked, , , , , 8, log8, is not equal to lg, Note,, log2, 2, , log10 41.15, = 0.50449, 3.2, Thus x = antilog 0.50449 =100.50449 = 3.195 correct to, 4 significant figures., Hence log10 x =, , Now try the following exercise, Exercise 13, , Problem 22. Solve the equation 2 x = 3, correct to, 4 significant figures., Taking logarithms to base 10 of both sides of 2 x = 3, gives:, log10 2x = log10 3, i.e., , x log10 2 = log10 3, log10 3 0.47712125 . . ., =, x=, log10 2 0.30102999 . . ., = 1.585, correct to 4 significant figures, , Indicial equations, , Solve the following indicial equations for x, each, correct to 4 significant figures:, 1. 3x = 6.4, , [1.690], , 2. 2 x = 9, , [3.170], , 3. 2 x−1 = 32x−1, , [6.058], , 5. 25.28 =4.2x, , [2.251], , 6. 42x−1 = 5x+2, , [3.959], , 7., , x −0.25 = 0.792, , 8. 0.027x = 3.26, equation 2 x+1 = 32x−5, , Problem 23. Solve the, correct to 2 decimal places., , Taking logarithms to base 10 of both sides gives:, , [−0.3272], , where P1 is the power input and P2 is the, P2, power output. Find the power gain, when, P1, n =25 decibels., [316.2], , (x + 1) log10 2 = (2x − 5) log10 3, x log10 2 + log10 2 = 2x log10 3 − 5 log10 3, , x(0.3010) + (0.3010) = 2x(0.4771) − 5(0.4771), i.e., , [2.542], , 9. The decibel gain n of an amplifier is given by:, , P2, n = 10 log10, P1, , log10 2x+1 = log10 32x−5, i.e., , [0.2696], , x 1.5 = 14.91, , 4., , Rearranging gives:, , 25, , 0.3010x + 0.3010 = 0.9542x − 2.3855, , Hence, , 3.4, 2.3855 + 0.3010 = 0.9542x − 0.3010x, 2.6865 = 0.6532x, , from which x =, , 2.6865, = 4.11, correct to, 0.6532, 2 decimal places, , Problem 24. Solve the equation x 3.2 = 41.15,, correct to 4 significant figures., Taking logarithms to base 10 of both sides gives:, log10 x 3.2 = log10 41.15, 3.2 log10 x = log10 41.15, , Graphs of logarithmic functions, , A graph of y = log10 x is shown in Fig. 3.1 and a graph, of y = loge x is shown in Fig. 3.2. Both are seen to be, of similar shape; in fact, the same general shape occurs, for a logarithm to any base., In general, with a logarithm to any base a, it is noted, that:, (i) loga1 = 0, Let loga = x, then a x = 1 from the definition of, the logarithm., If a x = 1 then x = 0 from the laws of indices., Hence loga 1 =0. In the above graphs it is seen, that log10 1 = 0 and loge 1 = 0
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26 Higher Engineering Mathematics, y, 2, , y, , 1.0, 1, , 0.5, 0, , 0, , 1, x, y 5 log10x, , 20.5, , 2, 3, , 3, 2, , 1, , 21, , x, 0.5, , 0.2, , 0.1, , 0.48 0.30 0 2 0.30 2 0.70 2 1.0, , 1, , 2, , 3, , 4, , 5, , 6, , x, , x, 6, 5, 4, 3, 2 1 0.5, 0.2, 0.1, y 5 loge x 1.79 1.61 1.39 1.10 0.69 0 20.69 21.61 22.30, , 22, , Figure 3.2, , 21.0, , Figure 3.1, , (ii) logaa = 1, Let loga a = x then a x = a from the definition of, a logarithm., If a x = a then x = 1., Hence loga a = 1. (Check with a calculator that, log10 10 = 1 and loge e = 1), , (iii) loga0 → −∞, Let loga 0 = x then a x = 0 from the definition of, a logarithm., If a x = 0, and a is a positive real number,, then x must approach minus infinity. (For, example, check with a calculator, 2−2 = 0.25,, 2−20 = 9.54 × 10−7, 2−200 = 6.22 × 10−61, and, so on), Hence loga 0 → −∞
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Chapter 4, , Exponential functions, 4.1 Introduction to exponential, functions, An exponential function is one which contains ex , e, being a constant called the exponent and having an, approximate value of 2.7183. The exponent arises from, the natural laws of growth and decay and is used as a, base for natural or Napierian logarithms., The most common method of evaluating an exponential function is by using a scientific notation calculator. Use your calculator to check the following, values:, e = 2.7182818, correct to 8 significant figures,, 1, , e−1.618 = 0.1982949, each correct to 7 significant, figures,, e0.12 = 1.1275, correct to 5 significant figures,, e−1.47 = 0.22993, correct to 5 decimal places,, e−0.431 = 0.6499, correct to 4 decimal places,, e, , 9.32, , , , 0.0256 e5.21 − e2.49 = 0.0256 (183.094058 . . ., − 12.0612761 . . .), = 4.3784, correct to 4, decimal places., Problem 2. Evaluate the following correct to 4, decimal places, using a calculator:, , , e0.25 − e−0.25, 5 0.25, e, + e−0.25, , 5, , e0.25 − e−0.25, e0.25 + e−0.25, , , , , , 1.28402541 . . . − 0.77880078 . . ., 1.28402541 . . . + 0.77880078 . . ., , , 0.5052246 . . ., =5, 2.0628262 . . ., , , , =5, , = 1.2246, correct to 4 decimal places., , = 11159, correct to 5 significant figures,, , e−2.785 = 0.0617291, correct to 7 decimal places., , Problem 1. Evaluate the following correct to 4, decimal places, using a calculator:, , , 0.0256 e5.21 − e2.49, , Problem 3. The instantaneous voltage v in a, capacitive circuit is related to time t by the, equation: v = V e−t /CR where V , C and R are, constants. Determine v, correct to 4 significant, figures, when t = 50 ms, C = 10 μF, R = 47 k, and V = 300 volts., v = V e−t /CR = 300e(−50×10, , −3)/(10×10−6 ×47×103)
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28 Higher Engineering Mathematics, Using a calculator,, v = 300e−0.1063829 ... = 300(0.89908025 . . .), = 269.7 volts, Now try the following exercise, Exercise 14 Further problems on, evaluating exponential functions, , (where 3! = 3 ×2 × 1 and is called ‘factorial 3’), The series is valid for all values of x., The series is said to converge, i.e. if all the terms are, added, an actual value for e x (where x is a real number), is obtained. The more terms that are taken, the closer, will be the value of ex to its actual value. The value of, the exponent e, correct to say 4 decimal places, may be, determined by substituting x = 1 in the power series of, equation (1). Thus,, e1 = 1 + 1 +, , 1. Evaluate the following, correct to 4 significant, figures: (a) e−1.8 (b) e−0.78 (c) e10, [(a) 0.1653 (b) 0.4584 (c) 22030], , +, , 2. Evaluate the following, correct to 5 significant, figures:, (a) e1.629 (b) e−2.7483 (c) 0.62e4.178, [(a) 5.0988 (b) 0.064037 (c) 40.446], In Problems 3 and 4, evaluate correct to 5 decimal, places:, 5e2.6921, 1, 3. (a) e3.4629 (b) 8.52e−1.2651 (c) 1.1171, 7, 3e, [(a) 4.55848 (b) 2.40444 (c) 8.05124], 5.6823, e2.1127 − e−2.1127, (b), e−2.1347, 2, −1.7295, − 1), 4(e, (c), e3.6817, [(a) 48.04106 (b) 4.07482 (c) −0.08286], , 4. (a), , 5. The length of a bar, l, at a temperature θ, is given by l = l0 eαθ , where l0 and α are, constants. Evaluate 1, correct to 4 significant figures, where l0 = 2.587, θ = 321.7 and, [2.739], α = 1.771 × 10−4., 6. When a chain of length 2L is suspended from, two points, 2D metres, hor apart,, √on2the2 same, L+ L +k, . Evalizontal level: D = k ln, k, uate D when k = 75 m and L = 180 m., [120.7m], , 4.2, , The power series for ex, , The value of e x can be calculated to any required degree, of accuracy since it is defined in terms of the following, power series:, ex = 1 + x +, , x2 x3 x4, +, +, +···, 2! 3!, 4!, , (1)2 (1)3 (1)4 (1)5, +, +, +, 2!, 3!, 4!, 5!, , (1)6 (1)7 (1)8, +, +, +···, 6!, 7!, 8!, , = 1 + 1 + 0.5 + 0.16667 + 0.04167, + 0.00833 + 0.00139 + 0.00020, + 0.00002 + · · ·, i.e., , e = 2.71828 = 2.7183,, correct to 4 decimal places, , The value of e0.05, correct to say 8 significant figures,, is found by substituting x = 0.05 in the power series for, e x . Thus, e0.05 = 1 + 0.05 +, , (0.05)2 (0.05)3, +, 2!, 3!, , (0.05)4 (0.05)5, +, +···, 4!, 5!, = 1 + 0.05 + 0.00125 + 0.000020833, +, , + 0.000000260 + 0.000000003, and by adding,, e0.05 = 1.0512711, correct to 8 significant figures, In this example, successive terms in the series grow, smaller very rapidly and it is relatively easy to determine the value of e0.05 to a high degree of accuracy., However, when x is nearer to unity or larger than unity,, a very large number of terms are required for an accurate, result., If in the series of equation (1), x is replaced by −x, then,, e−x = 1 + (−x) +, i.e. e−x = 1 − x +, , (−x)2 (−x)3, +, +···, 2!, 3!, , x2 x3, −, +···, 2! 3!, , In a similar manner the power series for e x may be used, to evaluate any exponential function of the form a ekx ,
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Exponential functions, where a and k are constants. In the series of equation (1),, let x be replaced by kx. Then,, , , (kx)2 (kx)3, kx, a e = a 1 + (kx) +, +, +···, 2!, 3!, , , (2x)2 (2x)3, 2x, Thus 5 e = 5 1 + (2x) +, +, +···, 2!, 3!, , , 4x 2 8x 3, = 5 1 + 2x +, +, +···, 2, 6, , , 4, i.e. 5 e2x = 5 1 + 2x + 2x 2 + x 3 + · · ·, 3, Problem 4. Determine the value of 5 e0.5 , correct, to 5 significant figures by using the power series, for ex ., ex = 1 + x +, Hence, , x2, 2!, , +, , x3, 3!, , +, , x4, 4!, , +···, , (0.5)2, (0.5)3, e0.5 = 1 + 0.5 +, +, (2)(1) (3)(2)(1), (0.5)4, , (0.5)5, , +, (4)(3)(2)(1) (5)(4)(3)(2)(1), (0.5)6, +, (6)(5)(4)(3)(2)(1), , +, , = 1 + 0.5 + 0.125 + 0.020833, + 0.0026042 + 0.0002604, + 0.0000217, i.e., , e0.5, , = 1.64872,, correct to 6 significant figures, , Hence 5e0.5 = 5(1.64872) = 8.2436,, correct to 5 significant figures, Problem 5. Expand ex (x 2 − 1) as far as the term, in x 5 ., The power series for ex is,, ex = 1 + x +, , x2 x3 x4 x5, +, +, +, +···, 2! 3!, 4! 5!, , Hence e x (x 2 − 1), , , x2 x3 x4 x5, = 1+x +, +, +, +, + · · · (x 2 − 1), 2!, 3! 4!, 5!, , , =, , x2 + x3 +, , x4 x5, +, +···, 2!, 3!, , 29, , , , , , x2 x3 x4 x5, +, +, +, +···, − 1+x +, 2!, 3! 4!, 5!, , , , Grouping like terms gives:, ex (x 2 − 1), , , , , x2, x3, = −1 − x + x 2 −, + x3 −, 2!, 3!, , , 4, , x, x4, x5 x5, +···, +, −, +, −, 2! 4!, 3!, 5!, , 1, 5, 11, 19 5, = − 1 −x + x2 + x3 + x4 +, x, 2, 6, 24, 120, when expanded as far as the term in x 5 ., Now try the following exercise, Exercise 15, series for ex, , Further problems on the power, , 1. Evaluate 5.6 e−1 , correct to 4 decimal places,, [2.0601], using the power series for e x ., 2. Use the power series for ex to determine, correct to 4 significant figures, (a) e2 (b) e−0.3 and, check your result by using a calculator., [(a) 7.389 (b) 0.7408], 3. Expand (1 − 2x) e2x as far as the term in x 4 ., , 8x 3, 1 − 2x 2 −, − 2x 4, 3, 2 1 , 4. Expand 2 ex, x 2 to six terms., ⎤, ⎡ 1, 5, 9, 1 13, 2x 2 + 2x 2 + x 2 + x 2, ⎥, ⎢, 3, ⎥, ⎢, ⎦, ⎣, 17, 21, 1, 1, + x2 + x2, 12, 60, , 4.3, , Graphs of exponential functions, , Values of ex and e−x obtained from a calculator,, correct to 2 decimal places, over a range x = −3, to x = 3, are shown in the following table.
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30 Higher Engineering Mathematics, x, ex, , y, , −3.0 −2.5 −2.0 −1.5 −1.0 −0.5 0, 0.05, , 0.08, , 0.14, , 0.22, , 0.37, , 0.61 1.00, , e−x 20.09 12.18, , 7.39, , 4.48, , 2.72, , 1.65 1.00, , 5, 3.87, , y5 2e0.3x, , 4, 3, , x, , 0.5, , 1.0, , 1.5, , 2.0, , 2.5, , 3.0, , 2, , ex, , 1.65, , 2.72, , 4.48, , 7.39, , 12.18, , 20.09, , 1, , e−x, , 0.61, , 0.37, , 0.22, , 0.14, , 0.08, , 0.05, , 1.6, , Figure 4.1 shows graphs of y = ex and y = e−x, , y 5 ex, 16, , −1.5 −1.0 −0.5, , x, 8, , −2x, e−2x, , 4, , 22, , 21, , 0, , 1, , 2, 2.2, , 3, , x, , A table of values is drawn up as shown below., , 12, , 23, , 21, 0, 20.74, , Problem 7. Plot a graph of y = 13 e−2x over the, range x = −1.5 to x = 1.5. Determine from the, graph the value of y when x = −1.2 and the value, of x when y = 1.4., , 20, y, , 22, , Figure 4.2, , y, , 5 e2x, , 23, , 1, , 2, , 3, , 3, , 2, , 0.5, , 1.0, , 1.5, , 0, , −1, , −2, , −3, , 20.086 7.389 2.718 1.00 0.368 0.135 0.050, , 1 −2x, 6.70, e, 3, , x, , 1, , 0, , 2.46 0.91 0.33 0.12 0.05 0.02, , A graph of 13 e−2x is shown in Fig. 4.3., Figure 4.1, , y, 1 e22x, , y 53, , 7, 6, , Problem 6. Plot a graph of y = 2 e0.3x over a, range of x = − 2 to x = 3. Hence determine the value, of y when x = 2.2 and the value of x when y = 1.6., , 5, 4, , 3.67, , 3, 2, , A table of values is drawn up as shown below., , 1.4, , 1, , x, , −3, , −2, , −1, , 0, , 1, , 2, , 3, , 21.5 21.0 20.5, , 0.5, , 1.0, , 1.5, , x, , 21.2 20.72, , 0.3x, , −0.9 −0.6 −0.3, , e0.3x, , 0.407 0.549 0.741 1.000 1.350 1.822 2.460, , 0, , 0.3, , 0.6, , 0.9, , 2 e0.3x 0.81 1.10 1.48 2.00 2.70 3.64 4.92, A graph of y = 2 e0.3x is shown plotted in Fig. 4.2., From the graph, when x = 2.2, y = 3.87 and when, y = 1.6, x = −0.74., , Figure 4.3, , From the graph, when x = −1.2, y = 3.67 and when, y = 1.4, x = −0.72., Problem 8. The decay of voltage, v volts, across, a capacitor at time t seconds is given by, v = 250 e, , −t, 3 ., , Draw a graph showing the natural
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Exponential functions, decay curve over the first 6 seconds. From the, graph, find (a) the voltage after 3.4 s, and (b) the, time when the voltage is 150 V., , of y when x = 1.4 and the value of x when, y = 4.5., [3.95, 2.05], 2. Plot a graph of y = 12 e−1.5x over a range, x = −1.5 to x = 1.5 and hence determine the, value of y when x = −0.8 and the value of x, when y = 3.5., [1.65, −1.30], , A table of values is drawn up as shown below., t, , 0, , e, , −t, 3, , −t, v = 250 e 3, , 2, , 3, , 1.00, , 0.7165 0.5134 0.3679, , 250.0, , 179.1, , t, e, , 1, , −t, 3, , −t, v = 250 e 3, , 128.4, 5, , 6, , 0.2636, , 0.1889, , 0.1353, , 65.90, , 47.22, , 33.83, , The natural decay curve of v = 250 e, Fig. 4.4., , −t, 3, , 3. In a chemical reaction the amount of starting, material C cm3 left after t minutes is given by, C = 40 e−0.006t . Plot a graph of C against t and, determine (a) the concentration C after 1 hour,, and (b) the time taken for the concentration to, decrease by half., [(a) 28 cm3 (b) 116 min], , 91.97, , 4, , is shown in, , 250, t, , y 5 250e2 3, , Voltage v (volts), , 200, , 150, , 4. The rate at which a body cools is given by, θ = 250 e−0.05t where the excess of temperature of a body above its surroundings at, time t minutes is θ ◦ C. Plot a graph showing, the natural decay curve for the first hour of, cooling. Hence determine (a) the temperature, after 25 minutes, and (b) the time when the, temperature is 195◦C., [(a) 70◦C (b) 5 min], , 4.4, , 100, 80, 50, , 0, , 1 1.5 2, 3 3.4 4, Time t(seconds), , 5, , 6, , Figure 4.4, , Napierian logarithms, , Logarithms having a base of ‘e’ are called hyperbolic,, Napierian or natural logarithms and the Napierian, logarithm of x is written as loge x, or more commonly, as ln x. Logarithms were invented by John Napier, a, Scotsman (1550–1617)., The most common method of evaluating a Napierian, logarithm is by a scientific notation calculator. Use your, calculator to check the following values:, , From the graph:, (a), , 31, , when time t = 3.4 s, voltage v = 80 V and, , (b) when voltage v = 150 V, time t = 1.5 s., Now try the following exercise, , ln 4.328 = 1.46510554 . . ., = 1.4651, correct to 4 decimal places, ln 1.812 = 0.59443, correct to 5 significant figures, ln 1 = 0, ln 527 = 6.2672, correct to 5 significant figures, ln 0.17 = −1.772, correct to 4 significant figures, , Exercise 16 Further problems on, exponential graphs, , ln 0.00042 = −7.77526, correct to 6 significant, figures, , 1. Plot a graph of y = 3 e0.2x over the range, x = −3 to x = 3. Hence determine the value, , ln e3 = 3, ln e1 = 1
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32 Higher Engineering Mathematics, From the last two examples we can conclude that:, loge ex = x, This is useful when solving equations involving exponential functions. For example, to solve e3x = 7, take, Napierian logarithms of both sides, which gives:, ln e3x = ln 7, i.e., from which, , 3x = ln 7, 1, x = ln 7 = 0.6486, correct to 4, 3, decimal places., , Problem 9. Evaluate the following, each correct, to 5 significant figures:, (a), (a), , (b), , (c), , 1, ln 7.8693, 3.17 ln 24.07, ., ln 4.7291 (b), (c), 2, 7.8693, e−0.1762, , 1, 1, ln 4.7291 = (1.5537349 . . .) = 0.77687,, 2, 2, correct to 5 significant figures, ln 7.8693 2.06296911 . . ., =, = 0.26215,, 7.8693, 7.8693, correct to 5 significant figures, 3.17 ln 24.07 3.17(3.18096625 . . .), =, e−0.1762, 0.83845027 . . ., = 12.027,, correct to 5 significant figures., , Problem 10., (a), , (a), , (b), , Evaluate the following:, , ln e2.5, 5e2.23 lg 2.23, (b), (correct to 3, 0.5, ln 2.23, lg 10, decimal places)., , , t, Problem 12. Given 32 = 70 1 − e− 2 determine, the value of t , correct to 3 significant figures., t, , Rearranging 32 = 70(1 − e− 2 ) gives:, t, 32, = 1 − e− 2, 70, t, 32 38, and, e− 2 = 1 −, =, 70 70, Taking the reciprocal of both sides gives:, t, 70, e2 =, 38, Taking Napierian logarithms of both sides gives:, , t, 70, ln e 2 = ln, 38, , t, 70, i.e., = ln, 2, 38, , 70, from which, t = 2 ln, = 1.22, correct to 3 signifi38, cant figures., , Problem 13., , ln e2.5, 2.5, =, =5, lg 100.5 0.5, , , , 4.87, Solve the equation: 2.68 = ln, x, , to find x., , 5e2.23 lg 2.23, ln 2.23, 5(9.29986607 . . .)(0.34830486 . . .), =, 0.80200158 . . ., = 20.194, correct to 3 decimal places., , Problem 11. Solve the equation: 9 = 4e−3x to, find x, correct to 4 significant figures., Rearranging 9 = 4e−3x gives:, , Taking the reciprocal of both sides gives:, 4, 1, = e3x, =, 9 e−3x, Taking Napierian logarithms of both sides gives:, , 4, = ln(e3x ), ln, 9, , 4, α, = 3x, Since loge e = α, then ln, 9, , 1, 4, 1, Hence, x = ln, = (−0.81093) = −0.2703,, 3, 9, 3, correct to 4 significant figures., , 9, = e−3x, 4, , From thedefinition, of a logarithm, since, 4.87, 4.87, then e2.68 =, 2.68 = ln, x, x, 4.87, Rearranging gives:, x = 2.68 = 4.87e−2.68, e, i.e., x = 0.3339, correct to 4, significant figures., 7, Problem 14. Solve = e3x correct to 4 signi4, ficant figures.
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Exponential functions, Taking natural logs of both sides gives:, , Since ln e = 1, , ln, , 7, = ln e3x, 4, , ln, , 7, = 3x ln e, 4, , ln, , 7, = 3x, 4, , −1 ±, , , , 12 − 4(1)(−10.953), 2, √, −1 ± 44.812 −1 ± 6.6942, =, =, 2, 2, , x = 2.847 or −3.8471, , i.e., , x = 0.1865, correct to 4, significant figures., , i.e., , Using the quadratic formula,, x=, , 0.55962 = 3x, , i.e., , Problem 15. Solve: e x−1 = 2e3x−4 correct to 4, significant figures., , x = −3.8471 is not valid since the logarithm of a, negative number has no real root., Hence, the solution of the equation is: x = 2.847, Now try the following exercise, , Taking natural logarithms of both sides gives:, , , , , ln e x−1 = ln 2e3x−4, , Exercise 17 Further problems on, evaluating Napierian logarithms, , and by the first law of logarithms,, , , , , ln e x−1 = ln 2 + ln e3x−4, , In Problems 1 and 2, evaluate correct to 5 significant figures:, , x − 1 = ln 2 + 3x − 4, , i.e., , Rearranging gives: 4 − 1 − ln 2 = 3x − x, 3 − ln 2 = 2x, , i.e., , 3 − ln 2, 2, = 1.153, , x=, , from which,, , Problem 16. Solve, correct to 4 significant, figures: ln(x − 2)2 = ln(x − 2) − ln(x + 3) + 1.6, Rearranging gives:, ln(x − 2)2 − ln(x − 2) + ln(x + 3) = 1.6, , 5e−0.1629, 1.786 ln e1.76, (b), lg 101.41, 2 ln 0.00165, ln 4.8629 − ln 2.4711, (c), 5.173, [(a) 2.2293 (b) −0.33154 (c) 0.13087], , 2. (a), , , , 3. ln x = 2.10, , [8.166], , 4. 24 + e2x = 45, , [1.522], , 5. 5 =, , and by the laws of logarithms,, , Cancelling gives:, , 1, ln 82.473, ln 5.2932 (b), 3, 4.829, 5.62 ln 321.62, (c), e1.2942, [(a) 0.55547 (b) 0.91374 (c) 8.8941], , 1. (a), , In Problems 3 to 7 solve the given equations, each, correct to 4 significant figures., , ln, , 33, , , (x − 2)2 (x + 3), = 1.6, (x − 2), , ln {(x − 2)(x + 3)} = 1.6, , and, , (x − 2)(x + 3) = e1.6, , i.e., , x 2 + x − 6 = e1.6, , or, , x 2 + x − 6 − e1.6 = 0, , i.e., , x 2 + x − 10.953 = 0, , e x+1 − 7, , 6. 1.5 = 4e2t, 7. 7.83 =, , 2.91e−1.7x, , , , t, −2, 8. 16 = 24 1 − e, x , 9. 5.17 = ln, 4.64, , , 1.59, = 2.43, 10. 3.72 ln, x, , [1.485], [−0.4904], [−0.5822], [2.197], [816.2], [0.8274]
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34 Higher Engineering Mathematics, , 11., 12., , y, , , , −x, 5 = 8 1−e 2, , [1.962], , ln(x + 3) − ln x = ln(x − 1), − 1)2, , − ln 3 = ln(x − 1), , y 5 Ae2kx, , [3], [4], , 13., , ln(x, , 14., , ln(x + 3) + 2 = 12 − ln(x − 2), , [147.9], , 15., , e(x+1), , [4.901], , 16., , ln(x + 1)2 = 1.5 − ln(x − 2), + ln(x + 1), , 17., , 18., , =, , 3e(2x−5), , 19., , If U2 = U1 e, , W, PV, , formula., 20., , A, , y 5 A(12e2kx ), , 0, , makeW the subject, of the, , U2, W = PV ln, U1, , Laws of growth and decay, , The laws of exponential growth and decay are of the, form y = A e−kx and y = A(1 − e−kx ), where A and k are, constants. When plotted, the form of each of these equations is as shown in Fig. 4.5. The laws occur frequently, in engineering and science and examples of quantities, related by a natural law include., l = l0 eαθ, , (ii) Change in electrical resistance, with temperature, Rθ = R0 eαθ, (iii) Tension in belts, , y, , , , The work done in an isothermal expansion of, a gas from pressure p1 to p2 is given by:, , p1, w = w0 ln, p2, , (i) Linear expansion, , x, , [3.095], , If the initial pressure p1 = 7.0 kPa, calculate, the final pressure p2 if w = 3 w0 ., [ p2 = 348.5 Pa], , 4.5, , 0, (a), , Transpose: b = ln t − a ln D to make t the, subject., a, [t = eb+a ln D = eb ea ln D = eb eln D, i.e. t = eb D a ], , R1, P, = 10 log10, find the value of R1, If, Q, R2, when P = 160, Q = 8 and R2 = 5., [500], , , A, , Figure 4.5, , (v) Biological growth, , y = y0 ekt, , (vi) Discharge of a capacitor q = Q e−t/CR, (vii) Atmospheric pressure, , p = p0 e−h/c, , (viii) Radioactive decay, , N = N0 e−λt, , (ix) Decay of current in an, inductive circuit, , i = I e− Rt /L, , (x) Growth of current in a, capacitive circuit, , i = I (1 − e−t/CR ), , Problem 17. The resistance R of an electrical, conductor at temperature θ ◦ C is given by, R = R0 eαθ , where α is a constant and, R0 = 5 × 103 ohms. Determine the value of α,, correct to 4 significant figures, when, R = 6 ×103 ohms and θ = 1500◦C. Also, find the, temperature, correct to the nearest degree, when the, resistance R is 5.4 ×103 ohms., R, = eαθ ., R0, Taking Napierian logarithms of both sides gives:, , Transposing R = R0 eαθ gives, , T1 = T0 eμθ, , (iv) Newton’s law of cooling θ = θ0 e−kt, , x, , (b), , ln, , R, = ln eαθ = αθ, R0
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Exponential functions, , , 6 × 103, 1, 1, R, Hence α = ln, =, ln, θ R0, 1500, 5 × 103, =, , 1, (0.1823215 . . .), 1500, , = 1.215477 · · ·× 10−4, Hence α = 1.215 × 10−4 ,, correct to 4 significant figures., R, = αθ, From above, ln, R0, θ=, , hence, , 1, R, ln, α R0, , When R = 5.4 × 103, α = 1.215477 . . . × 10−4 and, R0 = 5 ×103, , , 5.4 × 103, 1, θ=, ln, 1.215477 . . . × 10−4, 5 × 103, =, , 104, (7.696104 . . . × 10−2), 1.215477 . . ., ◦, , = 633 C, correct to the nearest degree., , 35, , Problem 19. The current i amperes flowing in a, capacitor at time t seconds is given by, −t, , i = 8.0(1 − e CR ), where the circuit resistance R is, 25 ×103 ohms and capacitance C is, 16 ×10−6 farads. Determine (a) the current i after, 0.5 seconds and (b) the time, to the nearest, millisecond, for the current to reach 6.0 A. Sketch, the graph of current against time., (a), , −t, , Current i = 8.0(1 − e CR ), −0.5, , = 8.0[1 − e (16 ×10−6 )(25 ×103 ) ] =8.0(1 − e−1.25), = 8.0(1 − 0.2865047 . . .) = 8.0(0.7134952 . . .), = 5.71 amperes, −t, , (b) Transposing i = 8.0(1 − e CR ), , gives, , −t, i, = 1 −e CR, 8.0, −t, , from which, e CR = 1 −, , i, 8.0 − i, =, 8.0, 8.0, , Taking the reciprocal of both sides gives:, Problem 18. In an experiment involving, Newton’s law of cooling, the temperature θ(◦ C) is, given by θ = θ0 e−kt . Find the value of constant k, when θ0 = 56.6◦ C, θ = 16.5◦ C and t = 83.0 seconds., Transposing, , θ = θ0 e−kt gives, θ, = e−kt, θ0, , from which, , θ0, 1, = −kt = ekt, θ, e, , Taking Napierian logarithms of both sides gives:, θ0, ln = kt, θ, from which,, , , 56.6, 1, 1 θ0, ln, k = ln =, t, θ, 83.0, 16.5, 1, =, (1.2326486 . . .), 83.0, Hence k = 1.485 × 10−2, , t, , e CR =, , 8.0, 8.0 − i, , Taking Napierian logarithms of both sides gives:, , , t, 8.0, = ln, CR, 8.0 − i, Hence, , , , 8.0, t = CRln, 8.0 − i, , , , = (16 × 10−6)(25 × 103 ) ln, , , , 8.0, 8.0 − 6.0, , , , when i = 6.0 amperes,, , , 8.0, 400, i.e. t = 3 ln, = 0.4 ln 4.0, 10, 2.0, = 0.4(1.3862943 . . .) = 0.5545 s, = 555 ms, to the nearest millisecond., A graph of current against time is shown in Fig. 4.6.
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36 Higher Engineering Mathematics, i (A), , Hence the time for the temperature θ 2 to be, one half of the value of θ 1 is 41.6 s, correct to 1, decimal place., , 8, 6, 5.71, , i 5 8.0 (12e2t/CR), , 4, , Now try the following exercise, , 2, , Exercise 18 Further problems on the laws, of growth and decay, , 0, , 0.5, 0.555, , 1.0, , 1.5, , t(s), , Figure 4.6, , Problem 20. The temperature θ2 of a winding, which is being heated electrically at time t is given, −t, θ2 = θ1 (1 − e τ ), , by:, where θ1 is the temperature (in, degrees Celsius) at time t = 0 and τ is a constant., Calculate,, (a), , θ1 , correct to the nearest degree, when θ2 is, 50◦ C, t is 30 s and τ is 60 s, , (b) the time t , correct to 1 decimal place, for θ2 to, be half the value of θ1 ., (a) Transposing the formula to make θ1 the subject, gives:, θ1 =, , θ2, , −t, (1 − e T ), , =, , 50, 1−e, , −30, 60, , 50, 50, =, =, 1 − e−0.5 0.393469 . . ., i.e. θ 1 = 127◦ C, correct to the nearest degree., (b) Transposing to make t the subject of the formula, gives:, −t, θ2, =1−e τ, θ1, −t, θ2, from which, e τ = 1 −, θ, , 1, t, θ2, Hence, − = ln 1 −, τ, θ, 1, , θ2, i.e., t = −τ ln 1 −, θ1, 1, Since, θ2 = θ1, 2, , , 1, t = −60 ln 1 −, 2, = −60 ln 0.5 = 41.59 s, , 1. The temperature, T ◦C, of a cooling object, varies with time, t minutes, according to the, equation: T = 150e−0.04t . Determine the temperature when (a) t = 0, (b) t = 10 minutes., [(a) 150◦ C (b) 100.5◦ C ], 2. The pressure p pascals at height h metres, −h, , above ground level is given by p = p0 e C ,, where p0 is the pressure at ground level, and C is a constant. Find pressure p when, p0 = 1.012 × 105 Pa, height h = 1420 m, and, C = 71500., [99210], 3. The voltage drop, v volts, across an inductor L henrys at time t seconds is given, − Rt, , by v = 200 e L , where R = 150 and, L =12.5 × 10−3 H. Determine (a) the voltage, when t = 160 ×10−6 s, and (b) the time for the, voltage to reach 85 V., [(a) 29.32 volts (b) 71.31 × 10−6 s], 4. The length l metres of a metal bar at temperature t ◦ C is given by l = l0 eαt , where, l0 and α are constants. Determine (a) the, value of α when l = 1.993 m, l0 = 1.894 m, and t = 250◦C, and (b) the value of l0 when, l = 2.416, t = 310◦C and α = 1.682 ×10−4., [(a) 2.038 × 10−4 (b) 2.293 m], 5. The temperature θ2◦ C of an electrical conductor at time t seconds is given by:, θ2 = θ1 (1 − e−t / T ), where θ1 is the initial, temperature and T seconds is a constant., Determine:, (a) θ2 when θ1 = 159.9◦C, t = 30 s and, T = 80 s, and, (b) the time t for θ2 to fall to half the value, of θ1 if T remains at 80 s., [(a) 50◦ C (b) 55.45 s ], 6. A belt is in contact with a pulley for a, sector of θ = 1.12 radians and the coefficient
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Exponential functions, , of friction between these two surfaces is, μ = 0.26. Determine the tension on the taut, side of the belt, T newtons, when tension, on the slack side T0 = 22.7 newtons, given, that these quantities are related by the law, T = T0 eμθ .Determine also the value of θ when, T = 28.0 newtons., [30.4 N, 0.807 rad], 7. The instantaneous current i at time t is given, −t, , by: i = 10 e CR when a capacitor is being, charged. The capacitance C is 7 ×10−6 farads, and the resistance R is 0.3 × 106 ohms. Determine:, (a) the instantaneous current when t is, 2.5 seconds, and, (b) the time for the instantaneous current to, fall to 5 amperes, Sketch a curve of current against time from, t = 0 to t = 6 seconds., [(a) 3.04 A (b) 1.46 s], 8. The amount of product x (in mol/cm3) found in, a chemical reaction starting with 2.5 mol/cm3, of reactant is given by x = 2.5(1 − e−4t ) where, t is the time, in minutes, to form product x. Plot, a graph at 30 second intervals up to 2.5 minutes, and determine x after 1 minute., [2.45 mol/cm3], 9. The current i flowing in a capacitor at time t, is given by:, −t, , be determined. This technique is called ‘determination, of law’., Graph paper is available where the scale markings, along the horizontal and vertical axes are proportional, to the logarithms of the numbers. Such graph paper is, called log-log graph paper., A logarithmic scale is shown in Fig. 4.7 where, the distance between, say 1 and 2, is proportional to, lg 2 − lg 1, i.e. 0.3010 of the total distance from 1 to 10., Similarly, the distance between 7 and 8 is proportional, to lg 8 − lg 7, i.e. 0.05799 of the total distance from 1 to, 10. Thus the distance between markings progressively, decreases as the numbers increase from 1 to 10., 1, , 2, , 3, , 4, , 5, , 6 7 8 910, , Figure 4.7, , With log-log graph paper the scale markings are from, 1 to 9, and this pattern can be repeated several times. The, number of times the pattern of markings is repeated on, an axis signifies the number of cycles. When the vertical axis has, say, 3 sets of values from 1 to 9, and the, horizontal axis has, say, 2 sets of values from 1 to 9,, then this log-log graph paper is called ‘log 3 cycle × 2, cycle’. Many different arrangements are available ranging from ‘log 1 cycle × 1 cycle’ through to ‘log 5, cycle × 5 cycle’., To depict a set of values, say, from 0.4 to 161, on an, axis of log-log graph paper, 4 cycles are required, from, 0.1 to 1, 1 to 10, 10 to 100 and 100 to 1000., Graphs of the form y = a ekx, , i = 12.5(1 − e CR ), , Taking logarithms to a base of e of both sides of y = a ekx, gives:, , where resistance R is 30 kilohms and the, capacitance C is 20 micro-farads. Determine:, , ln y = ln(a ekx ) = ln a + ln ekx = ln a + kx ln e, , (a), , the current flowing after 0.5 seconds, and, , (b) the time for the current to reach, 10 amperes., [(a) 7.07 A (b) 0.966 s], , 4.6 Reduction of exponential laws to, linear form, Frequently, the relationship between two variables, say, x and y, is not a linear one, i.e. when x is plotted against, y a curve results. In such cases the non-linear equation, may be modified to the linear form, y = mx + c, so that, the constants, and thus the law relating the variables can, , 37, , i.e. ln y = kx + ln a, , (since ln e = 1), , which compares with Y = m X + c, Thus, by plotting ln y vertically against x horizontally, a straight line results, i.e. the equation y = a ekx is, reduced to linear form. In this case, graph paper having a linear horizontal scale and a logarithmic vertical, scale may be used. This type of graph paper is called, log-linear graph paper, and is specified by the number, of cycles on the logarithmic scale., Problem 21. The data given below is believed to, be related by a law of the form y = a ekx , where a, and b are constants. Verify that the law is true and
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38 Higher Engineering Mathematics, The law of the graph is thus y = 18 e0.55x, , determine approximate values of a and b. Also, determine the value of y when x is 3.8 and the value, of x when y is 85., x −1.2 0.38, y, , 9.3, , 1.2, , 2.5, , 3.4, , 4.2, , When x is 3.8, y = 18 e0.55(3.8) = 18 e2.09, = 18(8.0849) = 146, , 5.3, , When y is 85, 85 = 18 e0.55x, , 22.2 34.8 71.2 117 181 332, , Since y = a ekx then ln y = kx + ln a (from above),, which is of the form Y = m X + c, showing that to produce a straight line graph ln y is plotted vertically against, x horizontally. The value of y ranges from 9.3 to 332, hence ‘log 3 cycle × linear’ graph paper is used. The, plotted co-ordinates are shown in Fig. 4.8 and since a, straight line passes through the points the law y = a ekx, is verified., Gradient of straight line,, k=, , AB, ln 100 − ln 10, 2.3026, =, =, BC, 3.12 − (−1.08), 4.20, , e0.55x =, , and, , 0.55x = ln 4.7222 = 1.5523, x=, , Hence, , Since ln y = kx + ln a, when x = 0, ln y = ln a, i.e. y = a, The vertical axis intercept value at x = 0 is 18, hence, a = 18, 1000, y, , 1.5523, = 2.82, 0.55, , Problem 22. The voltage, v volts, across an, inductor is believed to be related to time, t ms, by, t, , the law v = V e T , where V and T are constants., Experimental results obtained are:, v volts 883, , = 0.55, correct to 2 significant figures., , 85, = 4.7222, 18, , Hence,, , t ms, , 347, , 90, , 55.5 18.6, , 5.2, , 10.4 21.6 37.8 43.6 56.7 72.0, , Show that the law relating voltage and time is as, stated and determine the approximate values of V, and T . Find also the value of voltage after 25 ms, and the time when the voltage is 30.0 V., t, , Since v = V e T then ln v = T1 t + ln V which is of the, form Y = m X + c., Using ‘log3 cycle × linear’ graph paper, the points, are plotted as shown in Fig. 4.9., Since the points are joined by a straight line the law, , y 5a e kx, , 100, , 10, , t, , v = Ve T is verified., Gradient of straight line,, 1, AB, =, T, BC, ln 100 − ln 10, =, 36.5 − 64.2, , A, , B, , C, , =, , 2.3026, −27.7, , Hence T =, , −27.7, 2.3026, , = −12.0, correct to 3 significant figures., 1, 22, , 21, , Figure 4.8, , 0, , 1, , 2, , 3, , 4, , 5, , 6, , x, , Since the straight line does not cross the vertical axis, at t = 0 in Fig. 4.9, the value of V is determined, by selecting any point, say A, having co-ordinates, t, , (36.5,100) and substituting these values into v = V e T .
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Exponential functions, Now try the following exercise, , 1000, , v 5Ve, , Exercise 19 Further problems on reducing, exponential laws to linear form, , t, T, , 1. Atmospheric pressure p is measured at varying altitudes h and the results are as shown, below:, , (36.5, 100), , 100, , A, , Voltage, v volts, , Altitude, h m, , 10, , B, , C, , 1, 0, , 10, , 20, , 30, , 40, 50, Time, t ms, , 60, , 70, , 80, , Figure 4.9, , −36.5, , correct to 3 significant figures., , −t, , Hence the law of the graph is v = 2090 e 12.0 ., When time t = 25 ms,, −25, , v = 2090 e 12.0 = 260 V, −t, , When the voltage is 30.0 volts, 30.0 = 2090 e 12.0 ,, hence, , −t, , e 12.0 =, , 30.0, 2090, , t, , 2090, = 69.67, 30.0, Taking Napierian logarithms gives:, and, , 1500, , 68.42, , 3000, , 61.60, , 5000, , 53.56, , 8000, , 43.41, , Show that the quantities are related by the, law p =a ekh , where a and k are constants., Determine the values of a and k and state, the law. Find also the atmospheric pressure at, 10 000 m., , −5 h, , , 37.74 cm, , 2. At particular times, t minutes, measurements, are made of the temperature, θ ◦ C, of a, cooling liquid and the following results are, obtained:, , e 12.0, = 2090 volts,, , voltage, , 73.39, , p = 76 e−7×10, , 100, , V =, , 500, , a = 76, k = −7 × 10−5,, , 36.5, , Thus 100 = V e −12.0, i.e., , 90, , pressure, p cm, , e 12.0 =, , t, = ln 69.67 = 4.2438, 12.0, from which, time t = (12.0)(4.2438) = 50.9 ms, , Temperature θ ◦ C, , Time t minutes, , 92.2, , 10, , 55.9, , 20, , 33.9, , 30, , 20.6, , 40, , 12.5, , 50, , Prove that the quantities follow a law of the, form θ = θ0 ekt , where θ0 and k are constants,, and determine the approximate value of θ0, and k., [θ0 = 152, k = − 0.05], , 39
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Revision Test 1, This Revision Test covers the material contained in Chapters 1 to 4. The marks for each question are shown in, brackets at the end of each question., 1., , Factorise x 3 + 4x 2 + x − 6 using the factor theorem. Hence solve the equation, x 3 + 4x 2 + x − 6 =0, , 2., , (6), , Use the remainder theorem to find the remainder, when 2x 3 + x 2 − 7x − 6 is divided by, (a) (x − 2) (b) (x + 1), Hence factorise the cubic expression, 6x 2 + 7x − 5, by dividing out, 2x − 1, , 3., , Simplify, , 4., , Resolve the following into partial fractions, (a), (c), , 5., , x − 11, −2, , x2 − x, , (b), , (x 2, , (4), , 3−x, + 3)(x + 3), , x 3 − 6x + 9, x2 + x − 2, , 8., (24), , Evaluate, correct to 3 decimal places,, 5 e−0.982, 3 ln0.0173, , 6., , (7), , (2), , Solve the following equations, each correct to 4, significant figures:, x, , Solve the following equations:, , , (a) log x 2 + 8 − log(2x) = log 3, , (b) ln x + ln(x – 3) = ln 6x – ln(x – 2), (13), , R, U2, 9. If θ f − θi = ln, find the value of U2, J, U1, given that θ f = 3.5, θi = 2.5, R = 0.315, J = 0.4,, (6), U1 = 50, 10., , Solve, correct to 4 significant figures:, (a) 13e2x−1 = 7ex, , (a) ln x = 2.40 (b) 3x−1 = 5x−2, (c) 5 = 8(1 − e− 2 ), , 7. (a) The pressure p at height h above ground level is, given by: p = p0 e−kh where p0 is the pressure, at ground level and k is a constant. When p0, is 101 kilopascals and the pressure at a height, of 1500 m is 100 kilopascals, determine the, value of k., (b) Sketch a graph of p against h ( p the vertical, axis and h the horizontal axis) for values of, height from zero to 12 000 m when p0 is 101, kilopascals., (c) If pressure p = 95 kPa, ground level pressure, p0 = 101 kPa, constant k = 5 × 10−6, determine the height above ground level, h, in, kilometres correct to 2 decimal places., (13), , (10), , (b) ln (x + 1)2 = ln(x + 1) – ln(x + 2) + 2, , (15)
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Chapter 5, , Hyperbolic functions, (v) Hyperbolic secant of x,, , 5.1 Introduction to hyperbolic, functions, , sech x =, , Functions which are associated with the geometry of, the conic section called a hyperbola are called hyperbolic functions. Applications of hyperbolic functions, include transmission line theory and catenary problems., By definition:, (i) Hyperbolic sine of x,, ex − e−x, sinh x =, 2, , 1, 2, =, cosh x e x + e−x, , (5), , ‘sech x’ is pronounced as ‘shec x’, (vi) Hyperbolic cotangent of x,, coth x =, , e x + e−x, 1, = x −x, tanh x e − e, , (6), , ‘coth x’ is pronounced as ‘koth x’, (1), , Some properties of hyperbolic functions, Replacing x by 0 in equation (1) gives:, , ‘sinh x’ is often abbreviated to ‘sh x’ and is, pronounced as ‘shine x’, , sinh 0 =, , (ii) Hyperbolic cosine of x,, e x + e−x, cosh x =, 2, , Replacing x by 0 in equation (2) gives:, (2), , ‘cosh x’ is often abbreviated to ‘ch x’ and is, pronounced as ‘kosh x’, (iii) Hyperbolic tangent of x,, sinh x e x − e−x, =, tanh x =, cosh x e x + e−x, , (3), , ‘tanh x’ is often abbreviated to ‘th x’ and is, pronounced as ‘than x’, (iv) Hyperbolic cosecant of x,, cosech x =, , 1, 2, =, sinh x e x − e−x, , ‘cosech x’ is pronounced as ‘coshec x’, , e0 − e−0, 1−1, =, =0, 2, 2, , (4), , cosh 0 =, , e0 + e−0 1 + 1, =, =1, 2, 2, , If a function of x, f (−x) = − f (x), then f (x) is called, an odd function of x. Replacing x by −x in equation (1), gives:, e−x − e x, e−x − e−(−x), =, 2, 2, , x, −x, e −e, =−, = −sinh x, 2, , sinh(−x) =, , Replacing x by −x in equation (3) gives:, e−x − e−(−x), e−x − e x, =, e−x + e−(−x), e−x + e x, x, , e − e−x, =− x, = −tanh x, e + e−x, , tanh(−x) =
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42 Higher Engineering Mathematics, Hence sinh x and tanh x are both odd functions, , 1, and, (see Section 5.1), as also are cosech x =, sinh x, , , 1, coth x =, tanh x, If a function of x, f (−x) = f (x), then f (x) is, called an even function of x. Replacing x by −x in, equation (2) gives:, e−x + e−(−x), e−x + e x, =, 2, 2, = cosh x, , cosh(−x) =, , Hence cosh xis an evenfunction (see Section 5.2), as, 1, also is sech x =, cosh x, Hyperbolic functions may be evaluated easiest using a, calculator. Many scientific notation calculators actually, possess sinh and cosh functions; however, if a calculator, does not contain these functions, then the definitions, given above may be used., Problem 1. Evaluate sinh 5.4, correct to 4, significant figures., , Problem 3. Evaluate th 0.52, correct to 4, significant figures., Using a calculator with the procedure similar to that, used in Worked Problem 1,, th 0.52 = 0.4777, correct to 4 significant figures., Problem 4. Evaluate cosech 1.4, correct to 4, significant figures., cosech 1.4 =, , 1, sinh 1.4, , Using a calculator,, (i) press hyp, (ii) press 1 and sinh( appears, (iii) type in 1.4, (iv) press ) to close the brackets, (v) press = and 1.904301501 appears, (vi) press x −1, , Using a calculator,, (i) press hyp, (ii) press 1 and sinh( appears, , (vii) press = and 0.5251269293 appears, Hence, cosech 1.4 = 0.5251, correct to 4 significant, figures., , (iii) type in 5.4, (iv) press ) to close the brackets, (v) press = and 110.7009498 appears, Hence, sinh 5.4 = 110.7, correct to 4 significant figures., 1, Alternatively, sinh 5.4 = (e5.4 − e−5.4 ), 2, 1, = (221.406416 . . . − 0.00451658 . . .), 2, 1, = (221.401899 . . .), 2, = 110.7, correct to 4 significant figures., Problem 2. Evaluate cosh 1.86, correct to 3, decimal places., Using a calculator with the procedure similar to that, used in Worked Problem 1,, cosh 1.86 = 3.290, correct to 3 decimal places., , Problem 5. Evaluate sech 0.86, correct to 4, significant figures., sech 0.86 =, , 1, cosh 0.86, , Using a calculator with the procedure similar to that, used in Worked Problem 4,, sech 0.86 = 0.7178, correct to 4 significant figures., Problem 6. Evaluate coth 0.38, correct to 3, decimal places., coth 0.38 =, , 1, tanh 0.38, , Using a calculator with the procedure similar to that, used in Worked Problem 4,, coth 0.38 = 2.757, correct to 3 decimal places.
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Hyperbolic functions, , 43, , y, , Now try the following exercise, , 10, 8, 6, , Exercise 20 Further problems on, evaluating hyperbolic functions, , y 5sinh x, , 4, 2, , In Problems 1 to 6, evaluate correct to 4 significant, figures., , 23 22 21 0 1 2, 22, , 1. (a) sh 0.64 (b) sh 2.182, , 3 x, , 24, 26, , [(a) 0.6846 (b) 4.376], , 28, , 2. (a) ch 0.72 (b) ch 2.4625, , 210, , [(a) 1.271 (b) 5.910], Figure 5.1, , 3. (a) th 0.65 (b) th 1.81, [(a) 0.5717 (b) 0.9478], 4. (a) cosech 0.543 (b) cosech 3.12, [(a) 1.754 (b) 0.08849], 5. (a) sech 0.39 (b) sech 2.367, [(a) 0.9285 (b) 0.1859], , cosh x is an even function (as stated in Section 5.1)., The shape of y = cosh x is that of a heavy rope or chain, hanging freely under gravity and is called a catenary., Examples include transmission lines, a telegraph wire or, a fisherman’s line, and is used in the design of roofs and, arches. Graphs of y = tanh x, y = cosech x, y = sech x, and y = coth x are deduced in Problems 7 and 8., y, , 6. (a) coth 0.444 (b) coth 1.843, [(a) 2.398 (b) 1.051], , 10, , 7. A telegraph wire hangs so that its shape is, x, described by y = 50 ch . Evaluate, correct, 50, to 4 significant figures, the value of y when, x = 25., [56.38], 8. The length l of a heavy cable hanging under, gravity is given by l = 2c sh (L/2c). Find the, value of l when c = 40 and L =30., [30.71], 9., , V 2 = 0.55L tanh (6.3 d/L) is a formula for, velocity V of waves over the bottom of shallow water, where d is the depth and L is the, wavelength. If d = 8.0 and L =96, calculate, the value of V ., [5.042], , 6, 4, 2, 23 22 21 0, , Graphs of hyperbolic functions, , A graph of y = sinhx may be plotted using calculator, values of hyperbolic functions. The curve is shown in, Fig. 5.1. Since the graph is symmetrical about the origin,, sinh x is an odd function (as stated in Section 5.1)., A graph of y = cosh x may be plotted using calculator, values of hyperbolic functions. The curve is shown in, Fig. 5.2. Since the graph is symmetrical about the y-axis,, , 1 2, , 3, , x, , Figure 5.2, , Problem 7. Sketch graphs of (a) y = tanh x, and (b) y = coth x for values of x between, −3 and 3., A table of values is drawn up as shown below, −3, , x, , 5.2, , y 5cosh x, , 8, , sh x, , −10.02, , ch x, , 10.07, , y = th x =, , sh x, ch x, , y = coth x =, , ch x, sh x, , −2, , −1, , −3.63 −1.18, 3.76, , 1.54, , −0.995 −0.97 −0.77, −1.005 −1.04 −1.31
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44 Higher Engineering Mathematics, x, , 0, , 1, , 2, , 3, , sh x, , 0, , 1.18 3.63 10.02, , ch x, , 1, , 1.54 3.76 10.07, , 0, , 0.77 0.97, , A table of values is drawn up as shown below, −4, , x, , sh x, ch x, , 0.995, , cosech x =, , 1, sh x, , ch x, ch x, y = coth x =, sh x, , ±∞ 1.31 1.04, , 1.005, , A graph of y = tanh x is shown in Fig. 5.3(a), , sech x =, , −0.10 −0.28 −0.85, , 27.31, , 10.07, , 3.76, , 1.54, , 0.04, , 0.10, , 0.27, , 0.65, , 3, , 4, , 1, ch x, 0, , (b) A graph of y = coth x is shown in Fig. 5.3(b), , sh x, , 0, , Both graphs are symmetrical about the origin thus tanh x, and coth x are odd functions., , cosech x =, , 1, sh x, , ch x, Problem 8. Sketch graphs of (a) y = cosech x, and (b) y = sech x from x = −4 to x = 4, and, from, the graphs, determine whether they are odd or, even functions., , y 5 tanh x, , y, 1, 23 22 21, , sech x =, , 1, ch x, , 1, , 2, , 1.18 3.63 10.02 27.29, , ±∞ 0.85 0.28, , 0.10, , 2 3, , 1.54 3.76 10.07 27.31, , 1, , 0.65 0.27, , 0.10, , 3, , x, , 2, , y 5 cosech x, , 1, , (a), 232221, , y, , 01 2 3, 21, , y 5 cosech x, , 3, , x, , 22, y 5coth x, , 23, , 1, (a), 23 22 21 0, , 1, , 2 3, , x, , y, , 21, y 5 coth x, , 1, , 22, 23, , 232221 0, (b), , (b), , Figure 5.4, Figure 5.3, , 0.04, , A graph of y = cosech x is shown in Fig. 5.4(a)., The graph is symmetrical about the origin and is, thus an odd function., (b) A graph of y = sech x is shown in Fig. 5.4(b). The, graph is symmetrical about the y-axis and is thus, an even function., (a), , 21, , 2, , 0.04, , 1, , y, 0 1, , −1, , −0.04, , x, , (a), , −2, , −22.29 −10.02 −3.63 −1.18, , sh x, , y = th x =, , −3, , y 5 sech x, 1 2 3, , x
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45, , Hyperbolic functions, 5.3, , Hyperbolic identities, , For every trigonometric identity there is a corresponding hyperbolic identity. Hyperbolic identities may, be proved by either, (i) replacing sh x, , by, , e x + e−x, , e x − e−x, 2, , Problem 9. Prove the hyperbolic identities, (a) ch 2 x − sh2 x = 1 (b) 1 − th2 x = sech2 x, (c) coth 2 x − 1 =cosech2 x., , (a), and ch x, , by, , , or, 2, (ii) by using Osborne’s rule, which states: ‘the six, trigonometric ratios used in trigonometrical identities relating general angles may be replaced by, their corresponding hyperbolic functions, but the, sign of any direct or implied product of two sines, must be changed’., For example, since cos2 x + sin2 x = 1 then, by, Osborne’s rule, ch2 x − sh2 x = 1, i.e. the trigonometric functions have been changed to their corresponding, hyperbolic functions and since sin2 x is a product of two, sines the sign is changed from + to −. Table 5.1 shows, some trigonometric identities and their corresponding, hyperbolic identities., , x, , e − e−x, e x + e−x, +, = ex, ch x + sh x =, 2, 2, x, x, , e + e−x, e − e−x, ch x − sh x =, −, 2, 2, , , = e+−x, (ch x + sh x)(ch x − sh x) = (e x )(e−x ) = e0 = 1, i.e. ch2 x − sh2 x = 1, , (b) Dividing each term in equation (1) by ch2 x, gives:, ch2 x sh2 x, 1, −, = 2, ch2 x ch2 x, ch x, i.e. 1 −th2 x = sech2 x, , Table 5.1, Trigonometric identity, , Corresponding hyperbolic identity, , cos2 x + sin2 x = 1, , ch2 x − sh2 x = 1, , 1 + tan2 x = sec2 x, , 1 −th2 x = sech2 x, , cot 2 x + 1 =cosec 2 x, , coth2 x − 1 = cosech2 x, Compound angle formulae, , sin (A ± B) = sin A cos B ± cos A sin B, , sh (A ± B) = sh A ch B ± ch A sh B, , cos (A ± B) = cos A cos B ∓ sin A sin B, , ch (A ± B) = ch A ch B ± sh A sh B, , tan (A ± B) =, , tan A ± tan B, 1 ∓ tan A tan B, , th (A ± B) =, , th A ± th B, 1 ±th A th B, , Double angles, sin 2x = 2 sin x cos x, , sh 2x = 2 sh x ch x, , cos 2x = cos2 x − sin2 x, , ch 2x =ch2 x + sh2 x, , = 2 cos2 x − 1, , = 2 ch2 x − 1, , = 1 − 2 sin2 x, , = 1 + 2sh2 x, , tan 2x =, , 2 tan x, 1 − tan2 x, , (1), , th 2x =, , 2 th x, 1 + th2 x
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46 Higher Engineering Mathematics, (c), , Dividing each term in equation (1) by sh2 x, gives:, ch2 x sh2 x, 1, −, = 2, sh2 x sh2 x, sh x, , Problem 12., , Show that th2 x + sech2 x = 1., , L.H.S. = th2 x + sech2 x =, , i.e. coth2 x − 1 =cosech2 x, , =, , Problem 10. Prove, using Osborne’s rule, (a) ch 2 A = ch2 A + sh2 A, (b) 1 −th2 x = sech2 x., From trigonometric ratios,, cos 2 A = cos2 A − sin 2 A, , (1), , Osborne’s rule states that trigonometric ratios, may be replaced by their corresponding hyperbolic functions but the sign of any product, of two sines has to be changed. In this case,, sin2 A = (sin A)(sin A), i.e. a product of two sines,, thus the sign of the corresponding hyperbolic function, sh2 A, is changed from + to −. Hence, from, (1), ch 2A = ch2 A + sh2 A, (b) From trigonometric ratios,, 1 + tan2 x, , = sec2 x, , and tan2 x =, , sin2 x, cos2 x, , (2), =, , (sin x)(sin x), cos2 x, , i.e. a product of two sines., Hence, in equation (2), the trigonometric ratios, are changed to their equivalent hyperbolic function and the sign of th2 x changed + to −, i.e., 1 −th2 x = sech2 x, Problem 11., , Prove that 1 + 2 sh2 x = ch 2x., , Left hand side (L.H.S.), , , , 2, e x − e−x, = 1 + 2 sh x = 1 + 2, 2, 2x, , e − 2e x e−x + e−2x, = 1+2, 4, , e2x − 2 + e−2x, 2, 2x, , e + e−2x, 2, =1+, −, 2, 2, =1+, , =, , + e−2x, 2, , sh2 x + 1 ch2 x, = 2 = 1 = R.H.S., ch2 x, ch x, , Problem 13. Given Ae x + Be−x ≡ 4ch x−5 sh x,, determine the values of A and B., Ae x + Be−x ≡ 4 ch x − 5 sh x, x, , x, , e + e−x, e − e−x, −5, =4, 2, 2, 5, 5, = 2e x + 2e−x − e x + e−x, 2, 2, 1, 9, = − e x + e−x, 2, 2, Equating coefficients gives: A = −, , 1, 1, and B = 4, 2, 2, , Problem 14. If 4e x − 3e−x ≡ Psh x + Qch x,, determine the values of P and Q., 4e x − 3e−x ≡ P sh x + Q ch x, x, , x, , e − e−x, e + e−x, +Q, =P, 2, 2, P x P −x Q x Q −x, e − e + e + e, 2, 2, 2, 2, , , , , P+Q x, Q − P −x, e +, e, =, 2, 2, =, , 2, , e2x, , sh2 x + 1, ch2 x, , Since ch2 x − sh2 x = 1 then 1 + sh2 x = ch2 x, Thus, , (a), , 1, sh2 x, + 2, 2, ch x ch x, , = ch 2x = R.H.S., , Equating coefficients gives:, 4=, , P+Q, Q−P, and −3 =, 2, 2, , i.e. P + Q = 8, −P + Q = −6, , (1), (2), , Adding equations (1) and (2) gives: 2Q = 2, i.e. Q = 1, Substituting in equation (1) gives: P = 7.
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Hyperbolic functions, Now try the following exercise, Exercise 21 Further problems on, hyperbolic identities, In Problems 1 to 4, prove the given identities., 1. (a) ch (P − Q) ≡ ch P ch Q − sh P sh Q, (b) ch 2x ≡ ch2 x + sh2 x, 2. (a) coth x ≡ 2 cosech 2x + th x, (b) ch 2θ − 1 ≡2 sh2 θ, th A − th B, 1 −th A th B, (b) sh 2 A ≡ 2 sh A ch A, , 3. (a) th (A − B) ≡, , 4. (a) sh (A + B) ≡ sh A ch B + ch A sh B, (b), , sh2 x + ch2 x − 1, ≡ tanh4 x, 2ch2 x coth2 x, , 5. Given Pe x − Qe−x ≡ 6 ch x − 2 sh x, find P, and Q, [P = 2, Q =−4], 6. If 5e x − 4e−x ≡ A sh x + B ch x, find A and B., [A = 9, B = 1], , 5.4 Solving equations involving, hyperbolic functions, Equations such as sinh x = 3.25 or coth x = 3.478 may, be determined using a calculator. This is demonstrated, in Worked Problems 15 to 21., Problem 15. Solve the equation sh x = 3, correct, to 4 significant figures., If sinh x = 3, then x = sinh−1 3, This can be determined by calculator., (i) Press hyp, (ii) Choose 4, which is sinh−1, (iii) Type in 3, (iv) Close bracket ), (v) Press = and the answer is 1.818448459, i.e. the solution of sh x = 3 is: x = 1.818, correct to 4, significant figures., Problem 16. Solve the equation ch x = 1.52,, correct to 3 decimal places., , 47, , Using a calculator with a similar procedure as in Worked, Problem 15, check that:, x = 0.980, correct to 3 decimal places., With reference to Fig. 5.2, it can be seen that there, will be two values corresponding to y = cosh x =, 1.52. Hence, x = ±0.980, Problem 17. Solve the equation tanh θ = 0.256,, correct to 4 significant figures., Using a calculator with a similar procedure as in Worked, Problem 15, check that gives, θ = 0.2618, correct to 4 significant figures., Problem 18. Solve the equation sech x = 0.4562,, correct to 3 decimal places., sech, then, x = sech −10.4562 =, , x = 0.4562,, 1, 1, cosh−1, since cosh =, 0.4562, sech, , If, , i.e. x = 1.421, correct to 3 decimal places., With reference to the graph of y = sech x in Fig. 5.4, it, can be seen that there will be two values corresponding, to y = sech x = 0.4562, Hence, x = ±1.421, Problem 19. Solve the equation, cosech y = −0.4458, correct to 4 significant figures., −1, If cosechy = − 0.4458,, then y = cosech (−0.4458), 1, 1, since sinh =, = sinh−1, − 0.4458, cosech, i.e. y = −1.547, correct to 4 significant figures., , Problem 20. Solve the equation coth A = 2.431,, correct to 3 decimal places., coth, 2.431,, then, A = coth−1 2.431 =, A=, 1, 1, tanh−1, since tanh =, 2.431, coth, i.e. A= 0.437, correct to 3 decimal places., If, , Problem 21. A chain hangs in the form given by, x, y = 40 ch, . Determine, correct to 4 significant, 40, figures, (a) the value of y when x is 25, and (b) the, value of x when y = 54.30
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48 Higher Engineering Mathematics, (a), , x, , and when x = 25,, 40, 25, y = 40 ch, = 40 ch 0.625, 40, , y = 40 ch, , = 40(1.2017536 . . .) = 48.07, x, (b) When y = 54.30, 54.30 =40 ch , from which, 40, x, 54.30, ch =, = 1.3575, 40, 40, x, Hence,, = cosh−1 1.3575 =±0.822219 . . .., 40, (see Fig. 5.2 for the reason as to why the answer, is ±) from which, x = 40(±0.822219 . . ..) = ±32.89, , Following the above procedure:, (i) 2.6 ch x + 5.1 sh x = 8.73, x, , x, , e + e−x, e − e−x, i.e. 2.6, + 5.1, = 8.73, 2, 2, (ii) 1.3e x + 1.3e−x + 2.55e x − 2.55e−x = 8.73, i.e. 3.85e x − 1.25e−x − 8.73 =0, (iii) 3.85(e x )2 − 8.73e x − 1.25 =0, (iv) e x, , , −(−8.73) ± [(−8.73)2 − 4(3.85)(−1.25)], =, 2(3.85), √, 8.73 ± 95.463 8.73 ±9.7705, =, =, 7.70, 7.70, Hence e x = 2.4027 or e x = −0.1351, , Equations of the form a ch x + b sh x = c, where a, b and, c are constants may be solved either by:, (a), , plotting graphs of y = a ch x + b sh x and y = c, and noting the points of intersection, or more, accurately,, , (b) by adopting the following procedure:, , x, e − e−x, and ch x to, (i) Change sh x to, 2, , x, e + e−x, 2, (ii) Rearrange the equation into the form, pe x + qe−x +r = 0, where p, q and r are, constants., , (v), , Now try the following exercise, Exercise 22 Further problems on, hyperbolic equations, In Problems 1 to 8, solve the given equations, correct to 4 decimal places., 1., , 2., , (iv) Solve the quadratic equation p(e x )2 +re x +, q = 0 for e x by factorising or by using the, quadratic formula., , 3., , (b) sh A = −2.43, , (a) cosh B = 1.87 (b) 2 ch x = 3, [(a) ±1.2384 (b) ±0.9624], (a) tanh y = −0.76 (b) 3 th x = 2.4, [(a) −0.9962 (b) 1.0986], , 4., , (a) sech B = 0.235 (b) sech Z = 0.889, [(a) ±2.1272 (b) ±0.4947], , 5., , This procedure is demonstrated in Problem 22., , (a) cosech θ = 1.45 (b) 5 cosech x = 4.35, [(a) 0.6442 (b) 0.5401], , 6., Problem 22. Solve the equation, 2.6 ch x + 5.1 sh x = 8.73, correct to 4 decimal, places., , (a) sinh x = 1, , [(a) 0.8814 (b) −1.6209], , (iii) Multiply each term by e x , which produces, an equation of the form p(e x )2 +re x +, q = 0 (since (e−x )(e x ) = e0 = 1), , (v) Given e x = a constant (obtained by solving the equation in (iv)), take Napierian, logarithms of both sides to give, x = ln (constant), , x = ln 2.4027 or x = ln(−0.1351) which has no, real solution., Hence x = 0.8766, correct to 4 decimal places., , (a) coth x = 2.54 (b) 2 coth y = −3.64, [(a) 0.4162 (b) −0.6176], , 7., , 3.5 sh x + 2.5 ch x = 0, , [−0.8959]
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Hyperbolic functions, 8. 2 sh x + 3 ch x = 5, , [0.6389 or −2.2484], , 9. 4 th x − 1 = 0, , [0.2554], , 10. A chain hangs so, its shape is of the, xthat, . Determine, correct to, form y = 56 cosh, 56, 4 significant figures, (a) the value of y when, x is 35, and (b) the value of x when y is 62.35, [(a) 67.30 (b) ±26.42], , x3 x5, i.e. sinh x = x + + + · · · (which is valid for all, 3! 5!, values of x). sinh x is an odd function and contains only, odd powers of x in its series expansion., Problem 23. Using the series expansion for ch x, evaluate ch 1 correct to 4 decimal places., ch x = 1 +, Let, , 5.5 Series expansions for cosh x and, sinh x, , x = 1,, , then ch 1 = 1 +, , By definition,, x2 x3 x4 x5, +, +, +, +···, 2! 3! 4!, 5!, , from Chapter 4., Replacing x by −x gives:, e−x = 1 − x +, , x2 x3 x4 x5, −, +, −, +··· ., 2!, 3! 4! 5!, , 1, cosh x = (e x + e−x ), 2, , , x2 x3 x4 x5, 1, 1+x +, =, +, +, +, +···, 2, 2!, 3! 4! 5!, , , x2 x3 x4 x5, −, +, −, +···, + 1−x +, 2! 3!, 4! 5!, , , 2x 2 2x 4, 1, 2+, +, +···, =, 2, 2!, 4!, x2 x4, i.e. cosh x = 1 + + + · · · (which is valid for all, 2! 4!, values of x). cosh x is an even function and contains, only even powers of x in its expansion., 1, sinh x = (e x − e−x ), 2, , , x2 x3 x4 x5, 1, 1+x +, =, +, +, +, +···, 2, 2! 3!, 4! 5!, , , x2 x3 x4 x5, −, +, −, +···, − 1−x +, 2! 3!, 4! 5!, =, , x2 x4, +, + · · ·from above, 2! 4!, , 2x 3 2x 5, 1, 2x +, +, + ···, 2, 3!, 5!, , 14, 12, +, 2 × 1 4 ×3 × 2 × 1, , 16, + ···, 6 ×5 × 4 × 3 ×2 × 1, , +, , ex = 1 + x +, , 49, , = 1 + 0.5 + 0.04167 + 0.001389 + · · ·, i.e. ch 1 = 1.5431, correct to 4 decimal places,, which may be checked by using a calculator., Problem 24. Determine, correct to 3 decimal, places, the value of sh 3 using the series expansion, for sh x., sh x = x +, , x3 x5, +, + · · · from above, 3! 5!, , Let x = 3, then, 33 35 37 39 311, +, +, + +, +···, 3! 5! 7! 9! 11!, = 3 + 4.5 + 2.025 + 0.43393 + 0.05424, , sh 3 = 3 +, , + 0.00444 + · · ·, i.e. sh 3 = 10.018, correct to 3 decimal places., Problem, 25. Determine the power series for, θ, − sh 2θ as far as the term in θ 5 ., 2 ch, 2, In the series expansion for ch x, let x =, 2 ch, , θ, then:, 2, , , , θ, (θ/2)2 (θ/2)4, =2 1+, +, +···, 2, 2!, 4!, =2+, , θ2, θ4, +, +···, 4, 192
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50 Higher Engineering Mathematics, In the series expansion for sh x, let x = 2θ, then:, (2θ)3 (2θ)5, +, +···, 3!, 5!, 4, 4, = 2θ + θ 3 + θ 5 + · · ·, 3, 15, , sh 2θ = 2θ +, , Hence, , , , θ, θ2, θ4, ch, − sh 2θ = 2 +, +, +···, 2, 4, 192, , , 4, 4, − 2θ + θ 3 + θ 5 + · · ·, 3, 15, = 2 −2θ +, −, , θ2 4 3 θ4, − θ +, 4 3, 192, , 4 5, θ + · · · as far the term in θ 5, 15, , Now try the following exercise, Exercise 23 Further problems on series, expansions for cosh x and sinh x, 1. Use the series expansion for ch x to evaluate,, correct to 4 decimal places: (a) ch 1.5 (b) ch 0.8, [(a) 2.3524 (b) 1.3374], , 2. Use the series expansion for sh x to evaluate, correct to 4 decimal places: (a) sh 0.5, (b) sh 2, [(a) 0.5211 (b) 3.6269], 3. Expand the following as a power series as far, as the term in x 5 : (a) sh 3x (b) ch 2x, ⎡, ⎤, 9 3 81 5, (a), 3x, +, +, x, x, ⎢, 2, 40 ⎥, ⎣, ⎦, 2, (b) 1 + 2x 2 + x 4, 3, In Problems 4 and 5, prove the given identities,, the series being taken as far as the term in θ 5, only., 4. sh 2θ − sh θ ≡ θ +, , 5. 2 sh, , 31 5, 7 3, θ +, θ, 6, 120, , θ, θ, θ2 θ3, θ4, − ch ≡ − 1 + θ −, +, −, 2, 2, 8, 24 384, +, , θ5, 1920
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Chapter 6, , Arithmetic and geometric, progressions, 6.1, , Arithmetic progressions, , When a sequence has a constant difference between, successive terms it is called an arithmetic progression, (often abbreviated to AP)., Examples include:, , i.e., , For example, the sum of the first 7 terms of the series 1,, 4, 7, 10, 13, . . . is given by, 7, S7 = [2(1) + (7 − 1)3], since a = 1 and d = 3, 2, , (i) 1, 4, 7, 10, 13, . . . where the common difference, is 3 and, , 7, 7, = [2 + 18] = [20] = 70, 2, 2, , (ii) a, a + d, a + 2d, a + 3d,. . .where the common, difference is d., General expression for the n’th term of an AP, If the first term of an AP is ‘a’ and the common, difference is ‘d’ then, the n’th term is: a + (n − 1)d, In example (i) above, the 7th term is given by 1 +, (7 − 1)3 = 19, which may be readily checked., Sum of n terms of an AP, The sum S of an AP can be obtained by multiplying the, average of all the terms by the number of terms., a +l, , where ‘a’ is the, The average of all the terms =, 2, first term and l is the last term, i.e. l = a + (n − 1)d, for, n terms., Hence the sum of n terms,, , , a +l, Sn = n, 2, n, = {a + [a + (n − 1)d]}, 2, , n, S n = [2a + (n − 1)d], 2, , 6.2 Worked problems on arithmetic, progressions, Problem 1. Determine (a) the ninth, and (b) the, sixteenth term of the series 2, 7, 12, 17, . . ., 2, 7, 12, 17, . . . is an arithmetic progression with a, common difference, d, of 5., (a), , The n’th term of an AP is given by a + (n −1)d, Since the first term a = 2, d = 5 and n =9 then the, 9th term is:, 2 + (9 −1)5 = 2 + (8)(5) = 2 + 40 =42, , (b) The 16th term is:, 2 + (16 −1)5 = 2 +(15)(5) = 2 + 75 =77., Problem 2. The 6th term of an AP is 17 and the, 13th term is 38. Determine the 19th term.
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52 Higher Engineering Mathematics, The n’th term of an AP is a + (n −1)d, , The sum of the first 21 terms,, , a + 5d = 17, , (1), , The 13th term is: a + 12d= 38, , (2), , The 6th term is:, , Equation (2) −equation (1) gives: 7d = 21, from which,, 21, d = = 3., 7, Substituting in equation (1) gives: a + 15 =17, from, which, a = 2., Hence the 19th term is:, a + (n − 1)d = 2 + (19 − 1)3 = 2 + (18)(3) =, 2 + 54 = 56., , is, , an, , AP, , where, , a = 2 12, , 1. Find the 11th term of the series 8, 14, 20,, 26, . . ., [68], 2. Find the 17th term of the series 11, 10.7, 10.4,, 10.1, . . ., [6.2], and, , Hence if the n’th term is 22 then: a + (n − 1)d = 22, , i.e. 2 12 + (n − 1) 1 12 = 22, , (n − 1) 1 12 = 22 − 2 12 = 19 12 ., n −1 =, , 19 12, 1 12, , = 13 and n = 13 + 1 = 14, , i.e. the 14th term of the AP is 22., Problem 4. Find the sum of the first 12 terms of, the series 5, 9, 13, 17, . . ., 5, 9, 13, 17, . . . is an AP where a = 5 and d = 4. The sum, of n terms of an AP,, n, Sn = [2a + (n − 1)d], 2, Hence the sum of the first 12 terms,, S12 =, , Now try the following exercise, Exercise 24 Further problems on arithmetic, progressions, , Problem 3. Determine the number of the term, whose value is 22 in the series 2 12 , 4, 5 12 , 7, . . ., 2 12 , 4, 5 12 , 7, . . ., d = 1 12 ., , 21, [2a + (n − 1)d], 2, 21, 21, = [2(3.5) + (21 − 1)0.6] = [7 + 12], 2, 2, 399, 21, = 199.5, = (19) =, 2, 2, , S21 =, , 12, [2(5) + (12 − 1)4], 2, , = 6[10 + 44] = 6(54) = 324, Problem 5. Find the sum of the first 21 terms of, the series 3.5, 4.1, 4.7, 5.3, . . ., 3.5, 4.1, 4.7, 5.3, . . . is an AP where a = 3.5 and d = 0.6, , 3. The seventh term of a series is 29 and the, eleventh term is 54. Determine the sixteenth, term., [85.25], 4. Find the 15th term of an arithmetic progression, of which the first term is 2.5 and the tenth term, is 16., [23.5], 5. Determine the number of the term which is 29, in the series 7, 9.2, 11.4, 13.6, . . ., [11th ], 6. Find the sum of the first 11 terms of the series, 4, 7, 10, 13, . . ., [209], 7. Determine the sum of the series 6.5, 8.0, 9.5,, 11.0, . . . , 32, [346.5], , 6.3 Further worked problems on, arithmetic progressions, Problem 6. The sum of 7 terms of an AP is 35, and the common difference is 1.2. Determine the, first term of the series., n = 7, d = 1.2 and S7 = 35, Since the sum of n terms of an AP is given by, Sn =, , n, [2a + (n − 1)d], then, 2, , 7, 7, 35 = [2a + (7 − 1)1.2] = [2a + 7.2], 2, 2
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Arithmetic and geometric progressions, 35 × 2, = 2a + 7.2, 7, 10 = 2a + 7.2, 2a = 10 − 7.2 = 2.8,, 2.8, a=, = 1.4, 2, , Hence, , Thus, from which, , i.e. the first term, a = 1.4, , Problem 9. The first, twelfth and last term of an, arithmetic progression are 4, 31 12 , and 376 12, respectively. Determine (a) the number of terms in, the series, (b) the sum of all the terms and (c) the, ‘80’th term., (a), , Problem 7. Three numbers are in arithmetic, progression. Their sum is 15 and their product is 80., Determine the three numbers., , Let the AP be a, a +d, a +2d, . . . , a + (n − 1)d,, where a = 4, The 12th term is: a + (12 −1)d = 31 12, 4 + 11d = 31 12 ,, , i.e., , Let the three numbers be (a − d), a and (a + d), , from which, 11d = 31 12 − 4 = 27 12, , Then (a − d) + a + (a + d) = 15, i.e. 3a = 15, from, which, a = 5, , Hence d =, , 27 12, = 2 12, 11, The last term is a + (n − 1)d, , i.e. 4 + (n − 1) 2 12 = 376 12, , Also, a(a − d)(a + d) = 80, i.e. a(a 2 − d 2 ) = 80, Since a = 5, 5(52 − d 2 ) = 80, 125 − 5d 2 = 80, 125 − 80 = 5d 2, , (n − 1) =, , 376 12 − 4, 2 12, , 45 = 5d 2, √, 45, from which, d 2 = = 9. Hence d = 9 = ±3., 5, The three numbers are thus (5 − 3), 5 and (5 + 3), i.e., 2, 5 and 8., Problem 8. Find the sum of all the numbers, between 0 and 207 which are exactly divisible by 3., , =, , a + (n − 1)d = 207, , i.e., , 3 + (n − 1)3 = 207,, , n, [2a + (n − 1)d], 2, , , 150, 1, =, 2(4) + (150 − 1) 2, 2, 2, , , 1, = 75 8 + (149) 2, 2, , = 85[8 + 372.5], = 75(380.5) = 28537, , The sum of all 69 terms is given by, n, [2a + (n − 1)d], 2, 69, = [2(3) + (69 − 1)3], 2, 69, 69, = [6 + 204] = (210) = 7245, 2, 2, , S69 =, , = 149, , S150 =, , 207 − 3, = 68, 3, n = 68 + 1 = 69, , Hence, , 2 12, , (b) Sum of all the terms,, , (n − 1) =, , from which, , 372 12, , Hence the number of terms in the series,, n = 149 +1 =150, , The series 3, 6, 9, 12, . . ., 207 is an AP whose first term, a = 3 and common difference d = 3, The last term is, , 53, , (c), , 1, 2, , The 80th term is:, , a + (n − 1)d = 4 + (80 − 1) 2 12, , = 4 + (79) 2 12, = 4 + 197.5 = 201 12
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54 Higher Engineering Mathematics, Problem 10. An oil company bores a hole 80 m, deep. Estimate the cost of boring if the cost is £30, for drilling the first metre with an increase in cost of, £2 per metre for each succeeding metre., , 8. An oil company bores a hole 120 m deep. Estimate the cost of boring if the cost is £70 for, drilling the first metre with an increase in cost, of £3 per metre for each succeeding metre., [£29820], , The series is: 30, 32, 34, . . . to 80 terms, i.e. a = 30,, d = 2 and n = 80, Thus, total cost,, , n , Sn = 2a + (n − 1)d, 2, =, , 80, [2(30) + (80 − 1)(2)], 2, , = 40[60 + 158] = 40(218) = £8720, , 6.4, , Geometric progressions, , When a sequence has a constant ratio between successive terms it is called a geometric progression (often, abbreviated to GP). The constant is called the common, ratio, r., Examples include, (i) 1, 2, 4, 8, . . . where the common ratio is 2 and, , Now try the following exercise, , (ii) a, ar, ar 2 , ar 3 , . . . where the common ratio is r., General expression for the n’th term of a GP, , Exercise 25 Further problems on arithmetic, progressions, , If the first term of a GP is ‘a’ and the common ratio is, r, then, , 1. The sum of 15 terms of an arithmetic progression is 202.5 and the common difference is 2., Find the first term of the series., [−0.5], , the n’th term is: ar n−1, , 2. Three numbers are in arithmetic progression., Their sum is 9 and their product is 20.25., Determine the three numbers., [1.5, 3, 4.5], 3. Find the sum of all the numbers between 5 and, 250 which are exactly divisible by 4. [7808], 4. Find the number of terms of the series 5, 8,, 11, . . . of which the sum is 1025., [25], 5. Insert four terms between 5 and 22.5 to form, an arithmetic progression. [8.5, 12, 15.5, 19], 6. The first, tenth and last terms of an arithmetic, progression are 9, 40.5, and 425.5 respectively., Find (a) the number of terms, (b) the sum of, all the terms and (c) the 70th term., [(a) 120 (b) 26070 (c) 250.5], 7. On commencing employment a man is paid, a salary of £16000 per annum and receives, annual increments of £480. Determine his, salary in the 9th year and calculate the total, he will have received in the first 12 years., [£19840, £223,680], , which can be readily checked from the above examples., For example, the 8th term of the GP 1, 2, 4, 8, . . . is, (1)(2)7 = 128, since a = 1 and r = 2., Sum of n terms of a GP, Let a GP be a, ar, ar 2 , ar 3 , . . . , ar n−1, then the sum of n terms,, Sn = a + ar + ar 2 + ar 3 + · · · + ar n−1 · · ·, , (1), , Multiplying throughout by r gives:, r Sn = ar + ar 2 + ar 3 + ar 4, + · · · + ar n−1 + ar n + · · ·, , (2), , Subtracting equation (2) from equation (1) gives:, Sn − r Sn = a − ar n, i.e. Sn (1 − r) = a(1 − r n ), n, , −r ), Thus the sum of n terms, S n = a(1, (1 − r ) which is valid, when r < 1.
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Arithmetic and geometric progressions, Subtracting equation (1) from equation (2) gives, a(r n − 1), Sn =, which is valid when r > 1., (r − 1), For example, the sum of the first 8 terms of the GP 1, 2,, 1(28 − 1), 4, 8, 16, . . . is given by S8 =, , since a = 1 and, (2 − 1), r =2, i.e. S8 =, , 1(256 − 1), = 255, 1, , Sum to infinity of a GP, When the common ratio r of a GP is less than unity, the, a(1 −r n ), , which may be written, sum of n terms, Sn =, (1 −r), a, ar n, −, as Sn =, (1 −r) (1 −r), Since r < 1, r n becomes less as n increases, i.e. r n → 0, as n →∞. n, a, ar, Hence, → 0 as n →∞. Thus Sn →, as, (1 −r), (1 −r), n →∞., a, is called the sum to infinity, S∞,, The quantity, (1 −r), and is the limiting value of the sum of an infinite number, of terms,, a, i.e. S ∞ =, which is valid when −1 <r < 1., (1 − r), For example, the sum to infinity of the GP, 1 + 12 + 14 + · · · is, S∞ =, , 1, 1−, , 1, 2, , , since a = 1 and r = 12 , i.e. S∞ = 2., , 1, 2,, , 1 12 , 4 12 , 13 12 , . . . is a GP with a common ratio r = 3, , The sum of n terms, Sn =, Hence S7 =, , Problem 11. Determine the tenth term of the, series 3, 6, 12, 24, . . ., 3, 6, 12, 24, . . . is a geometric progression with a common ratio r of 2. The n’th term of a GP is ar n−1 ,, where a is the first term. Hence the 10th term is:, (3)(2)10−1 = (3)(2)9 = 3(512) = 1536., Problem 12. Find the sum of the first 7 terms of, the series, 12 , 1 12 , 4 12 , 13 12 , . . ., , 1 7, 2 (3 − 1), , (3 − 1), , =, , a(r n − 1), (r − 1), , 1, 2 (2187 −1), , 2, , = 546, , 1, 2, , Problem 13. The first term of a geometric, progression is 12 and the fifth term is 55. Determine, the 8’th term and the 11’th term., The 5th term is given by ar 4 = 55, where the first term, a = 12, Hence, , r4 =, , 55 55, =, a, 12, , , and, , r=, , 4, , 55, 12, , , = 1.4631719 . . ., , The 8th term is ar 7 = (12)(1.4631719 . . .)7 = 172.3, The 11th term is ar 10 = (12)(1.4631719 . . .)10 = 539.7, Problem 14. Which term of the series 2187, 729,, 243, . . . is 19 ?, 2187, 729, 243, . . . is a GP with a common ratio r = 13, and first term a = 2187, The n’th term of a GP is given by: ar n−1, n−1, 1, Hence, = (2187) 13, 9, , from which, , 6.5 Worked problems on geometric, progressions, , 55, , n−1, 1, 1, 1, =, =, 3, (9)(2187) 3237, 9, 1, 1, = 9=, 3, 3, , Thus (n − 1) = 9, from which, n =9 + 1 =10, i.e. 19 is the 10th term of the GP., Problem 15. Find the sum of the first 9 terms of, the series 72.0, 57.6, 46.08, . . ., The common ratio, r =, , ar 57.6, =, = 0.8, a, 72.0, , , , ar 2, 46.08, also, =, = 0.8, ar, 57.6
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56 Higher Engineering Mathematics, The sum of 9 terms,, a(1 − r n ) 72.0(1 − 0.89 ), S9 =, =, (1 − r), (1 − 0.8), =, , 72.0(1 − 0.1342), = 311.7, 0.2, , common ratio, (b) the first term, and (c) the sum of, the fifth to eleventh terms, inclusive., (a), , Problem 16. Find the sum to infinity of the, series 3, 1, 13 , . . ., 3, 1, 13 , . . . is a GP of common ratio, r = 13, The sum to infinity,, , (b) The sum of the 7th and 8th terms is 192. Hence, ar 6 + ar 7 = 192., Since r = 2, then 64a + 128a = 192, , a, 3, 9, 3, 1, S∞ =, = 2 = =4, =, 1, 1−r, 2, 2, 1− 3, 3, , 192a = 192,, from which, a, the first term, = 1., (c), , Now try the following exercise, Exercise 26 Further problems on geometric, progressions, 1. Find the 10th term of the series 5, 10, 20,, 40, . . ., [2560], 2. Determine the sum of the first 7 terms of the, [273.25], series 14 , 34 , 2 14 , 6 34 , . . ., 3. The first term of a geometric progression is 4, and the 6th term is 128. Determine the 8th and, 11th terms., [512, 4096], 4. Find the sum of the first 7 terms of the, series 2, 5, 12 12 , . . . (correct to 4 significant, figures)., [812.5], 5. Determine the sum to infinity of the series 4,, 2, 1, . . ., [8], 6. Find the sum to infinity of the series 2 12 , −1 14 ,, , 5, 1 23, 8, ..., , 6.6 Further worked problems on, geometric progressions, Problem 17. In a geometric progression the sixth, term is 8 times the third term and the sum of the, seventh and eighth terms is 192. Determine (a) the, , Let the GP be a, ar, ar 2 , ar 3 , . . . , ar n−1, The 3rd term = ar 2 and the sixth term = ar 5, The 6th term is 8 times the 3rd., √, 3, Hence ar 5 = 8ar 2 from which, r 3 = 8, r = 8, i.e. the common ratio r = 2., , The sum of the 5th to 11th terms (inclusive) is, given by:, S11 − S4 =, , a(r 11 − 1) a(r 4 − 1), −, (r − 1), (r − 1), , =, , 1(211 − 1) 1(24 − 1), −, (2 − 1), (2 − 1), , = (211 − 1) − (24 − 1), = 211 − 24 = 2048 − 16 = 2032, Problem 18. A hire tool firm finds that their, net return from hiring tools is decreasing by, 10% per annum. If their net gain on a certain tool, this year is £400, find the possible total of all future, profits from this tool (assuming the tool lasts for, ever)., The net gain forms a series:, £400 + £400 × 0.9 + £400 × 0.92 + · · · ,, which is a GP with a = 400 and r = 0.9., The sum to infinity,, a, 400, S∞ =, =, (1 − r), (1 − 0.9), = £4000 = total future profits, Problem 19. If £100 is invested at compound, interest of 8% per annum, determine (a) the value, after 10 years, (b) the time, correct to the nearest, year, it takes to reach more than £300.
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Arithmetic and geometric progressions, (a), , Let the GP be a, ar, ar 2 , . . . , ar n, The first term a = £100, The common ratio r = 1.08, Hence the second term is, ar = (100) (1.08) = £108,, which is the value after 1 year,, the third term is, , the fifth term is ar 4 = (50) (1.7188)4 = 436.39,, the sixth term is ar 5 = (50) (1.7188)5 = 750.06, Hence, correct to the nearest whole number, the 6 speeds, of the drilling machine are 50, 86, 148, 254, 436 and, 750 rev/min., , Now try the following exercise, , ar 2 = (100) (1.08)2 = £116.64,, which is the value after 2 years, and so on., Thus the value after 10 years, = ar 10 = (100) (1.08)10 = £215.89, (b) When £300 has been reached, 300 =ar n, i.e., , 300 = 100(1.08)n, , and, , 3 = (1.08)n, , Taking logarithms to base 10 of both sides gives:, lg 3 = lg(1.08)n = n lg(1.08),, by the laws of logarithms, lg 3, from which, n =, = 14.3, lg1.08, Hence it will take 15 years to reach more than, £300., Problem 20. A drilling machine is to have 6, speeds ranging from 50 rev/min to 750 rev/ min. If, the speeds form a geometric progression determine, their values, each correct to the nearest whole, number., Let the GP of n terms be given by a, ar, ar 2 , . . . , ar n−1 ., The first term a = 50 rev/min, The 6th term is given by ar 6−1 , which is 750 rev/min,, i.e., , ar 5 = 750, , 750 750, =, = 15, a, 50, √, 5, Thus the common ratio, r = 15 = 1.7188, from which r 5 =, , The first term is a = 50 rev/min, the second term is ar = (50) (1.7188) = 85.94,, the third term is ar 2 = (50) (1.7188)2 = 147.71,, the fourth term is ar 3 = (50) (1.7188)3 = 253.89,, , 57, , Exercise 27 Further problems on geometric, progressions, 1. In a geometric progression the 5th term is, 9 times the 3rd term and the sum of the 6th, and 7th terms is 1944. Determine (a) the common ratio, (b) the first term and (c) the sum, of the 4th to 10th terms inclusive., [(a) 3 (b) 2 (c) 59022], 2. Which term of the series 3, 9, 27, . . . is, 59049?, [10th], 3. The value of a lathe originally valued at, £3000 depreciates 15% per annum. Calculate, its value after 4 years. The machine is sold, when its value is less than £550. After how, many years is the lathe sold?, [£1566, 11 years], 4. If the population of Great Britain is 55 million, and is decreasing at 2.4% per annum, what, will be the population in 5 years time?, [48.71 M], 5. 100 g of a radioactive substance disintegrates, at a rate of 3% per annum. How much of the, substance is left after 11 years?, [71.53 g], 6. If £250 is invested at compound interest of, 6% per annum determine (a) the value after, 15 years, (b) the time, correct to the nearest, year, it takes to reach £750., [(a) £599.14 (b) 19 years], 7. A drilling machine is to have 8 speeds ranging from 100 rev/min to 1000 rev/min. If the, speeds form a geometric progression determine their values, each correct to the nearest, whole number., [100, 139, 193, 268, 373, 518,, 720, 1000 rev/min]
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Chapter 7, , The binomial series, 7.1, , Pascal’s triangle, , Table 7.1, (a 1 x)0, , A binomial expression is one which contains two terms, connected by a plus or minus sign. Thus ( p +q), (a +, x)2 , (2x + y)3 are examples of binomial expressions., Expanding (a + x)n for integer values of n from 0 to 6, gives the results as shown at the bottom of the page., From these results the following patterns emerge:, , (a 1 x)1, , (i) ‘a’ decreases in power moving from left to right., , (a 1 x)6, , 1, 1, 1, , (a 1 x), , 3, , (a 1 x), , 3, , 1, , 4, , 1, , (a 1 x), , 5, , 1, , 1, 3, , 4, , 1, , 6, , 5, 6, , 1, 2, , 1, , (a 1 x), , 2, , 10, 15, , 4, 10, , 20, , 1, 5, , 15, , 1, 6, , 1, , (ii) ‘x’ increases in power moving from left to right., (iii) The coefficients of each term of the expansions are, symmetrical about the middle coefficient when n, is even and symmetrical about the two middle, coefficients when n is odd., (iv) The coefficients are shown separately in Table 7.1, and this arrangement is known as Pascal’s triangle. A coefficient of a term may be obtained, by adding the two adjacent coefficients immediately above in the previous row. This is shown, by the triangles in Table 7.1, where, for example,, 1 + 3 = 4, 10 + 5 = 15, and so on., (v) Pascal’s triangle method is used for expansions of, the form (a + x)n for integer values of n less than, about 8., (a + x)0, (a + x)1, (a + x)2, (a + x)3, (a + x)4, (a + x)5, (a + x)6, , Problem 1. Use the Pascal’s triangle method to, determine the expansion of (a + x)7 ., From Table 7.1, the row of Pascal’s triangle corresponding to (a + x)6 is as shown in (1) below. Adding, adjacent coefficients gives the coefficients of (a + x)7, as shown in (2) below., 1, 1, , 6, 7, , 15, 21, , 20, 35, , 15, 35, , 6, 21, , 1, 7, , (1), 1, , (2), , The first and last terms of the expansion of (a + x)7 are, a 7 and x 7 respectively. The powers of ‘a’ decrease and, the powers of ‘x’ increase moving from left to right., , =, 1, = a+x, a+x, = (a + x)(a + x) =, a 2 + 2ax + x 2, = (a + x)2 (a + x) =, a 3 + 3a 2 x + 3ax 2 + x 3, 3, 4, = (a + x) (a + x) =, a + 4a 3 x + 6a 2 x 2 + 4ax 3 + x 4, 4, 5, = (a + x) (a + x) =, a + 5a 4 x + 10a 3 x 2 + 10a 2 x 3 + 5ax 4 + x 5, = (a + x)5 (a + x) = a 6 + 6a 5 x + 15a 4 x 2 + 20a 3 x 3 + 15a 2 x 4 + 6ax 5 + x 6
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The binomial series, of (a + x)n is given by:, , Hence, (a + x)7 = a 7 + 7a 6 x + 21a 5 x 2 + 35a 4 x 3, + 35a 3 x 4 + 21a 2 x 5 + 7ax 6 + x 7, Problem 2. Determine, using Pascal’s triangle, method, the expansion of (2 p − 3q)5 ., Comparing (2 p − 3q)5 with (a + x)5 shows that, a = 2 p and x = −3q., Using Pascal’s triangle method:, (a + x)5 = a 5 + 5a 4 x + 10a 3 x 2 + 10a 2 x 3 + · · ·, Hence, (2 p − 3q)5 = (2 p)5 + 5(2 p)4 (−3q), + 10(2 p)3 (−3q)2, + 10(2 p)2 (−3q)3, + 5(2 p)(−3q)4 + (−3q)5, i.e. (2p − 3q)5 = 32p 5 − 240p4 q + 720p3 q 2, − 1080p 2 q 3 + 810pq 4 − 243q 5, , Now try the following exercise, Exercise 28, triangle, , Further problems on Pascal’s, , 1. Use Pascal’s triangle to expand (x − y)7 ., , n(n − 1) n−2 2, x, a, 2!, n(n − 1)(n − 2) n−3 3, +, x, a, 3!, + ···, , (a + x)n = a n + na n−1 x +, , where 3! denotes 3 × 2 ×1 and is termed ‘factorial 3’., With the binomial theorem n may be a fraction, a, decimal fraction or a positive or negative integer., When n is a positive integer, the series is finite, i.e.,, it comes to an end; when n is a negative integer, or a, fraction, the series is infinite., In the general expansion of (a + x)n it is noted that the, n(n − 1)(n − 2) n−3 3, 4th term is:, a x . The number 3 is, 3!, very evident in this expression., For any term in a binomial expansion, say the r’th, term, (r − 1) is very evident. It may therefore be reasoned that the r’th term of the expansion (a + x)n is:, n(n − 1)(n − 2). . . to (r − 1) terms n−(r −1) r−1, x, a, (r − 1)!, If a = 1 in the binomial expansion of (a + x)n then:, n(n − 1) 2, x, 2!, n(n − 1)(n− 2) 3, +, x +···, 3!, , (1 + x)n = 1 + nx +, , which is valid for −1 < x < 1., When x is small compared with 1 then:, (1 + x)n ≈ 1 + nx, , x 7 − 7x 6 y + 21x 5 y 2 − 35x 4 y 3, + 35x 3 y 4 − 21x 2 y 5 + 7x y 6 − y 7, 2. Expand (2a + 3b)5 using Pascal’s triangle., , 7.3 Worked problems on the, binomial series, , 32a 5 + 240a 4 b + 720a 3 b2, + 1080a 2b3 + 810ab4 + 243b5, , Problem 3. Use the binomial series to determine, the expansion of (2 + x)7 ., The binomial expansion is given by:, , 7.2, , 59, , The binomial series, , The binomial series or binomial theorem is a formula, for raising a binomial expression to any power without, lengthy multiplication. The general binomial expansion, , n(n − 1) n−2 2, a x, 2!, n(n − 1)(n − 2) n−3 3, +, a x +···, 3!, , (a + x)n = a n + na n−1 x +
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60 Higher Engineering Mathematics, When a = 2 and n =7:, (7)(6) 5 2, (2) x, (2)(1), (7)(6)(5) 4 3 (7)(6)(5)(4) 3 4, +, (2) x +, (2) x, (3)(2)(1), (4)(3)(2)(1), , (2 + x)7 = 27 + 7(2)6 x +, , +, , (7)(6)(5)(4)(3) 2 5, (2) x, (5)(4)(3)(2)(1), , +, , (7)(6)(5)(4)(3)(2), (2)x 6, (6)(5)(4)(3)(2)(1), , +, , (7)(6)(5)(4)(3)(2)(1) 7, x, (7)(6)(5)(4)(3)(2)(1), , , , , , 1 5, 1, = c5 + 5c4 −, c−, c, c, , , (5)(4) 3, 1 2, +, c −, (2)(1), c, , , (5)(4)(3) 2, 1 3, +, c −, (3)(2)(1), c, , , (5)(4)(3)(2), 1 4, +, c −, (4)(3)(2)(1), c, , , 1 5, (5)(4)(3)(2)(1), −, +, (5)(4)(3)(2)(1), c, , 5, 1, 10 5, 1, i.e. c −, = c5 − 5c3 + 10c − + 3 − 5, c, c, c, c, , i.e. (2 + x)7 = 128 + 448x + 672x 2 + 560x 3, + 280x 4 + 84x 5 + 14x 6 + x 7, Problem 4. Use the binomial series to determine, the expansion of (2a − 3b)5 ., From equation (1), the binomial expansion is given by:, n(n − 1) n−2 2, a x, 2!, n(n − 1)(n − 2) n−3 3, +, x +···, a, 3!, , (a + x)n = a n + na n−1 x +, , When a = 2a, x = −3b and n = 5:, (2a − 3b)5 = (2a)5 + 5(2a)4 (−3b), +, , (5)(4), (2a)3 (−3b)2, (2)(1), , +, , (5)(4)(3), (2a)2 (−3b)3, (3)(2)(1), , +, , (5)(4)(3)(2), (2a)(−3b)4, (4)(3)(2)(1), , +, , (5)(4)(3)(2)(1), (−3b)5, (5)(4)(3)(2)(1), , i.e. (2a − 3b)5= 32a 5 −240a 4 b + 720a 3 b2, −1080a 2 b3 + 810ab4 −243b5, , Problem 5., series., , , , 1 5, Expand c −, using the binomial, c, , Problem 6. Without fully expanding (3 + x)7,, determine the fifth term., The r’th term of the expansion (a + x)n is given by:, n(n − 1)(n − 2) . . . to (r − 1) terms n−(r−1) r−1, x, a, (r − 1)!, Substituting n = 7, a = 3 and r − 1 = 5 −1 =4 gives:, (7)(6)(5)(4) 7−4 4, (3) x, (4)(3)(2)(1), i.e. the fifth term of (3 + x)7 = 35(3)3 x 4 = 945x 4, Problem 7. Find the middle term of, , , 1 10, ., 2p−, 2q, In the expansion of (a + x)10 there are 10 +1, i.e. 11, terms. Hence the middle term is the sixth. Using the, general expression for the r’th term where a = 2 p,, 1, x = − , n =10 and r − 1 = 5 gives:, 2q, , , 1 5, (10)(9)(8)(7)(6), 10–5, (2 p), −, (5)(4)(3)(2)(1), 2q, , , 1, = 252(32 p5) −, 32q 5, , , 1 10, p5, Hence the middle term of 2 p −, is −252 5, 2q, q, Problem 8. Evaluate (1.002)9 using the binomial, theorem correct to (a) 3 decimal places and (b) 7, significant figures.
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The binomial series, (1 + x)n = 1 + nx +, +, , n(n − 1) 2, x, 2!, n(n − 1)(n − 2) 3, x +···, 3!, , (1.002)9 = (1 + 0.002)9, , (9)(8), (1 + 0.002)9 = 1 + 9(0.002) +, (0.002)2, (2)(1), (9)(8)(7), (0.002)3 + · · ·, (3)(2)(1), , = 1 + 0.018 + 0.000144, + 0.000000672 + · · ·, = 1.018144672 . . ., Hence (1.002)9 = 1.018, correct to 3 decimal places, = 1.018145, correct to 7 significant, figures, Problem 9. Evaluate (0.97)6 correct to 4 significant figures using the binomial expansion., (0.97)6 is written as (1 − 0.03)6, Using the expansion of (1 + x)n where n = 6 and, x = −0.03 gives:, (1 − 0.03)6 = 1 + 6(−0.03) +, , (6)(5), (−0.03)2, (2)(1), , +, , (6)(5)(4), (−0.03)3, (3)(2)(1), , +, , (6)(5)(4)(3), (−0.03)4 + · · ·, (4)(3)(2)(1), , = 1 − 0.18 + 0.0135 − 0.00054, + 0.00001215 − · · ·, ≈ 0.83297215, i.e., , (3.039)4 may be written in the form (1 + x)n as:, (3.039)4 = (3 + 0.039)4, , , 0.039, = 3 1+, 3, , 4, , = 34 (1 + 0.013)4, , Substituting x = 0.002 and n = 9 in the general expansion for (1 + x)n gives:, , +, , 61, , (0.97)6 = 0.8330, correct to 4 significant, figures, , Problem 10. Determine the value of (3.039)4 ,, correct to 6 significant figures using the binomial, theorem., , (1 + 0.013)4 = 1 + 4(0.013), +, , (4)(3), (0.013)2, (2)(1), , +, , (4)(3)(2), (0.013)3 + · · ·, (3)(2)(1), , = 1 + 0.052 + 0.001014, + 0.000008788 + · · ·, = 1.0530228, correct to 8 significant figures, Hence (3.039)4 = 34 (1.0530228), = 85.2948, correct to 6 significant, figures, , Now try the following exercise, Exercise 29 Further problems on the, binomial series, 1. Use the binomial theorem to expand, (a + 2x)4 ., 4, a + 8a 3 x + 24a 2 x 2, + 32ax 3 + 16x 4, 2. Use the binomial theorem to expand (2 − x)6 ., , 64 − 192x + 240x 2 − 160x 3, + 60x 4 − 12x 5 + x 6, 3. Expand (2x − 3y)4 ., , 16x 4 − 96x 3 y + 216x 2 y 2, − 216x y 3 + 81y 4, , , 2 5, 4. Determine the expansion of 2x +, ., x, ⎡, ⎤, 320, 5, 3, ⎢ 32x + 160x + 320x + x ⎥, ⎢, ⎥, ⎣, ⎦, 160 32, + 3 + 5, x, x
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62 Higher Engineering Mathematics, 5., , Expand ( p + 2q)11 as far as the fifth term., ⎡, ⎤, p11 + 22 p10 q + 220 p 9q 2, ⎣, ⎦, + 1320 p8q 3 + 5280 p7q 4, , , 6., , Determine the sixth term of 3 p +, , q 13, 3, , 8., , Determine the middle term of (2a − 5b)8., [700000 a 4 b4 ], Use the binomial theorem to determine, correct to 4 decimal places:, (a) (1.003)8 (b) (1.042)7, [(a) 1.0243 (b) 1.3337], , 9., , Use the binomial theorem to determine, correct to 5 significant figures:, (a) (0.98)7 (b) (2.01)9, [(a) 0.86813 (b) 535.51], , 10., , Evaluate (4.044)6 correct to 3 decimal places., [4373.880], , 7.4 Further worked problems on the, binomial series, Problem 11., (a) Expand, , 1, in ascending powers of x as, (1 +2x)3, far as the term in x 3, using the binomial series., , (b) State the limits of x for which the expansion, is valid., (1 + x)n ,, , (a) Using the binomial expansion of, n = −3 and x is replaced by 2x gives:, 1, = (1 + 2x)−3, (1 + 2x)3, = 1 + (−3)(2x) +, +, , where, , (−3)(−4), (2x)2, 2!, , (−3)(−4)(−5), (2x)3 + · · ·, 3!, , = 1 − 6x + 24x 2 − 80x 3 + · · ·, , i.e. |x| <, , 1, 1, 1, or − < x <, 2, 2, 2, , Problem 12., (a) Expand, , 1, in ascending powers of x as, (4 − x)2, far as the term in x 3 , using the binomial, theorem., , ., , [34749 p8 q 5 ], 7., , (b) The expansion is valid provided |2x| < 1,, , (b) What are the limits of x for which the expansion in (a) is true?, 1, 1, 1, = , = , x 2, x 2, (4 − x)2, 42 1 −, 4 1−, 4, 4, , , −2, 1, x, =, 1−, 16, 4, Using the expansion of (1 + x)n, 1, 1 , x −2, =, 1, −, (4 − x)2 16, 4, , x, 1, =, 1 + (−2) −, 16, 4, , , (−2)(−3), x 2, +, −, 2!, 4, (−2)(−3)(−4) x 3, −, +, +···, 3!, 4, , , 1, x 3x 2 x 3, =, 1+ +, + +···, 16, 2, 16, 16, x, (b) The expansion in (a) is true provided, < 1,, 4, i.e. |x| < 4 or −4 < x < 4, (a), , Problem, 13. Use the binomial theorem to expand, √, 4 + x in ascending powers of x to four terms. Give, the limits of x for which the expansion is valid., , x , 4 1+, 4, , 1, , √, x 2, x, = 4 1+, = 2 1+, 4, 4, n, Using the expansion of (1 + x) ,, 1, , x 2, 2 1+, 4, , , 1, x, (1/2)(−1/2) x 2, = 2 1+, +, 2, 4, 2!, 4, √, 4+x =
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The binomial series, (1/2)(−1/2)(−3/2) x 3, +···, 3!, 4, , , x, x2, x3, =2 1+ −, +, −···, 8 128 1024, +, , x3, x x2, − +, −···, 4 64 512, x, <1,, This is valid when, 4, i.e. |x| <4 or −4 < x < 4, =2+, , 1, in ascending, (1 −2t ), powers of t as far as the term in t 3., , Problem 14. Expand √, , State the limits of t for which the expression, is valid., 1, √, (1 − 2t ), 1, , = (1 − 2t )− 2, , , 1, (−1/2)(−3/2), = 1+ −, (−2t ) +, (−2t )2, 2, 2!, +, , (−1/2)(−3/2)(−5/2), (−2t )3 + · · ·,, 3!, , when expanded by the binomial theorem as far as the x, term only,, , , , x, 3x, = (1 − x) 1 +, 1−, 2, 2, , , x 3x when powers of x higher than, = 1−x + −, unity are neglected, 2, 2, = (1 − 2x), √, (1 + 2x), Problem 16. Express √, as a power, 3, (1 − 3x), 2, series as far as the term in x . State the range of, values of x for which the series is convergent., √, 1, 1, (1 + 2x), 2 (1 − 3x)− 3, =, (1, +, 2x), √, 3, (1 − 3x), , 1, 1, (2x), (1 + 2x) 2 = 1 +, 2, (1/2)(−1/2), +, (2x)2 + · · ·, 2!, x2, =1+ x −, + · · · which is valid for, 2, 1, |2x| < 1, i.e. |x| <, 2, 1, , (1 − 3x)− 3 = 1 + (−1/3)(−3x), , using the expansion for (1 + x)n, , +, , 3, 5, =1+t + t2 + t3 +···, 2, 2, , |3x| < 1, i.e. |x| <, , 1, 1, 1, or − < t <, 2, 2, 2, , √, √, 3, (1 − 3x) (1 + x), Problem 15. Simplify, , x 3, 1+, 2, given that powers of x above the first may be, neglected., √, √, 3, (1 − 3x) (1 + x), , x 3, 1+, 2, 1, 1 , x −3, = (1 − 3x) 3 (1 + x) 2 1 +, 2, , , , , , x , 1, 1, ≈ 1+, (−3x) 1 +, (x) 1 + (−3), 3, 2, 2, , (−1/3)(−4/3), (−3x)2 + · · ·, 2!, , = 1 + x + 2x 2 + · · · which is valid for, , The expression is valid when |2t | <1,, i.e. |t| <, , 63, , Hence, , √, 1, 1, (1 + 2x), 2 (1 − 3x)− 3, √, =, (1, +, 2x), 3, (1 − 3x), , , x2, = 1+x −, + · · · (1 + x + 2x 2 + · · ·), 2, = 1 + x + 2x 2 + x + x 2 −, , x2, ,, 2, , neglecting terms of higher power than 2,, 5, = 1 +2x + x 2, 2, 1, 1, The series is convergent if − < x <, 3, 3, , 1, 3
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64 Higher Engineering Mathematics, Now try the following exercise, (c), Exercise 30 Further problems on the, binomial series, In problems 1 to 5 expand in ascending powers of x, as far as the term in x 3, using the binomial theorem., State in each case the limits of x for which the series, is valid., 1, 1., (1 − x), [1 + x + x 2 + x 3 + · · ·, |x| < 1], 2., , 1, (1 + x)2, [1 − 2x + 3x 2 − 4x 3 + · · ·, |x| < 1], , 3., , 4., , 5., , 1, (2 + x)3, , √, 2+x, , ⎡ , ⎤, 3x 3x 2 5x 3, 1, 1, −, +, −, +, ·, ·, ·, ⎣8, ⎦, 2, 2, 4, |x| < 2, , ⎤, √, x x2, x3, 2, 1, +, −, +, −, ·, ·, ·, ⎦, ⎣, 4 32 128, |x| < 2, , √, 19, 1 + 5x, ≈ 1+ x, √, 3, 6, 1 − 2x, , 8. If x is very small such that x 2 and higher powers may be, determine the power, √ neglected,, √, x +4 3 8−x, series for , 5, (1 + x)3, , 31, 4− x, 15, 9. Express the following as power series in, ascending powers of x as far as the term in, x 2 . State in each case the range of x for which, the series is valid., , , , 1−x, (1 + x) 3 (1 − 3x)2, , (a), (b), 1+x, (1 + x 2 ), ⎡, ⎤, 1, (a) 1 − x + x 2 , |x| < 1, ⎢, ⎥, 2, ⎢, ⎥, ⎣, ⎦, 1, 7 2, (b) 1 − x − x , |x| <, 2, 3, , ⎡, , 1, √, 1 + 3x, ⎤, ⎡, 27 2 135 3, 3, ⎢ 1 − 2 x + 8 x − 16 x + · · · ⎥, ⎢, ⎥, ⎣, ⎦, 1, |x| <, 3, , 7.5 Practical problems involving the, binomial theorem, Binomial expansions may be used for numerical approximations, for calculations with small variations and in, probability theory (see Chapter 57)., Problem 17. The radius of a cylinder is reduced, by 4% and its height is increased by 2%. Determine, the approximate percentage change in (a) its, volume and (b) its curved surface area, (neglecting, the products of small quantities)., , 6. Expand (2 + 3x)−6 to three terms. For what, values of x is the expansion valid?, ⎡ , ⎤, 189 2, 1, 1, −, 9x, +, x, ⎢ 64, ⎥, 4, ⎢, ⎥, ⎣, ⎦, 2, |x| <, 3, , Volume of cylinder =πr 2 h., Let r and h be the original values of radius and, height., The new values are 0.96r or (1 − 0.04)r and 1.02h or, (1 + 0.02)h., , 7. When x is very small show that:, , (a) New volume = π[(1 − 0.04)r]2 [(1 + 0.02)h], , (a), (b), , (1 − x)2, , 5, 1, √, ≈1+ x, 2, (1 − x), , (1 − 2x), ≈ 1 + 10x, (1 − 3x)4, , = πr 2 h(1 − 0.04)2 (1 + 0.02), Now (1 − 0.04)2 = 1 −2(0.04) + (0.04)2, = (1 − 0.08),, neglecting powers of small terms.
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The binomial series, Hence new volume, , ≈, , ≈ πr 2 h(1 − 0.08)(1 + 0.02), ≈ πr 2 h(1 − 0.08 + 0.02), neglecting, products of small terms, ≈ πr 2 h(1 − 0.06) or 0.94πr 2 h, i.e. 94%, of the original volume, Hence the volume is reduced by approximately 6%., (b) Curved surface area of cylinder =2πrh., New surface area, = 2π[(1 − 0.04)r][(1 + 0.02)h], = 2πrh(1 − 0.04)(1 + 0.02), ≈ 2πrh(1 − 0.04 + 0.02), neglecting, products of small terms, ≈ 2πrh(1 − 0.02) or 0.98(2πrh),, i.e. 98% of the original surface area, , 65, , bl 3, bl 3, (1 − 0.040) or (0.96), , i.e. 96%, 12, 12, of the original second moment of area, , Hence the second moment of area is reduced by, approximately 4%., Problem 19. The resonant frequency, of a, 1 k, vibrating shaft is given by: f =, , where k is, 2π I, the stiffness and I is the inertia of the shaft. Use the, binomial theorem to determine the approximate, percentage error in determining the frequency using, the measured values of k and I when the measured, value of k is 4% too large and the measured value, of I is 2% too small., Let f , k and I be the true values of frequency, stiffness, and inertia respectively. Since the measured value of, stiffness, k1 , is 4% too large, then, 104, k = (1 + 0.04)k, 100, The measured value of inertia, I1 , is 2% too small, hence, k1 =, , 98, I = (1 − 0.02)I, 100, The measured value of frequency,, , 1 k1, 1 12 − 12, =, f1 =, k I, 2π I1, 2π 1 1, I1 =, , Hence the curved surface area is reduced by, approximately 2%., Problem 18. The second moment of area of a, bl 3, rectangle through its centroid is given by, ., 12, Determine the approximate change in the second, moment of area if b is increased by 3.5% and l is, reduced by 2.5%., New values of b and l are (1 + 0.035)b and (1 − 0.025)l, respectively., New second moment of area, =, , 1, [(1 + 0.035)b][(1 − 0.025)l]3, 12, , =, , bl 3, (1 + 0.035)(1 − 0.025)3, 12, , ≈, , ≈, , =, , 1, 1, 1, [(1 + 0.04)k] 2 [(1 − 0.02)I ]− 2, 2π, , =, , 1, 1, 1 1, 1, (1 + 0.04) 2 k 2 (1 − 0.02)− 2 I − 2, 2π, , =, , 1, 1, 1 1 −1, k 2 I 2 (1 + 0.04) 2 (1 − 0.02)− 2, 2π, 1, , i.e., , 1, , f1 = f (1 + 0.04) 2 (1 − 0.02)− 2, , , , , 1, 1, ≈ f 1+, (0.04) 1 + − (−0.02), 2, 2, ≈ f (1 + 0.02)(1 + 0.01), , bl 3, (1 + 0.035)(1 − 0.075), neglecting, 12, powers of small terms, , Neglecting the products of small terms,, , bl 3, (1 + 0.035 − 0.075), neglecting, 12, products of small terms, , Thus the percentage error in f based on the measured, values of k and I is approximately [(1.03)(100) − 100],, i.e. 3% too large., , f1 ≈ (1 + 0.02 + 0.01) f ≈ 1.03 f
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66 Higher Engineering Mathematics, Now try the following exercise, 7., , The shear stress τ in a shaft of diameter, kT, D under a torque T is given by: τ =, ., π D3, Determine the approximate percentage error, in calculating τ if T is measured 3% too small, and D 1.5% too large., [7.5% decrease], , 8., , The energy W stored in a flywheel is given, by: W = kr 5 N 2 , where k is a constant, r, is the radius and N the number of revolutions. Determine the approximate percentage, change in W when r is increased by 1.3% and, N is decreased by 2%., [2.5% increase], , 9., , An error of +1.5% was made when measuring the radius of a sphere. Ignoring the, products of small quantities determine the, approximate error in calculating (a) the volume, and (b) the surface area., , (a) 4.5% increase, (b) 3.0% increase, , In a series electrical circuit containing inductance L and capacitance C the resonant fre1, √, . If the, quency is given by: fr =, 2π LC, values of L and C used in the calculation are, 2.6% too large and 0.8% too small respectively, determine the approximate percentage, error in the frequency., [0.9% too small], , 10., , The power developed by an engine is given, by I = k PLAN, where k is a constant. Determine the approximate percentage change in, the power when P and A are each increased, by 2.5% and L and N are each decreased, by 1.4%., [2.2% increase], , The viscosity η of a liquid is given by:, kr 4, η=, , where k is a constant. If there is, νl, an error in r of +2%, in ν of +4% and l of, −3%, what is the resultant error in η?, [+7%], , 11., , A magnetic pole, distance x from the plane, of a coil of radius r, and on the axis of the, coil, is subject to a force F when a current flows in the coil. The force is given by:, kx, , where k is a constant. Use, F=, 2, (r + x 2 )5, the binomial theorem to show that when x is, small compared to r, then, kx 5kx 3, F≈ 5 −, ., 2r 7, r, The flow, of water through a pipe is given by:, (3d)5 H, . If d decreases by 2% and H, G=, L, by 1%, use the binomial theorem to estimate, the decrease in G., [5.5%], , Exercise 31 Further practical problems, involving the binomial theorem, 1., , 2., , 3., , 4., , Pressure p and volume v are related by, pv 3 = c, where c is a constant. Determine the, approximate percentage change in c when, p is increased by 3% and v decreased by, 1.2%., [0.6% decrease], Kinetic energy is given by 12 mv 2 . Determine, the approximate change in the kinetic energy, when mass m is increased by 2.5% and the, velocity v is reduced by 3%., [3.5% decrease], , 5., , The radius of a cone is increased by 2.7%, and its height reduced by 0.9%. Determine, the approximate percentage change in its, volume, neglecting the products of small, terms., [4.5% increase], , 6., , The electric field strength H due to a magnet, of length 2l and moment M at a point on its, axis distance x from the centre is given by, , , 1, 1, M, −, H=, 2l (x − l)2 (x + l)2, Show that if l is very small compared with x,, 2M, then H ≈ 3 ., x, , 12.
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Revision Test 2, This Revision Test covers the material contained in Chapters 5 to 7. The marks for each question are shown in, brackets at the end of each question., 1., , Evaluate correct to 4 significant figures:, (a) sinh 2.47, , (6), , The increase in resistance of strip conductors, due to eddy currents at power frequencies is, given by:, , αt sinh αt + sin αt, λ=, 2 cosh αt − cos αt, Calculate λ, correct to 5 significant figures, when, α = 1.08 and t = 1., (5), , 3., 4., , Find the sum of the first eight terms of the series, 1, 2.5, 6.25, . . ., correct to 1 decimal place. (4), , 10., , Determine the sum to infinity of the series, (3), 5, 1, 15 , . . ., , 11., , A machine is to have seven speeds ranging from, 25 rev/min to 500 rev/min. If the speeds form a, geometric progression, determine their value, each, correct to the nearest whole number., (8), , 12., , Use the binomial series to expand (2a − 3b)6 ., , 13., , , , 1 18, ., Determine the middle term of 3x −, 3y, , (b) tanh 0.6439, , (c) sech 1.385 (d) cosech 0.874, 2., , 9., , If A ch x − B sh x ≡ 4ex − 3e−x determine the, values of A and B., (6), , (7), , (6), 14., , Solve the following equation:, , (a), , 3.52 ch x + 8.42 sh x = 5.32, correct to 4 decimal places., , Expand the following in ascending powers of t as, far as the term in t 3, , (7), , 5., , Determine the 20th term of the series 15.6, 15,, 14.4, 13.8, . . ., (3), , 6., , The sum of 13 terms of an arithmetic progression, is 286 and the common difference is 3. Determine, the first term of the series., (4), , 7., , An engineer earns £21000 per annum and receives, annual increments of £600. Determine the salary, in the 9th year and calculate the total earnings in, the first 11 years., (5), , 8., , Determine the 11th term of the series 1.5, 3, 6,, 12, . . ., (2), , 1, 1, (b) √, 1+t, (1 − 3t ), , For each case, state the limits for which the, expansion is valid., (12), 15., , When x is very small show that:, 1, 3, √, ≈ 1− x, 2, (1 + x)2 (1 − x), , 16., , (5), , R4 θ, The modulus of rigidity G is given by G =, L, where R is the radius, θ the angle of twist and, L the length. Find the approximate percentage, error in G when R is measured 1.5% too large,, θ is measured 3% too small and L is measured, 1% too small., (7)
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Chapter 8, , Maclaurin’s series, 8.1, , Introduction, , Some mathematical functions may be represented as, power series, containing terms in ascending powers of, the variable. For example,, ex = 1 + x +, sin x = x −, , x2 x3, +, +···, 2!, 3!, , x3 x5 x7, +, −, +···, 3! 5!, 7!, , x2 x4, and cosh x = 1 +, +, +···, 2! 4!, (as introduced in Chapter 5), Using a series, called Maclaurin’s series, mixed functions containing, say, algebraic, trigonometric and exponential functions, may be expressed solely as algebraic, functions, and differentiation and integration can often, be more readily performed., To expand a function using Maclaurin’s theorem,, some knowledge of differentiation is needed (More on, differentiation is given in Chapter 27). Here is a revision, or f (x), , y or f (x), , dy, dx, , ax n, , anx n−1, , sin ax, , a cos ax, , cos ax, , −a sin ax, , eax, , aeax, , ln ax, , 1, x, , sinh ax, , a cosh ax, , cosh ax, , a sinh ax, , of derivatives of the main functions needed in this, chapter., Given a general function f (x), then f (x) is the, first derivative, f (x) is the second derivative, and so, on. Also, f (0) means the value of the function when, x = 0, f (0) means the value of the first derivative when, x = 0, and so on., , 8.2, , Derivation of Maclaurin’s theorem, , Let the power series for f (x) be, f (x) = a0 + a1 x + a2 x 2 + a3 x 3 + a4 x 4, + a5 x 5 + · · ·, , (1), , where a0 , a1, a2, . . . are constants., When x = 0, f(0) = a0 ., Differentiating equation (1) with respect to x gives:, f (x) = a1 + 2a2 x + 3a3 x 2 + 4a4 x 3, + 5a5 x 4 + · · ·, , (2), , When x = 0, f (0) = a1 ., Differentiating equation (2) with respect to x gives:, f (x) = 2a2 + (3)(2)a3 x + (4)(3)a4 x 2, + (5)(4)a5 x 3 + · · · (3), f (0), 2!, Differentiating equation (3) with respect to x gives:, , When x = 0, f (0) = 2a2 = 2! a2 , i.e. a2 =, , f (x) = (3)(2)a3 + (4)(3)(2)a4 x, + (5)(4)(3)a5 x 2 + · · ·, , (4), , When x = 0, f (0) = (3)(2)a3 = 3! a3, i.e. a3 =, , f (0), 3!
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Maclaurin’s series, Continuing the same procedure gives a4 =, v, , f iv (0), ,, 4!, , f (0), , and so on., 5!, Substituting for a0 , a1, a2, . . . in equation (1) gives:, a5 =, , f (x) = f (0) + f (0)x +, +, , f(x) = f(0) + xf (0) +, i.e., , The values of f (0), f (0), f (0), . . . in the Maclaurin’s, series are obtained as follows:, , f (0) 3, x +···, 3!, , x3, + f (0) + · · ·, 3!, , 8.4 Worked problems on Maclaurin’s, series, Problem 1. Determine the first four terms of the, power series for cos x., , f (0) 2, x, 2!, , x2 , f (0), 2!, , 69, , (5), , Equation (5) is a mathematical statement called, Maclaurin’s theorem or Maclaurin’s series., , f (x) = cos x, , f (0) = cos 0 = 1, , f (x) = −sin x, , f (0) = −sin 0 = 0, , f (x) = −cos x, , f (0) = −cos 0 =−1, , f (x) = sin x, , f (0) = sin 0 =0, , f iv (x) = cos x, , f iv (0) = cos 0 = 1, , f v (x) = −sin x, , f v (0) = −sin 0 = 0, , f vi (x) = −cos x, , f vi (0) = −cos 0 =−1, , Substituting these values into equation (5) gives:, , 8.3, , Conditions of Maclaurin’s series, , f (x) = cos x = 1 + x(0) +, , Maclaurin’s series may be used to represent any function, say f (x), as a power series provided that at, x = 0 the following three conditions are met:, (a), , f(0) = ∞, For example, for the function f (x) = cos x,, f (0) = cos 0 =1, thus cos x meets the condition. However, if f (x) = ln x, f (0) = ln 0 =−∞,, thus ln x does not meet this condition., , (b) f (0), f (0), f (0), . . . = ∞, For example, for the function f (x) = cos x,, f (0) = −sin 0 =0, f (0) = −cos 0 = −1, and so, on; thus cos x meets this condition. However, if, f (x) = ln x, f (0) = 10 = ∞, thus ln x does not, meet this condition., (c), , +, , The resultant Maclaurin’s series must be, convergent, In general, this means that the values of the terms,, or groups of terms, must get progressively smaller, and the sum of the terms must reach a limiting, value., For example, the series 1 + 12 + 14 + 18 + · · · is convergent since the value of the terms is getting, smaller and the sum of the terms is approaching a, limiting value of 2., , i.e., , cos x = 1 −, , x2, x3, (−1) + (0), 2!, 3!, , x4, x5, x6, (1) + (0) + (−1) + · · ·, 4!, 5!, 6!, , x2 x4 x6, +, −, + ···, 2! 4! 6!, , Problem 2. Determine the power series for, cos 2θ., Replacing x with 2θ in the series obtained in, Problem 1 gives:, cos 2θ = 1 −, = 1−, , (2θ)2 (2θ)4 (2θ)6, +, −, +···, 2!, 4!, 6!, 4θ 2 16θ 4 64θ 6, +, −, +···, 2, 24, 720, , 2, 4, i.e. cos 2θ = 1 − 2θ 2 + θ 4 − θ 6 + · · ·, 3, 45, Problem 3. Using Maclaurin’s series, find the, first 5 (non zero) terms for the function, f (x) = sin x.
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70 Higher Engineering Mathematics, f (x) = sin x, , f (0) = sin 0 = 0, , f (x) = cos x, , f (0) = cos 0 = 1, , f (x) = −sin x, , f (0) = −sin 0 = 0, , f (x) = −cos x, , f (0) = −cos 0 = −1, , f iv (x) = sin x, , f iv (0) = sin 0 = 0, , f v (x) = cos x, , f v (0) = cos 0 = 1, , f vi (x) = −sin x, , f vi (0) = −sin 0 = 0, , f vii (x) = −cos x, , f vii(0) = −cos 0 = −1, , Problem 5. Determine the power series for tan x, as far as the term in x 3 ., f (x) = tan x, f (0) = tan 0 = 0, f (x) = sec2 x, f (0) = sec2 0 =, , Substituting the above values into Maclaurin’s series of, equation (5) gives:, sin x = 0 + x (1) +, , 2, , 3, , x5, x6, x7, +, (1) +, (0) +, (−1) + · · ·· · ·, 5!, 6!, 7!, i.e. sin x = x −, , f (0) = 2 sec2 0 tan 0 = 0, f (x) = (2 sec2 x)(sec 2 x), + (tan x)(4 sec x sec x tan x), by the, product rule,, = 2 sec4 x + 4 sec2 x tan 2 x, , f (0) = e = 1, , f (x) = 3 e3x, , f (0) = 3 e0 = 3, , f (x) = 9 e3x, , f (0) = 9 e0 = 9, , f (x) = 27 e3x, , f (0) = 27 e0 = 27, , f iv (x) = 81 e3x, , f iv (0) = 81 e0 = 81, , 0, , Problem 6., , x2, x3, = 1 + x (3) +, (9) +, (27), 2!, 3!, x4, +, (81) + · · ·· · ·, 4!, , e3x = 1 + 3x +, , i.e. e3x = 1 + 3x +, , 2!, , +, , 27x 3, 3!, , Substituting these values into equation (5) gives:, , +, , x2, x3, (0) + (2), 2!, 3!, , 1, i.e. tan x = x + x3, 3, , Substituting the above values into Maclaurin’s series of, equation (5) gives:, , 9x 2, , f (0) = 2 sec4 0 + 4 sec2 0 tan2 0 = 2, , f (x) = tan x = 0 + (x)(1) +, , f (x) = e, , e3x, , = 2 sec2 x tan x, , x3, x5, x7, +, −, + ···, 3!, 5!, 7!, , Problem 4. Using Maclaurin’s series, find the, first five terms for the expansion of the function, f (x) = e3x ., 3x, , f (x) = (2 sec x)(sec x tan x), , 4, , x, x, x, (0) + (−1) + (0), 2!, 3!, 4!, , 1, =1, cos2 0, , 81x 4, 4!, , f (x) = ln(1 + x), , f (0) = ln(1 +0) = 0, , f (x) =, , 1, (1 + x), , f (0) =, , 1, =1, 1+0, , f (x) =, , −1, (1 + x)2, , f (0) =, , −1, = −1, (1 + 0)2, , f (x) =, , 2, (1 + x)3, , f (0) =, , 2, =2, (1 + 0)3, , f iv (x) =, , −6, (1 + x)4, , f iv (0) =, , −6, = −6, (1 + 0)4, , f v (x) =, , 24, (1 + x)5, , f v (0) =, , 24, = 24, (1 + 0)5, , +···, , 9x2 9x3 27x4, +, +, + ···, 2, 2, 8, , Expand ln(1 + x) to five terms.
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Maclaurin’s series, Substituting these values into equation (5) gives:, x2, (−1), 2!, x4, x5, x3, (2) + (−6) + (24), +, 3!, 4!, 5!, , f (x) = ln(1 + x) = 0 + x(1) +, , x2 x3 x4 x5, +, −, +, − ···, i.e. ln(1 + x) = x −, 2, 3, 4, 5, Problem 7. Expand ln(1 − x) to five terms., Replacing x by −x in the series for ln(1 + x) in, Problem 6 gives:, ln(1 − x) = (−x) −, , (−x)2 (−x)3, +, 2, 3, (−x)4 (−x)5, −, +, −···, 4, 5, , i.e. ln(1 − x) = −x −, , x2 x3 x4 x5, −, −, −, − ···, 2, 3, 4, 5, , Problem, 8., , Determine the power series for, 1+x, ., ln, 1−x, , , , 1+x, = ln(1 + x) − ln(1 − x) by the laws of logln, 1−x, arithms, and from Problems 6 and 7,, , , 1+x, ln, 1−x, , , , , , , x2 x3 x4 x5, = x−, +, −, +, −···, 2, 3, 4, 5, , , x2 x3 x4 x5, −, −, −, −···, − −x −, 2, 3, 4, 5, , 2, 2, = 2x + x 3 + x 5 + · · ·, 3, 5, , , , , 1+x, x3, x5, i.e. ln, =2 x+, +, + ···, 1−x, 3, 5, Problem 9. Use Maclaurin’s series to find the, expansion of (2 + x)4 ., f (x) = (2 + x)4, , 24 = 16, , f (0) =, , f (x) = 4(2 + x)3, , f (0) = 4(2)3 = 32, , f (x) = 12(2 + x)2, , f (0) = 12(2)2 = 48, , f (x) = 24(2 + x)1, , f (0) = 24(2) = 48, , f iv (x) = 24, , f iv (0) = 24, , 71, , Substituting in equation (5) gives:, (2 + x)4, = f (0) + x f (0) +, = 16 + (x)(32) +, , x2, x3, x 4 iv, f (0) +, f (0) +, f (0), 2!, 3!, 4!, , x3, x4, x2, (48) + (48) + (24), 2!, 3!, 4!, , = 16 + 32x + 24x2 + 8x3 + x4, (This expression could have been obtained by applying, the binomial theorem.), x, , Problem 10. Expand e 2 as far as the term in x 4 ., x, , f (x) = e 2, , f (0) = e0 = 1, , 1 x, f (x) = e 2, 2, , 1, 1, f (0) = e0 =, 2, 2, , 1 x, f (x) = e 2, 4, , 1, 1, f (0) = e0 =, 4, 4, , 1 x, f (x) = e 2, 8, , 1, 1, f (0) = e0 =, 8, 8, , f iv (x) =, , 1 x, e2, 16, , f iv (0) =, , 1 0, 1, e =, 16, 16, , Substituting in equation (5) gives:, x, , e 2 = f (0) + x f (0) +, , x2, f (0), 2!, , x 4 iv, x3, f (0) +, f (0) + · · ·, 3!, 4!, , , , 1, x2 1, x3 1, +, +, = 1 + (x), 2, 2! 4, 3! 8, , x4 1, +···, +, 4! 16, +, , x, 1, 1, 1 4, 1, i.e. e 2 = 1 + x + x2 + x3+, x + ···, 2, 8, 48, 384
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72 Higher Engineering Mathematics, Problem 11. Develop a series for sinh x using, Maclaurin’s series., , f (x) = sinh x, f (x) = cosh x, , e0 − e−0, f (0) = sinh 0 =, =0, 2, 0, −0, e +e, f (0) = cosh 0 =, =1, 2, f (0) = sinh 0 = 0, f (0) = cosh 0 = 1, , f iv (x) = sinh x, f v (x) = cosh x, , f iv (0) = sinh 0 = 0, f v (0) = cosh 0 = 1, , f (x) = sinh x, f (x) = cosh x, , Substituting in equation (5) gives:, x2, x3, sinh x = f (0) + x f (0) +, f (0) +, f (0), 2!, 3!, x 4 iv, x5 v, +, f (0) +, f (0) + · · ·, 4!, 5!, x2, x3, x4, = 0 + (x)(1) + (0) + (1) + (0), 2!, 3!, 4!, x5, + (1) + · · ·, 5!, 3, 5, x, x, i.e. sinh x = x + +, + ···, 3! 5!, (as obtained in Section 5.5, page 49), Problem 12. Produce a power series for cos2 2x, as far as the term in x 6 ., , Now try the following exercise, Exercise 32 Further problems on, Maclaurin’s series, 1. Determine the first four terms of the power, series for sin 2x⎡using Maclaurin’s series. ⎤, 4 5, 4 3, ⎢sin 2x = 2x − 3 x + 15 x ⎥, ⎣, ⎦, 8 7, −, x +···, 315, 2. Use Maclaurin’s series to produce a power, series for cosh 3x as far as the term in x 6 ., , 9, 27, 81, 1 + x2 + x4 + x6, 2, 8, 80, 3. Use Maclaurin’s theorem to determine the first, x, three terms of the power series, for ln(1 + e2 )., x x, ln 2 + +, 2, 8, 4. Determine the power series for cos 4t as far as, the term in t 6 ., , 32, 256 6, t, 1 − 8t 2 + t 4 −, 3, 45, 3, , From double angle formulae, cos 2 A = 2 cos2 A − 1 (see, Chapter 17)., 1, from which,, cos2 A = (1 + cos 2 A), 2, 1, and, cos2 2x = (1 + cos 4x), 2, , 5. Expand e 2 x in a power, series as far as the term, 3, 9, 9, 3, 1 + x + x2 + x3, in x ., 2, 8, 16, 4, 6. Develop, as far as the term, in x , the power, 10, series for sec 2x., 1 + 2x 2 + x 4, 3, , From Problem 1,, x2 x4 x6, +, −, +···, 2! 4! 6!, (4x)2 (4x)4 (4x)6, hence, cos 4x = 1 −, +, −, +···, 2!, 4!, 6!, 32, 256 6, x +···, = 1 − 8x 2 + x 4 −, 3, 45, 1, Thus cos2 2x = (1 + cos 4x), 2, , , 32, 256 6, 1, 1 + 1 − 8x 2 + x 4 −, x +···, =, 2, 3, 45, cos x = 1 −, , i.e. cos2 2x = 1− 4x2 +, , 16 4 128 6, x −, x +···, 3, 45, , 7. Expand e2θ cos 3θ as far as theterm in θ 2 using, 5, Maclaurin’s series., 1 + 2θ − θ 2, 2, 8. Determine the first three terms of the series for, sin2 x by applying Maclaurin’s theorem., , 1, 2, x2 − x4 + x6 · · ·, 3, 45, 9. Use Maclaurin’s series to determine the expansion of (3 + 2t )4., , 81 + 216t + 216t 2 + 96t 3 + 16t 4
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74 Higher Engineering Mathematics, θ3, θ5, θ7, = θ−, +, −, +···, 18 600 7(5040), =1−, , 1, , 0, , 1, 1, 1, +, −, +···, 18 600 7(5040), , = 0.946, correct to 3 significant figures., 0.4, Problem 15. Evaluate 0 x ln(1 + x) dx using, Maclaurin’s theorem, correct to 3 decimal places., From Problem 6,, x2 x3 x4 x5, +, −, +, −···, ln(1 + x) = x −, 2, 3, 4, 5, ! 0.4, x ln(1 + x)dx, Hence, 0, , , , , x2 x3 x4 x5, x x−, +, −, +, − · · · dx, =, 2, 3, 4, 5, 0, , ! 0.4 , x3 x4 x5 x6, 2, x −, =, +, −, +, − · · · dx, 2, 3, 4, 5, 0, !, , x3 x4 x5 x6 x7, −, +, −, +, −···, 3, 8, 15 24 35, , =, , =, , 0.4, , 0.4, , 4. Use, Maclaurin’s theorem to, √, x ln(x + 1) as a power series., evaluate, correct to 3 decimal, 0.5 √, x ln (x + 1) dx., 0, , 8.6, , expand, Hence, places,, [0.061], , Limiting values, , It is sometimes, necessary to find limits of the form, , f (x), , where f (a) = 0 and g(a) = 0., lim, x→a g(x), For example,, , lim, , x→1, , , 1+3−4 0, x 2 + 3x − 4, =, =, x 2 − 7x + 6, 1−7+6 0, , and 00 is generally referred to as indeterminate., For certain limits a knowledge of series can sometimes, help., For example,, , 0, , (0.4)3 (0.4)4 (0.4)5 (0.4)6, −, +, −, 3, 8, 15, 24, +, , 1√, 3. Determine the value of 0 θ cos θ dθ, correct to 2 significant figures, using Maclaurin’s, series., [0.53], , , (0.4)7, − · · · − (0), 35, , = 0.02133 − 0.0032 + 0.0006827 − · · ·, = 0.019, correct to 3 decimal places., Now try the following exercise, Exercise 33 Further problems on, numerical integration using Maclaurin’s, series, 0.6, 1. Evaluate 0.2 3esin θ dθ, correct to 3 decimal, places, using Maclaurin’s series., [1.784], 2. Use Maclaurin’s theorem to expand cos2θ and, hence evaluate, correct to 2 decimal places,, ! 1, cos 2θ, dθ., [0.88], 1, 0, 3, θ, , , , , tan x − x, x→0, x3, ⎧, ⎫, 1, ⎪, ⎨ x + x3 + · · · − x ⎪, ⎬, 3, ≡ lim, from Problem 5, ⎪, x→0 ⎪, x3, ⎩, ⎭, lim, , ⎧, ⎫, 1, ⎪, , ⎨ x3 + · · ·⎪, ⎬, 1, 1, 3, = lim, =, = lim, 3, ⎪, x→0 ⎪, x→0, x, 3, 3, ⎩, ⎭, Similarly,, , sinh x, x→0, x, ⎫, ⎧, x3 x5 ⎪, ⎪, ⎪, ⎬, ⎨x +, +, +⎪, 3!, 5!, ≡ lim, from Problem 11, ⎪, x→0 ⎪, x, ⎪, ⎪, ⎭, ⎩, , , lim, , , , x2 x4, +, +··· = 1, = lim 1 +, x→0, 3!, 5!
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Maclaurin’s series, However, a knowledge of series does not help with, , 2, x + 3x − 4, examples such as lim, x→1 x 2 − 7x + 6, , Substituting x = 0 gives, , L’Hopital’s rule will enable us to determine such, limits when the differential coefficients of the numerator, and denominator can be found., , Applying L’Hopital’s rule again gives, , , , , cos x − 1, −sin x, lim, = lim, =0, x→0, x→0, 2x, 2, , L’Hopital’s rule states:, , , , f(x), f (x), lim, = lim , x→a g(x), x→a g (x), , , , , , Problem 18. Determine lim, , x→0, , provided g (a) = 0, , f (x), is still 00 ; if so, the, x→a g (x), numerator and denominator are differentiated again, (and again) until a non-zero value is obtained for the, denominator., The following worked problems demonstrate how, L’Hopital’s rule is used. Refer to Chapter 27 for methods, of differentiation., It can happen that lim, , , Problem 16. Determine lim, , x→1, , x 2 + 3x − 4, x 2 − 7x + 6, , , , The first step is to substitute x = 1 into both numerator and denominator. In this case we obtain 00 . It is, only when we obtain such a result that we then use, L’Hopital’s rule. Hence applying L’Hopital’s rule,, , , 2, , x + 3x − 4, 2x + 3, = lim, lim 2, x→1 x − 7x + 6, x→1 2x − 7, i.e. both numerator and, denominator have, been differentiated, =, , 5, = −1, −5, , , Problem 17. Determine lim, , x→0, , cos 0 − 1 1 − 1 0, =, =, again, 0, 0, 0, , sin x − x, x2, , , , Substituting x = 0 gives, , , sin x − x, sin 0 − 0 0, lim, =, =, 2, x→0, x, 0, 0, Applying L’Hopital’s rule gives, , , , , sin x − x, cos x − 1, lim, =, lim, x→0, x→0, x2, 2x, , x − sin x, x − tan x, , , , Substituting x = 0 gives, , , 0 − sin 0 0, x − sin x, lim, =, =, x→0 x − tan x, 0 − tan 0 0, Applying L’Hopital’s rule gives, , , , , x − sin x, 1 − cos x, = lim, lim, x→0 x − tan x, x→0 1 − sec2 x, Substituting x = 0 gives, , , 1 − cos 0, 1 − cos x, 1−1 0, =, lim, =, = again, x→0 1 − sec 2 x, 1 − sec2 0 1 − 1 0, Applying L’Hopital’s rule gives, , , , , 1 − cos x, sin x, =, lim, lim, x→0 1 − sec2 x, x→0 (−2 sec x)(sec x tan x), , , sin x, = lim, x→0 −2 sec2 x tan x, Substituting x = 0 gives, 0, sin 0, =, again, 2, −2 sec 0 tan 0 0, Applying L’Hopital’s rule gives, , , sin x, lim, x→0 −2 sec2 x tan x, ⎧, ⎫, ⎪, ⎪, ⎪, ⎪, ⎨, ⎬, cos x, = lim, 2, 2, x→0 ⎪, x) ⎪, ⎪, ⎪, ⎩ (−2 sec x)(sec, ⎭, 2, + (tan x)(−4 sec x tan x), using the product rule, Substituting x = 0 gives, 1, cos 0, =, −2 sec4 0 − 4 sec2 0 tan2 0 −2 − 0, =−, , 1, 2, , 75
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Chapter 9, , Solving equations by, iterative methods, 9.1, , Introduction to iterative methods, , Many equations can only be solved graphically or by, methods of successive approximations to the roots,, called iterative methods. Three methods of successive, approximations are (i) bisection method, introduced in, Section 9.2, (ii) an algebraic method, introduction in, Section 9.3, and (iii) by using the Newton-Raphson, formula, given in Section 9.4., Each successive approximation method relies on a, reasonably good first estimate of the value of a root, being made. One way of determining this is to sketch a, graph of the function, say y = f (x), and determine the, approximate values of roots from the points where the, graph cuts the x-axis. Another way is by using a functional notation method. This method uses the property, that the value of the graph of f (x) = 0 changes sign for, values of x just before and just after the value of a root., f(x), 8, , f(x)⫽x 2⫺x⫺6, , 4, , ⫺4, , ⫺2, , 0, ⫺4, ⫺6, , Figure 9.1, , 2, , 4, , x, , For example, one root of the equation x 2 − x − 6 = 0 is, x = 3. Using functional notation:, f (x) = x 2 − x − 6, f (2) = 22 − 2 − 6 = −4, f (4) = 42 − 4 − 6 = +6, It can be seen from these results that the value of f (x), changes from −4 at f (2) to +6 at f (4), indicating that, a root lies between 2 and 4. This is shown more clearly, in Fig. 9.1., , 9.2, , The bisection method, , As shown above, by using functional notation it is possible to determine the vicinity of a root of an equation by, the occurrence of a change of sign, i.e. if x 1 and x 2 are, such that f (x 1 ) and f (x 2 ) have opposite signs, there is, at least one root of the equation f (x) = 0 in the interval, between x 1 and x 2 (provided f (x) is a continuous function). In the method of bisection the mid-point of the, x1 + x2, interval, i.e. x 3 =, , is taken, and from the sign, 2, of f (x 3 ) it can be deduced whether a root lies in the, half interval to the left or right of x 3 . Whichever half, interval is indicated, its mid-point is then taken and the, procedure repeated. The method often requires many, iterations and is therefore slow, but never fails to eventually produce the root. The procedure stops when two, successive values of x are equal—to the required degree, of accuracy., The method of bisection is demonstrated in Problems 1 to 3 following.
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78 Higher Engineering Mathematics, Problem 1. Use the method of bisection to find, the positive root of the equation 5x 2 + 11x − 17 =0, correct to 3 significant figures., Let f (x) = 5x 2 + 11x − 17, then, using functional notation:, , Hence, f (1.25) = 5(1.25)2 + 11x − 17, = +4.5625, Since f (1) is negative and f (1.25) is positive, a root, lies between x = 1 and x = 1.25., 1 +1.25, Bisecting this interval gives, i.e. 1.125, 2, Hence, , f (0) = −17, f (1) = 5(1)2 + 11(1) − 17 = −1, , f (1.125) = 5(1.125)2 + 11(1.125) − 17, , f (2) = 5(2)2 + 11(2) − 17 = +25, , = +1.703125, , Since there is a change of sign from negative, to positive there must be a root of the equation between, x = 1 and x = 2. This is shown graphically in Fig. 9.2., f(x), , Since f (1) is negative and f (1.125) is positive, a root, lies between x = 1 and x = 1.125., 1 +1.125, Bisecting this interval gives, i.e. 1.0625., 2, Hence, f (1.0625) = 5(1.0625)2 + 11(1.0625) − 17, , 20, , = +0.33203125, , f (x) ⫽ 5x 2 ⫹ 11x ⫺ 17, 10, , ⫺4, , ⫺3 ⫺2, , ⫺1, , 0, , 1, , 2, , x, , ⫺10, ⫺17, ⫺20, , Figure 9.2, , Since f (1) is negative and f (1.0625) is positive, a root, lies between x = 1 and x = 1.0625., 1 +1.0625, Bisecting this interval gives, i.e. 1.03125., 2, Hence, f (1.03125) = 5(1.03125)2 + 11(1.03125) − 17, = −0.338867 . . ., Since f (1.03125) is negative and f (1.0625) is positive,, a root lies between x = 1.03125 and x = 1.0625., , The method of bisection suggests that the root is at, 1+2, = 1.5, i.e. the interval between 1 and 2 has been, 2, bisected., , Bisecting this interval gives, 1.03125 + 1.0625, i.e. 1.046875., 2, Hence, , Hence, f (1.5) = 5(1.5) + 11(1.5) − 17, 2, , = +10.75, Since f (1) is negative, f (1.5) is positive, and f (2) is, also positive, a root of the equation must lie between, x = 1 and x = 1.5, since a sign change has occurred, between f (1) and f (1.5)., 1 +1.5, i.e. 1.25 as the next, Bisecting this interval gives, 2, root., , f (1.046875) = 5(1.046875)2 + 11(1.046875) − 17, = −0.0046386. . ., Since f (1.046875) is negative and f (1.0625) is positive, a root lies between x = 1.046875 and x = 1.0625., Bisecting this interval gives, 1.046875 + 1.0625, i.e. 1.0546875., 2, The last three values obtained for the root are 1.03125,, 1.046875 and 1.0546875. The last two values are both
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Solving equations by iterative methods, 1.05, correct to 3 significant figure. We therefore stop, the iterations here., Thus, correct to 3 significant figures, the positive root, of 5x2 + 11x − 17 = 0 is 1.05, , Since f (1.75) is negative and f (1.5) is positive, a root, lies between x = 1.75 and x = 1.5., 1.75 + 1.5, Bisecting this interval gives, i.e. 1.625., 2, Hence, f (1.625) = 1.625 + 3 − e1.625, , Problem 2. Use the bisection method to determine the positive root of the equation x + 3 = e x ,, correct to 3 decimal places., , = −0.45341. . ., Since f (1.625) is negative and f (1.5) is positive, a root, lies between x = 1.625 and x = 1.5., 1.625 +1.5, Bisecting this interval gives, i.e. 1.5625., 2, Hence, , Let f (x) = x + 3 −ex, then, using functional notation:, f (0) = 0 + 3 − e0 = +2, f (1) = 1 + 3 − e1 = +1.2817 . . ., f (2) = 2 + 3 − e2 = −2.3890 . . ., , f (1.5625) = 1.5625 + 3 − e1.5625, , Since f (1) is positive and f (2) is negative, a root lies, between x = 1 and x = 2. A sketch of f (x) = x + 3 − ex ,, i.e. x + 3 =ex is shown in Fig. 9.3., f(x), , = −0.20823 . . ., Since f (1.5625) is negative and f (1.5) is positive, a, root lies between x = 1.5625 and x = 1.5., Bisecting this interval gives, , f(x) 5 x 1 3, , 1.5625 + 1.5, i.e. 1.53125., 2, , 4, , Hence, , 3, , 2, , f (1.53125) = 1.53125 + 3 − e1.53125, , f(x) 5 e x, , = −0.09270 . . ., Since f (1.53125) is negative and f (1.5) is positive, a, root lies between x = 1.53125 and x = 1.5., , 1, , 22, , 21, , 79, , 0, , 1, , 2 x, , Bisecting this interval gives, 1.53125 +1.5, i.e. 1.515625., 2, , Figure 9.3, , Bisecting the interval between x = 1 and x = 2 gives, 1 +2, i.e. 1.5., 2, Hence, f (1.5) = 1.5 + 3 − e1.5, = +0.01831. . ., Since f (1.5) is positive and f (2) is negative, a root lies, between x = 1.5 and x = 2., 1.5 + 2, Bisecting this interval gives, i.e. 1.75., 2, Hence, f (1.75) = 1.75 + 3 − e1.75, = −1.00460. . ., , Hence, f (1.515625) = 1.515625 + 3 − e1.515625, = −0.03664 . . ., Since f (1.515625) is negative and f (1.5) is positive, a, root lies between x = 1.515625 and x = 1.5., Bisecting this interval gives, 1.515625 + 1.5, i.e. 1.5078125., 2, Hence, f (1.5078125) = 1.5078125 + 3 − e1.5078125, = −0.009026 . . .
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80 Higher Engineering Mathematics, Since f (1.5078125) is negative and f (1.5) is positive,, a root lies between x = 1.5078125 and x = 1.5., , Hence the root of x + 3 =ex is x = 1.505, correct to 3, decimal places., , Bisecting this interval gives, , The above is a lengthy procedure and it is probably, easier to present the data in a table as shown in the, table., , 1.5078125 + 1.5, i.e. 1.50390625., 2, , x1, , Hence, , x2, , x3 =, , x1 + x2, 2, , f (x3 ), , f (1.50390625) = 1.50390625 + 3 − e1.50390625, , 0, , +2, , = +0.004676. . ., , 1, , +1.2817..., , Since f (1.50390625) is positive and f (1.5078125), is negative, a root lies between x = 1.50390625 and, x = 1.5078125., Bisecting this interval gives, , 2, , −2.3890..., , 1.50390625 + 1.5078125, i.e. 1.505859375., 2, Hence, f (1.505859375) = 1.505859375 + 3 − e, , 1.505859375, , = −0.0021666. . ., Since f (1.50589375) is negative and f (1.50390625), is positive, a root lies between x = 1.50589375 and, x = 1.50390625., Bisecting this interval gives, 1.505859375 + 1.50390625, i.e. 1.504882813., 2, Hence, f (1.504882813) = 1.504882813 + 3 − e1.504882813, , 1, , 2, , 1.5, , +0.0183..., , 1.5, , 2, , 1.75, , −1.0046..., , 1.5, , 1.75, , 1.625, , −0.4534..., , 1.5, , 1.625, , 1.5625, , −0.2082..., , 1.5, , 1.5625, , 1.53125, , −0.0927..., , 1.5, , 1.53125, , 1.515625, , −0.0366..., , 1.5, , 1.515625, , 1.5078125, , −0.0090..., , 1.5, , 1.5078125, , 1.50390625, , +0.0046..., , 1.50390625, , 1.5078125, , 1.505859375 −0.0021..., , 1.50390625, , 1.505859375 1.504882813 +0.0012..., , 1.504882813 1.505859375 1.505388282, , Problem 3. Solve, correct to 2 decimal places,, the equation 2 ln x + x = 2 using the method of, bisection., , = +0.001256. . ., Since f (1.504882813) is positive and, f (1.505859375) is negative,, a root lies between x = 1.504882813 and x =, 1.505859375., Bisecting this interval gives, 1.504882813 + 1.50589375, i.e. 1.505388282., 2, The last two values of x are 1.504882813 and, 1.505388282, i.e. both are equal to 1.505, correct to, 3 decimal places., , Let, , f (x) = 2 ln x + x − 2, f (0.1) = 2 ln(0.1) + 0.1 − 2 = −6.5051 . . ., (Note that ln 0 is infinite that is why, x = 0 was not chosen), f (1) = 2 ln 1 + 1 − 2 = −1, f (2) = 2 ln 2 + 2 − 2 = +1.3862 . . ., , A change of sign indicates a root lies between x = 1 and, x = 2., Since 2 ln x + x = 2 then 2 ln x = −x + 2; sketches of, 2 ln x and −x + 2 are shown in Fig. 9.4.
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Solving equations by iterative methods, , 81, , f(x), 2, , 1. Find the positive root of the equation, x 2 + 3x − 5 = 0, correct to 3 significant, figures, using the method of bisection. [1.19], , f(x)⫽⫺x⫹2, f(x)⫽ 2In x, , 1, , 2, , 1, , 0, , 3, , 2. Using the bisection method solve ex − x = 2,, correct to 4 significant figures., [1.146], 4, , x, , ⫺1, , 3. Determine the positive root of x 2 = 4 cos x,, correct to 2 decimal places using the method, of bisection., [1.20], , Figure 9.4, , 4. Solve x − 2 − ln x = 0 for the root near to 3,, correct to 3 decimal places using the bisection, method., [3.146], , As shown in Problem 2, a table of values is produced, to reduce space., , 5. Solve, correct to 4 significant figures,, x − 2 sin2 x = 0 using the bisection method., [1.849], , ⫺2, , x1, , x2, , x3 =, , x1 + x2, 2, , f (x 3 ), , 0.1, , −6.6051 . . ., , 1, , −1, , 2, , +1.3862 . . ., , 9.3 An algebraic method of successive, approximations, This method can be used to solve equations of the form:, , 1, , 2, , 1.5, , +0.3109 . . ., , 1, , 1.5, , 1.25, , −0.3037 . . ., , 1.25, , 1.5, , 1.375, , +0.0119 . . ., , 1.25, , 1.375 1.3125, , −0.1436 . . ., , where a, b, c, d, . . . are constants., Procedure:, , 1.3125, , 1.375 1.34375, , −0.0653 . . ., , 1.375 1.359375, , −0.0265 . . ., , First approximation, , 1.34375, 1.359375, , 1.375 1.3671875, , −0.0073 . . ., , 1.3671875 1.375 1.37109375, , a + bx + cx 2 + d x 3 + · · · = 0,, , (a), , +0.0023 . . ., , The last two values of x 3 are both equal to 1.37 when, expressed to 2 decimal places. We therefore stop the, iterations., , Using a graphical or the functional notation, method (see Section 9.1) determine an approximate value of the root required, say x 1., , Second approximation, (b) Let the true value of the root be (x 1 + δ1 )., (c), , Hence, the solution of 2 ln x + x = 2 is x = 1.37, correct to 2 decimal places., , Determine x 2 the approximate value of (x 1 + δ1 ), by determining the value of f (x 1 + δ1 ) = 0, but, neglecting terms containing products of δ1., , Third approximation, Now try the following exercise, Exercise 35 Further problems on the, bisection method, Use the method of bisection to solve the following, equations to the accuracy stated., , (d) Let the true value of the root be (x 2 + δ2 )., (e), , Determine x 3 , the approximate value of (x 2 + δ2 ), by determining the value of f (x 2 + δ2 ) = 0, but, neglecting terms containing products of δ2., , (f) The fourth and higher approximations are obtained, in a similar way., Using the techniques given in paragraphs (b) to (f),, it is possible to continue getting values nearer and
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82 Higher Engineering Mathematics, nearer to the required root. The procedure is repeated, until the value of the required root does not change on, two consecutive approximations, when expressed to the, required degree of accuracy., Problem 4. Use an algebraic method of, successive approximations to determine the value, of the negative root of the quadratic equation:, 4x 2 − 6x − 7 =0 correct to 3 significant figures., Check the value of the root by using the quadratic, formula., A first estimate of the values of the roots is made by, using the functional notation method, f (x) = 4x 2 − 6x − 7, f (0) = 4(0)2 − 6(0) − 7 = −7, f (−1) = 4(−1)2 − 6(−1) − 7 = 3, These results show that the negative root lies between 0, and −1, since the value of f (x) changes sign between, f (0) and f (−1) (see Section 9.1). The procedure given, above for the root lying between 0 and −1 is followed., , (a) Let a first approximation be such that it divides, the interval 0 to −1 in the ratio of −7 to 3, i.e. let, x 1 = −0.7, , The procedure given in (b) and (c) is now repeated, for x 2 = −0.7724., Third approximation, (d) Let the true value of the root, x 3 , be (x 2 + δ2 )., (e) Let f (x 2 + δ2 ) = 0, then, since x 2 = −0.7724,, 4(−0.7724 + δ2 )2 − 6(−0.7724 + δ2 ) − 7 = 0, 4[(−0.7724)2 + (2)(−0.7724)(δ2 ) + δ22 ], − (6)(−0.7724) − 6 δ2 − 7 = 0, Neglecting terms containing products of δ2 gives:, 2.3864 − 6.1792 δ2 + 4.6344 − 6 δ2 − 7 ≈ 0, , ≈, , −2.3864 − 4.6344 + 7, −6.1792 − 6, −0.0208, −12.1792, , ≈ +0.001708, , Second approximation, (b) Let the true value of the root, x 2 , be (x 1 + δ1 )., (c) Let f (x 1 + δ1 ) = 0, then, since x 1 = −0.7,, 4(−0.7 + δ1 )2 − 6(−0.7 + δ1 ) − 7 = 0, Hence, 4[(−0.7)2 + (2)(−0.7)(δ1 ) + δ12 ], − (6)(−0.7) − 6 δ1 − 7 = 0, Neglecting terms containing products of δ1, gives:, 1.96 −5.6 δ1 + 4.2 − 6 δ1 − 7 ≈ 0, , i.e., , i.e. x 2 = −0.7724, correct to 4 significant figures., (Since the question asked for 3 significant figure, accuracy, it is usual to work to one figure greater, than this)., , i.e. δ2 ≈, , First approximation, , i.e., , Thus, x 2 , a second approximation to the root is, [−0.7 +(−0.0724)],, , −5.6 δ1 − 6 δ1 = −1.96 − 4.2 + 7, δ1 ≈, , −1.96 − 4.2 + 7, −5.6 − 6, , 0.84, ≈, −11.6, ≈ −0.0724, , Thus x 3, the third approximation to the root is, (−0.7724 + 0.001708),, i.e. x 3 = − 0.7707, correct to 4 significant figures, (or −0.771 correct to 3 significant figures)., Fourth approximation, (f ) The procedure given for the second and third, approximations is now repeated for, x 3 = −0.7707, Let the true value of the root, x 4 , be (x 3 + δ3 )., Let f (x 3 + δ3 ) = 0, then since x 3 = −0.7707,, 4(−0.7707 + δ3)2 − 6(−0.7707, + δ3 ) − 7 = 0, 4[(−0.7707)2 + (2)(−0.7707) δ3 + δ32 ], − 6(−0.7707) − 6 δ3 − 7 = 0
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Solving equations by iterative methods, , 83, , f (1) = 3(1)3 − 10(1)2 + 4(1) + 7 = 4, , Neglecting terms containing products of δ3 gives:, , f (2) = 3(2)3 − 10(2)2 + 4(2) + 7 = −1, , 2.3759 − 6.1656 δ3 + 4.6242 − 6 δ3 − 7 ≈ 0, , Following the above procedure:, i.e. δ3 ≈, , −2.3759 − 4.6242 + 7, −6.1656 − 6, , First approximation, , −0.0001, ≈, −12.156, , (a), , ≈ +0.00000822, , Second approximation, , Thus, x 4, the fourth approximation to the root is, (−0.7707 + 0.00000822), i.e. x 4 = −0.7707, correct to 4 significant figures, and −0.771, correct to, 3 significant figures., , Let the first approximation be such that it divides, the interval 1 to 2 in the ratio of 4 to −1, i.e. let x 1, be 1.8., , (b) Let the true value of the root, x 2 , be (x 1 + δ1 )., (c), , Let f (x 1 + δ1 ) = 0, then since x 1 = 1.8,, 3(1.8 + δ1 )3 − 10(1.8 + δ1 )2, , Since the values of the roots are the same on, two consecutive approximations, when stated to, the required degree of accuracy, then the negative, root of 4x 2 − 6x − 7 = 0 is −0.771, correct to 3, significant figures., , + 4(1.8 + δ1 ) + 7 = 0, Neglecting terms containing products of δ1 and, using the binomial series gives:, 3[1.83 + 3(1.8)2 δ1 ] − 10[1.82 + (2)(1.8) δ1 ], , [Checking, using the quadratic formula:, , −(−6) ± [(−6)2 − (4)(4)(−7)], x=, (2)(4), 6 ± 12.166, =, = −0.771 and 2.27,, 8, correct to 3 significant figures], , + 4(1.8 + δ1 ) + 7 ≈ 0, 3(5.832 + 9.720 δ1) − 32.4 − 36 δ1, + 7.2 + 4 δ1 + 7 ≈ 0, 17.496 + 29.16 δ1 − 32.4 − 36 δ1, + 7.2 + 4 δ1 + 7 ≈ 0, , [Note on accuracy and errors. Depending on the, accuracy of evaluating the f (x + δ) terms, one or two, iterations (i.e. successive approximations) might be, saved. However, it is not usual to work to more than, about 4 significant figures accuracy in this type of calculation. If a small error is made in calculations, the only, likely effect is to increase the number of iterations.], , δ1 ≈, , −17.496 + 32.4 − 7.2 − 7, 29.16 − 36 + 4, , ≈−, , 0.704, ≈ −0.2479, 2.84, , Thus x 2 ≈ 1.8 −0.2479 =1.5521, Third approximation, , Problem 5. Determine the value of the, smallest positive root of the equation, 3x 3 − 10x 2 + 4x + 7 =0, correct to 3 significant, figures, using an algebraic method of successive, approximations., The functional notation method is used to find the value, of the first approximation., f (x) = 3x 3 − 10x 2 + 4x + 7, f (0) = 3(0)3 − 10(0)2 + 4(0) + 7 = 7, , (d) Let the true value of the root, x 3 , be (x 2 + δ2 )., (e), , Let f (x 2 + δ2 ) = 0, then since x 2 = 1.5521,, 3(1.5521 + δ2 )3 − 10(1.5521 + δ2 )2, + 4(1.5521 + δ2 ) + 7 = 0, Neglecting terms containing products of δ2 gives:, 11.217 + 21.681 δ2 − 24.090 − 31.042 δ2, + 6.2084 + 4 δ2 + 7 ≈ 0
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84 Higher Engineering Mathematics, δ2 ≈, ≈, , −11.217 + 24.090 − 6.2084 − 7, 21.681 − 31.042 + 4, −0.3354, −5.361, , ≈ 0.06256, Thus x 3 ≈ 1.5521 + 0.06256 ≈ 1.6147, , 9.4, , The Newton-Raphson formula, often just referred to as, Newton’s method, may be stated as follows:, If r1 is the approximate value of a real root of the, equation f (x) = 0, then a closer approximation to the, root r2 is given by:, , (f) Values of x 4 and x 5 are found in a similar way., f (x 3 + δ3 ) = 3(1.6147 + δ3)3 − 10(1.6147, + δ3 )2 + 4(1.6147 + δ3 ) + 7 = 0, giving δ3 ≈ 0.003175 and x 4 ≈ 1.618, i.e. 1.62, correct to 3 significant figures., , The Newton-Raphson method, , r2 = r1 −, , f(r1 ), f (r1 ), , The advantages of Newton’s method over the algebraic method of successive approximations is that it, can be used for any type of mathematical equation, (i.e. ones containing trigonometric, exponential, logarithmic, hyperbolic and algebraic functions), and it is, usually easier to apply than the algebraic method., , f (x 4 + δ4 ) = 3(1.618 + δ4 )3 − 10(1.618, + δ4 )2 + 4(1.618 + δ4 ) + 7 = 0, giving δ4 ≈ 0.0000417, and x 5 ≈ 1.62, correct to, 3 significant figures., Since x 4 and x 5 are the same when expressed to, the required degree of accuracy, then the required, root is 1.62, correct to 3 significant figures., , Problem 6. Use Newton’s method to determine, the positive root of the quadratic equation, 5x 2 + 11x − 17 =0, correct to 3 significant figures., Check the value of the root by using the quadratic, formula., The functional notation method is used to determine the, first approximation to the root., f (x) = 5x 2 + 11x − 17, , Now try the following exercise, , f (0) = 5(0)2 + 11(0) − 17 = −17, f (1) = 5(1)2 + 11(1) − 17 = −1, , Exercise 36 Further problems on solving, equations by an algebraic method of, successive approximations, Use an algebraic method of successive approximation to solve the following equations to the, accuracy stated., 1. 3x 2 + 5x − 17 = 0, correct to 3 significant, figures., [−3.36, 1.69], 2., , x 3 − 2x + 14 =0, correct to 3 decimal places., [−2.686], , 3., , x 4 − 3x 3 + 7x − 5.5 = 0, correct to 3 significant figures., [−1.53, 1.68], , 4., , x 4 + 12x 3 − 13 = 0, correct to 4 significant, figures., [−12.01, 1.000], , f (2) = 5(2)2 + 11(2) − 17 = 25, This shows that the value of the root is close to x = 1., Let the first approximation to the root, r1 , be 1., Newton’s formula states that a closer approximation,, f (r1 ), r2 = r1 −, f (r1 ), f (x) = 5x 2 + 11x − 17,, thus,, , f (r1 ) = 5(r1 )2 + 11(r1 ) − 17, = 5(1)2 + 11(1) − 17 = −1, , f (x) is the differential coefficient of f (x),, i.e., , f (x) = 10x + 11., , Thus f (r1 ) = 10(r1 ) + 11, = 10(1) + 11 = 21
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Solving equations by iterative methods, By Newton’s formula, a better approximation to the, root is:, , where sin2 means the sine of 2 radians, = 4 − 2.7279 + 2.1972 − 3.5, , −1, r2 = 1 −, = 1 − (−0.048) = 1.05,, 21, correct to 3 significant figures., A still better approximation to the root, r3 , is given by:, r3 = r2 −, , = −0.0307, f (x) = 2x − 3 cos x +, , = 5.9151, Hence, r2 = r1 −, , 0.0625, 21.5, , i.e. 1.05, correct to 3 significant figures., Since the values of r2 and r3 are the same when, expressed to the required degree of accuracy, the, required root is 1.05, correct to 3 significant figures., Checking, using the quadratic equation formula,, x=, , =, , [121 − 4(5)(−17)], (2)(5), , −11 ± 21.47, 10, , The positive root is 1.047, i.e. 1.05, correct to 3 significant figures (This root was determined in Problem 1, using the bisection method; Newton’s method is clearly, quicker)., Problem 7. Taking the first approximation as 2,, determine the root of the equation, x 2 − 3 sin x + 2 ln(x + 1) = 3.5, correct to 3, significant figures, by using Newton’s method., f (r1 ), Newton’s formula states that r2 = r1 −, , where, f (r1 ), r1 is a first approximation to the root and r2 is a better, approximation to the root., Since f (x) = x 2 − 3 sin x + 2 ln (x + 1) − 3.5, f (r1 ) = f (2) = 22 − 3 sin 2 + 2 ln 3 − 3.5,, , f (r1 ), f (r1 ), , −0.0307, 5.9151, = 2.005 or 2.01, correct to, 3 significant figures., , =2−, , = 1.05 − 0.003 = 1.047,, , √, , 2, 3, , = 4 + 1.2484 + 0.6667, , [5(1.05)2 + 11(1.05) − 17], = 1.05 −, [10(1.05) + 11], , −11 ±, , 2, x +1, , f (r1 ) = f (2) = 2(2) − 3 cos 2 +, , f (r2 ), f (r2 ), , = 1.05 −, , 85, , A still better approximation to the root, r3 , is given by:, r3 = r2 −, , f (r2 ), f (r2 ), , [(2.005)2 − 3 sin 2.005 + 2 ln 3.005 − 3.5], = 2.005 − , 2, 2(2.005) − 3 cos 2.005 +, 2.005 + 1, = 2.005 −, , (−0.00104), = 2.005 + 0.000175, 5.9376, , i.e. r3 = 2.01, correct to 3 significant figures., Since the values of r2 and r3 are the same when, expressed to the required degree of accuracy, then the, required root is 2.01, correct to 3 significant figures., Problem 8. Use Newton’s method to find the, positive root of:, x, (x + 4)3 − e1.92x + 5 cos = 9,, 3, correct to 3 significant figures., The functional notational method is used to determine, the approximate value of the root., f (x) = (x + 4)3 − e1.92x + 5 cos, , x, −9, 3, , f (0) = (0 + 4)3 − e0 + 5 cos 0 − 9 = 59
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86 Higher Engineering Mathematics, 1, − 9 ≈ 114, 3, 2, f (2) = 63 − e3.84 + 5 cos − 9 ≈ 164, 3, 3, 5.76, + 5 cos 1 − 9 ≈ 19, f (3) = 7 − e, 4, 3, 7.68, f (4) = 8 − e, + 5 cos − 9 ≈ −1660, 3, f (1) = 53 − e1.92 + 5 cos, , From these results, let a first approximation to the root, be r1 = 3., Newton’s formula states that a better approximation to, the root,, f (r1 ), r2 = r1 −, f (r1 ), f (r1 ) = f (3) = 7 − e, 3, , 5.76, , + 5 cos 1 − 9, , = 19.35, 5, x, sin, 3, 3, 5, f (r1 ) = f (3) = 3(7)2 − 1.92e5.76 − sin 1, 3, = −463.7, , 1., , x 2 − 2x − 13 =0, correct to 3 decimal, places., [−2.742, 4.742], , 2., , 3x 3 − 10x = 14, correct to 4 significant, figures., [2.313], , 3., , x 4 − 3x 3 + 7x = 12, correct to 3 decimal, places., [−1.721, 2.648], , 4., , 3x 4 − 4x 3 + 7x − 12 =0, correct to 3 decimal places., [−1.386, 1.491], , 5., , 3 ln x + 4x = 5, correct to 3 decimal places., [1.147], , 6., , x 3 = 5 cos 2x, correct to 3 significant figures., [−1.693, −0.846, 0.744], , 7., , f (x) = 3(x + 4)2 − 1.92e1.92x −, , Thus, r2 = 3 −, , 19.35, = 3 + 0.042, −463.7, , 8., , Solve the equations in Problems 1 to 5,, Exercise 35, page 81 and Problems 1 to, 4, Exercise 36, page 84 using Newton’s, method., , 9., , A Fourier analysis of the instantaneous value, of a waveform can be represented by:, , 1, π, + sin t + sin 3t, y= t+, 4, 8, , = 3.042 = 3.04,, correct to 3 significant figures., Similarly, r3 = 3.042 −, , f (3.042), f (3.042), , = 3.042 −, , (−1.146), (−513.1), , = 3.042 − 0.0022 = 3.0398 = 3.04,, correct to 3 significant figures., , Use Newton’s method to determine the value, of t near to 0.04, correct to 4 decimal places,, when the amplitude, y, is 0.880., [0.0399], 10., , Exercise 37 Further problems on Newton’s, method, In Problems 1 to 7, use Newton’s method to solve, the equations given to the accuracy stated., , A damped oscillation of a system is given by, the equation:, y = −7.4e0.5t sin 3t., , Since r2 and r3 are the same when expressed to the, required degree of accuracy, then the required root is, 3.04, correct to 3 significant figures., , Now try the following exercise, , θ, 300e−2θ + = 6, correct to 3 significant, 2, figures., [2.05], , Determine the value of t near to 4.2, correct, to 3 significant figures, when the magnitude, y of the oscillation is zero., [4.19], 11., , The critical speeds of oscillation, λ, of a, loaded beam are given by the equation:, λ3 − 3.250λ2 + λ − 0.063 = 0, Determine the value of λ which is approximately equal to 3.0 by Newton’s method,, correct to 4 decimal places., [2.9143]
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Chapter 10, , Binary, octal and, hexadecimal, 10.1, , Introduction, , All data in modern computers is stored as series of bits,, a bit being a binary digit, and can have one of two values,, the numbers 0 and 1. The most basic form of representing computer data is to represent a piece of data as a, string of 1’s and 0’s, one for each bit. This is called a, binary or base-2 number., Because binary notation requires so many bits to represent relatively small numbers, two further compact, notations are often used, called octal and hexadecimal. Computer programmers who design sequences of, number codes instructing a computer what to do would, have a very difficult task if they were forced to work, with nothing but long strings of 1’s and 0’s, the ‘native, language’ of any digital circuit., Octal notation represents data as base-8 numbers with, each digit in an octal number representing three bits., Similarly, hexadecimal notation uses base-16 numbers,, representing four bits with each digit. Octal numbers, use only the digits 0–7, while hexadecimal numbers, use all ten base-10 digits (0–9) and the letters A–F, (representing the numbers 10–15)., This chapter explains how to convert between the, decimal, binary, octal and hexadecimal systems., , 10.2, , Binary numbers, , The system of numbers in everyday use is the denary, or decimal system of numbers, using the digits 0 to 9., It has ten different digits (0, 1, 2, 3, 4, 5, 6, 7, 8 and 9), and is said to have a radix or base of 10., , The binary system of numbers has a radix of 2 and, uses only the digits 0 and 1., (a) Conversion of binary to decimal, The decimal number 234.5 is equivalent to, 2 × 102 + 3 × 101 + 4 × 100 + 5 × 10−1, i.e. is the sum of terms comprising: (a digit) multiplied, by (the base raised to some power)., In the binary system of numbers, the base is 2, so, 1101.1 is equivalent to:, 1 × 23 + 1 × 22 + 0 × 21 + 1 × 20 + 1 × 2−1, Thus the decimal number equivalent to the binary, number 1101.1 is 8 + 4 +0 + 1 + 12 , that is 13.5 i.e., 1101.12 = 13.510, the suffixes 2 and 10 denoting binary, and decimal systems of numbers respectively., Problem 1. Convert 110112 to a decimal number., From above: 110112 = 1 × 24 + 1 × 23 + 0 × 22, + 1 × 21 + 1 × 20, = 16 + 8 + 0 + 2 + 1, = 2710, Problem 2. Convert 0.10112 to a decimal, fraction., 0.10112 = 1 ×2−1 + 0 × 2−2 + 1 × 2−3 + 1 × 2−4, 1, 1, 1, 1, = 1× +0× 2 +1× 3 +1× 4, 2, 2, 2, 2
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88 Higher Engineering Mathematics, 1 1, 1, + +, 2 8 16, = 0.5 + 0.125 + 0.0625, , 2, 2, 2, 2, 2, 2, , =, , = 0.687510, , Problem 3., number., , Convert 101.01012 to a decimal, , 39, 19, 9, 4, 2, 1, 0, , Remainder, 1, 1, 1, 0, 0, 1, 1 0 0 1 1 1, , (most significant bit), , 101.01012 = 1 × 22 + 0 × 21 + 1 × 20 + 0 × 2−1, + 1 × 2−2 + 0 × 2−3 + 1 × 2−4, = 4 + 0 + 1 + 0 + 0.25 + 0 + 0.0625, = 5.312510, , The result is obtained by writing the top digit of the, remainder as the least significant bit, (a bit is a binary, digit and the least significant bit is the one on the right)., The bottom bit of the remainder is the most significant, bit, i.e. the bit on the left., Thus, , 3910 = 1001112, , The fractional part of a decimal number can be converted, to a binary number by repeatedly multiplying by 2, as, shown below for the fraction 0.625, , Now try the following exercise, Exercise 38 Further problems on, conversion of binary to decimal numbers, In Problems 1 to 5, convert the binary numbers, given to decimal numbers., 1. (a) 110 (b) 1011 (c) 1110 (d) 1001, [(a) 610 (b) 1110 (c) 1410 (d) 910 ], 2. (a) 10101 (b) 11001 (c) 101101 (d) 110011, [(a) 2110 (b) 2510 (c) 4510 (d) 5110 ], 3. (a) 101010 (b) 111000 (c) 1000001, (d) 10111000, [(a) 4210 (b) 5610 (c) 6510 (d) 18410 ], 4. (a) 0.1101, (d) 0.01011, , (b) 0.11001, (a) 0.812510, (c) 0.2187510, , (least significant bit), , (c) 0.00111, (b) 0.7812510, (d) 0.3437510, , 5. (a) 11010.11 (b) 10111.011 (c) 110101.0111, (d) 11010101.10111, (a) 26.7510, (b) 23.37510, (c) 53.437510 (d) 213.7187510, , (b) Conversion of decimal to binary, An integer decimal number can be converted to a corresponding binary number by repeatedly dividing by 2, and noting the remainder at each stage, as shown below, for 3910., , 0.625 3 2 5, , 1. 250, , 0.250 3 2 5, , 0. 500, , 0.500 3 2 5, , 1. 000, , (most significant bit) .1, , 0, , 1 (least significant bit), , For fractions, the most significant bit of the result is the, top bit obtained from the integer part of multiplication, by 2. The least significant bit of the result is the bottom, bit obtained from the integer part of multiplication by 2., Thus 0.62510 = 0.1012, Problem 4., , Convert 4710 to a binary number., , From above, repeatedly dividing by 2 and noting the, remainder gives:, 2 47, , Remainder, , 2 23, , 1, , 2 11, , 1, , 2, , 5, , 1, , 2, , 2, , 1, , 2, , 1, , 0, , 0, , 1, 1, , Thus 4710 = 1011112, , 0, , 1, , 1, , 1, , 1
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Binary, octal and hexadecimal, Problem 5. Convert 0.4062510 to a binary, number., , 1. (a) 5 (b) 15 (c) 19 (d) 29, (a) 1012, (b) 11112, (c) 100112 (d) 111012, , From above, repeatedly multiplying by 2 gives:, 0.40625 3 2 5, , 0. 8125, , 0.8125, , 325, , 1. 625, , 0.625, , 325, , 1. 25, , 0.25, , 325, , 0. 5, , 325, , 1. 0, , 0.5, , 1, , 1, , 0, , 1, , (a) 111112, , (b) 1010102, , (c) 1110012 (d) 1111112, 3. (a) 47 (b) 60 (c) 73 (d) 84, (a) 1011112, , (b) 1111002, , (b) 0.001112, , (c) 0.010012 (d) 0.100112, , Problem 6. Convert 58.312510 to a binary, number., The integer part is repeatedly divided by 2, giving:, Remainder, 0, 1, 0, 1, 1, 1, 1, , 4. (a) 0.25 (b) 0.21875 (c) 0.28125, (d) 0.59375, (a) 0.012, , i.e. 0.4062510 = 0.011012, , 58, 29, 14, 7, 3, 1, 0, , 2. (a) 31 (b) 42 (c) 57 (d) 63, , (c) 10010012 (d) 10101002, , .0, , 2, 2, 2, 2, 2, 2, , 89, , 1, , 5. (a) 47.40625 (b) 30.8125 (c) 53.90625, (d) 61.65625, ⎡, ⎤, (a) 101111.011012, ⎢, ⎥, ⎢ (b) 11110.11012 ⎥, ⎢, ⎥, ⎢ (c) 110101.11101 ⎥, 2⎦, ⎣, (d) 111101.101012, (c) Binary addition, Binary addition of two/three bits is achieved according, to the following rules:, , 1, , 0, , 1, , 0, , The fractional part is repeatedly multiplied by 2 giving:, 0.3125 3 2 5, 0.625 3 2 5, 0.25 3 2 5, 0.5, 325, , 0.625, 1.25, 0.5, 1.0, .0 1 0 1, , Thus 58.312510 = 111010.01012, Now try the following exercise, Exercise 39 Further problems on, conversion of decimal to binary numbers, In Problems 1 to 5, convert the decimal numbers, given to binary numbers., , sum, carry, sum, carry, 0+0 = 0, 0, 0+0+0 = 0, 0, 0+1 = 1, 0, 0+0+1 = 1, 0, 1+0 = 1, 0, 0+1+0 = 1, 0, 1+1 = 0, 1, 0+1+1 = 0, 1, 1+0+0 = 1, 0, 1+0+1 = 0, 1, 1+1+0 = 0, 1, 1+1+1 = 1, 1, These rules are demonstrated in the following worked, problems., Problem 7. Perform the binary addition:, 1001 + 10110, 1001, +10110, 11111
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90 Higher Engineering Mathematics, Problem 8. Perform the binary addition:, 11111 + 10101, , 10.3, , Octal numbers, , For decimal integers containing several digits, repeatedly dividing by 2 can be a lengthy process. In this case,, it is usually easier to convert a decimal number to a, binary number via the octal system of numbers. This, system has a radix of 8, using the digits 0, 1, 2, 3, 4,, 5, 6 and 7. The decimal number equivalent to the octal, number 43178 is:, , 11111, +10101, sum 110100, carry 11111, Problem 9. Perform the binary addition:, 1101001 + 1110101, , 4 × 83 + 3 × 82 + 1 × 81 + 7 × 80, , 1101001, +1110101, sum 11011110, carry 11, 1, , i.e. 4 × 512 + 3 × 64 + 1 × 8 + 7 × 1 or 225510, , Problem 10. Perform the binary addition:, 1011101 + 1100001 + 110101, , An integer decimal number can be converted to a corresponding octal number by repeatedly dividing by 8, and noting the remainder at each stage, as shown below, for 49310., , 1011101, 1100001, + 110101, sum 11110011, carry 11111 1, , 8 493, , Remainder, , 8 61, , 5, , 8, , 7, , 5, , 0, , 7, 7, , Now try the following exercise, , 5, , 5, , Thus 49310 = 7558, , Exercise 40 Further problems on binary, addition, , The fractional part of a decimal number can be converted, to an octal number by repeatedly multiplying by 8, as, shown below for the fraction 0.437510, , Perform the following binary additions:, 1. 10 + 11, , [101], , 2. 101 + 110, , [1011], , 3. 1101 + 111, , [10100], , 4. 1111 + 11101, 5. 110111 + 10001, 6. 10000101 + 10000101, , 0.4375 3 8 5, , 3. 5, , 385, , 4. 0, , 0.5, , .3, , [101100], [1001000], [100001010], , 7. 11101100 + 111001011, , [1010110111], , 8. 110011010 + 11100011, , [1001111101], , 9. 10110 + 1011 + 11011, , [111100], , 4, , For fractions, the most significant bit is the top integer, obtained by multiplication of the decimal fraction by, 8, thus,, 0.437510 = 0.348, , 10. 111 + 10101 + 11011, , [110111], , 11. 1101 + 1001 + 11101, , [110011], , The natural binary code for digits 0 to 7 is shown, in Table 10.1, and an octal number can be converted, to a binary number by writing down the three bits, corresponding to the octal digit., , 12. 100011 + 11101 + 101110, , [1101110], , Thus, , 4378 = 100 011 1112, , and 26.358 = 010 110.011 1012
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Binary, octal and hexadecimal, Table 10.1, , 0.5937510 = 0.468, , Thus, , Octal digit, , Natural, binary number, , 0, , 000, , 1, , 001, , 2, , 010, , 3, , 011, , 4, , 100, , 5, , 101, , 6, , 110, , 7, , 111, , The integer part is repeatedly divided by 8, noting the, remainder, giving:, , Problem 11. Convert 371410 to a binary number,, via octal., Dividing repeatedly by 8, and noting the remainder, gives:, Remainder, , 8 464, , 2, , 8, , 58, , 0, , 8, , 7, , 2, , 0, , 7, , 0.5937510 = 0.100 112, , Problem 13. Convert 5613.9062510 to a binary, number, via octal., , 8 5613, 8 701, 8 87, 8 10, 8, 1, 0, , The ‘0’ on the extreme left does not signify anything,, thus 26.358 = 10 110.011 1012, Conversion of decimal to binary via octal is demonstrated in the following worked problems., , 8 3714, , 0.468 = 0.100 1102, , From Table 10.1,, i.e., , Remainder, 5, 5, 7, 2, 1, 1, , 2, , 7, , 2, , 5, , 5, , This octal number is converted to a binary number,, (see Table 10.1)., 127558 = 001 010 111 101 1012, i.e., , 561310 = 1 010 111 101 1012, , The fractional part is repeatedly multiplied by 8, and, noting the integer part, giving:, 0.90625 3 8 5, 0.25, 385, , 7, , 91, , 0, , .7 2, , 2, , From Table 10.1, 72028 = 111 010 000 0102, i.e., 371410 = 111 010 000 0102, Problem 12. Convert 0.5937510 to a binary, number, via octal., , 7.25, 2.00, , This octal fraction is converted to a binary number,, (see Table 10.1)., 0.728 = 0.111 0102, i.e., , 0.9062510 = 0.111 012, , Thus, 5613.9062510 = 1 010 111 101 101.111 012, Multiplying repeatedly by 8, and noting the integer, values, gives:, 0.59375 3 8 5, 0.75, 385, , 4.75, 6.00, .4 6, , Problem 14. Convert 11 110 011.100 012 to a, decimal number via octal., Grouping the binary number in three’s from the binary, point gives: 011 110 011.100 0102
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92 Higher Engineering Mathematics, Using Table 10.1 to convert this binary number to an, octal number gives 363.428 and 363.428, = 3 × 82 + 6 × 81 + 3 × 80 + 4 × 8−1 + 2 × 8−2, = 192 + 48 + 3 + 0.5 + 0.03125, , pairs of hexadecimal digits RRGGBB, where RR is the, amount of red, GG the amount of green, and BB the, amount of blue., A hexadecimal numbering system has a radix of, 16 and uses the following 16 distinct digits:, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E and F, , = 243.5312510, Now try the following exercise, , ‘A’ corresponds to 10 in the decimal system, B to 11,, C to 12, and so on., (a) Converting from hexadecimal to decimal:, , Exercise 41 Further problems on, conversion between decimal and binary, numbers via octal, , For example, 1A16 = 1 × 161 + A × 160, = 1 × 161 + 10 × 1, , In Problems 1 to 3, convert the decimal numbers, given to binary numbers, via octal., 1. (a) 343 (b) 572 (c) 1265, , (a) 1010101112 (b) 10001111002, (c) 100111100012, , = 16 + 10 = 26, i.e., , 1A16 = 2610, , Similarly, 2E16 = 2 × 161 + E × 160, , 2. (a) 0.46875 (b) 0.6875 (c) 0.71875, , (a) 0.011112 (b) 0.10112, (c) 0.101112, , = 2 × 161 + 14 × 160, = 32 + 14 = 4610, 1BF16 = 1 × 162 + B × 161 + F × 160, , 3. (a) 247.09375 (b) 514.4375 (c) 1716.78125, ⎡, ⎤, (a) 11110111.000112, ⎢, ⎥, ⎣ (b) 1000000010.01112 ⎦, (c) 11010110100.110012, , and, , 4. Convert the binary numbers given to decimal, numbers via octal., , Table 10.2 compares decimal, binary, octal and hexadecimal numbers and shows, for example, that, 2310 = 101112 = 278 = 1716, , (a) 111.011 1 (b) 101 001.01, (c) 1 110 011 011 010.001 1, , (a) 7.437510 (b) 41.2510, (c) 7386.187510, , = 1 × 162 + 11 × 161 + 15 × 160, = 256 + 176 + 15 = 44710, , Problem 15. Convert the following hexadecimal, numbers into their decimal equivalents:, (a) 7A16 (b) 3F16, (a) 7A16 = 7 × 161 + A × 160 = 7 × 16 + 10 × 1, , 10.4, , Hexadecimal numbers, , The hexadecimal system is particularly important in, computer programming, since four bits (each consisting of a one or zero) can be succinctly expressed using, a single hexadecimal digit. Two hexadecimal digits represent numbers from 0 to 255, a common range used,, for example, to specify colours. Thus, in the HTML, language of the web, colours are specified using three, , = 112 + 10 = 122, Thus 7A16 = 12210, (b) 3F16 = 3 × 161 + F × 160 = 3 × 16 + 15 × 1, = 48 + 15 = 63, Thus 3F16 = 6310
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Binary, octal and hexadecimal, Table 10.2, , Problem 16. Convert the following hexadecimal, numbers into their decimal equivalents:, (a) C916 (b) BD16, , Decimal, , Binary, , Octal, , Hexadecimal, , 0, , 0000, , 0, , 0, , 1, , 0001, , 1, , 1, , 2, , 0010, , 2, , 2, , 3, , 0011, , 3, , 3, , 4, , 0100, , 4, , 4, , 5, , 0101, , 5, , 5, , 6, , 0110, , 6, , 6, , 7, , 0111, , 7, , 7, , 8, , 1000, , 10, , 8, , 9, , 1001, , 11, , 9, , 10, , 1010, , 12, , A, , 1A4E16 = 1 × 163 + A × 162 + 4 × 161 + E × 160, , 11, , 1011, , 13, , B, , = 1 × 163 + 10 × 162 + 4 × 161, , 12, , 1100, , 14, , C, , + 14 × 160, , 13, , 1101, , 15, , D, , = 1 × 4096 + 10 × 256 + 4 × 16 + 14× 1, , 14, , 1110, , 16, , E, , 15, , 1111, , 17, , F, , 16, , 10000, , 20, , 10, , 17, , 10001, , 21, , 11, , 18, , 10010, , 22, , 12, , 19, , 10011, , 23, , 13, , 20, , 10100, , 24, , 14, , 21, , 10101, , 25, , 15, , 22, , 10110, , 26, , 16, , 16 26 Remainder, 16 1 10 ; A16, 0 1 ; 116, , 23, , 10111, , 27, , 17, , most significant bit, , 24, , 11000, , 30, , 18, , 25, , 11001, , 31, , 19, , 26, , 11010, , 32, , 1A, , 27, , 11011, , 33, , 1B, , 28, , 11100, , 34, , 1C, , 16, , 27 15 ; F16, , 16, , 1 11 ; B16, , 29, , 11101, , 35, , 1D, , 30, , 11110, , 36, , 1E, , 31, , 11111, , 37, , 1F, , 32, , 100000, , 40, , 20, , 93, , (a) C916 = C × 161 + 9 × 160 = 12 × 16 + 9 × 1, = 192 + 9 = 201, Thus C916 = 20110, (b) BD16 = B × 161 + D × 160, = 11 × 16 + 13 × 1 = 176 + 13 = 189, Thus BD16 = 18910, Problem 17. Convert 1A4E16 into a decimal, number., , = 4096 + 2560 + 64 + 14 = 6734, Thus 1A4E16 = 673410, (b) Converting from decimal to hexadecimal, This is achieved by repeatedly dividing by 16 and noting, the remainder at each stage, as shown below for 2610 ., , 1 A, , least significant bit, , Hence 2610 = 1A16, Similarly, for 44710, 16 447 Remainder, , 0, , 1 ; 116, 1 B F, , Thus 44710 = 1BF16
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94 Higher Engineering Mathematics, Problem 18. Convert the following decimal, numbers into their hexadecimal equivalents:, (a) 3710 (b) 10810, (a) 16 37 Remainder, , In Problems 5 to 8, convert the given decimal, numbers into their hexadecimal equivalents., 5., , 5410 [3616], , 6., , 20010 [C816], , 7., , 9110 [5B16], , 8., , 23810 [EE16], , 16 2 5 5 516, 0 2 5 216, , (c) Converting from binary to hexadecimal:, 2, , 5, , most significant bit, , least significant bit, , Hence 3710 = 2516, (b) 16 108 Remainder, 16, , 6 12 5 C16, 0 6 5 616, , The binary bits are arranged in groups of four, starting from right to left, and a hexadecimal symbol is, assigned to each group. For example, the binary number 1110011110101001 is initially grouped in fours as:, 1110, ) *+ , 1010, ) *+ , 1001, ) *+ , and a hexadecimal symbol, ) *+ , 0111, E, 7, A, 9, assigned to each group as above from Table 10.2., Hence 11100111101010012 = E7A916, , 6 C, , Hence 10810 = 6C16, Problem 19. Convert the following decimal, numbers into their hexadecimal equivalents:, (a) 16210 (b) 23910, , Problem 20. Convert the following binary, numbers into their hexadecimal equivalents:, (a) 110101102 (b) 11001112, (a), , (a) 16 162 Remainder, 16 10 2 5 216, 0 10 5 A16, , Grouping bits in fours from the right gives:, 1101, ) *+ , 0110, ) *+ , and assigning hexadecimal symbols, D, 6, to each group gives as above from Table 10.2., Thus,, , A 2, , 110101102 = D616, , (b) Grouping bits in fours from the right gives:, 0110, ) *+ , 0111, ) *+ , and assigning hexadecimal symbols, 6, 7, to each group gives as above from Table 10.2., , Hence 16210 = A216, (b) 16 239 Remainder, 16 14 15 5 F16, 0 14 5 E16, , Thus,, , 11001112 = 6716, , E F, , Problem 21. Convert the following binary, numbers into their hexadecimal equivalents:, (a) 110011112 (b) 1100111102, , Hence 23910 = EF16, Now try the following exercise, , (a), Exercise 42 Further problems on, hexadecimal numbers, In Problems 1 to 4, convert the given hexadecimal, numbers into their decimal equivalents., 1., , E716 [23110], , 2., , 2C16, , [4410], , 3., , 9816 [15210], , 4., , 2F116 [75310], , Grouping bits in fours from the right gives:, 1100, ) *+ , 1111, ) *+ , and assigning hexadecimal symbols, C, F, to each group gives as above from Table 10.2., Thus, 110011112 = CF16, , (b) Grouping bits in fours from the right gives:, 0001, ) *+ , 1110, ) *+ , and assigning hexadecimal, ) *+ , 1001, 1, 9, E
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Binary, octal and hexadecimal, symbols to each group gives as above from, Table 10.2., Thus, 1100111102 = 19E16, , 95, , (a) Spacing out hexadecimal digits gives:, B, 7, + ,) * + ,) * and converting each into binary, 0111 1011, gives as above from Table 10.2., Thus, 7B16 = 11110112, , (d) Converting from hexadecimal to binary:, The above procedure is reversed, thus, for example,, 6CF316 = 0110 1100 1111 0011, from Table 10.2, , (b) Spacing out hexadecimal digits gives:, 7, D, 1, + ,) * + ,) * + ,) * and converting each into, 0001 0111 1101, binary gives as above from Table 10.2., Thus, 17D16 = 1011111012, , i.e. 6CF316 = 1101100111100112, Now try the following exercise, Problem 22. Convert the following hexadecimal, numbers into their binary equivalents:, (a) 3F16 (b) A616, (a) Spacing out hexadecimal digits gives:, F, 3, + ,) * + ,) * and converting each into binary, 0011 1111, gives as above from Table 10.2., Thus, 3F16 = 1111112, (b) Spacing out hexadecimal digits gives:, 6, A, + ,) * + ,) * and converting each into binary, 1010 0110, gives as above from Table 10.2., Thus, A616 = 101001102, Problem 23. Convert the following hexadecimal, numbers into their binary equivalents:, (a) 7B16 (b) 17D16, , Exercise 43 Further problems on, hexadecimal numbers, In Problems 1 to 4, convert the given binary, numbers into their hexadecimal equivalents., 1. 110101112, , [D716], , 2. 111010102, , [EA16], , 3. 100010112, , [8B16], , 4. 101001012, , [A516], , In Problems 5 to 8, convert the given hexadecimal, numbers into their binary equivalents., 5. 3716, , [1101112], , 6. ED16, , [111011012], , 7. 9F16, , [100111112], , 8. A2116, , [1010001000012]
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Revision Test 3, This Revision Test covers the material contained in Chapters 8 to 10. The marks for each question are shown in, brackets at the end of each question., 1., , Use Maclaurin’s series to determine a power series, for e2x cos 3x as far as the term in x 2 ., (9), , 2., , Show, using Maclaurin’s series, that the first four, terms of the power series for cosh 2x is given by:, 2, 4, cosh 2x = 1 + 2x + x 4 + x 6 ., 3, 45, 2, , 3., , 7., , (10), , Convert the following binary numbers to decimal, form:, (a) 1101 (b) 101101.0101, , Expand the function x ln(1 + sin x) using, Maclaurin’s series and hence evaluate:, ! 1, 2, x 2 ln(1 + sin x) dx correct to 2 significant, , (5), , 2, , 0, , figures., 4., , Use Newton’s method to determine the value of x,, correct to 2 decimal places, for which the value of, y is zero., (10), , 8., , Convert the following decimal number to binary, form:, (a) 27 (b) 44.1875, , (9), , (13), , Use the method of bisection to evaluate the root, of the equation: x 3 + 5x = 11 in the range x = 1 to, x = 2, correct to 3 significant figures., (11), , 5., , Repeat question 4 using an algebraic method of, successive approximations., (16), , 6., , The solution to a differential equation associated, with the path taken by a projectile for which the, resistance to motion is proportional to the velocity, is given by:, −x, , y = 2.5(e − e, x, , ) + x − 25, , 9., , Convert the following decimal numbers to binary,, via octal:, (a) 479 (b) 185.2890625, , (9), , 10. Convert, (a) 5F16 into its decimal equivalent, (b) 13210 into its hexadecimal equivalent, (c) 1101010112 into its hexadecimal equivalent, (8)
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Chapter 11, , Introduction to trigonometry, 11.1, , 169 = d 2 + 25, , Trigonometry is the branch of mathematics which deals, with the measurement of sides and angles of triangles, and their relationship with each other. There are, many applications in engineering where a knowledge, of trigonometry is needed., , 11.2, , 132 = d 2 + 52, , Hence, , Trigonometry, , The theorem of Pythagoras, , With reference to Fig. 11.1, the side opposite the right, angle (i.e. side b) is called the hypotenuse. The theorem, of Pythagoras states:, ‘In any right-angled triangle, the square on the, hypotenuse is equal to the sum of the squares on the, other two sides.’, Hence b2 = a2 + c2, , d 2 = 169 − 25 = 144, √, d = 144 = 12 cm, EF = 12 cm, , Thus, i.e., , Problem 2. Two aircraft leave an airfield at the, same time. One travels due north at an average, speed of 300 km/h and the other due west at an, average speed of 220 km/h. Calculate their distance, apart after 4 hours., After 4 hours, the first aircraft has travelled 4 × 300 =, 1200 km, due north, and the second aircraft has travelled 4 × 220 = 880 km due west, as shown in Fig. 11.3., Distance apart after 4 hours = BC., , A, N, , b, , c, , W, , B, E, , S, , B, , 1200 km, , C, , a, , Figure 11.1, , C, , Problem 1. In Fig. 11.2, find the length of EF., , 880 km, , A, , Figure 11.3, , D, f 5 5 cm, E, , e 513 cm, d, , Figure 11.2, , By Pythagoras’ theorem:, e2 = d 2 + f 2, , From Pythagoras’ theorem:, F, , BC 2 = 12002 + 8802 = 1 440 000 + 774 400, , and BC = (2 214 400), Hence distance apart after 4 hours = 1488 km.
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98 Higher Engineering Mathematics, Now try the following exercise, Exercise 44 Further problems on the, theorem of Pythagoras, , 7. Figure 11.5 shows a cross-section of a, component that is to be made from a round bar., If the diameter of the bar is 74 mm, calculate, the dimension x., [24 mm], , 1. In a triangle CDE, D = 90◦ , C D = 14.83 mm, and C E = 28.31 mm. Determine the length of, D E., [24.11 mm], , x, , 2. Triangle PQR is isosceles, Q being a right, angle. If the hypotenuse is 38.47 cm find (a), the lengths of sides P Q and Q R, and (b) the, value of ∠QPR. [(a) 27.20 cm each (b) 45◦ ], 3. A man cycles 24 km due south and then 20 km, due east. Another man, starting at the same, time as the first man, cycles 32 km due east and, then 7 km due south. Find the distance between, the two men., [20.81 km], 4. A ladder 3.5 m long is placed against a perpendicular wall with its foot 1.0 m from the wall., How far up the wall (to the nearest centimetre), does the ladder reach? If the foot of the ladder is now moved 30 cm further away from the, wall, how far does the top of the ladder fall?, [3.35 m, 10 cm], 5. Two ships leave a port at the same time. One, travels due west at 18.4 km/h and the other due, south at 27.6 km/h. Calculate how far apart the, two ships are after 4 hours., [132.7 km], , h, , 72 mm, , Figure 11.5, , 11.3 Trigonometric ratios of acute, angles, (a) With reference to the right-angled triangle shown, in Fig. 11.6:, (i), i.e., , sine θ =, , opposite side, hypotenuse, , sin θ =, , b, c, , cosine θ =, , (ii), , r 516 mm, , 6. Figure 11.4 shows a bolt rounded off at one, end. Determine the dimension h. [2.94 mm], , m, , 4m, , 7, , i.e., , cos θ =, tangent θ =, , (iii), , i.e., , sec θ =, cosecant θ =, , (v), Figure 11.4, , tan θ =, secant θ =, , (iv), , i.e., , a, c, , opposite side, adjacent side, , R 5 45 mm, , i.e., , adjacent side, hypotenuse, , cosec θ =, , b, a, hypotenuse, adjacent side, c, a, hypotenuse, opposite side, c, b
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Introduction to trigonometry, (vi), , adjacent side, opposite side, a, cot θ =, b, , cotangent θ =, i.e., , c, , 9, Since cos X = , then X Y = 9 units and, 41, X Z = 41 units., Using Pythagoras’, theorem: 412 = 92 + Y Z 2 from, , 2, which Y Z = (41 − 92 ) = 40 units., Thus, 40, 4, 40, , tan X =, =4 ,, 41, 9, 9, 41, 1, cosec X =, =1 ,, 40, 40, 41, 5, 9, sec X =, = 4 and cot X =, 9, 9, 40, sin X =, , b, , , a, , Figure 11.6, , (b) From above,, (i), , (ii), , (iii), (iv), , b, b, sin θ, = ac = = tan θ,, cos θ, a, c, sin θ, i.e. tan θ =, cos θ, a, cos θ, a, = c = = cot θ,, b b, sin θ, c, cos θ, i.e. cot θ =, sin θ, 1, sec θ =, cos θ, 1, cosec θ =, sin θ, (Note ‘s’ and ‘c’ go together), , 1, tan θ, Secants, cosecants and cotangents are called the, reciprocal ratios., (v), , cot θ =, , Problem 4. If sin θ = 0.625 and cos θ = 0.500, determine, without using trigonometric tables or, calculators, the values of cosec θ, sec θ, tan θ, and cot θ., 1, 1, =, = 1.60, sin θ, 0.625, 1, 1, =, = 2.00, sec θ =, cos θ, 0.500, 0.625, sin θ, =, = 1.25, tan θ =, cos θ, 0.500, 0.500, cos θ, =, = 0.80, cot θ =, sin θ, 0.625, , cosec θ =, , Problem 5. Point A lies at co-ordinate (2, 3) and, point B at (8, 7). Determine (a) the distance AB,, (b) the gradient of the straight line AB, and (c) the, angle AB makes with the horizontal., (a), , Points A and B are shown in Fig. 11.8(a)., In Fig. 11.8(b), the horizontal and vertical lines, AC and BC are constructed., , 9, Problem 3. If cos X =, determine the value of, 41, the other five trigonometry ratios., , Since ABC is a right-angled triangle, and, AC = (8 − 2) = 6 and BC = (7 − 3) =4, then by, Pythagoras’ theorem, , Fig. 11.7 shows a right-angled triangle X Y Z ., , AB2 = AC 2 + BC 2 = 62 + 42, , √, and AB = (62 + 42 ) = 52 = 7.211,, correct to 3 decimal places., , Z, , (b) The gradient of AB is given by tan A,, , 41, , i.e. gradient = tan A =, X, , Figure 11.7, , 99, , 9, , Y, , (c), , BC, 4 2, = =, AC, 6 3, , The angle AB makes with the horizontal is given, by tan−1 32 = 33.69◦.
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100 Higher Engineering Mathematics, f(x), 8, 7, 6, , B, , 4, 3, 2, , (a) sin α (b) cos θ (c) tan θ, , 15, 15, 8, (a), (b), (c), 17, 17, 15, , A, , 8, , ␣, , 17, , , 0, , 2, , 4, , 6, , 15, , 8, , Figure 11.10, , (a), f (x), 8, B, , 4. Point P lies at co-ordinate (−3, 1) and point, Q at (5, −4). Determine, (a) the distance PQ, (b) the gradient of the straight line PQ and, , 6, , (c), , 4, C, , A, , 2, , 0, , 2, , 4, , 6, , 8, , (b), , Now try the following exercise, Exercise 45 Further problems on, trigonometric ratios of acute angles, 1. In triangle ABC shown in Fig. 11.9, find, sin A, cos A, tan A, sin B, cos B and tan B., ⎤, ⎡, sin A = 35 , cos A = 45 , tan A = 34, ⎦, ⎣, sin B = 45 , cos B = 35 , tan B = 43, B, , A, , 3, C, , Figure 11.9, , 2. If cos A =, form., , Evaluating trigonometric ratios, , The easiest method of evaluating trigonometric functions of any angle is by using a calculator., The following values, correct to 4 decimal places,, may be checked:, , Figure 11.8, , 5, , 11.4, , the angle PQ makes with the horizontal., [(a) 9.434 (b) −0.625 (c) 32◦ ], , 15, find sin A and tan A, in fraction, 17, , 8, 8, sin A = , tan A =, 17, 15, , 3. For the right-angled triangle shown in, Fig. 11.10, find:, , sine 18◦ = 0.3090,, , cosine 56◦ = 0.5592, , cosine 115◦ = −0.4226,, sine 172◦ = 0.1392, sine 241.63◦ = −0.8799, cosine 331.78◦ = 0.8811, tangent 29◦ = 0.5543,, tangent 178◦ = −0.0349, tangent 296.42◦ = −2.0127, To evaluate, say, sine 42◦23 using a calculator, 23◦, means finding sine 42, since there are 60 minutes, 60, in 1 degree., 23, = 0.3833̇ thus 42◦ 23 = 42.383̇◦, 60, Thus sine 42◦23 = sine 42.383̇◦ = 0.6741, correct to 4, decimal places., 38◦, Similarly, cosine 72◦38 = cosine 72, = 0.2985,, 60, correct to 4 decimal places., Most calculators contain only sine, cosine and tangent functions. Thus to evaluate secants, cosecants and, cotangents, reciprocals need to be used. The following
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Introduction to trigonometry, values, correct to 4 decimal places, may be checked:, 1, secant 32◦ =, = 1.1792, cos 32◦, 1, = 1.0353, cosecant 75◦ =, sin 75◦, cotangent 41◦ =, secant 215.12◦ =, , 1, = −1.2226, cos 215.12◦, , cosecant 321.62◦ =, , 1, = −1.6106, sin 321.62◦, , cotangent 263.59◦ =, , 4. Press ◦ ”’, 5. Enter 12, 6. Press ◦ ”’, 7. Press), 8. Press = Answer = 2.798319…., Problem 8. Evaluate correct to 4 decimal places:, (a) sine 168◦14 (b) cosine 271.41◦, (c) tangent 98◦ 4, , If we know the value of a trigonometric ratio and need, to find the angle we use the inverse function on our, calculators., For example, using shift and sin on our calculator gives, sin−1 (, If, for example, we know the sine of an angle is 0.5 then, the value of the angle is given by:, sin−1 0.5 = 30◦ (Check that sin 30◦ = 0.5), does not mean, also be written as arcsin x), Similarly, if cos θ = 0.4371 then, θ = cos−1 0.4371 = 64.08◦, , 1, −1 x, sin x ; also, sin, , may, , each correct to 2 decimal places., Use your calculator to check the following worked, examples., Problem 6. Determine, correct to 4 decimal, places, sin 43◦ 39, 39 ◦, sin 43 39 = sin 43, = sin 43.65◦, 60, = 0.6903, ◦, , This answer can be obtained using the calculator as, follows:, 1. Press sin, 2. Enter 43, 3. Press ◦ ”’, 5. Press ◦ ”’, , (b), , 14◦, = 0.2039, 60, cosine 271.41◦ = 0.0246, , (c), , tangent 98◦ 4 = tan 98, , (a), , 6. Press ), , 7. Press = Answer = 0.6902512…., , sine 168◦14 = sine 168, , 4◦, = −7.0558, 60, , Problem 9. Evaluate, correct to 4 decimal places:, (a) secant 161◦ (b) secant 302◦29, 1, = −1.0576, cos 161◦, 1, 1, (b) sec 302◦29 =, =, 29◦, cos 302◦29, cos 302, 60, = 1.8620, (a), , and if tan A = 3.5984 then A = tan −1 3.5984, = 74.47◦, , 4. Enter 39, , 12◦, = 6 cos62.20◦, 60, = 2.798, , This answer can be obtained using the calculator as, follows:, 1. Enter 6, 2. Press cos, 3. Enter 62, , 1, = 0.1123, tan 263.59◦, , (Note that sin−1 x, , Problem 7. Determine, correct to 3 decimal, places, 6 cos 62◦12, 6 cos62◦ 12 = 6 cos62, , 1, = 1.1504, tan 41◦, , 101, , sec 161◦ =, , Problem 10. Evaluate, correct to 4 significant, figures:, (a) cosecant 279.16◦ (b) cosecant 49◦ 7, (a), (b), , 1, = −1.013, sin 279.16◦, 1, 1, cosec 49◦7 =, =, ◦, 7◦, sin 49 7, sin 49, 60, = 1.323, cosec 279.16◦ =, , Problem 11. Evaluate, correct to 4 decimal, places:, (a) cotangent 17.49◦ (b) cotangent 163◦ 52
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102 Higher Engineering Mathematics, 1, = 3.1735, tan 17.49◦, 1, 1, =, (b) cot 163◦52 =, ◦, 52◦, tan 163 52, tan 163, 60, = −3.4570, (a) cot 17.49◦ =, , Problem 12. Evaluate, correct to 4 significant, figures:, (a) sin 1.481 (b) cos(3π/5) (c) tan 2.93, (a) sin 1.481 means the sine of 1.481 radians. Hence, a calculator needs to be on the radian function., Hence sin 1.481 = 0.9960, (b) cos(3π/5) = cos 1.884955 · · · = −0.3090, (c) tan 2.93 = −0.2148, , cos−1 0.2437 means ‘the angle whose, cosine is 0.2437’, Using a calculator:, 1. Press shift 2. Press cos 3. Enter 0.2437, 4. Press ) 5. Press = The answer 75.894979…, is displayed, 6. Press ◦ ”’ and 75◦53 41.93 is displayed, Hence,, , cos−1 0.2437 = 75.89◦ = 77◦54, correct to the nearest minute., , Problem 16. Find the acute angle tan−1 7.4523 in, degrees and minutes, , Problem 13. Evaluate, correct to 4 decimal, places:, (a) secant 5.37 (b) cosecant π/4, (c) cotangent π/24, , tan−1 7.4523 means ‘the angle whose, tangent is 7.4523’, Using a calculator:, , (a) Again, with no degrees sign, it is assumed that, 5.37 means 5.37 radians., 1, = 1.6361, Hence sec 5.37 =, cos 5.37, 1, 1, (b) cosec (π/4) =, =, sin(π/4) sin 0.785398 . . ., = 1.4142, 1, 1, (c) cot(5π/24) =, =, tan(5π/24) tan 0.654498 . . ., = 1.3032, Problem 14. Find, in degrees, the acute angle, sin−1 0.4128 correct to 2 decimal places., sin−1 0.4128 means ‘the angle whose, sine is 0.4128’, Using a calculator:, 1. Press shift 2. Press sin 3. Enter 0.4128, 4. Press ) 5. Press = The answer 24.380848……, is displayed, Hence,, , sin−1 0.4128 = 24.38◦, , Problem 15. Find the acute angle cos−1 0.2437 in, degrees and minutes, , 1. Press shift 2. Press tan 3. Enter 7.4523, 4. Press ) 5. Press = The answer 82.357318…, is displayed, 6. Press ◦ ”’ and 82◦21 26.35 is displayed, Hence,, , tan−1 7.4523 = 82.36◦ = 82◦ 21, correct to the nearest minute., , Problem 17. Determine the acute angles:, (a) sec−1 2.3164 (b) cosec −11.1784, (c) cot −1 2.1273, (a) sec−1 2.3164 = cos−1, , , , 1, 2.3164, , , , = cos−1 0.4317 . . ., = 64.42◦ or 64◦ 25, or 1.124 radians, (b) cosec −11.1784 = sin−1, , , , 1, 1.1784, , , , = sin−1 0.8486 . . ., = 58.06◦ or 58◦4, or 1.013 radians
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103, , Introduction to trigonometry, (c), , cot −1 2.1273 = tan−1, , , , 1, 2.1273, , , Problem 20. In triangle E F G in Fig. 11.11,, calculate angle G., , = tan−1 0.4700 . . ., ◦, , ◦, , = 25.18 or 25 11, , E, , , or 0.439 radians, Problem 18. Evaluate the following expression,, correct to 4 significant figures:, 4 sec 32◦ 10 − 2 cot 15◦19, 3 cosec 63◦ 8 tan 14◦57, , F, , from which,, , 2.30, = 0.26406429 . . ., 8.71, G = sin−1 0.26406429 . . ., , i.e., , G = 15.311360 . . ., , sin G =, , i.e., , sec 32◦ 10 = 1.1813, cot 15◦ 19 = 3.6512, cosec 63◦ 8 = 1.1210, tan 14◦57 = 0.2670, 4 sec 32◦10 − 2 cot 15◦ 19, Hence, 3 cosec 63◦8 tan 14◦57, =, , 4(1.1813) − 2(3.6512), 3(1.1210)(0.2670), , =, , 4.7252 − 7.3024, 0.8979, , Hence,, , Now try the following exercise, , In Problems 1 to 8, evaluate correct to 4 decimal, places:, 1., , (a) sine 27◦ (b) sine 172.41◦, (c) sine 302◦52, , (a) 0.4540 (b) 0.1321, (c) −0.8399, , 2., , (a) cosine 124◦ (b) cosine 21.46◦, (c) cosine 284◦10, , (a) −0.5592 (b) 0.9307, (c) 0.2447, , 3., , (a) tangent 145◦ (b) tangent 310.59◦, (c) tangent 49◦ 16, (a) −0.7002 (b) −1.1671, (c) 1.1612, , 4., , (a) secant 73◦ (b) secant 286.45◦, (c) secant 155◦41, (a) 3.4203 (b) 3.5313, , correct to 4 significant figures., Problem 19. Evaluate correct to 4 decimal places:, (a) sec(−115◦ ) (b) cosec (−95◦ 47 ), Positive angles are considered by convention to be, anticlockwise and negative angles as clockwise., Hence −115◦ is actually the same as 245◦ (i.e., 360◦− 115◦), Hence sec(−115◦ ) = sec 245◦ =, , 1, cos 245◦, , = −2.3662, cosec (−95◦ 47 ) =, , ∠G = 15.31◦ or 15◦19, , Exercise 46 Further problems on, evaluating trigonometric ratios, , −2.5772, =, = −2.870,, 0.8979, , (b), , G, , Figure 11.11, , With reference to ∠G, the two sides of the triangle, given are the opposite side E F and the hypotenuse, E G; hence, sine is used,, , By calculator:, , (a), , 8.71, , 2.30, , 1, , = −1.0051, 47◦, sin −95, 60, , (c) −1.0974
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104 Higher Engineering Mathematics, 5., , (a) cosecant 213◦ (b) cosecant 15.62◦, (c) cosecant 311◦50, (a) −1.8361 (b) 3.7139, (c) −1.3421, , 6., , (a) cotangent 71◦ (b) cotangent 151.62◦, ◦, (c) cotangent 321, 23, (a) 0.3443 (b) −1.8510, (c) −1.2519, , 7., , (a) sine, , 8., , (a) sec, , In the triangle shown in Fig. 11.13, determine, angle θ in degrees and minutes., [20◦21 ], , , , 23, , 2π, (b) cos 1.681 (c) tan 3.672, 3, , (a) 0.8660 (b) −0.1010, (c) 0.5865, , π, (b) cosec 2.961 (c) cot 2.612, 8, , (a) 1.0824 (b) 5.5675, (c) −1.7083, , In Problems 9 to 14, determine the acute angle, in degrees (correct to 2 decimal places), degrees, and minutes, and in radians (correct to 3 decimal, places)., , 13.54◦, 13◦32 ,, 9. sin−1 0.2341, 0.236 rad, , 34.20◦ , 34◦12 ,, 10. cos−1 0.8271, 0.597 rad, , 39.03◦ , 39◦2 ,, 11. tan−1 0.8106, 0.681 rad, , 51.92◦, 51◦55 ,, 12. sec−1 1.6214, 0.906 rad, , 23.69◦, 23◦41 ,, 13. cosec−1 2.4891, 0.413 rad, , 27.01◦, 27◦1 ,, 14. cot −1 1.9614, 0.471 rad, 15., , 16., , In the triangle shown in Fig. 11.12, determine, angle θ, correct to 2 decimal places., [29.05◦ ], , 8, , Figure 11.13, , In Problems 17 to 20, evaluate correct to 4 significant figures., 17., , 4 cos 56◦19 − 3 sin 21◦57, , [1.097], , 18., , 11.5 tan 49◦ 11 − sin 90◦, 3 cos 45◦, , [5.805], , 19., , 5 sin 86◦3, 3 tan 14◦ 29 − 2 cos31◦ 9, , [−5.325], , 20., 21., , 22., , 23., , (sin 34◦27 )(cos 69◦2 ), (2 tan 53◦39 ), , 24., , 3 cot 14◦ 15 sec 23◦9, , 25., , cosec 27◦ 19 + sec 45◦29, 1 − cosec 27◦ 19 sec 45◦ 29, , 26., 9, , Figure 11.12, , If tan x = 1.5276, determine sec x, cosec x,, and cot x. (Assume x is an acute angle), [1.8258, 1.1952, 0.6546], , In Problems 23 to 25 evaluate correct to 4 significant figures, , 5, , , 6.4 cosec 29◦5 − sec 81◦, [0.7199], 2 cot 12◦, Determine the acute angle, in degrees and, minutes, , correct to the nearest, minute, given, 4.32 sin 42◦16, −1, [21◦42 ], by sin, 7.86, , [0.07448], [12.85], [−1.710], , Evaluate correct to 4 decimal places:, (a) sine (−125◦) (b) tan(−241◦), (c) cos(−49◦15, ), (a) −0.8192 (b) −1.8040, (c) 0.6528
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Introduction to trigonometry, , 27., , Evaluate correct to 5 significant figures:, (a) cosec (−143◦) (b) cot(−252◦ ), (c) sec(−67◦22 ), (a) −1.6616 (b) −0.32492, (c) 2.5985, , To ‘solve triangle ABC’ means ‘to find the length, AC and angles B and C’, sin C =, , sin B =, hence, , Problem 21. In triangle PQR shown in, Fig. 11.14, find the lengths of PQ and PR., P, , 388, Q, , 7.5 cm, , R, , PQ PQ, =, QR 7.5, PQ = 7.5 tan 38◦ = 7.5(0.7813), , = 5.860 cm, QR 7.5, cos 38◦ =, =, PR, PR, 7.5, 7.5, hence PR =, =, = 9.518 cm, ◦, cos 38, 0.7880, [Check: Using Pythagoras’ theorem, (7.5)2 + (5.860)2 = 90.59 = (9.518)2 ], , or, using Pythagoras’, theorem, 372 = 352 + AC 2 , from, , 2, which, AC = (37 − 352 ) = 12.0 mm., Problem 23. Solve triangle XYZ given, ∠X =90◦ , ∠Y = 23◦17 and Y Z = 20.0 mm., Determine also its area., It is always advisable to make a reasonably accurate, sketch so as to visualize the expected magnitudes of, unknown sides and angles. Such a sketch is shown in, Fig. 11.16., , sin 23◦ 17 =, , 37 mm, , B, , XZ, 20.0, Z, 20.0 mm, 238179, Y, , X, , Figure 11.16, , hence, , XZ = 20.0 sin 23◦17, = 20.0(0.3953) = 7.906 mm, , Problem 22. Solve the triangle ABC shown in, Fig. 11.15., 35 mm, , AC = 37 sin 18◦ 55 = 37(0.3242), , ∠Z = 180◦ − 90◦ − 23◦ 17 = 66◦43, , tan 38◦ =, , A, , AC, 37, , = 12.0 mm, , Figure 11.14, , hence, , 35, = 0.94595, 37, , hence ∠C = sin−1 0.94595 =71.08◦ = 71◦ 5 ., ∠B = 180◦ − 90◦− 71◦ 5 = 18◦55 (since angles in a, triangle add up to 180◦ ), , 11.5 Solution of right-angled, triangles, To ‘solve a right-angled triangle’ means ‘to find the, unknown sides and angles’. This is achieved by using, (i) the theorem of Pythagoras, and/or (ii) trigonometric, ratios. This is demonstrated in the following problems., , 105, , cos 23◦17 =, hence, , XY, 20.0, , XY = 20.0 cos 23◦17, = 20.0(0.9186) = 18.37 mm, , C, , Figure 11.15, , [Check: Using Pythagoras’ theorem, (18.37)2 + (7.906)2 = 400.0 = (20.0)2 ]
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106 Higher Engineering Mathematics, Area of triangle XYZ, =, , 1, 2, , 7. A ladder rests against the top of the perpendicular wall of a building and makes an angle of, 73◦ with the ground. If the foot of the ladder is, 2 m from the wall, calculate the height of the, building., [6.54 m], , (base) (perpendicular height), , = 12 (X Y )(X Z ) = 12 (18.37)(7.906), = 72.62 mm2, Now try the following exercise, , 11.6 Angles of elevation and, depression, , Exercise 47 Further problems on the, solution of right-angled triangles, 1. Solve triangle ABC in Fig. 11.17(i)., BC = 3.50 cm, AB = 6.10 cm,, ∠B = 55◦, D, , B, 4 cm, A, , 358, 5.0 cm, , G, , 3 cm, E, , A, , 418 15.0 mm, , C, , l, , F, , (i), , H, , (a) If, in Fig. 11.19, BC represents horizontal ground and AB a vertical flagpole, then the angle of, elevation of the top of the flagpole, A, from the, point C is the angle that the imaginary straight, line AC must be raised (or elevated) from the, horizontal CB, i.e. angle θ., , (ii), , (iii), , 2. Solve triangle DEF in Fig. 11.17(ii)., [F E = 5 cm, ∠E =53◦8 , ∠F = 36◦52 ], 3. Solve triangle GHI in Fig. 11.17(iii)., G H = 9.841 mm, GI = 11.32 mm,, ∠H = 49◦, 4. Solve thetriangle JKL in Fig. 11.18(i) and find, KL = 5.43 cm, JL = 8.62 cm,, its area., ∠J = 39◦, area = 18.19 cm2, 5. Solve the triangle MNO in Fig. 11.18(ii) and, find its area., , MN = 28.86 mm, NO= 13.82 mm,, ∠O = 64◦ 25 , area = 199.4 mm2, J, , M, , 258359, , 3.69 m, P, , 6.7 cm, K, , N, , 32.0 mm, 518, (i), , , , C, , Figure 11.17, , Q, , 8.75 m, R, , L, , O, (ii), , (iii), , Figure 11.18, , 6. Solve the triangle PQR in Fig. 11.18(iii) and, find its area., , PR = 7.934 m, ∠Q = 65◦ 3 ,, ∠R = 24◦ 57 , area = 14.64 m2, , B, , Figure 11.19, , (b) If, in Fig. 11.20, PQ represents a vertical cliff and, R a ship at sea, then the angle of depression of, the ship from point P is the angle through which, the imaginary straight line PR must be lowered, (or depressed) from the horizontal to the ship, i.e., angle φ., P, , , , Q, , R, , Figure 11.20, , (Note, ∠PRQ is also φ—alternate angles between, parallel lines.), Problem 24. An electricity pylon stands on, horizontal ground. At a point 80 m from the base of, the pylon, the angle of elevation of the top of the, pylon is 23◦ . Calculate the height of the pylon to the, nearest metre., Figure 11.21 shows the pylon AB and the angle of, elevation of A from point C is 23◦, tan 23◦ =, , AB, AB, =, BC, 80
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Introduction to trigonometry, Hence height of pylon AB, , Problem 26. The angle of depression of a ship, viewed at a particular instant from the top of a 75 m, vertical cliff is 30◦. Find the distance of the ship, from the base of the cliff at this instant. The ship is, sailing away from the cliff at constant speed and, 1 minute later its angle of depression from the top of, the cliff is 20◦. Determine the speed of the ship, in km/h., , = 80 tan 23◦ = 80(0.4245) = 33.96 m, = 34 m to the nearest metre., A, , C, , 23⬚, 80 m, , B, , Figure 11.21, , Problem 25. A surveyor measures the angle of, elevation of the top of a perpendicular building as, 19◦. He moves 120 m nearer the building and finds, the angle of elevation is now 47◦. Determine the, height of the building., , Figure 11.23 shows the cliff AB, the initial position, of the ship at C and the final position at D. Since the, angle of depression is initially 30◦ then ∠AC B = 30◦, (alternate angles between parallel lines)., AB, 75, =, BC, BC, 75, 75, BC =, =, = 129.9 m, tan 30◦ 0.5774, = initial position of ship from, base of cliff, , tan 30◦ =, hence, , The building PQ and the angles of elevation are shown, in Fig. 11.22., In triangle PQS,, , hence, , h, x + 120, h = tan 19◦(x + 120),, , i.e., , h = 0.3443(x + 120), , tan 19◦ =, , 308, A, 208, , (1), , 75 m, , P, , 208, , 308, B, , h, , C, , x, , D, , Figure 11.23, 478, , Q, , R, , 198, , S, , In triangle ABD,, , 120, , x, , AB, 75, =, BD BC + CD, 75, =, 129.9 + x, , tan 20◦ =, , Figure 11.22, , h, x, ◦, h = tan 47 (x), i.e. h = 1.0724x, , In triangle PQR, tan 47◦ =, hence, , 107, , Equating equations (1) and (2) gives:, , (2), Hence, , 75, 75, =, ◦, tan 20, 0.3640, = 206.0 m, , 129.9 + x =, , 0.3443(x + 120) = 1.0724x, 0.3443x + (0.3443)(120) = 1.0724x, (0.3443)(120) = (1.0724 − 0.3443)x, 41.316 = 0.7281x, 41.316, x=, = 56.74 m, 0.7281, From equation (2), height of building,, h = 1.0724x = 1.0724(56.74) = 60.85 m., , from which, , x = 206.0 − 129.9 = 76.1 m, , Thus the ship sails 76.1 m in 1 minute, i.e. 60 s, hence, speed of ship, distance 76.1, =, m/s, time, 60, 76.1 ×60 × 60, =, km/h = 4.57 km/h, 60 ×1000, , =
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108 Higher Engineering Mathematics, Now try the following exercise, Exercise 48 Further problems on angles of, elevation and depression, 1. If the angle of elevation of the top of a vertical, 30 m high aerial is 32◦, how far is it to the, aerial?, [48 m], 2. From the top of a vertical cliff 80.0 m high, the angles of depression of two buoys lying, due west of the cliff are 23◦ and 15◦ , respectively. How far are the buoys apart? [110.1 m], , trigonometric ratios and the theorem of Pythagoras may, be used for its solution, as shown in Section 11.5. However, for a non-right-angled triangle, trigonometric, ratios and Pythagoras’ theorem cannot be used. Instead,, two rules, called the sine rule and the cosine rule,, are used., , Sine rule, With reference to triangle ABC of Fig. 11.24, the sine, rule states:, a, b, c, =, =, sin A sin B sin C, , 3. From a point on horizontal ground a surveyor, measures the angle of elevation of the top of, a flagpole as 18◦ 40 . He moves 50 m nearer, to the flagpole and measures the angle of elevation as 26◦22 . Determine the height of the, flagpole., [53.0 m], 4. A flagpole stands on the edge of the top of a, building. At a point 200 m from the building, the angles of elevation of the top and bottom of the pole are 32◦ and 30◦ respectively., Calculate the height of the flagpole. [9.50 m], 5. From a ship at sea, the angles of elevation of, the top and bottom of a vertical lighthouse, standing on the edge of a vertical cliff are, 31◦ and 26◦ , respectively. If the lighthouse is, 25.0 m high, calculate the height of the cliff., [107.8 m], 6. From a window 4.2 m above horizontal ground, the angle of depression of the foot of a building, across the road is 24◦ and the angle of elevation, of the top of the building is 34◦. Determine,, correct to the nearest centimetre, the width of, the road and the height of the building., [9.43 m, 10.56 m], 7. The elevation of a tower from two points, one, due east of the tower and the other due west, of it are 20◦ and 24◦ , respectively, and the two, points of observation are 300 m apart. Find the, height of the tower to the nearest metre., [60 m], , 11.7, , Sine and cosine rules, , To ‘solve a triangle’ means ‘to find the values of, unknown sides and angles’. If a triangle is right angled,, , A, , c, , B, , b, , a, , C, , Figure 11.24, , The rule may be used only when:, (i) 1 side and any 2 angles are initially given, or, (ii) 2 sides and an angle (not the included angle) are, initially given., Cosine rule, With reference to triangle ABC of Fig. 11.24, the cosine, rule states:, a 2 = b2 + c 2 − 2bc cos A, or b2 = a 2 + c 2 − 2ac cos B, or c2 = a 2 + b 2 − 2ab cos C, The rule may be used only when:, (i) 2 sides and the included angle are initially given,, or, (ii) 3 sides are initially given., , 11.8, , Area of any triangle, , The area of any triangle such as ABC of Fig. 11.24 is, given by:
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Introduction to trigonometry, (i), (ii), (iii), , 1, 2 × base × perpendicular, 1, 2 ab sin C, , √, , height, or, , or 12 ac sin B or 12 bc sin A, or, , [s(s − a)(s − b)(s − c)], where, a+b+c, s=, 2, , 109, , vice-versa. In this problem, Y is the largest angle and, XZ is the longest of the three sides., Problem 28. Solve the triangle PQR and find its, area given that QR = 36.5 mm, PR = 29.6 mm and, ∠Q = 36◦ ., Triangle PQR is shown in Fig. 11.26., , 11.9, , Worked problems on the, solution of triangles and, finding their areas, , P, , r, , =51◦,, , Problem 27. In a triangle XYZ, ∠X, ∠Y = 67◦ and YZ = 15.2 cm. Solve the triangle and, find its area., , Q, , 36⬚, p ⫽ 36.5 mm, , q ⫽ 29.6 mm, , R, , Figure 11.26, , The triangle XYZ is shown in Fig. 11.25. Since, the angles in a triangle add up to 180◦, then, Z = 180◦ − 51◦− 67◦ = 62◦ . Applying the sine rule:, , Applying the sine rule:, 36.5, 29.6, =, ◦, sin 36, sin P, , 15.2, y, z, =, =, sin 51◦ sin 67◦, sin 62◦, 15.2, y, Using, =, and transposing gives:, sin 51◦ sin 67◦, 15.2 sin 67◦, = 18.00 cm =XZ, sin 51◦, z, 15.2, =, and transposing gives:, Using, ◦, sin 51, sin 62◦, y=, , z=, , 15.2 sin 62◦, = 17.27 cm =XY, sin 51◦, , X, , y, , 36.5 sin 36◦, = 0.7248, 29.6, , Hence P = sin−1 0.7248 =46◦ 27 or 133◦33 ., When P = 46◦27 and Q = 36◦ then, R = 180◦ − 46◦27 − 36◦ = 97◦ 33 ., When P = 133◦33 and Q =36◦ then, R = 180◦ − 133◦ 33 − 36◦ = 10◦27 ., Thus, in this problem, there are two separate sets of, results and both are feasible solutions. Such a situation, is called the ambiguous case., , 29.6, r, =, ◦, sin 97 33, sin 36◦, , 67⬚, Y, , sin P =, , Case 1. P = 46◦ 27 , Q =36◦ , R = 97◦33 ,, p =36.5 mm and q = 29.6 mm., From the sine rule:, , 51⬚, z, , from which,, , x ⫽15.2 cm, , Z, , Figure 11.25, , Area of triangle XYZ = 12 x y sin Z, = 12 (15.2)(18.00) sin 62◦ = 120.8 cm2 (or area, = 12 x z sin Y = 12 (15.2)(17.27) sin 67◦ = 120.8 cm2 )., It is always worth checking with triangle problems, that the longest side is opposite the largest angle, and, , from which,, 29.6 sin 97◦ 33, = 49.92 mm, sin 36◦, Area = 12 pq sin R = 12 (36.5)(29.6) sin 97◦ 33, r=, , = 535.5 mm2, Case 2. P = 133◦33 , Q = 36◦ , R = 10◦27 ,, p =36.5 mm and q = 29.6 mm.
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110 Higher Engineering Mathematics, From the sine rule:, 5., , r, 29.6, =, sin 10◦ 27, sin 36◦, from which,, r=, , 29.6 sin 10◦ 27, = 9.134 mm, sin 36◦, , Area = 21 pq sin R = 12 (36.5)(29.6) sin 10◦ 27, = 97.98 mm2 ., Triangle PQR for case 2 is shown in Fig. 11.27., , j = 3.85 cm, k = 3.23 cm, K = 36◦., ⎤, ⎡, J = 44◦29 , L = 99◦ 31 ,, ⎢ l = 5.420 cm, area = 6.132 cm2 or ⎥, ⎥, ⎢, ⎥, ⎢, ⎦, ⎣ J = 135◦31 , L = 8◦29 ,, 2, l = 0.811 cm, area = 0.917 cm, , 6. k = 46 mm, l = 36 mm, L =35◦ ., ⎤, ⎡, K = 47◦8 , J = 97◦ 52 ,, ⎢ j = 62.2 mm, area = 820.2 mm2 or ⎥, ⎥, ⎢, ⎥, ⎢, ⎦, ⎣ K = 132◦52 , J = 12◦ 8 ,, 2, j = 13.19 mm, area = 174.0 mm, , 133⬚33⬘, P, 9.134 mm, Q, , 11.10, , 29.6 mm, 36.5 mm, , 36⬚, , R, , Further worked problems on, solving triangles and finding, their areas, , 10⬚27⬘, , Figure 11.27, , Now try the following exercise, , Problem 29. Solve triangle DEF and find its area, given that EF = 35.0 mm, DE = 25.0 mm and, ∠E = 64◦., Triangle DEF is shown in Fig. 11.28., D, , Exercise 49 Further problems on solving, triangles and finding their areas, , f ⫽ 25.0 mm, 64⬚, , In Problems 1 and 2, use the sine rule to solve the, triangles ABC and find their areas., 1., , 2., , A = 29◦, B =68◦, b = 27 mm., C = 83◦ , a =14.1 mm,, c = 28.9 mm, area = 189 mm2, B = 71◦26 , C = 56◦32 , b = 8.60 cm., , A = 52◦2 , c = 7.568 cm,, a = 7.152 cm, area = 25.65 cm2, , In Problems 3 and 4, use the sine rule to solve the, triangles DEF and find their areas., 3. d = 17 cm, f = 22 cm, F = 26◦ ., , D = 19◦48 , E =134◦12 ,, e = 36.0 cm, area = 134 cm2, 4. d = 32.6 mm, e = 25.4 mm, D = 104◦22 ., , E = 49◦ 0 , F = 26◦ 38 ,, f = 15.09 mm, area = 185.6 mm2, In Problems 5 and 6, use the sine rule to solve the, triangles JKL and find their areas., , e, , E, , d ⫽ 35.0 mm, , F, , Figure 11.28, , Applying the cosine rule:, e2 = d 2 + f 2 − 2d f cos E, i.e., , e2 = (35.0)2 + (25.0)2, − [2(35.0)(25.0) cos 64◦], , = 1225 + 625 − 767.1 = 1083, √, from which, e = 1083 = 32.91 mm, Applying the sine rule:, 25.0, 32.91, =, ◦, sin 64, sin F, 25.0 sin 64◦, = 0.6828, from which, sin F =, 32.91, Thus, , ∠F = sin−1 0.6828, = 43◦4 or 136◦56
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Introduction to trigonometry, F = 136◦56 is not possible in this case since, 136◦56 + 64◦ is greater than 180◦ . Thus only, F = 43◦4 is valid, ∠D = 180◦ − 64◦ − 43◦4 = 72◦56, Area of triangle DE F = 12 d f sin E, = 12 (35.0)(25.0) sin 64◦ = 393.2 mm2., Problem 30. A triangle ABC has sides a =, 9.0 cm, b = 7.5 cm and c = 6.5 cm. Determine its, three angles and its area., Triangle ABC is shown in Fig. 11.29. It is usual first, to calculate the largest angle to determine whether the, triangle is acute or obtuse. In this case the largest angle, is A (i.e. opposite the longest side)., Applying the cosine rule:, a 2 = b2 + c2 − 2bc cos A, from which, 2bc cos A = b 2 + c2 − a 2, b 2 + c2 − a 2, 7.52 + 6.52 − 9.02, =, 2bc, 2(7.5)(6.5), = 0.1795, , and cos A =, , A, c 5 6.5 cm, B, , b 5 7.5 cm, , a 5 9.0 cm, , C, , Hence area, , = [11.5(11.5 − 9.0)(11.5 − 7.5)(11.5 − 6.5)], , = [11.5(2.5)(4.0)(5.0)] = 23.98 cm2, Alternatively, area = 12 ab sin C, = 12 (9.0)(7.5) sin 45◦16 = 23.98 cm2 ., Now try the following exercise, Exercise 50 Further problems on solving, triangles and finding their areas, In Problems 1 and 2, use the cosine and sine, rules to solve the triangles PQR and find their, areas., 1. q = 12 cm, r = 16 cm, P = 54◦ ., , p = 13.2 cm, Q = 47◦21 ,, R = 78◦39 , area = 77.7 cm2, 2. q = 3.25 m, r = 4.42 m, P = 105◦., , p = 6.127 m, Q = 30◦50 ,, R = 44◦10 , area = 6.938 m2, In problems 3 and 4, use the cosine and sine, rules to solve the triangles X Y Z and find their, areas., 3., , x = 10.0 cm, y = 8.0 cm, z =7.0 cm., , X = 83◦ 20 , Y = 52◦ 37 ,, Z = 44◦ 3 , area = 27.8 cm2, , 4., , x = 21 mm, y = 34 mm, z = 42 mm., , X = 29◦46 , Y = 53◦30 ,, Z = 96◦44 , area = 355 mm2, , Figure 11.29, , Hence A = cos−1 0.1795 = 79◦ 40 (or 280◦20 , which is, obviously impossible). The triangle is thus acute angled, since cos A is positive. (If cos A had been negative, angle, A would be obtuse, i.e. lie between 90◦ and 180◦)., Applying the sine rule:, 7.5, 9.0, =, sin 79◦40, sin B, from which,, 7.5 sin 79◦ 40, = 0.8198, 9.0, B = sin−1 0.8198 = 55◦4, , sin B =, Hence, , C = 180◦ − 79◦ 40 − 55◦ 4 = 45◦ 16, , Area = [s(s − a)(s − b)(s − c)],, a +b+c, 9.0 + 7.5 + 6.5, where, s=, =, 2, 2, = 11.5 cm, and, , 111, , 11.11 Practical situations involving, trigonometry, There are a number of practical situations where the, use of trigonometry is needed to find unknown sides and, angles of triangles. This is demonstrated in the following, problems., Problem 31. A room 8.0 m wide has a span, roof which slopes at 33◦ on one side and 40◦ on the, other. Find the length of the roof slopes, correct to, the nearest centimetre.
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112 Higher Engineering Mathematics, A section of the roof is shown in Fig. 11.30., , OA =, , B, , A, , 338, , The resultant, , , , (17257) = 131.4 V, , Applying the sine rule:, , 408, , C, , 8.0 m, , Figure 11.30, , Angle at ridge, B = 180◦ − 33◦− 40◦ = 107◦, From the sine rule:, 8.0, a, =, ◦, sin 107, sin 33◦, from which,, 8.0 sin 33◦, = 4.556 m, a=, sin 107◦, Also from the sine rule:, 8.0, c, =, sin 107◦ sin 40◦, from which,, , from which,, , 131.4, 100, =, sin 135◦ sin AO B, 100 sin 135◦, sin AOB =, 131.4, = 0.5381, , Hence angle AOB = sin−1 0.5381 =32◦ 33, 147◦27 , which is impossible in this case)., , (or, , Hence the resultant voltage is 131.4 volts at 32◦ 33, to V1 ., Problem 33. In Fig. 11.32, PR represents the, inclined jib of a crane and is 10.0 long. PQ is 4.0 m, long. Determine the inclination of the jib to the, vertical and the length of tie QR., R, , 8.0 sin 40◦, = 5.377 m, c=, sin 107◦, Q, , Hence the roof slopes are 4.56 m and 5.38 m, correct, to the nearest centimetre., , 120⬚, 4.0 m, , Problem 32. Two voltage phasors are shown in, Fig. 11.31. If V1 = 40 V and V2 = 100 V determine, the value of their resultant (i.e. length OA) and the, angle the resultant makes with V1 ., A, , 10.0 m, , P, , Figure 11.32, , Applying the sine rule:, , V2 ⫽100 V, , PR, PQ, =, sin 120◦ sin R, 45⬚, 0, V1 ⫽ 40 V B, , Figure 11.31, , Angle OBA = 180◦ − 45◦ = 135◦, Applying the cosine rule:, OA2 = V12 + V22 − 2V1 V2 cos OBA, = 402 + 1002 − {2(40)(100) cos 135◦}, = 1600 + 10000 − {−5657}, = 1600 + 10000 + 5657 = 17257, , from which,, PQ sin 120◦ (4.0) sin 120◦, =, PR, 10.0, = 0.3464, , sin R =, , Hence ∠R = sin−1 0.3464 = 20◦ 16 (or 159◦44 , which, is impossible in this case)., ∠P = 180◦ − 120◦ − 20◦ 16 = 39◦44 , which is the, inclination of the jib to the vertical., Applying the sine rule:, 10.0, QR, =, sin 120◦ sin 39◦44
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Introduction to trigonometry, , 113, , from which, length of tie,, QR =, , 28.5 m, , 10.0 sin 39◦44, = 7.38 m, sin 120◦, , 728, 34.6 m, 52.4 m, , 758, , Now try the following exercise, Exercise 51 Further problems on practical, situations involving trigonometry, 1. A ship P sails at a steady speed of 45 km/h in, a direction of W 32◦ N (i.e. a bearing of 302◦), from a port. At the same time another ship Q, leaves the port at a steady speed of 35 km/h in, a direction N 15◦ E (i.e. a bearing of 015◦)., Determine their distance apart after 4 hours., [193 km], 2. Two sides of a triangular plot of land are, 52.0 m and 34.0 m, respectively. If the area, of the plot is 620 m2 find (a) the length of, fencing required to enclose the plot and (b), the angles of the triangular plot., [(a) 122.6 m (b) 94◦49 , 40◦ 39 , 44◦ 32 ], 3. A jib crane is shown in Fig. 11.33. If the tie, rod PR is 8.0 long and PQ is 4.5 m long determine (a) the length of jib RQ and (b) the angle, between the jib and the tie rod., [(a) 11.4 m (b) 17◦33 ], , Figure 11.34, , 5. Determine the length of members BF and EB, in the roof truss shown in Fig. 11.35., [B F = 3.9 m, E B = 4.0 m], E, 4m, , 4m, , F, 2.5 m, A, , 50⬚, 5m, , D, 50⬚, 5m, , B, , 2.5 m, C, , Figure 11.35, , 6. A laboratory 9.0 m wide has a span roof, which slopes at 36◦ on one side and 44◦ on, the other. Determine the lengths of the roof, slopes., [6.35 m, 5.37 m], , 11.12 Further practical situations, involving trigonometry, Problem 34. A vertical aerial stands on, horizontal ground. A surveyor positioned due east, of the aerial measures the elevation of the top as, 48◦. He moves due south 30.0 m and measures the, elevation as 44◦ . Determine the height of the aerial., , R, , 130⬚ P, , Q, , Figure 11.33, , 4. A building site is in the form of a quadrilateral as shown in Fig. 11.34, and its area is, 1510 m2 . Determine the length of the perimeter of the site., [163.4 m], , In Fig. 11.36, DC represents the aerial, A is the initial, position of the surveyor and B his final position., From triangle ACD, tan 48◦ =, from which, , AC =, , DC, ,, AC, DC, tan 48◦, , Similarly, from triangle BCD,, BC =, , DC, tan 44◦
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114 Higher Engineering Mathematics, D, , (a) For the position shown in Fig. 11.37 determine, the angle between the connecting rod AB and, the horizontal and the length of OB., (b) How far does B move when angle AOB, changes from 50◦ to 120◦ ?, 488, , C, , (a) Applying the sine rule:, , A, , AB, AO, =, sin 50◦ sin B, , 30.0 m, , 448, , from which,, , B, , AO sin 50◦ 10.0 sin 50◦, =, AB, 30.0, = 0.2553, , sin B =, , Figure 11.36, , For triangle ABC, using Pythagoras’ theorem:, BC 2 = AB 2 + AC 2, , , DC, tan 44◦, , , DC 2, , 2, , , , DC, = (30.0) +, tan 48◦, , 2, , 2, , 1, tan2 44◦, , −, , 1, , 30.02, = 3440.4, 0.261596, , Hence, height of aerial,, DC =, , √, , Angle OAB = 180◦ − 50◦ − 14◦ 47 = 115◦ 13 ., 30.0, OB, =, ◦, sin 50, sin 115◦ 13, , = 30.02, , DC 2 (1.072323 − 0.810727) = 30.02, DC 2 =, , Hence the connecting rod AB makes an angle, of 14◦ 47 with the horizontal., , Applying the sine rule:, , , , tan 2 48◦, , Hence B = sin−1 0.2553 =14◦ 47 (or 165◦13 ,, which is impossible in this case)., , 3440.4 = 58.65 m, , from which,, OB =, , 30.0 sin 115◦13, = 35.43 cm, sin 50◦, , (b) Figure 11.38 shows the initial and final positions of, the crank mechanism. In triangle O A B , applying, the sine rule:, 30.0, 10.0, =, sin 120◦ sin A B O, from which,, sin A B O =, , Problem 35. A crank mechanism of a petrol, engine is shown in Fig. 11.37. Arm OA is 10.0 cm, long and rotates clockwise about O. The connecting, rod AB is 30.0 cm long and end B is constrained to, move horizontally., , Figure 11.37, , 50⬚, B⬘, , A⬘, , A, , 120⬚, 10.0 cm, O, , 10.0 cm, 508, , B, , 30.0 cm, , B, , A, , m, , 30.0 c, , 10.0 sin 120◦, = 0.2887, 30.0, , O, , Figure 11.38, , Hence A B O = sin−1 0.2887 =16◦ 47, which is impossible in this case)., , (or 163◦13
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Introduction to trigonometry, Angle OA B = 180◦ − 120◦ − 16◦ 47 = 43◦13 ., Phasor QR (which is joined to the end of PQ, to form triangle PQR) is 14.0 A and is at an, angle of 35◦ to the horizontal. Determine the, resultant phasor PR and the angle it makes with, phasor PQ., [32.48 A, 14◦19 ], , Applying the sine rule:, OB, 30.0, =, sin 120◦, sin 43◦ 13, from which,, 30.0 sin 43◦13, = 23.72 cm, OB =, sin 120◦, Since OB = 35.43 cm and OB = 23.72 cm then BB =, 35.43 − 23.72 = 11.71 cm., Hence B moves 11.71 cm when angle AOB changes, from 50◦ to 120◦ ., Problem 36. The area of a field is in the form of a, quadrilateral ABCD as shown in Fig. 11.39., Determine its area., B, , 42.5 m, , 2. Three forces acting on a fixed point are represented by the sides of a triangle of dimensions, 7.2 cm, 9.6 cm and 11.0 cm. Determine the, angles between the lines of action and the, three forces., [80◦ 25 , 59◦23 , 40◦12 ], 3. Calculate, correct to 3 significant figures, the, co-ordinates x and y to locate the hole centre, at P shown in Fig. 11.40., [x = 69.3 mm, y = 142 mm], P, , y, 568, , C, , 39.8 m, , 116⬚, 62.3 m, , A, , 1148, , 21.4 m, D, , Figure 11.39, , x, , 140⬚, , 100 mm, , Figure 11.40, , 4. An idler gear, 30 mm in diameter, has to be, fitted between a 70 mm diameter driving gear, and a 90 mm diameter driven gear as shown, in Fig. 11.41. Determine the value of angle θ, between the center lines., [130◦], , A diagonal drawn from B to D divides the quadrilateral, into two triangles., , 90 mm dia, , Area of quadrilateral ABCD, = area of triangle ABD + area of triangle BCD, = 12 (39.8)(21.4) sin 114◦ + 12 (42.5)(62.3) sin 56◦, = 389.04 + 1097.5 = 1487 m2, , 99.78 mm , , 30 mm dia, 70 mm dia, , Now try the following exercise, Exercise 52 Further problems on practical, situations involving trigonometry, 1. PQ and QR are the phasors representing the, alternating currents in two branches of a circuit. Phasor PQ is 20.0 A and is horizontal., , Figure 11.41, , 5. A reciprocating engine mechanism is shown, in Fig. 11.42. The crank AB is 12.0 cm long, and the connecting rod BC is 32.0 cm long., , 115
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116 Higher Engineering Mathematics, For the position shown determine the length, of AC and the angle between the crank and, the connecting rod., [40.25 cm, 126◦3 ], B, A, , 40⬚, C, , Figure 11.42, , 6. From Fig. 11.42, determine how far C moves,, correct to the nearest millimetre when angle, CAB changes from 40◦ to 160◦, B moving in, an anticlockwise direction., [19.8 cm], 25◦, , S of a tower mea7. A surveyor, standing W, sures the angle of elevation of the top of the, tower as 46◦30 . From a position E 23◦ S from, , the tower the elevation of the top is 37◦ 15 ., Determine the height of the tower if the, distance between the two observations is 75 m., [36.2 m], 8. An aeroplane is sighted due east from a radar, station at an elevation of 40◦ and a height, of 8000 m and later at an elevation of 35◦, and height 5500 m in a direction E 70◦ S. If, it is descending uniformly, find the angle of, descent. Determine also the speed of the aeroplane in km/h if the time between the two, observations is 45 s., [13◦57 , 829.9 km/h], 9. Sixteen holes are equally spaced on a pitch circle of 70 mm diameter. Determine the length of, the chord joining the centres of two adjacent, holes., [13.66 mm]
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Chapter 12, , Cartesian and polar, co-ordinates, 12.1, , From trigonometric ratios (see Chapter 11),, , Introduction, , There are two ways in which the position of a point in, a plane can be represented. These are, (a), , by Cartesian co-ordinates, i.e. (x, y), and, , (b) by polar co-ordinates, i.e. (r, θ), where r is a, ‘radius’ from a fixed point and θ is an angle from, a fixed point., , 12.2 Changing from Cartesian into, polar co-ordinates, In Fig. 12.1, if lengths x and y are known, then the, length of r can be obtained from Pythagoras’ theorem, (see Chapter 11) since OPQ is a right-angled triangle., Hence r 2 = (x 2 + y 2 ), , tan θ =, , y, x, , from which θ = tan−1, , y, x, , , y, r = x 2 + y 2 and θ = tan−1 are the two formulae we, x, need to change from Cartesian to polar co-ordinates. The, angle θ, which may be expressed in degrees or radians,, must always be measured from the positive x-axis, i.e.,, measured from the line OQ in Fig. 12.1. It is suggested, that when changing from Cartesian to polar co-ordinates, a diagram should always be sketched., Problem 1. Change the Cartesian co-ordinates, (3, 4) into polar co-ordinates., A diagram representing the point (3, 4) is shown in, Fig. 12.2., , , from which, r = x2 + y2, , P, , y, , y, P, , r, r, , , , , 0, , Figure 12.1, , 4, , y, , x, , Q, , 0, , x, , x, 3, , Figure 12.2
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118 Higher Engineering Mathematics, √, From Pythagoras’ theorem, r = 32 + 42 = 5 (note that, −5 has no meaning in this context). By trigonometric, ratios, θ = tan−1 43 = 53.13◦ or 0.927 rad., , y, , ␣, , [note that 53.13◦ = 53.13×(π/180) rad = 0.927 rad], Hence (3, 4) in Cartesian co-ordinates corresponds to (5, 53.13◦) or (5, 0.927 rad) in polar, co-ordinates., Problem 2. Express in polar co-ordinates the, position (−4, 3)., A diagram representing the point using the Cartesian, co-ordinates (−4, 3) is shown in Fig. 12.3., , y, , P, , 12, , ␣, , r, , P, , Figure 12.4, , Thus (−5, −12) in Cartesian co-ordinates corresponds to (13, 247.38◦) or (13, 4.318 rad) in polar, co-ordinates., Problem 4. Express (2, −5) in polar, co-ordinates., A sketch showing the position (2, −5) is shown in, Fig. 12.5., , , 0, , x, , 4, , , , 22 + 52 =, , Figure 12.3, , r=, , √, From Pythagoras’ theorem, r = 42 + 32 = 5., By trigonometric ratios, α = tan−1 34 = 36.87◦ or, 0.644 rad., Hence θ = 180◦ − 36.87◦ = 143.13◦ or, θ = π − 0.644 = 2.498 rad., Hence the position of point P in polar co-ordinate, form is (5, 143.13◦) or (5, 2.498 rad)., , α= tan−1, , 5, = 68.20◦ or 1.190 rad, 2, , θ = 2π − 1.190 = 5.093 rad, , y, , , , 2, 0, , A sketch showing the position (−5, −12) is shown in, Fig. 12.4., , and, , 12, 5, , = 67.38◦ or 1.176 rad, Hence θ = 180◦ + 67.38◦ = 247.38◦ or, θ = π + 1.176 = 4.318 rad, , ␣, , x, 5, , r, , P, , 52 + 122 = 13, , α= tan −1, , √, 29 = 5.385 correct to, 3 decimal places, , Hence θ = 360◦ − 68.20◦ = 291.80◦ or, , Problem 3. Express (−5, −12) in polar, co-ordinates., , , , x, , 0, , r, , 3, , r=, , , , 5, , Figure 12.5, , Thus (2, −5) in Cartesian co-ordinates corresponds, to (5.385, 291.80◦) or (5.385, 5.093 rad) in polar, co-ordinates.
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Cartesian and polar co-ordinates, Now try the following exercise, Exercise 53 Further problems on changing, from Cartesian into polar co-ordinates, In Problems 1 to 8, express the given Cartesian, co-ordinates as polar co-ordinates, correct to 2 decimal places, in both degrees and in radians., [(5.83, 59.04◦ ) or (5.83, 1.03 rad)], , (6.61, 20.82◦ ) or, 2. (6.18, 2.35), (6.61, 0.36 rad), , 1. (3, 5), , , 3. (−2, 4), , 4. (−5.4, 3.7), , 5. (−7, −3), , 6. (−2.4, −3.6), , 7. (5, −3), , 8. (9.6, −12.4), , If lengths r and angle θ are known then x =r cos θ and, y =r sin θ are the two formulae we need to change from, polar to Cartesian co-ordinates., Problem 5. Change (4, 32◦) into Cartesian, co-ordinates., A sketch showing the position (4, 32◦) is shown in, Fig. 12.7., Now x = r cos θ = 4 cos32◦ = 3.39, and, , y = r sin θ = 4 sin 32◦ = 2.12, , (4.47, 116.57◦) or, (4.47, 2.03 rad), , y, , (6.55, 145.58◦) or, (6.55, 2.54 rad), (7.62, 203.20◦), , 119, , r54, , y, , 5 328, , or, , (7.62, 3.55 rad), , 0, , x, , x, , (4.33, 236.31◦) or, (4.33, 4.12 rad), , Figure 12.7, , (5.83, 329.04◦) or, (5.83, 5.74 rad), , Hence (4, 32◦) in polar co-ordinates corresponds to, (3.39, 2.12) in Cartesian co-ordinates., , (15.68, 307.75◦) or, (15.68, 5.37 rad), , Problem 6. Express (6, 137◦) in Cartesian, co-ordinates., A sketch showing the position (6, 137◦) is shown in, Fig. 12.8., , 12.3 Changing from polar into, Cartesian co-ordinates, , x = r cos θ = 6 cos 137◦ = −4.388, which corresponds to length OA in Fig. 12.8., , From the right-angled triangle OPQ in Fig. 12.6., cos θ =, , Hence, , x, y, and sin θ = , from, r, r, trigonometric ratios, , x = r cos θ, , y = r sin θ = 6 sin 137◦ = 4.092, which corresponds to length AB in Fig. 12.8., y, , B, , y = r sin θ, , and, , r56, , y, , 5 1378, , P, A, r, , 0, , Figure 12.6, , 0, , x, , y, , Figure 12.8, x, , Q, , x, , Thus (6, 137◦) in polar co-ordinates corresponds to, (−4.388, 4.092) in Cartesian co-ordinates.
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120 Higher Engineering Mathematics, (Note that when changing from polar to Cartesian, co-ordinates it is not quite so essential to draw, a sketch. Use of x = r cos θ and y =r sin θ automatically, produces the correct signs.), Problem 7. Express (4.5, 5.16 rad) in Cartesian, co-ordinates., A sketch showing the position (4.5, 5.16 rad) is shown, in Fig. 12.9., x = r cos θ = 4.5 cos 5.16 = 1.948, , [(−2.615, −3.207)], , 6. (4, 4 rad), 7. (1.5, 300◦), , [(0.750, −1.299)], , 8. (6, 5.5 rad), , [(4.252, −4.233)], , 9. Figure 12.10 shows 5 equally spaced holes on, an 80 mm pitch circle diameter. Calculate their, co-ordinates relative to axes 0x and 0y in (a), polar form, (b) Cartesian form., Calculate also the shortest distance between, the centres of two adjacent holes., , y, , [(a) 40∠18◦, 40∠90◦, 40∠162◦,, 40∠234◦, 40∠306◦,, , 5 5.16 rad, A, x, , 0, r 5 4.5, , B, , (b) (38.04 + j12.36), (0 + j40),, (−38.04 + j12.36), (−23.51 − j32.36),, (23.51 − j32.36), 47.02 mm], , Figure 12.9, , y, , which corresponds to length OA in Fig. 12.9., y = r sin θ = 4.5 sin 5.16 = −4.057, which corresponds to length AB in Fig. 12.9., O, , Thus (1.948, −4.057) in Cartesian co-ordinates, corresponds to (4.5, 5.16 rad) in polar co-ordinates., , x, , Now try the following exercise, Exercise 54 Further problems on changing, polar into Cartesian co-ordinates, , Figure 12.10, , In Problems 1 to 8, express the given polar coordinates as Cartesian co-ordinates, correct to, 3 decimal places., 1. (5, 75◦), , [(1.294, 4.830)], , 2. (4.4, 1.12 rad), , [(1.917, 3.960)], , 3. (7, 140◦), , [(−5.362, 4.500)], , 4. (3.6, 2.5 rad), , [(−2.884, 2.154)], , 5. (10.8, 210◦ ), , [(−9.353, −5.400)], , 12.4 Use of Pol/Rec functions on, calculators, Another name for Cartesian co-ordinates is rectangular co-ordinates. Many scientific notation calculators, possess Pol and Rec functions. ‘Rec’ is an abbreviation of ‘rectangular’ (i.e., Cartesian) and ‘Pol’ is an, abbreviation of ‘polar’. Check the operation manual for, your particular calculator to determine how to use these
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Cartesian and polar co-ordinates, two functions. They make changing from Cartesian to, polar co-ordinates, and vice-versa, so much quicker and, easier., For example, with the Casio fx-83ES calculator, or, similar, to change the Cartesian number (3, 4) into polar, form, the following procedure is adopted:, 1. Press ‘shift’, 2. Press ‘Pol’, 3. Enter 3, 4. Enter ‘comma’ (obtained by ‘shift’ then )), 5. Enter 4, 6. Press ), 7. Press = The answer is: r = 5, θ = 53.13◦, , 121, , Similarly, to change the polar form number, (7, 126◦) into Cartesian or rectangular form, adopt the, following procedure:, 1. Press ‘shift’, 2. Press ‘Rec’ 3. Enter 7, 4. Enter ‘comma’, 5. Enter 126 (assuming your calculator is in, degrees mode), 6. Press ), 7. Press =, , Hence, (3, 4) in Cartesian form is the same as, (5, 53.13◦) in polar form., , The answer is: X = −4.11, and scrolling across,, Y = 5.66, correct to 2 decimal places., Hence, (7, 126◦) in polar form is the same as, (−4.11, 5.66) in rectangular or Cartesian form., , If the angle is required in radians, then before repeating, the above procedure press ‘shift’, ‘mode’ and then 4 to, change your calculator to radian mode., , Now return to Exercises 53 and 54 in this chapter and, use your calculator to determine the answers, and see, how much more quickly they may be obtained.
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Chapter 13, , The circle and its properties, 13.1, , Introduction, , A circle is a plain figure enclosed by a curved line, every, point on which is equidistant from a point within, called, the centre., , 13.2, , Properties of circles, , (i) The distance from the centre to the curve is, called the radius, r, of the circle (see OP in, Fig. 13.1)., , (vi) A quadrant is one quarter of a whole circle., (vii) A tangent to a circle is a straight line which, meets the circle in one point only and does not, cut the circle when produced. AC in Fig. 13.1 is, a tangent to the circle since it touches the curve, at point B only. If radius OB is drawn, then angle, ABO is a right angle., (viii) A sector of a circle is the part of a circle between, radii (for example, the portion OXY of Fig. 13.2, is a sector). If a sector is less than a semicircle it is called a minor sector, if greater than a, semicircle it is called a major sector., X, , Q, A, , Y, , O, , P, , O, S, , B, R, C, , T, R, , Figure 13.2, , Figure 13.1, , (ii) The boundary of a circle is called the circumference, c., (iii) Any straight line passing through the centre and, touching the circumference at each end is called, the diameter, d (see QR in Fig. 13.1). Thus, d = 2r., circumference, (iv) The ratio, = a constant for any, diameter, circle., This constant is denoted by the Greek letter π, (pronounced ‘pie’), where π = 3.14159, correct, to 5 decimal places., Hence c/d = π or c = πd or c = 2πr., (v) A semicircle is one half of the whole circle., , (ix) A chord of a circle is any straight line which, divides the circle into two parts and is terminated at each end by the circumference. ST, in, Fig. 13.2 is a chord., (x) A segment is the name given to the parts into, which a circle is divided by a chord. If the, segment is less than a semicircle it is called a, minor segment (see shaded area in Fig. 13.2)., If the segment is greater than a semicircle it is, called a major segment (see the unshaded area, in Fig. 13.2)., (xi) An arc is a portion of the circumference of a, circle. The distance SRT in Fig. 13.2 is called, a minor arc and the distance SXYT is called a, major arc.
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123, , The circle and its properties, (xii) The angle at the centre of a circle, subtended by, an arc, is double the angle at the circumference, subtended by the same arc. With reference to, Fig. 13.3, Angle AOC = 2 × angle ABC., (xiii) The angle in a semicircle is a right angle (see, angle BQP in Fig. 13.3)., , X, , Q, , A, , 3. A crank mechanism is shown in Fig. 13.5,, where XY is a tangent to the circle at point X. If, the circle radius OX is 10 cm and length OY is, 40 cm, determine the length of the connecting, rod XY., [38.73 cm], , B, O, O, , P, , Y, , 40 cm, , C, , Figure 13.3, , Figure 13.5, , Problem 1. If the diameter of a circle is 75 mm,, find its circumference., Circumference, c = π × diameter = πd, = π(75) = 235.6 mm., , 4. If the circumference of the earth is 40 000 km, at the equator, calculate its diameter., [12 730 km], 5. Calculate the length of wire in the paper clip, shown in Fig. 13.6. The dimensions are in, millimetres., [97.13 mm], , Problem 2. In Fig. 13.4, AB is a tangent to the, circle at B. If the circle radius is 40 mm and, AB = 150 mm, calculate the length AO., , 2.5 rad, , B, A, , r, O, , 12, , 2.5 rad, , Figure 13.4, , 32, , 6, , 3 rad, , A tangent to a circle is at right angles to a radius drawn, from the point of contact, i.e. ABO = 90◦ . Hence, using, Pythagoras’ theorem:, AO2 = AB2 + OB2, , AO = (AB2 + OB2 ) = [(150)2 + (40)2 ], = 155.2 mm, , Figure 13.6, , 13.3, , Radians and degrees, , One radian is defined as the angle subtended at the, centre of a circle by an arc equal in length to the radius., , Now try the following exercise, , s, r, , Exercise 55 Further problems on, properties of circles, 1. If the radius of a circle is 41.3 mm, calculate, the circumference of the circle., [259.5 mm], 2. Find the diameter of a circle whose perimeter, is 149.8 cm., [47.68 cm], , O, , , r, , Figure 13.7, , With reference to Fig. 13.7,, for arc length s,, θ radians =, , s, r
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124 Higher Engineering Mathematics, When s = whole circumference (= 2πr) then, s 2πr, θ= =, = 2π, r, r, i.e. 2π radians = 360◦ or π radians = 180◦, 180◦, Thus, 1 rad =, = 57.30◦, correct to 2 decimal, π, places., π, π, π, Since π rad = 180◦, then = 90◦, = 60◦ , = 45◦ ,, 2, 3, 4, and so on., Problem 3., (b) 69◦47 ., , Convert to radians: (a) 125◦, , (a) Since 180◦ = π rad then 1◦ = π/180 rad, therefore, π c, 125◦ = 125, = 2.182 rad, 180, (Note that c means ‘circular measure’ and indicates radian measure.), (b) 69◦ 47 = 69, , 47◦, = 69.783◦, 60, , π c, 69.783◦ = 69.783, = 1.218 rad, 180, Problem 4. Convert to degrees and minutes:, (a) 0.749 rad (b) 3π/4 rad., , Since 180◦ = π rad then 1◦ = 180/π, hence, π , 5π, (a) 150◦ = 150, rad =, rad, 180, 6, π , 3π, rad =, rad, (b) 270◦ = 270, 180, 2, π , 75π, 5π, (c) 37.5◦ = 37.5, rad =, rad =, rad, 180, 360, 24, , Now try the following exercise, Exercise 56 Further problems on radians, and degrees, 1. Convert to radians in terms of π: (a) 30◦, , π, 5π, 5π, ◦, ◦, (b) 75 (c) 225 . (a), (b), (c), 6, 12, 4, 2. Convert to radians: (a) 48◦ (b) 84◦51, (c) 232◦15 ., [(a) 0.838 (b) 1.481 (c) 4.054], 5π, 4π, 3. Convert to degrees: (a), rad (b), rad, 6, 9, 7π, (c), rad., [(a) 150◦ (b) 80◦ (c) 105◦ ], 12, 4. Convert to degrees and minutes: (a) 0.0125 rad, (b) 2.69 rad (c) 7.241 rad., [(a) 0◦ 43 (b) 154◦8 (c) 414◦53 ], , (a) Since π rad = 180◦ then 1 rad =180◦ /π, therefore, , 0.749 = 0.749, , 180, π, , ◦, , = 42.915◦, , 0.915◦ = (0.915 × 60) = 55 , correct to the nearest minute, hence, , , , 180, π, , 3π, 3π, rad =, 4, 4, , , , Arc length, From the definition of the radian in the previous section, and Fig. 13.7,, , 0.749 rad = 42◦ 55, (b) Since 1 rad =, , 13.4 Arc length and area of circles, and sectors, , ◦, , 180, π, , arc length, s = rθ where θ is in radians, , then, ◦, , Area of circle, 3, = (180)◦ = 135◦, 4, , For any circle, area = π × (radius)2, i.e., , Problem 5. Express in radians, in terms of π,, (a) 150◦ (b) 270◦ (c) 37.5◦ ., , Since r =, , area = πr 2, d, πd 2, , then area = πr 2 or, 2, 4
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The circle and its properties, Area of sector, Area of a sector =, , 125, , s 4.75, =, = 5.22 cm, θ 0.91, Diameter = 2 × radius= 2 × 5.22 =10.44 cm, Circumference, c = πd = π(10.44) = 32.80 cm, , Since s = rθ then r =, θ, (πr 2 ) when θ is in degrees, 360, 1, θ, (πr 2 ) = r 2 θ, =, 2π, 2, when θ is in radians, , Problem 6. A hockey pitch has a semicircle of, radius 14.63 m around each goal net. Find the area, enclosed by the semicircle, correct to the nearest, square metre., 1, Area of a semicircle = πr 2, 2, 1, When r = 14.63 m, area = π(14.63)2, 2, i.e. area of semicircle = 336 m2, Problem 7. Find the area of a circular metal, plate, correct to the nearest square millimetre,, having a diameter of 35.0 mm., πd 2, 4, π(35.0)2, When d = 35.0 mm, area =, 4, i.e. area of circular plate = 962 mm2, Area of a circle = πr 2 =, , Problem 8. Find the area of a circle having a, circumference of 60.0 mm., Circumference, c = 2πr, from which radius r =, , 60.0 30.0, c, =, =, 2π, 2π, π, , Area of a circle = πr 2, , , 30.0 2, i.e. area = π, = 286.5 mm2, π, , Problem 9. Find the length of arc of a circle of, radius 5.5 cm when the angle subtended at the, centre is 1.20 rad., Length of arc, s =rθ, where θ is in radians, hence, s = (5.5)(1.20) = 6.60 cm, , Problem 10. Determine the diameter and, circumference of a circle if an arc of length 4.75 cm, subtends an angle of 0.91 rad., , Problem 11. If an angle of 125◦ is subtended by, an arc of a circle of radius 8.4 cm, find the length of, (a) the minor arc, and (b) the major arc, correct to, 3 significant figures., (a), , π , Since 180◦ = π rad then 1◦ =, rad and, 180, π , 125◦ = 125, rad., 180, Length of minor arc,, π , s =rθ = (8.4)(125), = 18.3 cm,, 180, correct to 3 significant figures., , (b) Length of major arc, = (circumference − minor arc), = 2π(8.4) − 18.3 = 34.5 cm,, correct to 3 significant figures., (Alternatively, major arc =rθ, = 8.4(360 −125)(π/180) = 34.5 cm.), Problem 12. Determine the angle, in degrees and, minutes, subtended at the centre of a circle of, diameter 42 mm by an arc of length 36 mm., Calculate also the area of the minor sector formed., Since length of arc, s =rθ then θ = s/r, Radius, r =, , diameter 42, = = 21 mm, 2, 2, , s 36, hence θ = = = 1.7143 rad, r 21, 1.7143 rad = 1.7143 × (180/π)◦ = 98.22◦ = 98◦ 13 =, angle subtended at centre of circle., Area of sector, = 12 r 2 θ = 12 (21)2 (1.7143) = 378 mm2 ., Problem 13. A football stadium floodlight can, spread its illumination over an angle of 45◦ to a, distance of 55 m. Determine the maximum area that, is floodlit.
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126 Higher Engineering Mathematics, Floodlit area = area of sector, , π , 1, 1, = r 2 θ = (55)2 45 ×, 2, 2, 180, = 1188 m2, Problem 14. An automatic garden spray produces, a spray to a distance of 1.8 m and revolves through, an angle α which may be varied. If the desired, spray catchment area is to be 2.5 m2 , to what should, angle α be set, correct to the nearest degree., Area of sector = 12 r 2 θ, hence 2.5 = 12 (1.8)2 α, 2.5 × 2, = 1.5432 rad, from which, α =, 1.82, , , 180 ◦, 1.5432 rad = 1.5432 ×, = 88.42◦, π, Hence angle α = 88◦, correct to the nearest degree., Problem 15. The angle of a tapered groove is, checked using a 20 mm diameter roller as shown in, Fig. 13.8. If the roller lies 2.12 mm below the top of, the groove, determine the value of angle θ., 2.12 mm, 20 mm, 30 mm, , , Figure 13.8, , In Fig. 13.9, triangle ABC is right-angled at C (see, Section 13.2 (vii))., 2.12 mm, , 10, B mm, , 2, , 30 mm, C, , A, , Figure 13.9, , Length BC = 10 mm (i.e. the radius of the circle), and, AB = 30 −10 −2.12 = 17.88 mm from , Fig. 13.9., , 10, 10, θ, θ, −1, = 34◦, and = sin, Hence, sin =, 2 17.88, 2, 17.88, and angle θ = 68◦, , Now try the following exercise, Exercise 57 Further problems on arc, length and area of circles and sectors, 1. Calculate the area of a circle of radius 6.0 cm,, correct to the nearest square centimetre., [113 cm2 ], 2. The diameter of a circle is 55.0 mm. Determine, its area, correct to the nearest square, millimetre., [2376 mm 2 ], 3. The perimeter of a circle is 150 mm. Find its, area, correct to the nearest square millimetre., [1790 mm 2 ], 4. Find the area of the sector, correct to the, nearest square millimetre, of a circle having, a radius of 35 mm, with angle subtended at, [802 mm 2 ], centre of 75◦., 5. An annulus has an outside diameter of, 49.0 mm and an inside diameter of 15.0 mm., Find its area correct to 4 significant figures., [1709 mm 2 ], 6. Find the area, correct to the nearest square, metre, of a 2 m wide path surrounding a, circular plot of land 200 m in diameter., [1269 m2 ], 7. A rectangular park measures 50 m by 40 m. A, 3 m flower bed is made round the two longer, sides and one short side. A circular fish pond, of diameter 8.0 m is constructed in the centre, of the park. It is planned to grass the remaining, area. Find, correct to the nearest square metre,, the area of grass., [1548 m2 ], 8. Find the length of an arc of a circle of radius, 8.32 cm when the angle subtended at the centre, is 2.14 rad. Calculate also the area of the minor, sector formed., [17.80 cm, 74.07 cm2 ], 9. If the angle subtended at the centre of a circle, of diameter 82 mm is 1.46 rad, find the lengths, of the (a) minor arc (b) major arc., [(a) 59.86 mm (b) 197.8 mm], 10. A pendulum of length 1.5 m swings through, an angle of 10◦ in a single swing. Find, in, centimetres, the length of the arc traced by the, pendulum bob., [26.2 cm]
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127, , The circle and its properties, , 11. Determine the length of the radius and circumference of a circle if an arc length of 32.6 cm, subtends an angle of 3.76 rad., [8.67 cm, 54.48 cm], , 17. A 50◦ tapered hole is checked with a 40 mm, diameter ball as shown in Fig. 13.12. Determine the length shown as x., [7.74 mm], , 12. Determine the angle of lap, in degrees and minutes, if 180 mm of a belt drive are in contact, with a pulley of diameter 250 mm., [82◦30 ], 13. Determine the number of complete revolutions, a motorcycle wheel will make in travelling, 2 km, if the wheel’s diameter is 85.1 cm., [748], 14. The floodlights at a sports ground spread its, illumination over an angle of 40◦ to a distance, of 48 m. Determine (a) the angle in radians,, and (b) the maximum area that is floodlit., [(a) 0.698 rad (b) 804.1 m2], 15. Determine (a) the shaded area in Fig. 13.10, (b) the percentage of the whole sector that the, area of the shaded portion represents., [(a) 396 mm2 (b) 42.24%], , 12, , 70 mm, x, , m, , 40 m, 508, , Figure 13.12, , 13.5, , The equation of a circle, , The simplest equation of a circle, centre at the origin,, radius r, is given by:, x 2 + y2 = r 2, For example, Fig. 13.13 shows a circle x 2 + y 2 = 9., More generally, the equation of a circle, centre (a, b),, radius r, is given by:, (x − a)2 + ( y − b)2 = r 2, , mm, , (1), , Figure 13.14 shows a circle (x − 2)2 + ( y − 3)2 = 4., The general equation of a circle is:, 0.75, rad, , 50 mm, , x 2 + y 2 + 2ex + 2 f y + c = 0, , (2), , y, , Figure 13.10, , 3, x21y25 9, , 2, , 16. Determine the length of steel strip required to, make the clip shown in Fig. 13.11., [701.8 mm], , 1, 23 22 21 0, 21, , 100 mm, , 1, , 2, , 3, , x, , 22, 23, 1308, , 125 mm, rad, , Figure 13.13, 100 mm, , Figure 13.11, , Multiplying out the bracketed terms in equation (1), gives:, x 2 − 2ax + a 2 + y 2 − 2by + b 2 = r 2
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128 Higher Engineering Mathematics, y, , y, 4, , 3, , 5, , b53, , r5, , 2, , b 51, , 2, , 28, , 0, , 2, , 2, , r5, , 4, , x, , 4, , 26, , 24, , 22, , 0, , x, , a 524, , a52, , Figure 13.15, , Figure 13.14, , Alternatively, x 2 + y 2 + 8x − 2y + 8 = 0 may be rearranged as:, , Comparing this with equation (2) gives:, 2e = −2a, i.e. a = −, , (x + 4)2 + ( y − 1)2 − 9 = 0, , 2e, 2, , i.e., , 2f, and 2 f = −2b, i.e. b = −, 2, and c = a 2 + b2 − r 2 ,, , i.e., r = (a2 + b2 − c), , which represents a circle, centre (−4, 1) and radius 3,, as stated above., Problem 17. Sketch the circle given by the, equation: x 2 + y 2 − 4x + 6y − 3 = 0., , Thus, for example, the equation, x 2 + y 2 − 4x − 6y + 9 = 0, , (x + 4)2 + ( y − 1)2 = 32, , , , −4, a =−, ,, 2, , represents a circle with centre, , , −6, b=−, , i.e. at (2, 3) and radius, 2, , r = (22 + 32 − 9) = 2., Hence x 2 + y 2 − 4x − 6y + 9 =0 is the circle shown in, Fig. 13.14 (which may be checked by multiplying out, the brackets in the equation, (x − 2)2 + ( y − 3)2 = 4, Problem 16. Determine (a) the radius, and (b) the, co-ordinates of the centre of the circle given by the, equation: x 2 + y 2 + 8x − 2y + 8 =0., x 2 + y 2 + 8x − 2y + 8 =0 is of the form shown in equation (2),, , , , 8, −2, = −4, b = −, =1, where a = −, 2, 2, , √, and r = [(−4)2 + (1)2 − 8] = 9 = 3, Hence x 2 + y 2 + 8x − 2y + 8 =0 represents a circle centre (−4, 1) and radius 3, as shown in Fig. 13.15., , The equation of a circle, centre (a, b), radius r is, given by:, (x − a)2 + ( y − b)2 = r 2, The general equation of a circle is, x 2 + y 2 + 2ex + 2 f y + c = 0., 2e, 2f, From above a = − , b = −, and, 2, 2, , r = (a 2 + b2 − c)., Hence if x 2 + y 2 − 4x + 6y − 3 =0, , , , −4, 6, = 2, b = −, = −3, then a = −, 2, 2, , and r = [(2)2 + (−3)2 − (−3)], √, = 16 = 4, Thus the circle has centre (2, −3) and radius 4, as, shown in Fig. 13.16., Alternatively, x 2 + y 2 − 4x + 6y − 3 =0 may be rearranged as:, (x − 2)2 + ( y + 3)2 − 3 − 13 = 0, i.e., , (x − 2)2 + ( y + 3)2 = 42
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The circle and its properties, y, , v=, , i.e., , 4, , s, t, , 129, (1), , The unit of linear velocity is metres per second (m/s)., 2, , Angular velocity, 24, , 22, , 0, 22, 23, 24, , 2, , 4, r5, , 6 x, , 4, , 28, , The speed of revolution of a wheel or a shaft is usually, measured in revolutions per minute or revolutions per, second but these units do not form part of a coherent, system of units. The basis in SI units is the angle turned, through in one second., Angular velocity is defined as the rate of change, of angular displacement θ, with respect to time t ., For an object rotating about a fixed axis at a constant, speed:, , Figure 13.16, , angular velocity =, which represents a circle, centre (2, −3) and radius 4,, as stated above., Now try the following exercise, Exercise 58 Further problems on the, equation of a circle, 1. Determine the radius and the co-ordinates of, the centre of the circle given by the equation, x 2 + y 2 + 6x − 2y − 26 =0., [6, (−3, 1)], 2. Sketch the circle given by the equation, x 2 + y 2 − 6x + 4y − 3 =0., [Centre at (3, −2), radius 4], 3. Sketch the curve x 2 + ( y − 1)2 − 25 =0., [Circle, centre (0, 1), radius 5], , 4. Sketch the curve x = 6 1 − (y/6)2 ., [Circle, centre (0, 0), radius 6], , 13.6, , Linear and angular velocity, , Linear velocity, Linear velocity v is defined as the rate of change of, linear displacement s with respect to time t . For motion, in a straight line:, linear velocity =, , change of displacement, change of time, , i.e., , angle turned through, time taken, ω=, , θ, t, , (2), , The unit of angular velocity is radians per second, (rad/s). An object rotating at a constant speed of, n revolutions per second subtends an angle of 2πn, radians in one second, i.e., its angular velocity ω is, given by:, ω = 2πn rad/s, , (3), , From page 124, s =rθ and from equation (2) above,, θ = ωt, s = r(ωt ), s, = ωr, from which, t, s, However, from equation (1) v =, t, hence, , hence, , v = ωr, , (4), , Equation (4) gives the relationship between linear, velocity v and angular velocity ω., Problem 18. A wheel of diameter 540 mm is, 1500, rotating at, rev/min. Calculate the angular, π, velocity of the wheel and the linear velocity of a, point on the rim of the wheel., From equation (3), angular velocity ω = 2πn where n, is the speed of revolution in rev/s. Since in this case
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130 Higher Engineering Mathematics, n=, , 1500, 1500, rev/min =, = rev/s, then, π, 60π, , , 1500, = 50 rad/s, angular velocityω = 2π, 60π, , The linear velocity of a point on the rim, v = ωr, where, r is the radius of the wheel, i.e., 540, 0.54, mm =, m = 0.27 m., 2, 2, Thus linear velocity v = ωr = (50)(0.27), = 13.5 m/s, Problem 19. A car is travelling at 64.8 km/h and, has wheels of diameter 600 mm., (a) Find the angular velocity of the wheels in both, rad/s and rev/min., (b) If the speed remains constant for 1.44 km,, determine the number of revolutions made by, the wheel, assuming no slipping occurs., (a) Linear velocity v = 64.8 km/h, m, 1 h, km, × 1000, ×, = 18 m/s., = 64.8, h, km 3600 s, The radius of a wheel =, , 600, = 300 mm, 2, = 0.3 m., , From equation (5), v = ωr, from which,, v, 18, =, r, 0.3, = 60 rad/s, , angular velocity ω =, , From equation (4), angular velocity, ω = 2πn,, where n is in rev/s., Hence angular speed n =, , 60, ω, =, rev/s, 2π, 2π, , = 60 ×, , 60, rev/min, 2π, , = 573 rev/min, (b) From equation (1), since v = s/t then the time, taken to travel 1.44 km, i.e. 1440 m at a constant, speed of 18 m/s is given by:, time t =, , s, 1440 m, =, = 80 s, v, 18 m/s, , Since a wheel is rotating at 573 rev/min, then in, 80/60 minutes it makes, 573 rev/min ×, , 80, min = 764 revolutions, 60, , Now try the following exercise, Exercise 59 Further problems on linear, and angular velocity, 1. A pulley driving a belt has a diameter of, 300 mm and is turning at 2700/π revolutions, per minute. Find the angular velocity of the, pulley and the linear velocity of the belt, assuming that no slip occurs., [ω = 90 rad/s, v = 13.5 m/s], 2. A bicycle is travelling at 36 km/h and the diameter of the wheels of the bicycle is 500 mm., Determine the linear velocity of a point on the, rim of one of the wheels of the bicycle, and, the angular velocity of the wheels., [v = 10 m/s, ω = 40 rad/s], 3. A train is travelling at 108 km/h and has wheels, of diameter 800 mm., (a) Determine the angular velocity of the, wheels in both rad/s and rev/min., (b) If the speed remains constant for 2.70 km,, determine the number of revolutions, made by a wheel, assuming no slipping, occurs. , (a) 75 rad/s, 716.2 rev/min, (b) 1074 revs, , 13.7, , Centripetal force, , When an object moves in a circular path at constant, speed, its direction of motion is continually changing, and hence its velocity (which depends on both magnitude and direction) is also continually changing. Since, acceleration is the (change in velocity)/(time taken), the, object has an acceleration. Let the object be moving, with a constant angular velocity of ω and a tangential, velocity of magnitude v and let the change of velocity for a small change of angle of θ (=ωt ) be V in, Fig. 13.17. Then v2 − v1 = V . The vector diagram is, shown in Fig. 13.17(b) and since the magnitudes of v1, and v2 are the same, i.e. v, the vector diagram is an, isosceles triangle.
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The circle and its properties, v2, , Problem 20. A vehicle of mass 750 kg travels, around a bend of radius 150 m, at 50.4 km/h., Determine the centripetal force acting on the, vehicle., v1, , r, , , 2, , mv 2, The centripetal force is given by, and its direction, r, is towards the centre of the circle., , 5 t, 2v1, , r, , 131, , v2, , Mass m = 750 kg, v = 50.4 km/h, v, 2, , 50.4 × 1000, m/s, 60 × 60, = 14 m/s, , =, , V, (a), , (b), , Figure 13.17, , and radius r = 150 m,, , Bisecting the angle between v2 and v1 gives:, , thus centripetal force =, , θ, V, V /2, sin =, =, 2, v2, 2v, θ, 2, , i.e. V = 2v sin, , (1), , 750(14)2, = 980 N., 150, , Problem 21. An object is suspended by a thread, 250 mm long and both object and thread move in a, horizontal circle with a constant angular velocity of, 2.0 rad/s. If the tension in the thread is 12.5 N,, determine the mass of the object., , Since θ = ωt then, t=, , Centripetal force (i.e. tension in thread),, , θ, ω, , (2), , Dividing equation (1) by equation (2) gives:, V, 2v sin(θ/2) vω sin(θ/2), =, =, t, (θ/ω), (θ/2), For small angles, , sin(θ/2), ≈ 1,, (θ/2), , V, change of velocity, hence, =, t, change of time, = acceleration a = vω, However, ω =, thus vω = v ·, , v, (from Section 13.6), r, , v, v2, =, r, r, , v2, i.e. the acceleration a is, and is towards the centre of, r, the circle of motion (along V). It is called the centripetal, acceleration. If the mass of the rotating object is m, then, mv 2, by Newton’s second law, the centripetal force is, r, and its direction is towards the centre of the circle of, motion., , F=, , mv 2, = 12.5 N, r, , Angular velocity ω = 2.0 rad/s and, radius r = 250 mm = 0.25 m., Since linear velocity v = ωr, v = (2.0)(0.25), = 0.5 m/s., mv 2, Fr, , then mass m = 2 ,, r, v, (12.5)(0.25), i.e. mass of object, m =, = 12.5 kg, 0.52, Since F =, , Problem 22. An aircraft is turning at constant, altitude, the turn following the arc of a circle of, radius 1.5 km. If the maximum allowable, acceleration of the aircraft is 2.5 g, determine the, maximum speed of the turn in km/h. Take g as, 9.8 m/s2., The acceleration of an object turning in a circle is, v2, . Thus, to determine the maximum speed of turn,, r, 2, v, = 2.5 g, from which,, r
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132 Higher Engineering Mathematics, , , (2.5 gr) = (2.5)(9.8)(1500), √, = 36750 = 191.7 m/s, 60 × 60, km/h = 690 km/h, and 191.7 m/s= 191.7 ×, 1000, velocity, v =, , Now try the following exercise, Exercise 60 Further problems on, centripetal force, 1. Calculate the tension in a string when it is used, to whirl a stone of mass 200 g round in a horizontal circle of radius 90 cm with a constant, speed of 3 m/s., [2 N], , 2. Calculate the centripetal force acting on a, vehicle of mass 1 tonne when travelling around a bend of radius 125 m at 40 km/h. If this, force should not exceed 750 N, determine the, reduction in speed of the vehicle to meet this, requirement., [988 N, 5.14 km/h], 3. A speed-boat negotiates an S-bend consisting of two circular arcs of radii 100 m and, 150 m. If the speed of the boat is constant at, 34 km/h, determine the change in acceleration, when leaving one arc and entering the other., [1.49 m/s2]
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Revision Test 4, This Revision Test covers the material contained in Chapters 11 to 13. The marks for each question are shown in, brackets at the end of each question., 1. A 2.0 m long ladder is placed against a perpendicular pylon with its foot 52 cm from the pylon., (a) Find how far up the pylon (correct to the nearest mm) the ladder reaches. (b) If the foot of the, ladder is moved 10 cm towards the pylon how far, does the top of the ladder rise?, (7), 2. Evaluate correct to 4 significant figures:, (a) cos 124◦13, , (b) cot 72.68◦, , (4), , 3. From a point on horizontal ground a surveyor, measures the angle of elevation of a church spire, as 15◦. He moves 30 m nearer to the church and, measures the angle of elevation as 20◦. Calculate, the height of the spire., (9), 4. If secant θ = 2.4613 determine the acute, angle θ, (4), 5. Evaluate, correct to 3 significant figures:, 3.5 cosec 31◦ 17 − cot(−12◦ ), 3 sec 79◦ 41, , (5), , 6. A man leaves a point walking at 6.5 km/h in, a direction E 20◦ N (i.e. a bearing of 70◦). A, cyclist leaves the same point at the same time in a, direction E 40◦ S (i.e. a bearing of 130◦ ) travelling, at a constant speed. Find the average speed of the, cyclist if the walker and cyclist are 80 km apart, after 5 hours., (8), 7. A crank mechanism shown in Fig. RT4.1 comprises arm OP, of length 0.90 m, which rotates, anti-clockwise about the fixed point O, and, connecting rod PQ of length 4.20 m. End Q moves, horizontally in a straight line OR., (a) If ∠POR is initially zero, how far does end, Q travel in 14 revolution., , (b) If ∠POR is initially 40◦ find the angle, between the connecting rod and the horizontal and the length OQ., (c) Find the distance Q moves (correct to the, nearest cm) when ∠POR changes from 40◦, to 140◦., (16), 8. Change the following Cartesian co-ordinates into, polar co-ordinates, correct to 2 decimal places, in, both degrees and in radians:, (a) (−2.3, 5.4) (b) (7.6, −9.2), , (10), , 9. Change the following polar co-ordinates into, Cartesian co-ordinates, correct to 3 decimal, (6), places: (a) (6.5, 132◦) (b) (3, 3 rad), 10. (a), , Convert 2.154 radians into degrees and, minutes., (4), (b) Change 71◦17 into radians., , 11. 140 mm of a belt drive is in contact with a pulley of diameter 180 mm which is turning at 300, revolutions per minute. Determine (a) the angle, of lap, (b) the angular velocity of the pulley, and, (c) the linear velocity of the belt assuming that no, slipping occurs., (9), 12. Figure RT4.2 shows a cross-section through a, circular water container where the shaded area, represents the water in the container. Determine:, (a) the depth, h, (b) the area of the shaded portion,, and (c) the area of the unshaded area., (11), , 12 cm, , 608, , 12 cm, h, , P, , Figure RT4.2, O, , Figure RT4.1, , Q, , R, , 13. Determine, (a) the co-ordinates of the centre of, the circle, and (b) the radius, given the equation, x 2 + y 2 − 2x + 6y + 6 = 0, , (7)
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Chapter 14, , Trigonometric waveforms, (a), , 14.1 Graphs of trigonometric, functions, , y, 1.0, , y 5 sin A, , 0.5, , By drawing up tables of values from 0◦ to 360◦, graphs, of y = sin A, y = cos A and y = tan A may be plotted., Values obtained with a calculator (correct to 3 decimal places—which is more than sufficient for plotting, graphs), using 30◦ intervals, are shown below, with the, respective graphs shown in Fig. 14.1., , 0, , (b), , (a) y = sin A, 0, , 30◦, , 210◦, , 60◦, , 90◦, , 120◦, , 150◦ 180◦, , 240◦, , 270◦, , 300◦, , 330◦, , 0, , 210◦, , 60◦ 90◦ 120◦, , 150◦, , 180◦, , 240◦, , 270◦ 300◦, , 330◦, , 30◦, , A, , 21.0, (c), , y, 4, , y 5 tan A, , 0, , 30 60 90 120, , 330, 180 210 240 270 300, , 22, , 360, , A8, , 24, , Figure 14.1, , 0, , 0.500 0.866 1.000, From Fig. 14.1 it is seen that:, , 60◦, , 90◦, , 120◦, , 150◦, , tan A 0 0.577 1.732 ∞ −1.732 −0.577, 210◦, , A8, , 360◦, , (c) y = tan A, 0, , 30 60 90 120 150 180 210 240 270 300 330 360, , 150, , 30◦, , cos A −0.866 −0.500, , A, , y 5 cos A, , 2, , cos A 1.000 0.866 0.500 0 −0.500 −0.866 −1.000, A, , 0, , 360◦, , (b) y = cos A, 0, , A8, , y, , 20.5, , 0, , sin A −0.500 −0.866 −1.000 −0.866 −0.500, , A, , 210 240 270 300 330 360, , 0.5, , sin A 0 0.500 0.866 1.000 0.866 0.500, A, , 180, , 21.0, , 1.0, , A, , 30 60 90 120 150, , 20.5, , 240◦, , tan A 0.577 1.732, , 270◦, ∞, , 300◦, , 330◦, , −1.732 −0.577, , 180◦, 0, 360◦, 0, , (i) Sine and cosine graphs oscillate between peak, values of ±1., (ii) The cosine curve is the same shape as the sine, curve but displaced by 90◦., (iii) The sine and cosine curves are continuous and, they repeat at intervals of 360◦ ; the tangent
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Trigonometric waveforms, curve appears to be discontinuous and repeats at, intervals of 180◦., , 14.2, , Angles of any magnitude, , (i) Figure 14.2 shows rectangular axes XX’ and YY’, intersecting at origin 0. As with graphical work,, measurements made to the right and above 0 are, positive while those to the left and downwards, are negative. Let OA be free to rotate about 0., By convention, when OA moves anticlockwise, angular measurement is considered positive, and, vice-versa., 908, Y, Quadrant 2, , Quadrant 1, 1, , 1, , 1808, , 2, X9, , 0, , A, , 2, , 2, , Quadrant 3, , 1, , 08, , X, , 3608, , Quadrant 4, Y9, 2708, , (iii) Let OA be further rotated so that θ2 is any angle, in the second quadrant and let AC be constructed, to form the right-angled triangle OAC. Then:, sin θ2 =, , +, =+, +, , cos θ2 =, , tan θ2 =, , +, =−, −, , cosec θ2 =, , sec θ2 =, , +, =−, −, , cot θ2 =, , −, =−, +, +, =+, +, , −, =−, +, , (iv) Let OA be further rotated so that θ3 is any angle, in the third quadrant and let AD be constructed, to form the right-angled triangle OAD. Then:, sin θ3 =, , −, = − (and hence cosec θ3 is −), +, , cos θ3 =, , −, = − (and hence sec θ3 is +), +, , tan θ3 =, , −, = + (and hence cot θ3 is −), −, , (v) Let OA be further rotated so that θ4 is any angle, in the fourth quadrant and let AE be constructed, to form the right-angled triangle OAE. Then:, sin θ4 =, , −, = − (and hence cosec θ4 is −), +, , cos θ4 =, , +, = + (and hence sec θ4 is +), +, , tan θ4 =, , −, = − (and hence cot θ4 is −), +, , Figure 14.2, , (ii) Let OA be rotated anticlockwise so that θ1 is any, angle in the first quadrant and let perpendicular, AB be constructed to form the right-angled triangle OAB (see Fig. 14.3). Since all three sides, of the triangle are positive, all six trigonometric, ratios are positive in the first quadrant. (Note: OA, is always positive since it is the radius of a circle.), , (vi) The results obtained in (ii) to (v) are summarized, in Fig. 14.4. The letters underlined spell the word, CAST when starting in the fourth quadrant and, moving in an anticlockwise direction., 908, , 908, Quadrant 2, A, 1, , D, 1808, , Quadrant 1, , 1, , 1, , 2 2, , 1 1, , C, , 3, , 2, , 1, , Sine (and cosecant), positive, , A, , 0, , E B, , 1, , 2, , A, Quadrant 3, , 08, 3608, , 08, 3608, , 1808, Cosine, (and secant), positive, , Tangent, (and cotangent), positive, , A, Quadrant 4, , 2708, , 2708, , Figure 14.3, , All positive, , 1, , 4, , 135, , Figure 14.4
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136 Higher Engineering Mathematics, (vii) In the first quadrant of Fig. 14.1 all the curves, have positive values; in the second only sine is, positive; in the third only tangent is positive;, in the fourth only cosine is positive (exactly as, summarized in Fig. 14.4)., A knowledge of angles of any magnitude is needed, when finding, for example, all the angles between 0◦, and 360◦ whose sine is, say, 0.3261. If 0.3261 is entered, into a calculator and then the inverse sine key pressed, (or sin−1 key) the answer 19.03◦ appears. However, there is a second angle between 0◦ and 360◦ which the, calculator does not give. Sine is also positive in the second quadrant (either from CAST or from Fig. 14.1(a))., The other angle is shown in Fig. 14.5 as angle θ, where θ = 180◦ − 19.03◦ = 160.97◦. Thus 19.03◦ and, 160.97◦ are the angles between 0◦ and 360◦ whose, sine is 0.3261 (check that sin 160.97◦ = 0.3261 on your, calculator)., , S, , 19.038, , T, , 0, 20.4638, , 908 1808, , 2708, , 3328429, 3608 x, , 21.0, (a), 908, S, , 1808, , A, , , , , , T, , 08, 3608, , C, , 2708, (b), , Problem 2. Determine all the angles between 0◦, and 360◦ whose tangent is 1.7629, , A, , , 1808, , 2078389, , Figure 14.6, , 908, , 19.038, , y 5 sin x, , y, 1.0, , 08, 3608, , C, , A tangent is positive in the first and third quadrants (see Fig. 14.7(a)). From Fig. 14.7(b),, θ = tan −1 1.7629 =60◦26 . Measured from 0◦, the two, , 2708, , y 5 tan x, , y, , Figure 14.5, 1.7629, , Be careful! Your calculator only gives you one of these, answers. The second answer needs to be deduced from, a knowledge of angles of any magnitude, as shown in, the following problems., , 0, , 908, 608269, , 1808 2708, 2408269, , (a), , Problem 1. Determine all the angles between 0◦, and 360◦ whose sine is −0.4638, The angles whose sine is −0.4638 occurs in the, third and fourth quadrants since sine is negative in, these quadrants (see Fig. 14.6(a)). From Fig. 14.6(b),, θ = sin−1 0.4638 = 27◦ 38 ., Measured from 0◦, the two angles between 0◦ and, 360◦ whose sine is −0.4638 are 180◦ + 27◦ 38 , i.e., 207◦ 38 and 360◦ − 27◦38 , i.e. 332◦ 22 . (Note that, a calculator generally only gives one answer, i.e., −27.632588◦)., , 3608 x, , 908, A, , S, , , , 1808, , T, , C, , 2708, (b), , Figure 14.7, , 08, 3608, ,
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Trigonometric waveforms, angles between 0◦ and 360◦ whose tangent is 1.7629, are 60◦ 26 and 180◦ + 60◦ 26 , i.e. 240◦ 26 ., Problem 3. Solve sec−1 (−2.1499) =α for angles, of α between 0◦ and 360◦., Secant is negative in the second and third quadrants (i.e. the same as for, From Fig. 14.8,, cosine)., 1, −1, −1, = 62◦17 ., θ = sec 2.1499 =cos, 2.1499, Measured from 0◦, the two angles between 0◦ and 360◦, whose secant is −2.1499 are, α = 180◦ − 62◦ 17 = 117◦43 and, ◦, , ◦, , ◦, , α = 180 + 62 17 = 242 17, , A, , , , , 08, 3608, , T, , and, , α = 180◦ + 37◦ 20 = 217◦20, , Now try the following exercise, Exercise 61 Further problems on, evaluating trigonometric ratios of any, magnitude, 1. Find all the angles between 0◦ and 360◦ whose, sine is −0.7321., [227◦4 and 312◦56 ], , 3. If cotangent x = −0.6312, determine the values of x in the range 0◦ ≤ x≤ 360◦., [122◦16 and 302◦16 ], In Problems 4 to 6 solve the given equations., , 908, , 1808, , α = 37◦20, , 2. Determine the angles between 0◦ and 360◦, whose cosecant is 2.5317., [23◦16 and 156◦44 ], , , , S, , Hence, , 4. cos−1 (−0.5316) =t, 5. sec−1 2.3162 = x, , C, , 6. tan−1 0.8314 = θ, , 2708, , Figure 14.8, , Problem 4. Solve cot −1 1.3111 =α for angles of, α between 0◦ and 360◦., Cotangent is positive in the first and third quadrants (i.e. same as for tangent). From Fig. 14.9,, 1, = 37◦20 ., θ = cot −1 1.3111 = tan−1, 1.3111, 908, S, , C, , 2708, , Figure 14.9, , 08, 3608, , and cos 30◦ =, , T, , [x = 64◦25 and 295◦35 ], [θ = 39◦44 and 219◦44 ], , In Fig. 14.10, let OR be a vector 1 unit long and, free to rotate anticlockwise about O. In one revolution a circle is produced and is shown with, 15◦ sectors. Each radius arm has a vertical and, a horizontal component. For example, at 30◦, the, vertical component is T S and the horizontal component, is OS., From trigonometric ratios,, sin 30◦ =, , , , [t = 122◦ 7 and 237◦53 ], , 14.3 The production of a sine and, cosine wave, , A, , , , 1808, , 137, , TS, TS, =, , i.e. TS = sin 30◦, TO, 1, OS, OS, =, , i.e. OS = cos 30◦, TO, 1, , The vertical component TS may be projected across, to T S , which is the corresponding value of 30◦, on the graph of y against angle x ◦ . If all such, vertical components as TS are projected on to the
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138 Higher Engineering Mathematics, y, 908, 1208, , 1.0, , 608, , T 0.5, , 1508, , 0, , S 3608, 3308, , 2108, 2408, 2708, , Angle x 8, , S9, , R, , 1808, , y 5 sin x, , T9, , 308 608, , 1208, , 2108, , 2708, , 3308, , 20.5, , 3008 21.0, , Figure 14.10, , y, 158 08 R, T, 458, 608, , S, , 3308, 3158, , 1.0, , S9, , y 5 cos x, , 0.5, 2858, , 908, , 0, , 08, 2558, , O9, 308 608, , Angle x 8, 1208, , 1808, , 2408, , 3008, , 3608, , 20.5, , 1208, , 2258, 1508, 1808, , 2108, , 21.0, , Figure 14.11, , graph, then a sine wave is produced as shown in, Fig. 14.10., If all horizontal components such as OS are projected, on to a graph of y against angle x ◦ , then a cosine wave, is produced. It is easier to visualize these projections by, redrawing the circle with the radius arm OR initially in, a vertical position as shown in Fig. 14.11., From Figs. 14.10 and 14.11 it is seen that a cosine, curve is of the same form as the sine curve but is, displaced by 90◦ (or π/2 radians)., , 14.4, , Sine and cosine curves, , Graphs of sine and cosine waveforms, (i) A graph of y = sin A is shown by the broken line, in Fig. 14.12 and is obtained by drawing up a table, of values as in Section 14.1. A similar table may, be produced for y = sin 2 A., , A◦, , 2A, , sin 2 A, , A◦, , 2A, , sin 2 A, , 0, , 225, , 450, , 1.0, , 0, , 0, , 30, , 60, , 0.866, , 240, , 480, , 0.866, , 45, , 90, , 1.0, , 270, , 540, , 0, , 60, , 120, , 0.866, , 300, , 600, , −0.866, , 90, , 180, , 0, , 315, , 630, , −1.0, , 120, , 240, , −0.866, , 330, , 660, , −0.866, , 135, , 270, , −1.0, , 360, , 720, , 150, , 300, , −0.866, , 180, , 360, , 0, , 210, , 420, , 0.866, , 0, , A graph of y = sin 2 A is shown in Fig. 14.12.
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139, , Trigonometric waveforms, y, , y, y 5 sin A, , y 5 sin 2A, , 1.0, , y 5 cos 2A, , y 5 cos A, , 1.0, , 0, 0, , 90°, , 180°, , 270°, , 360°, , 908, , 1808, , 2708, , 3608, , A8, , A°, 21.0, , 21.0, , Figure 14.14, Figure 14.12, , (ii) A graph of y = sin 12 A is shown in Fig. 14.13 using, the following table of values., A◦, , 1, 2A, , (iii) A graph of y = cos A is shown by the broken line, in Fig. 14.14 and is obtained by drawing up a, table of values. A similar table may be produced, for y = cos 2 A with the result as shown., (iv) A graph of y = cos 12 A is shown in Fig. 14.15, which may be produced by drawing up a table, of values, similar to above., , sin 12 A, , 0, , 0, , 30, , 15, , 0.259, , 60, , 30, , 0.500, , 90, , 45, , 0.707, , 120, , 60, , 0.866, , 150, , 75, , 0.966, , 180, , 90, , 1.00, , 210, , 105, , 0.966, , 240, , 120, , 0.866, , 270, , 135, , 0.707, , 300, , 150, , 0.500, , 330, , 165, , 0.259, , Periodic functions and period, , 360, , 180, , 0, , (i) Each of the graphs shown in Figs. 14.12 to 14.15, will repeat themselves as angle A increases and, are thus called periodic functions., (ii) y = sin A and y = cos A repeat themselves every, 360◦ (or 2π radians); thus 360◦ is called the, period of these waveforms. y = sin 2 A and, y = cos 2 A repeat themselves every 180◦ (or, π radians); thus 180◦ is the period of these, waveforms., (iii) In general, if y = sin p A or y = cos p A (where p, is a constant) then the period of the waveform is, 360◦ / p (or 2π/ p rad). Hence if y = sin 3 A then, the period is 360/3, i.e. 120◦, and if y = cos 4 A, then the period is 360/4, i.e. 90◦., , y, , y 5 sin A, , 1.0, , 0, , 21.0, , Figure 14.13, , 90°, , 180°, , 0, , y, , 0, , 908, , 1808, , 2708, , 3608, , A8, , 21.0, , Figure 14.15, , y 5 sin 1 A, 2, , 270°, , y 5 cos 1 A y 5 cos A, 2, , 1.0, , 360°, , A°
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140 Higher Engineering Mathematics, Amplitude, Amplitude is the name given to the maximum or peak, value of a sine wave. Each of the graphs shown in, Figs. 14.12 to 14.15 has an amplitude of +1 (i.e. they, oscillate between +1 and −1). However, if y = 4 sin A,, each of the values in the table is multiplied by 4 and, the maximum value, and thus amplitude, is 4. Similarly, if y = 5 cos 2 A, the amplitude is 5 and the period is, 360◦/2, i.e. 180◦., , Problem 7., x = 360◦., , Sketch y = 4 cos2x from x = 0◦ to, , Amplitude= 4; period= 360◦/2 =180◦ ., A sketch of y = 4 cos2x is shown in Fig. 14.18., y, 4, , y 5 4 cos 2x, , Problem 5. Sketch y = sin 3 A between A = 0◦, and A = 360◦., 0, , Amplitude= 1; period= 360◦/3 =120◦., A sketch of y = sin 3 A is shown in Fig. 14.16., , 908, , 1808, , 2708, , 3608, , x8, , 24, , y, , Figure 14.18, , y 5 sin 3A, , 1.0, , 0, , 908, , 1808, , 2708, , 3608 A8, , 3, Sketch y = 2 sin A over one cycle., 5, , Amplitude= 2; period=, , 21.0, , 360◦ 360◦ × 5, = 600◦., =, 3, 3, 5, , 3, A sketch of y = 2 sin A is shown in Fig. 14.19., 5, , Figure 14.16, , y, 2, , Problem 6. Sketch y = 3 sin 2 A from A = 0 to, A = 2π radians., Amplitude= 3, period= 2π/2 = π rads (or 180◦)., A sketch of y = 3 sin 2 A is shown in Fig. 14.17., , 0, , y 5 2 sin, , 1808, , 3, A, 5, , 3608, , 5408, , 6008, , A8, , 22, , y, y 5 3 sin 2A, , 3, , 0, , Problem 8., , Figure 14.19, , 908, , 23, , Figure 14.17, , 1808, , 2708, , 3608, , A8, , Lagging and leading angles, (i) A sine or cosine curve may not always start at 0◦ ., To show this a periodic function is represented, by y = sin(A ± α) or y = cos(A ± α) where α, is a phase displacement compared with y = sin A, or y = cos A.
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141, , Trigonometric waveforms, (ii) By drawing up a table of values, a graph of, y = sin(A − 60◦ ) may be plotted as shown in, Fig. 14.20. If y = sin A is assumed to start at 0◦, then y = sin(A − 60◦ ) starts 60◦ later (i.e. has a, zero value 60◦ later). Thus y = sin(A − 60◦) is, said to lag y = sin A by 60◦ ., y 5 sin A, , y 5 sin(A 2 608), , 1.0, , 308, , y, 5, , 608, , y, , Amplitude= 5; period = 360◦/1 =360◦., 5 sin(A + 30◦ ) leads 5 sin A by 30◦ (i.e. starts 30◦, earlier)., A sketch of y = 5 sin(A + 30◦ ) is shown in Fig. 14.22., , y 5 5 sin A, y 5 5 sin(A 1 308), , 0, 0, , 908, , 1808, , 2708, , 3608, , A8, , 908, , 1808, , 2708, , 3608, , A8, , 308, 25, , 21.0, , Figure 14.22, , 608, , Figure 14.20, , (iii) By drawing up a table of values, a graph of, y = cos(A + 45◦ ) may be plotted as shown in, Fig. 14.21. If y = cos A is assumed to start at 0◦, then y = cos(A + 45◦ ) starts 45◦ earlier (i.e. has a, zero value 45◦ earlier). Thus y = cos(A + 45◦ ) is, said to lead y = cos A by 45◦ ., y, , Problem 10. Sketch y = 7 sin(2 A − π/3) in the, range 0 ≤ A ≤ 2π., Amplitude= 7; period = 2π/2 =π radians., In general, y = sin(pt − α) lags y = sin pt by α/p,, hence 7 sin(2 A − π/3) lags 7 sin 2 A by (π/3)/2,, i.e. π/6 rad or 30◦ ., A sketch of y = 7 sin(2 A − π/3) is shown in Fig. 14.23., , 458, y 5 cos A, y, , y 5 cos (A 1 458), , y 5 7 sin 2A, y 5 7 sin(2A 2 /3), , /6, 7, , 0, , 908, , 1808, , 2708, , 3608, , A8, 0, , 908, /2, , 21.0, 458, , Figure 14.21, , (iv) Generally, a graph of y = sin(A − α) lags, y = sin A by angle α, and a graph of, y = sin(A + α) leads y = sin A by angle α., (v) A cosine curve is the same shape as a sine curve, but starts 90◦ earlier, i.e. leads by 90◦ . Hence, cos A = sin(A + 90◦ )., Problem 9. Sketch y = 5 sin(A + 30◦ ) from, A = 0◦ to A = 360◦., , 1808, , , 2708, 3/2, , 3608, 2, , A8, , 7, /6, , Figure 14.23, , Problem 11. Sketch y = 2 cos(ωt − 3π/10) over, one cycle., Amplitude= 2; period = 2π/ω rad., 2 cos(ωt − 3π/10) lags 2 cos ωt by 3π/10ω seconds., A sketch of y = 2 cos(ωt − 3π/10) is shown in, Fig. 14.24.
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142 Higher Engineering Mathematics, y, , (ii) A graph of y = cos2 A is shown in Fig. 14.26, obtained by drawing up a table of values, similar, to above., , 3/10 rads, , 2, , y 5 2 cos t, y 5 2 cos(t 23/10), , y, /2, , 0, , /, , 3/2, , 2/, , t, , y 5 cos2 A, , 1.0, 0.5, , 22, , (i) A graph of y = sin2 A is shown in Fig. 14.25 using, the following table of values., A◦, , sin A, , 0, , 2708, , 3608 A8, , y = sin2 A and y = cos2 A are both periodic functions of period 180◦ (or π rad) and both contain, only positive values. Thus a graph of y = sin2 2 A, has a period 180◦ /2, i.e. 90◦ . Similarly, a graph, of y = 4 cos2 3 A has a maximum value of 4 and a, period of 180◦/3, i.e. 60◦., , (iii), , (sin A)2 = sin2 A, , 0, , 0, , 30, , 0.50, , 0.25, , 60, , 0.866, , 0.75, , 90, , 1.0, , 1.0, , 120, , 0.866, , 0.75, , 150, , 0.50, , 0.25, , 180, , 0, , 0, , 210, , −0.50, , 0.25, , 240, , −0.866, , 0.75, , 270, , −1.0, , 1.0, , 300, , −0.866, , 0.75, , 330, , −0.50, , 0.25, , 360, , 0, , Problem 12. Sketch y = 3 sin2 21 A in the range, 0 < A < 360◦., Maximum value = 3; period = 180◦/(1/2) = 360◦., A sketch of 3 sin2 12 A is shown in Fig. 14.27., y, y 5 3 sin2 1 A, 2, , 3, , 0, , 0, , y, , 908, , 1808, , 2708, , 3608, , Figure 14.27, y 5 sin2 A, , 1.0, , Problem 13. Sketch y = 7 cos2 2 A between, A = 0◦ and A = 360◦., , 0.5, , Figure 14.25, , 1808, , Figure 14.26, , Graphs of sin2 A and cos2 A, , 0, , 908, , 0, , Figure 14.24, , 908, , 1808, , 2708, , 3608, , A8, , Maximum value = 7; period = 180◦/2 = 90◦., A sketch of y = 7 cos2 2 A is shown in Fig. 14.28., , A8
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Trigonometric waveforms, y, , 14.5, y 5 7cos2 2A, , 7, , 0, , 908, , 1808, , 2708, , 3608, , A8, , Now try the following exercise, Exercise 62 Further problems on sine and, cosine curves, In Problems 1 to 9 state the amplitude and period, of the waveform and sketch the curve between, 0◦ and 360◦., , 3., 4., , y = 3 cos, , 2., , 5., , [1, 120◦], , y = cos 3A, 5x, y = 2 sin, 2, y = 3 sin 4t, , Sinusoidal form A sin (ωt ± α), , In Fig. 14.29, let OR represent a vector that is free to, rotate anticlockwise about O at a velocity of ω rad/s., A rotating vector is called a phasor. After a time, t seconds OR will have turned through an angle, ωt radians (shown as angle TOR in Fig. 14.29). If ST is, constructed perpendicular to OR, then sinωt = ST/ TO,, i.e. ST = TO sin ωt ., If all such vertical components are projected on to a, graph of y against ωt , a sine wave results of amplitude, OR (as shown in Section 14.3)., If phasor OR makes one revolution (i.e. 2π radians), in T seconds, then the angular velocity,, ω = 2π/ T rad/s, from which, T = 2π/ω seconds., T is known as the periodic time., The number of complete cycles occurring per second, is called the frequency, f, , Figure 14.28, , 1., , [2,, , Frequency =, , 144◦], , [3, 90◦], , θ, 2, 7, 3x, y = sin, 2, 8, , =, , [3, 720◦], , , 1, number of cycles, =, second, T, ω, ω, i.e. f =, Hz, 2π, 2π, , Hence angular velocity, ω = 2πf rad/s, , 7, , 960◦, 2, , 6., , y = 6 sin(t − 45◦), , [6, 360◦], , 7., , y = 4 cos(2θ + 30◦ ), , [4, 180◦], , 8., , y = 2 sin2 2t, , [2, 90◦], , 9., , 3, y = 5 cos2 θ, 2, , [5, 120◦], , Amplitude is the name given to the maximum or peak, value of a sine wave, as explained in Section 14.3. The, amplitude of the sine wave shown in Fig. 14.29 has an, amplitude of 1., A sine or cosine wave may not always start at 0◦., To show this a periodic function is represented by, y = sin (ωt ± α) or y = cos (ωt ± α), where α is a phase, displacement compared with y = sin A or y = cos A., A graph of y = sin (ωt − α) lags y = sin ωt by angle, y, , rads/s, , y ⫽ sin t, , 1.0, , T, t, 0, , S R, , 0, , ⫺1.0, , Figure 14.29, , 143, , t, , 908, , 1808, , 2708, , 3608, , /2, , , , 3/2, , 2, , t
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144 Higher Engineering Mathematics, α, and a graph of y = sin(ωt + α) leads y = sin ωt by, angle α., The angle ωt is measured in radians (i.e., rad, (t s) = ωt radians) hence angle α should also, ω, s, be in radians., The relationship between degrees and radians is:, ◦, , ◦, , 360 = 2π radians or 180 = π radians, 180, Hence 1 rad =, = 57.30◦ and, for example,, π, π, 71◦ = 71 ×, = 1.239 rad., 180, Given a general sinusoidal function, y = A sin(ω t ± α), then, (i), , A = amplitude, , (ii) ω = angular velocity = 2π f rad/s, 2π, = periodic time Tseconds, ω, ω, (iv), = frequency, f hertz, 2π, (v) α = angle of lead or lag (compared with, y = A sin ωt ), , (iii), , Problem 14. An alternating current is given by, i = 30 sin(100πt + 0.27) amperes. Find the, amplitude, periodic time, frequency and phase, angle (in degrees and minutes)., i= 30 sin(100πt + 0.27) A, hence amplitude =30 A, Angular velocity ω = 100π, hence, periodic time, T =, , 2π, 1, 2π, =, =, ω, 100π, 50, , Amplitude= maximum displacement = 2.5 m., Angular velocity, ω = 2π f = 2π(60) = 120π rad/s., Hence displacement = 2.5 sin(120πt + α) m., When t = 0, displacement = 90 cm = 0.90 m., Hence, , 0.90 = 2. sin(0 + α), , i.e., , sin α =, , Hence, , 0.90, = 0.36, 2.5, α = arcsin 0.36 = 21.10◦ = 21◦ 6, = 0.368 rad, , Thus displacement = 2.5 sin(120πt + 0.368) m, Problem 16. The instantaneous value of voltage, in an a.c. circuit at any time t seconds is given by, v = 340 sin(50πt − 0.541) volts. Determine:, (a), , the amplitude, periodic time, frequency and, phase angle (in degrees), , (b) the value of the voltage when t = 0, (c), , the value of the voltage when t = 10 ms, , (d) the time when the voltage first reaches, 200 V, and, (e), , the time when the voltage is a maximum., , Sketch one cycle of the waveform., (a), , Amplitude =340 V, Angular velocity, ω = 50π, Hence periodic time, T =, , = 0.04 s or 40 ms, , = 0.02 s or 20 ms, 1, 1, =, = 50 Hz, T, 0.02, , , 180 ◦, Phase angle, α = 0.27 rad = 0.27 ×, π, , 2π, 1, 2π, =, =, ω, 50π, 25, , Frequency, f =, , = 15.47◦ or 15◦28 leading, i = 30 sin(100πt), Problem 15. An oscillating mechanism has a, maximum displacement of 2.5 m and a frequency of, 60 Hz. At time t = 0 the displacement is 90 cm., Express the displacement in the general form, A sin(ωt ± α)., , Frequency, f =, , 1, 1, =, = 25 Hz, T, 0.04, , , , 180, Phase angle = 0.541rad = 0.541 ×, π, = 31◦ lagging v = 340 sin(50πt ), (b) When t = 0,, v = 340 sin(0 − 0.541) = 340 sin(−31◦), = −175.1 V
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Trigonometric waveforms, (c), , When t = 10 ms, , Now try the following exercise, , , , 10, then v = 340 sin 50π 3 − 0.541, 10, , Exercise 63 Further problems on the, sinusoidal form A sin(ωt ± α), , = 340 sin(1.0298) = 340 sin 59◦, , In Problems 1 to 3 find the amplitude, periodic, time, frequency and phase angle (stating whether, it is leading or lagging A sin ωt ) of the alternating, quantities given., , = 291.4 V, (d) When v = 200 volts, then 200 = 340 sin(50πt − 0.541), , 1. i = 40 sin(50πt + 0.29) mA, , 40, 0.04 s, 25 Hz, 0.29 rad, (or 16◦37 ) leading 40 sin 50 πt, , 200, = sin(50πt − 0.541), 340, Hence (50πt − 0.541) = arcsin, , 200, 340, , 2., , = 36.03◦ or 0.6288 rad, 50πt = 0.6288 + 0.541, , 3. v = 300 sin(200πt − 0.412) V, , 300 V, 0.01 s, 100 Hz, 0.412 rad, (or 23◦ 36 ) lagging 300 sin 200πt, , = 1.1698, Hence when v = 200 V,, 1.1698, = 7.447 ms, time, t =, 50π, (e), , When the voltage is a maximum, v = 340 V., Hence, , 340 = 340 sin(50πt − 0.541), 1 = sin(50πt − 0.541), , 50πt − 0.541 = arcsin 1, = 90◦ or 1.5708 rad, 50πt = 1.5708 + 0.541 = 2.1118, Hence time, t =, , 2.1118, = 13.44 ms, 50π, , A sketch of v = 340 sin(50πt − 0.541) volts is shown in, Fig. 14.30., Voltage V, 340, 291.4, 200, , 0, 2175.1, 2340, , Figure 14.30, , v 5340 sin(50 t 2 0.541), v 5340 sin 50 t, 20, 10, 7.447 13.44, , 30, , 40, , y = 75 sin(40t − 0.54) cm, , 75 cm, 0.157 s, 6.37 Hz, 0.54 rad, (or 30◦ 56 ) lagging75 sin 40t, , t (ms), , 4. A sinusoidal voltage has a maximum value of, 120 V and a frequency of 50 Hz. At time t = 0,, the voltage is (a) zero, and (b) 50 V., Express the instantaneous voltage v in the, form v = A sin(ωt ± α)., , (a) v = 120 sin 100πt volts, (b) v = 120 sin(100πt + 0.43) volts, 5. An alternating current has a periodic time of, 25 ms and a maximum value of 20 A. When, time t = 0, current i = −10 amperes. Express, the current i in the form i = A sin(ωt ± α)., , , , π, amperes, i = 20 sin 80πt −, 6, 6. An oscillating mechanism has a maximum displacement of 3.2 m and a frequency of 50 Hz., At time t = 0 the displacement is 150 cm., Express the displacement in the general form, A sin(ωt ± α)., [3.2 sin(100πt + 0.488) m], 7. The current in an a.c. circuit at any time, t seconds is given by:, i = 5 sin(100πt − 0.432) amperes, Determine (a) the amplitude, periodic time,, frequency and phase angle (in degrees) (b) the, value of current at t = 0 (c) the value of current, at t = 8 ms (d) the time when the current is first, a maximum (e) the time when the current first, , 145
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146 Higher Engineering Mathematics, reaches 3A. Sketch one cycle of the waveform, showing relevant points., ⎡, ⎤, (a) 5 A, 20 ms, 50 Hz,, ⎢ 24◦45 lagging, ⎥, ⎢, ⎥, ⎢ (b) −2.093 A, ⎥, ⎢, ⎥, ⎢ (c) 4.363 A, ⎥, ⎢, ⎥, ⎣ (d) 6.375 ms, ⎦, (e) 3.423 ms, , 14.6 Harmonic synthesis with, complex waveforms, A waveform that is not sinusoidal is called a complex, wave. Harmonic analysis is the process of resolving a, complex periodic waveform into a series of sinusoidal, components of ascending order of frequency. Many of, the waveforms met in practice can be represented by the, following mathematical expression., v = V1m sin(ωt + α1) + V2m sin(2ωt + α2 ), + · · · + Vnm sin(nωt + αn ), and the magnitude of their harmonic components, together with their phase may be calculated using, Fourier series (see Chapters 66 to 69). Numerical, methods are used to analyse waveforms for which, simple mathematical expressions cannot be obtained., A numerical method of harmonic analysis is explained, in the Chapter 70 on page 637. In a laboratory, waveform, analysis may be performed using a waveform analyser, which produces a direct readout of the component waves, present in a complex wave., By adding the instantaneous values of the fundamental and progressive harmonics of a complex wave for, given instants in time, the shape of a complex waveform, can be gradually built up. This graphical procedure is, known as harmonic synthesis (synthesis meaning ‘the, putting together of parts or elements so as to make up a, complex whole’)., Some examples of harmonic synthesis are considered in the following worked problems., Problem 17. Use harmonic synthesis to construct, the complex voltage given by:, v1 = 100 sin ωt + 30 sin 3ωt volts., The waveform is made up of a fundamental wave of, maximum value 100 V and frequency, f = ω/2π hertz, , and a third harmonic component of maximum value, 30 V and frequency = 3ω/2π(=3 f ), the fundamental, and third harmonics being initially in phase with each, other., In Fig. 14.31, the fundamental waveform is shown, by the broken line plotted over one cycle, the periodic, time T being 2π/ω seconds. On the same axis is plotted, 30 sin 3ωt , shown by the dotted line, having a maximum, value of 30 V and for which three cycles are completed, in time T seconds. At zero time, 30 sin 3ωt is in phase, with 100 sinωt ., The fundamental and third harmonic are combined by, adding ordinates at intervals to produce the waveform, for v1 , as shown. For example, at time T/12 seconds,, the fundamental has a value of 50 V and the third harmonic a value of 30 V. Adding gives a value of 80 V for, waveform v1 at time T/12 seconds. Similarly, at time, T/4 seconds, the fundamental has a value of 100 V and, the third harmonic a value of −30 V. After addition,, the resultant waveform v1 is 70 V at T/4. The procedure is continued between t = 0 and t = T to produce, the complex waveform for v1 . The negative half-cycle, of waveform v1 is seen to be identical in shape to the, positive half-cycle., If further odd harmonics of the appropriate amplitude, and phase were added to v1 a good approximation to a, square wave would result., Problem 18., given by:, , Construct the complex voltage, , , π, v2 = 100 sin ωt + 30 sin 3ωt +, volts., 2, , The peak value of the fundamental is 100 volts and the, peak value of the third harmonic is 30 V. However the, π, third harmonic has a phase displacement of radian, 2, π, leading (i.e. leading 30 sin 3ωt by radian). Note that,, 2, since the periodic time of the fundamental is T seconds,, the periodic time of the third harmonic is T/3 seconds,, π, 1, and a phase displacement of radian or cycle of the, 2, 4, third harmonic represents a time interval of (T/3) ÷ 4,, i.e. T/12 seconds., Figure, 14.32 shows graphs of 100 sin ωt and, , π, over the time for one cycle of the fun30 sin 3ωt +, 2, damental. When ordinates of the two graphs are added, at intervals, the resultant waveform v2 is as shown., If the negative half-cycle in Fig. 14.32 is reversed it, can be seen that the shape of the positive and negative, half-cycles are identical.
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Trigonometric waveforms, , 147, , Voltage v (V), 100, v15 100 sin t 1 30 sin 3t, 100 sin t, 50, , 30 sin 3t, , 30, T, 0, , T, 12, , T, 4, , T, 2, , 3T, 4, , Time t (s), , 230, 250, , 2100, , Figure 14.31, , Voltage v (V), , 100, , , v25 100 sin t 1 30 sin (3 t 1 2 ), 100 sin t, , 30 sin (3t 1 2 ), , 50, 30, , T, 4, , 0, , T, 2, , 3T, 4, , T, Time t (s), , 230, 250, , 2100, , Figure 14.32, , Problems 17 and 18 demonstrate that whenever, odd harmonics are added to a fundamental waveform,, whether initially in phase with each other or not, the, positive and negative half-cycles of the resultant complex wave are identical in shape. This is a feature, of waveforms containing the fundamental and odd, harmonics., , Problem 19. Use harmonic synthesis to construct, the complex current given by:, i1 = 10 sin ωt + 4 sin 2ωt amperes., Current i1 consists of a fundamental compon- ent,, 10 sin ωt , and a second harmonic component, 4 sin 2ωt ,
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148 Higher Engineering Mathematics, Current, i (A), i15 10 sin t 1 4 sin 2t, , 10, , 10 sin t, , 4 sin 2t, 4, 3T, 4, , T, 4, 0, , T, 2, , T, Time t (s), , 24, , 210, , Figure 14.33, , the components being initially in phase with each other., The fundamental and second harmonic are shown plotted separately in Fig. 14.33. By adding ordinates at, intervals, the complex waveform representing i1 is produced as shown. It is noted that if all the values in the, negative half-cycle were reversed then this half-cycle, would appear as a mirror image of the positive half-cycle, about a vertical line drawn through time, t = T/2., Problem 20., given by:, , Construct the complex current, , , π, i2 = 10 sin ωt + 4 sin 2ωt +, amperes., 2, , The fundamental component, 10 sin ωt , and the second, harmonic component, having an amplitude of 4 A and, π, a phase displacement of radian leading (i.e. leading, 2, π, 4 sin 2ωt by radian or T/8 seconds), are shown plotted, 2, separately in Fig. 14.34. By adding ordinates at intervals, the complex waveform for i2 is produced as shown., The positive and negative half-cycles of the resultant, waveform are seen to be quite dissimilar., From Problems 18 and 19 it is seen that whenever even harmonics are added to a fundamental, component:, (a), , if the harmonics are initially in phase, the negative, half-cycle, when reversed, is a mirror image of, , the positive half-cycle about a vertical line drawn, through time, t = T/2., (b) if the harmonics are initially out of phase with, each other, the positive and negative half-cycles, are dissimilar., These are features of waveforms containing the fundamental and even harmonics., Problem 21. Use harmonic synthesis to construct, the complex current expression given by:, , π, i = 32 + 50 sin ωt + 20 sin 2ωt −, mA., 2, The current i comprises three components—a 32 mA, d.c. component, a fundamental of amplitude 50 mA, and a second harmonic of amplitude 20 mA, lagπ, ging by radian. The fundamental and second har2, monic are shown separately in Fig. 14.35. Adding, ordinates at intervals, gives, the complex waveform, , π, ., 50 sin ωt + 20 sin 2ωt −, 2, This waveform is then added to the 32 mA d.c., component to produce the waveform i as shown., The effect of the d.c. component is to shift the whole, wave 32 mA upward. The waveform approaches that, expected from a half-wave rectifier.
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Trigonometric waveforms, Current, i (A), 10, , 10 sin t, i25 10 sin t 14 sin(2t 1 , 2), 4 sin(2t 1 , 2), , 4, , T, 0, , T, 4, , T, 2, , 3T, 4, , Time t (s), , 24, , 210, , Figure 14.34, , Current, i (mA), 100, i 32 50 sin t 20 sin(2t , , 50 sin t 20 sin(2t , , , ), 2, , , ), 2, , 50 sin t, , 50, , 32, , 20 sin(2t , , , ), 2, , 20, , T, 0, , 20, , 50, , Figure 14.35, , T, 4, , T, 2, , 3T, 4, , Time t (s), , 149
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150 Higher Engineering Mathematics, Voltage, v (V), , v 5 339.4 sin 100 t 1 67.9 sin(300 t 2, , 3, ), 4, , 339.4, 339.4 sin 100 t, 67.9 sin(300 t 2, 67.9, 267.9, , 15, 5, , 10, , 3, ), 4, 20, Time t (ms), , 2339.4, , Figure 14.36, , Problem 22. A complex waveform v comprises a, fundamental voltage of 240 V rms and frequency, 50 Hz, together with a 20% third harmonic which, has a phase angle lagging by 3π/4 rad at time t = 0., (a) Write down an expression to represent voltage, v. (b) Use harmonic synthesis to sketch the, complex waveform representing voltage v over one, cycle of the fundamental component., (a), , A fundamental voltage having an rms value of, 240, √ V has a maximum value, or amplitude of, 2 (240) i.e. 339.4 V., If the fundamental frequency is 50 Hz then, angular velocity, ω =2π f = 2π(50) = 100π rad/s., Hence the fundamental voltage is represented, by 339.4 sin 100πt volts. Since the fundamental frequency is 50 Hz, the time for one cycle, of the fundamental is given by T = 1/ f = 1/50 s, or 20 ms., The third harmonic has an amplitude equal to, 20% of 339.4 V, i.e. 67.9 V. The frequency of, the third harmonic component is 3 × 50 =150 Hz,, thus the angular velocity is 2π (150), i.e., 300π rad/s. Hence the third harmonic voltage, is represented by 67.9 sin(300πt − 3π/4) volts., Thus, voltage, v = 339.4 sin 100πt, + 67.9 sin (300πt−3π/4) volts, , (b) One cycle of the fundamental, 339.4 sin 100πt ,, is shown sketched in Fig. 14.36, together with, , three cycles of the third harmonic component, 67.9 sin(300πt − 3π/4) initially lagging by, 3π/4 rad. By adding ordinates at intervals,, the complex waveform representing voltage is, produced as shown., , Now try the following exercise, Exercise 64 Further problems on harmonic, synthesis with complex waveforms, 1. A complex current waveform i comprises, a fundamental current of 50 A rms and frequency 100 Hz, together with a 24% third, harmonic, both being in phase with each other, at zero time. (a) Write down an expression, to represent current i. (b) Sketch the complex, waveform of current using harmonic synthesis, over one cycle of the fundamental., (a) i = (70.71 sin 628.3t, + 16.97 sin 1885t ) A, 2. A complex voltage waveform v is comprised, of a 212.1 V rms fundamental voltage at a frequency of 50 Hz, a 30% second harmonic component lagging by π/2 rad, and a 10% fourth, harmonic component leading by π/3 rad., (a) Write down an expression to represent, voltage v. (b) Sketch the complex voltage
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Trigonometric waveforms, , waveform using harmonic synthesis over one, cycle of the fundamental waveform., ⎡, ⎤, (a) v = 300 sin314.2t, ⎢, ⎥, + 90 sin(628.3t − π/2) ⎦, ⎣, + 30sin(1256.6t + π/3) V, 3. A voltage waveform is represented by:, v = 20 + 50 sin ωt, + 20sin(2ωt − π/2) volts., Draw the complex waveform over one cycle of, the fundamental by using harmonic synthesis., 4. Write down an expression representing a, current i having a fundamental component of, amplitude 16 A and frequency 1 kHz, together, with its third and fifth harmonics being respectively one-fifth and one-tenth the amplitude, of the fundamental, all components being, in phase at zero time. Sketch the complex, , current waveform for one cycle of the fundamental using harmonic synthesis., i = 16 sin 2π103 t + 3.2 sin 6π103t, + 1.6 sin π104t A, 5. A voltage waveform is described by, , π, v = 200 sin 377t + 80 sin 1131t +, 4, , π, volts, + 20 sin 1885t −, 3, Determine (a) the fundamental and harmonic, frequencies of the waveform (b) the percentage third harmonic and (c) the percentage, fifth harmonic. Sketch the voltage waveform, using harmonic synthesis over one cycle of the, fundamental. ⎡, ⎤, (a) 60 Hz, 180 Hz, 300 Hz, ⎢ (b) 40%, ⎥, ⎣, ⎦, (c)10%, , 151
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Chapter 15, , Trigonometric identities, and equations, 15.1, , Hence, , Trigonometric identities, , cos2 θ + sin2 θ = 1, , (2), , Dividing each term of equation (1) by a 2 gives:, A trigonometric identity is a relationship that is true, for all values of the unknown variable., tan θ =, , sin θ, cos θ, 1, , cot θ =, , sec θ =, cos θ, sin θ, cos θ, , i.e., , 1, 1, cosec θ =, and cot θ =, sin θ, tan θ, , Hence, , are examples of trigonometric identities from, Chapter 11., Applying Pythagoras’ theorem to the right-angled, triangle shown in Fig. 15.1 gives:, a 2 + b 2 = c2, , (1), , 1 + tan2 θ = sec2 θ, , (3), , Dividing each term of equation (1) by b 2 gives:, , i.e., Hence, , c, , a 2 b2, c2, +, =, a2 a2, a2, 2 , b, c 2, 1+, =, a, a, , a 2 b2, c2, + 2= 2, 2, b, b, b, a 2, c 2, +1 =, b, b, cot2 θ + 1 = cosec2 θ, , (4), , Equations (2), (3) and (4) are three further examples, of trigonometric identities. For the proof of further, trigonometric identities, see Section 15.2., , b, , a, , Figure 15.1, , Dividing each term of equation (1), 2, , i.e., , 2, , gives:, , 15.2 Worked problems on, trigonometric identities, , 2, , a, b, c, + 2 = 2, 2, c, c, c, a 2 b 2, +, =1, c, c, (cos θ)2 + (sin θ)2 = 1, , by c2, , Problem 1. Prove the identity, sin2 θ cot θ sec θ = sin θ., With trigonometric identities it is necessary to start with, the left-hand side (LHS) and attempt to make it equal to
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154 Higher Engineering Mathematics, 3. 2 cos2 A − 1 = cos2 A − sin2 A, 4., , cos x − cos3 x, = sin x cos x, sin x, , 5. (1 + cot θ)2 + (1 − cot θ)2 = 2 cosec 2 θ, 6., , sin2 x(sec x + cosec x), = 1 + tan x, cos x tan x, , 15.3, , Trigonometric equations, , Equations which contain trigonometric ratios are called, trigonometric equations. There are usually an infinite, number of solutions to such equations; however, solutions are often restricted to those between 0◦ and 360◦ ., A knowledge of angles of any magnitude is essential, in the solution of trigonometric equations and calculators cannot be relied upon to give all the solutions (as, shown in Chapter 14). Fig. 15.2 shows a summary for, angles of any magnitude., 908, Sine, (and cosecant, positive), , 08, 3608, Tangent, (and cotangent, positive), , Cosine, (and secant, positive), , 2708, , Figure 15.2, , Equations of the type a sin2 A + b sin A + c = 0, (i) When a = 0, b sin A + c = 0, hence, c, c, sin A = − and A = sin−1 −, b, b, There are two values of A between 0◦ and, 360◦ which satisfy such an equation, provided, c, −1 ≤ ≤ 1 (see Problems 6 to 8)., b, (ii) When b = 0, a sin 2 A +c = 0, hence, c, c, sin2 A = − , sin A =, −, a , a, c, and A = sin−1 −, a, , (iii) When a, b and c are all non-zero:, a sin2 A + b sin A + c = 0 is a quadratic equation, in which the unknown is sin A. The solution of, a quadratic equation is obtained either by factorizing (if possible) or by using the quadratic, formula:, , −b ± (b2 − 4ac), sin A =, 2a, (see Problems 11 and 12)., (iv) Often the trigonometric identities, cos2 A + sin2 A = 1, 1 +tan2 A = sec2 A and, cot 2 A + 1 = cosec 2 A need to be used to reduce, equations to one of the above forms (see, Problems 13 to 15)., , 15.4 Worked problems (i) on, trigonometric equations, , All positive, , 1808, , If either a or c is a negative number, then the, value within the square root sign is positive., Since when a square root is taken there is a positive and negative answer there are four values, of A between 0◦ and 360◦ which satisfy such an, c, equation, provided −1 ≤ ≤ 1 (see Problems 9, a, and 10)., , Problem 6. Solve the trigonometric equation, 5 sin θ + 3 = 0 for values of θ from 0◦ to 360◦., 5 sin θ + 3 = 0, from which sinθ = − 35 = −0.6000, Hence θ = sin−1(−0.6000). Sine is negative in the third, and fourth quadrants (see Fig. 15.3). The acute angle, sin−1 (0.6000) = 36.87◦ (shown as α in Fig. 15.3(b))., Hence,, θ = 180◦ + 36.87◦ , i.e. 216.87◦ or, θ = 360◦ − 36.87◦ , i.e. 323.13◦, Problem 7. Solve 1.5 tan x − 1.8 =0 for, 0◦ ≤ x ≤ 360◦., 1.5 tan x − 1.8 =0, from which, 1.8, tan x =, = 1.2000., 1.5, Hence x = tan−1 1.2000
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Trigonometric identities and equations, y 5 sin , , y, 1.0, , Problem 8. Solve for θ in the range, 0◦ ≤ θ ≤ 360◦ for 2 sin θ = cos θ, , 216.878, 0, 20.6, , 908, , 323.138, , 2708, , 1808, , 3608, , , , 21.0, (a), 908, S, , 1808, , A, , a, , 08, 3608, , a, , T, , 2 sin θ, Dividing both sides by cos θ gives:, =1, cos θ, sin θ, From Section 15.1, tan θ =, ,, cos θ, hence 2 tan θ = 1, Dividing by 2 gives: tan θ = 12, from which, θ = tan−1 12, Since tangent is positive in the first and third quadrants,, θ = 26.57◦ and 206.57◦, Problem 9. Solve 4 sec t = 5 for values of t, between 0◦ and 360◦., 4 sec t = 5, from which sec t = 54 = 1.2500, Hence t = sec−1 1.2500, Secant = (1/cosine) is positive in the first and, fourth quadrants (see Fig. 15.5) The acute angle, sec−1 1.2500 =36.87◦ . Hence,, , C, , 2708, (b), , Figure 15.3, , Tangent is positive in the first and third quadrants (see, Fig. 15.4)., The acute angle tan−1 1.2000 =50.19◦ . Hence,, x = 50.19◦ or 180◦ + 50.19◦ = 230.19◦, , t = 36.87◦ or 360◦ − 36.87◦ = 323.13◦, 908, S, , A, , y 5 tan x, , y, , 36.878 08, , 1808, 1.2, , 36.878 3608, T, , 0, , 908, 50.198, , 1808, , 2708, , 3608 x, , C, 2708, , 230.198, , Figure 15.5, , Now try the following exercise, , (a), 908, S, , 1808, , Exercise 66 Further problems on, trigonometric equations, , A, 50.198, , 50.198, , 08, 3608, , In Problems 1 to 3 solve the equations for angles, between 0◦ and 360◦., [θ = 34.85◦ or 145.15◦], , 1. 4 −7 sin θ = 0, T, , C, 2708, (b), , Figure 15.4, , 155, , 2. 3 cosec A + 5.5 =0, , [A = 213.06◦ or 326.94◦], , 3. 4(2.32 − 5.4 cot t ) = 0, , [t = 66.75◦ or 246.75◦]
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156 Higher Engineering Mathematics, In Problems 4 to 6, solve for θ in the range, 0◦ ≤ θ ≤ 360◦., 4. sec θ = 2, , [60◦ , 300◦], , 5. cot θ = 0.6, , [59◦, 239◦], [41.81◦, 138.19◦], , 6. cosec θ = 1.5, , In Problems 7 to 9, solve for x in the range, −180◦ ≤ x ≤ 180◦ ., , y 5 cos A, , 1.0, 0.7071, , 0, , 1358, 2258, 1808, , 458, , 908, , [−30◦ , −150◦ ], , S, , In Problem 10 and 11, solve for θ in the range, 0◦ ≤ θ ≤ 360◦., 10. 3 sin θ = 2 cosθ, , [33.69◦, 213.69◦], , 11. 5 cos θ = − sin θ, , [101.31◦, 281.31◦], , A8, , (a), , [39.81◦, −140.19◦ ], , 9. cosec x = −2, , 3158 3608, , 20.7071, 21.0, , [±131.81◦ ], , 7. sec x = −1.5, 8. cot x = 1.2, , y, , 1808, , A, , 458, , 458, , 0, , 458, , 458, , 3608, , T, , C, 2708, (b), , Figure 15.6, , 15.5 Worked problems (ii) on, trigonometric equations, Problem 10. Solve 2 −4 cos2 A = 0 for values of, A in the range 0◦ < A < 360◦., 2 − 4 cos2, , A = 0, from which, A = = 0.5000, √, Hence cos A = (0.5000) = ±0.7071 and, A = cos−1(±0.7071)., Cosine is positive in quadrants one and four and negative in quadrants two and three. Thus in this case there, are four solutions, one in each quadrant (see Fig. 15.6)., The acute angle cos−1 0.7071 =45◦ . Hence,, cos2, , 2, 4, , Now try the following exercise, Exercise 67 Further problems on, trigonometric equations, In Problems 1 to 3 solve the equations for angles, between 0◦ and 360◦., 1. 5 sin2 y = 3, y = 50.77◦, 129.23◦,, 230.77◦ or 309.23◦, 2. cos2 θ = 0.25, [θ = 60◦, 120◦, 240◦ or 300◦ ], , A = 45◦, 135◦, 225◦ or 315◦, 3. tan2 x = 3, Problem 11. Solve 12 cot 2 y = 1.3 for, 0◦ < y < 360◦., cot 2 y = 1.3, from which, cot 2 y = 2(1.3) = 2.6, √, Hence cot y = 2.6 = ±1.6125, and y = cot −1, (±1.6125). There are four solutions, one in each, quadrant. The acute angle cot−1 1.6125 =31.81◦ ., , [θ = 60◦, 120◦, 240◦ or 300◦ ], 4. 5 + 3 cosec2 D = 8, [D = 90◦ or 270◦ ], , 1, 2, , Hence y = 31.81◦, 148.19◦, 211.81◦ or 328.19◦ ., , 5. 2 cot 2 θ = 5, θ = 32.32◦, 147.68◦,, 212.32◦ or 327.68◦
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Trigonometric identities and equations, 15.6 Worked problems (iii) on, trigonometric equations, , 3. 2 cosec 2 t − 5 cosec t =12, t = 14.48◦, 165.52◦,, 221.81◦ or 318.19◦, , Problem 12. Solve the equation, 8 sin2 θ + 2 sin θ − 1 = 0,, for all values of θ between 0◦ and 360◦., , 4. 2 cos2 θ + 9 cosθ − 5 = 0, , Factorizing 8 sin2 θ + 2 sin θ − 1 = 0 gives, (4 sin θ − 1) (2 sin θ + 1) = 0., Hence 4 sin θ − 1 = 0, from which, sin θ = 14 = 0.2500,, or 2 sin θ + 1 =0, from which, sin θ = − 12 = −0.5000., (Instead of factorizing, the quadratic formula can, of, course, be used)., θ = sin−1 0.2500 = 14.48◦ or 165.52◦ , since sine, is positive in the first and second quadrants, or, θ = sin−1(−0.5000) = 210◦ or 330◦, since sine is negative in the third and fourth quadrants. Hence, , Problem 14. Solve 5 cos2 t + 3 sin t − 3 =0 for, values of t from 0◦ to 360◦., Since cos2 t + sin2 t = 1, cos2 t = 1 − sin2 t . Substituting, for cos2 t in 5 cos2 t + 3 sin t − 3 = 0 gives:, 5(1 − sin2 t ) + 3 sin t − 3 = 0, 5 − 5 sin2 t + 3 sin t − 3 = 0, −5 sin2 t + 3 sin t + 2 = 0, , Problem 13. Solve 6 cos2 θ + 5 cosθ − 6 = 0 for, values of θ from 0◦ to 360◦., , θ = cos−1 0.6667 = 48.18◦ or 311.82◦, since cosine is positive in the first and fourth quadrants., , 5 sin2 t − 3 sin t − 2 = 0, Factorizing gives (5 sin t + 2)(sin t − 1) = 0. Hence, 5 sin t + 2 = 0, from which, sint = − 25 = −0.4000, or, sin t − 1 =0, from which, sin t = 1., t = sin−1(−0.4000) = 203.58◦ or 336.42◦, since sine, is negative in the third and fourth quadrants, or, t = sin−1 1 = 90◦. Hence t = 90◦ , 203.58◦ or 336.42◦, as shown in Fig. 15.7., y, y 5 sin t, , 1.0, , 203.588, , Now try the following exercise, Exercise 68 Further problems on, trigonometric equations, In Problems 1 to 3 solve the equations for angles, between 0◦ and 360◦., 1. 15 sin2 A + sin A − 2 =, 0, A = 19.47◦, 160.53◦,, 203.58◦ or 336.42◦, 2. 8 tan2 θ + 2 tan θ = 15 , , θ = 51.34◦, 123.69◦,, 231.34◦ or 303.69◦, , [θ = 60◦ or 300◦], , 15.7 Worked problems (iv) on, trigonometric equations, , θ = 14.48◦ , 165.52◦ , 210◦ or 330◦, , Factorizing 6 cos2 θ + 5 cos θ − 6 =0 gives, (3 cos θ − 2) (2 cos θ + 3) = 0., Hence 3 cosθ − 2 = 0, from which, cos θ = 23 = 0.6667,, or 2 cos θ + 3 =0, from which, cos θ = − 32 = −1.5000., The minimum value of a cosine is −1, hence the latter expression has no solution and is thus neglected., Hence,, , 157, , 0, 20.4, , 908, , 336.428, , 2708, , 3608 t 8, , 21.0, , Figure 15.7, , Problem 15. Solve 18 sec2 A − 3 tan A = 21 for, values of A between 0◦ and 360◦., 1 + tan2 A = sec2 A. Substituting, 18 sec2 A − 3 tan A = 21 gives, 18(1 + tan2 A) − 3 tan A = 21,, , for, , sec2 A, , in
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158 Higher Engineering Mathematics, i.e. 18 + 18 tan2 A − 3 tan A − 21 = 0, , Hence,, θ = 30.17◦ , 111.18◦, 210.17◦ or 291.18◦, , 18 tan2 A − 3 tan A − 3 = 0, Factorizing gives (6 tan A − 3)(3 tan A + 1) = 0., Hence 6 tan A − 3 =0, from which, tan A = 36 = 0.5000, or 3 tan A + 1 =0, from which, tan A = − 13 = − 0.3333., Thus A = tan−1 (0.5000) = 26.57◦ or 206.57◦, since, tangent is positive in the first and third quadrants, or, A = tan−1 (−0.3333) =161.57◦ or 341.57◦, since tangent is negative in the second and fourth quadrants., Hence,, A = 26.57◦ , 161.57◦ , 206.57◦ or 341.57◦, Problem 16. Solve, range 0 <θ < 360◦., cot 2 θ + 1 =, , 3 cosec 2 θ − 5 = 4 cot θ, , cosec 2 θ., , Substituting for, 3 cosec 2 θ − 5 =4 cot θ gives:, , Now try the following exercise, Exercise 69 Further problems on, trigonometric equations, In Problems 1 to 12 solve the equations for angles, between 0◦ and 360◦., 1., , 2 cos2 θ + sin θ = 1, , 2., , 4 cos2 t + 5 sin t = 3, , 3., , 2 cosθ − 4 sin2 θ = 0, , 4., , 3 cosθ + 2 sin2 θ = 3, [θ = 0◦ , 60◦, 300◦, 360◦], , 5., , 12 sin2 θ − 6 = cos θ , , 6., , 16 sec x − 2 = 14 tan2 x, [x = 52.53◦ or 307.07◦ ], , 7., , 4 cot2 A − 6 cosec A + 6 = 0, , cosec 2 θ, , in, , 8., , 5 sec t + 2 tan2 t = 3, , 9., , 2.9 cos2 a − 7 sin a + 1 =0, [a = 27.83◦ or 152.17◦ ], , 3 cot 2 θ + 3 − 5 = 4 cot θ, 3 cot 2 θ − 4 cot θ − 2 = 0, Since the left-hand side does not factorize the quadratic, formula is used. Thus,, , , [(−4)2 − 4(3)(−2)], 2(3), √, √, 4 ± (16 + 24) 4 ± 40, =, =, 6, 6, −(−4) ±, , [t = 190.1◦ , 349.9◦], , in the, , 3 (cot2 θ + 1) − 5 = 4 cot θ, , cot θ =, , [θ = 90◦ , 210◦, 330◦], , 10.3246, 2.3246, =, or −, 6, 6, Hence cot θ = 1.7208 or −0.3874, θ = cot −1, 1.7208 =30.17◦ or 210.17◦, since cotangent, is positive in the first and third quadrants, or, θ = cot −1(−0.3874) = 111.18◦ or 291.18◦, since, cotangent is negative in the second and fourth quadrants., , 10., , [θ = 38.67◦, 321.33◦], , θ = 48.19◦ , 138.59◦,, 221.41◦ or 311.81◦, , [ A = 90◦ ], , [t = 107.83◦ or 252.17◦ ], , 3 cosec2 β = 8 −7 cot β, β = 60.17◦, 161.02◦,, 240.17◦ or 341.02◦, , 11., , cot θ = sin θ, , 12., , tan θ + 3 cot θ = 5 sec θ, , [51.83◦, 308.17◦], [30◦ , 150◦]
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Chapter 17, , Compound angles, 17.1, , Compound angle formulae, , An electric current i may be expressed as i =, 5 sin(ωt − 0.33) amperes. Similarly, the displacement, x of a body from a fixed point can be expressed as, x = 10 sin(2t + 0.67) metres. The angles (ωt − 0.33) and, (2t + 0.67) are called compound angles because they, are the sum or difference of two angles. The compound, angle formulae for sines and cosines of the sum and, difference of two angles A and B are:, , (a), , sin(π + α) = sin π cos α + cos π sin α (from, the formula forsin(A + B)), = (0)(cos α) + (−1) sin α = −sin α, , (b) −cos(90◦ + β), = −[cos 90◦ cos β − sin 90◦ sin β], = −[(0)(cos β) − (1) sin β] = sin β, (c), , sin(A − B) − sin(A + B), = [sin A cos B − cos A sin B], , sin(A + B) = sin A cos B + cos A sin B, sin(A − B) = sin A cos B − cos A sin B, cos(A + B) = cos A cos B − sin A sin B, , − [sin A cos B + cos A sin B], = −2cos A sin B, , cos(A − B) = cos A cos B + sin A sin B, (Note, sin(A + B) is not equal to (sin A + sin B), and, so on.), The formulae stated above may be used to derive two, further compound angle formulae:, tan(A + B) =, tan(A − B) =, , tan A + tan B, 1 − tan A tan B, tan A − tan B, 1 + tan A tan B, , The compound-angle formulae are true for all values of, A and B, and by substituting values of A and B into the, formulae they may be shown to be true., Problem 1. Expand and simplify the following, expressions:, (a) sin(π + α) (b) −cos(90◦ + β), (c) sin(A − B) − sin(A + B), , Problem 2. Prove that, , π, = 0., cos(y − π) + sin y +, 2, cos(y − π) = cos y cos π + sin y sin π, = (cos y)(−1) + (sin y)(0), = −cos y, π, π, π, sin y +, = sin y cos + cos y sin, 2, 2, 2, = (sin y)(0) + (cos y)(1) = cos y, , π, Hence, cos(y − π) + sin y +, 2, , , = (−cos y) + (cos y) = 0, Problem 3. Show that, , π , π, tan x +, tan x −, = −1., 4, 4
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164 Higher Engineering Mathematics, , tan x + tan π4, π, tan x +, =, 4, 1 − tan x tan π4, from the formula fortan(A + B), , , 1 + tan x, tan x + 1, =, =, 1 − (tan x)(1), 1 − tan x, π, since tan = 1, 4, , , , tan x − tan π4, tan x − 1, π, tan x −, =, =, 4, 1 + tan x tan π4, 1 + tan x, , , , , π, π, Hence tan x +, tan x −, 4, 4, , , , 1 + tan x, tan x − 1, =, 1 − tan x, 1 + tan x, =, , tan x − 1 −(1 − tan x), =, = −1, 1 − tan x, 1 − tan x, , Problem 4. If sin P = 0.8142 and cos Q = 0.4432, evaluate, correct to 3 decimal places:, (a) sin(P − Q), (b) cos(P + Q) and, (c) tan(P + Q), using the compound-angle, formulae., Since sin P = 0.8142 then, P = sin−1 0.8142 =54.51◦ ., Thus cos P = cos 54.51◦ = 0.5806 and, tan P = tan 54.51◦ = 1.4025, Since cos Q = 0.4432, Q = cos−1 0.4432 =63.69◦ ., Thus sin Q = sin 63.69◦ = 0.8964 and, tan Q = tan 63.69◦ = 2.0225, (a) sin(P − Q), = sin P cos Q − cos P sin Q, = (0.8142)(0.4432) − (0.5806)(0.8964), = 0.3609 − 0.5204 = −0.160, (b) cos(P + Q), = cos P cos Q − sin P sin Q, = (0.5806)(0.4432) − (0.8142)(0.8964), = 0.2573 − 0.7298 = −0.473, (c) tan(P + Q), tan P + tan Q, (1.4025) + (2.0225), =, 1 − tan P tan Q, 1 − (1.4025)(2.0225), 3.4250, = −1.865, =, −1.8366, =, , Problem 5., , Solve the equation, , 4 sin(x − 20◦ ) = 5 cos x, for values of x between 0◦ and 90◦ ., 4 sin(x − 20◦ ) = 4[sin x cos 20◦ − cos x sin 20◦ ],, from the formula forsin(A − B), = 4[sin x(0.9397) − cos x(0.3420)], = 3.7588 sin x − 1.3680 cos x, Since 4 sin(x − 20◦ ) = 5 cos x then, 3.7588 sin x − 1.3680 cos x = 5 cos x, Rearranging gives:, 3.7588 sin x = 5 cos x + 1.3680 cos x, , and, , = 6.3680 cos x, sin x, 6.3680, =, = 1.6942, cos x, 3.7588, , i.e. tan x = 1.6942, and x = tan−1 1.6942 =59.449◦ or, 59◦ 27, [Check: LHS = 4 sin(59.449◦ − 20◦ ), = 4 sin 39.449◦ = 2.542, RHS = 5 cos x = 5 cos59.449◦ = 2.542], Now try the following exercise, Exercise 72 Further problems on, compound angle formulae, 1. Reduce the following to the sine of one, angle:, (a) sin 37◦ cos 21◦ + cos 37◦ sin 21◦, (b) sin 7t cos 3t − cos 7t sin 3t, [(a) sin 58◦ (b) sin 4t ], 2. Reduce the following to the cosine of one, angle:, (a) cos 71◦ cos 33◦ − sin 71◦ sin 33◦, π, π, π, π, (b) cos cos + sin sin, 3, 4, 3, 4, ⎡, ⎤, (a) cos 104◦ ≡ −cos 76◦, ⎣, ⎦, π, (b) cos, 12
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Compound angles, , 3. Show that:, , , , √, π, 2π, = 3 cos x, (a) sin x +, + sin x +, 3, 3, and , , 3π, − φ = cos φ, (b) − sin, 2, , same frequency (which is further demonstrated in, Chapter 25)., (iv) Since a = R cos α, then cos α = a/R, and since, b = R sin α, then sin α = b/R., , 4. Prove , that: , , , 3π, π, − sin θ −, (a) sin θ +, 4, 4, √, = 2(sin θ + cos θ), (b), , R, , cos(270◦ + θ), = tan θ, cos(360◦ − θ), , a, , In Problems 6 and 7, solve the equations for, values of θ between 0◦ and 360◦., , 7. 4 sin(θ − 40◦ ) = 2 sin θ, , b, , ␣, , 5. Given cos A = 0.42 and sin B = 0.73 evaluate, (a) sin(A − B), (b) cos(A − B), (c) tan(A+ B),, correct to 4 decimal places., [(a) 0.3136 (b) 0.9495 (c) −2.4687], , 6. 3 sin(θ + 30◦ ) = 7 cosθ, , 165, , [64.72◦ or 244.72◦], [67.52◦ or 247.52◦], , Figure 17.1, , If the values of a and b are known then the values, of R and α may be calculated. The relationship between, constants a, b, R and α are shown in Fig. 17.1., From Fig. 17.1, by Pythagoras’ theorem:, R = a 2 + b2, and from trigonometric ratios:, , 17.2 Conversion of a sin ωt + b cos ωt, into R sin(ωt + α), (i), , R sin(ωt + α) represents a sine wave of maximum value R, periodic time 2π/ω, frequency, ω/2π and leading R sin ωt by angle α. (See, Chapter 14)., , (ii), , R sin(ωt + α) may be expanded using the, compound-angle formula for sin(A + B), where, A = ωt and B = α. Hence,, R sin(ωt + α), = R[sin ωt cos α + cos ωt sin α], = R sin ωt cos α + R cos ωt sin α, = (R cos α) sin ωt + (R sin α) cos ωt, , (iii) If a = R cos α and b = R sin α, where a and, b are constants, then R sin(ωt + α) =a sin ωt +, b cos ωt , i.e. a sine and cosine function of the same, frequency when added produce a sine wave of the, , α = tan−1 b/a, Problem 6. Find an expression for 3 sin ωt + 4, cos ωt in the form R sin(ωt + α) and sketch graphs, of 3 sin ωt , 4 cosωt and R sin(ωt + α) on the, same axes., Let 3 sin ωt + 4 cosωt = R sin(ωt + α), then 3 sin ωt + 4 cosωt, = R[sin ωt cos α + cos ωt sin α], = (R cos α) sin ωt + (R sin α) cosωt, Equating coefficients of sin ωt gives:, 3 = R cos α, from which, cosα =, , 3, R, , Equating coefficients of cos ωt gives:, 4 = R sin α, from which, sin α =, , 4, R
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166 Higher Engineering Mathematics, There is only one quadrant where both sin α and cos α, are positive, and this is the first, as shown in Fig. 17.2., From Fig. 17.2, by Pythagoras’ theorem:, , R = (32 + 42 ) = 5, , Problem 7. Express 4.6 sin ωt − 7.3 cosωt in the, form R sin(ωt + α)., Let 4.6 sin ωt − 7.3 cos ωt = R sin(ωt + α)., then 4.6 sin ωt − 7.3 cos ωt, = R [sin ωt cos α + cos ωt sin α], = (R cos α) sin ωt + (R sin α) cos ωt, , R, , 4, , Equating coefficients of sin ωt gives:, , ␣, , 4.6 = R cos α, from which, cos α =, , 3, , Equating coefficients of cos ωt gives:, , Figure 17.2, , −7.3 = R sin α, from which, sin α =, , From trigonometric ratios: α = tan −1 43 = 53.13◦ or, 0.927 radians., Hence 3 sin ω t + 4 cos ωt = 5 sin(ω t + 0.927)., , By trigonometric ratios:, , , −1 −7.3, α = tan, 4.6, , Two periodic functions of the same frequency may be, combined by,, (a) plotting the functions graphically and combining, ordinates at intervals, or, , = −57.78◦ or −1.008 radians., , (b) by resolution of phasors by drawing or calculation., Hence, , Problem 6, together with Problems 7 and 8 following,, demonstrate a third method of combining waveforms., , 4.6 sin ω t − 7.3 cos ωt = 8.628 sin(ω t − 1.008)., , y, 0.927 rad, 5, , y 5 4 cos t, , 4, , y 5 3 sin t, , 3, , y 5 5 sin(t 1 0.927), , 2, 1, , 0.927 rad, , 0, 21, 22, 23, 24, 25, , /2, , −7.3, R, , There is only one quadrant where cosine is positive and, sine is negative, i.e. the fourth quadrant, as shown in, Fig. 17.4. By Pythagoras’ theorem:, , R = [(4.6)2 + (−7.3)2 ] = 8.628, , A sketch of 3 sin ωt , 4 cos ωt and 5 sin(ωt + 0.927) is, shown in Fig. 17.3., , Figure 17.3, , 4.6, R, , , , 3/2, , 2, , t (rad)
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Compound angles, , Hence α = 180◦ + 56.63◦ = 236.63◦ or 4.130 radians., Thus,, , 4.6, ␣, , −2.7 sin ω t − 4.1 cos ωt = 4.909 sin(ω t + 4.130)., , R, , An angle of 236.63◦ is the same as −123.37◦ or −2.153, radians., Hence −2.7 sin ωt − 4.1 cos ωt may be expressed also, as 4.909 sin(ω t − 2.153), which is preferred since it is, the principal value (i.e. −π ≤ α ≤ π)., , 27.3, , Figure 17.4, , Problem 8. Express −2.7 sin ωt − 4.1 cosωt in, the form R sin(ωt + α)., , Problem 9. Express 3 sin θ + 5 cos θ in the form, R sin(θ + α), and hence solve the equation, 3 sin θ + 5 cosθ = 4, for values of θ between 0◦ and, 360◦., Let 3 sin θ + 5 cos θ = R sin(θ + α), , Let −2.7 sin ωt − 4.1 cos ωt = R sin(ωt + α), = R[sin ωt cos α + cos ωt sin α], , = R[sin θ cos α + cos θ sin α], , = (R cos α)sin ωt + (R sin α)cos ωt, , = (R cos α)sin θ + (R sin α)cos θ, Equating coefficients gives:, , Equating coefficients gives:, −2.7, R, −4.1, −4.1 = R sin α, from which, sin α =, R, , −2.7 = R cos α, from which, cos α =, and, , 167, , There is only one quadrant in which both cosine and, sine are negative, i.e. the third quadrant, as shown in, Fig. 17.5. From Fig. 17.5,, , R = [(−2.7)2 + (−4.1)2 ] = 4.909, 4.1, and θ = tan −1, = 56.63◦, 2.7, , 3 = R cos α, from which, cos α =, , 3, R, , and 5 = R sin α, from which, sin α =, , 5, R, , Since both sin α and cos α are positive, R lies in the first, quadrant, as shown in Fig. 17.6., , R, , 5, , 908, ␣, 3, , ␣, , 22.7, , 1808, , 08, 3608, , u, 24.1, , Figure 17.6, , , From Fig. 17.6, R = (32 + 52) = 5.831 and, α = tan−1 53 = 59.03◦., Hence 3 sin θ + 5 cosθ = 5.831 sin(θ + 59.03◦), , R, , However, 2708, , Figure 17.5, , 3 sin θ + 5 cos θ = 4, , Thus 5.831 sin(θ + 59.03◦) = 4, from which, , , 4, ◦, −1, (θ + 59.03 ) = sin, 5.831
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168 Higher Engineering Mathematics, θ + 59.03◦ = 43.32◦ or 136.68◦, , i.e., , 6.5, , from which,, 6.774, 6.5, (A + 148.88◦ ) = sin−1, 6.774, = 73.65◦ or 106.35◦, , Hence sin(A + 148.88◦ ) =, , Hence θ = 43.32◦ − 59.03◦ = −15.71◦, θ = 136.68◦ − 59.03◦ = 77.65◦, , or, , Since −15.71◦ is the same as −15.71◦ + 360◦ , i.e., 344.29◦, then the solutions are θ = 77.65◦ or 344.29◦ ,, which may be checked by substituting into the original, equation., , Thus A = 73.65◦ − 148.88◦ = −75.23◦, ≡ (−75.23◦ + 360◦) = 284.77◦, A = 106.35◦ − 148.88◦ = −42.53◦, , or, Problem 10. Solve the equation, 3.5 cos A − 5.8 sin A = 6.5 for 0◦ ≤ A ≤ 360◦., , ≡ (−42.53◦ + 360◦ ) = 317.47◦, The solutions are thus A = 284.77◦ or 317.47◦ , which, may be checked in the original equation., , Let 3.5 cos A − 5.8 sin A = R sin(A + α), = R[sin A cos α + cos A sin α], = (R cos α) sin A + (R sin α) cos A, , Now try the following exercise, , Equating coefficients gives:, 3.5, 3.5 = R sin α, from which, sin α =, R, −5.8, and −5.8 = R cos α, from which, cosα =, R, There is only one quadrant in which both sine is positive and cosine is negative, i.e. the second, as shown in, Fig. 17.7., 908, , Exercise 73 Further problems on the, conversion of a sin ω t + b cos ω t into, R sin(ω t + α), In Problems 1 to 4, change the functions into the, form R sin(ωt ± α)., 1., , 5 sin ωt + 8 cosωt, , [9.434 sin(ωt + 1.012)], , 2., , 4 sin ωt − 3 cosωt, , [5 sin(ωt − 0.644)], , 3., , −7 sin ωt + 4 cos ωt, , 4., , 3.5, 1808, , R, , , Solve the following equations for values of θ, between 0◦ and 360◦ : (a) 2 sin θ + 4 cos θ = 3, (b) 12 sin θ − 9 cosθ = 7., , (a) 74.44◦ or 338.70◦, (b) 64.69◦ or 189.05◦, , 6., , Solve the following equations for, 0◦ < A < 360◦ : (a) 3 cos A + 2 sin A = 2.8, (b) 12 cos A − 4 sin A = 11., , (a) 72.73◦ or 354.63◦, (b) 11.15◦ or 311.98◦, , 7., , Solve the following equations for values of θ, between 0◦ and 360◦ : (a) 3 sin θ + 4 cosθ = 3, (b) 2 cosθ + sin θ = 2., [(a) 90◦ or 343.74◦ (b) 0◦ , 53.14◦ ], , 08, 3608, , 2708, , Figure 17.7, , , From Fig. 17.7, R = [(3.5)2 + (−5.8)2 ]= 6.774 and, 3.5, = 31.12◦ ., θ = tan−1, 5.8, Hence α = 180◦ − 31.12◦ = 148.88◦., Thus, 3.5 cos A − 5.8 sin A = 6.774 sin(A + 144.88◦ ) = 6.5, , [6.708 sin(ωt − 2.034)], , 5., ␣, , 25.8, , −3 sin ωt − 6 cos ωt, , [8.062 sin(ωt + 2.622)]
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Compound angles, , 169, , Also, for example,, 8., , Solve the following equations for values of, θ between 0◦ and 360◦: (a) 6 cos θ + sin θ =, √, 3 (b) 2 sin 3θ + 8 cos 3θ = 1., ⎡, ⎤, (a) 82.9◦ , 296◦, ⎣ (b) 32.36◦, 97◦, 152.36◦, 217◦, ⎦, 272.36◦ and 337◦, , 9., , cos 4 A = cos2 2 A − sin 2 2 A or, 1 − 2 sin2 2 A or, 2 cos2 2 A − 1, and cos 6 A = cos2 3 A − sin 2 3 A or, , The third harmonic of a wave motion is given, by 4.3 cos 3θ − 6.9 sin 3θ. Express this in the, form R sin(3θ ± α). [8.13 sin(3θ + 2.584)], , 10. The displacement x metres of a mass from, a fixed point about which it is oscillating is, given by x = 2.4 sin ωt + 3.2 cosωt , where t, is the time in seconds. Express x in the form, R sin(ωt + α). [x = 4.0 sin(ωt + 0.927)m], 11. Two voltages, v1 = 5 cos ωt and, v2 = −8 sin ωt are inputs to an analogue circuit. Determine an expression for the output, voltage if this is given by (v1 + v2 )., [9.434 sin(ωt + 2.583)], , 1 − 2 sin2 3 A or, 2 cos2 3 A − 1,, and so on., (iii) If, in the compound-angle, tan(A + B), we let B = A then, tan 2A =, , 2 tan A, 1 − tan2 A, , Also, for example,, tan 4 A =, and tan 5 A =, , 17.3, , Double angles, , (i) If, in the compound-angle, sin(A + B), we let B = A then, , formula, , for, , formula, , 2 tan 2 A, 1 − tan2 2 A, 2 tan 52 A, 1 − tan2 52 A, , and so on., , Problem 11. I3 sin 3θ is the third harmonic of a, waveform. Express the third harmonic in terms of, the first harmonic sin θ, when I3 = 1., , sin 2A = 2 sin A cos A, When I3 = 1,, , Also, for example,, , I3 sin 3θ = sin 3θ = sin(2θ + θ), , sin 4 A = 2 sin 2 A cos 2 A, , = sin 2θ cos θ + cos 2θ sin θ,, , and sin 8 A = 2 sin 4 A cos 4 A, and so on., (ii) If, in the compound-angle, cos(A + B), we let B = A then, , formula, , for, , = (2 sin θ cos θ) cos θ + (1 − 2 sin2 θ) sin θ,, from the double angle expansions, , cos 2A = cos2 A − sin2 A, Since cos2 A + sin2 A = 1, then, cos2 A = 1 − sin2 A, and sin2 A = 1 − cos2 A, and, two further formula for cos 2 A can be produced., Thus, i.e., and, i.e., , = 2 sin θ cos2 θ + sin θ − 2 sin3 θ, = 2 sin θ(1 − sin2 θ) + sin θ − 2 sin3 θ,, (since cos2 θ = 1 − sin2 θ), = 2 sin θ − 2 sin3 θ + sin θ − 2 sin3 θ, , cos 2 A = cos2 A − sin 2 A, 2, , from the sin(A + B) formula, , 2, , = (1 − sin A) − sin A, cos 2A = 1 − 2 sin2A, cos 2 A = cos2 A − sin 2 A, = cos2 A − (1 − cos 2 A), cos 2 A = 2cos2 A − 1, , i.e. sin 3θ = 3 sinθ − 4 sin3θ, , Problem 12. Prove that, , 1 − cos 2θ, = tan θ., sin 2θ, , for
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170 Higher Engineering Mathematics, 1 − cos 2θ, 1 − (1 − 2 sin2 θ), =, sin 2θ, 2 sin θ cos θ, 2, 2 sin θ, sin θ, =, =, 2 sin θ cos θ, cos θ, = tan θ = RHS, , LHS =, , Problem 13. Prove that, cot 2x + cosec 2x = cot x., LHS = cot 2x + cosec 2x =, , cos 2x, 1, +, sin 2x, sin 2x, , cos 2x + 1, sin 2x, (2 cos2 x − 1) + 1, =, sin 2x, 2, 2 cos x, 2 cos2 x, =, =, sin 2x, 2 sin x cos x, cos x, =, = cot x = RHS, sin x, =, , Problem 14. Solve the equation, cos 2θ + 3 sin θ = 2 for θ in the range 0◦ ≤ θ ≤ 360◦ ., Replacing the double angle term with the relationship, cos 2θ = 1 − 2 sin2 θ gives:, 1 − 2 sin θ + 3 sin θ = 2, , 2. Prove the following identities:, cos 2φ, (a) 1 −, = tan 2 φ, cos2 φ, (b), , 1 + cos 2t, = 2 cot2 t, sin2 t, , (tan 2x)(1 + tan x), 2, =, tan x, 1 − tan x, (d) 2 cosec 2θ cos 2θ = cot θ − tan θ, (c), , 3. If the third harmonic of a waveform is given by, V3 cos 3θ, express the third harmonic in terms, of the first harmonic cosθ, when V3 = 1., [cos 3θ = 4 cos3 θ − 3 cosθ], In Problems 4 to 8, solve for θ in the range, −180◦ ≤ θ ≤ 180◦, [−90◦ , 30◦, 150◦], , 4. cos 2θ = sin θ, , 5. 3 sin 2θ + 2 cosθ = 0, [−160.47◦, −90◦, −19.47◦ , 90◦], 6. sin 2θ + cos θ = 0, , [−150◦ , −90◦, −30◦ , 90◦], [−90◦ ], , 7. cos 2θ + 2 sin θ = −3, , 2, , −2 sin2 θ, , Rearranging gives:, + 3 sin θ − 1 = 0, or, 2 sin2 θ − 3 sin θ + 1 = 0, which is a quadratic in sin θ, Using the quadratic formula or by factorising gives:, (2 sin θ − 1)(sin θ − 1) = 0, from which, 2 sin θ − 1 = 0 or sin θ − 1 = 0, and, sin θ = 12 or sin θ = 1, from which,, θ = 30◦ or 150◦ or 90◦, Now try the following exercise, Exercise 74, angles, , Further problems on double, , 1. The power p in an electrical circuit is given by, v2, p = . Determine the power in terms of V ,, R, R and cos 2t when v = V cos t ., 2, V, (1 + cos 2t ), 2R, , 8. tan θ + cot θ = 2, , [45◦ , −135◦], , 17.4 Changing products of sines and, cosines into sums or differences, (i) sin(A + B) + sin(A − B) = 2 sin A cos B (from the, formulae in Section 17.1), i.e. sin A cos B, = 21 [sin(A + B) + sin(A − B)], , (1), , (ii) sin(A + B) − sin(A − B) = 2 cos A sin B, i.e. cos A sin B, = 21 [sin(A + B) − sin(A − B)], , (2), , (iii) cos(A + B) + cos(A − B) = 2 cos A cos B, i.e. cos A cos B, = 12 [cos(A + B) + cos(A − B)], , (3)
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172 Higher Engineering Mathematics, Solving the simultaneous equations gives:, A=, , From equation (7),, cos 6x + cos 2x = 2 cos 4x cos 2x, , X +Y, X −Y, and B =, 2, 2, , From equation (5),, , Thus sin(A + B) + sin(A − B) = 2 sin A cos B becomes,, , , , X +Y, X−Y, cos, (5), sin X + sin Y = 2 sin, 2, 2, , sin 6x + sin 2x = 2 sin 4x cos 2x, Hence, , Similarly,, , , , , X+Y, X−Y, sin X − sin Y = 2 cos, sin, (6), 2, 2, , , , X−Y, X+Y, cos, (7), cos X + cos Y = 2 cos, 2, 2, , , , X+Y, X−Y, cos X − cos Y = −2 sin, sin, (8), 2, 2, Problem 19., , Express sin 5θ + sin 3θ as a product., , 2 cos4x cos 2x, cos 6x + cos 2x, =, sin 6x + sin 2x, 2 sin 4x cos 2x, cos 4x, = cot 4 x, =, sin 4x, , Problem 23. Solve the equation, cos 4θ + cos 2θ = 0 for θ in the range 0◦ ≤ θ ≤ 360◦., From equation (7),, , 4θ + 2θ, cos 4θ + cos 2θ = 2 cos, 2, Dividing by 2 gives:, , , , , 5θ + 3θ, 5θ − 3θ, sin 5θ + sin 3θ = 2 sin, cos, 2, 2, , Hence, either, , Problem 20., , From equation (6),, , , , 7x + x, 7x − x, sin 7x − sin x = 2 cos, sin, 2, 2, = 2 cos 4x sin 3x, Problem 21., product., , Express cos 2t − cos 5t as a, , , , , , , , 2t + 5t, 2t − 5t, sin, 2, 2, , , 7, 3, 7, 3, = −2 sin t sin − t = 2 sin t sin t, 2, 2, 2, 2, , , , , 3, 3, since sin − t = −sin t, 2, 2, , cos 2t − cos 5t = −2 sin, , Problem 22., , Show that, , 4θ − 2θ, cos, 2, , , , cos 3θ cos θ = 0, cos 3θ = 0 or cos θ = 0, , from which, 3θ = 90◦ or 270◦ or 450◦ or 630◦ or, 810◦ or 990◦, and θ = 30◦ ,90◦ , 150◦ ,210◦ , 270◦ or 330◦, Now try the following exercise, Exercise 76 Further problems on changing, sums or differences of sines and cosines into, products, In Problems 1 to 5, express as products:, , From equation (8),, , , , , 3θ = cos−1 0 or θ = cos−1 0, , Thus,, , Express sin 7x − sin x as a product., , , , 2 cos3θ cos θ = 0, , Hence,, , From equation (5),, , = 2 sin 4θ cos θ, , , , cos 6x + cos 2x, = cot 4x., sin 6x + sin 2x, , 1., , sin 3x + sin x, , 2., , 1, 2 (sin 9θ − sin 7θ), , 3., , cos 5t + cos 3t, , 4., , 1, 8 (cos 5t − cos t ), , 5., , 1, 2, , 6., , Show that:, sin 4x − sin 2x, (a), = tan x, cos 4x + cos 2x, , [2 sin 2x cos x], [cos 8θ sin θ], [2 cos 4t cos t ], , − 14 sin 3t sin 2t, , 7π, π, cos, cos, 24, 24, , π, π, cos + cos, 3, 4, , (b), , 1, 2 {sin(5x − α) − sin(x, , + α)}, = cos 3x sin(2x − α)
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Compound angles, In Problems 7 and 8, solve for θ in the range 0◦ ≤, θ ≤ 180◦., 7., 8., , cos 6θ + cos 2θ = 0, [22.5◦, 45◦, 67.5◦, 112.5◦, 135◦, 157.5◦], sin 3θ − sin θ = 0, , 9., , [21.47◦, , cos 2x = 2 sin x, , or, , 158.53◦], , 10. sin 4t + sin 2t = 0, [0◦ , 60◦, 90◦, 120◦, 180◦, 240◦,, 270◦ , 300◦, 360◦], , 17.6, , Power waveforms in a.c. circuits, , (a) Purely resistive a.c. circuits, Let a voltage v = Vm sin ωt be applied to a circuit comprising resistance only. The resulting current, is i = Im sin ωt , and the corresponding instantaneous, power, p, is given by:, p = vi = (Vm sin ωt )(Im sin ωt ), i.e. p = Vm Im sin2 ωt, From double angle formulae of Section 17.3,, cos 2 A = 1 − 2 sin2 A, from which,, sin2 A = 12 (1 − cos 2 A) thus, sin2 ωt = 12 (1 − cos 2ωt ), Then power p = Vm Im, i.e., , , , 1, 2 (l, , , − cos 2ωt ), , p = 21 V m I m (1 − cos 2ω t), , The waveforms of v, i and p are shown in Fig. 17.8. The, waveform of power repeats itself after π/ω seconds and, hence the power has a frequency twice that of voltage, and current. The power is always positive, having a maximum value of Vm Im . The average or mean value of the, power is 12 Vm Im ., Vm, The rms value of voltage V = 0.707Vm , i.e. V = √ ,, 2, √, from which, Vm = 2 V ., , Average, power, , 1, , [0◦, 45◦, 135◦, 180◦], , In Problems 9 and 10, solve in the range, 0◦ to 360◦ ., , Maximum, power, , p, , p, i, v, , 173, , , , , 0, , 2, , , i, , t (seconds), , 2, v, , Figure 17.8, , Im, Similarly, the rms value of current, I = √ , from, 2, √, which, Im = 2 I . Hence the average power, P, developed in a purely, √ resistive, √ a.c. circuit is given by, P = 12 Vm Im = 12 ( 2V )( 2I ) = V I watts., Also, power P = I 2 R or V 2 /R as for a d.c. circuit,, since V = I R., Summarizing, the average power P in a purely, resistive a.c. circuit given by, P = V I = I 2R =, , V2, R, , where V and I are rms values., (b) Purely inductive a.c. circuits, Let a voltage v = Vm sin ωt be applied to a circuit containing pure inductance, case). The resulting, (theoretical, π, since current lags voltage, current is i = Im sin ωt −, 2, π, by radians or 90◦ in a purely inductive circuit, and, 2, the corresponding instantaneous power, p, is given by:, , π, p = vi = (Vm sin ωt )Im sin ωt −, 2, , π, i.e. p = Vm Im sin ωt sin ωt −, 2, However,, , , π, = −cos ωt thus, sin ωt −, 2, p = −Vm Im sin ωt cos ωt., , Rearranging gives:, p = − 12 Vm Im (2 sin ωt cosωt )., However, from double-angle formulae,, 2 sin ωt cos ωt = sin 2ωt., Thus, , power, p = − 21 V m I m sin 2ω t.
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174 Higher Engineering Mathematics, p, i, v, , p, v, , i, 1, , 0, , , , , 2, , , t (seconds), , 2, , Figure 17.9, , The waveforms of v, i and p are shown in Fig. 17.9., The frequency of power is twice that of voltage and, current. For the power curve shown in Fig. 17.9, the area, above the horizontal axis is equal to the area below, thus, over a complete cycle the average power P is zero. It, is noted that when v and i are both positive, power p is, positive and energy is delivered from the source to the, inductance; when v and i have opposite signs, power p, is negative and energy is returned from the inductance, to the source., In general, when the current through an inductance, is increasing, energy is transferred from the circuit to, the magnetic field, but this energy is returned when the, current is decreasing., Summarizing, the average power P in a purely, inductive a.c. circuit is zero., (c) Purely capacitive a.c. circuits, Let a voltage v = Vm sin ωt be applied to a circuit, containing, The resulting current is, pure capacitance., , i = Im sin ωt + π2 , since current leads voltage by 90◦, in a purely capacitive circuit, and the corresponding, instantaneous power, p, is given by:, , π, p = vi = (Vm sin ωt )Im sin ωt +, 2, , π, i.e. p = Vm Im sin ωt sin ωt +, 2, , π, However, sin ωt +, = cos ωt, 2, , thus, , p = Vm Im sin ωt cos ωt, , Rearranging gives p = 12 Vm Im (2 sin ωt cos ωt )., Thus power, p = 12 V m I m sin 2ω t., The waveforms of v, i and p are shown in Fig. 17.10., Over a complete cycle the average power P is zero., When the voltage across a capacitor is increasing,, energy is transferred from the circuit to the electric, field, but this energy is returned when the voltage is, decreasing., Summarizing, the average power P in a purely, capacitive a.c. circuit is zero., (d) R–L or R–C a.c. circuits, Let a voltage v = Vm sin ωt be applied to a circuit containing resistance and inductance or resistance and capacitance. Let the resulting current be, i = Im sin(ωt + φ), where phase angle φ will be positive for an R–C circuit and negative for an R–L circuit., The corresponding instantaneous power, p, is given by:, p = vi = (Vm sin ωt )Im sin(ωt + φ), i.e. p = Vm Im sin ωt sin(ωt + φ), Products of sine functions may be changed into differences of cosine functions as shown in Section 17.4,, i.e. sin A sin B = − 12 [cos(A + B) − cos(A − B)].
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Compound angles, p, i, v, , 175, , p, v, i, , 1, , , , , 0, , 2, , , t (seconds), , 2, , Figure 17.10, , p, i, v, , p, v, i, , 1, , 0, , , , , 2, , , t (seconds), , 2, , Figure 17.11, , Substituting ωt = A and (ωt + φ) = B gives:, power,, , p = Vm Im {− 12 [cos(ωt + ωt + φ), − cos(ωt − (ωt + φ))]}, , i.e., , p = 12 Vm Im [cos(−φ) − cos(2ωt + φ)], , However, cos(−φ) = cos φ, Thus p = 21 V m I m [cos φ − cos(2ω t + φ)], The instantaneous power p thus consists of, (i) a sinusoidal term, − 12 Vm Im cos(2ωt + φ) which, has a mean value over a cycle of zero, and, , (ii) a constant term, 12 Vm Im cos φ (since φ is constant, for a particular circuit)., Thus the average value of power, P = 12 Vm Im cos φ., √, √, Since Vm = 2 V and Im = 2 I , average power,, √, √, P = 12 ( 2 V )( 2 I ) cos φ, i.e., , P = V I cos φ, , The waveforms of v, i and p, are shown in Fig. 17.11, for an R–L circuit. The waveform of power is seen to
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176 Higher Engineering Mathematics, pulsate at twice the supply frequency. The areas of the, power curve (shown shaded) above the horizontal time, axis represent power supplied to the load; the small, areas below the axis represent power being returned to, the supply from the inductance as the magnetic field, collapses., A similar shape of power curve is obtained for an, R–C circuit, the small areas below the horizontal axis, representing power being returned to the supply from, the charged capacitor. The difference between the areas, , above and below the horizontal axis represents the heat, loss due to the circuit resistance. Since power is dissipated only in a pure resistance, the alternative equations, for power, P = I R2 R, may be used, where I R is the rms, current flowing through the resistance., Summarizing, the average power P in a circuit, containing resistance and inductance and/or capacitance, whether in series or in parallel, is given by, P = VI cos φ or P = I 2R R (V, I and I R being rms, values).
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Revision Test 5, This Revision Test covers the material contained in Chapters 14 to 17. The marks for each question are shown in, brackets at the end of each question., 1. Solve the following equations in the range 0◦, to 360◦., (a) sin−1(−0.4161) = x, (b) cot −1(2.4198) = θ, , (8), , 2. Sketch the following curves labelling relevant, points:, (a) y = 4 cos(θ + 45◦), (b) y = 5 sin(2t − 60◦ ), , (8), , 3. The current in an alternating current circuit at, any time t seconds is given by:, i = 120 sin(100πt + 0.274) amperes., , the amplitude, periodic time, frequency and, phase angle (with reference to 120 sin 100πt ), , , , 1 − cos2 θ, = tan θ, cos2 θ, , , 3π, (b) cos, + φ = sin φ, 2, (a), , sin2 x, = 1 tan 2 x, 1 + cos 2x 2, , 6. Solve the following trigonometric equations in the, range 0◦ ≤ x ≤ 360◦ :, , (b) 3.25 cosec x = 5.25, , the value of current when t = 6 ms, , (d) the time when the current first reaches 80 A, , (c) 5 sin2 x + 3 sin x = 4, , Sketch one cycle of the oscillation., , (d) 2 sec2 θ + 5 tan θ = 3, , (19), , 4. A complex voltage waveform v is comprised, of a 141.4 V rms fundamental voltage at a frequency of 100 Hz, a 35% third harmonic component leading the fundamental voltage at zero, time by π/3 radians, and a 20% fifth harmonic, component lagging the fundamental at zero time, by π/4 radians., (a), , (9), , (a) 4 cos x + 1 = 0, , (b) the value of current when t = 0, (c), , 5. Prove the following identities:, , (c), , Determine, (a), , (b) Draw the complex voltage waveform using, harmonic synthesis over one cycle of the, fundamental waveform using scales of 12 cm, for the time for one cycle horizontally and, 1 cm = 20 V vertically., (15), , Write down an expression to represent, voltage v., , (18), , 7. Solve the equation 5 sin(θ − π/6) = 8 cosθ for, values 0 ≤ θ ≤ 2π., (8), 8. Express 5.3 cos t − 7.2 sin t in the form, R sin(t + α). Hence solve the equation, 5.3 cos t − 7.2 sin t = 4.5 in the range, 0 ≤ t ≤ 2π., 9. Determine, , , , 2 cos3t sin t dt ., , (12), (3)
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Chapter 18, , Functions and their curves, y, , 18.1, , Standard curves, , 4, , When a mathematical equation is known, co-ordinates, may be calculated for a limited range of values, and, the equation may be represented pictorially as a graph,, within this range of calculated values. Sometimes it, is useful to show all the characteristic features of an, equation, and in this case a sketch depicting the equation can be drawn, in which all the important features, are shown, but the accurate plotting of points is less, important. This technique is called ‘curve sketching’, and can involve the use of differential calculus, with,, for example, calculations involving turning points., If, say, y depends on, say, x, then y is said to be a function of x and the relationship is expressed as y = f (x); x, is called the independent variable and y is the dependent, variable., In engineering and science, corresponding values are, obtained as a result of tests or experiments., Here is a brief resumé of standard curves, some of, which have been met earlier in this text., , 3, 2, 1, , 0, , 1, , 2, , 3, , x, , 3, , x, , (a), y, 5, 4, y 5 5 22x, , 3, 2, 1, , (i) Straight Line, The general equation of a straight, line is y = mx + c,, dy, and c is the y-axis, where m is the gradient i.e., dx, intercept., Two examples are shown in Fig. 18.1, , y 5 2x 1 1, , 5, , 0, , 1, , 2, (b), , Figure 18.1, , (ii) Quadratic Graphs, y, , The general equation of a quadratic graph is, y = ax 2 + bx + c, and its shape is that of a parabola., The simplest example of a quadratic graph, y = x 2 , is, shown in Fig. 18.2., , 8, 6, , y 5x 2, , 4, 2, , (iii) Cubic Equations, The general equation of a cubic graph is, y = ax 3 + bx 2 + cx + d., , 22 21 0, , Figure 18.2, , 1, , 2, , x
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Functions and their curves, The simplest example of a cubic graph, y = x 3 , is, shown in Fig. 18.3., y, 8, , (v) Circle (see Chapter 13, page 122), The simplest equation of a circle is x 2 + y 2 =r 2 ,, with centre at the origin and radius r, as shown in, Fig. 18.5., , y 5x 3, , 6, , y, , 4, , r, , 2, 22 21, , 179, , x21 y25 r 2, 1, , 2, , 22, , x, , 2r, , 24, , r, , O, , x, , 26, 28, , 2r, , Figure 18.3, , Figure 18.5, , (iv) Trigonometric Functions (see Chapter 14,, page 134), Graphs of y = sin θ, y = cos θ and y = tan θ are shown in, Fig. 18.4., , Figure 18.6 shows a circle, , y 5 sin , , 1.0, , 21.0, , (x − a)2 + (y − b)2 = r 2, , (x − 2)2 + (y − 3)2 = 4, , y, , 0, , More generally, the equation of a circle, centre (a, b),, radius r, is given by:, , , 2, , , , 3, 2, , y, , 2 , , (x 2 2)2 1 (y 2 3)2 5 4, , 5, 4, , (a), , r5, , 3, y, 1.0, , 0, 21.0, , b53, , y 5 cos , , 2, , , , 3, 2, , 2, , 2, , 0, , , 2, , 4, , x, , a52, , Figure 18.6, , (b), y, , 2, , (vi) Ellipse, , y 5 tan , , The equation of an ellipse is, 0, , , 2, , , , (c), , Figure 18.4, , 3, 2, , 2 , , x 2 y2, +, =1, a 2 b2, and the general shape is as shown in Fig. 18.7., The length AB is called the major axis and CD the, minor axis.
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180 Higher Engineering Mathematics, y, , (ix) Logarithmic Function (see Chapter 3, page 26), x2, , C, , a2, , 1, , y2, b2, , 51, , b, A, , B, O, , (x) Exponential Functions (see Chapter 4, page 30), x, , a, , y = ln x and y = lg x are both of the general shape shown, in Fig. 18.10., , y = ex is of the general shape shown in Fig. 18.11., , D, , y, , Figure 18.7, 3, c, y5x, , In the above equation, ‘a’ is the semi-major axis and, ‘b’ is the semi-minor axis., x 2 y2, (Note that if b = a, the equation becomes 2 + 2 = 1,, a, a, i.e. x 2 + y 2 = a 2 , which is a circle of radius a)., , 2, 1, 23, , 22, , 21, , (vii) Hyperbola, , 0, , 1, , 2, , 3, , x, , 21, , The equation of a hyperbola is, , 22, , x 2 y2, −, =1, a 2 b2, and the general shape is shown in Fig. 18.8. The, curve is seen to be symmetrical about both the, x- and y-axes. The distance AB in Fig. 18.8 is given, by 2a., , 23, , Figure 18.9, , y, , y, x2 y2, 51, 2, a2 b2, , A, , 0, , B, , O, , y 5 log x, , 1, , x, , x, , Figure 18.8, , Figure 18.10, , (viii) Rectangular Hyperbola, , (xi) Polar Curves, , The equation of a rectangular hyperbola is x y = c or, c, y = and the general shape is shown in Fig. 18.9., x, , The equation of a polar curve is of the form r = f (θ)., An example of a polar curve, r = a sin θ, is shown in, Fig. 18.12.
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Functions and their curves, y, , 181, , y, , 8, , 6, , y 5e x, , y 5 3(x 1 1), 4, 1, 2, , 0, , y5x11, , x, , 0, , Figure 18.11, , 1, , 2 x, , (a), y, 2, , a, , y 5 2 sin , , r 5a sin, , y 5 sin , 1, , O, , a, , 0, , , 2, , , , 3, 2, , 2, , , , (b), , Figure 18.13, Figure 18.12, , (ii) y = f (x) + a, , 18.2, , Simple transformations, , From the graph of y = f (x) it is possible to deduce the graphs of other functions which are transformations of y = f (x). For example, knowing the graph, of y = f (x), can help us draw the graphs of y = a f (x),, y = f (x) + a, y = f (x + a), y = f (ax), y = − f (x) and, y = f (−x)., (i) y = a f (x), For each point (x 1, y1 ) on the graph of y = f (x) there, exists a point (x 1, ay1 ) on the graph of y = a f (x)., Thus the graph of y = a f (x) can be obtained by, stretching y = f (x) parallel to the y-axis by a scale, factor ‘a’., Graphs of y = x + 1 and y = 3(x + 1) are shown in, Fig. 18.13(a) and graphs of y = sin θ and y = 2 sin θ are, shown in Fig. 18.13(b)., , The graph of y = f (x) is translated by ‘a’ units parallel to the y-axis to obtain y = f (x) + a. For example, if f (x) = x, y = f (x) + 3 becomes y = x + 3, as, shown in Fig. 18.14(a). Similarly, if f (θ) = cos θ,, then y = f (θ) + 2 becomes y = cos θ + 2, as shown in, Fig. 18.14(b). Also, if f (x) = x 2 , then y = f (x) + 3, becomes y = x 2 + 3, as shown in Fig. 18.14(c)., (iii) y = f (x + a), The graph of y = f (x) is translated by ‘a’ units parallel, to the x-axis to obtain y = f (x + a). If ‘a’ >0 it moves, y = f (x) in the negative direction on the x-axis (i.e. to, the left), and if ‘a’ <0 it moves y = f (x) in the positive, direction on the x-axis, to the right). For example,, if, , (i.e., π, π, becomes y = sin x −, f (x) = sin x, y = f x −, 3, 3, , π, as shown in Fig. 18.15(a) and y = sin x +, is shown, 4, in Fig. 18.15(b).
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182 Higher Engineering Mathematics, y, , y, , y 5sin x, , , 3, , 1, , 6, , y 5sin (x 2, , , ), 3, , y⫽x⫹3, 4, , 21, , y ⫽x, , 2, , , , , 2, , 0, , 3, 2, , 2, , x, , , 3, (a), , 4, , 2, , 0, , 6, , x, , y, , 4, , (a), , y 5sin x, , 1, , y 5 sin (x 1, 3, y ⫽ cos ⫹ 2, , 0, , 4, 21, , , 2, , , , , ), 4, , 3, 2, , 2, , x, , 1, y ⫽ cos , , 2, , 0, , , , 3, 2, , (b), , , , 2, , Figure 18.15, , Similarly graphs of y = x 2 , y = (x − 1)2 and, y = (x + 2)2 are shown in Fig. 18.16., , (b), , (iv) y = f (ax), y, , For each point (x 1, y1) on the graph of y = f (x), there, x1, exists a point, , y1 on the graph of y = f (ax). Thus, a, the graph of y = f (ax) can be obtained by stretching, 1, y = f (x) parallel to the x-axis by a scale factor, a, , 8, y ⫽ x2⫹ 3, 6, , y, y ⫽ x2, , 4, , y 5 x2, 6, , y 5 (x 1 2)2, , 4, , 2, , y 5 (x 2 1) 2, , 2, ⫺2, , ⫺1, , 0, , 1, , 2, , 22, , x, , (c), , Figure 18.14, , Figure 18.16, , 21, , 0, , 1, , 2, , x
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183, , Functions and their curves, y, , 1, For example, if f (x) = (x − 1)2 , and a = , then, 2, x, 2, f (ax) =, −1 ., 2, Both of these curves are shown in Fig. 18.17(a)., Similarly, y = cos x and y = cos 2x are shown in, Fig. 18.17(b)., , 8, y 5 x 21 2, , 4, , 22, , 21, , 0, , (v) y = − f (x), , 1, , x, , y 52(x 1 2 ), 2, , 24, , The graph of y = − f (x) is obtained by reflecting, y = f (x) in the x-axis. For example, graphs of y = ex, and y = −ex are shown in Fig. 18.18(a) and graphs of, y = x 2 + 2 and y = −(x 2 + 2) are shown in Fig. 18.18(b)., , 2, , 28, ( b), , Figure 18.18 (Continued), y, , 4, , The graph of y = f (−x) is obtained by reflecting, y = f (x) in the y-axis. For example, graphs of y = x 3, and y = (−x)3 = −x 3 are shown in Fig. 18.19(a), , x, y 5 ( 2 2 1)2, , 2, , 22, , (vi) y = f (−x), , y 5(x 21)2, , 0, , 2, , 4, , 6, , x, , y, , (a), , 20, , y 5 (2x )3, y, , y 5 cos x, , 1.0, , 10, , y 5 cos 2x, , 23, , 2, , 0, , , , y 5x3, , 3, 2, , 2, , 22, , 2, , 0, , 3 x, , 210, , x, , 220, , 21.0, (b), , (a), , Figure 18.17, y, y, , y 5 2In x, , y 5ex, , y 5 In x, , 1, 21, , 21, , x, , 0, , y 52ex, , ( b), , (a), , Figure 18.18, , Figure 18.19, , 1, , x
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184 Higher Engineering Mathematics, and graphs of y = ln x and y = −ln x are shown in, Fig. 18.19(b)., Problem 1. Sketch the following graphs, showing, relevant points:, (a) y = (x − 4)2 (b) y = x 3 − 8, (a) In Fig. 18.20 a graph of y = x 2 is shown by the broken line. The graph of y = (x − 4)2 is of the form, y = f (x + a). Since a = −4, then y = (x − 4)2 is, translated 4 units to the right of y = x 2 , parallel to, the x-axis., , Problem 2. Sketch the following graphs, showing, relevant points:, (a) y = 5 − (x + 2)3 (b) y = 1 + 3 sin 2x, (a) Figure 18.22(a) shows a graph of y = x 3 ., Figure 18.22(b) shows a graph of y = (x + 2)3 (see, f (x + a), Section (iii) above)., , y, , (See Section (iii) above)., 20, y ⫽x 3, , y, y ⫽x 2, , 10, , y ⫽ (x ⫺ 4)2, , 8, , 4, , ⫺4, , ⫺2, , ⫺2, , 2, , 0, , 4, , 6, , 0, , x, , 2, , –10, , x, , Figure 18.20, –20, , (b) In Fig. 18.21 a graph of y = x 3 is shown by the, broken line. The graph of y = x 3 − 8 is of the, form y = f (x) + a. Since a = −8, then y = x 3 − 8, is translated 8 units down from y = x 3 , parallel to, the y-axis., , (a), y, , (See Section (ii) above)., 20, y 5 (x 1 2)3, , y, 10, , 20, y ⫽x 3, y ⫽x 3 ⫺ 8, , 10, , ⫺3, , ⫺2, , ⫺1, , 0, , 1, , 2, , 3, , 24, , x, , 22, , 0, , 2, , –10, , –10, –20, , –20, –30, , (b), , Figure 18.21, , Figure 18.22, , x
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Functions and their curves, y, , y, , 1, , y 5 sin x, , 20, 0, y ⫽ ⫺(x ⫹ 2)3, , , 2, , ⫺2, , 3, 2, , x, , 21, , 10, , (a), y, , ⫺4, , , , y 5 sin 2x, , 1, 0, , 2, , x, 0, , –10, , , , , 2, , 3, 2, , 2 x, , 21, (b), , –20, y, , y 5 3 sin 2x, , 3, (c), y, , 2, , y ⫽ 5 ⫺ (x ⫹ 2)3, , 1, , 20, , 0, 10, , , 2, , , , 3, 2, , 2 x, , 21, ⫺4, , ⫺2, , 0, , 2, , x, , –10, , 22, 23, (c), y, , –20, 4, 3, , (d), , 2, , Figure 18.22 (Continued), , Figure 18.22(c) shows a graph of y = − (x + 2)3, (see − f (x), Section (v) above). Figure 18.22(d), shows the graph of y = 5 −(x + 2)3 (see, f (x) + a, Section (ii) above)., (b) Figure 18.23(a) shows a graph of y = sin x., Figure 18.23(b) shows a graph of y = sin 2x, (see f (ax), Section (iv) above)., Figure 18.23(c) shows a graph of y = 3 sin 2x (see, a f (x), Section (i) above). Figure 18.23(d) shows a, graph of y = 1 + 3 sin 2x (see f (x) + a, Section (ii), above)., , y ⫽1 ⫹ 3 sin 2x, , 1, , 0, , , 2, , , , ⫺1, ⫺2, (d), , Figure 18.23, , 3, 2, , 2 x, , 185
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186 Higher Engineering Mathematics, Now try the following exercise, Exercise 77 Further problems on simple, transformations with curve sketching, Sketch the following graphs, showing relevant, points:, (Answers on page 200, Fig. 18.39), 1., , y = 3x − 5, , 2., , y = − 3x + 4, , 3., , y = x2 + 3, , 4., , y = (x − 3)2, , 5., , y = (x − 4)2 + 2, , 6., , y = x − x2, , 7., , y = x3 +2, , 8., , y = 1 +2 cos 3x, , π, y = 3 −2 sin x +, 4, y = 2 ln x, , 9., 10., , 18.3, , also periodic of period 2π and is defined by:, 5, −1, when −π ≤ x ≤ 0, f (x) =, 1, when 0 ≤ x ≤ π, , 18.4 Continuous and discontinuous, functions, If a graph of a function has no sudden jumps or breaks it, is called a continuous function, examples being the, graphs of sine and cosine functions. However, other, graphs make finite jumps at a point or points in the interval. The square wave shown in Fig. 18.24 has finite, discontinuities as x = π, 2π, 3π, and so on, and is, therefore a discontinuous function. y = tan x is another, example of a discontinuous function., , 18.5, , Even and odd functions, , Even functions, A function y = f (x) is said to be even if f (−x) = f (x), for all values of x. Graphs of even functions are always, symmetrical about the y-axis (i.e. is a mirror image)., Two examples of even functions are y = x 2 and y = cos x, as shown in Fig. 18.25., , Periodic functions, , A function f (x) is said to be periodic if f (x + T ) =, f (x) for all values of x, where T is some positive, number. T is the interval between two successive repetitions and is called the period of the function f (x). For, example, y = sin x is periodic in x with period 2π since, sin x = sin(x + 2π) = sin(x + 4π), and so on. Similarly,, y = cos x is a periodic function with period 2π since, cos x = cos(x + 2π) = cos(x + 4π), and so on. In general, if y = sin ωt or y = cos ωt then the period of the, waveform is 2π/ω. The function shown in Fig. 18.24 is, , y, 8, 6, y 5x 2, , 4, 2, 23 22 21 0, , 1, , 2, , 3 x, , (a), y, , f (x), , y 5cos x, , 1, 2, ⫺2, , ⫺, , , , 0, , 2, , 2, 2, , 0, , x, , ⫺1, (b), , Figure 18.24, , Figure 18.25, , , 2, , x
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Functions and their curves, Odd functions, A function y = f (x) is said to be odd if f (−x) = − f (x), for all values of x. Graphs of odd functions are, always symmetrical about the origin. Two examples, of odd functions are y = x 3 and y = sin x as shown in, Fig. 18.26., Many functions are neither even nor odd, two such, examples being shown in Fig. 18.27., y, , y5x3, , Problem 3. Sketch the following functions and, state whether they are even or odd functions:, (a) y = tan x, ⎧, π, ⎪, 2, when 0 ≤ x ≤, ⎪, ⎪, 2, ⎪, ⎪, ⎨, π, 3π, ,, (b) f (x) = −2, when ≤ x ≤, ⎪, 2, 2, ⎪, ⎪, ⎪, ⎪, ⎩ 2, when 3π ≤ x ≤ 2π, 2, and is periodic of period 2π., , 27, , (a), 23, , 3 x, , 0, , A graph of y = tan x is shown in Fig. 18.28(a) and, is symmetrical about the origin and is thus an odd, function (i.e. tan(−x) = −tan x)., , (b) A graph of f (x) is shown in Fig. 18.28(b) and, is symmetrical about the f (x) axis hence the, function is an even one, ( f (−x) = f (x))., , 227, , (a), , y ⫽ tan x, , y, , y, 1, , y 5 sinx, , ⫺, 0 , 2, , 23 2 2, 2, 2, , , , 3, 2, , 0, , , , 2 x, , , , 2, , 2 x, , 21, (b), (a), , Figure 18.26, f(x ), 2, , y, y ⫽e x, , 20, , ⫺2, , 10, , ⫺, , 0, , x, , ⫺2, ⫺1 0, , 1 2 3 x, , ( b), , (a), , Figure 18.28, , y, , 0, (b), , Figure 18.27, , 187, , x, , Problem 4. Sketch the following graphs and state, whether the functions are even, odd or neither even, nor odd:, (a) y = ln x, (b) f (x) = x in the range −π to π and is, periodic of period 2π.
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188 Higher Engineering Mathematics, (a) A graph of y = ln x is shown in Fig. 18.29(a), and the curve is neither symmetrical about the, y-axis nor symmetrical about the origin and is thus, neither even nor odd., (b) A graph of y = x in the range −π to π is shown in, Fig. 18.29(b) and is symmetrical about the origin, and is thus an odd function., , , , 3. State whether the following functions, which, are periodic of period 2π, are even or odd:, 5, θ, when −π ≤ θ ≤ 0, (a) f (θ) =, −θ, when 0 ≤ θ ≤ π, ⎧, π, π, ⎨x, when − ≤ x ≤, 2, 2, (b) f (x) =, ⎩0, when π ≤ x ≤ 3π, 2, 2, , y, y ⫽ In x, , 1.0, , (a) odd (b) even, (c) odd (d) neither, , [(a) even (b) odd], , 0.5, , 1 2 3 4, , 0, , x, , 18.6, , ⫺0.5, , (a), y, , , ⫺2 ⫺, , y⫽x, , 0, , , , 2, , x, , ⫺, , If y is a function of x, the graph of y against x can be, used to find x when any value of y is given. Thus the, graph also expresses that x is a function of y. Two such, functions are called inverse functions., In general, given a function y = f (x), its inverse may, be obtained by interchanging the roles of x and y and, then transposing for y. The inverse function is denoted, by y = f −1 (x)., For example, if y = 2x + 1, the inverse is obtained by, (i) transposing for x, i.e. x =, , (b), , Figure 18.29, , Now try the following exercise, , Exercise 78 Further problems on even and, odd functions, In Problems 1 and 2 determine whether the given, functions are even, odd or neither even nor odd., 1. (a) x 4 (b) tan 3x (c) 2e3t (d) sin2 x, , (a) even, (b) odd, (c) neither (d) even, 2. (a) 5t 3 (b) ex + e−x (c), , Inverse functions, , cos θ, θ, , (d) ex, , y −1, y 1, = − and, 2, 2 2, , (ii) interchanging x and y, giving the inverse as, x 1, y= −, 2 2, x 1, Thus if f (x) = 2x + 1, then f −1 (x) = −, 2 2, A graph of f (x) = 2x + 1 and its inverse f −1 (x) =, x 1, − is shown in Fig. 18.30 and f −1 (x) is seen to be, 2 2, a reflection of f (x) in the line y = x., Similarly, if y = x 2 , the inverse is obtained by, √, (i) transposing for x, i.e. x = ± y and, (ii) interchanging, x and y, giving the inverse, √, y = ± x., Hence the inverse has two values for every value of x., Thus f (x) = x 2 does not have a single inverse. In, such a case the domain of the original function may, 2, the inverse is, be restricted, √ to y = x for x > 0. Thus, then y = + x. A graph of f (x) = x 2 and its inverse
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Functions and their curves, , 189, , Hence if f (x) = x − 1, then f−1(x) = x + 1, , y, y5 2x11, y5 x, , 4, , (b) If y = f (x), then y = x 2 − 4 √(x > 0), Transposing for x gives x = y +, √4, Interchanging x and y gives y = x + 4, Hence if √, f (x) = x 2 − 4 (x > 0) then, −1, f (x) = x + 4 if x > −4, , 2, 1, 21, , 1, , 0, , 2, , 3, , y5, , x, 1, 2, 2, 2, , 4, , x, , 21, , Figure 18.30, , y5 x 2, 4, y5 x, , 2, , y 5 Œ„, x, , 1, , 2, , 3, , x, , Figure 18.31, , √, , f −1 (x) = x for x > 0 is shown in Fig. 18.31 and, again,, f −1 (x) is seen to be a reflection of f (x) in the line y = x., It is noted from the latter example, that not all functions have an inverse. An inverse, however, can be, determined if the range is restricted., Problem 5. Determine the inverse for each of the, following functions:, (a) f (x) = x − 1 (b) f (x) = x 2 − 4 (x > 0), (c) f (x) = x 2 + 1, (a), , If y = f (x), then y = x 2 + 1 √, −1, Transposing for x gives x = y√, Interchanging x and y gives y = x − 1, which has, two values., Hence there is no inverse of f(x) = x2 + 1, since, the domain of f (x) is not restricted., , Inverse trigonometric functions, , y, , 0, , (c), , If y = f (x), then y = x − 1, Transposing for x gives x = y + 1, Interchanging x and y gives y = x + 1, , If y = sin x, then x is the angle whose sine is y., Inverse trigonometrical functions are denoted by prefixing the function with ‘arc’ or, more commonly,−1 ., Hence transposing y = sin x for x gives x = sin−1 y., Interchanging x and y gives the inverse y = sin−1 x., Similarly, y = cos−1 x, y = tan−1 x, y = sec−1 x,, y =cosec−1 x and y =cot −1 x are all inverse trigonometric functions. The angle is always expressed in, radians., Inverse trigonometric functions are periodic so it is, necessary to specify the smallest or principal value of the, angle. For sin−1 x, tan−1 x, cosec−1 x and cot −1 x, the, π, π, principal value is in the range − < y < . For cos−1 x, 2, 2, and sec−1 x the principal value is in the range 0 < y < π., Graphs of the six inverse trigonometric functions are, shown in Fig. 33.1, page 335., Problem 6. Determine the principal values of, (a) arcsin 0.5, √ , 3, (c) arccos −, 2, , (b) arctan(−1), √, (d) arccosec( 2), , Using a calculator,, (a) arcsin 0.5 ≡ sin−1 0.5 = 30◦, =, , π, rad or 0.5236 rad, 6, , (b) arctan(−1) ≡ tan−1 (−1) = −45◦, =−, , π, rad or −0.7854 rad, 4
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190 Higher Engineering Mathematics, √ , √ , 3, 3, −1, ≡ cos, = 150◦, −, (c) arccos −, 2, 2, 5π, rad or 2.6180 rad, 6, , , √, 1, (d) arccosec( 2) = arcsin √, 2, , , 1, = 45◦, ≡ sin−1 √, 2, =, , 4, , , or 0.7854 rad, , 8. cot −1 2, 9., , [0.4636 rad], , cosec−1 2.5, , [0.4115 rad], , [0.8411 rad], 10. sec−1 1.5, , , , , 1, π, or 0.7854 rad, 11. sin−1 √, 4, 2, 12. Evaluate x, correct to 3 decimal places:, x = sin−1, , π, = rad or 0.7854 rad, 4, Problem 7. Evaluate (in radians), correct to, 3 decimal places: sin−1 0.30 + cos−1 0.65., , π, , 7. tan −1 1, , 1, 4, 8, + cos−1 − tan−1, 3, 5, 9, [0.257], , 13. Evaluate y, correct to 4 significant figures:, √, √, y = 3 sec−1 2 − 4 cosec−1 2, + 5 cot−1 2, , sin−1 0.30 = 17.4576◦ = 0.3047 rad, , [1.533], , cos−1 0.65 = 49.4584◦ = 0.8632 rad, Hence sin−1 0.30 + cos−1 0.65, = 0.3047 +0.8632 = 1.168, correct to 3 decimal places., Now try the following exercise, , Determine the inverse of the functions given in, Problems 1 to 4., f (x) = x + 1, , 2., , f (x) = 5x − 1, , 3., , f (x) = x 3 + 1, , 4., , f (x) =, , 1, +2, x, , Asymptotes, , x +2, is drawn, x +1, up for various values of x and then y plotted against x,, the graph would be as shown in Fig. 18.32. The straight, lines AB, i.e. x = −1, and CD, i.e. y = 1, are known as, asymptotes., An asymptote to a curve is defined as a straight, line to which the curve approaches as the distance, from the origin increases. Alternatively, an asymptote can be considered as a tangent to the curve at, infinity., If a table of values for the function y =, , Exercise 79 Further problems on inverse, functions, , 1., , 18.7, , [ f −1(x) = x − 1], , , f −1 (x) = 15 (x + 1), √, [ f −1(x) = 3 x − 1], , 1, f −1(x) =, x −2, , Determine the principal value of the inverse functions in Problems 5 to 11., π, , 5. sin−1 (−1), − or −1.5708 rad, 2, π, , 6. cos−1 0.5, or 1.0472 rad, 3, , Asymptotes parallel to the x- and y-axes, There is a simple rule which enables asymptotes parallel to the x- and y-axis to be determined. For a curve, y = f (x):, (i) the asymptotes parallel to the x-axis are found by, equating the coefficient of the highest power of x, to zero., (ii) the asymptotes parallel to the y-axis are found by, equating the coefficient of the highest power of y, to zero.
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Functions and their curves, , 191, , y, , A, , 5, , 4, , 3, y5, 2, C, , x 12, x 11, , D, , 1, , 24, , 23, , 22, , 21, , 0, , 1, , 2, , 3, , 4, , x, , 21, 22, , y5, , x 12, x 11, , 23, 24, 25, B, , Figure 18.32, , With the above example y =, , x +2, , rearranging gives:, x +1, , y(x + 1) = x + 2, i.e., , yx + y − x − 2 = 0, , and, , x(y − 1) + y − 2 = 0, , (1), , The coefficient of the highest power of x (in this case x 1), is (y − 1). Equating to zero gives: y − 1 = 0, From which, y = 1, which is an asymptote of y =, as shown in Fig. 18.32., Returning to equation (1):, from which,, , x +2, x +1, , yx + y − x − 2 = 0, y(x + 1) − x − 2 = 0., , The coefficient of the highest power of y (in this case, y 1 ) is (x + 1). Equating to zero gives: x + 1 = 0 from, x +2, which, x = −1, which is another asymptote of y =, x +1, as shown in Fig. 18.32., , Problem 8. Determine the asymptotes for the, x −3, function y =, and hence sketch the curve., 2x + 1, Rearranging y =, , x −3, gives: y(2x + 1) = x − 3, 2x + 1, , i.e., 2x y + y = x − 3, or, 2x y + y − x + 3 = 0, and x(2y − 1) + y + 3 = 0, Equating the coefficient of the highest power of x to, zero gives: 2y − 1 = 0 from which, y = 12 which is an, asymptote., Since y(2x + 1) = x − 3 then equating the coefficient of, the highest power of y to zero gives: 2x + 1 = 0 from, which, x = − 12 which is also an asymptote., x − 3 −3, When x = 0, y =, =, = −3 and when y = 0,, 2x + 1, 1, x −3, 0=, from which, x − 3 = 0 and x = 3., 2x + 1, x −3, A sketch of y =, is shown in Fig. 18.33., 2x + 1
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192 Higher Engineering Mathematics, , y, , 6, , 4, y5, , x 23, 2x 11, , x 52, , 1, 2, , 2, y5, , 1, 2, , 2, 28, , 26, , 24, , 22, , 21, , 0, , 1, , 4, , y5, , 24, , 26, , Figure 18.33, , x 23, 2x 11, , 6, , 8, , x
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Functions and their curves, Problem 9. Determine the asymptotes parallel to, the x- and y-axes for the function, x 2 y 2 = 9(x 2 + y 2 )., Asymptotes parallel to the x-axis:, Rearranging x 2 y 2 = 9(x 2 + y 2 ) gives, , (iii) Equating the coefficient of the highest power, of x to zero gives m − 1 = 0 from which,, m = 1., Equating the coefficient of the next highest power, of x to zero gives m + c + 1 =0., and since m = 1, 1 + c + 1 = 0 from which,, c = −2., Hence y = mx + c = 1x − 2., , x 2 y 2 − 9x 2 − 9y 2 = 0, hence, , x 2 (y 2, , − 9) − 9y 2, , 193, , i.e. y = x − 2 is an asymptote., , =0, , To determine any asymptotes parallel to the x-axis:, Equating the coefficient of the highest power of x to zero, gives y 2 − 9 = 0 from which, y 2 = 9 and y = ±3., Asymptotes parallel to the y-axis:, Since x 2 y 2 − 9x 2 − 9y 2 = 0, then, , y 2 (x 2 − 9) − 9x 2 = 0, , Equating the coefficient of the highest power of y to zero, gives x 2 − 9 = 0 from which, x 2 = 9 and x = ±3., Hence asymptotes occur at y = ±3 and x = ±3., , Other asymptotes, To determine asymptotes other than those parallel to, x- and y-axes a simple procedure is:, (i) substitute y = mx + c in the given equation, (ii) simplify the expression, (iii) equate the coefficients of the two highest powers, of x to zero and determine the values of m and c., y = mx + c gives the asymptote., Problem 10. Determine the asymptotes for the, function: y(x + 1) = (x − 3)(x + 2) and sketch the, curve., , Rearranging y(x + 1) = (x − 3)(x + 2), yx + y = x 2 − x − 6, , gives, , The coefficient of the highest power of x (i.e. x 2 ) is 1., Equating this to zero gives 1 =0 which is not an equation, of a line. Hence there is no asymptote parallel to the, x-axis., To determine any asymptotes parallel to the y-axis:, Since y(x + 1) = (x − 3)(x + 2) the coefficient of, the highest power of y is x + 1. Equating this to, zero gives x + 1 = 0, from which, x = −1. Hence x = −1, is an asymptote., When x = 0, y(1) = (−3)(2), i.e. y = −6., When y = 0, 0 =(x − 3)(x + 2), i.e. x = 3 and x = −2., A sketch of the function y(x + 1) = (x − 3)(x + 2) is, shown in Fig. 18.34., Problem 11. Determine the asymptotes for the, function x 3 − x y 2 + 2x − 9 =0., Following the procedure:, (i) Substituting y = mx + c gives, x 3 − x(mx + c)2 + 2x − 9 =0., (ii) Simplifying gives, , Following the above procedure:, (i) Substituting y = mx + c into, y(x + 1) = (x − 3) (x + 2) gives:, (mx + c)(x + 1) = (x − 3)(x + 2), (ii) Simplifying gives, mx 2 + mx + cx + c = x 2 − x − 6, and (m − 1)x 2 + (m + c + 1)x + c + 6 =0, , x 3 − x[m 2 x 2 + 2mcx + c2 ] + 2x − 9 = 0, i.e., , x 3 − m 2 x 3 − 2mcx 2 − c2 x + 2x − 9 = 0, , and x 3 (1 − m 2 ) − 2mcx 2 − c2 x + 2x − 9 = 0, (iii) Equating the coefficient of the highest power of x, (i.e. x 3 in this case) to zero gives 1 −m 2 = 0, from, which, m = ±1., Equating the coefficient of the next highest power, of x (i.e. x 2 in this case) to zero gives −2mc = 0,, from which, c = 0.
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194 Higher Engineering Mathematics, y, , 6, , x2, , 2, , x 521, , y5, , 4, , 2, , 26, , 24, , 22, , 0, , 2, , 4, , y (x 11) 5 (x 23)(x 12), 22, , y (x 11) 5 (x 23)(x 12), , 24, , 26, , 28, , 210, , Figure 18.34, , 6, , x
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Functions and their curves, Hence y = mx + c = ±1x + 0, i.e. y = x and y = −x, are asymptotes., To determine any asymptotes parallel to the x- and, y-axes for the function x 3 − x y 2 + 2x − 9 =0:, Equating the coefficient of the highest power of x term, to zero gives 1 = 0 which is not an equation of a line., Hence there is no asymptote parallel with the x-axis., Equating the coefficient of the highest power of y term, to zero gives −x = 0 from which, x = 0., Hence x = 0, y = x and y = − x are asymptotes for the, function x3 − xy2 + 2x − 9 =0., Problem 12. Find the asymptotes for the function, x2 + 1, y=, and sketch a graph of the function., x, x2 + 1, gives yx = x 2 + 1., Rearranging y =, x, Equating the coefficient of the highest power x term to, zero gives 1 =0, hence there is no asymptote parallel to, the x-axis., , 1, Hence 1 = 2 and x 2 = 1, from which, x = ±1., x, When x = 1,, y=, , x2 + 1 1 + 1, =, =2, x, 1, , and when x = −1,, y=, , (−1)2 + 1, = −2, −1, , i.e. (1, 2) and (−1, −2) are the co-ordinates of the turning, d2 y, 2, d2 y, points. 2 = 2x −3 = 3 ; when x = 1, 2 is positive,, dx, x, dx, which indicates a minimum point and when x = −1,, d2 y, is negative, which indicates a maximum point, as, dx 2, shown in Fig. 18.35., Now try the following exercise, , Exercise 80 Further problems on, asymptotes, , Equating the coefficient of the highest power y term to, zero gives x = 0., , In Problems 1 to 3, determine the asymptotes, parallel to the x- and y-axes., , Hence there is an asymptote at x = 0 (i.e. the, y-axis)., , 1., , To determine any other asymptotes we substitute, y = mx + c into yx = x 2 + 1 which gives, , 2., , (mx + c)x = x 2 + 1, , 3., , x −2, x +1, x, y2 =, x −3, y=, , y=, , mx 2 + cx = x 2 + 1, , i.e., , and (m − 1)x 2 + cx − 1 = 0, , [y = 1, x = −1], [x = 3, y = 1 and y = −1], , x(x + 3), (x + 2)(x + 1), [x = −1, x = −2 and y = 1], , In Problems 4 and 5, determine all the asymptotes., , Equating the coefficient of the highest power x term to, zero gives m − 1 = 0, from which m = 1., Equating the coefficient of the next highest power x term, to zero gives c = 0. Hence y = mx + c = 1x + 0, i.e. y = x, is an asymptote., , 4. 8x − 10 + x 3 − x y 2 = 0, [x = 0, y = x and y = −x], , x2 + 1, is shown in Fig. 18.35., A sketch of y =, x, It is possible to determine maximum/minimum points, on the graph (see Chapter 28)., , In Problems 6 and 7, determine the asymptotes and, sketch the curves., , Since, then, , y=, , x2 + 1, x, , dy, 1, = 1 − x −2 = 1 − 2 = 0, dx, x, , for a turning point., , 5., , 6., , x2, , 1, =, + = x + x −1, x, x, , 195, , 7., , x 2 (y 2 − 16) = y, , y=, , x2 − x − 4, x +1, , [y = 4, y = −4 and x = 0], , , , x = −1, y = x − 2,, see Fig 18.40, page 202, , x y 2 − x 2 y + 2x − y = 5, , x = 0, y = 0, y = x,, see Fig. 18.41, page 202
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196 Higher Engineering Mathematics, y, , 5, , x, , 6, , y5, 4, , y, , x 211, x, , 2, , 24, , 22, , 2, , 0, , 4, , x, , 22, , y5, , x 211, x, , 24, , 26, , Figure 18.35, , 18.8, , Brief guide to curve sketching, , The following steps will give information from which, the graphs of many types of functions y = f (x) can be, sketched., (i) Use calculus to determine the location and nature, of maximum and minimum points (see Chapter 28), (ii) Determine where the curve cuts the x- and y-axes, (iii) Inspect the equation for symmetry., , (a), , If the equation is unchanged when −x is substituted for x, the graph will be symmetrical about, the y-axis (i.e. it is an even function)., , (b) If the equation is unchanged when −y is substituted for y, the graph will be symmetrical about, the x-axis., (c), , If f (−x) = − f (x), the graph is symmetrical about the origin (i.e. it is an odd, function)., , (iv) Check for any asymptotes.
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197, , Functions and their curves, y, , 18.9 Worked problems on curve, sketching, , 20, 15, , Problem 13. Sketch the graphs of, (a) y = 2x 2 + 12x + 20, , 10, , y 5 2x 2 1 12x 1 20, , (b) y = −3x 2 + 12x − 15, , 5, 2, , (a), , y = 2x 2 + 12x + 20 is a parabola since the equation is a quadratic. To determine the turning, point:, Gradient =, , 24, , 23, , 22, , 21, , 0, 23, 25, , 1, , 2, , 3, , x, , 210, , dy, = 4x + 12 = 0 for a turning point., dx, , y 5 23x 2 1 12x 2 15, 215, , Hence 4x = −12 and x = −3., , 220, , When x = −3, y = 2(−3)2 + 12(−3) + 20 =2., , 225, , Hence (−3, 2) are the co-ordinates of the turning, point, Figure 18.36, , d2 y, = 4, which is positive, hence (−3, 2) is a, dx 2, minimum point., When x = 0, y = 20, hence the curve cuts the, y-axis at y = 20., Thus knowing the curve passes through (−3, 2), and (0, 20) and appreciating the general shape, of a parabola results in the sketch given in, Fig. 18.36., (b), , Problem 14. Sketch the curves depicting the, following equations:, , (a) x = 9 − y 2 (b) y 2 = 16x, (c) x y = 5, (a), , y = −3x 2 + 12x − 15 is also a parabola (but, ‘upside down’ due to the minus sign in front of, the x 2 term)., Gradient =, , dy, = −6x + 12 = 0 for a turning point., dx, , Hence 6x = 12 and x = 2., When x = 2, y = −3(2)2 + 12(2) − 15 =−3., Hence (2, −3) are the co-ordinates of the turning, point, d2 y, = −6, which is negative, hence (2, −3) is a, dx 2, maximum point., When x = 0, y = −15, hence the curve cuts the axis, at y = −15., The curve is shown sketched in Fig. 18.36., , Squaring both sides of the equation and transposing gives x 2 + y 2 = 9. Comparing this with, the standard equation of a circle, centre origin and radius a, i.e. x 2 + y 2 = a 2, shows that, x 2 + y 2 = 9 represents a circle, centre origin and, radius 3. A sketch of this circle is shown in, Fig. 18.37(a)., , (b) The equation y 2 = 16x is symmetrical about the, x-axis and having its vertex at the origin (0, 0)., Also, when x = 1, y = ±4. A sketch of this, parabola is shown in Fig. 18.37(b)., (c), , a, represents a rectangular, The equation y =, x, hyperbola lying entirely within the first and third, 5, quadrants. Transposing x y = 5 gives y = , and, x, therefore represents the rectangular hyperbola, shown in Fig. 18.37(c).
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198 Higher Engineering Mathematics, y, , with the x- and y-axes of a rectangular co-ordinate, system, the major axis being 2(3), i.e. 6 units long, and the minor axis 2(2), i.e. 4 units long, as shown, in Fig. 18.38(a)., , 3, x, , y, , 4, (a), , x, , x 5 !(92y 2), 6, , y, 14, , (a) 4x 2 5 36 29y 2, y, , 1, , x, x, , 24, , 2Œ„3, (b) y 2 516x, , (b) 3y 2 11555x 2, , y, , Figure 18.38, , x, , (c) xy 5 5, , Figure 18.37, , Problem 15. Sketch the curves depicting the, following equations:, (a) 4x 2 = 36 −9y 2 (b) 3y 2 + 15 =5x 2, (a) By dividing throughout by 36 and transposing,, the equation 4x 2 = 36 − 9y 2 can be written as, x 2 y2, + = 1. The equation of an ellipse is of the, 9, 4, x 2 y2, form 2 + 2 = 1, where 2a and 2b represent the, a, b, x 2 y2, length of the axes of the ellipse. Thus 2 + 2 = 1, 3, 2, represents an ellipse, having its axes coinciding, , (b) Dividing 3y 2 + 15 = 5x 2 throughout by 15 and, x 2 y2, transposing gives, − = 1. The equation, 3, 5, 2, 2, y, x, − = 1 represents a hyperbola which is syma 2 b2, metrical about both the x- and y-axes, the distance, between the vertices being given by 2a., x 2 y2, − = 1 is as shown in, Thus a sketch of, 3, 5, √, Fig. 18.38(b), having a distance of 2 3 between, its vertices., Problem 16. Describe the shape of the curves, represented by the following equations:, , y 2, y2, = 2x, (b), (a) x = 2 1 −, 2, 8, 1/2, , x2, (c) y = 6 1 −, 16, , y 2, (a) Squaring the equation gives, 1−, 2, and transposing gives x 2 = 4 − y 2 , i.e., x2 =4
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199, , Functions and their curves, x 2 + y 2 = 4. Comparing this equation with, x 2 + y 2 = a 2 shows that x 2 + y 2 = 4 is the equation of a circle having centre at the origin (0, 0), and of radius 2 units., (b) Transposing, y2, , (c), , y2, = 2x, 8, , gives, , √, y = 4 x. Thus, , = 2x is the equation of a parabola having its, 8, axis of symmetry coinciding with the x-axis and, its vertex at the origin of a rectangular co-ordinate, system., 1/2, , y, x2, can be transposed to, =, y =6 1 −, 16, 6, 1/2, , x2, and squaring both sides gives, 1−, 16, y2, x2, x 2 y2, = 1 − , i.e., +, = 1., 36, 16, 16 36, This is the equation of an ellipse, centre at the origin of a rectangular co-ordinate system, the major, √, axis coinciding with the y-axis and being 2 36,, i.e. 12 units long. √, The minor axis coincides with, the x-axis and is 2 16, i.e. 8 units long., , Now try the following exercise, Exercise 81, sketching, , 1. Sketch the graphs of (a) y = 3x 2 + 9x +, , (a), , , y 2, x, Since =, 1+, 5, 2, y 2, x2, =1+, 25, 2, x 2 y2, i.e., −, =1, 25, 4, This is a hyperbola which is symmetrical about, √, both the x- and y-axes, the vertices being 2 25,, i.e. 10 units apart., (With reference to Section 18.1 (vii), a is equal, to ±5), , y, 15, a, (b) The equation =, is of the form y = , a =, 4 2x, x, 60, = 30., 2, This represents a rectangular hyperbola, symmetrical about both the x- and y-axis, and lying, entirely in the first and third quadrants, similar in, shape to the curves shown in Fig. 18.9., , 7, 4, , (b) y = −5x 2 + 20x + 50., ⎤, ⎡, (a) Parabola with minimum, , ⎥, ⎢, value at − 32 , −5 and , ⎥, ⎢, 3, ⎢, passing through 0, 1 4 . ⎥, ⎥, ⎢, ⎥, ⎢, ⎢(b) Parabola with maximum ⎥, ⎥, ⎢, ⎣, value at (2, 70) and passing⎦, through (0, 50)., In Problems 2 to 8, sketch the curves depicting the, equations given., 2., , x =4, , , y 2, 1−, 4, [circle, centre (0, 0), radius 4 units], , 3., Problem 17. Describe the shape of the curves, represented by the following equations:, , y 2, x, 15, y, (a) =, 1+, (b) =, 5, 2, 4 2x, , Further problems on curve, , 4., , 5., , 6., , 7., , √, , y, x=, 9, , y2 =, , , , parabola, symmetrical about, x-axis, vertex at (0, 0), , x 2 − 16, 4, ⎡, ⎤, hyperbola, symmetrical about, ⎢x- and y-axes, distance, ⎥, ⎢, ⎥, ⎣between vertices 8 units along ⎦, x-axis, , x2, y2, = 5−, 5, 2, ⎤, ⎡, ellipse, centre (0, 0), major axis, ⎣10 units along y-axis, minor axis⎦, √, 2 10 units along x-axis, , x = 3 1 + y2, ⎡, ⎤, hyperbola, symmetrical about, ⎢x- and y-axes, distance, ⎥, ⎢, ⎥, ⎣between vertices 6 units along ⎦, x-axis, x 2 y2 = 9, , , rectangular hyperbola, lying in, first and third quadrants only
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200 Higher Engineering Mathematics, 8., , 9., , , x = 13 (36 − 18y 2 ), ⎡, ⎤, ellipse, centre (0, 0),, ⎢major axis 4 units along x-axis,⎥, ⎢, ⎥, √, ⎣minor axis 2 2 units, ⎦, along y-axis, , ⎡, ⎤, hyperbola, symmetrical about x⎣and y-axes, vertices 2 units, ⎦, apart along x-axis, , Sketch the circle given by the equation, x 2 + y 2 − 4x + 10y + 25 =0., , 12., , √, y = 9 − x2, [circle, centre (0, 0), radius 3 units], , 13., , y = 7x −1, , 14., , y = (3x)1/2, , parabola, vertex at (0, 0), symmetrical about the x-axis, , 15., , y 2 − 8 =−2x 2, ⎡, ⎤, ellipse,, √ centre (0, 0), major, ⎢axis 2 8 units along the ⎥, ⎢, ⎥, ⎣ y-axis, minor axis 4 units ⎦, along the x-axis, , [Centre at (2, −5), radius 2], In Problems 10 to 15 describe the shape of the, curves represented by the equations given., 10., , 11., , , y = [3(1 − x 2 )], ⎡, ⎤, ellipse,, centre (0, 0), major axis, √, ⎣2 3 units along y-axis, minor ⎦, axis 2 units along x-axis, , y = [3(x 2 − 1)], , ⎡, ⎤, rectangular hyperbola, lying, ⎢in first and third quadrants, ⎥, ⎢, ⎥, ⎣symmetrical about x- and ⎦, y-axes, , Graphical solutions to Exercise 77, page 186, 1., , 2., , y, 10, , y, 4, , 5, , 2, , y 5 3x 25, , 0, , 1, , 2, , 3, , 0, , x, , 1, , 2, , 22, , 25, 3., , 4., , y, , 3, , x, , y 5 23x 14, , y, , 8, , 8, , y 5(x 23)2, , y 5 x 213, , 6, , 4, , 4, 2, 22, , Figure 18.39, , 21, , 0, , 1, , 2, , x, , 0, , 2, , 4, , 6, , x
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201, , Functions and their curves, 5., , 6., y, , y, 0.50, , 15, , y 5x 2x 2, , 0.25, , 10, , y 5(x24) 212, , 0, , 1, , x, , 5, , 2, , 0, , 4, , 6, , 8, , x, , 7., , 8., y, , y, 10, , y 5 11 2 cos 3x, , 3, 2, , y 5x 312, , 5, , 1, 22, , 21, , 0, , 2, , 1, , x, , , , 2, , 0, , 25, , 21, , 3, 2, , 210, 10., y, 9., , 3, , y, 6, , y 5 3 2 2 sin(x 1, , , ), 4, , 2, , y 5 2 ln x, , 4, 1, , 2, , 0, , p, 2, , p, , 3p, 2, , 2p, , x, , 0, 21, 22, , Figure 18.39 (Continued), , 1, , 2, , 3, , 4 x, , 2, , x
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202 Higher Engineering Mathematics, Graphical solutions to Problems 6 and 7, Exercise 80, page 195, y, 6, , 2, 2, , x 521, , y5, x, , 4, , 2, , 26, , 24, , 22, , x 2 2x2 4, y5, x 11, , 0, , 6 x, , 4, , 2, , x 2 2x 24, y5, x 11, , 22, 24, 26, , Figure 18.40, y, , xy 2 2 x 2y 1 2x 2y 5 5, 6, , y5, , x, , 4, , 2, , 26, , 24, , xy 2 2 x 2y 1 2x 2y 5 5, , 22, , 0, , 22, , 24, , 26, , Figure 18.41, , 2, , 4, , 6, , xy 2 2 x 2y 1 2x 2y 5 5, , x
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Chapter 19, , Irregular areas, volumes and, mean values of waveforms, 19.1, , Areas of irregular figures, , Areas of irregular plane surfaces may be approximately, determined by using (a) a planimeter, (b) the trapezoidal, rule, (c) the mid-ordinate rule, and (d) Simpson’s rule., Such methods may be used, for example, by engineers, estimating areas of indicator diagrams of steam engines,, surveyors estimating areas of plots of land or naval, architects estimating areas of water planes or transverse, sections of ships., (a), , A planimeter is an instrument for directly measuring small areas bounded by an irregular curve., , (iii) Areas PQRS, , y1 + y7, =d, + y2 + y3 + y4 + y5 + y6, 2, In general, the trapezoidal rule states:, Area =, ⎡ ⎛, ⎞, ⎤, , , first +, sum of, width of ⎣ 1 ⎝, ⎠ + remaining⎦, last, interval, 2 ordinate, ordinates, (c) Mid-ordinate rule, To determine the area ABCD of Fig. 19.2:, , (b) Trapezoidal rule, To determine the areas PQRS in Fig. 19.1:, , B, Q, y1, , y2, , y3, , y4, , y5, , y6, , R, y7, , C, y1, , y2, , y3, , y4, , y5, , y6, , d, , d, , d, , d, , d, , d, , D, , A, S, , P, d, , d, , d, , d, , d, , d, , Figure 19.2, , Figure 19.1, , (i) Divide base PSinto any number of equal intervals, each of width d (the greater the number, of intervals, the greater the accuracy)., (ii) Accurately measure ordinates y1 , y2 , y3 , etc., , (i) Divide base AD into any number of equal, intervals, each of width d (the greater the, number of intervals, the greater the accuracy)., (ii) Erect ordinates in the middle of each interval, (shown by broken lines in Fig. 19.2).
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204 Higher Engineering Mathematics, (iii) Accurately measure ordinates y1 , y2 , y3 , etc., (iv) Area ABCD = d(y1 + y2+ y3 + y4 + y5+ y6 ), , Area =, , , width of, sum of, interval, , 25, Speed (m/s), , In general, the mid-ordinate rule states:, , , mid-ordinates, , Graph of speed/time, , 30, , 20, 15, 10, , 0, , 1, , 2, 3, 4, Time (seconds), , 5, , 24.0, , 20.25, , 17.5, , 15.0, , 12.5, , 8.75, , 7.0, , 5.5, , 2.5, , 4.0, , (i) Divide base PS into an even number of intervals, each of width d (the greater the number, of intervals, the greater the accuracy)., , 1.25, , 5, , To determine the area PQRS of Fig. 19.1:, , 10.75, , (d) Simpson’s rule, , 6, , Figure 19.3, , (ii) Accurately measure ordinates y1 , y2 , y3, etc., (iii) Area PQRS =, , d, [(y1 + y7 ) + 4(y2 + y4 +, 3, y6 ) + 2(y3 + y5 )], , , , , first + last, 1 width of, Area =, 3 interval, ordinate, , +4, , 0 1, , 2, , sum of even, , , , ordinates, , , , sum of remaining, , 3, , 5, , = 58.75 m, (b) Mid-ordinate rule (see para. (c) above), The time base is divided into 6 strips each of width, 1 second., Mid-ordinates are erected as shown in Fig. 19.3 by, the broken lines. The length of each mid-ordinate, is measured. Thus, , odd ordinates, , 4, , , 0 + 24.0, + 2.5 + 5.5, 2, + 8.75 + 12.5 + 17.5, , Problem 1. A car starts from rest and its speed is, measured every second for 6 s:, Time t (s), , , area = (1), , In general, Simpson’s rule states:, , +2, , Thus, , area = (1)[1.25 + 4.0 + 7.0 + 10.75, + 15.0 + 20.25], , 6, , Speed v (m/s) 0 2.5 5.5 8.75 12.5 17.5 24.0, , = 58.25 m, , Determine the distance travelled in 6 seconds (i.e., the area under the v/t graph), by (a) the trapezoidal, rule, (b) the mid-ordinate rule, and (c) Simpson’s, rule., , (c) Simpson’s rule (see para. (d) above), , A graph of speed/time is shown in Fig. 19.3., , The time base is divided into 6 strips each of, width 1 s, and the length of the ordinates measured., Thus, area = 13 (1)[(0 + 24.0) + 4(2.5 + 8.75, + 17.5) + 2(5.5 + 12.5)], , (a) Trapezoidal rule (see para. (b) above), The time base is divided into 6 strips each of, width 1 s, and the length of the ordinates measured., , = 58.33 m
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Irregular areas, volumes and mean values of waveforms, , 205, , Problem 2. A river is 15 m wide. Soundings of, the depth are made at equal intervals of 3 m across, the river and are as shown below., Depth (m) 0, , 2.2 3.3, , 4.5 4.2 2.4, , 0, 140 160 200 190 180 130, , Calculate the cross-sectional area of the flow of, water at this point using Simpson’s rule., , 50, , From para. (d) above,, , = (1)[0 + 36.4 + 15] = 51.4 m2, , Width (m), , 50, , 50, , 0 2.8 5.2 6.5 5.8 4.1 3.0 2.3, , [143 m2 ], , Estimate the area of the deck., , Exercise 82 Further problems on areas of, irregular figures, , 2. Plot the graph of y = 2x 2 + 3 between x = 0, and x = 4. Estimate the area enclosed by the, curve, the ordinates x = 0 and x = 4, and the, x-axis by an approximate method., [54.7 square units], , 50, , 5. The deck of a ship is 35 m long. At equal, intervals of 5 m the width is given by the, following table:, , Now try the following exercise, , 1. Plot a graph of y = 3x − x 2 by completing, a table of values of y from x = 0 to x = 3., Determine the area enclosed by the curve, the, x-axis and ordinate x = 0 and x = 3 by (a) the, trapezoidal rule, (b) the mid-ordinate rule and, (c) by Simpson’s rule., [4.5 square units], , 50, , Figure 19.4, , Area = 13 (3)[(0 + 0) + 4(2.2 + 4.5 + 2.4), + 2(3.3 + 4.2)], , 50, , 19.2, , Volumes of irregular solids, , If the cross-sectional areas A1 , A2 , A3 , . . . of an irregular, solid bounded by two parallel planes are known at equal, intervals of width d (as shown in Fig. 19.5), then by, Simpson’s rule:, volume, V =, , d, [(A1 + A7 ) + 4(A2 + A4, 3, + A6) + 2 (A3 + A5)], , 3. The velocity of a car at one second intervals is, given in the following table:, time t (s) 0 1, velocity, v (m/s), , 2, , 3, , 4, , 5, , 6, , A1, , A2, , A3, , A4, , A5, , A6, , A7, , 0 2.0 4.5 8.0 14.0 21.0 29.0, , Determine the distance travelled in 6 seconds, (i.e. the area under the v/t graph) using, Simpson’s rule., [63.33 m], 4. The shape of a piece of land is shown in, Fig. 19.4. To estimate the area of the land,, a surveyor takes measurements at intervals, of 50 m, perpendicular to the straight portion, with the results shown (the dimensions being, in metres). Estimate the area of the land in, [4.70 ha], hectares (1 ha = 104 m2 )., , d, , d, , d, , d, , d, , d, , Figure 19.5, , Problem 3. A tree trunk is 12 m in length and has, a varying cross-section. The cross-sectional areas at, intervals of 2 m measured from one end are:, 0.52, 0.55, 0.59, 0.63, 0.72, 0.84, 0.97 m2, Estimate the volume of the tree trunk.
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206 Higher Engineering Mathematics, A sketch of the tree trunk is similar to that shown, in Fig. 19.5 above, where d = 2 m, A1 = 0.52 m2 ,, A2 = 0.55 m2 , and so on., Using Simpson’s rule for volumes gives:, Volume =, , 2, 3 [(0.52 + 0.97) + 4(0.55 + 0.63, , + 0.84) + 2(0.59 + 0.72)], = 23 [1.49 + 8.08 + 2.62] = 8.13 m3, , 1.76, 2.78, 3.10, 3.12, 2.61, 1.24, 0.85 m2, Determine the underwater volume if the, sections are 3 m apart., [42.59 m3 ], 2. To estimate the amount of earth to be removed, when constructing a cutting the crosssectional area at intervals of 8 m were estimated as follows:, 0, 2.8,, , Problem 4. The areas of seven horizontal, cross-sections of a water reservoir at intervals of, 10 m are:, 210, 250, 320, 350, 290, 230, 170 m2, Calculate the capacity of the reservoir in litres., Using Simpson’s rule for volumes gives:, , 3.7,, , 4.5,, , 4.1,, , 2.6,, , Estimate the volume of earth to be excavated., [147 m3], 3. The circumference of a 12 m long log of timber, of varying circular cross-section is measured, at intervals of 2 m along its length and the, results are:, Distance from, one end (m), , Circumference, (m), , 0, , 2.80, , 2, , 3.25, , 4, , 3.94, , 6, , 4.32, , = 16400 m3, , 8, , 5.16, , 16400 m3 = 16400 × 106 cm3 and since, 1 litre = 1000 cm3 ,, , 10, , 5.82, , 12, , 6.36, , Volume =, , 10, [(210 + 170) + 4(250 + 350, 3, + 230) + 2(320 + 290)], , =, , 10, [380 + 3320 + 1220], 3, , capacity of reservoir =, , 16400 × 106, litres, 1000, , 0 m3, , Estimate the volume of the timber in cubic, metres., [20.42 m3 ], , = 1 6400000, = 1.64 × 107 litres, Now try the following exercise, Exercise 83, Further problems on volumes, of irregular solids, 1. The areas of equidistantly spaced sections of, the underwater form of a small boat are as, follows:, , 19.3 The mean or average value of a, waveform, The mean or average value, y, of the waveform shown, in Fig. 19.6 is given by:, , y=, , area under curve, length of base, b
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Irregular areas, volumes and mean values of waveforms, , 207, , (iv) of a half-wave rectified waveform (see, Fig. 19.7(c)) is 0.318 × maximum value, or, (1/π) maximum value., Problem 5. Determine the average values over, half a cycle of the periodic waveforms shown in, Fig. 19.8., , y1 y2, d, , d, , y3, , y4, , y5, , y6, , y7, , d, , d, , d, , d, , d, , Voltage (V), , y, , b, , Figure 19.6, , 20, , 0, , If the mid-ordinate rule is used to find the area under the, curve, then:, , 1, , 2, , 3, , t (ms), , 4, , 210, , y=, , sum of mid-ordinates, number of mid-ordinates, , y1 + y2 + y3 + y4 + y5 + y6 + y7, =, 7, , Current (A), , (a), 3, 2, 1, 0, 21, 22, 23, , , , for Fig. 19.6, , 1, , 2, , 3, , 4, , 5 6, , t (s), , (b), Voltage (V), , For a sine wave, the mean or average value:, (i) over one complete cycle is zero (see Fig. 19.7(a)),, , V, Vm, , 10, , 0, , V, Vm, , 2, , 4, , 6, , 8, , t (ms), , 210, t, , 0, , (c), , t, , 0, , Figure 19.8, (a), , (b), , (a), , V, Vm, , Area under triangular waveform (a) for a half cycle, is given by:, Area =, , t, , 0, , (c), , Figure 19.7, , (ii) over half a cycle is 0.637 × maximum value, or, (2/π ) × maximum value,, (iii) of a full-wave rectified waveform (see Fig., 19.7(b)) is 0.637 × maximum value,, , 1, 2, , (base) (perpendicular height), , = 12 (2 × 10−3)(20), = 20 × 10−3 Vs, Average value of waveform, =, , area under curve, length of base, , =, , 20 × 10−3 Vs, 2 × 10−3 s, , = 10 V
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208 Higher Engineering Mathematics, (b) Area under waveform (b) for a half, cycle = (1 × 1) + (3 × 2) = 7 As., , (a) One cycle of the trapezoidal waveform (a) is, completed in 10 ms (i.e. the periodic time is, 10 ms)., , Average value of waveform, , Area under curve = area of trapezium, , area under curve, =, length of base, , =, , 1, 2, , (sum of parallel sides) (perpendicular, , distance between parallel sides), 7 As, =, 3s, , = 12 {(4 + 8) × 10−3}(5 × 10−3 ), = 30 × 10−6 As, , = 2.33 A, , Mean value over one cycle, , (c) A half cycle of the voltage waveform (c) is, completed in 4 ms., , =, , area under curve 30 × 10−6 As, =, length of base, 10 × 10−3 s, , = 3 mA, , Area under curve = 12 {(3 − 1)10−3 }(10), = 10 × 10−3 Vs, , (b) One cycle of the sawtooth waveform (b) is completed in 5 ms., , Average value of waveform, =, , area under curve, length of base, , =, , 10 × 10−3 Vs, 4 × 10−3 s, , Area under curve = 12 (3 × 10−3)(2), = 3 × 10−3 As, Mean value over one cycle, =, , = 2.5 V, , area under curve 3 × 10−3 As, =, length of base, 5 × 10−3 s, , = 0.6 A, , Current (mA), , Problem 6. Determine the mean value of current, over one complete cycle of the periodic waveforms, shown in Fig. 19.9., , 5, , 0, , 4, , 8, , 12, , 16, , 20, , 24, , 28 t (ms), , (a), Current (mA), , Problem 7. The power used in a manufacturing, process during a 6 hour period is recorded at, intervals of 1 hour as shown below., Time (h), , 0, , 1, , 2, , 3, , 4, , 5, , 6, , Power (kW), , 0, , 14, , 29, , 51, , 45, , 23, , 0, , Plot a graph of power against time and, by using the, mid-ordinate rule, determine (a) the area under the, curve and (b) the average value of the power., , 2, , The graph of power/time is shown in Fig. 19.10., (a), 0, , 2, , 4, , 6, , 8, (b), , Figure 19.9, , 10, , 12, , t (ms), , The time base is divided into 6 equal intervals, each of width 1 hour. Mid-ordinates are, erected (shown by broken lines in Fig. 19.10), and measured. The values are shown in, Fig. 19.10.
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Irregular areas, volumes and mean values of waveforms, , One cycle of the output voltage is completed in π radians, or 180◦ . The base is divided into 6 intervals, each of, width 30◦ . The mid-ordinate of each interval will lie at, 15◦, 45◦ , 75◦ , etc., At 15◦ the height of the mid-ordinate is, 10 sin 15◦ = 2.588 V., At 45◦ the height of the mid-ordinate is, 10 sin 45◦ = 7.071 V, and so on., The results are tabulated below:, , Graph of power/time, 50, 40, , Power (kW), , 209, , 30, 20, 10, 7.0, 0, , 21.5, 1, , 42.0, 2, , 49.5, , 37.0 10.0, , 3, 4, Time (hours), , 5, , 6, , Figure 19.10, , Area under curve = (width of interval), × (sum of mid-ordinates), = (1)[7.0 + 21.5 + 42.0, + 49.5 + 37.0 + 10.0], = 167 kWh (i.e. a measure, of electrical energy), , Mid-ordinate, , Height of mid-ordinate, , 15◦, , 10 sin 15◦ = 2.588 V, , 45◦, , 10 sin 45◦ = 7.071 V, , 75◦, , 10 sin 75◦ = 9.659 V, , 105◦, , 10 sin 105◦ = 9.659 V, , 135◦, , 10 sin 135◦ = 7.071 V, , 165◦, , 10 sin 165◦ = 2.588 V, sum of mid-ordinates =38.636 V, , Mean or average value of output voltage, , (b) Average value of waveform, , sum of mid-ordinates, number of mid-ordinates, 38.636, =, 6, = 6.439 V, =, , =, , area under curve, length of base, , =, , 167 kWh, = 27.83 kW, 6h, , (With a larger number of intervals a more accurate, answer may be obtained.) For a sine wave the actual, mean value is 0.637 ×maximum value, which in this, problem gives 6.37 V., , Alternatively, average value, =, , sum of mid-ordinates, number of mid-ordinates, , Voltage (V), , Problem 8. Fig. 19.11 shows a sinusoidal output, voltage of a full-wave rectifier. Determine, using, the mid-ordinate rule with 6 intervals, the mean, output voltage., , 10, , 0, , 308608908, , 2, , Figure 19.11, , 1808, , , 2708, 3, 2, , 3608, 2, , , , Problem 9. An indicator diagram for a steam, engine is shown in Fig. 19.12. The base line has, been divided into 6 equally spaced intervals and the, lengths of the 7 ordinates measured with the results, shown in centimetres. Determine (a) the area of the, indicator diagram using Simpson’s rule, and (b) the, mean pressure in the cylinder given that 1 cm, represents 100 kPa., , 3.6, , 4.0, , 3.5, , 2.9, , 12.0 cm, , Figure 19.12, , 2.2, , 1.7, , 1.6
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210 Higher Engineering Mathematics, , area =, , 12.0, cm. Using, 6, , Current (A), , (a) The width of each interval is, Simpson’s rule,, , 1, 3 (2.0)[(3.6 + 1.6) + 4(4.0, , 0, , + 2.9 + 1.7) + 2(3.5 + 2.2)], , Figure 19.13 (Continued ), , (b) Mean height of ordinates, =, , 30 t (ms), , (c), , area of diagram 34, =, length of base, 12, , = 2.83 cm, Since 1 cm represents 100 kPa, the mean pressure, in the cylinder, = 2.83 cm × 100 kPa/cm = 283 kPa., , 2. Find the average value of the periodic waveforms shown in Fig. 19.14 over one complete, cycle., [(a) 2.5 V (b) 3 A], Voltage (mV), , = 34 cm, , 15, , 25, , = 23 [5.2 + 34.4 + 11.4], 2, , 5, , 10, , 0, , 2, , 4, , 6, , 8, , 10, , t (ms), , 8, , 10, , t (ms), , Now try the following exercise, Exercise 84 Further problems on mean or, average values of waveforms, , Current (A), , (a), , 5, , 0, , Current (A), , 1. Determine the mean value of the periodic, waveforms shown in Fig. 19.13 over a half, cycle., [(a) 2 A (b) 50 V (c) 2.5 A], , 10, , 20, , t (ms), , Voltage (V), , (a), , (b), , Figure 19.14, , Time (ms), , 0 5, , 10, , 15, , 20, , 25, , 30, , Plot a graph of current against time and estimate the area under the curve over the 30 ms, period using the mid-ordinate rule and determine its mean value., [0.093 As, 3.1 A], , 100, , 5, , 10 t (ms), , 2100, (b), , Figure 19.13, , 6, , Current (A) 0 0.9 2.6 4.9 5.8 3.5 0, , 22, , 0, , 4, , 3. An alternating current has the following values, at equal intervals of 5 ms, , 2, , 0, , 2, , 4. Determine, using an approximate method, the, average value of a sine wave of maximum, value 50 V for (a) a half cycle and (b) a, complete cycle., [(a) 31.83 V (b) 0]
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Irregular areas, volumes and mean values of waveforms, , 5. An indicator diagram of a steam engine is, 12 cm long. Seven evenly spaced ordinates,, including the end ordinates, are measured as, follows:, 5.90, 5.52, 4.22, 3.63, 3.32, 3.24, 3.16 cm, , Determine the area of the diagram and the, mean pressure in the cylinder if 1 cm represents 90 kPa., [49.13 cm2 , 368.5 kPa], , 211
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Revision Test 6, This Revision Test covers the material contained in Chapters 18 and 19. The marks for each question are shown in, brackets at the end of each question., 1., , (a), , y = (x − 2)2, , (b), , y = 3 −cos 2x (d) 9x 2 − 4y 2 = 36, ⎧, π, ⎪, −1 −π ≤ x ≤ −, ⎪, ⎪, 2, ⎪, ⎪, ⎨, π, π, x − ≤x ≤, f (x) =, ⎪, 2, 2, ⎪, ⎪, ⎪, π, ⎪, ⎩ 1, ≤x ≤π, 2, , (e), , (c), , x 2 + y 2 − 2x + 4y − 4 = 0, , 2., , Determine the inverse of f (x) = 3x + 1, , 3., , Evaluate, correct to 3 decimal places:, 2 tan−1 1.64 + sec−1 2.43 − 3 cosec−1 3.85, , 4., , 6., , (x − 1)(x + 4), (x − 2)(x − 5), , A circular cooling tower is 20 m high. The inside, diameter of the tower at different heights is given, in the following table:, Height (m), , 0, , 5.0 10.0 15.0 20.0, , Diameter (m) 16.0 13.3 10.7, , (3), 7., (3), , (8), , Plot a graph of y = 3x 2 + 5 from x = 1 to x = 4., Estimate, correct to 2 decimal places, using 6 intervals, the area enclosed by the curve, the ordinates, , 8.6, , 8.0, , Determine the area corresponding to each diameter, and hence estimate the capacity of the tower in cubic, metres., (5), , (15), , Determine the asymptotes for the following, function and hence sketch the curve:, y=, , 5., , x = 1 and x = 4, and the x-axis by (a) the trapezoidal, rule, (b) the mid-ordinate rule, and (c) Simpson’s, rule., (11), , Sketch the following graphs, showing the relevant, points:, , A vehicle starts from rest and its velocity is, measured every second for 6 seconds, with the, following results:, Time t (s) 0, Velocity, v (m/s), , 1, , 2, , 0 1.2 2.4, , 3, , 4, , 3.7 5.2, , 5, , 6, , 6.0 9.2, , Using Simpson’s rule, calculate (a) the distance, travelled in 6 s (i.e. the area under the v/t graph), and (b) the average speed over this period., (5)
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Chapter 20, , Complex numbers, 20.1, , Cartesian complex numbers, , There are several applications of complex numbers, in science and engineering, in particular in electrical, alternating current theory and in mechanical vector, analysis., There are two main forms of complex number –, Cartesian form and polar form – and both are, explained in this chapter., If we can add, subtract, multiply and divide complex, numbers in both forms and represent the numbers on, an Argand diagram then a.c. theory and vector analysis, become considerably easier., (i) If the quadratic equation x 2 + 2x + 5 = 0 is, solved using the quadratic formula then,, , −2 ± [(2)2 − (4)(1)(5)], x=, 2(1), √, √, −2 ± [−16] −2 ± [(16)(−1)], =, =, 2, 2, √ √, √, −2 ± 16 −1 −2 ± 4 −1, =, =, 2, 2, √, = −1 ± 2 −1, √, It is not possible to evaluate −1 in real, terms., √ However, if an operator j is defined as, j = −1 then the solution may be expressed as, x = −1 ± j 2., (ii) −1 + j 2 and −1 − j 2 are known as complex, numbers. Both solutions are of the form a + jb,, ‘a’ being termed the real part and jb the, imaginary part. A complex number of the form, a + jb is called Cartesian complex number., , (iii) In pure √, mathematics the symbol i is used to, indicate −1 (i being the first letter of the word, imaginary). However i is the symbol of electric, current in engineering, and to avoid possible confusion the√, next letter in the alphabet, j , is used to, represent −1., Problem 1. Solve the quadratic equation, x 2 + 4 = 0., √, Since x 2 + 4 =0 then x 2 = −4 and x = −4., , , √, [(−1)(4)] = (−1) 4 = j (±2), √, = ± j2, (since j = −1), , i.e., x =, , (Note that ± j 2 may also be written ±2 j)., Problem 2. Solve the quadratic equation, 2x 2 + 3x + 5 = 0., Using the quadratic formula,, , −3 ± [(3)2 − 4(2)(5)], x=, 2(2), √, √, √, −3 ± −31 −3 ± (−1) 31, =, =, 4, 4, √, −3 ± j 31, =, 4, √, 3, 31, Hence x = − ± j, or −0.750 ± j1.392,, 4, 4, correct to 3 decimal places., (Note, a graph of y = 2x 2 + 3x + 5 does not cross, the x-axis and hence 2x 2 + 3x + 5 = 0 has no real, roots.)
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214 Higher Engineering Mathematics, Problem 3., (a), , j3, , (b), , Evaluate, j4, , j 23, , (c), , 20.2, , −4, (d) 9, j, , (a), , j 3 = j 2 × j = (−1) × j = − j, since j 2 = −1, , (b), , j 4 = j 2 × j 2 = (−1) × (−1) = 1, , (c), , j 23 = j × j 22 = j × ( j 2)11 = j × (−1)11, , (d), , j9=, , = j × (−1) = − j, j × j 8 = j × ( j 2)4 = j × (−1)4, = j ×1 = j, Hence, , The Argand diagram, , A complex number may be represented pictorially on, rectangular or cartesian axes. The horizontal (or x) axis is, used to represent the real axis and the vertical (or y) axis, is used to represent the imaginary axis. Such a diagram, is called an Argand diagram. In Fig. 20.1, the point A, represents the complex number (3 + j 2) and is obtained, by plotting the co-ordinates (3, j 2) as in graphical work., Figure20.1 also showstheArgand points B, C and D representing the complex numbers (−2 + j 4), (−3 − j 5), and (1 − j 3) respectively., , 4j, −4 −4 −4 − j, =, =, ×, =, j9, j, j, −j, −j2, 4j, =, = 4 j or j4, −(−1), , Imaginary, axis, B, , j4, j3, A, , j2, , Now try the following exercise, , j, , Exercise 85 Further problems on the, introduction to cartesian complex numbers, , 23, , 22 21 0, 2j, , [± j 5], , 2j 3, , x − 2x + 2 = 0, , [x = 1 ± j ], , 2j 4, , 3., , x 2 − 4x + 5 =0, , [x = 2 ± j ], , 4., , x 2 − 6x + 10 =0, , [x = 3 ± j ], , 5., , 2x 2 − 2x + 1 =0, , [x = 0.5 ± j 0.5], , 6., , x 2 − 4x + 8 =0, , 7., , 25x 2 − 10x + 2 = 0, , x 2 + 25 =0, , 2., , 2, , 3, , Real axis, , C, , D, , 2j 5, , Figure 20.1, , [x = 2 ± j 2], [x = 0.2 ± j 0.2], , 8. 2x 2 + 3x + 4 =0, , √, 23, 3, − ±j, or − 0.750 ± j 1.199, 4, 4, , 9. 4t 2 − 5t + 7 =0, , √, 87, 5, ±j, or 0.625 ± j 1.166, 8, 8, , 10. Evaluate (a) j 8, , 2, , 2j 2, , In Problems 1 to 9, solve the quadratic equations., 1., , 1, , 1, 4, (b) − 7 (c) 13, j, 2j, [(a) 1 (b) − j (c) − j 2], , 20.3 Addition and subtraction of, complex numbers, Two complex numbers are added/subtracted by adding/, subtracting separately the two real parts and the two, imaginary parts., For example, if Z 1 = a + jb and Z 2 = c + jd,, then, , Z 1 + Z 2 = (a + jb) + (c + j d), = (a + c) + j (b +d), , and, , Z 1 − Z 2 = (a + jb) − (c + j d), = (a − c) + j (b −d)
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Complex numbers, Thus, for example,, (2 + j 3) +(3 − j 4)= 2 + j 3 +3 − j 4, = 5 − j1, and (2 + j 3) −(3 − j 4)= 2 + j 3 −3 + j 4, = −1 + j7, The addition and subtraction of complex numbers may, be achieved graphically as shown in the Argand diagram, of Fig. 20.2. (2 + j 3) is represented by vector OP and, Imaginary, axis, , (3 − j 4) by vector OQ. In Fig. 20.2(a) by vector addition, (i.e. the diagonal of the parallelogram) OP + OQ = OR., R is the point (5, − j 1)., Hence (2 + j 3) +(3 − j 4) =5 − j1., In Fig. 20.2(b), vector OQ is reversed (shown as OQ ), since it is being subtracted. (Note OQ = 3 − j 4 and, OQ = −(3 − j 4) =−3 + j 4)., OP − OQ = OP + OQ = OS is found to be the Argand, point (−1, j 7)., Hence (2 + j 3) −(3 − j 4) =−1 + j 7, Problem 4. Given Z 1 = 2 + j 4 and Z 2 = 3 − j, determine (a) Z 1 + Z 2 , (b) Z 1 − Z 2 , (c) Z 2 − Z 1 and, show the results on an Argand diagram., , P (21j3), , j3, , 215, , j2, , (a) Z 1 + Z 2 = (2 + j 4) +(3 − j ), , j, 0, 2j, , 1, , 3, , 2, , 5 Real axis, R (5 2j ), , 4, , = (2 + 3) + j (4 −1) = 5 + j 3, (b) Z 1 − Z 2 = (2 + j 4) −(3 − j ), = (2 − 3) + j (4 −(−1)) = −1 + j 5, , 2j2, , (c) Z 2 − Z 1 = (3 − j ) −(2 + j 4), , 2j3, 2j4, , = (3 − 2) + j (−1 − 4) = 1 − j 5, , Q (3 2j 4), , Each result is shown in the Argand diagram of, Fig. 20.3., , (a), Imaginary, axis, S (211j7), , Imaginary, axis, , j7, (211 j 5), , j6, , j4, , j5, Q9, , j3, , j2, , P (21j3), , j, , j2, j, , 21 0, 2j, 1, , 2, , 3, , Real axis, , 1, , 2, , 2j 2, 2j 3, , 2j2, , 2j 4, , 2j3, , 2j 5, , Q (32j4), , 2j4, (b), , Figure 20.3, Figure 20.2, , ( 5 1j 3), , j3, , j4, , 23 22 21 0, 2j, , j5, , ( 12 j 5), , 3, , 4, , 5, , Real axis
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216 Higher Engineering Mathematics, 20.4 Multiplication and division of, complex numbers, , Problem 5. If Z 1 = 1 − j 3, Z 2 = −2 + j 5 and, Z 3 = −3 − j 4, determine in a + j b form:, , (i) Multiplication of complex numbers is achieved, by assuming all quantities involved are real and, then using j 2 = −1 to simplify., , (a) Z 1 Z 2, (c), , Hence (a + j b)(c + j d), = ac + a( j d) +( j b)c + ( j b)( j d), , Z1 Z2, Z1 + Z2, , (d) Z 1 Z 2 Z 3, , = −2 + j 5 + j 6 − j 215, = (−2 + 15) + j (5 + 6), since j 2 = −1,, , = (ac − bd) + j (ad + bc),, , = 13 + j11, , since j 2 = −1, (b), , = 12 − j 15 + j 8 − j 210, , Z1, 1 − j3, 1 − j3, −3 + j 4, =, =, ×, Z 3 −3 − j 4 −3 − j 4 −3 + j 4, , = (12 − (−10)) + j (−15 +8), = 22 − j 7, (ii) The complex conjugate of a complex number is obtained by changing the sign of the, imaginary part. Hence the complex conjugate, of a + j b is a − j b. The product of a complex, number and its complex conjugate is always a, real number., , Z1, Z3, , (a) Z 1 Z 2 = (1 − j 3)(−2 + j 5), , = ac + j ad + j bc + j 2bd, , Thus (3 + j 2)(4 − j 5), , (b), , =, , −3 + j 4 + j 9 − j 212, 32 + 42, , =, , 9 + j 13, 9, 13, =, + j, 25, 25, 25, or 0.36 + j0.52, , (c), , (1 − j 3)(−2 + j 5), Z1 Z2, =, Z 1 + Z 2 (1 − j 3) + (−2 + j 5), =, , 13 + j 11, , from part (a),, −1 + j 2, , =, , 13 + j 11 −1 − j 2, ×, −1 + j 2 −1 − j 2, , [(a + j b)(a − j b) may be evaluated ‘on sight’ as, a 2 + b2 ]., , =, , −13 − j 26 − j 11 − j 222, 12 + 22, , (iii) Division of complex numbers is achieved by, multiplying both numerator and denominator by, the complex conjugate of the denominator., , =, , 9 − j 37 9, 37, = −j, or 1.8 − j 7.4, 5, 5, 5, , For example,, (3 + j 4)(3 − j 4)= 9 − j 12 + j 12 − j 216, = 9 + 16 = 25, , For example,, , Z 1 Z 2 = 13 + j 11, from part (a), , 2 − j 5 2 − j 5 (3 − j 4), =, ×, 3 + j 4 3 + j 4 (3 − j 4), =, , (d) Z 1 Z 2 Z 3 = (13 + j 11)(−3 − j 4), since, , 6 − j 8 − j 15 + j 220, 32 + 42, , −14 − j 23 −14, 23, =, =, −j, 25, 25, 25, or −0.56 − j0.92, , = −39 − j 52 − j 33 − j 244, = (−39 + 44) − j (52 + 33), = 5 − j85, Problem 6., , Evaluate:, , , 1+ j3 2, 2, (b) j, (a), (1 + j )4, 1− j2
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Complex numbers, (a) (1 + j )2 = (1 + j )(1 + j ) =1 + j + j + j 2, , 4. (a) Z 1 + Z 2 − Z 3 (b) Z 2 − Z 1 + Z 4, , =1+ j + j −1= j2, (1 +, , j )4, , = [(1 +, , j )2]2 = (, , [(a) 7 − j 4 (b) −2 − j 6], , j 2)2 =, , j 24 = −4, , 5. (a) Z 1 Z 2 (b) Z 3 Z 4, [(a) 10 + j 5 (b) 13 − j 13], , 2, 2, 1, =, Hence, =−, 4, (1 + j ), −4, 2, (b), , =, , , 6. (a) Z 1 Z 3 + Z 4 (b) Z 1 Z 2 Z 3, [(a) −13 − j 2 (b) −35 + j 20], , 1 + j3 1 + j3 1 + j2, =, ×, 1 − j2 1 − j2 1 + j2, , 1+ j3, 1− j2, , Hence, , 2, , 1 + j2+ j3 +, 12 + 22, , j 26, , =, , 7. (a), , −5 + j 5, 5, , = −1 + j 1 = −1 + j, 8. (a), , = (−1 + j )2 = (−1 + j )(−1 + j ), = 1− j − j + j2 =− j2, , , 1+ j3 2, j, = j (− j 2) =− j 22 =2,, 1− j2, since j 2 = −1, , Now try the following exercise, , 1. Evaluate, (a), (3 + j 2) +(5 − j ) and, (b) (−2 + j 6) −(3 − j 2) and show the, results on an Argand diagram., [(a) 8 + j (b) −5 + j 8], 2. Write down the complex conjugates of, (a) 3 + j 4, (b) 2 − j ., [(a) 3 − j 4 (b) 2 + j ], 3. If z = 2 + j and w = 3 − j evaluate, (a) z + w (b) w − z (c) 3z − 2w (d), 5z + 2w (e) j (2w − 3z) (f ) 2 j w − j z, j 5 (d) 16 + j 3, , In Problems 4 to 8 evaluate in a + j b form, given Z 1 = 1 + j 2, Z 2 = 4 − j 3, Z 3 = −2 + j 3, and Z 4 = −5 − j ., , Z1, Z1 + Z3, (b), Z2, Z2 − Z4, , 11, −19, 43, −2, +j, (b), +j, (a), 25, 25, 85, 85, Z1 Z3, Z1, (b) Z 2 +, + Z3, Z1 + Z3, Z4, , 41, 45, 9, 3, + j, (b), − j, (a), 26, 26, 26, 26, , 1− j, 1, (b), 1+ j, 1+ j, , 1, 1, (a) − j (b) − j, 2, 2, , , −25 1 + j 2 2 − j 5, 10. Show that, −, 2, 3+ j4, −j, 9. Evaluate (a), , Exercise 86 Further problems on, operations involving Cartesian complex, numbers, , [(a) 5 (b) 1 − j 2 (c), (e) 5 (f ) 3 + j 4], , 217, , = 57 + j 24, , 20.5, , Complex equations, , If two complex numbers are equal, then their real parts, are equal and their imaginary parts are equal. Hence if, a + j b =c + j d, then a = c and b = d., Problem 7. Solve the complex equations:, (a) 2(x + j y) =6 − j 3, (b) (1 + j 2)(−2 − j 3) =a + j b, (a), , 2(x + j y) =6 − j 3 hence 2x + j 2y = 6 − j 3, Equating the real parts gives:, 2x = 6, i.e. x = 3, Equating the imaginary parts gives:, 2y = −3, i.e. y = − 32
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218 Higher Engineering Mathematics, (b) (1 + j 2)(−2 − j 3) =a + j b, −2 − j 3 − j 4 − j 26 = a + j b, , 2., , Hence 4 − j 7 =a + j b, , , , 2+ j, = j (x + j y), 1− j, , √, 3. (2 − j 3) = (a + j b), , Equating real and imaginary terms gives:, , 3, 1, x = , y =−, 2, 2, , [a = −5, b = −12], , a = 4 and b = −7, 4. (x − j 2y) −( y − j x) =2 + j, (a), , Solve the equations:, √, (2 − j 3) = (a + j b), , [x = 3, y = 1], , Problem 8., , 5. If Z = R + j ωL + 1/j ωC, express Z in, (a + j b) form when R = 10, L =5, C = 0.04, and ω = 4., [Z = 10 + j 13.75], , (b) (x − j 2y) +( y − j 3x) =2 + j 3, (a), , √, (2 − j 3) = (a + j b), (2 − j 3)2 = a + j b,, , Hence, i.e., , 20.6 The polar form of a complex, number, , (2 − j 3)(2 − j 3)= a + j b, , Hence 4 − j 6 − j 6 + j 29 = a + j b, , Thus a = −5 and b = −12, , (i) Let a complex number z be x + j y as shown in, the Argand diagram of Fig. 20.4. Let distance, OZ be r and the angle OZ makes with the positive, real axis be θ., , (b) (x − j 2y) +( y − j 3x) =2 + j 3, , From trigonometry, x = r cos θ and, , −5 − j 12= a + j b, , and, , Hence (x + y) + j (−2y − 3x) = 2 + j 3, , y = r sin θ, , Equating real and imaginary parts gives:, x+y=2, , Hence Z = x + j y = r cos θ + j r sin θ, (1), , and −3x − 2y = 3, , (2), , i.e. two simultaneous equations to solve., Multiplying equation (1) by 2 gives:, 2x + 2y = 4, , = r(cos θ + j sin θ), Z =r(cos θ + j sin θ) is usually abbreviated to, Z =r∠θ which is known as the polar form of, a complex number., , (3), , Imaginary, axis, , Adding equations (2) and (3) gives:, , Z, , −x = 7, i.e., x = −7, r, , From equation (1), y = 9, which may be checked, in equation (2)., , Now try the following exercise, Exercise 87, equations, , , O, , x, , A Real axis, , Figure 20.4, , Further problems on complex, , In Problems 1 to 4 solve the complex equations., 1. (2 + j )(3 − j 2) =a + j b, , jy, , [a = 8, b =−1], , (ii) r is called the modulus (or magnitude) of Z and, is written as mod Z or |Z |., r is determined using Pythagoras’ theorem on, triangle OAZ in Fig. 20.4,, , i.e., r = (x 2 + y 2 )
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Complex numbers, (iii) θ is called the argument (or amplitude) of Z and, is written as arg Z ., , Imaginary, axis, (23 1j4), , By trigonometry on triangle OAZ,, arg Z = θ = tan−1, , j3, r, , (iv) Whenever changing from cartesian form to polar, form, or vice-versa, a sketch is invaluable for, determining the quadrant in which the complex, number occurs., , 2j2, , 2, , 3, , Real axis, , r, , 2j4, , (3 2 j4), , (b) −3 + j 4 is shown in Fig. 20.6 and lies in the, second quadrant., Modulus, r = 5 and angle α = 53.13◦, from, part (a)., Argument =180◦ − 53.13◦ = 126.87◦ (i.e. the, argument must be measured from the positive real, axis)., , 2, , Hence −3 + j4 = 5∠126.87◦, , Real axis, , (c), Figure 20.5, , Hence the argument = 180◦ + 53.13◦ = 233.13◦,, which is the same as −126.87◦., , Argument, arg Z = θ = tan −1, , 3, 2, = 56.31◦ or, , Hence (−3 − j4) = 5∠233.13◦ or 5∠−126.87◦, , 56◦19, , (By convention the principal value is normally, used, i.e. the numerically least value, such that, −π < θ < π)., , In polar form, 2 + j 3 is written as 3.606∠56.31◦ ., Problem 10. Express the following complex, numbers in polar form:, , (d) 3 − j 4 is shown in Fig. 20.6 and lies in the fourth, quadrant., , (b) −3 + j 4, , Modulus, r = 5 and angle α = 53.13◦ , as above., Hence (3 − j4) = 5∠−53.13◦, , (c) −3 − j 4 (d) 3 − j 4, 3 + j 4 is shown in Fig. 20.6 and lies in the first, quadrant., , Modulus, r = (32 + 42 ) = 5 and argument, θ = tan −1 43 = 53.13◦., = 5∠53.13◦, , −3 − j 4 is shown in Fig. 20.6 and lies in the third, quadrant., Modulus, r = 5 and α = 53.13◦, as above., , , √, Modulus, |Z | =r = (22 + 32) = 13 or 3.606, correct, to 3 decimal places., , Hence 3 + j4, , , ␣1, , Figure 20.6, , , , (a), , j, ␣, 23 22 21 ␣, 2j, , (23 2 j4), , r, , (a) 3 + j 4, , r, , 2j3, , j3, , 0, , j2, , r, , Problem 9. Determine the modulus and argument, of the complex number Z = 2 + j 3, and express Z, in polar form., , Imaginary, axis, , (3 1j4), , j4, , y, x, , Z = 2 + j 3 lies in the first quadrant as shown in, Fig. 20.5., , 219, , Problem 11. Convert (a) 4∠30◦ (b) 7∠−145◦, into a + j b form, correct to 4 significant figures., (a), , 4∠30◦ is shown in Fig. 20.7(a) and lies in the first, quadrant.
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220 Higher Engineering Mathematics, Imaginary, axis, , Problem 12., (a), , 4, 308, 0, , (b) 3∠16◦ × 5∠−44◦ × 2∠80◦, , jy, Real axis, , x, , (a) 8∠25◦ ×4∠60◦ = (8 × 4)∠(25◦ +60◦) = 32∠85◦, (a), , (b) 3∠16◦ × 5∠ −44◦ × 2∠80◦, = (3 × 5 × 2)∠[16◦ + (−44◦ )+ 80◦ ] = 30∠52◦, , x, ␣, jy, , Real axis, 1458, , 7, , Problem 13., , Figure 20.7, , Using trigonometric ratios, x = 4 cos 30◦ = 3.464, and y = 4 sin 30◦ = 2.000., , (a), , Hence 4∠30◦ = 3.464 + j2.000, (b) 7∠145◦ is shown in Fig. 20.7(b) and lies in the, third quadrant., ◦, , Evaluate in polar form, , π, π, 10∠ × 12∠, 16∠75◦, 4, 2, (b), (a), π, 2∠15◦, 6∠−, 3, , (b), , ◦, , Angle α = 180 − 145 = 35, , ◦, , Hence x = 7 cos 35◦ = 5.734, and, , Determine, in polar form:, , 8∠25◦ × 4∠60◦, , y = 7 sin 35◦ = 4.015, , Hence 7∠−145◦ = −5.734 − j4.015, , (b), , 16∠75◦ 16, = ∠(75◦ − 15◦) = 8∠60◦, 2∠15◦, 2, π, π, × 12∠, , , , 4, 2 = 10 × 12 ∠ π + π − − π, π, 6, 4 2, 3, 6∠−, 3, 13π, 11π, = 20∠, or 20∠−, or, 12, 12, , 10∠, , 20∠195◦ or 20∠−165◦, , Alternatively, 7∠−145◦ = 7 cos(−145◦) + j 7 sin(−145◦), = −5.734 − j4.015, , Calculator, Using the ‘Pol’ and ‘Rec’ functions on a calculator, enables changing from Cartesian to polar and vice-versa, to be achieved more quickly., Since complex numbers are used with vectors and, with electrical engineering a.c. theory, it is essential that, the calculator can be used quickly and accurately., , 20.7 Multiplication and division in, polar form, If Z 1 =r1 ∠θ1 and Z 2 =r2 ∠θ2 then:, (i) Z1 Z2 = r1 r2 ∠(θ1 + θ2 ) and, (ii), , Z1 r1, = ∠(θ1 − θ2 ), Z2 r2, , Problem 14. Evaluate, in polar form, 2∠30◦ +5∠−45◦ − 4∠120◦., Addition and subtraction in polar form is not possible, directly. Each complex number has to be converted into, cartesian form first., 2∠30◦ = 2(cos 30◦ + j sin 30◦ ), = 2 cos 30◦ + j 2 sin30◦ = 1.732 + j 1.000, 5∠−45◦ = 5(cos(−45◦) + j sin(−45◦)), = 5 cos(−45◦) + j 5 sin(−45◦), = 3.536 − j 3.536, 4∠120◦ = 4( cos 120◦ + j sin 120◦ ), = 4 cos 120◦ + j 4 sin 120◦, = −2.000 + j 3.464, Hence 2∠30◦ + 5∠−45◦ − 4∠120◦
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Complex numbers, = (1.732 + j 1.000) +(3.536 − j 3.536), , 6. (a) 3∠20◦ × 15∠45◦, (b) 2.4∠65◦ × 4.4∠−21◦, [(a) 45∠65◦ (b) 10.56∠44◦], , − (−2.000 + j 3.464), = 7.268 − j 6.000, which lies in the fourth quadrant, , , , −6.000, = [(7.268)2 + (6.000)2 ]∠ tan−1, 7.268, , 7. (a) 6.4∠27◦ ÷ 2∠−15◦, (b) 5∠30◦ × 4∠80◦ ÷ 10∠−40◦, [(a) 3.2∠42◦ (b) 2∠150◦], π, π, 8. (a) 4∠ + 3∠, 6, 8, (b) 2∠120◦ + 5.2∠58◦ − 1.6∠−40◦, [(a) 6.986∠26.79◦ (b) 7.190∠85.77◦], , = 9.425∠−39.54◦, , Now try the following exercise, Exercise 88, form, , 221, , Further problems on polar, , 1. Determine the modulus and argument of, (a) 2 + j 4 (b) −5 − j 2 (c) j (2 − j )., ⎡, ⎤, (a) 4.472, 63.43◦, ⎢, ⎥, ⎣(b)5.385, −158.20◦⎦, (c) 2.236, 63.43◦, In Problems 2 and 3 express the given Cartesian, complex numbers in polar form, leaving answers, in surd form., 2. (a) 2 + j 3 (b) −4 (c) −6 + j, √, (a) 13∠56.31◦ (b)4∠180◦, √, (c) 37∠170.54◦, 3. (a) − j 3 (b) (−2 + j )3 (c) j 3(1 − j ), √, (a) 3∠−90◦ (b) 125∠100.30◦, √, (c) 2∠−135◦, In Problems 4 and 5 convert the given polar complex numbers into (a + j b) form giving answers, correct to 4 significant figures., 4. (a) 5∠30◦ (b) 3∠60◦ (c) 7∠45◦, ⎡, ⎤, (a) 4.330 + j 2.500, ⎢, ⎥, ⎣(b)1.500 + j 2.598⎦, (c) 4.950 + j 4.950, 5. (a) 6∠125◦ (b) 4∠π (c) 3.5∠−120◦, ⎡, ⎤, (a) −3.441 + j 4.915, ⎢, ⎥, ⎣(b) −4.000 + j 0, ⎦, , 20.8 Applications of complex, numbers, There are several applications of complex numbers, in science and engineering, in particular in electrical, alternating current theory and in mechanical vector, analysis., The effect of multiplying a phasor by j is to rotate, it in a positive direction (i.e. anticlockwise) on an, Argand diagram through 90◦ without altering its length., Similarly, multiplying a phasor by − j rotates the phasor through −90◦ . These facts are used in a.c. theory since certain quantities in the phasor diagrams, lie at 90◦ to each other. For example, in the R−L, series circuit shown in Fig. 20.8(a), V L leads I by, 90◦ (i.e. I lags V L by 90◦ ) and may be written as, j V L , the vertical axis being regarded as the imaginary axis of an Argand diagram. Thus V R + j V L = V, and since V R = IR, V = I X L (where X L is the inductive reactance, 2π f L ohms) and V = IZ (where Z is, the impedance) then R + j X L = Z ., , I, , VR, , VL, , I, , V, Phasor diagram, VL, , VR I, (a), , In Problems 6 to 8, evaluate in polar form., Figure 20.8, , VR, , VC, , V, Phasor diagram, VR, , V, , (c) −1.750 − j 3.031, , C, , R, , L, , R, , , , VC, V, (b), , I
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222 Higher Engineering Mathematics, Similarly, for the R−C circuit shown in Fig. 20.8(b),, VC lags I by 90◦ (i.e. I leads VC by 90◦) and, V R − j VC = V , from which R − j X C = Z (where X C, 1, is the capacitive reactance, ohms)., 2π fC, Problem 15. Determine the resistance and, series inductance (or capacitance) for each of the, following impedances, assuming a frequency of, 50 Hz:, (a) (4.0 + j 7.0) , , Problem 16. An alternating voltage of 240 V,, 50 Hz is connected across an impedance of, (60 − j 100) . Determine (a) the resistance (b) the, capacitance (c) the magnitude of the impedance and, its phase angle and (d) the current flowing., (a), , Impedance Z = (60 − j 100) ., Hence resistance = 60 , , (b) Capacitive reactance X C = 100 and since, 1, XC =, then, 2πf C, , (b) − j 20 , , (c) 15∠−60◦ , (a) Impedance, Z = (4.0 + j 7.0) hence,, resistance = 4.0 and reactance = 7.00 ., Since the imaginary part is positive, the reactance, is inductive,, , capacitance, C =, =, , i.e. X L = 7.0 , , 7.0, XL, =, = 0.0223 H or 22.3 mH, 2π f, 2π(50), , (b) Impedance, Z = j 20, i.e. Z = (0 − j 20) hence, resistance = 0 and reactance = 20 . Since the, imaginary part is negative, the reactance is cap1, acitive, i.e., X C = 20 and since X C =, 2πf C, then:, 1, 1, =, capacitance, C =, F, 2πf XC, 2π(50)(20), =, , 106, μF = 159.2 μF, 2π(50)(20), , (c) Impedance, Z, = 15∠−60◦ = 15[ cos (−60◦ ) + j sin (−60◦ )], , (c), , Magnitude of impedance,, |Z | =, , [(60)2 + (−100)2 ] = 116.6 , , (d) Current flowing, I =, , , , = −59.04◦, , V, 240∠0◦, =, Z 116.6∠−59.04◦, , Problem 17. For the parallel circuit shown in, Fig. 20.9, determine the value of current I and its, phase relative to the 240 V supply, using complex, numbers., XL 5 3 V, , R2 5 10 V, , 1, then capacitance,, 2πf C, , 1, =, μF, C=, 2πf XC, 2π(50)(12.99), , −100, 60, , The circuit and phasor diagrams are as shown in, Fig. 20.8(b)., , R1 5 4 V, , 106, , , , = 2.058 ∠59.04◦ A, , Hence resistance = 7.50 and capacitive reactance, X C = 12.99 , , = 245 μF, , , , Phase angle, arg Z = tan −1, , = 7.50 − j 12.99 , , Since X C =, , 106, μF, 2π(50)(100), , = 31.83 μF, , Since X L = 2πf L then inductance,, L=, , 1, 1, =, 2π f X C, 2π(50)(100), , R3 5 12 V, , I, , XC 5 5 V, , 240 V, 50 Hz, , Figure 20.9
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Complex numbers, V, Current I = . Impedance Z for the three-branch, Z, parallel circuit is given by:, , 10 N, , 8N, 210⬚, 120⬚, , 1, 1, 1, 1, +, +, ,, =, Z, Z1 Z2 Z3, , 45⬚, , where Z 1 = 4 + j 3, Z 2 = 10 and Z 3 = 12 − j 5, 1, 1, =, Z1, 4+ j3, 1, 4 − j3, 4− j3, =, ×, =, 4 + j 3 4 − j 3 42 + 32, , Admittance, Y1 =, , = 0.160 − j 0.120 siemens, , 15 N, , Figure 20.10, , The resultant force, , Admittance, Y2 =, , 1, 1, =, = 0.10 siemens, Z2, 10, , = f A + f B + fC, , Admittance, Y3 =, , 1, 1, =, Z3, 12 − j 5, , = 10(cos 45◦ + j sin 45◦) + 8(cos 120◦, , 1, 12 + j 5, 12 + j 5, =, ×, =, 12 − j 5 12 + j 5 122 + 52, , = 10∠45◦ + 8∠120◦ + 15∠210◦, + j sin 120◦) + 15(cos 210◦ + j sin 210◦ ), = (7.071 + j 7.071) + (−4.00 + j 6.928), , = 0.0710 + j 0.0296 siemens, Total admittance, Y = Y1 + Y2 + Y3, = (0.160 − j 0.120) + (0.10), + (0.0710 + j 0.0296), = 0.331 − j 0.0904, = 0.343∠−15.28◦ siemens, Current I =, , V, = VY, Z, , + (−12.99 − j 7.50), = −9.919 + j 6.499, Magnitude of resultant force, , = [(−9.919)2 + (6.499)2 ] = 11.86 N, Direction of resultant force, , , 6.499, = tan −1, = 146.77◦, −9.919, (since −9.919 + j 6.499 lies in the second quadrant)., , = (240∠0◦ )(0.343∠−15.28◦ ), = 82.32 ∠−15.28◦ A, Problem 18. Determine the magnitude and, direction of the resultant of the three coplanar, forces given below, when they act at a point., Force A, 10 N acting at 45◦ from the positive, horizontal axis., Force B, 87 N acting at 120◦ from the positive, horizontal axis., Force C, 15 N acting at 210◦ from the positive, horizontal axis., The space diagram is shown in Fig. 20.10. The forces, may be written as complex numbers., Thus force A, f A = 10∠45◦, force B, f B = 8∠120◦, and force C, fC = 15∠210◦., , Now try the following exercise, Exercise 89 Further problems on, applications of complex numbers, 1., , Determine the resistance R and series inductance L (or capacitance C) for each of the, following impedances assuming the frequency to be 50 Hz., (a) (3 + j 8) (b) (2 − j 3) , (c) j 14 , (d) 8∠−60◦ , ⎡, ⎤, (a) R = 3 , L = 25.5 mH, ⎢ (b) R = 2 , C = 1061 μF ⎥, ⎢, ⎥, ⎣ (c) R = 0, L = 44.56 mH ⎦, (d) R = 4 , C = 459.4 μF, , 223
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224 Higher Engineering Mathematics, 2. Two impedances, Z 1 = (3 + j 6) and, Z 2 = (4 − j 3) are connected in series to, a supply voltage of 120 V. Determine the, magnitude of the current and its phase angle, relative to the voltage., [15.76 A, 23.20◦ lagging], 3. If the two impedances in Problem 2 are connected in parallel determine the current flowing and its phase relative to the 120 V supply, voltage., [27.25 A, 3.37◦ lagging], 4. A series circuit consists of a 12 resistor, a, coil of inductance 0.10 H and a capacitance of, 160 μF. Calculate the current flowing and its, phase relative to the supply voltage of 240 V,, 50 Hz. Determine also the power factor of the, circuit., [14.42 A, 43.85◦ lagging, 0.721], 5. For the circuit shown in Fig. 20.11, determine, the current I flowing and its phase relative to, the applied voltage. [14.6 A, 2.51◦ leading], 6. Determine, using complex numbers, the magnitude and direction of the resultant of the, coplanar forces given below, which are acting at a point. Force A, 5 N acting horizontally,, Force B, 9 N acting at an angle of 135◦ to force, A, Force C, 12 N acting at an angle of 240◦ to, force A., [8.394 N, 208.68◦ from force A], XC 5 20 V, , R2 5 40 V, , R1 5 30 V, , XL 5 50 V, , R3 5 25 V, , I, V 5 200 V, , Figure 20.11, , 7. A delta-connected impedance Z A is given, by:, Z1 Z2 + Z2 Z3 + Z3 Z1, ZA =, Z2, Determine Z A in both Cartesian and polar, form given Z 1 = (10 + j 0) ,, Z 2 = (0 − j 10) and Z 3 = (10 + j 10) ., [(10 + j 20) , 22.36∠63.43◦ ], 8. In the hydrogen atom, the angular momentum, p, of the de Broglie wave is given, , jh, (±jmψ). Determine an, by: pψ = −, 2π, , mh, expression for p., ±, 2π, 9. An aircraft P flying at a constant height has, a velocity of (400 + j 300) km/h. Another aircraft Q at the same height has a velocity of, (200 − j 600) km/h. Determine (a) the velocity of P relative to Q, and (b) the velocity of, Q relative to P. Express the answers in polar, form, correct to the nearest km/h., (a) 922 km/h at 77.47◦, (b) 922 km/h at −102.53◦, 10. Three vectors are represented by P, 2∠30◦ ,, Q, 3∠90◦ and R, 4∠−60◦ . Determine in, polar form the vectors represented by (a), P + Q + R, (b) P − Q − R., (a) 3.770∠8.17◦, (b) 1.488∠100.37◦, 11. In a Schering bridge circuit,, Z X = (R X − j X C X ), Z 2 = − j X C2 ,, (R3 )(− j X C3 ), and Z 4 = R4, Z3 =, (R3 − j X C3 ), 1, where X C =, 2πf C, At balance: (Z X )(Z 3 ) = (Z 2 )(Z 4 )., C3 R4, Show that at balance R X =, C2, C2 R3, CX =, R4, , and
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Chapter 21, , De Moivre’s theorem, 21.1, , = 2197∠382.14◦(since 742.14, , Introduction, , ≡ 742.14◦ − 360◦ = 382.14◦), = 2197∠22.14◦ (since 382.14◦, , From multiplication of complex numbers in polar form,, , ≡ 382.14◦ − 360◦ = 22.14◦), , (r∠θ) × (r ∠θ) = r 2 ∠2θ, , or 2197∠22◦8, Similarly, (r∠θ)× (r∠θ)× (r∠θ) = r 3∠3θ, and so on., In general, De Moivre’s theorem states:, [r∠θ], , n, , = r n∠nθ, , Problem 2. Determine the value of (−7 + j 5)4,, expressing the result in polar and rectangular forms., , The theorem is true for all positive, negative and, fractional values of n. The theorem is used to determine, powers and roots of complex numbers., , 21.2, , Powers of complex numbers, ◦ 4, , ◦, , For example [3∠20 ] = 3 ∠(4 × 20 ) = 81∠80, De Moivre’s theorem., 4, , ◦, , by, , Problem 1. Determine, in polar form, (a) [2∠35◦ ]5 (b) (−2 + j 3)6., (a), , [2∠35◦]5 = 25 ∠(5 × 35◦),, from De Moivre’s theorem, , , 5, [(−7)2 + 52 ]∠ tan−1, −7, √, = 74∠144.46◦, , (−7 + j 5) =, , (Note, by considering the Argand diagram, −7 + j 5, must represent an angle in the second quadrant and not, in the fourth quadrant.), Applying De Moivre’s theorem:, √, (−7 + j 5)4 = [ 74∠144.46◦]4, √, = 744 ∠4 ×144.46◦, = 5476∠577.84◦, = 5476∠217.84◦, , = 32∠175◦, (b), , , 3, (−2 + j 3)= [(−2)2 + (3)2 ]∠ tan−1, −2, √, = 13∠123.69◦ , since −2 + j 3, lies in the second quadrant, √, (−2 + j 3)6 = [ 13∠123.69◦]6, √, = ( 13)6 ∠(6 × 123.69◦),, by De Moivre’s theorem, = 2197∠742.14◦, , or 5476∠217◦50 in polar form, Since r∠θ = r cos θ + j r sin θ,, 5476∠217.84◦ = 5476 cos217.84◦, + j 5476 sin217.84◦, = −4325 − j 3359, i.e., , (−7 + j5)4 = −4325 −j3359, in rectangular form
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226 Higher Engineering Mathematics, Now try the following exercise, Exercise 90 Further problems on powers, of complex numbers, , 13∠427.38◦. When the angle is divided by 2 an angle, less than 360◦ is obtained., Hence, , , , (5 + j 12) = [13∠67.38◦] and [13∠427.38◦], , 1. Determine in polar form (a) [1.5∠15◦]5, (b) (1 + j 2)6., [(a) 7.594∠75◦ (b) 125∠20.61◦], , 1, , =, , 2. Determine in polar and cartesian forms, (a) [3∠41◦]4 (b) (−2 − j )5., (a) 81∠164◦, −77.86 + j 22.33, (b) 55.90∠−47.18◦ , 38 − j 41, , [476.4∠119.42◦, −234 + j 415], , 4. (6 + j 5)3, 5. (3 − j 8)5, , [45530∠12.78◦, 44400 + j 10070], , 6. (−2 + j 7)4, , [2809∠63.78◦, 1241 + j 2520], , 7. (−16 − j 9)6, , 21.3, , =, , Roots of complex numbers, , , , , 1, ◦, × 67.38 and, 2, 1, × 427.38◦, 2, , , , √, √, 13∠33.69◦ and 13∠213.69◦, , Thus, in polar form, the two roots are, 3.61∠33.69◦ and 3.61∠−146.31◦., √, √, 13∠33.69◦ = 13(cos 33.69◦ + j sin 33.69◦ ), = 3.0 + j 2.0, √, √, 13∠213.69◦ = 13(cos 213.69◦ + j sin 213.69◦), = −3.0 − j 2.0, Thus, in cartesian form the two roots are, ±(3.0 + j2.0)., From the Argand diagram shown in Fig. 21.1 the two, roots are seen to be 180◦ apart, which is always true, when finding square roots of complex numbers., , The square root of a complex number is determined by, letting n =1/2 in De Moivre’s theorem,, , 1, 1 1, √ θ, i.e., [r∠θ] = [r∠θ] 2 = r 2 ∠ θ = r ∠, 2, 2, There are two square roots of a real number, equal in, size but opposite in sign., , , , = 3.61∠33.69◦ and 3.61∠213.69◦, , (38.27 × 106)∠176.15◦ ,, 106(−38.18 + j 2.570), , 1, 13 2 ∠, , 1, 13 2 ∠, , 3. Convert (3 − j ) into polar form and hence, evaluate (3 − j√, )7, giving the answer in polar, form., [ 10∠−18.43◦ , 3162∠−129◦ ], In problems 4 to 7, express in both polar and, rectangular forms., , 1, , = [13∠67.38◦] 2 and [13∠427.38◦] 2, , Imaginary axis, j2, 3.61, 213.698, , 33. 698, , 23, , 3, , Real axis, , 3.61, , Problem 3. Determine the two square roots of the, complex number (5 + j 12) in polar and cartesian, forms and show the roots on an Argand diagram., (5 + j 12) =, , , , [52 + 122 ]∠ tan−1, , , , 12, 5, , , , = 13∠67.38◦, When determining square roots two solutions result., To obtain the second solution one way is to, express 13∠67.38◦ also as 13∠(67.38◦ + 360◦ ), i.e., , 2j 2, , Figure 21.1, , In general, when finding the nth root of a complex, number, there are n solutions. For example, there are, three solutions to a cube root, five solutions to a fifth, root, and so on. In the solutions to the roots of a complex, number, the modulus, r, is always the same, but the
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De Moivre’s theorem, arguments, θ, are different. It is shown in Problem 3, that arguments are symmetrically spaced on an Argand, diagram and are (360/n)◦ apart, where n is the number, of the roots required. Thus if one of the solutions to the, cube root of a complex number is, say, 5∠20◦, the other, two roots are symmetrically spaced (360/3)◦ , i.e. 120◦, from this root and the three roots are 5∠20◦, 5∠140◦, and 5∠260◦ ., 1, , Problem 4. Find the roots of [(5 + j 3)] 2 in, rectangular form, correct to 4 significant figures., (5 + j 3) =, , √, 34∠30.96◦, , Applying De Moivre’s theorem:, (5 +, , 1, j 3) 2, , =, , 1, , 34 2 ∠ 12 × 30.96◦, , = 2.415∠15.48◦or 2.415∠15◦ 29, , (−14 + j 3) =, (−14 +, , −2, j 3) 5, , √, 205∠167.905◦, -, , =, , 205, , −2, 5 ∠, , , , 2, −, × 167.905◦, 5, , = 0.3449∠−67.164◦, or 0.3449∠−67◦ 10, There are five roots to this complex number,, , , −2, 1, 1, x 5 = 2 =√, 5 2, x, x5, The roots are symmetrically displaced from one, another (360/5)◦ , i.e. 72◦ apart round an Argand, diagram., Thus the required roots are 0.3449∠−67◦ 10 ,, 0.3449∠4◦ 50 , 0.3449∠76◦ 50 , 0.3449∠148◦ 50 and, 0.3449∠220◦50 ., Now try the following exercise, , The second root may be obtained as shown above, i.e., having the same modulus but displaced (360/2)◦ from, the first root., 1, , Thus, (5 + j 3) 2 = 2.415∠(15.48◦ + 180◦ ), = 2.415∠195.48◦, , Exercise 91 Further problems on the, roots of complex numbers, In Problems 1 to 3 determine the two square roots, of the given complex numbers in Cartesian form, and show the results on an Argand diagram., 1. (a) 1 + j (b) j, , In rectangular form:, , (a) ±(1.099 + j 0.455), (b) ±(0.707 + j 0.707), , 2.415∠15.48◦ = 2.415 cos 15.48◦, + j 2.415 sin15.48◦, = 2.327 + j0.6446, and, , 2. (a) 3 − j 4 (b) −1 − j 2, (a) ±(2 − j ), (b) ±(0.786 − j 1.272), , 2.415∠195.48◦ = 2.415 cos 195.48◦, + j 2.415 sin195.48◦, , 3. (a) 7∠60◦ (b) 12∠, , = −2.327 − j0.6446, [(5 + j 3)] 2 = 2.415∠15.48◦and, 2.415∠195.48◦or, ± (2.327 + j0.6446)., Problem 5. Express the roots of, (−14 + j 3), , −2, 5, , in polar form., , 3π, 2, (a) ±(2.291 + j 1.323), (b) ±(−2.449 + j 2.449), , 1, , Hence, , 227, , In Problems 4 to 7, determine the moduli and, arguments of the complex roots., 1, , 4. (3 + j 4) 3, Moduli 1.710, arguments 17.71◦ ,, 137.71◦ and 257.71◦
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228 Higher Engineering Mathematics, 1, , 5. (−2 + j ) 4, , 6. (−6 −, , ⎡, , ⎤, Modulus 1.223, arguments, ⎣ 38.36◦, 128.36◦,, ⎦, 218.36◦ and 308.36◦, , √, By definition, j = (−1), hence j 2 = −1, j 3 = − j ,, j 4 = 1, j 5 = j , and so on., θ2, θ3 θ4, θ5, Thus e j θ = 1 + j θ − − j + + j − · · ·, 2!, 3! 4!, 5!, Grouping real and imaginary terms gives:, , 1, j 5) 2, , e, , jθ, , Modulus 2.795, arguments, 109.90◦, 289.90◦, −2, , 7. (4 − j 3) 3, Modulus 0.3420, arguments 24.58◦,, 144.58◦ and 264.58◦, , However, from equations (2) and (3):, , , θ2 θ4, +, − · · · = cos θ, 1−, 2! 4!, , 8. For a transmission line, the characteristic, impedance Z 0 and the propagation coefficient, γ are given by:, , Z0 =, γ=, , R + j ωL, G + j ωC, , , , θ2 θ4, = 1−, +, −···, 2! 4!, , , θ3 θ5, +, −···, + j θ−, 3! 5!, , , , , and, , and, , , θ3 θ5, +, − · · · = sin θ, θ−, 3! 5!, , , [(R + j ωL)(G + j ωC)], , Given R = 25 , L =5 × 10−3 H,, G = 80 × 10−6 siemens, C = 0.04 × 10−6 F, and ω = 2000 π rad/s,, determine, in polar, , Z 0 = 390.2∠ −10.43◦ ,, form, Z 0 and γ ., γ = 0.1029∠61.92◦, , e jθ = cos θ + j sin θ, , Thus, , (4), , Writing −θ for θ in equation (4), gives:, e j (−θ) = cos(−θ) + j sin(−θ), However, cos(−θ) = cos θ and sin(−θ) = −sin θ, , 21.4 The exponential form of a, complex number, , Thus, , Certain mathematical functions may be expressed as, power series (for example, by Maclaurin’s series—see, Chapter 8), three examples being:, (i) ex = 1 + x +, , x2, 2!, , +, , x3, 3!, , +, , x4, 4!, , +, , x5, 5!, , x3 x5 x7, +, −, +···, 3!, 5! 7!, x2 x4 x6, +, −, +···, (iii) cos x = 1 −, 2! 4!, 6!, (ii) sin x = x −, , +···, , (1), (2), (3), , Replacing x in equation (1) by the imaginary number, j θ gives:, ( j θ)2 ( j θ)3 ( j θ)4 ( j θ)5, +, +, +, +· · ·, e j θ = 1+ j θ +, 2!, 3!, 4!, 5!, j 2θ 2, j 3θ 3, j 4θ 4, j 5θ 5, = 1 + jθ +, +, +, +, +···, 2!, 3!, 4!, 5!, , e −jθ = cos θ − j sin θ, , (5), , The polar form of a complex number z is:, z =r(cos θ + j sin θ). But, from equation (4),, cos θ + j sin θ = e jθ ., Therefore, , z = re jθ, , When a complex number is written in this way, it is said, to be expressed in exponential form., There are therefore three ways of expressing a complex number:, 1., , z =(a + j b), called Cartesian or rectangular form,, , 2., , z =r(cos θ + j sin θ) or r∠θ, called polar form, and, , 3., , z =re j θ called exponential form., , The exponential form is obtained from the polar form., π, , For example, 4∠30◦ becomes 4e j 6 in exponential, form. (Note that in re j θ , θ must be in radians.)
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De Moivre’s theorem, Problem 6. Change (3 − j 4) into (a) polar form,, (b) exponential form., (a), , (3 − j 4) = 5∠−53.13◦ or 5∠−0.927, in polar form, , (b) (3 − j 4) = 5∠−0.927 = 5e−j0.927, in exponential form, Problem 7. Convert 7.2e j 1.5 into rectangular, form., 7.2e j 1.5 = 7.2∠1.5 rad(= 7.2∠85.94◦) in polar form, , (a), , Thus if z =4e j 1.3 then ln z = ln(4e j1.3 ), = ln 4 + j1.3, (or 1.386 + j1.300) in Cartesian form., , (b) (1.386 + j 1.300) =1.90∠43.17◦ or 1.90∠0.753, in polar form., Problem 11. Given z = 3e1− j , find ln z in polar, form., If, , z = 3e1− j , then, , ln, , z = ln(3e1− j ), = ln 3 + ln e1− j, , = 7.2 cos 1.5 + j 7.2 sin1.5, , = ln 3 + 1 − j, , = (0.509 + j 7.182) in rectangular form, , = (1 + ln 3) − j, , Problem 8. Express, form., , π, z = 2e1+ j 3, , = 2.0986 − j 1.0000, in Cartesian, , π, z = (2e1 ) e j 3 by the laws of indices, π, (or 2e∠60◦ )in polar form, 3, , π, π, = 2e cos + j sin, 3, 3, = (2e1 )∠, , = (2.718 + j4.708) in Cartesian form, , = 2.325∠−25.48◦ or 2.325∠−0.445, Problem 12. Determine, in polar form, ln (3 + j 4)., ln(3 + j 4) = ln[5∠0.927] = ln[5e j 0.927], = ln 5 + ln(e j 0.927 ), = ln 5 + j 0.927, = 1.609 + j 0.927, = 1.857∠29.95◦ or 1.857∠0.523, , Problem 9. Change 6e2− j 3 into (a + j b) form., 6e2− j 3 = (6e2 )(e− j 3 ) by the laws of indices, , Exercise 92 Further problems on the, exponential form of complex numbers, , = 6e2 [cos (−3) + j sin (−3)], , 1. Change (5 + j 3) into exponential form., [5.83e j 0.54], , Problem 10. If z = 4e j 1.3 , determine ln z (a) in, Cartesian form, and (b) in polar form., , i.e., , Now try the following exercise, , = 6e2 ∠−3 rad (or 6e2 ∠−171.890 ), in polar form, , = (−43.89 − j6.26) in (a + jb) form, , If, , z = re j θ then ln z = ln(re j θ ), = lnr + ln e j θ, ln z = lnr + j θ,, , by the laws of logarithms, , 229, , 2. Convert (−2.5 + j 4.2) into exponential form., [4.89e j 2.11], 3. Change 3.6e j 2 into cartesian form., [−1.50 + j 3.27], π, , 4. Express 2e3+ j 6 in (a + j b) form., [34.79 + j 20.09], 5. Convert 1.7e1.2− j 2.5 into rectangular form., [−4.52 − j 3.38]
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230 Higher Engineering Mathematics, 6. If z = 7e j 2.1 , determine ln z (a) in Cartesian, form, and (b) in polar form., ⎤, ⎡, (a) ln 7 + j 2.1, ⎣(b) 2.86∠47.18◦or⎦, 2.86∠0.82, 7. Given z =4e1.5− j 2 , determine ln z in polar, form., [3.51∠−34.72◦ or 3.51∠−0.61], 8. Determine in polar form (a) ln (2 + j 5), (b) ln (−4 − j 3), ⎤, ⎡, (a) 2.06∠35.26◦or, ⎢ 2.06∠0.615 ⎥, ⎥, ⎢, ⎣(b) 4.11∠66.96◦or⎦, 4.11∠1.17, , 9. When displaced electrons oscillate about an, equilibrium position the displacement x is, given by the equation:, 5, 6, √, x = Ae, , ht, − 2m + j, , (4m f −h 2 ), t, 2m−a, , Determine the real part of x in terms of t ,, assuming (4m f − h 2 ) is positive., √, , , ht, (4m f − h 2 ), − 2m, cos, t, Ae, 2m −a
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Chapter 22, , The theory of matrices and, determinants, 22.1, , Matrix notation, , Matrices and determinants are mainly used for the solution of linear simultaneous equations. The theory of, matrices and determinants is dealt with in this chapter and this theory is then used in Chapter 23 to solve, simultaneous equations., The coefficients of the variables for linear simultaneous equations may be shown in matrix form., The coefficients of x and y in the simultaneous, equations, x + 2y = 3, , of the matrix. The number of rows in a matrix is usually, specified by m and the number of columns by n and, a matrix referred to as an ‘m by n’ matrix. Thus,, , 2 3 6, is a ‘2 by 3’ matrix. Matrices cannot be, 4 5 7, expressed as a single numerical value, but they can often, be simplified or combined, and unknown element values can be determined by comparison methods. Just as, there are rules for addition, subtraction, multiplication, and division of numbers in arithmetic, rules for these, operations can be applied to matrices and the rules of, matrices are such that they obey most of those governing, the algebra of numbers., , 4x − 5y = 6, , , 1, 2, become, in matrix notation., 4 −5, Similarly, the coefficients of p, q and r in the equations, 1.3 p − 2.0q + r = 7, 3.7 p + 4.8q − 7r = 3, 4.1 p + 3.8q + 12r = −6, ⎛, 1.3 −2.0, become ⎝3.7, 4.8, 4.1, 3.8, , ⎞, 1, −7⎠ in matrix form., 12, , The numbers within a matrix are called an array and the, coefficients forming the array are called the elements, , 22.2 Addition, subtraction and, multiplication of matrices, (i) Addition of matrices, Corresponding elements in two matrices may be added, to form a single matrix., Problem 1. Add the matrices, , , , , 2 −1, −3, 0, (a), and, and, −7, 4, 7 −4, ⎛, ⎞, ⎛, ⎞, 3 1 −4, 2 7 −5, (b) ⎝4 3, 1⎠ and ⎝−2 1, 0⎠, 1 4 −3, 6 3, 4
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232 Higher Engineering Mathematics, (a), , Adding the corresponding elements gives:, , , , , 2 −1, −3, 0, +, −7, 4, 7 −4, , , 2 + (−3) −1 + 0, =, −7 + 7, 4 + (−4), , , −1 −1, =, 0, 0, , (b) Adding the corresponding elements gives:, ⎛, ⎞ ⎛, ⎞, 3 1 −4, 2 7 −5, ⎝4 3, 1⎠ + ⎝−2 1, 0⎠, 1 4 −3, 6 3, 4, ⎛, ⎞, 3+2, 1 + 7 −4 + (−5), ⎠, = ⎝4 + (−2) 3 + 1, 1+0, 1+6, 4 + 3 −3 + 4, ⎞, ⎛, 5 8 −9, 1⎠, = ⎝2 4, 7 7, 1, (ii) Subtraction of matrices, If A is a matrix and B is another matrix, then (A − B), is a single matrix formed by subtracting the elements of, B from the corresponding elements of A., Problem, , −3, (a), 7, ⎛, 2, (b) ⎝−2, 6, , 2., , Subtract, , , , 0, 2 −1, from, and, −4, −7, 4, ⎞, ⎛, ⎞, 7 −5, 3 1 −4, 1, 0⎠ from ⎝4 3, 1⎠, 3, 4, 1 4 −3, , To find matrix A minus matrix B, the elements of B are, taken from the corresponding elements of A. Thus:, , , , 2 −1, −3, 0, (a), −, −7, 4, 7 −4, , , 2 − (−3) −1 − 0, =, −7 − 7, 4 − (−4), , , 5 −1, =, −14, 8, ⎞ ⎛, ⎞, ⎛, 2 7 −5, 3 1 −4, (b) ⎝, 1⎠ − ⎝−2 1, 0⎠, 4 3, 1 4 −3, 6 3, 4, , ⎛, 3−2, = ⎝4 − (−2), 1−6, ⎛, 1 −6, =⎝ 6, 2, −5, 1, , ⎞, 1 − 7 −4 − (−5), ⎠, 3−1, 1−0, 4 − 3 −3 − 4, ⎞, 1, 1⎠, −7, , Problem 3. If, , , , , −3, 0, 2 −1, A=, ,B=, and, 7 −4, −7, 4, , , 1, 0, C=, find A + B − C., −2 −4, , , −1 −1, A+ B =, 0, 0, (from Problem 1), , , , −1 −1, 1, 0, Hence, A + B − C =, −, 0, 0, −2 −4, , , −1 − 1, −1 − 0, =, 0 − (−2), 0 − (−4), , , −2 −1, =, 2, 4, Alternatively A + B − C, , , , , −3, 0, 2 −1, 1, 0, =, +, −, 7 −4, −7, 4, −2 −4, , , −3 + 2 − 1, 0 + (−1) − 0, =, 7 + (−7) − (−2) −4 + 4 − (−4), , , −2 −1, =, as obtained previously, 2, 4, , (iii) Multiplication, When a matrix is multiplied by a number, called scalar, multiplication, a single matrix results in which each, element of the original matrix has been multiplied by, the number., , , −3, 0, Problem 4. If A =, ,, 7 −4, ⎛, ⎞, , , 1, 0, 2 −1, ⎠ find, B=, and C = ⎝, −7, 4, −2 −4, 2 A − 3B + 4C.
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The theory of matrices and determinants, For scalar multiplication, each element is multiplied by, the scalar quantity, hence, , , , −3, 0, −6, 0, 2A = 2, =, 7 −4, 14 −8, , , , 2 −1, 6 −3, 3B = 3, =, −7, 4, −21 12, , , , 1, 0, 4, 0, and 4C = 4, =, −2 −4, −8 −16, Hence 2 A − 3B + 4C, , , , , −6 0, 6 −3, 4, 0, =, −, +, 14 −8, −21 12, −8 −16, , , −6 − 6 + 4, 0 − (−3) + 0, =, 14 − (−21) + (−8) −8 − 12 + (−16), , , −8, 3, =, 27 −36, , When a matrix A is multiplied by another matrix B,, a single matrix results in which elements are obtained, from the sum of the products of the corresponding rows, of A and the corresponding columns of B., Two matrices A and B may be multiplied together,, provided the number of elements in the rows of matrix, A are equal to the number of elements in the columns of, matrix B. In general terms, when multiplying a matrix, of dimensions (m by n) by a matrix of dimensions (n by, r), the resulting matrix has dimensions (m by r). Thus, a 2 by 3 matrix multiplied by a 3 by 1 matrix gives a, matrix of dimensions 2 by 1., Problem 5. If A =, find A × B., , , , , , 2 3, −5 7, and B =, 1 −4, −3 4, , Let A × B = C where C =, , , , C11 C12, C21 C22, , C11 is the sum of the products of the first row elements, of A and the first column elements of B taken one at a, time,, i.e. C11 = (2 × (−5)) + (3 × (−3)) = −19, C12 is the sum of the products of the first row elements, of A and the second column elements of B, taken one, at a time,, i.e. C12 = (2 × 7) + (3 × 4) = 26, , 233, , C21 is the sum of the products of the second row elements of A and the first column elements of B, taken, one at a time,, i.e. C21 = (1 × (−5)) + (−4 × (−3)) = 7, Finally, C22 is the sum of the products of the second, row elements of A and the second column elements of, B, taken one at a time,, i.e. C22 = (1 × 7) + ((−4) × 4) = −9, Thus, A × B =, , , , −19 26, 7 −9, , Problem 6. Simplify, ⎛, ⎞ ⎛ ⎞, 3, 4, 0, 2, ⎝−2, 6 −3⎠ × ⎝ 5⎠, 7 −4, 1, −1, The sum of the products of the elements of each row of, the first matrix and the elements of the second matrix,, (called a column matrix), are taken one at a time. Thus:, ⎛, , ⎞ ⎛ ⎞, 3, 4, 0, 2, ⎝−2, 6 −3⎠ × ⎝ 5⎠, 7 −4, 1, −1, ⎛, ⎞, (3 × 2) + (4 × 5) + (0 × (−1)), = ⎝(−2 × 2) + (6 × 5) + (−3 × (−1))⎠, (7 × 2) + (−4 × 5) + (1 × (−1)), ⎛ ⎞, 26, = ⎝ 29⎠, −7, ⎛, , ⎞, 3, 4, 0, Problem 7. If A = ⎝−2, 6 −3⎠ and, 7 −4, 1, ⎛, ⎞, 2 −5, B = ⎝ 5 −6⎠, find A × B., −1 −7, The sum of the products of the elements of each row of, the first matrix and the elements of each column of the, second matrix are taken one at a time. Thus:, ⎛, , ⎞ ⎛, ⎞, 3, 4, 0, 2 −5, ⎝−2, 6 −3⎠ × ⎝ 5 −6⎠, 7 −4, 1, −1 −7
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234 Higher Engineering Mathematics, ⎛, , ⎞, [(3 × 2), [(3 × (−5)), ⎜ + (4 × 5), +(4 × (−6)) ⎟, ⎜, ⎟, ⎜ + (0 × (−1))], +(0 × (−7))] ⎟, ⎜, ⎟, ⎜[(−2 × 2), ⎟, [(−2 × (−5)), ⎜, ⎟, ⎟, =⎜, +, (6, ×, 5), +(6, ×, (−6)), ⎜, ⎟, ⎜ + (−3 × (−1))], ⎟, +(−3, ×, (−7))], ⎜, ⎟, ⎜[(7 × 2), ⎟, [(7 × (−5)), ⎜, ⎟, ⎝ + (−4 × 5), +(−4 × (−6)) ⎠, + (1 × (−1))], +(1 × (−7))], ⎛, ⎞, 26 −39, = ⎝ 29 −5⎠, −7 −18, Problem 8. Determine, ⎛, ⎞ ⎛, ⎞, 1 0 3, 2 2 0, ⎝2 1 2⎠ × ⎝1 3 2⎠, 1 3 1, 3 2 0, , , , , 2 3, 2 3, A× B =, ×, 1 0, 0 1, , , [(2 × 2) + (3 × 0)] [(2 × 3) + (3 × 1)], =, [(1 × 2) + (0 × 0)] [(1 × 3) + (0 × 1)], , , 4 9, =, 2 3, , , , 2 3, 2 3, ×, B×A=, 0 1, 1 0, , , [(2 × 2) + (3 × 1)] [(2 × 3) + (3 × 0)], =, [(0 × 2) + (1 × 1)] [(0 × 3) + (1 × 0)], , , 7 6, =, 1 0, , , , 4 9, 7 6, =, , then A × B = B × A, Since, 2 3, 1 0, Now try the following exercise, , The sum of the products of the elements of each row of, the first matrix and the elements of each column of the, second matrix are taken one at a time. Thus:, ⎛, ⎞ ⎛, ⎞, 1 0 3, 2 2 0, ⎝2 1 2⎠ × ⎝1 3 2⎠, 1 3 1, 3 2 0, ⎛, ⎞, [(1 × 2), [(1 × 2), [(1 × 0), ⎜ + (0 × 1), + (0 × 3), + (0 × 2) ⎟, ⎜, ⎟, ⎜ + (3 × 3)] + (3 × 2)] + (3 × 0)]⎟, ⎜, ⎟, ⎜[(2 × 2), ⎟, [(2 × 2), [(2 × 0), ⎜, ⎟, ⎟, +, (1, ×, 1), +, (1, ×, 3), +, (1, ×, 2), =⎜, ⎜, ⎟, ⎜ + (2 × 3)] + (2 × 2)] + (2 × 0)]⎟, ⎜, ⎟, ⎜[(1 × 2), ⎟, [(1 × 2), [(1 × 0), ⎜, ⎟, ⎝ + (3 × 1), + (3 × 3), + (3 × 2) ⎠, + (1 × 3)], ⎞, 11 8 0, = ⎝ 11 11 2⎠, 8 13 6, , + (1 × 2)], , + (1 × 0)], , ⎛, , In algebra, the commutative law of multiplication states, that a × b = b × a. For matrices, this law is only true in, a few special cases, and in general A × B is not equal, to B × A., , , 2 3, If A =, and, 1 0, , Problem 9., , , 2 3, B=, show that A × B = B × A., 0 1, , Exercise 93 Further problems on addition,, subtraction and multiplication of matrices, In Problems 1 to 13, the matrices A to K are:, , , , , 3 −1, 5 2, A=, B=, −4, 7, −1 6, , , −1.3, 7.4, C=, 2.5 −3.9, ⎛, ⎞, 4 −7, 6, 4, 0⎠, D = ⎝−2, 5, 7 −4, ⎛, ⎞, 3, 6 2, E = ⎝ 5 −3 7⎠, −1, 0 2, ⎛, ⎞, , 3.1 2.4, 6.4, 6, F = ⎝−1.6 3.8 −1.9⎠ G =, −2, 5.3 3.4 −4.8, ⎛, ⎞, ⎛, ⎞, , 4, 1 0, −2, H=, J = ⎝−11⎠ K = ⎝0 1⎠, 5, 7, 1 0, Addition, subtraction and multiplication, In Problems 1 to 12, perform the matrix operation, stated., , , 8 1, 1. A + B, −5 13
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The theory of matrices and determinants, ⎡⎛, 2., , 7 −1, , ⎢⎜, ⎣⎝3, 4, , D+E, , 7 −2, , A− B, , 4., , 5., 6., , A+ B −C, , −7.5, , , 16.9, 45, , ⎡⎛, , A× B, , 17.2, , D× J, , 11., , E×K, , , , 16, , 0, , , , −27 34, , , 10., , A unit matrix, I, is one in which all elements of the, leading diagonal (\) have a value of 1 and all other elements have a value of 0. Multiplication of a matrix by, I is the equivalent of multiplying by 1 in arithmetic., , 22.4 The determinant of a 2 by 2, matrix, , , , a b, is defined, c d, , as (ad − bc)., The elements of the determinant of a matrix are, written, vertical lines. Thus, the determinant, between, , 3 −4, 3 −4, of, is written as, and is equal to, 1, 6, 1, 6, (3 × 6) − (−4 × 1), i.e. 18 −(−4) or 22. Hence the, determinant of a matrix can be expressed as a single, 3 −4, numerical value, i.e., = 22., 1, 6, , 43, , , A×C, , The unit matrix, , The determinant of a 2 by 2 matrix,, , ⎞⎤, , −11, , 22.3, , , , ⎟⎥, 28.6⎠⎦, , A× H, , 9., , 7, , −5.6 −7.6, , 4.6, , , , 8., , , , −26 71, , ⎢⎜, ⎣⎝ 17.4 −16.2, −14.2, 0.4, 7., , , , 1, , 9.3 −6.4, , 5 A + 6B, 2D + 3E −4F, , −2 −3, −3, , , , ⎞⎤, , ⎟⎥, 7⎠⎦, , 1, , , 3., , 8, , −6.4, , 26.1, , , , 22.7 −56.9, ⎡⎛, ⎞⎤, 135, ⎟⎥, ⎢⎜, ⎣⎝−52⎠⎦, −85, ⎞⎤, ⎡⎛, 5, 6, ⎟⎥, ⎢⎜, ⎣⎝12 −3⎠⎦, 1, 0, ⎞⎤, ⎡⎛, 55.4 3.4, 10.1, ⎟⎥, ⎢⎜, ⎣⎝−12.6 10.4 −20.4⎠⎦, −16.9 25.0, 37.9, , 12., , D× F, , 13., , Show that A ×, ⎤, ⎡C = C ×A, −6.4, 26.1, ⎢A × C =, ⎥, ⎢, 22.7 −56.9 ⎥, ⎢, ⎥, ⎢, ⎥, , ⎢, −33.5 −53.1 ⎥, ⎢, ⎥, ⎢C × A =, ⎥, 23.1 −29.8 ⎦, ⎣, Hence they are not equal, , 235, , Problem 10. Determine the value of, 3 −2, 7, 4, 3 −2, = (3 × 4) − (−2 × 7), 7, 4, = 12 − (−14) = 26, , Problem 11. Evaluate, , (1 + j ), j2, − j 3 (1 − j 4), , (1 + j ), j2, = (1 + j )(1 − j 4) − ( j 2)(− j 3), − j 3 (1 − j 4), = 1 − j 4 + j − j 24 + j 26, = 1 − j 4 + j − (−4) + (−6), since from Chapter 20, j 2 = −1, = 1− j4+ j +4 −6, = −1 − j 3, Problem 12. Evaluate, , 5∠30◦ 2∠−60◦, 3∠60◦ 4∠−90◦
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236 Higher Engineering Mathematics, 5∠30◦ 2∠−60◦, = (5∠30◦ )(4∠−90◦ ), 3∠60◦ 4∠−90◦, − (2∠−60◦ )(3∠60◦ ), = (20∠−60◦ ) − (6∠0◦ ), = (10 − j 17.32) − (6 + j 0), = (4 − j 17.32) or 17.78∠−77◦, Now try the following exercise, Exercise 94 Further problems on 2 by 2, determinants, , , 3 −1, 1. Calculate the determinant of, −4, 7, [17], 2. Calculate, the, , determinant of, −2, 5, 3 −6, 3. Calculate, the determinant, of, , , −1.3, 7.4, 2.5 −3.9, 4. Evaluate, , 5. Evaluate, , j2, −j3, (1 + j ), j, , [−3], , [−13.43], , [−5 + j 3], , 2∠40◦, , 5∠−20◦, , 7∠−32◦, , 4∠−117◦, (−19.75 + j 19.79), or, , 27.96∠134.94◦, , 22.5 The inverse or reciprocal of a, 2 by 2 matrix, The inverse of matrix A is A−1 such that A × A−1 = I ,, the unit matrix. , , 1 2, Let matrix A be, and let the inverse matrix, A−1, 3 4, , , a b, be, ., c d, Then, since A × A−1 = I ,, , , , , 1 2, a b, 1 0, ×, =, 3 4, c d, 0 1, , Multiplying the matrices on the left hand side, gives, , , , a + 2c b + 2d, 1 0, =, 3a + 4c 3b + 4d, 0 1, Equating corresponding elements gives:, b + 2d = 0, i.e. b = −2d, 4, and 3a + 4c = 0, i.e. a = − c, 3, Substituting for a and b gives:, ⎛, ⎞, 4, c, +, 2c, −2d, +, 2d, −, , ⎜, ⎟ , 3, 1 0, ⎜, ⎟, ⎜ , ⎟=, , 0 1, ⎝, ⎠, 4, 3 − c + 4c 3(−2d) + 4d, 3, ⎞, ⎛2, , , c 0, 1 0, ⎠, ⎝, 3, =, i.e., 0 1, 0 −2d, 2, 3, 1, showing that c = 1, i.e. c = and −2d = 1, i.e. d = −, 3, 2, 2, 4, Since b = −2d, b = 1 and since a = − c, a = −2., 3, , , , 1 2, a b, Thus the inverse of matrix, is, that is,, 3 4, c d, ⎞, ⎛, −2, 1, ⎝ 3, 1⎠, −, 2, 2, There is, however, a quicker method of obtaining the, inverse of a 2 by 2 matrix., , , p q, For any matrix, the inverse may be, r s, obtained by:, (i) interchanging the positions of p and s,, (ii) changing the signs of q and r, and, (iii) multiplying this new, matrix, by the reciprocal of, p q, the determinant of, r s, , , 1 2, Thus the inverse of matrix, is, 3 4, ⎞, ⎛, , , −2, 1, 4 −2, 1, =⎝ 3, 1⎠, 4 − 6 −3 1, −, 2, 2, as obtained previously., Problem 13. Determine the inverse of, , , 3 −2, 7, 4
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The theory of matrices and determinants, , , , p q, is obtained by interr s, changing the positions of p and s, changing the signs, of q and r and multiplying by the reciprocal of the, p q, determinant, . Thus, the inverse of, r s, , The inverse of matrix, , , , , , 1, 4 2, 3 −2, =, 7, 4, (3 × 4) − (−2 × 7) −7 3, ⎞, ⎛, 1, 2, , , ⎜ 13 13 ⎟, 1, 4 2, ⎟, =, =⎜, ⎝ −7 3 ⎠, 26 −7 3, 26 26, Now try the following exercise, Exercise 95 Further problems on the, inverse of 2 by 2 matrices, , , 3 −1, 1. Determine the inverse of, −4, 7, ⎞⎤, ⎡⎛, 7 1, ⎢⎜ 17 17 ⎟⎥, ⎟⎥, ⎢⎜, ⎣⎝ 4 3 ⎠⎦, 17 17, ⎛, ⎞, 1, 2, ⎜ 2, 3⎟, ⎟, 2. Determine the inverse of ⎜, ⎝ 1, 3⎠, −, −, 5, ⎞⎤, ⎡⎛3, 4, 5, 8 ⎟⎥, ⎢⎜ 7 7, 7 ⎟⎥, ⎢⎜, ⎣⎝ 2, 3 ⎠⎦, −6, −4, 7, 7, , , −1.3, 7.4, 3. Determine the inverse of, 2.5 −3.9, ⎡, , ⎤, 0.290 0.551, ⎣ 0.186 0.097, ⎦, correct to 3 dec. places, , 22.6 The determinant of a 3 by 3, matrix, (i) The minor of an element of a 3 by 3 matrix is, the value of the 2 by 2 determinant obtained by, covering up the row and column containing that, element., , 237, , ⎛, ⎞, 1 2 3, Thus for the matrix ⎝4 5 6⎠ the minor of, 7 8 9, element 4 is obtained ⎛by⎞covering the row, 1, (4 5 6) and the column ⎝4⎠, leaving the 2 by, 7, 2 3, , i.e. the minor of element 4, 2 determinant, 8 9, is (2 × 9) −(3 × 8) = −6., (ii) The sign of a minor depends on its position, within, ⎛, ⎞, + − +, the matrix, the sign pattern being ⎝− + −⎠., + − +, Thus, of element 4 in the matrix, ⎛ the signed-minor, ⎞, 1 2 3, ⎝4 5 6⎠ is − 2 3 = −(−6) = 6., 8 9, 7 8 9, The signed-minor of an element is called the, cofactor of the element., (iii) The value of a 3 by 3 determinant is the, sum of the products of the elements and their, cofactors of any row or any column of the, corresponding 3 by 3 matrix., There are thus six different ways of evaluating a 3 × 3, determinant—and all should give the same value., Problem 14. Find the value of, 3, 4 −1, 2, 0, 7, 1 −3 −2, The value of this determinant is the sum of the products, of the elements and their cofactors, of any row or of any, column. If the second row or second column is selected,, the element 0 will make the product of the element and, its cofactor zero and reduce the amount of arithmetic to, be done to a minimum., Supposing a second row expansion is selected., The minor of 2 is the value of the determinant remaining when the row and column containing the 2 (i.e., the second row and the first column), is covered up., 4 −1, Thus the cofactor of element 2 is, i.e. −11., −3 −2, The sign of element 2 is minus, (see (ii) above), hence, the cofactor of element 2, (the signed-minor) is +11., 3, 4, Similarly the minor of element 7 is, i.e. −13,, 1 −3, and its cofactor is +13. Hence the value of the sum of
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238 Higher Engineering Mathematics, the products of the elements and their cofactors is, 2 × 11 +7 × 13, i.e.,, , = j 2(9) − (1 − j )(5 − j 7), , 3, 4 −1, 2, 0, 7 = 2(11) + 0 + 7(13) = 113, 1 −3 −2, , = j 18 − [5 − j 7 − j 5 + j 27], , The same result will be obtained whichever row or, column is selected. For example, the third column, expansion is, (−1), , = j 2(5 − j 24) − (1 − j )(5 + j 5 − j 12) + 0, , 2, 0, 3, 4, 3 4, −7, + (−2), 1 −3, 1 −3, 2 0, , = j 18 − [−2 − j 12], = j 18 + 2 + j 12 = 2 + j 30 or 30.07∠86.19◦, , Now try the following exercise, , = 6 + 91 + 16 = 113, as obtained previously., , Problem 15., , Exercise 96 Further problems on 3 by 3, determinants, , 1, 4 −3, 2, 6, Evaluate −5, −1 −4, 2, , 1. Find the matrix of minors of, ⎛, ⎞, 4 −7, 6, ⎝−2, 4, 0⎠, 5, 7 −4, ⎡⎛, ⎞⎤, −16, 8 −34, ⎣⎝−14 −46, 63⎠⎦, −24, 12, 2, , 1, 4 −3, −5, 2, 6, Using the first row:, −1 −4, 2, =1, , 2 6, −5 6, −5, 2, −4, + (−3), −4 2, −1 2, −1 −4, , 2. Find the matrix of cofactors of, ⎛, ⎞, 4 −7, 6, ⎝−2, 4, 0⎠, 5, 7 −4, ⎡⎛, ⎞⎤, −16 −8 −34, ⎣⎝ 14 −46 −63⎠⎦, −24 −12, 2, , = (4 + 24) − 4(−10 + 6) − 3(20 + 2), = 28 + 16 − 66 = −22, 1, 4 −3, 2, 6, Using the second column: −5, −1 −4, 2, = −4, , −5 6, 1 −3, 1 −3, +2, −(−4), −1 2, −1, 2, −5, 6, , 3. Calculate the determinant of, ⎛, ⎞, 4 −7, 6, ⎝−2, 4, 0⎠, 5, 7 −4, , = −4(−10 + 6) + 2(2 − 3) + 4(6 − 15), = 16 − 2 − 36 = −22, Problem 16., , Determine the value of, , 8 −2 −10, 4. Evaluate 2 −3 −2, 6, 3, 8, , j2, (1 + j ) 3, (1 − j ), 1, j, 0, j4, 5, Using the first column, the value of the determinant is:, ( j 2), , 1, , j, , j4 5, , − (1 − j ), , (1 + j ) 3, j4, , 5. Calculate the determinant of, ⎛, ⎞, 3.1 2.4, 6.4, ⎝−1.6 3.8 −1.9⎠, 5.3 3.4 −4.8, , [−212], , [−328], , [−242.83], , 5, + (0), , (1 + j ) 3, 1, , j, , j2, 2, j, (1, +, j, ), 1, −3, 6. Evaluate, 5, −j4 0, , [−2 − j ]
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The theory of matrices and determinants, 3∠60◦, j2, 1, 7. Evaluate, 0, (1 + j ) 2∠30◦, 0, 2, j5, , 26.94∠−139.52◦ or, (−20.49 − j 17.49), 8. Find the eigenvalues λ that satisfy the following equations:, (a), , (2 − λ), 2, =0, −1, (5 − λ), , (b), , (5 − λ), 7, −5, 0, (4 − λ), −1, =0, 2, 8, (−3 − λ), , (You may need to refer to chapter 1, pages, 8–12, for the solution of cubic equations)., [(a) λ =3 or 4 (b) λ =1 or 2 or 3], , 22.7 The inverse or reciprocal of a, 3 by 3 matrix, The adjoint of a matrix A is obtained by:, (i) forming a matrix B of the cofactors of A, and, (ii) transposing matrix B to give B T , where B T is the, matrix obtained by writing the rows of B as the, columns of B T . Then adj A = BT ., The inverse of matrix A, A−1 is given by, A−1 =, , adj A, |A|, , where adj A is the adjoint of matrix A and |A| is the, determinant of matrix A., Problem 17. Determine the inverse of the matrix, ⎛, ⎞, 3, 4 −1, ⎜, ⎟, 0, 7⎠, ⎝2, 1 −3 −2, The inverse of matrix A, A−1 =, , adj A, |A|, , 239, , The adjoint of A is found by:, (i) obtaining the matrix of the cofactors of the elements, and, (ii) transposing this matrix., The cofactor of element 3 is +, , 0, 7, = 21., −3 −2, , 2, 7, = 11, and so on., 1 −2, ⎛, ⎞, 21, 11 −6, The matrix of cofactors is ⎝11 −5 13⎠, 28 −23 −8, , The cofactor of element 4 is −, , The transpose of the matrix of cofactors, i.e. the adjoint, of the matrix,, is obtained⎞by writing the rows as columns,, ⎛, 21 11, 28, and is ⎝ 11 −5 −23⎠, −6 13 −8, 3, 4 −1, 0, 7, From Problem 14, the determinant of 2, 1 −3 −2, is 113., ⎛, ⎞, 3, 4 −1, 0, 7⎠ is, Hence the inverse of ⎝2, 1 −3 −2, ⎞, ⎛, 28, 21 11, ⎝ 11 −5 −23⎠, ⎞, ⎛, 21 11, 28, −6 13 −8, 1 ⎝, 11 −5 −23⎠, or, 113, 113, −6 13 −8, Problem 18. Find the inverse of, ⎛, ⎞, 1, 5 −2, ⎜, ⎟, 4⎠, ⎝ 3 −1, −3, 6 −7, Inverse =, , adjoint, determinant, , ⎛, ⎞, −17, 9, 15, The matrix of cofactors is ⎝ 23 −13 −21⎠, 18 −10 −16, The transpose, ⎛ of the matrix⎞ of cofactors (i.e. the, −17, 23, 18, ⎝, 9 −13 −10⎠, adjoint) is, 15 −21 −16
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240 Higher Engineering Mathematics, ⎞, 1, 5 −2, The determinant of ⎝ 3 −1, 4⎠, −3, 6 −7, ⎛, , = 1(7 − 24) − 5(−21 + 12) − 2(18 − 3), = −17 + 45 − 30 = −2, ⎛, ⎞, 1, 5 −2, 4⎠, Hence the inverse of ⎝ 3 −1, −3, 6 −7, ⎛, ⎞, −17, 23, 18, ⎝ 9 −13 −10⎠, 15 −21 −16, =, −2, ⎛, ⎞, 8.5 −11.5 −9, 6.5, 5⎠, = ⎝−4.5, −7.5, 10.5, 8, Now try the following exercise, Exercise 97 Further problems on the, inverse of a 3 by 3 matrix, 1. Write down the transpose of, ⎞, ⎛, 4 −7, 6, ⎝−2, 4, 0⎠, 5, 7 −4, ⎡⎛, , ⎞⎤, 4 −2, 5, ⎣⎝−7, 4, 7⎠⎦, 6, 0 −4, , 2. Write down the transpose of, ⎞, ⎛, 3, 6 21, ⎝ 5 − 2 7⎠, 3, −1, 0 35, ⎡⎛, , ⎞⎤, 3, 5 −1, ⎣⎝ 6 − 2, 0⎠⎦, 3, 1, 3, 7, 2, 5, , 3. Determine the adjoint of, ⎞, ⎛, 4 −7, 6, ⎝−2, 4, 0⎠, 5, 7 −4, ⎞⎤, ⎡⎛, −16, 14 −24, ⎣⎝ −8 −46 −12⎠⎦, −34 −63, 2, 4. Determine the adjoint of, ⎛, ⎞, 3, 6 21, ⎜, ⎟, ⎝ 5 − 23 7⎠, −1, 0 35, ⎡⎛ 2, ⎞⎤, 42 13, − 5 −3 35, ⎢⎜, ⎟⎥, ⎢⎜−10 2 3 −18 1 ⎟⎥, 10, 2 ⎠⎦, ⎣⎝, − 23, −6 −32, 5. Find the inverse of, ⎞, ⎛, 4 −7, 6, ⎝−2, 4, 0⎠, 5, 7 −4, ⎞⎤, ⎛, −16, 14 −24, 1, ⎝ −8 −46 −12⎠⎦, ⎣−, 212, −34 −63, 2, ⎡, , ⎛, , ⎞, 3, 6 12, ⎜, ⎟, 6. Find the inverse of ⎝ 5 − 23 7⎠, 3, −1, 0 5, ⎛ 2, ⎞⎤, ⎡, − 5 −3 35, 42 13, ⎟⎥, ⎢ 15 ⎜, 3, −18 12 ⎠⎦, ⎝−10 2 10, ⎣−, 923, − 23, −6 −32
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Chapter 23, , The solution of simultaneous, equations by matrices and, determinants, (i) Writing the equations in the a1 x + b1 y = c form, gives:, , 23.1 Solution of simultaneous, equations by matrices, (a), , The procedure for solving linear simultaneous, equations in two unknowns using matrices is:, (i) write the equations in the form, a1 x + b1 y = c1, a2 x + b2 y = c2, (ii) write the matrix equation corresponding to, these, equations,, , a1 b1, x, c, i.e., ×, = 1, a2 b2, c2, y, , , a b, (iii) determine the inverse matrix of 1 1, a2 b2, , , 1, b2 −b1, i.e., a1, a1 b2 − b1 a2 −a2, (from Chapter 22), (iv) multiply each side of (ii) by the inverse, matrix, and, (v) solve for x and y by equating corresponding, elements., , Problem 1. Use matrices to solve the, simultaneous equations:, 3x + 5y − 7 = 0, 4x − 3y − 19 = 0, , (1), (2), , 3x + 5y = 7, 4x − 3y = 19, (ii) The matrix equation is, , , 3, 5, x, 7, ×, =, 4 −3, y, 19, , , 3, 5, (iii) The inverse of matrix, is, 4 −3, , , 1, −3 −5, 3, 3 × (−3) − 5 × 4 −4, ⎛3, 5 ⎞, ⎟, ⎜, i.e. ⎝ 29 29 ⎠, 4 −3, 29 29, (iv) Multiplying each side of (ii) by (iii) and, remembering that A × A−1 = I , the unit matrix,, gives:, ⎞, ⎛, 5, 3, , , , ⎜ 29 29 ⎟, 1 0 x, ⎟× 7, =⎜, ⎝ 4 −3 ⎠, 0 1, y, 19, 29, , 29
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242 Higher Engineering Mathematics, ⎞, ⎛, 21 95, , +, ⎜ 29 29 ⎟, x, ⎟, Thus, =⎜, ⎝ 28 57 ⎠, y, −, 29 29, , x, 4, i.e., =, y, −1, , (i) Writing the equations in the a1 x + b1 y + c1 z = d1, form gives:, x +y+z =4, 2x − 3y + 4z = 33, 3x − 2y − 2z = 2, , (v) By comparing corresponding elements:, , (ii) The matrix equation is, ⎛, ⎞ ⎛ ⎞ ⎛ ⎞, 1, 1, 1, x, 4, ⎝2 −3, 4⎠ × ⎝ y ⎠ = ⎝33⎠, 3 −2 −2, z, 2, , x = 4 and y = −1, Checking:, equation (1),, , (iii) The inverse matrix of, ⎛, ⎞, 1, 1, 1, 4⎠, A = ⎝2 −3, 3 −2 −2, , 3 × 4 + 5 × (−1) − 7 = 0 = RHS, equation (2),, 4 × 4 − 3 × (−1) − 19 = 0 = RHS, , is given by, (b) The procedure for solving linear simultaneous, equations in three unknowns using matrices is:, (i) write the equations in the form, a1 x + b1 y + c1 z = d1, a2 x + b2 y + c2 z = d2, a3 x + b3 y + c3 z = d3, (ii) write the matrix equation corresponding to, these equations, i.e., ⎛, ⎞ ⎛ ⎞ ⎛ ⎞, a1 b1 c1, x, d1, ⎝a2 b2 c2 ⎠ × ⎝ y ⎠ = ⎝d2 ⎠, a3 b3 c3, d3, z, (iii) determine the inverse matrix of, ⎞, ⎛, a1 b1 c1, ⎝a2 b2 c2 ⎠ (see Chapter 22), a3 b3 c3, , A−1 =, , adj A, |A|, , The adjoint of A is the transpose of the matrix of, the cofactors of the elements (see Chapter 22). The, matrix of cofactors is, ⎛, ⎞, 14 16 5, ⎝ 0 −5 5⎠, 7 −2 −5, and the transpose of this matrix gives, ⎛, ⎞, 14, 0, 7, adj A = ⎝16 −5 −2⎠, 5, 5 −5, The determinant of A, i.e. the sum of the products, of elements and their cofactors, using a first row, expansion is, , (iv) multiply each side of (ii) by the inverse, matrix, and, (v) solve for x, y and z by equating the corresponding elements., Problem 2. Use matrices to solve the, simultaneous equations:, x + y +z −4 = 0, , (1), , 2x − 3y + 4z − 33 = 0, 3x − 2y − 2z − 2 = 0, , (2), (3), , 1, , −3, 4, 2, 4, 2 −3, −1, +1, −2 −2, 3 −2, 3 −2, = (1 × 14) − (1 × (−16)) + (1 × 5) = 35, , Hence the inverse of A,, ⎛, ⎞, 14, 0, 7, 1, ⎝16 −5 −2⎠, A−1 =, 35, 5, 5 −5, (iv) Multiplying each side of (ii) by (iii), and remembering that A × A−1 = I , the unit matrix, gives
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The solution of simultaneous equations by matrices and determinants, ⎛, ⎞ ⎛ ⎞, 1 00, x, 1, ⎝0 1 0⎠ × ⎝ y ⎠ =, 35, 0 01, z, ⎛, ⎞, (14 × 4) + (0 × 33) + (7 × 2), × ⎝(16 × 4) + ((−5) × 33) + ((−2) × 2)⎠, (5 × 4) + (5 × 33) + ((−5) × 2), ⎛, ⎞, ⎛ ⎞, (14 × 4) + (0 × 33) + (7 × 2), x, 1, ⎝(16 × 4) + ((−5) × 33) + ((−2) × 2)⎠, ⎝y⎠ =, 35 (5 × 4) + (5 × 33) + ((−5) × 2), z, ⎛, , =, , ⎞, , 70, 1, ⎝−105⎠, 35, 175, ⎛, , ⎞, 2, = ⎝−3⎠, 5, (v) By comparing corresponding elements, x = 2,, y = −3, z = 5, which can be checked in the, original equations., Now try the following exercise, Exercise 98 Further problems on solving, simultaneous equations using matrices, , 243, , 6. In two closed loops of an electrical circuit, the, currents flowing are given by the simultaneous, equations:, I1 + 2I2 + 4 = 0, 5I1 + 3I2 − 1 = 0, Use matrices to solve for I1 and I2 ., [I1 = 2, I2 = −3], 7. The relationship between the displacement, s,, velocity, v, and acceleration, a, of a piston is, given by the equations:, s + 2v + 2a = 4, 3s − v + 4a = 25, 3s + 2v − a = −4, Use matrices to determine the values of s, v, and a., [s = 2, v = −3, a = 4], 8. In a mechanical system, acceleration ẍ ,, velocity ẋ and distance x are related by the, simultaneous equations:, 3.4 ẍ + 7.0 ẋ − 13.2x = −11.39, −6.0 ẍ + 4.0 ẋ + 3.5x = 4.98, 2.7 ẍ + 6.0 ẋ + 7.1x = 15.91, Use matrices to find the values of ẍ, ẋ and x., [ẍ = 0.5, ẋ = 0.77, x = 1.4], , In Problems 1 to 5 use matrices to solve the, simultaneous equations given., 1. 3x + 4y = 0, 2x + 5y + 7 = 0, , [x = 4, y = −3], , 2. 2 p +5q + 14.6 = 0, 3.1 p +1.7q + 2.06 =0, , (a), [ p =1.2, q = −3.4], , 3., , x + 2y + 3z =5, 2x − 3y − z = 3, −3x + 4y + 5z = 3, , When solving linear simultaneous equations in, two unknowns using determinants:, (i) write the equations in the form, a1 x + b1 y + c1 = 0, a2 x + b2 y + c2 = 0, , [x = 1, y = −1, z = 2], 4. 3a + 4b − 3c = 2, −2a + 2b + 2c = 15, 7a − 5b + 4c = 26, [a = 2.5, b = 3.5, c = 6.5], 5., , 23.2 Solution of simultaneous, equations by determinants, , p + 2q + 3r + 7.8 = 0, 2 p + 5q − r − 1.4 = 0, 5 p − q + 7r − 3.5 = 0, [ p = 4.1, q = −1.9, r = −2.7], , and then, (ii) the solution is given by, x, −y, 1, =, =, Dx, Dy, D, where Dx =, , b1 c1, b2 c2, , i.e. the determinant of the coefficients left, when the x-column is covered up,
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244 Higher Engineering Mathematics, a1 c1, , Dy =, , a2 c2, , i.e. the determinant of the coefficients left, when the y-column is covered up,, D=, , and, , find the values of u and a, each correct to 4, significant figures., Substituting the given values in v = u +at gives:, , a1 b1, , 21 = u + 3.5a, , (1), , a2 b2, , 33 = u + 6.1a, , (2), , i.e. the determinant of the coefficients left, when the constants-column is covered up., Problem 3. Solve the following simultaneous, equations using determinants:, 3x − 4y = 12, , (i) The equations are written in the form, i.e., and, , (ii) The solution is given by, u, −a, 1, =, =, Du, Da, D, , 7x + 5y = 6.5, Following the above procedure:, (i) 3x − 4y − 12 = 0, 7x + 5y − 6.5 = 0, (ii), , x, −y, 1, =, =, −4 −12, 3 −12, 3 −4, 5 −6.5, 7 −6.5, 7, 5, i.e., , x, (−4)(−6.5) − (−12)(5), =, =, , i.e., , where Du is the determinant of coefficients left, when the u column is covered up,, i.e., , Similarly, Da =, , i.e., Since, , x, 1, 86, =, then x =, =2, 86 43, 43, , and since, −y, 1, 64.5, = then y = −, = −1.5, 64.5 43, 43, Problem 4. The velocity of a car, accelerating at, uniform acceleration a between two points, is given, by v = u +at , where u is its velocity when passing, the first point and t is the time taken to pass, between the two points. If v = 21 m/s when t = 3.5 s, and v = 33 m/s when t = 6.1 s, use determinants to, , 3.5 −21, 6.1 −33, , 1 −21, 1 −33, , = (1)(−33) − (−21)(1), = −12, and, , x, −y, 1, =, =, 26 + 60 −19.5 + 84 15 + 28, x, −y, 1, =, =, 86 64.5 43, , Du =, , = (3.5)(−33) − (−21)(6.1), = 12.6, , −y, (3)(−6.5) − (−12)(7), 1, (3)(5) − (−4)(7), , a1 x + b1 y + c1 = 0,, u + 3.5a − 21 = 0, u + 6.1a − 33 = 0, , D=, , 1 3.5, 1 6.1, , = (1)(6.1) − (3.5)(1) = 2.6, Thus, i.e., and, , u, −a, 1, =, =, 12.6 −12 26, 12.6, = 4.846 m/s, 2.6, 12, a=, = 4.615 m/s2 ,, 2.6, each correct to 4 significant, figures., , u=, , Problem 5. Applying Kirchhoff’s laws to an, electric circuit results in the following equations:, (9 + j 12)I1 − (6 + j 8)I2 = 5, −(6 + j 8)I1 + (8 + j 3)I2 = (2 + j 4), Solve the equations for I1 and I2
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246 Higher Engineering Mathematics, Now try the following exercise, , 3 −4 −26, 87, D I1 = −5 −3, 2, 6 −12, = (3), , −3, 87, −5, 87, − (−4), 6 −12, 2 −12, + (−26), , −5 −3, 2, 6, , = 3(−486) + 4(−114) − 26(−24), , Exercise 99 Further problems on solving, simultaneous equations using determinants, In Problems 1 to 5 use determinants to solve the, simultaneous equations given., 1. 3x − 5y = −17.6, 7y − 2x − 22 = 0, [x = −1.2, y = 2.8], , = −1290, , 2. 2.3m − 4.4n = 6.84, 8.5n − 6.7m = 1.23, , 2 −4 −26, 1 −3, 87, D I2 =, −7, 6 −12, , [m = −6.4, n = −4.9], , = (2)(36 − 522) − (−4)(−12 + 609), + (−26)(6 − 21), = −972 + 2388 + 390, , 3. 3x + 4y + z = 10, 2x − 3y + 5z + 9 = 0, x + 2y − z = 6, [x = 1, y = 2, z = −1], , = 1806, , 4. 1.2 p − 2.3q − 3.1r + 10.1 = 0, , 2, 3 −26, 1, −5, 87, D I3 =, −7, 2 −12, , 4.7 p + 3.8q − 5.3r − 21.5 = 0, 3.7 p − 8.3q + 7.4r + 28.1 = 0, [ p = 1.5, q = 4.5, r = 0.5], , = (2)(60 − 174) − (3)(−12 + 609), + (−26)(2 − 35), = −228 − 1791 + 858 = −1161, D=, , and, , 2, 3 −4, 1 −5 −3, −7, 2, 6, , = (2)(−30 + 6) − (3)(6 − 21), + (−4)(2 − 35), = −48 + 45 + 132 = 129, Thus, I1, −I2, I3, −1, =, =, =, −1290 1806 −1161 129, giving, −1290, = 10 mA,, I1 =, −129, 1806, = 14 mA, 129, 1161, and I 3 =, = 9 mA, 129, I2 =, , 5., , x, y 2z, 1, − +, =−, 2 3, 5, 20, x 2y z, 19, +, − =, 4, 3, 2 40, 59, x +y−z =, 60, , 17, 5, 7, x = , y = ,z = −, 20, 40, 24, , 6. In a system of forces, the relationship between, two forces F1 and F2 is given by:, 5F1 + 3F2 + 6 = 0, 3F1 + 5F2 + 18 = 0, Use determinants to solve for F1 and F2 ., [F1 = 1.5, F2 = −4.5], 7. Applying mesh-current analysis to an a.c., circuit results in the following equations:, (5 − j 4)I1 − (− j 4)I2 = 100∠0◦, (4 + j 3 − j 4)I2 − (− j 4)I1 = 0, Solve the equations for I1 and I2., I1 = 10.77∠19.23◦ A,, I2 = 10.45∠−56.73◦ A
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248 Higher Engineering Mathematics, Hence, , Working backwards, from equation (3 ),, , x=, , Dx, 70, Dy, −105, =, = 2, y =, =, = −3, D, 35, D, 35, , z=, , −35, = 5,, −7, , 175, Dz, and z = D = 35 = 5, , from equation (2 ),, , Now try the following exercise, , from which,, , −5y + 2(5) = 25,, , y=, , Exercise 100 Further problems on solving, simultaneous equations using Cramers rule, , and from equation (1),, , 1. Repeat problems 3, 4, 5, 7 and 8 of Exercise, 98 on page 241, using Cramers rule., 2. Repeat problems 3, 4, 8 and 9 of Exercise 99, on page 244, using Cramers rule., , 23.4, , x + (−3) + 5 = 4,, from which,, x = 4+3−5 = 2, (This is the same example as Problems 2 and 7, and, a comparison of methods can be made). The above, method is known as the Gaussian elimination method., , Solution of simultaneous, equations using the Gaussian, elimination method, , We conclude from the above example that if, a11 x + a12 y + a13 z = b1, , Consider the following simultaneous equations:, x +y+z =4, , (2), , 3x − 2y − 2z = 2, , (3), , Leaving equation (1) as it is gives:, (1), , Equation (2) − 2 × equation (1) gives:, 0 − 5y + 2z = 25, , (2 ), , and equation (3) − 3 × equation (1) gives:, 0 − 5y − 5z = −10, , (3 ), , a31 x + a32 y + a33 z = b3, the three-step procedure to solve simultaneous equations in three unknowns using the Gaussian elimination method is:, a21, × equation (1) to form equa1. Equation (2) −, a11, a31, × equation (1) to, tion (2 ) and equation (3) −, a11, form equation (3 )., a32, × equation (2 ) to form equa2. Equation (3 ) −, a22, tion (3 )., 3., , Leaving equations (1) and (2 ) as they are gives:, x +y+z =4, , (1), , 0 − 5y + 2z = 25, , (2 ), , Equation (3 ) − equation (2 ) gives:, 0 + 0 − 7z = −35, , a21 x + a22 y + a23 z = b2, , (1), , 2x − 3y + 4z = 33, , x +y+z =4, , 25 − 10, = −3, −5, , (3 ), , By appropriately manipulating the three original equations we have deliberately obtained zeros in the positions shown in equations (2 ) and (3 )., , Determine z from equation (3 ), then y from, equation (2 ) and finally, x from equation (1)., , Problem 8. A d.c. circuit comprises three closed, loops. Applying Kirchhoff’s laws to the closed, loops gives the following equations for current flow, in milliamperes:, 2I1 + 3I2 − 4I3 = 26, I1 − 5I2 − 3I3 = −87, −7I1 + 2I2 + 6I3 = 12, , (1), (2), (3)
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The solution of simultaneous equations by matrices and determinants, Now try the following exercise, , Use the Gaussian elimination method to solve for, I1 , I2 and I3 ., (This is the same example as Problem 6 on page 243,, and a comparison of methods may be made), Following the above procedure:, 1. 2I1 + 3I2 − 4I3 = 26, 1, Equation (2) − × equation (1) gives:, 2, 0 − 6.5I2 − I3 = −100, −7, × equation (1) gives:, Equation (3) −, 2, 0 + 12.5I2 − 8I3 = 103, , (1), (2 ), , (3 ), (1), (2 ), , 12.5, × equation (2 ) gives:, −6.5, 0 + 0 − 9.923I3 = −89.308, (3 ), Equation (3 ) −, , 3. From equation (3 ),, −89.308, I3 =, = 9 mA,, −9.923, from equation (2 ), −6.5I2 − 9 =−100,, −100 +9, from which, I 2 =, = 14 mA, −6.5, and from equation (1), 2I1 + 3(14) − 4(9) = 26,, , from which, I 1 =, , 26 − 42 + 36 20, =, 2, 2, , = 10 mA, , 1. In a mass-spring-damper system, the acceleration ẍ m/s2 , velocity ẋ m/s and displacement, x m are related by the following simultaneous, equations:, 6.2 ẍ + 7.9 ẋ + 12.6x = 18.0, 7.5 ẍ + 4.8 ẋ + 4.8x = 6.39, , 2. 2I1 + 3I2 − 4I3 = 26, 0 − 6.5I2 − I3 = −100, , Exercise 101 Further problems on solving, simultaneous equations using Gaussian, elimination, , 13.0 ẍ + 3.5 ẋ − 13.0x = −17.4, By using Gaussian elimination, determine the, acceleration, velocity and displacement for the, system, correct to 2 decimal places., [ẍ = −0.30, ẋ = 0.60, x = 1.20], 2. The tensions, T1 , T2 and T3 in a simple framework are given by the equations:, 5T1 + 5T2 + 5T3 = 7.0, T1 + 2T2 + 4T3 = 2.4, 4T1 + 2T2, , = 4.0, , Determine T1 , T2 and T3 using Gaussian elimination., [T1 = 0.8, T2 = 0.4, T3 = 0.2], 3. Repeat problems 3, 4, 5, 7 and 8 of Exercise 98, on page 241, using the Gaussian elimination, method., 4. Repeat problems 3, 4, 8 and 9 of Exercise 99, on page 244, using the Gaussian elimination, method., , 249
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Revision Test 7, This Revision Test covers the material contained in Chapters 20 to 23. The marks for each question are shown in, brackets at the end of each question., 1. Solve the quadratic equation x 2 − 2x + 5 =0 and, show the roots on an Argand diagram., (9), 2. If Z 1 = 2 + j 5, Z 2 = 1 − j 3 and Z 3 = 4 − j determine, in both Cartesian and polar forms, the value, Z1 Z2, of, + Z 3 , correct to 2 decimal places., Z1 + Z2, (9), 3. Three vectors are represented by A, 4.2∠45◦ , B,, 5.5∠−32◦ and C, 2.8∠75◦. Determine in polar, form the resultant D, where D =B + C − A. (8), 4. Two impedances, Z 1 = (2 + j 7) ohms and, Z 2 = (3 − j 4) ohms, are connected in series to, a supply voltage V of 150∠0◦ V. Determine the, magnitude of the current I and its phase angle, relative to the voltage., (6), , 6., , Determine A × B., , (4), , 7., , Calculate the determinant of matrix C., , (4), , 8., , Determine the inverse of matrix A., , (4), , 9., , Determine E × D., , (9), , 10., , Calculate the determinant of matrix D., , (6), , 11., , Solve the following simultaneous equations:, 4x − 3y = 17, x + y+1 = 0, using matrices., , 12., , In questions 6 to 10, the matrices stated are:, , , , , −5, 2, 1, 6, A=, B=, 7 −8, −3 −4, , , j3, (1 + j 2), C=, (−1 − j 4) − j 2, ⎛, ⎞, ⎛, ⎞, 2 −1 3, −1, 3 0, D = ⎝−5, 1 0 ⎠ E = ⎝ 4 −9 2 ⎠, 4 −6 2, −5, 7 1, , Use determinants to solve the following simultaneous equations:, 4x + 9y + 2z = 21, , 5. Determine in both polar and rectangular forms:, (a) [2.37∠35◦]4 (b) [3.2 − j 4.8]5, √, (c) [−1 − j 3], , (6), , −8x + 6y − 3z = 41, 3x + y − 5z = −73, , (15), 13., , (10), , The simultaneous equations representing the currents flowing in an unbalanced, three-phase, starconnected, electrical network are as follows:, 2.4I1 + 3.6I2 + 4.8I3 = 1.2, −3.9I1 + 1.3I2 − 6.5I3 = 2.6, 1.7I1 + 11.9I2 + 8.5I3 = 0, Using matrices, solve the equations for I1 , I2, and I3 ., (10)
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Chapter 24, , Vectors, 24.1, , Introduction, , This chapter initially explains the difference between, scalar and vector quantities and shows how a vector is, drawn and represented., Any object that is acted upon by an external force will, respond to that force by moving in the line of the force., However, if two or more forces act simultaneously, the, result is more difficult to predict; the ability to add two, or more vectors then becomes important., This chapter thus shows how vectors are added and, subtracted, both by drawing and by calculation, and finding the resultant of two or more vectors has many uses in, engineering. (Resultant means the single vector which, would have the same effect as the individual vectors.), Relative velocities and vector i, j , k notation are also, briefly explained., , Now try the following exercise, Exercise 102 Further problems on scalar, and vector quantities, 1. State the difference between scalar and vector, quantities., In problems 2 to 9, state whether the quantities given are scalar (S) or vector (V) – answers, below., 2. A temperature of 70◦ C, 3. 5 m3 volume, 4. A downward force of 20 N, 5. 500 J of work, 6. 30 cm2 area, 7. A south-westerly wind of 10 knots, , 24.2, , Scalars and vectors, , The time taken to fill a water tank may be measured as,, say, 50 s. Similarly, the temperature in a room may be, measured as, say, 16◦C, or the mass of a bearing may, be measured as, say, 3 kg., Quantities such as time, temperature and mass are, entirely defined by a numerical value and are called, scalars or scalar quantities., Not all quantities are like this. Some are defined by, more than just size; some also have direction. For example, the velocity of a car is 90 km/h due west, or a force, of 20 N acts vertically downwards, or an acceleration of, 10 m/s2 acts at 50◦ to the horizontal., Quantities such as velocity, force and acceleration,, which have both a magnitude and a direction, are, called vectors., , 8. 50 m distance, 9. An acceleration of 15 m/s2 at 60◦ to the, horizontal, [Answers: 2. S 3. S 4. V 5. S 6. S 7. V, 8. S 9. V], , 24.3, , Drawing a vector, , A vector quantity can be represented graphically by a, line, drawn so that:, (a), , the length of the line denotes the magnitude of the, quantity, and, , (b) the direction of the line denotes the direction in, which the vector quantity acts.
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252 Higher Engineering Mathematics, An arrow is used to denote the sense, or direction, of the, vector., The arrow end of a vector is called the ‘nose’ and the, other end the ‘tail’., For example, a force of 9 N acting at 45◦ to the horizontal, is shown in Fig. 24.1., Note that an angle of + 45◦ is drawn from the horizontal, and moves anticlockwise., a, , 9N, 458, , 0, , Figure 24.1, , In this chapter a vector quantity is denoted by bold, print., , 24.4, , Addition of vectors by drawing, , Adding two or more vectors by drawing assumes that, a ruler, pencil and protractor are available. Results, obtained by drawing are naturally not as accurate as, those obtained by calculation., (a) Nose-to-tail method, Two force vectors, F1 and F2 , are shown in Fig. 24.3., When an object is subjected to more than one force,, the resultant of the forces is found by the addition of, vectors., , A velocity of 20 m/s at −60◦ is shown in Fig. 24.2., Note that an angle of −60◦ is drawn from the horizontal, and moves clockwise., , F2, , , 0, , F1, , 60⬚, , Figure 24.3, , 20 m/s, , b, , Figure 24.2, , Representing a vector, There are a number of ways of representing vector, quantities. These include:, 1., 2., , Using bold print, −→, AB where an arrow above two capital letters, denotes the sense of direction, where A is the, starting point and B the end point of the vector, , 3., , AB or a i.e. a line over the top of letters, , 4., , a i.e. an underlined letter, , The force of 9 N at 45◦ shown in Fig. 24.1 may be, represented as:, →, −, 0a or 0a or 0a, The magnitude of the force is 0a, Similarly, the velocity of 20 m/s at −60◦ shown in, Fig. 24.2 may be represented as:, →, −, 0b or 0b or 0b, The magnitude of the velocity is 0b, , To add forces F1 and F2 :, (i) Force F1 is drawn to scale horizontally, shown as, 0a in Fig. 24.4., (ii) From the nose of F1 , force F2 is drawn at angle, θ to the horizontal, shown as ab., (iii) The resultant force is given by length 0b, which, may be measured., This procedure is called the ‘nose-to-tail’ or ‘triangle’, method., , b, , 0, , , F1, , F2, , a, , Figure 24.4, , (b) Parallelogram method, To add the two force vectors, F1 and F2 , of Fig. 24.3:, (i) A line cb is constructed which is parallel to and, equal in length to 0a (see Fig. 24.5).
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253, , Vectors, (ii) A line ab is constructed which is parallel to and, equal in length to 0c., (iii) The resultant force is given by the diagonal of the, parallelogram, i.e. length 0b., This procedure is called the ‘parallelogram’ method., c, , , a, , F1, , (i) In Fig. 24.8, a line is constructed which is parallel, to and equal in length to the 8 N force, (ii) A line is constructed which is parallel to and equal, in length to the 5 N force, (iii) The resultant force is given by the diagonal, of the parallelogram, i.e. length 0b, and is, measured as 12 N and angle θ is measured, as 17◦ ., , b, , F2, 0, , (b) ‘Parallelogram’ method, , b, , Figure 24.5, , Problem 1. A force of 5 N is inclined at an angle, of 45◦ to a second force of 8 N, both forces acting at, a point. Find the magnitude of the resultant of these, two forces and the direction of the resultant with, respect to the 8 N force by:, (a) the ‘nose-to-tail’ method, and (b) the, ‘parallelogram’ method., The two forces are shown in Fig. 24.6. (Although the, 8 N force is shown horizontal, it could have been drawn, in any direction.), , 5N, , 5N, 458, 0, , , 8N, , Figure 24.8, , Thus, the resultant of the two force vectors in Fig. 24.6, is 12 N at 17◦ to the 8 N force., , Problem 2. Forces of 15 N and 10 N are at an, angle of 90◦ to each other as shown in Fig. 24.9., Find, by drawing, the magnitude of the resultant of, these two forces and the direction of the resultant, with respect to the 15 N force., , 458, 8N, , Figure 24.6, 10 N, , (a) ‘Nose-to tail’ method, (i) The 8 N force is drawn horizontally 8 units long,, shown as 0a in Fig. 24.7, (ii) From the nose of the 8 N force, the 5 N force, is drawn 5 units long at an angle of 45◦ to the, horizontal, shown as ab, (iii) The resultant force is given by length 0b and, is measured as 12 N and angle θ is measured, as 17◦., b, 5N, 0, , Figure 24.7, , 458, , , 8N, , a, , 15 N, , Figure 24.9, , Using the ‘nose-to-tail’ method:, (i) The 15 N force is drawn horizontally 15 units, long as shown in Fig. 24.10, (ii) From the nose of the 15 N force, the 10 N force, is drawn 10 units long at an angle of 90◦ to the, horizontal as shown, (iii) The resultant force is shown as R and is measured, as 18 N and angle θ is measured as 34◦ .
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254 Higher Engineering Mathematics, 195⬚, b, , Thus, the resultant of the two force vectors is 18 N at, 34◦ to the 15 N force., r, R, , 10 N, , , a, , 105⬚, , 15 N, , Figure 24.10, , 30⬚, , O, , Problem 3. Velocities of 10 m/s, 20 m/s and, 15 m/s act as shown in Fig. 24.11. Determine, by, drawing, the magnitude of the resultant velocity and, its direction relative to the horizontal., 2, , Figure 24.12, , Worked Problems 1 to 3 have demonstrated how, vectors are added to determine their resultant and their, direction. However, drawing to scale is time-consuming, and not highly accurate. The following sections demonstrate how to determine resultant vectors by calculation, using horizontal and vertical components and, where, possible, by Pythagoras’s theorem., , 20 m/s, , 10 m/s, , 1, , 24.5 Resolving vectors into horizontal, and vertical components, , 308, 158, 3, , 15 m/s, , Figure 24.11, , When more than two vectors are being added the ‘noseto-tail’ method is used., The order in which the vectors are added does not, matter. In this case the order taken is v1 , then v2 , then, v3 . However, if a different order is taken the same result, will occur., (i) v1 is drawn 10 units long at an angle of 30◦ to the, horizontal, shown as 0a in Fig. 24.12, (ii) From the nose of v1 , v2 is drawn 20 units long at, an angle of 90◦ to the horizontal, shown as ab, , A force vector F is shown in Fig. 24.13 at angle θ to the, horizontal. Such a vector can be resolved into two components such that the vector addition of the components, is equal to the original vector., F, , , , Figure 24.13, , The two components usually taken are a horizontal, component and a vertical component., If a right-angled triangle is constructed as shown in, Fig. 24.14, then 0a is called the horizontal component, of F and ab is called the vertical component of F., , (iii) From the nose of v2 , v3 is drawn 15 units long at, an angle of 195◦ to the horizontal, shown as br, , b, F, , (iv) The resultant velocity is given by length 0r and, is measured as 22 m/s and the angle measured to, the horizontal is 105◦., Thus, the resultant of the three velocities is 22 m/s at, 105◦ to the horizontal., , 0, , Figure 24.14, , , a
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Vectors, From trigonometry (see Chapter 11),, 0a, cos θ =, from which,, 0b, , 0a = 0b cos θ, , i.e., , the horizontal component of F = F cos θ, , and, , ab, from which,, sin θ =, 0b, , ab = 0b sin θ, = F sin θ, , the vertical component of F = F sinθ, , i.e., , 17.32 m/s, , = F cos θ, , Problem 4. Resolve the force vector of 50 N at an, angle of 35◦ to the horizontal into its horizontal and, vertical components., The horizontal component of the 50 N force,, 0a = 50 cos 35◦ = 40.96 N, The vertical component of the 50 N force,, ab = 50 sin 35◦ = 28.68 N, The horizontal and vertical components are shown in, Fig. 24.15., , 0, , 20, , 0, , a, 210 m/s, b, , Figure 24.16, , Problem 6. Resolve the displacement vector of, 40 m at an angle of 120◦ into horizontal and vertical, components., The horizontal component of the 40 m displacement,, 0a = 40 cos 120◦ = −20.0 m, The vertical component of the 40 m displacement,, ab = 40 sin 120◦ = 34.64 m, The horizontal and vertical components are shown in, Fig. 24.17., b, 40 N, 34.64 N, a, 220.0 N 0, , 28.68 N, , 358, , m/, , s, , b, 50 N, , 308, , 255, , 1208, , Figure 24.17, 40.96 N, , a, , Figure 24.15, , and, , √, 40.962 + 28.682, , = 50 N, , , 28.68, −1, θ = tan, = 35◦, 40.96, , Thus, the vector addition of components 40.96 N and, 28.68 N is 50 N at 35◦), Problem 5. Resolve the velocity vector of 20 m/s, at an angle of −30◦ to the horizontal into horizontal, and vertical components., , The horizontal component of the 20 m/s velocity,, 0a = 20 cos(−30◦) = 17.32 m/s, The vertical component of the 20 m/s velocity,, ab = 20 sin(−30◦) = −10 m/s, The horizontal and vertical components are shown in, Fig. 24.16., , 24.6 Addition of vectors by, calculation, Two force vectors, F1 and F2 , are shown in Fig. 24.18,, F1 being at an angle of θ1 and F2 being at an angle, of θ2 ., V, , F1 sin 1, F2 sin 2, , (Checking: by Pythagoras, 0b =, , F2, , F1, 1, , 2, , F1 cos 1, F2 cos 2, , Figure 24.18, , H
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256 Higher Engineering Mathematics, A method of adding two vectors together is to use, horizontal and vertical components., The horizontal component of force F1 is F1 cos θ1 and, the horizontal component of force F2 is F2 cos θ2, The total horizontal component of the two forces,, H = F1 cos θ1 + F2 cos θ2, The vertical component of force F1 is F1 sin θ1 and the, vertical component of force F2 is F2 sin θ2, The total vertical component of the two forces,, V = F1 sin θ1 + F2 sin θ2, Since we have H and V , the resultant of F1 and F2, is obtained by using the theorem of Pythagoras. From, H2 + V 2, Fig. 24.19,, 0b 2 = , 2, i.e., resultant = H 2 +, at an angle, V , V, −1, given by θ = tan, H, , The vertical component of the 8 N force is 8 sin 0◦, and the vertical component of the 5 N force is 5 sin 45◦, The total vertical component of the two forces,, V = 8 sin 0◦ + 5 sin 45◦ = 0 + 3.5355, = 3.5355, From Fig. 24.21, magnitude of resultant vector, √, = H2 + V 2, √, = 11.53552 + 3.53552 = 12.07 N, , R, , H ⫽11.5355 N, , nt, , lta, , R, , , 0, , H, , V ⫽ 3.5355 N, , , , b, , u, es, , nt, , lta, , u, es, , V, , Figure 24.21, , a, , The direction of the resultant vector,, , , , V, 3.5355, −1, −1, = tan, θ = tan, H, 11.5355, , Figure 24.19, , Problem 7. A force of 5 N is inclined at an angle, of 45◦ to a second force of 8 N, both forces acting at, a point. Calculate the magnitude of the resultant of, these two forces and the direction of the resultant, with respect to the 8 N force., , The two forces are shown in Fig. 24.20., , = tan −1 0.30648866 . . . = 17.04◦, Thus, the resultant of the two forces is a single vector, of 12.07 N at 17.04◦ to the 8 N vector., Perhaps an easier and quicker method of calculating, the magnitude and direction of the resultant is to use, complex numbers (see Chapter 20)., In this example, the resultant, = 8∠0◦ + 5∠45◦, = (8 cos 0◦ + j 8 sin0◦) + (5 cos 45◦ + j 5 sin45◦ ), , 5N, , = (8 + j 0) + (3.536 + j 3.536), , 458, , = (11.536 + j 3.536) N or 12.07∠17.04◦ N, , 8N, , Figure 24.20, , The horizontal component of the 8 N force is 8 cos 0◦, and the horizontal component of the 5 N force is, 5 cos 45◦, The total horizontal component of the two forces,, H = 8 cos 0◦ + 5 cos 45◦ = 8 + 3.5355, = 11.5355, , as obtained above using horizontal and vertical, components., Problem 8. Forces of 15 N and 10 N are at an, angle of 90◦ to each other as shown in Fig. 24.22., Calculate the magnitude of the resultant of these, two forces and its direction with respect to the, 15 N force.
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Vectors, , 257, , This is, of course, a special case. Pythagoras can only be, used when there is an angle of 90◦ between vectors., This is demonstrated in the next worked problem., 10 N, , Problem 9. Calculate the magnitude and, direction of the resultant of the two acceleration, vectors shown in Fig. 24.24., , 15 N, , Figure 24.22, , The horizontal component of the 15 N force is 15 cos0◦, and the horizontal component of the 10 N force is, 10 cos90◦, The total horizontal component of the two velocities,, , 28 m/s2, , H = 15 cos 0◦ + 10 cos 90◦ = 15 + 0 = 15, 15 sin 0◦, , The vertical component of the 15 N force is, and the vertical component of the 10 N force is 10 sin 90◦, The total vertical component of the two velocities,, V = 15 sin 0◦ + 10 sin 90◦ = 0 + 10 = 10, Magnitude of resultant vector, √, √, = H 2 + V 2 = 152 + 102 = 18.03 N, The direction of the resultant vector,, , , V, 10, −1, −1, θ = tan, = tan, = 33.69◦, H, 15, Thus, the resultant of the two forces is a single vector, of 18.03 N at 33.69◦ to the 15 N vector., , 15 m/s2, , Figure 24.24, , The 15 m/s2 acceleration is drawn horizontally, shown, as 0a in Fig. 24.25., From the nose of the 15 m/s2 acceleration, the 28 m/s2, acceleration is drawn at an angle of 90◦ to the horizontal,, shown as ab., b, , R, , There is an alternative method of calculating the resultant vector in this case., If we used the triangle method, then the diagram would, be as shown in Fig. 24.23., , 28, , , , ␣, a, , 15, , 0, , Figure 24.25, R, , 10 N, , The resultant acceleration, R, is given by length 0b., Since a right-angled triangle results, the theorem of, Pythagoras may be used., , , 15 N, , Since a right-angled triangle results then we could use, Pythagoras’s theorem without needing to go through, the procedure for horizontal and vertical components., In fact, the horizontal and vertical components are 15 N, and 10 N respectively., , and, , , , 152 + 282 = 31.76 m/s2, , −1 28, α = tan, = 61.82◦, 15, , 0b =, , Figure 24.23, , Measuring from the horizontal,, θ = 180◦ − 61.82◦ = 118.18◦
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258 Higher Engineering Mathematics, Thus, the resultant of the two accelerations is a single, vector of 31.76 m/s2 at 118.18◦ to the horizontal., , R, 21.118, , Problem 10. Velocities of 10 m/s, 20 m/s and, 15 m/s act as shown in Fig. 24.26. Calculate the, magnitude of the resultant velocity and its direction, relative to the horizontal., 2, , ␣, , , , 5.829, , Figure 24.27, , Measuring from the horizontal,, θ = 180◦ − 74.57◦ = 105.43◦, Thus, the resultant of the three velocities is a single, vector of 21.91 m/s at 105.43◦ to the horizontal., , 20 m/s, 1, , Using complex numbers, from Fig. 24.26,, , 10 m/s, 308, , resultant = 10∠30◦ + 20∠90◦ + 15∠195◦, , 158, 3, , = (10 cos 30◦ + j 10 sin30◦), , 15 m/s, , + (20 cos 90◦ + j 20 sin90◦ ), , Figure 24.26, , + (15 cos 195◦ + j 15 sin195◦), The horizontal component of the 10 m/s velocity =, 10 cos 30◦ = 8.660 m/s,, the horizontal component of the 20 m/s velocity is, 20 cos 90◦ = 0 m/s,, and the horizontal component of the 15 m/s velocity is, 15 cos195◦ = −14.489 m/s., The total horizontal component of the three velocities,, H = 8.660 + 0 − 14.489 = −5.829 m/s, The vertical component of the 10 m/s velocity =, 10 sin 30◦ = 5 m/s,, the vertical component of the 20 m/s velocity is, 20 sin 90◦ = 20 m/s,, and the vertical component of the 15 m/s velocity is, 15 sin 195◦ = −3.882 m/s., The total vertical component of the three forces,, V = 5 + 20 − 3.882 = 21.118 m/s, From Fig. 24.27, magnitude of resultant vector,, √, √, R = H 2 + V 2 = 5.8292 + 21.1182 = 21.91 m/s, The direction, the resultant, , of, vector,, V, 21.118, −1, −1, α = tan, = tan, = 74.57◦, H, 5.829, , = (8.660 + j 5.000) + (0 + j 20.000), + (−14.489 − j 3.882), = (−5.829 + j 21.118) N or, 21.91∠105.43◦ N, as obtained above using horizontal and vertical, components., The method used to add vectors by calculation will, not be specified – the choice is yours, but probably, the quickest and easiest method is by using complex, numbers., Now try the following exercise, Exercise 103 Further problems on, addition of vectors by calculation, 1., , A force of 7 N is inclined at an angle of 50◦, to a second force of 12 N, both forces acting at a point. Calculate magnitude of the
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Vectors, , resultant of the two forces, and the direction of the resultant with respect to the 12 N, force., [17.35 N at 18.00◦ to the 12 N force], , 8N, , 2. Velocities of 5 m/s and 12 m/s act at a point, at 90◦ to each other. Calculate the resultant, velocity and its direction relative to the 12 m/s, velocity., [13 m/s at 22.62◦ to the 12 m/s velocity], , 708, 5N, 608, , 3. Calculate the magnitude and direction of the, resultant of the two force vectors shown in, Fig. 24.28., [16.40 N at 37.57◦ to the 13 N force], 13 N, , Figure 24.30, 10 N, , 7. If velocity v1 = 25 m/s at 60◦ and v2 = 15 m/s, at −30◦ , calculate the magnitude and direction of v1 + v2 ., , 13 N, , [29.15 m/s at 29.04◦ to the horizontal], , Figure 24.28, , 4. Calculate the magnitude and direction of the, resultant of the two force vectors shown in, Fig. 24.29., [28.43 N at 129.30◦ to the 18 N force], , 8. Calculate the magnitude and direction of the, resultant vector of the force system shown in, Fig. 24.31., [9.28 N at 16.70◦], , 4 N 158, , 8N, , 22 N, 308, 18 N, , 608, , Figure 24.29, , 5. A displacement vector s1 is 30 m at 0◦. A second displacement vector s2 is 12 m at 90◦ ., Calculate magnitude and direction of the, resultant vector s1 + s2 ., [32.31 m at 21.80◦ to the 30 m, displacement], 6. Three forces of 5 N, 8 N and 13 N act as shown, in Fig. 24.30. Calculate the magnitude and, direction of the resultant force., [14.72 N at −14.72◦ to the 5 N force], , 6N, , Figure 24.31, , 9. Calculate the magnitude and direction of, the resultant vector of the system shown in, Fig. 24.32., [6.89 m/s at 159.56◦], , 259
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260 Higher Engineering Mathematics, Fig. 24.34(a) shows that the second diagonal of the, ‘parallelogram’ method of vector addition gives the, magnitude and direction of vector subtraction of oa, from ob., , 2 m/s, 3.5 m/s, 158, , b, , s, , d, , b, , 458, o, , ⫺a, , a, (a), , 308, , o, , a, , (b), , Figure 24.34, 4 m/s, , Problem 11. Accelerations of a1 = 1.5 m/s2 at, 90◦ and a2 = 2.6 m/s2 at 145◦ act at a point. Find, a1 + a2 and a1 − a2 (i) by drawing a scale vector, diagram, and (ii) by calculation., , Figure 24.32, , 10., , An object is acted upon by two forces of magnitude 10 N and 8 N at an angle of 60◦ to each, other. Determine the resultant force on the, object., [15.62 N at 26.33◦ to the 10 N force], A ship heads in a direction of E 20◦ S at a, speed of 20 knots while the current is 4 knots, in a direction of N 30◦ E. Determine the speed, and actual direction of the ship., [21.07 knots, E 9.22◦ S], , 11., , (i) The scale vector diagram is shown in Fig. 24.35., By measurement,, a1 + a2 = 3.7 m/s2 at 126◦, a1 − a2 = 2.1 m/s2 at 0◦, a1 ⫹ a2, 0, , 1, , 2, , 3, , Scale in m/s2, a1, a2, , 24.7, , Vector subtraction, , In Fig. 24.33, a force vector F is represented by oa., The vector (−oa) can be obtained by drawing a vector, from o in the opposite sense to oa but having the same, magnitude, shown as ob in Fig. 24.33, i.e. ob = (−oa), , 2.6 m/s2, , 1.5 m/s2, 126⬚, 145⬚, , a1 ⫺ a2, , ⫺a2, F, , 2F, , a, , o, , b, , Figure 24.33, , For two vectors acting at a point, as shown in, Fig. 24.34(a), the resultant of vector addition is:, os = oa + ob., Figure 24.33(b) shows vectors ob + (−oa), that is,, ob − oa and the vector equation is ob − oa = od. Comparing od in Fig. 24.34(b) with the broken line ab in, , Figure 24.35, , (ii) Resolving horizontally and vertically gives:, Horizontal component of a1 + a2 ,, H = 1.5 cos90◦ +2.6 cos 145◦ = −2.13, Vertical component of a1 + a2 ,, V = 1.5 sin90◦ + 2.6 sin145◦ = 2.99, FromFig. 24.36, magnitude of a1 + a2 ,, R = (−2.13)2 + 2.992 = 3.67 m/s2, , , 2.99, = 54.53◦ and, In Fig. 24.36, α = tan−1, 2.13, θ = 180◦ − 54.53◦ = 125.47◦, Thus,, , a1 + a2 = 3.67 m/s2 at 125.47◦
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Vectors, , 261, , The horizontal component of, v1 − v2 + v3 = (22 cos 140◦) − (40 cos 190◦), + (15 cos 290◦), , R, , 2.99, , = (−16.85) − (−39.39) + (5.13), = 27.67 units, , , , ␣, 22.13, , The vertical component of, , 0, , v1 − v2 + v3 = (22 sin 140◦ ) − (40 sin 190◦ ), + (15 sin 290◦ ), , Figure 24.36, , = (14.14) − (−6.95) + (−14.10), = 6.99 units, , Horizontal component of a1 − a2, = 1.5 cos90◦ − 2.6 cos 145◦ = 2.13, , The magnitude, of the resultant,, R = 27.672 + 6.992 = 28.54 units, , , 6.99, −1, The direction of the resultant R = tan, 27.67, = 14.18◦, Thus, v1 − v2 + v3 = 28.54 units at 14.18◦, Using complex numbers,, v1 − v2 + v3 = 22∠140◦ − 40∠190◦ + 15∠290◦, , Vertical component of a1 − a2, = 1.5 sin 90◦ − 2.6 sin 145◦ = 0, √, Magnitude of a1 − a2 = 2.132 + 02, = 2.13 , m/s2 , 0, Direction of a1 − a2 = tan −1, = 0◦, 2.13, a1 − a2 = 2.13 m/s2 at 0◦, , Thus,, , = (−16.853 + j 14.141), Problem 12. Calculate the resultant of (i), v1 − v2 + v3 and (ii) v2 − v1 − v3 when v1 = 22, units at 140◦ , v2 = 40 units at 190◦ and v3 = 15, units at 290◦ ., , − (−39.392 − j 6.946), + (5.130 − j 14.095), = 27.669 + j 6.992 =28.54∠14.18◦, (ii) The horizontal component of, , (i) The vectors are shown in Fig. 24.37., , v2 − v1 − v3 = (40 cos 190◦) − (22 cos 140◦), − (15 cos 290◦), , 1V, , = (−39.39) − (−16.85) − (5.13), = −27.67 units, The vertical component of, , 22, , v2 − v1 − v3 = (40 sin 190◦ ) − (22 sin 140◦), , 1408, 1908, 2H, 40, , 1H, , 2908, 15, , 2V, , Figure 24.37, , − (15 sin 290◦ ), = (−6.95) − (14.14) − (−14.10), = −6.99 units, From Fig., 24.38 the magnitude of the resultant,, R = (−27.67)2 + (−6.99)2 = 28.54 units, , , 6.99, = 14.18◦ , from which,, and α = tan −1, 27.67, θ = 180◦ + 14.18◦ = 194.18◦
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262 Higher Engineering Mathematics, 24.8, , 227.67, , ␣, , 0, , 26.99, R, , Relative velocity, , For relative velocity problems, some fixed datum point, needs to be selected. This is often a fixed point on the, earth’s surface. In any vector equation, only the start and, finish points affect the resultant vector of a system. Two, different systems are shown in Fig. 24.39, but in each, of the systems, the resultant vector is ad., b, , Figure 24.38, , b, c, , Thus, v2 − v1 − v3 = 28.54 units at 194.18◦, This result is as expected, since v2 − v1 − v3 =, − (v1 − v2 + v3 ) and the vector 28.54 units at, 194.18◦ is minus times (i.e. is 180◦ out of phase, with) the vector 28.54 units at 14.18◦, Using complex numbers,, v 2 − v 2 − v 3 = 40∠190◦ − 22∠140◦ − 15∠290◦, = (−39.392 − j 6.946), − (−16.853 + j 14.141), − (5.130 − j 14.095), = −27.669 − j 6.992, = 28.54∠ −165.82◦ or, 28.54∠194.18◦, , Now try the following exercise, Exercise 104, subtraction, , Further problems on vector, , 1. Forces of F1 = 40 N at 45◦ and F2 = 30 N at, 125◦ act at a point. Determine by drawing and, by calculation: (a) F1 + F2 (b) F1 − F2 ., [(a) 54.0 N at 78.16◦ (b) 45.64 N at 4.66◦ ], 2. Calculate the resultant of (a) v1 + v2 − v3, (b) v3 − v2 + v1 when v1 = 15 m/s at 85◦, v2 =, 25 m/s at 175◦ and v3 = 12 m/s at 235◦., [(a) 31.71 m/s at 121.81◦, (b) 19.55 m/s at 8.63◦ ], , a, , d, , a, d, (b), , (a), , Figure 24.39, , The vector equation of the system shown in Fig. 24.39(a), is:, ad = ab + bd, and that for the system shown in Fig. 24.39(b) is:, ad = ab + bc + cd, Thus in vector equations of this form, only the first and, last letters, ‘a’ and ‘d’, respectively, fix the magnitude, and direction of the resultant vector. This principle is, used in relative velocity problems., Problem 13. Two cars, P and Q, are travelling, towards the junction of two roads which are at right, angles to one another. Car P has a velocity of, 45 km/h due east and car Q a velocity of 55 km/h, due south. Calculate (i) the velocity of car P, relative to car Q, and (ii) the velocity of car Q, relative to car P., , (i) The directions of the cars are shown in, Fig. 24.40(a), called a space diagram. The velocity diagram is shown in Fig. 24.40(b), in which, pe is taken as the velocity of car P relative to, point e on the earth’s surface. The velocity of P, relative to Q is vector pq and the vector equation is pq = pe + eq. Hence the vector directions, are as shown, eq being in the opposite direction, to qe.
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Vectors, From the geometry√of the vector triangle,, 2, the magnitude of pq = 452 + 55, =71.06 km/h, 55, and the direction of pq = tan −1, = 50.71◦, 45, i.e. the velocity of car P relative to car Q is, 71.06 km/h at 50.71◦, , 263, , 3. A ship is heading in a direction N 60◦ E at a, speed which in still water would be 20 km/h., It is carried off course by a current of 8 km/h, in a direction of E 50◦ S. Calculate the ship’s, actual speed and direction., [22.79 km/h, E 9.78◦ N], , N, W, , q, , q, , E, S, P, , Q, , p, , e, , 45 km/h, (a), , i, j and k notation, , 24.9, , 55 km/h, , (b), , p, , e, (c), , Figure 24.40, , A method of completely specifying the direction of a, vector in space relative to some reference point is to use, three unit vectors, i, j and k, mutually at right angles, to each other, as shown in Fig. 24.41., z, , (ii) The velocity of car Q relative to car P is given by, the vector equation qp = qe + ep and the vector, diagram is as shown in Fig. 24.40(c), having ep, opposite in direction to pe., From the geometry, √ of this vector triangle, the mag= 71.06 m/s and the, nitude of qp = 452 +552 , 55, −1, = 50.71◦ but must, direction of qp = tan, 45, lie in the third quadrant, i.e. the required angle is:, 180◦ + 50.71◦ = 230.71◦, i.e. the velocity of car Q relative to car P is, 71.06 m/s at 230.71◦, , Now try the following exercise, Exercise 105, velocity, , k, i, , 0 j, , y, , x, , Figure 24.41, , Calculations involving vectors given in i, j k notation, are carried out in exactly the same way as standard, algebraic calculations, as shown in the worked example, below., , Further problems on relative, , 1. A car is moving along a straight horizontal, road at 79.2 km/h and rain is falling vertically, downwards at 26.4 km/h. Find the velocity of, the rain relative to the driver of the car., [83.5 km/h at 71.6◦ to the vertical], 2. Calculate the time needed to swim across a, river 142 m wide when the swimmer can swim, at 2 km/h in still water and the river is flowing, at 1 km/h. At what angle to the bank should, the swimmer swim?, [4 minutes 55 seconds, 60◦], , Problem 14. Determine:, (3i + 2j + 2k) − (4i − 3j + 2k), , (3i + 2j + 2k) − (4i − 3j + 2k) = 3i + 2j + 2k, − 4i + 3j − 2k, = −i + 5j, Problem 15. Given p = 3i + 2k,, q = 4i − 2j + 3k and r = −3i + 5j − 4k, determine:
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Chapter 25, , Methods of adding, alternating waveforms, 25.1 Combination of two periodic, functions, There are a number of instances in engineering and science where waveforms have to be combined and where, it is required to determine the single phasor (called, the resultant) that could replace two or more separate, phasors. Uses are found in electrical alternating current theory, in mechanical vibrations, in the addition of, forces and with sound waves., There are a number of methods of determining the, resultant waveform. These include:, (a) by drawing the waveforms and adding graphically, (b) by drawing the phasors and measuring the, resultant, (c) by using the cosine and sine rules, (d) by using horizontal and vertical components, (e) by using complex numbers, , 25.2, , Plotting periodic functions, , yR = 3 sin A + 2 cos A and obtain a sinusoidal, expression for this resultant waveform., y1 = 3 sin A and y2 = 2 cos A are shown plotted, in Fig. 25.1. Ordinates may be added at, say, 15◦, intervals. For example,, at 0◦, y1 + y2 = 0 + 2 = 2, at 15◦, y1 + y2 = 0.78 + 1.93 = 2.71, at 120◦, y1 + y2 = 2.60 + −1 = 1.6, at 210◦, y1 + y2 = −1.50 −1.73 = −3.23, and, so on., The resultant waveform, shown by the broken line,, has the same period, i.e. 360◦ , and thus the same frequency as the single phasors. The maximum value, or, , y, , 348, , 3.6, 3, , y1 5 3 sin A, y R 5 3.6 sin (A 1 34)8, , 2, , y2 5 2 cos A, , 1, , This may be achieved by sketching the separate functions on the same axes and then adding (or subtracting), ordinates at regular intervals. This is demonstrated in, the following worked problems., Problem 1. Plot the graph of y1 = 3 sin A from, A = 0◦ to A = 360◦ . On the same axes plot, y2 = 2 cos A. By adding ordinates, plot, , 0, 21, 22, 23, , Figure 25.1, , 908, , 1808, , 2708, , 3608, , A
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266 Higher Engineering Mathematics, amplitude, of the resultant is 3.6. The resultant waveπ, form leads y1 = 3 sin A by 34◦ or 34 ×, rad = 0.593, 180, rad., The sinusoidal expression for the resultant waveform is:, yR = 3.6 sin(A + 34◦ ) or, yR = 3.6 sin(A + 0.593), , y1 = 4 sin ωt and y2 = 3 sin(ωt − π/3) are shown plotted in Fig. 25.2., , 258, y1 5 4 sin t, , 4, , y25 3 sin(t 2 /3), , 2, 0, 22, 24, , y R 5 y1 1 y2, 908, /2, , 1808, , , 2708, 3/2, , 458, , 3.6, , y1 2 y2, y2, , y1, , 4, 2, , 0, 22, , 908, /2, , 1808, , , 2708, 3/2, , 3608, 2, , t, , 24, , Problem 2. Plot the graphs of y1 = 4 sin ωt and, y2 = 3 sin(ωt − π/3) on the same axes, over one, cycle. By adding ordinates at intervals plot, yR = y1 + y2 and obtain a sinusoidal expression for, the resultant waveform., , y, 6.1, 6, , y, , 3608, 2, , t, , 258, , 26, , Figure 25.3, , The amplitude, or peak value of the resultant (shown by, the broken line), is 3.6 and it leads y1 by 45◦ or 0.79, rad. Hence,, y1 − y2 = 3.6 sin(ωt + 0.79), Problem 4. Two alternating currents are given by:, and, i1 = 20 sin ωt, amperes, π, i2 = 10 sin ωt +, amperes., 3, By drawing the waveforms on the same axes and, adding, determine the sinusoidal expression for the, resultant i1 + i2 ., i1 and i2 are shown plotted in Fig. 25.4. The resultant, waveform for i1 + i2 is shown by the broken line. It has, the same period, and hence frequency, as i1 and i2 ., , Figure 25.2, , Ordinates are added at 15◦ intervals and the resultant is shown by the broken line. The amplitude, of the resultant is 6.1 and it lags y1 by 25◦, or 0.436 rad., Hence, the sinusoidal expression for the resultant waveform is:, yR = 6.1 sin(ωt − 0.436), , 30, 26.5, , y1 and y2 are shown plotted in Fig. 25.3. At 15◦ intervals, y2 is subtracted from y1. For example:, , , iR 5 20 sin t 110 sin (t 1 ), 3, , 20, , i1 5 20 sin t, i2 5 10 sin(t 1 ), 3, , 10, 908, 198, , 210, , Problem 3. Determine a sinusoidal expression, for y1 − y2 when y1 = 4 sin ωt and, y2 = 3 sin(ωt − π/3)., , 198, , , 2, , 1808, , 2708, , , , 3, 2, , 3608, 2 angle t, , 220, 230, , Figure 25.4, , ◦, , at 0 , y1 − y2 = 0 − (−2.6) = +2.6, at 30◦ , y1 − y2 = 2 − (−1.5) = +3.5, at 150◦ , y1 − y2 = 2 − 3 = −1, and so on., , The amplitude or peak value is 26.5 A., The resultant waveform leads the waveform of, i1 = 20 sin ωt by 19◦ or 0.33 rad
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Methods of adding alternating waveforms, , 267, , y1 5 4, , Hence, the sinusoidal expression for the resultant i1 + i2, is given by:, , 608 or /3 rads, , iR = i1 + i 2 = 26.5 sin(ωt + 0.33) A, y2 5 3, , Now try the following exercise, , Figure 25.5, y15 4, , Exercise 107 Further problems on plotting, periodic functions, , 3. Express 12 sin ωt + 5 cos ωt in the form, A sin(ωt ± α) by drawing and measurement., [13 sin(ωt + 0.395)], , 25.3 Determining resultant phasors, by drawing, The resultant of two periodic functions may be found, from their relative positions when the time is zero., For example, if y1 = 4 sin ωt and y2 = 3 sin(ωt − π/3), then each may be represented as phasors as shown in, Fig. 25.5, y1 being 4 units long and drawn horizontally, and y2 being 3 units long, lagging y1 by π/3 radians or, 60◦ . To determine the resultant of y1 + y2 , y1 is drawn, horizontally as shown in Fig. 25.6 and y2 is joined to the, end of y1 at 60◦ to the horizontal. The resultant is given, by yR . This is the same as the diagonal of a parallelogram, that is shown completed in Fig. 25.7., Resultant yR , in Figs. 25.6 and 25.7, may be determined, by drawing the phasors and their directions to scale and, measuring using a ruler and protractor., , 608, , 3, , 2. Two alternating voltages are given by, v1 = 10 sin ωt volts and v2 = 14 sin(ωt + π/3), volts. By plotting v1 and v2 on the same axes, over one cycle obtain a sinusoidal expression, for (a) v1 + v2 (b) v1 − v2 ., , (a) 20.9 sin(ωt + 0.63) volts, (b) 12.5 sin(ωt − 1.36) volts, , , , y 25, , 1. Plot the graph of y = 2 sin A from A = 0◦, to A = 360◦ . On the same axes plot, y = 4 cos A. By adding ordinates at intervals, plot y = 2 sin A + 4 cos A and obtain a sinusoidal expression for the waveform., [4.5 sin(A + 63.5◦ )], , 0, , yR, , Figure 25.6, y1 5 4, , , yR, y2 5 3, , Figure 25.7, , In this example, yR is measured as 6 units long and angle, φ is measured as 25◦., 25◦ = 25 ×, , π, radians = 0.44 rad, 180, , Hence, summarising, by drawing: y R = y 1 + y 2 =, 4 sin ωt + 3 sin(ωt − π/3) = 6 sin(ωt − 0.44), If the resultant phasor yR = y1 − y2 is required, then y2, is still 3 units long but is drawn in the opposite direction,, as shown in Fig. 25.8., yR, , , , 2y2 5 3, , 608, y1 5 4, , 608, , y2, , Figure 25.8
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268 Higher Engineering Mathematics, Problem 5. Two alternating currents are, given by: i1= 20 sinωt amperes and, π, amperes. Determine i1 + i2, i2 = 10 sin ωt +, 3, by drawing phasors., The relative positions of i1 and i2 at time t = 0 are shown, π, as phasors in Fig. 25.9, where rad = 60◦ ., 3, The phasor diagram in Fig. 25.10 is drawn to scale with, a ruler and protractor., , 10 A, , 0, , 608, , 20 A, a, , , , iR, , 210A, b, , Figure 25.11, , i2 5 10 A, , Now try the following exercise, 608, , Exercise 108 Further problems on, determining resultant phasors by, drawing, , i1 5 20 A, , Figure 25.9, iR, i2 5 10 A, , , , 608, i1 5 20 A, , Figure 25.10, , The resultant iR is shown and is measured as 26 A and, angle φ as 19◦ or 0.33 rad leading i1 . Hence, by drawing, and measuring:, , 1. Determine a sinusoidal expression for, 2 sin θ + 4 cos θ by drawing phasors., [4.5 sin(A + 63.5◦ )], 2. If v1 = 10 sin ωt volts and v2 = 14 sin(ωt + π/3), volts, determine by drawing phasors, sinusoidal expressions for (a) v1 + v2, (b) v1 − v2. , (a) 20.9 sin(ωt + 0.62) volts, (b) 12.5 sin(ωt − 1.33) volts, 3. Express 12 sin ωt + 5 cos ωt in the form, R sin(ωt ± α) by drawing phasors., [13 sin(ωt + 0.40)], , i R = i 1 + i 2 = 26 sin(ωt + 0.33)A, Problem 6. For the currents in Problem 5,, determine i1 − i2 by drawing phasors., At time t = 0, current i1 is drawn 20 units long horizontally as shown by 0a in Fig. 25.11. Current i2 is, shown, drawn 10 units long in broken line and leading by 60◦ . The current −i2 is drawn in the opposite, direction to the broken line of i2 , shown as ab in, Fig. 25.11. The resultant iR is given by 0b lagging by, angle φ., By measurement, iR = 17 A and φ = 30◦ or, 0.52 rad, Hence, by drawing phasors:, i R = i 1 −i2 = 17 sin(ωt − 0.52), , 25.4 Determining resultant phasors, by the sine and cosine rules, As stated earlier, the resultant of two periodic functions may be found from their relative positions when, the time is zero. For example, if y1 = 5 sin ωt and y2 =, 4 sin(ωt − π/6) then each may be represented by phasors as shown in Fig. 25.12, y1 being 5 units long and, drawn horizontally and y2 being 4 units long, lagging, y1 by π/6 radians or 30◦ . To determine the resultant of, y1 + y2 , y1 is drawn horizontally as shown in Fig. 25.13, and y2 is joined to the end of y1 at π/6 radians, i.e. 30◦, to the horizontal. The resultant is given by yR .
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Methods of adding alternating waveforms, yR, , y1 5 5, , y25 3, , /6 or 308, , y1 5 5 a, , , 308, , (b), , Figure 25.14, y2, , 54, , Using the sine rule:, , 3, 4.6357, =, sin φ, sin 135◦, , from which,, , sin φ =, , b, , Figure 25.13, , Using the cosine rule on triangle 0ab of Fig. 25.13 gives:, yR2 = 52 + 42 − [2(5)(4) cos 150◦], = 25 + 16 − (−34.641), = 75.641, √, from which, yR = 75.641 = 8.697, Using the sine rule,, , 458, , y1 5 2, , (a), , yR, , and, , 1358, , , , y1 5 2, , Figure 25.12, , from which,, , y25 3, , /4 or 458, , y2 5 4, , 0, , 8.697, 4, =, ◦, sin 150, sin φ, 4 sin 150◦, sin φ =, 8.697, = 0.22996, φ = sin−1 0.22996, = 13.29◦ or 0.232 rad, , 3 sin 135◦, = 0.45761, 4.6357, , φ = sin−1 0.45761, , Hence,, , = 27.23◦ or 0.475 rad., Thus, by calculation,, , y R = 4.635 sin(ωt + 0.475), , Problem 8. Determine, , π, 20 sin ωt + 10 sin ωt +, using the cosine, 3, and sine rules., From the phasor diagram of Fig. 25.15, and using the, cosine rule:, iR2 = 202 + 102 − [2(20)(10) cos 120◦], = 700, √, Hence, iR = 700 = 26.46 A, iR, , Hence, yR = y1 + y2 = 5 sin ωt + 4 sin(ωt − π/6), , i2 5 10 A, , = 8.697 sin(ωt − 0.232), Problem 7. Given y1 = 2 sin ωt and, y2 = 3 sin(ωt + π/4), obtain an expression, by, calculation, for the resultant, yR = y1 + y2 ., When time t = 0, the position of phasors y1 and y2, are as shown in Fig. 25.14(a). To obtain the resultant, y1 is drawn horizontally, 2 units long, y2 is drawn, 3 units long at an angle of π/4 rads or 45◦ and joined to, the end of y1 as shown in Fig. 25.14(b)., From Fig. 25.14(b), and using the cosine rule:, yR2 = 22 + 32 − [2(2)(3) cos 135◦], Hence,, , 269, , = 4 + 9 − [−8.485] = 21.49, √, yR = 21.49 = 4.6357, , 608, , , i1 5 20 A, , Figure 25.15, , Using the sine rule gives :, from which,, , 10, 26.46, =, sin φ sin 120◦, 10 sin 120◦, sin φ =, 26.46, = 0.327296, , and, , 0.327296 = 19.10◦, π, = 19.10 ×, = 0.333 rad, 180, , φ = sin, , −1
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270 Higher Engineering Mathematics, b, , Hence, by cosine and sine rules,, iR = i1 + i 2 = 26.46 sin(ωt + 0.333) A, , F, , F sin , , , , Now try the following exercise, , 0, , Exercise 109 Resultant phasors by the sine, and cosine rules, 1. Determine, using the cosine and sine rules, a, sinusoidal expression for:, y = 2 sin A + 4 cos A., [4.5 sin(A + 63.5◦ )], 2. Given v1 = 10 sin ωt volts and, v2 =14 sin(ωt + π/3) volts use the cosine and, sine rules to determine sinusoidal expressions, for (a) v1 + v2 (b) v1 − v2 ., , (a) 20.88 sin(ωt + 0.62) volts, (b) 12.50 sin(ωt − 1.33)volts, In Problems 3 to 5, express the given expressions, in the form A sin(ωt ± α) by using the cosine and, sine rules., 3. 12 sin ωt + 5 cos ωt, [13 sin(ωt + 0.395)], π, 4. 7 sin ωt + 5 sin ωt +, 4, [11.11 sin(ωt + 0.324)], , , , π, 5. 6 sin ωt + 3 sin ωt −, 6, [8.73 sin(ωt − 0.173)], , F cos , , a, , Figure 25.16, , i.e., , the horizontal component of F, H = F cos θ, , and sin θ =, , i.e., , ab, from which ab = 0b sin θ, 0b, = F sin θ, , the vertical component of F, V = F sin θ, , Determining resultant phasors by horizontal and vertical, components is demonstrated in the following worked, problems., Problem 9. Two alternating voltages are given by, v1 = 15 sin ωt volts and v2 = 25 sin(ωt − π/6), volts. Determine a sinusoidal expression for the, resultant vR = v1 + v2 by finding horizontal and, vertical components., The relative positions of v1 and v2 at time t = 0 are, shown in Fig. 25.17(a) and the phasor diagram is shown, in Fig. 25.17(b)., The horizontal component of vR ,, H = 15 cos0◦ + 25 cos(−30◦ ) = 0a + ab = 36.65 V, The vertical component of vR ,, V = 15 sin 0◦ + 25 sin(−30◦ ) = bc = −12.50 V, , vR = 0c = 36.652 + (−12.50)2, Hence,, by Pythagoras’ theorem, = 38.72 volts, , 25.5 Determining resultant phasors, by horizontal and vertical, components, If a right-angled triangle is constructed as shown in, Fig. 25.16, then 0a is called the horizontal component, of F and ab is called the vertical component of F., , tan φ =, , V, −12.50, =, = −0.3411, H, 36.65, , from which, φ = tan−1 (−0.3411) = −18.83◦, or − 0.329 radians., Hence,, , v R = v 1 + v2 = 38.72sin(ωt − 0.329)V, , From trigonometry (see Chapter 11),, 0a, from which,, 0b, 0a = 0b cos θ = F cos θ, cos θ =, , Problem 10. For the voltages in Problem 9,, determine the resultant vR = v1 − v2 using, horizontal and vertical components.
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Methods of adding alternating waveforms, v1 5 15 V, , 0, , /6 or 308, , v1 a, , , 271, , b, , 1508, , 308, , v2, v2 5 25 V, (a), , vR, , c, , (b), , Figure 25.17, i2 5 10 A, , The horizontal component of vR ,, H = 15 cos0◦ − 25 cos(−30◦ ) = −6.65V, 608, , The vertical component of vR ,, V = 15 sin0◦ − 25 sin(−30◦ ) = 12.50V, , Hence, vR = (−6.65)2 + (12.50)2, , i15 20 A, , Figure 25.19, , by Pythagoras’ theorem, = 14.16 volts, tan φ =, , V, 12.50, =, = −1.8797, H, −6.65, , from which, φ = tan −1(−1.8797) = 118.01◦, or 2.06 radians., Hence,, , Total vertical component,, V = 20 sin 0◦ + 10 sin 60◦ = 8.66 , By Pythagoras, the resultant, iR = 25.02 + 8.662, = 26.46 A, , 8.66, −1, Phase angle, φ = tan, = 19.11◦, 25.0, or 0.333 rad, Hence, by using horizontal and vertical components,, , π, 20 sin ωt + 10 sin ωt +, = 26.46 sin(ωt + 0.333), 3, , vR = v1 −v2 = 14.16 sin(ωt + 2.06)V, The phasor diagram is shown in Fig. 25.18., vR, , 2v2 5 25 V, , , , Now try the following exercise, Exercise 110 Further problems on, resultant phasors by horizontal and vertical, components, , 308, v1 5 15 V, , 308, , v2 5 25 V, , Figure 25.18, , Problem 11. Determine, , π, 20 sin ωt + 10 sin ωt +, using horizontal and, 3, vertical components., From the phasors shown in Fig. 25.19:, Total horizontal component,, H = 20 cos0◦ + 10 cos60◦ = 25.0, , In Problems 1 to 4, express the combination of, periodic functions in the form A sin(ωt ± α) by, horizontal and vertical components:, , π, 1. 7 sin ωt + 5 sin ωt +, 4, [11.11 sin(ωt + 0.324)], , π, 2. 6 sin ωt + 3 sin ωt −, 6, [8.73 sin(ωt − 0.173)], , π, 3. i = 25 sin ωt − 15 sin ωt +, 3, [i = 21.79 sin(ωt − 0.639)], , , , 3π, π, −7 sin ωt −, 4. x = 9 sin ωt +, 3, 8, [x = 14.38 sin(ωt + 1.444)]
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272 Higher Engineering Mathematics, 5. The voltage drops across two components when connected in series across, an a.c. supply are: v1 = 200 sin314.2t and, v2 = 120 sin(314.2t − π/5) volts respectively., Determine the:, (a) voltage of the supply (given by v1 + v2 ), in the form A sin(ωt ± α)., , the time is zero. For example, if y1 = 5 sin ωt and y2 =, 4 sin(ωt − π/6) then each may be represented by phasors as shown in Fig. 25.20, y1 being 5 units long and, drawn horizontally and y2 being 4 units long, lagging, y1 by π/6 radians or 30◦ . To determine the resultant of, y1 + y2 , y1 is drawn horizontally as shown in Fig. 25.21, and y2 is joined to the end of y1 at π/6 radians, i.e. 30◦, to the horizontal. The resultant is given by yR ., , (b) frequency of the supply., y1 5 5, , [(a) 305.3 sin(314.2t − 0.233)V, (b) 50 Hz], 6. If the supply to a circuit is v = 20 sin 628.3t, volts and the voltage drop across one of, the components is v1 = 15 sin(628.3t − 0.52), volts, calculate the:, (a) voltage drop across the remainder of, the circuit, given by v − v1 , in the form, A sin(ωt ± α)., , /6 or 308, , y2 5 4, , Figure 25.20, , (b) supply frequency., (c) periodic time of the supply., [(a) 10.21 sin(628.3t + 0.818)V, (b) 100 Hz (c) 10 ms], 7. The voltages across three components in a, series circuit when connected across an a.c., supply are:, , π, volts,, v1 = 25 sin 300 πt +, 6, , π, v2 = 40 sin 300 πt −, volts, and, 4, , π, volts., v3 = 50 sin 300 πt +, 3, Calculate the:, (a) supply voltage, in sinusoidal form, in the, form A sin(ωt ± α)., (b) frequency of the supply., (c), , periodic time., [(a) 79.83 sin (300 πt + 0.352)V, (b) 150 Hz (c) 6.667 ms], , 25.6 Determining resultant phasors, by complex numbers, As stated earlier, the resultant of two periodic functions may be found from their relative positions when, , 0, , y1 5 5 a, , , 308, , y2, , 54, , yR, , b, , Figure 25.21, , π, 6, = 5∠0◦ + 4∠ − 30◦, , In polar form, yR = 5∠0 + 4∠ −, , = (5 + j 0) + (4.33 − j 2.0), = 9.33 − j 2.0 = 9.54∠ − 12.10◦, = 9.54∠−0.21rad, Hence, by using complex numbers, the resultant in, sinusoidal form is:, y1 + y2 = 5 sin ωt + 4 sin(ωt − π/6), = 9.54 sin(ωt−0.21), Problem 12. Two alternating voltages are given, by v1 = 15 sin ωt volts and v2 = 25 sin(ωt − π/6), volts. Determine a sinusoidal expression for the, resultant vR = v1 + v2 by using complex numbers., The relative positions of v1 and v2 at time t = 0 are, shown in Fig. 25.22(a) and the phasor diagram is shown, in Fig. 25.22(b).
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Methods of adding alternating waveforms, v1 5 15 V, , 273, , v1, , , /6 or 308, , 1508, , v2 5 25 V, , vR, , (a), , (b), , Figure 25.22, , In polar form, vR = v1 + v2 = 15∠0 + 25∠ −, , π, 6, , = 15∠0◦ + 25∠ − 30◦, , From the phasors shown in Fig. 25.23, the resultant may, be expressed in polar form as:, i2 5 10 A, , = (15 + j 0) + (21.65 − j 12.5), = 36.65 − j 12.5 = 38.72∠ − 18.83◦, , 608, , = 38.72∠ − 0.329 rad, Hence, by using complex numbers, the resultant in, sinusoidal form is:, , i1 5 20 A, , Figure 25.23, , iR = 20∠0◦ + 10∠60◦, , vR = v1 + v2 = 15 sin ωt + 25 sin(ωt − π/6), = 38.72 sin(ωt − 0.329), , i.e., , = (25 + j 8.66) = 26.46∠19.11◦A or, , Problem 13. For the voltages in Problem 12,, determine the resultant vR = v1 − v2 using complex, numbers., π, In polar form, yR = v1 − v2 = 15∠0 − 25∠ −, 6, , iR = (20 + j 0) + (5 + j 8.66), , 26.46∠0.333 rad A, Hence, by using complex numbers, the resultant in, sinusoidal form is:, iR = i1 + i2 = 26.46 sin(ωt + 0.333)A, , = 15∠0◦ − 25∠ − 30◦, = (15 + j 0) − (21.65 − j 12.5), = −6.65 + j 12.5 = 14.16∠118.01◦, = 14.16∠2.06 rad, Hence, by using complex numbers, the resultant in, sinusoidal form is:, y1 − y2 = 15 sin ωt − 25 sin(ωt − π/6), = 14.16 sin(ωt − 2.06), Problem 14. Determine, , π, 20 sin ωt + 10 sin ωt +, using complex, 3, numbers., , Problem 15. If the supply to a circuit is, v = 30 sin 100 πt volts and the voltage drop across, one of the components is, v1 = 20 sin(100 πt − 0.59) volts, calculate the:, (a) voltage drop across the remainder of the, circuit, given by v − v1 , in the form, A sin(ωt ± α), (b) supply frequency, (c), , periodic time of the supply, , (d) r.m.s. value of the supply voltage, (a), , Supply voltage, v =v1 + v2 where v2 is the voltage, across the remainder of the circuit.
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274 Higher Engineering Mathematics, Hence, v2 = v − v1 = 30 sin 100 πt, − 20 sin(100 πt − 0.59), = 30∠0 − 20∠ − 0.59 rad, = (30 + j 0) − (16.619 − j 11.127), = 13.381 + j 11.127, = 17.40∠0.694 rad, Hence, by using complex numbers, the resultant, in sinusoidal form is:, v − v1 = 30 sin 100 πt − 20 sin(100 πt − 0.59), = 17.40 sin(ωt + 0.694) volts, ω, 100 π, =, = 50 Hz, 2π, 2π, 1, 1, (c) Periodic time, T = =, = 0.02 s or 20 ms, f, 50, , (b) Supply frequency, f =, , (d) R.m.s. value of supply voltage, = 0.707 × 30, = 21.21 volts, , Now try the following exercise, Exercise 111 Further problems on, resultant phasors by complex numbers, In Problems 1 to 4, express the combination of periodic functions in the form A sin(ωt ± α) by using, complex numbers:, , π, 1. 8 sin ωt + 5 sin ωt +, 4, [12.07 sin(ωt + 0.297)], , π, 2. 6 sin ωt + 9 sin ωt −, 6, [14.51 sin(ωt − 0.315)], , π, 3. v = 12 sin ωt − 5 sin ωt −, 4, [9.173 sin(ωt + 0.396)], , , , 3π, π, − 8 sin ωt −, 4. x = 10 sin ωt +, 3, 8, [16.168 sin(ωt + 1.451)], , 5. The voltage drops across two components when connected in series across, an a.c. supply are: v1 = 240 sin 314.2t and, v2 = 150 sin(314.2t − π/5) volts respectively., Determine the:, (a) voltage of the supply (given by v1 + v2 ), in the form A sin(ωt ± α)., (b) frequency of the supply., [(a) 371.95 sin(314.2t − 0.239)V, (b) 50 Hz], 6. If the supply to a circuit is v = 25 sin200πt, volts and the voltage drop across one of, the components is v1 = 18 sin(200πt − 0.43), volts, calculate the:, (a) voltage drop across the remainder of, the circuit, given by v − v1 , in the form, A sin(ωt ± α)., (b) supply frequency., (c) periodic time of the supply., [(a) 11.44 sin(200πt + 0.715)V, (b) 100 Hz (c) 10 ms], 7. The voltages across three components in a, series circuit when connected across an a.c., supply are:, , π, volts,, v1 = 20 sin 300πt −, 6, , π, volts, and, v2 = 30 sin 300πt +, 4, , π, volts., v3 = 60 sin 300πt −, 3, Calculate the:, (a) supply voltage, in sinusoidal form, in the, form A sin(ωt ± α)., (b) frequency of the supply., (c) periodic time., (d) r.m.s. value of the supply voltage., [(a) 79.73 sin(300π − 0.536) V, (b) 150 Hz (c) 6.667 ms (d) 56.37 V]
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Chapter 26, , Scalar and vector products, 26.1, , The unit triad, , When a vector x of magnitude x units and direction θ ◦, is divided by the magnitude of the vector, the result is a, vector of unit length at angle θ ◦ . The unit vector for a, 10 m/s at 50◦, velocity of 10 m/s at 50◦ is, , i.e. 1 at 50◦., 10 m/s, oa, In general, the unit vector for oa is, , the oa being, |oa|, a vector and having both magnitude and direction and, |oa| being the magnitude of the vector only., One method of completely specifying the direction of, a vector in space relative to some reference point is to, use three unit vectors, mutually at right angles to each, other, as shown in Fig. 26.1. Such a system is called a, unit triad., , r, k, , z, , j, , x, iO, a, , b, y, , Figure 26.2, , k, O, , j, , i, 4, , z, 3, 22, i, , k, o, , j, , P, , y, , (a), , x, , Figure 26.1, , In Fig. 26.2, one way to get from o to r is to move x, units along i to point a, then y units in direction j to get, to b and finally z units in direction k to get to r. The, vector or is specified as, or =xi + yj + zk, , k, O, r, , j, i, , 2, 5, , Problem 1. With reference to three axes drawn, mutually at right angles, depict the vectors, (i) op = 4i +3j −2k and (ii) or= 5i − 2j +2k., The required vectors are depicted in Fig. 26.3, op being, shown in Fig. 26.3(a) and or in Fig. 26.3(b)., , 22, (b), , Figure 26.3
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276 Higher Engineering Mathematics, b, , 26.2 The scalar product of two, vectors, v2, , When vector oa is multiplied by a scalar quantity, say k,, the magnitude of the resultant vector will be k times the, magnitude of oa and its direction will remain the same., Thus 2 ×(5 N at 20◦) results in a vector of magnitude, 10 N at 20◦ ., One of the products of two vector quantities is called the, scalar or dot product of two vectors and is defined as, the product of their magnitudes multiplied by the cosine, of the angle between them. The scalar product of oa and, ob is shown as oa • ob. For vectors oa = oa at θ1 , and, ob = ob at θ2 where θ2 > θ1 , the scalar product is:, , , , a, , O, , c, v2 cos , v1, (a), , v2, , oa • ob = oa ob cos(θ 2 − θ 1 ), , s, , v1, , For vectors v1 and v 2 shown in Fig. 26.4, the scalar, product is:, , co, , , , v 1 • v2 = v1 v2 cos θ, , v1, (b), v1, , Figure 26.6, , , , The projection of ob on oa is shown in Fig. 26.6(a) and, by the geometry of triangle obc, it can be seen that the, projection is v2 cos θ. Since, by definition, , v2, , Figure 26.4, , oa • ob = v1 (v2 cos θ),, The commutative law of algebra, a × b = b × a applies, to scalar products. This is demonstrated in Fig. 26.5. Let, oa represent vector v1 and ob represent vector v2 . Then:, oa • ob = v1 v2 cos θ (by definition of, a scalar product), , it follows that, oa • ob = v1 (the projection of v2 on v1 ), Similarly the projection of oa on ob is shown in, Fig. 26.6(b) and is v1 cos θ. Since by definition, ob • oa = v2 (v1 cos θ),, , b, , v2, , O, , it follows that, ob • oa = v2 (the projection of v1 on v2 ), , , , v1, , a, , Figure 26.5, , Similarly, ob • oa = v2 v1 cos θ = v1 v2 cos θ by the commutative law of algebra. Thus oa • ob = ob • oa., , This shows that the scalar product of two vectors, is the product of the magnitude of one vector and, the magnitude of the projection of the other vector on it., The angle between two vectors can be expressed in, terms of the vector constants as follows:, Because a • b = a b cos θ,, then, , cos θ =, , a•b, ab, , (1)
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Scalar and vector products, The direction cosines are:, cos α = , , The work done is F • d, that is F • AB in this case, , 3, = √ = 0.802, 2, 2, 2, 14, x +y +z, x, , y, , cos β = , , 2, = √ = 0.535, 14, x 2 + y2 + z2, , and cos γ = , , 1, = √ = 0.267, 2, 2, 2, 14, x +y +z, y, , (and hence α = cos−1 0.802 = 36.7◦, β = cos−1 0.535 =, 57.7◦ and γ = cos−1 0.267 =74.5◦)., Note that cos2 α + cos2 β + cos2 γ = 0.8022 + 0.5352 +, 0.2672 = 1., , Practical application of scalar product, Problem 6. A constant force of, F =10i + 2j −k newtons displaces an object from, A =i + j +k to B =2i − j +3k (in metres). Find the, work done in newton metres., , i.e. work done = (10i + 2j − k) • (i − 2j + 2k), But from equation (2),, a • b = a1 b1 + a2 b2 + a3 b3, Hence work done =, (10 × 1) + (2 × (−2)) + ((−1) × 2) = 4 Nm., (Theoretically, it is quite possible to get a negative, answer to a ‘work done’ problem. This indicates that, the force must be in the opposite sense to that given, in, order to give the displacement stated.), Now try the following exercise, Exercise 112, products, 1., , Further problems on scalar, , Find the scalar product a • b when, (i) a =i + 2j − k and b =2i + 3j +k, (ii) a =i − 3j +k and b = 2i + j +k, [(i) 7 (ii) 0], , One of the applications of scalar products is to the work, done by a constant force when moving a body. The work, done is the product of the applied force and the distance, moved in the direction of the force., i.e. work done = F • d, The principles developed in Problem 13, page 262,, apply equally to this problem when determining the, displacement. From the sketch shown in Fig. 26.8,, , Given p =2i − 3j, q = 4j −k and, r =i + 2j −3k, determine the quantities, stated in problems 2 to 8., 2., , (a) p • q (b) p • r, , [(a) −12 (b) −4], , 3., , (a) q • r (b) r • q, , 4., , (a) | p | (b) | r |, , [(a) 11 (b) 11], √, √, [(a) 13 (b) 14], , 5., , (a) p • (q + r) (b) 2r • (q − 2p), [(a) −16 (b) 38], , AB = AO+ OB = OB − OA, , 6., , (a) | p +r | (b) | p | +| r |, , that is AB = (2i − j + 3k) − (i + j + k), = i − 2j + 2k, B (2, 21, 3), , A (1,1,1), , [(a), , √, 19 (b) 7.347], , 7., , Find the angle between (a) p and q, (b) q and r., [(a) 143.82◦ (b) 44.52◦], , 8., , Determine the direction cosines of (a) p, (b) q (c) r., ⎡, ⎤, (a) 0.555, −0.832, 0, ⎣ (b) 0, 0.970, −0.243, ⎦, (c) 0.267, 0.535, −0.802, , 9., , Determine the angle between the forces:, F1 = 3i + 4j + 5k and, , O (0, 0, 0), , Figure 26.8, , 279, , F2 = i + j + k, , [11.54◦]
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Scalar and vector products, Squaring both sides of a vector product equation gives:, , (ii) From equation (7), , (|a × b|)2 = a 2 b2 sin2 θ = a 2 b2(1 − cos2 θ), , |a × b| =, , = a 2 b2 − a 2 b2 cos2 θ, , (6), , Now, , a a = a cos θ., 2, , = 14, and, , a•b, ., ab, , Multiplying both sides of this equation by, squaring gives:, a 2b2 cos2 θ =, , a • a = (1)(1) + (4 × 4) + (−2)(−2), b • b = (2)(2) + (−1)(−1) + (3)(3), , But θ = 0◦, thus a • a = a 2, Also, cos θ =, , [(a • a)(b • b) − (a • b)2 ], , = 21, , It is stated in Section 26.2 that a • b = ab cos θ, hence, •, , , , a 2 b2, , and, , a 2b2 (a • b)2, = (a • b)2, a 2b2, , Substituting in equation (6) above for a 2 = a • a, b2 = b • b, , Thus, , a • b = (1)(2) + (4)(−1) + (−2)(3), = −8, , |a × b| = (21 × 14 − 64), √, = 230 = 15.17, , Problem 8. If p = 4i + j −2k, q =3i − 2j + k and, r = i −2k find (a) ( p −2q) × r (b) p × (2r × 3q)., , and a 2 b2 cos2 θ = (a • b)2 gives:, , (a) ( p − 2q) × r = [4i + j − 2k, , (|a × b|)2 = (a • a)(b • b) − (a • b)2, , − 2(3i − 2j + k)] × (i − 2k), , That is,, , = (−2i + 5j − 4k) × (i − 2k), , |a × b| =, , [(a • a)(b • b) − (a • b) ], 2, , (7), , Problem 7. For the vectors a =i + 4j −2k and, b =2i − j +3k find (i) a × b and (ii) |a × b|., (i) From equation (5),, i, j k, a × b = a1 a2 a3, b1 b2 b3, , i, , j, , k, , = −2 5 −4, 1 0 −2, from equation (5), =i, , 5 −4, −2 −4, −j, 0 −2, 1 −2, +k, , −2 5, 1 0, , = i(−10 − 0) − j(4 + 4), , a a, a a, a a, = i 2 3 −j 1 3 +k 1 2, b2 b3, b1 b3, b1 b2, , + k(0 − 5), i.e., ( p − 2q) × r = −10i − 8j −5k, , Hence, i, j, k, 4 −2, a×b = 1, 2 −1, 3, =i, , 4 −2, 1 −2, 1, 4, −j, +k, −1, 3, 2, 3, 2 −1, , = i(12 − 2) − j(3 + 4) + k(−1 − 8), = 10i − 7j −9k, , (b) (2r × 3q) = (2i − 4k) × (9i − 6j + 3k), i j k, = 2 0 −4, 9 −6 3, = i(0 − 24) − j(6 + 36), + k(−12 − 0), = −24i − 42j −12k, , 281
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282 Higher Engineering Mathematics, The magnitude of M,, , Hence, , |M| = |r × F|, , = [(r • r)(F • F) − (r • F)2 ], , p × (2r × 3q) = (4i + j − 2k), × (−24i − 42j − 12k), , r • r = (1)(1) + (2)(2) + (3)(3) = 14, , i, j, k, =, 4, 1 −2, −24 −42 −12, , F • F = (1)(1) + (2)(2) + (−3)(−3) = 14, r • F = (1)(1) + (2)(2) + (3)(−3) = −4, , |M| = [14 × 14 − (−4)2 ], √, = 180 Nm = 13.42 Nm, , = i(−12 − 84) − j(−48 − 48), + k(−168 + 24), = −96i +96j − 144k, or −48(2i − 2j +3k), Practical applications of vector products, Problem 9. Find the moment and the magnitude, of the moment of a force of (i + 2j −3k) newtons, about point B having co-ordinates (0, 1, 1), when, the force acts on a line through A whose, co-ordinates are (1, 3, 4)., The moment M about point B of a force vector F which, has a position vector of r from A is given by:, , Problem 10. The axis of a circular cylinder, coincides with the z-axis and it rotates with an, angular velocity of (2i − 5j + 7k) rad/s. Determine, the tangential velocity at a point P on the cylinder,, whose co-ordinates are ( j + 3k) metres, and also, determine the magnitude of the tangential velocity., The velocity v of point P on a body rotating with angular, velocity ω about a fixed axis is given by:, v = ω × r,, where r is the point on vector P., v = (2i − 5j + 7k) × ( j + 3k), , Thus, M =r×F, , i, j k, = 2 −5 7, 0, 1 3, , r is the vector from B to A, i.e. r = BA., But BA = BO + OA = OA − OB (see Problem 13,, page 262), that is:, , = i(−15 − 7) − j(6 − 0) + k(2 − 0), = (−22i − 6j +2k) m/s, , r = (i + 3j + 4k) − ( j + k), = i + 2j + 3k, , The magnitude of v,, , |v| = [(ω • ω)(r • r) − (r • ω)2 ], , Moment,, M = r × F = (i + 2j + 3k) × (i + 2j − 3k), k, i j, 3, = 1 2, 1 2 −3, , r • r = (0)(0) + (1)(1) + (3)(3) = 10, ω • r = (2)(0) + (−5)(1) + (7)(3) = 16, Hence,, , = i(−6 − 6) − j(−3 − 3), + k(2 − 2), = −12i + 6j Nm, , ω • ω = (2)(2) + (−5)(−5) + (7)(7) = 78, , , (78 × 10 − 162 ), √, = 524 m/s = 22.89 m/s, , |v| =
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Scalar and vector products, , 283, , Now try the following exercise, Exercise 113, products, , Further problems on vector, , In problems 1 to 4, determine the quantities, stated when, p =3i +2k, q =i − 2j +3k and, r =−4i +3j − k., 1. (a) p × q (b) q × p, [(a) 4i − 7j −6k (b) −4i + 7j +6k], 2. (a) |p × r| (b) |r × q|, [(a) 11.92 (b) 13.96], 3. (a) 2p × 3r (b) (p +r) × q, (a) −36i −30j −54k, (b) 11i +4j −k, , magnitude about point Q having co-ordinates, (4, 0, −1) metres., M = (5i + 8j − 2k) Nm,, |M| = 9.64 Nm, 9. A sphere is rotating with angular velocity ω, about the z-axis of a system, the axis coinciding with the axis of the sphere. Determine the, velocity vector and its magnitude at position, (−5i +2j − 7k) m, when the angular velocity, is (i + 2j) rad/s., υ = −14i +7j +12k,, |υ|= 19.72 m/s, 10. Calculate the velocity vector and its magnitude for a particle rotating about the z-axis, at an angular velocity of (3i − j +2k) rad/s, when the position vector of the particle is at, (i − 5j +4k) m., [6i −10j −14k, 18.22 m/s], , 4. (a) p × (r × q) (b) (3p × 2r) × q, (a) −22i − j +33k, (b) 18i +162j +102k, 5. For vectors p =4i − j +2k and, q =−2i +3j − 2k determine: (i) p • q, (ii) p × q (iii) |p ×q| (iv) q × p and, (v) the angle between the vectors., ⎤, ⎡, (i) −15 (ii) −4i + 4j +10k, ⎥, ⎢, ⎣ (iii) 11.49 (iv) 4i −4j − 10k ⎦, , 26.4, , Vector equation of a line, , The equation of a straight line may be determined, given, that it passes through the point A with position vector, a relative to O, and is parallel to vector b. Let r be the, position vector of a point P on the line, as shown in, Fig. 26.10., , (v) 142.55◦, , b, P, , 6. For vectors a =−7i + 4j + 12 k and b =6i −, 5j −k find (i) a • b (ii) a × b (iii) |a ×b|, (iv) b ×a and (v) the angle between the, vectors., ⎤, ⎡, (i) −62 12 (ii) −1 12 i − 4j +11k, ⎥, ⎢, ⎣(iii) 11.80 (iv) 1 12 i +4j − 11k ⎦, (v) 169.31◦, 7. Forces of (i + 3j), (−2i − j), (i − 2j) newtons, act at three points having position vectors of, (2i + 5j), 4j and (−i + j) metres respectively., Calculate the magnitude of the moment., , A, , r, a, , O, , Figure 26.10, , [10 Nm], 8. A force of (2i − j + k) newtons acts on a line, through point P having co-ordinates (0, 3, 1), metres. Determine the moment vector and its, , By vector addition, OP = OA + AP,, i.e. r = a +AP., However, as the straight line through A is parallel to the, free vector b (free vector means one that has the same
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Scalar and vector products, is parallel to the vector 2i + 7j −4k. Determine, the point on the line corresponding to λ =2 in, the resulting equation., ⎡, ⎤, r = (5 + 2λ)i + (7λ − 2)j, ⎣, ⎦, + (3 − 4λ)k;, r = 9i + 12j − 5k, 2. Express the vector equation of the line in, problem 1 in standard Cartesian form., , x −5 y +2 3−z, =, =, =λ, 2, 7, 4, , In problems 3 and 4, express the given straight line, equations in vector form., 3., , 3x − 1 5y + 1 4 − z, =, =, 4, 2, 3, r = 13 (1 + 4λ)i + 15 (2λ − 1)j, + (4 − 3λ)k, , 4. 2x + 1 =, , 1 −4y 3z −1, =, 5, 4, r = 12 (λ − 1)i + 14 (1 − 5λ)j, + 13 (1 + 4λ)k, , 285
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Revision Test 8, This Revision Test covers the material contained in Chapters 24 to 26. The marks for each question are shown in, brackets at the end of each question., 1. State whether the following are scalar or vector, quantities:, (a) A temperature of 50◦C, (b) A downward force of 80 N, (c), , 70 m distance, , (f) An acceleration of 25 m/s2 at 30◦ to the, horizontal, (6), 2. Calculate the resultant and direction of the force, vectors shown in Fig. RT8.1, correct to 2 decimal, places., (7), 5N, , 7N, , 3. Four coplanar forces act at a point A as shown, in Fig. RT8.2 Determine the value and direction of the resultant force by (a) drawing (b) by, calculation using horizontal and vertical components., (10), 4N, A, 458, , Figure RT8.2, , 5. If velocity v1 = 26 m/s at 52◦ and v2 = 17 m/s, at −28◦ calculate the magnitude and direction, of v1 + v2 , correct to 2 decimal places, using, complex numbers., (10), , 7. If a = 2i + 4j −5k and b =3i − 2j +6k determine:, (i) a ·b (ii) |a +b| (iii) a × b (iv) the angle between, a and b., (14), 8. Determine the work done by a force of F newtons, acting at a point A on a body, when A is displaced, to point B, the co-ordinates of A and B being (2, 5,, −3) and (1, −3, 0) metres respectively, and when, F = 2i −5j + 4k newtons., (4), , 458, , 7N, 8N, , Plot the two voltages on the same axes to scales, π, of 1 cm = 50 volts and 1 cm = rad., 6, Obtain a sinusoidal expression for the resultant, v1 + v2 in the form R sin(ωt + α): (a) by adding, ordinates at intervals and (b) by calculation., (13), , 6. Given a = −3i + 3j + 5k, b = 2i − 5j + 7k and, c = 3i + 6j − 4k, determine the following:, (i) −4b (ii) a + b − c (iii) 5b − 3c., (8), , Figure RT8.1, , 5N, , v1 = 150 sin(ωt + π/3) volts and, v2 = 90 sin(ωt − π/6) volts, , 300 J of work, , (d) A south-westerly wind of 15 knots, (e), , 4. The instantaneous values of two alternating voltages are given by:, , 9. A force of F =3i −4j + k newtons acts on a line, passing through a point P. Determine moment M, and its magnitude of the force F about a point Q, when P has co-ordinates (4, −1, 5) metres and Q, has co-ordinates (4, 0, −3) metres., (8)
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Chapter 27, , Methods of differentiation, 27.1, , Introduction to calculus, , Calculus is a branch of mathematics involving or leading to calculations dealing with continuously varying, functions – such as velocity and acceleration, rates, of change and maximum and minimum values of, curves., Calculus has widespread applications in science and, engineering and is used to solve complicated problems, for which algebra alone is insufficient., Calculus is a subject that falls into two parts:, (i) differential calculus, or differentiation, which, is covered in Chapters 27 to 36, and, , f(x), B, , A, C, , f(x2), , f(x1), E, x1, , 0, , D, x2, , x, , Figure 27.2, , (ii) integral calculus, or integration, which is covered in Chapters 37 to 44., , 27.2, , For the curve shown in Fig. 27.2, let the points A and, B have co-ordinates (x 1 , y1) and (x 2 , y2), respectively., In functional notation, y1 = f (x 1 ) and y2 = f (x 2 ) as, shown., , The gradient of a curve, , If a tangent is drawn at a point P on a curve, then the, gradient of this tangent is said to be the gradient of the, curve at P. In Fig. 27.1, the gradient of the curve at P, is equal to the gradient of the tangent PQ., f (x), , The gradient of the chord AB, =, , BC BD − CD, f (x 2 ) − f (x 1 ), =, =, AC, ED, (x 2 − x 1 ), , For the curve f (x) = x 2 shown in Fig. 27.3., (i) the gradient of chord AB, , Q, , =, P, , f (3) − f (1) 9 − 1, =, =4, 3−1, 2, , (ii) the gradient of chord AC, 0, , Figure 27.1, , x, , =, , f (2) − f (1) 4 − 1, =, =3, 2−1, 1
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288 Higher Engineering Mathematics, y, , f(x), 10, , B, , f(x) 5 x 2, , 8, B (x 1 ␦x, y 1 ␦y), 6, ␦y, , 4, , C, , 2, , A, , f(x 1 ␦x), , A(x, y), , D, , ␦x, , f(x), 0, , 1, , 1.5, , 2, , 3, , x, x, , 0, , Figure 27.3, Figure 27.4, , (iii) the gradient of chord AD, f (1.5) − f (1) 2.25 − 1, =, = 2.5, =, 1.5 − 1, 0.5, (iv) if E is the point on the curve (1.1, f (1.1)) then, the gradient of chord AE, =, , f (1.1) − f (1) 1.21 − 1, =, = 2.1, 1.1 − 1, 0.1, , (v) if F is the point on the curve (1.01, f (1.01)) then, the gradient of chord AF, =, , f (1.01) − f (1) 1.0201 − 1, =, = 2.01, 1.01 − 1, 0.01, , Thus as point B moves closer and closer to point A the, gradient of the chord approaches nearer and nearer to the, value 2. This is called the limiting value of the gradient, of the chord AB and when B coincides with A the chord, becomes the tangent to the curve., , 27.3 Differentiation from first, principles, In Fig. 27.4, A and B are two points very close together, on a curve, δx (delta x) and δy (delta y) representing, small increments in the x and y directions, respectively., δy, Gradient of chord AB = ; however,, δx, δy = f (x + δx) − f (x)., δy, f (x + δx) − f (x), Hence, =, ., δx, δx, , δy, As δx approaches zero,, approaches a limiting value, δx, and the gradient of the chord approaches the gradient of, the tangent at A., When determining the gradient of a tangent to a curve, there are two notations used. The gradient of the curve, at A in Fig. 27.4 can either be written as, δy, or limit, δx→0 δx, δx→0, , , , limit, , In Leibniz notation,, , f (x + δx) − f (x), δx, , , , δy, dy, = limit, dx δx→0 δx, , In functional notation,, f (x) = limit, , , , δx→0, , f (x +δx) − f (x), δx, , , , dy, is the same as f (x) and is called the differential, dx, coefficient or the derivative. The process of finding the, differential coefficient is called differentiation., Problem 1. Differentiate from first principle, f (x) = x 2 and determine the value of the gradient, of the curve at x = 2., To ‘differentiate from first principles’ means ‘to find, f (x)’ by using the expression, , f (x) = limit, , δx→0, , f (x) = x 2, , f (x + δx) − f (x), δx
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Methods of differentiation, Substituting (x + δx) for x gives, f (x + δx) = (x + δx)2 = x 2 + 2xδx + δx 2 , hence, , 2, (x + 2xδx + δx 2 ) − (x 2 ), f (x) = limit, δx→0, δx, , = limit, , δx→0, , , + δx 2 ), , (2xδx, δx, , y, A, , (a), , 27.4 Differentiation of common, functions, From differentiation by first principles of a number of, examples such as in Problem 1 above, a general rule, for differentiating y = ax n emerges, where a and n are, constants., The rule is: if y = axn then, , dy, = anxn−1, dx, , f (x)= axn then f (x)= anxn−1 ) and is true for all, , (or, if, real values of a and n., For example, if y = 4x 3 then a = 4 and n =3, and, dy, = anx n−1 = (4)(3)x 3−1 = 12x 2, dx, If y = ax n and n =0 then y = ax 0 and, , dy, = (a)(0)x 0−1 = 0,, dx, i.e. the differential coefficient of a constant is zero., Figure 27.5(a) shows a graph of y = sin x. The gradient is continually changing as the curve moves from, dy, 0 to A to B to C to D. The gradient, given by, , may, dx, be plotted in a corresponding position below y = sin x,, as shown in Fig. 27.5(b)., , B, 0, ⫺, , D, , , , , 2, , 3, 2, , 2, , x rad, , C, , δx→0, , Differentiation from first principles can be a lengthy, process and it would not be convenient to go through this, procedure every time we want to differentiate a function., In reality we do not have to because a set of general, rules have evolved from the above procedure, which we, consider in the following section., , y ⫽ sin x, , ⫹, , = limit [2x + δx], As δx → 0, [2x + δx] →[2x + 0]. Thus f (x) = 2x, i.e., the differential coefficient of x 2 is 2x. At x = 2, the, gradient of the curve, f (x) = 2(2) = 4., , 289, , 0⬘, dy, dx, ⫹, (b), , 0, ⫺, , D⬘, , d, (sin x) ⫽ cos x, dx, A⬘, , 2, , C⬘, , , 3, 2, , 2, , x rad, , B⬘, , Figure 27.5, , (i) At 0, the gradient is positive and is at its steepest., Hence 0 is a maximum positive value., (ii) Between 0 and A the gradient is positive but is, decreasing in value until at A the gradient is zero,, shown as A ., (iii) Between A and B the gradient is negative but, is increasing in value until at B the gradient is at, its steepest negative value. Hence B is a maximum negative value., (iv) If the gradient of y = sin x is further investigated, dy, between B and D then the resulting graph of, dx, is seen to be a cosine wave. Hence the rate of, change of sin x is cos x,, i.e. if y = sin x then, , dy, = cos x, dx, , By a similar construction to that shown in Fig. 27.5 it, may be shown that:, if y = sin ax then, , dy, = a cos ax, dx, , If graphs of y = cos x, y = ex and y = ln x are plotted, and their gradients investigated, their differential coefficients may be determined in a similar manner to that, shown for y = sin x. The rate of change of a function is, a measure of the derivative.
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290 Higher Engineering Mathematics, The standard derivatives summarized below may be, proved theoretically and are true for all real values of x, , In general, the differential coefficient of a constant is always zero., (b) Since y = 6x, in the general rule a = 6 and n =1., , y or f (x), , dy, or f (x), dx, , ax n, , anx n−1, , sin ax, , a cos ax, , cos ax, , −a sin ax, , eax, , aeax, , ln ax, , 1, x, , The differential coefficient of a sum or difference is, the sum or difference of the differential coefficients of, the separate terms., , Hence, , In general, the differential coefficient of kx, where, k is a constant, is always k., Problem 4. Find the derivatives of, √, 5, (a) y = 3 x (b) y = √, 3 4, x, (a), , √, y = 3 x is rewritten in the standard differential, 1, , form as y = 3x 2 ., In the general rule, a = 3 and n =, , Thus, if f (x) = p(x) + q(x) − r(x),, (where f, p, q and r are functions),, then, , dy, = (6)(1)x 1−1 = 6x 0 = 6, dx, , , 1, 1, 3 1, dy, Thus, = (3), x 2 −1 = x − 2, dx, 2, 2, , f (x) = p (x) + q (x) − r (x), , Differentiation of common functions is demonstrated in, the following worked problems., Problem 2., , Find the differential coefficients of, 12, (a) y = 12x 3 (b) y = 3, x, , If y = ax n then, , dy, = anx n−1, dx, , (a) Since y = 12x 3 , a = 12 and n =3 thus, dy, = (12)(3)x 3−1 = 36x2, dx, 12, (b) y = 3 is rewritten in the standard ax n form as, x, y = 12x −3 and in the general rule a = 12 and, n = − 3., 36, dy, Thus, = (12)(−3)x −3−1 = −36x −4 = − 4, dx, x, Problem 3., (a), , Differentiate (a) y = 6 (b) y = 6x., , y = 6 may be written as y = 6x 0 , i.e. in the general, rule a = 6 and n =0., Hence, , dy, = (6)(0)x 0−1 = 0, dx, , 1, 2, , =, , 3, 2x, , (b), , 1, 2, , 3, = √, 2 x, , 4, 5, 5, = 4 = 5x − 3 in the standard differeny= √, 3 4, x, x3, tial form., In the general rule, a = 5 and n =− 43, , Thus, , , , dy, 4 − 4 −1 −20 − 7, = (5) −, x 3 =, x 3, dx, 3, 3, =, , −20, 7, 3x 3, , −20, = √, 3, 3 x7, , Problem 5., , Differentiate, with respect to x,, 1, 1, y = 5x 4 + 4x − 2 + √ − 3., 2x, x, y = 5x 4 + 4x −, , 1, 1, + √ − 3 is rewritten as, 2x 2, x, , 1, 1, y = 5x 4 + 4x − x −2 + x − 2 −3, 2, When differentiating a sum, each term is differentiated, in turn.
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Methods of differentiation, Thus, , dy, 1, = (5)(4)x 4−1 + (4)(1)x 1−1 − (−2)x −2−1, dx, 2, , 1 − 1 −1, + (1) −, x 2 −0, 2, 1 3, = 20x 3 + 4 + x −3 − x − 2, 2, , dy, 1, 1, i.e., = 20x3 + 4 + 3 − √, dx, x, 2 x3, Problem 6. Find the differential coefficients of, (a) y = 3 sin 4x (b) f (t ) = 2 cos3t with respect to, the variable., (a), , When y = 3 sin 4x then, , dy, = (3)(4 cos 4x), dx, = 12 cos 4x, , (b) When f (t ) = 2 cos 3t then, f (t ) = (2)(−3 sin 3t ) =−6 sin 3t, Problem 7. Determine the derivatives of, 2, (a) y = 3e5x (b) f (θ) = 3θ (c) y = 6 ln 2x., e, (a), , When y = 3e5x then, , (b), , f (θ) =, , dy, = (3)(5)e 5x = 15e5x, dx, , Problem 9. Determine the co-ordinates of the, point on the graph y = 3x 2 − 7x + 2 where the, gradient is −1., The gradient of the curve is given by the derivative., dy, = 6x − 7, dx, Since the gradient is −1 then 6x − 7 =−1, from which,, x =1, , When y = 3x 2 − 7x + 2 then, , When x = 1, y = 3(1)2 − 7(1) + 2 = −2, Hence the gradient is −1 at the point (1, −2)., , Now try the following exercise, Exercise 115 Further problems on, differentiating common functions, In Problems 1 to 6 find the differential coefficients of the given functions with respect to the, variable., 1., , f (θ) = (2)(−3)e−30 = −6e−3θ =, (c), , 2., , 3., , Problem 8. Find the gradient of the curve, y = 3x 4 − 2x 2 + 5x − 2 at the points (0, −2), and (1, 4)., The gradient of a curve at a given point is given by, the corresponding value of the derivative. Thus, since, y = 3x 4 − 2x 2 + 5x − 2, Then the gradient =, , 4., , dy, = 12x 3 − 4x + 5, dx, , At the point (0, −2), x = 0, Thus the gradient =12(0)3 − 4(0) + 5 =5, At the point (1, 4), x = 1, Thus the gradient =12(1)3 − 4(1) + 5 = 13., , (a) 5x 5 (b) 2.4x 3.5 (c), , 1, x, , , 1, (a) 25x 4 (b) 8.4x 2.5 (c) − 2, x, , 2, = 2e−3θ , thus, e3θ, , −6, e3θ, , 1, dy, 6, When y = 6 ln 2x then, =6, =, dx, x, x, , 291, , 5., , −4, (a) 2 (b) 6 (c) 2x, x, , , (a), , 8, (b) 0 (c) 2, x3, , √, √, 4, 3, (a) 2 x (b) 3 x 5 (c) √, x, , √, 1, 2, 3, (a) √ (b) 5 x 2 (c) − √, x, x3, −3, (a) √, (b) (x − 1)2 (c) 2 sin 3x, 3, x, ⎡, ⎤, 1, (a), √, (b), 2(x, −, 1), 3 4, ⎢, ⎥, x, ⎣, ⎦, (c) 6 cos 3x, 3, (a) −4 cos 2x (b) 2e6x (c) 5x, e, , −15, (a) 8 sin 2x (b) 12e6x (c) 5x, e
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294 Higher Engineering Mathematics, Note that the differential coefficient is not obtained by, merely differentiating each term in turn and then dividing the numerator by the denominator. The quotient, formula must be used when differentiating quotients., Problem 15. Determine the differential, coefficient of y = tan ax., , Let u = t e2t and v = 2 cos t then, du, dv, = (t )(2e2t ) + (e2t )(1) and, = −2 sin t, dt, dt, du, dv, dy v dx − u dx, =, Hence, dx, v2, , sin ax, . Differentiation of tan ax is thus, cos ax, treated as a quotient with u = sin ax and v = cos ax, , =, , (2 cos t )[2t e2t + e2t ] − (t e2t )(−2 sin t ), (2 cos t )2, , =, , 4t e2t cos t + 2e2t cos t + 2t e2t sin t, 4 cos2 t, , =, , 2e2t [2t cos t + cos t + t sin t ], 4 cos2 t, , y = tan ax =, , du, dv, dy v dx − u dx, =, dx, v2, (cos ax)(a cos ax) − (sin ax)(−a sin ax), =, (cos ax)2, a cos2 ax + a sin2 ax, a(cos2 ax + sin2 ax), =, (cos ax)2, cos2 ax, a, =, , sincecos2 ax + sin2 ax = 1, cos2 ax, (see Chapter 15), =, , dy, 1, Hence, = a sec2 ax since sec2 ax =, (see, dx, cos2 ax, Chapter 11)., Problem 16., , Find the derivative of y = sec ax., , 1, y = sec ax =, (i.e. a quotient). Let u = 1 and, cos, ax, v = cos ax, du, dv, v, −u, dy, = dx 2 dx, dx, v, , i.e., , =, , (cos ax)(0) − (1)(−a sin ax), (cos ax)2, , =, , , , , sin ax, a sin ax, 1, =, a, cos2 ax, cos ax, cos ax, , dy, = a sec ax tan ax, dx, , Problem 17., , Differentiate y =, , i.e., , dy, e2t, =, (2t cos t + cos t +t sin t), dx 2 cos2 t, , Problem 18. Determine the gradient, of the curve, √ , √, 3, 5x, ., 3,, at the point, y= 2, 2x + 4, 2, Let y = 5x and v = 2x 2 + 4, du, dv, v, −u, dy, (2x 2 + 4)(5) − (5x)(4x), dx, dx, =, =, 2, dx, v, (2x 2 + 4)2, 10x 2 + 20 − 20x 2, 20 − 10x 2, =, (2x 2 + 4)2, (2x 2 + 4)2, , √ , √, √, 3, , x = 3,, At the point, 3,, 2, √, dy, 20 − 10( 3)2, √, hence the gradient =, =, dx, [2( 3)2 + 4]2, =, , =, , 20 − 30, 1, =−, 100, 10, , Now try the following exercise, Exercise 117, Further problems on, differentiating quotients, , t e2t, 2 cost, , t e2t, The function, is a quotient, whose numerator is a, 2 cost, product., , In Problems 1 to 7, differentiate the quotients with, respect to the variable., , x cos x − sin x, sin x, 1., x, x2
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Chapter 28, , Some applications of, differentiation, 28.1, , Rates of change, , If a quantity y depends on and varies with a quantity, dy, x then the rate of change of y with respect to x is, ., dx, Thus, for example, the rate of change of pressure p with, dp, height h is, ., dh, A rate of change with respect to time is usually just, called ‘the rate of change’, the ‘with respect to time’, being assumed. Thus, for example, a rate of change of, di, current, i, is, and a rate of change of temperature,, dt, dθ, θ, is, , and so on., dt, Problem 1. The length l metres of a certain, metal rod at temperature θ ◦ C is given by, l = 1 + 0.00005θ + 0.0000004θ 2. Determine the, rate of change of length, in mm/◦ C, when the, temperature is (a) 100◦ C and (b) 400◦C., dl, The rate of change of length means, ., dθ, Since length, then, (a), , l = 1 +0.00005θ + 0.0000004θ 2,, dl, = 0.00005 + 0.0000008θ, dθ, , When θ = 100◦C,, dl, = 0.00005 + (0.0000008)(100), dθ, = 0.00013 m/◦C, = 0.13 mm/◦ C, , (b) When θ = 400◦C,, dl, = 0.00005 + (0.0000008)(400), dθ, = 0.00037 m/◦C, = 0.37 mm/◦ C, Problem 2. The luminous intensity I candelas, of a lamp at varying voltage V is given by, I = 4 ×10−4 V 2 . Determine the voltage at which the, light is increasing at a rate of 0.6 candelas per volt., The rate of change of light with respect to voltage is, dI, given by, ., dV, Since, , I = 4 × 10−4 V 2 ,, dI, = (4 × 10−4)(2)V = 8 × 10−4 V, dV, , When the light is increasing at 0.6 candelas per volt then, +0.6 = 8 × 10−4 V , from which, voltage, V=, , 0.6, = 0.075 × 10+4, 8 × 10−4, , = 750 volts, Problem 3. Newtons law of cooling is given by, θ = θ0 e−kt , where the excess of temperature at zero, time is θ0◦ C and at time t seconds is θ ◦ C. Determine, the rate of change of temperature after 40 s, given, that θ0 = 16◦C and k = −0.03
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300 Higher Engineering Mathematics, The rate of change of temperature is, Since, , θ = θ0 e−kt, dθ, = (θ0 )(−k)e−kt = −kθ0 e−kt, dt, , then, When, then, , dθ, dt, , θ0 = 16, k = −0.03 and t = 40, dθ, = −(−0.03)(16)e−(−0.03)(40), dt, = 0.48e1.2 = 1.594◦C/s, , Problem 4. The displacement s cm of the end, of a stiff spring at time t seconds is given by, s = ae−kt sin 2π f t . Determine the velocity of the, end of the spring after 1 s, if a = 2, k = 0.9 and, f = 5., ds, Velocity, v =, where s = ae−kt sin 2π f t (i.e. a, dt, product)., Using the product rule,, ds, = (ae−kt )(2π f cos 2π f t ), dt, + (sin 2π f t )(−ake−kt ), , rate of change of current when t = 20 ms, given, that f = 150 Hz., [3000π A/s], 2. The luminous intensity, I candelas, of a lamp, is given by I = 6 × 10−4 V 2 , where V is the, voltage. Find (a) the rate of change of luminous, intensity with voltage when V = 200 volts, and, (b) the voltage at which the light is increasing, at a rate of 0.3 candelas per volt., [(a) 0.24 cd/V (b) 250 V], 3. The voltage across the plates of a capacitor at, any time t seconds is given by v = V e−t /C R ,, where V , C and R are constants., Given V = 300 volts, C = 0.12 × 10−6 F and, R = 4 ×106 find (a) the initial rate of change, of voltage, and (b) the rate of change of voltage, after 0.5 s., [(a) −625 V/s (b) −220.5 V/s], 4. The pressure p of the atmosphere at height h, above ground level is given by p = p0e−h/c ,, where p0 is the pressure at ground level, and c is a constant. Determine the rate, of change of pressure with height when, p0 = 1.013 × 105 pascals and c = 6.05 × 104 at, 1450 metres., [−1.635 Pa/m], , When a = 2, k = 0.9, f = 5 and t = 1,, velocity, v = (2e−0.9 )(2π5 cos 2π5), + (sin 2π5)(−2)(0.9)e−0.9, = 25.5455 cos10π − 0.7318 sin 10π, = 25.5455(1) − 0.7318(0), = 25.55 cm/s, (Note that cos10π means ‘the cosine of 10π radians’,, not degrees, and cos 10π ≡ cos 2π = 1.), Now try the following exercise, Exercise 120, change, , Further problems on rates of, , 1. An alternating current, i amperes, is given by, i = 10 sin 2πf t , where f is the frequency in, hertz and t the time in seconds. Determine the, , 28.2, , Velocity and acceleration, , When a car moves a distance x metres in a time t seconds, along a straight road, if the velocity v is constant then, x, v = m/s, i.e. the gradient of the distance/time graph, t, shown in Fig. 28.1 is constant., If, however, the velocity of the car is not constant then, the distance/time graph will not be a straight line. It may, be as shown in Fig. 28.2., The average velocity over a small time δt and distance, δx is given by the gradient of the chord AB, i.e. the, δx, ., average velocity over time δt is, δt, As δt → 0, the chord AB becomes a tangent, such that, at point A, the velocity is given by:, v=, , dx, dt, , Hence the velocity of the car at any instant is given by, the gradient of the distance/time graph. If an expression
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Velocity, , Distance, , Some applications of differentiation, , x, , 301, , D, , ␦v, t, , C, ␦t, Time, , Time, , Figure 28.1, , Distance, , Figure 28.3, , The acceleration is given by the second differential, coefficient of distance x with respect to time t ., Summarizing, if a body moves a distance x metres, in a time t seconds then:, , B, , (i) distance x = f(t)., ␦x, A, ␦t, Time, , dx, , which is the gradient of, (ii) velocity v = f (t) or, dt, the distance/time graph., d2 x, dv, (iii) acceleration a = = f (t) or 2 , which is the, dt, dt, gradient of the velocity/time graph., , Figure 28.2, , for the distance x is known in terms of time t then the, velocity is obtained by differentiating the expression., The acceleration a of the car is defined as the rate, of change of velocity. A velocity/time graph is shown, in Fig. 28.3. If δv is the change in v and δt the, δv, corresponding change in time, then a = ., δt, As δt → 0, the chord CD becomes a tangent, such that, at point C, the acceleration is given by:, a=, , dv, dt, , Hence the acceleration of the car at any instant is, given by the gradient of the velocity/time graph. If an, expression for velocity is known in terms of time t, then the acceleration is obtained by differentiating the, expression., dx, dv, Acceleration a = . However, v = . Hence, dt, dt, , d2 x, d dx, = 2, a=, dt dt, dx, , Problem 5. The distance x metres moved, by a car in a time t seconds is given by, x = 3t 3 − 2t 2 + 4t − 1. Determine the velocity and, acceleration when (a) t = 0 and (b) t = 1.5 s., Distance, , x = 3t 3 − 2t 2 + 4t − 1 m, , Velocity, , v=, , Acceleration a =, (a), , dx, = 9t 2 − 4t + 4 m/s, dt, d2 x, = 18t − 4 m/s2, dx 2, , When time t = 0,, velocity v = 9(0)2 − 4(0) + 4 =4 m/s and, acceleration a = 18(0) − 4 = −4 m/s2 (i.e., deceleration), , (b) When time t = 1.5 s,, velocity v = 9(1.5)2 − 4(1.5) + 4 =18.25 m/s, and acceleration a = 18(1.5) − 4 =23 m/s2, , a
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302 Higher Engineering Mathematics, Problem 6. Supplies are dropped from a, helicoptor and the distance fallen in a time, t seconds is given by x = 12 gt 2, where g = 9.8 m/s2., Determine the velocity and acceleration of the, supplies after it has fallen for 2 seconds., 1, 1, x = gt 2 = (9.8)t 2 = 4.9t 2 m, 2, 2, dv, v=, = 9.8t m/s, dt, , Distance, Velocity, and acceleration, , a=, , d2 x, = 9.8 m/s2, dt 2, , When time t = 2 s,, velocity, v = (9.8)(2) = 19.6 m/s, and acceleration a = 9.8 m/s2, (which is acceleration due to gravity)., Problem 7. The distance x metres travelled by a, vehicle in time t seconds after the brakes are, applied is given by x = 20t − 53 t 2. Determine (a) the, speed of the vehicle (in km/h) at the instant the, brakes are applied, and (b) the distance the car, travels before it stops., (a) Distance, x = 20t − 53 t 2., , Problem 8. The angular displacement θ radians, of a flywheel varies with time t seconds and follows, the equation θ = 9t 2 − 2t 3 . Determine (a) the, angular velocity and acceleration of the flywheel, when time, t = 1 s, and (b) the time when the, angular acceleration is zero., (a) Angular displacement θ = 9t 2 − 2t 3 rad, Angular velocity ω =, , dθ, = 18t − 6t 2 rad/s, dt, , When time t = 1 s,, ω = 18(1) − 6(1)2 = 12 rad/s, Angular acceleration α =, When time t = 1 s,, , d2θ, = 18 − 12t rad/s2, dt 2, , α = 18 − 12(1) = 6 rad/s2, (b) When the angular acceleration is zero,, 18 − 12t = 0, from which, 18 =12t , giving time,, t = 1.5 s., Problem 9. The displacement x cm of the slide, valve of an engine is given by, x = 2.2 cos 5πt + 3.6 sin 5πt . Evaluate the, velocity (in m/s) when time t = 30 ms., , 10, dx, = 20 − t ., dt, 3, At the instant the brakes are applied, time = 0., , Displacement x = 2.2 cos 5πt + 3.6 sin 5πt, , Hence velocity, v = 20 m/s, , Velocity v =, , Hence velocity v =, , =, , 20 × 60 × 60, km/h, 1000, , = 72 km/h, (Note: changing from m/s to km/h merely involves, multiplying by 3.6.), (b) When the car finally stops, the velocity is zero, i.e., 10, 10, v = 20 − t = 0, from which, 20 = t , giving, 3, 3, t = 6 s., Hence the distance travelled before the car stops, is given by:, x = 20t − 53 t 2 = 20(6) − 53 (6)2, = 120 − 60 = 60 m, , dx, dt, , = (2.2)(−5π) sin 5πt + (3.6)(5π) cos 5πt, = −11π sin 5πt + 18π cos 5πt cm/s, When time t = 30 ms, velocity, , , , , 30, 30, = −11π sin 5π · 3 + 18π cos 5π · 3, 10, 10, = −11π sin 0.4712 + 18π cos 0.4712, = −11π sin 27◦ + 18π cos 27◦, = −15.69 + 50.39 = 34.7 cm/s, = 0.347 m/s
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303, , Some applications of differentiation, Now try the following exercise, , ⎡, , (a) 100 m/s (b) 4 s, (c) 200 m, , (d) −100 m/s, , 2. The distance s metres travelled by a car, in t seconds after the brakes are applied is, given by s = 25t − 2.5t 2. Find (a) the speed, of the car (in km/h) when the brakes are, applied, (b) the distance the car travels before it, stops., [(a) 90 km/h (b) 62.5 m], 3. The equation θ = 10π + 24t − 3t 2 gives the, angle θ, in radians, through which a wheel, turns in t seconds. Determine (a) the time, the wheel takes to come to rest, (b) the, angle turned through in the last second of, movement., [(a) 4 s (b) 3 rads], 4. At any time t seconds the distance x metres, of a particle moving in a straight line from, a fixed point is given by x = 4t + ln(1 − t )., Determine (a) the initial velocity and, acceleration (b) the velocity and acceleration, after 1.5 s (c) the time when the velocity is, zero., ⎤, ⎡, (a) 3 m/s; −1 m/s2, ⎥, ⎢, ⎢(b) 6 m/s; −4 m/s2⎥, ⎦, ⎣, (c), , (c) t = 6.28 s, 6., , 20t 3 23t 2, x=, −, + 6t + 5 represents the dis3, 2, tance, x metres, moved by a body in t seconds., Determine (a) the velocity and acceleration, at the start, (b) the velocity and acceleration, when t = 3 s, (c) the values of t when the, body is at rest, (d) the value of t when the, acceleration is 37 m/s2 and (e) the distance, travelled in the third second., ⎤, ⎡, (a) 6 m/s; −23 m/s2, ⎥, ⎢, ⎢(b) 117 m/s; 97 m/s2⎥, ⎥, ⎢, ⎥, ⎢(c) 3 s or 2 s, ⎥, ⎢, 4, 5, ⎥, ⎢, ⎦, ⎣(d) 1 12 s, (e) 75 16 m, , 28.3, , Turning points, , In Fig. 28.4, the gradient (or rate of change) of the, curve changes from positive between O and P to, negative between P and Q, and then positive again, between Q and R. At point P, the gradient is zero, and, as x increases, the gradient of the curve changes, from positive just before P to negative just after. Such, a point is called a maximum point and appears as the, ‘crest of a wave’. At point Q, the gradient is also zero, and, as x increases, the gradient of the curve changes, from negative just before Q to positive just after. Such, a point is called a minimum point, and appears as the, ‘bottom of a valley’. Points such as P and Q are given, the general name of turning points., y, R, , 3, 4s, , 5. The angular displacement θ of a rotating disc is, t, given by θ = 6 sin , where t is the time in sec4, onds. Determine (a) the angular velocity of the, disc when t is 1.5 s, (b) the angular acceleration, when t is 5.5 s, and (c) the first time when the, angular velocity is zero., , ⎤, , ⎥, ⎢, ⎣(b) α = −0.37 rad/s2 ⎦, , Exercise 121 Further problems on velocity, and acceleration, 1. A missile fired from ground level rises, x metres vertically upwards in t seconds and, 25, x = 100t − t 2. Find (a) the initial velocity, 2, of the missile, (b) the time when the height of, the missile is a maximum, (c) the maximum, height reached, (d) the velocity with which the, missile strikes the ground., , (a) ω = 1.40 rad/s, , P, Positive, gradient, , O, , Negative, gradient, , Positive, gradient, , Q, x, , Figure 28.4
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304 Higher Engineering Mathematics, It is possible to have a turning point, the gradient on, either side of which is the same. Such a point is given, the special name of a point of inflexion, and examples, are shown in Fig. 28.5., Maximum, point, , y, Maximum, point, , Problem 10. Locate the turning point on the, curve y = 3x 2 − 6x and determine its nature by, examining the sign of the gradient on either side., , dy, = 6x − 6., dx, dy, = 0. Hence 6x − 6 = 0,, (ii) At a turning point,, dx, from which, x = 1., (i) Since y = 3x 2 − 6x,, , x, , Minimum point, , Maximum and minimum points and points of, inflexion are given the general term of stationary, points., Procedure for finding and distinguishing between, stationary points:, (i) Given y = f (x), determine, , (iii) When x = 1, y = 3(1)2 − 6(1) = −3., Hence the co-ordinates of the turning point, are (1, −3)., , Figure 28.5, , (ii) Let, , positive to positive or negative to negative—, the point is a point of inflexion., , Following the above procedure:, Points of, inflexion, , 0, , (c), , dy, (i.e. f (x)), dx, , dy, = 0 and solve for the values of x., dx, , (iii) Substitute the values of x into the original, equation, y = f (x), to find the corresponding yordinate values. This establishes the co-ordinates, of the stationary points., , (iv) If x is slightly less than 1, say, 0.9, then, dy, = 6(0.9) − 6 = −0.6,, dx, i.e. negative., If x is slightly greater than 1, say, 1.1, then, dy, = 6(1.1) − 6 = 0.6,, dx, i.e. positive., Since the gradient of the curve is negative just, before the turning point and positive just after, (i.e. − ∨ +), (1, −3) is a minimum point., , To determine the nature of the stationary points:, Either, , Problem 11. Find the maximum and minimum, values of the curve y = x 3 − 3x + 5 by, , d2 y, and substitute into it the values of x, (iv) Find, dx 2, found in (ii)., If the result is:, (a) positive—the point is a minimum one,, (b) negative—the point is a maximum one,, (c) zero—the point is a point of inflexion,, or, , (a) examining the gradient on either side of the, turning points, and, , (v) Determine the sign of the gradient of the curve just, before and just after the stationary points. If the, sign change for the gradient of the curve is:, (a) positive to negative—the point is a maximum, one,, (b) negative to positive—the point is a minimum, one,, , (b) determining the sign of the second derivative., dy, = 3x 2 − 3, dx, dy, For a maximum or minimum value, =0, dx, Since y = x 3 − 3x + 5 then, , Hence 3x 2 − 3 = 0, from which, 3x 2 = 3 and x = ± 1, When x = 1, y = (1)3 − 3(1) + 5 =3, When x = −1, y = (−1)3 − 3(−1) + 5 =7, Hence (1, 3) and (−1, 7) are the co-ordinates of the, turning points.
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Some applications of differentiation, (a), , Considering the point (1, 3):, If x is slightly less than 1, say 0.9, then, dy, = 3(0.9)2 − 3,, dx, which is negative., If x is slightly more than 1, say 1.1, then, dy, = 3(1.1)2 − 3,, dx, which is positive., Since the gradient changes from negative to positive, the point (1, 3) is a minimum point., Considering the point (−1, 7):, If x is slightly less than −1, say −1.1, then, dy, = 3(−1.1)2 − 3,, dx, which is positive., If x is slightly more than −1, say −0.9, then, dy, = 3(−0.9)2 − 3,, dx, which is negative., , d2 y, , dy, = 3x 2 − 3, then 2 = 6x, dx, dx, d2 y, When x = 1,, is positive, hence (1, 3) is a, dx 2, minimum value., d2 y, is negative, hence (−1, 7) is, When x = −1,, dx 2, a maximum value., Thus the maximum value is 7 and the minimum value is 3., It can be seen that the second differential method of, determining the nature of the turning points is, in, this case, quicker than investigating the gradient., , Problem 12. Locate the turning point on the, following curve and determine whether it is a, maximum or minimum point: y = 4θ + e−θ ., y = 4θ + e−θ, dy, then, = 4 − e−θ = 0, dθ, for a maximum or minimum value., , Since, , 1, 4, , = eθ, giving θ = ln 14 = −1.3863 (see, , When θ = − 1.3863, y = 4(−1.3863) + e−(−1.3863), = 5.5452 +4.0000 = −1.5452, Thus (−1.3863, −1.5452) are the co-ordinates of the, turning point., d2 y, = e−θ ., dθ 2, When θ = −1.3863,, d2 y, = e+1.3863 = 4.0,, dθ 2, which is positive, hence (−1.3863, −1.5452) is a, minimum point., Problem 13. Determine the co-ordinates of the, maximum and minimum values of the graph, x3 x2, 5, y = − − 6x + and distinguish between, 3, 2, 3, them. Sketch the graph., Following the given procedure:, , Since the gradient changes from positive to negative, the point (−1, 7) is a maximum point., (b) Since, , Hence 4 = e−θ ,, Chapter 4)., , 305, , (i) Since y =, , 5, x3 x2, − − 6x + then, 3, 2, 3, , dy, = x2 − x −6, dx, dy, = 0. Hence, dx, x 2 − x − 6 = 0, i.e. (x + 2)(x − 3) = 0,, , (ii) At a turning point,, , from which x = −2 or x = 3., (iii) When x = −2,, y=, , 5, (−2)3 (−2)2, −, − 6(−2) + = 9, 3, 2, 3, , When x = 3,, y=, , 5, 5, (3)3 (3)2, −, − 6(3) + = −11, 3, 2, 3, 6, , Thus the co-ordinates, , of the turning points are, (−2, 9) and 3, −11 56 ., d2 y, dy, = x 2 − x − 6 then 2 = 2x−1., dx, dx, When x = −2,, , (iv) Since, , d2 y, = 2(−2) − 1 = −5,, dx 2, which is negative.
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306 Higher Engineering Mathematics, When x = 126.87◦,, , Hence (−2, 9) is a maximum point., When x = 3,, , y = 4 sin 126.87◦ − 3 cos126.87◦ = 5, , d2 y, = 2(3) − 1 = 5,, dx 2, which is positive., , , Hence 3, −11 56 is a minimum point., , When x = 306.87◦,, y = 4 sin 306.87◦ − 3 cos 306.87◦ = −5, , π , 126.87◦ = 126.87◦ ×, radians, 180, , Knowing (−2,, point (i.e. crest of, , 9) is a maximum, 5, a wave), and 3, −11 6 is a minimum point (i.e., bottom of a valley) and that when x = 0, y = 53 , a, sketch may be drawn as shown in Fig. 28.6., , = 2.214 rad, , π , 306.87◦ = 306.87◦ ×, radians, 180, = 5.356 rad, Hence (2.214, 5) and (5.356, −5) are, co-ordinates of the turning points., , y, 12, 8, , d2 y, = −4 sin x + 3 cos x, dx 2, , 9, 3 x2, y5 x, 2 2 26x 1 5, 3, 3, , When x = 2.214 rad,, , 4, , 22, , 21, , 0, , d2 y, = −4 sin 2.214 + 3 cos 2.214,, dx 2, 1, , 2, , 3, , x, , 24, , 2115, , the, , which is negative., Hence (2.214, 5) is a maximum point., When x = 5.356 rad,, , 28, , d2 y, = −4 sin 5.356 + 3 cos5.356,, dx 2, , 6, , 212, , Figure 28.6, , which is positive., Hence (5.356, −5) is a minimum point., A sketch of y = 4 sin x − 3 cos x is shown in Fig. 28.7., , Problem 14. Determine the turning points on the, curve y = 4 sin x − 3 cos x in the range x = 0 to, x = 2π radians, and distinguish between them., Sketch the curve over one cycle., , y, 5, , y ⫽ 4 sin x ⫺ 3 cos x, , Since y = 4 sin x − 3 cos x, dy, = 4 cos x + 3 sin x = 0,, dx, for a turning point, from which,, then, , 4 cos x = −3 sin x and, −4, sin x, =, = tan x, 3, cos x, , , −4, −1, = 126.87◦ or 306.87◦, since, Hence x = tan, 3, tangent is negative in the second and fourth quadrants., , 0, ⫺3, ⫺5, , Figure 28.7, , /2 2.214, , , , 5.356, 3/2, , x (rads), 2
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307, , Some applications of differentiation, Now try the following exercise, Exercise 122, points, , Further problems on turning, , 13. Show that the curve y = 23 (t − 1)3 + 2t (t − 2), has a maximum value of 23 and a minimum, value of −2., , In Problems 1 to 11, find the turning points and, distinguish between them., 1., , y = x 2 − 6x, , 2., , y = 8 + 2x − x 2, , [(1, 9) Maximum], , 3., , y = x 2 − 4x + 3, , [(2, −1) Minimum], , 4., , y = 3 + 3x 2 − x 3, , 5., , y = 3x 2 − 4x + 2, , Minimum at, , 6., , x = θ(6 − θ), , [Maximum at (3, 9)], , 7., , y = 4x 3 + 3x 2 − 60x − 12, , Minimum (2, −88);, Maximum(−2.5, 94.25), , [(3, −9) Minimum], , , , (0, 3) Minimum,, (2, 7) Maximum, 2, , 2, 3, 3, , , , 28.4 Practical problems involving, maximum and minimum values, There are many practical problems involving maximum and minimum values which occur in science and, engineering. Usually, an equation has to be determined, from given data, and rearranged where necessary, so, that it contains only one variable. Some examples are, demonstrated in Problems 15 to 20., Problem 15. A rectangular area is formed having, a perimeter of 40 cm. Determine the length and, breadth of the rectangle if it is to enclose the, maximum possible area., Let the dimensions of the rectangle be x and y. Then, the perimeter of the rectangle is (2x + 2y). Hence, 2x + 2y = 40,, , 8., , y = 5x − 2 ln x, , or, [Minimum at (0.4000, 3.8326)], , 9., , 10., , 11., , y = 2x − ex, , y =t3−, , x = 8t +, , [Maximum at (0.6931, −0.6136)], , t2, − 2t + 4, 2, ⎤, ⎡, Minimum at (1, 2.5);, ⎥, , ⎢, ⎣, 2 22 ⎦, Maximum at − , 4, 3 27, 1, 2t 2, , [Minimum at (0.5, 6)], , 12. Determine the maximum and minimum values, on the graph y = 12 cosθ − 5 sin θ in the range, θ = 0 to θ = 360◦. Sketch the graph over one, cycle showing relevant points., , Maximum of 13 at 337.38◦,, Minimum of −13 at 157.38◦, , x + y = 20, , (1), , Since the rectangle is to enclose the maximum possible, area, a formula for area A must be obtained in terms of, one variable only., Area A = x y. From equation (1), x = 20 − y, Hence, area A = (20 − y)y = 20y − y 2, dA, = 20 − 2y = 0, dy, for a turning point, from which, y = 10 cm, d2 A, = −2,, d y2, which is negative, giving a maximum point., When y = 10 cm, x = 10 cm, from equation (1)., Hence the length and breadth of the rectangle are, each 10 cm, i.e. a square gives the maximum possible, area. When the perimeter of a rectangle is 40 cm, the, maximum possible area is 10 × 10 = 100 cm2 ., Problem 16. A rectangular sheet of metal having, dimensions 20 cm by 12 cm has squares removed, from each of the four corners and the sides bent
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308 Higher Engineering Mathematics, upwards to form an open box. Determine the, maximum possible volume of the box., The squares to be removed from each corner are shown, in Fig. 28.8, having sides x cm. When the sides are bent, upwards the dimensions of the box will be:, length (20 − 2x) cm, breadth (12 − 2x) cm and height,, x cm., x, , x, x, , x, , Maximum volume = (15.146)(7.146)(2.427), = 262.7 cm 3, Problem 17. Determine the height and radius of a, cylinder of volume 200 cm3 which has the least, surface area., Let the cylinder have radius r and perpendicular, height h., Volume of cylinder,, V = πr 2 h = 200, , 12 cm, , Surface area of cylinder,, , (12 2 2x ), , x, , x, x, , x, 20 cm, , Figure 28.8, , A = 2πrh + 2πr 2, Least surface area means minimum surface area and a, formula for the surface area in terms of one variable, only is required., From equation (1),, , Volume of box,, V = (20 − 2x)(12 − 2x)(x), = 240x − 64x 2 + 4x 3, dV, = 240 − 128x + 12x 2 = 0, dx, for a turning point., Hence 4(60 − 32x + 3x 2 ) = 0,, i.e., , (1), , (20 2 2x ), , 3x 2 − 32x + 60 = 0, , Using the quadratic formula,, , 32 ± (−32)2 − 4(3)(60), x=, 2(3), = 8.239 cm or 2.427 cm., Since the breadth is (12 − 2x) cm then x = 8.239 cm is, not possible and is neglected. Hence x = 2.427 cm, d2 V, = −128 + 24x., dx 2, d2 V, When x = 2.427, 2 is negative, giving a maxdx, imum value., The dimensions of the box are:, length = 20 − 2(2.427) = 15.146 cm,, breadth = 12 − 2(2.427) = 7.146 cm,, and height = 2.427 cm, , h=, , 200, πr 2, , (2), , Hence surface area,, , , 200, A = 2πr, + 2πr 2, πr 2, 400, =, + 2πr 2 = 400r −1 + 2πr 2, r, d A −400, = 2 + 4πr = 0,, dr, r, for a turning point., Hence 4πr =, , 400, 400, and r 3 =, ,, r2, 4π, , from which,, , r=, , 3, , 100, π, , , = 3.169 cm, , d 2 A 800, = 3 + 4π., dr 2, r, d2 A, When r = 3.169 cm, 2 is positive, giving a mindr, imum value., From equation (2),, when r = 3.169 cm,, 200, = 6.339 cm, h=, π(3.169)2
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Some applications of differentiation, Hence for the least surface area, a cylinder of volume 200 cm3 has a radius of 3.169 cm and height of, 6.339 cm., Problem 18. Determine the area of the largest, piece of rectangular ground that can be enclosed by, 100 m of fencing, if part of an existing straight wall, is used as one side., Let the dimensions of the rectangle be x and y as shown, in Fig. 28.9, where P Q represents the straight wall., , P, y, , x, , Figure 28.10, , Surface area of box, A, consists of two ends and five, faces (since the lid also covers the front face.), Hence, , x, , Figure 28.9, , y=, From Fig. 28.9,, (1), , (2), , Since the maximum area is required, a formula for area, A is needed in terms of one variable only., From equation (1), x = 100 −2y, Hence area A =xy = (100 −2y)y = 100y −2y2, dA, = 100 − 4y = 0,, dy, for a turning point, from which, y = 25 m, d2 A, d y2, , 6 − 2x 2, 6, 2x, =, −, 5x, 5x, 5, , = −4,, , which is negative, giving a maximum value., When y = 25 m, x = 50 m from equation (1)., Hence the maximum possible area = x y = (50)(25) =, 1250 m2 ., Problem 19. An open rectangular box with, square ends is fitted with an overlapping lid which, covers the top and the front face. Determine the, maximum volume of the box if 6 m2 of metal are, used in its construction., A rectangular box having square ends of side x and, length y is shown in Fig. 28.10., , (2), , Hence volume, , V = x2 y = x2, , Area of rectangle,, A = xy, , (1), , Since it is the maximum volume required, a formula, for the volume in terms of one variable only is needed., Volume of box, V = x 2 y., From equation (1),, , y, , x + 2y = 100, , y, , x, , A = 2x 2 + 5x y = 6, , Q, , 309, , 6, 2x, −, 5x, 5, , , =, , 6x 2x 3, −, 5, 5, , dV, 6 6x 2, = −, =0, dx, 5, 5, for a maximum or minimum value., Hence 6 =6x 2 , giving x = 1 m (x = −1 is not possible,, and is thus neglected)., −12x, d2 V, =, 2, dx, 5, d2 V, When x = 1, 2 is negative, giving a maximum value., dx, From equation (2), when x = 1,, y=, , 2(1) 4, 6, −, =, 5(1), 5, 5, , Hence the maximum volume of the box is given by, , V = x 2 y = (1)2 45 = 45 m3, Problem 20. Find the diameter and height of a, cylinder of maximum volume which can be cut, from a sphere of radius 12 cm., A cylinder of radius r and height h is shown enclosed, in a sphere of radius R = 12 cm in Fig. 28.11.
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310 Higher Engineering Mathematics, d2 V, When h = 13.86, 2 is negative, giving a maximum, dh, value., From equation (2),, , r, , P, h, 2, h, , Q, , R, , 5, , 12, , cm, , r 2 = 144 −, , O, , h2, 13.862, = 144 −, 4, 4, , from which, radius r = 9.80 cm, Diameter of cylinder = 2r = 2(9.80) = 19.60 cm., Hence the cylinder having the maximum volume that, can be cut from a sphere of radius 12 cm is one in, which the diameter is 19.60 cm and the height is, 13.86 cm., , Figure 28.11, , Volume of cylinder,, V = πr 2 h, , Now try the following exercise, (1), , Using the right-angled triangle OPQ shown in, Fig. 28.11,, 2, h, r2 +, = R 2 by Pythagoras’ theorem,, 2, i.e., , r2 +, , h2, = 144, 4, , (2), , Since the maximum volume is required, a formula for, the volume V is needed in terms of one variable only., From equation (2),, r 2 = 144 −, , Exercise 123 Further problems on, practical maximum and minimum problems, 1., , The speed, v, of a car (in m/s) is related to, time t s by the equation v = 3 +12t − 3t 2., Determine the maximum speed of the car, in km/h., [54 km/h], , 2., , Determine the maximum area of a rectangular piece of land that can be enclosed by, 1200 m of fencing., [90000 m2], , 3., , A shell is fired vertically upwards and, its vertical height, x metres, is given by, x = 24t − 3t 2, where t is the time in seconds., Determine the maximum height reached., [48 m], , 4., , A lidless box with square ends is to be made, from a thin sheet of metal. Determine the, least area of the metal for which the volume, [11.42 m2 ], of the box is 3.5 m3., , 5., , A closed cylindrical container has a surface, area of 400 cm2 . Determine the dimensions, for maximum volume., radius = 4.607 cm;, , h2, 4, , Substituting into equation (1) gives:, , , h2, πh 3, V = π 144 −, h = 144πh −, 4, 4, dV, 3πh 2, = 144π −, = 0,, dh, 4, for a maximum or minimum value., Hence, 3πh 2, 4, , (144)(4), h=, = 13.86 cm, 3, , height = 9.212 cm, , 144π =, from which,, , −6πh, d2 V, =, 2, dh, 4, , 6., , Calculate the height of a cylinder of maximum volume which can be cut from a cone, of height 20 cm and base radius 80 cm., [6.67 cm]
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Some applications of differentiation, , 7., , 8., , The power developed in a resistor R by a, battery of emf E and internal resistance r is, E2 R, . Differentiate P with, given by P =, (R + r)2, respect to R and show that the power is a, maximum when R = r., Find the height and radius of a closed cylinder of volume 125 cm3 which has the least, surface area., height = 5.42 cm;, radius = 2.71 cm, , Problem 21. Find the equation of the tangent to, the curve y = x 2 − x − 2 at the point (1, −2)., Gradient, m, =, , 10., , 11., , Resistance to motion, F, of a moving vehicle, is given by F = 5x + 100x. Determine the, minimum value of resistance., [44.72], An electrical voltage E is given by, E =(15 sin 50πt + 40 cos 50πt ) volts,, where t is the time in seconds. Determine the, maximum value of voltage., [42.72 volts], The fuel economy E of a car, in miles per, gallon, is given by:, E = 21 + 2.10 × 10−2v 2, − 3.80 × 10−6v 4, where v is the speed of the car in miles per, hour., Determine, correct to 3 significant figures,, the most economical fuel consumption, and, the speed at which it is achieved., [50.0 miles/gallon, 52.6 miles/hour], , dy, = 2x − 1, dx, , At the point (1, −2), x = 1 and m = 2(1) − 1 =1., Hence the equation of the tangent is:, y − y1 = m(x − x 1), i.e. y − (−2) = 1(x − 1), i.e., , 9., , 311, , y+2 = x −1, y = x−3, , or, , The graph of y = x 2 − x − 2 is shown in Fig. 28.12. The, line AB is the tangent to the curve at the point C, i.e. (1,, −2), and the equation of this line is y = x − 3., , Normals, The normal at any point on a curve is the line which, passes through the point and is at right angles to the, tangent. Hence, in Fig. 28.12, the line CD is the normal., It may be shown that if two lines are at right angles, then the product of their gradients is −1. Thus if m is the, gradient of the tangent, then the gradient of the normal, 1, is −, m, Hence the equation of the normal at the point (x 1 , y1) is, given by:, y − y1 = −, , 1, (x − x1 ), m, , y, y ⫽ x 2 ⫺ x⫺ 2, , 2, , 28.5, , 1, , Tangents and normals, ⫺2, , Tangents, , ⫺1, , 2, , ⫺3 A, , Figure 28.12, , 3, B, , ⫺2, , y − y1 = m(x − x1), dy, = gradient of the curve at (x 1, y1)., dx, , 1, , ⫺1, , The equation of the tangent to a curve y = f (x) at the, point (x 1, y1) is given by:, , where m =, , 0, , C, D, , x
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312 Higher Engineering Mathematics, Problem 22. Find the equation of the normal to, the curve y = x 2 − x − 2 at the point (1, −2)., m = 1 from Problem 21, hence the equation of the, normal is, 1, y − y1 = − (x − x 1 ), m, 1, i.e. y − (−2) = − (x − 1), 1, i.e., or, , y + 2 = −x + 1, y = −x − 1, , Thus the line CD in Fig. 28.12 has the equation, y = −x − 1., Problem 23., , Determine the equations of the, x3, tangent and normal to the curve y =, at the point, 5, , , 1, −1, −, 5, , x3, Gradient m of curve y =, is given by, 5, d y 3x 2, =, m=, dx, 5, , , 3(−1)2 3, At the point −1, − 15 , x = − 1 and m =, =, 5, 5, Equation of the tangent is:, y − y1 = m(x − x 1 ), , , 3, 1, = (x − (−1)), i.e. y − −, 5, 5, i.e., or, or, , y+, , 1 3, = (x + 1), 5 5, , Hence equation of the normal is:, 15y + 25x + 28 = 0, Now try the following exercise, Exercise 124 Further problems on, tangents and normals, For the curves in problems 1 to 5, at the points, given, find (a) the equation of the tangent, and (b), the equation of the normal., (a) y = 4x − 2, (b) 4y + x = 9, , 1., , y = 2x 2 at the point (1, 2), , 2., , y = 3x 2 − 2x at the point (2, 8), (a) y = 10x − 12, , 3., , (b) 10y + x = 82, , , 1, x3, at the point −1, −, y=, 2, 2, (a) y = 32 x + 1, (b) 6y + 4x + 7 = 0, , 4., , y = 1 + x − x 2 at the point (−2, −5), (a) y = 5x + 5, , (b) 5y + x + 27 = 0, , , 1, 1, 5. θ = at the point 3,, t, 3, (a) 9θ + t = 6, (b) θ = 9t − 26 23 or 3θ = 27t − 80, , 5y − 3x = 2, , 1, y − y1 = − (x − x 1 ), m, , , 1, −1, i.e. y − −, =, (x − (−1)), 5, (3/5), , i.e., , 15y + 3 = −25x − 25, , 5y + 1 = 3x + 3, , Equation of the normal is:, , i.e., , Multiplying each term by 15 gives:, , 1, 5, = − (x + 1), 5, 3, 5, 5, 1, y+ =− x−, 5, 3, 3, y+, , 28.6, , Small changes, , If y is a function of x, i.e. y = f (x), and the approximate change in y corresponding to a small change δx in, x is required, then:, δy, dy, ≈, δx, dx, dy, and δy ≈, · δx or δy ≈ f (x) · δx, dx
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Some applications of differentiation, Problem 24. Given y = 4x 2 − x, determine the, approximate change in y if x changes from 1 to, 1.02., Since y = 4x 2 − x, then, dy, = 8x − 1, dx, , Percentage error, , , approximate change in T, 100%, =, original value of T, , , k, √ (−0.1), 2 l, =, √, × 100%, k l, , , , , −0.1, −0.1, 100% =, 100%, =, 2l, 2(32.1), = −0.156%, , Approximate change in y,, δy ≈, , 313, , dy, · δx ≈ (8x − 1)δx, dx, , When x = 1 and δx = 0.02, δy ≈ [8(1) − 1](0.02), ≈ 0.14, [Obviously, in this case, the exact value of dy, may be obtained by evaluating y when x = 1.02, i.e., y = 4(1.02)2 − 1.02 = 3.1416 and then subtracting from, it the value of y when x = 1, i.e. y = 4(1)2 − 1 = 3, giving, δy = 3.1416 −3 =0.1416., dy, Using δy =, · δx above gave 0.14, which shows that, dx, the formula gives the approximate change in y for a, small change in x.], , Hence the change in the time of swing is a decrease, of 0.156%., Problem 26. A circular template has a radius of, 10 cm (±0.02). Determine the possible error in, calculating the area of the template. Find also the, percentage error., Area of circular template, A = πr 2 , hence, dA, = 2πr, dr, Approximate change in area,, δA ≈, , dA, · δr ≈ (2πr)δr, dr, , When r = 10 cm and δr = 0.02,, Problem 25. The, √ time of swing T of a pendulum, is given by T = k l, where k is a constant., Determine the percentage change in the time of, swing if the length of the pendulum l changes from, 32.1 cm to 32.0 cm., , dT, 1 −1, =k, l 2, dl, 2, , , , δt ≈, , dT, δl ≈, dl, , , ≈, , , , , 0.4π, 100%, π(10)2, , = 0.40%, k, = √, 2 l, , Approximate change in T ,, , , i.e. the possible error in calculating the template area, is approximately 1.257 cm2., Percentage error ≈, , 1, √, If T = k l = kl 2 , then, , , , δ A = (2π10)(0.02) ≈ 0.4π cm 2, , , , k, √ δl, 2 l, , , k, √ (−0.1), 2 l, , (negative since l decreases), , Now try the following exercise, Exercise 125, changes, , Further problems on small, , 1. Determine the change in y if x changes from, 2.50 to 2.51 when, 5, (a) y = 2x − x 2 (b) y =, x, [(a) −0.03 (b) −0.008]
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314 Higher Engineering Mathematics, 2. The pressure p and volume v of a mass of gas, are related by the equation pv =50. If the pressure increases from 25.0 to 25.4, determine the, approximate change in the volume of the gas., Find also the percentage change in the volume, of the gas., [−0.032, −1.6%], 3. Determine the approximate increase in (a) the, volume, and (b) the surface area of a cube, of side x cm if x increases from 20.0 cm to, 20.05 cm., [(a) 60 cm3 (b) 12 cm2 ], 4. The radius of a sphere decreases from 6.0 cm, to 5.96 cm. Determine the approximate change, in (a) the surface area, and (b) the volume., [(a) −6.03 cm2 (b) −18.10 cm3 ], , 5. The rate of flow of a liquid through a, tube is given by Poiseuilles’s equation as:, pπr 4, Q=, where Q is the rate of flow, p, 8ηL, is the pressure difference between the ends, of the tube, r is the radius of the tube, L, is the length of the tube and η is the coefficient of viscosity of the liquid. η is obtained, by measuring Q, p, r and L. If Q can be, measured accurate to ±0.5%, p accurate to, ±3%, r accurate to ±2% and L accurate to, ±1%, calculate the maximum possible percentage error in the value of η., [12.5%]
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Chapter 29, , Differentiation of parametric, equations, 29.1 Introduction to parametric, equations, Certain mathematical functions can be expressed more, simply by expressing, say, x and y separately in terms, of a third variable. For example, y =r sin θ, x =r cos θ., Then, any value given to θ will produce a pair of values, for x and y, which may be plotted to provide a curve of, y = f (x)., The third variable, θ, is called a parameter and the, two expressions for y and x are called parametric, equations., The above example of y =r sin θ and x =r cos θ are, the parametric equations for a circle. The equation of, any point on a circle, centre at the origin and of radius, r is given by: x 2 + y 2 =r 2 , as shown in Chapter 13., To show that y =r sin θ and x =r cos θ are suitable, parametric equations for such a circle:, Left hand side of equation, , 29.2 Some common parametric, equations, The following are some of the most common parametric, equations, and Fig. 29.1 shows typical shapes of these, curves., (a), , Ellipse, , x = a cos θ, y = b sin θ, , (b) Parabola, , x = a t 2, y = 2a t, , (c), , x = a sec θ, y = b tan θ, c, x = c t, y =, t, , Hyperbola, , (d) Rectangular, hyperbola, (e), , Cardioid, , x = a (2 cosθ − cos 2θ),, y = a (2 sin θ − sin 2θ ), , (f ) Astroid, , x = a cos3 θ, y = a sin3 θ, , (g) Cycloid, , x = a (θ − sin θ ) , y = a (1− cos θ), , = x 2 + y2, = (r cos θ)2 + (r sin θ)2, = r 2 cos2 θ + r 2 sin2 θ, , , = r 2 cos2 θ + sin2 θ, = r = right hand side, 2, , (since cos2 θ + sin2 θ = 1, as shown in, Chapter 15), , 29.3, , Differentiation in parameters, , When x and y are given in terms of a parameter, say θ,, then by the function of a function rule of differentiation, (from Chapter 27):, dy, d y dθ, =, ×, dx, dθ dx
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Chapter 30, , Differentiation of implicit, functions, 30.1, , Implicit functions, , A simple rule for differentiating an implicit function, is summarised as:, , When an equation can be written in the form y = f (x), it is said to be an explicit function of x. Examples of, explicit functions include, y = 2x 3 − 3x + 4, y = 2x ln x, 3ex, and y =, cos x, In these examples y may be differentiated with respect, to x by using standard derivatives, the product rule and, the quotient rule of differentiation respectively., Sometimes with equations involving, say, y and x,, it is impossible to make y the subject of the formula., The equation is then called an implicit function and, examples of such functions include, y 3 + 2x 2 = y 2 − x and sin y = x 2 + 2x y., , 30.2 Differentiating implicit, functions, It is possible to differentiate an implicit function by, using the function of a function rule, which may be, stated as, du d y, du, =, ×, dx, d y dx, Thus, to differentiate y 3 with respect to x, the subdu, stitution u = y 3 is made, from which,, = 3y 2 . Hence,, dy, d 3, dy, (y ) = (3y 2 ) × , by the function of a function rule., dx, dx, , d, d, dy, [ f ( y)] = [ f ( y)] ×, dx, dy, dx, , (1), , Problem 1. Differentiate the following functions, with respect to x:, (a) 2y 4 (b) sin 3t ., (a) Let u =2y 4 , then, by the function of a function, rule:, du, dy, du dy, d, =, ×, =, (2y 4 ) ×, dx, dy dx, dy, dx, dy, = 8y3, dx, (b) Let u = sin 3t , then, by the function of a function, rule:, du, du dt, d, dt, =, ×, = (sin 3t ) ×, dx, dt, dx, dt, dx, dt, = 3 cos 3t, dx, Problem 2. Differentiate the following functions, with respect to x:, (a) 4 ln 5y, , 1, (b) e3θ−2, 5, , (a) Let u = 4 ln 5y, then, by the function of a function, rule:
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Chapter 31, , Logarithmic Differentiation, 31.1 Introduction to logarithmic, differentiation, With certain functions containing more complicated, products and quotients, differentiation is often made, easier if the logarithm of the function is taken before, differentiating. This technique, called ‘logarithmic, differentiation’ is achieved with a knowledge of (i) the, laws of logarithms, (ii) the differential coefficients of, logarithmic functions, and (iii) the differentiation of, implicit functions., , 31.2, , Laws of logarithms, , Three laws of logarithms may be expressed as:, (i) log(A × B) = log A + log B, , A, = log A − log B, (ii) log, B, (iii) log An = n log A, In calculus, Napierian logarithms (i.e. logarithms to a, base of ‘e’) are invariably used. Thus for two functions f (x) and g(x) the laws of logarithms may be, expressed as:, (i) ln[ f (x) · g(x)] = ln f (x) + ln g(x), , , f (x), = ln f (x) − ln g(x), (ii) ln, g(x), (iii) ln[ f (x)]n = n ln f (x), Taking Napierian logarithms of both sides of the equaf (x) · g(x), tion y =, gives:, h(x), , , f (x) · g(x), ln y = ln, h(x), , which may be simplified using the above laws of, logarithms, giving:, ln y = ln f (x) + ln g(x) − ln h(x), This latter form of the equation is often easier to, differentiate., , 31.3 Differentiation of logarithmic, functions, The differential coefficient of the logarithmic function, ln x is given by:, d, 1, (lnx) =, dx, x, More generally, it may be shown that:, d, f (x), [ln f (x)] =, dx, f (x), , (1), , For example, if y = ln(3x 2 + 2x − 1) then,, dy, 6x + 2, = 2, dx, 3x + 2x − 1, Similarly, if y = ln(sin 3x) then, dy 3 cos 3x, =, = 3 cot 3x., dx, sin 3x, Now try the following exercise, Exercise 131 Further problems on, differentiating logarithmic functions, Differentiate the following using the laws for, logarithms., , 4, 1. ln (4x − 10), 4x − 10
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326 Higher Engineering Mathematics, i.e. ln y = 2 ln(1 + x) +, 2. ln(cos 3x), , [−3 tan 3x], 2, 9x + 1, 3x 3 + x, , 3. ln(3x 3 + x), , 4., , ln(5x 2, , + 10x − 7), , 5. ln 8x, 6. ln(x 2 − 1), 7. 3 ln 4x, 8. 2 ln(sin x), , , , 9. ln(4x 3 − 6x 2 + 3x), , 10x + 10, 5x 2 + 10x − 7, , 1, x, , 2x, 2, x −1, , 3, x, , 1, 2, , − ln x − 12 ln(x + 2), by law (iii), of Section 31.2, (iii) Differentiate each term in turn with respect to x, using equations (1) and (2)., Thus, , 1, 1, 1 dy, 2, 1, 2, 2, =, +, − −, y dx, (1 + x) (x − 1) x (x + 2), , (iv) Rearrange the equation to make, Thus, , 31.4 Differentiation of further, logarithmic functions, As explained in Chapter 30, by using the function of a, function rule:, , 1 dy, d, (ln y) =, (2), dx, y dx, Differentiation, √of an expression such as, (1 + x)2 (x − 1), y=, √, may be achieved by using the, x (x + 2), product and quotient rules of differentiation; however the working would be rather complicated. With, logarithmic differentiation the following procedure is, adopted:, (i) Take Napierian logarithms of both sides of the, equation., √, , , (1 + x)2 (x − 1), Thus ln y = ln, √, x (x + 2), 6, 5, 1, (1 + x)2 (x − 1) 2, = ln, 1, x(x + 2) 2, (ii) Apply the laws of logarithms., 1, Thus ln y = ln(1 + x)2 + ln(x − 1) 2, 1, 2, , − ln x − ln(x + 2) , by laws (i), and (ii) of Section 31.2, , dy, the subject., dx, , , 2, 1, 1, dy, =y, +, −, dx, (1 + x) 2(x − 1) x, , [2 cot x], 12x 2 − 12x + 3, 4x 3 − 6x 2 + 3x, , ln(x − 1), , 1, −, 2(x + 2), (v) Substitute for y in terms of x., √, , dy, 2, (1 + x)2 (x − 1), Thus, =, √, dx, (1 + x), x (x + 2), +, , 1, 1, 1, − −, 2(x − 1) x 2(x + 2), , , , , , Problem 1., , Use logarithmic differentiation to, (x + 1)(x − 2)3, differentiate y =, (x − 3), , Following the above procedure:, (x + 1)(x − 2)3, (x − 3), , , (x + 1)(x − 2)3, then ln y = ln, (x − 3), , (i) Since, , y=, , (ii) ln y = ln(x + 1) + ln(x − 2)3 − ln(x − 3),, by laws (i) and (ii) of Section 31.2,, i.e. ln y = ln(x + 1) + 3 ln(x − 2) − ln(x − 3),, by law (iii) of Section 31.2., (iii) Differentiating with respect to x gives:, 1 dy, 1, 3, 1, =, +, −, ,, y dx, (x + 1) (x − 2) (x − 3), by using equations (1) and (2), (iv) Rearranging gives:, , , 1, 3, 1, dy, =y, +, −, dx, (x + 1) (x − 2) (x − 3)
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328 Higher Engineering Mathematics, (iv), (v), , , , 3, dy, 1, =y, +, − 1 − cot x, dx, x x ln 2x, , , dy, x3 ln 2x 3, 1, = x, +, − 1 − cot x, dx, e sin x x x ln 2x, , Now try the following exercise, , dy, when x = 1 given, dx, √, (x + 1)2 (2x − 1), y=, , (x + 3)3, , 7. Evaluate, , (x + 1)(2x + 1)3, y=, (x − 3)2 (x + 2)4, ⎤, ⎡, , 1, (x + 1)(2x + 1)3, 6, +, ⎥, ⎢, ⎥, ⎢ (x − 3)2 (x + 2)4 (x + 1) (2x + 1), ⎢, ⎥, ⎦, ⎣, 2, 4, −, −, (x − 3) (x + 2), √, (2x − 1) (x + 2), , 3. y =, (x − 3) (x + 1)3, √, ⎤, ⎡, , (2x − 1) (x + 2), 2, 1, , +, ⎥, ⎢, ⎥, ⎢ (x − 3) (x + 1)3 (2x − 1) 2(x + 2), ⎥, ⎢, , ⎦, ⎣, 1, 3, −, −, (x − 3) 2(x + 1), , 13, 16, , dy, , correct to 3 significant figures,, dθ, 2eθ sin θ, π, when θ = given y = √, 4, θ5, [−6.71], , 8. Evaluate, , Exercise 132 Further problems on, differentiating logarithmic functions, In Problems 1 to 6, use logarithmic differentiation, to differentiate the given functions with respect to, the variable., (x − 2)(x + 1), 1. y =, (x − 1)(x + 3), , ⎤, ⎡, 1, (x − 2)(x + 1), 1, +, ⎥, ⎢ (x − 1)(x + 3) (x − 2) (x + 1), ⎢, ⎥, ⎦, ⎣, 1, 1, −, −, (x − 1) (x + 3), , , , 31.5, , Differentiation of [ f (x)]x, , Whenever an expression to be differentiated contains a term raised to a power which is itself a function, of the variable, then logarithmic differentiation must be, used. For example, the, √ differentiation of expressions, such as x x , (x + 2)x , x (x − 1) and x 3x+2 can only be, achieved using logarithmic differentiation., , 2., , 4., , 5., , 6., , e2x cos 3x, y= √, (x − 4), , , 2x, e cos 3x, 1, 2 − 3 tan 3x −, √, 2(x − 4), (x − 4), y = 3θ sin θ cos θ, , , , 1, 3θ sin θ cos θ, + cot θ − tan θ, θ, , , 2x 4 tan x 4, 2x 4 tan x, 1, y = 2x, +, 2x, e ln 2x, e ln 2x x sin x cos x, , 1, −2 −, x ln 2x, , Problem 5., , Determine, , dy, given y = x x ., dx, , Taking Napierian logarithms of both sides of, y = x x gives:, ln y = ln x x = x ln x, by law (iii) of Section 31.2, Differentiating both sides with respect to x gives:, , 1, 1 dy, = (x), + (ln x)(1), using the product rule, y dx, x, i.e., , 1 dy, = 1 + ln x,, y dx, , from which,, , dy, = y(1 + ln x), dx, , i.e., , dy, = xx (1 + ln x), dx, , Problem 6., y = (x + 2)x ., , Evaluate, , dy, when x = −1 given, dx, , Taking Napierian logarithms of both sides of, y = (x + 2)x gives:, ln y = ln(x + 2)x = x ln(x + 2), by law (iii), of Section 31.2
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Logarithmic Differentiation, Differentiating both sides with respect to x gives:, , , 1 dy, 1, + [ln(x + 2)](1),, = (x), y dx, x +2, by the product rule., , , dy, x, Hence, =y, + ln(x + 2), dx, x +2, , , x, x, = (x + 2), + ln (x + 2), x+2, , , dy, −1 −1, When x = −1,, = (1), + ln 1, dx, 1, = (+1)(−1) = −1, Problem 7. Determine (a) the differential, √, dy, coefficient of y = x (x − 1) and (b) evaluate, dx, when x = 2., (a), , √, 1, y = x (x√− 1) = (x − 1) x , since by the laws of, m, indices n a m = a n, Taking Napierian logarithms of both sides gives:, 1, , ln y = ln(x − 1) x =, , 1, ln(x − 1),, x, , by law (iii) of Section 31.2., Differentiating each side with respect to x gives:, , , , , 1, 1, −1, 1 dy, =, + [ln(x − 1)], ,, y dx, x, x −1, x2, by the product rule., , , dy, 1, ln(x − 1), Hence, =y, −, dx, x(x − 1), x2, , , 1, ln(x − 1), dy √, x, = (x − 1), −, i.e., dx, x(x − 1), x2, (b) When x = 2,, , , , 1, dy √, ln(1), = 2 (1), −, dx, 2(1), 4, , , 1, 1, = ±1, −0 = ±, 2, 2, , Problem 8. Differentiate x 3x+2 with respect to x., , Let y = x 3x+2, Taking Napierian logarithms of both sides gives:, ln y = ln x 3x+2, i.e. ln y = (3x + 2) ln x, by law (iii) of Section 31.2., Differentiating each term with respect to x gives:, , 1 dy, 1, + (ln x)(3),, = (3x + 2), y dx, x, by the product rule., , , 3x + 2, dy, Hence, =y, + 3 ln x, dx, x, , , 3x + 2, = x 3x+2, + 3 ln x, x, , , 2, 3x+2, =x, 3 + + 3 ln x, x, Now try the following exercise, Exercise 133 Further problems on, differentiating [ f (x)]x type functions, In Problems 1 to 4, differentiate with respect to x., [2x 2x (1 + ln x)], , 1., , y = x 2x, , 2., , y = (2x − 1)x, (2x, √, x, , , − 1)x, , , 2x, + ln(2x − 1), 2x − 1, , (x+ 3), , √, x, (x + 3), , 3., , y=, , 4., , y = 3x 4x+1, , , 1, ln(x + 3), −, x(x + 3), x2, , , , 1, 4x+1, 3x, 4 + + 4 ln x, x, , 5. Show that when y = 2x x and x = 1,, 6. Evaluate, , dy, = 2., dx, , 4, d :√, x, (x − 2) when x = 3., dx, , , , 1, 3, , dy, 7. Show that if y = θ θ and θ = 2,, = 6.77,, dθ, correct to 3 significant figures., , 329
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Revision Test 9, This Revision Test covers the material contained in Chapters 27 to 31. The marks for each question are shown in, brackets at the end of each question., 1. Differentiate the following with respect to the, variable:, √, 1, (a) y = 5 +2 x 3 − 2 (b) s = 4e2θ sin 3θ, x, 3 ln 5t, (c) y =, cos 2t, 2, (d) x = , (13), 2, (t − 3t + 5), 2. If f (x) = 2.5x 2 − 6x + 2 find the co-ordinates at, the point at which the gradient is −1., (5), 3. The displacement s cm of the end of a stiff spring, at time t seconds is given by:, s = ae−kt sin 2π f t . Determine the velocity and, acceleration of the end of the spring after, 2 seconds if a = 3, k = 0.75 and f = 20., (10), 4. Find the co-ordinates of the turning points on, the curve y = 3x 3 + 6x 2 + 3x − 1 and distinguish, between them., (7), 5. The heat capacity C of a gas varies with absolute, temperature θ as shown:, C = 26.50 + 7.20 × 10, , −3, , θ − 1.20 × 10, , −6 2, , θ, , Determine the maximum value of C and the, temperature at which it occurs., (5), 6. Determine for the curve y = 2x 2 − 3x at the point, (2, 2): (a) the equation of the tangent (b) the, equation of the normal., (6), 7. A rectangular block of metal with a square crosssection has a total surface area of 250 cm2 . Find, the maximum volume of the block of metal. (7), , 8. A cycloid has parametric equations given by:, x = 5(θ − sin θ) and y = 5(1 − cos θ). Evaluate, d2 y, dy, when θ = 1.5 radians. Give, (b), (a), dx, dx 2, answers correct to 3 decimal places., (8), 9. Determine the equation of (a) the tangent, and (b), the normal, drawn to an ellipse x = 4 cos θ,, π, (8), y = sin θ at θ = ., 3, 10. Determine expressions for, , dz, for each of the, dy, , following functions:, (a) z =5y 2 cos x (b) z = x 2 + 4x y − y 2 ., , (5), , dy, 11. If x 2 + y 2 + 6x + 8y + 1 = 0, find, in terms of x, dx, and y., (3), 12. Determine the gradient of the tangents drawn to, (3), the hyperbola x 2 − y 2 = 8 at x = 3., 13. Use logarithmic, √ differentiation to differentiate, (x + 1)2 (x − 2), y=, , with respect to x., (6), (2x − 1) 3 (x − 3)4, 3eθ sin 2θ, √, and hence evaluate, θ5, dy, π, , correct to 2 decimal places, when θ = ., dθ, 3, (9), , 14. Differentiate y =, , , d √, t, (2t + 1) when t = 2, correct to 4, 15. Evaluate, dt, significant figures., (5)
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Chapter 33, , Differentiation of inverse, trigonometric and, hyperbolic functions, 33.1, , Inverse functions, , y +2, If y = 3x − 2, then by transposition, x =, . The, 3, y +2, function x =, is called the inverse function of, 3, y = 3x − 2 (see page 188)., Inverse trigonometric functions are denoted by prefixing the function with ‘arc’ or, more commonly, by, using the −1 notation. For example, if y = sin x, then, x = arcsin y or x = sin−1 y. Similarly, if y = cos x, then, x = arccos y or x = cos−1 y, and so on. In this chapter, the −1 notation will be used. A sketch of each of the, inverse trigonometric functions is shown in Fig. 33.1., Inverse hyperbolic functions are denoted by prefixing the function with ‘ar’ or, more commonly, by, using the −1 notation. For example, if y = sinh x, then, x = arsinh y or x = sinh−1 y. Similarly, if y = sech x,, then x = arsech y or x = sech−1 y, and so on. In this chapter the −1 notation will be used. A sketch of each of the, inverse hyperbolic functions is shown in Fig. 33.2., , 33.2 Differentiation of inverse, trigonometric functions, (i) If y = sin−1 x, then x = sin y., Differentiatingboth sides with respect to y gives:, dx, = cos y = 1 − sin2 y, dy, , since cos2 y + sin2 y = 1, i.e., However, , dx √, = 1 − x2, dy, , dy, 1, =, dx, dx, dy, , Hence, when y = sin−1 x then, dy, 1, =√, dx, 1 −x2, (ii) A sketch of part of the curve of y = sin−1 x is, shown in Fig. 33.1(a). The principal value of, sin−1 x is defined as the value lying between, −π/2 and π/2. The gradient of the curve between, points A and B is positive for all values of x, and thus only the positive value is taken when, 1, evaluating √, 1 − x2, x, (iii) Given y = sin−1, a, x = a sin y, Hence, , then, , x, = sin y, a, , and, , , dx, = a cos y = a 1 − sin2 y, dy, , 2, , x 2, a − x2, =a 1−, =a, a, a2, √, a a2 − x 2 √ 2, =, = a − x2, a
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Differentiation of inverse trigonometric and hyperbolic functions, y, , y, , y, 3/2, , 3/2, , y 5 sin21x, , , /2, , y 5cos21x, , /2, , B, , 21, 0, A 2/2, , y 5 tan21x, , , , D, , 11 x, , 21, , 11 x, , 0, 2/2, , 2, , 2, , 23/2, , 23/2, , (a), , /2, , C, , 2/2, , (b), , (c), y, , y, 3/2, , 3/2, , , /2, , , , y 5 sec21x, , y, , , y 5 cosec21x, , /2, , /2, , 21 0 11, 2/2, , x, , 21 0, 2/2, , 2, , 2, , 23/2, , 23/2, , (d), , x, , 0, , 11, , x, , y 5 cot21x, x, , 0, 2/2, , , (e), , (f), , Figure 33.1, , y, 3, 2, , y 5 sinh21x, , y 5cosh21x, , 2, , y 5 tanh21x, , 1, , 1, 01 2 3x, 23 22 21, 21, , 22 21 0, 21, , 22, , 22, , 23, , 23, , 1, , 21, , 2 3x, , (b), , (a), y, 3, , y, , y, 3, , (c), , y, y 5 sech21x, , y, y 5cosech21x, , 2, , 11 x, , 0, , y 5 coth21x, , 1, 0, 21, 22, , 1, , x, , 0, , x, , 21 0 11, , 23, (c), , Figure 33.2, , (e), , (f), , x, , 335
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336 Higher Engineering Mathematics, dy, 1, 1, =, =√, dx, dx, a2 − x 2, dy, x, dy, 1, i.e. when y = sin−1 then, =√, 2, a, dx, a − x2, , Table 33.1 Differential coefficients of inverse, trigonometric functions, , Thus, , y or f (x), , Since integration is the reverse process of differentiation then:, !, , 1, x, √, dx = sin−1 + c, 2, 2, a, a −x, , (i), , (iv) Given y = sin−1 f (x) the function of a function, dy, rule may be used to find, dx, , Then, , x, a, , 1, √, 2, a − x2, , sin−1 f (x), (ii), , Let u = f (x) then y = sin−1 u, , sin−1, , cos−1, , dy, du, 1, = f (x) and, =√, dx, du, 1 − u2, , tan −1, , x, a, , tan −1 f (x), , dy, dy du, 1, f (x), =, ×, =√, dx, du dx, 1 − u2, f (x), =, 1 −[ f (x)]2, , (iv), , Find, , dy, given y = sin−1 5x 2 ., dx, , (v), , Hence, if y = sin−1 5x 2, f (x) = 10x., Thus, , cosec−1, , (vi), , cot −1, , x, a, , x, a, , cot −1 f (x), dy, f (x), =, dx, 1 − [ f (x)]2, then, , f (x) = 5x 2, , and, , (b) Hence obtain the differential coefficient of, y = cos−1 (1 − 2x 2 )., , √, , a, x 2 − a2, , f (x), , f (x) [ f (x)]2 − 1, √, , −a, , x x 2 − a2, − f (x), , f (x) [ f (x)]2 − 1, −a, a2 + x 2, − f (x), 1 + [ f (x)]2, , (a) If y = cos−1 x then x = cos y., Differentiating with respect to y gives:, , dx, = −sin y = − 1 − cos2 y, dy, √, =− 1 − x2, , dy, 10x, 10x, =√, =, 2, 2, dx, 1 − (5x ), 1 −25x4, , Problem 2., (a) Show that if y = cos−1 x then, dy, 1, =√, dx, 1 − x2, , f (x), 1 + [ f (x)]2, x, , cosec−1 f (x), , From Table 33.1(i), if, y = sin−1 f (x) then, , x, a, , sec−1 f (x), , (v) The differential coefficients of the remaining, inverse trigonometric functions are obtained in, a similar manner to that shown above and a, summary of the results is shown in Table 33.1., Problem 1., , sec−1, , − f (x), , 1 − [ f (x)]2, a, a2 + x 2, , (see para. (i)), Thus, , f (x), , 1 − [ f (x)]2, −1, √, a2 − x 2, , x, a, , cos−1 f (x), (iii), , dy, or f (x), dx, , dy, 1, 1, =, =−√, dx, dx, 1 −x2, dy, The principal value of y = cos−1 x is defined as the, angle lying between 0 and π, i.e. between points C, and D shown in Fig. 33.1(b). The gradient of the curve, Hence
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338 Higher Engineering Mathematics, , Problem 6., , Differentiate y =, , cot −1 2x, 1 + 4x 2, , Using the quotient rule:, , , −2, − (cot −1 2x)(8x), (1 + 4x 2 ), dy, 1 + (2x)2, =, dx, (1 + 4x 2 )2, from Table 33.1(vi), =, , −2(1 +4x cot−1 2x), , −1, cos t − 1, =, (cos t − 1)2 + (sin t )2, (cos t − 1)2, , , , −1, (cos t − 1)2, =, cos t − 1 cos2 t − 2 cos t + 1 + sin2 t, =, , −(cos t − 1), 1 − cos t, 1, =, =, 2 − 2 cos t, 2(1 − cos t ) 2, , (1 +4x2 )2, Now try the following exercise, , Problem 7., , Differentiate y = x cosec, , −1, , x., , Using the product rule:, , −1, dy, = (x) √, + (cosec −1 x) (1), 2, dx, x x −1, from Table 33.1(v), −1, + cosec −1 x, =√, 2, x −1, Problem 8. Show that if, , , dy, sin t, 1, −1, then, y = tan, =, cos t − 1, dt, 2, , , If, , sin t, f (t ) =, cos t − 1, , then f (t ) =, =, , Exercise 135 Further problems on, differentiating inverse trigonometric, functions, In Problems 1 to 6, differentiate with respect to the, variable., x, 1. (a) sin−1 4x (b) sin−1, 2, , 4, 1, (a) √, (b) √, 2, 1 − 16x, 4 − x2, 2., , (a) cos−1 3x (b), , , , 3., , , (a), , 4., , −(cos t − 1), −1, =, (cos t − 1)2, cos t − 1, , −1, cos t − 1, , 2, sin t, 1+, cos t − 1, , 1, 6, (b) √, 2, 1 + 4x, 4 x (1 + x), 3, x, 4, , , 2, 4, (a) √, (b) √, t 4t 2 − 1, x 9x 2 − 16, 5., , Using Table 33.1(iii), when, , , sin t, −1, y = tan, cos t − 1, dy, then, =, dt, , √, 1, tan−1 x, 2, , (a) 2 sec−1 2t (b) sec−1, , since sin2 t + cos2 t = 1, =, , −3, −2, (a) √, (b) √, 1 − 9x 2, 3 9 − x2, , (a) 3 tan−1 2x (b), , (cos t − 1)(cos t ) − (sin t )(−sin t ), (cos t − 1)2, cos2 t − cos t + sin2 t, 1 − cos t, =, 2, (cos t − 1), (cos t − 1)2, , x, 2, cos−1, 3, 3, , 6., , θ, 5, cosec−1 (b) cosec−1 x 2, 2, 2, , −2, −5, (b) √, (a) √, 2, θ θ −4, x x4 − 1, √, (a) 3 cot −1 2t (b) cot −1 θ 2 − 1, , −1, −6, (b) √, (a), 2, 1 + 4t, θ θ2 − 1, (a)
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339, , Differentiation of inverse trigonometric and hyperbolic functions, , 7., , Showthat the differential coefficient of, x, 1 + x2, tan−1, ., is, 1 − x2, 1 − x2 + x4, , In Problems 8 to 11 differentiate with respect to, the variable., 8., , 9., , 10., , 11., , (a) 2x sin−1 3x (b) t 2 sec−1 2t, ⎤, ⎡, 6x, (a) √, + 2 sin−1 3x, ⎥, ⎢, 1 − 9x 2, ⎥, ⎢, ⎦, ⎣, t, + 2t sec−1 2t, (b) √, 4t 2 − 1, (a) θ 2 cos−1 (θ 2 − 1) (b) (1 − x 2 ) tan −1 x, ⎡, ⎤, 2, −1 (θ 2 − 1) − √ 2θ, (a), 2θ, cos, ⎢, ⎥, ⎢, 2 − θ2 ⎥, ⎢, ⎥, , , ⎣, ⎦, 1 − x2, −1, (b), − 2x tan x, 1 + x2, √, √, (a) 2 t cot −1 t (b) x cosec−1 x, √, ⎡, ⎤, 1, −2 t, −1, ⎢ (a) 1 + t 2 + √t cot t, ⎥, ⎢, ⎥, ⎣, ⎦, √, 1, −1, x− √, (b) cosec, 2 (x − 1), (a), , cos−1 x, sin−1 3x, (b), √, x2, 1 − x2, ⎡, ⎤, , 3x, 1, −1, − 2 sin 3x ⎥, ⎢ (a) x 3 √, 1 − 9x 2, ⎢, ⎥, ⎢, ⎥, ⎢, ⎥, x, ⎢, ⎥, ⎢, ⎥, cos−1 x, −1 + √, ⎣, ⎦, 2, 1−x, (b), 2, (1 − x ), , 33.3 Logarithmic forms of inverse, hyperbolic functions, Inverse hyperbolic functions may be evaluated most, conveniently when expressed in a logarithmic, form., x, x, For example, if y = sinh−1 then = sinh y., a, a, From Chapter 5, e y = cosh y + sinh y and, cosh 2 y −sinh2 y = 1, from which,, cosh y = 1 + sinh2 y which is positive since cosh y is, always positive (see Fig. 5.2, page 43)., , Hence e y =, , , , , =, , 1+, , 1 + sinh2 y + sinh y, , x 2, a, , +, , x, =, a, , , , a2 + x 2, a2, , , +, , x, a, , √, √, a2 + x 2 x, x + a2 + x 2, =, +, or, a, a, a, Taking Napierian logarithms of both sides gives:, 6, 5, √, x + a2 + x 2, y = ln, a, 6, 5, , 2 + x2, x, a, x, +, Hence, sinh−1 = ln, a, a, , (1), , 3, Thus to evaluate sinh−1 , let x = 3 and a = 4 in, 4, equation (1)., 6, 5, √, 3 + 42 + 32, −1 3, Then sin h, = ln, 4, 4, , , 3+5, = ln 2 = 0.6931, = ln, 4, By similar reasoning to the above it may be shown that:, 6, 5, √, x + x2 − a2, −1 x, cosh, = ln, a, a, , , a+x, x 1, and, tanh−1 = ln, a 2, a−x, Problem 9. Evaluate, correct to 4 decimal places,, sinh−1 2., 6, 5, √, 2 + x2, x, a, x, +, From above, sinh−1 = ln, a, a, With x = 2 and a = 1,, 6, 5, √, 2 + 12 + 22, −1, sinh 2 = ln, 1, √, = ln(2 + 5) = ln 4.2361, = 1.4436, correct to 4 decimal places, Using a calculator,, (i) press hyp, (ii) press 4 and sinh−1 ( appears, (iii) type in 2
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340 Higher Engineering Mathematics, (iv) press ) to close the brackets, (v) press = and 1.443635475 appears, Hence, sinh−1 2 = 1.4436, correct to 4 decimal places., Problem 10. Show that, , , a+x, x 1, tanh−1 = ln, and evaluate, correct, a 2, a−x, 3, to 4 decimal places, tanh −1, 5, If y = tanh−1, , x, x, then = tanh y., a, a, , If y = cosh−1, , x, x, then = cos y, a, a, , , e y = cosh y + sinh y = cosh y ± cosh2 y − 1, √, , x 2, x 2 − a2, x, −1 = ±, a, a, a, , x, = ±, a, =, , x±, , √, , x 2 − a2, a, , Taking Napierian logarithms of both sides gives:, 6, 5, √, x ± x 2 − a2, y = ln, a, , From Chapter 5,, 1 y, (e − e−y ) e2y − 1, sinh x, = 2y, = 21, tanh y =, y, −y, cosh x, e +1, 2 (e + e ), , by dividing each term by e−y, x e2y − 1, =, a e2y + 1, , Thus,, , from which, x(e2y + 1) = a(e2y − 1), Hence x + a = ae2y − xe2y = e2y (a − x), , , a+x, from which e2y =, a−x, Taking Napierian logarithms of both sides gives:, , Thus, assuming the principal value,, 6, 5, √, x + x2 − a2, −1 x, = ln, cosh, a, a, 14, 7, = cosh−1, 10, 5, x, −1, In the equation for cosh, , let x = 7 and a = 5, a, 6, 5, √, 7 + 72 − 52, −1 7, = ln, Then cosh, 5, 5, cosh−1 1.4 = cosh−1, , = ln 2.3798 = 0.8670,, correct to 4 decimal places., , , , and, , , a+x, 2y = ln, a−x, , , 1, a+x, y = ln, 2, a−x, , Now try the following exercise, , , , a+x, x 1, Hence, tanh−1 = ln, a 2, a−x, Substituting x = 3 and a = 5 gives:, tanh, , −1, , , , 5+3, 3 1, 1, = ln, = ln 4, 5 2, 5−3, 2, = 0.6931, correct to 4 decimal places., , Problem 11., , Prove that, 6, 5, √, x + x 2 − a2, −1 x, cosh, = ln, a, a, , and hence evaluate, 4 decimal places., , cosh −1 1.4, , correct to, , Exercise 136 Further problems on, logarithmic forms of the inverse hyperbolic, functions, In Problems 1 to 3 use logarithmic equivalents of, inverse hyperbolic functions to evaluate correct to, 4 decimal places., 1. (a) sinh−1, , 1, (b) sinh−1 4 (c) sinh−1 0.9, 2, [(a) 0.4812 (b) 2.0947 (c) 0.8089], , 2. (a) cosh−1, , 5, (b) cosh−1 3 (c) cosh−1 4.3, 4, [(a) 0.6931 (b) 1.7627 (c) 2.1380], , 3. (a) tanh−1, , 1, 5, (b) tanh−1 (c) tanh−1 0.7, 4, 8, [(a) 0.2554 (b) 0.7332 (c) 0.8673]
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Differentiation of inverse trigonometric and hyperbolic functions, 33.4 Differentiation of inverse, hyperbolic functions, x, x, If y = sinh−1 then = sinh y and x = a sinh y, a, a, dx, = a cosh y (from Chapter 32)., dy, Also cosh2 y − sinh2 y = 1, from which,, , x 2, , cosh y = 1 + sinh2 y =, 1+, a, √, a2 + x 2, =, a, √, a a2 + x 2 √ 2, dx, = a cosh y =, = a + x2, Hence, dy, a, dy, 1, 1, Then, =, =, dx, dx, a2 + x2, dy, x, [An alternative method of differentiating sinh−1, a, is to, the, 6 logarithmic form, 5 differentiate, √, 2, 2, x + a +x, with respect to x.], ln, a, −1, From the sketch of y = sinh, x shown, in Fig. 33.2(a), , , dy, is always positive., it is seen that the gradient i.e., dx, , It follows from above that, !, x, 1, dx = sinh−1 + c, √, 2, 2, a, x +a, 6, 5, √, x + a2 + x 2, +c, or, ln, a, It may be shown that, d, 1, (sinh−1 x)= , dx, x2 + 1, , Table 33.2 Differential coefficients of inverse, hyperbolic functions, dy, or f (x), dx, , y or f (x), (i) sinh−1, , x, a, , √, , sinh−1 f (x), (ii) cosh−1, , , , x, a, , (iii) tanh−1, , x, a, , f (x), [ f (x)]2 − 1, , √, , −a, , x a2 − x 2, − f (x), , f (x) 1 − [ f (x)]2, , sech−1 f (x), x, a, , √, , x, a, , coth−1 f (x), , −a, , x x 2 + a2, , cosech−1 f (x), (vi) coth−1, , , , 1, x 2 − a2, , f (x), 1 − [ f (x)]2, , x, a, , (v) cosech−1, , [ f (x)]2 + 1, , a, a2 − x 2, , tanh−1 f (x), (iv) sech−1, , f (x), , √, , cosh−1 f (x), , 1, x 2 + a2, , − f (x), , f (x) [ f (x)]2 + 1, a, a2 − x 2, f (x), 1 − [ f (x)]2, , Problem 12. Find the differential coefficient, of y = sinh−1 2x., , or more generally, d, f (x), [sinh−1 f (x)] = dx, [ f (x)]2 + 1, by using the function of a function rule as in, Section 33.2(iv)., The remaining inverse hyperbolic functions are differentiated in a similar manner to that shown above and, the results are summarized in Table 33.2., , From Table 33.2(i),, d, f (x), [sinh−1 f (x)] = , dx, [ f (x)]2 + 1, 2, d, (sinh−1 2x) = , Hence, dx, [(2x)2 + 1], 2, =, [4x2 + 1], , 341
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Differentiation of inverse trigonometric and hyperbolic functions, Hence, , d, cos x, [coth−1 (sin x)] =, dx, [1 − (sin x)2 ], , Since, , cos x, since cos2 x + sin2 x = 1, =, cos2 x, , then, , =, , 1, = sec x, cos x, , tanh−1, !, !, , x, a, , x, a, dx = tanh−1 + c, a2 − x 2, a, , x, 1, 1, dx = tanh−1 + c, a2 − x 2, a, a, !, !, 2, 1, Hence, dx, =, 2, , dx, (9 − 4x 2 ), 4 94 − x 2, , Using the product rule,, , , 1, dy, = (x 2 − 1), + (tanh−1 x)(2x), dx, 1 − x2, , 1, =, 2, , −(1 − x 2 ), =, + 2x tanh−1 x = 2x tanh−1 x − 1, (1 − x 2 ), !, , , Problem 19. Determine, , dx, (x 2 + 4), , ., , d , x, 1, sinh−1, =, dx, a, (x 2 + a 2), !, , , then, !, Hence, , dx, (x 2 + a 2 ), , = sinh−1, , 1, , dx =, (x 2 + 4), , !, , !, Problem 20. Determine, , x, +c, a, , 1, , dx, (x 2 + 22 ), , = sinh−1, , x, +c, 2, , !, , x, = 4 cosh−1 √ + c, 3, !, Problem 21. Find, , !, i.e., , 2, dx., (9 − 4x 2 ), , !, , 1, , dx, 3 2, 2, 2 −x, , x, 1 1, 3 tanh−1 3 + c, 2 2, 2, , 2, 2x, 1, dx = tanh−1, +c, 2, (9 − 4x ), 3, 3, , Now try the following exercise, , Exercise 137 Further problems on, differentiation of inverse hyperbolic, functions, In Problems 1 to 11, differentiate with respect to, the variable., x, 1. (a) sinh−1 (b) sinh−1 4x, 3, (a) , 2., , (a) 2 cosh −1, , 1, (x 2, , + 9), , (b) , , 4, (16x 2 + 1), , t, 1, (b) cosh −1 2θ, 3, 2, , 2, 1, (a) , (b) , 2, (t − 9), (4θ 2 − 1), , x, +c, a, , 1, , dx = cosh −1, , (x 2 − a 2 ), !, !, 4, 1, Hence, dx = 4 , dx, √, 2, (x − 3), [x 2 − ( 3)2 ], , then, , =, , 4, dx., , (x 2 − 3), , x, d , 1, cosh−1, =, dx, a, (x 2 − a 2 ), , Since, , a, a2 − x 2, , i.e., , Problem 18. Differentiate, y = (x 2 − 1) tanh−1 x., , Since, , =, , 3., , (a) tanh −1, , 2x, (b) 3 tanh−1 3x, 5, , 9, 10, (b), (a), 25 − 4x 2, (1 − 9x 2 ), , 4., , (a) sech−1, , 3x, 1, (b) − sech −1 2x, 4, 2, , −4, 1, (a) , (b) , x (16 − 9x 2 ), 2x (1 − 4x 2 ), , 343
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344 Higher Engineering Mathematics, , 5., , (a) cosech−1, , x, 1, (b) cosech−1 4x, 4, 2, , −4, −1, (a) , (b) , x (x 2 + 16), 2x (16x 2 + 1), 6., , 7., , 2x, 1, (b) coth−1 3t, 7, 4, , 14, 3, (a), (b), 49 − 4x 2, 4(1 − 9t 2), , (a) 2 sinh−1 (x 2 − 1), (a) coth−1, , (b), , , 1, cosh −1 (x 2 + 1), 2, (a) , , 8., , 9., , 10., , 2, , 1, (b) , (x 2 − 1), 2 (x 2 + 1), , (a) sech−1 (x − 1) (b) tanh−1(tanh x), , −1, (a), √, (b) 1, (x − 1) [x(2 − x)], , , t, −1, (b) coth−1 (cos x), (a) cosh, t −1, , −1, √, (b) −cosec x, (a), (t − 1) (2t − 1), √, (a) θ sinh−1 θ (b) x cosh−1 x, ⎤, ⎡, θ, + sinh−1 θ, (a) , ⎥, ⎢, (θ 2 + 1), ⎥, ⎢, ⎥, ⎢, √, −1, ⎣, x, cosh x ⎦, +, (b) , √, 2 x, (x 2 − 1), , 11. (a), , √, 2 sec h−1 t, tan h −1 x, (b), 2, t, (1 − x 2 ), , , ⎡, √ ⎤, 1, −1, −1, (a) 3 √, t, + 4 sech, ⎢, ⎥, t, (1 − t ), ⎢, ⎥, ⎢, ⎥, ⎣, ⎦, −1, 1 + 2x tanh x, (b), (1 − x 2 )2, , 12. Show that, , d, [x cosh−1 (cosh x)] = 2x., dx, , In Problems 13 to 15, determine the given, integrals., !, 1, 13. (a) , dx, 2, (x + 9), !, 3, (b) , dx, 2, (4x + 25), , x, 2x, 3, (a) sinh−1 + c (b) sinh−1, +c, 3, 2, 5, !, 1, 14. (a) , dx, 2, (x − 16), !, 1, (b) , dt, 2, (t − 5), , x, t, (a) cosh−1 + c (b) cosh−1 √ + c, 4, 5, !, !, dθ, 3, 15. (a) , (b), dx, 2, (16, −, 2x 2 ), (36 + θ ), ⎤, ⎡, θ, 1, (a) tan −1 + c, ⎥, ⎢, 6, 6, ⎥, ⎢, ⎦, ⎣, x, 3, −1, (b) √ tanh √ + c, 2 8, 8
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Chapter 34, , Partial differentiation, constant’. Thus,, , 34.1, , Introduction to partial, derivatives, , In engineering, it sometimes happens that the variation, of one quantity depends on changes taking place in, two, or more, other quantities. For example, the volume V of a cylinder is given by V = πr 2 h. The volume, will change if either radius r or height h is changed., The formula for volume may be stated mathematically, as V = f (r, h) which means ‘V is some function of r, and h’. Some other practical examples include:, , l, i.e. t = f (l, g)., (i) time of oscillation, t = 2π, g, (ii) torque T = I α, i.e. T = f (I, α)., (iii) pressure of an ideal gas p =, i.e. p = f (T, V )., (iv) resonant frequency fr =, , mRT, V, , 1, √, , 2π LC, i.e. fr = f (L , C), and so on., , When differentiating a function having two variables,, one variable is kept constant and the differential, coefficient of the other variable is found with respect, to that variable. The differential coefficient obtained is, called a partial derivative of the function., , 34.2, , First order partial derivatives, , A ‘curly dee’, ∂, is used to denote a differential coefficient in an expression containing more than one, variable., ∂V, means ‘the partial, Hence if V = πr 2 h then, ∂r, derivative of V with respect to r, with h remaining, , ∂V, d, = (πh) (r 2 ) = (πh)(2r) = 2πrh., ∂r, dr, ∂V, Similarly,, means ‘the partial derivative of V with, ∂h, respect to h, with r remaining constant’. Thus,, d, ∂V, = (πr 2 ) (h) = (πr 2 )(1) = πr 2 ., ∂h, dh, ∂V, ∂V, and, are examples of first order partial, ∂r, ∂h, derivatives, since n =1 when written in the form, ∂n V, ., ∂r n, First order partial derivatives are used when finding the, total differential, rates of change and errors for functions, of two or more variables (see Chapter 35), when finding, maxima, minima and saddle points for functions of two, variables (see Chapter 36), and with partial differential, equations (see Chapter 53)., Problem 1. If z = 5x 4 + 2x 3 y 2 − 3y find, ∂z, ∂z, (a), and (b), ., ∂x, ∂y, (a), , ∂z, , y is kept constant., ∂x, Since z = 5x 4 + (2y 2 )x 3 − (3y), then,, To find, , d, d, d, ∂z, =, (5x 4 ) + (2y 2 ) (x 3 ) − (3y) (1), ∂x, dx, dx, dx, = 20x 3 + (2y 2 )(3x 2 ) − 0., Hence, , ∂z, = 20x3 + 6x2 y2 ., ∂x
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346 Higher Engineering Mathematics, ∂z, , x is kept constant., ∂y, , (b) To find, , Problem 4., , Since z =(5x 4 ) + (2x 3 )y 2 − 3y, , 1, z= , 2, (x + y 2 ), , then,, ∂z, d, d, d, = (5x 4 ) (1) + (2x 3 ) (y 2 ) − 3 ( y), ∂y, dy, dy, dy, = 0 + (2x 3 )(2y) − 3, , Given y = 4 sin 3x cos 2t , find, , ∂y, , t is kept constant., To find, ∂x, Hence, , i.e., To find, , Hence, , i.e., , d, ∂y, = (4 cos 2t ) (sin 3x), ∂x, dx, = (4 cos 2t )(3 cos3x), ∂y, = 12 cos 3x cos 2t, ∂x, ∂y, , x is kept constant., ∂t, d, ∂y, = (4 sin 3x) (cos 2t ), ∂t, dt, = (4 sin 3x)(−2 sin 2t ), ∂y, = −8 sin 3x sin 2t, ∂t, , Problem 3., , If z =sin x y show that, 1 ∂z 1 ∂z, =, y ∂x x ∂ y, , ∂z, = y cos x y, since y is kept constant., ∂x, ∂z, = x cos x y, since x is kept constant., ∂y, , 1 ∂z, 1, ( y cos x y) = cos x y, =, y ∂x, y, , 1 ∂z, 1, (x cos x y) = cos x y., and, =, x ∂y, x, 1 ∂z, 1 ∂z, Hence, =, y ∂x x ∂y, , ∂z, ∂z, and, when, ∂x, ∂y, , −1, 1, = (x 2 + y 2 ) 2, z= , (x 2 + y 2 ), −3, ∂z, 1, = − (x 2 + y 2 ) 2 (2x), by the function of a, ∂x, 2, function rule (keeping y constant), , ∂z, Hence, = 4x3 y − 3., ∂y, Problem 2., ∂y, and, ∂t, , Determine, , ∂y, ∂x, , −x, , =, , 3, y2 ) 2, , (x 2 +, , =-, , −x, (x2 + y2 )3, , ∂z, −3, 1, = − (x 2 + y 2 ) 2 (2y), (keeping x constant), ∂y, 2, −y, =(x2 + y2 )3, Problem 5. Pressure p of a mass of gas is given, by pV = mRT, where m and R are constants, V is, the volume and T the temperature. Find expressions, ∂p, ∂p, for, and, ., ∂T, ∂V, Since pV = mRT then p =, , mRT, V, , ∂p, , V is kept constant., ∂T, , , mR d, mR, ∂p, =, (T ) =, Hence, ∂T, V, dT, V, , To find, , To find, , ∂p, , T is kept constant., ∂V, , d, ∂p, = (mRT), Hence, ∂V, dV, , , , 1, V, , , , = (m RT )(−V −2 ) =, Problem 6., , −mRT, V2, , The time of oscillation,, t , of, , l, where l is the, a pendulum is given by t = 2π, g, length of the pendulum and g the free fall, ∂t, ∂t, acceleration due to gravity. Determine, and, ∂l, ∂g
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Partial differentiation, To find, , ∂t, , g is kept constant., ∂l, t = 2π, , ∂t, Hence, =, ∂l, , , , , =, , To find, , 2π, √, g, 2π, √, g, , l, =, g, , , , , z =sin(4x + 3y) ⎡, , 5., , z = x 3 y2 −, , 1, √, 2 l, , , , , , 2π, √, g, , , , 1 −1, l 2, 2, , , , , , √, 1, l, = (2π l) √, g, g, , √ −1, = (2π l)g 2, , , √, ∂t, 1 −3, Hence, = (2π l) − g 2, ∂g, 2, , , √, −1, = (2π l) , 2 g3, √, −π l, = −π, = , g3, , l, g3, , 6., , z =cos 3x sin 4y, , 7., , The volume of a cone of height h and base, radius r is given by V = 13 πr 2 h. Determine, ∂V, ∂V, and, ∂h, ∂r , ∂V 1 2 ∂V 2, = πr, = πrh, ∂h 3, ∂r, 3, , 8., , The resonant frequency fr in a series electri1, cal circuit is given by fr = √, . Show, 2π LC, ∂ fr, −1, that, = √, ∂ L 4π C L 3, , 9., , An equation resulting from plucking a, string is:, , , , , nπ , nπb, nπb, t + c sin, t, x k cos, y = sin, L, L, L, ∂y, ∂y, Determine, and, ∂t, ∂x, , , ⎤, ⎡, nπb, ∂ y nπb nπ , t ⎥, ⎢ ∂t = L sin L x c cos L, ⎥, ⎢, ⎢, ⎥, , ⎥, ⎢, nπb, ⎢, t ⎥, − k sin, ⎥, ⎢, L, ⎥, ⎢, ⎥, ⎢, , , , nπ , ⎥, ⎢ ∂ y nπ, nπb, ⎢, t ⎥, ⎥, ⎢ ∂x = L cos L x k cos L, ⎥, ⎢, ⎢, ⎥, , ⎦, ⎣, nπb, t, + c sin, L, , Now try the following exercise, Exercise 138 Further problems on first, order partial derivatives, In Problems 1 to 6, find, 1., , z =2x y, , ∂z, ∂z, and, ∂x, ∂y, , ∂z, ∂z, = 2y, = 2x, ∂x, ∂y, ⎡, , 2., , 3., , z = x 3 − 2x y + y 2, , z=, , x, y, , ⎤, ∂z, 2 − 2y, =, 3x, ⎢ ∂x, ⎥, ⎢, ⎥, ⎣ ∂z, ⎦, = −2x + 2y, ∂y, ⎡ ∂z, 1 ⎤, =, ⎢ ∂x, y ⎥, ⎣ ∂z, −x ⎦, = 2, ∂y, y, , y, 1, +, 2, x, y, ⎡, , ⎤, ∂z, 2 y 2 + 2y, =, 3x, ⎢ ∂x, ⎥, x3, ⎢, ⎥, ⎣ ∂z, ⎦, 1, 1, 3, = 2x y − 2 − 2, ∂y, x, y, , π, =, lg, , ∂t, , l is kept constant., ∂g, t = 2π, , ⎤, ∂z, =, 4, cos(4x, +, 3y), ⎢ ∂x, ⎥, ⎢, ⎥, ⎣ ∂z, ⎦, = 3 cos(4x + 3y), ∂y, , , , , 2π, 1, 2π √, l, =, √, √ l2, g, g, , d 1, (l 2 ) =, dl, , , , 4., , 347, , ⎤, ∂z, = −3 sin 3x sin 4y, ⎥, ⎢ ∂x, ⎥, ⎢, ⎦, ⎣ ∂z, = 4 cos3x cos 4y, ∂y, ⎡
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348 Higher Engineering Mathematics, 10. In a thermodynamic system, k = Ae, where R, k and A are constants., , T S−H, RT, , as, , ,, , ∂A, ∂(S), ∂(H ), ∂k, (b), (c), (d), Find (a), ∂T, ∂T, ∂T, ∂T, ⎡, ⎤, AH T S−H, ∂k, RT, e, =, (a), ⎢, ⎥, ∂T, RT 2, ⎢, ⎥, ⎢, ⎥, ⎢, ⎥, ∂, A, kH, H −T S, ⎢ (b), ⎥, e RT, =−, ⎢, ⎥, 2, ∂T, RT, ⎢, ⎥, ⎢, ⎥, ⎢, ⎥, ∂(S), H, ⎢, ⎥, =− 2, ⎢ (c), ⎥, ⎢, ⎥, ∂T, T, ⎢, ⎥, ⎢, ⎥, ⎣, k ⎦, ∂(H ), (d), = S − R ln, ∂T, A, , ∂2V, . Thus,, ∂r∂h, ∂2V, ∂, =, ∂r∂h, ∂r, , , , ∂V, ∂h, , , =, , ∂, (πr 2 ) = 2π r., ∂r, , ∂V, with respect to h, keeping r, ∂r , , ∂ ∂V, , which is written as, constant, gives, ∂h ∂r, ∂2V, . Thus,, ∂h∂r, , , ∂, ∂2V, ∂ ∂V, =, =, (2πrh) = 2π r., ∂h∂r, ∂h ∂r, ∂h, , (iv) Differentiating, , (v), , ∂2V, ∂2V ∂2 V ∂2V, ,, ,, and, are examples of, 2, 2, ∂r, ∂h, ∂r∂h, ∂h∂r, second order partial derivatives., , 34.3, , Second order partial derivatives, , ∂2V, ∂2V, =, ∂r∂h ∂h∂r, and such a result is always true for continuous, functions (i.e. a graph of the function which has, no sudden jumps or breaks)., , (vi) It is seen from (iii) and (iv) that, , As with ordinary differentiation, where a differential, coefficient may be differentiated again, a partial derivative may be differentiated partially again to give higher, order partial derivatives., ∂V, (i) Differentiating, of Section 34.2 with respect, ∂r, , , ∂ ∂V, to r, keeping h constant, gives, which, ∂r ∂r, ∂2V, is written as, ∂r 2, Thus if V = πr 2 h,, then, , ∂2V, ∂, = (2πrh) = 2π h., 2, ∂r, ∂r, , ∂V, with respect to h, keeping, ∂h , , ∂ ∂V, r constant, gives, which is written, ∂h ∂h, ∂2V, as, ∂h 2, , (ii) Differentiating, , ∂2V, ∂, = (πr 2 ) = 0., 2, ∂h, ∂h, ∂V, (iii) Differentiating, with respect to r, keeping, ∂h , , ∂ ∂V, which is written, h constant, gives, ∂r ∂h, Thus, , Second order partial derivatives are used in the solution, of partial differential equations, in waveguide theory, in, such areas of thermodynamics covering entropy and the, continuity theorem, and when finding maxima, minima, and saddle points for functions of two variables (see, Chapter 36)., Problem 7. Given z =4x 2 y 3 − 2x 3 + 7y 2 find, ∂2z, ∂2 z, ∂2 z, ∂2z, (a) 2 (b) 2 (c), (d), ∂x, ∂y, ∂x∂ y, ∂ y∂x, (a), , ∂z, = 8x y 3 − 6x 2, ∂x, , ∂2 z, ∂ ∂z, ∂, =, =, (8x y 3 − 6x 2 ), ∂x 2, ∂x ∂x, ∂x, = 8y3 − 12 x, , (b), , ∂z, = 12x 2 y 2 + 14y, ∂y, , ∂2 z, ∂ ∂z, ∂, =, =, (12x 2 y 2 + 14y), 2, ∂y, ∂y ∂y, ∂y, = 24x2y + 14
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350 Higher Engineering Mathematics, , 2., , ⎡, ⎤, −2, −2, ⎢(a) x 2 (b) y 2 ⎥, ⎣, ⎦, (c) 0, (d) 0, , z = 2 ln x y, , ⎡, 3., , z=, , (x − y), (x + y), , ⎢, ⎢, ⎢, ⎣, , ⎤, 4x, −4y, (b), (x + y)3, (x + y)3 ⎥, ⎥, ⎥, 2(x − y), 2(x − y) ⎦, (c), (d), (x + y)3, (x + y)3, (a), , ⎡, 4., , z = sinh x cosh 2y, , (a) sinh x cosh 2y, , ⎢ (b) 4 sinh x cosh 2y, ⎢, ⎢, ⎢ (c) 2 cosh x sinh 2y, ⎣, (d) 2 cosh x sinh 2y, , 5. Given z = x 2 sin(x − 2y) find (a), (b), , ∂2 z, ∂ y2, , ⎤, ⎥, ⎥, ⎥, ⎥, ⎦, , ∂2z, and, ∂x 2, , ∂2 z, ∂2 z, =, ∂x∂ y ∂ y∂x, = 2x 2 sin(x − 2y) − 4x cos(x − 2y)., ⎡, ⎤, (a) (2 − x 2 ) sin(x − 2y), ⎢, ⎥, + 4x cos(x − 2y), ⎢, ⎥, ⎣, ⎦, (b) − 4x 2 sin(x − 2y), , Show also that, , ∂2z, , ∂2 z, , ∂2z, , ∂2z, , ,, and show that, =, ∂x 2 ∂ y 2, ∂x∂ y ∂ y∂x, x, when z = cos−1, y, , 6. Find, , ⎡, ⎤, −x, ∂2 z, ⎢(a) ∂x 2 = ( y 2 − x 2 )3 ,, ⎥, ⎢, ⎥, ⎢, ⎥, , , ⎢, ⎥, 2, 1, −x, 1, ∂ z, ⎢, ⎥, =, , +, ⎢(b), ⎥, 2 − x 2 ) y2, ⎢, ∂ y2, ( y2 − x 2 ) ⎥, (y, ⎢, ⎥, ⎢, ⎥, ⎣, ⎦, ∂2z, ∂2z, y, (c), =, =, ∂x∂ y ∂ y∂x, ( y 2 − x 2 )3, , 7. Given z =, , 3x, y, , , show that, , ∂2 z, ∂2 z, ∂2 z, =, and evaluate 2 when, ∂x∂ y ∂ y∂x, ∂x, , 1, 1, x = and y = 3., −√, 2, 2, 8. An equation used in thermodynamics is the, Benedict-Webb-Rubine equation of state for, the expansion of a gas. The equation is:, , , RT, C0 1, p=, + B0 RT − A0 − 2, V, T, V2, 1, Aα, + (b RT − a) 3 + 6, V, V, , γ , C 1+ 2 1 γ, −, V, +, e V2, T2, V3, Show that, =, , ∂2 p, ∂T 2, 6, V 2T 4, , , , , C, γ −γ, 1 + 2 e V 2 − C0 ., V, V
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Chapter 35, , Total differential, rates of, change and small changes, 35.1, , Problem 2. If z = f (u, v, w) and, z =3u 2 − 2v + 4w 3 v 2 find the total differential, dz., , Total differential, , In Chapter 34, partial differentiation is introduced for, the case where only one variable changes at a time,, the other variables being kept constant. In practice,, variables may all be changing at the same time., If z = f (u, v, w, . . .), then the total differential, dz,, is given by the sum of the separate partial differentials, of z,, i.e. dz =, , ∂z, ∂z, ∂z, du +, dv +, dw + · · ·, ∂u, ∂v, ∂w, , Problem 1. If z = f (x, y) and z = x 2 y 3 +, , (1), , 2x, + 1,, y, , determine the total differential, dz., , The total differential, ∂z, ∂z, ∂z, dz =, du +, dv +, dw, ∂u, ∂v, ∂w, ∂z, = 6u (i.e. v and w are kept constant), ∂u, ∂z, = −2 + 8w 3v, ∂v, (i.e. u and w are kept constant), ∂z, = 12w 2 v 2 (i.e. u and v are kept constant), ∂w, Hence, dz = 6u du + (8vw 3 − 2) dv + (12v2 w2 ) dw, , The total differential is the sum of the partial differentials,, i.e., , ∂z, ∂z, dx +, dy, dz =, ∂x, ∂y, 2, ∂z, = 2x y 3 +, (i.e. y is kept constant), ∂x, y, , 2x, ∂z, = 3x 2 y 2 2 (i.e. x is kept constant), ∂y, y, , , , , 2, 2x, dx + 3x2 y2 − 2 dy, Hence dz = 2xy3 +, y, y, , Problem 3. The pressure p, volume V and, temperature T of a gas are related by pV = kT ,, where k is a constant. Determine the total, differentials (a) dp and (b) dT in terms of p, V, and T ., (a), , ∂p, ∂p, dT +, dV ., ∂T, ∂V, kT, = kT then p =, V, k, ∂p, kT, = and, =− 2, V, ∂V, V, k, kT, = dT − 2 dV, V, V, , Total differential dp =, Since pV, hence, , ∂p, ∂T, , Thus, , dp
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Total differential, rates of change and small changes, dz, Hence the rate of change of z,, = 75.14 units/s,, dt, correct to 4 significant figures., Problem 5. The height of a right circular cone is, increasing at 3 mm/s and its radius is decreasing at, 2 mm/s. Determine, correct to 3 significant figures,, the rate at which the volume is changing (in cm3 /s), when the height is 3.2 cm and the radius is 1.5 cm., 1, Volume of a right circular cone, V = πr 2 h, 3, Using equation (2), the rate of change of volume,, dV, ∂V dr ∂V dh, =, +, dt, ∂r dt, ∂h dt, ∂V, 2, ∂V, 1, = πrh and, = πr 2, ∂r, 3, ∂h, 3, Since the height is increasing at 3 mm/s,, dh, i.e. 0.3 cm/s, then, = +0.3, dt, and since the radius is decreasing at 2 mm/s,, dr, i.e. 0.2 cm/s, then, = −0.2, dt, , , , 2, 1 2, dV, Hence, =, πrh (−0.2) +, πr (+0.3), dt, 3, 3, =, However,, Hence, , −0.4, πrh + 0.1πr 2, 3, , ∂A, 1, 1, = c sin B,, A = ac sin B,, 2, ∂a, 2, ∂A, 1, ∂A, 1, = a sin B and, = ac cos B, ∂c, 2, ∂B, 2, da, dc, = 0.4 units/s,, = −0.8 units/s, dt, dt, dB, = 0.2 units/s, and, dt, , , , , 1, 1, dA, =, c sin B (0.4) +, a sin B (−0.8), Hence, dt, 2, 2, , , 1, + ac cos B (0.2), 2, π, When a = 3, c = 4 and B = then:, 6, , , , , 1, 1, dA, π, π, =, (4) sin, (0.4) +, (3) sin, (−0.8), dt, 2, 6, 2, 6, , , 1, π, (3)(4) cos, (0.2), +, 2, 6, Since, , = 0.4 − 0.6 + 1.039 = 0.839 units2/s, correct, to 3 significant figures., Problem 7. Determine the rate of increase of, diagonal AC of the rectangular solid, shown in, Fig. 35.1, correct to 2 significant figures, if the sides, x, y and z increase at 6 mm/s, 5 mm/s and 4 mm/s, when these three sides are 5 cm, 4 cm and 3 cm, respectively., , h = 3.2 cm and r = 1.5 cm., , C, , dV, −0.4, =, π(1.5)(3.2) + (0.1)π(1.5)2, dt, 3, , b, B, , z 5 3 cm, , = −2.011 + 0.707 = −1.304 cm3 /s, Thus the rate of change of volume is 1.30 cm3/s, decreasing., Problem 6. The area A of a triangle is given by, A = 12 ac sin B, where B is the angle between sides a, and c. If a is increasing at 0.4 units/s, c is, decreasing at 0.8 units/s and B is increasing at 0.2, units/s, find the rate of change of the area of the, triangle, correct to 3 significant figures, when a is 3, units, c is 4 units and B is π/6 radians., Using equation (2), the rate of change of area,, d A ∂ A da ∂ A dc ∂ A dB, =, +, +, dt, ∂a dt, ∂c dt ∂ B dt, , 353, , y5, , 4 cm, , x5, , 5 cm, , A, , Figure 35.1, , (x 2 + y 2 ), , Diagonal AC = (BC 2 + AB 2 ), , = [z 2 + { (x 2 + y 2 )}2, = (z 2 + x 2 + y 2 ), , Diagonal AB =, , , Let AC = b, then b = (x 2 + y 2 + z 2 )
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354 Higher Engineering Mathematics, Using equation (2), the rate of change of diagonal b is, given by:, db ∂b dx ∂b dy ∂b dz, =, +, +, dt, ∂x dt, ∂ y dt, ∂z dt, , Since b = (x 2 + y 2 + z 2 ), , Exercise 141, change, , −1, x, ∂b, 1, = (x 2 + y 2 + z 2 ) 2 (2x) = , ∂x, 2, (x 2 + y 2 + z 2 ), y, ∂b, =, Similarly,, 2, ∂y, (x + y 2 + z 2 ), , ∂b, z, =, 2, ∂z, (x + y 2 + z 2 ), , and, , dx, = 6 mm/s = 0.6 cm/s,, dt, , dz, = 4 mm/s = 0.4 cm/s, dt, , Hence, , db, x, (0.6), = , 2, dt, (x + y 2 + z 2 ), + , , + , , y, (x 2 + y 2, , + z2), , z, (x 2 + y 2, , + z2), , (0.5), , (0.4), , When x = 5 cm, y = 4 cm and z = 3 cm, then:, 5, db, = , (0.6), 2, dt, (5 + 42 + 32 ), + , , + , , Further problems on rates of, , 1. The radius of a right cylinder is increasing at, a rate of 8 mm/s and the height is decreasing, at a rate of 15 mm/s. Find the rate at which the, volume is changing in cm3 /s when the radius, is 40 mm and the height is 150 mm., [+226.2 cm3 /s], 2. If z = f (x, y) and z = 3x 2 y 5 , find the rate of, change of z when x is 3 units and y is 2 units, when x is decreasing at 5 units/s and y is, increasing at 2.5 units/s., [2520 units/s], 3. Find the rate of change of k, correct to 4, significant figures, given the following data:, k = f (a, b, c); k = 2b ln a + c2 ea ; a is increasing at 2 cm/s; b is decreasing at 3 cm/s; c is, decreasing at 1 cm/s; a = 1.5 cm, b = 6 cm and, c = 8 cm., [515.5 cm/s], , dy, = 5 mm/s = 0.5 cm/s,, dt, and, , Now try the following exercise, , 4. A rectangular box has sides of length x cm,, y cm and z cm. Sides x and z are expanding at, rates of 3 mm/s and 5 mm/s respectively and, side y is contracting at a rate of 2 mm/s. Determine the rate of change of volume when x is, 3 cm, y is 1.5 cm and z is 6 cm., [1.35 cm3 /s], 5. Find the rate of change of the total surface area, of a right circular cone at the instant when the, base radius is 5 cm and the height is 12 cm if the, radius is increasing at 5 mm/s and the height, is decreasing at 15 mm/s., [17.4 cm2 /s], , 35.3, 4, , (52 + 42 + 32 ), 3, (52 + 42 + 32 ), , (0.5), , (0.4), , = 0.4243 + 0.2828 + 0.1697 = 0.8768 cm/s, Hence the rate of increase of diagonal AC is, 0.88 cm/s or 8.8 mm/s, correct to 2 significant figures., , Small changes, , It is often useful to find an approximate value for, the change (or error) of a quantity caused by small, changes (or errors) in the variables associated with the, quantity. If z = f (u, v, w, . . .) and δu, δv, δw, . . . denote, small changes in u, v, w, . . . respectively, then the corresponding approximate change δz in z is obtained from, equation (1) by replacing the differentials by the small, changes., Thus δz ≈, , ∂z, ∂z, ∂z, δu + δv +, δw + · · ·, ∂u, ∂v, ∂w, , (3)
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Total differential, rates of change and small changes, , Problem 8. Pressure p and volume V of a gas are, connected by the equation pV 1.4 = k. Determine, the approximate percentage error in k when the, pressure is increased by 4% and the volume is, decreased by 1.5%., , Hence δG ≈, , ∂k, ∂k, δp +, δV, ∂p, ∂V, , i.e., , Let p, V and k refer to the initial values., ∂k, Since, k = pV 1.4 then, = V 1.4, ∂p, ∂k, and, = 1.4 pV 0.4, ∂V, Since the pressure is increased by 4%, the change in, 4, pressure δp =, × p = 0.04 p., 100, Since the volume is decreased by 1.5%, the change in, −1.5, × V = −0.015V ., volume δV =, 100, Hence the approximate error in k,, δk ≈ (V ), , 1.4, , (0.04 p) + (1.4 pV, , 0.4, , )(−0.015V ), , ≈ pV 1.4[0.04 − 1.4(0.015)], ≈ pV 1.4[0.019] ≈, , 1.9, 1.9, pV 1.4 ≈, k, 100, 100, , i.e. the approximate error in k is a 1.9% increase., Problem 9. Modulus of rigidity G = (R 4 θ)/L,, where R is the radius, θ the angle of twist and L the, length. Determine the approximate percentage error, in G when R is increased by 2%, θ is reduced by, 5% and L is increased by 4%., , , , , , , R4, (−0.05θ), L, , , R4 θ, + − 2 (0.04L), L, , (0.02R) +, , R4 θ, R4 θ, [0.08 − 0.05 − 0.04] ≈ −0.01, ,, L, L, 1, G, δG ≈ −, 100, ≈, , Using equation (3), the approximate error in k,, δk ≈, , 4R 3 θ, L, , 355, , Hence the approximate percentage error in G is a, 1% decrease., Problem 10. The second moment of area of a, rectangle is given by I = (bl 3 )/3. If b and l are, measured as 40 mm and 90 mm respectively and the, measurement errors are −5 mm in b and +8 mm in, l, find the approximate error in the calculated value, of I ., Using equation (3), the approximate error in I ,, δI ≈, , ∂I, ∂I, δb + δl, ∂b, ∂l, , l3, ∂I, 3bl 2, ∂I, = and, =, = bl 2, ∂b, 3, ∂l, 3, δb = −5 mm and δl = +8 mm, , , , l3, (−5) + (bl 2 )(+8), 3, Since b = 40 mm and l = 90 mm then, Hence δ I ≈, , , δI ≈, , , 903, (−5) + 40(90)2 (8), 3, , ≈ −1215000 + 2592000, Using δG ≈, , Since, , and, , G=, , ∂G, ∂G, ∂G, δR +, δθ +, δL, ∂R, ∂θ, ∂L, 4R 3 θ ∂G, R4, R 4 θ ∂G, ,, =, ,, =, L ∂R, L, ∂θ, L, , −R 4 θ, ∂G, =, ∂L, L2, , 2, R = 0.02R, 100, Similarly, δθ = −0.05θ and δL =0.04L, Since R is increased by 2%, δ R =, , ≈ 1377000 mm4 ≈ 137.7 cm4, Hence the approximate error in the calculated value, of I is a 137.7 cm4 increase., Problem 11. The time of oscillation, t of a, , l, pendulum is given by t = 2π, . Determine the, g, approximate percentage error in t when l has an, error of 0.2% too large and g 0.1% too small.
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356 Higher Engineering Mathematics, Using equation (3), the approximate change in t ,, , H if the error in measuring current i is +2%,, the error in measuring resistance R is −3%, and the error in measuring time t is +1%., [+2%], , ∂t, ∂t, δt ≈ δl + δg, ∂l, ∂g, Since t = 2π, , and, , l ∂t, π, ,, =√, g ∂l, lg, , ∂t, = −π, ∂g, δl =, , 3., , l, (from Problem 6, Chapter 34), g3, , 0.2, l = 0.002 l and δg = −0.001g, 100, , π, hence δt ≈ √ (0.002l) + −π, lg, ≈ 0.002π, , l, + 0.001π, g, , , ≈ (0.001) 2π, , ≈ 0.0015t ≈, , l, g, , l, (−0.001 g), g3, l, g, , , + 0.0005 2π, , l, g, , 0.15, t, 100, , Hence the approximate error in t is a 0.15% increase., Now try the following exercise, Exercise 142, changes, , Further problems on small, , 1. The power P consumed in a resistor is given by, P = V 2 /R watts. Determine the approximate, change in power when V increases by 5% and, R decreases by 0.5% if the original values of V, and R are 50 volts and 12.5 ohms respectively., [+21 watts], 2. An equation for heat generated H is H = i 2 Rt ., Determine the error in the calculated value of, , fr =, , 1, √, , represents the resonant, 2π LC, frequency of a series connected circuit, containing inductance L and capacitance C., Determine the approximate percentage, change in fr when L is decreased by 3% and, C is increased by 5%., [−1%], , 4. The second moment of area of a rectangle, about its centroid parallel to side b is given by, I = bd 3/12. If b and d are measured as 15 cm, and 6 cm respectively and the measurement, errors are +12 mm in b and −1.5 mm in d,, find the error in the calculated value of I ., [+1.35 cm4 ], 5. The side b of a triangle is calculated using, b2 = a 2 + c2 − 2ac cos B. If a, c and B are, measured as 3 cm, 4 cm and π/4 radians respectively and the measurement errors, which occur are +0.8 cm, −0.5 cm and +π/90, radians respectively, determine the error in the, calculated value of b., [−0.179 cm], 6., , Q factor in, a resonant electrical circuit is given, 1 L, . Find the percentage change in, by: Q =, R C, Q when L increases by 4%, R decreases by 3%, and C decreases by 2%., [+6%], , 7. The rate, √ of flow of gas in a pipe is given by:, C d, , where C is a constant, d is the diamv= √, 6, T5, eter of the pipe and T is the thermodynamic, temperature of the gas. When determining the, rate of flow experimentally, d is measured and, subsequently found to be in error by +1.4%,, and T has an error of −1.8%. Determine the, percentage error in the rate of flow based on, the measured values of d and T ., [+2.2%]
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Chapter 36, , Maxima, minima and saddle, points for functions of two, variables, 36.1 Functions of two independent, variables, If a relation between two real variables, x and y,, is such that when x is given, y is determined, then, y is said to be a function of x and is denoted by, y = f (x); x is called the independent variable and y, the dependent variable. If y = f (u, v), then y is a function of two independent variables u and v. For example,, if, say, y = f (u, v) = 3u 2 − 2v then when u = 2 and, v = 1, y = 3(2)2 − 2(1) = 10. This may be written as, f (2, 1) = 10. Similarly, if u = 1 and v = 4, f (1, 4) = −5., , Consider a function of two variables x and y, defined by z = f (x, y) = 3x 2 − 2y. If (x, y) = (0, 0),, then f (0, 0) = 0 and if (x , y) =(2, 1), then f (2, 1)=10., Each pair of numbers, (x, y), may be represented, by a point P in the (x, y) plane of a rectangular, Cartesian co-ordinate system as shown in Fig. 36.1., The corresponding value of z = f (x, y) may be represented by a line PP drawn parallel to the z-axis., Thus, if, for example, z =3x 2 − 2y, as above, and P, is the co-ordinate (2, 3) then the length of PP is, 3(2)2 − 2(3) = 6. Figure 36.2 shows that when a large, number of (x, y) co-ordinates are taken for a function, z, , z, 6, p9, , o, , 3, , 0, , y, 2, p, , x, , Figure 36.1, , x, , Figure 36.2, , y
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358 Higher Engineering Mathematics, f (x, y), and then f (x, y) calculated for each, a large, number of lines such as P P can be constructed, and in, the limit when all points in the (x, y) plane are considered, a surface is seen to result as shown in Fig. 36.2., Thus the function z = f (x, y) represents a surface and, not a curve., , z, , Minimum, point, q, y, , 36.2 Maxima, minima and saddle, points, Partial differentiation is used when determining stationary points for functions of two variables. A function, f (x, y) is said to be a maximum at a point (x, y) if the, value of the function there is greater than at all points in, the immediate vicinity, and is a minimum if less than at, all points in the immediate vicinity. Figure 36.3 shows, geometrically a maximum value of a function of two, variables and it is seen that the surface z = f (x, y) is, higher at (x, y) = (a, b) than at any point in the immediate vicinity. Figure 36.4 shows a minimum value of a, function of two variables and it is seen that the surface, z = f (x, y) is lower at (x, y) = ( p, q) than at any point, in the immediate vicinity., , p, x, , Figure 36.4, z, t1, Maximum, point, t2, , b, , O, , y, , z, , Maximum, point, , a, x, , Figure 36.5, , b, y, a, x, , With functions of two variables there are three types, of stationary points possible, these being a maximum, point, a minimum point, and a saddle point. A saddle point Q is shown in Fig. 36.6 and is such that a, point Q is a maximum for curve 1 and a minimum for, curve 2., , Figure 36.3, Curve 2, , If z = f (x, y) and a maximum occurs at (a, b), the, curve lying in the two planes x = a and y = b must also, have a maximum point (a, b) as shown in Fig. 36.5. Consequently, the tangents (shown as t1 and t2) to the curves, at (a, b) must be parallel to Ox and Oy respectively., ∂z, ∂z, This requires that, = 0 and, = 0 at all maximum, ∂x, ∂y, and minimum values, and the solution of these equations, gives the stationary (or critical) points of z., , Q, , Curve 1, , Figure 36.6
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Maxima, minima and saddle points for functions of two variables, , 36.3 Procedure to determine, maxima, minima and saddle, points for functions of two, variables, Given z = f (x, y):, (i) determine, , ∂z, ∂z, and, ∂x, ∂y, , (ii) for stationary points,, , ∂z, ∂z, = 0 and, = 0,, ∂x, ∂y, , ∂z, = 0 and, (iii) solve the simultaneous equations, ∂x, ∂z, = 0 for x and y, which gives the co-ordinates, ∂y, of the stationary points,, (iv) determine, , ∂2 z, ∂2z ∂2z, ,, and, ∂x 2 ∂ y 2, ∂x∂ y, , (v) for each of the co-ordinates of the stationary, ∂2z ∂2z, points, substitute values of x and y into 2 , 2, ∂x ∂ y, ∂2 z, and, and evaluate each,, ∂x∂ y, , (vi) evaluate, , ∂2z, ∂x∂ y, , =, , ∂2z, ∂x∂ y, , 2, , (i), , ∂z, ∂z, = 2(x − 1) and, = 2(y − 2), ∂x, ∂y, , (ii) 2(x − 1) =0, , (1), , 2(y − 2) = 0, , (2), , (iii) From equations (1) and (2), x = 1 and y = 2, thus, the only stationary point exists at (1, 2)., (iv) Since, , , , ∂2 z, −, ∂x 2, , , , ∂2z, ∂ y2, , , , ∂ z, if < 0 and 2 < 0, then the stationary, ∂x, point is a maximum point,, , and, ∂2z, > 0, then the stationary, ∂x2, point is a minimum point., if < 0 and, , ∂2z, ∂z, = 2(x − 1) = 2x − 2, 2 = 2, ∂x, ∂x, ∂z, ∂2z, = 2(y − 2) = 2y − 4, 2 = 2, ∂y, ∂y, , ∂2z, ∂, and, =, ∂x∂ y ∂x, , ∂2z ∂2 z, ∂2 z, ,, and, ∂x 2 ∂ y 2, ∂x∂ y, , 2, , (c), , Following the above procedure:, , for each stationary point,, , and evaluate,, (viii) (a) if > 0 then the stationary point is a, saddle point., (b), , Problem 1. Show that the function, z =(x − 1)2 + (y − 2)2 has one stationary point only, and determine its nature. Sketch the surface, represented by z and produce a contour map in the, x-y plane., , and since, , into the equation, , , 36.4 Worked problems on maxima,, minima and saddle points for, functions of two variables, , 2, , (vii) substitute the values of, , 359, , (v), , , ∂z, ∂, = (2y − 4) = 0, ∂y, ∂x, , ∂2 z ∂2z, ∂2z, =, =, 2, and, =0, ∂x 2 ∂ y 2, ∂x∂ y, , , (vi), , , , 2, ∂2 z, =0, ∂x∂ y, , (vii) = (0)2 − (2)(2) = −4, ∂2z, (viii) Since < 0 and 2 > 0, the stationary point, ∂x, (1, 2) is a minimum., The surface z = (x − 1)2 + (y − 2)2 is shown in three, dimensions in Fig. 36.7. Looking down towards the, x-y plane from above, it is possible to produce a contour map. A contour is a line on a map which gives, places having the same vertical height above a datum, line (usually the mean sea-level on a geographical map).
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360 Higher Engineering Mathematics, z, , Problem 2. Find the stationary points of the, surface f (x, y) = x 3 − 6x y + y 3 and determine their, nature., y, 1, , Let z = f (x, y) = x 3 − 6x y + y 3, , 2, , Following the procedure:, (i), , o, 1, , ∂z, ∂z, = 3x 2 − 6y and, = −6x + 3y 2, ∂x, ∂y, , (ii) for stationary points, 3x 2 − 6y = 0, x, , −6x + 3y 2 = 0, , and, , Figure 36.7, , (iii) from equation (1), 3x 2 = 6y, , A contour map for z =(x − 1)2 + (y − 2)2 is shown in, Fig. 36.8. The values of z are shown on the map and these, give an indication of the rise and fall to a stationary point., , and, , y=, , 3x 2 1 2, = x, 6, 2, , y, , z51, , 2, , z54, , z59, , z 5 16, , 1, , 1, , Figure 36.8, , 2, , x, , (1), (2)
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Maxima, minima and saddle points for functions of two variables, and substituting in equation (2) gives:, , , 1 2 2, =0, x, −6x + 3, 2, 3, −6x + x 4 = 0, 4, , 3, x, 3x, −2 = 0, 4, from which, x = 0 or, , x3, − 2 =0, 4, , i.e. x 3 = 8 and x = 2, When x = 0, y = 0 and when x = 2, y = 2 from, equations (1) and (2)., Thus stationary points occur at (0, 0), and (2, 2)., , ∂2z, ∂2z, ∂ ∂z, ∂2z, =, 6x,, =, 6y, and, =, (iv), ∂x 2, ∂ y2, ∂x∂ y, ∂x ∂ y, =, , ∂, (−6x + 3y 2 ) = −6, ∂x, , ∂2 z, ∂2 z, = 0, 2 = 0, 2, ∂x, ∂y, ∂2 z, and, = −6, ∂x∂ y, ∂2 z, ∂2 z, = 12, 2 = 12, for (2, 2),, 2, ∂x, ∂y, ∂2 z, and, = −6, ∂x∂ y, 2 2, ∂ z, = (−6)2 = 36, (vi) for (0, 0),, ∂x∂ y, 2 2, ∂ z, for (2, 2),, = (−6)2 = 36, ∂x∂ y, (v), , for (0, 0), , , , ∂2z, (vii) (0, 0) =, ∂x∂ y, , , , 2, −, , ∂2z, ∂x 2, , , , ∂2 z, ∂ y2, , Now try the following exercise, Exercise 143 Further problems on, maxima, minima and saddle points for, functions of two variables, 1. Find the stationary point of the surface, f (x, y) = x 2 + y 2 and determine its nature., Sketch the surface represented by z., [Minimum at (0, 0)], 2. Find the maxima, minima and saddle points, for the following functions:, (a) f (x, y) = x 2 + y 2 − 2x + 4y + 8, (b) f (x, y) = x 2 − y 2 − 2x + 4y + 8, (c) f (x, y) = 2x⎡+ 2y − 2x y − 2x 2 − y 2 + 4.⎤, (a) Minimum at (1, −2), ⎣ (b) Saddle point at (1, 2) ⎦, (c) Maximum at (0, 1), 3. Determine the stationary values of the function f (x, y) = x 3 − 6x 2 − 8y 2 and distinguish, between them. Sketch an approximate contour, map to representthe surface f (x, y)., Maximum point at (0, 0),, saddle point at (4, 0), 4. Locate the stationary point of the function, z =12x 2 + 6x y + 15y 2 ., [Minimum at (0, 0)], 5. Find the stationary points of the surface, z = x 3 − x y + y 3 and distinguish between, them., , saddle point at, (0, 0),, minimum at 13 , 13, , , , = 36 − (0)(0) = 36, (2, 2) = 36 − (12)(12) = −108, (viii) Since (0, 0) > 0 then (0, 0) is a saddle point., ∂2 z, > 0, then (2, 2) is a, Since (2, 2) < 0 and, ∂x 2, minimum point., , 36.5 Further worked problems on, maxima, minima and saddle, points for functions of two, variables, Problem 3. Find the co-ordinates of the, stationary points on the surface, z = (x 2 + y 2 )2 − 8(x 2 − y 2 ), and distinguish between them. Sketch the, approximate contour map associated with z., , 361
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362 Higher Engineering Mathematics, (vii) (0, 0) = (0)2 − (−16)(16) = 256, , Following the procedure:, , (2, 0) = (0)2 − (32)(32) = −1024, , ∂z, (i), = 2(x 2 + y 2 )2x − 16x and, ∂x, ∂z, = 2(x 2 + y 2 )2y + 16y, ∂y, , (−2, 0) = (0)2 − (32)(32) = −1024, , (ii) for stationary points,, 2(x 2 + y 2 )2x − 16x = 0, i.e., , 4x 3 + 4x y 2 − 16x = 0, , and, , 2(x 2 + y 2 )2y + 16y = 0, , i.e., , 4y(x 2 + y 2 + 4) = 0, , (iii) From equation (1), y 2 =, Substituting, , y2 = 4 − x 2, , (1), (2), , 16x − 4x 3, =4 − x2, 4x, in equation (2) gives, , 4y(x 2 + 4 − x 2 + 4) = 0, i.e. 32y = 0 and y = 0, When y = 0 in equation (1),, , 4x 3 − 16x = 0, , i.e., , 4x(x 2 − 4) = 0, , from which, x = 0 or x = ±2, The co-ordinates of the stationary points are, (0, 0), (2, 0) and (−2, 0)., ∂2z, = 12x 2 + 4y 2 − 16,, (iv), ∂x 2, ∂2 z, ∂ y2, , = 4x 2 + 12y 2 + 16 and, , ∂2z, ∂x∂ y, , = 8x y, , (v) For the point (0, 0),, ∂2z, ∂2z, ∂2z, =, −16,, =, 16, and, =0, ∂x 2, ∂ y2, ∂x∂ y, For the point (2, 0),, ∂2z, ∂2z, ∂2z, =, 32,, =, 32, and, =0, ∂x 2, ∂ y2, ∂x∂ y, For the point (−2, 0),, ∂2z, ∂2z, ∂2z, = 32, 2 = 32 and, =0, 2, ∂x, ∂y, ∂x∂ y, , (vi), , ∂2 z, ∂x∂ y, , 2, = 0 for each stationary point, , (viii) Since (0, 0) > 0, the point (0, 0) is a saddle, point., 2 , ∂ z, > 0, the point, Since (0, 0) < 0 and, ∂x 2 (2, 0), (2, 0) is a minimum point., 2 , ∂ z, Since (−2, 0) < 0 and, > 0, the, ∂x 2 (−2, 0), point (−2, 0) is a minimum point., Looking down towards the x-y plane from above, an, approximate contour map can be constructed to represent the value of z. Such a map is shown in Fig. 36.9., To produce a contour map requires a large number of, x-y co-ordinates to be chosen and the values of z at, each co-ordinate calculated. Here are a few examples of, points used to construct the contour map., When z = 0, 0 =(x 2 + y 2 )2 − 8(x 2 − y)2, In addition, when, say, y = 0 (i.e. on the x-axis), 0 = x 4 − 8x 2 , i.e. x 2 (x 2 − 8) = 0, √, from which, x = 0 or x = ± 8, √, Hence the contour z = 0 crosses the x-axis at 0 and ± 8,, i.e. at co-ordinates (0, 0), (2.83, 0) and (−2.83, 0) shown, as points, S, a and b respectively., When z = 0 and x =2 then, 0 = (4 + y 2 )2 − 8(4 − y 2 ), i.e. 0 = 16 + 8y 2 + y 4 − 32 + 8y 2, i.e. 0 = y 4 + 16y 2 − 16, Let y 2 = p, then p2 + 16 p − 16 = 0 and, , −16 ± 162 − 4(1)(−16), p=, 2, −16 ± 17.89, =, 2, = 0.945 or −16.945, Hence y =, , √, , p=, , , , (0.945) or (−16.945), , = ± 0.97 or complex roots.
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Maxima, minima and saddle points for functions of two variables, , 363, , y, 4, , i, , z5, , 128, , 2, , z59, c, , g, , 0, z5, , S, f, , b, , 3, 22, , 3, 2, , a, , e, , x, , d, h, 22, , j, 24, , Figure 36.9, , Hence the z =0 contour passes through the co-ordinates, (2, 0.97) and (2, −0.97) shown as a c and d in Fig. 36.9., Similarly, for the z = 9 contour, when y = 0,, 9=, , (x 2, , + 02 )2, , i.e., , 9 = x 4 − 8x 2, , i.e., , x 4 − 8x 2 − 9 =0, , − 8(x 2 − 02 ), , Hence (x 2 − 9)(x 2 + 1) = 0., from which, x = ±3 or complex roots., Thus the z = 9 contour passes through (3, 0) and (−3, 0),, shown as e and f in Fig. 36.9., If z = 9 and x = 0, 9 = y 4 + 8y 2, i.e., , y 4 + 8y 2 − 9 = 0, , i.e., , (y 2 + 9)(y 2 − 1) = 0, , from which, y = ±1 or complex roots., Thus the z = 9 contour also passes through (0, 1) and, (0, −1), shown as g and h in Fig. 36.9., , When, say, x = 4 and y = 0,, z = (42 )2 − 8(42 ) = 128., when z = 128 and x = 0, 128 = y 4 + 8y 2, i.e., , y 4 + 8y 2 − 128 = 0, , i.e. (y 2 + 16)(y 2 − 8) = 0, √, from which, y = ± 8 or complex roots., Thus the z = 128 contour passes through (0, 2.83) and, (0, −2.83), shown as i and j in Fig. 36.9., In a similar manner many other points may be calculated, with the resulting approximate contour map shown in, Fig. 36.9. It is seen that two ‘hollows’ occur at the minimum points, and a ‘cross-over’ occurs at the saddle, point S, which is typical of such contour maps., Problem 4. Show that the function, f (x, y) = x 3 − 3x 2 − 4y 2 + 2, has one saddle point and one maximum point., Determine the maximum value.
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364 Higher Engineering Mathematics, , , Let z = f (x, y) = x 3 − 3x 2 − 4y 2 + 2., , (vi), , Following the procedure:, (i), , (ii) for stationary points, 3x 2 −6x = 0, , (1), , −8y = 0, , (2), , (iii) From equation (1), 3x(x − 2) = 0 from which,, x = 0 and x = 2., , , ∂2z, (viii) Since (0, 0) < 0 and, < 0, the, ∂x 2 (0, 0), point (0, 0) is a maximum point and hence the, maximum value is 0., Since (2, 0) > 0, the point (2, 0) is a saddle, point., , Hence the stationary points are (0, 0), and (2, 0)., (iv), , = 6x − 6,, , ∂x 2, , ∂2z, ∂ y2, , = −8 and, , = (0)2 = 0, , , , From equation (2), y = 0., , ∂2z, , 2, , (vii) (0, 0) = 0 −(−6)(−8) = −48, (2, 0) = 0 −(6)(−8) = 48, , ∂z, ∂z, = 3x 2 − 6x and, = − 8y, ∂x, ∂y, , and, , ∂2 z, ∂x∂ y, , ∂2z, ∂x∂ y, , The value of z at the saddle point is, 23 − 3(2)2 − 4(0)2 + 2 =−2., , =0, , An approximate contour map representing the surface, f (x, y) is shown in Fig. 36.10 where a ‘hollow effect’ is, seen surrounding the maximum point and a ‘cross-over’, occurs at the saddle point S., , (v) For the point (0, 0),, ∂2 z, ∂2 z, ∂2 z, =, −6,, =, −8, and, =0, ∂x 2, ∂ y2, ∂x∂ y, For the point (2, 0),, , Problem 5. An open rectangular container is to, have a volume of 62.5 m3 . Determine the least, surface area of material required., , ∂2z, ∂2z, ∂2 z, = 6, 2 = −8 and, =0, 2, ∂x, ∂y, ∂x∂ y, , y, 2, , z5, , 0, , MAX, , S, 2, 2, 52, , z, , z5, , 22, , Figure 36.10, , 24, , 21, , 21, , z5, , z5, , 22, , 3, , 2, , 4, , x
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Maxima, minima and saddle points for functions of two variables, From equation (1),, , (5) (5) z =62.5, z=, , from which,, , 365, , 62.5, = 2.5 m, 25, , ∂ 2 S 250 ∂ 2 S 250, ∂2 S, = 3 , 2 = 3 and, =1, 2, ∂x, x, ∂y, y, ∂x∂ y, When x = y = 5,, , y, z, , ∂2 S, ∂2 S, ∂2 S, =, 2,, =, 2, and, =1, ∂x 2, ∂ y2, ∂x∂ y, , = (1)2 − (2)(2) = −3, ∂2 S, > 0, then the surface area S is a, Since < 0 and, ∂x 2, minimum., , x, , Figure 36.11, , Hence the minimum dimensions of the container to have, a volume of 62.5 m3 are 5 m by 5 m by 2.5 m., Let the dimensions of the container be x, y and z as, shown in Fig. 36.11., , = (5)(5) + 2(5)(2.5) + 2(5)(2.5), , Volume, , V = x yz = 62.5, , (1), , Surface area,, , S = x y + 2yz + 2x z, , (2), , Exercise 144 Further problems on, maxima, minima and saddle points for, functions of two variables, , Substituting in equation (2) gives:, , , i.e., , S=xy +, , , , , 62.5, 62.5, + 2x, xy, xy, , 1. The function z = x 2 + y 2 + x y + 4x − 4y + 3, has one stationary value. Determine its, co-ordinates and its nature., [Minimum at (−4, 4)], , 125 125, +, x, y, , which is a function of two variables, ∂s, 125, = y − 2 = 0 for a stationary point,, ∂x, x, hence x 2 y =125, ∂s, 125, = x − 2 = 0 for a stationary point,, ∂y, y, hence x y 2 = 125, , (3), , (4), , Dividing equation (3) by (4) gives:, x2 y, x, = 1, i.e. = 1, i.e. x = y, x y2, y, Substituting y = x in equation (3) gives x 3 = 125, from, which, x = 5 m., Hence y = 5 m also, , = 75 m2, Now try the following exercise, , 62.5, From equation (1), z =, xy, , S = x y + 2y, , From equation (2), minimum surface area, S, , 2. An open rectangular container is to have a volume of 32 m3 . Determine the dimensions and, the total surface area such that the total surface, area is a minimum., , 4 m by 4 m by 2 m,, surface area = 48m2, 3. Determine the stationary values of the, function, f (x, y) = x 4 + 4x 2 y 2 − 2x 2 + 2y 2 − 1, and distinguish between them., ⎡, ⎤, Minimum at (1, 0),, ⎣ minimum at (−1, 0), ⎦, saddle point at (0, 0)
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366 Higher Engineering Mathematics, 4. Determine the stationary points of the surface, f (x, y) = x 3 − 6x 2 − y 2 ., , Maximum at (0, 0),, saddle point at (4, 0), 5. Locate the stationary points on the surface, f (x, y) = 2x 3 + 2y 3 − 6x − 24y + 16, and determine their nature., ⎡, ⎤, Minimum at (1, 2),, ⎣ maximum at (−1, −2),, ⎦, saddle points at (1, −2) and (−1, 2), , 6. A large marquee is to be made in the form, of a rectangular box-like shape with canvas, covering on the top, back and sides. Determine, the minimum surface area of canvas necessary, if the volume of the marquee is to the 250 m3., [150 m2 ]
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Revision Test 10, This Revision Test covers the material contained in Chapters 32 to 36. The marks for each question are shown in, brackets at the end of each question., 1., , (a) 5 ln (shx) (b) 3 ch3 2x, 2x, , (c) e, 2., , 6., , sech 2x, , (7), , Differentiate the following functions with respect, to the variable:, x, 1, (a) y = cos−1, 5, 2, (b) y = 3esin, , 2 sec−1 5x, x, , (d) y = 3 sinh−1 (2x 2 − 1), , 4., , ∂z ∂z, ,, ,, , ,, and, ., ∂x ∂ y ∂x 2 ∂ y 2 ∂x∂ y, ∂ y∂x, ∂2z, , 5., , 8., , The volume V of a liquid of viscosity coefficient, η delivered after time t when passed through a, tube of length L and diameter d by a pressure p, pd 4t, . If the errors in V , p and, is given by V =, 128ηL, L are 1%, 2% and 3% respectively, determine the, error in η., (8), , 9., , Determine and distinugish between the stationary, values of the function, , Evaluate the following, each correct to 3 decimal, places:, (6), , If z = f (x, y) and z = x cos(x + y) determine, ∂2z, , ∂2 z, , ∂2 z, , (12), , The magnetic field vector H due to a steady current I flowing around a circular wire of radius r, and at a distance x from its centre is given by, , , x, I ∂, √, H =±, 2 ∂x, r2 + x2, , (6), , An engineering function z = f (x, y) and, y, z = e 2 ln(2x + 3y). Determine the rate of, increase of z, correct to 4 significant figures,, when x = 2 cm, y = 3 cm, x is increasing at 5 cm/s, and y is increasing at 4 cm/s., (8), , (14), , (a) sinh−1 3 (b) cosh−1 2.5 (c) tanh−1 0.8, , If x yz = c, where c is constant, show that, , , dx d y, +, dz = −z, x, y, , (7), , 7., , −1 t, , (c) y =, , 3., , r2 I, Show that H = ± , 2 (r 2 + x 2 )3, , Differentiate the following functions with respect, to x:, , f (x, y) = x 3 − 6x 2 − 8y 2, and sketch an approximate contour map to represent the surface f (x, y)., (20), 10. An open, rectangular fish tank is to have a volume, of 13.5 m3 . Determine the least surface area of, glass required., (12)
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Chapter 37, , Standard integration, 37.1, , The process of integration, , The process of integration reverses the process of, differentiation. In differentiation, if f (x) = 2x 2 then, f (x) = 4x. Thus the integral of 4x is 2x 2 , i.e. integration is the process of moving from f (x) to f (x). By, similar reasoning, the integral of 2t is t 2., Integration is a process of summation oradding parts, together and an elongated S, shown as , is used to, replace, the words, ‘the integral of’. Hence, from above,, , 4x = 2x 2 and 2t is t 2., dy, In differentiation, the differential coefficient, indidx, cates that a function of x is being differentiated with, respect to x, the dx indicating that it is ‘with respect, to x’. In integration the variable of integration is shown, by adding d (the variable) after the function to be, integrated., , 37.2 The general solution of integrals, of the form ax n, , The general solution of integrals of the form ax n dx,, where a and n are constants is given by:, !, ax n dx =, , This rule is true when n is fractional, zero, or a positive, or negative integer, with the exception of n = −1., Using this rule gives:, !, 3x 4+1, 3, (i), 3x 4 dx =, + c = x5 + c, 4+1, 5, !, !, 2x −2+1, 2, −2, dx, =, 2x, dx, =, (ii), +c, x2, −2 +1, , !, Thus, , and, , 2t dt means ‘the integral of 2t, with respect to t ’., , As stated above, the differential coefficient of 2x 2 is, 4x, hence 4x dx = 2x 2 . However, the, differential coefficient of 2x 2 + 7 is also 4x. Hence 4x dx is also equal, to 2x 2 + 7. To allow for the possible presence of a constant, whenever the process of integration is performed,, a constant ‘c’ is added to the result., !, Thus, , =, , 4x dx means ‘the integral of 4x, with respect to x’,, !, , !, 4x dx = 2x 2 + c and, , 2t dt = t 2 + c, , ‘c’ is called the arbitrary constant of integration., , ax n+1, +c, n+1, , !, (iii), , √, , 2x −1, −2, +c=, + c, and, −1, x, !, , x dx =, , 1, , 1, x2, , 3, , x 2 +1, x2, dx =, +c=, +c, 1, 3, +1, 2, 2, , 2√ 3, x +c, 3, Each of these three results may be checked by differentiation., =, , (a), , The integral of a constant k is kx + c. For, example,, !, 8 dx = 8x + c, , (b) When a sum of several terms is integrated the result, is the sum of the integrals of the separate terms.
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Standard integration, For example,, !, (3x + 2x 2 − 5) dx, !, !, !, 2, = 3x dx + 2x dx − 5 dx, =, , 37.3, , 3x 2, 2x 3, +, − 5x + c, 2, 3, , Standard integrals, , Since integration is the reverse process of differentiation the standard integrals listed in Table 37.1 may be, deduced and readily checked by differentiation., Table 37.1 Standard integrals, !, , ax n+1, +c, n +1, (except when n =−1), , ax n dx =, , (i), !, , cos ax dx =, , (ii), !, (iii), !, , 1, sin ax dx = − cos ax + c, a, sec 2 ax dx =, , (iv), !, (v), !, (vi), !, (vii), !, !, , (b) When a = 2 and n = 3 then, !, 2t 3 dt =, , 1, cosec ax cot ax dx = − cosec ax + c, a, 1, sec ax tan ax dx = sec ax + c, a, 1 ax, e +c, a, , 1, dx = ln x + c, x, , , , Problem 1. Determine (a) 5x 2 dx (b) 2t 3 dt ., , ax n+1, The standard integral, ax n dx =, +c, n +1, (a) When a = 5 and n =2 then, !, 5x 2+1, 5x 3, 5x 2 dx =, +c=, +c, 2+1, 3, , 2t 3+1, 2t 4, 1, +c=, +c= t4 +c, 3+1, 4, 2, , Each of these results may be checked by differentiating, them., Problem 2. Determine, , !, 3, 4 + x − 6x 2 dx., 7, , (4 + 37 x − 6x 2 ) dx may be written as, , , , 4 dx + 37 x dx − 6x 2 dx, i.e. each term is integrated, separately. (This splitting up of terms only applies,, however, for addition and subtraction.), , !, 3, 2, Hence, 4 + x − 6x dx, 7, 1+1, 3 x, x 2+1, = 4x +, − (6), +c, 7 1+1, 2+1, 2, 3 x, x3, = 4x +, − (6) + c, 7 2, 3, , 1, tan ax + c, a, , 1, cosec 2 ax dx = − cot ax + c, a, , eax dx =, , (viii), (ix), , 1, sin ax + c, a, , 369, , = 4x +, , 3 2, x − 2x 3 + c, 14, , Note that when an integral contains more than one term, there is no need to have an arbitrary constant for each;, just a single constant at the end is sufficient., Problem 3. Determine, !, !, 2x 3 − 3x, (a), dx (b) (1 − t )2 dt., 4x, (a), , Rearranging into standard integral form gives:, !, 2x 3 − 3x, dx, 4x, !, ! 2, 2x 3 3x, x, 3, =, −, dx =, − dx, 4x, 4x, 2, 4, 2+1, 1 x, 3, =, − x +c, 2 2+1 4, 3, 1 x, 3, 3, 1, − x + c = x3 − x + c, =, 2 3, 4, 6, 4
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372 Higher Engineering Mathematics, !, 7., , (a), , !, 3 cos2x dx (b), , 7 sin 3θ dθ, ⎡, , ⎤, 3, ⎢ (a) 2 sin 2x + c ⎥, ⎢, ⎥, ⎣, ⎦, 7, (b) − cos 3θ + c, 3, !, !, 3, sec2 3x dx (b) 2 cosec 2 4θ dθ, 8. (a), 4, , 1, 1, (a) tan 3x +c (b) − cot 4θ +c, 4, 2, !, 9. (a) 5 cot 2t cosec 2t dt, !, 4, sec 4t tan 4t dt, (b), 3, ⎡, ⎤, 5, cosec, 2t, +, c, (a), −, ⎢, ⎥, 2, ⎢, ⎥, ⎣, ⎦, 1, (b) sec 4t + c, 3, !, !, 2 dx, 3 2x, e dx (b), 10. (a), 4, 3 e5x, , −2, 3, +c, (a) e2x + c (b), 8, 15 e5x, , !, ! 2, 2, u −1, 11. (a), du, dx (b), 3x, u, , 2, u2, (a) ln x + c (b), − ln u + c, 3, 2, !, 12., , (a), , (2+3x)2, √, dx (b), x, ⎡, , !, , !, , x3, +c, 3, , , , 3, , 33, 1, +c −, +c, 3, 3, 1, , , 1, 2, = (9 + c) −, + c =8, 3, 3, 3, , =, , Note that the ‘c’ term always cancels out when limits are, applied and it need not be shown with definite integrals., Problem 12. Evaluate, 3, 2, (a) 1 3x dx (b) −2 (4 − x 2 ) dx., !, , 2, , (a), 1, , , , 3x 2, 3x dx =, 2, , , , , , 3 2, 3 2, =, (2) −, (1), 2, 2, 1, 2, , 1, 1, =6 − 1 =4, 2, 2, !, (b), , , x3, (4 − x ) dx = 4x −, 3, −2, 3, , 3, , 2, , −2, , , , , (3)3, (−2)3, = 4(3) −, − 4(−2) −, 3, 3, , , −8, = {12 − 9} − −8 −, 3, , , 1, 1, = {3} − −5, =8, 3, 3, 4, , !, , √, , Definite integrals, , , x 2 dx =, , 1, , Problem 13., , Evaluate, , positive square roots only., 4, , !, 1, , 37.4, , 3, , 2, 1, + 2t dt, t, , ⎤, 18 √ 5, (a) 8 x + 8 x 3 +, x +c, ⎢, ⎥, 5, ⎢, ⎥, ⎣, ⎦, 3, 1, 4t, (b) − + 4t +, +c, t, 3, √, , limit and ‘a’ the lower limit. The operation of applying, the limits is defined as [x]ba = (b) − (a)., The increase in the value of the integral x 2 as x increases, 3, from 1 to 3 is written as 1 x 2 dx., Applying the limits gives:, , 1, , , θ +2, √, dθ, taking, θ, , , , ! 4, θ +2, θ, 2, √, +, dθ, dθ =, 1, 1, θ, 1, θ2, θ2, , ! 4 1, −1, θ 2 + 2θ 2 dθ, =, 1, , Integrals containing an arbitrary constant c in their, results are called indefinite integrals since their precise, value cannot be determined without further information., Definite integrals are those in which limits are applied., If an expression is written as [x]ba, ‘b’ is called the upper, , ⎡, ⎢θ, =⎣, , , 1, 2 +1, , 1, +1, 2, , , , +, , ⎤4, −1, 2 +1, , ⎥, ⎦, 1, − +1, 2, 1, , 2θ
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Standard integration, ⎡, =⎣, , 3, , θ2, 3, 2, , 1, , +, , 2θ 2, 1, 2, , ⎤4, ⎦ =, , , √, 2 3, θ +4 θ, 3, , 1, , Problem 16. Evaluate, !, ! 2, 4 e2x dx (b), (a), , 4, 1, , , , , √, 2, 2, 3, 3, =, (4) + 4 4 −, (1) + 4 (1), 3, 3, , , , 16, 2, =, +8 −, +4, 3, 3, , 1, , π, 2, , Problem 14. Evaluate, , !, , 2, , (a), , , 4 e2x dx =, , 1, , π, 2, , 2, , 4 2x, e, 2, , = 2[ e2x ]21 = 2[ e4 − e2 ], 1, , = 2[54.5982 −7.3891] =94.42, !, , 4, , (b), 1, , , 3, 3, du =, ln u, 4u, 4, , 3 sin 2x dx., , 4, , 3, = [ln 4 − ln 1], 4, 1, , 3, = [1.3863 −0] =1.040, 4, , 0, , !, , 1, , 3, du,, 4u, , each correct to 4 significant figures., , 2, 2, 1, = 5 +8− −4 = 8, 3, 3, 3, !, , 4, , 3 sin 2x dx, , Now try the following exercise, , 0, π, π, , , , 2, 2, 3, 1, = − cos 2x, = (3) −, cos 2x, 2, 2, 0, 0, , , , , π , 3, 3, − − cos 2(0), = − cos 2, 2, 2, 2, , , , 3, 3, = − cos π − − cos 0, 2, 2, , , , 3, 3 3, 3, = − (−1) − − (1) = + = 3, 2, 2, 2 2, , , , !, , Exercise 146, integrals, , In problems 1 to 8, evaluate the definite integrals, (where necessary, correct to 4 significant figures)., ! 1, ! 4, 3, 2, 5x dx (b), − t 2 dt, 1. (a), 1, −1 4, , 1, (a) 105 (b) −, 2, ! 2, ! 3, 2. (a), (3 − x 2 ) dx (b), (x 2 − 4x + 3) dx, , 2, , Problem 15. Evaluate, , Further problems on definite, , −1, , 4 cos 3t dt., , 1, , , 1, (a) 6 (b) −1, 3, , 1, , !, , , , 2, 2, 1, 4, 4 cos3t dt = (4), sin 3t =, sin 3t, 3, 3, 1, 1, , , , 4, 4, =, sin 6 −, sin 3, 3, 3, , 2, , Note that limits of trigonometric functions are always, expressed in radians—thus, for example, sin 6 means, the sine of 6 radians= −0.279415 . . ., ! 2, 4 cos 3t dt, Hence, 1, , , , , , 4, 4, = (−0.279415 . . .) −, (0.141120 . . .), 3, 3, = (−0.37255) − (0.18816) = −0.5607, , !, , 1, , π, , 3, cos θ dθ, 2, , 3. (a), 0, , !, , π, 2, , (b), , 4 cos θ dθ, , 0, , [(a) 0 (b) 4], !, 4. (a), , π, 3, π, 6, , !, , 2, , 2 sin 2θ dθ (b), , 3 sin t dt, 0, , [(a) 1 (b) 4.248], !, 5. (a), , !, , 1, , π, 6, , 5 cos3x dx (b), 0, , 3 sec2 2x dx, , 0, , [(a) 0.2352 (b) 2.598], , 373
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374 Higher Engineering Mathematics, !, , 2, , 6. (a), , 1 litre to 3 litres for a temperature rise from, 100 K to 400 K given that:, , cosec 2 4t dt, , 1, , !, (b), , π, 2, , π, 4, , (3 sin 2x − 2 cos3x) dx, [(a) 0.2527 (b) 2.638], , !, , 1, , 7. (a), , !, 3 e3t dt (b), , 0, , 2, , 2, dx, 2x, 3, e, −1, [(a) 19.09 (b) 2.457], , !, , 3, , 8. (a), 2, , 2, dx (b), 3x, , !, , 3, 1, , 2x 2 + 1, dx, x, [(a) 0.2703 (b) 9.099], , 9. The entropy change S, for an ideal gas is, given by:, ! V2, ! T2, dT, dV, Cv, −R, S =, T, T1, V1 V, where T is the thermodynamic temperature,, V is the volume and R = 8.314. Determine, the entropy change when a gas expands from, , Cv = 45 + 6 × 10−3 T + 8 × 10−6 T 2 ., [55.65], 10. The p.d. between boundaries a and b of an, ! b, Q, electric field is given by: V =, dr, 2πrε, 0 εr, a, If a = 10, b = 20, Q =2 × 10−6 coulombs,, ε0 = 8.85 ×10−12 and εr = 2.77, show that, V = 9 kV., 11. The average value of a complex voltage waveform is given by:, !, 1 π, (10 sin ωt + 3 sin 3ωt, V AV =, π 0, + 2 sin 5ωt) d(ωt), Evaluate V AV correct to 2 decimal places., [7.26]
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Chapter 38, , Some applications of, integration, 38.1, , Introduction, , There are a number of applications of integral calculus, in engineering. The determination of areas, mean and, r.m.s. values, volumes, centroids and second moments, of area and radius of gyration are included in this, chapter., , 38.2, , Areas under and between curves, , When y = 0, x = 0 or (x + 2) = 0 or (x − 4) = 0, i.e., when y = 0, x = 0 or −2 or 4, which means that the, curve crosses the x-axis at 0, −2, and 4. Since the curve, is a continuous function, only one other co-ordinate, value needs to be calculated before a sketch of the, curve can be produced. When x = 1, y = −9, showing that the part of the curve between x = 0 and x = 4, is negative. A sketch of y = x 3 − 2x 2 − 8x is shown in, Fig. 38.2. (Another method of sketching Fig. 38.2 would, have been to draw up a table of values.), y, , In Fig. 38.1,, !, total shaded area =, , b, , 10, , !, , c, , f (x)dx −, , a, , b, , f (x)dx, !, +, , 22, , d, , 21, , y 5 x 3 2 2x 2 2 8x, , 0, , 1, , 2, , 3, , 4, , f (x)dx, 210, , c, , y, 220, y 5 f (x), G, , Figure 38.2, , E, 0, , a, , b, , F, , c, , d, , x, , Shaded area, ! 0, ! 4, =, (x 3 − 2x 2 − 8x)dx − (x 3 − 2x 2 − 8x)dx, −2, , , , Figure 38.1, , =, , Problem 1. Determine the area between the curve, y = x 3 − 2x 2 − 8x and the x-axis., y = x 3 −2x 2 − 8x = x(x 2 −2x − 8) = x(x + 2)(x − 4), , x4, , 0, , −, , 2x 3, , −, , 0, 8x 2, , , −, , x 4 2x 3 8x 2, −, −, 4, 3, 2, , 4, 3, 2 −2, , , 2, 2, 1, = 6, − −42, = 49 square units, 3, 3, 3, , 4, 0, , x
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376 Higher Engineering Mathematics, , , 1, 1, − −13, = 7, 3, 2, , Problem 2. Determine the area enclosed between, the curves y = x 2 + 1 and y = 7 − x., At the points of intersection the curves are equal. Thus,, equating the y values of each curve gives:, x2 + 1 = 7 − x, x2 + x − 6 = 0, , from which,, , Factorizing gives (x − 2)(x + 3) = 0, from which x = 2 and x = −3, By firstly determining the points of intersection the, range of x-values has been found. Tables of values are, produced as shown below., x, , −3 −2 −1 0 1 2, , y = x2 + 1, , 10, , 5, , 2 1 2, , x, , −3, , 0 2, , y = 7−x, , 10, , 7, , 21, , 0, , !, , 2, −3, , !, =, , 2, −3, , !, =, , 2, −3, , , , y 5 3x, , 0, , 3y 5 x (or y 5 x3 ), , 1, , 2, , 3, , 4, , x, , Figure 38.4, , Shaded area, ! 1, ! 3 , x, x , =, dx +, 3x −, (4 − x) − dx, 3, 3, 0, 1, , y 5 x 2 11, , , 1, 3, 3x 2 x 2, x2 x2, + 4x −, −, −, 2, 6 0, 2, 6 1, , , , , 3 1, 9 9, − (0) + 12 − −, =, −, 2 6, 2 6, , , 1 1, − 4− −, 2 6, , , 1, 1, + 6−3, = 4 square units, = 1, 3, 3, , , =, , y572x, 1, , 2, , !, (7 − x)dx −, , y542x, , 2, , x, , Figure 38.3, , Shaded area =, , y, , 5, , 5, , 22, , Each of the straight lines are shown sketched in, Fig. 38.4., , 5, , y, , 23, , Problem 3. Determine by integration the area, bounded by the three straight lines y = 4 − x,, y = 3x and 3y = x., , 4, , A sketch of the two curves is shown in Fig. 38.3., , 10, , 5, = 20 square units, 6, , 2, −3, , (x 2 + 1)dx, , [(7 − x) − (x 2 + 1)]dx, Now try the following exercise, (6 − x − x 2 )dx, , x2 x3, = 6x −, −, 2, 3, , 2, −3, , , , , 9, 8, − −18 − + 9, = 12 − 2 −, 3, 2, , Exercise 147 Further problems on areas, under and between curves, 1. Find the area enclosed by the curve, y = 4 cos 3x, the x-axis and ordinates x = 0, π, [1 13 square units], and x =, 6
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Some applications of integration, , 377, , [Note that for a sine wave,, 2. Sketch the curves y = x 2 + 3 and y = 7 − 3x, and determine the area enclosed by them., [20 56 square units], 3. Determine the area enclosed by the three, straight lines y = 3x, 2y = x and y + 2x = 5., [2 12 square units], , In this case, mean value =, , , , Mean and r.m.s. values, =, , With reference to Fig. 38.5,, ! b, 1, mean value, y =, y dx, b −a a, 75, 6, 8, ! b, 8, 1, y2 dx, and, r.m.s. value = 9, b−a a, y, , 2, × 100 = 63.66 V], π, , (b) r.m.s. value, =, , 38.3, , 2, × maximum value, π, , mean value=, , , =, , !, , 1, π −0, !, , 1, π, , π, , π, , , v 2 d(ωt ), , 0, , , (100 sin ωt )2 d(ωt ), , 0, , 10000, π, , !, , π, , , sin2 ωt d(ωt ) ,, , 0, , which is not a ‘standard’ integral., It is shown in Chapter 17 that, cos 2 A = 1 − 2 sin2 A and this formula is used, whenever sin2 A needs to be integrated., , y 5 f(x), , Rearranging cos 2 A = 1 − 2 sin2 A gives, 1, sin2 A = (1 − cos 2 A), 2, y, , , Hence, , 0, , x5a, , x5b, , x, , Figure 38.5, , Problem 4. A sinusoidal voltage v = 100 sin ωt, volts. Use integration to determine over half a cycle, (a) the mean value, and (b) the r.m.s. value., (a), , Half a cycle means the limits are 0 to π radians., ! π, 1, Mean value, y =, v d(ωt ), π −0 0, !, 1 π, =, 100 sinωt d(ωt ), π 0, 100, =, [−cos ωt ]π0, π, 100, =, [(−cos π) − (−cos 0)], π, 200, 100, [(+1) − (−1)] =, =, π, π, = 63.66 volts, , , =, , =, , 10000, π, , 10000, π, , !, , !, , , sin2 ωt d(ωt ), , 0, π, , 0, , π, , , 1, (1 − cos 2ωt ) d(ωt ), 2, , , 10000 1, sin 2ωt, ωt −, π 2, 2, , π , 0, , 7⎧, , , 8, 10000 1, sin 2π, 8⎪, ⎪, 8⎨, π−, 8, π 2, 2, , =8, sin 0, 9⎪, ⎪, − 0−, ⎩, 2, , =, , =, , 10000 1, [π], π 2, , ⎫, ⎪, ⎪, ⎬, ⎪, ⎪, ⎭, , , , , 100, 10000, = √ = 70.71 volts, 2, 2, , [Note that for a sine wave,, 1, r.m.s. value= √ × maximum value., 2
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378 Higher Engineering Mathematics, y, , In this case,, , y 5 f (x), , 1, r.m.s. value = √ × 100 = 70.71 V], 2, A, , Now try the following exercise, x5a, , 0, , Exercise 148 Further problems on mean, and r.m.s. values, 1. The vertical height h km of a missile varies, with the horizontal distance d km, and is given, by h = 4d − d 2 . Determine the mean height of, the missile from d = 0 to d = 4 km., [2 23 km]., 2. The distances of points y from the mean value, of a frequency distribution are related to the, 1, variate x by the equation y = x + . Deterx, mine the standard deviation (i.e. the r.m.s., value), correct to 4 significant figures for, values of x from 1 to 2., [2.198], 3. A current i = 25 sin 100πt mA flows in an, electrical circuit. Determine, using integral, calculus, its mean and r.m.s. values each correct to 2 decimal places over the range t = 0, to t = 10 ms., [15.92 mA, 17.68 mA], , generated, V , is given by:, ! d, πx2 dy, V=, c, , Problem 5. The curve y = x 2 + 4 is rotated one, revolution about the x-axis between the limits x = 1, and x = 4. Determine the volume of solid of, revolution produced., Revolving the shaded area shown in Fig. 38.7, 360◦, about the x-axis produces a solid of revolution given by:, ! 4, ! 4, π y 2 dx =, π(x 2 + 4)2 dx, Volume =, !, , 1, , 1, 4, , =, , π(x 4 + 8x 2 + 16) dx, , 1, , x 5 8x 3, =π, +, + 16x, 5, 3, , v = E 1 sin ωt + E 3 sin 3ωt, , 4, , 1, , = π[(204.8 + 170.67 + 64), , where E 1 , E 3 and ω are constants., Determine the r.m.s. value of v over the, π, interval 0 ≤ t ≤ ., ω, ⎤, ⎡, E 12 + E 32, ⎦, ⎣, 2, , − (0.2 + 2.67 + 16)], = 420.6π cubic units, y, 30, , 20, , Volumes of solids of revolution, , With reference to Fig. 38.6, the volume of revolution,, V , obtained by rotating area A through one revolution, about the x-axis is given by:, ! b, πy2 dx, V=, , A, , 10, 5 D, 4, 0, , a, , If a curve x = f ( y) is rotated 360◦ about the y-axis, between the limits y = c and y = d then the volume, , x, , Figure 38.6, , 4. A wave is defined by the equation:, , 38.4, , x5b, , Figure 38.7, , y 5 x21 4, , B, , C, , 1, , 2, , 3, , 4, , 5, , x
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Some applications of integration, Problem 6. Determine the area enclosed by the, two curves y = x 2 and y 2 = 8x. If this area is, rotated 360◦ about the x-axis determine the volume, of the solid of revolution produced., , {(volume produced by revolving y 2 = 8x), − (volume produced by revolving y = x 2 )}, !, , 2, , i.e. volume =, , !, , x 4 − 8x = 0, , Hence, at the points of intersection, x = 0 and x = 2., When x = 0, y = 0 and when x = 2, y = 4. The points, of intersection of the curves y = x 2 and y 2 = 8x are, therefore at (0,0) and (2,4).√A sketch is shown in, Fig. 38.8. If y 2 = 8x then y = 8x., , Shaded area, !, , 2, , =, , , 1, 8 x 2 − x 2 dx, , !, √, , 8x − x 2 dx =, , 0, , 2 √, 0, , ⎤2 5 √ √, ⎡, 6, √ x 32, 3, 8 8 8, x, − {0}, −, =⎣ 8 3 − ⎦ =, 3, 3, 3, 2, 2, 0, , =, , 16 8 8, 2, − = = 2 square units, 3, 3 3, 3, y5x2, , y, , y 2 5 8x, (or y 5Œ(8x), , 4, , 2, , 0, , 2, , 8x 2 x 5, = π (8x − x )dx = π, −, 2, 5, 0, , 2, , 4, , 0, , = 9.6π cubic units, , x(x 3 − 8) = 0, , and, , π(x 4 )dx, , 0, , , , 32, − (0), = π 16 −, 5, , x 4 = 8x, from which,, , !, , 2, , π(8x)dx −, , 0, , At the points of intersection the co-ordinates of the, curves are equal. Since y = x 2 then y 2 = x 4 . Hence, equating the y 2 values at the points of intersection:, , Now try the following exercise, Exercise 149, , Further problems on volumes, , 1. The curve x y = 3 is revolved one revolution, about the x-axis between the limits x = 2 and, x = 3. Determine the volume of the solid, produced., [1.5π cubic units], y, 2. The area between 2 = 1 and y + x 2 = 8 is, x, rotated 360◦ about the x-axis. Find the volume produced., [170 23 π cubic units], 3. The curve y = 2x 2 + 3 is rotated about (a) the, x-axis between the limits x = 0 and x = 3,, and (b) the y-axis, between the same limits., Determine the volume generated in each case., [(a) 329.4π (b) 81π], 4. The profile of a rotor blade is bounded by the, lines x = 0.2, y = 2x, y = e−x , x = 1 and the, x-axis. The blade thickness t varies linearly, with x and is given by: t = (1.1 − x)K, where, K is a constant., (a) Sketch the rotor blade, labelling the limits., (b) Determine, using an iterative method, the, value of x, correct to 3 decimal places,, where 2x = e−x, , 1, , 2, , x, , Figure 38.8, , The volume produced by revolving the shaded area, about the x-axis is given by:, , 379, , (c) Calculate the cross-sectional area of the, blade, correct to 3 decimal places., (d) Calculate the volume of the blade in terms, of K, correct to 3 decimal places., [(b) 0.352 (c) 0.419 square units, (d) 0.222 K]
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380 Higher Engineering Mathematics, 38.5, , Centroids, , A lamina is a thin flat sheet having uniform thickness., The centre of gravity of a lamina is the point where, it balances perfectly, i.e. the lamina’s centre of mass., When dealing with an area (i.e. a lamina of negligible, thickness and mass) the term centre of area or centroid, is used for the point where the centre of gravity of a, lamina of that shape would lie., If x and y denote the co-ordinates of the centroid C, of area A of Fig. 38.9, then:, !, , !, , b, , 1, 2, , xy dx, x = !a, , b, , y2 dx, , y dx, , !, , 2, , =, , 0, ! 2, , !, , 1, 2, , y 2 dx, , 2, , (3x 2 )2 dx, , 0, , 8, , y dx, 0, , =, , =, , !, , 1, 2, , 2, , 9 x5, 2 5, , 4, , 9x dx, =, , 0, , , , 8, , 32, 5, 8, , 9, 2, , 0, , 8, , , =, , 2, , 18, = 3.6, 5, , Hence the centroid lies at (1.5, 3.6), , and y = ! ab, , b, , y=, , 1, 2, , y dx, , a, , Problem 8. Determine the co-ordinates of, the centroid of the area lying between the curve, y = 5x − x 2 and the x-axis., , a, , y, y 5 f(x), , y = 5x − x 2 = x(5 − x). When y = 0, x = 0 or x = 5., Hence the curve cuts the x-axis at 0 and 5 as shown, in Fig. 38.10. Let the co-ordinates of the centroid be, (x , y) then, by integration,, , Area A, C, , !, , x, x5b, , x, , x= !, , = !, , 0, , x(5x − x 2 ) dx, , 0, , 5, , 5, , y dx, , Figure 38.9, , 0, , Problem 7. Find the position of the centroid of, the area bounded by the curve y = 3x 2 , the x-axis, and the ordinates x = 0 and x = 2., If (x , y) are co-ordinates of the centroid of the given, area then:, !, , !, , 2, , 2, , x y dx, x = !0, , =, , 2, , x(3x 2 ) dx, , 0, , 2, , y dx, !, , = , (5x − x ) dx, 2, , 12, = 1.5, 8, , 5x 3, 3, , −, , 5x 2, 2, , −, , 0, , 5, x4, 4 0, 5, x3, 3 0, , y, , 6, , 2, , 4, , y 5 5x 2 x 2, , 3, , 0, , , , (5x 2 − x 3 ) dx, , = !0 5, , 3x dx, 0, , , , 5, , 2, , 3x 4, 3x dx, 4 0, =, = !0 2, [x 3 ]20, 3x 2 dx, 2, , !, , (5x − x 2 ) dx, , 0, , 8, , !, , 0, , =, , 5, , x y dx, , x5a, , 0, , !, , 5, , y, , C, , x, 2, , y, 0, , Figure 38.10, , 1, , 2, , 3, , 4, , 5, , x
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Some applications of integration, 625, −, = 3, 125, −, 2, , =, , y=, , 1, 2, , 625, 12, !, , 5, , 625, 625, 4 = 12, 125, 125, 3, 6, , , , 6, 125, , , , 2, , y dx, =, , 0, ! 5, , !, , =, , =, , =, , 5, , 0, ! 5, , y dx, 0, , 1, 2, , 4. Find the co-ordinates of the centroid of the area, which lies between the curve y/x = x − 2 and, the x-axis., [(1, −0.4)], 5. Sketch the curve y 2 = 9x between the limits, x = 0 and x = 4. Determine the position of the, centroid of this area., [(2.4, 0)], , 5, = = 2.5, 2, , 1, 2, , (5x − x 2 )2 dx, , (5x − x 2 ) dx, , 0, , !, , 5, , (25x − 10x + x ) dx, 3, , Theorem of Pappus, , 4, , ‘If a plane area is rotated about an axis in its own plane, but not intersecting it, the volume of the solid formed is, given by the product of the area and the distance moved, by the centroid of the area’., With reference to Fig. 38.11, when the curve y = f (x), is rotated one revolution about the x-axis between, the limits x = a and x = b, the volume V generated, is given by:, , 125, 6, 1 25x 3 10x 4 x 5, −, +, 2, 3, 4, 5, , 5, , 0, , 125, 6, 1, 2, , 38.6, , A theorem of Pappus states:, 2, , 0, , , , 381, , 25(125) 6250, −, + 625, 3, 4, 125, 6, , volume V = (A)(2π y ), from which, y =, , , , V, 2π A, , y, , = 2.5, , y 5 f(x), Area A, , Hence the centroid of the area lies at (2.5, 2.5)., , C, , (Note from Fig. 38.10 that the curve is symmetrical, about x = 2.5 and thus x could have been determined, ‘on sight’.), , y, x5a, , x5b x, , Figure 38.11, , Now try the following exercise, Exercise 150 Further problems on centroids, In Problems 1 and 2, find the position of the centroids of the areas bounded by the given curves, the, x-axis and the given ordinates., 1., , y = 3x + 2 x = 0, x = 4, , 2., , y=, , 5x 2, , x = 1, x = 4, , [(2.5, 4.75)], [(3.036, 24.36)], , 3. Determine the position of the centroid of a, sheet of metal formed by the curve, y = 4x − x 2 which lies above the x-axis., [(2, 1.6)], , Problem 9. (a) Calculate the area bounded by the, curve y = 2x 2 , the x-axis and ordinates x = 0 and, x = 3. (b) If this area is revolved (i) about the, x-axis and (ii) about the y-axis, find the volumes of, the solids produced. (c) Locate the position of the, centroid using (i) integration, and (ii) the theorem, of Pappus., (a), , The required area is shown shaded in Fig. 38.12., ! 3, ! 3, y dx =, 2x 2 dx, Area =, 0, , , =, , 0, , 3, 2x 3, , 3, , 0, , = 18 square units
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382 Higher Engineering Mathematics, y, , y 5 2x 2, , y=, , 18, , 1, 2, , !, , 2, , y dx, =, , 0, ! 3, , 1, 2, , !, , 3, , (2x 2 )2 dx, , 0, , 18, , y dx, , 12, , 0, , x, , 6, , y, 0, , 1, , 2, , =, , x, , 3, , Figure 38.12, , (b), , 3, , 1, 2, , !, , 3, , 4x 4 dx, =, , 0, , 18, , 1 4x 5, 2 5, , 3, , 0, , 18, , = 5.4, , (ii) using the theorem of Pappus:, , (i) When the shaded area of Fig. 38.12 is, revolved 360◦ about the x-axis, the volume, generated, !, , 3, , =, !, , !, , 0, , π(2x 2 )2 dx, , 81π = (18)(2π x ),, , i.e., from which,, , x=, , 0, 3, , =, , 3, , π y 2 dx =, , Volume generated when shaded area is, revolved about OY= (area)(2π x )., , 0, , 4π x 4 dx = 4π, , , , = 4π, , , , 3, , x5, 5, , 0, , 243, = 194.4πcubic units, 5, , Volume generated when shaded area is, revolved about OX = (area)(2π y)., 194.4π = (18)(2π y),, , i.e., , y=, , from which,, (ii) When the shaded area of Fig. 38.12 is, revolved 360◦ about the y-axis, the volume, generated, = (volume generated by x = 3), − (volume generated by y = 2x 2 ), ! 18, ! 18 , y, 2, =, π(3) dy −, π, dy, 2, 0, 0, , ! 18 , 18, y2, y, =π, dy = π 9y −, 9−, 2, 4 0, 0, = 81π cubic units, (c) If the co-ordinates of the centroid of the shaded, area in Fig. 38.12 are (x, y) then:, (i) by integration,, !, , !, , 3, , 3, , x y dx, x = !0, , =, , 3, , 0, , =, =, , 3, 0, , 18, 81, = 2.25, 36, , =, , Hence the centroid of the shaded area in, Fig. 38.12 is at (2.25, 5.4)., , Problem 10. A metal disc has a radius of 5.0 cm, and is of thickness 2.0 cm. A semicircular groove of, diameter 2.0 cm is machined centrally around the, rim to form a pulley. Determine, using Pappus’, theorem, the volume and mass of metal removed, and the volume and mass of the pulley if the density, of the metal is 8000 kg m−3., A side view of the rim of the disc is shown in Fig. 38.13., 2.0 cm, P, , x(2x 2 ) dx, , Q, , 18, , , 2x 3 dx, , 194.4π, = 5.4, 36π, , 0, , y dx, !, , 81π, = 2.25, 36π, , 5.0 cm, S, , 3, 2x 4, , 4, 18, , X, , 0, , Figure 38.13, , R, X
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Some applications of integration, When area PQRS is rotated about axis XX the volume generated is that of the pulley. The centroid of the, 4r, semicircular area removed is at a distance of, from its, 3π, diameter (see ‘Engineering Mathematics 6th edition’,, 4(1.0), Chapter 58), i.e., , i.e. 0.424 cm from PQ. Thus, 3π, the distance of the centroid from XX is 5.0 − 0.424,, i.e. 4.576 cm., The distance moved through in one revolution by the, centroid is 2π(4.576) cm., π(1.0)2, π, πr 2, =, = cm2, Area of semicircle =, 2, 2, 2, By the theorem of Pappus,, volume generated, = area × distance moved by, π , (2π)(4.576)., centroid =, 2, i.e. volume of metal removed = 45.16 cm3, Mass of metal removed = density × volume, 45.16 3, m, 106, = 0.3613 kg or 361.3 g, , = 8000 kg m−3×, , volume of pulley = volume of cylindrical disc, − volume of metal removed, = π(5.0)2 (2.0) − 45.16, = 111.9 cm3, Mass of pulley = density× volume, = 8000 kg m−3 ×, , 111.9 3, m, 106, , = 0.8952 kg or 895.2 g, , Now try the following exercise, Exercise 151 Further problems on the, theorem of Pappus, 1. A right angled isosceles triangle having a, hypotenuse of 8 cm is revolved one revolution, about one of its equal sides as axis. Determine the volume of the solid generated using, Pappus’ theorem., [189.6 cm3 ], 2. Using (a) the theorem of Pappus, and (b) integration, determine the position of the centroid, of a metal template in the form of a quadrant, , 383, , of a circle of radius 4 cm. (The equation of a, circle, centre 0, radius r is x 2 + y 2 = r 2 )., ⎡, ⎤, On the centre line, distance, ⎢ 2.40 cm from the centre, ⎥, ⎢, ⎥, ⎢, ⎥, ⎣ i.e. at co-ordinates, ⎦, (1.70, 1.70), 3., , (a) Determine the area bounded by the curve, y = 5x 2 , the x-axis and the ordinates, x = 0 and x = 3., (b) If this area is revolved 360◦ about (i) the, x-axis, and (ii) the y-axis, find the volumes of the solids of revolution produced, in each case., (c) Determine the co-ordinates of the centroid of the area using (i) integral calculus, and (ii) the theorem of Pappus., ⎡, ⎤, (a) 45 square units, ⎢(b) (i) 1215π cubic units ⎥, ⎢, ⎥, ⎢, ⎥, ⎣ (ii) 202.5π cubic units⎦, (c) (2.25, 13.5), , 4. A metal disc has a radius of 7.0 cm and is, of thickness 2.5 cm. A semicircular groove of, diameter 2.0 cm is machined centrally around, the rim to form a pulley. Determine the volume of metal removed using Pappus’ theorem, and express this as a percentage of the original volume of the disc. Find also the mass of, metal removed if the density of the metal is, 7800 kg m−3., [64.90 cm3 , 16.86%, 506.2 g], For more on areas, mean and r.m.s. values, volumes and, centroids, see ‘Engineering Mathematics 6th edition’,, Chapters 55 to 58., , 38.7 Second moments of area of, regular sections, The first moment of area about a fixed axis of a lamina, of area A, perpendicular distance y from the centroid, of the lamina is defined as Ay cubic units. The second, moment of area of the same lamina as above is given, by Ay 2 , i.e. the perpendicular distance from the centroid, of the area to the fixed axis is squared.
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384 Higher Engineering Mathematics, Second moments of areas are usually denoted by I and, have units of mm4 , cm4 , and so on., , limit, , Radius of gyration, , δx→0, , Several areas, a1 , a2, a3 , . . . at distances y1 , y2, y3 , . . ., from a fixed axis, may be replaced by a single area, + a3 + · · · at distance k from the, A, where A = a1 + a2 ;, axis, such that Ak 2 = ay 2 ., k is called the radius of ;, gyration of area A about the, given axis. Since Ak 2 = ay 2 = I then the radius of, gyration,, , k=, , It is a fundamental theorem of integration that, , I, A, , The second moment of area is a quantity much used in, the theory of bending of beams, in the torsion of shafts,, and in calculations involving water planes and centres, of pressure., The procedure to determine the second moment of, area of regular sections about a given axis is (i) to find the, second moment of area of a typical element and (ii) to, sum all such second moments of area by integrating, between appropriate limits., For example, the second moment of area of the rectangle shown in Fig. 38.14 about axis PP is found by, initially considering an elemental strip of width δx, parallel to and distance x from axis PP. Area of shaded, strip = bδx., , x=l, <, , !, , l, , x b δx =, 2, , x 2 b dx, , 0, , x=0, , Thus the second moment of area of the rectangle, about PP, 3 l, ! l, x, bl 3, 2, x dx = b, =, =b, 3 0, 3, 0, Since the total area of the rectangle, A = lb, then, 2, l, Al 2, =, I pp = (lb), 3, 3, l2, 3, i.e. the radius of gyration about axes PP,, I pp = Ak 2pp thus k 2pp =, , kpp =, , l2, l, =√, 3, 3, , Parallel axis theorem, In Fig. 38.15, axis GG passes through the centroid C, of area A. Axes DD and GG are in the same plane, are, parallel to each other and distance d apart. The parallel, axis theorem states:, IDD = IGG + Ad 2, Using the parallel axis theorem the second moment of, area of a rectangle about an axis through the centroid, , P, , G, l, d, , b, , x, , Area A, C, ␦x, P, , Figure 38.14, , Second moment of area of the shaded strip about, PP = (x 2 )(b δx)., The second moment of area of the whole rectangle about, PP is obtained by, all such strips between x =, ;summing, 2, 0 and x = l, i.e. x=l, x=0 x bδx., , G, D, , Figure 38.15, , D
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Some applications of integration, P, , G, l, 2, , A summary of derived standard results for the second, moment of area and radius of gyration of regular, sections are listed in Table 38.1., , l, 2, , C, , Problem 11. Determine the second moment of, area and the radius of gyration about axes AA, BB, and CC for the rectangle shown in Fig. 38.18., , b, , x, , l 5 12.0 cm, , ␦x, G, , P, , C, , A, , C, b 5 4.0 cm, , Figure 38.16, B, , may be determined. In the rectangle shown in Fig. 38.16,, bl 3, I pp =, (from above)., 3, From the parallel axis theorem, 2, 1, I pp = IGG + (bl), 2, , from which, IGG =, , B, A, , Figure 38.18, , From Table 38.1, the second moment of area about, axis AA,, , bl 3, bl 3, = IGG +, 3, 4, , i.e., , IAA =, , bl 3 bl 3, bl 3, −, =, 3, 4, 12, , bl 3, (4.0)(12.0)3, =, = 2304 cm4, 3, 3, , Perpendicular axis theorem, , 12.0, l, Radius of gyration,kAA = √ = √ = 6.93 cm, 3, 3, , In Fig. 38.17, axes OX , OY and OZ are mutually perpendicular. If OX and OY lie in the plane of area A then, the perpendicular axis theorem states:, , Similarly, IBB =, , IOZ = IOX + IOY, , and, Z, , Y, , O, , lb3 (12.0)(4.0)3, =, = 256 cm4, 3, 3, , 4.0, b, kBB = √ = √ = 2.31 cm, 3, 3, , The second moment of area about the centroid of a, bl 3, rectangle is, when the axis through the centroid is, 12, parallel with the breadth b. In this case, the axis CC is, parallel with the length l., Hence ICC =, , lb3 (12.0)(4.0)3, =, = 64 cm4, 12, 12, , Area A, X, , Figure 38.17, , 385, , and, , 4.0, b, kCC = √ = √ = 1.15 cm, 12, 12
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386 Higher Engineering Mathematics, Table 38.1 Summary of standard results of the second moments of areas of regular sections, Shape, , Position of axis, , Rectangle, , Second moment, , Radius of, , of area, I, , gyration, k, , bl 3, 3, , l, √, 3, , (2) Coinciding with l, , lb3, 3, , b, √, 3, , (3) Through centroid, parallel to b, , bl 3, 12, , l, √, 12, , (4) Through centroid, parallel to l, , lb3, 12, , b, √, 12, , (1) Coinciding with b, , bh 3, 12, , h, √, 6, , (2) Through centroid, parallel to base, , bh 3, 36, , h, √, 18, , (3) Through vertex, parallel to base, , bh 3, 4, , h, √, 2, , (1) Through centre, perpendicular to, , πr 4, 2, , r, √, 2, , (2) Coinciding with diameter, , πr 4, 4, , (3) About a tangent, , 5πr 4, 4, , r, 2, √, 5, r, 2, , Coinciding with diameter, , πr 4, 8, , r, 2, , (1) Coinciding with b, , length l, breadth b, , Triangle, Perpendicular height h,, base b, , Circle, , plane (i.e. polar axis), , radius r, , Semicircle, radius r, , Problem 12. Find the second moment of area and, the radius of gyration about axis PP for the, rectangle shown in Fig. 38.19., 40.0 mm, G, , G, 15.0 mm, , 25.0 mm, P, , Figure 38.19, , P, , IGG =, , lb3, where 1 = 40.0 mm and b = 15.0 mm, 12, , Hence IGG =, , (40.0)(15.0)3, = 11250 mm4, 12, , From the parallel axis theorem, I PP = IGG + Ad 2 ,, where A = 40.0 × 15.0 = 600 mm2 and, d = 25.0 +7.5 = 32.5 mm, the perpendicular, distance between GG and PP. Hence,, IPP = 11 250 + (600)(32.5)2, = 645000 mm4
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Some applications of integration, 2, IPP = AkPP, , from which,, , , kPP =, , , , IPP, =, area, , 387, , Problem 14. Determine the second moment of, area and radius of gyration of the circle shown in, Fig. 38.21 about axis YY ., , , 645000, = 32.79 mm, 600, , Problem 13. Determine the second moment of, area and radius of gyration about axis QQ of the, triangle BCD shown in Fig. 38.20., , r 5 2.0 cm, G, , G, , B, 3.0 cm, 12.0 cm, , G, , G, , Y, C, , 8.0 cm, , Figure 38.21, , D, 6.0 cm, , Q, , Y, , Q, , In Fig. 38.21, IGG =, , Figure 38.20, , Using the parallel axis theorem: I QQ = IGG +, where IGG is the second moment of area about the, centroid of the triangle,, Ad 2 ,, , bh 3, (8.0)(12.0)3, i.e., =, = 384 cm4 ,, 36, 36, A is the area of the triangle,, , Using the parallel axis theorem, IYY = IGG + Ad 2 ,, where d = 3.0 + 2.0 = 5.0 cm., IYY = 4π + [π(2.0)2 ](5.0)2, , Hence, , = 4π + 100π = 104π = 327 cm4, Radius of gyration,, , kYY =, , = 12 bh = 12 (8.0)(12.0) = 48 cm2, and d is the distance between axes GG and QQ,, = 6.0 + 13 (12.0) = 10 cm., , πr 4, π, = (2.0)4 = 4π cm4 ., 4, 4, , IY Y, =, area, , , , 104π, π(2.0)2, , , =, , √, 26 = 5.10 cm, , Problem 15. Determine the second moment of, area and radius of gyration for the semicircle shown, in Fig. 38.22 about axis XX ., , Hence the second moment of area about axis QQ,, G, , IQQ = 384 + (48)(10)2 = 5184 cm4, , B, , Radius of gyration,, , kQQ =, , IQ Q, =, area, , 10.0 mm, , G, B, , 15.0 mm, , , , , 5184, = 10.4 cm, 48, , X, , Figure 38.22, , X
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388 Higher Engineering Mathematics, 4r, The centroid of a semicircle lies at, from its, 3π, diameter., Using the parallel axis theorem:, IBB = IGG + Ad 2 ,, IBB =, , where, , =, , πr 4, (from Table 38.1), 8, π(10.0)4, = 3927 mm4,, 8, , π(10.0)2, πr 2, =, = 157.1 mm2, 2, 2, 4r, 4(10.0), d=, =, = 4.244 mm, 3π, 3π, , πr 4, The polar second moment of area of a circle=, 2, The polar second moment of area of the shaded area is, given by the polar second moment of area of the 7.0 cm, diameter circle minus the polar second moment of area, of the 6.0 cm diameter circle., Hence the polar second moment of area of the crosssection shown, , , , , π 7.0 4 π 6.0 4, =, −, 2, 2, 2, 2, = 235.7 − 127.2 = 108.5 cm4, , A=, and, Hence, , 3927 = IGG + (157.1)(4.244)2, , i.e., , 3927 = IGG + 2830,, , from which, IGG = 3927 − 2830 = 1097 mm4, , Problem 17. Determine the second moment of, area and radius of gyration of a rectangular lamina, of length 40 mm and width 15 mm about an axis, through one corner, perpendicular to the plane of, the lamina., The lamina is shown in Fig. 38.24., , Using the parallel axis theorem again:, I XX = IGG + A(15.0 + 4.244)2, i.e. IXX =, , Y, Z, , 1097 + (157.1)(19.244)2, , m, , 0m, , l54, , b 5 15 mm, X, , = 1097 + 58 179, = 59276 mm4 or 59280 mm4 ,, correct to 4 significant figures., , Radius of gyration, kXX =, , I XX, =, area, , , , 59 276, 157.1, , , , = 19.42 mm, , X, Z, , Y, , Figure 38.24, , From the perpendicular axis theorem:, I ZZ = I XX + IYY, , 7.0 cm, , 6.0 cm, , Problem 16. Determine the polar second moment, of area of the propeller shaft cross-section shown in, Fig. 38.23., , I XX =, , lb 3, (40)(15)3, =, = 45000 mm4, 3, 3, , and, , IYY =, , bl 3, (15)(40)3, =, = 320000 mm4, 3, 3, , Hence, , IZZ = 45 000 + 320 000, = 365000 mm4 or 36.5 cm4, , Radius of gyration,, , kZZ =, , Figure 38.23, , IZ Z, =, area, , , , 365 000, (40)(15), , , , = 24.7 mm or 2.47 cm
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Some applications of integration, Problem 18. Determine correct to 3 significant, figures, the second moment of area about axis XX, for the composite area shown in Fig. 38.25., , 389, , Problem 19. Determine the second moment of, area and the radius of gyration about axis XX for the, I -section shown in Fig. 38.26., S, 8.0 cm, , m, 0c, 4., , X, 1.0 cm, , 3.0 cm, , CE, , 7.0 cm, , X, 1.0 cm, , 3.0 cm, 8.0 cm, , 2.0 cm, , CD, , 2.0 cm, C, , C, y, , CT, T, , T, 6.0 cm, , X, , CF, , 4.0 cm, , 15.0 cm, , X, , S, , Figure 38.26, Figure 38.25, , The I -section is divided into three rectangles, D, E, and F and their centroids denoted by CD , CE and CF, respectively., , For the semicircle,, I XX =, , πr 4, π(4.0)4, =, = 100.5 cm4, 8, 8, , For the rectangle,, I XX =, , bl 3, 3, , =, , (6.0)(8.0)3, 3, , = 1024 cm4, , For the triangle, about axis TT through centroid C T ,, ITT =, , bh 3, (10)(6.0)3, =, = 60 cm4, 36, 36, , By the parallel axis theorem, the second moment of area, of the triangle about axis XX, , 2, = 60 + 12 (10)(6.0) 8.0 + 13 (6.0) = 3060 cm4 ., Total second moment of area about XX, = 100.5 + 1024 + 3060, = 4184.5, = 4180 cm4 , correct to 3 significant figures., , For rectangle D:, The second moment of area about C D (an axis through, CD parallel to XX ), =, , bl 3, (8.0)(3.0)3, =, = 18 cm4, 12, 12, , Using the parallel axis theorem:, I XX = 18 + Ad 2, where A = (8.0)(3.0) = 24 cm2 and d = 12.5 cm, Hence I XX = 18 + 24(12.5)2 = 3768 cm4., For rectangle E:, The second moment of area about CE (an axis through, CE parallel to XX ), =, , bl 3, (3.0)(7.0)3, =, = 85.75 cm4, 12, 12, , Using the parallel axis theorem:, I XX = 85.75 + (7.0)(3.0)(7.5)2 = 1267 cm4.
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390 Higher Engineering Mathematics, For rectangle F:, I XX, , E, , bl 3 (15.0)(4.0)3, =, =, = 320 cm4, 3, 3, , E, , Total second moment of area for the I-section about, axis XX,, , 9.0 cm, , I XX = 3768 + 1267 + 320 = 5355 cm4, D, , Total area of I -section, = (8.0)(3.0) + (3.0)(7.0) + (15.0)(4.0), , Figure 38.28, , = 105 cm2 ., Radius of gyration,, , I XX, k XX =, =, area, , D, , 12.0 cm, , , , 5355, 105, , 3. For the circle shown in Fig. 38.29, find the, second moment of area and radius of gyration, about (a) axis FF and (b) axis HH ., , , , , (a) 201 cm4 , 2.0 cm, (b) 1005 cm4, 4.47 cm, , = 7.14 cm, , H, , Now try the following exercise, H, , Exercise 152 Further problems on second, moment of areas of regular sections, , m, , 0c, , r5, , 1. Determine the second moment of area and, radius of gyration for the rectangle shown in, Fig. 38.27 about (a) axis AA (b) axis BB and, (c) axis CC., ⎡, ⎤, (a) 72 cm4 , 1.73 cm, ⎣(b) 128 cm4, 2.31 cm⎦, (c) 512 cm4 , 4.62 cm, B, , C, , 4., , F, , F, , Figure 38.29, , 4. For the semicircle shown in Fig. 38.30, find the, second moment of area and radius of gyration, about axis J J ., [3927 mm4 , 5.0 mm], , 8.0 cm, , A, , m, m, , A, , r5, , 10, , .0, , 3.0 cm, , J, B, , C, , J, , Figure 38.30, , Figure 38.27, , 2. Determine the second moment of area and, radius of gyration for the triangle shown in, Fig. 38.28 about (a) axis DD (b) axis EE and, (c) an axis through the centroid of the triangle, parallel to axis DD.⎡, ⎤, (a) 729 cm4 , 3.67 cm, ⎣(b) 2187 cm4 , 6.36 cm⎦, (c) 243 cm4, 2.12 cm, , 5. For each of the areas shown in Fig. 38.31 determine the second moment of area and radius of, gyration about axis LL, by using the parallel, axis theorem., ⎡, ⎤, (a) 335 cm4, 4.73 cm, ⎢, ⎥, ⎣(b) 22030 cm4, 14.3 cm⎦, (c) 628 cm4, 7.07 cm
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Some applications of integration, , 3.0 cm, 15 cm, , m, , .0 c, , 15 cm, , 4, ia 5, , D, , 5.0 cm, 2.0 cm, , 18 cm 10 cm, , 5.0 cm, , L, , L, (a), , (b), , (c), , 391, , 10. Determine the second moments of areas about, the given axes for the shapes shown in, Fig. 38.33. (In Fig. 38.33(b), the circular area, is removed.), ⎤, ⎡, I AA = 4224 cm4 ,, ⎣ I BB = 6718 cm4 , ⎦, ICC = 37300 cm4, , Figure 38.31, 3.0 cm, , 6. Calculate the radius of gyration of a rectangular door 2.0 m high by 1.5 m wide about a, vertical axis through its hinge., [0.866 m], , B, , 4.5 cm, 9.0 cm, , 16.0 cm, , m, , .0 c, , 7, ia 5, , 7. A circular door of a boiler is hinged so that, it turns about a tangent. If its diameter is, 1.0 m, determine its second moment of area, and radius of gyration about the hinge., [0.245 m4 , 0.559 m], 8. A circular cover, centre 0, has a radius of, 12.0 cm. A hole of radius 4.0 cm and centre X ,, where OX = 6.0 cm, is cut in the cover. Determine the second moment of area and the radius, of gyration of the remainder about a diameter, through 0 perpendicular to OX ., [14280 cm4 , 5.96 cm], 9. For the sections shown in Fig. 38.32, find, the second moment of area and the radius of, gyration about axis XX ., (a) 12190 mm4 , 10.9 mm, , D, , 4.0 cm, 15.0 cm, A, , 9.0 cm, (a), , A, C, B, , Figure 38.33, , 11. Find the second moment of area and radius, of gyration about the axis XX for the beam, section shown in Fig. 38.34., , 1350 cm4 ,, 5.67 cm, , 6.0 cm, , (b) 549.5 cm4 , 4.18 cm, 18.0 mm, , 2.0 cm, 8.0 cm, , 2.0 cm, , 12.0 mm, X, , 1.0 cm, , 6.0 cm, , 3.0 mm, 2.5 cm, 4.0 mm, , 3.0 cm, 2.0 cm, , X, , 2.0 cm, , X, (a), , Figure 38.32, , C, , 10.0 cm, (b), , X, , X, (b), , Figure 38.34, , 10.0 cm, , X
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Chapter 39, , Integration using algebraic, substitutions, 39.1, , 39.3 Worked problems on integration, using algebraic substitutions, , Introduction, , Functions which require integrating are not always in, the ‘standard form’ shown in Chapter 37. However, it is, often possible to change a function into a form which, can be integrated by using either:, (i) an algebraic substitution (see Section 39.2),, (ii) a trigonometric or hyperbolic substitution (see, Chapter 40),, (iii) partial fractions (see Chapter 41),, (iv) the t = tan θ/2 substitution (see Chapter 42),, (v) integration by parts (see Chapter 43), or, (vi) reduction formulae (see Chapter 44)., , Problem 1., , Determine, , , , cos(3x + 7) dx., , , , cos(3x + 7) dx is not a standard integral of the form, shown in Table 37.1, page 369, thus an algebraic, substitution is made., du, = 3 and rearranging gives, Let u = 3x + 7 then, dx, du, dx = . Hence,, 3, !, !, !, du, 1, cos(3x + 7) dx = (cos u), =, cos u du,, 3, 3, which is a standard integral, , 39.2, , Algebraic substitutions, , With algebraic substitutions, the substitution usually, made is to let u be equal to f (x) such that f (u) du, is a standard integral. It is found that integrals of the, forms,, !, , !, k, , [ f (x)] f (x) dx and k, n, , f (x), dx, [ f (x)]n, , (where k and n are constants) can both be integrated by, substituting u for f (x)., , =, , 1, sin u + c, 3, , Rewriting u as (3x + 7) gives:, !, 1, cos(3x + 7) dx = sin(3x + 7) + c,, 3, which may be checked by differentiating it., Problem 2., , , Find (2x − 5)7 dx., , (2x − 5) may be multiplied by itself 7 times and then, each term of the result integrated. However, this would
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Integration using algebraicsubstitutions, !, tan θ dθ = ln(sec θ)+ c,, , Hence, , (cos θ)−1 =, , since, , Let u =2x 2 + 1 then, , 1, = sec θ, cos θ, , !, , 2, , Hence, 0, , 39.5, , 3 , Problem 10. Evaluate 1 5x (2x 2 + 7) d x,, taking positive values of square roots only., du, du, Let u =2x 2 + 7, then, = 4x and dx =, dx, 4x, It is possible in this case to change the limits of integration. Thus when x = 3, u =2(3)2 + 7 =25 and when, x = 1, u = 2(1)2 + 7 = 9., , !, , 3x, , dx =, (2x 2 + 1), , x=3, , !, , , , u=25, , 5x (2x 2 + 7) dx =, , x=1, , u=9, , =, , =, , 5, 4, 5, 4, , √ du, 5x u, 4x, , x=3, , Thus, x=1, , u du, , =, , 6, , u3, , 9, , !, , 25, , 1, , u 2 du, 9, , 5 √ 3 √ 3 , =, 25 − 9, 6, , 5, 2, = (125 − 27) = 81, 6, 3, !, , 2, , Problem 11. Evaluate, , , , u, , −1, 2, , du, , x=0, , Since u = 2x 2 + 1, when x = 2, u =9 and when, x = 0, u =1., Thus, , 3, 4, , !, , x=2, , u, , −1, 2, , x=0, , du =, , 3, 4, , !, , u=9, , u, , −1, 2, , du,, , u=1, , i.e. the limits have been changed, ⎡, , =, , 1, 3 ⎢u2, , ⎤9, , √ , 3 √, ⎥, 9 − 1 = 3,, ⎣ 1 ⎦ =, 4, 2, 2 1, , taking positive values of square roots only., , In Problems 1 to 7, integrate with respect to the, variable., , 1, 1. 2x(2x 2 − 3)5, (2x 2 − 3)6 + c, 12, , , 3x, , dx,, , 5, − cos6 t + c, 6, , 2., , 5 cos5 t sin t, , 3., , 3 sec2 3x tan 3x, , 1, 1, sec2 3x + c or tan2 3x + c, 2, 2, , 4., , (2x 2 + 1), taking positive values of square roots only., 0, , x=2, , Exercise 154 Further problems on, integration using algebraic substitutions, , 9, , 9, , 25, , !, , 25 √, , ⎡ 3 ⎤25, , u2 ⎦, 5, 5x (2x 2 + 7) dx = ⎣, 4 3/2, , 5 , , 3, 4, , 3x du, √, u 4x, , Now try the following exercise, , !, , Thus the limits have been changed, and it is unnecessary, to change the integral back in terms of x., !, , x=2, x=0, , =, , Hence, !, , du, du, = 4x and dx =, dx, 4x, , Change of limits, , When evaluating definite integrals involving substitutions it is sometimes more convenient to change, the limits of the integral as shown in Problems 10, and 11., , 395, , 5., , , 2t (3t 2 − 1), ln θ, θ, , , 2, (3t 2 − 1)3 + c, 9, , , 1, (ln θ)2 + c, 2
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Revision Test 11, This Revision Test covers the material contained in Chapters 37 to 39. The marks for each question are shown in, brackets at the end of each question., !, ! , 2, theorem of Pappus to determine the volume of, 5, dx, 1. Determine: (a) 3 t dt (b) √, 3 2, material removed, in cm3 , correct to 3 significant, x, !, figures., (8), (c) (2 + θ)2 dθ, (9), 2., , 3., 4., , 5., , 6., , 7., , Evaluate the following integrals, each correct to, 4 significant figures:, , ! 2, ! π, 2, 3, 1 3, 3 sin 2t dt (b), +, +, dx, (a), x2 x 4, 1, 0, ! 1, 3, dt, (15), (c), 2t, e, 0, Calculate the area between the curve, y = x 3 − x 2 − 6x and the x-axis., , 400 mm, , 50 mm, 200 mm, , (10), , A voltage v = 25 sin 50πt volts is applied across, an electrical circuit. Determine, using integration,, its mean and r.m.s. values over the range t = 0 to, t = 20 ms, each correct to 4 significant figures., (12), Sketch on the same axes the curves x 2 = 2y and, y 2 = 16x and determine the co-ordinates of the, points of intersection. Determine (a) the area, enclosed by the curves, and (b) the volume of the, solid produced if the area is rotated one revolution, about the x-axis., (13), , Figure RT11.1, , 8., , A circular door is hinged so that it turns about, a tangent. If its diameter is 1.0 m find its second, moment of area and radius of gyration about the, hinge., (5), , 9., , Determine the following integrals:, !, !, 3 ln x, 7, dx, (a) 5(6t + 5) dt (b), x, !, 2, dθ, (c) √, (2θ − 1), , Calculate the position of the centroid of the, sheet of metal formed by the x-axis and the part of, the curve y = 5x − x 2 which lies above the x-axis., (9), A cylindrical pillar of diameter 400 mm has a, groove cut around its circumference as shown in, Fig. RT11.1. The section of the groove is a semicircle of diameter 50 mm. Given that the centroid, 4r, of a semicircle from its base is, , use the, 3π, , 10., , (9), , Evaluate the following definite integrals:, ! π, ! 1, , 2, 2, π, 2 sin 2t +, 3x e4x −3 dx, (a), dt (b), 3, 0, 0, (10)
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Integration using trigonometric and hyperbolic substitutions, Table 40.1 Integrals using trigonometric and hyperbolic substitutions, , f (x), Method, f (x)dx, , , sin 2x, 1, x+, +c, 1. cos 2 x, Use cos 2x = 2 cos 2 x − 1, 2, 2, , , sin 2x, 1, 2, x−, +c, 2. sin x, Use cos 2x = 1 − 2 sin 2 x, 2, 2, , See problem, 1, , 2, , 3. tan2 x, , tan x − x + c, , Use 1 + tan2 x = sec2 x, , 3, , 4. cot 2 x, , − cot x − x + c, , Use cot 2 x + 1 = cosec2 x, , 4, , 5., , cos m x, , sin n x, , (a) If either m or n is odd (but not both), use, cos 2 x + sin 2 x = 1, , 5, 6, , (b) If both m and n are even, use either, cos 2x = 2 cos 2 x − 1 or cos 2x = 1 − 2 sin 2 x, Use 12 [ sin(A + B) + sin(A − B)], , 6. sin A cos B, 7. cos A sin B, , Use, , 8. cos A cos B, , Use, , 9. sin A sin B, , Use, , 1, 10. , (a 2 − x 2 ), 11., , 12., , , (a 2 − x 2 ), 1, a2 + x 2, , 1, 13. , (x 2 + a 2 ), , 14., , , (x 2 + a 2 ), , 1, 15. , 2, (x − a 2 ), , 16., , , (x 2 − a 2 ), , sin−1, , x, +c, a, , 1, 2 [ sin(A + B) − sin(A − B)], 1, 2 [ cos(A + B) + cos(A − B)], − 12 [ cos(A + B) − cos(A − B)], , 7, 8, 9, 10, 11, 12, , Use x = a sin θ substitution, , 13, 14, , a 2 −1 x x 2, (a − x 2 ) + c, sin, +, 2, a 2, , Use x = a sin θ substitution, , 15, 16, , 1 −1 x, tan, +c, a, a, , Use x = a tan θ substitution, , 17–19, , sinh−1, , Use x = a sinh θ substitution, , 20–22, , a2, x x 2, (x + a 2 ) + c, sinh−1 +, 2, a 2, , Use x = a sinh θ substitution, , 23, , cosh−1, , Use x = a cosh θ substitution, , 24, 25, , x 2, x, a2, (x − a 2 ) − cosh−1 + c, 2, 2, a, , Use x = a cosh θ substitution, , 26, 27, , x, +c, a, 6, 5, , x + (x 2 + a 2 ), +c, or ln, a, , x, +c, a, 6, 5, , x + (x 2 − a 2 ), +c, or ln, a, , 399
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Integration using trigonometric and hyperbolic substitutions, !, , a cos θ dθ, , , since sin 2 θ + cos 2 θ = 1, (a 2 cos 2 θ), !, !, a cos θ dθ, =, = dθ = θ + c, a cos θ, x, x, Since x = a sin θ, then sin θ = and θ = sin−1 ., a, a, !, x, 1, dx = sin−1 + c, Hence , a, (a 2 − x 2 ), =, , !, , 3, , Problem 14. Evaluate, , , , 0, , !, , 3, , From Problem 13,, , , , 0, , 1, (9 − x 2 ), , 1, (9 − x 2 ), , dx., , dx, , , x 3, , since a = 3, = sin−1, 3 0, = (sin −1 1 − sin−1 0) =, , Problem 15. Find, , ! , , Since x = a sin θ, then sin θ =, , Also, cos 2 θ + sin 2 θ = 1, from which,, , , =, Thus, , a2 − x, a2, , cos θ dθ = a, , 2, , 1 + cos 2θ, 2, , , dθ, , (since cos 2θ = 2 cos 2 θ − 1), , , sin 2θ, a2, θ+, +c, 2, 2, , , a2, 2 sin θ cos θ, =, θ+, +c, 2, 2, since from Chapter 17, sin 2θ = 2 sin θ cos θ, =, , =, , a2, [θ + sin θ cos θ] + c, 2, , =, , a, , (a 2 − x 2 ), a, , a2, x x 2, (a − x2 ) + c, sin−1 +, 2, a 2, , Problem 16. Evaluate, , (a 2 − x 2 ) dx., , ! , 2, , , , x 2, , ! 4, , (16 − x 2 ) dx., , 0, , dx, Let x = a sin θ then, = a cos θ and dx = a cos θ dθ., dθ, ! , (a 2 − x 2 ) dx, Hence, ! =, (a 2 − a 2 sin 2 θ) (a cos θ dθ), ! =, [a 2 (1 − sin 2 θ)] (a cos θ dθ), ! , =, (a 2 cos 2 θ) (a cos θ dθ), !, = (a cos θ)(a cos θ dθ), !, , , 2, , 1−, , ! , a2, (a 2 − x 2 ) dx = [θ + sin θ cos θ], 2, , x (a 2 − x 2 ), a2, −1 x, sin, +c, +, =, 2, a, a, a, , From Problem 15,, , = a2, , , , cos θ = (1 − sin 2 θ) =, , =, , π, or 1.5708, 2, , x, x, and θ = sin−1, a, a, , , , ! 4, , (16 − x 2 ) dx, , 0, , 4, 16, x x, (16 − x 2 ), sin−1 +, 2, 4 2, 0, , , −1, −1, = 8 sin 1 + 2 (0) − [8 sin 0 + 0], π , = 8 sin −11 = 8, = 4π or 12.57, 2, , =, , Now try the following exercise, Exercise 158 Further problems on, integration using the sine θ substitution, !, 5, dt ., 1. Determine , (4 − t 2), , , x, 5 sin−1 + c, 2, !, 3, dx., 2. Determine , (9 − x 2 ), , , x, 3 sin−1 + c, 3, ! , (4 − x 2 ) dx., 3. Determine, , , x x, (4 − x 2 ) + c, 2 sin−1 +, 2 2, , 403
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404 Higher Engineering Mathematics, ! , , 4. Determine, , , !, (16 − 9t 2) dt ., , 8, 3t, t, (16 − 9t 2 ) + c, sin−1, +, 3, 4, 2, ! 4, π, , 1, 5. Evaluate, dx., , or 1.571, 2, 0, (16 − x 2 ), ! 1, (9 − 4x 2 ) dx., [2.760], 6. Evaluate, 0, , 40.6 Worked problems on integration, using tan θ substitution, !, Problem 17., , Determine, , 1, dx., 2, (a + x 2 ), , dx, = a sec 2 θ and dx = a sec2 θ dθ., Let x = a tan θ then, dθ, !, 1, Hence, dx, (a 2 + x 2 ), !, 1, (a sec2 θ dθ), =, 2, (a + a 2 tan2 θ), !, a sec2 θ dθ, =, 2, a (1 + tan 2 θ), !, a sec 2 θ dθ, =, , since 1+tan2 θ = sec 2 θ, a 2 sec2 θ, !, 1, 1, dθ = (θ) + c, =, a, a, x, Since x = a tan θ, θ = tan −1, a, !, x, 1, 1, Hence, dx = tan−1 + c, (a2 + x2 ), a, a, 2, , Evaluate, 0, , !, , 2, , 1, 0, , 0, , 5, dx, correct, (3 + 2x 2 ), , ! 1, 5, 5, dx =, dx, (3 + 2x 2 ), 2[(3/2), + x2], 0, !, 1, 5 1, dx, =, √, 2 0 [ (3/2)]2 + x 2, , 1, 1, 5, x, =, tan−1 √, √, 2, (3/2), (3/2) 0, =, , 5, 2, , , 2, tan−1, 3, , , 2, − tan −1 0, 3, , = (2.0412)[0.6847 − 0], = 1.3976, correct to 4 decimal places., Now try the following exercise, Exercise 159 Further problems on, integration using the tan θ substitution, , !, 3, t, 3, dt, ., tan −1 + c, 1. Determine, 4 + t2, 2, 2, !, 5, 2. Determine, dθ., 16 + 9θ 2, , 5, 3θ, tan −1, +c, 12, 4, !, , 1, , 3. Evaluate, 0, , !, , 3, , 4. Evaluate, , 3, dt ., 1 + t2, , [2.356], , 5, dx., 4 + x2, , [2.457], , 1, dx., (4 + x 2 ), , 1, dx, 2, 0 (4 + x ), 1 −1 x 2, since a = 2, tan, =, 2, 2 0, , 1 π, 1, −0, = (tan −1 1 −tan −1 0) =, 2, 2 4, π, = or 0.3927, 8, , From Problem 17,, , to 4 decimal places., !, , 1, , Evaluate, , 0, , !, Problem 18., , Problem 19., , 40.7 Worked problems on integration, using the sinh θ substitution, !, Problem 20., , Determine, , 1, dx., , 2, (x + a 2 ), , dx, Let x = a sinh θ, then, = a cosh θ and, dθ, dx = a cosh θ dθ
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Chapter 41, , Integration using, partial fractions, 41.1, , Introduction, , The process of expressing a fraction in terms of simpler, fractions—called partial fractions—is discussed in, Chapter 2, with the forms of partial fractions used being, summarized in Table 2.1, page 13., Certain functions have to be resolved into partial fractions before they can be integrated as demonstrated in, the following worked problems., , 41.2, , Worked problems on, integration using partial, fractions with linear factors, !, , Problem 1. Determine, , 11 −3x, dx., x 2 + 2x − 3, , (by algebraic substitutions — see Chapter 39), 6, 5, (x −1)2, + c by the laws of logarithms, or ln, (x +3)5, Problem 2. Find, !, 2x 2 − 9x − 35, dx., (x + 1)(x − 2)(x + 3), , It was shown in Problem 2, page 14:, 2x 2 − 9x − 35, 4, 3, 1, ≡, −, +, (x + 1)(x − 2)(x + 3) (x + 1) (x − 2) (x + 3), !, Hence, ! , , As shown in Problem 1, page 13:, , !, Hence, , 11 − 3x, 2, 5, ≡, −, x 2 + 2x − 3 (x − 1) (x + 3), , 11 − 3x, dx, + 2x − 3, , ! , 2, 5, =, −, dx, (x − 1) (x + 3), x2, , = 2 ln(x −1) − 5 ln(x + 3) + c, , ≡, , 2x 2 − 9x − 35, dx, (x + 1)(x − 2)(x + 3), , 3, 1, 4, −, +, dx, (x + 1) (x − 2) (x + 3), , = 4 ln(x+ 1) − 3 ln(x− 2) + ln(x+ 3) + c, 6, 5, (x + 1)4 (x + 3), +c, or ln, (x −2)3, !, Problem 3. Determine, , x2 + 1, dx., x 2 − 3x + 2
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410 Higher Engineering Mathematics, By dividing out (since the numerator and denominator are of the same degree) and resolving into partial, fractions it was shown in Problem 3, page 14:, x2 + 1, x 2 − 3x + 2, !, , ≡ 1−, , 2, 5, +, (x − 1) (x − 2), , ! , ≡, , , 5, 2, dx, +, 1−, (x − 1) (x − 2), , = (x −2) ln(x − 1) + 5 ln(x −2) + c, 5, or x + ln, , Problem 4., , !, , !, , 6, , 4(x − 4), dx, (x 2 − 2x − 3), ⎤, ⎡, 5 ln(x + 1) − ln(x − 3) + c, 6, ⎥, ⎢, 5, ⎥, ⎢, 5, ⎦, ⎣ or ln (x + 1) + c, (x − 3), , 2., , (x −2), +c, (x −1)2, 5, , Exercise 162 Further problems on, integration using partial fractions with, linear factors, In Problems 1 to 5, integrate with respect to x., !, 12, 1., dx, (x 2 − 9), ⎤, ⎡, 2 ln(x − 3) − 2 ln(x + 3) + c, ⎥, ⎢, , , ⎦, ⎣, x −3 2, +c, or ln, x +3, , x2 + 1, dx, x 2 − 3x + 2, , Hence, , Now try the following exercise, , Evaluate, 3, 2, , !, , x 3 − 2x 2 − 4x − 4, dx,, x2 + x − 2, , 3(2x 2 − 8x − 1), dx, (x + 4)(x + 1)(2x − 1), ⎡, ⎤, 7 ln(x + 4) − 3 ln(x + 1), ⎢, ⎥, ⎢ − ln(2x − 1) + c or, ⎥, ⎢, ⎥, , ⎢ , ⎥, ⎣, ⎦, (x + 4)7, ln, +c, (x + 1)3 (2x − 1), , 3., , correct to 4 significant figures., By dividing out and resolving into partial fractions it, was shown in Problem 4, page 15:, x 3 − 2x 2 − 4x − 4, 4, 3, ≡ x −3+, −, 2, x +x −2, (x + 2) (x − 1), !, 2, , x 3 − 2x 2 − 4x − 4, dx, x2 + x − 2, , !, , 3 , , 3, , Hence, , ≡, 2, , , =, , 9, − 9 + 4 ln 5 − 3 ln 2, 2, , x 2 + 9x + 8, dx, x2 + x − 6, , 4., , x + 2 ln(x + 3) + 6 ln(x − 2) + c, or x + ln{(x + 3)2 (x − 2)6 } + c, , , 3, 4, dx, −, (x + 2) (x − 1), , x2, − 3x + 4 ln(x + 2) − 3 ln(x − 1), 2, , , =, , x −3+, , !, , !, , 3x 3 − 2x 2 − 16x + 20, dx, (x − 2)(x + 2), ⎤, ⎡ 2, 3x, ⎣ 2 − 2x + ln(x − 2) ⎦, −5 ln(x + 2) + c, , 5., 3, 2, , , , In Problems 6 and 7, evaluate the definite integrals, correct to 4 significant figures., , − (2 − 6 + 4 ln 4 − 3 ln 1), = −1.687, correct to 4 significant figures., , !, , 4, , 6., 3, , x 2 − 3x + 6, dx, x(x − 2)(x − 1), , [0.6275]
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Integration usingpartial fractions, !, , 6, , 7., 4, , !, , x 2 − x − 14, dx, x 2 − 2x − 3, , [0.8122], , 8. Determine the value of k, given that:, !, , 1, 0, , (x − k), dx = 0, (3x + 1)(x + 1), , , , 1, 3, , 9. The velocity constant k of a given chemical, reaction is given by:, , ! , 1, dx, kt =, (3 − 0.4x)(2 − 0.6x), where x = 0 when t = 0. Show that:, , kt = ln, , 41.3, , 2(3 − 0.4x), 3(2 − 0.6x), , , , Worked problems on, integration using partial, fractions with repeated linear, factors, !, , Problem 5. Determine, , Problem 6. Find, , It was shown in Problem 6, page 16:, 5x 2 − 2x − 19, 2, 3, 4, ≡, +, −, 2, (x + 3)(x − 1), (x + 3) (x − 1) (x − 1)2, !, , 5x 2 − 2x − 19, dx, (x + 3)(x − 1)2, , Hence, , ! , ≡, , 3, 4, 2, +, −, (x + 3) (x − 1) (x − 1)2, , = 2 ln (x +3) + 3 ln (x −1) +, , , or ln (x +3)2 (x −1)3 +, , , dx, , 4, +c, (x − 1), , 4, +c, (x − 1), , Problem 7. Evaluate, ! 1 2, 3x + 16x + 15, dx,, (x + 3)3, −2, correct to 4 significant figures., It was shown in Problem 7, page 17:, , 2x + 3, dx., (x − 2)2, , 3, 6, 3x 2 + 16x + 15, 2, ≡, −, −, 3, 2, (x + 3), (x + 3) (x + 3), (x + 3)3, !, , It was shown in Problem 5, page 16:, Hence, 2x + 3, 2, 7, ≡, +, (x − 2)2, (x − 2) (x − 2)2, , !, ! , 2x + 3, 2, 7, Thus, dx ≡, +, dx, (x − 2)2, (x − 2) (x − 2)2, 7, +c, = 2 ln(x −2) −, (x −2), ⎡!, , 5x 2 − 2x − 19, dx., (x + 3)(x − 1)2, , ⎤, 7, dx, is, determined, using, the, algebraic, ⎦, ⎣ (x − 2)2, substitution u = (x − 2) — see Chapter 39., , 3x 2 + 16x + 15, dx, (x + 3)3, !, , ≡, , 1, −2, , , , 3, 6, 2, −, −, (x + 3) (x + 3)2 (x + 3)3, , , = 3 ln(x + 3) +, , 2, 3, +, (x + 3) (x + 3)2, , , dx, , 1, −2, , , , , 2 3, 2, 3, − 3 ln 1 + +, = 3 ln 4 + +, 4 16, 1 1, = −0.1536, correct to 4 significant figures., , 411
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412 Higher Engineering Mathematics, !, , Now try the following exercise, , 3 + 6x + 4x 2 − 2x 3, dx, x 2 (x 2 + 3), , Thus, , ! , , Exercise 163 Further problems on, integration using partial fractions with, repeated linear factors, In Problems 1 and 2, integrate with respect, to x., !, 4x − 3, dx, 1., (x + 1)2, , 7, 4 ln(x + 1) +, +c, (x + 1), !, , 5x 2 − 30x, , + 44, dx, (x − 2)3, ⎤, ⎡, 10, 5, ln(x, −, 2), +, ⎢, (x − 2) ⎥, ⎥, ⎢, ⎦, ⎣, 2, +, c, −, (x − 2)2, , 2., , In Problems 3 and 4, evaluate the definite integrals, correct to 4 significant figures., !, , 2, , 3., 1, , !, , 7, , 4., 6, , x 2 + 7x + 3, x 2 (x + 3), , [1.663], , 18 + 21x − x 2, dx, (x − 5)(x + 2)2, , [1.089], , ! 1, 5. Show that, 0, , , 4t 2 + 9t + 8, dt = 2.546,, (t + 2)(t + 1)2, , ≡, ! , =, !, , Worked problems on, integration using partial, fractions with quadratic factors, !, , Problem 8., , Find, , + 4x 2 − 2x 3, , 3 + 6x, x 2 (x 2 + 3), , dx., , It was shown in Problem 9, page 18:, 3 − 4x, 2, 1, 3 + 6x + 4x 2 − 2x 3, ≡ + 2+ 2, 2, 2, x (x + 3), x x, (x + 3), , dx, , !, x2, , 1, √, dx, + ( 3)2, , x, 3, = √ tan −1 √ , from 12, Table 40.1, page 399., 3, 3, !, , 4x, x2 + 3, , dx is determined using the algebraic substi-, , tution u =(x 2 + 3)., , ! , 2, 3, 1, 4x, dx, Hence, +, +, −, x x 2 (x 2 + 3) (x 2 + 3), 1, x, 3, + √ tan−1 √ − 2 ln(x 2 + 3) + c, x, 3, 3, , 2, x, x, 1 √, = ln 2, − + 3 tan−1 √ + c, x +3, x, 3, = 2 ln x −, , !, Problem 9., , Determine, , (x 2, , 1, dx., − a2), , A, B, 1, ≡, +, (x 2 − a 2 ) (x − a) (x + a), ≡, , 41.4, , , , , 2, 3, 1, 4x, dx, +, +, −, x x 2 (x 2 + 3) (x 2 + 3), , 3, dx = 3, (x 2 + 3), , Let, , correct to 4 significant figures., , 2, (3 − 4x), 1, + 2+ 2, x x, (x + 3), , A(x + a) + B(x − a), (x + a)(x − a), , Equating the numerators gives:, 1 ≡ A(x + a) + B(x − a), 1, Let x = a, then A = , and let x = −a, then, 2a, 1, B =−, 2a, !, 1, Hence, dx, (x 2 − a 2 ), !, ≡, , , 1, 1, 1, −, dx, 2a (x − a) (x + a)
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Integration usingpartial fractions, 1, [ln(x − a) − ln(x + a)] + c, 2a, , , x −a, 1, ln, +c, =, 2a, x +a, =, , 413, , Problem 12. Evaluate, ! 2, , 5, dx,, (9 − x 2 ), , 0, , correct to 4 decimal places., Problem 10. Evaluate, ! 4, 3, , 3, dx,, (x 2 − 4), , From Problem 11,, !, , 2, , correct to 3 significant figures., 0, , From Problem 9,, , , , ! 4, 1, x −2, 3, dx, =, 3, ln, 2, 2(2), x +2, 3 (x − 4), , 3, 2, 1, =, ln − ln, 4, 6, 5, =, , Problem 11. Determine, , (a 2, , 3, , 1, dx., − x 2), , Using partial fractions, let, 1, A, B, 1, ≡, ≡, +, (a 2 − x 2 ) (a − x)(a + x) (a − x) (a + x), ≡, , =, , 4, , 3 5, ln = 0.383, correct to 3, 4 3, significant figures., !, , , , , 1, 3+x, 5, dx, =, 5, ln, (9 − x 2 ), 2(3), 3−x, , A(a + x) + B(a − x), (a − x)(a + x), , Then 1 ≡ A(a + x) + B(a − x), 1, 1, Let x = a then A = . Let x = −a then B =, 2a, 2a, !, 1, Hence, dx, 2, (a − x 2 ), , !, 1, 1, 1, +, dx, =, 2a (a − x) (a + x), 1, [−ln(a − x) + ln(a + x)] + c, 2a, , , 1, a+x, +c, =, ln, 2a, a−x, , 2, 0, , , 5, 5, ln − ln 1, 6, 1, , = 1.3412, correct to 4 decimal places., , Now try the following exercise, Exercise 164 Further problems on, integration using partial fractions with, quadratic factors, !, x 2 − x − 13, 1. Determine, dx., (x 2 + 7)(x − 2), ⎤, ⎡, x, 3, 2, −1, ⎣ ln(x + 7) + √7 tan √7 ⎦, − ln(x − 2) + c, In Problems 2 to 4, evaluate the definite integrals, correct to 4 significant figures., !, , 6, , 2., 5, , !, , 2, , 3., 1, , !, , 5, , 4., 4, , 6x − 5, dx, (x − 4)(x 2 + 3), , [0.5880], , 4, dx, (16 − x 2 ), , [0.2939], , 2, dx, (x 2 − 9), , [0.1865], , =, , 2, , !, 5. Show, , that, 1, , , 2 +θ + 6θ 2 − 2θ 3, dθ, θ 2 (θ 2 + 1), , = 1.606, correct to 4 significant figures.
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Chapter 42, θ, The t = tan 2 substitution, 42.1, , Introduction, , sin θ =, , i.e., , 2t, (1 + t2 ), , (1), , !, , 1, dθ, where, a cos θ + b sin θ + c, a, b and c are constants, may be determined by using the, θ, substitution t = tan . The reason is explained below., 2, If angle A in the right-angled triangle ABC shown in, θ, Fig. 42.1 is made equal to, then, since tangent =, 2, opposite, θ, , if BC = t and AB = 1, then tan = t ., adjacent, 2, √, By Pythagoras’ theorem, AC = 1 +t 2, Integrals of the form, , C, 冪1 1 t 2, , A, , , 2, 1, , t, , θ, θ, Since cos 2x = cos2 − sin2, 2, 2, , =, , i.e., , 1, √, 1 + t2, , cos θ =, , 2, , , 2, t, − √, 1 + t2, , 1 −t 2, 1 +t 2, , (2), , θ, Also, since t = tan ,, 2, , , θ, θ, dt 1, 1, 2, 2, 1 + tan, from trigonometric, = sec =, dθ 2, 2 2, 2, identities,, , B, , Figure 42.1, , θ, θ, t, 1, and cos = √, Since, =√, 2, 2, 2, 1 +t, 1 +t 2, sin 2x = 2 sin x cos x (from double angle formulae,, Chapter 17), then, , i.e., , dt, 1, = (1 + t 2 ), dθ, 2, , Therefore sin, , θ, θ, sin θ = 2 sin cos, 2, 2, , , , t, t, =2 √, √, 1 + t2, 1 + t2, , from which,, , dθ =, , 2 dt, 1 +t 2, , (3), , Equations (1), (2) !and (3) are used to determine, 1, integrals of the form, dθ where, a cos θ + b sin θ + c, a, b or c may be zero.
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The t = tan θ2 substitution, When, , 42.2, , Worked problems on the, θ, t = tan substitution, 2, !, , Problem 1. Determine, , Hence, , t = −1, 2 = 2B, from which, B = 1, !, !, 2 dt, 1, 1, =, +, dt, 2, 1−t, (1 − t ) (1 + t ), = −ln(1 − t ) + ln(1 + t ) + c, , , (1 + t ), +c, = ln, (1 − t ), ⎧, ⎫, x⎪, ⎪, !, ⎨, ⎬, 1, +tan, dx, 2 +c, = ln, x, ⎪, cos x, ⎩ 1 −tan ⎪, ⎭, 2, , dθ, sin θ, , θ, 2 dt, 2t, and dθ =, from, then sin θ =, 2, 1+t2, 1 +t 2, equations (1) and (3)., , If t = tan, , !, , dθ, =, sin θ, , !, , 1, dθ, sin θ, 1 , , !, 2 dt, 2t, =, 1 + t2 1 + t2, !, 1, =, dt = ln t + c, t, , , !, dθ, θ, +c, = ln tan, sin θ, 2, , Thus, , Hence, , !, Problem 2. Determine, , 2 dt, x, 1 − t2, and dx =, from, then cos x =, 2, 2, 1+t, 1 + t2, equations (2) and (3)., , Thus, , dx, =, cos x, =, , !, !, , Thus, , π, Note that since tan = 1, the above result may be, 4, written as:, ⎧, ⎫, π, x ⎪, ⎪, !, ⎨, ⎬, tan, +, tan, dx, 4, 2, = ln, π, x +c, ⎪, cos x, ⎩ 1 − tan tan ⎪, ⎭, 4, 2, , π x , +, +c, = ln tan, 4 2, from compound angles, Chapter 17., , dx, cos x, , If tan, , !, , 1, , , 2 dt, 1 − t2, 1 + t2 1 + t2, , !, Problem 3. Determine, , 2 dt, x, 1 −t 2, and dx =, from, then cos x =, 2, 2, 1 +t, 1 +t 2, equations (2) and (3)., !, Thus, , 2, dt, 1 − t2, , 2, 2, =, 2, 1−t, (1 − t )(1 + t ), =, , B, A, +, (1 − t ) (1 + t ), , A(1 + t ) + B(1 − t ), =, (1 − t )(1 + t ), Hence, When, , 2 = A(1 + t ) + B(1 − t ), t = 1, 2 = 2 A, from which, A = 1, , dx, 1 +cos x, , If tan, , 2, may be resolved into partial fractions (see, 1 − t2, Chapter 2)., Let, , 415, , !, dx, 1, =, dx, 1 + cos x, 1 + cos x, , , !, 2 dt, 1, =, 1 − t2 1 + t2, 1+, 1 + t2, , , !, 2 dt, 1, =, (1 + t 2 ) + (1 − t 2 ) 1 + t 2, 1 +t 2, !, =, , !, , dt, , dx, x, = t + c = tan + c, 1 +cos x, 2, !, dθ, Problem 4. Determine, 5 +4 cos θ, , Hence
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416 Higher Engineering Mathematics, θ, 2 dt, 1 −t 2, and dx =, then cos θ =, 2, 1 +t 2, 1+t2, from equations (2) and (3)., , , 2 dt, !, !, dθ, 1 + t2, Thus, =, , , 5 + 4 cosθ, 1 − t2, 5+4, 1 + t2, , , 2 dt, !, 1 + t2, =, 2, 5(1 + t ) + 4(1 − t 2 ), (1 + t 2), !, !, dt, dt, =2, =, 2, 2, 2, t +9, t + 32, , , 1 −1 t, =2, tan, + c,, 3, 3, If t = tan, , from 12 of Table 40.1, page 399. Hence, , , !, 2, θ, dθ, −1 1, = tan, tan, +c, 5 +4 cos θ 3, 3, 2, Now try the following exercise, Exercise 165 Further problems on the, θ, t =tan substitution, 2, Integrate the following with respect to the variable:, ⎡, ⎤, !, dθ, ⎢ −2, ⎥, + c⎦, 1., ⎣, θ, 1 + sin θ, 1 + tan, 2, !, dx, 2., 1 − cos x + sin x, ⎫, ⎡ ⎧, ⎤, x ⎪, ⎪, ⎨ tan, ⎬, ⎢, ⎥, 2, ⎣ln, x ⎪ + c⎦, ⎪, ⎩ 1 + tan ⎭, 2, !, dα, 3., 3 + 2 cosα, , , , 2, 1, α, √ tan−1 √ tan, +c, 2, 5, 5, !, dx, 4., 3 sin x − 4 cos x, ⎧, ⎫, ⎤, ⎡, x, ⎪, ⎪, ⎨, ⎬, 2, tan, −, 1, ⎥, ⎢1, 2, + c⎦, ⎣ ln, x, ⎪, 5 ⎪, ⎩ tan + 2 ⎭, 2, , 42.3, , Further worked problems on the, θ, t = tan substitution, 2, !, , Problem 5., , Determine, , dx, sin x + cos x, , 1 − t2, x, 2t, ,, cos, x, =, and, then sin x =, 2, 1 + t2, 1 + t2, 2 dt, dx =, from equations (1), (2) and (3)., 1 + t2, Thus, 2 dt, !, !, dx, 1 + t2, = , , , sin x + cos x, 2t, 1 − t2, +, 1 + t2, 1 + t2, 2 dt, !, !, 2 dt, 1, + t2 =, =, 1 + 2t − t 2, 2t + 1 − t 2, 1 + t2, !, !, −2 dt, −2 dt, =, =, t 2 − 2t − 1, (t − 1)2 − 2, !, 2 dt, √, =, 2, ( 2) − (t − 1)2, 5√, 6, 2 + (t − 1), 1, +c, = 2 √ ln √, 2 2, 2 − (t − 1), , If tan, , (see Problem 11, Chapter 41, page 413),, !, i.e., , dx, sin x + cos x, ⎧√, ⎫, x⎪, ⎪, ⎨, ⎬, 2, −, 1, +tan, 1, 2 +c, = √ ln √, x, 2 ⎪, ⎩ 2 + 1 −tan ⎪, ⎭, 2, , Problem, 6. Determine, !, dx, 7 − 3 sin x + 6 cos x, From equations (1) and (3),, !, dx, 7 − 3 sin x + 6 cos x, !, =, , 2 dt, 1 + t2, , , , , 2t, 1 − t2, 7−3, +6, 1 + t2, 1 + t2
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The t = tan θ2 substitution, 2 dt, !, 1 + t2, =, 2, 7(1 + t ) − 3(2t ) + 6(1 − t 2 ), 1 + t2, !, 2 dt, =, 7 + 7t 2 − 6t + 6 − 6t 2, !, !, 2 dt, 2 dt, =, =, 2, t − 6t + 13, (t − 3)2 + 22, , , , 1 −1 t − 3, +c, =2, tan, 2, 2, from 12, Table 40.1, page 399. Hence, !, , dx, 7 − 3 sin x + 6 cos x, ⎞, ⎛, x, tan − 3, ⎟, ⎜, 2, = tan−1 ⎝, ⎠+c, 2, , , ⎫⎤, ⎧, 5, 3 ⎪, ⎪, ⎪ + t−, ⎪, ⎨, 1⎢ 1, 4, 4 ⎬⎥, ⎥+c, , , , , ln, = ⎢, ⎦, ⎪, 5, 5, 3 ⎪, 2⎣, ⎪, ⎪, ⎩, ⎭, 2, − t−, 4, 4, 4, ⎡, , from Problem 11, Chapter 41, page 413, ⎧, ⎫, 1, ⎪, ⎪, ⎨, ⎬, +, t, 1, 2, +c, = ln, 5 ⎪, ⎩ 2−t ⎪, ⎭, !, dθ, Hence, 4 cos θ + 3 sin θ, ⎧, ⎫, 1, θ⎪, ⎪, ⎨, ⎬, +, tan, 1, 2 +c, = ln 2, 5 ⎪, ⎩ 2 − tan θ ⎪, ⎭, 2, ⎧, ⎫, θ⎪, ⎪, ⎨, ⎬, 1, +2, tan, 1, 2 +c, or, ln, 5 ⎪, ⎩ 4 − 2 tan θ ⎪, ⎭, 2, Now try the following exercise, , !, Problem 7. Determine, , dθ, 4 cosθ + 3 sin θ, , From equations (1) to (3),, !, , dθ, 4 cos θ + 3 sin θ, 2 dt, 1 + t2, =, , , , , 1 − t2, 2t, 4, +, 3, 1 + t2, 1 + t2, !, !, dt, 2 dt, =, =, 2, 4 − 4t + 6t, 2 + 3t − 2t 2, !, , =−, , 1, 2, , 1, =−, 2, , =, , 1, 2, , !, , !, , !, , Exercise 166 Further problems on the, θ, t = tan substitution, 2, In Problems 1 to 4, integrate with respect to the, variable., !, dθ, 1., 5 + 4 sin θ, ⎡, ⎞, ⎤, ⎛, θ, 5, tan, +, 4, ⎟, ⎥, ⎢ 2 −1 ⎜, 2, ⎠ + c⎦, ⎣ tan ⎝, 3, 3, !, 2., , dt, , 2 , 5, 3 2, − t−, 4, 4, , dx, 1 + 2 sin x, ⎡, , ⎧, ⎤, √ ⎫, x, ⎪, ⎪, ⎨, ⎬, tan, 3, +, 2, −, ⎢ 1, ⎥, 2, ⎣ √ ln, √ ⎪ + c⎦, x, ⎪, 3 ⎩ tan + 2 + 3 ⎭, 2, , dt, 3, t2 − t − 1, 2, dt, , , 3 2 25, −, t−, 4, 16, , 417, , !, 3., , dp, 3 − 4 sin p +2 cos p, ⎧, ⎤, √ ⎫, p, ⎪, ⎪, ⎨, ⎬, tan, 11, −, 4, −, ⎢ 1, ⎥, 2, + c⎦, ⎣ √ ln, √, p, 11 ⎪, ⎩ tan − 4 + 11 ⎪, ⎭, 2, ⎡
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Revision Test 12, This Revision Test covers the material contained in Chapters 40 to 42. The marks for each question are shown in, brackets at the end of each question., 1. Determine the following integrals:, !, !, 2, 3, 2, (a) cos x sin x dx (b) , dx, (9 − 4x 2 ), !, 2, dx, (14), (c) , (4x 2 − 9), 2. Evaluate the following definite integrals, correct to, 4 significant figures:, ! π, ! π, 2, 3, 2, (a), 3 sin t dt (b), 3 cos5θ sin 3θ dθ, !, , 0, , 0, 2, , (c), 0, , 5, dx, 4 + x2, , (15), , 3. Determine:, !, x − 11, dx, (a), 2, x −x −2, !, 3−x, dx, (21), (b), (x 2 + 3)(x + 3), ! 2, 3, dx correct to 4 significant, 4. Evaluate, 2, 1 x (x + 2), figures., (12), !, dx, 5. Determine:, (8), 2 sin x + cos x, ! π, 2, dx, 6. Evaluate, correct to 3 decimal, π, 3 − 2 sin x, 3, places., (10)
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Chapter 43, , Integration by parts, 43.1, , Introduction, , 43.2 Worked problems on integration, by parts, , From the product rule of differentiation:, d, du, dv, (uv) = v, +u ,, dx, dx, dx, where u and v are both functions of x., dv, d, du, Rearranging gives: u, =, (uv) − v, dx, dx, dx, Integrating both sides with respect to x gives:, !, !, !, dv, d, du, u dx =, (uv) dx − v dx, dx, dx, dx, !, i.e., or, , !, dv, du, dx = uv− v, dx, dx, dx, !, !, u dv = uv − v du, , u, , Problem 1., , Determine, , x cos x dx., , From the integration by parts formula,, !, !, u dv = uv − v du, du, = 1, i.e. du = dx and let, dx , dv = cos x dx, from which v = cos x dx = sin x., Expressions for u, du and v are now substituted into, the ‘by parts’ formula as shown below., Let u = x, from which, , u, , dv, , x cos x dx, , This is known as the integration by parts formula, and provides a method of integrating, such, prod, x, dx,, t, sin, t dt ,, ucts, of, simple, functions, as, xe, , θ, e cos θ dθ and x ln x dx., Given a product of two terms to integrate the initial, choice is: ‘which part to make equal to u’ and ‘which, part to make equal to v’. The choice must be such that the, ‘u part’ becomes a constant after successive differentiation and the ‘dv part’ can be integrated from standard, integrals. Invariably, the following rule holds: If a product to be integrated contains an algebraic term (such as, x, t 2 or 3θ) then this term is chosen as the u part. The one, exception to this rule is when a ‘ln x’ term is involved;, in this case ln x is chosen as the ‘u part’., , , , , , , u, , v, , (x) (sin x), , , , , v, , du, , (sin x) (dx), , !, i.e., , x cos x dx = x sin x − (−cos x) + c, = x sin x +cos x + c, , [This result may be checked by differentiating the right, hand side,, i.e., , d, (x sin x + cos x + c), dx, = [(x)(cos x) + (sin x)(1)] − sin x + 0, using the product rule, = x cos x, which is the function, being integrated]
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Integration by parts, , 10. In determining a Fourier series to represent f (x) = x in the range −π to π, Fourier, coefficients are given by:, !, 1 π, x cos nx dx, an =, π −π, !, 1 π, x sin nx dx, and bn =, π −π, where n is a positive integer. Show by, using integration by parts that an = 0 and, 2, bn = − cos nπ., n, , !, , 1, , 11. The equation C =, , e−0.4θ cos 1.2θ dθ, , 0, , !, and, , 1, , S=, , e−0.4θ sin 1.2θ dθ, , 0, , are involved in the study of damped, oscillations. Determine the values of C, and S., [C = 0.66, S = 0.41], , 425
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Chapter 44, , Reduction formulae, !, , 44.1, , Introduction, , x n−1 ex dx = In−1, , then, !, , When using integration, by parts in Chapter 43, an, , integral such as x 2 e x dx requires integration by, parts twice. Similarly, x 3 e x dx requires integration, parts, three times. Thus, integrals such as, 5 by, x e x dx, x 6 cos x dx and x 8 sin 2x dx for example,, would take a long time to determine using integration by parts. Reduction formulae provide a quicker, method for determining such integrals and the method, is demonstrated in the following sections., , !, x n ex dx = x n ex − n, , Hence, , can be written as:, In = xn ex − nIn−1, , To determine, let, , , , dv = ex dx from which,, !, v = e x dx = ex, !, !, n x, n, x, Thus,, x e dx = x e − e x nx n−1 dx, , x 2 ex dx = I2 = x 2 ex − 2I1, and, , I1 = x 1 ex − 1I0, !, !, 0 x, I0 = x e dx = e x dx = ex + c1, , Hence, , I2 = x 2 ex − 2[xex − 1I0 ], = x 2 ex − 2[xex − 1(e x + c1 )], , !, i.e., , x2 ex dx = x 2 ex − 2xex + 2e x + 2c1, = ex (x2 − 2x +2) + c, , using the integration by parts formula,, !, n, x, = x e − n x n−1 ex dx, The integral on the far right is seen to be of the same, form as the integral on the left-hand side, except that n, has been replaced by n −1., Thus, if we let,, !, x n ex dx = In ,, , x 2 e x dx using a, , !, , du, = nx n−1 and du =nx n−1 dx, dx, and, , , , Using equation (1) with n = 2 gives:, , x n e x dx using integration by parts,, u = x n from which,, , (1), , Equation (1) is an example of a reduction formula since, it expresses an integral in n in terms of the same integral, in n −1., Problem 1. Determine, reduction formula., , 44.2 Using reduction formulae, n xfor, integrals of the form x e dx, , x n−1 e x dx, , (where c = 2c1 ), As with integration by parts, in the following examples, the constant of integration will be added at the last step, with indefinite integrals., Problem, 3 x 2., x e dx., , Use a reduction formula to determine
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427, , Reduction formulae, From equation (1), In = x n ex − n In−1, !, Hence, x 3 e x dx = I3 = x 3 ex − 3I2, , !, (sin x)nx n−1 dx, !, = x n sin x − n x n−1 sin x dx, , Hence In = x n sin x −, , I2 = x 2 e x − 2I1, 1 x, !I1 = x e − 1I, !0, 0 x, I0 = x e dx = e x dx = ex, , and, !, , x 3 e x dx = x 3 ex − 3[x 2e x − 2I1 ], , Thus, , = x 3 ex − 3[x 2e x − 2(xe x − I0 )], , Using integration by parts again, this time with, u = x n−1 :, du, = (n − 1)x n−2 , and dv = sin x dx,, dx, from which,, v=, , = x 3 ex − 3[x 2e x − 2(xe x − ex )], = x 3 ex − 3x 2 ex + 6(xe x − ex ), !, , = x 3 ex − 3x 2 ex + 6xe x − 6e x, , !, sin x dx = −cos x, , , Hence In = x sin x − n x n−1 (−cos x), n, , !, , x3ex dx = ex (x3 − 3x2 + 6x −6) + c, , i.e., , −, , (−cos x)(n − 1)x n−2 dx, , = x n sin x + nx n−1 cos x, , Now try the following exercise, , !, , − n(n − 1), Exercise 169 Further problems on using, reduction, formulae for integrals of the form, n x, x e dx, 1. Use, 4 xa reduction formula to determine, x e dx., [ex (x 4 − 4x 3 + 12x 2 − 24x + 24) + c], , 2. Determine t 3e2t dt using a reduction formula., , , , e2t 12 t 3 − 34 t 2 + 34 t − 38 + c, , i.e., , I n = xn sin x + nxn−1 cos x, , Problem, 3. Use a reduction formula to determine, 2, x cos x dx., Using the reduction formula of equation (2):, !, x 2 cos x dx = I2, = x 2 sin x + 2x 1 cos x − 2(1)I0, !, I0 = x 0 cos x dx, , and, , , , (a) xn cos x dx, , Let In = x n cos x dx then, using integration by parts:, du, if, u = x n then, = nx n−1, dx, and if dv = cos x dx then, !, v = cos x dx = sin x, , (2), , − n(n −1)In−2, , 3. Use, 1 3the2t result of Problem 2 to evaluate, 0 5t e dt, correct to 3 decimal places., [6.493], , 44.3 Using reduction formulae, n for, integrals, of, the, form, x cos x dx, n, and x sin x dx, , x n−2 cos x dx, , !, =, Hence, , cos x dx = sin x, , !, x2 cos x dx = x2 sin x +2x cos x − 2 sin x +c, , Problem 4. Evaluate, significant figures., Let, 3 us firstly, t cos t dt ., , find, , 2, 1, , a, , 4t 3 cos t dt , correct to 4, , reduction, , formula, , for
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428 Higher Engineering Mathematics, From equation (2),, !, t 3 cos t dt = I3 = t 3 sin t + 3t 2 cos t − 3(2)I1, , When n =2,, ! π, x 2 cos x dx = I2 = −2π 1 − 2(1)I0, , and, , and, , 0, , !, , I1 = t 1 sin t + 1t 0 cos t − 1(0)In−2, , !, , x 0 cos x dx, , 0, π, , =, , = t sin t + cos t, , cos x dx, 0, , Hence, !, t 3 cos t dt = t 3 sin t + 3t 2 cos t, − 3(2)[t sin t + cos t ], , = [sin x]π0 = 0, Hence, ! π, x 4 cos x dx = −4π 3 − 4(3)[−2π − 2(1)(0)], 0, , = −4π 3 + 24π or −48.63,, , = t sin t + 3t cos t − 6t sin t − 6 cost, 3, , π, , I0 =, , 2, , correct to 2 decimal places., , Thus, ! 2, 4t 3 cos t dt, 1, , = [4(t 3 sin t + 3t 2 cos t − 6t sin t − 6 cost )]21, = [4(8 sin 2 +12 cos 2 −12 sin 2 − 6 cos 2)], − [4(sin 1 +3 cos 1 − 6 sin 1 −6 cos 1)], , , (b) xn sin x dx, , Let In = x n sin x dx, Using integration by parts, if u = x n then, du, = nx n−1 and if dv = sin x dx then, dx , v = sin x dx = −cos x. Hence, , = (−24.53628) −(−23.31305), , !, x n sin x dx, , = −1.223, , !, , Problem, π 5. Determine a reduction formula, for 0 x n cos x dx and hence evaluate, π 4, 0 x cos x dx, correct to 2 decimal places., , = In = x n (−cos x) −, , (−cos x)nx n−1 dx, , !, = −x cos x + n, n, , x n−1 cos x dx, , From equation (2),, In = x n sin x + nx n−1 cos x − n(n − 1)In−2 ., ! π, x n cos x dx = [x n sin x + nx n−1 cos x]π0, hence, , Using integration by parts again, with u = x n−1 , from, du, which,, = (n − 1)x n−2 and dv = cos x, from which,, dx, v = cos x dx = sin x. Hence, , 0, , , In = −x n cos x + n x n−1 (sin x), , − n(n − 1)In−2, = [(π n sin π + nπ n−1 cos π), , !, −, , − (0 + 0)] − n(n − 1)In−2, = − nπ n−1 − n(n − 1)In−2, Hence, ! π, , = −x n cos x + nx n−1 (sin x), !, − n(n − 1) x n−2 sin x dx, , x 4 cos x dx = I4, , 0, , = −4π 3 − 4(3)I2 since n = 4, , (sin x)(n − 1)x n−2 dx, , i.e., , In = −xn cos x + nxn−1 sin x − n(n − 1)In−2 (3)
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Reduction formulae, Problem, 6. Use a reduction formula to determine, 3, x sin x dx., , Hence, !, 3, , π, 2, , 429, , θ 4 sin θ dθ, , 0, , Using equation (3),, !, x 3 sin x dx = I3, = −x 3 cos x + 3x 2 sin x − 3(2)I1, I1 = −x 1 cos x + 1x 0 sin x, , and, , = −x cos x + sin x, Hence, !, x 3 sin x dx = −x 3 cos x + 3x 2 sin x, − 6[−x cos x + sin x], = −x3cos x + 3x2 sin x, + 6x cos x − 6 sin x + c, !, , π, 2, , Problem 7. Evaluate, , 3θ 4 sin θ dθ, correct to 2, , 0, , decimal places., From equation (3),, , π, , In = [−x n cos x + nx n−1 (sin x)]02 − n(n − 1)In−2, , , π n−1, π n, π, π, = −, cos + n, sin, − (0), 2, 2, 2, 2, − n(n − 1)In−2, =n, , π n−1, 2, , − n(n − 1)In−2, , Hence, !, , π, 2, , !, 3θ sin θ dθ = 3, 4, , 0, , π, 2, , θ 4 sin θ dθ, , = 3I4, , , , π 3, π 1, =3 4, − 4(3) 2, − 2(1)I0, 2, 2, , , , π 1, π 3, − 4(3) 2, − 2(1)(1), =3 4, 2, 2, , , , π 3, π 1, =3 4, − 24, + 24, 2, 2, = 3(15.503 − 37.699 + 24), = 3(1.8039) = 5.41, Now try the following exercise, Exercise 170 Further problems on, reduction, formulae, for integrals of the form, n, x cos x dx and xn sin x dx, 1. Use, 5 a reduction formula to determine, x cos x⎡dx., ⎤, x 5 sin x + 5x 4 cos x − 20x 3 sin x, ⎢, ⎥, ⎣ − 60x 2 cos x + 120x sin x, ⎦, + 120 cos x + c, π 5, 2. Evaluate 0 x cos x dx, correct to 2 decimal, places., [−134.87], 3. Use, 5 a reduction formula to determine, x sin x dx., ⎡ 5, ⎤, −x cos x + 5x 4 sin x + 20x 3 cos x, ⎢, ⎥, ⎣ − 60x 2 sin x − 120x cos x, ⎦, + 120 sin x + c, π 5, 4. Evaluate 0 x sin x dx, correct to 2 decimal, places., [62.89], , 0, , = 3I4, , π 3, =3 4, − 4(3)I2, 2, π 1, − 2(1)I0 and, I2 = 2, 2, ! π, π, 2, θ 0 sin θ dθ = [−cos x]02, I0 =, 0, , = [−0 − (−1)] = 1, , 44.4 Using reduction formulae, nfor, integrals, of, the, form, sin x dx, , and cosn x dx, , (a) sinn x dx, , , Let In = sin n x dx ≡ sinn−1 x sin x dx from laws of, indices., Using integration by parts, let u = sinn−1 x, from which,
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432 Higher Engineering Mathematics, !, , Now try the following exercise, , cos4 x dx, , Hence, =, , 3, 1, cos3 x sin x +, 4, 4, , , , 1, 1, cos x sin x + x, 2, 2, , , , Exercise 171 Further problems on, formulae, for integrals of the form, , reduction, sinn x dx and cosn x dx, , 1, 3, 3, = cos3 x sin x + cos x sin x + x + c, 4, 8, 8, , 1. Use, 7 a reduction formula to determine, sin x dx., ⎤, ⎡, 6, 1, − sin6 x cos x − sin4 x cos x, ⎥, ⎢ 7, 35, ⎦, ⎣ 8, 16, − sin2 x cos x − cos x + c, 35, 35, π, 2. Evaluate 0 3 sin3 x dx using a reduction, formula., [4], , Problem 12. Determine a reduction formula, ! π, ! π, 2, 2, cosn x dx and hence evaluate, cos5 x dx, for, 0, , 0, , From equation (5),, !, 1, n −1, cosn x dx = cosn−1 x sin x +, In−2, n, n, , !, 0, , and hence, !, , π, 2, , 0, , , , 1, cos x dx =, cosn−1 x sin x, n, n, , +, , !, , π, 2, , formula., , π, 2, , 0, , n −1, In−2, n, , = [0 −0] +, , i.e., , cosn x dx = In =, , 0, , n −1, In−2, n, , n−1, In−2, n, , (6), , (Note that this is the same reduction formula as for, ! π, 2, sinn x dx (in Problem 10) and the result is usually, 0, , known as Wallis’s formula)., Thus, from equation (6),, !, , π, 2, , 0, , 4, cos5 x dx = I3 ,, 5, !, , π, 2, , I1 =, , and, , 2, I3 = I1, 3, 1, , cos x dx, π, 2, , = [sin x]0 = (1 − 0) = 1, π, 2, , Hence, , 44.5, , 0, , Further reduction formulae, , The following worked problems demonstrate further, examples where integrals can be determined using, reduction formulae., Problem, 13. Determine a reduction formula for, , tann x dx and hence find tan7 x dx., !, Let In =, , =, , !, tann x dx ≡, , !, , , , 4, 4 2, cos5 x dx = I3 =, I1, 5, 5 3, , 4 2, 8, =, (1) =, 5 3, 15, , sin5 x dx using a reduction, , 8, 15, , 4. Determine,, using a reduction formula,, !, 6, cos x dx., ⎡, ⎤, 1, 5, 5, 3, ⎢ 6 cos x sin x + 24 cos x sin x ⎥, ⎣, ⎦, 5, 5, + cos x sin x + x + c, 16, 16, , ! π, 2, 16, 7, cos x dx., 5. Evaluate, 35, 0, , 0, , !, , π, 2, , 3. Evaluate, , tann−2 x tan 2 x dx, by the laws of indices, , tan n−2 x(sec 2 x − 1) dx, , since 1 + tan2 x = sec2 x, !, !, = tan n−2 x sec2 x dx − tann−2 x dx
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Chapter 45, , Numerical integration, y f(x ), , y, , 45.1, , Introduction, , Even with advanced methods of integration there are, many mathematical functions which cannot be integrated by analytical methods and thus approximate, methods have then to be used. Approximate methods of, definite integrals may be determined by what is termed, numerical integration., It may be shown that determining the value of a definite, integral is, in fact, finding the area between a curve, the, horizontal axis and the specified ordinates. Three methods of finding approximate areas under curves are the, trapezoidal rule, the mid-ordinate rule and Simpson’s, rule, and these rules are used as a basis for numerical, integration., , y1 y2 y3 y4, , xa, , O, , x b, , d, , 45.2, , The trapezoidal rule, , b, Let a required definite integral be denoted by a y dx and, be represented by the area under the graph of y = f (x), between the limits x = a and x = b as shown in Fig. 45.1., Let the range of integration be divided into n equal, intervals each of width d, such that nd = b − a, i.e., b−a, d=, n, The ordinates are labelled y1 , y2, y3, . . . , yn+1 as, shown., An approximation to the area under the curve may be, determined by joining the tops of the ordinates by, straight lines. Each interval is thus a trapezium, and, since the area of a trapezium is given by:, , d, , x, , d, , Figure 45.1, , !, , b, a, , 1, 1, y dx ≈ (y1 + y2 )d + (y2 + y3 )d, 2, 2, 1, 1, + (y3 + y4 )d + · · · (yn + yn+1 )d, 2, 2, , 1, ≈ d y1 + y2 + y3 + y4 + · · · + yn, 2, 1, + yn+1, 2, , i.e. the trapezoidal rule states:, !, , b, a, , 1, area = (sum of parallel sides) (perpendicular, 2, distance between them) then, , yn1, , , y dx ≈, , , , 1 first + last, width of, interval, 2 ordinate, , , sum of remaining, +, ordinates, , (1)
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436 Higher Engineering Mathematics, Problem 1., , (a) Use integration to evaluate,, ! 3, 2, correct to 3 decimal places,, √ dx (b) Use the, x, 1, trapezoidal rule with 4 intervals to evaluate the, integral in part (a), correct to 3 decimal places., !, , 3, , (a), 1, , 2, √ dx =, x, , !, , 3, , Use the trapezoidal rule with 8, ! 3, 2, intervals to evaluate,, √ dx correct to 3, x, 1, decimal places., , 3−1, With 8 intervals, the width of each is, i.e. 0.25, 8, giving ordinates at 1.00, 1.25, 1.50, 1.75, 2.00, 2.25,, 2, 2.50, 2.75 and 3.00. Corresponding values of √ are, x, shown in the table below., , 1, , 2x − 2 dx, , 1, , ⎡, , Problem 2., , , , ⎤3, , −1, 2 +1, , , 1 3, ⎢ 2x, ⎥, =, 4x 2, =⎣, ⎦, 1, 1, − +1, 2, 1, √, √ , √ 3, = 4 x 1 = 4 3− 1, = 2.928, correct to 3 decimal places, (b) The range of integration is the difference between, the upper and lower limits, i.e. 3 − 1 = 2. Using, the trapezoidal rule with 4 intervals gives an inter3−1, val width d =, = 0.5 and ordinates situated, 4, at 1.0, 1.5, 2.0, 2.5 and 3.0. Corresponding values, 2, of √ are shown in the table below, each correct, x, to 4 decimal places (which is one more decimal, place than required in the problem)., x, , 2, √, x, , 1.0, , 2.0000, , 1.5, , 1.6330, , 2.0, , 1.4142, , 2.5, , 1.2649, , 3.0, , 1.1547, , x, , 2, √, x, , 1.00, , 2.0000, , 1.25, , 1.7889, , 1.50, , 1.6330, , 1.75, , 1.5119, , 2.00, , 1.4142, , 2.25, , 1.3333, , 2.50, , 1.2649, , 2.75, , 1.2060, , 3.00, , 1.1547, , From equation (1):, , ! 3, 1, 2, √ dx ≈ (0.25) (2.000 + 1.1547) + 1.7889, 2, x, 1, + 1.6330 + 1.5119 + 1.4142, , + 1.3333 + 1.2649 + 1.2060, = 2.932, correct to 3 decimal places., , From equation (1):, , ! 3, 2, 1, √ dx ≈ (0.5) (2.0000 + 1.1547), 2, x, 1, , , , + 1.6330 + 1.4142 + 1.2649, = 2.945, correct to 3 decimal places, This problem demonstrates that even with just 4 intervals a close approximation to the true value of 2.928, (correct to 3 decimal places) is obtained using the, trapezoidal rule., , This problem demonstrates that the greater the number, of intervals chosen (i.e. the smaller the interval width), the more accurate will be the value of the definite integral. The exact value is found when the number of, intervals is infinite, which is, of course, what the process, of integration is based upon., Problem 3. Use the trapezoidal rule to evaluate, ! π, 2, 1, dx using 6 intervals. Give the answer, 0 1 + sin x, correct to 4 significant figures.
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437, , Numerical integration, π, −0, With 6 intervals, each will have a width of 2, 6, π, i.e., rad (or 15◦) and the ordinates occur at, 12, π π π π 5π, π, 0, , , , ,, and, 12 6 4 3 12, 2, 1, Corresponding values of, are shown in the, 1 + sin x, table below., , !, , x, 0, , 1.0000, , π, (or 15◦), 12, , 0.79440, , π, (or 30◦ ), 6, , 0.66667, , π, (or 45◦), 4, , 0.58579, , π, (or 60◦ ), 3, , 0.53590, , 5π, (or 75◦), 12, , 0.50867, , π, (or 90◦ ), 2, , 0.50000, , 2, dx, 1 + x2, , (Use 8 intervals), , [1.569], , !, , 2 ln 3x dx, , (Use 8 intervals), , [6.979], , 0, 3, , 2., 1, , !, , π, 3, , 3., !, , 1, 1 + sin x, , 1, , 1., , , (sin θ) dθ, , (Use 6 intervals), , [0.672], , 0, 1.4, , 4., , e−x dx, 2, , (Use 7 intervals), , [0.843], , 0, , 45.3, , The mid-ordinate rule, , Let a required definite integral be denoted again, b, by a y dx and represented by the area under the graph, of y = f (x) between the limits x = a and x = b, as shown, in Fig. 45.2., y, y f(x), , From equation (1):, ! π, π 1, 2, 1, dx ≈, (1.00000 + 0.50000), 12, 2, 0 1 + sin x, , y1, , + 0.79440 + 0.66667, + 0.58579 + 0.53590, , O, , = 1.006, correct to 4, significant figures., , Now try the following exercise, Exercise 173 Further problems on the, trapezoidal rule, In Problems 1 to 4, evaluate the definite integrals, using the trapezoidal rule, giving the answers, correct to 3 decimal places., , y3, , yn, , a, , b x, d, , , , + 0.50867, , y2, , d, , d, , Figure 45.2, , With the mid-ordinate rule each interval of width d is, assumed to be replaced by a rectangle of height equal to, the ordinate at the middle point of each interval, shown, as y1 , y2, y3 , . . . , yn in Fig. 45.2., ! b, y dx ≈ d y1 + d y2 + d y3 + · · · + d yn, Thus a, ≈ d( y1 + y2 + y3 + · · · + yn ), i.e. the mid-ordinate rule states:, !, , b, a, , y dx ≈ (width of interval) (sum, of mid-ordinates), , (2)
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438 Higher Engineering Mathematics, From equation (2):, , Problem 4., , Use the mid-ordinate rule, ! 3 with (a) 4, 2, intervals, (b) 8 intervals, to evaluate, √ dx,, x, 1, correct to 3 decimal places., , 3−1, ,, (a) With 4 intervals, each will have a width of, 4, i.e. 0.5 and the ordinates will occur at 1.0, 1.5, 2.0,, 2.5 and 3.0. Hence the mid-ordinates y1 , y2, y3, and y4 occur at 1.25, 1.75, 2.25 and 2.75. Corre2, sponding values of √ are shown in the following, x, table., x, , 2, √, x, , 1.25, , 1.7889, , 1.75, , 1.5119, , 2.25, , 1.3333, , 2.75, , 1.2060, , From equation (2):, ! 3, 2, √ dx ≈ (0.5)[1.7889 + 1.5119, x, 1, + 1.3333 + 1.2060], = 2.920, correct to 3 decimal places., , !, , 3, 1, , 2, √ dx ≈ (0.25)[1.8856 + 1.7056, x, + 1.5689 + 1.4606 + 1.3720, + 1.2978 + 1.2344 + 1.1795], = 2.926, correct to 3 decimal places., , As previously, the greater the number of intervals, the nearer the result is to the true value (of 2.928, correct, to 3 decimal places)., !, Problem 5., , e, , −x 2, 3, , dx, correct to 4, , 0, , significant figures, using the mid-ordinate rule with, 6 intervals., 2.4 − 0, With 6 intervals each will have a width of, , i.e., 6, 0.40 and the ordinates will occur at 0, 0.40, 0.80, 1.20,, 1.60, 2.00 and 2.40 and thus mid-ordinates at 0.20, 0.60,, 1.00, 1.40, 1.80 and 2.20. Corresponding values of e, are shown in the following table., , (b) With 8 intervals, each will have a width of 0.25, and the ordinates will occur at 1.00, 1.25, 1.50,, 1.75, . . . and thus mid-ordinates at 1.125, 1.375,, 1.625, 1.875 . . ., 2, Corresponding values of √ are shown in the, x, following table., x, , 2.4, , Evaluate, , 2, √, x, , −x 2, 3, , −x 2, 3, , x, , e, , 0.20, , 0.98676, , 0.60, , 0.88692, , 1.00, , 0.71653, , 1.40, , 0.52031, , 1.80, , 0.33960, , 2.20, , 0.19922, , 1.125 1.8856, 1.375 1.7056, , From equation (2):, , 1.625 1.5689, , !, , 1.875 1.4606, , 2.4, , e, , −x 2, 3, , dx ≈ (0.40)[0.98676 + 0.88692, , 0, , 2.125 1.3720, , + 0.71653 + 0.52031, , 2.375 1.2978, , + 0.33960 + 0.19922], , 2.625 1.2344, 2.875 1.1795, , = 1.460, correct to 4 significant figures.
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Numerical integration, , 439, , y, , Now try the following exercise, , y a bx cx 2, , Exercise 174 Further problems on the, mid-ordinate rule, In Problems 1 to 4, evaluate the definite integrals, using the mid-ordinate rule, giving the answers, correct to 3 decimal places., y1, , !, , 2, , 3, dt, 1 + t2, , 1., 0, , !, , π, 2, , 2., 0, , (Use 8 intervals), , y3, , [3.323], d, , 1, dθ (Use 6 intervals), 1 + sin θ, , y2, , O, , d, , x, , [0.997], Figure 45.3, , !, , 3 ln x, , 3., , x, , 1, , !, , π, 3, , 4., , dx, , (Use 10 intervals) [0.605], , , (cos3 x) dx (Use 6 intervals) [0.799], , 0, , Since, , y = a + bx + cx 2 ,, , at, , x = −d, y1 = a − bd + cd 2, , at, , x = 0, y2 = a, , and at x = d, y3 = a + bd + cd 2, , 45.4, , Hence y1 + y3 = 2a + 2cd 2, , Simpson’s rule, , The approximation made with the trapezoidal rule is to, join the top of two successive ordinates by a straight, line, i.e. by using a linear approximation of the form, a + bx. With Simpson’s rule, the approximation made, is to join the tops of three successive ordinates by a, parabola, i.e. by using a quadratic approximation of the, form a + bx + cx 2 ., Figure 45.3 shows a parabola y = a + bx + cx 2 with, ordinates y1 , y2 and y3 at x = −d, x = 0 and x = d, respectively., Thus the width of each of the two intervals is d. The, area enclosed by the parabola, the x-axis and ordinates, x = −d and x = d is given by:, , d, bx 2 cx 3, (a + bx + cx )dx = ax +, +, 2, 3 −d, −d, , , bd 2 cd 3, = ad +, +, 2, 3, , , bd 2 cd 3, − −ad +, −, 2, 3, , !, , d, , 2, , 2, = 2ad + cd 3 or, 3, 1, d(6a + 2cd 2 ), 3, , y1 + 4y2 + y3 = 6a + 2cd 2, , And, , Thus the area under the parabola between x = −d, and x =d in Fig. 45.3 may be expressed as, 1, 3 d(y1 + 4y2 + y3 ), from equations (3) and (4), and the, result is seen to be independent of the position of the, origin., b, Let a definite integral be denoted by a y dx and, represented by the area under the graph of y = f (x), between the limits x = a and x = b, as shown in Fig. 45.4., The range of integration, b − a, is divided into an even, number of intervals, say 2n, each of width d., Since an even number of intervals is specified, an odd, number of ordinates, 2n + 1, exists. Let an approximation to the curve over the first two intervals be a parabola, of the form y = a + bx + cx 2 which passes through the, tops of the three ordinates y1, y2 and y3. Similarly, let, an approximation to the curve over the next two intervals be the parabola which passes through the tops of, the ordinates y3, y4 and y5 , and so on., !, , b, , Then, , y dx, a, , ≈, (3), , (4), , 1, 1, d(y1 + 4y2 + y3 ) + d(y3 + 4y4 + y5 ), 3, 3, 1, + d(y2n−1 + 4y2n + y2n+1 ), 3
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440 Higher Engineering Mathematics, y, , Thus, from equation (5):, !, y f(x), , 3, 1, , 2, 1, √ dx ≈ (0.5) [(2.0000 + 1.1547), 3, x, + 4(1.6330 + 1.2649) + 2(1.4142)], 1, = (0.5)[3.1547 + 11.5916, 3, , y2, , y1, , y3, , y4, , + 2.8284], , y2n1, , = 2.929, correct to 3 decimal places., , a, , O, , b, d, , d, , x, , d, , Figure 45.4, , ≈, , 1, d[(y1 + y2n+1 ) + 4(y2 + y4 + · · · + y2n ), 3, + 2(y3 + y5 + · · · + y2n−1 )], , (b) With 8 intervals, each will have a width of, 3−1, , i.e. 0.25 and the ordinates occur at 1.00,, 8, 1.25, 1.50, 1.75, . . . , 3.0. The values of the ordinates are as shown in the table in Problem 2,, page 436., Thus, from equation (5):, !, , 3, 1, , 2, 1, √ dx ≈ (0.25) [(2.0000 + 1.1547), x, 3, + 4(1.7889 + 1.5119 + 1.3333, , i.e. Simpson’s rule states:, !, , b, , y dx ≈, , a, , , , , 1 width of, first + last, ordinate, 3 interval, , , sum of even, +4, ordinates, , , sum of remaining, +2, odd ordinates, , + 1.2060) + 2(1.6330 + 1.4142, + 1.2649)], 1, = (0.25)[3.1547 + 23.3604, 3, , (5), , Note that Simpson’s rule can only be applied when an, even number of intervals is chosen, i.e. an odd number, of ordinates., Use Simpson’s rule with (a) 4, ! 3, 2, intervals, (b) 8 intervals, to evaluate, √ dx,, x, 1, correct to 3 decimal places., , + 8.6242], = 2.928, correct to 3 decimal places., It is noted that the latter answer is exactly the same as, that obtained by integration. In general, Simpson’s rule, is regarded as the most accurate of the three approximate, methods used in numerical integration., , Problem 6., , Problem 7., , Evaluate, !, , 3−1, ,, 4, i.e. 0.5 and the ordinates will occur at 1.0, 1.5,, 2.0, 2.5 and 3.0. The values of the ordinates are as, shown in the table of Problem 1(b), page 436., , (a) With 4 intervals, each will have a width of, , π, 3, 0, , , , 1, 1 − sin2 θ dθ,, 3, , correct to 3 decimal places, using Simpson’s, rule with 6 intervals.
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Numerical integration, π, −0, With 6 intervals, each will have a width of 3, 6, π, ◦, i.e., rad (or 10 ), and the ordinates will occur at, 18, π π π 2π 5π, π, 0, , , ,, ,, and, 18 9 6 9 18, 3, , , 1, 1 − sin2 θ are shown in, Corresponding values of, 3, the table below., θ, , π, 18, , 0, , π, 9, , , , 1 2, 1 − sin θ 1.0000 0.9950 0.9803 0.9574, 3, θ, , (or, , , 1, 1 − sin2 θ, 3, , 40◦), , (or, , π, 3, 0, , ≈, , (or, , 12.0, , 0, , Use Simpson’s rule to determine the approximate, charge in the 12 millisecond period., , From equation (5):, !, , 12.0, , Charge, q =, 0, , 1, i dt ≈ (2.0) [(0 + 0) + 4(3.5, 3, +10.0 + 2.0) + 2(8.2 + 7.3)], , = 62 mC, Now try the following exercise, , 0.9286, , 0.8969, , 0.8660, Exercise 175 Further problems on, Simpson’s rule, In Problems 1 to 5, evaluate the definite integrals, using Simpson’s rule, giving the answers correct, to 3 decimal places., , , , 1 2, 1 − sin θ dθ, 3, 1π , [(1.0000 + 0.8660) + 4(0.9950, 3 18, + 0.9574 + 0.8969), + 2(0.9803 + 0.9286)], , 1π , , [1.8660 + 11.3972 + 3.8178], 3 18, = 0.994, correct to 3 decimal places., =, , 2.0, , 60◦), , From Equation (5), !, , 10.0, , Charge, q, in millicoulombs, is given by, 12.0, q = 0 i dt., , π, 3, , 5π, 18, 50◦), , 7.3, , π, 6, , (or 10◦ ) (or 20◦) (or 30◦), , 2π, 9, , 8.0, , !, , π, 2, , 1., , [1.187], , 0, , !, , 1.6, , 1, dθ (Use 8 intervals), 1 + θ4, , [1.034], , sin θ, dθ, θ, , (Use 8 intervals), , [0.747], , x cos x dx, , (Use 6 intervals), , [0.571], , 2., 0, , !, , 1.0, , 3., 0.2, , Problem 8. An alternating current i has the, following values at equal intervals of, 2.0 milliseconds:, , , (sin x) dx (Use 6 intervals), , !, , π, 2, , 4., 0, , Time (ms), , Current i (A), , 0, , 0, , 2.0, , 3.5, , 4.0, , 8.2, , 6.0, , 10.0, , !, , π, 3, , 5., 0, , 2, , ex sin 2x dx (Use 10 intervals), [1.260], , In Problems 6 and 7 evaluate the definite integrals using (a) integration, (b) the trapezoidal rule,, , 441
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442 Higher Engineering Mathematics, (c) the mid-ordinate rule, (d) Simpson’s rule. Give, answers correct to 3 decimal places., ! 4, 4, dx (Use 6 intervals), 6., 3, 1 x, , (a) 1.875 (b) 2.107, (c) 1.765 (d) 1.916, !, , 6, , 7., 2, , 1, √, dx (Use 8 intervals), (2x − 1), , (a) 1.585 (b) 1.588, (c) 1.583 (d) 1.585, , In Problems 8 and 9 evaluate the definite integrals, using (a) the trapezoidal rule, (b) the mid-ordinate, rule, (c) Simpson’s rule. Use 6 intervals in each, case and give answers correct to 3 decimal places., ! 3, (1 + x 4 ) dx, 8., 0, , (a) 10.194 (b) 10.007, (c) 10.070, !, , 0.7, , 9., 0.1, , 10., , 1, dy, , (1 − y 2 ), , , , (a) 0.677 (b) 0.674, (c) 0.675, , A vehicle starts from rest and its velocity is, measured every second for 8 s, with values as, follows:, time t (s) velocity v (ms−1 ), 0, , 0, , 1.0, , 0.4, , 2.0, , 1.0, , 3.0, , 1.7, , 4.0, , 2.9, , 5.0, , 4.1, , 6.0, , 6.2, , 7.0, , 8.0, , 8.0, , 9.4, , The distance travelled in 8.0 s is given by, 8.0, 0 v dt, Estimate this distance using Simpson’s rule,, giving the answer correct to 3 significant, figures., [28.8 m], 11. A pin moves along a straight guide so that its, velocity v (m/s) when it is a distance x(m), from the beginning of the guide at time t (s) is, given in the table below., t (s), , v (m/s), , 0, , 0, , 0.5, , 0.052, , 1.0, , 0.082, , 1.5, , 0.125, , 2.0, , 0.162, , 2.5, , 0.175, , 3.0, , 0.186, , 3.5, , 0.160, , 4.0, , 0, , Use Simpson’s rule with 8 intervals to determine the approximate total distance travelled, by the pin in the 4.0 s period., [0.485 m]
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Revision Test 13, This Revision Test covers the material contained in Chapters 43 to 45. The marks for each question are shown in, brackets at the end of each question., !, , 3, , 1. Determine the following integrals:, !, !, (a) 5x e2x dx, (b) t 2 sin 2t dt, , (13), , 2. Evaluate correct to 3 decimal places:, ! 4, √, x ln x dx, , 5, dx using (a) integration (b) the, 2, x, 1, trapezoidal rule (c) the mid-ordinate rule, (d) Simpson’s rule. In each of the approximate methods use 8 intervals and give the answers, correct to 3 decimal places., (19), , (10), , 6. An alternating current i has the following values at, equal intervals of 5 ms:, , 5. Evaluate, , 1, , 3. Use reduction formulae to determine:, !, !, 3 3x, (a) x e dx, (b) t 4 sin t dt, !, , π, 2, , 4. Evaluate, formula., , 0, , cos6 x dx, , using, , a, , Time t (ms), , 0 5, , Current i(A) 0 4.8, (13), , reduction, (6), , 10, , 15, , 20, , 9.1 12.7, , 8.8, , 25, 3.5, , 30, 0, , Charge q, in coulombs, is given by, 30×10−3, i dt ., q= 0, Use Simpson’s rule to determine the approximate, charge in the 30 ms period., (4)
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Chapter 46, , Solution of first order, differential equations by, separation of variables, 46.1, , Family of curves, , dy, Integrating both sides of the derivative, = 3 with, dx, , respect to x gives y = 3 dx, i.e., y = 3x + c, where c, is an arbitrary constant., y = 3x + c represents a family of curves, each of the, curves in the family depending on the value of c., Examples include y = 3x + 8, y = 3x + 3, y = 3x and, y = 3x − 10 and these are shown in Fig. 46.1., y, , y 5 3x 1 3, , 12, , y 5 3x, , 8, y 5 3x 2 10, , 4, , 28, 212, 216, , Figure 46.1, , Problem 1., , Sketch the family of curves given by, dy, the equation, = 4x and determine the equation of, dx, one of these curves which passes through the point, (2, 3)., , y 5 3x 1 8, , 16, , 24 23 22 21 0, 24, , Each are straight lines of gradient 3. A particular curve, of a family may be determined when a point on the curve, is specified. Thus, if y = 3x + c passes through the point, (1, 2) then 2 = 3(1) + c, from which, c = −1. The equation of the curve passing through (1, 2) is therefore, y = 3x − 1., , 1, , 2, , 3, , 4, , x, , dy, = 4x with respect to x, Integrating both sides of, dx, gives:, !, !, dy, dx = 4x dx, i.e., y = 2x 2 + c, dx, Some members of the family of curves having, an equation y = 2x 2 + c include y = 2x 2 + 15,, y = 2x 2 + 8, y = 2x 2 and y = 2x 2 − 6, and these are, shown in Fig. 46.2. To determine the equation of the, curve passing through the point (2, 3), x = 2 and y = 3, are substituted into the equation y = 2x 2 + c., Thus 3 =2(2)2 + c, from which c = 3 −8 =−5., Hence the equation of the curve passing through the, point (2, 3) is y = 2x2 − 5.
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Solution of first order differential equations by separation of variables, , 6, , y, , 20, , 2x 2, , 30, , y, , y, y, 2, 2x 2, 2x 2 x 2 , 1, 8, 5, , y, , 10, , 4, , 3, , 2, , 1, , 0, , 1, , 2, , 3, , 4, , x, , 10, , Figure 46.2, , Now try the following exercise, Exercise 176, of curves, , Differential equations, , A differential equation is one that contains differential, coefficients., Examples include, (i), , The degree of a differential equation is that of the highest power of the highest differential which the equation, contains after simplification., 2 3 5, d x, dx, Thus, +2, = 7 is a second order differdt 2, dt, ential equation of degree three., Starting with a differential equation it is possible,, by integration and by being given sufficient data to, determine unknown constants, to obtain the original, function. This process is called ‘solving the differential equation’. A solution to a differential equation, which contains one or more arbitrary constants of integration is called the general solution of the differential, equation., When additional information is given so that constants may be calculated the particular solution of the, differential equation is obtained. The additional information is called boundary conditions. It was shown in, Section 46.1 that y = 3x + c is the general solution of, dy, = 3., the differential equation, dx, Given the boundary conditions x = 1 and y = 2, produces the particular solution of y = 3x − 1., Equations which can be written in the form, , Further problems on families, , 1. Sketch a family of curves represented by each, of the following differential equations:, dy, dy, dy, = 6 (b), = 3x (c), = x +2, (a), dx, dx, dx, 2. Sketch the family of curves given by the equady, tion, = 2x + 3 and determine the equation, dx, of one of these curves which passes through, the point (1, 3)., [ y = x 2 + 3x − 1], , 46.2, , 445, , dy, d2 y, dy, = 7x and (ii) 2 + 5, + 2y = 0, dx, dx, dx, , Differential equations are classified according to the, highest derivative which occurs in them. Thus example (i) above is a first order differential equation, and, example (ii) is a second order differential equation., , dy, dy, dy, = f (x),, = f ( y) and, = f (x) · f ( y), dx, dx, dx, can all be solved by integration. In each case it is possible, to separate the y’s to one side of the equation and the x’s, to the other. Solving such equations is therefore known, as solution by separation of variables., , 46.3, , The solution of equations of the, dy, form, = f (x), dx, , dy, A differential equation of the form, = f (x) is solved, dx, by direct integration,, !, i.e., y = f (x) dx, Problem 2. Determine the general solution of, dy, x, = 2 − 4x 3, dx, Rearranging x, , dy, = 2 − 4x 3 gives:, dx, , dy, 2 − 4x 3, 2 4x 3, 2, =, = −, = − 4x 2, dx, x, x, x, x
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446 Higher Engineering Mathematics, Integrating both sides gives:, , ! , 2, y=, − 4x 2 dx, x, 4 3, i.e. y = 2 ln x − x + c,, 3, which is the general solution., Find the particular solution of the, dy, differential equation 5 + 2x = 3, given the, dx, 2, boundary conditions y = 1 when x = 2., 5, , Problem 5., , The bending moment M of the beam, dM, is given by, = −w(l − x), where w and x are, dx, constants. Determine M in terms of x given:, M = 12 wl 2 when x = 0., dM, = −w(l − x) = −wl + wx, dx, , Problem 3., , d y 3 − 2x, 3 2x, dy, + 2x = 3 then, =, = −, dx, dx, 5, 5, 5, , ! , 3 2x, dx, −, Hence y =, 5, 5, 3x x 2, i.e., y=, −, + c,, 5, 5, which is the general solution., Substituting the boundary conditions y = 1 25 and x = 2, to evaluate c gives:, 1 25 = 65 − 45 + c, from which, c = 1, 3x x2, Hence the particular solution is y = − + 1., 5, 5, Since 5, , Problem, 4. Solve the equation, , dθ, = 5, given θ = 2 when t = 1., 2t t −, dt, Rearranging gives:, dθ, 5, dθ, 5, =, and, =t −, dt 2t, dt, 2t, Integrating gives:, , ! , 5, dt, θ=, t−, 2t, t2 5, i.e. θ = − ln t + c,, 2 2, which is the general solution., t−, , −, When θ = 2, t = 1, thus, c = 32 ., Hence the particular solution is:, 2 = 12, , t2 5, 3, − ln t +, 2, 2, 2, 1 2, i.e. θ = (t − 5 ln t + 3), 2, θ=, , 5, 2 ln, , 1 + c from which,, , Integrating with respect to x gives:, M = −wlx +, , wx 2, +c, 2, which is the general solution., , When M = 12 wl 2 , x = 0., w(0)2, 1, +c, Thus wl 2 = −wl(0) +, 2, 2, 1 2, from which, c = wl ., 2, Hence the particular solution is:, w(x)2 1 2, M = −wlx +, + wl, 2, 2, 1 2, 2, i.e. M = w(l − 2lx + x ), 2, 1, or M = w(l − x)2, 2, Now try the following exercise, Exercise 177 Further problems on, dy, equations of the form, = f (x)., dx, In Problems 1 to 5, solve the differential, equations., , sin 4x, dy, = cos 4x − 2x, y=, − x2 +c, 1., dx, 4, , dy, x3, 3, 2. 2x, =3 − x3, +c, y = ln x −, dx, 2, 6, 3., , dy, + x = 3, given y = 2 when x = 1., dx, , x2 1, y = 3x −, −, 2, 2, , 4. 3, , dy, 2, π, + sin θ = 0, given y = when θ =, dθ, 3, 3, , 1, 1, y = cos θ +, 3, 2
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Solution of first order differential equations by separation of variables, , 5., , 1, dy, + 2 = x − 3 , given y = 1 when x = 0., ex, dx, , , , 2, 1 2, x − 4x + x + 4, y=, 6, e, , 6. The gradient of a curve is given by:, dy x2, + = 3x, dx, 2, Find the equation of the curve if it passes, through the point 1, 13 ., , x3, 3, y = x2 − −1, 2, 6, 7. The acceleration, a, of a body is equal to its rate, dv, of change of velocity, . Find an equation for, dt, v in terms of t , given that when t = 0, velocity, v = u., [v = u +at], 8. An object is thrown vertically upwards with, an initial velocity, u, of 20 m/s. The motion, of the object follows the differential equation, ds, = u − gt , where s is the height of the object, dt, in metres at time t seconds and g = 9.8 m/s2 ., Determine the height of the object after 3, seconds if s = 0 when t = 0., [15.9 m], , 46.4, , The solution of equations of the, dy, form, = f ( y), dx, , dy, A differential equation of the form, = f ( y) is initially, dx, dy, rearranged to give dx =, and then the solution is, f ( y), obtained by direct integration,, !, !, dy, i.e., dx =, f ( y), Problem 6. Find the general solution of, dy, = 3 + 2y., dx, Rearranging, , dx =, , dy, = 3 + 2y gives:, dx, , dy, 3 + 2y, , 447, , Integrating both sides gives:, !, !, dy, dx =, 3 + 2y, Thus, by using the substitution u = (3 + 2y) — see, Chapter 39,, x = 12 ln (3 +2y) + c, , (1), , It is possible to give the general solution of a differential, equation in a different form. For example, if c = ln k,, where k is a constant, then:, x = 12 ln(3 + 2y) + ln k,, i.e., or, , 1, , x = ln(3 + 2y) 2 + ln k, , x = ln[k (3 +2y)], , by the laws of logarithms, from which,, , ex = k (3 + 2y), , (2), , (3), , Equations (1), (2) and (3) are all acceptable general, solutions of the differential equation, dy, = 3 + 2y, dx, Problem 7. Determine the particular solution of, dy, 1, ( y 2 − 1) = 3y given that y = 1 when x = 2, dx, 6, Rearranging gives:, , , , 2, y, y −1, 1, dy =, dy, dx =, −, 3y, 3 3y, Integrating gives:, , !, ! , y, 1, dx =, −, dy, 3 3y, y2 1, i.e., − ln y + c,, x=, 6, 3, which is the general solution., When y = 1, x = 2 16 , thus 2 16 = 16 − 13 ln 1 +c, from, which, c = 2., Hence the particular solution is:, x=, , y2, 1, − ln y + 2, 6, 3
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448 Higher Engineering Mathematics, Problem 8. (a) The variation of resistance,, R ohms, of an aluminium conductor with, dR, temperature θ ◦ C is given by, = α R, where α, dθ, is the temperature coefficient of resistance of, aluminium. If R = R0 when θ = 0◦C, solve the, equation for R. (b) If α = 38 ×10−4 /◦C, determine, the resistance of an aluminium conductor at 50◦ C,, correct to 3 significant figures, when its resistance, at 0◦C is 24.0 ., (a), , dy, dR, = α R is of the form, = f ( y), dθ, dx, dR, Rearranging gives: dθ =, αR, Integrating both sides gives:, !, !, dR, dθ =, αR, 1, θ = ln R + c,, i.e., α, which is the general solution., Substituting the boundary conditions R = R0 when, θ = 0 gives:, 1, ln R0 + c, α, 1, from which c = − ln R0, α, Hence the particular solution is, 0=, , 1, 1, 1, ln R − ln R0 = ( ln R − ln R0 ), α, α, α, , , R, 1, R, i.e. θ = ln, or αθ = ln, α, R0, R0, θ=, , Hence eαθ =, , R, from which, R = R0eαθ, R0, , (b) Substituting α = 38 ×10−4 , R0 = 24.0 and θ = 50, into R = R0 eαθ gives the resistance at 50◦ C, i.e., −4, R50 = 24.0 e(38×10 ×50) = 29.0 ohms, , , , 1., , dy, = 2 +3y, dx, , 2., , dy, = 2 cos2 y, dx, , 3. ( y 2 + 2), , 1, x = ln(2 + 3y) + c, 3, [tan y = 2x + c], , dy, 1, = 5y, given y = 1 when x =, dx, 2, 2, y, + 2 ln y = 5x − 2, 2, , 4. The current in an electric circuit is given by, the equation, Ri + L, , di, = 0,, dt, , where L and R are constants. Show that, − Rt, i = I e L , given that i = I when t = 0., 5. The velocity of a chemical reaction is given by, dx, = k(a − x), where x is the amount transdt, ferred in time t , k is a constant and a is, the concentration at time t = 0 when x = 0., Solve the equation and determine x in terms, of t ., [x = a(1 − e−kt )], 6., , (a), , Charge Q coulombs at time t seconds, is given by the differential equation, dQ Q, R, + = 0, where C is the capacidt, C, tance in farads and R the resistance in, ohms. Solve the equation for Q given, that Q = Q 0 when t = 0., , (b) A circuit possesses a resistance of, 250 ×103 and a capacitance of, 8.5 × 10−6 F, and after 0.32 seconds, the charge falls to 8.0 C. Determine, the initial charge and the charge after, 1 second, each correct to 3 significant, figures., −t, , [(a) Q = Q 0 e CR (b) 9.30 C, 5.81 C], Now try the following exercise, Exercise 178 Further problems on, dy, equations of the form, = f ( y), dx, In Problems 1 to 3, solve the differential, equations., , 7. A differential equation relating the difference, in tension T , pulley contact angle θ and coefdT, = μT . When θ = 0,, ficient of friction μ is, dθ, T = 150 N, and μ = 0.30 as slipping starts., Determine the tension at the point of slipping, when θ = 2 radians. Determine also the value, of θ when T is 300 N., [273.3 N, 2.31 rads]
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449, , Solution of first order differential equations by separation of variables, i.e. ( y 2 − 1)2 = Ax, 8. The rate of cooling of a body is given by, dθ, = kθ, where k is a constant. If θ = 60◦C, dt, when t = 2 minutes and θ = 50◦C when, t = 5 minutes, determine the time taken for θ, to fall to 40◦C, correct to the nearest second., [8 m 40 s], , 46.5 The solution of equations of the, dy, form, = f (x) · f ( y), dx, dy, A differential equation of the form, = f (x) · f ( y),, dx, where f (x) is a function of x only and f ( y) is a function, dy, = f (x) dx, and, of y only, may be rearranged as, f ( y), then the solution is obtained by direct integration, i.e., !, !, dy, = f (x) dx, f ( y), Problem 9. Solve the equation 4x y, , 2 ln ( y 2 − 1) = lnx + c, , the, , (1), , by the laws of indices., Separating the variables gives:, dθ, = 2e3t dt,, e−2θ, i.e. e2θ dθ = 2e3t dt, Integrating both sides gives:, !, !, 2θ, e dθ = 2e3t dt, , 1 2θ 2 3t 1, e = e −, 2, 3, 6, or, , 3e2θ = 4e3t − 1, , Problem 11. Find the curve which satisfies the, dy, equation x y = (1 + x 2 ), and passes through the, dx, point (0, 1)., (2), , If in equation (1), c = ln A, where A is a different, constant,, then ln( y 2 − 1)2 = ln x + ln A, i.e. ln( y 2 − 1)2 = ln Ax, , dθ, = 2e3t −2θ = 2(e3t )(e−2θ ),, dt, , general, , ln( y 2 − 1)2 − ln x = c, , 2, ( y − 1)2, from which, ln, =c, x, ( y2 − 1)2, = ec, x, , Problem 10. Determine the particular solution of, dθ, = 2e3t −2θ , given that t = 0 when θ = 0., dt, , 1 0 2 0, e = e +c, 2, 3, 1, 1 2, from which, c = − = −, 2 3, 6, Hence the particular solution is:, , or, , and, , dy, = y2 − 1, dx, , When t = 0, θ = 0, thus:, , Integrating both sides gives:, , ! , ! , 4y, 1, dy =, dx, y2 − 1, x, substitution u = y 2 − 1,, , 4x y, , 1 2θ 2 3t, e = e +c, 2, 3, , Separating the variables gives:, , , 4y, 1, d y = dx, y2 − 1, x, , Using the, solution is:, , Equations (1) to (3) are thus three valid solutions of the, differential equations, , Thus the general solution is:, , dy, = y2 − 1, dx, , (3), , Separating the variables gives:, dy, x, dx =, (1 + x 2 ), y, Integrating both sides gives:, 1, 2, , ln (1 + x 2 ) = ln y + c
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450 Higher Engineering Mathematics, When x = 0, y = 1 thus, c = 0., , 1, 2 ln 1 = ln 1 +c,, , from which,, , Hence the particular solution is 12 ln(1 + x 2 ) = ln y, 1, , 1, , i.e. ln(1 + x 2 ) 2 = ln y, from which, (1 + x 2 ) 2 = y, , Hence the equation of the curve is y = (1 +x2 )., Problem 12. The current i in an electric circuit, containing resistance R and inductance L in series, with a constant voltage source, Eis given by the, di, = Ri. Solve the, differential equation E − L, dt, equation and find i in terms of time t given that, when t = 0, i = 0., In the R − L series circuit shown in Fig. 46.3, the supply, p.d., E, is given by, E = V R + VL, V R = iR and V L = L, Hence, from which, , di, E = iR + L, dt, di, E − L = Ri, dt, , di, dt, , (by making, Chapter 39)., , VR, , substitution, , u = E − Ri,, , see, , 1, When t = 0, i = 0, thus − ln E =c, R, Thus the particular solution is:, −, , t, 1, 1, ln (E − Ri) = − ln E, R, L, R, , Transposing gives:, −, , 1, t, 1, ln (E − Ri) + ln E =, R, R, L, 1, t, [ln E − ln (E − Ri)] =, R, L, , , E, Rt, =, ln, E − Ri, L, , E, Rt, =e L, E − Ri, E − Ri, −Rt, Hence, and, =e L, E, −Rt, Ri = E − Ee L ., , from which, , E − Ri = Ee, , −Rt, L, , and, , Hence current,, i=, , L, , R, , a, , VL, , , , −Rt, E, 1−e L ,, R, , which represents the law of growth of current in an, inductive circuit as shown in Fig. 46.4., , i, E, , Figure 46.3, , i, E, R, , Most electrical circuits can be reduced to a differential, equation., di E − Ri, di, Rearranging E − L = Ri gives =, dt, dt, L, , i RE (1eRt/L ), , and separating the variables gives:, di, dt, =, E − Ri, L, , 0, , Integrating both sides gives:, !, !, di, dt, =, E − Ri, L, , Figure 46.4, , Hence the general solution is:, 1, t, − ln (E − Ri) = + c, R, L, , Time t, , Problem 13., Cv, , For an adiabatic expansion of a gas, , dp, dV, +Cp, = 0,, p, V
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Solution of first order differential equations by separation of variables, , where C p and Cv are constants. Given n =, show that pV n = constant., , Cp, ,, Cv, , Separating the variables gives:, , when y = 1., , Cv, , !, , dV, V, , Cv ln p = −C p ln V + k, , i.e., , Dividing throughout by constant Cv gives:, ln p = −, , Since, , Cp, k, ln V +, Cv, Cv, , Cp, = n, then ln p +n ln V = K ,, Cv, , where K =, i.e., logarithms., , [ y 2 = x 2 − 2 ln x + 3], or ln, , pV n = K ,, , by the laws of, , Hence pV n = e K , i.e. pV n = constant., , Now try the following exercise, Further problems on, dy, equations of the form, = f (x) · f ( y), dx, In Problems 1 to 4, solve the differential equations., dy, = 2y cos x, dx, , 8. The p.d., V , between the plates of a capacitor C charged by a steady voltage E, through a resistor R is given by the equation, dV, + V = E., CR, dt, (a), , Exercise 179, , 1., , dy, 6. Solve x y = (1 − x 2 ), for y, given x = 0, dx, when y = 1., 1, y= , (1 − x 2 ), 7. Determine the equation of the curve which, dy, satisfies the equation x y = x 2 − 1, and, dx, which passes through the point (1, 2)., , k, ., Cv, , ln p +ln V n = K, , dy, = 0, given x = 1, dx, [ln (x 2 y) = 2x − y − 1], , 5. Show that the solution of the equation, y2 + 1 y d y, =, is of the form, x 2 + 1 x dx, , 2, y +1, = constant., x2 +1, , Integrating both sides gives:, dp, = −C p, p, , dy, = e2x−y , given x = 0 when y = 0., dx, , 1, 1, e y = e2x +, 2, 2, , 4. 2y(1 − x) + x(1 + y), , dp, dV, Cv = −C p, p, V, , !, , 3., , [ln y = 2 sin x + c], , dy, 2. (2y − 1) = (3x 2 + 1), given x = 1 when, dx, y = 2., [ y 2 − y = x 3 + x], , Solve the equation for V given that at, t = 0, V = 0., , (b) Calculate V , correct to 3 significant, figures, when E =25V, C = 20 ×10−6 F,, R = 200 ×103 and t = 3.0 s., ⎤, ⎡, , −t, C, R, ⎦, ⎣(a) V = E 1 − e, (b) 13.2 V, 9. Determine the value of p, given that, dy, x 3 = p − x, and that y = 0 when x = 2 and, dx, when x = 6., [3], , 451
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Chapter 47, , Homogeneous first order, differential equations, 47.1, , Introduction, , Certain first order differential equations are not of the, ‘variable-separable’ type, but can be made separable by, changing the variable., dy, An equation of the form P, = Q, where P and Q are, dx, functions of both x and y of the same degree throughout,, is said to be homogeneous in y and x. For example, f (x, y) = x 2 + 3x y + y 2 is a homogeneous function, since each of the three terms are of degree 2. However,, x2 − y, is not homogeneous since the term, f (x, y) = 2, 2x + y 2, in y in the numerator is of degree 1 and the other three, terms are of degree 2., , (iv) Separate the variables and solve using the method, shown in Chapter 46., y, (v) Substitute v = to solve in terms of the original, x, variables., , 47.3 Worked problems on, homogeneous first order, differential equations, Problem 1. Solve the differential equation:, dy, y − x =x, , given x = 1 when y = 2., dx, Using the above procedure:, , 47.2 Procedure to solve differential, dy, equations of the form P, =Q, dx, (i) Rearrange P, , dy, dy Q, = Q into the form, = ., dx, dx P, , (ii) Make the substitution y = vx (where v is a funcdy, dv, tion of x), from which,, = v(1) + x, , by the, dx, dx, product rule., dy, in the equation, (iii) Substitute for both y and, dx, dy Q, = . Simplify, by cancelling, and an equation, dx P, results in which the variables are separable., , (i) Rearranging y − x = x, , dy, gives:, dx, , dy y − x, =, ,, dx, x, which is homogeneous in x and y., (ii) Let y = vx, then, , dy, dv, =v+x, dx, dx, , (iii) Substituting for y and, v+x, , dy, gives:, dx, , dv vx − x x(v − 1), =, =, =v − 1, dx, x, x
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Homogeneous first order differential equations, (iv) Separating the variables gives:, x, , dv, 1, = v − 1 − v = −1, i.e. dv = − dx, dx, x, , Integrating both sides gives:, !, , !, dv =, , y2, y, gives: 2 = ln x + c, which is, x, 2x, the general solution., , 2, = − ln 1 + c from, When x = 1, y = 2, thus:, 1, which, c = 2, y, Thus, the particular solution is: = − ln x + 2, x, or y = −x(ln x −2) or y = x(2 − ln x), Problem 2. Find the particular solution of the, d y x 2 + y2, equation: x, =, , given the boundary, dx, y, conditions that y = 4 when x = 1., Using the procedure of section 47.2:, d y x 2 + y2, =, gives:, dx, y, , d y x 2 + y2, =, which is homogeneous in x and y, dx, xy, since each of the three terms on the right hand, side are of the same degree (i.e. degree 2)., (ii) Let y = vx then, , dy, dv, =v+ x, dx, dx, , (iii) Substituting for y and, d y x 2 + y2, =, gives:, dx, xy, v+x, , dy, in the equation, dx, , dv, 1 + v2, x 2 + v2 x 2, x 2 + v2 x 2, =, =, =, dx, x(vx), vx 2, v, , (iv) Separating the variables gives:, x, , 1, Hence, v dv = dx, x, Integrating both sides gives:, !, !, 1, v2, v dv =, dx i.e., = ln x + c, x, 2, (v) Replacing v by, , 1, − dx, x, , Hence, v = −ln x + c, y, y, (v) Replacing v by gives: = −ln x + c, which is, x, x, the general solution., , (i) Rearranging x, , 453, , dv 1 + v 2, 1 + v2 − v2 1, =, −v=, =, dx, v, v, v, , When x = 1, y = 4, thus:, which, c = 8, , 16, = ln 1 + c from, 2, , Hence, the particular solution is:, or y2 = 2x2 (8 + lnx), , y2, = ln x + 8, 2x 2, , Now try the following exercise, , Exercise 180 Further problems on, homogeneous first order differential, equations, dy, 1. Find the general solution of: x 2 = y 2 ., dx, 3, , , x − y3, 1, = ln x + c, − ln, 3, x3, 2. Find the general solution of:, dy, = 0., [y = x(c − ln x)], x − y+x, dx, 3. Find the particular solution of the differential equation: (x 2 + y 2 )d y = x y dx, given that, x = 1 when y = 1., , , , 1, 2, 2, x = 2y ln y +, 2, x + y dy, = ., y − x dx, ⎤, ⎡, , , 1, 2y y 2, ⎣ − 2 ln 1 + x − x 2 = ln x + c ⎦, or x 2 + 2x y − y 2 = k, , 4. Solve the differential equation:, , 5. Find the particular, solution, of the differential, , , 2y − x d y, equation:, = 1 given that y = 3, y + 2x dx, when x = 2., [x 2 + x y − y 2 = 1]
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Homogeneous first order differential equations, y, gives:, x, , 2, y, ln 2 − 1 = ln x + c,, x, , (v) Replacing v by, , which is the general solution., , , 9, When y = 3, x = 1, thus: ln, − 1 = ln 1 +c, 1, from which, c = ln 8, Hence, the particular solution is:, , 2, y, ln 2 − 1 = ln x + ln 8 = ln 8x, x, by the laws of logarithms, 2, , y, y2, Hence,, −, 1, =, 8x, i.e., = 8x + 1 and, x2, x2, y 2 = x 2 (8x + 1), √, i.e., y = x (8x + 1), , Now try the following exercise, Exercise 181 Further problems on, homogeneous first order differential, equations, 1. Solve the differential equation:, x y 3 d y = (x 4 + y 4 )dx., , y 4 = 4x 4 (ln x + c), , dy, = 11y 2 − 16x y + 3x 2 ., dx, , , , , , y−x, 1 3, 13y − 3x, − ln, ln, 5 13, x, x, , 2. Solve: (9x y − 11x y), , = ln x + c, 3. Solve the differential equation:, dy, 2x, = x + 3y, given that when x = 1, y = 1., dx, , , (x + y)2 = 4x 3, 4. Show that the solution of the differential equady, = x 2 + y 2 can be expressed as:, tion: 2x y, dx, x = K(x 2 − y 2 ), where K is a constant., 5. Determine the particular solution of, d y x 3 + y3, , given that x = 1 when y = 4., =, dx, x y2, , , y 3 = x 3 (3 ln x + 64), 6. Show that the solution of the differential equad y y 3 − x y 2 − x 2 y − 5x 3, tion:, is of the, =, dx, x y 2 − x 2 y − 2x 3, form:, , , y2, 4y, y − 5x, = ln x + 42,, +, +, 18, ln, 2x 2, x, x, when x = 1 and y = 6., , 455
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Chapter 48, , Linear first order differential, equations, 48.1, , Integrating both sides gives:, , Introduction, , !, , dy, An equation of the form, + P y = Q, where P and, dx, Q are functions of x only is called a linear differential equation since y and its derivatives are of the first, degree., dy, + P y = Q is obtained by, (i) The solution of, dx, multiplying throughout by what is termed an, integrating factor., dy, (ii) Multiplying, + P y = Q by say R, a function, dx, of x only, gives:, R, , dy, + RPy = RQ, dx, , (1), , (iii) The differential coefficient of a product Ry is, obtained using the product rule,, dy, dR, d, (Ry) = R, +y, ,, i.e., dx, dx, dx, which is the same as the left hand side of, equation (1), when R is chosen such that, RP =, , dR, dx, , dR, = RP, then separating the variables, dx, dR, gives, = P dx., R, , (iv) If, , dR, =, R, , !, , !, P dx i.e. ln R =, , P dx + c, , from which,, , , R=e, , , i.e. R = Ae, , P dx+c, , P dx ,, , , , =e, , P dx c, , e, , where A = ec = a constant., , , (v) Substituting R = Ae P dx in equation (1) gives:, , , , , P dx d y, + Ae P dx P y = Ae P dx Q, Ae, dx, , , , , , dy, + e P dx P y = e P dx Q (2), i.e. e P dx, dx, (vi) The left hand side of equation (2) is, d P dx , ye, dx, which, may be checked by differentiating, , ye P dx with respect to x, using the product rule., (vii) From equation (2),, , d P dx , ye, = e P dx Q, dx, Integrating both sides gives:, ! , , ye P dx = e P dxQ dx, , , (viii) e, , P dx, , is the integrating factor., , (3)
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Linear first order differential equations, , 48.2 Procedure to solve differential, equations of the form, dy, + Py = Q, dx, (i) Rearrange the differential equation into the form, dy, + P y = Q, where P and Q are functions of x., dx, , (ii) Determine P dx., , , (iii) Determine the integrating factor e, , , (iv) Substitute e, , P dx, , P dx ., , into equation (3)., , (v) Integrate the right hand side of equation (3) to, give the general solution of the differential equation. Given boundary conditions, the particular, solution may be determined., , 48.3 Worked problems on linear first, order differential equations, 1 dy, Problem 1. Solve, + 4y = 2 given the, x dx, boundary conditions x = 0 when y = 4., Using the above procedure:, dy, (i) Rearranging gives, + 4x y = 2x, which is of the, dx, dy, form, + P y = Q where P = 4x and Q = 2x., dx, , , (ii), Pdx = 4xdx = 2x 2 ., , , (iii) Integrating factor e, , P dx, , (iv) Substituting into equation (3) gives:, !, 2, 2, ye2x = e2x (2x) dx, (v) Hence the general solution is:, 2, , 2, , ye2x = 12 e2x + c,, by using the substitution u = 2x 2 When x = 0,, y = 4, thus 4e0 = 12 e0 + c, from which, c = 72 ., Hence the particular solution is, 2, , 2, , ye2x = 12 e2x +, , 7, 2, , Problem 2. Show that the solution of the equation, dy, y, 3 −x2, + 1 =− is given by y =, , given, dx, x, 2x, x = 1 when y = 1., Using the procedure of Section 48.2:, , 1, dy, y = −1, which is, +, (i) Rearranging gives:, dx, x, dy, 1, of the form, + P y = Q, where P = and, dx, x, Q = −1. (Note that Q can be considered to be, −1x 0 , i.e. a function of x)., !, !, 1, (ii), P dx =, dx = ln x., x, , , (iii) Integrating factor e P dx = eln x = x (from the definition of logarithm)., (iv) Substituting into equation (3) gives:, !, yx = x(−1) dx, (v) Hence the general solution is:, yx =, , , , 2, 2, or y = 12 + 72 e−2x or y = 12 1 +7e−2x, , −x 2, +c, 2, , −1, + c, from, When x = 1, y = 1, thus 1 =, 2, 3, which, c =, 2, Hence the particular solution is:, yx =, , 2, , = e2x ., , 457, , i.e., , −x 2 3, +, 2, 2, , 2yx = 3 − x 2 and y =, , 3 − x2, 2x, , Problem 3. Determine the particular solution of, dy, − x + y = 0, given that x = 0 when y = 2., dx, Using the procedure of Section 48.2:, dy, (i) Rearranging gives, + y = x, which is of the, dx, dy, form, + P, = Q, where P = 1 and Q = x., dx, (In this case P can be considered to be 1x 0 , i.e. a, function of x)., , , (ii), P dx = 1dx = x., (iii) Integrating factor e, , , , P dx = e x .
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458 Higher Engineering Mathematics, (iv) Substituting in equation (3) gives:, !, ye x = e x (x) dx, (v), , Using the procedure of Section 48.2:, (4), , , , e x (x) dx is determined using integration by, parts (see Chapter 43)., !, xe x dx = xex − e x + c, Hence from equation (4): ye x = xe x − e x + c,, which is the general solution., When x = 0, y = 2 thus 2e0 = 0 − e0 + c, from, which, c = 3., Hence the particular solution is:, yex = xex − ex + 3 or y = x − 1 + 3e−x, , dy, (i) Rearranging gives, − (tan θ)y = sec θ, which is, dθ, dy, of the form, + P y = Q where P = −tan θ and, dθ, Q = sec θ., , , (ii), P dx = − tan θdθ = − ln(sec θ), = ln(sec θ)−1 = ln(cos θ)., , , (iii) Integrating, factor, e P dθ = eln(cosθ) = cos θ, (from the definition of a logarithm)., (iv) Substituting in equation (3) gives:, !, y cos θ = cos θ(sec θ) dθ, !, , Now try the following exercise, Exercise 182 Further problems on linear, first order differential equations, Solve the following differential equations., , dy, c , 1. x, =3− y, y =3+, dx, x, , , dy, 2, = x(1 − 2y), y = 12 + ce−x, 2., dx, , dy, 5t c, 3. t, −5t = −y, y= +, dt, 2 t, , , dy, + 1 = x 3 − 2y, given x = 1 when, 4. x, dx, , 47, x3 x, y =3, y= − +, 5 3 15x 2, , , 1 dy, 2, 5., + y =1, y = 1 +ce−x /2, x dx, , dy, x 1, 6., + x = 2y, y = + + ce2x, dx, 2 4, , 48.4 Further worked problems on, linear first order differential, equations, Problem 4. Solve the differential equation, dy, = sec θ + y tan θ given the boundary conditions, dθ, y = 1 when θ = 0., , i.e., , y cos θ =, , dθ, , (v) Integrating gives: y cos θ = θ + c, which is the, general solution. When θ = 0, y = 1, thus, 1 cos0 = 0 +c, from which, c = 1., Hence the particular solution is:, y cos θ = θ + 1 or y = (θ + 1) sec θ, Problem 5., (a) Find the general solution of the equation, (x − 2), (b), , (a), , d y 3(x − 1), +, y =1, dx, (x + 1), , Given the boundary conditions that y = 5 when, x = −1, find the particular solution of the, equation given in (a)., Using the procedure of Section 48.2:, (i) Rearranging gives:, dy, 3(x − 1), 1, +, y=, dx (x + 1)(x − 2), (x − 2), which is of the form, dy, 3(x − 1), + P y = Q, where P =, dx, (x + 1)(x − 2), 1, and Q =, (x − 2), !, !, 3(x − 1), (ii), P dx =, dx, which is, (x + 1)(x − 2), integrated using partial fractions.
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Linear first order differential equations, Let, , 3x − 3, (x + 1)(x − 2), , Now try the following exercise, , A, B, +, (x + 1) (x − 2), A(x − 2) + B(x + 1), ≡, (x + 1)(x − 2), ≡, , from which, 3x − 3 = A(x − 2) + B(x + 1), When x = −1,, −6 = −3 A, from which, A = 2, , 3 = 3B, from which, B = 1, Hence, , P dx, , = 2 ln (x + 1) + ln (x − 2), , 4. Show that the solution of the differential, equation, , = eln[(x+1), , 2 (x−2)], , = (x + 1)2 (x − 2), , (iv) Substituting in equation (3) gives:, 2, , y(x + 1) (x − 2), !, 1, dx, = (x + 1)2 (x − 2), x −2, !, = (x + 1)2 dx, (v) Hence the general solution is:, y(x + 1)2 (x − 2) = 13 (x + 1)3 + c, (b) When x = −1, y = 5 thus 5(0)(−3) = 0 + c, from, which, c = 0., Hence y(x + 1)2 (x − 2) = 13 (x + 1)3, i.e. y =, , (x + 1)3, 3(x + 1)2 (x − 2), , and hence the particular solution is, y=, , dy, 2, =, − y show, dx x + 2, 2, that the particular solution is y = ln (x + 2),, x, given the boundary conditions that x = −1, when y = 0., , 3. Given the equation x, , (iii) Integrating factor, , , dθ, + sec t (t sin t + cos t )θ = sec t , given, dt, , 1, t = π when θ = 1., θ = (sin t − π cos t ), t, , 3x − 3, dx, (x + 1)(x − 2), ! , 1, 2, dx, +, =, x +1 x −2, , = ln [(x + 1)2 (x − 2)], , e, , In problems 1 and 2, solve the differential equations., π, dy, = 1 − 2y, given y = 1 when x = ., 1. cot x, dx, 4, [y = 12 + cos2 x], 2. t, , When x = 2,, , !, , Exercise 183 Further problems on linear, first order differential equations, , (x + 1), 3(x − 2), , 4, dy, − 2(x + 1)3 =, y, dx, (x + 1), is y = (x + 1)4 ln (x + 1)2 , given that x = 0, when y = 0., 5. Show that the solution of the differential, equation, dy, + ky = a sin bx, dx, is given by:, , , a, y=, (k sin bx − b cos bx), k2 + b2, , 2, k + b2 + ab −kx, e ,, +, k2 + b2, given y = 1 when x = 0., dv, 6. The equation, = −(av + bt ), where a and, dt, b are constants, represents an equation of, motion when a particle moves in a resisting, medium. Solve the equation for v given that, v = u when t = 0., , , , b bt, b, v = 2 − + u − 2 e−at, a, a, a, , 459
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460 Higher Engineering Mathematics, 7. In an alternating current circuit containing, resistance R and inductance L the current, di, i is given by: Ri + L = E 0 sin ωt . Given, dt, i = 0 when t = 0, show that the solution of the, equation is given by:, , , E0, i=, (R sin ωt − ωL cosωt ), R 2 + ω2 L 2, , , E 0 ωL, e− Rt /L, +, R 2 + ω2 L 2, 8. The concentration, C, of impurities of an oil, purifier varies with time t and is described by, the equation, , dC, a, = b + dm − Cm, where a, b, d and m are, dt, constants. Given C = c0 when t = 0, solve the, equation and show that:, , , b, C=, + d (1 − e−mt /a ) + c0 e−mt /a, m, 9. The equation of motion of a train is given, dv, by: m = mk(1 − e−t ) − mcv, where v is the, dt, speed, t is the time and m, k and c are constants. Determine the speed, v, given v = 0 at, t = 0., , , , 1, e−t, e−ct, −, +, v=k, c c − 1 c(c − 1)
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Chapter 49, , Numerical methods for, first order differential, equations, y, , 49.1, , Introduction, , Not all first order differential equations may be solved, by separating the variables (as in Chapter 46) or by the, integrating factor method (as in Chapter 48). A number, of other analytical methods of solving differential equations exist. However the differential equations that can, be solved by such analytical methods is fairly restricted., Where a differential equation and known boundary, conditions are given, an approximate solution may be, obtained by applying a numerical method. There are, a number of such numerical methods available and the, simplest of these is called Euler’s method., , 49.2, , P, f (h), f (0), 0, , x, , h, , Figure 49.1, y, P, f (a 1 x), , f (a), , From Chapter 8, Maclaurin’s series may be stated as:, x2, 2!, , f (0) + · · ·, , Hence at some point f (h) in Fig. 49.1:, h2, f (0) + · · ·, f (h) = f (0) + h f (0) +, 2!, If the y-axis and origin are moved a units to the left,, as shown in Fig. 49.2, the equation of the same curve, , y 5 f (a 1 x), , Q, , Euler’s method, , f (x) = f (0) + x f (0) +, , y 5 f (x ), , Q, , 0, , x, a, , h, , Figure 49.2, , relative to the new axis becomes y = f (a + x) and the, function value at P is f (a)., At point Q in Fig. 49.2:, f (a + h) = f (a) + h f (a) +, , h2 , f (a) + · · ·, 2!, , (1)
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462 Higher Engineering Mathematics, which is a statement called Taylor’s series., If h is the interval between two new ordinates y0 and, y1 , as shown in Fig. 49.3, and if f (a) = y0 and y1 =, f (a + h), then Euler’s method states:, , y1 = y0 + h(y )0 , from equation (2), y1 = 4 + (0.2)(2) = 4.4, since h = 0.2, , Hence, , f (a + h) = f (a) + h f (a), y1 = y0 + h ( y )0, , i.e., , By Euler’s method:, , (2), , At point Q in Fig. 49.4, x 1 = 1.2, y1 = 4.4, and (y )1 = 3(1 + x 1 ) − y1, i.e. ( y )1 = 3(1 + 1.2) − 4.4 = 2.2, , y, , y 5 f (x), , Q, P, , y0, , y1 = y0 + h(y )0 from equation (2), , y1, , (a 1 h), , a, , 0, , If the values of x, y and y found for point Q are, regarded as new starting values of x 0, y0 and (y )0 , the, above process can be repeated and values found for the, point R shown in Fig. 49.5., Thus at point R,, , = 4.4 + (0.2)(2.2) = 4.84, x, , h, , When x 1 = 1.4 and y1 = 4.84,, ( y )1 = 3(1 + 1.4) − 4.84 = 2.36, , Figure 49.3, y, , The approximation used with Euler’s method is to take, only the first two terms of Taylor’s series shown in, equation (1)., Hence if y0 , h and (y )0 are known, y1 , which is an, approximate value for the function at Q in Fig. 49.3,, can be calculated., Euler’s method is demonstrated in the worked problems, following., , Q, , 4.4, P, , 4, , y0, , y1, , x0 5 1, , 0, , 49.3 Worked problems on Euler’s, method, , x1 5 1.2, , x, , h, , Figure 49.4, , Problem 1. Obtain a numerical solution of the, differential equation, , y, , dy, = 3(1 + x) − y, dx, , R, Q, P, , given the initial conditions that x = 1 when y = 4,, for the range x = 1.0 to x = 2.0 with intervals of 0.2., Draw the graph of the solution., dy, = y = 3(1 + x) − y, dx, With x 0 = 1 and y0, , = 4, ( y ), , 0 = 3(1 + 1) − 4 = 2., , y0, , 0, , 1.0, , y1, , x0 5 1.2, , x1 5 1.4, h, , Figure 49.5, , x
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463, , Numerical methods for first order differential equations, This step by step Euler’s method can be continued, and it is easiest to list the results in a table, as shown, in Table 49.1. The results for lines 1 to 3 have been, produced above., , y, , 6.0, , Table 49.1, x0, , (y )0, , y0, , 1., , 1, , 4, , 2, , 2., , 1.2, , 4.4, , 2.2, , 3., , 1.4, , 4.84, , 2.36, , 4., , 1.6, , 5.312, , 2.488, , 5., , 1.8, , 5.8096, , 2.5904, , 6., , 2.0, , 6.32768, , 5.0, , 4.0, 1.0, , 1.2, , 1.4, , 1.6, , 1.8, , 2.0, , x, , Figure 49.6, , For line 4, where x 0 = 1.6:, y1 = y0 + h( y )0, = 4.84 + (0.2)(2.36) = 5.312, and ( y )0 = 3(1 + 1.6) − 5.312 = 2.488, For line 5, where x 0 = 1.8:, y1 = y0 + h(y )0, = 5.312 + (0.2)(2.488) = 5.8096, and ( y )0 = 3(1 + 1.8) − 5.8096 = 2.5904, For line 6, where x 0 = 2.0:, y1 = y0 + h(y )0, , Problem 2. Use Euler’s method to obtain a, numerical solution of the differential equation, dy, + y = 2x, given the initial conditions that at, dx, x = 0, y = 1, for the range x = 0(0.2)1.0. Draw the, graph of the solution in this range., x = 0(0.2)1.0 means that x ranges from 0 to 1.0 in equal, intervals of 0.2 (i.e. h =0.2 in Euler’s method)., dy, + y = 2x,, dx, dy, = 2x − y, i.e. y = 2x − y, hence, dx, If initially x 0 = 0 and y0 = 1, then, ( y )0 = 2(0) − 1 = −1., Hence line 1 in Table 49.2 can be completed with, x = 0, y = 1 and y (0) = −1., , = 5.8096 + (0.2)(2.5904), = 6.32768, , Table 49.2, x0, , (As the range is 1.0 to 2.0 there is no need to calculate, (y )0 in line 6). The particular solution is given by the, value of y against x., dy, A graph of the solution of, = 3(1 + x) − y with initial, dx, conditions x = 1 and y = 4 is shown in Fig. 49.6., In practice it is probably best to plot the graph as, each calculation is made, which checks that there is a, smooth progression and that no calculation errors have, occurred., , y0, , (y )0, , 1., , 0, , 1, , −1, , 2., , 0.2, , 0.8, , −0.4, , 3., , 0.4, , 0.72, , 0.08, , 4., , 0.6, , 0.736, , 0.464, , 5., , 0.8, , 0.8288, , 0.7712, , 6., , 1.0, , 0.98304
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464 Higher Engineering Mathematics, dy, A graph of the solution of, + y = 2x, with initial, dx, conditions x = 0 and y = 1 is shown in Fig. 49.7., , For line 2, where x 0 = 0.2 and h = 0.2:, y1 = y0 + h(y ), from equation (2), = 1 + (0.2)(−1) = 0.8, , Problem 3., , , , and ( y )0 = 2x 0 − y0 = 2(0.2) − 0.8 = −0.4, , (a), , For line 3, where x 0 = 0.4:, y1 = y0 + h(y )0, = 0.8 + (0.2)(−0.4) = 0.72, , Obtain a numerical solution, using, Euler’s method, of the differential equation, dy, = y − x, with the initial conditions that at, dx, x = 0, y = 2, for the range x = 0(0.1)0.5. Draw, the graph of the solution., , (b) By an analytical method (using the integrating, factor method of Chapter 48), the solution of, the above differential equation is given by, y = x + 1 + ex ., , and ( y )0 = 2x 0 − y0 = 2(0.4) − 0.72 = 0.08, For line 4, where x 0 = 0.6:, , Determine the percentage error at x = 0.3, , y1 = y0 + h(y )0, = 0.72 + (0.2)(0.08) = 0.736, , (a), , and ( y )0 = 2x 0 − y0 = 2(0.6) − 0.736 = 0.464, For line 5, where x 0 = 0.8:, y1 = y0 + h(y )0, , dy, = y = y − x., dx, If initially x 0 = 0 and y0 = 2,, then (y )0 = y0 − x 0 = 2 − 0 =2., Hence line 1 of Table 49.3 is completed., , For line 2, where x 0 = 0.1:, , = 0.736 + (0.2)(0.464) = 0.8288, and ( y )0 = 2x 0 − y0 = 2(0.8) − 0.8288 = 0.7712, For line 6, where x 0 = 1.0:, , y1 = y0 + h(y )0 , from equation (2),, = 2 + (0.1)(2) = 2.2, and (y )0 = y0 − x 0, , y1 = y0 + h(y )0, , = 2.2 − 0.1 = 2.1, , = 0.8288 + (0.2)(0.7712) = 0.98304, As the range is 0 to 1.0, ( y )0 in line 6 is not needed., , For line 3, where x 0 = 0.2:, y1 = y0 + h(y )0, = 2.2 + (0.1)(2.1) = 2.41, , y, , and ( y )0 = y0 − x 0 = 2.41 − 0.2 = 2.21, 1.0, , Table 49.3, x0, 0.5, , 0, , Figure 49.7, , 0.2, , 0.4, , 0.6, , 0.8, , 1.0, , x, , y0, , ( y )0, , 1., , 0, , 2, , 2, , 2., , 0.1, , 2.2, , 2.1, , 3., , 0.2, , 2.41, , 2.21, , 4., , 0.3, , 2.631, , 2.331, , 5., , 0.4, , 2.8641, , 2.4641, , 6., , 0.5, , 3.11051
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Numerical methods for first order differential equations, For line 4, where x 0 = 0.3:, , 465, , Percentage error, , , actual − estimated, =, × 100%, actual, , , 2.649859 − 2.631, =, × 100%, 2.649859, , y1 = y0 + h(y )0, = 2.41 + (0.1)(2.21) = 2.631, and ( y )0 = y0 − x 0, = 2.631 − 0.3 = 2.331, , = 0.712%, , For line 5, where x 0 = 0.4:, Euler’s method of numerical solution of differential, equations is simple, but approximate. The method is, most useful when the interval h is small., , y1 = y0 + h(y )0, = 2.631 + (0.1)(2.331) = 2.8641, , , and ( y )0 = y0 − x 0, , Now try the following exercise, , = 2.8641 − 0.4 = 2.4641, For line 6, where x 0 = 0.5:, , Exercise 184, method, , y1 = y0 + h(y )0, = 2.8641 + (0.1)(2.4641) = 3.11051, dy, = y − x with x = 0, y = 2, A graph of the solution of, dx, is shown in Fig. 49.8., (b) If the solution of the differential equation, dy, = y − x is given by y = x + 1 +ex , then when, dx, x = 0.3, y = 0.3 + 1 +e0.3 = 2.649859., , Further problems on Euler’s, , 1. Use Euler’s method to obtain a numerical solution of the differential equation, dy, y, = 3 − , with the initial conditions that, dx, x, x = 1 when y = 2, for the range x = 1.0 to, x = 1.5 with intervals of 0.1. Draw the graph, of the solution in this range. [see Table 49.4], Table 49.4, x, , By Euler’s method, when x = 0.3 (i.e. line 4 in, Table 49.3), y = 2.631., , y, , 1.0, , 2, , 1.1, , 2.1, , 1.2, , 2.209091, , 1.3, , 2.325000, , 1.4, , 2.446154, , 1.5, , 2.571429, , y, , 3.0, , 2. Obtain a numerical solution of the differen1 dy, tial equation, + 2y = 1, given the initial, x dx, conditions that x = 0 when y = 1, in the range, x = 0(0.2)1.0, [see Table 49.5], , 2.5, , 3. (a), 2.0, 0, , Figure 49.8, , 0.1, , 0.2, , 0.3, , 0.4, , 0.5, , x, , y, dy, +1 = −, The differential equation, dx, x, has the initial conditions that y = 1 at, x = 2. Produce a numerical solution of, the differential equation in the range, x = 2.0(0.1)2.5
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466 Higher Engineering Mathematics, Table 49.5, , 49.4, , x, , An improved Euler method, , y, , 0, , 1, , 0.2, , 1, , 0.4, , 0.96, , 0.6, , 0.8864, , 0.8, , 0.793664, , 1.0, , 0.699692, , In Euler’s method of Section 49, the gradient ( y )0 at, P(x0 , y0 ) in Fig. 49.9 across the whole interval h is used, to obtain an approximate value of y1 at point Q. QR in, Fig. 49.9 is the resulting error in the result., y, , (b) If the solution of the differential equation by an analytical method is given, 4 x, by y = − , determine the percentage, x 2, error at x = 2.2, [(a) see Table 49.6 (b) 1.206%], Table 49.6, x, , Q, R, P, , y0, 0, , y, , x, , h, , Figure 49.9, , 2.0, , 1, , 2.1, , 0.85, , 2.2, , 0.709524, , 2.3, , 0.577273, , 2.4, , 0.452174, , 2.5, , 0.333334, , 4. Use Euler’s method to obtain a numerical soludy, 2y, tion of the differential equation, =x − ,, dx, x, given the initial conditions that y = 1 when, x = 2, in the range x = 2.0(0.2)3.0., If the solution of the differential equation is, x2, given by y = , determine the percentage, 4, error by using Euler’s method when x = 2.8, [see Table 49.7, 1.596%], Table 49.7, x, , x1, , x0, , In an improved Euler method, called the Euler-Cauchy, method, the gradient at P(x0 , y0 ) across half the interval, is used and then continues with a line whose gradient, approximates to the gradient of the curve at x 1, shown, in Fig. 49.10., Let y P1 be the predicted value at point R using Euler’s, method, i.e. length RZ, where, yP1 = y0 + h( y )0, , The error shown as QT in Fig. 49.10 is now less, than the error QR used in the basic Euler method and, the calculated results will be of greater accuracy. The, y, Q, T, , y, , 2.0, , 1, , 2.2, , 1.2, , 2.4, , 1.421818, , 2.6, , 1.664849, , 2.8, , 1.928718, , 3.0, , 2.213187, , (3), , R, , P, , S, , Z, 0, , x0, , x0 1 1 h, 2, h, , Figure 49.10, , x1, , x
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Numerical methods for first order differential equations, corrected value, yC1 in the improved Euler method is, given by:, yC1 = y0 +, , 1, 2 h[( y )0 + f (x1 , yP1 )], , (4), , The following worked problems demonstrate how, equations (3) and (4) are used in the Euler-Cauchy, method., Problem 4. Apply the Euler-Cauchy method to, solve the differential equation, dy, = y−x, dx, , For line 3, x 1 = 0.2, y P1 = y0 + h(y )0 = 2.205 + (0.1)(2.105), = 2.4155, yC1 = y0 + 12 h[(y )0 + f (x 1 , y P1 )], = 2.205 + 12 (0.1)[2.105 + (2.4155 − 0.2)], = 2.421025, ( y )0 = yC1 − x 1 = 2.421025 − 0.2 = 2.221025, For line 4, x 1 = 0.3, , in the range 0(0.1)0.5, given the initial conditions, that at x = 0, y = 2., dy, = y = y−x, dx, Since the initial conditions are x 0 = 0 and y0 = 2, then (y )0 = 2 − 0 = 2. Interval h = 0.1, hence, x 1 = x 0 + h = 0 + 0.1 = 0.1., From equation (3),, y P1 = y0 + h(y )0 = 2 + (0.1)(2) = 2.2, , y P1 = y0 + h(y )0, = 2.421025 + (0.1)(2.221025), = 2.6431275, yC1 = y0 + 12 h[(y )0 + f (x 1 , y P1 )], = 2.421025 + 12 (0.1)[2.221025, + (2.6431275 − 0.3)], = 2.649232625, (y )0 = yC1 − x 1 = 2.649232625 − 0.3, , From equation (4),, yC1 = y0 + 12 h[(y )0 + f (x 1 , y P1 )], = y0 +, , 467, , 1, 2 h[(y )0 + (y P1, , − x 1 )],, in this case, , = 2 + 21 (0.1)[2 + (2.2 − 0.1)] = 2.205, (y )1 = yC1 − x 1 = 2.205 − 0.1 = 2.105, If we produce a table of values, as in Euler’s method,, we have so far determined lines 1 and 2 of Table 49.8., The results in line 2 are now taken as x 0 , y0 and (y )0, for the next interval and the process is repeated., , = 2.349232625, For line 5, x 1 = 0.4, y P1 = y0 + h(y )0, = 2.649232625 + (0.1)(2.349232625), = 2.884155887, yC1 = y0 + 12 h[(y )0 + f (x 1 , y P1 )], = 2.649232625 + 12 (0.1)[2.349232625, + (2.884155887 − 0.4)], , Table 49.8, x, , y, , y, , 1., , 0, , 2, , 2, , 2., , 0.1, , 2.205, , 2.105, , 3., , 0.2, , 2.421025, , 2.221025, , 4., , 0.3, , 2.649232625, , 2.349232625, , 5., , 0.4, , 2.89090205, , 2.49090205, , 6., , 0.5, , 3.147446765, , = 2.89090205, (y )0 = yC1 − x 1 = 2.89090205 − 0.4, = 2.49090205, For line 6, x 1 = 0.5, y P1 = y0 + h(y )0, = 2.89090205 + (0.1)(2.49090205), = 3.139992255
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468 Higher Engineering Mathematics, Table 49.10, , yC1 = y0 + 12 h[(y )0 + f (x 1 , y P1 )], , x, , Error in, Euler method, , = 2.89090205 + 12 (0.1)[2.49090205, + (3.139992255 − 0.5)], = 3.147446765, Problem 4 is the same example as Problem 3 and, Table 49.9 shows a comparison of the results, i.e. it, compares the results of Tables 49.3 and 49.8., dy, = y − x may be solved analytically by the intedx, grating factor method of Chapter 48 with the solution, y = x + 1 +ex . Substituting values of x of 0, 0.1, 0.2, . . ., give the exact values shown in Table 49.9., The percentage error for each method for each value of, x is shown in Table 49.10. For example when x = 0.3,, % error with Euler method, , 0, , 0, , 0, , 0.1, , 0.234%, , 0.00775%, , 0.2, , 0.472%, , 0.0156%, , 0.3, , 0.712%, , 0.0236%, , 0.4, , 0.959%, , 0.0319%, , 0.5, , 1.214%, , 0.0405%, , Problem 5. Obtain a numerical solution of the, differential equation, dy, = 3(1 + x) − y, dx, , , actual − estimated, × 100%, =, actual, , , 2.649858808 − 2.631, × 100%, =, 2.649858808, , , in the range 1.0(0.2)2.0, using the Euler-Cauchy, method, given the initial conditions that x = 1 when, y = 4., , = 0.712%, , This is the same as Problem 1 on page 462, and a, comparison of values may be made., , % error with Euler-Cauchy method, , =, , Error in, Euler-Cauchy method, , , 2.649858808 − 2.649232625, × 100%, 2.649858808, , dy, = y = 3(1 + x) − y i.e. y = 3 + 3x − y, dx, x 0 = 1.0, y0 = 4 and h = 0.2, , = 0.0236%, , (y )0 = 3 + 3x 0 − y0 = 3 + 3(1.0) − 4 = 2, , This calculation and the others listed in Table 49.10, show the Euler-Cauchy method to be more accurate than, the Euler method., , x 1 = 1.2 and from equation (3),, , Table 49.9, x, 1., , 0, , 2., , Euler method, y, , Euler-Cauchy method, y, , Exact value, y = x + 1 + ex, , 2, , 2, , 2, , 0.1, , 2.2, , 2.205, , 2.205170918, , 3., , 0.2, , 2.41, , 2.421025, , 2.421402758, , 4., , 0.3, , 2.631, , 2.649232625, , 2.649858808, , 5., , 0.4, , 2.8641, , 2.89090205, , 2.891824698, , 6., , 0.5, , 3.11051, , 3.147446765, , 3.148721271
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Numerical methods for first order differential equations, (y )1 = 3 + 3x 1 − y P1, , y P1 = y0 + h(y )0 = 4 + 0.2(2) = 4.4, , = 3 + 3(1.6) − 5.351368, , yC1 = y0 + 12 h[(y )0 + f (x 1 , y P1 )], , = 2.448632, , = y0 + 12 h[(y )0 + (3 + 3x 1 − y P1 )], = 4 + 12 (0.2)[2 + (3 + 3(1.2) − 4.4)], = 4.42, (y )1 = 3 + 3x 1 − y P1 = 3 + 3(1.2) − 4.42 = 2.18, Thus the first two lines of Table 49.11 have been, completed., For line 3, x 1 = 1.4, , For line 5, x 1 = 1.8, y P1 = y0 + h(y )0 = 5.351368 + 0.2(2.448632), = 5.8410944, yC1 = y0 + 12 h[(y )0 + (3 + 3x 1 − y P1 )], = 5.351368 + 12 (0.2)[2.448632, + (3 + 3(1.8) − 5.8410944)], , y P1 = y0 + h(y )0 = 4.42 + 0.2(2.18) = 4.856, yC1 = y0 + 12 h[(y )0 + (3 + 3x 1 − y P1 )], = 4.42 +, , 469, , = 5.85212176, (y )1 = 3 + 3x 1 − y P1, , 1, 2 (0.2)[2.18, , = 3 + 3(1.8) − 5.85212176, , + (3 + 3(1.4) − 4.856)], = 4.8724, , = 2.54787824, For line 6, x 1 = 2.0, , (y )1 = 3 + 3x 1 − y P1 = 3 + 3(1.4) − 4.8724, = 2.3276, , y P1 = y0 + h(y )0, = 5.85212176 + 0.2(2.54787824), = 6.361697408, , For line 4, x 1 = 1.6, y P1 = y0 + h(y )0 = 4.8724 + 0.2(2.3276), , yC1 = y0 + 21 h[(y )0 + (3 + 3x 1 − y P1 )], = 5.85212176 + 12 (0.2)[2.54787824, , = 5.33792, , + (3 + 3(2.0) − 6.361697408)], , yC1 = y0 + 12 h[(y )0 + (3 + 3x 1 − y P1 )], , = 6.370739843, , = 4.8724 + 12 (0.2)[2.3276, + (3 + 3(1.6) − 5.33792)], = 5.351368, Table 49.11, x0, , y0, , y0, , 1., , 1.0, , 4, , 2, , 2., , 1.2, , 4.42, , 2.18, , 3., , 1.4, , 4.8724, , 2.3276, , 4., , 1.6, , 5.351368, , 2.448632, , 5., , 1.8, , 5.85212176, , 2.54787824, , 6., , 2.0, , 6.370739847, , Problem 6. Using the integrating factor, method the solution of the differential equation, dy, = 3(1 + x) − y of Problem 5 is y = 3x + e1 − x ., dx, When x = 1.6, compare the accuracy, correct to 3, decimal places, of the Euler and the Euler-Cauchy, methods., When x = 1.6, y = 3x + e1−x = 3(1.6) + e1−1.6 =, 4.8 + e−0.6 = 5.348811636., From Table 49.1, page 463, by Euler’s method, when, x = 1.6, y = 5.312, % error in the Euler method, , , 5.348811636 − 5.312, =, × 100%, 5.348811636, = 0.688%
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470 Higher Engineering Mathematics, From Table 49.11 of Problem 5, by the Euler-Cauchy, method, when x = 1.6, y = 5.351368, % error in the Euler-Cauchy method, , , 5.348811636 − 5.351368, × 100%, =, 5.348811636, = −0.048%, The Euler-Cauchy method is seen to be more accurate, than the Euler method when x = 1.6., , for the range x = 0 to x = 0.5 in increments of 0.1, given the initial conditions, that when x = 0, y = 1, (b) The solution of the differential equation, in part (a) is given by y = 2ex − x − 1., Determine the percentage error, correct to, 3 decimal places, when x = 0.4, [(a) see Table 49.13 (b) 0.117%], , Now try the following exercise, Table 49.13, Exercise 185 Further problems on an, improved Euler method, , x, 0, , 1, , 1, , 1. Apply the Euler-Cauchy method to solve the, differential equation, , 0.1, , 1.11, , 1.21, , 0.2, , 1.24205, , 1.44205, , 0.3, , 1.398465, , 1.698465, , 0.4, , 1.581804, , 1.981804, , 0.5, , 1.794893, , dy, y, = 3−, dx, x, for the range 1.0(0.1)1.5, given the initial, conditions that x = 1 when y = 2., [see Table 49.12], , y, , y, , Table 49.12, x, , y, , y, , 1.0, , 2, , 1, , 1.1, , 2.10454546, , 1.08677686, , 1.2, , 2.216666672, , 1.152777773, , 1.3, , 2.33461539, , 1.204142008, , 1.4, , 2.457142859, , 1.2448987958, , 1.5, , 2.583333335, , 2. Solving the differential equation in Problem 1 by the integrating factor method gives, 3, 1, y = x + . Determine the percentage error,, 2, 2x, correct to 3 significant figures, when x = 1.3, using (a) Euler’s method (see Table 49.4,, page 465), and (b) the Euler-Cauchy method., , 4. Obtain a numerical solution of the differential, equation, 1 dy, + 2y = 1, x dx, using the Euler-Cauchy method in the range, x = 0(0.2)1.0, given the initial conditions that, x = 0 when y = 1., [see Table 49.14], , Table 49.14, x, , y, , y, , 0, , 1, , 0, , 0.2, , 0.99, , −0.196, , [(a) 0.412% (b) 0.000000214%], , 0.4, , 0.958336, , −0.3666688, , 3. (a) Apply the Euler-Cauchy method to solve, the differential equation, dy, −x = y, dx, , 0.6, , 0.875468851, , −0.450562623, , 0.8, , 0.784755575, , −0.45560892, , 1.0, , 0.700467925
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Numerical methods for first order differential equations, 49.5, , The Runge-Kutta method, , The Runge-Kutta method for solving first order differential equations is widely used and provides a high degree, of accuracy. Again, as with the two previous methods, the Runge-Kutta method is a step-by-step process, where results are tabulated for a range of values of x., Although several intermediate calculations are needed, at each stage, the method is fairly straightforward., The 7 step procedure for the Runge-Kutta method,, without proof, is as follows:, dy, = f (x, y) given the, To solve the differential equation, dx, initial condition y = y0 at x = x 0 for a range of values of, x = x 0 (h)x n :, 1. Identify x 0 , y0 and h, and values of x 1 , x 2,, x 3 , . . .., , Using the above procedure:, 1., , 2. k1 = f (x 0 , y0 ) = f (0, 2);, dy, since, = y − x, f (0, 2) =2 − 0 = 2, dx, , , 3. k2 = f x 0 + h , y0 + h k1, 2, 2, , , 0.1, 0.1, = f 0+, ,2+, (2), 2, 2, = f (0.05, 2.1) = 2.1 − 0.05 = 2.05, , , 4., , , , h, h, 3. Evaluate k2 = f x n + , yn + k1, 2, 2, , = 2.1025 − 0.05 = 2.0525, 5., , = f (0.1, 2.20525), = 2.20525 − 0.1 = 2.10525, , h, yn+1 = yn + {k1 + 2k2 + 2k3 + k4 }, 6, 6., , Problem 7. Use the Runge-Kutta method to solve, the differential equation:, dy, =y−x, dx, in the range 0(0.1)0.5, given the initial conditions, that at x = 0, y = 2., , k4 = f (x 0 + h, y0 + hk3 ), = f (0 + 0.1, 2 + 0.1(2.0525)), , 6. Use the values determined from steps 2 to 5 to, evaluate:, , Thus, step 1 is given, and steps 2 to 5 are intermediate, steps leading to step 6. It is usually most convenient to, construct a table of values., The Runge-Kutta method is demonstrated in the following worked problems., , , h, h, k3 = f x 0 + , y0 + k2, 2, 2, , , 0.1, 0.1, = f 0+, ,2+, (2.05), 2, 2, = f (0.05, 2.1025), , , , h, h, 4. Evaluate k3 = f x n + , yn + k2, 2, 2, , 7. Repeat steps 2 to 6 for n = 1, 2, 3, . . ., , x 0 = 0, y0 = 2 and since h = 0.1, and the range is, from x = 0 to x = 0.5, then x 1 = 0.1, x 2 = 0.2, x 3 =, 0.3, x 4 = 0.4, and x 5 = 0.5, , Let n =0 to determine y1 :, , 2. Evaluate k1 = f(x n , yn ) starting with n =0, , 5. Evaluate k4 = f (x n + h, yn + hk3 ), , 471, , h, yn+1 = yn + {k1 + 2k2 + 2k3 + k4 } and when, 6, n = 0:, h, y1 = y0 + {k1 + 2k2 + 2k3 + k4 }, 6, = 2+, , 0.1, {2 + 2(2.05) + 2(2.0525), 6, + 2.10525}, , = 2+, , 0.1, {12.31025} = 2.205171, 6, , A table of values may be constructed as shown in, Table 49.15. The working has been shown for the first, two rows.
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472 Higher Engineering Mathematics, Table 49.15, n, , xn, , k1, , k2, , 0, , 0, , 1, , 0.1, , 2.0, , 2.05, , 2.0525, , 2.10525, , 2.205171, , 2, , 0.2, , 2.105171, , 2.160430, , 2.163193, , 2.221490, , 2.421403, , 3, , 0.3, , 2.221403, , 2.282473, , 2.285527, , 2.349956, , 2.649859, , 4, , 0.4, , 2.349859, , 2.417339, , 2.420726, , 2.491932, , 2.891824, , 5, , 0.5, , 2.491824, , 2.566415, , 2.570145, , 2.648838, , 3.148720, , dy, = y − x, f (0.1, 2.205171), dx, = 2.205171 − 0.1 = 2.105171, 3., , , , h, h, k2 = f x 1 + , y1 + k1, 2, 2, , , 0.1, 0.1, , 2.205171 +, (2.105171), = f 0.1 +, 2, 2, = f (0.15, 2.31042955), = 2.31042955 − 0.15 = 2.160430, , , 4., , , h, h, k3 = f x 1 + , y1 + k2, 2, 2, , , 0.1, 0.1, = f 0.1 +, , 2.205171 +, (2.160430), 2, 2, = f (0.15, 2.3131925) = 2.3131925 − 0.15, = 2.163193, , 5., , yn, , h, y2 = y1 + {k1 + 2k2 + 2k3 + k4 }, 6, , k1 = f (x 1 , y1 ) = f (0.1, 2.205171); since, , , , k4, 2, , Let n =1 to determine y2 :, 2., , k3, , k4 = f (x 1 + h, y1 + hk3 ), = f (0.1 + 0.1, 2.205171 + 0.1(2.163193)), , = 2.205171+, , 0.1, {2.105171+2(2.160430), 6, , + 2(2.163193) + 2.221490}, = 2.205171 +, , 0.1, {12.973907} = 2.421403, 6, , This completes the third row of Table 49.15. In a similar, manner y3 , y4 and y5 can be calculated and the results are, as shown in Table 49.15. Such a table is best produced, by using a spreadsheet, such as Microsoft Excel., This problem is the same as problem 3, page 459 which, used Euler’s method, and problem 4, page 461 which, used the improved Euler’s method, and a comparison of, results can be made., dy, = y − x may be solved, The differential equation, dx, analytically using the integrating factor method of, chapter 48, with the solution:, y = x + 1 +ex, Substituting values of x of 0, 0.1, 0.2, . . ., 0.5 will give, the exact values. A comparison of the results obtained, by Euler’s method, the Euler-Cauchy method and the, Runga-Kutta method, together with the exact values is, shown in Table 49.16., It is seen from Table 49.16 that the Runge-Kutta, method is exact, correct to 5 decimal places., , = f (0.2, 2.421490), = 2.421490 − 0.2 = 2.221490, 6., , h, yn+1 = yn + {k1 + 2k2 + 2k3 + k4 }, 6, and when n = 1:, , Problem 8. Obtain a numerical solution of the, dy, differential equation:, = 3(1 + x) − y in the, dx, range 1.0(0.2)2.0, using the Runge-Kutta, method, given the initial conditions that x = 1.0, when y = 4.0.
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Numerical methods for first order differential equations, , 473, , Table 49.16, , x, , Euler’s, method, y, , Euler-Cauchy, method, y, , Runge-Kutta, method, y, , Exact value, y = x +1 + e x, , 0, , 2, , 2, , 2, , 2, , 0.1, , 2.2, , 2.205, , 2.205171, , 2.205170918, , 0.2, , 2.41, , 2.421025, , 2.421403, , 2.421402758, , 0.3, , 2.631, , 2.649232625, , 2.649859, , 2.649858808, , 0.4, , 2.8641, , 2.89090205, , 2.891824, , 2.891824698, , 0.5, , 3.11051, , 3.147446765, , 3.148720, , 3.148721271, , Using the above procedure:, 1., , x 0 = 1.0, y0 = 4.0 and since h = 0.2, and the, range is from x = 1.0 to x = 2.0, then x 1 = 1.2,, x 2 = 1.4, x 3 = 1.6, x 4 = 1.8, and x 5 = 2.0, , Let n = 0 to determine y1 :, 2., , k1 = f (x 0 , y0 ) = f (1.0, 4.0); since, dy, = 3(1 + x) − y,, dx, , f (1.0, 4.0) = 3(1 + 1.0) − 4.0 = 2.0, , , h, h, 3. k2 = f x 0 + , y0 + k1, 2, 2, , , 0.2, 0.2, = f 1.0 +, , 4.0 +, (2), 2, 2, , 6., , , h, h, 4. k3 = f x 0 + , y0 + k2, 2, 2, , , 0.2, 0.2, , 4.0 +, (2.1), = f 1.0 +, 2, 2, , 2., , k1 = f (x 1 , y1 ) = f (1.2, 4.418733); since, dy, = 3(1 + x) − y, f (1.2, 4.418733), dx, = 3(1 + 1.2) − 4.418733 = 2.181267, , 3., , , , h, h, k2 = f x 1 + , y1 + k1, 2, 2, , , 0.2, 0.2, , 4.418733 +, (2.181267), = f 1.2 +, 2, 2, = f (1.3, 4.636860), = 3(1 + 1.3) − 4.636860 = 2.263140, , = 3(1 + 1.1) − 4.21 = 2.09, , = f (1.0 + 0.2, 4.1 + 0.2(2.09)), , when, , Let n = 1 to determine y2 :, , = f (1.1, 4.21), , 5. k4 = f (x 0 + h, y0 + hk3 ), , and, , h, y1 = y0 + {k1 + 2k2 + 2k3 + k4 }, 6, 0.2, {2.0 + 2(2.1) + 2(2.09) + 2.182}, = 4.0 +, 6, 0.2, {12.562} = 4.418733, = 4.0 +, 6, A table of values is compiled in Table 49.17. The, working has been shown for the first two rows., , = f (1.1, 4.2) = 3(1 + 1.1) − 4.2 = 2.1, , , h, yn+1 = yn + {k1 + 2k2 + 2k3 + k4 }, 6, n = 0:, , , , 4., , , h, h, k3 = f x 1 + , y1 + k2, 2, 2, , , 0.2, 0.2, , 4.418733 +, (2.263140), = f 1.2 +, 2, 2, , = f (1.2, 4.418), , = f (1.3, 4.645047) = 3(1 + 1.3) − 4.645047, , = 3(1 + 1.2) − 4.418 = 2.182, , = 2.254953
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474 Higher Engineering Mathematics, Table 49.17, , 5., , n, , xn, , k1, , k2, , 0, , 1.0, , 1, , 1.2, , 2.0, , 2.1, , 2.09, , 2.182, , 4.418733, , 2, , 1.4, , 2.181267, , 2.263140, , 2.254953, , 2.330276, , 4.870324, , 3, , 1.6, , 2.329676, , 2.396708, , 2.390005, , 2.451675, , 5.348817, , 4, , 1.8, , 2.451183, , 2.506065, , 2.500577, , 2.551068, , 5.849335, , 5, , 2.0, , 2.550665, , 2.595599, , 2.591105, , 2.632444, , 6.367886, , k4 = f (x 1 + h, y1 + hk3 ), = f (1.4, 4.869724) = 3(1 + 1.4) − 4.869724, = 2.330276, h, yn+1 = yn + {k1 + 2k2 + 2k3 + k4 }, 6, n =1:, , and, , when, , h, {k1 + 2k2 + 2k3 + k4 }, 6, 0.2, = 4.418733 +, {2.181267 + 2(2.263140), 6, + 2(2.254953) + 2.330276}, , y2 = y1 +, , = 4.418733 +, , k4, , yn, 4.0, , = f (1.2 + 0.2, 4.418733 + 0.2(2.254953)), , 6., , k3, , 0.2, {13.547729} = 4.870324, 6, , This completes the third row of Table 49.17. In a similar manner y3 , y4 and y5 can be calculated and the, , results are as shown in Table 49.17. As in the previous problem such a table is best produced by using a, spreadsheet., This problem is the same as Problem 1, page 462 which, used Euler’s method, and Problem 5, page 468 which, used the Euler-Cauchy method, and a comparison of, results can be made., dy, = 3(1 + x) − y may be, The differential equation, dx, solved analytically using the integrating factor method, of chapter 48, with the solution:, y = 3x +e1−x, Substituting values of x of 1.0, 1.2, 1.4, . . ., 2.0 will give, the exact values. A comparison of the results obtained, by Euler’s method, the Euler-Cauchy method and the, Runga-Kutta method, together with the exact values is, shown in Table 49.18., It is seen from Table 49.18 that the Runge-Kutta, method is exact, correct to 4 decimal places., , Table 49.18, , x, , Euler’s, method, y, , Euler-Cauchy, method, y, , Runge-Kutta, method, y, , Exact value, y = 3x + e1−x, , 1.0, , 4, , 4, , 4, , 4, , 1.2, , 4.4, , 4.42, , 4.418733, , 4.418730753, , 1.4, , 4.84, , 4.8724, , 4.870324, , 4.870320046, , 1.6, , 5.312, , 5.351368, , 5.348817, , 5.348811636, , 1.8, , 5.8096, , 5.85212176, , 5.849335, , 5.849328964, , 2.0, , 6.32768, , 6.370739847, , 6.367886, , 6.367879441
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Numerical methods for first order differential equations, The percentage error in the Runge-Kutta method, when, say, x = 1.6 is:, , , 5.348811636 − 5.348817, ×100% = −0.0001%, 5.348811636, From Problem 6, page 469, when x = 1.6, the percentage error for the Euler method was 0.688%, and, for the Euler-Cauchy method −0.048%. Clearly, the, Runge-Kutta method is the most accurate of the three, methods., Now try the following exercise, Exercise 186 Further problems on the, Runge-Kutta method, 1. Apply the Runge-Kutta method to solve the, dy, y, differential equation:, = 3 − for the range, dx, x, 1.0(0.1)1.5, given that the initial conditions, that x = 1 when y = 2., [see Table 49.19], Table 49.19, yn, , Table 49.20, n, , xn, , yn, , 0, , 0, , 1.0, , 1, , 0.2, , 0.980395, , 2, , 0.4, , 0.926072, , 3, , 0.6, , 0.848838, , 4, , 0.8, , 0.763649, , 5, , 1.0, , 0.683952, , dy, y, +1 = −, 3. (a) The differential equation:, dx, x, has the initial conditions that y = 1 at, x = 2. Produce a numerical solution of the, differential equation, correct to 6 decimal, places, using the Runge-Kutta method in, the range x = 2.0(0.1)2.5, (b) If the solution of the differential equation by an analytical method is given by:, 4 x, y = − determine the percentage error, x 2, at x = 2.2, [(a) see Table 49.21 (b) no error], , n, , xn, , 0, , 1.0, , 2.0, , 1, , 1.1, , 2.104545, , 2, , 1.2, , 2.216667, , n, , xn, , 3, , 1.3, , 2.334615, , 0, , 2.0, , 1.0, , 4, , 1.4, , 2.457143, , 1, , 2.1, , 0.854762, , 5, , 1.5, , 2.533333, , 2, , 2.2, , 0.718182, , 3, , 2.3, , 0.589130, , 4, , 2.4, , 0.466667, , 5, , 2.5, , 0.340000, , 2. Obtain a numerical solution of the differential, 1 dy, equation:, + 2y = 1 using the Rungex dx, Kutta method in the range x = 0(0.2)1.0, given, the initial conditions that x = 0 when y = 1., [see Table 49.20], , Table 49.21, yn, , 475
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Revision Test 14, This Revision Test covers the material contained in Chapters 46 to 49. The marks for each question are shown in, brackets at the end of each question., 1., 2., , 3., , Determine the equation of the curve which satisfies, dy, the differential equation 2x y = x 2 + 1 and which, dx, passes through the point (1, 2)., (5), , 6., , dV, +V = E, dt, , Solve the equation for V given that when time, t = 0, V = 0., , given the initial conditions that x = 1 when, y = 3, for the range x = 1.0 (0.1) 1.5, (b) Apply the Euler-Cauchy method to the differential equation given in part (a) over the same, range., (c), , (b) Evaluate voltage V when E =50 V, C =10 μF,, R = 200 k and t = 1.2 s., (14), 4., , 5., , Show that the solution to the differential equation:, d y x 2 + y2, =, is of the form, 4x, dx, y √ , √, 3y 2 = x 1 − x 3, given that y = 0 when, x = 1., (12), Show that the solution to the differential equation, dy, + (x sin x + cos x)y = 1, x cos x, dx, , (a) Use Euler’s method to obtain a numerical, solution of the differential equation:, y, dy, = + x2 − 2, dx, x, , A capacitor C is charged by applying a steady voltage E through a resistance R. The p.d. between the, plates, V , is given by the differential equation:, CR, , (a), , is given by: x y = sin x + k cos x where k is a, constant., (11), , dy, + x 2 = 5 given, Solve the differential equation: x, dx, that y = 2.5 when x = 1., (4), , Apply the integrating factor method to, solve the differential equation in part (a), analytically., , (d) Determine the percentage error, correct to 3 significant figures, in each of the two numerical, methods when x = 1.2, (30), 7., , Use the Runge-Kutta method to solve the difdy y, ferential equation:, = + x 2 − 2 in the range, dx x, 1.0(0.1)1.5, given the initial conditions that at, x = 1, y = 3. Work to an accuracy of 6 decimal, places., (24)
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Chapter 50, , Second order differential, equations of the form, d2 y, dy, a dx 2 + b dx + cy = 0, 50.1, , Introduction, , d2 y, dy, An equation of the form a 2 + b + cy = 0, where, dx, dx, a, b and c are constants, is called a linear second, order differential equation with constant coefficients. When the right-hand side of the differential, equation is zero, it is referred to as a homogeneous, differential equation. When the right-hand side is not, equal to zero (as in Chapter 51) it is referred to as a, non-homogeneous differential equation., There are numerous engineering examples of second, order differential equations. Three examples are:, (i), , dq, 1, d2q, + q = 0, representing an equaL 2 +R, dt, dt, C, tion for charge q in an electrical circuit containing, resistance R, inductance L and capacitance C in, series., , ds, d2 s, (ii) m 2 + a + ks = 0, defining a mechanical sysdt, dt, tem, where s is the distance from a fixed point, after t seconds, m is a mass, a the damping factor, and k the spring stiffness., P, d2 y, +, y = 0, representing an equation for the, (iii), 2, dx, EI, deflected profile y of a pin-ended uniform strut, , of length l subjected to a load P. E is Young’s, modulus and I is the second moment of area., d2, d, If D represents, and D2 represents 2 then the above, dx, dx, equation may be stated as, (aD2 + bD + c)y = 0. This equation is said to be in, ‘D-operator’ form., dy, d2 y, = Amem x and 2 = Am 2 em x ., If y = Aem x then, dx, dx, d2 y, dy, Substituting these values into a 2 + b + cy = 0, dx, dx, gives:, a(Am 2 em x ) + b(Amem x ) + c(Aem x ) = 0, i.e., , Aem x (am 2 + bm + c) = 0, , Thus y = Aem x is a solution of the given equation, provided that (am 2 + bm +c) = 0. am 2 + bm + c = 0 is, called the auxiliary equation, and since the equation is, a quadratic, m may be obtained either by factorizing or, by using the quadratic formula. Since, in the auxiliary, equation, a, b and c are real values, then the equation, may have either, (i) two different real roots (when b2 > 4ac) or, (ii) two equal real roots (when b 2 = 4ac) or, (iii) two complex roots (when b2 < 4ac).
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478 Higher Engineering Mathematics, Using the above procedure:, , 50.2, , Procedure to solve differential, equations of the form, d2 y, dy, a 2 + b + cy = 0, dx, dx, , (a) Rewrite the differential equation, dy, d2 y, a 2 +b, + cy = 0, dx, dx, as, , (aD2 + bD + c)y = 0, , (b) Substitute m for D and solve the auxiliary equation, am 2 + bm + c = 0 for m., (c) If the roots of the auxiliary equation are:, (i) real and different, say m = α and m = β,, then the general solution is, y = Aeαx + Beβx, (ii) real and equal, say m = α twice, then the, general solution is, y = (Ax + B)eαx, (iii) complex, say m = α ± jβ, then the general, solution is, y = eαx {A cosβx + B sinβx}, (d) Given boundary conditions, constants A and B,, may be determined and the particular solution, of the differential equation obtained., The particular solutions obtained in the worked problems of Section 50.3 may each be verified by substid2 y, dy, and 2 into the original, tuting expressions for y,, dx, dx, equation., , dy, d2 y, + 5 − 3y = 0 in D-operator form is, dx 2, dx, d, (2D2 + 5D − 3)y = 0, where D ≡, dx, (b) Substituting m for D gives the auxiliary equation, (a) 2, , 2m 2 + 5m − 3 = 0., Factorising gives: (2m − 1)(m + 3) = 0, from, which, m = 12 or m = −3., (c) Since the roots are real and different the general, 1, solution is y = Ae 2 x + Be−3x ., (d) When x = 0, y = 4,, hence, , 4= A+ B, , Since, , y = Ae 2 x + Be−3x, , (1), , 1, , dy 1 1 x, = Ae 2 − 3Be−3x, dx 2, dy, When, x = 0,, =9, dx, 1, (2), thus, 9 = A − 3B, 2, Solving the simultaneous equations (1) and (2), gives A = 6 and B = −2., then, , Hence the particular solution is, y = 6e 2 x − 2e−3x, 1, , Problem 2. Find the general solution of, d2 y, dy, 9 2 − 24 + 16y = 0 and also the particular, dt, dt, solution given the boundary conditions that when, dy, t = 0, y =, = 3., dt, Using the procedure of Section 50.2:, , 50.3, , Worked problems on, differential equations of, d2 y, dy, the form a 2 + b + cy = 0, dx, dx, , Problem 1. Determine the general solution of, d2 y, dy, 2 2 + 5 − 3y = 0. Find also the particular, dx, dx, dy, solution given that when x = 0, y = 4 and, = 9., dx, , d2 y, dy, − 24 + 16y = 0 in D-operator form is, dt 2, dt, d, 2, (9D − 24D +16)y = 0 where D ≡, dt, (b) Substituting m for D gives the auxiliary equation, 9m 2 − 24m + 16 =0., (a) 9, , Factorizing gives: (3m − 4)(3m − 4) = 0, i.e., m = 43 twice., (c) Since the roots are real and equal, the general, 4, solution is y = (At +B)e 3 t .
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2, , 479, , Second order differential equations of the form a ddxy2 + b dy, dx + cy = 0, (d) When t = 0, y = 3 hence 3 = (0 + B)e0, i.e. B = 3., , then, , 4, , Since y = (At + B)e 3 t, , , 4, dy, 4 4t, 3, then, = (At + B), e, + Ae 3 t , by the, dt, 3, product rule., dy, When t = 0,, =3, dt, 4, thus 3= (0 + B) e0 + Ae0, 3, , 4, , y = (−t + 3)e 3 t or y = (3 − t)e 3 t, Problem 3. Solve the differential equation, d2 y, dy, + 6 + 13y = 0, given that when x = 0, y = 3, 2, dx, dx, dy, and, = 7., dx, Using the procedure of Section 50.2:, (a), , − 3e−3x (A cos 2x + B sin 2x),, by the product rule,, −3x, , =e, , dy, d2 y, + 6 + 13y = 0 in D-operator form is, dx 2, dx, d, (D2 + 6D + 13)y = 0, where D ≡, dx, , [(2B − 3 A) cos 2x, − (2 A + 3B) sin 2x], , When x = 0,, , 4, i.e. 3 = B + A from which, A = −1, since, 3, B = 3., Hence the particular solution is, 4, , dy, = e−3x (−2 A sin 2x + 2B cos 2x), dx, , dy, = 7,, dx, , hence 7 =e0 [(2B − 3 A) cos 0 − (2 A + 3B) sin 0], i.e. 7 =2B − 3 A, from which, B = 8, since A = 3., Hence the particular solution is, y = e−3x(3 cos 2x + 8 sin 2x), Since, from Chapter 17, page 165,, a cos ωt + b sin ωt = R sin(ωt + α), where, , a, R = (a 2 + b2) and α = tan −1 then, b, 3 cos2x + 8 sin 2x, , = (32 + 82 ) sin(2x + tan−1 38 ), √, = 73 sin(2x + 20.56◦ ), √, = 73 sin(2x + 0.359), Thus the particular solution may also be, expressed as, √, y = 73 e−3x sin(2x + 0.359), , (b) Substituting m for D gives the auxiliary equation, m 2 + 6m + 13 =0., Now try the following exercise, Using the quadratic formula:, , −6 ± [(6)2 − 4(1)(13)], m=, 2(1), √, −6 ± (−16), =, 2, −6 ± j 4, = −3 ± j 2, i.e. m=, 2, (c), , Since the roots are complex, the general solution is, , Exercise 187 Further problems on, differential equations of the form, dy, d2 y, a 2 + b + cy = 0, dx, dx, In Problems 1 to 3, determine the general solution, of the given differential equations., 1. 6, , y = e−3x (A cos 2x + B sin 2x), (d) When x = 0, y = 3, hence, 3 =e0 (A cos 0 + B sin 0), i.e. A = 3., Since y = e−3x (A cos 2x + B sin 2x), , 2. 4, , d2 y d y, −, − 2y = 0, dt 2 dt, , d2θ, dθ, +4 +θ =0, 2, dt, dt, , , , y = Ae 3 t + Be− 2 t, 2, , 1, , , , , , 1, θ = (At + B)e− 2 t
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2, , Second order differential equations of the form a ddxy2 + b dy, dx + cy = 0, (c), , Since the roots are real and different, the general, solution is, , (d) When t = 0, V = 7 hence 7 = A + B, , t = 0,, , d2 i R di, 1, +, +, i = 0 in D-operator form is, dt 2 L dt LC, , , d, R, 1, 2, i = 0 where D ≡, D + D+, L, LC, dt, , (1), , dV, = Aωeωt − Bωe−ωt, dt, When, , Using the procedure of Section 50.2:, (a), , V = Aeωt + Be−ωt, , (b) The auxiliary equation is m 2 +, , dV, = 3ω,, dt, Hence m =, , 3 = A− B, , i.e., , (2), , Hence the particular solution is, V = 5eωt + 2e−ωt, , and, , m=, , sinh ωt = 12 (eωt − e−ωt ), cosh ωt =, , 1 ωt, 2 (e, , + e−ωt ), , then sinh ωt + cosh ωt = eωt, and, , cosh ωt − sinh ωt = e−ωt from Chapter 5., , Hence the particular solution may also be, written as, V = 5(sinh ωt + cosh ωt ), + 2(cosh ωt − sinh ωt ), i.e. V = (5 + 2) cosh ωt + (5 − 2) sinh ωt, i.e. V = 7 cosh ωt + 3 sinh ωt, Problem 6. The equation, d2i R di, 1, +, +, i =0, dt 2 L dt LC, represents a current i flowing in an electrical circuit, containing resistance R, inductance L and, capacitance C connected in series. If R = 200 ohms,, L =0.20 henry and C = 20 ×10−6 farads, solve the, equation for i given the boundary conditions that, di, when t = 0, i = 0 and = 100., dt, , 2, , When R = 200, L =0.20 and C = 20 ×10−6, then, , From equations (1) and (2), A = 5 and B = 2, , Since, , R, 1, m+, =0, L, LC, , 7, 8 2, , , 1, R, R 8, − 4(1), − ±9, L, L, LC, , 3ω = Aω − Bω,, , thus, , 481, , =, , (c), , 7, 8 , , 200 8, 4, 200 2, 9, −, −, ±, 0.20, 0.20, (0.20)(20 × 10−6 ), 2, −1000 ±, 2, , √, , 0, , = −500, , Since the two roots are real and equal (i.e. −500, twice, since for a second order differential equation there must be two solutions), the general, solution is i = (At +B)e−500t ., , (d) When t = 0, i = 0, hence B = 0, di, = (At + B)(−500e−500t ) + (e−500t )(A),, dt, by the product rule, di, = 100, thus 100 =−500B + A, dt, i.e. A = 100, since B = 0, When t = 0,, , Hence the particular solution is, i = 100te−500t, Problem 7. The oscillations of a heavily damped, pendulum satisfy the differential equation, dx, d2 x, + 6 + 8x = 0, where x cm is the, dt 2, dt, displacement of the bob at time t seconds., The initial displacement, is equal to +4 cm and the, , dx, is 8 cm/s. Solve the, initial velocity i.e., dt, equation for x.
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482 Higher Engineering Mathematics, Using the procedure of Section 50.2:, , from, , 2. A body moves in a straight line so that its, distance s metres from the origin after time, d2 s, t seconds is given by 2 + a2 s = 0, where a, dt, is a constant. Solve the equation for s given, ds, 2π, that s = c and, = 0 when t =, ., dt, a, [s = c cos at ], , (c) Since the roots are real and different, the general, solution is x =Ae−2t + Be−4t ., , 3. The motion of the pointer of a galvanometer, about its position of equilibrium is represented, by the equation, , dx, d2 x, + 6 + 8x = 0 in D-operator form is, (a), 2, dt, dt, d, 2, (D + 6D + 8)x = 0, where D ≡ ., dt, (b) The auxiliary equation is m 2 + 6m + 8 =0., Factorising gives: (m + 2)(m + 4) = 0,, which, m = −2 or m = −4., , (d) Initial displacement means that time t = 0. At this, instant, x = 4., Thus 4 = A + B, , I, , If I , the moment of inertia of the pointer about, its pivot, is 5 ×10−3, K , the resistance due to, friction at unit angular velocity, is 2 × 10−2, and F, the force on the spring necessary to, produce unit displacement, is 0.20, solve the, equation for θ in terms of t given that when, dθ, t = 0, θ = 0.3 and, = 0., dt, [θ = e−2t (0.3 cos 6t + 0.1 sin 6t )], , (1), , Velocity,, dx, = −2 Ae−2t − 4Be−4t, dt, dx, = 8 cm/s when t = 0,, dt, thus, , 8 = −2 A − 4B, , (2), , From equations (1) and (2),, A = 12 and B = −8, Hence the particular solution is, x = 12e−2t − 8e−4t, , 4. Determine an expression for x for a differential, dx, d2 x, equation 2 + 2n + n 2 x = 0 which repredt, dt, sents a critically damped oscillator, given that, dx, at time t = 0, x = s and, = u., dt, [x = {s + (u + ns)t }e−nt ], 5., , i.e. displacement, x = 4(3e−2t − 2e−4t ) cm, , Now try the following exercise, Exercise 188 Further problems on second, order differential equations of the form, dy, d2 y, a 2 + b + cy = 0, dx, dx, 1. The charge, q, on a capacitor in a certain, electrical circuit satisfies the differential equadq, d2 q, tion 2 + 4 + 5q = 0. Initially (i.e. when, dt, dt, dq, t = 0), q = Q and, = 0. Show that the, dt, charge, √ in the circuit can be expressed as:, q = 5 Qe−2t sin(t + 0.464)., , dθ, d2θ, +K, + Fθ = 0., 2, dt, dt, , di 1, d2i, L 2 + R + i = 0 is an equation repredt, dt C, senting current i in an electric circuit. If, inductance L is 0.25 henry, capacitance C, is 29.76 ×10−6 farads and R is 250 ohms,, solve the equation for i given the boundary, di, conditions that when t = 0, i = 0 and = 34., , dt, , 1 −160t, − e−840t, e, i=, 20, , 6. The displacement s of a body in a damped, mechanical system, with no external forces,, satisfies the following differential equation:, 2, , ds, d2 s, + 6 + 4.5s = 0, 2, dt, dt, , where t represents time. If initially, when, ds, t = 0, s = 0 and, = 4, solve the differential, dt, 3, equation for s in terms of t ., [s = 4t e− 2 t ]
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Chapter 51, , Second order differential, equations of the form, d2 y, dy, a dx 2 + b dx, 51.1 Complementary function and, particular integral, If in the differential equation, a, , d2 y, dy, +b, + cy = f (x), 2, dx, dx, , (1), , the substitution y = u + v is made then:, a, , + cy = f (x), The general solution, u, of equation (3) will contain two, unknown constants, as required for the general solution, of equation (1). The method of solution of equation (3), is shown in Chapter 50. The function u is called the, complementary function (C.F.)., If the particular solution, v, of equation (2) can be determined without containing any unknown constants then, y = u +v will give the general solution of equation (1)., The function v is called the particular integral (P.I.)., Hence the general solution of equation (1) is given by:, , d(u + v), d2(u + v), +b, + c(u + v) = f (x), dx 2, dx, , y = C.F. + P.I., , Rearranging gives:, 2, 2, , du, dv, d u, d v, a 2 +b, + cu + a 2 + b +cv, dx, dx, dx, dx, , 51.2, , = f (x), If we let, a, , d2 v, dx 2, , +b, , dv, + cv = f (x), dx, , (i) Rewrite the given differential equation as, (aD2 + bD+ c)y = f (x)., (2), , then, du, d2 u, a 2 +b, + cu = 0, dx, dx, , Procedure to solve differential, equations of the form, d2 y, dy, a 2 + b + cy = f (x), dx, dx, , (3), , (ii) Substitute m for D, and solve the auxiliary, equation am 2 + bm +c = 0 for m., (iii) Obtain the complementary function, u, which, is achieved using the same procedure as in, Section 50.2(c), page 478.
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484 Higher Engineering Mathematics, Table 51.1 Form of particular integral for different functions, Type, , Straightforward cases, ‘Snag’ cases, Try as particular integral: Try as particular integral:, , (a) f (x) = a constant, , v=k, , (b) f (x) = polynomial (i.e., , v = a + bx + cx 2 + · · ·, , f (x) = L + M x + N x 2 +, , v = kx (used when C.F., contains a constant), , See, problem, 1, 2, 3, , ···, , where any of the coefficients, may be zero), (c) f (x) = an exponential function, (i.e. f (x) =, , v = keax, , (i) v = kxeax (used when eax, , Aeax ), , 4, 5, , appears in the C.F.), (ii) v = kx 2 eax (used when eax, and xeax both appear in, the C.F.), , (d) f (x) = a sine or cosine function v = A sin px + B cos px, , v = x(A sin px + B cos px), , (i.e. f (x) = a sin px + b cos px,, , (used when sin px and/or, , where a or b may be zero), , cos px appears in the C.F.), , 6, , 7, 8, , (e) f (x) = a sum e.g., (i), , f (x) = 4x 2 − 3 sin 2x, , (ii), , f (x) = 2 − x + e3x, , 9, (i), , v = ax 2 + bx + c, + d sin 2x + e cos 2x, , (f ) f (x) = a product e.g., f (x) = 2ex, , (ii) v = ax + b + ce3x, v = ex (A sin 2x + B cos 2x), , 10, , cos 2x, , (iv) To determine the particular integral, v, firstly, assume a particular integral which is suggested by f (x), but which contains undetermined coefficients. Table 51.1 gives some, suggested substitutions for different functions, f (x)., (v) Substitute the suggested P.I. into the differential equation (aD2 + bD +c)v = f (x) and, equate relevant coefficients to find the constants, introduced., (vi) The general solution is given by, y = C.F. + P.I., i.e. y = u +v., (vii) Given boundary conditions, arbitrary constants, in the C.F. may be determined and the particular, solution of the differential equation obtained., , 51.3, , Worked problems on, differential equations of the, d2 y, dy, form a 2 + b, + cy = f (x), dx, dx, where f (x) is a constant or, polynomial, , Problem 1. Solve the differential equation, d2 y d y, +, − 2y = 4., dx 2 dx, Using the procedure of Section 51.2:, (i), , d2 y d y, +, − 2y = 4 in D-operator form is, dx 2 dx, (D2 + D − 2)y = 4.
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2, , Second order differential equations of the form a ddxy2 + b dy, dx + cy = f (x), (ii) Substituting m for D gives the auxiliary equation m 2 + m − 2 = 0. Factorizing gives: (m − 1), (m + 2) = 0, from which m = 1 or m = −2., (iii) Since the roots are real and different, the C.F.,, u = Aex + Be−2x ., (iv) Since the term on the right hand side of the, given equation is a constant, i.e. f (x) = 4, let, the P.I. also be a constant, say v = k (see, Table 51.1(a))., (v) Substituting v = k into (D2 + D − 2)v = 4, gives (D2 + D − 2)k = 4. Since D(k) = 0 and, D2 (k) = 0 then −2k = 4, from which, k = −2., Hence the P.I., v = −2., (vi) The general solution is given by y = u + v, i.e., y = Aex + Be−2x − 2., Problem 2. Determine the particular solution of, d2 y, dy, the equation 2 − 3 = 9, given the boundary, dx, dx, dy, conditions that when x = 0, y = 0 and, = 0., dx, , d2 y, dy, − 3 =9, 2, dx, dx, (D2 − 3D)y = 9., , in, , D-operator, , Hence the particular solution is, y = −1 + 1e3x − 3x,, i.e. y = e3x − 3x − 1, , Problem 3. Solve the differential equation, d2 y, dy, 2 2 − 11 + 12y = 3x − 2., dx, dx, Using the procedure of Section 51.2:, dy, d2 y, (i) 2 2 − 11 + 12y = 3x − 2, dx, dx, form is, , in, , D-operator, , (2D2 − 11D + 12)y = 3x − 2., (ii) Substituting m for D gives the auxiliary, equation 2m 2 − 11m + 12 =0. Factorizing gives:, (2m − 3)(m − 4) = 0, from which, m = 32 or, m = 4., (iii) Since the roots are real and different, the C.F.,, 3, , u =Ae 2 x + Be4x, (iv) Since f (x) = 3x − 2 is a polynomial, let the P.I.,, v = ax + b (see Table 51.1(b))., , Using the procedure of Section 51.2:, (i), , 485, , form, , is, , (ii) Substituting m for D gives the auxiliary equation, m 2 − 3m =0. Factorizing gives: m(m − 3) = 0,, from which, m = 0 or m = 3., , (v) Substituting v = ax + b into, (2D2 − 11D +12)v = 3x − 2 gives:, (2D2 − 11D + 12)(ax + b) = 3x − 2,, i.e. 2D2 (ax + b) − 11D(ax + b), + 12(ax + b) = 3x − 2, , (iii) Since the roots are real and different, the C.F.,, u = Ae0 + Be3x , i.e. u = A +Be3x ., , i.e., , (iv) Since the C.F. contains a constant (i.e. A) then let, the P.I., v = kx (see Table 51.1(a))., , Equating the coefficients of x gives: 12a = 3,, from which, a = 14 ., , (v) Substituting v = kx into (D2 − 3D)v = 9 gives, (D2 − 3D)kx = 9., D(kx) = k and D2 (kx) = 0., Hence (D2 − 3D)kx = 0 −3k = 9, from which,, k = −3., Hence the P.I., v = −3x., (vi) The general solution is given by y = u + v, i.e., y = A +Be3x −3x., (vii) When x = 0, y = 0, thus 0 = A + Be0 − 0, i.e., 0= A+ B, (1), dy, d, y, = 3Be3x − 3;, = 0 when x = 0, thus, dx, dx, 0 = 3Be0 − 3 from which, B = 1. From equation (1), A = −1., , 0 − 11a + 12ax + 12b = 3x − 2, , Equating the constant terms gives:, −11a + 12b = −2., , i.e. −11 14 + 12b = −2 from which,, 1, 11 3, = i.e. b =, 4 4, 16, 1, 1, Hence the P.I., v = ax + b = x +, 4, 16, (vi) The general solution is given by y = u + v, i.e., 12b = −2 +, , 3, 1, 1, y = Ae 2 x + Be4x + x +, 4, 16
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490 Higher Engineering Mathematics, Now try the following exercise, given by:, Exercise 191 Further problems on, differential equations of the form, dy, d2 y, a 2 + b + cy = f (x) where f (x) is a sine or, dx, dx, cosine function, , y = e−4t (A cos 2t + B sin 2t ), 15, +, (sin 4t − 8 cos4t ), 13, 7., , In Problems 1 to 3, find the general solutions of the, given differential equations., 1. 2, , d2 y d y, −, − 3y = 25 sin 2x, dx 2 dx, 3, , y = Ae 2 x + Be−x, − 15 (11 sin 2x − 2 cos 2x), 2., , + 0.024 sin 200t − 0.010 cos 200t, , d2 y, dy, − 4 + 4y = 5 cos x, dx 2, dx, y = (Ax + B)e2x − 45 sin x + 35 cos x, , 3., , dq 1, d2q, L 2 + R + q = V0 sin ωt represents the, dt, dt C, variation of capacitor charge in an electric circuit. Determine an expression for, q at time t seconds given that R = 40 ,, L =0.02 H, C = 50 × 10−6 F, V0 = 540.8 V, and ω = 200 rad/s and given the boundary, dq, conditions that when t = 0, q = 0 and, = 4.8, dt, q = (10t + 0.01)e−1000t, , , , d2 y, + y = 4 cos x, dx 2, , 51.6, , [ y = A cos x + B sin x + 2x sin x], 4. Find the particular solution of the differend2 y, dy, tial equation 2 − 3 − 4y = 3 sin x; when, dx, dx, dy, x = 0, y = 0 and, = 0., dx, ⎤, ⎡, 1, 4x − 51e−x ), (6e, y, =, ⎥, ⎢, 170, ⎥, ⎢, ⎦, ⎣, 1, − (15 sin x − 9 cos x), 34, 5. A differential equation representing the, d2 y, + n 2 y = k sin pt ,, motion of a body is, dt 2, where k, n and p are constants. Solve the equation (given n = 0 and p2 = n 2) given that when, dy, t = 0, y =, = 0., dt, , , , k, p, y= 2, sin, nt, sin, pt, −, n − p2, n, 6. The motion of a vibrating mass is given by, d2 y, dy, + 8 + 20y = 300 sin4t . Show that the, 2, dt, dt, general solution of the differential equation is, , Worked problems on, differential equations of the, d2 y, dy, form a 2 + b, + cy = f (x), dx, dx, where f (x) is a sum or a product, , Problem 9. Solve, d2 y d y, +, − 6y = 12x − 50 sin x., dx 2 dx, Using the procedure of Section 51.2:, (i), , d2 y d y, +, − 6y = 12x − 50 sin x in D-operator, dx 2 dx, form is, (D2 + D − 6)y = 12x − 50 sin x, , (ii) The auxiliary equation is (m 2 + m − 6) = 0, from, which,, (m − 2)(m + 3) = 0,, i.e. m = 2 or m = −3, (iii) Since the roots are real and different, the C.F.,, u = Ae2x + Be−3x ., (iv) Since the right hand side of the given differential, equation is the sum of a polynomial and a sine, function let the P.I. v = ax + b + c sin x + d cos x, (see Table 51.1(e)).
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Chapter 52, , Power series methods, of solving ordinary, differential equations, 52.1, , Introduction, , Second order ordinary differential equations that cannot be solved by analytical methods (as shown in, Chapters 50 and 51), i.e. those involving variable coefficients, can often be solved in the form of an infinite series, of powers of the variable. This chapter looks at some of, the methods that make this possible—by the Leibniz–, Maclaurin and Frobinius methods, involving Bessel’s, and Legendre’s equations, Bessel and gamma functions and Legendre’s polynomials. Before introducing, Leibniz’s theorem, some trends with higher differential, coefficients are considered. To better understand this, chapter it is necessary to be able to:, (i) differentiate standard functions (as explained in, Chapters 27 and 32),, (ii) appreciate the binomial theorem (as explained in, Chapters 7), and, (iii) use Maclaurins theorem (as explained in Chapter 8)., , 52.2 Higher order differential, coefficients as series, The following is an extension of successive differentiation (see page 296), but looking for trends, or series,, , as the differential coefficient of common functions, rises., dy, d2 y, = a 2 eax , and so, = aeax ,, (i) If y = eax , then, 2, dx, dx, on., If we abbreviate, , dy, d2 y, as y , … and, as y ,, dx, dx 2, , dn y, as y (n) , then y = aeax , y = a 2eax , and the, dx n, emerging pattern gives:, , y(n) = an eax, , For example, if y = 3e2x , then, d7 y, = y (7) = 3(27 ) e2x = 384e2x, dx 7, (ii) If y = sin ax,, , π, y = a cos ax = a sin ax +, 2, y = −a 2 sin ax = a 2 sin(ax + π), , , 2π, 2, = a sin ax +, 2, y = −a 3 cos x, , , 3π, and so on., = a 3 sin ax +, 2, , (1)
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494 Higher Engineering Mathematics, , , In general, y(n) = an sin ax + nπ, 2, , (2), , For example, if, d5 y, y = sin 3x, then 5 = y (5), dx, , , , 5π, π, 5, = 3 sin 3x +, = 35 sin 3x +, 2, 2, , y(n) =, , y = sinh 2x, then, , (iii) If y = cos ax,, , π, y = −a sin ax = a cos ax +, 2, , , 2π, y = −a 2 cos ax = a 2 cos ax +, 2, , , 3π, 3, 3, and so on., y = a sin ax = a cos ax +, 2, , nπ , y(n) = an cos ax +, 2, , (3), , (5), , (iv) If y = x a, y = a x a−1 , y = a(a − 1)x a−2 ,, y = a(a − 1)(a − 2)x a−3 ,, and y(n) = a(a − 1)(a − 2) . . . . . (a − n + 1) x a−n, , (v) If y = sinh ax, y = a cosh ax, y = a sinh ax, 2, , y = a 3 cosh ax, and so on, , 25, {[0] sinh 2x + [2] cosh 2x}, 2, = 32 cosh 2x, , =, , (vi) If y = cosh ax,, , Since cosh ax is not periodic (see graph on page, 43), again it is more difficult to find a general, statement for y (n) . However, this is achieved with, the following general series:, , = −256 cos 2x, , d4 y, For example, if y = 2x6 , then 4 = y (4), dx, 6!, = (2), x 6−4, (6 − 4)!, 6 × 5 × 4× 3 × 2× 1 2, = (2), x, 2×1, = 720x2, , + [1 − (−1)5 ] cosh 2x}, , y = a 3 sinh ax, and so on, , = 4(26 ) cos (2x + π), , where a is a positive integer., , 25, {[1 + (−1)5 ] sinh 2x, 2, , y = a 2 cosh ax, , , , 6π, (6), 6, then 6 = y = 4(2 ) cos 2x +, dx, 2, 6, = 4(2 ) cos (2x + 3π), d6 y, , a!, xa−n, (a − n)!, , =, , d5 y, = y (5), dx 5, , y = a sinh ax, , For example, if y = 4 cos 2x,, , or y(n) =, , an, {[1 +(−1)n ] sinh ax, 2, + [1 −(−1)n ] cosh ax}, , For example, if, , = 243 cos 3x, , In general,, , Since sinh ax is not periodic (see graph on page, 43), it is more difficult to find a general statement for y (n) . However, this is achieved with the, following general series:, , (4), , y(n) =, , an, {[1 − (−1)n ] sinh ax, 2, + [1 + (−1)n ] cosh ax}, , (6), , 1, For example, if y = cosh 3x,, 9 7, 1 3, d7 y, (7), (2 sinh 3x), then 7 = y =, dx, 9 2, = 243 sinh 3x, 1, 1, 2, (vii) If y = ln ax, y = , y = − 2 , y = 3 , and so, x, x, x, on., In general, y(n) = (−1)n−1, , (n − 1)!, xn, , For example, if y = ln 5x, then, , d6 y, (6) = (−1)6−1 5! = − 120, =, y, dx 6, x6, x6, , (7)
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Power series methods of solving ordinary differential equations, 1, Note that if y = ln x, y = ; if in equation (7),, x, 0 (0)!, n = 1 then y = (−1) 1, x, 1, (−1)0 = 1 and if y = then (0)!= 1 (Check that, x, (−1)0 = 1 and (0)! = 1 on a calculator)., , 52.3, , Leibniz’s theorem, y = uv, , If, , 495, , (8), , where u and v are each functions of x, then by using the, product rule,, y = uv + vu, , (9), , y = uv + v u + vu + u v, , Now try the following exercise, , = u v + 2u v + uv, , Exercise 193 Further problems on higher, order differential coefficients as series, , 1. (a) y (4) when y = e2x (b) y (5) when y, , = u v + 3u v + 3u v + uv, , t, = 8e2, , 1 t, (b) e 2 ], 4, , + 4u (1)v (3) + uv (4), , [(a) 81 sin3t (b) −1562.5 cos5θ], 3. (a) y (8) when y = cos 2x, 2, (b) y (9) when y = 3 cos t, 3, , 2, 29, (a) 256 cos2x (b) − 8 sin t, 3, 3, t7, 8, (b) 630 t ], , 4. (a) y (7) when y = 2x 9 (b) y (6) when y =, [(a) (9! )x 2, , (11), , 1, 5. (a) y (7) when y = sinh 2x, 4, (b) y (6) when y = 2 sinh 3x, [(a) 32 cosh 2x (b) 1458 sinh 3x], 6. (a) y (7) when y = cosh 2x, 1, (b) y (8) when y = cosh 3x, 9, [(a) 128 sinh 2x (b) 729 cosh 3x], , (12), , From equations (8) to (12) it is seen that, (a), , 2. (a) y (4) when y = sin 3t, 1, (b) y (7) when y =, sin 5θ, 50, , 7. (a) y (4) when y = 2ln 3θ, 1, (b) y (7) when y = ln 2t, 3, , , y = u v + vu + 2u v + 2v u + uv + v u, y (4) = u (4)v + 4u (3)v (1) + 6u (2)v (2), , Determine the following derivatives:, , [(a) 16 e2x, , (10), , the n’th derivative of u decreases by 1 moving, from left to right,, , (b) the n’th derivative of v increases by 1 moving from, left to right,, (c), , the coefficients 1, 4, 6, 4, 1 are the normal binomial, coefficients (see page 58)., , In fact, (uv)(n) may be obtained by expanding (u + v)(n), using the binomial theorem (see page 59), where the, ‘powers’ are interpreted as derivatives. Thus, expanding, (u + v)(n) gives:, y(n) = (uv)(n) = u(n) v + nu(n−1) v (1), n(n− 1) (n−2) (2), v, u, 2!, n(n− 1)(n −2) (n−3) (3), +, v +···, u, 3!, +, , (13), , Equation (13) is a statement of Leibniz’s theorem,, which can be used to differentiate a product n times., The theorem is demonstrated in the following worked, problems., Problem 1. Determine y (n) when y = x 2 e3x ., For a product y = uv, the function taken as, , (a) −, , 240, 6, (b) 7, θ4, t, , (i) u is the one whose nth derivative can readily be, determined (from equations (1) to (7)),, (ii) v is the one whose derivative reduces to zero after, a few stages of differentiation.
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496 Higher Engineering Mathematics, Thus, when y = x 2 e3x , v = x 2 , since its third derivative, is zero, and u = e3x since the nth derivative is known, from equation (1), i.e. 3n eax, Using Leinbiz’s theorem (equation (13),, y, , (n), , =u, , (n), , n(n − 1) (n−2) (2), v + nu, v +, v, u, 2!, n(n − 1)(n − 2) (n−3) (3), +, v + ···, u, 3!, , By Leibniz’s equation, equation (13),, , , n(n − 1) (n), y (2)+ 0, y (n+2)(1 + x 2 ) + n y (n+1)(2x)+, 2!, + 2{y (n+1) (x) + n y (n) (1) + 0} − 3{y (n) } = 0, , (n−1) (1), , i.e. (1 + x 2 )y (n+2) + 2n x y (n+1) + n(n − 1)y (n), + 2x y (n+1) + 2 ny (n) − 3y (n) = 0, (1 + x 2 )y (n+2) + 2(n + 1)x y (n+1), , where in this case v = x 2 , v (1) = 2x, v (2) = 2 and, v (3) = 0, , or, , Hence, y (n) = (3n e3x )(x 2 ) + n(3n−1 e3x )(2x), , i.e. (1 + x2 )y(n+2) + 2(n + 1)xy(n+1), , n(n − 1) n−2 3x, (3 e )(2), 2!, n(n − 1)(n − 2) n−3 3x, (3 e )(0), +, 3!, = 3n−2 e3x (32 x 2 + n(3)(2x), , + (n 2 − n + 2n − 3)y (n) = 0, , + (n2 + n − 3)y(n) = 0, , +, , Problem 4., , + n(n − 1) + 0), i.e., , y(n) = e3x 3n−2 (9x2 + 6nx + n(n− 1)), , Problem 2. If x 2 y + 2x y + y = 0 show that:, x y (n+2) + 2(n + 1)x y (n+1) + (n 2 + n + 1)y (n) = 0, Differentiating each term of x 2 y + 2x y + y = 0, n times, using Leibniz’s theorem of equation (13),, gives:, , , y (n+2) x 2 + n y (n+1) (2x) +, , n(n − 1) (n), y (2) + 0, 2!, , , , + {y (n+1) (2x) + n y (n) (2) + 0} + {y (n) } = 0, i.e. x 2 y (n+2) + 2n x y (n+1) + n(n − 1)y (n), + 2x y (n+1) + 2n y (n) + y (n) = 0, i.e. x 2 y (n+2) + 2(n + 1)x y (n+1), + (n 2 − n + 2n + 1)y (n) = 0, or, , Find the 5th derivative of y = x 4 sin x., , If y = x 4 sin x, then using Leibniz’s equation with, u = sin x and v = x 4 gives:, , nπ 4 , y (n) = sin x +, x, 2, , , (n − 1)π, + n sin x +, 4x 3, 2, , , (n − 2)π, n(n − 1), sin x +, 12x 2, +, 2!, 2, , , n(n − 1)(n − 2), (n − 3)π, +, sin x +, 24x, 3!, 2, , n(n − 1)(n − 2)(n − 3), sin x, +, 4!, , (n − 4)π, +, 24, 2, , , 5π, + 20x 3 sin(x + 2π), and y (5) = x 4 sin x +, 2, , , (5)(4), 3π, 2, +, (12x ) sin x +, 2, 2, , x2 y(n+2) + 2(n + 1) x y(n+1), + (n + n + 1)y, 2, , (n), , +, , =0, , , π, (5)(4)(3)(2), (24) sin x +, (4)(3)(2), 2, , , , 5π, π, sin x +, ≡ sin x +, ≡ cos x,, 2, 2, +, , Problem 3. Differentiate the following, differential equation n times:, (1 + x 2 )y + 2x y − 3y = 0., , Since, , (5)(4)(3), (24x) sin (x + π), (3)(2)
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497, , Power series methods of solving ordinary differential equations, , , 3π, sin(x + 2π) ≡ sin x, sin x +, ≡ −cos x,, 2, and, , 52.4 Power series solution by the, Leibniz–Maclaurin method, , sin (x + π) ≡ −sin x,, , then y (5) = x 4 cos x + 20x 3 sin x + 120x 2 (−cos x), + 240x(−sin x) + 120 cos x, i.e. y(5) = (x4 − 120x2 + 120)cos x, + (20x3 − 240x) sin x, , (i) Differentiate the given equation n times, using, the Leibniz theorem of equation (13),, , Now try the following exercise, , (ii) rearrange the result to obtain the recurrence, relation at x = 0,, , Exercise 194 Further problems on, Leibniz’s theorem, Use the theorem of Leibniz in the following, problems:, 1. Obtain the n’th derivative of: x 2 y., x 2 y (n) + 2n x y (n−1) + n(n − 1)y (n−2), 2. If, ⎡, ⎢, ⎣, , y = x 3 e2x, , find, , y (n), , and hence, , y (3) ., , , , ⎤, , y (n) = e2x 2n−3 {8x 3 + 12nx 2, , ⎥, + n(n − 1)(6x) + n(n − 1)(n − 2)} ⎦, , y (3) = e2x (8x 3 + 36x 2 + 36x + 6), 3. Determine the 4th derivative of: y = 2x 3 e−x ., [ y (4) = 2e−x (x 3 − 12x 2 + 36x − 24)], 4. If y = x 3 cos x determine the 5th derivative., [ y (5) = (60x − x 3 ) sin x +, (15x 2 − 60) cos x], 5. Find an expression for y (4) if y = e−t sin t ., [ y (4), , =, , −4 e−t sin t ], , 6. If y = x 5 ln 2x find y (3) ., [ y (3) = x 2 (47 + 60 ln 2x)], 7. Given 2x 2 y + x y + 3y = 0 show that, 2x 2 y (n+2) + (4n + 1)x y (n+1) + (2n 2 − n +, 3)y (n) = 0., 8. If y = (x 3 + 2x 2 )e2x determine an expansion, for y (5)., [ y (5), , =, , e2x 24 (2x 3, , For second order differential equations that cannot be, solved by algebraic methods, the Leibniz–Maclaurin, method produces a solution in the form of infinite, series of powers of the unknown variable. The following simple 5-step procedure may be used in the, Leibniz–Maclaurin method:, , + 19x 2 + 50x, , + 35)], , (iii) determine the values of the derivatives at x = 0,, i.e. find ( y)0 and ( y )0 ,, (iv) substitute in the Maclaurin expansion for, y = f (x) (see page 69, equation (5)),, (v) simplify the result where possible and apply, boundary condition (if given)., The Leibniz–Maclaurin method is demonstrated, using, the above procedure, in the following worked problems., Problem 5. Determine the power series solution, of the differential equation:, dy, d2 y, + x + 2y = 0 using Leibniz–Maclaurin’s, 2, dx, dx, method, given the boundary conditions that at, dy, = 2., x = 0, y = 1 and, dx, Following the above procedure:, (i) The differential equation is rewritten as:, y + x y + 2y = 0 and from the Leibniz theorem, of equation (13), each term is differentiated n, times, which gives:, y (n+2) +{y (n+1) (x)+n y (n) (1)+0}+2 y (n) = 0, i.e., , y (n+2) + x y (n+1) + (n + 2) y (n) = 0, (14), , (ii) At x = 0, equation (14) becomes:, y (n+2) + (n + 2) y (n) = 0, from which, y (n+2) = −(n +2) y (n)
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498 Higher Engineering Mathematics, This equation is called a recurrence relation, or recurrence formula, because each recurring, term depends on a previous term., (iii) Substituting n =0, 1, 2, 3, … will produce, a set of relationships between the various, coefficients., For n =0,, , ( y )0 = −2( y)0, , n =1, ( y )0 = −3( y )0, , (v) Collecting similar terms together gives:, , 2x 2 2 × 4x 4, y = ( y)0 1 −, +, 2!, 4!, 2 × 4 × 6x 6 2 × 4 × 6 × 8x 8, +, 6!, 8!, 5, , 3x 3 3 × 5x 5, +, − · · · + ( y )0 x −, 3!, 5!, , −, , n =2, ( y (4) )0 = −4( y )0 = −4{−2( y)0 }, = 2 × 4( y)0, n =3,, , ( y (5) )0 = −5( y, , )0 = −5{−3( y )0 }, , = 3 × 5( y )0, , −, , n =5, ( y (7) )0 = −7( y (5) )0 = −7{3×5( y )0 }, , +, 5, + ( y )0 ×, , x7, −, +···, 2×4×6, , n =6, ( y (8) )0 = −8( y (6) )0 =, , (iv) Maclaurin’s theorem from page 69 may be, written as:, y = ( y)0 + x( y )0 +, , x2, x3, ( y )0 + ( y )0, 2!, 3!, +, , x 4 (4), ( y )0 + · · ·, 4!, , Substituting the above values into Maclaurin’s, theorem gives:, y = ( y)0 + x( y )0 +, , x2, {−2( y)0 }, 2!, , x4, x3, +, {−3( y )0 } + {2 × 4( y)0 }, 3!, 4!, +, , x6, x5, {3 × 5( y )0 } + {−2 × 4 ×6( y)0 }, 5!, 6!, , +, , x7, {−3 × 5 × 7( y )0 }, 7!, +, , x8, 8!, , {2 × 4 × 6 × 8( y)0 }, , x8, − ···, 3×5×7, , , , x, x3, x5, −, +, 1 1×2 2×4, , = −3 × 5 × 7( y )0, , −8{−2 × 4 × 6( y)0}= 2 × 4 × 6×8(y)0, , , , , x4, x6, x2, i.e. y = ( y)0 1 −, +, −, 1, 1×3 3×5, , n =4, ( y (6) )0 = −6( y (4) )0 = −6{2 × 4( y)0 }, = −2 × 4 × 6( y)0, , 3 × 5 × 7x 7, + ···, 7!, , 6, , The boundary conditions are that at x = 0, y = 1, dy, = 2, i.e. ( y)0 = 1 and ( y )0 = 2., and, dx, Hence, the power series solution of the differendy, d2 y, tial equation: 2 + x, + 2y = 0 is:, dx, dx, , x2, x4, x6, y = 1− +, −, 1 1 ×3 3 ×5, , , x, x8, x3, +, −··· +2, −, 3 ×5 × 7, 1 1×2, , 5, 7, x, x, +, −, +···, 2×4 2×4×6, Problem 6. Determine the power series solution, of the differential equation:, d2 y d y, +, + x y = 0 given the boundary conditions, dx 2 dx, dy, that at x = 0, y = 0 and, = 1, using, dx, Leibniz–Maclaurin’s method., Following the above procedure:, (i) The differential equation is rewritten as:, y + y + x y = 0 and from the Leibniz theorem of
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Power series methods of solving ordinary differential equations, equation (13), each term is differentiated n times,, which gives:, y, , (n+2), , i.e., , +y, , (n+1), , +y, , (n), , (x) + n y, , (n−1), , (1) + 0 = 0, , y (n+2) + y (n+1) + x y (n) + n y (n−1) = 0, (15), , (ii) At x = 0, equation (15) becomes:, y (n+2) + y (n+1) + n y (n−1) = 0, from which, y (n+2) = −{y (n+1) + n y (n−1) }, This is the recurrence relation and applies for, n ≥1, (iii) Substituting n = 1, 2, 3, . . . will produce a set of, relationships between the various coefficients., For n = 1, ( y )0 = −{( y )0 + ( y)0 }, n = 2, ( y (4) )0 = −{( y )0 + 2( y )0 }, n = 3, ( y (5) )0 = −{( y (4) )0 + 3( y )0 }, n = 4, ( y (6) )0 = −{( y (5) )0 + 4( y )0 }, n = 5, ( y (7) )0 = −{( y (6) )0 + 5( y (4) )0 }, n = 6, ( y (8) )0 = −{( y (7) )0 + 6( y (5) )0 }, From the given boundary conditions, at x = 0,, dy, y = 0, thus ( y)0 = 0, and at x = 0,, = 1, thus, dx, ( y )0 = 1, From the given differential equation,, y + y + x y = 0, and, at x = 0,, ( y )0 + ( y )0 + (0)y = 0 from which,, ( y )0 = −( y )0 = −1, Thus, ( y)0 = 0, ( y )0 = 1, ( y )0 = −1,, ( y )0 = −{( y )0 + ( y)0 } = −(−1 +0) = 1, ( y (4) )0 = −{( y )0 + 2( y )0 }, = −[1 + 2(1)] = −3, ( y (5) )0 = −{( y (4) )0 + 3( y )0 }, = −[−3 +3(−1)] =6, ( y (6) )0 = −{( y (5) )0 + 4( y )0 }, = −[6 + 4(1)] = −10, ( y (7) )0 = −{( y (6) )0 + 5( y (4) )0 }, = −[−10 +5(−3)] =25, , 499, , ( y (8) )0 = −{( y (7) )0 + 6( y (5) )0 }, = −[25 +6(6)] = −61, (iv) Maclaurin’s theorem states:, x2, x3, y = ( y)0 + x( y )0 + ( y )0 + ( y )0, 2!, 3!, x 4 (4), ( y )0 + · · ·, 4!, and substituting the above values into, Maclaurin’s theorem gives:, +, , y = 0 + x(1) +, , x2, x3, x4, {−1} + {1} + {−3}, 2!, 3!, 4!, , +, , x6, x7, x5, {6} + {−10} + {25}, 5!, 6!, 7!, , x8, {−61} + · · ·, 8!, (v) Simplifying, the power series solution of, d2 y d y, +, the differential equation:, + x y = 0 is, dx 2 dx, given by:, +, , y = x−, , x2 x3 3x4 6x5 10x6, + −, +, −, 2! 3! 4!, 5!, 6!, +, , 25x7 61x8, −, +···, 7!, 8!, , Now try the following exercise, Exercise 195 Further problems on power, series solutions by the Leibniz–Maclaurin, method, 1. Determine the power series solution of the difdy, d2 y, ferential equation: 2 + 2x, + y = 0 using, dx, dx, the Leibniz–Maclaurin method, given that at, dy, x = 0, y = 1 and, = 2., dx, ⎤, ⎡, , x 2 5x 4 5 × 9x 6, ⎥, ⎢ y = 1 − 2! + 4! − 6!, ⎥, ⎢, ⎥, ⎢, , , 3, ⎢ 5 × 9 × 13x 8, 3x ⎥, ⎥, ⎢ +, −··· +2 x −, ⎢, 8!, 3! ⎥, ⎥, ⎢, ⎢, ⎥, ⎥, ⎢, ⎦, ⎣ 3 × 7x 5 3 × 7 × 11x 7, +, −, +···, 5!, 7!
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500 Higher Engineering Mathematics, 2. Show that the power series solution of the difd2 y, dy, ferential equation: (x + 1) 2 + (x − 1) −, dx, dx, 2y = 0, using the Leibniz–Maclaurin method,, is given by: y = 1 + x 2 + ex given the boundary, dy, conditions that at x = 0, y =, = 1., dx, 3. Find the particular solution of the differd2 y, dy, − 4y = 0, ential equation: (x 2 + 1) 2 + x, dx, dx, using the Leibniz–Maclaurin method, given, the boundary conditions that at x = 0, y = 1, dy, and, = 1., dx, x3 x5 x7, y = 1 + x + 2x 2 +, −, +, +···, 2, 8, 16, 4. Use the Leibniz–Maclaurin method to determine the power series solution for the differend2 y d y, + x y = 1 given that, tial equation: x 2 +, dx, dx, dy, at x = 0, y = 1 and, = 2., dx, , ⎤, ⎡, x4, x6, x2, ⎢ y = 1 − 22 + 22 × 42 − 22 × 42 × 62 ⎥, ⎥, ⎢, 5, , ⎥, ⎢, x3, x5, ⎥, ⎢, ⎥, ⎢, + ··· +2 x − 2 + 2, 2, ⎥, ⎢, 3, 3 ×5, ⎥, ⎢, , ⎥, ⎢, 7, x, ⎦, ⎣, − 2, +, ·, ·, ·, 3 × 52 × 72, , (iv) equate coefficients of corresponding powers of, the variable on each side of the equation;, this enables index c and coefficients a1 , a2 ,, a3 , … from the trial solution, to be determined., This introductory treatment of the Frobenius method, covering the simplest cases is demonstrated, using the, above procedure, in the following worked problems., Problem 7. Determine, using the Frobenius, method, the general power series solution of the, d2 y d y, differential equation: 3x 2 +, − y = 0., dx, dx, The differential equation may be rewritten as:, 3x y + y − y = 0., (i) Let a trial solution be of the form, , y = x c a0 + a1 x + a2 x 2 + a3 x 3 + · · ·, 4, + ar x r + · · ·, , (16), , where a0 = 0,, i.e. y = a0 x + a1 x, c, , c+1, , + a2 x, , c+2, , + a3 x c+3, , + · · · + ar x c+r + · · ·, , (17), , (ii) Differentiating equation (17) gives:, y = a0cx c−1 + a1 (c + 1)x c, + a2(c + 2)x c+1 + · · ·, + ar (c + r)x c+r−1 + · · ·, and, , y = a0c(c − 1)x c−2 + a1 c(c + 1)x c−1, + a2 (c + 1)(c + 2)x c + · · ·, , 52.5 Power series solution by the, Frobenius method, A differential equation of the form y + P y + Qy = 0,, where P and Q are both functions of x, such that the, equation can be represented by a power series, may be, solved by the Frobenius method., The following 4-step procedure may be used in the, Frobenius method:, (i) Assume, a trial solution of the form y4 =, :, xc a0 + a1 x + a2 x2 + a3 x3 + · · · + ar xr + · · ·, , + ar (c + r − 1)(c + r)x c+r−2 + · · ·, (iii) Substituting y, y and y into each term of the, given equation 3x y + y − y = 0 gives:, 3x y = 3a0 c(c − 1)x c−1 + 3a1 c(c + 1)x c, + 3a2(c + 1)(c + 2)x c+1 + · · ·, + 3ar (c + r − 1)(c+r)x c+r−1 +· · · (a), y = a0 cx c−1 +a1 (c + 1)x c +a2 (c + 2)x c+1, + · · · + ar (c + r)x c+r−1 + · · ·, , (b), , (ii) differentiate the trial series,, (iii) substitute the results in the given differential, equation,, , −y = −a0 x c − a1 x c+1 − a2 x c+2 − a3 x c+3, − · · · − ar x c+r − · · ·, , (c)
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Power series methods of solving ordinary differential equations, (iv) The sum of these three terms forms the left-hand, side of the equation. Since the right-hand side is, zero, the coefficients of each power of x can be, equated to zero., For example, the coefficient of x c−1 is equated, to zero giving: 3a0 c(c − 1) + a0 c = 0, or a0 c[3c − 3 + 1] = a0 c(3c − 2) = 0, , (18), , The coefficient of x c is equated to zero giving:, 3a1c(c + 1) + a1 (c + 1) − a0 = 0, i.e., , a1 (3c2 + 3c + c + 1) − a0, = a1(3c2 + 4c + 1) − a0 = 0, , or, , a1 (3c + 1)(c + 1) − a0 = 0, , 501, , a1, a0, =, (2 × 4) (2 × 4), since a1 = a0, a2, a0, when r = 2, a3 =, =, (3 × 7) (2 × 4)(3 × 7), a0, or, (2 × 3)(4 × 7), a3, when r = 3, a4 =, (4 × 10), a0, =, (2 × 3 × 4)(4 × 7 × 10), and so on., , Thus, when r = 1, a2 =, , From equation (16), the trial solution was:, (19), , In each of series (a), (b) and (c) an x c term is, involved, after which, a general relationship can, be obtained for x c+r , where r ≥ 0., In series (a) and (b), terms in x c+r−1 are present;, replacing r by (r + 1) will give the corresponding, terms in x c+r , which occurs in all three equations, i.e., in series (a), 3ar+1 (c + r)(c + r + 1)x c+r, in series (b), ar+1 (c + r + 1)x c+r, in series (c), −ar x c+r, Equating the total coefficients of x c+r to zero, gives:, 3ar+1 (c + r)(c + r + 1) + ar+1 (c + r + 1), − ar = 0, , y = x c {a0 + a1 x + a2 x 2 + a3 x 3 + · · ·+ ar x r + · · ·}, Substituting c = 0 and the above values of a1 , a2 , a3, …, into the trial solution gives:, , , y = x a0 + a0 x +, , , a0, x2, (2 × 4), , , a0, x3, +, (2 × 3)(4 × 7), , , , a0, x4 + · · ·, +, (2 × 3 × 4)(4 × 7 × 10), , x3, x2, i.e. y = a0 1 + x +, +, (2 × 4) (2 × 3) (4 × 7), , x4, +, +···, (21), (2 × 3 × 4)(4 × 7 × 10), 0, , which simplifies to:, ar+1 {(c + r + 1)(3c + 3r +1)} − ar = 0, , (20), , Equation (18), which was formed from the coefficients of the lowest power of x, i.e. x c−1, is called, the indicial equation, from which, the value of, c is obtained. From equation (18), since a0 = 0,, 2, then c = 0 or c =, 3, , (a) When c = 0:, From equation (19), if c = 0, a1 (1 × 1) − a0 = 0,, i.e. a1 = a0, From equation (20), if c = 0,, ar+1 (r + 1)(3r + 1) − ar = 0,, ar, i.e. ar+1 =, r ≥0, (r + 1)(3r + 1), , 2, (b) When c = :, 3, , , 5, 2, − a0 = 0, i.e., From equation (19), if c = , a1(3), 3, 3, a0, a1 =, 5, 2, From equation (20), if c =, 3, , , 2, ar+1, + r + 1 (2 + 3r + 1) − ar = 0,, 3, , , 5, (3r + 3) − ar, i.e. ar+1 r +, 3, = ar+1 (3r 2 + 8r + 5) − ar = 0,, ar, i.e. ar+1 =, r ≥0, (r + 1)(3r + 5)
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502 Higher Engineering Mathematics, a1, a0, =, (2 × 8) (2 × 5 × 8), a0, since a1 =, 5, a2, when r = 2, a3 =, (3 × 11), a0, =, (2 × 3)(5 × 8 × 11), a3, when r = 3, a4 =, (4 × 14), a0, =, (2×3×4)(5×8×11×14), and so on., , Thus, when r = 1, a2 =, , y = x c {a0 + a1 x + a2 x 2 + a3 x 3 + · · ·+ ar x r + · · ·}, 2, Substituting c = and the above values of a1 , a2 ,, 3, a3 , … into the trial solution gives:, , , , a , 2, a0, 0, y = x 3 a0 +, x+, x2, 5, 2×5×8, , , a0, +, x3, (2 × 3)(5 × 8 × 11), , , , a0, +, x4 + · · ·, (2 × 3 × 4)(5 × 8 × 11 × 14), , 2, x2, x, 3, i.e. y = a0 x 1 + +, 5 (2 × 5 × 8), x3, (2 × 3)(5 × 8 × 11), , +, , x4, + ···, (2 × 3 × 4)(5 × 8 × 11 × 14), , x3, (2 × 3)(5 × 8 × 11), , +, , x4, +···, (2 × 3 × 4)(5 × 8 × 11 × 14), , , , Problem 8. Use the Frobenius method to, determine the general power series solution of the, differential equation:, d2 y, dy, 2x 2 2 − x, + (1 − x)y = 0., dx, dx, The differential equation may be rewritten as:, 2x 2 y − x y + (1 − x)y = 0., (i) Let a trial solution be of the form, , From equation (16), the trial solution was:, , +, , +, , y = x c {a0 + a1 x + a2 x 2 + a3 x 3 + · · ·, + ar x r + · · ·}, , (23), , where a0 = 0,, i.e. y = a0 x c + a1 x c+1 + a2 x c+2 + a3 x c+3, + · · · + ar x c+r + · · ·, , (24), , (ii) Differentiating equation (24) gives:, y = a0 cx c−1 + a1 (c + 1)x c + a2 (c + 2)x c+1, + · · · + ar (c + r)x c+r−1 + · · ·, and y = a0 c(c − 1)x c−2 + a1 c(c + 1)x c−1, + a2(c + 1)(c + 2)x c + · · ·, + ar (c + r − 1)(c + r)x c+r−2 + · · ·, (iii) Substituting y, y and y into each term of, the given equation 2x 2 y − x y + (1 − x)y = 0, gives:, , , (22), , Since a0 is an arbitrary (non-zero) constant in each, solution, its value could well be different., Let a0 = A in equation (21), and a0 = B in equation (22)., Also, if the first solution is denoted by u(x) and the, second by v(x), then the general solution of the given, differential equation is y = u(x) + v(x). Hence,, , x3, x2, +, y = A 1 +x +, (2 × 4) (2 × 3)(4 × 7), , x4, +, +···, (2 ×3 × 4)(4 × 7 × 10), , 2, x2, x, +Bx3 1+ +, 5 (2 × 5 ×8), , 2x 2 y = 2a0 c(c − 1)x c + 2a1 c(c + 1)x c+1, + 2a2 (c + 1)(c + 2)x c+2 + · · ·, + 2ar (c + r − 1)(c + r)x c+r + · · ·, (a), −x y = −a0 cx c − a1 (c + 1)x c+1, − a2 (c + 2)x c+2 − · · ·, − ar (c + r)x c+r − · · ·, , (b), , (1 − x)y = (1 − x)(a0 x c + a1 x c+1 + a2 x c+2, + a3 x c+3 + · · · + ar x c+r + · · ·), = a0 x c + a1 x c+1 + a2 x c+2 + a3 x c+3, + · · · + ar x c+r + · · ·
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Power series methods of solving ordinary differential equations, − a0 x c+1 − a1 x c+2 − a2 x c+3, − a3 x, , c+4, , − · · · − ar x, , c+r+1, , −···, , (c), , (iv) The indicial equation, which is obtained by, equating the coefficient of the lowest power of, x to zero, gives the value(s) of c. Equating the, total coefficients of x c (from equations (a) to (c)), to zero gives:, i.e., , 2a0c(c − 1) − a0 c + a0 = 0, a0 [2c(c − 1) − c + 1] = 0, , i.e., , a0 [2c2 − 2c − c + 1] = 0, , i.e., i.e., , a0 [2c2 − 3c + 1] = 0, a0 [(2c − 1)(c − 1)] = 0, c = 1 or c =, , 2ar (c + r − 1)(c + r) − ar (c + r), + ar − ar−1 = 0, from which,, ar [2(c + r − 1)(c + r) − (c + r) + 1] = ar−1, and ar =, , ar−1, 2(c +r −1)(c +r)−(c +r) +1, , when r = 4,, a3, a3, =, a4 =, 4(8 + 1) 4 × 9, a0, =, (1 × 2 × 3 × 4) × (3 × 5 × 7 × 9), and so on., From equation (23), the trial solution was:, , y = x c a0 + a1 x + a2 x 2 + a3 x 3 + · · ·, + ar x r + · · ·, , 1, 2, The coefficient of the general term, i.e. x c+r ,, gives (from equations (a) to (c)):, from which,, , , i.e. y = a0 x 1 1+, , (25), , +, , ar−1, 2(r)(1 + r) − (1 + r ) +1, ar−1, =, 2, 2r + 2r − 1 − r + 1, ar−1, ar−1, = 2, =, 2r + r, r (2r + 1), , when r = 2,, a1, a1, =, 2(4 + 1) (2 × 5), a0, a0, =, or, (1 × 3)(2 × 5), (1 × 2) × (3 × 5), , a2 =, , when r = 3,, a2, a2, a3 =, =, 3(6 + 1) 3 × 7, a0, =, (1 × 2 × 3) × (3 × 5 × 7), , 4, , Substituting c = 1 and the above values of a1 , a2 ,, a3, … into the trial solution gives:, , a0, a0, 1, y = x a0 +, x+, x2, (1×3), (1×2)×(3×5), a0, +, x3, (1 × 2 × 3) × (3 × 5 × 7), a0, +, x4, (1×2×3×4)×(3×5×7×9), , + ···, , (a) With c = 1, ar =, , Thus, when r = 1,, a0, a0, a1 =, =, 1(2 + 1) 1 × 3, , 503, , +, , (b) With c =, , x2, x, +, (1×3) (1 × 2) × (3 × 5), , x3, (1 × 2 × 3) × (3 × 5 × 7), x4, (1×2×3×4)×(3×5×7×9), , + ···, (26), , 1, 2, , ar−1, 2(c + r − 1)(c + r) − (c + r) + 1, from equation (25), ar−1, , , , i.e. ar = , 1, 1, 1, 2 +r −1, +r −, + r +1, 2, 2, 2, ar−1, , , = , 1, 1, 1, 2 r−, r+, − −r +1, 2, 2, 2, ar−1, , = , 1, 1, 2 r2 −, − −r +1, 4, 2, ar =
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504 Higher Engineering Mathematics, ar−1, ar−1, = 2, 1, 1, 2r, −r, 2r 2 − − − r + 1, 2 2, ar−1, =, r(2r − 1), , solution of the given differential equation is, y = u(x) + v(x),, , =, , , i.e. y = A x 1 +, , a0, a0, =, 1(2 − 1) 1 × 1, a1, a1, when r = 2, a2 =, =, 2(4 − 1) (2 × 3), a0, =, (2 × 3), a2, a2, when r = 3, a3 =, =, 3(6 − 1) 3 × 5, a0, =, (2 × 3) × (3 × 5), a3, a3, when r = 4, a4 =, =, 4(8 − 1) 4 × 7, a0, =, (2×3×4)×(3×5×7), , Thus, when r = 1, a1 =, , and so on., From equation (23), the trial solution was:, , y = x c a0 + a1 x + a2 x 2 + a3 x 3 + · · ·, 4, + ar x r + · · ·, , x4, (2 × 3 × 4) × (3 × 5 × 7), , + ···, , x4, (1 × 2 × 3×4)×(3×5×7×9), , , 1, x2, +··· +Bx2 1+x+, (2 × 3), , +, , +, , x3, (2 × 3) × (3 × 5), , +, , x4, +···, (2 × 3 × 4) ×(3 × 5 ×7), , , , Problem 9. Use the Frobenius method to, determine the general power series solution of the, d2 y, differential equation: 2 − 2y = 0., dx, , (i) Let a trial solution be of the form, , y = x c a0 + a1 x + a2 x 2 + a3 x 3 + · · ·, + ar x r + · · ·, , 4, , (28), , where a0 = 0,, i.e. y = a0 x c + a1 x c+1 + a2 x c+2 + a3 x c+3, + · · · + ar x c+r + · · ·, , x2, (2 × 3), , (29), , (ii) Differentiating equation (29) gives:, , x3, +, (2 × 3) × (3 × 5), +, , x3, (1 × 2 × 3) × (3 × 5 ×7), , The differential equation may be rewritten as:, y − 2y = 0., , 1, Substituting c = and the above values of a1 , a2 ,, 2, a3 , … into the trial solution gives:, , 1, a0 2, a0, 2, x +, x3, y=x a0 +a0 x +, (2×3), (2×3)×(3×5), , a0, +, x4 + · · ·, (2 × 3 × 4) × (3 × 5 × 7), , 1, i.e. y = a0 x 2 1 + x +, , +, , x2, x, +, (1 × 3) (1 × 2) × (3 × 5), , y = a0cx c−1 + a1 (c + 1)x c + a2 (c + 2)x c+1, + · · · + ar (c + r)x c+r−1 + · · ·, (27), , Since a0 is an arbitrary (non-zero) constant in, each solution, its value could well be different., Let a0 = A in equation (26), and a0 = B in equation (27). Also, if the first solution is denoted by, u(x) and the second by v(x), then the general, , and y = a0 c(c − 1)x c−2 + a1 c(c + 1)x c−1, + a2(c + 1)(c + 2)x c + · · ·, + ar (c + r − 1)(c + r)x c+r−2 + · · ·, (iii) Replacing r by (r + 2) in, ar (c + r − 1)(c + r) x c+r−2 gives:, ar+2 (c + r + 1)(c + r + 2)x c+r
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Power series methods of solving ordinary differential equations, , 505, , , , 2x 2 4x 4, = a0 1 +, +, +···, 2!, 4!, 5, 6, 2x 3 4x 5, + a1 x +, +, +···, 3!, 5!, , Substituting y and y into each term of the given, equation y − 2y = 0 gives:, y − 2y = a0 c(c − 1)x c−2 + a1 c(c + 1)x c−1, + [a2(c+1)(c + 2)−2a0 ]x c +· · ·, , Hence, a0 c(c − 1) =0 from which, c = 0 or, c = 1 since a0 = 0, , Since a0 and a1 are arbitrary constants, depending on boundary conditions, let a0 = P, and a1 = Q, then:, , , 2x2 4x4, +, +···, y=P 1 +, 2!, 4!, , , 3, 4x5, 2x, +, +···, (33), +Q x+, 3!, 5!, , For the term in x c−1 , i.e. a1c(c + 1) = 0, , (b) When c =1: a1 = 0, and from equation (31),, , + [ar+2 (c + r + 1)(c + r + 2), − 2ar ] x c+r + · · · = 0, , (30), , (iv) The indicial equation is obtained by equating, the coefficient of the lowest power of x to zero., , With c = 1, a1 = 0; however, when c = 0, a1 is, indeterminate, since any value of a1 combined, with the zero value of c would make the product, zero., , a2 =, Since, , For the term in x c ,, a2 (c + 1)(c + 2) − 2a0 = 0 from which,, 2a0, a2 =, (31), (c + 1)(c + 2), For the term in x c+r ,, , (32), , (a) When c = 0: a1 is indeterminate, and from, equation (31), 2a0, 2a0, a2 =, =, (1 × 2), 2!, 2ar, and, (r + 1)(r + 2), 2a1, 2a1, 2a1, =, =, when r = 1, a3 =, (2 × 3) (1 × 2 × 3), 3!, 2a2, 4a0, =, when r = 2, a4 =, 3×4, 4!, , 2a0 2 2a1 3, Hence, y = x 0 a0 + a1 x +, x +, x, 2!, 3!, , 4a0 4, +, x + ···, 4!, In general, ar + 2 =, , from equation (28), , 2ar, (c + r + 1)(c + r + 2), 2ar, =, (r + 2)(r + 3), , ar+2 =, , from equation (32) and when r = 1,, a3 =, , a4 =, , from which,, 2ar, (c + r + 1)(c + r + 2), , c = 1,, , 2a1, = 0 since a1 = 0, (3 × 4), , when r = 2,, , ar+2 (c + r + 1)(c + r + 2) − 2ar = 0, , ar+2 =, , 2a0, 2a0, =, (2 × 3), 3!, , 2a2, 2, 2a0 4a0, =, ×, =, (4 × 5) (4 × 5), 3!, 5!, , when r = 3,, a5 =, , 2a3, =0, (5 × 6), , Hence, when c = 1,, , , 2a0 2, 4a0 4, x +, x +···, y = x 1 a0 +, 3!, 5!, from equation (28), 5, 6, 2x 3 4x 5, i.e. y = a0 x +, +, + ..., 3!, 5!, Again, a0 is an arbitrary constant; let a0 = K ,, , , 2x3 4x5, +, +···, then, y=K x+, 3!, 5!, However, this latter solution is not a separate solution,, for it is the same form as the second series in equation, (33). Hence, equation (33) with its two arbitrary constants P and Q gives the general solution. This is always
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506 Higher Engineering Mathematics, the case when the two values of c differ by an integer (i.e., whole number). From the above three worked problems,, the following can be deduced, and in future assumed:, (i) if two solutions of the indicial equation differ by, a quantity not an integer, then two independent, solutions y = u(x) + v(x) result, the general solution of which is y = Au + Bv (note: Problem 7, 1, 2, had c = 0 and and Problem 8 had c = 1 and ;, 3, 2, in neither case did c differ by an integer), (ii) if two solutions of the indicial equation do differ by, an integer, as in Problem 9 where c = 0 and 1, and, if one coefficient is indeterminate, as with when, c = 0, then the complete solution is always given, by using this value of c. Using the second value, of c, i.e. c = 1 in Problem 9, always gives a series, which is one of the series in the first solution., Now try the following exercise, Exercise 196 Further problems on power, series solution by the Frobenius method, 1. Produce, using Frobenius’ method, a power, series solution for the differential equation:, d2 y d y, 2x 2 +, − y = 0., dx, dx, ⎤, ⎡, , 2, x, ⎥, ⎢y = A 1 + x +, ⎥, ⎢, (2 × 3), ⎥, ⎢, ⎢, ⎥, 3, ⎥, ⎢, x, ⎢, +, +··· ⎥, ⎥, ⎢, (2 × 3)(3 × 5), ⎥, ⎢, ⎥, ⎢, , ⎥, ⎢, 1, ⎥, x2, x, ⎢, ⎥, ⎢ +Bx2 1+, +, ⎥, ⎢, (1, ×, 3), (1, ×, 2)(3, ×, 5), ⎥, ⎢, ⎥, ⎢, , 3, ⎥, ⎢, x, ⎣, +, + ··· ⎦, (1 × 2 × 3)(3 × 5 × 7), 2. Use the Frobenius method to determine the, general power series solution of the differend2 y, tial equation: 2 + y = 0., dx, ⎡, , ⎤, x2 x4, ⎢ y = A 1 − 2! + 4! − · · ·, ⎥, ⎢, ⎥, ⎥, , ⎢, 3, 5, ⎢, ⎥, x, x, ⎢, + B x − + − ··· ⎥, ⎢, ⎥, 3! 5!, ⎣, ⎦, = P cos x + Q sin x, , 3. Determine the power series solution of the, d2 y, dy, differential equation: 3x 2 + 4 − y = 0, dx, dx, using the Frobenius method., ⎡, ⎤, , x, x2, ⎢y = A 1 + (1 × 4) + (1 × 2)(4 × 7) ⎥, ⎢, ⎥, ⎢, ⎥, x3, ⎢, ⎥, ⎢, +, +··· ⎥, ⎢, ⎥, (1 × 2 × 3)(4 × 7 × 10), ⎢, ⎥, , ⎢, ⎥, 2, 1, x, x, ⎢, ⎥, −3, +, 1+, ⎢ + Bx, ⎥, ⎢, (1 × 2) (1 × 2)(2 × 5)⎥, ⎢, ⎥, ⎢, ⎥, x3, ⎣, +, + ··· ⎦, (1 × 2 × 3)(2 × 5 × 8), 4. Show, using the Frobenius method, that, the power series solution of the differential, d2 y, − y = 0 may be expressed as, equation:, dx 2, y = P cosh x + Q sinh x, where P and Q are, constants. [Hint: check the series expansions, for cosh x and sinh x on page 47], , 52.6 Bessel’s equation and Bessel’s, functions, One of the most important differential equations in, applied mathematics is Bessel’s equation and is of the, form:, d2 y, dy, + (x 2 − v 2 )y = 0, x2 2 + x, dx, dx, where v is a real constant. The equation, which has, applications in electric fields, vibrations and heat conduction, may be solved using Frobenius’ method of the, previous section., Problem 10. Determine the general power series, solution of Bessels equation., d2 y, dy, +x, + (x 2 − v 2 )y = 0 may, 2, dx, dx, be rewritten as: x 2 y + x y + (x 2 − v 2 )y = 0, Bessel’s equation x 2, , Using the Frobenius method from page 500:, (i) Let a trial solution be of the form, y = x c {a0 + a1 x + a2 x 2 + a3 x 3 + · · ·, + ar x r + · · ·}, where a0 = 0,, , (34)
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507, , Power series methods of solving ordinary differential equations, i.e. y = a0 x c + a1 x c+1 + a2 x c+2 + a3 x c+3, + · · · + ar x, , c+r, , +···, , (35), , Similarly, if c = −va1[1 − 2v] = 0, The terms (2v + 1) and (1 − 2v) cannot both be, zero since v is a real constant, hence a1 = 0., , (ii) Differentiating equation (35) gives:, y = a0cx c−1 + a1 (c + 1)x c, + a2 (c + 2)x c+1 + · · ·, + ar (c + r)x c+r−1 + · · ·, and y = a0 c(c − 1)x c−2 + a1 c(c + 1)x c−1, + a2 (c + 1)(c + 2)x c + · · ·, + ar (c + r − 1)(c + r)x c+r−2 + · · ·, (iii) Substituting y, y and y into each term of, the given equation: x 2 y + x y + (x 2 − v 2 )y = 0, gives:, , Since a1 = 0, then from, a3 = a5 = a7 = . . . = 0, , + a2(c + 1)(c + 2)x c+2 + · · ·, , a2 =, , a0, v 2 − (c + 2)2, , a4 =, , a0, 2, 2, [v − (c + 2) ][v 2 − (c + 4)2 ], , a6 =, , a0, [v 2 − (c + 2)2 ][v 2 −(c + 4)2 ][v 2 − (c + 6)2 ], and so on., , a2 =, , + a1(c + 1)x c+1 + a2 (c + 2)x c+2 + · · ·, + ar (c + r)x c+r + · · · + a0 x c+2 + a1 x c+3, + a2 x, , − a1 v x, , + · · · + ar x, , 2 c+1, , 2 c, , + · · · − a0 v x, , − · · · − ar v x, , 2 c+r, , +··· = 0, (36), , (iv) The indicial equation is obtained by equating, the coefficient of the lowest power of x to zero., Hence,, , a0 [c2 − c + c − v 2 ] = 0, , from which, c = +v or c = −v since a0 = 0, For the term in x c+r ,, − ar v 2 = 0, , ar [(c + r)2 − v 2 ] =−ar−2, , For the term in x c+1 ,, a1[c(c + 1) + (c + 1) − v 2 ] = 0, i.e., , a1 [(c + 1)2 − v 2 ] = 0, , but if c = v, , a1 [(v + 1)2 − v 2 ] = 0, , a6 =, , =, , ar [(c + r − 1)(c + r) + (c + r) − v 2 ] =−ar−2, i.e. ar [(c + r)(c + r − 1 + 1) − v 2 ] =−ar−2, , i.e. the recurrence relation is:, ar−2, ar =, for r ≥ 2, v2 − (c + r)2, , =, , =, , ar (c + r − 1)(c + r) + ar (c + r) + ar−2, , i.e., , a4 =, , =, , a0 [c2 − v 2 ] = 0, , i.e., , =, , =, , a0c(c − 1) + a0 c − a0 v 2 = 0, , from which,, , (37), , When c = +v,, , + ar (c + r − 1)(c + r)x c+r + · · · + a0 cx c, , c+r+2, , equation, , and, , a0 c(c − 1)x c + a1 c(c + 1)x c+1, , c+4, , a1[2v + 1] = 0, , i.e., , (37), , =, , a0, 2, v − (v + 2)2, , =, , a0, 2, 2, v − v − 4v − 4, , −a0, −a0, = 2, 4 + 4v, 2 (v + 1), a0, , , v 2 − (v + 2)2 v 2 − (v + 4)2, a0, [−22 (v + 1)][−23(v + 2)], a0, 5, 2 (v + 1)(v + 2), a0, 24 × 2(v + 1)(v + 2), a0, 2, 2, 2, [v −(v+2) ][v −(v+4)2 ][v 2−(v+6)2 ], a0, 4, [2 × 2(v + 1)(v + 2)][−12(v + 3)], −a0, 24 × 2(v + 1)(v + 2) × 22 × 3(v + 3), −a0, and so on., 26 × 3! (v + 1)(v + 2)(v + 3), , The resulting solution for c = +v is given by:, y=u=, , A x v 1−, , x4, x2, +, 22 (v +1) 24 × 2! (v +1)(v +2), , x6, − 6, +···, 2 × 3! (v +1)(v + 2)(v + 3), (38)
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508 Higher Engineering Mathematics, which is valid provided v is not a negative integer, and where A is an arbitrary constant., When c = −v,, a0, a0, = 2, a2 = 2, v − (−v + 2)2, v − (v 2 − 4v + 4), −a0, −a0, =, = 2, 4 − 4v, 2 (v − 1), a0, a4 = 2, [2 (v − 1)][v 2 − (−v + 4)2 ], a0, = 2, [2 (v − 1)][23 (v − 2)], a0, = 4, 2 × 2(v − 1)(v − 2), a0, Similarly, a6 = 6, 2 × 3! (v−1)(v−2)(v−3), , upper case Greek letter gamma, and the gamma function, (x) is defined by the integral, ! ∞, t x−1 e−t dt, (40), (x) =, 0, , and is convergent for x > 0, , ∞, , t x e−t dt, , 0, , and by using integration by parts (see page 420):, , , , x e−t ∞, (x + 1) = t, −1 0, ! ∞ −t , e, x t x−1 dx, −, −1, 0, ! ∞, e−t t x−1 dt, = (0 − 0) + x, 0, , Hence,, , = x(x) from equation (40), , y =w=, , , , x4, x2, B x −v 1 + 2, + 4, 2 (v−1) 2 ×2! (v−1)(v−2), , x6, + 6, +···, 2 × 3! (v − 1)(v − 2)(v − 3), which is valid provided v is not a positive integer, and where B is an arbitrary constant., The complete solution of Bessel’s equation:, x2, , !, , From equation (40), (x + 1) =, , , d2 y, dy 2, +x, + x − v 2 y = 0 is:, dx 2, dx, , y= u +w =, , x4, x2, + 4, A xv 1 − 2, 2 (v + 1) 2 × 2!(v + 1)(v + 2), , x6, − 6, +···, 2 × 3!(v + 1)(v + 2)(v + 3), , x2, −v, +Bx, 1+ 2, 2 (v − 1), x4, + 4, 2 × 2!(v − 1)(v − 2), +, , x6, +· · ·, 6, 2 × 3!(v−1)(v−2)(v−3), , , (39), , The gamma function, The solution of the Bessel equation of Problem 10 may, be expressed in terms of gamma functions. is the, , This is an important recurrence relation for gamma, functions., Thus, since, then similarly,, , (x + 1) = x(x), (x + 2) = (x + 1)(x + 1), = (x + 1)x(x), , and, , (41), , (x + 3) = (x + 2)(x + 2), = (x + 2)(x + 1)x(x),, and so on., , These relationships involving gamma functions are used, with Bessel functions., , Bessel functions, The power series solution of the Bessel equation may be, written in terms of gamma functions as shown in worked, problem 11 below., Problem 11. Show that the power series solution, of the Bessel equation of worked problem 10 may, be written in terms of the Bessel functions Jv (x), and J−v (x) as:, AJv (x) + BJ −v (x), x v , 1, x2, =, − 2, 2, (v + 1) 2 (1! )(v + 2), x4, + 4, −···, 2 (2! )(v + 4)
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Power series methods of solving ordinary differential equations, , +, , x −v , 2, , This is called the Bessel function of the first order kind,, of order v, and is denoted by Jv (x),, x v , 1, x2, i.e. Jv (x) =, − 2, 2, (v + 1) 2 (1!)(v + 2), , x4, + 4, −···, 2 (2!)(v + 3), , 1, x2, − 2, (1 − v) 2 (1! )(2 − v), , x4, + 4, −···, 2 (2! )(3 − v), , From Problem 10 above, when c = +v,, −a0, a2 = 2, 2 (v + 1), If we let a0 =, , provided v is not a negative integer., , 1, 2v (v + 1), , For the second solution, when c = −v, replacing v, by −v in equation (42) above gives:, , then, −1, −1, =, 22 (v + 1) 2v (v + 1) 2v+2 (v + 1)(v + 1), −1, = v+2, from equation (41), 2 (v + 2), , a2k =, , a2 =, , Similarly, a4 =, , a2, 2, v − (c + 4)2, , from equation (37), , from, =, , (−1)k, 22k−v (k! ) (k − v + 1), , which,, , when, , a2, a2, =, (v − c − 4)(v + c + 4) −4(2v + 4), since c = v, −1, −a2, −1, = 3, =, 2 (v + 2) 23 (v + 2) 2v+2 (v + 2), , =, when k = 2, a4 =, , 1, , =, , 2v+4 (2! )(v + 3), , when k = 3, a6 =, =, , The recurrence relation is:, ar =, , (−1)r/2, , r , r, 2v+r, ! v + +1, 2, 2, , (−1)1, −1, 22−v (1! )(2 − v), (−1)2, 24−v (2! )(2 − v + 1), 1, 24−v (2! )(3 − v), , a2k =, , (−1)k, 2v+2k (k!)(v + k + 1), , (42), for k = 1, 2, 3, . . ., , Hence, it is possible to write the new form for equation, (38) as:, , 1, x2, v, y = Ax, −, 2v (v + 1) 2v+2 (1! )(v + 2), , x4, + v+4, −···, 2 (2! )(v + 3), , (−1)3, 26−v (3! )(3 − v + 1), 1, 26−v (3! )(4 − v), , and so on., , , , 1, x2, −, 2−v (1 − v) 22−v (1! )(2 − v), , x4, + 4−v, −···, 2 (2! )(3 − v), , x −v, 1, x2, − 2, i.e. J−v (x)=, 2, (1 −v) 2 (1!)(2 − v), , x4, −· · ·, + 4, 2 (2!)(3 −v), Hence, y = Bx −v, , And if we let r = 2k, then, , (−1)0, 2−v (0! )(1 − v), , 22−v (1! )(1 − v + 1), , since (v + 2)(v + 2) = (v + 3), −1, and a6 = v+6, and so on., 2 (3! )(v + 4), , k = 0, a0 =, , 1, since 0! = 1 (see page 495), 2−v (1 − v), , when k = 1, a2 =, , =, , =, , 509, , provided v is not a positive integer., Jv (x) and J−v (x) are two independent solutions of the, Bessel equation; the complete solution is:, y = AJ v (x) + B J −v (x) where A and B are constants
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510 Higher Engineering Mathematics, i.e. y = AJ v (x)+ BJ −v (x), x v , 1, x2, − 2, =A, 2, (v + 1) 2 (1!)(v + 2), +, +B, , x −v , 2, , From this series two commonly used function are, derived,, , x4, − ···, 24 (2!)(v + 4), , 1, x2, − 2, (1 −v) 2 (1!)(2 − v), x4, + 4, − ···, 2 (2!)(3 −v), , In general terms: Jv (x) =, , x v ;, ∞, , i.e. J0(x) =, , , , 1, 1 x 4, 1 x 2, +, −, (0! ) (1! )2 2, (2! )2 2, 1 x 6, −, +···, (3! )2 2, , = 1−, , , , (−1)k x 2k, 22k (k! )(v+k+1), , 2 k=0, x −v ;, ∞, (−1)k x 2k, and J−v (x) =, 2k, 2, k=0 2 (k! )(k − v + 1), , x2, 22 (1!)2, , x4, 24 (2!), , −, 2, , =, , x 2, x n 1, 1, −, 2, n! (n + 1)! 2, , x 4, 1, +, − ···, (2! )(n + 2)! 2, , , , 26 (3!)2, , +···, , x, x3, x5, − 3, + 5, 2 2 (1!)(2!) 2 (2!)(3!), −, , It may be shown that another series for Jn(x) is given by:, , x6, , , x 2, 1, x, 1, and J1(x) =, −, 2 (1! ) (1! )(2! ) 2, , x 4, 1, +, −···, (2! )(3! ) 2, , Another Bessel function, , Jn (x) =, , +, , x7, +···, 27 (3!)(4!), , Tables of Bessel functions are available for a range of, values of n and x, and in these, J0 (x) and J1(x) are most, commonly used., Graphs of J0 (x), which looks similar to a cosine, and, J1 (x), which looks similar to a sine, are shown in, Figure 52.1., , y, 1, y ⫽ J0(x), , 0.5, y ⫽ J1(x), , 0, , ⫺0.5, , Figure 52.1, , 2, , 4, , 6, , 8, , 10, , 12, , 14, , x
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511, , Power series methods of solving ordinary differential equations, i.e. y = a0 x c + a1 x c+1 + a2 x c+2 + a3 x c+3, Now try the following exercise, Exercise 197 Further problems on Bessel’s, equation and Bessel’s functions, 1. Determine the power series solution of Besd2 y, dy, sel’s equation: x 2 2 + x, + (x 2 −v 2 )y = 0, dx, dx, when v = 2,up to and including, the term in x 6 ., , 2, x4, x, − ···, y = Ax 2 1 − +, 12 384, , + · · · + ar x c+r + · · ·, , (44), , (ii) Differentiating equation (44) gives:, y = a0 cx c−1 + a1 (c + 1)x c, + a2 (c + 2)x c+1 + · · ·, + ar (c + r)x c+r−1 + · · ·, and y = a0 c(c − 1)x c−2 + a1 c(c + 1)x c−1, + a2 (c + 1)(c + 2)x c + · · ·, , 2. Find the power series solution, of, , the, Bessel function: x 2 y + x y + x 2 − v 2 y = 0, in terms of the Bessel function J3(x) when, v = 3. Give the answer up to and including the, term in x 7 ., ⎡, ⎤, x 3 1, x2, −, ⎢ y = AJ3 (x) = 2, 4 22 5 ⎥, ⎢, ⎥, ⎣, ⎦, x4, + 5, −···, 2 6, 3. Evaluate the Bessel functions J0 (x) and J1 (x), when x = 1, correct to 3 decimal places., [J0(x) = 0.765, J1(x) = 0.440], , + ar (c + r − 1)(c + r)x c+r−2 + · · ·, (iii) Substituting y, y and y into each term of the, given, equation:, , , 1 − x 2 y − 2x y + k(k + 1)y = 0 gives:, a0 c(c − 1)x c−2 + a1 c(c + 1)x c−1, + a2 (c + 1)(c + 2)x c + · · ·, + ar (c + r − 1)(c + r)x c+r−2 + · · ·, − a0 c(c − 1)x c − a1 c(c + 1)x c+1, − a2 (c + 1)(c + 2)x c+2 − · · ·, , 52.7 Legendre’s equation and, Legendre polynomials, Another important differential equation in physics, and engineering applications is Legendre’s equation, d2 y, dy, of the form: (1 − x 2 ) 2 − 2x, + k(k + 1)y = 0 or, dx, dx, 2, (1 − x )y − 2x y + k(k + 1)y = 0 where k is a real, constant., Problem 12. Determine the general power series, solution of Legendre’s equation., To solve Legendre’s equation, (1 − x 2 )y − 2x y + k(k + 1)y = 0 using the Frobenius, method:, (i) Let a trial solution be of the form, , y = x c a0 + a1 x + a2 x 2 + a3 x 3, , 4, + · · · + ar x r + · · ·, (43), where a0 = 0,, , − ar (c + r − 1)(c + r)x c+r − · · · − 2a0 cx c, − 2a1 (c + 1)x c+1 − 2a2 (c + 2)x c+2 − · · ·, − 2ar (c + r)x c+r − · · · + k 2 a0 x c, + k 2 a1 x c+1 + k 2 a2 x c+2 + · · · + k 2 ar x c+r, + · · · + ka0 x c + ka1 x c+1 + · · ·, + kar x c+r + · · · = 0, , (45), , (iv) The indicial equation is obtained by equating the, coefficient of the lowest power of x (i.e. x c−2 ) to, zero. Hence, a0c(c − 1) = 0 from which, c = 0 or, c = 1 since a0 = 0., For the term in x c−1 , i.e. a1 c(c + 1) = 0 With, c = 1, a1 = 0; however, when c = 0, a1 is indeterminate, since any value of a1 combined with, the zero value of c would make the product zero., For the term in x c+r ,, ar+2 (c + r + 1)(c + r + 2) −ar (c + r − 1), (c + r) − 2ar (c + r) + k 2 ar + kar = 0
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512 Higher Engineering Mathematics, from which,, ar (c+r−1)(c+r)+2(c+r)−k 2 −k, ar+2 =, (c+r+1)(c+r +2), , , , ar [(c + r)(c + r + 1) − k(k + 1)], =, (c + r + 1)(c + r + 2), (46), When c = 0,, ar+2 =, , ar [r(r + 1) − k(k + 1)], (r + 1)(r + 2), , =, , a0 [−k(k + 1)], (1)(2), a1[(1)(2) − k(k + 1)], (2)(3), −a1 [k 2 + k − 2] −a1 (k − 1)(k + 2), =, 3!, 3!, , a2 [(2)(3) − k(k + 1)] −a2 k 2 + k − 6, =, a4 =, (3)(4), (3)(4), =, , −a2 (k + 3)(k − 2), (3)(4), , =, , −(k + 3)(k − 2) a0 [−k(k + 1)], ., (3)(4), (1)(2), , , , a0 k(k + 1)(k + 3)(k − 2), =, 4!, For r = 3,, , =, , a3[(3)(4) − k(k + 1)] −a3 [k 2 + k − 12], =, (4)(5), (4)(5), −a3 (k + 4)(k − 3), (4)(5), , −(k + 4)(k − 3) −a1 (k − 1)(k + 2), =, ., (4)(5), (2)(3), =, , , k(k + 1) 2, i.e. y = a0 1 −, x, 2!, , +, , For r = 2,, , a5 =, , + ···, , k(k +1)(k − 2)(k + 3) 4, x −···, 4!, , (k − 1)(k + 2) 3, + a1 x −, x, 3!, , For r = 1,, a3 =, , a1(k − 1)(k − 3)(k + 2)(k + 4), and so on., 5!, , Substituting values into equation (43) gives:, , a0 k(k + 1) 2, 0, y = x a0 + a1 x −, x, 2!, −, , a1 (k − 1)(k + 2) 3, x, 3!, , , , , , +, , For r = 0,, a2 =, , a0k(k + 1)(k − 2)(k + 3) 4, x, 4!, a1 (k − 1)(k − 3)(k + 2)(k + 4) 5, +, x, 5!, , +, , , (k − 1)(k − 3)(k + 2)(k + 4) 5, x − · · · (47), 5!, , From page 506, it was stated that if two solutions of, the indicial equation differ by an integer, as in this case,, where c = 0 and 1, and if one coefficient is indeterminate, as with when c = 0, then the complete solution is, always given by using this value of c. Using the second, value of c, i.e. c = 1 in this problem, will give a series, which is one of the series in the first solution. (This may, be checked for c = 1 and where a1 = 0; the result will be, the first part of equation (47) above)., , Legendre’s polynomials, (A polynomial is an expression of the form:, f (x) = a + bx + cx 2 + d x 3 + · · ·). When k in equation, (47) above is an integer, say, n, one of the solution series, terminates after a finite number of terms. For example,, if k = 2, then the first series terminates after the term in, x 2 . The resulting polynomial in x, denoted by Pn (x), is, called a Legendre polynomial. Constants a0 and a1 are, chosen so that y = 1 when x = 1. This is demonstrated, in the following worked problems., Problem 13., P2 (x)., , Determine the Legendre polynomial, , Since in P2 (x), n =k = 2, then from the first part of, equation (47), i.e. the even powers of x:, , , 2(3) 2, y = a0 1 −, x + 0 = a0 {1 − 3x 2 }, 2!, a0 is chosen to make y = 1 when x = 1, i.e. 1 = a0 {1 −3(1)2 } = −2a0 , from which, a0 = −, , 1, 2
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Power series methods of solving ordinary differential equations, Hence, P2 (x)= −, , 1, 1, 1 − 3x 2 = (3x2 − 1), 2, 2, , Problem 14. Determine the Legendre polynomial P3 (x)., Since in P3 (x), n =k = 3, then from the second part of, equation (47), i.e. the odd powers of x:, , (k − 1)(k + 2) 3, x, y = a1 x −, 3!, , (k − 1)(k − 3)(k + 2)(k + 4) 5, +, x − ···, 5!, , , (2)(5) 3 (2)(0)(5)(7) 5, i.e. y = a1 x −, x +, x, 3!, 5!, , , 5 3, = a1 x − x + 0, 3, a1 is chosen to make y = 1 when x = 1., , , , 3, 5, 2, from which, a1 = −, i.e. 1 = a1 1 − = a1 −, 3, 3, 2, , , 5, 3, 1, Hence, P3 (x) =− x− x 3 or P3 (x) = (5x3− 3x), 2, 3, 2, , Rodrigue’s formula, An alternative method of determining Legendre polynomials is by using Rodrigue’s formula, which states:, n, , 1 dn x2 − 1, Pn (x)= n, (48), 2 n!, dxn, This is demonstrated in the following worked problems., Problem 15. Determine the Legendre polynomial, P2 (x) using Rodrigue’s formula., n, , 1 dn x 2 − 1, In Rodrigue’s formula, Pn (x) = n, 2 n!, dx n, and when n =2,, P2 (x) =, =, , 1 d 2 (x 2 − 1)2, 22 2!, dx 2, 1 d2 (x 4 − 2x 2 + 1), 23, dx 2, d 4, (x − 2x 2 + 1), dx, , = 4x 3 − 4x, , 513, , , , d2 x 4 − 2x 2 + 1, d(4x 3 − 4x), =, and, = 12x 2 − 4, dx 2, dx, , , , 1 d2 x 4 −2x 2 +1, 1, = 12x 2 − 4, Hence, P2 (x) = 3, 2, 2, dx, 8, , 1 2, i.e. P2 (x) = 3x − 1 the same as in Problem 13., 2, Problem 16. Determine the Legendre polynomial, P3 (x) using Rodrigue’s formula., n, , 1 dn x 2 − 1, and, In Rodrigue’s formula, Pn (x) = n, 2 n!, dx n, when n = 3,, 3, , 1 d3 x 2 − 1, P3 (x) = 3, 2 3!, dx 3, , , , 1 d3 x 2 − 1 x 4 − 2x 2 + 1, = 3, 2 (6), dx 3, , , 1 d3 x 6 − 3x 4 + 3x 2 − 1, =, (8)(6), dx 3, , , d x 6 −3x 4 +3x 2 −1, = 6x 5 − 12x 3 + 6x, dx, , , d 6x 5 −12x 3 +6x, = 30x 4 − 36x 2 + 6, dx, , , d 30x 4 − 36x 2 + 6, and, = 120x 3 − 72x, dx, , , 1 d3 x 6 − 3x 4 + 3x 2 − 1, Hence, P3 (x) =, (8)(6), dx 3, 1, , 1 , =, 120x 3 − 72x = 20x 3 − 12x, (8)(6), 8, , 1, i.e. P3 (x)= 5x3 − 3x the same as in Problem 14., 2, Now try the following exercise, Exercise 198 Legendre’s equation and, Legendre polynomials, 1. Determine the power series solution of, the Legendre equation:, , , 1 − x 2 y − 2x y + k(k + 1)y = 0 when, (a) k = 0 (b) k = 2, up to and including the
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Chapter 53, , An introduction to partial, differential equations, 53.1, , Introduction, , A partial differential equation is an equation that, contains one or more partial derivatives. Examples, include:, ∂u, ∂u, +b, =c, (i) a, ∂x, ∂y, (ii), , (iii), , ∂ 2u, 1 ∂u, = 2, 2, ∂x, c ∂t, (known as the heat conduction equation), ∂ 2u ∂ 2 u, +, =0, ∂x 2 ∂ y 2, (known as Laplace’s equation), , Equation (i) is a first order partial differential equation, and equations (ii) and (iii) are second order, partial differential equations since the highest power, of the differential is 2., Partial differential equations occur in many areas of, engineering and technology; electrostatics, heat conduction, magnetism, wave motion, hydrodynamics and, aerodynamics all use models that involve partial differential equations. Such equations are difficult to solve,, but techniques have been developed for the simpler, types. In fact, for all but for the simplest cases, there are, a number of numerical methods of solutions of partial, differential equations available., To be able to solve simple partial differential equations knowledge of the following is required:, (a), , partial integration,, , (b) first and second order partial differentiation — as, explained in Chapter 34, and, (c), , the solution of ordinary differential equations —, as explained in Chapters 46–51., , It should be appreciated that whole books have been, written on partial differential equations and their solutions. This chapter does no more than introduce the, topic., , 53.2, , Partial integration, , Integration is the reverse process of differentiation., ∂u, Thus, if, for example,, = 5 cos x sin t is integrated par∂t, tially with respect to t , then the 5 cosx term is considered, as a constant,, !, !, and u = 5 cos x sin t dt = (5 cos x) sin t dt, = (5 cos x)(−cos t ) + c, = −5 cos x cos t + f (x), ∂2u, = 6x 2 cos 2y is integrated partially, ∂x∂ y, with respect to y,, !, , !, ∂u, then, cos 2y d y, = 6x 2 cos 2y d y = 6x 2, ∂x, , , 1, 2, = 6x, sin 2y + f (x), 2, , Similarly, if, , = 3x 2 sin 2y + f (x)
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516 Higher Engineering Mathematics, and integrating, !, u=, , ∂u, partially with respect to x gives:, ∂x, , [3x 2 sin 2y + f (x)] dx, , = x3 sin 2y + (x)f (x) + g(y), f (x) and g(y) are functions that may be determined, if extra information, called boundary conditions or, initial conditions, are known., , 53.3, , From the boundary conditions, when x = 0,, u = cos y, hence, cos y =, , from which, F( y) = cos y, ∂2u, Hence, the solution of 2 = 6x 2 (2y − 1) for the given, ∂x, boundary conditions is:, u=, , Solution of partial differential, equations by direct partial, integration, , The simplest form of partial differential equations, occurs when a solution can be determined by direct partial integration. This is demonstrated in the following, worked problems., Problem 1. Solve the differential equation, ∂2u, = 6x 2 (2y − 1) given the boundary conditions, ∂x 2, ∂u, that at x = 0,, = sin 2y and u =cos y., ∂x, ∂ 2u, Since 2 = 6x 2 (2y − 1) then integrating partially with, ∂x, respect to x gives:, !, !, ∂u, = 6x 2 (2y − 1)dx = (2y − 1) 6x 2 dx, ∂x, = (2y − 1), , 6x 3, + f ( y), 3, , = 2x 3 (2y − 1) + f ( y), where f (y) is an arbitrary function., From the boundary conditions, when x = 0,, ∂u, = sin 2y., ∂x, sin 2y = 2(0)3 (2y − 1) + f ( y), , Hence,, from which,, , f ( y) = sin 2y, , ∂u, = 2x 3 (2y − 1) + sin 2y, ∂x, Integrating partially with respect to x gives:, !, u = [2x 3 (2y − 1) + sin 2y]dx, Now, , =, , 2x 4, (2y − 1) + x(sin 2y) + F( y), 4, , (0)4, (2y − 1) + (0)sin 2y + F( y), 2, , x4, (2y − 1) + x sin y + cos y, 2, , Problem 2. Solve the differential equation:, ∂2u, ∂u, = cos(x + y) given that, = 2 when y = 0,, ∂x∂ y, ∂x, and u = y 2 when x = 0., ∂ 2u, = cos(x + y) then integrating partially with, ∂x∂ y, respect to y gives:, !, ∂u, = cos(x + y)d y = sin(x + y) + f (x), ∂x, Since, , ∂u, = 2 when y = 0,, From the boundary conditions,, ∂x, hence, 2 = sin x + f (x), from which, f (x) = 2 − sin x, i.e., , ∂u, = sin(x + y) + 2 − sin x, ∂x, , Integrating partially with respect to x gives:, !, u = [sin(x + y) + 2 − sin x]dx, = −cos(x + y) + 2x + cos x + f (y), From the boundary conditions, u = y 2 when x = 0, hence, y 2 = −cos y + 0 + cos 0 + f ( y), = 1 − cos y + f ( y), from which, f (y) = y 2 − 1 + cos y, Hence, the solution of, , ∂ 2u, = cos(x + y) is given by:, ∂x∂ y, , u = −cos(x + y) + 2x + cos x + y2 − 1 + cos y
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517, , An introduction to partial differential equations, ⎞, ⎛ 2, 3x − (x 2 + y 2 + z 2 ), ⎟, ⎜, ⎜ + 3y 2 − (x 2 + y 2 + z 2 )⎟, ⎠, ⎝, , Problem 3. Verify that, 1, φ(x, y, z) = , satisfies the partial, x 2 + y2 + z2, ∂ 2φ ∂ 2φ ∂ 2φ, differential equation: 2 + 2 + 2 = 0., ∂x, ∂y, ∂z, , =, Thus, , , The partial differential equation, ∂ 2φ ∂ 2φ ∂ 2φ, +, +, = 0 is called Laplace’s equation., ∂x 2 ∂ y 2 ∂z 2, If φ(x, y, z) = , , 1, , 1, , then differentiating partially with respect to x gives:, 3, ∂φ, 1, = − (x 2 + y 2 + z 2 )− 2 (2x), ∂x, 2, , = −x(x 2 + y 2 + z 2 )− 2, , 5, ∂ 2φ, 3, = (−x) − (x 2 + y 2 + z 2 )− 2 (2x ), 2, ∂x, 2, 2, , 2, , =, , 3x 2, , −, , 2 − 32, , (3x 2 ) − (x 2, , y2 + z2 ), , +, , (x 2 + y 2 + z 2 ) 2, , (3y 2 ) − (x 2 + y 2 + z 2 ), ∂ 2φ, =, 5, ∂ y2, (x 2 + y 2 + z 2 ) 2, , of, , [u =2t y 2 +, , f (t )], , ∂u, = 2t cos θ given that u = 2t when, ∂t, θ = 0., [u =t 2 (cos θ − 1) + 2t ], , Verify that u(θ, t ) =θ 2 + θt is a solution of, ∂u, ∂u, −2, =t., ∂θ, ∂t, , 4., , Verify that u = e−y cos x is a solution of, ∂ 2u ∂ 2u, +, = 0., ∂x 2 ∂ y 2, , 5., , Solve, , ∂ 2φ, (3z 2 ) − (x 2 + y 2 + z 2 ), =, 5, ∂z 2, (x 2 + y 2 + z 2 ) 2, , ∂2u, = 8e y sin 2x given that at y = 0,, ∂x∂ y, ∂u, π, = sin x, and at x = , u =2y 2 ., ∂x, 2, u = −4e y cos 2x − cos x + 4 cos 2x, , + 2y 2 − 4e y + 4, , Thus,, ∂ 2 φ ∂ 2 φ ∂ 2 φ (3x 2 ) − (x 2 + y 2 + z 2 ), + 2+ 2 =, 5, ∂x 2, ∂y, ∂z, (x 2 + y 2 + z 2 ) 2, , +, , solution, , 3., , 5, , +, , general, , Solve, , y2 + z2 ) 2, , Similarly, it may be shown that, , and, , the, , 3, , (x 2 + y 2 + z 2 ), +, , Determine, ∂u, = 4t y., ∂y, , 2., , 1, , 5, 2, , (x 2, , satisfies the Laplace equation, , Now try the following exercise, , 1., , + (x + y + z ) (−1), by the product rule, =, , 1, x 2 + y2 + z2, , =0, , Exercise 199 Further problems on the, solution of partial differential equations by, direct partial integration, , 3, , and, , 5, , (x 2 + y 2 + z 2 ) 2, , ∂ 2φ ∂ 2φ ∂ 2φ, +, +, =0, ∂x 2 ∂ y 2 ∂z 2, , = (x 2 + y 2 + z 2 )− 2, , x 2 + y2 + z2, , + 3z 2 − (x 2 + y 2 + z 2 ), , 6., , (3y 2 ) − (x 2 + y 2 + z 2 ), 5, , (x 2 + y 2 + z 2 ) 2, (3z 2 ) − (x 2 + y 2 + z 2 ), 5, , (x 2 + y 2 + z 2 ) 2, , ∂2u, = y(4x 2 − 1) given that at x = 0,, ∂x 2, ∂u, u =sin y and, = cos 2y., ∂x, , , 4, x, x2, + x cos 2y + sin y, −, u=y, 3, 2, , Solve
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518 Higher Engineering Mathematics, (c), 7., , 8., , 9., , 10., , ∂2u, ∂u, Solve, = sin(x + t) given that, =1, ∂x∂t, ∂x, when t = 0, and when u =2t when x = 0., [u =−sin(x + t) + x + sin x + 2t + sin t ], x, Show that u(x, y) = x y + is a solution of, y, ∂2u, ∂2u, 2x, + y 2 = 2x., ∂x∂ y, ∂y, Find the particular solution of the differential, ∂ 2u, equation, = cos x cos y given the ini∂x∂ y, ∂u, = x, and, tial conditions that when y =π,, ∂x, when x = π, u =2 cos y., , π2, x2, u = sin x sin y + + 2 cos y −, 2, 2, Verify that φ(x, y) = x cos y + e x sin y satisfies the differential equation, ∂2φ ∂2φ, +, + x cos y = 0., ∂x 2 ∂ y 2, , 53.4 Some important engineering, partial differential equations, There are many types of partial differential equations. Some typically found in engineering and science, include:, (a), , Laplace’s equation, used extensively with electrostatic fields is of the form:, ∂ 2u ∂ 2u ∂ 2u, +, +, = 0., ∂x 2 ∂ y 2 ∂z 2, , (d) The transmission equation, where the potential u in a transmission cable is of the form:, ∂ 2u, ∂2 u, ∂u, =, A, +B, + Cu where A, B and C are, 2, 2, ∂x, ∂t, ∂t, constants., Some of these equations are used in the next sections., , 53.5, , Let u(x, t ) = X (x)T (t ), where X (x) is a function of x, only and T (t ) is a function of t only, be a trial solution to, ∂ 2u 1 ∂ 2u, the wave equation 2 = 2 2 . If the trial solution is, ∂x, c ∂t, ∂ 2u, ∂u, =X T., simplified to u = XT, then, = X T and, ∂x, ∂x 2, ∂ 2u, ∂u, Also, = XT and 2 = XT ., ∂t, ∂t, ∂ 2u, Substituting into the partial differential equation 2 =, ∂x, 1 ∂ 2u, gives:, c2 ∂t 2, 1, X T = 2 XT, c, Separating the variables gives:, X , 1 T , = 2, X, c T, , The wave equation, where the equation of motion, is given by:, 1 ∂2u, ∂ 2u, =, ∂x 2, c2 ∂t 2, T, , with T being the tension in a string, ρ, and ρ being the mass/unit length of the string., where c2 =, , (b) The heat conduction equation is of the form:, ∂ 2 u 1 ∂u, =, ∂x 2 c2 ∂t, h, , with h being the thermal conducσρ, tivity of the material, σ the specific heat of the, material, and ρ the mass/unit length of material., where c2 =, , Separating the variables, , X, 1 T, = 2, where μ is a constant., X, c T, X, Thus, since μ =, (a function of x only), it must be, X, 1 T, independent of t ; and, since μ = 2, (a function of t, c T, only), it must be independent of x., Let μ =, , If μ is independent of x and t , it can only be a conX, then X = μX or X − μX = 0 and if, stant. If μ =, X, 1 T, μ= 2, then T = c2 μT or T − c2 μT = 0., c T, Such ordinary differential equations are of the form, found in Chapter 50, and their solutions will depend, on whether μ > 0, μ = 0 or μ < 0.
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An introduction to partial differential equations, , Problem 4. Find the general solution of the, following differential equations:, (a) X − 4X =0, , y, u 5 f (x, t ), , Worked Problem 4 will be a reminder of solving, ordinary differential equations of this type., , P, u(x, t ), , (b) T + 4T = 0., 0, , (a), , If X − 4X =0 then the auxiliary equation (see, Chapter 50) is:, m 2 − 4 = 0 i.e. m 2 = 4 from which,, m = +2 or m = −2, Thus, the general solution is:, X = Ae2x + Be−2x, , (b) If T + 4T = 0 then the auxiliary equation is:, m 2 + 4 = 0 i.e. m 2 = −4 from which,, √, m = −4 = ± j 2, Thus, the general solution is:, T = e0 {A cos 2t + B sin 2t } =A cos2t + B sin2t, Now try the following exercise, , 519, , L, , x, , x, , Figure 53.1, , fixed. The position of any point P on the string depends, on its distance from one end, and on the instant in time., Its displacement u at any time t can be expressed as, u = f (x, t ), where x is its distance from 0., The equation of motion is as stated in Section 53.4 (a),, 1 ∂2u, ∂2u, i.e. 2 = 2 2 ., ∂x, c ∂t, The boundary and initial conditions are:, (i) The string is fixed at both ends, i.e. x = 0 and, x = L for all values of time t ., Hence, u(x, t ) becomes:, , u(0, t ) = 0, for all values of t ≥ 0, u(L , t ) = 0, (ii) If the initial deflection of P at t = 0 is denoted by, f (x) then u(x, 0) = f (x), , Exercise 200 Further problems on revising, the solution of ordinary differential equation, 1. Solve T = c2 μT given c = 3 and μ = 1., [T = Ae3t + Be−3t ], 2. Solve T − c2 μT = 0 given c = 3 and μ = −1., [T = A cos 3t + B sin 3t], 3. Solve X = μX given μ = 1., , X = Aex + Be−x, 4. Solve X − μX = 0 given μ = −1., [X = A cos x + B sin x], , (iii) Let the initial velocity of P be g(x), then, , ∂u, = g(x), ∂t t =0, Initially a trial solution of the form u(x, t ) = X (x)T (t ), is assumed, where X (x) is a function of x only and T (t ), is a function of t only. The trial solution may be simplified to u = XT and the variables separated as explained, in the previous section to give:, X, 1 T, = 2, X, c T, When both sides are equated to a constant μ this results, in two ordinary differential equations:, T − c2 μT = 0 and X − μX =0, , 53.6, , The wave equation, , An elastic string is a string with elastic properties, i.e., the string satisfies Hooke’s law. Figure 53.1 shows a, flexible elastic string stretched between two points at, x = 0 and x = L with uniform tension T . The string will, vibrate if the string is displace slightly from its initial, position of rest and released, the end points remaining, , Three cases are possible, depending on the value, of μ., , Case 1: μ > 0, For convenience, let μ = p2 , where p is a real constant., Then the equations, X − p 2 X = 0 and T − c2 p2 T = 0
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520 Higher Engineering Mathematics, have solutions: X = Ae px + Be− px and, T = Cecpt + De−cpt where A, B, C and D are constants., But X =0 at x = 0, hence 0 = A + B i.e. B = −A and, X = 0 at x = L , hence, 0 = Ae p L + Be− p L = A(e p L − e− p L )., Assuming (e p L – e− p L ) is not zero, then A = 0 and since, B = −A, then B = 0 also., This corresponds to the string being stationary; since it, is non-oscillatory, this solution will be disregarded., , nπ, Thus sin pL =0 i.e. pL =nπ or p =, for inteL, ger values of n., Substituting in equation (4) gives:, , , , cnπt, cnπt, nπ x , C cos, + D sin, u = B sin, L, L, L, , , nπ x, cnπt, cnπt, i.e. u = sin, An cos, + Bn sin, L, L, L, , In this case, since μ = p2 = 0, T = 0 and X = 0. We, will assume that T (t ) = 0. Since X = 0, X = a and, X = ax + b where a and b are constants. But X =0 at, x = 0, hence b = 0 and X = ax and X =0 at x = L, hence, a = 0. Thus, again, the solution is non-oscillatory and is, also disregarded., , (where constant An = BC and Bn = B D). There, will be many solutions, depending on the value of, n. Thus, more generally,, , ∞ , <, cnπ t, nπx, un (x, t) =, An cos, sin, L, L, n=1, , cnπt, + Bn sin, (5), L, , Case 3: μ < 0, , To find An and Bn we put in the initial conditions, not yet taken into account., , Case 2: μ = 0, , For convenience,, let μ = − p2 then X + p 2 X =0 from which,, X = A cos px + B sin px, and T + c2 p2 T = 0, , (i) At t = 0, u(x, 0) = f (x) for 0 ≤ x ≤ L, Hence, from equation (5),, (1), u(x, 0) = f (x) =, , from which,, , n=1, , T = C cos cpt + D sin cpt, , (2), , , (ii) Also at t = 0,, , (see worked Problem 4 above)., Thus, the suggested solution u = XT now becomes:, u = {A cos px + B sin px}{C cos cpt + D sin cpt }, (3), Applying the boundary conditions:, (i) u = 0 when x = 0 for all values of t ,, thus 0 = {A cos 0 + B sin 0}{C cos cpt, i.e., , ∂u, ∂t, , t =0, , An sin, , nπx , L, , (6), , = g(x) for 0 ≤ x ≤ L, , Differentiating equation (5) with respect to t gives:, , , ∞ , ∂u <, cnπt, nπ x, cnπ, An −, =, sin, sin, ∂t, L, L, L, n=1, , , cnπ, cnπt, + Bn, cos, L, L, and when t = 0,, , + D sin cpt }, , 0 = A{C cos cpt + D sin cpt }, , i.e. g(x) =, , + D sin cpt } = 0), , ∞, cπ < , nπx , Bn n sin, L, L, , (7), , n=1, , u = {B sin px}{C cos cpt, + D sin cpt }, , ∞ , <, nπ x cnπ , sin, g(x) =, Bn, L, L, n=1, , from which, A = 0, (since {C cos cpt, Hence,, , ∞ , <, , (4), , (ii) u = 0 when x = L for all values of t, Hence, 0 = {B sin pL}{C cos cpt + D sin cpt }, Now B = 0 or u(x, t ) would be identically zero., , From Fourier series (see page 638) it may be shown that:, nπ x, between, An is twice the mean value of f (x) sin, L, x = 0 and x = L, !, 2 L, nπ x, f (x)sin, dx, i.e., An =, L 0, L, for n = 1, 2, 3, . . . (8)
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An introduction to partial differential equations, and Bn, , cnπ , , is twice the mean value of, L, nπ x, g(x)sin, between x = 0 and x = L, L, ! L, 2, L, nπ x, i.e., Bn =, g(x)sin, dx, cnπ L, L, 0, , or, , 521, , Bn =, , 2, cnπ, , !, , L, , g(x)sin, 0, , u(x, 0 ), , y, 4, u 5 f (x ), , 2, , 0, , nπ x, dx, L, , 50 x (cm), , 25, , Figure 53.2, , (9), , Summary of solution of the wave equation, The above may seem complicated; however a practical problem may be solved using the following 8-point, procedure:, 1. Identify clearly, conditions., , the, , initial, , and, , boundary, , 2. Assume a solution of the form u = XT and express, the equations in terms of X and T and their, derivatives., 3. Separate the variables by transposing the equation, and equate each side to a constant, say, μ; two, separate equations are obtained, one in x and the, other in t ., 4. Let μ = − p 2 to give an oscillatory solution., 5. The two solutions are of the form:, X = A cos px + B sin px, and T = C cos cpt + D sin cpt., Then u(x, t ) = {A cos px + B sin px}{C cos cpt +, D sin cpt }., 6. Apply the boundary conditions to determine constants A and B., 7. Determine the general solution as an infinite sum., 8. Apply the remaining initial and boundary conditions and determine the coefficients An and Bn, from equations (8) and (9), using Fourier series, techniques., Problem 5. Figure 53.2 shows a stretched string, of length 50 cm which is set oscillating by, displacing its mid-point a distance of 2 cm from its, rest position and releasing it with zero velocity., ∂ 2u 1 ∂ 2u, Solve the wave equation: 2 = 2 2 where, ∂x, c ∂t, c2 = 1, to determine the resulting motion u(x, t )., , Following the above procedure,, 1. The boundary and initial conditions given are:, 6, u(0, t ) = 0, i.e. fixed end points, u(50, t ) = 0, u(x, 0) = f (x) =, , 2, x 0 ≤ x ≤ 25, 25, , =−, , 100 −2x, 2, x +4 =, 25, 25, 25 ≤ x ≤ 50, , (Note: y = mx + c is a straight line graph, so the gradient, m, between 0 and 25 is 2/25 and the y-axis, 2, intercept is zero, thus y = f (x) = x + 0; between, 25, 25 and 50, the gradient =−2/25 and the y-axis, 2, intercept is at 4, thus f (x) = − x + 4)., 25, , ∂u, = 0 i.e. zero initial velocity., ∂t t =0, 2. Assuming a solution u = XT , where X is a function of x only, and T is a function of t only,, ∂ 2u, ∂u, ∂u, then, = X T and 2 = X T and, = XT and, ∂x, ∂x, ∂y, ∂2u, = XT . Substituting into the partial differential, ∂ y2, ∂2u, 1 ∂2u, equation, 2 = 2 2 gives:, ∂x, c ∂t, 1, X T = 2 XT i.e. X T = XT since c2 = 1., c, X, T, 3. Separating the variables gives:, =, X, T, Let constant,, μ=, , X, T, X, T, =, then μ =, and μ =, X, T, X, T
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522 Higher Engineering Mathematics, from which,, , Each integral is determined using integration by, parts (see Chapter 43, page 420) with the result:, , X − μX = 0 and T − μT = 0., 4., , Letting μ = − p 2 to give an oscillatory solution, gives:, X + p 2 X = 0 and T + p2 T = 0, , From equation (9),, , The auxiliary equation, for each is: m 2 + p 2 = 0, , from which, m = − p2 = ± j p., 5., , 6., , 16, nπ, sin, 2, 2, n π, 2, , An =, , Bn =, , , !, , L, , g(x) sin, 0, , nπ x, dx, L, , Solving each equation gives:, X = A cos px + B sin px, and, T = C cos pt + D sin pt ., Thus,, u(x, t ) ={A cos px+B sin px}{C cos pt +D sin pt }., , Substituting into equation (b) gives:, , Applying the boundary conditions to determine, constants A and B gives:, , u n (x, t ) =, , ∂u, ∂t, , = 0 = g(x) thus, Bn = 0, , t =0, , =, , u(x, t ) = B sin px{C cos pt + D sin pt } (a), , or, more generally,, u n (x, t ) =, , ∞, <, n=1, , nπ x, sin, 50, , , , From equation (8),, 2, L, , !, , L, , nπ x, dx, L, 0, , ! 25 , 2, 2, nπ x, =, x sin, dx, 50 0, 25, 50, , ! 50 , nπ x, 100 − 2x, sin, dx, +, 25, 50, 25, , An =, , Hence,, , sin, , nπ x, 50, , , An cos, , , , nπt, 50, nπt, + Bn sin, 50, , 16, n2π 2, , sin, , nπ, nπt, cos, 2, 50, , , , , , ∞, , u(x, t) =, , nπx, 16 < 1, nπ, nπ t, sin, sin, cos, π2, n2, 50, 2, 50, n=1, , For stretched string problems as in problem 5 above, the, main parts of the procedure are:, , 2., , Determine An from equation (8)., !, 2 L, nπ x, Note that, f (x) sin, dx is always equal, L 0, L, nπ, 8d, (see Fig. 53.3), to 2 2 sin, n π, 2, Determine Bn from equation (9), , 3., , Substitute in equation (5) to determine u(x, t ), , (b), , where An = BC and Bn = B D., , ∞, <, , nπ x, 50, , nπt, + (0) sin, 50, , 1., nπt, An cos, 50, , nπt, + Bn sin, 50, , sin, , n=1, , (ii) u(50, t ) = 0, hence, 0 = B sin 50 p{C cos pt + D sin pt }. B = 0,, hence sin 50 p =0 from which, 50 p =nπ and, nπ, p=, 50, 7. Substituting in equation (a) gives:, , , nπ x, nπt, nπt, u(x, t ) = B sin, C cos, + D sin, 50, 50, 50, , ∞, <, n=1, , (i) u(0, t ) =0,hence 0 = A{C cos pt + D sin pt }, from which we conclude that A = 0., Therefore,, , 8., , 2, cnπ, , y, , f (x) sin, , y 5 f (x ), d, , 0, , Figure 53.3, , L, 2, , L, , x
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An introduction to partial differential equations, y, u 5 f (x, t ), , Now try the following exercise, Exercise 201 Further problems on the, wave equation, 1. An elastic string is stretched between two, points 40 cm apart. Its centre point is displaced, 1.5 cm from its position of rest at right angles, to the original direction of the string and then, released with zero velocity. Determine the subsequent motion u(x, t ) by applying the wave, 1 ∂ 2u, ∂2u, equation 2 = 2 2 with c2 = 9., ∂x, c ∂t, ∞, , u(x, t ) =, , 12 < 1, nπ, nπ x, sin, sin, π2, n2, 2, 40, n=1, , cos, , 3nπt, 40, , 2. The centre point of an elastic string between, two points P and Q, 80 cm apart, is deflected, a distance of 1 cm from its position of, rest perpendicular to P Q and released initially with zero velocity. Apply the wave, ∂ 2u 1 ∂ 2u, equation 2 = 2 2 where c = 8, to deter∂x, c ∂t, mine the motion of a point distance x from P at, time t ., ∞, , u(x, t ) =, , 523, , 8 < 1, nπ, nπ x, nπt, sin, sin, cos, 2, 2, π, n, 2, 80, 10, n=1, , P, u (x, t ), , 0, , L, , x, , x, , Figure 53.4, , Fig. 53.4, where the bar extends from x = 0 to x = L, the, temperature of the ends of the bar is maintained at zero,, and the initial temperature distribution along the bar is, defined by f (x)., Thus, the boundary conditions can be expressed as:, 6, u(0, t ) = 0, for all t ≥ 0, u(L , t ) = 0, and, , u(x, 0) = f (x) for 0 ≤ x ≤ L, , As with the wave equation, a solution of the form, u(x, t ) = X (x)T (t ) is assumed, where X is a function of, x only and T is a function of t only. If the trial solution, is simplified to u = XT , then, ∂u, ∂ 2u, ∂u, = X T and, =XT, = XT, ∂x, ∂x 2, ∂t, Substituting into the partial differential equation,, ∂ 2u, 1 ∂u, = 2, gives:, ∂x 2, c ∂t, 1, X T = 2 XT, c, Separating the variables gives:, , 53.7, , The heat conduction equation, , ∂ 2 u 1 ∂u, =, is solved, ∂x 2 c2 ∂t, in a similar manner to that for the wave equation; the, equation differs only in that the right hand side contains, a first partial derivative instead of the second., The conduction of heat in a uniform bar depends on, the initial distribution of temperature and on the physical properties of the bar, i.e. the thermal conductivity,, h, the specific heat of the material, σ , and the mass, per unit length, ρ, of the bar. In the above equation,, h, c2 =, σρ, With a uniform bar insulated, except at its ends, any heat, flow is along the bar and, at any instant, the temperature, u at a point P is a function of its distance x from one, end, and of the time t . Consider such a bar, shown in, The heat conduction equation, , X , 1 T, = 2, X, c T, Let − p2 =, , X, 1 T, = 2 where − p2 is a constant., X, c T, , X, If − p2 =, then X = − p2 X or X + p 2 X = 0,, X, giving X = A cos px + B sinpx, 1 T, T, then, = − p2 c2 and integrating, and if − p 2 = 2, c T, T, with respect to t gives:, !, !, T, dt = − p2 c2 dt, T, from which, ln T = − p2 c2 t + c1, The left hand integral is obtained by an algebraic, substitution (see Chapter 39).
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524 Higher Engineering Mathematics, If ln T = − p2c2 t + c1 then, 2 2, 2 2, 2 2, T = e− p c t +c1 = e− p c t ec1 i.e. T = k e−p c t (where, constant k = ec1 )., 2 2, Hence, u(x, t ) = XT = {A cos px + B sin px}k e− p c t, 2 2, i.e. u(x, t ) = {P cos px + Q sin px}e− p c t where, P = Ak and Q = Bk., Applying the boundary conditions u(0, t ) =0 gives:, 2 2, 2 2, 0={P cos 0+ Q sin 0}e− p c t = P e− p c t from which,, 2, 2, P = 0 and u(x, t ) = Q sin px e− p c t ., 2 2, Also, u(L , t ) =0 thus, 0 = Q sin pL e− p c t and since, nπ, Q = 0 then sin pL =0 from which, pL =nπ or p =, L, where n =1, 2, 3, . . ., There are therefore many values of u(x, t )., Thus, in general,, u(x, t ) =, , ∞ , <, , Q n e− p, , 2 c2 t, , sin, , n=1, , nπ x , L, , ∞ , <, n=1, , Q n sin, , nπ x , L, , From Fourier series, Q n = 2 × mean, nπ x, from x to L., f (x) sin, L, !, 2 L, nπ x, f (x) sin, dx, Hence,, Qn =, L 0, L, Thus, u(x, t ) =, , u (x, 0 ), u (x, t ), , 0, , 1, , x (m ), , 1, , x (m ), , P, u (x, t ), 0, x, , Figure 53.5, , Applying the remaining boundary condition, that when, t = 0, u(x, t ) = f (x) for 0 ≤ x ≤ L, gives:, f (x) =, , 15, , Assuming a solution of the form u = XT , then, from, above,, X = A cos px + B sin px, and T = k e− p, , value, , of, , , , ∞ ! L, 2 2, nπ x, nπ x, 2<, f (x) sin, dx e− p c t sin, L, L, L, 0, n=1, , This method of solution is demonstrated in the following, worked problem., Problem 6. A metal bar, insulated along its sides,, is 1 m long. It is initially at room temperature of, 15◦ C and at time t = 0, the ends are placed into ice, at 0◦C. Find an expression for the temperature at a, point P at a distance x m from one end at any time, t seconds after t = 0., The temperature u along the length of bar is shown in, Fig. 53.5., ∂ 2 u 1 ∂u, and the, The heat conduction equation is 2 = 2, ∂x, c ∂t, given boundary conditions are:, u(0, t ) = 0, u(1, t ) = 0 and u(x, 0) = 15, , 2 c2 t, , ., , Thus, the general solution is given by:, u(x, t ) = {P cos px + Q sin px}e− p, u(0, t ) = 0 thus 0 = P e− p, , 2 c2 t, , 2 c2 t, , from which, P = 0 and u(x, t ) ={Q sin px}e− p, , 2 c2 t, , ., , 2 2, p}e− p c t ., , Also, u(1, t ) =0 thus 0 = {Q sin, Since Q = 0, sin p = 0 from which, p = nπ where, n = 1, 2, 3, . . ., ∞ , , <, 2 2, Q n e− p c t sin nπ x, Hence, u(x, t ) =, n=1, , The final initial condition given was that at t = 0,, u = 15, i.e. u(x, 0) = f (x) = 15., ∞, <, {Q n sin nπ x} where, from Fourier, Hence, 15 =, n=1, , coefficients, Q n = 2 × mean value of 15 sin nπ x from, x = 0 to x = 1,, i.e., , Qn =, , 2, 1, , =−, , !, , 1, 0, , cos nπ x 1, 15 sin nπ x dx = 30 −, nπ, 0, , 30, [cos nπ − cos 0], nπ
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525, , An introduction to partial differential equations, =, , 30, (1 − cos nπ), nπ, , = 0 (when n is even) and, , 60, (when n is odd), nπ, , Hence, the required solution is:, ∞ , , <, 2 2, u(x, t) =, Q n e− p c t sin nπ x, , conduction equation to be, ⎡, , take c2 = 1., , ⎣u(x, t ) = 320, π2, , ∞, <, n(odd)=1, , ∂ 2 u 1 ∂u, =, and, ∂x 2 c2 ∂t, , , 1, nπ, nπ x −, sin, sin, e, 2, n, 2, 20, , n2 π 2 t, 400, , ⎤, , ⎦, , n=1, , 60, =, π, , ∞, <, n(odd)=1, , 1, 2 2 2, (sin nπ x)e−n π c t, n, , Now try the following exercise, Exercise 202 Further problems on the heat, conduction equation, 1. A metal bar, insulated along its sides, is 4 m, long. It is initially at a temperature of 10◦C, and at time t = 0, the ends are placed into ice at, 0◦C. Find an expression for the temperature, at a point P at a distance x m from one end at, any time t seconds after t = 0., ⎤, ⎡, ∞, nπ x, 40 < 1 − n2 π 2 c2 t, ⎦, ⎣u(x, t ) =, 16, sin, e, π, n, 4, , 53.8, , Laplace’s equation, , The distribution of electrical potential, or temperature,, over a plane area subject to certain boundary conditions,, can be described by Laplace’s equation. The potential, at a point P in a plane (see Fig. 53.6) can be indicated, by an ordinate axis and is a function of its position, i.e., z = u(x, y), where u(x, y) is the solution of the Laplace, ∂2u ∂ 2u, two-dimensional equation 2 + 2 = 0., ∂x, ∂y, The method of solution of Laplace’s equation is similar, to the previous examples, as shown below., Figure 53.7 shows a rectangle OPQR bounded by, the lines x = 0, y = 0, x = a, and y = b, for which, we are required to find a solution of the equation, ∂ 2u ∂ 2 u, +, = 0. The solution z =(x, y) will give, say,, ∂x 2 ∂ y 2, , n(odd)=1, , 2. An insulated uniform metal bar, 8 m long,, has the temperature of its ends maintained at, 0◦C, and at time t = 0 the temperature distribution f (x) along the bar is defined by, f (x) = x(8 − x). If c2 = 1, solve the heat con∂ 2 u 1 ∂u, =, duction equation, to determine, ∂x 2 c2 ∂t, the temperature u at any point in the bar at, time t ., ⎤, ⎡, 3 <, ∞, 1 − n2 π 2 t, nπ x, 8, ⎦, ⎣u(x, t ) =, e 64 sin, π, n3, 8, , P, , x, , 0, , Figure 53.6, z, y, R, y5b, , Q, , u (x, y ), , n(odd)=1, , 3. The ends of an insulated rod PQ, 20 units long,, are maintained at 0◦ C. At time t = 0, the temperature within the rod rises uniformly from, each end reaching 4◦ C at the mid-point of, PQ. Find an expression for the temperature, u(x, t ) at any point in the rod, distant x from, P at any time t after t = 0. Assume the heat, , y, , z, , 0, , Figure 53.7, , P, x5a, , x
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526 Higher Engineering Mathematics, the potential at any point within the rectangle OPQR., The boundary conditions are:, , Since there are many solutions for integer values of n,, u(x, y) =, , u = 0 when x = 0 i.e. u(0, y) = 0, , for 0 ≤ y ≤ b, , u = 0 when x = a i.e. u(a, y) =0, , for 0 ≤ y ≤ b, , u = 0 when y = b i.e. u(x, b) =0, , for 0 ≤ x ≤ a, , u = f (x) when y = 0 i.e. u(x, 0) = f (x), for 0 ≤ x ≤ a, As with previous partial differential equations, a solution of the form u(x, y) = X (x)Y (y) is assumed, where, X is a function of x only, and Y is a function of, y only. Simplifying to u = X Y , determining partial, ∂2u ∂2u, derivatives, and substituting into 2 + 2 = 0 gives:, ∂x, ∂y, X Y + XY =0, X, Y, Separating the variables gives:, =−, X, Y, Letting each side equal a constant, − p2 , gives the two, equations:, X + p 2 X = 0 and Y − p 2 Y = 0, from which, X = A cos px + B sin px and, Y = C e py + D e− py or Y = C cosh py + D sinh py (see, Problem 5, page 480 for this conversion)., This latter form can also be expressed as:, Y = E sinh p( y + φ) by using compound angles., Hence u(x, y) = X Y, = {A cos px + B sin px}{E sinh p( y + φ)}, or u(x, y), = {P cos px + Q sin px}{sinh p( y + φ)}, where P = AE and Q = B E., The first boundary condition is: u(0, y) = 0, hence, 0 = P sinh p(y + φ) from which, P = 0. Hence,, u(x, y) = Q sin px sinh p(y + φ)., The second boundary condition is: u(a, y) = 0,, hence, 0 = Q sin pa sinh p(y + φ), from which,, nπ, for, sin pa = 0, hence, pa = nπ or p =, a, n = 1, 2, 3, . . ., The third boundary condition is: u(x, b) = 0,, hence, 0 = Q sin px sinh p(b + φ) from which,, sinh p(b + φ) = 0 and φ = −b., Hence, u(x, y) = Q sin px sinh p(y − b) =, Q 1 sin px sinh p(b − y) where Q 1 = −Q., , ∞, <, , Q n sin px sinh p(b − y), , n=1, , =, , ∞, <, , Q n sin, , n=1, , nπ x, nπ, sinh, (b − y), a, a, , The fourth boundary condition is: u(x, 0) = f (x),, hence,, , f (x) =, , ∞, <, , Q n sin, , n=1, , i.e., , f (x) =, , ∞ , <, n=1, , nπ x, nπb, sinh, a, a, , , nπ x, nπb, sin, Q n sinh, a, a, , From Fourier series coefficients,, , , nπb, = 2 × the mean value of, Q n sinh, a, nπ x, f (x) sin, from x = 0 to x = a, a, ! a, nπ x, f (x) sin, i.e., =, dx from which,, a, 0, Q n may be determined., This is demonstrated in the following worked, problem., Problem 7. A square plate is bounded by the, lines x = 0, y = 0, x = 1 and y = 1. Apply the, ∂ 2u ∂ 2u, Laplace equation 2 + 2 = 0 to determine the, ∂x, ∂y, potential distribution u(x, y) over the plate, subject, to the following boundary conditions:, u = 0 when x = 0 0 ≤ y ≤ 1,, u = 0 when x = 1 0 ≤ y ≤1,, u = 0 when y = 0 0 ≤ x ≤ 1,, u = 4 when y = 1 0 ≤ x ≤ 1., Initially a solution of the form u(x, y) = X (x)Y (y) is, assumed, where X is a function of x only, and Y is a, function of y only. Simplifying to u = X Y , determining, ∂ 2u ∂ 2u, partial derivatives, and substituting into 2 + 2 = 0, ∂x, ∂y, gives:, X Y + XY =0, X, Y, Separating the variables gives:, =−, X, Y, Letting each side equal a constant, − p 2 , gives the two, equations:, X + p2 X = 0 and Y − p2 Y = 0
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An introduction to partial differential equations, 16, (for odd values of n), nπ, 16, 16, Hence, Q n =, =, cosech nπ, nπ(sinh nπ) nπ, , from which, X = A cos px + B sin px, , =, , and Y = Ce py + De− py, or Y = C cosh py + D sinh py, or Y = E sinh p(y + φ), , Hence, from equation (a),, ∞, <, Q n sin nπ x sinh nπ y, u(x,y) =, , Hence u(x, y) = X Y, = {A cos px + B sin px}{E sinh p(y + φ)}, , n=1, , or u(x, y), = {P cos px + Q sin px}{sinh p(y + φ)}, , =, , where P = AE and Q = BE., The first boundary condition is: u(0, y) = 0, hence, 0 = P sinh p(y + φ) from which, P = 0., Hence, u(x, y) = Q sin px sinh p(y + φ)., The second boundary condition is: u(1, y) = 0, hence, 0 = Q sin p(1) sinh p(y + φ) from which,, sin p =0, hence, p =nπ for n =1, 2, 3, . . ., The third boundary condition is: u(x, 0) = 0, hence,, 0 = Q sin px sinh p(φ) from which,, sinh p(φ) = 0 and φ =0., Hence, u(x, y) = Q sin px sinh py., Since there are many solutions for integer values of n,, u(x, y) =, , ∞, <, , Q n sin px sinh py, , =, , Q n sin nπ x sinh nπ y, , ∞, <, n(odd)=1, , 1, (cosech nπ sin nπ x sinhnπy), n, , Exercise 203 Further problems on the, Laplace equation, 1. A rectangular plate is bounded by the, lines x = 0, y = 0, x = 1 and y = 3. Apply the, ∂ 2u ∂ 2u, Laplace equation 2 + 2 = 0 to determine, ∂x, ∂y, the potential distribution u(x, y) over the plate,, subject to the following boundary conditions:, u =0 when x = 0, u =0 when x = 1, u =0 when y = 2, u =5 when y = 3, , (a), , n=1, , The fourth boundary condition is: u(x, 1) = 4 = f (x),, ∞, <, Q n sin nπ x sinh nπ(1)., hence, f (x) =, , 16, π, , Now try the following exercise, , n=1, ∞, <, , 527, , ⎡, ⎣u(x, y) = 20, π, , ∞, <, n(odd)=1, , 0 ≤ y ≤ 2,, 0 ≤ y ≤ 2,, 0 ≤ x ≤ 1,, 0 ≤ x ≤ 1., , ⎤, 1, cosechnπ sin nπ x sinh nπ(y −2)⎦, n, , n=1, , From Fourier series coefficients,, Q n sinh nπ = 2 × the mean value of, f (x) sin nπ x from x = 0 to x = 1, i.e. =, , 2, 1, , !, , 1, , 4 sin nπ x dx, 0, , cos nπ x 1, =8 −, nπ, 0, 8, =−, (cos nπ − cos 0), nπ, 8, =, (1 −cos nπ), nπ, = 0 (for even values of n),, , 2. A rectangular plate is bounded by the, lines x = 0, y = 0, x = 3, y = 2. Determine the, potential distribution u(x, y) over the rectangle using the Laplace equation, ∂2u ∂ 2u, +, = 0, subject to the following, ∂x 2 ∂ y 2, boundary conditions:, u(0, y) = 0, u(3, y) = 0, u(x, 2) = 0, u(x, 0) = x(3 − x), , ⎡, , ⎣u(x, y) = 216, π3, , ∞, <, n(odd)=1, , 0 ≤ y ≤ 2,, 0 ≤ y ≤ 2,, 0 ≤ x ≤ 3,, 0 ≤ x ≤ 3., , ⎤, 1, nπ x, 2nπ, nπ, cosech, sin, sinh, (2 − y)⎦, 3, 3, 3, n3
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Revision Test 15, This Revision Test covers the material contained in Chapters 50 to 53. The marks for each question are shown in, brackets at the end of each question., 1., , d2 y, dy, (b), + 2 + 2y = 10ex given that when x = 0,, dx 2, dx, dy, y = 0 and, = 1., (20), dx, 2., , u (x,0), , Find the particular solution of the following differential equations:, d2 y, (a) 12 2 − 3y = 0 given that when t = 0, y = 3, dt, dy 1, and, =, dt 2, , 1, 0, , 40 x (cm), , Figure RT15.1, , 6., , In a galvanometer the deflection θ satisfies the, differential equation:, , Determine the general power series solution of, Bessel’s equation:, x2, , dθ, d2 θ, +2 +θ = 4, dt 2, dt, , d2 y, dy, +x, + (x 2 − v 2 )y = 0, dx 2, dx, , and hence state the series up to and including the, term in x 6 when v = +3., (26), , Solve the equation for θ given that when t = 0,, dθ, = 0., (12), θ = 0 and, dt, 3., , Determine y (n) when y = 2x 3 e4x ., , 4., , Determine the power series solution of the differend2 y, dy, tial equation: 2 + 2x, + y = 0 using Leibnizdx, dx, Maclaurin’s method, given the boundary conditions, dy, that at x = 0, y = 2 and, = 1., (20), dx, , 5., , 20, , (10), , Use the Frobenius method to determine the general power series solution of the differential, d2 y, equation: 2 + 4y = 0., (21), dx, , 7., , 8., , 9., , Determine the general solution of, , ∂u, = 5x y, ∂x, , (2), , ∂ 2u, = x 2 (y − 3), Solve the differential equation, ∂x 2, given the boundary conditions that at x = 0,, ∂u, = sin y and u =cos y., (6), ∂x, Figure RT15.1 shows a stretched string of length, 40 cm which is set oscillating by displacing its, mid-point a distance of 1 cm from its rest position and releasing it with zero velocity. Solve the, 1 ∂2u, ∂ 2u, =, where c2 = 1, to, wave equation:, ∂x 2, c2 ∂t 2, determine the resulting motion u(x, t )., , (23)
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Chapter 54, , Presentation of, statistical data, 54.1, , Some statistical terminology, , The relative frequency with which any member of a, set occurs is given by the ratio:, , Data are obtained largely by two methods:, (a), , by counting—for example, the number of stamps, sold by a post office in equal periods of time, and, , (b) by measurement—for example, the heights of a, group of people., When data are obtained by counting and only whole, numbers are possible, the data are called discrete. Measured data can have any value within certain limits and, are called continuous (see Problem 1)., A set is a group of data and an individual value within, the set is called a member of the set. Thus, if the masses, of five people are measured correct to the nearest 0.1 kg, and are found to be 53.1 kg, 59.4 kg, 62.1 kg, 77.8 kg and, 64.4 kg, then the set of masses in kilograms for these five, people is:, {53.1, 59.4, 62.1, 77.8, 64.4}, and one of the members of the set is 59.4, A set containing all the members is called a population., Some members selected at random from a population, are called a sample. Thus all car registration numbers, form a population, but the registration numbers of, say,, 20 cars taken at random throughout the country are a, sample drawn from that population., The number of times that the value of a member occurs, in a set is called the frequency of that member. Thus, in the set: {2, 3, 4, 5, 4, 2, 4, 7, 9}, member 4 has a frequency of three, member 2 has a frequency of two and, the other members have a frequency of one., , frequency of member, total frequency of all members, For the set: {2, 3, 5, 4, 7, 5, 6, 2, 8}, the relative frequency of member 5 is 29, Often, relative frequency is expressed as a percentage and the percentage relative frequency is: (relative, frequency × 100)%., Problem 1. Data are obtained on the topics given, below. State whether they are discrete or continuous, data., (a) The number of days on which rain falls in a, month for each month of the year., (b) The mileage travelled by each of a number of, salesmen., (c), , The time that each of a batch of similar, batteries lasts., , (d) The amount of money spent by each of, several families on food., (a), , The number of days on which rain falls in a given, month must be an integer value and is obtained by, counting the number of days. Hence, these data, are discrete., , (b) A salesman can travel any number of miles (and, parts of a mile) between certain limits and these, data are measured. Hence the data are continuous.
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530 Higher Engineering Mathematics, (c), , The time that a battery lasts is measured and can, have any value between certain limits. Hence these, data are continuous., , (d) The amount of money spent on food can only be, expressed correct to the nearest pence, the amount, being counted. Hence, these data are discrete., , Now try the following exercise, Exercise 204 Further problems on discrete, and continuous data, In Problems 1 and 2, state whether data relating to, the topics given are discrete or continuous., 1. (a), , The amount of petrol produced daily, for, each of 31 days, by a refinery., , (b) The amount of coal produced daily by, each of 15 miners., (c), , The number of bottles of milk delivered, daily by each of 20 milkmen., , (b) horizontal bar charts, having data represented, by equally spaced horizontal rectangles (see Problem 3), and, (c), , vertical bar charts, in which data are represented by equally spaced vertical rectangles (see, Problem 4)., , Trends in ungrouped data over equal periods of time, can be presented diagrammatically by a percentage, component bar chart. In such a chart, equally spaced, rectangles of any width, but whose height corresponds, to 100%, are constructed. The rectangles are then subdivided into values corresponding to the percentage, relative frequencies of the members (see Problem 5)., A pie diagram is used to show diagrammatically the, parts making up the whole. In a pie diagram, the area of, a circle represents the whole, and the areas of the sectors, of the circle are made proportional to the parts which, make up the whole (see Problem 6)., Problem 2. The number of television sets, repaired in a workshop by a technician in six,, one-month periods is as shown below. Present these, data as a pictogram., , (d) The size of 10 samples of rivets produced, by a machine., , (a) continuous (b) continuous, (c) discrete, (d) continuous, , Month, , Number repaired, , January, , 11, , February, , 6, , (a) The number of people visiting an exhibition on each of 5 days., , March, , 15, , April, , 9, , (b) The time taken by each of 12 athletes to, run 100 metres., , May, , 13, , June, , 8, , 2., , (c) The value of stamps sold in a day by, each of 20 post offices., (d) The number of defective items produced in each of 10 one-hour periods, by a machine., , (a) discrete (b) continuous, (c) discrete (d) discrete, , Each symbol shown in Fig. 54.1 represents two television sets repaired. Thus, in January, 5 12 symbols are used, to represent the 11 sets repaired, in February, 3 symbols, are used to represent the 6 sets repaired, and so on., Month, January, February, March, , 54.2, , Presentation of ungrouped data, , April, May, , Ungrouped data can be presented diagrammatically in, several ways and these include:, (a), , pictograms, in which pictorial symbols are used, to represent quantities (see Problem 2),, , June, , Figure 54.1, , Number of TV sets repaired, , ; 2 sets
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Presentation ofstatistical data, Problem 3. The distance in miles travelled by, four salesmen in a week are as shown below., Salesmen, , P, , Q, , R, , S, , Distance travelled miles 413 264 597 143, Use a horizontal bar chart to represent these data, diagrammatically., Equally spaced horizontal rectangles of any width, but, whose length is proportional to the distance travelled,, are used. Thus, the length of the rectangle for salesman, P is proportional to 413 miles, and so on. The horizontal, bar chart depicting these data is shown in Fig. 54.2., , Salesmen, , S, , 531, , Problem 5. The numbers of various types of, dwellings sold by a company annually over a, three-year period are as shown below. Draw, percentage component bar charts to present these, data., Year 1 Year 2 Year 3, 4-roomed bungalows, , 24, , 17, , 7, , 5-roomed bungalows, , 38, , 71, , 118, , 4-roomed houses, , 44, , 50, , 53, , 5-roomed houses, , 64, , 82, , 147, , 6-roomed houses, , 30, , 30, , 25, , A table of percentage relative frequency values, correct, to the nearest 1%, is the first requirement. Since,, , R, Q, , percentage relative frequency, , P, 0, , 100, , 200, 300, 400, 500, Distance travelled, miles, , 600, , =, , frequency of member × 100, total frequency, , then for 4-roomed bungalows in year 1:, , Figure 54.2, , Problem 4. The number of issues of tools or, materials from a store in a factory is observed for, seven, one-hour periods in a day, and the results of, the survey are as follows:, Period, , 1, , 2 3 4, , 5, , 6 7, , Number of, issues, 34 17 9 5 27 13 6, , percentage relative frequency, =, , 24 × 100, = 12%, 24 + 38 + 44 + 64 + 30, , The percentage relative frequencies of the other types, of dwellings for each of the three years are similarly, calculated and the results are as shown in the table, below., , Present these data on a vertical bar chart., , Number of issues, , In a vertical bar chart, equally spaced vertical rectangles, of any width, but whose height is proportional to the, quantity being represented, are used. Thus the height of, the rectangle for period 1 is proportional to 34 units,, and so on. The vertical bar chart depicting these data is, shown in Fig. 54.3., 40, 30, 20, 10, 1, , Figure 54.3, , 2, , 3, , 4 5, Periods, , 6, , 7, , Year 1 Year 2 Year 3, (%), (%), (%), 4-roomed bungalows, , 12, , 7, , 2, , 5-roomed bungalows, , 19, , 28, , 34, , 4-roomed houses, , 22, , 20, , 15, , 5-roomed houses, , 32, , 33, , 42, , 6-roomed houses, , 15, , 12, , 7, , The percentage component bar chart is produced by, constructing three equally spaced rectangles of any, width, corresponding to the three years. The heights of, the rectangles correspond to 100% relative frequency,, and are subdivided into the values in the table of percentages shown above. A key is used (different types, of shading or different colour schemes) to indicate
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532 Higher Engineering Mathematics, corresponding percentage values in the rows of the table, of percentages. The percentage component bar chart is, shown in Fig. 54.4., , Research and, development, Labour, 728 368, 188 Materials, 1268, 1088, Overheads, , Percentage relative frequency, , Key, 100, , 6-roomed houses, , 90, , 5-roomed houses, , 80, , 4-roomed houses, , 70, , 5-roomed bungalows, , 60, , 4-roomed bungalows, , 50, , Profit, , Ip ⬅ 1.88, , Figure 54.5, , 40, 30, , (b) Using the data presented in Fig. 54.4,, comment on the housing trends over the, three-year period., , 20, 10, 1, , 2, Year, , 3, , Figure 54.4, , (c) Determine the profit made by selling 700, units of the product shown in Fig. 54.5., (a), , Problem 6. The retail price of a product costing, £2 is made up as follows: materials 10 p, labour, 20 p, research and development 40 p, overheads, 70 p, profit 60 p. Present these data on a pie diagram., , £413 × 37, , i.e. £152.81, 100, Similarly, for salesman Q, the miles travelled are, 264 and his allowance is, , A circle of any radius is drawn, and the area of the circle, represents the whole, which in this case is £2. The circle, is subdivided into sectors so that the areas of the sectors, are proportional to the parts, i.e. the parts which make, up the total retail price. For the area of a sector to be, proportional to a part, the angle at the centre of the circle, must be proportional to that part. The whole, £2 or 200 p,, corresponds to 360◦. Therefore,, , £264 × 37, , i.e. £97.68, 100, Salesman R travels 597 miles and he receives, £597 × 37, , i.e. £220.89, 100, , 10, degrees, i.e. 18◦, 200, 20, 20 p corresponds to 360 ×, degrees, i.e. 36◦, 200, 10 p corresponds to 360 ×, , and so on, giving the angles at the centre of the circle, for the parts of the retail price as: 18◦, 36◦ , 72◦, 126◦, and 108◦, respectively., The pie diagram is shown in Fig. 54.5., Problem 7., (a) Using the data given in Fig. 54.2 only,, calculate the amount of money paid to each, salesman for travelling expenses, if they are, paid an allowance of 37 p per mile., , By measuring the length of rectangle P the, mileage covered by salesman P is equivalent to, 413 miles. Hence salesman P receives a travelling, allowance of, , Finally, salesman S receives, £143 × 37, , i.e. £52.91, 100, (b) An analysis of Fig. 54.4 shows that 5-roomed, bungalows and 5-roomed houses are becoming, more popular, the greatest change in the three, years being a 15% increase in the sales of, 5-roomed bungalows., (c), , Since 1.8◦ corresponds to 1 p and the profit occupies 108◦ of the pie diagram, then the profit per, unit is, 108 × 1, , that is, 60 p, 1.8
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Presentation ofstatistical data, The profit when selling 700 units of the product is, £, , ⎡, , ⎤, 6 equally spaced horizontal, ⎢ rectangles, whose lengths are ⎥, ⎢, ⎥, ⎣ proportional to 35, 44, 62, ⎦, 68, 49 and 41, respectively., , 700 × 60, , that is, £420, 100, , Now try the following exercise, , 4., , Present the data given in Problem 2 above on, a horizontal bar chart., ⎡, ⎤, 5 equally spaced, ⎢ horizontal rectangles, whose ⎥, ⎢, ⎥, ⎢ lengths are proportional to ⎥, ⎢, ⎥, ⎣ 1580, 2190, 1840, 2385 and ⎦, 1280 units, respectively., , 5., , For the data given in Problem 1 above,, construct a vertical bar chart., ⎡, ⎤, 6 equally spaced vertical, ⎢ rectangles, whose heights ⎥, ⎢, ⎥, ⎢ are proportional to 35, 44, ⎥, ⎢, ⎥, ⎣ 62, 68, 49 and 41 units, ⎦, respectively., , 6., , Depict the data given in Problem 2 above on, a vertical bar chart., ⎡, ⎤, 5 equally spaced vertical, ⎢ rectangles, whose heights are ⎥, ⎢, ⎥, ⎢ proportional to 1580, 2190, ⎥, ⎢, ⎥, ⎣ 1840, 2385 and 1280 units, ⎦, respectively., , 7., , A factory produces three different types of, components. The percentages of each of, these components produced for three, onemonth periods are as shown below. Show this, information on percentage component bar, charts and comment on the changing trend, in the percentages of the types of component, produced., , Exercise 205 Further problems on, presentation of ungrouped data, 1., , The number of vehicles passing a stationary, observer on a road in six ten-minute intervals, is as shown. Draw a pictogram to represent, these data., Period of, Time, , 1, , 2, , 3, , 4, , 5, , 6, , Number of, Vehicles, 35 44 62 68 49 41, ⎡, ⎢, ⎢, ⎢, ⎢, ⎢, ⎢, ⎣, 2., , ⎤, , If one symbol is used to, ⎥, represent 10 vehicles,, ⎥, ⎥, working correct to the, ⎥, ⎥, nearest 5 vehicles,, ⎥, gives 3 12 , 4 12 , 6, 7, 5 and 4 ⎦, symbols respectively., , The number of components produced by a, factory in a week is as shown below:, Day, , Number of Components, , Mon, , 1580, , Tues, , 2190, , Wed, , 1840, , Thur, , 2385, , Fri, , 1280, , Show these data on a pictogram., ⎡, ⎤, If one symbol represents, ⎢ 200 components, working ⎥, ⎢, ⎥, ⎢ correct to the nearest, ⎥, ⎢, ⎥, ⎢ 100 components gives: ⎥, ⎢, ⎥, ⎣ Mon 8, Tues 11, Wed 9, ⎦, Thurs 12 and Fri 6 12 ., 3., , For the data given in Problem 1 above, draw, a horizontal bar chart., , Month, , 1, , 2, , 3, , Component P, , 20, , 35, , 40, , Component Q, , 45, , 40, , 35, , Component R, , 35, , 25, , 25, , ⎡, , ⎤, Three rectangles of equal, ⎢ height, subdivided in the ⎥, ⎢, ⎥, ⎢ percentages shown in the ⎥, ⎢, ⎥, ⎢ columns above. P increases ⎥, ⎢, ⎥, ⎣ by 20% at the expense, ⎦, of Q and R, , 533
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534 Higher Engineering Mathematics, 8., , A company has five distribution centres and, the mass of goods in tonnes sent to each, centre during four, one-week periods, is as, shown., Week, , 1, , 2, , 3, , 4, , Centre A, , 147, , 160, , 174, , 158, , Centre B, , 54, , 63, , 77, , 69, , Centre C, , 283, , 251, , 237, , 211, , Centre D, , 97, , 104, , 117, , 144, , Centre E, , 224, , 218, , 203, , 194, , Use a percentage component bar chart to, present these data and comment on any, trends., ⎡, ⎤, Four rectangles of equal, ⎢ heights, subdivided as follows: ⎥, ⎢, ⎥, ⎢ week 1: 18%, 7%, 35%, 12%, ⎥, ⎢, ⎥, ⎢ 28% week 2: 20%, 8%, 32%, ⎥, ⎢, ⎥, ⎢ 13%, 27% week 3: 22%, 10%, ⎥, ⎢, ⎥, ⎢ 29%, 14%, 25% week 4: 20%, ⎥, ⎢, ⎥, ⎢ 9%, 27%, 19%, 25%. Little, ⎥, ⎢, ⎥, ⎢ change in centres A and B, a, ⎥, ⎢, ⎥, ⎢ reduction of about 8% in C, an ⎥, ⎢, ⎥, ⎣ increase of about 7% in D and a ⎦, reduction of about 3% in E., 9., , The employees in a company can be split, into the following categories: managerial 3,, supervisory 9, craftsmen 21, semi-skilled 67,, others 44. Show these data on a pie diagram., ⎤, ⎡, A circle of any radius,, ⎥, ⎢ subdivided into sectors, ⎥, ⎢, ⎢ having angles of 7 1 ◦ , 22 1 ◦, ⎥, 2, 2 ⎥, ⎢, ◦, ◦, ⎦, ⎣ 52 1 , 167 1 and110◦,, 2, 2, respectively., , 10., , The way in which an apprentice spent his, time over a one-month period is as follows:, drawing office 44 hours, production 64 hours,, training 12 hours, at college 28 hours., Use a pie diagram to depict this information., ⎤, ⎡, A circle of any radius,, ⎢ subdivided into sectors ⎥, ⎥, ⎢, ⎢ having angles of 107◦, ⎥, ⎥, ⎢, ⎦, ⎣ 156◦, 29◦and 68◦ ,, respectively., , 11., , (a) With reference to Fig. 54.5, determine, the amount spent on labour and materials to produce 1650 units of the product., (b) If in year 2 of Fig. 54.4, 1% corresponds, to 2.5 dwellings, how many bungalows, are sold in that year. [(a) £ 495, (b) 88], , 12., , (a) If the company sell 23500 units per, annum of the product depicted in, Fig. 54.5, determine the cost of their, overheads per annum., (b) If 1% of the dwellings represented in, year 1 of Fig. 54.4 corresponds to, 2 dwellings, find the total number of, houses sold in that year., [(a) £ 16450, (b) 138], , 54.3, , Presentation of grouped data, , When the number of members in a set is small, say ten or, less, the data can be represented diagrammatically without further analysis, by means of pictograms, bar charts,, percentage components bar charts or pie diagrams (as, shown in Section 54.2)., For sets having more than ten members, those members, having similar values are grouped together in classes, to form a frequency distribution. To assist in accurately counting members in the various classes, a tally, diagram is used (see Problems 8 and 12)., A frequency distribution is merely a table showing, classes and their corresponding frequencies (see Problems 8 and 12)., The new set of values obtained by forming a frequency, distribution is called grouped data., The terms used in connection with grouped data are, shown in Fig. 54.6(a). The size or range of a class, is given by the upper class boundary value minus, the lower class boundary value, and in Fig. 54.6 is, 7.65 − 7.35, i.e. 0.30. The class interval for the class, shown in Fig. 54.6(b) is 7.4 to 7.6 and the class mid-point, value is given by,, , , upper class, boundary value, , , , , +, , lower class, boundary value, , 2, and in Fig. 54.6 is, , 7.65 +7.35, , i.e. 7.5., 2, ,
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Presentation ofstatistical data, (a), , 535, , Class interval, , 81 83 87 74 76 89 82 84, Lower, class, boundary, , Class, mid-point, , Upper, class, boundary, , 86 76 77 71 86 85 87 88, 84 81 80 81 73 89 82 79, 81 79 78 80 85 77 84 78, 83 79 80 83 82 79 80 77, , (b), , to 7.3, , 7.35, , 7.4 to 7.6, , 7.5, , 7.7 to, , 7.65, , Figure 54.6, , One of the principal ways of presenting grouped, data diagrammatically is by using a histogram, in, which the areas of vertical, adjacent rectangles are, made proportional to frequencies of the classes (see, Problem 9). When class intervals are equal, the heights, of the rectangles of a histogram are equal to the frequencies of the classes. For histograms having unequal, class intervals, the area must be proportional to the frequency. Hence, if the class interval of class A is twice, the class interval of class B, then for equal frequencies,, the height of the rectangle representing A is half that, of B (see Problem 11). Another method of presenting, grouped data diagrammatically is by using a frequency, polygon, which is the graph produced by plotting frequency against class mid-point values and joining the, co-ordinates with straight lines (see Problem 12)., A cumulative frequency distribution is a table showing the cumulative frequency for each value of upper, class boundary. The cumulative frequency for a particular value of upper class boundary is obtained by adding, the frequency of the class to the sum of the previous frequencies. A cumulative frequency distribution is formed, in Problem 13., The curve obtained by joining the co-ordinates of cumulative frequency (vertically) against upper class boundary (horizontally) is called an ogive or a cumulative, frequency distribution curve (see Problem 13)., , Problem 8. The data given below refer to the gain, of each of a batch of 40 transistors, expressed, correct to the nearest whole number. Form a, frequency distribution for these data having seven, classes., , The range of the data is the value obtained by taking, the value of the smallest member from that of the, largest member. Inspection of the set of data shows, that, range = 89 −71 = 18. The size of each class is, given approximately by range divided by the number of, classes. Since 7 classes are required, the size of each, class is 18/7, that is, approximately 3. To achieve seven, equal classes spanning a range of values from 71 to 89,, the class intervals are selected as: 70–72, 73–75, and, so on., To assist with accurately determining the number in each, class, a tally diagram is produced, as shown in, Table 54.1(a). This is obtained by listing the classes, in the left-hand column, and then inspecting each of the, 40 members of the set in turn and allocating them to, the appropriate classes by putting ‘1s’ in the appropriate rows. Every fifth ‘1’ allocated to the particular row, is shown as an oblique line crossing the four previous, ‘1s’, to help with final counting., A frequency distribution for the data is shown in, Table 54.1(b) and lists classes and their corresponding frequencies, obtained from the tally diagram. (Class, mid-point value are also shown in the table, since they, are used for constructing the histogram for these data, (see Problem 9))., , Problem 9. Construct a histogram for the data, given in Table 54.1(b)., , The histogram is shown in Fig. 54.7. The width of, the rectangles correspond to the upper class boundary values minus the lower class boundary values and, the heights of the rectangles correspond to the class, frequencies. The easiest way to draw a histogram is, to mark the class mid-point values on the horizontal, scale and draw the rectangles symmetrically about the, appropriate class mid-point values and touching one, another.
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536 Higher Engineering Mathematics, Table 54.1(a), Class, , Tally, , 70–72, , 1, , 73–75, , 11, , 76–78, , 1111 11, , 79–81, , 1111 1111 11, , 82–84, , 1111 1111, , 85–87, , 1111 1, , 88–90, , 111, , 80, , 130 170, , 80 100, , 90 120, , 80 120 100 110, , 50 100 110, , Table 54.1(b), , Frequency, , 90 110, , 90 100, , Frequency, , 70–72, , 71, , 1, , 73–75, , 74, , 2, , 76–78, , 77, , 7, , Table 54.2, , 79–81, , 80, , 12, , Class, , Frequency, , 82–84, , 83, , 9, , 20–40, , 2, , 85–87, , 86, , 6, , 50–70, , 6, , 88–90, , 89, , 3, , 80–90, , 12, , 100–110, , 14, , 120–140, , 4, , 150–170, , 2, , 74, , 77, , 80, , 83, , 86, , 89, , 80, , Inspection of the set given shows that the majority of the members of the set lie between £80, and £110 and that there are a much smaller number of extreme values ranging from £30 to £170., If equal class intervals are selected, the frequency, distribution obtained does not give as much information as one with unequal class intervals. Since, the majority of members are between £80 and £100,, the class intervals in this range are selected to be, smaller than those outside of this range. There is no, unique solution and one possible solution is shown in, Table 54.2., , Class mid-point, , 71, , 40 110, , 70 110, , Class, , 16, 14, 12, 10, 8, 6, 4, 2, , 70, , Problem 11. Draw a histogram for the data given, in Table 54.2., , Class mid-point values, , Figure 54.7, , Problem 10. The amount of money earned, weekly by 40 people working part-time in a factory,, correct to the nearest £10, is shown below. Form a, frequency distribution having 6 classes for these, data., 80, , 90, , 70 110, , 140, , 30, , 90, , 90 160 110, , 50 100 110, , 80, , 60 100, , When dealing with unequal class intervals, the histogram must be drawn so that the areas, (and not the, heights), of the rectangles are proportional to the frequencies of the classes. The data given are shown in, columns 1 and 2 of Table 54.3. Columns 3 and 4 give, the upper and lower class boundaries, respectively. In, column 5, the class ranges (i.e. upper class boundary minus lower class boundary values) are listed. The, heights of the rectangles are proportional to the ratio, frequency, , as shown in column 6. The histogram is, class range, shown in Fig. 54.8.
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Presentation ofstatistical data, , 537, , Table 54.3, 2, Frequency, , 3, Upper class boundary, , 4, Lower class boundary, , 5, Class range, , 20–40, , 2, , 45, , 15, , 30, , 2, 1, =, 30 15, , 50–70, , 6, , 75, , 45, , 30, , 6, 3, =, 30 15, , 80–90, , 12, , 95, , 75, , 20, , 12 9, =, 20 15, , 100–110, , 14, , 115, , 95, , 20, , 14 10 12, =, 20, 15, , 120–140, , 4, , 145, , 115, , 30, , 4, 2, =, 30 15, , 150–170, , 2, , 175, , 145, , 30, , 2, 1, =, 30 15, , Frequency per unit, class range, , 1, Class, , 6, Height of rectangle, , The size of each class is given approximately by, , 12/15, 10/15, 8/15, 6/15, 4/15, 2/15, , range, number of classes, , 30, , 60, 85 105 130, Class mid-point values, , 160, , Figure 54.8, , Problem 12. The masses of 50 ingots in, kilograms are measured correct to the nearest 0.1 kg, and the results are as shown below. Produce a, frequency distribution having about 7 classes for, these data and then present the grouped data as, (a) a frequency polygon and (b) a histogram., 8.0 8.6 8.2 7.5 8.0 9.1 8.5 7.6 8.2 7.8, , Since about seven classes are required, the size of each, class is 2.0/7, that is approximately 0.3, and thus the, class limits are selected as 7.1 to 7.3, 7.4 to 7.6, 7.7 to, 7.9, and so on., The class mid-point for the 7.1 to 7.3 class is, 7.35 +7.05, , i.e. 7.2, for the 7.4 to 7.6 class is, 2, 7.65 +7.35, , i.e. 7.5, and so on., 2, To assist with accurately determining the number in, each class, a tally diagram is produced as shown, in Table 54.4. This is obtained by listing the classes, in the left-hand column and then inspecting each of the, 50 members of the set of data in turn and allocating it, Table 54.4, , 8.3 7.1 8.1 8.3 8.7 7.8 8.7 8.5 8.4 8.5, , Class, , 7.7 8.4 7.9 8.8 7.2 8.1 7.8 8.2 7.7 7.5, , 7.1 to 7.3, , 111, , 8.1 7.4 8.8 8.0 8.4 8.5 8.1 7.3 9.0 8.6, , 7.4 to 7.6, , 1111, , 7.4 8.2 8.4 7.7 8.3 8.2 7.9 8.5 7.9 8.0, , 7.7 to 7.9, , 1111 1111, , 8.0 to 8.2, , 1111 1111 1111, , 8.3 to 8.5, , 1111 1111 1, , 8.6 to 8.8, , 1111 1, , 8.9 to 9.1, , 11, , The range of the data is the member having the largest, value minus the member having the smallest value., Inspection of the set of data shows that:, range = 9.1 − 7.1 = 2.0, , Tally
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Class, , 8.7, , 9.0, , 9.15, , 8.4, , 8.85, , 8.1, , 8.55, , 7.8, , Class mid-point values, , Class mid-point Frequency, Figure 54.10, , 7.1 to 7.3, , 7.2, , 3, , 7.4 to 7.6, , 7.5, , 5, , 7.5 to 7.9, , 7.8, , 9, , 8.0 to 8.2, , 8.1, , 14, , 7.1 to 7.3, , 8.1 to 8.5, , 8.4, , 11, , 8.0 to 8.2 14,, , 8.2 to 8.8, , 8.7, , 6, , 8.9 to 9.1, , 8.9 to 9.1, , 9.0, , 2, , Form a cumulative frequency distribution for these, data and draw the corresponding ogive., , Problem 13. The frequency distribution for the, masses in kilograms of 50 ingots is:, , A frequency polygon is shown in Fig. 54.9, the, co-ordinates corresponding to the class midpoint/frequency values, given in Table 54.5. The, co-ordinates are joined by straight lines and the polygon is ‘anchored-down’ at each end by joining to the, next class mid-point value and zero frequency., , Frequency, , 7.5, , 8.25, , 7.2, , Table 54.5, , 7.95, , Histogram, , 7.65, , 14, 12, 10, 8, 6, 4, 2, 0, , 7.35, , to the appropriate class by putting a ‘1’ in the appropriate row. Each fifth ‘1’ allocated to a particular row is, marked as an oblique line to help with final counting., A frequency distribution for the data is shown in, Table 54.5 and lists classes and their corresponding frequencies. Class mid-points are also shown in this table,, since they are used when constructing the frequency, polygon and histogram., , Frequency, , 538 Higher Engineering Mathematics, , 14, 12, 10, 8, 6, 4, 2, 0, , Frequency polygon, , 3,, , 7.4 to 7.6, , 5,, , 7.7 to 7.9 9,, , 8.3 to 8.5 11,, , 8.6 to 8.8, 6,, , 2,, , A cumulative frequency distribution is a table giving values of cumulative frequency for the value of, upper class boundaries, and is shown in Table 54.6., Columns 1 and 2 show the classes and their frequencies. Column 3 lists the upper class boundary values, for the classes given in column 1. Column 4 gives, the cumulative frequency values for all frequencies, less than the upper class boundary values given in, column 3. Thus, for example, for the 7.7 to 7.9 class, Table 54.6, 1, Class, , 7.2, , 8.4, 7.5, 7.8, 8.1, 8.7, Class mid-point values, , 9.0, , 2, 3, 4, Frequency Upper Class Cumulative, boundary, frequency, Less than, , Figure 54.9, , 7.1–7.3, , 3, , 7.35, , 3, , A histogram is shown in Fig. 54.10, the width of, a rectangle corresponding to (upper class boundary, value—lower class boundary value) and height corresponding to the class frequency. The easiest way to draw, a histogram is to mark class mid-point values on the, horizontal scale and to draw the rectangles symmetrically about the appropriate class mid-point values and, touching one another. A histogram for the data given in, Table 54.5 is shown in Fig. 54.10., , 7.4–7.6, , 5, , 7.65, , 8, , 7.7–7.9, , 9, , 7.95, , 17, , 8.0–8.2, , 14, , 8.25, , 31, , 8.3–8.5, , 11, , 8.55, , 42, , 8.6–8.8, , 6, , 8.85, , 48, , 8.9–9.1, , 2, , 9.15, , 50
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Presentation ofstatistical data, shown in row 3, the cumulative frequency value is, the sum of all frequencies having values of less than, 7.95, i.e. 3 +5 + 9 =17, and so on. The ogive for the, cumulative frequency distribution given in Table 54.6, is shown in Fig. 54.11. The co-ordinates corresponding to each upper class boundary/cumulative frequency, value are plotted and the co-ordinates are joined by, straight lines (—not the best curve drawn through, the co-ordinates as in experimental work.) The ogive, is ‘anchored’ at its start by adding the co-ordinate, (7.05, 0)., , 40.1, , 39.7, , 40.5, , 40.5, , 39.9, , 40.8, , 40.0, , 40.2, , 40.0, , 39.9, , 39.8, , 39.7, , 39.5, , 40.1, , 40.2, , 40.6, , 40.1, , 39.7, , 40.2, , 40.3, , ⎡, , ⎤, There is no unique solution,, ⎢ but one solution is:, ⎥, ⎢, ⎥, ⎢ 39.3−39.4 1; 39.5−39.6 5; ⎥, ⎢, ⎥, ⎢ 39.7−39.8 9; 39.9−40.0 17; ⎥, ⎢, ⎥, ⎣ 40.1−40.2 15; 40.3−40.4 7; ⎦, , 50, Cumulative frequency, , 40.5−40.6 4; 40.7−40.8 2, 40, , 2. Draw a histogram for the frequency distribution given in the solution of Problem 1., , 30, , ⎡, , ⎤, Rectangles, touching one another,, ⎢ having mid-points of 39.35,, ⎥, ⎢, ⎥, ⎣ 39.55, 39.75, 39.95, . . . and, ⎦, heights of 1, 5, 9, 17, . . ., , 20, 10, , 7.05, , 7.35 7.65 7.95 8.25 8.55 8.85 9.15, Upper class boundary values in kilograms, , Figure 54.11, , Now try the following exercise, , 3. The information given below refers to the, value of resistance in ohms of a batch of, 48 resistors of similar value. Form a frequency distribution for the data, having about 6, classes, and draw a frequency polygon and histogram to represent these data diagramatically., 21.0 22.4 22.8 21.5 22.6 21.1 21.6 22.3, , Exercise 206 Further problems on, presentation of grouped data, , 22.9 20.5 21.8 22.2 21.0 21.7 22.5 20.7, , 1. The mass in kilograms, correct to the nearest, one-tenth of a kilogram, of 60 bars of metal, are as shown. Form a frequency distribution, of about 8 classes for these data., , 23.2 22.9 21.7 21.4 22.1 22.2 22.3 21.3, 22.1 21.8 22.0 22.7 21.7 21.9 21.1 22.6, 21.4 22.4 22.3 20.9 22.8 21.2 22.7 21.6, 22.2 21.6 21.3 22.1 21.5 22.0 23.4 21.2, , 39.8, , 40.3, , 40.6, , 40.0, , 39.6, , 39.6, , 40.2, , 40.3, , 40.4, , 39.8, , 40.2, , 40.3, , 39.9, , 39.9, , 40.0, , 40.1, , 40.0, , 40.1, , 40.1, , 40.2, , 39.7, , 40.4, , 39.9, , 40.1, , 39.9, , 39.5, , 40.0, , 39.8, , 39.5, , 39.9, , 40.1, , 40.0, , 39.7, , 40.4, , 39.3, , 40.7, , 39.9, , 40.2, , 39.9, , 40.0, , ⎡, , ⎤, There is no unique solution,, ⎢ but one solution is:, ⎥, ⎢, ⎥, ⎢ 20.5–20.9 3; 21.0–21.4 10; ⎥, ⎢, ⎥, ⎣ 21.5–21.9 11; 22.0–22.4 13; ⎦, 22.5–22.9 9; 23.0–23.4 2, 4. The time taken in hours to the failure of 50, specimens of a metal subjected to fatigue failure tests are as shown. Form a frequency distribution, having about 8 classes and unequal, class intervals, for these data., , 539
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540 Higher Engineering Mathematics, , 28 22 23 20 12 24 37 28 21 25, , 2.10, , 2.29, , 2.32, , 2.21, , 2.14, , 2.22, , 21 14 30 23 27 13 23, , 7 26 19, , 2.28, , 2.18, , 2.17, , 2.20, , 2.23, , 2.13, , 3 21 24 28 40 27 24, , 2.26, , 2.10, , 2.21, , 2.17, , 2.28, , 2.15, , 20 25 23 26 47 21 29 26 22 33, , 2.16, , 2.25, , 2.23, , 2.11, , 2.27, , 2.34, , 27, , 2.24, , 2.05, , 2.29, , 2.18, , 2.24, , 2.16, , 2.15, , 2.22, , 2.14, , 2.27, , 2.09, , 2.21, , 2.11, , 2.17, , 2.22, , 2.19, , 2.12, , 2.20, , 2.23, , 2.07, , 2.13, , 2.26, , 2.16, , 2.12, , 24 22 26, , 9 13 35 20 16 20 25 18 22, ⎡, , ⎤, There is no unique solution,, ⎢ but one solution is: 1–10 3; ⎥, ⎢, ⎥, ⎣ 11–19 7; 20–22 12; 23–25 11; ⎦, 26–28 10; 29–38 5; 39–48 2, 5. Form a cumulative frequency distribution and, hence draw the ogive for the frequency distribution given in the solution to Problem 3., , 20.95 3; 21.45 13; 21.95 24;, 22.45 37; 22.95 46; 23.45 48, 6. Draw a histogram for the frequency distribution given in the solution to Problem 4., ⎡, ⎤, Rectangles, touching one another,, ⎢ having mid-points of 5.5, 15,, ⎥, ⎢, ⎥, ⎢ 21, 24, 27, 33.5 and 43.5. The, ⎥, ⎢, ⎥, ⎢ heights of the rectangles (frequency ⎥, ⎢, ⎥, ⎣ per unit class range) are 0.3,, ⎦, 0.78, 4. 4.67, 2.33, 0.5 and 0.2, 7. The frequency distribution for a batch of, 50 capacitors of similar value, measured in, microfarads, is:, ⎡, ⎤, 10.5–10.9, 2, 11.0–11.4 7,, ⎣ 11.5–11.9 10, 12.0–12.4 12, ⎦, 12.5–12.9 11, 13.0–13.4 8, Form a cumulative frequency distribution for, these data., , (10.95 2), (11.45 9), (11.95 11),, (12.45 31), (12.95 42), (13.45 50), 8. Draw an ogive for the data given in the solution, of Problem 7., 9. The diameter in millimetres of a reel of wire, is measured in 48 places and the results are as, shown., , (a), , Form a frequency distribution of diameters having about 6 classes., , (b) Draw a histogram depicting the data., (c), , Form a cumulative frequency distribution., , (d) Draw an ogive for the data., ⎡, ⎤, (a) There is no unique solution,, ⎢, ⎥, but one solution is:, ⎢, ⎥, ⎢, ⎥, 2.05–2.09 3; 2.10–21.4 10; ⎥, ⎢, ⎢, ⎥, 2.15–2.19 11; 2.20–2.24 13; ⎥, ⎢, ⎢, ⎥, 2.25–2.29 9; 2.30–2.34 2 ⎥, ⎢, ⎢, ⎥, ⎢, ⎥, ⎢, ⎥, ⎢ (b) Rectangles, touching one, ⎥, ⎢, ⎥, another, having mid-points of ⎥, ⎢, ⎢, ⎥, 2.07, 2.12 . . .and heights of ⎥, ⎢, ⎢, ⎥, 3, 10, . . ., ⎢, ⎥, ⎢, ⎥, ⎢, ⎥, ⎢, ⎥, ⎢ (c) Using the frequency, ⎥, ⎢, ⎥, distribution given in the, ⎢, ⎥, ⎢, ⎥, solution to part (a) gives:, ⎢, ⎥, ⎢, ⎥, ⎢, ⎥, 2.095 3; 2.145 13; 2.195 24; ⎥, ⎢, ⎢, 2.245 37; 2.295 46; 2.345 48 ⎥, ⎢, ⎥, ⎢, ⎥, ⎢, ⎥, ⎢ (d) A graph of cumulative, ⎥, ⎢, ⎥, ⎢, ⎥, frequency against upper, ⎢, ⎥, ⎢, ⎥, class boundary having, ⎢, ⎥, ⎣, ⎦, the coordinates given, in part (c).
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Chapter 55, , Measures of central tendency, and dispersion, 55.1, , Measures of central tendency, , A single value, which is representative of a set of values,, may be used to give an indication of the general size of, the members in a set, the word ‘average’ often being, used to indicate the single value., The statistical term used for ‘average’ is the arithmetic, mean or just the mean., Other measures of central tendency may be used and, these include the median and the modal values., , 55.2 Mean, median and mode for, discrete data, Mean, The arithmetic mean value is found by adding together, the values of the members of a set and dividing by the, number of members in the set. Thus, the mean of the set, of numbers: {4, 5, 6, 9} is:, 4+5+6+9, , i.e. 6, 4, In general, the mean of the set: {x 1 , x 2, x 3, . . ., x n } is, x=, , ;, x, x1 + x2 + x3 + · · · + xn, , written as, n, n, , ;, where is the Greek letter ‘sigma’ and means ‘the sum, of’, and x (called x-bar) is used to signify a mean value., , Median, The median value often gives a better indication, of the general size of a set containing extreme values. The set: {7, 5, 74, 10} has a mean value of 24,, which is not really representative of any of the values of the members of the set. The median value is, obtained by:, (a), , ranking the set in ascending order of magnitude, and, , (b) selecting the value of the middle member for sets, containing an odd number of members, or finding, the value of the mean of the two middle members, for sets containing an even number of members., For example, the set: {7, 5, 74, 10} is ranked as, {5, 7, 10, 74}, and since it contains an even number of, members (four in this case), the mean of 7 and 10 is, taken, giving a median value of 8.5. Similarly, the set:, {3, 81, 15, 7, 14} is ranked as {3, 7, 14, 15, 81} and the, median value is the value of the middle member, i.e. 14., , Mode, The modal value, or mode, is the most commonly, occurring value in a set. If two values occur with, the same frequency, the set is ‘bi-modal’. The set:, {5, 6, 8, 2, 5, 4, 6, 5, 3} has a model value of 5, since the, member having a value of 5 occurs three times., Problem 1. Determine the mean, median and, mode for the set:, {2, 3, 7, 5, 5, 13, 1, 7, 4, 8, 3, 4, 3}
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542 Higher Engineering Mathematics, The mean value is obtained by adding together the, values of the members of the set and dividing by the, number of members in the set., Thus, mean value,, 2 + 3 + 7 + 5 + 5 + 13 + 1, +7 + 4 + 8 + 3 + 4 + 3, 65, x=, =, =5, 13, 13, To obtain the median value the set is ranked, that is,, placed in ascending order of magnitude, and since the, set contains an odd number of members the value of the, middle member is the median value. Ranking the set, gives:, {1, 2, 3, 3, 3, 4, 4, 5, 5, 7, 7, 8, 13}, The middle term is the seventh member, i.e. 4, thus the, median value is 4. The modal value is the value of, the most commonly occurring member and is 3, which, occurs three times, all other members only occurring, once or twice., Problem 2. The following set of data refers to the, amount of money in £s taken by a news vendor for, 6 days. Determine the mean, median and modal, values of the set:, {27.90, 34.70, 54.40, 18.92, 47.60, 39.68}, 27.90 + 34.70 + 54.40, + 18.92 + 47.60 + 39.68, Mean value =, = £37.20, 6, The ranked set is:, {18.92, 27.90, 34.70, 39.68, 47.60, 54.40}, Since the set has an even number of members, the mean, of the middle two members is taken to give the median, value, i.e., Median value =, , 34.70 + 39.68, = £37.19, 2, , Since no two members have the same value, this set has, no mode., , Now try the following exercise, Exercise 207 Further problems on mean,, median and mode for discrete data, In Problems 1 to 4, determine the mean, median, and modal values for the sets given., 1. {3, 8, 10, 7, 5, 14, 2, 9, 8}, [mean 7 13 , median 8, mode 8], 2. {26, 31, 21, 29, 32, 26, 25, 28}, [mean 27.25, median 27, mode 26], 3. {4.72, 4.71, 4.74, 4.73, 4.72, 4.71, 4.73, 4.72}, [mean 4.7225, median 4.72, mode 4.72], 4. {73.8, 126.4, 40.7, 141.7, 28.5, 237.4, 157.9}, [mean 115.2, median 126.4, no mode], , 55.3 Mean, median and mode for, grouped data, The mean value for a set of grouped data is found by, determining the sum of the (frequency × class mid-point, values) and dividing by the sum of the frequencies,, f1 x 1 + f2 x 2 + · · · + fn x n, f1 + f2 + · · · + fn, ;, ( f x), = ;, f, , i.e. mean value x =, , where f is the frequency of the class having a mid-point, value of x, and so on., Problem 3. The frequency distribution for the, value of resistance in ohms of 48 resistors is as, shown. Determine the mean value of resistance., 20.5–20.9 3, 21.0–21.4 10,, 21.5–21.9 11, 22.0–22.4 13,, 22.5–22.9 9, 23.0–23.4 2, The class mid-point/frequency values are:, 20.7 3, 21.2 10, 21.7 11, 22.2 13,, 22.7 9 and 23.2 2, For grouped data, the mean value is given by:, ;, ( f x), x= ;, f
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Measures of central tendency and dispersion, where f is the class frequency and x is the class midpoint value. Hence mean value,, , Mean, , Median, Mode, 16, , Y, , B, , A, , 14, 5.6, , 12, Frequency, , (3 × 20.7) + (10 × 21.2) + (11 × 21.7), + (13 × 22.2) + (9 × 22.7) + (2 × 23.2), x=, 48, 1052.1, =, = 21.919., 48, , 543, , i.e. the mean value is 21.9 ohms, correct to 3 significant, figures., , C, 24, , 10, , D, , 8, , 32, 16, , 6, 4, , 12, , 10, , 2, , E, , F, , 6, , 14 15 16 17 18 19 20 21 22 23 24 25 26 27, Time in minutes, , Histogram, The mean, median and modal values for grouped data, may be determined from a histogram. In a histogram,, frequency values are represented vertically and variable values horizontally. The mean value is given by, the value of the variable corresponding to a vertical, line drawn through the centroid of the histogram. The, median value is obtained by selecting a variable value, such that the area of the histogram to the left of a vertical, line drawn through the selected variable value is equal, to the area of the histogram on the right of the line. The, modal value is the variable value obtained by dividing, the width of the highest rectangle in the histogram in, proportion to the heights of the adjacent rectangles. The, method of determining the mean, median and modal, values from a histogram is shown in Problem 4., Problem 4. The time taken in minutes to, assemble a device is measured 50 times and the, results are as shown. Draw a histogram depicting, this data and hence determine the mean, median, and modal values of the distribution., 14.5–15.5, , 5, 16.5–17.5, , 8,, , 18.5–19.5 16, 20.5–21.5 12,, 22.5–23.5, , 6, 24.5–25.5, , 3, , The histogram is shown in Fig. 55.1. The mean value, lies at the centroid of the histogram. With reference to, any arbitrary axis, say YY shown at a time of 14 minutes,, the position of the horizontal value of the, ; centroid can be, obtained from the relationship AM = (am), where A, is the area of the histogram, M is the horizontal distance, of the centroid from the axis YY , a is the area of a rectangle of the histogram and m is the distance of the centroid, of the rectangle from YY . The areas of the individual, rectangles are shown circled on the histogram giving a, , Y, , Figure 55.1, , total area of 100 square units. The positions, m, of the, centroids of the individual rectangles are 1, 3, 5, . . .units, from YY . Thus, 100M = (10 × 1) + (16 × 3) + (32 × 5), + (24 × 7) + (12 × 9) + (6 × 11), i.e., , M=, , 560, = 5.6 units from YY, 100, , Thus the position of the mean with reference to the time, scale is 14 + 5.6, i.e. 19.6 minutes., The median is the value of time corresponding to a vertical line dividing the total area of the histogram into two, equal parts. The total area is 100 square units, hence the, vertical line must be drawn to give 50 units of area on, each side. To achieve this with reference to Fig. 55.1,, rectangle ABFE must be split so that 50 −(10 + 16) units, of area lie on one side and 50 − (24 +12 + 6) units of, area lie on the other. This shows that the area of ABFE, is split so that 24 units of area lie to the left of the line, and 8 units of area lie to the right, i.e. the vertical line, must pass through 19.5 minutes. Thus the median value, of the distribution is 19.5 minutes., The mode is obtained by dividing the line AB, which, is the height of the highest rectangle, proportionally to, the heights of the adjacent rectangles. With reference to, Fig. 55.1, this is done by joining AC and BD and drawing, a vertical line through the point of intersection of these, two lines. This gives the mode of the distribution and is, 19.3 minutes.
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544 Higher Engineering Mathematics, Now try the following exercise, , is the root-mean-square value of the members of the set, and for discrete data is obtained as follows:, , Exercise 208 Further problems on mean,, median and mode for grouped data, 1. The frequency distribution given below refers, to the heights in centimetres of 100 people., Determine the mean value of the distribution,, correct to the nearest millimetre., 150–156, , (b) calculate the deviation of each member of the set, from the mean, giving, (x 1 − x), (x 2 − x ), (x 3 − x), . . . ,, , 5, 157–163 18,, , 164–170 20, 171–177 27,, 178–184 22, 185–191, , (c) determine the squares of these deviations, i.e., , 8, [171.7 cm], , 2. The gain of 90 similar transistors is measured, and the results are as shown., 83.5–85.5, 95.5–97.5, , 3, , 3. The diameters, in centimetres, of 60 holes, bored in engine castings are measured and, the results are as shown. Draw a histogram, depicting these results and hence determine, the mean, median and modal values of the, distribution., 7, 2.016–2.019 16,, , 2.021–2.024 23, 2.026–2.029, 2.031–2.034, , 5, , (d) find the sum of the squares of the deviations, that is, , (e) divide by the number of members in the set, n,, giving, , By drawing a histogram of this frequency distribution, determine the mean, median and, modal values of the distribution., [mean 89.5, median 89, mode 88.2], , 2.011–2.014, , (x 1 − x)2 , (x 2 − x )2 , (x 3 − x)2 , . . .,, , (x 1 − x)2 + (x 2 − x )2 + (x 3 − x)2 , . . .,, , 6, 86.5–88.5 39,, , 89.5–91.5 27, 92.5–94.5 15,, , 55.4, , (a) determine the measure of central tendency, usually, the mean value, (occasionally the median or modal, values are specified),, , 9,, , ⎤, mean 2.02158 cm,, ⎣ median 2.02152 cm, ⎦, mode 2.02167 cm, ⎡, , Standard deviation, , (x 1 − x)2 + (x 2 − x )2 + (x 3 − x)2 + · · ·, n, (f) determine the square root of (e)., The standard deviation is indicated by σ (the Greek, letter small ‘sigma’) and is written mathematically as:, 75, 6, 8 ;, 2, 8, (x, −, x, ), Standard deviation, σ = 9, n, where x is a member of the set, x is the mean value of, the set and n is the number of members in the set. The, value of standard deviation gives an indication of the, distance of the members of a set from the mean value., The set: {1, 4, 7, 10, 13} has a mean value of 7 and a, standard deviation of about 4.2. The set {5, 6, 7, 8, 9}, also has a mean value of 7, but the standard deviation is, about 1.4. This shows that the members of the second, set are mainly much closer to the mean value than the, members of the first set. The method of determining the, standard deviation for a set of discrete data is shown in, Problem 5., , (a) Discrete data, The standard deviation of a set of data gives an indication, of the amount of dispersion, or the scatter, of members, of the set from the measure of central tendency. Its value, , Problem 5. Determine the standard deviation, from the mean of the set of numbers:, {5, 6, 8, 4, 10, 3} correct to 4 significant figures.
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Measures of central tendency and dispersion, The arithmetic mean,, ;, x, 5 + 6 + 8 + 4 + 10 + 3, x=, =, =6, n, 6, , ;, (x − x )2, Standard deviation, σ =, n, , 545, , From Problem 3, the distribution mean value,, x = 21.92, correct to 4 significant figures., The ‘x-values’ are the class mid-point values, i.e. 20.7,, 21.2, 21.7, . . ., , (8 − 6)2 ,, , Thus the (x − x )2 values are (20.7 − 21.92)2 ,, (21.2 − 21.92)2 , (21.7 − 21.92)2 , . . ., , The sum of the (x − x )2 values,, <, i.e., (x − x )2 = 1 + 0 + 4 + 4 + 16 + 9 = 34, , and the f (x − x)2 values are 3(20.7 − 21.92)2 ,, 10(21.2 − 21.92)2 , 11(21.7 −21.92)2 , . . ., ;, The, f (x − x )2 values are, , (x − x)2, , (5 − 6)2 ,, , The, values are:, (4 − 6)2 , (10 − 6)2 and (3 − 6)2 ., , (6 − 6)2 ,, , ;, , (x − x )2, 34, =, = 5.6̇, n, 6, since there are 6 members in the set., Hence, standard deviation,, , ;, √, (x − x )2, = 5.6, σ=, n, , 4.4652 + 5.1840 + 0.5324 + 1.0192 + 5.4756, , and, , + 3.2768 = 19.9532, 4, ;:, f (x − x)2, 19.9532, ;, =, = 0.41569, f, 48, , = 2.380, correct to 4 significant figures., , (b) Grouped data, , = 0.645, correct to 3 significant figures., , For grouped data, standard deviation, 75, 6, 8 ;, 8, { f (x − x)2 }, 9, ;, σ=, f, where f is the class frequency value, x is the class midpoint value and x is the mean value of the grouped data., The method of determining the standard deviation for a, set of grouped data is shown in Problem 6., Problem 6. The frequency distribution for the, values of resistance in ohms of 48 resistors is as, shown. Calculate the standard deviation from the, mean of the resistors, correct to 3 significant figures., 20.5–20.9, , 3, 21.0–21.4 10,, , 21.5–21.9 11, 22.0–22.4 13,, 22.5–22.9, , and standard deviation,, 75, 46, 8 ;:, 2, 8, √, x), f, (x, −, ;, = 0.41569, σ =9, f, , 9, 23.0–23.4, , Now try the following exercise, Exercise 209 Further problems on, standard deviation, 1. Determine the standard deviation from the, mean of the set of numbers:, {35, 22, 25, 23, 28, 33, 30}, correct to 3 significant figures., , [4.60], , 2. The values of capacitances, in microfarads, of, ten capacitors selected at random from a large, batch of similar capacitors are:, 34.3, 25.0, 30.4, 34.6, 29.6, 28.7, 33.4,, , 2, 32.7, 29.0 and 31.3, , The standard deviation for grouped data is given by:, ;, σ=, , { f (x − x)2 }, ;, f, , , , Determine the standard deviation from the, mean for these capacitors, correct to 3 significant figures., [2.83 μF]
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546 Higher Engineering Mathematics, 3. The tensile strength in megapascals for 15, samples of tin were determined and found, to be:, 34.61, 34.57, 34.40, 34.63, 34.63,, 34.51, 34.49, 34.61, 34.52, 34.55,, 34.58, 34.53, 34.44, 34.48 and 34.40, Calculate the mean and standard deviation, from the mean for these 15 values, correct to, 4 significant figures., , mean 34.53 MPa, standard, deviation 0.07474 MPa, 4. Determine the standard deviation from the, mean, correct to 4 significant figures, for the, heights of the 100 people given in Problem 1, of Exercise 208, page 544., [9.394 cm], 5. Calculate the standard deviation from the, mean for the data given in Problem 3 of, Exercise 208, page 544, correct to 3 significant, figures., [0.00544 cm], , number of members. These ten parts are then called, deciles. For sets containing a very large number of members, the set may be split into one hundred parts, each, containing an equal number of members. One of these, parts is called a percentile., Problem 7. The frequency distribution given, below refers to the overtime worked by a group of, craftsmen during each of 48 working weeks in a, year., 25–29, , 5, 30–34, , 4, 35–39 7,, , 40–44 11, 45–49 12, 50–54 8,, 55–59, , 1, , Draw an ogive for this data and hence determine, the quartile values., The cumulative frequency distribution (i.e. upper class, boundary/cumulative frequency values) is:, 29.5, , 5, 34.5, , 9, 39.5 16, 44.5 27,, , 49.5 39, 54.5 47, 59.5 48, , 55.5 Quartiles, deciles and, percentiles, Other measures of dispersion which are sometimes, used are the quartile, decile and percentile values., The quartile values of a set of discrete data are, obtained by selecting the values of members which, divide the set into four equal parts. Thus for the set:, {2, 3, 4, 5, 5, 7, 9, 11, 13, 14, 17} there are 11 members, and the values of the members dividing the set into four, equal parts are 4, 7, and 13. These values are signified by Q 1 , Q 2 and Q 3 and called the first, second and, third quartile values, respectively. It can be seen that the, second quartile value, Q 2 , is the value of the middle, member and hence is the median value of the set., For grouped data the ogive may be used to determine the, quartile values. In this case, points are selected on the, vertical cumulative frequency values of the ogive, such, that they divide the total value of cumulative frequency, into four equal parts. Horizontal lines are drawn from, these values to cut the ogive. The values of the variable, corresponding to these cutting points on the ogive give, the quartile values (see Problem 7)., When a set contains a large number of members, the, set can be split into ten parts, each containing an equal, , The ogive is formed by plotting these values on a graph,, as shown in Fig. 55.2. The total frequency is divided, into four equal parts, each having a range of 48/4, i.e., 12. This gives cumulative frequency values of 0 to 12, corresponding to the first quartile, 12 to 24 corresponding to the second quartile, 24 to 36 corresponding to the, third quartile and 36 to 48 corresponding to the fourth, quartile of the distribution, i.e. the distribution is divided, into four equal parts. The quartile values are those of the, variable corresponding to cumulative frequency values, of 12, 24 and 36, marked Q 1 , Q 2 and Q 3 in Fig. 55.2., These values, correct to the nearest hour, are 37 hours,, 43 hours and 48 hours, respectively. The Q 2 value is, also equal to the median value of the distribution. One, measure of the dispersion of a distribution is called the, semi-interquartile range and is given by (Q 3 − Q 1 )/2,, and is (48 −37)/2 in this case, i.e. 5 12 hours., Problem 8. Determine the numbers contained in, the (a) 41st to 50th percentile group, and (b) 8th, decile group of the set of numbers shown below:, 14 22 17 21 30 28 37, , 7 23 32, , 24 17 20 22 27 19 26 21 15 29
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Measures of central tendency and dispersion, , Cumulative frequency, , 50, , are as shown. Determine the median and first, and third quartile values for this data., , 40, , 27 37 40 28 23 30 35 24 30 32 31 2, 30, , [30, 25.5, 33.5 days], 20, , 2. The number of faults occurring on a production line in a nine-week period are as shown, below. Determine the median and quartile, values for the data., , 10, , 25, , 55, 30, 35Q1 40 Q2 45 Q3 50, Upper class boundary values (hours), , 60, , Figure 55.2, , The set is ranked, giving:, 7 14 15 17 17 19 20 21 21 22 22 23, 24 26 27 28 29 30 32 37, (a), , There are 20 numbers in the set, hence the first, 10% will be the two numbers 7 and 14, the second 10% will be 15 and 17, and so on. Thus the, 41st to 50th percentile group will be the numbers, 21 and 22., , (b) The first decile group is obtained by splitting the, ranked set into 10 equal groups and selecting the, first group, i.e. the numbers 7 and 14. The second, decile group are the numbers 15 and 17, and so on., Thus the 8th decile group contains the numbers, 27 and 28., , Now try the following exercise, Exercise 210 Further problems on, quartiles, deciles and percentiles, 1. The number of working days lost due to, accidents for each of 12 one-monthly periods, , 30 27 25 24 27 37 31 27 35, [27, 26, 33 faults], 3. Determine the quartile values and semiinterquartile range for the frequency distribution given in Problem 1 of Exercise 208,, page 544., Q 1 = 164.5 cm, Q 2 = 172.5 cm,, Q 3 = 179 cm, 7.25 cm, 4. Determine the numbers contained in the 5th, decile group and in the 61st to 70th percentile, groups for the set of numbers:, 40 46 28 32 37 42 50 31 48 45, 32 38 27 33 40 35 25 42 38 41, [37 and 38; 40 and 41], 5. Determine the numbers in the 6th decile group, and in the 81st to 90th percentile group for the, set of numbers:, 43 47 30 25 15 51 17, 36 44 33 17 35 58 51, , 21, 35, , 37 33 44 56 40 49 22, 44 40 31 41 55 50 16, [40, 40, 41; 50, 51, 51], , 547
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Chapter 56, , Probability, 56.1, , Introduction to probability, , The probability of something happening is the likelihood or chance of it happening. Values of probability lie, between 0 and 1, where 0 represents an absolute impossibility and 1 represents an absolute certainty. The probability of an event happening usually lies somewhere, between these two extreme values and is expressed, either as a proper or decimal fraction. Examples of, probability are:, that a length of copper wire, has zero resistance at 100◦C, , 0, , that a fair, six-sided dice will, stop with a 3 upwards, , 1, 6, , or 0.1667, , that a fair coin will land with, 1, a head upwards, 2 or 0.5, that a length of copper wire has, 1, some resistance at 100◦C, If p is the probability of an event happening and q is the, probability of the same event not happening, then the, total probability is p + q and is equal to unity, since it is, an absolute certainty that the event either does or does, not occur, i.e. p + q = 1, , Expectation, The expectation, E, of an event happening is defined, in general terms as the product of the probability p of, an event happening and the number of attempts made,, n, i.e. E = pn., Thus, since the probability of obtaining a 3 upwards, when rolling a fair dice is 16 , the expectation of getting, a 3 upwards on four throws of the dice is 16 × 4, i.e. 23, Thus expectation is the average occurrence of an event., , Dependent event, A dependent event is one in which the probability of, an event happening affects the probability of another, event happening. Let 5 transistors be taken at random, from a batch of 100 transistors for test purposes, and the, probability of there being a defective transistor, p1 , be, determined. At some later time, let another 5 transistors, be taken at random from the 95 remaining transistors in, the batch and the probability of there being a defective, transistor, p2, be determined. The value of p2 is different, from p1 since batch size has effectively altered from 100, to 95, i.e. probability p2 is dependent on probability p1 ., Since 5 transistors are drawn, and then another 5 transistors drawn without replacing the first 5, the second, random selection is said to be without replacement., , Independent event, An independent event is one in which the probability, of an event happening does not affect the probability, of another event happening. If 5 transistors are taken at, random from a batch of transistors and the probability of, a defective transistor p1 is determined and the process, is repeated after the original 5 have been replaced in, the batch to give p2 , then p1 is equal to p2 . Since the, 5 transistors are replaced between draws, the second, selection is said to be with replacement., , Conditional probability, Conditional probability is concerned with the probability of say event B occurring, given that event A has, already taken place., If A and B are independent events, then the fact that, event A has already occurred will not affect the probability of event B., If A and B are dependent events, then event A having, occurred will effect the probability of event B.
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Probability, 56.2, , thus the total probability,, , Laws of probability, , 20 33, +, =1, 53 53, hence no obvious error has been made)., p+q =, , The addition law of probability, The addition law of probability is recognized by the, word ‘or’ joining the probabilities. If pA is the probability of event A happening and pB is the probability, of event B happening, the probability of event A or, event B happening is given by pA + pB (provided events, A and B are mutually exclusive, i.e. A and B are events, which cannot occur together). Similarly, the probability, of events A or B or C or . . . N happening is given by, pA + pB + pC + · · · + pN, , 549, , Problem 2. Find the expectation of obtaining a 4, upwards with 3 throws of a fair dice., Expectation is the average occurrence of an event and is, defined as the probability times the number of attempts., The probability, p, of obtaining a 4 upwards for one, throw of the dice is 16, Also, 3 attempts are made, hence n =3 and the, expectation, E, is pn, i.e. E = 16 × 3 = 12 or 0.50, , The multiplication law of probability, The multiplication law of probability is recognized by, the word ‘and’ joining the probabilities. If pA is the, probability of event A happening and pB is the probability of event B happening, the probability of event A, and event B happening is given by pA × pB . Similarly,, the probability of events A and B and C and . . .N happening is given by, pA × pB × pC × · · · × pN, , 56.3, , Worked problems on probability, , Problem 1. Determine the probabilities of, selecting at random (a) a man, and (b) a woman, from a crowd containing 20 men and 33 women., (a), , The probability of selecting at random a man, p, is, given by the ratio, number of men, number in crowd, i.e. p =, , 20, 20, =, or 0.3774, 20 + 33 53, , (b) The probability of selecting at random a women,, q, is given by the ratio, number of women, number in crowd, i.e. q =, , 33, 33, =, or 0.6226, 20 + 33 53, , (Check: the total probability should be equal to 1;, p=, , 20, 33, and q =, 53, 53, , Problem 3. Calculate the probabilities of, selecting at random:, (a) the winning horse in a race in which 10 horses, are running,, (b) the winning horses in both the first and second, races if there are 10 horses in each race., (a), , Since only one of the ten horses can win, the probability of selecting at random the winning horse is, 1, number of winners, , i.e., or 0.10, number of horses, 10, (b) The probability of selecting the winning horse in, 1, . The probability of selecting, the first race is 10, 1, the winning horse in the second race is 10, . The, probability of selecting the winning horses in the, first and second race is given by the multiplication, law of probability, i.e., probability =, =, , 1, 1, ×, 10 10, 1, or 0.01, 100, , Problem 4. The probability of a component, failing in one year due to excessive temperature is, 1, 1, , due to excessive vibration is, and due to, 20, 25, 1, excessive humidity is . Determine the, 50, probabilities that during a one-year period a, component: (a) fails due to excessive temperature, and excessive vibration, (b) fails due to excessive, vibration or excessive humidity, and (c) will not fail, because of both excessive temperature and, excessive humidity.
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550 Higher Engineering Mathematics, Let pA be the probability of failure due to excessive, temperature, then, pA =, , 1, 19, and pA =, 20, 20, , (where pA is the probability of not failing)., Let pB be the probability of failure due to excessive, vibration, then, pB =, , 1, 24, and pB =, 25, 25, , Let pC be the probability of failure due to excessive, humidity, then, pC =, (a), , 1, 49, and pC =, 50, 50, , The probability of a component failing due to, excessive temperature and excessive vibration is, given by:, pA × pB =, , 1, 1, 1, ×, =, or 0.002, 20 25 500, , (b) The probability of a component failing due to, excessive vibration or excessive humidity is:, pB + pC =, (c), , 1, 1, 3, +, =, or 0.06, 25 50 50, , The probability that a component will not fail due, to excessive temperature and will not fail due to, excess humidity is:, pA × pC =, , 19 49, 931, ×, =, or 0.931, 20 50 1000, , Problem 5. A batch of 100 capacitors contains, 73 which are within the required tolerance values,, 17 which are below the required tolerance values,, and the remainder are above the required tolerance, values. Determine the probabilities that when, randomly selecting a capacitor and then a second, capacitor: (a) both are within the required tolerance, values when selecting with replacement, and (b) the, first one drawn is below and the second one drawn, is above the required tolerance value, when, selection is without replacement., (a), , The probability of selecting a capacitor within the, 73, . The first capacrequired tolerance values is, 100, itor drawn is now replaced and a second one is, drawn from the batch of 100. The probability of, , this capacitor being within the required tolerance, 73, ., values is also, 100, Thus, the probability of selecting a capacitor within, the required tolerance values for both the first and, the second draw is, 73, 5329, 73, ×, =, or 0.5329, 100 100 10000, (b) The probability of obtaining a capacitor below the, 17, required tolerance values on the first draw is, ., 100, There are now only 99 capacitors left in the batch,, since the first capacitor is not replaced. The probability of drawing a capacitor above the required tol10, erance values on the second draw is , since there, 99, are (100 −73 − 17), i.e. 10 capacitors above the, required tolerance value. Thus, the probability of, randomly selecting a capacitor below the required, tolerance values and followed by randomly selecting a capacitor above the tolerance’ values is, 17, 10, 170, 17, ×, =, =, or 0.0172, 100 99 9900 990, Now try the following exercise, Exercise 211 Further problems on, probability, 1. In a batch of 45 lamps there are 10 faulty, lamps. If one lamp is drawn at random, find, the probability of it being (a) faulty and, (b) satisfactory., ⎤, ⎡, 2, or, 0.2222, (a), ⎥, ⎢, 9, ⎥, ⎢, ⎦, ⎣, 7, (b) or 0.7778, 9, 2. A box of fuses are all of the same shape and, size and comprises 23 2 A fuses, 47 5 A fuses, and 69 13 A fuses. Determine the probability, of selecting at random (a) a 2 A fuse, (b) a 5 A, fuse and (c) a 13 A fuse., ⎡, ⎤, 23, (a), or 0.1655, ⎢, ⎥, 139, ⎢, ⎥, ⎢, ⎥, 47, ⎢ (b), or 0.3381 ⎥, ⎢, ⎥, 139, ⎢, ⎥, ⎣, ⎦, 69, or 0.4964, (c), 139
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Probability, , 3. (a) Find the probability of having a 2 upwards, when throwing a fair 6-sided dice. (b) Find, the probability of having a 5 upwards when, throwing a fair 6-sided dice. (c) Determine, the probability of having a 2 and then a 5, on two successive throws of a fair 6-sided, dice., , 1, 1, 1, (a) (b) (c), 6, 6, 36, 4. Determine the probability that the total score, is 8 when two like dice are thrown., , 5, 36, 5. The probability of event A happening is 35 and, the probability of event B happening is 23 . Calculate the probabilities of (a) both A and B, happening, (b) only event A happening, i.e., event A happening and event B not happening, (c) only event B happening, and (d) either, A, or B, or A and B happening., , 1, 4, 13, 2, (d), (a) (b) (c), 5, 5, 15, 15, 6. When testing 1000 soldered joints, 4 failed, during a vibration test and 5 failed due to, having a high resistance. Determine the probability of a joint failing due to (a) vibration,, (b) high resistance, (c) vibration or high, resistance and (d) vibration and high, resistance., ⎤, ⎡, 1, 1, (b), (a), ⎢, 250, 200 ⎥, ⎥, ⎢, ⎣, 9, 1 ⎦, (c), (d), 1000, 50000, , 56.4 Further worked problems on, probability, , neither of the components is defective when drawn, (a) with replacement, and (b) without replacement., (a) With replacement, The probability that the component selected on the first, 35, 7, draw is satisfactory is , i.e. . The component is now, 40, 8, replaced and a second draw is made. The probability, 7, that this component is also satisfactory is . Hence, the, 8, probability that both the first component drawn and the, second component drawn are satisfactory is:, 7 7 49, × =, or 0.7656, 8 8 64, (b) Without replacement, The probability that the first component drawn is sat7, isfactory is . There are now only 34 satisfactory, 8, components left in the batch and the batch number is 39., Hence, the probability of drawing a satisfactory compo34, nent on the second draw is . Thus the probability that, 39, the first component drawn and the second component, drawn are satisfactory, i.e. neither is defective, is:, 7 34 238, ×, =, or 0.7628, 8 39 312, Problem 7. A batch of 40 components contains, 5 which are defective. If a component is drawn at, random from the batch and tested and then a second, component is drawn at random, calculate the, probability of having one defective component,, both with and without replacement., The probability of having one defective component can, be achieved in two ways. If p is the probability of drawing a defective component and q is the probability of, drawing a satisfactory component, then the probability, of having one defective component is given by drawing, a satisfactory component and then a defective component or by drawing a defective component and then a, satisfactory one, i.e. by q × p + p ×q, With replacement:, , Problem 6. A batch of 40 components contains, 5 which are defective. A component is drawn at, random from the batch and tested and then a second, component is drawn. Determine the probability that, , 551, , 1, 5, =, 40 8, 35 7, q=, =, 40 8, p=, , and
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552 Higher Engineering Mathematics, Hence, probability of having one defective component is:, 1 7 7 1, × + ×, 8 8 8 8, i.e., 7, 7, 7, +, =, or 0.2188, 64 64 32, Without replacement:, 1, 7, p1 = and q1 = on the first of the two draws. The, 8, 8, batch number is now 39 for the second draw, thus,, p2 =, p1 q2 + q1 p2 =, , 5, 35, and q2 =, 39, 39, 1 35 7, 5, ×, + ×, 8 39 8 39, , =, , 35 + 35, 312, , =, , 70, or 0.2244, 312, , Problem 8. A box contains 74 brass washers,, 86 steel washers and 40 aluminium washers. Three, washers are drawn at random from the box without, replacement. Determine the probability that all, three are steel washers., Assume, for clarity of explanation, that a washer is, drawn at random, then a second, then a third (although, this assumption does not affect the results obtained)., The total number of washers is 74 + 86 + 40, i.e. 200., The probability of randomly selecting a steel washer on, 86, . There are now 85 steel washers in, the first draw is, 200, a batch of 199. The probability of randomly selecting a, 85, steel washer on the second draw is, . There are now, 199, 84 steel washers in a batch of 198. The probability of, randomly selecting a steel washer on the third draw is, 84, . Hence the probability of selecting a steel washer, 198, 84, on the third draw is, . Hence the probability of select198, ing a steel washer on the first draw and the second draw, and the third draw is:, 86, 85, 84, 614040, ×, ×, =, = 0.0779, 200 199 198 7880400, , Problem 9. For the box of washers given in, Problem 8 above, determine the probability that, there are no aluminium washers drawn, when three, washers are drawn at random from the box without, replacement., The probability of not, an aluminium washer on, , drawing, 160, 40, , i.e., . There are now 199, the first draw is 1 −, 200, 200, washers in the batch of which 159 are not aluminium, washers. Hence, the probability of not drawing an alu159, . Similarly,, minium washer on the second draw is, 199, the probability of not drawing an aluminium washer on, 158, the third draw is, . Hence the probability of not draw198, ing an aluminium washer on the first and second and, third draws is, 160 159 158 4019520, ×, ×, =, = 0.5101, 200 199 198 7880400, Problem 10. For the box of washers in, Problem 8 above, find the probability that there are, two brass washers and either a steel or an, aluminium washer when three are drawn at random,, without replacement., Two brass washers (A) and one steel washer (B) can be, obtained in any of the following ways:, 1st draw, , 2nd draw, , 3rd draw, , A, , A, , B, , A, , B, , A, , B, , A, , A, , Two brass washers and one aluminium washer (C) can, also be obtained in any of the following ways:, 1st draw, , 2nd draw, , 3rd draw, , A, , A, , C, , A, , C, , A, , C, , A, , A, , Thus there are six possible ways of achieving the, combinations specified. If A represents a brass washer,
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Probability, B a steel washer and C an aluminium washer, then the, combinations and their probabilities are as shown:, Draw, , Probability, , First Second Third, A, , A, , B, , 73, 86, 74, ×, ×, = 0.0590, 200 199 198, , A, , B, , A, , 86, 73, 74, ×, ×, = 0.0590, 200 199 198, , B, , A, , A, , 74, 73, 86, ×, ×, = 0.0590, 200 199 198, , A, , A, , C, , 73, 40, 74, ×, ×, = 0.0274, 200 199 198, , A, , C, , A, , 40, 73, 74, ×, ×, = 0.0274, 200 199 198, , C, , A, , A, , 74, 73, 40, ×, ×, = 0.0274, 200 199 198, , are visited, calculate the probabilities that, (a) they both have a telephone and (b) one, has a telephone but the other does not have, telephone., [(a) 0.64 (b) 0.32], 3. Veroboard pins are packed in packets of 20, by a machine. In a thousand packets, 40 have, less than 20 pins. Find the probability that if 2, packets are chosen at random, one will contain, less than 20 pins and the other will contain 20, pins or more., [0.0768], 4. A batch of 1 kW fire elements contains 16, which are within a power tolerance and 4, which are not. If 3 elements are selected at, random from the batch, calculate the probabilities that (a) all three are within the power, tolerance and (b) two are within but one is not, within the power tolerance., [(a) 0.4912 (b) 0.4211], , The probability of having the first combination or the, second, or the third, and so on, is given by the sum of, the probabilities,, i.e. by 3 × 0.0590 +3 × 0.0274, that is, 0.2592, , Now try the following exercise, Exercise 212, probability, , Further problems on, , 1. The probability that component A will operate, satisfactorily for 5 years is 0.8 and that B will, operate satisfactorily over that same period of, time is 0.75. Find the probabilities that in a, 5 year period: (a) both components operate satisfactorily, (b) only component A will operate, satisfactorily, and (c) only component B will, operate satisfactorily., [(a) 0.6 (b) 0.2 (c) 0.15], 2. In a particular street, 80% of the houses have, telephones. If two houses selected at random, , 5. An amplifier is made up of three transistors,, A, B and C. The probabilities of A, B or C, 1, 1 1, ,, and, , respecbeing defective are, 20 25, 50, tively. Calculate the percentage of amplifiers, produced (a) which work satisfactorily and, (b) which have just one defective transistor., (a) 89.38%, (b) 10.25%, 6. A box contains 14 40 W lamps, 28 60 W, lamps and 58 25 W lamps, all the lamps being, of the same shape and size. Three lamps are, drawn at random from the box, first one, then a, second, then a third. Determine the probabilities of: (a) getting one 25 W, one 40 W and one, 60 W lamp, with replacement, (b) getting one, 25 W, one 40 W and one 60 W lamp without, replacement, and (c) getting either one 25 W, and two 40 W or one 60 W and two 40 W lamps, with replacement., [(a) 0.0227 (b) 0.0234 (c) 0.0169], , 553
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Revision Test 16, This Revision Test covers the material contained in Chapters 54 to 56. The marks for each question are shown in, brackets at the end of each question., 1., , A company produces five products in the following, proportions:, , Class, intervals (mm), , Product A 24 Product B 16 Product C 15, Product D 11 Product E 6, Present these data visually by drawing (a) a vertical, bar chart, (b) a percentage component bar chart,, (c) a pie diagram., (13), 2., , The following lists the diameters of 40 components, produced by a machine, each measured correct to, the nearest hundredth of a centimetre:, 1.39, 1.40, 1.36, 1.38, 1.37, 1.41, , 3., , 1.36, 1.24, 1.36, 1.35, 1.34, 1.35, , 1.38, 1.28, 1.35, 1.42, 1.34, 1.38, , 1.31, 1.42, 1.45, 1.30, 1.32, 1.27, , 1.33, 1.34, 1.29, 1.26, 1.33, 1.37, , 1.40, 1.43, 1.39, 1.37, 1.30, , 1.28, 1.35, 1.38, 1.33, 1.38, , 1.24–1.26, , 2, , 2, , 1.27–1.29, , 4, , 6, , 1.30–1.32, , 4, , 10, , 1.33–1.35, , 10, , 20, , 1.36–1.38, , 11, , 31, , 1.39–1.41, , 5, , 36, , 1.42–1.44, , 3, , 39, , 1.45–1.47, , 1, , 40, (10), , 6., , Determine the probabilities of:, (a) drawing a white ball from a bag containing, 6 black and 14 white balls,, (b) winning a prize in a raffle by buying 6 tickets, when a total of 480 tickets are sold,, , Determine for the 10 measurements of lengths, shown below:, , (c) selecting at random a female from a group of, 12 boys and 28 girls,, , (a) the arithmetic mean, (b) the median, (c) the, mode, and (d) the standard deviation., , (d) winning a prize in a raffle by buying 8 tickets, when there are 5 prizes and a total of 800 tickets, are sold., (8), 7., , The heights of 100 people are measured correct to, the nearest centimetre with the following results:, 150–157 cm, 5 158–165 cm, 166–173 cm 42 174–181 cm, 182–189 cm, 8, , 18, 27, , Draw an ogive for the data of component measurements given below, and hence determine the, median and the first and third quartile values for, this distribution., , The probabilities of an engine failing are given by:, p1, failure due to overheating; p2 , failure due to, ignition problems; p3 , failure due to fuel blockage., 1, 1, 2, When p1 = , p2 = and p3 = , determine the, 8, 5, 7, probabilities of:, (a) all three failures occurring,, (b) the first and second but not the third failure, occurring,, (c) only the second failure occurring,, (d) the first or the second failure occurring but not, the third., (12), , Determine for the data (a) the mean height and, (b) the standard deviation., (12), 5., , Cumulative, frequency, , (a) Using 8 classes form a frequency distribution, and a cumulative frequency distribution., (b) For the above data draw a histogram, a frequency polygon and an ogive., (21), , 28 m, 20 m, 32 m, 44 m, 28 m, 30 m, 30 m, 26 m,, 28 m and 34 m, (10), 4., , Frequency, , 8., , In a box containing 120 similar transistors 70 are, satisfactory, 37 give too high a gain under normal
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Revision Test 16, operating conditions and the remainder give too low, a gain., Calculate the probability that when drawing two, transistors in turn, at random, with replacement,, of having, (a) two satisfactory,, , 555, , (b) none with low gain,, (c) one with high gain and one satisfactory,, (d) one with low gain and none satisfactory., Determine the probabilities in (a), (b) and (c), above if the transistors are drawn without, replacement., (14)
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Chapter 57, , The binomial and, Poisson distributions, 57.1, , The binomial distribution, , The binomial distribution deals with two numbers only,, these being the probability that an event will happen, p,, and the probability that an event will not happen, q., Thus, when a coin is tossed, if p is the probability of the, coin landing with a head upwards, q is the probability of, the coin landing with a tail upwards. p + q must always, be equal to unity. A binomial distribution can be used, for finding, say, the probability of getting three heads in, seven tosses of the coin, or in industry for determining, defect rates as a result of sampling. One way of defining, a binomial distribution is as follows:, ‘if p is the probability that an event will happen and, q is the probability that the event will not happen,, then the probabilities that the event will happen, 0, 1, 2, 3, . . . ,n times in n trials are given by the, successive terms of the expansion of (q + p)n, taken, from left to right’., , The binomial expansion of (q +, is:, n(n − 1) n−2 2, p, q n + nq n−1 p +, q, 2!, n(n − 1)(n − 2) n−3 3, +, p +···, q, 3!, from Chapter 7., This concept of a binomial distribution is used in, Problems 1 and 2., p)n, , Problem 1. Determine the probabilities of having, (a) at least 1 girl and (b) at least 1 girl and 1 boy in a, , family of 4 children, assuming equal probability of, male and female birth., The probability of a girl being born, p, is 0.5 and the, probability of a girl not being born (male birth), q,, is also 0.5. The number in the family, n, is 4. From, above, the probabilities of 0, 1, 2, 3, 4 girls in a family, of 4 are given by the successive terms of the expansion, of (q + p)4 taken from left to right. From the binomial, expansion:, (q + p)4 = q 4 + 4q 3 p + 6q 2 p2 + 4q p 3 + p 4, Hence the probability of no girls is q 4,, 0.54 = 0.0625, , i.e., the probability of 1 girl is 4q 3 p,, i.e., , 4 × 0.53 × 0.5 = 0.2500, , the probability of 2 girls is 6q 2 p2 ,, i.e., , 6 × 0.52 × 0.52 = 0.3750, , the probability of 3 girls is 4q p3,, i.e., , 4 × 0.5 × 0.53 = 0.2500, , the probability of 4 girls is p4,, i.e., , 0.54 = 0.0625, Total probability, (q + p)4 = 1.0000, , (a), , The probability of having at least one girl is the, sum of the probabilities of having 1, 2, 3 and 4, girls, i.e., 0.2500 + 0.3750 + 0.2500 + 0.0625 = 0.9375
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The binomial and Poisson distributions, (Alternatively, the probability of having at least, 1 girl is: 1 − (theprobability of having no girls), i.e., 1 − 0.0625, giving 0.9375, as obtained previously.), (b) The probability of having at least 1 girl and, 1 boy is given by the sum of the probabilities of, having: 1 girl and 3 boys, 2 girls and 2 boys and 3, girls and 2 boys, i.e., , Industrial inspection, In industrial inspection, p is often taken as the, probability that a component is defective and q is the, probability that the component is satisfactory. In this, case, a binomial distribution may be defined as:, ‘the probabilities that 0, 1, 2, 3,… , n components, are defective in a sample of n components, drawn, at random from a large batch of components, are, given by the successive terms of the expansion of, (q + p)n , taken from left to right’., , 0.2500 + 0.3750 + 0.2500 = 0.8750, (Alternatively, this is also the probability of having, 1 − (probability of having no girls + probability of, having no boys), i.e., 1 −2 × 0.0625 =0.8750, as obtained previously.), Problem 2. A dice is rolled 9 times. Find the, probabilities of having a 4 upwards (a) 3 times and, (b) less than 4 times., Let p be the probability of having a 4 upwards. Then, p = 1/6, since dice have six sides., Let q be the probability of not having a 4 upwards., Then q = 5/6. The probabilities of having a 4 upwards, 0, 1, 2, . . ., n times are given by the successive terms of, the expansion of (q + p)n , taken from left to right. From, the binomial expansion:, (q + p)9 = q 9 + 9q 8 p + 36q 7 p2 + 84q 6 p3 + · · ·, The probability of having a 4 upwards no times is, q 9 = (5/6)9 = 0.1938, , 557, , This definition is used in Problems 3 and 4., Problem 3. A machine is producing a large, number of bolts automatically. In a box of these, bolts, 95% are within the allowable tolerance values, with respect to diameter, the remainder being, outside of the diameter tolerance values. Seven, bolts are drawn at random from the box. Determine, the probabilities that (a) two and (b) more than two, of the seven bolts are outside of the diameter, tolerance values., Let p be the probability that a bolt is outside of the, allowable tolerance values, i.e. is defective, and let q be, the probability that a bolt is within the tolerance values,, i.e. is satisfactory. Then p = 5%, i.e. 0.05 per unit and, q = 95%, i.e. 0.95 per unit. The sample number is 7., The probabilities of drawing 0, 1, 2, . . . , n defective, bolts are given by the successive terms of the expansion, of (q + p)n , taken from left to right. In this problem, , The probability of having a 4 upwards once is, (q + p)n = (0.95 + 0.05)7, , 9q 8 p = 9(5/6)8(1/6) = 0.3489, , = 0.957 + 7 × 0.956 × 0.05, , The probability of having a 4 upwards twice is, 36q 7 p2 = 36(5/6)7(1/6)2 = 0.2791, The probability of having a 4 upwards 3 times is, , + 21 × 0.955 × 0.052 + · · ·, Thus the probability of no defective bolts is, , 84q 6 p3 = 84(5/6)6(1/6)3 = 0.1302, (a) The probability of having a 4 upwards 3 times is, 0.1302, (b) The probability of having a 4 upwards less than 4, times is the sum of the probabilities of having a 4, upwards 0, 1, 2, and 3 times, i.e., 0.1938 + 0.3489 + 0.2791 + 0.1302 = 0.9520, , 0.957 = 0.6983, The probability of 1 defective bolt is, 7 × 0.956 × 0.05 = 0.2573, The probability of 2 defective bolts is, 21 × 0.955 × 0.052 = 0.0406, and so on., (a), , The probability that two bolts are outside of the, diameter tolerance values is 0.0406
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558 Higher Engineering Mathematics, (b) To determine the probability that more than, two bolts are defective, the sum of the probabilities of 3 bolts, 4 bolts, 5 bolts, 6 bolts and 7 bolts, being defective can be determined. An easier way, to find this sum is to find 1 − (sum of 0 bolts, 1 bolt, and 2 bolts being defective), since the sum of all, the terms is unity. Thus, the probability of there, being more than two bolts outside of the tolerance, values is:, 1 − (0.6983 + 0.2573 + 0.0406), i.e. 0.0038, , probabilities of 0, 1, 2, . . ., 10 students successfully, completing the course in three years., Let p be the probability of a student successfully completing a course of study in three years and q be the, probability of not doing so. Then p = 0.45 and q =, 0.55. The number of students, n, is 10., The probabilities of 0, 1, 2, . . ., 10 students successfully, completing the course are given by the successive terms, of the expansion of (q + p)10 , taken from left to right., (q + p)10 = q 10 + 10q 9 p + 45q 8 p2 + 120q 7 p3, , The probability of a component being damaged, p,, is 4 in 50, i.e. 0.08 per unit. Thus, the probability of a, component not being damaged, q, is 1 − 0.08, i.e. 0.92., The probability of there being 0, 1, 2, . . ., 6 damaged, components is given by the successive terms of, (q + p)6 , taken from left to right., (q + p)6 = q 6 + 6q 5 p + 15q 4 p2 + 20q 3 p3 + · · ·, (a) The probability of one damaged component is, 6q 5 p = 6 × 0.925 × 0.08 = 0.3164, (b) The probability of less than three damaged components is given by the sum of the probabilities of, 0, 1 and 2 damaged components., q 6 + 6q 5 p + 15q 4 p2, = 0.926 + 6 × 0.925 × 0.08, + 15 × 0.924 × 0.082, = 0.6064 + 0.3164 + 0.0688 = 0.9916, , + 210q 6 p4 + 252q 5 p5 + 210q 4 p6, + 120q 3 p7 + 45q 2 p8 + 10q p9 + p 10, Substituting q = 0.55 and p = 0.45 in this expansion, gives the values of the successive terms as: 0.0025,, 0.0207, 0.0763, 0.1665, 0.2384, 0.2340, 0.1596, 0.0746,, 0.0229, 0.0042 and 0.0003. The histogram depicting, these probabilities is shown in Fig. 57.1., 0.24, 0.22, 0.20, Probability of successfully completing course, , Problem 4. A package contains 50 similar, components and inspection shows that four, have been damaged during transit. If six, components are drawn at random from the contents, of the package determine the probabilities that in, this sample (a) one and (b) less than three are, damaged., , 0.18, 0.16, 0.14, 0.12, 0.10, 0.08, 0.06, 0.04, , Histogram of probabilities, The terms of a binomial distribution may be represented pictorially by drawing a histogram, as shown in, Problem 5., Problem 5. The probability of a student, successfully completing a course of study in three, years is 0.45. Draw a histogram showing the, , 0.02, 0, , Figure 57.1, , 0 1 2 3 4 5 6 7 8 9 10, Number of students
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The binomial and Poisson distributions, , 559, , Now try the following exercise, Exercise 213 Further problems on the, binomial distribution, 1. Concrete blocks are tested and it is found, that, on average, 7% fail to meet the required, specification. For a batch of 9 blocks, determine the probabilities that (a) three blocks and, (b) less than four blocks will fail to meet the, specification., [(a) 0.0186 (b) 0.9976], 2. If the failure rate of the blocks in Problem 1, rises to 15%, find the probabilities that (a) no, blocks and (b) more than two blocks will fail, to meet the specification in a batch of 9 blocks., [(a) 0.2316 (b) 0.1408], 3. The average number of employees absent, from a firm each day is 4%. An office within, the firm has seven employees. Determine the, probabilities that (a) no employee and (b) three, employees will be absent on a particular day., [(a) 0.7514 (b) 0.0019], 4. A manufacturer estimates that 3% of his output, of a small item is defective. Find the probabilities that in a sample of 10 items (a) less, than two and (b) more than two items will be, defective., [(a) 0.9655 (b) 0.0028], 5. Five coins are tossed simultaneously. Determine the probabilities of having 0, 1, 2, 3, 4, and 5 heads upwards, and draw a histogram, depicting the results., ⎤, ⎡, Vertical adjacent rectangles,, ⎥, ⎢, ⎢ whose heights are proportional to⎥, ⎥, ⎢, ⎣ 0.0313, 0.1563, 0.3125, 0.3125, ⎦, 0.1563 and 0.0313, 6. If the probability of rain falling during a particular period is 2/5, find the probabilities of, having 0, 1, 2, 3, 4, 5, 6 and 7 wet days in a, week. Show these results on a histogram., ⎤, ⎡, Vertical adjacent rectangles,, ⎥, ⎢, ⎢ whose heights are proportional⎥, ⎥, ⎢, ⎥, ⎢ to 0.0280, 0.1306, 0.2613,, ⎥, ⎢, ⎦, ⎣ 0.2903, 0.1935, 0.0774,, 0.0172 and 0.0016, 7. An automatic machine produces, on average, 10% of its components outside of the, , tolerance required. In a sample of 10 components from this machine, determine the probability of having three components outside of, the tolerance required by assuming a binomial, distribution., [0.0574], , 57.2, , The Poisson distribution, , When the number of trials, n, in a binomial distribution, becomes large (usually taken as larger than 10), the calculations associated with determining the values of the, terms becomes laborious. If n is large and p is small,, and the product np is less than 5, a very good approximation to a binomial distribution is given by the corresponding Poisson distribution, in which calculations, are usually simpler., The Poisson approximation to a binomial distribution, may be defined as follows:, ‘the probabilities that an event will happen 0, 1, 2,, 3, … , n times in n trials are given by the successive, terms of the expression, , , , λ2 λ3, e−λ 1 + λ +, +, +···, 2! 3!, taken from left to right’., , The symbol λ is the expectation of an event happening, and is equal to np., Problem 6. If 3% of the gearwheels produced, by a company are defective, determine the, probabilities that in a sample of 80 gearwheels, (a) two and (b) more than two will be defective., The sample number, n, is large, the probability of a, defective gearwheel, p, is small and the product np is, 80 × 0.03, i.e. 2.4, which is less than 5., Hence a Poisson approximation to a binomial distribution may be used. The expectation of a defective, gearwheel, λ = np = 2.4, The probabilities of 0, 1, 2, . . . defective gearwheels, are given by the successive terms of the expression, , , λ2 λ3, +, +···, e−λ 1 + λ +, 2! 3!
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560 Higher Engineering Mathematics, taken from left to right, i.e. by, λ2 e−λ, ,..., e−λ , λe−λ ,, 2!, Thus probability of no defective gearwheels is, e−λ = e−2.4 = 0.0907, probability of 1 defective gearwheel is, λe−λ = 2.4e−2.4 = 0.2177, probability of 2 defective gearwheels is, λ2e−λ 2.42 e−2.4, =, = 0.2613, 2!, 2×1, (a) The probability of having 2 defective gearwheels, is 0.2613, (b) The probability of having more than 2 defective, gearwheels is 1 − (the sum of the probabilities of, having 0, 1, and 2 defective gearwheels), i.e., 1 − (0.0907 + 0.2177 + 0.2613),, , (a) one, and (b) less than three machines breaking, down in any week., Since the average occurrence of a breakdown is known, but the number of times when a machine did not break, down is unknown, a Poisson distribution must be used., The expectation of a breakdown for 35 machines is, 35 × 0.06, i.e. 2.1 breakdowns per week. The probabilities of a breakdown occurring 0, 1, 2, . . . times are, given by the successive terms of the expression, , , λ2 λ3, −λ, +, +··· ,, 1+λ+, e, 2! 3!, taken from left to right., Hence probability of no breakdowns, e−λ = e−2.1 = 0.1225, probability of 1 breakdown is, λe−λ = 2.1e−2.1 = 0.2572, probability of 2 breakdowns is, , that is, 0.4303, , The principal use of a Poisson distribution is to determine the theoretical probabilities when p, the probability of an event happening, is known, but q, the, probability of the event not happening is unknown. For, example, the average number of goals scored per match, by a football team can be calculated, but it is not possible to quantify the number of goals which were not, scored. In this type of problem, a Poisson distribution, may be defined as follows:, ‘the probabilities of an event occurring 0, 1, 2, 3, …, times are given by the successive terms of the, expression, , , , λ2 λ3, e−λ 1 + λ +, +, +··· ,, 2! 3!, taken from left to right’, , The symbol λ is the value of the average occurrence of, the event., Problem 7. A production department has 35, similar milling machines. The number of, breakdowns on each machine averages 0.06 per, week. Determine the probabilities of having, , (a), , λ2 e−λ 2.12 e−2.1, =, = 0.2700, 2!, 2×1, The probability of 1 breakdown per week is, 0.2572, , (b) The probability of less than 3 breakdowns per, week is the sum of the probabilities of 0, 1, and, 2 breakdowns per week,, i.e., , 0.1225 + 0.2572 + 0.2700, i.e. 0.6497, , Histogram of probabilities, The terms of a Poisson distribution may be represented pictorially by drawing a histogram, as shown in, Problem 8., Problem 8. The probability of a person having an, accident in a certain period of time is 0.0003. For a, population of 7500 people, draw a histogram, showing the probabilities of 0, 1, 2, 3, 4, 5 and 6, people having an accident in this period., The probabilities of 0, 1, 2, . . . people having an accident are given by the terms of expression, , , λ2 λ3, +, +··· ,, e−λ 1 + λ +, 2! 3!, taken from left to right.
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The binomial and Poisson distributions, , Probability of having an accident, , 0.28, , Use a Poisson distribution to determine the, probability of more than two employees going, to hospital during this period of time if there, are 2000 employees on the payroll., [0.5768], , 0.24, 0.20, 0.16, , 3. When packaging a product, a manufacturer, finds that one packet in twenty is underweight., Determine the probabilities that in a box of, 72 packets (a) two and (b) less than four will, be underweight., [(a) 0.1771 (b) 0.5153], , 0.12, 0.08, 0.04, 0, , 0, , 1, , 2, 3, 4, 5, Number of people, , 6, , Figure 57.2, , The average occurrence of the event, λ, is, 7500 × 0.0003, i.e. 2.25, , 4. A manufacturer estimates that 0.25% of his, output of a component are defective. The components are marketed in packets of 200. Determine the probability of a packet containing, less than three defective components., [0.9856], , The probability of no people having an accident is, e−λ = e−2.25 = 0.1054, The probability of 1 person having an accident is, λe−λ = 2.25e−2.25 = 0.2371, The probability of 2 people having an accident is, λ2 e−λ 2.252 e−2.25, =, = 0.2668, 2!, 2!, and so on, giving probabilities of 0.2001, 0.1126,, 0.0506 and 0.0190 for 3, 4, 5 and 6 respectively having an accident. The histogram for these probabilities is, shown in Fig. 57.2., , Now try the following exercise, Exercise 214 Further problems on the, Poisson distribution, 1. In problem 7 of Exercise 213, page 559,, determine the probability of having three components outside of the required tolerance using, the Poisson distribution., [0.0613], 2. The probability that an employee will go to, hospital in a certain period of time is 0.0015., , 5. The demand for a particular tool from a store, is, on average, five times a day and the demand, follows a Poisson distribution. How many of, these tools should be kept in the stores so that, the probability of there being one available, when required is greater than 10%?, ⎡, ⎤, The probabilities of the demand, ⎢, ⎥, ⎢ for 0, 1, 2, . . . tools are, ⎥, ⎢, ⎥, ⎢ 0.0067, 0.0337, 0.0842, 0.1404,⎥, ⎢, ⎥, ⎢ 0.1755, 0.1755, 0.1462, 0.1044,⎥, ⎢, ⎥, ⎢ 0.0653, . . . This shows that the ⎥, ⎢, ⎥, ⎢ probability of wanting a tool ⎥, ⎢, ⎥, ⎢ 8 times a day is 0.0653, i.e., ⎥, ⎢, ⎥, ⎣ less than 10%. Hence 7 should ⎦, be kept in the store, 6. Failure of a group of particular machine, tools follows a Poisson distribution with a, mean value of 0.7. Determine the probabilities of 0, 1, 2, 3, 4 and 5 failures in a week and, present these results on a histogram., ⎤, ⎡, Vertical adjacent rectangles, ⎥, ⎢, ⎢ having heights proportional⎥, ⎥, ⎢, ⎣ to 0.4966, 0.3476, 0.1217, ⎦, 0.0284, 0.0050 and 0.0007, , 561
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Chapter 58, , The normal distribution, 58.1 Introduction to the normal, distribution, , Frequency, , When data is obtained, it can frequently be considered, to be a sample (i.e. a few members) drawn at random, from a large population (i.e. a set having many members). If the sample number is large, it is theoretically, possible to choose class intervals which are very small,, but which still have a number of members falling within, each class. A frequency polygon of this data then has a, large number of small line segments and approximates, to a continuous curve. Such a curve is called a frequency, or a distribution curve., An extremely important symmetrical distribution curve, is called the normal curve and is as shown in Fig. 58.1., This curve can be described by a mathematical equation and is the basis of much of the work done in more, advanced statistics. Many natural occurrences such as, the heights or weights of a group of people, the sizes, of components produced by a particular machine and, the life length of certain components approximate to a, normal distribution., , Variable, , Figure 58.1, , Normal distribution curves can differ from one another, in the following four ways:, (a) by having different mean values, (b) by having different values of standard deviations, , (c) the variables having different values and different, units and, (d) by having different areas between the curve and, the horizontal axis., A normal distribution curve is standardized as follows:, (a) The mean value of the unstandardized curve is, made the origin, thus making the mean value,, x , zero., (b) The horizontal axis is scaled in standard deviax −x, tions. This is done by letting z =, , where, σ, z is called the normal standard variate, x is the, value of the variable, x is the mean value of the, distribution and σ is the standard deviation of the, distribution., (c) The area between the normal curve and the horizontal axis is made equal to unity., When a normal distribution curve has been standardized, the normal curve is called a standardized normal, curve or a normal probability curve, and any normally, distributed data may be represented by the same normal, probability curve., The area under part of a normal probability curve is, directly proportional to probability and the value of the, shaded area shown in Fig. 58.2 can be determined by, evaluating:, !, , , , 1, √, e, (2π), , z2, 2, , , , dz, where z =, , x −x, σ, , To save repeatedly determining the values of this function, tables of partial areas under the standardized normal curve are available in many mathematical formulae, books, and such a table is shown in Table 58.1, on, page 564.
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The normal distribution, , 563, , Probability, density, , 22.22, , 0, (a), , z-value, , 0, (b), , 2.78 z-value, , 0, (c), , 2.78 z-value, , z1, z-value, 0 z2, Standard deviations, , Figure 58.2, , Problem 1. The mean height of 500 people is, 170 cm and the standard deviation is 9 cm., Assuming the heights are normally distributed,, determine the number of people likely to have, heights between 150 cm and 195 cm., The mean value, x , is 170 cm and corresponds to a, normal standard variate value, z, of zero on the standardized normal curve. A height of 150 cm has a z-value, x −x, 150 −170, given by z =, standard deviations, i.e., σ, 9, or −2.22 standard deviations. Using a table of partial areas beneath the standardized normal curve (see, Table 58.1), a z-value of −2.22 corresponds to an area, of 0.4868 between the mean value and the ordinate, z = −2.22. The negative z-value shows that it lies to, the left of the z = 0 ordinate., This area is shown shaded in Fig. 58.3(a). Similarly,, 195 −170, 195 cm has a z-value of, that is 2.78 standard, 9, deviations. From Table 58.1, this value of z corresponds, to an area of 0.4973, the positive value of z showing, that it lies to the right of the z = 0 ordinate. This area, is shown shaded in Fig. 58.3(b). The total area shaded, in Figs. 58.3(a) and (b) is shown in Fig. 58.3(c) and is, 0.4868 +0.4973, i.e. 0.9841 of the total area beneath, the curve., However, the area is directly proportional to probability., Thus, the probability that a person will have a height, of between 150 and 195 cm is 0.9841. For a group of, 500 people, 500 ×0.9841, i.e. 492 people are likely to, have heights in this range. The value of 500 × 0.9841 is, 492.05, but since answers based on a normal probability, distribution can only be approximate, results are usually, given correct to the nearest whole number., Problem 2. For the group of people given in, Problem 1, find the number of people likely to have, heights of less than 165 cm., , 22.22, , Figure 58.3, , 165 −170, A height of 165 cm corresponds to, i.e., 9, −0.56 standard deviations., The area between z = 0 and z = −0.56 (from Table 58.1), is 0.2123, shown shaded in Fig. 58.4(a). The total area, under the standardized normal curve is unity and since, the curve is symmetrical, it follows that the total area, to the left of the z = 0 ordinate is 0.5000. Thus the area, to the left of the z =−0.56 ordinate (‘left’ means ‘less, than’, ‘right’ means ‘more than’) is 0.5000 − 0.2123,, i.e. 0.2877 of the total area, which is shown shaded in, Fig 58.4(b). The area is directly proportional to probability and since the total area beneath the standardized, normal curve is unity, the probability of a person’s height, being less than 165 cm is 0.2877. For a group of 500 people, 500 × 0.2877, i.e. 144 people are likely to have, heights of less than 165 cm., Problem 3. For the group of people given in, Problem 1 find how many people are likely to have, heights of more than 194 cm., 194 −170, that is,, 9, 2.67 standard deviations. From Table 58.1, the area, , 194 cm corresponds to a z-value of
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564 Higher Engineering Mathematics, Table 58.1 Partial areas under the standardized normal curve, , 0, , x −x, σ, 0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, 3.0, 3.1, 3.2, 3.3, 3.4, 3.5, 3.6, 3.7, 3.8, 3.9, , z=, , z, , 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 0.0000, 0.0398, 0.0793, 0.1179, 0.1554, 0.1915, 0.2257, 0.2580, 0.2881, 0.3159, 0.3413, 0.3643, 0.3849, 0.4032, 0.4192, 0.4332, 0.4452, 0.4554, 0.4641, 0.4713, 0.4772, 0.4821, 0.4861, 0.4893, 0.4918, 0.4938, 0.4953, 0.4965, 0.4974, 0.4981, 0.4987, 0.4990, 0.4993, 0.4995, 0.4997, 0.4998, 0.4998, 0.4999, 0.4999, 0.5000, , 0.0040, 0.0438, 0.0832, 0.1217, 0.1591, 0.1950, 0.2291, 0.2611, 0.2910, 0.3186, 0.3438, 0.3665, 0.3869, 0.4049, 0.4207, 0.4345, 0.4463, 0.4564, 0.4649, 0.4719, 0.4778, 0.4826, 0.4864, 0.4896, 0.4920, 0.4940, 0.4955, 0.4966, 0.4975, 0.4982, 0.4987, 0.4991, 0.4993, 0.4995, 0.4997, 0.4998, 0.4998, 0.4999, 0.4999, 0.5000, , 0.0080, 0.0478, 0.0871, 0.1255, 0.1628, 0.1985, 0.2324, 0.2642, 0.2939, 0.3212, 0.3451, 0.3686, 0.3888, 0.4066, 0.4222, 0.4357, 0.4474, 0.4573, 0.4656, 0.4726, 0.4783, 0.4830, 0.4868, 0.4898, 0.4922, 0.4941, 0.4956, 0.4967, 0.4976, 0.4982, 0.4987, 0.4991, 0.4994, 0.4995, 0.4997, 0.4998, 0.4999, 0.4999, 0.4999, 0.5000, , 0.0120, 0.0517, 0.0910, 0.1293, 0.1664, 0.2019, 0.2357, 0.2673, 0.2967, 0.3238, 0.3485, 0.3708, 0.3907, 0.4082, 0.4236, 0.4370, 0.4484, 0.4582, 0.4664, 0.4732, 0.4785, 0.4834, 0.4871, 0.4901, 0.4925, 0.4943, 0.4957, 0.4968, 0.4977, 0.4983, 0.4988, 0.4991, 0.4994, 0.4996, 0.4997, 0.4998, 0.4999, 0.4999, 0.4999, 0.5000, , 0.0159, 0.0557, 0.0948, 0.1331, 0.1700, 0.2054, 0.2389, 0.2704, 0.2995, 0.3264, 0.3508, 0.3729, 0.3925, 0.4099, 0.4251, 0.4382, 0.4495, 0.4591, 0.4671, 0.4738, 0.4793, 0.4838, 0.4875, 0.4904, 0.4927, 0.4945, 0.4959, 0.4969, 0.4977, 0.4984, 0.4988, 0.4992, 0.4994, 0.4996, 0.4997, 0.4998, 0.4999, 0.4999, 0.4999, 0.5000, , 0.0199, 0.0596, 0.0987, 0.1388, 0.1736, 0.2086, 0.2422, 0.2734, 0.3023, 0.3289, 0.3531, 0.3749, 0.3944, 0.4115, 0.4265, 0.4394, 0.4505, 0.4599, 0.4678, 0.4744, 0.4798, 0.4842, 0.4878, 0.4906, 0.4929, 0.4946, 0.4960, 0.4970, 0.4978, 0.4984, 0.4989, 0.4992, 0.4994, 0.4996, 0.4997, 0.4998, 0.4999, 0.4999, 0.4999, 0.5000, , 0.0239, 0.0636, 0.1026, 0.1406, 0.1772, 0.2123, 0.2454, 0.2760, 0.3051, 0.3315, 0.3554, 0.3770, 0.3962, 0.4131, 0.4279, 0.4406, 0.4515, 0.4608, 0.4686, 0.4750, 0.4803, 0.4846, 0.4881, 0.4909, 0.4931, 0.4948, 0.4961, 0.4971, 0.4979, 0.4985, 0.4989, 0.4992, 0.4994, 0.4996, 0.4997, 0.4998, 0.4999, 0.4999, 0.4999, 0.5000, , 0.0279, 0.0678, 0.1064, 0.1443, 0.1808, 0.2157, 0.2486, 0.2794, 0.3078, 0.3340, 0.3577, 0.3790, 0.3980, 0.4147, 0.4292, 0.4418, 0.4525, 0.4616, 0.4693, 0.4756, 0.4808, 0.4850, 0.4884, 0.4911, 0.4932, 0.4949, 0.4962, 0.4972, 0.4980, 0.4985, 0.4989, 0.4992, 0.4995, 0.4996, 0.4997, 0.4998, 0.4999, 0.4999, 0.4999, 0.5000, , 0.0319, 0.0714, 0.1103, 0.1480, 0.1844, 0.2190, 0.2517, 0.2823, 0.3106, 0.3365, 0.3599, 0.3810, 0.3997, 0.4162, 0.4306, 0.4430, 0.4535, 0.4625, 0.4699, 0.4762, 0.4812, 0.4854, 0.4887, 0.4913, 0.4934, 0.4951, 0.4963, 0.4973, 0.4980, 0.4986, 0.4990, 0.4993, 0.4995, 0.4996, 0.4997, 0.4998, 0.4999, 0.4999, 0.4999, 0.5000, , 0.0359, 0.0753, 0.1141, 0.1517, 0.1879, 0.2224, 0.2549, 0.2852, 0.3133, 0.3389, 0.3621, 0.3830, 0.4015, 0.4177, 0.4319, 0.4441, 0.4545, 0.4633, 0.4706, 0.4767, 0.4817, 0.4857, 0.4890, 0.4916, 0.4936, 0.4952, 0.4964, 0.4974, 0.4981, 0.4986, 0.4990, 0.4993, 0.4995, 0.4997, 0.4998, 0.4998, 0.4999, 0.4999, 0.4999, 0.5000
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The normal distribution, (a), , 565, , the number of bottles likely to contain less, than 750 ml,, , (b) the number of bottles likely to contain, between 751 and 754 ml,, 20.56 0, (a), , z-value, , (c), , the number of bottles likely to contain more, than 757 ml, and, , (d) the number of bottles likely to contain, between 750 and 751 ml., , (a), 20.56 0, , z-value, , (b), , Figure 58.4, , between z = 0, z = 2.67 and the standardized normal, curve is 0.4962, shown shaded in Fig. 58.5(a). Since, the standardized normal curve is symmetrical, the total, area to the right of the z =0 ordinate is 0.5000, hence the, shaded area shown in Fig. 58.5(b) is 0.5000 − 0.4962,, i.e. 0.0038. This area represents the probability of a person having a height of more than 194 cm, and for 500, people, the number of people likely to have a height of, more than 194 cm is 0.0038 ×500, i.e. 2 people., , 0, (a), , 2.67, , z-value, , 0, (b), , 2.67, , z-value, , Figure 58.5, , Problem 4. A batch of 1500 lemonade bottles, have an average contents of 753 ml and the standard, deviation of the contents is 1.8 ml. If the volumes of, the contents are normally distributed, find, , The z-value corresponding to 750 ml is given, 750 −753, x −x, i.e., = −1.67 standard deviby, σ, 1.8, ations. From Table 58.1, the area between z = 0, and z = −1.67 is 0.4525. Thus the area to the, left of the z =−1.67 ordinate is 0.5000 −0.4525, (see Problem 2), i.e. 0.0475. This is the probability of a bottle containing less than 750 ml., Thus, for a batch of 1500 bottles, it is likely that, 1500 ×0.0475, i.e. 71 bottles will contain less, than 750 ml., , (b) The z-value corresponding to 751 and 754 ml, 754 −753, 751 −753, and, i.e. −1.11 and, are, 1.8, 1.8, 0.56 respectively. From Table 58.1, the areas, corresponding to these values are 0.3665 and, 0.2123 respectively. Thus the probability of a, bottle containing between 751 and 754 ml is, 0.3665 +0.2123 (see Problem 1), i.e. 0.5788. For, 1500 bottles, it is likely that 1500 ×0.5788, i.e., 868 bottles will contain between 751 and 754 ml., 757 −753, ,, (c) The z-value corresponding to 757 ml is, 1.8, i.e. 2.22 standard deviations. From Table 58.1,, the area corresponding to a z-value of 2.22 is, 0.4868. The area to the right of the z =2.22, ordinate is 0.5000 −0.4868 (see Problem 3), i.e., 0.0132. Thus, for 1500 bottles, it is likely that, 1500 ×0.0132, i.e. 20 bottles will have contents, of more than 757 ml., (d) The z-value corresponding to 750 ml is −1.67, (see part (a)), and the z-value corresponding to, 751 ml is −1.11 (see part (b)). The areas corresponding to these z-values are 0.4525 and 0.3665, respectively, and both these areas lie on the left, of the z = 0 ordinate. The area between z =−1.67, and z = −1.11 is 0.4525 −0.3665, i.e. 0.0860 and, this is the probability of a bottle having contents, between 750 and 751 ml. For 1500 bottles, it is
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566 Higher Engineering Mathematics, likely that 1500 ×0.0860, i.e. 129 bottles will be, in this range., Now try the following exercise, Exercise 215 Further problems on the, introduction to the normal distribution, , 7. The intelligence quotients of 400 children, have a mean value of 100 and a standard deviation of 14. Assuming that I.Q.’s are normally, distributed, determine the number of children, likely to have I.Q.’s of between (a) 80 and 90,, (b) 90 and 110 and (c) 110 and 130., [(a) 65 (b) 209 (c) 89], , 1. A component is classed as defective if it has a, diameter of less than 69 mm. In a batch of 350, components, the mean diameter is 75 mm and, the standard deviation is 2.8 mm. Assuming, the diameters are normally distributed,, determine how many are likely to be classed, as defective., [6], , 8. The mean mass of active material in tablets, produced by a manufacturer is 5.00 g and the, standard deviation of the masses is 0.036 g. In, a bottle containing 100 tablets, find how many, tablets are likely to have masses of (a) between, 4.88 and 4.92 g, (b) between 4.92 and 5.04 g, and (c) more than 5.04 g., , 2. The masses of 800 people are normally distributed, having a mean value of 64.7 kg and a, standard deviation of 5.4 kg. Find how many, people are likely to have masses of less than, 54.4 kg., [22], , [(a) 1 (b) 85 (c) 13], , 3. 500 tins of paint have a mean content of, 1010 ml and the standard deviation of the contents is 8.7 ml. Assuming the volumes of the, contents are normally distributed, calculate the, number of tins likely to have contents whose, volumes are less than (a) 1025 ml (b) 1000 ml, and (c) 995 ml., [(a) 479 (b) 63 (c) 21], 4. For the 350 components in Problem 1, if those, having a diameter of more than 81.5 mm are, rejected, find, correct to the nearest component, the number likely to be rejected due to, being oversized., [4], 5. For the 800 people in Problem 2, determine, how many are likely to have masses of more, than (a) 70 kg and (b) 62 kg., [(a) 131 (b) 553], 6. The mean diameter of holes produced by a, drilling machine bit is 4.05 mm and the standard deviation of the diameters is 0.0028 mm., For twenty holes drilled using this machine,, determine, correct to the nearest whole number, how many are likely to have diameters, of between (a) 4.048 and 4.0553 mm and, (b) 4.052 and 4.056 mm, assuming the diameters are normally distributed., [(a) 15 (b) 4], , 58.2, , Testing for a normal distribution, , It should never be assumed that because data is continuous it automatically follows that it is normally, distributed. One way of checking that data is normally, distributed is by using normal probability paper, often, just called probability paper. This is special graph, paper which has linear markings on one axis and percentage probability values from 0.01 to 99.99 on the, other axis (see Figs. 58.6 and 58.7). The divisions on the, probability axis are such that a straight line graph results, for normally distributed data when percentage cumulative frequency values are plotted against upper class, boundary values. If the points do not lie in a reasonably, straight line, then the data is not normally distributed., The method used to test the normality of a distribution is, shown in Problems 5 and 6. The mean value and standard, deviation of normally distributed data may be determined using normal probability paper. For normally distributed data, the area beneath the standardized normal, curve and a z-value of unity (i.e. one standard deviation) may be obtained from Table 58.1. For one standard, deviation, this area is 0.3413, i.e. 34.13%. An area of, ±1 standard deviation is symmetrically placed on either, side of the z = 0 value, i.e. is symmetrically placed, on either side of the 50% cumulative frequency value., Thus an area corresponding to ±1 standard deviation, extends from percentage cumulative frequency values of, (50 + 34.13)% to (50 − 34.13)%, i.e. from 84.13% to, 15.87%. For most purposes, these values are taken as, 84% and 16%. Thus, when using normal probability
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99.99, , 99.99, , 99.9, 99.8, , 99.9, , 99, 98, , 99, 98, , 95, , 95, Percentage cumulative frequency, , Percentage cumulative frequency, , The normal distribution, , 90, Q, , 80, 70, 60, 50, 40, 30, 20, , P, , R, , 10, 5, 2, 1, 0.5, 0.2, 0.1, 0.05, , 567, , 90, B, 80, 70, 60, 50, 40, 30, 20, , A, , C, , 10, 5, 2, 1, 0.5, 0.2, 0.1, 0.05, , 0.01, 30, , 32, 34, 36, 38, Upper class boundary, , 40, , 42, , Figure 58.6, , 0.01, , 10 20 30 40 50 60 70 80 90 100 110, Upper class boundary, , Figure 58.7, , paper, the standard deviation of the distribution is, given by:, , , variable value for 84% cumulative frequency −, variable value for 16% cumulative frequency, 2, Problem 5. Use normal probability paper to, determine whether the data given below, which, refers to the masses of 50 copper ingots, is, approximately normally distributed. If the data is, normally distributed, determine the mean and, standard deviation of the data from the graph drawn., Class mid-point value (kg), , Frequency, , 29.5, , 2, , 30.5, , 4, , 31.5, , 6, , 32.5, , 8, , 33.5, , 9, , 34.5, , 8, , Class mid-point value (kg), , Frequency, , 35.5, , 6, , 36.5, , 4, , 37.5, , 2, , 38.5, , 1, , To test the normality of a distribution, the upper class, boundary/percentage cumulative frequency values are, plotted on normal probability paper. The upper class, boundary values are: 30, 31, 32, …, 38, 39. The corresponding cumulative frequency values (for ‘less than’, the upper class boundary values) are: 2, (4 + 2) = 6,, (6 + 4 +2) = 12, 20, 29, 37, 43, 47, 49 and 50. The corresponding percentage cumulative frequency values are, 6, 2, × 100 =4,, × 100 = 12, 24, 40, 58, 74, 86, 94, 98, 50, 50, and 100%., The co-ordinates of upper class boundary/percentage, cumulative frequency values are plotted as shown
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568 Higher Engineering Mathematics, in Fig. 58.6. When plotting these values, it will always, be found that the co-ordinate for the 100% cumulative, frequency value cannot be plotted, since the maximum, value on the probability scale is 99.99. Since the points, plotted in Fig. 58.6 lie very nearly in a straight line,, the data is approximately normally distributed., The mean value and standard deviation can be determined from Fig. 58.6. Since a normal curve is symmetrical, the mean value is the value of the variable, corresponding to a 50% cumulative frequency value,, shown as point P on the graph. This shows that the, mean value is 33.6 kg. The standard deviation is determined using the 84% and 16% cumulative frequency, values, shown as Q and R in Fig. 58.6. The variable values for Q and R are 35.7 and 31.4 respectively; thus two, standard deviations correspond to 35.7 − 31.4, i.e. 4.3,, showing that the standard deviation of the distribution, 4.3, i.e. 2.15 standard deviations., is approximately, 2, The mean value and standard deviation of the distribution can be calculated using, ; , fx, mean, x = ; , f, and standard deviation,, 75, 6, 8 ;, 8, [ f (x − x̄ )2 ], 9, ; , σ=, f, where f is the frequency of a class and x is the class midpoint value. Using these formulae gives a mean value, of the distribution of 33.6 (as obtained graphically) and, a standard deviation of 2.12, showing that the graphical, method of determining the mean and standard deviation, give quite realistic results., Problem 6. Use normal probability paper to, determine whether the data given below is normally, distributed. Use the graph and assume a normal, distribution whether this is so or not, to find, approximate values of the mean and standard, deviation of the distribution., Class mid-point values, , Frequency, , 5, , 1, , 15, , 2, , 25, , 3, , 35, , 6, , Class mid-point values, , Frequency, , 45, , 9, , 55, , 6, , 65, , 2, , 75, , 2, , 85, , 1, , 95, , 1, , To test the normality of a distribution, the upper class, boundary/percentage cumulative frequency values are, plotted on normal probability paper. The upper class, boundary values are: 10, 20, 30, …, 90 and 100. The corresponding cumulative frequency values are 1, 1 +2 = 3,, 1 + 2 +3 = 6, 12, 21, 27, 29, 31, 32 and 33. The per1, centage cumulative frequency values are, × 100 =3,, 33, 3, × 100 =9, 18, 36, 64, 82, 88, 94, 97 and 100., 33, The co-ordinates of upper class boundary values/percentage cumulative frequency values are plotted as, shown in Fig. 58.7. Although six of the points lie approximately in a straight line, three points corresponding to, upper class boundary values of 50, 60 and 70 are not, close to the line and indicate that the distribution is, not normally distributed. However, if a normal distribution is assumed, the mean value corresponds to, the variable value at a cumulative frequency of 50%, and, from Fig. 58.7, point A is 48. The value of the, standard deviation of the distribution can be obtained, from the variable values corresponding to the 84% and, 16% cumulative frequency values, shown as B and C in, Fig. 58.7 and give: 2σ = 69 −28, i.e. the standard deviation σ = 20.5. The calculated values of the mean and, standard deviation of the distribution are 45.9 and 19.4, respectively, showing that errors are introduced if the, graphical method of determining these values is used, for data which is not normally distributed., Now try the following exercise, Exercise 216 Further problems on testing, for a normal distribution, 1. A frequency distribution of 150 measurements, is as shown:
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569, , The normal distribution, ⎡, Class mid-point value Frequency, , Graphically,, , x = 27.1, σ = 0.3;, , ⎢, ⎣ by calculation, x = 27.079,, , ⎤, ⎥, ⎦, , 26.4, , 5, , 26.6, , 12, , 26.8, , 24, , 27.0, , 36, , 27.2, , 36, , Load (kN) 17 19 21 23 25 27 29 31, , 27.4, , 25, , Frequency, , 27.6, , 12, , Use normal probability paper to show that this, data approximates to a normal distribution and, hence determine the approximate values of the, mean and standard deviation of the distribution. Use the formula for mean and standard, deviation to verify the results obtained., , σ = 0.3001, 2. A frequency distribution of the class mid-point, values of the breaking loads for 275 similar, fibres is as shown below:, , 9 23 55 78 64 28 14, , 4, , Use normal probability paper to show that this, distribution is approximately normally distributed and determine the mean and standard, deviation of the distribution (a) from the graph, and (b) by calculation., (a) x = 23.5 kN,, , σ = 2.9 kN, , (b) x = 23.364 kN, σ = 2.917 kN
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Chapter 59, , Linear correlation, y, , 59.1 Introduction to linear correlation, Correlation is a measure of the amount of association, existing between two variables. For linear correlation,, if points are plotted on a graph and all the points lie on, a straight line, then perfect linear correlation is said, to exist. When a straight line having a positive gradient can reasonably be drawn through points on a graph, positive or direct linear correlation exists, as shown, in Fig. 59.1(a). Similarly, when a straight line having, a negative gradient can reasonably be drawn through, points on a graph, negative or inverse linear correlation exists, as shown in Fig. 59.1(b). When there is no, apparent relationship between co-ordinate values plotted on a graph then no correlation exists between the, points, as shown in Fig. 59.1(c). In statistics, when two, variables are being investigated, the location of the coordinates on a rectangular co-ordinate system is called, a scatter diagram—as shown in Fig. 59.1., , Positive linear correlation, , x, , (a), y, , 59.2 The product-moment formula, for determining the linear, correlation coefficient, , Negative linear correlation, , x, , (b), y, , The amount of linear correlation between two variables, is expressed by a coefficient of correlation, given the, symbol r. This is defined in terms of the deviations of, the co-ordinates of two variables from their mean values, and is given by the product-moment formula which, states:, coefficient of correlation,, , ;, xy, r = -:; ; 4, x2, y2, , (1), , where the x-values are the values of the deviations of coordinates X from X, their mean value and the y-values, , No correlation, (c), , Figure 59.1, , x
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Linear correlation, are the values of the deviations of co-ordinates Y from Y ,, their mean value. That is, x = (X − X) and y = (Y − Y )., The results of this determination give values of r lying, between +1 and −1, where +1 indicates perfect direct, correlation, −1 indicates perfect inverse correlation and, 0 indicates that no correlation exists. Between these values, the smaller the value of r, the less is the amount of, correlation which exists. Generally, values of r in the, ranges 0.7 to 1 and −0.7 to −1 show that a fair amount, of correlation exists., , 59.3 The significance of a coefficient, of correlation, When the value of the coefficient of correlation has been, obtained from the product moment formula, some care, is needed before coming to conclusions based on this, result. Checks should be made to ascertain the following, two points:, (a), , that a ‘cause and effect’ relationship exists, between the variables; it is relatively easy, mathematically, to show that some correlation exists, between, say, the number of ice creams sold in a, given period of time and the number of chimneys, swept in the same period of time, although there, is no relationship between these variables;, , (b) that a linear relationship exists between the, variables; the product-moment formula given in, Section 59.2 is based on linear correlation. Perfect, non-linear correlation may exist (for example, the, co-ordinates exactly following the curve y = x 3 ),, but this gives a low value of coefficient of correlation since the value of r is determined using, the product-moment formula, based on a linear, relationship., , Let X be the variable force values and Y be the, dependent variable extension values. The coefficient of, correlation is given by:, ;, r = -:;, , Force (N), , 10, , 20, , 30, , 40, , 50, , 60, , y2, , 4, , X, , Y, , x = (X − X), , y = (Y − Y ), , 10, , 0.22, , −30, , −0.699, , 20, , 0.40, , −20, , −0.519, , 30, , 0.61, , −10, , −0.309, , 40, , 0.85, , 0, , −0.069, , 50, , 1.20, , 10, , 0.281, , 60, , 1.45, , 20, , 0.531, , 70, , 1.70, , 30, , 0.781, , ;, , 280, = 40, 7, ;, 6.43, Y = 6.43, Y =, = 0.919, 7, X =280, X =, , xy, , x2, , y2, , 20.97, , 900, , 0.489, , 10.38, , 400, , 0.269, , 3.09, , 100, , 0.095, , 0, , 0.005, , 2.81, , 100, , 0.079, , 10.62, , 400, , 0.282, , 0, , ;, , 23.43, x y = 71.30, , ;, , 900, x 2 = 2800, , ;, , 0.610, y 2 = 1.829, , 70, , Thus, , Extension, (mm), , x2, , xy, ;, , where x = (X − X ) and y = (Y − Y ), X and Y being, the mean values of the X and Y values respectively., Using a tabular method to determine the quantities of, this formula gives:, , 59.4 Worked problems on linear, correlation, Problem 1. In an experiment to determine the, relationship between force on a wire and the, resulting extension, the following data is obtained:, , 571, , 0.22 0.40 0.61 0.85 1.20 1.45 1.70, , Determine the linear coefficient of correlation for, this data., , 71.3, r=√, = 0.996, [2800 ×1.829], , This shows that a very good direct correlation exists, between the values of force and extension.
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572 Higher Engineering Mathematics, xy, , x2, , y2, , −46.5, , 3.7, , 581, , −242.1, , 24.3, , 2411, , Problem 2. The relationship between expenditure, on welfare services and absenteeism for similar, periods of time is shown below for a small company., Expenditure, (£ 000), , 3.5 5.0 7.0, , 10, , 12, , 15 18, , Days lost, , 241 318 174 110 147 122 86, , Determine the coefficient of linear correlation for, this data., Let X be the expenditure in thousands of pounds and Y, be the days lost., The coefficient of correlation,, ;, xy, r = -:; ; 4, x2, y2, , ;, , −674.8, x y = −2172, , Thus, r=√, , ;, , 62.9, x 2 = 169.2, , ;, , 7242, y 2 = 40441, , −2172, = −0.830, [169.2 ×40441], , This shows that there is fairly good inverse correlation, between the expenditure on welfare and days lost due, to absenteeism., Problem 3. The relationship between monthly, car sales and income from the sale of petrol for a, garage is as shown:, , where x = (X − X ) and y = (Y − Y ), X and Y being the, mean values of X and Y respectively. Using a tabular, approach:, , Cars sold, , 2 5 3 12 14 7 3 28 14 7 3 13, , Income from, petrol sales 12 9 13 21 17 22 31 47 17 10 9 11, (£ 000), , X, , Y, , x = (X − X ), , y = (Y − Y ), , 3.5, , 241, , −6.57, , 69.9, , 5.0, , 318, , −5.07, , 146.9, , 7.0, , 174, , −3.07, , 2.9, , 10, , 110, , −0.07, , −61.1, , 12, , 147, , 1.93, , −24.1, , X, , Y, , x = (X − X), , y = (Y − Y ), , 15, , 122, , 4.93, , −49.1, , 2, , 12, , −7.25, , −6.25, , 18, , 86, , 7.93, , −85.1, , 5, , 9, , −4.25, , −9.25, , 3, , 13, , −6.25, , −5.25, , 12, , 21, , 2.75, , 2.75, , 14, , 17, , 4.75, , −1.25, , ;, ;, , X =70.5, X =, , 70.5, = 10.07, 7, , 1198, Y = 1198, Y =, = 171.1, 7, , Determine the linear coefficient of correlation, between these quantities., Let X represent the number of cars sold and Y the, income, in thousands of pounds, from petrol sales. Using, the tabular approach:, , 7, , 22, , −2.25, , 3.75, , xy, , x2, , y2, , 3, , 31, , −6.25, , 12.75, , −459.2, , 43.2, , 4886, , 28, , 47, , 18.75, , 28.75, , −744.8, , 25.7, , 21580, , 14, , 17, , 4.75, , −1.25, , −8.9, , 9.4, , 8, , 7, , 10, , −2.25, , −8.25, , 3733, , 3, , 9, , −6.25, , −9.25, , 4.3, , 0
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Linear correlation, X, , Y, , x = (X − X ), , y = (Y − Y ), , 13, , 11, , 3.75, , −7.25, , ;, , 111, = 9.25, 12, ;, 219, Y = 219, Y =, = 18.25, 12, , ;, , X = 111, X =, , xy, , x2, , y2, , 45.3, , 52.6, , 39.1, , 39.3, , 18.1, , 85.6, , 32.8, , 39.1, , 27.6, , 7.6, , 7.6, , 7.6, , −5.9, , 22.6, , 1.6, , −8.4, , 5.1, , 14.1, , −79.7, , 39.1, , 162.6, , 539.1, , 351.6, , 826.6, , −5.9, , 22.6, , 1.6, , 18.6, , 5.1, , 68.1, , 57.8, , 39.1, , 85.6, , −27.2, x y = 613.4, , ;, , 14.1, x 2 = 616.7, , ;, , 52.6, y 2 = 1372.7, , The coefficient of correlation,, ;, xy, r = -:; ; 4, x2, y2, 613.4, =√, = 0.667, {(616.7)(1372.7)}, Thus, there is no appreciable correlation between, petrol and car sales., Now try the following exercise, Exercise 217, correlation, , Further problems on linear, , In Problems 1 to 3, determine the coefficient of, correlation for the data given, correct to 3 decimal, places., , 1., , X, Y, , 14, 900, , 18, 1200, , 23, 1600, , 30, 2100, , 2., , X, Y, , 2.7, 11.9, , 4.3, 7.10, , 1.2, 33.8, , 1.4, 25.0, , 3., , X, Y, , 24, 39, , 41, 46, , 9, 90, , 18, 30, , 50, 3800, [0.999], 4.9, 7.50, [−0.916], , 73, 98, [0.422], , 4. In an experiment to determine the relationship, between the current flowing in an electrical, circuit and the applied voltage, the results, obtained are:, Current, (mA), Applied, , 5 11 15 19 24 28, , voltage (V) 2, , 4, , 6, , 33, , 8 10 12 14, , Determine, using the product-moment formula, the coefficient of correlation for these, results., [0.999], 5. A gas is being compressed in a closed cylinder, and the values of pressures and corresponding, volumes at constant temperature are as shown:, Pressure (kPa) Volume (m3 ), 160, , 0.034, , 180, , 0.036, , 200, , 0.030, , 220, , 0.027, , 240, , 0.024, , 260, , 0.025, , 280, , 0.020, , 300, , 0.019, , Find the coefficient of correlation for these, values., [−0.962], 6. The relationship between the number of miles, travelled by a group of engineering salesmen, in ten equal time periods and the corresponding value of orders taken is given below., Calculate the coefficient of correlation using, the product-moment formula for these values., , 573
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574 Higher Engineering Mathematics, , Miles, travelled, , Orders taken, (£ 000), , 1370, 1050, 980, 1770, 1340, 1560, 2110, 1540, 1480, 1670, , 23, 17, 19, 22, 27, 23, 30, 23, 25, 19, , 7. The data shown below refers to the number, of times machine tools had to be taken out of, service, in equal time periods, due to faults, occurring and the number of hours worked by, maintenance teams. Calculate the coefficient, of correlation for this data., Machines, out of, service:, 4 13 2 9 16 8 7, Maintenance, hours:, 400 515 360 440 570 380 415, [0.632], , [0.937]
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Chapter 60, , Linear regression, y, , 60.1, , (Xn, Yn ), , Introduction to linear regression, , Q, , Dn, , Regression analysis, usually termed regression, is used, to draw the line of ‘best fit’ through co-ordinates on, a graph. The techniques used enable a mathematical equation of the straight line form y = mx + c to, be deduced for a given set of co-ordinate values, the, line being such that the sum of the deviations of, the co-ordinate values from the line is a minimum,, i.e. it is the line of ‘best fit’. When a regression analysis is made, it is possible to obtain two lines of best fit,, depending on which variable is selected as the dependent variable and which variable is the independent, variable. For example, in a resistive electrical circuit, the, current flowing is directly proportional to the voltage, applied to the circuit. There are two ways of obtaining experimental values relating the current and voltage., Either, certain voltages are applied to the circuit and the, current values are measured, in which case the voltage is, the independent variable and the current is the dependent, variable; or, the voltage can be adjusted until a desired, value of current is flowing and the value of voltage is, measured, in which case the current is the independent, value and the voltage is the dependent value., , 60.2 The least-squares regression, lines, For a given set of co-ordinate values, (X 1, Y1),, (X 2, Y2 ), . . . , (X n , Yn ) let the X values be the independent variables and the Y -values be the dependent values., Also let D1, . . . , Dn be the vertical distances between the, line shown as PQ in Fig. 60.1 and the points representing the co-ordinate values. The least-squares regression, line, i.e. the line of best fit, is the line which makes the, value of D12 + D22 + · · · + Dn2 a minimum value., , H4, , H3, (X1, Y1 ), D1, , D2, (X2, Y2 ), , P, , x, , Figure 60.1, , The equation of the least-squares regression line is, usually written as Y = a0 + a1 X , where a0 is the, Y -axis intercept value and a1 is the gradient of the line, (analogous to c and m in the equation y = mx + c). The, values of a0 and a1 to make the sum of the ‘deviations squared’ a minimum can be obtained from the two, equations:, <, <, Y = a0 N + a1, X, (1), <, <, <, (X Y ) = a0, X + a1, X2, (2), where X and Y are the co-ordinate values, N is the, number of co-ordinates and a0 and a1 are called the, regression coefficients of Y on X . Equations (1) and (2), are called the normal equations of the regression lines, of Y on X . The regression line of Y on X is used to estimate values of Y for given values of X . If the Y -values, (vertical-axis) are selected as the independent variables,, the horizontal distances between the line shown as PQ
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576 Higher Engineering Mathematics, in Fig. 60.1 and the co-ordinate values (H3, H4 , etc.), are taken as the deviations. The equation of the regression line is of the form: X =b0 + b1Y and the normal, equations become:, <, <, X = b0 N + b1, Y, (3), <, <, <, (XY) = b0, Y + b1, Y2, (4), where X and Y are the co-ordinate values, b0 and b1, are the regression coefficients of X on Y and N is the, number of co-ordinates. These normal equations are of, the regression line of X on Y , which is slightly different, to the regression line of Y on X . The regression line of, X on Y is used to estimated values of X for given values, of Y . The regression line of Y on X is used to determine, any value of Y corresponding to a given value of X . If, the value of Y lies within the range of Y -values of the, extreme co-ordinates, the process of finding the corresponding value of X is called linear interpolation. If, it lies outside of the range of Y -values of the extreme, co-ordinates than the process is called linear extrapolation and the assumption must be made that the line of, best fit extends outside of the range of the co-ordinate, values given., By using the regression line of X on Y , values of X, corresponding to given values of Y may be found by, either interpolation or extrapolation., , Determine the equation of the regression line of, inductive reactance on frequency, assuming a linear, relationship., Since the regression line of inductive reactance on frequency is required, the frequency is the independent, variable, X , and the inductive reactance is the dependent variable, Y . The equation of the regression line of, Y on X is:, Y = a 0 + a1 X, and the regression coefficients a0 and a1 are obtained, by using the normal equations, ;, ;, Y = a0 N + a1 X, ;, ;, ;, and, XY = a0 X + a1 X 2, (from equations (1) and (2)), A tabular approach is used to determine the summed, quantities., Frequency, X, , 60.3 Worked problems on linear, regression, Problem 1. In an experiment to determine the, relationship between frequency and the inductive, reactance of an electrical circuit, the following, results were obtained:, , ;, , 50, , 30, , 2500, , 100, , 65, , 10000, , 150, , 90, , 22500, , 200, , 130, , 40000, , 250, , 150, , 62500, , 300, , 190, , 90000, , 350, X =1400, , 30, , 100, , 65, , 150, , 90, , 200, , 130, , 250, , 150, , 300, , 190, , 350, , 200, , ;, , 200, Y = 855, , ;, , ;, , 122500, X 2 = 350000, , Y2, , XY, , Frequency Inductive reactance, (Hz), (ohms), 50, , X2, , Inductive, reactance, Y, , 1500, , 900, , 6500, , 4225, , 13500, , 8100, , 26000, , 16900, , 37500, , 22500, , 57000, , 36100, , 70000, XY = 212000, , ;, , 40000, Y 2 = 128725
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Linear regression, The number of co-ordinate values given, N is 7., Substituting in the normal equations gives:, 855 = 7a0 + 1400a1, 212000 = 1400a0 + 350000a1, 1400 ×(1) gives:, 1197000 = 9800a0 + 1960000a1, , Solving these equations in a similar way to that in, Problem 1 gives:, b0 = −6.15, , (1), (2), , (3), , and b1 = 1.69, correct to 3 significant figures., Thus the equation of the regression line of frequency on, inductive reactance is:, X = −6.15 + 1.69 Y, , 7 × (2) gives:, 1484000 = 9800a0 + 2450000a1, , (4), , (4) − (3) gives:, 287000 = 0 + 490000a1, from which, a1 =, , 287000, = 0.586, 490000, , Substituting a1 = 0.586 in equation (1) gives:, 855 = 7a0 + 1400(0.586), i.e., , a0 =, , 577, , 855 −820.4, = 4.94, 7, , Thus the equation of the regression line of inductive, reactance on frequency is:, Y = 4.94 + 0.586 X, Problem 2. For the data given in Problem 1,, determine the equation of the regression line of, frequency on inductive reactance, assuming a linear, relationship., In this case, the inductive reactance is the independent, variable X and the frequency is the dependent variable, Y . From equations 3 and 4, the equation of the regression, line of X on Y is:, X = b0 + b1 Y, and the normal equations are, <, <, X = b0 N + b1, Y, <, <, <, Y + b1, Y2, and, XY = b0, From the table shown in Problem 1, the simultaneous, equations are:, 1400 = 7b0 + 855b1, 212000 = 855b0 + 128725b1, , Problem 3. Use the regression equations, calculated in Problems 1 and 2 to find (a) the value, of inductive reactance when the frequency is 175 Hz, and (b) the value of frequency when the inductive, reactance is 250 ohms, assuming the line of best fit, extends outside of the given co-ordinate values., Draw a graph showing the two regression lines., (a), , From Problem 1, the regression equation of inductive reactance on frequency is, Y = 4.94 + 0.586 X . When the frequency, X , is, 175 Hz, Y = 4.94 +0.586(175) = 107.5, correct to, 4 significant figures, i.e. the inductive reactance is, 107.5 ohms when the frequency is 175 Hz., , (b) From Problem 2, the regression equation of frequency on inductive reactance is, X = −6.15 +1.69 Y . When the inductive reactance, Y , is 250 ohms,, X = −6.15 +1.69(250) = 416.4 Hz, correct to 4, significant figures, i.e. the frequency is 416.4 Hz, when the inductive reactance is 250 ohms., The graph depicting the two regression lines is shown, in Fig. 60.2. To obtain the regression line of inductive reactance on frequency the regression line equation, Y = 4.94 +0.586X is used, and X (frequency) values of, 100 and 300 have been selected in order to find the corresponding Y values. These values gave the co-ordinates, as (100, 63.5) and (300, 180.7), shown as points A, and B in Fig. 60.2. Two co-ordinates for the regression, line of frequency on inductive reactance are calculated, using the equation X =−6.15 +1.69Y , the values of, inductive reactance of 50 and 150 being used to obtain, the co-ordinate values. These values gave co-ordinates, (78.4, 50) and (247.4, 150), shown as points C and D, in Fig. 60.2., It can be seen from Fig. 60.2 that to the scale drawn, the, two regression lines coincide. Although it is not necessary to do so, the co-ordinate values are also shown, to indicate that the regression lines do appear to be the
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578 Higher Engineering Mathematics, y, , Using a tabular approach to determine the values, of the summations gives:, , Inductive reactance in ohms, , 300, , Radius, X, , 250, 200, , B, D, , 150, , X2, , Force, Y, , 55, , 5, , 3025, , 30, , 10, , 900, , 16, , 15, , 256, , 12, , 20, , 144, , 11, , 25, , 121, , 9, , 30, , 81, , 7, , 35, , 49, , 100, A, 50, , C, , 0, , 100, , 200 300 400, Frequency in hertz, , 500, , x, , ;, , Figure 60.2, , 5, X = 145, , lines of best fit. A graph showing co-ordinate values is, called a scatter diagram in statistics., , 5 10 15 20 25 30 35 40, , Radius (cm) 55 30 16 12 11, , 9, , 7, , 5, , Determine the equations of (a) the regression line of, force on radius and (b) the regression line of radius, on force. Hence, calculate the force at a radius of, 40 cm and the radius corresponding to a force of, 32 newtons., , ;, , Thus, Let the radius be the independent variable X , and the, force be the dependent variable Y . (This decision is, usually based on a ‘cause’ corresponding to X and an, ‘effect’ corresponding to Y .), (a) The equation of the regression line of force on, radius is of the form Y = a0 + a1 X and the constants a0 and a1 are determined from the normal, equations:, ;, ;, Y = a0 N + a1 X, ;, ;, ;, and, XY = a0 X + a1 X 2, (from equations (1) and (2)), , 40, Y = 180, , and, , ;, , 25, X 2 = 4601, , Y2, , XY, , Problem 4. The experimental values relating, centripetal force and radius, for a mass travelling at, constant velocity in a circle, are as shown:, Force (N), , ;, , 275, , 25, , 300, , 100, , 240, , 225, , 240, , 400, , 275, , 625, , 270, , 900, , 245, , 1225, , 200, XY= 2045, , ;, , 1600, Y 2 = 5100, , 180 = 8a0 + 145a1, 2045 = 145a0 + 4601a1, , Solving these simultaneous equations gives, a0 = 33.7 and a1 = −0.617, correct to 3 significant figures. Thus the equation of the regression, line of force on radius is:, Y = 33.7 − 0.617X, (b) The equation of the regression line of radius on, force is of the form X =b0 + b1Y and the constants b0 and b1 are determined from the normal, equations:
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Linear regression, ;, , and, , ;, X = b0 N + b1 Y, ;, ;, ;, X Y = b0 Y + b1 Y 2, (from equations (3) and (4)), , The values of the summations have been obtained, in part (a) giving:, 145 = 8b0 + 180b1, and 2045 = 180b0 + 5100b1, Solving these simultaneous equations gives, b0 = 44.2 and b1 = −1.16, correct to 3 significant, figures. Thus the equation of the regression line of, radius on force is:, , 4. The data given in Problem 2, [X = −0.056 +4.56Y ], 5. The relationship between the voltage applied, to an electrical circuit and the current flowing, is as shown:, Applied voltage (V), , 2, , 5, , 4, , 11, , 6, , 15, , 8, , 19, , i.e. the force at a radius of 40 cm is 9.02 N., , 10, , 24, , The radius, X , when the force is 32 newtons is, obtained from the regression line of radius on, force, i.e. X = 44.2 −1.16(32) = 7.08,, , 12, , 28, , 14, , 33, , The force, Y , at a radius of 40 cm, is obtained, from the regression line of force on radius, i.e., y = 33.7 −0.617(40) = 9.02,, , i.e. the radius when the force is 32 N is 7.08 cm., Now try the following exercise, Exercise 218, regression, , Further problems on linear, , In Problems 1 and 2, determine the equation of the, regression line of Y on X , correct to 3 significant, figures., X, , 14, , 18, , 23, , 30, , 50, , Y, , 900, , 1200, , 1600, , 2100, , 3800, , [Y = −256 +80.6X ], 2., , [X = 3.20 + 0.0124Y ], , Current (mA), , X = 44.2 − 1.16Y, , 1., , 3. The data given in Problem 1, , X, Y, , 6, , 3, , 9, , 15, , 2, , 14, , 21 13, , 1.3 0.7 2.0 3.7 0.5 2.9 4.5 2.7, [Y = 0.0477 +0.216X ], , In Problems 3 and 4, determine the equations of, the regression lines of X on Y for the data stated,, correct to 3 significant figures., , Assuming a linear relationship, determine, the equation of the regression line of applied, voltage, Y , on current, X , correct to 4 significant figures., [Y = 1.142 + 2.268X ], 6. For the data given in Problem 5, determine the, equation of the regression line of current on, applied voltage, correct to 3 significant figures., [X = −0.483 +0.440Y ], 7. Draw the scatter diagram for the data given, in Problem 5 and show the regression lines, of applied voltage on current and current on, applied voltage. Hence determine the values, of (a) the applied voltage needed to give a, current of 3 mA and (b) the current flowing, when the applied voltage is 40 volts, assuming, the regression lines are still true outside of the, range of values given., [(a) 7.92 V (b) 17.1 mA], 8. In an experiment to determine the relationship, between force and momentum, a force X , is, applied to a mass, by placing the mass on an, inclined plane, and the time, Y , for the velocity, , 579
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580 Higher Engineering Mathematics, , to change from u m/s to v m/s is measured. The, results obtained are as follows:, Force (N), , Time (s), , 11.4, , 0.56, , 18.7, , 0.35, , 11.7, , 0.55, , 12.3, , 0.52, , 14.7, , 0.43, , 18.8, , 0.34, , 19.6, , 0.31, , Determine the equation of the regression line, of time on force, assuming a linear relationship, , between the quantities, correct to 3 significant, figures., [Y = 0.881 −0.0290X ], 9. Find the equation for the regression line of, force on time for the data given in Problem 8,, correct to 3 decimal places., [X =30.194 −34.039Y ], 10. Draw a scatter diagram for the data given in, Problem 8 and show the regression lines of, time on force and force on time. Hence find, (a) the time corresponding to a force of 16 N,, and (b) the force at a time of 0.25 s, assuming, the relationship is linear outside of the range, of values given., [(a) 0.417 s (b) 21.7 N]
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Revision Test 17, This Revision Test covers the material contained in chapters 57 to 60. The marks for each question are shown in, brackets at the end of each question., 1. A machine produces 15% defective components., In a sample of 5, drawn at random, calculate,, using the binomial distribution, the probability, that:, , Torque X, , Current Y, , 0, , 3, , 1, , 5, , (a) there will be 4 defective items,, , 2, , 6, , (b) there will be not more than 3 defective items,, , 3, , 6, , (c) all the items will be non-defective., , 4, , 9, , 5, , 11, , 6, , 12, , 7, , 12, , 8, , 14, , 9, , 13, , Draw a histogram showing the probabilities of 0, 1,, 2, . . . , 5 defective items., (20), 2. 2% of the light bulbs produced by a company are, defective. Determine, using the Poisson distribution, the probability that in a sample of 80 bulbs:, (a) 3 bulbs will be defective, (b) not more than, 3 bulbs will be defective, (c) at least 2 bulbs will be, defective., (13), 3. Some engineering components have a mean length, of 20 mm and a standard deviation of 0.25 mm., Assume that the data on the lengths of the components is normally distributed., In a batch of 500 components, determine the, number of components likely to:, , Determine the linear coefficient of correlation for, this data., (18), 6. Some results obtained from a tensile test on a steel, specimen are shown below:, Tensile force (kN) Extension (mm), , (a) have a length of less than 19.95 mm,, , 4.8, , 3.5, , (b) be between 19.95 mm and 20.15 mm,, , 9.3, , 8.2, , (15), , 12.8, , 10.1, , 4. In a factory, cans are packed with an average of, 1.0 kg of a compound and the masses are normally, distributed about the average value. The standard, deviation of a sample of the contents of the cans, is 12 g. Determine the percentage of cans containing (a) less than 985 g, (b) more than 1030 g,, (c) between 985 g and 1030 g., (10), , 17.7, , 15.6, , 21.6, , 18.4, , 26.0, , 20.8, , (c) be longer than 20.54 mm., , 5. The data given below gives the experimental values, obtained for the torque output, X , from an electric, motor and the current, Y , taken from the supply., , Assuming a linear relationship:, (a) determine the equation of the regression line, of extension on force,, (b) determine the equation of the regression line, of force on extension,, (c) estimate (i) the value of extension when the, force is 16 kN, and (ii) the value of force, when the extension is 17 mm., (24)
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Chapter 61, , Introduction to Laplace, transforms, 61.1, , 61.3 Linearity property of the, Laplace transform, , Introduction, , The solution of most electrical circuit problems can, be reduced ultimately to the solution of differential, equations. The use of Laplace transforms provides an, alternative method to those discussed in Chapters 46 to, 51 for solving linear differential equations., , From equation (1),, !, L{k f (t )} =, , ∞, , !, , ∞, , =k, , 61.2, , Definition of a Laplace transform, , The Laplace transform, of the function f (t ) is defined, ∞, by the integral 0 e−st f (t ) dt , where s is a parameter, assumed to be a real number., , Common notations used for the Laplace, transform, , e−st f (t ) dt, , 0, , i.e. L{k f (t )} = kL{ f (t )}, , (2), , where k is any constant., Similarly,, , !, , ∞, , L{a f (t ) + bg(t )} =, , e−st (a f (t ) + bg(t )) dt, , 0, , !, , There are various commonly used notations for the, Laplace transform of f (t ) and these include:, , ∞, , =a, , e−st f (t ) dt, , 0, , !, , ∞, , +b, , (i) L{ f (t )} or L{ f (t )}, , e−st g(t ) dt, , 0, , (ii) L( f ) or L f, (iii), , e−st k f (t ) dt, , 0, , i.e. L{a f (t ) + bg(t )} = aL{ f (t )} + bL{g(t )},, , f (s) or f (s), , Also, the letter p is sometimes used instead of s as, the parameter. The notation adopted in this book will, be f (t ) for the original function and L{ f (t )} for its, Laplace transform., Hence, from above:, !, , ∞, , L{ f (t)} =, 0, , e−st f (t) dt, , (1), , (3), , where a and b are any real constants., The Laplace transform is termed a linear operator, because of the properties shown in equations (2) and (3)., , 61.4 Laplace transforms of, elementary functions, Using the definition of the Laplace transform in equation (1) a number of elementary functions may be, transformed. For example:
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Introduction to Laplace transforms, , , (a) f (t)= 1. From equation (1),, !, , ∞, , L{1} =, , −st, , e, 0, , =, , , , e−st, (1) dt =, −s, , ∞, , t e−st e−st, − 2, −s, s, , ∞, 0, , by integration by parts,, , e0, ∞e, e, −, 0, −, =, −, −s, s2, s2, , , 1, = (0 − 0) − 0 − 2, s, , 0, , −s(∞), , 1, 1, = − [e−s(∞) − e0 ] = − [0 − 1], s, s, 1, = (provided s > 0), s, , −s(∞), , since (∞ × 0) = 0,, , (b) f (t)= k. From equation (2),, =, , L{k} = kL{1}, , k, 1, = , from (a) above., Hence L{k} = k, s, s, (c), , (where a is a real constant = 0)., From equation (1),, !, , ∞, , e−st (eat ) dt =, , 0, , !, , ∞, , e−(s−a)t dt,, , 0, , from the laws of indices,, e−(s−a)t, =, −(s − a), , ∞, , 0, , 1, =, (0 − 1), −(s − a), , 1, =, s−a, (provided (s − a) > 0, i.e. s > a), (d) f (t) = cos at (where a is a real constant)., From equation (1),, !, , ∞, , L{cos at } =, , e−st cos at dt, , 0, , , =, , e−st, (a sin at − s cos at ), s2 + a2, , ∞, 0, , by integration by parts twice (see page 423),, =, , e−s(∞), (a sin a(∞) − s cos a(∞)), s2 + a2, −, , =, , s, s2 + a2, , s2, , 1, (provided s > 0), s2, , (f) f (t)= t n (where n =0, 1, 2, 3, …)., , f (t) = eat, , L{eat } =, , 583, , e0, (a sin 0 − s cos 0), + a2, , By a similar method to (e) it may be shown, 2, (3)(2) 3!, that L{t 2 } = 3 and L{t 3} =, = 4 . These, s, s4, s, results can be extended to n being any positive, integer., n!, Thus L{t n } = n+1 provided s > 0), s, (g) f (t)= sinh at. From Chapter 5,, 1, sinh at = (eat − e−at ). Hence,, 2, , , 1, 1, L{sinh at } = L eat − e−at, 2, 2, 1, 1, = L{eat } − L{e−at }, 2, 2, from equations (2) and (3),, , , 1, 1, 1, 1, −, =, 2 s −a, 2 s +a, from (c) above,, , 1, 1, 1, =, −, 2 s −a s +a, a, = 2, (provided s > a), s − a2, A list of elementary standard Laplace transforms are, summarized in Table 61.1., , 61.5 Worked problems on standard, Laplace transforms, , ( provided s > 0), , (e) f (t) = t. From equation (1),, −st ! −st, ! ∞, te, e, −st, e t dt =, −, dt, L{t } =, −s, −s, 0, , ∞, 0, , Problem 1. Using a standard list of Laplace, transforms, determine, , the following:, 1 4, (a) L 1 + 2t − t, (b) L{5e2t − 3e−t }., 3
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584 Higher Engineering Mathematics, =, , Laplace transforms, ∞, L{ f (t)}= 0 e−st f (t) dt, , 5, 3, −, s −2 s +1, , =, , 1, s, , 5(s + 1) − 3(s − 2), (s − 2)(s + 1), , =, , Table 61.1 Elementary standard Laplace transforms, Function, f (t), (i), , 1, , (ii), , k, , (iii), , eat, , (iv), , sin at, , (v), , cos at, , (vi), , t, , (vii), , t2, , (viii), , t n (n = 1, 2, 3, . . .), , (ix), , cosh at, , (x), , sinh at, , (a), , k, s, 1, s −a, a, 2, s + a2, s, 2, s + a2, 1, s2, 2!, s3, n!, s n+1, s, s2 − a2, a, s2 − a2, , , , 1 4, L 1 + 2t − t, 3, 1, = L{1} + 2L{t } − L{t 4},, 3, from equations (2) and (3), , , , 4!, 1, 1, 1, = +2 2 −, ,, s, s, 3 s 4+1, from (i), (vi) and (viii) of Table 61.1, , , 1 4.3.2.1, 2, 1, = + 2−, s s, 3, s5, 8, 1, 2, = + 2− 5, s s, s, , (b) L{5e2t − 3e−t } = 5L(e2t ) − 3L{e−t },, from equations (2) and (3), , , , , 1, 1, =5, −3, ,, s −2, s − (−1), from (iii) of Table 61.1, , 2s + 11, s2 − s − 2, , Problem 2. Find the Laplace transforms of:, (a) 6 sin 3t − 4 cos5t (b) 2 cosh 2θ − sinh 3θ., (a), , L{6 sin 3t − 4 cos5t }, = 6L{sin 3t } − 4L{cos5t }, , , , , 3, s, =6 2, −4 2, ,, s + 32, s + 52, from (iv) and (v) of Table 61.1, =, , 18, s2 + 9, , −, , 4s, s2 + 25, , (b) L{2 cosh 2θ − sinh 3θ}, = 2L{cosh 2θ} − L{sinh 3θ}, , , , s, 3, =2 2, −, s − 22, s 2 − 32, from (ix) and (x) of Table 61.1, =, , 2s, 3, −, s2 − 4 s2 − 9, , Problem 3., , Prove that, 2, a, (b) L{t 2} = 3, (a) L{sin at } = 2, s + a2, s, s, (c) L{cosh at } = 2, s − a2, (a) From equation (1),, !, , ∞, , L{sin at } =, , e−st sin at dt, , 0, , , =, , e−st, (−s sin at − a cos at ), s2 + a2, , ∞, 0, , by integration by parts,, =, , 1, [e−s(∞) (−s sin a(∞), s2 + a2, − a cos a(∞)) − e0 (−s sin 0, −a cos 0)]
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Introduction to Laplace transforms, =, , 1, [(0) − 1(0 − a)], 2, s + a2, , =, , a, (provided s > 0), s2 + a2, , =, , !, , ∞, , L{t } =, , (s 2 + 4) − s 2, 4, =, 2, 2, 2s(s + 4), 2s(s + 4), 2, = 2, s(s + 4), , e−st t 2 dt, , 0, , , , t 2 e−st 2t e−st 2e−st, =, − 3, −, −s, s2, s, , ∞, 0, , by integration by parts twice,, , , , 2, = (0 − 0 − 0) − 0 − 0 − 3, s, =, , (b) Since cosh 2x = 2 cosh2 x − 1 then, 1, cosh2 x = (1 + cosh 2x) from Chapter 5., 2, 1, Hence cosh2 3x = (1 + cosh 6x), 2 , , 1, Thus L{cosh 2 3x} = L (1 + cosh 6x), 2, , 2, (provided s > 0), s3, , 1, 1, = L{1} + L{cosh 6x}, 2, 2, , , , 1, s, 1 1, +, =, 2 s, 2 s 2 − 62, , (c) From equation (1),, , L{cosh at } = L, , , 1 at, (e + e−at ) ,, 2, , =, from Chapter 5, , 1, 1, = L{eat } + L{e−at },, 2, 2, equations (2) and (3), , , , , 1, 1, 1, 1, =, +, 2 s −a, 2 s − (−a), from (iii) of Table 61.1, , 1, 1, 1, +, =, 2 s −a s +a, , 1 (s + a) + (s − a), =, 2 (s − a)(s + a), =, , s, s2 − a2, , (provided s > a), , Problem 4. Determine the Laplace transforms of:, (a) sin2 t (b) cosh2 3x., (a), , , , , 1, s, 1, −, s, 2 s 2 + 22, from (i) and (v) of Table 61.1, , =, , (b) From equation (1),, 2, , 1, 2, , 585, , Since cos 2t = 1 −2sin2 t then, 1, sin2 t = (1 − cos2t ). Hence,, 2, , , 1, 2, L{sin t } = L (1 − cos 2t ), 2, 1, 1, = L{1} − L{cos 2t }, 2, 2, , 2s 2 − 36, s2 − 18, =, 2s(s 2 − 36) s(s2 − 36), , Problem 5. Find the Laplace transform of, 3 sin(ωt + α), where ω and α are constants., Using the compound angle formula for sin(A + B),, from Chapter 17, sin(ωt + α) may be expanded to, (sin ωt cos α + cos ωt sin α). Hence,, L{3sin (ωt + α)}, = L{3(sin ωt cos α + cos ωt sin α)}, = 3 cosαL{sin ωt } + 3 sin αL{cosωt },, since α is a constant, , , , , s, ω, + 3 sin α 2, = 3 cosα 2, s + ω2, s + ω2, from (iv) and (v) of Table 61.1, 3, = 2, (ω cos α + s sin α), (s + ω 2 ), Now try the following exercise, Exercise 219 Further problems on an, introduction to Laplace transforms, Determine the Laplace transforms in Problems, 1 to 9.
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Chapter 62, , Properties of Laplace, transforms, 62.1, , The Laplace transform of eat f (t), , From Chapter 61, the definition of the Laplace transform, of f (t ) is:, ! ∞, L{ f (t )} =, e−st f (t ) dt, (1), 0, , !, , ∞, , Thus L{eat f (t )} =, , ∞, , =, , e−st (eat f (t )) dt, e, , f (t ) dt, , (2), , Hence the substitution of (s − a) for s in the transform, shown in equation (1) corresponds to the multiplication, of the original function f (t ) by eat . This is known as a, shift theorem., , 62.2 Laplace transforms of the form, eat f(t), From equation (2), Laplace transforms of the form, eat f (t ) may be deduced. For example:, (i) L{eat t n }, , page 584., , n!, , then L{eat t n } =, , ω, from (iv) of Table, s 2 + ω2, , ω, from equa(s −a)2 + ω2, tion (2) (provided s > a)., then L{eat sin ωt} =, , 61.1, page 584., −(s−a), , (where a is a real constant), , s n+1, , 61.1, page 584., , Since L{cosh ωt } =, , 0, , Since L{t n } =, , Since L{sin ωt } =, , (iii) L{eat cosh ωt}, , 0, , !, , (ii) L{eat sin ωt}, , from (viii) of Table 61.1,, n!, , from equation (2), , from (ix) of Table, , s−a, from equa(s − a)2 − ω2, tion (2) (provided s > a)., then L{eat cosh ωt} =, , A summary of Laplace transforms of the form, eat f (t ) is shown in Table 62.1., Table 62.1 Laplace transforms of the form, eat f (t ), Function eat f (t ), (a is a real constant), (i) eat t n, (ii) eat sin ωt, (iii) eat cos ωt, (iv) eat sinh ωt, , (s − a)n+1, above (provided s > a)., , s, s 2 − ω2, , (v) eat cosh ωt, , Laplace transform, L{eat f (t )}, n!, (s − a)n+1, ω, (s − a)2 + ω2, s −a, (s − a)2 + ω2, ω, (s − a)2 − ω2, s −a, (s − a)2 − ω2
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588 Higher Engineering Mathematics, =, , Problem 1. Determine (a) L{2t 4e3t }, (b) L{4e3t cos 5t }., , =, (a) From (i) of Table 62.1,, , , 4!, L{2t e } = 2L{t e } = 2, (s − 3)4+1, 4 3t, , , , 4 3t, , = 8L{e3t cos 2t } − 10L{e3t sin 2t }, =, , (b) From (iii) of Table 62.1,, , =, , 4(s − 3), s2 − 6s +34, , =, , 1, Since cos 2x = 1 −2 sin2 x, sin2 x = (1 − cos 2x)., 2, Hence,, , , 1, L 3e− 2 x sin2 x, , (a) From (ii) of Table 62.1,, L{e, , 3, 3, sin 3t }=, =, 2, 2, (s − (−2)) + 3, (s +2)2 + 9, =, , 3, 3, =, s 2 + 4s + 4 + 9 s2 + 4s + 13, , (b) From (v) of Table 62.1,, L{3eθ cosh 4θ} = 3L{eθ cosh 4θ}=, =, , 3(s − 1), s 2 −2s +1−16, , 8s − 44, 8(s − 3) − 10(2), = 2, (s − 3)2 + 22, s − 6s + 13, , Problem 4. Show that, , , 1, 48, −2x, 2, sin x =, L 3e, (2s + 1)(4s 2 + 4s + 17), , Problem 2. Determine (a) L{e−2t sin 3t }, (b) L{3eθ cosh 4θ}., , −2t, , 10(2), 8(s − 3), −, 2, 2, (s − 3) + 2, (s − 3)2 + 22, from (iii) and (ii) of Table 62.1, , L{4e3t cos 5t } = 4L{e3t cos 5t }, , , s −3, =4, (s − 3)2 + 52, 4(s − 3), s 2 − 6s + 9 + 25, , 10, s2 + 6s + 5, , (b) L{2e3t (4 cos 2t − 5 sin 2t )}, , 2(4)(3)(2), 48, =, =, 5, (s − 3), (s − 3)5, , =, , 10, 10, =, (s + 3)2 − 22 s 2 + 6s+9 − 4, , 3(s − 1), (s − 1)2 − 42, , =, , , , 1 1, = L 3e− 2 x (1 − cos 2x), 2, , , , , 1, 1, 3, 3, = L e− 2 x − L e− 2 x cos 2x, 2, 2, ⎛ , , ⎛, ⎞, , ⎞, 1, s− −, ⎟, ⎟ 3⎜, 3⎜, 1, 2, ⎟, ⎟− ⎜, , , = ⎜, ⎜, ⎟, , , ⎝, ⎠, 2, 1, 2, 2⎝, ⎠, 1, s− −, +22, s− −, 2, 2, , 3(s − 1), s2 − 2s −15, , Problem 3. Determine the Laplace transforms of, (a) 5e−3t sinh 2t (b) 2e3t (4 cos 2t − 5 sin 2t )., (a) From (iv) of Table 62.1,, L{5e−3t sinh 2t } = 5L{e−3t sinh 2t }, , , 2, =5, (s − (−3))2 − 22, , from (iii) of Table 61.1 (page 584) and (iii), of Table 62.1 above,, , , 1, 3 s+, 3, 2, −, = , , , 1, 1 2, 2 s+, 2, + 22, s, +, 2, 2, =, , 3, 6s + 3, , − , 1, 2s + 1, 4 s2 + s + + 4, 4
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590 Higher Engineering Mathematics, (b) Second derivative, , Substituting into equation (3) gives:, , Let the second derivative of f (t ) be f (t ), then from, equation (1),, !, , ∞, , L{ f (t )} =, , i.e., , e−st f (t ) dt, , Hence, , 0, , (c) Let f (t ) = e−at then f (t ) = −ae−at and f (0) = 1., , Integrating by parts gives:, !, , ∞, , −st, , e, , f (t ) dt = e, , 0, , −st, , ∞, f (t ) 0 + s, , !, , ∞, , e, , Substituting into equation (3) gives:, −st, , f (t ) dt, , L{−ae−at } = sL{e−at } − 1, , 0, , −aL{e−at } = sL{e−at } − 1, , = [0 − f (0)] + sL{ f (t )}, , 1 = sL{e−at } + aL{e−at }, , assuming e−st f (t ) → 0 as t → ∞, and f (0) is the, value of f (t ) at t = 0. Hence, { f (t )} = − f (0) + s[s( f (t )) − f (0)], from equation (3),, ⎫, ⎪, ⎪, ⎪, ⎪, = s2 L{ f (t)} − sf (0) − f (0) ⎪, ⎪, ⎬, 2 , d y, ⎪, or L, ⎪, ⎪, dx2, ⎪, ⎪, ⎪, ⎭, 2, , = s L{ y} − sy(0) − y (0), L{ f (t)}, , i.e., , (4), , dy, at x = 0., where y (0) is the value of, dx, Equations (3) and (4) are important and are used in, the solution of differential equations (see Chapter 64), and simultaneous differential equations (Chapter 65)., Problem 5. Use the Laplace transform of the first, derivative to derive:, (a) L{k} =, , k, 2, (b) L{2t } = 2, s, s, , 1, (c) L{e−at } =, s +a, From equation (3), L{ f (t )} = sL{ f (t )} − f (0)., (a) Let f (t ) = k, then f (t ) = 0 and f (0) = k., Substituting into equation (3) gives:, L{0} = sL{k} − k, k = sL{k}, k, Hence L{k} =, s, (b) Let f (t ) = 2t then f (t ) = 2 and f (0) = 0., i.e., , L{2} = sL{2t } − 0, 2, = sL{2t }, s, 2, L{2t}= 2, s, , 1 = (s + a)L{e−at }, Hence L{e−at } =, , 1, s+a, , Problem 6. Use the Laplace transform of the, second derivative to derive, s, L{cos at } = 2, s + a2, From equation (4),, L{ f (t )} = s 2 L{ f (t )} − s f (0) − f (0), Let f (t ) = cos at , then f (t ) = −a sin at and, f (t ) = −a 2 cosat , f (0) = 1 and f (0) = 0, Substituting into equation (4) gives:, L{−a 2 cos at } = s 2 {cos at } − s(1) − 0, i.e., , −a 2 L{cos at } = s 2 L{cos at } − s, s = (s 2 + a 2 )L{cos at }, , Hence, , from which, L{cos at } =, , s, s2 + a2, , Now try the following exercise, Exercise 221 Further problems on the, Laplace transforms of derivatives, 1. Derive the Laplace transform of the first, derivative from the definition of a Laplace, transform. Hence derive the transform, L{1} =, , 1, s
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Properties of Laplace transforms, Let, 2. Use the Laplace transform of the first derivative to derive the transforms:, 1, 6, (b) L{3t 2} = 3, (a) L{eat } =, s −a, s, 3. Derive the Laplace transform of the second, derivative from the definition of a Laplace, transform. Hence derive the transform, a, L{sin at } = 2, s + a2, 4. Use the Laplace transform of the second, derivative to derive the transforms:, a, (a) L{sinh at } = 2, s − a2, s, (b) L{cosh at } = 2, s − a2, , f (t ) = 5 + 2 cos3t, , L{ f (t )} = L{5 + 2 cos3t } =, , 5, 2s, + 2, s s +9, , from (ii) and (v) of Table 61.1, page 584., By the initial value theorem,, limit[ f (t )] = limit [sL{ f (t )}], t →0, , s→∞, , , , 5, 2s, i.e. limit[5 + 2 cos 3t ]= limit s, +, s→∞, t →0, s s2 + 9, , 2s 2, = limit 5 + 2, s→∞, s +9, 2∞2, = 5+2, ∞2 + 9, i.e. 7 = 7, which verifies the theorem in this case., 5 + 2(1) = 5 +, , i.e., , The initial value of the voltage is thus 7 V., , 62.4 The initial and final value, theorems, There are several Laplace transform theorems used to, simplify and interpret the solution of certain problems., Two such theorems are the initial value theorem and the, final value theorem., , (a) The initial value theorem states:, , Problem 8. Verify the initial value theorem for, the function (2t − 3)2 and state its initial value., Let, Let, , f (t ) = (2t − 3)2 = 4t 2 − 12t + 9, L{ f (t )} = L(4t 2 − 12t + 9), , 2, 12 9, =4 3 − 2 +, s, s, s, , from (vii), (vi) and (ii) of Table 61.1, page 584., limit [ f (t)]= limit [sL{ f (t)}], s→∞, , t→0, , For example, if f (t ) = 3e4t then, L{3e4t } =, , 3, s −4, , from (iii) of Table 61.1, page 584., By the initial value theorem,, , , 3, limit[3e4t ] = limit s, s→∞, t →0, s −4, , , 3, i.e., 3e0 = ∞, ∞−4, i.e., , 3 =3, which illustrates the theorem., , Problem 7. Verify the initial value theorem for, the voltage function (5 + 2 cos3t ) volts, and state its, initial value., , By the initial value theorem,, , , 8, 12 9, limit[(2t − 3)2 ] = limit s 3 − 2 +, s→∞, t →0, s, s, s, , 8, 12, = limit 2 −, +9, s→∞ s, s, 8, 12, +9, i.e., (0 − 3)2 = 2 −, ∞, ∞, i.e. 9 = 9, which verifies the theorem in this case., The initial value of the given function is thus 9., , (b) The final value theorem states:, limit [f (t)]= limit [sL{ f (t)}], t→∞, , s→0, , For example, if f (t ) = 3e−4t then:, , , 3, limit[3e−4t ] = limit s, t →∞, s→0, s +4, , 591
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592 Higher Engineering Mathematics, i.e., , 3e−∞ = (0), , , , 3, 0+4, , , , i.e. 0 = 0, which illustrates the theorem., Problem 9. Verify the final value theorem for the, function (2 + 3e−2t sin 4t ) cm, which represents the, displacement of a particle. State its final steady, value., f (t ) = 2 + 3e−2t sin 4t, , Let, , L{ f (t )} = L{2 + 3e−2t sin 4t }, , , 2, 4, = +3, s, (s − (−2))2 + 42, 12, 2, = +, s (s + 2)2 + 16, from (ii) of Table 61.1, page 584 and (ii) of, Table 62.1 on page 587., By the final value theorem,, t →∞, , s→0, , limit[2 + 3e−2t sin 4t ], t →∞, , = limit s, s→0, , 2, 12, +, s (s + 2)2 + 16, , , = limit 2 +, s→0, , Now try the following exercise, Exercise 222 Further problems on initial, and final value theorems, 1. State the initial value theorem. Verify the theorem for the functions (a) 3 −4 sin t (b) (t − 4)2, and state their initial values., [(a) 3 (b) 16], 2. Verify the initial value theorem for the voltage, functions: (a) 4 +2 cos t (b) t − cos 3t and state, their initial values., [(a) 6 (b) −1], , limit[ f (t )] = limit[sL{ f (t )}], i.e., , The initial and final value theorems are used in pulse, circuit applications where the response of the circuit, for small periods of time, or the behaviour immediately, after the switch is closed, are of interest. The final value, theorem is particularly useful in investigating the stability of systems (such as in automatic aircraft-landing, systems) and is concerned with the steady state response, for large values of time t , i.e. after all transient effects, have died away., , , , 12s, (s + 2)2 + 16, , i.e. 2 + 0 = 2 +0, i.e. 2 = 2, which verifies the theorem in this case., The final value of the displacement is thus 2 cm., , 3. State the final value theorem and state a practical application where it is of use. Verify the, theorem for the function 4 +e−2t (sin t + cos t ), representing a displacement and state its final, value., [4], 4. Verify the final value theorem for the function, 3t 2e−4t and determine its steady state value., [0]
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Chapter 63, , Inverse Laplace transforms, 63.1 Definition of the inverse Laplace, transform, If the Laplace transform of a function f (t ) is F(s),, i.e. L{ f (t )} = F(s), then f (t ) is called the inverse, Laplace transform of F(s) and is written as, f (t ) = L−1{F(s)}., , 1, −1 1, = 1., For example, since L{1} = then L, s, s, a, Similarly, since L{sin at } = 2, then, s + a2, −1, , L, , , , a, s2 + a2, , F(s) = L{ f (t)}, , = sin at, and so on., , Tables of Laplace transforms, such as the tables in, Chapters 61 and 62 (see pages 584 and 587) may be, used to find inverse Laplace transforms., However, for convenience, a summary of inverse, Laplace transforms is shown in Table 63.1., , Problem 1. Find the following inverse Laplace, transforms:, , , , , 1, 5, (a) L−1 2, (b) L−1, s +9, 3s − 1, From (iv) of Table 63.1,, , , a, −1, L, = sin at,, s2 + a2, , L−1 {F(s)} = f (t), , (i), , 1, s, , 1, , (ii), , k, s, , k, , (iii), , 1, s −a, , eat, , (iv), , a, s 2 +a 2, , sin at, , (v), , s, s 2 +a 2, , cosat, , (vi), , 1, s2, , t, , (vii), , 2!, s3, , t2, , (viii), , n!, s n+1, , tn, , (ix), , a, s 2 −a 2, , sinh at, , (x), , s, s 2 −a 2, , cosh at, , (xi), , n!, (s − a)n+1, , eat t n, , (xii), , ω, (s − a)2 + ω2, , eat sinωt, , (xiii), , s−a, (s − a)2 + ω2, , eat cosωt, , (xiv), , ω, (s − a)2 − ω2, , eat sinhωt, , (xv), , s−a, (s − a)2 − ω2, , eat coshωt, , , , 63.2 Inverse Laplace transforms of, simple functions, , (a), , Table 63.1 Inverse Laplace transforms
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594 Higher Engineering Mathematics, Hence L−1, , , , , , , 1, 1, −1, =, L, s2 + 9, s 2 + 32, , , 3, 1 −1, = L, 3, s 2 + 32, =, , (b) L−1, , , , 1, sin 3t, 3, , ⎧, ⎪, ⎪, ⎨, , , , ⎫, ⎪, ⎪, ⎬, , 5, 5, , , = L−1, 1 ⎪, ⎪, 3s − 1, ⎪, ⎪, ⎩3 s −, ⎭, 3, ⎫, ⎧, ⎪, ⎪, ⎪, ⎪, ⎬ 5 1, 1, 5 −1 ⎨, = e3t, , = L, ⎪, 1 ⎪, 3, 3, ⎪, ⎪, ⎭, ⎩ s−, 3, , (b) L−1, , , , , , , 4s, s, −1, =, 4L, s 2 − 16, s 2 − 42, = 4 cosh 4t,, from (x) of Table 63.1, , Problem 4. Find, , , , , 3, 2, (b) L−1, (a) L−1 2, s −7, (s − 3)5, (a) From (ix) of Table 63.1,, , , a, −1, L, = sinh at, s2 − a2, Thus, L−1, , from (iii) of Table 63.1, Problem 2. Find the following inverse Laplace, transforms:, , , 6, 3, −1, −1, (a) L, (b) L, 3, s, s4, , 2, (a) From (vii) of Table 63.1,, =t2, s3, , , 6, 2, −1, −1, Hence L, = 3L, = 3t 2 ., 3, s, s3, L−1, , , , (b) From (viii) of Table 63.1, if s is to have a power, of 4 then n = 3., , , 3!, 6, −1, 3, −1, Thus L, = t i.e. L, = t3, s4, s4, Hence, , L−1, , , , , , 3, 1 −1 6, 1, = L, = t3 ., 4, 4, s, 2, s, 2, , Problem 3. Determine, , , , , 7s, 4s, (a) L−1 2, (b) L−1 2, s +4, s − 16, (a) L−1, , , , , , , 7s, s, −1, =, 7L, = 7 cos 2t,, s2 + 4, s 2 + 22, from (v) of Table 63.1, , , , , , , 3, 1, −1, =, 3L, √, s2 − 7, s 2 − ( 7)2, 5, 6, √, 7, 3 −1, √, =√ L, 7, s 2 − ( 7)2, √, 3, = √ sinh 7t, 7, , (b) From (xi) of Table 63.1,, , , n!, −1, L, = eat t n, (s − a)n+1, , , 1, 1, Thus L−1, = eat t n, n+1, (s − a), n!, , , 2, −1, and comparing with L, shows that, (s − 3)5, n = 4 and a = 3., Hence, L, , −1, , , , 2, (s − 3)5, , , , , 1, = 2L, (s − 3)5, , , 1 3t 4, 1, =2, e t = e3t t 4, 4!, 12, , Problem 5. Determine, , , 3, −1, (a) L, s 2 − 4s + 13, , , 2(s + 1), −1, (b) L, s 2 + 2s + 10, , −1
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Inverse Laplace transforms, (a) L, , −1, , , , , , , 3, 3, −1, =L, s 2 − 4s + 13, (s − 2)2 + 32, = e2t sin 3t,, , (b) L−1, , , , from (xii) of Table 63.1, , , 2(s + 1), 2(s + 1), −1, =L, s 2 + 2s + 10, (s + 1)2 + 32, , , = 2e−t cos 3t,, , Now try the following exercise, Exercise 223 Further problems on inverse, Laplace transforms of simple functions, Determine the inverse Laplace transforms of the, following:, 1. (a), , 7, 2, (b), s, s −5, , [(a) 7 (b) 2e5t ], , from (xiii) of Table 63.1, 2. (a), , Problem 6. Determine, , , 5, (a) L−1 2, s + 2s − 3, , , 4s − 3, −1, (b) L, s 2 − 4s − 5, (a) L, , −1, , , , 3. (a), , , , , 5, 5, −1, =L, s 2 + 2s − 3, (s + 1)2 − 22, ⎧, ⎫, 5, ⎪, ⎪, ⎨, ⎬, (2), −1, 2, =L, 2, 2, ⎪, ⎩ (s + 1) − 2 ⎪, ⎭, , 2s, 3, (b) 2, 2s + 1, s +4, , 3 1, (a) e− 2 t (b) 2 cos2t, 2, 1, s 2 + 25, , (b), , from (xiv) of Table 63.1, , , 4s − 3, 4s − 3, −1, −1, =L, (b) L, s 2 − 4s − 5, (s − 2)2 − 32, , , 4(s − 2) + 5, = L−1, (s − 2)2 − 32, , , 4(s − 2), = L−1, (s − 2)2 − 32, , , 5, + L−1, (s − 2)2 − 32, ⎧, ⎫, 5, ⎪, ⎪, ⎨, ⎬, (3), 2t, −1, 3, = 4e cosh 3t + L, 2, 2, ⎪, ⎩ (s − 2) − 3 ⎪, ⎭, , , from (xv) of Table 63.1, = 4e2t cosh 3t +, , 5 2t, e sinh 3t,, 3, from (xiv) of Table 63.1, , , , (a), , 4. (a), , 5s, 2s 2 + 18, , 5, = e−t sinh 2t,, 2, , , 4, s2 + 9, , 5. (a), , 6. (a), , 7. (a), , 8. (a), , 5, 8, (b) 4, s3, s, , (b), , 6, s2, , 1, 4, sin 5t (b) sin 3t, 5, 3, , , 5, (a) cos 3t (b) 6t, 2, , (a), , 4, 5 2, t (b) t 3, 2, 3, , 3s, 7, (b) 2, 1 2, s − 16, s −8, 2, , 7, (a) 6 cosh 4t (b) sinh 4t, 4, 4, 15, (b), 3s 2 − 27, (s − 1)3, , 5, (a) sinh 3t (b) 2 et t 2, 3, 1, 3, (b), (s + 2)4, (s − 3)5, , 1, 1, (a) e−2t t 3 (b) e3t t 4, 6, 8, , 595
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596 Higher Engineering Mathematics, s +1, 3, 9. (a) 2, (b) 2, s + 2s + 10, s + 6s + 13, , 3, (a) e−t cos 3t (b) e−3t sin 2t, 2, 10. (a), , 2(s − 3), s 2 − 6s + 13, , (b), , 7, s 2 − 8s + 12, , , 7, (a) 2e3t cos 2t (b) e4t sinh 2t, 2, , 11. (a), , 2s + 5, 3s + 2, (b) 2, s 2 + 4s − 5, s − 8s + 25, ⎤, ⎡, 1, (a) 2e−2t cosh 3t + e−2t sinh 3t, ⎥, ⎢, 3, ⎦, ⎣, 14 4t, 4t, (b) 3e cos 3t + e sin 3t, 3, , When s = 2, 3 =3 A, from which, A = 1., When s = −1, −9 = −3B, from which, B = 3., , , 4s − 5, −1, Hence L, s2 − s − 2, , , 1, 3, −1, ≡L, +, s −2 s +1, , , , , 1, 3, = L−1, + L−1, s −2, s +1, = e2t + 3e−t , from (iii) of Table 63.1, Problem 8., , Find L−1, , , , 3s 3 + s 2 + 12s + 2, (s − 3)(s + 1)3, , , , 3s 3 + s 2 + 12s + 2, (s − 3)(s + 1)3, A, D, B, C, +, +, +, 2, s − 3 s + 1 (s + 1), (s + 1)3, , , A(s + 1)3 + B(s − 3)(s + 1)2, + C(s − 3)(s + 1) + D(s − 3), ≡, (s − 3)(s + 1)3, ≡, , 63.3 Inverse Laplace transforms using, partial fractions, Sometimes the function whose inverse is required is not, recognisable as a standard type, such as those listed in, Table 63.1. In such cases it may be possible, by using, partial fractions, to resolve the function into simpler, fractions which may be inverted on sight. For example,, the function,, F(s) =, , 2s − 3, s(s − 3), , Problem 7., , Determine L−1, , 4s − 5, 2, s −s −2, , , , 4s − 5, 4s − 5, A, B, ≡, ≡, +, s2 − s − 2, (s − 2)(s + 1) (s − 2) (s + 1), A(s +1) + B(s −2), ≡, (s − 2)(s + 1), Hence 4s − 5 ≡ A(s + 1) + B(s − 2)., , 3s 3 + s 2 + 12s + 2 ≡ A(s + 1)3 + B(s − 3)(s + 1)2, + C(s − 3)(s + 1) + D(s − 3), When s = 3, 128 =64 A, from which, A = 2., When s = −1, −12 =−4D, from which, D = 3., , cannot be inverted on sight from Table 63.1. However,, 2s − 3, 1, 1, by using partial fractions,, ≡ +, which, s(s − 3) s s − 3, may be inverted as 1 + e3t from (i) and (iii) of Table 61.1., Partial fractions are discussed in Chapter 2, and a summary of the forms of partial fractions is given in Table 2.1, on page 13., , , Hence, , Equating s 3 terms gives: 3 = A + B, from which, B = 1., Equating constant terms gives:, 2 = A − 3B − 3C − 3D,, i.e., , 2 = 2 − 3 − 3C − 9,, , from which, 3C = −12 and C = − 4, Hence, 3 2, , 3s + s + 12s + 2, L−1, (s − 3)(s + 1)3, , , 2, 3, 1, 4, +, ≡ L−1, +, −, s − 3 s + 1 (s + 1)2 (s + 1)3, 3, = 2e3t + e−t − 4e−t t + e−t t 2 ,, 2, from (iii) and (xi) of Table 63.1
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Inverse Laplace transforms, , , 7s + 13, s(s 2 + 4s + 13), , , −s + 3, −1 1, ≡L, +, s s 2 + 4s + 13, , , , 1, −s + 3, + L−1, ≡ L−1, s, (s + 2)2 + 32, , , , −1 1, −1 −(s + 2) + 5, ≡L, +L, s, (s + 2)2 + 32, , , , s +2, −1 1, −1, −L, ≡L, s, (s + 2)2 + 32, , , 5, + L−1, (s + 2)2 + 32, , Hence L−1, , Problem 9. Determine, , , 5s 2 + 8s − 1, L−1, (s + 3)(s 2 + 1), 5s 2 + 8s − 1, A, Bs + C, ≡, + 2, 2, (s + 3)(s + 1) s + 3 (s + 1), ≡, , A(s 2 + 1) + (Bs + C)(s + 3), (s + 3)(s 2 + 1), , Hence 5s 2 + 8s − 1 ≡ A(s 2 + 1) + (Bs + C)(s + 3)., When s = −3, 20 =10 A, from which, A = 2., Equating s 2 terms gives: 5 = A + B, from which, B = 3,, since A = 2., Equating s terms gives: 8 = 3B + C, from which,, C = −1, since B = 3., , , 5s 2 + 8s − 1, −1, Hence L, (s + 3)(s 2 + 1), , , , , 2, 3s − 1, + 2, s +3 s +1, , , , , 2, 3s, + L−1 2, ≡ L−1, s +3, s +1, ≡ L−1, , − L−1, , 5, ≡ 1 − e−2t cos 3t + e−2t sin 3t, 3, from (i), (xiii) and (xii) of Table 63.1, , Now try the following exercise, , 1, , , , s2 + 1, , Use partial fractions to find the inverse Laplace, transforms of the following functions:, 1., , 11 −3s, s 2 + 2s − 3, , 2., , 2s 2 − 9s − 35, (s + 1)(s − 2)(s + 3), , [4e−t − 3e2t + e−3t ], , 3., , 5s 2 − 2s − 19, (s + 3)(s − 1)2, , [2e−3t + 3et − 4et t ], , 4., , 3s 2 + 16s + 15, (s + 3)3, , [e−3t (3 − 2t − 3t 2)], , = 2e−3t + 3 cost − sin t,, from (iii), (v) and (iv) of Table 63.1, Problem 10. Find, , L−1, , , , 7s + 13, s(s 2 + 4s + 13), , , , 7s + 13, A, Bs + C, ≡ + 2, s(s 2 + 4s + 13), s, s + 4s + 13, A(s 2 + 4s + 13) + (Bs + C)(s), ≡, s(s 2 + 4s + 13), , 5., , Hence 7s + 13 ≡ A(s 2 + 4s + 13) + (Bs + C)(s)., When s = 0, 13 =13 A, from which, A = 1., Equating, B = −1., , s2, , , , Exercise 224 Further problems on inverse, Laplace transforms using partial fractions, , , terms gives: 0 = A + B, from which,, , Equating s terms gives: 7 =4 A + C, from which, C = 3., , 597, , 6., , [2et − 5e−3t ], , 7s 2 + 5s + 13, (s 2 + 2)(s + 1), , √, √, 3, 2 cos 2t + √ sin 2t + 5e−t, 2, 3 +6s + 4s 2 − 2s 3, s 2 (s 2 + 3), , √, √, √, [2 + t + 3 sin 3t − 4 cos 3t ]
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598 Higher Engineering Mathematics, , 7., , 26 −s 2, s(s 2 + 4s + 13), , 2, 2 − 3e−2t cos 3t − e−2t sin 3t, 3, , 63.4, , Poles and zeros, , It was seen in the previous section that Laplace transφ(s), forms, in general, have the form f (s) =, . This is, θ(s), the same form as most transfer functions for engineering systems, a transfer function being one that relates, the response at a given pair of terminals to a source or, stimulus at another pair of terminals., Let a function in the s domain be given by:, φ(s), f (s) =, where φ(s) is of less, (s − a)(s − b)(s − c), degree than the denominator., Poles: The values a, b, c, … that makes the denominator zero, and hence f (s) infinite, are called, the system poles of f (s)., If there are no repeated factors, the poles are, simple poles., If there are repeated factors, the poles are, multiple poles., Zeros: Values of s that make the numerator φ(s) zero,, and hence f (s) zero, are called the system, zeros of f (s)., s −4, has simple poles at s = −1, (s + 1)(s − 2), s +3, has a, and s = +2, and a zero at s = 4, (s + 1)2 (2s + 5), 5, simple pole at s = − and double poles at s = −1, and, 2, s +2, a zero at s = −3 and, has simple, s(s − 1)(s + 4)(2s + 1), 1, poles at s = 0, +1, −4, and − and a zero at s = −2, 2, For example:, , The location of a pole in the s-plane is denoted by a, cross (×) and the location of a zero by a small circle, (o). This is demonstrated in the following examples., From the pole-zero diagram it may be determined that, the magnitude of the transfer function will be larger, when it is closer to the poles and smaller when it is close, to the zeros. This is important in understanding what the, system does at various frequencies and is crucial in the, study of stability and control theory in general., Problem 11., R(s) =, , Determine for the transfer function:, , 400 (s + 10), s (s + 25)(s 2 + 10s + 125), , (a) the zero and (b) the poles. Show the poles and, zero on a pole-zero diagram., (a) For the numerator to be zero, (s + 10) = 0., Hence, s = −10 is a zero of R(s)., (b) For the denominator to be zero, s = 0 or s = −25, or s 2 + 10s + 125 =0., Using the quadratic formula., , √, −10 ± 102 −4(1)(125) −10 ± −400, =, s=, 2, 2, =, , −10 ± j 20, 2, , = (−5 ± j 10), Hence, poles occur at s = 0, s =−25, (−5 + j10), and (−5 −j10), The pole-zero diagram is shown in Figure 63.1., j, , j10, , 225, , 220, , 215, , 210, , 25, , 0, , Pole-zero diagram, The poles and zeros of a function are values of complex, frequency s and can therefore be plotted on the complex, frequency or s-plane. The resulting plot is the pole-zero, diagram or pole-zero map. On the rectangular axes, the, real part is labelled the σ -axis and the imaginary part, the jω-axis., , 2j10, , Figure 63.1, ,
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Inverse Laplace transforms, Now try the following exercise, , Problem 12. Determine the poles and zeros for, the function: F(s) =, , (s + 3)(s − 2), (s + 4)(s 2 + 2s + 2), , Exercise 225, and zeros, , Further problems on poles, , and plot them on a pole-zero map., 1. Determine, For the numerator to be zero, (s + 3) =0 and (s − 2) = 0,, hence zeros occur at s = −3 and at s = +2 Poles occur, when the denominator is zero, i.e. when (s + 4) = 0, i.e., s = −4, and when s 2 + 2s + 2 = 0,, i.e. s =, , −2 ±, , , , √, 22 − 4(1)(2), − 2 ± −4, =, 2, 2, , −2 ± j2, =, = (−1 +j) or (−1 −j), 2, The poles and zeros are shown on the pole-zero map of, F(s) in Figure 63.2., , j, , 23, , 22, , 21, , 0, , function:, , (a) the zero and (b) the poles. Show the poles, and zeros on a pole-zero diagram., , (a) s = −4 (b) s = 0, s = −2,, s = 4 + j 3, s= 4 − j 3, 2. Determine the poles and zeros for the function:, (s − 1)(s + 2), F(s) =, and plot them on, (s + 3)(s 2 − 2s + 5), a pole-zero map., , poles at s = −3, s = 1 + j 2, s = 1 − j 2,, zeros at s = +1, s = −2, s −1, (s + 2)(s 2 + 2s + 5), determine the poles and zeros and show them, on a pole-zero diagram., ⎡, ⎤, poles at s = −2, s = −1 + j 2,, ⎣, ⎦, s = −1 − j 2,, zero at s = 1, , 3. For the function G(s) =, , j, , 24, , for the transfer, 50 (s + 4), R(s) =, s (s + 2)(s 2 − 8s + 25), , 1, , 2, , 3, , , , 2j, , Figure 63.2, , It is seen from these problems that poles and zeros, are always real or complex conjugate., , 4. Find the poles and zeros for the transfer funcs 2 − 5s − 6, tion: H (s) =, and plot the results in, s(s 2 + 4), the s-plane., , poles at s = 0, s = + j 2, s = − j 2,, zeros at s = −1, s = 6, , 599
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Chapter 64, , The solution of differential, equations using Laplace, transforms, 64.1, , Introduction, , An alternative method of solving differential equations, to that used in Chapters 46 to 51 is possible by using, Laplace transforms., , 64.2, , Procedure to solve differential, equations by using Laplace, transforms, , (i) Take the Laplace transform of both sides of the, differential equation by applying the formulae, for the Laplace transforms of derivatives (i.e., equations (3) and (4) of Chapter 62) and, where, necessary, using a list of standard Laplace transforms, such as Tables 61.1 and 62.1 on pages 584, and 587., (ii) Put in the given initial conditions, i.e. y(0), and y (0)., (iii) Rearrange the equation to make L{y} the subject., (iv) Determine y by using, where necessary, partial, fractions, and taking the inverse of each term by, using Table 63.1 on page 593., , 64.3, , Worked problems on solving, differential equations using, Laplace transforms, , Problem 1. Use Laplace transforms to solve the, differential equation, dy, d2 y, − 3y = 0, given that when x = 0,, 2 2 +5, dx, dx, dy, y = 4 and, = 9., dx, This is the same problem as Problem 1 of Chapter 50,, page 478 and a comparison of methods can be made., Using the above procedure:, , , dy, d2 y, − 3L{y} = L{0}, + 5L, (i) 2L, dx 2, dx, , , 2[s 2 L{y} − sy(0) − y (0)] + 5[sL{y}, − y(0)] − 3L{y} = 0,, from equations (3) and (4) of Chapter 62.
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The solution of differential equations using Laplace transforms, (ii), , y(0) = 4 and y (0) = 9, Thus 2[s 2 L{y} − 4s − 9] + 5[sL{y} − 4], − 3L{y} = 0, i.e., , This is the same as Problem 3 of Chapter 50, page 479., Using the above procedure:, 2 , , d x, dy, (i) L, + 13L{y} = L{0}, +, 6L, d y2, dx, , 2s 2 L{y} − 8s − 18 + 5sL{y} − 20, , + 6[sL{y} − y(0)] + 13L{y} = 0,, , (iii) Rearranging gives:, , from equations (3) and (4) of Chapter 62., , (2s 2 + 5s − 3)L{y} = 8s + 38, 8s + 38, i.e. L{y} = 2, 2s + 5s − 3, y = L−1, , , , 8s + 38, 2s 2 + 5s − 3, , [s 2 L{y} − sy(0) − y (0)], , Hence, , − 3L{y} = 0, , (iv), , 601, , (ii), , y(0) = 3 and y (0) = 7, Thus s 2 L{y} − 3s − 7 + 6sL{y}, , , , − 18 + 13L{y} = 0, (iii) Rearranging gives:, , 8s + 38, 8s + 38, ≡, + 5s − 3 (2s − 1)(s + 3), , (s 2 + 6s + 13)L{y} = 3s + 25, , 2s 2, , L{y} =, , A, B, ≡, +, 2s − 1 s + 3, , i.e., , A(s + 3) + B(2s − 1), ≡, (2s − 1)(s + 3), , y = L−1, , Hence 8s + 38 = A(s + 3) + B(2s − 1)., 1, 1, When s = , 42 =3 A, from which, A = 12., 2, 2, When s = −3, 14 =−7B, from which, B = −2., , , 8s + 38, −1, Hence y = L, 2s 2 + 5s − 3, , , 12, 2, −1, =L, −, 2s − 1 s + 3, 6, 5, , , 12, 2, −1, −1, =L, , −L, s +3, 2 s − 12, 1, , Hence y = 6e 2 x − 2e−3x , from (iii) of, Table 63.1., Problem 2. Use Laplace transforms to solve the, differential equation:, dy, d2 y, +6, + 13y = 0, given that when x = 0, y = 3, 2, dx, dx, dy, and, = 7., dx, , (iv), , , , = L−1, =L, , −1, , 3s + 25, s 2 + 6s + 13, , , , , 3s + 25, + 6s + 13, , , , 3s + 25, (s + 3)2 + 22, 3(s + 3) + 16, (s + 3)2 + 22, , , , , , 3(s + 3), =L, (s + 3)2 + 22, , , 8(2), + L−1, (s + 3)2 + 22, −1, , , , s2, , = 3e−3t cos2t + 8e−3t sin 2t, from (xiii), and (xii) of Table 63.1, Hence y = e−3t (3 cos 2t + 8 sin 2t), Problem 3. Use Laplace transforms to solve the, differential equation:, d2 y, dy, −3, = 9, given that when x = 0, y = 0 and, 2, dx, dx, dy, = 0., dx, This is the same problem as Problem 2 of Chapter 51,, page 485. Using the procedure:
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602 Higher Engineering Mathematics, , , , , d2 y, dy, (i) L, = L{9}, − 3L, dx 2, dx, Hence [s 2 L{y} − sy(0) − y (0)], , Using the procedure:, , 2 , dy, d y, (i) L, + 10L{y} = L{ e2x + 20}, −, 7L, dx 2, dx, Hence [s 2 L{y} − sy(0) − y (0)] − 7[sL{y}, 1, 20, − y(0)] + 10L{y} =, +, s −2, s, , 9, −3[sL{y} − y(0)] =, s, y(0) = 0 and y (0) = 0, , (ii), , (ii), Hence s 2 L{y} − 3sL{y} =, , 9, s, , (iii) Rearranging gives:, 9, (s 2 − 3s)L{y} =, s, 9, 9, i.e. L{y} =, =, s(s 2 − 3s) s 2 (s − 3), y = L−1, , (iv), , , , 9, 2, s (s − 3), , + 10L{y} =, (iii) (s 2 − 7s + 10)L{y} =, , , , 9, C, A, B, ≡ + 2+, − 3), s, s, s −3, , s 2 (s, , ≡, , A(s)(s − 3) + B(s, s 2 (s − 3), , When s = 0, 9 =−3B, from which, B = −3., When s = 3, 9 =9C, from which, C = 1., Equating s 2 terms gives: 0 = A + C, from which,, A = −1, since C = 1. Hence,, L−1, , , , , 9, 1, 3, 1, −1, =L, − − 2+, s 2 (s − 3), s s, s −3, = −1 − 3x + e3x , from (i),, , (iv), , ≡, , 3(21s − 40) − s(s − 2), 3s(s − 2), , =, , −s 2 + 65s − 120, 3s(s − 2), , A, B, C, D, +, +, +, s, s − 5 s − 2 (s − 2)2, , A(s − 5)(s − 2)2 + B(s)(s − 2)2, , ≡, , , , + C(s)(s − 5)(s − 2) + D(s)(s − 5), s(s − 5)(s − 2)2, , Hence, , d2 y, , −s 2 + 65s − 120, , dy, + 10y = e2x + 20, given that when, dx, dx, dy, 1, x = 0, y = 0 and, =−, dx, 3, , =, , −s 2 + 65s − 120, s(s − 5)(s − 2)2, , Problem 4. Use Laplace transforms to solve the, differential equation:, −7, 2, , 21s − 40 1, −, s(s − 2) 3, , −s 2 + 65s − 120, 3s(s − 2)(s 2 − 7s + 10), , −s 2 + 65s − 120, 1, =, 3 s(s − 2)(s − 2)(s − 5), , 1 −s 2 + 65s − 120, =, 3 s(s − 5)(s − 2)2, 2, , −s + 65s − 120, 1, y = L−1, 3, s(s − 5)(s − 2)2, , (vi) and (iii) of Table 63.1., i.e. y = e3x − 3x −1, , 21s − 40, s(s − 2), , Hence L{y} =, , − 3) + Cs 2, , Hence 9 ≡ A(s)(s − 3) + B(s − 3) + Cs 2 ., , , , 1, 3, , , 1, 2, Hence s L{y} − 0 − −, − 7sL{y} + 0, 3, y(0) = 0 and y (0) = −, , ≡A(s − 5)(s − 2)2 + B(s)(s − 2)2, + C(s)(s − 5)(s − 2) + D(s)(s − 5)
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The solution of differential equations using Laplace transforms, When s = 0, −120 = − 20 A, from which, A = 6., , Hence, , E = A(R + Ls) + Bs, , When s = 5, 180 =45B, from which, B = 4., , When, , s = 0, E = AR,, , When s = 2, 6 =−6D, from which, D = −1., , from which,, , A=, , Equating s 3 terms gives: 0 = A + B + C, from, which, C = −10., 2, , −s + 65s − 120, 1, Hence L−1, 3, s(s − 5)(s − 2)2, , , 1 −1 6, 4, 10, 1, = L, +, −, −, 3, s s − 5 s − 2 (s − 2)2, , Problem 5. The current flowing in an electrical, circuit is given by the differential equation, Ri + L(di/dt ) = E, where E, L and R are, constants. Use Laplace transforms to solve the, equation for current i given that when t = 0,, i = 0., Using the procedure:, , , di, = L{E}, (i) L{Ri} + L L, dt, , from which,, , B =−, , L−1, , , , Hence current i =, E, s, , (ii) i(0) = 0, hence RL{i} + LsL{i} =, , E, s, , (iii) Rearranging gives:, E, s, , E, s(R + Ls), , , E, (iv) i = L−1, s(R + Ls), i.e. L{i} =, , E, A, B, ≡ +, s(R + Ls), s, R + Ls, ≡, , , , R, R, s =− , E = B −, L, L, , A(R + Ls) + Bs, s(R + Ls), , EL, R, , , E, s(R + Ls), , , −E L/R, −1 E/R, =L, +, s, R + Ls, , , E, EL, = L−1, −, Rs, R(R + Ls), ⎧, ⎛, ⎞⎫, ⎪, ⎪, ⎨ E 1 E, ⎬, ⎜ 1 ⎟, = L−1, − ⎝, ⎠, ⎪, ⎪, R R, ⎩R s, +s ⎭, L, ⎧, ⎫, ⎪, ⎪, ⎪, ⎪, ⎬, E −1 ⎨ 1, 1, , = L, −, ⎪, R ⎪, R, s, ⎪, ⎪, ⎩, ⎭, s+, L, , 10, x, 4, Thus y = 2 + e5x − e2x − e2x, 3, 3, 3, , (R + Ls)L{i} =, , When, , Hence, , 1, = [6 + 4 e5x − 10 e2x − x e2x ], 3, , i.e. RL{i} + L[sL{i} − i(0)] =, , E, R, , , , Rt, E, 1 − e− L, R, , Now try the following exercise, Exercise 226 Further problems on solving, differential equations using Laplace, transforms, 1., , A first order differential equation involving, current i in a series R − L circuit is given by:, di, E, + 5i = and i = 0 at time t = 0., dt, 2, Use Laplace transforms to solve for i, when (a) E = 20 (b) E = 40 e−3t and, (c) E = 50 sin 5t ., ⎤, ⎡, (a) i = 2(1 − e−5t ), ⎥, ⎢(b) i = 10( e−3t − e−5t ), ⎥, ⎢, ⎦, ⎣, 5 −5t, (c) i = ( e − cos 5t + sin 5t ), 2, , 603
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604 Higher Engineering Mathematics, In Problems 2 to 9, use Laplace transforms to solve, the given differential equations., 2., , 9, , 7., , dy, d2 y, − 24, + 16y = 0, given y(0) = 3 and, dt 2, dt, , , y (0) = 3., , 4, , y = (3 − t ) e 3 t, 8., , 3., , 4., , 5., , 6., , d2 x, + 100x = 0, given x(0) = 2 and, dt 2, [x = 2 cos10t ], x (0) = 0., d2 i, di, + 1000 + 250000i = 0, given, 2, dt, dt, i(0) = 0 and i (0) = 100. [i = 100t e−500t ], d2 x, , dx, +6, + 8x = 0, given x(0) = 4 and, dt 2, dt, x (0) = 8., [x = 4(3e−2t − 2e−4t )], dy, 2, d2 y, −2, + y = 3 e4x , given y(0) = −, dx 2, dx, 3, 1, and y (0) = 4, 3, , 1, y = (4x − 1) e x + e4x, 3, , d2 y, + 16y = 10 cos4x, given y(0) = 3 and, dx 2, y (0) = 4., , 5, y = 3 cos4x + sin 4x + x sin 4x, 4, d2 y dy, +, − 2y = 3 cos3x − 11 sin 3x, given, dx 2 dx, y(0) = 0 and y (0) = 6, [ y = ex − e−2x + sin 3x], , 9., , d2 y, dy, −2, + 2y = 3 e x cos 2x, given, 2, dx, dx, y(0) = 2 and y (0) = 5, y = 3e x (cos x + sin x) − ex cos 2x, , , , 10. Solve, using Laplace transforms, Problems 4, to 9 of Exercise 187, page 480 and Problems, 1 to 5 of Exercise 188, page 482., 11. Solve, using Laplace transforms, Problems 3, to 6 of Exercise 189, page 486, Problems 5, and 6 of Exercise 190, page 488, Problems 4, and 7 of Exercise 191, page 490 and Problems, 5 and 6 of Exercise 192, page 492.
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Chapter 65, , The solution of simultaneous, differential equations using, Laplace transforms, 65.1, , Introduction, , It is sometimes necessary to solve simultaneous differential equations. An example occurs when two electrical, circuits are coupled magnetically where the equations, relating the two currents i1 and i2 are typically:, , L1, , di1, di2, +M, + R1 i1 = E 1, dt, dt, , L2, , di2, di1, +M, + R2 i2 = 0, dt, dt, , where L represents inductance, R resistance, M mutual, inductance and E 1 the p.d. applied to one of the circuits., , 65.2 Procedure to solve simultaneous, differential equations using, Laplace transforms, (i) Take the Laplace transform of both sides of each, simultaneous equation by applying the formulae for the Laplace transforms of derivatives (i.e., equations (3) and (4) of Chapter 62, page 589) and, using a list of standard Laplace transforms, as in, Table 61.1, page 584 and Table 62.1, page 587., , (ii) Put in the initial conditions, i.e. x(0), y(0), x (0),, y (0)., (iii) Solve the simultaneous equations for L{y} and, L{x} by the normal algebraic method., (iv) Determine y and x by using, where necessary,, partial fractions, and taking the inverse of each, term., , 65.3 Worked problems on solving, simultaneous differential, equations by using Laplace, transforms, Problem 1. Solve the following pair of, simultaneous differential equations, dy, +x =1, dt, dx, − y + 4et = 0, dt, given that at t = 0, x = 0 and y = 0., Using the above procedure:, , dy, + L{x} = L{1}, (i) L, dt, , (1)
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606 Higher Engineering Mathematics, , L, , , dx, − L{y} + 4L{et } = 0, dt, , (2), , Hence, −4s 2 + s − 1 = A(s − 1)(s 2 + 1) + Bs(s 2 + 1), , Equation (1) becomes:, , + (Cs + D)s(s − 1), , 1, [sL{y} − y(0)] + L{x} =, s, , (1 ), , from equation (3), page 589 and Table 61.1,, page 584., , When s = 0, −1 = −A, , hence A = 1, , When s = 1, −4 = 2B, , hence B =−2, , Equating s 3 coefficients:, , Equation (2) becomes:, [sL{x} − x(0)] − L{y} = −, (ii), , 4, s −1, , 0 = A + B + C hence C = 1, (2 ), , x(0) = 0 and y(0) = 0 hence, , (since A = 1 and B = −2), Equating s 2, , −4 = −A + D − C hence D =−2, , Equation (1 ) becomes:, sL{y} + L{x} =, , 1, s, , (since A = 1 and C = 1), , (1 ), , and equation (2 ) becomes:, , Thus L{x} =, , 4, s −1, 4, or −L{y} + sL{x} = −, s −1, sL{x} − L{y} = −, , =, (2 ), , (iii) 1 × equation (1 ) and s × equation (2 ) gives:, 1, sL{y} + L{x} =, s, −sL{y} + s 2 L{x} = −, , 4s, s −1, , (4), , =L, , −1, , +s −1, s(s − 1)(s 2 + 1), , (5), , −4s 2 + s − 1, A, B, Cs + D, ≡ +, + 2, 2, s(s − 1)(s + 1), s, (s − 1) (s + 1), , , A(s − 1)(s 2 + 1) + Bs(s 2 + 1), s(s − 1)(s 2 + 1), , 1, 2, s −2, −, +, s (s − 1) (s 2 + 1), , , , 1, 2, s, 2, −, +, −, s (s − 1) (s 2 + 1) (s 2 + 1), , , , x = 1 −2et + cos t − 2 sin t,, , y=, , Using partial fractions, , + (Cs + D)s(s − 1), , , , 2, s −2, 1, −, +, s (s − 1) (s 2 + 1), , dx, − y + 4 et = 0, dt, from which,, , −4s 2 + s − 1, s(s − 1), −4s 2, , , , −4s 2 + s − 1, s(s − 1)(s 2 + 1), , from Table 63.1, page 593, From the second equation given in the question,, , 1, 4s, −, s s −1, (s − 1) − s(4s), =, s(s − 1), , =, , x =L, , −1, , i.e., , (s 2 + 1)L{x} =, , from which, L{x} =, , (iv) Hence, , (3), , Adding equations (3) and (4) gives:, , =, , coefficients:, , =, , dx, + 4 et, dt, d, (1 − 2 et + cos t − 2 sin t ) + 4 et, dt, , = −2 et − sin t − 2 cos t + 4 et, i.e. y = 2et − sin t − 2 cos t, [Alternatively, to determine, equations (1 ) and (2 )], , y,, , return, , to
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607, , The solution of simultaneous differential equations using Laplace transforms, and equation (2 ) becomes, , Problem 2. Solve the following pair of, simultaneous differential equations, , (2s − 1)L{y} − 2(3) − sL{x }, , dx, dy, − 5 + 2x = 6, dt, dt, dy dx, −, − y = −1, 2, dt, dt, , +8=−, , 3, , i.e. (3s + 2)L{x} − 5sL{y} =, , given that at t = 0, x = 8 and y = 3., , (3s + 2)L{x} − 5sL{y}, , Using the above procedure:, , , dy, dx, (i) 3L, − 5L, + 2L{x} = L{6}, dt, dt, , , dy, dx, 2L, −L, − L{y} = L{−1}, dt, dt, , 6, +9, s, − sL{x} + (2s − 1)L{y}, =, , (1), , 1, = − −2, s, , (2), , 3[sL{x} − x(0)] − 5[sL{y} − y(0)], 6, s, , from equation (3), page 589, and Table 61.1,, page 584., , i.e. (3s + 2)L{x} − 3x(0) − 5sL{y}, 6, + 5y(0) =, s, Equation (2) becomes:, , 6, s, (1 ), , (A), , , , 1, = (3s + 2) − − 2 (4), s, i.e. s(3s + 2)L{x} − 5s 2 L{y} = 6 + 9s, , (3 ), , −s(3s + 2)L{x} + (6s 2 + s − 2)L{y}, = −6s −, , 2[sL{y} − y(0)] − [sL{x } − x(0)], , 2, −7, s, , (4 ), , Adding equations (3 ) and (4 ) gives:, , 1, − L{y} = −, s, , (s 2 + s − 2)L{y} = −1 + 3s −, , from equation (3), page 589, and Table 61.1,, page 584,, , + x(0) − L{y} = −, , 1, s, , i.e. (2s − 1)L{y} − 2y(0) − sL{x}, , 2, s, , =, , −s + 3s 2 − 2, s, , from which, L{y} =, , 3s 2 − s − 2, s(s 2 + s − 2), , i.e. 2sL{y} − 2y(0) − sL{x}, , + x(0) = −, , ⎪, ⎪, ⎪, ⎪, ⎪, ⎪, ⎪, ⎪, (2 ) ⎭, , (1 ), , −s(3s + 2)L{x} + (3s + 2)(2s − 1)L{y}, , i.e. 3sL{x} − 3x(0) − 5sL{y}, + 5y(0) + 2L{x} =, , 6, +9, s, ⎫, ⎪, ⎪, ⎪, ⎪, ⎪, ⎪, ⎪, (1 ) ⎪, ⎬, , (2 ), , (iii) s × equation (1 ) and (3s + 2) × equation (2 ), gives:, , , 6, +9, (3), s(3s + 2)L{x} − 5s 2 L{y} = s, s, , Equation (1) becomes:, + 2L{x} =, , 1, s, , Using partial fractions, 1, s, , (2 ), , (ii) x(0) = 8 and y(0) = 3, hence equation (1 ), becomes, (3s + 2)L{x} − 3(8) − 5sL{y}, + 5(3) =, , 6, s, , (1 ), , 3s 2 − s − 2, s(s 2 + s − 2), ≡, , A, B, C, +, +, s, (s + 2) (s − 1), , =, , A(s + 2)(s − 1) + Bs(s − 1) + Cs(s + 2), s(s + 2)(s − 1)
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608 Higher Engineering Mathematics, i.e. 3s 2 − s − 2 = A(s + 2)(s − 1), , Using partial fractions, , + Bs(s − 1) + Cs(s + 2), , 8s 2 − 2s − 6, s(s + 2)(s − 1), , When s = 0, −2 = −2 A, hence A =1, When s = 1, 0 = 3C, hence C = 0, , ≡, , A, B, C, +, +, s, (s + 2) (s − 1), , =, , A(s + 2)(s − 1) + Bs(s − 1) + Cs(s + 2), s(s + 2)(s − 1), , When s = −2, 12 =6B, hence B =2, Thus L{y} =, , 3s 2 − s − 2, 1, 2, = +, s(s 2 + s − 2), s (s + 2), , (iv) Hence y = L−1, , , , i.e. 8s 2 − 2s − 6 = A(s + 2)(s − 1), , , , 1, 2, = 1 +2e−2t, +, s s +2, , + Bs(s − 1) + Cs(s + 2), , Returning to equations (A) to determine L{x} and, hence x:, (2s − 1) × equation (1 ) and 5s × (2 ) gives:, (2s − 1)(3s + 2)L{x} − 5s(2s − 1)L{y}, , , 6, = (2s − 1), +9, s, and −s(5s)L{x} + 5s(2s − 1)L{y}, , , 1, = 5s − − 2, s, , 6, −9, s, , (5), Thus L{x} =, , (6), , and − 5s 2 L{x} + 5s(2s − 1)L{y}, (6 ), , Adding equations (5 ) and (6 ) gives:, (s 2 + s − 2)L{x} = −2 + 8s −, =, , 6, s, , −2s + 8s 2 − 6, s, , from which, L{x} =, =, , 8s 2 − 2s − 6, s(s 2 + s − 2), 8s 2 − 2s − 6, s(s + 2)(s − 1), , 8s 2 − 2s − 6, 3, 5, = +, s(s + 2)(s − 1) s (s + 2), , Hence x = L−1, , , , , 3, 5, = 3 + 5e−2t, +, s s +2, , Therefore the solutions of the given simultaneous differential equations are, (5 ), , = −5 − 10s, , When s = 1, 0 =3C, hence C = 0, When s = −2, 30 = 6B, hence B = 5, , i.e. (6s 2 + s − 2)L{x} − 5s(2s − 1)L{y}, = 12 + 18s −, , When s = 0, −6 = −2 A, hence A = 3, , y = 1 +2e−2t and x = 3 +5e−2t, (These solutions may be checked by substituting the, expressions for x and y into the original equations.), Problem 3. Solve the following pair of, simultaneous differential equations, d2 x, −x = y, dt 2, d2 y, + y = −x, dt 2, dx, =0, given that at t = 0, x = 2, y = −1,, dt, dy, and, = 0., dt
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The solution of simultaneous differential equations using Laplace transforms, Equation (7) −equation (8) gives:, , Using the procedure:, (i), , [s 2 L{x} − sx(0) − x (0)] − L{x} = L{y}, , [−1 − (s 2 − 1)(s 2 + 1)]L{y}, , (1), , = 2s + s(s 2 − 1), , [s 2 L{y} − sy(0) − y (0)] + L{y} = −L{x} (2), , i.e., , −s 4 L{y} = s 3 + s, , and, , L{y} =, , from equation (4), page 590, (ii) x(0) = 2, y(0) = −1, x (0) = 0 and y (0) = 0, hence s 2 L{x} − 2s − L{x} = L{y}, s 2 L{y} + s + L{y} = −L{x}, , 1, 1, s3 + s, =− − 3, 4, −s, s s, , , 1, 1, −1, y=L, − − 3, s s, , from which,, , (1 ), (2 ), , 1, y = −1 − t 2, 2, , i.e., , (iii) Rearranging gives:, (s 2 − 1)L{x} − L{y} = 2s, 2, , L{x} + (s + 1)L{y} = −s, , (3), (4), , Equation (3) ×(s 2 + 1) and equation (4) ×1, gives:, (s 2 + 1)(s 2 − 1)L{x} − (s 2 + 1)L{y}, = (s 2 + 1)2s, L{x} + (s 2 + 1)L{y} = −s, , Now try the following exercise, Exercise 227 Further problems on solving, simultaneous differential equations using, Laplace transforms, , (5), , Solve the following pairs of simultaneous differential equations:, , (6), , 1., , Adding equations (5) and (6) gives:, [(s 2 + 1)(s 2 − 1) + 1]L{x} = (s 2 + 1)2s − s, i.e. s 4 L{x} = 2s 3 + s = s(2s 2 + 1), 2., , s(2s 2 + 1) 2s 2 + 1, =, from which, L{x} =, s4, s3, =, , (iv), , Hence x = L, , −1, , , , 2s 2, 1, 2, 1, + 3 = + 3, s3, s, s s, , 2, 1, + 3, s s, , x = 5 cos t + 5 sin t − e2t − et − 3 and, y = e2t + 2et − 3 − 5 sin t, 3., , Returning to equations (3) and (4) to determine y:, 1 × equation (3) and (s 2 − 1) × equation (4) gives:, 2, , (s − 1)L{x} − L{y} = 2s, , (7), , (s − 1)L{x} + (s − 1)(s + 1)L{y}, = −s(s 2 − 1), , (8), , 2, , 2, , dy, dx, −y+x +, − 5 sin t = 0, dt, dt, dx, dy, 3 + x − y + 2 − et = 0, dt, dt, given that at t = 0, x = 0 and y = 0., 2, , , , 1, x = 2 + t2, 2, , i.e., , dx dy, +, = 5 et, dt, dt, dy, dx, −3, =5, dt, dt, given that when t = 0, x = 0 and y = 0., [x = et − t − 1 and y = 2t − 3 + 3et ], 2, , 2, , d2 x, + 2x = y, dt 2, d2 y, + 2y = x, dt 2, given that at t = 0, x = 4, y = 2,, and, , dy, = 0., dt, , , , dx, =0, dt, , √, x = 3 cos t + cos(√3 t ) and, y = 3 cos t − cos( 3 t ), , 609
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Revision Test 18, This Revision Test covers the material contained in Chapters 61 to 65. The marks for each question are shown in, brackets at the end of each question., 1., , Find the Laplace transforms of the following, functions:, (a) 2t 3 − 4t + 5 (b) 3e−2t − 4 sin 2t, , 2., , (c) 3 cosh 2t, , (d) 2t 4e−3t, , (e) 5e2t cos 3t, , (f) 2e3t sinh 4t, , 4., (16), , (c), (e), (g), , 12, 5, (b) 5, 2s + 1, s, 4s, 5, (d) 2, 2, s +9, s −9, s −4, 3, (f) 2, (s + 2)4, s − 8s − 20, 8, s 2 − 4s + 3, , , , 13 − s 2, s(s 2 + 4s + 13), , 5., , (24), , In a galvanometer the deflection θ satisfies the, differential equation:, , Use Laplace transforms to solve the equation for θ, dθ, = 0. (13), given that when t = 0, θ = 0 and, dt, Solve the following pair of simultaneous differential equations:, 3, , dx, = 3x + 2y, dt, , 2, , dy, + 3x = 6y, dt, , (17), , Use partial fractions to determine the following:, , , 5s − 1, (a) L−1 2, s −s −2, , 2, 2s + 11s − 9, (b) L−1, s(s − 1)(s + 3), , , , d2 θ, dθ, +2 +θ = 4, 2, dt, dt, , Find the inverse Laplace transforms of the following functions:, (a), , 3., , (c) L−1, , given that when t = 0, x = 1 and y = 3., 6., , (20), , Determine the poles and zeros for the transfer func(s + 2)(s − 3), tion: F(s) =, and plot them on, (s + 3)(s 2 + 2s + 5), a pole-zero diagram., (10)
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Chapter 66, , Fourier series for periodic, functions of period 2π, f (x), , 66.1, , Introduction, 1, , Fourier series provides a method of analysing periodic, functions into their constituent components. Alternating currents and voltages, displacement, velocity and, acceleration of slider-crank mechanisms and acoustic, waves are typical practical examples in engineering and, science where periodic functions are involved and often, requiring analysis., , 66.2, , Periodic functions, , A function f (x) is said to be periodic if, f (x + T ) = f (x) for all values of x, where T is, some positive number. T is the interval between two, successive repetitions and is called the period of, the functions f (x). For example, y = sin x is periodic in x with period 2π since sin x = sin(x + 2π), = sin(x + 4π), and so on. In general, if y = sin ωt then, the period of the waveform is 2π/ω. The function, shown in Fig. 66.1 is also periodic of period 2π and is, defined by:, , −1, when −π < x < 0, f (x) =, 1, when, 0<x <π, If a graph of a function has no sudden jumps or breaks, it is called a continuous function, examples being the, graphs of sine and cosine functions. However, other, graphs make finite jumps at a point or points in the, interval. The square wave shown in Fig. 66.1 has finite, , 22, , 2, , , , 0, , 2, , x, , 21, , Figure 66.1, , discontinuities at x = π, 2π, 3π, and so on. A great, advantage of Fourier series over other series is that it, can be applied to functions which are discontinuous as, well as those which are continuous., , 66.3, , Fourier series, , (i) The basis of a Fourier series is that all functions of practical significance which are defined in, the interval −π ≤ x ≤ π can be expressed in, terms of a convergent trigonometric series of the, form:, f (x) = a0 + a1 cos x + a2 cos 2x, + a3 cos 3x + · · · + b1 sin x, + b2 sin 2x + b3 sin 3x + · · ·, when a0 , a1, a2, . . . b1, b2, . . . are real constants, i.e.
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612 Higher Engineering Mathematics, f (x) = a0 +, , ∞, ;, , (an cos nx + bn sinnx), , f (x), 8, , (1), , n=1, , where for the range −π to π:, ! π, 1, a0 =, f (x) dx, 2π −π, !, 1 π, f (x)cos nx dx, an =, π −π, (n = 1, 2, 3, . . .), , and, , bn =, , 1, π, , !, , 2 2/2, , /2, , 0, , , , 3/2 x, , 23, , Figure 66.2, , π, −π, , f (x)sin nx dx, , (n = 1, 2, 3, . . .), , Fig. 66.2, the sum of the Fourier series at the points of, π, discontinuity (i.e. at , π, . . . is given by:, 2, , (ii) a0 , an and bn are called the Fourier coefficients, of the series and if these can be determined, the, series of equation (1) is called the Fourier series, corresponding to f (x)., (iii) An alternative way of writing the series is by, using the a cos x + b sin x = c sin(x + α) relationship introduced in Chapter 17, i.e., , 1, 8 + (−3) 5, = or 2, 2, 2, 2, , 66.4, , Worked problems on Fourier, series of periodic functions of, period 2π, , f (x) = a0 + c1 sin(x + α1 ) + c2 sin(2x + α2 ), + · · · + cn sin(nx + αn ),, where a0 is a constant,, c1 = (a12 + b12 ), . . .cn = (an2 + bn2 ), are the amplitudes of the various components,, and phase angle, αn = tan−1, , an, bn, , (iv) For the series of equation (1): the term, (a1 cos x + b1 sin x) or c1 sin(x + α1) is called the, first harmonic or the fundamental, the term, (a2 cos 2x + b2 sin 2x) or c2 sin(2x + α2 ) is called, the second harmonic, and so on., For an exact representation of a complex wave, an infinite number of terms are, in general, required. In many, practical cases, however, it is sufficient to take the first, few terms only (see Problem 2)., The sum of a Fourier series at a point of discontinuity, is given by the arithmetic mean of the two limiting values, of f (x) as x approaches the point of discontinuity from, the two sides. For example, for the waveform shown in, , Problem 1. Obtain a Fourier series for the, periodic function f (x) defined as:, 5, −k, when −π < x < 0, f (x) =, +k, when, 0<x <π, The function is periodic outside of this range with, period 2π., The square wave function defined is shown in Fig. 66.3., Since f (x) is given by two different expressions in the, two halves of the range the integration is performed in, two parts, one from −π to 0 and the other from 0 to π., , f (x), k, , 22, , 2, , , , 0, 2k, , Figure 66.3, , 2, , x
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Fourier series for periodic functions of period 2π, 5, , 5, =, 2π, , , , −cos 2π 12 + n, , 1, 2 +n, , , cos 2π 12 − n, −, , 1, 2 −n, , 5, =, 2π, , −cos 0, cos 0, − 1, − 1, , 2 +n, 2 −n, , 6, , When n is both odd and even,, 5, 1, 1, 5, an =, + 1, , 1, 2π, 2 +n, 2 −n, − 1, , −1, +n, , 2, , 5, , − 1, 2, , 1, −n, 6, , 5, , 1, , , , , −n, sin 2π 12 + n, −, , , 1, 2 −n, 2 +n, 6, sin 0, sin 0, − 1, − 1, , 2 −n, 2 +n, , sin 2π, 1, , 2, , When n is both odd and even, bn = 0 since sin(−π),, sin 0, sin π, sin 3π, . . . are all zero. Hence the Fourier, series for the rectified sine wave,, θ, i = 5 sin is given by:, 2, f (θ) = a0 +, , 6, , ∞, <, (an cos nθ + bn sin nθ), n=1, , , i.e. i = f (θ) =, , 2, 2, 5, + 1, , 1, 2π, 2 +n, 2 −n, 5, 6, 1, 1, 5, =, + 1, , 1, π, 2 +n, 2 −n, , =, , 10 20, 20, −, cos θ −, cos 2θ, π, 3π, (3)(5)π, −, , 20, cos 3θ − · · ·, (5)(7)π, , , , 20 1 cos θ cos 2θ cos 3θ, −, −, −, − ···, i.e. i =, π 2, (3), (3)(5) (5)(7), , Hence, 5, a1 =, π, , 1, , 5, a2 =, π, , 1, , 5, a3 =, π, , 1, , 1, bn =, π, 5, =, π, , 3, 2, , 5, 2, , 7, 2, , +, , +, , +, , !, , − 12, , − 32, , − 52, , 2π, , , , Now try the following exercise, Exercises 228 Further problems on Fourier, series of periodic functions of period 2π, , 2 2, −20, =, −, 5 3, (3)(5)π, , 1. Determine the Fourier series for the periodic, function:, 5, −2, when −π < x < 0, f (x) =, +2, when, 0<x <π, , , 5 2 2, −20, =, −, =, π 7 5, (5)(7)π, and so on, , 1, , 0, , 0, , 5, =, π, , 1, , 2π, , !, , , −20, 5 2 2, =, =, −, π 3 1, 3π, , 1, , which is periodic outside this range of, period 2π., , ⎡, ⎤, 8, 1, ⎢ f (x) = π sin x + 3 sin 3x ⎥, ⎢, ⎥, , ⎣, ⎦, 1, + sin 5x + · · ·, 5, , θ, 5 sin sin nθ dθ, 2, , , 1, 1, − cos θ, +n, 2, 2, , , 1, − cos θ, −n, dθ, 2, from Chapter 40, , 5, =, 2π, , sin θ, 1, , 1, , −n, , 2 −n, 2, , , , sin θ, −, 1, , 1, , +n, , 2 +n, , , , 615, , 2π, , 2, , 0, , 2. For the Fourier series in Problem 1, deduce a, π, π, series for at the point where x =, 4, 2, , π, 1 1 1, = 1 − + − +···, 4, 3 5 7
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616 Higher Engineering Mathematics, 3. For the waveform shown in Fig. 66.6 determine (a) the Fourier series for the function and, (b) the sum of the Fourier series at the points, of discontinuity., , ⎤, ⎡, 1, 1 2, cos x − cos 3x, (a) f (x) = +, ⎥, ⎢, 2 π, 3 , ⎥, ⎢, ⎥, 1, ⎢, ⎥, ⎢, + cos 5x − · · ·, ⎥, ⎢, 5, ⎦, ⎣, 1, (b), 2, f (x), , 0, , ⎧, 0, when −π < t < 0, ⎪, ⎪, ⎪, ⎨, π, 1, when, 0<t <, f (t ) =, 2, ⎪, ⎪, π, ⎪−1, when, ⎩, <t <π, 2, The function has a period of 2π., ⎡, , ⎛, , ⎢, ⎜, ⎢, ⎜, ⎢, ⎜, ⎢, ⎜, 2, ⎢ f (t ) = ⎜, ⎢, π⎜, ⎢, ⎜, ⎢, ⎜, ⎣, ⎝, , 1, , 23 2 2, 2, 2, , 6. Determine the Fourier series for the periodic, function of period 2π defined by:, , , 2, , , , 3, 2, , x, , Figure 66.6, , 4. For Problem 3, draw graphs of the first three, partial sums of the Fourier series and show that, as the series is added together term by term the, result approximates more and more closely to, the function it represents., 5. Find the term representing the third harmonic, for the periodic function of period 2π given by:, , 0, when −π < x < 0, f (x) =, 1, when, 0<x <π, , 2, sin 3x, 3π, , cos t −, , 1, cos 3t, 3, , 1, + cos 5t − · · ·, 5, 1, + sin 2t + sin 6t, 3, 1, + sin 10t + · · ·, 5, , ⎞⎤, ⎟⎥, ⎟⎥, ⎟⎥, ⎟⎥, ⎟⎥, ⎟⎥, ⎟⎥, ⎟⎥, ⎠⎦, , 7. Show that the Fourier series for the periodic, function of period 2π defined by, 5, f (θ) =, , 0,, , when −π < θ < 0, , sin θ, when, , 0<θ <π, , is given by:, 2, f (θ) =, π, , , , 1 cos 2θ cos 4θ, −, −, 2, (3), (3)(5), , cos 6θ, −, −···, (5)(7)
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Chapter 67, , Fourier series for a, non-periodic function, over range 2π, 67.1 Expansion of non-periodic, functions, If a function f (x) is not periodic then it cannot be, expanded in a Fourier series for all values of x. However,, it is possible to determine a Fourier series to represent, the function over any range of width 2π ., Given a non-periodic function, a new function may, be constructed by taking the values of f (x) in the, given range and then repeating them outside of the, given range at intervals of 2π. Since this new function is, by construction, periodic with period 2π,, it may then be expanded in a Fourier series for, all values of x. For example, the function f (x) = x, is not a periodic function. However, if a Fourier, series for f (x) = x is required then the function, is constructed outside of this range so that it is, periodic with period 2π as shown by the broken lines in, Fig. 67.1., For non-periodic functions, such as f (x) = x, the sum, of the Fourier series is equal to f (x) at all points in the, given range but it is not equal to f (x) at points outside, of the range., For determining a Fourier series of a non-periodic function over a range 2π, exactly the same formulae for the, Fourier coefficients are used as in Section 66.3(i)., , f (x), 2, , f (x)5x, , 0, , 2, , 22, , 4, , x, , Figure 67.1, , 67.2, , Worked problems on Fourier, series of non-periodic functions, over a range of 2π, , Problem 1. Determine the Fourier series to, represent the function f (x) = 2x in the range, −π to +π., The function f (x) = 2x is not periodic. The function is, shown in the range −π to π in Fig. 67.2 and is then, constructed outside of that range so that it is periodic of, period 2π (see broken lines) with the resulting saw-tooth, waveform., For a Fourier series:, f (x) = a0 +, , ∞, <, n=1, , (an cos nx + bn sin nx)
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621, , Fourier series for a non-periodic function over range 2π, When θ = π, f (θ) = π 2, , 4π 2, 1, Hence π 2 =, + 4 cos π + cos 2π, 3, 4, 1, 1, cos 4π + · · ·, + cos 3π +, 9, 16, , 1, − 4π sin π + sin 2π, 2, , , , 1, + sin 3π + · · ·, 3, , i.e. π 2 −, , , cos 3t, 4, π, ⎢ f (t ) = 2 + 1 − π cos t + 32, ⎢, , ⎣, cos 5t, + 2 +···, 5, ⎡, , , , , 4π 2, 1 1, = 4 −1 + −, 3, 4 9, , −, , π2, 3, π2, 3, , , 1, +, − · · · − 4π(0), 16, , , 1 1, 1, = 4 −1 + − +, −···, 4 9 16, , , 1 1, 1, = 4 1− + −, +···, 4 9 16, , Hence, , π2, 1 1 1, = 1− + − +· · ·, 12, 4 9 16, , or, , 1, 1, π2, 1, = 1− 2 + 2 − 2 +· · ·, 12, 2, 3, 4, , Now try the following exercise, Exercise 229 Further problems on Fourier, series of non-periodic functions over a range, of 2π, 1. Show that the Fourier series for the function, f (x) = x over the range x = 0 to x = 2π is, given by:, , f (x) = π − 2 sin x + 12 sin 2x, , + 13 sin 3x + 14 sin 4x + · · ·, 2. Determine the Fourier series for the function, defined by:, 5, 1 − t, when −π < t < 0, f (t ) =, 1 + t, when 0 < t < π, Draw a graph of the function within and, outside of the given range., , ⎤, ⎥, ⎥, ⎦, , 3. Find the Fourier series for the function, f (x) = x ⎡, + π within the range, −π < x < π.⎤, 1, ⎢ f (x) = π + 2 sin x − 2 sin 2x ⎥, ⎢, ⎥, , ⎣, ⎦, 1, + sin 3x − · · ·, 3, 4. Determine the Fourier series up to and, including the third harmonic for the, function defined by:, 5, x,, when 0 < x < π, f (x) =, 2π − x, when π < x < 2π, Sketch a graph of the function within and, outside of the given range, assuming the period, is 2π., , ⎤, ⎡, 4, cos3x, π, cos x +, f (x) = −, ⎢, 2 π, 32 ⎥, ⎥, ⎢, , ⎦, ⎣, cos 5x, +, +···, 2, 5, 5. Expand the function f (θ) = θ 2 in a Fourier, series in the range −π < θ < π., Sketch the function within and outside of the, given range., ⎤, , ⎡, 1, π2, ⎢ f (θ) = 3 − 4 cos θ − 22 cos 2θ ⎥, ⎥, ⎢, ⎥, ⎢, , ⎦, ⎣, 1, + 2 cos 3θ − · · ·, 3, 6. For the Fourier series obtained in Problem 5,, ∞ 1, ;, let θ = π and deduce the series for, 2, n=1 n, , 1, 1, 1, π2, 1, 1 + 2 + 2 + 2 + 2 +··· =, 2, 3, 4, 5, 6, 7. Show that the Fourier series for the triangular, waveform shown in Fig. 67.5 is given by:, , 8, 1, 1, y = 2 sin θ − 2 sin 3θ + 2 sin 5θ, π, 3, 5, , 1, − 2 sin 7θ + · · ·, 7
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622 Higher Engineering Mathematics, in the range 0 to 2π., , 8. Sketch the waveform defined by:, ⎧, 2x, ⎪, ⎪, ⎨ 1 + , when −π < x < 0, π, f (x) =, ⎪, 2x, ⎪, ⎩ 1 − , when 0 < x < π, π, , y, 1, , 0, 21, Figure 67.5, , , , 2, , , , Determine, the Fourier, series in this range. ⎤, ⎡, 1, 8, ⎥, ⎢ f (x) = π 2 cos x + 32 cos 3x, ⎢, ⎥, ⎦, ⎣, 1, 1, + 2 cos 5x + 2 cos 7x + · · ·, 5, 7, 9. For the Fourier series of Problem 8, deduce a, π2, series for, 28, π, 1, 1, 1, 1, =1 + 2 + 2 + 2 + 2 + ···, 8, 3, 5, 7, 9
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Chapter 68, , Even and odd functions and, half-range Fourier series, 68.1, , Even and odd functions, , ∞, <, , Hence f (x) = a0 +, , an cos nx, , n=1, , Even functions, A function y = f (x) is said to be even if f (−x) = f (x), for all values of x. Graphs of even functions are always, symmetrical about the y-axis (i.e. is a mirror image)., Two examples of even functions are y= x 2 and y = cos x, as shown in Fig. 18.25, page 186., , Odd functions, A function y = f (x) is said to be odd if f (−x) =, − f (x) for all values of x. Graphs of odd functions are, always symmetrical about the origin. Two examples, of odd functions are y = x 3 and y = sin x as shown in, Fig. 18.26, page 187., Many functions are neither even nor odd, two such, examples being shown in Fig. 18.27, page 187., See also Problems 3 and 4, page 187., , 68.2 Fourier cosine and Fourier, sine series, , where, , and, , ! π, 1, f (x) dx, 2π −π, !, 1 π, f (x) dx, =, π 0, , a0 =, , (due to symmetry), !, 1 π, an =, f (x) cos nx dx, π −π, !, 2 π, f (x) cos nx dx, =, π 0, , (b) Fourier sine series, The Fourier series of an odd periodic function f (x), having period 2π contains sine terms only (i.e. contains, no constant term and no cosine terms)., Hence f (x) =, , ∞, <, , bn sin nx, , n=1, , (a) Fourier cosine series, The Fourier series of an even periodic function, f (x) having period 2π contains cosine terms only, (i.e. contains no sine terms) and may contain a constant, term., , where, , bn =, , 1, π, , =, , 2, π, , !, , π, , −π, , !, , f (x) sin nx dx, , π, , f (x) sin nx dx, 0
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624 Higher Engineering Mathematics, 2, an =, π, , Problem 1. Determine the Fourier series for the, periodic function defined by:, ⎧, π, ⎪, −2,, when −π < x < −, ⎪, ⎪, ⎪, 2, ⎪, ⎨, π, π, 2,, when − < x <, f (x) =, ⎪, 2, 2, ⎪, ⎪, π, ⎪, ⎪, ⎩ −2,, when, < x < π., 2, and has a period of 2π., , 2, =, π, 4, =, π, =, , 4, π, , =, , 4, π, , The square wave shown in Fig. 68.1 is an even function, since it is symmetrical about the f (x) axis., Hence from para. (a) the Fourier series is given by:, f (x) = a0 +, , ∞, <, , and, f (x), , 2, , /2, , 0, , , , From para. (a),, , =, , 1, π, , π, , f (x) dx, , 0, , !, , π/2, 0, , !, 2 dx +, , π, π/2, , , −2 dx, , , 1, π/2, =, [2x]0 + [−2x]ππ/2, π, =, , π/2, , !, 2 cos nx dx +, , 0, , 5, , π, π/2, , , −2 cos nx dx, , 6, −sin nx π, +, n, 0, π/2, , , sin(π/2)n, −0, n, , , −sin(π/2)n, + 0−, n, , , , , 2 sin(π/2)n, 8, nπ, =, sin, n, πn, 2, sin nx, n, , , , π/2, , 3/2 2, , x, , an =, , 8, for n = 1, 5, 9, . . ., πn, −8, for n =3, 7, 11, . . ., πn, , 8, −8, 8, , a5 =, , and so on., Hence a1 = , a3 =, π, 3π, 5π, Hence the Fourier series for the waveform of Fig. 68.1, is given by:, , 1, 1, 8, cos x − cos 3x + cos 5x, f (x) =, π, 3, 5, , 1, − cos 7x +· · ·, 7, , When x = 0, f (x) = 2 (from Fig. 68.1)., , Figure 68.1, , !, , !, , Problem 2. In the Fourier series of Problem 1 let, x = 0 and deduce a series for π/4., , ⫺2, , 1, a0 =, π, , f (x) cos nx dx, , 0, , When n is odd, an =, , (i.e. the series contains no sine terms.), , ⫺ ⫺/2, , π, , When n is even, an = 0, , an cos nx, , n=1, , ⫺3/2, , !, , 1, [(π) + [(−2π) − (−π)] = 0, π, , Thus, from the Fourier series,, , 8, 1, 1, 2=, cos 0 − cos 0 + cos 0, π, 3, 5, , 1, − cos 0 + · · ·, 7, 1 1 1, 2π, = 1 − + − +···, Hence, 8, 3 5 7, i.e., , 1 1 1, π, = 1− + − +···, 4, 3 5 7, , Problem 3. Obtain the Fourier series for the, square wave shown in Fig. 68.2.
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626 Higher Engineering Mathematics, Hence the Fourier series is:, , π2, 1, 1, f (θ) = θ 2 =, − 4 cos θ − 2 cos 2θ + 2 cos 3θ, 3, 2, 3, , , 1, 1, − 2 cos 4θ + 2 cos 5θ − · · ·, 4, 5, , Problem 5., , For the Fourier series of Problem 4,, ∞ 1, ;, π2, =, let θ = π and show that, 2, 6, n=1 n, , When θ = π, f (θ) = π 2 (see Fig. 68.3). Hence from the, Fourier series:, , π2, 1, 1, − 4 cos π − 2 cos 2π + 2 cos3π, π2 =, 3, 2, 3, , 1, 1, − 2 cos 4π + 2 cos 5π − · · ·, 4, 5, i.e., , , , π2, 1, 1, 1, 1, π −, = −4 −1 − 2 − 2 − 2 − 2 − · · ·, 3, 2, 3, 4, 5, , , 2π 2, 1, 1, 1, 1, = 4 1 + 2 + 2 + 2 + 2 +···, 3, 2, 3, 4, 5, 2, , i.e., , 2π 2, 1, 1, 1, 1, = 1 + 2 + 2 + 2 + 2 +···, (3)(4), 2, 3, 4, 5, π2, 1, 1, 1, 1, 1, = 2 + 2 + 2 + 2 + 2 +···, 6, 1, 2, 3, 4, 5, , i.e., Hence, , ∞, <, 1 π2, =, 6, n2, n=1, , Now try the following exercise, Exercise 230 Further problems on Fourier, cosine and Fourier sine series, 1. Determine the Fourier series for the function, defined by:, ⎧, π, ⎪, ⎪, ⎪ −1, −π < x < − 2, ⎪, ⎨, π, π, 1, − < x <, f (x) =, ⎪, 2, 2, ⎪, ⎪, ⎪, ⎩ −1, π < x < π, 2, , which is periodic outside of this range of, period 2π., , ⎤, ⎡, 1, 4, ⎥, ⎢ f (x) = π cos x − 3 cos 3x, ⎥, ⎢, ⎥, ⎢, 1, ⎥, ⎢, + cos 5x, ⎥, ⎢, 5, ⎥, ⎢, ⎢, ⎥, ⎦, ⎣, 1, − cos 7x + · · ·, 7, 2. Obtain the Fourier series of the function, defined by:, 5, t + π,, −π < t < 0, f (t ) =, t − π,, 0<t <π, which is periodic of period 2π. Sketch the, given function., ⎤, ⎡, f (t ) = −2(sin t + 12 sin 2t, ⎥, ⎢, ⎥, ⎢, + 13 sin 3t, ⎥, ⎢, ⎦, ⎣, 1, + 4 sin 4t + · · ·), 3. Determine the Fourier series defined by, 5, 1 − x, −π < x < 0, f (x) =, 1 + x, 0 < x < π, which is periodic of period 2π., ⎡, π, f (x) = + 1, 2 , ⎢, ⎢, 1, 4, ⎢, cos x + 2 cos 3x, −, ⎢, π, 3, ⎢, , ⎣, 1, + 2 cos 5x + · · ·, 5, , ⎤, ⎥, ⎥, ⎥, ⎥, ⎥, ⎦, , 4. In the Fourier series of Problem 3, let x = 0, and deduce a series for π 2 /8., 2, π, 1, 1, 1, =1 + 2 + 2 + 2 + ···, 8, 3, 5, 7, , 68.3, , Half-range Fourier series, , (a) When a function is defined over the range say 0, to π instead of from 0 to 2π it may be expanded, in a series of sine terms only or of cosine terms, only. The series produced is called a half-range, Fourier series.
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Even and odd functions and half-range Fourier series, f (x), B, , calculated as in Section 68.2(b), i.e., f(x)5x, , , , f (x) =, , A, 2, , 22, , , , 2 x, , Figure 68.4, , (b) If a half-range cosine series is required for the, function f (x) = x in the range 0 to π then an, even periodic function is required. In Figure 68.4,, f (x) = x is shown plotted from x = 0 to x = π., Since an even function is symmetrical about the, f (x) axis the line AB is constructed as shown., If the triangular waveform produced is assumed, to be periodic of period 2π outside of this range, then the waveform is as shown in Fig. 68.4. When, a half-range cosine series is required then the, Fourier coefficients a0 and an are calculated as, in Section 68.2(a), i.e., ∞, <, , an cos nx, , n=1, , where, , an =, , and, (c), , 1, a0 =, π, 2, π, , !, , π, , f (x) dx, 0, , !, , π, , f (x) cos nx dx, 0, , If a half-range sine series is required for the function f (x) = x in the range 0 to π then an odd periodic function is required. In Figure 68.5, f (x) = x, is shown plotted from x = 0 to x = π. Since an, odd function is symmetrical about the origin the, line CD is constructed as shown. If the sawtooth, waveform produced is assumed to be periodic of, period 2π outside of this range, then the waveform, is as shown in Fig. 68.5. When a half-range sine, series is required then the Fourier coefficient bn is, f (x), , f (x)5x, , , C, 22, , 2, D, , Figure 68.5, , 0, 2, , ∞, <, , bn sin nx, , n=1, , 0, , f (x) = a0 +, , 627, , , , 2, , 3 x, , 2, where bn =, π, , !, , π, , f (x) sin nx dx, 0, , Problem 6. Determine the half-range Fourier, cosine series to represent the function f (x) = 3x in, the range 0 ≤ x ≤ π., From para. (b), for a half-range cosine series:, f (x) = a0 +, , ∞, <, , an cos nx, , n=1, , When f (x) = 3x,, !, !, 1 π, 1 π, a0 =, f (x)dx =, 3x dx, π 0, π 0, , π, 3π, 3 x2, =, =, π 2 0, 2, ! π, 2, an =, f (x) cos nx dx, π 0, !, 2 π, =, 3x cos nx dx, π 0, , 6 x sin nx cos nx π, =, by parts, +, π, n, n2 0, , , , π sin nπ cos nπ, cos 0, 6, +, − 0+ 2, =, π, n, n2, n, , , cos nπ cos0, 6, 0+, − 2, =, π, n2, n, =, , 6, (cos nπ − 1), πn 2, , When n is even, an = 0, 6, −12, When n is odd, an = 2 (−1 −1) =, πn, πn 2, −12, −12, −12, Hence a1 =, , a5 =, , and so on., , a3 =, π, π32, π52, Hence the half-range Fourier cosine series is given by:, , 3π 12, 1, f (x) = 3x =, −, cos x + 2 cos 3x, 2, π, 3, , 1, + 2 cos 5x + · · ·, 5
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628 Higher Engineering Mathematics, f (x ), , Problem 7. Find the half-range Fourier sine, series to represent the function f (x) = 3x in the, range 0 ≤ x ≤ π., 2, , From para. (c), for a half-range sine series:, f (x) =, , ∞, <, , , , 2 x, , Figure 68.6, , bn =, =, , 2, π, 2, π, , 2, =, π, , !, !, , π, , f (x) sin nx dx, , 0, π, , cos x sin nx dx, !, , 0, π, 0, , 1, [sin(x + nx) − sin(x − nx)] dx, 2, , , , 1 −cos[x(1 + n)] cos[x(1 − n)] π, +, π, (1 + n), (1 − n), 0, , , −cos[π(1 + n)] cos[π(1 − n)], 1, =, +, π, (1 + n), (1 − n), , , −cos 0, cos 0, +, −, (1 + n) (1 − n), , 6, = − cos nπ, n, , =, , 6, n, 6, 6, 6, Hence b1 = , b3 = , b5 = and so on., 1, 3, 5, 6, When n is even, bn = −, n, 6, 6, 6, Hence b2 = − , b4 = − , b6 = − and so on., 2, 4, 6, Hence the half-range Fourier sine series is given by:, , 1, 1, f (x) = 3x = 6 sin x − sin 2x + sin 3x, 2, 3, When n is odd, bn =, , 1, 1, − sin 4x + sin 5x −· · ·, 4, 5, , , , Problem 8. Expand f (x) = cos x as a half-range, Fourier sine series in the range 0 ≤ x ≤ π, and sketch, the function within and outside of the given range., When a half-range sine series is required then an, odd function is implied, i.e. a function symmetrical, about the origin. A graph of y = cos x is shown in, Fig. 68.6 in the range 0 to π. For cos x to be symmetrical, about the origin the function is as shown by the broken, lines in Fig. 68.6 outside of the given range., From para. (c), for a half-range Fourier sine series:, , n=1, , 0, 21, , When f (x) = 3x,, !, !, 2 π, 2 π, bn =, f (x) sin nx dx =, 3x sin nx dx, π 0, π 0, , 6 −x cos nx sin nx π, by parts, =, +, π, n, n2 0, , , −π cos nπ sin nπ, 6, =, +, − (0 + 0), π, n, n2, , f (x) =, , y 5 cos x, , bn sin nx, , n=1, , ∞, <, , 1, , bn sin nx dx, , When n is odd,, , , −1, 1, 1, +, bn =, π, (1 + n) (1 − n), , , −1, 1, −, +, =0, (1 + n) (1 − n), When n is even,, , , 1, 1, 1, −, bn =, π, (1 + n) (1 − n), , , −1, 1, −, +, (1 + n) (1 − n), , , 2, 1, 2, =, −, π (1 + n) (1 − n), , , 1 2(1 − n) − 2(1 + n), =, π, 1 − n2, , , 1 −4n, 4n, =, =, 2, π 1−n, π(n 2 − 1), Hence b2 =, , 8, 16, 24, , b4 =, , b6 =, and so on., 3π, 15π, 35π
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629, , Even and odd functions and half-range Fourier series, Hence the half-range Fourier sine series for f (x) in the, range 0 to π is given by:, f (x) =, , 8, or f (x)=, π, , 8, 16, sin 2x +, sin 4x, 3π, 15π, 24, sin 6x + · · ·, +, 35π, , , 1, 2, sin 2x +, sin 4x, 3, (3)(5), +, , , 3, sin 6x +· · ·, (5)(7), , Now try the following exercise, Exercise 231 Further problems on, half-range Fourier series, 1. Determine the half-range sine series for the, function defined by:, ⎧, π, ⎨ x, 0 < x <, 2, f (x) =, ⎩ 0, π < x < π, 2, ⎤, ⎡, π, 2, sin x + sin 2x, f (x) =, ⎥, ⎢, π, 4, ⎥, ⎢, ⎥, ⎢, 1, ⎥, ⎢, −, sin, 3x, ⎥, ⎢, 9, ⎦, ⎣, , π, − sin 4x + · · ·, 8, 2. Obtain (a) the half-range cosine series and, (b) the half-range sine series for the function, ⎧, π, ⎪, ⎨ 0, 0 < t <, 2, f (t ) =, π, ⎪, ⎩ 1,, <t <π, 2, , ⎡, 1 2, −, cos t, (a), f, (t, ), =, ⎢, 2 π, ⎢, ⎢, 1, ⎢, − cos 3t, ⎢, 3, ⎢, ⎢, , ⎣, 1, + cos 5t − · · ·, 5, , ⎤, ⎥, ⎥, ⎥, ⎥, ⎥, ⎥, ⎥, ⎦, , , 2, ⎢ (b) f (t ) = π sin t − sin 2t, ⎢, ⎢, 1, 1, ⎢, + sin 3t + sin 5t, ⎢, ⎢, 3, 5, ⎢, , ⎣, 1, − sin 6t + · · ·, 3, ⎡, , ⎤, ⎥, ⎥, ⎥, ⎥, ⎥, ⎥, ⎥, ⎦, , 3. Find (a) the half-range Fourier sine series and, (b) the half-range Fourier cosine series for the, function f (x) = sin2 x in the range 0 ≤ x ≤ π., Sketch the function within and outside of the, given range., , ⎤, ⎡, sin 3x, 8 sin x, −, (a), f, (x), =, ⎢, π (1)(3) (1)(3)(5) ⎥, ⎥, ⎢, ⎥, ⎢, sin 5x, ⎥, ⎢, −, ⎥, ⎢, (3)(5)(7), ⎥, ⎢, ⎥, ⎢, , ⎥, ⎢, sin, 7x, ⎥, ⎢, −, −···, ⎥, ⎢, (5)(7)(9), ⎥, ⎢, ⎦, ⎣, 1, (b) f (x) = (1 − cos 2x), 2, 4. Determine the half-range Fourier cosine series, in the range x = 0 to x = π for the function, defined by:, ⎧, π, ⎪, 0<x <, ⎪, ⎨ x,, 2, f (x) =, π, ⎪, ⎪, ⎩ (π − x), 2 < x < π, ⎡, ⎢, ⎢, ⎢, ⎢, ⎢, ⎢, ⎣, , f (x) =, , , 2, π, −, cos 2x, 4 π, cos 6x, +, 32, , cos 10x, +, +···, 52, , ⎤, ⎥, ⎥, ⎥, ⎥, ⎥, ⎥, ⎦
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Chapter 69, , Fourier series over any range, 69.1 Expansion of a periodic function, of period L, (a) A periodic function f (x) of period L, repeats itself when x increases by L, i.e., f (x + L) = f (x). The change from functions, dealt with previously having period 2π to functions having period L is not difficult since it may, be achieved by a change of variable., (b) To find a Fourier series for a function f (x) in, L, L, the range − ≤ x ≤ a new variable u is intro2, 2, duced such that f (x), as a function of u, has, 2π x, L, period 2π. If u =, then, when x = − ,, L, 2, L, u = −π and when x = , u =+π. Also, let, 2, , Lu, f (x) = f, = F(u). The Fourier series for, 2π, F(u) is given by:, F(u) = a0 +, 1, where a0 =, 2π, , ∞, <, , (an cos nu + bn sin nu),, , L, L, and the limits of integration are − to +, 2, 2, instead of from −π to +π. Hence the Fourier, series expressed in terms of x is given by:, , , , ∞, ;, 2πnx, f (x)= a0 +, an cos, L, n=1, , , 2πnx, + bn sin, L, where, in the range −, , a0 =, , 1, L, , 2, an =, L, and, , bn =, , 2, L, , !, , L, L, to + :, 2, 2, , L, 2, −L, 2, , f (x) dx,, , !, , , , 2πnx, dx, f (x) cos, −L, L, 2, , !, , , , 2π nx, f (x) sin, dx, −L, L, 2, , L, 2, , L, 2, , n=1, π, , !, !, , −π, , F(u) du,, , 1 π, F(u) cos nu du, π −π, !, 1 π, bn =, F(u) sin nu du, π −π, , The limits of integration may be replaced by any interval, of length L, such as from 0 to L., , an =, , and, , (c) It is however more usual to change the formula of, 2π x, , then, para. (b) to terms of x. Since u =, L, du =, , 2π, dx,, L, , Problem 1. The voltage from a square wave, generator is of the form:, 5, v(t ) =, , 0, −4 < t < 0, 10, 0 < t < 4, and has a period of 8 ms., , Find the Fourier series for this periodic function.
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Fourier series over any range, , 10, 24, , , ⎤4, πnt, −10 cos, 1⎢, 4 ⎥, ⎥, πn , = ⎢, ⎣, ⎦, 4, 4, 0, ⎡, , v (t), , 28, , 631, , 4, , 0, , 8, , 12 t (ms), , =, , Period L 5 8 ms, , −10, [cos πn − cos0], πn, , Figure 69.1, , When n is even, bn = 0, The square wave is shown in Fig. 69.1. From para. (c),, the Fourier series is of the form:, , v(t ) = a0 +, , ∞ , <, n=1, , a0 =, , 1, L, , =, , 1, 8, , an =, =, =, , =, , !, , L, 2, −L, 2, , 0, , −10, 20, (−1 − 1) = ,, π, π, , b3 =, , −10, 20, (−1 − 1) =, ,, 3π, 3π, , b5 =, , 20, , and so on., 5π, , , , , , 2πnt, 2πnt, + bn sin, an cos, L, L, , v(t ) dt =, , 1, 8, , !, , 4, −4, , v(t ) dt, , Thus the Fourier series for the function v(t ) is given by:, , , 1, 10 dt = [10t ]40 = 5, 8, −4, 0, , , L, !, 2πnt, 2 2, dt, v(t ) cos, −L, L 2, L, , , !, 2πnt, 2 4, dt, v(t ) cos, 8 −4, 8, , ! 0, , πnt, 1, 0 cos, dt, 4 −4, 4, , , ! 4, πnt, dt, 10 cos, +, 4, 0, , ⎤4, ⎡, πnt, 10 sin, 1⎢, 4 ⎥, ⎢, ⎥ = 10 [sin πn − sin 0], πn , ⎣, ⎦, 4, πn, 4, 0, !, , When n is odd, b1 =, , !, , 0 dt +, , 4, , = 0 for n =1, 2, 3, . . ., , , ! L, 2πnt, 2 2, v(t ) sin, dt, bn =, L −L, L, 2, , , !, 2πnt, 2 4, v(t ) sin, dt, =, 8 −4, 8, , ! 0, , πnt, 1, dt, 0 sin, =, 4 −4, 4, , , , , πt, 1, 3πt, 20, sin, + sin, v(t) = 5 +, π, 4, 3, 4, , , 5πt, 1, + sin, + ···, 5, 4, Problem 2. Obtain the Fourier series for the, function defined by:, ⎧, ⎪, ⎨0, when −2 < x < −1, f (x) = 5, when −1 < x < 1, ⎪, ⎩0, when 1 < x < 2, The function is periodic outside of this range of, period 4., The function f (x) is shown in Fig. 69.2 where period,, L = 4. Since the function is symmetrical about the f (x), axis it is an even function and the Fourier series contains, no sine terms (i.e. bn = 0)., , f (x), 5, , 25 24 23 22 21 0, , !, , 4, , +, 0, , , , , πnt, 10 sin, dt, 4, , L54, , Figure 69.2, , 1, , 2, , 3, , 4, , 5, , x
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632 Higher Engineering Mathematics, Hence the Fourier series for the function f (x) is, given by:, , Thus, from para. (c),, f (x) = a0 +, , ∞, <, n=1, , a0 =, , 1, L, , =, , 1, 4, , !, , L, 2, , f (x) dx =, , −L, 2, , !, , , , 2πnx, an cos, L, , −1, −2, , !, 0 dx +, , 1, 4, , !, , 2, , −2, , f (x) =, f (x) dx, !, , 1, −1, , , , 2, , 5 dx +, , 0 dx, 1, , 1, 1, 10 5, = [5x]1−1 = [(5) − (−5)] =, =, 4, 4, 4, 2, 2, an =, L, 2, =, 4, =, , 1, 2, , !, , , , , 2πnx, f (x) cos, dx, L, , L, 2, −L, 2, , !, , 2, −2, , !, , The function f (t ) =t in the interval 0 to 3 is shown in, Fig. 69.3. Although the function is not periodic it may, be constructed outside of this range so that it is periodic, of period 3, as shown by the broken lines in Fig. 69.3., , , 2πnx, dx, f (x) cos, 4, −1, , −2, , f (t ), f (t ) 5t, , πnx , dx, 0 cos, 2, !, , 23, , πnx , 5 cos, dx, 2, −1, , 0, , 2, , +, 1, , πnx , 0 cos, dx, 2, , ⎡, , πnx ⎤1, sin, 5⎢, ⎥, = ⎣ πn2 ⎦, 2, 2, −1, , , , πn, −πn, 5, =, sin, − sin, πn, 2, 2, When n is even, an = 0, When n is odd,, 5, 10, (1 − (−1)) =, π, π, 5, −10, (−1 − 1) =, 3π, 3π, , 5, 10, a5 =, (1 − (−1)) =, and so on., 5π, 5π, , 3, , 6, , t, , Period L 5 3, , 1, , !, , a3 =, , Problem 3. Determine the Fourier series for the, function f (t ) = t in the range t = 0 to t = 3., , , , +, , a1 =, , , , , πx, 5, 10, 1, 3πx, cos, +, − cos, 2, π, 2, 3, 2, , , , , 5πx, 7πx, 1, 1, + cos, − cos, + ···, 5, 2, 7, 2, , Figure 69.3, , From para.(c), the Fourier series is given by:, f (t ) = a0 +, , , , , , ∞ , <, 2πnt, 2πnt, + bn sin, an cos, L, L, n=1, , 1, a0 =, L, 1, =, 3, an =, , 2, L, , 2, =, L, 2, =, 3, , !, , L, 2, −L, 2, , !, , 3, , 0, , !, , , 1 t2, t dt =, 3 2, , L, 2, −L, 2, , !, , L, 0, , !, , 3, 0, , 1, f (t ) dx =, L, , !, , L, , f (t ) dx, , 0, 3, , =, 0, , 3, 2, , , , 2πnt, dt, f (t ) cos, L, , , 2πnt, t cos, dt, L, , , , 2πnt, t cos, dt, 3
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Fourier series over any range, ⎡, , , , 2πnt, t sin, 2⎢, 3, ⎢, , = ⎢ , 2πn, 3⎣, 3, , , , , , 2πnt, cos, 3, + , , 2πn 2, 3, , ⎤3, ⎥, ⎥, ⎥, ⎦, 0, , by parts, ⎫, ⎡⎧, ⎪, ⎪, ⎪, ⎪, ⎪, ⎪, ⎨ 3 sin 2πn, ⎬, cos, 2πn, 2⎢, ⎢, +, = ⎢ , 2, 2πn, 3 ⎣⎪, 2πn ⎪, ⎪, ⎪, ⎪, ⎪, ⎩, ⎭, 3, 3, ⎫⎤, ⎧, ⎪, ⎪, ⎪, ⎪, ⎪, ⎪, ⎨, cos 0 ⎬⎥, ⎥, − 0+ , 2 ⎪⎥ = 0, ⎪, ⎦, 2πn, ⎪, ⎪, ⎪, ⎪, ⎭, ⎩, 3, bn =, , 2, L, , !, , L, 2, −L, 2, , , f (t ) sin, , , 2πnt, dt, L, , , , , 2πnt, 2 L, =, t sin, dt, L 0, L, , , !, 2πnt, 2 3, dt, =, t sin, 3 0, 3, ⎡, , , , ⎤3, 2πnt, 2πnt, −t cos, sin, ⎥, 2⎢, 3, 3, ⎢, ⎥, , , = ⎢, + , ⎥, 2πn, 3⎣, 2πn 2 ⎦, 3, 3, !, , 0, , by parts, ⎫, ⎡⎧, ⎪, ⎪, ⎪, ⎪, ⎪, ⎪, ⎨ −3 cos 2πn, ⎬, sin, 2πn, 2⎢, ⎢, +, , = ⎢, , 2πn, 3 ⎣⎪, 2πn 2 ⎪, ⎪, ⎪, ⎪, ⎪, ⎩, ⎭, 3, 3, ⎫⎤, ⎧, ⎪, ⎪, ⎪, ⎪, ⎪, ⎪, ⎨, sin 0 ⎬⎥, ⎥, − 0+ , 2 ⎪⎥, ⎪, ⎦, 2πn, ⎪, ⎪, ⎪, ⎪, ⎭, ⎩, 3, ⎡, ⎤, =, , cos 2πn ⎥, −3, −3, 2⎢, ⎢ −3, ⎥, , ⎦ = πn cos 2πn = πn, 2πn, 3⎣, 3, , Hence b1 =, , −3, −3, −3, , b2 =, , b3 =, and so on., π, 2π, 3π, , 633, , Thus the Fourier series for the function f (t ) in the range, 0 to 3 is given by:, , , , , 2πt, 4πt, 3, 1, 3, sin, + sin, f (t) = −, 2, π, 3, 2, 3, , , 6πt, 1, + ···, + sin, 3, 3, , Now try the following exercise, Exercise 232 Further problems on Fourier, series over any range L, 1. The voltage from a square wave generator is, of the form:, , 0, −10 < t < 0, v(t ) =, 5,, 0 < t < 10, and is periodic of period 20. Show that the, Fourier series for the function is given by:, , , , πt, 3πt, 5 10, 1, v(t ) = +, sin, + sin, 2, π, 10, 3, 10, , , 5πt, 1, +···, + sin, 5, 10, 2. Find the Fourier series for f (x) = x, range x = 0 to x = 5., , , ⎡, 2π x, 5 5, ⎢ f (x) = 2 − π sin 5, ⎢, , , ⎢, 1, 4π x, ⎢, + sin, ⎢, ⎢, 2, 5, ⎢, , , ⎣, 6π x, 1, + sin, +···, 3, 5, , in the, ⎤, ⎥, ⎥, ⎥, ⎥, ⎥, ⎥, ⎥, ⎦, , 3. A periodic function of period 4 is defined by:, 5, −3, −2 < x < 0, f (x) =, +3,, 0<x <2, Sketch the function and obtain the Fourier, series for the function., , ⎤, ⎡, πx , 12, sin, f (x) =, ⎥, ⎢, π, 2, ⎥, ⎢, , , ⎥, ⎢, 1, 3π x, ⎥, ⎢, + sin, ⎥, ⎢, 3, 2, ⎥, ⎢, ⎢, , ⎥, , ⎦, ⎣, 1, 5π x, +···, + sin, 5, 2
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634 Higher Engineering Mathematics, , 4. Determine the Fourier series for the half, wave rectified sinusoidal voltage V sin ωt, defined by:, ⎧, π, ⎪, ⎨V sin ωt, 0 < t < ω, f (t ) =, π, 2π, ⎪, ⎩, 0,, <t <, ω, ω, which is periodic of period, ⎡, ⎢, ⎢, ⎢, ⎢, ⎢, ⎢, ⎣, , 2π, ω, , V, V, + sin ωt, π 2, 2V cos 2ωt, −, π, (1)(3), , cos 4ωt cos 6ωt, +, +, +···, (3)(5), (5)(7), , f (t ) =, , Problem 4. Determine the half-range Fourier, cosine series for the function f (x) = x in the range, 0 ≤ x ≤ 2. Sketch the function within and outside of, the given range., A half-range Fourier cosine series indicates an even, function. Thus the graph of f (x) = x in the range 0 to, 2 is shown in Fig. 69.4 and is extended outside of this, range so as to be symmetrical about the f (x) axis as, shown by the broken lines., , ⎤, ⎥, ⎥, ⎥, ⎥, ⎥, ⎥, ⎦, , f (x ), f (x ) 5 x, 2, 24, , 22, , 0, , 2, , 4, , 6, , x, , Figure 69.4, , 69.2 Half-range Fourier series for, functions defined over range L, πx, (a) By making the substitution u =, (see, L, Section 69.1), the range x = 0 to x = L corresponds to the range u = 0 to u = π. Hence a, function may be expanded in a series of either, cosine terms or sine terms only, i.e. a half-range, Fourier series., (b) A half-range cosine series in the range 0 to L can, be expanded as:, f (x) = a0 +, , ∞, <, , an cos, , nπ x , L, , n=1, , where, , !, , 1, a0 =, L, , L, , f (x) dx and, 0, , !, , 2, an =, L, , L, , f (x) cos, , nπ x , L, , 0, , dx, , (c) A half-range sine series in the range 0 to L can, be expanded as:, f (x)=, , ∞, <, n=1, , 2, where bn =, L, , nπx , bn sin, L, , !, , L, , f (x) sin, 0, , nπ x , L, , dx, , From para. (b), for a half-range cosine series:, , f (x) = a0 +, 1, a0 =, L, , ∞, <, , an cos, , n=1, , !, , L, 0, , nπ x , L, , 1, f (x) dx =, 2, , !, , 2, , x dx, 0, , , 2, 1 x2, =1, =, 2 2 0, !, nπ x , 2 L, an =, f (x) cos, dx, L 0, L, !, nπ x , 2 2, x cos, dx, =, 2 0, 2, ⎡, nπ x ⎤2, nπ x , cos, x sin, ⎢, ⎥, + 22 ⎦, = ⎣ nπ 2, nπ, ⎡⎛, , 2, , 2, ⎞, , ⎛, , 0, , ⎞⎤, , cos nπ ⎟ ⎜, cos 0 ⎟⎥, ⎢⎜ 2 sin nπ, = ⎣⎝ nπ + 2 ⎠ − ⎝0 + 2 ⎠⎦, nπ, nπ, 2, 2, 2, ⎤, ⎡, 1 ⎥, ⎢ cos nπ, = ⎣ 2 − 2 ⎦, nπ, nπ, 2, 2, 2, 2, =, (cos nπ − 1), πn
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Fourier series over any range, When n is even, an = 0, −8, −8, −8, a1 = 2 , a3 = 2 2 , a5 = 2 2 and so on., π, π 3, π 5, Hence the half-range Fourier cosine series for f (x) in, the range 0 to 2 is given by:, , , , 3πx, πx, 8, 1, f (x) = 1 − 2 cos, + 2 cos, π, 2, 3, 2, , , 5πx, 1, + ···, + 2 cos, 5, 2, Problem 5. Find the half-range Fourier sine, series for the function f (x) = x in the range, 0 ≤ x ≤ 2. Sketch the function within and outside of, the given range., A half-range Fourier sine series indicates an odd function. Thus the graph of f (x) = x in the range 0 to 2 is, shown in Fig. 69.5 and is extended outside of this range, so as to be symmetrical about the origin, as shown by, the broken lines., f (x ), f (x ) 5 x, , 22, , 0, , 2, , 4, , 6 x, , 22, , bn sin, , nπ x , , n=1, , −4, −4, (1) =, 2π, 2π, , b3 =, , −4, 4, (−1) =, and so on., 3π, 3π, , Thus the half-range Fourier sine series in the range 0 to, 2 is given by:, , , , πx, 2πx, 4, 1, f (x)=, sin, − sin, π, 2, 2, 2, , , , , 1, 1, 3πx, 4πx, + sin, − sin, +···, 3, 2, 4, 2, , Now try the following exercise, Exercise 233 Further problems on, half-range Fourier series over range L, , ⎢, ⎢, ⎢, ⎢, ⎢, ⎢, ⎢, ⎢, ⎢, ⎣, , From para. (c), for a half-range sine series:, ∞, <, , b2 =, , ⎡, , Figure 69.5, , f (x) =, , −4, 4, (−1) =, π, π, , 1. Determine the half-range Fourier cosine series, for the function f (x) = x in the range, 0 ≤ x ≤ 3. Sketch the function within and outside of the given range., , 2, 24, , Hence b1 =, , L, , !, , nπ x , 2 L, f (x) sin, dx, L 0, L, !, nπ x , 2 2, x sin, =, dx, 2 0, L, ⎡, nπ x ⎤2, nπ x , sin, −x cos, ⎥, ⎢, nπ 2, + 22 ⎦, =⎣, nπ, , , πx, 3 12, f (x) = − 2 cos, 2 π, 3, , , 3π x, 1, + 2 cos, 3, 3, , , , 5π x, 1, +···, + 2 cos, 5, 3, , ⎤, ⎥, ⎥, ⎥, ⎥, ⎥, ⎥, ⎥, ⎥, ⎥, ⎦, , bn =, , ⎡⎛, , 2, , 2⎞, , ⎛, , 0, , ⎞⎤, , sin nπ ⎟ ⎜, sin 0 ⎟⎥, ⎢⎜ −2 cos nπ, = ⎣⎝ nπ + 2 ⎠ − ⎝0 + 2 ⎠⎦, nπ, nπ, 2, 2, 2, −4, −2 cos nπ, =, cos nπ, =, nπ, nπ, 2, , 635, , 2. Find the half-range Fourier sine series, for the function f (x) = x in the range, 0 ≤ x ≤ 3. Sketch the function within and outside of the given range., , , , ⎡, πx 1, 2π x ⎤, 6, sin, − sin, f (x) =, ⎢, ⎥, π, 3, 2, 3, ⎢, ⎥, ⎢, ⎥, , , ⎢, ⎥, 3π, x, 1, ⎢, ⎥, + sin, ⎢, ⎥, 3, 3, ⎢, ⎥, ⎢, ⎥, , , , ⎣, ⎦, 4π x, 1, − sin, +···, 4, 3
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636 Higher Engineering Mathematics, 3. Determine the half-range Fourier sine series, for the function defined by:, , t, 0 < t < 1, f (t ) =, (2 − t ), 1 < t < 2, ⎤, , πt, 8, sin, f, (t, ), =, ⎥, ⎢, π2, 2, ⎥, ⎢, ⎥, ⎢, , , ⎥, ⎢, 3πt, 1, ⎥, ⎢, − 2 sin, ⎥, ⎢, 3, 2, ⎥, ⎢, , , ⎥, ⎢, ⎦, ⎣, 5πt, 1, −···, + 2 sin, 5, 2, ⎡, , 4. Show that the half-range Fourier cosine series, for the function f (θ) = θ 2 in the range 0 to 4, is given by:, , πθ, 16 64, f (θ) =, − 2 cos, 3, π, 4, , , 2πθ, 1, − 2 cos, 2, 4, , , , 3πθ, 1, + 2 cos, −···, 3, 4, Sketch the function within and outside of the, given range.
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Chapter 70, , A numerical method of, harmonic analysis, 70.1, , bn =, , Introduction, , Many practical waveforms can be represented by simple mathematical expressions, and, by using Fourier, series, the magnitude of their harmonic components, determined, as shown in Chapters 66 to 69. For waveforms not in this category, analysis may be achieved by, numerical methods. Harmonic analysis is the process, of resolving a periodic, non-sinusoidal quantity into a, series of sinusoidal components of ascending order of, frequency., , 70.2 Harmonic analysis on data given, in tabular or graphical form, The Fourier coefficients a0 , an and bn used in Chapters 66 to 69 all require functions to be integrated, i.e., 1, a0 =, 2π, , !, , π, −π, , 1, f (x)dx =, 2π, , !, , 2π, , 1, π, , 1, =, π, , !, , π, , −π, , !, , 2π, , f (x) sin nx dx, f (x) sin nx dx, , 0, , = twice the mean value of f (x) sin nx, in the range 0 to 2π, However, irregular waveforms are not usually defined, by mathematical expressions and thus the Fourier coefficients cannot be determined by using calculus. In these, cases, approximate methods, such as the trapezoidal, rule, can be used to evaluate the Fourier coefficients., Most practical waveforms to be analysed are periodic., Let the period of a waveform be 2π and be divided into, p equal parts as shown in Fig. 70.1. The width of each, f(x), y0 y1 y2 y3 y4, , f (x) dx, , 0, , = mean value of f (x), in the range−π to π or 0 to 2π, ! π, 1, f (x) cos nx dx, an =, π −π, !, 1 2π, f (x) cos nx dx, =, π 0, = twice the mean value of f (x) cos nx, in the range 0 to 2π, , yp, 0, , 2/p, , , , Period 5 2, , Figure 70.1, , 2, , x
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638 Higher Engineering Mathematics, 2π, . Let the ordinates be labelled y0 ,, p, y1 , y2 , . . . y p (note that y0 = y p ). The trapezoidal rule, states:, , 1, Area = (width of interval) (first + last ordinate), 2, interval is thus, , Draw the graph of voltage V against angle θ and, analyse the voltage into its first three constituent, harmonics, each coefficient correct to 2 decimal, places., , + sum of remaining ordinates, , , 2π 1, (y0 + y p ) + y1 + y2 + y3 + · · ·, p 2, , Voltage (volts), , ≈, , 1, Since y0 = y p , then (y0 + y p ) = y0 = y p, 2, p, <, 2π, Hence area ≈, yk, p, , ≈, , area, length of base, p, 1 2π <, 2π, , p, , k=1, , 1<, yk ≈, yk, p, p, , k=1, , 1<, yk, p, p, , (1), , k=1, , Similarly, an = twice the mean value of f (x) cos nx in, the range 0 to 2π,, 2<, yk cos nxk, p, p, , thus an ≈, , (2), , k=1, , and bn = twice the mean value of f (x) sin nx in the, range 0 to 2π,, , 90, , 180, , y3 y4 y5 y6, , y11 y12, , y7, , 270, , 360 degrees, , The graph of voltage V against angle θ is shown in, Fig. 70.2. The range 0 to 2π is divided into 12 equal, 2π, π, intervals giving an interval width of, , i.e., rad, 12, 6, ◦, or 30 . The values of the ordinates y1 , y2 , y3 , . . . are, 62, 35, −38, . . . from the given table of values. If, a larger number of intervals are used, results having, a greater accuracy are achieved. The data is tabulated in the proforma shown in Table 70.1, on, page 639., From equation (1), a0 ≈, , p, 1 ;, 1, yk = (212), p k=1, 12, , = 17.67 (since p = 12), , 2<, thus bn ≈, yk sinnxk, p, p, , (3), From equation (2), an ≈, , k=1, , Problem 1. The values of the voltage v volts at, different moments in a cycle are given by:, θ ◦ (degrees) V (volts), , y9, y8, , Figure 70.2, , However, a0 = mean value of f (x) in the range 0 to 2π., Thus a0 ≈, , 80, 60, 40 y, 1, y2, 20, , 0, 220, 240, 260, 280, , k=1, , Mean value =, , y10, , hence, , a1 ≈, , 2, (417.94) = 69.66, 12, , a2 ≈, , 2, (−39) = −6.50, 12, , a3 ≈, , 2, (−49) = −8.17, 12, , θ ◦ (degrees) V (volts), , 30, , 62, , 210, , −28, , 60, , 35, , 240, , 24, , 90, , −38, , 270, , 80, , 120, , −64, , 300, , 96, , 150, , −63, , 330, , 90, , 180, , −52, , 360, , 70, , and, , From equation (3), bn ≈, hence, , p, 2 ;, yk cos nx k, p k=1, , b1 ≈, , p, 2 ;, yk sin nx k, p k=1, , 2, (−278.53) = −46.42, 12
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Table 70.1, , Ordinates, , θ◦, , V, , cos θ, , V cos θ, , sin θ, , V sin θ, , cos 2θ, , V cos 2θ, , sin 2θ, , V sin 2θ, , cos 3θ V cos 3θ, , sin 3θ, , y1, , 30, , 62, , 0.866, , 53.69, , 0.5, , 31, , 0.5, , 31, , 0.866, , 53.69, , 0, , 0, , 1, , 62, , y2, , 60, , 35, , 0.5, , 17.5, , 0.866, , 30.31, , −17.5, , 0.866, , 30.31, , −1, , −35, , 0, , 0, , y3, , 90 −38, , 0, , 0, , 0, , 0, , −1, , 38, , y4, , 120 −64, , −0.5, , y5, , 150 −63, , y6, , −0.5, , V sin 3θ, , −38, , −1, , 38, , 32, , 0.866, , −55.42, , −0.5, , 32, , −0.866, , 55.42, , 1, , −64, , 0, , 0, , −0.866, , 54.56, , 0.5, , −31.5, , 0.5, , −31.5, , −0.866, , 54.56, , 0, , 0, , 1, , −63, , 180 −52, , −1, , 52, , 0, , 0, , 1, , −52, , 0, , −1, , 52, , 0, , 0, , y7, , 210 −28, , −0.866, , 24.25, , −0.5, , 14, , 0.5, , −14, , 0.866 −24.25, , 0, , 0, , −1, , 28, , y8, , 240, , 24, , −0.5, , 1, , 24, , 0, , 0, , y9, , 270, , 80, , 0, , 0, , 0, , 1, , 80, , y10, , 300, , 96, , 0.5, , y11, , 330, , 90, , y12, , 360, , 70, , 12, ;, k=1, , yk = (212), , 0, , 0, , −0.866, , −20.78, , −0.5, , −12, , 0.866, , −1, , −80, , −1, , −80, , 0, , 48, , −0.866, , −83.14, , −0.5, , −48, , −0.866 −83.14, , −1, , −96, , 0, , 0, , 0.866, , 77.94, , −0.5, , −45, , 0.5, , 45, , −0.866 −77.94, , 0, , 0, , −1, , −90, , 1, , 70, , 1, , 70, , 1, , 70, , 0, , 0, , 12, ;, k=1, , −12, , 0, , 0, , yk cos θk, = 417.94, , 0, , 0, 12, ;, k=1, , yk sin θk, = −278.53, , 12, ;, k=1, , yk cos 2θk, = −39, , 0, 12, ;, k=1, , 20.78, 0, , 0, , yk sin 2θk, = 29.43, , 12, ;, k=1, , yk cos 3θk, = −49, , 12, ;, k=1, , yk sin 3θk, = 55, , A numerical method of harmonic analysis, , 1, , 639
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640 Higher Engineering Mathematics, , and, , b2 ≈, , 2, (29.43) = 4.91, 12, , b3 ≈, , 2, (55) = 9.17, 12, , Hence equation (4) may be re-written as:, v = 17.67 + 83.71 sin(θ + 2.16), + 8.15 sin(2θ − 0.92), + 12.28 sin(3θ − 0.73) volts, which is the form used in Chapter 25 with complex, waveforms., , Substituting these values into the Fourier series:, f (x) = a0 +, , ∞, <, , Now try the following exercise, , (an cos nx + bn sin nx), , n=1, , Exercise 234 Further problems on, numerical harmonic analysis, , gives: v = 17.67 + 69.66 cos θ − 6.50 cos 2θ, − 8.17 cos 3θ + · · · − 46.42 sin θ, + 4.91 sin 2θ + 9.17 sin 3θ + · · ·, , (4), , Note that in equation (4), (−46.42 sin θ + 69.66 cosθ), comprises the fundamental, (4.91 sin 2θ − 6.50 cos 2θ), comprises the second harmonic and (9.17 sin 3θ −, 8.17 cos 3θ) comprises the third harmonic., It is shown in Chapter 17 that:, , Determine the Fourier series to represent the periodic functions given by the tables of values in, Problems 1 to 3, up to and including the third harmonic and each coefficient correct to 2 decimal, places. Use 12 ordinates in each case., 1. Angle θ ◦, , Displacement y 40 43 38, , a sin ω t + b cos ω t = R sin(ω t + α), √, where a = R cos α, b = R sin α, R = a 2 + b2 and, b, α = tan−1, a, , For the fundamental, R = (−46.42)2 + (69.66)2, , Angle θ ◦, , a = R cos α, then cos α =, , and if, , −46.42, a, =, R, 83.71, which is negative,, , and the third harmonic, (9.17sin3θ −8.17cos3θ)=12.28sin(3θ −0.73), , 0, , 30, , 60, , 90, , Angle θ ◦ 180 210 240 270, , 120 150, , 300, , 330, , Voltage v 15.0 12.5 6.5 −4.0 −7.0 −7.5, ⎤, ⎡, v = 5.00 − 10.78 cos θ + 6.83 sin θ, ⎦, ⎣, − 1.96 cos 2θ + 0.80 sin 2θ, + 0.58 cos 3θ − 1.08 sin 3θ, , Hence α = tan−1, , (4.91 sin 2θ − 6.50 cos 2θ) = 8.15 sin(2θ − 0.92), , 17, , Voltage v −5.0 −1.5 6.0 12.5 16.0 16.5, , The only quadrant where cos α is negative and sin α is, positive is the second quadrant., , Thus (−46.42 sin θ + 69.66 cosθ ), = 83.71 sin(θ + 2.16), By a similar method it may be shown that the second, harmonic, , 23, , 210 240 270 300 330 360, , 2. Angle θ ◦, , 69.66, b, b = R sin α, then sin α = =, R, 83.71, which is positive., , b, 69.66, = tan−1, a, −46.42, = 123.68◦ or 2.16 rad, , 30, , Displacement y 11 9 10 13 21 32, ⎤, ⎡, y = 23.92 + 7.81 cos θ + 14.61 sin θ, ⎦, ⎣, + 0.17 cos 2θ + 2.31 sin 2θ, − 0.33 cos 3θ + 0.50 sin 3θ, , = 83.71, If, , 30 60 90 120 150 180, , 3., , Angle θ ◦ 30 60, , 90, , 120 150, , 180, , Current i 0 −1.4 −1.8 −1.9 −1.8 −1.3, Angle θ ◦ 210 240 270 300 330 360, Current i, ⎡, , 0, , 2.2, , 3.8, , 3.9, , 3.5 2.5, , ⎤, i = 0.64 + 1.58 cosθ − 2.73 sin θ, ⎣, − 0.23 cos 2θ − 0.42 sin 2θ ⎦, + 0.27 cos 3θ + 0.05 sin 3θ
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A numerical method of harmonic analysis, Problem 2. Without calculating Fourier, coefficients state which harmonics will be present, in the waveforms shown in Fig. 70.4., , 70.3 Complex waveform, considerations, It is sometimes possible to predict the harmonic content of a waveform on inspection of particular waveform, characteristics., , f(x), 2, , (i) If a periodic waveform is such that the area above, the horizontal axis is equal to the area below, then the mean value is zero. Hence a0 = 0 (see, Fig. 70.3(a))., , 2, , f (x) = f (x + π) represents a waveform which, repeats after half a cycle and only even, harmonics are present (see Fig. 70.3(d))., , (v), , f (x) = − f (x + π) represents a waveform for, which the positive and negative cycles are, identical in shape and only odd harmonics are, present (see Fig. 70.3(e))., , f (x), , , , 0, , 2, , 2 x, , (a) a0 5 0, , 2 x, , (b) Contains no sine terms, f (x), , f (x), , 22 2, , , , 0, , 0, , , , 22 2 0, , 2 x, , 2 x, , (c) Contains no cosine terms (d) Contains only even harmonics, , , , 2, , x, , f(x), 5, , (iii) An odd function is symmetrical about the origin, and contains no cosine terms (see Fig. 70.3(c))., (iv), , 0, 22, , (a), , (ii) An even function is symmetrical about the, vertical axis and contains no sine terms (see, Fig. 70.3(b))., , f (x), , (b), , 2, , 0, , , , 2, , x, , Figure 70.4, , (a), , The waveform shown in Fig. 70.4(a) is symmetrical about the origin and is thus an odd, function. An odd function contains no cosine, terms. Also, the waveform has the characteristic f (x) = − f (x + π), i.e. the positive and negative half cycles are identical in shape. Only, odd harmonics can be present in such a waveform. Thus the waveform shown in Fig. 70.4(a), contains only odd sine terms. Since the area, above the x-axis is equal to the area below,, a0 = 0., , (b) The waveform shown in Fig. 70.4(b) is symmetrical about the f (x) axis and is thus an, even function. An even function contains no sine, terms. Also, the waveform has the characteristic f (x) = f (x + π), i.e. the waveform repeats, itself after half a cycle. Only even harmonics, can be present in such a waveform. Thus the, waveform shown in Fig. 70.4(b) contains only, even cosine terms (together with a constant, term, a0 )., , f (x), , 2, , 0, , , , 2, , (e) Contains only odd harmonics, , Figure 70.3, , 641, , x, , Problem 3. An alternating current i amperes is, shown in Fig. 70.5. Analyse the waveform into its, constituent harmonics as far as and including the, fifth harmonic, correct to 2 decimal places, by, taking 30◦ intervals.
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642 Higher Engineering Mathematics, Investigating waveform characteristics has thus saved, unnecessary calculations and in this case the Fourier, series has only odd sine terms present, i.e., , i, 10, , 5, 2180, , 2120, , 2150, , 260, , 290, , 230 0, , i = b1 sin θ + b3 sin 3θ + b5 sin 5θ + · · ·, , y5, 180, , y1 y2 y 3 y4, 30 60 90 120 150, , 25, , 240, , 300, , 360, , 330, y11, y10, , 210 270, y7, y, y8 9, , 8, , 210, , Figure 70.5, , With reference to Fig. 70.5, the following characteristics, are noted:, , A proforma, similar to Table 70.1, but without the, ‘cosine terms’ columns and without the ‘even sine, terms’ columns is shown in Table 70.2 up to, and, including, the fifth harmonic, from which the Fourier, coefficients b1, b3 and b5 can be determined. Twelve, co-ordinates are chosen and labelled y1 , y2 , y3 , . . . y12, as shown in Fig. 70.5, From equation (3), Section 70.2,, , (ii) Since the waveform is symmetrical about the origin the function i is odd, which means that there, are no cosine terms present in the Fourier series., (iii) The waveform is of the form f (θ) = − f (θ + π), which means that only odd harmonics are, present., , 2<, ik sin nθk , where p = 12, p, p, , bn =, , (i) The mean value is zero since the area above the, θ axis is equal to the area below it. Thus the, constant term, or d.c. component, a0 = 0., , k=1, , Hence b1 ≈, , 2, (48.24) = 8.04,, 12, , b3 ≈, , 2, (−12) = −2.00,, 12, , b5 ≈, , 2, (−0.24) = −0.04, 12, , and, , Table 70.2, Ordinate, , θ, , sin θ, , i, , i sin θ, , sin 3θ, , i sin 3θ, , sin 5θ, , i sin 5θ, , y1, , 30, , 2, , 0.5, , 1, , 1, , 2, , 0.5, , 1, , y2, , 60, , 7, , 0.866, , 6.06, , 0, , 0, , −0.866, , −6.06, , y3, , 90, , 10, , −1, , −10, , y4, , 120, , 7, , 0.866, , 6.06, , 0, , 0, , −0.866, , −6.06, , y5, , 150, , 2, , 0.5, , 1, , 1, , 2, , 0.5, , 1, , y6, , 180, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , y7, , 210, , −2, , −0.5, , 1, , −1, , 2, , −0.5, , 1, , y8, , 240, , −7, , −0.866, , 6.06, , 0, , 0, , y9, , 270, , −10, , 1, , −10, , y10, , 300, , −7, , −0.866, , 6.06, , 0, , 0, , y11, , 330, , −2, , −0.5, , 1, , −1, , 2, , −0.5, , 1, , y12, , 360, , 0, , 0, , 0, , 0, , 0, , 0, , 0, , 1, , 10, , −1, , 12, ;, k=1, , 10, , yk sin θk = 48.24, , 12, ;, k=1, , yk sin 3θk = −12, , 1, , 10, , 0.866, −1, , 10, , 0.866, , 12, ;, k=1, , −6.06, −6.06, , yk sin 5θk = −0.24
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643, , A numerical method of harmonic analysis, Thus the Fourier series for current i is given by:, i = 8.04 sin θ − 2.00 sin 3 θ − 0.04 sin 5 θ, , y, 40, 30, 20, 10, , Now try the following exercise, , 1. Without performing calculations, state which, harmonics will be present in the waveforms, shown in Fig. 70.6, , Current /amperes, , Exercise 235 Further problems on a, numerical method of harmonic analysis, , 0, ⫺10, ⫺20, , (a) only odd cosine terms present, (b) only even sine terms present, , 90⬚, , 180⬚, , 270⬚, , 360⬚ ⬚, , 3/2, , 2 rads, , (a), 10, 5, /2, , 0, , , (b), , Figure 70.7, f(t), 4, , 22 2 0 2, , 4, , t, , 24, , (a), , y, , 4. Determine the Fourier series as far as the third, harmonic to represent the periodic function y, given by the waveform in Fig. 70.8. Take 12, intervals when analysing the waveform., , 10, , 2, , 0, , 3. For the waveform of current shown in, Fig. 70.7(b) state why only a d.c. component and even cosine terms will appear in the, Fourier series and determine the series, using, π/6 rad intervals, up to and including the sixth, harmonic., , I = 4.00 − 4.67 cos2θ + 1.00 cos 4θ, − 0.66 cos 6θ, , 2, x, , y, 100, 80, 60, 40, 20, , 210, (b), , Figure 70.6, 2908, , 2. Analyse the periodic waveform of displacement y against angle θ in Fig. 70.7(a) into, its constituent harmonics as far as and, including the third harmonic, by taking 30◦, intervals., ⎡, ⎤, y = 9.4 + 13.2 cos θ − 24.1 sin θ, ⎢, ⎥, + 0.92 cos 2θ − 0.14 sin 2θ ⎦, ⎣, + 0.83 cos 3θ + 0.67 sin 3θ, , 0 220 908 1808, 240, 260, 280, 2100, , Figure 70.8, , ⎡, , 2708, , 3608, , 8, , ⎤, y = 1.83 − 27.77 cosθ + 83.74 sin θ, ⎢, ⎥, − 0.75 cos 2θ − 1.59 sin 2θ, ⎣, ⎦, + 16.00 cos3θ + 11.00 sin 3θ
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Chapter 71, , The complex or, exponential form, of a Fourier series, 71.1, , Introduction, , The form used for the Fourier series in Chapters 66 to, 70 consisted of cosine and sine terms. However, there is, another form that is commonly used—one that directly, gives the amplitude terms in the frequency spectrum and, relates to phasor notation. This form involves the use of, complex numbers (see Chapters 20 and 21). It is called, the exponential or complex form of a Fourier series., , e j θ − e− j θ, 2j, , sin θ =, , from which,, , (4), , Thus, from page 630, the Fourier series f (x) over, any range L,, , , , , ∞ , <, 2πnx, 2πnx, f (x) = a0 +, + bn sin, an cos, L, L, n=1, , may be written as:, , 71.2 Exponential or complex, notation, , f (x) = a0 +, , , , ∞, <, , an, , ej, , 2πn x, L, , n=1, , , , It was shown on page 226, equations (4) and (5) that:, e j θ = cos θ + j sin θ, , (1), , and e− j θ = cos θ − j sin θ, , (2), , e j θ + e− j θ = 2 cosθ, e j θ + e− j θ, 2, , Similarly, equation (1) – equation (2) gives:, e j θ − e− j θ = 2 j sin θ, , 2πn x, L, , ej, , 2πn x, L, , , , 2πn x, L, , − e− j, 2j, , , , Multiplying top and bottom of the bn term by − j (and, remembering that j 2 = −1) gives:, , Adding equations (1) and (2) gives:, , from which, cos θ =, , + bn, , + e− j, 2, , (3), , f (x) = a0 +, , ∞, <, , , an, , ej, , 2πn x, L, , n=1, , , − j bn, , + e− j, 2, , ej, , 2πn x, L, , 2πn x, L, , , , − e− j, 2, , 2πn x, L, ,
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The complex or exponential form of a Fourier series, Rearranging gives:, , 71.3, , , ∞ , <, 2πn x, an − j bn, ej L, f (x) = a0 +, 2, n=1, , , an + j bn − j 2πn x, L, +, e, 2, , (5), , The Fourier coefficients a0 , an and bn may be replaced, by complex coefficients c0 , cn and c−n such that, c0 = a0, an − j bn, 2, , (7), , an + j bn, =, 2, , (8), , cn =, and c−n, , (6), , The complex coefficients, , From equation (7), the complex coefficient cn was, an − j bn, defined as: cn =, 2, However, an and bn are defined (from page 630) by:, 2, an =, L, 2, bn =, L, , !, , L, 2, , − L2, , !, , L, 2, , − L2, , ∞, <, , cn e j, , 2πn x, L, , +, , ∞, <, , c−n e− j, , 2πn x, L, , Since e = 1, the c0 term can be absorbed into the summation since it is just another term to be added to the, summation of the cn term when n =0. Thus,, cn e, , x, j 2πn, L, , +, , n=0, , ∞, <, , c−n e− j, , 2πn x, L, , (10), , n=1, , The c−n term may be rewritten by changing the limits, n =1 to n =∞ to n = −1 to n =−∞. Since n has been, 2πn x, made negative, the exponential term becomes e j L, and c−n becomes cn . Thus,, f (x) =, , ∞, <, , cn e j, , 2πn x, L, , n=0, , +, , −∞, <, , , 2πnx, f (x) sin, dx, L, ⎞, 2πnx , f (x) cos L dx, ⎟, 2πnx ⎠, L, 2 2, − j L L f (x) sin L dx, , , , cn e j, , −2, , Thus, cn =, 1, =, L, , cn e j, , 2π nx, L, , !, , (11), , Equation (11) is the complex or exponential form of, the Fourier series., , 2, , , 2πnx, f (x) cos, dx, L, − L2, , , ! L, 2πnx, 1 2, f (x) sin, dx, − j, L − L2, L, , from which,, 1, cn =, L, , !, , L, 2, , − L2, , 2πn x, L, , n=−∞, , , , L, 2, , From equations (3) and (4),, , 2πn x, 2πn x, ! L, 1 2, e j L + e− j L, dx, f (x), cn =, L − L2, 2, , 2πn x, 2πn x, ! L, 1 2, e j L − e− j L, dx, − j, f (x), L − L2, 2j, , f (x), , n=−1, , ∞, <, , L, , 2 2, L −L, 2, , −, , Since the summations now extend from −∞ to −1 and, from 0 to +∞, equation (10) may be written as:, f (x) =, , , , (9), , n=1, , 0, , f (x) =, , , 2πnx, f (x) cos, dx and, L, , ⎜, ⎝, , n=1, , ∞, <, , , , ⎛, , where c−n represents the complex conjugate of cn (see, page 216)., Thus, equation (5) may be rewritten as:, f (x) = c0 +, , 645, , i.e., , cn =, , 1, L, , 1, L, !, , !, , ej, L, 2, , − L2, L, 2, , − L2, , , 2πn x, + e− j L, dx, 2, , 2πn x, 2πn x, e j L − e− j L, dx, f (x), 2, , 2πn x, L, , f (x) e−j, , 2π nx, L, , dx, , (12), , Care needs to be taken when determining c0 . If n appears, in the denominator of an expression the expansion can, be invalid when n =0. In such circumstances it is usually, simpler to evaluate c0 by using the relationship:, 1, c0 = a0 =, L, , !, , L, 2, , − L2, , f (x)dx (from page 630)., , (13)
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646 Higher Engineering Mathematics, Problem 1. Determine the complex Fourier series, for the function defined by:, ⎧, ⎨0, when −2 ≤ x ≤ −1, f (x) = 5, when −1 ≤ x ≤1, ⎩, 0, when 1 ≤ x ≤ 2, , Hence, from equation (11), the complex form of the, Fourier series is given by:, ∞, <, , f (x) =, , cn e, , x, j 2πn, L, , =, , n=−∞, , ∞, <, , 5, πn j π nx, sin, e 2, πn, 2, n=−∞, (14), , The function is periodic outside this range of, period 4., This is the same Problem as Problem 2 on page 631 and, we can use this to demonstrate that the two forms of, Fourier series are equivalent., The function f (x) is shown in Figure 71.1, where the, period, L = 4., From equation (11), the complex Fourier series is, given by:, ∞, <, , f (x) =, , cn e, , Let us show how this result is equivalent to the result, involving sine and cosine terms determined on page 632., From equation (13),, 1, c0 = a0 =, L, =, , x, j 2πn, L, , n=−∞, , where cn is given by:, ! L, 2πn x, 1 2, f (x) e− j L dx (from equation 12)., cn =, L, L −2, With reference to Figure 71.1, when L = 4,, , ! −1, ! 1, ! 2, x, 1, − j 2πn, 4, 0 dx +, 5e, dx +, 0 dx, cn =, 4 −2, −1, 1, 1, =, 4, , !, , 1, , −1, , 5 e−, , j πn x, 2, , 1, , j πn x, , 5 e− 2, dx =, 4 − j πn, 2, , 5, πn, sin, (from equation (4))., πn, 2, , L54, , Figure 71.1, , 3, , 4, , 1, −1, , 5 dx, , 5, πn, sin, , then, πn, 2, , 5, sin π = 0, 2π, , (in fact, all even terms will be zero since, sin nπ = 0), c3 =, , 5, πn, 5, 3π, 5, sin, =, sin, =−, πn, 2, 3π, 2, 3π, , By similar substitution,, c5 =, , 5, 5π, , c7 = −, , 5, , and so on., 7π, , Similarly,, , 5, , 2, , !, , 5, 5 1, 5, = [1 − (−1)] =, x, 4 −1 4, 2, , c2 =, , c−1 =, , 1, , − L2, , 1, f (x)dx =, 4, , 5, π, 5, sin =, π, 2, π, , f (x), , 25 24 23 22 21 0, , L, 2, , c1 =, , −1, , , j πn, −5 − j πn x 1, −5 − j πn, =, =, e 2, e 2 − e 2, −1, j 2πn, j 2πn, πn, , πn, 5 e j 2 − e− j 2, =, πn, 2j, =, , Since cn =, , !, , 5, , x, , 5, −π, 5, sin, =, −π, 2, π, , c−2 = −, , 5, −2π, sin, = 0 = c−4 = c−6 , and so on., 2π, 2, , c−3 = −, , 5, −3π, 5, sin, =−, 3π, 2, 3π, , c−5 = −, , 5, −5π, 5, sin, =, , and so on., 5π, 2, 5π
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647, , The complex or exponential form of a Fourier series, Hence, the extended complex form of the Fourier series, shown in equation (14) becomes:, 5 j 3π x, 5 j 5π x, 5 5 j πx, + e 2 −, e 2 +, e 2, 2 π, 3π, 5π, πx, 5 j 7π x, 5, −, e 2 + · · · + e− j 2, 7π, π, 5 − j 3π x, 5 − j 5π x, 2 +, 2, −, e, e, 3π, 5π, 5 − j 7π x, 2 + ···, −, e, 7π, , πx, 5 5 j πx, = +, e 2 + e− j 2, 2 π, , 5 j 3π x, 3π x, −, e 2 + e− j 2, 3π, , , 5π x, 5π x, 5, +, e 2 + e− j 2 − · · ·, 5π, , πx, πx, e j 2 + e− j 2, 5 5, = + (2), 2 π, 2, , 3π x, 3π x, 5, e j 2 + e− j 2, −, (2), 3π, 2, , f (x) =, , , 5π x, 5π x, 5, e j 2 + e− j 2, − ···, +, (2), 5π, 2, , , π x 10, 3π x, 5 10, = +, cos, −, cos, 2 π, 2, 3π, 2, , , 5π x, 10, +, cos, − ···, 5π, 2, (from equation (3)), , , , πx, 1, 3πx, 5 10, cos, − cos, i.e. f (x) = +, 2 π, 2, 3, 2, , , 5πx, 1, −···, + cos, 5, 2, which is the same as obtained on page 632., ∞, <, , 5, nπ, sin, e, Hence,, πn, 2, n=−∞, , j πnx, 2, , is equivalent to, , , , , πx, 3πx, 5 10, 1, +, cos, − cos, 2 π, 2, 3, 2, , , 1, 5πx, −···, + cos, 5, 2, , Problem 2. Show that the complex Fourier series, for the function f (t ) =t in the range t = 0 to t = 1,, and of period 1, may be expressed as:, f (t ) =, , ∞, 1, j < e j 2πnt, +, 2 2π n=−∞ n, , The saw tooth waveform is shown in Figure 71.2., , f (t), f (t) 5 t, , 21, , 0, , 1, , 2, , t, , Period L 5 1, , Figure 71.2, , From equation (11), the complex Fourier series is, given by:, f (t ) =, , ∞, <, , 2πnt, L, , cn e j, , n=−∞, , and when the period, L =1, then:, f (t ) =, , ∞, <, , cn e j 2πnt, , n=−∞, , where, from equation (12),, cn =, , 1, L, , !, , L, 2, , − L2, , f (t ) e− j, , 2πnt, L, , dt =, , !, , 1, L, , L, , f (t ) e− j, , 2πnt, L, , dt, , 0, , and when L =1 and f (t ) =t , then:, cn =, , 1, 1, , !, 0, , 1, , t e− j, , 2πnt, 1, , !, , 1, , dt =, , t e− j 2πnt dt, , 0, , Using integration by parts (see Chapter 43), let u = t ,, du, from which,, = 1, and dt = du, and, dt, let dv = e− j 2πnt , from which,, !, e− j 2πnt, v = e− j 2πnt dt =, − j 2πn
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649, , The complex or exponential form of a Fourier series, Now try the following exercise, , when −1 < t < 1 and has period 2., , , ∞, 1 < e(2− j πn) − e−(2− j πn), e j πnt, f (t ) =, 2 n=−∞, 2 − j πn, , Exercise 236 Further problems on the, complex form of a Fourier series, 1. Determine the complex Fourier series for the, function defined by:, , 0, when −π ≤ t ≤ 0, f (t ) =, 2, when 0 ≤ t ≤ π, The function is periodic outside of this range, of period 2π., ∞, <, , j, (cos nπ − 1) e j nt, nπ, n=−∞, , , 2, 1 j 3t 1 j 5t, jt, =1− j, e + e + e +···, π, 3, 5, , , 2 − j t 1 − j 3t 1 − j 5t, + j, + e, + e, +···, e, π, 3, 5, , f (t ) =, , 71.4, , If even or odd symmetry is noted in a function, then time, can be saved in determining coefficients., The Fourier coefficients present in the complex Fourier, series form are affected by symmetry. Summarising, from previous chapters:, An even function is symmetrical about the vertical axis, and contains no sine terms, i.e. bn = 0., For even symmetry,, a0 =, , 2. Show that the complex Fourier series for the, waveform shown in Figure 71.3, that has, period 2, may be represented by:, f (t ) = 2 +, , ∞, <, , n=−∞, (n= 0), , Symmetry relationships, , 2, an =, L, , j2, (cos nπ − 1) e j πnt, πn, , =, , L, , f (x)dx, , and, , 0, , !, , L, 0, , !, , L, 2, , , , , 2πnx, f (x) cos, dx, L, , f (x) cos, , 0, , , 2πnx, dx, L, , For odd symmetry,, , 4, , bn =, 0, , 4, L, , !, , An odd function is symmetrical about the origin and, contains no cosine terms, a0 = an = 0., , f (t), , 21, , 1, L, , 1, , 2, , t, , Period L 5 2, , Figure 71.3, , 3. Show that the complex Fourier series of, Problem 2 is equivalent to:, , 1, 8, sin πt + sin 3πt, f (t ) = 2 +, π, 3, , 1, + sin 5πt + · · ·, 5, 4. Determine the exponential form of the Fourier, series for the function defined by: f (t ) = e2t, , =, , 2, L, 4, L, , !, !, , L, , , , , 2πnx, dx, L, , , 2πnx, dx, f (x) sin, L, , f (x) sin, , 0, L, 2, , 0, , an − j bn, From equation (7), page 645, cn =, 2, Thus, for even symmetry, bn = 0 and, cn =, , an, 2, =, 2, L, , !, , L, 2, , 0, , , , 2πnx, dx, f (x) cos, L, , (15), , For odd symmetry, an = 0 and, − j bn, 2, cn =, = −j, 2, L, , !, , L, 2, , 0, , , , , 2πnx, dx, f (x) sin, L, , (16), , For example, in Problem 1 on page 646, the function, f (x) is even, since the waveform is symmetrical about
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652 Higher Engineering Mathematics, i.e., f (x) =, , , 1, 1, 8, sin x + sin 3x + sin 5x, π, 3, 5, , f (x) =, , , 1, + sin 7x + · · ·, 7, , Hence,, f (x) =, , ∞, <, n=−∞, , 8, ≡, π, , −j, , 3. Determine the complex Fourier series to represent the function f (t ) = 2t in the range −π, , ∞ , <, j2, cos nπ e j nt, to +π., f (t ) =, n, n=−∞, , 2, (1 −cos n π) e jnx, nπ, , , 1, 1, sinx + sin 3x + sin 5x, 3, 5, 1, + sin7x + · · ·, 7, , 4. Show that the complex Fourier series in, problem 3 above is equivalent to:, , 1, 1, f (t ) = 4 sin t − sin 2t + sin 3t, 2, 3, , 1, − sin 4t + · · ·, 4, , , , Now try the following exercise, Exercise 237 Further problems on, symmetry relationships, , 71.5, , 1. Determine the exponential form of the Fourier, series for the periodic function defined by:, ⎧, π, ⎪, −2, when −π ≤ x ≤ −, ⎪, ⎪, ⎪, 2, ⎪, ⎪, ⎨, π, π, 2, when − ≤ x ≤ +, f (x) =, ⎪, 2, 2, ⎪, ⎪, ⎪, ⎪, π, ⎪, ⎩−2, when + ≤ x ≤ + π, 2, , The frequency spectrum, , In the Fourier analysis of periodic waveforms seen in, previous chapters, although waveforms physically exist, in the time domain, they can be regarded as comprising, components with a variety of frequencies. The amplitude and phase of these components are obtained from, the Fourier coefficients an and bn ; this is known as a, frequency domain. Plots of amplitude/frequency and, phase/frequency are together known as the spectrum, of a waveform. A simple example is demonstrated in, Problem 6 following., , and has a period of 2π., f (x) =, , , 1, 8, 1, cos x − cos 3x + cos5x, π, 3, 5, , 1, − cos 7x + · · ·, 7, , , ∞ , <, 4, nπ, sin, e j nx, nπ, 2, n=−∞, , Problem 6. A pulse of height 20 and width 2 has a, period of 10. Sketch the spectrum of the waveform., , 2 Show that the exponential form of the Fourier, series in problem 1 above is equivalent to:, , The pulse is shown in Figure 71.5., , f (t ), 20, , 21, , 0, , 1, , t, L 5 10, , Figure 71.5
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The complex or exponential form of a Fourier series, The complex coefficient is given by equation (12):, cn =, , 1, L, , !, , L, 2, , − L2, , 1, =, 10, , !, , f (t )e− j, , − j 2πnt, 10, , 20e, , =, , 20, 10, , =, , 20, πn, , i.e. cn =, , dt, πnt, , 1, −1, , 2πnt, L, , 20 e− j 5, dt =, 10 − j πn, 5, , , , 1, −1, , , , 5, πn, πn, e− j 5 − e j 5, − j πn, ej, , πn, 5, , − e− j, 2j, , πn, 5, , 20, nπ, sin, πn, 5, , A graph of |cn | plotted against the number of the, harmonic, n, is shown in Figure 71.6., Figure 71.7 shows the corresponding plot of cn, against n., Since cn is real (i.e. no j terms) then the phase must be, either 0◦ or ±180◦, depending on the sign of the sine,, as shown in Figure 71.8., When cn is positive, i.e. between n =−4 and n = +4,, angle αn = 0◦ ., When cn is negative, then αn = ±180◦; between n =+6, and n =+9, αn is taken as +180◦, and between n =−6, and n = −9, αn is taken as −180◦., Figures 71.6 to 71.8 together form the spectrum of the, waveform shown in Figure 71.5., , 71.6, , from equation (4), page 644., From equation (13),, ! L, !, 1 2, 1 1, f (x) dx =, 20 dt, c0 =, L − L2, 10 −1, 1, 1, [20t ]1−1 =, [20 − (−20)] = 4, 10, 10, 20, π, c1 =, sin = 3.74 and, π, 5, 20 π , = 3.74, c−1 = − sin −, π, 5, Further values of cn and c−n , up to n = 10, are calculated, and are shown in the following table., =, , n, , cn, , c−n, , 0, , 4, , 4, , 1, , 3.74, , 3.74, , 2, , 3.03, , 3.03, , 3, , 2.02, , 2.02, , 4, , 0.94, , 0.94, , 5, , 0, , 0, , 6, , −0.62, , −0.62, , 7, , −0.86, , −0.86, , 8, , −0.76, , −0.76, , 9, , −0.42, , −0.42, , 10, , 0, , 0, , 653, , Phasors, , Electrical engineers in particular often need to analyse alternating current circuits, i.e. circuits containing, a sinusoidal input and resulting sinusoidal currents and, voltages within the circuit., It was shown in chapter 14, page 143, that a general, sinusoidal voltage function can be represented by:, v = Vm sin (ωt + α) volts, , (19), , where Vm is the maximum voltage or amplitude of the, voltage v, ω is the angular velocity ( = 2π f , where f is, the frequency), and α is the phase angle compared with, v = Vm sin ωt ., Similarly, a sinusoidal expression may also be expressed, in terms of cosine as:, v = Vm cos(ωt + α)volts, , (20), , It is quite complicated to add, subtract, multiply and, divide quantities in the time domain form of equations, (19) and (20). As an alternative method of analysis a, waveform representation called a phasor is used. A phasor has two distinct parts—a magnitude and an angle;, for example, the polar form of a complex number, say, 5∠π/6, can represent a phasor, where 5 is the magnitude, or modulus, and π/6 radians is the angle or argument., Also, it was shown on page 228 that 5∠π/6 may be, written as 5 e j π/6 in exponential form., In chapter 21, equation (4), page 228, it is shown that:, e j θ = cos θ + j sin θ, which is known as Euler’s formula., , (21)
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654 Higher Engineering Mathematics, |cn|, 4, 3, 2, 1, 210 29 28 27 26 25 24, , 23 22 21, , 0, , 1, , 2, , 3, , 4, , 5, , 1, , 2, , 3, , 4, , 5, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 6, , 7, , 8, , 9, , 10, , n, , Figure 71.6, cn, 4, 3, 2, 1, 210, , 29 28, , 27 26 25, , 24, , 23 22 21, , 0, , 10 n, , Figure 71.7, ␣n, 1808, , 908, , 210 29 28 27, , 26, 25 24, , 23 22, , 21 0, , 2908, , 21808, , Figure 71.8, , 6, , 7, , 8, , 9, , 10, , n
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The complex or exponential form of a Fourier series, Imaginary axis, , From equation (21),, e j (ωt +α) = cos(ωt + α) + j sin(ωt + α), and, , Vm e j (ωt +α) = Vm cos(ωt + α), + j Vm sin(ωt + α), , Thus a sinusoidal varying voltage such as in equation (19) or equation (20) can be considered to be either, the real or the imaginary part of Vm e j (ωt +α), depending on whether the cosine or sine function is being, considered., Vm e j (ωt + α) may be rewritten as Vm e j ωt e j α since, a m+n = a m × a n from the laws of indices, page 1., The e j ωt term can be considered to arise from the fact, that a radius is rotated with an angular velocity ω, and, α is the angle at which the radius starts to rotate at time, t = 0 (see Chapter 14, page 143)., Thus, Vm e j ωt e j α defines a phasor. In a particular circuit the angular velocity ω is the same for all the, elements thus the phasor can be adequately described, by Vm ∠α, as suggested above., Alternatively, if, , and, , v = Vm cos(ωt + α) volts, , 1 jϑ, cos θ =, e + e− j θ, 2, from equation (3), page 644, , then, , , , 1 j (ωt +α), v = Vm, + e− j (ωt +α), e, 2, , i.e., , 1, 1, v = Vm e j ωt e j α + Vm e− j ωt e− j α, 2, 2, , Thus, v is the sum of two phasors, each with half the, amplitude, with one having a positive value of angular, velocity (i.e. rotating anticlockwise) and a positive value, of α, and the other having a negative value of angular, velocity (i.e. rotating clockwise) and a negative value of, α, as shown in Figure 71.9., 1, 1, The two phasors are Vm ∠α and Vm ∠−α., 2, 2, From equation (11), page 645, the Fourier representation of a waveform in complex form is:, cn e, , and, , j 2πnt, L, , = cn e, , j ωnt, , for positive values of n, , , 2π, since ω =, L, , cn e−j ωnt for negative values of n., , 655, , , 1 Vm, 2, , ␣, ␣, , 0, 2, , Real axis, , 1, , V, , m, , , , Figure 71.9, , It can thus be considered that these terms represent phasors, those with positives powers being phasors rotating, with a positive angular velocity (i.e. anticlockwise), and, those with negative powers being phasors rotating with, a negative angular velocity (i.e. clockwise)., In the above equations,, n =0 represents a non-rotating component, since e0 = 1,, n =1 represents a rotating component with angular, velocity of 1ω,, n =2 represents a rotating component with angular, velocity of 2ω, and so on., Thus we have a set of phasors, the algebraic sum of, which at some instant of time gives the magnitude of, the waveform at that time., Problem 7. Determine the pair of phasors that, can be used to represent the following voltages:, (a) v = 8 cos 2t, (b) v = 8 cos (2t − 1.5), (a), , From equation (3), page 644,, cos θ =, , 1 jθ, (e + e−jθ ), 2, , Hence,, v = 8 cos2t = 8, , , , 1 j2t, e + e−j2t, 2, , = 4e j2t + 4e−j2t, This represents a phasor of length 4 rotating anticlockwise (i.e. in the positive direction) with an, angular velocity of 2 rad/s, and another phasor, of length 4 and rotating clockwise (i.e. in the
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Imaginary axis, , 656 Higher Engineering Mathematics, Problem 8. Determine – the pair of phasors that, can be used to represent the third harmonic, ⫽ 2 rad/s, , 0, , 2, , 4, , Real axis, , ⫽ 2 rad/s, , negative direction) with an angular velocity of, 2 rad/s. Both phasors have zero phase angle., Figure 71.10 shows the two phasors., (b) From equation (3), page 644,, , 1 jθ, e + e− j θ, cos θ =, 2, Hence, v = 8 cos(2t − 1.5), , , 1 j (2t −1.5), + e− j (2t −1.5), =8, e, 2, = 4e j (2t −1.5) + 4e− j (2t −1.5), v = 4e2t e−j 1.5 + 4e−j2t e j1.5, , Imaginary axis, , This represents a phasor of length 4 and phase, angle −1.5 radians rotating anticlockwise (i.e. in, the positive direction) with an angular velocity of, 2 rad/s, and another phasor of length 4 and phase, angle +1.5 radians and rotating clockwise (i.e. in, the negative direction) with an angular velocity of, 2 rad/s. Figure 71.11 shows the two phasors., , 5 2 rad/s, , , 1 jt, e − e− j t from page 644, 2j, , gives: v = 8 cos 3t − 20 sin 3t, , , 1 j 3t, =8, e + e− j 3t, 2, , − 20, , , 1 j 3t, e − e− j 3t, 2j, , = 4e j 3t + 4e− j 3t −, , 10 j 3t 10 − j 3t, e + e, j, j, , = 4e j 3t + 4e− j 3t −, , 10( j ) j 3t 10( j ) − j 3t, e +, e, j( j), j( j), , = 4e j 3t + 4e− j 3t + 10 j e j 3t − 10 j e− j 3t, since j 2 = −1, = (4 + j 10)e j 3t + (4 − j 10) e− j 3t, (4 + j 10) =, , , , 42 + 102 ∠ tan−1, , , , 10, 4, , , , and (4 − j 10), , 1.5 rad, , Real axis, , = 10.77∠−1.19, Hence, v = 10.77 ∠1.19 +10.77 ∠ −1.19, , 4, , 5 2 rad/s, , Figure 71.11, , sin t =, , = 10.77∠1.19, 4, 1.5 rad, , 0, , , 1 jt, e + e− j t, 2, , Using cos t =, and, , Figure 71.10, , i.e., , v = 8 cos3t − 20 sin 3t., , Thus v comprises a phasor 10.77∠1.19 rotating anticlockwise with an angular velocity if 3 rad/s, and a, phasor 10.77∠−1.19 rotating clockwise with an angular, velocity of 3 rad/s.
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The complex or exponential form of a Fourier series, Now try the following exercise, , Exercise 238, , Further problems on phasors, , 1. Determine the pair of phasors that can be used, to represent the following voltages:, (a) v = 4 cos4t, , (b) v = 4 cos(4t + π/2)., , [(a) 2e j 4t + 2e− j 4t , 2∠0◦ anticlockwise, 2∠0◦ clockwise, each with, ω = 4 rad/s, (b) 2e j 4t e j π/2 + 2 e− j 4t e− j π/2 , 2∠π/2, anticlockwise, 2∠ −π/2 clockwise,, each with ω = 4 rad/s], , 2. Determine the pair of phasors that can represent the harmonic given by:, v = 10 cos 2t − 12 sin 2t ., [(5 + j 6)e j 2t + (5 − j 6)e− j 2t ,, 7.81∠0.88 rotating anticlockwise,, 7.81 ∠−0.88 rotating clockwise, each, with ω = 2 rad/s], 3. Find the pair of phasors that can represent the, fundamental current: i = 6 sin t + 4 cost ., [(2 − j 3) e j t + (2 + j 3) e− j t ,, 3.61∠−0.98 rotating anticlockwise,, 3.61 ∠ 0.98 rotating clockwise, each, with ω = 1 rad/s], , 657
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Revision Test 19, This Revision Test covers the material contained in Chapters 66 to 71. The marks for each question are shown in, brackets at the end of each question., 1., , Obtain a Fourier series for the periodic function, f (x) defined as follows:, , −1, when − π ≤ x ≤ 0, f (x) =, 1, when, 0≤x ≤π, , 210, , 0.27, , 240, , 0.13, , 270, , 0.45, , The function is periodic outside of this range with, period 2π., (13), , 300, , 1.25, , 330, , 2.37, , 2., , Obtain a Fourier series to represent f (t ) = t in the, range −π to +π., (13), , 360, , 3.41, , 3., , Expand the function f (θ) = θ in the range, 0 ≤θ ≤ π into (a) a half range cosine series, and, (b) a half range sine series., (18), , 4., , (a) Sketch the waveform defined by:, ⎧, ⎨0, when −4 ≤ x ≤ −2, f (x) = 3, when −2 ≤ x ≤ 2, ⎩, 0, when 2 ≤ x ≤ 4, and is periodic outside of this range of period 8., (b) State whether the waveform in (a) is odd, even, or neither odd nor even., (c) Deduce the Fourier series for the function, defined in (a)., (15), , 5., , Displacement y on a point on a pulley when, turned through an angle of θ degrees is given by:, θ, , y, , 30, , 3.99, , 60, , 4.01, , 90, , 3.60, , 120, , 2.84, , 150, , 1.84, , 180, , 0.88, , Sketch the waveform and construct a Fourier series, for the first three harmonics, (23), 6., , A rectangular waveform is shown in Fig. RT19.1., (a) State whether the waveform is an odd or even, function., (b) Obtain the Fourier series for the waveform in, complex form., (c) Show that the complex Fourier series in (b) is, equivalent to:, , 20, 1, 1, f (x) =, sin x + sin 3x + sin 5x, π, 3, 5, , 1, + sin 7x + · · ·, 7, (18), f (x ), 5, , 22, , 2, , , , 0, , 25, , Figure RT19.1, , 2, , 3, , x
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Essential formulae, Number and Algebra, , (ax 2, , Laws of indices:, a m × a n = a m+n, m, , an =, , √, n, , am, , am, an, , f (x), + bx + c)(x + d), , ≡, = a m−n (a m )n = a mn, , a −n =, , 1, an, , a0 = 1, , Definition of a logarithm:, If y = a x then x = loga y, , Quadratic formula:, If ax 2 + bx + c = 0 then x =, , C, Ax + B, +, (ax 2 + bx + c) (x + d), , −b ±, , √, , Laws of logarithms:, b2 − 4ac, , log(A × B) = log A + log B, , A, log, = log A − log B, B, , 2a, , Factor theorem:, , log An = n × log A, , If x = a is a root of the equation f (x) = 0, then (x − a), is a factor of f (x)., , Exponential series:, Remainder theorem:, If (ax 2 + bx + c) is divided by (x − p), the, remainder will be: ap 2 + bp +c., , ex = 1 + x +, , or if (ax 3 + bx 2 + cx + d) is divided by (x − p), the, remainder will be: ap3 + bp2 + cp + d., , x2 x3, +, +···, 2! 3!, (valid for all values of x), , Hyperbolic functions:, Partial fractions:, Provided that the numerator f (x) is of less degree than, the relevant denominator, the following identities are, typical examples of the form of partial fractions used:, f (x), (x + a)(x + b)(x + c), B, C, A, +, +, ≡, (x + a) (x + b) (x + c), f (x), (x + a)3 (x + b), A, C, D, B, ≡, +, +, +, (x + a) (x + a)2 (x + a)3 (x + b), , sinh x =, , e x − e−x, 2, , cosech x =, , 1, 2, = x, sinh x, e − e−x, , cosh x =, , e x + e−x, 2, , sech x =, , 1, 2, = x, cosh x, e + e−x, , tanh x =, , e x − e−x, e x + e−x, , coth x =, , e x + e−x, 1, = x, tanh x, e − e−x, , cosh2 x − sinh2 = 1 1 − tanh 2 x = sech2 x, coth2 x − 1 = cosech2 x
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660 Higher Engineering Mathematics, Arithmetic progression:, , Boolean algebra:, , If a = first term and d = common difference, then the, arithmetic progression is: a, a + d, a + 2d, . . ., , Laws and rules of Boolean algebra, , The n’th term is: a + (n − 1)d, n, Sum of n terms, Sn = [2a + (n −1)d], 2, , Geometric progression:, If a = first term and r = common ratio, then the geometric progression is: a, ar, ar 2 , . . ., The n’th term is: ar n−1, a(1 −r n ), a(r n − 1), or, (1 −r), (r − 1), a, If −1 <r < 1, S∞ =, (1 −r), , Sum of n terms, Sn =, , Binomial series:, (a + b)n = a n + na n−1 b +, +, , n(n − 1) n−2 2, a b, 2!, , Commutative Laws: A + B = B + A, A· B = B · A, Associative Laws:, A + B + C = (A + B) + C, A · B · C = (A · B) · C, Distributive Laws: A · (B + C) = A · B + A · C, A + (B · C) = (A + B) · (A+C), Sum rules:, A+ A =1, A+1=1, A+0= A, A+ A = A, Product rules:, A· A =0, A·0=0, A·1= A, A· A = A, Absorption rules:, A+ A· B = A, A · (A + B) = A, A+ A· B = A+ B, A+ B = A· B, De Morgan’s Laws:, A· B = A+ B, , n(n − 1)(n − 2) n−3 3, a b +···, 3!, , (1 + x)n = 1 + nx +, +, , n(n − 1) 2, x, 2!, , n(n − 1)(n − 2) 3, x +···, 3!, , Geometry and Trigonometry, Theorem of Pythagoras:, b 2 = a 2 + c2, A, , Maclaurin’s series:, x2, f (x) = f (0) + x f (0) +, f (0), 2!, x3, +, f (0) + · · ·, 3!, , Newton Raphson iterative method:, If r1 is the approximate value for a real root of the equation f (x) = 0, then a closer approximation to the root,, r2 , is given by:, r2 = r1 −, , f (r1 ), f (r1 ), , c, , b, , B, , a, , cosec θ =, , 1, sin θ, , C, , Figure FA1, , Identities:, sec θ =, , 1, cos θ, , 1, sin θ, tan θ =, tan θ, cos θ, cos2 θ + sin2 θ = 1 1 + tan2 θ = sec2 θ, , cot θ =, , cot 2 θ + 1 = cosec2 θ
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Essential formulae, Triangle formulae:, , Products of sines and cosines into sums or, differences:, , With reference to Fig. FA2:, Sine rule, , a, b, c, =, =, sin A sin B, sin C, , sin A cos B = 12 [sin(A + B) + sin(A − B)], cos A sin B = 12 [sin(A + B) − sin(A − B)], , a 2 = b2 + c2 − 2bc cos A, , Cosine rule, , 661, , cos A cos B = 12 [cos(A + B) + cos(A − B)], A, , sin A sin B = − 12 [cos(A + B)−cos(A − B)], c, , b, , B, , a, , Sums or differences of sines and cosines, into products:, , C, , , , Figure FA2, , Area of any triangle, (i), (ii), (iii), , 1, 2, , × base × perpendicular height, , 1, 2 ab sin C, , √, , , , x+y, x−y, cos, 2, 2, , , , x−y, x+y, sin, sin x − sin y = 2 cos, 2, 2, , , , x+y, x−y, cos x + cos y = 2 cos, cos, 2, 2, , , , x−y, x+y, sin, cos x − cos y = −2 sin, 2, 2, sin x + sin y = 2 sin, , or 21 ac sin B or 12 bc sin A, , [s(s − a)(s − b)(s − c)] where s =, , a +b+c, 2, , Compound angle formulae:, sin(A ± B) = sin A cos B ± cos A sin B, cos(A ± B) = cos A cos B ∓ sin A sin B, tan(A ± B) =, , tan A ± tan B, 1 ∓ tan A tan B, , If R sin (ωt + α) = a sin ωt +b cos ωt,, then a = R cos α, b = R sin α,, , b, R = (a 2 + b2 ) and α = tan −1, a, , For a general sinusoidal function, y = A sin(ωt ±α), then:, A = amplitude, ω = angular velocity = 2π f rad/s, 2π, = periodic time T seconds, ω, ω, = frequency, f hertz, 2π, α = angle of lead or lag (compared with, y = A sin ωt ), , Double angles:, sin 2 A = 2 sin A cos A, cos 2 A = cos2 A − sin 2 A = 2 cos2 A − 1, = 1 − 2 sin A, 2, , tan 2 A =, , 2 tan A, 1 − tan2 A, , Cartesian and polar co-ordinates:, , , If co-ordinate (x, y) = (r, θ) then r = x 2 + y 2 and, y, θ = tan−1, x, If co-ordinate (r, θ) = (x, y) then x =r cos θ and, y =r sin θ.
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662 Higher Engineering Mathematics, The circle:, With reference to Fig. FA3., Area = πr 2 Circumference = 2πr, π radians = 180◦, , Equation of an ellipse, centre at origin, semi-axes a, x 2 y2, and b:, + =1, a 2 b2, Equation of a hyperbola:, , x 2 y2, − =1, a 2 b2, , Equation of a rectangular hyperbola: x y = c2, s, , r, , Irregular areas:, , , , Trapezoidal rule, , r, , , , , width of 1 first + last, Area ≈, interval, 2 ordinates, , , sum of remaining, +, ordinates, , Figure FA3, , For sector of circle:, s = rθ (θ in rad), , Mid-ordinate rule, , , , width of, sum of, Area ≈, interval, mid-ordinates, , shaded area = 21 r 2 θ (θ in rad), Equation of a circle, centre at (a, b), radius r:, (x − a)2 + (y − b)2 = r 2, , Linear and angular velocity:, , Simpson’s rule, , , , 1 width of, first + last, Area ≈, ordinate, 3 interval, , If v = linear velocity (m/s), s = displacement (m),, t = time (s), n =speed of revolution (rev/s),, θ = angle (rad), ω = angular velocity (rad/s),, r = radius of circle (m) then:, v=, , θ, s, ω = = 2πn v = ωr, t, t, , centripetal force =, , mv 2, r, , where m = mass of rotating object., , Graphs, Equations of functions:, Equation of a straight line: y = mx + c, Equation of a parabola:, y = ax 2 + bx + c, Circle, centre (a, b), radius r:, (x − a)2 + (y − b)2 =r 2, , +4, , , , sum of even, ordinates, , +2, , , , sum of remaining, odd ordinates, , Vector Geometry, If a = a1i + a2 j+ a3 k and b = b1 i + b2 j+ b3 k, a · b = a1 b1 + a2 b2 + a3 b3, a·b, |a | = a12 + a22 + a32 cos θ =, |a| |b|, i, j k, a × b = a1 a2 a3, b1 b2 b3, , |a × b | = [(a · a)(b · b) − (a · b)2 ]
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664 Higher Engineering Mathematics, y or f (x), , dy, or f (x), dx, , cos−1 f (x), , , , tan−1, , x, a, , tan−1 f (x), sec−1, , x, a, , sinh−1 f (x), cosh−1, , −a, , f (x), , x, a, , x, a, , −a, a2 + x 2, , , , 1, x2, , + a2, , [, , √, , +1, , 1, x 2 − a2, f (x), , cosh−1 f (x), , , , x, tanh −1, a, , a, 2, a − x2, , tanh −1 f (x), , f (x), 1 − [ f (x)]2, , sech −1, , x, a, , sech −1 f (x), , [ f (x)]2 − 1, , √, , a, a2 − x 2, f (x), 1 − [ f (x)]2, , coth−1 f (x), , Product rule:, When y = uv and u and v are functions of x then:, dy, dv, du, =u +v, dx, dx, dx, , When y =, , u, and u and v are functions of x then:, v, du, dv, dy v dx − u dx, =, dx, v2, , Function of a function:, , f (x), f (x)]2, , − f (x), , f (x) [ f (x)]2 + 1, , Quotient rule:, , − f (x), 1 + [ f (x)]2, √, , x, a, , a, , f (x), , f (x) [ f (x)]2 − 1, , −a, , x x 2 + a2, , coth−1, , − f (x), , f (x) [ f (x)]2 − 1, , cot −1 f (x), , √, , f (x), 1 + [ f (x)]2, , √, x x 2 − a2, , x, a, , x, a, , cosech −1 f (x), , a, , cosec−1, , cosech −1, , a, a2 + x 2, , x x 2 − a2, , x, cosec−1, , sinh−1, , 1 − [ f (x)]2, , √, , sec−1 f (x), , cot −1, , − f (x), , dy, or f (x), dx, , y or f (x), , If u is a function of x then:, dy dy du, =, ×, dx du dx, , Parametric differentiation:, If x and y are both functions of θ, then:, , d dy, dy, dy, d2 y, dθ dx, =, = dθ and, 2, dx, dx, dx, dx, dθ, dθ, , −a, , x a2 − x 2, − f (x), , f (x) 1 − [ f (x)]2, , Implicit function:, d, d, dy, [ f (y)] = [ f (y)] ×, dx, dy, dx
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Essential formulae, Maximum and minimum values:, dy, = 0 for stationary points., dx, dy, Let a solution of, = 0 be x = a; if the value of, dx, 2, d y, when x = a is: positive, the point is a minimum,, dx 2, negative, the point is a maximum., If y = f (x) then, , Velocity and acceleration:, If distance x = f (t ), then, velocity, , v = f (t ) or, , acceleration a = f (t ) or, , d2 x, dt 2, , Tangents and normals:, Equation of tangent to curve y = f (x) at the point, (x 1 , y1) is:, y − y1 = m(x − x 1 ), where m = gradient of curve at (x 1, y1 )., Equation of normal to curve y = f (x) at the point, (x 1 , y1) is:, 1, y − y1 = − (x − x 1 ), m, , Partial differentiation:, Total differential, If z = f (u, v, . . .), then the total differential,, dz =, , Small changes, If z = f (u, v, . . .) and δx, δy, … denote small changes, in x, y, … respectively, then the corresponding change,, , δz ≈, , ∂z, ∂z, du + dv + . . . ., ∂u, ∂v, , Rate of change, du dv, If z = f (u, v, . . .) and, ,, , … denote the rate of, dt dt, change of u, v, … respectively, then the rate of change, of z,, dz, ∂z du ∂z dv, =, ·, +, ·, + ..., dt, ∂u dt, ∂v dt, , ∂z, ∂z, δx + δy + . . . ., ∂x, ∂y, , To determine maxima, minima and saddle points for, functions of two variables: Given z = f (x, y),, (i) determine, , dx, and, dt, , 665, , ∂z, ∂z, and, ∂x, ∂y, , (ii) for stationary points,, , ∂z, ∂z, = 0 and, = 0,, ∂x, ∂y, , ∂z, = 0 and, (iii) solve the simultaneous equations, ∂x, ∂z, = 0 for x and y, which gives the co-ordinates, ∂y, of the stationary points,, (iv) determine, , ∂2z, ∂2z ∂2 z, ,, and, ∂x 2 ∂ y 2, ∂x∂ y, , (v) for each of the co-ordinates of the stationary points, substitute values of x and y into, ∂2z, ∂2 z ∂2z, ,, and, and evaluate each,, ∂x 2 ∂ y 2, ∂x∂ y, , (vi) evaluate, , ∂2z, ∂x∂ y, , 2, for each stationary point,, , ∂2 z ∂2z, ∂2z, (vii) substitute the values of 2 , 2 and, into, ∂x ∂ y, ∂x∂ y, 2 2 2 2 , ∂ z, ∂ z, ∂ z, −, the equation =, ∂x∂ y, ∂x 2, ∂ y2, and evaluate,, (viii) (a) if > 0 then the stationary point is a saddle, point, ∂2z, < 0, then the stationary, ∂x2, point is a maximum point, and, , (b) if < 0 and, , ∂2z, > 0, then the stationary, (c) if < 0 and, ∂x2, point is a minimum point
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666 Higher Engineering Mathematics, , 1, (a 2 + x 2 ), , Standard integrals:, , , y, , , , y dx, , 1, , 1, sin ax + c, a, , sin ax, , 1, − cos ax + c, a, , sec2 ax, cosec 2 ax, , ln, , , sec ax tan ax, , 1, sec ax + c, a, , eax, , 1 ax, e +c, a, , 1, x, , ln x + c, , cos2 x, sin2 x, , 1, ln(sec ax) + c, a, , , sin 2x, 1, x+, +c, 2, 2, , , sin 2x, 1, x−, +c, 2, 2, , tan 2 x, , tan x − x + c, , cot 2 x, , −cot x − x + c, , 1, (a 2 − x 2 ), , (a 2 − x 2 ), , cosh−1, , ln, , , 1, cosec ax cot ax − cosec ax + c, a, , tan ax, , 1, (x 2 − a 2 ), , 1, tan ax + c, a, 1, − cot ax + c, a, , sin−1, , x, +c, a, , a 2 −1 x x 2, (a − x 2 ) + c, sin, +, 2, a 2, , x+, , , , (x 2 + a 2 ), +c, a, , a2, x x 2, (x + a 2 ) + c, sinh−1 +, 2, a 2, , (x 2 + a 2 ), , , , x, + c or, a, , sinh−1, , (x 2 + a 2 ), , a, , cos ax, , , , y dx, , 1 −1 x, tan, +c, a, a, , x n+1, +c, n +1, (except where n = −1), , ax n, , , , , , y, , Integral Calculus, , x, + c or, a, , x+, , , , (x 2 − a 2 ), +c, a, , x 2, x, a2, (x − a 2 ) − cosh−1 + c, 2, 1, a, , (x 2 − a 2 ), , θ, t = tan substitution, 2, To determine, , , , sin θ =, dθ =, , 1, dθ let, a cos θ + b sin θ + c, 1 − t2, 2t, and, cos, θ, =, (1 + t 2 ), 1 + t2, 2 dt, (1 + t 2 ), , Integration by parts:, If u and v are both functions of x then:, !, !, dv, du, u dx = uv − v, dx, dx, dx, , Reduction formulae:, !, , x n ex dx = In = x n ex − n In−1, !, x n cos x dx = In = x n sin x + nx n−1 cos x, −n(n − 1)In−2
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Essential formulae, !, !, , π, , x n cos x dx = In = −nπ n−1 − n(n − 1)In−2, , 0, , 667, , Centroids:, With reference to Fig. FA5:, , n, , n, , x sin x dx = In = −x cos x + nx, , n−1, , sin x, , !, , −n(n − 1)In−2, , !, , !, , b, , x y dx, , 1, n −1, sinn x dx = In = − sinn−1 x cos x +, In−2, n, n, !, 1, n −1, cosn x dx = In = cosn−1 sin x +, In−2, n, n, ! π/2, ! π/2, n −1, n, sin x dx =, cosn x dx = In =, In−2, n, 0, 0, !, tann−1 x, tann x dx = In =, − In−2, n −1, !, (ln x)n dx = In = x(ln x)n − n In−1, , a, , x̄ = !, , b, , y 2 dx, , a, ! b, , y dx, , y dx, , a, , a, , y, y 5 f(x), Area A, C, , x, 0, , With reference to Fig. FA4., , and ȳ =, , b, , 1, 2, , y, x5a, , x5b, , x, , Figure FA5, , y, , Theorem of Pappus:, , y 5 f(x), , With reference to Fig. FA5, when the curve is rotated one, revolution about the x-axis between the limits x = a and, x = b, the volume V generated is given by: V = 2πAȳ., , A, , 0, , x5a, , x5b, , Parallel axis theorem:, , x, , If C is the centroid of area A in Fig. FA6 then, , Figure FA4, , Area under a curve:, , 2, 2, + Ad 2 or k 2B B = kGG, + d2, Ak 2B B = AkGG, , !, , b, , area A =, , y dx, a, , G, , Mean value:, mean value =, , 1, b−a, , !, , B, , b, , y dx, a, , C, , R.m.s. value:, , Area A, , , r.m.s. value =, , 1, b−a, , !, , , , b, , d, , y 2 dx, a, , G, , Volume of solid of revolution:, !, , b, , volume =, a, , π y 2 dx about the x-axis, , Figure FA6, , B
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668 Higher Engineering Mathematics, Second moment of area and radius of gyration:, Shape, , Position of axis, , Rectangle, length l, breadth b, , (1) Coinciding with b, , Second moment, of area, I, , (2) Coinciding with l, (3) Through centroid,, parallel to b, (4) Through centroid,, parallel to l, , Triangle, (1) Coinciding with b, Perpendicular, height h, (2) Through centroid,, base b, parallel to base, (3) Through vertex,, parallel to base, , Circle, radius r, , (1) Through centre,, perpendicular to plane, (i.e. polar axis), (2) Coinciding with diameter, (3) About a tangent, , Semicircle, , Coinciding with, , radius r, , diameter, , Perpendicular axis theorem:, then, , =, , Ak 2O X, , +, , Ak 2OY, , or, , k 2O Z, , bl 3, 3, lb 3, 3, bl 3, 12, , 1, √, 3, b, √, 3, 1, √, 12, , lb 3, 12, , b, √, 12, , bh 3, 12, bh 3, 36, , h, √, 6, h, √, 18, , bh 3, 4, , h, √, 2, , πr 4, 2, , r, √, , πr 4, 4, 5πr 4, 4, , r, √2, 5, r, 2, , πr 4, 8, , r, 2, , 2, , Numerical integration:, , If OX and OY lie in the plane of area A in Fig. FA7,, Ak 2O Z, , Radius of, gyration, k, , =, , k 2O X, , Z, , + k 2OY, , Trapezoidal rule, !, , , , , , 1 first + last, width of, ydx ≈, interval, 2 ordinates, , , Area A, , +, O, , ordinates, , Mid-ordinate rule, , X, Y, , Figure FA7, , sum of remaining, , !, , , , , width of, sum of, ydx ≈, interval, mid-ordinates, ,
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Essential formulae, Linear first order:, , Simpson’s rule, !, , 669, , , , , 1 width of, first + last, ydx ≈, ordinate, 3 interval, , , sum of even, +4, ordinates, , , sum of remaining, +2, odd ordinates, , dy, If, + P y = Q, where P and Q are functions of x, dx, only (i.e. a linear first order differential equation), then, , , (i) determine the integrating factor, e, , P dx, , (ii) substitute the integrating factor (I.F.) into, the equation, !, y (I.F.) = (I.F.) Q dx, , (iii) determine the integral (I.F.) Q dx, , Differential Equations, Numerical solutions of first order, differential equations:, , First order differential equations:, Separation of variables, !, dy, If, = f (x) then y =, f (x) dx, dx, !, !, dy, dy, = f (y) then, dx =, If, dx, f (y), dy, If, = f (x) · f (y) then, dx, , !, , dy, =, f (y), , Euler’s method:, y1 = y0 + h(y )0, Euler-Cauchy method: y P1 = y0 + h(y )0, , !, f (x) dx, , Homogeneous equations:, dy, If P, = Q, where P and Q are functions of both x and, dx, y of the same degree throughout (i.e. a homogeneous, first order differential equation) then:, (i) Rearrange P, , 1, yC1 = y0 + h[(y )0 + f (x 1 , y p1 )], 2, Runge-Kutta method:, dy, To solve the differential equation, = f (x, y) given, dx, the initial condition y = y0 at x = x 0 for a range of, values of x = x 0(h)x n :, and, , dy, dy Q, = Q into the form, =, dx, dx P, , (ii) Make the substitution y = vx (where v is a function of x), from which, by the product rule,, dy, dv, = v(1) + x, dx, dx, dy, in the equation, (iii) Substitute for both y and, dx, dy Q, =, dx P, (iv) Simplify, by cancelling, and then separate the, dy, variables and solve using the, = f (x) · f (y), dx, method, y, (v) Substitute v = to solve in terms of the original, x, variables., , 1. Identify x 0 , y0 and h, and values of x 1 , x 2 , x 3 , . . ., 2. Evaluate k1 = f (x n , yn ) starting with n = 0, , , h, h, 3. Evaluate k2 = f x n + , yn + k1, 2, 2, , , h, h, 4. Evaluate k3 = f x n + , yn + k2, 2, 2, 5. Evaluate k4 = f(x n + h, yn + hk3 ), 6. Use the values determined from steps 2 to 5 to, evaluate:, h, yn+1 = yn + {k1 + 2k2 + 2k3 + k4 }, 6, 7. Repeat steps 2 to 6 for n = 1, 2, 3, . . ., , Second order differential equations:, d2 y, dy, If a 2 + b, + cy = 0 (where a, b and c are condx, dx, stants) then:, (i) rewrite the differential equation as, (aD2 + bD +c)y = 0, (ii) substitute m for D and solve the auxiliary equation, am 2 + bm + c = 0
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670 Higher Engineering Mathematics, (iii) if the roots of the auxiliary equation are:, (a) real and different, say m = α and m = β, then the general solution is, , y, , y(n), , cos ax, , , nπ , a n cos ax +, 2, , xa, , a!, x a−n, (a − n)!, , y = Aeαx + Beβx, (b) real and equal, say m = α twice, then the, general solution is, y = (Ax + B) eαx, (c), , sinh ax, , complex, say m = α ± jβ, then the general, solution is, y = eαx (A cosβx + B sin βx), , cosh ax, , (iv) given boundary conditions, constants A and B, can be determined and the particular solution, obtained., ln ax, d2 y, dy, + b + cy = f (x) then:, dx2, dx, (i) rewrite the differential equation as, (aD2 + bD +c)y = 0., , , an :, 1 + (−1)n sinh ax, 2, , 4, + 1 − (−1)n cosh ax, , an :, 1 − (−1)n sinh ax, 2, , 4, + 1 + (−1)n cosh ax, (−1)n−1, , (n − 1)!, xn, , If a, , (ii) substitute m for D and solve the auxiliary equation, am 2 + bm + c = 0., , Leibniz’s theorem:, To find the n’th derivative of a product y = uv:, y (n) = (uv)(n) = u (n) v + nu (n−1) v (1), , (iii) obtain the complimentary function (C.F.), u, as, per (iii) above., (iv) to find the particular integral, v, first assume a particular integral which is suggested by f (x), but, which contains undetermined coefficients (See, Table 51.1, page 484 for guidance)., (v) substitute the suggested particular integral into, the original differential equation and equate, relevant coefficients to find the constants, introduced., (vi) the general solution is given by y = u +v., , +, , n(n − 1) (n−2) (2), v, u, 2!, , +, , n(n − 1)(n − 2) (n−3) (3), v +···, u, 3!, , Power series solutions of second order, differential equations:, (a), , Leibniz-Maclaurin method, , (i) Differentiate the given equation n times,, using the Leibniz theorem,, (ii) rearrange the result to obtain the recurrence, relation at x = 0,, , (vii) given boundary conditions, arbitrary constants in, the C.F. can be determined and the particular, solution obtained., , (iii) determine the values of the derivatives at, x = 0, i.e. find (y)0 and (y )0 ,, , Higher derivatives:, , (iv) substitute in the Maclaurin expansion for, y = f (x),, , y, , y(n), , eax, , a n eax, , nπ , a n sin ax +, 2, , sin ax, , (v) simplify the result where possible and apply, boundary condition (if given)., (b) Frobenius method, (i) Assume a trial solution of the form:, y = x c {a0 + a1 x + a2 x 2 + a3 x 3 + · · · +, a0 = 0,, ar x r + · · ·}
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671, , Essential formulae, (ii) differentiate the trial series to find y, and y ,, (iii) substitute the results in the given differential, equation,, (iv) equate coefficients of corresponding powers, of the variable on each side of the equation: this enables index c and coefficients, a1 , a2, a3, . . . from the trial solution, to be, determined., , and in particular:, x n 1, x 2, 1, Jn (x) =, −, 2, n! (n + 1)! 2, , x 4, 1, +, −···, (2! )(n + 2)! 2, J0 (x) = 1 −, , x2, x4, +, 22 (1! )2 24 (2! )2, −, , Bessel’s equation:, The solution of x 2, , y = Ax v 1 −, +, , d2 y, dx 2, , +x, , and J1 (x) =, , dy, + (x 2 − v 2 )y = 0 is:, dx, , −, , x, 22 (v + 1), x4, , or, in terms of Bessel functions and gamma functions:, y = AJv (x) + B J−v (x), x v , 1, x2, − 2, =A, 2, (v + 1) 2 (1! )(v + 2), +, x −v , 2, , x4, −···, 24 (2! )(v + 4), , In general terms:, , and J−v (x) =, , (−1)k x 2k, + k + 1), , 22k (k! )(v, , ∞, x −v <, , 2, , is:, , , k(k + 1) 2, x, y = a0 1 −, 2!, k(k + 1)(k − 2)(k + 3) 4, +, x −···, 4!, , , (k − 1)(k + 2) 3, + a1 x −, x, 3!, , , , , , Rodrigue’s formula:, Pn (x) =, , 1 d n (x 2 − 1)n, dxn, , 2n n!, , , , Statistics and Probability, Mean, median, mode and standard, deviation:, , ∞, x v <, k=0, , d2 y, dy, − 2x, + k(k + 1)y = 0, 2, dx, dx, , , , 1, x2, − 2, (1 − v) 2 (1! )(2 − v), , 2, , The solution of (1 − x 2 ), , (k − 1)(k − 3)(k + 2)(k + 4) 5, +, x −···, 5!, , x4, + 4, −···, 2 (2! )(3 − v), , Jv (x) =, , x7, +···, 27(3! )(4! ), , Legendre’s equation:, , + 2), , , x6, +···, − 6, 2 × 3! (v + 1)(v + 2)(v + 3), , x4, x2, −v, + Bx, + 4, 1+ 2, 2 (v − 1) 2 × 2! (v − 1)(v − 2), , x6, + 6, +···, 2 × 3! (v − 1)(v − 2)(v − 3), , +B, , x, x3, x5, − 3, + 5, 2 2 (1! )(2! ) 2 (2! )(3! ), , 2, , 24 × 2! (v + 1)(v, , x6, +···, 26 (3! )2, , k=0, , (−1)k x 2k, 22k (k! )(k − v + 1), , If x = variate and f = frequency then:, ;, fx, mean x̄ = ;, f, The median is the middle term of a ranked set of data.
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672 Higher Engineering Mathematics, The mode is the most commonly occurring value in a, set of data., , Standard deviation:, 7, 4, 8 ;:, 8, f (x − x̄)2, 9, ;, for a population, σ=, f, , Binomial probability distribution:, If n =number in sample, p =probability of the occurrence of an event and q = 1 − p, then the probability of, 0, 1, 2, 3, . . . occurrences is given by:, n(n − 1) n−2 2, p ,, q, 2!, n(n − 1)(n − 2) n−3 3, p ,..., q, 3!, , Normal approximation to a binomial, √ distribution:, Mean = np Standard deviation σ = (npq), , Poisson distribution:, If λ is the expectation of the occurrence of an event then, the probability of 0, 1, 2, 3, . . . occurrences is given by:, e−λ, e−λ, e−λ , λe−λ , λ2, , λ3, ,..., 2!, 3!, , x2, , Percentile values (χ 2p ) for the Chi-square distribution, with ν degrees of freedom—see Table 76.1, page 60, on, the website., , , ; (o − e)2, 2, where o and e are the observed and, χ =, e, expected frequencies., , y2, , Sample, number of members N , mean x, standard deviation s., Sampling distributions, mean of sampling distribution of means μx, standard error of means σx, standard error of the standard deviations σs ., , Standard error of the means:, Standard error of the means of a sample distribution, i.e., the standard deviation of the means of samples, is:, , Product-moment formula for the linear, correlation coefficient:, xy, ;, , Chi-square distribution:, , Population, number of members N p , mean μ, standard deviation σ ., , (i.e. successive terms of the (q + p)n expansion)., , ;, , Percentile values (t p ) for Student’s t distribution with ν, degrees of freedom — see Table 74.2, page 38, on the, website., , Symbols:, , q n , nq n−1 p,, , Coefficient of correlation r = - ;, , Student’s t distribution:, , , , σ, σx = √, N, , , , Np − N, Np − 1, , , , for a finite population and/or for sampling without, replacement, and, σ, σx = √, N, , where x = X − X and y = Y − Y and (X 1 , Y1),, (X 2 , Y2 ), . . . denote a random sample from a bivariate, normal distribution and X and Y are the means of the, X and Y values respectively., , for an infinite population and/or for sampling with, replacement., , Normal probability distribution:, , The relationship between sample mean, and population mean:, , Partial areas under the standardized normal curve — see, Table 58.1 on page 564., , μx = μ for all possible samples of size N are drawn, from a population of size N p .
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Essential formulae, Estimating the mean of a population, (σ known):, , page 33, on the website. The confidence limits of a, population mean based on sample data is given by:, tc s, x±√, (N − 1), , The confidence coefficient for a large sample size,, (N ≥ 30) is z c where:, Confidence Confidence, level %, coefficient z c, , Laplace Transforms, , 99, , 2.58, , 98, , 2.33, , Function, , 96, , 2.05, , f (t ), , 95, , 1.96, , 90, , 1.645, , 80, , 1.28, , 50, , 0.6745, , zcσ, x±√, N, , , , Np − N, Np − 1, , 1, s, , k, , k, s, 1, s−a, , sin at, , a, s 2 +a 2, , cos at, , s, s 2 +a 2, , , , for a finite population of size N p , and by, zcσ, x ± √ for an infinite population, N, , Laplace transforms, ∞, L{ f (t )} = 0 e−st f (t ) dt, , 1, , eat, The confidence limits of a population mean based on, sample data are given by:, , t, t n (n = positve integer), , 1, s2, n!, s n+1, , Estimating the mean of a population, (σ unknown):, , cosh at, , s, s 2 −a 2, , The confidence limits of a population mean based on, sample data are given by: μx ± z c σx ., , sinh at, , a, s 2 −a 2, , e−at t n, , n!, (s+a)n+1, , Estimating the standard deviation of a, population:, The confidence limits of the standard deviation of a population based on sample data are given by:, s ± z c σs ., , Estimating the mean of a population based, on a small sample size:, The confidence coefficient for a small sample size, (N < 30) is tc which can be determined using Table 74.1,, , 673, , e−at sin ωt, , ω, (s+a)2 +ω 2, , e−at cos ωt, , s+a, (s+a)2 +ω 2, , e−at cosh ωt, , s+a, (s+a)2 −ω 2, , e−at sinh ωt, , ω, (s+a)2 −ω 2
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Index, Adding alternating waveforms, 265, Adjoint of matrix, 239, Algebra, 1–12, Algebraic method of successive, approximations, 81–84, substitution, integration, 392–395, Amplitude, 140, 143, Angle between two vectors, 276, of any magnitude, 135–137, of depression, 106–108, of elevation, 106–108, Angular velocity, 129, Applications of complex numbers, 221, differentiation, 299, rates of change, 299–300, small changes, 312–313, tangents and normals, 311–312, turning points, 303–307, velocity and acceleration,, 300–303, integration, 375, areas, 375–376, centroids, 380–381, mean value, 377–378, r.m.s. value, 377–378, second moment of area, 383–391, volumes, 378–379, Arc, 122, length, 124, Area of circle, 124, sector, 125, triangle, 108–111, irregular figures, 203–205, under curve, 375–376, Argand diagram, 214, Argument, 219, Arithmetic mean, 541, progression, 51, Astroid, 315, Asymptotes, 190–196, Auxiliary equation, 477, Average, 541, value of waveform, 206–210, Base, 87, Bessel functions, 508, Bessel’s equation, 506–511, , Binary addition, 89, numbers, 87–90, Binomial distribution, 556–559, expression, 58, series/theorem, 59–66, practical problems, 64–66, Bisection method, 77–81, Bits, 87, Boundary conditions, 445, 516, Brackets, 2, Calculus, 287, Cardioid, 315, Cartesian complex numbers, 213–218, co-ordinates, 117–120, Catenary, 43, Centre of area, 380, gravity, 380, mass, 380, Centripetal acceleration, 131, force, 130–132, Centroids, 380–381, Chain rule, 295, Change of limits, 395–396, Chord, 122, Circle, 122, 179, area, 124, equation of, 127–129, 179, Circumference, 122, Class interval, 534, Coefficient of correlation, 570–571, Cofactor, 237, Combination of periodic functions,, 265–274, Common difference, 51, logarithms, 20, ratio, 54, Complementary function, 483, Complex numbers, 213–224, applications of, 221–224, Cartesian form, 213, coefficients, 645–649, conjugate, 216, equations, 217–218, exponential form, 228–230, form of Fourier series, 644–649, polar form, 218–221, , powers of, 225–226, roots of, 226–228, Complex wave, 146–151, considerations, 641–643, Compound angles, 163–176, Computer numbering systems, 87, Conditional probability, 548, Continuous data, 529, function, 186, 611, Contour map, 359, Conversion of a sin ωt + b cosωt into, R sin(ωt + α), 165–169, Correlation, linear, 570–574, Cosecant, 98, Cosh, 41, series, 49–50, Cosh θ substitution, 406–408, Coshec, 41, Cosine, 98, curves, 138–143, rule, 108, 268–270, wave production, 137–138, Cotangent, 99, Coth, 41, Cramer’s rule, 247–248, Cross product, 280, Cubic equations, 9, 178, Cumulative frequency distribution, 535, Curve sketching, 196–202, Cycloid, 315, , Deciles, 546–547, Decimal numbers, 87, Definite integrals, 372–374, Degree of differential equation, 445, De Moivre’s theorem, 225–230, Dependent event, 548, Depression, angle of, 106–108, Derivatives, 288, Laplace transforms of, 589, Determinant, 235, 237–239, to solve simultaneous equations,, 241–244, Determination of law, 37, Diameter, 122
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676 Index, Differential coefficient, 288, Differential equations, 445, a, , dy, d2 y, + cy = 0 type,, +b, dx, dx 2, 477–482, , a, , d2 y, dy, +b, + cy = f(x) type,, dx 2, dx, 483–492, , dy, = f(x) type, 445–447, dx, dy, = f(y) type, 447–449, dx, dy, = f(x) · f(y) type, 449–451, dx, dy, + P y = Q type, 456–460, dx, degree of, 445, first order, separation of variables,, 444, homogeneous first order, 452–455, numerical methods, 461, partial, 515, power series method, 493, simultaneous, using Laplace, transforms, 605–609, using Laplace transforms, 600–604, Differentiation, 68, 287, applications, 299–314, from first principles, 288, function of a function, 295–296, 320, implicit, 320–324, inverse hyperbolic function,, 341–344, trigonometric function, 334–339, logarithmic, 325–329, methods of, 287–298, of common functions, 289–292, of hyperbolic functions, 331–332, of parametric equations, 315–319, partial, 345, first order, 345–348, second order, 348–350, product, 292–293, quotient, 293–295, successive, 296–298, Direction cosines, 278, Discontinuous function, 186, Discrete data, 529, Dividend, 7, Divisor, 7, D-operator form, 477, , Dot product, 276, Double angles, 45, 169–170, Elastic string, 519, Elevation, angle of, 106–108, Ellipse, 179, 199, 315, Equations, 3, Bessel’s, 506–511, circle, 126, complex, 217–218, heat conduction, 518, 523–525, hyperbolic, 47–48, indicial, 24–25, 501, 503, 507, Laplace, 515, 517, 518, 525–527, Legendre’s, 511–513, Newton-Raphson, 84, normal, 575, of circle, 127–129, 179, quadratic, 5–6, simple, 3, simultaneous, 4–5, 241–247, solving by iterative methods, 77–86, tangents, 311–312, transmission, 518, trigonometric, 154–158, wave, 518–523, Euler-Cauchy method, 466, Euler’s formula, 653, Euler’s method, 461–470, Even function, 42, 186–188, 623, Expectation, 548, Exponential form of complex number,, 228–230, Fourier series, 644, Exponential function, 27–39, graphs of, 29–31, power series, 28–29, Extrapolation, 576, Factorisation, 2, Factor theorem, 8–10, Family of curves, 444, Final value theorem, 591–592, First moment of area, 383, Formulae, 4, Fourier coefficients, 612, Fourier series, 146, cosine, 623–626, exponential form, 645, half-range, 626–629, 634, non-periodic over range 2π,, 617–622, over any range, 630–636, periodic of period 2π, 611–616, sine, 623–626, , Frequency, 143, 529, curve, 562, distribution, 534, 538, domain, 652, polygon, 535, 538, relative, 529, spectrum, 652–653, Frobenius method, 500–506, Functional notation, 288, Function of a function, 295–296, 320, Functions of two variables, 357–366, Fundamental, 612, Gamma function, 508, Gaussian elimination, 248–249, General solution of a differential, equation, 445, 447, Geometric progression, 54–57, Gradient of a curve, 287–288, Graphs of exponential functions, 29–31, hyperbolic functions, 43–44, logarithmic function, 25–26, trigonometric functions, 134, Grouped data, 534–539, 545, Growth and decay laws, 34–37, Half range Fourier series, 624–629,, 634, Half-wave rectifier, 148, Harmonic analysis, 146, 637–643, Harmonic synthesis, 146–151, Heat conduction equation, 518,, 523–525, Hexadecimal number, 92–95, Higher order differentials, 493–495, Histogram, 535, 538, 543, of probabilities, 558, 560, Homogeneous, 452, 477, Homogeneous first order differential, equations, 452–460, Horizontal bar chart, 530, component, 254, 270, Hyperbola, 180, rectangular, 180, 199, 315, Hyperbolic functions, 41–50, 159, differentiation of, 331–332, graphs of, 43–44, inverse, 334–344, solving equations, 47–48, Hyperbolic identities, 45–47, 160–161, logarithms, 20, 31–34, 325, Hypotenuse, 97, Identities, hyperbolic, 45–47, 160–161, trigonometric, 45, 152–154, i, j,k notation, 263
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Index, Imaginary part, 213, Implicit differentiation, 320–324, Implicit function, 320, Independent event, 548, Indices, laws of, 1, 2, Indicial equations, 24–25, 501, 503,, 507, Industrial inspection, 557–558, Initial conditions, 516, Initial value theorem, 591, Integrating factor, 456, Integration, 368, algebraic substitution, 392–396, applications of, 375–391, areas, 375–376, centroids, 380–381, mean value, 377–378, r.m.s. value, 377–378, second moment of area, 383–391, t = tan θ/2 substitution, 414, volumes, 378–379, by partial fractions, 409–413, by parts, 420–425, change of limits, 395–396, coshθ substitution, 406–408, definite, 372–374, hyperbolic substitutions, 399,, 404–408, numerical, 73–74, 435–442, reduction formulae, 426–434, sineθ substitution, 402–404, sinh θ substitution, 404–406, standard, 368–374, tan θ substitution, 404, t = tan θ/2 substitution, 414–418, trigonometric substitutions,, 398–404, Interpolation, 576, Inverse functions, 101, 188–190, 334, hyperbolic, 334, differentiation of, 341–344, trigonometric, 189, 334, differentiation of, 334–339, Inverse Laplace transforms, 593–597, using partial fractions, 596–597, Inverse matrix, 236, Irregular areas, 203, volumes, 205, Iterative methods, 77, Lagging angle, 140, Lamina, 380, Laplace’s equation, 515, 517, 518,, 525–527, Laplace transforms, 582–586, common notations, 582, , definition, 582, derivatives, 589–591, for differential equations, 600–604, for simultaneous differential, equations, 605–609, inverse, 593–597, using partial fractions, 596–597, linearity property, 582, of elementary functions, 582–585, properties of, 587, Laws of, growth and decay, 34–37, indices, 1, 2, logarithms, 22–24, 325, probability, 548, Leading angle, 140, Least-squares regression lines,, 575–580, Leibniz, notation, 288, theorem, 495–497, Leibniz-Maclaurin method, 497–500, Legendre polynomials, 512–514, Legendre’s equation, 511–514, L’Hopital’s rule, 75, Limiting values, 74–76, Linear, correlation, 570–574, first order differential equation,, 456–460, regression, 575–580, second order differential equation,, 477, velocity, 129–130, Logarithmic, differentiation, 325–329, forms of inverse hyperbolic, functions, 339–340, scale, 37, Logarithms, 20–26, graphs of, 25–26, laws of, 22–24, 325, Log-linear graph paper, 37, Log-log graph paper, 37, Lower class boundary value, 534, Maclaurin’s series/theorem, 68–76, numerical integration, 73–74, Matrices, 231–240, adjoint, 239, determinant of, 235–236, 237–239, inverse, 236, 239–240, reciprocal, 236, 239–240, to solve simultaneous equations,, 241–247, , 677, , transpose, 239, unit, 235, 236, Maximum point, 303, practical problems, 307–311, Mean value, 377–378, 541–543, 562, of waveform, 206–211, Measures of central tendency, 541, 544, Median, 541–543, Mid-ordinate rule, 203–204, 437–439, Minimum point, 303, practical problems, 307–311, Mode, 541–543, Modulus, 218, 277, Moment of a force, 282, Napierian logarithms, 20, 31–34, 325, Natural logarithms, 20, 31–34, 325, Newton-Raphson method, 84–86, Non-homogeneous differential, equation, 477, Non-right angled triangles, 108, Norm, 277, Normal, 311–312, distribution, 562–569, equations, 575, probability curve, 562, probability paper, 566, Nose-to-tail method, 252, Numerical integration, 73–74, 435–442, methods for first order differential, equations, 461, Numerical methods, 146, for first order differential equations,, 461–475, of harmonic analysis, 637–643, Octal numbers, 87, 90–92, Odd function, 41, 43, 187–188, 623,, 641, 649, Ogive, 535, 539, 546, Order of precedence, 2, Osborne’s rule, 45, 46, 161, Pappus theorem, 381–383, Parabola, 178, 197, 315, 439, Parallel axis theorem, 384–385, Parallelogram method, 252, Parameter, 315, Parametric equations, 315–319, Partial differential equations, 515, Partial differentiation, 345–350, equations, 515–527, Partial integration, 515, Partial fractions, 13, 409, inverse Laplace transforms,, 596–597
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678 Index, integration, using, 409–413, linear factors, 13–15, 409–411, quadratic factors, 16–17, 412–413, repeated linear factors, 17–19,, 411–412, Particular solution of differential, equation, 445, 478, Particular integral, 483, 484, Pascal’s triangle, 58–59, Percentage component bar chart, 530, Percentile, 546–547, Period, 139, Periodic function, 139, 186, 611, 630, combination of, 265–274, Periodic time, 143, 144, Perpendicular axis theorem, 385, Phasor, 143, 221, 267–274, 653–657, Pictogram, 530, Pie diagram, 530, 532, Planimeter, 203, Point of inflexion, 304, Poisson distribution, 559–561, Polar, co-ordinates, 117–121, curves, 180, form, 120, 121, 213, 218–221, 228, Poles, 598–599, Pole-zero diagram, 599, Pol/Rec function, 120, 220, Polynomial division, 6–8, Polynomial, Legendre’s, 512–514, Population, 529, Power series for e x , 28–29, cosh x and sinh x, 49–50, Power series methods of solving, differential equations, 493–514, by Frobenius’s method, 500–506, by Leibniz-Maclaurin method,, 497–500, Power waveforms, 173–176, Powers of complex numbers, 225–226, Practical trigonometry, 111–116, Precedence, 2, 3, Principal value, 219, Probability, 548–553, laws of, 549, paper, 566, Product rule of differentiation,, 292–293, Product-moment formula, 570–573, Pythagoras, theorem of, 97–98, Quadrant, 122, Quadratic equations, 5–6, graphs, 178, , Quartiles, 546, Quotient rule of differentiation,, 293–294, Radian, 123, 144, Radius, 122, of curvature, 318, 319, of gyration, 384, Radix, 87, Rates of change, 299–300, 352–354, Reciprocal matrix, 236–237, 239–240, ratios, 99, Rectangular hyperbola, 180, Recurrence formula, 498, relation, 498, 507, Reduction formulae, 426–434, of exponential laws to linear form,, 37–39, Regression, coefficients, 575, linear, 575, Relation between trigonometric and, hyperbolic functions, 159–162, Relative frequency, 529, velocity, 262–263, Remainder theorem, 10–12, Resolution of vectors, 254, Resultant phasor by complex numbers,, 272, horizontal and vertical components,, 270, phasor diagrams, 267, plotting, 265, sine and cosine rules, 268, Right-angled triangles, 105–108, R.m.s. values, 377–378, Rodrigue’s formula, 513, Roots of complex numbers, 226–228, Runge-Kutta method, 471–476, Saddle point, 357–366, Sample, 529, Scalar product, 276–280, application of, 279, Scalar quantity, 251, Scatter diagram, 570, 578, Secant, 99, Sech, 41, Second moment of area, 383–391, Second order differential equations,, 445, 477–492, Sector, 122, area of, 124–127, Segment, 122, Semicircle, 122, Semi-interquartile range, 546, Separation of variables, 445, , Series, binomial, 59–66, exponential, 28–29, Maclaurin’s, 68–74, sinh and cosh, 49–50, Set, 529, Simple equations, 3, Simpson’s rule, 204, 439–442, Simultaneous differential equations by, Laplace transforms, 605–609, Simultaneous equations, 4–5, by Cramers rule, 245, by determinants, 243–247, by Gaussian elimination, 246–247, by matrices, 241–243, Sine, 98, curves, 138–143, rule, 108, wave, 207, wave production, 137–138, Sine θ substitution, 402–403, Sinh, 41, series, 49–50, Sinh θ substitution, 404–406, Sinusoidal form, A sin(ωt ± α),, 143–145, Small changes, 312–314, 354–356, Solution of any triangle, 109–116, right-angled triangles, 105–108, Space diagram, 262, Square wave, 146, Spectrum of waveform, 652–653, Standard curves, 178–181, derivatives, 290, deviation, 544–546, integration, 368, Stationary points, 304, Statistical tables, normal curve, 564, Straight line, 178, Sum to infinity, 55, Successive differentiation, 296–298, Symmetry relationships, 649–652, Tables, statistical, normal curve, 564, Tally diagram, 534, 535, 537, Tangent, 98, 311–312, Tangential velocity, 282, Tanh, 41, Tanθ substitution, 404, Taylor’s series, 462, Testing for a normal distribution,, 566–569, Theorems, binomial, 59–66, Maclaurin’s, 68–76, Pappus, 381–383
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Index, parallel axis, 384–385, perpendicular axis, 385, Pythagoras, 97–98, Total differential, 351–352, Transfer function, 598, Transformations, 181–186, Transmission equation, 518, Transposition of formulae, 4, Trapezoidal rule, 203, 204, 435–437, Trial solution, 519, Triangle, area of, 108–111, Trigonometric ratios, 98–100, equations, 154–158, evaluation of, 100–105, functions, 134, and hyperbolic substitutions,, integration, 398–408, identities, 152–154, , inverse function, 189, 334, waveforms, 134–151, Trigonometry, 97, practical situations, 111–116, t = tan θ/2 substitution, 414–418, Turning points, 303–307, Ungrouped data, 530–534, Unit matrix, 235, Unit triad, 275, Upper class boundary value, 534, Vector addition, 255–260, nose-to-tail method, 252, 253, parallelogram method, 252–253, Vector drawing, 251, Vector equation of a line, 283–285, , Vector products, 280–283, applications of, 282, Vector quantities, 251, Vector subtraction, 260–262, Vectors, 251, Velocity and acceleration, 300–303, Vertical bar chart, 530, component, 254, 270, Volumes, of irregular solids, 205–206, of solids of revolution, 378–379, Wallis’s formula, 432, Wave equation, 519–523, Waveform analyser, 146, Work done, 279, Zeros (and poles), 598, , 679
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