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Calculus, - - - - - - - - - - - - - - ' - - - - - - - Fifth Edition, , Frank Ayres, Jr., PhD, . Formerly Professor and Head of the Department of Mathematics, Dickinson College, , Elliott Mende/son, PhD, Professor of Mathematics, Queens College, , Schaum's Outline Series, , New York Chicago San Frnncisco Lisbon London, Madrid Mexico City Milan New Delhi San Juan, Seoul Singapore Sydney Toronto
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Til, McGlOw Htff,, , 1'''1' 1/I1l", , FRANK AYRES, Jr., PhD, was fonnerly Professor and Head of the Department at Dickinson College, Carlisle., Pennsylvania. He is the coauthor of Schaum's Outline ofTrigorwmetry and Schaum's Outline of College Mathematics., ELLIOTT MENDELSON, PhQ, is Professor of Mathematics at Queens College. He is the author of Scliaum 's, Outline of Begin"ing Calculus., Schaum's Outline of CALCULUS, Copyright e 2009, 1999, 1990, 1962 by The McGraw-Hill Companies, Inc. All rights reserved. Printed in the, United States of America. Except as permitted under the Copyright Act of 1976. no part of this publication may, be reproduced or distributed in any fonns or by any means, or stored in a data base or retrieval system, without, the prior written pennission of the publisher., 4567891011CUSCUS0143210, MHID 0-07-150861-9, ISBN 978-0-07-150861-2, Sponsoring Editor: Charles Wall, Production Supervisor: Tama Harris McPhatter, Editing Supervisor: Maureen B. Walker, Interior Designer: Jane Tenenbaum, Project Manager: Madhu Bhardwaj, Library of Conl~ress Cataloging-in-Publication Data Is on file with the Library of Congress.
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Preface, The purpose of this book is to help students understand and use the calculus. Everything has been aimed, toward making this easier, especially for students with limited· background in mathematics or for readers who, have forgotten their earlier training in mathematics. The topics covered include all the material of standard, courses in elementary and intermediate calculus. The direct and concise exposition typical of the Schaum, Outline series has been amplified by a large number of examples, followed by many carefully solved problems. In choosiag these problems, we have attempted to anticipate the difficulties that normally beset the, beginner. In addition, each chapter concludes with a collection of supplementary exercises with answers., This fifth edition has enlarged the number of solved problems and supplementary exercises. Moreover, we, have made a great effort to go over ticklish points of algebra or geometry that are likely to confuse the student, The author believes that most of the mistakes that students make in a calculus course are not due to a deficient, comprehension of the principles of calculus, but rather to their weakness in high-school algebra or geometry., Students are urged to continue the study of each chapter until they are confident about their mastery of the, material. A good test of that accomplishment would be their ability to answer the supplementary problems., The author would like to thank many people who have written to me with corrections and suggestions, in, particular Danielle Cinq-Mars, Lawrence Collins, L.D. De longe, Konrad Duch, Stephanie Happ, Lindsey Oh,, and Stephen B. Soffer. He is also grateful to his editor, Charles Wall, for all his patient help and guidance., ELLIOTT MENDELSON
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Contents, CHAPTER 1 Linear Coordinate Systems. Absolute Value. Inequalities, Linear Coordinate System, , Finite Intervals, , Infinite Intervals, , 1, , Inequalities, , CHAPTER 2 Rectangular Coordinate Systems, , 9, , Coordinate Axes Coordinates Quadrants The Distance Formula, . Midpoint Formulas Proofs of Geometric Theorems, , The, , 18, , CHAPTER 3 Lines, The Steepness of a Line The Sign of the Slope Slope and Steepness, Equations of Lines A Point-Slope Equation Slope-Intercept Equation, Parallel Lines Perpendicular Lines, , CHAPTER 4 Circles, , 29, , Equations of Circles, , The Standard Equation of a Circle, , 37, , CHAPTER 5 Equations and Their Graphs, The Graph of an Equation, Sections, , Parabolas, , Ellipses, , HyperbOlas, , Conic, , CHAPTER 6 Functions, , 49, , CHAPTER 7 Limits, , 56, , Limit of a Function, , Right and Left Limits, , Theorems on Limits, , Infinity, , 66, , CHAPTER 8 Continuity, Continuous Function, , 73, , CHAPTER 9 The Derivative, l', , Delta Notation, , The Derivative, , Notation for Derivatives, , Differentiability, , 79, , CHAPTER 10 Rules for Differentiating Functions, Differentiation Composite Functions. The Chain Rule Alternative Formulation of the Chain Rule Inverse Functions Higher Derivatives, , -
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Contents, 90, , CHAPTER 11 Implicit Differentiation, Implicit Functions, , Derivatives of Higher Order, , CHAPTER 12 Tanglmt and Normal Lines, , 93, , The Angles of Intersection, , CHAPTER 13 Law Ilf the Mean. Increasing and Decreasing Functions, Relative Maximum and Minimum, , 98, , Increasing and Decreasing Functions, , CHAPTER 14 Maximum and Minimum Values, , 105, , Critical Numbers, Second Derivative Test for Relative Extrema, First Derivative Test Absolute Maximum and Minimum Tabular Method for Finding the Absolute Maximum and Minimum, , CHAPTER IS Curve Sketching. Concavity. Symmetry, Concavity, ymptotes, Functions, , Vertical Asymptotes, Points of Inflection, Symmetry, Inverse Functions and Symmetry, Hints for Sketching the Graph of y =f (x), , 119, Horizontal AsEven and Odd, , CHAPTER 16 Review of Trigonometry, Angle Measure, , Directed Angles, , 130, Sine and Cosine Functions, , CHAPTER 17 Differentiation of Trigonometric Functions, , 139, , Continuity of cos x and sin x, Graph of sin x Graph of cos x Other Trigonometric Functions, Derivatives, Other Relationships, Graph of y =, tan x, Graph of y = sec x Angles Between Curves, , 152, , CHAPTER 18 Invel'se Trigonometric Functions, The Derivative of sin-I x, gent Function, , The Inverse Cosine Function, , The Inverse Tan-, , CHAPTER 19 Rectilinear and Circular Motion, Rectilinear Motion, , Motion Under the Influence of Gravity', , 161, Circular Motion, , CHAPTER 20 Related Rates, , 167, , CHAPTER 21 Diffe!rentials. Newton's Method, , 173, , The Differential, , Newton's Method, , CHAPTER 22 Antiderivatives, Laws for Antiderivatives, , 181
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Contents, 190, , CHAPTER 23 The Definite Integral. Area Under a Curve, Sigma Notation, , Area Under a Curve, , Properties of the Definite Integral, , 198, , CKAPTER 24 The Fundamental Theorem of Calculus, Mean-Value Theorem for Integrals Average Value of a Function on a Closed, Interval Fundamental Theorem of Calculus Change of Variable in a Definite Integral, , 206, , CHAPTER 25 The Natural Logarithm, The Natural Logarithm • Properties of the Natural Logarithm, , CHAPTER 26 Exponential and Logarithmic Functions, , 214, , Properties of e' The General Exponential Function, Functions, , CHAPTER 27, , ~Hopital's, , General Logarithmic, , Rule, , 222, , L'H6pital's Rule Indeterminate Type 0'00, Indeterminate Types 00, 00°, and 1-, , 'Indeterminate Type ·00-00, , CHAPTER 28 Exponential Growth and Decay, , 230, , Half-Life, , CHAPTER 29 Applications of Integration I: Area and Arc Length, Area Between a Curve and the y Axis, , Areas Between Curves, , 235, Arc Length, , CHAPTER 30 Applications of Integration II: Volume, Disk Formula Washer Method Cylindrical Shell Method, of Shells Formula Cross-Section Formula (Slicing Formula), , 244, Difference, , CHAPTER 31 Techniques of Integration I: Integration by Parts, , 259, , CHAPTER 32 Techniques of Integration II:Trigonometric Integrands and, Trigonometric Substitutions, , 266, , Trigonometric Integrands, , Trigonometric Substitutions, , CHAPTER 33 Techniques of Integration III: Integration by Partial Fractions, , 279, , Method of Partial Fractions, , CHAPTER 34 Techniques of Integration IV: Miscellaneous Substitutions, , 288
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Contents, CHAPTER 35 Improper Integrals, , 293, , Infinite Limits of Integration, , Discontinuities of the Integrand, , CHAPTER 36 Applilcations of Integration III: Area of a Surface of Revolution, , 301, , CHAPTER 37 Parametric Representation of Curves, , 307, , Parametric Equations, , Arc Length for a Parametric Curve, , CHAPTER 38 Curvature, , 312, , Derivative of Arc Length, Curvature, The Radius of Curvature, Circle of Curvature, The Center of Curvature, The Evolute, , CHAPTER 39, , Planl~, , The, , Vectors, , 321, , Scalars and Vectors, Sum and Difference of Two Vectors, Components of, a Vector Scalar Product (or Dot Product) Scalar and Vector Projections, Differentiation of Vector Functions, , CHAPTER 40 Curvilinear Motion, , 332, , Velocity in Curvilinear Motion, Acceleration in Curvilinear Motion, Tangential and Normal Components of Acceleration, , CHAPTER 41 Polar Coordinates, , 339, , Polar and Rectangular Coordinates, Inclination, Points of Intersection, of the Arc Length, Curvature, , Some Typical Polar Curves, Angle of, Angle ofIntersection The Derivative, , 352, , CHAPTER 42 Infinite Sequences, Infinite Sequences, , Limit of a Sequence, , Monotonic Sequences, , CHAPTER 43 Infinite Series, , 360, , Geometric Series, , CHAPTER 44 Series with Positive Terms. The Integral Test. Comparison Tests, , 366, , Series of Positive Terms, , CHAPTER 45 Altel'nating Series. Absolute and Conditional Convergence., The Ratio Test, , 375, , Alternating Series, , CHAPTER 46 Power Series, Power Series, , 383, Uniform Convergence
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Contents, , CHAPTER 47 Taylor and Maclaurin Series. Taylor's Formula with Remainder, Taylor and Maclaurin Series, , Applications of Taylor's Fonnula with Remainder, , CHAPTER 48 Partial Derivatives, , 405, , Functions of Several Variables Limits, Partial Derivatives of Higher Order, , Continuity, , Partial Derivatives, , 414, , CHAPTER 49 Total Differential.Differentiability.Chain Rules, Total Differential, , 396, , Differentiability, , Chain Rules, , Implicit Differentiation, , 426, , CHAPTER 50 Space Vecturs, Vectors in Space Direction Cosines of a Vector, Detenninants Vector, Perpendicular to Two Vectors, Vector Product of Two Vectors Triple Scalar Product Triple Vector Product, The Straight Line The Plane, , 441, , CHAPTER 51 Surfaces and Curves in Space, Planes Spheres Cylindrical Surfaces Ellipsoid Elliptic Paraboloid, Elliptic Cone Hyperbolic Paraboloid Hyperboloid of One Sheet Hyperboloid of1Wo Sheets Tangent Line and Nonnal Plane to a Space Curve Tangent, Plane and Nonnal Line to a Surface· Surface of Revolution, , CHAPTER 52 Directional Derivatives. Maximum and Minimum Values, Directional Derivatives Relative Maximum and Minimum Values, Maximum and MininlU~ Values, , 452, Absolute, , CHAPTER 53 Vector Differentiation and Integration, , 460, , Vector Differentiation Space Curves Surfaces The Operation V, Divergence and Curl Integration Line Integrals, , 474, , CHAPTER 54 Double and Iterated Integrals, The Double Inte~l, , The Iterated Integral, , CHAPTER 55 Centroids and Moments of Iriertia of Plane Areas, Plane Area by Double Integration, , Centroids, , 481, , Moments of Inertia, , CHAPTER 56 Double Integration Applied to Volume Under a, , Surface and the Area of a Curved Surface, , 489, , CHAPTER 57 Triple Integra.ls, , 498, , Cylindrical and Spherical Coordinates The Triple Integral, Triple Integrals Centroids and Moments of Inertia, , CHAPTER 58 Masses of Variable Density, , Evaluation of, , 510
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.., , ...---, , Contents, , CHAPTER 59 Diffe.'ential Equations of First and Second Order, Separable Differential Equations, Second-Order Equations, , Homogeneous Functions, , 516, Integrating Factors, , Appendix A, , 527, , Appendix B, , 528, , Index, , 529
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Linear Coordinate Systems., Absolute Value. Inequalities, Linear Coordinate System, A linear coordinate system is a graphical representation of the real numbers as the points of a straight line. To, each number corresponds one and only one point, and to each point corresponds one and only one number., To set up a linear coordinate system on a given line: (1) select any point of the line as the origin and let, that point correspond to the number 0; (2) choose a positive direction on the line and indicate that direction, by an arrow; (3) choose a fixed distance as a unit of measure. If x is a positive number, find the point corresponding to x by moving a distance of x units from the origin in the positive direction. If x is negative,, find the point corresponding to x by moving a distance of -x units from the origin in the negative direction., (For example, if x =-2, then -x = 2 and the corresponding point lies 2 units from the origin in the negative, direction.) See Fig. 1-1., -4, , I, , I, , I, , I, , I, , -3 -512 -2 -3/2 -I, , o, 112, I' 4Vi, , 2, , II, 4, , Fig. 1-1, , The number assigned to a point by a coordinate system is called the coordinate of that point. We often, will talk as if there is no distinction between a point and its coordinate. Thus, we might refer to "the point 3", rather than to "the point with coordinate 3.", The absolute value Ixl of a number x is defined as follows:, if x is zero or a positive number, IXI={ x, , -x if x is a negative number, , For example, 141 = 4,1-31:::; -(-3):::; 3, and 101 = O. Notice that, if x is a negative number, then -x is positive., Thus, Ixl ~ 0 for all x., ", The following properties hold for any numbers x and y., (1.1), , (1.2), , (1.3), , (1.4), , I-xl =Ixl, , When x =0, I-xl =1-01 =101 =Ixl., When x >D, -x < 0 and I-xl =-(-x) =x =Ixl., When x < 0, -x> 0, and I-xl =-x =Ixl., Ix-yJ= Iy-xl, This follows from (1.1), since y - x =-(x - y)., Ixl = c implies that x = ±c., For example, if Ixl =2, then x =±2. For the proof, assume Ixl =c., If x ~ 0, x =Ixl =c. If x < 0, -x =Ixl =c; then x =-(-x) =-c., IxF =xl, , Ifx ~ 0, Ixl :::; x and 1x12 =x2• If x$; 0, Ixl =-x and IxF =(_X)2 =xl., . (1.5), , lxyl =Ixl . Iyl, , By (1.4), lxyl2 =(xy)2 =x2y2 = Ixl21yl2 = (lxl . lyl)2. Since absolute values are nonnegative, taking, square roots yields Ixyl = Ixl . Iyl.
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CHAPTER 1 Linear Coordinate Systems, , o, a, , •, b •, , •a, , c., b, , Open interval (a, b): a <)C < b, , Closed interval [a, b): a ~ x ~ b, , Fig. 1-4, , By a half-open interval we mean an open interval (a, h) together with one of its endpoints. There are two, such intervals: [a, h) is the set of all x such that a ~ x < h, and (a, h] is the set of all x such that a < x ~ h., , Infinite Intervals, Let (a, 00) denote the set of all x such that a < x., Let [a, 00) denote the set of all x such that a ~ x., Let (-co, h) denote the set of all x such that x < h., Let (-co, ~l denote the set of all x such that x ~ h., , Inequalities, Any inequality, such as 2x - 3 > 0 or 5 < 3x + 10 ~ 16, determines an interval. To solve an inequality means, to determine the corresponding interval of numbers that satisfy the inequality., Solve 2x - 3 > O., , EXAMPLE 1.1:, , 2x-3>O, 2x>3, , (Adding 3), , x> t, , (Dividing by, , 2:, , Thus, the corresponding interval is (t,oo)., Solve 5 < 3x + to ~ 16., , EXAMPLE 1.2:, , 5<3x+1O~16, , -5 < 3x ~ 6 (Subtracting 10), , -t < x ~ 2, , (Dividing by 3), , Thus, the corresponding interval is (-t, 2]., Solve -2x + 3 < 7., , EXAMPLE 1.3:, , -2x+3 < 7, -2x < 4, x> -2, , (Subtracting 3), (Dividing by - 2), , (Recall that dividing by a negative number reverses an inequality.) Thus, the corresponding interval is (-2, 00)., , ,, SOLVED PROBLEMS, , 1., , Describe and diagram the following intervals, and write their interval notation, (a) -3 < x < 5; (b) 2 ~ x::; 6;, (c) -4 < x::; 0; (d) x > 5; (e) x::; 2; (f) 3x - 4::; 8; (g) 1 < 5 - 3x < II., , (a) All numbers greater than -3 and less than 5; the interval notation is (-3, 5):, , o, , -3, , o, , •
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CHAPTER 1 Linear Coordinate Systems, , (e) Ix + 21 < 3 is equivalent to -3 < x + 2 < 3. Subtracting 2, we obtain -5 < x < 1, which defines the open, interval (-5, 1):, , o, , o, , -s, , •, , °, , (f) The inequality Ix - 41 < 0 detennines the interval 4 - 0< x < 4 + O. The additional condition < Ix - 41 tells, , x'", , us that, 4. Thus, we get the union of the two intervals (4 - 0, 4) and (4, 4 + 0). The result is called the, deleted S-neighborhood of 4:, , o, , o, , 3., , •, , o, , ., , 4-8, , Describe and diagram the intervals detennined by the following inequalities, (a) 15 - xl, (c) II - 4xl < t., , ., , ~, , 3; (b) 12x - 31 < 5;, , (a) Since 15 - xl =Ix - 51, we have Ix - 51 ~ 3, which is equivalent to -3 ~ x - 5 ~ 3. Adding 5, we get 2 ~ x S 8,, which defines the closed interval [2, 8]:, , •, , .., , :, , :8, , 2, , (b) 12x - 31 < 5 is equivalent to -5 < 2x - 3 < 5. Adding 3, we h~ve -2 < 2x < 8; then dividing by 2 yields, -I < x < 4, which defines-the open interval (-I, 4):, ------~o~------------<o~--------~., -1, , 4, , (c) Since 11 - 4x1 =14x - 11, we have 14x - 11< t, which is equivalentto -t < 4x - 1 < t. Adding 1, we get, t < 4x < t. Dividing by 4, we obtain t < x < t. which defines the open imel"'aill. i):··, , o, , o, , 1/8, , 4., , •, , 3/8, , Solve the inequalities: (a) 18x - 3.il> 0; (b) (x + 3)(x - 2)(x - 4) < 0; (c) (x + 1)2(x - 3) > 0, and diagram the solutions., (a) Set 18x - 3.il = 3x(6 - x) = 0, obtaining x = 0 and x = 6. We need to detennine the sign of 18x - 3x2on each, of the intervals x < O. 0 < x < 6, and x > 6. to detennine where 18x - 3.il> O. Note that it is negative when, x < 0 (since x is negative and 6 - x is positive). It becomes positive when we pass from left to right through, o(since x changes sign but 6 - x remains positive), and it becomes negative when we pass through 6 (since x, remains .positive but 6 - x changes to negative). Hence, it is positive when and only when 0 < x < 6., , o, , o, o, , •, , 6, , (b) The crucial points are x =-3, x = 2, and x =4. Note that (x + 3)(x - 2)(x - 4) is negative for x < -3 (since, each of me factors is negative) and that it changes sign when we pass through each of the crucial points., Hence, it is negative fof x < -3 and for 2 < x < 4:, , o, , o, , -3, , o, 4, , •, , (c) Note tpat (x + I) is always positive (except at x = -1, where it is 0). Hence (x + 1)2 (x - 3) > 0 when and only, when x - 3 > 0, that is, for x > 3:, , o, S., , Solve 13x - 71 = 8., By (1.3). 13x - 71 = 8 if and only if 3x - 7 =±8. Thus, we need to solve 3x - 7 =8 and 3x - 7 =-8. Hence, we, get x = 5 or x = -t.
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CHAPTER 1 Linear Coordinate Systems, , (d), (e), , ~-~~4, 12+~1>1., , (f), , Ans., , .,-,.. ,:1,, .'1'", , ~1<3, (a)t <x < 3; (b)x~t or x~ 0; (c) -6 ~x ~ 18; (d)x~ -t or x~t; (e)x> 0 or x< -lor -t <x < 0;, (f)x>torx<-t, , 12. Describe and diagram the set determined by each of the following conditions:, (a) x(x - 5) < 0, (b) (x-2)(x-6»0, , (c) (x + I)(x - 2) < 0, (d) x(x-2)(x+3»0, , (e) (x+2)(x+3)(x+4)<0, (f) (x-l)(x+ 1)(x-2)(x+3»0, (g) (X_I)2(X+4»0, (h), (i), (j), (k), , (x - 3)(x + 5)(x - 4)2 < 0, (x - 2)3 > 0, (x + 1)3 < 0, , (x - 2)3(x + I) < 0, (1) (X-I)3 (x+ 1),,<0, (m) (3x - 1)(2x + 3) > 0, (n) (x - 4)(2x - 3) < 0, Ans., , (a) O<x< 5; (b)x> 60r x< 2; (c) -I <x< 2; (d)x> 2 or-3 <x<O; (e)-3 <x< -2 or x<-4;, (f}x>2or-1 <x< 1 orx<-3;(g)x>-4andx;t 1;(h)-5<x<3;(i)x>2;(j)x<-I;, (k)-I <x<2;(l)x< 1 andx;t-I;(m)x>torx<-t;(n>!<x<4, , 13. Describe and diagram the set determined by each of the following conditions:, (a) xl <4, (b) xl ~ 9, (c) (X_2)2~ 16, , (d) (2x+ If> I, (e) xl+3x-4>O, (f) xl + 6x + 8 ~ 0, (g) xl < 5x+ 14, (h), , 2x2 >x+ 6, , (i) fu2+13x<5, (j) xl+ 3xl > lOx, , Ans., , (a) -2 < x < 2; (b) x ~ 3 or x ~ -3; (c) -2 ~ x:5 6; (d) x > 0 or x < -I; (e) x > I or x < -4; (f) -4:5 x:5 -2;, (g)L2 <x< 7; (h)x> 2 or x <-t; (i) -t<x<t; (j) -5 <x< 0 or x> 2, , 14. Solve: (a) -4 < 2 - x < 7, , 3x-1, (d) 2x+3>3, , (b), , 2x-1, x, , --<3, , (e), , _x_<l, , (f), , IX:21, , x+2, , ~2, , Ans. (a) -5 <x < 6; (b) x>O or x < -I; (c) x > -2; (d) -~<x <t; (e) x < 0 or 0 <x < t; (f) x:5 -4 or x ~-I
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Rectangular Coordinate, Systems, Coordinate Axes, In any plane rJ}, choose a pair of perpendicular lines. Let one of the lines be horizontal. Then the other line, must be vertical. The horizontal line is called the x axis, and the vertical line the y axis. (See Fig. 2-1.), y, I, , b -, , --, , ---1 p(a. b), -------,----I, I, , I, I, , I, I, , I, I, I, I, --~--~~--~~--~~--~~------x, , -2, , -\, , Ia, , 0, , _\, , I, , I, I, , Fig. 2-1, , Now choose linear coordinate systems on the x axis and the y axis satisfying the following conditions:, The origin for each coordinate system is the point 0 at which the axes intersect. The x axis is directed from, left to right, and the y axis from bottom to .top. The part of the x axis with positive coordinates is called the, positive x axis. and the part of the y axis with positive coordinates is called the positive y axis., We shall establish a correspondence between the points of the plane <!JI and pairs of real numbers., , Coordinates, Consider any t>oint P of the plane (Fig. 2-1). The vertical line through P intersects the x axis at a unique, point; let a be the coordinate of this point on the x axis. The number a is called the x coordinate of P (or the, abscissa of P). The horizontal line through P intersects the y axis at a unique point; let b be the coordinate, of this point on the y axis. The number b is called the y coordinate of P (or the ordinate of P)~ In this way., every point P has a unique pair (a, b) of real numbers associated with it. Conversely. every pair (a. b) of real, numbers is associated with a unique point in the plane., The coordinates of several points are shown in Fig. 2-2. For the sake of simplicity. we have limited them, to integers.
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CHAPTER 2 Rectangular Coordinate Systems, Quadrant II consists of all points with negative x coordinate and positive ycoordinate. Quadrants mand IV, are also shown in Fig. 2-4., 1, , U, , I, , (-.+), , (+.+), , (-1.2)., , 2, , ----~--~~--~--~--~~---------x, , III, , IV, , (-.-), , (+.-), , Rg.2-4, , The points on the x axis have coordinates of the form (a, 0). The y axis consists of the points with coordinates of the form (0, b)., ., Given a coordinate system, it is customary to refer to the point with coordinates (a, b) as "the point, (a, b)." For example, one might say, "'.fhe point (0, 1) lies on the y axis.", , The Distance Formula, The distance Pl2between poinits PI and P2with coordinates (XI' YI) and (x2' Y2) in a given coordinate system, (see Fig. 2-5) is given by the following distance formula:, (2.1), 1, , 1,, I, , I, I, I, , I, - - - - - - ~ R(x,. YI), P1(x .. '1) I, I, I, , I, , I, , I, , I, , I, IA, , A I, , ____~----~I~--------~~2~----X, , Fig. 2·5, , To see this, let R be the point wh~re the vertical line through P2 intersects the horizontal line through PI' The, X coordinate of R is x2', , rean theorem, (~P2)2, , the same as that of P2' The Ycoordinate of R is YI' the same as that of PI' By the Pythago=(~R)2 + (~R)2. IfAl andA 2 are the projections of PI and P2on the x axis, the segments, , PIR and AIA2 are opposite sides of a rectangle, so that ~R =AI~' But AIA2 = IXI - x21 by property (1.12)., 'I 1 -~R=IYI-Y21· Hence, (~P2), 2, 2, S0, -~R=lxl-X21· S'Imlary,, =lxt -x2'2 +IYt-Y2' 2 = (X I -X2)2 +(Yt-Y2)'
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-, , CHAPTER 2 Rectangular Coordinate Systems, , Proofs of Geometric Theorems, Proofs of geometric theorems can often be given more easily by use of coordinates than by deductions from, axioms and previously derived theorems. Proofs by means of coordinates are called analytic, in contrast to, so-called synthetic proofs from axioms., EXAMPLE 2.2: Let us prove analytically that the segment joining the midpoints of two sides of a triangle is one-half, the length of the third side. Construct a coordinate system so that the third side AB lies on the positive x axis, A is the, origin, and the third vertex C lies above the x axis, as in Fig. 2-7., , y, C(u, II), , ------~~~--------~-------------x, , A, , B, , Fig. 2-7, Let b be the x coordinate of B. (In other words, let b = AB.) Let C have coordinates (u, v). Let MI and M2 be the, midpoints of sides AC and BC, respectively. By the midpoint formulas (2.2), the coordinates of MI are, coordinates of M2 are, , (u; b , ~). By the distance formula (2.1),, , (~, ~), and the, , which is half the length of side AB., , SOLVED PROBLEMS, , 1., , Show that the distance between a point P(x, y) and the origin is ~X2 +T., Since the origin has coordinates (0, 0), the distance formula yields ~r-(x-_-O""")2=-+-(-y-_-0-:"')2 =~X2 +T., , 2., , Is the triangle with vertices A(l, 5), B(4, 2), and C(5, 6) isosceles?, AB=~(l-4)2 +(5-2)2 =~(-3)2 +(3)2 =.J9+9 =./f8, I', , AC=~(l-5)2 +(5-6)2 =~(-4)2 +(_1)2, , =.Ji6+1 =$.7, , BC = ~(4 -5)2 + (2- 6)2 =~(_1)2 + (-4)2 =.JI +16 = ffi, , Since AC = BC, the triangle is isosceles., 3., , Is the triangle with vertices A(-5, 6), B(2, 3), and C(5, 10) a right triangle?
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CHAPTER 2 Rectangular Coordinate Systems, , segments are in proportion.) But A,Q' =x -I, and Q'~ = 6 - x. So, 3x - 3 = 12 - 2x. Hence 5x = 15, whence x = 3. By similar reasoning,, , ., 6=! = ~,and cross-multiplying yields, =, , ~ ~ =~, from which it follows that y =4., , y, P2(6,7), , I, , I, , I, , 2, , I, I, I, , PI, , I, , I, , AI, , I AJ, , I, , .t, , .t, , 6, , Fig. 2-9, , 6., , In Fig. 2-10, find the coordinates of points A, B, C, D, E, and F., y, , E., , 4, , 3, , c., • F, , 2, , A., , -\ B, -2, , Fig. 2-10, Ans., , (A) = (-2, I); (B) =(0, -I): (C) =(1,3); (0) = (-4, -2); (E) =(4, 4); (F) = (7,2)., , 7., , Draw a coordinate system and show the points having the following coordinates: (2, -3), (3, 3), (-I, I), (2, -2),, (0,3), (3, 0), (-2, 3)., , 8., , Find the distances between the following pairs of points:, (a) (3,4) and (3, 6), (b) (2,5) and (2, -2), (d) (2,3) 6nd (5, 7), (e) '(-2,4) and (3,0), Am., , 9., , (a) 2; (b)7; (c) I; (d) 5; (e), , (c) (3, l) and (2, l), and (4, -I), , (f) (-2,, , .J41: (f) tffi, , Draw the triangle with vertices A(2, 5), B(2, -5), and C(-3, 5). and find its area., Ans., , Area =25, , t)
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CHAPTER 2 Rectangular Coordinate Systems, , 22. Prove analytically that the sum of the squares of the medians of a triangle is equal to three-fourths the sum of the, squares of the sides., 23., , ~~e, , analytically that the line segments joining the midpoints of opposite sides of a quadrilateral bisect each, , other., 24. Prove that the coordinates (x, y) of the point Q that divides the line segments from PI(x\, y) to P 2(X2• Y2) in the, ratio '1:'2 are determined by the formulas, X=, , 'i~+'2XI, , Ij+'i, (Hint: Use the reasoning of Problem 5.), , 25. Find the coordinates of the point Q on the segment Pl2 such that ~QIQP2 =t, jf (a) PI =(0. 0), P2 =(7. 9);, (b) PI =(-1, 0), P 2 = (0, 7); (c) PI = (-7, -2), P2 = (2, 7); (d) PI = (1, 3), P 2 = (4. 2)., Ans. (a) (.Jt-.2);(b), , (', , (-t ..If); (c), , (-5.t);(d), , (.If,-V-)
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.., , -~, , CHAPTER 3 Lines, , Y, , Y, , --~----~~---------------x, , Rg.3-3, , Rg.3-2, , Now let the line :i be horizontal, as in Fig. 3-4(c). Here YI =Y2, so that Y2 - YI =O. In addition, x2 -, , XI:;:', , O., , Hence, m =_0_ =O. The sLope of:i is zero., x2 -XI, Line!£ is vertical in Fig. 3-4(d), where we see that Y2 - YI > 0 while x2 -, , XI, , =O. Thus, the expression, , Y2 - YI is undefined. The sLope is not defined for a verticaL line ~. (Sometimes we describe this situation by, X 2 -XI, , saying that the slope of!£ is "infinite."), Y, , x, , --~~--~--------x, , !l, (a), , (b), Y, , Y, , P2(X 2, Y2), , - ....-+---...- ! l, , p.(x .. Y.), , --------~------------x, , x, , (c), , (d), , Rg.3-4, , Slope and Steepness, Consider any line;;e with positive slope, passing through a point PI (xl' YI }; such a line is shown in Fig. 3-5., Choose the point P2(X2, Y2} on ~ such that x 2 - XI = 1. Then the slope m of ~ is equal to the distance AP2 •, As the steepness of the line increases, AP2 increases without limit, as shown in Fig. 3-6(a). Thus, the slope, of!£ increases without bound from 0 (when :i is horizontal) to +00 (when the line is vertical). By a similar, argument, using Fig. 3-6(b), we can show that as a negatively sloped line becomes steeper, the slope steadily, decreases from 0 (when the line is horizontal) to -00 (when the line is vertical).
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CHAPTER 3 Lines, , A Point-Slope Equation, A point-slope equation of the line ~ is any equation of the form (3.1). If the slope m of ~ is known, then, each point (x.' y.) of!£ yields a point-slope equation of~. Hence, there is an infinite number of point-slope, equations for~. Equation (3.1) is equivalent to y - y. = m(x - xJ, EXAMPLE 3.2:, , (a) The line passing through the point (2, 5) with slope 3 has a point-slope equation, , (b) Let!£ be the line through the points (3, -1) and (2, 3). Its slope is m = 3, CD, v+l, y-3, of OL are ~3 =-4 and --2 =-4., , x-, , ~ =~ =3., , 2~31) =~1 =-4. Two point-slope equations, , x-, , Slope-Intercept Equation, If we multiply (3.1) by x - xl' we obtain the equation y - y. =m(x- x.), which can be reduced first to y - y. =, mx - mx., and then to y =mx + (y. - mx.). Let b stand for the number y. - mx •. Then the equation for line, ~becomes, , y=mx+b, , (3.2), , Equation (3.2) yields the value y =b when x = 0, so the point (0, b) lies on~. Thus, b is the y coordinate, of the intersection of ~ and the y axis, as shown in Fig. 3-8. The number b is called the y intercept of ~, and, (3.2) is called the slope-intercept equation for !e., y, , --~~----~-----------------x, , Fig. 3-8, , EXAMPLE 3.3:, , The line through the points (2,3) and (4,9) has slope, , Its slope-intercept equation has the fonn y = 3x + b. Since the point (2, 3) lies on the line, (2, 3) must satisfy this, equation. Substitution yields 3 = 3(2) + b, from which we find b = -3. Thus, the slope-intercept equation is y = 3x - 3,, , Another method for finding this equation is to write a point-slope equation of the line, say y - ~ =3. Then, multiplying by x - 2 and adding 3 yields y = 3x - 3,, x-, , Parallel Lines, Let !e. and ~2 be parallel non vertical lines, and let A. and A2 be the points at which ;£. and!e 2 intersect the, y axis, as in Fig. 3-9(a). Further, let B. be one unit to the right of AI' and B2 one unit to the right of A2, Let, C. and C2 be the intersections of the verticals through B. and B2 with ~. and ~2' Now, triangle A.B.C. is, congruent to triangle A2BP2 (by the angle-side-angle congruence theorem), Hence, B.C. = B2C2 and, BC_. __, BC, Slopeof!e =_., 2_2 =slopeof;£, •, 1 - 1, 2
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CHAPTER 3 Lines, y, y, , 2, , ----~~--_+--~--~--~--~x, -2, -1, 0, 4, , -1, , Rg.3-11, , Rg.3-10, , 2., , Let;e be the perpendicular bisector of the line segment joining the points A(-1, 2) and B(3, 4), as shown in, Fig. 3-11. Find an equation for ;e., , ;e passes through the midpoint M of segment AB. By the midpoint formulas (2.2), the coordinates of Mare (1, 3)., , i 1-, , The slope of the line through A and B is ~, = = Let m be the slope of ;e. By Theorem 3.2, t m =-1,, whence m = -2., The slope-intercept equation for;e has the form y =-2x + b. Since M (l, 3) lies on;e, we have 3 =-2(1) + b., Hence, b =5, and the slope-intercept equation of;e is y =-2x + 5., , 3 (:1), , 3., , Determine whether the points A( I, -1), B(3, 2), and C(7, 8) are collinear, that is, lie on the same line., A, B, and C are collinear if and only if the line AB is identical with the line AC, which is equivalent to the, 2-(-1) 3, 8-(-1) 9 3, slope of AB being equal to the slope of AC. The slopes of AB and AC are 3=t ='2 and ~ =6" ='2', , Hence, A, B, and C are collinear., 4., , Prove analytically that the figure obtained by joining the midpoints of consecutive sides of a quadrilateral is a, parallelogram., Locate a quadrilateral with consecutive vertices, A, B, C, and D on a coordinate system so that A is the origin, B, lies on the positive x axis, and C and D lie above the x axis. (See Fig. 3-12.) Let b be the x coordinate of B, (II, v) the, coordinates of C, and (x, y) the coordinates of D. Then, by the midpoint formula (2.2), the midpoints M" M2, M3, and, , M4 of sides AB,BC, CD, and DA have coordinates, , (!, 0), (, , II ;, , b,, , ~), ( x; II., , y; v). and (I' ~Jrespectively., , We must show that M,M;t3M4 is a parallelogram. To do this, it suffices to prove that lines M,M2 and M3M4 are, parallel and that lines M#3 and M,M4are parallel. Let us calculate the slopes of these lines:, y y+v, slope(M M )= '2--23 4, X, x+u, , v, , -'2 =E., II, II, '2--2- -'2, , I', , y, , -L=_y_, x-b, , -2-, , x-b, , 1-0, , slope(M 1M4 )=L- =-Lb, x b x-, , '2-'2, , Since slope(M,M2) = slope(M3M4), M)M2 and M)M4 are parallel. Since slope(M2M3) =slope(M)M4)' M#) and, M)M4 are paralic\, Thus, M,M#)M4 is a parallelogram.
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CHAPTER 3 Lines, , Now, conversely, we assume that m.m2 = -I, where m. and m2are the slopes of nonverticallines ~. and ~2', Then ~. is not parallel to ~2' (Otherwise, by Theorem 3.1, m. =m2and, therefore, m~ = -1, which contradicts the, fact that the square of a real number is never negative.) We must show that~. and ~2 are perpendicular. Let P be, the intersection of~. and:£2 (see Fig. 3-14). Let:£3 be the line through P that is perpendicular to ~ •. If m3 is the, slope of ~3' then, by the first part of the proof, m.m3 = -1 and, therefore, m1m3= m.m2' Since m.m3 = -1, m. '# 0;, therefore, m3 = m2. Since ~2 and ~3 pass through the same point P and have the same slope, they must coincide., Since ~. and ~3 are perpendicular, :£. and :£2 are also perpendicular., , ~~~--~~---------x, , Fig. 3-14, , 6., , 7., , Show that~ if a and b are not both zero, then the equation ax + by =c is the equation of a line and, conversely,, every line has an equation of that form., Assume b '# 0. Then, if the equation ax + by =c is solved for y, we obtain a slope-intercept equation, y = (-alb) x + clb of a line. If b = 0, then a '# 0, and the equation ax +by = c reduces to ax = c; this is equivalent, to x =cia, the equation of a vertical line., Conversely, every non vertical line has a slope-intercept equation y =mx + b, which is equivalent to -mx + y =b,, an equation of the desired form. A vertical line has an equation of the form x =c, which is also an equation of the, required form with a = 1 and b =O., Show that the line y = x makes an angle of 45° with the positive x axis (that is, that angle BOA in Fig. 3-15, contains 45°)., y, , ----~~----~~--------x, , Fig. 3-15, , Let A be the point on the line y = x with coordinates (1, 1). Drop a perpendicular AB to the positive x axis., Then AB = 1 and OB = 1. Hence, angle OAB = angle BOA, since they are the base angles of isosceles triangle, BOA. Since angle OBA is a right angle,, Angle OAB + angle BOA = 1800, , -, , angle OBA = 1800, , Since angle BOA = angle OAB, they each contain 45°., , -, , 900 = 90 0, , : :,
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CHAPTER 3 Lines, , 11. (a) Describe the lines having equations of the form i = a., (b) Describe the lines having equations of the form y = b., , (c) Describe the line having the equation y = -x., 12. (a) Find the slopes and y intercepts of the lines that have the following equations: (i) y = 3x - 2; (ii) 2x - 5y = 3;, , (iii)y=4x- 3; (iv)y= -3; (v) ~+1= 1., (b) Find the coordinates of a point other than (0, b) on each of the lines of part (a)., Ans. (a)(i) m = 3, b = -2; (ii) m = t b = -t (iii) m = 4, b = -3; (iv) m = 0, b = -3; (v) m = -t; b = 2;, (b) (i) (1, 1); (ii) (-6, -3); (iii) (1, I); (iv) (1, -3); (v) (3, 0), 13. If the point (3, k) lies on the line with slope m = -2 passing through the point (2, 5), find k., , Ans. k=3, 14. Does the point (3, -2) lie on the line through the points (8, 0) and (-7, -6)?, , Ans., , Yes, , 15. Use slopes to determine whether the points (7, -I), (10, I), and (6, 7) arc the vertices of a right triangle., , Ans. They are., 16. Use slopes to determine whether (8, 0). (-1, -2), (-2, 3). and (7, 5) are the vertices of a parallelogram., , Ans. They are., 17. Under what conditions are the points (II, v + w). (v,, , Ans., , II, , + w). and (w. II + v) collinear?, , Always., , 18. Determine k so that the points A(7, 3), B( -1,0), and C(k, -2) are the vertices of a right triangle with right angle at B., , Ans., , k= I, , 19. Determine whether the following pairs of lines are parallel, perpendicular. or neither:, (a) y=3x+2andy=3x-2, , y = 2x - 4 and y = 3x + 5, 3x-2y=5and2x+3y=4, 6x + 3y, = 1 and 4x + 2y = 3, x=3 andy=-4, (f) 5x + 4y = 1 and 4x + 5y = 2, (g) x = -2 and x = 7, , (b), (c), (d), (e), , Ans. (a) Parallel; (b) neither; (c) perpendicular; (d) parallel; (e) perpendicular; (f) neither; (g) parallel, 20. Draw the line determined by the equation 2x + 5y = 10. Determine whether the points (10, 2) and (12. 3) lie on, this line., , ~j -!.'.:'~~ ,, , . ;1.1..,-, , :tf~;~~':", ...;, , ,, , ', , -:
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Circles, Equations of Circles, For a point P(x, y) to lie on the circle with center C(a,b) and radius r, the distance PC must be equal to r, (see Fig. 4-1). By the distance formula (2.1),, PC =~(x-a)2 +(y-b)2, , Thus, P lies on the circle if and only if, , (x-a)2+(y-b)2 =r2, , (4.1), , Equation (4.1) is called the standard equation of the circle with center at (a, b) and radius r., , Rg.4-1, , EXAMPLE 4.1:, (a) The circle with center (3,1) and radius 2 has the equation (x - 3)2 + (y - 1)2 = 4., (b) The circle with center (2, -1) and radius 3 has the equation (x - 2)2 + (y + 1)2 =9., (c) What is the set of points satisfying the equation (x - 4)2 + (y - 5)2 =25?, By (4.1), this is the equation of the circle with center at (4, 5) and radius 5. That circle is said to be the graph of the, given equation, that is, the set of points satisfying the equation., (d) The graph of the equation (x + 3)2 + y2 =2 is the circle with center at (-3, 0) and radius .fi., , The Standard Equation of a Circle, The standard equation of a circle with center at the origin (0, 0) and radius r is, x2+y2 = r2, , (4.2), , For example, x2 + f = 1 is the equation of the circle with center at the origin and radius 1. The graph of, =5 is the circle with center at the origin and radius $., The equation of a circle sometimes appears in a disguised form. For example, the equation, , xl + f, , x2+y2+8x-6y+21=0, , (4.3)
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---4_, , CHAPTER 4 Circles, , SOLVED PROBLEMS, , 1., , Identify the graphs of (a) J.x2 + 2'1 - 4x + y + 1 = 0; (b) r + 'I - 4y + 7 = 0; (c) r + 'I - 6x - 2y + 10 = O., (a) First divide by 2, obtaining x 2 + yl - 2x + t y + t = O. Then complete the squares:, (x 2, , -, , 2x + 1) + (yl + t y + fr;) + t = I + -k =-it, (x-l)2 +(y+ t)2 =-\t-t=-it-'\-=i\-, , Thus, the graph is the circle with center (1, -t) and radius t., (b) Complete the square:, , x2 +(y_2)2 +7=4, x 2 + (y - 2)2 = -3, , Because the right side is negative, there are no points in the graph., (c) Complete the square:, , (X-3)2 +(y-l)2 +10=9+1, (x - 3)2, , +(y - 1)2 =0, , The only solution is the point (3, 1)., , 2., , Find the standard equation of the circle with center at C(2, 3) and passing through the point P( -1, 5)., The radius of the circle is the distance, CP=~(5-3)2+(-1-2)2 =~22+(_3)2 =~4+9=J13, , So the standard equation is (x - 2)2 + (y - 3)2 = 13., 3., , Find the standard equation of the circle passing through the points P(3, 8), Q(9, 6), and R(13, -2)., First method: The circle has an equation of the form r +'I + Ax + By + C = O. Substitute the values of x and y, at point P, to obtain 9 + 64 + 3A + 8B + C = 0 or, 3A + 8B+ C =-73, , (I), , A similar procedure for points Q and R yields the equations, , 9A+6B+C=-1l7, , (2), , I3A - 2B+ C =-173, , (3), , Eliminate C from (I) and (2) by subtracting (2) from (I):, , -6A+2B=44 or -3A+B=22, , (4), , Eliminate C from (1) and (3) by subtracting (3) from (1):, , .,..lOA+IOB=l00 or -A+B=lO, , (5), , Eliminate B from (4) and (5) by subtracting (5) from (4), obtaining A =-6. Substitute this value in (5) to find that, B =4. Then solve for C in (I): C =-87., Hence, the original equation for the circle is xl +Y - 6x + 4y - 87 =O. Completing the squares then yields, (x- W +(y+ 2)2 =87 +9+4= 100, , Thus, the circle has center (3, -2) and radius 10., Second method: The perpendicular bisector of any chord of a circle passes through the center of the circle., Hence, the perpendicular biseCtor ;e of chord PQ will intersect the perpendicular bisector .M. of chord QR at the, center of the circle (see Fig. 4-2).
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CHAPTER 4 Circles, , The perpendicular bisector .At of chord PQ in Fig. 4-3 also passes through the center of the circle, so the, intersection of ~ and .M. will be the center of the circle. The slope of PQ is I. Hence. by Theorem 3.2, the slope, of.M. is -1. So .M. has the slope-intercept equation y =-x + h'. Since the midpoint (2, 2) of chord PQ is a point on, .M., we have 2 = -(2) + h'; hence, h' =4, and the equation of.M. is y =-x + 4. We must find the common solution, ofy =-x+4 and y =-tx+!. Setting, , ,", , -x+4=-tx+t, ,', , yields x = -1. Therefore, y = -x + 4 = - (-I) + 4 = 5, and the center C of the circle is (-I, 5). The radius is, the distance PC= ~(-I- 3)2 +(5- 3)2 =./16+ 4 =flO. The standard equation of the circle is then, (x + 1)2 + (y - 5)2 = 20., 5., , 1, , .. ',.,, , ;.~:r;-?:, . ,.<~ ~~, . .~~~;., , "'~", , Find the standard equation of every circle that passes through the points P(l. -I) and Q(3. I) and is tangent to the, line y = -3x: •, Let C(c, d) be the center of one of the circles, and let A be the point of tangency (see Fig. 4-4). Then, because, CP =CQ, we have, Cp2 = CQ 2, , or (c-l)2 +(d+ 1)2 =(c-3)2 +(d _1)2, , Expanding and simplifying. we obtain, c+d=2, , (I), , y= -3x, , ------~~--~--~--------~---------x, , Fig. 4-4, ., 3c+d, -2-2, In addition, CP = CA. and by the formula of Problem 8 in Chapter 3. CA = JW . Setting CP =CA thus yields, (c - 1)2 +(d + 1)2 = (3c;0d)2 . Substituting (1) in the right-hand side and, , ,., , mUlti~?Ying by 10 then yields, , .• "'jJo;, j, , 1O[(c - 1)2 + (d + 1)2] =(2c + 2)2, , from which, , 3(.'2 + 5d 2 - 14(.' + lOci + 8 = 0, , By (l), we can replace d by 2 - c, obtaining, , 2c 2 -llc+12=O, , or, , (2c-3)(c-4)=O, , Hence, c =t or c =4. Then (I) gives us the two solutions c =t, d =t and c =4. cI = -2. Since the radius, CA =3c: / ' these solutions produce radii of, , viO, , b =JW210 and vlO~ =JW. Thus, there are two such circles, and, , vlO, , • ~, •, , , ',' ~J~ , .
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.....----, , I', , CHAPTER 4 Circles, , their standard equations are, , (x-~Y +{y-{Y =~, , 6., , and, , Find the standard equation~ of the circles satisfying the following conditions:, (a) center at (3, '5) and radius 2, (c) center at (5,0) and radius J3, (e) center at (-2, 3) and passing through (3, -2), Ans., , 7., , Identify the graphs of the following equations:, , Ans., , (b) r+y1-"4x+5y+ 10=0, (d) 4X2+4y2+8y-3=0, (f), , x 2 + i + J2x - 2 = 0, , (a) circle, <:enter at (-8.6). radius 3JW : (b) circle, center at (2, -t), radius f: (c) circle, center at (-f .f)., radius J2/2; (d) circle, center at (0, -1), radius t; (e) empty graph; (f) circle. center at (-J2I2,O),, radius .J5fi, , Find the standard equations of the circles through (a) (-2, I), (l, 4), and (-3, 2); (b) (0, 1), (2, 3), and (1, 1+.J3);, (c) (6, I), (2, -5), and (1, -4); (d) (2, 3), (-6, -3), and (l, 4)., Ans., , 9., , (b) center at (4, -1) and radius 1, (d) center at (-2, -2) and radius 5J2, (f) center at (6, 1) and passing through the origin, , (a) (x- 3)2 + (y - 5)2= 4; (b) (x- 4)2 + (y+ 1)2= 1; (c) (x- 5)2+ i= 3; (d) (X+2)2+ (y + 2)2 = 50;, (e) (x+ 2)2 + (y- 3)2= 50; (f) (x - 6)2 +(y-l)2 = 37, , (a) x 2 +i+16x-12y+IO=0, (c) x2 + y2 + x - Y = 0, (e) r + y2 - x - 2y + 3 = 0, , 8., , (x-4)2+(y+2)2=10, , (a) (x+ 1)2+ (y- 3)2= 5; (b) (x- 2)2+ (y-l)2=4; (c) (X-4)2+(y +2)2 = 13; (d) (X+2)2+ y2 = 25, , For what values of k does the circle (x + 2k)2 + (y - 3k)2 = 10 pass through the point (1, O)?, Ans., , k=rr ork=-1, , 10. Find the standard equations of the circles of radius 2 that are tangent to both the lines x = 1 and y =3., Ans . . (x+ 1)2 + (y_I)2=4; (x+ 1)2 + (y- 5)2=4; (x- 3)2+ (y_I)2 =4; (x- 3)2+ (v"- 5)2 = 4, , 11. Find the value of k so that r + y2 + 4x - 6y + k = 0 is the equation of a circle of radius 5., Ans., , k =-12, , 12. Find the standard equation of the circle having as a diameter the segment joining (2, -3) and (6, 5)., , }}~:~'~j, , 13. Find the standard equation of every circle that passes through the origin. has radius 5. and is such that the y, coordinate of its center is -4., , ',"'t'l,\~'~:'~,", .'. t, , I (, , ,-~.; :'.", , ,;::, , Ans., , (x - 3)2 + (y + 4)2 = 25 or (x + 3)2 + 'y + 4)2 = 25
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CHAPTER 4 Circles, , 14. Find the standard equation of the circle that passes through the points (8, -5) and (-1, 4) and has its center on the, line 2x + 3y =3., , 15. Find the standard equation of the circle with center (3, 5) that is tangent to the line 12x - 5y + 2 = O., Ans., , (x - 3)2 + (y - 5)2 = 1, , 16. Find the standard equation of the circle tl)at passes through the point (1,3 +../2) and is tangent to the line x + y =2, , at (2,0)., , 17. Prove analytically that an angle inscribed in a semicircle is a right angle. (See Fig. 4-5.), , 18. Find the length of a tangent from (6, -2) to the circle (x - 1)2 + (y - 3)2 = 1. (See Fig. 4-6.), Ans., , 7, y, y, , ---4~------~------~~x, , (-r,O), , ---+--------~--------x, , (r,O), , (6, -2), , Fig. 4·5, , Fig. 4-6, , 19. Find the standard equations of the circles that pass through (2, 3) and are tangent to both the lines 3x - 4y = -I, and 4x + 3y = 7., , 6)2 + ( Y -"5, 12)2 = I, Ans. (x - 2)2 + (y - 8)2 = 25 and ( x - '5, 20. Find the standard equations of the circles that have their centers on the line 4x + 3y =8 and are tangent to both, the lines x + y =-2 and 7x - y =-6., , ,, , 21. Find the standard equation of the circle that is concentric with the circle x2 + y2 - 2x - 8y + I = 0 and is tangent to, the line 2x - Y =3., , 22. Find the standard equations of the circles that have radius 10 and are tangent to the circle xl +y2 =25 at the point (3. 4).
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CHAPTER 4 Circles, 23. Find the longest and shortest distances from the point (7, 12) to the circle xl + f + 2x + 6y·- 15 = O., Ans., , 22 and 12, , 24. Let ~I and ~2 be two intersecting circles determined by the equations xl + f + Alx + Bly + CI = 0 and x2 + f + ., A~ + B~ + C2 = O. For any number k *- -I, show that, , ,, , x 2 + y2, , + Alx + Bly + CI + k(x + y2 + A 2x + B2y + C 2 ) = 0, 2, , is the equation of a circle through the intersection points of ~ I and ~2' Show, conversely. that every such circle, may be represented by such an equation for a suitable k., 25. Find the standard equation of the circle passing through the point (-3, 1) and containing the points of intersection, , of the circles xl + y2 + 5x = 1 and x2+ f + y =7., , Am., , (Use Problem 24.), , 2, (, 3)2 569, (x+1) + Y+1O =100, , 26. Find the standard equations of the circles that have centers on the line 5x - 2y = -21 and are tangent to both, coordinate axes., Ans., , (x, , + 7)2 + (y + 7)2 =49 and (x + 3)2 + (y - 3)2 =9, , 27. (a) If two circles xl + f + Alx + Bly + CI = 0 and xl + l- + A~ + B~ + C2 = 0 intersect at two points. find an, equation of the line through their points of intersection., (b) Prove that if two circles intersect at two poinL~. then the line through their points of intersection is, perpendicular to the line through their centers., , 28. Find the points of intersection of the circles xl + f + 8y - 64 = 0 and xl + f, Ans., , (8, 0) and, , - 6x -, , 16 =O., , (-it ,Jf ), , 29. Find the equations of the lines through (4. 10) and tangent to the circle x2 + y2 - 4y - 36 =O., Ans., , y=-3x+22andx-3y+26=0, , 30. (GC) Use a graphing calculator to draw the circles in Problems 7(d). 10, 14. and 15. (Note: It may be necessary, to solve for y.), , 31. (GC) (a) Use a graphing calculator to shade the interior of the circle with center at the origin and radius 3., (b) Use a graphing calculator to shade the exterior of the circle xl + (y - 2)2 = l., , 32. (GC) Use a graphing calculator to graph the following inequalities: (a) (x - 1)2 + y2 < 4; (b) xl + f, , - 6x -, , 8y > O.
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.Equations and Their Graphs, . The Graph of an Equation, The graph of an equation involving x and y as its only variables consists of all points (x, y) satisfying the, equation., EXAMPLE 5.1: (a) What is the graph of the equation 2x - Y = 31, The equation is equivalent to y = 2x - 3, which we know is the slope-intercept equation of the line with slope 2, and y intercept -3., (b) What is the graph of the equation xl +y2 - 2x + 4y - 4 =01, Completing the square shows that the given equation is equivalent to the equation (x - 1)2 + (y + 2)2 =9. Hence, its, graph is the circle with center (I, -2) and radius 3., , Parabolas, Consider the equation y =r. If we substitute a few values for x and calculate the associated values of y, we, obtain the results tabulated in Fig. 5-1. We can plot the corresponding points, as shown in the figure. These, points suggest the heavy curve, which belongs to a family of curves called parabolas. In particular, the, graphs of equations of the formy =cx2, where c is a nonzero constant, are parabolas, as are any other curves, obtained from them by translations and rotations., y, , x, , y, , 3, , 9, 4, , 2, I, , 0, , I, 0, , -), , -2, , -3, , 4, 9, -) -2 -I, , o, , x, 1, , 2, , ), , Rg.5-1, , In Fig. 5-1, we note that the graph of y =:Xl contains the origin (0, 0) but all its other points lie above the, x axis, since r is positive except when x =O. When x is positive and increasing, y increases without bound., , Hence, in the first quadrant, the graph moves up without bound as it moves right. Since (_X)2 =x 2, it follows, that, if any point (x, y) lies on the graph in the first quadrant, then the point (-x, y) also lies on the graph in, the second quadrant. Thus, the graph is symmetric with respect to the y axis. The y axis is called the aXis of, symmetry of this parabola.
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CHAPTER 5 Equations and Their Graphs, , Ellipses, ., , 2, , 2, , To construct the graph of the equation ~ + ~ = 1, we again compute a few values and plot the corresponding points, as shown in Fig. 5-2. The graph suggested by these points is also drawn in the figure; it is a, 2, , 2, , member of a family of curves called ellipses. In particular, the graph of an equation of the form x 2 + ~2 = I, is an ellipse, as is any curve obtained from it by translation or rotation., 2, a2, Note that, in contrast to parabolas, ellipses are bounded. In fact, if (x, y) is on the graph of xl), + Y4 = 1, then, 2, , 2, , 2, , "7, , .~ ~ ~ + ~ =I, and, therefore,r~ 9. Hence, -3 ~x~ 3. So, the graph lies between the vertical lines x =-3, and x = 3. Its rightmost point is (3, 0), and its leftmost point is (-3, 0). A similar argument shows that the, , graph lies between the horizontal lines y = -2 and y =2, and that its lowest point is (0. -2) and its highest point, is (0, 2). In the first quadrant, as x increases from 0 to 3, y decreases from 2 to O. If (x, y) is any point on the, graph, then (-x, y) also is on the graph. Hence, the graph is symmetric with respect to the y axis. Similarly, if, (x, y) is on the graph. so is (x, -y). and therefore the graph is symmetric with respect to the x axis., y, , x, , y, , 3, 2, , 0, ~lvs .. :t1.S, dv'2 .. :t1.9, , (J, , ±2, :t~V2, , -I, , (x. y), , (-x. y), , x, , -3, , (x. -y), , ±jV3, , -2, -3, , -2, , 0, , Fig. 5-2, 2, , 2, , When a =b, the ellipse Q\ + Yb 2 =1 is the circle with the equation, r + y2 =a2 , that is, a circle with center, ., at the origin and radius a. Thus, circles are special cases of ellipses., , Hyperbolas, 2, , 2, , Consider the graph of the equation ~ - ~ = 1. Some of the points on this graph are tabulated and plotted in, Fig. 5-3. These points suggest the curve shown in the figure, which is a member of a family of curves called, x 2 y2, hyperbolas. The graphs of equations of the form - 2 - b 2' = 1 are hyperbolas, as are any curves obtained, from them by translations and rotations., a, y, , x, , y, , ~3, , 0, , ±4, ±5, , ±, ± 1.76, ±, :t2.67, :t2v'3 ... ±3.46, , ~6, , "-, , "-, , h''''''', t ..., , " ......, , "- ......, , ......, , "-2, ;', , /', , "", , "", , ", , 2, , ......, ;', , ", , -2, , -4, , Rg.5-3, , x 2 y2, x2, y2, Let us look at the hyperbola 9 - "4 = I in more detail. Sin.:e 9 = 1+ "4 ~ I, it follows that r ~ 9. an I, therefore, Ixl ~ 3. Hence, there are no points on the graph between the vertical lines x =-3 and x = 3. If (x, y)
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CHAPTER 5 Equations and Their Graphs, , is on the graph, so is (-x, y); thus, the graph is symmetric with respect to the y axis. Similarly, the graph is, symmetric with respect to the x axis. In the first quadrant, as x increases, y increases without bound., , Rg.5-4, , Note the dashed lines in Fig. 5-3; they are the lines y =tx and y = -tx, and they are called the asymptotes of the hyperbola: Points on the hyperbola get closer and closer to these asymptotes as they recede from, the origin. In general, the asymptotes of the hyperbola x: - Yb:, a, , =1 are the lines y =!!..a x, , and y =-!!.. x., a, , Conic Sections, Parabolas, ellipses, and hyperbolas together make up a class of curves called conic sections. They can be, defined geometrically as the intersections of planes with the surface of a right circular cone, as shown in, Fig. 5-4., , SOLVED PROBLEMS, , 1., , Sketch the graph of the cubic curve y = xl., The graph passes through the origin (0, 0). Also, for any point (x, y) on the graph, x and y have the same sign;, hence, the graph lies in the first and third quadrants. In the first quadrant, as x increases, y increases without, bound. Moreover. if (x, y) lies on the graph, then (-x, -y) also lies on the graph. Since the origin is the midpoint, of the segment connecting the points (x, y) and (-x, -y), the graph is symmetric with respect to the origin. Some, points on the graph are tabulated and shown in Fig. 5-5; these points suggest the heavy curve in the figure., y, , x, , y, , 0, , 0, , 1/2, , 1/8, , I, , I, , 312, , 27/8, , 2, , 8, , -112, , -1/8, , -I, , -I, , -312, -2, , -27/8, , x, , -8, Rg.~5
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CHAPTER 5 Equations and The;r Graphs, , 2., , -r., , Sketch the graph of the equation y =, If (x. y) is on the graph of the parabola y = (Fig. 5-1). then (x. -y) is on the graph of y =, and vice, versa. Hence. the graph of y =, is the reflection in the x axis of the graph of y = The result is the parabola, in Fig. 5-6., , -r, , 3., , .., , r., , -r., , Sketch the graph of x = )'2., This graph is obtained from the parabola y =x2 by exchanging the roles of x and y. The resl!lting curve is a, parabola with the x axis as its axis of symmetry and its "nose" at the origin (see Fig. 5-7). A point (x. y) is on, the graph of x =y2 if and only if (y. x) is on the graph of y =r. Since the segment connectinjthe points (x. y), and (y., , {;;.;), , r, , x) is perpendicular to the diagonal line y =x(why?). and the midpoint ( x; y. x; Y, , of that segment, , is on the line y = x (see Fig. 5-8). the parabola x = y2 is obtained from the parabola y =r by reflection in the, \', , line y=x., , :~ :~-: -, , y, , -3 -2 -\, 0 \, 2, 3, ---'--r-~~~-r-'----------X, , Fig. 5-6, y, , y, , ,,,, (x. y), , ________, , ~~---------------------x, , Fig. 5-7, Fig. 5-8, , 4., , Let :£ be a line. and let F be a point not on :£. Show that the set of all points equidistant from F and :£ is a, parabola., Construct a coordinate systlem such that F lies on the positive y axis. and the x axis is parallel to ~ and, halfway between F and ~. (See Fig. 5-9.) Let 2p be the distance between F and~. Then ~ has the equation y =-po, and the coordinates of F arc (0. pl.
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CHAPTER 5 Equations and Their Graphs, , Consider an arbitrary point P(x, y). Its distance from ;e is Iy +pi, and its distance from F is ~X2 + (y _ p)2 •, Thus, for the point to be equidistant from F and;e, we must have Iy + pi =~r + (y - p)2 • Squaring yields, (y +p)2, + (y _p)2, from which we find that 4py, This is the equation of a parabola with the y axis as its, axis of symmetry. The point F is called the focus of the parabola, and the line!£ is called its directrix. The chord, AB through the focus and parallel to;e is called the latus rectum. The "nose" of the parabola at (0, 0) is called its, vertex., , =r, , =r., , y, , ---------+----~_+~-----------------------x, , Fig. 5-9, , 5., , Find the length of the latus rectum of a parabola 4py =r., The y coordinate of the endpoints A and B of the lactus rectum (see Fig. 5-9) is p. Hence, at these points. 4p2 =r, and, therefore, x =±2p. Thus, the length AB of the latus rectum is 4p., , 6., , Find the focus, directrix, and the length of the latus rectum of the parabola y = f x 2, and draw its graph., The equation of the parabola can be written as 2y =r. Hence, 4p =2 and p =f. Therefore, the focus is, at (0, f), the equation of the directix is y =-f. and the length of the latus rectum is 2. The graph is shown in, Fig. 5-10., , ~----------~,,~__~--~--___, -3, -2, -\, , x, , Rg.5-10, , 7., , Let F and F' be two distinct points at a distance 2c from each other. Show that the set of all points P(x, y) such, that PF + PF' =2a. a> c is an ellipse.
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CHAPTER 5 Equations and Their Graphs, , r,, , Construct a coordinate system such that the x axis passes through F and, the origin is the midpoint of the, segment Fr, and F lies on the positive x axis. Then the coordinates of F and rare (c, 0) and (-c, 0)., y, , B(O. b), , -----t--. .----~~----~~~------x, , B'(O.-b), , Rg.5-11, , (See Fig. 5-11.) Th~s, the condition PF + PF' = 2a is equivalent to J(x - c)2 + y2 + J(x + C)2 + y2 = 2a., After rearranging and squaring twice (to eliminate the square roots) and performing indicated operations. we, obtain, , (1), , Since a > c, a2 - c' > O. Let b = ../a 2 - c'. Then (1) becomes blxl + alyl = a2b2• which we may rewrite as, , x2, , y2, , (j2 + bf = I, the equation of an ellipse., When y = 0, xl = a2; hence. the ellipse intersects the x axis at the points A'(-a, 0), and A(a. 0), called the, vertices of the ellipse (Fig. 5-11). The segment A'A is called the major axis; the segment OA is called the, semimajor axis and has length a. The origin is the center of the ellipse. F and F' are called the foci (each is, afocus). When x = 0, yl = bl. Hence. the ellipse intersects the y axis at the points 8'(0, -b) and 8(0, b). The, segment B' B is called the millor axis; the segment 08 is called the semimillor axis and has length b. Note that, b = ../a 2 - c2 < J;;2 = a. Hence. the semiminor axis is smaller than the semimajor axis. The basic relation among, a, b, and c is a2 = b2 + c2., The eccentricity of an ellipse is definec. tv be e=da. Note thatO< e < 1. Moreover. e= ../a1 - b 2 /a =Jl-(bla)2 ., Hence. when e is very small. bta is very close to I. the minor axis is close in size to the major axis, and the ellipse, is close to being a circle. On the other hand, when e is close to I, bta is close to zero, the minor axis is very small in, comparison with the major axis. and the ellipse is very "flat.", 8., , Identify the graph of the equation 9xl + 16y2 = 144., The given equation is equivalent to x 2116 + yl/9 = I. Hence, the graph is an ellipse with semimajor axis, of length a =4 and semiminor axis of length b =3. (See Fig. 5-12.) The vertices are (-4, 0) and (4.0). Since, c = ../a 2 - b 2 =../16 - 9 =J7, the eccentricity e is cia = J714 '" 0.6614., , 9., , Identify the graph of the equation 25xl + 4y2 = 100., The given equation is equivalent to xl/4 +yl125 = I, an ellipse. Since the denominator under y2 is larger, than the denominator under xl, the graph is an ellipse with the major axis on the y axis and the minor axis on, the xaxis (see Fig. 5-13). The vertices are at (0. -5) and (0, 5). Since c =../a2 - b 2 =..fil, the eccentricity is, J2I15", 0.9165.
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CHAPTER 5 Equations and Their Graphs, y, , --_~2+----r--~~-X, , -5, , -3, , Rg.5-12, , Fig. 5-13, , 10. Let F and F' be distinct points at a distance of 2c from each other. Find the set of all points P(x, y) such that, IpF - PF'I = 2a, for a < c., Choose a coordinate system such that the x axis passes through F and F', with the origin as the midpoint of, the segment FF' and with F on the positive x axis (see Fig. 5-14). The coordinates of F and F' are (c, 0) and, (-c, 0). Hence, the given condition is equivalent to ~(x - C)2 + y2 - ~(x + C)2 + y2 = ±2a. After manipulations, required to eliminate the square roots, this yields, (I), , Since c > a, c2 - a1 > O. Let b =.Jc2 - a2 • (Notice that a2 + b2 =c2.) Then (I) becomes b2x2 - a2f, x2, , y2, , =a1b2, which, , we rewrite as a2 - b2 = I, the equation of a hyperbola., When y =0, x =±a. Hence, the hyperbola intersects the x axis at the points A'(-a, 0) and A(a, 0), which are, called the vertices of the hyperbola. The asymptotes are y =±!!.x. The segment A'A is called the transverse axis., a, , The segment connecting the points (0, -b) and (0, b) is called the conjugate axis. The center of the hyperbola is, the origin. The points F and F' are called the foci. The eccentricity is defined to be e =, , J., , ~ = ~ ~1 +(%, , Since c > a, e > I. When e is close to 1, b is very small relative to a, and the hyperbola has a very pointed "nose";, when e is very large, b is very large relative to a, and the hyperbola is very "flat.", y, , ---------*~~~~~~~-------x, , Fig. 5-14
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CHAPTER 5 Equations and Their Graphs, 11. Identify the graph of the equation 25xl - 16f =400., The given equation is equivalent to xl/16 - fl25 =1. This is the equation of a hyperbola with the x axis as its, transverse axis, vertices (-4, 0) and (4, 0), and asymptotes y= ±tx. (See Fig. 5.15.), y, /, , I, I, /, , I, /, , /, , ------~~~~, , __, , /, , /, , ----~-----------x, , -s, , Rg.5-15, , 12. Identify the graph of the equation f, , - 4xl2 = 4. 2, , The given equation is equivalent to Y4 - ~ = 1. This is the equation of a hyperbola, with the roles of x and, )' interchanged. Thus, the transverse axis is the y axis, the conjugate axis is the x axis, and the vertices are (0, -2), and (0, 2). The asymptotes are x = ±t y or, equivalently, y =±2t. (See Fig. 5-16.), 13. Identify the graph of the equation y = (x - If., A point (u, v) is on the graph of y = (x - 1)2 if and only if the point (u - 1, v) is on the graph of y =xl. Hence,, the desired graph is obtained from the parabola y = xl by moving each point of the latter one unit to the right., (See Fig. 5-17.), (x - 1)2, , (y - 2)2, , 14. Identify the graph of the equation ---4-- + 9, 1., A point (u, v) is on the graph if and only if the point (u - I, v - 2) is on the graph of the equation xl/4 + yl9 = 1., Hence, the desired graph is obtained by moving the ellipse x2/4 + il9 = lone unit to the right and two units, upward. (See Fig. 5-18.) The center of the ellipse is at (I, 2), the major axis is along the line x = 1, and the minor, axis is along the line y =2., , \, , \, , \, \, \, , I, , \, -I, , /, /-1, /, , /, /, /, Fig. 5-16
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CHAPTER 5 Equations and Their Graphs, y, , y, , (1,5), , (-1,2), , (3,2), , (1,-1), , Ag.5-18, -3, , -2, , -I, , 0, , 3, , 4, , Ag.5-17, , 15. How is the graph of an equation F(x - a, Y - b) = 0 related to the graph of the equation F(x, y) = O?, A point (u, v) is on the graph of F(x - a, y - b) =0 if and only if the point (u - a, v - b) is on the graph of, F(x, y) = O. Hence, the graph of F(x - a, y - b) = 0 is obtained by moving each point of the graph of F(x, y) = 0, by a units to' the right and b units upward. (If a is negative, we move the point lal units to the left. If b is negative,, we move the point Ibl units downward.) Such a motion is called a lranslalion., 16. Identify the graph of the equation y =xl - lx., Competing the square in x, we obtain y + 1 =(x - I f Based on the results of Problem 15, the graph is, obtained by a translation of the parabola y =xl so that the new vertex is (I, -I). [Notice that y + I is)' - (-I }.J It, is shown in Fig. 5-19., y, , -, '-.-, , ,, ____, , ~, , __, , ~, , __, , ~~~~, , __, , ~, , ____ x, , ,":", , Ag.5-19
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Equations and Their Graphs, , CHAPTER 5, , 17. Identify the graph of 4,x2 - 9f - 16x + 18y - 29 =0., Factoring yields 4(x2 - 4x) - 9(f - 2y) - 29 = 0, and then 2completiny the square in x and y produces, 4(x - 2)2 - 9(y - 1)2= 36. Dividing by 36 then yields, , (x~ 2), , (y ~ 1) = 1. By the results of Problem IS, the, 2, , 2, , graph of this equation is obtained by translating the hyperbola x9 - Y4 = 1 two units to the right and one unit, upward, so that the new center of symmetry of the hyperbola is (2, I). (See Fig. 5-20.), 18. Draw the graph of the equation xy = I., Some points of the graph are tabulated and plotted in Fig. 5-21. The curve suggested by these points is shown, dashed as well. It can be demonstrated that this curve is a hyperbola with the line y = x as transverse axis. the line, y =-x as converse axis, vertices (-1, -1) and (1, I), and the x axis and y axis as asymptotes. Similarly. the graph, of any equation .xy == d, where d is a positive constant, is a hyperbola with y = x as transverse axis and y = -x as, converse axis. and with the coordinate axes as asymptotes. Such hyperbolas are called equilateral hyperbolas., They can be shown to be rotations of hyperbolas of the form x2la 2 -y2la2 = I., y, ",, , /', , ,/, , ./, , ,/, /', , ", , ,//', , ;.~.I), , "-, , /', , --------~--~~/'~------~~--~-------x, /', , /', , /', , /', /', , Rg., , ~20, , y, , x, 3, 2, , y, , ,, , 1/3, , \, , 112, , 2, , 112, 1/3, , 114, -1/4, -113, -1/2, , -1, -2, -3, , ~, , \, , 1, , 4, -4, , ----- ... - ......, , \, , "" .......- -- ...., , 2, 3, , ', , ----~--~--~--~--~--~, , -3, -2, , .... ", \, , -1, -112, , ,, \, , -113, , ,,, •, , -2, , -3, , I, I, •, I, , Fig., , ~21, , -4, , __, , ~, , __, , ~, , ________ x
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CHAPTER 5 Equations and Their Graphs, , 19. (a) On the same sheet of paper, draw the graphs of the following parabolas:, (i) y = 2r;, (ii) y = 3r;, (iv) y=tx2;, (v) y=tx2., (b) (Ge) Use a graphing calculator to check your answers to (a)., , (iii) y = 4r;, , 20. (a) On the same sheet of paper, draw the graphs of the following parabolas and indicate points of intersection:, y =r;, (ii) y =-r;, (iv) x=-f., (b) (GC) Use a graphing calculator to check your answers to (a)., , (i), , (iii) x= f;, , 21. Draw the graphs of the following equations:, (a) y=x3-1, (d) y=-x3, , (b) y= (x-2)3, (e) y=-(x-l)3, , (c) y=(x+l)3-2, (f) y=-(x-I)3+2, , 22. (Ge) Use a graphing calculator to answer Problem 21., , 23. Identify and draw the graphs of the following equations:, (a) f-xi=1, (d) xy=4, , lOy =r, (j) 3f, =9, , (g), , -r, , Ans., , (b) 25r + 36f =900, (e) 4r+4f= I, (h) 4r+9f= 16, , (c) 2r-y2=4, , (f) 8x= f, (i) xy=-I, , (a) hyperbola. y axis as transverse axis, vertices (0, ±I), asymptotes y =±r, (b) ellipse, vertices (±6. 0), foci (±./IT, 0); (c) hyperbola, x axis as transverse axis, vertices (±J2,0), asymptotes y =±xJ2x;, (d) hyperbola, y =x as transverse axis, vertices (2, 2) and (-2, -2), x and y axes as asymptotes; (e) circle,, center (0, 0), radius t; (f) parabola, vertex (0, 0), focus (2, 0), directrix x = -2; (g) parabola, vertex (0,0),, focus (0, t), directrix y =-t; (h) ellipse, vertices (±2, 0), foci (±tv'5',O); (i) hyperbola, y =-x as transverse, axis, vertices (-I, 1) and (1, -I), x and y axes as asymptotes; (j) hyperbola. y axis as transverse axis, vertices, (0, ±.j3), asymptotes y =±X.J3x/3, , 24. (GC) Use a graphing calculator to draw the graphs in Problem 23., , 25. Identify and draw the graphs of the following equations:, , r-, , (b) 5r+y2-2Ox+6y+25=0, (c), 6x-4y+ 5 =0, (e) 3r+2f+ 12x-4y+ 15=0, (f) (x-l)(y+2)= I, (g) xy - 3x - 2y + 5 =0 [Hint: Compare (f).], (h) 4r+f+8x+4y+4=0, (i) 2x2-8x-y+ll=O, (j) 25r+16y2-IOOx-32y-284=0, , (a) 4r-3f+8x+12y-4=0, (d) 2r+y-4x+4y+6=0, , Ans., , (a) empty graph; (b) ellipse. center at (2, -3); (c) parabola, vertex at (3, -1); (d) single point (l, -2);, (e) empty graph; (f) hyperbola, center at (1, -2); (g) hyperbola, center at (2,3); (h) ellipse, center at, (-1,2); (i) parabola, vertex at (2,3); (j) ellipse, center at (2, I), , 26. (GC) Use a graphing calculator to draw the graphs in Problem 25.
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CHAPTER 5 Equations and Their Graphs, , ··\"i~if, ., t!;1"', , I, , :~, , ;;, , ~, , JJ~, , 27. Find the focus, directIix, and length of the latus rectum of the following parabolaS: (a) lor =3y; (b) 21 =3x;, (c) 4y =x2 + 4x + 8; (d) 8y =, , -r., , i),, , (t, , =-i., , =-to, , directrix y, latus rectum,b-; (b) focus at ,0), directrix X, latus rectum, (c) focus at (-2. 2), directrix y = O. latus rectum 4; (d) focus at (0. -2). directrix y = 2, latus rectum 8, , Ails. (a) focus al(O,, , ~, , t;, , 28. Find an equation for each parabola satisfying the following conditions:, , =, , Focus at (0, -3). directrix y =3, (b) Focus at (6, 0), directrix x 2, Foclls lit (1. 4), directrix y =0, (d) Vertex at (1. 2). focus at (I. 4), Vertex at (3, 0), directrix x = 1, Vertex at the origin, y axis as axis of symmetry, contains the point (3, 18), Vertex at (3, 5), axis of symmetry parallel to the y axis, contains the point (5, 7), Axis of symmetry parallel to the x axis, contains the points (0, 1), (3, 2), (1, 3), Latus rectum is the segment joining (2, 4) and (6, 4), contains the point (8, 1), (j) Contains the points (1. 10) and (2. 4), axis of symmetry is vertical. vertex is on the line 4x - 3y = 6, , (a), (c), (e), (f), (g), (h), (i), , ~~, , j.<--~-;, , -r;, , (j) Y - 2, , ;".,:, , =, , (a) 12y =, (b) 8(x - 4) I; (c) 8(y - 2) = (x - 1)2; (d) 8(y - 2) =(x - 1)2; (e) 8(x - 3), (f) y= 2il; (g) 2(y - 5) =(x - W; (h) 2(x- l1J)=-5(y-#Y; (i)4(y- 5) = -(x- 4)2;, , Ans., , .~~?~~ +, , =2(x -, , =f;, , 3)2 or y-i\= 26(x-f!Y, , 29. Find an equation for each ellipse satisfying the following conditions:, (a), (b), (c), (d), (e), (I), , Center at the origin, one focus at (0, 5). length of semimajor axis is 13, Center at the origin, major axis on the y axis, contains the points (I. 2J3) and, Center at (2, 4), focus at (7. 4), contains the point (5,8), Center at (0, 1). one vertex at (6, 1), eccentricity t, Foci at (0. ±t). contains (t, 1), Foci (U, ±9), semiminor axis of length 12, , x2, (a) 144, , AilS., , x2, (f) 144, , x2, , y2, , y2, , + 169 = 1; (b) "4 + 16 =1; (c), , (x - 2)2, 45, , +, , (t. Jf5), , (y - 4)2, x 2 (y -1)2, 20 -1; (d) 36 +, , --w- =1; (e) x, , 2, , 9y2, , + 25 = 1;, , y2, , + 225 = 1, , 30. Find an equation for each hyperbola satisfying the following conditions:, 1· ", , :~~~~, , (a), (b), (c), (d), , Center at the origin, transverse axis the x axis, contains the points (6, 4) and (-3, 1), Center at the origin, one vertex at (3, 0), one asymptote is y = t x, Has asymptotes y ±fix, contains the point (1, 2), Center at the origin, one focus at (4. 0), one vertex at (3, 0), , =, , i'!'Y-~;i~, , '1, ,.,-, , 31. Find an equation of the hyperbola consisting of all points P(x, y) such that IPF - PF'I= 2fi. where F =(fi,fi), and F' =(-fi,-fi)., AilS., , .\}'=, , 1, x2, , y2, , 32. (GC) Use a graphing calculator to draw the hyperbola, , '9 - "4 =1 and its asymptotes y =±t x., , 33. (GC) Use a graphing calculator to draw the ellipses r, graph obtained from the former one?, , + 4f = 1 and (x - 3)2 + 4(y - 2)2 = 1. How is the latter
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Functions, We say that a quantity y is afunction of some other quantity x i.f the value of y is determined by the value of x., Iffdenotes the function, then we indicate the dependence ofy onx by means of the fonnulay =f(x). The letter, x is called the independent variable, and the letter y is called the dependent variable. The independent variable, is also called the argument of the function, and the dependent variable is called the value of the function., For example, the area A of a square is a function of the length s of a side of the square, and that function can, be expressed by the fonnula A =sl. Here, s is the independent variable and A is the dependent variable., The domain of a function is the set of numbers to which the function can be applied, that is, the set of, numbers that are assigned to the independent variable. The range of a function is the set of numbers that the, function associates with the numbers in the domain., EXAMPLE 6.1: The formula/(x) =Xl detennines a function/that assigns to each real number x its square. The domain consists of all real numbers. The range can be seen to consist of all nonnegative real numbers. (In fact, each value, Xl is nonnegative. Conversely, if r is ~ry nonnegative real number, then r appears as a value when the function is applied, to.Jr. since r=(.Jr)2.), EXAMPLE 6.2:, , Let g be the function defined by the formula g(x) =Xl - 4x + 2 for all real numbers. Thus,, g(l) = (1)2 - 4(1) + 2 = 1- 4 + 2 = -1, , and, g(-2) = (_2)2 -4(-2)+ 2=4 +8 + 2= 14, , Also, for any number a, g(a + I) = (a + 1)2 - 4(a + 1) + 2 = a2 + 2a + 1 - 4a - 4 + 2 = a2 - 2a - 1., , EXAMPLE 6.3: (a) Let the function hex) = 18x - 3Xl be defined for all real numbers x. Thus. the domain is the set of, all real numbers. (b) Let the area A of a certain rectangle, one of whose sides has length x, be given by A = 18x - 3Xl., Both x and A must be positive. Now, by completing the square, we obtain, A =-3(X2 - 6x) =-3[(x - 3)2 - 9] =27 - 3(x - 3)2, , Since A> 0, 3(x - 3)2 < 27, (x - 3)2 < 9. Ix - 31 < 3. Hence, -3 < x - 3 < 3. 0 < x < 6. Thus, the function detemlining A has the open interval (0, 6) as its domain. The graph of A = 27 - 3(x - 3)2 is the parabola shown in Fig. 6-1., From the graph, we see that the range of the function is the half-open interval (0, 27)., Notice that the function of part (b) is given by the same formula as the function of part (a), but the domain of the, former is a proper subset of the domain of the latter.
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CHAPTER 6, , Functions, , A, 27, , --O~------~3-------6L-----X, , Fig. 6-1, , The graph of a functionJis defined to be the graph of the equation y =j{x)., EXAMPLE 6.4: (a) Consider the functionJtx) = Lt!. Its graph is the graph of the equation y = Ixl. and is indicated in, Fig. 6-2. Notice thatf (x) =x when x ~ 0, whereas f (x) = -x when x ~ O. The domain off consists of all real numbers., (/n general. if afurlctioll is given by means of aformula, then. if nothing is said to the cofltrary. we shall asswne that the, domain consists of all numbers for which the fonnula is defined.) From the graph in Fig. 6-2. we see that the range of, the function consists of all nonnegative real numbers. (/11 general. the range of aflme/ioll is the set ofy coordinates of, all points in the graph oflhefunction.) (b) The formula g(x) = 2x +3 defines a function g. The graph of this function is, the graph of the equation y =2x + 3, which is the straight line with slope 2 and y intercept 3. The set of all real numbers, , is both the domain and range of g., y, , Fig. 6-2, , EXAMPLE 6.5:, , Let a function g be defined as follows:, , X2, , g(x) =, , {, , if2~x~4, , x+l, , A function defined in this way is said to be dqined by cases. Notice that the domain of g is the closed interval rI. 4]., , In a rigorous development of mathemaJics, a function J is defined to be a set of ordered pairs such lha t. if, =z. However, such a definition obscures the intuitive meaning of the, notion of function., , (x, y) and (x, z) are in the setf, then y
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CHAPTER 6 Functions, , 'SOLVED PROBLEMS, , 1., , x-I, , Given f(x) =x 2 + 2' find (a)f(O); (b)f(-I); (c)f(2a); (d)f(lIx); (e)f(x + h)., , 0-1, , (a) f(O) = 0+2, , 1, , -1-1, , =-"2, , (b) f(-I)= 1+2, , lIx-1, x-x 2, (d) f(1/x) = lIx2 + 2 = 1+ 2x2, 2., , x+h-l, , 2a-1, , = 4a2 + 2, , x+h-l, , 2z- 1 = 2z (23 -, , ~~: ~ ~~, , f(4)., , t) = Jf f(x), , (b) f(x + 3), , f(x-l), , Detennine the domains of the functions, (n) y='/4-x 2, , (b) y=./x 2 -16, , I, (d) y= x 2 -9, , (e) y= x 2 +4, , (a), (b), (c), (d), (e), 4., , -, , (c) f(2a), , (e) f(x+h)=(x+W+2=x 2 +2hx+h 2 +2, , Iff(x) =2', show that (a) f(x + 3) - f(x -1) =Jf f(x) and (b), (a) f(x + 3) - f(x -1) = 2H ), , 3., , 2, , =-"3, , I, x-2, , (c) y = -, , x, , Since y must be real, 4 - xl ~ 0, or xl ~ 4. The domain is the interval -2 ~ x ~ 2., Here, xl - 16 ~ 0, or xl ~ 16. The domain consists of the intervals x ~ -4 and x ~ 4., The function is defined for every value of x except 2., The function is defined for x ~ ±3., Since xl + 4 ~ 0 for all x, the domain is the set of all real numbers., , Sketch the graph of the function defined as follows:, f(x) =5 when 0 < x ~ I, , f(x) = lOwhen I < x ~ 2, , f(x) =15when 2<x ~3, , f(x) =20 when 3<x ~ 4, , etc., , Detennine the domain and range of the function., The graph is shown in Fig. 6-3. The domain is the set of all positive real numbers, and the range is the set of, integers. 5, 10, 15.20, ...., y, 0---, , 25, ~, , IS, , 10, , o~----, , 0----o~----, , --+-____-L______L-____-L______L-____L-________ X, , o, , 3, , 4, , 5, , Fig. 6-3, 5., , A rectangular plot requires 2000 ft of fencing to enclose it. If one of its dimensions is x (in feet), express its area, y (in square feet) as a function of x, and determine the domain of the function., Since one dimension is x, the otheris t(2ooo - 2x) = 1000 - x. The area is then y =x(lOOO - x), and the, domain of this function is 0 < x < 1000., , 6., , Express the length I of a chord of a circle of radius 8 as a function of its distance x from the center of the circle., Determine the domain of the function.
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CHAPTER 6 Functions, From Fig. 6-4 }Ve see that t/=../64 - x 2 , so that 1= 2"64 - x 2 • The' domain is the interv;U O~ x < 8., , Fig. 6-4, , 7., , From each comer of a square of tin, 12 inches on a side, small squares of side x (in inches) are removed, and, the edges are turned up to form an open box (Fig. 6-5). Express the volume V of the box (in cubic inches) as a, function of x, and determine the domain of the function., , I, I, IJI, II, I~, , I, I, , ',',, , 12 - 2:, , ., '~., , ', , Fig. 6-5, ~, , ......", , :', , =, , The box has a square base of side 12 - 2x and a height of x. The volume of the box is then V x(12 - 2x)2 =, 4x(6 - X)2. The domain is the interval 0 < x < 6., , As x increases over its domain, V increases for a time and then decreases thereafter. Thus, among such boxes, that may be constructed, there is one of greatest volume, say M. To determine M. it is necessary to locate the, precise value of x at which V ceases to increase. This problem will be studied in a later chapter., , 8., , Iff(x) =x2 + 2x, find f(a+~- f(a) and interpret the result., f(a+h)- f(a) = [(a+h)2 +2(a+h)]-(a 2 +2a), h, h, , 2a+2+h, , On the graph of the function (Fig. 6-6), locate points P and Q whose respective abscissas are a and a + h., The ordinate of P is./ta), and that of Q is./ta + h). Then, f(a+h)- f(a) difference of ordinates, I, f PQ, It, = difference of abscissas s ope 0, , I(.H)-/(o), , Fig. 6-6
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CHAPTER 6 Functions, , 9., , Letf(x) =Xl- 2x + 3. Evaluate (a)f(3); (b)f(-3); (c)f(-x); (d)f(x + 2); (e)f(x -2); (f)f(x+ h); (g)f(x+ h)f(x); (h) f(x + hk - f(x), , =, , =, , =, , (a) f(3) 32 - 2(3) + 3 9 - 6 + 3 6, (b) f(-3)=(-3)2-2(-3)+3=9+6+3=18, (c) f(-x)=(-X)2_2(-x)+3=Xl+2x+3, , (d) f (x + 2) =(x + 2)2 - 2(x + 2) + 3 =Xl + 4x + 4 - 2x - 4 + 3 =Xl + 2x + 3, (e) f(x- 2)=(x- 2)2 - 2(x- 2)+ 3 =Xl-4x+4 - 2x+4+ 3 =Xl- 6x+ 11, (f) f(x+ h) =(x+ h)2 - 2(x+ h) + 3 =Xl +2hx+ h2 - 2x- 2h+ 3 =Xl + (2h - 2)x+ W - 2h + 3), (g) f(x+h) -f(x) - [Xl + (2h - 2)x+(h2- 2h+3») - (Xl- 2x+3)=2hx+h2_2h =h(2x+h - 2), (h), , - f(x), k, , f(x+h, , h(2x~h-2) =2x+h-2, , 10. Draw the graph of tlie 'function f(x) =../4 - x 2, and find the domain and range of the function., The graph off is the graph of the equation y =../4 - x 2. For points on this graph, f =4 - Xl; that is, Xl + f =4., The graph of the last equation-is the circle with center at the origin and radius 2, Since y =../4 - x2 ;::: 0, the, desired graph is the upper half of that circle. Fig. 6-7 shows that the domain is the interval - 2 ~ x ~ 2, and the, range is the interval 0 ~ y ~ 2., , y, , -2, , 0, , Fig. 6-7, , 11. Iff(x) =Xl- 4x+ 6, find (a)f(O); (b)f(3); (c)f(-2). Show that fm= f{f) andf(2 - h) =f(2 + h)., , Ans., , (a) -6; (b) 3; (c) 18, , x-I fmd(a)f(O); (b)f(l); (c)f(-2). Show thatf (, 12. If f(x) =. x+l', , Ans. (a) -1; (b) 0; (c) 3, 13. Iff(x) =Xl- x, show thatf(x + 1) =f(-x)., , a::, , 14. Iff(x) =1/x, show that f(a) - f(b) =f( b a)., ., 5x+3, 15. If y =f(x) = 4x _ 5' show that x =f(y)·, , 1, :x1)=-f(x) andf.(-:x1) =- f(xl'
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CHAPTER 6 Functions, , 16. Detennine the domain of each of the following functions:, (a) y=x2+4, , 2x, y= (x-2)(x+ I), , (e), , Ans., , •, , (b) y=.JX2 + 4, (f), , (a), (b), (g) all values ofx; (c), , 1, , y= .J9-x2, Ld~2;, , (c) y=.JXl -4, , (d), , x 2 -1, (g) y= x 2 + 1, , (h), , x, y=x+3, y=, g, 2-x, , (d)x;t:-3; (e)x;t:-l, 2; (f) -3 <x< 3; (h)0$x<2, , 17. Compute l(a +'1z - f(o) in the following cases:, , .~ I ,, , I, , f(x) = x _ 2 when a ;t: 2 and a + h ;t: 2, , (a), , (b) f(x)=.Jx-4 wh~na~4anda+h~4, (c) f(x), ,":'1, , x, , = x+ I, , when a;t: -1 anda+h;t:-l, --I, , Ans., , I, , 1, , (a) (a-2)(a+h-2); (b) .Ja+h-4 +.Ja-4; (c) (a+l)(a+h+l), , 18•. Draw the graphs of the following functions, and find their domains and ranges:, (a) f(x)=-x 2 + 1, , (c), , (b) f(x) =, , j, , X-l, , ifO<x<l, , 2x, , if I $x, , f (x) = [xl =the greatest integer less than or equal to x, , (d), , x 2 -4, f(x)=-x-2, , (g) f(x) = Lx - 31, , (j) f(x), , Ans., , , ~;:>, , ~.}-<~, , =x -Ixl, , (a), (b), (c), (d), (e), (f), (g), (h), (i), (j), (k), , (e) f(x)=5-x 2, (h), , (f), , f (x) = 4/x, , (k) f(x) =, , f(x) = -4£, , (i) f(x) = lxI/x, , G, , ifx~O, , ifx<O, , domain, all numbers; range, y $ 1, domain, x> 0; range, -I < y < 0 or y ~ 2, domain, all numbers; range, all integers, domain, x ;t: 2; range, y ;t: 4, domain, all numbers; range, y $ 5, domain, x ~ 0; range, y $ 0, domain, all numbers; range, y ~ 0, domain, x ;t: 0; range, y ;t: 0, domain, x;t: 0; range, {-I, I}, domain, all numbers; range, y $ 0, domain, all numbers; range, y ~ 0, , 19. (GC) Use a graphing calculator to verify your answers to Problem 18., , 20. Evaluate the expression f(x + hl - f(x) for the following functionsf:, (a) f(x)=3x-r, , (b), , (c) !(x)=3x-5, , (d) l(x)=x3 -2, , Ans., , (a) 3 - 2x -- II, , f(x)=fb, , (b) J2(x +~) + fb, , (c) 3, , (d) 3r + 3xh + h, , 2, , 21. Find a formula for the function f whose graph consists of all points satisfying each of the following equations., (In plain language, solve each equation for y.), (a) xSy+4x-2=:O, , An.\'., , (b) x= 2+),, , 2-),, , 2-4x, 2(x-l), (a) f(x)=-xr; (b) f(x)=--.x+l; (c)f(x) = 2x, , (c) 4r-4xy+y2=0
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CHAPTER 6 Functions, , 22. Graph the following functions and find their domain and range:, if-l<x<O, ifO<x<2, (a) I(x), , X+2, !x, , =, , (b) g(x), , ifOSx<1, , !2-X, , =, , x-I, , J -4-;, , (c) h(x) = x;, , 1, , if3Sx<4, , if x :;t2, ifx=2, , Ans. (a) domain = (-1, I], range = [0, 2), (b) domain = union of (0,2) and [3, 4), range = (0, 3), (c) domain and range = set of all real numbers, 23. (GC) Verify your answers to Problem 22 by means of a graphing calculator., 24. In each of the following cases, defme a function that has the given set!1lJ as its domain and the given set ~ as its, range: (a)!1lJ =(0, 2) and ~ =(1,7); (b)!1lJ = (0, 1) and ~ =(1. 00)., Ans., , (a) One such function isf(x) =3x + I. (b) One such function is f(x) =-I1_., , -x, , 25. (a) Prove the vertical line test: A set of points in the xy plane is the graph of a function if and only if the set, intersects every vertical line in at most one point., (b) Determine whether each set of points in Fig. 6-8 is the graph of a function., Ans., , Only (b) is the graph of a function., , (a), , (b), , (c), , (d), , Rg.6-8
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Limits, Limit of a Function, Iffis a function, then we say:, A is the limit off(x) as x approaches a, , if the value of f(x) gets arbitrarily close to A as x approaches a. This is written in mathematical notation as:, limf(x) = A, X-+a, , For example, lim i'- = 9, since i'- gets arbitrarily close to 9 as x approaches as close as one wishes to 3. The, definition of Ifffi3 f(x) =A was stated above in ordinary language. The definition can be stated in more precise, mathematica(fa"nguage as follows: lim f(x) = A if and only if. for any given positive number E, however, small, there exists a positive numbe(8~uch that, 'whenever 0 Ix - al < 0, then If(x) - AI < E., The gist of the definition is illustrated in Fig. 7-1. After E has been chosen [that is, after interval (ii), has been chosen], then 0 can be found [that is, interval (i) can be determined] so that, whenever x :F- a is, on interval (i), say at xo' then f(x) is on interval (ii), at f(x o)' Notice the important fact that whether or not, lim f(x) = A is true does not depend upon the value off(x) when x = a. In fact,f(x) need not even be defined, , <, , x-+", , whenx=a., Xo, , ---~O~---<O)-+-I- - - ( 0 ) - - - - -, , a-S, , a, , a+S, , x, , f(x o ), ---0......-----+1-1-1, A-f, A, , ----0-!(x), A+f, , (ii), , (i), , Fig. 7-1, , EXAMPLE 7.1:, , 2, 2, lim x - 4 _ 4 although x -24 is not defined when x =2. Since, .-+2 x - 2 ,, x-, , x2 - 4, x-2, , = (x -, , 2)(x + 2), x-2, , =x + 2, , 2-4, , we see that ~2 approaches 4 as x approaches 2., xEXAMPLE 7.2:, , Let us use the precise definition to show that lim (4x - 5) = 3. Let E> 0 be chosen. We must produce, , .... 2, , some 8> 0 such that, whenever 0 < Ix - 21< 8, then I(4x - 5) - 31 < E ., First we note that 1(4x - 5) - 31 = 14x - 81 =41x - 21·, If we take 8to be E 14, then, whenever 0 < Ix - 21 < 8, 1(4x- 5) - 31 =41x - 21 < 48 =E·
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CHAPTER 7 Limits, , Right and Left Limits, Next we want to talk about one-sided limits of f(x) as x approaches a from the right-hand side or from the, left-hand side. By lim f(x) =A we mean thatfis defined in some open interval (c, a) andf(x) approaches A, -'-+0·, as x approaches a through values less than a, that is, as x approaches a from, the left. Similarly, lim f(x) =A, ., means thatfis defined in some open interval (a, d) andf(x) approaches A as x approaches afrom the right., If f is defined in an interval to the left of a and in an interval to the right of a, then the statement lim f(x) =A is, x-+a, equivalent to the conjunction of the two statements lim f(x) =A and lim f(x) =A. We shall see by examples, x.....a·, x-+a·, below that the existence of the limit from the left does not imply the existence of the limit from the right, and, conversely., When a function is defined only on one side of a point a, then we shall identify limf(x) with the one-sided, x-+a, limit, if it exists. For example, iff(x) = JX, thenfis defined only at and to the right of O. Hence, since lim, JX =0, we will also write lim JX =O. Of course, lim JX does not exist, since JX is not defined when x <, O. This is an example where the existence of the limit from one side does not entail the existence of the limit, from the other side. As another interesting example, consider the function g(x) =$x, which is defined only, for x> O. In this case, lim $x does not exist, since IIx gets larger and larger without bound as x approaches, ofrom the right. Therefore, lim $x does not exist., x~a·,, , 1, , x~, , x~O., , x~-, , .r~O·, , x-+o, , EXAMPLE 7.3: The functionf(x) = ~9-X2 has the interval-3'~ x ~ 3 as its domain. If a is any number on the interval (-3, 3), then lim ~9-X2 exists and is equal to ~9-a2. Now consider a = 3. Let x approach 3 from the left; then, rn-:2, x ..... a, lim_ ,,9-x2 = O. For x > 3, ~9-X2 is not defined, since 9 - xl is negative. Hen~e, lim ~9-X2 = lim ~9-X2 =O., %-+3, rn-:2, rn-:22, .....3, x-+3Similarly, X~~3 ,,9-X2 =x~~+ ,,9-x =O., , Theorems on Limits, The following theorems are intuitively clear. Proofs of some of them are given in Problem 11., Theorem 7.1:, , Iff(x) = c, a constant, then limf(~)=c., ' ....a, , For the next five theorems, assume lim f(x) =A and lim g(x) = B., x~a, , x~a, , Theorem 7.2:, , lim c· f(x) = c Iimf(x) = cA., , Theorem 7.3:, , lim [f(x)±g(x)] =limf(x)± limg(x) =A ± B., , ,....., , x-+a, , Theorem 7.4:, , .I-+cJ, , lim [f(x)g(x)] = limf(x)· limg(x) = A· B., x-+a, , Theorem 7.5:, , x-+o, , . (f(X»), , hm, ...... -(), gx, , x-+a, , ~i.T.f(x), , x-+a, , A ., If B O., , = rlmg(), x =-B', , *, , X-+II, , Theorem 7.6:, , lim~f(x) ="limf(x) =ifA, if !ifA is defined., .r-.a, , .r-+a, , Infinity, Let, limf(x) =+00, x-+a, , mean that, as x approaches a, f(x) eventually becomes greater than any preaSsigned positive number, however, large. In such a case, we say thatf(x) approaches +00 asx approaches a. More precisely, limf(x) =+00 if and only, ., , x-M, , if, for any positive number M, there exists a positive number S such that, whenever 0 < Ix - al < S, thenf(x) > M.
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CHAPTER 7 Limits, , Similarly, let, limf(x) =-0<:, X-HI, , mean that, as x approaches aJ(x) eventually becomes less than any preassigned number. In that case, we say, thatf(x) approaches - 0 0 as x approaches a., Let, limf(x) = oc, x--+a, , mean that, as x approaches a, It(x) Ieventually becomes greater than any preassigned positive number. Hence,, = 00 if and only if lim It(x)1 =+00., , x_, limf(x), , _., , These defmitions can be extended to one-sided limits in the obvious way., EXAMPLE 7.4:, , (a), , -I, ·, (b) I<-+1, 1m 1)2 =-oc:, x-(, , lim~=+oc, , >-+0, , x, , (c) liml=oc, .-+0 X, , EXAMPLE 7.5:, , (a), , lim 1 = +00. As x approaches 0 from the right (that is, through positive numbers), l/x is positive and eventu., , \~o+x, , ally becomes larger than any preassigned number., (b), , lim 1 = -00 since, as x approaches.O from the left (that is, through negative numbers), l/x is negative and, , .-+0- X, , eventually becomes smaller than any preassigned number., , The limit concepts already introduced can be extended in an obvious way to the case in which the variable, approaches +00 or -00. For example,, ', lim f(x)=A, , x--++oo, , means that f(x) approaches A as x -7 +00, or, in more precise terms, given any positive E, there exists a, number N such that, whenever x> N, then If(x) - AI < E. Similar definitions can be given for the statements, lim f(x)=A, lim f(x) =+00, lim f(x) =-00, limf(x)=-oo, ahdlim f(x) =+00., x~~, , x~, , EXAMPLE 7.S:, , Caution:, , ~ =0, , lim, , >-+_ X, , and, , .r-+--oo, , X~Q, , x-+-OO, , lim, , .-+_, , (2+~)=, 2., x, , When limf(x}=±oc and limg(x) = ±oo, Theorems 7.3-7.5 do not make sense and cannot be used., ~~a, , x~o, , J., , For example, lim ~:= +oc and lim = +00, but, <-+0 x, .-+0 X, lim IIIIX: = lim x 2 = 0, .-+0, , Note:, , X, , .-+0, , We say that a limit, such as limf(x) or lim f(x) exists when the.limit is a real number, but not when the, x-kr, x-HOO, 2, 2, limit is +00 or -00 or 00. For example, since lim x -24 = 4, we say that lim x -24 exists. However, although lim ~ == +oc,, I, <.... 2 x,.... 2 x.-+0 X, we do not say that lim 2" exists., , ....0 x, , SOLVED PROBLEMS, , 1., , Verify the following limit computations:, (a), , lim 5x =5 lim x =5·2 =10, x-+2, , l-+2
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CHAPTER 7 Limits, , (b) lim (2x+3)= 2limx+ lim 3 = 2, 2+3= 7, .1.... 2, , .t~2, , ..... 2, , (c) Iim(x2-4x+I)=4-8+1=-3, .... 2, , Iim(x-2), lim x-2 _ ....3, 1, (d) ......, 3 x+2 -lim(x+2), 5", .... 3, , ....., , [Note: Do not assume from these problems that Iimf(x) is invariably f(a),], , (g), , 2., , lim.xl - 25 = lim (x- 5) =-10, x+ 5 .... -5, , .... -5, , Verify the following limit computations:, (a), , r, , x-4, }l!]x2-x-12, , r, , x-4, , .~(x+3)(x-4), , lim_I_=l, ....... x+3 7, , The division by x - 4 before passing to the limit is valid since x*,4 as x -+ 4; hence, x - 4 is never zero,, I'1m x 2 +3x+9 -9, ....3 X+3, 2, , ,Im--=, r-27 I'Im-(X-3)(X2+3x+9), (b) I...., 2, 3 x - 9, ....3 (x - 3)(x +3), , ------, , Here, and again in Problems 4 and 5, h is a variable, so that it might be thought that we are dealing with, functions of two variables. However, the factthafxls a variable plays no role in these problems; for the moment,, x can be considered a constant., "., - c__ c- -, , , X2 + X - 2 I' (x -l)(x + 2L I' x + 2 - 00 ', ", ,, (e) I,1m, ( 1)2, 1m (X - 1)2 - ...., 1m1 x- 1 - , no IlDllt eXIsts,, ...2 x .-+1, , 3., , In the following problems (aHc), you can interpret .r-+too, lim as either 1-..., lim·00 or ......., lim-00 ; it will not matter which, Verify, the limit computations,, (a), , lim 3x-2 = lim 3-2/x = 3-0 =!, 9x+7 ....too 9+7/x 9+0 3, r 6X2+2x+1 r 6+2/x+l/x 2, , ....too, , (b) .~'!!.5X2-3X+4 .~'!!.5-3/x+4Ix2, (c), , (d), (), e, (f), , 2, 2, x + x - 2 r l/x + l/x - 2/r, %~'!!. 4r -1 = .~.. .4- l/x3, , r, , 6+0+0 6, 5-0+05", , Q- 0, 4-, , ;.r, , lim, = lim ~ =-00, .... -.. ,x + I .... -... 1+ l/x, 2X3, I', 2x, ,, I1m -2- = 1m 'i7'i'T.:r = +00, , .....oox + I, ...._1+ I/x, , lim (r-7x4-2x+5)= lim r(I-].-4+, .1....+00, , lim, (_,,00, , , .... +00, , (1-].- : +J.)=, X, , A-, , r, , x, , X, , _~5)=+OC since, , A, , (1- 0 - 0+0)= I and lim r = +00, , (g) ...., lim_ (r-7x4-2x+5)= __, lim X5, , .r.........., , x, (1-].-:, + ~)=-oo since, A, , ~
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CHAPTER 7 Limits, , 4., , . f(x+h)- [(x), Givenf(x) =xl - 3.(, find hm, h, h...O, Sincef(x) = xl - 3x, we havef(x + h) = (x + h)2 - 3(x + h) and, , . f(x + h) - f(x), \1m, h, , h ... O, , (x, = I'1...., 1m, 0, , 2, , + 21u + h2- 3x - 311) - (x 2 - 3x), h, , .::;21u~+'-ihf-2__-=3=h, = \.hUll...o, h, , = lim (2x+ h- 3)= 2x- 3., h...O, , 5., , ~. f(x+h)- f(x), I, Given f(x) = ..,5x+ I, find lim, h, when x> --5', h...O, , r, , h~, , f(x+h)- f(x), h, , r, , = A~, , .J5x+5h+l-$X+I, h, , .J5x + 5h + 1 - .J5x + 1 .J5x + 5h + 1 + .J5x + 1, - \.Im~~~-.--~---~~~~~~~~, - h...O, h, .J5x + 5h + 1 + .J5x + 1, ., , (5x+5h+I)-(5x+l), , =h\1111, ~r;::==::=:===='="""""'r.:=====-., ...O h(.J5x + 5h + 1 + .J5x+ l), ·, = IUII, h-+O, , 6., , (a) In each of the following, (a) to (e), detennine the points x = a for which each denominator is zero. Then see, -'Wh:Ll-h:.ppens to y as x -+ a- and as x -+ a+, and verify the given solutions., (b) (GC) Check the answers in (a) with a graphing calculator., (a) y, , - . I I...... ', , (b), , ", , =f(x) =2/x; The denominator is zero when x = O. As x -+ 0-, y -+ -00; asx -+ 0+, y -+ +00., , Y = I(x) = (x ... ~)(; _ 2) : The denominator is zero for x =-3 and x = 2. As x -+ -3-, y -+ -00; as x -+ -3+,, Y -+ +00. As x -+ 2-, y -+ -00; as x, , (c), ... _\-, , 5, 5, =---,:::=:=, .J5x + 5h + 1 + .J5x + 1 2.J5x + 1, , -+ 2+, y -+ +00., , (x /2)(; _I) ; The denominator is zero for x = -2 and x =1. As x -+ -2-, y -+ -00; as x -+ -2+,, , y = f(x), , y -+ +00. As x -+ 1-, y -+ +00; as x -+ 1+, y -+ -00 ., , ~ ,-. I, , V:~~l~, , (x + 2)(x - 1), .•, (x _ 3)2 ; The denommator IS zero for x, , t}~, , (d) y = f(x) =, (e), , 7., , Y = f(x) =, , =3. As x -+ 3-, y -+ +00; as x -+ 3+, y -+ +00., , (x+ 2)(\- x), .., x_3, ; The denommator IS zero for x = 3. As x -+ 3-, y -+ +00; as x -+ 3+, y -+, -00., , For each of the functions of Problem 6, determine what happens to y as x -+ -00 and x -+ +00., (a) As x -+ ±oo, y:: 2/x -+ O. When x < 0, y < O. Hence, as x -+ -00, y -+ 0-. Similarly, as x -+ +00, y -+ 0·., (b) Divide numerator and denominator of (x+, obtaining, , 3)(;- 2) by xl (the highest power of x in the denominator),, IIx-l/x 2, (1 + 3Ix)(\ - 2/x), , Hence, as x ~ ±oo,, , 0-0, , y~ (1+0)(\-0), , 0, , =T=O, , As x -+ -00, the factors x - I, x + 3. and x - 2 are negative, and, therefore, y -+ 0-. As x, are positive, and, therefore, y ~ 0+., (c) Similar to (b)., , ~, , +00, those factors
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---~CI», , CHAPTER 7 Limits, , 2, , 2, , (d) (x+ 2)(x-l) = x2 + X - 2. 1+ lIx- 21x 2 , after dividing numerator and denominator by r (the highest, (x- 3)2, x - 6x+9 1-6Ix+9Ix, power of x in the denominator). Hence, as x ~ ±co, Y~ ~ ~ ~ ~ ~, = I. The denominator (x - 3)2 is, , t, , always nonnegative. As x ~ -00, both x + 2 and x-I are negative and their product is positive; hence,, y ~ 1+. As x ~ +00, both x + 2 and x-I are positive, as is their product; hence, y ~ 1+., (e) . (x+2)(1- x), , 2, , -x -x+ 2 = -x-l + 2/x , after dividing numerator and denominator byx (the highest, x-3, x-3, 1-3~, power of x in the denominator). As x ~ ±oo, 2/x and 3/x approach 0, and - x - I approaches ± 00. Thus, the, denominator approaches 1 and the numerator approaches ± 00. As x ~ -00, X + 2 and x - 3 are negative and, 1 - x is positive; so, y ~ +00. As x ~ +00, x + 2 and x - 3 are positive and I - x is negative; so, y ~ -00., , 8., , Examine the function of Problem 4 in Chapter 6 as x ~ a- and as x ~ a+ when a is any positive integer., Consider, as a typical case, a = 2: As x ~ 2-,f(x) ~ 10. As x ~ 2+,f(x) ~ 15. Thus, limf(x), does not exist:, ....2, In general, the limit fails to exist for all positive integers. (Note, however, that limf(x), =, lim, f(x) = 5, since f(x), 1'-100, x-+O·, is not defined for x ~ 0.), , 9., , Use the precise definition to show that lim, (x2+ 3x) = 10., ....2, Let E> 0 be chosen. Note that (x- 2)2 =r - 4x+ 4, and so, r + 3x - 10 = (x - 2)2 + 7x -14 = (x - 2)2 +, 7(x - 2). Hence I(r + 3x) - 101 = I(x - 2)2 + 7(x - 2)1 ~ ~ - 212 + 71x - 21. If we choose ~ to be the minimum of I, and E18, then ~2 ~ ~, and, therefore, 0 < ~ - 21 < ~ impli~s l(x 2+ 3x) - 101< ~ + 7~ ~ ~ + 7~ = 8~ ~ E., , 10. If limg(x) = B ¢ 0, prove that there exists a positive number ~ such that 0 < ~ - al < ~ implies Ig(x)l> mJ. ., x->a, 2, Letting E = IBV2 we obtain a positive ~ such that 0 < ~ - al < ~ implies Ig(x) - BI < IB1/2. Now, if 0 < ~ - al < ~,, then IBI = Ig(x) + (B - g(x»1 ~ Ig(x)1 + IB - g(x)1 < Ig(x)1 + IBV2 and, therefore, IBI/2 < Ig(x)l., 11. Assume (I) limf(x) = A and (n) limg(x) = B. Prove:, ,f-+a, , ;c~a, , (a) lim [f(x) + g(x)] = A + B, x-+a, , (b) limf(x)g(x)=AB, x...., , (a) Let E > 0 be chosen. Then E 12 > O. By (I) and (II), there exist positive ~I and ~2 such that 0 < ~ - al < ~I, implies If(x) - AI < E12 and 0 < Ix - al < ~2 implies Ig(x) - Bj < E 12. Let ~ be the minimum of ~I and ~2', Thus, for 0 < Ix - al <~, If(x) - AI < E12 and Ig(x) - BI < E12. Therefore, for 0 < ~ - al < ~,, l(f(x) + g(x)-(A + B)I =l(f(x)- A)+(g(x)- B)I, , ~lf(x)-AI+I8(x)- BI < I+I=E, (b) Let E> 0 be chosen. Choose E' to be the minimum of E 13 and 1 and E1(3IBI) (if B ¢ 0), and E 1(31.41) (if, A ¢ 0). Note that (E')2 ~ E' since E' ~ I Moreover, IBI E' ~ E13 and k41 E' ~ E13. By (I) and (II), there exist, , positive ~I and O2 such that 0 < Ix - al < ~I implies If(x) - AI < E * and 0 < ~ - al < ~2 implies Ig(x) - BI < E'., Let obe the minimum of ~I and ~2' Now, for 0 < ~ - al < ~,, If(x)g(x) - ABI, , =l(f(x) - A)(g(x) - B) +B(f(x) - A) +A(g(x) - B)I, ~ If(x) - AXg(x) -, , 8)1 + IB(f(x) - A)I + !A(g(x) - 8)1, , = If(x) - Allg(x) - BI + 1B11f(x) - AI + !A118(x) - BI, , ~(E')2 +IBIE' +IAIE'~E' +1+1~1+1+1=E, 1, (c) Bypart(b),itsufficestoshowthat lim-(l)=-B, .-+. g x . Let E>O be chosen. Then B2E/2>0. Hence,there, exists a positive ~I such that 0 < ~ - al < ~I implies Ig(x) - BI < IB~E .
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CHAPTER 7 Limits, By Problem 10, there exists a positive O2 such that 0 < Ix - al < 02 implies Ig(x)1 > IBV2. Let 0 be the minimum, , of, , o. and or Then 0 < Ix - al < 0 implies that, _I, , Ig(x), , ., , -.11B = lBlI8Wf, 1!! ___8{.t)1 < ~. 2 - e, 2 lBP-, , 12. Prove that, for any polynomial function, , li,mf(x) = f(a), , ......, , This follows from Theorems 7.1-7.4 and the obvious fact that lim x =a ., , ......., , 13. Prove the following generalizations of the results of Problem 3. Let f(x) = a.x· + a._1x"-· + ... + a.x + a o and, g(x) = b1xl + bHxH + ... + b.x + bo be two polynomials., , (a), , lim f(x), .....:toog(x), , =5L, hl, , if 11 =k, , (b), , lim f(x), , ....:toog(x), , =0, , if 11 < k, , (c), , lim f(x), ._g(x), , =±oo, , if n > k. (It is +00 if and only if a. and bi have the same sign.), , (d), , lim f(x) =, ....._g(x), , ±oo, , if n > k. (The correct sign is the sign of a.bt ( -I).-t.), 2, , x, I, Ii, x, I, I', I,, 14. Prove (a) .~~(X_2)3=-oc;(b), .~'!x+l= ; (c) .....'! x-I =+oa., , °, , (a) Let M be any negative number. Choose positive and equal to the minimum of 1 and, and 0 < Ix -, , 21 < O. Then Ix- 21 3 < 83 S; 8 S;, , 1, Therefore, (x _ 2)3 =, , WI' Hence, R> IMI, , = -M. But (x - 2)3'< O., , I, , -1X=2P < M., , (b) Let e be any positive number, and let M, , I II I, , =lIe. Assume x > M. Then, , x, 1 =x+l<x<M=e, I, I, I, 1 = x+l, x+lx2, , .., , (c) Let M be any positive number. Assume x > M + 1. Then x-I ~, , 15. Evaluate: (a) limhl; (b) limhl; (c) limhl, 1-+0" X, , .x-+O~, , X, , 1-+0, , X, , (a) When x > 0, Ixl = x. Hence, ~-+o;, lim hl, = ~-tO", lim 1=1., X, = - x. Hence , .l~Olim hl, = x-tOlim -1 =-1., (b) When x < 0 , I-I, JA, X, , (c), , lim hl, does not exist' since ....., lim0" hl, x, X, , • ....0, , ;~~-~~~.-, , ~~ ::,-~--:~.:, , 16. Evaluate the following limits:, (a), (b), , lim (x 2 - 4x), ..... 2, , lim, , x-+-I, , (Xl, , + 2X2 -, , 3x-4), , *' lim hl ., .-+0' X, , xx = x> M., 2, , WI' Assume, , x <2
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CHAPTER 7 Limits, , (g), , +, , x~, , Ans., , ., , 3- x, , 3', , lim 3' - 3-x, , (a), , ., , 19. Fmd ~~~, Ans., , -!-; (b) -t; (c) 0; (d) +00; (e) 0; (f), , 1; (g)-1, , f(a+h)- f(a)., .., It, for the functlOnsfm Problems 11, 12, 13, 15, and 16 (a, b, d, g) of Chapter 6., , (4}2 5)2; (16)(a) 2a, (b), , (11) 2a-4: (12) (aJ1)2;(13)2a-I;(15), 4a, , ~,(d), , (a13)2', , 20. (GC) Investigate the behavior of, X, if x>O, f(x)= { x+ 1, if x~O, as x -+ O. Draw a graph and verify it with a graphing calculator., lim I(x) = 0; lim f(x) = 1; limf(x) does not exist., , Ans., , ;;/'., , %-+0-, , .t~O·, , .(-.0, , 21. Use Theorem 7.4 and mathematical induction to prove lirnx" = an for all positive integers n., x-+a, , 22. For f(x) = 5x - 6, find 0> 0 such that, whenever 0 < ~ - 41 < 0, then If(x) - 141 < e, when (a) e=, , t, , and, , (b) e=O.OOI., , ;i:;j~, , Ans., , (a), , I~; (b) 0.0002, , 23. Use the precise definition to prove: (a) lim5x=15; (b) lirnx2=4; (c) lim(x2 -3x+5)=3., x-.), , x~2, , x~2, , 24. Use the precise definition to prove:, (a), , lim 1= DC, , x-+I, , x-+o X, ,~, , x2, (d) .1-+-00, lim +1=-00, X, , ~1=1, X~+OO x-, , (c) lim, , (b) lim -=:Ll = 00, , x-, , ,, , 25. Let/(x), g(x), and hex) be such that (l)f(x) ~ g(x):$; hex) for all values in certain intervals to the left and right of, a, and (2) lim f(x) = lim hex) = A. Prove lim g(x), x-+a.r-+tl, , X-+d, , =A., , (Hint: For e> 0, there exists 0> 0 such that, whenever 0 < ~ therefore, A - e < f(x) :$; g(x) :$; hex) < A + e.), 26. Prove: Iff(x), , (Hillt:, ..,., , ~{~, .'., , ~, , al < 0, then If(x) - AI < e and Ih(x) - AI < e, , and,, , M for all x in an open interval containing a and if limf(x) = A, then A ~ M., •, , x-+a, , Assume A > M. Choose e =, , teA - M) and derive a contradiction.), , 27. (GC) Use a graphing calculator to confirm the limits found in Problems led, e,j), 2(a, b, d), 16, and 18., , 28. (a) Show that x-+_, Iim(x-~x2-1)=O., (Hillt: Multiply and divide by x + ~.), , x2, (b) Show that the hyperbola a2, , 29. (a), , y2, , -, , ••, , b 2 = I gets arbltranly close to the asymptote y =, , Find lim.JX+3 -$., X-+O, , X, , (Hint: Multiply the numerator and denominator by -Ix + 3 +.fj.), (b) (GC) Use a graphing calculator to confirm the result of part (a)., , ab x as x approaches, , 00.
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••, , CHAPTER 7' Limits, 30. Letf(x) = .JX, (a) lim f(x), x-+4 +, , - 1 if x> 4 andf(x) =xl - 4x + 1 ifx < 4. Find:, (b) lim f(x), ....4 -, , (c) lim f(x), ......, , Ans. (a) l;.(b) 1; (c) 1, 31. Let g(x) = lOx -7 if x> 1 and g(x) =3x + 2 ifx < 1. Find:, (a) lim g(x), ...... 1., , (b) lim g(x), x-+I-, , (c) lim g(x), .. -+1, , Ans. (a) 3; (b) 5~ (c) It does not exist.
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,, , Continuity, Continuous Function, . A functionJis defined to be continuous at Xo if the following three conditions hold:, (i) J(xo) is defined;, (ii) lim J(x) exists;, .1-+.10, , (iii) lim J(x) =J(x o)', .1-+,10, , For example,f(.t) =xl + I is continuous at 2, since lim J(x) =5 =J(2). Condition (i) implies that a func...... 2, , tion can be continuous only at points of its domain. Thus. J(x) =.J4 - x 2 is not continuous at 3 because J(3), is not defined., LetJbe a function that is defined on an interval (a, xJ to the left of Xo and/or on an interval (xo' b) to the, right of xo' We say thatJis discontinuous at.to ifJis not continuous at xo' that is, if one or more of the conditions (i)-(iii) fails., ', EXAMPLE 8.1:, ,, , (a), , f(x) = ~2 is discontinuous at 2 because f(2) is not defined and also because limf(x) does not exist, X-., _2, (since lim f(x) = 00). See Fig. 8-1., x-+2, , 1/, , o, , !L, 12, , 1, , I, I, I, , Fig. 8-1, 1, , x - 4, ..., 2 bec:tllse f ( 2) IS, . not de fillIed. However, I'lln f(), (b) f(x) =--2IS dlscontllluoliS at, x, ~_2, lim(x + 2) =4 so that condition (ii) holds., , =(._2, 1m (x + 2)(x2x-, , 2), , , .... 2, , The discontinuity at 2 in Example 8.I(b) is said to be removable because, if we extended the functionJ, by defini ng its value at x =2 to be 4, then the extended function g would be continuous at 2. Note that g(x) ==, x + 2 for all x. The graphs of f(x) = : ~; and g(x) = x + 2 are identical except at x = 2, where the former has, a "hole." (See Fig. 8-2.) Removing the discontinuity consists simply of filling the "hole."
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CHAPTER 8 Continuity, , Rg.8-2, , The discontinuity at 2 in Example 8.1 (a) is not removable. Redefining the value ofjat 2 cannot change, the fact that lim ~2 does not exist., x-+2, , x-, , We also call a discontinuity of a functionj at Xo removable whenj(xo) is defined and changing the value, of the function at Xo produces a function that is continuous at xo', Define a functionfas follows:, , EXAMPLE 8.2:, , X2, , f(x)= { 0, , if x*2, ifx=2, , Here Iimf(x}, == 4, butf(2} = O. Hence, condition (iii) fails, so thatfhas a discontinuity at 2. But if we change the, x.... 2, value offat 2 to be 4, then we obtain a function h such that h(x} = xl for all x, and h is continuous at 2. Thus, the, discontinuity off at 2 was removable., , *, , EXAMPLE 8.3: Letfbe the function such that f(x) = Ixl for all x O. The graph of/is shown in Fig. 8-3. fis discontinuous at 0 because f(O) is not defined. Moreover, x, , limf(x)=lim.:!.=1 and lim f(x) = Iim-X =-1, , x-.o+, , Thus, lim f(x}, x-Hl"", , ,l'-+O· X, , .t-tO-, , x-t-O-, , X, , * limo+ f(x}. Hence, the discontinuity offat 0 is not removable., x...., , - - - - - - - ( ) -1, , I', , Fig. 8-3, , The kind of discontinuity shown in Example 8.3 is called ajllmp discontinuity. In general, a functionj, has a jump discontinuity atxo if lim j(x) and lim f(x} both exist and lim f(x) lim f(x). Such a discontinuity, is not removable., x....x., x.... .,;, x-txo, x-tx~, , *, , EXAMPLE 8.4:, , The function of Problem 4 in Chapter 6 has a jump discontinuity at every positive integer., , Properties of limits lead to corresponding properties of continuity.
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CHAPTER 8 Continuity, , Theorem 8.1: Assume thatf and g are continuous at xO' Then:, (a) The constant function h(x) =c for all x is continuous at every xo', (b) cfis continuous at xo' for any constant c. (Recall that cfhas the value c . f (x) for each argument x.), (c) f + g is continuous at xO', (d) f - g is continuous at xo', (e) fg is continuous at xo', (f) fig is continuous at Xo if g(x():f:. O., (g), , ifl is continuous at, , Xo, , if ~ /(xo) is defined., , These results follow immediately from Theorems 7.1-7.6. For example, (c) holds because, lim (f(x) + g(x», , x-+.to, , Theorem 8.2:, , ., , =x-+.to, lim f(x) + lim g(x) = f(x o) + g(xo), x-+.to, , The identify function /(x) = x is continuous at every xo', , This follows fwm the fact that lim x = xo', x~, , ..., , We say that a functionfis continuous on a set A iffis continuous at every point of A. Moreover, if we just, say thatf is c01ltinuolls, we mean thatfis continuous at every real number., The original intuitive idea behind the notion of continuity was that the graph of a continuous function was, supposed to be "continuous" in the intuitive sense that one could draw the graph without taking the pencil off, the paper. Thus, the graph would not contain any "holes" or "jumps." However, it turns out that our precise, definition of continuity goes well beyond that original intuitive notion; there are very complicated continuous, functions that could certainly not be drawn on a piece of paper., Theorem 8.3:, , Every polynomial function, , is continuous., , This is a consequence of Theorems 8.1 (a-e) and 8.2., EXAMPLE 8.5: As an instance of Theorem 8.3, considerthe function xl - 2x+ 3. Note that, by Theorem 8.2, the identity, function x is continuous and, therefore, by Theorem 8.1 (e), xl is continuous, and, by Theorem 8.1 (b), -2x is continuous. By, Theorem 8.1(a). the constant function 3 is continuous. Finally, by Theorem 8.l(c), xl - 2x + 3 is continuous., Theorem 8.4: Every rational/unction H(x) = f~x~, wheref(x) and g(x) are polynomial functions. is continuous on, the set of all points at which g(x) :f:. o., gx, , This follows from Theorems 8.l(f) and 8.3. As examples, the function H (x) =-::f--,1 is continuous at all, x 7, points except I and -I, and the function G(x) = x2- 1 is continuous at all points (since x2 + 1 is never 0)., x +, We shall use a special notion of continuity with respect to a closed interval [a, b). First of all, we say that, a function/is c01ltilluouS Oil the right at a if/(a) is defined and lim f(x) exists, and lim /(x) = f(a). We say, x~at, X-+l'+, that/is continuous 011 the left at b iff(b) is defined and lim f(x) exists, and lim f(x) =/(b)., x-+b-, , x-+b-, , Definition: fis continuous on [a, b) if/is continuous at each point on the open interval (a, b),fis continuous on the, right at a, andfis continuous on the left at b., , Note that whether fis continuous on [a, b] does not depend on the values off, if any, outside of [a, b]. Note, also that every continuous function (that is, a function continuous at all real numbers) must be continuous on, any closed interval. In particular, every polynomial fUdction is continuous on any closed interval.
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CHAPTER 8 Continuity, , We want to discuss certain deep properties of continuous functions that we shall use but whose proofs are, beyond the scope of this book., Theorem 8.5 (Intermediate Value Theorem): Ifjis continuous on [a, b] andf(a):# f(b), then, for any number c, betweenf(a) andf(b), there is at least one number Xo in the open interval (a, b) for whichf(xJ = c., , Figure 8-4(a) is an illustration of Theorem 8.5. Fig. 8-5 shows that continuity throughout the interval, is essential for the validity of the theorem. The following result is a special case of the Intermediate Value, Theorem., , 1(·) - , - - - - - - - - - - - - -, , I, I, , I, , I, , I, I, I :a:, ~, , 0", , •, , (a), , (b) [(x), , =0 has three roots, , hetween :r = a and, , I, , = b., , Rg.8-4, , 'II, , I(b) - - - - - - - -, , I, I, , I, , I, , I(a), , I, I, I, I, I, 11, , b, (b) ((x) =0 has no root, between x = a and x = b,, , (a), , Rg.8-5, , Corollary 8.6: Iffis continuous on [a, b] andf(a) andf(b) have opposite signs, then the equationf(x) =0 has at least, one root in the open interval (a, b), and, therefore, the graph off crosses the x-axis at least once between a and b. (See, Fig.8-4(b).), , ,, TheoremS.7 (Extreme Value Theorem):, value M on the interval., , Iffis continuous on [a, b], thenftakes on a least value m and a greatest, , As an illustration of the Extreme Value Theorem, look at Fig. 8-6(a), where the minimum value m occurs, at x = c and the maximum value M occurs at x = d. In this case, both c and d lie inside the interval. On, the other hand, in Fig. 8-6(b), the minimum value m occurs at the endpoint x =a and the maximum value, M occurs inside the interval. To see that continuity is necessary for the Extreme Value Theorem to be true,, consider the function whose graph is indicated in Fig. 8-6(c). There is a discontinuity at c inside the interval;, the function has a minimum value at the left endpoint x =a but the function has no maximum value.
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CHAPTER 8 Continuity, , '1/, , 11, , I, I, I, I, , 1M, , I, I, I, , 1m, 0, , a, , I, I, d, , c, , 1m, I, , x, , b, , 0, , X, , a, , (a), , 0, , (b), '1/, , I, , I, I, , I, , 1m, I, , I, I, , I, , 0, , a, , c, , x, , b, , (c), , Fig. 8-6, , Another useful property of continuous functions is given by the following result., Theorem 8.8: If/is continuous at c andf(c) > 0, then there is a positive number 0 such that, whenever, c - 8 < x < c + 8, then/(x) > O., , This theorem is illustrated in Fig. 8-7. For a proof, see Problem 3., '1/, , I(c+o) ---------1(0) - - - - - - - - 1(0- 0) - - - - - - -, , o, , c+o, , Fig. 8-7, , SOLVED PROBLEMS, , 1., , Find the discontinuities of the following functions. Determine whether they are removable. If not removable,, determine whether they are jump discontinuities. (GC) Check your answers by showing the graph of the function, on a graphing calculator., (a) f(x) = 1., x, , Nonremovable discontinuity at x =O.
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CHAPTER 8 Continuity, , x-I, , (b), , f (x) =(x+3)(x-2)', , (c), , f() - (x+ 2)(x -I), , x -, , Nonremovable discontinuities at x =-3 and x =2., Nonremovable discontinuity at x =3., , (X-3)2, , 3, (d) f(x) = X 2- 297 ., x -, , Has a removable discontinuity at x =3. (Note that xl - 27 =(x - 3)(r, , + 3x +9).) Also has a nonremovable discontinuity at x = -3., , 4-x2, f(x) = 3-.JX2 +5', , (e), , Has a removable discontinuity at x = ±2. Note that, 2, 4-r 3+ x +S =3+.JX2+S., . 3-.JX2+S 3+ x 2+S, , (f) f(x) = x~x+_Xl)2 ., , Has a nonremovable discontinuity at x = I., , (g) f(x) = [xJ = the greatest integer ~ x., (h) f(x) =x - [x)., (i) f(x) = 3xl - 7r + 4x - 2., , Has a jump discontinuity at every integer., Has a nonremovable discontinuity at every integer., A polynomial has no discontinuities., , 2, , 2., , (j), , f(X)={~ ~::~, , (k), , . lX, f(x) = x, , ....., , ~". ~, , Removable discontinuity at x =O., , .':~~~~?~, ~, , if X~O., if 0 < x < I, 2-x if x~ 1., 2, , ~, , :: ", '~5., , ...::'t[-", , No discontinuities., , Show that the existence of lim f(a + hh - f(a) implies thatfis continuous at x =a., h...O, , lim(f(a+h)- f(a»=lim(f("a+hl- f(a) .h)=, h...O, , h...O, , lim f(a+h)- f(a) . lim h = lim f(a+h)- f(a) ·0= 0, h...O, h, h...O, h...O, h, But, , ",, , ~~tf:J, ., , lim (f(a+h)- f(a» = limf(a+h)-limf(a)= limf(a + h)- f(a), h...o, , h...o, , h...o, , h...o, , ~:::·-.--'j.i, , :., .~ ~:~~~:-;~, :~, , ., , : • t",, , Hence, limf(a + h) = f(a). Note that limf(a + h) =limf(x). So, limf(x) =f(a)., It~O, , 3., , x~a, , h-.O, , x-+a, , Prove Theorem 8.8., By the continuity offat c, limf(x) = f(c). If we let E= f(c)/2 > 0, then there exists a positive 8 such that 0 <, ,, Ix - cI < 8 implies that !f(x) - f(c)1 <f(c)l2. The latter inequality also holds when x =c. Thus, Ix - cI < 8 implies, !f(x) - f(c) I <f(c)l2. The latter implies -f(c)/2 <f(x) - f(c) <f(c)l2. Addingf(c) to the left-hand inequality, we, obtainf(c)/2 <f(x)., , ...., , ilj, , "i, j, , 1 l\~", I, , ", , ,, , -1 'I, , 1, , -, , ",*, , '~\I[.~', , t, , l', , .. .., , '", , ''!, , .\~~<~~ • •, , -, , ~"1, , •, , ', , ,, ,, , " , '.,..., , ,, , ~, , f., , "'T.l/.':;, , ".",~,,,,,;,;,,;., •• ,, , ", , , .... ~, I, , ,, 4., , Determine the discontinuities of the following functions and state why the function fails to be continuous at those, points. (GC) Check your answers by graphing the function on a graphing calculator., (a) f(x) =Xl -3x-1O, , (b), , f(x) ={x+ 3 ~f x ~ 2, x2 +1 ifx<2, , (c) f(x)=Ixt-x, , (d), , 4 x if x<3, f(x)= x=2 ifO<x<3, x-I if x~O, , x+2, , (e), , f(x) = ~: =~, , 1, , "", , .~
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CHAPTER 8 Continuity, , (g) f(i:) = xl- 7x, ( i), , (k), , ~", , x, , (a) Removable discontinuity at x =-2. (NOlc that x2 - 3x - 10 =(x + 2)(x - 5).), (b, c, g) None., (d) Jump discontinuity at x = o., (e) Rcmovablc discontinuities at x =± 1., (f) Removable discontinuities at x =3, x =-5. (Note that xl + 2x - 5 =(x + 5)(x - 3) and xl + xl - 17x +, 15 =(x + 5)(x - ~)(x - 1).), (h) Removable discontinuity at x =2 and nonremovable discontinuity at x =3., (i) Removable discontinuity at x =-1 and nonremovable discontinuity at x = -3., (j) Removable discontinuity at x =2 and nonremovable discontinuity at x =-2., (k) Removable discontinuity at x = 1 and non removable discontinuity at x = -1., , Ails., , .~~,~~~~ '~, , 2, , + 3x + 2, ~ Xl +4x+3, x-I, f(x) =.JX2 +3-2, , f(x), , 5., , Show thatf(x) = Ixl is continuous., , 6., , If Fig. 8-5(a) is the graph of f(x) =x - 4x 21 , show that there is a removable discontinuity at x = 7 and that, xc = 10 there., , 7., , Prove: Iffis continuous on the interval [a, b] and c is a number in (a, b) such thatf(c) < 0, then there exists a, positive number 0 such that, whenever c - 0 < x < c + 0, thenf(x) < O., , 2, , i, , (Hi1lt: Apply Theorem 8.8 to-f), , 8., , Sketch the graphs of the following functions and determine whether they are continuous on the closed interval, [0, I]:, , ::~~'j ..., , /", , ~ -'t'<,·, ;·;',.i!: .,, , (c), , I ifx<O, f(x)= -0 ifOSxSI, { 1 ifx>1, X2, ifxSO, f(x) ={ - x2 ifx > 0, , (e), , f(x)= 0, , (a), , (b) f(x) =, , l±, I, , ifx> 0, ifxSO, , (d) f(x)=lifO<xSI, , ifxSO, ifO<x<1, x ifx~ 1 ., , X, :.,i, ., ~, , ., , {, , .-.", Ans., , (a) Yes. (b) No. Not continuous on the right at O. (c) Yes. (d) No. Not defined at O. (e) No. Not continuous, on the left at I.
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The Derivative, Delta Notation, Let f be a function. As usual, we let x stand for any argument of J, and we let y be the corresponding value, of f Thus, y =f(x). Consider any number Xo in the domain off Let Llx (read "delta x") represent a small, change in the value of x, from Xo to Xo + Llx, and then let l:1y (read "delta y") denote the corresponding change, in the value of y. So, l:1y = f(x o+ Llx) - f(xJ. Then the ratio, l:1y _ changeiny _ f(xo+Llx)- f(x o), I:1x - change in x I:1x, , is called the average rate of change of the functionf on the interval between Xo and Xo + Llx., Lety= I(x) =r+2x. Starting atxo=1, changexto 1.5. Then III = 0.5. The corresponding change in, , EXAMPLE 9.1:, , y is fly =1(l.5) - 1(1) = 5.25 - 3 =2.25. Hence, the average rate of change of yon the interval between x = 1 and x = 1.5, . fly _ 2.25 - 4 5, IS III - 0.5 - . ., , The Derivative, If y = f(x) and Xo is in the domain off, then by the instantaneous rate of change off at Xo we mean the limit, of the average rate of change between Xo and Xo + I:1x as I:1x approaches 0:, , provided that this limit exists. This limit is also called the derivative off at Xo', , Notation for Derivatives, Let us consider the derivative off at an arbitrary point x in its domain:, , Jim l:1y, I:1x, , 6<-+0, , = lim f(x+l:1x)- f(x), AHO, , I:1x, , I', , The value of the derivative is a function of x, and will be denoted by any of the following expressions:, D Y = dy, x, , dx, , = y' = f'(x) =..4.., y =..4.. f(x) = lim l:1y, dx, dx, 6<-+0 I:1x, , The valuef'(a) of the derivative off at a particular point a is sometimes denoted by, , ilx;a
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CHAPTER 9 The Derivative, , Differentiability, A function is said to be differentiable at a point Xo if the derivative of the function exists at that point., Problem 2 of Chapter 8 shows that differentiability implies continuity. That the converse is false is showI:\, in Problem II., , "~:fJt:~,~~', ., , ', , :,', , SOLVED PROBLEMS, , <[~'t~~,, , 1., , . ---:.: ~~-,, , ~:-;i~·~f, , ;- -~., , :~, , .., , Given y =I(x) = xl + 5x - 8, find Ily and Ilylllx as x changes (a) from Xo = 1 to XI = Xo + Ilx = 1.2 and (b) from, xo= I tox i =0.8., (a) !!.x =XI - Xo = 1.2 - 1 = 0.2 and Ily =J(xo+ Ilx) - J(xo) =J(1.2) - J(l) = --0.56 - (-2) = 1.44., Ily _ 1.44_, So !!.x - 0.2 - 7.2., , ;~:;~;Ki, , (b) Ilx = 0.8 - I = --0.2 and Ily = J(0.8) - J(I) = -3.36 - (-2) = -1.36. So, , ~, , J--~i~~¢:~, , ~ = "jj~t = 6.8., , Geometrically. Ilylllx in (a) is the slope of the secant line joining the points (I, -2) and (1.2. --0.56) of the, parabola y =xl + 5x - 8, and in (b) is the slope of the secant line joining the points (0.8, -3.36) and 0, -2) of the, same parabola., , ~ __ ,:. ~_.~~I ~, , 2., , -::-:'::,, , ., , ,~, , If a body (that is, a material object) starts out at rest and then falls a distance of s feet in 1 seconds, then physical, laws imply that s = 16t2. Find !!.slllt as 1 changes from to to to + M. Use the result to find !!.sIM as 1 changes:, (a) from 3 to 3.5, (b) from 3 to 3.2, and (c) from 3 to 3.1., , (a) Here 10 = 3,M = 0.5, and !!.slllt = 32(3) + 16(0.5) = 104 ft/sec., (b) Here 10 = 3.!!.t = 0.2. and !!.slllt = 32(3) + 16(0.2) = 99.2 ft/sec., (c) Here '0= 3.M =0.1. and !!.slllt=97.6 ftlsec., , ., , ~k"$i~,, , Since !!.s is the displacement of the body from time t =to to t =to + Ilt., , ~ "".:.. -.i".-:, , !!.s, At, , 3., , ' 0f the bodyover th", al, = displacement, time, = average veIoclty, e bme Interv, , Find dyldx. given y =,;1- xl - 4. Find also the value of dyldx when (a) X = 4, (b) x = O. (c) X = -I., , y + Ily = (x + IlX)3 - (x + IlX)2 - 4, , =Xl + 3x 2(llx) + 3X(Il~)2 + (IlX)3 - x 2- 2x(llx) - (IlX)2 - 4, Ily =(3x 2 - 2x)llx + (3x -1)(llx)2 + (ll.tV, , ~~ =3x 2-, , 2x + (3x -1)llx + (IlX)2, , dy, = lim l3x 2 - 2x +(3x -l)llx + (IlX)2] = 3x 2- 2x, dx AI-+O, , (a), , !il, , x=4, , =3(4)2-2(4)=40;, , (b), , :tl.r-O = 3(W - 2(0) = 0;, , (c), , :tl, , _, xa 1, , = 3(-1)2 - 2(-1) =5
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",, , CHAPTER 9 The Derivative, , 4., , Find the derivative of y = I(x) = r + 3x + 5., 6.y =I(x+ 6.x) - I(x) =[(x+ 6.X)2 + 3(x+ 6.x) + 5)]- [x 2+ 3x+ 5], = [x 2+ 2x6.x+ (6.X)2 + 3x+ 36.x + 5] - [x 2+ 3x+ 5] = 2x6.x+ (6.X)2 + 36.x, , =(2x +6.x +3)6.x, , ~~ =2x+6.x+3, d, So, !!ldx = lim (2x + 6.x + 3) = 2x + 3., Ar->O, , 5., , Find the derivative of y = I(x) = ~2 at x = 1 and x = 3., x6.y= I(x+ 6.x)- I(x) =(x+, , L)- 2, , 1, (x-2)-(x+6.x-2)., x- 2 = (x- 2)(x+ 6.x- 2) ., , - (x- 2)(x+ 6.x- 2), ~_, , -1, 6.x - (x- 2)(x+ 6.x- 2), , dv - I'1m ~~,......:.....,....--::.-I, So :::.Jf...., , dx - Ar ....O (x - 2)(x +6.x - 2), , -I, , (x - 2)2 ., , dy_ -I - 1 Atx-3 dy- -I - 1, Atx --I , dx, - (1- 2)2 - - ., - , dx - (3 - 2)2 - - ., , 6., , Find the derivative of I(x) =, , i: ~ 1., I(x + 6.x) = 2(x +6.x) - 3, 3(x+6.x)+4, , 2x+26.x-3, l(x+6.x)- I(x) = 3x+36.x+4, , 2x-3, 3x+4, , (3x +4)[(2x - 3) + 26.x]- (2x - 3)[(3x + 4) + 36.x], (3x +4)(3x+ 36.x +4), , =, , _ (6x+8-6x+9)6.x, - (3x + 4)(3x + 36.x+ 4), l(x+6.x)- I(x), 6.x, , 176.x, (3x + 4)(3x + 36.x+4), , 17, (3x +4)(3x + 36.x+4), , f'(x) =l!~(3X+4)(3~:36.x+4), , 7., , l, , (3x ;4)2, , Find the derivative of y = I(x) = .J2x +1., y+ 6.y =(2x+ 26.x + 1)"2, 6.y=(2x+26.x+ 1)1/2 -(2x+ 1)112, =[(2 +26. +1)1/2_(2 1)"2](2x+26.x+I)1/2+(2x+I)112, x, x, x+, (2x+26.x+I)lh+(2x+I)I12, , (2x+ 26.x + 1)- (2x+ I), , 26.x, , =(2x + 26x + 1)lh + (2x + 1)112 = (2x + 26.x + 1)1/2 + (2x + 1)112
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• .0, 8., , CHAPTER <}, , Tile Derivative, , Find the derivative off(x) =XI/3. Examinef(O)., f(x+ 6x), , =(x+ Ax)11l, , f(x + Ax) - f(x) = (x + Ax)1I3 -, , Xl/l, , [(x + AxY/3 -, , XIl3][(X, , (x + ~X)213, , =, , + ~X)lI3 + X 1/3 (X + Ax)1I3 + x2/3], + XI/3(X + Ax)W + X U3, , x+Ax-x, , I, - (x + Ax)213 + Xltl(X + ~t)I/3 + Xli], , f(x+&)- f(x) _, , Ax, , The derivative does not exist at x = 0 because the denominator is zero there. Note that the function/is, continuous at x =o., 9., , Interpret dyldx geometrically., From Fig. 9-1 we see that ~yl6x is the slope of the secant line joining an arbitrary but fixed point P(x, y) and, a nearby point Q(x + &, y + 6)') of the curve. As Ax ~ 0, P remains fixed while Q moves along the curve toward, p, and the line PQ revolves about P toward its limiting position, the tangent line PTmoves to the curve at P., Thus, dyldx gives the slope of the tangent line at P to the curve y =f(x)., 'Y, , T, , o, Rg.9-1, For example, from Problem 3, the slope of the cubic y =,xl the point x = 9; and it is m = 5 at the P!1:.1t x = -1., , r - 4 is m = 40 at the point x = 4; it is m = 0 at, , 10. Find dsldt for the function of Problem 2 and interpret the result., , !; = 32to + 16&., ~ .~ :::r'{ ~.,~.~, , :~~.~~>~, , Hence,, , dds, t, , =limo (32to+16~t)=32to, AI ..., , As ~t ~ 0, As/ill gives the average velocity of the body for shorter and shorter time intervals ~t. Then we can, consider dsldt to be the instantalleous velocity v of the body at time toFor example, at t = 3, v = 32(3) = 96 ft/sec. In general, if an object is moving on a straight line. and its, position on the line has coordinate s at time t, then its instantaneous velocity at time t is ds/dt. (See Chapter 19.), , 11. Findf(x) whenf(x) = lxl., The function is continuous for all values of x. For x < O,f(x) = -x and, f'(x), , = lim -(x + Ax)-(-x) = lim -Ax = lim-l=-l, At ... o, , &, , <11 ...0, , Ax, , At... o
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CHAPTER 9 The Derivative, , Similarly, for x > O,f(x) = x and, f'(x) =lim (X+Ax)-X, Ar...o, Ax, , = lim Ax = lim 1= 1, Ar ...o Ax, , /(0) -lim, IAxI, - , x)-0, - an d I'1m f(O+Ax)A_, A_ •, At x -0/(, Ar-+O, LU, Ar ...o LU, , ItI -t;, , Ar-+O, , ItI t 1, , As Ax ~ 0-,, = = -I ~ -I. But, as Ax ~ ()+,, = = ~ I. Hence, the derivative does not exist, atx= O., Since the function is continuous at 0, this shows that continuity does not imply differentiability., , 12. Compute E=, , ~- ~, , for the function of (a) Problem 3 and (b) Problem 5. Verify that, , E~ 0 as Ax ~ O., , E= [3x 2- 2x+(3x-I)Ax+ (Ax)2]-(3x 2- 2x) = (3x-1 + Ax) Ax, , (a), , (b) E=, , -1, __-_1_= -(x-2)+(x+Ax-2) =, 1, Ax, . (x-2Xx+Ax-2) (X-2)2, (x-2)2(x+Ax-2), (x-2)2(X+Ax-2), , Both obviously go to zero as Ax ~ O., , i, , 13. Interpret ~y = Ax + E Ax of Problem 12 geometrically., In Fig. 9-1, ~y =RQ and, Ax= PRtanLTPR= RS; thus, EAx= SQ. For a change Ax in x from P(x, y), ~y, is the corresponding change in y along the curve while, Ax is the corresponding change in y along the tangent, , i, , i, , line PT. Since their difference E Ax is a mUltiple of (Ax)2, it goes to zero faster than Ax; and :, , ~x can be used as, , an approximation of ~y when 1&1 is small., , 14. Find ~y and ~y/&, given, (a) y =2x - 3 and x changes from 3.3 to 3.5., (b) y == xl +4x and x changes from 0.7 to 0.85., (c) Y =21x and x changes from 0.75 to 0.5., Ans. (a) 0.4 and 2; (b) 0.8325 and 5.55; (c) t and -.If, 15. Find ~y, given y = xl - 3x + 5, x == 5, and &, Ans., , ~y=-0.0699;, , =-0.01. What then is the value of y when x =4.99?, , y= 14.9301, , 16. Find the average velocity (see Problem 2), given: (a) s == (3t 2 + 5) feet and t changes from 2 to 3 seconds., (b) s =(2t2 +'St - 3) feet and t changes from 2 to 5 seconds., Ans. (a) 15 ftlsec; (b) 19 ftlsec, 17. Find the increase in the volume of a spherical balloon when its radius is increased (a) from' to, + M inches;, (b) from 2 to 3 inches. (Recall that volume V = 4",3.)
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CHAPTER 9 The Derivative, , 18. Find the derivative of each of the following:, (a) y=4x-3, (d) y = l/r, (g), , y=JX, , (j), , y = 11../2+ X, , AilS., , (b) y=4-3x, (e) y=(2x-l)/(2x+ 1), (h) y= I/JX, , (c) y=r+2x-3, (f) y (1 + 2x)/(1 - 2x), (i) . y = + 2x, , =, ../1, , i, , (a) 4; (b) -3; (c) 2(x + 1); (d) -2/xl; (e) (2 4 1)2; (f) (1, )2; (g) Ie ; (h) 1e ;J'(i), 1, x+, - x, 2vx, 2x"x, (J.) _ 2(2+x)3/2, , A;, 1+2x, , 19. Find the slope of the tangent line to the following curves at the point x = 1 (see Problem 9): (a) y = 8 - 5xl;, (b) y= x!l;(c), , Ans., , x~3', , (a) -10; (b) -1; (c), , ., , -t, , 20. (GC) Use a graphing calculator to verify your answers in Problem 19. (Graph the curve and the tangent line that, you found.), , 21. Find the coordinates of the vertex (that is, the tuming point) of tilt! parabola y = xl - 4x + 1 by making use of, the fact that, at the vertex, the slope of the tangent line is zero. (See Problem 9.) (GC) Check your answer with a, graphing calculator., Ans., , (2, -3), , 22. Find the slope m of the tangent lines to the parabola y = -xl + 5x - 6 at its points of intersection with the x axis., AIlS., , At x = 2, m = 1. At x =3, m =-1., , 23. When an object is moving on a straight line and its coordinate on that line is s at time t (where s is measured in, feet and t in seconds), find the. velocity at time t = 2 in the following cases:, (a) s=f+3t, , (c), , s="/1+2, , (See Problem 10.) ,, Ans., , (a) 7 ftlsec; (b) 0 ftlsec; (c), , t ftlsec, , 24. Show that the instantaneous rate of change of the volume V of a cube with respect to its edge x (measured in, inches) is 12 in 3/in when x =' 2 in.
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Rules for Differentiating, Functions, Differentiation, Recall that a function/is said to be differentiable at Xo if the derivativef'(Xo) exists. A function is said to be, differentiable on a set if the function is differentiable at every point of the set. If we say that a function is, differentiable, we mean that it is differentiable at every real nun,lber. The process of finding the derivative, of a function is called differentiation., Theorem 10.1 (Differentiation Formulas): In the following formulas, it is assumed that u, v, and, that are differentiable at x; c and m are assumed to be constants., (I), , (2), , ix, it, , I\', , are functions, , (c) = 0 (The derivative of a constant function is zero.), (x) = I (The derivative of the identity function is I.), , dll, (3) .!L(cu)-c, dx, -dx, , d (u+v+ ... ) = du, (4) dx, dx + dv, dx +..., , (5) .!L(u _ v), dx, , (6), , =dlldx _ dvdx, , ix (/Iv) = ct + v :, II, , du, , v, , 1) =--:!rx, ix (xm ) =, , (8) .!L(, , dx x, , (9), , (Difference Rule), (Product Rule), , dv, , dx, (7) .!L(~)= v dx -/I, 2, dx v, , (Sum Rule), , IIlX,"-1, , provided that V:F- 0, , (Quotient Rule), , provided that x :F- 0, (Power Rule), , Note that formula (8) is a special case of formula (9) when m = -I. For proofs, see Problems 1-4., EXAMPLE 10.1:, D, (x 3 + 7x+ 5) = D, (x 3 )+ D, (7x) + D,(5), , =3x' + 7 D, (x) + 0, = 3x2 + 7, , (Sum Rule), , (Power Rule, fonnulas (3) and (I», , (formula (2», , Every polynomial is differentiable, and its derivative can be computed by using the Sum Rule, Power Rule, and, formulas (1) and (3)., , --
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CHAPTER 10, , Rules for Differentiating Functions, , Composite Functions. The Chain Rule, The composite junction /og offunctions g and/is defined as follows: (f og)(x) =/(g(x». The function g, is applied first and then/. g is called the inner junction, and/is called the outer junction. /0 g is called the, composition of g andf', EXAMPLE 10.2:, , Letf(x) =.xl and g(x) = x + I. Then:, , ,, , (f og)(x) = f(g(x» = f(x+ I) = (x + 1)2 = x 2 + 2x+ 1, (go flex) = g(f(x» = g(X2) = x 2 + I, , Thus, in this case, fog:F- go f., , When / and g are differentiable. then so is their composition /0 g. There are two procedures for finding, the derivative of /0 g. The first method is to compute an explicit formula for /(g(x» and differentiate., EXAMPLE 10.3:, , If f(x) =.xl + 3 and g(x) = 2x + 1, then, , y= f(g(x» = f(2x+ I) =(2x+ 1)2 +3=4x 2+4x+4, , and, , i=8x+4, , Thus, Dx (fog)=8x+4., , The second method of computing the derivative of a composite function is based on the following rule., , Chain Rule, Dif(g(x» = f'(g(x»· g'(x), , Thus, the derivative off 0 g is the product of the derivative of the outer functionf(evaluated at g(x» and the derivative, of the inner function (evaluated at x). It is assumed that g is differentiable at x and thatfis differentiable at g(x)., EXAMPLE 10.4:, , In Example lO.3,f'(x), , =2x and g'(x) = 2. Hence, by the Chain Rule,, , Dx(f(g(x» = f'(g(x»· g'(x) = 2g(x)· 2 = 4g(x) = 4(2x + 1) = 8x +4, , Alternative Formulation of the Chain Rule, Let u =g(x) and y =/(u). Then the composite function of g and/is y =/(u) =/(g(x», and we have the fonnula:, , Ex- dy du, , dx - du dx, , EXAMPLE 10.5:, , (Chain Rule), , Let)' = u3 and u = 4.xl - 2x + 5. Then the composite function y = (4.xl - 2x + 5)3 has the derivative, dy = dy dll = 3u2(8x _ 2) = 3(4x 2 - 2x + 5)2(8x - 2), dx dll dx, , t, , :y, , Warning. In the Alternative Formulation of the Chain Rule., =, du, the yon the left denotes the composite function of x, whereas the)' on the right denotes the ongin~ %nction of u. Likewise, the two occurrences of u have different meanings. This notational confusion is made up for by the simplicity of the, alternative formulation.
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CHAPTER 10 Rules for Differentiating Functions, , Inverse Functions, Two functions f and g such that g(f(x» =x andf(g(y» =y are said to be inverse functions. Inverse functions, reverse the effect of each other. Given an equation y = f(x), we can find a formula for the inverse of fby, solving the equation for x in terms of y., EXAMPLE 10.6:, , (a) Letf(x) =x+ t Solving' the equationy =x + 1 for x, we obtain x = y - 1. Then the inverse g off is given by the, fonnula g(y) =y - 1. Note that g reverses the effect off andfreverses the effect of g., (b) Letf(x) =-x. Solving y =-x for x, we obtain x =-yo Hence, g(y) =-y is the inverse off. In this case, the inverse, off is the same function as f., (c) Let f(x) =.,fX. fis defined only for nonnegative numbers, and its range is the set of nonnegative numbers. Solving y =JX for x, we get x =y, so that g(y) =y. Note that, since g is the inverse off, g is only defined for nonnegative numbers, since the values off are the nOMegative numbers. (Since y = f(g(y», then, if we allowed g to, be defined for negative numbers, we would have -1 =f(g(-I» = f(1) = I, a contradiction.), (d) The inverse of/(x) =lx-I is the function g(y)= y; I., , Notation, The inverse off will be denotedf-I., Do not confuse this with the exponential notation for raising a number to the power -1. The context will, usually tell us which meaning is intended., Not every function has an inverse function. For example, the functionf(x) = i'- does not possess an inverse. Sincef(l) 1 f(-I), an inverse function g would have to satisfy g(l) 1 and g(l) == -I, which is, impossible. (However, if we restricted the functionf(x) =i'- to the domain x ~ 0, then the function g(y) =, would be an inverse function off.), The condition that a functionJmust satisfy in order to have an inverse is thatfis one-to-one, that is, for, any XI and x2, if XI X2' thenJ(xl}:f:. J(x2). Equivalently,fis one-to-one if and only if, for any XI and x2, if, J(x l} =!(X2)' then XI = X2', , = =, , =, , JY, , *, , Let us show that the functionf(x) = 3x + 2 is one-to-one. Assumef(xl) = f(X2)' Then 3xt + 2 =, 2, 3X2 + 2, 3x t =3~, XI =.12. Hence,fis one-to-one. To fmd Ule inverse, solve y = 3x + 2 for x, obtaining X =y; . Thus,, 2, f-I(y) =y; . (In general, if we can solve y =f(x) for Xin tenns ofy, then we know thatfis one-to-one.), , EXAMPLE 10.7:, , Theorem 10.2 (Differentiation Formula for Inverse Functions): Letfbe one-to-one and continuous on an interval (a, b). Then:, (a) The range offis an interval I (possibly infinite) andfis either increasing or decreasing. Moreover,f-t is continuous on I., (b) Iffis differentiable at Xo andf'(Xo):f:. 0, thenf-t is differentiable at Yo = f{Xo) and (f-I)'(yO) =, , rlxo)", , The latter equation is sometimes written, , where x =fll(y)., For the proof, see Problem 69., EXAMPLE 10.8:, , (a) Lety=f{x)=,rfor x>O. Then x= r'{y)=Ji. Since : =2x,, this is a special case of Theorem 8.l(9) when m =t.), , ~ =2~ =2$' Thus, D,(Ji) =2$' (Note that
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CHAPTER 10 Rules for Differentiating Functions, , (b) Let y =f(x), , =x3 for all x. Then x = f-I(y) =</y =yll3, , for all y. Since, , all y ¢ O. (Note thatf-I(O) = 0 andf'(O) = 3(0)2 = 0.), , '*, , =3x2 ,, , ':yt =pix =J=2l31, This holds for, y, ., 2, , •, , Higher Derivatives, If y =f(x) is differentiable. its derivative y' is also called the first derivative of f If y' is qifferentiable, its, derivative is called the second derivative off If this second derivative is differentiable. then its derivative is, called the third derivative off, and so on., , Notation, First derivative:, , y',, , f'(x),, , Second derivative:, , y",, , f"(x),, , Third derivative:, , yIf' , f"'(x),, , nth derivative:, , In), pn),, , dy, dx' DxY, d 2y, dx 2 , D;y, d 3y, dx 3 ' D~y, dny, dx n ' D;y, , SOLVED PROBLEMS, 1., , ., , Remember that, (I), , (2), ( 3), , .!L f(x) = lim, dt, , d, , d, , ~, , .u-.o, , f(x + llx) - f(x), llx, , d c = lim c:: c = lim 0 = 0, dx, 4,-+0 uX, , d (), , r, , dx x = 4!~, d (, , )_, , 4x-+O, , (x + ax) - x·, , r, , dx cu - 4~1, , r, , ax, , r llx r I I, 4~ ax =4~O =, , cu(x + llx) - cu(x), llx, , =c 4!~O, 2., , d, , Prove Theorem 10.1, (1 )-(3): (1) dx (c) = 0; (2) dx (x) = 1; (3) dx (cu) = c dx', , u(x + ax) - u(x), , ax, , r, , ~c, , u(x + ax) - u(x), ax, , du, = c dx, , Prove Theorem 10.1, (4), (6), (7):, (4), , tx(u+v+ .. ')=:+~+'", , (6), , dx (uv) = II dx, , dv, , + v dx, , dll, , dv, , (7), , d, , dll, , tx(%) = VdX~ udi, , v, , provided that ¢O, , (4) It suffice to prove this for just two summands. II and v. Letf(x) =II + v. Then, f(x + ax) - f(x), , ax, , _ u(x + llx) llx, -, , :':j;~ft::, , {I, , u(x + llx) + v(x + ax) - u(x) - v(x), , Taking the limit as llx --+ 0 yields, , tx (u + v) = :, , ax, u(x) + v(x + ax) -, , + ~., , llx, , v(x)
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CHAPTER 10 Rules for Differentiating Functions, (6) Letf(x) =uv. Then, f(x + ax) - lex), ax, , u(x + ax)v(x +ax) - u(x)v(x), ~x, , _ [u(x+ ax)v(x+ ax)-v(x)u(x+ ~x)]+[v(x)u(x +~x)- u(x)v(x)], , -, , Ax, , =u(x+ Ax) v(x+ ~~- vex) + v(x)u(x+ ~-u(x)., Taking the limit as ~x -+ 0 yields, !(uv) = u(x) !v(x)+v(x) !U(x)=u~ +v:, , Note that lim u(x + ~x) =u(x) because the differentiability of u implies its continuity., 6HO, , =vx, , (7) Set f(x) =!! u«x», then, v, , I(x + ax) - lex), ax, , u(x+~x), , ~, , vex + ~x), , vex), , U(X + ax)v(x) -, , u(x)v(x +ax), , Ax( v(x)v(x + ~x»), , Ax, , _ [u(x +~x)v(x)- u(x)v(x)]- [u(x)v(x+ ~x)-u(x)v(x»), Ax[v(x)v(x+ Ax)], ( ), =vx, , d, , u(x+ax)-u(x), , Ax, , v(x)v(x +~x), , _ d (u)_ v(x)-ixu(x)-u(x)-ixv(x), , and for ~x -+ 0, dx ~(x) -dx -; 3., , ()v(x+ax)-v(x), ux, ax, , [V(X)]2, , v~-u~, v2, , Prove Theorem 10.1 (9): DxCXM) =rnxm-I, when m is a nonnegative integer., Use mathematical induction. When m =G,, , Assume the formula is true for m. Then, by the Product Rule,, , =XM, , +rnx M = (m+ l)xM, , Thus, the formula holds for m + 1., 4., , Prove Theorml 10.1(9): D.(x M ) = rnx m- I, when m is a negative integer., Let m = -k, where k is a positive integer. Then, by the Quotient Rule and Problem 3,, , k-I, , ., , ., , x - kx- l - I -- rnxm- I, -- -k 7--
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CHAPTER 10 Rules for Differentiating Functions, , f i ' y ~ 3-lx, 14• D1' ferentiate, 3 +lx.·, , Use the Quotient Rule:, , ,, , d, , d, , ,_ (3 +2x)di(3-2x)-(3-2x)t1X(3 +2x) _ (3+2xX-2)-(3-2x)(2), Y(3+lx)2, (3+lx)2, , .., 15. Differentiate y =, dy, dx, , =, , -12 ., (3+2x)2, , J4x_x =(4 _xX2)1!2 •, 2, , 2, , 2, , (4 - X2)'I2-9x(X2)-X2i(4 4-x2, , (4 -X2)U2(lx)-(x1XtX4 _ x2tl12(-2x), 4-x1, , r)112, , _ (4-x2t2(2x)+r(4-x2t'2 (4_X 2)/2, 4-x 2, (4-X 2 )112, , 16. Find ~, given x =, , yJl- y2 •, , By the Product Rule,, , By Theorem 10.2,, , 17. Find the slope of the tangent line to the curve x = y2 - 4y at the points where the curve crosses the y axis., The intersection points are (0, 0) and (0, 4). We have : = 2y-4 and so :, At (0,0) the 'slope is -t, and at (0, 4) the slope is t., 18•. Derive the Chain Rule: DJJ(g(x» = f'(g(x» . g'(x»., ., Let H=fog. Lety=g(x) andK';'g(x+h)-g(x).Also, let, , = dx}dy = 2 ~4', , F(t)=f(y+t~-f(Y) f'(y) fort¢O., , Since lim F(t) =0, let F(O) =O. Thenf(y + t) - f(y) =t(F(t) +f'(y» for all t. When t:::; K,, , .-.0, , f(y+K)- f(y) =K(F(K) + f'(y», f(g(x +h» - f(g(x» =K(F(K) + f'(y», H(X+hl-H(x), , Hence,, , r, , Now,, , =~(F(K)+ f'(y», , K _ r g(x+h)- g(x) _ '( ), h, -g x, , 12J7I- h~, , Since lim K =0, lim F(K) =O. H!!nce,, ", , h...O, , h-tO, , H'(x) = f'(y)g'(x) = f'(g(x»g'(x)., 19. Find dydx' given y = u~ -11 and u =~X2 +2., , ., , u +, , El...-, , 4u, du - (u 2 +1)2, , and, , y, , ..:',
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CHAPTER 10 Rules for Differentiating Functions, dy _ Ex. du _, dx - du dx -, , Then, , 4u, (U 2, , 2x _, , 8x, , + 1)2 3u 2 - 3u(u 2 + 1)2, , 20. A point moves along the curve y =xl - 3x + 5 so that x.= tJi + 3, where 1 is time. At what rate is y changing, when 1=47', We must find the value of dy/dl when 1=4. First, dyldx = 3(xl - I) and dx/dl, , =1I(4Ji). Hence,, , dy _ dy dx _ 3(X2 -1), dl - dx dt - 4Ji, When 1 =4, x =, , tJ4 + 3 =4, and, , 1, = 3(~~; I), , ~, , units per unit of time., , 21. A point moves in the plane according to equations x =t2 + 21 and y = 21 3 - 61. Find dyldx when 1=0, 2, and 5., Since the first equation may be solved for 1 and this result substituted for 1 in the second equation, y is a, function of x. We have dyldl =6t2- 6. Since dxld! =21 + 2, Theorem 8.2 gives us dtldx = 1/(21 + 2). Then, , dy _ dy dl _ 2, 1_, dx - dt dx - 6(1 -I) 2(1 + 1) - 3(1 -I)., The required values of dyldx are -3 at 1=0,3 at 1=2, and 12 at t =5., 22. If y = xl - 4x and x = ../212 + 1, find dyldt when 1=,fi., , i, , and, , dx, 21, dl = (212 + 1)112, , dy _ dy dx _ 41(x- 2), dt - dx dt - (21 2 + 1)112, , So, "'2, When t --o,J~,, , =2(x-2), , -, , ~5, , X-o,J;), , ~5), and dy, dl -- 4,fi(J5 - 2) - 4,fi, 5 (5- - 2 o,J;)., , J5, , -, , 23. Show that the function/(x) =xl + 3x2 - 8x + 2 has derivatives of all orders and find them., f'(x) =3xl + 6x - 8,J"(x) =6x + 6.f'"(X) =6, and all derivatives of higher order are zero., , 24. Investigate the successive derivatives of f(x) = xA'3 at x = O., , f'(x) = tx"3, 213, 4, / "(x) -- .!., v X- -- 9x2Jl, , pn)(o) does not exist for n, , ~, , and, , 1'(0) = 0, , and, , /,,(0) does not exist, , 2., , 25. If /(x) = I': x = 2(\ - xtl, find a formula for /(n)(x)., , f'(x) =2(-1)(1- xt2(-I) =2(1- xt2, , =2(1 !)(I- xt2, , /"(x) =2(l!)(-2)(1- xt3 (-I) = 2(2!)(1- xt3, f'''(x) = 2(2!)(-3)(1- x)-4(-I) = 2(3!)(\ - x)-4, , =, , which suggestpn)(x) 2(n!)(1 - x)-(n+ I). This result may be established by mathematical induction by showing, that if PA)(X) = 2(k!)( I - x)-(A+ I), then, JlA+I)(X) = -2(k!)(k + 1)(1- xt(k+2)(-I) = 2[(k + 1)1](1- xt(k+2)
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CHAPTER 10 Rules for Differentiating Functions, , 62. f(x)=r+2, , Ans., , no inverse function, , 63. f(x)=x', , Ans., , x= I, , ,, 2x-l, 64. f(x)= x+2, , Ans., , X, , -I, , (y) ,T:. dx, I, I -VJ, ="y. dy =3x 2 ='3 y, , -I-I(y)- 2y+l. dx _ _5~, -, , - - y-2' dy - (y_2)2, , 65. Find the points, at which the function/(x) =Ix + 21 is differentiable., Ans., , All points except x =-2, , 66. (GC) Use a graphing calculator to draw the graph of the parabola y =xl - 2x and the curve y = !x2 - hi. Find all, points of discontinuity of the latter curve., Ans. x=Oandx=2, 67. Find a fonnula for the nth derivative of the following functions: (a) f(x) = x~ 2; (b) f(x) =.JX., , Ans., , (a) ,CO'(x) = (-1)"+1, , 2n!, (x+ 2)0+1, , (b) t<"'(x) = (_1)"+1 3·5· 7· ... . (2n - 3) X-{2n-I)ll, , 2n, , 68. Find the second derivatives of the following functions:, (a) I(x) = 2x - 7, (b) I(x) = 3xl + 5:c - 10, (c) I(x) =x!4, (d) l(x)=.,J7-x, Ans., , 2, , (a) 0; (b) 6; (c) (x + 4)3; (d), , 1, , 1, , 4 (7 _ X)312, , 69. Prove Theorem 10.2., Ans., , Hints: (a) Use the intennediate value theorem t9 show that the range is an interval. That/is 'increasing, or decreasing follows by an argument that uses the extreme value and intennediate value theorems. The, continuity ofI-I is then derived easily., f-I(y) - r<Y ), 1, 1, (b), y_ Yo 0 - l(f-I(y»- /(r<Yo» I(x)- I(xo), ri(y)- 1-I(yO), x-~o, By the continuity Of/-I. as y -+ Yo. x -+ xo. and we get (f-I)'(yO)= f'(~o)', , I'
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Implicit Differentiation, Implicit Functions, An equationf(x, y) =0 defines y implicitly as a function of x. The domain of that implicitly defined function, consists of those x for which there is a unique y such thatf(x. y) = o., EXAMPLE 11.1:, , *", , (a) The equation xy + x - 2)' - I = 0 can be solved for y, yielding y = 1- x . This function is defined for x 2., x- 2, (b) The equation, + 9y2 - 36 =0 does not determine a unique function y. If we solve the equation for)" we, obtain y =, x 2 • We shall think of the equation as implicitly defining two functions. y =, x 2 and, 2, y=, x • Each of these functions is defined for Ixl ~ 3. The ellipse determined by the original equation is, the union of the graphs of the two functions., , 4r, , ±t,./9 -t../9 -, , t../9 -, , If y is a function implicitly defined by an equalionf(x. y), ferent ways:, I., , 2., , =0, the derivative y' can be found in two dif-, , Solve the equation for y and calculate y' directly. Except for very simple equations, this method is usually, impossible or impractical., Thinking of y as a function of x. differentiate both sides of the original equationf(x, y) = 0 and solve the, resulting equation for y'. This differentiation process is known as implicit differentiation., , EXAMPLE 11.2:, (a) Find y', given x)' + x - 2)' - I = O. By implicit differentiation, xy' + y D,(x) - 2y' - Dz<!) = Dx(O). Thus, xy' + y-, , 2y' = O. Solve for y': y' = 21 + )' . In this case, Example I 1.1 (a) shows that we can replace y by 1- x and find y', -x, x- 2, in tenns of x alone. We see that it would have been just as easy to differentiate y = \- x by the Quotient Rule., , x- 2, , However, in most cases, we cannot solve for y or for y' in terms of x alone., + 91 - 36 =0, find y' when x = $. By implicit differentiation, 4Dx(r) + 9D.(f) - DP6) = D/O)., (b) Given, Thus, 4{2x) + 9(2yy') =O. (Note that D,<f) = 2yy' by the Power Chain Rule.) Solving for y', we get y' =-4x19y., When x =. . ./5, y =, For the function y corresponding to the upper arc of the ellipse (see Example I 1.1 (b»,, y=, and y' =-$/3. For the function y corresponding to the lower arc of the ellipse, y =, and y' =-$/3., , 4r, , -t, , it., , -t, , Derivatives of Higher Order, Derivatives of higher order may be obtained by implicit differentiation or by a combination of direct and, implicit differentiation., EXAMPLE 11.3: In Example Il.2(a), y' = 21 + Y . Then, , -x, , ,If, , .\, , =D r,,')= D, , (I, , + y)= (2-x)y' -(1 + y)(-I), x v , 2-x, (2_X)2, , _(2-x)y'+I+y_ (, (2 - X)2, -, , y), , 1+- +1+, 2-x ( ) 2-x, Y _ 2+2y, (2 - X)2, - (2 - X)2
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CHAPTER 11 Implicit Differentiation, , 4., , Find y", given: (a) x + xy + y =2: (b) x3 - 3xy +1, , ., 2(1 + y), Ans. (a) y " =, - - ' (b) y"=, (1 + X)2 ', , =1., , 4X)', , (y2 -, , X)3, , ,, 5., , Find y', y", and)'111 at: (a) the point (2, 1) on xl -l- x = 1; (b) the point (1, 1) on x, , Ans., 6., , (a), , 3, , + 3x y - 6xy2 + 21 =o., 2, , t, --t, t; (b) 1,0,0, , Find the slope of the tangent line at a point (xo' Yo) of: (a) Irxl + aVo = a2lJ2; (b) b2xl - a2f = a2lJ2; (c) x3 + 1 -, , 6.ry= O., , 7., , Prove that the lines tangent to the curves 5y - 2x + yl- xly = 0 and 2y + 5x + x4 - xV = 0 at the origin intersect, at right angles., , 8., , (a) The total surface area of a closed rectangular box whose base is a square with side y and whose height is x is, given by S = 21 + 4xy. If S is constant, fmd dy/dx without solving for y., (b) The total surface area of a right circular cylinder of radius r and height h is given by S = 2m.2 + 21trh, If S is, constant, find dr/dh., , Ans.· (a) 9., , Xry:, , (b), , -2r~h, , ., , I, , y" ]312, For the cucle x2 + y2 = r, show that [l + (y')2, , I=r', 1, , 10. Given S = 1tX(x+ 2y) and V= 7tx2y, show that dS/dx = 21t(x - y) when Vis a constant, and dV/dx =-1tX(x - y), when S is a constant., 11. Derive the formula Dx(x m) = mt m- I of Theorem 10.1(9) when m = p/q, where p and q are nonzero integers. You, may assume that xplq is differentiable. (Hint: Let y =xplq. Then yq =x p• Now use implicit differentiation.), 12. (GC) Use implicit differentation to find an equation of the tangent line to, answer on a graphing calculator., , Ans., , y=-x+8, , JX + JY = 4, , at (4,4), ailct verify your
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Tangent and Normal Lines, An example of a graph of a continuous function/is shown in Fig. 12-1(a). If Pis a point of the graph having abscissa x, then the coordinates of Pare (x,J(x». Let Q be a nearby point having abscissa x + ~x. Then, the coordinates of Q are (x + &,j(x + ~x». The line PQ has slope /(x + ~ -/(x) . As Q approaches P, along the graph, the lines PQ get closer and closer to the tangent line fJ to the graph at P. (See Fig. 12-1 (b).), Hence, the slope of PQ approaches the slope of the tangent line. Thus, the slope of the tangent line is, lim /(x+ &) - /(x) , which is the derivative f'(x)., &, , dx->O, , y, , y, , ,, "" ...., ,, ........ ........, ", ,, ........ efT, ,,, ,,, ,,, ,, , " ........ ........, " " ........, "" ,, ", , ", ,, ,, ', Q" ,, ,,, ~", , "", ", , Q(x+~,f(x+~», , ,, --r------------------------+x, , ... ·efT, , /I., , --r-----------------------~x, , (a), , (b), , Rg.12-1, , If the slope m of the tangent line at a point of the curve y =/ (x) is zero, then the cU'Vi: has a horizontal, tangent line at that point, as at points A, C, and E of Fig. 12-2. In general, if the derivative of/is m at a point, (xo' Yo)' then the point-slope equation of the tangent line is y - Yo =m(x - xo)' If/is continuous at xo' but, lim f'(x) = 00, then the curve has a vertical tangent line atxo' as at points Band D of Fig. 12-2., , X-+X(I, , •, , JI, , Rg.12-2
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CHAPTER 12, , Tangent and Normal Lines, , The normal line to a curve at one of its points (xo' Yo) is the line that passes through the point and is perpendicular to the tangent line at that point Recall that a perpendicular to a line with nonzero slope m has slope -11m., Hence, if• m ¢ 0 is the slope of the tangent line, then y - y0 =-( lIm)(x - x 0) is a point-slope equation of the, ,, normallme. If the tangent line is horizontal, then the normal line is vertical and has equation x = x . If the, tangent line is vertical, then the normal line is horizontal and has equation y = )'0', 0, , The Angles of Intersection, The angles of intersection of two curves are defined as the angles between the tangent lines to the curves at, their point of intersection., To determine the angles of intersection of the two curves:, 1., , Solve the equations of the curves simultaneously to find the points of intersection., , 2., , Find the slopes 111( and 1112 of the tangent lines to the two curves at each point of intersection., , 3., , If 111 ( =111 2, the angle of intersection is 0°, and if 111 ( =-VI1I 2• the angle of intersection is 90°; otherwise,, the angle of intersection i/J can be found from the fOImula, , i/J is the acute angle of intersection when tan i/J > 0, and 180° -i/J is the acute angle of intersection when, tan i/J < O., , ,-'., , SOLVED PROBLEMS, , 1., , Find equations of the tangent and normal lines to y = f(x) =xl - 2.x2 + 4 at (2,4)., /,(x) = 3xl - 4x. Thus, the slope of the tangent line at (2,4) is m =/,(2) =4. and an equation of the tangent, line is y - 4 = 4(x - 2). The slope-intercept equation is y = 4x - 4., An equation of the normal line at (2,4) is y - 4 =-t(x - 2). Its slope-intercept equation is y = -t x + t., , 2., , Find equations of the tangent and normal lines to xl + 3xy + y2 = 5 at (I, I)., By implicit differentiation, 2x + 3xy' + 3y + 2yy' = O. So, y' = - ; ; : ;;. Then the slope of the tangent line, at (1, 1) is -I. An equation of the tangent line is y - I = -(x - 1). Its slope-intercept equation is y = -x + 2. An, equation of the normal line is y - 1 = x-I, that is, y = x., , 3., , Find the equations of the tangent lines with slope m =-t to the ellipse 4x2 + 9y2 = 40., By implicit differentiation. y' =-4x/9y. So, at a point of tangency (xo' Yo)' m =-4 xo 19yo =- t. Then Yo = 2xo', Since the point is on the ellipse, 4x~ + 9y~ = 40. So, 4x~ +9(2xo)2 = 40. Therefore, x~ = I, and Xo = ± 1. The, required points are (1, 2) and (-I, -2)., At (1, 2), an equation of the tangent line is y - 2 =-i(x -1)., At (-1, -2), an equation of the tangent line is y+2=-t(x+ 1)., , 4., , Find an equation of the tangent lines to the hyperbola).,J. - y2 =16 that pass through the point (2, -2)., By implicit differentiation, 2x - 2yy' = 0 and, therefore, y' = xly. So, at a point of tangency (xo' Yo)' the slope of, the tangent line must be x/Yo' On the other hand, since the tangent line must pass through (xo' Yo) and (2, -2), the, . Yo+2, siopels - - ., xo-2, )'0 + 2 Hence, Xo2 - 2Xo =Yo22Th', . Id'lIlg Xo + Yo -- 8,an, . d., + Yo' us, 2Xo + 2Yo -- x()2 - Yo2 -- 16• Yle, Thus, -Xo = --2', Yo, , xo -, , therefore, Yo = 8 - xo', If we substitute 8 - Xo for Yo in x~ - y~ = 16 and solye for ro' we get Xo = 5. Then Yo = 3. Hence, an equation of, the tangent line is y - 3 = t(x - 5).
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Tangent and Normal Lines, , CHAPTER 12, , 9., , One of the points of intersection of the curves (a) y2 = 4x and (b)2x2 = 12 - 5y is (I, 2). Find the acute angle of, intersection of the curves at that point., For (<I), y' = 2Iy. For (b), y' = -4xI5. Hence, at (I, 2), m. = 1 and m2 = -t. So,, tanq>=, Then, , fjJ '" 83°, , m -m, 1+.i, •, 2 =~=9, l+m.m2 1-5', , ,, , 40' is the acute angle of intersection., , 10. Find the angles of intersection of the curves (a) 2x2 + y2 = 20 and (b) 4y2 - xl = 8., Solving simultaneously, we obtain y2 = 4, Y = ±2. Then the points of intersection are (±2.J2, 2) and (±2.J2, - 2)., For (a), y'= -2x/y, anMor (b), y'=xI4y. At the point (2.J2, 2), m. = -2.J2 and11tz:.= t.J2. Since m.m2 = -I, the, angle of intersection)s 90° (that is, the curves are orthogonal). By symmetry, the curves are orthogonal at each of, their points of intersection., 11. A cable of a certain suspension bridge is attached to supporting pillars 250 ft apart. If it hangs in the form of a, , parabola with the lowest point 50 ft below the point of suspension, find the angle between the cable and the pillar., Take the origin at the vertex of the parabola, as in Fig. 12-4. The equation of the parabola is y = 6is x 2 and, y'= 4x1625., At (125, 50), m = 4(125)/625 = 0.8000 and e= 38°40'. Hence, the required angle is IP = 90° - e = 51 ° 20'., Y, , __, , ~, , __, , ~~~~~, , __+-_______ x, , /, , /, , Fig. 12-4, , 12. Examine xl + 4xy + 16y2 = 27 for horizontal and vertical tangent lines., AilS., , Horizontal tangents at(3, -, , t) and, , (-3,, , t). Vertical tangents at (6,, , -, , t) and (-6,, , -, , t)·, , 13. Find equations of the tangent and normal lines to x 2 - y2 = 7 at the point (4, -3)., Ans., , 4x + 3y = 7 and 3x - 4y = 24, , 14. At what points on the curve y = x3 + 5 is its tangent line: (a) parallel to the line 12x - y = 17; (b) perpendicular to, the line x + 3y = 2?, Ans., , (a) (2, 13), (-2, -3); (b) (1. 6). (-1,4), , 15. Find equations of the tangent lines to 9r + 16y2 = 52 that are parallel to the line 9x - 8y = I., AilS., , 9x - 8y = ±26
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CHAPTER 12 Tangent and Normal Lines, 16. Find equations of the tangent lines to the hyperbola xy = 1 that pass through the point (-1, 1)., Ans., , y, , =(2.[i - 3)x +2.[i - 2; y =-(2.[i +3)x ~ 2.[i - 2, , 17. For the parabola y2 = 4px, show that an equation of the tangent line at one of its points P(xo' yo> is YoY = 2p(x +xo)', , r, , 18. For the ellipse b2 + a'f = aW. show that the equations of its tangent lines of slope mare, y= mX± .Ja 2m2 +b2, , -, , 19. For the hyperbola ~r a2y2 = a2~. show that (a) an equation of the tangent line at one of its points P(xo' Yo) is, b2xrf - a2yoY = aW; and (b) the equations of its tangent lines of slope mare y =mt± .Ja2m~ -; b2 •, , )';, , 20. Show that the normal line to a parabola at one of its points P bisects the angle included between the focal radius, of P and the line through P parallel to the axis of the parabola., , 21. Prove: Any tangent line to a parabola, except at the vertex, intersects the directrix and the latus rectum (produced, if necessary) in points equidistant from the focus., , , 22. Prove: The cbord joining the points of contact of the tangent lines to a parabola from, passes through the focus., , ~y, , point on its directrix, , 23. Prove: The normal line to an ellipse at any of its points P bisects the angle included between the focal radii of P., , +.JY, , 24. Prove: (a) The sum of the intercepts on the coordinate axes of any tangent line to JX, =Ja is a constant., (b) The sum of the squares of the intercepts on the coordinat~ axes of any tangent line to r/3 + y213 =a'll3 is a, constant., , 25. Find the acute angles of intersection of the circles, , r - 4x +y2 =0 and r +y2 =8., , 26. Show that the curves y = xl + 2 and y =2r + 2 have a common tangent line at the point (0, 2) and intersect at the, point (2, 10) at an angle , such that tan ~ =J.i., , 27. Show that the ellipse 4i + 9y2 =45 and the hyperbola r, , - 4y2 =5 are orthogonal (that is, intersect at a right angle)., , 28.' Find equations of the tangent and normal lines to the parabola y =4r at the point (-1, 4),, Am., , y + 8x + 4 =0; 8y .:. x - 33 = 0, , 29. At what points on the curve y =2x3 + l3x2 + 5x + 9 does its tangent line pass through the origin?, Ans. x=-3, -1,, , t, , ,i, , : ..
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Law of the Mean. Increasing and, Decreasing Functions, Relative Maximum and Minimum, A functionJis said to have a relative maximum at Xo ifJ(xo) ?J(x) for all x in some open interval containing, Xo (and for whichJ(x) is defmed). In other words, the value ofJ at Xo is greater than or equal to all values of, J at nearby points. Similarly, J is said to have a relative minimum at Xo if J(xo) ~ J(x) for all x in some open, interval containing Xo (and for whichJ(x) is defined). In other words, the value ofJ at Xo is less than or equal, to all values ofJat nearby points. By a relative extremum ofJwe mean either a relative maximum or a relative minimum off., Theorem 13.1: If/has a relative extremum at a pointxo at whichj'(xo) is defined, thenj'(xo) =O., Thus, if/is differentiable at a point at which it has a relative extremum, then the graph of/has a horizontal, tangent line at that point. In Fig. 13-1, there are horizontal tangent lines at the points A and B where / attains a relative, maximum value and a relative minimum value, respectively. See Problem 5 for a proof of Theorem 13.1., , __ __-+______________________, ~, , ~x, , Rg.13-1, , Theorem 13.2 (Rolle's Theorem): Let/be continuous on the closed interval [a, b] and differentiable on the open, interval (a, b). Assume that/(a) =/(b) =O. Thenj'(xO> =0 for at least one pointxo in (a, b)., , This means that, if the graph of a continuous function intersects the x axis at x =a and x =b, and the function is differentiable between a and b, then there is at least one point on the graph between a and b where, the tangent line is horizontal. See Fig. 13-2, where there is one such point. For a proof of Rolle's Theorem,, see Problem 6.
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CHAPTER 13 Law of the Mean, II, , Fig. 13-2, , Corollary 13.3 (Generalized Rolle's Theorem): Let g be continuous on the closed interval (a, b] and differentiable, on the open interval (a, b). Assume that g(a) =g(b). Then g'(xJ =0 for at least one point Xo in (a, b)., See, Fig. 13-3 for an example in which there is exactly one such point. Note that Corollary 13.3 follows, from Rolle's Theorem if we letf(x) = g(x) - g(a)., II, , x, , x., , a, , Fig. 13-3, , Theorem 13.4 (Law of the Mean)t: Let/be continuous on the closed interval [a, b] and differentiable on the open, interval (a, b). Then there is at least one point Xo in (a. b) for which, /(b) - /(a), b-a, , = f'(x, , ), 0, , See Fig. 13-4. For a proof, see Problem 7. Geometrically speaking, the conclusion says that there is some, point inside the interval where the slope f(:rJ of the tangent line is equal to the slope (J(b) - f(a»/(b - a), of the line Pl2 connecting the points (a'/(a» and (b,f(b» of the graph. At such a point, the tangent line is, parallel to Pl2, since their slopes are equal., II, , ~--------~----~--~%, o, a, %0, b, , Fig. 13-4, , The Law of the Mean is also called the Mean-Value Theorem for Derivatives.
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CHAPTER 13, , Law of the Mean, , Theorem 13.5 (Extended Law of the Mean): Assume that/(x) and g(x) are continuous on la, b], anddifferentiable, on (a, b). Assume also that g'(x) :F- for all x in (a. b). Then there exists at least one point Xo in (a, b) for which, , °, , I(b)- I(a) _ /'(xo ), g(b) - g(a) - g'(xo ), , For a proof. see Problem 13. Note that the Law of the Mean is the special case when g(x} =x., Theorem 13.6 (Higher-Order Law of the Mean): If I and its first" - I derivatives are continuous on [a, b] and, jI"'(X) exists on (a. b), then there is at least one Xo in (a. b) such that, , /,(a), , rea), , I(b) = l(a)+-I!-(b-a)+~(b-a)2 + ..., <n-I)( ), a (b-a),,-I, (n -I)!, , +I, , (I), , j<")(x ), , +__, 0 (b-a)", n!, , (For a proof. see Problem 14.), When b is replaced by x, formula (I) becomes, , I(x) = I(a) + I'I(~) (x - a) + I';\a) (x - a)2, <n-I)( ), +I, a (x-a)"-I, (n-I)!, , +, , I(nl(x ), , n!, , 0, , + ..., (2), , (x-a)n, , for some Xo between a and x., In the special case when a = O. formula (2) becomes, , I(x)= 1(0)+ 1'1\0) x+ 1';\0) x 2 + ..., (3), J!"-l) (0) n-I j<n)(xo), + (n, -1)', . x +--,-x", 11., , for some Xo between 0 and x., , Increasing and Decreasing Functions, A functionJis said to be increasing on an interval if u < v impliesJ(u) <J(v) for all u and v in the interval., Similarly,fis said to be decreasing on an interval if u < v impliesJ(u) > J(v) for all u and v in the interval., Theorem 13.7: (a) Iff is positive on an interval, thenlis increasing on that interval. (b), terval, thenlis decreasing on that interval., , Iff is negative on an in-, , For a proof, see Problem 9., , SOLVED PROBLEMS, , 1., , 2., , Find the value of Xo prescribed in Rolle's Theorem for I(x) =.x3 - 12x on the interval 0 ~ x ~213., Note that 1(0) = 1(213) =O. If j'(x) = 3x2 - 12 = O. then x =± 2. Then Xo = 2 is the prescribed value., , =, , Does Rolle's Theorem apply to the functions (a) I(x) = x: ix, and (b) I(x) = x:~ i x on the interval (0, 4)?, (a) I(x) =0 when x = 0 or x =4. Sincelhas a discontinuity at x =2, a point on [0,4]. the theorem does not, apply.
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CHAPTER 13 Law of the Mean, (b) /(x) = 0 whenx=O or x= 4./has a discontinuity atx= -2, a point not on [0, 4]. In a~dition., f(x) =(xl + 4x - 8)1(x+ 2'f exists everywhere except atx= -2. So. the theorem appbes and Xo = 2(..fj -1)., , the positive root of xl + 4x - 8 = O., 3., , Find the value of Xo prescribed by the law of the mean when/(x) = 3xl + 4x - 3 ~g a 4= I. b = 3., /(a) =/(1) = 4,f(b) =/(3) = 36.f(xJ =6xo +4, and b - a = 2. So. 6Xo +4=-r =16. Thenxo=2., , 4., , Find a valuexo prescribed by the extended law of the mean when/(x) = 3x+ 2 and 8(X) = xl + 1. on [1.4]., We have to find Xo so that, /(b)- /(a) _ /(4)- /(1), g(b)-g(a) - 8(4)-g(l), , Then, , 5., , Xo, , = 14-5 =l=.l$?=-..L, 17-2, , 5 8 (xo), , 2Xo, , =t·, , Prove Theorem 13.1: If/has a relative extremum at a pointxo at whichf(xJ is defined, thenf(xJ=O., Consider the case of a relative maximum. Since /has a relative maximum at xo' then, for sufficiently small, /(xo+ ax) - /(xo), IL\x!,f(xo+ L\x) </(xJ, and so/{xo+ L\x) - /~J < O. Thus, when tlx < 0,, tlx, > O., So,, , = lim /(xo+ fu) - /(Xu) ~ 0, , tlx, , Ar....O-, , " 0, /(xo+fu)/(xo) < 0. Hence,, When uX>, fu, , Sincef(xJ ~ 0 andf(xJ $ 0, it follows thatf(xJ= O., 6., , Prove Rolle"s Theorem (Theorem 13.2): If/is continuous on the closed interval [a. b) and differentiable on the, open interval (a. b). and if/(a) =/(b) =0, thenf(xo-> == 0 for some point Xo in (a, b)., If/(x) = 0 throughout [a. b], thenf(x) = 0 for ali x in (a, b). On the other hand. if/(x) is positive (negative), somewhere in (a, b), then, by the Extreme Value Theorem (Theorem 8.7),fhas a maximum (minimum) value at, some point Xo on [a. b). That maximum (minimum) value must be positive (negative), and, therefore. Xo lies on, (a, b), sincef(a) =/(b) =O. Hence,fhas a relative maximum (minimum) at xO' By Theorem 13.l,f(xO-> =o., , 7., , Prove the Law of the Mean (Theorem 13.4): Let/be continuous on the closed interval [a, b) and differentiable on, the open interval (a, b). Then there is at least one point Xo in (a, b) for which (f(b) - /(a»/(b - a) ={(xo)', Let F(x) = /(x)- /(a)- /(b)- /(a) (x- a)., b-a, Then F(a) =0 = F(b). So, Rolle's Theorem applies to F on [a, b1. Hence, for some Xo in (a, b), P(xJ = O., But F'(x)=/'(x) /(b)- /(a). Thus, /'(x), ., b-a, 0, , 8., , /(b)- /(a) =0., , b-a, , Show that, if g is increasing on an interval, then -g is decreasing on that interval., Assume u < v. Then g(u) < g(v). Hence, -g(u) > -g(v).
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CHAPTER 13, , 9., C, , ,, , •• -,, , Law of the Mea.n, , Prove Theorem 13.7: (a) Iff' is positive on an interval, thenfis increasing on that interval, (b) Iff' ·is negative on, an interval, thenfis decreasing on that interval., (a) Let a and b be any two points on the interval with a < b. By the Law of the Mean. (f(b) -f(a»/(b - a) =, f'(x o) for some point Xo in (a, b). Since Xo is in the interval,f'(xo) > O. Thus, (f(b) - f(a»/(b - a) > O. But,, a < b and, therefore. b - a > O. Hence,f(b) - f(a) > o. So,f(a) <feb)., (b) Let g =-.f. So, g' is positive on the interval. By part (a), g is increasing on the interval. So,fis decreasing on, the interval., ,/ ., , lO. Show thatf(x) =x5 + 20x - 6 is an increasing function for all values of x., f'(x) = 5x 4 + 20 > 0 for all x. Hence, by Theorem 13.7(a).fis increasing everywhere., , 11. Show thatf(x) = I - x3 - x 7 is a decreasing function for all values of x., f'(x) = -3,r - 7x!' < 0 for all x:t O. Hence, by Theorem 13.7(b).fis decrea~ing on any interval not, containing O. Note that, if x < O,f(x) > I = f(O), and, if x> 0,f(0) = I >f(x). So,fis decreasing for all real, numbers., 12. Show thatf(x) = 4x3 + X - 3 = 0 has exactly one real solution., f(O) = -3 andf(l) = 2. So, the intermediate value theorem tellsuli thatf(x) = 0 has a solution in (0, I). Since, f'(x) = 12x2 + I > O.fis an increasing function. Therefore, there cannot be two values of x for whichf(x) = O., , 13. Prove the Extended Law of the Mean (Theorem 13.5): Iff(x) and g(x) are continuous on [a, b), and differentiable, on (a, b), and g' (x) :t 0 for all x in (a, b), then there exists at least one point Xo in (a, b) for which, f(b)- f(a) _ /'(x o), g(b)-g(a) - g'(xo)', Suppose that g(b) = g(a). Then, by the generalized Rolle's Theorem, g'(x) = 0 for some x in (a, b),, contradicting our hypothesis. Hence, g(b) :t g(a)., , Let F(x) = f(x) - feb), , feb) - f(a) (g(x)- g(b»., g(b)- g(a), , F(a), , Then, , =0 = F(b), , and, , F'(x) =/,(x) - feb) - f(a) g'(x), g(b)- g(a), , .., . h f'( ) f(b)- f(a) '( )-0, By Rolle's Theorem, there eXists Xo III (a, b) for whlc, Xo - g(b)- g(a) g Xo •, 14. Prove the Higher-Order Law of the Mean (Theorem 13.6): I~ f and its first n - 1 derivatives are continuous on, [a, b] andj<n)(x) exists on (a, b), then there is at least one Xo in (a, b) such that, , f(b)=f(a)+, , f', , 1(~, , ), , f"( ), j</I-I) (a), (b-a)+T(b-a)2+ ... + (n-I)! (b-a)(n-I), , +, , jI")(x ), , n!, , 0, , (b-a)n, , (1), , Let a constant K be defined by, , (2), , and consider, , '( ), , 1"( ), , j<"-I) (x), , f, F(x) = f(x)- f(b)+-rr-(b-x)+-i-(b-x)2, + ... + (n-I)! (b-X)/I-I + K(b-x.,/I
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CHAPTER 13 Law of the Mean, Now F(a) =0 by (2), and F(b) =O. By Rolle's Theorem, there existsxo in (a, b) suchthat, F'(xo) =/'(xo)+[f"(xoXb-Xo)- /'(X o)]+[, , !"i~o) (b- xo)2 - !"(XoXb-Xo)1, , ~, ~~~\~', , +" .+[PO)(Xo) (b- x )11-1 _ j<"-I) (xo) (b - x )0-2]_ Kn(b- x )0-1, (n -1)!, , 0, , (n - 2)!, , .0, , ., , 0, , - P")(xo) (b - x )0-1 - Kn(b - x )"-1 =0, , - (n-l)!, , 0, , 0, , p")(x ), , Then K = - - I0_, and (2) becpmes (1)., , n., , 15. Iff(x) =0 for all x on (a. b), thenfis constant on (a, b)., Let u and v be any two points in (a, b) with u < v. By the Law of the Mean, there exists Xo in (u, v) for which, ·f(v)-/(u) /'(Xo). By hypothesis,f(xO> =O. Hence,f(v) - f(u) = 0, and, therefore,f(v) = f(u)., v-u, , 16. Iff(x), , =r - 4x+ 3 on [1, 3]. find a value prescribed by Rolle's Theorem., , Ans. xo=2, , 17. Find a value prescribed by the Law of the Mean. given:, (a) y =xlon [0, 6], (b) y=ar+bx+con [x"x2], , Ans., Ans., , Xo =2../3, xo =t(x, +x2 ), , ......, , 18. Iff(x) = g'(x) for all xin (a, b), prove that there exists a constant K such thatf(x) = g(x) + K for all x in (a. b)., (Hint: D.(f(x) - g(x» =0 in (a. b). By Problem 15, there is a constant K such thatf(x) - g(x) =Kin (0, b).), , 19. Find a value Xo precribed by the extended law of the mean when f(x) = xl + 2x - 3, g(x) = r, interval [0, 1J., , Ans., , - 4x +'6 on the, , t, , 20. Show that xl + px + q = 0 has: (a) one real root if p > 0, and (b) three real roots if 4pl + 27q2 < O., 21. Show that f(x) = ~!~ has neither a relative maximum nor a relative minimum. (Hint: Use Theorem 13.1.), I', , 22. Show thatf(x) =5xl + llx - 20 =0 has exactly one real solution., ~}t.?~~-~', , 23. (a) Where are the following functions (i)-(vii) increasing and where are they decreasing? Sketch the graphs., (b) (GC) Check your answers to (a) by means of a graphing calculator., (i) f(x)=3x+5, (ii) f(x) = -7x + 20, , Ans., Ans., , (iii)f(x)=r+6x-ll, , Ans., , Increasing everywhere, Decreasing everywhere, Decreasing on (-00, -3), increasing on (-3, +00), , :!.•:--~'..
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CHAPTER 13, (iv) f(x) = S + 8.x-xl, (v) f(x) =.J4 - x 2, (vi) f(x) = Ix - 21 + 3, (vii) f(x) =-::rL-:r4, x -, , Ans., Ans., Ans., Ans., , Law of the Mean, , Increasing on (-00, 4), decreasing on (4, +00), Increasing on (-2, 0), decreasing on (0, ,2), Decreasing on (-00, 2), increasing on (2. +00), Decreasing on (-00, -2), (-2. 2). (2, +00); never increasing, , 24. (GC) Use a graphing calculator to estimate the intervals on whichf(x) = r + 2x3 - 6x + I is increasing. and the, ,, intervals on which it is decreasing., , 25. For the following functions, determine whether Rolle's Theorem is applicable. If it is. find the prescribed values., (a), (b), (c), (d), , f(x) =x3'4 - 2 on [-3, 3], f(x) = Ixl - 41 on [0. 8], f(x) = ~ - 41 on [0. 1], 2, f(X)=x -3x -4 on[-1,4], x- S, , Ans., Ans., Ans., Ans., , No. Not differentiable at x = O., No. Not differentiable at x = 2., No·f(O) f(l), Yes. Xo =S - J6, , *
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Maximum and Minimum Values, Critical Numbers _., A number Xo in the domain of/such that either f(xJ =0 or f'(xJ is not defined is called a critical number, off, Recall (Theorem, that, if/has a relative extremum at Xo andf'(xJ is defined, thenf(xo) = 0 and,, therefore, x is a critical number off Observe, however, that the condition thatf(xJ =0 does not guarantee, thatfhas a ~lative extremum at xO' For example, if/(x) =x3, thenf(x) =3r, and therefore, 0 is a critical, number off; butfhas neither a relative maximum nor a relative minimum at O. (See Fig. 5-5)., , 13.n, , EXAMPLE 14.1:, , r..., , (a) Let/(x) = 7x2 - 3x + 5. ThenJ'(x) = 14x -_3. Setf'(x) = 0 and solve. The only critical number of / is, (b) Letflx) =~ - 2il +x+ t~'(x) = Jr - 4x+T:SOrvingftx}~O, we find that the critical numhers are I and t., -O:L-.I.CI/(X), numt..:r off, -______=.X2l3. TI1ellf'(x) = 1, 3 x-In =2, 3x";,. Sincef'(O) is not defined, 0 is tilt: only critical, -, , We shall find some conditions under which we can conclude that a function/has a relative maximum or, a relative minimum at a given critical number., , Second Derivative Test for Relative Extrema, Assume thatf'(xo) =0 and thatj"(xo) exists. Then:, (i) if f"(xJ < 0, then/has a relative maximum at xo;, (ii) iff"(xJ > 0, then/has a relative minimum atxo;, (iii) if f"(xJ =0, we do not know what is happening at xo', A proof is given in Problem 9. To see that (iii) holds, consider the three functionsf(x) =.t, g(x) =-.t, and, h(x) =x3. Sincef'(x) =: 4x\ g'(x) =-4x3, and h'(x) =3r, 0 is a critical number of all three functions. Since, rex) = 12.x2. g"(x) =-12r, and h"(x) =6.x, the second derivative of all three functions is 0 at O. However,, /has a relative minimum at 0, g has a relative maximum at 0, and h has neither a relative maximum nor a, relative minimum at O., EXAMPLE 14.2:, , (a) Consider the functionj{x) =7x2 - 3x + 5 of Example I(a). The only critical number was k Sincef"(x):= 14,, /"(17) =14,> O. So, the second derivative test tells us that/has a relative minimum at,t., (b) Considerthe functionj{x) =x3 - 2il +x + I of Example I(b). Note thatf"(x) = 6x - 4. At the critical numbers I, and t,f"(l) =2 > 0 and f"W =-2 < O. Hence /has a relative minimum at I and a relative maximum at t·, (c) In Example I(c),f(x) =x"l and f'(x) = ix-ln. The only critical number is 0, wheref' is not defined. Hence,f"(O), is not defined and the second derivative test is not applicable., , If the second derivative test is not usable or convenient, either because the second derivative is 0, or does, not exist, or is difficult to compute, then the following test can be applied. Recall thatf'(x) is the slope of, the tangent line to the graph of/ at x.
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CHAPTER 14, , Maximum and Minimum Values, , First. Derivative Test, Assume/'(xo}, , O., , Case, , {+, -}, If/' is ~ositive in an open interval immediately to the left of xo' and negative in an open interval immediately, to the nght of xu' then/has a relative maximum atxo' (See Fig. 14-1(a).}, , ", , Case {-, +}, , Iff' is ~egati~e in an open interval. immediately to the left of xO' and positive in an open interval immediately, to the nght of xO' then/has a relatIve minimum at xO' (See Fig. 14-1 (b).), ., , Cases {+, +} and {-, -}, If /' has the same sign in open intervals immediately to the left and to the right of x • then / has neither a, relative maximum nor a relative minimum at xO' (See Fig. 14-I(c. d).), 0, For a proof of the first derivative test, see Problem 8., , ---, , ----'-, , (D), , ~---./, , ~-(b), , (d), , (e), Fig. 14-1, , EXAMPLE 14.3: Consider the three functions/(x) = .0, g(x) = -x'. and hex) = i' discussed above. At their critical, number n, the second derivative test was not applicable because the second derivative was O. Let us try the first derivative test., (a) f'(x) = 4i'. To the left of 0, x < 0, and so,f'(x) < O. To the right of 0, x > 0, and so,f'(x) > O. Thus, we have the, case {-, +} and / must have a relati ve minimum at O., (b) g' (x) = -4x'. To the left of 0, x < 0, and so, g '(x) > O. To the right of O. x > 0, and so, g '(x) < O. Thus, we have the, case {+, -} and g must have a relative maximum at O., (c) h '(x) = 3x2. h '(x) > 0 on both sides of O. Thus, we have the case {+. +} and h has neither a relative maximum nor, a relative minimum at O. There is an inflection point at x = O., These results can be verified by looking at the graphs of the functions.
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••, , CHAPTER 14 Maximum and Minimum Values, , Absolute Maximum and Minimum, , ., , An absolute maximum of a function/on a set S occurs at Xo in S if/(x) S/(xJ for all x in S. An absolute, minimum of a fundion / on a set S Occurs at Xo in S if /(~) ~ /(xJ for all x in S., , Tabular Method for Rnding the Absolute Maximum and Minimum, Let/be continuous on [a, b] and differentiable on (a, b). By the Extreme Value Theorem, we know that/has, an absolute maximum and minimum on [a, b]. Here is a tabular method for determining what they are and, where they occur. (See Fig. 14-2.), x, , J(x), , Fig. 14-2, , First, find the critical numbers (if any) cI ' c2' ••• of/in (a, b). Second, list these numbers in a table, along, with the endpoints a and b of the interval. Third, calculate the value offfor all the numbers in the table., Then:, 1. The largest of these values is the absolute maximum of/on [a, b]., 2. The smallest of these values is the absolute minimum of/on [a, b]., EXAMPLE 14.4:, , Let us find the absolute maximum and minimum ofj{x) =r -:xl - x + 2 on [0, 2]., , f'(x) = 3:xl - 2x-l = (3x+ l)(x-l).Hence, the critical numbers arex=-t andx= 1. The only critical number in, , [0, 2] is 1. From the table in Fig. 14-3, we see that the maximum value off on [0, 2] is 4, which is attained at'the right, endpoint 2, and the minimum value is I, attained at 1., , x, , J(x), , I, , I, , o, , 2, , 2, , 4, , Fig. 14-3, , Let us see why the method works. By the Extreme Value Theorem,fachieves maximum and minimum, values on the closed interval [a, b]. If either of those values occurs at an endpoint, that value will appear in, the table and, since it is actualiy a maximum or minimum, it will show up as the largest or smallest value. If, the maximum 9r minimum is assumed at a point Xo inside the interval,fhas a relative maximum or minimum, at Xo and, therefore, by Theorem 13.1,j'(xo) =O. Thus, Xo will be a critical number and will be listed in the, table, so that the corresponding maximum or minimum value /(xo) will be the largest or smallest value in, the right-hand column., Assume thatfis a continuous function defined on an interval 1. The interval 1 can be a finite or, infinite interval. If/has a unique relative extremum within 1, then that relative extremum is also an absolute extremum, onl., , Theorem 14.1:, , To see why this is so, look at Fig. 14-4, 'vhere / is assumed to have a unique extremum, a relative maximum at c. Consider any other num,ber d in J. The graph moves downward on both sides of c. S,o, if f(d)
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CHAPTER 14 Maximum and Minimum Values., were greater than/(c), then, b~ ~e Extreme Val~e Theorem for the closed interval with endpoints c and, d, / would have an absolute nummum at some point u between c and d. (u could not be equal t, d), Then/would have a r~lative minimum at u, contradicting our hypothesis that/has a relative extre~u~o~nl~, at c. e can e~tend thIS argument to the case where/has a relative minimum at c by applying the result we, ., have Just obtamed to -f, , V:, , ----r----------r---------r--~--~x, , c, , II, , d, , Rg.14-4 ', , SOLVED PROBLEMS, , 1., , Locate the absolute maximum or minimum of the following functions on their domains:, (a) y =-x 2; (b) y = (x- 3)2; (c) y =-./25-4x 2 ; (d) y= .Jx-4., (a) y =-xl has an absolute maximum (namely, 0) when x =0, since y < 0 when x :t:. O. It has no relative, minimum, since its range is (-00, 0). The graph is a parabola opening downward, with vertex at (0, 0)., , .,, , ~.~:, , (b) y = (x -3)2 has an absolute minimum, 0, when x =3, since y > 0 when x:t:. 3. It has no absolute maximum,, since its range is (0, +00). The graph is a parabola opening upward, with vertex at (3, 0)., (c) y = -./25 - 4x 2 has 5 as its absolute maximum, when x = 0, since 25 - 4x2 < 25 when x:t:. O. It has 0 as its, absolute minimum, when x = t. The graph is the upper half of an ellipse., (d) y = .Jx - 4 has 0 as its absolute minimum when x = 4. It has no absolute maximum. Its graph is the upper, half of a parabola with vertex at (4, 0) and the x axis as its axis of symmetry., , J'.", , i:-·t/:'':.;t~, , ,., , .., 2., , t, , Let f(x) = t x 3 + x 2 - 6x + 8. Find: (a) the critical numbers off, (b) the points at whichfhas a relative maximum, or minimum; (c) the intervals on whichfis increasing or decreasing., (a) rex) =x 2 + X - 6 = (x + 3)(x - 2). Solvingr(x) =0 yields the critical numbers -3 and 2., (b) r(x) = 2x + 1. Thus,f"(-3) = -5 < 0 andr(2) =5. Hence, by the second derivative test,fhas a relative, maximum at x =-3, where f(-3) = -¥. By the second derivative test,fhas a relative minimum at x = 2,, where f(2) =, (c) Look atf'(x) = (x+ 3)(x- 2). When x > 2,f'(x) >0. For-3 <x< 2,f'(x) < O. For x<-3, f'(x) > O. Thus, by, Theorem l3.,7,fis increasing for x < -3 and 2 < x, and decreasing for -3 < x < 2., , t-, , A sketch of part of the graph offis shown in Fig. 14-5. Note thatfhas neither absolute maximum nor absolute, minimum.
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CHAPTER 14 Maximum and Minimum Values, (-3.4312~, , V, , f:i~,?, ~;~;", , ",I -, , Fig. 14-5, , 3., , Let/(x)·=.t + 2r- 3.r - 4x + 4. Find: (a) the critical numbers off, (b) the points at which/has a relative, extremum; (c) the intervals on which/is increasing or decreasing., (a) f'(x) = 4x3 + fu2 - 6x - 4. It is clear that x = I is a zero of f'(x). Dividingf'(x) by x - I yields 4.r + lOx + 4,, which factors into 2(2r + 5x+ 2) = 2(2%+ 1)(x+ 2). Thus,!,(x) = 2(x- 1)(2% + l)(x+ 2), and the critical, numbers are 1, -t, and -2., (b) /"(x) = 12r + 12% - 6 = 6(2r + 2x - 1). Using the second derivative test, we find: (i) at x =1,/"0) = 18> 0,, and there is a relative minimum: (ii) at x =-t, I"(-t) =-9 < 0, so that there is a relative maximum: (iii) at, x =-2,/"(-2) =18> O. so that there is a relative minimum., (c) f'(x) > 0 when x > 1,f'(x) < 0 when -t < x < 1, f'(x) > 0 when -2 < x < -t, andf'(x) < 0 when x < -2., Hence, lis increasing when x > 1 or -2 < x < -t, and decreasing when -t < x < I ?C x < -2., The graph is sketched in Fig. 14-6., , Fig. 14-6, , 4., , Examine /(x) =x ~ 2 for relative extrema, and find the intervals on which/is increasing or decreasing., / (x) = ~x - 2t l , so that f'(x) =-(x - 2t2 =- (x~ 2)1' Thus,f' is never 0, and the only number where f' is not, , defined is the number 2. which is not in the domain off. Hence,fhas no critical numbers. So,fhas no relative, extrema. Note thatf'(x) < 0 for x t:. 2. Hence,fis decreasing for x < 2 and for x> 2. There is a nonremovable, discontinuity at x = 2. The graph is shown in Fig. 14-7., , ,, , V.i"Fig. 14-7
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CHAPTER 14, 5., , Maximum and Minimum Values, , Locate the relative extrema of j(x) = 2 +xW and the intervals on whichfl'S l'ncreasl'n, d, ., , . -1 -113 _ 2, _", ., ' ., g or ecreas1I1g., f (.l) - 3 x - 3X ll3 • Then x - 0 IS a cntlcal number, smcef'(O) IS not defined (but 0 is in the domain off)., , ~ote th;tf~(x) ap?~a~hes as x appr~a:hes O. ~hen x < O,/'(x) is negative and, therefore./is decreasing. When, ~t: ~~.(x) IS posItive and, therefore./Is mcreas1l1g. The graph is sketched in Fig. 14-8.fhas an absolute minimum, 00, , ,, II, , (0.2), , o, Fig. 14-8, 6., , Use the second derivative test to examine the relative extrema of the following functions: (a) j(x) = x(12 - 2x)1;, 250, ., (b) f(x) = x 2 +-., x, (a) f'(x) =x(2)(12 - 2x)( -2) + (12 - 2x)2 =(12 - 2x)(I2 - 6x) = 12(x - 6)(x - 2). So, 6 and 2 are the critical, numbers.f"(x) 12(2x - 8) = 24(x - 4). So,/"(6) 48> 0, and 1"(2) = -48 < O. Hence./has a relative, minimum at x = 6 and a relative maximum at x =2., , =, , =, , 3, , (b), , I'(x) = 2x- 2;20 = 2( x, , ~}25). So, the only critical number is 5 (where xl -125 = O).r(x) = 2 + 500lxl., , Since 1"(5) = 6·> 0, fhas a relative minimum at x = 5., , 7., , Determine the relative extrema ofj(x), , =(x -, , 2)213., , I'(x) = 3(x!2)213' So. 2 is the only critical number. Sincel'(2) is not defined.f"(2) will be undefined., Hence, we shall try the first derivative test. For x < 2, f'(x) < 0, and, for x> 2, I'(x) > O. Thus, we have the case, {-, +} of the first derivative test, andfhas a relative minimum at x = 2., , 8., , Prove the first derivative test., Assumef'(xo> = O. Consider the case {+, -I: Iff' is positive in an open interval immediately to the left of xo', and negative in an open interval immediately to the right of xu' thenfhas a relative maximum at xO' To see this,, notice that, by Theorem 13.8, sinceI' is positive in an open interval immediately to the left of xo' fis increasing, in that interval. and, since f' is negative in an open interval immediately to the right of xo' fis decreasing in that, interval. Hence, fhas a relative maximu~at xo' The case {-, +} follows from the case {+, -} applied to - fIn, the case {+. +}, fwill be increasing'in an interval around xo' and, in the case {-, -}, fwill be decreasing in an, interval around xo' So, in both cases, fhas neither a relative maximum nor minimum at xo', , 9., , Prove the second derivative test: Iff(x) is differentiable on an open interval containing a critical value Xo off, and, f"(xo> exists andf"(xo> is positive (negative), thenfhas a relative minimum (maximum) at xo', Assume r(xO> > O. Then, by Theorem 13.8, I' is increa~ing at xo' Since I'(xo) =0, this implies that I' is, negative nearby and to the left of XO' and I' is positive nearby and to the right of xo' Thus, we have the case {-, +}, of the nrst derivative test and, therefore, fhas a relative minimum at xo' In the opposite situation, where f,,(xo) < 0,, the result we have just prqved is applicable to the function g(x) =- j(x). Then g has a relative minimum at Xo and,, therefore, fhas a relative maximum at xo', , 10. Among those positive real numbers II and II whose sum is 50, find that choice of u and v that makes their product, P as large as possible., P =11(50 - u). Here, u is any positive number less than 50. But we also car allow u to be 0 or 50, since, in, those cases, P = 0, which will certainly not be the largest possible value. So, P is a continuous function u(50 - u).
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CI:IAPTER 14 Maximum and Minimum Values, defined on [0; 50]. P =SOu - u2 is also differentiable everywhere. and dPldu =50 - 2u. Setting dPldu =0 yields, a unique critical number u =25. By the tabular method (Fig. 14-9). we see that the maximum value of Pis 625., when u = 25 (and, therefore, v = 50 - u = 25)., , u, , P, , 25, , 625, , o, , 0, , 50, , 0, , Fig. 14-9, 11. Divide the number 120 into two parts such that the product P of one part and the square of the other is a maximum., Let x be one part and 120 -x the other part.1)ten P =(120 - x)r and 0 ~x ~ 120. Since dPldx= 3x(80 - x),, the critical numbers are 0 and 80. Using the tabular method, we find P(Ol =0, P(80) = 256,000 and P(120) = O., So, the maximum value occurs when x =80, and the required parts are 80 and 40., 12. A sheet of paper for a poster is to be 18 ft2 in area. The margins at the top and bottom are to be 9 inches, and the, margins at the sides 6,inches. What should be the dimensions of the sheet to maximize the printed area?, Let x be one dimension. measured in feet. Then 18/x is the other dimension. (See Fig. 14-10.) The only, restriction o~ x is thatx> O. The printed area in square feet is A =(x-l>(lli _1 ), and dA = ~ _~., ", x 2, dx xl 2, 3/4, , 1/2, , 18/20, , Rg.14-10, Solving dAldx =0 yields the critical number x =2$. Since d2AIdx2 =-361xl is negative when x =2J3, the, second derivative test tells us that A has a relative maximum at x = 2$ . Since 2$ is the only critical number in, the interval (0. +00), Theorem 14.1 tells us that A has an absolute maximum at x=2.J3. Thus, one side is 2$ ft, and the other side is 18/(2$) =3$ ft., 13. At 9 A.M., ship 8 is 65 miles due east of another ship A. Ship 8 is then sailing due west at 10 milh,'l~.d A is, sailing due south at 15 milh. If they continue on their respective courses, when will they be nearest one another,, and how near? (See Fig. 14-11.), Let Ao and 8 0 be the positions of the ships at 9 A.M., and A, and 8 , their positions t hours later. The distance, covered in (hours by A is 1St miles; by B, lOt miles. The distance D between the ships is determined by D2 =, (15t)2 + (65 ~ IOt)2. Then, , ,, , ,, , 2D, , £If, =2(l5tX15) +2(65 -lOt)(-IO); he~ce, dfr,; 325tD650 ', , Fig. 14-11, , c,!
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CHAPTER 14, , Maximum and MinimumVa/ues, I, , ~?lving dD/~ = 0, yields the critical number t = 2. Since D > 0 and 325t - 650 is negative to the left of 2 and ., positIve to the nght of 2, the case (-, +) Qf the first derivative test tells us that t = 2 yields a relative minimum for D, Since t = 2 is the only critical number, Theorem 14.1 implies that there is an absolute minimum at t = 2., ., Setting t = 2 in D2 = (15t)2 + (65 - IOt)2 yields D = 15.jIT miles. Hence, the ships are nearest at II A.M., at, which time they are 15.jIT miles apart., , 14. A cylindrical container with circular base is to hold 64 in 3. Find its dimensions so that the amoun; (surface area), of metal required is a minimum when the container is (a) an open can and (b) a closed can., Let rand h be, respectively, the radius of the base and the height in inches, A the amount of metal, and V the, volume of the container., (a) Here V = 1trh = 64, and A = 21trh + 1tr. To express A as a function of one variable, we solve for h in the first, relation (because it is easier) and substitute in the second, obtaining, , and the critical number is r=4/~. Then h = 641rcr2 =4/~. Thus, r=h = 4/~ in., Now dA/dr > 0 to the right of the critical number, and dA/dr <: 0 to the left of the critical number. So, by, the first derivative test, we have a relative minimum. Since there is no other critical number, that relative, minimum is an absolute minimum., (b) Here again V= nrh = 64, but A = 21trh + 2nr = 2nr(64/1tr) + 2nr = 128/r+ 21tr. Hence, •, , dA _ 128 + 4 _ 4(rcr 3 - 32), dr --7 rcrr2, and the critical number is r = 2~41rc . Then h = 641rcr 2 = 4~41rc . Thus, h = 2r = 4V41rc in. That we have, found an absolute minimum can be shown as in part (a)., , 15. The total cost of producing x radio sets per day is $( tx 2 + 35x + 25), and the price per set at which they may be, sold is $(50-tx)., (a) What should be the daily output to obtain a maximum total profit?, (b) Show that the cost of producing a set is a relative minimum at that output., (a) The profit on the sale of x sets per day is P = x(50 - t x) - (t x 2 + 35x + 25). Then dPldx = 15 - 3x12; solving, dPldx = 0 gives the critical number x = 10., Since d 2 Pldx 2 -t < 0, the second derivative test shows that we have found a relative maximum. Since, x = 10 is the only critical number, the relative maximum is an absolute maximum. Thus, the daily output that, maximizes profit is 10 sets per day., ., ., t x 2 + 35x, 1, 25, dC I 25 ., . dC'ldx 0, (b) The costofproducmg a set IS C=, x + 25 =4x+35+x·Then, dX=4-Xl,solvmg I =, gives the critical number x 10., Since dlCldx 2 = 501x3 > 0 when x = 10, we have found a relative minimum. Since there is only one, critical number, this must be an absolute minimum., , =, , =, , 16. The cost of fuel to run a locomotive is proportional to the square of the speed and $25 per hour for a speed of, 25 miles per hour. Other costs amount to $100 per hour, regardless of the speed. Find the speed that minimizes, the cost per mile., Let v be the required speed, and let C be the total cost per mile. The fuel cost per hour is kif, where k is a, constant to be determined. When v= 25 mi/h, kif = 625k = 25; hence, k = 1125., , c-, , costin$1h _ v 2 /25+ 100 = ~+ 100, - speed in milh v, 25, v
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CHAPTER 14 Maximum and Minimum Values, , Then, Since v> 0, the only relevant critical number is v =50. Since tPCldJ =2001'; > 0 wh~~ v =50, the ~econd, derivative test tells us that C has a relative minimum at v 50. Since v 50 is the only cntlcal number III (0, f<x»,, Theorem 14.1 tells us that C has an absolute minimum at v = 50. Thus, the most economical speed is 50 milh., , =, , =, , 17. A lIlan in a rowboat at P in Fig. 1+ 12,5 miles from the nearest point A on a straight shore, wishes to reach a point, B, 6 mi les from A along the shore, in the shortest time. Where should he land if he can row 2 milh and walk 4 milh?, p, , C, , B, 6-%, , Fig. 14-12, Let C be the point between A and B at which the man lands, and let AC = x. The distance rowed is, PC =../25 +Xl , and the rowing time required is, , '1, , =di~ce = ~ . The distance walked is CB =6 - x,, , and the walking time required is '2 =(6 - x)/4. Hence, the total time required is, , °, , The critical number obtained from setting 2x-../25 + x 2 = is x~ t../3 - 2.89. Thus, he should ,land at a, point about 2.89 miles from A toward B. (How do we know that this point yields the shortest time?), 18. A given rectangular area is to be fenced off in a field that lies along a straight river. If no fencing is needed along, the river, show that the least amount of fencing will be required when the length of the field is twice its width., Let x be the length of the field, and y its width. The area of the field is A = xy. The fencing required is, F= x+2y, anddFldx= I +2 dyldx. When dFldx=O, dyldx=-1., Also, dAldx = = y + x dyldx. Then y - t x =0, and x = 2y as required., To see that F has been minimized, note that dyldx = - rIA and, , °, , Now use the second derivative test and the uniqueness of the critical number., 19. Find the dimensions of the right circular cone of minimum volume V that can be circumscribed about a sphere of, radi us 8 indites., Let x be the radius of the base of the cone, and y + 8 the height of the cone. (See Fig. 14-13.) From the similar, right triangles ABC and AED, we have, , x_, , y +8, , 8- ~y2-64, , and .therefore, , 2 _, , X -, , 64(y+8)2, 2, y-64, t;...·, , ,;, ~, , Also,, , ,So,, , dV, , 4y, , 64n(y+8Xy-24), = 3(y- 8)2, , f ~, , ~,, , .
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i, I, I, , I, I, , I, , -I-, , CHAPTER 14 Maximum and Minimum Values, , I, , A, , I, I, , I, I, , "1-"::::""--1, , C, , Fig. 14-13, The relevant critical number is y = 24. Then the height of the cone is y + 8 = 32 inches, and the radius of the base, is 8.[i inches. (How do we know that the volume has been minimized?), , 20. Find the dimensions of the rectangle of maximum area A that can be inscribed in the portion of the parabola, yz = 4px intercepted by the line x = a., Let PBB'P' in Fig, 14-14 be the rectangle, and (x, y) the coordi~ates of p, Then, , Rg.14-14, , A=2 y(a-X)=2 y (a-, , X~ )=2ay - ~:, , and, , dA, 3/, -=2a--, , dy, , 2p, , Solving dAld)' = 0 yields the critical number y = ~4apI3. The dimensions of the rectangle are 2y = t ~3ap and, a - x = a - (y2/4p) =2a/3,, Since d 2Aldyl = -3ylp < 0, the second derivative test an~ t'"le uniqueness of the critical number ensure that we, have found the maximum area,, , 21. Find the height of the right circular cylinder of maximum volume V that can be inscribed in a sphere of radius R., (See Fig. 14-15.), , Fig. 14-15
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CHAPTER 14 Maximum and Minimum Values, Let r be the radius of the base, and 2h the height of the cylinder. From the geometry, V =21tilh and ,:z + Ir =R2., Then, , . dh, dV =21C (- II, r3 +2~., h) When V .is a malumum,, ., dV -,, - 0 fro m whiCh..2, 2hl, From the last relatIOn,, dr =-Iir ' so dr, dr, r = ., Then R2 =,:z + h2=2h2 +h2, so that h =RI.J3 and the height of the cylinder is 2h =2R/.J3. The secondderivative test can be used to verify that we have found a maximum value of V., 2~. A wall of a building is to be braced by a beam that must pass over a parallel wall 10ft high and 8 ft from the, building. Find the length L of the shortest beam that can be used. ., See Fig. 14-16. Let be the distance from the foot of the beam to the foot of the parallel wall, and let y be the, , x, , distance (in feet) from the ground to'the top of the beam. Then L =J(x +8)2 + y2 •, , Rg.14-16, "1, . I, Y x+8, 1O(x+8), x . Hence,, Al so, from sImI ar tnang es, 10 = -x- and, therefore, Y, , =, , The relevant critical nu~ber is x = 2~. The length of the shortest beam is, , 2~8 ~4~1O,OOO +100 =({IiOO +4)3/2 ft, The first derivative test and Theorem 14.\ guarantee that we really have found the shortest length., , "II\"~' ~ l"I\\I~;l\"1, , ", , \, , I", , \, , 1:'I\',~~1-/':, , ", ', , ':'!',~, ~, , ,, , -., , ., , ,'", , ':, , ,,(, , .. ', , I, , ~'I, , ~,'.., , 23. Examine each of the following for relative maximum and minimum values, using the first deri-.:ative test., (a) !(x)=r+2x-3, (b) !(x)=3+2x-r, , (c) !(x)=.~+2x2-4x-8, , (d) !(x), , =r - 6x2 +9x- 8, , (e) f(x), , =(2 -, , x)', , Ans.. x = -\ yields relative minimum-4, Ans. x = I yields relative maximum 4, Ans. x, yields relative minimum -1ft: x=-2 yields relative, maximum 0, Ans. x = 1 yields relative maximum -4: x =3 yields relative, minimum -8 ', Ans. neither relative maximum nor relative minimum, , =t, , ., , ';: .',, , .
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CHAPTER 14 Maximum and Minimum Values, , (g) I(x) = (x - 4)4(X + 3)3, (h) I(x) = x 3 + 48/x, , Ans., , x = 0 yields relative.maximum 16; x = ±2 yields relative, minimum 0, , Ails., , x = 0 yields relative maximum 6912; x = 4 yields relative, minimum 0; x = -3 yields neither, x = -2 yields relative maximum -32; x = 2 yields relative, minimum 32, , Ails., , x = -2 yields relative maximum 0; x = 0 yields-relative, minimum -</4; x = I yields neither, , Ails., , 24. Examine the functions of Problem 23 (a - f) for relative maximum and minimum values, using the second, derivative test., ., , 25. Show that y = (a l -, , X)2, , + (a 2 -, , X)2, , + ... +(all -, , X)2, , has an absolute minimum when x = a l + a z + ... + a •., n, , 26. Examine the following for absolute maximum and minimum values on the given interval., (a) y=-ron-2<x<2, , (b) y = (x - 3)2 on 0 ~ x ~ 4, , Ans., Ans., , maximum(=O)atx=O, maximum (= 9) at x = 0; minimum (= 0) at x = 3, , (c) y = .,/25 - 4x 2 on -2 ~ x ~ 2, , AIlS., , maximum (= 5) at x = 0; minimum (= 3) at x = ±2, , (d) y=.Jx-4 on4~x~29, , AilS., , maximum (= 5) at x = 29; minimum (= 0) at x = 4, , 27. The sum of two positive numbers is 20. Find the numbers if: (a) their product is a maximum; (b) the sum of their, squares is a minimum: (c) the product of the square of one and the cube of the other is a maximum., Ans., , (a) 10, \0; (b) 10,10; (c) 8,12, , 28. The product of two positive numbers is 16. Find the numbers if: (a) their sum is least; (b) the sum of one and the, square of the other is least., Ans., , (a) 4, 4; (b) 8, 2, , 29. An open rectangular box with square ends is to be built to hold 6400 ft3 at a cost of $0.75/ft2 for the base and, $0.25/ft2 for the sides. Find the most economical dimensions., Ans., , 20x20x 16, , 30. A wall 8 ft high is 3ift from a house. Find the shortest ladder that will reach from the ground to the house when, leaning over the wall., AilS., , 15ift, , 31. A company olTers the following schedule of charges: $30 per thousand for orders of 50,000 or less, with the, charge decreased by 37ft for each thousand above 50,000. Find the order size that makes the company's, receipts a maximum., AilS., , 65,000, , 32. Find an equation of the line through the point (3, 4) that cuts from the first quadrant a triangle of minimum area., AilS., , 4x+.)y-24=0
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CHAPTER 14 Maximum and Minimum Values, 33. At what point in the first quadrant on the parabola y =4 - 1!- does the tangent line, together with the coordinate, axes, determine a triangle of minimum area?, Ans. (2.[513,8/3), 34. Find the minimum distance from the point (4, 2) to the parabola y2 =8x., Ans., , 2./2, , 35•. (a) Examine 2r - 4xy + 3y2 - 8x + 8y - 1 = o for maximum and minimum values of y. (b) (GC) Check your, answer to (a) on a graphing calculator., Ans. (a) Maximum at (5, 3); (b) minimum at (-1, -3), 36. (GC) Find the absolute maximum and minimum off(x) =~ - 31!- - 8x - 3 on [-1, 2) to three-decimal-place, accuracy., Ans. Maximum 1.191 atx= -{).866; minimum -14.786 atx = 1.338, 37. An elec~c current, when flowing ina circular coil of radius r, exerts a force F = (x 2, , :'2, , !, , )St2, , on a small magnet, , located at a distance x a~ve the center of the coil. Show that F is greatest when x = r ., 38. The work done ,by a voltaic cell of constant electromotive force E and constant internal resistance r in passing, a steady current through an external resistance R is proportional to WR/(r + R)2. Show that the work done is, greatest when R =r., , y2, , 2, , 39. A tangent line is drawn to the ellipse ~5 + 16 = 1 so that the part intercepted by the coordinate axes is a, minimum. Show that its length is 9., 2, , 2, , i25, , 40. A rectangle is inscribed in the ellipse 4~ +, = 1 with its sides parallel to the axes of the ellipse. Find the, dimensions of the rectangle of (a) maximum area and (b) maximum perimeter that can be so inscribed., Ans. (a) 20./2 x 15./2; (b) 32 X 18, 41. Find the radius R of the right circula.r cone of maximum volume that can be inscribed in a sphere of radius r., (Recall that the volume of a right circular cone lif radius R and height h is t TrR 2h.), , Ans., , R=tr./2, , 42. A right circular cylinder is inscribed in a right circular cone of radius r. Find the radius R of the cylinder if:, (a) its volume is a maximum; (b) its lateral area is a maximum. (Recall that the volume of a right circular cylinder, of radius R and height h is 1fR2h, and its lateral area is 27tRh.), AilS., , (a) R=tr; (b) R=tr, , 43. Show that a conical tent of given volume will require the least amount of material when its height h is./2 times, the radius r of the base. [Note first that the surface area A =7t(r +h2).], 44. Show that the equilateral triangle of altitude 3r is the isosceles triangle of least area circumscribing a circle of, radius r., , :, , ., , ,"
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CHAPTER 14 Maximum and Minimum Values, , 45. Detennine the dimensions of the right circular cylinder of maximum lateral surface area that can be inscribed in a, sphere of radius 8., , Ans., , h=2r=8J2, , 46. Investigate the possibility of inscribing a right circular cylinder of maximum total area (including its top and, bottom) in a right circular cone of radius r and height h., , Ans., , If II > 2r. radius of cylinder, , =-t( h/~ r)·, , "
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Curve S/cetching., Concavity. Symmetry, Concavity, From an intuitive standpoint, an arc of a curve is said to be concave upward if it has the shape of a cup (see, . Fig. 15-1(a» and is said to be concave downward if it has the shape of a cap (see Fig. 15-1(b». Note that a, more precise definition is available. An arc is concave upward if, for each xO' the arc lies above the tangent, line at Xo in some open interval around xO' Similarly, an arc is concave downward if, for each x O' the arc lies, below the tangent line at Xo in some open interval around xO' Most curves are combinations of concave upward and concave downward. For example, in Fig. 15-1 (c), the curve is concave downward from A to Band, from C to D, but concave upward from B to C., ~, , D, , A, (a), , ,(b), , Concave upward, , Concave downward, , (c), , Fig. 15-1, , The second derivative of/tells us about the concavity of the graph off, Theorem 15.1:, , (a) Iff"(x) > 0 for x in (a, b), then the graph of/is concave upward for a <x < b., (b) Iff"(x) < 0 for x in (a, b), then the graph of/is concave downward for a < x < b., , For the proof, see Problem 17., EXAMPLE 15.:11:, , (a) Let/(x) =x2• Thenj'(x) =2x,f"(x) =2. Since /,,(x) > 0 for all x, the graph of/is always concave upward. This, was to be expected, since the graph is a parabola that opens upward., (b) Let /(x) =y =.)1- x2 • Then f =1 - xl, xl + f =1. So, the graph is the upper half of the unit circle with, the center at the origin. By implicit differentiation, we obtain x + yy' = 0 and then I + yy" + (y')2 = O. So,, y" = -[1 + (y')2Jly. Since y> 0 (except at x = I). y" < O. Hence, the graph is always concave downward., whicJt is what we would expect., , -Iga
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CHAPTER 15, , Curve Sketching. Concavity. Symmetry, , Points of Inflection, A point of inflection on a curve y -/(x) is a point at which the concavity changes, that is, the curve is concave, upward on one side and concave downward on the other side of the point. So, if y" exists in an open interval, containing' x o' then y" < 0 on one side of Xo and y" > 0 on the other side of xo' Therefore, if y" is continuous, at x o' then y = 0 at xo' Thus, we have:, II, , If the graph of/has an inflection point at Xo andr exists in an open interval containing x andf" is, continuous at xo' then/"(xO> = O., 0, , Theorem 15.2:, , EXAMPLE 15.2:, (a) Let lex) = xl. Then/'(x) = 3r..r(x) = 6.x. Thus,j"(x) < 0 for x < O. andf"(x) > 0 for x < O. Hence. the graph of, / has an inflection point at x = O. (See Fig. 5-5.) Note thatr(O} = 0, as predicted by Theorem 15.2., (b) Let lex) = .x-4. Thenf'(x) = 4xl. and rex) = 12x2. Solvingf"(x) = 0 yields x = O. However. the graph of/ does not, have an inflection point at x = O. It is concave upward everywhere. This example shows that/" (xo) = 0 does not, necessarily imply that there is an intlection point at xo', (c) Let/(x)=tx3 +tx2 -6x+8. Solvingf"(x) = 2x+ 1 =0, we find that the graph has an inflection point at (-t, W)., Note that this is actually an inflection point. since/'(x) < 0 for x <, and rex) > 0 for x > -to See Fig. 14-5., , -t, , Vertical Asymptotes, A vertical line x =Xo such thatJ(x) approaches + 00 or -00 as x approaches Xo either from the left or the right, is called a vertical asymptote of the graph off. If J(x) has the form g(x)lh(x), where g and h are continuous, functions, then the graph of/has a vertical asymptote x =Xo for every Xo such that h(xO> =0 (and g(xo) 0)., , '*, , Horizontal Asymptotes, A horizontal line y= Yo is called a horizontal asymptote of the graph ofJif either lim J(x) = Yo or lim J(x) = Yo', x-+Thus, a horizontal asymptote is approached by the graph as one moves further and further to the left or further, and further to the right., .r~+-, , EXAMPLE 15.3:, (a) Let /(x) = 1. Then the graph of/has a vertical asymptote at x = 0, which is approached both from the left and the, x, right. The line y = 0 (that is, the x axis) is a horizontal asymptote both on the left and the right See Fig. 5-21., (b) Let lex) = -L . Then x = 2 is a vertical asymptote of the graph off, which is approached both from the left and, x- 2, the right. The line y = 0 is a horizontal asymptote, which is approached both on the left and the right. See Fig. 14-7., (c) Let /(x) = (x -1)(';+ 3)' Then the graph of/has vertical asymptotes at x = I and x = -3. The line y = 0 is a horizontal asymptote, which is approached both on the left and the right., (d) Let lex) =, , ~ ~ ~. Then the graph of/has a vertical asymptote at x = 3, which is approached both from the left, , and the right. The line y = I is a horizontal asymptote, which is approached both on the left and the right., , Symmetry, We say that two points P and Q are symmetric with respect to a lille I if I is the perpendicular bisector of the line, segment connecting P and Q. [See Fig. 15-2(a).], We say that two pointe; P and Q are symmetric with respect to a point B if B is the midpoint of the segment connecting P and Q., A curve is said to be symmetric with respect to a line I (respectively, point B) if, for any point P on the curve, there, is another point Q on the curve such that P and Q are symmetric with respect to I (respectively, B). [See Fig. l5-2(b, c).1, , If a curve is symmetric with respect to a line I, then I is called an axis oj symmetry of I. For example, any, line throufh the center of a circle is an axis of symmetry of that circle.
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CHAPTER 15 Curve Sketching. Concavity. Symmetry, , Q, , ,,, , \, , \, , \, , ,,, , ,, ,,, , \, , \, , \, , ,,, , \, \, , \, , ,Q, , \, , plt\, , ,, , \, , ", , ,, , \, , \, \, \, , \, , I, , \, , \',, , /, , I, , I, , ./, , p, , (c), , (b), , (II), , Fig. 15-2, , .-------- ----------., (xJ'), , (-xJ'), , .---- ., , (x.-y), , (II), , (b), , "", , .--, , (-x,-y), , (c), , Fig. 15-3, , Points (x, y) and (-x, y) are symmetri.c with respect to the y axis, and points (x, y) and (x, -y) are symmetric with respect to the x axis. Points (x, y) and (-x, -y) are symmetric with respect to the origin. See, Fig. 15-3(a-c)., Consider the graph of an equation F(x, y) =O. Then:, (i) The graph is symmetric with respect to the y axis if and only if F(x, y) =0 implies F(-x, y) =O., (ii) The graph is symmetric with respect to the x axis if and only if F(x, y) = 0 implies F(x. -y) = O., (iii) The graph is symmetric with respect to the origin if and only if F(x. y) =0 implies F(-x. -y) =O., EXAMPLE 15.4, (a) The parabola y =xl is symmetric with respect to the y axis., (b) The parabohl x = f is symmetric with respect to the x axis., (c) A circle xl + y2 = an ellipse ~~ + ~:, y axis. the x axis. and the origin., , r., , =1. and a hyperbola ~~ - ~: =1are symmetric with respect to the, , EXAMPLE 15.5: A point P(a. b) is symmetric to the point Q(b. a) with respect to the line y =x. To see this. note first, that the line PQ'has slope -I. Since the line y =x has slope I, the line PQ is perpendicular 10 the line y =x. In addition,, the midpoint of the segment connecting P and Qis ( a; b •b ; a ). which is on the line y =x. Hence. the line y =x is the, perpendicular bisector of that segment., '
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CHAPTER 15, , Curve Sketching. Concavity. Symmetry, , Inverse Functions and Symmetry, We say that two curves C I and C2 are symmetric to each other with respect to a line I if, for any point P on, one of the curves, the point Q that is symrrietric to P with respect to I is on the other curve. (In other words,, if we "reflect" one of the curves in the line I, the result is the other curve.), Theorem 15.3: Consider any one-to-one functionJand its inverse functionf-I. Then the graphs of fandJ-I are symmetric to each other with respect to the line y = x., ", , To see this, assume that (a, b) is on the graph off ThenJ(a) =b. Hence,f-I(b) =a. that is, (b, a) is on the, graph of J- I • By Example 5, (a, b) and (b, a) are symmetric with respect to the line y =x., EXAMPLE 15.6:, , t.\:., , tx, , (a) [ffix) = 2t. then f-I(x) =, Hence. the lines y =2x and)' = are symmetric with respect to the line y =x., (b) Let C 1 be the parabola that is the graph of the equation y = x 2, and let C2 be the parabola that is the graph of the, equation x =y2. Then C 1 and C 2 are sYlllmetric with respect to the line y =x. since the equation x =l results from, the equation y =xl by interchanging x and y., , Even and Odd Functions, A functionJis said to be evell if. for any x in its domain, -x is also in its domain andJ(-x) =J(x).fis said to, be an odd function if, for any x in its domain, -x is also in its domain andJ(-x) =-f(x)., EXAMPLE 15.7: Any polynomial. such as 3x" - Sx4 + 7, that involves only even powers of x determines an even, function. Any polynomial. such as 5x9 + 2r - 4xl + 3x, that involves only odd powers of x determines an odd function., , A function J is even if and only if its graph is symmetric with respect to the y axis. In fact. assume J is, even and (x. y) is on its graph. Then y =f(x). Hence, y =f( -x) and, therefore, (-x, y) is on the graph. Thus., the graph is symmetric with respect to the y axis. The converse is left as Problem 16(a)., A functionJis odd if and only if its graph is symmetric with respect to the origin. In fact, assumeJis odd, and (x, y) is on its graph. Then y =J(x). Hence, -y =J(-x) and, therefore, (-x, -y) is on the graph. Thus, the, graph is symmetric with respect to the origin. The converse is left as Problem 16(b)., , Hints for Sketching the Graph G of y, I., 2., , 3., 4., , 5., 6., , =/(X), , Calculate y'. and. if convenient, y"., Use y' to find any critical numbers (where y' =O. or y' is undefined and y is defined). Determine whether, these critical numbers yield relative maxima or minima by using the second derivative test or the first, derivative test., Use y' to determine the intervals on which y is increasing (when y' > 0) or decreasing (when y' < 0)., Use y" to determine where G is concave upward (when}!" > 0) or concave downward (when, y" < 0). Check points where y" =0 to determine whether they are inflection points (if y" > 0 on one side, and y" < 0 on the other side of the point)., Look for vertical asymptotes. If y =g(x)/h(x), there is a vertical asymptote x =Xo if h(x(J) = 0 and g(xo) O., Look for horizontal asymptotes. If lim J(x) = Yo' then y = Yo is a horizontal asymptote on the right. If, x-++lim J(x) =Yo. then y =Yo is a horizontal asymptote on the left., , *, , x-+-, , Determine the behavior "at infinity." If lim J(x) =+00 (respectively, -00 ), then the curve moves upward, x-++(respectively. downward) without bound to the right. Similarly, if lim J(x) =+00 (respectively, -00), then, the curve moves upward (respectively, downward) without bounCtto the left., 8. Find the y intercepts (where the curve cuts the y axis, that is, where x = 0) and the x intercepts (where, the curve cuts the x axis. that is, where y = 0)., 9. Indicate any comer points, where y' approaches one value from the left and another value from the right., An example is the origin on the graph of y =Ixl., to. Indicate any cusps, where y' approaches +00 from both sides or y' approaches -00 from both sides. An, example is the origin on the graph of y = v'iXi., 11. Find any oblique asymptotes y =nu+ b such that lim(J(x)-(nu+b) =0 or lim(j(x)-(nu+b) =O., An oblique asymptote is an asymptote that is neither~erticaI nor horizontal. H 7.
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CHAPTER 15 Curve Sketching. Concavity. Symmetry, , SOLVED PROBLEMS, , 1., , Examine y =3x4 - lOi' - 12.x2 + 12x - 7 for concavity and points of inflection., We have, y' = 12i' - 30x2 - 24x+ 12, )''' = 36x 2 - 60x - 24 = 12(3x + 1)(x - 2), , Set y" = 0 an~ solve to obtain the possible points of inflection x = - t and 2. Then:, When x < -to, When-t<x<2., Whenx>2., , y" =+, and the arc is concave upward., y" =-, and the arc is concave downward., y" =+, and the arc is concave upward., , The points ofinflection are (-t,-W) and (2, -63), since y" changes sign at x ='-t and x =2. See Fig. 15-4., 'JJ, , Rg.15-4, , 2., , Examine y =x4 - 6x + 2 for concavity and points of inflection, and sketch the graph., We have y" =12.x2. By Theorem 15.2, the possible point of inflection is at x =O. On the intervals x < 0 and, x> 0, y" is positive, and the arcs on both sides of x =0 are concave upward. The point (0, 2) is not a point of, inflection. Setting y' =4i' - 6 =0, we find the critical number x =if3fi. At this point, y" = 12.x2 > 0 and we have, a relative minimum by the second derivative test. Since there is only one critical number, there is an absolute, minimum at this point (where x - 1.45 and y - - 3.15). See Fig. 15-5., J".', , Rg.15-5, , 3., , Examine y = 3x + (x + 2)315 for concavity and points of inflection, and sketch the graph,, 3 + 5(x +32)215 an d", . 0 f'III fl ection, . .IS at x =-., 2 When x> - 2", ., y' =, y = 25(x-6, + 2)1" . The POSSI'bl e POlllt, ,y IS, negative and the arc is concave downward. When x< - 2, y" is positive and the arc is concave upward. Hence,, there is an inflection point at x =-2, .vhere y =-6. (See Fig. 15-6.) Since y' > 0 (except at x =-2), y is an, increasing function, and there are no relative extrema.
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CHAPTER 15, , Curve Sketching. Concavity. Symmetry, , Rg.15-6, , 4., , Iff"(xo) = 0 andf''' (xo) 7; 0, then there is an inflection point at xo', , Since f"'(xJ = O,f"'(xo} is either positive or negative. Hence.!" is either increasing or decreasing at xo' Since, f"(xJ = 0, f" ha.c; opposite signs to the left and right of xo' So, the curve will have opposite concavity on the two, sides of xo' and there is an inflection point at xo', , 5., , Find equations of the tangent lines at the points of inflection of y =f(x) =.x4 - 6x3 + 12x2 - 8x., A point of inflection exists at x = Xu whenf"(xo} = 0 andf'''(x()} 7; O. Here,, f'(x) = 4x 3 -I8x 2 + 24x - 8, /,,(x) =12x 2, , -, , 36x+ 24 = I2(x -I)(x- 2), , f'''(x) = 24x - 36 = 12(2x - 3), , The possible points of inflection are at x = 1 and 2. Since 1'''(1) 7; 0 and /,,'(2) 7; 0, the points (I, -1) and (2. 0), are points of inflection., At (1, -1), the slope of the tangent line is m =1'(1) = 2, and its equation is, y=y,=m(x-x,), , or, , y+I=2(x'-1), , or, , y=2x-3, , At (2. 0). the slope is 1'(2) = 0, an'd the equation of the tangent line is y = O., , 6., , Sketch the graph of)' =f(x) = 2tl- 5x2 + 4x - 7., f'(x) = 6x2 - lOx + 4,f"(x) = 12x - 10, and /"'(x) = 12. Now, 12x - 10 > 0 when x> i and I2x - 10 < 0 when, x < i. Hence, the graph offis concave upward when x > t, and concave downward whenx < i. Thus, there is an, inflection point at x = i. Since f"(x) = 2(3x2 - 5x + 2) = 2(3x - 2)(x - I), the critical numbers are x = t and x = I., Since /"(t) = -2 < 0 andf"(I) = 2, there is a relative maximum at x = t (where)' = -W - -5.96) and a relative, minimum at x = I (where y = - 6). See Fig. 15-7., ., , 7, , Sketch the graph of y = f(x) = ~2', , 2, , x-, , 2, , 2, , 4 Th' I, - 4 +4, x - 4 +--2, 4, 2, 8(, y= x x2 =--2, en y = - ( x-4 2)2 an d", y =x-2)3', xx- =x+ +--2', x-, , Solving y' = 0, we obtain the critical numbers x = 4 and x = O. Sincef"(4) = 1 > 0 and/,,{O) = -1 < 0, there, is a relative minimum at x = 4 (where y = 8) and a relative maximum at x = 0 (where y = 0). Since y" is never O., there are no inflection points. The line x = 2 is a vertical asymptote. The line y = x + 2 is an oblique asymptote on, both sides, since. on the CJrve, y - (x + 2) = ~2 -t 0 as x -t ±oo. See Fig. 15-8., , x-
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CHAPTER 15 Curve Sketching,. Concavity. Symmetry, , Rg.15-7, , 8, 1, , ,, , ,, ,,, ,,, , ,,, , ,, ,,, , ,,, , ,, , ,, ,,, , ,, ,,, , ,, ,,, , ,,, , ,,, , I,', I,, , ~(, , , 1, , ,, , ,,, ,, ,,, , ,, ,,, ,,, , ,, , ,", , ":, , 1, , ~],i;:,i, , 11, , ; --.': r·, , 4, , ,,, , ,,, , Rg.l5-8, , ,., , 8., , Sketch the graph of g(x) =2xl - 9.\.2 + 36., g'{x) =6x2 - 18x = 6x(x - 3) and g"(x) = 12x - 18 6(2x - 3). So, the critical numbers are x= 0 (where y =36), and x = 3 (where y = 9). Since gl/(O) =-18 < 0 and g"(3) = 18> 0, there is a relative maximum at x = and a, relative minimum atx =3. Setting gl/(x) = yields x =t, where there is an inflection point, since g"(x) =6(2x - 3), changes sign at x =t., g(x) -+ +00 as x -+ +00, and g(x) -+ - 00 as x -+ - 00. Since g(-I) =29 and g(-2) =-16, the intermediate, value theorem implies that there is a zero Xo of g between -1 and -2. (A graphing calculator shows Xo - -1.70.), That is the only zero because g is increasing up to the point (0, 36), decreasing from (0, 36) to (3, 9), and then, increasing from (3, 9). See Fig. 15-9., , =, , °, , °
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Curve Sketching. Concavity. Symmetry, , CHAPTER 15, , ;:~ ~!', , ', , ",,',1-, , Fig. 15-9, , 9., , x2, , Sketch the graph of )' = (x _ 2)(x _ 6) ., There are vertical asymptotes at x = 2 and x = 6 ., .'_ 2x(x-2)(x-6)-2x 2(x-4) _, 8x(3-x), ) (X_2)2(X-6)Z, - (X_2)2(X_W, , ¥" = (x - 2)2(X- 6)2(24 -16x) - 8x(3- x)(2)(x- 2)(x- 6)(2x - 8), (x - 2)4(X - 6)4, , ., , _ 8(2x 3 - 9x 2 + 36), - (x-2)j(x-W, , The critical numbers are x =0 (where y =0) and x = 3 (where y =-3). Calculation shows that y"(O) > 0 and, y"(3) < O. Hence, there is a relative minimum at x = 0 and a relative maximum at x = 3. Since y -+ I when x-+, ±oo, the line y = I is a horizontal asymptote on both the left and the right. Setting y" =0 yields g(x) =2.x3- 9x2 +, 36 O. By Problem 8. we see that we have a unique inflection point Xo - -1.70 (where y - 0.10). See Fig. 15-10., , =, , r, , ,~.,, , •, , I, I, I, I, , I, I, , I, I, I, I, , I, I, I, , I, , I, I, I, I, I, I, I, I, , ~---------1---------------------------I, I, I, , 3, , 6, , Fig. 15·10, , 10. Sketch the graph of y(r - 4) = x4., , -t-. Then y = ± ~. The curve exists only for r > 4, that is, for x > 2 or x < -2, plus the isolated, x - 4, x -4, 4, , y2 =, , point (0, 0)., , 2, , 2
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••, , CHAPTER 15 Curve Sketching. Concavity. Symmetry, , The curve is symmetric with respect to both coordinate axes and the origin. So, from now on, we shall, consider only the first quadrant Then, , The only critical number is 2fi (where y = 4). Since y" > 0, the graph is concave upward and there is a, relative minimum at (2fi, 4). The lines x = 2 and x = -2 are vertical asymptotes. The rest of the graph in the other, quadrants is obtained by reflection in the axes and origin. Note that there is also an oblique asymptote y =x, since, y'- - r = x4/(r - 4) - r = 4/(r - 4) ~ 0 as x ~ :too. By symmetry, y = -x is also an asymptote. See Fig. 15-11., , 'V/', , 11, , 1, , I, , /, , /', , I /, I /, 1/, , /(, , /, , /, , I /0',, , 1/, , V, , n/, , / I, , 1, , 2, , 1, , 'I, , ', , /1, , // r, , I', , /, , /, , ~", 1, , 1 ", , 1, 1, 1, I, , I", , I~', I, ', I',, , Rg.15-11, , 11 •. Examine the functions of Problem 23(a-j) of Chapter 14., Ans., , (a), (b), (c), (d), (e), , No inflection point, concave upward everywhere, No inflection point, concave downward everywhere, Inflection point atx=-t, concave upward for x> -t, concave downward for x<-t, Inflection point at x = 2, concave upward for x > 2, concave downward for x < 2, Inflection point atx = 2, concave downward for x> 2, concave upward for x < 2, , (f) Inflection point at x =±, , 2f ', , concave upward for x >, , _M<x<2.J3., 3, , 3, , 2f, , and x < -, , 2f, concave downward for·, ., , 12. Prove: Ifj{x) = a,il + br + ex + d has two critical numbers, their average is the abscissa at the point of inflection., If there is just one critical number, it is the abscissa at the point of inflection., , 13. Discuss and ~ketch the graphs of the following equations:, (a) xy, , =(r - 9)2, , Ans. Symmetric with respect to the origin, vertical asymptote x = 0, relative minimum at (3, 0), relative, maximum at (-3, 0), no inflection points, concave upward for x> 0, X4, , (b) y= l-x 2, , Ans. Symmetric with respect to the y axis, vertical asymptotes x = ± I, relative minimum at (0, 0), relative., maxima at (±fi, - 4), no inflection points, concave upward for Lli < 2
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CHAPTER 15 Curve Sketching. Concavity. Symmetry, , Ans., , Vertical asymptote x =0, relative minimum at (1,3), inflection point at (-if}., 0), concave upward for, x < -if}. and x > 0, ,, , Ails., , Relative maximum at (4, 2if4), relative minimum at (0, 0), where there is a "cusp," inflection point, (6,0), concave upward for x> 6. oblique asymptote y = -x + 2 to the left and the right, , (e), , Ails., , ", , y=l+L, x-I, , =, , Vertical asymptote x I. relative maximum at (0, I). relative minimum at (2. 5). concave upward for x >, I and downward for x < 1, no inflection points. increasing for x < 0 and x > 2, decreasing for 0 < x < I and, I < x < 2, oblique asymptote y x + 2, , =, , (f) ), _ _, x_, , - x2 + I, , Ans., , Symmetric with respect to the origin, relative maximum at (1, t), relative minimum at (-I. -t ),, increasing on -I < x < I. concave upward on -J3 < x < 0 and x> J3. concave downward on x < -J3, and 0 < x < J3. inflection points at x =0 and x =±J3, horizontal asymptote y =0 on both sides, , (g) y = x..r;-::T, , Ans., , Defined for x ~ I. increasing. concave upward for x> t. and downward for x < t. infl'ection point, (t, ~J3), , (h) y= x~2-x, , Ans., (i), , AilS., , Relative maximum at x = t, increasing for x < t, concave downward for x < 3, inflection point (3, -3), , v=x+2 1, x, , =, , Vertical asymptote x 0, horizontal asymptote y = 0 on both sides, relative minimum (-2, -t), increasing, for -2 < x < 0, concave upward for -3 < x < 0 and x > 0" inflection point at (-3, -t), y ~ + 00 as x ~ 0, , 14. Show that any function F(x) that is defined for all x may be expressed in one and only one way as the sum of an, even and an odd function. [Hint: Let E(x)=t(F(x)+F(-x».j, 15. Find an equation of the new curve C 1 that is obtained when the graph of the curve C with the equation xl - 3xy +, 21 = I is reflected in: (a) the x axis: (b) the)' axis: (c) the origin., ', , Ans., , (a) x2 + 3xy + 21 = I; (b) same as (a); (c) C itself, , 16. (a) If the graph off is symmetric with respect to the y axis, show thatfis even. (b) If the graph of fis symmetric, with respect to the origin, then show thatfis odd. [Him: For (a). if x is in the domain off, (x,J(x» is on the graph, and, therefore. (-x,f(x» is on the graph. Thus,f(-x) = j{x).j, 17. Prove Theorem 15.1: (a) If f"(x) > 0 for x in (a, h). then the graph of fis concave upward for a < x < h. (b) If, f"(x) < 0 for x in (a. b), then the graph off is concave downward, for a < x < h.
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CHAPTER 15 Curve Sketching, Concavity. Symmetry, , -,-, , [For (a),let Xo belong to (a, b). Sincej"(xo> > 0, /' is increasing in some open interval I containing xO', Assume x in I and x > xO' By the law of the mean,f(x) - /(xO> =/,(x')(x - xo> for some x' with Xo < x' < x. Since, /' is increasing,/,(xO> </,(X). Then/(x) =/,(x')(x - xo> +/(xo> >/,(xO>(x - xo> +/(xO>. But y =/,(xO><x - xo> +, /(xo) is an equation of the tangent line at xo' A similar argument works when x < xo' Thus, the curve lies above the, tangent line and, therefore, is concave upward.], , 3r, , 18. (GC) Use a graphing calcolator to draw the graph of/(x) =,;1- +4x - 2. Show analytically that/is, increasing and that there is an inflection point at (-I, 3). Use the calculator to draw the graph ofI-I and y = x, and, observe that the graphs of/andf' are symmetric with respect to y = x., 2, , 19. (Ge) Try to sketch the graph of y = 1 ~ 2 5 by standard methQds and then use a graphing calculator for, x - x +, additional information (such as the l~ation of any vertical asymptotes)., , ,
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Review of Trigonometry, Angle Measure, The traditional unit for measuring angles is the degree. 360 degrees make up a complete rotation. However,, it turns out that a different unit, the radian, is more useful in calculus. Consider a circle of radius 1 and, with center at a point C. (See Fig. 16-1.) Let CA and CB be two radii for which the arc AB of the circle has, length 1. Then one radian is taken to he the measure of the central angle ACB., , /, , I, , /, , '", , .",..----........, , ....., , A, , I, , I, I, I, \, , ,, , C, , ,8, /, , ,, , \, , I, , "", , ./, ........., , _--",.., ....-, , //, , Rg.16-1, , If u is the number of d~rees in angle ACB, then the ratio of u to 3600 is equal to the ratio of AB to the, circumference 21t. Since AB =1, ul360 = II21t and, therefore. u = 180/1t. So,, , 1 radian = 1!O degrees, , (I), , If 1t is approximated as 3.14, then I radian is about 57.3 degrees. Multiplying equation (1) by 7tlI80. we, obtain:, I degree = I~O radians, , (2), , The table in Fig. 16-2 shows the radian measure of some important degree measures., Now take any circle of radius r with center O. (See Fig. 16-3.) Let LDOE contain (J radians and let s be, the length of arc DE. The ratio of (J to the number 21t of radians in a complete rotation is equal to the ratio, of s to the entire circumference 21tr. So, (J/21t = s/21tr. Therefore., s= r(J, , (3)
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-I-, , CHAPTER 16 Review of Trigonometry, , Degrees, , Radians, 1t, , 30 1, , 6, 1t, , 45, , 4, 1t, , 60, , -, , 3, , 1t, , 90, , 2, , 180, , 1t, , 270, , -, , 360, , 21t, , E, , 31t, 2, , Rg.16-2, , Rg.16-3, , Directed Angles, Ifan angle is thought of as being generated by a rotation, then its measure will be counted as positive if, the rotation is counterclockwise and negative if the rotation is clockwise. See, for example, angles ,of 1t12, radians and -1tI2 radians in Fig. 16-4. We shall also allow angles of more than one complete rotation. For, example, Fig. 16-5 shows a counterclockwise angle generated by a complete rotation plus another quarter of, a complete rotation, yielding an angle of 21t + 1tI2 =51t12 radians, and an angle of 31t radians generated by, one and a half turns in the counterclockwise direction., , +11 turns, , - !': radians, 2, , 'TT2 ra d'lans, , (-90"), , (90"), , + I~ turns, , or, , or, , + 5"., 2" rad'lans, , +3". radians, , Rg.16-4, , Rg.16-5, , Sine and Cosine Functions, . Consider a coordinate system with origin at 0 and point A at (1, 0). Rotate the arrow OA through an angle, of 9 degrees to a new position OB. Then (see Fig. 16-6):, 1. cos 9 is defined to be the x coordinate of the point B., 2. sin 9 is defined to be the y coordinate of the point B., y, 8(C05 8. sin 8), , o, , Fig. 16-6, , A(l.O), , :l
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Review of Trigonometry, , CHAPTER 16, EXAMPLE 16.1:, , (a) If e:::: rt/2, the final position B is (0, 1). Hence, cos(rt/2) = 0 and sin(rt/2) = 1., (b) If, 1t, then B is (-1. 0). So, cos 1t = -1 and sin 1t = O., (c) If, 31t12, then B is (0, -1). So, cos(37t/2) = 0 and sin(37t/2) =-1., (d) If = 0 or = 21t, then B is (1,0). Hence, cos 0 = 1 and sin 0 = 0, and cos 21t = 1 and sin 21t = O., , e=, e=, e, , e, , Let us see that our definitions coincide with the traditional definitions in the case of an acute angle of, a triangle. Let 0 be an acute angle of a right triangle DEF and let !l.OBG be a similar triangle with hypotenuse I. (See Fig. 16-7.) Since the triangles are similar. BG I BO = EF I ED, that is, BG =ble. and, likewise,, OG = ale Hence, cos 0 =ale and sin 0 = ble. This is the same as the traditional definitions:, adjacent side, cos O=-;-"--hypotenuse, , . 0 opposite side, sm =~'----, hypotenuse, , and, , y, , 8(cos 8, sin 8), , opposite:, side. b, , ....., , ,, , ", ,, \, , ,, , ,,, , \, , D, , adjace:nt, side:. a, , o, , F, , ~, , G, , A(l,O), , x, , c, , Rg.16-7, TABLE, , 16.1, 9, cos 9, , sin 9, , 0, , I, , 0, , nl6, , 30, , nl4, , 45, , -F3/2, fil2, , 112, fil2, , nl3, , 60, , 112, , -F312, , n/2, , 90, , 0, , 1, , n, , 180, , -1, , 0, , 3n/2, , 270, , 0, , -\, , RADIANS, , 0, , DEGREES, , We now can use the values obtained from high-school trigonometry. [See Problem 22(a-e).] Table 16-1, lists the most useful values., Let us first collect some simple consequences of the definitions., (16.1), , (16.2), , cos(O+ 27t) =cos Oand sin(O+21t)=sin 0, This holds because an additional complete rotation of 21t radians brings us back to the same point., cos(-O) = cos oand sin(-O) = -sin o(see Fig. 16-8)
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CHAPTER 16 Review of Trigonometry, sin28 + cos2 8 = 1 [~accordance with tradiational notation, sin2 8 and cos2 8 stand for (sin 8)2 and, (cos 8)2.], ,, In Fig. 16-6, 1= OB =.Jcos28 + sin 28 by Problem 1 of Chapter 2. (16.3) implies sin2 8 = 1 cos2 8 and cos2 8 = 1 - sin2 8., , (16.3), , y, , y, , -- -, , .........., , ....., , ,, \, , \, , (+,+), , (-,+), , \, , \, \, \, , ,, , , c,, \\, , ', , \, , .r, , --------------4-----------~x, , I, , I, , I, , I, , I, , /, , ,I, , (-,-), , I, , (+,-), , /, -;', , "", , Rg.16-8, , (16.4), (16.5), , Rg.16-9, , In the four quadrants, the sine and cosine have the signs shown in Fig. 16-9., , For any point A(x, y) different from the origin 0, let r be its distance from the origin, and let 8 be, the radian measure of the angle from the positive x axis to the arrow ~A. (See Fig. 16-10.) The pair, (r, 9) are called polar coordinates of A. Then x =r cos 8 and y =r sin 8. (See Problem 8.), , For the derivation of more complicated formulas, we shall depend on the following result., (16.6) cos(u - v) = cos u cos v+ sin u sin v, , For the proof, see Problem II., cos(u + v) = cos u cos v- sin u sin v, Replace vby - vin (16.6) and use (16.2)., (16.8), cos(1tI2 - v) =sin v and sin(1tI2 - v) = cos v, Replace u by rtl2 in (16.6) and use cos(rtl2) = 0 and sin(1tI2) =1. This yields cos(1t12 - v) = sin v. In, this formula, replace v by (1tI2 - v) to obtain cos v =sin(1tI2 - v)., (16.9) sin(u + v) = sin 1/ cos v+ cos u sin v, By (16.6) and (16.8),, (16.7), , sin(1I + v) = cos[(nl2) - (u + v)] =cos[(n/2 -II) - v], , =cos(nl2 - u)cosv+ sin(n/2 -1I)sinv =sinucosv+ cos II sinv, (16.10) sin(u - v) = sin u cos v - cos u sin v, , Replace vby - vin (16.9) and use (16.2)., (16.11) cos 2u ='cos 2 Ii - sin 2 11 =2 cos 2 U - 1 = I - 2 sin 2 u, Replace v by 1/ in (16.7) to get cos 2u-= cos 2 II - sin2 Ii. Use sin2 1/ = I - cos2 1/ and cos2 1/ =1 - sin 2, u to obtain the other two forms., (16.12) sin 2u = 2 sin II cos U, Replace vby Ii in (16.9)., (16.13), , COS2(~)= 1+~osu, cosu = cos(, , 2'~) = 2COS ~ )-1, 2, , (
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CHAPTER 16 Review of Trigonometry, , SOLVED PROBLEMS, , 1., , Translate the following degree measures into radian measures: (a) 54°; (b) 120°., (a), , 54°=54(1~0, , radians) = l~n radians, , /, , (b) 120° = 120( I~O radians) = 2; radians, 2., , Translate the following radian measures into degree measures: (a) 2; radians; (b) 5; radians; (c) 2 radians., (a), , 2n radians = 2n (180 degrees) = 720, , 5, , 5, , n, , (b) 5; radians = 5;(I!0 degrees)=1500, (c) 2 radians=2(I!0, 3., , degrees)=(3~r, , (a) In a circle of radius r = 3 'Centimeters, what arc length s along the circumference corresponds to a central, angle eof 1f16 radians?, (b) In a circle of radius 4 feet, what central angle corresponds to an arc length of 8 feet?, We know that s = rO, where 0 is measured in radians., (a), , S=3(~)=~centimeters, , (b) O=(f)=!=2radians, 4., , What rotations between 0 and 21t radians have the same effect as the rotations with the following measures?, (a) lin radians; (b) 405°; (c) - j radians; (d) -Sit radians., (a) lin = 2n + 3: So, the equivalent rotation is 3: radians., (b) 405° = (360 + 4St. So, the equivalent rotation is 4So., (c) - j + 2n = Sf. So, the equivalent rotation is Sf radians., (d) -Sit + 61t = It. So, the equivalent rotation is It radians., , 5., , Find sin eif eis an acute angle such that cosO = t., 2, By (16.3), t + sin 2 0= 1. So, sin 2 0= i and, therefore, sinO = it. Since Ois acute, sin 0 is positive. So,, sine= t., , 6., , Show that sin (It - 6) = sin 0 and cos (It - 6) = -cos e., By (16.1 0), sin (It - 6) = sin It cos e- cos It sin 0 = (0) cos e- (-l)sin e= sin, cos It cos 0 + sin It sin e= (-I) cos e+ (0) sin e= -cos O., , 7., , e. By (16.6), cos (It -, , Calculate. the following values: (a) sin 27r13; (b) sin 7; ; (c) cos 91t; (d) sin 420°; (e) cos 37r14;, (g) sin 7rl8;~) sin 19°., (a) By Problem 6, sin 2f =Sin(n-j)=sinj=, (b) By (16.1), sin 7; = sin(2tr+j}= sinj=, , f, , 4'-, , (c) By (16.1), cos 91t = cos (7t + 87t) = cos It =-1, (d) By (16.1), sin 3900 = sin(30 + 360)° = sin 30° =, (e) By Problem 6, cos 3: =cos(n-~)= -cos~, , t, , =-1, , 6) =, , (D cos !tI12;, , \, , ...•, , ,
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CHAPTER 16 Review of Trigonometry, , (f), , =12 .J2, + .J3 .J2 = .J2 +.J6, 2, 2 2, 4, , cos lL =cos( 1C _1C) =cos 1C cosK + sin Ksin K, 12, 3 4, 3, 4, 3, 4, , . 2(1C)_I-COS(1C/4) _1-(.,fi12), , (g) By (16.14). Sin g -, , 2, , -, , 2, , 2-.J2, , ., , 1C _, , - 4 - ' Hence. Slng-±, , ., , ~2-.J2., 2, .Smce, , 1C, , 1C, , 0<g<2', , .1C, . posItive, .. an dh, &, .1C.[i:Jf, Sin, g IS, . t erelore,, Sln, 2, g =, (h) 19° cannot be expressed in terms of more fami liar angles (such as 30°. 45°, or 60°) in such.Q way that any of, our formulas are applicable. We must then use the sine table in Appendix A, which gives 0.3256; this is an, approximation correct to four decimal places., , 8., , Prove the result of (16.5): If (r, 0) are polar coordinates of (x, y), then x = r cos e and y =r sin e., Let D be the foot of the perpendicular from A(x, y) to the x axis (see Fig. 16-12). Let Fbe the point on the ray, OA at a unit distance from the origin. Then F = (cos sin 0). If E is the foot of the perpendicular from F to the x, axis. then OE = cos and FE = sin Since MDO is similar to MEO (by the AA criterion). we have:, , e, , e,, , e, , OD, OE, , = OA = AD, OF, , FE', , that is _x_=!.=+, • cose 1 sine, , Hence, x = r cos e and y = r sin e. When A(x, y) is in one of the. other quadrants, the proof can be reduced to, the case where A is in the first quadrant. The case when A is on the x axis or the y axis ,is very easy., y, A(,t, y), , o, , E, , D, , Fig. 16-12, , 9., , Find rectangular coordinates of the point with !ldar coordinates r = 3., By (16.5), x = rcose =3cos-K = 31. and y = rsine =, , J3)., r =r + y2 = 1 + 3 = 4. Then r=, , 10. Find polar coordinates ofthe point (l,, By (16.5)., , e= 7tl6., , 3sin~ = 3( -!) = l, , J3, , 2. So, cose= f=-!, and sine= ~= 2' Thus, e=, , ~., , 11. (a) Prove the law of cosines (16.15(a». (b) Prove the law of sines (l6.l5(b»., (a) See Fig. 16-11. Take a coordinate system with C as origin and B on the positive x axis. Then B has, coordinates (a, 0). Let (x, y) be the coordinates of A. By (16.5), x = b cos and y = b sin By the distance, formula (2.1),, , e, , e.
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-., , CHAPtER 16 Review of Trigonometry, , Therefore., , =b2COS 2 0 - 2abcosO+a 2+b2sin 2 0 (Algebra: (u - 11)2 =u2- 2uII+ 112), , =a2 + b2(cOs 20+ sin 2 0)- 2ab cos 0, , (b) See Fig. 16-13. Let D be the foot of the perpendicular from A to side BC. and let h = AD. Then, sinB=ADI AB=hlc. Thus. h = c sinB and.so the area of MBC = t(base xheight)=tah =tacsinB., (Verify that this also holds when LB is obtuse.) Similarly, tbcsinA = the area of MBC = tabsinC., Hence. tacsinB =tbcsinA =tabsinC. Dividing by tabc, we obtain the law of sines., A, , .~, D, , B, , a, , C, , Fig. 16-13, 12. Prove the identity (16.6): cos(u - v) = cos u cos v+ sin u sin v., Consider the case where 0 ~ v< u < v+ 1t. (See Fig. 16-14.) By the law of cosines., BC2 = P + F - 2(1)(I)cos(LBOC). Thus,, (cosu - cos 11)2 + (sinu - sin 11)2 = 2- 2cos(u - 11), , cos 2 U - 2cosucoslI+ cos 2 11+ sin 2 U - 2sinusinll+sin 2 v = 2 - 2cos(u - 11), (COS, , 2, , 1l+ sin 2 u)+ (cos1 11+ sin 2 11)- 2(cosucosv+ sin II sin 11) = 2- 2coS(u - 11), 1+ 1- 2(cosucosv+ sinusinv) = 2 - 2cos(u - v), , cosucosv+sinusinll = cos(u - 11), All the other cases can be derived from the case above., y, C(COI U., , I, I, , /, , ,in u), ~--, , I, , /, , I, , \, \, \, , \, , .\, \., , "' ....., , -Fig. 16-14
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Differentiation of, Trigonometric Functions, Continuity of cos x and sin x, It is clear that cos x and sin x are contmuous functions, that is, for any 0,, , limcos(O+h)=cosO, h~O, , limsin(O+h)=sinO, , and, , h~O, , To see this, observe that, in Fig. 17-1, as h approaches 0, point C approaches point B. Hence, the x coordinate of C (namely, cos (0+ h» approaches the x coordinate of B (namely, cos 0), and the y coordinate of C, (namely, sin (0 + h» approaches the y coordinate of B (namely, sin 0)., , ,,"'--- c. . ., , (cos (9 + h), sin (8 + II», , /, , I, , I, , I, I, , \, \, , I, I, , \, \, , I, , ,, "- ....., , ....., , _..... .,,-, , I, /, ,/, , Rg.17-1, , To find the derivative of sin x and cos x, we shall need the following limits., (17.1), (17.2), , lim sinO = 1, 9~0 0, I·1m 1- cosO, Ll, =0, 6-+0, , (], , For a proof of (17.1), see Problem 1. From (17.1), (17.2) is derived as follows:, , 1- cosO, , o, , 1- cosO 1+cosO 1- cos 2 0, = 0 . 1+ cos 0 = 0(1 +cos 0), , =, , sin 2 0 = sinO. sinO, 0(1 + cosO), 0 1+cosO'
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CHAPTER 17, , Differentiation of Trigonometric Functions, , Hence,, , r, , J!!J, , l-cos8, 8, , (17.3), , D/sinx)=cosx, , (17.4), , D.(cosx) =-sinx, , lim sin8 .lim sin8 ~1. sinO =1.~=1.0=0, 9-+0, 8 8-+0 1+ cos8, 1+ cosO, 1+ 1, , For a proof of (17.3), see Problem 2. From (17.3) we can derive (17.4), with the help of the chain rule, and (16.8), as follows:, Dx(cosx) = DA (sin(, , ~ -x)) =cos( ~ -x ).(-1) =-sinx, , Graph of sin x, Since sin (x + 2lr) =sin x, we need only construct the graph for 0 ~ x ~ 2lr. Setting Dx(sin x) =cos x =0 and, noting that cos x =0 in [0, 2lr] when and only when x =lr/2 or x =3lr/2, we find the critical numbers lr/2 and, 3lrl2. Since D;(sinx) = Dx(cosx) = -sinx, and -sin(lr/2) = -1 < 0 and -sin(3nt2) = 1 > 0, the second derivative test implies that there is a relative maximum at (lr/2, 1) and a relative minimum at (3lrl2, -1). Since, DA(sin x) = cos x is positive in the first and fourth quadrants, sin x is increasing for 0 < x < lr/2 and for, 3lr/2 < x < 2lr. Since D;(sinx) =-sinx is positive in the third and fourth quadrants, the graph is concave, upward for lr < X < 2lr. Thus, there will be an inflection point at (lr, 0), as well as at (0, 0) and (2lr, 0). Part, of the graph is shown in Fig. 17-2., , Graph of cosx, Note that sin (lrl2 + x) = sin (lr/2) cos x + cos (lr/2) sin x = 1· cos x + 0 . sin x = cos x. Thus, the graph of, cos x can be drawn by moving the graph of sin x by lrl2 units to the left, as shown in Fig. 17-3., y, , ,. ,. ,., 31t, 2, , -It, , ,,, , "",--- ..... ,, , "", , ,., , ,, , ......, , x, , 'y= sin x, , Fig. 17-2, y, , ...., , "", , "", , \, \, , , ...., , ,,, , ...., , "-, , '-, , y =cosx, , Fig. 17-3, , ,. ,..,.._ ..... - ......... ...., , ...., , ,,, , ,, , \
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CHAPTER 17, , Differe~tiation, , of Trigonometric Functions, , The graphs of y = sin x and y = cos x consist of repeated waves, with each wave extending over an interval, of length 2n. The length (period) and height (amplitude) of the waves can be changed by multiplying the, argument and the value, respectively, by constants., EXAMPLE 17.1: Let y = cos 3x. The graph is sketched in Fig. 17-4. Because cos 3(x + 2rc/3) = cos (3x + 2rc) =, cos 3x, the function is of period p = 21C13. Hence, the length of each wave is 21C13. The number of waves over an interval, of length 2rc (corresponding to one complete rotation of the ray determining the angle x) is 3. This number is called, the frequency f of cos 3x. In general, pf = (length of each wave) x (number of waves in an interval of 21C) = 21C. Hence,, f= 2rclp., y, , x, , Fig. 17-4, , For any b > 0, the functions sin bx and cos bx have frequency b and period 2mb., , y = 2 sin x. The graph of this function (see Fig. 17-5) is obtained from that of y = sin x by doubling, the y values. The period and frequency are the same as those of)' = sin x, that is, p = 21C and f = 1. The amplitude, the, maximum height of each wave, is 2., EXAMPLE 17.2:, , y, , x, , Fig."17-5, , In general, if b > 0, theny=A sin bx andy=A cos.bxhave period 2mb, frequency b, and amplitude, IA I. Figure 17-6 shows the graph of y = 1.5 sin 4x., , EXAMPLE 17.3:
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CHAPTER 17 Differentiation of Trigonometric Functions, , Graph of y, , =ll~n x, , Since tan x has period Tr, it suffices to detennine the graph in (-Tr/2, TrI2). Since tan (-x) = -tan x, we need, only draw the graph in (0, TrI2) and then reflect in the origin. Since tan x =(sinx)/(cos x), there will be vertical, asymptotes at x = TrI2'and x =-Tr/2. By (17.5), D.(tan x) > 0 and, therefore, tan x is increasing., D;(tanx) =D,(sec 2 x) =2sec x(tan x sec x) =2 tan X sec 2 x., , Thus, the graph is concave upward when tan x > 0, that is, for 0 < x < Tr/2, and there is an inflection point, at (0, 0). Some special values of tan x are given in Table 17-1, and the graph is shown in Fig. 17-7., TABLE, , 17-1, , x, , tan x, , 0, Tr, , 0, , 6", Tr, , 1/--, , 0.58, , 4", , I, , Tr, , .J3 -1.73, , 3", , ,, I., , -I, , x, , I, I, I, I, I, , I, , I, I, I, , I, I, , ,., , I, , I, , I, , I, Rg.17-7, , For an acute angle 0 of a right triangle,, tanO= sinO = opposite + adjacent _ opposite, cosO hypotenuse. hypotenuse - adjacent
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· CHAPTER 17 Differentiation of Trigonometric Functions, , Graph of y = sec x, Since sec x = lI(cos x), the graph will have a vertical asymptote x =Xo for all Xo for which cosxo = 0, that is,, for x =(2n + 1)Trl2, where n is any integer. Like cos x. sec x has a period of 2n, and we can confine our attention to (-n, n). Note that Isec xl ~ 1, since Icos xl ~ 1. Setting Dx(sec x) = tan x sec x = 0, we find critical, numbers at x =0 and x = n, and the first derivative test tells us that there is a relative minimum at x = 0 and, a relative maximum at x = n., Since, D;(secx), , =Dx(tan .\'secx) =tan x(tanxsecx) + secx(sec 2 x) =secx(tan 2 x+ sec 2 x), , there are no inflection points and the curve is concave upward for -n/2 < x < n/2. The graph is shown in, Fig. 17-8., y, , 31t,, , --', 2:,, , ,,, ,,, ,, , -1t, , -1t,, , 0, , 2', , ]t,, , 2, , -1, , 1t, , 31t', ;.......,, , X, , 2:,, ,, , -2, ,,,, Fig. 17-8, , Angles Between Curves, , By the angle of inclination of a nonverticalline ~, we mean the smaller counterclockwise ang~e a. from, the positive x axis to the line. (See Fig. 17-9.) If m is the slope of~, then m = tan a.. (To see thiS, look at, Fig. 17-10. where the line ~' is assumed to be paran~l to ~ and, therefore, has the same slope m. Then m =, (sin a. - O)/(cos a. - 0) = (sin a.)/(cos a.) = tan a..), y, :[', , P(cos Q, sin 0), , "\ \, x, , Rg.17-9, , x, , ,/, , /, , /, , \, , I, , x, , _/, , Fig. 17-10, , By the angle between two curves at a point of intersection P, we mean the smaller of the two angles between the tangent lines to the curves at P. (See Problems 17 and 18.)
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CHAPTER 17 Differentiation of Trigonometric Functions, 14. Sketch the graph off(x) = sin x + cos x., f(x) has a period of 2n. Hence, we need only consider the interval [0, 2n].f(x) = cos x - sin x, and r(x) =, -(sin x + cos x). The critical numbers occur where cos x =sin x or tan x = I, x =1CI4 or x =51C14., , f"(nI4) = -(.fin + .fi12) =-.fi < O. So, there is a relative maximum at x =n14, y = J2., f"{5nI4) = -(-.fin - .fin) =.fi > O. Thus, there is a relative minimum at x= 5nI4,y= -Ji. The, inflection points occur wheref"{x) = -(sin x + cos x) = 0, sin x = -cos x, tan x =-1, x =31C14 or x = 71C14,), =O., See Fig. 17-13., ~, y, , x, , Fig. 17-13, , 15. Sketch the graph of f(x) = cos x - cos 2 x., f'(x) = -sinx - 2(cosx)(-sinx) = (sinxX2cosx -I), and, f"(x) = (sin x)(-2 sin x) +(2cosx -I)(cosx), , = 2(cos 2 X -, , sin 2 x) - cos x = 4cos 2 X - cosx - 2, , Since/has period 2n, we need only consider [-n, nl, and sincefis even, we have to look at only [0, n]. The, critical numbers are the solutions in [0, nJ of sin x = 0 or 2 cos x - I = O. The first equation has solutions 0 and n,, and the second is equivalent to cosx = t, which has the solution n/3.j"(0) = 1 > 0; so there is a relative minimum, at (0, O).j"(n) =3 > 0; so there is a relative minimum at (n, -2). f"(n/3) =-f < 0; hence there is a relative, maximum at (n/3, t). There are inflection points between 0 and 1CI3 and between 1CI3 and 1r, they can be found by, using the quadratic formula to solve 4 cos 2 x - cos x - 2 = 0 for cos x and then using a cosine table or a calculator, to approximate x. See Fig. 17-14., y, 2, , 1, x, , Rg.17-14, 16. Find the absolute extrema off (x) =sin x + x on [0, 21rJ., f'(x) = cos x + l. Settingf'(x) = 0, we get cos x =-I and, therefore, the only critical number in [0, 2nJ is, x = n. We list n and the two endpoints 0 and 2n and compute the values off(x)., x, , f(x), , o, , 0, , 2n, , 2n, , Hence, the absolute maximum 2n is achieved at x =2n, and the absolute minimum 0 at x =O.
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CHAPTER 17 Differentiation of Trigonometric Functions, , 17. Find the angle at which the lines ..!£.: y= x+ 1 and !£2: y= -3x+5 intersect, Let a. and ~ be the angles of inclination of !£. and!£l (see Fig. 17-15), and let m. and m1be the, respective slopes. Then tan a. =ml =1and tan ~ = m2 =-3. ~ - a l is the angle of intersection. Now, by, Problem 5,, tana2 - tana., I+tanatana, I, 2, , lan(a -a), 2, , •, , ~-m., , _, , -3-1, , l+m.m 2 -1+(-3)(1), , =-4 =2, , -2, , From a graphing calculator, ~ - a l - 63.4°., , !£I, , ---------r~~r-----~-------------------.x, , Rg.17-15, , 18. Find the angle a between the parabolas y =xl and x =y2 at (I. I),, Since D.(xl) =2x and D.(/i) = 11 (2/i). the slopes al(l, 1) are 2 and, Thus. using a graphing calculator. we approximate a by 36,9°,, , \", 1.....-",, , ., , I, , ~, , . - ,, , I, , t. Hence. lana = 2-it » =! =i1+ (t, , ', , ', , ~, , ', , '~', , 19. Show that cot(x + 1r) =cot x, sec(x + 21r) =sec x, and csc(x + 21r) =csc x,, 20. Find the period p. frequency J, and amplitude A of 5 sin(x/3) and sketch its graph., , Ans., , p:z:61r,f=t.A=5, , '21. Find all solutions of cos x =O., , Ans., , x =(2n + I)~ for all int~gers n, , 22. Find all solutions oflan x= 1., , Ans., , x =(4n + l)t for all integers n, , ,, , ', , ,.. ,, , '/;'., , .. r, I:~r..
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CHAPTER 17 Differentiation of Trigonometric Functions, , 23. Sketch the graph of I(x) = 2 sinx ., -cosx, , Ans., , See Fig. 17-16., , V3), , ( _!!3' _ 3, , Fig. 17-16, , 24. Derive the fonnula tan(u + v) = tan u + tan v ., I-tanutanv, 25. Find l., , (a) y = sin 3x + cos 2x, (b) y = tan(.r), (c) y = tan 2 x, (d) y = cot(1 - 2r), (e) y =x~ sin x, (f) y= cosx, x, . sinax, , AIlS., AIlS., Ans., Ans., Ans., Ans., , r, , y' = 3 cos 3x - 2 sin 2x, y' = 2x sec2 (xl), , y' =2 tan x sec 2 x, y' =4x csc2 (1 - 2r), y' =xl cos x + 2x sin x, ' -xsinx-cosx, y =, Xl, , sin J (2x), , 26. Evaluate: (a) hm-.-- ; (b) 1m . 2(3 ), .....0 sm bx, x-+o xsm, x, , 27, If x = A sin kt + B cos kt. show that, , ~;~ = -k 2x., , 28. (a) If y = 3 sin(2x + 3), show that y"+ 4y = O. (b) If y = sin x + 2cos x, show that ylll + y" + y' + Y = O., 29. (i) Discuss and sketch the following on the interval 0 :S x < 2H. (ii) (GC) Check your answers to (i) on a graphing, calculator., (a) y=tsin2x, (b) y = cos 2x - cos X, , (c) y=x-2sinx, (d) y = sin x(l + cos x), , (e) y = 4cos 3 X - 3 cos x, AilS., , (a) maximum at x = 1r/4, 51r/4; minimum at x = 31r/4, 71r/4; inflection point at x = 0, 1r/2, H, 31r/2, (b) maximum at x = 0, 1C', minimum at x 1r/3, 51r/3; inflection point at x = 32°32', 126°23', 233°37',, , =, , 327°28', (e) maximum at x = 51r/3; minimum at x = 1r/3; inflection point at x = 0, H, (d) maximum at x = 1r/3; minimum at x =51r/3; inflection point at x =0, tr, 104°29',255°31', (e) maximum at x =0, 21r/3, 41r/3; minimum at x =1r/3. tr, 51r/3; inflection point at x =7rI2, 31r/2, 1r/6,, 57d6,71r/6. 111r/6
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CHAPTER 17 Differentiation of Trigonometric Functions, 30. If the aJ;lgle of elevation of the sun is 45° and is decreasing by t radians per hour, how fast is the shadow cast on, the ground by a pole 50 ft tall lengthening?, , Ans. 25 ftlh, 31. Use implicit differentiation to find y': (a) tan y = r: (b) cos (xy) = 2y., , ,, Ans. (a) y" = 2x cos 2 y; (b) y = -, , y sin(xy), .., 2+x sm(xy)
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,, , Inverse Trigonometric Functions, The sine and cosine functions and the other trigonometric functions are not one-to-one and. therefore, do, not have inverse functions. However, it is possible to restrict the domain of trigonometric functions so that, they become one-to-one., Looking at the graph of y sin x (see Fig. 17-2), we note that on the interval - nl2 ~ x ~ nl2 the restriction of sin x is one-to-one. We then define sin-I x to be the corresponding inverse function. The domain of, this function is [-1, 1], which is the range of sin x. Thus,, , =, , I., 2., 3., , sin-I (x) =y if and only if sin y =x., The domain of sin-I x is [-1, 1]., The range of sin-I x is [-nl2, nl2]., The graph of sin-I x is obtained from the graph of sin x by reflection in the line y =x. See Fig. 18-1., y, , .!!., 2, , 1C, , -2"", y, , =sin-I x, , Rg.18-1, EXAMPLE 18.1: In general, sin-I x = the number y in [-7t/2, 7tI2] such that sin y = x. In particular, sin-I 0 = 0,, sin-I I = 7t12, sin-I (-1) = -7tI2. sin-I (t) = Jr/6. sin-I (.fi/2) = Jr/4. sin-I (,,[3/2) = Jr/3. Also. sin-'(-t) = Jr 16. In general., sin-I (-x) = -sin-I x. because sin (-y) =-sin y., , The Derivative of sln-1 x, Let y = sin-I x. Since sin x is differentiable, sin-I x is differentiable by Theorem 10.2. Now, sin y = x and., therefore. by implicit differentiation. (cos y) y' = 1. Hence. y' = l/(cos y). But cos 2 y = I - sin 2 y = I - r., So. cos y =±.Jl- x 2, , •, , By definition of sin-I x, y is in the interval [-Tt/2, Tt/2] and, therefore, cos y ~ O.
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CHAPTER 18 Inverse Trigonometric Functions, , r.--?, ', 1, Hence, cos y =,,1- x2 • Thus, Y = .Jl- x2 • So, we have shown that, , (18.1), , 1, . -I), Dx (sm, x = 1f1""":22, , .', , f'/1-x, , The Inverse Cosine Function, If we restrict the domain of cos x to [0, n], we obtain a one-to-one function (with range [-1, 1D. SO we can, define COS-I x to be the inverse of that restriction., , 1. COS-I (x) =y if and only if cos y =x., 2. The domain of COS-I x is [-I, I]., 3. The range of COS-I x is [0, n]., 'The graph of COS-I x is shown in Fig. 18-2. It is obtained by reflecting the graph of y =cos x in the line, y=x., y, , -I, , o, Y =.cos-1 X, , Fig. 18-2, , An argument similar to the one above for (18.1) shows that, (18.2), , 1, , D (COS-I x) =--==, , .JI- x2, , x, , The Inverse Tangent Function, Restricting the domain of tan x to the interval (-nl2, nl2), we obtain a one-to-one function (with range the, set of all real numbers), whose inverse we take to be tan-I x. Then:, , 1. tan-I(x) =y if and only if tan Y=x., 2. The domain of tan-I x is (-00, +00)., 3: The range of tan-I x is (-nl2, nl2)., , ,, , EXAMPLE 18.2: In general, tan-I x =the number yin (-1t/2. 1t/2) such that tany =X. In particular, lan-I 0 =0, tan-II =, lt/4, tan-I (./3) =tr/3, tan-I (.J3/3) =tr/6, Since tan (-x) =-tan x, it follows that tan-I (-x) =-tan-I x. For example, tan-I, (-1) = -lt/4., , The graph of y = tan-I x is shown in Fig. 18-3. It is obtained from the graph of y = tan x by reflection in, the line y =x. 'Note that y =nl2 is a horizontal asymptote on the right and y =-n/2 is a horizontal asymptote, on the left.
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CHAPTER 18, , In~erse, , Trigonometric Functions, , )', , 2"", ,, , 2, , y = lan-I, , X, , Fig. 18-3, , I, D (tan-' x) = - x, 1+X2, In fact, if y = tan-' x, tan y = x and, by implicit differentiation, (sec 2 y) y' = 1. Hence,, , (18.3), , 1, 1, I, y' = sec 2 y = 1+ tan 2 y = 1+ x 2, , •, , Inverses of cot x, sec x, and csc x are defined in similar fashion., cot-' x. Restrict cot x to (0, 1t). Then the domain of coe' x is (-00, +00) and, , y =coC' x if and only if cot y =x, (18.4), , 1, Dx (cot-'x)=-I+x 2, , The proof is similar to that of (18.3). The graphs Qf cot x and coe' x are shown in Fig. 18-4., y, , y, , ------------, , ", , -------------, , o, , JC, , (b) Y = coe l x, , (tI) Y = COl X, , Fig. 18-4
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CHAPTER 18 Inverse Trigonometric Functions, sec-1 x. Restrict sec x to the union of [0, rcl2) and [rc, 3rcl2). Then the domain of see- I x consists of all y such, that Iyl ~ I and, y =see-I x if and only if sec y =x, (18.5), , I, D (sec- t x) =--===, x, ., x.Jx2-1, , For the proof, see Problem 1. The graph of sec x appeared in Fig. 17-8, and that of sec-I x is shown in, Fig. 18-5., y, , -I, , y =sec-I X, , Rg.18-5, , csc-1 x. Restrict csc x to the union of (0, rc/2] and (rc, 3rcl2]. Then the domain of csc- I x consists of all y such, that Iyl ~ 1 and, y = csc- I x if and only if esc y = x, (18.6), , 1, D (esc-I x)= ----,,==, x, X.JX2 -I, , The proof is similar to that of (18.5). The graphs of esc x and esc-I x are shown in Fig. 18-6., 'I, , y, , ,, ,, ,, , :n, , ,: I, , ., .., , ::, , •, :', , .., , ,, , 2"", , 1, , I, , I, I, , I, I, I, I, , "1, , I, I, I, , -I, , I, , I, I, , I, , I, , ,, 1, , I, I, (a), , x, , -I, , ,,, , (b) y, , ., ··, ., \, , y =esc x, , Rg.18-6, , =csc- l X
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CHAPTER 18 Inverse Trigonometric Functions, , The apparently arbitrary choices of ~e domains for the inverse trigonometric functions were made in, order to obtain simple formulas for the derivatives .., Do not confuse the notation for the inverse trigonometric functions with exponential notation. For example, sin-I x is not the same as (sin xtl. To avoid the possibility of such confusion, one can use the following, alternative notation for the inverse trigonometric functions:, arcsin x =sin-1 x, arccos x =COS-I x, etc., , SOLVED PROBLEMS, 1., , Prove (18.5): D.(sec- ' x) = J 12 -1, Let y = sec-I x. Then sec, and, by implicit differentiation. tan y sec Y (y') I. Now tan 2 Y sec2 Y - I i" - 1;, hence, tan y =±Jx 2 -I. By definition of sec-I X, y is in [0, rrJ2) or [n, 3rrJ2), and, therefore, tan y is positive., Thus, tany"= JX2 -1 So,, , !=:, , =, , 1, 1, Y = tanysecy = xJx2-1, I, , In Problems 2-8, find the first derivative y'., , 2., , Y = sin- ' (2x - 3)., By (l8. t) and the Chain Rule,, , )" =, , 4., , 5., , I, D (2x _ 3) =, 2, ==, I, JI-(2x-3)2 •, .J12x-4x2-8 J3x-x 2-2, , )' = tan- I(3i")., By (18.3) and the Chain Rule, )" =, , 1+, , dx2)2 D.(3xl) = -I :;x4 ., , )'=cot-IO:~)., By (18.4) and the Chain Rule,, , ,_, 1, D (I+x)_, 1, x (I-x)-(l+x)(-I), y --I+(I+x)2 • I-x - 1+(1+x)2, (l_X)2, I-x, , I-x, , =, , =
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CHAPTER 18 Inverse Trigonometric Functions, , 7., , Y=XCSC-I(±)+.Jl~x2, , Y' =..i[, ab, , forO<x< 1., , (£tanx)] = ab1 a +ba2tan 2x ba sec2 x= a, , 1, D, (b, )2 z a, 1+ -tanx, , 2, , 2, , 2, , 2, , sec X, + b2 tan 2x, , a, , _, , I, , - a2 cos 2x + b2 sin 2 x, 9., , If y2 sin x + y = tan-I x, find y'., By implicit differentiation, 2yy' sinx +y2 cosx+ y' = '[73"1 Hence,, +x, y'(2ysinx + I) =-II2 - y2 cos x and. therefore., +x, , r,. •, , , 1- (1 + x 2 )y2 COSX, y = (1 + x2 )(2ysinx + I), , m; (e) tan- I(-.J3); (f) sec-I (2); (g) sec-I (-2), , 10. Evaluate: (a) sin-I (-fil2) (b) cos-I(I); (c) COS-I(O): (d) cos- 1, , (a) sin-I (-fil2) = -sin-I (fi 12 =-re14, (b) cos- l(1) = 0, since cos(O) = I and 0 is in [0, xl, (c) COS-I(O) = 7tI2, since cos(x/2) = 0 and (1tI2) is in [0. x], (~) cos-I<t>:re/3, (e) tan-I (-../3) = -Ian-I (../3) = -re/3, (f) sec- I(2) =1t/3, since, , r~:·~-:-f~, , It";:-:', , J;::,; ,, i,'\', , 1.. ' - , - ,, , ", , _\ = -2 and 41t13 is in [x, 31t12)', , 11. Show that sin-I x + COS-I X =~., , D.(sin- ' x + COS-I x) = ~2 - ~, = O. Then, by Problem 15 of Chapter 13. sin-I x + COS-I X is a, vI-x, vI-x 2, constant. Since sin-I 0 +COS-I 0 = 0 + = ~. that constant is ~., , I, , 12. (a) Prove: sin (sin-I(y» = y; (b) find sin-I(sin x); (c) prove that sin-'(sin x) = x if and only if x is in [-x12. 7tl2l., (a) This follows directly from the definition of sin-I(y)., (b) sin-I(sin x} = sin-IQ = O., (c) sin-I y is equal to that number x in [-xl2, 1t/2] such that sin x =y. So, if x is in [-1tI2, xl2], sin-I (sin x) = x., If x is not in [-xl2, x/2], then sin-I (sin x) :I: x, since, by definition, sin-I (sin x) must be in [-1tI2, xl2]., 13. Evaluate: (a) cos(2sin-'(t»; (b) sin(cos-I(-t»., , (a) By (16.11), cos(2sin-l(t», , =1- 2sin 2(sin- ' (t», , = 1- 2(t)2, , =I - ! =, , :, , tL~', ~~::,;~'~'", , =COS(~/3) =t=2, , sec(f), (g) sec- I(-2) = 4x/3, since sec (4re 13) = coS(lre/3), , ', , r, L, , *.
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CHAPTER 18 Inverse Trigonometric Functions, , (b) sin2(cos-1 (-t», , =1- cos1 (COS-I (-t» =1- (-t)2 = tr., , Hence, sin(cos-I(-t» =±Jf/4. Since cos-I(-f) is in the second quadrant, sin(cos-I(-t» > O. So,, sin(cos- I(-t» = Jf/4., , 14. The lower edge of a mural. 12 ft high, is 6 ft above an observer's eye. Under the assumption'that the most, favorable view is obtained when the angle subtended by the mural al the eye is a maximum, at what distance, from the wall should the observer stand?, Let denote the subtended angle, and let x be the distance from the wall. From Fig. IS·7, tan (0 + ~)= IS/x,, lan~ =6/x, and, , o, , tan 0 = tan [(0 + 41) -, , tan(O+I/')-tan~, , 1/'] = 1 + tan(O +I/')tan~, , (\S/x):....(6/x) _ 12x, 1 + (1S/xX6/x) - Xl + lOS, , Rg.18-7, Then, 0= tan, , -I(, x2, , l2X), + lOS, , and, , dO, dx, , 2, , 12(-x +lOS), , = X4 + 360x2 + 11664, , The critical number x =6.J3 - 10.4. By the first derivative test, this yields a relative maximum. The observer, should stand about 10.4 ft in front of the wall., , , j, , 15. Evaluate: (a) sin- I(-./3I2); (b), , 005- 1(./312):, , (c), , co~-I(-./3I2); (d) tan- I(-./3/3): (e) sec-I(.fi):, , (f) sec-I (-.fi)., Alls., , (a), , -~; (b) ~; (c) 5:; (d), , -.g: (e) %; (f) 5:, , 16. Prove: tan-I x + cot-I X =~., In Problems 17-24, find y'., , 3, , AlIS., , 19. y= tan-I, , (~), , 1, , Ans., , _, , Ans., , - .'(2+9, , Ans., , .J4_X2, '3
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CHAPTER 18 Inverse Trigonometric Functions, , Ans., , 2X(COS-1, , (1)x + Jx2-4, 1 ), , Ans., , Ans., , Ans., , 8, x3 Jx 2-4, , 25. Prove fonnulas (18.2), (18.4), and (18.6)., , m. Find: (a) sin 9; (b) cos 9; (c) tan 9; (d) cot 9; (e) sec 9; (f) csc 9; (g) cos 29; (h) sin 29., , 26. Let () =cos- 1, , Ans., , 3$., , 2., , (a) -7-' (b)"f, (c), , 3$., , 2$., , 7., , 7$., , 2 ' (d) 15' (e) 2' (f) 15' (g) -, , 41., , 12$, , 49' (h) 49, , 27. Let () =sin-I(-t). Find: (a) sin 9; (b) cos 9; (c) tan 9; (d) cot 9; (e) sec 9; (f) csc 9; (g) cos 29; (h) sin 29., , Ans., , (a), , -t; (b) 2f; (c) - ~; (d) -216; (e), , 5ff; (f) -5; (g) j; (h) - 4:, , 29. Evaluate: (a) cos(sin-I(-(r»; (b) tan(sec-I(t»; (c) sin(cos-I(t) + sec-I 4); (d) COS-I (cos, , 4.fJ., , 216 ., , Jif -/f5 ., , Ans. (a) 0 ' (b) -7-' (c) 20+10' (d), , 3:)., , n, "2, , 30. Find the domain and range of the function/(x) =sin(sec- I x)., , Ans. Domain lxf ~ 1; range (-1, 1), 31. (a) For which values of x is tan-I(tan x) = x true?, , (b) (OC) Use a graphing calculator to draw the graph of y = tan-I(tan x) - x to verify your answer to (a)., , 32. A light is to be placed directly above the center of a circular plot of radius 30 ft, at such a height that the edge, of the plot will get maximum illumination. Find the height if the intensity I at any point on the edge is directly, proportion'll to the cosine of the angle of incidence (angle between the ray of light and the vertical) and inversely, proportional to the square of the distance from the source., (Hillt: Let x be the required height, y the distance from the light to a point on the edge, and 9 the angle of, "d, cos() = (x2 +kx, lllCI ence. Then I =k 7, 900)3/2 .), , Ans., , Isfi ft, , 33. Show that sin-I x =tan-{, , JI ~ x, , 2, , ), , for Ixl < 1. Examine what happens when lxf = 1.
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CHAPTER 18 Inverse Trigonometric Functions, , 34. (GC) Evalute sin-I, Ans., , 0.6435, , 35. (a) Find sec(tan-'(, calculator., , Am., , m by using a graphing calculator., , t». (b) Find an algebraic fonnula for sec (tan-I (2.x». (c) (GC) Verify (a) and (b) on a graphing, , (a).J7ji; (b) ../1 + 4x 2, , (±) for x ~ 1; (b) sec-I x = 21t' - COS-I (±) for x ::;; - I., , 36. Prove: (a) sec-Ix = cos.- I, , ., , I, , (The formula of part (a) would hold in general for !xl ~ 1 if we had defined sec-I x to be the inverse of the, restriction of sec x to (-1tI2, 1tI2). However, if we had done that, then the fonnula for Dx(sec-Ix) would have been, 1/(1 x 1../x 2 - I ) instead of the simpler formula l/(x../ x 2 -1 ).)
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Rectilinear and Circular Motion, \, , Rectilinear Motion, Rectilinear motion is motion of an object on a straight line. If there is a coordinate system on that line, and, s denotes the coordinate of the object at any time t, then the position of the object is given by a function, s =f(t). (See Fig. 19-1.), • s, -2, , -1, , 0, , 2, , 3, , Fig. 19-1, , The position at a time t + flt, very close to t, is f(t + flt). The "distance" the object travels between time, t and time t + flt is f(t + flt) - f(t). !he time the object has traveled is flt. So. the average velocity over this, period of time is, f(t + flt) - /(t), flt, , (Note that the "distance" can be negative when the object is moving to the left along the s axis. So the average, velocity can be positive or negative or zero,), As flt approaches zero. this average velocity approaches what we think of as the instantaneolls velocity v, at time t. So,, v = lim f(t + flt) - f(t), &--+0, , flt, , =f'(t), , Hence, the instantaneous velocity v is the derivative of the position function s. that is, v =dsldt., The sign of the instantaneous velocity v tells us in which direction the object is moving along the line. If, v =ds/dt > 0 on an interval of time, then by Theorem 13.7(a), we know that s must be increasing, that is, the, object is moving in the direction of increasing s along the line. If v = ds/dt < O. then the object is moving in, the direction of decreasing s., The instantaneous speed of the object is defined as the absolute value of the velocity. Thus. the speed, indicates how fast the object is moving, but not its direction. In an automobile, the speedometer tells us the, instantaneous speed at which the car is moving., The acceler9tioll a of an object moving on a straight line is defined as the rale at which the velocity is, changing, that is, the derivative of the velocity:, , EXAMPLE 19.1: Let the position of an automobile on a highway be given by the equation .I' = f(t) = t 2 - 5t, where s, is measured in miles and t in hours. Then its velocity v = 21 - 5 milh and its acceleration Cl = 2 milh 2• Thus, its velocity, is increasing at the rate of 2 miles per hour per hour. .
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CHAPTER 19, , Rectilinear and Circular Motion, , When an object moving along a straight line changes direction, its velocity v =O. For, a change in direc. tion occurs when the position s reaches a relative extremum, and this occurs only when dsldt = O. (However,, the converse is false; dsldt = 0 does not always indicate a relative extremum. An example is s = t 3 at 1=0.), EX~MPLE 19.2:, Assume that an object moves along a straight line according to the equation s = /(t) = (t - 2)2,, where s is measured in feet and t in seconds. (The graph of/is shown in Fig. 19-2.) Then v = f'(t) = 2(t - 2) ftlsee and, a = 2 ftlsec 2. For t < 2, v < 0 and the object is moving to the left. (See Fig. 19-3.) For t> 2, v > 0 and the object is moving to the right. The object changes direction at (= 2, where v O. Note that, although the velocity v is 0 at time t = 2,, the object is moving at that time; it is not standing still. When we say that an object is stalldillg still, we mean that its, position is constant over a whole interval of time., , =, , s, , [----.--~----~-----.• s, , Fig. 19-2, , Fig. 19-3, , Motion Under the Influence of Gravity, If an ohject has heen thrown slmight up or down. or just stalts from rest, and the only force acting upon it is, the gravitational pull of the earth, then the resulting rectilinear motion is referred to as free fall., Put a coordinate system on the vertical line on which the object is moving. Assume that this s axis is directed upward (see Fig. 19-4), and that ground level (the surface of the earth) corresponds to S = O. It is a fact, of physics that the acceleration a is a constant approximately equal to - 32 ftlsec 2. (In the metric system, this, constant is -9.8 mlsec2• The symbol "m"stands for "meters.") Note that the acceleration is negative because, the pull of the earth causes the velocity to decrease., Since, , ~; =a =-32, we have:', , earth, , /, , Fig. 19-4, , (19.1), (19.2), , v=vo-321, d, where Vo is the initial velocity when t =O.t Now, v = d~' Hence,, 2, S = So + '1lt - 161, where So is the initial position. the value of s when t =O.i, , In fact, lJ,(Vo - 321) = - 32 = D,v. So, by Chapter 13, Problem 18, vand "0 - 321 differ by a constant. Since vand "0 - 321 are equal, when ( = O. that constant difference is O., , In fact, lJ,(so + 'I. - 161') = "0 - 321 =D,r. So! by Chapter 13, Problem 18, S and So + I\)t - 161' differ by a constant. Since sand, So + I\)t - 161 2 are equal when 1= 0, that constant difference is O., t
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-'1-, , CHAPTER 19 Rectilinear and Circular Motion, , Circular Motion, The motion of a particle P along a circle is completely described by an equation 8 =/(1), where 8 is the, central angle (in radians) swept over in time 1 by a line joining P to the center of the circle. The x and y coordinates of P are given by x =r cos 8 and y =r sin 8., By the angular velocity m of P at time t, we mean ~~., 2, , . t, we mean dt, dm =ddt 8 •, · a 0 f P at time, By the angu Iar acee leratlOn, 2, , I, , SOLVED PROBLEMS, , 1., , !t, , A body moves along a straight line according to the law s =, , 3, , -, , Illl: end of 2 seconds., , v = ~:, , =tt 2 - 2: hence, when 1=2,, , a = C;;;, 2., , 2t. Determine its velocity and acceleration at, ., , v = t(2)2 - 2 =4 ftlsee·., , =31; hence, when 1= 2, a =3(2) =6 ftlsec 2•, , The path of a particle moving in a straight line is given by s =t3 - 61 2 + 9t + 4., (a), (b), (c), (d), (e), , Find s and a when v = O., Find s and v when a =O., When is s increasing?, When is v increasing?, When does the direction of motion change?, We have, v, , (a), (b), (c), (d), (e), 3., , =~ =3t2 -121 + 9 =3(t -1)(1- 3),, , a, , =~~ =6(1 - 2), , When v= 0.1 = I and 3. When I =I, s = 8 and a =-6. When 1= 3, s = 4 and a =6., When a = 0, 1= 2. At I = 2, s = 6 and v = - 3., s is increasing when v> O. that is. when I < I and I> 3., v is increasing when a > 0, that is, when t> 2., The direction of motion changes when v= 0 and a O. From (a), the direction changes when 1=1 and 1= 3., , *', , A body moves along a horizontal line according to s =/(1) =13 - 91 2 + 241., (a) When is s increasing, and when is it decreasing?, (b) When is v increasing, and when is it decreasing?, (c) Find the total distance traveled in the first 5 seconds of motion., We have, v = ~~, , =3t 2 -181+ 24 = 3(1 - 2)(t-4)., , a = ~~ =6(t - 3), , (a) s is increasing when v> 0, that is, when 1 < 2 and t> 4., s is decreasing when v < 0, that is, when 2 < 1 < 4., (b) v is increasing when'a > 0, that is, when I> 3., v is decreasing when a < 0, that is, when 1<3,, (c) When 1= 0, s =0 and the body is at O. The initial motion is to the right (v> 0) for the first 2 seconds; when, 1= 2, the body is s = /(2) =20 ft from O., During the next 2 seconds, it moves to the left. after which it is s =/(4) = 16 ft from O., , ~:, , ..., , ,.,.
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CHAPTER 19 Rectilinear and Circular Motion, , It then moves to the right, and after 5 seconds of motion in all, it is s =/(5) = 20 ft from O. The total, distance traveled is 20 + 4 + 4 =28 ft (see Fig. 19-5)., , o, , 20, , I, , 1, , 4, , . -------., , Fig. 19-5, , 4., , A particle 1II0ves in a horiz.ontalline according to s =/(1) = 14 -, , ,, 61), , + 12t 2 - lOt + 3., , (a) When is the speed increasing, and when decreasing?, (b) When does the direction of motion change?, (c) Find the total distance traveled in the first 3 seconds of motion., Here, v =:, , =4t 3 -, , 181 2 + 241 - 10 =2(t -1)2(21 - 5),, , a= dv =12(/-1)(1-2), dl, , (a) v changes sign att = 2.5, and a changes sign att = 1, t =2., For t < I, v < 0 and a > O. Since a > 0, v is increasing. Since v < 0, the speed Ivl =- v is decreasing., For I < t < 2, v < 0 and a < O. Since a < 0, v is decreasing. Since v < 0, the speed Ivl =- v is increasing., For 2 < t < 2.5, v< 0 and a > O. As in the first case, the speed is decreasing., For t > 2.5, v> 0 and a > O. v is increa~ing. Since v> 0, the speed Ivl = v is increasing., (b) The direction of motion changes at 1= 2.5, since, by the second derivative test, s has a relative extremum, there., (c) When 1 = 0, S =3 and the particle is 3 units to the right of O. The motion is to the left until 1= 2.5, at which, time the particle is i units to the left of 0, When 1 = 3, s =0; the particle has moved i units to the right., The total distance traveled is 3 + + =11- units. (See Fig. 19-6.), , '*, , *, o, , 3, , 1------ -------- ------------...,,, ,", ,....--------_, .. - -- -27/16, , --., , Rg.l9-6, 5., , A stone, projected vertically upward with initial velocity 112 ftlsee, moves according to s = 112/- 16t 2, where s, is the distance from the starting point. Compute (li) the velocity and acceleration when t = 3 and when t =4, and, (h) when the greatest height reached. (c) When will its height be 96 ft?, We have v= dsldl = 112 -32t and a = dvldt= -32., (a) At t = 3. v = 16 and a = - 32. The stone is rising at 16 ftlsec., At t =4, v =- 16 and a =- 32. 'Pl\; stone is falling at 16 ftlsec., (b) At the highest point of the motion, v = O. Solving v= 0 = 112 - 321 yields 1 =3.5. At this time, s = 196 ft., (c) Letting 96 = 112t - 16t 2 yields t 2 - 71 + 6 =0, from which I = I and 6. At the end of I seeond of motion, the, stone is at a height of 96 ft and is rising, since v> O. At the end of 6 seconds, it is at the same height but is, falling since v < O., , 6., , A particle rotates counterclockwise from rest according to e= 13/50 - I, where eis in radians and t in seconds., Calculate the angubr displacement e. the angular velocity m, and the angular acceleration a at the end of, 10 seconds,, m = de = 31, dt, , 7., , 2, , SO, , -, , 1= 5 radlsec, , ', , a = ~~ = ~b =~ radlsec 2, , At t =0, a stone is dropped from the top of a building 1024 ft high. When does it hit the ground, and with what, speed? Find the speed also in miles per hour.
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CHAPTER 19 Rectilinear and Circular Motion, , Since So = 1024 and '" = 0, equation (19.2) becomes s = 1024 - 161 2, and the time of hitting the ground is the, solution of 1024 - 161 2 = O. This reduces to 12 = 64, yielding 1= ±8. Since the motion occurs when I ~ 0, 1=8., The equation (19.1) is v = - 321, yielding v = - 32(8) = - 256 ftlsec when 1= 8, that is, when the stone hits the, ground. (The velocity is negative because the stone is moving downward.) The speed is 256 fllsee. To change to, miles per hour, note the following:, x feet per second = 60x feet per minute = 60(60x) feet per hour, , 15 x nu'1es per hoUf., = 3600x'l, 5280 nu es per hour = 22, Thus,, (19.3), , x feet per second=i x miles per hour., , In particular, when x = 256, we get 174,\ miles per hour., , 8., , If a rocket is shot vertically upward from the ground with an initial velocity of 192 ftlsec, when does it reach its, maximum height above the ground, and what is that maximum height? Also find how long it takes to reach the, ground again and with what speed it hits the ground., Equations (19.1) and (19.2) are v= 192 - 321 and s = 1921 - 1612. At the maximum height, v= 0, and,, therefore, 1= 6. So, it takes 6 seconds to reach the maximum height, which is 192(6) - 16(6)2 = 576 ft. The rOCket, returns to ground level when 0 = 1921 - 161 2, that is, when 1= 12. Hence, it took 6 seconds to reach the ground, again. exactly the same time it took to reach the maximum height. The velocity when I = 12 is 192 - 32( 12) =, -192 fllsec. Thus. its final speed is the same as its initial speed., , 9., , Show that, if an object is moving on a straight line, then its speed is increasing when its velocity v and its, acceleration a have the same sign, and its speed is decreasing when v and a have opposite sign. (Hint: The speed, S = luI. When v> 0, S = v and dSldt = didt =a. When v< O. S = -v and dSldt =-didt = -a.), , 10. An object moves in a straight line according to the equation s =13 - 61 2 + 9t, the units being feet and seconds., Find its position. direction. and velocity, and determine whether its speed is increasing or decreasing when (a), t=t;(b) t=t;(c) t=t;(d)t=4., , s =l(-ft: moving to the right with v =J]-ft/sec: speed decreasing, s = f ft; moving to the left with v = -t ft/sec; speed increasing, s =t ft; moving to the left with v = -t ft/sec; speed deereasing, s = 4 ft; moving to the right with v =9 ft/sec; speed increasing, , Ans. (a), (b), (c), (d), , 11. The distance of a locomotive from a fixed point on a straight track at time tis 31 4 - 44t 3 + 14412. When is it in, reverse?, , Ans. 3 < 1< 8, , ,., , 12. Examine, as in Problem 2, each of the following straight line motions: (a) s = 13 - 9t 2 + 24t: (b) .\' = t 3 - 3t 2 + 3t + 3;, (c)s=2t 3 -12t 2 + 18t-5; (d)s=3t 4 -28t 3 +90t 2 -108t,, , Ans., , The changes of direction occur at 1=2 and t =4 tn (a), not at all in (b), at t = I and 1=3 in (c). and at I = I, in (d)., , 13. An object moves vertically upward from the earth according to the equation s = 64t - 16t 2• Show that it has lost, one-half its velocity in its first 48 ft of rise,
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CHAPTER 19 Rectilinear and Circular Motion, , 14. A baIl is thrown vertically upward ~om the edge of a roof in such a manner that it eventually falls to the street, 112 ft below. If it moves so that its distance s from the roof at time t is given by s = 94t - 16t 2, find (a) the', position of the ball, its velocity, and the direction of motion when t =2, and (b) its velocity when it strikes the, street (s in feet, and t in seconds)., Ans., , ~",, , ~., , ., , (a) 240 ft above the street, 32 ftlsec upward; (b) -128 ftlsec, , 15. A wheel turns through an angle of oradians in time I seconds so that 0= 128t - 12t 2• Find the angular velocity, and acceleration at the end of 3 seconds., Ans., , (0, , = 56 rad/sec; a =- 24 rad/sec2, , 16. A stone is dropped down a well that is 144 ft deep. When will it hit the bottom of the well?, Ans., , After 3 seconds, , 17. With what speed in miles per hour does an object dropped from the top o(a 10-story building hit the ground?, Assume that each story of the building is 10ft high., Ans., , 54fr mlh, , 18. An automobile moves along a straight road. If its position is given by s = 8t 3 - 12t 2 + 6t - I, with s in miles and t, in hours, what distance does it travel from t = 0 to (= I?, Ans., , 2 miles, , 19. Answer the same question as in Problem 18, except that s = 51 Ans., , (2, , and the car operates from 1=0 to t = 3., , 6.5 miles, , velocity in feet per second if it hit the, 20. A stone was thrown straight up from the ground. What was its initial, I, ground after 15 seconds?, Ans., , 240 ftlsec, , 21. (GC) Let the position s of an object moving on a ~;trlight line be given by s = t 4 - 3(2 + 2t. Use a graphing, calculator to estimate when the object changes direction, when it is moving to the right, and when it is moving to, the left. Try to find corresponding exact formulas., Ans., , Change of direction at (= -1.3660, 0.3660, and l. The object moves left for t < -1.3660 and for, 0.3660 < t .( I. The exact values of t at which the object changes direction are 1 and -1 ~.J3 ., , 22. (GC) An object is moving along a straight line according to the equation s =3t - t 2• A second object is moving, along the same line according to the equation s = t) - (2 + I. Use a graphing calculator to estimate (a) when they, occupy the same position and (b) when they have the same velocity. (c) At the time(s) when they have the same, position, are they moving in the same direction?, Ans., , (a) 0.3473 and 1.5321; (b) (= ±l; (c) opposite directions at both intersections.
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Related Rates, If a quantity y is a function of time t, the rate of change of y with respect to time is given by dyldt. When two, or more quantities, all functions of the time t, are related by an equation, the relation of their rates of change, lIIay be found by differentiating both sides of the equation., EXAMPLE 20.1: A 25-foot ladder rests against a vertical wall. (See Fig. 20-1.) If the bottom of the ladder is sliding, away from the base of the wall at the rate of 3 ft/sec, how fast is the top of the ladder moving down the wall when the, bottom of the ladder is 7 feet from the base?, , -, , x, , Rg.20-1, , Let x be the distance of the bottom of the ladder from the base of the wall, and let y be the distance of the, top of the ladder from the base of the wall. Since the bottom of the ladder is moving away from the base of, the wall at the rate of 3 ftlsec, dxldt =3. We have to find dyldt when x =7. By the Pythagorean Theorem,, (20.1), , This is the relation between x and y. Differentiating both sides with respect to t. we get, , ,., Since d.'ddt =3, 6x + 2y dyldt =0, whence, (20.2), , This is the desired equation for dyldt. Now. for our particular problem, x =7. Substituting 7 for.t in equation, (20.1). we get 49 + f =625, f =576, y =24. In equation (20.2). we replace x and y by 7 and 24, obtaining, 21 + 24 dyldt =O. Hence, dyldt =- t. Since dyldt < 0, we conclude that the top of the ladder is sliding down, the wall at the rate of t ft/sec when the bottom of the ladder is 7 ft from the base of the wall., --."~.=• •
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CHAPTER 20, , Related Rates, , SOLVED PROBLEMS, , 1., , Gas is escaping from a spherical balloon at the rate of 2 ft3/ min. How fast is the surface area shrinking when the, radius is 12 ft?, A sphere of radius r has volume V =11£r3 and surface area S =41£r. By hypothesis. dVldt =-2. Now,, dVldt = 41£r drldt. So, -2= 4m.2 dr/dt and, therefore. drldt = -1I(21£r). Also, dSldt = 8rtr dr/dt. Hence,, dSldt = -81£r/21£r =--4//". So, when': = 12, dS/dt =-li =-to Thus, the surface area is shrinking at the nite of, , t fe/min., 2., , '1',·, , Water is running out of a conical funnel at the rate of I inl/sec. If the radius of the base of the funnel is 4 in and, the height is 8 in, find the rate at which the water level is dropping when it is 2 in from the top. (The formula for, the volume V of a cone is t1£r2 h, where r is the radius of the base and h is the height.), Let r be the radius and II the height of the surface of the water at time t, and let V be the volume of the water, in the cone. (See Fig. 20-2.) By similar triangles, r/4 = h/8, whence r = th ., , Fig. 20-2, , Then, So, , dV, dt, , =~"'h2, dh, 4", dt·, , By hypothesis, dV/dt =-I. Thus, ., 2, -1 =t"'h, ,. dlz, dt', , . Id'mg Tt=, dh Trh2', -4, Yle, , Now, when the water level is 2 in from the top, h = 8 - 2 = 6. Hence, at that time, dhldt =-1/(91£), and so the, water level is dropping at the rate of 11(91£) in/sec., ,, , .1.'., , 3., , Sand falling from a chute forms a conical pile whose altitude is always equal to t the radius of the base. (a), How fast is the volume increasing when the radius of the base is 3 ft and is increasing at the rate of 3 in/min?, (b) How fast is the radius increasing when it is 6 ft and the volume is increasing at the rate of 24 ft3/min?, Let r be the radius of the base, and h the height of the pile at time t. Then, , (a) When r = 3 and dr/dt = t, dVldt = 31£ ft3/ min., (b) When r =6 and dVldt =24, drldt = 1/(21£) ftlmin., 4., , Ship A is sailing due south at 16 milh, and ship B, 32 miles south of A, is sailing due east at 12 milh. (a) At what, rate are they approaching or separating at the end of I hour? (b) At the end of 2 hours? (c) When do they cease to, approach each other, an~ how far apart are they at that time?
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••, , CHAPTER 20 Related Rates, , tet Ao and Bo be the initial positions of the ships, and A, and B, their positions I hours later. Let D be the, distance between them 1 hours later. Then (see Fig. 20-3):, D2= (32 -161)2 +(121)2, , and, , 2D t1fr, , ;: : 2(32 -161)(-16) + 2(121)(12) =2(4001 - 512)., , Ao, ;..., , A,, , Rg.20-3, , Hence dD = 4001-512, , dt, , D, , (a) When t = 1, D =20 and, , df, =-5.6. They are approaching at 5.6 milh., , t1fr =12. They are separating at 12 milh., They cease.to approach each other when, =0, that is. when 1 =~ = 1.28 h. at which time they are, , (b) When 1 =2. D =24 and, , (c), , ttt,, , D = 19.2 mIles apart., , S., , Two parallel sides of a rectangle are being lengthened at the rate of 2 in/sec, while the other two sides are, shortened in such a way that the figure remains a rectangle with constant area A =50 in2• What is the rate, of change of the perimeter P when the length of an increasing side is (a) 5 in? (b) 10 in? (c) What are the, dimensions when the perimeter ceases to decrease?, Let x be the length of the sides that are being lengthened, and y the length of the other sides. at time I., Then, , '-, , -".-, , ;f~~i, , ;~~~, ::~.tf·~, , (a) When x = 5, y = 10 and tWdl = 2. Then, 5 ~ + 10(2) = O. So, , ~ = -4, , and, , : = 2(2 - 4) = -4 in/sec (decreasing), , (b) When x = 10. y = 5 and tWdl = 2. Then, 10, , 7, +5(2) = O. So ~. = -I, , and, , :, , =2(2 -I) = 2 in/sec (decreasing), , (c) The perimeter will cease to decrease when dP/dt = 0, that is, when dy/dl = -dx/dt =-2. Then, x(-2) +,y(2) =0, and the rectangle is a square of side x =y =5./2 in., 6., , The radius of a sphere is r when the time is t seconds. Find the radius when the rate of change of the surface area, and the rate of change of the radius are equal., The surface area S =41t2; ~ence. dS/dl =81tr dr/dr. When dS1d1 =dr/dl. 81tr =1 and the radius r = 1I81t., , 7., , A weight W is attached to a rope 50 ft long that passes over a pulley at a point P. 20 ft above the ground. The, other end of the rope is attached to a truck at a point A, 2 ft above the ground. as shown in Fig. 20-4. If the truck, moves away at the rate of 9 ft/sec, how fast is the weight rising when it is 6 ft above the ground?
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CHAPTER 20, , Related Rates, , p, .".-.. ,., , Fig. 20-4, Let x denote the distance the weight has been raised, and y the horizontal distance from point A1 where the, rope is attached to the tnlck, to the vertical line passing through the pulley. We must find dxldt when dyldt = 9, andx= 6., Now, y2=(30+X)2_(lS)2, , When x, , 8., , dy 30+x d"C, di=, y dt, , and, , =6, y =ISJ3 and dyldt =9. Then 9 = 30 +J dx, d, IS,,3, , t, , ,from which dx, , dI, , =2.2 J3 ft/sec .., , A light L hangs H ft above a street. An object h ft tall at 0, directly under the light. moves in a straight line along, the street at v ft/sec. Find <I formula for the velocity V of the tip of the shadow cast by the object on the street at, t seconds. (See Fig. 20-5:), , L, , Fig. 20-5, After t seconds, the object has moved a distance vI. Let y be the distance of the tip of the shadow from O. By, similar triangles, (y - VI)ly =hlH. Hence,, y= Hvt, H-h, , and, therefore,, , v = dy =..JiJ!.....=, dt, , H- h, , 1, v, 1- (hlH), , Thus. the velocity of the tip of the shadow is proportional to the velocity of the object, the factor of, proportionality depending upon the ratio hill. As II ~ 0, V ~ v. while as h ~ H. V ~ + 00., , SUPPLEMENTARY PROBLEMS, .:, - . .-- __ r9., , A rectangular trough is 8 ft long. 2 ft across the top, and 4 ft deep. If water flows in at a rate of 2 ftl/min, how fast, is the surface rising when the water is 1 ft deep?, Ans., , t ft/min
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CHAPTER 20 Related Rates, , 10. A liquid is flowing into a vertical cylindrical tank of radius 6 ft at the rate of 8 ft3/min. How fast is the surface, rising?, , Ans. 2J91t ftlmin, 11. A man S ft tall walks at a rate of 4 ftlsec directly away from a street light that is 20 ft above the street. (a) At what, rate is the tip of his shadow moving? (b}At what rate is the length of his shadow changing?, , Ans. (a) .Jf-ft/sec;'(b) t ft/sec, 12. A balloon is rising vertically over a point A on the ground at the rate of IS ftlsec. A point B on the ground is level, with and 30 ft from A. When the balloon is 40 ft from A, at what rate is its distance from B changing?, , Ans., , 12 ftlsec, , 13. A ladder 20 ft long leans against a house. If the foot of the ladder is moving away from the house at the rate of, 2 ft/sec, find how fast (a) the top of the ladder is moving downward, and (b) the slope of the ladder is decreasing,, when the foot of the ladder is 12 ft from the house., , Ans. (a) t ft/sec; (b) i per second, 14. Water is being withdrawn from a conical reservoir 3 ft in radius and 10 ft deep at 4 ft3/min. How fast is the, surface falling when the depth of the water is 6 ft? How fast is the radius of this surface diminishing?, , Ans., , 100/811t ftlmin; 1O/271t ftlmin, , 15. A barge. whose deck is 10 ft below the level of a dock, is being drawn in by means of a cable attached to the deck, and passing through a ring on the dock. When the barge is 24 ft away and approaching the dock at ft/sec, how, fast is the cable being pulled in? (Neglect any sag in the cable.), , t, , Ans., , nftlsec, , 16. A boy is flying a kite at a height of 150 ft. If the kite moves horizontally away from the boy at 20 ftlsec, how fast, is the string being paid out when the kite is 250 ft from him?, , Ans., , 16 ft/sec, , 17. One train. starting at II A.M., travels east at 45 milh while another. starting at noon from the same point, travels, south at 60 milh. How fast are they separating at 3 P.M.?, , Ans., , IOS/i/2 miIh, , 18. A light .is at the top of a pole 80 ft high. A ball is dropped at the same heighl'from a point 20 ft from the light., Assuming that the ball falls according to s = 16t2• how fast is the shadow of the ball moving along the ground I, second later1, AilS., , 200ftlsec, , 19. Ship A is 15 miles east of 0 and moving west at 20 miIh; ship B is 60 mi south of 0 and moving north at 15 milh., (a) Are they approaching or separating after I h and at what rate? (b) After 3 h? (c) When are they nearest one, another?, , Ans. (a) approaching.IIS/J8'i miIh; (b) separating, 9$012 miIh; (c) I h S5 min
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CHAPTER 20 Related Rates, , 20. Water. at a rate of 10 fi3/min. is pouring into a leaky cistern whose shape is a cone 16 ft deep and 8 ft in diameter, at the top. At the time the water is 12 ft deep, the water level is observed to be rising at 4 in/min. How fast is the, water leaking away?, AilS., , (10 - 37t) ft3/ min, , 21. A solution is passi~g through a conical filter 24 in deep and 16 in across the top. into a cylindrica( vessel of, diameter 12 in. At what rate is the level of the solution in the cylinder rising if. when the depth of the solution in, the filter is 12 in. its level is falling at the rate I in/min?, Ans., , tin/min, , 22. Oil from a leaking oil tanker radiates outward in the form of a circular filmon the surface of the water. If the, radius of the circle increases at the rate of 3 meters per minute, how fast is the area of the circle increasing when, the radius is 200 meters?, Ans., , 12007t m 2/min, , 23. A point moves on the hyperbola Xl - 4y2 = 36 in such a way that the x coordinate increases at a constant rate of, 20 units per second. How fast is the y coordinate changing at the point (10. 4)?, Ans., , 50 units/sec, , 24. If a point moves along the curve y =Xl - lx. at what point is the y coordinate changing twice as fast as the, x coordinate?, Ans., , (2.0)
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Differentials., Newton's Method, If a functionjis differentiable at x, then j'(x) = lim !!:.yl/u, where!!:.y =j(x + !!:.x) - j(x). Hence, for values, <1x....o, , of ~ close to 0, !!:.y/~ will be close tof(x). This is often written Ay/~ - f(x). whence, !!:.y - j'(x)!!:.x, , (21.1 ), , j(x+!!:.x) - j(x) + j'(x)!!:.x, , (21.2), , This implies, , Formula (21.2) can be used to approximate values of a func'tion., EXAMPLE 21.1:, , Let us estimate '/16.2. Let f(x) = f;,x= 16, and !!:.x = 0.2. Thenx+~x= 16.2, f(x + ~x) = '/16.2,, , and f(x) = Jf6 = 4. Since f'(x) = D,(X" 2 ) = !X-1I2 = 1I(2f;) = 11(2$6)= t, formula (21.2) becomes, , '/16.2 - 4 + *(0.2) = 4.025, , (This approximation turns out to be correct to three decimal places. To four decimal places, the correct value is, 4.0249, which can be checked on a graphing calculator.), , EXAMPLE 21.2: Let us estimate sin (0.1). Here,f(x) = sin x, x = 0, and ~x = 0.1. Then x + ~x = 0.1,, f(x + ~x) = sin (0.1), andf(x) = sin 0 = O.Sincef(x) = cos x = cos 0 = 1, formula (21.2) yields, sin(O.l) - 0 + 1(0.1) = 0.1, ,, The actual value turns out to be 0.0998, correct to four decimal places. Note that the method lIsed for this problem, shows that sin u can be approximated by u for values of 1/ close to O., , A limitation of formula (21.2) is that we have no information about how good the approximation is. For, example, if we want the approximation to be correct to four decimal places. we do not know how small !!:.X, should be chosen., , -D
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CHAPTER 21, , Differentia/s. Newton's Method, , The Differential, . The product on the right side of equation (21.1) is called the differential of/and is denoted by df, , Definition, The differential dJ ofJis defined by, dJ =f'(X)tlX, , Note that df is a function of two variables, x and Ax. If Ilx is small, then formula (21.1) becomes, f(x+ Ilx)- f(x) - df, , (21.3), , rex)., , This fonnula is illustrated in Fig. 21-1. Line :£ is tangent to the graph off at P; so its slope is, Hence,, f'(x) = RTIPR = RTlllx. Thus, RT = f'(x)llx = df. For Ax small, Q is close to P on the graph and, therefore RT - RQ, that is, df - f(x + Ilx) - f(x), which is formula (21.3)., , I df, I, , ------~R, , P(:c. f(:c», , x +~:c, , Rg.21-1, , When the functionfis given by a formula. say f(x) = tan x. then we often will write df as d (tan x). Thus,, d(tan x) = df = f'(X)!lX, Similarly, d(x3 - 2x) =(3r - 2) Ilx. In particular, if f(x), , = sec 2 x Ilx, , =x,, , dx = df = f'(x) Ilx = (I) Ilx = Ilx, , Since dx =Ilx, we obtain df =j'(x) dx. When Ilx;:j:. 0, division by Ilx yields dfl dx =j'(x). Whenf(x) is written, as y, then dfis written dy and we get the traditional notation dyldx for the derivative., If u and v are functions and c is a constant, then the following formulas are easily derivable:, d(c) =0, , d(cu) =cdu, , d(uv) =Ii dv+v du, , d(u+v)=du+dv, , d(~)= vdu-udv, 2, v, , v
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CHAPTER 21 Differentials. Newton's Method, , Newton's Method, Assume that we know that Xo is close to a solution of the equation, (21.4), , J(X)=O, , whereJis a differentiable function. Then the tangent line '!J to the graph ofJat the point with x coordinate, XQ will ordinarily intersect the x axis at a point whose x coordinate XI is closer to the solution of (21.4) than, is Xo. (See Fig. 21-2.), , y. •, , o, , Fig. 21·2, , 9ne point-slope equation of the line '!J is, , since f(xo> is the slope of ?J. If ?J intersects the x axis at (XI' 0), then, , Hence,, , XI, , J(Xo), , = Xo - f'(x ), o, , Now carry out the same reasoning, but beginning with XI instead of Xu- The result is a number X2 that, should be closer to the solution of (21.4) than XI' where x2 =XI - J(XI)!f(X I). If we keep on repeating this, procedure, we would obtain a sequence of numbers Xo, Xl' X2, ••• , X., ..• determined by the formula, J(X.), , X.+l, , =, , (21.5), , x. - f'(x.), , This is known as Newton's method for finding better and better approximations to a solution of the equationJ(x) =O. However, the method does not always work. (Some examples of the troubles that can arise are, shown in Problems 23 and 24.), EXAMPLE 21.3:, , We can approximate J3 by applying Newton's method to the functionj(x), , =r - 3. Here,f(x) =2x, , and (21.5) reads, x~, , +3, , ~, , (21.6)
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CHAPTER 21, , Differentials. Newton's Method, , Let the tirst approximation Xo be I, since we know that 1 < $ < 2. Successively substituting n = 0, 1, 2, ... in, ., ., (21.6),t we get, , .2 _, , .\ -, , 22 + 3 _7 - 1 75, 2(2) - 4 - . -, , . _ (1.75)2 + 3, .\) - 2(1.75), , 1.732142857, , X4, , =, , (1.732142857)2 + 3, 2(1.732142857), , Xs, , =, , (1.73205081)2 + 3, 2(1.73205081) = 1.732050808, , X6, , =, , (1.732050808)2 + 3, 2(1.732050808), , 1.73205081, , 1. 732050 808, , Since our calculator yielded X6 = X s' we can go no further, and we have obtained the approximation .[3 -1.732 050 808,, which is, in fact, correct to the indicated number of decimal places., , SOLVED PROBLEMS, , 1., , Use formula (21.2) to approximate: (a) !J124 ; (b) sin 61°., (a) Let f(x) = $, x = 125, and ~x =-I. Then x + III = 124, f(x + t1x) =!J124, and f(x) =Vf2S =5., Since, , )- D ( 1/3)_.1 X -2/3 _l_I__ L!"_...L, f '( xxX, -3, -3(125)2/3-35 2 -75, formula (21.2) yields !./f24 - 5 + (*)(-1) =5-* =-¥f - 4.9867. (To four decimal places, the correct answer, can be shown to be 4.9866.), (b) Letf(x) =sin X, X =rri3. and t1x =rri180. Then x + III =61 °,f(x + Ill) =sin 61°. and f(x) =.[312., Since rex) =cos x = cOS(Jr /3) =t, formula (21.2) yields, sin 61 0, , -1 +H)( ~O) 1, , 0.8660 + 0.0087 = 0:8747, , (To four decimal places, the correct answer can be shown to be 0.8746.), , 2., , Approximate the change in the volume V of a cube of side X if the side is increased by 1%.., Here,, is O.Olx,J(x) = V =x3. andfex) = 3r. By fonnula (21.1), the increase is approximately (3r)(O.OIx), O.03xl. (Thus, the volume increases by roughly 3%.), , ax, , 3., , Find dy for each of the following functions y =f(x):, , (a), , y=xl+4r-5x+6., dy = d(x 3 ) + d(4x 2 ) - d(5x) +d(6) = (3x 2+8x - 5)dx, , (b), , t, , Y =(2x3 + 5)312., dy = t(2x1 + 5) 1/2 d(2x 3 + 5) =f(2xl + 5)1/2(6x 2dx) ~ 9x 2(2x 3 + 5) I12 dx, , The computations are so tedious that a calculator, preferably a programmable calculator, should be used., , =
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CHAPTER 21, , 6., , Differentials. Newton's Method, , =, , Approximate the roots of2 cos x - Xl O., ., ', Drawing the graphs of y = 2 cos x and y = Xl, we see that there are two real roots, close to 1 and -1. (Since the, function 2 cos x - xl is even, if r is one root, the other root is -r.) Apply Newton 's method with Xo 1. Then, f(x) 2 cos x - xl andf(x) = -2 sin x - 2x = -2(x + sin x). Equation (21.5) becomes, , =, , =, , <', , j,, , X, n+l, , =x +, n, , I 2cosx - Xl, ': n, 2 x + smx, ", , I!, , x; + 2(xn sinxn + cos xu), 2(x" +sinxJ, , Then, XI -, , 1.02188593, , Xl, , -1.02168997, , Xl, , -1.021689954, , x 4 -1.021689954, A graphing calculator produces 1.021 69, which is correct to the indicated number of places. Thus, the answer, obtained by Newton's method is accurate to at least five places., , ...., , ,":.1, ', ', , ', , ~~'~~/':':~.';, 7., , Use equation (21.2) to approximate: (a) ~; (b) ~1020; (c) cos 59°; (d) tan 44°., , Ans., , 8., ....., , (a) 2.03125; (b) 3.996 88; (c) 0.5151; (d) 0.9651, , Use equation (2l.l) to approximate the change in (a) x' as X changes from 5 to 5.01; (b), to 0.98 ., , 1X, , as X changes from 1, , ,',':, , Ans., , 9., , (a) 0.75; (b) 0.02, , A circular plate expands under the influence of heat so that its radius increases from 5 to 5.06 inches. Estimate, the increase in area ., , . Ans., , 0.61t in2 - 1.88 inl, , 10. The radius of ~ ball of ice shrinks from 10 to 9.8 inches. Estimate the decrease in (a) the volume; (b) the surface area., , Ans., , (a) 80n inl; (b) 16n in2, , 11. The velocity attained by an object falling freely a distance h feet from rest is given by v = -./64.4h ft/sec., Estimate the error in vdue to an error 0[0.5 ft when h is measured as 100 ft., , Ans., , 0.2 ft/sec, , 12. If an aviator flies around the world at a distance 2 miles above the equator, estimate how many more miles he, will travel than a person who travels along the equator., AilS., , 12.6 miles
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CHAPTER 21, , Differentials. Newton's Method, , 13. The radius of a circle is to be measured and its area computed. If the radius can be measured to an accuracy of 0.00 1, in and the area must be accurate to 0.1 inches 2, estimate the maximum radius for which this process can be used., , Ans., , 16 in, , 14. If pV =20 and p is measured as 5 ± 0.02, estimate V., , Ans., , V = 4 ± 0.016, , 15. If F = IIr and F is measured as 4 ± 0.05, estimate r., , Ans., , 0.5 ± 0.003, , 16. Estimate the change in the total surface of a right circular cone when (a) the radius r remains constant while, the height h changes by a small amount MI; (b) the height remains constant while the radius changes by a small, , amountl::1r., Ans., , 2, r2 +h2, , (a) rcrhMlI.Jr2 +h2 ; (b) rc(/J+2r +2r)l::1r, , 17. Find dy for each of the following:, , (a) y =(5 _X)3, sinx, (b) y = x, , AilS., AilS., , (c) Y =cos- t (2x), , AilS., , (d) y =cos (br), , AilS., , X)2 dx, xcosx-sinx dx, x2, , -3{5 -, , -2 dx, JI-4x 2, -2bx sin (br) dx, , 18. Find dy/dx in the following examples by using differentials:, (a) 2xyl+3ry= I, , AilS., , (b) xy=sin(x-y), , Ans., , 2y(yl +3x), 3x(2y2+x), cos{x-y)-y, ) +x, cos (, x-y, , 19. (GC) Use Newton's method to find the solutions of the following equations. to four decimal places:, (a) xl+3x+ I =0, , (b) x - cos x =0, (c) xl + 2x2 - 4 = 0, , AilS., AilS., AilS., , ~.3222, , 0.7391, 1.1304, , 20. (GC) Use Newton's method to approximate the following to four decimal places:, (a), , :J3, , (b), , :.jill, , ,., , AilS., AilS., , 1.3161, 3.0098, , 21. (a) Verify that Newton's method for calculating ..r, yields the equation xn • t, (b) (GC) Apply part (a) to approximate J5 to four decimal places., , AilS., , (b) 2.2361, , =!(x. + ;. J
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CHAPTER 21, , Differentials. Newton's Method, , 22. (GC) Show that x 3 + r - 3 = 0 has a unique solution in (1,2) and use Newton's method to approximate it to four, decimal places., , Ans., , 1.1746, , 23. Show that Newton's method does not work if it is applied to the equation X" 3 = 0, with Xo = 1., , ,, , 24. Show that Newton's method does not give approximations to the solutions of the following equations. starting, with the given initial values, and explain why it docs lIot work in those cases., (a) .xl - 3r + 3x + 2 = O. with Xo = I., (b) x3 - 3x2 + x-I = O. with Xo = I., (c), , f(x), , ={, , Jx-2, , forx~2, , -J2-x, , for x < 2, , ., , with -"0 = 3, , 25. (GC) Approximate 1t by using Newton's method to find a solution of cos x + 1 = O., Ans., , 3.141592654. (Note how long it takes for the answer to stabilize.), , 26. (GC) Use Newton's method to esti~ate the unique positive solution of cos x = ~., , Ans., , 1.029866529
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Antide riva tives, If F'(X) =f(x~, then F is called an antiderivative off., EXAMPLE 22.1:, , xl is an antiderivative of 3r, since D,(xl) =3r. But xl + 5 is also an antiderivative of 3r, since, , D.(5) =O., , In general, if F(x) is an antiderivative off(x), then F(x) + C is also an anti derivative ofj{x), where C, is any constant., (II) On the other hand, if F(x) is an antiderivative off(x), and if G(x) is any other antiderivative ofj{x), then, G(x) =F(x) + C, for some constant C., (I), , Property (II) follows from Problem 13 of Chapter 18, since F'(x) = f(x) =G'(x)., From Properties (I) and (II) we see that, if F(x) is an antiderivative off(x), then the antiderivatives off(x), are precisely those functions of the form F(x) + C. for an arbitrary constant C., , ff(x)dx will denote any antiderivative off(x). In this notation,f(x) is called the integrand., , Notation:, , Terminology:, , An antiderivative f f(x)dx is also called an indefinite integral., , An explanation of the peculiar notation f f(x)dx (including the presence of the differential dx) will be, given later., EXAMPLE 22:2:, , (a) fxdx=tx2+C;(b) f-sinxdx=cosx+C., , Laws for Antlderlvatives, Law 1., , f Odx= C., , Law 2., , fldx =x+c., , Law 3., , f adx =ax+C., , Law 4., , f, , X,+l, , x'dx = r + I +C for any rational number r'* -I., , (4) follows from the fact that D. ( ::~ ) =x' for r,* -I., , LawS., , faf(.t)dx=aff(x)dx., , Note that D. (a f f(x)dx) =aD. (f f(x)dx) =af(x)., Law 6., , f (J(x) + g(x»dx = f f(x)+dx+ f g(x)dx., , Note that D., Law 7., , (f f(x)dx+ f g(x)dx) =D. (f f(x)dx) + D. (f g(x)dx) =f(x)+ g(x)., , f (J(x)- g(x»dx= f f(x)dx- f g(x)dx., , Note that D.(f f(x)dx- f g(X)dT) = D.(f f(x)dx)- D.(f g(X)dT) = f(x)- g(x)., , ----~.lI!m!:m-.
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CHAPTER 22 Antiderivatives, , 17., , J sec (4x, 2, , X, , 2, , -, , 5)dx., , Let u =4x2 - 5. Then du =8x dx, tdu =x dx. Thus,, , Jxsec (4x, 2, , 18., , 2, , -, , J, , 5) dx =t sec 2 u du =ttanu +C =ttan(4x 2 - 5) +C, , Jx .Jx+ ldx., 2, , Let u =x + 1. Then du =dx and x =u - 1. Thus,, , Jx .Jx +1dx = J(u -1)2.JU du = J(u, 2, , =J(U 512 -, , 2, , -, , 2u + I)U 1l2 du, , 2U 312 + U1l2 ) du =tu 112 - 2(-t)U 512 + tu 3/2 + C, , =2U 312 (tu 2 -, , tu + t>+ C, , =2(x + 1)3/2 [+(x + 1)2 - t(x + 1) + t1 +C, 19. A stone is thrown straight up from the ground with an initial velocity of 64 ftlsec. (a) When does it reach its, maximum height? (b) What is its maximum height? (c) When does it hit the ground? (d) What is its velocity, when it hits the ground?, , J, , J, v = J-32.dl = -321 + C, , In free-fall problems, v = a dl and $ = v dl because a = ~~ and v =:. Since a =-32 ft/sec 2,, , 1, , Letting 1= 0, we see that C1 = 110, the initial velocity_atl =0. Thus, v = -321 + 110. Hence,, , Letting 1= 0, we see that C2 =$0' the initial position at I = 0. Hence, $, , =-16/ 2 + vol + $0, , In this problem, $0 = 0 and 110 = 64. So,, v=-32/+64. $=-16t 2 +64t, , (a), (b), (c), (d), , °, , At the maximum height, ~: =v =0. So, -32t + 64 = and, therefore, 1=2 seconds., When I =2, $ = -16(2)2 + 64(2) = 64 ft, the maximum height., When the stone hits the ground, 0 =$ =-16t2 + 64t. Dividing by I, =-16/+ 64 and, therefore, t= 4., When 1=4, v=-32(4)+64=-64ft/sec., , °, , 20. Find an equation of the curve passing through the point (3, 2) and having slope 5x2 - x + 1 at every point (x, y)., Since the slope is the derivative, dyldx = 5x2 - x + I.. Hence,, , J, , y= (5x 2 -x+ l)dx =t x3 -tx2 +x+C, , Since (3, 2) is on the curve, 2 =t(3)l- t(3)2 + 3+ C =45 curve is ~, , y = t x3 21. Justify the Substitution Method:, , t +3 +C. So, C =-¥. Hence, an equation of the, , t x2 +X - ¥, , Jf(g(x»g'(x)dx = Jfeu) duo, , Here, u = g(x) and duldx= g'(x). By the Chain Rule,, D., , (J feu) dU)= D. (J feu) dU). 'i: =f(u)· 'i: =f(g(x»· g'(x)
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CHAPTER 22, , Antiderivatives, , In Problems 53-64, use any method., , 55., , (x+l)dx, , fiF:==:==, "x +2x-4, 2, , Ans., , tx4 -tx3 +tx 2 +C, , Ails., , !ex! _x)S +C, , Ails., , JX2 +2x-4 +C, , Ans., , t<1+Ji)3+C, , 58., , fsec 3x tan 3x dx, , Ans., , tsec3x+C, , 59., , fcsc (2x) dx, , Ails., , -tcot2x+ C, , 60., , fx sec, , Ans., , tanx-x+C, , 2, , 2, , 2, (X ), , dx, , 61. ftan2~dx, , 62., , fcos x sinx dx, , Ails., , -tcossx+C, , 63., , f J5-x, dx, , Ans., , sin-I ( x f ) + C, , 64., , 2, f l-4tan, sec xdx, x, , Ails., , tsin- ' (2tanx)+C, , 4, , 2, , 2, , 65.. A stone is thrown straight up from a building ledge that is 120 ft above the ground, with an initial velocity of, 96 ft/sec. (a) When will it reach its maximum height? (b) What will its maximum height be? (c) When will it hit, the ground? (d) With what speed will it hit the ground?, Ans., , (a) t, , =3 sec; (b) 264 ft; (c), , 6+:f6 -7.06sec; (d) -129.98 ft/sec ., , 66. An object moves on the x axis with acceleration a = 3t - 2 ftlsec 2. At time t =0, it is at the origin and moving, with a speed of 5 ft/sec in the negative direction. (a) Find a formula for its velocity v. (b) Find a formula for its, position x. (c) When and where does it change direction? (d) At what times is it moving toward the right?, Ans., , (a)v=ft2-2t-5;(b)x=tt3-t2-5t;(c), , 2+..J34, 2-..J34, 3 ; (d) t> 2+..J34, 3, or t<, 3, , 67. A rocket shot straight up from the ground hits the ground 8 seconds later. (a) What was its initial velocity?, (b) What was its maximum height?, , Ans., , (a) 128 ft/sec; (b) 256 ft
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CHAPTER 22 Antiderivatives, , 68. A driver applies the brakes on a car going at 55 miles per hour on a straight road. The brakes cause a constant, deceleration of 11 ft/sec2• (a) How soon will the car stop? (b) How far does the car move after the brakes were, applied?, ', , Ans. (a) 5 sec; (b) 137.5 ft, 69. Find the equation of a curve going through the point (3,7) and having slope 4x2 - 3 at (x, y).
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The Definite Integra/., Area Under a Curve, Sigma Notation, The Greek capital letter L denotes repeated addition., EXAMPLE 23.1:, s, , (a), , L,j = 1+2+3+4+5= 15., i"1, J, , (b) L,(2i+l)=1+3+5+7., 10, , (c) L,i~ =22+3 2+ ... +(10)2, 4, , (d), , L,, , COS j1C, , = COS1C + cos21l' + cos31l' + COS41C, , i=1, , In general, if fis a function defined on the integers, and if nand k are integers such that II. ~ k, then:, n, , Lf(j) = f(k)+ f(k+ 1)+···+ f(lI), j-k, , Area Under a Curve, Assume thatfis a function such thatf(x) ~ 0 for all x in a closed interval [a, b). Its graph is a curve that, lies on or above the x axis. (See Fig. 23-1.) We have an intuitive idea of the area A of the region ~ under, the graph, above the x axis, and between the vertical lines x = a and x = b. We shall specify a method for, evaluating A., Choose points XI' x2, ••• , Xn-I between a and b. Let Xo = a and Xn = b. Thus (see Fig. 23-2),, , The interval [a, b) is divided into n subintervals [xo, XI], [XI' X2], ... , [xn _ l • xn). Denote the lengths of these, subintervals by ~IX, ~~.... , ~,,x. Hence, if I :5 k:5 n,
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CHAPTER 23 The Definite Integral. Area Under a Curve, , Fig. 23-1, , .to-I, , x.., , ;r, , b, , Fig. 23-2, , Draw vertical line segments x =Xl from the x axis up to the graph. This divides the region minto n strips., Letting L\~ denote the area of the kth strip, we obtain, , We can approximate the area L\kA in the following manner. Select any point xl in the kth subinterval, [xl-\, Xk]' Draw the vertical line segment from the point Xl on the x axis up to the graph (see the dashed lines, in Fig. 23-3); the length of this segment is !(x;). The rectangle with base ~~ and height !(x;) has area, !(xi) L\~, which is approximately the area L\kA of the kth strip. Hence, the total area A under the curve is, approximately the sum, , •, , I,!(XZ) L\k x =!(Xj) L\IX+ !(x~) L\2X+'", l=1, , +!(x~) L\.x, , (23.1)
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The Definite Integral. Area Under a Curve, , CHAPTER 23, , y, , x~ b, , x, , Fig. 23-3, , The approximation becomes better and better as we divide the interval [a, b) into more and more subintervals and as we make the lengths of these subintervals smaller and smaller. If successive approximations, can be made as close as one wishes to a specific number, then that number will be denoted by, , I: /(x)dx, and will be called the definite integral of/from a to b. Such a number does 1I0t exist in all cases, but it does, exist, for example, when the functionfis continuous on [a, b). When a f(x) dx exists, its value is equal to, the area A under the curve. t, In the notation /(x) dx, b is called the upper limit and a is called the lower limit of the definite, a, integral., For any (not necessarily nonnegative) function/on [a, b], sums of the form (23.1) can be defined, without, using the notion of area. If there is a number to which these sums can be made as close as we wish, as n gets, , J", , r, , larger and larger and as the maximum of the lengths ~ approaches 0, then that number is denoted JI> f(x) dx, , J, b, , a, , and is called the definite integral of/on [a, b]. When /(x) dx exists, we say that/is integrable on [a, b]., b, a, We sha!l.assume without proof that fa /(x) dx exists for every function/that is continuous on [a, b]. To, evaluate /(x) dx, it suffices to find the limit of a sequence of sums (23.1) for which the number n of subintervals approaches infinity and the maximum lengths of the subintervals approach O., , fa, , EXAMPLE 23.2:, , Let us show that, , I:ldx=b-a, Let a =Xo < XI < X 2 <...< X,,_I < x", , =b, , (23.2), , be a subdivision of [a, b]. Then a corresponding sum (23.1) is, , 11, , n, , L!(XI)L1 I X= LL1 k x ', i.1, , k=1, , (because !(x) = I for all x), , =b-a, Since every approximating sum is b - a,, , I: I dx =b - a., , tThe definite integral is also called the Riemann integral of/on [a., , bl. and the sum (23.1) is called a Riemann sum tor/on [a. bl.
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CHAPTER 23 The Definite Integral. Area Under a Curve, , An alternative argument would use the fact that the region under the graph of the constant function 1 and, above the x axis, between x =a and x =b, is a rectangle with base b - a and height 1 (see Fig. 23-4). So,, 1dx, being the area of that rectangle, is b - a., , I:, , y, , ,, , .,:, , a, , b, , x, , Fig. 23-4, EXAMPLE 23.3:, , Let us calculate, , I: x, , dx., , Let a = Xo < XI < x2<,,,<x._1 <x. = b be a subdivision of [a, b] into n equal subintervals. Thus, each 8~=, (b - a)ln. Denote (b - a)ln by 8x. Then XI = a + 8x, X:! = a + 28x, and, in general, Xi = a + k 8x. In the kth subinterval,, [xH Xt], choose x: to be the right-hand endpoint Xi' Then the approximating sum (23.1) has the form, •, , n, , f(X,)8 k X= LX,8 k X= L(a+k .it).it, k=1, , k-I, •, , •, , n, , = L(a.it+k(8x)2)= La.it+ Lk(8x)2, .1=1, , .1=1, , t=1, , +, , = n(a .it) (.it)2 t.k =n(a b~ a, , )+( b~a, , n, , n(n +, 2, , 1»), , =a(b-a)+!(b-a)2 n;1, ~, n(n+ 1), , Here we have used the fact that .4.J k =- 2 - ' (See Problem 5.), .1=1, , Now, as n -7 00, (n + 1)/n =1 + lin -7 1 + 0 = 1. Hence, the limit of our approximating sums is, , In the next chapter, we will fmd a method for calculating, computation used in this example., , r, , f(x) dx that will avoid the kind of tedious, , Q, , Properties of the 'Definite Integral, , I: c f(x) dx =cI: f(x) dx, n, , This follows from the fact that an approximating sum, , proximating sum, , n, , b, , k-I, , a, , (23.3), b, , L cf(x: ) Ai; for fa cf(x) dx is equal to c times the api=1, , L f(xI) 8,tX for Jf(x) dx, and that the same relation holds for the corresponding limits., , I: -f(x) dx =- 1: f(x) dx, This is the special case of (23.3) when c =-1., , (23.4)
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CHAPTER 23 The Definite Integral. Area Under a Curve, , (23.5), , I:(f(x)+g(x»dx= I:f(x)dx+ J:g(x)dx, , J(f(x) +g(x» dx is equal, /I", to the sum L.J(xn ~kX +2, g(xO ~kX of approximating sums for J f(x) dx and Jg(x) dx., n ., , This follows from the fact that an approximating sum 2, (f(xZ) +g(xz)) ~kX for, l=1, , k=1, , k=l, , b, , b, , ab, , a, , a, , I:(f(x)-g(x»dx= I:f(x)dx- J:g(x)dx, , (23.6), , Sincef(x) - g(x) =f(x) + (-g(x), this follows from (23.5) and (23.4)., If a < c < b, thenfis integrable on [a, b] if and only if it is integrable on [a, e] and [e, b]. Moreover, iff, is integrable on [a, b],, , I:, , f(x) dx =, , r, , f(x) dx+, , r, , f(x) dx, , (23.7), , This is obvious whenf(x) ~ 0 and we interpret the integrals as areas. The general result follows from looking, at the corresponding approximating sums, although the case where one of the· subintervals of [a. b] contains, c requires some extra thought., ., We have defined, follows:, , 1:, , f(x) dx only when a < b. We can extend the definition to' all possible cases as, , (i), , J: f(x) dx =0, , (ii), , Jab f(x) dx =_Jb f(x) dx when a < b, II, , In particular, we always have:, , 1:, , f(x) dx =-, , I: f(x) dx, , for any c and d, , (23.8), , It can readily be verified that the laws (23.2)-(23.6), the equation in (23.7), and the result of Example 23.3, all remain valid for arbitrary upper and lower limits in the integrals., , SOLVED PROBLEMS, , 1., , r, , Assume I(x) ~ 0 for all x in [a, b]. Let A be the area between the graph ofI and the x axis, from x = a to x, (See Fig. 23-5.) Show that, , I(x) dx = -J ., , x, , a, , y=j{x), , Fig. 23-5, , =b.
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CHAPTER 23, , f, , Let B be the. area between the graph of -f and the x axis, from x =a to x =b. By symmetry, B =A. But,, f(x) dx =f(x) dx by (23.4)., , 1:-, , f, , Since, 2., , ••, , The Definite Integral. Area Under a Curve, , f, , -f(x) dx =B,, , f(x) dx =-B =-A, , r, , Consider a functionfthat, between a and b, assumes both positive and negative values. For example, let its graph, be as in Fig. 23-6. Then, , f(x) dx is the difference between the sum of the areas above the x axis and below the, , graph and the sum of the areas below the x axis and above the graph. In the case of the graph shown in Fig. 23-6,, , b ."',,1"'--', , ",, , 'J, , Rg.23-6, , To see this. apply (23.7) and Problem 1:, , b, Iq, f~, J~, f~, Jb, Ja f(x)dx = " f(x) dx+ c, f(x)dx+ f(x) dx+ c, f(x) dx+ c, f(x) dx=A , CI, , 3., , A2 + A) - A4 + As, , Assume thatfand g are integrable on [a, b). Prove:, (a) Iff(x)~Oon [a, b], then ff(x)dx~O., (b) Iff(x), , r, , r, , ~ g(x) on [a. b), then f(x) dx ~ g(x) dx., , (c) If m ~f(x) S; M for all x in [a, b), then m(b - a) ~, , r, , f(x) dx S; M(b - a)., t',J, , (a) Since every approximating sum I,f(x;) dkx ~ 0, it follows that, k=l, , ff(x)dx~O, (b) g(x) - f~x) ~ 0 on [a. b). So, by (a),, , r, , (g(x) - f(x» dx ~ O. By (23.6),, , f g(x) dx - rf(x) dx ~ O. Hence,, , 1: f(x) dT::; 1: g(x) dx, (c) By (b),, b, , J: m dx~ 1:f(x)S; J: M dx, , J• Mdx=M, , . But. by (23.2) and (23.3),, , 1: mdx= mJ:ldx=m(b-a)and, , Jb"ldT=M(b-a).Hence., ., m(b-a)~ rf(x)dx~M(b-a), , ,
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CHAPTER 23, , 4., , I;, , Evaluate, , The Definite Integral. Area Under a Curve, , x 2 dx., , '*], , This is the area under the parabolay = xl from x = 0 tox:::; I. Divide [0, 1] into n equal subintervals. Thus,, , each !;.,.x c 1/n. In the kth subinterval [ k ~ I, , ,let x; be the right endpoi~t kin. Thus, the approximating sum, , (23.1) is', , ', , ,, , Now,, , ~, , £,.; k 2, , =, , n(n + 1)(2n + I), , 6, , (see Problem 12)., , k~1, , Hence,, , ~ f(x')!;. x=...L3 n(n+l)(2n+1) l(n+l)(2n+l), £,.;, k=1, , k, , n, , k, , 6, , 6, , n, , n, , 2, So, the approximating sums approach t(1 + 0)(2 + 0) = t as n ~ 00. Therefore,, ,, Jro x dx =, will derive a simpler method for obtaining the same result, l, , 5,, , ·;', , t. In the next chapter, 'we, , ~, n(n+ I), ., Prove the formula £,.;k =- 2 - used 10 Example 23.3., k-I, , Reversing the order of the summands in, n, , ~) = I + 2+3+"'+(n-2)+(n-I)+n, , ~'f¥~~'", :;, ,, , k=1, , we get, n, , '-, , ., , .., , ~) =n + (n - I) + (n - 2) + ... + 3 + 2 + 1., -, , hi, , ,-, , ~-':~-t, , ~:,;-d.:;:-;., , ,~;~1~>:~, ·..... id, , Adding the two equations yields, , e, ,, , ••, , ", , :~~., , 2~ k =(n + l) + (n + l) + (n + I) + .. , + (n + lJ +(n + 1) + (n + 1) = n(n + I), k-I, , ~ii', , since the sum in each column is n + I. Hence, dividing by 2, we get, , .'••• to,,,,:", , -,"-, , .~, , -k··-, , ., , 6., , Calculate: (a), AIlS., , 7., , (b), , JS-2 xdx:, , (c) rl3x2dx., , Jo, , (a) 3(4-1)=9; (b) t(52_(-2)2)=1l; (c) 3(t)=1, , Find the area under the parabola y =xl - 2x + 2, above the x axis, and between x = 0 and x = I., Ans., , 8,, , J4I 3 dr., , +-2[tW-0 2 »)+2(1-0)=t, , Evaluate, , J:(3x+ 4) dx.
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CHAPTER 23, , 9., , The Definite Integral. Area Under a Curve, , For the function! graphed in Fig. 23-7, express, , 10. Show that 3 ~, , f:, , f(x) dx in terms of the areas AI' A2, and A3•, , r, , x 3 dx ~ 192. [Hint: Problem 3(c).], , 11. Evaluate f~ ~1- x 2 dx. (Hint: Find the correspon~ng area by geometric reasoning.), , ;iJ.:;~i, , iii, , Ans. 7tl4, y, , ----~----------~------_7------~-X, , Fig. 23-7, 12. Use mathematical induction to prove the formula ~)2, , =n(n + 1~2n + 1) of Problem 4. (Verify it when n = 1, and, , kal, , then show that, if it holds for n, then it holds for n + 1.), ,, , 2', , 2, , n, , 100, , 18, , 13. Evaluate (a) Lcos 1 ; (b) L(4j+l); (c) L4j; (d) L2P., 6, j=O, j=O, j=1, j=1, , Ans. (a) 3 \$; (b) 15; (c) 20200; (d) 4218, 14. Let the graph ofJbetween x = 1 and x = 6 be as in Fig. 23-8. Evaluate, , r, , J(x) dx., , Ans. 1-3+t=-t, y, , 2, , ----~o~--~--+---~--~-,~~-------+x, , -\, , -2, , Fig. 23-8, 15. IfJis continuous on [a, b],f(x) ~ 0 on [a, b), andJ(xo) > 0 for some Xo in [a, b), prove that, [Hint: By the continuity off, f(x) > t J(xo) > 0 for all x in some subinterval [c,, , J: f(x) dx > O., , dJ. Use (23.7) and Problem 3(a, c).]
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The Fundamental Theorem, of Calculus, Mean-Value Theorem for Integrals, Letfbe continuous on [a, b]. Then there exists c in [a, b] such that, , J: f(x)dx= (b-a)f(c), , (24.1 ), , To see this,let m and M be the minimum and maximum values offin [a, b], and apply Problem 3(c) of, Chapter 23 to obtain, ., m(b-a) $, , r, , I fb, 'm$ b-a "f(x)dx$M, , and, therefore,, , f(x) dx $ M(b-a), , So, by the intemlediatc value theorm, - 1b fb f(x) dx = f(c) for some c in [a, b]., -a a, , Average Value of a Function on a Closed Interval, Letfbe defined on [a, b]. Sincefmay assume infinitely many values on [a, b], we cannot talk about the, average of all of the values off. Instead, divide [a, b] into n equal subintervals, each of Il.x = b-a. Select an, n, , arbitrary point x; in the kth subinterval. Then the average of the n values f(x;), f(xi), ... ,f(x;) is, f(x;) + f(xi) + ... + f(x;), n, , =.!. ~ f(x'), n£.J, 1:=1, , I:, , When n is large, this value is intuitively a good estimate of the "average value offon [a, .~J." However, since, I, I, -=--Il.x,, 11, b-a, , I n, I n, - Lf(x;)=-b_ Lf(x;)ll.x, , n 1:=1, , As 11 ~ 00, the sum on the right approaches, Definition:, , a 1:=1, , 1:, , f(x) dx. This suggests the following definition., , The average value of/on [a, b] is - b, I Jb f(x) dx., , -a, , a, , Letfbe continuous on [a, b]. If x is in [a, b], then, Dx, , For a proof, see Problem 4., , U:, , r, , f(t) dt is a function of x, and:, , /(t) dt) = /(x), , (24.2)
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CHAPTER24 The Fundamental Theorem of Calculus, , Fundamental Theorem of Calculus, , f, , Letfbe continuous on [a, b], and let F(x) = f(x) dx, that is, F is an antiderivative ofJ. Then, , I: f(x)dx =F(b)- F(a), To see this, note that, by (24.2),, , r, , f(t) dt and F(x) have the same derivative,f(x). Hence, by Problem 18, x, , Q, , of Chapter 13, there is a constant K such that, , fa f(t) dt =F(x) + K. When x =a, we get, , F(a)+K= f:f(t)dt=O, , Hence,, , (24.3), , So,, , K==I-F(aY, , f: f(t) dt =F(x)- F(a). When x =b, this yields, , I: f(t) dt =F(b)- F(a), , r, , Equation (24.3) provides a simple way of computing f(x) dx when we can find an antiderivative F ofJ., The expression F(b) - F(a) on the right side of (24.3) is often abbreviated as F(x)]:. Then the fundamental, theorem of calculus can be written as follows:, , I: f(x) dx ff(x) dx (, =, , EXAMPLE 24.1:, , (i) The complicated evaluation of, one:, , 1:xdx in Example 23.3 of Chapter 23 can be replaced by the following simple, , (ij) The very tedious computation of f>2dx in Problem 4' of Chapter 23 can be replaced by, , (iii) In general, Jb x'dx= _1_Xr+I]b =_l_(b,+1 _a r +l), a, r+ 1, a, r+ 1, , for r *-1, , Change of Variable in a Definite Integral, , J, , In the computation of a definite integral by the fundamental theorem, an antiderivative f(x) dx is required., In Chapter 22, we saw that substitution of a new variable II is sometimes useful in finding f(x) dx. When, the substitution also is made in the definite integral, the limits of integration must be replaced by the corresponding values of u., , J
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The Fundamental Theorem of Calculus, , CHAPTER 24, , Evaluate, , EXAMPLE 24.2:, , r, , -/5x+4 dx., , Let u = 5x + 4. Then du = 5, , dx. When x =1,, , U, , = 9, and when x = 9,, , U, , =49. Hence,, , For justification of this method, see Problem 5., , SOLVED PROBLEMS, I., , rl(12, , Evaluate J, , o, , sin 2 xcosx dx., , Jsin xcosx d.x =tsin x by Quick Formula I. Hence, by the fundamental theorem,, 2, , 3, , r, , J:'2 sin 2xcosx d.x= tsin 3X]:12 = t[(sin~, 2., , Find the area under the graph of f(x) = :tb ' above the x axis, and between 0 and 1., The area is, , 3., , -(SinO»)] =HP - 0 3)=t, , J~ h, , d.x =sin-I (, , t)l =, , sin-I (t)-sin-I(O) =t, , - 0 =t·, , Find the average value of f(x) = 4 - xl on [0, 2]., The average value is, , b~a f f(x)dr= t S: (4-X2)dx=t(4X-~)I =t[(8-t)-(0-0)]= t, 4., , Prove fonnula (24.2): D. ([ f(t) dl) = f(x), Let hex) =, , 1: f(l) dl. Then:, , C, I1, , h(x+l1x)-h(x) =, =, , 'f(l)dl- J:f(t)dt, , s: f(t) dl + [+'''f(1) dt - s: f(t) dt, , (by 23.7), , ··11x, =", S f(l) dt, = Ilx· f(x'), , - hex), Thus, hex + l1x), Ilx, , = f('), x, , for some x' between x and x + Ilx (by the mean value, theorem for integrals), , t', an d th erelore,, , D, (IX fU) dt)= D.(h(x)) = lim h(x+ Il;) - hex) = lim f(x'), II, , A.t-tO, , X, , ,0,,, ~O, , But, as Ilx -+ 0, x + Ilx -+ x and so, x· -+ x (since x· is belweenx and x + Ilx). Sincefis continuous,, lim f(x') =f(x)., , 11.....0
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CHAPTER 24 The Fundamental Theorem of Calculus, , S., , Justify a change of variable in a definite integral in the following precise sense. Given, , J: I(x) dx. let x =g(u), , where. as x varies from a to b. u increases or decreases from c to d. (See Fig. 24-1 for the case where u is, increasing.) Show that I, , r, , f(x) dx =, , r, , f(g(u»g'(u) du, , (The right side is obtained by substituting g(u) for x, g'(u) du for dx. and changing the limits of integration from a, and b to c and d.), b, , Q, , c, , d, , &I, , Fig. 24-1, , J, , Let F(x) = I(x) dx. that is. rex) = j(x). By the Chain Rule., ", , Thus., , D.(F(g(u» = F'(g(u»· g'(u) = f(g(u»g'(u), , Jf(g(u»g'(u) du =F(g(u», , So, by the fundamental theorem., d, , J, <, , d, , I(g(u»g'(u) du =F(g(u}}],, , =F(g(d»- F(g(c», =F(b}':' F(a} =I:f(x)dx, , 6., , (a), , J:, , Iffis an even function. show that. for a> O. f/(x) dx =2 f(x) dx:, , (b) Iffis an odd function. show that. for a> O. f/(x) dx =O., Let u = -x. Then du = -dx, and, , 1 f(x)dx=Jof(-u)(-I)du=-Jof(-u)du= Jr"f(-u)du, 0, , -d, , o, , G, , U, , Rewriting u as x in the last integral. we have:, , (*), , f/(x)dx= f:I(-x)dx, , Thus., , J:, , [/(x) dx =f/(x) dx + f(x) dx (by (23.7», , J:, J:, , =, =, (a) Iffis even,f(-x) +f(x), , =2f(x). whence, , (b) Iffisodd,f(-'~)+f(x}=O, whence, , I(-x) dx+, , f:, , I(x) dx (by, , f(-x) + f(x)dx, , f:, , (.», , (by 23.5», , f:, , [/(x) dx = 2f(x) dx =2 f(x) dx., , rQ, f"-" f(x)dx= Jr"Odx=O, ldx=O., o, Jo
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CHAPTER 24, , 7., , The Fundamental Theorem of Calculus, , Trapezoidal Rule, (a) Letf(x) ~ 0 on [a, b). Divide [a, b) into n equal parts, each of length!u, , xn_ l • (See Fig. 24-2(a).) Prove the following trapezoidal rule:, (b) Use the trapezoidal rule with n = 10 t~ approximate, , r, , =b ~ a, by means of poi,nts X I ,.\2, ••• ,, , f(x) dx -, , ~X(f(a)+ 2~f(Xt)+ f(b»), , f>2 dx., , (a) The area of the strip, over [XH Xl], is approximately the area of trapezoid ABeD (in Fig. 24-2(b»:,, ~X (f(x t _ l ) + f(x k »t (Remember that Xo = a and X. = b.) So, the area under the curve is approximated by, the sum of the trapezoidal areas,, , t, , .-1, , ~{ ([f(xo) + f(x l ») + [f(x t ) + f(x 2 )] + ... + [f(x._ t ) + f(x.)]} = ~x[f(a) + 2IJ(xk ) + f(b)], ., , r-t, , y, , .x, , (b), , (a), , Fig. 24-2, , (b) With, , ;;L~, , <'"If-~:, , /I, , = 10, a =0, b = I, !u =i, , and X t, , =kIlO, we get, , P+12)=201(29, ), 1x dx- 20I(02 +2I, 100, lOOLP+l, t, , I, , 2, , 9, , o, , . ....;.: .., ~·{~I, :', , t-I, , =, , ~i~~;, , do [I ~ (285) + I], , pi, , (by Problem 12 of Chapter 23), , =0.335, The exact value is t (by Example 24.1 (ii»", , In Problems 8-22, use the fundamental theorem of calculus to evaluate the definite integral., 8., , r, , 9., , r(-x1- -x1) dx, , Ans. .Il", 9, , 10., , fdx, ITx, , Ans. 2, , -I, , -3, , (2x 2 -, , 2, , Xl), , dx, , 3, , t Recall that the area of a trapezoid of height h and bases b l and bz is th(b l + b2 )., , AilS., , t
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CHAPTER 24 The Fundamental Theorem of Calculus, , 11., , 1, , 3, "'., , 1(12, , sin x cit, , Ans., , 12. J:(2+x).cit, , I, , 13. J:(2-X)2 cit, 14., , J: (3 - 2x+ x, , Ans. 6, Ans., , 2, , ) cit, , :Ii, 2, , t, , Ans. 9, , 16., , t, f, , 17., , r../1, , 18., , J: 2 + I) cit, , 19., , f "Ji+;, 1 cit, , Ans. 2, , 20., , J>(I- JX)2 cit, , Ans., , 21., , t ../x x-IS dx, , Ans. 6, , 22., , J:I( sin!dt, , Ans. 4, , 15., , (1- t 2)t dt ., , Ans., , -t, , (1- u)Jii du, , Ans., , -W-, , + 3x cit, , X (X, , 3, , o, , 2, , 4, , Ans. 26, Ans., , ~, , 3, , I, , 10, , In Problems 23-26, use Problem 6(a, b)., 23., , r, , --dx, cit, , _2X1 +4, , 24. t(x3 -xS)dx, , Ans., , n, , '4, , Ans. 0, , 25., , f3 sin jcit, , Ans. 0, , 26., , f"2, -1(/2 cosxcit, , Ans. 2, , 27. Prove: D.(f f(t)dt)=-f(X)., 28. Prove. D, •, , (('(.) f(t) dt )=f(g(x»g'(x) - f(h(x»h'(x)., h(.), , In Problems 29-32, use Problems 27-28 and (24.2) to find the given derivative., 29. D., , U: sint dt), , 30. D.(I:t2 dt), , Ans. sinx, Ans. -xl
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CHAPTER 24 .The Fundamental Theorem of Calculus, , -31. D.(J;in'tJdt), 32., , Ans. sin] x cos x ., , Dx(L~' cost dt), , Ans. 4 cos 4x - 2x cos xl, , 33. Compute the average value of the following functions on the indicated intervals., (a) f(x) = <IX on [0, 1), , Ails., , (b), , Ans., , f(x) = sec 2 x on, , [0,1'], , =3x 2 -Ion [-I, 4], f(x) =sinx-cosxon[O,1t'], , f, , Hi, 1t', , (c) f(x), , Ans. 12, , (d), , Ans., , 34. Use the change-oC-variables method to find, Ans., , 11t', , J3 ../2x +3 x dx., 112, , Jf, , 35. An object moves along the x axis fC?r a period of time T. If its initial position is XI and its final position is x2, show, ', that .Its average veIoclty, was x2 T-XI ., ., 36 Let f(x), •, , AilS., , ={cos, x, I-x, , Cor x < 0. Evaluate, , for x~O, , r, , f(x) dx., , -1(/2, , t, , I, 37. Evaluate lim-,, h-iO I, , fl." ' -5+7 dx., J, , X, , ·,:·,"t, , Ans., , 14, , 38. (Midpoint Rule) In an approximating sum (23.1) iJ(x;)t1 l x, if we select, , x; to be the midpoint of the kth, , lal, , subinterval, then the sum is said to be obtained by the midpoint rule. Apply the midpoint rule to approximate, , f~ x 2 dx, using a division into five equal subintervals, and compare with the exact result of t., , :!0t~i, , :f~~', '", '~., , ~, , Ans., , 0.33, , 39. (Simpsoll's Rule) If we divide, sum for f(x) dx,, , J:, , la, b] into" equal subintervals, where" is even, the following approximating, , I •• .;.,;: • ., ;p~, , -, , is said to be obtained by Simpson's rule. Except for the first and last terms, the coefficients consist of alternating, 4s and 2s. (The basic idea is to use parabolas as approximating arcs instead of line segments as in the trapezoidal, rule. Simpson's rule is usually much more accurate than the midpoint or trapezoidal rule.), Apply Simpson's rule to approximate (a) J~ x 2 dx and (b), the answers obtained by the fundamental theorem., , Ans., , (a), , J: sinx dx with n = 4, and compare the results with, , t, which is the exact answer; (b) 7;(2..[i + I) - 2.0046 as compared to 2
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CHAPTER 24, , The Fundamental Theorem of Calculus, , J>3, , 40. Consider, fix. (a) Show that the fundamental theorem yields the answer t. (b) (Ge) With n = 10,, approximate (to four decimal places) the integral by the trapezoidal, midpoint, and Simpson's rules., Ans., , Trapezoidal 0.2525; midpoint 0.2488; Simpson's 0.2500, , 41. Evaluate:, (a), , lim l(cosl[ +cos 2n +... +cos nn), ........ n, n, n, n, , (b), , ~ tn [sec (tn)+ sec (2 tn)+ .. ·+sec (n -1) tn)+t ], , Ans., , 2, , (a)t, , J:, , 2, , 2, , cosxfix=O;(b) 1:'6 sec2xdx=, , 42. (a) Use a substitution to evaluate, , 1, , r ~ dx (to eight decimal places)., 2, , J. "x+l, , (b) (Ge) Use a graphing calculator to estimate the integral of (a)., , Ans., , (a) t(2 -.,fi) - 0.39052429; (b) 0.39052429, , 43. (Ge) Estimate, Am., , ·/4, 0, , 1 xsin (tanx) dx (to four decimal places)., 3, , 0.0262, , r, , 44. (Ge) Consider x</ x~ + 2X2 -1 dx. Estimate (to six decimal places) its value using the trapezoidal and, Simpson's rule (both with n = 4), and compare with the value given by a graphing calculator., Ans., , trapezoidal 3.599492; Simpson's 3.571557: graphing calculator 3.571639
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The Natural Logarithm, The traditional way of defining a logarithm, loga b, is to define it as that number u such that aU = b. For, example, loglo 100 2 because 102 100. However, this definition has a theoretical gap. The flaw is that, we have not yet defined li' when II is an irrational number, for example, J2 or 1t. This gap can be filled in,, but that would require an extensive and sophisticated detour. t Instead, we take a different approach that will, eventually provide logically unassailable definitions of the logarithmic and exponential functions. A temporary disadvantage is that the motivation for our initial definition will not be obvious., , =, , =, , The Natural Logarithm, We are already familiar with the formula, , f, , xr+1, , x'dx= r+l +C, , (rt:-l), , The problem remains of finding out what happens when r =- 1, that is, of finding the antiderivative of X-I., The graph of y = lit, for t > 0, is shown in Fig. 25-1. It is one branch of a hyperbola. For x> 1, the definite, integral, d, , f -dt, t, I, , is the value of the area under the curve y = lit and above the t axis, between t = 1 and t =x., , Definition, Inx = JX !dt for x>O, I t, The function In x is called the natural logarithm. The reasons for referring to it as a logarithm will be made, clear later. By (24.2),, , (25.1), , I, D x On x) =-X, , for x > 0, y, y, , = III, , 4, , x, , Fig. 25-1, , Some calculus telttbooks just ignore the difficulty. They assume that Ii' is defined when a > 0 and u is any real number and that the, usual laws for cltponents are valid ., , ..4'»
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CHAPTER 25, , The Natural Logarithm, , Hence, the natural logarithm is the antiderivative of x- I, but only on the interval (0, +00). An antiderivative, for all x 0 will be constructed below in (25.5)., , *, , Properties of the Natural Logarithm, (25.2) In I = 0, since In 1= JI !dt = O., I t, (25.3) If x> 1, then In x > O., This is true by virtue of the fact that, , r, , }dt represents an area, or by Problem 15 of Chapter 23., , (25.4) IfO<x<l,thenlnx<O., , I, , Inx= JIx t dt = -, , J1It':( by (23.8). Now, for 0 < x < I, if x S t S I, then lit> 0 and, therefore, by, x, , Problem 15 of Chapter 23, JI !dt > O., x t, 1, (25.5) (a) Dx(lnLd) = for x 0, , *, , x, , (b), , J~dx=ln Ixl+C, , forx*O, , The argument is simple. For x> 0, Ixl = x, and so DrOn Ixl) = Dx(ln x) = l/x by (25.1). For x < 0, Ixl = -x,, and so, Dx(ln Ixl) = DxOn (-x» = DuOn u)Dx(u), , (Chain Rule, with u = -x> 0), , =(~)(-I)=_1, =~, u, -u x, I, , EXAMPLE 25.1: Dx (inI3x+ 21) = 3x+ 2 Dx(3x+2), , (Chain Rule), , 3, = 3x+2, , (25.6) In uv= In u + In v, Note that, Dx (In (ax» = ~ Dx(ax), ax, , (by the Chain Rule and (25.1», , 1, 1, =-(a)=-=D, (lnx), ax, x, x, , Hence, In (ax) = In x + K for some constant K (by Problem 18 of Chapter 13). When x = I, In a =, In 1 + K = 0 + K = K. Thus, In (ax) = In x + In a. Replacing a and x by II and v yields (25.6)., , (25.7) In(';) = Inu-Inv, u, In (25.6), replace u by -., v, , (25.8), , 1, In-=-Inv, v, , In (25.7), replace u by 1 and use (25.2).
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CHAPTER 25, , (25.9), , In (xr) = r In x for any rational number r and x> O., , The Natural Logarithm, , ., , By the Chain Rule, Dx(ln (xr» = -.!,(rx r- I ) = !... = Dx(rlnx). So, by. Problem 18 ofChap~er 13, In (x) =, x, x, r In x + K for some constant K. When x = 1, In 1 =r In 1 + K. Since In 1 =0, K =0, yielding (25.9)., EXAMPLE 25.2:, , In <J2x - 5 =In (2x - 5)1/3 =tIn (2x - 5)., , (25.11), , In x is an increasing function., D x (lnx) = ~, x > 0 since x > O. Now lise Theorem 13.7., In u = In v implies u = v., This is a direct consequence of (25.10). For, if u i:-. v, then either u < vor v< u and, therefore, either, In u < In v or In v < In u., , (25.12), , !<ln2< 1, , (25.10), , y, I, I, I, I, I, I, , I, I, I, , I, \, \, \, , \, \, , --- ---------2, , 3, , Rg.25-2, , The area under the graph of y = lit, between t = 1 and t = 2, and ~~ove the t axis, is greater than the, area t of the rectangle with base [1,2] and height t. (See Fig. 25-2.) It is also less than the area, 1 of the rectangle with base [1, 2] and height 1. (A more rigorous argument would use Problems, 3(c) and 15 of Chapter 23.), (25.13), , lim In x = +00, Let k be any positive integer. Then, for x > 22k,, In x> In (22k) =2k In 2> 2k( t) = k, , (25.14), , by (25.19) and (25.9). Thus, as x --7 + 00, In x eventually exceeds every positive integer., lim Inx=-oo, Let u = lIx. As x --7 0+,, , U--7+oo., , Hence,, , lim Inx = lim In(.!.) = lim-lnu, , (by (25.8», , =-limlnu=-oo, , (by (25.13», , .r~", , u~+oo, , U, , u~-+oo, , u->+-, , (25.15), , Quick Formula II:, , Jg'(x), g(x) dx = In Ig(x)I+C, ., , 1, , By the Chain Rule and (25.5) (a), D/lnlg(x)l) = g(x) g'(x).
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CHAPTER 25 The Natural Logarithm, , For 0 < x < 1, -} is increasing on [x, I]. Then, by Problems 3(c) and 15 of Chapter 23,, , I, , --(I-x)<lnx=, x, , izl-dt=, f'(--I) dt<-I(l-x), txt, I, , 1, , Hence, 1- - < In x < x-I. When x = I, the three tenns are all equal to O., x, , • ·3., , Find the derivatives of the following functions., (a) y = In (x + 3)2 =2 In (x + 3)., , ,, 2, y = x+3, , Ans., (b) y, , Ans., , =(In (x+ 3»2, y, , ' 21 ( 3) I, = n x + x + 3 =2 Inx(x+ 3+3), , (c) y =In [(,il + 2)(x2+3)] = In (,il + 2) + In (x2+3), 2, , Ans., , 1 3x 2) +-2-(2x)=-)-+-2I, 3x, 2x, y' =~(, .,\- + 2, x +3, x +2 x +3, X4, , (d) y= In (3x _ 4)2, Ans., , Inx 4 -In (3x - 4)2, , =4Inx- 2In (3;X - 4), , , 4, 2, 4, 6, y =----(3)=---x 3x-4, x 3x-4, , (e) y = In sin 5x, Ans., , (f), , Ans., , Ans., , (h), , 1, , .y' = -'-5-cos(5x)(5) = 5cot5x, Sill x, y = In (x +, , -it +x, , 2), , ,. 1+ t(l + x2t"2(2x), y = X+(l+Xl)"1, , y,=.!._l_(_2x)= _ _, x_, ,. 2 3-Xl, 3- x2, , y=xlnx-x, , Ans. y'=Inx, (i), , Ans., , 1+ x(l + x2tlll (I + X2)"1, x + (l + x2)ln (1 + X 2)"2, , y = In (In (tan x», , tanx+cotx, y =. In (tan x), , I, , Jj;;2
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CHAPTER 25 The Natural Logarithm, , (c), , y= ..Jx2+3cosx, (3x-5)3, x, I. ), y=y ---tanx--x +3, 3x-5, , Ans., , (d), , I, , (, , 2, , y=4;;~;, , Ans., , 3y, y =- 4r -9, I, , 11. Express in terms of In 2 and In 3: (a) InW); (b) In, , 17., , Ans. (a) 7 In 3; (b) In 2-3 In 3, 12. Express in tennsofln 2 and In 5: (a) In 50; (b) Ini; (c) In$; (d) In, , Ans. (a) In 2 +2In5; (b) -21n 2; (c), , 10., , -t InS; (d) - (3 In 2 + In 5), , 13. Find the area under the curve y = ~ and above the x axis, between x = 2 and x = 4., , Ans., , In 2, , 14. Find the average value of.!. on [3, 5)., , x, , Ans., , 12 In .2.3, , 15. Use implicit differentiation to find y': (a) yl = In (xl + yl); (b) 3y - 2x = I + In xy., , 16. Evaluate lim 1 in 2 + h ., ~....o h, 2, 1, , Ans., , 2", , f, , ,17. Check the formula cscxdx = In Icscx- cotxl+C., , f, , 18. (GC) Approtimate In2= tdl to six decimal places by (a) the trapezoidal rule; (b) the midpoint rule;, (c) Simpson's rule, in each case with" = 10., ., Ans., , (a) 0.693771; (b) 0.692835; (c) 0.693147, , 19. (GC) Use Newton's method to approximate the root of xl + In x = 2 to four decimal places., Ans., , 1.3141, , ,., , _L_i, , >~;~.
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Exponential and Logarithmic, Functions, From Chapter 25, we know that the natural logarithm In x is an increasing differentiable function with domain the set of all positive real numbers and range the set of all real numbers. Since it is increasing, it is a, one-to-one function and, therefore, has an inverse function, which we shall denote bye'., Definition, e' is the inverse of In x., , It follows that the domain of e' is the set of all real numbers and its range is the set of all positive real, numbers. Since e' is the inverse of In x, the graph of e' can be obtained from that of In x by reflection in the, line y =x. See Fig. 26-1., y, , .t, , Rg26-1, , Our notation may be confusing. It should not be assumed from the notation that e' is an ordinary power of, base e with exponent x. Later in this chapter, we will find out that this is indeed true, but we do not know it yet., , Properties of el(, , (26.2), , e' > 0 for all x, The range of e' is the set of positive real numbers., In (e') = x, , (26.3), , e1nx =x, , (26.1), , Properties (26.2) and (26.3) follow from the fact that e' and In x are inverses of each other.
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CHAPTER 26 Exponential and Logarithmic Functions, , (26.4) et is an increasing function., Assume u < v. Si~ce u =In (eM) and v= In (e"), In (eM) < In (e'). But, since In x is increasing, eM < eO., [For, if e"S eM, then In (e") S In (e").], (26.5), , Dx(e') =e', , Let y =e'. Then In y =x. By implicit differentiation, ..!.. y' =1 and, therefore, y' =y =et. For a more, 1, , y, , rigorous argument, let/ex) =In x and/-'(y) =eY. Note that f'(x) =-. By Theorem 1O.2(b),, x, (f-I)'(y), , =f'(/I(y»',, , EXAMPLE 26.1: Dx(e',nx) =D.(e")Dx(u), , that is, D,(eY ) =lI~Y, , =e', , (Chain Rule, with u =sin x), , =e· (cos x) =emX(cosx), (26.6), , feX dx =~ +C, , EXAMPLE 26.2:, , J, , To find xe" dx, letu =x 2 , du = 2xdx. Then, , Jxex'dx=tJelidu=te" +C=tez' +C, (26.7), , fe-xdx =-e- +C, X, , f, , f, , ', , Let u = -x, du =-dx. Then e-xdx = - eMdu = -e" + C = _e- + C., (26.8), , X, , eO = I, , By (26.3), ) = e'n I =eO., , (26.9), , eu+' =eOe', In (e"+» =u + v= 'In (e") + In (e") =In (eMe") by (25.6). Hence, e"+' =eWe" because In x is a one-to-one, function., , (26.10) eM-. = eM, e", , By (26.9). elt-'e" =e<u-v)+' =e". Now divide bye"., (26.11) e-', , =-e'I, , Replace Il by 0 in (26.10) and use (26.8)., (26.12) x < ~ for all x, By Problem 7 of Chapter 25, In x Sx-) <x. By (26.3) and (26.4), x=eI"x <~., , =+00, , (26.13) lim e', x-+...., , This follows from (26.4) and (26.12)., (26.14) lim eX =0, x-+-, , 1, , Let u =-x. As x -7 00 , U -7 +00 and. by (26.13), e" -7 +00. Then, by (26.11), eX = e- U= u, e, , -7, , O., , The mystery of the letter e in the expression et can now be cleared up., Definition, , ', , Let e be the number such that In e = I., Since In x is a one-to-one function from the set of positive real numbers onto the set of all real numbers, there must, be exactly one number x such that In x = I. That number is designated e., , Since, by (25.12). In 2 < 1 < 2 In 2 =In 4. we know thilt 2 < e < 4., (26.15) (GC), e- 2.718281828, This estimate can be obtained from a graphing calculator. Later we will find out how to approximate e to, any degree of accuracy.
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CHAPTER 26 Exponential and Logarithmic Functions, , Now we can show that the notation e< is not misleading, that is, that tr actually is a power of e. First of, all, this can be proved for positive integers x by mathematical induction. [In fact, by (26.3), e ~ e lne = el. So,, by (26.9), en+ l = ene' = e"e for any positive integer n and therefore, if we assume by inductive hypothesis that, en represents the produce of e by itself n times, then e"+l is the product of e by itself n + 1 times.] By (26.8), eO = 1, which corresponds to the standard definition of eO. If n is a positive integer, e-n would ordinarily be, defined by lien and this is identical to the function value given by (26.11). If k and n are positive integers,, then the power e k1n is ordinarily defined as tfik. Now, in fact, by (26.9), the product ekillek/~ • .. e k1n , where, there are II factors, is equal to ekhr+k!II+" +klll =e k • Thus, the function value e k1n is identical to the nth root, of ek • For negative fractions, we again apply (26.11) to see that the function value is identical to the value, specified by the usual definition. Hence, the function value e' is the usual power of e when x is any rational, number. Since our function tr is continuous, the value of if when x is irrational is the desired limit of er for, rational numbers r approaching x., The graph of y = tr is shown in Fig. 26-2. By (26.13), the graph rises without bound on the right and, by, (26.14), the negative x axis is a horizontal asymptote on the left. Since D;(e X ) =Dx(eX ) =eX> O. the graph is, concave upward everywhere. The graph of)' = e- is also shown in Fig. 26-2. It is obtained from the graph, of y =tr by reflection in the y axis., X, , n....- (1 +*f, For a proof, see Problem 5., , (26.16) e' = lim, (26.17) e = lim, , 11-++-, , (1 + t, , r, , This is a special case of (26.16) when x = 1. We can use this formula to approximate e, although the, convergence to e .is rather slow. For example, when 11 = 100, we get 2.7169 and, when n = 10 000,, we get 2.7181, which is correct only to three decimal places., y, , Rg.26-2, , The General Exponential Function, Let a > O. Then we can define a' as follows:, Definition, Note thallhis is consistent with the definition of e' since, when a =e, In a = 1., (26.18) Dx (aX), , = (In a)a', , In fact,, (chainrulewithu =xlna), , =e"(lna) =ex1na(lna) =a'(lna), EXAMPLE, 26.3:, ., , D,(2') = (.n 2)2'.
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CHAPTER 26 Exponential and Logarithmic Functions, , (26.19) faxtlx=, , I~a aX +C, , This is a direct consequence of (26.18)., EXAMPLE 26.4:, , JlOx ~ In \0 lOx + C, , We can derive the usual properties of powers., (26.20) If = 1, , If= t!' Ina = t!'= 1, , =aUa', all+. =e(II+" In a =e" In a + "In a =e" In ae,ln a =aUa", aU, , (26.21) 0"*", (26.22) a, , U, -', , =li', , By (26.21), a"-;a'= at. . ..,.. =aU. Now divide by a'., 1, , (26.23) a-' =li', , Replace U by 0 in (26.22) and use (26.20)., (26.24) aU' =(au)", (aU)", , (26.25) (ab)", , =e,ln(a') =, , e,(u(1na)), , =, , e(u.)lna, , =aU', , =aMbo, , Recall that we know that Dx (x) = rx""'1 for rational numbers r. Now we are able to prove that fonnula, for any real number r., (26.26) DX (x') =rx""'1, Since x =e' In x,, D,(x r ) = D,(e rInX ) = D.(eU)D,(u), =, , e (r( ~ ))=r(x ~, r, , U, , {, , ), , (Chain Rule with U = rlnx), , =r ~~ =rx r - I, , General Logarithmic Functions, Let a > O. We want to defme a function logax that plays the role of the traditional logarithm to the base a. If, ,, Inx, y =loga x, then a' =x and, therefore, In (a Y ) =Inx, y Ina =1nx, y =-1-., na, , Definition, Inx, loga x=-lna, -., , (26.27) y =loga X is equivalent to a', ,, , y, , =x, , Inx, , =log" x ¢::> y = In a ¢::> y In a =In x, ¢::>, , In(a') = In x ¢::> aY = x (The symbol ¢::> is the symbol for equivalence,, that is, if and only if.), , Thus, the general logarithmic function with base a is the inverse of the general exponential function with, base a., (26.28) a108, > =x
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••, , CHAPTER 26, , Exponential and Logarithmic Functions, , (26.29) logo (a') == x, These follow from (26.27). See Problem 6., The usual properties of logarithm can easily, , be derived. See Problem 7., , Inx = In x. Th us, the naturaIogant, I ' hm turns out to be a loganthm, ., . t he, · that. Iog. x = Inx, N otlce, In e = -110, usual sensei With base c., SOLVED PROBLEMS, 1., , Evaluate: (a) In (e l ); (b) e71 "!; (c), , n, , e(l 3)-,;, , (e) 1"., , (a) In (e l ) = 3 by (26.2), (b) e7In2=(eln2)'=27= 128 by (26.24) and (26.3), lnl, , (c) eOlll )-2, , 1u = euln 1= ell!0) = eO = I by (26.8), , (d), , 2., , e, 3, =-2, =..,, by (26.10), e, e", , Find the derivatives of: (a) enll ; (b)5 n ; (c)3.xK; (d) ile'., (a) D, (elx+l) = el<+1 (3) = 3el o+ I by the Chain Rule, (b) D,(5 h )=Du (5")D(u), , (chain rule with u=3x), , = (In 5)5" (3), , by (26.18), , = 3(1n5)51', , D, (3x") =3(nx"-1 ) =3nx·- 1 by (26.26), , (c), , (d) D.(x2e') = x 2D,(e') + e' D, (X2), , =x 2e' + e' (2x) = Xl", 3., , f, , J, , Find the following antiderivative: (a) 3(2') tit; (b) x 2e'> dL, , f3(2') tit -- 3f2' tit-2' + C -ln2, -~2x +C, - 3In2, , 1, , (a), , (b) Let u =Xl, du, , 4., , by the product rule, , (x + 2), , 1 + C =je'", I +C, =3x 2 dx. Then fX2~ tit = j1 fe"du = je", , Solve the following equations for x: (a) Inxl =2; (b) In (lnx) =0; (c) r,-I =3; (d) e' - 3e-' = 2., In general, In A = B is equivalent to A = eI, and eC = D is equivalent to C = In D., (a) . In.xl = 3 In x. Hence, In.xl = 2 yields 3 In x = 2, Inx = t, x = e2l3., (b) In (In x) = 0 is equivalent to In x = eO = 1, which, in tum, is equivalent to x, , =e l = e., , =In 3, and then to x = In 32+ 1., ., Multiply both sides bye': ell - 3 =2e', e2x - 2e' - 3 =O. Letting u =e' yields the quadratic equation u2 - 2u3 =0; (u - 3) (u + I) = 0, with solutions u =3 and u =-I. Hence, e' = 3 or e' =-I. The latter is impossible, , (c) e1r-1 = 3 is equivalent to 2x·-·1, (d), , since e' is always positive. Hence, e' = 3 and, therefore, x = In 3., , 5., , Prove (26.16): e" = lim (I +, 'I-++-, , Lel an, , =(1 +, , *J., , ~)", ., II, , Then, , u), , (In(l+u/n)-Inl)', Ina =n 1n (I +- =u, /, n, nun, A, . (In(l+ulll)-1111)., ....lor D(I), 11Ie expression, I, IS a d'«, iuerence quotIent, n x at x = I > WI'th uX, = U / II. A S II, , U II, , U/II, , ., , ~ O. So, that differencc quotient approaches D, (Inx)I';1, , lim Ina n = u(l) =u. So, lim an ~ lim e, , ,,~+-, , '., , n-++-, , n-++-, , ln, , ••, , = eU., , =(lIx)I'.1 = I. Hence,, , ~, , +00,
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CHAPTER 26 Exponential and Logarithmic Functions, l, , r12, 15. (GC) Use Simpson's, . rule with n = 4 to approximate Jr e- dx to four decimal places., o, , Ans., , 0.8556, , P, , 16. If interest is paid at r percent per year and is compounded n times per year, then dollars become, dollars after 1 year. If n , +00, then the interest is said to be compounded continuously., , P(l + 1O~, , )", n, , (a) If compounded continuously at r percent per year, show that P dollars becomes Pe"IOO dollars after 1 year,, and Pen'loodollars after t years., (b) At r percent compounded continuously, how many years does it take for a given amount of money to double?, (c) (GC) Estimate to two decimal places how many years it would take to double a given amount of money, compounded continuously at 6% per year?, (d) (GC) Compare the result of compounding continuously at 5% with that obtained by compounding once a, year., Ans., , (b) IOO~n2) _ 69;31; (c) about 11.55 years;, , (d) After 1 year, $1 becomes $1.05 when compounded once a year, and about $1.0512 when compounded, continuously., 17. Find (log,oe)·ln 10., Ans., , I, , 18. Write as a single logarithm with ba~e a: 3 log. 2 + log. 40 -log. 16, Ans., , log. 20, , 19. (GC) Estimate log27 to eight decimal places., Ans., , 2.80735492, , 20. Show that 10gb x =(log. X)(logb a)., 21. (GC) Graph y =e- r12 . Indicate absolute extrema, inflection points. asymptotes, and any symmetry., Ans., , Absolute maximum at (0, I). inflection points at x =±I. x axis is a horizontal asymptote on the left and, right, symmetric with respect to the y axis., , 22. Given eX' - x + y2 = I. find, , t, , by implicit differentiation., , _1_-..:..yeX', ___, Ans• =2, y+xe'7, e' - e-', , 23. (GC) Grapb y =sinhx = - 2 - ', , e' - e-', , Je' +e-', , 24. Evaluate ---dx., , Ans. In (e'+ e-,) + C, 25. Use logarithmic differentiation to find the derivative of y = Xli, •, Ans., , 3y(I-lnx), x2
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L'Hopifal's Rule, Limits of the form lim, , ~g~ can be evaluated by the following theorem in the indeterminate cases where f(x), , and g(x) both approach 0 or both approach ±oo., , L'Hopital's Rule, Iff(x} and g(x) either both approach 0 or both approach ±ce, then, · f(x) -I' r(x)·, 11m ( ) - 1m '(), gx, g x, , Here, "lim" stands for any of, lim,, , xlim,, ... _, , lim,, x...a, , lim., , x-+a~, , For a sketch of the proof, see Problems I, II, and 12. It is assumed. in the case of the last three types of, limits, that g'(x) 'i: 0 for x sufficiently close to a, and in the case of the first two limits, that g'(x) 'i: 0 for, sufficiently large or sufficiently small values of x. (The corresponding statements about g(x) 'i: 0 follow by, Rolle's Theorem.), EXAMPLE 27.1:, , Since In x approaches +00 as x approaches +00, L'Hopita\'s Rule implies that, lim Inx = lim l/x = lim 1=0, x, ,(~"" 1, .t~ X, , -l~t_, , EXAMPLE 27.2:, , Since eX approaches +00 as x approaches +00, L'Hopital's Rule implies that, lim, , .r~+-, , EXAMPLE 27.3:, , ~ =Iim J.r=0, e, , e, , .r~+-, , We already know from Problem l3(a) of Chapter 7 that, lim 3x 2 +5x-8, ,-1'- 7x 2 - 2x + I, , 3, , "7, , Since both 3x2 + 5x - 8 and 7x2 - 2t + I approach +00 as x approaches +00, L'H6pital's Rule tells us that, lim 3x~ + 5x - 8 = lim 6x +5, 7x -2x+1 H+- 14x-2., , ,-I+-
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CHAPTER 27 L'Hopifal's Rule, , and another application of the rule tells us that, , lim 6x+5 =lim -h-='\=t, 14x - 2 x........, , x-+...., , EXAMPLE 27.4:, , Since tan x approaches 0 as x approaches 0, L'HopitaI's Rule implies that, 2, , lim tanx = lim sec x = lim _,_12- = pI = I, 1, • ...0, X, .-+0, .-+0 cos X, ,, , Indeterminate Type O· 00, Iff(x) approaches 0 and g(x) approaches ±oo, we do not know how to find limf(x)g(x). Sometimes such a, problem can be transformed into a problem to which L'Hopiial's Rule is applicable., EXAMPLE 27.5: As x approaches 0 from the right, In x approaches -00. So, we do not know how to find lim x In x., • ...0', But as x approaches 0 from the right, lIx approaches +co. So, by L'Hopital's Rule,, , In x= lim -11, In x = lim -1/, lIx = lim -X= 0, ,, IImx, .-+0', X, x-+O' X 2 x-+O', , .-+0', , Indeterminate Type 00 - 00, If f(x) and g(x) both approach, , we do not know what happens to lim(f(x) - g(x». Sometimes we can, transform the problem into a L'Hopital's-type problem., EXAMPLE 27.6:, , Iim(cscx, x...., , o, , 00,, , _1)x is a problem of this kind. But,, , lim (cscx _1) = lim (_._1__ 1) = lim x- ~inx, x •...o, x...o xsmx, .....0, , SIflX, , X, , Since x - sin x and x sin x both approach 0, L'H6pitaI's Rule applies and we get lim 1- c~s~ , Here both, ', •...0 xcosx smx, numerator an d denommator, approac hOd, an L'H'Oplt'aI' s Rule yel'IdS, lim, .-.0, , ,sinx, = __, O_=Q=O, -xsmx+cosx+cosx 0+ 1+ I 2, , Indeterminate Types 0°, 00°, and 1, If lim y is of one of these types, then lim (In y) will be of type O· 00., 00, , In Iimx'"u,y=x"uu is of type 00 and we do not know what happens in the limit. But, In x .-+0', In y = sinx Inx = csc x and In x and,csc x approach:too. So, by L'Hopital's Rule,, EXAMPLE 27.7:, , , Iny= I'1m, I1m, , x-+o', , ..... 0· -, , I/x, csc x cot x, , • 2, , ", , I'Im---=sm x, I'1m smx, smx, -x cos x, x-+O' X, cos x, , x-+o', , =-lim, sinx lim tanx =-(1)(0) = 0, .-+0' X x-+O+, Here. we used the fact that Iim«sinx)/x) = I (Problem 1 of Chapter 17), Now, since lim In y = 0,, x-+O+, , • ...0, , Iimy= Iime iny =eo = 1, , x-+O', , x-+O'
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CHAPTER 27, , In!~, , EXAMPLE 27.8:, , IlnxlX, y = IInxlx is of type, , 00, , 0, , L'Hopifal's Rule, , and it is not clear what happens in,' the limit. But, , ,, , Iny=xlniin xl= In~~: xl and both In'lIn xl and 1/x approach +00. So L'Hopital's Rule yields, , I·1m In '= I'lin ( -I-)/(- 1) = lim--=, x, 0, ) .....0' x In x, x 2 x->o' In x, ', , x....o·, , .;, , since, lim _1_ = O., In x, , lim y = lim eln), = eO = 1, , lienee,, , x ...o·, , X~O·, , x-+O·, , In limxll{x-I) , y = X ll{x-l) is of type 1~ and we cannot see what happens in the limit. But In y = In Xl, and both the numerator, the denominator approach O. So by L'Hopital' s Rule, we get, xEXAMPLE 27.9:, , and, , lim In y = lim 111x = 1., x-+l, , lim y =lim elny = e l = e, , Hence", , x-+1, , .1-+1, , x-+I, , SOLVED PROBLEMS, 1., , Prove the following, , ~ form of L'Hopital's Rule. Assumef(x) and g(x) are differentiable and g'{x} '*, , 0, , °in some, , open interval (a, b) and lim f(x) = = lim g(x). Then, if lim f;«X» exists,, .1'-+,,+, , x-ta·, , .:r-+a:+, , lim f(x), Ha', , g(x), , g, , X, , =lim f:(x), ' ....a· g (x), , Since lim I(x) = 0 = lim g(x), we may assume thatf(a) and g(a) are defined and thatf(a) = g(a) = O., .1'-+a', Replacing b by x in the Extended Law of the Mean (Theorem 13.5), and using the fact thatf(a) = g(a) = 0, we, J, , -+(1+, , obtain, f(x), g(x), , for some, , Xo, , with a<xo <x. So,, , Xo ~a+, , as, , f(x)- f(a), , !'(xo), , = g(x) - g(a) = g'(xo), , x ~, , a+. Hence, ,, , lim I(x) = lim f:(x), g(x) x-;a· g (x), , Ha+, , We also can obtain the Q fonn of L'Hopital's Rule for lim (simply let II = -x). and then the results for lim and, O, 6, ~, I~, }~~ yield the 0' form of L'Hopital's Rule ~~., ', 2., , We already know by Examples 1 and 2 that lim Inx = 0 and lim ~ = O. Show further that lim (lnx)" = 0 and, x", .. ., X-+_ X, x....- e, x...._, x, lim -;- = 0 for all poslttve tntegers n., x.... _ e, Use mathematical induction. Assume these results for a given 11 ~ 1. Uy L'Hopital's Rule,, lim (Inxr, x ...._, x, , l, , lim (n+l)(lnx)"(lIx), =,......., I, , (n+l)lim (In x)" =(11+1)(0)=0, x ...._, x, , Likewise,, I'm x, , x~_, , n, , 1, , I' (n+l)x", ex+ =x~'!, eX, , (n+l)lim, , x. . _ Xx"e =(n+I)(O)=O
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CHAPTER 27, , So, , L'Hopifal's Rule, , lim Iny= lim lntanx = lim (sec 2 xltanx)/(secxtanx)= lim ~~x =Ql =1,, z-o(.I2)- sec X, z-o(.I2)z-o(./2,. sm x, , z-o(.I2)", , (1), , lim, , Z-+~, , .fi+Xi., x, , .J x 2 = x-++lim, x->... 2 + x, , We get lim, , J2+ii, 2, , 2+ x and we are going around in a circle. So, L'Hopital's Rule is of no use. But,, x', , ="0+1 = I, , 4., , Criticize the following use of L'Hopital's Rule:, , r, , 2, Xl - x - X - 2, r 3x 2 - 2x - I, ,I~ Xl - 3x 2 + 3x - 2 = ,I~~ 3Xl - 6x + 3, , 6x2 I' -6=, 6 I, · I1m, z ... l, x6, - 6= x1m, ... l, , The second equation is an incorrect use of L'Hopital's Rule, since lim (3x 2 - 2x -1) =7 and lim (3x 2 - 6x + 3) = 3., z-+l, .1'-+2, So, the correct limit should be, , t., , s., , (GC) Sketch the graph of y = xe- = :. ., , .,, ., y, =, O., So., the, positive, x, axis, is, a, horizontal, asyomptote., Since, See Fig 27 -I. By Example 2, xlim, ......, lim e- X = t-, lim y = - . y' = e- X (I - x) and y" =e-' (x - 2). 1ben x = 1 is a critical number. By the second, aerivative tes{lhere is a relative maximum at (I. lie) since y" < 0 at x = O. The graph is concave downward for, x < 2 (where y" < 0) and concave upward for x> 2 (where yIP> 0). (2. 2Iel) is an inflection point. The graphing, calculator gives us the estimates lIe - 0.37 and 21e2 - 0.27., X, , x, , Rg.27-1, , 6., , (GC) Sketch the graph y = x Inx., See Fig. 27-2.1l1e graph is defined only for x>O. Clearly, lim y =t-. By Example 5, lim y =O. Since y' =1+ Inx, x-++".... 0·, and y" =l/x > O. the critical number at x = lie (where y' =0) yields. by the second derivative test, a relative, minimum at (lIe. -lIe). The graph is concave upward everywhere., y, , x, , o, , Fig. 27-2
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CHAPTER 27, , (y), (b'), , e- WU sec 2 x = 0, , x!$1_)=0, lim (1 __._, smx, .--+0, , (e') lim, .--+0, , (h') lim, , X, , (~_, 2, x, , (COSX)"x, , (k'), , )-_13, , x-++-, , r1m (.smx-cosx )""" = 1/e, , (1'), , s, , lim (1- e- r )e<, , =lie, , ~, , r1m (tanx t't = I, , (m') lim xta"t R ' = e-2/r<, , ., , (p'), , .t-tl, , 2', , 1, , -3", , lim ~=O, X, ., , .r-t+O+, , 1000, , lim _n_=O, S, , (r'), , x-++-, , X, , lim, eX(1-e X), x....o (l + x)ln (1- x), , =I, , X~+-, , (0') hm V=O, x--+ ...., , (q') lim In / =0, x--+...., , (i'), , x--+t,,-, , Y= e, , XX, , X~O+, , (i') lim (eX + 3x)l/· = e4, , =1, , 1, I, x-2 =""'4, ), , -0, , (_IInx___x-I, X_) =_12, , (g') lim, , JX, , X, , .....0, , (n') lim (1 + I1x, , (s'), , ..... 1, , ", , (f) lim (lnx __1_)= 0, , x--+t", x--+ ...., , (d') lim, , (e') l~ (sec l x- tan 3 x)= 00, , 2, I-cosx -, , x-+o, , (a') ~ x 2 -4, , x--+o, , x, , 0, , .(4, , lim (x-sin- I x)csc 3 x=-i, , (z), , L'Hopifa/'s Rule, , X, , lim -Llim l-e =1, .....0 1 + X .--+0 (1:- x), , 11. Verify the sketch of the proof of the following ~ form of L'H6pital's Rule at +00. Assumef(x) and g(x) are, differentiable and g'(x) *- 0 for all x ~ c. and lim f(x) = 0 = lim g(x). Then,, x~, , I'(x), ·m, -II, g'(x), , if, , X~+-, , lim f(x) = lim f:(x), , exists,, , x-+ ...., , x-+_, , g(x), , x ........, , g (x), , Proof Let F(u) = f(l/u) and G(u) = g(l/u). Then, by Problem I for a ~ O+, and with F and G instead off and g,, ., , f(x), , ., , hm g X(), t .... _, , F(u), , ., , F'(u), , =0--+0+, hm G(OU) = 0--+0', hm G'(U ), 2, , = lim (f'(I1u)· (_111/ », 2, .....0· (g'{l/u)· (-lIu », , = lim f;(I/u) = lim, .--+0', , g (lIu), , x ........, , f;(x), g (x), , ., 12. Fill in the gaps in the proof of the following;; form of L'H6pital's Rule in the lim case. (The other cases follow, easy as in the ~ form.) Assume f(x) and ~(x) ~re differentiable and g'(x} *- 0 i;;~+me open interval (a, b) and, limf{x}=±oo=, limg{x}. Then,, t, , x-+a, , ., , , x-+a-+, , if, , K = lim f:(x), X--+Q', , exists,, , g (x), , lim f(x) = lim f:(x), .-ta', , g(x), , x-tQ', , g (x), , Proof Assume E> 0 and choose c so that IK - (l'(x)/g'(x»1 < E/2 for a < x < c. Fix d in (a. c). Let a < y < d. By, the extended mean value theorem, there exists x' such that, , y<x' <d, , f(d)- fey), g(d)- g(y), , and, , I'(x'), g'(x'), , Then, K - fed) - f(Y)1 < §., , I, , 2, , g(d)- g(y), , and so, , K _[(fey) _ f(d»)/(I_ g(d), g(y) g(y), g(y), , I, , )]1 < §.2, , Now we let y ~ a+. Since g(y) ~ ±CO and fed) and g(d) are constant, f(d)/g(y) ~ 0 and 1- g(d)/g(y) ~ 1. So,, for y close to a,, , I, , I, , K - f(y) < E ., g(y), , Hence,, , lim fey) =K, y....a·, , g(y), , 0
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CHAPTER 27 L'Hopital's Rule, , 13. (GC) In the following cases, try to find the limit by analytic methods, and then check by estimating the limit on a, graphing calculator: (a) lim X 1/s ; (b) lim X 1/s ; (c) lim (1- cosxY; (d) lim (.JX2 + 3x - x)., 1....0+, , Ans., , .I.........,f~, , .r-++-, , (a) 0; (b) 1; (c) 1; (d>t, , 14. The current in a coil containing a resistance R, an inductance, L, and a constant electromotive force, E, at time tis, given by i =~ (1- e- R11L ). Obtain a formula for estimating i when R is very close to O., Ans., , Et, , T, , ,.
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,, , Exponential Growth and Decay, Assume that a quantity )' varies with time and that, dy _ky, , (28.1), , dx-, , for some nonzero constant k. Let F(t) =yle u . Then, by the Quotient Rule,, , Hence, F(t) must be a constant C. (Why?) Thus, yle U = C and, tnerefore, y = Cekt • To evaluate C, let t = O., Then y(O) = Ceo = C(l) = C. If we designate y(O) by Yo, then C = Yo and we have obtained the general form, of the solution of equation (28.1):, (28.2), , If k > 0, we say that y grows exp()nentially and k is called the growth constant. If k < 0, we say that y decays, exponentially, and k is called the decay constant. The constant Yo is called the initial value., , ., , ~, , r, , From Problem 2 of Chapter 27, we know that lim --;- = O. So, when k > 0, lim i t = O. Thus, a quantity, u_e, t_e, that grows exponentially grows much more rapidly than any power of t. There are many natural processes,, such as bacterial growth or radioactive decay, in which quantities increase or decrease at an exponential, rate., , Half-Life, Assume that a quantity y of a certain substance decays exponentially, with decay constant k. Let Yo be the, quantity at time t =O. At what time Twill only half of the original quantity remain?, , By (28.2), we get the equation y =yoe u . Hence, at time T,, .1)', 2 (), , =,fOelT, \J, , In (t) = In (e lT ) = kT, (28.3), , -ln2=kT, T __ In2, , -, , k, , (28.4
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..,-, , CHAPTER 28 Exponential Growth and Decay, , Note that the same value T is obtained for any original amount Yo' T is called the half-life of the substance., It is related to the decay constant k by the equation (28.3). So, if we know the value of either k or T, we can, compute the value of the other. Also observe that, in (28.4), k < 0, so that T> O., The value of k can be obtained by experiment. For a given initial value Yo and a specific positive time to', we observe the value of y, substitute in the equation (28.2), and solve for k., ,SOLVED PROBLEMS, , 1., , Given that the half-life Tofradium is 1690 years, how much will remain of one gram of radium after 1000 years?, From (28.3), k = _1~2 = and the quantity of radium is given by y = yoe-(ln2)1I1690. Noting that Yo = 1, and substituting 1000 for I, we get the quantity, , l2io, , y =e-(ln2)lroJ/l690 _ e-693.111690 _ e-{l·4101 - 0.6636 grams, Thus, about 663.6 milligrams are left after 1000 years., 2., , If 20% of a radioactive substance disappears in one year, find its half-life T. Assume exponential decay., By (28.2), 0.8yo = YOel(l) = yoel. So, 0.8 = el whence, k= In (0.8) = InW = In 4 -In 5. From (28.4),, , -.lnl102 - 3. I063 years., T -k -ln5-ln4, 3., , Assume that the number of bacteria in a culture grows exponentially with a growth constant of 0.02, time being, measured in hours. (Although the number of bacteria must be a nonnegative integer, the assumption that the, number is a continuous quantity always seems to lead to results that are experimentally verified.), (a) How many bacteria will be present after 1 hour if there are initially 1000?, (b) Given the same initial 1000 bacteria, in how many hours will there be 100 000 bacteria?, (a) From (28.2), y = IOOOeo o2 -1000(1.0202) = 1020.2 -1020, (b) From (28.2),, 100 000 = 1000e0 021, 100= eO.021, , In 100 = 0.021, 21n1O= 0.02t (since InlOO = In (lW = 21n1O), 1=100 In 10 -100(2.0326) = 203.26 hours, , Note: Sometimes. instead of giving the growth constant, say k = 0.02, one gives a corresponding rate of, increase per unit time (in our case, 2% per hour.) This is not quite accurate. A rate of increase of r% per unit time, is approximately the same as a value of k= O.Or when r is relatively small (say, r ~ 3). In fact, with an r% rate of, growth, y =yo( I + O.Or) after one unit of time. Since y = yoel when t = 1, we get I + O.Or = ek and, therefore,, k =In (1 + O.Or). This is close to O.Or, since In (I + x) - x for small positive x. (For example, In 1.02 - 0.0198, and In 1.03 - 0.02956.) For that reason, many textbooks often interpret a rate of increase of r% to mean that, k =O.Or. ', 4., , If a quantity y increases or decreases exponentially, find a formula for the average value of y over a time interval [0, bl., , By definition, the average value y••, (28.1), ky:;, , t, , and, therefore, Y.. = b~, , = b ~ 0 S: y dt =, , lk S:, , ky dt (where k is the growth or decay constant). By, , S: t dt. By the Fundamental Theorem of Calculus., , S:~ dt=y(b)-y(O)=y(b)-yo', , Thus,, , Y•• = :k(y(b)-yo)
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CHAPTER 28, , Exponential Growth and Decay, , 5., , If the population of a country is 100 million people and the population is increasing exponentially,with a growth, constant k = In 2, calculate precisely the population after 5 years., By (28.2), the population y = yoe tl =101 e(la2)5 =101 (e ln2 )5 = IO'(25 ) = 32(108 ). Thus, the population will reach, 3.2 billion people in 5 years., , 6., , Carbon-Dating. A certain isotope 14C of carbqn occurs in living organisms in a fixed proportion to ordinary, carbon. When that organism dies, its 14C decays exponentially, and its half-life is 5730 years. AsSume that a piece, of charcoal from a wood fire was found in cave and contains only 9% of the I·C expected in a corresponding, piece of wood in a live tree. (This figure is obtained by measuring the amount of ordinary carbon in the piece of, charcoal.) How long ago was the wood burned to form that charcoal?, If y is the amount of 14C present in the piece of charcoal, we have y = yoe kl• The present quantity O.09Yo = yoe kr,, where't is the elapsed time. Thus, 0.09= eAr, In (0.09) = kf, f= (In (O.09»/k. Since the half-life T= 5730 and, k = -(In 2)IT = -(In 2)/5730, we obtain, , ·~~~",, , "8-11;~., , __ 5730 In(0.09) _ 5730 (In lOO-ln9) 19906, In2, In2, years, , f -, , 7., , .... :'., , -:;;~'"{!;",.,, • ". I~, , ,-:, , Newton's Law of Cooling: The rate of change of the temperature of an object is proportional to the difference, between the object's temperature and the temperature of the surrounding medium., Assume that a refrigerator is maintained at a constant temperature of 45°F and that an object having a, temperature of 800F is placed inside the refrigerator. If the temperature of the object drops from 800F to 700F in, 15 minutes, how long will it take for the object's temperature to decrease to 6O"F?, Let u be the temperature of the object. Then, by Newton's Law of Cooling, duldt = k(u -45), for some, (negative) constant k. Let y = u - 45. Then dy/dt = du/dt = kyo Thus, by (28.2), y = yoe kl • Since u is initially 8ooF,, Yo = 80 - 45 = 35. So, y = 35e H • When t = 15, u = 70 and y = 25. Hence, 25 = 35e l5k , 5 = 7e lSk and, therefore,, 15k = In <t)= In5-ln7. Thus, k =-ts(ln5-ln7). When the object's temperature is 60"F,), = 15. So, 15 = 35ekt ,, 3 = 7ekt and therefore, kt = In(t) = In3-ln7. Thus,, , t=, , In3-ln7, In3-In7, ., k, =15In5_In7-37.7727mmutes, , Hence, it would take about 22.7727 minutes for the object's temperature to drop from 70" to 60°., 8., , r, , Compound Interest. Assume that a savings account earns interest at a rate of r% per year. So, after one year,, an aIllount of P dollars would become, , p( 1+ 160) dollars and, after t years, it would become p( 1+ 160, , dollars. However, if the interest is calculated n times a year instead of once a year, then in each period the, , r, , interest rate would be (rln)%; after t years, ti>.'!re would have been nt such periods and the final amount would be, , P(I + tOOn, , If we let n -+ +00, then we say thatlhe interest is compounded colltinuou.I'ly. In such a case, the, , final amount would be, , Let $100 be deposited in a savings account paying an interest rale of 4% per year. Arter 5 years, how much, would be in the account if:, (a) The interest is calculated once a year?, , (b) The interest is calculated quarterly (that is, four times per year)?, (c) The interest is compounded continuously?, (a) lOO( 1.04)5 - 121.6653 rlolJar., (b) 100( 1.0 1)20 - 122.0190 dollar., (c) 100eo.04(S) = 100eo.2 - 122.1403 dollar.
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CHAPTER 28 Exponential Growth and Decay, , 9., , Assume that, in a chemical reaction, a certain substance decomposes at a rate proportional to the amount present., Assume that an initial quantity of 10,000 grams is reduced to 1000 grams in 5 hours. How much would be left of, an initial quantity of 20,000 grams after 15 hours?, , Ans., , 20 grams, , 10. A container with a maximum capacity of 25,000 fruit flies initially contains 1000 fruit flies. If the population, grows exponentially with a growth constant of (In 5)110 fruit flies per day, in how many days will the container, be full?, , Ans. 20 days, 11. The half-life of radium is 1690 years. How much will be left of 32 grams of radium after 6760 years?, , Ans. 2grams, 12. If a population grows exponentially and increases at the rate of 2.5% per year, find the growth constant k., , Ans., , In 1.025 - 0.0247, , 13. A saltwater solution initially contains 5 Ib of salt in 10 gal of fluid. If water flows in at the rate of t gal/min, and the mixture flows out at the same rate, how much salt is present after 20 min?, , Ans., , ~~ =-~( 1~ ). At I =20, S =5e-, , 1, , -, , 1.8395 lb., , 14. Fruit flies in an enclosure increase exponentially in such a way that their population doubles in 4 hOllrs. How, many times the initial number will there be after 12 hours?, , Ans. 8, 15. (GC) If the world population in 1990 was 4.5 billion and it is growing exponentially with growth constant, k =(In 3)18, estimate the world population in the years (a) 2014; (b) 2020., , Ans. (a) 111.5 billion;'(b) 277.0 billion, 16. (GC) If a thermometer with a reading of 65°F is taken into the outside air where the temperature is a constant, 25°F, the thermometer reading decreases to 500F in 2.0 minutes., (a) Find the thermometer reading after one more minute., (b) How much longer (after 3.0 minutes) will it take for the thermometer reading to reach 32°F?, Use Newton's Law of Cooling., (', , AilS., , (a) 45°F; (b) about 4.4 minutes more, , 17. (GC) Under continuous compounding at a rate of r% per year:, (a) How'long does it take for a given amount of money P to double?, (b) If a given amount P doubles in 9 years, what is r?, (c) If r= 8, how much mllst be deposited now to yield $100,000 in 17 years?, , Ans. (a) 100 In2 _ 69.31; (b) about 7.7: (c) about $25,666, r, r
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CHAPTER 28, , Exponential Growth and Decay, , 18. An object cools from 120"F to 95°F in half an hour when surrounded by air whose temperature is 70"F. Use, Newton's Law of Cooling to find its temperature at the end of another half an hour., , ~;, ~~~z;~, .:~!., , -.. -, , 19. If an amount of money receiving interest of 8% per year is compounded continuously, what is thy equivalent, yearly rate of return?, Am., , about 8.33%, , 20. How long does it take for 90% of a given quantity of the radioactive element cobalt-60 to decay, given that its, half-life is 5.3 years?, Ans., , about 17.6 years, , 21. A radioactive substance decays exponentially. If we start with an initial quantity of Yo, what is the average, ., ., quantity present over the first half-life?, Am., , ..lL, 21n2
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Applications of Integration I:, Area and Arc Length, Area Between a Curve and the y Axis, We already know how to find the area of a region like that shown in Fig. 29-1, bounded below by the x axis,, above by a curve y =f(x) , and lying between x =a and x =h. The area is the definite integral f(x)dx., , I:, , y, , Fig. 29-1, , Now consider a region like that shown in Fig. 29-2, bounded on the left by the y axis, on the right by a, curve x =g(y), and lying between y =c and y =d. Then, by an argument similar to that for the case shown in, Fig. 29-1, the area of the region is the definite integral, , r, , g(y)dy., , y, , +---'----'---'-----'----.x, , Fig. 29-2
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Applications of Integration I, , CHAPTER 29, , EXAMPLE 29.1:, , Consider, the region bounded on the right, by the parabola x =4 -2 i, on the left by. the y axis. and, ., ., , above and below by y =2 and y =-1. See Fig. ?9-3. Then the area of this region is, Theorem of Calculus, this is, (4y- ty3)1~, = (8 -, , t)- (-4- (-t» = 12 - t = 12 -, , J (4 - y2)dy., , By the Fundamental, , -I, , 3= 9, , y, , ~--'---"x, , Fig. 29-3, , Areas Between Curves, Assume thatf and g are continuous functions such that g(x) '5.f(x) for (l '5. x '5. b. Then the curve y =f(x) lies, above the curve y =g(x) between x = a and x =b. The area A of the region between the two curves and lying, between x =a and x = b is given by the formula, A=, , J: (f(x) - g(x»dx, , (29.1 ), , To see why this formula holds, first look at the special case where 0 '5. g(x) '5. f(x) for a '5. x '5. b. (See, Fig. 29-4.) Clearly, the area is the· difference between two areas, the area AI of the region under the curve, y = f(x) and above the x axis, and the area A of the region under the curve y = g(x) and above the x axis., b, , Since A, =, , b, , B, , J. f(x)dx and At = fa g(x)dx,, , A =A, - A,, , =J: f(x)dx- J: g(x)dx, , ,.:., , =L(f(x) -, , g(x»dx by (23.6), , y, , ....., , ", , ~, , --, , ........, , ...,..... ........ y = g(x), , ----~----~--------------L----x, , a, , o, , b, , Fig. 294
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CHAPTER 29 Applications of Integration I, Now look at the general case (see Fig. 29-5), when one or both of the curves y =I(x) and y =g(x) may lie, below the x axis. Let m < 0 be the absolute minimum of g on [a, b]. Raise both curves by Iml units. The new, graphs, shown in Fig. 29-6, are on or above the x axis and enclose the same area A as the original graphs., The upper curve is the graph of y =I(x) + Iml and the lower curve is the graph of y = g(x) + Iml. Hence, by, the special case above,, , 1:, , 1:, , A = «f(x) +1m I-(g(x) + I m I»dx = (/(x) - g(x»dx, y, y, , .t, , Fig. 29-5, , Fig. 29-6, , EXAMPLE 29.2: Find the area A of the region Wl under ~e line y;;;; t x + 2. above the parabola y ;;;;, the y axis and x;;;; 1. (See the shaded region in Fig. 29-7.) By (29.1)., , A = fl((1 x + 2)-X2, Jo 2, , r. and between, , )dx=(14 x +2x- 1r)1, =(1+, 2- 1)- (0+0 - 0);;;; 2-+ 24 _3.-= 23, 3, 4, 3, 12 12 12 12, 2, , -2, , ,, , -I' 0, , x, , Fig. 29-7, , Arc Length, Let/be differentiable on [a, b]. Consider the part of the graph of/from (a, I(a)) to (b,f(b)). Let us find a, formula for the length L of this curve. Divide [a, b] into n equal subintervals, each of length ~x. To each, point xk in this subdivision there corresponds, a point Pk(Xk,f(Xk» on the curve. (See Fig. 29-8.) For large n,, o, the sum Po~ + ~ P2 + ... + Pn-IP", to the length of the curve., , =L Pk- I~, l-I, , of the lengths of the line segments Pk-lk is an approximation
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CHAPTER 29 Applications of Integration I, y, , x, , Fig. 29-8, , By the distance formula (2.1),, , Now, xk - Xk_1 =!::.x and, by the law of the mean (Theorem 13.4),, , f(x k) - f(x k_l ) = (x k - xH )f'(x:) = (!::.x)f'(x;), for some x; in (~k-I' xk). Thus,, , PH~ = ~(!::.X)2 + (/::.X)2 (f'(X;»2 = ~ (1 +(f'(X;»2)(!::'X)2, , = ~l + (f'(x; »2 ~(/::.X)2 =~l + (f'(X;»2!::.X, , :tPk-l~ =:t~1+(f'(x:»2!::.x, , So,, , k=1, , k=1, , r, , The right-hand sum is an approximating sum for the ~efinite integral, . ~, n --7 +00, we get the arc length formula:, , ~l + (f'(X»2 dx. Therefore, letting, , (29.2), , EXAMPLE 29.3:, , Find the arc length L of the curve y =x3f2 from x = 0 to x = 5., , By (29.2), since y'=tx l12, , L=, , J: ~l +, , =t, , f:, , (I, , (y')2 dx=, , f: ~l, , =tJX:, +txdx·, , +tX)"2 (t)dx =H(l + tX)~121, , (by Quick Formula I and the Fundamental Theorem of Calculus)
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CHAPTER 29 Applications of Integration I, , SOLVED PROBLEMS, , I., , Find the area bounded by the parabola x = 8 +2y - y. the y axis, and the lines y = -I and y = 3., Note, by completing the square, that x =-(y2 - 2y - 8) = -«(y - 1)2 - 9) = 9 - (y - 1)2 = (4 - y)(2 + y). Hence,, the vertex of, the parabola is (9, 1) and the parabola cuts the y axis at y =4 and y = -2. We want the area of the, shaded region in Fig. 29-9, which is given by, , 3, , L(s+ 2y-y2)dy=(Sy+ y2 -, , tl) ]3_I = (24+9-9)- (-S+ 1- t)='i-, , y, , y), , Fig. 29-9, , 2., , Find the area of the region between the curves y = sin x and y = cos x from x = 0 to x = 7tl4., The curves intersect at (TC/4, ../212), and 0 ~ sin x < cos x for 0 ~ x < 1t/4. (See Fig. 29-10.) Hence, the area is, , In, , ](/4, , ( ../2 ../2), (cosx-sinx)dx=(sinx+cosx) ]"/4, 0 = T+T -(0+1)=../2-1, , Vi, T, 11', , 4, , Fig. 29-10, , 3., , Find the area of the region bounded by the parabolas.), =6x - xl and y =xl - 2x., By solving 6x - xl =xl - 2x, we see that the parabolas intersect when x =0 and x =4, that is, at (0, 0) and (4, 8),, (See Fig. 29-11.) By completing the square, the fIrst parabola has the equation y = 9 - (x - W; therefore, it has its, vertex at (3, 9) and opens downward. Likewise. the second parabola has the equation y =(x _1)2 - I; therefore,, its vertex is at (I, -I) and it opens upward. Note that the first parabola lies above the second parabola in the given, region. By (29.1). the required area is, 4, , 4, , 4, , o, , o, , 0, , 1«6x-x2)-(x2 -2x»dx= Jr (8x-2x 2)dx=(4x2 -iX3)], , =(64-.qt)=¥
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CHAPTER 29, , 11, , Applications of Integration I, , 1/= %1-2%, , Rg.29-11, , 4., , Find the area of the region bounded by the parabola i = 4x and the line y = 2x - 4., Solving the equations simultaneously. we get (2x - 4)2 = 4x. Xl - 4x + 4 = x. Xl - 5x +4 = O. (x - I)(x - 4) = O., Hence, the curves intersect when x = 1 or x = 4. that is, at (I, -2) and (4,4). (See Fig. 29-12.) Note that neither, curve is above the other throughout the region. Hence, it is better to take y as the independent variable and, rewrite the curves as x =t)'2 and x = t (y + 4). The line is always to the right of the parabola., The area is obtained by integrating along the y axis:, , t:(t<y+4)- tl)dy= t t:(2y+8- yl)dy, , =Hl + 8y- tl)t =t«16+ 32- ,)- (4 -16 + t» =9, , ..... 1',, , Rg.29-12, , 5., , Find the area of the region between the curve y = x 3 - 6x 2 + 8x and the x axis., Since Xl - 6x 2 + 8x =x(x 2 - 6x +8) =x(x - 2)(x - 4), the curve crosses the x axis at x =0, x =2, and x =4., The graph looks like the curve shown in Fig. 29-13. (By applying the quadratic fonnula to y', we find that the, maximum and minimum values occur at x = 2 ± tJi.) Since the part of the region with 2 S x S 4 lies below the, x axis, we must calculate two separate integrals, one with respect to y between x = 0 and x = 2, and the other with, respect to -y between x =2 and x =4. Thus, the required area is
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CHAPTER 29 Applications of Integration I, II, , o, , 2, , Fig. 29-13, , Note that, if we had made the mistake of simply calculating the integral!·, (xl- 6x 2 +8x)dx, we would have, ., 0, got the incorrect answer O., 6., , Find the area enclosed by the curve y = xl - x4., The curve is symmetric with respect to the coordinate axes. Hence ~ired area is four times the portion, lying in the first quadrant. (See Fig. 29-14.) In the first quadrant, y = .,Jx2 - X4 = x.JI - x 2 and the curve intersects the x axis at x =0 and x = 1. So, the required area is, 4, , f>Jl=?' dx = -2f~ (1- X2)112(-2x)dx, = -2( t)(\ - x 2 )l121, =, , -teo _1, , 312, , (by Quick Formula I), , )= -f(-I)=, , t, , II, , Fig. 29-14, , 7., , Find the arc length of the curve x = 3yl'2 -1 from y = 0 to y = 4., We can reverse the roles of x and y in the arc length formula (29.2): L =, , ~dx)2, , dx, , d, , L=f:';1 + Jf y dy = Tr f: (1 + .y.y)'I2(.y.)dy =n(t)(1 + Jf y)J/2 I =~«82)JI2 - Il'l) =~(82.J82 -1), 8., , Find the arc length of the curve 24xy = x4 + 48 from x =2 to x =4., y = Xl + 2X-', Hence, y' =t x 2 - 2/x 2 , Thus,, , *, , (y,)2 =irx., I + (y/)2 = ir x·, , -t+, , 4., , '4 =(t x2+ Xl2)2, , + t + X4, , 9, , I VI + l dY) dy. Since dy = '2 ill,, , '
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CHAPTER 29 Applications of Integration I, , So,, = (14 X 3 - 2x- 1 ), , 9., , J: = (t - t) - (t - I) =If, , ., . a., ., Fmd the arc length of the catenary y ="l(e x1• + e- x1a ) from x =0 to x = a., y' =t(e x1a + e-·· 1a ) and, therefore,, , So,, , 10. Find the area of the region lying above the x axis and under the parabola y = 4x - xl., , ¥, , Ans., , 11. Find the area of the region bounded by the parabola y =xl - 7x + 6, the x axis, and the lines x =2 and x =6., Ans., , .?f, , 12. Find the area of the region bounded by the given curves., , t ... :, , ', , (a) y = x2, y = 0, x =2. x =5, (b) y = xl, y =0, x = I. x =3, (c) y=4x-xl.y=O,x= l,x=3, (d) x = 1 + ),2, x = 10, (e) x = 3y2 - 9, x = O. Y = 0, Y = 1, (f) x = y2 + 4y. x = 0, (g) y =9 - xl, y =x + 3, (h) y=2-ry=-x, (i) y =xl - 4, Y = 8 - 2x2, G) Y =xl - 4xl. y =4x2, , Ans., Ans., Ans., Ans., Ans., Ans., Ans., Ans., Ans., Ans., , (k), , y =e. y =e-X, x =0, x =2, , Ans., , e +1, e2 -2, , (I), , y = el"+e-xl", y = 0, x = ±a, , Ans., , 2a(e~1), , =0, x = I. x = e, I, (n) y =, - - y=O, x=+l, 1+x2', -, , Ans., , 24, , (0) y=tanx,x=O,x=4, , Ans., , (p) y=25-x2,256x=3y2.16y=9r, , Ans., , ,, , Ans., , (104JIT - 125) 127, , 2, , (m) xy = 12, )', , ,., , Ans., , 39, 20, , 22, , T, , 36, , 8, , 1f, , J¥, .!, 2, , 32, , W-fi, 2, , ,., , 2", , tln2, , 13. Find the length of the indicated arc of the given curve., (a) y3 = 8x2 frornx= I tox= 8, (b) 6xy = xl + 3 from x = 1 to x = 2, (c) 27y2 = 4(x - 2)3 from (2, 0) to (11, 6.[3, , AilS., , tt, , Ans., , 14
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CHAPTER 29 Applications of Integration I, , (d) y=tx2 -tlnx fromx= 1 tox=e, 1C, , \, , 1C, , (e) y =In cos x from x ='6 to x ='4, , (0, , X 2/l, , +y2ll =4 from x =I to x =8, , Ans., Ans., Ans., , te -t, 2, , InC::), 9, , 14. (GC) Estimate the arc length of y =sin x from x =0 to x =1t to an accuracy of four decimal places. (Use, Simpson's Rule with n = 10.), , Ans. 3.8202, , ,.
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Applications of, .Integration II: Volume, A solid of revolution is obtained by revolving a region in a plane about a line t~at does not intersect the, region. The line about which the rotation takes place is called the axis of revolution., Letfbe a continuous function suchthatf(x) ;:: 0 for a:::; x:::; b. Consider the region r& under the graph off,, above the x axis, and between x =a and x =b. (See Fig. 30-1.) If r& is revolved about the x axis, the resulting, solid is a solid of revolution. The generating regions r& for some familiar solids are shown in Fig. 30-2., y, , ., , a, , Fig. 30-1, , Disk Formula, The volume V of the solid of revolution obtained by revolving the region r& of Fig. 30-1 about the x axis is, given by, (disk formula), , y, , y, , rl-------,, , -~-------~-~,x, , (a) Cone, , --r---------~~-x, , h, , (b) Cylinder, , Fig. 30·2, , (e) Sphere
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CHAPTER 30 Applications of I~tegration II: Volume, See Problem 9 for a sketch of the proof of this formula., Similarly, when the axis of rotation is the y axis and the region that is revolved lies between the y axis, and a curve x =g(y) and between y =c and y =d (see Fig. 30-3), then the volume Vof the resulting solid of, revolution is given by the formula, (disk formula), , y, , d, , ...., , ,, , " x=g(y), \, , ,, , I, , I, I, \, , ", , \., , x, , \, , \, \, , --'•, , c, , Rg.30-3, EXAMPLE 30.1: Consider the solid of revolution obtained by revolving about the x axis the region in the first quadrant bounded by the parabola y2 = 8x and the line x = 2. (See Fig. 30-4.) By the disk fomlUla. the volume is, , x, , Rg.30-4, , EXAMPLE 30.2: Consider the solid of revolution obtained by revolving about the y axis the region bounded by the, parabolay=4r and the lines x = 0 andy= 16. (See Fig. 30-5.) To find its volume. we use the version of the disk formula, in which we integrate along the y axis. Thus,
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CHAPTER 30 Applications of Integration II: Volume, y, , --~~--~--~------~x, , Rg.3(}'5, , Washer Method, Assume that 0 ~ g(x) ~f(x) for a ~ x ~ b. Consider the region between x = a and x =b and lying between, y =g(x) and y =f(x). (See Fig. 30-6.) Then the volume Vof the solid of revolution obtained by revolving this, region ~bout the x axis is given by the formula, (washer formula)t, y, , t"-oY = f(x) _""",,r-, , ~~-----+------------~--------z, a, b, , Fig. 30-6, , 1:, , The justification is clear. The desired volume is the difference of two volumes, the volumes 1r (f(X»2dx, of the solid of revolution generated by revolving about the x axis the region under y =f(x) and the volume, 1r (g(x»2dx of the solid of revolution generated by revolving about the x axis the region under y == g(x)., , J:, , A similar formula, , v == 1r, , J: [(f(y»2 - (g(y»2 ]dy, , (washer formula), , holds when the region lies between the two curves x = f(y) and x == g(y) and between y == c and y = d, and it, is revolved about the y axis. (It is assumed that 0 ~ g(y) ~f(y) for c ~ y ~ d.), tThe word "washer" is used because each t" ,in vertical strip of the region being revolved produces a solid that resembles a plu'l1bing, part called a washer (a small cylindrical disk with a hole in the middle).
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CHAPTER 30 Applications of Integration II: Volume, y, , a, , b, , x, , Fig. 30-7, EXAMPLE 30.3: Consi.tier the solid of revolution obtained by revolving about the x axis the region bounded by the, curves, Y = 4x2, x = 0, and y = 16. (The same region as in Fig. 30-5.) Here the upper curve is y = 16 and the lower curve, is y = 4x2. Hence, by the washer formula,, , Cylindrical Shell Method, Consider the solid of revolution obtained by revolving about the y axis the region (JJt in the first quadrant between the x axis and the curve y =f(x), and lying between x =a and x =b. (See Fig. 30-7.) Then the volume, of the solid is given by, , v =2n, , 1:, , xf(x) dx =2n, , 1: xy dx, , (cylindrical shell formula), , See Problem 10 for the justification of this formula., A similar formula holds when the roles of x and yare reversed, that is, the region (JJt in the first quadrant, between the y axis and the curve x =f(y), and lying between y =c and y =d, is revolved about the x axis, V =2n, , f yf(y)dy= 2nf yxdy, , EXAMPLE 30.4: Revolve about the yaxis the region above the x axis and belowy=2x2, and betweenx=O andx=5. By the, cylindrical shell formula, the resulting solid has volume, , Note that the volume could also have been computed by the washer formula, but the calculation would, have been somewhat more complicated., ,., , Difference of Shells Formula, Assume that 0 ~ g(x) ~f(x) on an interval [a, b) with a ~ O. Let (JJt be the region in the first quadrant between, the curves y =f(x) and y = g(x) and between x = a and x = b. Then the volume of the solid of revolution, obtained by revolving (JJt about the y axis is given by, I, , v =2n s: x(j(x)- g(x))dx, , (difference of shells formula)
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CHAPTER 30 Applications of Integration II: Volume, , This obviously follows from the cylindrical shells formula because the required volume is the difference of, two volumes obtained by the cylindrical shells formula. Note that a similar.formula holds when the roles of, x and yare reversed., EXAMPLE 30.5: Consider the region in the first quadrant bounded above by y = xl, below by y =xl, and lying, between x = 0 and x = 1. When revolved aboUl the y axis, this region generates a solid of revolution whose volume, according to the difference of shells fomlUla, is, J, , Cross-Section Formula (Slicing Formula), Assume that a solid lies entirely between the plane perpendicular to the x axis at x =a and the plane perpendicular to the x axis at x =b. For each x such that a ~ x S; b, assume that the plane perpendicular to the x, axis at that value of x intersects the solid in a region of area A(x). (See Fig. 30-8.) Then the volume Vof the, solid is given by, , V=, , r, , A(x)dx, , (cross-section formula)t, , For justification, see Problem 11., , a, , Fig. 30-8, , EXAMPLE 30.6: Assume that half of a salami of length h is such that a cross-section perpendicular to the axis of the, salami at a distance x from the end 0 is a circle of radius ../X. (See Fig. 30-9.) Hence, the area A(x) of the cross-section, is 1C(../X)2 = 1CX. SO, the cross-section formula yields, ., , ----t-~x, , o, , Fig.3()'9, , tThis formula is also called the slicing formula because each cross-sectional area A(x) is obtained by slicing through the solid.
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CHAPTER 30 Applications of Integration II: Volume, , SOLVED PROBLEMS, , 1., , Find the volume of a cone that has height h and whose base has radius r., The cone is generated by revolving about the x axis the region between the line y = x and the x axis,, between x =0 and x =h. (See Fig. 30-2(a).) By the disk formula, the volume of the cone is, , 2., , Fin~, , 3., , Find the volume of a sphere of radius r., The sphere is generated by revolving about the x axis the region between the semicircle y =.jr2 - x 2 and, the x axis, between x =-r and x =r. (See Fig. 30-2(c).) By the symmetry with respect to the y axis, we can use, the part of the given region between x = 0 and x =r and then double the result. Hence. by the disk formula, the, volume of the sphere is, , 4., , Let !'A be the region between the x axis. the curve y =il. and the line x =2. (See Fig. 30-10.), , f, , the volume of the cylinder of height h and radius r., The cylinder is generated by revolving about the x axis the region between the line y =r and the x axis,, between x =0 and x =h. (See Fig. 30-2(b).) By the disk formula, the volume of the cylinder is, , (a) Find the volume of the solid obtained by revolving ~ about the x axis., (b) Find the volume of the solid obtained by revolving ~ about the y axis., y, , 8, , 4, , :---+----~, , Rg.30-10, , (a) The disk fonnula yields the volume, , x
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CHAPTER 30 Applications of Integration II: Volume, (b) (First solution) The cylindrical shells formula yields the votume, , (Second solution) Integrating along the y axis and using the washer formula yields the volume, , S., , Find the volume of the solid' obtained by revolving about the y axis the region in the first quadrant inside the, circle Xl + Y. = r, and between y = a and y = r (where 0 < a < r), See Fig. 30-11. (The solid is a "polar cap" of a, sphere of radius r.), y, , r, , ---O~--------~r--~~X, , Rg.30-11, Integrating along the y axis, the disk formula yields the volume, , 6., , Find the volume of the solid obtained by revolving about the y axis the region in the first quadrant bounded, above by the parabola y = 2 - Xl and below by the parabola)' == xl. (See Fig. 30-12.), y, , "-------~------., , Fig. 30-12, , x
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CHAPTER 30 ,Applications of Integration II: Volume, The curves intersect at (1.1). By the difference of cylindrical shells formula. the volume is, , 7,, , Consider the region ~ bounded by the parabola y = 4x2 and the lines x = 0 and y = 16. (See Fig. 30-5.) Find the, volume of the solid obtained by revolving ~ about the line y =-2., To solve this problem. we reduce it to the Case of a revolution about the x axis. Raise the region ~ vertically, upward through a distance of 2 units. This changes ~ into a region ~' that is bounded below by the parabola, y = 4x2 + 2. on the left by the y axis. and above by the line y = 18. (See Fig. 30-13.) Then the original solid of, revolution has the same volume as the solid of revolution obtained by revolving R* around the x axis. The latter, volume is obtained by the washer formula:, , v =rr, , f:, , f:, , (18 2 -(4x 2 + 2)2) dx = rr (256-16x 4 -16x 2 - 4) dr, , =1r(252x-.lfr -Jtr}I =rr(504- 5~2 -.!f}= 53~t1r, , 8,, , (\s in Problem 7. consider the region ~ bounded by the parabola y = 4x2 and the lines x = 0 and y = 16., (See Fig. 30-5.) Find the volume of the solid obtained by revolving ~ about the line x = -I., Y, , --~----~--~--~--~x, , Fig. 30-13, , To solve this problem. we reduce it to the case of a revolution about the y axis. Move the region '!A to the right, through a distance of 1 unit. This changes ~ into a region ~' that is bounded on the right by the parabola, y =4(x - 1)2, above by y = 16. and on the left by x = I. (See Fig. 30-14.) The desired volume is the same as that, obtained wIlen we revolve '?)t' about the y axis. The latter volume is got by the difference of cylindrical shells, formula:, , v = 2rr f x(16-4(x-l)2) dx= 21r f x(16- 4x 2+8x - 4) dl, , f, , = 21r (16x- 4x 3 + 8x 2- 4x) dx = 21r(8x2 -, , X4, , = 21r[(72-81 + 72-18) -(8 -1 +t- 2}J= If, , + tx) - 2x 2)]:
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. CHAPTER 30 Applications ·of Integration II: Volume, y, , 16, , 4, , Rg.30-14, , 9. Justify the disk fOmlula: V =TC, , 1:(j(x»2dx., , ., , Divide the interval [a, b) into n equal subintervals, each of length Ax = b - a . (See Fig. 30-15.), Consider the volume V; obtained by revolving the region, , ~j, , n, , above the ith subinterval about the x axis., , If nil and Mj are the absolute minimum and absolute maximum of/on the ith subinterval, then V; lies, between the volume of a cylinder of radius mj and height !:u and the volume of a cylinder of radius M,, and height, , ~x. Thus, TCnI/tu S ~ s TCM/~ and, therefore, m/ S TCl S M/. (We have assumed that the, , volume of a cylinder of radius r and height h is ltrh.) Hence, by the intermediate value theorem for the, continuous function, , t, , (f(X»2,, , there exists, , x;, , in the ith subinterval such that, , ~ =(j(xn)2, , and. therefore,, , Y; =TC(j(X;) ~x. Thus,, , ., , V=, , ~:~' ,, •,-04 •.'_ .... ,, ~",:,~~:, , ;..:;, , :t Y;, , = TC :t(j(X;»)2 Ax, , Letting n -+ -i-, we obtain the disk formula., , 1-1, , y, , a, , '--------'--.... x, , Rg.30-15
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CHAPTER 30 Applications of Integration II: Volume, , r, , 10. Justify the cylindrical shells fonnula: V =2lr xf(x) dx., Divide [a, b] into n equal subintervals, each of length tlx. (See Fig. 30-16.) Let f1ii be the region above the, ith subinterval. Let, , x; be the midpoint, , XI-!; Xi, , of the ith interval. The solid obtained by revolving the region, , f1i, about the y axis is approximately the solid obtained by revolving the rectangle with base tlx and height, , i, =, . !(x~)., , The latter solid is a cylindrical shell, that is, it lies between the cylinders obtained by revolving the, rectangles with the same height !(x;) and with bases [0, xi-!] and [0, Xi]' Hence, it has volume, ,, , ,, , ", , ', , =, , lrXU(x;>-lrX~.f(x:> lr!(x;)(x; - xU, = lr!(x;)(xi - XH)(Xi + x/-I)= lr!(x;)(2x;)(tlx)= 2lrx;!(x;Xtlx), , J:, , Thus, the total Vis approximated by 2lri;:!(x;)tlX which approaches 2lr x!(x)dx as n ~ +00., i=1, , 11. Justify the cross-section fonnula: V =, , 1:, , A(x) dx., , Divide [a, b) into n equal subintervals [xI-!' Xi], and choose a point x; in [XI-!' xJ. If n is large, tlx is small, and the piece of the solid between xi-! and Xi' will be close to a (noncircular) disk of thickness tlx and base area, , ., , ,, , A(xn. (See Fig. 30-17.) This disk has volume A(x;)tlx. So V is approximated by LA(x;>tlx, which approaches, b, , f.• A(x) dx as n -+ +00., , ,. l, y, , ------'---~, , a, , x, , Fig. 30-16, , I', , :::t--r---+----. x, a, ",', , Fig. 30-17
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CHAPTER 30 Applications of Integration II:' Volume, , ~~.~:.;-, , ~-~:~;~:;, , //.'., ','\', , ., , 12. A solid has a circular base of radius 4 units. Find the volume of the solid if every plane section perpendicular to a, particular fixed diameter is an equilateral triangle., , Take the circle as iri Fig. 30-18. with the fixed diameter on the x axis. The equation of the circle is r + i = 16., The cross-section ABC of the solid is an equilateral triangle of side 2y and area A(x) =J3f =J3(l6 - r). Then,, by the cross-section formula,, , v = J3 J: (16 - x") (t>: =..J3(16x-tx3)]: = 2~6..J3, 13. A solid has a base in the form of an ellipse with major axis 10 and minor axis 8. Find its volume if every section, ', , .... ,. ,, , perpendicular to the major axis is an isosceles triangle with altitude 6., Take the ellipse· as in Fig. 30-19, with equation, , ~~ + i~ = 1. The section ABC is an isosceles triangle of base, , 2),. altitude 6, and area A(x) =6)' =6{ ~../25 - x 2 ). Hence,, , V =¥f5 ../25-x1dx = 601r, -5, , z, , x, , y, Fig. 30-18, , (Note that, , t "/25- x1dx is the area of the upper half of, , the circler +f, , =25 and, therefore, is,equal to 25m2.), , z, , c, , '----I--x, , y, , Fig. 30-19, , SUPPLEMENTARY PROBLEMs, '•••, , "", , ,, , "',, , .., , I, , ', , 14. Consider the region ~ bounded by the parabola f, , =8x and the line x =2. (See Fig. 30-4.), , (a) Find the voluille of the solid generated hy revolving ~ about the)' axis., (b) Find the volume of the solid generated by revolving ~ about the line x = 2., , Ans., , (a), , 12~1r;, , (b), , 2~~1r
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CHAPTER 30 Applications of Integration II: Volume, , 15. Find the volume of the solid generated by revolving the region between the x axis and the parabola y = 4x - xl, about the line y = 6., Ans., , 14081t', -15-, , 16. Find the volume of the torus (doughnut) generated by revolving the circle (x - a)2 +f, where 0 < b < a., , =b2 about the y axis,, , Ans. 2x2a1J2 ., 17. Consider the region ~ bounded by y= -xl - 3x+6 and x + y = 3. Find the volume of the solid generated by, revolving ~ about:, (a) the x axis; (b) the line x =3., Ans. (a) 17f;1t'; (b) 25f1t', In Problems 18-26, find the volume generated when the given region is revolved about the given line. Use the disk, formula., 18. The region bounded by y =2x2. y =O. x == O. x =5, about the x axis., Ans. 25001t, 19. The region bounded by xl AilS., , f, , = 16. Y = O. x = 8, about the x axis., , 2561t', -3-, , 20. The region bounded by y = 4xl. x = 0, y = 16. about y =16. (See Fig. 30-5.), Ans., , 40961t', -15-, , 21. The region bounded by f = xl. y =O. x = 2, about the x axis., Ans. 41t, 22. The region bounded by y =xl. y = O. x == 2. about x = 2., AilS., , 161t', -5-, , 23. The region within the curve f = x'(1 - xl). about the x axis., AilS., , 24. The region within the ellipse 4xl + 9f = 36, about the x axis., Ans., , 161t, , 25. The region within the ellipse 4xl + 9f =36, about the y axis., Ans. 241t
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CHAPTER 30, ~":\'i, , 26. The region within the parabola x =9 -, , Ans., , Applications of Integration H: Volume, , r and between y =x - 7 and the y axis, about the y axis., , 96311', , -5-, , In Problems 27-32. find the volume of the solid generated by revolving the given region about the given line. Use the, washer fonnula., 27. The region bounded by y = lr, y = 0, x = 0, x =5, about the y axis., , Ans. 6251t, 28. The region bounded by xl -, , Ans., , r = 16, y =0, x = 8, about the y axis:, , I 28.J31t, , 29. The region bounded by y = xl, x =0, y =8. about x = 2., , Ans., , 14411', , S-, , 30. The region bounded by y =xl, y = 4x.,... xl, about the x axis., , Ans., , 3211', , -3-, , 31. The region bounded by)' =xl, Y =4x - xl, about y = 6., , Ans., , 6411', -3-, , 32. The region bounded by x =9 -, , Ans., , r. Y= x - 7, about x = 4., , 15311', -5-, , In Problems 33-37, find the volume of the solid generated by revolving the given region about the given line. Use the, cylindrical shells formula., , =, , 33. The region bounded by y lr, y =0, x =0, x = 5. about x =6., , Ans. 3751t, 34. The region bounded by )' = xl, y =0, x =2, about y =8., , Ans., , 32011', -7-, , 35. The region bounded by )' = xl. y = 4x - x2, about x = 5., , Ans., , 6411', -3-, , 36. The region bounded by )', , Ans., , =xl :- 5x + 6 and)' =0, about the)' axis.
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CHAPTER 30 Applications of Integration II: Volume, 37. The region bounded by x =9 - f, y =x -7, x =0, about y =3., Am., , 369n, , -2-, , In Problems 38-42, find the volume generated by revolving the given region about the given line. Use any appropriate, method., 38. The region bounded by y =e- z ', y = O. x =0, x = I, about the y axis., Am. 1t(1 - e- I ), 39. The region bounded by y =2r, y = 2x + 4, about x = 2., , Ans. 271t, 40. The region bounded by y =2x, y =0, x =0, x = I, about the y axis., , Ans., , 4n, T, , 41. The region bounded by y =xl, x= f, about the x axis., , Ans., , 3n, 10, , 42. The region bounded by xy = 4, Y = (x - 3)2, about the x axis., , 43. Find the volume of the frustum of a cone whose lower base is of radius R. upper base is of radius r, and, altitude is h., , 44. A solid has a circular base of radius 4 units. Find the volume of the solid if every plane perpendicular to a, fixed diameter (the x axis of Fig. 30-18) is: (a) a semicircle; (b) a square; (c) an isosceles right triangle with the, hypotenuse in the plane of the base., , Ans., , (a) 12:n; (b), , lOr;, , (c) 2~6, , 45. A solid has a base in the form of an ellipse with major axis 10 and minor axis 8. Find its volume if every section, perpendicular to the major axis is an isosceles right triangle with one leg in the plane of the base., , Ans., , 640, -3-, , 46. The base ora solid is the first-quadrant region bounded by the line 4x + 5y =20 and the coordinate axes. Find its, volume If every plane section perpendicular to the x axis is a semicircle., , Ans., , IOn, , -3-, , 47. The base of a solid is the circle xl +f = l6x, and every plane section perpendicular to the x axis is a rectangle, whose height is twice the distance of the plane ofthe section from the origin. Find its volume., A,lS., , 10241t, , ~', , ~;-
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CHAPTER 30 Applications of Integration II: Volume, 48. The section of a certain solid cut by any plane perpendicular to the x axis is a circle with the ends of a diameter, ., lying on the parabolas = 4x and xl = 4y. Find its volume., , r, , Ans., , 65611r, 280, , 49. The section of a certain solid cut by any plane perpendicular to the x axis is a square with the ends of a diagonal, lying on the parabolas y2 =4x and xl =4)'. Find its volume., .,, , Ans., , 144, , 35, , 50. A hole cif radius 1 unit is bored through a sphere of radius 3 units, the axis of the hole being a diameter of th~, sphere. Find the volume of the remaining part of the sphere., , Ans., , 641r,fi, , -3-
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CHAPTER 32 Techniques of Integration II, By definition of the inverse tangent, -7tl2 < () < 7tI2. So, cos () > 0 and, therefore, sec () > O. Thus,, , sec() =lsec()l= "4 + x 2/2. Hence,, , I, , dx, , 2sec 2 () d(), 4tan2 ()(2sec(), , I z2"4+x2 =, = II sec() d() II cos() d(), 4 tan, 4 sin, 2 (), , 2 (), , -41 I(sin()t2 cos() d(), , =t<-(sin()t ' )+C=-4 ~ ()+C, sm, Now we must evaluate sin ()., sin()=tan()=, , Analytic method:, , sec(), , x/2, = x ., "4+x2/2 "4+X2, , Geometric method: Draw the right triangle shown in Fig. 32-1. From this triangle we see that sin () = xl "4 + x 2 •, (Note that it follows also for () < 0.), , ..J4+;2 +C, , Hence,, , 4x, , Fig. 32-1, , This example illustrates the following general rule:, , Strategy I., , If, , .Ja2 +x2 occurs in an integrand, try the substitution x = a tan ()., , EXAMPLE 32.12:, , Find, , I x Jdx9- x ., 2, , 2, , Let x = 3 sin (), that is, () = sin-I (xl3). Then dx = 3 cos () d() and, "9 - x 2 = "9 - 9sin 2 () = 3"sin 2 () = 3"cos 2 6 = 31cos61, By definition of the inverse sine, -7tl2 < ()< 7tI2 and, therefore, cos, , (», , O. Thus, cos() = Icos()1 = "9 - x 2 13, Now,, , =lIcsc 8d8, I x "9dx-x =1 9 sin3cos()d(), 8 (3 cos 8) 9, 2, , 2, , ,, , 2, , 2, , = _1 cot 8 + C = _1 C?S 8 + C = _ 1 "9 - x 13 + C = _1 ~ + C, 9, 9 sm 8, 9, xl3, 9, x, 2, , This example illustrates the following general method:, , Strategy II., EXAMPLE 32.13:, , If "a 2 - x2 occurs in an integrand, try the substitution x = a sin, , e.
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CHAPTER 32 Techniques of Integration II, , x-2 +C, 4.J4x-X2, , In Problems 72 and 73, first apply integration by parts., , 72., , Jx sin-I x dx == t<2X2 -1) sin-I x + t x.Jl- x2 + C
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Techniques of Integration III:, Integration by Partial Fractions, We shall give a general method for finding antiderivatives of the form, , J~~~~ dx, where, , N(x) and D(x) are, , polynomials. A function of the form ~~~~ is called a rational function. (N(x) is the numerator and D(x) is, the denominator.) As examples, consider, x-I, , Jx +8 dx, , X 3 -x, --dx, J x+2, , and, , 3, , Two restrictions will be assumed, neither of which limits the applicability of our method: (i) the leading, coefficient (the coefficient of the highest power of x) in D(x) is +1; (ii) N(x) is of lower degree than D(x). A, quotient N(x)lD(x) that satisfies (ii) is called a proper rational function. Let us see that the restrictions (i)-(ii), are not essential., EXAMPLE 33.1:, note that, , Consider the case where, , f, , ~~~~ is 5x ~~~, 8, , _, , 4' Here, our first restriction is not satisfied. However,, , 2X3, dx-1f 2x3, 5x + 3x - 4 - 5 x 8 + t x 8, , t, , dx, , The integral on the right side satisfies restrictions (i) and (ii)., , EXAMPLE 33.2: Consider the case where, divide N(x) by D(x):, , ~~x~, x, , is, , J. Here, our second restriction is not satisfied. But we can, , ~s +, x, , +, , Hence,, , f, , J, , and the problem i$ reduced' to evaluating l~;: dx, which satisfies our restrictions., , A polynomial is said to be irredllcible if it is not the product of two polynomials of lower degree., Any linear polynomialf(x) =ax + b is automatically irreducible, since polynomials of lower degree than, f(x) are constants andf(x) is not the product of two constants., ., Now consider any quadratic polynomial g(x) = or + bx + c. Then, g(x), , is irreducible if and only if b2 - 4ac < 0
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CHAPTER 33 Techniques of Integration III, , To see why this is so, assume that g(x) is reducible. Then g(x) :;:; (Ax, x :;:; -DIC are roots of g(x). The quadratic formula, , + B)(Cx + D). Hence, x:;:; -BIA and, , should yield these roots. Therefore, b 2 - 4ac cannot be negative. Conversely, assume b2 - 4ac ~ O. Then the, quadratic formula yields two roots of g(x). But, if r is a root of g(x), then g(x) is divisible by x - r. tHence,, g(x) is reducible., EXAMPLE 33.3:, (a) r + 4 is irreducible, since IJ2 - 4ac =0 - 4(1)(4)::= -16 < O., (b) r + x - 4 is reducible, since IJ2 - 4ac = I - 4(1)(-4) = 17 ~ O., , We will assume without proof the following fairly deep property of polynomials with real coefficients., THEOREM 33.1: Any polynomial D(x) with leading coefficient I can be expressed as a product of linear factors of, the form x - a and of irreducible quadratic factors of the form r + bx + c. (Repetition of factors is permitted.), EXAMPLE 33.4:, (a) x 3 - 4x=x(x2 -4) =x(x- 2)(x+ 2), (b) x 3 + 4x =x(r + 4), (xl + 4 is irreducible.), , (c) x4 - 9 =(x2 - 3)(r + 3) =(x (d), , Xl -, , J3 )(x + J3 )(r + 3)', , (xl + 3 is irreducible.), , 3x2 -x + 3 = (x+ I)(x- 2)2, , Method of Partial Fractions, Assume that we wish to evaluate, , J~~~~, , dx, where, , ~~~~, , is a proper rational function and D(x) has, , leadin~, , coefficient 1. First, write D(x) as a product of linear and irreducible quadratic factors., Our method will depend on this factorization. We will consider various cases and, in each case, we will, first explain the method by means of an example and then state the general procedure., , Case I, D(x) is a product of distinct linear factors., EXAMPLE 33.5:, , Find, ;, , In this case, D(x) = r, , Jx!:4', , - 4 = (x -, , 2)(x + 2). Write, I, =~+_B_, (x-2)(x+2) x.-2 x+2, , It is assumed that A and B are certain constants, that we must now evaluate. Clear the denominators by multiplying, both sides by (x - 2)(x + 2):, 1 =A(x + 2) + R(x - 2), , (I), , First, substitute -2 for x in (I): 1 =A(O) + B(-4) =-48. Thus, B =-t., Second, substitute 2 for x in (I): 1 =A(4) + B(O) ::: 4A. Thus, A = t. Hence,, 1, (x-2)(x+2), t, , 1, , 1, , 1, , I, , 4" x-2 -4" x+2, , In general, if a polynomial h(x) has r as a root, then h(x) must be divisible by x - r.
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CHAPTER 33 Techniques of Integration III, , Jx2~4 = Hi x~2 -i X!2)dt=tlnlx-21-tlnlx+21+C, , So,, , =t(lnlx-21-lnlx+21)+C, x 2, =l.lnl, - 1+ C, 4, x+2, , EXAMPLE 33.S:, , F' d, , Jx3 +x2+ 1)-6x', dt, (x, , •, Factoring the denominator yields x(xl + x - 6) = x(x - 2)(x + 3). The integrand is x(x.:2}(~ + 3) ., Represent it in the following form:, In, , x+ 1, =A.+.JL.+-.L, x(x-2)(x+3) x x-2 x-3, , Clear the denominators by mUltiplying by x(x - 2)(x + 3):, , x + 1 = A(x - 2)(x + 3) + Bx(x + 3) + Cx(x - 2), , (2), , Let x be 0 in (2): I = A(-2)(3) +B(0)(3) + C(O)(-2) = -6A. So, A = -to, Let x be 2 in (2): 3 = A(O)(5) + B(2)(5) + C(2)(0) = lOB. So, B =m-., Let x be -3 in (2): -2 =A(-5)(0) + B(-3)(0) + C(-3)(-5) = 15C. So, C =-k, , (x+l)dt, J x 3 + x2 - 6x, , Hence,, , J(_11+l_1__ .2_I_)dt, 6 x 10 x + 2 15 x + 3, = -tin Ixl +ib-In Ix + 21-iln Ix + 31 +C, , A, , General Rule for Case I, , Represent the integrand as a sum of terms of the form - - for each linear factor x - a of the denominax-a, tor, where A is an unknown constant. Solve for the constants. Integrating yields a sum of terms of the form, A In Ix- al., , Remark.' We assume without proof that the integrand always has a representation of the required kind. For, every particular problem, this can be verified at the end of the calculation., Case II, D(x) is a product of linear factors, some of which occur more than once., EXAMPLE 33.7:, , F' d, In, , J (3x+5)dt, , X 3 _X 2 _X+1', , First factor the denominator: t, , x3- xl - x + I =(x + l)(x - \)2, Then represent the integrand, , t, , 3, , x -, , 3; + 5 I as a sum of the following form:, -x+, , In trying to find linear factors~of a denominator that is a polynomial with integral coefficients, test each of the divisors r of the constant, term to see whether it is a root of the polynomial. If it is, then x - r is a factor of the polynomial. In the given example, the constant, term is 1. Both of its divisors. 1 and -1, tum out to be roots.
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CHAPTER 33 Techniques of Integration III, Note that, for the factor (x - 1) that occurs twice, there are terms with both (x - 1) and (x...,. 1)2 in the denominator., . Now clear the denominators by multiplying both sides by (x + l)(x - 1)2:, , 3x + 5 =A(x _1)2+ B(x+ l)(x- I) + C(x +1), Let x = 1. Then 8 = (O)A + (2)(0)B + (2)C = 2C. Thus, C = 4., Let x = -I. Then 2 = (4)A + (0)(-2)B + (O)C = 4A. Thus, A =t., To find B. compare the co~fficients of x 2 on both sides of (l). On the left it is 0, and on the right it is A + B., Hence, A + B = O. Since A = t, B =, Thus,, , -to, , 3x+5, , 1 -1- - 1 -1- + 4 1- 2 x + 1 2 x-I, (x - 2)2, , Therefore,, , Jx-dx(, 1)2, Jx (3x+5)dx, -x2 -x+ 1-i'lnlx+ll-tlnlx-ll+4 I, , 3, , By Quick Fonnula I,, J(x-lt2dx=-(x-1)-1 = __1J ~=, (x _1)2, x-I, , Jx, , So,, , (3x+5)dx, 1, - x2 -x+ 1 =tlnlx+ll-tlnlx-II-4-=-1+C, . x, , 3, , =tln lx + lI __, 4_+C, lx-II x-I, , EXAMPLE 33.8:, , .J(x+l)dx, , Find x3(x _ 2)2', , Represent the integrand, , x3~:~IJ)2, , in the following fooo:, , Clear denominators by mUltiplying by x3(x - 2)2:, , x + 1 =Ax2(x - 2)2 + Bx(x - 2)2 + C(x - 2)2 + Dx3(x - 2) + Er, Letx=O. Then 1 =4C. So, C=t., Let x = 2. Then 3 = 8E. So, E = t., Compare coefficients of X. Then 1 = 4B - 4C. Since C = t, B = t., 2, , I, , Compare coefficients of x • Then 0 = 4A - 4B +4C. Since B=t andC=t, A =t., Compare coefficients of .0.0 =A + D. Since A =, , So,, , and, , t, D = -to, , (1)
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.-, , CHAPTER 33 Techniques of Integration III, A, , General Rule for Case II, , A, , Foreachrepeatedlinearfactor(x-r)thatoccursktimesinthedenominator,use-1-+ ( ~ )2 + ... + ( k )k, x-r x-r, x-r, as part of the representation of the integrand. Every linear factor that occurs only once is handled as in Case I., , Case III, D(x) is a product of one or more distinct irreducible quadratic factors and possibly also some linear factors, (that may occur more tl)an once)., t., General Rule for Case III, , Linear factors are handled as in Cases I-II. For each irreducible quadratic factor x2 + bx + c, place a term, . the representation, . 0 fth', x2Ax+B, +bx +C 10, e 10tegrand., EXAMPLE 33.9:, , .f, , (x-I)dx, FInd x(x2 + 1)(x2 + 2)', , Represent the integrand as follows:, , Clear the denominators by mUltiplying by x(,Xl + I)(r + 2)., x - I =A(r + l)(r '+ 2) + (Bx + C)x(x2+ 2) + (Dx + E)x(r + 1), , Multiply out on the right:, , x-I, , =(A + B + D)x4 + (B+ E)r + (3A + C + D)r + (2C + E)x + 2A, , Comparing coefficients, we get:, 2A =-1,, , 2C+E=I,, , 3A+C+D=0,, , B+E=O,, , A+B+D=O, , Hence, A = -t and, therefore, C + D = t. B + D = t. From the latter two equations, C - B = I. From 2C + E = I and, B + E = 0, we get 2C - B = I. Now, fromC - B = 1 and 2C - B = I, we get C = 0. Hence, from C - B= I, B = -1., Then, from B + D = t, it follows that D =t. Finally, from B + E =0, E =1., Thus,, , Hence,, , Now,, , f x/+l dx=!f }~I dx=tln(x, , Also,, , 3x + 2, , x 2 + 2', , f, , 2, , (by Quick Formula 11), , +1), , dx -- f--.lL, dx + f_2dx, x +2, Xl + 2, 2, , =~ f x?: 2 dx+Ji tan-I (12 )=t ln, Therefore,, , - I) dx, +2), f x(x2(x+1)(x, 2, , t, , 2, , Ji, , -tlnlxl+ In(x +1)+T tan, , (X, , -I (, , 2, , +2)+ Ji tan-I, , X ), Ji, +C, , (12)
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CHAPTER 33 Techniques of Integration III, Case IV, D(x) is a product of zero or more linear factors and one or more irreducible quadratic factors., General Rule for Case IV, , Linear factors are handled as in Cases I-II. For each irreducible quadratic facto,r xl + bx + c that occurs to, the kth power, insert as part of the representation of the integrand., , EXAMPLE 33.10:, Let, , Find, , J(x2x ++1)23 dx., 2, , 2, , 2X2 + 3 = Ax + B + Cx + D . Then, (X2+1)2, x 2 +1, (x2+1)2, , 2x2 + 3 = (Ax + B)(x2 + 1) + Cx + D = Axl + Bx2 + (A + C)x + (B + D), , Compare coefficients: A, , = O. B = 2, A + C =O., , B + D = 3. Hence. C =0, D = I. Thus., , =2 tan-l x + J(x2 ~ 1)2 dx, In the second integral. let x =tan 8. Then, , J, , (x2, , ~ 1)2 dx =J~~ :8 =Jcos 8 d8 =H8 + sin8cos8), 2, , =~(8+ tan~8+ 1)= ~(tan-I x + x/+ I), Thus,, , SOLVED PROBLEMS, , 1•, , F111dJx, , 4, -, , xl -x- 1 dx, , Xl, , -x2, , ., , The integrand is an improper fraction. By division,, , We write, , x+l, X2(X-', , I), , = A +JL+~and, obtain, 2, x, , x, , x-I, , x+ I, , =Ax{x -, , I) + B(x - I) +, , Cr
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CHAPTER 33 Techniques of Integration III, Forx= 0, 1= -B and B = -I. For x= 1,2 = C. Forx= 2,3 = 2A + B +4C and A =-2. Thus., , 2., , Find, , f (x +~:; + 3)', , Let (x +2)(x + 3) = x ~ 2 + x! 3' Clear the denominators:, x = A(x + 3) + B(x + 2), , Let x = -2. Then -2 = A. Let x = -3. Then -3 = -8. So, B = 3., , xdx, f (x+ 2)(x+ 3), , f I, f I, -2 x+2 dx+3 x+3 dx, , = -2 In Ix+ 21 + 31n Ix+ 31+ C = -In«x + 2)2)+ 1n(lx + 31)3 + C, (X +, , I, , 3)31, , =In (x + 2)2 + C, , 3., , Find, , f x(x + 2)(x, +2, dx, -I) ., Xl, , 2, , Let, , x +2, x(x + 2)(x - I), , A. + ~ + .....£... Clear the denominators:, x, , x +2, , x- I, , r+2=A(x+2)(x-l)+Bx(x-l)+Cx(x+2), , Letx = O. Then 2 = -2A. So. A = -I. Let x = -2. Then 6 = 6B. So. B = I. Let x = l. Then 3 = 3C., So. C = I. Hence., x2+2, dx=-f1dx+f-l_dx+f-l- dx, f x(x +, 2)(x -I), x, x +2, x- I, , = -In Ixl +In Ix+21 +In lx-II + C= In I(X+2)(X-I)1, x, +C, 4., , Find, , .1, , 3, , x +I, dx, (x + 2)(x - 1)3 ., , Let, , 3, , x +I, =..A.... +...1L + _C_ + _D_ Clear the denominators:, (x+2)(x-l)) x+2 x-I (X_l)2 (x-I)l, , ,, , xl + I = A(x - I)l + B(x + 2)(x - 1)2 + C(x + 2)(x - I) + D(x + 2), , Let x = -2. Then -7 = -27A. So. A = 1r. Let x = I. Then 2 = 3D. So. D = t. Compare coefficients of Xl. Then I = A + B., Since A = i, B = -W. Compare coefficients of r. 0 = -3A + C. Since A = t" C = t., Thus,, , +1, dx=~f-I-dx+ 2°f-l_ dx +lf-l_ dx + l f - l - dx, f (x +x), 2)(x -I)), 27 x + 2, 27 x - I, 9 (x - 1)2, 3 (x - I»), 7, 20, 71, II, = 27 In Ix + 21 + 27 1n Ix - II -"9 x _ I -"3 (x _1)2 + C, , 'I
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Techniques of Integration IV:, Miscellaneous Substitutions, I., , Assume that, in a rational function, a variable is replaced by one of the following radicals., 1. ~. Then the substitution ax + b =t' will produce a rational function. (See Pro~lems 1-3.), 2. ~ q + px + x 2 • Then the substitution q + px + r =(z - X)2 will yield a rational function. (See Problem, 4.), ~r-q-+-P-X-_-X-2 =~(a+ x)(f3-x). Then the substitution q +px-x2 =(<X+X)2 Z2will produce a rational, , 3., , function. (See Problem 5.), II., , Assume that, in a rational function, some variables are replaced by sin x and/or cos x. Then the substitution x = 2 tan- 1 Z will produce an integral of a rational function of z., The reason that this will happen is that, ., , 1- Z2, , 2z, , slOx=-1-2 ', , +z, , cos x = 1+ Z2, , ,, , (34.1 ), , (See Problem 6 for a derivation of the first two equations.), In the final result, replace z by tan (X/2). (See Problems 7-10.), , WJ."~I;, ."J •., , ,-,-'c, , !~.!!$i, ~ . -~."- -. ., , SOLVED PROBLEMS, , 1., , Find, , -, , Jx ~., I-x, , Let 1- x = Z2. Then x= I -, , Z2,, , dx=-2z dz, and, , :.", , dx, J(l-Z2)Z, -2z dz -2J..AL, Jx-/l-x, l-z2, , By integration by partial fractions, one obtains, ln l l + z l+c. Hence,, -2f..AL=I - Z2, 1- z, 2., , Find, , f (x-, , f x~=Inll-~I+c, 1- x, I + 1- x, , ~., , 2) x+2, , Let x + 2 = Z2. Then x = Z2 - 2, dx = 2z dz, and, , dx, f 2z dz 2f dz, f (x-2)../x+2, z(z2-4), z2-4, , tfa:..--,~--
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Improper Integrals, 1, b, , ', , For a definite integral f(x)dx to be defined, it suffice~ that a and b are real numbers and thatf(x) is continuous on [a, b]. We shall now study two different kinds of integrals that we shall call improper integrals., , Infinite Limits of Integration, , See Problems 1-3,5, and 6., (b), , J:, , f(x)dx, , =}~~r f(x)dx, , See Problem 4,, (c), , r:, , f(x)dx, , = J~ f(x)dx+, , J: f(x)dx, , provided that both limits on the right exist. See Problem 7., , Discontinuities of the Integrand, (a) If fis continuous on [a, b] except that it is not continuous from the right at a, then, , r, , f(x)dx =~i~, , r, , f(x)dx, , See Problem 16., (b) Iffis continuous on [a, b] except that it is not continuous from the left at b, then, , J f(x)dx =Ii"! JU f(x)dx, b, , a, , u-+b, , a, , See Problems 9. 10, 12, 14, and 15., (c) Iffis continuous on fa, b] except at a point c in (a. b), then, , J ~ f(x)dx =.= Ii"! J" f(x)dx+ lim Jb f(x)dx, a, , lI-+C, , a, , u-+c+ u, , provided that both integrals on the right exist. See Problems 11 and 13., When the limit defining an improper integral exists, we say that the integral is convergent. In the opposite case, we say that the integral is divergent. If the integral is divergent, we say that it is equal to +00, (respectively -00) if the limit defining the improper integral approaches +00 (respectively -00).
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CHAPTER 35 Improper Integrals, , 10. Evaluate, , 12o 2dx-x ., , The integrand is discontinuous at x = 2., , f2 2dx = lim, , Jo -, , X, , .-+2-, , r" 2dxX = .-+2lim -In(2 - X)]", 0, , Jo -, , = lim - (In(2 - u) - In 2», "-.2-, , J, , =+00, , Hence, the integral diverges to +00., , r4 dx, 11. Evaluate Jo (x _ 1)2 ., The integrand is discontinuous at x = I, which is inside (0, 4), (See Fig. 35-1.), , ], , ' "dx, I'1m--1, .....I1m, ,- fa ---=, (x - 1)2 ....., x-I, 1-, , ", , 0, , (-L.I - (-1») = .....lim -(-Ll + 1) =+00, , = ....., lim,- - u -, , Hence,, , 1-, , U -, , ., , 4, 4, J4o (xdx )2 is divergent. (We do not have to consider lim r ( dx )2 at all. For r ( dx )2 to be, .....I+JO x- 1, Jo x- l, 1, , convergent, both lim, "~I-, , 1" (x-dxl)2 and ..lim J4 (x-dx1)2, ~1", , 0, , u, , mu~t exist.), , 11, , :, '~. ~ c:, , '.~~\'~~'., , i~~;~:'~, , ~}:.~\~i+., , Rg.35-2, , Fig. 35-1, , 12. Find the area of the region between the curve y =, , Jl-x, x, , 2, , , the x axis, and x = 0 and x = I. (See Fig, 35-2.), , The area is, x dx= lim 1" x, 1o ~, ....., .Jl- x, 1, , 1- 0, , 2, , dx, , = lim - (1- x 2 )/2, ..... 1-, , J., , 0, , (by Quick Formula I), , =lim -[.Jl.=U2 -1]= 1, .. ~l-, , ,
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CHAPTER 35 Improper Integrals, , 22. Show that the areas of the following regions are infinite:, (a) Above the x axis, under y =, , I, , 4"3"', , -x, , from x = -2 tox = 2., , (b) Above the x ax is, under xy = 9 and to the right of x = I., 23. Show that the area of the region in the first quadrant under y = e-2x is t, and that the volume generated by, revolving that region about the x axis is ~., 24. Find the length of the indicated arc: (a) 9y2 = x(3 - x)2, a loop; (b) x2l.l + y2l.l = a2l.l, entire length;, (c) 91 = .x2(2x + 3), a loop, , Ans. (a) 4./3 units; (b) 6a units; (c) 2./3 units, , f (X~b)P converges for p < I and diverges to +, , 25. Show that, , 00, , for p ~ I., , 26. Let 0 5,ft..x) 5, g(x) for a 5, x < b. Assume that lim f(x) = +00 and lim g(x) = +00. (See Fig. 35-3.) It is not hard to, show that, if, , r, , r, , neither does, , fba g(x)dx. A similar result also holds for a < x 5, b. with, , r-.b-, , g(x)dx converges, then so does, , .l-fb-, , f(x)dx and. equ~valently. if, , r, , f(x)dx does not converge, then, , lim replacing lim., .r-+b-, , l-+cr t, , y, , o, , %, , Fig. 35-3, As an example, consider, , l, , r I dx, , Jo -x, , 4 •, , For 0 5, x < 1, ., , l-x4=(I-x)(I+x)(I+x2)<4(1-x) and 1_1_<_'_4, 4 I-x, , l, , l-x, , l, , Since -4' r ,dx does not converge, neither does r I dx 4 •, Jo -x, Jo -x, Now consider, , r 2dx C, • For 0 < x ~ I., Jo x +vx, l, , 2 I JX < ~. Since, x + x vX, , f vX~ dx converges., so does r dxrx', ,, J x + x, l, , 0, , o, , 2, , Determine whether each of the following converges:, (a), , Ans., , rl e'~ (b) Jr,,'4 cosx, dx; (c) r"'· cosx dx, x, J Tx, , Jo Xli), , o, , o, , (a) and (c) converge, , 27. Assume that 0 5,f(x) ~ g(x) for x ~ a. Assume also that lim f(x) = lim g(x) = O. (See Fig. 35-4.) It is not hard to, z-t+-, , show that, if, , .... ..-, , J.-g(x)dx converges, so does f~f(x)dx (and, equivalently, that, if f~f(x)dx does not converge,, , then neither does f~g(x)dt)., , .
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CHAPTER35 Improper Integrals, , o, , :r, , Fig. 35-4, dx, . For x ~ 1, J 1, <~. Since f-~ converges, so does, -As an dxexample, consider J-' .Jx4+2x+6, x4+2x+6 x, , x, J, .Jx· + 2x +6 ., Determine whether or not each of the following converges:, , (a), , J- .Jx3+2x, dx, ; (b) r-e-xldx; (c) J,- dx, J,, .JX+X4, 0, , 2, , AI1.L, , all converge, , 28. Define the gamma function r(t) =, , J;, , xt-ie-'dx for t > O. It can be proved that r(t) is convergent (Ibis is left as a, , project for the student.), (a), (b), (c), (d), , Show that f( I) = 1., Show that f(2) = 1. (Hint: Use integration by parts.), Prove that r(t + I) =tf(t) for all t ::- O. (Hint: Use integration by parts.), Usc part (c) to show that f(1l + I) =II! for all positive integers n. (Recall that II! =1· 2·3·4 .... · n.)
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Applications of Integration III:, Area. of a Surface of Revolution, If an arc of a curve is revolved about a line that does not intersect the arc, then the resulting surface is called, a surface of revoLution. By the suiface area of that surface, we mean the area of its outer surface., Letfbe a continuous function on [a, b] that is differentiable in (a, b) and such thatf(x) ~ 0 for a ~ x ~ b., Then the surface area S of the surface of revolution generated by revolving the graph off on [a, b] about the, x axis is given by the formula, (36.1), , For a justification of this formula, see Problem II., There is another formula like (36.1) that is obtained when we exchange the roles of x and y. Let g be a, continuous function on [c. d] that is differentiable on (c, d) and such that g(y) ~ 0 for c ~ y ~ d. Then the, surface area S of the surface of revolution generated by revolving the graph of g on [c.d] about the y axis is, given by the formula:, S =2n, , d, Ldx 1+ (dx)2, dy dy =2n J.. g(y)~l + (g'(y»2 dy, , (36.2), , Similarly, if a curve is given by parametric equations x = f(u). y = g(ll) (see Chapter 37). and, if the arc, from u =U I to u = U2 is revolved about the x axis, then the surface area of the resulting surface of revolution, is given by the formula, S =2n, , t. ,, , y, , (dX, )2 ( d' )2, dll + d~ du, , (36.3), , Here, we have assumed thatfandg are continuous on [11 1.112] and differentiable on (U 1.1l2)' and thaty= g(ll) ~ 0, on lUI. 112]' Another such formula holds in the case of a revolution around the)' axis., , SOLVED PROBLEMS, 1., , Find the area S of the surface of revolution generated by revolving about the x axis the arc of the parabola i, from x = 0 to x =3., , By implicit differentiation,, , and, , 1+ (, , +36, Ixd)2 =7, 2, , = 12T
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CHAPTER 36 Applications of Integration III, 6., , Find the area of the surface of revolution generated by revolving about the x axis the hypocycloid x =a cos3 (),, y =a sin3 ()., The required surface is generated by revolving the arc from () = 0 to () = 7t. We have, 2 2 2, = -3acos 2()sin(),, = 3asin2()cos(), and, = 9a cos ()sin () . Then, , ~, , ~, , (~, , r+( ~~I, , dx)2 +(dy)2, . ~ (asin 3 B)3acosBsinBdO, S=2(27r)fo~ Y dB, d() d()=2(27r)fo, = 12~27r, 7., , (square units), , Find the area of the surface of revolution generated by revolving about the x axis the cardioid, =cos 3()- cos 20, y =2 sin 0- sin 20., The required surface is generated by revolving the arc from 0 = 0 to 0 = 7t. (See Fig. 36-2.) We have, , x, , <, , I, , ~ = 2co~0 - 2cos28,, , ~ = -2sinO+ 2sin 20,, , v, , Fig. 36-2, and, , (~r +(~J ~S(1-sinOsin20-cosOcos20)=S(I-cosO), Then, S = 27r f: (2sinO - sin20)(2fi.jl- cosO) dO, , f:, , = Sfi7r sinO(I- cos 0)% dO = ( 16f 7r(I - cos 0)% ), = 12~7r, 8., , I, , (square units), , Show that the surface area of a cylinder of radius r and height h is 27trh., The surface is generated by revolving about the x axis the curve y =r from x =0 to x =h. Since, , +( i I = 1. Then, by (3~.1),, , i=, , O., , 1, , h, , I, , h, , S=27r rdx=27r(rx)] =27rrh, o, , 9., , 0, , Show that the surface area of a sphere of radius r is 41tr., The surface area is genemted by revolving about the x axis the semicircle y = .jr2- x2 from x =- r to x = r., By symmetry, this is double the surface area from x = 0 to x = r. Since y2 =r - r., , 2y: = -2x, , and therefore, , dy, x, dx=-y, , and
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CHAPTER 36 Applications of Integration III, Hence, by (36.1),, , S =2.2nf y, o, , ~dx = 4nrl' Idx=4nrxJ' =4nr2, '/7, 0, , 0, , 10. (a) Show that the surface area of a cone with base of radius r and with slant height s (see Fig. 36-3) is 7trs., , ,, , (b) Show that the surface area of a frustum of a cone having bases of radius 1', and 1'2 and slant height it (see, Fig. 36-4) is 1t(r, + 1'2)11. (Note that the/rits/1I11l is obtained by revolving the right-hand segment of the slant, height around the base of the triangle.), ;.",, , ;';.., , ,, , Fig. 36-3, , Fig. 36-4, , (a) Cut open the cone along a slant height and open it up as part of a cirCle of radius s (as shown in Fig. 36-5)., Note that the portion of the circumference cut off by this region is 2nr (the circumference of the base of the, cone.) Now the desired area S is the difference between nsl (the area of the circle in Fig. 36-5) and the area, , 1-, , A, of the circular sector with central angle 8. This area A, is :n (ns2) ::; 8s 2• Since the arc cut off by 8 is, 2ns - 2m; we get 8 = 2ns - 2nr. Thus, A, = n(s - r)s. Hence, S =.nsl - n(s - r)s =nrs square units., s, , \, \, \, , \, \, \, \, , ,,, ,, ,,, , \, , I, I, I, , I, I, , I, I, , I, , I, , I, I, , ,, ,,, , I, , I, , ", Fig. 36-5, , !i = u, +u. Then r2 u, = r, u, + ri"' So, u, = ,,'~ r.' Now, by, ", 1'2, 2, ,, nr2(u, + u) - nr, u, = n(r2- r,)u, + nr2 u = nr, u + nr2 u = nCr, + r2)u, , (b) From the similar triangles in Fig. 36-4, we get, part (a), the surface area of the frustum, square units., , •, IS
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CHAPTER 36 Applications of Integration III, 11. Sketch a derivation of fonnula (36.1)., Assume that [a, b] is divided into n equal subintervals, [XH> Xl], each of length !:u =b ~ a . The total surface, area S is the sum of the surface areas Sl generated by the arcs between the points (Xl_I' f(x k-I » and (xl,f(xJ), each, of which is approximated by the surface area generated by the line segment between (Xl-I.!(Xk-I» and (Xl,J(Xk»', The latteris the area of a frustum of a cone. In the notation of Fig. 36-6, this is, by virtue of Problem lO(b):, , r, , n(J(xk_I )+ f(xk))~{!:u)2 +(~yy = 2n f(Xk_ ) ; f(xk ) 1~{!:u)2 +(~y)2, ,, , Now, f(x l_,)+ f(xll, being the average off(xk_,) andf(xk), is between those two values and, by the, 2, ~, intennediate value theorem, is equal to f(x;) for some x; in (Xk.I.XJ. Also, J{!:u)2 (L\y)2 = VI ~) !:U. By, the mean value, theorem ~, , +, , +(, , =f' (x:) for some x: in (Xk_I' xJ. Thus, S is approximated by the sum, , r, , and it can be shown that this sum can be made arbitrarily close to 2n f{x)Jl + {J/{X))2 tb. t Hence, the latter is, equal to S., (x/c>f(xJ), , ,,, , (Xk-I.!(Xk-I)) f - - - - - - - t, , ,,, , I, I, , I, I, , I, I, , ,,,, ,,,,,, , xk_1, , Rg.36-6, , In Problems 12-20, find the area of the surface of revolution generated by revolving the given arc about the given axis:, , 12. y= mx fromx=O, tox=2; x axis, ,., 13. y=, , txl from =°to x =3; x axis, X, , Ans., , 4mn.JI +m2, , Am., , n(82m -1)/9, , tIn general, the following result can be proved:, Bliss's 1;beorem: Assume/and g are continuous on [a, bl. Divide [a, bl into subintervals, and let iikx =x. - Xi-l' In each [Xk-lo xJ. choose x; and x:' Then the approximating sum, to, , ., , •, , with a = Xo < XI < ... < x. < b,, , IJ( x;)g(X:)~t. can be made arbitrarily close, W, , b, , [Xt-I' Xt ], , J f{x)g{x)dx by letting n -4 +00 and making the maximum lengths of the subintervals approach O., ,
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CHAPTER 36 Applications of Integration III, , ! tr[9-.182 +In (9 +~)], , 14. y=tx 3 frornx=Otox=3;yaxis, , AIlS., , 15. One loop of 8f = x2(l - X2); x axis, , Ans., , In, , 16. y=x3/6+1/2xfromx= 1 tox=2;yaxis, , AIlS., , (11+ln2)n, , 4, , .-, , 17. y=lnxfromx= 1 tox=7;yaxis, , Ans. [34..{i + In (3 + 2-/2), , 18. One loop of 91 = x(3 - X)2; Y axis, , AIlS., , 28n..J3/5, , 19. An arch of x = a( 0 - sin e). )' = a(l - cos e); x axis, , Ans., , 641C.a 2/3, , 20. x = el cost, y = el sin I from I ~ 0 to t = ~ n; x axis, , Ans., , 2n..fi(2elf + 1)/5, , Jtr, , 21. Find the surface area of a zone cut from a sphere of radius r by two parallel planes, each at a distance ~a from, the center., , Ans., , bear, , 22. Find the surface area of a torus (doughnut) generated by revolving the circle xl + (y -b)2 = a2 about the x axis., Assume 0 < a < b.
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Parametric Representation, of Curves, Parametric Equations, If the coordinates (x, y) of a point P on a curve are given as functions x = f(u), y = g(u) of a third variable or, parameter, u, the equations x =f(u) and y =g(u) are called parametric equations of the curve., EXAMPLE 37.1:, (a) x =cos (), y =4 sin2 () are parametric equations, with parameter (), of the parabola 4r + y =4. since, 4r+ y =4 cos2 ()+ 4 sin) ()= 4., (b), , x:= ft,), =4 -, , f is another parametric representation, with parameter t. of the same curve., , It should be noted that the first set of parametric equations represents only a portion of the parabola (Fig. 37-1 (a»,, whereas the second represents the entire curve (Fig. 37-/(b))., , II, t=O, , 8_=_n~__~__~~x, , o, , 8=0, , (b), , (a), , Fig. 37-1, EXAMPLE 37.2:', (a) The equations x =r cos B, y =r sin () represent the circle of radius, with center at the origin, since, x 2 + y2 =r cos2 B+ r sin 2 B= r(cos2 ()+ sin 2 (}) =r2. The parameter Bcan be thought of as the angle from the, positive x axis to the segment from the origin to the point P on the circle (Fig. 37-2)., (b) The equations x = a +, cos e, y = b +, sin erepresents the circle of radius, with center at (a. b). since, (x - a)2 + (y - b)2 =,2 cos' e+,2 sin 2 B= ,2(COS 2 B+ sin' e) = /".
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CHAPTER 37, , Parametric Representation of Curves, , y, , ____+-______, , ~-L----~----, , __ x, , Rg.37-2, , Assume that a curve is specified by means of a pair of parametric equations x =f(lI) and y =g(u). Then, the first and second derivatives : and ~ are given by the following formulas., , (37.1) First Derivative, , d)' (d\')j(dX), dll, , d~ = d;/, , ~~ =: .~ ., , This follows from the Chain Rule fonnula, , (37.2) Second Derivative, , This folIows from the Chain ~ule formula, , ~l ( : ) = ~ . ~ ., , Arc Length for a Parametric Curve, If a curve is given by parametric equations x =f(t), y =get), then the length of the arc of the curve between, the points corresponding to parameter values t I and 12 is, , L =1'2, ", , (dx)2, +(dy )2 dt, dt, dt, , This formula can be derived by an argument similar to that for the arc length fonnula (29.2)., , SOLVED PROBLEMS, , I•, , · I -I', dv an d -d', d~ Y .I t' x = I - SII1, ,, FIll(, tX, .\", d.x, ., - = I - cos 1 and, ~, , t,)', , = I - cos I,, , dy., dy, -d =Sill t. By (37.1), d.x, t, , A..(dY)= (I-cos, dt d.x, , sin I, = I -~t, . Then, , t)(cos t)-(sin t)(sin t), (I - cos 1)2, , cOS/-(cos 2t+sin 2 /), , =, , (1- cos /)2, , cost-I, 1, = (1- cos if = cos /- 1
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Parametric Representation of Curves, , CHAPTER 37, , Hence, by (37.2),, 2, , d y, , I, /, I, d.x 2 = cos I - I (1- cos 1) = (1- cos 1)2, , 2., , ~, , Find: and, , ifx=e'cosl,y=e'sint., , d.x, •, -dI = e'(cos t - sm I). and, , dy, •, dy cos I + sin I, -dI = e'(cos I +sm t). By (37.1), d.x = cos-sm, I . I' Then,, , !L(dy)= (cos I-sin 1)2 -(cos t+sin t)(-sin I-COS I), (cos t - sin 1)2, , dl d.x, , =, , (cos I - sin 1)2 + (cos I + sin 1)2, (cos I - sin 1)2, , 2(cos 21+ sin 2n·, (cos I - sin 1)2, , 2, - (cos I - sin 1)2, So, by (37.2),, 2, , d y=, 2, !e'(COst-sinl)=, 2, dx 2 (cos 1- sin t)2, e' (cos I - sin t), , 3., , Find- an equation of the tangent line to the curve x = Jr, y = t - ~ at the point where t = 4., , L·., , dy, dx = ~ and !!l..d = I + 2, By (37.1). dx = 2Jr + 1. So. the slope of the tangent line when 1=4 is, dI 2", t, I, I, I, 2-/4 + t = Jf. When t = 4, x = 2 and y = An equation of the tangent line is y - = Jf(x - 2)., , t., , 4., , t, , The position of a particle that is moving along a curve is given at time t by the parametric equations x =, 2 - 3 cos I, Y = 3 + 2 sin t. where x and yare measured in feet and I in seconds. (See Fig. 37-3.) Note that, t(x - 2)2 + t<y - 3)2 = I, so that the curve is an ellipse. Find: (a) the time rate of change of x when 1= 1tI3;, (b) the time rate of change of y when 1= 51t13; (c) the time rate of change of the angle of inclination Oof the, tangent line when 1= 21t13., , ~=3sint, , and, , 7r=2cosl.ThentanO=:=tcOlt., , Ir dx 3./3, (a) When 1=3' dt =-2-ft/sec, , 51r dy, (b) When 1=3' dl = 2m = I ft/sec, _, _I, dO _ -t csc 2I _ -6csc 2I, (c) 0 - tan (tcott). So, dt - 1+ t cott 2 I - 9+4.cot2 l', , I', , t=O, , x, , o, Fig. 37-3
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CHAPTER 37. Parametric Representation of Curves, , ;i~~1, , ':";~1r~, , When t, , = 27r,, 3, , dO = --6(2/13)2, = _ 24 , Thus, the angle of inclination of the tangent line is decreasing at the, 31, dt 9+4(-1/13)2, , rate of 1+ radians per second., , S., , Find the arc length of the curve x = t2, )' = t 3 from t = 0 to t = 4., , ~~ =2t.1r =3t, , 2, , and, , (~;r + (~J =4t, , 2, , +9t4 =4t 2 (I+lt 2 )., , Then, , L= J>tJl+ft2dt=tJ:(I+7t2)"2(tt)dt, = H(l + ft 2)3n, , 6., , k= -/r(37.[fi -1), , Find the length ofan arch of the cycloid x = 0 - sin 0, y = I - cos 0 between 0 = 0 and 0 = 2n., , ~~ =sin8, , %, , = I-cosO,, , and, , (%f, , +, , (~~r =(l-cOS8)2 +sin10=2(l-COS8)=4sin 1(!). Then, , L= 2J: sin(!)d8 = -4COS(!), , ., , El., , r, , = -4(COS7r - cosO) = 8, , d 2y, , In Problems 7-11. find. (a) dx' (b) dx 2 ', , 7., , x = 2 + t, Y = 1 + t2, , Ans., , (a) 2t; (b) 2, , 8., , x = t + lit, y = t + 1, , Ans., , (a) PI(P - 1); (b) - 2t3/(t2- 1)3, , 9., , x = 2 sin t, y = cos, , 2t, , Ans., , (a) -2 sint; (b)-I, , 10. x = cos3 0, Y = sin 3 0, , Ans., , (a) -tan 9; (b) 1/(3 cos· Osin 0,1, , Ans., , (a) tan ~; (b) l/(a~cos3~), , 11., , x = a(cos, , tP + tPsin, , IP), y = a(sin ~ - ~cos ~), , 12. Find the slope of the curve x = e-' cos 2t. y = e-2J sin 2t at the point t = O., Ans., , -2, , 13. Find the rectangular coordinates of the highest point of the curve x = 96t, Y = 96t - 16P. (Hint: Find t for, maximum y.), , AIlS., , (288,144), , 14. Find equations of the tangent line and normal line to the following curves at the points determined by the given, value of the parameter:, , 5r, , (a) x = 3e'. Y =, at t =0, (b) x=acos 4 0,y=asin 4 0at 8=f, , Ans., , (a) 3y+5x= 30, 5y-3x= 16; (b) 2x+2y=a,y=x
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CHAPTER 37, , Parametric Representation of Curves, , 15. Find an equation of the tangent line at any point P(x, y) of the curve x = a cos3 1, y = a sin3 1. Show that the length, of the segment of the tangent line intercepted by the coordinate axes is a., Ans., , xsin 1+ ycos 1=~sin21, , 16. For the curve x = (1- 1, y = f - I, locate the points where the tangent line is (a) horizontal, and (b) vertical. Show, that, at the point where the curve crosses itself. the two tangent lines are mutually perpendicular., Ans., , (a) I=±, , f;, , (b) 1=0, , In Problems 17-20, find the length of the specified arc of the given curve., , 17. The circle x =a cos 0, y= a sin ofrom 0= 0 to 0= 21t., Ans., , 2M, , 18. x = e cos t. y = e sin 1from 1=0 to 1= 4., , Ans., , fi(e 4 - I), , 19. x= In~, y= tan-I tfrom 1 = 0 to t = l., , Ans., , In(1 + fi), , 20. x =2 cos 0+ cos20+ I. y =2 sin 0+ sin20., , Ans., , 16, , 21. The position of a point at time 1 is given as x =tt 2• Y =t(6t + 9)3n. Find the distance the point travels from, t =0 to I = 4., Ans., , 20, , 22. Identify the curves given by the following parametric equations and write equations for the curves in terms of x and y:, (a), (b), (c), (d), , x = 31 + 5. y = 41 - 1, x= t+2. y= f, x=I-2 y=_t_, ., 1-2, x = 5 cos I, Y = 5 sin 1, , Ans., Ans., Ans., Ans., , Straight line: 4x - 3y =23, Parabola: y = (x - 2)2, Hyperbola: y = 1 + I, Circle: xl + y2 = is, , 23. (GC) Use a graphing calculator to find the graphs of the following parametric curves:, (a) x= O+sin O,y= I-cos 0, (b) x =3 cos3 0, Y=3 sin3 0, (c) x = 2 cot e, y = 2 sin20, 30, 30 2, (d) x= (1+03)' y= (1+0 3), , (cycloid), (hypocycloid), (witch of Agnesi), (folium of Descartes)
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Curvature, Derivative of Arc Length, Let y == f(x) have a continuous first derivative. Let A(XO, Yo) be a fixed point on its graph (see Fig. 38-1) and denote, by s the arc length measured from A to any other point P(x, y) on the curve. We know that, by fonnula (29.2)., , if s is chosen so as to increase with x. Let Q(x + ~, y + ~y) be a point on the curve near P. Let, the arc length from P to Q. Then, , I", , ~s, , denote, , Jl + dx )2, , ds _ 1m I!.s, _+, A, - uX, dx - t.x--+O, , (dy, , ., , and. similarly,, , FOO), , ds = lim ~s = + 1+ dx, dy ay->O ~y dy, , 2, , The plus or minus sign is to be taken in the first formula according as s increases or decreases as x increases,, and in the second formula according as s increases or decreases as y increases., y, , x, , o, Fig. 38-1, , When a curve is given by parametric equations x =f(u), y =g(u),, , ds, . I:!.s, ( -dx, -=hm-=+, du /1." .... 0 ~u du, , )2 +( dy- )2, du, , Here the plus or minus sign is to be taken according as s increases or decreases as, , ••, , .'!, , increases .
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CHAPTER 38 Curvature, To avoid the repetition of ambiguous signs, we shall assume hereafter that the direction on each arc has, been established so ~at the derivative of arc length will be positive., , Curvature, The curvature K of a curve y = f(x) at any point P on it is defined to be the rate of change of the direction of, the curve at P, that is, of the angle of inclination of the tangent line at P, with respect to the arc length s., (See Fig. 38-2.) Intuitively, the curvature tells us how fast the tangent line is turning. Thus, the curvature is, large when the curve bends sharply., y, , x, , Fig. 38-2, , As formulas for the curvature, we get:, (38.1 ), , or, in terms of y,, (38.2), , For a derivation, see Problem 13., K is sometimes defined so as to be positive. If this j's assumed, then the sign of K should be ignored in, what f o l l o w s . ·, ., , The Radius of Curvature, The radius of curvl;lture R at a point P on a curve is defined by R =, , Ii ~, , provided that K ~ o., , The Circle of Curvature, The circle. of curvature, or osculating circle of a curve at a point P on it, is the circle of radius R lying on the, concave Side of ilie curve and tangent to it at P. (See Fig. 38-3.), , Fig. 38-3
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G-, , CHAPTER 38, , Curvature, , To construct the circle of curvature, on: the concave side of the curve, construct the normal line at P and, on it layoff a segment PC of length R. The point C is the center of the required circle., , The Center of Curvature, The center of curvature for a point P(x, y) of a curve is the center C of the circle of curvature at P. The coordinates ( , ) of the center of curvature are given by, J, , or by, , See Problem 9 for details., , The Evolute, The evolute of a curve is the locus of the centers of curvature of the given curve. (See Problems 11-12.), , ,":", , SOLVED PROBLEMS, , 1., , Find, , %. at P(x, y) on the parabola y = 3x2., , ~~ =Jl+(~y =~1+(6x)2 =.jl+36x, 2., , Find, , 2, , j~ and ~;, at P(x, y) on the ellipse x2 + 4)'2 = 8., , ,;.,, , dy, , dy, , Since 2x + 8y- =0 dx, 'dx, , x, =--4, .. y, , d')", , I+(~, , dy, , 3., , Find, , ~~ at P(, , dx, 4y, and -d = - - . Then, y, x, , l6y2, , = 1+ - , - =, , ) on the curve x, , x-, , = sec, , x 2 +16y" 2+3i, ds_~2+3y22, 2, = -2--2 and -d' - -_2, x, -Y, J, Y, , ,y = tan .
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CHAPTER 38 Curvature, , 4., , The coordinates (x, y) in feet of a moving particle P are given by x = cos t - 1, Y = 2 sin t + 1, where t is the time, in seconds. At what rate is P moving along the curve when (a) t= 57f16, (b) t = 57f13, and (c) P is moving at its, fastest and slowest?, , (a) When t = 57t16, dsldt = ~I + 3(t) = JOI2 ft/sec., (b) When t = 57f13, dsldt = ~I + 3(t) = .fi/2 ft/sec., , ..I',", , (c) Let S = ~: = ...JI + 3cos t. Then c;% = -3co~tsint. Solving dS/dt = 0 gives the critical numbers t = 0,, 7t/2, 7t, 37t/2., , _,~~t:'~, , 2, , ", , When t = 0 and 7t, the rate dsldt = ~l + 3(1) = 2 ft/sec is fastest. When t = 7tI2 and 3rri2, the rate, dsldt = ~l + 3(0) = I ft/sec is slowest. The curve is shown in Fig. 38-4., ""-", .-;,.,-1", ,, , >1, , :..;.'.', , ',', , Fig. 38-4, , 5., , Find the curvature of the parabola f, , !!l=§.., dx, , (a), , y,, , so 1+(dy )2 =1+ 36 and dly =_~dy __ 36, dx, y2, dx 2, y2 dx - y3, , :ii, , !!2, , A, . (ely)2 _, I, _ -1/6, t(3,6).1+ dx -2 and dx2=-"6,soK- 23/2 =-24', , (b) At (t, , ,, , - 3):, , 2, , 1+ (ely)2 = 5 and d y2 =i so K = 4/3, dx, dx, 3', 53/2, , (c) At(O,O),;Z is undefined. But, 6., , =12x at the points: (a) (3, 6); (b) ( 1, -3); (c) (0, 0)., , =M, 75', , t=i=o, l+(tY =1, ~=i,and K=-i., , Find the CUnKlture of the cycloid x = - sin ,)' = I - cos at the highest point of all arch. (See Fig. 38-5.), 'I, , ;,:,:;., , \\:~'~~'!;, , ;-~'~~~~, ~~Ji~:~., , Fig. 38-5
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CHAPTER 38, , Curvature, , To find the highest point on the interval 0 < x < 21t; dyld = sin , so that the critical number on the interval is, 2 = cos, < 0 when =1t, the point =1t is a relative maximum point and is the highest point, of the curve on the interval., , x =1t. Since tPyld, , To find the curvature,, , dx, , dO, , At, , 7., , dy, . 0, dO = sm ,, , =1- cosO,, =1t, dyldx =0,, , dy _ sinO, dx -I-cosO', , 2, , -.!L(, , d y=, sinO ) dO _, dx 2 de 1-cosO dx -, , 1, (1'-cOSO)2, , d"yldx 2 =-t, and K =-t., , Find the curvature of the cissoid l(2 - x), , =.x3 at the point (1, 1). (See Fig. 38-6.), , Fig. 38-6, Differentiating the given equation implicitly with respect to x, we obtain, , -yl + (2 - x)2yy' = 3x2, , (1), , -2)')" +(2- x)2yy" + (2.,... x)2(),')2 - 2yy' =6x, , (2), , and, , From (I). for x = y = 1. -1 + 2y' = 3 and y' =2. Similarly, from (2), for x =y = I and y' = 2, we find y" = 3., Then K = 3/(1 + 4)3/2 = 3$/25., , 8., , Find the point of greatest curvature on the curve y, , dy, , 1, , dx=-X, , and, , So,, , " l, ber'IS, tI1cref ore, x, num, The cnhca, , 9., , =In x., , I The reqUIre, . d pomt, . IS, . (1, . In 2 ) ., =J2', J2 '-2, , Find the coordinates of the center of curvature C of the curve y = f(x) at a point P(x, y) at which y' ~ O. (See Fig. 38-3.), The center of curvature C( , ) lies: (I) on the normal line at P and (2) at a distance R from P measured, toward the concave side of the curve. These conditions give, respectively,, , From the first,, , - x, , =- y'(, , - y). Substitution in the second yields, , and, therefore,, , f3 - y = ± 1+y(y')2, "
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CHAPTE;R 38 Curvature, , To detennine the correct sign, note that, when the curve is concave upward. y" > 0 and, since C then lies above, P, - y > O. Thus, the proper sign in this case is +. <rou should show that the sign is also + when y" < 0.) Thus,, , 1+(V')2, y, , f3=y+~, , and a=x, , 1'[1 +(y')l], Ii, , Y, , 10. Find the equation of the circle of curvature of 2xy + x + y =4 at the point (I, 1)., , Differentiating yields 2y + 2xy' + 1 +y' =O. At (I, 1), y' =-1 and 1 + (y')2 = 2. Differentiating again yields, y" =~. Then, , 41 + 2xy" +y" = O. At (I, 1),, K=, , i!h ', , R =3f ', , a =1-, , -lg) =l, , f3 =1+ 473, , t, , =, , ., , -,,~~,, , ·1 ";,...-, , The required equation is (x - )l +(y -, , )2 =R2 or (x -, , 11. Find the equation of the evolute of the parabola f, , W +(y - W =l., , = 12x., , At P(x, y):, , ., , -:>, , Then, , ~(1+3lt), 3J2x3l1, , ", , -~' ~."~,, , =x+ 2.J3!Jt 3) =3x+6, 3, , and, , "--'i,~:-t!~, , ~t-ttj~:, _, 1+ 361l, f3- y+ -36/yl, , The equations =3x + 6, =- yl/36 may be regarded as parametric equations of the evolute with x and, y, connected by the equation of the parabola, as parameters. However, it is relatively simple in this problem, to eliminate the parameters. Thus, x = ( - 6)13, y = -~36f3, and substituting in the equation of the parabola,, we have, (36f3)211 = 4(a - 6) or 81f32 = 4(a - 6)3, , The parabola and its evolute are shown in Fig. 38-7., , ., ", , '-.';", '1'- "", , {,;,~;;,f.i, , "-;~;;~~, "-:~i-~_, , Fig. 38-7
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CHAPTER 38, + sin, , 12. Find the equation of the evolute of the curve x = cos, , ,y = sin, , -, , cos ., , At P(x, y):, , ~=(kosO,, , dy, dO, , dy, dx, , =O'sm 0,, , =tanO,, ,, , Then, 2, , a = x - tanOscc 0 = x - OsinO= cosO, (sec) 0)/0, and, 2, , sec 0, f3 -- y+ (sec~, 0)/0, and, , = cos ,, , = sin, , )'+ OcosO= sinO, , are parametric equations of the evolute (see Fig. 38-8)., , 11, , Evolute, (Circle), Fig. 38-8, , 13. Derive formula (38.1)., tan, , is the slope of the tangent line and, therefore,, , Hence, , This yields, , from which, , Curvature
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CHAPTER 38 Curvature,, , tis, In Problems 14-16, find tis, dx and dy", 14. r+y2=25, , 15. y2 =x3, , 5, ds_, -/25-x 2 ' dy -, , Ans., , ds, dx, , Ans., , : =t-/4+9x,, , d, , d~ =, , 5, , Ji5-Y2, ~4+9y213, 3y1/l, , Ans., ds, In Problems 17 and 18, find dx', , Ans., , 18. 27ay2=4(x-a)l, , Ans., , : =J(x+2a)/3a, , Ans., , t-/4+91 2, , 21. x = cos t. Y = sin t, , Ans., , I, , 22. x = cos3 t, Y = sinl t, , Ans., , fsin2t, , In Problems 19-22. find ~., 19. x="',y=tl, , 20. x = 2 cos t, Y = 3 sin t, , 23. Find the curvature of each curve at the, given points:, (a) y=x3/3atx=O,x= 1,x=-2, , (c) y=sinxatx=O,, , X=t7C, , (b) r=4ayatx=O.x=2a, (d) y=e-,l atx=O, , Ans. (a) 0, 1i/2, -4v'f71289; (b) 1I2a, Ii/Sa; (c) 0, -1; (d)-2, 24. Show (a) the curvature of a straight line is 0; (b) the curvature of a circle is numerically the reciprocal, of its radius., , 25. Find the points of maximum curvature of (a) y = e'; (b) y = txl., , ,, , r', , Ans. (a) x = -t In 2; (b) x =5~/4, 26. Find the radius of curvature of, (a) x3 + xy2 - 61 =0 at (3. 3)., (b) x=2atan ,y=atan2 at (x.y)., (c) x::: a cos· ,y = a sin4 at(x. y)., , Ans. (a) 5$; (b) 2alsecl 81; (c) 2a(sin4 + cos·, , )312
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CHAPTER 38, , Curvature, , 27. Find the center of curvature of (a) Problem 26(a); (b) y = sin x at a maximum point., , Ans., , (a) C(-7. 8): (b)C(-I'O), , 28. Find the equation, Am., , or the circle of curvature of the parabola f = 12x at the points (0. 0) and (3,6)., , (x - 6)= + y2 = 36: (x - 15)2 + (y + 6): = 288, , 29. Find the equation of the evolute of (a) b2)..2 + (/2y2 = a 2b2: (b) xU3 + jU3 + aU,I; (l:) x = 2 cost + cos 2t,, y =2 sin t + sin 2t., AilS., , (a)(a )~J+(b )213= (a2_b2)2J3; (b) ( + )213+( f3 = t(2sin t - sin2t), , )2J3=2a 2J3;(c) a=f(2cost-cos2t), ,
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Plane Vectors, Scal~rs, , and Vectors, , Quantities such as time, temperature, and speed, which have magnitude only, are called scalars. Quantities, such as force, velocity, and acceleration, which have both magnitude and direction, are called vectors. Vectors are represented geometrically by directed line segments (arrows). The direction of the arrow (the angle, that it inakes with some fixed directed line of the plane) is the direction of the vector, and the length of the, arrow represents the magnitude of the vector., Scalars will be denoted by letters a, b, c, ... in ordinary type; vectors will be denoted in bold type by letters a, b, C, ••. , or by an expression of the form OP (where it is assumed that the vector goes from 0 to P., (See Fig. 39-1(a).) The magnitude (length) of a vector a or OP will be denoted lal or IOPI., B, , p, , o, , a=b, , (a), , (c), , (6), , (d), , Fig. 39-1, , 1\vo vectors a andb are said to be equal (and we write a =b) if they have the same direction and magnitude. A vector whose magnitude is that of a, but whose direction is opposite that of a, is called the negative, of a and is denoted -a. (See Fig. 39-1 (a).), If a is a vector and k is a positive scalar, then ka is defined to be a vector whose direction is that of a and, whose magnitude is k times that of a. If k is a negative scalar, then ka has direction opposite that of a and, has magnitude Ikl times that of a., ', We also assume a zero vector 0 with magnitude 0 and no direction. We define -0 =0, Oa =0, and kO =O., Unless indicated otherwise, a given vector has no fixed position in the plane and so may be moved under, parallel displacement at will. In particular, if a and b are two vectors (Fig. 39-1(b», they may be placed so, as to have a common initial or beginning point P (Fig. 39-1(c» or so that the initial point of b coincides with, the terminal or endpoint of a (Fig. 39-1 (d»., , Sum and Difference of Two Vectors, If a and b are the vectors of Fig. 39-I(b), their sum a + b is to be found in either of two equivalent ways:, , By placing the vectors as in Fig. 39-1(c) and completing the parallelogram PAQB of Fig. 39-2(a). The, vector PQ is the required sum., 2. By placing the vectors as in Fig. 39-1(d) and completing the triangle PAB of Fig. 39-2(b). Here, the, vector PB is the required sum., 1., , It follows from Fig. 39-2(b) that three vectors may be displaced to form a tria1gle, provided that one of, them is either the sum or the negative of the sum of the other two.
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CHAPTER 39, , Plane Vectors, , B, , Q, , i>, b, , A, , a,, , P, P, , (a), , (b), , (d), , (c), , Rg.39-2, , If a and b are the vectors of Fig. 39-1 (b), their difference a - b is to be found in either of two equivalent, ways:, 1., 2., , From the relation a - b =a + (- b) as in Fig. 39-2(c)., By placing the vectors as in Fig. 39-1(c) and completing the triangle. In Fig. 39-2(d), the vector, , BA =a- b., If a, b, and c are vectors, the following laws are valid., , PROPERTY (39.1) (Commutative Law), PROPERTY (39.2) (Associative Law), PROPERTY (39.3) (Distributive Law), , a+b=b+a, a + (b + c) = (a + b) + c, k(a + b) =ka + kb, , See Problems 1 to 4., , Components of a Vector, In Fig. 39-3(a), let a =PQ be a given vector, and let PM and PN be any two other directed lines through P., Construct the parallelogram PAQB. Then, , a=PA+PB, and a is said to be resolved in the directions PM and PN. We shall call PA and PB the vector components of, a in the pair of directions PM and PN., Consider next the vector a in a rectangular coordinate system (Fig. 39-3(b», having equal units of measure on the two axes. Denote by i the vector from (0, 0) to (1, 0), and by j the vector from (0, 0) to (0, 1)., The direction of i is that of the positive x axis, the direction of j is that of the positive y axis, and both are, unit vectors, that. is, vectors of magnitude 1., From the imtial point P and the terminal point Q of a, drop perpendiculars to the x axis, meeting it at M, and N, respectively, and to the y axis, meeting it at Sand T, respectively. Now, MN = ali, with a l positive,, and ST= a2 j, with a2 negative. Then: MN = RQ = ali, ST = PR = a2 j, and, (39.1), , 01, . (a), , M, , N, (b), , Fig. 39-3
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CHAPTER 39 Plane Vectors, , Let us call ali and ~j the vector components of a. t The scalars a l and ~ will be called the scalar components (or the x component and y component, or simply the components) of a. Note that 0 = Oi + OJ., Let the direction of a be given by the angle 8, with 0 ~ 8 < 2n, measured counterclockwise from the positive x axis to the vector. Then·, , lal=~a; +ai, , (39.2), , and, (39.3), , with the quadrant of (J being detennined by, al = lal cos 8, a2 ;' lal sin 8, If a =ali + a2j and b =bd + b2j, then the following hold., , PROPERTY (39.4), PROPERTY (39.5), PROPERTY (39.6), PROPERTY (39.7), , a =b if and only if a l = bl and a2 = b2, ka = /cali + /ca2j, a + b =(a l + bl)i + (a2 + b:zlj, a - b = (a l - bl)i + (a2 - b:zlj, , Scalar Product (or Dot Product), The scalar product (or dot product) of vectors a and b is defined by, a· b = lallbl cos 8, , (39.4), , where 8 is the smaller angle between the two vectors when they are drawn with a common initial point (see, Fig. 39-4). We also define: a • 0 = 0 • a = o., B, , A, p, Rg.394, , From the definitions, we can derive the following properties of the scalar product., PROPERTY (39;8) (Commutative Law), PROPERTY (39.9), PROPERTY (39.10), PROPERTY (39.11), PROPERTy(39.12), PROPERTY (39.13) (Distributive Law), PROPERTY (39.14), , ,, , j., , [, , t A pair of directions, , (such as OM and, , a·b=b·a, , a . a =lal 2 and lal =M, a . b = 0 if and only if (a =0 or b = 0 or a is perpendicular to b), i·i=j·j=landi·j=O, a . b = (a.i + ~j) . (bli + b2j) =albl + a2b2, a.(b+c)=a.b+a.c, (a +b).(c+d) = a ·c+a· d +b· c+b· d, , an need not be mentioned, since they are detennined by the coordinate system.
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a-, , CHA'PTER 39, , Plane Vectors, , Scalar and Vector Projections, In equation (39.1), the scalar a l is called the scalar projection of a on any vector whose direction is that of, the positive x axis, while the vector ali is called the vector projection of a on any vector whose direction is, that of the positive x axis. In general, for any nonzero vector b and any vector a, we define we define a . ~, to be the scalar projection of a on b, and ( a . ~) 1:1 to be the vector projection of a on b., Note that. when b has the direction of the positive x axis,, , (~ee Problem 7.), , ~ = i., , PROPERTY (39.15) a . b is the product of the length of a and the scalar projection of bon a. Likewise, a . b, is the product of the length of b and the scalar projection of a on b. (See Fig. 39-5.), , Fig. 39-5, , Differentiation of Vector Functions, Let the curve of Fig. 39-6 be given by the parametric equations x =f(u) and y = g(u). The vector, r, , =xi + yj = f(u)i + g(ll)j, , joining the origin to the point P(x. y) of the curve is called the position vector or the radius vector of P. It is a, function of u. (From now on, the letter r will be used exclusively to denote position vectors. Thus, a =3i + 4j, is meant to be a "free" vector, whereas r = 3i + 4j is meant to be the vector joining the origin to P(3, 4).), · . -d, dr 0 f the f, ', . h respect to u IS, . de fime d to be l'1m r(u + ~u), - r(u), The denvattve, unctIOn, r Wit, A .., , u, , ~~, , ~, , Straightforward computation yields:, , dr, du, , = dx i+ dy ., du, , du J, , (39.5), , Let s denote the arc length measured from a fixed point Po of the curve so that s increases with u. If 't'is, the angle that drldu makes with the positive x axis, then, tan't' =, , (~~; )/(:) = ~ =the slope of the curve at P, , Fig. 39-6
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CHAPTER 39 Plane Vectors, , ' de, Moreover, dr., du IS a vector 0 f, magmtu, , 2(, (, ), II, dr, du, , =, , dx, du, , dy )2, , + du, , ds, , =du, , (39.6), , whose direction is that of the tangent line to the curve at P.1t is customary to show this vector with P as its, initial point., If now the scalar variable It is taken to be the arc length s, then equation (39.5) becomes, (39.7), , :J, , Z)2 ,, , !he direction of tis T, while its magnitude is (, +(, which is equal to I. Thus, t =drlds is the, Unit tangent vector to the curve at P., Since t is a unit vector, t and dtlds are perpendicular. (See Problem 10.) Denote by n a unit vector at P, having the direction of dtlds. As P moves along the curve shown in Fig. 39-7, the magnitude of t remains, constant; hence dtlds measures the rate of change of the direction of t. Thus, the magnitude of dtlds at P is, the absolute value of the curvature at P, that is, Idtldsl =1K1, and, dt, ds =IKI n, , (39.S), , 1/, , Fig. 39-7, , SOLVED PROBLEMS, , 1., , Prove a + b = b + a., From Fig. 39-S, a + b = PQ = b + n., Q, , Fig. 39-8, , 2., , Prove(a+b)+c=a+(b+c)., From Fig. 39-9, PC = PB + BC = (a + b) + c. Also, PC = PA + AC = a + (b + c).
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CHAPTER 39, , Plane Vectors, , c, B, , p, , Fig. 39·9, , 3., , Let a, b, and c be three vectors issuing from P such that their endpoints A, B, and C lie on a line, as shown in, Fig. 39-10. If C divides BA in the ratio x: y, where x + y = 1, show that c =xa + yb., Just note that, c = PB+ BC = b+x(a - b) = xa +(J-x)b =xa + yb, As an example, if C bisects BA, then c = t(a + b) and BC = Ha - b)., , B, A, , P, Fig. 39-10, , 4., , Prove: The diagonals of a parallelogram bisect each other., Let the diagonals intersect at Q, as in Fig. 39-11. Since PB = PQ + QB = PQ - BQ, there are positive numbers x and y such that b =x(a + b) - y(a - b) =(x - y)a + (x + y)b. Then x + y = 1 and x - y = O. Hence, x = y = t,, and Q is the midpoint of each diagonal., , c, B, , p, Fig. 39-11, , 5., , For the vectors a = 3i + 4j and b = 2i - j, find the magnitude and direction of (a) a and b; (b) a, (a) For a = 3i + 4j: lal = Ja~ + ai, angle and is 53°8'., For b, , + b; (c) b -, , a., , =./32 + 4 2 = 5; tan 0 =a2 /a l =t and cosO =a/lal =t; then 0 is a first quadrant, , =2i - j: Ibl =.J4+T =$; tan 0 =-t and cosO = 2/ /5; 0= 360° -, , 26°34' = 333°26'.
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CHAPTER 39 Plane Vectors, , (b) a + b =(3i +4j)+ (21 - j) = 5i + 3j. Then la + bl= .JS1 + 31 =./34. Since tan 0 = t and cosO= 5/../34,, , 0= 30°58'., (c) b-a=(2i- j)-(31 +4j)=-I- 5j. Then Ib-al=.j26. Since tan 0=5 andcosO=-l/.j26, 0= 258°41'., 6., , Prove: The median to the base of an isosceles triangle is perpendicular to the base. (See Fig. 39-12, where lal = Ibl.), , .;• ",'_0-, , .,.~~, , ., , ~' ', ..... .,1 •• -", , ( ,,!, , • ~ . ., , <,f~::~t~':' ',~, , i,:~~L,, , Rg.39-12, , From Problem 3, since m bisects the base, m = t(a +b). Then, m·(b - a)=t(a + b)· (b-a), =t(a· b- a·a + b·b - b· a)= t(b·b- a ·a)=O, Thus, the median is perpendicular (0 the base., 7., , If b is a nonzero vector, resolve it vector 8 into components a. and 8 2, respectively parallel and perpellolcular to b., , In Fig. 39-13, we have 8 = 8. + 8 2, 8. = cb, and 8 2 ' b = O. Hence, 82 = 8, a·b, (8 - cb) . b = a . b - dbl 2 = 0, whence c = Ibl2 . Thus,, , - 8., , = a - cb. Moreover., , 82 ., , b=, , and, The scalar a . 1:1 is the scalar projection of a on b. The vector ( a . 1:1 ) 1:1 is the vector projection of 8 on b., , Rg.39-13, , 8., , Resolve 8 = 4i +3j into components a. and a2, parallel and perpendicular, respectively, to b = 31 +j., ., a·b 12+3 3, From Problem 7, c =Ibj2 =\ 0 = 2"' Then, , ,j:
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Plane Vectors, , CHAPTER 39, , At ()= 7tl4,, t=, , ., , 1, 1, -JiI+Ji, j,, , Ji. Ji, , Idtl, , dt, 2, ds = 3a 1+ 3a j, IKI= cis = 3a, , ., , 1 dt, and n = iKi ds, , 1, 1., =.fi, i + .fi J, , I, , 13. Show that the vector a =ai + bj is perpendicular to the line ax t by +c =O., I, , Let P,(x,. y,) and P2(~' Y2) be two distinct points on the line. Then ax, + by, + c = 0 and ax2 + bY2 + c = o., Subtracting the first from the second yields, (I), , Now, , By (1), the left side is zero. Thus, a is perpendicular (normal) to the line., 14. Use vector methods to find:, , (a) The equation of the line through P,(2, 3) and perpendicular to the line x + 2y + 5 = O., (b) The equation of the line through, P,(2, 3) and P2(5, -I)., Take P(x, y) to be any other point on the required line., (a) By Problem 13, the vector a =i + 2j is normal to the line x + 2y + 5 =O. Then PIP =(x - 2)i + (y - 3)j is, parallel to a if (x - 2)i + (y - 3).i =k(i + 2j) for some scalar k. Equating components, we get x - 2 =k and, y - 3 =2k. Eliminating k, we obtain the required equation y - 3 = 2(x - 2), or, equivalently, 2x - y - t =O., (b) We have PIP =(x - 2)i + (y - 3)j and P IP2 =3i - 4j. Now a =4i + 3j is perpendicular to PIP z and, hence, to, PIP. Thus, 0 = a· PIP = (4i + 3j)· [(x - 2)i + (y - 3)j] and, equivalently, 4x + 3y - 17 = O., , IS. Use vector methods to find the distance of the point PI (2, 3) from the line 3x + 4y - 12 =O., At any convenient point on the line, say A(4, 0), construct the vector a = 3i + 4j perpendicular to the line. The, required distance is d = IAPII cos () in Fig. 39-14. Now, a . API = la11AP11 cos ()= lal d. Hence,, d=a.AP,, lal, , = (3i+4j)·(-2i+3j), 5, , -6+12_6, -5--'5, , Fig. 39-14, , 16. The work done by a force expressed as a vector b in moving an object along a vector a is defined as the product, , of the magnitude of b in the direction of a and the distance moved. Find the work done in moving an object along, the vector a =3i + 4j if the force applied is b =21 +J., The work done is, (magnitude of b in the direction of a) . (distance moved) = (lbl cos() lal = b . a =(2i + j). (3i +4j) = 10
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CHAPTER 39, , Plane Vectors, , 17. Given the vectors a. b. c in Fig. 39-15. construct (a) 2a; (b) -3b; (c) a + 2b; (d) a + b - c; (e) a - 2b + 3c., 18. Prove: The line joining the midpoints of two sides of a triangle is parallel to and one-half the length of the third, side. (See Fig. 39-16.), , 19. If a, b, c, d arc consecutive sides of a quadrilateral (s~e Fig. 39-17). show that a, Q be two nonconsecutive vertices.) Express PQ in two ways., , + b + c + d =O. (Hint: Let P and, , p, Fig. 39-15, , Fig. 39-16, , Fig. 39-17, , 20. Prove: If the midpoints of the consecutive sides of any quadrilateral are joined, the resulting quadrilateral is a, parellelogram. (See Fig. 39-18.), , c, , r---:;j;.z--_B, , p, , a, , Fig. 39-18, 21. Using Fig. 39-19, in which lal = Ibl is the radius of a c,ircle. prove that the angle inscribed in a semicircle is a right, angle., , Fig. 39-19, , 22. Find the length of each of the following vectors and the angle it makes with the positive x axis: (a) i, (c) i + fjj; (d) i - fjj., , An.\'., , (a)./2. fJ =, , + j; (b) -i + j;, , t,.; (b) ./2, fJ = 3,./4; (c) 2, fJ = 7tl3; (d) 2, fJ= 57t13, , 23. Prove: If u is obtained by rotating the unit vector i counterclockwise about the origin tht"cIugh the angle 6, then, u = i cos 6+ j sin fJ.
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••, , CHAPTER 39 Plane Vectors, 24. Use the law of cosines for triangles to obtain a . b =lallbl cosO = t(l al 2 + Ibl 2 -.lcF)., , 25. Write each of the following vectors in the form at + bj., (a) The vector joining the origin to P(2. -3); (b) The vector joining P,(2, 3) to P2(4, 2);, (c) The vector joining P2(4, 2) to P, (2. 3); (d) The unit vector in the direction of 31 + 4j;, (e) The vector having magnitude 6 and direction 120°, , Ans. (a) 21- 3j; (b) 21 - j; (c)-2i+ j; (d) ti+tj; (e) -31+3./3j, , 26. Using vector methods, derive the formula for the distance between P,(x"y,) and P2(X2' Y2)', . 27. ,Given 0(0, 0), A(3, 1), and B(I, 5) as vertices of the parallelogram OAPB, find the coordinates of P., , Ans. (4,6), 28. (a) Find k so that a = 3i + 2j and b = i + kj are perpendicular. (b) Write a vector perpendicular to a = 2i + 5j., , 29. Prove Properties (39.8) to (39.15)., 30. Find the vector projection and scalar projection of b on a, given: (a) a = i - 2j and b = -31 + j; (b) a = 2i + 3j a~d, b =lOi + 2j., Ans., , (a) -I + 2j, -$.; (b) 4i + 6j, 2J13, , 31. Prove: Thr~ vectors a, b, c will, after parallel displacement, form a triangle provided (a) one of them is the sum, of the other two or (b) a + b + c =0., 32. Show that a = 31 - 6j, b = 4i + 2j, and c = -71 + 4j are the sides of a right triangle. Verify that the midpoint of the, hypotenuse is equidistant from the vertices., 33. Find the unit tangent vector t = dr/ds, given: (a) r = 41 cos 0+ 4j sin 0; (b) r, , =eel + e-ej; (c) r = Oi + 02j., , 8, , Ans., , 1+28j, (), a -I.., sm O', + J cos 0; (b) eli18- e- 1 ; (c) r.-:-;;;:;, e + e- •, vI +40 2, , 34. (a) Find n for the curve of Problem 33(a); (b) Find n for the curve of Problem 33(c); (c) Find t and n given x =, cos 0 + 0 si n 0, y = sin 0 - 0 cos 8., , Ans., , (a)icosO-jsin8;(b), , "1~:02, , i+, , ~j;(C)t=icos8+jSin8'D=-isinO+jCOSO, , ~ ',: '~',, , ", , "
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Curvilinear Motion, Velocity in Curvilinear Motion, Consider a point P(x, y) moving along a curve with the equations x, ferentiating the position vector, , =f(t). y =g(t), where t is time. By dif-, , r=xi+yj, , (40.1), , with respect to t, we obtain the velocity vector, (40.2), , where Vx, , d.x, , = dt, , dy, and Vy = dt', , The magnitude of v is called the speed and is given by, Ivl, , =.JV:V =~V2x + v2 =ds, dt, y, , The direction of v at P is along the tangent line to the curve at p. as shown in Fig. 40-1. If T denotes the, direction of v (the angle between v and the positive x axis), then tan T= v/vx • with the quadrant being determined by Vx = Ivl cos T and Vy = Ivl sin T., , Fig. 40-1, , Acceleration in Curvilinear Motion, Differentiating (40.2) with respect to t, we obtain the acceleration vector, (40.3)
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CHAPTER 40, , Curvilinear Motion, , y, , o, Fig. 40-3, , SOLVED PROBLEMS, , 1., , Discuss the motion given by the equations x = cos 2m, y = 3 sin 2m. Find the magnitude and direction of the, velocity and acceleration vectors when: (a) t = t; (b) t = t., The motion is along the ellipse 9,r + y2 = 9. Beginning (at t = 0) at (I, 0), the moving point traverses the curve, counterclockwise., r = xi + yj = (cos2m)i + (3sin2m)j, , v = ~~ = v) + v),j = -(2nsin2m)i + (6ncos2m)j, a= ~~ =a,i+ayj=-(4n 2 cos2m)i-(12n 2 sin2m)j, , (a) Att =, , t:, , a, , a, , 1, , tanl/>=-L=3$, cos I/> =-1, XI =M, ax, a, 2,,7, So, I/> = 259 0 6'., , (b) At t = t:, , v = $ni - 3nj, , Ivl= 2$n,, , 5n, , So, r=""3', , tan r, , and, , a = 2n 2j + 6$n2 j, , = -$ cosr = t
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CHAPTER 40, , Curvilinear Motions, , y, , Fig. 40-4, , 2., , A point travels counterclockwise about the circle:xl + y2 = 625 at a rate Ivl = 15. Find r. lal. and; at (a) the point, (20. 15) and (b) the point (5,-IOJ6). Refer to Fig. 40-4., Using the parametric equations x = 25 cos 8. y = 25 sin 8. we have at P(x. y):, r = (25 cos 8)i + (25 sin 8)j, v = ~~ = [(-25 sin 8)i + (25 cos 8)jI ~~, = (-15 sin 8)i + 15 cos 8)j, a=, , c:; =[(-15COs9)i-(15sin8)j]~~, , = (-9 cos 8)i - (9 sin 9)j, since Ivl = 15 is equivalent to a constant angular speed of ~~ =, , l, , (a) At the point (20. 15). sin9 = t and cos9 = t. Thus., , \, v=-9i+12j, tanr=-t. cosr=-t. So r=126° 52', a=-Jji--¥j. lal=9. tan;=t. cos;=-t. So ;=216°52', (b) At the point (5. -IOJ6~ sin 8 = -tJ6 and cos9 = t. Thus., , v'=6J6i+3j, tanr=J61l2, cosr=tJ6. So r=W 32', a=-ti+.1I-/6j, 181=9. tanl/J=-2J6, cos;=-t. So 1/J=1010 32', , 3., , A particle moves on the first-quadrant arc of r = 8y so that v, = 2. Find Ivl. r. lal. and, Using the parametric equations x = 49, Y = 282• we have, , I/>, , at the point (4. 2).
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CHAPTER 40, , ., , dO, , dO, , Curvilinear Motion, , 1, , Smce, Vy = 40Tt = 2 and Tt = 20' we have, , e, , v = 2.I + 2'J an d a = - 0I3 .I, , At the point (4, 2), 0 = 1. Then, v=2i+2j,, a=-i,, , 4., , Ivl=2J2,, , lal= 1,, , tan~=I, COH=tJ2. So 'r=t1t', , tanl{> = 0,, , cosl{>=-1. So, , l{>=", , Find the magnitudes of the tangential and normal components of acceleration for the motion x = e' cos t,, y = e' sin t at any time t., We have:, r = xi + yj = (e' cost)i + (e' sint)j, , v = e' (cost - sint)i + e' (sin t + cost)j, a = -2e' (sint)i + 2e' (cost)j, , Then lal = 2e' . Also,, , S., , ~: = Ivl = J2e' and la, I= I~;~I = J2e'. Finally,, , A particle moves from left to right along the parabola y = r with constant speed 5. Find the magnitude of the, tangential and normal components of the acceleration at (1, 1)., Since the speed is constant, la,I=I~;~I=o.At (l,l),y'= 2x= 2 andyH= 2. The radius of curvature at(I, 1) is, (1 + (y')2 )3/2 5$, IV 12 ., then R=, Iy"l, -2-' Hence, la n l=T=2$., , 6., , The centrifugal force F (in pounds) exerted by a moving particle of weight W (in pounds) at a point in its path is, given by the equation F =, , ~ la"l. Find the centrifugal force exerted by a particle, weighing 5 pounds, at the ends, , of the major and minor axes as it traverses the elliptical path x = 20 cos t, Y = 15 sin t, the measurements being in, feet and seconds. Use g = 32 ftlsec 2•, We have:, r = (20cost)i + (15sint)j, , v = (-20sint)i + (l5cost)j, a = -20(cost)i -15(sint)j, Then, , ~: = Ivl = .J400sin 2 t + 225cos2 t, , and, , 175sintcost, d 2s, 2, dt = .J400sin2t+225cos2t, , At the ends of the major axis (t = 0 or t = 1t'):, 2, , lal=20 ,, , 2, 2, laI 1=14, d,!'i1=0 ' Ian 1=.J20 -0 =20, , and, , F=t(20)=tpound~
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CHAPTER 40 Curvilinear Motions, , (t = I or t= 3;):, , At the ends of the minor axis, , lal= 15, la,l=O, la.l= 15, , 7,, , and, , F = 1\(15)";, , '* pounds, , Assuming the equations of motion of a projectile to be x = ttl cos 1/1. Y= vot sin 1/1 - t gt 2• where t.b is the initial, velocity, 1/Iis the angle of projection, g = 32 ftlsec 2, and x and y are measured in feet and t in seconds, find: (a) the, equation of motion in rectangular coordinates; (b) the range; (c) the angle of projection for maximum range; and, (d) the speed and direction of the'projectile after 5 sec of flight if Ib = 500 ftlsec and 1/1= 45°, (See Fig. 40-5.), , Rg.40-5, , (a) We solve the first of the equations for t = __, x_ and substitute in the second:, Vo cos 1/1, , y, , x·, =v ---sm, cos 1/1, , 0 Vo, , 2, , X - =xtan -1/1 t g cos 1/1 ), 1/1, , -, , ( Vo, , P, , 2, , 2v~ cos2 1{1, , (b) Solving y= voIsin I{I- t gt 2 = 0 for t. we get t = 0 and t = (2tb sin !lNg. For (he latter. we have, Range = x = vocos I{I, , 2v sin I{I, 0, , g, , v2 sin 21{1, = .....0'--"""'"-, , g, , ., dx 2v 2 cos2yr, (c) Forxamaxlmum'- = 0, 0; hence cos 21/1= 0 and l{I=tlr., dI{I, , (d) For Ib =500 and, , I{I, , =t lr,, , g, , = 250..{it and y = 250..{it - 161 2• Then, , X, , ,, , ';,'-, , ,:;l~;f;~, , v, = 250..{i and v, =250..{i ..:. 321, , ,, , :,:~;~!~,, ,': ~~ ,:- :, , When t = 5, v, =250.,fi and v, =250..{i -160. Then, v, , tan r =...L = 0.5475. So r = 28° 42',, v,, , 8., , and, , ,, , Ivl = Jv,2 + v2 = 403 ftlsec, , A point P moves on a circle x =r cos 13, y = r sin /3 with constant speed v. Show that, if the radius vector to P, moves with angular velocity wand angular acceleration a, (a) v = rwand (b) a = r.Jw4 +a 2 •, (a) vt, , =-riin {31; = -rwsin /3, , and, , v, = rcos /31; = rwcos{3, , Then, (b) a, =, , a,=, , d;,' = -rwcosf31; - rsin/3~~ =-rw cos{3 - rasin{3, 2, , d: =-rwsinfj7,+rcos/3Tt=-rw sin/3+racos/3, , dv, , Then, , dR, , dw, , 2, , l,:,
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CHAPTER 40, , 9., , ·, , Curvilinear Motion, , Find the magnitude and direction of velocity and acceleration at time t, given, (a) x= e', y=e 2'-4e'+ 3; at t= 0, , Ans., , (a)lvl=$,f;;;296°34';lal=l, ~=O, , (b) x = 2 - t, Y = 2t3 - t; aU = I, , Ails., , (b)lvl;;;,J26, f= 101°19'; lal= 12, ~=tn, , (c) x=cos3t,y=sint;at t=tn, , Ans., , (c)lvl=$,, , (d) x = e' cos t, y, = e' sin t;. atl = 0, , Ails., , (d) Ivl =.,J2,, , "-:,, , 10. A particle moves on the first-quadrant arc of the parabola y2 = 12x with, , Vx, , and ~ at (3, 6)., Ans., , Vy, , = 15, Ivl = 15.,J2,, , f, , = tn;, , ax, , f, , 161°34'; lal;;;.J4j, ~;;;353° 40', , = tn; lal = 2, ~ = tn, , = 15. Find vY' lvi, and f; and ax, a y, lal,, ., , = 0, ay = -7512, lal = 75/2, ~ = 3n12, , 11. A particle moves along the curve y = xl/3 with, and acceleration when x = 3., , Ans., , f;;;, , Vx, , = 2 at all times. Find the magnitude and direction of the velocity, , Ivl=2..j82, f=83°40'; lal=24, ~=tn, , 12. A particle moves around a circle of radius 6 ft at the constant speed of 4 ftlsec. Determine the magnitude of its, acceleration at any position., Ans., , la, I = 0, lal = lanl = 8/3 ftlsec 2, , 13. Find the magnitude and direction of the velocity and acceleration, and the magnitudes of the tangential and, normal components of acceleration at time t, for the motion:, (a) x = 3t, Y = 9t - 3t2; att = 2, (b) x=cost+.tsint,y=sint-tcost;att= I, , Ans., , (a) Ivl = 3.,J2, f= 7n14; lal = 6, ~ = 3n 12; la, I= Ian 1= 3.,J2, (b) Ivl = 1, f= I; lal =.,J2, ~= 102° 18'; la,l = lanl = 1, , 14. A particle moves along the curve y = tx 2 la,l and lanl when t = 1., , Ans., , Vx, , t in x so that x = tt 2, for t> O. Find vx' vy' lvi, and 't'; ax, ay' lal, and~;, , = 1, vy= 0, Ivl = 1, f=O; ax = 1, ay= 2, lal = $, ~=63° 26'; la,l = 1, la,.l = 2, , 15. A particle moves along the path y = 2x - xl with Vx =4 at all times. Find the magnitudes of the tangential and, normal components of acceleration at the position (a) (1, 1) and (b) (2, 0)., , Ans., , (a) la,l = 0, lanl = 32; (b) la,l = 641$, la.l = 32$, , 16. If a particle moves on a circle according to the equations x = r cos OJt, y = r sin OJt, show that its speed is OJr., , 17. Prove that if a particle moves with constant speed, then its velocity and acceleration vectors are perpendicular;, and, conversely, prove that if its velocity and acceleration vectors are perpendicular, then its speed is constant.
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Polar Coordinates, The position of a point Pin a plane may be described by its coordinates (x, y) with respect to a given rectangular coordinate system. Its position may also be described by choosing a fixed point 0 and specifying the, directed distance p =OPand the angle 8 that OPmakes with a fixed half-line OX. (See Fig. 41-1.) This is, the polar coordinate system. The point 0 is called the pole, and OX is called the polar axis., P(P • • ), , O'-~, , __________~Z, Fig. 41-1, , To each number pair (p, B) there corresponds one and only one point. The converse is not true. For example, (1,0) and (1, 2Jr) describe the same point, on the polar axis and at a distance I from the pole. That, same point also corresponds to (-I, n). (When p is negative, the point corresponding to (p, B) is obtained as, follows: Rotate the polar axis OX through 8 radians (counterclockwise if 8 is positive and clockwise if 8 is, negative) to a new position OX' and then move IPI units on the half-line opposite to OX'.), In general, a point Pwith polar coordinates (p. 8) also can be described by (p, 8 ±2n1r) and (-p, 8 ±(2n + 1)1l),, where n is any nonnegative integer. In addition, the pole itself corresponds to (0, 8), with arbitrary 8., EXAMPLE 41.1:, , In Fig. 41-2, several points and their polar coordinates are shown. Note that point C has polar co-, , ordinates (I, 3;)., , ., , A polar equation of the form p =f( 8) or F(p, 8) =0 determines a curve, consisting of those. points corresponding to pairs (p, () that satisfy the equation. For example, the equation p = 2 determines the circle, with center at the pole and radius 2. The equation p =-2 determines the same set of points. In general, an, equation p =c, where c is a constant, determines the circle with center at the pole and radius lei. An equation, 8= c determines the line through the pole obtained by rotating the polar axis through c radians. For example,, 8= n/2 is the line through the pole and perpendicular to the polar axis., (2'I), , ,, , (I.~), , .-----~~--------~x, , (1.0), , Fig. 41-2
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41··, , CHAPTER 41, , Polar Coordinates, , Polar and Rectangular Co.ordinates, Given a pole and polar axis, set up a rectangular coordinate system by letting the polar axis be the positive, x axis and letting the y axis be perpendicular to the x axis at the pole. (See Fig. 41-3.) Then the pole is the origin, of the rectangular system. If a point P has rectangular coordinates (x, y) and polar coordinates (p, 8), then, , x = p cos e and, , y= p sin e, , (41.1), , These equations entail, (41.2), , ,,,P(x.y), ,, ,,,, ,, ,,, ,,, , +, , ________________ .J--.::;..J..-_.l-_ _ _ _ _, , 0:,, , x, , ,,, ,, , ,,, ,, Fig. 41-3, , EXAMPLE 41.2: Consider the polar curve p = cos O., Multiplying by p, we get p2 = p cos O. Hence, x2 + y2 =x holds for the rectangular coordinates of points on the, curve. That is equivalent to x?- - x + y2 = 0 and completion of the square with respect to x yields (X_t)2 + y2 = t., Hence, the curve is the circle with center at (t, 0) and radius t. Note that, as Ovaries from 0 to n/2, the upper semicircle is traced out from (I, 0) to (0, 0), and then, as 0 varies from ~ to n, the lower semicircle is traced out from (0, 0), back to (1, 0). This whole path is retraced once more as 0 varies from n to 2n. Since cos 0 has a period of 2n, we have, completely described the curve., EXAMPLE 41.3: Consider the parabola y = x 2• In polar coordinates, we get p sin 0 =, p = tan 0 sec 0, which is a polar equation of the parabola., , rr cos, , 2, , 0, and, therefore,, , Some Typical Polar Curves, (a), (b), (c), (d), , Cardioid: p = 1 + sin 8. See Fig. 41-4(a)., Lima~on: p = 1 + 2 cos 8. See Fig. 41-4( b)., Rose with three petals: p = cos 38. See Fig. 41-4(c)., Lemniscate: p2 = cos 28. See Fig. 41-4(d)., , At a point POll a polar curve, the angle ",from the radius vector OPto the tangent PTto the curve (see, Fig. 41-5) is given by, where, , ,, , P, , dp, , = de, , (41.3), , For a proof of this equation, see Problem I. Tan '" plays a role in polar coordinates similar to that of the, slope of the tangent line in rectangular coordinates., , i, ~
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CHAPTER 41, , Polar Coordinates, , k---~-----~3~---x, , k---'--+----x, , (a), , (b), , \, \, \, , ·~~---+'---x, , ,, ,,, (c), , -;--------*-------~--x, , (d), , Rg.41-4, , Angle of Inclination, The angle of inclination 't of the tangent line to a curve at a point P(p, fJ) on it (see Fig. 41-5) is given by, , pcos (J +p' sin (J, tanf= -psm, . (J + pI cos (J, , (41.4), , For a proof of tliis equation, see Problem 4., , Points of Intersection, Some or all of the points of intersection of two polar curves p = !t«(]) and p= h(fJ) (or equivalent equations), may be found by solving, (41.5)
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CHAPTER 41, , O'-~, , ____________, , Polar Coordinates, , ~~X, , Fig. 41·5, , EXAMPLE 41.4: Find the points of intersection of p = 1 + sin {} and p = 5 - 3 sin {}., Setting 1 + sin {} = 5 - 3 sin {}, we obtain sin {} = 1. Then p = 2 and {} = n/2. The only point of intersection is, (2, nl2). Note that we need not indicate the infinite number of other pairs that designate the same point., , Since a point may be represented by more than one pair of polar coordinates, the intersection of two curves, may contain points that no single pair of polar coordinates satisfies (41.5)., EXAMPLE 41.5: Find the points of intersection of p = 2 sin 2{} and p = I., Solution of the equation 2 sin 2{}= 1 yields sin2{} =t, and, therefore, within [0, 2n), {}= n1l2, ·5nIl2, 13n/12,, 17n112. We have found four points of intersection: (I, nIl 2), (1, 5nIl2), (1, 13n/12), and (1, 17,./12). But the circle, p = 1 also can be represented as p = -1. Now solving 2 sin 2{} = -1,. we get sin 2{} = - t and, therefore, {} = 7n112,, Ilnll2, 19Jr1l2, and 23Jr112. Hence we get four more points of intersection (-I, 7JrI12), (-I, lln/I2), (-1, 19JrI12),, and (-I, 23Jr/l2)., , When the pole is a point of intersection, it may not appear among the solutions of (41.5). The pole is a, point of intersection when there exist (}I and (}2 such thatfl( (}I) =0 =f2( ~)., EXAMPLE 41.6: Find the points of intersection of p = sin {} and p = cos {}., From the equation sin {}= cos {}, we obtain the points of intersection (.[iI2, Jrl4) and (-.[iI2, 5nI4). However, both, curves contain the pole. On p = sin {}, the pole has coordinates (0, 0), whereas, on p = cos {}, the pole has coordinates, (0, Jr/2)., EXAMPLE 41.7: Find the points of intersection of p = cos 2{} and p = cos {}., Setting cos 2{} = cos {}and noting that cos 2{}= 2 cos2 {}- 1, we get 2 cos2 {}- cos {}- 1 = 0 and, therefore, (cos {}- 1), (2 cos 9+ 1) = o. So, cos 9= 1 or cos{} = -to Then 9= 0, 2JrI3, 4Jr/3, yielding points intersection (1, 0), (-t, 2Jr/3),, and (-t, 4Jr/3). But the pole is also an intersection point, appearing as (0, Jr14) on p = cos 29 and as (0, nl2) on p = cos 9., , of, , Angle of Intersection, The angle of intersection, lP, of two curves at a common point P(p, (J), not the pole, is given by, tanlP =, , tan III - tan III, 'rI, ""2, l+ tan 1l', tan 11'2, , (;+ i .6), , where 'IIi and '1'2 are the angles from the radius vector OPto the respective tangent lines to the curves at P., (See Fig. 41-6.) This formula follows from the trigonometric identity for tan( 'III - '1'2), since lP = 11', - 11'2', , c., , O'-------------------x, Fig. 41·6
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CHAPTER 41, , Polar Coordinates, , EXAMPLE 41.8: Find the (acute) angles of intersection of p = cos 28 and p = cos 8., The points of intersection were found in Example 7. We also need tan 'I't and tan '1'2' For p = cos 8, formula (41.3), yields tan '1'1 =-cot 8. For p=cos 28, formula (41.3) yields tan "'2 =-tcot28., At the point (I, 0), !an "'I = -cot 0 = 00 and, likewise, tan "'2 = 00. Then "'I = "'2 = lrl2 and, therefore, tfJ = O., At the point, , (--!, 2f). tan "'I = 1313 and tan '1'2 = -1316. So, by (41.6),, t/I-, , tan.,.-, , (1313)+(1316) _ 313, 1-(1/6), -.5, , and. therefore, the acute angle of intersection tfJ "" 46° 6'. By symmetry, this is also the acute angle of intersection at, the point (-t, 4lr/3)., At the pole, on p = cos 8, the pole is given by 8= lrfl. On P = cos 28, the pole is given by 8 = lr/4 and 8 = 3lr/4., Thus, at the pole there are two intersections, the acute angle being lr/4 for each., , The Derivative of the Arc Length, The derivative of the arc length is given by, , ~ =~p2+(pl)2, , (41.7), , where pi =~~ and it is understood that s increases with 8., For a proof. see Problem 20., , Curvature, The curvature of a polar curve is given by, (41.8), , For a proof, see Problem 17., , SOLVED PROBLEMS, , 1., , Derive formula (41.3): tan'll = P ~~ = :" where pi ~., In Fig. 41-7, Q(p+ tlp, 8+ MJ) is a point on the curve near P. From the right triangle PSQ., sintl8, tan A. = SP = SP =, psinM, =, psinM, =, P-wSQ OQ-OS p+tlp-pcosM p(1-cosM)+tlp, l-cosM+~, , p, , Now as Q --+ P along the curve, tl8 --+ 0, OQ --+ OP, PQ --+ PT, and LA. --+ Lyt., , ,, , o, Rg.41-7, , M, , M
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CHAPTER 41, , ;\~~1, ":, , As M} --+ 0,, , Polar Coordinates, , S~~6 --+ 1 and 1- ~J .:l6 --+ O. Thu~,, , ,.;-:,.~, , ·, , .,:.~~, , :-.., , P, , 1, , dO, , 1 tan/\. = dp/dO = P dp, tan '" = J!!1,, , i*, "<'~, , ~,~~:"'ry, , In Problems 2 and 3. use fonnula (41.3) to find tan ",for the given curve at the given point., , . -~ ;II., , 2., , At, -', , ,, , p= 2+cos8 at 0= ~. (See Fig. 41-8.), , ",, , e=}-. p= 2+t=t. p' = -Sine=-lj. and tan",=f=-jr·, II, , Fig. 41-8, , 3., , P = 2 sin 3eat 0= ~. (See Fig. 41-9.), At, , e=, , *., , p= 2i=.J2. p' = 6 cOS3e=6(- ~ )=-3.J2 and tan", = :' =...:t·, T, , Rg.41-~, , ., , '.' ~, , 4., , ., , ., , pcosO+ p'sine, . 0, 0., -psm + pcos, From Fig.41-7. 'f= ¥'+ eand, , Denve fonnula (41.4): tan 'f =, , dO + sinO, ., tan¥'+ tan 6, Pdp cos6, tan'f='~dn(¥,+O)= 1 ta, t 0=, dO' 0, - n'l' an, I_p_~, , dp cosO, , pcose + ~~ sinO, = -:-----::~-, , ~~ cos6- psinO, , 5., , pcosO+ p' sinO, -psinO+ p'cosO, , Show that. if p = f( fJ) passes through the pole and 0, is such thatf( 0,) = O. then the direction of the tangent line to, the curve at the pole (0.0,) is, (See Fig. 41-10.), , e,., , 1/, , Fig. 41-10
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CHAPTER 41, , 9., , Investigate p = 1 + sin, , Polar Coordinates, , ofor horizontal and verticaJ tangents. (See Fig. 41-13.), , .'~~l~, , ;.~~, Rg.41-13, At P(p. 6):, , tan - (1 + sin9)cos9 + cos 9 sin 9, t' -(1 + sin O)sin 9 + cos 2 0, ", , cos 9(1 + 2sin8), (sinO+ 1)(2sin9-1), , We set cos 0 (1 + 2 sin 6) = 0 and solve. obtaining (} =n/2. 3TCI2. 7n16. and Ilnl6. We also set (sin (J + I), (2 sin ()- I) = 0 and solve. obtaining ()= 3nl2. n16. and 5n16., , ..,.', , For, For, For, For, , 0 =nl2: There is a horizontal tangent at (2. nfl)., () =7nl6 and 11n16: There are horizontal tangents at (t. 7TCI6) and (t, IITCI6)., ()= nl6 and 5n16: There are vertical tangents at (t. n16) and (t. 5nI6)., 0 = 3nl2: By Problem 5, there is a vertical tangent at the pole., , 10. Show that the angle that the radius vector to any point of the cardioid p = a( 1 - cos 6) makes with the curve is, one-half that which the radius vector makes with the polar axis., At any point P(p, (}) on the cardioid,, , p' = asinO, , and, , tanlJl, , =..e.pi =1-.cosO, = tan.Q.., smO, 2, , So lJI=tO., In Problems 11-13, find the a!1gies of intersection of the given pair of curves., , 11., , p= 3 cos O. p= 1 + cos O. (See Fig. 41-14.) ., , Fig. 41-14, Solve 3 cos 0= 1 + cos 0 for the points of intersection. obtaining (312, n/3) and (3/2, 5n/3). The curves also, intersect at the pole., For p= 3 cos 0:, For p= 1 + cos, , 0:, , pi =-3 sinO, p'=-sinO, , and, and, , tan lJI. = -cotO, , tanY' ;:: _1 +.CO! ~, 2, , smO
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CHAPTER 41, , Polar Coordinates, , At 0= n/3. tan"'l = -1./3. tan '1'2 = -./3. and tantP= 11./3. The acute angle of intersection at (t. n/3) and. by, symmetry. at (t. 5n/3) is n16., At the pole. either a diagram or the result of Problem 5 shows that the curves are orthogonal., 12. p=sec 2 tO. p=3csc 2tO., Solve sec 2 to = 3csc 2 to for the points of intersection. obtaining (4. 2n/3) and (4. 4n/3)., For p=sec2tO:, For p = 3 csc2to:, , p' = sec 2 t°tanto, p' = -3 csc 2tOcotto, , and, and, , tan '1'1 = cotto, tan '1'2 =-tanto, , At 0=2n/3. tan'l'l = 11./3. and tann2 =-./3. and tP=tn; the curves are orthogonal. Likewise. the curves are, orthogonal at 0= 4n/3., , 13. p= sin 20,p=cos 0. (See Fig. 41-15.), The curves intersect at the points (./312. n16) and (-./3/2. 5n/6) and the pole., For p = sin 20:, For p=cos 0:, , p' = 2cos20, p' =-sinO, , and, and, , tan '1'1 = t tan 20, tan "'2 =-cotO, , At 0= n/6. tan"'l = ./312. tan '1'2 = -./3. and tantPl = -3./3. The acute angle of intersection at the point (./312., n/6) is ,= tan- 13J) = 79° 6'. Similarly. at 0= 5n/6. tan '1'1 = -./312. tan '1'2 =./3. and the angle of intersection, is tan~1 3./3., At the pole, the angles of intersection are 0 and nl2., y, , Fig. 41·15, , In Problems 14-16, find, , :0 at the point P(p,, , 0)., , 14. p= cos 20. ', , 15. P(I + cos 0) = 4., Differentiation yields -p sin 0 + Ii( I + cos 0) = O. Then, , ,_ psinO, , 4sinO, , p - I + cosO, , (1 + cos 0)2, , an, , d, , ds -.J 2 (')2 _, 4.[i, dO - P + P - (l+cosO)JI2
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CHAPTER 41, , Polar Coordinates, •, , ,, , ~t~, , t, , I, , !, , 1\, - '. I, , '", , /1'", , ,~..., , ~. ~, , .. '.... -0', , •, , ... l, , ", , ... ,t,., , • ~(, , ,, , -, , fA.,, , -, , I., , t ., , _, , I , .\, , ,., , In Problems 22-25, find tan ",for the given curve at the given points., , 22. p=3-sinOatO=O, 0=3,,/4, , Ans., , -3; 3.fi -I, , 23. P = a(l - cos lJ) at 0= "/4, 0= 3"fl, , Ans., , .fi -1:-1, , 24. p(1 - cos 0) = a at 0= "/3,0= 5,,/4, , Ans., , -13/3: I+.fi, , 25. ti = 4 sin 20at 0= 5"112,0= 2,,/3, , Ans., , -113: 13, , In Problems 26-29, find tan t for the given curve at the given point., 26. p= 2 + sin Oat 0= ,,/6, , Ans., , -313, , 27. ti = 9 cos 20; at 0= ,,/6, , Ans., , 0, , 28. P = sin3 «(j3) at 0= ,,/2, , AIlS., , -13', , 29. 2P(1 - sin lJ) = 3 at 0= ,,/4, , Ans., , I+fi, , 30. Investigate p = sin 20 for horizontal and vertical tangents., Ans., , horizontal tangents at 0= 0, ",54°44', 125°16',234°44',305° 16'; vertical tangents at 0= "/2, 3"12,, 35°16', 144°44',215°16',324°44', , In Problems 31-33, find the acute angles of intersection of each pair of curves., 31. p= sin 0, p= sin 20, , Ans., , tP = 79°6' at 0= ,,/3 and 5,,/3: tP = 0 at the pole, , 32. p= .fisin8, ti = cos 20, , Ans., , tP = ,,/3at 0= "/6, 5,,/6: tP = ,,14 at the pole, , , 33. ti= 16 sin 20, ti =4csc 20, , Ans., , tP = ,,/3 at each intersection, , 34. Show that each pair of curves intersects at right angles at all points of intersection., (a) p = 4 cos 0, p = 4 sin 0, (c) ti cos 20= 4, p2 sin 20= 9, , (b) p= e9• p= e- 9, (d) p=1+cosO,p=l-cosO, , 35. Find the angle of intersection of the tangents to p = 2 - 4 sin 8 at the pole., Ans., , 2,,/3
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Polar Coordinates, , CHAPTER 41, , 36. Find the curvature of each o~ these curves at P(p, 8): (a), 4 cos 0,, (a) I/(fie 8 ); (b) 2; (c) +"cos28; (d), , Ans., , p = ell; (b) P = sin 0; (c) tY = 4 cos 20; (d) p,= 3 sin 0 +, ', , t, , 37. Find ~~ for the curve p = a cos 8., Ans., , a, , 38. Find, , ~~, , for the curve p = a(l, , ,, , + cos 8)., , a"2+ 2cos8, , Ans., , 39. Suppose a particle moves along a curve, (a) Multiply the equation, 2, , v =(, , ~~, , r, , p = f( 8) with its position at any time I given by p.= g(t), 8= h(/)., , (~ =p2 + (p/)2 obtained in Problem 20 by (~~, , r= ~~ r 1:r, , r, , to obtain, , +(, , p2 (, , de, d81dl, '., P d8, . I dp, (b) From tan'l'=p- =P, Id ,0btall1 sIIlV/=--d andcos'l'=--d ., dP, dpI, vI, vI, , In Problems 40-43, find all points of intersection of the given equations., , 40. p= 3 cos 8. p= 3 sin 0, , Ans., , (0,0), (3fi/2, lCf4), , 41. P = cos 0, p = I - cos 0, , AilS., , (0.0)., , 42. p= 8, p= lC, , Ans., , (lC, 1t), (-lC,-1t), , 43. P =sin 20, p = cos 2fJ, , AlIS., , (0, 0),, , <t, lCf3), (t, -1f13), , (21l+ l)lC), 2', 6, for n =0, I, 2, 3, 4, 5, (11., , 44. (Ge) Sketch the curves in Problems 40-43, find their graphs on a graphing calculator, and check your a,nswers to, Problems 40-43., , 45. (GC) Sketch the graphs of the following equations and then check your answers on a graphing calculator:, (a) p = 2 cos 4fJ, , (b) p = 2 sin 50, , (c), , tY = 4 sin uj, , (f), , p2, , (d) p=2(I-cos8), , (e), , P=I+~OSO, , (g) P = 2 - sec fJ, , (h), , p=, , =~, , i, , (In parts (g) and (II). look for asymptotes.), , 46. Change the following rectangular equations to polar equations and sketch the graphs:, (b) 4x= y2, (e) y =b, , (a) r-4x+y2=O, (d) x = a, , Ans., , (a), , (c) xy= I, , (0, , y = IIU + b, , p =4 cos fJ; (b) p = 4 cot Ocsc fJ; (c) tY =sec Ocsc 8; (d) p = a sec 0; (e) p = b csc 8;, b, , (f)p= sinO-mcosO
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CHAPTER 41, , Polar Coordinates, , 47. (GC) Change the following polar equations to rectangular coordinates and then sketch the graph. (Verify on a, graphing calculator.) (a) p= 2c sin 0; (b) p= 0; (c) p= 7 sec 0, , ·48. (a) Show that the distance between two points with polar coordinates (p" 0,) and (Pl., , Oz) is, , (b) When 0, = Oz. what does the distance simplify to? Explain why this is so., Ans., , IP, - 1'2/, , (c) When 0, - O2 =, Ans., , I' what does the formula yield? Explain the signficance of the result., , ~p:+p~, , (d) Find the distance between the points with the polar coordinates (I. 0) and, , Ans., , (I. t)., , ~2-J2, , 49. (a) Letfbe a continuous function such thatf( 0) ~ 0 for a < 0 < p. Let A be the area of the region bounded by, , =t 1:, , =tf:, , the lines 0= a and 0= p. and the polar curve p= f(O)· Derive the formula A, (f(0»2dO, p2dO., (Hint: Divide [a. PI into n equal parts. each equal to dO. Each resulting subregion has area approximately, equal to t dO(f(0;»2. where 0; is in the ith subinterval.), (b) Find the area inside the cardioid p = 1 + sin O., (c) Find the area of one petal of the rose with three petals, p =cos 30. (Hint: Integrate from, to, , -i i·)
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Infinite Sequences, Infinite Sequences, An infinite sequence (s) is a function whose domain is the set of positive integers; Sn is the value of this, function for a given positive integer n. Sometimes we indicate (sn) just by writing the first few tenns of the, sequence s" S2' S3' .•• , .'In' •••• We shall consider only sequences where the values sn are real numbers., EXAMPLE 42.1:, (a) (*)istheseQuencel,t,1, ... ,*, ...., (b) ((tr) is the sequence t,, , i, i, ... ,in' .. ., , (c) (n 2 ) is the sequence of squares 1,4,9, 16, . .. ,11 2, . . . ., (d) (2n) is the sequence of positive even integers 2, 4, 6, 8, ... , 2n, ...., (e) (2n -I) is the sequence of positive odd integers 1,3,5, 7, .. .., , Limit of a Sequence, If (s) is an infinite sequence and L is a number, then we say that lim sn = L if Sn gets arbitrarily close to L, as n increases without bound., n-+tFrom a more precise standpoint, lim sn = L means that, for any positive real number E > 0, there exists, a positive integer no such that, whenever n ~ no, we have Is. - LI < E. To illustrate what this means, place, the points L, L - E, and L + E on a coordinate line (see Fig. 42-1), where E is some positive real number., Now, if we place the points s" S2' S3' ••• on the coordinate line, there will eventually be an index no such that, s, ",S, +" S +,. S +3' . . . and all subsequent terms of the sequence will lie inside the interval (L - E, L+ E)., ','", nG -', nG, Sm+ ,, , • I, L-E, , L, , •, , L+E, , Fig. 42-1, , If lim s = L, then we say that the sequence (s.> converges to L. If there is a number L such that (s >, _", n, converges to L, then we say that (s.) is convergent. When (s.) is not convergent, then we say that (s.> is, , divergent., EXAMPLE 42.2:, , (1) is convergent, since lim 1=O. To see this, observe thall/II can be made arbitrarily close to, II, , " ....... It, , o by making n large enough. To get an idea of why this is so, note that 1110 =0.1, 11100 =0.01, 111000 =0.001, and, so on. To check that the precise definition is satisfied, let E be any positive number. Take no to be the smallest positive, integer greater than liE. So, liE < 110. Hence, if n ~ no, then n > 1/ E and, therefore, lin < E. Thus, if n ~ no, Ilin - 01 < E., This proves lim 1=0., 11, ,, , 11 ..... . . -, , _, , EXAMPLE. 42.3: (2n) is a divergent sequence, since lim.... 2n, large as n lIlcreases., , *' L for each real number L. In fact, 2n gets arbitrarily
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CHAPTER 42 Infinite Sequences, We write lim S =+00 if s. gets arbitrarily large as n increases. In such a case, we say that (sn> diverges to, n-++- n, More precisely, lim s =+00 if and only if, for any number c, no matter how large, there exists a positive, n~"'" ,n, integer no such that, whenever n ~ no, we have Sn > c., Likewise, we write n-+too, lim sn =-00 if Sn gets arbitrarily small as n increases. In such a case, we say that (sn>, diverges to -00. More precisely, lim s. =-00 if and only if, for any number c, no matter how small, there, n-+...., exists a positive integer no such that, whenever n ~ no, we have Sn < c., We shall write lim sn =oc: if lim Isnl =foe, that is, the magnitude of Sn gets arbitrarily large as n increases., , +00., , n-++-, , n-+of-, , EXAMPLE 42.4: (a) lim 2n=+oo;(b) lim (l-n)3=~;(c) lim (_l)n(n2) =00. Note that, incase(c),the sequence, •, ,,~, "~, II~, converges neither to +00 nor to ~., EXAMPLE 42.5: The sequence «-I)") is divergent, but it diverges neither to +00, nor to --, nor to 00. Its values, oscillate between 1 and-I., , A sequence (s.> is said to be bounded above if there is a number c such that S. S c for all n, and (s.) is said, to be bounded below if there is a number b such that b S Sn for all n. A sequence (s.) is said to be bounded if, it is bounded both above and below. It is clear that a sequence (s.> is bounded if and only if there is a number, d such that ~nl S d for all n., EXAMPLE 42.6: (a) The sequence (2n) is bounded below (for example, by 0) but is not bounded above. (b) The, sequence «-I)"} is bounded. Note that «_I)n) is -1, 1, -1, ... So, 1(-1)·1 S I for all n., , Theorem 42.1:, , Every convergent sequence is bounded., , For a proof, "See Problem 5., The converse of Theorem 42.1 is false. For example, the sequence «-I)n > is bounded but not convergent., Standard arithmetic operations on convergent sequences yield convergent sequences, as the following, intuitively obvious results show., Theor~m 42.2:, , .,,n_, .n_, -, , Assume !~~ sn = c and !~ In = d. Then:, , (a) lim k = k, where k is a constant., (b) nlim, ...._ ks n = kc, where k is a constant., (c), , (d), (e), (f), , lim, lim, lim, lim, , (sn +I.)=c+d., (sn -I.)=c-d., (s.I.)=cd ., (S.lI.) = c/d provided that d;t 0 and In;t 0 for all n., , For proofs of parts (c) and (e), see Problem 10., The following facts about sequences are intuitively clear., Theorem 42.3:, , If lim s. = 00 and s. ;t 0 for all n, then lim, ,,~+-, , ..L = O., , n~Sn, , For a proof, see Problem 7., Theorem 42.4:, , ..., , lim a· =00., ,,-., In particular, if a > I, then lim a' = +00., n......., (b) If Irl < 1. then lim r' = O., , (a) If, , lal > I, then, ,, , -, , ....., , For proofs, see Problem 8., Theorem 42.5 (Squeeze Theorem):, then lim In =L., fI~-+-, , For a proof, see Problem 11., , If lim, , n~+-, , S, II, , = L = lim, , U, , n-.+- ", , ,and there is an integer msuch that s. S In S lin for all n ~ m.
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••, Corollary 42.6:, , CHAPTER 42 Infinite Sequences, , I lu I for all n ~ ni, then lim t, = 0 is equivalent to lim la I = O., n-+.... ,., , If lim u. = 0 and there is an integer nI such that, ,.~...., , This is a consequence of Theorem 42.S and the fact that lim a, n-+...., , EXAMPLE 42. 7:, , lil~, , " .... ~, , II, , It,. S;, , n, , II ....... ,., , ~ O., , (-I)· ~ = O. To see Ihis, use Corollary 42.6, noting that 1(_1)" ~I S;! and lim ! = o., 11, n, n, .....~ 11, , Theorem 42.7: Assume that/is a function that is continuous at c, and assume that lim s", s. are in the domain off Then lim /(s,,) = fCc)., II-'too·, , =c, wh!1re all the terms, , Il-t+-, , See Problem 33., It is clear that whether or not a sequence converges would not be affected by deleting, adding, or altering a finite, number of terms at the beginning of the sequence. Convergence depends on what happens "in the long run.", We shall extend the notion of infinite sequence to the case where the domain of a sequence is allowed to, be the set of nonnegative integers or any set consisting of all integers greater than or equal to a fixed Integer., For example, if we take the domain to be the set of nonnegative integers, then (2n + I) would denote the, sequence of positive odd integers. and (112") would denote the sequence I. t, t. t ....., , Monotonic Sequences, (a)·, (b), (c), (d), (e), , A sequence (s") is said to be nondecreasing if s" ::; S"+I for all n., A sequence (s") is said to be increasing if Sn < $"+1 for all n., A sequence (s") is said to be Tlonillcreasil1g if s" ~ S"+I for all n., A sequence (s.) is said to be decreasing if s" > Sn+1 for alll!., A sequence is said to be monotonic if it is either nondecreasing or nonincreasing., , Clearly, every increasing sequence is nondecreasing (but not conversely), and every decreasing seql,lence, is nonincreasing (but not conversely)., EXAMPLE 42.8: (a) The sequence I, I, 2, 2, 3, 3,4, 4, ... is nondecreasing, but not increasing. (b) -I -I, -2, -2,, -3, -3, -4, -4, ... is nonincrcasing, but not decreasing., I, , An important basic property of the real number system is given by the following result. Its proof is beyond, the scope of this book., Theorem 42.8:, , Every bounded monotonic sequence is convergent., , There are several methods for showing that a given sequence (s.) is nondecreasing, increasing, nonin-., creasing. or decreasing. Let us concentrate on the property that (s.) is increasing., Method I: Show that S.+I, , - S., , > O., , ., 3n, EXAMPLE 42.9: Consider s. = 4n + 1 . Then, , 3(n+l), , 3n+3 S, , = 4(n + 1) + 1 = 4n + S', , S.+1, , 0,, , 3n + 3, 3n, (12n 2 + ISn + 3) - (I2n 2 + ISn), sn+l - s" =-4-n-+-S - -4-n-+-1 =..:......;;-'-~(4;;-n'-+'-:S"f)(-:-:4,--'n'-+-'-:I')-..:.:.......:.., , 3, , =(4n+S)(4n+l) >0, since 411 + 5 > 0 and 4" + 1 > 0, , Method 2: When all .1'" > 0, show that s"+/s,, > I., EXAMPLE 42.10:, , Using the same example sn =, , 4~: 1 as above,, , ~ = ( 3n + 3 )/(~) = 3n + 3 4n + I = 12n 2 +2 ISn + 3 > 1, s., 4n+5, 4n+1, 3n 4n+S, 12n +15n, ', , since 12n 2 + 15n + 3 > 12n 2 + ISn > O.
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CHAPTER 42 Infinite Sequences, , Method 3: Find a differentiable function f(x) such that f(n) =sJor all n, and show that j'(x) > 0 for all x ~ 1, (and, hence, theit f is an increasing function for x ~ 1)., EXAMPLE 42.11:, , Consider s. = 4~: 1 again. Let f(x) =, , 4~~ 1 . Then, , /,(x) = (4X! 1)2 > 0 for all x., , SOLVED PROBLEMS, , 1., , For each of the following sequences, write a formula for t~e nth term and determine the limit (if it exists). It is, assumed that n = 1, 2, 3, ...., 1 1 1 I, , (a), , 2' 4' 6' S' ...., I I, , (), e, , 1 I, , 1, , -2' 3' -4' 5' -6'''', • IC·, . 31C, • 2, . 5IC, sm'2 ' sm ~ sm 2 ,sm ~ sm 2 ' ..., , (c) 1,, , (d), , 0.9,0.99,0.999.0.9999, ..., , (f), , t, Hf. (4r. (~r, ..., , 1, 1, (a) s. = -2 ; lim -2 = O., II~ n, II, , (b) s"= n+ ; lim n+l=lim (1- +11)=I-lim +11=1-0=1., n 1 "....... n, II-++n, n-++- n, , (c) s. =, , (-1>-+', , (-1)"+', , ; lim - - =O. This is intuitively clear, but one can also apply Theorem 42.3 to the sequence, n, "....... n, «-I)M' n), since lim (-1>-+'" = 00., "......., , 1- I ~. ; !~~ (1- I ~. ) =1- !~l. 1~" =1- 0 =1., , (d) sn =, , Note that lim 101", 11 .... ....,, , =0 by virtue of Theorem 42.4(b)., , (e) s. =sin ": . Note that the sequence consists of repetitions of the cycle 1,0, -1,0 and has no limit., (f) sn =, , 2., , (n n+ I)·; "........, lim (n + 1)" = lim (I + 1)" = eby (26.17)., n, n.....n, , ........., , Evaluate lim s in the following cases:, 2, (a) s = 5,,2 -4n+ 13, (b) s = 8n -3, • 3,,2 - 95n - 7, "2n +5, , c, 3n+7, ( ) n3 - 211-9, , 2, R allth r 5x -4x+13 5 b Ch, 7 Pr bl, 13 Th ~, r 5,,2-4n+13· 5 A 'mil, ec, at x~ 3x2 -95x-7 3 y apter, 0 em . ere ore, .~ 3n2 -95n -7 3' SI ar, result holds whenever s" is a quotient of polynomials of the same degree., 2, 2, (b) Recall that lim 8x +- 3 =+00 by Chapter 7, Problem 13. Therefore, lim 811 -53 =+00. A similar result holds, ,....... 2x 5, "....... 211+, whenever s. is a rational function whose numerator has greater degree than the denominator (and whose, leading coefficients have the same sign)., (c) Recall that lim 33x; 7 9 0 by Chapter 7, Problem 13. Therefore, lim 3 3n + 7 9 O. The same result, ........ x - x "....... " - 211holds whenever s" is a rational function whose denominator has greater degree than the numerator., , (a), , 3., , For each of the following sequencies, determine whether it is nondecrcasing, increasing, nonincreasing,, d~creasing, or none of these. Then determine its limit, if it exists., 5" - 2, , (a) s" = 7" +3, , n, , (b) s" = F, , 1, , (c) 3", , - 2 Th f'() = (7 x + 3)(5) - (5x - 2)(7) :::: 29 > 0, (a) Let f( x) = 5x, 7x+3' en, x, (7x+3)2, (7x+3)2', Hence,Ax) is an, , increa~ing, , function and, therefore, (s,,) is an increasing sequence., , (b) Let f(x) = ~. Then j'(x)= 2' -.;~~X)2', , 1- x(ln2), , 2'
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CHAPTER 42 Infinite Sequences, , Since In 2 > t (by (25.12», x(ln2) > xl2 ~ 1, when x ~ 2. Thus, I - x(1n2) < 0 when x ~ 2 and, therefore,, rex) < 0 when x ~ 2. So,Ax) is decreasing for x ~ 2 and this implies that 5. is decreasing for 11 ~ 2. Note that, 5. = t = 52' Hence, (5.) is nonincreasing. Now let us find the limit. By L'HOpital's Rule., , x, lim 2 ,, , r~i-«>, , (c), , t, , ~;:I =(3!+1 )/(1.) =, , =X-.....,, lim (I n 21)2X =0, , 2~ =0, , < I. Hence, (s.) is decreasing., , 1, Theorem 42.4(b) tells us that lim 3• = lim (-31)", n~"", , 4., , and, therefore, 'I~+lim, , "~, , =O., , 1 ·3·5·7 ... (211 - 1) ., Show that the sequence s" = 2.4.6.8 ... (211) IS convergent., , Let us use Theorem 42.8. (s.) is bounded, since 0 < s. < I. Let us show that (s.> is decreasing. Note that, I· 3 . 5 . 7 ... (211 + 1), 211 + I, s".1 = 2.4.6·8 ... (2n+2) =s"211+2, --<s•, , 5., , Prove Theorem 42.1: Every convergent sequence (sJ is boundl.:d., Let lim s = L. Take E = I. Then there exists a positive integer 110 such that, whenever fI, •, Is" - LI <"-HI . Hence. for 11 ~ 11o, using the triangle inequality, we get, ~"I, , Show that the sequence, , (~:), , we have, , = Ks. -L)+LI s~" -LI+ILI <l + ILl, , So, if we take M to be the maximum of I + ILl and, bounded., 6., , ~ 1/ 0,, , I,r, I, IS21.ls31, ... ,lsJ, then IsJ S M for all n. Thus, (s,,) is, , is divergent., , •, 1I! _ I· 2·3 ... /I 1 3 4, 11 11 l', •, bo ded S b Th, . 42 1, Smce "F- 2.2.2 ... 2 222'" 2 >2 lor 11 > 4, the sequence IS not un . 0, y eorem .,, the sequence cannot be ~onvergent., 7., , Prove Theorem 42.3: If lim s" = 00 and s. '" 0 for all n, then lim -'. =o., ., n ..... +n .....+- S, Consider any E > o. Since lim s. =00, there exists some positive integer m such that, whenever 11 ~ m,, •, , ~"I > ~, , 8., , Prove Theorem 42.4: (a) if, , 11 ......., , and, therefore,, , lal> 1. then, , -.L - 01 =l-.Ll, < E., S", I.f",, , lim an = oc; (b) If, , II-++-, , So,, , Irl < J. then, , lim, n........, , -.L, =n, ., sri·, , lim r' = O., , II-++-, , (a) LetM>O.andlet 1a1=I+b.So,.h>O.Now, 1a1"=(l+b)"=I+nh+ ... >1+nh>M when 11~~., (b) Let a = IIr. Since Irl < 1. lal > I. By part (a), lim a" =00. Hence, lim (lIr") =00. So, by Theorem 42.3., II-++lim r" =0., ,,~, , fl-++-, , 9., , 1, Prove: lim 2" =O., n-+"*-, , lim 2· = 00 by Theorem 42.4(a). Hence, lim 21" = 0 by Theorem 42.3., n.....+-, , 11-++.0, , 10. Prove Theorem 42.2(c) and (e)., Assume !~ s" = c and !~!!. In, , =d.
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CHAPTER 42 Infinite Sequences, , .-, , (c) lim (s. +t.) =c +d. Let E>O. Then there exist integers m, and ~ such that ~. It. - dl < EI2 for n ~ ~. Let m be the maximum of m, and ~. So; for n ~ m.ls. Hence. for n ~ m., , cl < EI2 for n ~ m, and, cl < EI2 and It. - dl < EI2., , I(s. +t.)-(c+d)l= I(s. -c)+(t. -d)1 ~ Is. -cl+lt. -dl <I+I= E, (e), , lim, (s n", t ) =cd. Since (s ) is convergent. it is bounded. by Theorem 42.1 and. therefore. there is a positive, ~....., number M such that ~.I ~ M for all n. Let E> O.lf d;:/: O. there exists an integer m, such that Is. -cl < El2ldl, for n ~ m, and. therefore. Idlls. - cl < EI2 for n ~ mi' If d =O. then we can choose m, =1 and we would, still have Idlls. - cl < EI2 for n ~ mi' There also exists ~ such that It. - dl < EI2M for n ~ ~. Let m be the, maximum of m, and~. If n ~ m., II, , lSi. -cdl, , =Is.(/. - d)+ d(s. - cli ~ Is.(t. - d)l + Id(s. -c)1, =Is. lit. -dl+ldlls -cl ~M(2~ )+I=E, n, , 11. Prove the Squeeze Theorem: If lim Sn =L = lim u., and there is an integer m such that Sn ~ tn ~ un for all n ~ m,, •, then lim t. = L:, Let E> O. There is an integer m, ~ m such that Is. - LI < E/4 and lu. - LI < E/4 for n ~ mi' Now assume, n ~ mi' Since s. ~ t. ~ u•. lt. -sJ ~ lu. - sJ. But, , .-, , II~, , II ........, , lu. -s.1 =I(u. -L)+(L-sj ~ lu. - LI + IL-s.1 < t+t=I, Thus, Itn - s" I< E/2. Hence,, , In each of Problems 12-29, determine for each given sequence (s.) whether it is bounded and whether it is, nondecreasing. increasing. nonincreasing. or decreasing. Also determine whether it is convergent and, if possible, find, its limit. (Note: If the sequence has a finite limit, it must be bounded. If it has an intinite limit, it must be unbounded.), 12., , (n+,~), , Ans. nondecreasing; increasing for n ~ 2; limit +00, , 13. (Sin n:), , Ans. bounded; no limit, , 14. (~), , Ans. increasing; limit +00, , 15. ( n! ), 10', 16., , ,, , (l~n), , 17. (t(1 + (_l)n+l», , 18., , (lnn~l), , Ans. increasing for n ~ 10; limit +00, , Ans. decreasing for 11 ~ 3; limit 0, , Ans. bounded: no limit, , Ans. decreasing: limit 0
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CHAPTER 42 Infinite Sequences, , (~~), , Ails., , nonincreasing; decreasing for n ~ 2; limit 0, , 20. (!ifn), , Ails., , decreasing for n ~ 3; limit 1, , 19., , 21., , ( ~), 11+2, , AilS., , increasing; limit 3, , 22., , (COs~), , AilS., , increasing, limit 1, , AilS., , decreasing; limit 0, , AilS., , limit 0, , 25. (J;+l- Fn), , AIlS., , decreasing; limit 0, , 26.., , (3,,2~4), , AilS., , decreasing; limit 0, , 27., , (II sin ~), , AilS., , increasing, limit 1t, , AilS., , increasing; limit +00, , AilS., , increasing; li~it +00, , 23., , 24., , (Si~n), , ! ",', , Tn each of Problems 30-32, find a plausible formula for a sequence whose first few terms are given. Find the limit, (if it exists) of your sequence., , 3 9 27 81, 30. I, 2' 4' 6' 8' ..., , AilS., , 31. -1,1,-1,1,-1,1, ..., , AilS., , 32., , 3 7 II 3 19, , T' 4' 7' 2' IT' ..., , AilS., , 3"-1, , S", , ..., , = 2(n -I) ; Inmt IS +00, , Sn = (_1)n; no limit, , S", , =, , i ;decreasing, limit is t, , j~ =, , 33. Prove Theorem 42.7. (Hillt: Let E> O. Choose &> 0 such that, for x in the domain off for which, have If(x) - f(c)1 < E· Choose m so that n ~ m implies Is" - cl < 0.), , 34. Show that lim ~l/nP = I for p > O. (Hint: nP''' = e(P In ")I,,.), "-+...., , Ix -cl < 0, we
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CHAPTER 42 Infinite Sequences, , /2, , 35. (GC) Use a graphing calculator to investigate s =, +5 for n = 1 to n = 5. Then determine analytically the, •, 4, 4 +, behavior of the sequence.,, n n, , Ans. decreasing; limit is, , t, , ., n5, 36. (GC) Use a graphing calculator to investigate s. = 2. for n = 1 to n = 10. Then determine analytically the, behavior of the sequence., Ans. decreasing for n ~ 7; limit is 0, , 37. Prove that It-Hlim a. = 0 is equivalent to n-Hlim lan 1=0., , 38. If s. > 0 for all n and lim, , ,If""'", , S2, , = c. prove that lim s. =, , It, , .........., , 39. (GC) Define s. by recursion as follows: SI =2 and, , Fc., , S.+1, , t(, , =, , s. + :.) for n ~ 1., , (a) Use a graphing calculator to estimate s. for n =2•.. , • 5., (b) Show that. if lim s. exists, then lim s. =.J2., II ....+., II ........., (c) Prove that hm sexIsts ., , ....... ., I, , 40. Define s. by recursion as follows: SI = 3, and, (a) Prove s. < 6 for all n., (b) Show that < s. > is increasing., (c) Prove that lim sexists., "........ " ., (d) Evaluate lim s ., , ....... ., , Ans. (d) 6, , 41. Prove Theorem 42.2, parts (a), (b). (d), (f)., , ,, , S.+1, , = 1:.<s. + 6) for n~ 1.
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Infinite Series, Let (s,,) be an infinite sequence. We can fonn the infinite sequence of partial sums (S,,) as follows:, SI =Sl, , We usually will designate the sequence (Sn) by the notation, ~s, L.." =s I +s 2 +···+s + ..., 1/, , The numbers Sl' S2' ••• ' S", ••. will be called the terms of the series., If S is a number such that lim S. =S, then the series, is said to converge and S is called the sum of, , n........., , the series. We usually designate S by, , LS", , +-, , LS"', "~, , ., , =S, then, the series £..J, ~ s is said to diverge. If lim S =-toe. then, +It-++the series is said to diverge to +00 and we write LS" =+00. Similarly. if lim S" =, then the series is said, +II-++to diverge to, and we write L s" =, "=1, If there is no number S such that n-++lim S, , It, , II, , II, , -00,, , -00, , -DC., , "=1, , EXAMPLE 43.1: Consider the sequence «_1)"+1). The terms are SI = I, S2 = -I, S3 = 1. S4 = -1. and so on. Hence., the partial sums begin with SI = 1. S2 = 1 + (-1) = O. S3 = 1 + (-1) + 1 = I, S4 = 1 + (-1) + (1) + (-1) =O. and continue, with alternating Is and Os. So, lim S" does not exist and the series diverges (but not to +00 or -(0)., ., , -, , "..., , Geometric Series, Consider the sequence (ar,,-I), which consists of the tenus a, ar, ar, ar3, ...., The series, , L ar"-I is called a geometric series with ratio rand first tenn a. Its nth partial sum S" is given by, S" =a+ar+ar 2+"'+ar,,-1, , Multiply by 1':, , rS", , =ar +ar2 +... +ar/l- I + ar", , Subtract:, , S" -rS" =a-ar", , Hence,, , (1- r)S" = a(l- r"), , a, , S = a(l- r"), ", l-r, l•
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CHAPTER 43 Infinite Series, , Everything now depends on the ratio r. If I~ < 1, then lim r' =0 (by Theorem 42.4(b» and, therefore,, .--HoM, lim S =a/(I- r). If I~ > I, then lim r" =00 (by Theorem 42.4(a» and, therefore, lim Sn =00. (A trivial, n-Hn~+eo, II-Hexception occurs when a =O. In that case, all terms are 0, the series converges, and its sum is 0.) These results, are summarized as follows:, II, , Theorem 43.1:, , Given a geometric series, , Lar, , n I, - :, , (a) If I~ < 1, the series converges and has sum I ~ r', (b) If I~ > I and a 0, the series diverges to 00., , *, , EXAMPLE 43.2:, , Take the geometric series, , L<t>a-I with ratio r = t and first term a = I:, I+++t+t+···, , ..., , m= t = 2. Thus, ~<t>'-I =2., , 1, By Theorem 43.1(a), the series converges and has sum I_, , Ls., L(s. +1.)., , We can multiply a series, by a constant c to obtain a new series, and, to obtain a new series, , LI., , *, , If C 0, then, , Theorem 43.2:, , Lcs., and we can add two series Ls., , Lcs. converges if and only if Ls. converges. Moreover. in the case of convergence., , ..., , ..., , ,,=1, , /1=1, , Lcsn=cLs., To obtain this result, denote by 1'. =CS I +CS2 +.. ·+cs" the nth partial sum of the series Ics". Then, T. =cS., where Sn is the nth partial sum of ~>". So, lim T" exists if and only if lim S" exists, and when the, n-++limits exist, lim Tn =C lim S•. This yields Theorem 43.2., n~+oo, , 1'1-+...., , Theorem 43.3:, , "-++00, , Assume that two series, , Ls. and Lt. both converge. Then their sum L(s. +I,,) also converges and, ..., , ..., , too, , n=l, , ,.=), , L(s.+ ''>=Ls.+ LI., To see this. let S. and T. be the nth partial sums of Is" and ~>., respectively. Then the nth partial sum, Un of I(s. +1.) is easily seen to be S. + T•. So, lim U. = lim S. + lim 1'.. This yields Theorem 43.3., n-++-, , n~+-, , Corollary 43.4: Assume that two series, converges and, , I'J~, , Ls. and LI. both converge. Then their difference L(SII-III) also, , ..., ... ..., L(s. -( )= Ls. -L'., 11, , n=1, , ,, , n = l ' n~1, , This follows directly from Theorems 43.2 and 43.3. Just note that, series I(-l)t n •, , I, , (s" -I,,) is the sum of Is" and the, , Theorem 43.5: If ~ s converges. then lim s =O., .L.J n, n-+.... ", , ...., , To see this, assume that, , Is. =S. This means that lim S. =S, where, as usual. S" is the nIh partial sum of, "=1, , n-++-, , the series. We also have lim S._I = S. But, s. =Sn - Sn-I' So. lim sn = lim Sn -lim S,,_I, n .... +-, , 11-+...., , "-++00, , 11-+...., , =S - S =O.
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CHAPTER 43 Infinite Series, Corollary 43.6 (The Dive~ence Theorem):, , If lim s. does not exist or lim s :1= 0, th~n ~ s diverges. ,, n-++-, , k, , n-+-t- n, , n, , This is an immediate logical consequence of Theorem 43.5., , t + t + t + t + '" diverges., lim 211n+ 1 =-:-21 :1= 0, the Divergence Theorem implies that the series, di~erges., Here, .1'" = "L.1ln+ l' Since II......., ., ,, , EXAMPLE 43,.3:, , The series, , The converse of Theorem 43.5 is not valid: lim, s = 0 does not imply that", -)+00 n, L.J s n converges. This is shown, by the following example., •, , L, , 1t t, , EX~MPLE 43.4:, C~nsider the so-called harmonic series ~ = 1+ + + + ~ + .... Let us look at the following, ' ., partial sums of thiS senes:, S2 =, , 1+1, 111111112, , S4 =1+'2+'3+4'> 1+'2+4'+4'= 1+'2+'2= 1+'2, , 1111, 1111, 4, 1, ~=~+3+~+~+i>~+i+i+i+i=~+i=~+'2, >1+1, 2, , 1111111, , I, , SI6 =Sg +'9+lO+TI+ 12 + 13 + 14 +15+16, , 11111111, 1, >~+16+16+16+16+16+16+16+16=~+'2, >1+1, 2, , Continuing in this manner, we would obtain SJ2 > 1+ t, S64 > 1+ t, and, in general, S2k > 1+ k12 when k > 1. This implies, that lim S" =-too and, therefore, the harmonic series diverges. But notice that lim s" = lim lin = O., fI-++-, , ,,~+-, , n-+"-, , Remark: Convergence or divergence is not affected by the addition or deletion of a finite number of, terms at the beginning of a series. For example, if we delete the first k terms of a series 'and the sum of the, deleted terms is c, then each new partial sum Tn has the form SII+k - c. (For example, TI is Sk+1 - c.) But, lim (S.+k - c) exists if and only if lim S k exists, and lim S,.+k exists if and only if lim Sn exists., n~+-, , n-++oo, , n~, , n+, , n-io+-., , It will often be useful to deal with series in which the terms of (Sll) are indexed by the n011- ., negative integers: So- SI' S2. S3' •.•• Theli the partial'sums SII would also begin with So =S(h and the sum of a, , Notation.', , ...., , convergent series would be written as, , Is"., 11=0, , SOLVED PROBLEMS, 1., , ' I IlC scncs, ' 3, I + 52, I + 53, I + ' " t'or convergence., EX<lllllllC, , *. Since Irl, = I_I [ilS) i~~ t·, , This is a geometric series with ratio r = t and the first term a =, that the series converges and that its sum is 1~ r, 2•, , ', h, ., Examllle, I c sencs, , The nth term is, , =, , ItI< I, Theorem 43.1 (a) te1\s us, , =, , I, I, I, 1, T2, + 2.3, + 3.4, + 4.5, + . .. ..lor convergence., n. (~ +1) . This is equal to ~ - 11 ~ 1 . Hence, the 11th 'lartial sum
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CHAPTER 43 Infinite Series, , S =_1_+_1_+_1_+_1_+.+, 1, • I· 2 2·3 3·4 4·5, n' (n + I), , =1 __1_, n+l, , Thus, lim S = lim (1- -L ) = 1- 0 = 1. Hence, the series converges and its sum is 1., .......... ........., n+ I, 3., , t, , We know that the geometric series 1+t + + t + ,t + ... converges to S = 2. Examine the series that results, when: (a) its first four terms are dropped; (b) the terms 3, 2, and 5 are added to the beginning of the series., , l., , (a) The resulting series is a geometric series ,t+i +... with ratio t· It converge~ to 12IA~2) YJ~ = Note, that this is the same as S - (1 + t+t+ t) = 2-(.y.) = t., .., (b) The new series is 3 + 2 + 5 + I + t + + t +,t + .... The new partial sums are the old ones plus (3 + 2 + 5)., Since the old partial sums converge to 2, the new ones converge to 2 + 10 =12. Thus, the new series is, convergent and its sum is 12., , t, , 4., , Show that the series, , t +t +t +, , *, , + .. , diverges., Here, s. = 2·2~ 1= 1- 21" . Since lim 21" = 0, it follows that lim sn = 1- 0 = 1 O. So, by the Divergence, •, , n~+-, , Theorem, the senes dIverges., , *, , -¥ +.,. - ... for convergence., This is a geometric series with ratio, = -t. Since I" = t > 1, Theorem 43.1 (b) tells us that the series diverges., , 5., , Examine the serie-s 9 - 12 + 16 -, , 6., , Evaluate ~. 2", , ~ (-I)", , =1- 2'1 + 4'I - '8I - \6I -. "., , • -0, , This is a geometric series with ratio, = ., a, 1, 1, 2, sumls 1-,=1-(-112) 3/2=3', 7., , n~, , t, , and first term a = 1. Since, , 1,1 = t < 1, the series converges and its, , Show that the infinite decimal 0.999 ... is equal to 1., 0.999 ... =, , ?o + 1~ + 1~ + .... This is a geometric series with first term a = i, , ., I, th, a, 9/10, Hence, It, converges to e sum 1-, = 1- (1/1 0), , and ratio, = i1r., , 9/10 1, 9/10 = ., , ' th, ' f.3+n+n+n+···, 1111, 8, • Exarrune, e senes, , Here,, , S II, , So, lim S, tI-++-, , 9., , n, , = (2n _ d(2n + 1), Note that (2n _ 1hn + 1) =~( 2n1_ 1 - 2,/+ 1) . Hence, the nth partial sum S" is, , =t. Thus, the series converges to t., ,., , Examine the series 3 + .J3 + if] + if3 + ...., S, n, , *, , = if] = 31/• = ton 3)1• • Then lim s = eO = 1 O. By the Divergence Theorem, the series diverges., n-++- n, , 10. Examine the series i1r+ir+rr+fJ+···., , This series is obtained from the harmonics series by deleting the first nine terms. Since the harmonic series, diverges, so does this series., , ., , '~,
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CHAPTER 43 Infinite Series, , 11. (Zeno's Paradox) Achilles (A) and a tortoise (T) have a race. T gets a 1000 ft head start, but A runs at 10 ft/sec,, whereas Tonly does 0.01 ftlsec. When A reaches T's starting point, Thas moved a short distance ahead. When A, reaches that point, T again has moved a short distance ahe~d, etc. Zeno claimed that A would never catch T. Show, that this is not so., When A reaches T's starting point, 100 seconds have passed and T has moved 0.01 (100) = 1 ft. A covers that, additional I ft in 0.1 seconds, but Thas moved 0.01(0.1) =0.001 ft further. A needs 0.0001 seconds to cover that, distance, but T meanwhile has moved 0.01(0.0001) = 0.000001 ft, etc. The limit of the distance batween A and, T approaches O. The time involved is 100 + 0.1 +0.0001 + 0.000000 1+ .. " which is a geOliletric series with first, term a = 100 and ratio r= 111000. Its sum is, , _a_=, 100, 1- r 1- (1/1000), , 100, 999/1000, , 100000, , 999", , which is a little more than 100 seconds. The seeming paradox arises from the artificial division of the event into, infinitely many shorter and shorter steps., , 12. Examine each of the following geometric series. If the series converges, find its sum., (a), , AilS., AIlS., Ans., AilS., , 4-1++-*+· .., l+t+t+Jt+ .. ·, (c) 1-t+t-*+ .. ·, (d) 1+ e- I + e-2 + e-3 + ..., , (b), , S=, , Jt-, , Diverges, , s-1., -4, S=_e_, e-I, , 13. A rubber ball is dropped from a height of 10 ft. Whenever it hits the ground, it bounces straight up three-fourths, , of the previous height. What is the total distance traveled by the ball before it stops?, 70 ft, , Ans., , 14. Examine the series "", 4-J n(n1+ 4), , nI, , 1 + 3.7, 1 + ..., + 2.6, , S =-:it, , Ans., , . I, . "", I, 15• Exauune, t le senes "'" n(n + lXn + 2), , 1, 1, =1· 21. 3 + 2·3·4, + 3.4.5 + ..., , .l, S -- 4, , Ans., , ...., , 16. Evaluate ~>" when Sn is the following:, (a) 3-n, Ans., , (a), , (b), , n(n~2), , t: (b) t; (c), , *:, , (c), , n(n~ 3), , (d) (n ~ I)!, , (d) I, , 17. Show that each of the following series diverges:, (a), , 3+f+t+t+ .. ·, , (d), , e+f+~+£+'", , 1, I, 1, I, (c) -+~+-+-+, ..., , 2 ,,2, , ~, , 18. Evaluate the following, (a), , I, 1), L+-( 'F+7n, n=O, , (b), , fin, , n=1, , (c), , ~ 2n+l, , ~n2(n+l)2, , ifi
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CHAPTER 43 Infinite Series, , (d), , f, , 2";3·, , (e), , .... 2"-1, Ly, , (f), , .=1, , .=1, , .=0, , (h), , ...., , .... 5·, L¥, , f.-1 (~!t, , . .... 23•, (1), , L 32•, • =1, , (j)~), "=1, , Ans., , (a) Jt; (b) +00; (c) I; (d) t; (e) I; (f) +00; (g) +00; (h) -i; (i) 8; (j) +00, , 19. (GC) In Problems 1 and 6" use a calculator to compute the first'1O partial sums and determine to hqw many, decimal places the 10th piutial sum is a correct estimate of the sum of the series•, , 20. (GC) (a) If Ixl < I, what function is represented by, , ...., , LX" =I +x +x +x +... ?, 2, , 3, , .=0, 2, 3, , (b) Use a graphing calculator to graph I + x + x + x +... + x 9 on the interval (-I, I) and compare the graph, with that of the function in (a)., , Ans., , 1, , (a) -1-, , -x, , 21. In each of the following, find those values of x for which the given series converges, and then find the function, represented by the sum of the series for those values of x., (a) f(3x)n, .=0, , AlIS., , (b) f(X-2)', .=0, , (c), , f(1f, ",,0, , (d), , f(x21)", n=0, , 1, 2, (a)lxl<-31'-113 ;(b)l<x<3'-3 ;(c)lxl<2'-22 ;(d)-I<x<3'-3, , - x, , -x, , -x, , -x
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Series with Positive Terms. The, Integral Test. Comparison Tests, Series of Positive Terms, If all the terms of a series, , II SII are positive, then the series is called a positive series., , For a positive series LS,,' the sequence of partial sums (S,> is an increasing sequence, since, S,,+I =Sn + sn+1 > SIlo This yields the following useful result., Theorem 44.1:, , 2., , A positive series, , Sn, , converges if and only if the sequence of partial sums (sn) is bounded., , L, , To see this, note first that, if sII converges, then, by definition, (S.) converges and, therefore, by Theorem 42.1,, (Sn> is bounded. Conversely, if (Sn> is bounded, then, since (S,.) is increasing, Theorem 42.8 implies that, (S) converges, that is, Ls. converges., , L, , Theorem 44.2 (Integral Test): Let, sn be a positive series and letf(x) be a continuous, positive decreasing function on [1, + 00) sHc,h that f(n) = Sll for all positive integers n. Then:, , LSn converges if and only if r-f(x}dxconverges, From Fig. 44,-1 we see that, so,, , r, , f(x)dx < SI + S2 + ... + S._I, , = SII_I' If ~>n converges, then (S,> is bounded;, , f f(x) dx will be bounded for all u ~ 1 and, therefore, r-f(x)dx converges. Conversely, from Fig. 44-1, , r-, , we have S2 +S3 +"'+SII <, , Sn <, , r, , f(x)dx and, therefore, Sn <, , r, , f(x)dx+s l • Thus, if r-f(x)dxconverges, then, , f(x) dx + s, and so (SII) will be bounded. Hence, by Theorem 44.1,, , Theorem 44.2., , S,', 2, , 3, , 4, , 5, , Fig. 44-1, , 2>11 converges. This proves
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CHAPTER 44, , EXAMPLE 44.1:, , Series with Positive Terms, , Inn., dIverges., , L- n, , Inx, Let f(x) =-. Now., , x, , -Inx dx= lim I"lnx dx= lim t(lnx)2]· = lim t«InU)2 -0)=+00, II x, ........ I X, u......., 1, • ....-, , Hence. by the integral test,, EXAMPLE 44.2:, , L I~n diverges., , ~...L converges., ~n2, , Let f(x) = ~. Now., x, , III, , (, , +- 1, JU -:rdx=lim-- =lim---l=l, 1 ), JI -:rdx=lim, x, U~+.. I X, u-HX 1, u-++_, U, , nr, , Hence. by the .mtegral test. '", £.J I converges., , Remark: The integral test can be easily be extended to the case where the lower limit of the integral is, changed from I to any positive integer., , La" and Lb b~ two positive series such that there is a positive integer m, , Theorem 44.3 (Comparison Test): Let, for which at :s; b. for all integers k ~ m. Then:, (I), , (2), , If, If, , n, , Lb. converges. so does La.:, La. diverges. so does Lb•., , We may assume in the derivation of Theorem 44.3 that m = 1, since convergence is not affected by deletion of a finite number of terms at the beginning of a series. Note also that (2) is a logical conse'C!,ence of, (1). To prove (I). assume that Lb. converges. Let B. =bl +b2 +· .. +b. be the nth partial sum for ~b. and, let An =a l + (/2 + ... + an be the 11th partial slim Lan' Then A. ~ Bn, since at ~ bi for all k. From the fact, that Lbn converges, it follows, by Theorem 44.1, that the sequence (8,) is bounded. Since An ~ Bn for all, n, it follows that the sequence (An> is bounded. So, by Theorem 44.1, La. converges. This proves Theorem 44.3., EXAMPLE 44.3:, , ~, ~5 converges., ~II+, , Let a• = ~5, and b. = ~., Then a. < b" for all n. By Example 2., n +, n, , L ~ converges. So, by the comparison test,, , L n2~ 5 converges., EXAMPLE 44.4:, , 11, , L 3n ~ 5 diverges., , 1 and b = - 3, Let a. = -4, I 5' Now, a. :s; b. for 11 ~ 5. (To see this, observe that .l.. S _1- is equivalent to 311 +, II, •, n+, 411, 3n + 5, 5 :s; 411. which is equivalent to 5 :s; II.) Recall that the harmonic series ~ diverges (by Chapter 43, Example 4)., Hence. ~ .l.. diverges by Theorem 43.2. The comparison test implies that ~_l_ diverges., , L, , ~4n, , ~3n+5, , Sometimes. as in Example 4, complicated maneuvers are needed in order to apply the com'parison test., The following result offers a much more flexible tool., II, , Theorem 44.4 (LImit Comparison Test): Let ~, a and ~, b be two positive series such that L = .....lim -b" exists, ~", ~., and 0 < L < +00. Then, converges if and only if, converges., •, , La", , Lb"
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CHAPTER 44, , Series with Positive Terms, , Assume that Lbn converges. Let c be a positive number such that L < c. Then there exists a positive integer rn such that an/b" < c for all n ~ m. Hence, an < cb" for alln ~ m. But, since L bn converges, so does L cb"., Therefore, by the comparison test, Lan converges. Conversely, if Lan converges, then Lbn converges., (In fact, lim bn = ~ > 0 and we can use the same kind of argument that was just given.), n~f<oo (I", , 2, , - 511 + 4 diverges., 7111+2, ', ", When dealing with quotients of polynomials, a good rule of thumb is to ignore everything except the leading, 31 . Let us try a limit comparison with 1. Now, terms. In this case, we have 3 /: =-7, ,, , EXAMPLE 44.5:, , ,,311, L.J, , 7 11, , /I, , /I, , lim [(3n, n......., , 2, , +, , ..,. 5n, 7n 3 +2, , 4)/1], , = lim 3n, , 12, , n-->+-, , ~n, , ", , ~, , J11 6 5114 22, - 11 +7, , -, , ~n2 + 4n 3, 7', , 7n +2, , L 311;11-) 511;, 4 diverges., +, , Since" 1 diverges, the limit comparison test tells us that, EXAMPLE 44.S:, , 3, , converges., , . Us\ng the rule of thumb given in Example 5, we should look at, with 2"":, , 17= ~~. = ;2 ., , So, let us try a limit comparison, , 11, , Let us divide the numerator and denominator by /1 3• Note that, in the denominator, wewould get, , So, the result would be, , ., . companson, ., " J 6 5n 4- 22 7, Hence, smce, we know, by Examp12th, e , at ", ~I, 2"" converges, the I'mllt, test .ImpI'les t h, at ~, n, n - n +, converges., , SOLVED PROBLEMS, , 1., , L ~, yvhere p is constant. This is called a p-series. Then:, (a) If p> 1, the series L nIp converges., (b) If p ::;; I, the series L ~ diverges., We may assume that p :f. I, since we already know that the harmonic series L 1 diverges. We may also assume, I, n, , Consider'the series, , 11, , 11, , thai p > 0; if p::;; 0, lim, , °, , and the Divergence Theorem implies that the series diverges. Let us apply the, n, p, integral test with f(x) = l/x • (f(x) is positive and decreasing in [1, +00).) Now,, tI-++oot, , -:f., P, , f,, , ~- -dx=, 1, lim, xP, u-->+-lim, - u-->+-, , f,, , u, , I, '-P ]", -dx=, lim _x_, xP, u-->+- I - P ,, , (E2.., __, I - P 1- p ., 1 )
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CHAPTER 44 Series with Positive Terms, lim _I_I =O. So, lim (I -' .-++_ Ul-~ =.-++_, uP.-++_ - p, , I., , UI, , (a) p> Then p -I > 0 and lim, , L nI,, , converges., , (h) p < 1. Then 1- p > 0 and lim ul- p, , ., , .-+...., , dIverges., , =+00. So, lim (lu-p, '-P U-HN, , -I-_1_), =_I_I', By the integral test,, p, p-, , _1_1_) = +00 and, by the integral test,, -p, , L~n, , In Problems 2-7, examine the given series for convergence., 2., , I, , 1, , I, , 1 + - + - + - + .. ·., , J3 J5 J7, So, , =~.Let f(x)=~., v2n-1, , v2x-1, , On [I,+OO),f(x) > 0 andfisdecreasing., , r+- $=t, dx = lim r· rf?.-. = lim -21 r· (2x 2x -I, u-++- J, ",2x - 1 u-++_, J,, , J,, , = lim t(2)(2x _1)"2, ,,-++-, , I, , 1, , o-"2(2)dx, , = lim «2u _1)"2 - I) =+00, .~, , Hence, the series diverges by the integral test., , 3., , I, , I, , I, , I, , 3" + TO + 29 + ... + n3 + 2 + .. '., n3 ~ 2 < ~3', , L ~3 is convergent, since it is a p-series with p =3 > 1. Thus. by the comparison test. L n3 ~ 2, , is convergent., , 4., , I, , 1, , I, , 1+ 2! + 3! + 4! + ...., , 1, s. =.1,., ~ 2~-' is a convergent geometric series (with, n. Note that .1,, n. = n(n- 1) ..... 3 . 2 ~ 2~-1 for n ~ 2. Since ~, ratio t),, is convergent by the comparison test., n., , L -\, , 345, 5.' 2 +2f+y+'4J+, ...., sn = n ~ 1. Use limit comparison with -!!r =-\-., n, n n, , We know that, , L nl2 converges. So, by the limit comparison test, L n; 1 converges., , 6• 1+ -L+.l+-L+···, 22 33 44, S., , =~., I ·n ~ 2~', n ,Now, ~, n =n·n ...., , comparison test,, , L nln converges., , and ~, ~ 2!-' is a convergent geometric series (r =t). So, by the, , ., , ,, , ,"'.,",, :~.:- ~::..~~, , ~"1.,,_- .~;~~~, Sn, , ~ ~ ',~ ; '~~'(", , =n:n' ++11 . Use limit comparison with n:1/ =1n :, , 1)/lJ=, lim n + n =, 1, n, n +I, , +, .-++_ [(n2, n3 + 1, lim, , ","1', , 3, , .-++-, , 3, , '.'>;:~":, ,--', , ""
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We know that the hannonic series, , R., , Positive Terms, , Series with, , CHAPTER 44, , L 1n diverges. So, by the limit comparison test, £oJn+1, ~ n: + 1 diverges., ., , 1, 1, I, 21n2 + 31n3 + 41n4 + ..., So, , = -11- is detined for n ~ 2., , n nn, , dx, J2'- -.-I-=lim, .\ n x, , u-tt-, , J" -I-=Iim, dx, In(lnll) ]" = lim (In(1nu)-ln(ln2))=+oo., n x ....... 2 X, , 2, , " ..... _, , Hence, the series diverges by the integral test., 9., , How many terms of, , L -4n suffice to obtain two-decimal place accuracy (that is, an error < S/l03)?, , If we use k tenns, then we require that the error, , I, I, I, ~, I, L.~ 2"L2"=, £oJ 2"~, "_I n, "_I ", "-t+1 n, , I., , f+- 2"dx=, I, . J" 2"dx=, I·, 11m, hm--I, I, , x, , 101 ...., , +_, , l X, , ............, , =hm - (I---:I), x......... u k, , Hence, 200 <k. Thus, it suffices to use 201 terms of the series. (The graphing calculator can be used to find, , L::2I .. 1.64.), 201, , ._1 11, , Ls., , 10. Assume, converges by virtue of the integral test applied to/ex) and, for each n, the error (or remainder) Rk, after k tenns is defined to be, . +-, , Then, , Rk =, , L s. < 1:-/(x)dx., , lI=k+1, , Find a bound on the error when, , f.-1 :2 is approximated by the first five, , I, The error Rs < 2"dx = t = 0.2., s x, , J, , +-, , 11. Assume, , tenus: t +., , t, , + ~ + I~ +, , is, , =, , ~~~ .. 1.4636., , Ls. and Lc. are positive series, Lc. converges, and s. ~c. for all n. Then the error R1 after k tenns is, , +-, , L, ~I with an error < 0.0000 I?, .~I n +, I, I, I f+- I, I, = -5-. and c. =:3"', It suffices to have L :3" < 0.00001. Now, ~ s <, sdx, = 4k4 ., • n +, n, n, £oJ n, x, , At least how many terms will suffice to estimate, In this case,, , S, , I, , +-, , I ., , .-1+1, , •, , +-, , .=k+l, , So, we need 4k4 < 0.00001 = 100,000' EqUivalently, 100,000 <.w, 25,000 <~, k ~ 13., , k
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CHAPTER 44, , Series with Positive Terms, , For Problems 12-43. determine whether the series converges., , 12., , 13., , L~, n, , 3, L n(n +1), , Ans. converges; comparison with, , L(n+I)(n+2), , Ans. diverges; limit comparison with, , L~, , 14. Ln2n+l, , · · companson, . WI'th k.J, ~ 11, 1, Ans. diverges; IlImt, , 15. L;, , Ans. converges; integral test, , 16. L, , Ans. converges; limit comparison with, , 2n, (n + I)(n + 2)(n + 3), , L~, n, , 17., , L(2n~I)2, , Ans., , · . comparison, ., ~ 1, converges; IImlt, WI'th k.J, , 18., , Ln3~1, , Ans., , ., · . comparison, converges; IImlt, Wit'h~, k.J I, , nr, , nr, , 19. L n - 2, , ., Ans. converges; I·umt. comparIson, Wit'h~, k.J I, , 20. Linn, 2, , Ans. converges; limit comparison with, , nr, , 7, , n +2, , Lnsin(~), , Ans. diverges; Divergence Theorem, , Li, , Ans. diverges; p-series, p = t < 1, , 23. L)-i, , Ans. converges; comparison with, , 24. Linn, , Ans., , diverges; comparison with, , L I~n, , Ans., , diverges; comparison with, , L~, , Ans., , diverges; limit comparison with, , (for n ~ 3), , Ans., , diverges; integral test, , (for n ~ 3), , Ans. converges, integral test, , 21., , 22., , ..rn, , 25., , L, , I+~nn, , I', , 26. L, , n+l, n./3n-2, , 27., , nlnnl~ (Inn), , 28, , •, , L, L, , I, , nlnn(\n(lnn»2, , L -k, n, , L 2LI ,n ~ 2, , L In, , .~, , :".
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CHAPTER 44, , Series with Positive Terms, , 1 + 1 +_1_+ .. ,, 29• 412 + 72, ', W 132, I'IInit, , companson, ' . WIt, 'h £..J"2, '" 1, = (3 n+1 1)2 ; converges;, ", n, , A ns, s,, , 3 + 3 +_3_+.,., 30. 3 +¥, 3iir 4"3, •, .I, , Ans,, , s,, , =~; diverges: p-series, p = t < 1, , 1 1 1, 31. 1+5+"9+13+···., Ans,, , sn =!.....--4, 1 3: diverges: limit comparison with '", n£..I 1, , n, , 1, 1, 11, 32. 2 + 3.4 + 4.5,6 + ~ , ~, , 33., , n, , + ...., , Ans., , sn, , = III + 1)(11: 2) ... (211): converges; limit comparison with L ;2, , 2, , 3, , 4, 5, + 3. 33 + 4. 34 + ...., , 3" + 2. 32, Ans., , sn, , = ~ ~! ;converges; limit comparison with L jn, , 1, 1, 1, 1, 34. 2 + 2.2 2 + 3.2 3 + 4.24 + ... ,, An.\'., , I, = n2', ; converges;., companson WIt'1",1, 1 £..J 2", , s., , 2, 3, 4, 5, 35. D+ 2·4 + 3·5 + 4·6 + ...., Ans., , ., 1, =n(nn ++12): d'Iverges; 1"ImIt companson, WI'th '", £..J n, , s,, , 1 2, 3, 4, 36. 2+32'+'43+34+ .. ·., A '1.1:, 37. 1+, , s. = (n ~ I)' : converges; comparison with, , iz +, , A ns., , 3L2, , s. =, , +, , L 2!-1, , 13 + ...., , l', , ., , 'th, , n(.+2)/2; converges; companson WI, , L "2, 1, n, , 38 1+ 1 + 3. + ..i + .. '., •, 5 10 17, Ans., , sn, , =n+, n2+ I) ; diverges; limit comparison with'", 1, £..J1t, , 2 2·4 2·4·6, 2·4·6·8, 39. 5' + 5.8 + s:s.TI + 5·8 ·)1·14 + ...., Ans., , s. =, , 2·4 ..... (2n), C;. Q .. • /')..L, , 'tM\;, , .•, , converges;companson WIth, , '", £..J<t)'
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••, , CHAPTER 44 Series with Positive Terms, , 42., , s, , Ans., , s = n4 +2, ; diverges; limit comparison with ~ 1, • n +n, L. n, , o, , 3, , 1, 22, , 234, -1 + 32 _ 2 + 42 - 3 + 52 - 4 + ...., s =(, • n+, , Ans., , 43., , =n~+n, +1; converges, limit comparison with L., ~~, n, , Ans., , 1, 2) _, , ;)2 -n ;diverges;, limit comparison with ~ 1, ', L. n, , 1, 1, 1, F + 33 - 22 + 43 - 32 + 5) - 42 +...., , s. = (, , Ans., , n+, , 1, 1)3, , -n, , . companson, . WI'th, 2; converges; I'Imlt, , L3n1, , 44. (GC) Estimate the error when:, , f 3. ~ 1 is approximated by the sum of its first six terms., , (a), , n=1, , (b), , -f, , 40, , ~ 3 is approximated by the sum of its first six terms., , n-=I, , Ans., , (a) 0.0007; (b) 0.00009, , 45. (GC) (a) Estimate the error when the geometric series, , L ]. is approximated by the sum of its first six terms., , (b) How many terms suffice to compute the sum if the allowable error is 0.00005?, , Ans., , (a) 0.047; (b) 16, , f -'n . with an error < 0.00 I?, Find a bound on the error if we approximate f -'. by the sixth partial sum., , 46. (GC) (a) How many terms suftice to approximate, , 11=1, , (b), , ,, , _, , (c) What is your approximation to, , ~n, , n=1, , Ans., , (a) 7; (b) 0.0015; (c) 1.0811, , -t, , 47. (GC) Let S. be the nth partial sum 1+ + ... + ~ of the divergent harmonic series., (a), (b), (c), (d), Ans., , Prove In (n + 1) S S. S 1+ Inn., Let En =S. -Inn. Prove that (E.) is bounded and decreasing., Prove that (Eo> converges. Its limit is denoted 'Yand is called Euler's constant., Use a graphing calculator to approximate E999 to eight decimal places., (d) 0.57771608 (in fact,, , r- 0,57721566.), , 48. (Extension of the limit comparison test.) Assume Ls. and Lt. are positive series. Prove:, , =0 and L'. converge, so does Ls•., If lim ~ =+00 and Ltn diverges, so does LSn', n~+- In, , (a) If lim Stn, n~", , (b), , ,., , ,, , L -'., by the sixth partial sum, correct to four decimal places?, n
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CHAPTER 44, , 49. Use the extension of the limit comparison test to determine whether, , AIlS., , converges; use, , Series with Positive Terms, , L (lnn7)4 converges., , L;!r and Problem 48(a), , LSn, , 50. Assume, is a positive series and lim ns" exists and is positive. Prove that Ls.diverges. (Hint: Limit, "-HM, .,, comparison with L(1ln).), , 51. Assume, , Ls. and Lt. are convergent positive series. Prove that Ls.t. converges.
Page 385 : Alternating Series. Absolute, and Conditional Convergence., The Ratio Test, Alternating Series, A series whose terms are alternately positive and negative is said to be an alternating series. It can be written in the form, , where an are all positive., Theorem 45.1 (Alternating Series Theorem):, quence (an) is decreasing; (2) lim an =0, Then:, , Let L(_l)n+l an be an alternating series. Assume that: (1) the se-, , n~+oo, , (I), , L, , (_l)n+1 an, , converges to a sum A, and, , (II), , If An is the nth partial sum and Rn =A magnitude than the first term omitted),, , (I), , Since, , (an), , An, , is the corresponding error, then IR.I < an+l (that is, the error is less in, , is decreasing, a 2n+1 > a 2n+2 and, therefore, a 2n+1 -, , a 2,,+2, , > 0, Hence,, , So, the sequence (A 2) is increasing, Also,, A, , .I~/I, , =aJ-, , (a 2 -, , a3 ) -, , (a 4 - a S) - .. , -(a2n-2 -, , a211-1 ) -, , a 2/1, , <a I, , Hence, (~J is boun~ed. Therefore, by Theorem 42,8, (A 2n ) converges to a limit L, Now, 'A 2n+1 = A 2n + a2n +I ,, Hence,, , ,., lim A2n+1 = lim ~n + lim a2n+1 =L + 0 =L, , 11-'+-, , Thus, lim An = L and, therefore, L, , n~"", , (_l)n+1 an, , n~+-, , converges,, , n-++-, , (II), , =(a2n+1 - a 2n+2 ) + (a 2n+3 - a 2n +4 ) +.. ,> 0, and R 2n =a2n+1 Hence, IR2nl< a2n+l' For odd indices, R2,,+I =-(a2n+2 - a2n+3 ) R 2n, , (a 2n +) -, , (a 2n+2 - a2n +3 ) - (a 2n +4 - a2n+S ) -, , , .., , < a2n+l', , < 0 and R 2n+1 =-(12,,+2 +, a2n+4) +(~n+S - ~n+li)+'" > -a 2n +2 , Hence, l~n+11 < a2n+2 , Thus, for all k, IRk I< ak+I ,, (a 2n +4, , -, , a2,,+s) -'"
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..,,-----, , CHAPTER 45 Alternating Series, , Thealtemating harmonic seri~, , EXAMPLE 45.1:, , 1 1 1 1 1 .. ·, 1--+---+---+, , 2, , 345, , 6, , converges by virtue of the Alternating Series Theorem. By part (II) of that theorem, the magnitude IR"I of the error, after II terms is less than n ~ l' If we want an error less than 0.1, then it suffices to take 11 ~ 1 ~ 0.1, equivalent to I0 ~ 11 + 1. So, n ~ 9. Thus, we must use, , = I~' which is, , Definition, Consider an arbitrary series ~>"., , LS" is said to be absolutely convergent if Lis,.! is convergent., LS" is said to be conditionally convergent if it is convergent but not absolutely convergent., EXAMPLE 45.2:, , The alternating hannonic series I(-lrl -k is conditionally convergent., , EXAMPLE 45.3:, , The series I(-I)"+1 ~2 is absolutely convergent., , We shall state without proof two significant results about absolute and conditional convergence. In wh!!t follows,, by a rearrangement of a series we mean a series obtained from the given series by rearranging its terms (that is, by, changing the order in which the terms occur)., , LS" is absolutely convergent, then every rearrangement of LS" is convergent and has the same slim as Is,;., If L s" is conditionally convergent, then if c is any real number or, or --, there is a rearrangement of L s", , (I), , If, , (2), , +00, , with sum c., Theorem 45.2: If a series is absolutely convergent, then it is convergent., For a proof, see Problem 1., , Note that a positive series is absolutely convergent if and only if it is convergent., The following test is probably the most useful of all convergence tests., Theorem 45.3 (The Ra~lo Test):, (1), (2), , ,, , LS" be any series., , If lim, , 1.1'S,.' +1\= r <'1, the9 LS" is absolutely convergent., , If lim, , ISn+I\= rand (r> 1or r=+o<:), then Is" diverges., , ,,-++-, , n--+-+-, , (3), , Let, , S", , If lim \.1'"+1\ = I, then we can draw no conclusion about the convergence or divergence of, , L, , S •, , . ", , ~~, , Problem 14., Theorem 45.4 (The Root Test):, , Let, , LS" be any series., , (2), , "fSJ =r < 1. then Is" is absolutely convergent., If lim ,,1ol,.1, "fSJ =r and (r > 1 or r = +0<:), then L srI diverges., , (3), , If lim • Ii;,), =1, then we can draw no conclusion about the convergence or divergence of, n, , (1), , If n-Jo+oG'IloJ"', lim, ,,~+-, , n-++-, , ".Il, , ', , For a proof, see ProLlem 15., , I, , Sn', , For a proof, see
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.., , CHAPTER 45 Alternating Series, , EXAMPLE 45.4:, absolutely., , Consider the series, , L 22: .Then lim ~=lim i, II, , n-++- II, , 11-++-, , =O. So, by the root test, the series converges, , SOLVED PROBLEMS, , 1. Show that, if, , Ls. is absolutely convergent, then it is convergent., , o~ S. + Is.1 ~ 2Is.l. Since Lls.1 converges, so does L21s"i. Then, by the comparison test, L(s. + Is"l) converges., Hence, Ls. = L«s. + Is.l) -Is.b converges by Corollary 43.4., In Problems 2-13, determine whether the given series converges absolutely, conditionally, or not at all., , 2., , I 1 I I, 2"-5"+10-17+'", , L, , s. =(-1),,+1 +1', +1 converges by comparison with the convergent p-series, n+, n+, I, ', L(-l)""tl n21+ 1 is absolutely convergent., 3., , I 2 3 4, ---+---+, ee 2 e3 e4 ... •, , sn = (_1)·+1 ;. The series, absolutely convergent., , 4., , L; converges by the integral test ( using f(x) = :. ). Hence, L (-I )n+l;, , I, I, 1, I, 1--+---+--·, .., , JiJ3J4J5', , sn = (_1)"+1, test. But, , 5., , L ~., So,, n, , 1, , In., , L"*, , 1, , Since (, , is, , ", , "*), , is a decreasing sequence, the series converges by virtue of the alternating series, , is divergent, since it is a p-series with p =t < I., , 1, , 1-2"+4-8+ .. ··, The series I + t +f - t + ... is a geometric series with ratio r = t. Since Irl < I, it converges and, therefore, the, given series is absolutely convergent., , 6., , 1_1+l.._-±.+, ..., 3 32 33, S. = (_1)·+1 3!1 . Let us apply the ratio test:, lim IS.+II=n+I/..1L=n+l 1, s", 3" 3,,-1, II 3 ., , ,I-t+-, , So,, , Hence, the given series is absolutely convergent., , 7., , 1, , 21, , 31, , 41, , 2"-323+433-5 43 +...., , s. =(-lr~lI:, , convergent p-series, , 8., , 2, , 31, , 41, , ~3' Look at L~J Is..l =, , I, , L ~3, , • Hence,, , n:, , 1 ~3 < ~3' So, Lls.1 converges by comparison with the, , the given series is absolutely convergent., , 51, , 3-42"+53-64+· ..., s. =(_1)"+1 ~ ~, , i kNote that (~ ~ i Mis a decreasing sequence ( since, , series is convergent by the Alternating Series Theorem. However,, with £"/1, ~ 1. Thus, the given series is conditionally convergent., , D, ( (x\\\x ) < 0). Hence, the given, , ISIII > -t~. So, Lls,,1 diverges by comparison, , "'
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CHAPTER 45 Alternating Series, , 9., , 23 25 27, 2 - - + - - - + .. ·, 3! 5! 7!, ., , s = (-1)"+1, ", , 0", , 22q- 1, . Apply the ratio test:, (2n-l)!, , si 2, l;:1 =(211 +1)!, , :<, , 2, ,,+1, , ..., , Hence, lim, 11 ..... . . -, , 10., , 1, , "2 -, , /, , 22,,-1, , (211-1)!, , 4, , =(211 + 1)(211), , IS"+II, =0 and, therefore, the series is absolutely convergent., sn, , 4, , 9, , 16, , 23 + 1 + 33 + 1 - 4 3 + 1 + ..., , n~: 1. Since ( / : 1) is a decreasing sequence for II ~ 2, the given series converges by the, Alternating Series Theorem. The series L ISql is divergent by limit comparison with L 1. Hence, the given series, , s" =, , (_1)"+1, , is conditionally convergent., , 11., , 1, , 2-, , 2, , 3, , ', , n, , 4, , 23 + I + 33 + 1 - 4 3 + 1 + ...., , -4-. Hence, the given series is absolutely, , s = (_I)"+I_Il_., ~ Is"1 is convergent by limit comparison with ~, ", n3 + 1 L..., L... II, convergent., , 12., , 1, , 1, 1, 2· 22 + 3· 23, , N -, , -, , 1, 4.24 + ...., , s" = (-Iy+1 n1" . Apply the ratio test:, S,,+II = (n +1)2,,+1, 1, / 1, 1, s:, n2" = n + 12, II, , 1, , Thus, lim, , ,,~-, , IS"+11 = -21 < 1. So the given series is absolutely, ~onvergent., ., ~, , L (-1),,+1 (n: 1)!,, 3, , 13., , Apply the ratio test:, , Is"+II_ (n + 1)3 /~, _ (ill)) (_1), (n+l)! n, n+2, , . IS" - (11+2)!, , Isn, , So, lim SM11 = O. Hence, the given series is absolutely convergent., n ..... ..-, , 14. Justify the ratio test (Theorem 45.3)., (a) Assume lim, , fI-+-+-, , 11, , I~I, = r < 1, Choose t such that r < t < I. Then there exists a positive integer m such thaI, if, sit, , ~ m, IS;:11 ~ t. Hence,, , But,, , Lrlsml is a convergent geometric series (with ratio t < 1). So, by the comparison test, Lls,,1 converges., , Hence,, , LSq is absolutely convergent.
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CHAPTER 45 Alternating Series, , ISn+'I= rand (r > I or r= +00). Choose t so that 1 < t < r. There exists a positive integer m such, , (b) Assume, lim s., ., ._, , that, if, , n~ m, Is;:' I~ t. Hence,, , Therefore, lim s. = 00 and, by the divergence theorem, LSo diverges., (c) Consider, , 1-., , ~, , iT, , lim, n ........, , k n2 ', , I~I = lim [(--Ll, )/1]n =._, lim -ILl =I. In this case, the series diverges. Now consider, n+, n+, ., So, , ........., , _, , ._IS.I- 0- ( + nr)-.2.. ( + ), lim s.+, - lim, , (n, , 1, , 1)2 /, , 1 - li, , n, , n, , 2, , -1, 1 -, , In this case, the series converges., 15. Justify the root test (Theorem 45.4)., , n_ vfsJ';, , r < 1. Choose t so that r < t < 1. Then, there exists a positive integer m such that, .~ ::;; t for n ~ m. Hence, ISol : ; for n ~ m. Therefore, Lisol converges by comparison with the convergent, geometric series, SO,, is absolutely convergent., (b) Assume!!.!! ~ = r and (r> I or r =+00). Choose t so that 1 < t < r. For some positive integer m, ~ ~ t, , (a) Assume lim, , Lt". LSn, , t", , =+00, lim s. = So, by the Divergence Theorem, Ls. diverges., Consider ~ 1 and ~ ~. In both ca~;'iim .fsl ~1~(Note that lim n-' =lim e-(lno)/. =I.), LJ n, k n, n~+- VI.)nl, n~+OO, for n ~ m. Then Is.1 ~ r for n ~ m. Since lim In, , (c), , 00., , n-H'M, , In Problems 16-22, use the ratio test to test the series for convergence., , So,, So, the series converges by the ratio test., , So,, Hence, lim, , Is'+'1 =+oc and the series diverges by the ratio test., , II...., ..... s", , 1. 2 1. 2 . 3 1· 2 ·3 ·4 +..., 18 1+-+--+, ., 1·3 1·3·5 1·3·5·7, S, , •, , =, , I, , n!, 1· 3 . 5 ...... (2n - 1),, , Then, , /, n!, n+l, s:- = 1·3·5·(n+I)!, .... (2n +1) I· 3 ·5· ..... (2n - I) = 2n +1., S,,+,, , So, lim s,,+rl =-21 < 1. Hence, the series converges by the ratio test., n~+-, , sn, , s., So, lim, , n-++-, , n+ 1 1, =-n40-' ., , Then, , ~_(n+21-)/(n+ 1_1_)_1 n(n+2), SIt n + 1 4", n 4, 4 (n + 1)2 ., , 1, Is"+', sn I= -4 < l. Hence, the series converges by the ratio test., , 0, , -', , -, , 7' ",
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CHAPTER 45, , Alternating Series, , Then, , Then lim, , n-+-+-, , I~I, s. =I. So the ratio test yields no conclusion. However, limit comparison with k'" 1n shows that the, , series diverges., , ", , 113", 21. '", k (11+, l)!', So+1 _ (n + 1)3"+1 /~ _ n + 1_3_, S., , n n+2', , So,, , (n+l)o+I/~=(n+I)" =(1+1)"., , So., , -, , (1I+2)!, , (n+l)l -, , lim, , IS"+II, sIt =0, , lim, , 1~I=e>1, s", , ,.~, , Hence. the series converges by the ratio test., , 22. "'~, k n.,., So+1 =, S., , (n+I)!, , n!, , 11, , n, , "~...., , Hence, the series diverges by the ratio test., , In Problems 23-40. determine whether the given alternating series is absolutely convergent, cqnditionally convergent., or divergent., 23. ~(_1)"+11.., nl, , Ans., , 24. ~(_I)"+I_l, , Ans. conditionally convergent, , Inn, , ~,- ~"""'",, , .', , 25., , L (_1)"+1 _n_, +1, Il, , absolutely convergent, , Ans. divergent, , 26. ~ (_l)"+I..l!!!L., , Ans., , condi!ionally convergent, , 27. ~(-I)"+I_I-, , Ans., , conditionally convergent, , 3n+ I, , 2n-l, , 28., , ~(-I)"+'*, , Ans. divergent, , 29., , L(- 1),,+1 (21l-1)2, 1, , AilS., , absolutely convergent, , Ans., , conditionally convergent, , Ans., , absolutely convergent, , 30. L(-I)"+I, , 31., , 1, , .jn(n + 1), , ~(-I)·+' (n11)2
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CHAPTER 45 Alternating Series, , 32. L (-I)It+' _1_, n2 +2, 33. L(-Ir', , 1, (n!)2, , 34. L(-lr,_nn2 +1, 2, , Ans. absolutely convergent, , absolutely convergent, , Ans., , Ans. conditionally convergent, '., , 35. L(-Ir,_nn4+2, , Ans. absolutely convergent, , 36. L(-Ir'n{lf, , \4, , Ans. absolutely convergent, , 2, , 37. L(-I)n+' 2n -3, n +n+2, 38., , L(-Ir' n;'1, , 39., , L(-Ir' 2':...2, , 3, , 40. L~, 2, , Ans. divergent, , Ans. absolutely convergent, , Ans., , absolutely convergent, , Ans. absolutely convergent, , n, , 41'. (GC) How many terms of L(-I)-+'-\ will suffice to get an approximation within 0.0005 of the actual sum?, 11., Find that approximation., Ans., , n =6; 1~ - 0.632, , 42. (GC) How many terms of L (_I)n+' (2n ~ I)! will suffice to get an approximation of the actual sum with an error, <0.001? Find that approximation., Ans. n =.3; 0.842., , 43. (GC) How many terms of L(-I)n+,, <0.001? Find that approximation., Ans., , *, , will suffice to get an approximation of the actual sum with an error, , n = 1000; 0.693, , In Problems 44-49, determine whether the series converges., , ,., , ~, , (n!)2, , 44. ~ (2n)!, , Ans. convergent, , Ans. divergent, , Ans. divergent
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CHAPTER 45 Alternating Series, , ~ll';", , 47., , L~, n!, , Ans., , convergent, , 48., , L (n +4"2)", , Ans., , convergent, , 49., , L(ll~lr, , Ans., , divergent, , ,, , 50. Determine whether L(-l)"+I("n+ 1- Jti} is absolutely convergent, conditionally convergent, or divergent., Ans., , ;. c. ", , conditionally convergent, , In Problems 51 and 52, lind the number of terms that suffice to approximate the sum of the given series to fourdecimal-place accuracy (that is, with ari error < SIlOS) and compute the approximation., , 52. (GC) f(-I)'H', , (211~1)!, , Anso, , n = 6; 0.9721, , Ails., , II, , =4; 0.8415, , '1=1, , 53. Let Irt < 1, (a) Prove that Lnr" = r+2r2 +3r3 +4r 4 + ... converges ., , .-, , (b) Show that ~ IIr", , = (1 ~ r)2 . (Hillt: Let S =r + 2r2 + 3r2+ 4r4 +"', multiply this equation by r, and subtra~t, , the result from the original equation.), <-, , (c) Show that, , L ;" =2., rt=1
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·Power Series., Power Se'rles, An infinite series, , ...., Ian(x- C)", , =ao+al(x- c)+a2(x- C)2 + ..., , (46.1), , 11=0, , is called a power series in x about c with coefficients (a,,). An important special case, , ...., , Ia"x n =ao + alx + a2x 2 +..., , (46.2), , 11=0, , is a power series about O., For a given value of x, the series (46.1) either converges or diverges. Hence, (46.1) determines a function, fwhose domain is the set of all x for which (46.1) converges and whose corresponding valuef(x) is the sum, of the series., Note that (46.1) converges when x =c., EXAMPLE 46.1:, , The power series about 0, ioo, , LX' = I + x + x" +..., ,.=0, , is a geometric series with ratio x. Thus, it converges for Ixl < 1, and its sum iS~. So, the domain of the, corresponding function is an interval around O., -x, ioo, , Theorem 46.1:, , Assume that the power series La.(x - c)' converges for Xo:t c. Then it converges absolutely for all, /lVO, , x such that Ix - cl < Ixo - cl (that is, for all x that are closer to c than Xo)., , For a proof, see Problem 4., ioo, , For a power series La.(x- c)n, one of the following three cases holds:, (a) it converges (or all x; or, ••0, (b) it converges for all x in an open interval (c - RI , C + RI ) around c, but not outside the closed interval, [c - R I , C + Rtl; or, (c) it converges only for x =c., , Theorem 46.2:, , ...., , By the interval of convergence of :~>n(x- c)n we mean:, , In case (a): (-00, +00), In case (b): (c - RI , C + RI ), In case (c): {c}, , ,,=0
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CHAPTER 46, , Power Series, , ....., , By the radius of convergence of ~>"(x- c)" we mean:, In case (a):, "=0, In case (b): R,, 00, , In case (c):, , o·, , Note: In case (b), whether the power series converges at neither endpoint of its interval of convergence or, at one or both of those endpoints depends upon the given series., ", ., For a sketch of a proof of Theorem 46.2. see Problem 5., EXAMPLE 46.2:, , The power series, , f-,-(X_-_2t-)' = (x _ 2) + (x - 2)2 + (x - 2)3 + ..., • al, , 2, , n, , 3, , is a power series about 2. Let us use the ratio test to find the interval of convergence., , S.+II = Ix - 21"+1, s" ' n + 1, , fix - 21", Il, , 1, , = _n_lx - 21. Thus,, 11 + I, , lim 1S'+11 == Ix - 21., , 11___ s., , So, by the ratio test, the series converges absolutely for Ix - 21 < I. The latter inequality is equivalent to -I < x - 2 < I,, , -, , which, in turn, is equivalent to I <x< 3. Hence, the interval of convergence is (1, 3) and the radius of convergence is 1., At the endpoint x = 1, the series becomes ~)(-l)lIlfl], which converges by the Alternating Series Theorem. At the endpoint x = 3, the series becomes I,(lIn), ttle1divergent hannonic series. Thus, the power series converges for 1 $x< 3., • =1, , EXAMPLE 46.3:, , The power series, , .... x', x2 x3, I, --I+x+-+-+'", 1, 2'. ., 31, n., .=0, , is a power series about O. (Recall that O! = I.) Let us use the ratio test:, , Hence, by the ratio test, the series converges (absolutely) for all x. Jts interval of convergence is (.-" +00) and its, radius of convergence is 00., •, , EXAMPLE 46.4:, , The power series, , ...., , Ln!x" = I + x+ 2!x 2 +3!x3 + ..., • =0, , is a power series about O. Let us use the ratio test again:, , S.+II=, I s., , (n+,?\~xl·+1, n.x, , =(n+ 1)lxl. So,, , lim IS.+II=+oo., ........ s", , except when x = O. Thus, the series converges only for x = O.lts (degenerate) "interval" of convergence is {O) and its, radius of convergence is 0
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CHAPTER 46, , Power Series, , Uniform Convergence, Let (I.) be a sequence of functions, all defined on a setA. Let/be a function defined on A. Then (I.) is said, to converge uniformly to / on A if, for every E > 0, there exists a positive integer m such that, for each x in A, and every n ~ m, !fn(x) - fix) I< E •, , ..., , *, , If a power series ~ an(x - c)· converges for Xo c and d < Ixo - cI, then the sequence of partial sums, .t.i.J, ..., (St (x)), where St (x) =La. (x - c)·, 'C"gnverges uniformly to La" (x - c)" on the interval consisting of all x such that, , Theorem 46.3:, , Ix - cI < d. Hence, the co~vergence is uniform on any interval'St}jctly inside the interval of convergence., , The reader is referred to more advanced books on analysis for a proof of this result., If (f.) converges uniformly to/ on a set A and each/" is continuous on A, then/is continuous on A., , Theorem 46.4:, , For a proof, see Problem 6., , ...., Corollary 46.5: The function defined by a power series La.(x - c)" is continuous at all points within its interval of, convergence., , This follows from Theorems 46.3 and 46.4., , -, , ..., , Theorem 46.6 (Integration of Power Series): Let/be the function defined by a power series La,,(x-c)· on, its interval of convergence (with radius of convergence R,). Then:, ,,:0, , (46.3), , (a), , where the interval of convergence of the power series on the right side of formula (46.3) is the same as that of the, original series. K is an arbitrary constant of integration. Note that the antiderivative of/is obtained by term-byterm integration of the given power series., (b) If a and b are in the interval of convergence, then:, , b, , _~( (X_C),,+I)]b, , fJ(x)dx-.~ an, Thus., , 1:, , (46.4), , n+1, , n=O, , d, , /(x)dx is obtained by term-by-term integration., , For a proof of Theorem 46.6. the reader should consult a more advanced book on analysis ., , ...., Theorem 46.7 (Differentiation of Power Series):, , Let/be the function defined by a power series Lan(x-c)", , on its interval of convergence (with radius of convergence R1). Then/is differentiable in that interval and, , ,., , ..., , !'(x) =Lna,,(x-c)"-1, nEO, , for, , Ix-cl<R1, , r, , ._0, , (46.5), , Thus, the derivative is obtained by term-by-term differentiation of the power series. The interval of convergence of, the power series on the right side of formula (46.5) will be the same as for the original power series., , For a proof, the reader is referred to more advanced texts in analysis.
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CHAPTER 46, , EXAMPLE 46.5:, , Power Series, , We already know by Example 1 that, for Ixl < I,, , 1, I-x, , ..., , =Lx" =1+X+X2 +Xl +"'+x" + ..., , (46.6), , .-0, , Now. D, (1 ~, , x) = 0_1, , X)2', , SO, by Theorem 46.7., __1 - =1+ 2x + 3x 2 + ... + IU"-I + . . . for !xl < I, (1- X)2, , ..., , ..., , 11:1, , ncO, , =Lnx·-' = L(n + I)x·, EXAMPLE 46.6:, , We know already that, , 1, ..., - - = ~ x" =1+x+ x 2 +Xl + ... + x" + ..., I-x £.J, .=0, , forW<1, , Replace x by -x. (This is permissible, since I-xl = lxl < 1.) The result is, , ~ (-x)" =~ (-l)"x" = 1- X+X2 _Xl + ..., , _1_=, , ~, , l+x, , (46.7), , ~, , By Theorem 46.6(a), we can integrate tenn by tenn:, , ..., , .+1, , dx=, J-I+x, ~, , ..., , ", , f::, , n, , ~(_I)·_x_+K= ~(_\)"-'L+K, , In II + xl, , n+1, , ..., , ", , =L(-I)"-':, , for !xl < 1, , for Ixt < I, , +K, , 1t=1, , Letting x = 0 and noting that In 1 = 0, we find that K = O., Note also that, for !xl < I, we have -I < x < I, 0 < 1 + x < 2, and, therefore, 11 + xl = 1 + x. Hence,, In (1 + x), , ..., , =L(-Ir, , l, , ., , :, , for !xl, , <I, , .=1, , (46.8), The ratio test shows that this series converges., If we replace x by x - I, we ob~a;n:, , Inx = f (_1),,-1 (x -I)", ~, , .-1, , n, , for lx-II < 1, , (46.9), , Note that lx - 11 < I is equivalent to 0 < x < 2., Thus, In x is definable by a power series within (0, 2)., , ..., , Theorem 46.8 (Abel's Theorem):, , Assume that the power series La,,(x - c)" has a finite interval of convergence, .-0, , lx - cl < R, and let/be a function whose values in that interval are given by thatpower series. If the power series also, converges at the right-hand endpoint b = c + R, of the interval of convergence, then lim lex) exists and is equal to the, :t-tb-, , sum of the series at b. The analogous result holds at the left-hand endpoint a = c ~ R I •, , The reader is referred to advanced books on analysis for a proof.
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CHAPTER 46 Power Series, , This is a continuation of Example 6. By formula (46.8),, , EXAMPLE 46.7:, , ~, , In (I + x) = L(_l)n-1 :, , ., , for Ixl < I, , n=1, , At the right-hand endpoint x = I of the interval of convergence, the power series becomes the convergent, alternating harmonic series, ~, , I, , ...1, , n, , L(-W- I-= I-t+t+t+···, By Abel's Theorem, thi~ series is equal to x-w, liin In(1 + x) _= •In2. So,, In 2= I-t+t.-t+ ..., , (46.10), , Start again with, , EXAMPLE 46.8:, , _1_ = ~ x. = I + x + x 2 + xl + ... + x" + for Ixl < I, I-x ~, , .-0, , Replace x by -_-c-, obtaining, , (46.11), , Since !-xll < I is equivalentto Ixl < I, (46.11) holds for Ixl < 1., Now, by Theorem 46.6(a), the antiderivative tan-I x of -1I2 can be obtained by term-by-term integration:, +x, ~, , 2.+1, , ,=0, , n+, , x I +K forlxl< I, tan-I x= L(-l)'-2, , Here K is the constant of integration. If we let x = 0 and note that tan-I 0 = 0, it follows that K = O. Hence,, ..., , 2n+1, , tan-I x= L(-l)"_x-=x-txl +tx' -txl + ..., n=O, 2n+l, , (46.12), , At the right-hand endpoint x = I of the interval of convergence, the series in (46.12) becomes, I', , ..., 1, ~(-l)n-=I-t+t-t+···, ~, 2n+l, , which converges by virtue of the Alternating Series Theorem. So, by Abel's Theorem,, , I-t+t-t+···= lim tan-I (x) = tan-II =1!.4, .r~I", , (46.13)
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CHAPTER 46, ., , L ~ converges for all x. Let /(x) = L x", ~, , We know already, by Example 3, that, By term-by-term differentiation (Theorem 46.7),, EXAMPLE 46.9:, , ...., , f'(x) =, , .aO, , <-, , 1, , L (:~-1)! = I, 0=1, , Not~, , Power Series, , "=0, , ., , ~, , ,,!, , 0-0, , for all x., , n!, , .~'; = f(x), , lhat/(O) = 1. Therefore, by formula (28.2),f(x) = tf. Thus,, <-, , ez=L~, 0=0 n., , for all x, , (46.14), , SOLVED PROBLEMS, , 1., , Find the interval of convergence of the power series, , and identify the function represented by this power series., Use the ratio test:, 0, , Ss.+.lj = Ix - 21"+1 / Ix - 21 = _n_1x - 21., ,,+ I, n, n+ I, j, , S, , 0,, , lim jSoHj = Ix - 21, , n-.+- S", , Hence, the interval of convergence is Ix - 21 < 1. (This is equivalent to -I < x - 2 < l. which, in tum, is, equivalent to 1 < x < 3.) At the right endpoint x = 3, the series is the divergent hannonic series, and, at the left, endpoint x = 1, the series is the negative of the convergent alternating hannonic series. So, the series converges, for 1 ~x< 3., Let h(x) =, , ., L (2)0, x~, . By Theorem 46.7, h'(x) =I(x- 2)0-1. This series is a geometric series with, <-, , <-, , I, , 0.1, , •. 1 I, , I, , first term I and ratio (x - 2); so its sum is 1- (x- 2) -3-', -x Thus, h'(x) = -3-', -x Hence,, h(x) = 3~x =-lnl3-xl+C. Now,, , J, , - (2 2)", , h(2) = I~=o, , and -lnl3-21+C=O. So,, , 0=1, , c=o, , ., , Moreover, since x < 3 in the interval of convergence, 3 - x > 0 and, therefore, 13 - xl = 3 - x. Thus,, h(x) =-In (3 - x)., , In Problems 2 and 3, find the interval of convergence of the given series and the behavior at the endpoints (if any)., , Use the ratio test:, Ixl., + 1)2 ,,2, 11+1, j ~j= ~/~=(_n_)2, S", , (/I, , Hence,, , lim jSn+lj=lxl., .... _ s", , Hence, the interval of convergence is Ixl < I. The radius of convergence is 1. At x = 1, we obtain the convergent, p-series with p = 2. At x = -I, the series converges by the alternating seri ..$ test. Thus, the series converges for, A, , -l~x~l.
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CHAPTER 46 Power Series, , 3,, , f(X+l)" =(x+l)+ (x+I)2 + (X+I)l + ...., ,=, In, fi.J3, Use the ratio test:, , Sn+ll= Ix+ IIMI / Ix+ lin = ~ n Ix+ 11. Hence,, .In+t In, n +1, I s,, , lim ISM'I=Ix+ 11., sn, , .~, , Hence, the interval of convergence is Ix + II < I. This is equivalent to -I < x + I < I, which, in tum. is ~ivalent, to -2 < x < O. The radius of convergence is I. At the right endpoint x =0, we get the divergent p-series, ioo, , (with p =t). At the left endpoint x =-2, we get the alternating series ~, , (-I)., , In, , L~, n=1, , ,which converges by the, , "n, , Alternating Series Theorem. Thus, the series converges for -2 ~ x < O., , 4., , Prove Theorem 46.1., Since La. (Xo - c)' converges,, , !!!!!.an (Xo - c)' =0 by Theorem 43.5. Hence, there is a positive number M, , such that la.llxo - cI' < M for all n, by Theorem 42.1. Assume Ix - cI < lxo - cl. Let, , Ix-d, , r= Ix _ d < 1., o, , Then,, , .'., , Ia.llx - d"= 1a,llxo - d" rn < Mr"., , Therefore, LIa,(x - c)"1 is convergent by comparison with the convergent geometric series LMr". Thus,, a. (x - c)' is absolutely convergent., _, , L, , 5., , Prove Theorem 46.2., Only a very intuitive argument is possible here. Assume that neither case (a) nor case (c) holds. Since case (a), does not hold. the power series does not converge for some x c. Since case (c) does not hold, the series does, converge for some x c. Theorem 46.1 implies that there is an interval (c - K, c + K) around c in which the series, converges. The interval of convergence is the maximal such interval. (Using Theorem 46.1. one takes the "least, upper bound" R, of all K such that the series converges in (c - K, c + K). Then. (c - RI , C + R,) is the desired, interval.), , *', , *', , 6., , Prove Theorem 46.4. ., Assume x is in A and E > O. Since (f.) converges uniformly to I on A. there is a positive integer m such that,, if n ~ m, then If. (y) - 1(y)I< El3 for all y in A. Since I., is continuous at x. there exists 8> 0 such that, for any x·, in A, if lx' - xl < 8, then 1/.,(x') - 1.,(x)1 < E/3. Hence if lx' - xl < 8,, , .---"~.;;:c~, ~'c'__, , .-', , I/(x') - I(x) I=l(f(x') - 1m'x'» +(f.. (x') - 1m (x» + (fm (x) - l(x»1, ~1/(x')- 1... (x')I+I/.. (x·)- 1,.(x)1 + 1/.. (x)-, , I(x) I, , ,, , This proves the continuity ofI at x., • •• - t, , 7., , ., . Jba f.(x)dr., If (f.) converges uniformlyto/on, [a, b) and each/. is continuous on [a, b]. then J6"/(x)dx= ~, Assume E > O. There is a positive integer m such that, if n ~ m. then II,• (x) - l(x)l< --a, E, b for all x in [a, b]., Therefore, f~ If. (x) - I(x) I dx < E. Then, , If I(x) dx- f f.(X)dxl = If(I"(x) - I(x» dxl ~ flf.(x) - l(x)1 d'( <, , E, , for n ~ In, , .~J} ., , .~~;~lt~, ~, , ., , :~::;.. ~, , ". -,.-' ~ -'!
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CHAPTER 46, , 8., , Power Series, , Prove that the functionJ defined by a power series is continuous within its interval of convergence (Corollary 46.5)., J(x) = lim S.(x) and the convergence is unifonn by Theorem 46.3. Each S.(x). being a polynomial. is, "-++-, , continuous. Hence,fis continuous by Theorem 46.4., , 9., , Find a power series about 0 that represents the function -xI2' In what interval is the representation valid?, I, ...., +x, By fonnula (46.11), -I- . 2 = 2,(-1)" x 2" for Ix! < l. Hence, ', +.\, .~, +-, , _x_ = ~ (-l)" X 2"+1, I+X2 k.-0, ., , ~.~., , ., , ~l~.-< ~, :.:--;.."-;',",:, , ~t-·", , -, , for Ix! < 1, , The series diverges at both endpoints x = I and x = -1., , ~.,.- ~.-, , In Problems 10 and II. use the ratio test to find the interval of convergence and indicate what happens at the, endpoints (if any)., 10. 2, 1~n x·., , =, s", I~I, , (/I + \) WeI+1 /, , \0"+1, , 1IW", , 10', , =(n + I)M, n, , 10 ., , Hence, , Iiml~I;;;;J&·, 10, , ,, , .-++- S., , We get convergence when IxtlIO < I, that is. when !xl < 10. That is the interval of convergence. The series, diverges at both endpoints ± 10., 11. ~.fl.. (x - n)' ., ~3", , l, , IS;:II, , (n+I)Ix-mo+ /nlx-mn =n+1 Ix-m, . 3"+1, 3·, n, 3,, , Hence,, , limIS.+II= Ix-m., s., 3 ., , .-++-, , So. the interval Qf convergence is Ix - nl < 3. The series diverges at both endpoints., 12. Find the interval of convergence of 2, !!!It, (~~)!x"., Apply the ratio test:, , S.+I \= «11 + 1)!)2 Ixl"+1 /(11!)2 Ixl", (n + 1)2, (2n + 2)!, (2n)! = (2n + 2)(2n + I) Ixl, \ S., , Hence,, , lim IS.+I\ = J&., s., 4, , "-++-, , So. the interval of convergence is Ix! < 4., , 13. Find a power series a~ut 0 that represents 1:r', Start with I ~ x = 2,X" for Ix! < 1. Replace x by, , r:, , .-0, , for, , Ixl < I, , for, , W< 1, , (since Jx31 < I is equivalent to Ixl < 1). Multiply by x:, +-, , _x_= ~ X~"+I, I_Xl, , ~, , ,..0, , In Problems 14-16. find simple fonnulas for the functionJ(x) represented by the given power series.
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CHAPTER 46, , Power Series, , ..., , 4+1, , ..., , xf(x) = L XI)', ( = L ~ -1- x = e' -1- X, .=1 n + . nz(), Hence, I(x) = e' -1- x ., x, , 15. t x3 + t x6 + t x 9 + .. ' ., ..., , 3., , ..., , Let I(x) = L ~n' Then, , /'(x) = Lx3.-1 = x 2 + x 5 + Xl + ...., , 11-1, , pi, , This is a geometric series with ratio xl. So, it converges for (xlI < 1, which is equivalent to txt < 1. Hence,, 2, , •, , 2, , /'(x) =-x I, '3 for Ixl < I. Therefore, I(x) =I x I3 dx =-t In 11- x31+ C. But/(O) =O. Hence, C =O. Also,, , -x, , -x, , I - xl >0 for Ixl < 1. Therefore,, , ,. '.,1.,, , ..... '., , . ,., , ;1 ",.,', , IxI < 1., , for, 16. x+2x 3 +3r+4x7 +...., , "(f~;), , The ratio test shows that the series converges for Ixl < I. Let, , ..., , g(x)=x+2x +3x +4x + ... =Lnx, 3, , 5, , 7, , ;~~.;, 2 1, .-, , n=1, , ...., , Then 2g(x) =L 2nx, , 2 1, .-, , Hence, taking antiderivatives., , "=1, , I, , ...., , 2, , 2 g(x)dx = K + Lx2• = K + I: x 2, "=1, , ..., , (since LX24 is a geometric series with ratio x 2 ), ,..1, , .",', , ~, , .;.-", , .. ··t!·.-., , Now differentiate:, for, , txt <1, , 17. (GC) Approximate rIll In (I + x) dx to two-decimal-place accuracy (that is, with an error < 5/I()l)., Jo, x, By formula (46.8), In(1 + x)= x - tx2 + tx 3- tx4 + ... for txt < 1, So, , In (1+x), x, , l-tx+.lx2 -tx3 + ... = ~ (-I)"x", 3, ~ n+1, ...0, , ·11, J!~, ,';~ ".~:, , ,->.j', '~',, , By Theorem 46.6(b),, I2, , I, , rlllln(1+x)dx=~(-I)' x·+ 1, 10, x, ~ n + 1 n + 1., n=O, , =~ (-1)' _1_, ~ (n + 1)2, , n=O, , 2n• 1, , which is a convergent alternating series., In order to get an approximation with an error less than 5/103, we must find" such that the first omitted tenn, (n 11)2 2LI is ~ 1~3 = 260· So, we must have 200 ~ (n + 1)2 2n+l. Trial and error shows that n ~ 3. Hence, we, can use the tenns corresponding to n = 0, 1. 2:, I, I, J, 65, 2-16+ 72 =144 - 0.45, , This answer can be confirmed by a graphing calculator (which yields 0.44841421 as an approximation).
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CHAPTER 46, , power Series, , +-, , 18. Find the function defined by, , 1)' x" •, ,.0, , This is a geometric series with ratio 2x and first term 1. Hence, it converges for 12.xl < 1, that is, for, ., , ,1, , IxI < t, and, , Its sum IS 1- 2x', +-, , 19. Find the interval of convergence of, Apply the ratio test:, , :~~~~, , I, , '1~~~, , i;~~, , By L'Hopital's rule, lim, 11-+....., , L in (~'+ I),, .;1, , ~I_, s., , Ixt"+', , /, , - In (n + 2), , Ixl", _ In (n + I), In (n + 1) - In (n + 2), , IxI, , ISn+', I=IxI. Hence, the interval of convergence is given by Ix! < 1. (For X= 1; we get, sn, , f In (,! + 1)' which we know is divergent. For x, , = -I, we get the convergent alternating series, , f In«(,:~ 1)'), ".1, , ",_I, , 20. Approximate 1 with an error less than 0.000 I., e, By formula (46.14),, +-, , e' -, , x", L, .. -n,., , for all x., , O, , I, , +- (_I\n, , Hence - = e- I = '" L.:.L, 'e, ~ n!, , By the Alternating Series Theorem, we seek the least n such that lin!!!> 0.0001 = lIlO,OOO, that is, IO,()()()!!> nL, Trial and error shows that n ~ 8. So, we must use the terms corresponding to n = 0, 1, ... ,7:, I, , I - I + 2" -, , 1, , 1, , 1, , 1, , I, , 103, , '6 + 24 - 120 + 720 - 5040 = 280 - 0.3679, , (A graphing calculator yields the answer 0.3678794412, correct to 10 decimal places.), 21. Approximate, , J~ e- xl dx to two-decimal-place accuracy, that is, with an error less than 5/I()3 =0.005., , By formula (46.14),, +-, , eX =, , n, , L~, n., , Hence,, , for all x., , for all x, , .=0, , By Theorem 46.6(b),, , .., , ,',, , J,r e_ •1dx l, , o, , I, , 2n, (_I\n X +1, ,i.; L.:.L__, , '+", , - •.0 n!, , 2n+1, , '", , +0-,i.;, , -.-0, , (-1)", , I, , --, , n! 2n+1, , We can apply the Alternating Series Theorem. The magnitude of the first term omitted (2n 1I)n! should be, !!> 0.005 = 11200. So, 200 ~ (2n + l)n!. Trial and ~rror shows that n ~ 4. Hence, we should use the first four terms,, that is, those corresponding to n = 0, 1,2,3:, I I, 1 26, 1-'3+10- 42 = 35 -0.743, , (A graphing calculator yields the approximation 0.74682413, correct to eight decimal places.), 22. Find a power series expansion for x ~ 3 about O., x~3, , =~, , (xli) + I· By formula.(46.7), 11 x, , +-, , =L(-I)" x" =1- x+ x, a=O, , 2, , - Xl, , + .... for Ix! < 1.
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CHAPTER 46, , Power Series, , Hence,, , t, , ..., ()' =L(-I)", ..., (x/3)1 + 1 - L(-l)', ~"•, ,,-0, , 1'", •, x+3 =L(-l)' 3:+1, , Thus,, , for, , n.O, , IIt, , <1, , for W<3, , ,=0, , The series diverges at x = ± 3., , 23. Find a power series expansion for, , 1x, , about 1., , ..., , 1 = 1 (I 1)' By formula (46.7), - 11 = L(-I)' x' for Ixl < 1. Hence,, x, + x+x ...0, , ±= 1+(!-l) - f(-I)'(x-I)", , ~:j.~:;~;, ", , i;!~.:·, , for lx-II < 1, , .-..0, , :1, ", (, , j, , \~ ~, , ,--, , In Proplems 24-31, find the interval of convergence of the given power series ., 24., , .I.1IX', , Ans. -1 <x< 1, , 25., , I. n(nx"+ 1), , Ans., , -I~x~I, , 26., , I.;;., , Ans., , -5~x<5, , 27., , I. n(n + I)(n + 2), , Ans., , -I~x~1, , 28., , L (In (n + 1»2, , Ans. -1, , ~x<, , I, , 29., , I. T+ij1", x', , Ans. -I, , ~x~, , 1, , 30., , I. (x-4)", , Ans., , 3~x~5, , 31., , I. (3x-2)", , Ans., , -I <x<t, , 211, , X, , x-+I, , n1, , 5", , I', , 32. Express e-2Jc as a power series about O., Ans., , f (-1)'2', x', n!, ...0, , 33. Represent e~fl as a power series about 2., , ...., , Ans., , I. 211(,J!) (x- 2)', ,DO, , :~-:
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••, , CHAPTER 46, , Power Series, , 34. Represent In x as a power series about 2., ... (-1\10+1, , Ans., , In2+ LT(x-2)", 0-1, , 35. (GC) Find In (0.97) with seven-decimal-place accuracy. (Hint: Use the power series for In (1- x) about 0.), , ,, , Ans., .,.\,, 1,'_, , "i', , ;:"., , .:}:~~~, , ::'~~~~), , -0,0304592, , 36. How many terms in the power series for In (I + x) about 0 must be used to find In 1.02 with an error, ~ 0.00000005?, Three, , Ans., , 37. (Ge) Use a power series to compute e-2 to four-decimal-place accuracy., Ans., , 0.1353, , 38. (GC) Evaluate, Ans., , dx, , "2, , to four-decimal-place accuracy., 1o -1-+x, 4, , 0.4940, , In Problems 39 and 40, find the interval of convergence of the given series ., ... ", 39. L~, ,,=1, -, , Ans., , (-00, +00), , Ans., , x= 0, , n, , -.-, , ~'.-~~~~~', , +-, , 40., , ::'::i~~':, , ,, , L, I'h~x", ".0, ., , _; ~·tj""J •., , --.',', , • + -., Lfas a power series about O., , 41. Represent cosh x =, Ans., , fL, 0=0, , (2n)!, , 42. Find a power series about 0 for the normal distribution function, Ans,, , ~ (-1)0, , X 2,,+1, , ~ n !(2") 2n + 1, , . expansIOn, 'b, x, · d a power senes, 43 • FIII, a out 0 for III II +, _ x', , 44. (GC) Approximate tan-I t to two-decimal-place accuracy., Ans., , 0.46, , J:, , e- t 'lldt.
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Power Series, , CHAPTER 46, , ...., , 45. Show that the conve~e of Abel's Theorem is not valid, that is, if f(x) =La.x· for Ixl < r, where r is the radius, ...., , of convergence of the power series, and Ii"! f(x) exists, then, I, f(x) = I+x'), , ,..0, , L a,," need not converge. (Hirlt: Look at, , - ' -, , ...., 46. Find a simple formula for the functionf(x) represented by Lrl 2x"., "=\, , Ans., , M, (I-x), ..., , 47. Find'a simple formula for the functionf(x) represented by L (n :I)n', n=2, , Ans. x + (1 - x) In (1 - x), , ..., , 48. (a) Show that (1-\)2 = Lnx'" for lxI < 1. (Hint: Use Example 5.), , 2, , .-\, , 2, , (b) Show that (1 ':X)3, , ..., , =Ln(n -1)x· for lxI < 1. (Hint: First 4.ivide the series by x. integrate. factor out x. use, •• 2, , part (a). and differentiate.) ,, (c) Show that, , x· for, , x(f~:;l = fn 2, ...., ...., n3\, , (d) Evaluate, , L;' and L~', ,..1, , Arls. (d) 2 and 6, , ,, , n-I, , Ixl < 1.
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"CHAPTER 4',7:, ;'·"31~·:t'·, , ., , ";, , ,.:?~~"'!~!':';, , -~-';il:t', , Taylor and Maclaurin Series., Taylor's Formula with Remainder, Taylor and Maclaurin Series, Letfbe a function that is infinitely differentiable at x =c, that is, the derivatives [<n)(c) exist for all positive, integers n,, The Taylor series for f about c is the power series, +-, , ~:a.(x-c)·, , =a o +a1(x-c)+a 2(x-c)2 +..., , n;O, , P")(c), , where an =- -n.I- for all n. Note thatf(O) is taken to mean the functionfitself, so 'that ao=f(c)., The Maclaurin series for fis the Taylor series for f about 0, that is, the power series, +-, , Ia.x", , =a o +al.x+~.x2 +..., , • ;0, , P")(O), , where a =- - I- for all n., •, n., EXAMPLE 47.1: The Maclaurin series fOI" sin x, Letf(x) = sin x. Then, f'(x) = cos x,, f"(x) = -sinx,, , f"'(x) = -cosx,, Since f )(x), , = sin x, further derivatives repeal this cycle of four functions. Since sin 0 =0 and cos 0 = 1./(2k)(0) =0 and, P2k+I)(O) = (-I tHence, a2k = 0 and a 21:+1 = dk~-I)r So, the Maclaurin series for sin x is, 4, , ~, , (-I)-, , "-(2k+I)'x, 1=0', , 21+1 _, , Xl, , x~, , x1, , -x---7' +..., , +-5', 3•, .•, , An application of the ratio test shows that this series converges for all x. We do not know that sin x is equal, to its Maclaurin series. This will be proved later., EXAMPLE 47.2:, , Let us find the Maclaurin series for f(x) = -I1_., , -x, , f '( x ) = (1_IX)2, , •, , 4·3·2, f 4( x )= (1X)5, , ', , f"( ), X, , = (1_2X)l •, , f~()= 5·4·3·2, X, (1- X)6, , f"'(), , x, , 3 . X)4, 2 •, =(1-
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CHAPTER 47 Taylor and Maclaurin Series, , n', , r)(O), , ., , ., , 1·, , We, can see the pattern:, r)(x) = (1 'rl ' Hence, a = - - I- = 1for all n, and the Maclaunn senes lor -1-IS, +_, ,, -x, II, n., -x, c, , ~x·. In this case, we already know that I ~ x is equal to its Maclaurin series for lxl < 1., 11=0, +00, , Theorem 47.1: If f(x) = ~>n(x - e)n for somex;t: e, then this series is the Taylor series forf, that is, h. =, , p.l( ), ----nffor, , .=0, +00, , all n. In particular, if f(x) = L hnx" for some x ;t: 0, then this series is the Maclaurin series for f., .=0, +-, , Assume f(x) =~ h. (x - e)n for some x ;t: c. Thenf( c) =boo By term-by-term differentiation (Theorem 46.7),, +00, , +_, , 11=0, , f' (x) = ~ nh. (x - e)n-I inthe interval of convergence of ~ hn(x - e)n . Hence fcc) =b,. Differentiating again, we, , get, , !"<;; =fn(n -l)bll(x - e)"-2. So,f'(c) =~bz and, th:refore, b2=f;~c) ., ..0, , •, , +00, , Differentiating again, we get flll(X) = Ln(n -I)(n - 2)bn(x-: e)"-3. So, f"'(e) = 3!b) and, therefore, b3=, Iterating this procedure, we obtain, ncO, ., b = p"l(e), n, n!, , M), T', •, , for all n;;:: 0, , Thus, the series is the Taylor series for f, EXAMPLE 47.3:, , We already know by formula (46.8) that, ...., , n, , In(l+x)=L(-l)"-lx, n=1, , ., , +...., , n, , for Ixl<1, , 11, , Hence, by Theorem 47.1, the series k.J, ~ (_1)·-1 £ must be the Maclaurin series for In (1 + x). It is not necessary to, 11, , go through the laborious process of c~lllputing the Maclaurin series for In (1 + x) directly from the definition of, Maclaurin series., , EXAMPLE 47.4:, , If f(x) = I~x' findj<47) (0)., , 2, , '~M, , +00, , We know that - 1 = Lx n for lxl < I. Hence, by Theorem 47.1, thecoefficient of x", namely I, is equal to--,-., x .=0, n., j<47l(0), , So, for n = 47, 1= (47)! and, therefore, jC47)(O) = (47)!., , Theorem 47.2 (Taylor's Formula with Remainder): Letfby a function such that its (n + l)st derivativepn+11 exists, in (a. {3). Assume also that c and x are in (a, {3). Then there is some x· between e and x such that, ,, !"(c), P"I(C), j!"+I)(x'), f(x) = f(c) + f'(e)(x - c) + 2'!(x - C)2 + ... +---,;y-(x - e)" + (n + I)! (x - C)II+I, n, , pkl(c), , = L-k,-(x-C)k +RII(x), PO, , ., , •, , p.+II(X·), , Here, RII (x) = (n + I)! (x - e)n+l is called the remainder term or the error., Theorem 47.2 can be derived from Theorem 13.6 (the Higher-Order Law of the Mean)., , (47.1)
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CHAPTER 47 . Taylor and Maclaurin Series, , .Applications of Taylor's Formula with Remainder, (I) Showing that certain functions are represented by their Taylor series by proving that, lim Rn(x) = 0, n-++-, , From Taylor's formula (47.1),, R,,(x) = f(x)-, , ,, , j<k)(e), L -k'-(xe)k, n, , k=O, , •, , If lim Rn (x) = 0 then, n-+_, , that is,J(x) is equal to its Taylor series., , Remark:, , dn, lim -, =0 for any d. To see this, recall that, , n n-+_, , n., , 43.5. nlim, .... _ ~, n. =0 for any x., , xn, L I", converges for all x. Hence, by Theorem, n., +-, , IPO, , ., , EXAMPLE 47 .S: sin x is equal to its Maclaurin series., Whenj(x) = sin x, then every derivative f·'(x) is either sin x. cos x, -sin x. or -cos x. and, therefore, If··(x)1 :::; I., So,, , _Ij(n+l'(x'), "+11< I(x-e)nfll, IR.(x)l- (lI+l)! (x-c) - (n+1)!, , By the Remark above. lim I(x - e)"tI I =O. Hcnce, lim R (x) =O. Thercfore. sin x is equal to its Maclaurin series:, " ...+_, , (n, , + I)!, , "......., , (47.2), , (II) Approximating values of functions or integrals, Use a bound on Rix) to get a bound on the error when we approximate the sum of an infinite series by, a partial sum., EXAMPLE 47.6, , Let us approximate e to four decimal places, that is. with an error less than 0.00005., ., , +-, , Preliminary result: e < 3. To see this, note that. since e', , =I x~,, ..0, , n., , _I_~I_II 1 1 . 1 1, e-e - ~, , - + +-2'., +-3', +-4' +-5' + ..., naO n., ..., , Now, for the functionj(x) =e, we wish to make the magnitude of the error R,,( I) < 0.00005. By Taylor's formula, with remainder. with x = 1,, , P"+I)(XO)I, I, , IR" (1)1 = (n + I)!', , 0, , where 0 < x <l
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CHAPTER 47 Taylor and Maclaurin Series, Since D.(ez) = ez, pw+l)(x) = ez for all x. So, p.+I)(X·) = ez'. Since ez is an increasing function, ez' < e' = e < 3. Thus,, IR.(l)I< (n 11)!' Since we wish to make the error <0.00005, it suffices to have, (n 1l)! ~ 0.00005,, , that is,, , (n 11)! $;, , 20.~ ,, , 60,000 $; (n + I)!., , Trial and error shows that this holds for n ~ 8. So, we can use the partial sum, , r ~n., 8, , -1.7183 ., , • 00, , Theorem 47,3 (The Binomial Series): Assume r t: O. Then, ~ r(r-IXr-2) .. ·(r-n+ I), , (1 + x)' = 1+ £oJ, , I, , n., , /'I_I, , x·, , ', , for Ixl< 1, , r(r - I), r(r - l)(r - 2) 3, =1+rx+--x2+, x + ..., , 2!, , 3!, , (47.3), , Apply the ratio test to the given series:, , SIO+III s. -, , r(r -1)(r - 2) .. ·(r - n)X-+ ' jr(r -I)(r - 2) .. ·(r -n + 1)x" 1, (II + I)!, n!, , So,, , -1I)XI, , · ISII+II t'Im--=x, I(r, I, IIm-=, s. 11-+.... n + 1, , 11-+...., , ', , Hence, the series converges for Ixl < 1. For a sketch of the proof that this series is equal to (I + x)" see, Problem 31., Note that, if r is a positive integer k. then the coefficients of X' for n > k are 0 and we get the binomial, formula, (l+X)k, , .r, , =, , 4, , ~, , EXAMPLE 47.7:, , 0, , k', , 11.'(k~, , n ),xn, ., , Let us expand "I + x as a power series about O. This is the binomial series for r = t., , + (1/2)(-112) 2 + (112)(-1/2)(-3/2) 3, F+ =1+111, I!X, 2!, x, 3!, x, , ,/l't'X, , + (1/2)(-1/2)(-3/2)(-5/2) X4 + ..., 4!, , +1, , =1, , x, , -t, , x2, , + I~r -1~8 X4 +..., , (47.4), , Let us find a power series expansion about 0 for -r,!...;1- x, Take the binomial series for r= -t, and then replace x by -x:, , EXAMPLE 47.8:, , +- 1. 3 . 5 ... (2n-l), --1+ .=1 2·4· 6 .. · (2n) x", , r, , (47.5)
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.,,----Theorem 47.4:, , CHAPTER 47 Taylor and Maclaurin Series, +-, , +-, , +-, , n=O, , n=O, , ,.-0, , If f(x) =Lanx" for Ix! < R. and g(x) =Lbnx nfor ,Ix! < Rz, then f(x)g(x) =Lc.x" for'lx! < minimum, rI, , (R I , Rz), where cn =Lakb,,_I', , The reader is referred to more advanced treatments of analysis for a proof. Theorem 47.4 guarantees that,, iff and g have power series expansions, then so does their product., , ,, , SOLVED PROBLEMS, 1., , Find a power series expansion about 0 for cos x., We know by Example 5 that, , Then, by Theorem 46.7, we can differentiate term by term:, , 2., , Find a power series about, , f, , x, , Use the identity sin = cos, , for sin x., , (x - f). Then, by Problem 1,, , ,.)2, , ,.)4, , +- (-1)1 (,.)21, 1(, 1(, sinx=t;(2k)!X- 2 =1- 2!x-"2 +4!x-"2,-'", , 3., , Iff(x) =tan-I x, evaluatef38)(O)., We know by formula (46.12) that, , :n + 1 =x-tx + tr -tx + .. ,, , +-, , 2n+1, , tan-I x= L(-I)·, , 7, , 3, , for Ixl< 1, , .=0, , Hence, by Theorem 47.1, the coefficient of x38 in this power series is equal to f~;8~W) . But the coefficient of,iIK is O., So, f(38)(0) = o., 4., , Find power series expansions about 0 for the following functions:, (c) 1I~, , (b) xe-lx, , (a) cos (xl), +-, , (-1)1, , +-, , (_l)k, , (a) cosx = L (2k)! XU by Problem 1. Therefore, COS(X2) =L (2k)! X41., 1.0, , (b) We know that e' =, , +-, , k, , +-, , L ~!" So, e- = L, lx, , 1-0, , (_1)12k, , k!, , Xl,, , Hence,, , .~O, , 1=0, , (c) This is the binomial series for r=-t., 3~1 -1 1, (-1/3)(-4/3) 2 (-1/3)(-4/3)(-7/3) 3, 1/\11+X, - -'3 x +, 2!, x +, 3!, x, , + (-1/3)(-4/3)(-7/3)(-10/3) X4 + .. ,, 4!, , =1+, , ·(311- 2», L- (-1)"(1·4·7··, 3"n!, 0=1, , ~
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Taylor and Maclaurin Series, , CHAPTER 47, , 5., , Find the first five tenns of the Maclaurin series for e'(sin x)., Method 1: Letf(x) = e'(sin x). Then, f'(x) = e'(sinx.+ cos x),, , f"(x) = 2e'(cosx),, , r)(x)=-4e'(sinx),, , and, , f"'(x) = 2e'(cosx':'" sinx), , f')(x)=-4e'(sinx+cosx), , 1'")(0), , Hence, since a. = n! ,we get tlo = 0, a l = 1, ~ = 1, ~ =, , t. a, , 4, , = O. and a, = -i. Thus, , ~~')',,;, , ;.;',,", f····, , Method 2: eX (sin x) = + x + ~~ + ~ + ...)( x -, , (1, , ~+~, , _ ...). If we multiply out according to the rule in, , Theorem 47.4. we get the same result as above. For example, c, = ..g-rr+-clo = -i., 6., , We know that sinx = x -, , ~~ + ~ - .... For what values of x will approximating sin x by x produce an error < 0.005?, , 1~(x)l= Ir~~x') x31~ 1~3 . (Here, If(3)(x)l~ 1since!3) is ~os x.) So, we require Ixll/6 < 0.005, which is, , :~.-~~'<, ·C.:I/', A,, , ••, , i :~;Il;!, , equivalent to Ixil < 0.03. So, we want Ixl < ~0.03 - 0.31., 3, , 7., , If we approximate sin x by x - ~! for Ixl < 0.5. what is a bound on the error?, Since sin x is equal to an alternating series for any x. the error will be less than the magnitude of the first tenn _, omitted, in this case 1xl'/5!. When Ixl < 0.5, the error will be less than lio (O.W - 0.00026., , 8., , Approximate, , 1 •, , 1x, o, , Sin X dx, , with an' error less than 0.005., , Hence,, I, , (I, , Therefore., , sinx, , Jo x, , ~ (-I)k f.1 2k, ~ (-I)k X2k+I ]1, dx, =~(2k+ I)! ox dx= ~(2k+ 1)! 2k+ 1 0, _~, , (_I)k, , 1, , -to (2k + 1)1 2k +1, This is an alternating series. We must find k so that (2k ~ I)! 2k1+ I ~ 0.005. or, equivalently. 200 ~ (2k + l) !(2k + 1)., It is true for k ~ 2. So, we need 1- iT = H- 0.944., 9., , Find a pow~r series about 0 for sin-I x., By fonnula (47.5),, 1, ~ 1·3·5 .. · (2n - 1), "/1-x =1+ £... 2.4.6 ... (2n) x", •• 1, , for Ixl<l, , Replace x by fl., 1, ~ I ·3·5 .. · (2n - 1), r,--:,=I+ £... 246 (2) (2. for Itl<1, .... 1-t 2, n = I · · .. •, n, , ~, , i, , :>*
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Tay/or and Maclaurin Series, , CHAPTER 47, , So, for Ixl < I,, , . _IX= IX ---dt=x+, I, L,+w \ ·3·5 .. · (2n -I), , sm, , o, , .J1=t2, , x~·+1, , 2·4·6 .. · (2n) .~n + 1, , 0=1, , JO. Find Maclaurin series for the following functions: (a) sin(xl ); (b) sin2 x., Recall that, if a fUllction has a power series expansion in an interval about 0, then that power s~ries is the, Maclaurin series of the function., , f, , (a) sin x =, , 1-0, , (_\)1 , X21tl, , for all x. Hence, sin(x l ) =, , (2k + I)., , 1=0, , ~~., , . 2, , (b) sm x, , 1- cos(2x) _ I (, , -"2, , 2, , +- (-I )1+1 2, . 2 . ~, sm x IS .£.J (2k)', , ., , (-1)1221, 1- L, (2k)! x, +-, , 1=0, , 21 - 1, , 1-', , I, , (-1)1 , X 61+3 and this series is the Maclaurin series for, , (2k + I)., , 2k) _ +- (_I)hI22k-1, - L, , (2k)!, , 1=1, , ., , x, , 2k, , ., , ., , by Problem I. So, the Maclaunn senes for, , •, ., , 2k, x., , 11. Find the first four nonzero terms of the Maclaurin series for f(x) = sec x., It would be very tedious to compute the successive derivatives. Instead, since sec x cos x =I, we can proceed, , .-, , differently. We assume sec x =La"x". Then, .-0, , X2- +, X4, x-6 + .. ·) =1, ...)(1 (ao +ax+ax2+n.x3+, 1, 2, -:l, 2 24- -720., , We then "multiply Ollt," compare coefficients on the two sides of the equation, and solve for the a•., , Thus., , 2, , An alternative method would be to carry out a "long division" of 1 by 1- x, 2, , 4, , 12. Find the Maclaurin series for the following functions:, 2, (a) sin (xS); (b) -I I, 5; (c) cos X., , +X, , +AilS., , (-1)1, , (a) L(2k+I)!X, , IOI 5, ' ;, , .-, , (b) L(-I)·x, , 1-0, , ho{), , 5, ,,;, , +~, , (c) 1+ L, , (-1)1 22k - 1 21, , (2k)!, , X, , I-I, , 13. Find the Taylor 'series for In x about 2., Ans., , 6, , + ~4 - 7~0 + ..., , In2+ ~(_I)"-I (x-2)", £.J, 112", , "-I, , 14. Find the first three nonzero terms of the Maclaurin series for (a) si~x; (b) e' cos x., e
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CHAPTER 47, , Taylor and Maclaurin Series, , 15. Compute the first three nonzero terms of the Maclaurin series for tan x., , Ans., , x + t X 3 +1r x' +..., , 16. Compute the first three nonzero terms of the Maclaurin series for sino, x., , Ans., , X+t x3+ix' +..., , x 1- [Hint:, , 17. Find the Taylor series for cos about, I, , Ans., , 1C )2t, , +.. (_l)k (, , 2~ (2k)!, , x- 3, , J3, , +00, , (-I)k, , -T ~(2k+l)!, , 18. (GC) Use power series to approximate, , (, , Use an identity for cos, , (t+(x- t )).], , 1C )2k+l, , x- 3, , 2, r", tan-I x dx, Jo, x, , Ans. 0.4872, 19. (GC) Use power series to approximate, , 112ln(1 +x), dx correctly to four decimal places., x, , 1o, , Ans. 0.4484, 20. (GC) Use power series to approximate J~ ~1 + x 2 dx correctly to four decimal places., , Ans., , 1.0948, , 21. (GC) What is a bound on the error if we approximate e' by I + x +t x 2 for Ixl ~ 0.05? (You may use eO·ns < 1.06.), , ,, , .", , Ans. 0.0000221, 22. (GC) What is a bound on the error if we approximate In (1 +x) by x for lxi ~ 0.05?, , Ans., , 0.00 125, , .., , 23. (GC) Use the Taylor series for sin x about, , t to approximate sin 62° correctly to five decimal places., , Ans. 0.88295, 24. (GC) In what interval can you choose the angle if the values of cos x are to be computed using three terms of its, Taylor series about and the error must not exceed 0.000 05?, , t, , Ans., , Ix -tl ~ 0.0669, , 25. (GC) Use power series to compute to four-decimal-place accuracy: (a) e-2; (b) sin 32°: (c) cos 36°., , Ans. (a) 0.1353; (b) 0.5299; (c) 0.8090, 26. (GC) For what range of x can:, (a) e' be replaced by I + x +t x 2 if the allowable error is 0.0005?, (b) sin x be replaced by x - tx J + Tilrx' if the allowable error is 0.00005?, , Ans. (a) lxi < 0.1: (b) Ix1 < 47°, , ;',:'
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CHAPTER 47, , Taylor and Maclaurin Series, , ., e esin.t, e eros \27. Use power series to evaluate: (a) lim--=,---; (b) lim - 2 ., x-+O, , Am., , (a), , i;, , (b), , X, , x-+o, , X, , I, , 28. (GC) Use power series to evaluate:, (a), , Jr (1- tsin2 xtl12 dx (to three-decimal-place accuracy)., , (b), , f~ cosJX dx (to five-decimal-place accuracy)., , (c), , (to four-decimal-place accuracy)., 1o -1-+x, ., , lr12, , o, , dx, , 112, , 4, , Ans. (a) 1.854; (b) 0.76355; (c) 0.4940, , 29. (GC) Use power series to approximate the length of the curve y =t Xl from x =Oto x =.5, with four-decimalplace accuracy., , Ans., , 0.5031, , 30. (GC) Use power series to approximate the area between the curve y = sin (xl) and the x axis from x = 0 to x = 1,, ii,, , ,, , ~fi~t~, , with four-decimal-place accuracy., AilS., , 0.3103, , 31. Prove that the binomial series expansion in Theorem 47.3 is correct., , ,, ~ ,(,-1)(,- 2)"'('-11 + I), . , .lor d., dy and, Hmt:, Let y =1+ £.J', ,, x". U se tenn-by-term d'N'", luerentlatlon to fiIJ1d the senes, "=1, n., x, , show that, ,, , i, , = 1~x· Then derive y = (l, , + x)'. [Use "separation ofvariables";, , J~ = J;::.J, , .;'", , ~'fH/~, , 32. Expand the polynomialf(x) =x4 - llxl + 43xl - 60x + 14 as a power series about 3, and find, AilS., , 1.185, , f:'2 f(x)dx~
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Partial Derivatives, Functions of Several Variables, If a real number z is assigned to each point (x, y) of a part of the xy plane, then z is said to be given as a, , function, z =/(x, y), of the independent variables x and y. The set of all points (x, y, z) satisfying Z =/(x, y), is a surface in three-dimensional space. In a similar manner, functions w =/(x, y, z, ...) of three or more, independent variables may be defined, but no geometric picture is available., There are a number of differences between the calculus of one and two variables. However, the calculus, of functions of three or more variables differs only slightly from that of functions of two variables. The study, here will be limited largely to functions of two variables., , Limits, By an open disk with center (a, b) we mean the set of points (x, y) within some fixed distance 0 from (a, b),, that is, such that ~(x-a)2 +(y- b)2 <0. By a deleted disk around (a, b) we mean an open disk"without its, center (a, b)., Let/be a function of two variables and assume that there are points in the domain of/arbitrarily close to, (a, b). To say that/(x, y) has the limit L as (x, y) approaches (a, b) means intuitively that/(x, y) can be made, arbitrarily close to L when (x, y) is sufficiently close to (a, b). Mote precisely,, lim /(x,y) = L, , (x.yH(a,b), , if, for any E > 0, there exists 0> 0 such that~ for any (x, y) in the domain of/ and in the deleted disk of radius, <5 around (a, b), I/(x, y) - LI < E. This is equivalent to saying that, for any E> 0, there exists 0> 0 such that, 0< ~(x- a)2 +(y - b)2 < 0 implies I/(x, y) - LI < E for any (x, y) in the domain off. Note that it is not assumed that/(a, b) is defined., Laws for limits analogous to those for functions of one variable (Theorems 7.1-7.6) also hold here and, with similar proofs., EXAMPLE 48.1:, , Using these standard laws for limits, we see that, lim, , ,, EXAMPLE 48.2:, , Let us show that, , (.(.y)-+(3.1), , (3XyZ +1 XY) = 3(3)(1) +t(3)(1) == t +t = 11, 7 +y 2, 7 +1, , In some cases, these standard laws do not suffice., lim, , (x.y)-+(O.O), , ~xy2, 2 = O. Our usual limit rules would yield QO' which is indeterminate. So, we need a, +y, , x, , more involved argument. Assume E > O. Now,, , ;xy22 -01=1 ;xy2 21=3Ix'I~I~3IX'~3~X2 + y2 <3c5=E, x+y, x+y, 1x+y, . if we choose 15 =E /3 and we assume that 0 < ~x2 + yZ < 15 .
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CHAPTER 48 Partial Derivatives, , ,, , Let us show that, , lim, , x 2 _yl, , + y 2 does not exist., Let (x, y) -+ (0, 0) along the x axis, where y = O. Then, -+ y:, = x~, y x, , EXAMPLE 48.3:, , (X.1)-+(0,0), , 2, , X, , $, , 2 2, , = I. So, the limit along the x axis is I. Now, , ', , 2, , let (x. y) -+ (0,0) along the y axis, where x = O. Then x 2 +- y2 = - Y2 = -\. So, the limit along the y axis is -I. Hence,, , x Y, Y, there can be no common limit as one approaches (0, 0), and the limit does not exist., , EXAMPLE 48.4:, , Let us show that, , 2)2 does not exist., , 2, , lim, (' • .1')-+(0.0), , (, , \ - Y2, x, , +Y, , Here, we cannot use the same argument as in Exampfe 3, since, , (2 2)2, , ;2 ~ ~2, , approaches I as (x, y) approaches, , (0.0) along both the x axis and the y axis. However, we can let (x, y) approach (0, 0) along the line y = x. Then, , 2)2 = O. So, (2, 2)2 ~ 0 along y = x. Since this is different from the limit 1 approached along, 22 - ) .22 ): = (2, X2 - X2, X 2 - Y2, (, Xx+y, x+x, x+y, ,, the x axis, there is no limit as (x. y) ~ (0. 0)., , Continuity, Let I be a function of two variables and assume that there are points in the domain of I arbitrarily close, to (a, b). Then I is continuous at (ll. b) if and only if I is defined at (a. b), lim I(x. y) exists, and, lim I(x. y) = lea, b), ,, (x.yH(u.b), (x.y)-+(a.b), , We say that/is continuous on a set A if/is continuous at each point of A., This is a generalization to two variables of the definition of continuity for functions of one variable. The, basic properties of continuous functions of one variable (Theorem 8.1) carry over easily to two variables., In addition. every polynomial in two variables, such as 7X S - 3xy3 - y4 + 2xy2 +5, is continuous at all points., Every continllolls function of one variable is also continuous as a function of two variables., The notions of limit and continuity have obvious generalizations to functions of three or more variables., , Partial Derivatives, Let Z =lex, y) be a function of two variables. If x varies while y is held fixed, z becomes a function of x. Then, its derivative with respect to x, lim /(x+flx.y)- /(x.),), flx, , &x-iO, , is called the (first) partial derivative of/with respect to x and is deno~e1h(x, y) or, , ~~ or ~., , Similarly, if y varies while x is held fixed, the (first) partial derivative of/with respect to y is, /, (x, y) =, y, , az = a/ = lim lex, y + fly) - lex, y), , dy, , dy, , &y-+O, , fly, , Lctf(x. y) = x2 sin y. Thenix(x, y) = 2x sin y andJ,,(x, y) = xl cos y., Note that, whenix is computed. y is temporarily treated like a constant, and. when!, is computed. x is temporarily, treated like a constant., The partial derivatives have simple geometric interpretations. Consider the surface z =f(x, y) in Fig. 48-1. Through, the point P(x. y. z). there is a curve APB that is the intersection with the surface of the plane through P parallel to the, xz plane (the plane determined by the x axis and the z axis). Similarly. CPD is the curve through P that is the intersection with the surface z =f(x,y) of the plane thro~h P that is parallel to the yz plane. As x varies while y is held fixed., Pmoves along the curve APB, and the value of, at (x, y) is the slope of the tangent line ~ the curve APB at P., Similarly. as y varies while x is held fixed, P mov~s along the curve CPD, and the value of ~ at (x. y) is the slopr ' If, the tangent line to the curve CPD at P., EXAMPLE 48.5:, , f
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CHAPTER 48 Partial Derivatives, z, B, , --'-D, -/P(X.;), , C, , A, , --------------------y, Rg.48-1, , Partial Derivatives of Higher Order, , aazx ,yielding, az=i;u(X,y)= axa(az), az =iyx(x,y) =dya(az), ax2, ax and ayax, ax, , We can take the partial derivatives with respect to x and y of, , ,, , 2, , Similarly, from, , ~, , 2, , we obtain, , ady22z= ~'Y(x, y) = dya(az), a2z, a(az), dy and axdy = ixy(x, y) = ax dy, Assume that IX), and /,x exist and are continuous in an open disk. Then/>y =/,x at every point of the disk., , Theorem 48.1:, , For a proof, see Problem 30., Let us verify Theorem 48.1 for/(x, y) =r(sin yx)., , EXAMPLE 48.6:, , f.(x,y) =x 2 (cos yx)(y) + 2x(sinyx) =x[xy(cosyx) + 2sinyx], , J, (x, y) =X2(cosyX)x + x3 (cosyx), ~'X(x,y), , =x[x(y(-sinyx)(x) +cosyx) + 2(cosyx)(x)], = x 2 [-xy sinyx + 3cosyx], , l>y(x,y) = x3 (-sinyx)(y) + 3x 2 cosyx = X2[_xy sinyx + 3cos yx], , Partial derivatives also can be defined for functions of three or more variables. An analogue of Theorem 48.1, holds for any two orderings of given subscripts., Note that partial derivativeS may fail to exist when the required limits do not exist., , SOLVED PROBLEMS, 1., , y, , lim xcos(x- )., 4, Since the standard limit laws apply, the limits are:, , Evaluate: (a), , lim (2xy4_7x2y2);(b), , (x.y)-+(3.2), , (X.y)-+(lf,O), , (a) 2(3)(2)4 -7(3)2(2)2 =96 - 252 = -156; (b), , 2., , Evaluate, , ., ltm, , 7rCOS~ = 7rf, , x, , 2, -2---2', , (x,y)-+(O,O) X, , +Y, , As (x, y) ~ (0, 0) along the y axis, x =0 and ~ =0 ~ o., x +y
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CHAPTER·4a, , Partial Derivatives, , As (x, y) -+ (0, 0) along the x axis, y = 0 and ~ == x~ = I -+ 1., . does not eXIst., .', x +y x, Hence, the Ii~t, , 3., , Evaluate, , lim, (x.r)-+(O.O), , Since Ixl= U, , ~.%)I, •, x 2 + y2, , :5~x2, + )'2, 1~1:5IYI-+ 0 as (x,y) -+ (0,0). So,, ,, x- + )/2, lim, (x. r)-+(O.O), , 4., , .%)I, , ~X2, , + y2, , O., , The function f(x,y) = sin(x+ y) is continuous everywhere except at (0,0) and on the line y = -x. where it is not, x+y, defined. Canf(O, 0) be defined so that the new function is continuous?, As (x, y) -+ (0, 0), x + y -+ 0 and. therefore. sin(x + y) -+ 1. since lim sinu = 1. So. if we letf(O, 0) = 1. the, x+y, •...0 u, new function will be continuous at (0,0). Thus, the original discontinuity wac; removable., In Problems 5-9, find the first partial derivatives., , 5., , Z=, , 2x2 - 3.%)1 + 4f·, Treating y as a constant and differentiating with respect to x yields, , f = 4x - 3y., , Treating x a'i a constant and differentiating with respect to y yields, , ~~, = -3x + 8y., , x 2 y2, y x, . 2, Treating y as a constant and differentiating with respect to x yields ik = 2x -.;., dX y x, ., d, d'l'l", ., ., ,, h, ', Id, k, x 2 +-., 2y, T reatmg x as a constant an I lerenltatlllg Wit respect to y Yle S J = - '2, ., oy, y, x, , 6., , Z=-+-., , 7., , Z=, , 8., , Z=, , sin (2x + 3y)., , ~~ =3cos(2x + 3y), , g~ = 2cos(2x + 3y), , and, , 2xv• + .,......'-:r7', y2, Zax -1+x4 y2 l+x2y4, , and, , tan-I (x 2y) + tan-I (.%)12)., , a, , g~ =e, , x1, , +X)'(2x + y), , and, , 10. The area of a triangle is given by K = tabsin C. When a = 20. b = 30, and C = 30°, find:, (a) The rate of change of K with respect to a, when band C are constant., (b) The rate of change of K with respect to C, when a and b are constant., (c) The rate of change of b with respect to a. when K and C are constant., (a), , ~~ ='tbsinC=t(30)(siri300)=.!f, , (b), , ~~ =tabcosC=t(20X30)(cos300)= l50fj, , (c) b=~ and ab =_~= 2(tabsinC) =_£=_1, asinC, da, a 2 sinC, a 2 sinC, a, 2
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CHAPTER 48 Partial Derivatives, , 19. Detennine whether the following functions are solutions of Laplace's equation, , =, , (b) Z=t(e'+l), , (a) Z e' cos Y, , (c), , £i-, , + ~ = 0:, , z=x2-1, , az = it cosy -a, a2z.2 = it cosy, -a, x, '!X., , (a), , ~ =-e'siny, ~~ =-e'cosy, Then ~~, 2 + dy2 =o., , ax, , az =1(ex+1), ll-l( . . .+l), ax, 2, 'ax 2 -2", , (b), , ~ = t<e'+l), ~ =1(tr+1 ), so,£t+~=e'.':f.O., , a, , ~ =2x, ax2z =2, 2, , ax, , (c), , ~=-2Y, ~=-2, So, n~, i)x 2 + dy2 =O., , In Problems 20-24, evaluate the given limit., , 20., , 21., , 22., , (.,yHH,2), , r1m, , 2-, , x-2y, x +y, , Ans., , -t, , Iim, , -2--2, , x- Y, X +y, , AlIS., , no limit, , 3xy, 2X2 + y2, , Ans., , no limit, , (x.,)->(o.o), , lim, , (.,yH(O,O), , 2, , 23., , ~, 2, ('.,)->(0.0) x + y., , AlIS., , no limit, , 24., , x2 + y2, 2, (•• ,H(O.O) Jx + y2 + 4 - 2, , Ans., , 4, , lim, , lim, , 25. Detennine whether each of the following functions can be defined at (0, 0) so as to be continuous:, (a), , Ans., , y2, , xr+yr, , (b), , x- y, x+y, , (a) no; (b) no; (c) yes; (d) no, , (c), , xl +I, , x2+y2, , (d), , ..!..:!L., X2+y2
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CHAPTER 48 Partial Derivatives, By a similar argument using ~ = (f(a + h. b + h) - f(a, b + h» - (f(a + h. b) - f(a. b» and the Mean-Value, , Theorem, we get, , t~~~ =fp(a.b), 31. Show that Theorem 48.1 no longer holds if the continuity assumption for f", and fp is dropped. Use the following, , function:, ., f( x, y) =, , XY(X2 -, , !o, , 2, , i), , x +y, , *, , 2, , ., , *, , if (x. y) (0, 0), if (x, y) = (0, 0), , [Find formulas for.!.(x, y) andJ,,(x, y) for (x, y) (0, 0); evaluate!.(O, 0) andJ,,(O, 0), and thenf",(O. 0) andf,,;~(j. 0).], , ,
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CHAPTER 49 Total Differential, , Differentiability, A function z =!(x, y) is said to be differentiable at (a, b) if functions, , EJ, , and E2 exist such that, (49.4), , and, , lim, , (dr.6yH(O,Q), , =, , EJ, , lim, , E2, , (dr.6yH(O.O), , =0, , Note that formula (49.4) can be written as, (49.5), , . We say that z =!(x, y) is differentiable on a set A if it is differentiable at each point of A., As in the case of one variable. differentiability implies continuity. (See Problem 23.), EXAMPLE 49.2:, !,(x, y) = 4y. Then, , Let us see that z = I(x, y) = x + 2f is differentiable at every point (a, b). Note that.f.(x, y) = I and, ~z=, , I(a + ~x, b+ ~y)- I(a, b)=a+ ~x+ 2(b+ ~y)2 -a - 2b 2, , = ~x + 4b~y+ 2(~y)2 = !.(a, b) ~x + I,(a, b) ~y+ (2~y) ~y, , Definition: By an open set in a plane, we mean a set A of points in the plane such that every point of A belongs to an, open disk that is included in A., , Examples of open sets are an open disk and the interior of a rectangle., Assume that/(x. y) is such that/. and h. are continuous in an open setA. Then 1 is differentiable in A;, , Theorem 49.1:, , For the proof. see Problem 43., Letz= I(x, y)=J9-x 2 - yl. Then!. = J, , EXAMPLE 49.3:, , -x, , 9-x2 -yl, , and!, = ~, , -y, . So. by Theorem 49.1 ,, 9-x 2-yl, , 1 is differentiable in the open disk of radius 3 and center at the origin (0, 0) (where the denominators of.f. and!, exist, and are continuous). In that disk, r +y2 < 9. Take the point (a. b) = (1, 2) and let us evaluate the change ~z as we move, from (1,2) to (1.03,2.01). So, ~ =0.03 and ~y =0.01. Let us approximate ~z by, , -1, , -2, , dz =1.(1, 2) ~x+ !,(1, 2) ~y =2"(0.03) + 2"(0.01) =-0.025, , The actual difference ~z is J9 - (1.03)2 - (2.01)2 .- .J9 - 1- 4 - 1.9746 - 2 = -0.0254., , ,, Chain Rules, Chain Rule (2 --+ 1), Letz =!(x, y), wherefis differentiable, and letx= g(t) andy = h(t), where g and h are differentiable functions, of one variable. Then z =f(g(t), h(t» is a differentiable function of one variable. and, dz, , az dx az dy, , ----+-dt - ax dt ay dt, , (49.6)
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CHAPTER 49, , 4., , Total Differential, , Approximate the area of a rectangle of dimensions 35.02 by 24.97 units., For dimensions x by y, the area isA =xyso that dA =, , ~~.d.t+ ~ dy= yd.t + xdy., , Withx= 35, d.t= 0.02,, , Y = 25, and dy = - 0.03, we have A = 35(25) = 875 and dA = 25(0.02) + 35(- 0.03) = - 0.55. The area is, approximately A + dA = 874.45 square units. The actual area is 874. 4494., S., , Approximate the change in the hypotenuse of a right triangle of legs 6 and 8 inches when the shofter leg is, lengthened by t inch and the longer leg is shortened by t inch., Let x, y. and z be the shorter leg, the longer leg. and the hypotenuse of the triangle. Then, , y, , *, , When x = 6, = 8. d.t =, approximately, inch., , 6., , t. and dy= -to then dz = 6~);!(~t) = do inch. Thus the hypotenuse is lengthened by, 6, , 8, , The power consumed in an electrical resistor is given by P = £2IR (in watts). If E = 200 volts and R = 8 ohms. by, how much does the power change if E is decreased by 5 volts and R is decreased by 0.2 ohm?, We have, , When E = 200. R = 8, dE = -5. and dR =- 0.2. then, , r, , dP = 2(200) (-5) - (2r (-0.2) =-250 + 125 =-125, 8, , The power is reduced by approximately 125 watts., , 7., , The dimensions of a reCtangular block of wood were f~und to be 10. 12, and 20 inches, with a possible error of, 0.05 in each of the measurements. Find (approximately) the greatest error in the surface area of the block and the, percentage error in the area caused by the errors in the individual measurements., The surface area is S = 2(X)' + yz + zx): then, , dS=, , ~~ d.t+ ~~ dy+ ~~ dz = 2(y+ z) dx+ 2(x+ z) dy+ 2(y+ x)dz, , The greatest error in S occurs when the errors in the lengths are of the same sign, say positive. Then, , dS =2(12+ 20)(0.05)+ 2(10+ 20XO.05)+ 2(12+ 10)(0.05) = 8.4in 2, The percentage error is (error/area)( I00) = (8.411120)(100), , 8., , =, , =0.75%., , For the fonnula R ElC, find the maximum error and the percentage error if C = 20 with a possible error of 0.1, and E = 120 with a possible error of 0.05., Here
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CHAPTER 49 Total Differential, , 28. Find duldt, given:, (a) u=x2yl;x=2t',y=3t2, , Ans., , 6xft(2yt + 3x), , (b) u = x cos y +y sin x; x = sin 21, y =cos 2t, , Am., , 2(cos y + y cos x) cos 2t - 2(-x sin y + sin x) sin 21, , (c) u =xy + yz + zx; x = e' , y = e-l, Z=e' + e-l, , Ans., , (x + 2y + z)e' - (2x + y + z)e-l, , ., , ,, , 29. At a certain instant, the radius of a right circular cylinder is 6 inches and is increasing at the rate 0.2 in/sec, while, the altitude is 8 inches and is decreasing at the rate 0.4 in/sec. Find the time rate of change (a) of the volume and, (b) of the surface at that instant., Am., , (a) 4.8nin3/sec; (b) 3.2n in2/sec, , 30. A particle moves in a plane so that at any time 1 its abscissa and ordinate are given by x =2 + 3t, y =t2 + 4 with x, and y in feet and t in minutes. How is the distance of the particle from the origin changing when t = I?, , Ans., , 5/../2 ftlmin, , 31. A point is moving along the curve of intersection of x2 + 3xy + 31 =Z2 and the plane x - 2y + 4 =O. When x =2, and is increasing at 3 units/sec, find (a) how y is changing, (b) how z is changing. and (c) the speed of the point., , Ans., , (a) increasing 3/2 units/sec; (b) increasing 75114 units/sec at (2.3, 7) and decreasing 75114 units/sec at, (2,3, -7); (c) 6.3 units/sec, , 32. Find dZ/aS and dZ/dt, given:, (a) Z=x2 - 2y2; X =3s +21, y =3s - 2t, (b) Z= x2 + 3xy + I; x = sin s + cos t, y =sin s - cos t, (c) Z= x2 + 21; x = e' - e', y =e' + e', (d) Z = sin (4x+ 5y); x=s +t, y= s- t, (e) Z =e'Y; x = S2 + 2st, y =2st + t2, , 33. (a) If II =j(x, y) and x = r cos, , Ans., Ans., Ans., Ans., Ans., , 6(x - 2y); 4(x + 2y), 5(x + y) cos s; (x - y) sin t, 2(x + 2y)e'; 2(2y - x)e', 9 cos (4x+ 5y); - cos (4x+ 5y), 2e')'[tx + (s + t)y); 2eX)'[(s + t)x + sy], , e, y =r sin e, show that, =(dU)2 +-1(dU)2, ( dU)2, ax +(dU)2, dy, or r2 de, , (b) If II =j(x, y) and x = r cosh s, y =r sinh s, show that, , 34. (a) Ifz=j(x+ ay)+ g(x -ay), show that, , ~= d2~' (Hint: Write z=j(I/)+ g(v), U=x+ ay, v=x -ay.), , (b) If z =x"f(ylx), show that XdZ/dX + ydZ/dy = nz., , ;)~fj~f::, Lr\,L~~>, ,_, , ,.,1", , ", , :;:~tt~i::'~~
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CHAPTER 49, , Total Differential, , 42. If u =x+ y+z, v=r + f+t, and w=.xl+ yl +t, show that, , ax, , yz, ()y, x+z, au = (x - y)(x - z) , av = 2(x - y)(y - z) ,, , S-, , I, , dw - 3(x - z)(y - z), , 43. Fill in the gaps in the following sketch of a proof of Theorem 49.1. Assume thatf(x, y) is such that/. and.t,'are, continuous in an open setA. We must prove thatfis differentiable in A., There exists x' between a and a + ax such that, f(a + dx, b) - f(a, b) = f.(x·, b) dx, , and there exists y' between b and b + dy such that, f(a + dx, b + dy) - f(a + dx, b) =f,(a + dx, y*) dy., , Then, dz=f(a+dx.b+dy)- f(a.b), = [J(a + dx. b) - f(a, b)] + [J(a + dx, b + dy) - f(a + dx, b)], , =f.(x·, b)dx + f,(a + dx, y') dy, Let EI = f.(x· ,y)- f.(a,b) and E2 = f/a + dx.y')- f,(a, b). Then, , ,::~~., , :~,;::--~~, , To show that EI, , ~, , 0 andE2 ~ O. use the continuity of/. and!,., , <;i)~, , ;;n'1:j, 44. Show that continuity off(x. y) does not imply differentiability. even when!. and!, both exist. Use the function, 2 xy ,,2, f( x.y) = x + J, , 1o, , [Hint: Show thatfis not continuous at (0.0) and., !,(O, 0) by a direct computation.], , if (x. y) ~ (0. 0), if (x. y) =(0.0), , therefor~., , not differentiable. Show the existence of/.(O. 0) and, , 45. Find a functionf(x. y) such thatfjO. 0) =!,(O. 0) = 0, andfis not continuous at (0. 0). This shows that existence of the, first partial derivatives does not imply continuity. [Hint: Defme f(x. y) = 2xy ,,2 for (x. y) ~ (0. 0) andf(O, 0) = 0.], x, , ,, , +J
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Space Vectors, Vectors in Space, As in the plane (see Chapter 39), a vector in space is a quantity that has both magnitude and direction. Three, vectors a, b, and c, not in the same plane and no two parallel, issuing from a common point are said to fonn, a right·handed system or triad if c has the direction in which the right-threaded screw would move when, rotated through the smaller angle in the direction from a to b, as in Fig. 50-1. Note that, as seen from a point, on c, the rotation through the smaller angle from a to b is counterclockwise., z, , y, , x, , Rg.50-1, , Rg.50-2, , We choose a right-handed rectangular coordinate system in space and let i, j, and k be unit vectors along, the positive x, y and z axes, respectively, as in Fig. 50-2. The coordinate axes divide space into eight parts,, called octants. The first octant, for example, consists of all points (x, y, z) for which x> 0, y > 0, z > O., As in Chapter 39, any vector a may be written as, , If P (x, y, z) is apoint in space (Fig. 50-2), the vector r from the origin 0 to P is called the position vector, of P and may be written as, r= OP = 08 + BP = OA +AB+ BP =xi + yj+zk, , (50.1)
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CHAPTER SO Space Vectors, , The algebra of vectors developed in Chapter 39 holds here with only such changes as the difference in dimensions requires. For example, if a=a.i+azj+a3k and b=b.i+bzj+b3k. then, ka = ka. i + ka2 j + ka3k for k any scalar, a = b.if and only if a. = bl'aZ = b2 ,and a3 = b3, a ± b= (a. ±b.)i + (a z ±b2 )j + (a 3 ±b3 )k, , a . b = lallbl cos 0, where Ois the smaller angle between a and b, i . i = j . j = k . k = 1 and i . j = j . k = k . i =0, lal = Ja:a = ~a; + a; + a~, a . b = 0 if and only if a = 0, or b = 0, or a and b are perpendicular, From (50.1), we have, (50.2), , as the distance of the point P (x, y. z) from the origin. Also, if ~(xl'Yl'z.) and P2(X2'Y2,Z2) are any two points, (see Fig. 50-3), then, , and, , (50.3), , is the familiar formula for the distance between two points. (See Problems 1-3.), z, z, , x, , I', , Rg.50-3, , Fig. 50-4, , Direction Cosines of a Vector, Let a=a.i+a2 j+a3k make, Fig. 50-4. From, , angle~, , a, /3, and y, respectively, with the positive x, y. and z axes, as in, , i . a = lillal cosa = lalcosa,, , ., j . a = lal cos/3, k . a = lal cosy
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CHAPTER 50, , Space Vectors, , we have, cos a, , i. 8, , ~, , R., j.a ~, cosp = 1iI = Iif', , =1iI =fai ', , k.8 ~, cos r = Tal = Iif, , These are the direction cosines of a. Since, , the vector u =i cosa+ j cos{3+ k cosr is a unit vector parallel to a., , Determinants, We shall assume familiarity with 2 x 2 and 3 x 3 determinants. In particular,, , That expansion of the 3 x 3 determinant is said to be "along the first row." It is equal to suitable expansions along the other rows and down the columns., , Vector Perpendicular to Two Vectors, Let, , be two nonparallel vectors with common initial point P. By an easy computation, it can be shown that, , all· la) all. lal, e= a, lb b + b b J + b, 1, , I, , 2), , ), , I, , i, , ~I, b k = al, , 12, , j, , k, , a 1 a), b, b, b), I • 2, , (50.4), , is perpendicular to (normal to) both a and b and, hence, to the plane of these vectors., In Problems 5 and 6, we show that, lei =lallbl sin 0 =area of a parallelogram with nonparallel sides a and b, , (50.5), , If a and b are parallel, then b =lea, and (50.4) shows that e =0; that is, C is the zero vector. The zero vector, by definition, has magnitude 0 but no specified direction., , Vector Product of Two Vectors, Take, , with initial point P and denote by n the unit vector normal to the plane of a and b, so directed that a, b, and, n (in that order) form a right-handed triad at P. as in Fig. 50-5. The vectorproduct or cross product of a and, b is defined as, a x b =lallbl sinO n, , (50.6)
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CHAPTER 50 Space Vectors, , where 8 is again the smaller angle between a and b. Thus, a x b is a vector perpendicular to both a and b., We show in Problem 6 that la x bl = lal Ibl sin8 is the area of the parallelogram having a and b as nonparallel sides., If a and b are parallel, then 8 = 0 or nand a x b = O. Thus,, ixi=jxj=kxk=O, , (50.7), , aXb, , Fig. 50-5, , In (50.6). if the order of a and b is reversed, then 0 must be replaced by -0; hence,, bxa=-(axb), , (50.8), , Since the coordinate axes were chosen as a right-handed system, it follows that, i xj, , jx i, , = k,, =-k,, , jxk =i,, , k xi, , =-i,, , k xj, , =j, , i x k =-j, , (50.9), , In Problem 8, we prove for any vectors a, b, and c, the distributive law, (a + b) x c = (a x c) + (b x c), , (50.10), , Multiplying (50.10) by -1 and using (50.8). we have the companion distributive law, cx(a + b)=(cxa) + (cxb), , (50.11 ), , (a+b)x(c+d) = aXc+axd+bxc+bxd, , (50.12), , Then, also,, , i, , and, , (See Problems 9 and 10.), , a xb = a l, bl, , j, , k, , a2, , a), , b2, , b), , (50.13)
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CHAPTER SO, , Space Vectors, , Triple Scalar Product, In Fig. 50-6, let 0 be the smaller angle between b and c and let ¢ be the smaller ~gle between a and b x c. Let h, denote the height and A the area of the base of the parallelepiped. Then the triple scalar product is by definition, a· (b x c) = a ·Ibllcl sinO n = lallbllcl sinO cos¢ = (Ial cos¢)(lbllci sinO) = hA, = volume of para\lelepiped, It may be shown (see Problem II) that, , al, , a2 a3, b2 b3 =(axb)· c, c2 c3, , a . (bxc)= bl, , cI, , (50.14), , Fig. 50-6, , I, , Also, , c·(axb)=, , ~, , bl, , whereas, , bl, b· (a x c) = a l, , C2, , ~, , b2, , b2, a2, , c3, , al, ~ = bl, b3, cl, , b3, a3, , =-, , al, bl, , a2, b2, c2, , ~, , b3, , a2 a3, b2, , ·, , =a -(b xc), , c), , b3, , =-a . (b x c), , Similarly, we have, , a·(b xc) = c·(a xb) = b·(cx a), , (50.15), , a ·(b xc) = -b ·(a xc) = -c ·(b x a) =- a ·(c x b), , (50.16), , and, , From the definition of a· (b x c) as a volume, it follows that if a, b, and c are coplanar, then a . (b x c) =0,, and conversely., The parentheses in a ·(b x c) and (a x b)·c are not necessary. For example, a . b x c can be interpreted, only as a· (b x c) or (a· b) x c. But a·b is a scalar, so (a· b) x c is without meaning. (See Prob''!m 12.)
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CHAPTER 50 Space Vectors, , Triple Vector Product, In Problem 13, we show that, , Similarly,, , aX (b xc)= (a· c)b -(a· b)c, , (50.17), , (a X b) x c = (a . c)b - (b . c)a, , (50.18), , Thus, except when b is perpendicular to both a and c, a x (b x c) ~ (a x b) x c and the use of parentheses, is necessary., , The Straight Line, A line in space through a given point PrJ..x() y() Zo) may be defined as the locus of all points P(x, y. z) such, that PoP is parallel to a given direction a=ati+a2j+~k. Let roand r be the position vectors of Po and P, (Fig. 50-7). Then, r - r0 =ka where k is a scalar variable, , (50.19), , is the vector equation of line PPo. Writing (50.19) as, , then separating components to obtain, /, , Fig. 50-7, , and eliminating"', we have, x-xo = y-Yo, al, a2, , = z-zo, a3, , (50.20), , as the equations of the line in rectangular coordinates. Here, [at, a2• a3l is a set of direction nllmbers for the, line and [ ~I' ~21 ' ~I] is a set of directioll cpsines of the line., , i, , 1
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CHAPTER 50, , Space Vectors, , If anyone of the numbers a" a2' o~ a3 is zero, the corresponding numerator in (50.20) must be zero. For, exarriple, if at =0 but a2 , a3 :t 0, the equations of the line are, , x-xo=O, , and, , The Plane, A plane in space through a given point Po(xo, Yo. Zo) can be defined as the locus of all lines through Po and, perpendicular (nonnal) to a given line (direction) a =Ai + Bj + Ck (Fig. 50-8). Let P(x, y, z) be any other, point in the plane. Then r - r 0 = PuP is perpendicular to a, and the equation of the plane is, (r - ro)' a, , =0, , (50.21), , o, Fig.5~, , In rectangular coordinates, this becomes, [(x - xo)i +(y - yo)j + (z- Zo)k)· (Ai+ Bj+ Ck) =0, A(x - xo)+,B(y- Yo)+ C(z- Zo) =0, , or, or, , Ax, , + By +Cz + D=O, , (50.22), , where D = -(Axo + Byo + CZo)·, Conversely, let Po(xo• Yo. zo) be a point on the surface Ax + By + C'l, + D = O. Then also, Axo + Byo + CZo + D = O. Subtracting the second of these equations from the first yields A(x - ·"Co) +, B(y - Yo) + C(z - Zo) =(Ai + Bj + Ck)- [(x - xo)i + (y - yo)j + (z - Zo)k) =0 and the constant vector Ai +Bj +, Ck is nonnal to the surface at each of its points. Thus, the surface is a plane., , SOLVED PROBLEMS, , 1., , Find the distance of the point P,(I, 2,3) from (a) the origin, (b) the x axis, (c) the z axis. (d) the xy plane, and (e), the point P2(3. -1,5),, In Fig. 50-9,, , .IF + 22 +32 = M., , (a) r= OPt = i + 2j + 3k; hence, Irl =•, , =AB + BP, = 2j + 3k; hence, IAP,I =.J4+9 =Jf3., DP, = DE + Ell, =2j + i; hence, IDll,1 =$., , (b) AP,, (c), , (d) BP, = 3k, so IBP,I =3., (e) P,P2 =(3 - J)i + (-1 -2)j + (5 - 3)k =2i - 3j + 2k; hence,IP,P21=.J4 +9+4, , =.Ji7.
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CHAPTER 50 Space Vectors, z, , ........., , ,."-' ., , ---y, , x, , Fig. 50-9, , 2., , Find the angle obetween the vectors joining 0 to PI(l, 2, 3) and P2(2, -3, -I)., Let r, = OPt = i +2j+ 3k and r 2 =OP2 =2i - 3j -k. Then, cosO=!L5..=1(2)+2(-*3(-I)=_1, Ir,llrzl, 14, 2, , Ji4, , and, , 0=120°., , 'J';/',:./':;"'.,......., 3., , Find the angle a = LBAC of the tri.angle ABC (Fig. 50-10) whose vertices are A( 1, 0, I), B(2, -1, I). C(-2, I, 0)., C, , B, , Fig. 50-10, , Let a = AC = -31 + j - k and b = An = 1- j. Then, a·b -3-1, cos a = lallbl = Jfi -~.85280 and a - 148"31'., , 4., , Find the direction cosines of a = 31 + 12j + 4k., ··, The d·, Ireclton cosmes are cos, , 5., , I· a 3, f.I, i. a, 12, k· a 4, a = Iilf, = 13' cos f' = laf = 13' cos r = 181 = 13', , If a = ali + aJ + a)k and b = b,i + bJ + b)k are two vectors issuing from a point P and if, , ,, show that lei = lalbl sin 0, where 0 is the smaller angle between a and b., , B·b, , We have cos 0 = lallbl and, lei, lallbl, Hence, lei = lallbl sin 0 as required.
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. CHAPTER 50, , Space Vectors, , 6., , Find the area of the parallelogram whose nonparallel sides are a and h., From Fig. 50-11, h = Ibl sin 0 and the area is hlal = lallblsin O., , 7., , Let 8 1 and 3 2, respectively, be the components of a parallel and perpendicular to b, as in Fig. 50-12. Show that, a 2 x b = a x b and a I x b = O., If 0 is the angle between a and b, then lall= lal cos eand la 21= lal sinO. Since a, a 2• and b are coplanar,, , ., , J, , a 2 x b = la 211bl sin, , ~, , n = lal sin 0 Ihl n = lallbl sin 0 n '" a x b, , Since 8 1 and b are parallel, a l x b = O., , 8., , Prove: (a + b) x c = (a x c) + (b x c)., In Fig. 50-13, the initial point P of the vectors a, b, and c is in the plane of the paper, while their endpoints are, above this plane, The vectors 8 1 and b l are, respectively, the components of a and b perpendicular to c. Then a.,, b .. 8. + b •• 8. X c, b. x c, and (a. + b.) x c all lie in the plane of the paper., , n, , Rg.50-12, , Rg.50-11, B, , J ••_, , Q, , Fig. 50-13, In triangles PRS and PMQ,
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CHAPTER 50 Space Vectors, , when k = 1. Similarly B is on the line because, -ti+tJ+k =k(2i- J-4k), when k = - t. The point C is not on the line because, 1- 2j - 2k = k(2i - j - 4k), for no value of k., , 16. Write the equation of the plane:, (a) Passing through Po(l, 2, 3) and parallel to 3x - 2y + 4z - S = O..•, (b) Passing through Po(l. 2. 3) and P.(3. -2. I). and perpendicular to the plane 3x - 2y + 4z - S = O., (c) Through PO<I. 2. 3). P.(3, -2, I) and P2(S, 0, -4)., , Let P(x, y, z) be a general point in the required plane., (a) Here a = 31 - 2j + 4k is normal to the given plane and to the required plane. The vector equation of the latter, I is (r - ra)' a = 0 and the rectangular equation is, 3(x-I)-2(y-2)+4(z-3)=0, 3x - 2y + 4z - II =0, (b) Here r. - ra = 2i - 4j - 2k and a = 3i - 2j + 4k are parallel to the required plane; thus, (r. - ro) x a is, n9rmal to this plane. Its vector equation is (r - ro) . [(r l - ro) x a] =O. The rectangular equation is, , or, , (r- ro)' 2, , ~ .~4~1 = (x- I)i +(y- 2)j + (z- 3)k]· [-201 -14j + 8k], , 3 -2, , ~, , = -20(x -I) -14(y -2) + 8(z - 3) =0, , or 20x+ 14y-8z-24=0., (c) Here r. - ro = 2i - 4j - 2k and r2 - ro = 4i = 2j - 7k are parallel to the required plane, so that (rl - ro) x, (r2 - ra> is normal to it. The vector equation is (r - ro) . [(r. - ro) x (r2 - roll =0 and the rectangular, equation is, j, , k, , (r-ro)' 2 -4 -2 =[(x-l)i+(y-2)j+(z-3)k]·[-i41+6j+12k], , 4 -2 -7, =24(x-I)+6(y..:.2)+ 12(z-3)=O, or 4x + y + 2z - 12 =O., , ,, , n, , 17. Find the shortest distance d between the point Po(1, 2, 3) and the plane given by the equation, 3x - 2y + 5z - 10 = O., :, A normal to the plane is a = 31 - 2j + Sk. Take PI (2. 3, 2) as a convenient point in n. Then. apart from sign. d, is the scalar projection of PoP I on a. Hence., , d, , =/(r. - ro)' a/=I(I + j- k) ·(31- 2J+5k)l= 1-/38, lal, , /38, , 19
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CHAPTER 50, , Space Vectors, , SUPPLEMENTARY PROBLEMS, 18. Find the length of (a) the vector a = 2i, P,(3, 4, 5) to P2(1, -2,3)., , Ans. (a), , M; (b), , + 3j + k; (b) the vector b = 3i -, , 5j, , + 9k; and (c) the vector c, joining, , .J[i5; (c) 2JIT, , 19. For the vectors of Problem 18:, (a) Show that a and b are perpendicular., (b) Find the smaller angle between a and c, and that between band c., (c) Find the angles that b makes with the coordinate axes., , :, , .., 20. Prove: i . i, , =j . j =k . k = I, , and, , i .j, , =j, , .k, , =k . i = O., , 21. Write a unit vector in the direction of a and a unit vector in the direction of b of Problem 18 ., , •I, , Arts., , (a) ../14 i + 3../14 . + Jf4 k' (b) _3_ i _ _, 5_. + _9_ k, 7, 14 J 14 ', .J[i5 Jli5 J Jli5, , 22. Find the interior angles ~ and y of the triangle of Problem 3., , -;~.', ,~, , y'--., , 23. For the unit cube in Fig. 50-14, tind (a) the angle between its diagonal and an edge, and (b) the angle between its, diagonal and a diagonal of a face., , z, , y, , '."'J, , ~~~l~J~~ ~, •, , Fig. 50-14, , >', , 24. Show that the scalar projection of b onto a is given by, , al~~', , 25. Show that the vector c of (50.4) is perpendicular to both a and b.
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-D, , CHAPTER 50 Space Vectors, , 26. Given a = 1+ j. b = 1- 2k. and e = 21 + 3j + 4k. confinn the following equations:, (a), (c), (e), (g), , =, =, , a x b -21 + 2j - k, e x a -41 + 4j - k, , a·(axb)=O, ax(bxe)=3i-3j-14k, ., , (b) b x e = 61 - 8j + 3k, (d) (a + b) x (a - b) =41-4j +2k, (f) a· (b x c) =-2, (h) eX(axb)=-lli-6j+IOk, , f, , 27. Find the area of the triangle whose vertices are A(1. 2. 3). B(2. -I. I). and C(-2, I, -1). (Hint: lAB x ACI =twice, the area.), , Ans. 5./3, 28. Find the volume of the parallelepiped whose edges are OA. OB, and OC. for A(1, 2. 3), B(I. I, 2). and C(2. 1, I)., , Ans. 2, , 29. If u = a x b, v = b xc, W = e x a, show that:, (a) u· e = v . a = W • b, (b) a· u = b . u = 0, b . v = e . v = 0, e . W = a . w =0, (c) u· (v x w) =[a . (b x e)F, , 30. Show that (a + b)· [(b +e}x (e+ a}) =2a· (b x c)., , 31. Find the smaller angle of intersection of the planes 5x - 14y + 2z - 8 = 0 and lOx - II Y + 2z + 15 = O. (Hint: Find, the angle between their normals.), , 32. Write the vector equation of the line of intersection of the planes x +y - z - 5 = 0 and 4x - y - z + 2 =O., , Ans. (x - l)i + (y - 5)j + (z - l)k =k(-21 - 3j - 5k), where Po(1. 5, 1) is a point on the line., 33. Find the shortest distance between the line through A(2, -1. -I} and B(6, -8. 0) and the line through C(2, 1, 2), and D(O, 2, -1)., , Ans. .j6/6, , 34. Define a line through Po(xOo Yo. Zo) as the locus of all points P(x, y. z) such that PoP and OPo are perpendicular., Show that i~ vector equation is (r - ro) . ro =O., , 35. Find the rectangular equations of the line through P0(2, -3, 5) and, (a) Perpendicular to 7x - 4y + 2z - 8 =O., (b) Parallel to the line x - y + 2z + 4 = 0, 2x + 3y + 6z - 12 = O., (c) Through P1(3, 6. -2)., , Ans., , x-2_y+3_z-5.(b}x-2=y+3=z-5, (}x-2=y+3_d, (a) -7- -4 - 2', 12, 2, -5' c I, 9 - -7, , ~, , ., , ~, , .....1,,,.~: '.
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CHAPTER SO, , Space Vectors, , 36. Find the equatioIi of the plane:, (a) Through Po(1, 2, 3) and parallel to a =2i + j - k and b =3i + 6j - 2k., (b) Through Po(2, -3, 2) and the line 6x + 4y + 3z + 5 =0, 2x +y + z - 2 =O., (c) Through P o(2, -I, -I) and PI (1,2,3) and perpendicular to 2x + 3y"": 5z - 6:= O., AIlS. (a) 4x + y + 9z - 33 = 0; (b) 16x + 7}' + 8z - 27 = 0; (c)'9x - y + 3z - 16 = 0, , 37. If ro = i + j + k, r l = 2i + 3j + 4k, and r 2=3i + 5j + 7k are three position vectors, show th~t ro x r l + r l x r 2+ r 2, x r0 =O. What can be said of the tenninal points of these vectors?, Ans., , They are collinear., , 38. If Po. PI' and P2 are three noncollinear points and r o, rio and r 2 are their position vectors, what is the position of, ra x r l + r l X r2 + r2 X ro with respect to the plane PoPIP2?, Ans., , nonnal, , 39. Prove: (a) a x (b x c) + b x (c x a) + c x (a, , X, , b) =0; (b) (a x b) . (c x d) = (a· c)(b . d) - (a· d)(b . c)., , 40. Prove: (a) The perpendiculars erected at the midpoints of the sides of a triangle meet in a point; (b) the, perpendiculars dropped from the vertices to the opposite sides (produced if necessary) of a triangle meet in, a point., , 41. Let A( I. 2, 3). B(2, -1. 5). and C(4. 1.3) be three vertices of the parallelogram ABeD. Find (a) the coordinates of D;, (b) the area of ABCD; and (c) the area of the orthogonal projection of ABCD on each of the coordinate planes., AilS., , (a) D(3, 4, 1); (b) 2fi6 ; (c) 8, 6, 2, , 42. Prove that the area of a parallogram in space is the square rool of the sum of the squares of the areas of, projections of the parallelogram on the coordinate planes.
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Surfaces and Curves in Space, Planes, We already know (fonnula (50.22» that the equation of a plane has the fonn Ax + By + Cz + D =0, where, Ai + Bj + Ck is a nonzero vector perpendicular to the plane. The plane passes through the origin (0, 0, 0), when and only when D =O., , Spheres, From the distance fonnula (50.3), we see that an equation of the sphere with radius r and center (a. b. c) is, , So a sphere with center at the origin (0, O. 0) and radius r has the equation, , Cylindrical Surfaces, An equation F(x. y) = 0 ordinarily defines a curve '€ in the xy plane. Now, if a point (x. y) satisfies this, equation, then, for any z. the point (x, y, z) in space also satisfies the equation. So, the equation F(x, y) =0, detennines the cylindrical surface obtained by moving the curve C(6 parallel to the z axis. For example, the, equation x2 +f =4 detennines a circle in the xy plane with radius 2 and center at the origin. If we move this, circle parallel to the z axis, we obtain a right circular cylinder. Thus, what we ordinarily call a cylinder is a, special case of a cylindrical surface., Similarly. an equation F(y, z) =0 detennines the cylindrical surface obtained by moving the curve in the yz, plane defined by F(y, z) =0 parallel to the x axis. An equation F(x, z) =0 detennines the cylindrical surface, obtained by moving the curve in the xz plane defined by F(x, z) =0 parallel to the y axis., More precisely, the cylindrical surfaces defined above are called right cylindrical surfaces. Other cylindrical surfaces can be obtained by moving the given curve parallel to a line that is not perpendicular to the, plane of the curve., ,., EXAMPLE 51.1: The equation z =r determines a cylindrical surface generated by moving the parabola z =rlying, in the xz plane parallel to the y axis., , Now we shall look at examples of surfaces detennined by equations of the second degree in x, y, and z., Such surfaces are called quadric surfaces. Imagining what they look like is often helped by describing their, intersections with planes parallel to the coordinate planes. Such intersections are called traces.
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CHAPTER 51, , Surfaces and Curves in Space, , Ellipsoid, , The nontrivial traces are ellipses. See Fig. 51-1. In general, the equation of an ellipsoid has the fonn, x2, a, , y2, , Z2, , 2" + -I2 +""""2, , '), , c, , =I, , (a > 0, b > 0,, , C>, , 0), , When a =b =c, we obtain a sphere., z, , y, , x, , Rg.51-1, , Elliptic Paraboloid, , The surface lies on or above the xy plane. The traces parallel to the xy plane (for a fixed positive z) are, circles. The traces parallel to the xz or yz plane are parabolas. See Fig. 51-2. In general, the equation of an, elliptic paraboloid has the form, Z, , x2, , y2, , C=(if + 17, , (a > 0, b > 0, c> 0), , z, , Rg.51-2, , and the traces parallel to the xy plane are ellipses. When a = h, we obtain a circular paraboloid, as in the, given ex: mple.
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CHAPTER 51 Surfaces and Curves in Space, , Elliptic Cone, , See Fig. 51-3. This is a pair of ordinary cones, meeting at the origin. The traces parallel to the xy plane, are circles. The traces parallel to thexz or yz plane are hyperbolas. In general, the equation of an elliptic, cone has the form, , y2, .,.=.,.+, b, c a, Z2, , x2, , (a>O, b>O, c>O), , 2, , and the traces parallel to the xy plane are ellipses. When a =b, we obtain a right circular cone, as in the given, example., z, , --------~---------y, , Fig. 51-3, , Hyperbolic Paraboloid, , z= 2y2-il, See Fig. 51-4. The surface resembles a saddle. The traces parallel to the xy plane are hyperbolas. The other, traces are parabolas. In general, the equation of a hyperbolic paraboloid has the form, , In the given example, c = I, a =1, and b =1I.J2., , Hyperboloid of One Sheet, , See Fig. 51 ~5. The traces parallel to the xy plane are circles and the other traces are hyperbolas. In general,, a hyperboloid of one sheet has an equation of the form
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CHAPTER 51, , Surfaces and Curves in Space, , and the traces parallel to the xy plane are ellipses., z, , y, , Rg.51-4, l, , H-UW-H1t--- y, x, , Rg.51-5, , Hyperboloid of Two Sheets, , See Fig. 51-6. The traces parallel to the xy plane are circles, and the other traces are hyperbolas. In general,, a hyperboloid of two sheets has an equation of the form, (a, , > 0, b > 0, c > 0), , and the traces parallel to the xy plane are ellipses., In general equations given above for various quadric surfaces, permutation of2the variables x, y, Z is, , y2, c, , Z2, , x, , understood to produce quadric surfaces of the same type. For example, """'2 - """'2 - -b2 =1 also detennines a, , ., , hyperboloid of two sheets., , a
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CHAPTER 51 Surfaces and Curves in Space, , Tangent Line and Normal Plane to a Space Curve, A space curve may be defined parametrically by the equations, , =, , x /(t),, , y = g(t),, , (51.1), , Z = h(t), , x, , y, , Rg.51-6, , At the point Po(Xo, Yo, ZQ) of the curve (determined by t =to), the equations of the tangent line are, x-xo = Y-Yo, dx I dt dy I dt, , = z-Zo, dz I dt, , (51.2), , and the equations of the normal plane (the plane through Po perpendicular to the tangent line there) are, dx, dy, dz, -(x-x, )+-(y-y )+-(Z-7_)=O, dt, 0, dt, 0, dt, '1l, , (51.3), , See Fig. 51-7. In both (51.2) and (51.3), it is understood that the derivative has been evaluated at the point, Po. (See Problems 1 and 2.), Tangent line, Normal Plane) P(;, ~X,y,Z), Po(Xo + Ax, Yo + lly. Zo + llz), , ...., , Rg.51-7, , Tangent Plane and Normal Line to a Surface, The equation of the tangent plane to the surface F(x, y, z) =0 at one of its points Po(xo, Yo, ZQ) is, , of, , of, , of, , -(x-x, )+-(y-y )+-(Z-7_)=O, oX, 0, i)y, 0, oz '1l, , (51.4)
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CHAPTER 51, , Surfaces and Curves in Space, , and the equations of the normal line at Po are, (51.5), , with the understanding that the partial derivatives have been evaluated at the point Po. See Fig. 51-8., (See Problems 3-9.), ", Nonnalline, , Rg.51-8, , A space curve may also be defined by a pair of equations, F(x, y, z) = 0, G(x, y, z) =, , °, , (51.6), , At the point Po(xo. Yo, 20) of the curve, the equations of the tangent line are, , x-xo, , aF aF, , Y-Yo, , =, , z-Zo, , = aF, , aF aF, aF, ax ay, aG aG aG aG de aG, ay Tz Tz ax ax' ay, dy, , Tz, , Tz ox, , (51.7), , and the equation of the normal plane is, , (51.8), , In (51.7) and (51.8), it is understood that ali partial derivatives have been evaluated at the point Po. (See, Problems 10 and 11.), , Surface of Revolution, Let the graph of y =f(x) in the xy plane be revolved about the x axis. As a point (xo, Yo) on the graph revolves,, a resulting point (x o, y, z) has the distance )'0 from the point (xo' 0, 0). So, squaring that distance, we get, , Then, the equation of the surface of revolution is, (51.9)
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4., , CHAPTER Sl Surfaces and Curves in Space, , SOLVED PROBLEMS, , 1., , Derive (51.2) and (51.3) for the tangent line and normal plane to the space curve x =/(t), y = g(t), z = h(t) at the, point Po(xo. Yo> Zo) determined by the value t = 10. Refer to Fig. 51-7., Let P'o(Xo+ ~x, Yo +~y, Zo + ~z), determined by t= to + ~t, be another point on the curve. As Po~ Po along, the curve, the chord PoP'oapproaches the tangent line to the curve at Po as the limiting position., A simple set of direction numbers for the chord PoP'o is, Po ~ Po, &, , ~ 0 and [ t, , [~x, ~y, ~z], but we shall use [t 't ,~ ].Then as, , 't ,~]~ [~ ,!frt,1ftJ. a s~t of direction numbers of the tangent line at Po· Now if ., , P(x, y, z) is an arbitrary point on this tangent line, then [x - Xo, y....: Yo, z - zo] is a set of direction numbers of Pop., Thus, since the sets of direction numbers are proportional, the equations of the tangent line at Po are, , x-xo = Y-Yo = z-Zo, dx I dt dy I dt dz I dt, If R(x, y, z) is an arbitrary point in the normal plane at Po, then, since PoR and PaP are perpendicular, the, equation of the normal plane at Po is, , 2. -Find the equations of the tangent line and normal plane to:, (a) The curve x =t, Y =f. z =(J at the point t =1., (b) The curve x =t - 2, Y =3f + 1, z =2t3 at the point where it pierces the yz plane., (a) At the point t = I or (I, I, 1), dxldt = I, dyldt =2t =2, and dz/dt =3t2 =3. Using (51.2) yields, for the, equations of the tangent line, x II =Y ~ I = z;1; using (51.3) gives the equation of the normal plane as, (x-l)+2(y-I)+3(z-I)=x+2y+3z-6=0., (b) The given curve pierces the yz plane at the point where x =t - 2 =0, that is, at the point t =2 or (0, 13,, 16). At this point, dxldt = 1, dyldt =6t = 12, and dzldt =6f =24. The equations of the tangent line are, = y ~213 = z 6, and the equation of the normal plane is x + 12(y - 13) +24(z - 16) =x + 12y +, , T, , 21, , 24z - 540 =O., , 3., , Derive (51.4) and (51.5) for the tangent plane to the surface F(x, y, z) =0 at the point Po(xo• Yo. zo). Refer to, Fig.51-8., Let x =/(t), y =g(t}, z =h(t) be the parametric equations of any curve on the surface F(x. y, z) =0 and passing, through the point Po. Then, at Po., , with the understanding that all derivatives have been evaluated at Po., This relation expresses the fact that the line through Po with direction numbers, , 1, , [t, 1r ,1ft] is perpendicular, , to the line through Po having direction numbers [~~ , ac:;, ~~ The first set of direction numbers belongs to the, tangent to the curve which lies in the tangent plane of the surface. The second set defines the normal line to the, surface at Po. The equations of this normal are
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CHAPTER 51 Surfaces and Curve,s in Space, , and the equation of the tangent plane at Po is, , In Problems 4 and 5, lind the equations of the tangent plane and nonnalline to the given surface at the given point., , z =; 3.r+ 2)'2 -, , 4., , II; (2, 1,3)., , =O. At (2, 1,3), aaFex = 6x ,, =12,, aaF =4y = 4, and ~ =-I. The equation of, Y, oz, 2) + 4(y - I) - (z - 3) = 0 or 12x + 4y -z =25., , Put F(x, y, z) =3.r+ 2)'2_ z - 11, i, , the tangent plane is 12(x -, , _• ', , ':.~~{~;, , The equations of the norrnalline are x 2 = y ~ I = ,z~13 ., , 12, , "t:·::~'?t, , F(x,)" z) = r+ 31- 4z 2 + 3xy - lO)'z + 4x - 5z - 22 = 0; (1, -2. I)., , 5., , ~~ = 2x + 3y +4 =0,, , At (I, -2. 1),, , *" =, , 6)' + 3x - lOz =-19, and, , *, , =-8z -, , lOy - 5 = 7. The equation of, , the tangent plane is O(X - I) -19(y + 2) + 7(z - I) =0 or 19y -7z + 45 =0., The equations of the normal line are x - I = 0 and, ,, , 6., , ~~; = z 7I or x;' 1, 7y + 19z - 5 =O., , 2, y2, 2, Show that the equation of the tangent plane to the surface ~ - j;r -If = 1at the point PO<Xo, Yo, Zo) is, a, c, , 'uo _ )'Yo _, , a2, , b2, , 5.. - 1, c2, , aF, , -, , •, , 2xo aF, , AtPo'-a =-2, x II, , 2yo, '-ay =--b2', , and, , aF, 2"0, -a, =--2· The equationofthetangentplaneis, Z, C, , 2xo, ' 2y, 2"0, -a2( x - x0 )-~()'-y, )--(Z-7_)=0., b1, 0, c2, '1l, •, , 'uo, , )'Yo, , ZZo, , x~, , y~;;, , ., , ., , This becomes -a2 - -b, - 2 = -:r - -b, = I. SInce Po IS on the surface., 2 - ,c, 2 - 2", a, c, 7., , I, , Show that the surfaces F(x, y, z) =r + 41- 4z2 - 4 =0 and G(x, y. z) =r + y2 + Z2 - 6x - 6y + 2z + 10 =0 are, tangent at the point (2, 1, I)., It is to be shown that the two surfaces have the same tangent plane at the given point. At (2,1, 1),, , ",1,', , and, , of, , aaFIX = 2x-4•, , aF, -a, =8y=8,, )', , ~=-8z=-8, oZ ', , ~~ =2x-6=-2,, , aG, ay =2),-6=-4,, ,, , Tz=2z+2=4, , dG, , Since the sets of direction numbers [4, 8, -8) and [-2, -4. 4) of the normal lines of the two surfaces are, proportional, the surfaces have the common tangent plane, l(x-2)+2(y-I)-2(z-1)=0, , 8., , or, , x+2y-2z=2, , Show that the surfaces F(x. y, z) = x). + yz - 4zx =0 and G(x. y. z) =3z2 - 5x + y = 0 intersect at right angles at the, point( I, 2, I)., It is to be shown that the tangent planes to the surfaces at the point are perpendicular or, what is the same, that, the normal lines at the point are perpendicular. At (1, 2, 1)., , 1Ii, , = y- 4z = -2,, , aF _, _, -a, y -x+z-2 ,, , aF, Tz=y-4x=-2
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4D, , CHAPTER 51 Surfaces and Curves in Space, , A set of direction numbers for the normal line to F(x, y, z) = 0 is [II' ml' nl] = [1, -I, 1]. At the same point,, , t~ =-5,, , oG, Oy -I, - ,, , Pfz=6Z=6, , A set of direction numbers for the normal line to G(x, y, z) = 0 is [/2 '~' n2] = [-5, 1, 6]., Since 11/2 + mlm2 + n1n2= 1(-5) + (-1)1 + 1(6) = 0, these directions are perpendicular., 9., , Show that the surfaces F(x, y. z) = 3xl + 4y2 + Si - 36 = 0 and G(x, y. z) = xl + 2y2 - 4Z2 - 6 = 0 intersect at right, angles., , ., , ., , of, , of, , of, , ., , At any pomt Po(Xo, Yo, Zo) on the two surfaces, ox = 6xo' Oy = SYo ,and Tz = 16Zo: hence [3.to, 4y() SZol IS a, set of direction numbers for the normal to the surface F(x, y. z) = 0 at Pr; Similatly, [.to, 2yOt -41.01 is a set of, direction numbers for the normal line to G(x, y. z) = 0 at Prj Now, since, 6(x~, , + 2y~ -4~)- (3x~ +4y~ +S~)=6(6)- 36= 0,, , these directions are perpendicular., 10. Derive (51.7) and (51.S) for the tangent line and normal plane to the space curve C: F(x, y, z) = 0, G(x, y, z) =0, at one of its points Po(xo, Yo> Zo>., , At Po, the directions, , [~~ ,~ , arz], , and, , [~~ , t~ ,t~], , are normal, respectively, to the tangent planes of the, , surfaces F(x, y, z) = 0 and G(x, y, z) = O. Now the direction, , aF I Oy, [laGIOy, , aF I ozl, dG/az', , OF I dZ of I oxl, dGldx', , IdGldz, , dn dX dF I dylJ, dGIOy, , IdGldx, , being perpendicular to each of these directions, is that of the tangent line to Cat Pr; Hence, the equations of the, tangent line are, x-xo, , _, , y- Yo, , dF I Oy of I dz -ldF I dz dF I dxl, oGIOy oGloz dGldz dGldx, , I, , and the equation of the normal plane is, , 0F! dy of I 01.1, IdF I 01. of I dxl, IdF I dx of I Oyl, oGIOy dGloz(x-xo)+ oGldz oGlox(y-yo)+ dGldx dGIOy(z-Zo)=O, , 1, , 11. Find the equations of the tangent line and the normal plane to the curve xl +y2 + ·c = 14, x + Y+ 1. = 6 at the point, (1,2,3)., , Set F(x,y,z)=x 2 + y2 +1. 2 -14=0 and G(x,t,z)=x+y+z-6=0. At(I,2,3),, (', , I~~~~ ~~~~~1=12[ 2tl=l~ ~1=-2, dFIOZ of,ax'=16 21=4 IdF/dX aFIOyI=12 41=-2, dG I o~ 1 1 'oG I dx dG I Oy 1 1, , IaG I dz, , '1, , 2, With [1, -2, 1] as a set of direction numbers of the tangent, its equations are x ')1 = Y~2 = 1. 3 . The equation, of the normal plane is (x - 1) - 2(y - 2) + (z - 3) = x - 2y + z = O.
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CHAPTER 51, , Surfaces and Curves in Space, , 12. Find equations of the surfaces of revolution generated by revolving the given ~urve about the given axis: (a) y :::;, xl about the x axis; (b) y =1 about the y axis; (c) z =4y about the y axis., In each case, we use an ippropriate fonn of (51.9): (a) + t = xA; (b) xl + t =~; (c) xl + t = 16y2., , r, , Y, , 13. Identify the locus of all points (x, y, z) that are equidistant from the point (0, -I, 0) and the plane y = 1., Squaring the distances, we get r + (y + 12) + Z2 =(y - 1)2, whence.r + Z2 =-4y, a circular paraboloid., ,, , 14. Identify the surface 4.r We have, , + Z2 - 8x + 2y + 2z + 3 = 0 by completing the squares., , )'2, , 4(r - 2x) - (y2 - 2y) + (t + 2z) + 3 = 0, , This is a hyperboloid of one sheet, centered at (1, 1, -1)., , ~ '~•.!, , r, , _, , \, , I, , -, , I, , .., , r. I ', , ., •, , -, , I,.:, , •, , ,, , .-, , ••, , 15. Find the equations of the tangent line and the nonnal plane to the given curve at the given point:, (a) x=2/,y=r,z=/ 3;1=1, , AilS., , x22=y~I=Z31;2x+2Y+3z-9=0, , (b) x=/e',y=eI,Z=I;I=O, , AilS., , f=XY!=f;x+y+z-I=O, , (c) x=/cos/,y=tsint,z=t;t=O, , AIlS., , x=z.y=O;x+z=O, , 16. Show that the curves (a) x = 2 - t, Y = -1//, Z = 2fl and (b) x = I + e, y =sin e- 1, z =2 cos eintersect at right, angles at P( I, -1, 2). Obtain the equations of the tangent line and nonnai plane of each curve at P., , AIlS., , (a) x.:-/ = y;1 =, , Z4 2 ;X-y-4z+6=0;, , (b) x- y =2.,z=2; x+y=O, , 17. Show that the tangent lines to the helix x = a cos t, y = a sin I, Z = bt meet the xy plane at the same angle., , 18. Show that the length of the curve (51.1) from the point I, , =10 to the point t =t I is given by, , Find the length of the helix of Problem 17 from 1=0 to I =II', ", , I., , ', , 19. Find the equations of the tangent line and the normal plane to the given curve at the given point:, , (a) r + 2y2 + 2z2 = 5, 3x - 2y - z = 0; (I, I, I)., (b) 9x2 + 4y2 - 36z =0, 3x+ y + z -Z2- 1 =0; (2, -3, 2)., (,,) 4z2 =xy,r+r=8z; (2,2,1).
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CHAPTER 51 Surfaces and Curves in Space, , Ans., , (a), , X21=y~l=z~l;, , 2x+7y-8z-I=0; (b), , XI"2=y~2.y+3=0;X+Z-4=0;, , 2, (e) XI"2=y-=-1 ,z_1 =O;x-y=O, , 20. Find the equations of the tangent plane and normal line to the given surface at the given point:, , (e) r + 2t + 3y2; (2, -2, -2), , x-I ~, Ans. x-2y+3z= 14;-1-=, -2 = 8 3, x-x y+y z-z, Ans. XIX +YIY + ZIZ = .::......::L =__I =__I, XI, YI, z,, x-2 ~ i l l, Ans. x+3y-2z=0;-I-= 3 = -2, , (d) 2r+2xy+y2+z+I=0;(I,-2,-3}, , Ans., , (e) Z =xy; (3, -4, -12), , Ans., , (a) r+y2+t=14;(I,-2,3}, (b) r+y2+t=r;(x l'YI,zl), , r;, , Z-2Y=I;X-I=0,y~2=Z~13, 4x-3y+z=12; x 3 = y~34 =, 4, , Zi l2, , 21. (a) Show that the sum of the intercepts of the plane tangent to the surface xlr.! + ylr.!+ zll2 = aIr.! at any of its points is a., (b) Show that the square root of the sum of the squares of the intercepts of the plane tangent to the surface, yW+ z'1l3 = a'1l3 at any of its points is a., , rn +, , 22. Show that each pair of surfaces are tangent at the given point:, (a) r +y2 + t =18, xy =9; (3, 3, 0)., (b) r+ y2+ t -8x- 8y - 6z + 24 = 0, r + 3y2+2i= 9; (2, I, I)., , 23. Show that each pair of surfaces are perpendicular at the given point:, (a) r+ 2y2- 4z2 = 8,4 r - y2+2z2 = 14; (2, 2,1)., (b) r+ f+ t = 50. r+ f- IOz + 25 =0; (3, 4, 5)., 24. Show that each of the surfaces (a) 14r + III + 8z2 =66, (b) 3z2 - 5x + Y = 0, and (c) xy + yz - 4zx = 0 is, perpendicular to the other two at the point (I, 2, I)., 25. Identify the following surfaces., (a) 36y2-r+36z2 =9., (b) 5y =-Z2+ r., , (c) r+41-4t-6x-16y-16z+5=0., Ans., , (a) hyperboloid of one sheet (around the X axis); (b) hyperbolic paraboloid; (c) hyperboloid of one, sheet. centered at (3. 2, -2), , 26. Find an equation of a curve that, when revolved about a suitable axis, yields the paraboloid f + Z2 - 2x = O., AilS., , Y = .ffi or z = fiX, about the x axis, , 27. Find an equation of the surface obtained by revolving the given curve about the given axis. Identify the type of, surface: (a) x = y2 about the X axis; (b) x = 2y about the x axis., Ans., , (a), , X, , = f + t (circular paraboloid); (b), , f, , 2, , + Z2 = ~ (right circular cone)
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Directional Derivatives., Maximum and Minimum Values, Directional Derivatives, Let P(x, y, z) be a point on a surface z = I(x, y). Through p. pass planes parallel to the xz and yz planes,, cutting the surface in the arcs PR and PS, and cutting the xy plane in the lines P*M and P*N, as shown in, Fig. 52-1. Note that p* is the foot of the perpendicular frol11 P to the xy plane. The partial derivatives azl, and azlay, evaluated at P*(x, y), give, respectively, the rates of change of z = p*p when y is held fixed and, when x is held fixed. In other words, they give the rates of change of z in directions parallel to the x and y, axes. These rates of change are the slopes of the tangent lines of the curveS PR and PS at P., , ax, , Rg.52·1, , Consider next a plane through P perpendicular to the xy plane and making an angle e with the x axis. Let, it cut the surface in the curve PQ and the xy plane in the IineP*L. The directional derivative of/(x, y) at p*, in the direction 8 is given by, , az, , az ., , dz, -=-cos8+-sm8, ds ax, ay, , (52.1), , The direction eis the direction of the vector (cos 8)i + (sin 8)j., The directional derivative gives the rate of change of z = p*p in the direction of P*L; it is equal to the, slope of the tangent line of the curve PQ at P. (See Problem I.)
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CHAPTER 52, , Directional Derivatives, , The directional derivative at a point p* is a function of (J. We shall see that there is a direction, determined, by a vector called the gradient of I at p* (see Chapter 53), for which the directional derivative at p* has, a maximum value. That maximum value is the slope of the steepest tangent line that can be drawn to the, surface at P., For a function w = F(x, y, z), the directional derivative at P(x, y, z) in the direction determined by the, angles a, /3, r is given by, dF, ds, , By the direction determined by a,, (cos r)k,, ', , of, of, of, = ox cosa+ oy cos/3+azcos r, , /3, and r. we mean the direction of the vector (cos, , a)i + (cos, , /3)j +, , Relative Maximum and Minimum Values, Assume that z = fix. y) has a relative maximum (or minimum) value at Po(xo, Yo, Zo). Any plane through Po, perpendicular to the xy plane will cut the surface in a curve having a relative maximum (or minimum) point, , ~ cosO+ ~ sinO of z = I(x, y) must equal zero at Po. In particular,, when 0=0, sin 8 = 0 and cos 8 = 1, so that ~ = O. When 8 = ~, sin 8 = 1 and cos 8 = 0, so that ~ = O., , at Po. Thus, the directional derivative, , Hence, we obtain the following theorem., Theorem 52.1:, , If z = f(x, y) has a relative extremum at Po(x:o, y() Zo) and, , ~ and ~ exist at (.lO, Yo). then ~ = 0 and, , af, ay =0 at (xo, Yo)', We shall cite without proof the following sufficient conditions for the existence of a relative maximum, or minimum., Theorem 52.2:, , Let z = f(x, y) have fIrst and second partial derivatives in an open set including a point (xo, Yo) at, , 2f, . af, ,, ax = 0 and dfay =O. Define 11 =( axaay )2 - (a2f)(a2f), ax, dy2' Assume 11 < 0 at (xo, Yo)· Then:, , whIch, , 2, , z = f(x, y) has, , a relative m1nimum at (xo' Yo), , l, , a relative maximum at (xo', Yo), , If 11 > 0, there is neither a relative maximum nor a relative minimum at (xo, Yo)., If 11 =0, we have no information., , Absolute Maximum and Minimum Values, Let A be a set of points in the xy plane. We say that A is bounded if A is included in some disk. By the compLement of A in the xy plane, we mean the set of all points in the xy plane that are not in A. A is said to be closed, if the complement of A is an open set., Example 1:, , The following are instances of closed and bounded sets., , (a) Any closed disk D, that is, the set of all points whose distance from a fixed point is less than or equal to some, fixed positive number r. (Note that the complement of D is open because any point not in D can be surrounded by, an open disk having no points in D.), (b) The inside and boundary of any rectangle. More generally, the inside and boundary of any "simple closed curve,", that is, a curve that does not interset itself except at its initial and terminal point.
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CHAPTER 52, , Directional Derivatives, , Theorem 52.3: Let f(x, y) be a function that is continuous on a closed. bounded set A. Then f has an abs,olute, maximum and an absolute minimum value in A., , The reader is referred to more advanced texts for a proof of Theorem 52.3. For three or more variables,, an analogous result can be derived., ,, SOLVED PROBLEMS, 1., , Derive fonnula (52.1)., In Fig. 52-I, let P**(x + tu, y + toy) be a second point on P*L and denote by lls the distance p*p**;, Assuming that.z = f(x. y) possesses continuous first partial derivatives, we have, by Theorem 49.1,, , where EI and E2 ~ 0 as tox and toy ~ O. The average rate of change between points p* and'p** is, , t.z _ dZ, , tu dZ toy, tu, toy, ax t::.s, + dY lls + EI lls + E2 lls, oz· 8 +E1COS 8 +E 2 sm. 8, =~, ox cosu + dysm, , lls -, , Ll, , where 0 is the angle that the line p*p** makes with the x axis. Now let p** ~ p* along P*L The directional, derivative at P*. that is, the instantaneous rate of change of z. is then, , 2., , Find the directional derivative of z=:r - 6y2 at P*(7, 2) in the direction: (a) 0= 45°; (b) 0 = 135°., The directional derivative at any point P*(x, y) in the direction 0 is, , ~ = ~; cos 8 + ~sin(J= 2xcos8-12ysin8, (a) At P*(7, 2) in the direction 8= 45°,, , ~ =2(7)(t!i)-12(2)(t.fi)=-s!i, (b) At P*(7, 2) in the direction 8=135°,, *= 2(7X-t!i)-12(2Xt!i)= -19!i, , 3., , Find the directional derivative of z = ye' at P*(O, 3) in the direction (a) 8= 30°; (b) 8= 120°., Here, dzlds = ye' cos 8+ e' sin 8., (a) At (0, 3) in the direction 8= 30°, ddds= 3(1)(-tJ3)+t=t(3J3 +1)., (b) At (0,3) in the direction 8= 120°, ddds = 3(1)(-t) + tJ3 = t(-3+ J3)., , 4., , The temperature T of a heated circular plate at any of its points (x. y) is given by T =, , 2, , 64, , x +y2 +, , 2' the origin being, , at the center of the plate. At the point (1; 2), find the rate of change of T in the direction 8 = n/3.
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CHAPTER 52, , -., , Directional Derivatives, , We have, dT, , ds =, , 64(2x), + 2)2 cos9 -, , (x 2 + y2, , •, ••, 1t dT, 128 1, At (1, 2) m the dIrection 0= 3' ds =-49 "2, , 5., , (X2, , 64(2y)., + y2 + 2)Z sm9, , 256.J3, (1 2 "'3), - 49, T =- 64, 49 + V:J·, , The electrical potential Vat any point (x, y) is given by V =In ~X2 + y2 . Find the rate of change of V at the point, (3,4) in the direction toward the point (2, 6)., Here,, dV = - 2 x- - 2 COSu£I +- ZY, '£1, -ds, - - 2 sm u, X +Y, x +y, , Since Ois a second-quadrant angle and tan 9= (6 - 4)/(2 - 3) = -2. cos 0= -11$ and sin 0= 21 $., ' the m, . d'Icated d'rrectlon,, . ds, dV = 25, 3 (Hence, at (3 , 4) m, 6., , 1), , $, , 4 $2 = 25', $, + 25, , Find the maximum directional derivative for the surface and point of Problem 2., At P*(7, 2) in the direction 0, dz/ds = 14 cos 9- 24 sin 9., To find the value of o for which, , ~;, , is a maximum, set fo (-~) = -14 sin 0- 24 cos 0= O. Then tan 9=, , -* =-4, , and 0 is either a second- or fourth-quadrant angle. For the second-quadrant angle, sin 0= 121M and cos = -7/J193., For the fourth-quadrant angle. sin 0= -121M and cos 0= 7/Jf93., , (~ ) = fo (-14 sin 0 -, , Since' f;2, , 24 cos 0) = -14 cos 0 + 24 sin 9 is negative for the fourth-quadrant angle,, , the maximum directional derivative is, 7., , ~~ = 14(, , 24( -, , = 2Jf93, and the direction is 9='300°15'., , Find the maximum directional derivative for the function and point of Problem 3., At P*(O, 3) in the direction 9. dzlds = 3 cos 0+ sin 9., ., To find the value of 9 for which ~ is a maximum, set fo (~;) = -3 sin 9 + cos 0 = O. Then tan 9 = t and 0, is either a first- or third-quadrant angle., Since f;2, , (~) =fo, , (-3 sin 0+ cos 9) = -3 cos 0- sin Ois negative for the first-quadrant angle, the, , maximum directional derivative is ddsz = 3, ., , 8., , k )- Jh), , b + vlO, b = JfO, and the direction is 0= 18°26'., , vl0, , In Problem 5. show that V changes most rapidly along the set of radial lines through the origin., At any point (Xl' YI) in the direction 8, ddV =~cosO+~sin9. Now V changes most rapidly, S, , when, , XI, , +YI, , Xl, , +YI, , dd9(ddV)=-~Sin9+~cosO=0,, and then tan 9= YI ~~x~ + Y~~ = YI . Thus, Ois the angle of, S, Xl +YI, Xl +Y, Xl, Xl +YI, Xl, I, , i~clination of the line joining the origin and the point (Xl' YI)', , ., , l', , 9., , Find the directional derivative of F(x, y, z) = xy + 2xz - yZ + ZZ at the point (I, -2, I) along the curve X = t,, Y= t - 3, z = f in the direction of increasing z., A set of direction numbers of the tangent to the curve at (1. -2, 1) is [I, 1,2]; the direction cosines are [1IJ6,, 1/../6, 21J6]. The directional derivative is
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·8·, , CHAPTER 52, , Directional Derivatives, , 10. Examine f(x, y) = i'- + y2 - 4x + 6y + 25 for maximum and minimum values., The conditions, , ~ = 2x - 4 = 0 and ~ = 2y + 6 = 0 are satisfied when x = 2, y = -3. Since, , f(x, y) =(i'-- 4x+ 4) + & + 6y+9)+25 -4 -9= (x- 2)2+ (y+ 3)2 + 12, , it is evident thatf(2, -3) = 12 is the absolute minimum value of the function. Geometrically, (2, -3, 12) is the, lowest point on the surface z = i'- + y2 - 4x + 6y + 25. Clearly,f(x, y) has no absolute maximum :alue., 11. Examine f(x,y), , =xl + yl + 3xy for maximum and minimum values., , df = 3(i'- + y) = 0 and df =3& + x) =0 are satisfied when x = 0,, , We shall use Theorem 52.2. The conditions, ., y =0 and when x =-I, Y =-1., , ~, , aaxf =6x = 0, axafay = 3, and aay2f =6y =O. Then, 2, , At (0, 0),, , ~, , 2, , 2, , 2, , 2f, 2f, dXdY _(aax2 )(a()y2 )_- 9 > 0, (lY)', 2, , and (0, 0) yields neither a relative maximum nor minimum., , af, 2, , af =3, and ady2f =-6. Then, 2, , 2, , At (-1, -I), ax 2 =-6, ax ay, , (iiir -(~)(~)=-27<0, , and, , ~+~<O, , Hence,f(-l, -1) = 1 is a relative maximum value of the function., Clearly, there are no absolute maximum or minimum values. (When y = O,f(x, y) = xl can be made arbitrarily, large or small.), , 12. Divide 120 into three nonnegative parts such that the sum of their products taken two at a time is a maximum., Let x, y, and 120 - (x + y) be the three parts. The function to be maximized is S = xy + (x + y){l20 - x - y)., Since 0 ~ x + y ~ 120, the domain of the function consists of the solid triangle shown in Fig. 52-2. Theorem 52.3, guara,ntees an absolute maximum., y, 120, , 120, , x, , Fig. 52-2, , Now,, , ~~ =y+(I20-x-y)-(x+y)= 120-2x-y, and, , ~~, , =x+(120-r-y)-(x+y)= 120-x-2y
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CHAPTER S2 Directional Derivatives, , Setting dS/dX = dS/dy = 0 yields 2x + y = 120 and x + 2y = 120., Simultaneous solution gives x = 40, y = 40, and 120 - (x + 4) = 40 as the three parts, and S = 3(4Q2) = 4800., So, if the absolute maximum occurs in the interior of the triangle, Theorem 52.1 tells us we have found it. It is, still necessary to check the boundary of the triangle. When y = 0, S = x( 120 - x). Then dSldx = 120 - 2x, and the, critical number is x = 60. The corresponding maximum value of Sis 60(60) = 3600, which is <4800. A similar, result holds when x = O. Finally, on the hypotenuse, where y = 120 - x, S = x(120 - x) and we again obtain a, maximum of 3600. Thus, the absolute maximum is 4800. and x = y = z = 40., , 13. Find the point in the plane 2x - y + 2z = 16 nearest the origin., , Let (x, y, z) be the required point; then the square of its distance from the origin is D = i'- + y + Z2. Since also, 2x - y+ 2z = 16, we havey= 2x+ 2z -16 and D =i'- +(2x+ 2z - 16)2 + Z2., Then the conditions dD/dX = 2x + 4(2x + 2z - 16) = 0 and dD/dZ = 4(2x + 2z - l6) + 2z = 0 are equivalent, to 5x + 4z = 32 and 4x + 5z = 32. and x = z = f. Since it is known that a point for which D is a minimum exists,, (f,--'l-,f) is that point., , 14. Show that a rectangular parallelepiped of maximum volume V with constant surface area S is a cube.·, Let the dimensions be x, y, and z. Then V = xyz and S = 2(xy + yz + zx)., The second relation may be solved for z and substituted in the first, to express Vas a function of x and y. We, prefer to avoid this step by simply treating z as a function of x and y. Then, , dV =yz+xy~, dX, dX', , a;; =xz+xy~~, , From the latter two equations, adZ = - y+ z and ~ = -~. Substituting in the first two yields the conditions, x, x+y, vy, x+y, dV, xy(y + z), dV, xy(x + z), ., dX =yz- x+y, oand dy =xz- x+y =O.whichreducetoY(z-x)=Oandi'-(z-y)=O.Thusx=y=z,, as required., , y: -4, , 15. Find the volume V of the largest rectangular parallelepiped that can be inscribed in the ellipsoid x~ + + = I., Let P(x. y, z) be the vertex in the first octant. Then V = 8xyz. Consider z to be defined as a fungtion ~f th~, independent variables x and y by the equation of the ellipsoid. The necessary conditions for a maximum are, , (I), , From the equation of the ellipsoid, obtain 2; + 2i ~z = 0 and, between these relations and (I) to obtain a c x, , ~~ + 2i ~ = O. Eliminate dZ/iJx and dZ/dy, c, , and, finally., , (2), Combine (2) with the equation of the ellipsoid to get x = afj/3, y = bfj/3. and z = c fj /3., Then V = 8xyz = (8fjl9)abc cubic units.
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CHAPTER 52, , Directional Derivatives, , 16. Find the directional derivatives of the given function at the given point in the indicated direction., (a), , (b), (c), , (d), , z =r+xy +T, (3, I), 9=l, z = x3 - 3xy + I, (2, I), 9 = tan- 1m., z = y +x cosxy. (0, 0),9=, z =2\"2 + 3xy - y2, (1, -I), toward (2,, , Ails., , f., , I)., , (a) +<7 + 5Jj); (b) 2 1.Ji31l 3; (c), , t(l + Jj);, , (d) 11$/5, , 17. Find the maximum directional derivative for each of the functions of Problem 16 at the given point, , AilS., , (a), , ffi:, , (b), , 3M:, , (c) fl.: (d), , J26, , 18. Show that the maximal directional derivative of V = In, , Jx, , 1, , + y2 of Problem ~ is constant a.ong any circl~ r + T = r., , 19. On a hill represented by z = 8 - 4xl - 2y~. find (a) the direction of the steepest grade at (1. 1.2) and (b) the, direction of the contour line (the direction for which z = constant). Note that the directions are mutually, perpendicular., , AilS., , (a) tan-let>. third quadrant; (b) tan -1(-2), , 20. Show that the sum of the squares of the directional derivatives of z = I(x. y) at any of its points is constant for any, two mutually perpendicular directions and is equal to the square of the maximum directional derivative., 21. Given z =I(x, y) and w =g(x, y) such that az/i)x =aw/'dy and az!'dy= -aw/ax. If 91 and ~ are two mutually, perpendicular directions. show that at any point P(x, y), azlas l = aw/as2 and azlas2 = -aw/asl., 22. Find the directional derivative of the given function at the given point in the indicated direction:, (a) xy2z. (2. 1. 3), [1, -2. 2J., (b) xl + T + Z2. (I, 1. 1). toward (2.3,4)., , (c) xl + yl-2xz. (1. 3. 2), along xl + yl AIlS., , 2xz = 6, 3xl - T + 3z = 0 in the direction of increasing z., , (a) -Jf; (b) 6MI7; (c) 0, , 23. Examine each of the following functions for relative maximum and minimum values., (a) z =2x + 4y - xl - T - 3, (b) z = x3 +1- 3xy, (c) z =xl+ 2xy+ 2y2, (d) z = (x- y)(1-xy), (e) z=2r+T+6xy+ IOx-6y+5, (f) z = 3x - 3y - 2x3 - xy2 + 2ry + )'3, (g) z=xy(2x+4y+ 1), , Ans. maximum = 2 when x = I. Y = 2, AIlS. minimum =-1 whenx= l,y= 1, Ans. minimum = 0 when x = 0, y = 0, AIlS. neither maximum nor minimum, Ans. neither maximum nor minimum, Ans. minimum = --/6 when x =--/6/6. y= -/6/3;, maximum-/6 when x = -/6/6, y = --/6/3, Ans. maximum m-when x = -t, y =-n-, , 24. Find positive numbers x, y, Z such that, (a) x + y + z = 18 and xyz is a maximum, (c) x + y + z = 20 and xyi is a maximum, , Ans., , (b) xyz = 27 andx + y + z is a minimum, (d) x + y + z = 12 and xyl z3 is a maximum, , (a) x = y = z = 6; (b) x = y = z = 3; (c) x = y = 5, z = 10; (d) x = 2,, , y = 4, z = 6
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CHAPTER 52 Directional Derivatives, 25. Find the minimum value of the square of the distance from the origin to the plane Ax + By + Cz + D =O., Ans, , IJ2/(A2+ 8 2 + ('2), , 26. (a)' The surface area of a rectangular box without a top is to be 108 ft2. Find the greatest possible volume., (b) The volume of a rectangular box without a top is to be 500 ft3• Find the minimum surface area., , Ans., , (a) 108 ft3; (b) 300 ft2, , 27. Find the point on z =xy - I nearest the origin., Ans., , (0,0, -1), , 28. Find the equation of the plane through (I, I, 2) that cuts off the least volume in the first octant., Ans., , 2x + 2y + z = 6, , 29. Determine the values of p and q so that the sum S of the squares of the vertical distances of the points (0, 2), (1, 3),, and (2, 5) from the line y = px + q is a minimum. (Hint: S = (q - 2)2 +(p + q - 3)2 + (2p + q - 5)2.), Ans. p=t:, , q=t
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CHAPTER 53 Vector Differentiation and Integration, , cm-, , Space Curves, Consider the space curve, , x= /(t),, , y, , =g(t),, , z =h(t), , (53.5), , where /(t), g(t), and h(t) have continuous first and second derivatives. Let the position vector of a general, variable point P(x, y, z) of the curve be given by, r=xi +yj +zk, As in Chapter 39, t = dr/ds is the unit tangent vector to the curve. If R is the position vector of a point, (X, Y, Z) on the tangent line at P, the vector equation of this line is (see Chapter 50), R - r =kt for k a scalar variable, , (53.6), , and the equations in rectangular coordinates are, X-x Y-y, dxlds =dylds, , where [ : ' : ', , Z-z, , =dzlds, , ~~ ] is a set of direction cosines of the line. In the corresponding equation (51.2), a set of, , direction numbers [ : ', , 1r, ~~] was used., , The vector equation of the normal plane to the curve at P is given by, (R - r)' t =0, , (53.7), , where R is the position vector of a general point of the pl~ne., Again, as in Chapter 39, dtlds is a vector perpendicular to t. If D is a unit vector having the direction of, dtlds, then, dt, ds, , =IKln, , where IKI is the magnitude of the curvature at P. The unit vector, (53.8), is called the principal normal to the curve at P., The unit vector b at P, defined by, b=txn, , (53.9), , is called the binfJrmal at P. The three vectors t, n, b form at P a right-handed triad of mutually orthogonal, vectors. (See Problems I and 2.), At a general point P of a space curve (Fig. 53-1), the vectors t, n, b determine three mutually perpendicular planes:, 1., , The osculatin'g plane, containing t and D, having the equation (R - r) • b =0, , 2., , The normal plane, containing nand b, having the equation (R - r) • t =0, , 3., , The rectifying plane, containing t and b, having the equation (R - r)' n =0
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CHAPTER 53, , Vector Diff~rentiation and Integration, , z, , ,, , y, , x, Fig. 53-1, , In each equation, R is the position vector of a general point in the particular plane., , Surfaces, Let F(x, y, z) =0 be the equation of a surface. (See Chapter 51.) A parametric representation results when x,, y, and z are written as functions of two independent variables or parameters u and v, for example, as, , =, , x J.(u, v),, , y=Nu, v),, , z = f3(u, v), , (53.10), , z =J3(Uo, v), , (53.11 ), , When u is replaced with uo, a constant, (53.10) becomes, x, , =J.(Uo, v),, , the equation of a space curve (u curVe) lying on the surface. Similarly, when v is replaced with "b, a constant,, (53.10) becomes, x =J.(u, vo),, , y =J;(u, 110),, , (53.12), , the equation of another space curve (v curve) on the surface. The two curves intersect in a point of tl1e surface, obtained by setting u = Uo and v = 110 simultaneously in (53.10)., The position vector of a general point P on the surface is given by, , r = xi + yj + zk = iJ.(u, v) + jf2(U, v) + kf3(U, v), Suppose (53.11) and (53.12) are the II and v curves through P. Then. at P., , is a vector tangent to the u curve, and, , (53.13)
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CHAPTER 53, , Vector Differentiation and Integration, , is a vector tangent to the v curve. The two tangents detennine a plane that is the tangent plane to the surface, at P (Fig. 53-2). Clearly, a normal to this plane is given by, x The unit no~al to the surface at P is, defined by, dU dV, , dr dr., , dr dr, -xn= dU dV, 'dr drl, . IdU x dV, , (53.14), , o, Fig. 53-2, , If R is the position vector of a general point on the normal 10 the surface at P, its vector equation is, , drxdr), (R-r)=k ( -, , ., , dU, , dV, , (53.15), , If R is the position vector of a general point on the tangent plane to the surfa~e at P, its vector equation is, (R-r)., , - =0, (-dUdrxdVdr), , (53.16), , (See Problem 3.), , The Operation V, In Chapter 52, the directional derivative of z =f(x, y) at an arbitrary point (x, y) and in a direction making an, angle () with the positive x axis is given as, , L~t, , us write, (53.17), , Now a =i cos ()+ j sin () is a unit vector whose direction makes the angle () with the positive x axis. The other, factor on the right of (53.17), when written as, , (i ! + ~ f., j, , ), , suggests the definition of a vector differential
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CHAPTER 53 Vector Differentiation and Integration, , operator V (del), defined by, (53.18), , r, , is called the gradient offor grad! From (53.17), we see thatthe comIn vector analysis, Vf= i ~ + j, ponent of Vfin the direction of a ~nit vector a is the directional derivative offin the directi6n of a., Let r =xi + yj be the position vector to P(x, y). Since, , dr, , = Vf· ds, , I~I =IVi1 cos;, , and, , where; is the angle between the vectors Vfand drlds. we see thatdflds is maximal when cos; = 1. that, is. when Vf and dr/ds have the same direction. Thus. the maxiJilUm value·of the directional derivative at, P is IV!I; and its direction is that of Vf (Compare the 'discussion of maximum directional derivatives in, Chapter 52.) (See Problem 4.), For w = F(x, y, z), we define, , t'7F _ . aF, v, , -, , I, , . aF k aF, , ax + J ay + Tz, , and the directional derivative of F(x, y, z) at an arbitrary point P(x, y, z) in the direction a, dF, , di =VF'a, , =ali + aJ + a 2k is, (53.19), , As in the case of functions of two variables, IVF1 is the maximum value of the directional derivative of, F(x, y, z) at P(x, y, z), and its direction is that of VF. (See Problem 5.), Consider now the surface F(x, y, z) = O. The equation of the tangent plane to the surface at one of its, points Po(xo, Yo. Zo) is given by, , aF, , aF, , aF, , (x-x )-+(y-y )-+(Z-7_)o, 0 i)y, .. "1l, , ax, , ., , az, , =[(x-xo)i+(y~yo)j+(z-Zo)k] .[i~~ +j~ +k~]=O, , (53.20), , with the understanding that the partial derivatives are evaluated at P(}' The first factor is an arbitrary vector, through Po in the tangent plane; hence the second factor VF, evaluated at Po, is normal to the tangent plane,, that is. is normal to the surface at Po. (See Problems 6 and 7.), , Divergence and Curl, The divergence of a vector function F =ifl(x. y. z) + jf2(X. y, z) + kf3(X, y, z), sometimes called del dot F. is, defined by, (53.21 )
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CHAPTER 53, , Vector Differentiation and Integration, , The curl of the vector function F, or del cross F, is defined by, i, , j, , k, , a a a, curl F =V x F = ax, , ay az, It fz h, (53.22), , (See Problem 8.), , Integration, Our discussion of integration here will be limited to ordinary integration of vectors and to so-called "line, integrals." As an example of the former, let, F(u) =i cos u + j sin u + auk, , be a vector depending upon the scalar variable u. Then, F'(u) =-i sin u + j cos u + ak, , JF'(u)du =J(-isinu + jcos u + ak) du, , and, , =i, , J-sin u du + j Jcos u du + kJa du, , =icosu + jsinu +auk +c, =F(u)+c, where c is an arbitrary constant vector independent of u. Moreover,, , j, , uab, U=<J, , F'(u) du = [F(u)+c]::: = F(b)-F(a), , (See Problems 9 and 10.), , Line Integrals, Consider two points Po and PI in space. joined by an arc C. The arc may be a segment of a straight line or, a portion of a space curve x = g\(t), y = g2(t), z = git), or it may consist of several subarcs of curves. In any, case, Cis assumyd to be continuous at each of its points and not to intersect itself. Consider further a vector, function, F =F(x, y, z) =iJ;(x, y, z) + Jt;(x, y, z) + k.h(x, y, z), , which at every point in a region about C and, in particular, at every point of C, defines a vector of known, magnitude and direction. Denote by, r =xi +yj +zk, , (53.23)
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CHAPTER 53, , Vector Differentiation and Integration, , the position vector of P(x, y, z) on C. The integral, , I, , II (, , cPo, , III, , F. dr), ds ds = cPo F. dr, , (53.24), , is called a line integral, that is, an integral along a given path C., As an example, let F denote a force. The work done by it in moving a particle over dr IS given by (see, Problem 16 of Chapter 39), IFlldrlcos, , e =F • dr, , and the work done in moving the particle from Po to PI along the arc C is given by, , From (53.23),, , dr;:;idx+jdy+kck, and (53.24) becomes, (53.25), (See Problem 11.), , I .•..;, , SOLVED PROBLEMS, , ... --, , .. . -....., , ..r,~~~,, , ~, , 1., , A particle moves along the curve x = 4 cos 1, y = 4 sin 1, Z = 61. Find the magnitude of its velocity and acceleration, attimes I =0 and I = 11r., Let P(x, y, z) be a point on the curve, and, r = xi + yj + zk = 4i cos I + 4j sin I + 6kl, , be its position vector. Then, v = d1r = -4i sin t + 4j cos t + 6k, £1, , At t= 0:, , Att=1 1r :, , and, , a=~, , =-4i cos t -, , v =4j +6k, , Ivl = ./16 + 36 = 2Jl3, , a=-4i, , lal=4, , v=-4i +6k, , Ivl = ./16+36 = 2Jl3, , a=-4j, , lal=4, , 4j sin t
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CHAPTER 53, , 2., , Vector Differentiation and Integration, , At the point (I, I, I) or t = 1 of the space curve x = t, Y=, , r, Z = tl, find:, , (a) The equations of the tangent line and normal plane., (b) The ,unit tangent. principal normal. and binormal., (c ) The equations of the principal normal and binormal., We have, r= Ii +rj +fk, , ~ == i + 2tj + 3rk, ds, dt, , 2, 4, ==Idrl==, dt .J! +4t +9t, , t == dr, , ds, , Att = I. r = i + j + k and t =, , 2, , =dr dt = i +2, dt ds, , 3t k, I + 4t 2 +9t 4, , +., (i + 2j + 3k)., ,,14, , (a) If R is the position vector of a general point (X. Y, Z) on the tangent line, its vector equation is R - r = kt or, (X - l)i + (Y - l)j + (Z - I)k =, , h (i + 2j + 3k), , ,,14, , and its rectangular equations are, X-I Y-l Z-l, -1- =2-: == -3If R is the position vector of a general point (X, Y, Z) on the normal plane, its vector equation is, (R - r) • t = 0 or, [(X -I)i +(Y -I)j +(Z-I)k].-b (i + 2j + 3k)=O, , ,,14, , and its rectangular equation is, (X-I)+2(Y-I)+3(Z-I)=X+2Y+3Z-6=O, (See Problem 2(a) of Chapter 51.), , ,, , ,, , At t = I,, , dt _ -lli-8j+9k, ds 98, , Then, , n, , 1 dt, , =lKf ds =, , d, an, , 1~I==t~=IKI., , -lli-8j+9k, , .fi66, , I ;•, , and, , ;~, , .' ', , j k, b=txn= Jf4~ 1 2 3 = i.n(3i-3J+k), 14 266 -II -8 9 ,,19, , (c) If R is the position vector of a geneml point (X. Y, Z) on the principal normal, its vector equation is, R-r=knor, , ., , ,, , ~, , --~, , ",, , ,, , (X-l)i+(Y-I)j+(Z-I)k= k -lli-8j+9k, , m, , ;, , -, , ,, .... ,, , ~, .~
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Vector Differentiation and Integration, , CHAPTER 53, , and the equations in rectangular coordinates are, , X-I, , Y-I, , Z-l, , -=tT=~=-9-, , If R is the position vector of a general point (X, y, Z) on the binonnal, its vector equation is R - r = kb or, , ,, (X-I)i+(Y-l)j+(Z-I)k= k, , 3i, , $+k, 19, , and the equations in rectangular coordinates are, X -I Y -1 Z-I, -3- = --==r- = -1-, , 3,, , Find the equations of the tangent plane and normal line to the surface x =2(u + v), y =3(u - v)., P(u =2, v= I)., Here, r, , =2(/1 + v)i + 3(/1 -, , v)j + /wk., , ~: =2i + 3j + vk,, , ~~ = 2i -, , z =uv at the point, , 3j + uk, , and at the point P,, r, , ar = 2'1+ 3'J + k ,, au, , =6i + 3j + 2k., , ar x ar = 9i - 2J -, , and, , au av, , ~: = 2i -, , 3j + 2k, , 12k, , The vector and rectangular equations of the nonnalline are, , (X - 6)i + (Y - 3)j + (Z- 2)k =k(9i - 2j - 12k), , or, , X-6+Y-3=Z-2, 9, -2, -12, The vector and rectangular equations of the tangent plane are, and, , m-. ar) =0, (au, av, , (R - r). -'- x -, , [(X - 6)i + (Y - 3)j + (Z - 2)k] • [9i - 2j - 12k] = 0, , or, , 9X-2Y-12Z-24=O, , and, , 4,, , :~, , (a) Find the direc~ional derivative off(x. y) = xl - 6y2 at the point (7, 2) in the direction, (b) Find the maximum value of the directional derivative at (7, 2)., , (a) Vf=, and, , (i tx + j! }Xl, , -6y2) =, , itx (x, , 2, , _6y2)+, , j! (x, , 2, , -6y2)= 2xi-12yj, , e = tn-.
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CHAPTER S3, , Vector Differentiation and Integration, , At (7,2), Vf= 141 - 24j, and, , Vf· a =(141 -24j)·, , (*I+*j, , )=7../2 -12../2 =-5../2, , .., , is the directional derivative., (b) At (7,2), with Vf = 141 - 24j, IV!I = ./142 + 242 = 2M is the maximum directional derivative. Since, .1\1>'-', , ~=, , :/kJI- :Ah, , j = icos8+ jsin8, , the direction is e = 300°15'. (See Problems 2 and 6 of Chapter 52.), S,, , (a) Find the directional derivative of F(x, y, z) = X- - 2y2 +4z2 at P(1, I, -1) in the direction a = 21 + j - k., (b) Find the maximum value of the directional derivative at P., Here, , and at (J, 1, -1), VF =21 - 4j - 8k., , =(2i - 4j - 8k) • (2i +j - k) = 8, (b) At P, IVFI = $4 =2.fif. The direction is a =2i - 4j - 8k., , (a) VF' a, , 6., , Given the surface F(x. y. z) =xl + 3xyz + 2yl- Z3 - 5 =0 and one of its points Po (1. I, 1). find (a) a unit normal, to the surface at Po; (b) the equations of the nonnalline at Po; and (c) the equation of the tangent plane at Po ., . Here, VF =(3x- + 3yz)1 + (3xz + 6y2)j + (3xy - 3z2)k, andatPo(l,l, 1), VF=61+9j., , VF =.Jf3, 2 1+ Jij, 3 j'IS a UnIt. normaI at Po; th e other --.:Ji312. -.:Ji3, 3 J.., (), a IVFI, (b) The equations of the normal line are X I = Y 3' I , Z =I., , 2, , (c) The equation of the tangent plane is 2(X -1) + 3(Y - 1) = 2X + 3Y - 5 = O., 7., , Find the angle of intersection of the surfaces, , ,, , at the point (2, I, -2)., We have, , ;~(~>, VF. = V(x- +1+ Z2 - 9) = 2xi + 2yj + 2zk, , and, , VF2 = V(X- + 2y2 - z - 8) =2xi + 4yj - k, , ,.~., , ~, , .;.... :., , ;i:~p{;, :~~;-};;,
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CHAPTER S3, , Vector Differentiation and Integration, , At t=O, r = i + t j +~ =0, from which c2 =- i -t j. Thus,, r=(e' -1)1+(te2t +tl-t)J+tt2k, , 11. Find the work done by a force F =(x + yz)1 + (y + xz)j + (z + xy)k in moving a particle from the origin 0 to, C(I, I, I), (a) along the straight line OC: (b) along the curve x = I, Y= fl, z = tl: and (c) along the straight lines, from 0 toA(I, 0, 0), A to B(I, 1,0), and B to C., , F· dr= [(x+yz)i + (y+xy)j +(z+xy)k]' [i dx+ j dy+k dz], = (x + yz) dx + (y + xz) dy + (z;t .l)') dz, , (a) Along the line OC, x = y = z and dx = dy = dz. The integral to be evaluated becomes, , W=, , f, , (1,I.1), , c (0.0,0), , F.dr=3 J(I,(x+x2 )dx= [( tx 2 +X3, 0, , )1 =t, 1, , (b) Along the given curve, x = 1and dx = dt: y = fl and dy = 21 dl: z = tl and dz = 3fl dl. At O. 1= 0: at C, t = 1., , Then, , (c) From 0 to A: y =z = 0 and dy = dz = 0, and x varies from 0 to 1., From A to B: x = 1. z = 0, dx = dz = 0, and y varies from 0 to 1., From B to C: x = y = I and dx = dy = 0, and z varies from 0 to I., Now, for the distance from 0 to A, WI = =, , ., , Jo(I xdx = t: for the distance from A to B, W2 = J(Ioydy =t: and for, , the distance from B to C, W3 = (I(Z+ I) dz=t. Thus, W= WI + W2+ W3= 1Jo, ., In general, the value of a line integral depends upon the path of integration. Here is an example of one which, does not, that is, one which is independent of the path. It can be shown that a line integral, , f (f.dx + hdy + hdz), , is independent of the pat~ if there exists a function l/i,.x, y, z) such that d; =t.dx +/2 dy + 13 dz. In this problem, the, integrand is, , (x+ yz) dx + (y+xz) dy + (z+xy) dz = d [t (xl + i+ r}+xyz], , 12. Find ds/dt and d 2s/dI 2• given (a) s = (I + I)i + (fl + t + l)j + (t3 + fl + t + l)k and (b) s = Ie' cos 21 + je' sin 2t + flk., , Ans., , (a) 1+ (21+ I)j + (3fl + 21+ I)k, 2j + (61 + 2)k;, (b) el(cos 21-2 sin 21)i + e'(sin 21 + 2 cos 21)j + 21k,, e'(-4 sin 21 - 3 cos 2t)1 + e'(-3 sin21 + 4 cos 2t)j + 2k
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CHAPTER 53, , Vector Differentiation and Integration, , 13. Given a = ui + u2j + u3k, b = i cos u + j sin u, and c = 3u2i - 4uk. First compute a' b, a x b, a • (b x c), and, a x (b x c), and find the derivative of each. Then find the derivatives using the fonnulas., Fin~ (a) the magnitudes of its velocity, and acceleration at time ( = I; (b) the components of velocity and acceleration at time (= 1 in the direction, a = 4i - 2j + 4k., , 14. A particle moves along the curve x = 3(2, Y = fl- 2t, Z = 13, where I is time., , AIlS., , (a) Ivl = 3$,lal = 2Jl9: (b) 6, 1f, , :;"".,'., , ,'<;" ., ~:,,~~~~,,~, .>;i~t':t, , ~t~~~, , 15. Using vector methods, find the equations of the tangent line and normal plane to the curves of Problem 15 of, Chapter 51., , 16. Solve Problem 16 of Chapter 51 using vector methods., 17. Show that the surfaces x = u. y = 5u - 3 v 2, Z = v and x = u, y = v, Z = -1411, 4 are perpendicular at P( 1, 2, 1)., U-II, , 18. Using vector methods, find the equations of the tangent plane and normal line to the surface:, (a) x =, (b) x =, AIlS., , = v, Z = uv at the point (II, v) = (3, ~ 4)., Y = v, Z = u2 - v 2 at the point (u. v) = (2, I)., , II,),, II,, , (a) 4X-3Y+Z-12=0, X-4 3 =, ( b) 4X - 2Y - Z - 3 =, , °' - 4, , Yr4=Z~l2, , X - 2 = Y - 1= Z - 3, , 2, , 1, , • 19. (a) Find the equations of the osculating and rectifying planes to the curve of Problem 2 at the given point., (b) Find the equations of the osculating, nonnal, and rectifying planes to x = 21 - fl, y = 12, Z = 2t + t2 at t = 1., '--<.,:, , ., , ~, , -, , AilS., , (a) 3X - 3Y+Z- 1 =0, llX + 8Y - 9Z-1O=0, (b) X+2Y-Z=O, Y+2Z-7=0,5X-2Y+Z-6=0, , ,..,-, , -~~~-:~,;' ., , ., , ;;~"., , 20. Show that the equation of the osculating plane to a space curve at P is given by, , d2r), , dr x - =0, (R-r). (lit dt 2, 21. Solve Problems 16 and 17 of Chapter 52, using vector methods., , 22. Find J:F(u) du, given, , (a) F(u) = u3i + (3u 2 - 2u)j + 3k; a = 0, b = 2; (b) F(u) = e"i + e-2Mj + uk; a = 0, b = 1, All.\'., , (a)4i+4j+6k;(h)(e-l)i+, , ~(I-e-2)j+~k, , 23. The acceleration of a particle at any time t is given by a'; dvldt = (t +l)i + 12j + (/ 2 - 2)k. If at t =0, the, displacement is r = 0 and the velocity is v = i - k, find v and r at any time t., AilS., , V, , =(tfl + I + I)i + ttJj + (tf - 2t -, , I)k; r, , =(ttl + tfl + t)i + -rrt4j + (*t 4 -, , 12 - t)k
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CHAPTER 53, , Vector Differentiation and Integration, , -., , 24. In each of the following, find the work done by the given force F in moving a particle from 0(0, 0, 0) to C( I, 1, 1), along (I) the straight line x = y = z, (2) the curve x = t, Y =fl, z =tl, and (3) the straight lines from 0 to A( 1,0,0),, A to B(I, 1,0), and B to C., (a) F =xi + 2yj;- 3xk., (b) F=(y+z)i+(x+z)J+(x+y)k., (c) F =(x +xyz)i +(y +xlz)j +(z +xly)k., Ans.· (a)3;(b)3;(c)t,H,t, , 25. If r =xi +yj + zk, show that (a) div r =3 and (b) curl r =O., 26. If/ =/(x, y, z) has partial derivatives of order at least two, show that (a) V x V/= 0; (b) V • (V xf) =, , a2 a2 el2), (c) V' V/= (dXf+ayr+dzf /., , O;f
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Double and Iterated Integrals, The Double Integral, Consider a function z = f(x, y) that is continuous on a bounded region R of the xy plane. Define a partition rtP of R by drawing a grid of horizontal and vertical lines. This divides the region into n subregions, RI , R2, ••• ,R" of areas ~IA, ~~, ... , ~"A, respectively. (See Fig. 54-1.) In each subregion, RkJ choose a, point Pk(Xk' Yk) and fonn the sum, , ", LJ(Xk'Yk)~kA = f(xI'YI)~IA + ... + f(x",y")~"A, , (54.1), , k=1, , Define the diameter of a subregion to be the greatest distance between any two points within or on its boundary, and denote by d~ the maximum diameter of the subregions. Suppose that we select partitions so that, d~ ~ 0 and II ~ +00. (In other words, we choose more and more subregions and we make their diameters, smaller and smaller.) Then the double illtegral of I(x, y) over R is defined as, (54.2), y, , -+------------------------x, , o, , Rg.54-1, This is not a genuine limit statement. What (54.2) really says is that II f(x,y)dA is a number such that, for, any E > 0, there exists a positive integer, , R, , 110, , such that, for any n ~ 110 and any partition with d~ < IIno' and any, , ", , corresponding approximating slim Lf(Xk'Yk)~kA, we have, k=1, , ", , Lf(Xk'Yk)~kA- IIf(x,y)dA < E, k=1, , R, , When z =f(x, y) is nonnegative on the region R, as in Fig. 54-2, the double integral (54.2) may be interpreted as a volume. Any tennftxk, y.)~.A of (54.1) gives t~e volume of a vertical column whose base is of, , -fl-
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CHAPTER 54 Double and Iterated Integrals, , 'l, , . area M and whose altitude is the distance =!(xv yJ measured along the vertical from the selected point, Pt(xt, yJ to the surface, =!(x. y). This. in tum, may be taken as an approximation of the volume of the, vertical column whose lower base is the subregion Rl and whose upper base is the projection of Rt on the, surface. Thus, (54.1) is an approximation of the volume "under the surface" (that is, the volume with lower, base R and upper base the surface cut off by moving a line parallel to the , axis along the boundary of R). It, is intuitively clear that (54.2) is the measure of this volume., The evaluation of even the simplest double integral by direct summation is usually very difficult., , Rg.54-2, , The Iterated Integral, Consider a volume defmed as above. and assume that the boundary of R is such that no line parallel to the x, axis or to the y axis cuts it in more than two points. Draw the tangent lines x =a and x =b to the boundary, with points of tangency K and L, and the tangent lines y =c and y =d with points of tangency M and N. (See, Fig. 54-3.) Let the equation of the plane arc LMK be y =gl(x), and that of the plane arc LNK be y = g2(X),, Divide the interval a ~ x ~ b into m subintervals hi' h2••••• hm of respeCtive lengths dlX, d~.••• dmX by, the insertion of points ~,. ~2' •••• ~"""I so that a =~ < ~I < ~ < ... < ~m-I < ~m =b. Similarly. divide the, interval c ~ y ~ d into n subintervals k" ~, ...• kn of respective lengths dIY, ~•...• d~ by the insertion, points 11,. 112' •.• , 11~, so that c = 110 < 111 < 112 < ... < 1J.-1< 11. =d. Let Il.. be the greatest d;X and let Jl.. be, the greatest dJy. Draw the parallel lines x =~I' X = ~2' ... ,X =~m-I and the parallel lines y =171' Y = 172' ..• ,, Y = 17~1t thus dividing the region R into a set of rectangles RjJ of areas djX dJY' plus a set of nonrectangles, along the boundary (whose areas will be small enough to be safely ignored). In each subinterval hi select a, point x =Xj and, in each subinterval kj select a point Y =YJ. thereby determining in each subregion Rij a point, PIj{xj , Yj)' With each subregion Rij associate, by means of the equation of the surface. a number zij =ft..x;, Yj)', and form the sum, , L, i.l.2~"jm, , !(Xpyj)djXdjy, , (54,3), , j.I,2,....., , Now, (54.3) is merely a special case of (54.1). So. if the number of rectangles is indefinitely increased, in such a manner that both Il.. ~ 0 and Jl.. ~ 0, the limit of (54.3) should be equal to the double integral, (54.2)., In effecting ,this limit, let us first choose one of the subintervals. say hj. and form the sum, , of the contributions of all rectangles having hj as one dimension. that is. the contributions of all rectangles, lying on the ith column. When n ~ +00. Jl.n ~ 0,
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CHAPTER 54, , Double and Iterated Integrals, , z, , y, , Rg.54-3, , Now summing over the, , 111, , columns and lelling /1/, , ~, , +00, we have, , (54.4), , Although we shall not use the brackets hereafter, it must be clearly understood that (54.4) caUs for the, evaluation of two simple definite integrals in a prescribed order: first. the integral off(x, y) with respect to y, (considering x as a constant) from y =g.(x), the lower boundary of R, to y =82(X), the upper boundary of R,, and then the integral of this result with respect,to x from the abscissa x =a of the leftmost point of R to the, abscissa x =b of the rightmost point of R.The integral (54.4) is called an iterated or repealed integral., It will be left as an exercise to sum first for the contributions of the rectangles lying in each row and then, over all the rows to obtain the equivalent iterated integral, , f, , d Jh2(y), , c, , . I(x,y)dxdy, , ~(y), , ,, , (54.5), , where x = h.(y) and x =h2(y) are the equations of the plane arcs MKN and MLN, respectively., In Problem 1, it is shown by a different procedure that the ite(ated integral (54.4) measures the volume, under discussion. For the evaluation of iterated integrals, see Problems 2 to 6., The principal difficulty in setting up the iterated integrals of the next several chapters will be that of, inserting the limits of integration to cover the region R. The discussion here assumed the simplest of regions:, more complex regions are considered in Problems 7 to 9., , SOLVED PROBLEMS, , 1., , Let z =f(x, y) be nonnegative and continuous over the region R of the xy plane whose boundary consists of the, arcs of two curves y = gl(x) and y =g2(X) intersecting at the points K and L, as in Fig. 51.-4. Find a fonnula for the, volume V under the surface z = f(x, y).
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CHAPTER 54 Double and Iterated Integrals, z, , y, , x, Fig, 54-4, ~.!., , Let the section of this volume cut by a plane x = Xi' where a < Xi < b, meet the boundary of R at the points, and T(xi, g2(Xi and let it meet the surface z =fix, y) in the arc UValong which Z =fixi, y). The area, of this section STUV is given by, , », , »,, , S(Xi' gl(Xi, , Thus, the areas of cross sections of the volume cut by planes parallel to the yz plane are known functions, A(x) =r(XI) f(x,y)dy of x, where x is the distance of the sectioning plane from the origin. By the cross-section, R,(XI), , formula of Chapter 30, the required volume is given by, , v = fb A(x)dx = Jb[Jg,(XI), f(x,y)dyJdx, B,(XI), o, , 0, , This is the iterated integral of (54.4)., In Problems 2-6, evaluate the integral on the left., , .3., , 4., , 2, J2J1Y (x+ y)dxdy= J2 [tx +xyPYdy =J2 6y2dy =[2yl]2 =14, 1, , Y, , 1, , Y, , 1, , 1JX'H xdydx =12 [xyt,+x dx =12, 2, , -I, , -I 2x'-2, , 2xl-2, , -1, , (Xl, , + x2 -, , 1, , 2Xl, , +2x)dx =t, , s., 6., , J, , r"/2 4COS 9, , Jo, , 2, , (JrI2[ 1, , p1dpd8= Jo, , '4 P4, , ]4COSB, 2, , (Jr/2, , d8= Jo (64cos 4 8-4)d8, , =[64( 38 + sin 8 + sin 48) _ 40J"12 -IOn, 8, , 7., , Evaluate, )', , 4, , 32, , 0, , -, , Jf dA, where R is the region in the first quadrant bounded by the semicubical parabola y2 = xl and the, , mey=x. R, , '
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CHAPTER 54, , Double and Iterated Integrals, , The line and parabola intersect in the points (0, 0) and (1, 1), which establish the extreme values of x and yon, the region R., Solution 1 (Fig. 54-5): Integrating first over a horizontal strip, that is, with respect to x from x y (the line), to x =y'll3 (the parabola), and then with respect to y from y =0 to Y 1, we get, , =, , =, , Rg.54-5, Solution 2 (Fig. 54-6): Integrating first over a vertical strip, that is, with respect to y from y =x;3fl (the, parabola) to y =x (the line), and then with respect to x from x =0 to x = 1, we obtain, , If dA= J~D"dydx = f~(X_X3/2)dx=[tx2 -t rS'2 n= f0R, , Rg.54-6, , 8., , Evaluate, , If dA where R is the region between y = 2x and y = xl lying to the left of x = 1., R, , Integrating first over the vertical strip (see Fig. 54-7), we have, , If dA =JJx:x dydx =S; (2x - x 2)dx = !, R, , Fig. 54-7
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••, , CHAPTER 54 Double and Iterated Integrals, , When horizontal strips are used (see Fig. 54-8), two iterated integrals are necessary. Let RI denote the part of, R lying below AB, and R2 the part above AB. Then, , IIdA= IIdA+ If dA =I~J,:dxdy+ fI~l2dxdy=-&+t=t, R, , II,, , R,, , Fig. 54-8, , 9., , Evaluate II x2dA where R is the region in the' first quadrant bounded by the hyperbola xy = 16 and the lines, R, , y = x, y = 0, and x = 8. (See Fig. 54-9.), , o, , x, , Rg.54-9, It is evident from Fig. 54-9 that R must be separated into two regions, and an iterated integral evaluated for, each. Let RI denote the part of R lying above the line y =2, and R2 the part below that line. Then, , As an exercise, you might separate R with the line x =4 and obtain, , JroI f33y ex2 dx dy by first reversing the order of integration., The given integral cannot be evaluated directly, since ex> dx is not an elementary function. The region R of, , 10. Evaluate, , f, , integration (see Fig. 54-10) is bounded by the lines x =3y, x =3, and y = O. To reverse the order of integration,, first integrate with respect to y from y =0 to y = x/3, and then with respect to x from x =0 to x =3. Thus., , 1o f3 ex> dxdy =I I, , 3x, , 1, , 3,, , 0 0, , /J eX1, , 13[ex2 y]x/3 dx, , dydx =, , 0, , 0
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Double and Iterated Integrals, , CHAPTER 54, y, , o, , x, , Rg.54-10, , ,~l, 'c', , I, , Cc'", , 11. Evaluate the iterated integral at the left:, (a), , J~r dxdy= I, , (b), , (c), , rr(X 2+ y2)dydx=Jf-, , (d), , (e), , fr'"0 xly2dxdv. = 4, , (f), , (g), , f~Io, , ", , .,, , ~iL~,', ,-:, ., , ., , ;~f~f~>, , xl, , xe1 dydx=te-l, , (i), , r, , (k), , 1'o,,41'..,01«0 p3COS, , an, , -', , O /2), , o, , 1, 2, , 0, , 0, , ""'8, , pdpd8=3, 2, , 8dpd8=*, , (h), (j), (I), , (x+ y)dxdy = 9, , 1'r xy2dydx=*, 1'r';(x+yl)dydx=i, rt' ydxdy=Jj, o, , xl, , ., , o •, 2 ,, , f'2r, rn 1'0--0, o, , o, , oP2cos8dpd8=t, pJ cos 28dpd8 = -it 1r, , 12. Using an iterated integral, evaluate each of the following double integrals. When feasible, evaluate the iterated, integrals in both orders., (a), (b), (c), (d), (e), , ~f:, , J., , I, , J: J:, , (f), , x over the region bounded by y =xl and y =.x3, y over the region of part (a), xl over the region bounded by y =x, y =lx, and x =2, lover each first-quadrant region bounded by 2y =xl, y = 3x, and x + y =4, y over the region above y = 0 bounded by yl = 4x and f =5 - x, , ~ over the region in the first quadrant bounded by xl =4 - 2y, 2y_)'2, , I, Ans. 1if, Ans. 1r, Ans. 4, Ans., Ans. 5, , t;,, , Ans. 4, , 13. In Problems II (a) to (h), reverse the o~d.:r of integration and evaluate the resulting iterated integral.
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Centroids and Moments of, Inertia of Plane Areas, Plane Area by Double Integration, If f(x. y) = I. the double integral of Chapter 54 becomes JI dA. In cubic units. this measures the volume of, , a cylinder of unit height; in square units. it measures the a:ea A of the region R., In polar coordinates, , A=, , If dA= f, , fJfP2(8), , R, , a, , pdpd8, , p!(9), , where 8 = a, 8 =P. p =PI (8), and p =Pz( 8) are chosen as boundaries of the region R., , Centroids, The centroid (x,y) of a plane region R is intuitively thought of in the following way. If R is supposed to, have a uniform unit density, and if R is supported from below at the point (x.y). then R will balance (that, is, R will not rotate at all)., To locate (x.y), first consider the vertical line x =x . If we divide R into subregions RI • . . . , R., of areas, ~IA .... , ~"A as in Chapter 54, and if we select points (xv yJ in each Rk• then the moment (rotational force), of Rk about the line x =x is approximately (Xi - x)dkA . So, the moment of R about x =x is approximately, , L•, , (Xi -, , x)dkA. Making the partition of R finer and finer, we get JI (x - x) dA as the moment of R about, , =X. In order to have no rotation about x =x. we must have If (x-:-x) dA =O. But, R, , ~, , X, , R, , JI(x-x)dA= JIxdA- JIxdA= JIxdA-xJIdA, R, , R, , R, , R, , R, , Hence, we must have JI x dA =x JI dA. Similarly. we get JI y dA =yJI dA. So, the centroid is determined by, the equations, R, R, R, R, , If x dA = xIf dA, R, , a'nd, , R, , Note that JI dA is equal to the area A of the region R., R, , If y dA =yIf dA, R, , R
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Centroids and Moments of Inertia, , CHAPTER 55, , Moments of Inertia, The moments of inertia of a plane region R with respecllo the coordinate axes are given by, , Ix =, , Hy2 dA, , and, , I)' =, , R, , Hx dA, 2, , R, , The polar moment of inel1ia (the lilOment of inertia with respect to a line through the origin and perpendicular, to the plane of the area) of a plane region R is given by, , JJ, , 10 =Ix +Iy = (X2+y2)dA, R, , SOLVED PROBLEMS, 1., , Find the area bounded by the parabola y =xl and the line y =2x + 3., Using vertical strips (see Fig. 55-0, we have, , 3f2>+3, , f3, , A = f-I.' dy dx = )2x + 3 - x 2 )dx =32/3 square units, , Rg.55-1, , 2., , =, , Find the area bounded by the parabolas y2 4 - x and y2 = 4 - 4x., Using horizontal strips (Fig. 55-2) and taking advantage of symmetry, we have, , =6, , f:, , (1- t, , r) dy = 8 square units, , Rg.55-2, , 3., , Find the area outside the circle p =2 and inside the cardiod p =2( I + cos 8)., Owing to symmetry (see Fig. 55-3), the required area is twice that swept over as Ovaries from 0= 0 to, O=tn. Thus,
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CHAPTER 55 Centroids and Moments of Inertia, , 12. Find /.. 1" and 10 for the area of the circle p, , =2(sin (J+ cos 8). (See Fig. 55-12.), , Since xl+y2 = p2,, , =4[t(J - cos2(J- tsin4(J]~~:.. ::= 61t' =3A, It is evident from Fig. 55-12 that /x =I,. Hence, /., , r:'~.:"::, , =/, =t /0 =t A., , ·~;:!:t, ..:~'{, , -f!'i¥', ;_r •..i', -~., , c,', , ,, , Fig. 55-12, , ,.'.,, , 13. Use double integration to find the area., , Bounded by 3x + 4y =24, x =0, y =0, Bounded by x + y =2, 2y =x + 4, y =0, Bounded by xl =4y, 8y =xl + 16, Within p =2(1 - cos 8), Bounded by p =tan (J sec (J and (J =rcJ3, (0 Outside p =4 and inside (J =8 cos (J, , Afls. 24 square units, Afls. 6 square units, Ans. -¥ square units, Ans. 61t square units, Ans. t.-!i square units, Ans. 8(i 1t' + J3) square units, , (a), (b), (c), (d), (e), , '~, , .. t, , 14. Locate the ~entroid of each of the following areas., (a), (b), (c), (d), (e), , The area of Problem 13(a), The first-quadrant area of Problem l3(c), The first-quadrant area bounded by y2 =6x, y =0, x =6, The area bounded by y2 =4x, xl =5 - 2y, x =0, The first-quadrant area bounded by xl - 8y + 4 =0, xl =4y, x =0, (f) The area of Problem 13(e), , Ans., Afls., Ans., Afls., Afls., Ans., , (t. 2), G,t), , Ans., , C61t'+6J3, 22 ), 21t'+3J3 '21t'+3J3, , (If,t), , (i,tJ), (t,t), , (tJ3,*), , ~, , .", .~,, , (g) The first-quadrant area of Problem 13(f), IS. Verify that, , tr, a, , [g;«(J) -, , r, , g~«(J)]d(J =, , ,, , r(8) pdp d(J, a .,(8), , :;~~'-- ,~, ." .., :, , ..., , =IfR dA; then infer that, , ,~, , . ..;, , ._', , It·~''1:. ~, , tw~~~,:", ,:;f'", , ··:tWi~,~, 'it:. :<,r:, ~, , If f(x,y)dA =If f(pCos(J,psin()p dp d(J, R, , ., , I, , 'ji;~t:;e, , R, , :~;~!':, , ",, , .~~~~.~., if;',,;;', ":~~H":;, , 16. Find /. and I, for each of the following areas:, , (a) The area of Problem 13(a), , Ans. /x =6k' y, I =-¥A, , (b) The area cut from y2 =8x by its latus rectum, , Ans., , 1.=1/-A: 1,=j,fA, , Ans., , I z =.l.A', 14, , IY =iA, , Ans., , / :r =~A', 70, , IY =iA, , (c) The, , area bounded by y =xl and y =x, , (d) The area bounded by y =4x - xl and y =x, , ~!;:~~~~, , .!""_'.J,, .:~t~t; ~, _'''Df, , ,.:~:o, , 3tr,i.;, ';-.';<
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CHAPTER 55, , 17. Find Ix and Iy for one loop of, , Ans. I, =(l~, , rr =cos 2 B., , -i)A; l,=(~ +i)A, , 18. Find 10 for (a) the loop of e= sin 2 Band (b) the area enclosed by, , Ans., , Centroids and Moments· of Inertia, , (a) fA ; (b), , e= I + cos e., ,, , itA, , 19. (a) Let the region R shown in Fig. 55-13 have area A and centroid (x. y) . If R is revolved about the x axis, show, that the volume Vof the resulting solid of revolution is equal to 21t'~ . (Hint: Use the method of cylindrical, shells.), (b) Prove the Theorem of Pappus: If d is the distance traveled by the centroid during the revolution (of part (a»),, show that V = Ad., (c) Prove that the volume of the torus generated by revolving the disk shown in Fig. 55-14 about the x axis is, 2n2a2b. (It is assumed that 0 < a < b.), y, , R, , b, , a, Fig. 55-13, , b, , x, , x, Rg.55-14
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Double Integration Applied to, Volume Under a Surface and the, Area of a Curved Surface, Let z = f(x, y) or z = f(P, fJ) define a surface., The volume V under the surface, that is, the volume of a vertical column whose upper base is in the surface, and whose lower base is in the xy plane, is given by the double integral, , v=JJzdA, , (56.1 ), , R, , where R is the region forming the lower base., The area S of the portion R* of the surface lying above the region R is given by the double integral, , IJ, , S=, , dz)2 +(az)2, ay dA, ( ax, , (56.2), , 1+, , If the surface is given by x =f(Y, z) and the region R lies in the yz plane. then, , IJ, , S=, , (ax)2, ay + dz dA, (ax)2, , (56.3), , I+, , If the surface is given by y =f(x, z) and the region R lies in the xz plane, then, (56.4), , SOLVED PROBLEMS, , ., , ,, , 1., , Find the volume in the first octant between the planes z =0 and z =x + y + 2, and inside the cylinder .r +y2 = 16., From Fig. 56-1, it is evident that z = x + y + 2 is to be integrated over a quadrant of the circle r + l = 16 in, the xy plane. Hence,, v=, , II zdA = f: foJl6:! (x + y + 2)dy dx = f: (x.J16 - x +8 -! x + 2.J16 - x )dx, 2, , R, , 2, , 2
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CHAPTER 56 Double Integration Applied to Volume, , x, , (4,0,0), , Fig. 56-1, '/",, , 2., , Find the volume bounded by the cylinder x2 + y2 =4 and the planes y + z =4 and z =O., From Fig. 56-2, it is evident that z =4 - Y is to be integrated over the circle r + y2 =4 in the xy plane. Hence,, , V=, , 3., , r.-::;-(4 - y)dx dy= 2 J2J,H (4- y)dxdy= 1611" cubic units, J2JH, -2 -,,4-,', , -2 0, , Find the volume bounded above by the paraboloid r + 4y2 =z, below by the plane z =0, and laterally by the, cylinders y2 = x and r = y. (See Fig. 56-3.), The required volume is obtained by integrating Z =r + 4y2 over the region R common to the parabolas f =x, and xI =y in the xy plane. Hence,, V=, , J~t~(x2+4l)dydx= J~[x2Y+4y3r dx=tcubicunits, , Fig. 56·2, , 4., , Fig. 56·3, , Find the volume of one of the wedges cut from the cylinder 4xI + f =a2 by the planes z = 0 and z =my. (See, Fig. 56-4.), The volume is obtained by integrating z = my over half the ellipse, 4xI + f =a2• Hence,, ,, , a12 J,J.'-4X', ial2 ~<' dx = rna, io 0 my dy dx = m 0 [y2 ]".2_, 0, 3, , V=2, , 4, , 3, , cubic units
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CHAPTER 56 Double Integration Applied to Volume, , Fig. 56-4, S., , Find the volume bounded by the paraboloid xl +y2 =4z, the cylinder xl +y2 =8y, and the plane z =O. (See, Fig. 56-5.), The required volume is obtained by integrating z=t<x2+ y2) over the circle xl + y2 =8y. Using cylindrical, coordinates (see Chapter 57), the volume is obtained by integrating z = t p2 over the circle p = 8 sin fJ. Then,, , V=, , If, R, , = 1~, 6., , flf f hnB, , zdA=JoJo, , f:, , I flf flolnB, , zpdpdO='4JoJo, , p1dpdO, , f:, , [p4 ~~n9 dO =256 sin 4 0 dO =961r cubic units, , Find the volume removed when a hole of radius a is bored through a sphere of radius 2a, the axis of the hole, being a di'ameter of the sphere. (See Fig. 56-6.), z, , z, , .., , ,-', , =::-:+---- y, 'i'i'f---y, , x, , Fig. 56-5, , Fig. 56-6, , From the figure, it is obvious that the required volume is eight times the volume in the first octant bounded, by the cylinder p2 =a2, tl1e sphere p2 + Z2 =4a 2, and the plane Z=O. The latter volume is obtained by integrating, 2- p2 over a quadrant of the circle p =a. Hence,, Z=, , J4a, , 7., , Derive formula (56.2)., Consider a region R' of area S on the surface z =f(x, y). Through the boundary of R' pass a vertical cylinder, (see Fig. 56-7) cutting thexy plane in the region R. Now divide R into n subregions RI , ••• ,Rn of areas M I , ••., , ,
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CHAPTER 56 Double Integration Applied to Volume, , ~A",, , and denote by ~sl' the area of the projection of ~A, on R'. In that ith subregion of R', choose a point P, and, draw there the tangent plane to the surface. Let the area of the projection of R, on this tangent plane be denoted, by ilI;. We shall use aT; as an approximation of the corresponding surface area as,.., , ,, , :':~0, t.. ", ~ ~, , ·.fr:t~, , Rg.56-7, , ,, Now the angle between the xy plane and the tangent plane at Pi is the angle "I, between the z axis with, , ~z~~, , direction numbers [0, 0, 1] and the normal, [ -, , ;(fi~~), , I], , ~ ,- ~, =[- ~~ ,- ~~, 1]. to the surface at p~ Thus,, , Then (see Fig. 56-8), , Rg.56-8, , Hence, an approximation of S is, , 8., , n, , ", , i.1, , i-I, , LilT; = Lsecr;M" and, , Find the area of the portion of the cone x 2+ y2 =3z1 1ying above the xy plane and inside the cylinder x2+ y2 =4)'., Solution 1: Refer to Fig. 56-9. The projection of the required area on the xy plane is the region R enclosed by, the circle xl + y2 = 4y. For the cone,, So, , 2 2, 1+(aZ)2, = 9z +X +y2 = 12~2 =1, ax +(az)2, dy, 9z, 9z, 3, 2, , .
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CHAPTER 56 Double Integration Applied to Volume, z, , Z2 + 6y, , = 36,, , x=o, , Fig. 56-13, , '., ~-., , "-' I ~ I ' :., , ':, , t, , ~., , ,, , '.", , ., , 12. Find the volume cut from 9xl + 4f + 36z =36 by the plane z =O., , Ans. 31t cubic units, 13. Find the volume under z =3x and above the first-quadrant area bounded by x =0, y =0, x =4, and xl + f, , Ans. 98 cubic units, 14. Find the volume in the first octant bounded by xl + Z =9, 3x + 4y = 24, x =0, y =0, and z = O., , Ans., , 1485116 cubic units, , 15. Find the volume in the first octant bounded by xy = 4z, y = x, and x = 4., , Ans. 8 cubic units, 16. Find the volume in the first octant bounded by xl +f, , Ans., , =25 and z =y., , 125/3 cubic units, , 17. Find the volume common to the cylinders xl +f = 16 and xl + z2 = 16., , Ans., , 1024/3, , cubic units, , 18. Find the volume in the first octant inside f + Z2 =9 and outside y2 = 3x., , Ans. 277t116 cubic units, 19. Find the volume in the first octant bounded by xl + Z2 = 16 and x - y =O., , Ans. 64/3 cubic units, , =25.
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CHAPTER 56 Double Integration Applied to Volume, 20. Find the volume in front of x =0 and common to f + t, Ans., , =4 and f + t + 2x = 16., , 281t cubic units, , 21. Find the volume inside p = 2 and outside the cone Z2 = p2., Ans., , 22. Find the volume inside y2 + t, Ans., , ,, , 321t/3 cubic units, , =2 and outside xl - f - t =2., , 8n(4 - .fi)/3 cubic units, , 23. Find the volume COllllllon to p2 + Z2 =a2 and p, Ans., , 2(31t -4)a2/9 cubic units, , 24. Find the volume inside xl + f, , r~>'.'i'.> '., , Ans., , =a sin B., , =9; bounded below by xl +f + 4z = 16 and above by z =4., , 817tl8 cubic units, , t \', , •, ~~:, , "., , 25. Find the volume cut from the paraboloid 4xl + f = 4z by the plane z - y = 2., , .', , Ans., , 91t cubic units, , 26. Find the volume generated by revolving the cardiod p = 2( 1 - cos 8) about the polar axis., Ans., , V =2n, , If yp dpd8 =64n/3 cubic units, , 27. Find the volume generated by revolving a petal of p = sin 2B about either axis., Ans., , 327t1105 cubic units, , 28. Find the area of the portion of the cone xl +yl = Z2 inside the vertical prism whose base is the triangle bounded by, the lines y =x, x =0, and y = 1 in the xy plane., Ans., , t.fi square units, , 29. Find the area of the portion of the plane x + y + z = 6 inside the cylinder x2 + f = 4., , ;., , ", , t, , -, , Ans., , 4fjn square units, , 30. Find the area of the portion of the sphere xl + f + Z2 =36 inside the cylinder xl + y2 =6y., Ans., , n(1t - 2) square units, , 31. Find the area of the portion of the sphere xl + f + Z2 = 4z inside the paraboloid xl + f = z., Ans., , 41t square units
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CHAPTER 56 Double Integration Applied to Voluf!'e, 32. Find the area of the portion of the sphere r, , +y2 +t =25 between the planes z =2 and z =4., , Ans. 20lt square units', 33. Find the area of the portion of the surface z = xy inside the cylinder r +y2 = 1., Ans., , 2,,(2./2 -1)/3 square units, , 34. Find the area of the surface of the cone r +y2 - 9t =0 above the plane z =0 and inside the cylinder r + y2 =6y., Ans., , 3M" square units, , 35. Find the area of that part of the sphere r + y2 + Z2 = 25 that is within the elliptic cylinder 2i'- + yl = 25., Ans., , 50lt square units, , 36. Find the area of the surface of r + y2 - az = 0 which lies directly above the lemniscate 4p2 = a2 cos 2()., Ans., , S =~, , JJ J4 p +a1p dp dO =;2 (i -t) square units, 2, , 37. Find the area of the surface of r +y2 + t, , =4 which lies directly above the cardioid p =1 - cos ()., , Ans. 8[" -./2 .... In(./2 + 1)] square units
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Triple Integrals, Cylindrical and Spherical Coordinates, Assume that a point P has coordinates (x, y, z) in a right-handed rectangular coordinate system. The corresponding cylindrical coordinates of P are (r, 8, z), where (r., 8) are polar coordinates for the point (x, y), in the xy plane. (Note the notational change here from (p, 8) to (r, 8) for the polar coordinates of (x. y): see, Fig. 57-1.) Hence. we have the relations, x, , = rcose,, , tane = I, x, , y=rsine,, , In cylindrical coordinates, an equation r = c represents a right cylinder of radius c with the z axis as its axis, of symmetry. An equation 8= crepresents a plane through the z axis., A point P with rectangular coordinates (x. y, z) has the spherical coordinates (p, 8, iP), where p = lOP!,, eis the same as in cylindrical coordinates, and iP is the directed angle from the positive z axis to the vector OP., (See Fig. 57-2.) In spherical coordinates, an equation p = c represents a sphere of radius c with center at the, origin. An equation iP =c represents a cone with vertex at the origin and the z axis as its axis of symmetry., z, z, , p(r, 6, z), , 11, , 1, , ~--~I~-------- Y, , 'A------+------ y, , I, ~,I, , 6', x, , 'I, , -----~, x, , Y, , Fig. 57·2, , Rg.57-1, , The following additional relations, easily deduced from Fig. 57-2 and the equations above, hold among, spherical. cylindrical, and rectangular coordinates:, r=, , psin iP,, , x = p sin iP cos e,, , y = p sin iP sh, , e,, , z =p cos iP
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CHAPTER 57 Triple Integrals, , The Triple Integral, Letf(x, y, z) be a continuous function on a three-dimensional region R. The definition of the double integral, , can be extended in an obvious way to obtain the definitie;m of the triple integral, , IfIf(x, y, z) dV., R, , Iff(x, y, z) =1, then, , IfIf(x, y, z) dV may be interpreted as measuring the volume of the region R., R, , Evaluation of Triple Integrals, As in the case of double integrals, a triple integral can be evaluated in terms of iterated integrals., In rectangular coordinates,, , fIf f(x, y, z) dV =I lYle..), , Itl("'Y), , b, , a, , R, , ,,(x), , l,eX.Y), , Ill, =Jcd IXzCy), x,Cy), , ex •y ), , ll(X.)'), , f(x, y, z). dz dy dx, •, , f(x, y, z) dz dx dy, etc., , where the limits of integration are chosen to cover the region R., In cylindrical coordinates,, fJ fr2(s) f ll (r.8), , fIf f(r. 8. z) dV =f, R, , a, , r,(8), , l,(r.81, , f(r, 8. z)r dz dr d8, , where the limits of integration are chosen to cover the region R. (See Problem 23.), In spherical coordinates,, , iIf f(p,, R, , ~, 8) dV, , Pz (;.8), •, =JafJ JM6) fpM.8), f(p, ~, 8)p2 sm ~ dp d~ d8, ~(s), , where the limits of integration are chosen tocovei the region R. (See Problem 24.), Discussion of the definitions: Consider the functionf(x, y, z), continuous over a region R of ordinary, space. After slicing R with planes x =;1 and y =7]. as in Chapter 54, let these subregions be further sliced by, planes z = ~. The region R has now been separat~d into a number of rectangular parallelepipeds of volume, .1~jk = .1XI.1Yj.1Zk and a number of partial parallelepipeds which we shall ignore. In each complete parallelepiped. select a point Pijk(x,., yp Zk); then compute f(x;, Yj' Zk) and form the sum, , I, , f(x j • Yj' z.).1Vijk, , =I, , I~I."•.,, , I~I.",..., J~I,,,, ••, , j=I.",n, , f(xj! Y)' Zk).1X/.1y j .1Zk, , (57.1), , k=I., ..p, , k31 ....p, , The triple integral of f(x, y, z) over the region R is defined to be the limit of (57.1) as the number of parallelepipeds is indefinitely increased in such a manner that all dimensions of each go to zero., In evaluating this limit, we may sum first each set of parallelepipeds having .1 j X and .1/. for fixed i andj,, as two dimensions and consider the limit as each .1kz ~ O. We have, , ,, , Now these are the columns, the basic subregions, of Chapter 54; hence,, , ~~'!, , L, , .-..... 1.1......, 1'-....., ..., j~I, , ,n, , k=I ...p, , f(x j , Yl' Zk).1V;jk =, , Iff f(x, Y, z)dz dx dy = Iff f(x, y, z)dz dy dx, R, , R, ,, , .
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CHAPTER 57 Triple Integrals, , '~, , ;;~JR~~,t:, .., ;t;l~, , ,,/:=:_, , T>, , --~-+;1:.', -,',",,'0, , '. ~ :1~'~iFi~·, , -.-;:-~~:., , Fig. 57·3, , 3., , Compute the triple integral of f(r, (J, z) =r over the region R bounded by the paraboloid r = 9 - z and the plane, , z =O. (See Fig. 57-4.), , :- .;,~;. ~~t~'i, , ';%4!J);, '~,, , .; .~ ";'.: '!., ~.'-:~ ... "~, , -<, , ~)fMi;, x, , Fig. 57-4, , Integrate first with respect to z from z =0 to Z =9 with respect to (J from (J= 0 to 8= 21t. This yields, , r, then with respect to r from r =0 to r =3, and finally, , r, , Iff r 2dV = J: J;,z r2(r dz dr dO) =J:-J: ,-3(9 - r2)dr dO, R, , 4., , io4i.flkl i4, , Show that the following integrals give the same volume: (a) 4, and (c), , 4(4J4, Jo, , )"14, , (.f4HT dx dz ely., , Jo, , 0, , (,'+r)/4, , dz ely dt, (b), , i4 i2 i~ ely dtdz:, ,[i, , 0 0, , 0
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CHAPTER 57 Triple Integrals, (a) Here z ranges from z =t(x 2 + )'2) to Z =4, that is, the volume is bounded below by the paraboloid, 4z = x2 + y2 and above by the plane z = 4. The ranges of y and x cover a quadrant of the circle x2 + y2 = 16, z, = 0, the projection of the curve of intersection of the paraboloid and the plane z = 4 on the xy plane. Thus,, the integral gives the volume cut from the paraboloid by the plane Z=4., (b) Here y ranges from y = 0 to Y =,j4z - x 2 , that is, the volume is bounded on the left by the xz plane and on, the right by the paraboloid y2 = 4z The ranges of x and z cover one-half the area cut from the parabola, x 2 = 4z, Y = 0, the curve of intersection of the paraboloid and the xz plane, by the plane z =4'. The region R is, that of (a)., (c) Here the volume is bounded behind by the yz plane and in front by the paraboloid 4z = x2 + yl. The ranges, of z and y cover one-half the area cut from the parabola 1 = 4z, x = 0, the curve of intersection of the, paraboloid and the yz plane, by the plane z =4. The region R is that of (a)., , r., , S., , Compute the triple integral of F(p. 0, tfJ) = 1/P over the region R in the first octant bounded by the cones tfJ = ~, and tfJ = tan-I 2 and the sphere p =J6. (See Fig. 57-5.), , y, , j:4~1, '.", , Fig. 57-5, Integrate first with respect to p from p= 0 to p=$, then with respect to ~ from, finally with respect to ofrom 0 to~. This yields, , 1, r J, fIJ -dV=, p, Jo, , 1112 lJIn-'21 J6, 1<14, , R, , 1112 J","o'2, , = 3Jo, , ,,14, , 0, , .= t ,=, to, , tan-I 2, and, , 1, _p2sin,dpdtfJdO, p', -, , sin,dtfJ dO, /, , 6., , Find the volume bounded by the paraboloid z = 2x2 +1 and the cylinder z = 4 -I. (See Fig. 57-6.), Integrate first with respect to z from z =2x2 +1 to Z =4 - y, then with respect to y from y =0 to y = ../2 - x2, (obtain x 2 + i = 2 by eliminating x between the equations of the two surfaces), and finally with respect to x from, x = 0 to x =.fi (obtained by setting y = 0 in x2 + y2 = 2) to obtain one-fourth of the required volume. Thus,
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CHAPTER 57 Triple Integrals, , Fig. 57-6, 7., , Find the volume within the cylinder r =4 cos 0 bounded above by the sphere r + t, z =O. (See Fig. 57-7.), , = 16 and below by the plane, , z, , y, , Fig. 57-7, Integrate first with respect to z from z =0 to Z =-/16 - r2 , then with respect to r from r =0 to r =4 cos 0, and, finally with respect to 8 from 0= 0 to 0= 1t to obtain the required volume. Thus,, (Ir, , (4 .... 9, , V = Jo Jo, , =- 6f, 8., , J:, , (~, , Jo, , (n, , (4<018, , r dz dy d8 =Jo Jo, , ~, , rvl6 - r2 drd8, , (sin 3 8 -1)d8 = ~ (31r - 4) cubic units, , Find the coordinates of the centroid of the volume within the cylinder r =2 cos 0, bounded above by the, paraboloid l= r and below by the plane z =O. (See Fig. 57-8.), (K12 r2eo08, , ("'2 (200.9 (,1, , V =2Jo Jo, , ·M,., , Jo rdzdrdO =2Jo Jo, , (KI2 (2e0l9 (", , =III xdV =2Jo, /I, , Jo, , r2 drd(), , Jo (rcos())rdzdrd()
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'CHAPTER 57 Triple Integrals, z, , ,, , x, , y, , Fig. 57-8, Then, , :x = M ,./V = t. By symmetry. y = 0; Also., Mry =, , 2c059 f", f"l2 f 2oos 9, Jo zr dz dr dO= Ju Jo r drdO, , IfIzdV =2Jof"'2 Jfo, R, , and Z = M ",.IV = JB-. Thus, the centroid has coordinates (t. 0, JB-)., , 9., , For the right circular cone of radius a and height h, find (a) the centroid: (b) the moment of inertia with respect to, its axis: (c) the moment of inertia with respect to any line through its vertex and perpendicular to its axis: (d) the, moment of inertia with respect to any line through its centroid and perpendicular to its axis: and (e) the moment, of inertia with respect to any diameter of its base., Take the cone as in Fig. 57-9. so that its equation is r = Z, Then:, , X, , V=4, , ., 1"'21"(hr--rl, h) drdO., l"l2j" fhlor'. rdzdrdO=4, o, , 0, , 0, , (0.0. h), , 0, , a, , d, , (0.0, JAh), , c, , r.O) y, , (r. O.0), , x, , Rg.57-9
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CHAPTER 57, , Triple Integrals, , 11. Locate the centroid of the volume cut from one nappe of a cone of vertex angle 60° by a sphere of radius 2 whose, center is at the vertex of the cone., Take the surface as in Fig. 57-11, so that x =y =0. In spherical coordinates, the equation of the cone is, I/J =Ir 16, and the equation of the sphere is 'p =2. Then, , ,, "t' ,, , ". i, , .. ~,~~\,;, , r'r12, , rJfl6, , = 8 Jo Jo sin 21/J di/J dO = Ir, , -"'I, , z, , y, , x, , \, \, , Rg.57-11, , ...., , ,, , 12. Find the moment of inertia with respect to the z axis of the volume of Problem II., , 12, , = 128 r.. r"\inJI/J di/J dO= 128(1_1$)r· dO= 81r (16-9$)= 5-2$ V, 5 Jo Jo, 5 3 8, Jo 15, 5, f2, , 13. Describe the curve detennined by each of the following pairs of equations in cylindrical coordinates:, (a) r = 1. z = 2; (b) r =2, z = (J; (c) 0= 7tl4, r =.,fi; (d) (J~ 7tl4, z = r., , Ans., , (a) circle of radius I in plane z = 2 with center having.rectangular coordinates (0, 0, 2); (b) helix on right, circular cylinder r = 2; (c) vertical line through point having rectangular coordinates (I, 1,0); (d) line, through origin in plane (J = 7tl4, making an angle of 45° with xy plane
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CHAPTER 57 Triple Integrals, , 14. Describe the curve determined by each of the following pairs of equations in spherical coordinates:, 1C, 1C, 2"'1C, (a)p= I. O=lt;(b) e='4"=6";(c)p=, • .,,='4', , Ans., , (a) circle of mdius 1 in Xl. plane with center at origin; (b) halfline on intersection of plane 0 = 7tl4 and, cone ,= n/6; (c) circle of radius J2 in plane z= J2 with center on z axis, , 15. Transfonn each of the following equations in either rectangular. cylindrical, or spherical coordinates into, equivalent equations in the two other coordinate systems:, (a) p = 5; (b) Z2 = r; (c) r +y2 + (z -1)2 = I, Ans., , ia-;r,),·~.,''':, , J, , ~~'.;i:, , (a)r+f+z2=25, r+z2=25; (b) z2 =xl+f. cos ,=t (that is. ;=n/4 or ~=3n/4); (c) r+z2=2z,, p=2cos;, 2, , 16. Evaluate the triple integml on the left in each of the following:, , (a), (b), (c), , J: dz dxdy=, , J~f, , I, , frr dzdydx= 24I, J,6 f 2-2, J,4-2Y xdz dx dy = 144, o ..., , 0, , /J-X/J, , o, , 0, , 0, , [ r t'l2t, =, , o, , (d), , r2rJ,~ (16 - r2 )II217.drdO =-n, 256, o 0 0, 5, , (e), , r·rf, o, , 0, , 0, , 0, , :.'",, , 21 3 3, / -./, , xdzdydx, , ], , o·p4sin~dpd~de=2500n, , 17. Evaluate the integral of Problem J6(b) after changing the order to dz dx dy., , 18. Evalute the integml of Problem J6( c). changing the order to dx dy dz and to dy dz dx., , 19. Find the following volumes. using integrals in rectangular coordinates:, (a) Inside r + f = 9. above z = O. and below x + z = 4, Am., (b) Bounded by the coordinate planes and 6x + 4y + 3z = 12, Ans., (c) Inside xl + f =4x. above z = 0, and below xl + f = 4z, Ans., , 361t cubic units, 4 cubic units, 67t cubic units, , 20. Find the following volumes. using triple integrals in cylindrical coordinates:, (a) The volume of Problem 4., (b) The volume of Problem J9(c)., (c) That inside r = 16, above z = O. and below 2z = y, , Ans., , 64/3 cubic units, , Ans., Am., , (3.t.f), , (c) The first-octant volume of Problem 19(a), , Ans., , (d) That of Problem 19( c), (e) That of Problem 20(c), , Ans., Ans., , 64-91C, 23, 73n-128), ( 16(n -I)' 8(n - I), 32(n -I), (t, 0.-11-), (0, 37t14, 37t116), , 21. Find the centroid of each of the following volumes:, (a) Under Z2 = xy and above the triangle y = x. y =O., x = 4 in the plane z = 0, (b) That of Problem 19(b), , <t.t.I)
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.T~iple, , CHAPTER 57, , Integrals, , 22. Find the moments of inertia I~. I y • I z of the following volumes:, (a) That of Problem 4, , Ans., , ! ~ =!1 =.3l.V·, I % =l6.V, T, T, , (b) That of Problem 19(b), , Ans., , (c) That of Problem 19(c), , Ans., , =tv-" ! =2V''r:l1f, I =Jl V, ! x =.1iV'!, mV', T I I " '=, yI, T . ! z =.&!.V, 9, , (d) That cut from;: = r2 by the plane;: = 2, , AT/s., , I \", , t, , I~, , =!1 =tv'!, -tv, 'z.-"', , 23. Show that. in cylindrical coordinates. the triple integral of a functionj{r. B. z) over a region R may be represented by, ll, , f, , a, , J,,(B) JZ'«,II) f(r. B, z)r liz dr dB, ,,(B), , :.(,,8), , [Hint: Consider. in Fig. 57-12. a representative subregion of R bounded by two cylinders having the z axis as, axis and of radii rand r + Art respectively. cut by two horizontal planes through (0, 0, z) and (0, O. z + Az)., respectively, and by two vertical planes through the z llJt,.is making angles Band B+ AB. respectively. with the xz, plane. Take IlV = (r MJ) Ar Az as an approximation of its volume.], , y, , x, , Fig. 57-12, 24. Show that, in spherical coordinates, the triple integral of a function !(p, 1ft. ') over a region R may be represented by, II, , Ia, , I"(B) JP,(•.81!(p.l/J. B)pl sm,, •, dp dl/J dB, fI(B), , 1'1( ••8), , [Hint: Consider. in Fig. 57-13. a representative subregion of R bounded by two spheres centered at O. of radii, p and p + Ap. respectively. by two cones having 0 as vertex. the z axis as axis. and semiverticaI angles l/Jand, q,+ 1l4J. respectively. and by two vertical planes through the z axis making angles Band B+ /lB. respectively., with the yz plane. Take AV =(pAl/J)(psinq,1l8)(llp) = pl sinq,ApAl/JA8 as an approximation of its volume.], 25. Change the following points from rectangular to cylindrical coordinates: (a) (1, 0, 0); (b) (fi. fit 2);, (c) (-$,1,5)., , Ans., , (a) (1, 0, 0); (b), , (2, ll); (c) (2, 5:,5)
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Triple Integrals, , CHAPTER 57, , J, , x, , Fig. 57·13, 26. Change the following points from cylindrical to rect~gular coordinates: (a) (5, ~. 1); (b) (2, Ans., , (a) (, , l5f 'I): (b)(v'3. -I, 0);, , ~, 0); (c) (0. 7, 1)., , (c)(O, 0, 1), , 27. Change the following points from rectangular to spherical coordinates: (a) (I, 0, 0); (b) (.[i,.[i, 2);, (c)(l. -I,-.[i)., Ans., , (a), , (I, 0, ~); (b) (2.[i. l t)= (c) (~, 7:. 3:), , 28. Change the following points from spherical to rectangular coordinates: (a)(l, O. 0); (b) (2, 0, 1t); (c), Ans., , (4, t. ~)., , (a) (0, O. I); (b) (0. O. -2): (c) (Ii,.[i, 2 v'3), , 29. Describe the surfaces determined by the following equations:, , -r; (f) 0 = t; (g) p = 2sinq,, , . (a) z = r; (b) r= 4 cos 0; (c) pcosq, =4; (d) psinq, = 4; (e) q, =, AlIS., , (a) circular paraboloid; (b) right circular cylinder (x - 2)2 +Y =4; (c) plane z = 4; (d) right circular, cylinder r + f = 16; (e) the xy plane; (f) right circular cone with the z axis as its axis; (g) right circular, cylinder r + f =4, , ,
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j, , Masses of Variable Density, Homogeneous masses can be treated as geometric figures with density 0 = 1. The mass of a homogeneous, body of volume V and density ois m = oV., For a non homogenous mass whose density 0 varies continuously, an element of mass dm is given by, (1), (2), (3), , O(x, y) ds for a planar material curve (~.g., a pieco of fine wire);, o(x, y) dA for a material two-dimensional plate (e.g., a thin sheet of metal);, O(x, y, z) dV for a material body., , The center of mass (x, y) of a planar plate that is distributed over a region R with density O(x, y) is determined by the equations, , mx = My, , and, , my = Mx', , where, , My, , =, , H, , O(x,y)xdA, , R, , and, , Mx ~, , H, , O(x,y)ydA, , R, , An analogous result holds for the center of mass of a three-dimensional body. The reasoning is similar to, that for centroids in Chapter 55., The moments of inertia of a planar mass with respect to the x axis and the y axis are Ix = o(x, y)y2 dA, , H, , = If o(x,y)x 2 dA. Similar fonnulas with triple integrals hold for three-dimensional bodies. (For exR, , and I,, , R, , ample, Ix, , =, , HI, , O(x, y, Z)(y2, , + Z2) dA.), , R, , SOLVED PROBLEMS, , 1., , Find the mass of a semicircular wire whose density varies as the distance from the diameter joining the ends., Take the wire as in Fig. 58-1, so that O(x, y) =kyo Then, fromr +y2 = r2., , ~l +( t Ydx=~dx, . m = ft5(x,y) ds =J' kyLdx =krf' dx =2kr2 units, -, y, -,, ds=, , and, , y, , p(x,y), , ~~----~r---~~~-x, , (-r,O), , 0, , Fig. 58-1
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CHAPTER 58 Masses of Variable Density, , 2., , Find the mass of a square plate of side a if the density varies as the square of the distance from a vertex., Take the square as in Fig. 58-2, and let the vertex from which distances are measured be at the origin. Then, 6(x, y) =k(r +y2) and, , f: f: k(x 2+ y2) dx dy=kf: (ta +af) dy =tka units, , If, , m = 6(x, y)dA =, , 4, , 3, , R, , y, , x, , Rg. SS:2, , 3., , Find the mass of a circular plate of radius r if the density varies as the square of the distance from a point on the, circumference., Take the circle as in Fig. 58-3 and let A(r. 0) be the fixed point on the circumference.. Then 6(x, y) =, k[(x - r)2 + f] and, , f', , m = 6(x,y) dA ;;;; 2 _, f.~, 0, k[(x - r)2 + y2] dy dx =t klfr4 unIts, , If, R, , y, , A(r.O), , x, , Rg.58-3, , 4., , Find the center of mass of a plate in the form of the segments cut from the parabola f = 8x by its latus rectum, x =2 if the density varies as the distance from the latus rectum. (See Fig. 58-4.), , Here, 6(x.,y) = 2 - x and, by symmetry, Y= O. For the upper half of the plate., , My, , 'f.4[4 t, y6], 128, = R 6(x,y)xdA= f.04f2yl/(2-x)xdxdy=k, 0 3'- 64 + (24)(64) dY=3fk, , If, , and x =M, 1m = t. The center of mass has coordinates (t, 0).
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CHAPTER 58, , Masses of Variable Density, , 4), , x, , J, , Fig. 58-4, 5., , Find the center of mass of a plate in the form of the upper half of the cardioid r = 2( 1 + cos 0) if the density, varies as the distance from the pole. (See Fig. 58-5.), , r" r2(1+cos9) (kr)r dr dO ':= tk Jr" (1 + cos 0)3 dO = "kn, , Jf 15(r, 0) cIA = Jo Jo, 'Jf, r, Mx = 15{r, O)y cIA = Jo, m=, , o, , R, , lr, , J,2(1+<O'0), 0, , (kr)(rsinO)r dr dO, , R, , = 4k Jr" (I + COSO)4 sinO dO =1¥k, ., o, , My =, , r" r2(1+ros8) (kr)(rcosO)r dr dO = 14kn, , If 15(r, O)x cIA = J, 'J,, o, , R, , Then x = My, m, , = 21, Y= Mx = 96, 10, , m, , 25n, , 0, , ,, , ,and the center of mass has coordinates (2101 296 )., ' 5n, , x, · ;';, , :"'.;."',:i;, , ", , Fig. 58-5, , 6., , Find the moment of inertia with respect to the x axis of the plate having for edges one arch of the curve y = sin x, and the x axi.s :f its density varies as the distance from the x axis. (See Fig. 58-6.), , r" r,jnx ky dy dx = tk Jr" sin 2 x dx =tkn, , m = 15(x,y) cIA =Jo Jo, , Jf, , o, , R, , y, , ""T----X, , I-lg.58-6
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••, , CHAPTER 58 Masses of Variable Density, , 7., , Find the mass of a sphere of radius a if the density varies inversely as the square of the distance from the center., k, _ k, Take the sphere as in Fig. 58-7. Then o(x,y,z) X2 + y2 + Z2 - p2 and, , m=, , IIf O(x,y,z)dV=8f, , KI2 rfl2, , 0, , Jo, , f" pzp2sm,dpd,dO, k, ., 0, , R, , (KI2 (.12, , (lfl2, , = 8ka Jo Jo sin 4' d4' dO =8ka Jo dO =4k1ra' units, z, , ,,',, , y, ~, , "", , , -'-, , ;,~, , , ',',, , .., , ,~,, , Rg.58-7, , 8., , Find the center of mass of a right circular cylinder of radius a and height h if the density varies as the distance, from the base., Take the cylinder as in Fig. 58-8, so that its equation is r = a and the volume in question is that part of the, cylinder between the planes z = and z =h. Clearly, the center of mass lies on the z axis. Then, , °, , m=, , flf, R, , r·/2 r" rh, , (.,2, , r", , O(z,r,O) dV = 4Jo Jo Jo(fez)r dz dr dO = 2kh2 Jo Jor dr dO, I,'"~, , ", , IIf 8(Z,r,O)z dV= 4Jor,,/2 Jr"oJor (kz 2)r dz dr dO = tkh Jor"'2 Jor r dr dO, h, , M.., =, , a, , J, , R, , and Z =M xylm =tho Thus the center of mass has coordinates (0, 0, th)., z, , y, , Fig. 58-8, , -
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CHAPTER 58, , 9., , Masses of Variable Density, , Find the mass of, (a) A straight rod of length a whose density varies as the square of the distance from one end, Ails., , tka l, , units, , (b) A plate in the fonn of a right triangle with legs a and b, if the density varies as the sum of the distance from, the legs, Ans., , tkab(a+b)units, , (c) A circular plate of radius II whose density varies as the distance from the center, , Ans., , tka 31t' units, , (d) A pla~e in the form of an ellipse b2x2 + a'y'- = Q1b2, if the density varies as the sum of the distances from its axes, AilS., , tkab(a + b) units, , (e) A circular cylinder of height b and radius of base a, if the density varies as the square of the distance from, its axis, , (f) A sphere of radius a whose denliity varies as the distance from a fixed diametral plane, , (g) A circular cone of height b and radius of base a whose density varies as the distance from its axis, , (h) A spherical surface whose density varies as the distance from a fixed diametral plane, AilS., ':, , ~"!~' n•, , i{t;·';'t, , 2kal rt units, , 10. Find the center of mass, , J", , (a) One quadrant of the plate of Problem 9(c}, , Ans., , (3a12rt, 3a12rt), , (b) One quadrant of a circular plate of radius a, if the density varies as the distance from a bounding radius, (the x axis), , Ans., , (3aJ8,3an:116), , (c) A cube of edge a. if the density varies as the sum of the distances from three adjacent edges (on the, coordinate axes), , Ans., , (5aI9. 5a19. 5a19)
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CHAPTER 58, , Masses of Variable Density, , (d) An octant of a sphere of radius a, if the density varies as the distance from one of the plane faces, Ans., , (l6al151t, 16a1157t, 8al15), , (e) A right circular cone of height b and radius of base a, if the density varies as the distance from its base, Ans. . (0,0, 2b15), , 11. Find the moment of inertia of:, (a) A square plate of side a with respect to a side, if the density varies as the square of the distance from an, extremity of that side, , (b) A plate in the form of a circle of radius a with respect to its center, if the density varies as the square of the, distance from the center, , (c) A cube of edge a with respect to an edge, if the density varies as the square of the distance from one, extremity of that edge, , (d) A right circular cone of height b and radius of base II with respect to its axis. if the density varies as the, distance from the axis, , (e) The cone of (d), if the density varies as the distance from the base, , ,
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Differential Equations of First, and Second Order, A differential equatipn is an equation that involves a functiOJ1, say y, of one variable, say x, and derivatives, , ~ + 2 ~ +3y-7sinx+4x =0, equation also can be written as y" + 2y' + 3y - 7 sin x + 4x = o., ofy or differentials ofx andy. Examples are, , and dy= (x+ 2y) dx. The first, , The order of a differential equation is the order of the derivative of highest order appearing in it. The first, of the above equations is of order two, and the second is of order one., A solution of a differential equation is a function y that satisfies the equation. A general solution of an, equation is a fonnula that describes all solutions of the equation. It turns out that a gener:al solution of a differential equation of order 11 will contain n arbitrary constants., , Separable Differential Equations, A separable differential equation is a first-order equation that can be put in the fonn, · h', , f( x ) d r+g (), 0, y d y=,, , W hIC, , . I, , dy, , f(x), , IseqUlvaentto dx=- g(y), , A separable equation·can be solved by taking antiderivatives, , Jf(x) dx+ Jg(y) dy =, , C, , The result is an equation involving x and y that detennines y as a function of x. (See Problems 4-6, and for, justification, see Problem 61.), , Homogeneous Functions, A function f(x, y) is said to be homogeneous of degree n if f(Ax, Aj) = A"/(X, y). The equation M(x, y) dx +, = 0 is said to be homogeneous if M(x, y) and N(x, y) are homogeneous of the same degree. It is, easy to verify that the substitution, , N(x, y) dy, , y = tJX,, , dy= tJ dx+x dtJ, , will transfonn a homogeneous equation into a separable equation in the variables x and v., , Integrating Factors, Certain differential equations may be solved after multiplication by a suitable function of x and y produces an, integrable combination of tenns. Such a function is called an integrating factor of the equations. In looking, for integrable combinations, note that:, (i), , (iii), , d(xy) =xdy+ ydx, d(ln xy) = xdy+ ydx, xy, , (ii), (iv), , = xdy~ydx, x, d(_I_u k+l ) =Uk du, d(y/x), , k+1
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.8, , Differential Equations, , CHAPTER 59, , Moreover, d(F) + d(G) + ... =0 yields F + G +. =constant. ~ee Problems 10-14.), The so-called linear differential equations 0/ the first order., + Py =Q , where P and Q are functions, , Ix, , of x alone, have the function ~(x) =elf," as integrating factor. (See Problems 15-17.), An equation of the form, , t+, , Py = Qyn, where n ;#: 0, 1 and where P and Q are functions of x alone, can, , be reduced to the linear fonn by the substitution, , yl-n =z,, , dy, y-n dx, , 1 dz, , =I - n dx, , (See Problems 18-19)., , Second-Order ..::E:.:!qu=a:..:t:.:..::io:.:..:n~s_ _ _ _ _ _ _ _ _---::----::-:-:---:-_ _ _ _ _ __, The second-order equations that will be solved in this chapter are of the following types:, , d2y, , dx 2 =/(x), 2, , ddxy = /, 2, , d2y, dx l, , (See Problem 23.), , (dy), ~x, dx, , =/(y), , (See Problems 24 and 25.), (See Problems 26 and 27.), , dy, dy, dx 2 + P dx +Qy = R,, 2, , where P and Q are constants and R is a constant or function of x only., , (See Problems 28-33.), If the equation m2 + Pm + Q =0 has two distinct roots 1111 and~, then y =Cle m1x + C2e""-' is the general, , solution of the equation, , ~ + P ~ +Qy =O. If the two roots are identical so that ml =1n2 =In, then, y =Cle +C2xe mx =e""(CI +C2x), lfU, , is the general solution., d2, d, The general solution of, + P +Qy =0 is called the complementary function of the equation, , J Ix, , d 2y, dy, dxl + P dx +Qy =R(x), , (59.1 ), , If/(x) satisfies (59.1 ). then the general solution of (59.1) is, y = complementary function, , +/(x), , The function/(x) is called a particular solution of (59.1)., , SOLVED PROBLEMS, , 1., , Show that (a) y = 2e x, (b) y = 3x. and (c) y = Cle x + C1x, where CI and C2 are arbitrary constants, are solutions of, the differential equation y"( 1 - x) +y'x - y =O., , ,, , (a) Differentiate y = 2e' twice to obtain y' = 2e' and y" = 2e '. Substitute in the differential equation to obtain, the identity 2e X( 1 - x) + 2e 'x - 2e x = O., (b) Differentiate y = 3x twice to obtain y' = 3 and y" = O. Substitute in the differential equation to obtain the, identity 0( 1 - x) + 3x - 3x =O., (c) Differentiate y = Cle' + C2x twice to obtain y' = Cle' + C1and y" = Cle·. Substitute in the differential, equation to obtain the identity Cle'(l - x) + (Cle X + C2 )x - (Cle' + C2x) =O., Solution (c) is the general solution of the differential equation because it satisfies the equation and contains, the proper number of essential arbitrary constants. Solutions (a) and (b) are called particular soll/tions because, each may be obtained by assigning particular values to the arbitrary constants of the general solution.
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CHAPTER S9 Differential Equations, , 19. Solve: +ytanx=rsecx., Wrilthe equation in the form y-3£+,-2 tanx=secx. Then use the substitution,-2 = z, y-3* =-l~ to, obtain dx -2ztanx=-2secx., ., The integiating factor is ~(x) =e-2J .... "'" ='cos1 x. It gives cos2x dz - 2z cosx sin x dx= - 2cos x dx' from, which, , zcos 2x=-2sinx+C,, , 2, , or, , T=-2sinx+C, , 20. When a bullet is fired into a sand bank. its retardation is assumed equal to the square root of its velocity on, entering. For how long will it travel if its velocity on entering the bank is 144 ft/sec?, ., Let Ifrepresent thebullefs velocity 1 seconds after striking the bank. Then the retardation is =.r;" so, =-dt and 2.[; =-I +C., , 't, , -j;, , When 1 =0, If= 144 and C = 2.Ji44 = 24. Thus, 2.[; = -I +24 is the law governing the motion of the bullet., When If= 0,1 = 24; the bullet will travel for 24 seconds before coming to rest., , 21. A tank contains 100 gal of brine holding 200 Ib of salt in solution. Water containing 1 Ib of salt per gallon flows, into the tank at the rate of 3 gal/min, and the mixture, kept uniform by stirring, flows out at the same rate. Find, the amount of salt at the end of 90 min., Let q denoce the number of pounds of salt in the tank at the end of t minutes. Then, is the rate of change of, the amount of salt at time I., 1, Three pounds of salt enter the tank each minute, and 0.03q-pounds are removed. Thus, ~ = 3 - 0.03q., , !ft, , Rearranged, this becomes 3- ~~03q = dl, and integration yields, , ., , In(0.03q - 3), C, 0.03, =-1+., When 1 =0, q = 200 and C = J~~ so that In(0.03q - 3) =~.031 + In3. Then O.Olq - 1 =e-O,03" and q = 100 +, l()()e-O-Ol'. When 1 = 90, q = 100 + lOOc2.1 - 106.72 lb., , 22. Under certain conditions, cane sugar in water is converted into dextrose at a rate proportional to the amount that, is unconverted at any time. If, of 75 grams at time 1= 0, 8 grams are converted during the first 30 min, find the, amount converted in 1 hours., Let q denote the amount converted in 1 minutes. Then, = k(75 - q), from which 7;:" = k dl, and integration gives In (75 - q) =-kt + C., 1, q, When 1 =0, q = 0 and C = In 75, so that In (75 - q) =-kt + In 75., When 1 =30 and q =8, we have 30k =In75 -In 67; hence, k =0.0038, and q = 75( I - e-O,oo~)., When 1 =90, q =75( 1 - e-O·34 ) - 21.6 grams., , t, , !ft, , d2y, 23. Solve dx 2 = xe' + cosx., Here, , i (:Z )= xe' +cosx. Hence, ~ J(xe' +cosx)dx, =, , = xe' - e' + sin x +CI' and another integration yields, , y =xe' - 2e'<- cosx + Clx + Cz•, , Jd +x dydx =a., 2, , 24. Solve x2, , Let p=, , Z:, , then, , ~{ = i and the given equation becomes x2i+xp= a or xdp+ pdx=~dx. Then integra-, , tion yields xp = a In Ix\+CI' or x!!1.. = a In Ixl + C. When this is written as dy = a In Ixl dx + CI dx, integration gives, dx, x, x, y= taln2lxl+ Ciin Ixl+ Cr
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CHAPTER 59 Differential Equations, To find a particular solution of the equation, we note that the right-hand member is R(x) =~. This suggests, that the particular solution will contain a term in ~ and perhaps other terms obtained by successive differentiation. We assume it to be of the form y = Ar + Bx + C, where the constants A, B, C are to be determined. Hence, we substitute y =Ar + Bx + C, y' =lAx + B, and y" =2A in the differential equation to obtain, 2A +3(2Ax + B)-4(Axl+Bx+C)=Xl, , or, , -4Ax2 +(6A-4B)x+(2A +3B-4C)=xl, , Since this latter equation is an identity in x, we have -4A = I, 6A - 4B =0, and 2A + 3B - 4C =O. These yield, A, , =-t. B=-j. C = - ~~, and y=-tx2 --Ix - ~~ is a particular solution. Thus, the general solution is, , Xl _1 x - 11, Y -- CI eX +C1e-4x _1, 4, 8 32', , 33., , SOlve~-2.£-3y=cosx., Here m2 ina - 3 = 0, from which m = -I, 3; the complementary function is y = Cle-x + C,Ix. The right~hand, !.., , member of the differential equation suggests that a particular solution is of the form A cos x + B sin x. Hence, we, substitute y = A cos x + B sin x, y'= B cos x - A sin x, and y" = - A cos x - B sin x in the differential equation to, obtain, (-Acosx- Bsinx)- 2(Bcosx-Asinx)- 3(A cos x + Bsinx)= cosx, , or, , -2(2A + B)cosx + 2(A - 2B)sinx =cosx, , -!, B =- I~ . The general, , The latter equation yields - 2(2A + B) = I and A - 2B =0, from which A =, . CIe-, + C2"_3x -scosx-TO, I, 1 smx, . ., · IS, so Iutton, , 34. A weight attached to a spring moves up and down so that the equation of motion is ~ + 16s =0, where s is the, stretch of the spring at time t. If s = 2 and ~ =I when 1 =0, find s in terms of I., 1, Here ml + 16 = 0 yields m =±4i, and the general solution is s =A cos 41 T B sin41. Now when 1=0, S = 2 =A,, so that s = 2 cos 4t + B sin4t., Also when 1= 0, dsldt = I =-8 sin 41 + 4B cos 4t =4B, so that B = Thus, the required equation is, s = 2cos4t + tsin4t., , t-, , 2, 35. The electric current in a certain circuit is given by dd + 4 dd1 + 25041 =ItO. If 1=0 and dd1 = 0 when t = 0, find, · terms 0 f t., t, t, t, I In, 2, Here m + 4m + 2504 =0 yields m = -2 + 50i. -2 - 50;; the complementary function is e-21 (A cos 50t + B sin, SOt). Because the right-hand member is a constant, we find that the particular solution is 1 = 110/2504 =0.044., Thus, the general solution is 1 =e-2, (A cos SOt + B sin SOt) + 0.044., Also when t =0, dlldt =0 =e-21[(-2A + SOB) cos SOt - (28 + 50A) sin 50t] =-2A + SOB. Then B = ~.0018,, and the required relation is 1 = -e-21(0.044 cos SOt + 0.00 18 sin 50t) + 0.044., , !, , 36. A chain 4 ft long starts to slide off a flat roof with 1 ft hanging over the edge. Discounting friction, find (a) the, velocity with which it slides off and (b) the time required to slide off., Let s denote the length of the chain hanging over the edge of the roof at time t., (a) The force F causing the chain to slide off the roof is the weight of the part hanging over the edge. That, weight is mgs/4. Hence,, F = mass x acceleration, , =ms" =t mgs, , or, , s" =t gs, , MUltiplying by 2s' yields 2s's" =t gss' and integrating once gives (S')2 =t gs2 + CI •, When t =0, s = I and s' =O. Hence, CI =-tg and s' =tJi.Js 2 -I. When s =4, s' =, , t.Ji5i ft/sec.
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CHAPTER 59, , (b) Since.Js~s_ 1 = t, , Ji dt, integration yields In Is + .J, , S2 -, , Differential Equations, , 11 = tJii + C2" When t = 0, s = 1. Then, , 2, , C2 =Oand In(s+.Js -1)=tJii., When s = 4, t = ~ In(4 +.Ji5)seconds., , "g, , ., , 37. A boat of mass 1600 Ib has a speed of 20 ft Isec when its engine is suddenly stopped (at t = 0). 1ge resistance, of the water is proportional to the speed of the boat and is 200 lb when t = O. How far will the boat have moved, when its speed is reduced to 5 ft/sec?, Let s denote the distance traveled by the boat t seconds after the engine is stopped. Then the force F on the, boat is, , F = ms" =- Ks' from which s" = -ks', g, , To detemune k, we note that at t =0, s' =20 and s" = force =- 200 =-4. Then k =-s" I s' =t. Now, 1600, mass, s" = = -~, and integration gives lnv =-tt + Cr or v= C."e-1I5 •, ., r, When t =0, v = 20. Then C. = 20 and v = ~ = 20e- '5. Another integration yields s = -10Qe-ffS + C2•, , 'Z, , When t =0, s = 0; then C2 = 100 and s = 100(1 - e-ffS). We require the value of s when v= 5 = 20e-li5, that is,, when e- r/5 =t. Then s::;: 100(1- t) = 75 ft., .. ,, .., , ' .. ~, ,;, , ~~, ~.~, , 38. Form the differential equation whose general solution is:, , c,'; + I, Cr+ C!, ~.:) )' = C. + C:.\ + C}r, (g) y = C. sin x + C2 cos X, (:\), , (':J, , .Y, , ~, , \'=, , Ans., , {OJ J = Clx of C, ~J), , xy=x'-C, (f) Y = C.e' + C2ih, (h) y =C.e' cos(3x + CJ, , (a) xy' = 2(y - 1); (b) y' = (y - xy')2; (c) 4ry = 'lxl/ + (yJ2; (d) xy' + y = 3r; (e) y'" = 0;, (f) y" - 3/ + 2y = 0; (g) y" + y =0; (h) y" - 2y' + lOy = 0, , 39. Solve:, , ydy-4xdx=0, l dy - 3x5 dx=O, ry'::;: l(x - 4), (x-2y)dy+(y+4x)dx=0, (2y + I)y' =3ry, xy'(2y-I)=y(I-x), (r + l) dx = 2xy dy, (x + y) dy =(x - y) dx, x(x + y) dy - Y dx = 0, (j) x dy - y dx + xe-yf· dx =0, (k) dy =(3y + eh) dx, (I) x 2i dy =(I -.xyl) dx, , (a), (b), (c), (d), (e), (f), (g), (h), (i), , Ans., Ans., Ans., Ans., Ans., Ans., Ans., Ans., Ans., AilS., Ans., , y=4r+C, 21= 3X'+ C, r-xy+2y=Cry, xy-y+2r=C, y + In ~II = r + C, In IxY = x + 2y + C, r-y2=Cx, r.- 2x)' - y = C, y= Ce-J/·, &"-' + In ICx! = 0, y = (Ce' - l)e2>, Am. 2x3y3=3r+ C, , 40. The tangent and normal to a curve at a point P(x, y) meet the x axis in T and N, respectively, and the y axis in S, and M, respectively. Determine the family of curves satisfying the conditions:, (a) TP = PS; (b) NM =MP; (c) TP = OP; (d) NP, , Ans., , =OP, , (a) xy= C; (b) 2r + y2= C; (c)xy = C, Y = Cx; (d)r±y= C
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Index, A, Abel's theorem. 386, Abscissa, 9, Absolute maximum alld minimum, 107,453, Absolute value, I, Absolutely convergent series, 376, Acceleration:, angUlar. 163, in curvilinear motion, 332, in rectilinear motion, 161, tangential and normal components of, 333, vector, 332, Alternating:, harmonic series, 376, series, 375, theorem. 375, Amplitude. 141, Analytic proofs of geometric theorems. 13, Angle:, between two curves, 144.342, measure, 130, of inclination, 144,341, Angular velocity and acceleration, 163, Antiderivative. 181, Approximation by differentials, 174, Approximation by series, 398, Arclength.237,308, derivative of, 312, 343, formula, 238, Area:, between curves, 236, by integration, 190. 481, in polar coordinates, 351, 520, of a curved surface, 489, of a surface of revolution, 30 I, under a curve, 190, Argument, 49, Asymptote, 120, of hyperbola. 39, Average rate of change, 73, Average value of a function, 198, Average velocity, 161, Axis of revolutiQll, 244, Axis of symmetry, 120, of a parabola, 37, B, Binomial series, 399, Binormal vector, 461, Bliss's theorem, 305, Bounded sequence, 353, Bounded set in a plane, 453, , C, Carbon dating, 232, Cardioid, 340, Catenary, 220, Center of curvature, 314, Center of mass, 510, Center:, of a hyperbola. 43, of an ellipse, 42, Centroid:, of a plane region. 481, of a volume. 500, Chain rule. 80,415, Change of variables in an integral, 199, Circle, 29, equation of, 29, of curvature, 313, osculating, 313, Circular motion, 163, Closed interval, 2, Closed set. 453, Comparison test, 367, Complement. 453, Complementary function. 517, Completing the square. 30, Components of a vector, 322, Composite function, 80, Composition, 80, Compound interest. 221. 232, Concave upward, downward, 119, Concavity, 119, Conditionally convergent series, 376, Cone. elliptic, 443, Conic sections, 39, Conjugate axis of a hyperbola, 43, Continuous function, 66, 68, 405, on [a,bJ, 68, on the left (right), 68, Convergence of series. 360, absolute, conditional, 376, Convergence, uniform, 385, Convergent sequence, 352, Coordinate, I, axes, 9, Coordinate system:, cylindrical and spherical, 498, linear, 1, rectangular, 9, right-handed, 426, polar. 133, 339, Cosecant, 142, Cosine, 131
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Index, Cosine (ConI.):, direction cosines, 428, Cotangent, 142, Critical numbers, 105, Cross product of vC?ctors, 428, Cross-section formula, 248, Cubic curve, 39, Curl, 465, Curvature, 313, of a polar curve. 343, Curve sketching, 122, Curvilinear motion, 332, Cycloid, 315, Cylindrical coordinates, 498, Cylindrical shell formula, 247, Cylindrical surfaces, 441, D, Decay constant, 230, Decreasing:, function. 100, sequence, 354, Definite integral, 192, Degree, 130, Del. 464, Deleted disk, 405, Delta neighborhood, 4, Delta notation, 73, Density, 510, Dependent variable. 49, Derivative. 73, directional. 452, first. 62, higher order, 82, 90, of a vector function, 324, of arc length, 312, 343, of inverse functions, 81, partial, 405, second. 82, third, 82, Determinants, 428, Difference of shells formula. 247, Difference rule for derivdves. 7!;, Differentiability. 74, 415, Differential, 174, total, 414, Differential equation, 516, linear. of the first order, 517, order of a, 516, second order. 517, separable. 516, solution (general) of a, 516, Differentiation, 79, formulas, 79, implicit, 90, 417, logarithmic, 210, of inverse functions, 81, of power series, 385, of trigonometric functions. 139, of vector functions, 324. 460, , Directed angles; 131, Direction cosines, 428, Direction numbers, 431, Directional derivative, 452, Directrix of a parabola, 41, Discontinuity, 66, jump, 67, removable, 66, Disk:, deleted. 405, open, 405, Disk formula. 244, Displacement, 74, Distance formula, 11, for polar coordinates, 351, Divergence (div):, of a sequence, 352, of a series, 360, of a vector function, 464, Divergence theorem, 362, Domain of a function, 49, Dol product of vectors, 323, Double integral. 474, 489, E, , e,215, e,214, , Eccentricity, of an ellipse. 42, of a hyperbola. 43, Ellipses, 38, center, eccentricity, foci, major axis,, minor axis of; 42, Ellipsoid. 442, Elliptic:, cone. 443, paraboloid. 442, Equations, graphs of, 37, Equilateral hyperbola. 46 ., Even functions, 122, Evolute,314, Exponential functions, 214. 216, Exponential growth and decay. 230, Extended law of the mean, 100, Extreme Value Theorem, 69, Extremum, relative, 98, , F, First derivative, 62, First derivative test, 106, First octant, 426, Foci:, of an ellipse, 42, of a hyperbola. 43, Focus of a parabola, 41, Free fall, 162, Frequency; 141, Function, 49, compcsite, 80, continuous, 405, , ,
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Index, Function (Coni.), decreasing, 100, differentiable, 74, 415, domain of a, 49, even, 122, exponential, 214, 216, homogeneous, 516, hyperbolic,220, implicit, 90, increasing, 100, integrable, 192, inverse, 81, inverse trigonometric, 152, logarithmic, 206, odd, 122, of several variables, 405, one-to-one, 81, range of a, 49, trigonometric, 139, Fundamental Theorem of Calculus, 199, G, Gamma function, 300, General exponential function. 216, General logarithmic functions, 217, Generalized Rolle'$ theorem. 99, Geometric series, 360, Gradient. 453. 464, Graphs of equations, 20, 37, Graphs of functions. 122, Gravity. 162, Growth constant, 230, , H, Half-life, 231, Half-open interval, 3, Hannonic series, 362, Higher order:, derivatives, 90, partial derivatives, 407, Higher order law of the mean, 100, Homogeneous:, bodies, 510, equation, ?7?, function, 516, Horizontal asymptote, 120, Hyperbola, 38, 43, asymptotes of, 39, center, conjugate axes, eccentricity,, foci, tranverse axes, vertices, equilateral, 43, 46, Hyperbolic functions, 220, Hyperbolic paraboloid, 443, Hyperboloid:, of one sheet, 443, of two sheets, 444, , Improper integra1s. 293, Increasing, function, 100, sequence, 354, Indefinite integral, 181, Independent variable, 49, Inde~natetypes,223, , Inequalities, 3, Infinite intervals, 3, Infinite limit, 57, of integration, 293, Infinite sequence, 352, limit of, 352, Infmite series, 360, Inflection point, 120, Initial position, 162, Initial velocity, 162, Instantaneous rate of change, 73, Instantaneous velocity, 161, Integrable, 192, Integral:, definite, 192, double, 474, improper. 293, indefinite, 181, iterated. 475, line. 466, Riemann. 192, test for convergence. 366, triple. 499, Integrand. 181, Integrating factor. 516, Integration:, by miscellaneous substitutions. 288, by partial fractions. 279, by parts. 259, by substitution. 182, by trigonometric substitution, 268, of power series. 385, plane area by double. 481, Intercepts. 21, Intermediate Value Theorem. 69, Interval of convergence. 383, Intervals. 2, Inverse cosecant, 155, Inverse cosine. 153, Inverse cotangent, 154, Inverse function. 81, Inverse secant. 155, Inverse sine. 152, Inverse tangent. 153, Inverse trigonometric functions, 152, Irreducible polynomial. 279, Iterated integral. 475, , J, Jump discontinuity. 67, , I, , Implicit differentiation. 90. 417, Implicit functions. 90, , L, , Latus rectum of a parabola, 41
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Index, Law of cosines, 134, Law of sines, 134, Law of the mean, 99, Extended, 100, Higher-order, 100, Lemniscate, 340, Length of are, 130, L"H6pital's Rule, 222, Lima~on, 340, Limit:, infinite, 57, of a function, 56, 405, of a sequence, 352, right and left, 57, Limit comparison test, 367, Line, 18, equation of a, 20, in space, 431, slope of a, 18, Line integral, 466, Linear coordinate system, I, Linear differential equation of the, first order, 517, Logarithm, natural, 206, Logarithmic differentiation, 2 t 0, Logarithmic functions, 217, Lower limit of an integral, t 92, M, Maclaurin series, 396, Major axis of an ellipse, 42, Mass, 510, Maximum and minimum:, absolute, 107, relative, 98, Mean-Value theorem for derivatives, 99, Mean-Value theorem for integrals, 198, Midpoint formulas, 12, Minor axis of an ellipse, 42, Midpoint rule for integrals, 204, Moment of inertia:, of planar mass, 510, of planar region, 482, of a volume, 500, Monotonic sequence, 354, Motion:, circular, 163, curvilinear, 332, rectilinear, 161, Motion under the influence of gravity, 162, , N, Natural logarithm, 206, Newton's law of cooling, 232, Newton's method, 175, Nondecreasing (nonincreasing) sequence, 354, Normal component of acceleration, 333, Normal line to a plane curve, 94, Normal line to a surface, 445, Normal plane to a space curve, 445, 461, , o, Octants, 426, One-to-one function, 81, Open disk, 405, Open intervaJ, 2, Open set, 415, Ordinate, 9, Origin, I, Osculating circle, 313, Osculating plane, 461, , ,, , p, Pappus, theorem of, 488, Parabola, 37, focus, directrix, latus rectum, vertex, 4 i, Paraboloid:, elliptic, 442, hyperbolic, 443, Paradox, Zeno's, 364, Parallel lines, slopes of, 22, Parameter, 307, Pam metric equations, 307 ., for surfaces, 462, Partial derivative, 405, higher order, 407, Partial fractions, 279, Partial sums of a series, 360, Particular solution, 517, Period, 141, Perpendicular lines, slopes of, 22, Plane, 432, vectors, 321, Point of inflection, 120, Point-slope equation of a line, 21, Polar axis, 339, Polar coordinates, 133,339,340, Polar curves, 340, Polar equation, 339, Pole, 339, Position vector, 324,426, Positive series, 366, Positive x axis, y axis, 9, Power c"hain rule, 84, Power rule for derivatives, 79, Power series, 383, differentiation of, 385, integration of, 385, interval of convergence of, 383, radius of convergence of, 384, unifonn convergence of, 385, p-series, 368, Principal normal, 461, Product rule for derivatives, 79, , Q, Quadr.tnts, 10, Quickfonnula I, 182, Quick fonnula II, 208, Quotient rule for derivatives, 79
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Index, R, Radian measure, 130, Radius of convergence, 383, Radius of curvature, 313, Radius vector, 324, Range of a function, 49, Rate of change, 73, Ratio of a geometric series, 360, Ratio test, 376, Rational function, 68, 279, Rectangular coordinate system, 9, Rectifying plane, 461, Rectilinear motion, 161, Reduction formulas, 263-264, Related rates, 167, Relative extrema (maximum and minimum), 98, 105,, 106,453, Remainder term, 397, Removable discontinuity, 70, Riemann integral, 192, Riemann sum, 192, Right-handed system, 426, Rolle's theorem, 98, generalized, 99, Root test, 376, Rose with three petals, 340, S, Scalar product of vectors, 323, Scalars, 321, Secant function, 142, Second derivative, 82, Second derivative test, 105, Semimajor (semiminor) axis of an ellipse, 42, Separable differential equation, 516, Sequences:, bounded,353, convergent and divergent, 352, limit of, 352, decreasing, increasing, nondecreasing,, nonincreasing, monotonic, 354, Series, infinite, 360, absolutely convergent, 376, alternating, 375, binomial, 399, conditionally convergent, 376, convergent and divergent, 360, geometric, 360, harmonic, 362,, Maclaurin, 396, partial sums of, 360, positive, 366, power, 383, p-series, 368, remainder after n terms of a, 397, Taylor, 396, sum of, 360, terms of, 360, with positive terms, 366, , Sigma notation, 190, Simpson's rule, 204, Sine, 131, Slicing formula, 248, Slope of a line, 18, Slopes:, of parallel lines, 22, of perpendicular lines, 22, Slope-intercept equation of a line, 21, Solid of revolution, 244, Space curve, 445, 461, Space vectors, 426, Speed,332, Sphere, 441, Spherical coordinates, 498, Squeeze theorem, 353, Standard equation of a circle, 29, Standing still, 162, Substitution method, 182, Sum of a series, 360, Sum rule for derivatives, 79, Surfaces, 462, cylindrical,441, Surface of revolution, 301, 446, Symmetry, 120, 122, axis of, 37, 120, T, Tabular method for absolute extrema, 107, Tangent function, 142, Tangent line to a plane curve, 93, Tangent line to a space curve, 445, Tangent plane to a surface, 445, Tangential component of acceleration, 333, Taylor series, 396, Taylor's formula with remainder, 397, Terms of a series, 360, Third derivative, 82, Total differential, 414, Transverse axis of a hyperbola, 43, Trapezoidal rule, 202, Triangle inequality, 2, Trigonometric functions, 131, 140, 142, Trigonometric integrands, 266, Trigonometric substitutions, 268, Trigonometry review, 130, Triple integral, 499, Triple scalar product, 430, Triple vector product, 431, U, Uniform convergence, 385, Unit normal to a surface, 463, Unit tangent vector, 325, Upper limit of an integral, 192, V, , Vector:, equation of a line, 431, equation of a plane, 432
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Index, Vector (Cont.):, position, 324, 426, product, 428, projections, 324, radius, 324, unit, 322 ., unit tangent, 325, velocity, 332, zero, 321, Vector functions, 324, curl of, 465, differentiation of, 324. 460, divergence of, 464, integration of, 46S, Vectors, 321, acceleration, 332, addition of, 321, components of, 322, cross product of, 428, difference of, 322, direction cosines of, 428, dot product of, 323, magnitude of, 321, plane, 321, scalar product of, 323, scalar projection of, 324, space, 426, SUIll of. 321, triple scalar product of, 430, . triple vector product of, 431, vector product of, 428, vector projl!ction of, 324, Velocity:, angular, 163, average, 161, , Vetocity (Cont.):, in curvilinear motion, 332, in rectilinear motion, 161, initial, 162, instantaneous, 161, vector, 332, Vertex of a parabola, 41, Vertical asymptote, 120, Vertices:, of a hyperbola, 43, of an ellipse, 42, Volume:, given by an itt;rated integral, 475, of solids of revolution, 244, under a surface, 489, with area of cross section given, 248, , I, , W, Washer formula, 246, Work done by a force, 329, X, x axis, 9, , positive, 9, x coordinate, 9, , y, y cui.:>, 9, , positive, 9, y coordinate, 9, y intercept, 21, Z, Zeno's paradox, 364, Zero vector, 321
