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Contents, Chapter 1. Limit and Continuity of Multivariable Functions, 1.1. Functions of Several Variables, 1.2. Limit and Continuity, , 1, 1, 5, , Chapter 2. Partial Derivatives, 2.1. Definition and Examples, 2.2. Second Order Partial Derivative, 2.3. Higher Order Partial Derivative, , 13, 13, 19, 22, , Chapter 3., , Differentiability, , 25, , Chapter 4., , Extreme Values, , 47, , Chapter 5. Multiple Integrals, 5.1. Double Integrals over Rectangles, , 60, 60, , i

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Preface, This text book is an initiative by the BOS Mathematics, Savitribai Phule Pune University., The syllabus of Savitribai Phule Pune University is always considered commendable, with a list, of reputed books in the subject recommended for reference. Many times teachers face difficulty, in doing justice to every aspect covered collectively by all these books. So, while preparing new, syllabus for S.Y.B.Sc./S.Y.B.A. it was thought the University should prepare textbooks for Mathematics with the following objectives:, 1. Uniformising notations, definitions and to focus on important aspects of revised syllabus which, need to be stressed and understood., 2. Collecting all relevant topics, problems, questions prescribed in syllabus., 3. Providing a ready reference to teachers and students., The book is written in accordance with the new prescribed syllabus of S.Y.B.Sc., S.Y.B.A. Mathematics (2013 Pattern), for the Paper I - MT 211: Multivariable Calculus I (First Term). The, syllabus deals with important topics in Mathematics, Vector Calculus., This book consists of a detail introduction at the beginning of each chapter, several illustrative, examples and problems for practice with hints and solutions given at the end of each section. A, proper understanding of all the illustrative examples would go a long way in making the subject, fully comprehensible., In case of queries/suggestions, send an email to: sumati.munot @ gmail.com, We hope our endeavour will benefit both students and teachers., -Authors, , ii

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Syllabus:- Paper I MT 211: Multivariable Calculus I, (1) Limit and Continuity of Multivariable functions: . . . [06], 1.1 Functions of several variables, graphs and level curves of function of two variables ., 1.2 Limit and continuity in higher dimensions ., (2) Partial Derivatives: . . . [04], 2.1 Definition and examples ., 2.2 Second order partial derivative, the mixex derivative theorem ., 2.3 Partial derivatives of higher order., (3) Differentiability: . . . [12], 3.1 Differentiability, the increment theorem for functions of two variables(without proof)., 3.2 Chain rules for composite function., 3.3 Directional derivatives, gradient vectors., 3.4 Tangent planes, normal lines and differentials., (4) Extreme Values: . . . [10], 4.1 Extreme values, First derivative test and Second derivative test for local extreme, values., 4.2 Lagrange’s multipliers method for finding extreme values of constraint function (One, Constraint), 4.3 Taylors Formula for two variables., (5) Multiple Integrals: . . . [16], 5.1 Double Integral over rectangles, Fubini’s theorem for calculating double integrals, (Without proof), 5.2 Double integrals in polar form., 5.3 Triple integral in cylindrical and spherical coordinates., 5.4 Triple integral in cylindrical and spherical coordinates., 5.5 Substitution in multiple integrals, Application to are and volumes., Text book: Prepared by the BOS Mathematics, Savitribai Phule Pune University, Pune., Recommended Book: Thomas’ Calculus’ 11th Edition, G.B. Thomas. Revised by Maurice, D. Weir, Joel Hass and Frank R. Giordano. Pearson Education 2012. Articles: 14.1 to 14.10,, 15.1,15.3,15.4,15.6,15.7, Reference Books:, (1) Basic Multivariable Calculus, J.E. Marsden, A.J. Tromba, A. Weinstein. Springer Verlag, (Indian Edition)., (2) Shanti Narayan, R.K. Mittal, A Text-book of Vector Calculus, S. Chand and Company., (3) D.V.Widder, Advanced Calculus 2 (2 nd Edition), (4) T.M. Apostol. Calculus Vol.2 (2 nd Edition), John Wiley, Newyork (1967)., , iv

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CHAPTER 1, , Limit and Continuity of Multivariable Functions, 1.1. Functions of Several Variables, Recall for the function of one variable, the underlying set was R, the set of real numbers. For, two variables, corresponding role is played by the set R2 , the cartesian product of R with itself., R2 = {(x, y)/x, y ∈ R} where R2 is represented by points in the plane. Similarly, points in space, are represented by R3 = {(x, y, z)/x, y, z ∈ R} and Rn = {(x1 , x2 , . . . , xn )/x1 , x2 , . . . , xn ∈ R}., Definition 1.1. Real Valued Functions of Two Independent Variables:, Let D = {(x, y)/x, y ∈ R} ⊂ R2 . A real valued function f on D is a rule that assigns a unique, single real number w = f (x, y) to each element in D., Definition 1.2. Real Valued Functions of Three Independent Variables:, Let D = {(x, y, z)/x, y, z ∈ R} ⊂ R3 . A real valued function f on D is a rule that assigns a unique, single real number w = f (x, y, z) to each element in D., Definition 1.3. Real Valued Functions of n Independent Variables:, Let D = {(x1 , x2 , · · · xn )/x1 , x2 , · · · xn ∈ R} ⊂ Rn .A real valued function f on D is a rule that, assigns a unique single real number w = f (x1 , x2 , · · · xn ) to each element in D., The set D is the domain of function. The set f (D) = {w ∈ R/w = f (x1 , x2 , · · · xn )} is range, of function. w is dependent variable (output variable) and x1 , x2 , · · · xn are independent variables, (input variables)., p, Example 1.1. Sketch the domain of f (x, y) = y − x2 . What is the range of function?, p, Solution: The domain D is the set of all pairs (x, y) in the plane for which y − x2 is real, ∴ y − x2 ≥ 0. ∴ y ≥ x2 ., Therefore, domain D is the set of points that lie above and on the parabola y = x2 ., The range of function is the set of all non-negative numbers w ≥ 0., Example 1.2. What are the domain and range of f (x, y) =, , 1, ., xy, , Solution: The domain D is the set of all pairs in the plane for which xy 6= 0., The range of function is the set {w ∈ R/w ∈ (−∞, 0) ∪ (0, ∞)} ., xy, Example 1.3. What are the domain and range of f (x, y) = 2, ., (x − y 2 ), Solution: The domain of function consists of all points in the plane except the points for which, x2 − y 2 = 0., i.e. all points in plane except points on the lines y = x and y = −x., The range of function is entire set of real numbers i.e. −∞ < w < ∞., Example 1.4. Find the domain and range of f (x, y) =, , (x2, , 1, ., + y2 + z2), , Solution: The domain D is the set of points in the space for whichx2 + y 2 + z 2 6= 0., i.e. the domain is set of points (x, y, z) 6= (0, 0, 0)., The range of function is the set of all positive numbers i.e. 0 < w < ∞. Interior and Boundary, points, Definition 1.4. Interior point: A point (x0 , y0 ) in a region R in the xy plane is an interior, point of R if every disk centered at (x0 , y0 ) of positive radius lies entirely in R., Definition 1.5. Boundary point: A point (x0 , y0 ) is boundary point of R if every disk, centered at (x0 , y0 ) of positive radius contains points that lie in R as well as points lie outside of, R., 1

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Chapter 1. Limit and Continuity of Multivariable Functions § 1.1. Functions of Several Variables, Note: The boundary point itself need not belong to R., Definition 1.6. Open Region: A region is open if it consists entirely of interior points., Definition 1.7. Closed Region: A region is closed if it contains all its interior and boundary, points., Definition 1.8. Bounded Region: A region in the plane is bounded if it lies inside a disk of, fixed radius., Definition 1.9. Unbounded Region: A Region in the plane is unbounded if it is not bounded., For example:, (i) Bounded regions: Triangles, rectangles, circles,disks., (ii) Unbounded regions: lines, coordinate axes, quadrants, half planes., In Fig.1.1,function is defined only where (y − x2 ) ≥ 0., i.e.domain is closed, unbounded region., √, Example 1.5. If f (x, y) = y − x then, (1) find the boundary of domain of function., (2) determine whether domain is an open region,closed region or neither, (3) decide whether domain is bounded or unbounded., √, Solution: Let f (x, y) = y − x, The domain of f (x, y) is set of all pairs in the plane for which (y − x) ≥ 0, i.e. y ≥ x and range of f (x, y) is setpof all non negative numbers i.e. w ≥ 0, Boundary of domain of function is (y − x) = 0, ⇒y=x, i.e. The straight line passing through origin., Therefore, domain is closed region, which is unbounded., 1.1.1. Graphs and Level curves of function of two variables., Definition 1.10. Level Curve: The set of all points in the plane where a function f (x, y), has a constant value ’c’ is called a level curve of f ., i.e.{(x, y) ∈ D/f (x, y) = c}, Definition 1.11. Graph of Function: The set of all points {x, y, f (x, y)} in a space , for, (x, y) in domain of f is called the graph of the function f ., The graph of the function f is also called surface z = f (x, y)., Example 1.6. Plot the level curves for the function f (x, y) =, , (x + y), , x 6= y, if c = 0, 1., (x − y), , Solution: For Level curve, f (x, y) = c, c ∈ R, (x + y), =c, (x − y), , , c−1, y=, x, c+1, i.e.The level curves are lines passing through origin with slope, , (c − 1), (c + 1), , (1) For c = 0, slope is -1, ∴ The level curve is the line y = −x., (2) For c = 1, slope is 0, ∴ The level curve is y = 0 i.e. x axis., (x + y), F ig.1.2 Level curves for f (x, y) =, , x 6= y, (x − y), Example 1.7. Plot the graph and level curves of the function f (x, y) = 100 − x2 − y 2 for, c = 0, 51, 75., Page|2

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§ 1.1. Functions of Several Variables Chapter 1. Limit and Continuity of Multivariable Functions, Solution: Let f (x, y) = 100 − x2 − y 2 The domain D of f is entire xy plane and range of f is, the set of real numbers less than or equal to 100., For Level curves:, f (x, y) = c, c ∈ R, 100 − x2 − y 2 = c, x2 + y 2 = 100 − c, √, Therefore, level curves are circles centered at origin with radius 100 − c., (1) For c = 0, x2 + y 2 = 100 Therefore, level curve is circle with center origin and radius 10, (2) For c = 51, x2 + y 2 = 49 The level curve is circle with center origin and radius 7, (3) For c = 75, x2 + y 2 = 25 The level curve is circle with center origin and radius 5, F ig.1.3 Level curves for f (x, y) = 100 − x2 − y 2 For Graph of f (x, y):, The graph is the paraboloid z = 100 − x2 − y 2 . as shown in Fig.1.4, also the level curve f (x, y) = 100 consists of the origin alone., Example 1.8. Sketch the level curves of f (x, y) = −(x − 1)2 − y 2 + 1, for f (x, y) = 1, 0, −3, −8., Solution: Let f (x, y) = −(x − 1)2 − y 2 + 1, For level curves, f (x, y) = c, c ∈ R, (x − 1)2 + y 2 = 1 − c, √, i.e.level curves are circles centered at (1,0)and radius 1 − c, here,c = 1, 0, −3, −8, i.e.level curves are circles centered at (1,0) and respective radii are 0,1,2,3., 2, 2, Example, √ 1.9., √ Find an equation of the level curve of f (x, y) = 16 − x − y that passes through, the point (2 2, 2)., √ √, Solution: Let level curves f (x, y) = 16 − x2 − y 2 that passes through the point (2 2, 2)., √, √ 2, ∴ f (x, y) = 16 − (2 2)2 − ( 2 ), =6, , 16 − x2 − y 2 = 6, x2 + y 2 = 10, √, Level curve is a circle with center at origin and radius is 10. EXERCISE 1.1 In example 1-12, Q1: Find domain and range of functions., Q2: Describe the level curves of functions., Q3: Find the boundary of domain of functions., Q4: Determine whether the domain is an open region, closed region or neither., Q5: Decide whether domain is bounded or unbounded., (1) f (x, y) = yp− x., (2) f (x, y) = (y − x., (3) f (x, y) = 4x2 + 9y 2 ., (4) f (x, y) = x2 − y 2 ., (5) f (x, y) = xy., y, (6) f (x, y) = 2 ., x, 1, ., (7) f (x, y) = p, 2, 2, p 16 − x − y, (8) f (x, y) = 9 − x2 − y 2 ., (9) f (x, y) = ln(x2 + y 2 )., 2, 2, (10) f (x, y) = e−(x +y ) ., (11) f (x, y) = sin−1 (y − x)., Page|3

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Chapter 1. Limit and Continuity of Multivariable Functions § 1.1. Functions of Several Variables, y, (12) f (x, y) = tan−1 ( ). In example 13-16. Find an equation for the level curve of the function, x, f (x, y) that passes through√the√given point., 2, 2, (13) f (x, y) = 16, √ − x − y , (2 2, 2)., (14) f (x, y) = x2 − 1 , (1, 0)., √ √, R y dt, , (− 2, 2)., (15) f (x, y) = x, 2, 1+t, ANSWERS 1.1, (1), , (2), , (3), , (4), , (5), , (6), , (7), , (8), , Page|4, , (i) Domain :- all points in xy Plane. Range :- all real numbers., (ii) level curves are straight lines y − x = c, i.e. parallel to y = x, (iii) No boundary points., (iv) Both open and closed., (v) Unbounded., (a) Domain :-set of all (x, y) such that y ≥ x, Range :- z ≥ 0., (b) straight lines√y − x = c, c ≥ 0, (c) boundary is y − x = 0 i.e.straight lines y = x., (d) closed., (e) Unbounded., (a) Domain :-all points in xy Plane., Range :- z ≥ 0., (b) f (x, y) = 0 , the origin. f (x, y) = c > 0,ellipses with center(0,0), major and minor, axis along X andY axis respectively., (c) No boundary points, (d) Both open and closed., (e) Unbounded., (i) Domain :-all points in xy Plane., Range :- all real Numbers., (ii) Level curves:For f (x, y) = 0 ,the union of lines y = ±x for f (x, y) = c 6= 0 ,, hyperbolas centered at (0,0)with foci on the x axis if c > 0 and on the Y axis if, c<0., (iii) No boundary points., (iv) Both open and closed., (v) Unbounded., (i) Domain :-all points in xy Plane., Range :- all real Numbers., (ii) Level curves:For f (x, y) = 0 ,the X and Y axis for f (x, y) 6= 0 , hyperbolas with X, and Y axis as asymptotes ., (iii) No boundary points., (iv) Both open and closed., (v) Unbounded., (i) Domain :-all (x, y) 6= (0, y), Range :- all real Numbers., (ii) Level curves:For f (x, y) = 0 ,the X axis without origin for f (x, y) 6= 0, parabolas, minus origin., (iii) boundary is line x = 0., (iv) open region., (v) Unbounded., (i) Domain :-all pairs (x,y) satisfying x2 + y 2 < 16., 1, Range :- z ≥ ., 4, (ii) Level curve :are circles centered at origin with radii r < 4., (iii) boundary is the circle x2 + y 2 = 16., (iv) open region., (v) bounded., (i) Domain :-all pairs (x,y) satisfying x2 + y 2 ≤ 9., Range :- 0 ≤ z ≤ 3., (ii) Level curves are circles centered at origin with radii r ≤ 3

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§ 1.2. Limit and Continuity, , Chapter 1. Limit and Continuity of Multivariable Functions, , boundary is the circle x2 + y 2 = 9., closed region., bounded., Domain :-(x, y) 6= (0, 0)., Range :-all real Numbers., (ii) Level curves are circles centered at origin with radii r > 0., (iii) boundary is the single point (0,0)., (iv) open region., (v) unbounded., (i) Domain :-all points in xy Plane., Range :-0 < z < 1., (ii) Level curves are origin itself and the circles centered at origin with radii r > 0., (iii) No boundary points., (iv) both open and closed region., (v) unbounded., (i) Domain :-all (x,y) satisfying −1 ≤ (y − x) ≤ 1., π, π, Range :-− ≤ z ≤ ., 2, 2, (ii) Level curves are straight lines of the form y − x = c , −1 ≤ c ≤ 1., (iii) boundary are two straight lines y = 1 + x and y = −1 + x., (iv) closed region., (v) unbounded., (i) Domain :-all (x, y) , x 6= 0., π, π, Range :-− < z <, 2, 2, (ii) Level curves are straight lines of the form y − x = c , −1 ≤ c ≤ 1., (iii) boundary are two straight lines y = 1 + x and y = −1 + x., (iv) closed region., (v) unbounded., 2, x + y 2 = 10, x = 1 or x = −1, √, tan−1 y − tan−1 x = 2tan−1 2, , (iii), (iv), (v), (9), (i), , (10), , (11), , (12), , (13), (14), (15), , 1.2. Limit and Continuity, Limit of Function If the values of a function z = f (x, y) can be made as close as we like to a, fixed number L by taking the point (x, y) close to the point (x0 , y0 ) but not equal to (x0 , y0 ) , We, say that L is the limit of f (x, y) as (x, y) approaches to (x0 , y0 )., Since,, p, p, |x − x0 | = (x − x0 )2 ≤ (x − x0 )2 + (y − y0 )2 and, p, p, |y − y0 | = (y − y0 )2 ≤ (x − x0 )2 + (y − y0 )2, p, then (x − x0 )2 + (y − y0 )2 < δ, ⇒|x − x0 | < δ and |y − y0 | < δ, conversely , if for some δ > 0 , p, both |x − x0 | < δ and |y − y0 | < δ, p, √, 2, 2, 2, (x − x0 ) + (y − y0 ) <, δ2 +, ⇒, pδ = 2δ thus , the open√ square |x − x0 | < δ and |y −, yp, (x − x0 )2 + (y − y0 )2 < 2δ and contains the open disc, 0 | < δ lies inside the open disc, (x − x0 )2 + (y − y0 )2 < δ, Definition 1.12. Limit of Function of two variables: A function f (x, y) approaches the, limit L ,as (x, y) approaches (x0 , yo ) , if there exists a corresponding number δ > 0 such that for, all (x, y) in the domain of f ,, p, |f (x, y) − L| <  whenever 0 < (x − x0 )2 + (y − y0 )2 < δ, or 0 < |x − x0 | < δ and 0 < |y − y0 | < δ, In symbols , we can write as, lim, , f (x, y) = L, , (x,y)→(x0 ,y0 ), , Page|5

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Chapter 1. Limit and Continuity of Multivariable Functions, , If, , § 1.2. Limit and Continuity, , Theorem 1.1. Properties of Limits:, lim, f (x, y) = L ,, lim, g(x, y) = M , L, M ∈ R then, (x,y)→(x0 ,y0 ), , (i), , lim, , (x,y)→(x0 ,y0 ), , [f (x, y) ± g(x, y)] = L ± M, , (x,y)→(x0 ,y0 ), , (ii), , lim, , [f (x, y).g(x, y)] = L.M, , (x,y)→(x0 ,y0 ), , (iii), , lim, , Kf (x, y) = KL ,K, L ∈ R and, , (x,y)→(x0 ,y0 ), , (iv), , lim, (x,y)→(x0 ,y0 ), , f (x, y), L, =, , M 6= 0., g(x, y), M, , Note:-If a function f (x, y) has different limits along two different paths as (x, y) → (x0 , y0 ), then, lim, f (x, y) does not exists. continuous Function:, (x,y)→(x0 ,y0 ), , Definition 1.13. continuous Function of two variables: A function f (x, y) is said to be, continuous at the point (x0 , yo ) , if, (i) f (x, y) is defined at (x0 , yo ), i.e. f (x0 , yo ) exists., (ii), lim, f (x, y) exists., (x,y)→(x0 ,y0 ), , (iii), , lim, (x,y)→(x0 ,y0 ), , f (x, y) = f (x0 , yo ), , Definition 1.14. A function f (x, y) is said to be continuous in a region R if it is continuous, at every point of its domain., Theorem 1.2. Properties of continuous function:, If f (x, y) and g(x, y) are both continuous at point (x0 , y0 ) then, (i) f (x, y) ± g(x, y) are continuous at point (x0 , y0 ), (ii) Kf (x, y) is continuous at point (x0 , y0 ), (iii) f (x, y).g(x, y) is continuous at point (x0 , y0 ), f (x, y), is continuous at point (x0 , y0 ) , g(x, y) 6= 0, (iv), g(x, y), (v) |f (x, y)| is continuous at point (x0 , y0 ), Theorem 1.3. Let f (x, y) be continuous function at point (x0 , y0 ) then f (x, y0 ) is continuous, function at x = x0 and f (x0 , y) is continuous function at y = y0 ,where f (x, y0 ) and f (x0 , y) being, function of one variable., Proof. Since,f (x, y) is continuous function at point (x0 , y0 ), lim, f (x, y) = f (x0 , yo ), , (x,y)→(x0 ,y0 ), , For given  > 0 , there exists δ > 0 such that, |x − x0 | < δ and |y − y0 | < δ ⇒ |f (x, y) − f (x0 , y0 )| < , |x − x0 | < δ and |y0 − y0 | < δ ⇒ |f (x, y0 ) − f (x0 , y0 )| < , |x − x0 | < δ ⇒ |f (x, y0 ) − f (x0 , y0 )| < , ∴ f (x, y0 )is continuous function at x = x0, Similarly, |y − y0 | < δ ⇒ |f (x0 , y) − f (x0 , y0 )| < , ∴ f (x0 , y)is continuous function at y = y0, The converse of above theorem, need not be true., ,  2xy, , (x, y) 6= (0, 0);, For example: Let f (x, y) =, x2 + y 2,  0,, (x, y) = (0, 0)., lim f (x, 0) = lim 0 = 0, , x→0, , x→0, , lim f (0, y) = lim 0 = 0, , y→0, , Page|6, , x→0

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§ 1.2. Limit and Continuity, , Chapter 1. Limit and Continuity of Multivariable Functions, , ∴ f (x, y) is continuous function at (0, 0) when considered as a single variable x or y, Now ,considered along path y = mx, lim, , f (x, y) = lim f (x, mx), x→0, , (x,y)→(0,0), , = lim, , x→0 x2, , =, , 2mx2, + m 2 x2, , 2m, 1 + m2, , ∴ this limit depends on m., ∴ limit does not exists at (0, 0), i.e. Function f (x, y) of two variable is not continuous at (0, 0), Definition 1.15. continuity of composite Function: If f (x, y) is continuous at (x0 , y0 ), and g is single variable function continuous at f (x0 , y0 ) then the composite function h = gof, defined by h(x, y) = g(f (x, y)) is continuous at (x0 , y0 )., xy, for example: e(x−y) , cos( 2, ) , ln(1 + x2 y 2 ), x +1, Example 1.10. Evaluate the limits , if they exists, 3x2 − y 2 + 5, (a), lim, (x,y)→(0,0) x2 + y 2 + 2, cosy + 1, (b), lim, π y − sinx, (x,y)→( ,0), 2, p, (c), lim, ln x2 + y 2 + z 2, (x,y,z)→(0,−2,0), , Solution:, 3x2 − y 2 + 5, 3(0)2 − (0)2 + 5, =, (0)2 + (0)2 + 2, (x,y)→(0,0) x2 + y 2 + 2, 5, =, 2, , (a), , lim, , cosy + 1, cos(0) + 1, lim, =, π, π y − sinx, (0) − sin( ), (x,y)→( ,0), 2, 2, 2, =, −1, = −2, p, p, (c), lim, ln x2 + y 2 + z 2 = ln (0)2 + (−2)2 + (0)2, (x,y,z)→(0,−2,0), √, = ln 4, = ln2, √, 2x − y − 2, Example 1.11. Find, lim, , 2x − y 6= 4, (x,y)→(2,0) 2x − y − 4, (b), , Solution:, , √, lim, (x,y)→(2,0), , √, 2x − y − 2, 2x − y − 2, √, =, lim, 2x − y − 4, (x,y)→(2,0) ( 2x − y)2 − (2)2, √, 2x − y − 2, √, √, =, lim, (x,y)→(2,0) [ 2x − y + 2][ 2x − y − 2], 1, √, =, lim, 2x − y + 2, (x,y)→(2,0), 1, =√, 4+2, 1, =, 4, Page|7

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Chapter 1. Limit and Continuity of Multivariable Functions, , § 1.2. Limit and Continuity, , x+y, Example 1.12. Let f (x, y) =, and  = 0.02 Show that there exists δ > 0 such that for, 2 + cosx, p, all (x, y) , x2 + y 2 < δ ⇒ |f (x, y) − f (0, 0)| < ., Solution:, f (x, y) =, , x+y, ,  = 0.02, 2 + cosx, Since, −1 ≤ cosx ≤ 1, ∴ (−1 + 2) ≤ (2 + cosx) ≤ (2 + 1), , ∴ 1 ≤ (2 + cosx) ≤ 3, 1, 1, ∴ ≤, ≤1, 3, 2 + cosx, |x + y|, x+y, ∴, ≤|, | ≤ |x + y| ≤ |x| + |y|, 3, 2 + cosx, x+y, | ≤ |x| + |y|, ∴|, 2 + cosx, x+y, consider, |f (x, y) − f (0, 0)| = |, − 0|, 2 + cosx, x+y, =|, | ≤ |x| + |y|, 2 + cosx, ∴ if |x| < δ, |y| < δ then |f (x, y) − f (0, 0)| ≤ |x| + |y| < 2δ = , , 0.02, = 0.01 such that, ∴ there exists δ = =, 2, 2, |x| < δ, |y| < δ ⇒ |f (x, y) − f (0, 0)| < , Example 1.13. Let f (x, y, z) = tan2 x + tan2 y + tan2 z and  = 0.03., Show that there exists a , δ > 0 such that for all (x, y, z),, p, x2 + y 2 + z 2 < δ ⇒ |f (x, y, z) − f (0, 0, 0)| < ., Solution: f (x, y, z) = tan2 x + tan2 y + tan2 z,  = 0.03, Consider,, |f (x, y, z) − f (0, 0, 0)| = |tan2 x + tan2 y + tan2 z|, ≤ |tan2 x| + |tan2 y| + |tan2 z|, = tan2 x + tan2 y + tan2 z, Now if |x| < δ, |y| < δ, |z| < δ, ⇒ |f (x, y, z) − f (0, 0, 0)| ≤ tan2 x + tan2 y + tan2 z, ≤ tan2 δ + tan2 δ + tan2 δ, = 3tan2 δ = , , 0.03, ∴ there exists tan2 δ =, =, = 0.01, 3, 3, ∴ tanδ = 0.1, ∴ there exists δ = tan−1 (0.1), such that |x| < δ, |y| < δ, |z| < δ ⇒ |f (x, y, z) − f (0, 0, 0)| < , Example 1.14. Show that f (x, y) =, , x2, has no limit as (x, y → (0, 0) by considering, x2 − y, , different paths., Solution: Along X-axis (i.e.y = 0):, lim, (x,y)→(0,0), , Page|8, , x2, x→0 x2, =1, , f (x, y) = lim f (x, 0) = lim, x→0

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§ 1.2. Limit and Continuity, , Chapter 1. Limit and Continuity of Multivariable Functions, , Along Y -axis (i.e.x = 0):, lim, , 0, y→0 0 − y, =0, , f (x, y) = lim f (0, y) = lim, , (x,y)→(0,0), , y→0, , Along y = kx2 , (k 6= 1):, f (x, y) = lim f (x, kx2 ) = lim, , lim, , x→0 x2, , x→0, , (x,y)→(0,0), , x2, − kx2, , x2, x→0 x2 (1 − k), 1, =, 1−k, = lim, , ∴ different limits for different paths., ∴ limit does not exist., Example 1.15. Show that, , lim, , tan−1 (, , (x,y)→(0,0), , π, |x| + |y|, ) = by using polar coordinates., 2, 2, x +y, 2, , Solution: by polar coordinates,, x = rcosθ, y = r sin θ and r → 0 as (x, y) → (0, 0), lim, (x,y)→(0,0), , tan−1 (, , |x| + |y|, |rcosθ| + |rsinθ|, ) = lim tan−1 (, ), 2, 2, r→0, x +y, r2, |r|(|cosθ| + |sinθ|), = lim tan−1 (, ), r→0, r2, , consider,, lim tan−1 (, , r→0+, , (|cosθ| + |sinθ|), |r|(|cosθ| + |sinθ|), ) = lim tan−1 (, ), r2, r, r→0+, = lim tan−1 (∞), r→0+, , =, , π, 2, , consider,, |r|(|cosθ| + |sinθ|), −(|cosθ| + |sinθ|), ) = lim tan−1 (, ), 2, −, r, r, r→0, π, =, 2, |x|, +, |y|, π, ∴, lim tan−1 ( 2, ) =, x + y2, 2, (x,y)→(0,0), , lim tan−1 (, , r→0−, , Example 1.16. At what points the function f (x, y) = sin, , 1, is continuous ?, xy, , Solution:, 1, f (x, y = sin, is continuous at all points(x, y) except x = 0 or y = 0., xy, Example 1.17. At what points(x,y,z) in a space the function f (x, y, z) =, , 1, is contin|xy| + |z|, , uous ?, Solution:, f (x, y, z) =, , 1, is continuous at all points(x, y, z) except (0, y, 0) or (x, 0, 0)., |xy| + |z|, , Example 1.18. Define f (0, 0) , so that f (x, y) = ln(, , 3x2 − x2 y 2 + 3y 2, ) is continuous at origin., x2 + y 2, Page|9

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§ 1.2. Limit and Continuity, , xy − y − 2x + 2, ,x 6= 1, x−1, (x,y)→(1,1), y+4, (iv), lim, (x,y)→(2,−4) x2 y − xy + 4x2 − 4x, √, √, x−y+2 x−2 y, √, (v), lim, ,x 6= y., √, x− y, (x,y)→(0,0), x+y−4, √, (vi), lim, (x,y)→(2,2), √x + y − 2, 2x − y − 2, (vii), lim, (x,y)→(2,0) 2x − y − 4, √, √, x− y+1, (viii), lim, x−y−1, (x,y)→(4,3), Each of following example gives a function f (x, y) and a positive number , show that, there exists a δ > 0 such that for all (x, y),, p, x2 + y 2 < δ ⇒ |f (x, y) − f (0, 0)| < , (i) f (x, y) = x2 + y 2 ,  = 0.01, y, (ii) f (x, y) = 2, ,  = 0.05, x +1, x+y, ,  = 0.01, (iii) f (x, y) = 2, x +1, x+y, (iv) f (x, y) =, ,  = 0.02, 2 + cosx, Each of following example gives a function f (x, y, z) and a positive number , show that, there, exists a δ > 0 such that for all (x, y, z),, p, x2 + y 2 + z 2 < δ ⇒ |f (x, y, z) − f (0, 0, 0)| < , (i) f (x, y, z) = x2 + y 2 + z 2 ,  = 0.015, (ii) f (x, y, z) = xyz,  = 0.008, x+y+z, (iii) f (x, y, z) = 2, ,  = 0.015, x + y2 + z2 + 1, (iv) f (x, y) = tan2 x + tan2 y + tan2 z,  = 0.03, Show that following functions have no limit as(x, y) → (0, 0), by considering different, paths., x, (i) f (x, y) = − p, x2 + y 2, 4, x, (ii) f (x, y) = 4, x + y2, x4 − y 2, (iii) f (x, y) = 4, x + y2, xy, (iv) f (x, y) =, |xy|, x−y, (v) f (x, y) =, x+y, x+y, (vi) f (x, y) =, x−y, x2 + y, (vii) f (x, y) =, y, x2, (viii) f (x, y) = 2, x −y, By using Polar coordinates,find the limit of following functions as (x, y) → (0, 0) if exists., x3 − xy 2, (i) f (x, y) = 2, x + y2, x3 − y 3, (ii) f (x, y) = cos( 2, ), x + y2, 2, y, (iii) f (x, y) = 2, x + y2, 2x, (iv) f (x, y) = 2, x + x + y2, |x| + |y|, (v) f (x, y) = tan−1 ( 2, ), x + y2, (iii), , (3), , (4), , (5), , (6), , Chapter 1. Limit and Continuity of Multivariable Functions, , lim, , Page|11

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Chapter 1. Limit and Continuity of Multivariable Functions, , § 1.2. Limit and Continuity, , x2 − y 2, x2 + y 2, (7) At what points the following functions are continuous?, (i) f (x, y) = sin(x + y), (ii) f (x, y) = ln(x2 + y 2 ), x+y, (iii) f (x, y) =, x−y, y, (iv) f (x, y) = 2, x +1, 1, (v) f (x, y) = sin, xy, x+y, (vi) f (x, y) =, 2 + cosx, x2 + y 2, (vii) f (x, y) = 2, x − 3x + 2, 1, (viii) f (x, y) = 2, x −y, 2, 2, 2, (ix) f (x, y, z) = x, p + y − 2z, (x) f (x, y, z) = x2 + y 2 − 1, (xi) f (x, y, z) = lnxyz, (xii) f (x, y, z) = ex+y cosz, 1, (xiii) f (x, y, z) = xysin, z, 1, (xiv) f (x, y, z) = 2, x + z2 − 1, 1, (xv) f (x, y, z) =, |y| + |z|, 1, (xvi) f (x, y, z) =, |xy| + |z|, (8) Define f (0, 0), so that function f (x, y) be continuous at the origin ?, 3x2 − x2 y 2 + 3y 2, (i) f (x, y) = ln(, ), x2 + y 2, 3x2 y, (ii) f (x, y) = 2, x + y2, (9) Show that function f (x, y, z) = x + y − z is continuous at every point (x0 , y0 , z0 )., (10) Show that function f (x, y, z) = x2 + y 2 + z 2 is continuous at origin., ANSWERS 1.2, √, 5, 1, 1, 19, (1)(i) (ii) 0 (iii)2 6 (iv), (v) 1 (vi) 1 (vii) (viii) ln2 (ix) 1 (x) 1 (xi) 0 (xii) −2 (xiii), (xiv), 2, 36, 2, 12, −1, −π, (xv) 2 (xvi) tan−1 (, ) (xvii) 3 (xviii) ln2, 2, 4, 1, 1, 1, (2)(i)]0 (ii) 2 (iii) −1 (iv) (v) 2 (vi) 4 (vii) (viii), 2, 4, 4, (3) (i)δ√= 0.1 (ii) δ = 0.05 (iii) δ = 0.005 (iv) δ = 0.01, (4) (i) 0.015 (ii) δ = 0.2 (iii) δ = 0.005 (iv) δ = tan−1 (0.1), π, (6) (i)0 (ii) 1 (iii) does not exist (iv) does not exist (v) (vi) does not exists, 2, (7) (i)All (x, y) (ii)All (x, y) except (0, 0) (iii)All (x, y) so that x 6= y (iv) All (x, y) (v) All (x, y), except x = 0 or y = 0 (vi)All (x, y) (vii)All (x, y) so that x 6= 2 and x 6= 1 (viii)All (x, y) so that, y 6= x2 (ix)All (x, y, z) (x)All (x, y, z) except interior of cylinder x2 + y 2 = 1 (xi)All (x, y, z) so that, xyz > 0 (xii)All (x, y, z) (xiii)All (x, y, z) with z 6= 0 (xiv)All (x, y, z) with x2 + y 2 6= 1 (xv)All, (x, y, z) except (x, 0, 0) (xvi)All (x, y, z) except (0, y, 0) or (x, 0, 0), (8) (i)ln3 (ii)0, (vi) f (x, y) =, , Page|12

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CHAPTER 2, , Partial Derivatives, 2.1. Definition and Examples, If z = f (x, y) is a function of two variable x and y.It is possible to consider the rate of change of, z w.r.t. x ,keeping y as a constant and also rate of change of z w.r.t. y ,keeping x as a constant,We, get a partial derivative., To distinguish partial derivative from ordinary derivative we use the symbol ’∂’rather than d which, is previously used., Definition 2.1. Partial derivative of function w.r.t.x: The partial derivative of function, f (x, y) w.r.t.x at point(x0 , y0 ) is, f (x0 + h, y0 ) − f (x0 , y0 ), ∂f, |(x0 ,y0 ) = lim, , if limit exists., h→0, ∂x, h, It is denoted by fx (x0 , y0 ), Note:, d, f (x, y0 ) at x = x0, dx, (2) The slope of the curvez = f (x, y0 ) at the point P (x0 , y0 , f (x0 , y0 )) in the plane y = y0 is, the value of the partial derivative of f w.r.t. x at (x0 , y0 ), The tangent line to the curve at point P is the line in the plane y = y0 that passes through, P with this slope., (1) An equivalent expression for the partial derivative is, , Definition 2.2. Partial derivative of function w.r.t.y: The partial derivative of function, f (x, y) w.r.t.y at point(x0 , y0 ) is, f (x0 , y0 + k) − f (x0 , y0 ), ∂f, |(x0 ,y0 ) = lim, , if limit exists., k→0, ∂y, k, d, It is denoted by fy (x0 , y0 ) or, f (x0 , y)|y=y0, dy, Note:The slope of the curvez = f (x0 , y) at the point P (x0 , y0 , f (x0 , y0 )) in the plane x = x0 is the, value of the partial derivative of f w.r.t. y at (x0 , y0 ), The tangent line to the curve at point P is the line in the plane x = x0 that passes through P with, this slope., ∂f, ∂f, and, of following functions, ∂x, ∂y, (a) f (x, y) = 5xy − 7x2 − y 2 + 3x − 6y + 2 at (2, −3), (b) f (x, y) = sin2 (x − 3y), , Example 2.1. Find partial derivative, , Solution:, (a) Let f (x, y) = 5xy − 7x2 − y 2 + 3x − 6y + 2, ∂f, = 5y − 14x + 3, ∂x, ∂f, |, = 5(−3) − 14(2) + 3, ∂x (2,−3), = −40, ∂f, = 5x − 2y − 6, ∂y, ∂f, |, = 5(2) − 2(−3) − 6, ∂y (2,−3), = 10, 13

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§ 2.1. Definition and Examples, , Chapter 2. Partial Derivatives, , Example 2.3. By using limit definition of partial derivative,, ∂f, ∂f, compute the partial derivative, and, of f (x, y) = 4 + 2x − 3y − xy 2 at (−2, 1), ∂x, ∂y, Solution:f (x, y) = 4 + 2x − 3y − xy 2, f (x0 + h, y0 ) − f (x0 , y0 ), ∂f, |(x0 ,y0 ) = lim, h→0, ∂x, h, ∂f, f (−2 + h, 1) − f (−2, 1), |, = lim, h→0, ∂x (−2,1), h, [4 + 2(−2 + h) − 3 − (−2 + h)] − [4 − 4 − 3 + 2], = lim, h→0, h, [4 − 4 + 2h − 3 + 2 − h] − [−1], = lim, h→0, h, h−1+1, = lim, h→0, h, =1, Similarly,, f (x0 , y0 + k) − f (x0 , y0 ), ∂f, |(x0 ,y0 ) = lim, k→0, ∂y, k, ∂f, f (−2, 1 + k) − f (−2, 1), |(−2,1) = lim, k→0, ∂y, k, [4 + 2(−2) − 3(1 + k) − (−2)(1 + k)2 ] − [−1], = lim, k→0, k, [4 − 4 − 3 − 3k + 2 + 4k + 2k 2 + 1], = lim, k→0, k, 2, 2k + k, = lim, k→0, k, = lim 2k + 1, k→0, , =1, ∂f, at (x0 , y0 , z0 ), Example 2.4. Write the definition of partial derivative, ∂z, ∂f, at (1, 2, 3) for f (x, y, z) = x2 yz 2, and use this to find, ∂z, Solution: Let f (x, y, z) = x2 yz 2 ,, ∂f, f (x0 , y0 , z0 + h) − f (x0 , y0 , z0 ), |(x0 ,y0 ,z0 ) = lim, h→0, ∂z, h, ∂f, f (1, 2, 3 + h) − f (1, 2, 3), |, = lim, h→0, ∂z (1,2,3), h, 2, ∂f, 2(3 + h) − 2(9), |, = lim, h→0, ∂z ((1,2,3)), h, 18 + 12h + 2h2 − 18, = lim, h→0, h, = lim (12 + 2h), h→0, , = 12, ∂x, Example 2.5. Find, at (1, −1, −3), if the equation xz + ylnx − x2 + 4 = 0. defines x as a, ∂z, function of y and z and partial derivative exists., Page|15

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§ 2.1. Definition and Examples, , Chapter 2. Partial Derivatives, Solution:Let xz + ylnx − x2 + 4 = 0, , differentiating partially w.r.t.’z’, ∂x y ∂x, ∂x, x+z, +, − 2x, =0, ∂z, x ∂z, ∂z, ∂x, y, =0, ∴ x + (z + − 2x), x, ∂z, ∂x, −x2, ∴, =, ∂z, zx + y − 2x2, , at point (1,-1,-3), , ∂x, −(1)2, =, ∂z, −3 − 1 − 2, 1, =, 6, Example 2.6. Find the slope of the tangent to the parabola at (1, 2, 5), if the plane x = 1, intersects the paraboloid z = x2 + y 2 in a parabola., Solution: Equation of paraboloid is given by z = x2 + y 2 ., Plane x = 1 intersects this paraboloid into parabola as shown in fig.2.1. The Slope is partial, ∂z, derivative, at point (1, 2), ∂y, , ∴, , ∂z, ∂(x2 + y 2 ), =, ∂y, ∂y, = 2y, , ∂z, |, =4, ∂y (1,2), Example 2.7. If resistors of R1 , R2 ,and R3 ohms are connected in parallel to make an R-ohms, 1, 1, 1, 1, =, resistors , the value of R as, +, +, ., R, R1 R2 R3, ∂R, ,when R1 = 30, R2 = 45,and R3 = 90 ohms., Find, ∂R2, Solution: Since,, , 1, R, ∂ 1, ∴, ( ), ∂R2 R, 1 ∂R, ∴− 2, R ∂R2, 1 ∂R, ∴ 2, R ∂R2, ∂R, ∴, ∂R2, Page|16, , 1, 1, 1, +, +, R1 R2 R3, ∂, 1, 1, 1, =, (, +, +, ), ∂R2 R1 R2 R3, 1, =0−, +0, (R2 )2, 1, =, (R2 )2, R, = ( )2 · · · · · · · · · · · · (a), R2, =

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§ 2.1. Definition and Examples, , Chapter 2. Partial Derivatives, , If R1 = 30, R2 = 45, R3 = 90 then, 1, 1, 1, 1, +, +, =, R, R1 R2 R3, 1, 1, 1, =, +, +, 30 45 90, 1, =, 15, ∴ R = 15, ∴ equation (a) becomes, ∂R, 15, = ( )2, ∂R2, 45, 1, =, 9, EXERCISE 2.1, ∂f, ∂f, (1) Find, and, of following functions., ∂x, ∂y, (i) f (x, y) = x2 + 3xy + y − 1, (ii) f (x, y) = ysin(xy), 2y, (iii) f (x, y) =, y + cosx, (iv) f (x, y) = 2x2 − 3y − 4, (v) f (x, y) = x2 − xy + y 2, x2 − 1, (vi) f (x, y) =, y+2, (vii) f (x, y) = 5xy − 7x2 − y 2 + 3x − 6y + 2, (viii) f (x, y) = (xy − 1)2, 3, (ix) f (x, y) = (2x, p − 3y), 2, 2, (x) f (x, y) = x + y, 2, y, 3, (xi) f (x, y) = [x + ] 3, 2, 1, (xii) f (x, y) =, x+y, x, (xiii) f (x, y) = 2, x + y2, x+y, (xiv) f (x, y) =, xy − 1, y, (xv) f (x, y) = tan−1 ( ), x, (xvi) f (x, y) = ex+y+1, (xvii) f (x, y) = e−x .sin(x + y), (xviii) f (x, y) = ln(x + y), (xix) f (x, y) = exy lny, (xx) f (x, y) = sin2 (x − 3y), (xxi) f (x, y) = cos2 (3x − y 2 ), (xxii) f (x, y) = xy, (xxiii) f (x, y) = logy x, Ry, (xxiv) f (x, y) = g(t)dt , (g is continuous for all t), x, , (2), (i), (ii), (iii), (iv), (v), (vi), (vii), , Find fx , fy and fz of following functions., f (x, y, z) = 1 + xy 2 − 2z 2, f (x, y, z) = xy +pyz + xz, f (x, y, z) = x − y 2 + z 2, 1, −, 2, 2, 2, f (x, y, z) = (x + y + z ) 2, f (x, y, z) = sin−1 (xyz), f (x, y, z) = sec−1 (x + yz), f (x, y, z) = ln(x + 2y + 3z), Page|17

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Chapter 2. Partial Derivatives, (viii), (ix), (x), (xi), (xii), (xiii), , § 2.1. Definition and Examples, , f (x, y, z) = yz ln(xy), 2, 2, 2, f (x, y, z) = e−(x +y +z ), f (x, y, z) = e−(xyz), f (x, y, z) = tanh(x + 2y + 3z), f (x, y, z) = sinh(xy − z 2 ), f (x, y, z) = x sin(y + 3z), , (3) Find the first order partial derivative of the following functions., (i) f (t, α) = cos(2πt − α), 2u, 2, (ii) g(u, v) = v e v, (iii) h(ρ, φ, θ) = ρ sin φ cos θ, (iv) g(r, θ, z) = r(1 − cos θ) − z, (4) By using limit definition of partial derivative, Compute the partial derivative, , ∂f, ∂f, and, ∂x, ∂y, , of, f (x, y) = 1 − x + y − 3x2 y at (1, 2), (5) Write the definition of partial derivative, , ∂f, ∂f, at (x0 , y0 , z0 ) use this to find, at (−1, 0, 3), ∂y, ∂y, , for f (x, y, z) = −2xy 2 + yz 2 ., ∂z, (6) Find, at (1, 1, 1), if the equation xy + z 3 x − 2yz = 0 defines z as a function of x and y, ∂x, and partial derivative exists., ∂z, , if the equation yz − ln z = x + y defines z as a function of x and y and partial, (7) Find, ∂x, derivative exists., ∂f, ∂f, ∂f, ∂f, ANSWERS 2.1 (1) (i), = 2x + 3y,, = 3x + 1 (ii), = y 2 cos xy,, = xy cos(xy) + sin(xy), ∂x, ∂y, ∂x, ∂y, ∂f, 2y sin x, ∂f, 2 cos x, ∂f, ∂f, ∂f, ∂f, (iii), =, ,, =, (iv), = 4x,, = −3 (v), = 2x − y,, = −x + 2y, 2, 2, ∂x, (y + cosx) ∂y, (y + cosx), ∂x, ∂y, ∂x, ∂y, ∂f, ∂f, ∂f, ∂f, ∂f, = 2x(y + 2),, = x2 − 1 (vii), = 5y − 14x + 3,, = 5x − 2y − 6 (viii), =, (vi), ∂x, ∂y, ∂x, ∂y, ∂x, ∂f, ∂f, ∂f, ∂f, x, ∂f, 2y(xy − 1),, = 2x(xy − 1) (ix), = 6(2x − 3y)2 ,, = −9(2x − 3y)2 (x), =p, ,, =, ∂y, ∂x, ∂y, ∂x, x2 + y 2 ∂y, y, 2x2, ∂f, 1, ∂f, −1, ∂f, −1, ∂f, ∂f, p, = r, ,, = r, (xii), =, ,, =, (xiii), =, (xi), 2, 2, 2, 2, ∂x, ∂y, ∂x, (x, +, y), ∂y, (x, +, y), ∂x, y, y, x +y, 3, x3 +, 3 3 x3 +, 2, 2, y 2 − x2 ∂f, −2xy, ∂f, −y 2 − 1 ∂f, −x2 − 1, ∂f, −y, ∂f, ,, =, (xiv), =, ,, =, (xv), = 2, ,, =, 2, 2, 2, 2, 2, 2, 2, 2, 2, (x + y ) ∂y, (x + y ), ∂x, (xy − 1) ∂y, (xy − 1), ∂x, x + y ∂y, x, ∂f, ∂f, ∂f, ∂f, (xvi), = ex+y+1 ,, = ex+y+1 (xvii), = −e−x sin(x + y) + e−x cos(x + y),, =, 2, 2, x +y, ∂x, ∂y, ∂x, ∂y, xy, 1, ∂f, 1, ∂f, ∂f, e, ∂f, e−x cos(x + y) (xviii), =, ,, =, (xix), = yexy ln y,, = xexy ln y +, ∂x, (x + y) ∂y, (x + y), ∂x, ∂y, y, ∂f, ∂f, ∂f, (xx), = 2 sin(x − 3y) cos(x − 3y),, = −6 sin(x − 3y) cos(x − 3y) (xxi), = −6 cos(3x −, ∂x, ∂y, ∂x, ∂f, ∂f, ∂f, ∂f, y 2 ) sin(3x − y 2 ),, = 4y cos(3x − y 2 ) sin(3x − y 2 ) (xxii), = yxy−1 ,, = xy . ln x (xxiii), =, ∂y, ∂x, ∂y, ∂x, 1, ∂f, ln x, ∂f, ∂f, ∂f, y, ∂f, x, ,, =−, (xxiv), = −g(x),, = g(y) (xxv), =, ,, =, x ln y ∂y, y(ln y)2, ∂x, ∂y, ∂x, (1 − xy)2 ∂y, (1 − xy)2, (2)(i)fx = 1 + y 2 , fy = 2xy, fz = −4z (ii)fx = y + z, fy = x + z, fz = y + x (iii)fx = 1, fy =, −3, −3, −y, −z, 2, 2, 2, 2, 2, 2, p, , fz = p, (iv)fx = −x(x +y +z ) 2 , fy = −y(x +y +z ) 2 , fz = −z(x2 +y 2 +, 2, 2, 2, 2, y +z, y +z, −3, yz, xz, xy, 1, p, z 2 ) 2 (v)fx = p, , fy = p, , fz = p, (vi)fx =, , fy =, 2, 2, 2, 2, 2, 2, 2, 2, 2, |x + yz| (x + yz)2 − 1, 1−x y z, 1−x y z, 1−x y z, z, y, 1, 2, p, p, , fz =, (vii)fx =, , fy =, , fz =, 2, 2, x, +, 2y, +, 3z, x, +, 2y, + 3z, |x + yz| (x + yz) − 1, |x + yz| (x + yz) − 1, Page|18

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Chapter 2. Partial Derivatives, , § 2.2. Second Order Partial Derivative, , y, Solution: Let f (x, y) = tan−1 ( ), x, , ∂ y, ∂f, 1, =, ( ), y, ∂x, 1 + ( )2 ∂x x, x, 2, −y, x, (, ), = 2, x + y 2 x2, −y, = 2, x + y2, ∂f, ∂ y, 1, also,, =, y 2 ∂y ( x ), ∂y, 1+( ), x, 1, x2, ( ), = 2, 2, x +y x, x, = 2, x + y2, ∂2f, ∂ ∂f, =, ( ), ∂x2, ∂x ∂x, ∂, −y, =, ( 2, ), ∂x x + y 2, ∂, 1, = −y ( 2, ), ∂x x + y 2, −1, ∂, = −y[ 2, ] (x2 + y 2 ), (x + y 2 )2 ∂x, y, = 2, (2x), (x + y 2 )2, 2xy, = 2, (x + y 2 )2, ∂2f, ∂ ∂f, =, ( ), 2, ∂y, ∂y ∂y, ∂, x, =, ( 2, ), ∂y x + y 2, ∂, 1, ), = x. ( 2, ∂y x + y 2, −1, ∂, = x[ 2, ] (x2 + y 2 ), (x + y 2 )2 ∂y, −2xy, = 2, (x + y 2 )2, ∂2f, ∂ ∂f, =, ( ), ∂x∂y, ∂x ∂y, ∂, x, =, ( 2, ), ∂x x + y 2, (x2 + y 2 ) − x(2x), =, (x2 + y 2 )2, y 2 − x2, = 2, (x + y 2 )2, Page|20

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§ 2.2. Second Order Partial Derivative, , Chapter 2. Partial Derivatives, , ∂2f, ∂ ∂f, =, ( ), ∂y∂x, ∂y ∂x, ∂, −y, =, ), ( 2, ∂y x + y 2, (x2 + y 2 )(−1) − (−y)2y, =, (x2 + y 2 )2, y 2 − x2, = 2, (x + y 2 )2, Theorem 2.1. The Mixed Derivative Theorem (Clairaut’s Theorem):, Statement:If f (x, y) and its partial derivatives fx , fy , fxy , fyx are defined throughout an open region containing a point (a, b) and are all continuous at (a, b) then, fxy (a, b) = fyx (a, b), Proof. Given:Functions f, fx , fy , fxy , fyx are all defined on an open region containing point, (a, b).Also all are continuous at point (a, b)., Claim:fxy (a, b) = fyx (a, b)., Since,f, fx , fy , fxy , fyx are all defined in the interior of a rectangle R in the xy plane containing, point (a, b)., Let h and k be the numbers such that the point (a + h, b + k) also lies in R. Consider,, (1), , ∆ = F (a + h) − F (a), , (2), , Where, F (x) = f (x, b + k) − f (x, b), , apply the Mean Value theorem to F on (a, a + h),which is continuous because it is differentiable., ∴ Equation (1) becomes,, ∆ = hF 0 (c1 ), c1 ∈ (a, a + h), , (3), From Equation (2), , F 0 (x) = fx (x, b + k) − fx (x, b), ∴ Equation (3) becomes,, ∆ = h[fx (c1 , b + k) − fx (c1 , b)], , (4), , apply Mean Value theorem to function g(y) = fx (c1 , y), ∴ g(b + k) − g(b) = kg 0 (d1 ), d1 ∈ (b, b + k), ∴ fx (c1 , b + k) − fx (c1 , b) = kfxy (c1 , d1 ), Equation (4) becomes ,, (5), , ∆ = hkfxy (c1 , d1 ), f or some point(c1 , d1 ) ∈ R0, , Now,By using equation (2),equation (1)becomes, ∆ = f (a + h, b + k) − f (a + h, b) − f (a, b + k) + f (a, b), ∆ = [f (a + h, b + k) − f (a, b + k)] − [f (a + h, b) − f (a, b)], (6), , ∆ = φ(b + k) − φ(b), , (7), , where, φ(y) = f (a + h, y) − f (a, y), , apply Mean Value theorem to equation (6) we get,, ∆ = kφ0 (d2 ), d2 ∈ (b, b + k), , (8), from equation (7), , φ0 (y) = fy (a + h, y) − fy (a, y), , (9), ∴ Equation (8) becomes ,, , ∆ = k[fy (a + h, d2 ) − fy (a, d2 )], apply Mean Value theorem to function fy (x, d2 ),we get, fy (a + h, d2 ) − fy (a, d2 ) = hfyx (c2 , d2 ), c2 ∈ (a, a + h), Page|21

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§ 2.3. Higher Order Partial Derivative, , Chapter 2. Partial Derivatives, , Solution: Let f (x, y, z) = 2z 3 − 3(x2 + y 2 )z, ∂f, = −6xz, ∂x, ∂2f, = −6z, ∂x2, ∂f, = −6yz, also,, ∂y, ∂2f, = −6z, ∂y 2, ∂f, also,, = 6z 2 − 3(x2 + y 2 ), ∂z, ∂2f, = 12z, ∂z 2, ∂2f, ∂2f, ∂2f, +, +, = −6z − 6z + 12z, ∴, ∂x2, ∂y 2, ∂z 2, ∂2f, ∂2f, ∂2f, ∴, +, +, =0, ∂x2, ∂y 2, ∂z 2, ∴ f (x, y, z) satisfies Laplace equation. EXERCISE 2.2, (1) Find all second order partial derivatives of the following functions., (i) f (x, y) = x + y + xy, (ii) f (x, y) = sin xy, (iii) g(x, y) = x2 y + cos y + y sin x, (iv) h(x, y) = xey + y + 1, (v) r(x, y) = ln(x + y), y, (vi) s(x, y) = tan−1 ( ), x, (2) Verify that Wxy = Wyx for the following functions., (i) W = ln(2x + 3y), (ii) W = ex + x ln y + y ln x, (iii) W = xy 2 + x2 y 3 + x3 y 4, (iv) W = x sin y + y sin x + xy, (v) W = log(x − 2y), y, (vi) W = tan−1 ( ), x, (vii) W = ax2 + 2hxy + by 2, x, (viii) W = sin−1 ( ), y, (ix) W = log(x sin y + y sin x), y, (x) W = log tan( ), x, 1, (xi) W = x2 y 2 cos( ), x2, (xii) W = x3 y + exy, (3) Show that each of following function satisfies a Laplace equation., (i) f (x, y, z) = x2 + y 2 − 2z 2, (ii) f (x, y, z) = 2z 3 − 3(x2 + y 2 )z, (iii) f (x, y) = e−2y, p cos(2x), (iv) f (x, y) = ln x2 + y 2, −1, 2, 2, 2, (v) f (x, y, z) = (x + y + z ) 2, (vi) f (x, y, z) = e3x+4y cos(5z), (vii) f (x, y) = log(x2 + y 2 ), y, (viii) f (x, y) = a log(x2 + y 2 ) + b tan−1 ( ), x, (ix) f (x, y) = 2(ax + by)2 − (x2 + y 2 ) , a2 + b2 = 1, (4) Show that each of following function are solutions of the wave equation, (i) W = sin(x + ct), , 2, ∂2W, 2∂ W ., =, c, ∂t2, ∂x2, , Page|23

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Chapter 2. Partial Derivatives, (ii), (iii), (iv), (v), (vi), (vii), , = cos(2x + 2ct), = sin(x + ct) + cos(2x + 2ct), = ln(2x + 2ct), = tan(2x − 2ct), = 5 cos(3x + 3ct) + ex+ct, = f (u) , f is differentiable function of u and u = a(x + ct) , a is constant., x, ∂2f, x2 − y 2, y, = 2, (5) If f (x, y) = x2 tan−1 ( ) − y 2 tan−1 ( ) then prove that, x, y, ∂y∂x, x + y2, 3, 3, 2, 2, (6) If f (x, y) = log(x + y − x y − xy )then show that, ∂f, ∂f, 2, +, =, ∂x ∂y, x+y, (7) If f (x, y) = log(tan x + tan y + tan z)then show that, ∂u, ∂u, ∂u, sin 2x, + sin 2y, + sin 2z, =2, ∂x, ∂y, ∂z, x2 + y 2, then show that, (8) If f (x, y) =, x+y, ∂u ∂u 2, ∂u ∂u, (x − y)2, [, −, ] = 4[1 −, −, ]=4, ∂x ∂y, ∂x ∂y, (x + y)2, 1, −, 2, 2, 2, (9) If u = (x + y + z ) 2 then show that, ∂u, ∂u, ∂u, ( )2 + ( )2 + ( )2 = u4, ∂x, ∂y, ∂z, (10) If z = f (x + ay) + φ(x − ay) then prove that, 2, ∂2z, 2∂ z, =, a, ∂y 2, ∂x2, 1 ∂2z, 1 ∂2z, (11) Evaluate 2 2 + 2 2 if,a2 x2 + b2 y 2 − c2 z 2 = 0, a ∂x, b ∂y, ∂2f, (12) If f (x, y, z) = log(x2 + y 2 + z 2 ) Find, ∂y∂z, 3f, ∂, (13) If f (x, y, z) = exyz then show that, = (1 + 3xyz + x2 y 2 z 2 )exyz, ∂x∂y∂z, 1, 2, 2, 2, (14) If u = (x + y + z ) 2 then prove that, ∂2u ∂2u ∂2u, 2, + 2 + 2 =, 2, ∂x, ∂y, ∂z, u, ANSWERS 2.2 (1) (i)fxx = 0, fyy = 0, fxy = 1, fyx = 1 (ii)fxx = −y 2 sin xy, fyy = −x2 sin xy, fxy =, fyx = cos xy − xy sin xy (iii)gxx = 2y − y sin x, gyy = − cos y, gxy = gyx = 2x + cos x (iv)hxx =, −1, −1, −1, 0, hyy = xey , hxy = hyx = ey (v)rxx =, , ryy =, , rxy = ryx =, (vi)sxx =, (x + y)2, (x + y)2, (x + y)2, 2xy, −2xy, y 2 − x2, 1, ,, s, =, ,, s, =, s, =, (11) p, yy, xy, yx, 2, 2, 2, 2, 2, 2, 2, 2, 2, (x + y ), (x + y ), (x + y ), [c a2 x2 + b2 y 2 ], −4yz, (12)] 2, (x + y 2 + z 2 )2, , Page|24, , W, W, W, W, W, W, , § 2.3. Higher Order Partial Derivative

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Chapter 3. Differentiability, Similarly, fy (0, 0) = 0 · · · (ii), From (i) and (ii) both the first partial derivatives of f (x, y) exists at (0, 0)., Now, suppose that f is differentiable at (0, 0) then by the definition of differentiability, f (∆x, ∆y) − f (0, 0) = fx (0, 0)∆x + fy (0, 0)∆y + ε1 ∆x + ε2 ∆y;, p, p, ∴ |∆x · ∆y| − |0| = 0 · ∆x + 0 · ∆y + ε1 ∆x + ε2 ∆y · · · (iii), ε1 , ε2 → 0 as (∆x, ∆y) → (0, 0)., Since (iii) holds for all small values of ∆x and ∆y, put ∆y = ∆x in (iii), we get, p, |(∆x)2 | = ε1 ∆x + ε2 ∆x, ∴ |∆x| = ∆x(ε1 + ε2 ), |∆x|, ∴, = ε1 + ε2, ∆x, Taking limit as ∆x → 0 of both sides., |∆x|, ∴ lim, = lim (ε1 + ε2 ), ∆x→0 ∆x, ∆x→0, ∴ ±1 = 0;, which is absurd. Hence f is not differentiable at (0, 0)., Moreover,, For continuity of f (x, y) at (0, 0). Consider, p, |f (x, y) − f (0, 0)| = | |xy||, √ √, x· y, =, ≤ x2 + y 2 < ε, p, √, ..., x ≤, x2 + y 2, p, √, y ≤, x2 + y 2, p, p, √, ⇒ x2 + y 2 < ε(= δ) Thus, |f (x, y) − f (0, 0)| < ε, whenever x2 + y 2 < δ, ⇒ f (x, y) is continuous at (0, 0)., example 2: Show that the function f (x, y) = |x|(1 + y) is not differentiable at (0, 0) but is, continuous at (0, 0)., Solution : Given that, f (x, y) = |x|(1 + y)., (x0 , y0 ) = (0, 0), ∴ f (x0 , y0 ) = f (0, 0), = 0, Consider, f (x0 + ∆x, y0 ) − f (x0 , y0 ), ∆x→0, ∆x, , f (∆x, 0) − f (0, 0), ∆x→0, ∆x, |∆x|(1 + 0) − 0, = lim, ∆x→0, ∆x, |∆x|, = lim, ∆x→0 ∆x, =, , lim, , lim, , Now, |∆x|, lim, +, ∆x, ∆x→0, |∆x|, lim, ∆x→0− ∆x, , , , , ∆x, =, lim, = 1 · · · (i), ∆x→0+ ∆x, , , −∆x, =, lim, = −1 · · · (ii), ∆x, ∆x→0−, , , , , |∆x|, .., does not exist : ( . by (i) and (ii)), ∴ lim, ∆x→0, ∆x, f (∆x, 0) − f (0, 0), i.e. lim, does not exist, which means that fx (0, 0) does not exist., ∆x→0, ∆x, Since existence of fx (0, 0) and fy (0, 0) is a necessary condition for differentiability, therefore f is, not differentiable at (0, 0)., Page|27

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Chapter 3. Differentiability, To show that f (x, y) is continuous at (0, 0) we will use ε − δ definition., Let ε > 0. Consider, |f (x, y) − f (0, 0)| =, =, ≤, ∴ |f (x, y) − f (0, 0)| ≤, , |f (x, y) − 0| = ||x|(1 + y)|, |x||1 + y|, 2|x|, if |y| < 1, 2|x| < ε, ε, ∴ |f (x, y) − f (0, 0)| < ε, if |x| < = δ, 2, nε, , o, , then |f (x, y) − f (0, 0)| < ε when |x| < δ, |y| < δ, 2, ∴ lim f (x, y) = 0 = f (0, 0) ⇒ f (x, y) is continuous at (0, 0)., Take δ = min, , ,1, , ∆x→0, , example 3: Let, 2xy, f (x, y) =, , if (x, y) 6= (0, 0), 2, x + y2, =, 0,, if (x, y) = (0, 0), Show that f (x, y) is not differentiable at (0, 0) even though fx (0, 0) and fy (0, 0) exists., Solution : First let us show that fx (0, 0) & fy (0, 0) exist, f (∆x, 0) − f (0, 0), ∆x→0, ∆x, 0−0, = lim, ∆x→0 ∆x, = 0, , fx (0, 0) =, , lim, , Similarly fy (0, 0) = 0 i.e. both fx (0, 0) & fy (0, 0) exist., Now, we will find the limit of f (x, y) along a path y = mx, m 6= 0., ∴, , lim, , f (x, y) =, , (x,y)→(0,0), , lim, , , f (x, mx), , 2x · mx, x2 + m2 x2, , (x,mx)→(0,0), , =, , lim, , x→0, , 2m, 1 + m2, f (x, y) does not exist., =, , which depends upon the path. i.e., , lim, (x,y)→(0,0), , Hence, f is not continuous at (0, 0)., Therefore f is not differentiable at (0, 0)., example 4: Let, x2 − y 2, , (x, y) 6= (0, 0), x2 + y 2, =, 0,, (x, y) = (0, 0), Show that f (x, y) is differentiable at (0, 0)., f (x, y) = 2xy, , Solution :, f (∆x, 0) − f (0, 0), ∆x→0, ∆x, 0−0, = lim, ∆x→0 ∆x, = 0, , fx (0, 0) =, , lim, , Similarly fy (0, 0) = 0., ∴ fx (0, 0) and fy (0, 0) both exist., Now,, ∆f, ∆f, , = f (x0 + ∆x, y0 + ∆y) − f (x0 , y0 ), = f (∆x, ∆y) − f (0, 0), , ∴ f (∆x, ∆y) − f (0, 0) = 0 · ∆x + 0 · ∆y + ε1 ∆x + ε2 ∆y; where, Page|28

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Chapter 3. Differentiability, 2(∆x)2 ∆y, , if (∆x, ∆y) 6= (0, 0), (∆x)2 + (∆y)2, 0,, if (∆x, ∆y) = (0, 0), , ε1 =, =, and, , −2(∆x)(∆y)2, , if (∆x, ∆y) 6= (0, 0), (∆x)2 + (∆y)2, =, 0,, if (∆x, ∆y) = (0, 0), Here as (∆x, ∆y) → (0, 0), ε1 → 0, ε2 → 0. ∴ f (∆x, ∆y) − f (0, 0) = fx (0, 0)∆x + fy (0, 0)∆y +, ε1 ∆x + ε2 ∆y; ε1 → 0, ε2 → 0 as (∆x, ∆y) → (0, 0), Hence by the definition, f (x, y) is differentiable at (0, 0)., ε2 =, , Theorem 3: (Sufficient Conditions for Differentiability :-), If f (x, y) is a function of two variables x and y such that, (i) fx (a, b) and fy (a, b) exist, (ii) One of the first partial derivatives fx , fy is continuous at (a, b)., Then f (x, y) is differentiable at (a, b)., Proof : Suppose fy is continuous at (a, b) ⇒ fy exist in the neighbourhood of (a, b), (say square, δ- neighbourhood of (a, b)) i.e ∃δ > 0 so that the point (a + h, b + k) lies in the δ-neighbourhood, of (a, b) where |h| < δ, |k| < δ., Now, ∆f, , = f (a + h), b + k) − f (a, b), = f (a + h, b + k) − f (a + h, b) + f (a + h, b) − f (a, b) · · · ?, , Dfine the function g(y) as g(y) = f (a + h, y), Here g is derivable in (b, b + k) and we have g 0 (y) = fy (a + h, y)., Also g is continuous in [b, b + k]., Hence by LM V T (IInd form), g(b + k) − g(b) = kg 0 (b + kθ); 0 < θ < 1., i.e. f (a + h, b + k) − f (a + h, b) = kfy (a + h, b + kθ) · · · (1), Since fy is continuous at (a, b), lim, fy (a + h, b + kθ) = fy (a, b), , (h,k)→(0,0), , lim, , [fy (a + h, b + kθ) − fy (a, b)] = 0, , (h,k)→(0,0), , If we put fy (a + h, b + kθ) − fy (a, b) = ψ(h, k) then, , lim, , ψ(h, k) = 0., , (h,k)→(0,0), , With this equation (1) becomes, f (a + h, b + k) − f (a + h, b) = k(fy (a, b) + ψ(h, k)), f (a + h, b + k) − f (a + h, b) = kfy (a, b) + kψ(h, k) · · · (2), Now we hare, , f (a + h, b) − f (a, b), fx (a, b) = lim, h→0, h, , , f (a + h, b) − f (a, b), − fx (a, b) = 0., ∴ lim, h→0, h, f (a + h, b) − f (a, b), Put φ(h) =, − fx (a, b) then lim φ(h) = 0 i.e. φ(h) → 0 as (h, k) → (0, 0)., h→0, h, ∴ f (a + h, b) − f (a, b) = hfx (a, b) + hφ(h, k) · · · (3), Putting (2), (3) and (1) in ? we get, ∆f = f (a + h, b + k) − f (a, b) = hfx (a, b) + kfy (a, b) + hφ(h, k) + kψ(h, k); where φ(h, k) → 0 and, ψ(h, k) → 0 as (h, k) → (0, 0)., Hence, by the definition of differentiability, f (x, y) is differentiable at (a, b)., Differentials : Let z = f (x, y) be a differentiable function of two variables x and y. The differential, ∂z, ∂z, or total differential of z; denoted by dz; is defined as dz =, dx +, dy; where dx and dy (are, ∂x, ∂y, called the differentials of x and y) are two new independent variables., Page|29

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Chapter 3. Differentiability, ∴ df, ∴ df, , = fx (x0 , y0 )∆x + fy (x0 , y0 )∆y, 1, 1, =, (0 · 1) −, × (0 · 01), 20, 125, = 0 · 005 − 0 · 00008, = 0 · 00492., , ∴ df, Hence,, f (x0 + r, ∆x, y0 + ∆y) ≈ f (x0 , y0 ) + df, 4·1, 2, ≈, + 0 · 00492, ∴, 25 · 01, 3, = 0 · 4 + 0 · 00492, =, 0 · 40492., Exercise (A), (1) Show that the function, xy, f (x, y) = p, , if x2 + y 2 6= 0, 2, 2, x +y, =, 0,, if x = y = 0., is continuous at the origin but is not differentiable at the origin., (2) If,  ,  , 1, 1, f (x, y) = x sin, + y sin, , y 6= 0, x, y, =, 0,, x, y = 0., 4, 4, x +y, (3) If f (x, y) = 2, , x2 + y 2 6= 0 and f (0, 0) = 0. Show that f (x, y) is differentiable at, x + y2, (0, 0)., x3 − y 3 2, (4) Show that f (x, y) = 2, , x + y 2 6= 0 is continuous at (0, 0) but not differentiable at, x + y2, (0, 0)., (5) Using differentials, find the approximate value of, (i) [3p, · 8)2 + 2(2 · 1)3 ]1/5 (ii) f (5 · 12, 6 · 85), (iii) (5 · 98)2 + (8 · 01)2 (iv) f (5 · 04 × 9 · 98)., (6) If f (x, y) = x3 − xy + 3y 2 then find f (4 · 98, 4 · 1)., (7) If f (x, y, z) = (x2 + y 2 + z 2 )1/2 , find f (0 · 1, 4 · 01, 3 · 1)., (8) Let z = x2 − xy + 3y 2 then find dz if x = 5, y = 4, ∆x = −2, ∆y = 1., Answers, (5) (i) 2 · 01 (ii) 5 · 10 (iii) 9 · 996 (iv) 50 · 3 (6) 140 · 7, (7) 5 · 068 (8) −123., 3.2 : Composite Function : Chain Rule., For function of one variable, y = f (x) and x = φ(t) then y = f (φ(t)) is called a composite function, of t., dy, dy dx, Its derivative w.r.t. 0 t0 is given by, =, ·, which is known as Chain Rule., dt, dx dt, For function of two variables also we will have composite function and Chain Rule., 1. Suppose u = f (x, y) is a function of two independent variables x, y and x, y are themselves, function of single variable, 0 t0 i.e. x = φ(t), y = ψ(t) then, u = f [φ(t), ψ(t)] = F (t) is called a composite function of single variable0 t0 ., e.q.:(1) u = f (x, y) = x + y and x = at, y = bt2, then u = f (at, bt2 ) = at + bt2 is a composite function of single variable 0 t0 ., (2) u = sin(x + y 2 ) and x = cost, y = t2 + 1 then, u = sin[cost + (t2 + 1)2 ], a composite function of 0 t0 ., 2. Suppose W = f (u, v) is a function of two variables u, v and u, v are functions of two, variables x and y, i.e. u = φ(x, y), v = ψ(x, y) then, W = f [φ(x, y), ψ(x, y)] = F (x, y) is called a composite function of two variables x and y., Page|31

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Chapter 3. Differentiability, e.q.: W = f (u, v) and u = x + y, v = x − y then, W = f (x + y, x − y) is a composite function of x and y., 3. Suppose Z = f (x) is a function of one variable x and x is itself a function of two variables, u and v i.e. x = φ(u, v) then, Z = f (φ(u, v)) is a composite function two variables u and v., e.q.: Z = f (u); u = ax + by then Z = f (ax + by) is a composite function of x and y., Theorem 4: Chain Rule (I):, If u = f (x, y) is a differentiable function of x and y, x = φ(t), y = ψ(t) are themselves functions of, single variable 0 t0 . Then the composite function u = f [φ(t), ψ(t)] is differentiable function of single, du, ∂u dx ∂u dy, variable 0 t0 and its total derivative is given by, =, ·, +, · ., dt, ∂x dt, ∂y dt, Proof : We have given that u = f (x, y) and x = φ(t), y = ψ(t). Let ∆x = φ(t + ∆t) − φ(t) and, ∆y = ψ(t + ∆t) − ψ(t) be the increments in x and y respectively corresponding to an increment, ∆t in t., ∂u, ∂u, ∆x+ ∆y+ε1 ∆x+ε2 ∆y · · · (1), Since u = f (x, y) is differentiable, by increment theorem ∆u =, ∂x, ∂y, where ε1 → 0, ε2 → 0 as (∆x, ∆y) → (0, 0)., , , , , ∂u, ∂u, + ε1 ∆x +, + ε2 ∆y, ∴ ∆u =, , ,  ∂x, ∂y, ∆u, ∆x, ∆y, ∂u, ∂u, ∴, =, + ε1, +, + ε2, · · · (2), ∆t, ∂x, ∆t, ∂y, ∆t, As x = φ(t), y = ψ(t) are differentiable functions at t, ∴ they are continuous at t and hence ∆x, ∆y → 0 as ∆t → 0., ∴ ε1 → 0, ε2 → 0 as ∆t → 0., dx, ∆y, dy, ∆x, =, and lim, =, ., Also lim, ∆t→0 ∆t, ∆t→0 ∆t, dt, dt, Taking limit as ∆t → 0 of equation (2), , ,  , , ,  ,  , ∆u, ∂u, ∂u, ∆x, ∆y, lim, = lim, + ε1, + lim, + ε2, lim, lim, ∆t→0 ∆t, ∆t→0 ∂x, ∆t→0 ∂y, ∆t→0 ∆t, ∆t→0 ∆t, du, ∂u dx ∂u dy, ⇒, =, ·, +, ·, dt, ∂x dt, ∂y dt, which proves Chain Rule (I)., Theorem 5: Chain Rule (II):, If W = f (u, v) is a differentiable function of two variables u and v., u = φ(x, y), v = ψ(x, y) are differentiable functions of x and y then the composite function, W = f [φ(x, y), ψ(x, y)] = F (x, y) is also, differentiable and, ∂w, ∂x, ∂w, ∂y, , =, =, , ∂w, ∂u, ∂w, ∂u, , ∂u ∂w ∂v, +, ·, ,, ∂x, ∂v ∂x, ∂u ∂w ∂v, ·, +, ·, ∂y, ∂v ∂y, ·, , Proof : Since u, v and w are differentiable functions, by Chain Rule (I), ∆u =, ∆v =, ∆w =, , ∂u, ∂u, ∆x +, ∆y + ε1 ∆x + ε2 ∆y · · · (1), ∂x, ∂y, ∂v, ∂v, ∆x +, ∆y + ε3 ∆x + ε4 ∆y · · · (2), ∂x, ∂y, ∂w, ∂w, ∆u +, ∆v + ε5 ∆u + ε6 ∆v · · · (3);, ∂u, ∂v, , where ε1 , ε2 , ε3 , ε4 → 0 as (∆x, ∆y) → (0, 0) and, ε5 , ε6 → 0 as (∆u, ∆v) ⇒ (0, 0)., Page|32

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Chapter 3. Differentiability, , Now ∆w =, , , , , ∂w, ∂w, + ε5 ∆u +, + ε6 ∆v ∴ by (3), Vsing (1) and (2) in above we get, ∂u, ∂v, , , , ∂w, ∂u, ∂u, ∆w =, + ε5, ∆x +, ∆y + ε1 ∆x + ε2 ∆y, ∂u, ∂x, ∂y, , , , ∂v, ∂v, ∂w, + ε6, ∆x +, ∆y + ε3 ∆x + ε4 ∆y, +, ∂v, ∂x, ∂y, , , , , ∂w ∂u ∂w ∂v, ∂w ∂u ∂w ∂v, ∆x +, ∆y, ∴ ∆w =, ·, +, ·, ·, +, ·, ∂u ∂x, ∂v ∂x, ∂u ∂y, ∂v ∂y, +α1 ∆x + α2 ∆y · · · (4);, , where α1 and α2 are sum of terms each containing the factors ε1 , ε2 , · · · ε6 ., ∴ α1 → 0, α2 → 0 as (∆x, ∆y) → (0, 0)., From (4) w = F (x, y) is differentiable at (x, y)., ∆w, ∂w ∂u ∂w ∂v, Now put ∆y = 0 and divide by ∆x; Equation (4) becomes, =, ·, +, ·, + α1 Taking, ∆x, ∂u ∂x ∂v ∂x, ∂w, ∂w ∂u ∂w ∂v, limit as ∆x → 0., =, ·, +, ·, Similarly put ∆x = 0 and divide by ∆y to equation, ∂x, ∂u ∂x, ∂v ∂x, ∂w, ∂w ∂u ∂w ∂v, (4); taking limit as ∆y → 0 we get, =, ·, +, ·, which proves the theorem., ∂y, ∂u ∂y, ∂v ∂y, # The Chain Rule for functions of three variables:If W = f (x, y, z) is a differentiable function of three variables x, y, z and x, y, z are differentiable, functions of single variable 0 t0 then the composite function w = f (t) is also differentiable function, dw, ∂w dx ∂w dy ∂w dz, of 0 t0 and its derivative is, =, ·, +, ·, +, · . # Chain Rule for functions on, dt, ∂x dt, ∂y dt, ∂z dt, surfaces:If W = f (x, y, z) is a differentiable function of three variables x, y, z and x = g(r, s), y = h(r, s), z =, k(r, s) are differentiable functions of two variables r and s. Then the composite function, ∂w ∂x, ∂w, =, ·, +, w = f [g(r, s), h(r, s), k(r, s)] = F (r, s) is differentiable function of r and s; and, ∂r, ∂x ∂r, ∂w ∂y ∂w ∂z ∂w, ∂w ∂x ∂w ∂y ∂w ∂z, ·, +, · ,, =, ·, +, ·, +, · , # Chain Rule for functions of many, ∂y ∂r ∂z ∂r ∂s, ∂x ∂s ∂y ∂s ∂z ∂s, variables:If W = f (x1 , x2 , x3 , · · · , xn ) is a differentiable function of a finite set of varibles x1 , x2 , x3 , · · · , xn, and each x1 , x2 , x3 , · · · xn is a differentiable function of a finite set of variables p1 , p2 , p3 , · · · , pr ., Then W = f [p1 , p2 , p3 , · · · , pr ] is differentiable function of finite set of variables p1 , p2 , p3 , · · · , pr, and we have, ∂w, =, ∂p1, ∂w, =, ∂p2, so on, ∂w, =, ∂pr, , ∂w, ∂x1, ∂w, ∂x1, , ∂x1, ∂w ∂x2, ∂w ∂x3, +, ·, +, ·, + ··· +, ∂p1, ∂x2 ∂p1, ∂x3 ∂p1, ∂x1, ∂w ∂x2, ∂w ∂x3, ·, +, ·, +, ·, + ··· +, ∂p2, ∂x2 ∂p2, ∂x3 ∂p2, ·, , ∂w, ∂xn, ∂w, ∂xn, , ∂xn, ,, ∂p1, ∂xn, ·, ,, ∂p2, ·, , ∂w ∂x1, ∂w ∂x2, ∂w ∂x3, ∂w ∂xn, ·, +, ·, +, ·, + ··· +, ·, ∂x1 ∂pr, ∂x2 ∂pr, ∂x3 ∂pr, ∂xn ∂pr, , ∂w, ∂w, −a, = 0., ∂x, ∂y, Solution : We have given that w = f (ax + by) put u = ax + by then w = f (u)., ∴ By Chain Rule, example 1: If w = f (ax + by) then show that b, , ∂w, ∂x, ∂w, ∴b, ∂x, ∂w, ∂y, ∂w, ∴a, ∂y, , dw ∂u, dw, ·, =a·, du ∂x, du, dw, = ab, · · · (1), du, =, , dw ∂u, dw, ·, =b·, du ∂y, du, dw, = ab, · · · (2), du, =, , Page|33

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Chapter 3. Differentiability, ∂w, ∂w, −a, = 0., ∂x, ∂y, example 2: If z = f (y + ax) + g(y − ax) prove that zxx = a2 zyy , assuming that second order, partial derivative of f, g exist and a is constant., From (1) and (2) b, , Solution : Put u = y + ax, v = y − ax. Hence, z = f (u) + g(v); where, u = φ(y, x) = y + ax, v = ψ(y, x) = y − ax, ∴ By Chain Rule, zx =, , ∂z, ∂x, , ∂z ∂u ∂z ∂v, ·, +, ·, ∂u ∂x ∂v ∂x, = f 0 (u)a + g 0 (v) · (−a), , zx =, , ∂z, ∂x, , = a(f 0 (u) − g 0 (v)) · · · (1), , =, , Again differentiate w.r.t. x., ∂, ∂u, ∂, ∂v, ∂2z, =, [a(f 0 (u) − g 0 (v))] ·, +, [a(f 0 (u) − g 0 (v))] ·, 2, ∂x, ∂u, ∂x ∂v, ∂x, 2 00, 2 00, = a f (u) + a g (v) · · · (2), , zxx =, zxx, Now zy =, , ∂z, ∂z ∂u, ∂z ∂v, =, ·, +, ·, = f 0 (u) + g 0 (v) · · · (3) Differentiate w.r.t. y again, we get, ∂y, ∂u ∂y, ∂v ∂y, , ∂2z, = f 00 (u) + g 00 (v) · · · (4) from (2) and (4) zxx = a2 zyy proved., ∂y 2, y, example 3: If u = xy 2 log, then find du., x, ∂u, ∂u, dx +, dy · · · (1) Now, Solution : We know that du =, ∂x, ∂y,  , y, ∂u, −1, 2 1, 2, + xy · y, = y log, ·y, ∂x, x, x2, x, y, 2, 2, = y log, − y · · · (2), x, zyy =, , 1 1, + xy 2 · y ·, x, x, x, y, = 2xy log, + xy · · · (3), x, h, y, i, h, y, i, From (2) and (3) equation (1) becomes du = y 2 log, − y 2 dx + 2xy log, + xy dy exx, x, du, 3, 3, ample 4: If u = x + y ; x = a cos t, y = b sin t then find, . Also verify the result., dt, Solution : We have given that u = x3 + y 3 ; x = a cos t, y = b sin t., ∴ u is a composite function of single variable t., ∂u, ∂y, , du, dt, du, dt, , = 2xy log, , =, , y, , ∂u dx ∂u dy, ·, +, ·, ∂x dt, ∂y dt, , = 3x2 (−a sin t) + 3y 2 (b cos t), , du, = −3a3 cos2 t sin t + 3b3 sin2 t cos t · · · (1) Verification :, dt, We have u = x3 + y 3 putting values of x and y u = a3 cos3 t + b3 sin3 t. Differentiate w.r.t. t., du, ∴, = −3a3 cos2 t · sin t + 3b3 sin2 t · cos t which is equation (1)., dt, , , y−x z−x, ∂u, ∂u, ∂u, example 5: If u = u, ,, , show that x2, + y2, + z2, = 0 Solution : We have, xy, xz, ∂x, ∂y, ∂z, Putting values of x and y we get, , Page|34

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Chapter 3. Differentiability, , given that u = u, , y−x z−x, ,, xy, xz, , , , put, , y−x, 1 1, = − and, xy, x y, z−x, 1 1, s =, = −, xz, x z, ∴ u = u(r, s) i.e. u is composite function of x and y., ∂u ∂r, ∂u ∂s, ∂r, 1 ∂r, 1 ∂r, ∂u, =, ·, +, ·, · · · (1) Here, = − 2,, = 2,, = 0. and, ∴ Chain Rule, ∂x, ∂r ∂x ∂s ∂x, ∂x, x ∂y, y ∂z, ∂s, 1 ∂s, ∂s, 1, = − 2,, = 0,, = 2., ∂x, x ∂y, ∂z, z, ∂u, ∂u ∂u, ∴ Equation (1) becomes x2, =−, −, · · · (2) Again, ∂x, ∂r, ∂s, ∂u, ∂u ∂r ∂u ∂s, =, ·, +, ·, ∂y, ∂r ∂y, ∂s ∂y, ∂u, ∂u, ∴ y2, =, · · · (3), ∂y, ∂r, r =, , Also, ∂u, ∂u ∂r ∂u ∂s, =, ·, +, ·, ∂z, ∂r ∂z, ∂s ∂z, ∂u, ∂u, ∴ z2, =, · · · (4), ∂z, ∂s, ∂u, ∂u, ∂u, + y2, + z2, = 0. example 6: If ϕ(x − az, cy − bz) = 0, Adding (2), (3) and (4) we get x2, ∂x, ∂y, ∂z, ∂z, ∂z, show that ap + bq = c; where p =, ,q =, ., ∂x, ∂y, Solution : We have ϕ(cx − az, cy − bz) = 0. Put r = cx − az, s = cy − bz so that ϕ(r, s) = 0., Differentiate r and s partially w.r.t. x, y., ∴, , ∂r, ∂x, ∂s, ∂x, , ∂z ∂r, ∂z, ,, = −a, and, ∂x ∂y, ∂y, ∂z ∂s, ∂z, = −b ,, =c−b ., ∂x ∂y, ∂y, = c−a, , ∂ϕ, ∂ϕ ∂r, ∂ϕ ∂s, =, ·, +, ·, ∂x, ∂r ∂x, ∂s ∂x, , , , , ∂z, ∂ϕ, ∂z, ∂ϕ, ∴0 =, c−a, +, −b, ∂r, ∂x, ∂s, ∂x, , , ∂ϕ ∂z, ∂ϕ, ∂ϕ, 0 = c, +, −a, −b, ∂r, ∂x, ∂r, ∂s, ∂ϕ, ac, ∂z, ∂r, ⇒a, =, · · · (1), ∂ϕ, ∂ϕ, ∂x, a, +b, ∂r, ∂s, , Now by chain rule to ϕ we have, , Also, ∂ϕ, ∂y, , ∂ϕ ∂r ∂ϕ ∂s, ·, +, ·, ∂r ∂y, ∂s ∂y, , , , , ∂ϕ, ∂z, ∂ϕ, ∂z, ∴0 =, −a, +, c−b, ∂r, ∂y, ∂s, ∂y, , , ∂ϕ ∂z, ∂ϕ, ∂ϕ, 0 = c, −, a, +b, ∂s, ∂y, ∂r, ∂s, ∂ϕ, bc, ∂z, ∂s, ⇒b, =, · · · (2), ∂ϕ, ∂ϕ, ∂y, a, +b, ∂r, ∂s, =, , Page|35

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Chapter 3. Differentiability, ∂ϕ, ∂ϕ, ac, + bc, ∂z, ∂z, ∂r, ∂s = c = R.H.S. example 7: If w = xy and, Adding (1) and (2) a, +b, =, ∂ϕ, ∂ϕ, ∂x, ∂y, a, +b, ∂r, ∂s, dw, π, x = cos t, y = sin t find, at t = ., dt, 2, Solution : We have w = xy, 1, M-(I) : w = cos t · sin t = sin 2t, 2, , , dw, d 1, 1, ∴, =, sin 2t = · 2 cos 2t = cos 2t, 2, dt dt 2, π, dw, = cos 2 · = −1., ∴, dt t= π, 2, 2, M-(II) : By using Chain Rule, dw, ∂w dx ∂w dy, =, ·, +, ·, dt, ∂x dt, ∂y dt, dw, ∴, = y(− sin t) + x(cos t), dt, dw, = − sin2 t + cos2 t = cos 2t, dt,  , dw, π, = cos 2 ·, ∴, π, dt t=, 2, 2, = −1., u+v, example 8: If z = f (x, y); x = uv, y =, ., u−v, ∂z, ∂z, ∂z, Then prove that 2x, =u, +v ., ∂x, ∂u, ∂v, u+v, ., Solution : z = f (x, y); x = uv, y =, u−v, ∴ z = F [u, v] ∴ By Chain Rule., ∂z, ∂z ∂x ∂z ∂y, =, ·, +, ·, ∂u, ∂x ∂u ∂y ∂u, , , ∂z, ∂z u − v − (u + v), =, (v) +, ∂x, ∂y, (u − v)2, ∂z, ∂z, 2v, ∂z, = v, −, ∂u, ∂x (u − v)2 ∂y, ∂z, ∂z, 2uv ∂z, ⇒u, = uv, −, · · · (1), ∂u, ∂x (u − v)2 ∂y, Again, ∂z, ∂z ∂x ∂z ∂y, =, ·, +, ·, ∂v, ∂x ∂v, ∂y ∂v, , , ∂z u − v − (u + v)(−1), ∂z, =, (u) +, ∂x, ∂y, (u − v)2, ∂z, ∂z, 2u ∂z, = u, −, ∂v, ∂x (u − v)2 ∂y, ∂z, ∂z, 2uv ∂z, ⇒v, = uv, −, · · · (2), ∂v, ∂x (u − v)2 ∂y, Adding (1) and (2), ∂z, ∂z, +v, ∂x, ∂y, ∂z, ∂z, u, +v, ∂x, ∂y, u, , Hence the result is proved., Page|36, , ∂z, ∂x, ∂z, = 2x, ∂x, = 2uv

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Chapter 3. Differentiability, , , ∂u, ∂r, , 2, , 1, + 2, r, , , , ∂u, ∂θ, , 2, Solution : We have, u = f (x, y), u = f (r cos θ, r sin θ) = F (r, θ), , ∴ By Chain Rule, ∂u, ∂r, ∂u, ∴, ∂r, , =, =, , ∂u, ∂x, ∂u, ∂x, , ∂x ∂u ∂y, +, ·, ∂r, ∂y ∂r, ∂u, · cos θ +, sin θ · · · (1), ∂y, ·, , and, , ∴, ∴, , 1, ·, r, , ∂u, ∂θ, ∂u, ∂θ, ∂u, ∂θ, , ∂u ∂x ∂u ∂y, ·, +, ·, ∂x ∂θ, ∂y ∂θ, ∂u, ∂u, =, (−r sin θ) +, (r cos θ), ∂x, ∂y, ∂u, ∂u, = −, sin θ +, cos θ · · · (2), ∂x, ∂y, =, , Squaring and adding (1) and (2) we get,  ,  2,  2,  2, 1 ∂u 2, ∂u, ∂u, ∂u, 2, 2, + 2, =, (cos θ + sin θ) +, (sin2 θ + cos2 θ), ∂r, r, ∂θ, ∂x, ∂y,  2,  ,  2  2, ∂u, 1 ∂u 2, ∂u, ∂u, + 2, =, +, ., ∂r, r, ∂θ, ∂x, ∂y, ∂u, =, example 12: Show that u = f (x2 y); where f is differentiable function of x and y satisfies x, ∂x, ∂u, 2y ., ∂y, du ∂v, ∂u, ∂u, =, ·, = f 0 (v) · 2xy ∴ x, = f 0 (v) ·, Solution : Put x2 y = v then u = f (v) so that, ∂x, dv ∂x, ∂x, ∂u, du ∂v, ∂u, 2x2 y · · · (1) and, =, ·, = f 0 (v) · x2 ∴ 2y, = f 0 (v) · 2x2 y · · · (2) From (1) and (2) we, ∂y, dv ∂y, ∂y, ∂u, ∂u, = 2y ., observe that x, ∂x, ∂y, Exercise (B), ∂z, ∂z, 1. If z is a function of x and y if x = eu + e−v , y = e−u − e+v then prove that, −, =, ∂u ∂v, ∂z, ∂z, x, −y ., ∂x, ∂y, 2. If z = f (x, y); x = eu cos v, y = eu sin v then show that, ∂z, ∂z, ∂z, y, +x, = e2u ., ∂u, ∂v, ∂y, ∂z, ∂z, ∂z, 3. If z = f (u, v); u = x + y, v = x − y then show that, +, =2 ., ∂x ∂y, ∂u, ∂u, 4. If u = x3 + y 3 ; x = a cos t, y = b sin t. Find the value of, ., ∂t,  , x, 5. If z = tan−1, ; x = u + v, y = u − v then show that, y, ∂z, ∂z, u−v, +, = 2, ., ∂u ∂v, u + v2, 6. If w = f (x, y); x = u + v, y = uv then show that, ∂w, ∂w, ∂w, (i) (u − v), =u, −v, ., ∂x, ∂u, ∂v, ∂w, ∂w ∂w, (ii) (u − v), =, −, ., ∂y, ∂u, ∂v, u+v, ∂z, ∂z, ∂z, 7. If z = f (x, y); x = uv, y =, then show that 2x, =u, +v ., u−v, ∂x, ∂u, ∂v, Page|38

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Chapter 3. Differentiability, Solution :, (fx )(0,0) = (cos y − yex )(0,0), = 1, (fy )(0,0), = (−x sin y − ex )(0,0), = −1., ∴ Equation of tangent plane is, 1(x − 0) − 1(y − 0) − (z − 0) = 0 i.e. x − y − z = 0., example 2: Find the equation of tangent plane to the surface, z = 4x2 + y 2 at the point (1, 1, 5)., Solution : z = f (x, y) = 4x2 + y 2, ∴ (fx )(1,1,5) = (8x)(1,1,5) = 8, (fy )(1,1,5) = (2y)(1,1,5) = 2, ∴ Equation of tangent plane is, 8(x − 1) + 2(y − 1) − (z − 5) = 0 i.e. 8x + 2y − z − 5 = 0., # Increment / Change and Distance :, To estimate the change in the value of function f when we move a small distance ds from a point, ·, ds, example 1. Estimate how, p0 in a perticular direction of ū. df =, (Of )p · ū, |{z}, | {z0 }, xey, , directional derivative distanceincrement, , much the value of f (x, y, z) =, + yz will change if the point p(x, y, z) moves 0.1 unit from, p0 (2, 0, 0) straight forward to p1 (4, 1, −2)., Solution : We have p0 (2, 0, 0) and p1 (4, 1, −2), ∴ The direction vector p0 p1 = 2ī + j̄ − 2k̄ ≡ ū, 2, 1, 2, so that û = ī + j̄ − k̄., 3, 3, 3, Now,, (Of )(2,0,0) = (ey ī + (xey + z)j̄ + y k̄)(2,0,0) = ī + 2j̄, , , 2, 1, 2, 4, ∴ (Of )p0 · û = (ī + 2j̄) ·, ī + j̄ − k̄ =, 3, 3, 3, 3, Hence, the change in the value of, f = df, , = ((Of )p0 · û) · ds,  , 4, =, · (0 · 1), 3, ≈ 0 · 13., , Exercise (C), 1. Find the directional derivative of functions at the given point in the given direction., (i) f (x, y) = 2x2 + y 2 at the, √ point (−1, 1) in the direction 3ī − 4j̄., (ii) h(x, y) = tan−1 (y/x) +√ 3 sin−1 (xy/2), at the point (1, 1) in the direction 3ī − 2j̄., (iii) g(x, y) = x − (y 2 /x) + 3 sec−1 (2xy), at the point (1, 1) in the direction 12ī + 5j̄., (iv) f (x, y, z) = xy + yz + zx, at the point (1, −1, 2) in the direction 3ī + 6j̄ − 2k̄., (v) f (x, y, z) = x2 + 2y 2 − 3z 2 , at the point (1, 1, 1) in the direction ī + j̄ + k̄., (vi) g(x, y, z) = 3ex cos(yz), at the point (0, 0, 0) in the direction 2ī + j̄ − 2k̄., 2. Find the directions in which the function (a) increase most rapidly (b) decrease most, rapidly at a given point., (i) f (x, y) = x2 + xy + y 2 at the point (−1, 1)., (ii) g(x, y) = x2 y + exy sin y at the point (1, 0)., x, (iii) f (x, y, z) = − yz at the point (4, 1, 1)., y, (iv) g(x, y, z) = xey + z 2 at the point (1, log 2, 1/2)., (v) h(x, y, z) = log(xy) + log(yz) + log(xz) at the point (1, 1, 1)., (vi) w(x, y, z) = log(x2 + y 2 − 1) + y + 6z at the point (1, 1, 0)., 3. Find the equation of tangent plane and normal line of the surfaces, Page|45

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Chapter 3. Differentiability, f (x, y, z) = x2 + y 2 + z 2 = 3 at the point (1, 1, 1)., g(x, y, z) = x2 + y 2 − z 2 = 18 at the point (3, 5, −4)., h(x, y, z) = 2z − x2 = 0 at the point (2, 0, 2)., w(x, y, z) = x2 − xy − y 2 − z = 0 at the point (1, 1, −1)., p(x, y, z) = x + y + z = 1 at the point (0, 1, 0)., x2, Find on equation for the tangent to the ellipse, + y 2 = 2 at the point (−2, 1). (Ans:, 4, x − 2y = −4), Find parametric equations for the tangent line to the curve of intersection of surfaces at, the given point, (i) f (x, y, z) = x + y 2 + xz = 4, g(x, y, z) = x = 1 at the point (1, 1, 1)., (ii) f (x, y, z) = xyz = 1, g(x, y, z) = x2 + 2y 2 + 3z 2 = 6 at the point (1, 1, 1)., (iii) f (x, y, z) = x2 + 2y + 2z = 4, g(x, y, z) = y = 1 at the point (1, 1, 1/2)., (iv) f (x, y, z) = x + y 2 + z = 2, g(x, y, z) = y = 1 at the point (1/2, 1, 1/2)., (v) f (x, y, z) = x3 + 3x2 y 2 + y 3 + 4xy − z 2 = 0, g(x, y, z) = x2 + y 2 + z 2 = 11 at the, point (1, 1, 3)., Find the plane tangent to the level surfaces at the given point, (i) z = log(x2 + y 2 ) at the point (1, 0, 0)., 2, 2, (ii) z = e√(x +y ) at the point (0, 0, 1)., (iii) z = y − x at the point (1, 2, 1). p, By about how much will f (x, y, z) = log x2 + y 2 + z 2 change if the point p(x, y, z) moves, from p0 (3, 4, 12) a distance of ds = 0 · 1 units in the direction of 3ī + 6j̄ − 2k̄?, Estimate how much value of f (x, y, z) = ex cos(yz) will change if the point p(x, y, z) moves, from the origin a distance of ds = 0 · 1 units in the direction of 2ī + 2j̄ − 2k̄?, How much value of g(x, y, z) = x + x cos z − y sin z + y change if the point p(x, y, z) moves, from p0 (2, −1, 0) a distance of ds = 0 · 2 units towards the point p1 (0, 1, 2)?, Estimate how much the value of h(x, y, z) = cos(πxy) + xz 2 will change if the point, p(x, y, z) moves from p0 (−1, −1, −1) a distance of ds = 0 · 1 units towards the origin?, Is there a direction A in which the rate of change of f (x, y) = x2 − 3xy + 4y 2 at point, p(1, 2) equals 14? Give reason for your answer., In what two directions is the derivative of f (x, y) = x2 − y 2 /x2 + y 2 at point p(1, 1) equal, to zero?, In what direction is the derivative of f (x, y) = xy + y 2 at point p(3,, √ 2) equal to zero?, The derivative of f (x, y) at p0 (1, 2) in the direction of ī + j̄ is 2 2 and in the direction, of −2j̄ is −3. What is the derivative of f in the direction of ī − 2j̄? Give reason for your, answer., (i), (ii), (iii), (iv), (v), , 4., 5., , 6., , 7., 8., 9., 10., 11., 12., 13., 14., , Page|46

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CHAPTER 4, , Extreme Values, Introduction :, A continuous function defined on closed bounded region in XY -plane take absolute maximum and, minimum values on the domains. It is important to find these values and to know where they, occur., For function of one variable, to find local extreme points we see the points where the graph has, horizontal tangent line. At such points we look for local maxima, local minima and point of, inflection., For function of two variables, we see the points where the surface z = f (x, y) has a horizontal, tangent plane. At such points we look for local maxima, local minima and saddle points., 4.1: Local (Relative) Maxima and Minima of a Function of Two Variables:Definition 1: Local (Relative) Maximum Value :Suppose f (x, y) is defined on region R. (a, b) is a point in R and in domain of f (x, y). f (a, b) is, called a local (relative) maximum value of function f (x, y) if there exists some neighbourhood of, (a, b) such that for every point (a + b, b + k) of this neighbourhood f (a, b) ≥ f (a + h, b + k)., The point (a, b) is called Local (Relative) Maximum Point., Definition 2: Local (Relative) Minimum Value :Suppose f (x, y) is defined on a region R. (a, b) is a point in R and in the domain of f (x, y), f (a, b), is called a local (relative) minimum value of function f (x, y) if there exists some neighbourhood of, (a, b) such that for every point (a + h, b + k) of this neighbourhood f (a, b) ≤ f (a + h, b + k). The, point (a, b) is called Local (Relative) Minimum Point., Definition 3: f (a, b) is said to be a local (relative) extreme value of the function f (x, y) if it is, either a local (relative) maximum or local (relative) minimum value., ? First Derivative Test : (Necessary condition for extremum):Theorem 1: If f (x, y) has a local maximum or minimum value at an interior point (a, b) of it’s, domain and if the first partial derivatives fx (x, y) and fy (x, y) exists in a neighbourhood of (a, b), then fx (a, b) = 0 and fy (a, b) = 0., Proof: We shall prove this result by taking one variable constant., Take y = constant = b (say). Then the function of two variables f (x, y) becomes f (x, b) which is, of one variable., Here f (x, b) will have an extreme value of (a, b) if its derivatives w.r.t. x at (a, b) (i.e. at x = a) is, zero., i.e. (fx (x, b))(a,b) = 0 or (fx (x, b))x=a = 0 ⇒ fx (a, b) = 0., Similarly we can prove that fy (a, b) = 0 by taking x constant., The theorem is proved., Remark : The converse of the above theorem is not true e.g. consider f (x, y) = x2 − y 2 ., Here fx (x, y) = 2x, fy (x, y) = −2y., Take the point (a, b) = (0, 0). Then fx (0, 0) = 0 and fy (0, 0) = 0., This shows that both first order partial derivatives at (0, 0) vanish but f has neither maxima nor, minima at (0, 0)., For consider any neighbourhood of (0, 0) for small values of h both (2h, h) and (h, 2h) points are, in neighbourhood of (0, 0) and we have f (2h, h) > 0, f (h, 2h) < 0 ⇒ f has neither maxima nor, minima at (0, 0)., Thus, it is clear that vanishing of the first order partial derivatives is a necessary condition but not, a sufficient condition., Definition 4: A point (a, b) is said to be a critical point or a stationary point of a function f (x, y), if fx (a, b) = 0 = fy (a, b)., Note that: Every point of extremum is a stationary point but every stationary point may not be, 47

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Chapter 4. Extreme Values, a point of extremum., Definition 5: A point (a, b) is said to be a saddle point of a function f if in every neighbourhood, of (a, b) there are points (x, y) for which f (x, y) > f (a, b) and also the points for which f (x, y) <, f (a, b)., ? Second Derivative Test For Extrema :Theorem 2: Suppose f (x, y) is a function of two variable x and y defined in a region R such that, its first and second order partial derivatives are continuous in some neighbourhood of (a, b) of the, region R and fx (a, b) = 0 = fy (a, b) then, (i) f has local maximum at (a, b) if, 2 (a, b) > 0., fxx (a, b) < 0, fxx (a, b) · fyy (a, b) − fxy, (ii) f has local minimum at (a, b) if, 2 (a, b) > 0., fxx (a, b) > 0, fxx (a, b) · fyy (a, b) − fxy, (iii) f has saddle point at (a, b) if, 2 (a, b) < 0., fxx (a, b) · fyy (a, b) − fxy, (iv) Test is inconclusive at (a, b) if, 2 (a, b) = 0., fxx (a, b) · fyy (a, b) − fxy, 2 is called descriminant of f and, The expression fxx · fyy − fxy, 2 = fxx fxy, fxx · fyy − fxy, fxy fyy, example 1: exampleine for extreme values the function, f (x, y) = 2(x2 − y 2 ) − x4 + y 4 ., Solution :, fx = 4x − 4x3 ,, fxx = 4 − 12x2 ,, 3, fy = −4y + 4y , fyy = −4 + 12y 2 , fxy = 0, For extremum we have fx = 0, fy = 0., ∴ 4x − 4x3 = 0 and −4y + 4y 3 = 0, ∴ 4x(1 − x2 ) = 0 and 4y(−1 + y 2 ) = 0, ⇒ x = 0, 1 − x2 = 0 and y = 0, −1 + y 2 = 0, ⇒ x = 0, x = ±1 and y = 0, y = ±1., ∴ The critical /stationary points are (0, 0), (0, ±1), (±1, 0), (1, 1) and (−1, −1)., [At (0, 0) :], fxx (0, 0) = 4, fyy (0, 0) = −4, fxy (0, 0) = 0., 2 = 4(−4) − 02 = −16 < 0, ∴ fxx · fyy − fxy, , ⇒ At (0, 0)f has a saddle point., , [At (0, ±1) :], , fxx (0, ±1) = 4 > 0, fyy (0, ±1) = 8, fxy = 0, , ∴ fxx · fyy −, , 2, fxy, , 2, fxy, , = (−8)(−4) − 0 = 32 > 0 ⇒ f has maximum at (±1, 0) and f (±1, 0) = 1., fxx (1, 1) = −8, fyy (1, 1) = 8, fxy = 0., , [At (1, 1) :], ∴ fxx · fyy −, , ⇒ f has minimum at (0, ±1) and f (0, ±1) = −1., fxx (±1, 0) = −8 > 0, fyy (±1, 0) = −4, fxy = 0, , [At (±1, 0) :], ∴ fxx · fyy −, , = (4)(8) − 0 = 32 > 0, , 2, fxy, , [At (−1, −1) :], , = (−8)(8) − 0 = −64 < 0 ⇒ f has saddle point at (1, 1)., fxx (−1, −1) = −8, fyy (−1, −1) = 8, fxy = 0., , 2, fxy, , ∴ fxx · fyy −, = (−8)(8) − 0 = −64 < 0 ⇒ f has saddle point at (−1, −1)., example 2: Investigate the maximum and minimum values of, f (x, y) = (x + y − 1)(x2 + y 2 )., Solution : fx = x2 + y 2 + 2x(x + y − 1), fy = x2 + y 2 + 2y(x + y − 1),, fxx = 6x + 2y − 2, fyy = 2x + 6y − 2, fxy = 2y + 2x., For extremum, we have fx = 0 = fy, ∴ x2 + y 2 + 2x(x + y − 1) = 0 · · · (i) and, x2 + y 2 + 2y(x + y − 1) = 0 · · · (ii), subtracting (i) and (ii), we get (x + y − 1)(x − y) = 0., ⇒ x = y or x = 1 − y., Case (1): With x = y (i) becomes, x2 + x2 + 2x(x + x − 1) =0, ∴ 6x2 − 2x =0, i.e. 2x(3x − 1) =0, Page|48

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Chapter 4. Extreme Values, ⇒ x = 0 or x =, , 1, 3, , , 1 1, ., As x = y ∴ we get the points as (0, 0) and, ,, 3 3, Case (2): With x = 1 − y in (i) we get 1 − 2y + 2y 2 = 0., This has imaginary roots., , , 1 1, ∴ The stationary points are (0, 0) and, ., ,, 3 3, [At (0, 0) :], fxx = −2 < 0, fyy = −2, fxy = 0, , , 2 = (−2) · (−2) − 0 = 4 > 0, ∴ fxx · fyy − fxy, ⇒ f has maximum at (0, 0) and f (0, 0) = 0., , , 2, 2, 4, 1 1, :], fxx = > 0, fyy = , fxy =, ,, [At, 3 3, 3, 3,   ,  3, 2, 2, 16, −4, 1 1, 2 =, ∴ fxx · fyy − fxy, −, =, < 0 ⇒ f has saddle point at, ,, 3, 3, 9, 3, 3 3, example 3: Find extreme value of a function., f (x, y) = xy − x2 − y 2 − 2x − 2y + 4., Solution : fx = y − 2x − 2, fy = x − 2y − 2, fxx = −2 < 0, fyy = −2, fxy = 1., For extremum we have fx = 0 = fy, ∴ y − 2x − 2 = 0 and x − 2y − 2 = 0 solving these for x and y we get x = y = −2., ∴ The point (−2, −2) is the only point where f may have extreme values., 2 at (−2, −2) = (−2)(−2) − (1)2 = 3 > 0., Now fxx · fyy − fxy, ⇒ f has local maximum at (−2, −2), and f (−2, −2) = 8., example 4: Find and classify the extreme points of the function, f (x, y) = x4 − 3x2 y + y 3 ., Solution : fx = 4x3 − 6xy, fy = −3x2 + 3y 2, fxx = 12x2 − 6y, fyy = 6y, fxy = −6x, For extremum, we have fx = 0 = fy, ∴ 4x3 − 6xy = 0, 3x2 + 3y 2 = 0, 2x(2x2 − 3y) = 0, ∴ y = ±x, −3, 3, ., ⇒ x = 0 or 2x2 = 3y ∴ x = or x =,  2   2 , 3 3, −3 3, ∴ The critical points are, ,, ,, ,, and (0, 0)., 2 2, 2 2, , ,  2,  ,  ,  , 3 3, 3, 3, 3, 3, [At, ,, :], fxx = 12, −6, = 18 > 0 fyy = 6, = 9 fxy = −6, = −9, 2 2, 2, 2, 2, 2, , , 3 3, 2, 2, ∴ fxx · fyy − fxy = (18)(9) − (−9) = 81 > 0., ⇒ f has local minimum at, ,, ., 2 2, , , −3 3, Similarly, f has also local minimum at, ,, 2 2, , ,  2,  ,  , −3 3, −3, 3, 3, −3, [At, ,, :], fxx = 12, −6, = 18 > 0 fyy = (6) = 9 fxy = −6, =9, 2 2, 2, 2, 2, 2, , , −3 3, 2, 2, ∴ fxx · fyy − fxy = (18)(9) − (9) = 81 > 0., ⇒ f has local minimum at, ,, ., 2 2, [At (0, 0) :], fxx = 0, fyy = 0, fxy = 0., 2 = 0 ∴ Test Fails., ∴ fxx · fyy − fxy, But f (x, x) = x3 (1 − 2x), 1, −1, ∴ For 0 < x < , f (x, x) > 0 and for, < x < 0, f (x, x) < 0., 2, 2, ⇒ f has saddle point at (0, 0)., example 5: Find the extreme values of the function, 50 20, f (x, y) = xy +, +, x, y, 20, 50, Solution : fx = y − 2 , fy = x − 2, x, y, , Page|49

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Chapter 4. Extreme Values, 100, 40, , fyy = 3 , fxy = 1, 3, x, y, For extremum, we have fx = 0 = fy, 50, 20, ∴ y − 2 = 0 and x − 2 = 0., x, y, 50, 20, 200x4, ⇒ y = 2 with this x − 2 = 0 becomes x −, =0, y, 2500, ,  x, x3, x3, = 0 ⇒ x = 0 or 1 −, = 0., ⇒x 1−, 2500, 2500, 50, ∴ x = 5. putting this in y − 2 = 0 gives y = 2, x, ∴ (5, 2) is the only point where f take extreme value, 4, fxx (5, 2) = > 0, fyy (5, 2) = 5, fxy (5, 2) = 1., 5,  , 4, 2, (5) − (1)2 = 3 > 0, ∴ fxx · fyy − fxy =, 5, ⇒ f has minimum at (5, 2) and f (5, 2) = 30., fxx =, , example 6: Find extreme values of the function, f (x, y) = 3x2 (y − 1) + y 2 (y − 3) + 1., Solution : fx = 6x(y − 1), fy = 3(x2 − 2y + y 2 ), fxx = 6(y − 1), fyy = 6(y − 1), fxy = 6x, For extremum, we have fx = 0 = fy, ∴ 6x(y − 1) = 0 and 3(x2 − 2y + y 2 ) = 0 ⇒ x = 0 or y = 1., When x = 0, x2 − 2y + y 2 = 0 ⇒ y = 0 or y = 2, when y = 1, x2 − 2y + y 2 = 0 ⇒ x = ±1., ∴ The stationary points are (0, 0), (0, 2), (1, 1), (−1, 1)., [At (0, 0) :], fxx = −6 < 0, fyy = −6, fxy = 0., 2 = (−6)(−6) − 0 = 36 > 0, ∴ fxx · fyy − fxy, , ⇒ f has maximum at (0, 0) and f (0, 0) = 1., , [At (0, 2) :], , fxx = 6 > 0, fyy = 6, fxy = 0, , ∴ fxx · fyy −, ∴ fxx · fyy −, 2, fxy, , 2, fxy, , 2, fxy, , = 36 − 0 = 36 > 0. ⇒ f has minimum at (0, 2) and f (0, 2) = −3., [At (1, 1) :], , fxx = 0, fyy = 0, fxy = 6, , = 0 − 36 = −36 < 0., , ⇒ f has saddle point at (1, 1)., , [At (−1, 1) :], , fxx = 0, fyy = 0, fxy = −6, , (−6)2, , ∴ fxx · fyy −, =0−, = −36 < 0., ⇒ f has saddle point at (−1, 1)., example 7: A rectangular box open at the top is to have a volume of 32m3 . What must be the, dimensions so that the total surface area is minimum?, Solution: Let the length, breadth and height of the rectangular box be x, y, z respectively, with, surface S and volume V., Here, V = 32m3 ⇒ xyz = 32 · · · (i), We want to minimize the surface area of the rectangular box., 32, But surface area = S is given by S = 2z(x + y) + xy. But from (i) z =, xy, , , 1 1, +, = f (x, y) say,, ∴ S = xy + 64, x y, 64, 64, 128, 128, Now Sx = y − 2 , Sy = x − 2 , Sxx = 3 , Syy = 3 , Sxy = 1., x, y, x, y, For extremum, Sx = 0 = Sy, 64, 64, ∴ y − 2 = 0 and x − 2 = 0, x, y, ∴ x2 y = 64 and y 2 x = 64, 64, ⇒ x2 y = y 2 x ⇒ x = y ∴ y − 2 = 0, x, 64, ⇒ x − 2 = 0 ⇒ x3 = 64 ⇒ x = 4 and hence y = 4., x, ∴ The point (4, 4) is only point at which S may take extreme value. [At (4, 4) :], Sxx (4, 4) =, 128, = 2 > 0, Syy (4, 4) = 2, Sxy = 1, 64, Page|50

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Chapter 4. Extreme Values, 2 = (2)(2) − (1)2 = 3 > 0, ∴ Sxx · Syy − Sxy, , we have V = xyz = 32, ∴ At (4, 4, 2) :, ∴ (S)min =, =, (S)min =, Hence, Length, , ⇒ S has minimum at (4, 4)., , ∴ (4)(4)z = 32 ⇒ z = 2., , S has minimum value, 2z(x + y) + xy, 2(2)(4 + 4) + (4)(4), 32 + 16 = 48, = 4m, breadth = 4m, Height = 2m., , In example (4), we have obtained the minimum of the function x4 − 3x2 y + y 3 and in example (7),, we have found the minimum of the function 2z(x + y) + xy subject to the condition xyz = 32., Here we observe that these two problems are of different types. example (4) is a problem of free, extrema where as example (7) we have an additional condition called constraint or side condition, i.e. problem is of constrained extrema., To solve example (7) we have obtained the function S in terms of two variables x and y by replacing the value of z from the side condition. Another method to solve the problems of constrained, extrema is given by ’Lagrange’. The method is known as ’Lagrange’s method of multipliers., 4.2: Lagrange’s Method of undetermined multiplier(s) :M-(1): Let f (x, y, z) be a function of three variables x, y, z which is to be exampleined for extremum and let the variables x, y, z are connected by the relation φ(x, y, z) = 0 · · · (1) Since, ∂f, ∂f, ∂f, ∂f, ∂f, ∂f, f (x, y, z) is to have extremum ∴, = 0,, = 0,, = 0 so that, dx +, dy +, dz =, ∂x, ∂y, ∂z, ∂x, ∂y, ∂z, ∂φ, ∂φ, ∂φ, dx +, dy +, dz = 0 · · · (3) Multiply equa0 · · · (2) Differentiating the relation (1) we have, ∂x, ∂y, ∂z, , , , ∂f, ∂f, ∂φ, ∂φ, dx+, dy +, tion (3) by a parameter λ and adding in equation (2) we get, +λ, +λ, ∂x, ∂x, ∂y, ∂y, , , ∂f, ∂φ, +λ, dz = 0. This equation will be satisfied identically if coefficients of dx, dy, dz are 0., ∂z, ∂z, i.e. if, , ∂φ, ∂f, +λ, ∂x, ∂x, ∂f, ∂φ, +λ, ∂y, ∂y, ∂φ, ∂f, +λ, ∂z, ∂z, , = 0 · · · (4), = 0 · · · (5), = 0 · · · (6), , The equation (1), (4), (5) and (6) will determine the values of x, y, z and λ for which f (x, y, z) is, stationary., M-(2): Suppose f (x, y, z) and g(x, y, z) are differentiable functions. To find the local maximum, and minimum values of f (x, y, z) subject to the constraint g(x, y, z) = 0, find the values of x, y, z, and λ that satisfy the edquations Of = λOg and g(x, y, z) = 0 simulteneously., Remark :, 1. Inspite of usefulness in applications, Lagrange’s method fails to determine the nature of, the stationary point., 2. It is used in finding extreme values of constrained functions., example 1: Divide the number 36 into three parts so that the continued product of the first,, square of the second and cube of the third may be maximum., Solution : Let the numbers be x, y, z respectively., Let f (x, y, z) = xy 2 z 3 and g(x, y, z) ≡ x + y + z = 36., Construct the auxiliary function F as, F, , = f (x, y, z) + λg(x, y, z), , F, , = xy 2 z 3 + λ(x + y + z − 36), , Differentiating F partially w.r.t. x, y, z and λ, and then equating to 0, we get, Fx = y 2 z 3 + λ, = 0 · · · (1), Fy = 2xyz 3 + λ, = 0 · · · (2), Fz = 3xy 2 z 2 + λ, = 0 · · · (3), Fλ = x + y + z − 36 = 0 · · · (4), Page|51

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Chapter 4. Extreme Values, Now multiply equation (1) by x, (2) by y, (3) by z and adding,we get, , 6xy 2 z 3 + λ(x + y + z) = 0, ∴ 6xy 2 z 3 + 36λ = 0, −xy 2 z 3, ⇒λ =, 6, Putting this value in, xy 2 z 3 + λx, xy 2 z 3, ∴ xy 2 z 3 −, ·x,  6 x, ∴ xy 2 z 3 1 −, 6, x, ⇒1−, 6, ⇒x, , = 0, = 0, = 0, .., = 0, . xy 2 z 3 6= 0, , = 6., , Similarly, putting the value of λ in 2xy 2 z 3 + λy = 0 and 3xy 2 z 3 + λz = 0 respectively we get y = 12, and z = 18., ∴ The three numbers are 6, 12, & 18., and f (6, 12, 18) = 6(12)2 (18)3 = 5038848., example 2: Obtain the shortest distance of the point (1, 2, −3) from the plane 2x − 3y + 6z = 20., Solution : Suppose A(1, 2, −3) and let p(x, y, z) be any point on the plane (say) φ(x, y, z) =, 2x − 3y + 6z − 20 = 0., The distance = d2 = Ap = (x − 1)2 + (y − 2)2 + (z + 3)2 ≡ f (x, y, z). Which is to be minimize., Construct the auxiliary function, F, ∴F, , = f (x, y, z) + λφ(x, y, z), = (x − 1)2 + (y − 2)2 + (z + 3)2 + λ(2x − 3y + 6z − 20)., , =, 2(x − 1) + 2λ, = 0 · · · (1), =, 2(y − 2) − 3λ, = 0 · · · (2), =, 2(z + 3) + 6λ, = 0 · · · (3), = 2x − 3y + 6z − 20 = 0 · · · (4), 2(x − 1)2 + 2λ(x − 1) = 0 · · · (5), Multiply equation (1) by (x−1), (2) by (y−2), (3) by (z+3) we get 2(y − 2)2 − 3λ(y − 2) = 0 · · · (6), 2(z + 3)2 + 6λ(z + 3) = 0 · · · (7), 2, 2, 2, Adding (5), (6) and (7) we get 2[(x − 1) + (y − 2) + (z + 3) ] + λ(2x − 3y + 6z) + 22λ = 0, −d2, . Taking value of λ in equation (1), (2) and (3) we get, ∴ 2d2 + 42λ = 0 ⇒ λ =, 21, d2, −d2, d2, x−1= , y−2=, , z+3=, 21, 14, 7,  2 2  2 2  2 2, d, −d, d, ∴ d2 =, +, +, 21, 14, 7, ⇒ d = 6., Fx, Fy, Differentiating F w.r.t. x, y, z and λ, equate to zero, Fz, Fλ, , ∴ The shortest distance is 6 unit., example 3: Show that the greatest value of 8xyz under the condition, Solution : Let f (x, y, z) = 8xyz, g(x, y, z) =, Page|52, , y2, z2, x2, +, +, − 1 = 0., 9, 16, 4, , x2, y2, z2, 64, +, +, = 1 is √ ., 9, 16, 4, 3

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Chapter 4. Extreme Values, f (x, y) = 49 − x2 − y 2 on the line x + 3y = 10., f (x, y) = x2 y on the line x + y = 3., f (x, y) = x2 + y 2 , g(x, y) = x2 − 2x + y 2 − 4y = 0., f (x, y) = 3x − y + 6, g(x, y) = x2 + y 2 − 4 = 0., f (x, y, z) = x + 2y + 3z on x2 + y 2 + z 2 = 25., a) f (x, y) = x + y, g(x, y) = xy − 16 = 0, x > 0, y > 0, b) f (x, y) = xy, on the line x + y = 16., (viii) f (x, y) = xy, g(x, y) = x2 + y 2 − 10 = 0., (ix) Find the three real numbers whose sum is 9 and the sum of whose squares is as, small as possible., (x) Find the largest product of positive numbers, x, y, z if x + y + z 2 = 16., Find the maximum value of xyz subject to the condition, x + y + z = 1., Find the extreme values of the function x2 +y 2 +z 2 subject to the condition ax+by+cz = p., 5xyz, is maximum., If xyz = 8, find the values of x, y, z for which, x + 2y + 4z, If x2 y 2 (x + y + 1) then show that origin is a point of minima., Determine the shortest distance from the point (1, 0) to the parabola y 2 = 4x., Show that the shortest distance from the point (a, b, c) to the plane Ax + By + Cz + D = 0, Aa + Bb + Cc + D, ., is √, A2 + B 2 + C 2, If a, b, c are positive numbers, find the extreme value of, f (x, y, z) = xa y b z c subject to the condition x + y + z = 1., 4, 9, 16, If + +, = 25. Obtain the values of x, y, z which makes x + y + z minimum. Use, x, y, z, Lagrange’s Method., (ii), (iii), (iv), (v), (vi), (vii), , 3., 4., 5., 6., 7., 8., , 9., 10., , Page|55

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Chapter 4. Extreme Values, 4.3: Taylor’s Formula For Functions of Two Variables:, Theorem 3: If f (x, y) and its partial derivatives of order (n + 1) are continuous in neighbourhood, of a point (a, b) and if (a + h, b + k) is any point in this neighbourhood then there exists a positive, number c, 0 < c < 1 such that, , , , , , , ∂, 1, ∂, ∂ 2, 1, ∂, ∂ n, ∂, +k, f (a, b)+, h, +k, f (a, b)+· · ·+, h, +k, f (a, b), f (a+h, b+k) = f (a, b)+ h, ∂x, ∂y, 2!, ∂x, ∂y, n!, ∂x, ∂y, , , 1, ∂, ∂ n+1, +, h, +k, f (a + ch, b + ck)., (n + 1)!, ∂x, ∂y, Proof: Let x = a + ht, y = b + kt; where 0 ≤ t ≤ 1 is a parameter, ∴ f (x, y) = f (a + ht, b + kt) = F (t)., Since f (x, y) possesses continuous partial derivatives of order (n+1) in any neighbourhood of point, the (a, b)., F (t) is continuous in [0, 1] and, ∂f dx ∂f dy, ·, +, ·, F 0 (t) =, ∂x dt, ∂y dt, , ∂f, ∂f, ∂, ∂, 0, f, F (t) = h, +k, = h, +k, ∂x, ∂y, ∂x, ∂y, ∂f 0 dx ∂f 0 dy, ·, +, ·, ∂x dt, ∂y dt, , , , , ∂, ∂f, ∂f, ∂, ∂f, ∂f, =, h, +k, ·h+, h, +k, ·k, ∂x, ∂x, ∂y, ∂y, ∂x, ∂y,  2, , , , ∂ f, ∂2f, ∂2f, ∂2f, = h 2 +k, h+ h, +k 2 k, ∂x, ∂x∂y, ∂y∂x, ∂y, 2, 2, 2, ∂ f, ∂ f, ∂ f, =h2 2 + 2hk, + k2 2, ∂x, ∂x∂y, ∂y, , 2, ∂, ∂, = h, +k, f, ∂x, ∂y, , , ∂, ∂ n+1, 000, n+1, F (t), · · · , F, (t) = h, +k, f, ∂x, ∂y, 1, 1, 1, By Maclaurin’s theorem, we have F (1) = F (0)+F 0 (0)+ F 00 (0)+· · ·+ F n (0)+, F n+1 (c) · · · ?, 2!, n!, (n + 1)!, But, ∴ F 00 (t) =, , F (1) = f (a + h, b + k),, F (0) = f (a, b),, , , ∂, ∂, 0, F (0) =, h, +k, f (a, b),, ∂x, ∂y, , , ∂, ∂ 2, 00, F (0) =, h, +k, f (a, b), · · · ,, ∂x, ∂y, , , ∂, ∂ n, n, F (0) =, h, +k, f (a, b),, ∂x, ∂y, , , ∂, ∂ n+1, n+1, F, (c) =, h, +k, f (a + ch, b + ck), ∂x, ∂y, Putting all these values in equation ? we get, , , ∂, ∂, f (a + h, b + k) = f (a, b) + h, +k, f (a, b), ∂x  ∂y, , 1, ∂, ∂ 2, +, h, +k, f (a, b), 2!, ∂x  ∂y, , 1, ∂, ∂ n, +··· +, h, +k, f (a, b), n!  ∂x, ∂y , 1, ∂, ∂ n+1, +, h, +k, f (a + ch, b + ck);, (n + 1)!, ∂x, ∂y, 0 < c < 1., Page|56

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Chapter 4. Extreme Values, First n derivative terms are evaluated at point (a, b). The last term is evaluated at some point, (a + ch, b + ck) on the line segment joining (a, b) and (a + h, b + k)., Remark :, 1, The last term is called the remainder and the theorem is called Taylor’s expansion about, the point (a, b)., 2. Another form of Taylor’s Formula / theorem is, , , ∂, ∂, f (a, b), + (y − b), f (x, y) = f (a, b) + (x − a), ∂x, , 2∂y, ∂, ∂, 1, (x − a), + (y − b), f (a, b), +, 2!, ∂x , ∂y, , 1, ∂, ∂ n+1, +··· +, (x − a), + (y − b), f (a + (x − a)c, b + (y − b)c),, (n + 1)!, ∂x, ∂y, 0 < c < 1., This is called Taylor’s expansion of f (x, y) about the point (a, b) in powers of (x − a) and, (y − b)., 3. If a = 0, b = 0 and h, k are independent variables i.e. h = x, k = y then we get, , , ∂, ∂, f (x, y) = f (0, 0) + x, f (0, 0), +y, ∂x, ∂y, , , 1, ∂, ∂ 2, +, f (0, 0), x, +y, 2!, ∂x, ∂y, , , 1, ∂, ∂ n+1, +··· +, x, +y, f (cx, cy), 0 < c < 1., (n + 1)!, ∂x, ∂y, This is called Maclaurin’s expansion., 4. Taylor’s Formula gives polynomial approximation of two variables. First three terms give, the functions linearization. The last term gives approximation error., Working Rule : To workout the examples, we will be interested in the expansion of some function, f (x, y) upto and including certain terms., For this we need to write first few terms of the expansion:, f (x, y) = f (a, b) + (hfx (a, b) + kfy (a, b)), 1, + [h2 fxx (a, b) + 2hkfxy (a, b) + k 2 fyy (a, b)], 2!, , 1  3, +, h fxxx (a, b) + 3h2 kfxxy (a, b) + 3hk 2 fxyy (a, b) + k 3 fyyy (a, b), 3!, OR, f (x, y) = f (a, b) + [(x − a)fx (a, b) + (y − b)fy (a, b)], 1, + [(x − a)2 fxx (a, b) + 2(x − a)(y − b)fxy (a, b) + (y − b)2, 2!, fyy (a, b)] + · · ·, example 1: Expand f (x, y) = x3 + xy 2 in powers of (x − 2) and (y − 1)., Solution : We have f (x, y) = x3 + xy 2 a = 2, b = 1., ∴ f (2, 1) = 10., fx = 3x2 + y 2 ∴ fx (2, 1) = 13,, , fy = 2xy ∴ fy (2, 1) = 4,, , fxx = 6x, , ∴ fxx (2, 1) = 12, fyy = 2x, , ∴ fyy (2, 1) = 4,, , fxy = 2y, , ∴ fxy (2, 1) = 2,, , ∴ fyx (2, 1) = 2,, , fxx = 6, , ∴ fxxx (2, 1) = 6, fyyy = 0, , ∴ fyyy (2, 1) = 0,, , fxxy = 0, , ∴ fxxy (2, 1) = 0,, , ∴ fxyy (2, 1) = 2., , fyx = 2y, fxyy = 2, , Page|57

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CHAPTER 5, , Multiple Integrals, 5.1. Double Integrals over Rectangles, We consider a function f (x, y) defined on rectangular region R R : a ≤ x ≤ b, c ≤ y ≤ d., We sub divide R into small rectangles using a network of lines parallel to the x-axis and y-axis as, shown in figure. The lines divide R into n rectangular pieces, where the number of such pieces n, gets large as the width and height of each piece gets small. These rectangles form a partion of R. A, small rectangular piece of width 4x and height 4y has area 4A = 4x4y. If we number the small, pieces partitioning R in some order, the their areas are given by numbers 4A1 , 4A2 , 4A3 , · · · 4An ,, where 4Ak is the area of the k th small rectangle., To form a Riemann sum over R, we choose a point (xk , yk ) in the k th small rectangle, multiply the, n, X, value of f at the point by the area 4Ak , and add togehter the products: Sn =, f (xk , yk )4Ak ., k=1, , Depending on how we take (xk , yk ) in k th small rectangle, we may get different values for Sn ., We are interested in what happens to these Riemann sums as the widths and heights of all the, small rectangles in the partition of R approach zero. The norm of a partition P written kP k, is, the largest width or height of any rectangle in the partition. If kP k = 0 · 1, then all the rectangles, in the partion of R have width at most 0 · 1 and height at most 0 · 1. Sometimes the Riemann, sums converge as the norm of P goes to zero, written kP k → 0. The resulting limits is written as, n, X, f (xk , yk )4Ak . As kP k → 0 and the rectangles get narrow and short, their number n, lim, kP k→0, , k=1, , increases, so we can also write this limit as lim, , n→∞, , n, X, , f (xk , yk )4Ak . with the understanding that, , k=1, , 4Ak → 0 as n → ∞ and kP k → 0., If f is continuous throughout R, then as we refine the mesh width to make both 4x and 4y go to, zero, the sums inZ (1), Z approach a limit, Z Zcalled the double integral of f over R., The notation is, , f (x, y)dA or, R, , Z Z, f (x, y)dA = lim, R, , n→∞, , f (x, y)dxdy.Thus, R, , n, X, , f (xk , yk )4Ak = lim sn ., n→∞, , k=1, , Fubini’s theorem for calculating double integrals :Theorem 1 Fubini’s Theorem (First Form) :Z Z, If f (x, y) is continuous on the rectangular region R : a ≤ x ≤ b, c ≤ y ≤ d, then, f (x, y)dA =, R, Z dZ b, Z bZ d, f (x, y)dxdy =, f (x, y)dydx. Fubini’s theorem state that double integrals over rectc, , a, , a, , c, , angles can be calculated as iterated integrals. This means that we can evaluate a double integral, by integrating w.r.t. one variable at a time., Z Z, Example 1: Calculate, f (x, y)dA for, R, , f (x, y) = 4 − y 2 and R : 0 ≤ x ≤ 3, 0 ≤ y ≤ 2., , 60

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§ 5.1. Double Integrals over Rectangles, , Chapter 5. Multiple Integrals, , Solution : By Fubini’s theorem, Z Z, , 3Z 2, , Z, f (x, y)dA =, , R, , 0, , (f − y 2 )dydx, , 0, 3, , Z, , y3, 4y −, 3, , =, 0, 3, , Z, , 2, dx, y=0, , 16, dx, 3, , =, 0, , 16 3, =, [x], 3 0, = 16., Reversing the order of integration, Z Z, Z 2Z 3, f (x, y)dA =, (4 − y 2 )dxdy, R, 0, 0, Z 2, (4 − y 2 )[x]3x=0 dy, =, 0, , , 2, y3, (4 − y )dy = 3 4y −, = 3, 3 0, 0, (16), = 3, 3, = 16., Z, , 2, , 2, , By reversing the order of integration we sec that the value of integral remains the same., Z 3Z 0, Example 2: Evaluate, (x2 y − 2xy) dydx., −2, , 0, , Solution : by Fubini’s Theorem,, Z 3Z 0, Z, (x2 y − 2xy) dydx =, 0, ,  Z 0, , Z 0, 2, x, ydy − 2x, ydy dx, x=0, y=−2, y=−2, ),  2 0, Z 3 (  2 0, y, y, x2, =, − 2x, dx, 2 y=−2, 2 y=−2, x=0, , , , Z 3  , 0−4, 2 0−4, − 2x, dx, =, x, 2, 2, x=0, Z 3, =, {−2x2 + 4x} dx, , −2, , 3, , 0, , 3, x3, 2, = −2 + 2x, 3, x=0, = [−12 + 18] = 6., , , Reversing the order of integration, Z 0Z 3, Z, (x2 y − 2xy) dxdy =, −2, , 0, , −2, Z 0, , 0, , , , d3, 2x2, y−, y, 3, 2, , 3, dy, x=0, , (6y − 9y) dy, , =, −2, Z 0, , −3ydy, , =, −2, , , 0, y2, = −3, 2 y=−2, = [0 + 6], = 6., , Page|61

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§ 5.1. Double Integrals over Rectangles, , Chapter 5. Multiple Integrals, , Theorem 2: Fubini’s Theorem (Stronger Form) :Let f (x, y) be continuous on a region R., (1). If R is defined by a ≤ x ≤ b, g1 (x) ≤ y ≤ g2 (x), with g1 and g2 are continuous on [a, b], then, Z b Z g2 (x), Z Z, f (x, y) dydx., f (x, y) dA =, a, , R, , g1 (x), , (2). If R is defined by c ≤ y ≤ d, h1 (y) ≤ x ≤ h2 (y). with h1 and h2 are continuous on [c, d], then, Z d Z h2 (y), Z Z, f (x, y) dxdy., f (x, y) dA =, c, , R, , h1 (y), , R R sin x, Example 1: Evaluate, R x dA where R is the triangle in the xy-plane bounded by the x-axis,, the line y = x and the line x = 1., Solution: The region of integration is shown in figure we integrate first w.r.t. y and then with, respect to x., , Z Z, , sin x, dA =, R x, , Z Z, , Z, , 1, , x=0, 1, , Z, , Z, , x, , y=0, , sin x, dxdy =, x, , Z, , 1, , x=0, 1, , Z, , Z, , x, , y=0, , sin x, dy, x, , , , Z, , 1, , dx =, x=0, , sin x x, [y]y=0 dx, x, , sin x, sin x, dA =, (x − 0) dx =, sin xdx = [− cos x]10 = − cos 1 + cos 0 = 1 − cos 1 ., R x, x=0 x, 0, :- Sketch the region of integration for the integral, RExample, 2 R 2x, 0 x2 (4x + 2) dydx and write an equivalent integral with the order of integration reversed., Solution : The region of integration is given by the inequalities x2 ≤ y ≤ 2x and 0 ≤ x ≤ 2., Therefore the region is bounded by the curve parabola y = x2 and the lines y = 2x, x = 0, x = 2., , Page|62

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§ 5.1. Double Integrals over Rectangles, Z, , Chapter 5. Multiple Integrals, , 2 Z 2x, , 2, , Z, , (4x + 2)[y]2x, y=x2 dx, , (4x + 2) dydx =, 0, , x2, , 0, 2, , Z, , (4x + 2)(2x − x2 ) dx, , =, 0, 2, , Z, , (6x2 − 4x3 + 4x) dx, , =, 0, , = [2x3 − x4 + 2x2 ]20, = 24 − 24 + 23 = 8., To find limits for integrating in the reverse order, we consider a horizontal line passing from left to, right through the region., , y, √, This line enters at x = and leaves at x = y. To cover all such lines we let y run from y = 0 to, 2, y = 4. Now by changing the order of integration., , Z Z, Z 2 Z 2x, Z 4 Z √y, Z 4, Z 4, √, sin x, y2, √, y, 2, [2x, +2x], 2y, +, 2, d, dA, (4x+2) dydx =, (4x+2), dxdy, =, dy, =, y, −, −, y, y, y, 2, R x, 0, x2, 0, y=0, 0, x, 2, 2, ", #4, , Z 4, Z Z, Z 2 Z 2x, 2, 2, 3/2, sin x, y, 2y, y3, √ −y, (4x + 2) dydx =, y+2 y, dA, dy =, +, −, = 8., 2, 2, 3/2, 6, x2, 0, 0, R x, 0, RR, Example : Evaluate, R f (x, y) dA where f (x, y) = y cos xy over the rectangle 0 ≤ x ≤ π, 0 ≤, y ≤ 1., Solution :, Z Z, Z π Z 1, f (x, y) dA =, y cos xy dydx., R, , x=0 y=0, 1 Z π, , Z, =, , y cos xy dxdy, y=0, Z 1, , x=0, , , , sin xy, =, y, y, y=0, Z 1, =, sin πy dy, , π, dy, x=0, , y=0, , =[− cos πy]1y=0, = − cos π + cos 0, = − (1) + 1, =2., Example : Integrate f over the given region f (u, v) = v −, from the first quadrant of uv plane by the line u + v = 1., Solution :, , √, , u over the triangular region cut off, , Page|63

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§ 5.1. Double Integrals over Rectangles, , Chapter 5. Multiple Integrals, , Consider the strip parallel to v-axis. This strip varies from u = 0 to u = 1. The limits of u are, u = 0 to u = 1. Now the values of v interms of u are v = 0 to v = 1 − u. The limits of v are v = 0, to v = 1 − u., Z Z, Z Z, √, (v − u) dA, f (u, v) dA =, R, R, Z 1 Z 1−u, √, =, (v − u) dvdu, u=0 v=0, 1  2, , Z, =, , u=0, Z 1 , , √, v, − uv, 2, , 1−u, du, v=0, , , (1 − u)2 √, =, − u(1 − u) du, 3, u=0, , Z 1 , (1 − u)2, 1/2, 3/2, −u +u, du, =, 2, u=0, ", #1, (1 − u)3 u3/2 u5/2, =, −, +, 6, 3/2, 5/2, u=0, , 2 2 1, =0 − + −, 3 5 6, −10 + 6 1, =, −, 15, 6, −4 1, =, −, 15, 6, −8 − 5, =, 30, −13, =, ., 30, Example : Evaluate, Solution :, , R 1 R √1−s2, 0, , 0, , Z, , 8tdtds (the st plane)., 1Z, , √, , 1−s2, , Z, , 1, , 8tdtds =, 0, , 0, , s=0, 1, , Z, =, , √, , 4(1 − s2 ) ds, , s=0, , 1, s3, =4 s −, 3 0, , , 1, =4 1 −, 3, 8, = ., 3, , , Page|64, , 2, , 1−s, [4t2 ]t=0, ds

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§ 5.1. Double Integrals over Rectangles, Example : Evaluate, , R 0 R −v, −2 v, , Chapter 5. Multiple Integrals, , 2dpdv (the pv plane), , Solution : We have the limits of pare p = v and p = −v, v = −2, v = 0., , Z, , 0, , Z, , −v, , Z, , 0, , 2dpdv =, −2, , v, , −2, , Example : Evaluate, , 2[p]−v, p=v, , 0, , 0, , −4v 2, dv =, 2[−v − v] dv =, −4vdv =, 2, −2, −2, Z, , Z, , , , 0, = 8., −2, , RR, , f (x, y) dA where R is the region in the first quadrant bounded by the, x, lines y = x, y = 2x, x = 1, x = 2 and f (x, y) = ., y, Solution : Draw the lines y = x, y = 2x, x = 1, x = 2., R, , Page|65

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§ 5.1. Double Integrals over Rectangles, , Chapter 5. Multiple Integrals, Z Z, , Z, , 2, , Z, , 2x, , f (x, y) dA =, R, , x=1, 2, , Z, , y=x, , x, dydx, y, , x[log y]2x, y=x dx, , =, x=1, Z 2, , x[log 2x − log x] dx, , =, x=1, Z 2, , =, , x log 2dx, x=1, , x2, = log 2, 2, 3, =, log 2., 2, , , Example : Evaluate, , RR, R, , 2, 1, , 1, dxdy where R is the square 1 ≤ x ≤ 2, 1 ≤ y ≤ 2., xy, , Solution :, Z Z, R, , dxdy, xy, , Z, , 2, , =, x=1, 2, , Z, =, , x=1, , Z, , 2, , 1, dydx, xy, y=1, Z 2, 1, 1, dx, dy, x, y=1 y, , = [log x]2x=1 [log y]2y=1, = [log 2 − log 1][log 2 − log 1], = [log 2]2 ., RR, 2, 2, Example : Evaluate, R f (x, y) dA where f (x, y) = x + y and R is triangular region with, vertices (0, 0), (1, 0) and (0, 1)., Solution : Let O(0, 0), A(1, 0), B(0, 1)., , y2 − y1, 1−0, =, = −1. Equation of AB is, x2 − x1, 0−1, y − 0 = m(x − 1), y, = −1(x − 1), y, = −x + 1., , Slope of AB = m =, , Consider the strip parallel to y axis. This strip varies from x = 0 to x = 1., The limits of x are from x = 0 to x = 1. Now the values of y interms of x at the end point of the, Page|66

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§ 5.1. Double Integrals over Rectangles, are y = 0 and y =, , Chapter 5. Multiple Integrals, , 1, 1, 1, . Draw the curve x4 = y and the lines x = , y = 0, y = ., 16, 2, 16, , 1, Consider the strip parallel to y-axis. This strip varies from x = 0 to x = . The limits of x are, 2, 1, 4, x = 0 to x = . Now the values of y interms of x are y = 0 to y = x, 2, ∴ The limits of y are y = 0 to y = x4 ., Z, , 1/16 Z 1/2, , Z, , 5, , 1/2 Z x4, , cos(16yx ) dxdy =, 0, , y 1/4, , x=0, Z 1/2, , =, , cos(16πx5 )x4 dx, , x=0, π/2, , Z, =, , cos t, 0, , =, =, =, , cos(16πx5 ) dydx, , y=0, , dt, 80π, , 1, π/2, [sin t]0, 80π, 1, π, [sin − sin 0], 80π, 2, 1, ., 80π, , Put 16πx5 = t, 80πx4 dx = dt., 1, π, when x = , t =, 2, 2, when x = 0, t = 0., R 8 R 2 dydx, ., Example : Change the order of integration and evaluate 0 √, 3x, y4 + 1, √, Solution : We have the limits of y are y = 3 x to y = 2 that is y 3 = x to y = 2 and limits of x, are x = 0 to x = 8., Draw the curve y 3 = x and the straight lines y = 2, x = 0, x = 8., , Page|69

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§ 5.1. Double Integrals over Rectangles, , Chapter 5. Multiple Integrals, , Consider the strip parallel to x-axis. This strip varies from y = 0 to y = 2. Therefore limits of y, are y = 0 to y = 2. Now the values of x interms of y at the end points on strip are x = 0 to x = y 3 ., Z 2 Z y3, Z 2, Z 2, Z 8Z 2, dydx, dxdy, 1, y3, y3, =, =, [x], dy, =, dy., x=0, √, 4, 4, 4, 3 x y4 + 1, y=0 x=0 y + 1, y=0 y + 1, y=0 y + 1, 0, Substitute, y4 + 1 =, Z t17, Z 8Z 2, dt, 1, dydx, =, = log 17., √, 4, 3x y + 1, 4t, 4, 1, 0, R1R1, Example : Change the order of integration and evaluate 0 y x2 exy dxdy., Solution : We have the limits of x are x = y to x = 1 and the limits of y are y = 0 to y = 1., Draw the lines y = 0, y = 1, x = 1, y = x., , Consider the strip parallel to y-axis. This strip varies from x = 0 to x = 1., ∴ The limits of x are x = 0 to x = 1. Now the values of y interms of x at the end point of strip are, y = 0 to y = x., ∴ The limits of y are y = 0 to y = x., By changing order of integration, Z, 0, , 1Z 1, , x2 exy dxdy =, , y, , Z, , 1, , Z, , x=0, 1, , x, x=0, Z 1, , =, 0, , Z, , 1, , =, , x2 exy dydx, , y=0, , Z, =, , x, , 2, , , , exy, x, , =, =, =, =, , dx, y=0, , 2, , x[ex − e0 ] dx, Z 1, x2, xe dx −, xdx, , 0, , Z, , x, , 1, , , , 0, 1, 2, x, , dt, −, 2, 2, 0, 1 t1 1, [e ] − + 0, 2 0 2, 1, 1, [e − 1] −, 2, 2, e, − 1., 2, et, , 0, , dt, ., 2, R3R1, 3, Example : Change the order of integration and evaluate it 0 √x/3 ey dydx., r, x, Solution : We have the limits of y are y =, and y = 1. The limits of x are x = 0 and x = 3., 3, Put x2 = t, xdx =, , Page|70

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§ 5.1. Double Integrals over Rectangles, Draw the curve y 2 =, , Chapter 5. Multiple Integrals, , x, , y = 1, x = 0, x = 3., 3, , Consider the strip parallel to x-axis. This strip varies from y = 0 to y = 1., ∴ The limits of y are y = 0 to y = 1. Now the values of x in terms of y at the end point of strip, are x = 0 and x = 3y 2 ., ∴ The limits of x are x = 0 to x = 3y 2 ., By changing order of integration, Z 3Z 1, Z 1 Z 3y2, 3, y3, s, ey dxdy, x e dydx =, 0, , y=0, , 3, , Z, , 1, , x=0, 3, , =, y=0, 1, , Z, , 2, , 3y, dy, ey [x]x=0, 3, , ey (3y 2 ) dy, , =, y=0, 1, , Z, , et dt, , =, 0, [et ]10, , =, = e − 1., Example : Evaluate, , R ln 8 R ln y, 1, , 0, , ex+y dxdy., , Solution :, Z, 1, , ln 8 Z ln y, , ex+y dxdy =, , 0, , Z, , ln 8 Z ln y, , 1, , Z, , , , ey dy, , x=0, ln 8, , =, 1, , Z, , ex dx, , y y, [ex ]ln, x=0 e dy, , ln 8, , =, , (y − 1)ey dy, , 1, , = [(y −, =, =, =, =, , 8, 1)ey ]ln, 1, , Z, , ln 8, , −, , ey dy, , 1, y ln 8, [e ]1, , (ln 8 − 1)8 −, (ln 8 − 1)8 − [8 − e], 8 ln 8 − 8 − 8 + e, 8 ln 8 − 16 + e., , Example : Sketch the region of integration, determine the order of integration and evaluate the, R π R π sin y, integral. 0 x, dydx., y, Solution : We have the limits of y are y = x and y = π and the limits of x are z = 0 to z = π., Draw the lines x = 0, x = π, y = x, y = π., Page|71

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§ 5.1. Double Integrals over Rectangles, , Chapter 5. Multiple Integrals, , R π sin y, dy w.r.t. y. Therefore we change the order of integration., x, y, Consider the strip parallel to x-axis. This strip varies from y = 0 to y = π. The limits of y are, y = 0 to y = π. Now the values x interms of y at the end point of strip are x = 0 to x = y., ∴ The limits of x are x = 0 to x = y. Hence by changing order of integration we get, We cannot integrate, , Z, 0, , π, , Z, x, , π, , sin y, dydx =, y, , Z, , π, , Z, , y=0, π, , Z, =, , y=0, π, , yπ, , x=0, , sin y, dxdy, y, , sin y y, [x]x=0 dy, y, , Z, =, , sin y dy, 0, , = [− cos y]π0, = − cos π + cos 0, = 2., , : Change the order of integration and evaluate, RExample, 2R2, 2, 0 x 2y sin xy dydx., Solution : We have the limits of y are y = x and y = 2 and the limits of x are x = 0 and x = 2., Draw the lines x = 0, x = 2, y = x, y = 2., , Consider the strip parallel to x-axis. This trip varies from y = 0 to y = 2. The limits of y are y = 0, to y = 2. Now the values of x interms of y are x = 0 to x = y., ∴ The limits of x are x = 0 to x = y., Page|72

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§ 5.1. Double Integrals over Rectangles, , Chapter 5. Multiple Integrals, , ∴ By changing order of integration., Z, Z 2Z 2, 2, 2y sin xy dydx =, 0, , 2, , y=0, Z 2, , x, , =, , Z, , y, , 2y, y=0, Z 2, , =, , 2y 2 sin xy dxdy, , x=0, 2, , , , − cos xy, y, , y, dy, x=0, , 2y[− cos y 2 + cos 0] dy, , y=0, 2, , Z, =, , − cos y 2 (2ydy) +, , y=0, Z 4, , Z, , 2, , 2dydy, 0, , − cos t dt + [y 2 ]20, 0, Z 4, +4, = − sin t, =, , 0, , = 4 − sin 4, = 4 − (−0 · 7568), = 4 · 7568., Definition : The Jacobian determinant or Jacobian of the co-ordinates transformation, ∂x ∂x, ∂u, ∂v = ∂x ∂y − ∂x ∂y . (1) The Jacobian is also, x = g(u, v) y = h(u, r) is J(u, v) = ∂y ∂y, ∂u ∂v, ∂v ∂u, ∂u ∂v, ∂(x, y), denoted by J(u, v) =, . e.g. for polar co-ordinates we have, ∂(u, v), x = r cos θ, y = r sin θ., The Jacobian is, ∂(x, y), J(r, θ) =, ∂(r, θ), , ∂x ∂x, cos θ −r sin θ, ∂r ∂θ, =, =, ∂y ∂y, sin θ r cos θ, ∂r ∂θ, = 4 cos θ2 + r sin2 θ, = r, , The Jacobian of co-ordinate transformation, , ∂x ∂x ∂x, ∂u ∂v ∂w, ∂(x, y, z), ∂y ∂y ∂y, =, x = g(u, v, w), y = h(u, v, w), z = k(u, v, w). is defined by J(u, v, w) =, ∂(u, v, w), ∂u ∂v ∂w, ∂z ∂z ∂z, ∂u ∂v ∂w, e.g. for cylindtrical co-ordinates we have x = r cos θ, y = r sin θ, z = z The Jacobian transformation is, , J(r, θ, z) =, , ∂(x, y, z), ∂(r, θz), , ∂x ∂x ∂x, ∂r ∂θ ∂z, cos θ −r sin θ 0, ∂y ∂y ∂y, =, = sin θ r cos θ 0, ∂r ∂θ ∂z, 0, 0, g, ∂z ∂z ∂z, ∂r ∂θ ∂z, = cos θ[r cos θ − 0] + r sin θ[sin θ − 0] + 0, = r cos2 θ + r sin2 θ, = r(cos2 θ + sin2 θ), = r., , ∂(x, y), for the transformation, ∂(u, v), (a) x = u cos v, y = u sin v (b) x = u sin v, y = u cos v., Example : Find the Jacobian, , Page|73

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§ 5.1. Double Integrals over Rectangles, , Chapter 5. Multiple Integrals, Solution : (a) Since x = u cos v, y = u sin v., , ∂(x, y), ∂(u, v), , ∂x ∂x, ∂u ∂v = cos v −u sin v, =, ∂y ∂y, sin v u cos v, ∂u ∂v, = u cos2 v + u sin2 v, = u., , (b) Since we have, x = u sin v, y = u cos v, , ∂(x, y), ∂(u, v), , ∂x ∂x, ∂u ∂v = sin v u cos v, =, ∂y ∂y, cos v −u sin v, ∂u ∂v, = −u sin2 v − u cos2 v, = −u., , Example : Solve the system u = x − y, v = 2x + y for x and y interms of u and v. Then find the, ∂(x, y), value of the Jacobian, ., ∂(u, v), Solution : We have, u = x − y (1), v = 2x + y (2), u+v, v − 2u, Solving (1) and (2) we get x =, , y=, ., 3, 3, , ∂(x, y), J=, ∂(u, v), , =, , =, =, , ∂x, ∂u, ∂y, ∂u, 1 2, +, 9 9, 1, ., 3, , ∂x, ∂v, ∂y, ∂v, , =, , 1, 3, −2, 3, , 1, 3, 1, 3, , ∂(x, y, z), of the transformation, ∂(u, v, w), (a) x = u cos v, y = u sin v, z = w, 1, (b) x = 2u − 1, y = 3v − 4, z = (w − 4)., 2, Solution : (a), Example : Find the Jacobian, , J=, , ∂(x, y, z), ∂(u, v, w), , ∂x ∂x ∂x, ∂u ∂v ∂w, cos v −u sin v 0, ∂y ∂y ∂y, =, = sin v u cos v 0, ∂u ∂v ∂w, 0, 0, 1, ∂z ∂z ∂z, ∂u ∂v ∂w, = cos v[u cos v − 0] + u sin v[sin v − 0] + 0[0 − 0], = u cos2 v + u sin2 v, = u, , Page|74

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§ 5.1. Double Integrals over Rectangles, , Chapter 5. Multiple Integrals, , (b), , ∂(x, y, z), J=, ∂(u, v, w), , ∂x ∂x, ∂u ∂v, ∂y ∂y, =, ∂u ∂v, ∂z ∂z, ∂u ∂v, , 3, = 2, −0, 2, = 3., , ∂x, ∂w, ∂y, ∂w, ∂z, ∂w, , =, , 2 0, 0 3, 0 0, , 0, 0, 1, 2, , Example : Solve the system u = x + 2y, v = x − y for x and y in terms of u and v. Then find, ∂(x, y), the value of the Jacobian, ., ∂(u, v), u−v, u+v, Solution : (1) and (2) y =, , x=, ., 3, 3, ∂(x, y), J=, ∂(u, v), , =, , =, =, , −1, 3, 1, 3, 1 1, +, 9 9, 2, 9, 1, 3, 1, 3, , y, R 4 R 2 +1 2x − y, 2x − y, y, Example : Evaluate 0, dxdy by applying the transformation u =, , v =, y, 2, 2, 2, x=, 2, and integrating over an appropriate region in the uv-plane., Solution : We sketch the region R of integration in the xy-plane and identify its boundaries. We, 2x − y, y, solve two equations u =, , v = for x and y in terms of u and v. x = u + v, y = 2v. We, 2, 2, find the boundaries of G by substituting these expressions into the equations for boundaries of R., xy equations for Corresponding uv-equations, Simplified, the boundary of R, for the boundary of G, uv-equations, y, 2v, =v, x=, u+v =, u=0, 2, y2, x= +1, u+v =v+1, u=1, 2, y=0, 2v = 0, v=0, y=4, 2v = 4, v = 2., , WeZuse the formula inZ change, of variable, Z, Z, f (x, y) dxdy =, f (g(u, v), h(u, v)) |J| dudv, R, , G, , Page|75

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§ 5.1. Double Integrals over Rectangles, , Chapter 5. Multiple Integrals, y, Z 1 Z 2, +1, 2x − y, 2, u dvdu = 4., dxdy, =, 2, y, 2, u=0 v=0, 0, x=, 2, R 1 R 1−x √, Example : Evaluate 0 0, x + y(y − 2x)2 dydx., , Z, , 4Z, , Solution : We sketch the region R of integration in the xy plane and identify its boundaries. The, integrand suggest the transformation u = x + y and v = y − 2x. Solving these two equation for x, 2u v, u v, + We can find the boundaries of the uv-region G., and y in terms of u and v. x = − , y =, 3 3, 3, 3, xy equations for Corresponding uv-equations, Simplified, the boundary of R, for the boundary of G, uv-equations, u v, u v, y =1−x, 2 + =1− +, u=1, 3 u, 3 v 3 3, y=0, 2 + =0, 2u + v = 0 i.e. v = −2u, u3 v3, x=0, v=u, − =0, 3 v3, u, − =1, x=1, u − v = 3, v = u − 3, 3 3, , J=, , 1, 3, 2, 3, , 1, −, 3, 1, 3, , =, Z, 0, , 3, 1, 1 2, + = = . By using change of variable in double integration, 9 9, 9, 3, 1 Z 1−x √, , x + y(y − 2x) dydx =, , 0, , =, , =, , =, , =, , 1, , 2, =, , Page|76, , u, , √, ,  , 1, uv, dvdu, 3, u=0 v=−2u, 1  3 u, Z, 1 1, v, u2, du, 3 u=0, 3 v=−2u, 1, Z, 1 1, u 2 [u3 + 8u3 ] du, 9 u=0, 7, Z, 1 1, 9u 2, 9 0,  9 1, u2 , , ,  9 , Z, , 2, , 2, ., 9, , Z, , 0, , 2

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§ 5.1. Double Integrals over Rectangles, , Chapter 5. Multiple Integrals, , Area in Polar Co-ordinates :RR, The area of a closed and bounded region R in the polar co-ordinate plane is A =, R rdrdθ., 2, Example : Find the area enclosed by the lemniscate r = 4 cos 2θ., Solution : We sketch the graph of the lemniscate to determine the limits of integration. We see, that the total area is 4 times the first-quadrant portion., Z π Z, A = 4 4, θ=0, , √, , 4 cos 2θ, , rdrdθ, , r=0, , Z π  2 √4 cos 2θ, r, = 4 4, dθ, θ=0 2 r=0, Z π, = 2 4 4 cos 2θ dθ, 0, , π, sin 2θ 4, = 8, 2, 0, , , sin π, = 4, − sin 0, 2, = 4., , , Changing Cartesian Integrals into Polar Integrals, :RR, The procedure for changing a Cartesian integral, f, (x,, y) dxdy into a polar integral has two, R, steps., Step 1: Substitute x = r cos θ and y = r sin θ and replace dxdy by rdrdθ in the cartesian integral., Step 2: Supply polar limits of integration, R R for boundary ofR R., R, The cartesian integral then becomes, R f (x, y) dxdy =, G f (r cos θ, r sin θ) rdrdθ. where G G, denotes the region of integration in polar co-ordinates., R2Rx, Example : Evaluate 0 0 ydydx., Solution :, Z, , 2Z x, , Z, ydydx =, , 0, , 0, , 0, , Z, =, 0, , 2  2 x, y, , 2, , 2  2, x, , 2, , dx, , y=0, , dx, , 2, x3, 6 x=0, 8−0, 6, 4, ., 3, , , =, =, =, , Page|77

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§ 5.1. Double Integrals over Rectangles, , Chapter 5. Multiple Integrals, , If we divide R into the region R1 and R2 shown in figures (A), Z Z, Z Z, dA, da +, A =, R1, , Z, , 1, , =, y=0, , Z, , R2, y, , Z, , √, x≡ y, , y=0, , y, [x]x≡√y, , =, =, , √, , y, , dxdy, , Z, , x=y−2, √, , 4, , dy +, y=1, , y=0, , y, , [x]x=y−2 dy, 4, , 1, (y 2 − y + 2) dy, , y=1, , 3 1, , 4,  3, 2, ,  2y 2 ,  2,  +  y − y + 2y , , ,  3 ,  3, 2, 2 0, 2, 1, 4 14 15, +, −, +6, 3, 3, 2, 12 − 15, 2, 9, ., 2, , , =, , Z, , y=1, , 1, 1, Z, 2, 2, (y + y ) dy +, , 1, , =, , =, , 4, , dxdy +, , √, , 1, , =, Z, , √, , Z, , On the other hand, reverting the order of integration figure B gives., Z 2 Z x+2, A =, dydx, −1, 2, , Z, =, , −1, Z 2, , =, , x2, , [y]x+2, dx, y=x2, [x + 2 − x2 ] dx, , −1, , x2, x3, + 2x −, 2, 3, , , , 4−1, (8 + 1), + x(2 + 1) −, 2, 3, , =, =, =, =, , 2, , , , −1, , , , 3, +3, 2, 9, ., 2, , Example : Find the area bounded by the co-ordinate axes and the line x + y = 2., Solution : DrawZthe Z, co-ordinate axes and the line x + y = 2., 2, x−2, Required Area =, dydx = −2., x=0, , y=0, , Evaluating Polar Integrals :Example : Change the cartesian integral into an equivalent polar integral. Then evaluate the, polar integral, Page|80

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§ 5.1. Double Integrals over Rectangles, , Chapter 5. Multiple Integrals, , x = r cos θ, y = r sin θ (0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π), Z, , 1, , −1, , Z √1−y2, √, , −, , 2, , Z, , 2, , 1, , Z, , 2π, , log(r2 + 1) rdr dθ, , log(x + y + 1) dxdy =, 1−y 2, , r=0, , Z, , θ=0, , 1, , log(r2 + 1) rdr[θ]2π, θ=0, , =, r=0, , Z, , 1, , log(r2 + 1) rdr, , = 2π, r=0, 2, , Put r + 1 = t, 2rdr = dt, when r = 0, t = 1, when r = 1, t = 2, Z 2, = π, log t dt, t=1, , , Z 2, 1, t dt, = π [t log t]2t=1 −, t, 1, = π{2 log 2 − 0 − [t]21 }, = π(2 log 2 − 1), = π(log 4 − 1)., Part 3, Triple Integrals in Rectangular Co-ordinates :We use triple integrals to find the volumes of three dimentional shapes, moments of solids and the, masses and the average value of functions of three variables., Triple Integrals :If F (x, y, z) is a function defined on a closed bounded region D in space - the region occupied by, a solid ball, for example, or a lump of clay - then the integral of F over D may be defined in, the following way. We partition a rectangular region containing D into rectangular cells by planes, parallel to the co-ordinate planes. We number the cells that lie inside D from 1 to n in some order, a, typical cell having dimensions 4xk by 4yk by 4zk and volume 4Vk . We choose a point (xk , yk , zk ), n, X, F (xk , yk , zk ) 4Vk (1) If F is continuous and the bounding, in each cell and form the sum Sn =, k=1, , surface of D is made of smooth surfaces joined along continuous curves, then, Z ZasZ4xk , 4yk and 4zk, approach zero independently the sums Sn approach a limit lim Sn =, n→∞, , F (x, y, z) dv. We, D, , call this limit the triple integral of F over D. The limit also exists for some discontinuous functions., Properties of Triple Integrals :If F = F (x, y, z) and G = G(x, y, z) are continuous, then, RRR, RRR, 1. R R RD kF dv = k, R DR FR dv (any number, R R R k)., 2. R R RD (F ± G) dv =, D F dv ±, D Gdv., 3. R R RD F dv ≥ 0, if, F, ≥, 0, on, D., RRR, 4., F, dv, ≥, D, D Gdv if F ≥ G on D., 5. If the domain D of a continuous function F is partitioned by smooth surfaces into a finite, number of nonoverlappingRcells, RR, RRR, RRR, RRR, D1 , D2 , D3 , · · · , Dn , then, D F dv =, D1 F dv +, D2 F dv + · · · +, Dn F dv., RRR, Definitions :- The volume of a closed, bounded region D in space is V =, D dv., Example : Find the volume of the region D enclosed by the surfaces z = x2 + 3y 2 and z =, 8 − x2 − y 2 ., RRR, 2, 2, 2, 2, Solution : The volume is v =, D dzdydx z = x + 3y = 8 − x − y, 2, 2, 2x + 4y = 8, x2 + 2y 2 = 4., Page|83

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§ 5.1. Double Integrals over Rectangles, , Chapter 5. Multiple Integrals, 2y 2 =, y2, y, , 4 − x2, 4 − x2, =, r2, 4 − x2, = ±, 2, Z, , 2, , Required volume V =, x=−2, v, u, u4, t, , v, u, u4, t, , − x2, Z 8−x2 −y2, 2, v, =, u, u 4 − x2 z=x2 +3y 2, t, y=−, 2, v, , Z, , 2, u, − x2, Z 2 Z t4 − x, 2, 2, 2 (8 − x2 − 4y 2 ) dydx, v, (8 − x2 − 4y 2 ) dydx = 4, u, u 4 − x2, y=0, x=−2, x=0, t, y=−, 2, Areas, Moments and Centers of Mass :Area of Bounded Regions in the Plane :If we take f (x, y) = 1 in the definition of the double integral over a region R, then the partial, n, n, X, X, sums reduces to Sn =, f (xk , yk )4Ak =, 4Ak (1) As 4x and 4y approach zero, the coverage, , Z, , Z, , u, , k=1, , k=1, , of R by the 4Ak ’s becomes increasingly complete and we define the area of R to be the limit., Z Z, n, X, dA. (2), 4Ak =, Area = lim, n→∞, R, k=1, RR, Definition :- The area of a closed, bounded plane region R is A =, R dA. (3) To evaluate the, integral in (3), we integrate the constant function f (x, y) = 1 over R., Example : Find the area of the region R bounded by y = x and y = x2 in the first quadrant., 2, Solution : DrawZ the, line y = x and, Z 1parabola y =Zx 1., 1Z x, 1, [y]xy=x2 dx =, [x − x2 ] dx = ., Required area =, dydx =, 6, 0, 0, y=x2, 0, Example : A thin plate covers the triangular region bounded by the x-axis and the lines x = 1, and y = 2x in the first quadrant. The plate’s density at the point (x, y) is δ(x, y) = 6x + 6y + 6., Find the plate’s mass first moments, center of mass, moments of intertia, and radii of gyration, about the co-ordinate axes., , Solution : We sketch the plate and put in enough detail to determine the limits of integration for, the integrals we have to evaluate., The Z, plate’s, Z mass is, Z, Z, 1, , 2x, , M=, , 1, , 2x, , x=0, , y=0, , δ(x, y) dydx =, 0, , y=0, , (6x + 6y + 6) dydx. = 14., , The first, about the x-axis, Z 1moment, Z 2x, Z 1 Zis2x, Mx =, yδ(x, y) dydx =, (6xy + 6y 2 + 6y) dydx11., 0, , 0, , 0, , 0, , 0, , 0, , Similarly, Z 1 the, Z 2xfirst moment about the y-axis is, My =, xδ(x, y) dydx = 10., The moment, is, Z 1 Zx-axis, Z 1 Z 2x of intertia about the, 2x, 2, Ix =, y δ(x, y) dydx =, (6xy 2 + 6y 3 + 6y 2 ) dydx = 12., 0, , 0, , 0, , 0, , Similarly,, moment of inertia about the y-axis is, Z 1 Z the, 2x, 39, Iy =, x2 δ(x, y) dydx =, 5, 0, 0, 39, 60 + 39, 99, ∴ I0 = Ix + Iy = 12 +, =, = ., 5, 5, 5, The three radii of gyration are, , Page|84

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§ 5.1. Double Integrals over Rectangles, , Chapter 5. Multiple Integrals, , r, , p, 12, =, 6/7, r 14, r, p, 39, 39, Iy /M =, Ry =, /14 =, 5, r, r 70, p, 99, 99, I0 /M =, R0 =, /14 =, ., 4, 70, Z 1Z 1Z 1, (x2 + y 2 + z 2 ) dzdydx., Example : (1) Evaluate, Rx =, , p, , Ix /M, , =, , 0, , 0, , 0, , Solution :, 1Z 1Z 1, , Z, , 2, , 2, , 1, z3, (x + y )z +, dydx, 3 z=0, x=0 y=0, , Z 1 Z 1 , 1, (x2 + y 2 ) +, dydx, 3, x=0 y=0, , 1, Z 1 , 1, y3, 2, x +, y+, dx, 3, 3 y=0, x=0, , Z 1 , 1, 2, 2, dx = ., x +, 3, 3, x=0, Z, , 2, , (x + y + z ) dzdydx =, 0, , 0, , 0, , =, =, =, Z, , eZ eZ e, 1, , Z, , 1, , , , 2, , 2, , 1, dxdydz., xyz, , (2) Evaluate, 1, , 1, , 1, , Solution :, eZ eZ e, , Z, 1, , 1, , 1, , eZ e, , Z, , 1 1, dxdydz =, yz x, , Z1 e, , =, 1, 1Z, , Z, , πZ, , 1, [log x]e1 dydz, yz, 1, Z, 1 e 1, dydz = 1, z 1 y, , π, , Example : Evaluate, , y sin z dxdydz., 0, , 0, , 0, , Solution :, Z, , 1, , Z, , π, , π, , Z, , Z, , 1, , Z, , π, , y sin z dxdydz =, z=0, , y=0, , x=0, , y sin z dydz, z=0 y=0, Z 1  2 π, y, , = π, , z=0, , Z, , 1Z, , 2−x Z, , dzdydx., 0, , 2−x−y, , y=0, , π3, [1 − cos 1]., 2, , 2−x−y, , Example : Evaluate, Solution :, Z 1 Z 2−x Z, , 2, , sin z dz =, , 0, , 0, , Z, , 1 Z 2−x, , [z]2−x−y, z=0, , Z, , 1 Z 2−x, , dzdydx =, dydx =, (2 − x − y) dydx, 0, 0, 0, y=0, , , , Z 1, Z, Z, 2−x, (2 − x)2, 1 1, 7, y2, =, (2 − x)y −, dx =, (2 − x2 ) −, dx =, (2 − x)2 dx = ., 2 y=0, 2, 2 x=0, 6, x=0, x=0, Example : Find the volume of the region D enclosed by the surfaces, z = x2 + 3y 2 and z = 8 − x2 − y 2 ., Z Z Z, Solution : The volume is V =, dzdydx., 0, , 0, 1, , 0, , , , D, , z = x2 + 3y 2 and z = 8 − x2 − y 2, The limits of z. are z = x2 + 3y 2 to z = 8 − x2 − y 2 ., x2 + 3y 2 = 8 − x2 − y 2, x2 + 2y 2 = 4, 4 − x2, y2 =, r2, 4 − x2, y=± (, ), 2, Page|85

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§ 5.1. Double Integrals over Rectangles, , Chapter 5. Multiple Integrals, r, The limits of y are from y = −, , 4 − x2, to y =, 2, , r, , 4 − x2 = 0 ∴ x = ±2, The limits of x are from x = −2 to x = 2., , Z, V, , 2, , x=−2, , Z, , v, u, u4, t, , − x2, Z 8−x2 −y2, 2, v, dzdydx, u, u 4 − x2 z=x2 +3y 2, t, y=−, 2, , Z, , =, , 2, , = 2, , Z, , v, u, u4, t, , 2, x=0, , Z, , 2, , Z, , 2, , − x2 Z, 2, , y=0, v, u, u4 −, t, , Z, , dzdydx, , x2, (8 − 2x2 − 4y 2 ) dydx, , y=0, , , , 2 (4 − x2 )y −, , = 4, , 8−x2 −y 2, , z=x2 +3y 2, , 2, , = 4, x=0, , 4 − x2, ., 2, , 2y 3, , x=0, , 3, , , Z, , 2, , = 8, x=0, , , 2, (4 − x ), , , , 4−, 2, , x2, , v, u, u, t 4, , − x2, 2, , y=0, , 1, 2, , −, , 2, 3, , , , 4−, 2, , x2, , 3, 3, 2 , 2 2, x2 ) 2 ,  dx,  (4 −√x ) − 2(4 − √, = 8, , 2, 3·2 2 , x=0, , , Z, , =, , =, , =, , =, , =, , =, =, =, , 3, Z, (8)2 2, 2 2, √, (4 − x ) dx, 3 2 0, Put x = 2 sin θ dn = 2 cos θdθ, when x = 0, θ = 0, π, when x = 2, θ =, 2, Z π, 16, 2 8 cos3 θ2 cos θdθ., √, 3 2 0, √ Z π, , , 8 2 2, 1 + cos 2θ 2, 16, dθ, 3 0, 2, √ Z π, 8 2 2, 4(1 + 2 cos 2θ + cos2 2θ) dθ, 3 0, √ Z π , , 8 2 2, 1 + cos 4θ, 4 1 + 2 cos 2θ +, dθ, 3 0, 2, √ , π, 8 2, 2 sin 2θ θ sin 4θ 2, 4 θ+, + −, 3, 2, 2, 8, √, √  0, , , 8 2, pi π, 8 2, 3π, ·4, +, =, 4, 3, 2, 4, 3, 4, √, 8 2π., , Example : Evaluate the cylindrical co-ordinate integrals, R 2π R 1 R √2−r2, dz rdrdθ., 0, 0 r, Page|86, , , 3, 2,  dx

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§ 5.1. Double Integrals over Rectangles, , Chapter 5. Multiple Integrals, Solution :, Z, , 1, , Z, , 1, , Z, , 1, , 1, , Z, , Z, , 1, , Z, , 1, , (x + y + z) dydxdz =, −1, , −1, , (x + y + z) dydxdz, , −1, , z=−1, 1, , Z, , x=−1, , Z, , 1, , y=−1, , , , =, z=−1 x=−1, Z 1 Z 1, , y2, (x + z)y +, 2, , 1, dxdz, y=−1, , [2(x + z) + 0] dxdz, , =, z=−1, Z 1, , =, , x=−1, , [x2 + xz]1x=−1 dz, , z=−1, 1, , Z, , 2zdz, , =, z=−1, , z2, = 2, 2, = 0., , , Example : Evaluate, Solution :, Z eZ eZ, , ReReRe, 1, , 1, , 1, , 1, , z=−1, , ln r ln s ln t dt drds., , e, , Z, , e, , Z, , e, , ln r ln s ln t dt drds =, 1, , 1, , 1, , , ln r ln s, , r=1, Zs=1, e Z e, , =, , [t log t]e1, , Z, , e, , −, 1, , , 1, t · dt drds, t, , ln r ln s {e log e − 0 − [t]e1 } drds, , r=1, Zs=1, e Z e, , ln r ln s {e log e − e + 1} drds, , , Z e, 1, e, =, ln s [r ln r]r=1 −, r dr (e log e − e + 1) ds, r, 1, Zs=1, e, =, ln s[e log e − 0 − e + 1](e log e − e + 1) ds, s=1, , Z e , = (e log e − e + 1)2 [s log s]es=1 −, ds, =, , Zs=1, e, , r=1, , 1, , = (e log e − e + 1)2 {e log e − e + 1}, = (e log e − e + 1)3 ., , Page|88