SECOND EDITION A Textbook of ENGINEERING MATHEMATICS-I H.S. Gangwar l Prabhakar Gupta A Textbook of ENGINEERING MATHEMATICS-I This page intentionally left blank A Textbook of ENGINEERING MATHEMATICS-I (SECOND EDITION) Prabhakar Gupta H.S. Gangwar M.Sc. (Math.), M.Tech., Ph.D. Dean Academics SRMS College of Engineering and Technology, Bareilly (U.P.) M.Sc., Ph.D. Lecturer Deptt. of Mathematics SRMS College of Engineering and Technology, Bareilly (U.P.) UPTU An Imprint of Copyright © 2010, 2009 New Age International (P) Ltd., Publishers Published by New Age International (P) Ltd., Publishers All rights reserved. No part of this ebook may be reproduced in any form, by photostat, microfilm, xerography, or any other means, or incorporated into any information retrieval system, electronic or mechanical, without the written permission of the publisher. All inquiries should be emailed to rights@newagepublishers.com ISBN (13) : 978-81-224-2847-6 PUBLISHING FOR ONE WORLD NEW AGE INTERNATIONAL (P) LIMITED, PUBLISHERS 4835/24, Ansari Road, Daryaganj, New Delhi - 110002 Visit us at www.newagepublishers.com Preface to the Second Revised Edition This book has been revised exhaustively according to the global demands of the students. Attention has been taken to add minor steps between two unmanageable lines where essential so that the students can understand the subject matter without mental tire. A number of questions have been added in this edition besides theoretical portion wherever necessary in the book. Latest question papers are fully solved and added in their respective units. Literal errors have also been rectified which have been accounted and have come to our observation. Ultimately the book is a gift to the students which is now error free and user- friendly. Constructive suggestions, criticisms from the students and the teachers are always welcome for the improvement of this book. AUTHORS This page intentionally left blank Some Useful Formulae 1. 2. sin ix = i sin hx 3. sin x = e ix − e − ix 2i 4. cos x = e ix + e − ix 2 5. Sin h2x = 1 (cosh 2x – 1) 2 6. cos h2x = 1 (cosh 2x + 1) 2 7. z a x dx = ax a ≠ 1, a > 0 log a z z z sin haxdx = 1 cos hax a cos haxdx = 1 sin hax a tan haxdx = 1 log cos hax a z z z z z z z 1 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. cos ix = cos hx a2 − x2 1 2 x −a 1 2 x +a 2 2 dx = sin −1 x x = arc sin a a dx = log x + x 2 − a 2 dx = 1 x x tan −1 = arc tan a a a a2 x sin −1 a 2 a 2 − x 2 dx = x 2 e ax sin bx dx = e ax (a sin bx – b cos bx) a2 + b2 e ax cos bx dx = e ax (a cos bx + b sin bx) a 2 + b2 sec ax dx = a2 − x2 + 1 log |sec ax + tan ax| a 1 log |cosec ax – cot ax| a 18. z 19. sin x = x − x 3 x5 + ...... 3 5 20. cos x = 1 − x2 x 4 + ........ 2 4 21. tan x = x + x3 2 5 17 7 + x − x + ........ 3 15 315 22. log (1 + x) = x − 23. log (1 – x) = − x − 24. sin hx = x + 25. d x a = ax loge a dx 26. 1 d cos −1 x = − dx 1 − x2 27. 1 d cot −1 x = − dx 1 + x2 28. 29. 30. cosec ax dx = d dx d dx x2 x3 − − ......... 2 3 x 3 x5 + + ......... 3 5 cosec −1 x = − log a x = x2 x3 x 4 + − + ........ 2 3 4 1 x 1 x x2 − 1 log a e d x a tan −1 . = 2 dx a a + x2 Contents PREFACE TO THE SECOND REVISED EDITION SOME USEFUL FORMULAE U NIT I. Differential Calculus-I 1.0 Introduction 1.1 nth Derivative of Some Elementary Functions Exercise 1.1 1.2 Leibnitz’s Theorem Exercise 1.2 Exercise 1.3 Partial Differentiation 1.3 Function of Two Variables 1.4 Partial Differential Coefficients Exercise 1.4 1.5 Homogeneous Function 1.6 Euler’s Theorem on Homogeneous Functions Exercise 1.5 1.7 Total Differential Coefficient Exercise 1.6 Curve Tracing 1.8 Procedure for Tracing Curves in Cartesian Form Exercise 1.7 1.9 Polar Curves Exercise 1.8 1.10 Parametric Curves Exercise 1.9 Expansion of Function of Several Variables 1.11 Taylor’s Theorem for Functions of Two Variables Exercise 1.10 Objective Type Questions Answers to Objective Type Questions U NIT II. Differential Calculus-II 2.1 Jacobian Exercise 2.1 194 1 1 6 7 13 19 20 20 21 33 35 36 47 48 62 63 64 71 73 77 78 80 81 81 89 90 94 95150 95 109 2.2 Approximation of Errors Exercise 2.2 2.3 Extrema of Function of Several Variables Exercise 2.3 2.4 Lagrange’s Method of Undetermined Multipliers Exercise 2.4 Objective Type Questions Answers to Objective Type Questions U NIT III. Matrices 3.0 Introduction 3.1 Definition of Matrix 3.2 Types of Matrices 3.3 Operations on Matrices 3.4 Trace of Matrix 3.5 Properties of Transpose 3.6 Properties of Conjugate Matrices 3.7 Singular and Non-Singular Matrices 3.8 Adjoint of a Square Matrix 3.9 Inverse of a Matrix (Reciprocal) Exercise 3.1 3.10 Elementary Row and Column Transformations 3.11 Method of Finding Inverse of a Non-Singular Matrix by Elementary Transformations Exercise 3.2 3.12 Rank of a Matrix Exercise 3.3 3.13 System of Linear Equations (Non-Homogeneous) 3.14 System of Homogeneous Equations 3.15 Gaussian Elimination Method Exercise 3.4 3.16 Linear Dependence of Vectors Exercise 3.5 3.17 Eigen Values and Eigen Vectors Exercise 3.6 3.18 Cayley-Hamilton Theorem Exercise 3.7 3.19 Diagonalization of a Matrix 3.20 Application of Matrices to Engineering Problems Exercise 3.8 Objective Type Questions Answers to Objective Type Questions 111 119 121 134 135 145 147 150 151257 151 151 152 155 156 156 156 163 163 163 165 167 168 174 175 186 188 197 200 206 210 214 214 230 232 238 239 249 253 255 257 U NIT IV. Multiple Integrals 4.1 Multiple Integrals 4.2 Double Integrals 4.3 Working Rule 4.4 Double Integration for Polar Curves Exercise 4.1 4.5 Change of the Order of Integration 4.6 Change of Variables in a Multiple Integral Exercise 4.2 4.7 Beta and Gamma Functions 4.8 Transformations of Gamma Function 4.9 Transformations of Beta Function 4.10 Relation between Beta and Gamma Functions 4.11 Some Important Deductions 4.12 Duplication Formula 4.13 Evaluate the Integrals Exercise 4.3 4.14 Application to Area (Double Integrals) Exercise 4.4 4.15 Triple Integrals Exercise 4.5 4.16 Application to Volume (Triple Integrals) Exercise 4.6 4.17 Dritchlet’s Theorem Exercise 4.7 Objective Type Questions Answers to Objective Type Questions U NIT V. Vector Calculus Vector Differential Calculus 5.1 Vector Function 5.2 Vector Differentiation 5.3 Some Results on Differentiation Exercise 5.1 5.4 Scalar Point Function 5.5 Vector Point Function 5.6 Gradient or Slope of Scalar Point Function 5.7 Geometrical Meaning of Gradient, Normal 5.8 Directional Derivative 5.9 Properties of Gradient Exercise 5.2 5.10 Divergence of a Vector Point Function 5.11 Physical Interpretation of Divergence 5.12 Curl of a Vector 258332 258 258 259 259 266 268 274 281 283 285 286 287 287 289 294 299 300 311 312 314 315 322 323 329 330 332 333418 333 333 333 334 336 337 337 337 338 338 339 350 351 352 353 5.13 Physical Meaning of Curl 5.14 Vector Identities Exercise 5.3 5.15 Vector Integration 5.16 Line Integral 5.17 Surface Integral 5.18 Volume Integral Exercise 5.4 5.19 Green’s Theorem Exercise 5.5 5.20 Stoke’s Theorem 5.21 Cartesian Representation of Stoke’s Theorem Exercise 5.6 5.22 Gauss’s Divergence Theorem 5.23 Cartesian Representation of Gauss’s Theorem Exercise 5.7 Objective Type Questions Answers to Objective Type Questions 353 354 363 365 365 366 367 374 376 384 386 388 399 400 401 413 414 418 Unsolved Question Papers (20042009) 419431 Index 433434 UNIT 1 Differential Calculus-I 1.0 INTRODUCTION Calculus is one of the most beautiful intellectual achievements of human being. The mathematical study of change motion, growth or decay is calculus. One of the most important idea of differential calculus is derivative which measures the rate of change of a given function. Concept of derivative is very useful in engineering, science, economics, medicine and computer science. dy d2 y , second order derivative, denoted by dx dx 2 d3 y third order derivative by and so on. Thus by differentiating a function y = f(x), n times, dx 3 dny or Dny or yn(x). Thus, the process successively, we get the nth order derivative of y denoted by dx n of finding the differential co-efficient of a function again and again is called Successive Differentiation. The first order derivative of y denoted by 1.1 nth DERIVATIVE OF SOME ELEMENTARY FUNCTIONS 1. Power Function (ax + b)m Let y = (ax + b)m y1 = ma (ax + b)m–1 y2 = m (m–1)a2 (ax + b)m–2 ..... ...... ................................... ..... ...... ................................... yn = m(m–1) (m–2) ... (m – n – 1) an (ax + b)m–n Case I. ⇒ When m is positive integer, then m (m – 1)...(m – n + 1)(m – n)...3 ⋅ 2 ⋅ 1 n yn = a (ax + b)m–n (m – n)...3 ⋅ 2 ⋅ 1 yn = m dn m ( + ) = ax b a n ( ax + b) m− n m−n dx n 1 2 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Case II. When m = n = +ve integer yn = n n a (ax + b)0 = 0 n an ⇒ dn ( ax + b)n = n a n n dx Case III. When m = –1, then y = (ax + b)–1 = ∴ 1 ( ax + b) yn = (–1) (–2) (–3) ... (–n) an (ax + b)–1–n ⇒ RS T dn 1 n dx ax + b UV = (−1) n a W (ax + b) n n n+1 Case IV. Logarithm case: When y = log (ax + b), then a ax + b Differentiating (n–1) times, we get y1 = yn = an d n −1 ( ax + b) −1 dx n−1 Using case III, we obtain ⇒ l n −1 q = (−1) ( ax +(nb)− 1) a dn log( ax + b) dx n n n 2. Exponential Function (i) Consider y = amx y1 = mamx. loge a y2 = m2amx (loge a)2 ................................... ................................... yn = mn amx (loge a)n (ii) Consider Putting y = emx a = e in above yn = mnemx 3. Trigonometric Functions cos (ax + b) or sin (ax + b) Let y = cos (ax + b), then F ax + b + π I H 2K 2π I F y = – a cos (ax + b) = a cos ax + b + H 2 K 3π I F y = + a sin (ax + b) = a cos ax + b + H 2 K y1 = – a sin (ax + b) = a cos 2 2 3 3 2 3 ................................................................................. ................................................................................. 3 DIFFERENTIAL CALCULUS-I F I H K d nπ I F y = sin( ax + b) = a sin ax + b + H 2K dx yn = nπ dn cos ( ax + b) = a n cos ax + b + n 2 dx n Similarly, n n n 4. Product Functions eax sin (bx + c ) or e ax cos (bx + c ) Consider the function y = eax sin (bx + c) y1 = eax·b cos (bx + c) + aeax sin (bx + c) = eax [b cos (bx + c) + a sin (bx + c)] To rewrite this in the form of sin, put a = r cos φ, b = r sin φ, we get y1 = eax [r sin φ cos (bx + c) + r cos φ sin (bx + c)] y1 = reax sin (bx + c + φ) Here, r = a 2 + b 2 and φ = tan (b/a) –1 Differentiating again w.r.t. x, we get y2 = raeax sin (bx + c + φ) + rbeax cos (bx + c + φ) Substituting for a and b, we get y2 = reax. r cos φ sin (bx + c + φ) + reax r sin φ cos (bx + c+ φ) y2 = r2eax [cos φ sin (bx + c + φ) + sin φ cos (bx + c + φ)] ∴ = r2 eax sin (bx + c + φ + φ) y2 = r2 eax sin (bx + c+ 2φ) Similarly, y3 = r3eax sin (bx + c + 3φ) ...................................................... yn = dn dx n e ax sin( bx + c ) = r n e ax sin (bx + c + nφ) In similar way, we obtain d n ax e cos(bx + c) = r ne ax cos (bx + c + nφ) yn = dx n 1 1 − 5x + 6x 2 1 = ( 2 x − 1)( 3 x − 1) Example 1. Find the nth derivative of Sol. Let or ∴ 1 1 − 5x + 6x 2 2 3 − y = 2x − 1 3x − 1 y = yn = 2 (By Partial fraction) dn dn –1 (2x – 1) – 3 (3x – 1)–1 dx n dx n 4 A TEXTBOOK OF ENGINEERING MATHEMATICS—I L(−1) n 2 OP LM(−1) n 3 OP (−1) n a d As ( ax + b) = = 2 M –3 MN (2x − 1) PQ MN (3x − 1) PQ dx ( ax + b) L 2 − 3 OP . y = ( −1) n M N (2x − 1) ( 3x − 1) Q n n n n+1 n+1 n or n n+1 n n+1 Example 2. Find the nth derivative of eax cos2 x sin x. (1 + cos 2 x) sin x Sol. Let y = eax cos2 x sin x = eax 2 1 ax 1 ax e sin x + e ( 2 cos 2x sin x) = 2 2× 2 1 ax 1 e sin x + e ax sin( 3x) − sin x = 2 4 1 ax 1 ax e sin x + e sin 3x y = 4 4 1 n ax 1 ∴ yn = r e sin (x + nφ) + r1 n e ax sin ( 3x + nθ) . 4 4 where and n n +1 l or n −1 n+1 n +1 n n r = q a 2 + 1 ; tan φ = 1/a a 2 + 9 ; tan θ = 3/a. 2x , find yn. Example 3. If y = tan–1 1 − x2 2x Sol. We have y = tan −1 1 − x2 Differentiating y w.r.t. x, we get r1 = y1 = 1 1+ ⋅ 2 F 2x I H1−x K (U.P.T.U., 2002) FG H d 2x dx 1 − x 2 2 2(1 + x 2 ) = 2 = (1 − x ) 2(1 − x ) + 4x IJ = K e1 + x − 2x + 4x j ⋅ (1 − x ) 2 2 4 (1 + x 2 ) 2 y1 = 1 1 1 − , (by Partial fractions) i x−i x+i LM N = LM MN n −1 n −1 1 ( −1) ( n − 1) ( −1) ( n − 1) − i ( x − i) n ( x + i) n ( −1)n−1 ( n − 1) i = 2 2 OP Q Differentiating both sides (n–1) times w.r. to ‘x’, we get yn = 2 2 2 ( x + i)( x − i ) y1 = (1 + x 2 ) 2 2 ( x − i ) − n − ( x + i ) −n ( −1)n−1 ( n − 1) i OP PQ r − n (cos θ − i sin θ) − n − r − n (cos θ + i sin θ) − n (where x = r cos θ, 1 = r sin θ) 5 DIFFERENTIAL CALCULUS-I = ( −1) n − 1 ( n − 1) r − n i n − 1 r–n sin nθ, where r = yn = 2(–1)n–1 Sol. We have x2 + 1 F 1I. H xK θ = tan–1 Example 4. If y = x log cos nθ + i sin nθ − cos nθ + i sin nθ x −1 . Show that x +1 LM x − n − x + n OP N (x − 1) (x + 1) Q yn = (–1)n–2 n−2 y = x log x −1 = x [log(x – 1) – log (x + 1)] x +1 n (U.P.T.U., 2002) n Differentiating w.r. to ‘x’, we get LM 1 − 1 OP N x − 1 x + 1Q F 1 I + F−1 + 1 I = log (x – 1) – log (x + 1) + 1 + H x − 1K H x + 1 K y1 = log (x – 1) – log (x + 1) + x or y1 = log (x –1) – log (x + 1) + 1 1 + x −1 x +1 Differentiating (n–1) times with respect to N, we get yn = = = = d n−1 d n−1 d n −1 d n −1 dx dx dx dx n −1 log (x − 1) − n −1 d n− 2 log (x + 1) + n −1 (x − 1) −1 + n −1 RS d log(x − 1)UV − d RS d log(x + 1)UV + (−1) n − 1 + (−1) n − 1 Tdx W dx Tdx W (x − 1) (x + 1) n− 2 dxn−2 (x + 1) −1 n −1 n− 2 F H I K F H n− 1 n n I K ( −1)n −1 n − 1 ( −1) n−1 n − 1 d n− 2 d n− 2 1 1 − + + (x − 1) n ( x + 1) n dx n − 2 x − 1 dx n − 2 x + 1 ( −1)n− 2 n − 2 (x − 1) n−1 − ( −1) n− 2 n − 2 (x + 1)n−1 + (x − 1)n + ( −1)n − 1 (n − 1) n − 2 LM x − 1 − x + 1 − (n − 1) − (n − 1) OP N (x − 1) (x + 1) (x − 1) (x + 1) Q L x − n − x + n OP . n−2M N (x − 1) (x + 1) Q n−2 n−2 = ( −1) n n n −2 = ( −1) n n Example 5. Find yn (0) if y = Sol. We have ( −1)n−1 (n − 1) n − 2 y = x3 x −1 2 n n . x3 x 3 − 1 + 1 ( x − 1)( x 2 + x + 1) 1 = + 2 = 2 2 ( x − 1)( x + 1) x −1 x −1 x −1 (x + 1)n 6 A TEXTBOOK OF ENGINEERING MATHEMATICS—I or y = x2 − 1 + 1 1 +1+ x +1 x−1 x+1 a fa f 1 1L 1 1 O + M − y = x+ x + 1 2 N x − 1 x + 1 PQ 1L 1 1 O + y = x+ M 2 N x − 1 x + 1 PQ (−1) n O 1 L ( −1) n P + y = 0+ M 2 NM ( x − 1) ( x + 1) QP ( −1) n L 1 1 y = MN (x − 1) + (x + 1) OPQ 2 ( −1) n L 1 1 O + x = 0, y (0) = M 2 N (−1) (1) PQ y = or x2 + x + 1 1 + ( x + 1) ( x − 1)( x + 1) n ∴ n n +1 n n+1 n or n +1 n n +1 n At n +1 n When n is odd, yn(0) = When n is even, yn(0) = ( −1) n n 2 ( −1) n n 2 n +1 1+1 = − n −1 + 1 = 0 . EXERCISE 1.1 1 . If y = x2 , find nth derivative of y. ( x − 1)2 ( x + 2) (U.P.T.U., 2002) LMAns. y = (−1) n + 1 + 5(−1) n + 4(−1) n OP 3( x − 1) 9( x − 1) 9( x + 2) PQ MN LMAns. (−1) n L a − b OOP x 2 . Find the nth derivative of . M P ( a − b) N (x − a) ( x − a)( x − b) (x − b) Q PQ MN L 1 + x OP . 3 . Find the nth derivative of tan M N1 − x Q n n 2 n n+2 n n +1 n n +1 2 2 n +1 n +1 –1 [Ans. ( −1)n −1 n − 1 sin n θ sin nθ where θ = cot–1x ] 4 . If y = sin3 x, find yn. 5 . Find nth derivative of tan–1 LMAns. 3 sinF x + n π I − 1 .3 . sinF 3x + n π I OP H 2KQ N 4 H 2K 4 Ans. a −1f n − 1 a sin θ sin nθ n F xI . H aK n− 1 −n n 7 DIFFERENTIAL CALCULUS-I a f LMAns. 2(−1) n OP MN (x + 1) PQ Ans. a −1f n − 1 ⋅ 2x LMAns. e 1 − 5 cos(2x + n tan ) OP 2 N Q Ans. e x x + n 6 . Find yn, where y = ex.x. n 1− x 7 . Find yn, when y = . 1+ x n+1 n− 1 8 . Find nth derivative of log x2. x 9 . Find yn, y = ex sin2 x. −n −1 n/2 2 1 0 . If y = cos x · cos 2x · cos 3x find yn. 1 (cos 6 x + cos 4x + cos 2x + 1) [Hint: cos x · cos 2x · cos 3x = 4 LMAns. 1 L6 cosF 6x + n π I + 4 cosF 4x + n π I + 2 cosF 2x + n π I OOP H 2K H 2 K PQQ N 4 MN H 2 K n n n 1.2 LEIBNITZ'S* THEOREM Statement. If u and v be any two functions of x, then Dn (u.v) = nc0 Dn (u).v + nc1Dn–1(u). D(v) + nc2 Dn–2(u).D2 (v) + ... + ncr Dn–r (u).Dr(v) + ... + ncn u. Dn v ...(i) (U.P.T.U., 2007) Proof. This theorem will be proved by Mathematical induction. D (u.v) = D (u).v + u.D(v) = 1c0 D (u).v + 1c1 u.D(v) Now, ...(ii) This shows that the theorem is true for n = 1. Next, let us suppose that the theorem is true for, n = m from (i), we have Dm (u.v) = m Differentiating w.r. to x, we have m c0 Dm(u).v + mc1 Dm–1 (u) D (v) + mc2 Dm–2(u) D2 (v) + ... + mcr Dm–r(u) Dr (v) + ... + mcm u Dm(v) r m r Dm+1 (uv) = mc0 D m+1 ( u) ⋅ v + D m ( u) ⋅ D( v) + mc1 D m ( u) D( v) + Dm−1 ( u) D 2 ( v) + mc2 mD m− 1 r m r ( v )r ( u) D 2 ( v ) + D m − 2 ( u). D 3 ( v) +...+ m c r Dm − r +1 ( u) Dr v + D m − r ( u) Dr +1 ( v) m + ..... + mcm D( u) ⋅ D ( v ) + uD m m +1 But from Algebra we know that mcr + mcr+1 = m+1cr+1 and mc0 = m+1c0 = 1 c h c h ∴ Dm+1(uv) = m +1 c 0 D m +1 ( u) ⋅ v + m c 0 + m c1 D m ( u) ⋅ D( v) + m c1 + m c 2 D m −1 u ⋅ D 2 v + ... + c c + c hD m r m r +1 m−r ( u) ⋅ D r +1 ( v ) +...+ m +1 c m +1 u ⋅ D m +1 ( v ) c As c = m m m +1 cm +1 = 1 h * Gottfried William Leibnitz (16461716) was born Leipzig (Germany). He was Newton’s rival in the invention of calculus. He spent his life in diplomatic service. He exhibited his calculating machine in 1673 to the Royal society. He was linguist and won fame as Sanskrit scholar. The theory of determinants is said to have originated with him in 1683. The generalization of Binomial theorem into multinomial theorem is also due to him. His works mostly appeared in the journal ‘Acta eruditorum’ of which he was editor-in-chief. 8 A TEXTBOOK OF ENGINEERING MATHEMATICS—I ⇒ Dm+1(uv) = m+1 c0 D m+1 (u) ⋅ v + m+1 c1D m (u) ⋅ D(v) + m +1 c2 D m−1 (u) ⋅ D 2 ( v)+... + m +1 c r +1 D m− r (u) ⋅ D r +1 (v)+...+ m+1 cm+1u ⋅ D m +1 (v) ...(iii) Therefore, the equation (iii) shows that the theorem is true for n = m + 1 also. But from (2) that the theorem is true for n = 1, therefore, the theorem is true for (n = 1 + 1) i.e., n = 2, and so for n = 2 + 1 = 3, and so on. Hence, the theorem is true for all positive integral value of n. Example 1. If y1/m + y–1/m = 2x, prove that (x2 – 1) yn+2 + (2n + 1) xyn+1 + (n2 – m2) yn = 0. Sol. Given ⇒ y1/m + 1 y 1/ m (U.P.T.U., 2007) = 2x y2/m – 2xy1/m + 1 = 0 (y ) – 2x(y1/m) + 1 = 0 (y1/m = z) z2 – 2xz + 1 1/m 2 or ⇒ ∴ z = 2x ± 4x 2 − 4 =x ± 2 x2 − 1 x ± x2 − 1 ⇒ y = x ± Differentiating equation (i) w.r.t. x, we get ⇒ y1 = ⇒ or x2 − 1 y1/m = m x ± x −1 y1 = 2 my m −1 ⇒ y1 m LM1 ± 2x OP = m x ± x − 1 x −1 N 2 x −1 Q 2 2 ⇒ 2 m 2 x 2 − 1 = my x −1 y (x – 1) = m y Differentiating both sides equation (ii) w.r.t. x, we obtain 2y1y2(x2 – 1) + 2xy21 = 2m2 yy1 2 1 2 ...(i) 2 2 ...(ii) y2 (x2 – 1) + xy1 – m2y = 0 Differentiating n times by Leibnitz's theorem w.r.t. x, we get Dn (y2) · (x2 – 1) + nc1 Dn–1y2·D2(x2 – 1) + nc2 Dn−2 y2 D2 (x2 − 1) + Dn (y1)x + nc1 Dn–1 (y1) Dx–m2yn =0 ⇒ yn+2 (x2 – 1) + nyn+1· 2x + n(n − 1) yn · 2 + yn+1 · x + nyn – m2yn = 0 2 ⇒ (x2 – 1)yn+2 + (2n + 1) xyn+1 + (n2 – n + n – m2)yn = 0 ⇒ (x2 – 1) yn+2 + (2n + 1) xyn+1 (n2 – m2) yn = 0. Hence proved. Example 2. Find the nth derivative of ex log x. Sol. Let u = ex and v = log x n x n Then D (u) = e and D (v) = ( −1) n−1 n − 1 xn D n ( ax + b) −1 = (−1) n n ( ax + b)n+1 9 DIFFERENTIAL CALCULUS-I By Leibnitz’s theorem, we have Dn (ex log x) = Dnex log x + nc1 Dn–1 (ex) D(log x) + nc2 Dn–2 (ex)D2 (log x) + ... + ex Dn (log x) FG H IJ + ... + e (−1) n − 1 K x L ( −1) n − 1O n n( n − 1) PP . +...+ D (e log x) = e M log x + − MN x 2x x Q n−1 1 n(n − 1) x 1 e − 2 = e · log x + ne · + x 2 x x x x n n −1 ⇒ n x x 2 n Example 3. Find the nth derivative of x2 sin 3x. Sol. Let u = sin 3x and v = x2 ∴ F 3x + n π I H 2K Dn(u) = Dn (sin 3x) = 3n sin D(u) = 2x, D2 (v) = 2, D3 (v) = 0 By Leibnitz’s theorem, we have Dn (x2 sin 3x) = Dn (sin 3x)x2 + nc1 Dn–1 (sin 3x) · D (x2) + nc2 Dn–2(sin 3x) · D2(x2) I · x + n3 sin FG 3x + n − 1π IJ · 2x H K 2 K n(n − 1) F n − 2π IJ ⋅ 2 + · 3 sin G 3x + H 2 2 K F nπ I + 2nx · 3 sin FG 3x + n − 1π IJ = 3 x sin 3x + H H 2K 2 K F n − 2π IJ . + 3 n(n−1) · sin G 3x + H 2 K F H = 3n sin 3x + nπ 2 2 n–1 n–2 n 2 n-1 n−2 Example 4. If y = x log (1 + x), prove that yn = ( −1) n − 2 n − 2 (x + n) (x + 1) n . Sol. Let u = log (1 + x), v = x Dn (u) = = ⇒ dn d n −1 log ( 1 + ) x = dx n dx n−1 d n −1 dx ⋅ n −1 (U.P.T.U., 2006) F d log (1 + x)I H dx K d n −1 1 ( x + 1) −1 = x+1 dx n −1 ( −1) n−1 n − 1 n D (u) = ( x + 1) n and D(v) = 1, D2(v) = 0 By Leibnitz’s theorem, we have yn = Dn (x log (1 + x) = Dn (log (1 + x)) x + nc1 Dn–1 (log (1 + x)) Dx = x ( −1) n −1 n − 1 ( x + 1) n + n( −1) n −2 n − 2 ( x + 1) n−1 10 A TEXTBOOK OF ENGINEERING MATHEMATICS—I LM − x(n − 1) + n(x + 1) OP N (x + 1) (x + 1) Q L − xn + x + xn + n OP n−2 M N (x + 1) Q L x + n OP . Hence proved. n−2 M N (x + 1) Q yn = (–1)n–2 n − 2 ⇒ = (–1)n–2 = (–1)n–2 n n n n Example 5. If y = a cos (log x) + b sin (log x). Show that x2y2 + xy1 + y = 0 x2yn+2 + (2n +1) xyn+1 + (n2 + 1) yn = 0. y = a cos (log x) + b sin (log x) and Sol. Given ∴ or y1 = – a sin (log x) (U.P.T.U., 2003) F 1 I + b cos (log x) F 1 I H xK H xK xy1 = – a sin (log x) + b cos (log x) Again differentiating w.r.t. x, we get xy2 + y1 = – a cos (log x) ⇒ ⇒ F 1 I – b sin (log x) F 1 I H xK H xK x2y2 + xy1 = – {a cos (log x) + b sin (log x)} = – y x2y2 + xy1 + y = 0. Hence proved. ...(i) Differentiating (i) n times, by Leibnitz’s theorem, we have yn+2 · x2 + n yn+1 · 2x + n(n − 1) yn · 2 + yn+1 · x + nyn + yn = 0 2 ⇒ x2yn+2 + (2n + 1) xyn+1 + (n2 – n + n + 1) yn = 0 ⇒ x2yn+2 + (2n + 1) xyn+1 + (n2 + 1) yn = 0. Hence proved. Example 6. If y = (1 – x)–α e–αx, show that (1 – x)yn+1 – (n + αx) yn – nαyn–1 = 0. Sol. Given y = (1 – x)–α. e–αx Differentiating w.r.t. x, we get y1 = α (1 – x)–α–1 e–αx – (1 – x)–α e–αx·α y1 = (1 – x)–α e–αx ·α = y1 (1 – x) = αxy LM 1 − 1OP = yα LM x OP N1 − x Q N1 − x Q Differentiating n times w.r.t. x, by Leibnitz’s theorem, we get yn+1 (1 – x) – nyn = αyn· x + nαyn–1 ⇒ (1 – x)yn+1 – (n + αx) yn – nαyn–1 = 0. Hence proved. FG y IJ = log F x I , prove that x y + (2n + 1)xy + (n + m ) y = 0. H bK H mK F yI F x I = m log x Sol. We have cos GH JK = log H mK b m m Example 7. If cos–1 2 m –1 2 n+2 n+1 2 n 11 DIFFERENTIAL CALCULUS-I ⇒ y = b cos F m log x I H mK On differentiating, we have F m log x I · m ⋅ 1 H mK x m xI F xy = – mb sin H m log K m 2 y1 = – b sin ⇒ 1 Again differentiating w.r.t. x, we get F H xy2 + y1 = – mb cos m log x (xy2 + y1) = – m2b cos I K 1 1 x m· · x m m m F m log x I = –m y H mK 2 x2y2 + xy1 + m2y = 0 or Differentiating n times with respect to x, by Leibnitz’s theorem, we get yn+2 · x2 + nyn+1 · 2x + n(n − 1) · 2yn + xyn+1 + nyn + m2yn = 0 2 ⇒ x2yn+2 + (2n + 1) xyn+1 + (n2 – n + n + m2)yn = 0 ⇒ x2yn+2 + (2n + 1) xyn+1 + (n2 + m2) yn = 0. Hence proved. Example 8. If y = (x2 – 1)n, prove that (x2 – 1)yn+2 + 2xyn+1 – n (n+1)yn = 0 (U.P.T.U., 2000, 2002) UV W RS T dP d dn (1 – x 2 ) n + n(n + 1) Pn = 0. (x2 – 1)n, show that n dx dx dx 2 n Sol. Given y = (x – 1) Differentiating w.r. to x, we get Hence, if Pn = y1 = n(x2 – 1)n–1. 2x = ⇒ 2nx (x 2 − 1)n (x 2 − 1) (x2 – 1) y1 = 2nxy Again differentiating, w.r.t. x, we obtain (x2 – 1) y2 + 2xy1 = 2nxy1 + 2ny Now, differentiating n times, w.r.t. x by Leibnitz's theorem (x2 – 1)yn+2 + 2nxyn+1 + or or 2n(n − 1) yn + 2xyn+1 + 2nyn = 2nxyn+1 + 2n2yn + 2nyn 2 (x2 – 1)yn+2 + 2xyn+1 (n+1 – n) + (n2 – n + 2n – 2n2 – 2n)yn = 0 (x2 – 1) yn+2 + 2xyn+1 – (n2 + n)yn = 0 ⇒ (x2 – 1) yn + 2 + 2xyn+1 − n(n + 1) yn = 0. Hence proved. ...(i) 12 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Second part: Let y = (x2 – 1)n ∴ Now Pn = RS T d d yn (1 − x 2 ) dx dx dn y = yn dx n UV = d n(1 − x )y s W dx 2 n +1 = (1 – x2)yn+2 – 2xyn+1 = – (x 2 − 1) y n+ 2 + 2xyn +1 ⇒ RS UV = – n(n + 1)y T W d R S(1 − x ) dPdx UVW + n (n + 1)y = 0. Hence proved. dx T d d (1 − x 2 ) Pn dx dx 2 or [Using equation (i)] n n n Example 9. Find the nth derivative of y = xn–1 log x at x = 1 . 2 Sol. Differentiating y1 = (n – 1) xn–1–1 log x + xn–1 or 1 x (n − 1)x n −1 ⋅ log x x n−1 + ⇒ xy1 = (n – 1)y + xn–1 x x Differentiating (n–1) times by Leibnitz's theorem, we get y1 = xyn + n–1c1 yn-1 = (n – 1)yn–1 + n−1 ⇒ xyn + (n – 1)yn–1 = (n – 1)yn–1 + n−1 ⇒ xyn = At x = yn n − 1 i.e. y = d n −1 n−1 = ( n − 1)( n − 2)...2.1 = n − 1 x dx n −1 n−1 n x 1 2 FG 1 IJ = 2 n − 1 . H 2K Example 10. If y = (1 – x2)–1/2 sin–1x, when –1 < x < 1 and − π π < sin–1x < , then show 2 2 that (1 – x2)yn+1 – (2n + 1) xyn – n2 yn–1 = 0. Sol. Given y = (1 – x2)–1/2 sin–1x Differentiating y1 = – y1 = ⇒ 1 (1 – x2)–3/2 (–2x) sin–1x + (1 – x2)–1/2· 2 1 2 −2 x(1 − x ) sin −1 x (1 − x ) y1 (1–x ) = xy + 1 2 2 + xy + 1 1 = 2 (1 − x 2 ) (1 − x ) 1 1 − x2 13 DIFFERENTIAL CALCULUS-I Differentiating n times w.r.t. x, by Leibnitz's theorem, we get n(n − 1) yn+1 (1 – x2) + nyn (–2x) + yn–1 · (–2) = xyn + nyn–1 2 (1 – x2)yn+1 – (2n + 1)xyn – (n2 – n + n)yn–1 = 0 ⇒ Hence proved. (1 – x2)yn+1 – (2n + 1)xyn – n2yn–1 = 0. Example 11. If y = xn log x, then prove that (i) yn+1 = n (ii) yn = nyn–1 + x Sol. (i) We have y = xn log x an − 1f . Differentiating w.r. to x, we get y1 = nxn–1 · log x + ⇒ xn x xy1 = nxn . log x + xn xy1 = ny + xn ...(i) Differentiating equation (i) n times, we get n xyn+1 + nyn = nyn + ⇒ yn+1 = (ii) yn = n Proved. x FG d x . log xIJ H dx K F x + nx . log xI d = GH x JK dx d d = n x .log xj + .x e dx dx dn d n−1 x n .log x = n dx dx n −1 e j n−1 n n−1 n −1 n−1 n−1 n −1 n −1 n n −1 n −1 an − 1f . Proved. = nyn–1 + e j dn x n log x n dx d n −1 ∴ yn −1 = n−1 x n−1 log x dx As yn = e EXERCISE 1.2 Find the nth derivative of the following: 1 . ex log x. 2 . x2 ex. LMAns. e Llog x + c ⋅ 1 − c ⋅ 1 + 2 c ⋅ 1 +...+(−1) MN x x x N x n 1 n 2 n 2 3 3 n−1 n − 1 n c n x −n OP OP QQ Ans. e x x 2 + 2nx + n(n − 1) j 14 A TEXTBOOK OF ENGINEERING MATHEMATICS—I LMAns. 6(−1) n − 4 OP x MN PQ LMAns. 2 (−1) n OP MN (1 + x) PQ n 3 . x log x. 3 4. n− 3 1− x . 1+ x n n+1 5 . x2 sin 3x. LMAns. 3 x sinF 3x + nπ I + 2nx ⋅ 3 Lsin{3x + 1 (n − 1)π}O + 3 n(n − 1) sin RS3x + π (n − 2)UVOP MN PQ H 2K 2 T 2 WQ N 6 . e (2x + 3) . Ans. e o( 2x + 3) + 6n( 2x + 3) + 12(n − 1)( 2x + 3) + 8n(n − 1)(n − 2)t n −1 2 n x n− 2 x 3 2 2 7 . If x = tan y, prove that (1 + x2) yn+1 + 2nxyn + n(n–1)yn–1 = 0. (U.P.T.U., 2006) 8 . If y = ex sin x, prove that y′′–2y′ + 2y = 0. 9 . If y = sin (m sin–1x), prove that (1 – x2)yn+2 – (2n + 1)x · yn+1 + (m2 – n2)yn = 0. 1 0 . If x = cos h 2 2 (U.P.T.U., 2004, 2002) LMF 1 I log yOP , prove that (x – 1)y + xy – m y = 0 and (x – 1)y + (2n + 1)xy NH m K Q 2 2 2 2 1 n+2 n+1 + (n – m )yn = 0. FG y IJ = log F x I , prove that x y + (2n + 1)xy + 2n y = 0. H bK H nK n 1 1 . If cos–1 2 2 n+2 n+1 n 1 2 . If y = etan–1x, prove that (1 + x2)yn+2 + {2(n + 1) x –1}yn+1 + n(n+1)yn = 0. 1 3 . If sin–1y = 2 log (x + 1), show that (x + 1)2yn+2 + (2n + 1)(x + 1)yn+1 + (n2 + 4)yn = 0. d 1 4 . If y = C1 x + x 2 − 1 i + C dx − x − 1 i , prove that (x – 1)y + (2n + 1)xy = 0. n 2 2 n 2 n+2 n+1 1 5 . If x = cos [log (y )], then show that (1 – x )yn+2 – (2n + 1) xyn+1 – (n + a ) yn = 0. 1/a 2 2 2 1.2.1 To Find (yn)0 i.e., nth Differential Coefficient of y, When x = 0 Sometimes we may not be able to find out the nth derivative of a given function in a compact form for general value of x but we can find the nth derivative for some special value of x generally x = 0. The method of procedure will be clear from the following examples: –1 Example 1. Determine yn(0) where y = em.cos x. Sol. We have –1 y = em.cos x Differentiating w.r.t. x, we get –1 y 1 = em cos x m or FG IJ ⇒ H 1 − x K 1 − x . y = – me −1 2 2 2 2 2 1 − x 2 y1 = – my ⇒ (1 – x )y1 = m y 2 1 ...(i) m cos–1x 15 DIFFERENTIAL CALCULUS-I Differentiating again (1 – x2) 2y1 y2 – 2xy21 = 2m2yy1 ⇒ (1 – x2)y2 – xy1 = m2y ...(ii) Using Leibnitz's rule differentiating n times w.r.t. x 2n(n − 1) yn – xyn+1 – nyn = m2yn (1 – x2)yn+2 – 2nxyn+1 – 2 (1 – x2)yn+2 – (2n +1)xyn+1 – (n2 + m2)yn = 0 or Putting x = 0 yn+2 (0) – (n2 + m2)yn (0) = 0 ⇒ replace n by (n – 2) yn+2 (0) = (n2 + m2)yn(0) ...(iii) yn (0) = {(n – 2)2 + m2} yn–2 (0) replace n by (n – 4) in equation (iii), we get yn–2 (0) = {(n – 4 )2 + m2} yn–4 (0) ∴ yn(0) = {(n – 2)2 + m2} {(n – 4)2 + m2} yn–4 (0) Case I. When n is odd: yn(0) = {(n – 2)2 + m2} {(n – 4)2 + m2} .... (12 + m2)y1 (0) [The last term obtain putting n = 1 in eqn. (iii)] 1 m cos −1 x m⋅ Now we have y1 = − e 1 − x2 mπ ...(iv) ...(v) As cos −1 0 = At x = 0, y1(0) = −me 2 Using (v) in (iv), we get yn(0) = − π 2 mπ 2 {an − 2f + m } {an − 4f + m } .... e1 + m j me . 2 2 2 2 2 2 Case II. When n is even: yn(0) = {(n – 2)2 + m2} {(n – 4)2 + m2} .... (22 + m2)y2 (0) [The last term obtain by putting n = 2 in (iii)] From (ii), y2(0) = m2(y)0 ∴ y2(0) = m2 emπ/2 ...(vi) −1 As y = e m cos x ...(vii) −1 ∴ y(0) = e m cos 0 = emπ/2 From eqns. (vi) and (vii), we get yn(0) = {(n – 2)2 + m2} {(n – 4)2 + m2} .... (22 + m2) m2emπ/2. Example 2. If y = (sin–1x)2. Prove that yn(0) = 0 for n odd and yn(0) = 2.22.42....(n – 2)2, n ≠ 2 for n even. (U.P.T.U., 2005, 2008) –1 2 Sol. We have y = (sin x) ...(i) 1 −1 On differentiating y 1 = 2 sin–1 x · ⇒ y1 1 − x 2 = 2 y , As y = sin x 2 1− x Squaring on both sides, y21 (1 – x2) = 4y e j 16 or A TEXTBOOK OF ENGINEERING MATHEMATICS—I Again differentiating 2(1 – x2) y1 y2 – 2xy12 = 4y1 (1 – x2) y2 – xy1 = 2 ...(ii) Differentiating n times by Leibnitz’s theorem 2n n − 1 (1 – x2) yn+2 – 2nxyn+1 – yn – xyn+1 – nyn = 0 2 a f (1 – x2) yn+2 – (2n + 1) xyn+1 – n2yn = 0 Putting x = 0 in above equation yn+2 (0) – n2 yn (0) = 0 ⇒ yn+2 (0) = n2 yn(0) replace n by (n – 2) yn(0) = (n – 2)2 yn –2 (0) or ...(iii) Again replace n by (n – 4) in (iii) and putting the value of yn–2 (0) in above equation yn (0) = (n – 2)2 (n – 4)2 yn–4 (0) Case I. If n is odd, then yn(0) = (n – 2)2 (n – 4)2 (n – 6)2 ... 12·y1 (0) 1 But y1(0) = 2 sin–1 0· =0 1–0 ∴ yn(0) = 0. Hence proved. Case II. If n is even, then yn(0) = (n – 2)2 (n – 4)2 ... 22· y2 (0) From (ii) y2 (0) = 2 Using this value in eqn. (iv), we get yn(0) = (n – 2)2 (n – 4)2 .... 22·2 yn(0) = 2.22.42 ... (n – 2)2, n ≠ 2 otherwise 0. Proved. or m Example 3. If y = x + 1 + x2 Sol. Given y = x + 1 + x2 ∴ 2 y1 = m x + 1 + x = y1 1 + x 2 or Squaring m m x + 1 + x2 1+ x , find yn 0). m −1 LM1 + x OP N 1+x Q ...(i) 2 m = 2 my 1 + x2 = my y21(1+x2) = m2y2 2 1 or ...(iv) ...(ii) 2 2 Again differentiating, y (2x) + (1 + x )2y1y2 = m ·2yy1 y2 (1 + x2) + xy1 – m2y = 0 Differentiating n times by Leibnitz’s theorem 2n(n − 1) (1 + x2)yn+2+ 2nxyn+1 + yn + xyn+1 + nyn – m2yn = 0 2 ...(iii) 17 DIFFERENTIAL CALCULUS-I (1 + x2)yn+2 + (2n + 1)xyn+1 + (n2 – m2)yn = 0 Putting x = 0, we get yn+2 (0) + (n2 – m2) yn(0) = 0 ⇒ yn+2 (0) = – (n2 – m2) yn (0) replace n by n – 2 yn (0) = – { (n – 2)2 – m2} yn–2 (0) Again replace n by (n – 4) in (iv) and putting yn–2 (0) in above equation or ...(iv) yn(0) = (–1)2 {(n – 2)2 – m2} {(n – 4}2 – m2} yn–4 (0) Case I. If n is odd But or ⇒ Case II. If n is even yn(0) = – {(n – 2)2 – m2} {(n – 4)2 – m2} ... {12 – m2} y1 (0) y1(0) = my(0) y1(0) = m (As y (0) = 1) yn(0) = {m2 – (n – 2)2}{m2 – (n – 4)2} ... (m2 – 12)·m. yn (0) = {m2 – (n – 2)2} {m2 – (n – 4)2}... (m2 – 22) y2 (0) ⇒ yn (0) = {m2 – (n – 2)2} {m2 – (n – 4)2} ... (m2 – 22)·m2. (As y2 (0) = m2). –1 Example 4. Find the nth differential coefficient of the function on cos (2 cos x) at the point x = 0. or Sol. Let y = cos (2 cos–1 x) On diffentiating, y1 = – sin (2 cos–1 x) LM −2 OP MN 1 − x PQ ...(i) 2 –1 1 − x 2 = 2 sin (2 cos x) Squaring on both sides, we get y1 e −1 = 4 sin2 2 cos x y21 (1 – x2) { e j j} 2 −1 = 4 1 − cos 2 cos x or 2 1 2 2 y (1 – x ) = 4 (1 – y ) Again differentiating w.r.t. x, we get 2y1y2 (1 – x2) – 2xy21 = – 8yy1 or y2 (1 – x2) – xy1 + 4y = 0 ...(ii) Differentiating n times by Liebnitz’s theorem 2n n − 1 (1 – x2) yn+2 – 2nxyn+1 – yn – xyn+1 – nyn + 4yn = 0 2 a f (1 – x2) yn+2 – (2n + 1) xyn+1 – (n2– 4) yn = 0 Putting x = 0 in above equation, we get yn+2 (0) – (n2 – 4) yn (0) = 0 or yn+2 (0) = (n2 – 4) yn(0) Replace n by n – 2, we get yn (0) = {(n – 2)2 – 4} yn–2 (0) ...(iii) 18 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Again replace n by (n – 4) in (iii) and putting yn –2 (0) in above then, we get yn (0) = {(n – 2)2 – 4} {(n – 4)2 – 4} yn – 4 (0) Case I. If n is odd yn (0) = {(n – 2)2 – 4} {(n – 4)2 – 4} .......... (12 – 4) y1 (0) But y1 (0) = 2 sin (2 cos–1 0) = 2 sin (π) = 0 ∴ yn (0) = 0. Case II. If n is even yn (0) = {(n – 2)2 – 4}{(n – 4)2 – 4} .......... {22 – 4} y2 (0) yn (0) = 0 Hence for all values of n, even or odd, yn (0) = 0. Example 5. Find the nth derivative of y = x2 sin x at x = 0. Sol. We have (U.P.T.U., 2008) y = x2 sin x = sin x. x2 ...(i) Differentiate n times by Leibnitz’s theorem, we get yn = nC Dn (sin x). x2 + nC Dn–1 (sin x) D(x2) + nC Dn–2 (sin x) D2 (x2) + 0 FG H IJ + 2nx . sin FG x + n − 1 πIJ + n(n – 1) sin FG x + n − 2 πIJ K H 2 K H 2 K F nπ IJ + 2nx sin FG x + nπ − π IJ + n(n – 1) sin FG x + nπ − πIJ = x sin G x + H 2K H 2 K H 2 2K F nπ IJ – 2nx cos FG x + nπ IJ – n(n – 1) sin FG x + nπ IJ = x sin G x + H 2K H 2K H 2K F nπ IJ – 2nx cos FG x + nπ IJ y = (x – n + n) sin G x + H 2K H 2K 0 nπ = x2 . sin x + 2 1 2 2 2 2 2 n Putting x = 0, we obtain nπ . 2 Example 6. If y = sin (a sin–1 x), prove that yn (0) = (n – n2) sin (1 – x2)yn+2 – (2n + 1)xyn+1 – (n2 – a2)yn = 0 Also find nth derivative of y at x = 0. Sol. We have [U.P.T.U. (C.O.), 2007] y = sin(a sin–1 x) Differentiating w.r.t. x, we get y1 = cos(a sin–1 x). or or a 1 − x2 ...(i) y1 1 − x 2 = a cos(a sin–1 x) Squaring on both sides, we obtain y12 (1 – x2) = a2 cos2 (a sin–1 x) = a2 [1 – sin2 (a sin–1 x)] y12 (1 – x2) = a2 (1 – y2) ...(ii) 19 DIFFERENTIAL CALCULUS-I Differentiating again, we get 2y1 y2 (1 – x2) – 2xy12 = – 2 a2yy1 y2 (1 – x2) – xy1 = – a2y or ...(iii) Differentiating n times by Leibnitz’s theorem a f y – xy – ny = – a y (1 – x2)yn+2 – 2nxyn+1 − 2n n − 1 2 or 2 n n+1 n n (1 – x2)yn+2 – (2n + 1)xyn+1 – (n2 – a2)yn = 0 ...(iv) Putting x = 0 in relation (iv), we get yn+2 (0) – (n2 – a2)yn (0) = 0 yn+2 (0) = (n2 – a2)yn (0) or ...(v) Replace n by (n – 2) in relation (v), we get yn (0) = {(n – 2)2 – a2}yn–2 (0) Again replace n by (n – 4) in equation (v) and putting yn–2 (0) in above relation, we get yn (0) = {(n – 2)2 – a2} {(n – 4)2 – a2}yn–4 (0) Case I. When n is odd: yn (0) = {(n – 2)2 – a2} {(n – 4)2 – a2} ..... {12 – a2}y1 (0) ...(vi) [The last term in (vi) obtain by putting n = 1 in equation (v)] Putting x = 0, in equation (i), we get y1 (0) = cos(a sin–1 0). a = cos 0 . a ⇒ y1 (0) = a yn (0) = {(n – 2)2 – a2} {(n – 4)2 – a2} ..... {12 – a2}. a Hence, Case II. When n is even: yn (0) = {(n – 2)2 – a2} {(n – 4)2 – a2} ..... {22 – a2}y2 (0). Putting x = 0 in (iii), we get [The last term obtain by putting n = 2 in equation (v)] y2 (0) = – a2 y(0) = – a2 x0 = 0 Hence, (As y(0) = 0) yn (0) = 0. EXERCISE 1.3 1 . If y = tan–1 x, find the value of y7 (0) and y8 (0). a sin–1 x [Ans. 6 and 0.] 2 . If y = e , prove that (1 – x ) yn + 2 – (2n + 1)xyn + 1 – (n + a ) yn = 0 and hence find the value of yn when x = 0. Ans. n is odd, yn (0) = 2 2 2 {an − 2f + a } {an − 4f + a } ... (3 + a ) (1 + a )a n is even, y (0) = {an − 2f + a } {an − 4f + a } ...... (4 + a )(2 + a ) a . 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 n 3 . If log y = tan–1x, show that (1 + x2) yn + 2 + {2(n + 1)x – 1} yn + 1 + n (n + 1) yn = 0 and hence find y3, y4 and y5 at x = 0. [U.P.T.U. (C.O.), 2003] Ans. y3 (0) = – 1, y4 (0) = – 1, y5 (0) = 5 20 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 4 . If f (x) = tan x, then prove that fn (0) – nc2 fn – 2 (0) + nc4 fn – 4 (0) – ....... = Sin 5 . If y = sin–1x, find yn (0). FG nπ IJ . H 2K Ans. n is odd, yn (0) = (n – 2)2 (n – 4)2 .... 52 . 32 . 1 n is even, yn (0) = 0. 6 . Find yn (0) when y = sin (m sin h–1 x). a fFG IJ {an − 2f + m } {an − 4f + m } .... (1 + m ) m · n n−1 Ans. n is odd, yn (0) = –1 H 2 K 2 2 2 2 2 2 is even, yn = (0) = 0. LM { N 7 . If y = log x + 1 + x 2 }OPQ , show that 2 yn + 2 (0) = – n2 yn (0) hence find yn (0). a f Ans. n is odd, yn (0), = 0 n is even yn(0) = −1 n− 2 2 (n – 2)2 (n – 4)2 .... 42 22.2. PARTIAL DIFFERENTIATION Introduction Real world can be described in mathematical terms using parametric equations and functions such as trigonometric functions which describe cyclic, repetitive activity; exponential, logrithmic and logistic functions which describe growth and decay and polynomial functions which approximate these and most other functions. The problems in computer science, statistics, fluid dynamics, economics etc., deal with functions of two or more independent variables. 1.3 FUNCTION OF TWO VARIABLES If f (x, y) is a unique value for every x and y, then f is said to be a function of the two independent variables x and y and is denoted by z = f (x, y) Geometrically the function z = f (x, y) represents a surface. The graphical representation of function of two variables is shown in Figure 1.1. y ¦ (x,y) z O z = ¦ (x,y) Fig. 1.1 x 21 DIFFERENTIAL CALCULUS-I 1.4 PARTIAL DIFFERENTIAL COEFFICIENTS The partial derivative of a function of several variables is the ordinary derivative with respect to any one of the variables whenever, all the remaining variables are held constant. The difference between partial and ordinary differentiation is that while differentiating (partially) with respect to one variable, all other variables are treated (temporarily) as constants and in ordinary differentiation no variable taken as constant, Definition: Let z = f (x, y) Keeping y constant and varying only x, the partial derivative of z w.r.t. ‘x’ is denoted by and is defined as the limit ∂z ∂x = Lim b ∂z ∂x g b g f x + δx, y − f x, y δx→0 δx ∂z Partial derivative of z, w.r.t. y is denoted by ∂y and is defined as ∂z ∂y = Lim δy→0 Notation: The partial derivative b g b g. f x, y + δy − f x, y δy ∂z ∂z ∂f is also denoted by or fx similarly ∂y is denoted ∂x ∂x ∂f by ∂y or fy. The partial derivatives for higher order are calculated by successive differentiation. Thus, ∂2 f ∂ 2z = ∂x 2 ∂x 2 = fxx, ∂2z ∂2 f = = fyy ∂y 2 ∂y 2 ∂2z ∂2 f = ∂x∂y ∂x∂y = fxy, ∂2 f ∂2z = = fyx and so on. ∂y∂x ∂y∂x ∂z ∂z and : ∂ y ∂x Let z = f(x, y) represents the equation of a surface in xyzcoordinate system. Suppose APB is the curve which a plane through any point P on the surface O to the xz-plane, cuts. As point P moves along this curve APB, its coordinates z and x vary while y remains constant. The slope of the tangent line at P to APB represents the rate at which z-changes w.r.t. x. Geometrical interpretation of Hence, and Z A P O ∂z = tan θ (slope of the curve APB at the point P) ∂x ∂z = tan φ (slope of the curve CPD at point P) ∂y Y f q B D X Fig. 1.2 22 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 1. Verify that ∂ 2u ∂ 2u = where u(x, y) = loge ∂x∂y ∂y∂x Fx + y I GH xy JK 2 Fx + y I GH xy JK 2 2 (U.P.T.U., 2007) 2 Sol. We have u(x, y) = loge ⇒ u(x, y) = log (x2 + y2) – log x – log y ...(i) Differentiating partially w.r.t. x, we get ∂u ∂x = 2x 1 – x2 + y2 x Now differentiating partially w.r.t. y. ∂ 2u ∂y∂x = – 4xy ...(A) ex + y j 2 2 2 Again differentiate (i) partially w.r.t. y, we obtain ∂u ∂y = 2y ex + y j 2 2 1 y – Next, we differentiate above equation w.r.t. x. ∂ 2u 4xy = – 2 2 ∂x∂y x + y2 e ...(B) j Thus, from (A) and (B), we find ∂ 2u ∂ 2u = . ∂y∂x ∂x∂y Hence proved. FG y IJ , verify that ∂ f = ∂ f . H xK ∂y∂x ∂x∂y F yI f = tan G J H xK 2 Example 2. If f = tan–1 Sol. We have 2 –1 ...(i) Differentiating (i) partially with respect to x, we get ∂f = ∂x FG –y IJ = F –y I F y I H x K GH x + y JK 1+G J H xK 1 2 2 2 ...(ii) 2 Differentiating (i) partially with respect to y, we get ∂f 1 1 x = = 2 2 x ∂y x + y2 y 1+ x Differentiating (ii) partially with respect to y, we get FG IJ H K F GH ∂2 f −y ∂ = ∂y∂x ∂y x 2 + y 2 ...(iii) I = ex + y ja–1f − b– ygb2yg JK ex + y j 2 2 2 2 2 23 DIFFERENTIAL CALCULUS-I = y2 − x2 ...(iv) ex + y j 2 2 2 Differentiating (iii) partially with respect to x, we get F GH ∂2 f x ∂ = 2 ∂x∂y ∂x x + y 2 = I = ex + y ja1f − xa2xf JK ex + y j 2 2 3 2 2 y2 − x2 ...(v) ex + y j 2 2 2 ∂2 f ∂2 f = . Hence proved. ∂y∂x ∂x∂y ∴ From eqns. (iv) and (v), we get F ∂u ∂u I F ∂u ∂u I Example 3. If u(x + y) = x + y , prove that G − J = 4 G 1 − ∂x − ∂y J . H K H ∂x ∂y K 2 2 Sol. Given ∴ 2 u = x2 + y2 x+y bx + yga2xf − ex + y ja1f = x + 2xy − y b x + yg bx + y g bx + ygb2yg − ex + y ja1f = y + 2xy − x ∂u = ∂y bx + yg bx + yg ∂u ∂x 2 = 2 2 2 2 2 and 2 2 2 2 2 ∴ ∂u ∂u + ∂x ∂y ∂u ∂u 1– – ∂x ∂y or ∂u ∂u – ∂y ∂x and 4xy = b x + yg = 1− 2 4xy ( x + y) = 2 (x − y) 2 ...(i) (x + y) 2 ex + 2xy − y j − e y + 2xy − x j = bx + y g 2 ex − y j 2 bx − yg = = bx + yg bx + y g 2 2 2 2 2 2 2 2 ∴ 2 ...(ii) From (ii), we get F ∂u − ∂u I GH ∂x ∂y JK 2 = b g = 4 F1 − ∂u − ∂u I , from (i). Hence proved. GH ∂x ∂y JK bx + yg 4 x− y 2 2 24 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 4. If u = sin–1 Sol. Given and from (i), y ∂u ∂x = ∂u ∂y = ∂u ∂y = x Example 5. If f(x, y) = x2 tan–1 2 yx ...(ii) x + y2 2 F– x I + 1 ⋅ 1 R| F x I U| GH y JK 1 + FG y IJ x S|1 − GH y JK V| H xK T W 1 2 2 x − + ey − x j 2 2 2 xy ...(iii) x + y2 2 ∂u ∂u +y = 0. ∂y ∂x Hence proved. FG y IJ – y tan FG x IJ then prove that ∂ f = ∂ f . H xK H yK ∂x∂y ∂y∂x 2 2 2 –1 FG y IJ + x · 1 × FG − y IJ – y · 1 · FG 1 IJ H xK F yI H x K F xI H yK 1+G J 1+G J H xK H yK ∂f F y I yx – y = 2x tan FG y IJ – y = 2x · tan G J – H xK ∂x H xK x + y x + y ∂f = 2x · tan–1 ∂x 2 2 2 2 2 2 3 –1 or ...(i) 2 2 − ey − x j 2 Adding (ii) and (iii), we get x Sol. –1 2 x or –1 –1 ∴ or F x I + tan FG y IJ , show that x ∂u + y ∂u = 0. GH y JK ∂y H xK ∂x F xI F yI u = sin G J + tan G J H xK y H K ∂u F yI 1 1 1 = . + . G− J ∂x R| F x I U| y 1 + FG y IJ H x K S|1 − GH y JK V| H xK W T –1 2 2 2 2 Differentiating both sides with respect to y, we get FG 1 IJ –1 = 2x –1 = x − y ...(i) H xK yI x +y x +y F 1+G J H xK F xI F xI ∂f 1 1 1 = x · · – 2y tan G J – y · · G− J ∂y H yK F yI x F xI H y K 1+G J 1+G J H xK H yK F x I xy ∂f x = – 2y tan G y J + H K x +y ∂y x +y ∂2 f = 2x · ∂y∂x Again 2 1 2 2 · 2 2 –1 2 2 2 2 2 2 –1 2 2 2 3 or 2 2 2 2 25 DIFFERENTIAL CALCULUS-I j – 2y tan FG x IJ = x – 2y tan FG x IJ . H yK H yK x +y e x x2 + y2 = 2 –1 2 –1 Differentiating both sides with respect to x, we get ∂2 f = 1 – 2y ∂x ∂y F 1I 2y x −y G J =1– = y H K x +y x + y F xI 1+G J H yK 2 1 2 2 2 2 2 2 2 ...(ii) Thus, from (i) and (ii), we get ∂2 f ∂2 f = . Hence proved. ∂x∂y ∂y∂x Example 6. If V = (x2 + y2 + z2)–1/2, show that ∂ 2V ∂ 2V ∂ 2V + + = 0. 2 ∂y ∂x 2 ∂z 2 Sol. Given V = (x2 + y2 + z2)–1/2. ∴ ∂V ∂x = – ∴ ∂ 2V ∂x 2 = – x – ...(i) 1 (x2 + y2 + z2)–3/2 2x = – x (x2 + y2 + z2)–3/2 2 LM RS 3 ex + y + z j NT 2 2 2 2 –5/2 UV e W ⋅ 2x + x 2 + y 2 + z 2 = 3x2 (x2 + y2 + z2)–5/2 – (x2 + y2 + z2)–3/2 = (x2 + y2 + z2)–5/2 [3x2 – (x2 + y2 + z2)] or and ∂ 2V = (x2 + y2 + z2)–5/2 (2x2 – y2 – z2) ∂x 2 Similarly from (i), we can find j –3/2. ⋅1 OP Q ...(ii) ∂ 2V ∂y 2 = (x2 + y2 + z2)–5/2 (2y2 – x2 – z2) ...(iii) ∂ 2V ∂z 2 = ( x2 + y2 + z2)–5/2 (2z2 – x2 – y2) ...(iv) Adding (ii), (iii) and (iv), we get ∂ 2V ∂ 2V ∂ 2V + + 2 ∂y ∂x 2 ∂z 2 = (x2 + y2 + z2)–5/2 [(2x2 – y2 – z2) + (2y2 – x2 – z2) + (2z2 – x2 – y2)] = (x2 + y2 + z2)–5/2 [0] = 0. Hence proved. Example 7. If u = f (r), where r2 = x2 + y2, show that ∂ 2u ∂ 2u 1 f ′ (r). 2 + ∂y 2 = f ′′ (r) + r ∂x (U.P.T.U., 2001, 2005) 26 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Sol. Given r2 = x2 + y2 ...(i) Differentiating both sides partially with respect to x, we have 2r Similarly, ∂r ∂x = 2x or ∂r ∂y = Now, ∂r x = ∂x r ...(ii) y r ...(iii) u = f(r) ∂u ∂r x = f ′ (r). = f ′ (r) · , from (ii) ∂x ∂x r Again differentiating partially w.r.to x, we get ∴ LM f ′ arfx OP = r f ′ arf.1 + xf ′′arfb∂r / ∂xg – xf ′ arfb∂r / ∂xg r N r Q O x 1 L ∂ u rf ′ ar f + x f ′′ar f − f ′ arfP , from (ii). M = r r N ∂x Q ∂ u O 1 L y rf ′ ar f + y f ″ ar f – f ′ arfP Similarly, = M ∂y r MN r PQ L ex + y j f ′ arfOP ∂ u ∂ u 1 M 2rf ′ ar f + ex + y j f ″ arf − Adding, + = PQ r ∂x ∂y r M N ∂ 2u ∂x 2 = ∂ ∂x 2 2 or 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 = 1 [2rf ′(r) + r2f ″(r) – rf ′(r)], from (i) r2 = 1 f ′(r) + f ″(r). Hence proved. r Example 8. If u = log (x3 + y3 + z3 – 3xyz); show that F ∂ + ∂ + ∂I u=– 9 . GH ∂x ∂y ∂z JK bx + y + zg 2 2 Sol. Given and (U.P.T.U., 2003) u = log (x3 + y3 + z3 – 3xyz). ∴ ∂u ∂x = x 3 + y 3 + z 3 − 3xyz Similarly, ∂u ∂y = x + y 3 + z 3 – 3xyz ∂u ∂z = 3x 2 − 3 yz ...(i) 3 y 2 − 3xz ...(ii) 3 3z 2 − 3xy x 3 + y 3 + z 3 – 3xyz . ...(iii) 27 DIFFERENTIAL CALCULUS-I Adding eqns. (i), (ii) and (iii), we get ∂u ∂u ∂u + ∂y + ∂x ∂z = e 3 x 2 + y 2 + z 2 – xy − yz − zx x + y + z − 3xyz 3 3 3 j e 3 x 2 + y 2 + z 2 – xy − yz − zx = j bx + y + zgex + y + z − xy − yz − zxj 2 2 2 a f As a 3 + b 3 + c 3 – 3abc = a + b + c ( a 2 + b 2 + c 2 − ab − bc − ca) ∂u ∂u ∂u + + ∂x ∂y ∂z or = 3 . x+y+z ...(iv) F ∂ ∂ + ∂ I u = FG ∂ + ∂ + ∂ IJ FG ∂ + ∂ + ∂ IJ u Now, G + H ∂x ∂y ∂z K H ∂x ∂y ∂z K H ∂x ∂y ∂z JK F ∂ ∂ ∂ I F ∂u ∂u ∂u I F ∂ ∂ ∂ I F 3 I = G ∂x + ∂y + ∂z J G ∂x + ∂y + ∂z J = G ∂x + ∂y + ∂z J G x + y + z J , from (iv) H KH K H KH K L ∂ F 1 I ∂ F 1 I ∂ F 1 IO = 3 M ∂x G x + y + z J + ∂y G x + y + z J + ∂z G x + y + z J P K H K H K PQ MN H L 1 OP 1 1 −9 − − = 3 M− MN bx + y + zg bx + y + zg bx + y + zg PQ = bx + y + zg . 2 2 2 2 2 Hence proved. Example 9. If u = tan–1 ∂ 2u ∂x∂y Sol. Given ∴ e xy 1 + x2 + y2 = 1 e1 + x + y j 2 3/2 2 ⋅ xy u = tan–1 ∂u ∂y j , show that e1 + x + y j 2 2 ⋅ 1 LM1 + {x y /e1 + x + y j}OP N Q LM √ e1 + x + y j.1 − y 1 e1 + x + y j 2 ×x M + 1 x e +y j MN e1 + x + y j − y x = . √ e1 + x + y j 1+ x + y + x y = 2 2 2 2 2 2 2 2 2 2 2 −1/2 2 2 2 2 2 2 2 2 2y OP PP Q 28 A TEXTBOOK OF ENGINEERING MATHEMATICS—I ∂u ∂y or = x 1 + x2 x = e1 + x je1 + y j e1 + x + y j e1 + y j e1 + x + y j . 2 2 2 2 2 2 2 Again differentiating partially w.r.to x LM OP L x OP ∂ M x 1 = MMe1 + y j e1 + x + y j PP = e1 + y j ∂x MM e1 + x + y j PP N Q N Q LM e1 + x + y j1 − x 1 e1 + x + y j 2x OP 1 2 = M PP e1 + y j MN e1 + x + y j Q e1 + x + y j – x = 1 1 . = . Hence proved. e1 + y j e1 + x + y j e1 + x + y j ∂ 2u ∂x ∂y ∂ ∂x 2 2 2 2 2 2 2 2 2 2 2 2 3/2 2 Example 10. If 2 2 2 −1/2 2 2 3/2 2 2 y2 x2 z2 + + = 1, show that a2 + u c2 + u b2 + u FG ∂u IJ + FG ∂u IJ + FG ∂u IJ = 2 FG x ∂u + y ∂u + z ∂u IJ . H ∂x K H ∂y K H ∂z K H ∂x ∂y ∂z K 2 2 2 2 (U.P.T.U., 2002) Sol. We have y2 x2 z2 + + = 1 a2 + u b2 + u c2 + u where u is a function of x, y and z Differentiating (i) partially with respect to x, we get ...(i) LM x OP ∂u y z + + MMe a + uj eb + uj e c + uj PP ∂x = 0 N Q 2x / e a + uj 2x / e a + uj ∂u = = ∂x LMx / ea + uj + y / eb + uj + z / e c + uj OP ∑ LMx / e a + uj OP Q N N Q 2 y / e b + uj 2z / e c + uj ∂u ∂u = ; = ∂y ∑ LMNx / e a + uj OPQ ∂z ∑ LMNx / ea + uj OPQ = 2 2 2x − a2 + u 2 2 2 2 2 2 2 2 2 or 2 2 2 2 2 2 2 Similarly, 2 2 2 2 2 2 2 2 2 2 2 Adding with square L O FG IJ + FG ∂u IJ + FG ∂u IJ = 4 MNx / e a + uj + y / eb + uj + z / e c + uj PQ H K H ∂y K H ∂z K LM∑ RSx / e a + uj UVOP WQ N T ∂u ∂x 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 29 DIFFERENTIAL CALCULUS-I FG ∂u IJ + FG ∂u IJ + FG ∂u IJ = H ∂x K H ∂y K H ∂z K 2 2 or Also, x 2 ∂u ∂u ∂u +y +z = ∂x ∂z ∂y = 4 ∑ LMNx / e a + uj OPQ 2 ...(ii) 2 2 LM 2x O 2y 2z P + + ∑ LMNx / e a + uj OPQ MN e a + uj eb + uj e c + uj PQ 2 2 2 2 ∑ LMNx 2 / e a 2 + uj OPQ 2 2 2 1 2 2 2 2 [1], from (i) ...(iii) From (ii) and (iii), we have FG ∂u IJ + FG ∂u IJ + FG ∂u IJ = 2 FG x ∂u + y ∂u + z ∂u IJ . Hence proved. H ∂x K H ∂y K H ∂z K H ∂x ∂y ∂z K 2 2 2 Example 11. If xx yy zz = c, show that at x = y = z, ∂ 2z = – (x log ex)–1. Where z is a function of x and y. ∂x ∂y Sol. Given xx.yy.zz = c, where z is a function of x and y. Taking logarithms, x log x + y log y + z log z = log c. ...(i) Differentiating (i) partially with respect to x, we get LMx FG 1 IJ + blog xg1OP + LMz FG 1 IJ + blog zg1OP ∂z = 0 N H xK Q N H zK Q ∂x b1 + log xg ∂z = − b1 + log zg ∂x ∂z b1 + log yg Similarly, from (i), we have ∂y = – b1 + log zg ∂ F ∂z I ∂ z ∂ L F 1 + log y I O ∴ = ∂x G ∂y J = ∂x∂y H K ∂x MMN−GH 1 + log z JK PPQ , from (iii) = or 2 ∂2z ∂ –1 ∂x∂y = – (1 + log y) · ∂x [(1 + log z) ] or LM– b1 + log zg ⋅ 1 ⋅ ∂z OP z ∂x Q N ∂ z b1 + log yg · R|S−F 1 + logxIU|V , from (iii) = ∂x∂y zb1 + log zg T| GH 1 + log zJKW| ∂ z b1 + log xg At x = y = z, we have =– ∂x∂y x b1 + log xg −2 = – (1 + log y) · 2 or 2 2 ∴ 2 3 ...(ii) ...(iii) 30 A TEXTBOOK OF ENGINEERING MATHEMATICS—I ∂2z 1 1 ∂x∂y = − x 1 + log x = – x log e e + log x = (As log e e = 1) b ⇒ g b g a f = – mx log aexfr . Hence proved. 1 x log ex =– −1 Example 12. If u = log (x3 + y3 – x2y – xy2) then show that ∂ 2u 4 ∂ 2u ∂ 2u . + 2 + = − 2 2 2 ∂x∂y ∂x ∂y x+y b Sol. We have 3 3 2 g 2 u = log (x + y – x y – xy ) ∂u 3x 2 − 2xy − y 2 = 3 ∂x x + y 3 − x 2 y − xy 2 e 3 y − x − 2xy ∂u = 3 ∂y x + y 3 − x 2 y − xy 2 2 e ...(i) j ...(ii) 2 e Adding (i) and (ii), we get j j e 3x 2 − 2xy − y 2 + 3 y 2 − x 2 − 2xy ∂u ∂u + = ∂y ∂x x 3 + y 3 − x 2 y − xy 2 e b g j 2 x− y 2 2 2 j bx + yg ex + y – 2xyj 2 bx − yg ∂u ∂u 2 + = = ∂x ∂y bx + ygbx − yg bx + yg F∂ ∂I ∂ u ∂ u ∂ u Now, +2 + = G + J u H ∂x ∂y K ∂x ∂x∂y ∂y F∂ ∂I F∂ ∂I = G ∂x + ∂y J G ∂x + ∂y J u H KH K F ∂ ∂ I 2 (from iii) = G + J . H ∂x ∂y K x + y ∂ F 1 I ∂ F 1 I G J + 2 =2 ∂y GH x + y JK ∂x H x + y K = 2 ... (iii) 2 2 2 2 2 2 2 =– 2 b x + yg 2 Example 13. If u = exyz, show that ∂ 3u = (1 + 3xyz + x2y2z2)exyz. ∂x∂y∂z – 2 b x + yg 2 =– 4 bx + yg 2 . Hence proved. 31 DIFFERENTIAL CALCULUS-I u = exyz ∴ Sol. We have or ∂ xyz xyz 2 xyz ∂y (e ·xy) = e x yz + e .x ∂ 2u ∂y∂z = ∂ 2u ∂y∂z = (x2yz + x) exyz ∂ 3u ∂x∂y∂z Hence ∂u = exyz.xy ∂z = (2xyz + 1) exyz + (x2yz + x) exyz·yz = (1 + 3xyz + x2y2z2) exyz. Hence proved. Example 14. If u = log r, where r2 = (x – a)2 + (y – b)2 + (z – c)2, show that ∂ 2u ∂ 2u ∂ 2u 1 + + = 2. 2 2 2 ∂x ∂z ∂y r Sol. Given r2 = (x – a)2 + (y – b)2 + (z – c)2, ...(i) Differentiating partially with respect to x, we get 2r Similarly, FG x − a IJ . H r K ∂r b y − bg and ∂r = az − cf = ∂r ∂x ∂y or ...(ii) r IJ K FG H ∂u ∂x = 1 ∂r 1 x−a = , from (ii) r r ∂x r ∂u ∂x = x−a r2 ∂ 2u ∂x 2 = FG x − a IJ = r a1f − ax − af 2rb∂r / ∂xg Hr K r ∂ u r − 2 ax − af = , from (ii) ∴ ∂x 2 Similarly, ∂ 2u ∂y 2 ∂ 2u ∂ 2u ∂ 2u + + ∂y 2 ∂x 2 ∂z 2 ∂ ∂x 2 2 4 2 2 2 or ∴ ∂y r u = log r. Now, ∴ ∂r = ∂x = 2 (x – a) or r4 = = = b g ; ∂ u = r − 2 ( z − c) . r2 − 2 y − b 2 2 2 ∂z 2 r4 2 r4 {a f + b y − bg + az − cf } 3r 2 − 2 x − a 2 2 2 r4 1 3r 2 − 2r 2 , from (i) = 2 . Hence proved. 4 r r 32 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 15. If u = x2 tan–1 (y/x) – y2 tan–1 (x/y), prove that ∂ 2u x2 − y2 = 2 . ∂y∂x x + y2 Sol. Given u = x2 tan–1 (y/x) – y2 tan–1 (x/y). Differentiating partially with respect to x, we get ∂u ∂x = FG y IJ + 2x tan FG y IJ − y ⋅ 1 ⋅ 1 H xK 1 + b y / xg H x K 1 + bx / y g y 1 x2 = – = – 2 x2 y x +y 2 – 2 ⋅ − −1 y3 x +y 2 2 2 2 2 + 2x tan–1 y x j + 2x tan FG y IJ = – y + 2x tan FG y IJ H xK H xK ex + y j e y x2 + y2 2 –1 –1 2 Again differentiating partially with respect to y, we get ∂ 2u ∂y∂x = ∂ ∂y RS– y + 2x tan FG y IJ UV = – 1 + 2x 1 ⋅ 1 H xKW F yI x T 1+G J H xK = –1+ –1 2 x2 − y2 2x 2 = . Hence proved. x2 + y2 x2 + y2 Example 16. If z = f (x – by) + φ (x + by), prove that ∂ 2z ∂2z . b = ∂y 2 ∂x 2 z = f (x – by) + φ (x + by) ∂z = f ′ (x – by) + φ′ (x + by) ∂x ∂2z = f ″ (x – by) + φ″ (x + by). ∂x 2 ∂z ∂y = – bf ′ (x – by) + bφ′ (x + by) 2 Sol. Given ∴ and Again from (i), ...(i) ...(ii) ∂2z ∂ 2z 2 2 2 from (ii). Hence proved. 2 = b f ″ (x – by) + b φ′′ (x + by) = b ∂y ∂x 2 ′ and Example 17. If u (x, y, z) = log (tan x + tan y + tan z). Prove that sin 2x ∂u ∂u ∂u + sin 2y ∂y + sin 2z = 2. ∂x ∂z Sol. ∂u ∂x = sec 2 x tan x + tan y + tan z (U.P.T.U., 2006) 33 DIFFERENTIAL CALCULUS-I ∂u ∂y = sec 2 y tan x + tan y + tan z sec 2 z ∂u = tan x + tan y + tan z ∂z ∂u ∂u ∂u ∴ sin 2x + sin 2y ∂y + sin 2z ∂x ∂z = = sin 2x sec 2 x + sin 2y sec 2 y + sin 2z sec 2 z tan x + tan y + tan z b 2 tan x + tan y + tan z btan x + tan y + tan zg g = 2. Hence proved. EXERCISE 1.4 ∂ 3u 2 2 2 1. Find ∂x∂y∂z if u = ex +y +z . Ans. 8xyzu 2. Find the first order derivatives of LMAns. ∂u = x b y log x + yg; ∂u = x log xOP ∂y N ∂x Q F I LMAns. ∂u = 1 ∂u = − y ex − y j FH x + x − y IK OP (ii) u = log H x + x − y K MN ∂x x − y ∂y PQ F x − y I , show that ∂u = – y ∂u . 3. If u = sin G ∂x x ∂y H x + y JK xy + 1 xy (i) u = xxy. 2 2 2 2 2 –1 4. If u = ex (x cos y – y sin y), prove that ∂ 2u ∂ 2u + = 0. ∂x 2 ∂y 2 FG y IJ , prove that ∂ u + ∂ u = 0. H xK ∂x ∂y 2 5. If u = a log (x2 + y2) + b tan–1 2 6. If z = tan (y – ax) + (y + ax)3/2, prove that 7. If u = x2 x y2 y z2 z 1 1 1 , prove that 2 2 2 ∂ 2z 2 ∂ z a – = 0. ∂x 2 ∂y 2 ∂u ∂u ∂u + ∂y + = 0. ∂x ∂z 2 −1 2 2 2 −1 34 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 8. If u = 2(ax + by)2 – (x2 + y2) and a2 + b2 = 1, find the value of ∂ 2u ∂ 2u . 2 + ∂x ∂y 2 LM OP −4 yz MMAns. ex + y + z j PP N Q ∂ 2u . 9. If u = log (x2 + y2 + z2), find the value of ∂y∂z e j 1 10. If u = x 2 + y 2 + z 2 2 , then prove that ∂2u ∂x 2 11. If z = f (x + ay) + φ (x – ay), prove that 12. If u = cos–1 + ∂2u ∂x 2 16. If u = r , where r = x +y +z 2 2 17. If u = (x2 + y2 + z2)–1, prove that 18. If θ = t e ∂ 2u ∂2u 2 = . 2 + 2 ∂y u ∂z ∂ 2u = 0. ∂y 2 + 2 ∂2 f ∂x 2 = b2 ∂2 f ∂t 2 . ∂ 2u ∂ 2u ∂ 2u . find 2 + ∂y 2 + ∂x ∂z 2 [Ans. m (m + 1)rm−2] ∂ 2u ∂ 2u ∂2u + + = 2 (x2 + y2 + z2)–2. ∂x 2 ∂z 2 ∂y 2 r2 4t , find the value of n, when 1 ∂ r 2 ∂r FG r ∂θ IJ = ∂θ . H ∂r K ∂t 2 LMAns. n = − 3 OP 2Q N 19. For n = 2 or – 3 show that u = rn (3 cos2 θ – 1) satisfies the differential equation FG r ∂u IJ + 1 ∂ FG sin θ ∂u IJ = 0. H ∂r K sin θ ∂θ H ∂θ K F ∂ u I + 1 ∂u + 1 ∂ u = 0. 20. If u = e cos (a log r), show that G H ∂r JK r ∂r r ∂θ ∂ ∂r 2 2 2 aθ − 2 2 2 z ∂z x −y j ∂z 21. If e e = (x – y), show that y +x = x2 – y2. 2 2 ∂x 2 ∂y 2 [Hint: Solve for z = (y – x ) log (x – y)]. 22. If u = 2 2 ∂2z ∂2z . 2 = a ∂x 2 ∂y 2 15. Prove that f (x, t) = a sin bx. cos bt satisfies − 2 ∂u ∂u ∂u + ∂y + = (x + y + z)2. ∂x ∂z 14. If u = x2y + y2z + z2x, show that n 2 bx − yg / bx + yg , prove that x ∂∂ux + y ∂∂uy = 0. 13. If u = log (x2 + y2), show that m [Ans. 0] ∂ 2u 1 ∂ 2u ∂2u and r2 = (x – a)2 + (y – b)2 + (z – c)2, prove that + + = 0. ∂y 2 r ∂x 2 ∂z 2 35 DIFFERENTIAL CALCULUS-I ∂ 2u ∂ 2u ∂2u 23. If u (x, y, z) = cos 3x cos 4y sin h 5z, prove that = 0. 2 + ∂y 2 + ∂x ∂z 2 FG ∂u IJ FG ∂x IJ = 1 FG ∂v IJ FG ∂y IJ . H ∂x K H ∂u K 2 H ∂y K H ∂v K F ∂x I F ∂x I F ∂θI F ∂θ I F ∂y I 25. If x = r cos θ, y = r sin θ, find G J , G J G J G J G J . H ∂r K H ∂θ K H ∂xK H ∂y K H ∂x K 24. If x2 = au + bv; y2 = au – bv, prove that y θ v r, x y, u r x, [Ans. cos θ, – r sin θ, –r –1 sin θ, r –1 cos θ, – cot θ.] 1.5 HOMOGENEOUS FUNCTION A polynomial in x and y i.e., a function f (x, y) is said to be homogeneous if all its terms are of the same degree. Consider a homogeneous polynomial in x and y f (x, y) = a0xn + a1 xn−1 y + a2 xn-2y2 + ..... + an yn LMa + a F y I + a F y I +...+ a F y I OP GH x JK P MN GH x JK GH x JK Q F yI f (x, y) = x F G J H xK n 2 = xn 0 1 n 2 n or Hence every homogeneous function of x and y of degree n can be written in above form. NOTE: Degree of Homogeneous function = degree of numerator – degree of denominator. Remark 1: If f (x, y) = a0xn + a1xn+1 . y–1 + a2xn+2 . y–2 + ...... + anxn+n y –n R| F F x xI xI U + + + a a .... S| GH y JK GH y JK |V| y T W F xI f (x, y) = x F G J ; degree = n H yK f (x, y) = a y + a y . x + ...... + a y . x R| F x I F x I U| = y Sa + a G J + .... + a G J V H y K |W |T H y K F xI f (x, y) = y F G J ; degree = –n H yK n 2 = x n a0 + a1 ⇒ Remark 2: If n 2 n –n –n–1 0 –n–n 1 n n n ⇒ 0 n 1 –n Another forms are also possible i.e., f (x, y) = yn F F x I ; f (x, y) = y F(y/x) GH y JK n n 36 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 1.6 EULER’S THEOREM ON HOMOGENEOUS FUNCTIONS Statement: If f is a homogeneous function of x, y of degree n then x ∂f ∂f + y ∂y = nf. ∂x (U.P.T.U., 2006) Proof. Since f is a homogeneous function ∴ FG y IJ H xK f (x, y) = xn F ...(i) Differentiating partially w.r.t. x and y, we get FG y IJ + x F′ FG y IJ FG − y IJ H xK H xK H x K ∂f FG y IJ FG 1 IJ = x F′ ∂y H xK H xK ∂f = nxn−1 F ∂x n 2 n ...(ii) ...(iii) Multiplying (ii) by x and (iii) by y and adding, we have x FG y IJ – x y F′ FG y IJ + x y F′ FG y IJ H xK H xK H xK F yI = nx F G J H xK ∂f ∂f +y = nxn F ∂x ∂y n−1 n−1 n ⇒ x ∂f ∂f +y = nf ∂y ∂x (from (i)). In general if f (x1, x2, ..., xn) be a homogeneous function in x1, x2, ..., xn then x1 ∂f + ... + xn ∂x n ∂f ∂f + x2 ∂x1 ∂x 2 = nf. Corollary 1. If f is a homogeneous function of degree n, then 2 x ∂2 f ∂x 2 ∂2 f ∂2 f 2 + 2xy +y = n (n –1)f. ∂x∂y ∂y 2 Proof. We have ∂f ∂f + y ∂y = n f ∂x Differentiating (i) w.r.t. x and y respectively, we get ...(i) ∂f ∂2 f ∂2 f ∂f +x + y = n ∂x ∂x∂y ∂x ∂x 2 ...(ii) x and x ∂f ∂2 f ∂2 f + ∂y + y ∂y∂x ∂y 2 = n ∂f ∂y ...(iii) Multiplying (ii) by x and (iii) by y and adding, we have x2 ∂2 f ∂x 2 + 2xy ∂2 f ∂2 f ∂f ∂f + y2 +y =n 2 + x ∂x∂y xy ∂y ∂x F x ∂f + y ∂t I GH ∂x ∂y JK 37 DIFFERENTIAL CALCULUS-I ⇒ x 2 ∂2 f ∂x 2 ∂2 f ∂2 f 2 + 2xy +y ∂x∂y ∂y 2 = n2f − nf = n (n – 1) f Example 1. Verify Euler’s theorem for the function f (x, y) = ax2 + 2hxy + by2. Sol. Here the given function f (x, y) is homogeneous of degree n = 2. Hence the Euler’s theorem is ∂f ∂f x +y = 2f ...(i) ∂y ∂x Now, we are to prove equation (i) as follows: ∂f ∂f = 2 ax + 2 hy, ∂y = 2hx + 2by ∂x ∴ x ∂f ∂f + y ∂y ∂x = 2ax2 + 2hxy + 2hxy + 2by2 = 2(ax2 + 2 hxy + by2) = 2f ⇒ x ∂f ∂f +y ∂x ∂y = 2f, which proves equation (i). Example 2. Verify Euler’s theorem for the function u = xn sin FG y IJ . H xK Sol. Since u is homogeneous function in x and y of degree n, hence we are to prove that x ∂u ∂u + y ∂y ∂x = nu FG y IJ H xK ∂u F yI F yI F y I = nx sin G J + x cos G J G − J H xK H xK H x K ∂x F yI F yI ∂u x = nx sin G J – x y cos G J H H xK K x ∂x F yI ∂u = yx cos G J y H xK ∂y u = xn sin We have ∴ or n–1 n n n–1 n–1 Similarly, Adding (ii) and (iii), we get x ∂u ∂u + y ∂y ∂x = nxn sin ∂u ∂u + y ∂y = nu ∂x which verifies Euler’s theorem. ⇒ ...(i) x FG y IJ H xK 2 ...(ii) ...(iii) 38 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 3. If u = sin–1 w x − yz xy x + y {| , show by Euler’s theorem that y ∂u ∂u = – · ∂x x ∂y w x − y z ⇒ sin u = x − y xy x + y {| x+ y Sol. We have u = sin–1 Let f = sin u = x− y x+ y Here, f is a homogeneous function in x and y where, degree n = 1 1 – =0 2 2 ∴ By Euler’s theorem, we have x or or ∂f ∂f +y = 0. f = 0 ∂x ∂y ∂ ∂ (sin u) + y ∂y (sin u) = 0 ∂x ∂u ∂u x cos u · + y cos u ∂y = 0 ∂x x ⇒ x As f = sin u ∂u ∂u +y = 0 ∂x ∂y y ∂u ∂u = − . ∂x x ∂y ⇒ Hence proved. Example 4. If u = log [(x4 + y4)/ (x + y)], show that x ∂u ∂u +y = 3. ∂x ∂y Sol. We have x4 + y4 x4 + y4 u u = loge ⇒e = x+ y x+y Let f = eu = (U.P.T.U., 2000) x4 + y4 x+y Here the function f is a homogeneous function in x and y of degree, n = 4 – 1 = 3 ∴ By Euler’s theorem x ⇒ x ∂f ∂f + y ∂y = nf = 3f ∂x ∂ ∂ (eu) + y ∂y (eu) = 3f ∂x DIFFERENTIAL CALCULUS-I ⇒ F x ∂u + y ∂u I GH ∂x ∂y JK e = 3e ⇒ u x u ∂u ∂u +y = 3. Hence proved. ∂x ∂y Example 5. Verify Euler’s theorem for u = sin–1 F x I + tan FG y IJ . GH y JK H xK –1 (U.P.T.U., 2006) Sol. Here u is a homogeneous function of degree, n = 1 – 1 = 0; hence by Euler’s theorem x ∂u ∂u +y = 0 ∂x ∂y ∂u = ∂x Now, or x ∂u = ∂x ∂u ∂y = and or y ∂u = ∂y 1 1− FG x IJ H yK 2 x 2 y −x ⋅ – 2 FG FG IJ H H K y 1 2 − 2 x y 1+ x 1 + y IJ K xy ...(i) 2 x + y2 F − x I + 1 . FG 1IJ GH y2 JK yI 2 H xK F F xI + 1 GH x JK 1−G J H yK 1 2 −x 2 y −x 2 . + xy ...(ii) ex + y 2 j 2 Adding (i) and (ii), we get x ∂u ∂u +y = 0, hence Euler’s theorem is verified. ∂x ∂y Example 6. If u = x sin–1 x2 Sol. We have F xI GH y JK + y sin FGH xy IJK , find the value of –1 ∂ 2u ∂ 2u 2 + 2xy ∂x∂y ∂x ∂ 2u . ∂y2 (U.P.T.U., 2007) F xI GH y JK + y sin FGH xy IJK F xI F 1 I + y · 1 FG − y IJ ∂u x = sin G J + H xK ∂x H y K 1 − x GH y JK y 1− u = x sin–1 ∴ + y2 –1 –1 2 2 y2 x2 2 40 A TEXTBOOK OF ENGINEERING MATHEMATICS—I ⇒ x 2 F xI GH y JK + 2x 2 – x y− y y −x F − x I + sin FG y IJ + y 1 FG 1 IJ 1 ∂u = x· GH y2 JK H xK ∂y y H xK x 1− 1− 2 ∂u = x sin–1 ∂x 2 2 ...(i) –1 and ∂u y ∂y = – ⇒ 2 y2 x2 y F yI + y sin G J + H xK x − y y2 − x2 2 x2 –1 2 Adding (i) and (ii), we get x 2 ∂u ∂u + y ∂y = x sin–1 ∂x 2 F x I + y sin FG y IJ = u GH y JK H xK –1 ...(ii) ...(iii) Differentiating (iii) partially w.r.t. x and y respectively. and ∂u ∂u ∂ 2u ∂ 2u ∂ 2u ∂ 2u +x + y = ⇒ x + y =0 ∂x ∂x ∂x2 ∂x2 ∂x∂y ∂x∂y ...(iv) ∂ 2u ∂ 2u ∂ 2u ∂u ∂u ∂ 2u + +y = ⇒ y + x =0 ∂x∂y ∂y∂x ∂y ∂y ∂y 2 ∂y2 ...(v) x Multiplying equation (iv) by x and (v) by y and adding, we get x2 Example 7. If u = tan–1 x2 ∂ 2u ∂ 2u ∂ 2u 2 + 2xy + y = 0. ∂y2 ∂x∂y ∂x2 x 3 + y3 ∂u ∂u , prove that x +y = sin 2u and evaluate x− y ∂x ∂y ∂ 2u ∂ 2u + 2xy + y2 ∂x∂y ∂x2 ∂ 2u . ∂y2 Sol. We have u = tan –1 x 3 + y3 x 3 + y3 ⇒ tan u = x− y x− y x 3 + y3 Let f = tan u = x− y ∴ Since f (x, y) is a homogeneous function of degree n = 3–1=2 By Euler’s theorem, we have x ⇒ x ∂f ∂f ∂ ∂ +y = nf ⇒ x (tan u) + y (tan u) = 2 tan u ∂x ∂x ∂y ∂y ∂u ∂u · sec2 u + y · sec2 u = 2 tan u ∂y ∂x DIFFERENTIAL CALCULUS-I or x ∂u tan u ∂u +y = 2 = sin 2u. Proved. ∂x sec 2 u ∂y ...(i) Differentiating (i) partially w.r.t. x, we get x or ∂u ∂ 2u ∂u ∂ 2u + + y = 2 cos 2u . 2 ∂x∂y ∂x ∂x ∂x ∂ 2u ∂u ∂ 2u + y = (2 cos 2u –1) 2 ∂ ∂ x y ∂x ∂x Multiplying by x, we obtain x ∂ 2u ∂u ∂ 2u = x (2 cos 2u –1) 2 + xy ∂x ∂x∂y ∂x Again differentiating equation (i) partially w.r.t. y, we get x2 x ...(ii) ∂ 2u ∂u ∂ 2u ∂u +y + = 2 cos 2u 2 ∂y∂x ∂y ∂y ∂y or y or y2 ∂u ∂ 2u ∂ 2u + x = (2 cos 2u – 1) 2 ∂y ∂ ∂ y x ∂y ∂u ∂ 2u ∂ 2u 2 + xy ∂x∂y = y ( 2 cos 2u – 1) ∂y (multiply by y) ∂y ...(iii) Adding (ii) and (iii), we get x2 ∂ 2u ∂ 2u ∂ 2u 2 = ( 2cos 2u – 1) 2 + 2xy ∂x∂y + y ∂x ∂y2 F x ∂u + y ∂u I GH ∂x ∂y JK = (2cos 2u – 1) sin 2u, (from (i)) = (2sin 2u cos 2u – sin 2u) = sin 4u – sin 2u = 2 cos FG 4u + 2uIJ . cos FG 4u − 2uIJ . H 2 K H 2 K ∂ 2u ∂ 2u 2 Hence, x + 2xy +y ∂y2 = 2 cos 3u · cos u. ∂x∂y ∂x2 2 ∂ 2u Example 8. If z = xm f x2 Sol. Let then ∂ 2z ∂x 2 FG y IJ + x g FG x IJ , prove that H xK H yK n + 2xy ∂ 2z ∂ x∂ y + y2 u = xm f ∂ 2z ∂y 2 + mnz = (m + n – 1) F x ∂z + y ∂z I GH ∂x ∂y JK · FG y IJ , v = x g FG x IJ H xK H yK n z = u+v ...(i) Now, u is homogeneous function of degree m. Therefore with the help of (Corollary 1, on page 36), we have 42 A TEXTBOOK OF ENGINEERING MATHEMATICS—I ∂ 2u ∂ 2u ∂ 2u 2 x + 2xy +y = m (m – 1) u ∂y2 ∂x∂y ∂x2 2 F x I , we have GH y JK Similarly for v = xn g x ∂ 2v ∂ 2v 2 + 2xy +y ∂ x∂ y ∂y 2 ∂x 2 ∂ 2v 2 ...(ii) = n (n – 1) v ...(iii) Adding (ii) and (iii), we get ∂ ∂2 ∂2 2 x 2 (u + v) + 2xy ∂x∂y (u + v) + y ∂y 2 (u + v) = m (m – 1) u + n (n – 1) v ∂x 2 ⇒ x2 ∂ 2z ∂2z ∂ 2z 2 + 2xy + y ∂y2 = m (m – 1) u + n (n – 1) v (As z = u + v). ∂x ∂y ∂x2 ...(iv) Again from Euler’s theorem, we get x Adding x ∂u ∂v ∂u ∂v + y ∂y = m u and x + y ∂ y = nv ∂x ∂x ∂ ∂ (u + v) + y ∂y (u + v) = mu + nv ∂x ∂z ∂z + y ∂y = mu + nv ...(v) ∂x m (m –1) u + n (n – 1) v = (m2 u + n2v) – (mu + nv) = m (m + n) u + n (m + n) v – mn (u + v) – (mu + nv) = (mu + nv) (m + n) − (mu + nv) − mnz = (mu + nv) (m + n – 1) – mnz ⇒ x Now, = (m + n –1) F x ∂z + y ∂z I GH ∂x ∂y JK – mnz, from (v) Putting this value in equation (iv), we get F x ∂z + y ∂z I GH ∂x ∂y JK – mnz F x ∂z + y ∂z I ∂ 2z ∂ z ∂ 2z ⇒ x 2 + 2xy ∂x∂y + y ∂y2 + mnz = (m + n –1) GH ∂x ∂y JK . Hence proved. ∂x L x + 2y + 3z OP Example 9. If u = sin M MN x8 + y8 + z8 PQ , show that x2 ∂ 2z ∂2z ∂ 2z 2 + 2xy + y = (m + n –1) ∂x ∂y ∂y2 ∂x2 2 2 2 –1 x ∂u ∂u ∂u + y ∂y + z + 3 tan u = 0. ∂x ∂z (U.P.T.U., 2003) DIFFERENTIAL CALCULUS-I LM x + 2y + 3z OP MN x + y + z PQ L x + 2y + 3z OP sin u = M MN x + y + z PQ , let f = sin u u = sin–1 Sol. We have ⇒ 8 8 8 8 8 8 ∴ Degree of homogeneous function f, n = 1 – 4 = – 3, from Euler’s theorem, we have x ⇒ x ∂f ∂f ∂f + y ∂y + z = – 3 sin u ∂x ∂z ∂ ∂ ∂ (sin u) + y ∂y (sin u) + z (sin u) = – 3 sin u ∂x ∂z F x ∂u + y ∂u + z ∂u I GH ∂x ∂y ∂z JK cos u = – 3 sin u or or x ∂u ∂u ∂u + y ∂y + z + 3 tan u = 0. ∂x ∂z Hence proved. R| π 2x 2 + y 2 + zx 12 U| j |V , find the value of | e Example 10. If V = log sin S || 2ex 2 + xy + 2yz + z 2 j 13 || T W e x Sol. We have ∂V ∂V ∂V + y ∂y + z , when x = 0, y = 1, z = 2. ∂x ∂z R| π 2x 2 + y 2 + zx 12 U| j |V | e V = log sin S || 2ex 2 + xy + 2yz + z 2 j 13 || T W R π 2x + y + zx U j |V | e e = sin S || 2 ex + xy + 2yz + z j || T W R| π 2x 2 + y 2 + zx 12 U| j |V | e sin (e ) = S || 2 ex 2 + xy + 2yz + z 2 j 13 || T W R| π 2x 2 + y 2 + zx 12 U| j |V | e f = sin (e ) = S || 2 ex 2 + xy + 2yz + z 2 j 13 || T W 2 ∴ V 2 or Let –1 V –1 v Since f is a homogeneous function ∴ n = 1 – 2 1 = 3 3 2 1 2 1 2 3 44 A TEXTBOOK OF ENGINEERING MATHEMATICS—I By Euler’s theorem, we get ∂f ∂f ∂f x +y +z = ∂x ∂z ∂y ⇒ x 1 f 3 1 ∂ ∂ ∂ (sin–1 eV) + y (sin–1 eV) + z (sin–1 eV) = sin–1 eV 3 ∂x ∂z ∂y Fx ∂V + y ∂V + z ∂V I FG 1 × eV IJ 1 GH ∂x ∂y ∂z JK H 1 − e 2V K = 3 sin e ⇒ –1 ⇒ x ∂V ∂V ∂V +y +z = ∂y ∂x ∂z ee j V Now, esin e j −1 and V 1 1 − e 2V × sin–1 eV V 3 e = x = 0 y =1 z =2 ...(i) RS π UV = 1 , e = 1 T 2 × 2W 2 2 2V = sin x = 0 y =1 z =2 V π 4 Putting all these values in equation (i), we get ⇒ x ∂V ∂V ∂V +y +z = ∂x ∂y ∂z 1− 1 2 1 π × × = 1 3 4 2 x ∂V ∂V ∂V +y +z = ∂x ∂y ∂z π Example 11. If u = x3y2 sin–1 x 12 . FG y IJ , show that H xK ∂u ∂ 2u ∂u ∂ 2u ∂ 2u 2 + y ∂y = 5u and x2 + 2xy + y = 20u. ∂y2 ∂x ∂x2 ∂x∂y u = x3y2 sin–1 Sol. FG y IJ H xK FG yIJ sin FG y IJ = x F FG y IJ F FG y IJ = FG y IJ sin y H xK H xK H xK H xK H xK x 2 = x5 2 –1 5 −1 ∴ u is a homogeneous function of degree 5 i.e., n = 5 By Euler’s theorem, we get x ∂u ∂u + y ∂y = 5u. Proved. ∂x Next, we know that (from Corollary 1, on page 36) or 12 π × 12 12 x2 ∂ 2u ∂ 2u ∂ 2u 2 + 2xy +y ∂y2 = n (n – 1) u = 5 (5 – 1) u ∂x∂y ∂x2 x2 ∂ 2u ∂ 2u ∂ 2u 2 + 2xy + y = 20 u. Hence proved. ∂x∂y ∂y2 ∂x2 45 DIFFERENTIAL CALCULUS-I FG y IJ + f FG y IJ , prove that H xK H xK Example 12. If u = x f1 x2 2 2 ∂ 2u ∂ 2u 2 ∂ u = 0. + 2 xy + y ∂x∂y ∂x 2 ∂y 2 u1 = Sol. Let FG y IJ and u = x f FG y IJ , then u = u + u H xK H xK x f1 0 2 2 1 2 Since u1 is a homogeneous function of degree one. So by Corollary 1 on page 36, we get x2 ∂2u1 2 + 2xy ∂ u1 ∂ 2 u1 + y2 ∂x ∂y ∂y2 ...(i) = 1 (1 – 1) = 0 ∂x2 and u2 is also a homogeneous function of degree 0 2 ∂ u2 ∂ 2 u2 ∂ 2 u2 + 2xy + y2 ∴ 2 ∂y 2 ∂x∂y ∂x Adding (i) and (ii), we get x2 x2 ∂2 ∂x 2 (u1 + u2) + 2xy ⇒ x2 ...(ii) = 0 ∂2 ∂2 (u1 + u2) + y2 (u1 + u2) = 0 ∂x∂y ∂y 2 ∂ 2u ∂ 2u ∂ 2u 2 + 2xy +y ∂x∂y ∂x 2 ∂y 2 = 0. Hence proved. Example 13. If u = sin–1 (x3 + y3)2/5, evaluate 2 ∂ 2u x 2∂ 2u 2 ∂ u . + + xy y 2 ∂x∂y ∂x 2 ∂y 2 u = sin–1 (x3 + y3)2/5 Sol. Given F1 + y I GH x JK 3 sin u = (x + y ) 3 Let f = sin u ∴ f = x 6/5 which is homogeneous of degree n = x ⇒ or 3 2/5 x =x F1 + y I GH x JK 3 6/5 25 3 25 3 ...(i) 6 . By Euler’s theorem 5 ∂f ∂f 6 +y = f ∂x ∂y 5 ∂ ∂ sin u + y sin u ∂x ∂y b x g b g = 65 sin u ∂u ∂u cos u + y cos u = 6 sin u ∂x ∂y 5 x 6 ∂u ∂u tan u = +y 5 ∂x ∂y ...(ii) 46 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Differentiating (ii) w.r. to ’x‘, we get 6 ∂u ∂u ∂ 2u ∂ 2u sec2 u. +x 2 +y = 5 ∂x ∂x ∂x∂y ∂x Multiplying by x x ∂u ∂ 2u ∂ 2u 6 ∂u + x 2 2 + xy = sec2 u. x ∂x ∂x∂y ∂x 5 ∂x ...(iii) Differentiating (ii) w.r. to ‘y’, we get x 6 ∂ 2u ∂u ∂ 2u ∂u sec2 u + +y 2 = ∂y 5 ∂y∂x ∂y ∂y Multiplying by y y ∂u ∂ 2u ∂ 2u 6 ∂u + y 2 2 + xy sec2 u. y = x y ∂y ∂ ∂ 5 ∂y ∂y ...(iv) Adding (iii) and (iv), we get F x ∂u + y ∂u I FG 6 sec u − 1IJ GH ∂x ∂y JK H 5 K ∂u ∂u 6 ∂ 2u ∂ 2u ∂ 2u 6 6 As x +y = tan u x 2 2 + 2xy + y2 2 = tan u FG sec2 u − 1IJ K H ∂ x ∂y 5 ∂x∂y 5 5 ∂x ∂y F xI F yI Example 14. If u = 3x cot G J + 16y cos G J then prove that H xK H yK x2 or 2 ∂ 2u ∂ 2u 2∂ u = 2 + 2xy ∂x∂y + y ∂x ∂y 2 4 x –1 2 4 –1 2 ∂ 2u ∂ 2u 2∂ u xy 2 + + y = 12u. ∂x∂y ∂x2 ∂y 2 Sol. The given function is homogeneous function of degree 4. By Euler’s theorem, we have x ∂u ∂u +y = 4u ∂x ∂y ...(i) Differentiating (i) w.r. to x, we get ∂u ∂u ∂ 2u ∂ 2u +x 2 +y 2 = 4 ∂x ∂x ∂x ∂y or x ∂u ∂u ∂ 2u ∂ 2u + x2 2 + xy 2 = 4x ∂x ∂x ∂x ∂y ...(ii) Differentiating (i) w.r. to y, we get x or xy ∂u ∂ 2u ∂u ∂ 2u + +y 2 = 4 ∂y ∂x∂y ∂y ∂y ∂u ∂ 2u ∂u ∂ 2u +y + y 2 2 = 4y ∂y ∂x∂y ∂y ∂y ...(iii) 47 DIFFERENTIAL CALCULUS-I Adding (ii) and (iii), we get or F GH I JK x2 2 ∂u ∂u ∂ 2u ∂ 2u 2∂ u +y = 3 x xy y 2 + + 2 2 ∂x ∂y ∂x∂y ∂x ∂y x2 2 ∂ 2u ∂ 2u 2∂ u = 3 × 4u = 12u. Hence proved. xy y 2 + + ∂x∂y ∂x 2 ∂y 2 EXERCISE 1.5 Verify Euler’s theorem 1. 2 2 x −y . 1 4 4. x 3 y − 3 tan–1 FG y IJ. H xK 2. Fx + y I Fx + y I. GH JK GH JK 5. b x + yg . 1 4 1 4 1 5 1 5 xy F xI GH y JK + cot FGH yx IJK. F xI F yI 6. cos G y J + cot GH JK . H K x 3. cos–1 –1 –1 –1 F x + y I , show that x ∂u + y ∂u = 3. (U.P.T.U., 2000) GH x + y JK ∂y ∂x ∂u 9 ∂u I F I F 8. If u = H x + y K H x + y K , show that x + y ∂y = u [U.P.T.U. (AG), 2005] 20 ∂x F xI F xI ∂z ∂z 9. If z = x y sin G y J + log x – log y, show that x +y = 6x y sin G y J . H K H K ∂x ∂y 4 4 7. If u = loge 1 4 1 4 4 2 –1 1 5 1 5 4 2 –1 [U.P.T.U. (C.O.)., 2003] ∂ ∂ ∂ u u u 10. If u = x3 + y3 + z3 + 3xyz; show that x +y +z = 3u. ∂x ∂y ∂z 11. If u = cos–1 LM x − y OP , prove that x ∂u + y ∂u = 0. ∂y ∂x Nx + y Q 12. If u = log [(x2 + y2)/(x + y)], prove that x ∂u ∂u + y ∂y = 1. ∂x 13. If u = sin–1 {(x2 + y2)/(x + y)}, show that x ∂u ∂u + y ∂y = tan u. ∂x 14. If u = sin–1 {(x + y)/( x + y )}. Show that x (U.P.T.U., 2008) ∂u 1 ∂u + y ∂y = tan u. 2 ∂x ∂u 1 ∂u + y ∂y = sin 2u. 2 ∂x 16. If u is a homogeneous function of degree n, show that 15. If u = tan–1 [(x2 + y2)/(x + y)], then prove that x ∂ 2u ∂ 2u ∂u . (a) x 2 + y ∂x∂y = (n – 1) ∂x ∂x ∂ 2u ∂u ∂ 2u . (b) y 2 + x ∂x∂y = (n – 1) ∂y ∂y 48 A TEXTBOOK OF ENGINEERING MATHEMATICS—I FG y IJ , show that x ∂u + y ∂u = 0. H xK ∂x ∂y LM x + y OP ∂ u 1 ∂ 2u ∂ 2u , prove that x + 2xy + y 1 8 . If u = sin M P 2 = 144 tan u [tan u – 11]. 2 ∂ ∂ x y ∂y ∂x MN x + y PQ F x +y I 5 ∂u ∂u 1 9 . Prove that x +y = tan u if u = sin G 2 ∂x ∂y H x + y JK . 17. If u = f –1 1 4 1 4 1 6 1 6 2 2 2 3 2 3 –1 2 0 . Verify Euler’s theorem for f = z x+ y + y x + . z+x y+ z 2 F xI GH y JK , show that x ∂∂ux + y ∂∂uy = 2u and x ∂∂xu2 + 2xy ∂∂x∂uy + 2 21. If u = y2 y2 e y x + x2 tan–1 2 ∂ 2u = 2u. ∂y2 2 2 2 e x + y j sin u = x + y , prove that x ∂∂xu2 + 2xy ∂∂x∂uy + y ∂∂yu2 = tanu 12 F 13 + tan u I GH 12 12 JK . F y2 I ∂ 2u 2 3 . If u = tan G x J , show that x H K ∂x2 2 22. If 1 3 1 3 2 2 2 –1 2 4 . If f (x, y) = 2 1 x 2 + ∂f ∂f log x − log y 1 + , show that x + y ∂y + 2f (x, y) = 0. 2 2 ∂x xy x +y 2 5 . If u = sec–1 {(x3 + y3) / (x + y)}, show that x 1.7 Let ∂u ∂u + y ∂y = 2 cot u. ∂x TOTAL DIFFERENTIAL COEFFICIENT z = f (x, y) ...(i) where x = φ (t) and y = ψ (t), then z can be expressed as a function of t alone by substituting the values of x and y in terms of t from the last two equations in equation (i). And we can find the ordinary differential coefficient dz , which is called total differential dt coefficient of z with respect to t. Since it is very difficult sometimes to express z in terms of t alone by eliminating x and y. So we are now to find and y in terms of t in z = f (x, y). dz without actually substituting the values of x dt 49 DIFFERENTIAL CALCULUS-I Let δx, δy and δz be the increaments in x, y and z corresponding to a small increament δt in the value of t. then z + δz = f (x + δx, y + δy) ...(ii) where x + δx = φ (t + δt), y + δy = ψ (t + δt) Now, b g f x + δx, y + δy − f (x, y) δz dz = Lim = Lim (from ii) δt→0 δt δt→0 dt δt = Lim δt→0 b g f x + δx, y + δy − f (x + δx, y) + f (x + δx, y) − f (x, y) δt b {Adding and subtracting f (x + δx, y)} b g g f x + δx, y + δy − f (x + δx, y) Lim f x + δx, y − f (x, y) = Lim + 0 δt→ δt→0 δt δt LM f bx + δx, y + δyg − f bx + δx, yg ⋅ δy OP Lim LM f bx + δx, yg − f bx, yg ⋅ δx OP δy δt PQ + t 0 MN δx δt PQ MN Also, as δt → 0, δx → 0, δy → 0 L ∂f bx, yg δy OP + Lim LM ∂f bx, yg δx OP dz ∴ = Lim M t 0 M ∂y t 0 M ∂x δt PQ δt PQ dt N N ∂f bx, yg dx ∂f bx, yg dy = ⋅ + = Lim δt→0 δ→ δ→ ∂y ∴ dz dt = δ→ dt ∂x ∂z dx ∂z dy + ∂y dt ∂x dt dt ...(iii) (As z = f (x, y)) ∂z dx1 ∂z dx2 ∂z dxn + +....+ ∂x1 dt ∂x2 dt ∂xn dt The above relation can be also written as In general dz dt = dz = ∂z ∂z dx + ∂y dy , which is called total differential of z. ∂x Corollary: If z = f (x, y) and suppose y is the function of x, then f is a function of one independent variable x. Here y is intermediate variable. Identifying t with x in (iii), we get dz dx = ∂z dy ∂z dx ∂z ∂z dy dz + ∂y ⇒ = + . ∂x dx dx ∂x ∂y dx dx 1.7.1 Change of Variables Let z = f (x, y) where x = φ (s, t) and y = ψ (s, t) then z is considered as function of s and t. Now the derivative of z with respect s is partial but not total. Keeping t constant the equation (iii) modified as ∂z ∂s = ∂f ∂f ∂y ∂x · + · ∂x ∂y ∂s ∂s ...(A) 50 A TEXTBOOK OF ENGINEERING MATHEMATICS—I In a similar way, we get ∂z ∂t = ∂f ∂f ∂y ∂x · + · ∂x ∂y ∂t ∂t ...(B) The equations (A) and (B) are known as chain rule for partial differentiation. Example 1. Find the total differential coefficient of x2 y w.r.t. x when x, y are connected by x2 + xy + y2 = 1. Sol. Let z = x2 y ...(i) Then the total differential coefficient of z dz = dx ∂z dy ∂z dx ∂z dy ∂z · + · = + ∂y · ∂x dx ∂x dx ∂x dx ∂z ∂z = 2xy, ∂y = x2 and we have ∂x x2 + xy + y2 = 1 Differentiating w.r.t. x, we get ...(ii) From (i) 2x + dy dy x + y + 2y dx dx = 0 dy dx = 0 dy dx = – (2x + y) + (x + 2y) ⇒ ...(iii) 2x + y x + 2y Putting these values in equation (ii), we get F GH 2x + y dz = 2xy + x2 − x + 2y dx I = 2xy – x b2x + yg . JK bx + 2 y g 2 Example 2. If f (x, y) = 0, φ (y, z) = 0, show that ∂f ∂φ dz ∂y ∂z dx = Sol. We have From (i) From (ii) ∂f ∂φ . · ∂x ∂y f (x, y) = 0 φ (y, z) = 0 FG ∂f IJ H ∂x K ∂f ∂f dy dy = + =0⇒ = − ⋅ F ∂f I ∂x ∂y dx dx GH ∂y JK F ∂φ I GH ∂y JK ∂φ ∂φ dz dφ dz ⋅ = + =0⇒ = − dy dy ∂y ∂z dy FG ∂φ IJ H ∂z K df dx ...(i) ...(ii) 51 DIFFERENTIAL CALCULUS-I Multiplying these two results, we get FG ∂f IJ FG ∂φ IJ H ∂x K × H ∂y K dz = dx F ∂f I FG ∂φ IJ GH ∂y JK H ∂z K ∂f ∂φ dz ∂y ∂z dx or Example 3. If u = u x2 ∂f ∂φ · ⋅ ∂x ∂y Hence proved. F y − x , z − x I , show that GH xy xz JK ∂u ∂u ∂u + y2 + z2 = 0. y ∂ ∂x ∂z Sol. Let s = ∂s ∂x So (U.P.T.U., 2005) y−x 1 1 1 1 z− x = – and t = = – xy x y x z zx 1 = – , x2 ∂s ∂t 1 ∂t 1 ∂t 1 = 2 , = − 2, = 2, =0 ∂y ∂x y y ∂ ∂ z x z ∂s = 0 ∂z u = u(s, t) Since ∴ ∂u ∂x = ∂u ∂s ∂u ∂t ⋅ ⋅ + ∂s ∂x ∂t ∂x ⇒ ∂u ∂x = ∂u ∂s x2 or Next, ∂u ∂x ∂u ∂y ∂u ∂y or and = FG – 1 IJ + ∂u FG − 1 IJ H x K ∂t H x K 2 2 ∂u ∂u – ∂s ∂t ∂u ∂s ∂u ∂t + ∂s ∂y ∂t ∂y = – = ∂u ∂u ∂u 2 + 0 ⇒ y y ∂y = ∂s ∂s 1 ∂u ∂u ∂s ∂u ∂t + =0+ 2 z ∂s ∂z ∂t ∂z ∂t ∂u ∂t 1 = 2 ∂u = ∂z ∂u ⇒ z2 = ∂z Adding (i), (ii) and (iii), we get ∂u ∂u ∂u ∂u ∂u ∂u ∂u x2 + y2 ∂y + z2 = – – + + = 0. Hence proved. ∂x ∂z ∂s ∂t ∂s ∂t Example 4. Prove that ...(i) ∂ 2u ∂ 2u ∂ 2u ∂ 2u + = + , where ∂ξ 2 ∂η2 ∂y2 ∂x2 x = ξ cos α – η sin α, y = ξ sin α + η cos α. ...(ii) ...(iii) 52 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Sol. We have ∴ Now, Let ⇒ x = ξ cos α – η sin α, y = ξ sin α + η cos α. ∂y ∂y ∂x ∂x = cos α, = – sin α and = sin α, = cos α ∂ξ ∂η ∂ξ ∂η u = u (x, y) ∂u ∂u ∂x ∂u ∂y ∂u ∂u = · + · = cos α + sin α ∂ξ ∂x ∂ξ ∂y ∂ξ ∂x ∂y ∂u ∂η = ∂u ∂x ∂u ∂u ∂y ∂u · + · = – sin α + cos α ∂η ∂x ∂x ∂y ∂η ∂y Again ∂ 2u ∂ξ 2 = ∂ ∂ξ ⇒ ∂ 2u ∂ξ 2 = cos2α ∂ 2u ∂η2 = ∂ ∂η and and = ...(ii) FG ∂uIJ = F cos α ∂ + sin α ∂ I F cos α ∂u + sin α ∂u I H ∂ξ K GH ∂x ∂y JK GH ∂x ∂y JK ∂ 2u ∂ 2u ∂ 2u 2 + 2 sin α cos α + sin α ...(iii) ∂y2 ∂x∂y ∂x 2 ∂u ∂ ∂ ∂u ∂u + cos α – sin α + cos α = – sin α ∂η ∂x ∂y ∂x ∂y FG IJ F H K GH IF JK GH I JK ∂ 2u ∂ 2u ∂ 2u ∂ 2u 2α = sin2α – 2 sin α cos α + cos 2 ∂x∂y ∂y2 ∂η ∂x2 Adding (iii) and (iv), we get ∂ 2u ∂ 2u + ∂ξ 2 ∂η2 ...(i) ...(iv) ∂ 2u ∂ 2u 2 α + sin2α) + (cos2α + sin2α) · (cos ∂y2 ∂x2 ∂ 2u ∂ 2u . Hence proved. 2 + ∂x ∂y2 Example 5. If u = f (y – z, z – x, x – y), show that ∂u ∂u ∂u + + = 0. ∂ ∂z y ∂x Sol. Let r = y – z, s = z – x, t = x – y Then u = f (r, s, t) = (U.P.T.U., 2003) ∴ ∂u ∂x = ∂f ∂t ∂f ∂r ∂f ∂s · + · + · ∂t ∂x ∂r ∂x ∂s ∂x ⇒ ∂u ∂x = ∂r ∂s ∂t ∂f ∂f ∂f = 0, = −1, =1 ×0+ (–1) + (1) As ∂x ∂x ∂x ∂r ∂s ∂t ⇒ ⇒ ∂u ∂f ∂f = – + ∂s ∂t ∂x ∂u ∂f ∂r ∂f ∂s ∂f ∂t ∂y = ∂r · ∂y + ∂s · ∂y + ∂t · ∂y ∂u ∂y = ∂f ∂f ∂f ∂r ∂s ∂t = 1, = 0, = −1 (1) + (0) + (–1) As ∂y ∂y ∂y ∂r ∂s ∂t = ∂f ∂f – ∂r ∂t ...(i) ...(ii) 53 DIFFERENTIAL CALCULUS-I and ∂u ∂z = ∂f ∂r ∂f ∂s ∂f ∂t · + · + · ∂r ∂z ∂s ∂z ∂t ∂z ∂r ∂s ∂t ∂f ∂f ∂f = −1, = 1, = 0 (–1) + (1) + (0) As ∂z ∂z ∂z ∂r ∂s ∂t ∂u ∂f ∂f ⇒ = – + ...(iii) z ∂ ∂s ∂r Adding equations (i), (ii) and (iii), we get ∂u ∂u ∂u + ∂y + = 0. Hence proved. ∂x ∂z = Example 6. If x = r cos θ, y = r sin θ, show that ∂r = ∂x ∂x ∂x ∂θ ∂ 2θ ∂ 2θ ; =r and find the value of + · r∂θ ∂r ∂x ∂y2 ∂x2 Sol. We have x = r cos θ, y = r sin θ y ⇒ r2 = x2 + y2 and θ = tan–1 x ∂r ∂r x = r cos θ = cos θ ∴ 2r · = 2x ⇒ = ...(i) ∂ x r ∂x r ∂x and we have x = r cos θ ⇒ = cos θ ...(ii) ∂r From equations (i) and (ii), we get ∂r ∂x = . Hence proved. ∂x ∂r 1 ∂x ∂x Now, = – r sin θ ⇒ = – sin θ ...(iii) r ∂θ ∂θ y ∂θ 1 y r sin θ − 2 =– and = =– 2 2 2 x ∂x y r2 x +y 1+ 2 x 1 ∂θ = – sin θ ⇒ r = – sin θ ...(iv) ∂x r From equations (iii) and (iv), we obtain 1 ∂x ∂θ = r . Hence proved. ∂x r ∂θ 2 xy ∂θ ∂ 2θ ∂ 2θ y × 2x y Since = – 2 ⇒ = ⇒ 2 ...(v) 2 2 = 2 2 2 ∂x ∂x ∂x x +y x + y2 x2 + y2 FG IJ H K FG H IJ K e and ∂θ ∂y 1 = y2 x2 Adding equations (v) and (vi), we get ∂ 2θ ∂ 2θ + ∂y2 ∂x2 1+ = 0. e j FG 1IJ = x ∴ ∂ θ = – 2xy H xK ex + y j ∂y ex + y j j 2 2 2 2 2 2 2 ...(vi) 54 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 7. If φ (x, y, z) = 0, show that FG ∂y IJ FG ∂z IJ FG ∂x IJ = – 1. H ∂z K H ∂xK y H ∂y K x (U.P.T.U., 2004) z Sol. We have φ (x, y, z) = 0 Keeping y as constant, differentiate partially w.r.t. x, we get FG ∂φ IJ + FG ∂φ IJ · FG ∂z IJ H ∂xK H ∂z K H ∂xK FG ∂z IJ H ∂xK ⇒ = 0 y FG ∂φ IJ H ∂x K = – FG ∂φ IJ y H ∂z K ...(i) Next, keeping z as constant, differentiate partially w.r.t. y, we obtain ∂φ ∂x ⇒ F ∂x I + ∂φ = 0 GH ∂y JK ∂y z F ∂φ I F ∂x I = − GH ∂y JK GH ∂y JK FG ∂φ IJ H ∂x K FG ∂φ IJ ∂ y FG IJ = – H ∂z K H ∂z K x F ∂φ I GH ∂y JK ...(ii) z Similarly, ...(iii) Multiplying equations (i), (ii) and (iii), we get FG ∂y IJ FG ∂z IJ FG ∂x IJ H ∂z K H ∂xK y H ∂y K x = –1. Hence proved. z Example 8. If x + y = 2eθ cos φ and x – y = 2i eθ sin φ, show that ∂ 2V ∂ 2V ∂ 2V · 2 = 4xy 2 + ∂x ∂y ∂φ ∂θ Sol. We have x + y = 2eθ cos φ x – y = 2ieθ sin φ Adding 2x = 2 (eθ) (cos φ + i sin φ) x = eθ + i φ and subtracting, we get y = eθ – i φ Let ∴ (U.P.T.U., 2001) ...(i) ...(ii) V = V (x, y) ∂V ∂V ∂x ∂V ∂y = · + · ∂θ ∂x ∂θ ∂y ∂θ ∂V ∂V ∂V = x +y ∂θ ∂y ∂x As ∂y ∂x = e θ+iφ = x, = e θ − iφ = y ∂θ ∂θ 55 DIFFERENTIAL CALCULUS-I FG ∂VIJ = FG x ∂ + y ∂ IJ FG x ∂V + y ∂V IJ H ∂θ K H ∂x ∂y K H ∂x ∂y K F ∂V ∂V I ∂ V ∂ V ∂ V ∂ V = x +y + 2xy + G x ∂x + y ∂y J H K ∂θ ∂y ∂x ∂y ∂x ∂ ∂ 2V 2 = ∂θ ∂θ ∴ 2 ⇒ 2 2 2 2 2 2 2 ...(iii) 2 ∂V ∂y ∂V ∂x ∂V ∂y ∂V ∂V ∂x = ix = = –iy = · + = (ix) + (–iy) As ∂φ ∂φ ∂x ∂y ∂φ ∂x ∂y ∂φ ∂φ Now F x ∂V − y ∂V I GH ∂x ∂y JK F ∂ ∂ I F ∂V ∂V I ∂ V ∂ F ∂V I G J = = i G x ∂x − y ∂y J i G x ∂x − y ∂y J ∂φ H ∂φ K H K H K ∂φ F ∂ 2V ∂ 2V ∂ 2V + x ∂V + y ∂VI = − G x2 2 + y2 2 − 2xy ∂x∂y ∂x ∂y JK H ∂x ∂y ∂V = i ∂φ ⇒ 2 Next 2 2 ∂ 2v ∂ 2v ∂v ∂v ∂ 2V 2 ∂ v 2 y 2 xy −x −y − + = –x 2 2 2 ∂x ∂y ∂x ∂y ∂x ∂y ∂φ Adding equations (iii) and (iv), we get ∂ 2V ∂ 2V + ∂φ 2 ∂θ 2 = 4xy ...(iv) ∂ 2V . Hence proved. ∂x ∂y Example 9. If x = r cos θ, y = r sin θ, z = f (x, y), prove that 1 ∂z 1 ∂z ∂z ∂z ∂z ∂z = cos θ – sin θ ; = sin θ + cos θ. r ∂θ r ∂θ ∂x ∂r ∂y ∂r e ∂ 2 r n .cos nθ j = – n (n – 1) r n – 2 · sin (n – 2) θ. ∂x ∂y Sol. Here z is a function of x and y where x and y are functions of r and θ. Prove also that ∴ We have and Now, ∂z ∂z ∂r ∂z ∂θ = · + · ∂x ∂r ∂x ∂θ ∂x ...(i) ∂z ∂z ∂r ∂z ∂θ = · + · ∂y ∂r ∂y ∂θ ∂y ...(ii) x = r cos θ, y = r sin θ, so r2 = x2 + y2 and θ = tan–1 FG y IJ . H xK ∂r ∂θ sin θ cos θ ∂r ∂θ = cos θ, ∂y = sin θ, =– and ∂y = r r ∂x ∂x Substituting these values in equations, (i) and (ii), we have Then, and 1 ∂z ∂z ∂z . Hence proved. = cos θ – sin θ r ∂x ∂r ∂θ ...(iii) ∂z ∂y ...(iv) = sin θ 1 ∂z ∂z . Hence proved. + cos θ r ∂r ∂θ 56 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Again substituting rn cos nθ for z in (iv), we get ∂ cos θ ∂ n ∂ n n (r cos nθ) ∂y (r cos nθ) = sin θ · ∂r (r cos nθ) + r ∂θ cos θ = sin θ (nrn – 1 cos nθ) + (– rnn sin nθ) r = nrn−1(cos nθ sin θ − cos θ sin nθ) e ∂ r n cos nθ or ∂y Now, e j = nr ∂ 2 r n cos nθ ∂x∂y n–1 sin (θ – nθ) ...(v) j = ∂ LM ∂ er n cos nθ j OP ∂x M N ∂y PQ ∂ ∂ nr sin a1 − nfθ , from (v) = n r sin a1 − nfθ = ∂x ∂x sin θ ∂ n−1 L ∂ O r sina1 − nfθ}P = n Mcos θ {rn−1 sina1 − nfθ} − r ∂θ { N ∂r Q sin θ n−1 L r a1 − nf cos a1 − nf θOP = n Mcos θ an − 1f rn − 2 sin a1 − nf θ – r N Q n– 1 n– 1 = – n (n –1) rn–2 [sin (n – 1)θ cos θ – cos (n – 1) θ sin θ] = – n (n – 1) rn-2 sin (n – 2) θ. Hence proved. Example 10. If u = f (x, y) and x = r cos θ, y = r sin θ, prove that ∂ 2u ∂ 2u ∂ 2u 1 2 = 2 + 2 + 2 ∂y ∂x ∂r r 1 ∂u ∂ 2u + · 2 r ∂r ∂θ Or F ∂ u I F ∂ 2uI Transform G H ∂x JK + GH ∂y2 JK = 0 into polars and show that u = (Ar + Br ) sin nθ satisfies 2 n 2 the above equation. Sol. We know x = r cos θ, y = r sin θ ∴ r2 = From (ii), we get 2r or Similarly, ...(ii) FG y IJ H xK ...(iii) ∂r ∂r x r cos θ = 2x or = = , from (i) r r ∂x ∂x ∂r = cos θ ∂x ∂r r sin θ y = = = sin θ r ∂y r Also from (iii), ...(i) x2 + y2 θ = tan–1 and –n ∂θ = ∂x FG FG IJ H H K y 1 − 2 2 · x y 1+ x ...(iv) ...(v) IJ = – y K x +y 2 2 57 DIFFERENTIAL CALCULUS-I r sin θ sin θ ∂θ = – =– 2 r r ∂x ∂θ 1 x r cosθ 1 cos θ = 2 2 · x = x2 + y2 = ∂y = r r y 1+ or FG IJ FG IJ H K H xK and Now, we know ∂u ∂u ∂r ∂u ∂θ = · + · ∂x ∂r ∂x ∂θ ∂x ...(vii) IJ K FG H sin θ ∂u ∂u − (cos θ) + , from (iv), (vi) r ∂r ∂θ sin θ ∂ ∂ ∂ (u) = cos θ (u) – (u) r ∂θ ∂x ∂r ∂u Replacing u by ∂x 2 ∂ u sin θ ∂ ∂u ∂ u ∂ ∂ ∂u = cos θ. – , 2 = ∂x ∂x r ∂r ∂x ∂θ ∂x ∂x = or ...(vi) ...(viii) FG IJ H K FG IJ FG IJ H K H K ∂ L ∂u sin θ ∂u O sin θ ∂ L ∂u sin θ ∂u O cos θ − cos θ − − = cos θ ⋅ M P M r ∂θ Q r ∂θ N r ∂θ PQ ∂r N ∂r ∂r 2 sin θ LMF – sin θ ∂u + cos θ ⋅ ∂ u I − 1 ∂ FG sin θ ⋅ ∂u IJ OP L ∂ 2u ∂ F 1 ∂u I O G = cos θ Mcos θ 2 − sin θ ⋅ G ⋅ J P – ∂θ ∂r JK r ∂θ H ∂θ K PQ ∂r r MNH ∂r H r ∂θ K PQ MN ∂r or LM MN RS T ∂ 2u 1 ∂ 2u 1 ∂u ∂ 2u cos sin θ ⋅ − θ ⋅ − = cos θ r ∂r ∂θ r 2 ∂θ ∂r 2 ∂x2 – or LM MN UVOP WPQ F GH 2 2 sin θ – sin θ ∂u + cos θ ⋅ ∂ u − 1 sin θ ∂ u + cos θ ⋅ ∂u ∂θ ∂r ∂θ r ∂r ∂θ 2 r I OP JK PQ 2sin θ cos θ ∂ 2u sin 2 θ ∂ 2u ∂ 2u ∂ 2u sin 2 θ ∂u 2 θ = cos – + + 2 2 2 2 ∂r ∂θ r ∂r ∂x ∂r ∂θ r r + Similarly, 2 cos θ sin θ ∂u ...(ix) r2 ∂θ ∂ 2u 2sin θ cos θ ∂ 2u cos2 θ ∂ 2u ∂ 2u + + 2 = sin2 θ ∂y r r2 ∂r 2 ∂r ∂θ ∂θ 2 2 cos θ sin θ cos 2 θ ∂u – r2 ∂r r Adding equations (ix) and (x), we get + ∂u ∂θ ...(x) ∂ 2u 1 ∂ 2u ∂ 2u ∂ 2u 2 θ + sin2 θ) 2 θ + cos2 θ) + = (cos + (sin 2 ∂y r2 ∂x2 ∂r 2 ∂θ 2 + 1 ∂ 2u 1 1 ∂u ∂u ∂ 2u (sin2 θ + cos2 θ) = + 2 2 2 + r ∂r . r r ∂r ∂r ∂θ Hence proved. 58 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Or If u = and ∴ 1 ∂ 2u 2 + r2 ∂r (Arn + Br–n) sin nθ, then ∂u ∂r = n (Arn–1 – Br–n – 1) sin nθ; ∂ 2u ∂r 2 = n [A(n – 1)rn – 2 + B (n + 1) r–n –2] sin nθ; ∂ 2u ∂θ 2 = ∂ ∂θ ∂u = (Arn + Br–n) n cos nθ ∂θ FG ∂uIJ = – (Ar + Br ) n sin nθ H ∂θ K n –n 2 1 ∂u ∂ 2u n–2 + B (n + 1) r–n–2] sin nθ 2 + r ∂r = n[A (n – 1) r ∂θ 1 1 (Arn + Br–n)n2 sin n θ + n(Arn–1 – Br–n–1) sin nθ r r2 = [A{n2–n–n2 + n}rn–2 + B(n2 + n – n2 –n)r–n – 2] sin nθ – = 0. Hence proved. Example 11. If x = r cos θ, y = r sin θ or r2 = x2 + y2, prove that ∂ 2r 1 ∂ 2r + 2 = 2 ∂y r ∂x Sol. Since x = r cos θ and y = r sin θ ∂r ∂x = ∂ 2r ∂ 2r + ∂y2 ∂x2 = ∴ Adding ∂ 2r ∂ 2r + ∂y2 ∂x2 or Also, 1 r R|F ∂r I F ∂r I U| S|GH ∂x JK + GH ∂y JK V| · T W 2 2 2 y ∂ 2r x ∂r r 2 − x2 ∂ r r2 − y2 ; = ; = ; = , ∂y2 r ∂y r ∂x2 r3 r3 er − x j + er − y j = 2r − ex + y j 2 2 2 r3 2 2 r3 2 2 r3 = 2r 2 − r 2 , ä x2 + y2 = r2 r3 = 1 r ...(i) R|F ∂r I F ∂r I U| 1 R|F x I F y I U| 1 F x2 + y2 I S|GH ∂x JK + GH ∂y JK V| = r S|GH r JK + GH r JK V| = r G r2 J K T W H T W 1 Fr I = G J,äx +y =r r Hr K 2 2 2 2 2 2 2 2 2 2 = ∂ r 1 ∂ 2r = + , from (i) Hence proved. 2 ∂y2 r ∂x Example 12. If V = f (2x – 3y, 3y – 4z, 4z – 2x), compute the value of 6Vx + 4Vy + 3Vz. (U.P.T.U., 2008) Sol. Let r = 2x – 3y, s = 3y – 4z, t = 4z – 2x ∴ V = f(r, s, t) 59 DIFFERENTIAL CALCULUS-I Vx = = ⇒ ∂f ∂r ∂f ∂s ∂f ∂t ∂V = ⋅ + ⋅ + ⋅ ∂x ∂r ∂x ∂s ∂x ∂t ∂x a f ∂f ∂f ∂f ⋅2 + ⋅0 + −2 ∂r ∂s ∂t As = ∂r ∂s ∂t = 2, = 0, = −2 ∂x ∂x ∂x Vx = 2fr – 2ft Vy = = ⇒ ...(i) ∂f ∂r ∂f ∂s ∂f ∂t ∂V = ⋅ + ⋅ + ⋅ ∂y ∂r ∂y ∂s ∂y ∂t ∂y a f af ∂f ∂f ∂f −3 + 3 + ⋅0 ∂r ∂s ∂t As ∂r ∂s ∂t = − 3, = 3, = 0 ∂y ∂y ∂y Vy = – 3fr + 3fs ...(ii) Vz = 4fr – 4fs ...(iii) Similarly Multiplying (i), (ii) and (iii) by 6, 4, 3 respectively and adding, we get 6Vx + 4Vy + 3Vz = 12fr – 12ft – 12fr + 12fs + 12fr – 12fs ⇒ 6Vx + 4Vy + 3Vz = 0. Hence Proved. Example 13. If u = x log xy, where x3 + y3 + 3xy = 1, find du . dx [U.P.T.U. (C.O.), 2005] Sol. By total differentiation, we know that ∂u dx ∂u dy ∂u ∂u dy du + ⋅ ⋅ + ⋅ = = ∂x dx ∂y dx ∂x ∂y dx dx ...(i) 1 ∂u = log xy + ⋅ y = log xy + 1 y ∂x ...(ii) we have u = x log xy af ∂u x x x = = xy y ∂y ...(iii) Also, given that x3 + y3 + 3xy = 1 Differentiating w.r.t. ‘x’, we get 3x2 + 3y2 ⇒ or dy dy + 3y + 3x = 0 dx dx (x2 + y) + (x + y2) dy = 0 dx e e x2 + y dy = − dx x + y2 Using (ii), (iii) and (iv) in (i), we get j j ...(iv) F GH x x2 + y du − = (1 + log xy) y x + y2 dx I JK 60 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 14. If x = u + v + w, y = vw + wu + uv, z = uvw and f is a function of x, y, z, show ∂f ∂f ∂f ∂f ∂f ∂f that u . +v +w = x + 2y + 3z ∂u ∂v ∂w ∂x ∂y ∂z Sol. Let f = f(x, y, z) ∂f ∂f ∂x ∂f ∂y ∂f ∂z ⋅ + ⋅ + = ∂u ∂x ∂u ∂y ∂u ∂z ∂u = or a f ∂f ∂f ∂f + w+v + vw ∂x ∂y ∂z a As a f ∂y ∂x ∂z = 1, = w+v , = vw ∂u ∂u ∂u f ∂f ∂f ∂f ∂f + u w+v + uvw = u ∂x ∂y ∂z ∂u u ...(i) ∂f ∂f ∂x ∂f ∂y ∂f ∂z ⋅ + ⋅ + ⋅ = ∂v ∂x ∂v ∂y ∂v ∂z ∂v = or v a a f ∂y ∂f ∂f ∂f ∂x ∂z As = 1, = u+w , = uw + u+w + uw ∂v ∂v ∂v ∂z ∂x ∂y f a ∂f ∂f ∂f ∂f = v + v u+w + uvw ∂ ∂ ∂z x y ∂v f ...(ii) a f ...(iii) Similarly w ∂f ∂f ∂f ∂f + w u+v + uvw = w ∂x ∂y ∂z ∂w Adding (i), (ii) and (iii), we get or au + v + wf ∂∂xf + 2avw + wu + uvf ∂∂yf + 3uvw ∂∂fz u ∂f ∂f ∂f +v +w = ∂u ∂v ∂w u ∂f ∂f ∂f ∂f ∂f ∂f +v +w = x + 2y + 3z . Proved. ∂u ∂v ∂w ∂x ∂y ∂z Example 15. If by the substitution u = x2 – y2, v = 2xy, f(x, y) = φ(u, v) show that Sol. We have ∂2 f ∂2 f ∂x ∂y 2 + 2 F ∂ φ + ∂ φI . GH ∂u ∂v JK 2 = 4(x2 + y2) 2 2 2 f(x, y) = φ(u, v) Differentiating partialy w.r.t. ‘x’. ∂φ ∂φ ∂f ∂φ ∂u ∂φ ∂v = ⋅ + ⋅ = 2x + 2y ∂u ∂x ∂v ∂x ∂u ∂v ∂x ∂2 f ∂x 2 As ∂u ∂v = 2x, = 2y ∂x ∂x FG IJ = ∂ FG 2x ∂φ + 2y ∂φ IJ H K ∂x H ∂u ∂v K ∂ F ∂φ ∂φ I ∂u ∂ F ∂φ ∂φ I ∂v 2x + 2y J 2x + 2y J ⋅ + = G G H K H ∂u ∂u ∂v ∂x ∂v ∂u ∂v K ∂x = ∂ ∂f ∂x ∂x 61 DIFFERENTIAL CALCULUS-I F 2x ∂ φ + 2y ∂ φ I .2x + F 2x ∂ φ + 2y ∂ φ I .2y GH ∂u ∂u∂v JK GH ∂v∂u ∂v JK 2 = or 2 2 2 2 2 2 ∂ 2φ ∂ 2φ 2∂ φ 8 4 xy y + + ∂u∂v ∂x 2 ∂u2 ∂v2 Again differentiating f(x, y) partially w.r. to y ∂2 f = 4x2 ...(i) ∂f ∂u ∂v ∂φ ∂φ ∂φ ∂u ∂φ ∂v As = − 2 y, = 2x = ⋅ + = − 2y + 2x ∂y ∂y ∂y ∂u ∂y ∂v ∂y ∂u ∂v ∂2 f ∂y 2 = FG H ∂ ∂φ ∂φ −2 y + 2x ∂y ∂u ∂v IJ K FG IJ IJ H K K F ∂ φ + 2x ∂ φ I b−2yg + F −2y ∂ φ + 2x ∂ φ I a2xf = G −2 y GH ∂u∂v ∂v JK H ∂u ∂u∂v JK = FG H ∂ ∂φ ∂φ ∂u ∂ ∂φ ∂φ ∂v + 2x −2 y + 2x + −2 y ∂u ∂u ∂v ∂y ∂v ∂u ∂ v ∂y 2 2 2 2 2 ∂2 f or ∂y 2 = 4y2 2 2 ∂ 2φ ∂ 2φ 2∂ φ 8 4 xy x − + ∂u∂v ∂u2 ∂v2 ...(ii) Adding (i) and (ii), we get ∂2 f ∂x 2 + ∂2 f ∂y 2 2 2 e j ∂∂uφ + 4 ex + y j ∂∂vφ e j FGH ∂∂uφ + ∂∂vφ IJK . Hence Proved. = 4 x2 + y 2 = 4 x2 + y2 2 2 2 2 2 2 2 2 Example 16. If x2 + y2 + z2 – 2xyz = 1, show that dx 1−x 2 + dy 1− y 2 + dz 1 − z2 = 0. Sol. We have x2 + y2 + z2 – 2xyz = 1 x2 – 2xyz + y2 z2 = 1 – y2 – z2 + y2 z2 or (x – yz)2 = (1 – y2) (1 – z2) or (x – yz) = Again e1 − y j e1 − z j 2 (y – zx)2 = (1 – x2) (1 – z2) or (y – zx) = Let ...(i) y2 – 2xyz + z2 x2 = 1 – x2 – z2 + z2 x2 or Similarly 2 (z – xy) = e1 − x j e1 − z j e1 − x j ⋅ e1 − y j 2 2 ...(ii) 2 2 ...(iii) u h x2 + y2 + z2 – 2xyz – 1 = 0 62 A TEXTBOOK OF ENGINEERING MATHEMATICS—I By total differentiation, we get ∂u ∂u ∂u dx + dy + dz = 0 ∂x ∂y ∂z du = As or (2x – 2yz)dx + (2y – 2zx)dy + (2z – 2xy)dz = 0 ∂u = 2x − 2 yz ∂x ∂u = 2 y − 2zx ∂y ∂u = 2z − 2xy ∂z or (x – yz)dx + (y – zx)dy + (z – xy)dz = 0 ...(iv) Putting (i), (ii) and (iii) in equation (iv), we get e1 − y j e1 − z j . dx + e1 − x j e1 − z j dy + e1 − x j e1 − y j dz = 0 2 2 Dividing by dx 1−x 2 + 2 2 2 2 e1 − x j e1 − y j e1 − z j , we get 2 dy 1− y 2 + 2 dz 1 − z2 2 = 0. Hence proved. EXERCISE 1.6 LMA ns. dy = − L yx + y log y OOP MN dx MMN x log x + xy PPQPQ du LA ns. du = 1 + log xy + x L− (x + y) O O b g y MM ( y + x) PPPP 2 . If u = x log xy, where x + y + 3xy = 1, find .M dx dx MN N QQ LMA ns. du = 2xy − x L b2x + yg OOP du MM bx + 2yg PPP 3 . If u = x y, where x + xy + y = 1, find . MN dx N QQ dx 4 . If V is a function of u, v where u = x – y and v = x – y, prove that F ∂ V + xy ∂ V I . ∂ 2V ∂ 2V x + y J 2 2 = (x + y) G ∂x ∂y ∂v K H ∂u y −1 dy 1 . Find if xy + yx = c. dx x x -1 y 2 3 2 2 3 2 2 2 2 2 2 5 . Transform the Laplacian equation 2 ∂ 2u ∂ 2u + ∂y2 = 0 by change of variables from x, y to r, θ ∂x2 when x = er cos θ, y = er sin θ. 6 . Find du , if u = x log xy where x3 + y3 + 3xy = 1. dx LMA ns. e FG ∂ u + ∂ uIJ = 0OP H ∂r ∂θ K PQ MN LMA ns. du = 1 + log xy − x . x + y OP y x + y QP NM dx −2r 2 2 2 2 2 2 63 DIFFERENTIAL CALCULUS-I 7 . If the curves f (x, y) = 0 and φ (x, y) = 0 touch, show that at the point of contact ∂f ∂φ ∂f ∂φ · = · . ∂ ∂x y ∂y ∂x dy , when (cos x)y = (sin y)x. dx 9 . If x = r cos θ, y = r sin θ, prove that 8 . Find F GH I JK LMA ns. dy = y tan x + log sin y OP N dx log cos x − x cot y Q 2 2 ∂ 2r ∂ 2r ∂ r · . 2 = ∂y∂x ∂x2 ∂y 1 0 . If z is a function of x and y and x = eu + e–v, y= e–u –ev prove that ∂z ∂z ∂z ∂z . – =x – y ∂v ∂y ∂u ∂x 1 1 . If u = log (tan x + tan y + tan z) prove that ∂u ∂u ∂u + (sin 2y) ∂y + (sin 2z) = 2. ∂x ∂z 1 2 . If u = 3 (lx + my + nz)2 – (x2 + y2 + z2) and l2 + m2 + n2 = 1, show that (sin 2x) ∂ 2u ∂ 2u ∂ 2u + + = 0. ∂x2 ∂y2 ∂z2 1 3 . If u = x2 + 2xy – y log z, where x = s + t2, y = s – t2, z = 2t, find LMA ns. ∂u = 8, ∂u = 8t – 4OP N ∂s ∂t Q ∂u ∂u , at (1, 2, 1). ∂s ∂t 1 4 . If u = x2 – y2 + sin yz, where y = ex and z = log x, find du . dx LMA ns. 2ex – e j + e cosee log xjFG log x + 1 IJ OP H xK Q N ∂z ∂z 1 5 . If z = z(u, v), u = x – 2xy – y and v = y, show that (x + y) + bx − yg = 0 is equivalent ∂x ∂y x 2 to x x 2 ∂z = 0. ∂v CURVE TRACING Introduction It is analytical method in which we draw approximate shape of any curve with the help of symmetry, intercepts, asymptotes, tangents, multiple points, region of existence, sign of the first and second derivatives. In this section, we study tracing of standard and other curves in the cartesian, polar and parametric form. 64 1.8 A TEXTBOOK OF ENGINEERING MATHEMATICS—I PROCEDURE FOR TRACING CURVES IN CARTESIAN FORM The following points should be remembered for tracing of cartesian curves: 1.8.1 Symmetry (a) Symmetric about x-axis: If all the powers of y occurring in the equation are even then the curve is symmetrical about x-axis. Example. y2 x2 + = 1, y2 = 4ax. a2 b2 (U.P.T.U., 2008) (b) Symmetric about y-axis: If all the powers of x occurring in the equation are even then the curve is symmetrical about y-axis. Example. x2 = 4ay, x4 + y4 = 4x2y2. (c) Symmetric about both x- and y-axis: If only even powers of x and y appear in equation then the curve is symmetrical about both axis. Example. x2 + y2 = a2. (U.P.T.U., 2008) (d) Symmetric about origin: If equation remains unchanged when x and y are replaced by – x and – y. Example. x5 + y5 = 5a2x2y. Remark: Symmetry about both axis is also symmetry about origin but not the converse (due to odd powers). (e) Symmetric about the line y = x : A curve is symmetrical about the line y = x, if on interchanging x and y its equation does not change. Example. x3 + y3 = 3axy. (U.P.T.U., 2008) (f) Symmetric about y = – x : A curve is symmetrical about the line y = – x, if the equation of curve remains unchanged by putting x = – y and y = – x in equation. Example. x3 – y3 = 3axy. (U.P.T.U., 2008) 1.8.2 Regions (a) Region where the curve exists: It is obtained by solving y in terms of x or vice versa. Real horizontal region is defined by values of x for which y is defined. Real vertical region is defined by values of y for which x is defined. (b) Region where the curve does not exist: This region is also called imaginary region, in this region y becomes imaginary for values of x or vice versa. 1.8.3 Origin and Tangents at the Origin If there is no constant term in the equation then the curve passes through the origin otherwise not. If the curve passes through the origin, then the tangents to the curve at the origin are obtained by equating to zero the lowest degree terms. Example. The curve a2y2 = a2x2 – x4, lowest degree term (y2 – x2) equating to zero gives y = ± x as the two tangents at the origin. 1.8.4 Intercepts (a) Intersection point with x- and y-axis: Putting y = 0 in the equation we can find points where the curve meets the x–axis. Similarly, putting x = 0 in the equation we can find the points where the curve meets y-axis. 65 DIFFERENTIAL CALCULUS-I (b) Points of intersection: When curve is symmetric about the line y = ± x, the points of intersection are obtained by putting y = ± x in given equation of curve. (c) Tangents at other points say (h, k) can be obtained by shifting the origin to these points (h, k) by the substitution x = x + h, y = y + k and calculating the tangents at origin in the new xy plane. Remark. The point where dy/dx = 0, the tangent is parallel to x-axis. And the point where dy/dx = ∞, the tangent is vertical i.e., parallel to y-axis. 1.8.5 Asymptotes If there is any asymptotes then find it. (a) Parallel to x-axis: Equate the coefficient of the highest degree term of x to zero, if it is not constant. (b) Parallel to y-axis: Equate the coefficient of the highest degree term of y to zero, if it is not constant. Example. x2y – y – x = 0 highest power coefficient of x i.e., x2 = y Thus asymptote parallel to x-axis is y = 0 Similarly asymptote parallel to y-axis are x2 – 1 = 0 ⇒ x = ± 1. (c) Oblique asymptotes (not parallel to x-axis and y-axis): The asymptotes are given by y = mx + c, where m = lim x→∞ FG y IJ and c = lim (y – mx). H xK x→∞ (d) Oblique asymptotes (when curve is represented by implicit equation f (x, y) = 0): The asymptotes are given by y = mx + c where m is solution of φn(m) = 0 and c is the solution of cφ′n(m) + φn–1(m) = 0 or c = af af − φ n−1 m . Here φn(m) and φn–1(m) are obtained by putting x = 1 and φ ′n m y = m in the collection of highest degree terms of degree n and in the collection of the next highest degree terms of degree (n – 1). 1.8.6 Sign of First Derivatives dy/dx (a ≤ x ≤ b) dy > 0 , then curve is increasing in [a, b]. dx dy < 0 , then curve is decreasing in [a, b]. (b) dx dy = 0 , then the point is stationary point where maxima and minima can occur. (c) If dx (a) 1.8.7 Sign of Second Derivative (a) (b) d2 y dx 2 d2 y dx 2 £ P´ (a ≤ x ≤ b) £³P > 0 , then curve is convex or concave upward (holds water). < 0 , then the curve is concave or concave downward (spills water). 66 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 1.8.8 Point of Inflexion A point where d2y/dx2 = 0 is called an inflexion point where the curve changes the direction of concavity from downward to upward or vice versa. w Example 1. Trace the curve y = x3 – 3ax2. Sol. 1. Symmetry: Here the equation of curve do not hold any condition of symmetry. So there is no symmetry. 2. Origin: Since there is no constant add in equation so the curve passes through the origin. The equation of tangent at origin is y = 0 i.e., x-axis (lowest degree term). 3. Intercepts: Putting y = 0 in given equation, we get x3 – 3ax2 = 0 ⇒ x = 0, 3a Thus, the curve cross x-axis at (0, 0) and (3a, 0) . 4. There is no asymptotes. dy 5. = 3x2 – 6ax. dx dy For stationary point = 0 = 3x2 – 6ax = 0 ⇒ x = 0, 2a. dx 6. d2y FG d y IJ H dx K P v> FQ J>NG Q FP J>R G> Fig. 1.3 FG d y IJ H dx K 2 2 = 6x – 6a, > m = – 6a < 0 (concave) and ymax = 0 and 2 dx 2 x=0 = 6a > 0 (convax) and ymin. = – 4a3. 2 x=2a = 12a – 6a d2y = 0 ⇒ 6x – 6a = 0 ⇒ x = 0, a. dx 2 8. Region: – ∞ < x <∞ since y is defined for all x. 7. Inflexion point: 9. Sign of derivative Interval Sign of y Quadrant Sign of y′ Nature of curve –∞<x<0 y<0 III y′ > 0 increasing 0 < x < 2a y<0 IV y′ < 0 decreasing 2a < x < 3a y<0 IV y′ > 0 increasing 3a < x < ∞ y>0 I y′ > 0 increasing Using the above calculations. We draw the graph in Figure 1.3. Example 2. Trace the curve y2 (a – x) = x3, a > 0. Sol. 1. Symmetry: Since y has even power so the curve is symmetric about x-axis. 2. Origin: The curve passes through the origin. 3. Tangent at origin: The coefficient of lowest degree term is y2 = 0 or y = 0 and y = 0 i.e., there is a cusp at the origin. 4. Intercepts: Putting x = 0, then y = 0 ⇒ origin is the only point where the curve meets the co-ordinate axes. 5. Asymptotes: Asymptotes parallel to y-axis obtained by equating to zero the highest degree term of y i.e., (x – a) = 0 ⇒ x = a. (U.P.T.U., 2006) w ³>[> b¬°¡©¤>>>¯ «¦¤«¯> m F J>NG Fig. 1.4 v> 67 DIFFERENTIAL CALCULUS-I x3 this shows that the values of x > a, y is imaginary. So the curve does a−x not exist x > a. Similarly, the curve does not exist when x < 0. Here the curve only for 0 ≤ x < a. 7. Sign of derivation: 6 . Region: y2 = 1 F H I K 3a −x dy dy 2 =± , in the first quadrant for 0 ñ x < a, > 0, curve increasing in the first dx dx ( a − x) a − x quadrant. The shape of figure is shown in the (Fig. 1.4). x2 Example 3. Trace the following curve and write its asymptotes. x3 + y3 = 3axy. (U.P.T.U., 2003) Sol. 1. Symmetry: Interchage x and y. Then equation of curve remain unchanged. ∴ The curve is symmetric about the line y = x. 2. Origin: The curve passes through origin. 3. Tangent at origin: The coefficient of lowest degree term is x = 0 or y = 0 ∴ x = 0 and y = 0 are tangents at origin. 4. Intercepts: Putting y = x, we get w 2x3 = 3ax2 ⇒ 2x3 – 3ax2 = 0 >³ ´> [ ⇒ x2 (2x – 3a) = 0 3a 2 Q JQ Å Å P P ⇒ x = 0 and x = At x = 0, y = 0 and at x = 3a 3a ,y= 2 2 Thus, points of intersection are (0, 0) and F 3a , 3a I H 2 2K along the line y = x. 5. Asymptotes: Since the coefficients of highest powers of x and y are constants, there are no asymptotes parallel to x-and y-axis. Putting x = 1 and y = m in highest degree term (x3 + y3), we get OQSÝ > m>FNJ>NG ³> I> m = –1 (Real solution) Again putting x = 1 and y = m in next highest degree term (– 3axy), we get c = At a−3amf = a 3m2 I> Fig. 1.5 φ3(m) = 1 + m3 = 0 ⇒ ´> m m = –1, c = – a. Asymptotes y = mx + c ⇒ y = –x–a or y + x + a = 0. >[ >N v 68 A TEXTBOOK OF ENGINEERING MATHEMATICS—I dy ay − x 2 = 2 ⇒ dx y − ax an angle 135° with x-axis. 6 . Derivative: FG dy IJ H dx K FH 3a 3a , 2 2 F 3a 3a I I = – 1 i.e., the tangent at H 2 , 2 K making K 7 . Region: When both x and y are negative simultaneously equation of curve is not satisfied (–x)3 + (–y)3 = 3a (–x) (–y) ⇒ – (x3 + y3) = 3axy ⇒ There is no part of the curve exists in 3rd quadrant The shape of the curve is shown in Fig. 1.5. 2 2 2 Example 4. Trace the curve x 3 + y 3 = a 3 (Astroid). 2 2 2 F x I + FG y IJ = 1 H aK H aK FG x2 IJ 1 3 + FG y2 IJ 1 3 = 1 H a2 K H a 2 K Sol. We have x 3 + y 3 = a 3 ⇒ or 23 23 ...(i) 1 . Symmetry: Since there are even powers of x and y so the curve is symmetric about both axis. 2. Origin: Since there is constant term so the curve does not passes through the origin. 3. Intercept: Putting y = 0, we get x = ± a. ⇒ The curve cross x-axis at (a, 0) and (–a, 0) Similarly, putting x = 0 then y = ± a i.e., the curve cross y-axis at (0, a) and (0, –a). FG yIJ ⇒ FGH dyIJK = 0 i.e., tangent dx H xK F dyI = – ∞ = tan F − π I i.e., at (a, 0) is along x-axis and GH dx JK H 2K dy 4. Derivative: =– dx 1 3 w ( a, 0) `> FNJ G (0, a) tangent at (0, a) is y-axis. Fy I F x2 I 3 5. Region: G J = 1 – G 2 J Ha K Ha K 2 1 3 1 a F J>NG 2 ∴ if y2 x > 1 i.e., x > a, we have 2 < 0 i.e., y2 < 0 = y is a a _> m F J>NG v b FNJ> G Fig. 1.6 imaginary, so the curve does not exist for x > a. Similarly, the curve does not exist for x < – a and y > a, y < – a. 6. Asymptotes: No asymptotes. The shape of the figure is shown in the Fig. 1.6. Example 5. Trace the curve x2y2 = a2 (y2 – x2). Sol. 1. Symmetry: The curve is symmetric about both axis. 2. Origin: The curve passes through origin. 69 DIFFERENTIAL CALCULUS-I 3 . Tangents: The tangents at the origin y2 – x2 = 0. i.e., y = ± x. 4. Intercepts: Cross the line at origin i.e., (0, 0). 5. Region: y2 = a2x2/(a2 – x2) this shows that the curve does not exist for x2 > a2 i.e., for x > a and x < – a. 6. As x → a, y2 → ∞ and as x → – a, y2 → ∞ The shape of the curve is shown in Fig. 1.7. w ³>[> m Example 6. Trace the curve y2 (a + x) = x 2 (b – x), (Strophoid). Sol. 1. Symmetry: There is only even power of y so the curve is symmetric about x-axis. 2. Origin: The curve passes through the origin. 3. Tangents at origin: The lowest degree term is ay2 – bx2 b ∴ tangents are ay2 – ax2 = 0 ⇒ y = ± x. a 4. Intercepts: Putting y = 0, so x2 (b – x) = 0 ⇒ x = 0, b The curve meets x-axis at (0, 0) and (b, 0). y-intercept: put x = 0 then y = 0. So (0, 0) is the y-intercept. 5. Asymptotes: Asymptotes parallel to y-axis is x + a = 0 ⇒ x = – a. ³>[> w¢ Fig. 1.7 w ³>[>¡ m ³>[>> F¡J>NG b−x , y becomes imaginary a+x when x > b and x < – a. Thus curve exist only in the region – a < x < b. 6. Region: y = ± x 7. Derivative: ( −2 x 2 − 3 ax + bx + 2 ab) dy = ⇒ 2( a + x) 3/2 (b − x ) dx FG dyIJ H dxK v Fig. 1.8 (b, 0) = ∞ = tan π 2 Thus, the tangent at (b, 0) is parallel to y-axis. Therefore from above calculations, we draw the curve (Fig. 1.8) of given equation. Example 7. Trace the curve y2 (x2 + y2) + a2 (x2 – y2) = 0. Sol. 1. Symmetry: The curve is symmetric about both axis. 2. Origin: Curve passes through origin. 3. Tangents at origin: The tangents at the origin are x2 – y2 = 0 or y = ± x. 4. Intercepts: At x-axis = (0, 0) At y-axis ⇒ y4 – a2y2 = 0 (put x = 0) y2 (y2 – a2) = 0 ⇒ y = ± a, 0 Thus, the curve cross y-axis at (0, a) and (0, – a) and (0, 0). v 70 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 5 . Tangent at new point (0, – a) and (0, a) Putting y = y + a, we get (y + a)2 x 2 + ( y + a) 2 + a 2 x 2 − ( y + a) 2 FNJ> G w ´>[>³ ´>[>³ =0 ³>[>> ³>[> ∴ The tangent at the new points are y = 0, y = 0 v i.e., parallel to x-axis. m 2 2 2 2 2 2 6 . Region: x = y (a – y )/(a + y ) The curve does not exist when y2 > a2 or y > a and y < – a FNJ> G ∴ The curve exist in the region when – a < y < a. 7 . No asymptotes: The shape of the curve shown in Fig. 1.9. Fig. 1.9 Example 8. Trace the curve x2y2 = a2 (x2 + y2). Sol. 1. Symmetry: Curve is symmetric about both axis. 2. Origin: Curve passes through the origin. Tangnt at (0, 0) are x2 + y2 = 0 ⇒ y = ± ix, which give imaginary tangents. So (0, 0) is a conjugate point. w 3. The curve does not cross the axis. 4. Asymptotes: Asymptotes are x = ± a and y = ± a. 5. Region: y2 = a2x2/(x2 – a2). If x2 < a2 i.e., x < ± a ´>[> then y is imaginary i.e., the curve does not exist when x < a and x < – a. v¢ v m Similarly does not exist when y < a and y < – a ´>[>> ´>[>> 6. y2 → ∞ as x → a and x2 → ∞ as y → a ∴ The shape of the curve is shown in Fig. 1.10. Example 9. Trace the curve (x2 – a2) (y2 – b2) = a2b2. w¢ Sol. The given curve is (x2 – a2) (y2 – b2) = a2b2 or Fig. 1.10 x2y2 – b2x2 – a2y2 = 0 or x2y2 = b2x2 + a2y2 1. Symmetry about both the axes. w ³>[> 2. (0, 0) satisfies the equation of the curve. Tangents at the origin are a2y2 + b2x2 = 0, which give imaginary tangents. So (0, 0) is a conjugate point. ´>[>¡ 3. The curve does not cross the axes. 4. Equating the coefficients of highest powers of x and y we find that x = ± a and y = ± b are the asymptotes. v¢ v m ´>[>>¡ 5. Solving for y, we get y2 = b x . x 2 − a2 ∴ If x2 < a2 i.e., x is numerically less than a, y2 is negative i.e., y is imaginary i.e., the curve does not exist between the lines x = − a and x = a. ³>[>> 2 2 w¢ Fig. 1.11 Similarly arguing we find that the curve does not exist between the lines y = – b and y = b. 6. y2 → ∞ as x → a and x2 → ∞ as y → a. With the above data, the shape of the curve is as shown in Fig. 1.11. 71 DIFFERENTIAL CALCULUS-I Example 10. Trace the curve y2 (a – x) = x2 (a + x). Sol. 1. Symmetry about x-axis. 2. Passes through (0, 0), the tangents are y2 = x2 or y = ±x. Tangents being real and distinct, node is expected at the origin. 3. Curve crosses the x-axis at (– a, 0) and (0, 0). Shifting the origin to (–a, 0) and equating the lowest degree terms to zero, we get new y-axis as the tangent at the new origin. 4. x = a is the asymptote. 5. For x < – a, the curve does not exist. Similarly for x > a, the curve does not exist. 6. As x → a, y2 → ∞. 7. No point of inflexion. ∴ Shape of the curve is as shown in Fig. 1.12. x=a Y (–a, 0) (a, 0) O X¢ X Y¢ Fig. 1.12 EXERCISE 1.7 1 1 1 2. x 2 + y 2 = a 2 1 . Trace the curve y2 (2a – x) = x3 (U.P.T.U. (C.O.), 2003) (U.P.T.U., 2004) Y Y O X Asymptote Ans. = (0, a) y Ans. x Tangent a/4 O a/4 (a, 0) X 72 A TEXTBOOK OF ENGINEERING MATHEMATICS—I F xI F yI 3 . H K + GH JK = 1 a b 2 3 2 3 4 . x6 + y6 = a2x2y2 Y Y y = –x (0, b) y=x b Ans. –a a Ans. X O –(a, 0) x (a, 0) –b (0, –b) 5 . a2y2 = x3 (2a – x) 6 . 9ay2 = x(x – 3a)2 Y Y O Ans. (2a, 0) 7 . a4y2 = a2 x4 – x6 O Ans. X X 8 . xy2 = 4a2 (2a – x) Y Ans. 3a Y (–a, 0) –a O a Ans. X 2a O (a, 0) 9 . y3 = x(a2 – x2) X (2a, 0) 1 0 . y = 8a3/(x2 + 4a2) Y Y A Ans. (a, 0) (–a, 0) X O (0 , 2a) y = 2a Ans. x1 x2 O Asymptote X 73 DIFFERENTIAL CALCULUS-I 1.9 POLAR CURVES The general form (explicit) of polar curve is r = f(θ) or θ = f(r) and the implicit form is F(r, θ) = 0. Procedure: 1. Symmetry: (a) If we replace θ by −θ, the equation of the curve remains unchanged then there is symmetry about initial line θ = 0 (usually the positive x-axis in cartesian form). Example. r = a (1 ± cos θ). (b) If we replace θ by π − θ, the equation of the curve remains unchanged, then there is a π symmetry about the line θ = (passing through the pole and z to the initial line) which is usually 2 the positive y-axis in cartesian). Example. r = a sin 3θ. (c) There is a symmetry about the pole (origin) if the equation of the curve remains unchanged by replacing r into – r. Example. r2 = a cos 2θ. (d) Curve is symmetric about pole if f(r, θ) = f(r, θ + π) Example. r = 4 tan θ. (e) Symmetric about θ = F H I K π π i.e., (y = x), if f(r, θ) = f r, − θ 4 2 F H I K 3π 3π i.e., (y = – x), if f (r, θ) = f r, −θ 4 2 2 . Pole (origin): If r = f(θ1) = 0 for some θ = θ1 = constant then curve passes through the pole (origin) and the tangent at the pole (origin) is θ = θ1. (f) Symmetric about θ = Example. r = a (1 + cos θ) = 0, at θ = π. 3 . Point of intersection: Points of intersection of the curve with initial line and line θ= π π are obtained by putting θ = 0 and θ = . 2 2 4 . Region: If r2 is negative i.e., imaginary for certain values of θ then the curve does not exist for those values of θ. 5 . Asymptote: If lim r = ∞ then an asymptote to the curve exists and is given by equation θ→ α r sin (θ – α) = f ′(α) where α is the solution of 1 = 0. f(θ) 6 . Tangent at any point (r, θ ): Tangent at this is obtained from tan φ = rdθ , where dr φ is the angle between radius vector and the tangent. 7. Plotting of points: Solve the equation for r and consider how r varies as θ varies from 0 to ∞ or 0 to – ∞. The corresponding values of r and θ give a number of points. Plot these points. This is sufficient for tracing of the curve. (Here we should observe those values of θ for which r is zero or attains a minimum or maximum value). 74 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 1. Trace the curve r2 = a2 cos 2θ (U.P.T.U., 2000, 2008) q = 3 p/4 Y q= π Hence, the straight lines θ = ± are the tangents 4 at origin to the curve. 3 . Intersection: Putting θ = 0 ∴ r2 = a2 ⇒ r = ± a the curve meets initial line to the points (a, 0) and (–a, π). 4 . As θ varies from 0 to π, r varies as given below: θ = 0 30 45 90 135 150 180 2 2 2 2 2 r = a a /2 0 – a 0 a /2 a2 ←imaginary→ p/4 π π or θ = ± 2 4 q= i.e., cos 2θ = 0 ⇒ 2θ = ± p/2 Sol. 1. Symmetry: Since there is no change in the curve when θ replace by – θ. So the curve is symmetric about initial line. 2 . Pole: Curve passes through the pole when r2 = a2 cos 2θ = 0 (–a, p) X (a, 0) –3 p/4 O q= q = – p/4 Fig. 1.13 5 . Region: The above data shows that curve does not exist for values of θ which lying between 45º and 135º. Example 2. Trace the curve r = a sin 3θ (U.P.T.U., 2002) Sol. 1. Symmetry: The curve is not symmetric about the initial line. 2 . Origin: Curve passes through the origin when r = 0 ⇒ ⇒ a sin 3θ = 0 3θ = 0, π, 2π, 3π, 4π, 5π ⇒ θ = 0, π 2π 4 π 5π , , π, , 3 3 3 3 are the tangents at the pole. 3 . Asymptote: No asymptote since r is finite for any value of θ. 4 . Region: Since the maximum value of sin 3θ is 1. π 3 π 5π , , , etc. 2 2 2 π 3 π 5π or θ= , , , etc. 6 6 6 for which r = a (maximum value). ∴ The curve exist in all quadrant about the lines θ = q q = p/2 q = 2p/3 = 5p / q = p/3 p/6 3θ = 6 q= So, a a (q = p ) π 3π 5 π , , at distance r = a. q = 7p/6 6 6 6 The shape of the curve is given in the (Fig. 1.14). q = 4 p/3 q=0 O q = 11p/4 q = 3 p/2 q = 5 p/3 Fig. 1.14 75 DIFFERENTIAL CALCULUS-I Example 3. Trace the curve r = aeθ Sol. 1. No symmetry about the initial line. 2 . As θ → ∞ , r → ∞ and r always positive. 3. Corresponding values of θ and r are given below: θ = r = π/2 aeπa/2 0 a π aeπa 3π/2 ae3π/2 Y q = p/2 q=0 X q = p/2 2π ae2πa O with the above data the shape of the curve is shown below: Example 4. Trace the curve r2 cos 2θ = a2 (Hyperbola) 1 Sol. 1. Symmetry about pole and about the line θ = π . 2 2 . Changing to cartesian the equation becomes q = 3 p/2 Fig. 1.15 q = p/2 Y x2 – y2 = a2. ∴ The equation of the asymptotes are y = ± x or 1 π are its polar asymptotes. 4 3 . When θ = 0, r2 = a2 or r = ±a i.e., the points (a, 0) and (– a, 0) lie on the curve. (Here co-ordinates of the points are polar coordinates). 4 . Solving for r we get r2 = a2/cos 2θ. This shows 1 that as θ increases from 0 to π , r increases from a to ∞. 4 1 3 5 . For values of θ lying between π and π , r2 is 4 4 negative i.e., r is imaginary. So the curve does not exist for 1 3 π< θ < π. 4 4 q = p/4 θ=± (a = p) Fig. 1.16 p/4 q = p/2 q= q = 3p/4 a a π i.e., y-axis. 2 a q=0 q=p 30° 45° 60° 90° 120° q= 5p /4 a 1 π or 2 . Putting r = 0, cos 2θ = 0 or 2θ = ± 2 1 1 θ = ± π , i.e., the straight lines θ = ± π are the tangents 4 4 to the curve at the pole. 3 . Corresponding values of θ and r are given below: θ = 0° X q = – p/4 q = 3p/2 Example 5. Trace the curve r = a cos 2θ. Sol. 1. Symmetry about the initial line and the line θ= (a = 0) O q=p q = 7 p/4 q = 3p/2 Fig. 1.17 135° 150° 180° 1 1 1 1 − a − a a a 0 – a 0 a 2 2 2 2 Plot these points and due to symmetry about the initial line the other portion can be traced. r =a 76 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 6. Trace the curve r = a (1 – cos θ) (cardoid). Sol. 1. Symmetry: No change in equation when we replace θ by – θ. So the equation of curve is symmetric about initial line. 2 . Pole (region): If the curve passes through origin then r = 0 ⇒ a (1 – cos θ) = 0 cos θ = 1 = cos 0 ⇒ θ = 0. ⇒ Here, the straight line θ = 0 is tangent at origin. 3 . Intersection: Putting θ = 0 then r = 0 and putting θ = π, then r = 2a ∴ Intersection points on initial line = (0, 0) and (2a, π). 4 . Region: It exists in all quadrant. 5 . Asymptotes: No. asymptotes. 6 . As θ increases from 0 to π, r also increases from 0 to 2a. The corresponding values of r and θ given below: θ= 0 60° 90° 120° 0 = π/2 (2a, π) (0, 0) θ=0 180° a 3a Fig. 1.18 r =a a 2a 2 2 With the above data the shape of the curve is given in (Fig. 1.18). Example 7. Trace the curve r = a + b cos θ, when a < b. Sol. 1. Symmetry: The curve is symmetric about initial line. 2 . Origin: For origin r = 0 ⇒ a + b cos θ = 0 ⇒ cos θ = − 3 . Corresponding values of θ and r are given below: π 4 θ= 0 r =a+b a+ FG b IJ H 2K π 3 π 2 a+ a F bI H 2K 2π 3 Fa − bI H 2K a b 3π 4 FG a − b IJ H 2K π (a – b) b 2π then r is positive for all values of θ from 0 to but r is negative when θ 2 3 3 2π 3π = π or π. Here, r must vanish somewhere between θ = and . Let θ = α (lying between 4 3 4 2π 3π and ). 3 4 q = 2 p/3 q = p/2 q = p/3 For which r = 0 then θ = α is the straight line which is θ= α tangent to the curve at the pole and for values of θ lying between α and π, r is negative and points corresponding to such values of θ will be marked in the opposite direction on these lines as r is θ= π O negative for them. A Thus in this case a < b, the curve passes through the origin Let a > RS−F a I UV and form two loops, one inside the T H bK W other as shown in the Fig. 1.19. when θ = α = cos–1 Fig. 1.19 77 DIFFERENTIAL CALCULUS-I EXERCISE 1.8 1 . Trace the curve r = a (1 + cos θ) 2 . Trace the curve r = a sin 2θ Y (a, p/2) q = 3 p/4 (a, p) Ans. X O (2a, 0) q = p/2 q = p/4 q=p Ans. q= 0 0 q = 5 p/4 3 . Trace the curve q = 7 p/4 4 . Trace the curve r = 2a cos θ r= a sin 2 θ cos θ Y Y P (r, q) x=0 r Ans. O q a a C (a, 0) A X Ans. A (a, 0) X O [Hint: Change it in cartesian coordinates, y2 (x – a) = – x3] 5 . Trace the curve r = 2 (1 – 2 sin θ) 6 . Trace the curve r = a cos 2θ q = 3 p/4 q = p/2 q=p O q = p/2 q = p/4 X Ans. Ans. (U.P.T.U., 2003) q=p q=0 q = 3p/2 q = 5 p/4 q = 3p/2 x 78 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 8 . Trace the curve r2 cos 2θ = a2 7 . Trace the curve r = a (1 + sin θ) q = p/2 q = p/2 Y q = p/4 (0, 2a) Ans. Ans. O (–a, 0) q=0 X (a, 0) q=p [Hint: Change in cartesian form x2 – y2 = a2] 1.10 O (a, 0) (a, p) q = 3p/2 q = – p/4 PARAMETRIC CURVES Let x = f1(t) and y = f2(t) be the parametric equations of a curve where t is a parameter. Method I: Eliminate the parameter ‘t’ if possible and we shall get the cartesian equation of curve which can easily traced. Example. x = a cos t, y = a sin t ⇒ x2 + y2 = a2. Method II: When the parameter ‘t’ cannot be eliminated. 1 . Symmetry: If x = f1(t) is even and y = f2(t) is odd then curve is symmetric about x-axis: Similarly, if x = f1(t) is odd and y = f2 (t) is even then curve is symmetric about y-axis. 2 . Origin: Find ‘t’ for which x = 0 and y = 0. 3 . Intercept: x-intercept obtained for values of t for which y = 0, y-intercept for values of t for which x = 0. 4 . Determine least and greatest values of x and y. lim lim 5 . Asymptotes: t → t x(t) = ∞, t → t1 y(t) = ∞, then t = t1 is asymptote. 1 6 . Tangents: dy dy = ∞ (vertical tangent) and = 0 (horizontal tangent). dx dx Remark. If the given equations of the curves are periodic functions of t having a common period, then it is enough to trace the curve for one period. Example 1. Trace the curve x = a (t – sin t), y = a (1 – cos t) (Cycloid) Sol. 1. Symmetry: Since x is odd and y is even so the curve is symmetric about y-axis. 2 . Origin: Putting x = 0 and y = 0, we get and 0 = a (t – sin t) ⇒ t = 0 0 = a (1 – cos t) ⇒ cos t = 1 ⇒ t = 0, 2π, 4π, etc. 79 DIFFERENTIAL CALCULUS-I 3. dx dy dy/dt dy a sin t = a (1 – cos t), = a sin t ∴ = = dt dx/dt dt dx a (1 − cos t) dy t 2a sin t/2.cos t/2 ⇒ = = cot 2 2 dx a(1 − 1 + 2 sin t/2) dy Corresponding values of x, y, for different values of t are given below: dx 3π π π 2π t = 0 2 2 π 3π x = 0 a aπ a 2aπ −1 +1 2 2 y = 0 a 2a a 0 F H I K F H I K dy = ∞ 1 0 – 1 –∞ dx 4 . Tangents at y = 0 are vertical and at y = 2a is horizontal. Curve is periodic for period 2π in the interval [0, 2π]. Curve repeats over intervals of [0, 2 aπ] refer Fig. 1.20. Y 2a – 4ap – 2ap 2ap O 4ap Fig. 1.20 Y t = p/2 B (0, a) Example 2. Trace the curve x = a cos3 t, y = a sin3 t. (Astroid) Sol. 1. dy/dt = 3a sin2 t cos t. dx/dt = –3a cos2 t sin t 2 A t=0 X¢ (–a, 0) dy dy dt 3 a sin t cos t = = dx dt dx −3 a cos2 t sin t ∴ a A¢ x= a y= 0 π 4 a 2 2 a 2 2 π 2 0 a 3π 2 −a 2 2 a 2 2 π – a 0 a (a, 0) X t = 3 p/2 B¢ (0, –a) Y¢ dy or = – tan t. dx 2 . Corresponding values of x, y and dy/dx for different values of t are given below: t = 0 O 5π 4 a 2 2 −a 2 2 Fig. 1.21 3π 2 0 – a 7π 4 a 2 2 −a 2 2 2π a 0 dy = 0 – 1 –∞ 1 0 –∞ 1 1 0 dx Plotting the above points and observing the inclinations of the tangents at these points the shape of the curve is as shown in Fig. 1.21. 80 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 3. Sol. 1. x = a (t + sin t), Y y = a (1 – cos t) dy/dt = a (sin t); B t = –p ap dx/dt = a (1 + cos t) ∴ t = –p/2 t = p/2 2 a(2 cos 2 1 2 X X¢ 1 O t=0 2 Y¢ t) Fig. 1.22 2 a sin t cos t = t=p 2a dy a(sin t ) dy dt = = dx a(1 + cos t ) dx dt 1 C ap 1 t. 2 2 . Corresponding values of x, y and dy/dx for different values of t are given below: or t = – π x = – πa y = 2a dy/dx = – ∞ dy|dx = tan 1 − π 2 1 – a ( π – 1) 2 a – 1 1 π 2 0 0 1 π + 1) 2 a 1 a( 0 0 EXERCISE 1.9 Trace the following curves: 1 . x = a sin 2t (1 + cos 2t), y = a cos 2t (1 – cos 2t) Y C A O Ans. 2 . x = a (t – sin t), y = a (1 + cos t) X B Y C B ap ap 2a t = p/2 t = 3p/2 Ans. X¢ A t=p O Y¢ t = 2p X π aπ 2a ∞ 81 DIFFERENTIAL CALCULUS-I 3 . x = a cos t + 1 a log tan2 2 F t I , y = a sin t (Tractrix) H 2K Y B (0, a) t = p/2 a t=0 X¢ Ans. t = –p X O a (0, –a) B¢ 4. x = a (t + t 3 ) 1 + t4 ,y= a (t − t 3 ) 1 + t4 Y y=x Ans. X (–a, 0) (a, 0) O y = –x EXPANSION OF FUNCTION OF SEVERAL VARIABLES 1.11 TAYLOR'S THEOREM FOR FUNCTIONS OF TWO VARIABLES Let f(x, y) be a function of two independent variables x and y. If the function f(x, y) and its partial derivatives up to nth order are continuous throughout the domain centred at a point (x, y). Then LM ∂f (a, b) + k ∂f (a, b) OP ∂y Q N ∂x f(a + h, b + k) = f(a, b) + h LM OP MN PQ ∂ f ( a, b) ∂ f ( a, b) ∂ f ( a, b) O 1 L ∂ f ( a, b) + 3h k + 3hk +k PP+... + 3 Mh ∂y ∂x ∂y ∂x∂y MN ∂x Q + 2 ∂ 2 f ( a, b) ∂ 2 f ( a, b) 1 2 ∂ f ( a, b) h2 2 hk + k + 2 ∂x∂y ∂y 2 ∂x 2 3 3 2 3 3 2 2 3 3 2 3 3 Or LM ∂ + k ∂ OP f (a, b) + 1 LMh ∂ + k ∂ OP f(a, b) ∂y Q 2 N ∂x N ∂x ∂y Q ∂O 1 L ∂ + 3 Mh ∂x + k ∂y P f ( a, b)+... N Q 2 f(a + h, b + k) = f(a, b) + h 3 82 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Proof. Suppose P(x, y) and Q (x + h, y + k) be two neighbouring points. Then f (x + h, y + k), the value of f at Q can be expressed in terms of f and its derivatives at P. Here, we treat f(x + h, y + k) as a function of single variable x and keeping y as a constant. Expanded as follows using Taylor's theorem for single variable.* ∂f ( x , y + k ) h 2 ∂ 2 f ( x , y + k ) + +... ∂x 2 ∂x 2 f(x + h, y + k) = f(x, y + k) + h ...(i) Now expanding all the terms on the R.H.S. of (i) as function of y, keeping x as constant. LM f (x, y) + k ∂f (x, y) + k ∂ f (x, y) +...OP 2 ∂y ∂y MN PQ O ∂ f ( x , y ) k ∂ f ( x , y) ∂ L + + …P + h M f ( x , y) + k 2 ∂y ∂x M ∂y PQ N O ∂f ( x, y ) k ∂ f ( x, y ) h ∂ L f ( x, y ) + k + +...P+... + M ∂y 2 ∂x M 2 ∂y PQ N L ∂f (x, y) + k ∂f (x, y) OP f(x + h, y + k) = f(x, y) + M h ∂y Q N ∂x ∂ f ( x , y) O ∂ f ( x , y) 1 L ∂ f ( x , y) h +k + 2hk + M PP + ... 2 MN ∂x∂y ∂y ∂x Q 2 2 f(x + h, y + k) = 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 For any point (a, b) putting x = a, y = b in above equation then, we get LM ∂f (a, b) + k ∂f (a, b) OP ∂y Q N ∂x f (a + h, b + k) = f (a, b) + h + LM MN OP PQ 2 ∂ 2 f ( a, b) 1 2 ∂ 2 f ( a, b) 2 ∂ f ( a, b) + ... h 2 hk k + + 2 ∂x 2 ∂y 2 ∂y 2 Or LM ∂ + k ∂ OP f (a, b) + 1 LMh ∂ + k ∂ OP f (a, b) ∂y Q 2 N ∂x N ∂x ∂y Q ∂O 1 L ∂ h + k P f ( a, b) + ... + Hence proved. M ∂y Q 3 N ∂x 2 f (a + h, b + k) = f (a, b) + h 3 Alternative form: Putting a+h = x⇒h=x–a b+k = y⇒k=y–b LM N then f (x, y) = f (a, b) + ( x − a) ∂ ∂ + ( y − b) ∂x ∂y OP f (a, b) + 1 L(x − a) ∂ + (y − b) ∂ O f (a, b) + ... ...(ii) 2 MN ∂x ∂y PQ Q 2 * Taylor's theorem for single variable f (x + h) = f (x) + h ∂f h 2 ∂2 f h 3 ∂ 3 f + ... + + ∂x 2 ∂x 2 3 ∂x 3 83 DIFFERENTIAL CALCULUS-I 1.11.1 Maclaurin's Series Expansion It is a special case of Taylor's series when the expansion is about the origin (0, 0). So, putting a = 0 and b = 0 in equation (2), we get LM ∂ + y ∂ OP f (0,0) + 1 LMx ∂ + y ∂ OP f (0,0) + ... 2 N ∂x ∂y Q N ∂x ∂y Q π Example 1. Expand e cos y about the point F 1, I (U.P.T.U., 2007) H 4K Sol. We have f (x, y) = e cos y ...(i) π π e π F I = e cos = , f 1, and a = 1, b = H 4K 4 4 2 F πI ∂f 1, ∂f H 4 K = e cos π = e ∴ From (1) = e cos y ⇒ 2 f (x, y) = f (0, 0) + x x x x ∂x 4 ∂x 2 F I H K = – e sin π = – e π ∂f 1, ∂f 4 x ∂y = – e sin y ⇒ ∂y ∂2 f ∂x 2 4 2 F πI H 4 K = e , ∂ f = – e sin y ∂ 2 f 1, = ex cos y ⇒ 2 ∂x 2 2 x ∂x∂y F πI H 4 K = – e , ∂ f = – e cos y = – e . ∂ 2 f 1, 2 ∂x∂y By Taylor's theorem, we have F H π f 1 + h, + k 4 2 x ∂y 2 2 F ∂f F1, π I ∂f F1, π I I I = f F1, π I + GG h H 4 K + k H 4 K JJ + ... K H 4 K H ∂x ∂y K π π +k=y⇒k=y– , equation (2) reduce in the form 4 4 e e e 1 π e − ( x − 1) 2 ⋅ + (x – 1) · + O− + 2 2 4 2 2 2 Let 1 + h = x ⇒ h = x – 1 and f ((x, y) = ex cos y = ...(ii) F H I FG K H IJ K LM N F y − π I FG − e IJ + F y − π I FG − e IJ OP + ... H 4 K H 2 K H 4 K H 2 K PQ e L F π I + (x − 1) − (x − 1) F y − π I − F y − π I + ...OP . ⇒ f (x, y) = 1 + (x − 1) − y − M H 4K 2 H 4 K H 4K Q 2N 2 + 2 (x – 1) 2 2 Example 2. Expand f (x, y) = ey log (1 + x) in powers of x and y about (0, 0) Sol. We have f (x, y) = ey log (1 + x) Here, a = 0 and b = 0, then f (0, 0) = e0 log 1 = 0 Now, ∂f ey = ∂x 1+ x ∂f y ∂y = e log (1 + x) ⇒ ∂f ( 0, 0 ) = 1 ∂x ⇒ ∂f (0,0) = 0 ∂x 84 A TEXTBOOK OF ENGINEERING MATHEMATICS—I ∂2 f ey = ∂x∂y 1+ x ⇒ ∂2 B ∂N 2 ey (1 + x ) 2 ⇒ = ey log (1 + x) ⇒ ∂2 f ∂y 2 = − ∂2 f (0,0) ∂x∂y = 1 ∂2 f (0,0) = – 1 ∂x2 ∂2 f (0,0) = 0 ∂y2 Now, applying Taylor's theorem, we get FG h ∂f (0,0) + k ∂f (0,0) IJ + 1 FG h ∂ f (0,0) ∂y K 2H ∂x H ∂x ∂ f (0, 0) ∂ f (0, 0) I +k + 2hk JK + ... ∂x∂y ∂y 2 2 2 2 f (0 + h, 0 + k) = f (h, k) = f (0, 0) + 2 2 2 Let h = x, k = y, then, we get f (x, y) = ey log (1 + x) = f (0,0) + = 0 + (x × 1 + y × 0) + ⇒ FG x ∂f (0, 0) + y ∂f (0, 0) IJ + ... ∂y K H ∂x 1 [x2 (–1) + 2xy × 1 + y2 × 0] + ... 2 x2 + xy + ... . 2 ey log (1 + x) = x – Example 3. Find Taylor’s series expansion of function f (x, y) = e − x − y . cos xy about the point x0 = 0, y0 = 0 up to three terms. (U.P.T.U., 2006) 2 2 Sol. We have f (x, y) = e − x − y cos xy. Now, we get the following terms f (0, 0) = 1. ∂f ∂ f ( 0, 0 ) −x 2 − y 2 = – e (2x cos xy + y sin xy) ⇒ =0 ∂x ∂x ∂f ∂ f ( 0, 0 ) −x 2 − y 2 (2y cos xy + x sin xy) ⇒ =0 ∂y = – e ∂y 2 Similarly, ∂ 2 f (0, 0) = 0, ∂x∂y ∂3 f (0,0) ∂x ∂y 2 ∂ f ( 0, 0 ) = 0, 2 ∂ 2 f ( 0 ,0 ) ∂x 2 3 ∂ f (0, 0) ∂x∂y 2 = −2, = 0, ∂ 2 f (0, 0) ∂y 2 =–2 ∂ 3 f ( 0, 0 ) =0 ∂x 3 3 ∂y 3 = 0 Applying Taylor's theorem FG h ∂ + k ∂ IJ f (0, 0) + 1 FG h ∂ + k ∂ IJ f (0, 0) ∂y K 2 H ∂x H ∂x ∂y K ∂I 1 F ∂ h + k J f (0, 0) + ... + G ∂y K 3 H ∂x 2 f (h, k) = f (0, 0) + 3 85 DIFFERENTIAL CALCULUS-I Putting h = x, k = y and all values then, we get f (x, y) = e − x − y · cos xy = 1 + (x × 0 + y × 0) 1 2 x ( −2) + 2 xy × 0 + y 2 ( −2) + 2 2 2 + ⇒ 2 1 3 x × 0 + 3x 2 y × 0 + 3xy 2 × 0 + y 3 × 0 3 2 A − N − O . cos xy = 1 – x2 – y2 .... . Example 4. Find Taylor's expansion of f (x, y) = cot–1 xy in powers of (x + 0. 5) and (y – 2) up to second degree terms. Hence compute f (– 0.4, 2.2) approximately. Sol. Here f (x, y) = cot–1 xy Now f (– 0.5, 2) = cot–1(–1) = 3π 4 −y ∂f = ∂x 1 + x2y2 ⇒ ∂f ( −0.5, 2 ) = – 1 ∂x −x ∂f = 1 + x2y2 ∂y ⇒ ∂f ( −0.5, 2) = 1/4 ∂y ∂2 f ( x 2 y 2 − 1) = ∂x∂y (1 + x 2 y 2 )2 ⇒ ∂ 2 f ( −0.5, 2) = 0 ∂x∂y ∂2 f ∂x 2 ⇒ ∂2 f ∂y 2 = = 2 xy 3 (1 + x 2 y 2 )2 2x 3 y ⇒ (1 + x y ) 2 2 2 ∂ 2 f ( −0.5, 2) ∂x 2 ∂ 2 f ( −0.5, 2) ∂y 2 = – 2 = – 1 8 Now applying Taylor's series expansion, we get FG ∂ + k ∂ IJ ⋅ f (−0.5, 2) + 1 FG h ∂ + k ∂ IJ f (–0.5, 2) +… ∂y K H ∂x ∂y K 2 H ∂x 2 f (– 0.5 + h, 2 + k) = f (–0.5, 2) + h Let ∴ or – 0.5 + h = x ⇒ h = x + 0.5 2+k = y⇒k=y–2 |As a = – 0.5, b = 2 π 3 1 f (x, y) = cot–1 xy = + (x + 0.5) (–1) + (y – 2) × 4 4 1 1 (x + 0.5) 2 ( −2) + 2 (x + 0.5)( y − 2) × 0 + ( y − 2) 2 − . + 2 8 LM N I OP KQ 3π 1 1 – (x + 0.5) + (y – 2) – (x + 0.5)2 – (y – 2)2 + .... 4 4 16 x = – 0.4 and y = 2.2 f (x, y) = Putting F H 3π 0.2 1 – (0.1) + – (0.1)2 – (0.2)2 4 4 16 = 2.29369. f (– 0.4, 2.2) = 86 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 5. Calculate log (1.03)1/3 + ( 0.98 ) 1/ 4 − 1 approximately by using Taylor's expansion up to first order terms. FH x + y − 1IK 1 3 Sol. Let f (x, y) = log 1 4 f (1, 1) = log 1 = 0 ∂f = ∂x Now, − 3 FH 2 1× x 3 IK 1 1 x3 + y4 −1 ∂f (1,1) = ∂x ⇒ 1 3 Taking a = 1, b = 1 ∂f ∂y = 3 − 1× y 4 F I 4 H x + y − 1K 1 3 1 4 ∂f (1,1) = ∂x ⇒ 1 4 Now, applying Taylor's theorem FG h ∂ + k ∂ IJ f (1,1) + ... H ∂x ∂y K L O f (1 + h, 1 + k) = log M(1 + h) + (1 + k ) − 1P N Q 1 1 L O f (1 + h, 1 + k) = log M(1 + h) + (1 + k ) − 1P = 0 + h × +k× N Q 3 4 f (1 + h, 1 + k) = f (1, 1) + But ∴ 1 3 1 4 1 3 1 4 ...(i) Putting h = 0.03 and k = – 0.02 in equation (i) then, we get LM N 1 1 OP = 0.03 × 1 + (– 0. 02) × 1 Q 3 4 log (1.03 ) 3 + ( 0.98 ) 4 − 1 = 0.005. Example 6. Expand xy in powers of (x – 1) and (y – 1) up to the third degree terms. (U.P.T.U., 2003) Sol. Here f (x, y) = xy Now ∂f = yxy–1 ∂x ⇒ ∂f (1, 1) = 1 ∂x ∂f y ∂y = x log x ⇒ ∂f (1, 1) = 0 ∂y ∂2 f = xy–1 + yxy–1 · log x ∂x∂y ⇒ ∂ 2 f (1, 1) = 1 ∂x∂y ∂2 f ∂x 2 = y (y – 1) xy–2 ⇒ = xy · (log x)2 ⇒ ∂2 f ∂y 2 f (1, 1) = 1 ∂2 f (1, 1) ∂x2 ∂ 2 f (1, 1) ∂y 2 = 0 = 0 87 DIFFERENTIAL CALCULUS-I ∂3 f ∂x∂y 2 ∂3 f ∂x 2 ∂y ∂3 f ∂x 3 ∂3 f ∂y 3 = yxy–1(log x)2 + 2xy–1 · log x ⇒ = (y – 1)xy–2 + y(y – 1) xy–2 · log x + yxy–2 ⇒ = y(y – 1) (y – 2)xy–3 ⇒ = xy(log x)3 ⇒ ∂ 3 f (1, 1) ∂x∂y 2 ∂3 f (1, 1) ∂x 2∂y ∂3 f (1,1) ∂x3 ∂3 f (1, 1) ∂y3 = 0 = 1 = 0 = 0 Now applying Taylor's theorem, we get FG h ∂ + k ∂ IJ f (1, 1) + 1 FG h ∂ + k ∂ IJ f (1, 1) H ∂x ∂y K ∂y K 2 H ∂x 1 F ∂ + G h + k ∂∂y IJK · f (1, 1) + ... 3 H ∂x 2 f (1 + h, l + k) = f (1, 1) + 3 Let ⇒ 1 + h = x and 1 + k = y h = x – 1 and k = y – 1 ∂ ∂O ∂ ∂O 1 L L f (x, y) = f (1, 1) + M( x − 1) ∂x + ( y − 1) ∂y P f + 2 M( x − 1) ∂x + ( y − 1) ∂y P f N Q N Q ∂ ∂O 1 L + 3 M( x − 1) ∂x + ( y − 1) ∂y P f + ... N Q 2 ∴ 3 Using all values in above equation, we get f (x, y) = xy = 1 + (x – 1) + 0 + + 1 0 + 2 ( x − 1)( y − 1) + 0 2 1 0 + 3 ( x − 1)2 ( y − 1) + 0 + 0 3 = 1 + (x – 1) + (x – 1) (y – 1) + Example 7. Obtain Taylor's expansion of tan–1 second degree terms. Hence compute f (1.1, 0.9). Sol. Here f (x, y) = tan–1 Now, y x ∂f y = – 2 ∂x (x + y 2 ) 1 ( x − 1)2 ( y − 1) . 2 y about (1, 1) up to and including the x (U.P.T.U., 2002, 2005) ∴ f (1, 1) = π 4 ⇒ ∂f (1, 1) = ∂x − 1 2 88 A TEXTBOOK OF ENGINEERING MATHEMATICS—I b g = 12 ∂f x ∂y = ( x 2 + y 2 ) ⇒ ∂f 1, 1 ∂y ∂2 f ∂x 2 ⇒ ∂2 f 1, 1 ∂x 2 ∂2 f ∂y 2 = = 2 xy (x + y ) 2 2 2 −2 xy ∂2 f ⇒ (x + y ) 2 b g = 12 2 2 2 ∂ B y2 − x2 = ∂N∂O ( x 2 + y 2 )2 By Taylor’s theorem ∂y 2 (1, 1) = − 1 2 ∂2 f = 0 ∂x∂y ⇒ LMh ∂ + k ∂ OP f (1, 1) + 1 LMh ∂ + k ∂ OP f (1,1) + ... ∂y Q 2 N ∂x N ∂x ∂y Q 2 f (1 + h, 1 + k) = f (1, 1) + Let l+h = x⇒ h= x–1 1+k = y⇒k =y–1 ∴ f (x, y) = f (1, 1) + ( x − 1) LM OP f + ... N Q π 1 L F 1 I + 2(x − 1)( y − 1) 1 1 ( x − 1) = + (x – 1) F − I + (y – 1) F I + M H 2K H K H K 2 4 N 2 2 F 1IO × 0 + ( y − 1) − P H 2KQ LM N OP Q ∂ ∂ 1 ∂ ∂ ( x − 1) + ( y − 1) + ( y − 1) f+ ∂x ∂y 2 ∂x ∂y 2 2 2 or 1 π 1 1 1 – (x – 1) + (y – 1) + (x – 1)2 – (y – 1)2 + … . 4 4 2 2 4 Putting x = 1.1, y = 0.9, we get f (x, y) = π 1 1 1 1 – · (1.1 – 1) + (0.9 – 1) + (1.1 – 1)2 – (0.9 – 1)2 4 2 2 4 4 = 0.785 – 0.05 – 0.05 + 0.0025 – 0.0025 = 0.685. f (1.1, 0. 9) = Example 8. Find Taylor's expansion of Sol. Here ∴ f (x, y) = 1 + x + y 2 in power of (x – 1) and (y – 0). 1 + x + y2 f (1, 0) = 2 ∂f 1 = ∂x 2 1 + x + y2 ⇒ ∂ B (1, 0) = 1/2 2 ∂N ∂f = ∂y 1+ x+ y ⇒ ∂ B (1, 0) = 0 ∂O ∂2 f ∂x 2 1 4 (1 + x + y 2 ) 3/2 ⇒ ∂2 B 1 1, 0 = – 2 4.2 3/2 ∂N ∂2 f ∂y 2 = – = y 2 1 1 + x + y2 – y2 (1 + x + y ) 2 3/2 ⇒ b g ∂ 2 f (1, 0) ∂y 2 = 1 2 89 DIFFERENTIAL CALCULUS-I y ∂2 f = – 2(1 + x + y 2 ) 3/2 ∂x∂y By Taylor's theorem, we get ∂ 2 f (1, 0) = 0. ∂x∂y ⇒ LM ∂ + k ∂ OP f (1, 0) + 1 Lh ∂ + k ∂ O f (1, 0) + ... ∂y PQ 2 MN ∂x N ∂x ∂y Q 2 f (1 + h, 0 + k) = f (1, 0) + h Let ⇒ f (x, y) = 1 + h = x ⇒ h = x – 1, k = y 1 1 1 1 (x − 1)2 − + ( x − 1)y × 0 + y2. + ... 2 + ( x − 1) ⋅ + y×0 + 3/2 2 4.2 2 2 2 2 = LM N 2 1+ FG H IJ K OP Q x − 1 ( x − 1) 2 y 2 − + + ... . 4 32 4 EXERCISE 1.10 1 . Expand f (x, y) = x2 + xy + y2 in powers of (x – 1) and (y – 2). [Ans. f (x, y) = 7 + 4 (x – 1) + 5 (y – 2) + (x – 1)2 + (x – 1) (y – 2) + (y – 2)2 + ...] 2 . Evaluate tan–1 F 0.9 I . H 1.1 K [Ans. 0.6904] 3 . Expand f (x, y) = sin (xy) about the point (1, π/2) up to and second degree term. LMAns. f (x, y) = 1 − π (x − 1) − π (x − 1)F y − π I − 1 F y − π I OP + ... H 2K 2 H 2K Q 8 2 N 2 2 2 4 . Obtain Taylor's expansion of x2y + 3y – 2 in powers of (x – 1) and (y + 2). [Ans. f (x, y) = – 10 – 4 (x – 1) + 4 (y + 2) – 2(x – 1)2 + ...] 5 . Expand exy in powers of (x – 1) and (y – 1). LMAns. eRS1 + (x − 1) + ( y − 1) + (x − 1) + (x − 1)(y − 1) + (y − 1) + ...UVOP 2 2 T WQ N 2 2 6 . Expand cosx cosy in powers of x and y. LMAns. f (x, y) = 1 − 1 (x + y ) + 1 (x + 6 x y + y ) + ...OP 2 24 N Q 2 7 . Expand f (x, y) = e2x cos 3y up to second degree. 2 4 2 2 [Ans. 1 + 2x + 2x2 – 4 9 2 y + ...] 2 8 . Obtain Taylor's series expansion of (x + h) (y + k) /(x + h + y + k) LMAns. xy + hy + kx − h y + 2hkxy − k x + ...OP MN x + y (x + y) (x + y) (x + y) (x + y) (x + y) PQ 2 2 2 2 2 2 2 2 3 3 3 9 . Obtain Taylor's expansion of (1 + x – y)–1 in powers of (x – 1) and (y – 1). [Ans. 1 – x + y + x2 – 2xy + y2 + ...] LMAns. y + xy − y + (x y − xy ) + y + ...OP 2 2 3 N Q 2 1 0 . Find Machaurin's expansion of ex log (1 + y). 2 2 3 90 A TEXTBOOK OF ENGINEERING MATHEMATICS—I OBJECTIVE TYPE QUESTIONS A. Pick the correct answer of the choices given below: ∂ 3u is ∂x∂y∂z 1 . If u = exyz, then (i) (x2yz + x) (ii) (x2yz – x)exyz (iii) exyz.xy (iv) (1 + 3xyz + x2y2z2)exyz x− y 2 . If u = sin–1 x+ y , then ∂u is equal to ∂x (i) − x ∂u y ∂y (ii) (iii) − y ∂u x ∂y (iv) xy 3 . If z = log x (i) − (iii) x 2 + y 2 then x +y 2 2 , x +y (ii) 2 y x , x2 + y2 x2 + y2 4 . If u = ex2 + y2 + z2, then (iv) x x2 + y2 x x −y 2 ∂ 3u is ∂x∂y∂z (i) 7xy (ii) 6xyzu (iii) – 8xyzu (iv) 8xyzu 5 . If z = xn–1 y f(y/x) then x ∂ u ∂ u +y is 2 ∂ y∂x ∂x 2 2 ∂z ∂x (ii) nz (iii) n(n – 1)z (iv) n (i) n 6 . If f(x, y, z) = 0 then ∂u ∂y ∂z ∂z and are ∂y ∂x y 2 ∂u ∂y ∂z ∂y ∂x ∂y ∂z is ⋅ ⋅ ∂y ∂z ∂x (i) –1 (ii) 1 (iii) 0 (iv) 2 7 . If u = x3 y2 sin–1 (y/x), then x ∂u ∂u +y is: ∂x ∂y (i) 0 (ii) 6u (iii) – 8u (iv) 5u 2 ,− , y x2 + y2 y x − y2 2 91 DIFFERENTIAL CALCULUS-I 8 . If u = xy f(y/x), then x ∂u ∂u +y is ∂x ∂y (i) yu (ii) xu (iii) u (iv) 2u 9 . The degree of u = [z2/(x4 + y4)]1/3 is 2 1 (i) (ii) 3 3 −2 (iii) (iv) 3 3 (i) t22 dz is dt (ii) 0 (iii) 23t17 (iv) 23t22 1 0 . If z = x4 y5 where x = t2 and y = t3 then ∂y ∂z and at (1, –1, 2) are ∂x ∂x 1 1 (i) − , 1 (ii) − , − 1 2 2 1 1 , −1 ,1 (iii) (iv) 2 2 1 2 . The equation of curve x3 + y3 = 3axy inersects the line y = x at the point 1 1 . If x2 + y2 + z2 = a2 then FG 3a, 3a IJ H 2K F 3a 3a I (iii) G , J H 2 2K (i) FG 3a , − 3a IJ H 2 2K F 3a , − 3a IJ (iv) G − H 2 2K (ii) 1 3 . The equation of the curve x2y – y – x = 0 has maximum (i) One asymptote (ii) Four asymptotes (iii) Three asymptotes (iv) None of these 1 4 . The curve r2 cos 2θ = a2 is symmetric about the line π π (i) θ = (ii) θ = − 2 2 3π (iii) θ = − (iv) θ = π 2 1 5 . The parametric form of the curve x = a cost, y = a sin t is symmetric about (i) x-axis (ii) y-axis (iii) both axis (iv) about y = x B. Fill in the blanks: 1 . The nth derivative of y = xn–1 log x at x = 2 . The nth derivative of y = x sin x is .......... 3 . If y = ex sin x, then y″ – 2y′ + 2y = .......... 4 . If y = sin 2x then y6 (0) = ........... 1 is .......... 2 92 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 5 . If y = sin hx, then y2n (x) = .......... 6 . If y = ex.x, then yn(x) = ........... 7 . If y = sin3 x, then yn (x) = .......... 8 . The nth derivative (yn) of y = x2 sin x at x = 0 is .......... 9 . If y = tan–1 x and n is even, then yn (0) = .......... 1 0 . If y = cos(m sin–1 x) and n is odd then yn (0) = .......... 1 1 . If y = (sin–1 x)2 and n is odd then yn (0) = .......... 1 2 . If z = f(x – by) + φ(x + by), then b 2 1 3 . If u = log(2x + 3y), then ∂2z = .......... ∂x 2 ∂ 2u = ........... ∂x∂y 1 4 . If u = eax + by f(ax – by), then b ∂u ∂u +a = .......... ∂x ∂y 1 5 . If u = x2 + y2, x = s + 3t, y = 2s – t, then ∂u = .......... ∂s 1 6 . If f(x, y) be a homogeneous function of degree n, then x 2 1 7 . If log u = ∂x + 2xy 2 ∂2 f ∂2 f + y 2 2 = .......... ∂x∂y ∂y ∂u ∂u x y , then x = .......... +y ∂y ∂x x+y ∂u ∂u ∂u +y +z = .......... ∂x ∂y ∂z 1 9 . If f(x, y) = 0, φ(y, z) = 0, then ∂f ∂φ dz = .......... ∂y ∂z dx 2 0 . If x = er cos θ, y = er sin θ then x 3 − x 2 y + xy 2 + y 3 x − xy − y 2 2 2 . If u = log ∂2 f 2 2 1 8 . If u = (x2 + y2 + z2)1/2, then x 2 1 . If u = (U.P.T.U., 2008) 2 x4 − y4 x +y 3 3 , then ∂θ ∂θ = .......... and = .......... ∂y ∂x , then x x ∂u ∂u +y = .......... ∂x ∂y ∂u ∂u +y = .......... ∂x ∂y 2 3 . If tlim x(t) = ∞, tlim y(t) = ∞, then t = .......... is asymptote. →t →t 1 1 2 4 . The curve r = a (1 – cos θ) is symmetric about the .......... dy = .......... dx 2 6 . The curve x2y2 = a2 (y2 – x2) has tangents at origin .......... 2 5 . The tangent is parallel to x-axis if 2 7 . The curve y2 (a – x) = x2(a + x) exists if ..... x ..... 2 8 . The curve y2 (x2 + y2) + a2 (x2 – y2) = 0, cross y-axis at ..... and ..... 2 9 . If f(x, y) = ey log(1 + x), then expansion of this function about (0, 0) up to second degree term is .......... 3 0 . tan–1 {(0.9) (–1.2)} = .......... 93 DIFFERENTIAL CALCULUS-I 3 1 . If f(x, y) = ex . sin y, then b g = .......... ∂ 3 f 0, 0 ∂x 3 3 2 . Expansion of exy up to first order tem is .......... 3 3 . f(x, y) = f(1, 2) + ......... C. Indicate True or False for the following statements: 1 . If u, v are functions of r, s are themselves functions of x, y then a f = ∂au, vf × ∂bx, yg b g ∂b r , s g ∂a r , s f ∂ u, v ∂ x, y 2 . Geometrically the function z = f(x, y) represents a surface in space. 3 . If f(x, y) = ax2 + 2hxy + by2 then x ∂f ∂f +y = 2f. ∂x ∂y 4 . If z is a function of two variables then dz is defined as dz = ∂z ∂z dx + dy . ∂x ∂y 5 . If f(x 1, x 2, ..., x n ) be a homogeneous function then x1 ∂f ∂f ∂f + x2 + ... + xn ∂x1 ∂x 2 ∂xn = n (n – 1) f. 6 . If u = x2 + y2 x −y 2 2 + 4 , then x ∂u ∂u +y = 4. ∂x ∂y 7 . When the function z = f(x, y) differentiating (partially) with respect to one variable, other variable is treated (temporarily) as constant. 8 . To satisfied Euler’s theorem the function f(x, y) should not be homogeneous. ∂z ∂z 9 . The partial derivatives and are interpreted geometrically as the slopes of the ∂y ∂x tangent lines at any point. 10. (i) The curve y2 = 4ax is symmetric about x-axis. (ii) The curve x3 + y3 = 3axy is symmetric about the line y = – x. (iii) The curve x2 + y2 = a2 is symmetric about both the axis x and y. (iv) The curve x3 – y3 = 3axy is symmetric about the line y = x. 11. (i) If there is no constant term in the equation then the curve passes through the origin otherwise not. d2 y ≠ 0 is called an inflexion point. dx 2 (iii) If r2 is negative i.e., imaginary for certain values of θ then the curve does not exist for those values of θ. (ii) A point where π . 2 (i) Maclaurin’s series expansion is a special case of Taylor’s series when the expansion is about the origin (0, 0). (ii) Taylor’s theorem is important tool which provide polynomial approximations of real valued functions. (iv) The curve r = aeθ is symmetric about the line θ = 12. 94 A TEXTBOOK OF ENGINEERING MATHEMATICS—I (iii) Taylor’s theorem fail to expand f(x, y) in an infinite series if any of the functions fx(x, y), fxx(x, y), fxy(x, y) etc., becomes infinite or does not exist for any value of x, y in the given interval. (iv) f(x, y) = f(a, b) + LMax − af ∂ − ax − bf ∂ OP f(a, b) + ..... . ∂y Q N ∂x ANSWERS TO OBJECTIVE TYPE QUESTIONS A. Pick the correct answer: 1 . (iv) 2 . (iii) 4 . (iv) 5 . (i) 7 . (iv) 8 . (iv) 1 0 . (iv) 1 1 . (i) 1 3 . (iii) 1 4 . (i) B. Fill in the blanks: 1. 2 n −1 2. 3 . Zero a 6 . ex x + n 9 . Zero 12. ∂ 2z FG H x sin x + 4 . Zero f 3 . (iii) 6 . (i) 9 . (iii) 1 2 . (iii) 1 5 . (iii) IJ K FG H en − n2 j sin n2π 7 . Try yourself 8. 1 0 . Zero 1 1 . Zero ∂ 2u ∂y∂x 1 6 . n(n – 1)f 13. ∂y 2 1 5 . 2x + 4y IJ K nπ nπ − n cos x + 2 2 5 . sin hx 1 4 . 2 abu 1 7 . 3u log u 18. u ∂f ∂φ 1 9 . ∂x ⋅ ∂y 20. − 21. u 2 4 . initial line 22. 1 25. 0 2 3 . t1 26. y = ± x 27. – a < x < a 2 8 . (0, a) and (0, – a) 3 0 . – 0.823 31. 0 sin 2 θ cos 2 θ , y x x2 + xy 2 3 2 . e{1 + (x – 1) + (y – 1)} 29. x − L R ∂f O ∂f U 1 R ∂f U ∂f x − 1f + b y ⋅ 2g V +......P a S 3 3 . MSaz − 1f + b y ⋅ 2g V + ∂y W 2 T ∂y W ∂x MNT ∂x PQ 2 C. True or False: 1. F 5. F 9. T 1 0 . (i) T 1 1 . (i) T 1 2 . (i) T 2. T 6. F 3. T 7. T 4. T 8. F (ii) F (iii) T (iv) F (ii) F (ii) T (iii) T (iii) T (iv) F (iv) F GGG UNIT 11 Differential Calculus-II 2.1 JACOBIAN The Jacobians* themselves are of great importance in solving for the reverse (inverse functions) derivatives, transformation of variables from one coordinate system to another coordinate system (cartesian to polar etc.). They are also useful in area and volume elements for surface and volume integrals. 2.1.1 Definition If u = u (x, y) and v = v (x, y) where x and y are independent, then the determinant ∂u ∂x ∂v ∂x ∂u ∂y ∂v ∂y is known as the Jacobian of u, v with respect to x, y and is denoted by a f b g ∂ u, v or J (u, v) ∂ x, y as Similarly, the Jacobian of three functions u = u (x, y, z), v = v (x, y, z), w = w (x, y, z) is defined J (u, v, w) = a b f g ∂ u, v, w = ∂ x, y , z ∂u ∂x ∂v ∂x ∂w ∂x * Carl Gustav Jacob Jacobi (1804–1851), German mathematician. 95 ∂u ∂y ∂v ∂y ∂w ∂y ∂u ∂z ∂v ∂z ∂w ∂z 96 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 2.1.2 Properties of Jacobians 1. If u =u (x, y) and v = v (x, y), then a f × ∂ bx, yg = 1 or JJ′′ = 1 ∂ a u , vf ∂ b x , yg ∂ bx, yg ∂ au, vf and J′ = J = ∂ au, vf ∂ bx, yg ∂ u, v where, Proof: Since u = u (x, y) v = v (x, y) Differentiating partially equations (i) and (ii) w.r.t. u and v, we get ∂y ∂u ∂u ∂x ∂u =1 = · + × ∂u ∂x ∂u ∂y ∂u Now, a f b g ∂u =0 = ∂v ∂y ∂u ∂x ∂u · + × ∂v ∂x ∂v ∂y ∂v =0 = ∂u ∂v =1 = ∂v ∂v ∂v ∂x ∂y · + ∂y × ∂x ∂u ∂u ∂v ∂x ∂ y ∂v · + × ∂x ∂v ∂v ∂y b g = a f ∂ x, y ∂ u, v × ∂ x, y ∂ u, v ∂u ∂x ∂v ∂x ∂u ∂y ∂v ∂y × ∂x ∂u ∂y ∂u (U.P.T.U., 2005) ...(i) ...(ii) ...(iii) ...(iv) ...(v) ...(vi) ∂x ∂v ∂y ∂v = ∂u ∂u ∂y ∂x ∂x ∂y ∂ ∂ u u ∂v ∂v × ∂x ∂y ∂x ∂y ∂v ∂v (By interchanging rows and columns in II determinant) = ∂u ∂x ∂u ∂y + ∂x ∂u ∂y ∂u ∂v ∂x ∂v ∂y + . ∂x ∂u ∂y ∂u ∂u ∂x ∂u ∂y + ∂x ∂v ∂y ∂v ∂v ∂x ∂v ∂y . + ∂x ∂v ∂y ∂v (multiplying row-wise) Putting equations (iii), (iv), (v) and (vi) in above, we get a f b g b g = a f 1 0 0 =1 1 JJ′ = 1. Hence proved. ∂ x, y ∂ u, v × ∂ x, y ∂ u, v or 2. Chain rule: If u, v, are function of r, s and r, s are themselves functions of x, y i.e., u = u (r, s), v = v (r, s) and r = r (x, y), s = s (x, y) a f = b g ∂ u, v ∂ x, y then Proof: Here and u = r = a f a f a f b g ∂ u, v ∂ r, s ⋅ ∂ r, s ∂ x, y u (r, s), v = v (r, s) r (x, y), s = s (x, y) 97 DIFFERENTIAL CALCULUS-II Differentiating u, v partially w.r.t. x and y ∂u ∂u ∂r ∂u ∂s = · + · ∂x ∂r ∂x ∂s ∂x ∂u ∂u ∂r ∂u ∂s = · + · ∂y ∂r ∂y ∂s ∂y ∂v ∂x ∂v ∂y Now, = = ...(ii) ∂v ∂s ∂v ∂r · + · ∂s ∂x ∂r ∂x ∂v ∂r ∂v ∂s · + · r y ∂ ∂ ∂s ∂y a f · ∂ar, sf = a f ∂bx, yg ∂u ∂r ∂v ∂r ∂u ∂s ∂v ∂s = ∂u ∂r ∂v ∂r ∂u ∂s ∂v ∂s ∂ u, v ∂ r, s ...(i) ...(iii) ...(iv) . ∂r ∂x ∂s ∂x ∂r ∂y ∂s ∂y . ∂r ∂x ∂r ∂y ∂s ∂x ∂s ∂y By interchanging the rows and columns in second determinant = ∂u ∂r ∂u ∂s + ∂r ∂x ∂s ∂x ∂v ∂r ∂v ∂s + ∂r ∂x ∂s ∂x ∂u ∂r ∂u ∂s + ∂r ∂y ∂s ∂y ∂v ∂r ∂v ∂s + ∂r ∂y ∂s ∂y |multiplying row-wise Using equations (i), (ii), (iii) and (iv) in above, we get a f a f a f = b g ∂ u, v ∂ r, s . ∂ r, s ∂ x, y ∂u ∂x ∂v ∂x ∂u ∂y ∂v ∂y = a f b g ∂ u, v . ∂ x, y Or a f = ∂ au, vf ⋅ ∂ ar, sf . Hence proved. ∂ ar, sf ∂ b x, yg b g ∂ au, vf Example 1. Find , when u = 3x + 5y, v = 4x – 3y. ∂ bx, yg ∂ u, v ∂ x, y Sol. We have ∴ Thus, u = 3x + 5y v = 4x – 3y ∂u ∂u ∂v ∂v = 3, = 5, = 4 and =–3 ∂x ∂y ∂x ∂y a f = b g ∂ u, v ∂ x, y ∂u ∂x ∂v ∂x ∂u ∂y ∂v ∂y = 3 4 5 = – 9 – 20 = – 29. −3 98 A TEXTBOOK OF ENGINEERING MATHEMATICS—I a b Sol. We have ∴ f g ∂ u, v, w of the following: ∂ x, y, z Example 2. Calculate the Jacobian u = x + 2y + z, v = x + 2y + 3z, w = 2x + 3y + 5z. u = x + 2y + z v = x + 2y + 3z w = 2x + 3y + 5z ∂u ∂u ∂u ∂v ∂v ∂v = 1, = 2, = 1, = 1, = 2, = 3, ∂x ∂y ∂z ∂x ∂y ∂z (U.P.T.U., 2007) ∂w ∂w ∂w = 2, ∂y = 3 and = 5. ∂x ∂z Now, a b ∂u ∂x ∂v ∂x ∂w ∂x f g ∂ u, v, w = ∂ x, y, z ∂u ∂y ∂v ∂y ∂w ∂y ∂u ∂z ∂v ∂z ∂w ∂z 1 = 1 2 2 2 3 1 3 5 = 1 (10 – 9) – 2 (5 – 6) + 1 (3 – 4) = 2. Example 3. Calculate b g if u = 2yz , v = 3zx , w = 4xy . y x z ∂ au, v, wf Sol. Given 3zx 2yz 4xy ,v= ,w= y x z u = ∂ x, y, z 3zx ∂v 3x −2 yz ∂u 2z ∂u 3z ∂v 2y ∂v ∂u = , = = , = , = , =− 2 , 2 , ∂z y x y y z x y y ∂x ∂ ∂ ∂ ∂ x x ∴ ∂w 4xy 4x 4y ∂w ∂w = , = and =– 2 ∂x z ∂y ∂z z z Now, a b f g ∂ u, v, w = ∂ x, y, z = – ∂u ∂y ∂v ∂y ∂w ∂y ∂u ∂z ∂v ∂z ∂w ∂z – = 2 yz 2 x 3z y 4y z 2z x −3 zx y2 4x z 2y x 3x y –4 xy z2 LM 2y L 12xz 12xyz O 2z L −12xyz − 12xy O 12x2 O − M P – + P M + zy2 PP 2 2 2 yz PQ x MN yz yz PQ x MN yz x MN y z Q 2 2 yz 12x yz 2 f g ∂ bx, y, zg ∂ au, v, wf But, we have × = 1 (Property 1) ∂ au, v, wf ∂ bx, y, zg ∂ bx, y, zg 1 . ∴ = 96 ∂ au, v, wf ⇒ a b ∂u ∂x ∂v ∂x ∂w ∂x ∂ u, v, w = 0 + 48 + 48 = 96. ∂ x, y, z 99 DIFFERENTIAL CALCULUS-II Example 4. If u = xyz, v = x2 + y2 + z2, w = x + y + z find J (x, y, z). Sol. Here we calculate J(u, v, w) as follows: J(u, v, w) = ∂u ∂x ∂v ∂x ∂w ∂x ∂u ∂y ∂v ∂y ∂w ∂y ∂u ∂z ∂v ∂z ∂w ∂z yz = 2x 1 zx 2y 1 (U.P.T.U., 2002) xy 2z 1 = yz (2y – 2z) – zx (2x – 2z) + xy (2x – 2y) = 2 [y2z – yz2 – zx2 + z2x + xy (x – y)] = 2 [– z (x2 – y2) + z2 (x – y) + xy (x – y)] As x 2 – y 2 = ( x – y )( x + y) = 2 (x – y)[– zx – zy + z2 + xy] But = 2 (x – y)[z (z – x) – y (z – x)] = 2 (x – y) (z – y) (z – x) = – 2 (x – y) (y – z) (z – x) J(x, y, z) · J(u, v, w) = 1 ∴ J (x, y, z) = Example 5. If x = vw , y = uv and u = r sin θ cos φ, wu , z = v = r sin θ sin φ, w = r cos θ, calculate b g. ∂ br, θ, φg ∂ x, y, z Sol. Here x, y, z are functions of u, v, w and u, v, w are functions of r, θ, φ so we apply IInd property. b b g ∂ bx, y, zg ∂ au, v, wf g = ∂ au, v, wf · ∂ br, θ, φg ∂ x, y, z ...(i) ∂ r, θ, φ Consider b g = ∂ au, v, wf ∂ x, y, z ∂x ∂u ∂y ∂u ∂z ∂u ∂x ∂v ∂y ∂v ∂z ∂v 1 2 0 = = 1 2 w u 1 2 v u 1 8 ∂x ∂w ∂y ∂w ∂z ∂w w v 0 1 2 u v 1 2 v w 1 2 u w 0 LM w v u + v w u OP 1 MN v u w w u v PQ = 8 1+ 1 = 2 = 1 8 4 100 A TEXTBOOK OF ENGINEERING MATHEMATICS—I b a g = f f = g ∂u ∂r ∂v ∂r ∂w ∂r ∂ x, y, z ⇒ Next ...(ii) ∂ u, v, w a b ∂ u, v, w ∂ r , θ, φ ∂u ∂θ ∂v ∂θ ∂w ∂θ ∂u sin θ cos φ ∂φ ∂v = sin θ sin φ ∂φ ∂w cos θ ∂φ r cos θ cos φ −r sin θ sin φ r cos θ sin φ r sin θ cos φ −r sin θ 0 = sinθ cos φ (r2 sin2 θ cos φ) – r cos θ cos φ (–r2 sin θ cos θ cos φ) + r2 sin θ sin φ (sin2 θ sin φ + cos2 θ sin φ) = r2 sin θ cos2 φ (sin2 θ + cos2 θ) + r2 sin θ sin2 φ a b f g ∂ u, v, w = r2 sin θ cos2 φ + r2 sin θ sin2 φ = r2 sin θ ∂ r, θ, φ Using (ii) and (iii) in equation (i), we get ⇒ ...(iii) b g = 1 × r sin θ = r sin θ . 4 ∂ br, θ, φg 4 ∂ x, y, z 2 2 Example 6. If u = x (1 – r2)−1/2, v = y (1 – r2)−1/2, w = z (1 – r2)−1/2 where Sol. Since a b f g ∂ u, v, w x 2 + y 2 + z 2 , then find ∂ x, y, z . r = r2 = x2 + y2 + z2 ∂r x ∂r y ∂r z = , = , = r ∂y r ∂x r ∂z 2 –1/2 Differentiating partially u = x (1 – r ) w.r.t. x, we get ∴ ∂u ∂x FG −1IJ (– 2r) e1 − r j . ∂r e j H 2K ∂x 1 x = e1 − r j + rx e1 − r j . = + r 1−r = 1 − r2 2 ⇒ ∂u ∂x = −1 2 2 +x −1 2 2 −3 2 −3 2 2 e j 3 1 − r2 2 and ∂u ∂y = x ∂u ∂y = ∂u ∂z = FG −1IJ e1 − r j · (– 2r) ∂r = xr · y H 2K ∂y e1 − r j r 2 xy 3 2 2 e1 − r j xz 3 2 2 e1 − r j −3 2 3 2 2 3 2 2 e1 − r j 1 − r2 + x2 Differentiating partially u w.r.t. y, we get ⇒ x2 101 DIFFERENTIAL CALCULUS-II ∂v ∂x Similarly, ∂w ∂x Thus, a b = yx e1 − r j yz 1 − r 2 + y 2 ∂v ∂v = 3 3 , ∂z = ∂y 1 − r2 2 1 − r2 2 3 , e 2 2 = zx e1 − r j ∂w = ∂y 3 , 2 2 e1 − r j xy 3 1 − r2 2 j yx f g = e1 − r j ∂ u, v, w ∂ x, y, z 3 2 2 e j 1−r + y e 2 3 2 2 1 9 1 − r2 2 j j 3 j 3 yz j e 1 − r2 2 1 − r2 + z2 zy e1 − r j e 1 − r2 2 3 1 − r2 2 e 2 1−r + z ∂w = 3 . ∂z 2 2 1−r xz 2 zx = 3 , 2 2 3 1 − r2 2 j 2 zy 1 − r 2 + x2 e e j 3 2 2 3 2 2 e1 − r j e1 − r j 1 − r2 + x2 xy yx 2 xz 1−r + y 2 yz 1−r + z zx zy e j 9 2 2 [(1 – r + x ) {(1 – r + y )(1 – r + z )– y z } 1 − r e j 2 − = 2 2 2 2 2 2 2 2 2 – xy {xy (1 – r2 + z2) – xyz2} + xz{xy2z – zx(1 – r2 + y2)}] = −9 1 − r2 2 e j [(1 – r2 + x2) (1 – r2 + y2) (1 – r2 + z2) – (1 – r2) (y2z2 + x2y2 + x2z2) – x2y2z2] −9 2 e1 − r j [(1 – r ) + (1 – r ) (x + y + z )] = e1 − r j [(1 – r ) + (1 – r ) r ] = e1 − r j . (1 – r ) [1 – r + r ] = e1 − r j . = 2 2 2 2 3 −9 2 2 3 −9 2 2 2 2 2 2 2 2 2 2 2 2 2 2 −5 2 Example 7. Verify the chain rule for Jacobians if x = u, y = u tan v, z = w. (U.P.T.U., 2008) Sol. We have x = u ⇒ y = u tan v ⇒ z = w ⇒ ∂x ∂x ∂x = 1, = = 0 ∂u ∂v ∂w ∂y ∂y ∂y = tan v, = u sec2 v, =0 ∂u ∂v ∂w ∂z ∂z ∂z = =1 = 0, ∂u ∂v ∂w 1 0 0 b g = tan v u sec2 v 0 = u sec v J = ∂ bu, v, wg 0 0 1 ∂ x, y, z 2 ...(i) 102 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Solving for u, v, w in terms of x, y, z, we have u = x v = tan–1 y y = tan −1 u x w = z y ∂u ∂u ∂u ∂v ∂v x ∂v ∂w ∂w ∴ , , = 1, = = 0, = − 2 = 2 = 0, = = 0 and 2 2 ∂x ∂y ∂z ∂x ∂y ∂z ∂x ∂y x +y x +y ∂w = 1 ∂z 1 0 x b g= − y J′ = ∂ bx, y, zg x2 + y2 ∂ u, v, w 2 x +y 0 0 0 0 = 2 x 2 x +y 2 = 1 1 F yI xG 1 + 2 J H xK 2 = 1 u sec2 v ...(ii) Hence from (i) and (ii), we get J.J′ = u sec2 v. 1 u sec 2 v = 1. 2.1.3 Jacobian of Implicit Functions If the variables u, v and x, y be connected by the equations f1 (u, v, x, y) = 0 f2 (u, v, x, y) = 0 i.e., u, v are implicit functions of x, y. Differentiating partially (i) and (ii) w.r.t. x and y, we get ...(i) ...(ii) ∂f1 ∂f1 ∂f1 ∂u ∂v + · + · = 0 ∂x ∂x ∂x ∂u ∂v ...(iii) ∂f1 ∂f1 ∂f1 ∂u ∂u ∂ y + ∂u · ∂ y + ∂v · ∂ y = 0 ...(iv) ∂f 2 ∂f 2 ∂f 2 ∂u ∂v + · + · = 0 ∂x ∂x ∂x ∂u ∂v ...(v) ∂f 2 ∂f ∂u ∂f ∂v + 2 · + 2 · = 0 ∂y ∂y ∂y ∂u ∂v ...(vi) Now, ∂ f1 b g × ∂ au, vf = ∂ u ∂f ∂ bx, yg ∂ au, vf ∂ f1 , f2 2 ∂u = ∂ f1 ∂v ∂ f2 ∂v ∂u ∂x ∂v ∂x ∂u ∂y ∂v ∂y ∂ f1 ∂ u ∂ f1 ∂v + ∂u ∂x ∂v ∂x ∂f1 ∂u ∂ f1 ∂ v + ∂ u ∂ y ∂v ∂ y ∂ f2 ∂ u ∂ f 2 ∂ v + ∂u ∂ x ∂ v ∂ x ∂f2 ∂u ∂ f2 ∂ v + ∂u ∂y ∂v ∂y 103 DIFFERENTIAL CALCULUS-II Using (iii), (iv), (v) and (vi) in above, we get ∂f ∂f − 1 − 1 ∂x ∂y = ∂ f2 ∂ f2 − − ∂x ∂y Thus, ⇒ = (– 1)2 b g × ∂ au, vf = (– 1) ∂ b f , f g ∂ bx, yg ∂ bx, yg ∂ au, vf ∂bf , f g ∂ bx, yg ∂ au, vf = (– 1) ∂ bx, yg ∂bf , f g ∂ au, vf ∂ f1 , f2 2 1 ∂ f1 ∂x ∂ f1 ∂y ∂ f2 ∂x ∂ f2 ∂y 2 1 2 1 2 2 Similarly for three variables u, v, w a b f = (– 1) g ∂ u, v, w ∂ x, y, z b ∂ f1 , f 2 , f 3 3 b ∂ x , y, z b g ∂ f1 , f 2 , f 3 a ∂ u, v, w f g g and so on. Example 8. If u3 + v3 = x + y, u2 + v2 = x3 + y3, show that a f b g y2 − x2 ∂ u, v . = 2uv u − v ∂ x, y Sol. Let f 1 ≡ u3 + v3 − x − y = 0 f 2 ≡ u2 + v2 − x3 − y3 = 0 ∂ f1 ∂ f1 –1 ∂x ∂y ∂ f1 , f2 Now, = = ∂ ∂ f f ∂ x, y 2 2 –3x 2 ∂x ∂y a f b g b g b ∂ f1 , f2 and a f ∂ u, v g= ∂ f1 ∂u ∂ f1 ∂v ∂ f2 ∂u ∂ f2 ∂v 3u2 = 2u –1 –3 y 3v 2 2v 2 = 3 (y2 – x2) = 6 uv (u – v) b g a f = ∂ bx, yg = 3 ey − x j = ( y − x ) . Hence Proved. b g ∂ b f , f g 6uv au – vf 2uv (u − v) ∂ au, vf ∂ f1 , f2 Thus, (U.P.T.U., 2006) ∂ u, v ∂ x, y 1 2 2 2 2 Example 9. If u, v, w are the roots of the equation in k, b g = – au − vfav − wfaw − uf . that aa − bfab − cfac − af ∂ au, v, wf ∂ x, y, z 2 x a+ k + y z + = 1, prove c+ k b+ k 104 A TEXTBOOK OF ENGINEERING MATHEMATICS—I x y z + =1 a+ k c+ k b+ k x (b + k) (c + k) + y (a + k) (c + k) + z (a + k) (c + k) = (a + k) (b + k) (c + k) k3 + k2 (a + b + c – x – y – z) + k{bc + ca + ab – (b + c) x – (c + a) y – (a + b)z} + (abc – bcx – cay – abz) = 0 Since its roots are given to be u, v, w, so we have u + v + w = – (a + b + c – x – y – z) uv + vw + wu = bc + ca + ab – (b + c) x – (c + a)y – (a + b)z uvw = – (abc – bcx – cay – abz) Let f1 ≡ u + v + w + a + b + c – x – y – z = 0 f 2 ≡ uv + vw + wu – bc – ca – ab + (b + c) x + (c + a) y + (a + b)z = 0 f 3 ≡ uvw + abc – bcx – cay – abz = 0 Sol. We have or or Now, b + g= ∂ f1 , f2 , f3 b ∂ x, y , z g ∂ f1 ∂x ∂ f2 ∂x ∂ f3 ∂x ∂ f1 ∂y ∂ f2 ∂y ∂ f3 ∂y ∂ f1 ∂z ∂ f2 ∂z ∂ f3 ∂z 1 0 b + c a − b = bc c a − b –1 b+ c –bc b ∂ f1 , f 2 , f 3 and a f ∂ u, v, w g= ∂ f1 ∂v ∂ f2 ∂v ∂ f3 ∂v 0 a−c b a−c ∂ f1 ∂w ∂ f2 ∂w ∂ f3 ∂w 1 = v+w vw = = (a – b)(a – c)(b – c) 1 1 1 vw wu uv a v + w f a w + uf au + v f 0 0 u−v u−w a w u− v –1 a+ b –ab a f a f a f = a f a f ∂ f1 ∂u ∂ f2 ∂u ∂ f3 ∂u –1 c+ a –ca f va u − w f (C2 → C2 – C1, C3 → C3 – C1) = (u – v)(u – w)(v – w) Thus, ∴ b ∂ f1 , f2 , f3 g = – aa − bf aa − cfab − cf b g au − vfau − wf av − wf a f ∂ bx, y, zg au − vfav − wfau − wf . Hence proved. =– ∂ au, v, wf aa − bfab − cfaa − cf a b f g b g ∂ x, y , z ∂ u, v, w = (–1)3 ∂ f1 , f2 , f3 ∂ x, y, z ∂ u, v , w As JJ ′ = 1. Example 10. If u = 2axy, v = a (x2 – y2) where x = r cosθ, y = r sin θ, then prove that a f a f ∂ u, v = – 4a2r3. ∂ r, θ 105 DIFFERENTIAL CALCULUS-II Sol. We have u = 2axy, v = a (x2 – y2) ∂u ∂u 2ay ∂ u, v ∂x ∂y = ∂v = ∂v 2ax ∂ x, y ∂x ∂y 2ax a f = – 4a (x + y ) −2ay b g ∂ au, vf As x + y = r = − 4a r ∂ bx, yg ∂x ∂x cos θ –r sin θ ∂ bx, yg ∂r ∂θ = ∂y = =r y ∂ sin θ r cos θ ∂ ar, θf ∂r ∂θ ∂ au, vf ∂ bx, yg ∂ au, vf = · = (− 4a r ).r = − 4a r . Hence proved. ∂ ar, θf ∂ bx, yg ∂ ar, θf Now, 2 2 2 2 2 2 or and 2 2 Hence 2 2 2 3 Example 11. If u3 + v3 + w3 = x + y + z, u2 + v2 + w2 = x3 + y3 + z3, u + v + w = x2 + y2 + z2, then show that f b gb ga f g a fa fa f a b x− y y− z z− x ∂ u, v, w ⋅ = ∂ x, y, z u− v v− w w− u f 1 ≡ u3 + v3 + w3 – x – y – z = 0 f 2 ≡ u2 + v2 + w2 – x3 – y3 – z3 = 0 f 3 ≡ u + v + w – x2 – y2 – z2 = 0 Sol. Let Now, b ∂ f1 , f2 , f3 b ∂ x, y , z g g = = ∂f 1 ∂f 1 ∂f 1 ∂x ∂f 2 ∂y ∂f 2 ∂z ∂f 2 ∂x ∂f 3 ∂y ∂f 3 ∂z ∂f 3 ∂x ∂y ∂z –1 0 –3x2 3 x2 − y2 e –2x 2 ⇒ and b g = 0 3 x2 − z2 j e a f –1 –3y2 –2y j –1 –3z2 –2z C2 → C2 − C1 , C3 → C3 − C1 2 x− z = – 6[(x – y ) (x – z) – (x – z2) (x – y)] = – 6 (x – y) (x – z) [(x + y) – (x + z)] ∂ f1 , f2 , f3 b ∂ x, y , z b b 2 x− y –1 –3x2 –2x g ∂ f1 , f2 , f3 a f ∂ u, v, w 2 2 g = 6 (x – y) (y – z) (z – x) g = ∂f1 ∂u ∂f 2 ∂u ∂f 3 ∂u ∂f 1 ∂v ∂f 2 ∂v ∂f 3 ∂v ∂f1 ∂w ∂f 2 ∂w ∂f 3 ∂w = 3u 2 3v 2 3w 2 2u 2v 2w 1 1 1 106 A TEXTBOOK OF ENGINEERING MATHEMATICS—I b g 3 bw 2 − u2 g 2a v − u f 2a w − u f 3u 2 = 3 v2 − u2 2u 1 0 C2 → C2 − C1 , C3 → C3 − C1 0 Expand it with respect to third row, we get = 6[(v2 – u2) (w – u) – (w2 – u2) (v – u)] = 6 (v – u) (w – u) [(v + u) – (w + u)] b ∂ f1 , f2 , f3 ⇒ a f ∂ u, v, w Hence a b g = – 6 (u – v) (v – w) (w – u) b ∂ f1 , f2 , f3 g f = (–1) ∂ bx, y, zg = + 6 bx − ygby − zgaz − xf ∂ bf , f , f g 6 au − vfav − wfaw − uf g ∂ au, v, w f bx − ygby − zgaz − xf · Hence proved. = au − vfav − wfaw − uf ∂ u, v, w ∂ x, y, z 3 1 2 3 Example 12. u, v, w are the roots of the equation (x – a)3 + (x – b)3 + (x – c)3 = 0, find b g. ∂ ba, b, cg ∂ u, v, w Sol. We have (x – a)3 + (x – b)3 + (x – c)3 = 0 x3 – a3 – 3xa (x – a) + x3 – b3 – 3xb (x – b) + x3 – c3 – 3xc (x – c) = 0 or 3x3 – 3x2 (a + b + c) + 3x (a2 + b2 + c2) – (a3 + b3 + c3) = 0 Since u, v, w are the roots of this equation, we have u+v+w = a+b+c uv + vw + wu = a2 + b2 + c2 α +β+ γ = −b a As αβ + βγ + γα = c a a 3 + b3 + c 3 3 f1 ≡ u + v + w – a – b – c = 0 αβγ = − d a uvw = Let f 2 ≡ uv + vw + wu – a2 – b2 – c2 = 0 f 3 = uvw – Now a 3 + b3 + c 3 3 g = −−21a −−21b −−21c = −−21a 2aa0− bf 2aa0− cf , FG c2 → c2 − c1IJ H c3 → c3 − c1K ∂b a, b, cg − a 2 − b2 − c 2 − a 2 e a 2 − b2 j e a 2 − c 2 j b ∂ f1 , f 2 , f 3 = – 2{(a – b) (a2 – c2) – (a – c) (a2 – b2)} = – 2(a – b) (b – c) (c – a) and b ∂ f1 , f 2 , f 3 b ∂ u, v , w g g = v +1 w u +1 w v +1 u = v +1 w vw wu uv FG c → c − c IJ Hc → c − c K vw wau − vf vau − wf 0 0 u−v u−w , 2 2 1 3 3 1 107 DIFFERENTIAL CALCULUS-II = (u – v) v(u – w) – (u – w) w(u – v) = – (u – v) (v – w) (w – u) Thus b ∂ f1 , f 2 , f 3 a a g f = a−1f ∂aa, b, cf = − −2aa − bf ab − cf ac − af f ∂b f , f , f g − au − vfa v − wfaw − uf ∂au, v, wf aa − bfab − cf ac − af . = −2 au − vfav − wf aw − uf ∂ u, v, w ∂ a, b, c 3 1 2 3 2.1.4 Functional Dependence Let u = f1 (x, y), v = f2 (x, y) be two functions. Suppose u and v are connected by the relation f (u, v) = 0, where f is differentiable. Then u and v are called functionally dependent on one another ∂v ∂u ∂u ∂v (i.e., one function say u is a function of the second function v) if the , , and ∂y are not ∂x ∂y ∂x all zero simultaneously. Necessary and sufficient condition for functional dependence (Jacobian for functional dependence functions): Let u and v are functionally dependent then f (u, v) = 0 ...(i) Differentiate partially equation (i) w.r.t. x and y, we get ∂f ∂v ∂f ∂u · + ∂v ∂x ∂u ∂x ∂f ∂v ∂f ∂u · + · ∂v ∂y ∂u ∂y = 0 ...(ii) = 0 ...(iii) There must be a non-trivial solution for Thus, or or ∂f ∂f ≠ 0, ≠ 0 to this system exists. ∂u ∂v ∂u ∂x ∂u ∂y ∂v ∂x ∂v ∂y = 0 For non-trivial solution xAx = 0 ∂u ∂x ∂v ∂x ∂u ∂y ∂v ∂y = 0 Changing all rows in columns a f = 0 b g ∂ u, v ∂ x, y Hence, two functions u and v are “functionally dependent” if their Jacobian is equal to zero. 108 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Note: The functions u and v are said to be “functionally independent” if their Jacobian is not equal to zero i.e., J(u, v) ≠ 0 Similarly for three functionally dependent functions say u, v and w. J(u, v, w) = a f = 0. ∂ bx, y, zg ∂ u, v, w Example 13. Show that the functions u = x + y – z, v = x – y + z, w = x2 + y2 + z2 – 2yz are not independent of one another. Also find the relation between them. Sol. Here u = x + y – z, v = x – y + z and w = x2 + y2 + z2 – 2yz a b f = g ∂u ∂x ∂v ∂x ∂w ∂x ∂u ∂y ∂v ∂y ∂w ∂y = 1 1 2x 1 –1 2y − 2z ∂ u, v, w ∂ x, y, z Now, ∂u ∂z ∂v ∂z ∂w ∂z = 1 1 2x 1 –1 2y − 2z –1 1 2z − 2y 0 0 (C3 → C3 + C2) 0 = 0. Hence u, v, w are not independent. Again or u + v = x + y – z + x – y + z = 2x u – v = x + y – z – x + y – z = 2 (y – z) ∴ (u + v)2 + (u – v)2 = 4x2 + 4 (y – z)2 = 4(x2 + y2 + z2 – 2yz) = 4w 2 2 ⇒ (u + v) + (u – v) = 4w 2(u2 + v2) = 4w or u2 + v2 = 2w. Example 14. Find Jacobian of u = sin–1 x + sin–1 y and v = x 1 − y 2 + y 1 − x 2 . Also find relation between u and v. u = sin–1 x + sin–1 y, v = x 1 − y 2 + y 1 − x 2 Sol. We have a f = ∂bx, yg ∂ u, v Now, = − Next, xy 1−x 2 1− y 2 ∂u ∂x ∂v ∂x +1−1+ ∂u ∂y ∂v ∂y 1 1− x = 1 − y2 − xy 1−x 2 1 1 − y2 1 − y2 2 xy 1− x 2 − xy 1− y 2 + 1 − x2 = 0. Hence u and v are dependent. { 2 2 u = sin–1 x + sin–1 y ⇒ u = sin–1 x 1 − y + y 1 − x { } As sin −1 A + sin −1 B = sin −1 A 1 − B 2 + B 1 − A 2 ⇒ or x 1 − y 2 + y 1 − x2 = v v = sin u. sin u = } 109 DIFFERENTIAL CALCULUS-II Example 15. Show that ax2 + 2hxy + by2 and Ax2 + 2Hxy + By2 are independent unless a h b = = . A H B Sol. Let u = ax2 + 2hxy + by2 v = Ax2 + 2Hxy + By2 If u and v are not independent, then ∂u ∂x ∂v ∂x or ∂u ∂y ∂v ∂y = a f =0 b g ∂ u, v ∂ x, y 2ax + 2hy 2 Ax + 2Hy 2hx + 2by 2Hx + 2By =0 ⇒ (ax + hy) (Hx + By) – (hx + by) (Ax + Hy) = 0 ⇒ (aH – hA) x2 + (aB – bA) xy + (hB – bH) y2 = 0 But variable x and y are independent so the coefficients of x2 and y2 must separately vanish and therefore, we have aH – hA = 0 and hB – bH = 0 i.e., a A i.e, = h b a h = and = H B A H h b = · Hence proved. H B Example 16. If u = x2 e–y cos hz, v = x2 e–y sin hz and w = 3x4 e–2y then prove that u, v, w are functionally dependent. Hence establish the relation between them. Sol. We have u = x2 e–y cos hz, v = x2 e–y sin hz, w = 3x4 e–2y a b ∂u ∂x ∂u ∂y ∂u ∂z 2xe − y cos hz −x 2 e − y cos hz x 2 e − y sin hz f = ∂v ∂x ∂v ∂y ∂v ∂z = 2xe sin hz −x e sin hz x e cos hz g ∂w ∂x ∂w ∂y ∂w ∂z 12x e −6x e 0 ∂ u, v, w ∂ x, y, z −y 3 −2 y 2 −y 2 −y 4 −2 y = 2x e–y cos hz{0 + 6x6 e–3y cos hz} + x2 e–y cos hz{0 – 12x5 e–3y cos hz} + x2 e–y sin hz{–12x5 e–3y sin hz + 12x5 e–3y sin hz} = 12x7 e–4y cos h2 z – 12x7 e–4y cos h2 z = 0 Thus u, v and w are functionally dependent. Next, 3u2 – 3v2 = 3(x4 e–2y cos h2 z – x4 e–2y sin h2 z) = 3x4 e–2y (cos h2 z – sin h2 z) = 3x4 e–2y ⇒ 3u2 – 3v2 = w. EXERCISE 2.1 1 . If x = r cos θ, y = r sin θ find b g. ∂ ar, θf ∂ x, y LM Ans. 1 OP rQ N x 2x 3 x x xx , y2 = 3 1 , y3 = 1 2 show that the Jacobian of y1, y2, y3 with respect to x1, x1 x2 x3 x2, x3 is 4. (U.P.T.U., 2004(CO), 2002) 2 . If y1 = 110 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 3 . If x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ, show that b g = r sin θ. ∂ br, θ, φg ∂ x, y, z 2 4 . If u = x + y + z, uv = y + z, uvw = z, evaluate e j b g b g (U.P.T.U., 2000) b g. ∂ au, v, wf ∂ x, y, z 2 (U.P.T.U., 2003) Ans. u v LMAns. − y OP 2x Q N ∂ u, v x2 + y2 y2 5 . If u = ,v= , find ∂ x , y . 2x 2x 6 . If x = a cos h ξ cos η, y = a sin h ξ sin η, show that b g = 1 a (cos h 2ξ – cos 2η). 2 ∂ bξ, ηg ∂ x, y 2 7 . If u3 + v + w = x + y2 + z2, u + v3 + w = x2 + y + z2, u + v + w3 = x2 + y2 + z, then evaluate a b LM b1 − 4xy − 4yz − 4zx + 6xyzg OP Ans. MM 27u2v2w2 + 2 − 3 eu2 + v2 + w2 j PP Q N f g ∂ u, v, w . ∂ x, y, z f g L −2(x − y) ( y − z) ( z − x) O (U.P.T.U., 2001) MAns. au − vfav − wfaw − uf P N Q 8 . If u, v, w are the roots of the equation (λ – x)3 + (λ – y)3 + (λ – z)3 = 0 in λ, find a b ∂ u, v, w . ∂ x, y, z 9 . If u = x1 + x2 + x3 + x4, uv = x2 + x3 + x4, uvw = x3 + x4 and uvwt = x4, show that b ∂ x1 , x2 , x3 , x4 g = u v w. a f ∂ bx, yg ∂ au, vf 1 0 . Calculate J = and J′ = . Verify that JJ′ = 1 given ∂ bx, yg ∂ au, vf ∂ u, v, w, t (i) u = x + 3 2 y2 y2 . ,v= x x Ans. J = e 2 u , J ′ = e −2 u . (ii) x = eu cos v, y = eu sin v. 1 1 . Show that a f a f LMAns. J = 2 y , J ′ = x OP x 2y Q N ∂ u, v = 6r3 sin 2θ given u = x2 – 2y2, v = 2x2 – y2 and x = r cos θ, y = r sin θ. ∂ r, θ 1 2 . If X = u2v, Y = uv2 and u = x2 – y2, v = xy, find a fL b g MN j OPQ ∂ X, Y 2 2 2 2 2 2 2 . Ans. 6x y x + y x – y ∂ x, y e je a f Ans. – 6 e x cos y b g ∂ au, v, wf Ans. – 2 1 4 . Find , if u = 3x + 2y – z, v = x – y + z, w = x + 2y – z. ∂ bx, y, zg Ans. bx − ygb y − zga z − xf 1 5 . Find J (u, v, w) if u = xyz, v = xy + yz + zx, w = x + y + z. 1 3 . Find ∂ u, v, w , if u = x2, v = sin y, w = e–3z. ∂ x, y, z −3 z 111 DIFFERENTIAL CALCULUS-II 1 6 . Prove that u, v, w are dependent and find relation between them if u = xey sin z, v = xey Ans. dependent, u 2 + v 2 = w cos z, w = x2e2y. b g b g f g 2 y+z x− y 3x 2 Ans. dependent, uvw 2 = 1 ,v= . 2 , w = x 2 y+z 3 x−y 1 8 . If X = x + y + z + u, Y = x + y – z – u, Z = xy – zu and U = x2 + y2 – z2 – u2, then show ∂ X, Y, Z, U that J = = 0 and hence find a relation between X, Y, Z and U. ∂ x, y, z, u b g a b 17. u = Ans. XY = U + 2Z 1 9 . If u = x y− z ,v= y z ,w= , then prove that u, v, w are not independent and also x− y z−x Ans. uv + vw + wu + 1 = 0 find the relation between them. 2 0 . If u = x + 2y + z, v = x – 2y + 3z, w = 2xy – xz + 4yz – 2z2, show that they are not Ans. 4w = u 2 − v 2 independent. Find the relation between u, v and w. a f b g ∂ u, v x+y and v = tan–1 x + tan–1 y, find . Are u and v functionally related? If ∂ x, y 1 − xy yes find the relationship. [Ans. yes, u = tan v] 2 1 . If u = g = u v. f ∂au, vf x −y 2 3 . If x + y + u – v = 0 and uv + xy = 0 prove that . = ∂bx, yg u +v 2 2 . If u = x + y + z, uv = y + z, uvw = z, show that 2 2 2 b a ∂ x, y, z 2 ∂ u, v, w 2 2 2 2 2 2 4 . Find Jacobian of u, v, w w.r.t. x, y, z when u = yz xy zx . ,v = ,w = z x y [Ans. 4] 2.2 APPROXIMATION OF ERRORS Let u = f (x, y) then the total differential of u, denoted by du, is given by du = ∂f ∂f dx + ∂y dy ∂x ...(i) If δx and δy are increments in x and y respectively then the total increment δu in u is given by or But δu = f (x + δx, y + δy) = δu ≈ ∴ From (i) δu ≈ f (x + δx, y + δy) – f (x, y) f (x, y) + δu du, δx ≈ dx and δy ≈ dy ...(ii) ∂f ∂f δx + δy ∂x ∂y ...(iii) 112 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Using (iii) in (ii), we get the approximate formula f (x + δx, y + δy) ≈ f (x, y) + ∂f ∂f δx + ∂y δy ∂x ...(iv) Thus, the approximate value of the function can be obtained by equation (iv). Hence δx or dx = Absolute error δx x and 100 × or dx = x Proportional or Relative error dx dx or 100 × = x x Percentage error in x. 1 2 Example 1. If f (x, y) = x y 10 , compute the value of f when x = 1.99 and y = 3.01. (U.P.T.U., 2007) 1 Sol. We have f (x, y) = ∴ Let ∂f ∂x x 2 y 10 9 ∂f 1 2 − 10 = x y ∂y 10 1 = 2xy 10 , As x = 2, δx = – 0.01 a f a f x + δx = 2 + –0.01 = 1.99 y + δy = 1 + 2.01 = 3.01 y = 1, δy = 2.01 Now, f (x + δx, y + δy) = f (x, y) + ∂f ∂f δx + ∂y δy ∂x af 1 ⇒ f{2 + (– 0.01), 1 + 2.01} = f (2, 1) + 2 × 2 1 10 × (– 0.01) + af 9 1 − (2)2. 1 10 × (2.01) 10 1 ⇒ f (1.99, 3.01) ≈ 2 × 1 10 + (– 0.04) + 0.804 ≈ 4 – 0.04 + 0.804 = 4.764. 2 Example 2. The diameter and height of a right circular cylinder are measured to be 5 and 8 cm. respectively. If each of these dimensions may be in error by ± 0.1 cm, find the relative percentage error in volume of the cylinder. Sol. Let diameter of cylinder = x cm. height of cylinder = y cm. then V = πx2 y 4 (radius = ∴ πxy ∂V ∂V πx2 = , = ∂x 2 4 ∂y ⇒ dV = ∂V ∂V dx + dy ∂x ∂y x ) 2 113 DIFFERENTIAL CALCULUS-II ⇒ dV = 1 1 π xy. dx + πx2. dy 2 4 1 1 2 πxy. dx πx dy dV dy dx 2 + 4 2 = = 2. + 2 V x y x y π π x y or or 100 × FG H 4 dV dx = 2 100 × V x 4 IJ + 100 × dy y K Given x = 5 cm., y = 8 cm. and error dx = dy = ± 0.1. So 100 × IJ K FG H 01 . 01 . dV + = ± 100 2 × 5 8 V = ± 5.25. Thus, the percentage error in volume = ± 5.25. Example 3. A balloon is in the form of right circular cylinder of radius 1.5 m and length 4 m and is surmounted by hemispherical ends. If the radius is increased by 0.01 m and the length by 0.05 m, find the percentage change in the volume of the balloon. [U.P.T.U., 2005 (Comp.), 2002] Sol. Let radius = r = 1.5 m, δr = 0.01 m height = h = 4 m, δh = 0.05 m OLS 2 3 2 3 4 3 πr + πr = πr2h + πr 3 3 3 ∂V ∂V = 2πrh + 4πr2, = πr2 ∂r ∂h ∂V ∂V dV = dr + dh = (2πrh + 4πr2) dr + πr2dh ∂r ∂h dV 2πr h + 2r πr 2 = dr + dh V 4r 4r 2 2 OLS>ª volume (V) = πr2 h + ∴ R>ª OLS a f F I F I Fig. 2.1 πr h + πr h + H 3K H 3K 3 × 2a h + 2r f 3 3 = dr + dh = [2(h + 2r) dr +rdh] ra 3h + 4r f ra 3h + 4rf a3h + 4rf or = 3 [2(4 + 3)(0.01) + 1.5 (0.05)] 15 . 12 + 6 a f 1 0.215 [0.14 + 0.075] = 9 9 0.215 dV 100 × = 100 × = 2.389% 9 V = ⇒ Thus, change in the volume = 2.389%. δr = dr δh = dh 114 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 4. Calculate the percentage increase in the pressure p corresponding to a reduction 1 of % in the volume V, if the p and V are related by pV1.4 = C, where C is a constant. 2 Sol. We have pV1.4 = C Taking log of equation (i) log p + 1.4 log V = log C Differentiating ...(i) dp 1 1.4 dV · dp + · dV = 0 ⇒ = – 1.4 p p V V or 100 × FG H dp dV = −1.4 100 × V p = −1.4 × IJ K FG −1IJ H 2K 1 dV 1 = %= 2 × 100 V 2 = 0.7 Hence increase in the pressure p = 0.7%. Example 5. In estimating the cost of a pile of bricks measured as 6′ × 50′ × 4′, the tape is stretched 1% beyond the standard length. If the count is 12 bricks to one ft3, and bricks cost Rs. 100 per 1000, find the approximate error in the cost. (U.P.T.U., 2004) Sol. Let length (l) = x ...(i) breadth (b) = y height (h) = z ∴ V = lbh = xyz or log V = log x + log y + log z On differentiating or 1 dV = V 100 × ⇒ 100 × 1 1 1 dx + y y + dz x z dy dV dx dz = 100 × + 100 × + 100 × y V x z dV = 1+1+1=3 V 3V 3 6 × 50 × 4 dV = = = 36 cube fit 100 100 a ∴ f Number of bricks in dV = 36 × 12 = 432 Hence, the error in cost = 432 × .100 = Rs. 43.20. 1000 Example 6. The angles of a triangle are calculated from the sides a, b, c of small changes δa, δb, δc are made in the sides, show that approximately δA = a [δa – δb cos C – δc. cos B] 2∆ where ∆ is the area of the triangle and A, B, C are the angles opposite to a, b, c respectively. Verify that δA + δB + δC = 0. (U.P.T.U., 2001) 115 DIFFERENTIAL CALCULUS-II Sol. From trigonometry, we have b 2 + c2 − a2 ⇒ b2 + c2 – a2 = 2bc cos A 2bc = b2 + c2 – 2bc cos A cos A = ⇒ a2 On differentiating, we get a · da ⇒ bc sin A · dA ⇒ 2∆ · dA ⇒ 2 ∆ · dA = = = = b · db + c · dc – db · c cos A – b · dc cos A + bc sin A · dA a · da – b · db – c · dc + db · c cos A + b · dc cos A a · da – (b – c cos A) · db – (c – b cos A) dc a da – (a cos C + c cos A – c cos A) db – (a cos B + b cos A – b cos A) dc 1 bc sin A 2 a = b cos C + c cos B As ∆ = 2 ∆ · dA = a da – db.a cos C – dc.a cos B or ⇒ δA = a [δa – δb. cos C – δc. cos B] 2∆ As δA ≈ dA δa ≈ da, δb ≈ db δc ≈ dc Hence proved. Similarly, δB = δC = and Adding δA, δB and δC, we get δA + δB + δC = b [δb – δc. cos A – δa. cos C] 2∆ c [δc – δa. cos B – δb. cos A] 2∆ 1 [(a – b cos C – c cos B) δa + (b – a cos C – c cos A) δb 2∆ + (c – a cos B – b cos A) δc] = ⇒ 1 [(a – a) δa + (b – b) δb + (c – c) δ c] 2∆ δA + δB + δC = 0. Verified. Example 7. Show that the relative error in c due to a given error in θ is minimum when θ = 45° if c = k tan θ. Sol. We have c = k tanθ ...(i) On differentiating, we get dc = k sec2 θ dθ ...(ii) 2 2 d θ dc sec θdθ From (i) and (ii), we get = = c sin 2θ tan θ Thus, dc will be minimum if sin2θ is maximum c i.e., sin 2θ = 1 = sin 90 ⇒ 2θ = 90 ⇒ θ = 45°. Hence proved. Since sin θ lies between – 1 and 1 116 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 8. Find approximate value of a0.98f + a2.01f + a1.94f . Sol. Suppose f (x, y, z) = e x + y + z j ∂f ∂f ∴ = x ex + y + z j , = y ex + y + z j ∂y ∂x 2 2 12 2 2 2 Now 2 12 2 2 −1 2 2 2 2 2 −1 2 , ∂f 2 2 2 −1 2 = z x +y +z ∂z e j ∂f ∂f ∂f δx + ∂y δy + δz ∂x ∂z x = 1, y = 2, z = 2, δx = – .02, δy = .01, δz = –.06. f (x + δx, y + δy,z + δz) = f (x, y, z) + Let ⇒ e f (0.98, 2.01, 1.94) = 12 + 2 2 + 2 2 j 12 + (9) −1 2 af a −1 2 (.01) +2 9 −1 2 −0.06 × ( −0.02) + 2 (9) f 1 (0.02 – 0.02 + 0.12) 3 = 3 – 0.04 = 2.96. = 3– Example 9. Prove that the relative error of a quotient does not exceed the sum of the relative errors of dividend and the divisor. Sol. Let x = dividend y = divisor z = quotient x Then = z y Taking log on both sides log x – log y = log z dy dz dx – = z y x ⇒ the relative error in quotient is equal to difference of the relative errors of dividend and divisor. dz Hence (relative error) in quotient does not exceed the sum of relative errors of dividend z and the divisor dz dx dy i.e., < + . Hence proved. z x y Differentiating Example 10. The work that must be done to propel a ship of displacement D for a distance 2 S2 D 3 . Find approximately the increase of work necessary when the ‘S in time ‘t’ proportional to t2 displacement is increased by 1%, the time diminished by 1% and the distance diminished by 2%. Sol. Let the work = W 2 then S2D 3 W ∝ t2 117 DIFFERENTIAL CALCULUS-II ⇒ W = K· S 2 D −2 3 , where K is proportional constant. t2 Taking log on both sides 2 log D – 2 log t 3 dW dS 2 dD dt On differentiating = 2 + –2 W S 3 D t dt dW dS 2 dD 100 × ⇒ 100 × = 2 100 × + – 2 100 × t S D W 3 −4 2 = 2 (–2) + (1) – 2 (–1) = 3 3 −4 ∴ Approximate increase of work = %. 3 Example 11. The height h and semi-vertical angle α, of a cone are measured from there A, the total area of the cone, including the base, is calculated. If h and α are in error by small quantities δh and δα respectively, find the corresponding error in the area. Show further that, if π , an error of + 1 per cent in h will be approximately compensated by an error of –19.8 α= 6 minutes in α. Sol. Total area of the cone A = πr2 + πrl log W = log K + 2 log S + IJ K FG H IJ K FG H IJ K FG H A = πh2 tan2 α + π (h tan α) (h sec α) or = πh2 (tan2α + tan α sec α) r = tan α ⇒ r = h tan α h l = sec α ⇒ l = h sec α h δA = 2πh δh (tan2 α + tan α sec α) Differentiating Fig. 2.2 + πh2 (2tan α sec2 α. δα + sec3α δα + tan2 α · sec α · δα) δA = 2πh tan α.δh (tan α + sec α) + πh2 (sec2 α + 2.tan α sec α or + tan2 α) sec α . δα = 2πh tan α. δh (tan α + sec α) + πh2 (sec α + tan α)2 sec α.δα LM N δA = πh2 (tan α + sec α) 2tan α. Now, putting 1 3 . 1 + 100 f OP Q π δh , × 100 = 1 and δA = 0 in above. 6 h π π π δh 1 π π π + tan + sec . sec . δα 2 tan 100 × . 0 = πh2 tan + sec 6 h 100 6 6 6 6 6 α= FG H ⇒ 2· a δh + tan α + sec α sec α. δα . h IJ LM KN FG 1 + 2 IJ · 2 · δα = 0 H 3 3K 3 FG H IJ K FG H IJ K OP Q 118 ⇒ A TEXTBOOK OF ENGINEERING MATHEMATICS—I 1 100 3 + FG 3IJ δα= 0 ⇒ δα = – 1 radian. H 3K 100 3 δα = – or =– =– 1 100 3 9 5π 3 × 180 degree π × 60 minutes (1° = 60 minutes) 9 × 60 = – 19.858 minutes. Hence proved. 5 × 3.14 × 1.732 Example 12. If the sides and angles of a plane triangle vary in such a way that its circum da db dc + + = 0 where da, db, dc are small cos A cos B cos C increaments in the sides a, b, c respectively. C Sol. Let R be the circum radius. a We know that R = R 2sin A ∂R 1 a × cos A ∂R ∴ = , =– O ∂ A 2sin A 2 sin 2 A ∂a ∂R ∂R ⇒ dR = da + dA ∂a ∂A B A a cos A 1 da – . dA = 2 sin 2 A 2sin A Fig. 2.3 1 a cos A or 0 = da − . dA As R = constant 2sin A 2 sin A radius remains constant, prove that RS T ⇒ da − UV W a cosA da a dA = 0 ⇒ = . dA cosA sinA sinA da = 2R dA cosA db Similarly, = 2R dB cosB dc = 2R dC cosC Adding these equations, we get da db dc + + = 2R (dA + dB + dC) cosA cosB cosC A+ B+ C = π ∴ dA + dB + dC = 0 From (i) and (ii), we get da db dc + + = 0. Hence proved. cosA cosB cosC ⇒ ...(i) ...(ii) 119 DIFFERENTIAL CALCULUS-II Example 13. Considering the area of a circular ring as an increment of area of a circle, find approximately the area of the ring whose inner and outer radii are 3 cm. and 3.02 cm. respectively. Sol. Let area of a circle = πr2 or A = πr2 ⇒ dA = 2πr dr ...(i) Now area of circular ring δA is the difference between A1 of outer ring and A2 of inner sphere choose r = 3 cm. and δr = 0.02, then ⇒ ⇒ A1 – A2 = A (r + δr) – A (r) = δA ≈ dA A1 – A2 = dA = 2πr.dr (from i) A1 – A2 = 2 π × 3 × .02 = .12 π cm2. O Example 14. Calculate (2.98)3 Sol. Let f (x, y) = yx ⇒ Now, Let ⇒ ∂f ∂x 3.02 3 cm 0.02 ∂f = yx. log y, ∂y = xyx–1 ∂f ∂f f (x + δx, y + δy) = f (x, y) + δx + ∂y δy ∂x Fig. 2.4 x = 3, δx = – 0.02, y = 3, δy = 0 f (2.98, 3) = f (3, 3) + 33. log 3 × (– 0.02) + 0 = 33 – 33 × (.4771213) × 0.02 log 3 = .4771213 = 27 (1 – 0.00954) = 26.74. EXERCISE 2.2 1. The period T of a simple pendulum is T = 2π possible errors up to 1% in l and 2.5 % in g. l . Find the maximum error in T due to g (U.P.T.U., 2003) Ans. 1.75% 2. In the estimating the number of bricks in a pile which is measured to be (5 m × 10 m × 5 m), count of bricks is taken as 100 bricks per m3. Find the error in the cost when the tape is stretched 2% beyond its standard length. The cost of bricks is Rs. 2,000 per thousand bricks. (U.P.T.U., 2000) Ans. Rs. 3000 3. Find the approximately value of f(0.999) where f(x) = 2x4 + 7x3 – 8x2 + 3x + 1. LMHint: f a0.999f = f bx + δxg = f a x f + ∂f . δx = f a1f + f ′ a1fa –0.001fOP ∂x N Q Ans. 4.984 120 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 1 mv2e find approximate change in T as the mass 2 m changes from 49 to 49.5 and the velocity v changes from 1600 to 1590. 4. If the kinetic energy T is given by T = Ans. 144000 units 5. Find the approximate value of (1.04)3.01. Ans. 1.12 6. If ∆ be the area of a triangle, prove that the error in ∆ resulting from a small error in c is given by 1 1 1 ∆ 1 – + + δ∆ = δc 4 s s−a s−b s− c 7. Considering the volume of a spherical shell as an increment of volume of a sphere, calculate approximately the volume of a spherical shell whose inner diameter is 8 inches 1 inch. and whose thickness is Ans. 4π cubic inches 6 8. A diameter and altitude of a can in the form of right circular cylinder are measured as 4 cm. and 6 cm. respectively. The possible error in each measurement is 0.1 cm. Find approximately the maximum possible error in the value computed for the volume and LM N OP Q Ans. 5.0336 cm 3 , 3.146 cm 2 lateral surface. x2 y2 9. Find the percentage error in calculating the area of ellipse 2 + 2 = 1, when error of a b Ans. 2% + 1 % is made in measuring the major and minor axis. 5 10. The quantity Q of water flowing over a v-notch is given by the formula Q = cH 2 where H is the head of water and c is a constant. Find the error in Q if the error in H is 1.5%. Ans. 3.75% 11. Find the percentage error in calculated value of volume of a right circular cone whose altitude is same as the base radius and is measured as 5 cm. with a possible error of 0.02 cm. Ans. 1.2% 12. Find possible percentage error in computing the parallel resistance r of three resistance 1 1 1 1 r1, r2, r3 from the formula = + + if r1, r2, r3 are each in error by plus 1.2%. 2 r r r r 1 3 Ans. 12% 13. The diameter and the height of a right circular cylinder are measured as 4 cm. and 6 cm. respectively, with a possible error of 0.1 cm. Find approximately the maximum possible error in the computed value of the volume and surface area. Ans. 1.6 π cu. cm, π sq. cm 14. Find a3.82f + 2a2.1f 2 1 3 5 Ans. = 2.012 . 15. Show that the acceleration due to gravity is reduced nearly by 1% at an altitude equal to 0.5 % of earth’s radius. Given that an external point x kilometers from the earth’s centre, F rI such an acceleration is given by g GH JK , where r is the radius of the earth. x 2 16. Calculate the error in R if RI = E and possible errors in E and I are 20% and 10% respectively. Ans. 10% 121 DIFFERENTIAL CALCULUS-II 17. In the manufacture of closed cylindrical boxes with specified sides a, b, c (a ≠ b ≠ c) small changes of A %, B %, C % occurred in a, b, c respectively from box to box from the specified dimension. However, the volume and surface area of all boxes were according to specification, show that A B C = = a b−c b c−a c a−b a f a f a f a f 1 18. Find 83.7 4 . Ans. 3.025 19. Find y (1.997) where y(x) = x4 – 2x3 + 9x + 7. Ans. 24.949 20. The time T of a complete oscillation of a simple pendulum of length l is governed by T l = 2π g where g is a constant. (a) Find approximate error in the calculated value of T corresponding to an error of 2% in the value of L. (U.P.T.U., 2008) Ans. 1% (b) By what percentage should the length be changed in order to correct a loss of 2 Ans. – 0.278% minutes per day? 2.3 EXTREMA OF FUNCTION OF SEVERAL VARIABLES Introduction In some practical and theoretical problems, it is required to find the largest and smallest values of a function of two variables where the variables are connected by some given relation or condition known as a constraint. For example, if we plot the function z = f(x, y) to look like a mountain range, then the mountain tops or the high points are called local maxima of f(x, y) and valley bottoms or the low points are called local minima of f(x, y). The highest mountain and lowest valley in the entire range are said to be absolute maximum and absolute minimum. The graphical representation is as follows. Absolute maximum Z Local maximum Y O X Absolute minimum Local minimum Fig. 2.5 Definition Let f (x, y) be a function of two independent variables x, y such that it is continuous and finite for all values of x and y in the neighbourhood of their values a and b (say) respectively. 122 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Maximum value: f(a, b) is called maximum value of f(x, y) if f (a, b) > f(a + h, b + k). For small positive or negative values of h and k i.e., f(a, b) is greater than the value of function f (x, y,) at all points in some small nbd of (a, b). Minimum value: f(a, b) is called minimum value of f(x, y) if f (a, b) < f(a + h, b + k). Note: f(a + h, b + k) – f (a, b) = positive, for Minimum value. f(a + h, b + k) – f (a, b) = negative, for Maximum value. Extremum: The maximum or minimum value of the function f(x, y) at any point x = a and y = b is called the extremum value and the point is called “extremum point”. Geometrical representation of maxima and minima: The function f(x, y) represents a surface. The maximum is a point on the surface (hill top). The minimum is a point on the surface (bottom) from which the surface ascends (climbs up) in every direction. Z = f (x, y) Z = f (x, y) P (maximum) Q (minimum) Y Y X X (a) (b) Fig. 2.6 Saddle point: It is a point where function is neither maximum nor minimum. At such point f is maximum in one direction while minimum in another direction. Example: z = xy, hyperbolic paraboloid has a saddle point at the origin. Remark 1 : If f(x, y) ≤ f(a, b) where (x, y) is a neighbourhood of (a, b). The number f(a, b) is called local maximum value of f(x, y). Remark 2 : If f(x, y) ≥ f(a, b) where (x, y) is a neighbourhood of f(a, b). The number f(a, b) is called local minimum value of f(x, y). 2.3.1 Condition for the Existence of Maxima and Minima (Extrema) By Taylor’s theorem f (a + h, b + k) = f (a, b) + F h ∂f + k ∂f I GH ∂x ∂y JK + ( a, b ) 1 2 F h ∂ f + 2 hk ∂ f + k ∂ f I GH ∂x J ∂x∂y ∂y K 2 2 2 2 2 2 +.... ...(i) 2 ( a, b ) Neglecting higher order terms of h2, hk, k2, etc. Since h, k are small, the above expansion reduce to ∂f a, b ∂f a, b f (a + h, b + k) = f (a, b) + h + k ∂y ∂x a f ∂f a a, bf ∂f a a, bf ⇒ f (a + h, b + k) – f (a, b) = h + k ∂x ∂y a f ...(ii) 123 DIFFERENTIAL CALCULUS-II The necessary condition for a maximum or minimum value (L.H.S. of eqn (ii) negative or positive) is h a f + k ∂f a a, bf = 0 ∂f a, b ∂x ∂y a f = 0, ∂f a a, bf = 0 h and k can take both + ve and – ve value ...(iii) ∂f a, b ⇒ ∂y ∂x The conditions (iii) are necessary conditions for a maxmium or a minimum value of f(x, y). Note: The conditions given by (iii) are not sufficient for existence of a maximum or a minimum value of f(x, y). 2.3.2 Lagrange’s Conditions for Maximum or Minimum (Extrema) Using the conditions (iii) in equation (i) (2.3.1) and neglecting the higher order term h3, k3, h2 k etc. we get f (a + h, b + k) – f (a, b) = Putting ∂2 f ∂x 2 = f (a + h, b + k) – f (a, b) = = LM MN OP PQ ∂2 f ∂2 f ∂2 f 1 + k2 2 h 2 2 + 2hk ∂x∂y 2 ∂y ( a, b ) ∂x r, ∂2 f ∂2 f = s, = t, then ∂x∂y ∂y 2 1 [h2r + 2hks + k2 t] 2 LM h r + 2hkrs + k tr OP r N Q L O 1 M a hr + ksf + k ert − s j P PQ r 2M N 1 2 2 2 2 2 ⇒ f (a + h, b + k) – f (a, b) = 2 2 ...(iv) If rt – s2 > 0 then the numerator in R.H.S. of (iv) is positive. Here sign of L.H.S. = sign of r. Thus, if rt –s2 > 0 and r < 0, then f (a + h, b + k) – f (a, b) < 0 if rt – s2 > 0 and r > 0, then f (a + h, b + k) – f (a, b) > 0. Therefore, the Lagranges conditions for maximum or minimum are: (U.P.T.U., 2008) 1. If rt –s2 > 0 and r < 0, then f (x, y) has maximum value at (a, b). 2. If rt –s2 > 0 and r > 0, then f (x, y) has minimum value at (a, b). 3. If rt –s2 < 0, then f (x, y) has neither a maximum nor minimum i.e., (a, b) is saddle point. 4. If rt –s2 = 0, then case fail and here again investigate more for the nature of function. 2.3.3 Method of Finding Maxima or Minima 1. Solve ∂f ∂f = 0 and ∂y = 0, for the values of x and y. Let x = a, y = b. ∂x The point P (a, b) is called critical or stationary point. 124 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 2. Find r, s and t at x = a, y = b. 3. Now check the following conditions: (i) If rt – s2 > 0 and r < 0, f (x, y) has maximum at x = a, y = b. (ii) If rt – s2 > 0 and r > 0, f (x, y) has minimum at x = a, y = b. (iii) If rt – s2 < 0, f(x, y) has neither maximum nor minimum. (iv) If rt – s2 = 0, case fail. Example 1. Find the maximum and minimum of u = x3 + y3 – 63(x + y) + 12xy. Sol. ∂u ∂u = 3x2 – 63 + 12y; ∂y = 3y2 – 63 + 12x ∂x r = ∂ 2u ∂ 2u ∂ 2u = 12, t = = 6y 2 = 6x, s = ∂x ∂y 2 ∂x∂y Now for maximum and minimum value ⇒ and ∂u ∂u = 0 and ∂y = 0 ∂x 2 3x + 12y – 63 = 0 ⇒ x2 + 4y – 21 = 0 3y + 12x – 63 = 0 ⇒ y + 4x – 21 = 0 Subtracting (i) and (ii), we get 2 2 ...(i) ...(ii) (x – y) (x + y) – 4 (x – y) = 0 ⇒ (x – y) (x + y – 4) = 0 ⇒ x = y and x + y = 4 Putting x = y in (i), we get x2 + 4x – 21 = 0 ⇒ (x + 7) (x – 3) = 0 ∴ Again putting x = 3, x = – 7 y = 3, y = – 7 y = 4 – x in eqn (i), we get ∴ x2 + 4 (4 – x) – 21 = 0 ⇒ x2 – 4x – 5 = 0 ⇒ (x – 5) (x + 1) = 0 ⇒ x = 5, x = – 1 ∴ y = 4 – 5 = – 1, y = 4 – (–1) = 5. Hence (3, 3), (5, –1), (–7, –7) and (–1, 5) may be possible extremum. At x = 3, y = 3, we have r = 18; s = 12, t = 18 ∴ rt – s2 = 18 × 18 – (12)2 > 0 and r = 18 > 0. So there is minima at x = 3, y = 3, and the minimum value of u is (3)3 + (3)3 – 63 (3 + 3) + 12 (3) (3) = – 216. At x = 5, y = – 1, we have r = 30; s = 12, t = – 6 ∴ rt – s2 = 30 (– 6) – (12)2 < 0, so there is neither maxima nor minima at x = 5, y = – 1. At x = – 7, y = – 7, we have r = – 42; s = 12, t = – 42 ∴ rt – s2 = (– 42) (– 42) – (12)2 > 0 and r < 0, so there is maxima at x = – 7, y = – 7 and its maximum value is (– 7)3 + (– 7)3 – 63 (– 7 – 7) + 12 (– 7) (– 7) = 784. At x = – 1, y = 5, we have r = – 6 ; s = 12, t = 30 ∴rt – s2 = (– 6) (30) – (12)2 < 0, so there is neither maxima nor minima at x = – 1, y = – 5. 125 DIFFERENTIAL CALCULUS-II Example 2. Show that minimum value of u = xy + Sol. a3 a3 + is 3a2. x y ∂u ∂u ∂2u = y − a 3 x −2 ; = x − a 3 y −2 , r = 2 = 2 a 3 x −3 ; ∂x ∂y ∂x s = ∂2u ∂2u = 1; t = = 2 a 3 y −3 . ∂ x∂ y ∂y 2 Now for maximum or minimum we must have So from ∂u ∂u =0, =0 ∂x ∂y ∂u = 0, we get y – a3x–2 = 0 or x2y = a3 ∂x ∂u = 0, we get x – a3y−2 = 0 or xy2 = a3 ∂y Solving (i) and (ii), we get x2y = xy2 or xy (x – y) = 0 and from or ...(i) ...(ii) x = 0, y = 0 and x = y. From (i) and (ii), we find that x = 0 and y = 0 do not hold as it gives a = 0, which is against hypothesis. ∴ We have x = y and from (i) we get x3 = a3 or x = a and therefore, we have x = a = y. This satisfies (ii) also. Hence it is a solution. at At x = a = y, we have r = 2a3a–3 = 2, s = 1, t = 2 ∴ rt – s2 = (2) (2) – 12 = 3 > 0 Also r = 2 > 0. Hence, there is minima at x = a = y ∴ The minimum value of u = xy + (a3/x) + (a3/y) x = a=y = a.a + (a3/a) + (a3/a) = a2 + a2 + a2 = 3a2. Hence proved. Example 3. Discuss the maximum or minimum values of u when u = x3 + y3 – 3axy. (U.P.T.U., 2004) 2 ∂ u ∂u ∂u = 3x2 – 3ay; ∂y = 3y2 – 3ax; r = = 6x; Sol. ∂x ∂x2 ∂2u ∂2u s = ∂x∂y = −3a, t = 2 = 6 y. ∂y ∂u ∂u Now for maximum or minimum, we must have ∂x = 0, ∂y = 0 ∂u So from = 0, we get x2 – ay = 0 ∂y ∂u and from = 0, we get y2 – ax = 0 ∂y Solving (i) and (ii), we get (y2/a)2 – ay = 0 or y4 – a3y = 0 or y (y3 – a3) = 0 or y = 0, a. Now from (i), we have when y = 0, x = 0, and when y = a, x = ± a. ...(i) ...(ii) 126 A TEXTBOOK OF ENGINEERING MATHEMATICS—I But x = – a, y = a, do not satisfy (ii), here are not solutions. Hence the solutions are x = 0, y = 0; x = a, y = a; At x = 0, y = 0, we have r = 0, s = – 3a, t = 0. ∴ rt – s2 = 0 – (–3a)2 = negative and there is neither maximum nor minimum at x = 0, y = 0. At x = a, y = a, we get r = 6a, s = – 3a, t = 6a ∴ rt – s2 = (6a)(6a) – (–3a)2 = 36a2 – 9a2 > 0 Also r = 6a > 0 if a > 0 and r < 0 if a < 0. Hence there is maximum or minimum according as a < 0 or a > 0. The maximum or minimum value of u = – a3 according as a < 0 or a > 0. Example 4. Determine the point where the function u = x2 + y2 + 6x + 12 has a maxima or minima. ∂u ∂x Sol. = 2x + 6; r ≡ ∂u = 2y ∂y ∂2 u ∂2u ∂2u = 2; s ≡ = 0; t ≡ =2 ∂x∂y ∂x 2 ∂y 2 ∂u ∂u = 0, = 0. ∂x ∂y Now for maxima or minima we must have From ∂u = 0, we get 2x + 6 = 0 ∂x or x = – 3 ∂u From ∂y = 0, we get 2y = 0 or y = 0 Also at x = – 3, y = 0, r = 2, s = 0, t = 2 ∴ rt – s2 = 2 (2) – (0)2 = 4 > 0 and r = 2 > 0 Hence, there is minima at x = –3, y = 0. Example 5. A rectangular box, open at the top, is to have a volume of 32 c.c. Find the dimensions of the box requiring least material for its construction. (U.P.T.U., 2005) Sol. V = 32 c.c. Let length = l, breadth = b and height = h Total surface area S = 2lh + 2bh + lb ...(i) S = 2(l + b)h + lb Now volume V = lbh = 32 ⇒ b = 32 lh Putting the value of ‘b’ in equation (i) S = 2 F l + 32 I h + l F 32 I H lh K H lh K 64 32 + l h 64 ∂S 32 = 2l − 2 = 2h – 2 , ∂h l h S = 2lh + ∴ ∂S ∂l ...(ii) h b l Fig. 2.7 ...(iii) 127 DIFFERENTIAL CALCULUS-II For minimum S, we get ∂S ∂l 64 32 =0 ⇒h= 2 2 l l ∂S 32 16 = 0 ⇒ 2l – 2 = 0 ⇒ l = 2 and ∂h h h From (iv) and (v), we get = 0 ⇒ 2h – Putting h = 2, in equation (v), we get l = Now, and ∂l 2 ∂ 2S = 2 ⇒ s = 2 and ∂ l ∂h ∴ rt – s2 16 =4 4 32 =4 4×2 b = ∂2S ...(v) 32 × h 4 ⇒ h3 = 8 ⇒ h = 2 256 h = From (ii) ...(iv) 128 128 = = 2 ⇒r=2>0 64 l3 = ∂2S 64 64 = =8 ⇒t = 8 8 h3 ∂h = 2 × 8 – 4 = 12 > 0 = 2 ⇒ rt – s2 > 0 and r > 0 Hence, S is minimum, for least material l = 4, b = 4, h = 2. Example 6. Examine the following surface for high and low points z = x2 + xy + 3x + 2y + 5. Sol. ∂z ∂x = 2x + y + 3; r = ∂z = x+2 ∂y ∂2 z ∂2 z ∂2 z = = = = =0 s t 2 ; 1 ; ∂x∂y ∂x 2 ∂y 2 For the maximum or minimum we must have ∂z ∂x From From ∂z ∂x ∂z ∂y = 0, ∂z = 0. ∂y = 0, we get 2x + y + 3 = 0 ...(i) = 0, we get x + 2 = 0. ...(ii) From (ii), we get x = – 2 and ∴ from (i) y = – 3 + 4 = 1. Hence, the solution is x = – 2, y = 1 and for these values we have r = 2, s = 1, t = 0. ∴ rt – s2 = (2) (0) – (1)2 = – 1 < 0. ∴ There is neither maximum nor minimum at x = – 2, y = 1. 128 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 7. Discuss the maximum and minimum values of 2 sin + cos (x + y). Sol. Let ∴ 1 1 (x + y) cos (x – y) 2 2 1 1 (x + y) cos (x – y) + cos (x + y) 2 2 = sin x + sin y + cos (x + y) ∂u = cos x – sin (x + y); = cos y – sin (x + y) ∂y u = 2 sin ∂u ∂x r = ∂2 u ∂2u = – sin x – cos (x + y); s = = – cos (x + y); ∂x 2 ∂x∂y t = ∂2u = – sin y – cos (x + y) ∂y 2 For maximum or minimum, we must have ∂u ∂u = 0, =0 ∂x ∂y ∂u = 0, we get cos x – sin (x + y) = 0 ∂x ∂u From ∂y = 0, we get cos y – sin (x + y) = 0. Solving (i) and (ii), we get cos x = cos y which gives x = 2nπ ± y, where n is any integer. In particular x = y. From ...(i) ...(ii) When x = y, from (i), we get cos x – sin 2x = 0 or cos x (1 – 2 sin x) = 0 which gives cos x = 0, sin x = 1 1 , then x = nπ + (–1)n. π = y, 2 6 and for these value of x and y, we have If sin x = 1 2 x= y LM N 1 1 – cos 2 nπ + ( −1) n . π 3 2 Similarly, t < 0 and s < 0 and r > s, t > s. ∴ rt – s2 > 0. Also r < 0. r = – Hence, there is a maximum when x = nπ + (–1)n. OP < 0. (Note) Q 1 π = y. 6 1 π = y, x = y 2 1 From here, we get x = y = ± π , (3/2)π, (5/2)π etc. 2 1 If x = π = y, then r = – 1 + 1 = 0, s = 1, t = 0 2 If cos x = 0, then x = 2nπ + ∴ rt – s2 < 0. Hence, there is neither maximum nor minimum at x = 1 π = y. 2 129 DIFFERENTIAL CALCULUS-II 1 If x = − π = y, then r = 1 + 1 = 2 = t, s = 1 2 ∴ rt – s2 = (2 × 2) –1 = 3 > 0. Also r > 0 1 π = y. In a similar way we can discuss other values too. 2 Example 8. Show that the distance of any point (x, y, z) on the plane 2x + 3y – z = 12, from the origin is given by Hence, there is minimum at x = – l = x 2 + y 2 + ( 2x + 3 y − 12) 2 . Hence find a point on the plane that is nearest to the origin. Sol. l = distance between (x, y, z) and (0, 0, 0) = ( x − 0 ) 2 + ( y − 0) 2 + ( z − 0 ) 2 = x 2 + y 2 + (2x + 3 y − 12) 2 , z = 2x + 3y – 12 x2 + y 2 + z2 = l2 = x2 + y2 + (2x + 3y – 12)2 = u (say) l2 = u = 5x2 + 10y2 + 12xy – 48x – 72y + 144 or or ∴ ∂u ∂x = 10x + 12y – 48; ...(i) ∂u = 20y + 12x – 72 ∂y ∂2 u ∂ 2u ∂2u = 10; s = = 12 ; t = = 20 ∂ y∂ x ∂x 2 ∂y 2 = (10) (20) – (12)2 = 56 > 0 and r > 0, so there is a minimum value of l. r = ∴ Also and rt – s2 ∂u ∂x ∂u ∂y = 0 ⇒ 10x + 12y – 48 = 0 or 5x + 6y = 24 ...(ii) = 0 ⇒ 20y + 12x – 72 = 0 or 5y + 3x = 18 ...(iii) F 12 I , y = F 18 I H7K H7K F 12 I + 3 F 18 I – z = 12 2x + 3y – z = 12 or 2 H7K H7K Solving (ii) and (iii), we get x = Also or 24 + 54 – 7z = 84 or 7z = 24 + 54 – 84 = – 6 ∴ The required point is or z = − F 12 , 18 , − 6 I . H 7 7 7K Example 9. Discuss the maximum and minimum values of Sol. Let Then x4 + 2x2y – x2 + 3y2. u = x4 + 2x2y – x2 + 3y2 ∂u ∂x = 4x3 + 4xy – 2x ; ∂u = 2x2 + 6y; ∂y 6 7 130 A TEXTBOOK OF ENGINEERING MATHEMATICS—I ∴ ∂2 u ∂2u ∂ 2u = 12x2 + 4y – 2; s = = 4x; t = =6 2 ∂x ∂y 2 ∂ x∂ y r = ∂u ∂u = 0, = 0. ∂y ∂x For maximum and minimum, we must have From ∂u = 0, we get 2x(2x2 + 2y – 1) = 0 ∂x or 2x2 + 2y – 1 = 0 ...(i) ∂u From ∂y = 0, we get 2x2 + 6y = 0 or x2 + 3y = 0 Solving (i) and (ii), we get 4y + 1 = 0 or y = – ∴ From (ii), we get x2 = – 3y = ∴ The solutions are x = When x = ...(ii) 1 4 F 3 I or x = ± 1 3 H 4K 2 1 1 1 1 3,y=– 3,y=– and x = – 2 4 2 4 1 1 3,y=– , we get 2 4 r = 12 ∴ rt – s2 = 6 × 6 – F 3 I + 4F − 1 I − 2 = 6 , s = 4 FH 1 3 IK = 2 3 , t = 6 H 4K H 4K 2 e2 3 j > 0. Also r > 0 2 ∴ There is a minimum when x = Again when x = – 1 1 3, y =– 2 4 1 1 3, y=– , we have r = 6, s = – 2 3 , t = 6 2 4 ∴ rt – s2 = (6) (6) – e−2 3 j > 0. Also r > 0. 2 Hence as before there is a minimum when 1 1 3,y=– x = . 2 4 Example 10. Find the shortest distance between the lines x − 3 y − 5 z −7 x +1 y +1 z+1 = = . = = and 1 −2 1 7 −6 1 x − 3 y − 5 z −7 = = Sol. Let = λ ⇒ x = λ + 3, y = 5 – 2λ, z = 7 + λ 1 −2 1 Thus any point P on the line is (3 + λ, 5 – 2λ, 7 + λ) x +1 y +1 z+1 = = µ ⇒ x = – 1 + 7µ, y = – 1 – 6µ , z = – 1 + µ = 7 −6 1 The point Q is (– 1 + 7 µ, – 1 – 6µ, – 1 + µ) ∴ Distance between these two lines is and let D = ⇒ b3 + λ + 1 − 7µg + ( 5 − 2λ + 1 + 6µ) + (7 + λ + 1 − µ) 2 D2 = u(Say) = 6λ2 +86µ2 – 40λµ + 105. 2 2 131 DIFFERENTIAL CALCULUS-II ∂u ∂u = 12λ – 40µ, = 172µ – 40λ. ∂λ ∂µ ∂u ∂u = 0 and = 0 ⇒ 12λ – 40µ = 0, 172µ – 40λ = 0. But ∂λ ∂µ Solving these equations, we get λ = 0, µ = 0 ∴ ∂2u ∂2 u ∂2u = 172 = 12, t = , = s = – 40 ∂λ∂µ ∂λ2 ∂µ 2 Now, rt – s2 = (12)·(172) – (– 40)2 = 464 > 0 ⇒ rt – s2 > 0 and r > 0 Hence u occurs minimum value at λ = 0 and µ = 0. The shortest distance is given by r = D = 4 2 + 62 + 82 = 116 = 2 29 . Example 11. The temperature T at any point (x, y, z) in space is T (x, y, z) = Kxyz2 where K is a constant. Find the highest temperature on the surface of the sphere x2 + y2 + z2 = a2. (U.P.T.U., 2008) Sol. T = Kxyz2 ...(i) 2 2 2 2 2 2 2 2 x + y + z = a ⇒z = a – x – y From (i) T = Kxy (a2 – x2 – y2) ∂T ∂x ∴ = Ky (a2 – x2 – y2) – 2 Kx2y = Ky(a2 – 3x2 – y2) ∂T 2 2 2 ∂y = Kx(a – x – 3y ) for maximum and minimum value ∂T ∂T = 0 and = 0 ⇒ x = 0 and y = 0 ∂x ∂y or 3x2 + y2 = a2 x2 + 3y2 = a2 Similarly, Solving x=y=± r = a 2 ∂ 2T ∂2T = – 6 Kxy, s = = K (a2 – 3x2 – 3y2) ∂x 2 ∂ x∂ y ∂ 2T t = ∂y 2 = – 6Kxy and At (0, 0) r = 0, s = Ka2 and t = 0 ∴ rt – s2 = 0.0 – Ka2 = – Ka2 < 0 So, there is neither maximum nor minimum at x = 0 and y = 0 At a a a a ,y= and x = – ,y=– 2 2 2 2 3 a 2K 6 2 3 2 r = – Ka = – Ka < 0, t = − Kxy, s = − 2 2 4 2 x = 132 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 9 2 4 a4K2 K a – = 2K2a4 > 0 4 4 ∴ rt – s2 > 0 and r < 0 Hence, T has a maximum value at x = + a/2 and y = ± a/2 rt – s2 = The maximum value of T = K · FG IJ = Ka . H K 8 4 a2 a2 4 2 Example 12. Find the maximum and minimum values of the function z = sin x sin y sin (x + y). Sol. Given z = sin x sin y sin (x + y) or = 1 [2 sin x sin y] sin (x + y) 2 = 1 [cos (x – y) – cos (x + y)] sin (x + y) 2 = 1 [2 sin (x + y) cos (x – y) – 2 sin (x + y) cos (x + y)] 4 z = ∴ 1 [sin 2x + sin 2y – sin(2x + 2y)] 4 1 ∂z = [cos 2x – cos (2x + 2y)] 2 ∂x ∂z 1 = [cos 2y – cos (2x + 2y)] ∂y 2 r = ∂2 z = – sin 2x + sin (2x + 2y) ∂x 2 ...(A) s = ∂2z = sin (2x + 2y) ∂x∂y ...(B) t = ∂2 z = – sin 2y + sin (2x + 2y) ∂y 2 ...(C) For maximum or minimum, we must have ∂z ∂z = 0, =0 ∂y ∂x From ∂z = 0, we get cos 2x – cos (2x + 2y) = 0 ∂x ...(i) From ∂z = 0, we get cos 2y – cos (2x + 2y) = 0 ∂y ...(ii) Solving (i) and (ii), we get cos 2x = cos 2y which gives 2x = 2nπ ± 2y. In particular 2x = 2y or x = y When x = y, from (i), we get cos 2x – cos 4x = 0 or cos 2x –(2 cos2 2x – 1) = 0 or 2 cos2 2x – cos 2x – 1 = 0 cos 2θ = 2 cos2 θ – 1 133 DIFFERENTIAL CALCULUS-II 1 ± (1 + 8 ) or cos 2x = or 2x = 2nπ ± 0, 2mπ ± or x = nπ, mπ ± 4 π 3 In particular x = = 1± 3 1 = 1, – 4 2 2π , where m, n are zero or any integers 3 π 3 π π , we have y = x = 3 3 2π 4π and then r = – sin + sin , from (A) 3 3 When x = 3 3 – =– 3; 2 2 4π s = sin , from (B) 3 = – 3 2 or s = – and t = – sin = – 2π 4π + sin , from (C) 3 3 3 – 2 3 =– 3 2 e− 3 j e− 3 j – FGH − 23 IJK = 3 – 34 = 94 2 ∴ rt – s2 = = positive. π π = y, rt – s2 > 0 , r < 0, so there is a maximum at x = = y. Thus at x = 3 3 π π π π + Hence, maximum value = sin . sin . sin 3 3 3 3 F H 3 3 3 3 3 ⋅ ⋅ = . 2 2 2 8 π π If we take x = – , then y = x = – 3 3 1 3,t= 3 and r = 3, s= 2 9 ∴ rt – s2 = >0, r>0 4 π There is a minimum at x = – = y. 3 I K = Hence, the minimum value = sin F − π I sin F − π I sin RSF − π I + F − π I UV H 3 K H 3 K TH 3 K H 3 K W 134 A TEXTBOOK OF ENGINEERING MATHEMATICS—I = – sin = − π π 2π · sin sin 3 3 3 3 3 3 3 3 ⋅ ⋅ = − · 2 2 2 8 Example 13. Find the local minima or local maxima of the function f(x, y) = 4x2 + y2 – 4x + 6y – 15. Sol. We have f(x, y) = 4x2 + y2 – 4x + 6y – 15 ∂f = 8x – 4 ∂x For stationary point and ∂f = 2y + 6. ∂y ∂f ∂f = 0 = 0 and ∂y ∂x 1 and 2y + 6 = 0 ⇒ y = – 3 2 Now, the given function can be written as f(x, y) = (2x – 1)2 + (y + 3)2 – 25 ...(i) 2 2 Since (2x – 1) ≥ 0 and (y + 3) ≥ 0. From (i), we get f(x, y) ≥ – 25 for all values of x and y. Hence f(1/2, – 3) = – 25 is a local minimum. ∴ 8x – 4 = 0 ⇒ x = Example 14. Find the local minima or local maxima of the function f(x, y) = – x2 – y2 + 2x + 2y + 16. ∂f = – 2x + 2 = 0 ⇒ x = 1 ∂x Sol. ∂f = – 2y + 2 = 0 ⇒ y = 1 ∂y f(x, y) = 18 – {(x – 1)2 + (y – 1)2} Now, ...(i) Since (x – 1) ≥ 0 and (y – 1) ≥ 0, from (i) we get f(x, y) ≤ 18 for all values of x and y. 2 2 Hence f(1, 1) = 18 is a local maxima. EXERCISE 2.3 1. Discuss the maximum values of u, where LMAns. x = a , y = a OP 2 2Q N 2. Find the points (x, y) where the function u = xy (1 – x – y) is maximum or minimum. LMAns. Maxima at x = 1 , y = 1 OP 3 3Q N u = 2a2xy – 3ax2y – ay3 + x3y + xy3 3. Find the extrema of f (x, y) = (x2 + y2) e(6x + 2x2). [Ans. minima at (0, 0) minimum value = 0 and at (–1, 0) min. value = e–4, 1 saddle point (– , 0)]. 2 135 DIFFERENTIAL CALCULUS-II 4. If the perimeter of a triangle is constant prove that the area of this triangle is maximum when the triangle is equilateral. [Hint: 2s = a + b + c, ∆ = s( s − a)( s − b)( s − c ) ] 2s ] 3 5. Show that the rectangular solid of maximum volume that can be inscribed in a sphere is a cube. [Ans. Maximum when a = b = c = [Hint: V = xyz, diagonal of cubic = V = xy x2 + y 2 + z2 = d ⇒ z = d2 − x2 − y2 OP Q d 2 − x 2 − y 2 so 6. Find the shortest distance from origin to the surface xyz2 = 2. [Ans. 2] 7. In a plane triangle ABC find the maximum value of cos A cos B cos C. [Ans. 1/8] 8. Discuss the maximum or minimum values of u given by u = x3y2 (1 – x – y). [Ans. Maximum at x = 0, y = 1/3] 9. Find the maximum and minimum values of u = 6xy + (47 – x – y) (4x + 3y). [Ans. Max. value of u = 3384] 10. Discuss the maxima and minima of the function f (x, y) = cos x, cos y cos (x + y) (U.P.T.U., 2007) 11. Divide 24 into three parts such that the continued product of the first, square of the second and the cube of the third may be maximum. [Ans. 4, 8, 12] 12. Examine for extreme values: u (x, y) = x2 + y2 + 6x + 12. [Ans. Min. value = 3] 2.4 LAGRANGE'S* METHOD OF UNDETERMINED MULTIPLIERS Let φ (x, y, z) is a function of three independent variables, where x, y, z are related by a known constraint g(x, y, z) = 0 Thus the problem is Extrema of u = f(x, y, z) ...(i) Subject to g (x, y, z) = 0 ...(ii) For stationary point ∂f ∂f ∂f = = =0 ∂x ∂z ∂y ∴ df = ∂f ∂f ∂f dx + dy + dz = 0 ∂x ∂z ∂y ∂g ∂g ∂g dx + dy + dz = 0 ∂z ∂y ∂x Multiplying eqn. (iv) by λ and adding to (iii), we obtain From (ii) dg = FG ∂f + λ ∂g IJ dx + FG ∂f + λ ∂g IJ dy + FG ∂f + λ ∂g IJ dz = 0 H ∂x ∂x K H ∂y ∂y K H ∂z ∂z K *Joseph Louis Lagrange (1736–1813). ...(iii) ...(iv) ...(v) 136 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Since x, y, z are independent variables ∴ ∂g ∂f +λ = 0 ∂x ∂x ∂f ∂g +λ ∂y ∂y = 0 ...(vi) ...(vii) ∂g ∂f +λ = 0 ...(viii) ∂z ∂z On solving (ii), (vi), (vii) and (viii), we can find x, y, z and λ for which f (x, y, z) has maximum or minimum. Notes: 1. The Lagrange's method of undetermined multiplier's introduces an additional unknown constant λ known as Lagrange's multiplier. 2. Nature of stationary points cannot be determined by Lagranges method. Example 1. Determine the maxima and minima of x2 + y2 + z2 when ax2 + by2 + cz2 = 1. Sol. Let f (x, y, z) = x2 + y2 + z2 ...(i) and g(x, y, z) ≡ ax2 + by2 + cz2 – 1 = 0 ...(ii) ∂f ∂f ∂f = 2x, ∂y = 2y, = 2z ∂x ∂z ∂g ∂g ∂g From (ii) = 2ax, ∂y = 2by, = 2cz. ∂x ∂z Now from Lagrange's equations, we get From (i) ∂f ∂g = 2x + λ . 2ax = 0 ⇒ 2x (1 + λa) = 0 ⇒ x (1 + λa) = 0 ...(iii) +λ ∂x ∂x ∂f ∂g = 0 ⇒ 2y + λ · 2by = 0 ⇒ 2y (1 + λb) = 0 ⇒ y ( 1 + λb) = 0 ...(iv) + λ ∂y ∂y and or ∂f ∂g = 0 ⇒ 2z + λ · 2cz = 0 ⇒ 2z (1 + λc) = 0 ⇒ z ( 1 + λb) = 0 ...(v) + λ ∂z ∂z Multiplying these equations by x, y, z respectively and adding, we get x2 (1 + λa) + y2 (1 + λb) + z2 (1 + λc) = 0 (x2 + y2 + z2) + λ (ax2 + by2 + cz2) = 0 ...(vi) Using (i) and (ii) in above equation, we get f+ λ = 0⇒λ=– f Putting λ = – f in equations (iii), (iv) and (v), we get ⇒ i.e., x (1 – fa) = 0, y (1 – fb) = 0, z (1 – fc) = 0 1 – fa = 0, 1 – fb = 0, 1 – fc = 0 f = 1 1 1 , , . These give the max. and min. values of f. a b c Example 2. Find the extreme value of x2 + y2 + z2, given that ax + by + cz = p. (U.P.T.U., 2007) Sol. Let u = x2 + y2 + z2 ...(i) Given ax + by + cz = p. ...(ii) 137 DIFFERENTIAL CALCULUS-II For max. or min. from (i), we have du = 2x dx + 2y dy + 2z dz = 0. ...(iii) Also from (ii), a dx + b dy + c dz = 0. ...(iv) Multiplying (iv) by λ and adding in (iii), we get (x dx + y dy + z dz) + λ (a dx + b dy + c dz) = 0. Equating the coefficients of dx, dy and dz to zero, we get x + λa = 0, y + λb = 0, z + λc = 0 These are Lagrange's equations. Multiplying these by x, y, z respectively and adding, we get ...(v) x (x + λa) + y (y + λb) + z (z + λc) = 0 or or (x2 + y2 + z2) + λ (ax + by + cz) = 0 u + λp = 0 or λ = – u/p. ∴ From (v), we get x– FG au IJ = 0, y – FG bu IJ = 0, z – FG cu IJ = 0 HpK HpK HpK x y u z = = = a p b c or x y z = = a b c or F x I + b FG y IJ + c F z I = p H aK H bK H cK F xI F xI F xI a H K + b H K + c H K = p, from (vi) a a a F x I = p or x = ap (a + b + c ) H aK a +b +c From (ii), we get a2 2 2 2 or 2 or 2 2 2 2 Similarly, bp y = a +b +c These give the minimum value of u. Hence minimum value of u is u = = 2 2 2 2 2 , z= a2 p2 (a 2 + b2 + c 2 )2 ( a 2 + b 2 + c 2 )p 2 ( a2 + b2 + c 2 ) 2 + = 2 cp a + b2 + c 2 2 b2 p2 = 1 and lx + my + nz = 0. Sol. Let Given u = x 2 y2 z2 + + a4 b4 c 4 x2 y2 z2 + + = 1 a2 b2 c2 lx + my + nz = 0 + ( a2 + b2 + c 2 )2 p2 (a2 + b2 + c 2 ) Example 3. Find the maximum and minimum values of and ...(vi) c 2 p2 (a 2 + b2 + c 2 )2 · x 2 y2 z2 x2 y2 z2 + + + + where a4 b4 c 4 a2 b2 c2 ...(i) ...(ii) ...(iii) 138 A TEXTBOOK OF ENGINEERING MATHEMATICS—I From (i), (ii) and (iii), we get FG x IJ dx + FG y IJ dy + FG z IJ dz = 0 Ha K Hb K Hc K FG x IJ dx + FG y IJ dy + FG z IJ dz = 0 Ha K Hb K Hc K and 4 4 4 ...(iv) 2 2 2 ...(v) l dx + m dy + n dz = 0 Multiplying (v) and (vi) by λ1, λ2 and adding, we get ...(vi) FG x + λ x + λ lIJ dx + FG y + λ y + λ mIJ dy + FG z + λ z + λ nIJ dz = 0 Ha a K Hb b K K Hc c 1 2 4 2 1 2 4 2 1 2 4 2 Equating to zero the coefficients of dx, dy and dz, we get x λ1x y λ y z λ z + 2 + λ 2 l = 0 , 4 + 12 + λ 2 m = 0 , 4 + 12 + λ 2 n = 0 a4 a c c b b These are Lagrange’s equations. Multiplying these by x, y, z and adding, we get ...(vii) FG x + y + z IJ + λ FG x + y + z IJ + λ (lx + my + nz) = 0 Ha b c K Ha b c K or or 2 2 2 4 4 4 or or or or or 2 2 2 2 2 2 u + λ1 (1) + λ2 (0) = 0, using (i), (ii) and (iii) u + λ1 = 0 or λ1 = – u ∴ From (vii), we have y uy x ux z uz − 2 + λ 2 l = 0, − + λ 2n = 0 − 2 + λ 2 m = 0, 4 4 a a c4 c2 b b x (1 – a2u) = – la4 λ2, y (1 – b2u) = – mb4 λ2, z (1 – c2u) = – nc4λ2 or or 1 2 −la 4 λ 2 − mb 4 λ 2 − nc 4 λ 2 , z= 2 , y = 2 1− a u 1−b u 1 − c2u Substituting these values in (iii), we get x = l2 a 4 m2 b 4 n2 c 4 + + =0 1 − a2 u 1 − b 2 u 1 − c 2 u Σl2 a4 (1 – b2u) (1 – c2u) = 0 Σl2 a4 {b2c2u2 – (b2 + c2) u + 1} = 0 u2 (Σl2a4b2c2) – u {Σl2a4 (b2 + c2)} + Σl2a4 = 0 RSΣl a FG 1 + 1 IJ UV u + Σl a = 0 T H c b KW R| F a a I U| F a l + b m + c n IJ = 0, (l a + m b + n c ) u – SΣl G + J V u + GH b c c a a b K |T H c b K |W 2 2 a2b2c2 (l2a2 + m2b2 + n2c2)u2 – a2b2c2 2 2 2 2 2 2 2 2 2 2 4 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 which gives the max. and min. values of u. Example 4. Find the minimum value of x2 + y2 + z2 when yz + zx + xy = 3a2. Sol. Let u = x2 + y2 + z2 Given yz + zx + xy = 3a2. For max. or min. from (i), we have ...(i) ...(ii) du = 2x dx + 2y dy + 2z dz = 0 ...(iii) 139 DIFFERENTIAL CALCULUS-II Also from (ii), (y dz + z dy) + (z dx + x dz) + (x dy + y dx) = 0 (z + y) dx + (x + z) dy + (y + x) dz = 0. ...(iv) Multiplying equation (iv), by λ and adding in (iii), we get [x + λ (z + y)] dx + [y + λ (x + z)] dy + [z + λ (y + x)] dz = 0 Equating the coefficients of dx, dy, dz to zero, we get x + λ (z + y) = 0, y + λ (x + z) = 0, z + λ (x + y) = 0 ...(v) These are Lagrange’s equations Multiplying these by x, y, z respectively and adding, we get [x2 + λx (z + y)] + [y2 + λy (x + z)] + [z2 + λz (y + x)] = 0 (x2 + y2 + z2) + 2λ (xy + yz + zx) = 0 or u + 2λ (3a2) = 0 or λ = − or or 6a2 u ( z + y) ∴ From (v), we get x = or u 6a 2 . ,y= u ( x + z) 6a 2 ,z= u ( y + x) 6 a2 x z u y = 2 = = z+y x + y 6a x+z 2 – 6a x + uy + uz = 0, ux – 6a2y + uz = 0, ux + uy – 6a2z = 0 Eliminating x, y, z, we get or −6 a 2 u u −6 a = 0, which gives the max. and min. values of u. u −6 a 2 u u −6 a 2 u + 6a 2 u + 6 a2 u −6 a 2 − u 0 0 −6 a − u −6 a 2 1 1 u −1 0 0 −1 u (u + 6a2)2 or u 2 2 u 2 2 = 0, C2 → C2 – C1 C3 → C3 – C1 = 0 or (u + 6a2)2 −6 a 2 1 1 u −1 0 0 1 −1 = 0, R3 → R3 – R2 2 or (u + 6a ) [– 6a – u(–1–1)] = 0 |Expand with respect to first column 2 2 2 2 2 or (u + 6a ) [– 6a + 2u] = 0 or u = – 6a , 3a . But u cannot be equal to – 6a2, since sum of three squares (viz. x2, y2, z2) from (i), cannot be negative. Hence u = 3a2 gives max. or min. value of u. Example 5. Find the minimum distance from the point (1, 2, 0) to the cone z2 = x2 + y2. (U.P.T.U., 2006) Sol. Let (x, y, z) be any point on the cone then distance from the point (1, 2, 0) is D2 = (x – 1)2 + (y – 2)2 + (z – 0)2 u = (x – 1)2 + (y – 2)2 + z2 Let Subject to 2 2 2 x + y – z = 0 ...(i) ...(ii) 140 A TEXTBOOK OF ENGINEERING MATHEMATICS—I for minimum, from (i) and (ii), we get du = (x – 1)dx + (y – 2)dy + zdz = 0 xdx + ydy – zdz = 0 ...(iii) ...(iv) Multiplying equation (iv) by λ and adding in (iii), we get (x – 1)dx + (y – 2)dy + zdz + λ (xdx + ydy – zdz) = 0 ⇒ ⇒ {x (1 + λ) –1} dx + {y (1 + λ) – 2} dy + {z (1 – λ)} dz = 0 x (1 + λ) –1 = 0, y ( 1 + λ) –2 = 0, z (1 – λ) = 0 ⇒ x = 1 , 1+λ y= 2 , λ=1 1+λ ...(v) 1 1 2 = , y= =1 2 1+1 1+1 Putting the value of x and y in equation (ii), we get ∴ x = 1 5 5 + 1 – z2 = 0 ⇒ z2 = ⇒z=± 4 4 2 Hence, the minimum distance from the point (1, 2, 0) is F 1 − 1I + (1 − 2) + FG 5 IJ = 1 + 1 + 5 = 10 D = H2 K H2K 4 4 4 2 2 2 2 D2 = or 5 ⇒ D= 2 5 . 2 Example 6. Show that the rectangular solid of maximum volume that can be inscribed in a sphere is a cube. (U.P.T.U., 2003) Sol. Let the length, breadth and height of solid are Z l = 2x b = 2y h = 2z ∴ Volume of the solid V = lbh = 2x·2y·2z ⇒ V = 8xyz Equation of the sphere ⇒ x2 + y2 + z2 = R2 x + y2 + z2 – R2 = 0 2 ...(i) ...(ii) For maximum differentiating (i), (ii), we get dV = 8yzdx + 8xzdy + 8xydz = 0 and 2xdx + 2ydy + 2zdz = 0 Multiplying (iv) by λ and adding in (iii), we get Y X Fig. 2.8 ...(iii) ...(iv) 8yzdx + 8xzdy + 8xydz + λ (2xdx + 2ydy + 2zdz) = 0 ⇒ (2λx + 8yz)dx + (2λy + 8xz)dy + (2λz + 8xy)dz = 0 Equating the coefficient of dx, dy and dz to zero, we get ⇒ λx = – 4yz , λy = – 4xz, λz = – 4xy These are Lagrange’s equations ...(v) 141 DIFFERENTIAL CALCULUS-II Multiplying equation (v) by x, y, z respectively, we get λx2 = – 4xyz, λy2 = – 4xyz, λz2 = – 4xyz From these, we get λx2 = λy2 = λz2 ⇒ x2 = y2 = z2 ⇒ x = y= z Thus, the rectangular solid is a cube. Proved. Example 7. Use the method of Lagrange's multiplier to find the volume of the largest rectangular parallelopiped that can be inscribed in the ellipsoid Sol. Let l = 2x b = 2y h = 2z V = 8xyz ∴ ...(i) 2 2 and x 2 y2 z2 + + =1. a2 b2 c2 (U.P.T.U., 2002, 2000) 2 y x z + 2 + 2 −1= 0 2 a b c For largest volume, from (i) and (ii), we get ...(ii) dV = yz·dx + xz·dy + xy·dz = 0 and ...(iii) y x z dx + 2 dy + 2 dz = 0 2 a b c Now, equation (iii) + λ × equation (iv), we get ...(iv) λ λ λ z) dz = 0 2 x) dx + (xz + 2 y) dy + (xy + a b c2 λ λ λ ⇒ yz + 2 x = 0, xz + 2 y = 0, xy + 2 z = 0 a b c Multiplying (v) with x, y, z respectively and adding then, we get (yz + L x y + z OP = 0 3xyz + λ M + Na b c Q 2 2 2 2 2 2 ⇒ 3xyz + λ = 0 (using ii) ⇒ λ = – 3xyz Putting the value of λ in any one of equation (v), we get ⇒ Similarly, 1– FG H x a2 = 0 ⇒ yz 1 − 3x2 a2 = 0⇒ x= yz – 3xyz · y = Hence, the largest volume V = 8 · b abc 3 3 3 ⋅ , z= 3x 2 a2 a , 3 c 3 IJ = 0 K ...(v) 142 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 8. Determine the point on the paraboloid z = x2 + y2 which is closest to the point (3, – 6, 4). Sol. Let (x, y, z) be any point on paraboloid nearest to the point (3, – 6, 4) ∴ D = ⇒ Let D2 = (x – 3)2 + (y + 6)2 + (z – 4)2 u = (x – 3)2 + (y + 6)2 + (z – 4)2 2 ( x − 3)2 + ( y + 6)2 + ( z − 4) 2 ...(i) 2 Subject to x + y – z = 0 For minimum distance, differentiating (i) and (ii), we get du = 2(x – 3) dx + 2(y + 6) dy + 2(z – 4) dz = 0 and 2xdx + 2ydy – dz = 0 ...(ii) ...(iii) ...(iv) Now, equation (iii) + λ × (iv), we get {2x (1 + λ) – 6} dx + {2y (1 + λ) + 12} dy + {2z – λ – 8) dz = 0 ⇒ x(1 + λ) – 3 = 0, y (1 + λ) + 6 = 0, 2z – (λ + 8) = 0 3 6 (λ + 8) , y=– ,z= 1+λ 1+λ 2 Putting the values of x, y, z in equation (ii), we get ⇒ x= ...(v) 9 36 (λ + 8) – = 0 2 + (1 + λ ) (1 + λ ) 2 2 45 (λ + 8) – = 0 (1 + λ )2 2 or ⇒ 90 – (λ + 1)2 (λ + 8) = 0 ⇒ ⇒ λ3 + 10λ2 + 17λ – 82 = 0 λ = 2 Hence x = 1, y = – 2, z = 5. Example 9. Find the maximum and minimum distances from the origin to the curve 3x2 + 4xy + 6y2 = 140. Sol. Let (x, y) be any point on the curve ∴ distance from (0, 0) is given by D2 = x2 + y2 = u (say) Subject to 3x2 + 4xy + 6y2 – 140 = 0 ...(i) ...(ii) For maximum and minimum from (i) and (ii), we get du = 2xdx + 2ydy = 0 6xdx + 4dx y + 4xdy+ 12ydy = 0 ...(iii) ...(iv) Now, equation (iii) + λ × (iv), we get {x (2 + 6λ) + 4yλ} dx + {y (2 + 12λ) + 4xλ}dy = 0 ⇒ 2x + λ (6x + 4y) = 0 ...(v) 2y + λ (12y + 4x) = 0 ...(vi) 143 DIFFERENTIAL CALCULUS-II Multiplying the above equations by x, y respectively and adding, we get 2 (x2 + y2) + 2λ (3x2 + 4xy + 6y2) = 0 ⇒ 2u + 2λ (140) = 0 (using (i) and (ii)) u 140 Putting the value of λ in eqns. (v) and (vi), we get ∴ λ = – u (6x + 4y) = 0 ⇒ (140 – 3u)x – 2uy = 0 140 u (12y + 4x) = 0 ⇒ – 2ux + (140 – 6u)y = 0 and 2y – 140 This system has non-trivial solution if 2x – 140 − 3u −2u −2u (140 − 6u) = 0 ⇒ (140 – 3u) (140 – 6u) – 4u2 = 0 ⇒ 14u2 – 1260u + (140)2 = 0 u2 – 90u – 1400 = 0 ⇒ (u – 70) (u – 20) = 0 u = 70, 20 Thus, the maximum and minimum distances are 70 , 20 . | As D2 = u Example 10. A wire of length b is cut into two parts which are bent in the form of a square and circle respectively. Find the least value of the sum of the areas so found. Sol. Let part of square = x and part of circle = y ⇒ x + y = b ∴ side of square = x , 4 radius of circle = y |As 2πr = y 2π area of square = x2 16 area of circle = πy 2 4π 2 = y2 4π 2 y x2 + 16 4π Subject to b =x + y ⇒ x + y – b = 0 For minimum from (i) and (ii), we get Here, let u = sum of areas = y x dx + dy = 0 8 2π dx + dy = 0 du = and ...(i) ...(ii) ...(iii) ...(iv) 144 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Now, (iii) + λ × (iv), we get xdx ydy + + λ (dx + dy) = 0 8 2π y x + λ dy = 0 + λ dx + ⇒ 8 2π y x + λ = 0, +λ = 0 ⇒ 8 2π ⇒ x = – 8λ, y = – 2πλ Putting x and y in equation (ii), we get b – 8λ – 2πλ = b ⇒ λ = – 8 + 2π 8b 2πb Thus x = – 8λ = , y = – 2πλ = 8 + 2π 8 + 2π ∴ The least value of areas is, from (i) F H I K IJ K FG H ...(v) x2 y 2 + 16 4π 64b 2 4π 2 b 2 + = 16(8 + 2π) 2 4π (8 + 2π) 2 u = = b 2 ( π + 4) 4( π + 4) 2 = b2 . 4( π + 4) Example 11. Find the dimension of rectangular box of maximum capacity whose surface area is given when (a) box is open at the top (b) box is closed. (U.P.T.U., 2008) Sol. Let the length, breadth and height of box are x, y, z respectively. So volume V = xyz ...(i) There will be two surface area one for open and one for closed box ∴ or Here nxy + 2yz + 2zx = S (say) ...(ii) g(x, y, z) ≡ nxy + 2yz + 2zx – S = 0 n = 1, when the box is open on the top n = 2, when the box is closed. ...(iii) The Lagrange’s equations are ∂g ∂V = yz + λ(ny + 2z) = 0 +λ ∂x ∂x ...(iv) ∂g ∂V = xz + λ(nx + 2z) = 0 +λ ∂y ∂y ...(v) ∂g ∂V = xy + λ(2y + 2x) = 0 +λ ∂z ∂z Multiplying (iv), (v), (vi) by x, y, z respectively and adding, we get ...(vi) 3xyz + λ [2(nxy + 2yz + 2zx)]= 0 or 3V + λ[2S] = 0 ⇒ λ= − 3V 2S ...(vii) 145 DIFFERENTIAL CALCULUS-II Putting value of λ from (vii) in (iv), (v) and (vi), we get 3xyz 3V (ny + 2z) = 0 ⇒ yz − (ny + 2z) = 0 2S 2S 2S or nxy + 2xz = 3 2S Similarly nxy + 2yz = 3 2S 2yz + 2zx = 3 From (viii) and (ix), we get x = y and from (ix), (x), we get yz − ny nx = 2 2 Putting (xi) and (xii) in equation (ii), we have ny = 2z ⇒ z = ...(viii) ...(ix) ...(x) ...(xi) ...(xii) nx nx +2· · x = S ⇒ 3nx2 = S 2 2 S or x2 = 3n (a) When box is open n = 1 nx · x + 2 · x · ∴ x2 = S 3 ⇒ x= S 3 Hence, the dimensions of the open box are x = y = 1 S S and z = 2 3 3 S 6 S ⇒ x= 6 Hence, the dimensions of the closed box are ∴ x2 = x = y= S 6 (b) When box is closed n = 2 and z= S · 6 EXERCISE 2.4 1. Find the maximum and minimum distances of the point (3, 4, 12) from the sphere, x2 + y2 + z2 = 1. (U.P.T.U., 2001) Ans. Dmin = 12, Dmax = 14 2. Find the maximum and minimum distances from the origin to the curve x2 + 4xy + 6y2 = 140. [U.P.T.U. (C.O.), 2003] Ans. Dmin = 4.5706, Dmax. = 21.6589 3. The temperature T at any point (x, y, z) in space is T = 400 xyz2. Find the highest temperature at the surface of a sphere x2 + y2 + z2 = 1. Ans. T = 50 146 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 4. Find the maxima and minima of x2 + y2 + z2 subject to the conditions : ax2 + by2 + cz2 LMAns. l + m + n = 0OP N ( au − 1) (bu − 1) (cu − 1) Q 2 = 1, lx + my + nz = 0. 2 2 5. Find the maximum value of u = xpyqzr when the variables x, y, z are subject to the LMAns. u = FG p IJ FG q IJ F r I OP H a K H bK H cK Q N p condition ax + by + cz = p + q + r. q r 6. A rectangular box, which is open at the top has a capacity of 256 cubic feet. Determine the dimensions of the box such that the least material is required for the construction of the box. Use Lagrange's method of multipliers to obtain the solution. Ans. length = breadth = 8′, height = 4′ 7. Find the minimum value of x2 + y2 + z2 subject to condition 1 1 1 + + = 0. x y z Ans. Minimum value = 27 8. Determine the point in the plane 3x – 4y + 5z = 50 nearest to the origin. Ans. (3, – 4, 5) 9. Find the length and breadth of a rectangle of maximum area that can be inscribed in the 3 2 , b = 2 , area = 12 2 10. Divide 24 into three parts such that the continued product of the first square of the second and the cube of the third may be maximum. ellipse 4x2 + 9y2 = 36. Ans. l = Ans. 4, 8, 12, maximum value = 4·82·123 11. Find the volume of the largest rectangular parallelopiped that can be inscribed in the ellipsoid of revolution 4x2 + 4y2 + 9z2 = 36. Ans. Maximum volume = 16 3 12. Using the Lagrange's method of multipliers, find the largest product of the numbers x, y and z when x2 + y2 + z2 = 9. Ans. 3 3 13. A torpedo has the shape of a cylinder with conical ends. For given surface area, show 2 that the dimensions which give maximum value are l = h = r, where l is the length 5 of the cylinder, r is the radius and h is the altitude of cone. 14. Find the dimensions of a rectangular box, with open top of given capacity (volume) such that the sheet metal (surface area) required is least. Ans. x = y = 2z = (2V)1/3, V = volume 15. Find the maximum and minimum values of x 2 + y 2 when 13x2 – 10xy + 13y2 = 72. Ans. Maximum = 3, minimum = 2 147 DIFFERENTIAL CALCULUS-II 16. A tent of given volume has a square base of side 2a and has its four sides of height b vertical and is surmounted by a pyramid of height h. Find the values of a and b in terms of h so that the canvas required for its construction be minimum. 1 h 5 h, b = (4a2)h, S = 8ab + 4a a 2 + h 2 ; a = ]. 3 2 2 17. If x and y satisfy the relation ax2 + by2 = ab, prove that the extreme values of function u = x2 + xy + y2 are given by the roots of the equation 4(u – a) (u – b) = ab. 18. Find the maximum value of xmynzp when x + y + z = a. [Hint: V = 4a2b + Ans. am+n+b· mm·nn·pp/(m + n + p)m + n + p 19. Find minimum distance from the point (1, 2, 2) to the sphere x2 + y2 + z2 = 30. Ans. 3 20. Determine the perpendicular distance of the point (a, b, c) from the plane lx + my + nz LMAns. Minimum distance = la + mb + nc OP l +m +n Q N = 0 by the Lagrange’s method. 2 2 2 OBJECTIVE TYPE QUESTIONS A. Pick the correct answer of the choices given below: b g for the function u = e sin y, v = (x + log sin y) is (U.P.T.U., 2008) ∂b x , y g ∂ u, v 1. The Jacobian x (i) 1 (ii) 0 (iii) sin x sin y – xy cos x cos y (iv) a f for the function u = 3x + 5y, v = 4x – 3y is b g ∂ u, v ∂ x, y 2. The Jacobian (i) 29 2 (ii) xy 2 3 (iii) x y – y (iv) – 29 3. If x = r cos θ, y = r sin θ, z = Z, then (i) 2r (iii) ex x 5 r 4. If u = x sin y, v = y sin x, then b a ∂ x, y, z g is f ∂ r , θ, Z (ii) r2 – 5 (iv) r a f is b g ∂ u, v ∂ x, y (i) sin x sin y (ii) sin x sin y – xy cos x cos y (iii) cos x cos y – xy sin y (iv) 0 148 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 5. If u = 3x + 2y – z, v = x – y + z, w = x + 2y – z, then J (u, v, w) is (i) 5 (ii) 0 (iii) – 2 (iv) xy + 3x3 6. If in the area of an ellipse if a two per cent error is made in measuring the major and minor axis then the percentage error in area is (i) 2% (ii) 3% (iii) 5% (iv) 4% 7. The value of (1.05)3.02 is (i) 2.35 (ii) 1.25 (iii) 1.15 (iv) 0.57 8. If an error of 2% is made in measuring the sides of a rectangle, then what is the percentage of error in calculating its area? (i) 1% (ii) 2% (iii) 4% (iv) 8% 9. The relation among relative error of quotient, relative errors of dividend and the divisor (Take x = dividend, y = divisor, z = quotient) is (i) dz dx dy = + z x y (ii) dy dz dx < +2 z x y (iii) dz dx dy < − z x y (iv) dz dx dy = − z x y 10. If u = x2 + y2 + 6x + 12 then the stationary point is (i) (– 3, 0) (ii) (– 3, 5) (iii) (– 2, 6) 11. The extreme values of (iv) (5, 6) f(x, y) = x2 + 2y2 on the circle x2 + y2 = 1 are (i) fmax = 2, fmin = 1 (ii) fmax = 0, fmin = – 2 (iii) fmax = 7, fmin = – 5 (iv) None of these 12. If u = x4 + 2x2y – x2 + 3y2 then rt – s2 is equal to (i) 24 (ii) 36 (iii) – 58 (iv) 14 B. Fill in the blanks: 1. If u = u (r, s), v = v (r, s) and r = r (x, y), s = s (x, y) then b g is .......... a f ∂au, vf 3. If u = x (1 – y), v = xy then = .......... ∂bx, yg ∂au, vf ∂bx, yg × 4. = .......... ∂bx, yg ∂au, vf 2. If x = r cos θ, y = r sin θ then ∂ x, y ∂ r, θ a f = .......... b g ∂ u, v ∂ x, y 149 DIFFERENTIAL CALCULUS-II C. Indicate True or False for the following statements: 1. (i) The functions u and v are said to be functionally independent if their Jacobian is not equal to zero. (ii) If f1 (u, v, x, y) = 0 and f2 (u, v, x, y) = 0 then u, v are said to be implicit functions. (iii) m functions of n variables are always functionally dependent when m > n. (iv) If x = r cos θ, y = r sin θ then 2. (i) b g = r – 2r + 1. ∂ar , θf ∂ x, y 2 dx represents relative error. x (ii) If f(a, b) < f(a + h, b + k) then f(a, b) is a maximum value. (iii) If f(a, b) > f(a + h, b + k) then f(a, b) is a minimum value. (iv) A point where function is neither maximum nor minimum is called saddle point. 3. (i) Nature of stationary points cannot be determined by Lagrange’s method. (ii) Solving fx = 0 and fy = 0 for stationary point. (iii) Extremum is a point which is either a maximum or minimum. (iv) Extrema occur only at stationary points. However, stationary points need not be extrema. D. Match the Following: 1. (i) Maxm (a) rt – s2 = 0 (ii) Minm (b) rt – s2 < 0 (iii) Saddle point (c) rt – s2 > 0, r > 0 (iv) Failure case (d) rt – s2 > 0, r < 0 2. (i) δx or dx (ii) δx dx or x x dx x (iv) f(x, y) ≤ f(a, b) (iii) 100 × (a) Local maximum (b) Absolute error (c) Relative error (d) Percentage error (U.P.T.U., 2008) 150 A TEXTBOOK OF ENGINEERING MATHEMATICS—I ANSWERS TO OBJECTIVE TYPE QUESTIONS A. Pick the correct answer: 1. (ii) 4. (ii) 7. (iii) 2. (iv) 5. (iii) 8. (iii) 3. (iv) 6. (v) 9. (iv) 10. (i) 11. (i) 12. (i) 2. r 3. x 1. (i) T (ii) T (iii) T (iv) F 2. (i) T (ii) F (iii) F (iv) T 3. (i) T (ii) T (iii) T (iv) T B. Fill in the blanks: 1. 4. 1 a f a f a f b g ∂ u, v ∂ r , s ⋅ ∂ r , s ∂ x, y C. True or False: D. Match the following: 1. (i) → (d) (ii) → (c) (iii) → (b) (iv) → (a) 2. (i) → (b) (ii) → (c) (iii) → (d) (iv) → (a) GGG UNIT III Matrices 3.0 INTRODUCTION The term matrix was apparently coined by sylvester about 1850, but introduced first by Cayley in 1860. In this unit, we focus on matrix theory itself which theory will enable us to obtain additional important results regarding the solution of systems of linear algebraic equations. One way to view matrix theory is to think in terms of a parallel with function theory. In mathematics, we first study numbers—the points on a real number axis. Then we study functions, which are mappings or transformations, from one real axis to another. For example, f (x) = x2 maps the point x = 2, say on x-axis to the point f = 4 on f or y-axis. Just as functions act upon numbers, we shall see that matrices act upon vectors and are mappings from one vector space to another. 3.1 DEFINITION OF MATRIX A matrix is a collection of numbers arranged in the form of a rectangular array. These numbers known as elements or entries are enclosed in brackets [ ] or ( ) or O O. Therefore a matrix A may be expressed as LM a a A = MM MNa M OP P M M P P L a Q 11 a12 L a1n 21 a22 L a2n m1 M am2 ...(i) mn The horizontal lines are called rows and vertical lines are called columns. The order of matrix A is m × n and is said to be a rectangular matrix. 3.1.1 Notation Elements of matrix are located by the double subscript ij where i denotes the row and j the column. In view of subscript notation in (1), one also writes A = [aij], where i = 1, 2, ...,m and j = 1, 2, ... n 151 152 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 3.2 TYPES OF MATRICES There are following types of matrices: (a) Rectangular matrix: A matrix in which the number of rows and columns are not equal, i.e., (m ≠ n) is called a rectangular matrix, e.g., LM1 N3 OP Q 2 2 5 . 1 2×3 (b) Square matrix: A matrix in which the number of rows and columns are equal, i.e., (m = n) is called a square matrix, e.g., LM 5 MM 3 N0 OP P 2PQ 2 1 1 0 1 . 3× 3 (c) Row matrix: A matrix which has only single row and any number of columns is called a row matrix, e.g., [1 2 0 5]1 × 4 (d) Column matrix: A matrix which has only single column and any number of rows, LM 1 OP 2 i.e., (m × 1) order is called column matrix, e.g., MM PP 0 MN 5 PQ . 4×1 (e) Null matrix or zero matrix: Any m × n matrix is called a null matrix if each of its elements is zero and is denoted by Om × n or simply by O simply, e.g., LM0 0 0OP . N0 0 0Q (f) Diagonal matrix: A square matrix A = [aij] is called diagonal matrix, if all the elements except principal diagonal are zero. Thus, for diagonal matrix aij ≠ 0, i = j and aij = 0, i ≠ j, e.g., LM1 MM0 N0 0 2 0 0 0 5 OP PP Q (g) Scalar matrix: Any diagonal matrix in which all its diagonal elements are equal to a scalar, say (K) is called a scalar matrix LM 5 MM0 N0 Thus, i.e., 0 5 0 OP P 5PQ 0 0 A = [aij]n × n is a scalar matrix if aij = RS 0 TK when i ≠ j when i = j (h) Identity matrix (or unit matrix): Any diagonal matrix is called an identity matrix, if each of its diagonal elements is unity. Thus a matrix A = {aij}n × n is called identity matrix iff aij = RS0, T1, i≠ j . An identity matrix of order n is denoted by I or In. i =j 153 MATRICES Thus, I4 = LM1 MM0 N0 0 1 0 0 0 0 0 0 1 OP PP Q (i) Symmetric matrix: A square matrix A = [aij]n × n is said to be symmetric matrix if its (i, j)th element is equal to (j, i)th element. Thus matrix A is said to be symmetric matrix if aij = aji ∀ i, j. e.g., LM 1 MM−3 N−4 −3 2 −5 LM 0 MM−2 N−5 2 0 −3 −4 −5 3 OP PP Q (j) Skew symmetric matrix: A sqaure matrix A = [aij] is said to be skew symmetric if aij = – aji ∀ i, j. But for diagonal elements aii = – aii ⇒ 2aii = 0 ⇒ aii = 0. This proves that every leading diagonal element of a skew symmetric matrix is zero. Thus, OP PP Q 5 3 is a skew symmetric matrix. 0 (k) Triangular matrix: If every element above or below the leading diagonal of a square matrix is zero, the matrix is called a triangular matrix. It has the following two forms: (i) Upper triangular matrix: A square matrix in which all the elements below the leading diagonal are zero i.e., aij = 0; i > j is called an upper triangular matrix. LM 2 MM0 N0 e.g., 3 5 0 1 2 4 OP PP Q (ii) Lower triangular matrix: A square matrix in which all the elements above the leading diagonal are zero i.e., aij = 0; i < j is called a lower triangular matrix. LM5 MM 3 N6 e.g., 0 2 3 0 0 1 OP PP Q (l) Transpose of a matrix: The matrix is obtained by interchange the rows and columns of a given matrix A, is called the transpose of A and is denoted by A′ or AT e.g., If LM 2 3 A = M1 2 MN5 1 LM 2 OP 3 3 0P , then A′ = M MM5 2 1PQ N6 5 6 OP P 3 2P P 0 1Q 1 2 5 1 (m) Conjugate of a matrix: The matrix obtained by replacing each element by its conjugate complex number of a given matrix A, is called the conjugate of A and is denoted by A . Thus, if A = LM1 + 2i −4i OP , N 3 1 − 2iQ then A = LM1 − 2i 4i OP N 6 1 + 2iQ 154 A TEXTBOOK OF ENGINEERING MATHEMATICS—I (n) Tranjugate or conjugate transpose of a matrix: The conjugate of a transposed matrix A or transpose of a conjugate matrix A is called a tranjugate matrix of A and is denoted by A*. Thus, if A = LM 2 + 3i −5i OP , N 6 1 − 2iQ then A* = LM 2 − 3i 6 OP N 5i 1 + 2iQ (o) Hermitian matrix: A square matrix A = [aij] is said to be “Hermitian matrix” if its conjugate transpose matrix A* is equal to itself i.e., A* = A. Thus A = [aij] is hermitian matrix if aij = a ji ∀ i and j. Hence for diagonal elements, aii = a ii i.e., every leading diagonal element in a hermitian matrix is wholly real. LM 1 3 + i i OP MM3 − i 2 5iPP N −i −5i 0 Q e.g., (p) Skew hermitian matrix: Any square matrix A = [aij] is said to be a skew hermitian matrix if A* = – A −a ji ∀ i, j Thus A is skew hermitian if aii = For diagonal elements and aii = – a ii ⇒ aii + a ii = 0 aii = x + iy a ii = x – iy then a ii + a ii = x + iy + x – iy = 2x If or x = 0 Hence all the diagonal elements of a skew hermitian matrix are either zero or pure imaginary. LM i e.g., −3 + i MM N−4 + 5i 3+i 0 2i 4 + 5i 2i 0 OP PP Q (q) Nilpotent matrix: A square matrix A is said to be nilpotent of index p if p is the least positive integer such that Ap = 0. Thus, a square matrix A is said to be nilpotent of index 2, if LM0 0OP is a nilpotent matrix N1 0 Q L0 0 O L0 0 O L0 0 O A = M1 0 P M1 0 P = M0 0 P = 0 N Q N Q N Q A2 = 0; e.g., As 2 (r) Idempotent matrix: A square matrix A is said to be periodic of period p if p is the least positive integer such that Ap+1 = A. If p = 1 so that A2 = A, then A is called idempotent. Thus a square matrix A is said to be “idempotent” (or of period 1) if e.g., A2 = A. LM1 0OP , N0 1 Q L1 0OP LM1 0OP = LM1 0OP = A A = M N0 1Q N0 1Q N0 1Q A = 2 155 MATRICES (s) Involutory matrix: A square matrix A is said to be involutory if A2 = I, I being the identity matrix. e.g., A = LM1 0OP , A = LM1 0OP LM1 0OP = LM1 0OP = I N0 1 Q N0 1 Q N0 1 Q N0 1 Q 2 (t) Orthogonal matrix: A square matrix A is said to be an orthogonal matrix, if A′A = AA′ = I (u) Unitary matrix: A square matrix A is said to be unitary matrix if AA* = A*A = I. (U.P.T.U., 2001, 2005) 3.3 OPERATIONS ON MATRICES 3.3.1 Scalar Multiple of a Matrix Consider a matrix A = [aij]m × n. Let k be any scalar belonging to a field over which A is defined. The scalar multiple of k and A, denoted by kA, is defined as kA = [kaij]m × n i.e., each element of A is multiplied by k. If k = –1, then (–1) A = [–aij] (–1) A is denoted by –A and is called the negative of matrix A. –A is also called “additive inverse of A”. Thus, if A = LM 1 2OP , then 3A = L 1 ⋅ 3 MN –1 ⋅ 3 N –1 4Q 2⋅ 3 4⋅ 3 OP = LM 3 6 OP Q N –3 12Q 3.3.2 Addition of Matrices Any two matrices can be added if they are of the same order. If A = [aij]m × n, B = [bij]m × n then A + B = [aij + bij]m × n e.g., Let A = LM 2 5OP , B = LM5 N 3 1Q N2 OP Q –4 0 , then A + B = LM7 1OP N5 1Q 3.3.3 Subtraction of Matrices Any two matrices can be subtracted if they are of the same order. If A = [aij]m × n, B = [bij]m × n; then A – B = [aij – bij]m × n 3.3.4 Properties of Addition of Matrices (i) Commulative law A + B =B+ A (ii) Associative law (A + B) + C = A + (B + C) (iii) Each matrix has an additive inverse (iv) Cancellation law A + B =A+ C⇒ B= C 156 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 3.3.5 Multiplication of Matrices The product AB of two matrices A and B is only possible if the number of columns of A = number of rows of B. In the product AB, A is called the ‘prefactor’ and B is called the ‘post factor’. La Let A = M a N 11 e.g., 21 a12 a 22 OP Q a13 a 23 , B = Lb a a a O M L AB = M a N a a PQ MMNbb La b + a b + a b = Ma b + a b + a b N 11 12 13 21 22 23 11 21 31 11 11 12 21 13 31 21 11 22 21 23 31 LMb MMb Nb 11 21 31 b12 b22 b32 OP PP Q b12 b22 b32 OP PP Q a11b12 + a12 b22 + a13 b32 a 21b12 + a 22 b22 + a 23 b32 OP Q 3.3.6 Properties of Multiplication of Matrices (i) Associative law: (AB) C = A (BC) (ii) Distributive law: A(B + C) = AB + AC. 3.4 TRACE OF MATRIX If A = [aij]n × n be a square matrix, then the sum of its diagonal elements is defined as the trace of the matrix, hence ∞ trace of A = ∑ aii i= 1 e.g., A= LM 1 0 5OP MM 0 2 3PP , the trace of A = 1 + 2 + (– 2) = 1. N –1 1 2Q 3.5 PROPERTIES OF TRANSPOSE (i) (A′)′ = A (ii) (KA)′ = KA′, K being scalar (iii) (iv) (A + B)′ = A′ + B′ (AB)′ = B′A′. 3.6 PROPERTIES OF CONJUGATE MATRICES (i) (ii) (iii) dKAi = K A , K being a scalar, d A + Bi = A + B, d A B i = A B. 157 MATRICES Example 1. Show that every square matrix can be uniquely expresses as the sum of a symmetric and a skew symmetric matrix. Sol. Let A be any square matrix 1 1 (A + A′) + (A – A′) 2 2 1 1 Taking P = (A + A′), Q = (A – A′), we get 2 2 A = P+ Q ...(i) 1 1 Now P′ = A′ + A′ ′ A + A′ ′ = 2 2 1 = (A′ + A) = P 2 1 1 and Q′ = [(A–A′)]′ = [A′ – (A′)′] 2 2 1 = (A′ – A) = – Q 2 ⇒ P′ = P, Q′ = – Q Hence P is symmetric and Q is skew symmetric. This shows that a square matrix A is expressible as a sum of a symmetric and skew symmetric matrix. Deduction: To prove that this representation is unique, let if possible A = R + S be another representation of A, where R is symmetric and S is skew symmetric, ⇒ R = R′, S′ = – S Now A = R+ S ⇒ A′ = (R + S)′ ⇒ A′ = R′ + S′ Evidently A = b ⇒ g LM b g OP N Q A′ = R – S As R = R ′, S ′ = – S 1 (A + A′) = P 2 1 Also A – A′ = R + S – (R – S) = 2S or S = (A – A′) = Q 2 Thus R = P, S = Q This proves that the representation is unique. ∴ A + A′ = R + S + R – S = 2R or R = Example 2. Every square matrix can be uniquely expressed as P + iQ, where P and Q are hermitian. Sol. Let A be square matrix. Evidently A = = Taking 1 1 (A + A* ) + (A – A∗) 2 2 1 1 A− A* (A + A*) + i 2i 2 RS a T fUVW 1 1 (A + A*), Q = (A – A*), we get 2i 2 A = P + iQ P = ...(i) 158 A TEXTBOOK OF ENGINEERING MATHEMATICS—I L1 O 1 (A* + A ) = 1 (A* + A) = P, P* = M a A + A *fP = 2 2 N Q 2 L1 O F 1 IJ A* – A** = – 1 (A* – A) = Q Q* = M a A – A *fP = G 2i 2i N Q H −2i K * ∗∗ so that * Thus P* = P, Q* = Q ⇒ P and Q are hermitian. In this event (i) proves that A is expressible as P + iQ where P and Q are hermitian. Deduction: To prove that this representation is unique, let if possible. A = R + iS be another representation where R and S are hermitian so that R* = R, S* = S A* = (R + iS)* = R* + i S* = R + (–i)S = R – iS or A + A* = (R + iS) + (R – iS) = 2R 1 R = (A + A*) = P 2 A – A* = (R + iS) – (R – iS) = 2iS 1 (A + A*) = Q ⇒ S = 2i Finally, R = P, S = Q. Hence the representation is unique. Example 3. If A is unitary matrix, show that A′ is also unitary. Sol. ⇒ AA* = A*A = I (AA*)* = (A*A)* = I* = I (I* = I) (A*)* A* = A* (A*)* = I AA* = A* A = I (AA*)′ = (A*)′ A′ = ⇒ (A′)* · A′ = Hence, A′ is a unitary matrix. LM1 Example 4. If A = M 2 MN 2 Sol. Here and 2 1 2 OP PP Q a f As A * * = A (A*A)′ = (I)′ A′ (A*)′ = I A′ (A′)* = I Proved. 2 2 show that A2 – 4A – 5I = 0. 1 LM1 2 2OP L1 2 2O = M 2 1 2P M 2 1 2P MN 2 2 1PQ MMN 2 2 1PPQ LM1 + 4 + 4 2 + 2 + 4 2 + 4 + 2OP L 9 M = M 2 + 2 + 4 4 + 1 + 4 4 + 2 + 2P = M 8 MN 2 + 4 + 2 4 + 2 + 2 4 + 4 + 1PQ MN8 LM1 2 2OP L 4 8 8O M 4 8PP 4A = 4 M 2 1 2P = M 8 MN 2 2 1PQ MN8 8 4PQ A2 8 9 8 8 8 9 OP PP Q 159 MATRICES LM 9 8 8OP LM 4 8 8OP LM5 0 0OP 4 8 P – M 0 5 0P ∴ A – 4A − 5I = M 8 9 8P – M 8 MN8 8 9PQ MN8 8 4PQ MN0 0 5PQ LM 9 – 9 8 – 8 8 – 8OP LM0 0 0OP = M 8 – 8 9 – 9 8 – 8P = M 0 0 0P = 0. Proved. MN8 – 8 8 – 8 9 – 9PQ MN0 0 0PQ L 3 –4OP prove that A = L1 + 2K – 4K O , K being any positive integer. Example 5. If A = M MN K 1 − 2KPQ N1 –1Q L 3 –4OP and K be any positive integer Sol. Let A = M N1 –1Q L1 + 2K – 4K OP , we see that To prove A = M N K 1 − 2KQ L 3 −4OP = LM1 + 2 ⋅1 −4⋅ 1 OP A = A = M N1 –1Q N 1 1 – 2⋅1Q 2 K K 1 This proves that result is true for K = 1 LM 3 –4OP LM 3 –4OP = LM 9 – 4 N1 –1Q N1 –1Q N 3 – 1 L1 + 2 ⋅ 2 –4⋅ 2 OP = M 2 1 – 2 ⋅ 2Q N A2 = Now OP = LM5 Q N2 –12 + 4 –4 + 1 –8 –3 OP Q This proves that the result is true for K = 2. Let us suppose that the result is true for K = n, so that LM1 + 2n – 4n OP . N n 1 − 2n Q L1 + 2n – 4n OP LM 3 – 4OP = A A = M N n 1 − 2n Q N1 – 1Q L 3 + 6n − 4n –4 − 8n + 4nOP = LM1 + 2an + 1f –4an + 1f OP = M 3n + 1 − 2n –4n − 1 + 2n Q N n + 1 1 − 2an + 1fQ N An = Now, An + 1 n This proves that the result is true for K = n + 1, if it is true for K = n. Also, we have shown that the result is true for K = 1, 2. Hence by mathematical induction the required result follows. Example 6. Find the nature of the following matrices A + A*, AA* and A – A*. Sol. and (U.P.T.U., 2001) z A* = A → Hermitian matrix, (A + A*)* = A* + A { Hermitian (AA*)* = (A*)*·A* = A·A* ⇒ Hermitian matrix. (A – A*)* = A* – (A*)* = A* – A = – (A – A*) ⇒ skew symmetric matrix. 160 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 7. Show that the matrix A = LMα + iγ β + iδ OP is a unitary matrix if Nβ + iδ α iγ Q α2 + β2 + γ2 + δ2 = 1. Sol. A = (U.P.T.U., 2005) LMα + iγ β + iδOP ∴ A = LM α iγ β iδ OP N β + iδ α iγ Q N β i δ α + i γ Q * But for unitary matrix AA* = I LMα + iγ β + iδOP LM α iγ β iδ OP = LM1 0OP N β + i δ α i γ Q N β i δ α + i γ Q N 0 1 Q ∴ L α +γ +β +δ ⇒ M αβ βγ + αδ + γδ – αβ – iαδ + iβγ – δγ i i – MN 2 2 2 αβ – iαδ + iβγ + γδ – αβ – iβγ + iαδ – δγ β2 + δ 2 + α 2 + γ 2 2 LM1 0OP N 0 1Q OP L1 0O 0 = M α + β + γ + δ PQ N0 1PQ OP PQ = LMα + β + γ + δ 0 MN 2 ⇒ ⇒ 2 2 2 2 2 2 α2 + β2 + γ2 + δ2 = 1. Proved. LM 1 1 + iOP is unitary. (U.P.T.U., 2001) –1 Q 3 N1 – i 1 + iO 1 L 1 A = M 1 – i –1 PQ 3 N 1 + iO 1 L 1 A* = M –1 PQ 3 N1 – i 1 + iO 1 L 1 1 + iO 1 L 1 × A*·A = M P M –1 Q –1 PQ 3 N1 – i 3 N1 – i (1 + i ) – (1 + i)O 1 L 3 0O L1 0O 1 L 1 + (1 + 1) = = =I = M (1 + 1) + 1 PQ 3 MN0 3PQ MN0 1PQ 3 N(1 – i ) – (1 – i ) Example 8. Prove that the matrix Sol. Let 2 1 Therefore, A is a unitary matrix. Example 9. Define a unitary matrix. If N = LM 0 1 + 2iOP is a matrix, then show that N –1 + 2i 0 Q (I – N) (I + N)–1 is a unitary matrix where I is an identity matrix. Sol. A square matrix A is said to be unitary if A*A = I. Now I– N = LM1 0OP – LM 0 1 + 2iOP = LM 1 N0 1Q N–1 + 2i 0 Q N1 – 2i (U.P.T.U., 2000) OP 1 Q –1 – 2i 161 MATRICES and I+ N = LM1 0OP + LM 0 1 + 2iOP = LM 1 1 + 2iOP N0 1Q N –1 + 2i 0 Q N–1 + 2i 1 Q |I + N| = 1 – (– 1 – 4) = 6 Let ∴ ⇒ ⇒ A = A11 = 1, A12 = – (– 1 + 2i) = 1 – 2i A21 = – (1 + 2i) = – 1 – 2i, A22 = 1 B = LM 1 N –1 – 2i OP Q LM N 1 – 2i 1 or B′ = 1 1 – 2i Adj A = Adj (I + N) = B′ = (I + N)–1 = and LM 1 1 + 2iOP N–1 + 2i 1 Q LM N LM 1 N1 – 2i Adj ( I + N) 1 1 = |I + N| 6 1 – 2i –1 – 2i 1 –1 – 2i 1 OP 1 Q OP Q OP Q –1 – 2i –1 – 2iO L 1 –1 – 2iO LM P M 1 Q N1 – 2i 1 PQ N –2 – 4i O 1 L –4 = = C (say) M –4 PQ 6 N 2 – 4i 2 + 4i O 1L –4 ∴ C* = M – 4 PQ 6 N –2 + 4i 2 + 4i O L – 4 –2 – 4i O 1 L 36 0O 1 L –4 ⇒ C*C = = M P M P M PQ i i 4 4 2 4 4 –2 + – – – 0 36 36 N QN Q 36 N L1 0OP = I . Hence Proved. = M N0 1Q LM 1 2 3 OP 2 3 is nilpotent of index 2. Example 10. Show that A = M 1 MN–1 –2 –3 PPQ LM 1 2 3OP LM 1 2 3OP 2 3 1 2 3 Sol. Here A = M 1 MN–1 –2 –3PPQ MMN–1 –2 –3PPQ LM 1 + 2 – 3 2 + 4 – 6 3 + 6 – 9OP LM0 0 0OP 2+ 4– 6 3 + 6 – 9P = M0 0 0P = 0 = M 1+2– 3 MN–1 – 2 + 3 –2 – 4 + 6 –3 – 6 + 9PQ MN0 0 0PQ Now (I – N) (I + N)–1 = 2 Here A is nilpotent of index 2. 1 1 6 1 – 2i 162 A TEXTBOOK OF ENGINEERING MATHEMATICS—I LM a h gOP LMxOP f P, C = M yP then find out ABC. Example 11. If A = [x y z], B = M h b MNg f c PQ MNzPQ LMa h gOP Sol. AB = [x y z] M h b f P = [ax + hy + gz hx + by + fz gx + fy + cz] MNg f cPQ LM xOP and ABC = (AB) C = [ax + hy + gz hx + by + fz gx + fy + cz] M yP MN zPQ = [(ax + hy + gz)x + (hx + by + fz)y + z(gx + fy + cz)] = [ax2 + by2 + cz2 + 2fy + 2zx + 2hxy]. A2 A3 + +..., show that 2 3 Example 12. If eA is defined as I + A + LMcos hx sin hx OP, where A = LMx xOP. N sin hx cos hxQ N x xQ Lx xOP LMx xOP = LM2x 2x OP = 2x LM1 1OP A = M Nx xQ Nx xQ MN2x 2x PQ N1 1Q F L1 1OI = 2x B G say B = M1 1PJ H N QK eA = ex Sol. 2 2 2 2 2 2 2 Similarly, A3 = 22x3 B, A4 = 23x4 B, ... etc. ∴ eA = 1 + A + = I + xB + = 1 2 A2 A3 + + ..... 2 3 2x 2 B 2 2 x 3 B + + ..........|As A = xB 2 3 F I GH 2 I + ( 2x )B + ( 2 x2) B + ( 2 x3) B + . . . . .JK 2 3 LM 2 + 2x + (2x) + (2x) +... 0 + 2x + ( 2x) + ( 2x) +...OP 2 3 2 3 1 M PP = ( 2x ) ( 2x ) ( 2x) ( 2x ) 2M MM0 + 2x + 2 + 3 +... 2 + 2x + 2 + 3 +...PP N Q LM 1 ee + e j 1 ee – e jOP 1 L e + 1 e – 1O 2 = M P = e M 12 1 2 MN e – 1 e + 1PQ MN 2 ee – e j 2 ee + e jPPQ Lcos hx sin hxOP . Hence Proved. = e M N sin hx cos hxQ x 2 3 2 3 2 3 2 3 2x 2x 2x 2x x –x x –x x –x x –x x 163 MATRICES 3.7 SINGULAR AND NON-SINGULAR MATRICES A square matrix A is said to be singular, if |A| = 0. If |A|≠ 0, then A is called a non-singular matrix or a regular matrix. (U.P.T.U., 2008) 3.8 ADJOINT OF A SQUARE MATRIX Adjoint of A is obtained by first replacing each element of A by its cofactor in |A| and then taking transpose of the new matrix or by first taking transpose of A and then replacing each element by its cofactor in the determinant of A. LM a A = Ma MN a Let OP P a PQ 21 a12 a 22 a13 a 23 , then 31 a 32 33 11 a11 A = a 21 a12 a 22 a13 a 23 a 31 a 32 a 33 The cofactors of A in |A| are as follows: A11 = a 22 a 32 A21 = – A31 = a12 a 32 a12 a 22 LM A B = MA MN A a 23 a 21 , A13 = a 33 a 31 a13 a11 , A22 = a 33 a 31 a13 a11 , A23 = – a 33 a 31 a13 a11 , A32 = – a 23 a 21 a13 a11 , A33 = a 23 a 21 A12 A22 A32 11 Let a 23 a 21 A12 = – a 33 a 31 21 31 OP PP Q LM MM N A13 A11 A23 then Adj A = B′ = A12 A33 A13 a 22 a 32 a12 a 32 a12 a 22 A21 A22 A23 A31 A32 A33 OP PP Q 3.8.1 Properties of Adjoint If A = [aij] is a square matrix of order n then (i) adj A′ = (adj A)′ (ii) adj A* = (adj A)* (iii) adjoint of a symmetric (Hermitian) matrix is symmetric (Hermitian). 3.9 INVERSE OF A MATRIX (RECIPROCAL) Consider only square matrices. Inverse of a n-square matrix A is denoted by A–1 and is defined as follows: where I is n × n unit matrix AA–1 = A–1A = I adj A or A–1 = · A 3.9.1 Properties of Inverse (i) Inverse of A exists only if |A| ≠ 0 i.e., A is non-singular. (ii) The inverse of a matrix is unique. If B and C are two inverses of the same matrix A then (CA) B = C (AB), IB = CI i.e., B = C, so inverse is unique. 164 A TEXTBOOK OF ENGINEERING MATHEMATICS—I (iii) Inverse of a product is the product of inverses in the reverse order i.e., (AB)–1 = B–1 A–1, since (AB) (B–1 A–1) = A (BB–1) A–1 = AIA–1 = AA–1 = I. (iv) For a diagonal matrix D with dii as diagonal elements, D–1 is a diagonal matrix with 1 reciprocal d as the diagonal elements. ii (v) Transportation and inverse are commutative i.e., (A–1)T = (AT)–1, taking transpose of AA–1 = A–1A = In (A–1)T AT = AT (A–1)T = IT = I i.e., (A–1)T is the inverse of AT or (A–1)T = (AT)–1. (vi) (A–1)–1 = A Taking inverse of (AA–1) = I, (AA–1)–1 = (A–1)–1 A–1 = I–1 = I = AA–1. Thus, A = (A–1)–1. Example 13. Find the inverse of matrix A, where LM–1 1 2OP A = M 3 –1 1P. MN–1 3 4PQ Sol. Here we find A11, A12......... cofactors of A as follow: A11 = –1 3 A21 = – A31 = 1 3 1 –1 LM –7 Let B = M 2 MN 3 ∴ A–1 = 1 = – 7, A12 = – 4 3 –1 3 1 = – (12 + 1) = – 13, A13 = –1 4 –1 2 = – (4 – 6) = 2, A22 = –1 4 2 –1 = 1 + 2 = 3, A32 = – 1 3 –13 –2 7 LM MM N OP PP Q LM MM N –7 adj A 1 = –13 A 10 10 2 –2 2 OP LM PP MM Q N 2 –1 = – 4 + 2 = – 2, A23 = – 4 –1 2 –1 = – (– 1 – 6) = 7, A33 = 1 3 10 –7 T 2 ∴ adj A = (B) = –13 –2 10 3 –0 ⋅ 7 7 = –1 ⋅ 3 –2 1 –1 = 10 3 2 –2 2 0⋅2 –0 ⋅ 2 0⋅2 OP PP Q 1 =2 3 1 = 1– 3 = –2 –1 3 7 , A = 10 –2 OP PP Q 0⋅3 0⋅7 . –0 ⋅ 2 Example 14. If A and B are n-rowed orthogonal (unitary) matrices, then AB and BA are also orthogonal (unitary). Sol. (i) Let A and B be n-rowed orthogonal matrices then A′A = AA′ = I and B′B = BB′ = I To prove AB and BA are orthogonal we have (AB)′ (AB) = (B′A′) (AB) = B′ (A′A) B = B′IB = B′B = I (BA)′ (BA) = (A′B′) (BA) = A′ (B′B) A = A′IA = A′A = I Thus (AB)′ (AB) = I and (BA)′ (BA) = I This ⇒ AB and BA are orthogonal. (ii) Let A and B be unitary, then A*A = AA* = I and B*B = BB* = I. To prove AB and BA are unitary complete the proof yourself. 165 MATRICES EXERCISE 3.1 1. Find the values of p, q, r, s, t and u if LM1 2OP LM–3 –2OP LM p qOP = = , B –5 and C r s P so that A + B – C = 0. 3 4 1 A= M P P M M MN5 6PQ MN 4 3PQ MN t uPQ [Ans. p = – 2, q = 0, r = 4, s = – 1, t = 9, u = 9] LM0 1 0OP 2. If A = M0 0 1P , then show that A = pI + qA + rA . MNp q rPQ L2 3 1OP, B = LM1 2 –6OP , find 3A – 4B. 3. If A = M N0 1 5Q N0 –1 3 Q LM1 –2 3 OP 4. If A = M2 3 –1P , show that 6A + 25A – 42I = 0. MN3 1 2 PQ LM1 –3 2OP LM1 4 1 0OP LM2 1 –1 –2OP 1 –3 , B = 2 1 1 1 , C = 3 –2 –1 –1 . 5. If A = M 2 MN4 –3 –1PPQ MMN1 –2 1 2PPQ MMN2 –5 –1 0PPQ 3 2 LMAns. L2 N MN0 OPOP 3 QQ 1 27 1 2 Show that (i) AB = AC, (ii) (B + C) A = BA + CA. LM 0 6. If A = M MNtan α2 OP P prove that 0 P Q Lcos α – sin αOP · I + A = (I – A) M Nsin α cos α Q – tan α 2 7. Find the product of the matrices LM 2 –1 0OP L2 1 2 1OP, B = M 0 4 1P . A= M N1 1 1 1Q MM–2 1 0PP N 1 3 2Q 8. If Aα = LMAns. L1 N MN1 1 1 OP OP 3Q Q 3 LM cos α sin αOP , then show that N– sin α cos αQ L cos nα sin nα OP = A , (A ) = M N– sin nα cos nαQ α n nα where n is any positive integer. Also prove that Aα and Aβ commute and that Aα Aβ = Aα + β. Also prove that Aα A–α = I. 166 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 9. If A = LM0 1OP , and B = LM 1 OP , find BA and AB if they exist. N2Q N1 2Q Ans. AB = LM3 4OP L2 1 2O 10. If A = M1 1P, B = M P , find (AB)′. MN2 0PQ N1 2 4Q Hence verify that (AB)′ = B′A′. LM 2OP ; BA does not exist N 5Q LM LM10 3 MMAns. ( AB)′ = M11 3 MN 22 6 N 4 2 4 OPOP PPPP QQ LM 2 –2 –4OP MM–1 3 4 PP is idempotent. N 1 –2 3 Q L 1 2 2 OP 1 M 2 1 –2 is orthogonal. 12. Show that the matrix A = 3 M MN–2 2 1 PPQ 1 L1 + i i – 1O 13. Prove that the matrix A = M P is unitary. 2 N1 + i 1 – i Q LM–2 + 3i 1 – i 2 + i OP 4 – 5i 5 P as sum of hermitian and skew hermitian matrices. 14. Express M 3 MN 1 1 + i –2 + 2iPQ LM 1 L –4 4 – i 4 O 1 L 6i –2 – i 2iO OP MMAns. 2 MM 4 + i 8 10PP + 2 MM 2 – i –10i 0 PP PP MN 3 – i 2 –4PQ MN–1 + i 2i 4iPQ Q N LM 1 2 3OP 2 3P is nilpotent. 15. Show that M 1 MN–1 –2 –3PQ 16. Express given matrix A as sum of a symmetric and skew symmetric matrices. LM L6 6 7O L 0 2 –2OOP LM6 8 5OP MMAns. MM6 2 5PP + MM –2 0 –2PPPP A = M4 2 3P . N MN7 5 1PQ MN 2 2 0PQQ MN1 7 1PQ LM1 0 0 0OP 1 –1 0 0 17. Show that A = M MM1 –2 1 0PPP is involutory. N1 –3 3 –1Q 11. Show that the matrix A = 167 MATRICES 18. Find the inverse of the matrix A. LM 1 1 3 OP 3 –3 . A = M1 MN–2 –4 4 PPQ Lcos α sin α OP = sec 2α LM cos α 19. Prove that M Nsin α cos αQ N– sin α –1 –1 – sin α cos α 20. If ω is one the imaginary cube roots of unity and if LM1 1 A = M1 ω MN1 ω 2 OP PP Q LM L12 1M = Ans. A –5 MM 4M MN –1 N LM MM N 1 1 1 –1 ω , then show that A = 1 3 2 ω 1 1 ω2 ω OP. Q 4 –1 –1 OPOP PP –1PQ PQ 6 –3 OP PP Q 1 ω . ω2 3.10 ELEMENTARY ROW AND COLUMN TRANSFORMATIONS The following transformations on a given matrix are defined as elementary transformations: (i) Inter-change of any two rows (columns). (ii) Multiplication of any row (column) by any non-zero scalar k. (iii) Addition to one row (column) of another row (column) multiplied by any non-zero scalar. We will use the following notations to represent the elementary row (column) operations: (a) Rij (Cij) or Ri ↔ Rj (Ci ↔ Cj) is used for the inter-change of ith and jth rows (columns). (b) Ri (K) [Ci (K)] or Ri → KRi (Ci → KCi) will denote the multiplication of the elements of the ith row (column) by a non-zero scalar K. (c) Rij (K) [Cij (K)] or Ri → Ri + KRj (Ci → Ci + KCj) is used for addition to the elements of ith row (column) the elements of jth row (column) multiplied by the constant K. 3.10.1 Elementary Matrices The square matrices obtained from an identity or unit matrix by any single elementary transformation (i), (ii) or (iii) are called “elementary matrices”. 3.10.2 Properties of Elementary Transformations (i) Every elementary row (column) transformation on a matrix can be effected by pre-post multiplication by the corresponding elementary matrix of an appropriate order. (ii) The inverse of an elementary matrix is an elementary matrix. 3.10.3 Equivalent Matrices Two matrices A and B are said to be equivalent, denoted by A ~ B, if one matrix say A can be obtained from B by the application of elementary transformations. 3.10.4 Properties of Equivalent Matrices (i) If A and B are equivalent matrices, then there exist non-singular matrices R and C such that B = RAC. 168 A TEXTBOOK OF ENGINEERING MATHEMATICS—I (ii) Every non-singular square matrix can be expressed as the product of elementary matrices. (iii) If there exist a finite system of elementary matrices R1, R2,....., Rm such that (Rm... R2 R1) A = I and A is non-singular, than A−1 = (Rm ... R2 R1) I. 3.11 METHOD OF FINDING INVERSE OF A NON-SINGULAR MATRIX BY ELEMENTARY TRANSFORMATIONS The property III gives a method of finding the inverse of a non-singular matrix A. In this method, we write A = IA. Apply row transformations successively till A of L.H.S. becomes identity matrix I. Therefore, A reduces to I, I reduces to A–1. LM0 1 2 2OP 1 1 2 3P Example 1. Find the inverse of A = M MM2 2 2 3PP . N2 3 3 3Q Sol. Let A = IA ⇒ R1 ↔ R2 LM0 1 2 2OP LM1 0 0 0OP MM12 12 22 33PP = MM00 10 01 00PP A MN2 3 3 3PQ MN0 0 0 1PQ LM1 1 2 3OP LM0 1 0 0OP MM0 1 2 2PP = MM10 00 01 00PP A MN22 23 23 33PQ MN0 0 0 1PQ Applyirg R3 → R3 – 2R1, R4 → R4 – 2R1, we get LM1 1 2 3 OP LM0 1 0 0OP MM00 10 –22 –32 PP = MM10 –20 10 00PP A MN0 1 –1 –3PQ MN0 –2 0 1PQ Applying R2 → R2 + R3, we have LM1 1 2 3 OP L0 MM00 10 –20 –1–3PP = MM1 MN0 1 –1 –3PQ MMN00 1 –2 –2 –2 0 1 1 0 OP PP PQ 0 0 A 0 1 169 MATRICES Applying R1 → R1 – R2, R4 → R4 – R2, we have LM1 0 2 4OP LM–1 3 –1 0OP MM00 01 –20 –1–3PP = MM 01 –2–2 11 00PP A MN0 0 –1 –2PQ MN–1 0 –1 1PQ Applying R1 → R1 + R3, R4 → 2R4 – R3, we get LM1 0 0 1 OP L–1 1 0 0O MM0 1 0 1PP = MM 1 –2 1 0PP A P MMN–20 –22 –31 02PPQ MN00 00 20 3 1Q Applying R1 → R1 + R4, R2 → R2 – R4, R3 → R3 – 3R4 LM1 0 0 0 OP L–3 3 –3 2 O MM0 1 0 0 PP = MM 3 –4 4 –2PP A MN00 00 20 10 PQ MM–26 –82 10–3 –62 PP Q N Applying R3 → – Hence 1 R , R → (– 1) R4, we obtain 2 3 4 LM1 0 0 0OP L–3 3 –3 2 O MM0 1 0 0PP = MM 3 –4 4 –2PP A MN00 00 10 10PQ MM–32 –24 –53 –23 PP Q N LM–3 3 –3 2OP 3 –4 4 –2 A = M MM–3 4 –5 3PPP . N 2 –2 3 –2Q –1 Example 2. Find by the elementary row transformation inverse of the matrix. Sol. Let Now ∴ LM0 1 2OP MM1 2 3PP. N 3 1 1Q LM0 1 2OP A = M1 2 3P MN3 1 1PQ A = IA LM0 1 2OP LM1 0 0OP MM1 2 3PP = MM0 1 0PP A N3 1 1Q N0 0 1Q (U.P.T.U., 2000, 2003) 170 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Applying R1 ↔ R2, we get R3 → R3 – 3R1 LM1 2 3OP LM0 1 0OP MM0 1 2PP = MM1 0 0PP A N3 1 1Q N0 0 1Q LM1 2 3OP LM0 1 0OP MM0 1 2PP = MM1 0 0PP A N0 –5 –8Q N0 –3 1Q R1 → R1 – 2R2, R3 → R3 + 5R2 LM1 0 –1OP LM–2 1 0OP MM0 1 2 PP = MM 1 0 0PP A N0 0 2 Q N 5 –3 1Q R1 → R1 + R3 → 1 R 2 3 Therefore, 1 R , R → R2 – R3, we get 2 3 2 1 0 0 1 2 0 1 0 = –4 0 0 2 5 LM MM N OP PP Q LM MM N LM1 0 0OP LM1 2 MM0 1 0PP = MM –4 N0 0 1Q N5 2 LM1 2 A = M –4 MN5 2 –1 –1 2 3 –3 –1 2 3 –3 2 –1 2 3 –3 2 OP PP Q 1 2 –1 A 1 OP PP Q 1 2O P –1 P . 1 2PQ 1 2 –1 A 1 2 Example 3. Find the inverse of the following matrix employing elementary transformations: Sol. Let ∴ LM3 –3 4OP MM2 –3 4PP · N0 –1 1Q A = IA LM3 –3 4OP LM1 0 0OP MM2 –3 4PP = MM0 1 0PP A N0 –1 1Q N0 0 1Q Applying R1 → R1 – R2, we get LM1 0 0OP LM1 –1 0OP MM2 –3 4PP = MM0 1 0PP A N0 –1 1Q N0 0 1Q [U.P.T.U. (C.O.), 2002] 171 MATRICES R2 → R2 – 2R1 LM1 0 0OP LM 1 –1 0OP MM0 –3 4PP = MM–2 3 0PP A N0 –1 1Q N 0 0 1Q R3 → 3R3 – R2 R2 → R2 + 4R3 R2 → – Hence LM1 0 0 OP LM 1 –1 0OP MM0 –3 4 PP = MM–2 3 0PP A N0 0 –1Q N 2 –3 3Q LM1 0 0 OP LM1 MM0 –3 0 PP = MM6 N0 0 –1Q N 2 –1 –9 –3 OP PP Q 0 12 A 3 1 R , R → (– 1) R3, we obtain 3 2 3 LM1 0 0OP LM 1 –1 0 OP MM0 1 0PP = MM–2 3 –4PP A N0 0 1Q N–2 3 –3Q LM 1 –1 0OP 3 –4 . A = M –2 MN –2 3 – 3PPQ –1 Example 4. Find the inverse of the matrix Sol. Let ⇒ LM–1 –3 3 –1OP M 1 1 –1 0P A = M 2 –5 2 –3 P. MN–1 1 0 1PQ A = IA LM–1 –3 3 –1OP LM1 0 0 0OP MM 12 –51 –12 –30PP = MM00 10 01 00PP A MN–1 1 0 1PQ MN0 0 0 1PQ Applying R2 → R2 + R1, R3 → R3 + 2R1, R4 → R4 – R1, we get LM–1 –3 3 –1OP LM 1 0 1 1 –2 2 –1 P MM 00 –11 = M M2 0 MN 0 4 –38 –52 PPQ MN−1 0 0 0 1 0 OP PP PQ 0 0 A 0 1 172 A TEXTBOOK OF ENGINEERING MATHEMATICS—I R2 → – 1 R 2 2 LM –1 MM 0 MN 00 –3 1 –11 4 3 –1 8 –3 OP PP PQ LM MM MM N 1 –1 1 – 12 2 = –5 2 2 –1 R3 → R3 + 11R2, R4 → R4 – 4R2 LM–1 –3 3 –1 OP L 1 MM 00 10 –1–3 11 22PP = MM – 1 2 MN 0 0 1 0 PQ MM – 71 2 N R3 ↔ R4 OP 0 0P A 1 0P P 0 0 1PQ 0 1 – 2 0 0 0 –1 2 – 11 2 2 0 0 0 1 0 LM–1 –3 3 –1 OP LM 1 0 0 0OP MM 00 10 –11 102PP = MM– 11 2 – 12 2 00 01PP A MN 0 0 –3 1 2PQ MMN – 72 – 112 1 0PPQ R4 → 2R4 LM –1 –3 3 –1 OP LM 1 0 0 0OP MM 0 1 –1 1 2PP = MM– 1 2 – 1 2 0 0PP MN 00 00 –61 10 PQ MN –71 –112 02 01PQ Applying R4 → R4 + 6R3, we get R2 → R2 + R3 R2 → R2 – OP PP PQ 0 0 A 0 1 LM–1 –3 3 –1 OP LM 1 0 0 0OP MM 0 1 –1 1 2PP = MM– 1 2 – 1 2 0 0PP A MN 00 00 10 01 PQ MN –11 12 02 61PQ LM–1 –3 3 –1 OP LM 1 0 0 0OP MM 0 1 0 1/2PP = MM1 2 3 2 0 1PP A MN 00 00 10 01 PQ MN –11 12 02 61PQ 1 R 2 4 LM–1 –3 3 –1OP L 1 0 0 0 O MM 0 1 0 0 PP = MM 1 1 –1 –2PP A MN 00 00 10 01 PQ MM–11 12 02 61 PPQ N 173 MATRICES Applying R1 → R1 + 3R2, we get LM–1 0 3 –1OP LM 4 3 –3 –6OP MM 0 1 0 0 PP = MM 1 1 –1 –2PP A MN 00 00 10 01 PQ MN–11 12 02 16 PQ Applying R1 → R1 – 3R3, we obtain LM–1 0 0 –1OP LM 1 –3 –3 –9OP MM 00 10 01 00 PP = MM 11 12 –10 –21 PP A MN 0 0 0 1 PQ MN–1 1 2 6 PQ Applying R1 → R1 + R4 LM–1 0 0 0OP LM 0 –2 –1 –3OP MM 00 10 01 00PP = MM 11 12 –10 –21 PP A MN 0 0 0 1PQ MN–1 1 2 6 PQ In the last, we apply R1 → (– 1) R1 Therefore LM1 0 0 0OP LM 0 2 1 3 OP MM00 10 01 00PP = MM 11 12 –10 –21 PP A MN0 0 0 1PQ MN–1 1 2 6 PQ LM 0 2 1 3 OP 1 1 –1 –2 A = M MM 1 2 0 1 PPP . N–1 1 2 6 Q –1 Example 5. With the help of elementary operations, find the inverse of Sol. Since LM1 2 1OP A = M 3 2 3P. MN1 1 2PQ A = IA LM1 2 1OP LM1 0 0OP MM3 2 3PP = MM0 1 0PP A N1 1 2Q N0 0 1Q Applying R2 → R2 – 3R1, R3 → R3 – R1, we get LM1 2 1OP LM 1 0 0OP MM0 –4 0PP = MM–3 1 0PP A N0 –1 1Q N–1 0 1Q 174 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Applying R2 → 1 R2 , we get 4 LM1 2 1OP LM 1 MM0 1 0PP = MM 3 4 N0 –1 1Q N –1 0 –1 4 0 LM1 0 1OP LM – 1 2 MM0 1 0PP = MM 3 4 N0 0 1Q N – 1 4 12 –1 4 –1 4 OP PP Q 0 0 A 1 Applying R3 → R3 + R2, R1 → R1 – 2R2, we have OP PP Q 0 0 A 1 Now applying R1 → R1 – R3, we get Hence LM1 0 0OP LM – 1 4 3 4 –1OP MM0 1 0PP = MM 3 4 – 1 4 0 PP A N0 0 1Q N – 1 4 – 1 4 1 Q LM– 1 4 3 4 –1OP –1 4 0 P. A = M34 MN– 1 4 – 1 4 1 PQ –1 EXERCISE 3.2 1. Find the inverse of the following matrices: LM1 –1 0 2 OP MM02 11 12 –11 PP. MN 3 2 1 6 PQ LM 7 6 2OP 2. M –1 2 4P. MN 3 3 8PQ LM 1 1 3 OP 3 –3 . 3. M 1 MN–2 –4 –4PPQ LM2 4 3 OP 3 6 5P 4. M MM2 5 2 PP. N 4 5 14Q LM L 2 –1 1 –1OOP MMAns. MM –5 –3 1 1 PPPP MM MMN 23 –13 –10 10 PPQPP N Q LM L 4 –42 20 OP OP 1 M MMAns. 130 M 20 50 –30P PP MN –9 –3 20 PQ Q N LM 1 L12 4 6 O OP MMAns. 4 MM–5 –1 –3PP PP MN–1 –1 –1PQ Q N LM L–23 29 – 64 5 – 18 5O OP MMAns. MM 10 –12 26 5 7 5 PP PP 65 25 P MM MMN 12 –2 PP 3 5 1 5 Q PQ –2 N 175 MATRICES LM L 3 –1 5 OP OP 1 M MMAns. 14 M 5 3 –1P PP MN –1 5 3 PQ Q N LM L –1 2 1OOP MMAns. MM –7 11 6PPPP N MN –2 3 2PQQ LM L 0 1 4 – 1 2 OOP MMAns. MM –1 b3 4g i b1 2g iPPPP N MN 0 1 4 1 2 PQQ LM L 1 –2 1 0 OOP MMAns. MM 1 –2 2 –3PPPP MM MMN –20 13 –2–1 13 PPQPP N Q LM L–2 1 0 1 OOP MMAns. MM 1 0 2 –1PPPP 1 –3 1P MM MMN–4 P N –1 0 –2 2 PQPQ LM 1 L 1 –1 1 O OP MMAns. 2 MM–8 6 –2PP PP MN 5 –3 1 PQ Q N LM 1 L 2 2 –3OOP MMAns. 5 MM–3 2 2 PPPP MN 3 –3 2 PQQ N LM L 3 2 6O OP MMAns. MM1 1 2PP PP N MN 2 2 5PQ Q LM 1 2 –1OP 5. M –1 1 2 P. MN 2 –1 1 PQ LM4 –1 1 OP 6. M 2 0 –1P. MN1 –1 3 PQ LM i –1 2iOP 7. M 2 0 2 P. MN –1 0 1 PQ LM1 2 3 1OP M1 3 3 2 P 8. M2 4 3 3P. MN1 1 1 1PQ LM 2 –6 –2 –3 OP M 5 –13 –41 –72 PP. 9. M –1 4 MN 0 1 0 1 PQ LM0 1 3OP 10. M 1 2 3P. MN3 1 1PQ LM2 1 2OP 11. M2 2 1P. MN1 2 2PQ LM 1 2 –2OP 12. M –1 3 0 P. MN 0 –2 1 PQ 3.12 RANK OF A MATRIX A positive number r is said to be the rank of matrix A if matrix A satisfies the following conditions. (i) There exists at least one non-zero minor of order r. (ii) Every minor of order (r + 1) and higher, if any, vanishes. The rank of matrix A is denoted by ρ(A) or r (A). Or The rank of a matrix or a linear map is the dimension of the image of the matrix or the linear map corresponding to the number of linearly independent rows or columns of the matrix or to the number of non-zero singular values of the map. 176 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Result: (a) Rank of A and A′ is same. (b) For a rectangular matrix A of order m × n, rank of A ≤ min (m, n) i.e., rank cannot be exceed the smaller of m and n. Minor of ) matrix. The determinant corresponding to any r × r submatrix of m × n matrix. A is called a minor of order r of the matrix A of order m × n. Example. If A = LM2 3 5OP , then 2 1 N1 2 3Q 3 , 2 3 2 5 2 , 3 1 5 are all minors of order 2 of A. 3 3.12.1 Normal Form The normal form of matrix A of rank r is one of the forms Ir, LMI 0OP, I N 0 0Q r r 0, LMI OP N 0Q r where Ir is an identity matrix of order r. This form can be obtained by the application of both elementary row and column operations on any given matrix A. 3.12.2 Procedure to Obtain Normal Form Consider Am × n = Im × m· Am × n· In × n Apply elementary row operations on A and on the prefactor Im × m and apply elementary column operations on A and on the postfactor In × n, such that A on the L.H.S. reduces to normal form. Then Im × m reduces to Pm × m and In × n reduces to Qn × n; resulting in N = PAQ. Here P and Q are non-singular matrices. Thus for any matrix of rank r, there exist nonsingular matrices P and Q such that PAQ = N = LM I 0OP . N 0 0Q r 3.12.3 Echelon Form A matrix A = [aij] is an echelon matrix or is said to be in echelon form, if the number of zeros preceding the first non-zero entry (known as distinguished elements) of a row increases row by row until only zero rows remain. In row reduced echelon matrix, the distinguished elements are unity and are the only non-zero entry in their respective columns. “The number of non-zero rows in an Echelon form is the rank”. Example 1. Find the rank of matrix Sol. Let LM 2 3 –2 4OP MM 3 –2 1 2PP. MN –23 42 03 45PQ LM 2 3 –2 4OP 3 –2 1 2 A = M MM 3 2 3 4PPP N–2 4 0 5Q (U.P.T.U., 2006) 177 MATRICES R1 → R1 + R4, R2 → R2 – R3, we get LM 0 7 –2 9 OP 0 –4 –2 –2 A ~ M MM 3 2 3 4 PPP N–2 4 0 5 Q R3 → R3 + R4 LM 0 7 –2 9 OP 0 –4 –2 –2 ~ M MM 1 6 3 9 PPP N–2 4 0 5 Q Q R1 ↔ R3 R4 → R4 + 2R1 Now, LM 1 6 3 9 OP 0 –4 –2 –2P ~ M MM 0 7 –2 9 PP N–2 4 0 5 Q LM1 6 3 9 OP 0 –4 –2 –2 A ~ M MM0 7 –2 9 PPP N0 16 6 23Q |A| = –4 7 16 –2 –2 6 –2 –9 23 (Expanded w.r. to first column) = – 4 (–46 –54) + 2 (161 – 144) –2 (42 + 32) = 400 + 34 – 148 = 286 ⇒ |A| ≠ 0 Thus there is a non-singular minor of order 4. Hence ρ(A) = 4. Example 2. Find the rank of matrix A by echelon form. LM2 3 –1 –1OP 1 –1 –2 –4 A = M MM 3 1 3 –2PPP. N6 3 0 –7 Q Sol. Applying R1 ↔ R3, we get LM1 –1 –2 –4OP M2 3 –1 –1PP A ~ M3 1 MN6 3 03 −–27PQ (U.P.T.U., 2005) 178 A TEXTBOOK OF ENGINEERING MATHEMATICS—I R2 → R2 – 2R1, R3 → R3 – 3R1, R4 → R4 – 6R1 LM1 –1 –2 –4OP M0 5 3 7 PP A ~ M0 4 MN0 9 129 1710PQ Applying R3 → R3 – 4 9 R , R → R4 – R , we get 5 2 4 5 2 1 –1 –2 –4 0 5 3 7 ~ 0 0 33 5 22 5 0 0 33 5 22 5 LM MM MN In the last, we apply R4 → R4 – R3, we get LM1 0 A ~ M MM0 N0 –1 5 0 0 –2 3 33 5 0 OP PP PQ OP PP PQ –4 7 22 5 0 Here the number of non-zero rows = 3 Therefore ρ(A) = 3 Example 3. Find the rank of following matrix: LM 3 4 5 6 7 OP MM 45 65 76 78 89 PP A = MM10 11 12 13 14PP. MN15 16 17 18 16PQ Sol. Applying R1 → R1 – R2 and then again R1 → – R1 LM 1 1 1 1 1 OP MM 54 65 67 78 89 PP A ~ MM10 11 12 13 14PP MN15 16 17 18 16PQ R2 → R2 – 4R1, R3 → R3 – 5R1, R4 → R4 – 10R1, R5 → R5 – 15R1 LM1 1 1 1 1OP MM00 11 22 33 44PP ~ MM0 1 2 3 4PP MN0 1 2 3 4PQ 179 MATRICES Applying R3 → R3 – R2, R4 → R4 – R2, R5 → R5 – R2, we get LM1 1 1 1 1OP MM00 10 02 03 04PP A ~ MM0 0 0 0 0PP MN0 0 0 0 0PQ Here number of non-zero rows = 2 Therefore ρ(A) = 2. Example 4. Reduce the matrix A to its normal form, where LM0 1 –3 –1OP 1 0 1 1 A = M MM3 1 0 2 PPP and hence find the rank of A. N1 1 –2 0 Q Sol. Applying R1 ↔ R2, we have LM1 0 1 1 OP 0 1 –3 –1P A ~ M MM3 1 0 2 PP N1 1 –2 0 Q R3 → R3 – 3R1, R4 → R4 – R1 LM1 0 1 1 OP 0 1 –3 –1 ~ M MM0 1 –3 –1PPP N0 1 –3 –1Q R3 → R3 – R2, R4 → R4 – R2 C3 → C3 – C1, C4 → C4 – C1 LM1 0 1 1 OP 0 1 –3 –1P ~ M MM0 0 0 0 PP N0 0 0 0 Q LM1 0 0 0 OP 0 1 –3 –1 A ~ M MM0 0 0 0 PPP N0 0 0 0 Q 180 A TEXTBOOK OF ENGINEERING MATHEMATICS—I C3 → C3 + 3C2, C4 → C4 + C2, we get LM1 0 0 0OP 0 1 0 0 LI 0O P A ~ M =M MM0 0 0 0PP N 0 0PQ N0 0 0 0Q 2 Hence ρ (A) = 2. Example 5. Reduce the matrix A to its normal form, when LM 1 2 –1 4 OP 2 4 3 4 A = M MM 1 2 3 4 PPP. N–1 –2 6 –7 Q Hence, find the rank of A. Sol. LM 1 2 –1 4 OP 2 4 3 4 A = M MM 1 2 3 4 PPP N–1 –2 6 –7Q Applying R2 → R2 – 2R1, R3 → R3 – R1, R4 → R4 + R1 LM1 2 –1 4 OP 0 0 5 –4 ~ M MM0 0 4 0 PPP N0 0 5 –3Q Applying C2 → C2 – 2C1, C3 → C3 + C1, C4 → C4 – 4C1 C3 ↔ C2 R3 → R3 – LM1 0 0 0 OP 0 0 5 –4 ~ M MM0 0 4 0 PPP N0 0 5 –3Q LM1 0 0 0 OP 0 5 0 –4 ~ M MM0 4 0 0 PPP N0 5 0 –3Q 4 R , R → R4 – R2 5 2 4 LM1 0 0 0 OP 0 5 0 –4 ~ M MM0 0 0 16 5PPP N0 0 0 1 Q (U.P.T.U., 2001, 2004) 181 MATRICES C3 ↔ C4 R2 → R2 + 4R4, Applying R4 → R4 – LM1 0 0 0OP 0 5 –4 0 ~ M MM0 0 16 5 0PPP N0 0 1 0Q LM1 0 0 0OP 0 5 0 0 ~ M MM0 0 16 5 0PPP N0 0 1 0Q 16 1 5 R3, then R2 → R and R3 → R. 5 5 2 16 3 1 0 0 0 0 1 0 0 I3 0 ~ = 0 0 1 0 0 0 0 0 0 0 OP PP LM PQ N LM MM MN ρ (A) = 3. OP Q Example 6. Find the non-singular matrices P and Q such that the normal form of A is PAQ where LM1 3 6 –1OP A = M1 4 5 1 P MN1 5 4 3 PQ . Hence, find its rank. 3×4 Sol. Here we consider A3 × 4 = I3 × 3· A3 × 4· I4 × 4 LM1 3 6 –1OP L1 MM1 4 5 1 PP = MM0 N1 5 4 3 Q MN0 0 1 0 As Am × n = Im × m· Am × n· In × n OP LM10 0P A M M0 1PQ M N0 0 1 0 0 OP LM10 0P A M M0 1PQ M N0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 1 Applying R2 → R2 – R1, R3 → R3 – R1 (pre), we get LM1 3 6 –1OP LM 1 0 MM0 1 –1 2 PP = MM –1 1 N0 2 –2 4 Q N –1 0 R3 → R3 – 2R1 (pre) 0 OP PP PQ 0 0 1 0 LM1 3 6 –1OP LM 1 0 0OP LM10 01 00 00OP MM0 1 –1 2 PP = MM–11 –21 01PP A MM0 0 1 0PP Q MN0 0 0 1PQ N0 0 0 0 Q N OP PP PQ 182 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Applying C2 → C2 – 3C1, C3 → C3 – 6C1, C4 → C4 + C1 (post), we get LM1 0 0 0OP LM 1 0 0OP LM10 –31 –60 01OP MM0 1 –1 2PP = MM–11 –21 10PP × A MM0 0 1 0PP Q MN0 0 0 1PQ N0 0 0 0Q N C3 → C3 + C2, C4 → C4 – 2C2 (post) LM1 0 0 0OP LM 1 0 0OP LM10 –31 –91 –27 OP MM0 1 0 0PP = MM–11 –21 10PP × A MM0 0 1 0 PP Q MN0 0 0 1 PQ N0 0 0 0Q N Therefore, I2 = N = PAQ, where LM 1 0 0OP LM10 –31 –91 –27 OP P P = M–1 1 0P, Q = M M 0 0 1 0P MN 1 –2 1PQ M0 0 0 1 P Q N and Rank of A = 2. Example 7. Find non-singular matrices P and Q such that PAQ is the normal form where LM 1 –1 2 –1 OP A = M 4 2 –1 2 P MN 2 2 –2 0 PQ . 3 × 4 Sol. Here we consider A3 × 4 = I3 × 3· A3 × 4· I4 × 4 LM1 –1 2 –1OP L1 MM4 2 –1 2 PP = MM0 N2 2 –2 0 Q MN0 0 1 0 OP LM PP MM Q MN 1 0 0 0 A 0 1 0 As Am × n = Im × m Am × n In × n 0 1 0 0 0 0 1 0 0 0 0 1 Applying R2 → R2 – 4R1, R2 → R2 – 2R1 (pre), we get LM1 MM0 N0 –1 6 4 2 –9 –6 OP LM 1 0 0OP LM10 PP = MM – 4 1 0PP A MM0 Q N – 2 0 1Q MN0 –1 6 2 0 1 0 0 0 0 1 0 OP PP PQ 0 0 0 1 Applying C2 → C2 + C1, C3 → C3 – 2C1, C4 → C4 + C1 (post) LM1 0 0 0OP LM 1 0 0OP LM10 11 –20 01OP MM0 6 –9 6PP = MM–4–2 10 01PP A MM0 0 1 0PP Q MN0 0 0 1PQ N0 4 –6 2Q N OP PP PQ 183 MATRICES R2 → 1 1 R2, R3 → R (pre) 3 2 3 LM1 0 0 0OP LM 1 0 0 OP LM10 11 –20 10OP MM0 2 –3 2PP = MM– 4 3 1 3 0 PP A MM0 0 1 0PP N0 2 –3 1Q N –1 0 1 2Q MN0 0 0 1PQ R3 → R3 – R2 (pre) –2 LM1 0 0 0 OP LM 14 10 0 OP LMM10 11 0 10OPP MM0 2 –3 2 PP = MM– 3 3 0 PP A MM0 0 1 0PP N0 0 0 –1Q N1 3 – 1 3 1 2Q N0 0 0 1Q C3 ↔ C4 (post) LM1 0 0 0 OP LM 1 0 0 OP LM10 11 10 –20 OP MM0 2 2 3PP = MM– 4 3 1/ 3 0 PP A MM0 0 0 1 PP N0 0 1 0 Q N 1 3 – 1 3 1 2Q MN0 0 1 0 PQ C3 → C3 – C2, C4 → C4 + 3 C (post) 2 2 LM1 0 0 0OP LM 1 0 0 OP LM 10 11 –10 –31 22 OP MM0 2 0 0PP = MM – 4 3 1 3 0 PP A MM 0 0 0 1 PP N0 0 –1 0Q N 1 3 – 1 3 1 2 Q MN 0 0 1 0 PQ Applying R2 → 1 R , R → (– 1) R3 (pre) 2 2 3 LM1 0 0 0OP LM 1 0 MM0 1 0 0PP = MM– 2 3 1 6 N0 0 1 0Q N – 1 3 1 3 ⇒ Therefore, And OP LM PP MM Q MN 1 0 0 0 A 0 –1 2 0 1 1 0 0 0 –1 0 1 OP 32 P 1 P P 0 Q –1 2 I3 = PAQ LM 1 0 0 OP LM10 11 –10 0 P, Q = M P = M– 2 3 1 6 MN – 1 3 1 3 – 1 2PQ MM00 00 01 N ρ (A) = 3. Note: In such problems other Ans. is possible. –1 2 32 1 0 OP PP PQ 184 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 8. Find the rank of LM 6 1 3 8 OP 16 4 12 15 P A= M MM 5 3 3 8 PP. N 4 2 6 –1Q Sol. Applying C1 ↔ C2, C3 → 1 C 3 3 LM1 6 1 8 OP 4 16 4 15P A ~ M MM3 5 1 8 PP N2 4 2 –1Q By C2 → C2 – 2C1, C3 → C3 – C1, we have LM1 4 0 8 OP M4 8 0 15P ~ M3 –1 –2 4 P MN2 0 0 –1PQ By R2 → R2 – 2R1 LM1 4 0 8 OP 2 0 0 1P ~ M MM3 1 2 4 PP N2 0 0 1Q R3 → R3 – R2, R4 → R4 – R2, we get LM1 4 0 8 OP 2 0 0 –1 ~ M MM1 –1 –2 5 PPP N0 0 0 0 Q ⇒ |A| = 0 i.e., minor of order 4 = 0 Next, we consider a minor of order 3 ∴ ∴ 1 2 1 4 0 –1 0 0 –2 = 1 (0 – 0) – 4 (– 4 – 0) + 0 = 16 ≠ 0 ρ (A) = 3. Example 9. Find the value of a such that the rank of A is 3, where LM1 1 –1 0OP 4 4 –3 1 P A= M MM a 2 2 2PP. N 9 9 a 3Q 185 MATRICES LM1 1 –1 0OP 4 4 –3 1P A = M MMa 2 2 2PP N9 9 a 3Q Sol. Applying R2 → R2 – 4R1, R3 → R3 – 2R1, R4 → R4 – 9R1, we have LM 1 1 –1 0OP 0 0 1 1 A ~ M MMa – 2 0 4 2PPP N 0 0 a + 9 3Q Again R3 → R3 – 4R2, R4 → R4 – 3R2 LM 1 1 –1 0 OP 0 0 1 1P ~ M MMa – 2 0 0 –2PP N 0 0 a+6 0 Q R4 ↔ R3 LM 1 1 –1 0 OP 0 0 1 1 ~ M MM 0 0 a + 6 0 PPP Na – 2 0 0 –2Q Cases: (i) If a = 2, |A| = 1·0·8. (– 2) = 0, rank of A = 3. (ii) If a = – 6, no. of non-zero rows is 3, rank of A = 3. Example 10. For which value of ‘b’ the rank of the matrix. LM1 5 4 OP A = M0 3 2 P is 2. MN b 13 10PQ (U.P.T.U., 2008) Sol. Since the rank of matrix A is 2 so the minor of 3rd order must be zero i.e., |A| = 0. Thus 1 5 4 0 3 2 b 13 10 = 0 (30 – 26) – 5 (0 – 2b) + 4 (0 – 3b) = 0 ⇒ Hence 4 + 10b – 12b = 0 ⇒ b = 2 4 – 2b = 0 186 A TEXTBOOK OF ENGINEERING MATHEMATICS—I EXERCISE 3.3 Find the rank of the following matrix by reducing normal form: 1 2 –3 1 2 –1 3 1 0 –1 4 1 2 1 . 1. (U.P.T.U., 2001) 2. 3 –1 1 3 –1 1 2 –1 1 0 1 2 0 1 3 1 0 [Ans. ρ(A) = 3] LM MM MM MN OP PP PQ LM MM MN OP PP PP PQ 9 1 –1 . 9 9 LM1 2 3OP LM9 7 3 6OP 3. M1 4 2P. [Ans. 2] 4. M 5 –1 4 1 P. MN2 6 5PQ MN6 8 2 4PQ LM0 0 0 0 0OP LM 1 2 –1 3OP M0 1 2 3 4 P M4 1 2 1P 5. M 3 –1 1 2 P. [Ans. 3] 6. M0 2 3 4 1 P. MN0 3 4 1 2PQ MN 1 2 0 1PQ (U.P.T.U. special exam., 2001) LM1 2 3 4 5OP LM –1 2 –2OP 2 3 4 5 6 P M 7. M 3 4 5 6 7 P. [Ans. 3] 8. M 1 2 1 P. MN –1 –1 0 PQ MN 4 5 6 7 8PQ LM1 2 1 0 OP LM1 1 2 OP MM23 –12 21 25 PP . [Ans. 3] 10. M1 2 3 P. 9. M P 5 6 3 2 MN0 –1 –1PQ PP MM N1 3 –1 –3Q [Ans. 4] [Ans. 3] [Ans. 3] [Ans. 3] [Ans. 2] Find the rank of the following matrix. LM1 2 3 4 OP 2 3 4 5 11. M MM 3 4 5 6 PPP. MN 4 5 6 7 PQ LM 3 –2 0 –1 –7 OP M 0 2 2 1 –5P 13. M 1 –2 –3 –2 1 P. MN 0 1 2 1 –6PQ LM 1 2 –1 4 OP 5 . 15. M 2 4 3 MN–1 2 6 –7PPQ 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 LM2 1 3 4OP M0 3 4 1 P [Ans. 3] 12. M2 3 7 5P. MN2 5 11 6PQ LM1 2 3OP [Ans. 4] 14. M2 3 1P. MN3 1 2PQ LM 0 i –iOP [Ans. 2] 16. M – i 0 – i P. MN–3 1 0 PQ [Ans. 3] [Ans. 3] [Ans. 3] 187 MATRICES LM 3 –1 –2OP 17. M –6 2 –4 P. MN 3 1 –2PQ LM1 2 3 1OP 19. M2 4 6 2P. MN1 2 3 2PQ LM0 2 3OP [Ans. 2] 18. M0 4 6P. MN0 6 9PQ LM 4 2 3 OP [Ans. 2] 20. 8 MM 4 6 PP. N –2 –1 –15Q LM1 –2 3OP 21. M 2 –1 2P. MN3 1 2PQ LM 3 4 5 6 7 OP MM 45 65 76 78 98 PP 23. MM10 11 12 13 14PP. MN15 16 17 18 19PQ LM 5 6 7 8 OP M6 7 8 9P 25. M11 12 13 14 P. MN16 17 18 19PQ LM 5 3 14 4OP 27. M 0 1 2 1 P. MN1 –1 2 0PQ LM 1 2 –5OP [Ans. 3] 22. M –4 1 –6P. MN 6 3 –4PQ [Ans. 1] [Ans. 1] Find the Echelon form of the following matrix and hence find the rank. LM2 3 –1 –1OP M1 –1 –2 –4P [Ans. 2] 24. M 3 1 3 –2 P. MN6 3 0 –7 PQ LM–2 –1 –3 –1OP M 1 2 –3 –1P [Ans. 3] 26. M 1 0 1 1 P. MN 0 1 1 –1PQ LM0 1 3 –2OP M0 4 –1 3 P [Ans. 3] 28. M 0 0 2 1 P. MN0 5 –3 4 PQ [Ans. 2] [Ans. 3] [Ans. 3] [Ans. 2] 29. Determine the non-singular matrices P and Q such that PAQ is in the normal form for A. Hence find the rank of A. LM3 2 –1 5 OP –2 P. A = M5 1 4 MN1 –4 11 –19PQ LM LM1 MM LM 0 0 1 OP MM0 MMAns. P = M 01 1 3 – 5 3P Q = M MM – 1 3 1 6 PP MM0 MM N2 Q M MN MN0 4 17 17 0 0 9 119 –1 7 –1 17 0 OP OP PP PP PP PP rank = 2 PP 1 PP PP 31 Q Q 9 217 –1 7 0 (other forms are also possible) 188 A TEXTBOOK OF ENGINEERING MATHEMATICS—I LM2 1 –3 –6OP 30. A = M 3 –3 1 2 P. (U.P.T.U., 2002) MN1 1 1 2 PQ LM OP LM 0 0 1 OP LM1 –1 4 0 OP –5 0 1 0 MMAns. P = M –1 0 2 P, Q = M PP and rank = 3PP M –2 0 0 1 MM –3 1 9 PP M MM PP N 14 28 28 Q N0 0 0 1 PQ N Q LM LM 1 0 0OP LM1 –1 –1OPOP LM1 1 2 OP MMAns. P = M–1 1 0P, Q = M0 1 –1PPP 31. A = M1 2 3 P. MN–1 1 1PQ MN0 0 1 PQQ N MN0 –1 –1PQ LM LM1 1 3 –4/ 3 – 1 OP OP 1 0 0 L O 3 LM 1 2 3 –2OP M M M P 0 –1 6 –5 6 7 6P P rank = 2 –2 1 0 , = Ans. P = Q M PP PP 32. A = M 2 –2 1 3 P. MM PP MM0 0 M 1 0 MN 3 0 4 1 PQ M N–1 –1 1Q M0 0 P 0 1 PQ Q N N 3.13 SYSTEM OF LINEAR EQUATIONS (NON-HOMOGENEOUS) Let us consider the following system of m linear equations in n unknowns x1, x2, ....., xn: a11 x1 + a12 x 2 + .... + a1n xn = b1 a 21 x1 + a 22 x 2 + .... + a 2n xn = b2 = ... .... .... .... a m1 x1 + am 2 x 2 + .... + amn x n = bn U| |V || W ...(i) In matrix notation these equations can be put in the form AX = B ...(ii) LM a a A = M MM .... Na 21 a12 a22 m1 .... am2 11 where OP P .... P P a Q a1n a2n .... .... mn LM x OP LM b OP b x X = M P and B = M P MM M PP MM M PP Nx Q Nb Q 1 1 2 2 n n 189 MATRICES Augmented matrix: The augmented matrix [A : B] or à of system (i) is obtained by augmenting A by the column B. LM a a à = [A : B] = M MM L Na 11 21 i.e., m1 OP P L L L M LP P M b Q a a a12 L a1n M b1 a22 L a2n M b2 m2 mn n [U.P.T.U., (C.O.), 2003] 3.13.1 Conditions for Solution of Linear Equations Consistent: If the ranks of A and augmented matrix [A : B] are equal, then the system is said to be consistent otherwise inconsistent. There are following conditions for exist the solution of any system of linear equations : (i) If ρ(A) = ρ[A : B] = r = n (where n is the number of variables) then the system has a unique solution. (ii) If ρ(A) = ρ[A : B] = r < n. then the system has infinitely many solutions in terms of remaining n – r unknowns which are arbitrary. If n – r = 1 (then solution is one variable independent solution and let equal to K). n – r = 2 (then solution is two variable independent solution and let variables equal to K1, and K2) and so on. Trivial solution: It is a solution where all xi are zero i.e., x1 = x2 ... = xn = 0. Example 1. Check the consistency of the following system of linear nonhomogeneous equations and find the solution, if exists: (U.P.T.U., 2007) 7x1 + 2x2 + 3x3 = 16 2x1 + 11x2 + 5x3 = 25 x1 + 3x2 + 4x3 = 13. LM7 2 3OP LMx OP LM16OP A = M 2 11 5P , X = Mx P , B = M 25P MN1 3 4PQ MNx PQ MN13PQ LM7 2 3 M 16OP [A : B] = M2 11 5 M 25P MN1 3 4 M 13PQ 1 Sol. Here, 2 3 The augmented matrix Applying R1 ↔ R3, we get [A : B] ~ Again R2 → R2 – 2R1, R3 → R3 – 7R1 LM1 3 4 M 13OP MM 2 11 5 M 25PP N7 2 2 M 16Q LM1 3 4 M 13 OP ~ 0 MM 5 –3 M –1 PP N0 –19 –25 M –75Q 190 A TEXTBOOK OF ENGINEERING MATHEMATICS—I R3 → R3 + 19 R2 5 LM1 3 4 M 13 OP M –3 M –1 PP ~ M0 5 MM0 0 –182 M – 394 PP N 5 5 Q ⇒ ρ (A) = ρ [A : B] = 3 ⇒ r = n = 3. ∴ The system is consistent. The given system has a unique solution. Now from AX = B LM1 MM0 MN0 OP Lx O LM 13 OP M P –3 P Mx P = M –1 P MM –394 PP –182 P M P P P M x 5 Q N Q N 5 Q 3 4 5 2 0 3 LM x + 3x + 4x OP L 1 3 O MM 5x – 3x PP = MMM – 1 PPP MM –182 x PP MM – 39 4 PP Q N 5 Q N 5 1 ⇒ 1 2 2 3 3 3 ⇒ x1 + 3x2 + 4x3 = 13 ...(i) 5x2 – 3x3 = –1 ...(ii) 182 394 x3 = 5 5 On solving these equations, we get the final solution ...(iii) 95 100 197 , x2 = , x3 = . 91 91 91 Example 2. Test the consistency of following system of linear equations and hence find the solution. (U.P.T.U., 2005) 4x1 – x2 = 12 – x1 + 5x2 – 2x3 = 0 – 2x2 + 4x3 = – 8 x1 = LM 4 –1 0 M 12OP Sol. The augmented matrix [A : B] = M –1 5 –2 M 0 P MN 0 –2 4 M –8PQ Applying R1 ↔ R2, we get LM –1 5 –2 M 0 OP [A : B] ~ M 4 –1 0 M 12 P MN 0 –2 4 M –8PQ 191 MATRICES R2 → R2 + 4R1 LM –1 5 –2 M 0 OP ~ 0 19 –8 M 12 P MM N 0 –2 4 M –8PQ R1 → – R1, R3 → R3 + 2 R, 19 2 LM1 ~ M0 MM N0 –5 2 M 19 –8 60 19 M 0 M OP 12 P –128 P P 19 Q 0 ∴ ρ(A) = ρ [A : B] = 3 i.e., r = n = 3 (The system is consistent). Hence, there is a unique solution ⇒ ⇒ LM –1 MM 0 MM 0 N –5 19 0 OP L x O L 0 O PM P M P –8 P M x P = M 12 P MM –128 PP 60 P MM PP x N 19 Q 19 PQ N Q 2 1 2 3 x1 – 5x2 + 2x3 = 0 ...(i) 19x2 – 8x3 = 12 60 x 3 –128 = 19 19 32 ⇒ x3 = – 15 putting the value of x3 in equation (ii), we get 32 19x2 – 8 – = 12 15 256 76 =– ⇒ 19x2 = 12 – 15 15 –76 4 =– ⇒ x2 = 15 × 19 15 and putting the values of x1, x2 in equation (i), we get 4 32 +2 − x1 – 5 − = 0 15 15 ...(ii) FG H IJ K FG IJ FG IJ H K H K x1 + ⇒ ⇒ Hence, 20 64 – = 0 15 15 44 x1 – = 0 15 44 x1 = 15 4 32 44 x1 = , x2 = – and x3 = – . 15 15 15 ...(iii) 192 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 3. Solve 2x1 – 2x2 + 4x3 + 3x4 = 9 x1 – x2 + 2x3 + 2x4 = 6 2x1 – 2x2 + x3 + 2x4 = 3 x1 – x2 + x4 = 2 Sol. The augmented matrix is LM 2 –2 4 3 M 9OP 1 –1 2 2 M 6 [A : B] = M MM 2 –2 1 2 M 3PPP N1 –1 0 1 M 2Q R1 ↔ R2 LM1 –1 2 2 M 6OP 2 –2 4 3 M 9 P ~ MM MN21 –2–1 10 21 MM 32PPQ R2 → R2 – 2R1, R4 → R4 – R1, R3 → R3 – 2R1 LM1 −1 2 2 M 6 OP M0 0 0 –1 M –3PP ~ M MN00 00 –3–2 –2–1 MM –4–9PQ R2 → (– 1) R2, R3 → (– 1) R3, R4 → (– 1) R4 LM1 –1 2 2 M 6 OP M0 0 0 1 M 3 PP ~ M MN00 00 23 21 MM 94 PQ R3 → R3 – R4 R3 ↔ R2 LM1 –1 2 2 M 6 OP 0 0 0 1 M 3P ~ MM MN00 00 12 11 MM 45PPQ LM1 –1 2 2 M 6 OP 0 0 1 1 M 5P ~ MM MN00 00 02 11 MM 34PPQ 193 MATRICES R4 ↔ R3 LM1 –1 2 2 M 6 OP 0 0 1 1 M 5 ~ M MM0 0 2 1 M 4PPP N0 0 0 1 M 3 Q R3 → R3 – 2R2 LM1 –1 2 2 M 6 OP LM1 –1 2 2 M 6OP 0 0 1 1 M 5 P M0 0 1 1 M 5 P ~ M MM0 0 0 −1 M –6PP ~ MM0 0 0 1 M 6PP bR → – R g N 0 0 0 1 M 3 Q N 0 0 0 1 M 3Q 3 3 R4 → R4 – R3 and then R4 → (– 1) R4 LM1 –1 2 2 M 6 OP M0 0 1 1 M 5PP ~ M MN00 00 00 10 MM 63 PQ Hence, ρ (A) = 3 and ρ [A : B] = 4 ⇒ ρ (A) ≠ ρ [ A : B]. So the given system is inconsistent and therefore it has no solution. Example 4. Investigate for what values of λ, µ the equations x + y + z = 6, x + 2y + 3z = 10, x + 2y + λz = µ have (i) no solution (ii) a unique solution (iii) an infinity of solutions. Sol. The augmented matrix LM1 1 1 M 6 OP [A : B] = M1 2 3 M 10 P MN1 2 λ M µ PQ R2 → R2 – R1, R3 → R3 – R1 R3 → R3 – R1 LM1 1 1 M 6 OP 2 M 4 P ~ M0 1 MN0 1 λ – 1 M µ – 6PQ LM1 1 1 M 6 OP 2 M 4 ~ M0 1 MN0 0 λ – 3 M µ – 10PPQ (i) For no solution ρ (A) ≠ ρ [A; B] it is only possible when λ = 3. (U.P.T.U., 2001) 194 A TEXTBOOK OF ENGINEERING MATHEMATICS—I (ii) For unique solution ρ (A) = ρ [A ® B] it is only possible when λ – 3 ≠ 0 i.e., λ ≠ 3 and µ ≠ 10. (iii) For infinite number of solutions ρ (A) = ρ [A : B] = r < n it is only possible when λ = 3 and µ = 10. Example 5. Show that the equations x+ y+ z=6 x + 2y + 3z = 14 x + 4y + 7z = 30 are consistent and solve them. Sol. The augmented matrix is LM1 1 1 M 6 OP [A : B] = M1 2 3 M 14 P MN1 4 7 M 30PQ R2 → R2 – R1, R3 → R3 – R1 R3 → R3 – 3R2 LM1 1 1 M 6 OP ~ M0 1 2 M 8 P MN0 3 6 M 24PQ LM1 1 1 M 6OP ~ M0 1 2 M 8 P MN0 0 0 M 0PQ Hence, i.e., ρ (A) = ρ [A : B] = 2 r = 2 < 3 (n = 3) ∴ n – r = 3 – 2 = 1 (one variable independent solution). The system is consistent and have infinitely solutions. Now ∴ AX = B LM1 1 1OP LMx OP LM6OP MM0 1 2PP MM y PP = MM8PP N0 0 0 Q N z Q N0Q x+y+z = 6 ...(i) y + 2z = 8 ...(ii) Let z = k Putting z = k in (ii), we get y + 2k = 8 ⇒ y = 8 – 2k From (i) x + 8 – 2k + k = 6 ⇒ x = k – 2 Therefore, x = k – 2, y = 8 – 2k and z = k. Ans. 195 MATRICES Example 6. Solve 3x + 3y + 2z = 1 x + 2y = 4 10y + 3z = – 2 2 x – 3y – z = 5 Sol. The augmented matrix is 3 3 1 2 [A : B] = 0 10 2 −3 R1 ↔ R3 1 2 3 3 ~ 0 10 2 −3 LM MM MN LM MM MN OP PP PQ 0 : 4O 2 : 1P P 3 : −2P P −1 : 5Q 2 0 3 −1 : : : : 1 4 −2 5 0 2 3 −1 : : : : 4 −11 −2 −3 R2 → R2 – 3R1, R4 → R4 – 2R1 LM1 0 ~ M MM0 N0 R3 → R3 + OP PP PQ 10 7 R , R → R4 – R2 3 2 4 3 LM1 M0 ~ M MM0 MM0 N R3 → 2 −3 10 −7 3 3 R3, R4 → R 29 17 4 R4 → R4 + R3 LM1 0 ~ M MM0 N0 LM1 0 ~ M MM0 N0 2 0 : −3 2 29 3 −17 3 : 0 0 2 −3 0 0 0 2 1 −1 2 −3 0 0 0 2 1 0 : : : : : : : : : : OP −11 P −116 P P 3 P 68 P 3 PQ 4 OP PP PQ 4 −11 −4 4 OP PP PQ 4 −11 −4 0 196 A TEXTBOOK OF ENGINEERING MATHEMATICS—I ⇒ i.e., ρ (A) = ρ [A : B] = 3 r = 3 = n = number of variables. Hence, the system is consistent and has unique solution. Now, AX = B ⇒ LM1 2 0OP LxO LM 4OP MM00 −30 21PP MMyPP = MM−−114PP MN0 0 0PQ MNz PQ MN 0PQ ⇒ x + 2y = 4 ...(i) – 3y + 2z = – 11 z = –4 ...(ii) ...(iii) On solving (i) and (ii), we get x = 2, y = 1. Hence, x = 2, y = 1 and z = – 4. Example 7. Apply the matrix method to solve the system of equations x + 2y – z = 3 3x – y + 2z = 1 2x – 2y + 3z = 2 x– y +z = – 1 [U.P.T.U., 2003; U.P.T.U. (C.O.), 2003] Sol. The augmented matrix is LM1 2 −1 : 3OP 3 −1 2 : 1P [A : B] = M MM2 −2 3 : 2PP N1 −1 1 : −1Q R2 → R2 – 3R1, R3 → R3 – 2R1, R4 → R4 – R1 LM1 2 0 −7 ~ M MM0 −6 N0 −3 R3 → R3 – −1 5 5 2 OP PP PQ : : : : 3 −8 −4 −4 6 3 R2, R4 → R4 – R 7 7 2 LM1 MM0 M0 ~ M MM0 N 2 −1 : −7 5 5 7 1 − 7 : 0 0 : : OP −8 P 20 P P 7 P 4 − PP 7Q 3 197 MATRICES R4 → R4 + ⇒ 1 R 5 3 LM1 2 −1 : 3OP M0 −7 55 : 20−8PP ~ M MM0 0 7 : 7 PP 0Q N0 0 0 ρ (A) = ρ [A : B] = 3 = number of variables. Hence, the system is consistent and has a unique solution. Now, AX = B ⇒ LM 3 OP LM1 2 −1OP MM0 −7 55PP LMxyOP = MM −208 PP MM0 0 7 PP MMNz PPQ MM 7 PP MN 0 PQ N0 0 0Q ⇒ x + 2y – z = 3 – 7y + 5z = – 8 ...(i) ...(ii) 5 20 z = ⇒z=4 7 7 From (i) and (ii), we get, x = – 1, y = 4 ⇒ x = – 1, y = 4, z = 4. 3.14 SYSTEM OF HOMOGENEOUS EQUATIONS If in the set of equations (1) of (3.13), b1 = b1 = ... = bn = 0, the set of equation is said to be homogeneous. Result 1: If r = n, i.e., the rank of coefficient matrix is equal to the number of variables, then there is always a trivial solution (x1 = x2 = ... = xn = 0). Result 2: If r < n, i.e., the rank of coefficient matrix is smaller than the number of variables, then there exist a non-trivial solution. Result 3: For non-trivial solution always |A| = 0. Example 8. Solve the following system of homogeneous equations: x + 2y + 3z = 0 3x + 4y + 4z = 0 7x + 10y + 12z = 0 Sol. Here, A = LM1 2 3OP MM3 4 4PP N7 10 12Q 198 A TEXTBOOK OF ENGINEERING MATHEMATICS—I LM1 2 3OP ~ M0 −2 −8P MN0 −4 −9PQ R3 → R3 – 2R2 LM1 2 3OP A ~ M0 −2 −5P MN0 0 1PQ This shows rank (A) = 3 = number of unknowns. Hence, the given system has a trivial solution i.e., x = y = z = 0. Example 9. Solve x + y – 2z + 3w = 0 x – 2y + z – w = 0 4x + y – 5z + 8w = 0 5x – 7y + 2z – w = 0 Sol. The coefficient matrix A is LM1 1 −2 3OP 1 −2 1 −1 A = M MM4 1 −5 8PPP N5 −7 2 −1Q R2 → R2 – R1, R3 → R3 – 4R1, R5 → R5 – 5R1 R3 → R3 – R2, R4 → R4 – 4R2 LM1 1 −2 −3 0 3 ~ M MM0 −3 3 N0 −12 12 3 −4 −4 −16 OP PP PQ LM1 1 −2 0 −3 3 ~ M MM0 0 0 N0 0 0 3 −4 0 0 OP PP PQ ⇒ ρ (A) = 2 < 4 (n = 4), so there exist a non-trivial solution. Now, LM1 1 −2 0 −3 3 ⇒ M MM0 0 0 N0 0 0 AX = B OP LMx OP LM 0 OP PP MMyz PP = MM 00 PP PQ MNwPQ MN 0 PQ 3 −4 0 0 199 MATRICES ⇒ x + y – 2z + 3w = 0 – 3y + 3z – 4w = 0 ...(i) ...(ii) Choose z = k1, w = k2, then from (ii) and (i), we get 4 k 3 2 – 3y + 3k1 – 4k2 = 0 ⇒ y = k1 – and x + k1 – 4 5 k2 – 2k1 + 3k2 = 0 ⇒ x = k1 – k. 3 3 2 where k1 and k2 are arbitrary constants. Example 10. Find the value of λ such that the following equations have unique solution. λx + 2y – 2z – 1 = 0, 4x + 2λy – z – 2 = 0, 6x + 6y + λz – 3 = 0 (U.P.T.U., 2003) Sol. We have λx + 2y – 2z = 1 4x + 2λy – z = 2 6x + 6y + λz = 3 The coefficient matrix A is LMλ 2 A = M 4 2λ MN 6 6 OP PP Q −2 −1 λ For unique solution |A| ≠ 0 ∴ ⇒ LMλ 2 −2OP MM 4 2λ −1PP = λ + 11λ – 30 ≠ 0 N 6 6 λQ 3 (λ – 2) (λ2 + 2λ + 15) ≠ 0 ⇒ λ ≠ 2. Example 11. Determine b such that the system of homogeneous equations (U.P.T.U., 2008) 2x + y + 2z = 0 x + y + 3z = 0 4x + 3y + bz = 0 has (i) trivial solution and (ii) non-trivial solution. Sol. (i) For trivial solution, |A| ≠ 0 ∴ ⇒ ⇒ |A| = 2 1 4 1 1 3 2 3 ≠0 b 2 (b – 9) – (b – 12) + 2 (3 – 4) ≠ 0 2b – 18 – b + 12 – 2 ≠ 0 ⇒ b – 8 ≠ 0 ⇒ b ≠ 8. (ii) For non-trivial solution, |A| = 0 ⇒ b – 8 = 0 ⇒ b = 8. 200 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 3.15 GAUSSIAN ELIMINATION METHOD Gaussian elimination method is an exact method which solves a given system of equations in n unknowns by transforming the coefficient matrix into an upper triangular matrix and then solve for the unknowns by back substitution. Example 12. Solve the system of equations: 2x1 + 3x2 + x3 = 9 x1 + 2x2 + 3x3 = 6 3x1 + x2 + 2x3 = 8 (U.P.T.U., 2006) by Gaussian elimination method. Sol. The augmented matrix is: LM2 MM N3 [A : B] = 1 R1 ↔ R2 3 2 1 1 3 2 : : : OP PP Q 9 6 8 LM1 2 3 : 6OP ~ M 2 3 1 : 9P MN3 1 2 : 8PQ R2 → R2 – 2R1, R3 → R3 – 3R1 LM1 2 3 : 6OP ~ M0 −1 −5 : −3 MN0 −5 −7 : −10PPQ R3 → R3 – 5R2 LM1 2 3 : 6OP ~ M0 −1 −5 : −3P MN0 0 18 : 5PQ which is upper triangular form ∴ and 18x 3 = 5 ⇒ x3 = 5 18 x1 + 2x2 + 3x3 = 6 – x2 – 5x3 = – 3 From (ii) x2 + 25 25 29 = 3 ⇒ x2 = 3 – = 18 18 18 again from (i), we have x1 + 2 × ⇒ Hence, 29 5 73 +3× = 6 ⇒ x1 + =6 18 18 18 73 35 x1 = 6 – = 18 18 35 29 5 . x1 = , x2 = and x3 = 18 18 18 ...(i) ...(ii) 201 MATRICES Example 13. Solve by Gaussian elimination method 10x1 – 7x2 + 3x3 + 5x4 = 6 – 6x1 + 8x2 – x3 – 4x4 = 5 3x1 + x2 + 4x3 + 11x4 = 2 5x1 – 9x2 – 2x3 + 4x4 = 7 Sol. The augmented matrix is LM 10 −7 3 5 : 6OP −6 8 −1 −4 : 5P [A : B] = M MM 3 1 4 11 : 2PP N 5 −9 −2 4 : 7Q R3 ↔ R2 LM 10 −7 3 5 : 6OP M 3 81 −41 −114 :: 25PP ~ M−6 MN 5 −9 −2 4 : 7PQ R2 → R2 – 3 6 1 R , R → R3 + R , R → R4 – R 10 1 3 10 1 4 2 1 LM10 MM 0 M ~ M MM 0 MM MN 0 R2 → − −7 3 5 : 31 10 31 10 19 2 : 19 5 4 5 −1 : 7 2 3 2 : 11 2 − OP 1P P 5P 43 PP 5P P 4P PQ 6 10 5 2 R2, R3 → R2, R3 → − R3 31 19 11 LM10 MM 0 M ~ M MM 0 MM MN 0 −7 3 5 : 1 1 95 31 : 1 4 19 − 5 19 : 1 7 11 − 3 11 : OP 2P P 31 P 43 PP 19 P P −8 P 11 PQ 6 202 A TEXTBOOK OF ENGINEERING MATHEMATICS—I R3 → R3 – R2, R4 → R4 – R2 LM10 MM 0 ~ M MM 0 MM MN 0 R3 → − −7 3 1 1 0 0 5 : 95 31 1960 − 589 1138 − 341 15 19 4 − 11 − : : : OP 2 P P 31 P 1295 P P 589 P 270 P − 341 PQ 6 19 11 R3, R4 → − R4 15 4 LM10 MM 0 ~ M MM 0 MM MN 0 R4 → R4 – R3 LM10 MM 0 M ~ M MM 0 MM 0 N −7 3 1 1 0 1 0 1 −7 3 1 1 0 1 0 0 5 95 31 392 93 569 62 5 95 31 392 93 923 186 : : : : : : : : OP 2 P P 31 P 259 P − P 93 P 135 P 62 PQ 6 OP 2 P P 31 P 259 P − P 93 P 923 P 186 PQ 6 Hence, the coefficient matrix is an upper triangular form ∴ or and ⇒ 923 923 x4 = ⇒ x4 = 1 186 186 392 259 392 259 x3 + x = − ⇒ x3 + = − 93 4 93 93 93 259 392 651 − x3 = − = − =–7 93 93 93 95 2 95 2 x2 + x3 + x = ⇒ x2 – 7 + = 31 4 31 31 31 x2 – 122 2 2 122 124 + = = ⇒ x2 = =4 31 31 31 31 31 Again 10x1 – 7x2 + 3x3 + 5x4 = 6 ⇒ 10x1 – 7 × 4 + 3 × (–7) + 5 × 1 = 6 ⇒ 10x1 – 28 – 21 + 5 = 6 203 MATRICES ⇒ 10x1 – 44 = 6 ⇒ 10x1 = 50 ⇒ x1 = 5 Therefore, x1 = 5, x2 = 4, x3 = – 7 and x4 = 1. 3.15.1 Gauss-Jordan Elimination Method Apply elementary row operations on both A and B such that A reduces to the normal form. Then the solution is obtained. Example 14. Solve by Gauss-Jordan elimination method: 2x1 + x2 + 3x3 = 1 4x1 + 4x2 + 7x3 = 1 2x1 + 5x2 + 9x3 = 3 LM2 1 3OP LM1OP A = M 4 4 7P , B = M1P MN2 5 9PQ MN3PQ Sol. Here ∴ Augmented matrix LM2 1 3 : 1OP [A : B] = M4 4 7 : 1P MN2 5 9 : 3PQ R2 → R2 – 2R1, R3 → R3 – R1 R3 → R3 – 2R2 R1 → LM2 1 3 : 1 OP ~ M0 2 1 : −1P MN0 4 6 : 2 PQ LM2 1 3 : 1 OP ~ 0 2 1 : −1 MM0 0 4 : 4 PP N Q 1 1 1 R , R → R2 , R3 → R3 2 1 2 2 4 LM1 M ~ M0 MM MN0 R1 → R1 − 1 R 2 2 LM1 M ~ M0 MM MN0 1 2 1 0 0 1 0 3 2 1 2 1 5 4 1 2 1 : : : : : : OP PP PP 1 PQ 1 2 1 − 2 OP PP PP 1 PQ 3 4 1 − 2 204 A TEXTBOOK OF ENGINEERING MATHEMATICS—I R1 → R1 – 5 1 R3 , R2 → R2 – R3 4 2 LM1 0 0 : − 1 OP 2 ~ M0 1 0 : − 1 P MM PP N0 0 1 : 1 Q Hence, the matrix A is in normal form ∴ x1 = – 1 , x = – 1, x3 = 1. 2 2 Example 15. Solve by Gauss-Jordan elimination method: 2x1 + 5x2 + 2x3 – 3x4 = 3 3x1 + 6x2 + 5x3 + 2x4 = 2 4x1 + 5x2 + 14x3 + 14x4 = 11 5x1 + 10x2 + 8x3 + 4x4 = 4 Sol. LM2 5 2 −3 : 3 OP 3 6 5 2 : 2P [A : B] = M MM4 5 14 14 : 11PP N5 10 8 4 : 4 Q R2 → R2 – R1, R3 → R3 – R2, R4 → R4 – R3 R1 ↔ R4, R2 ↔ R3 LM2 5 2 −3 : 3 OP 1 1 3 5 : −1 ~ M MM1 −1 9 12 : 9 PPP N1 5 −6 −10 : −7Q LM1 5 −6 −10 : −7OP 1 −1 9 12 : 9 P ~ M MM1 1 3 5 : −1PP N2 5 2 −3 : 3 Q R2 → R2 – R1, R3 → R3 – R1, R4 → R4 – 2R1 LM1 5 −6 −10 : −7OP 0 −6 15 22 : 16 P ~ M MM0 −4 9 15 : 6 PP N0 −5 14 17 : 17Q R2 → R2 – R4, then R2 → – R2, R3 → R3 – 4R2, then again R3 → –R3 , R4 → R4 – 5R2 , then also again R4 → –R4 LM1 5 −6 −10 : −7 OP 0 1 −1 −5 : 1 P ~ M MM0 0 −5 5 : −10PP N0 0 9 −8 : 22 Q 205 MATRICES R3 → – 1 R , then R4 → R4 – 9R3 5 3 LM1 5 −6 −10 : −7OP 0 1 −1 −5 : 1 ~ M MM0 0 1 −1 : 2 PPP N0 0 0 1 : 4 Q R3 → R3 + R4, R2 → R2 + 5R4, R1 → R1 + 10R4 LM1 5 −6 0 : 33OP 0 1 −1 0 : 21P ~ M MM0 0 1 0 : 6 PP N0 0 0 1 : 4 Q R2 → R2 + R3, R1 → R1 + 6R3 LM1 5 0 0 : 69OP M0 1 0 0 : 27PP ~ M MM0 0 1 0 : 26PP N0 0 0 1 : 4 Q R1 → R1 – 5R2 LM1 0 0 0 : −66OP 0 1 0 0 : 27 P ~ M MM0 0 1 0 : 6 PP N0 0 0 1 : 4 Q Hence, the coefficient matrix is in normal form ∴ x1 = – 66, x2 = 27, x3 = 6 and x4 = 4. Example 16. Find the values of λ for which the equations 3x + y – λz = 0; 4x – 2y – 3z = 0; 2 λx + 4y + λz = 0 have a non-trivial solution. Obtain the most general solutions in each case. Sol. The coefficient matrix is LM 3 1 −λOP A = M 4 −2 −3 P MN2λ 4 λ PQ for non-trivial solution A = 0 ∴ LM 3 1 −λOP A = M 4 −2 −3 P = 0 ⇒ 3 (–2λ + 12) – (4λ + 6λ) – λ (16 + 4λ) = 0 NM2λ 4 λ PQ ⇒ – 6λ + 36 – 10λ – 16λ – 4λ2 = 0 ⇒ – 4λ2 – 32λ + 36 = 0 ⇒ λ2 + 8λ – 9 = 0 206 A TEXTBOOK OF ENGINEERING MATHEMATICS—I ⇒ ⇒ λ2 + 9λ – λ – 9 = 0 ⇒ (λ – 1) (λ + 9) = 0 λ = 1, – 9. Case I. If λ = 1, R2 → R2 – LM 3 1 −1 OP A = M 4 −2 −3 P MM 2 4 1 PP N Q 4 2 R , R → R3 – R1 3 1 3 3 LM3 ~ M0 MM MN0 R3 → R3 + R2 LM3 ~ M0 MM N0 1 −1 −10 3 10 3 5 − 3 5 3 1 − 10 3 0 OP PP PP Q OP 5 − P 3P 0 PQ −1 This shows ρ (A) = 2 < 3 (n = 3) ∴ Let z = k 10 y 5z k − − = 0 ⇒ 2y + k = 0 ⇒ y = – 2 3 3 and 3x + y – z = 0 Case II. If λ = – 9 ⇒ 3x – k k –k=0⇒x= . Ans. 2 2 LM 3 1 9 OP A = M 4 −2 −3P MN−18 4 −9PQ Solve itself like case I. x = − 3k 9k ,y= − , z = k. 2 2 EXERCISE 3.4 Examine whether the following systems of equations are consistent. If consistent solve. 1. x+y+z = 6 2x + 3y – 2z = 2 5x + y + 2z = 13 (U.P.T.U., 2000) [Ans. x = 1, y = 2, z = 3] 2. 3x + 3y + 2z = 1 x + 2y = 4 207 MATRICES 3. 4. 5. 6. 10y + 3z = – 2 2x – 3y – z = 5 [Ans. x = 2, y = 1 and z = – 4] x1 + 2x2 − x3 = 3 3x1 – x2 + 2x3 = 1 2x1 – 2x2 + 3x3 = 2 x1 – x2 + x3 = – 1 (U.P.T.U., 2002) [Ans. x1 = –1, x2 = 4, x3 = 4] x+y+z = 7 x + 2y + 3z = 8 y + 2z = 6 [Ans. inconsistent; no solution] 5x + 3y + 7z = 4 3x + 26y + 2z = 9 7x + 2y + 10z = 5 [Ans. x = (7 – 6k)/11, y = (3 + k)/11, z = k] – x1 + x2 + 2x3 = 2 3x1 – x2 + x3 = 6 – x1 + 3x2 + 4x3 = 4 [Ans. x1 = 1, x2 = – 1, x3 = 2] 7. Find the values of a and b for which the system has (i) no solution, (ii) unique solution and (iii) infinitely many solution. 2x + 3y + 5z = 9 (i) a = 5, b ≠ 9 7x + 3y – 2z = 8 Ans. (ii) a ≠ 5, b any value 2x + 3y + az = b (iii) a = 5, b = 9 8. Discuss the solutions of the system of equations for all values of λ. x + y + z = 2, 2x + y – 2z = 2, λx + y + 4z = 2. [Ans. Unique solution if λ ≠ 0; infinite number of solutions if λ = 0] 9. 10. x+y+z = 6 x + 2y + 3z = 14 x + 4y + 7z = 30 [Ans. x = k – 2, y = 8 – 2k, z = k] 2x – y + z = 7 3x + y – 5z = 13 x+y+z = 5 [Ans. x = 4, y = 1, z = 0] Solve the following homogeneous equations: 11. x + 2y + 3z = 0 2x + y + 3z = 0 3x + 2y + z = 0 12. 13. [Ans. x = y = z = 0] x + y + 3z = 0 x–y+z = 0 x – 2y = 0 x–y+z = 0 [Ans. x = – 2k, y = –k, z = k] x + 2y + 3z = 0 3x + 4y + 4z = 0 7x + 10y + 12z = 0 [Ans. x = y = z = 0] 208 14. A TEXTBOOK OF ENGINEERING MATHEMATICS—I 4x + 2y + z + 3w = 0 6x + 3y + 4z + 7w = 0 2x + y + w = 0 [Ans. x = k1, y = – 2k1 – k2, z = – k2, w = k2] 15. For what values of λ the given equations will have a non-trivial solution. x + 2y + 3z = λx 2x + 3y + z = λx 3x + y + 2z = λy [Ans. λ = 6] 16. Find the value of ‘a’ so that the following system of homogeneous equations have exactly 2 linearly independent solutions. ax1 – x2 – x3 = 0 – x1 + ax2 – x3 = 0 [Ans. a = – 1] – x1 – x2 + ax3 = 0 17. Apply the test of rank to examine if the following equations are consistent: 2x – y + 3z = 8 – x + 2y + z = 4 3x + y – 4z = 0 and if consistent, find the complete solution. [Ans. x = y = z = 2] 18. Show that the equations – 2x + y + z = a x – 2y + z = b x + y – 2z = c have no solutions unless a + b + c = 0, in which case they have infinitely many solutions. Find their solutions a = 1, b = 1 and c = – 2. [Ans. x = k – 1, y = k – 1, z = k] 19. Show that the system of equations x + 2y – 2w = 0, 2x – y – w = 0, x + 2y – w = 0. – y + 3z – w = 0 do not have a non-trivial solution. 4x 20. Show that the homogeneous system of equations x + y cos γ + z cos β = 0, x cos γ + y z cos α = 0, x cos β + y cos γ + z = 0, has non-trivial solution if α + β + γ = 0. + Solve the following system of equations by Gaussian elimination method: 21. x1 + 2x2 – x3 = 3 2x1 – 2x2 + 3x3 = 2 3x1 – x2 + 2x3 = 1 x1 – x2 + x3 = – 1 [Ans. x1 = 1, x2 = 4, x3 = 4] 22. 23. 2x1 + x2 + x3 = 10 3x1 + 2x2 + 3x3 = 18 x1 + 4x2 + 9x3 = 16 [Ans. x1 = 7, x2 = – 9, x3 = 5] 2x1 + x2 + 4x3 = 12 8x1 – 3x2 + 2x3 = 20 4x1 + 11x2 – x3 = 33 [Ans. x1 = 3, x2 = 2, x3 = 1] 209 MATRICES 24. 25. 26. x1 + 4x2 – x3 = –5 x1 + x2 – 6x3 = – 12 3x1 – x2 – x3 = 4 LMA ns. x = 117 , x = − 81 , x = 148 OP 71 71 71 Q N 1 2 2x1 + x2 + 2x3 + x4 = 6 6x1 – 6x2 + 6x3 + 12x4 = 36 4x1 + 3x2 + 3x3 – 3x4 = –1 2x1 + 2x2 – x3 + x4 = 10 2x1 – 7x2 + 4x3 = 9 x1 + 9x2 – 6x3 = 1 – 3x1 + 8x2 + 5x3 = 6 [Ans. x1 = 2, x2 = 1, x3 = – 1, x4 = 3] [Ans. x1 = 4, x2 = 1, x3 = 2] Solve by Gauss-Jordan method: 27. x – 3y – 8z = – 10 3x + y = 4 2x + 5y + 6z = 13 28. 29. 3 [Ans. x = y = z = 1] x+y+z = 6 2x + 3y – 2z = 2 5x + y + 2z = 13 [Ans. x = 1, y = 2, z = 3] 3x + y + 2z = 3 2x – 3y – z = – 3 x + 2y + z = 4 [Ans. x = 1, y = 2, z = – 1] 4x + 3y + 3z = – 2 x+z = 0 4x + 4y + 3z = – 3 [Ans. x = 1, y = 1, z = – 1] 31. Test the consistency and solve 2x – 3y + 7z = 5, 3x + y – 3z = 13, 2x + 19y – 47z = 32. [Ans. Inconsistent] Solve the following system by any method: 32. 2x + 6y + 7z = 0 6x + 20y – 6z = – 3 6y – 18z = – 1 [Ans. Inconsistent] 30. 33. 34. 4x – y + 6z = 16 x – 4y – 3z = – 16 2x + 7y + 12z = 48 5x – 5y + 3z = 0 2x1 + x2 + 5x3 + x4 = 5 x1 + x2 – 3x3 – 4x4 = – 1 3x1 + 6x2 – 2x3 + x4 = 8 2x1 + 2x2 + 2x3 – 3x4 = 2 35. 5x + 3y + 7z = 4 3x + 26y + 2z = 9 7x + 2y + 11z = 5 LMA ns. x = −9k + 16 , y = −6k + 16 , z = kOP 5 3 5 3 N Q LMA ns. x = 2, x = 1 , x = 0, x = 4OP 5 5Q N 1 2 3 4 LMA ns. x = 7 , y = 3 , z = 0OP 11 11 N Q 210 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 3.16 LINEAR DEPENDENCE OF VECTORS The set of vectors* (row or column matrices) X1, X2, ... Xn is said to be linearly dependent if there exist scalars a1, a2,... an not all zero such that a1X1 + a2X2 + ... + anXn = O [O is null matrix] 3.16.1 Linear Independence of Vectors If the set of vectors is not linearly dependent then it is said to be linearly independent. i.e., if every relation of the type a1X1 + a2X2 + ... + anXn = O ⇒ a1 = a2 = ... = an = 0. Example 1. Show that the vectors (3, 1, – 4), (2, 2, – 3) and (0, – 4, 1) are linearly dependent. Sol. Let X1 = (3, 1, – 4), X2 = (2, 2, – 3), X3 = (0, – 4, 1) Now, a1X1 + a2X2 + a3X3 = O |linear dependence a1(3, 1, – 4) + a2 (2, 2, – 3) + a3 (0, – 4, 1) = (0, 0, 0) ⇒ (3a1 + 2a2, a1 + 2a2 – 4a3, – 4a1 – 3a2 + a3) = (0, 0, 0) ⇒ 3a1 + 2a2 = 0, a1 + 2a2 – 4a3 = 0, – 4a1 – 3a2 + a3 = 0 The system of equations is homogeneous. Now the coefficient matrix is R1 ↔ R2 LM 3 2 0OP 2 −4 A = M 1 MN− 4 −3 1PPQ R2→ R2 – 3R1, R3 → R3 + 4R1 R2 → – 1 1 R, R → R 4 2 3 5 3 LM 1 2 − 4OP ~ M 3 2 0 MN− 4 −3 1PPQ LM1 2 − 4OP ~ M0 − 4 12 MN0 5 −15PPQ LM1 2 −4OP ~ M0 1 −3P MN0 1 −3PQ LMx OP x * An ordered set of n numbers belonging to a field F and denoted by X = [x , x , ... x ] or M P is called MM : PP an n-dimensional vector over F. Nx Q 1 2 1 2 n n 211 MATRICES LM1 2 −4OP ~ M0 1 −3P MN0 0 0 PQ R3 → R3 – R2 and ⇒ ρ(A) = 2 and n = 3 Let a3 = k a2 – 3a3 = 0 ⇒ a2 – 3k = 0 ⇒ a2 = 3k Again a1 + 2a2 – 4a3 = 0 ⇒ a1 + 6k – 4k = 0 ⇒ a1 = – 2k a1 = – 2k, a2 = 3k, a3 = k. Therefore, Hence, a1, a2 and a3 cannot be zero otherwise there is a trivial solution which is impossible. So the vectors X1, X2 and X3 are linearly dependent and the relation is – 2kX1 + 3kX2 + kX3 = O ⇒ 2X1 – 3X2 – X3 = O. Example 2. Find the value of λ for which the vectors (1, – 2, λ), (2, – 1, 5) and (3, – 5, 7λ) are linearly dependent. (U.P.T.U., 2006) Sol. Let X1 = (1, – 2, λ), X2 = (2, – 1, 5), X3 = (3, – 5, 7λ) Now a1X1 + a2X2 + a3X3 = 0 |For linear dependence ⇒ a1 (1, – 2, λ) + a2 (2, – 1, 5) + a3 (3, – 5, 7λ) = (0, 0, 0) ⇒ U| V λa + 5a + 7λa = 0 |W a1 + 2a2 + 3a3 = 0 – 2a1 – a2 – 5a3 = 0 1 2 ...(i) 3 The system is homogeneous ∴ For non-trivial solution* |A| = 0 ⇒ |A| = 1 2 3 −2 −1 −5 = 0 λ 5 7λ = (– 7λ + 25) – 2 (– 14λ + 5λ) + 3 ( – 10 + λ) = 0 = – 7λ + 25 + 18 λ – 30 + 3 λ = 0 = 14λ – 5 = 0 ⇒ λ = 5 . 14 Example 3. Show that the vectors [0, 1, – 2], [1, – 1, 1] [1, 2, 1] form a linearly independent set. Sol. Let X1 = [0, 1, – 2], X2 = [1, – 1, 1] X3 = [1, 2, 1] Also suppose a1X1 + a2X2 + a3X3 = O ⇒ a1 [0, 1, –2] + a2 [1, – 1, 1] + a3 [1, 2, 1] = O ⇒ {a2 + a3, a1 – a2 + 2a3, – 2a1 + a2 + a3] = [0, 0, 0] ⇒ a2 + a3 = 0 a1 – a2 + 2a3 = 0 * For linear dependence the solution must be non-trivial otherwise it will be linear independence. 212 A TEXTBOOK OF ENGINEERING MATHEMATICS—I – 2a1 + a2 + a3 = 0 LM 0 1 1OP A = M 1 −1 2P NM−2 1 1PQ Now R1 ↔ R2 LM 1 −1 2OP 0 1 1 A ~ M MN−2 1 1PPQ R3 → R3 + 2R1 R3 → R3 + R2 LM1 −1 2OP ~ M0 1 1P NM0 −1 5PQ LM1 −1 2OP ~ M0 1 1P = ρ (A) = 3 MN0 0 6PQ LM1 −1 2OPLM a OP LM0OP MM00 10 16PPMMaa PP = MM0PP QN Q N0Q N 1 Now 2 3 ⇒ a1 – a2 + 2a3 = 0 a2 + a3 = 0 a3 = 0 ⇒ a1 = a2 = a3 = 0 All a1, a2 and a3 are zero. Therefore, they are linearly independent. Example 4. Examine the vectors X1 = [1, 1, 0]T, X2 = [3, 1, 3]T, X3 = [5, 3, 3]T are linearly dependent. LM1OP LM3OP LM5OP Sol. Here X = M1P , X = M1P , X = M 3P MN0PQ MN3PQ MN3PQ 1 Now ⇒ ⇒ ⇒ 2 3 a1X1 + a2X2 + a3X3 = O LM1OP LM3OP LM5OP L0O a M1P + a M1P + a M 3P = M0P MN0PQ MN3PQ MN3PQ MMN0PPQ LMa + 3a + 5a OP LM0OP MM 0a ++3aa ++ 33aa PP = MM00PP Q NQ N 1 2 3 1 2 3 1 2 3 2 3 a1 + 3a2 + 5a3 = 0 a1 + a2 + 3a3 = 0 0.a1 + 3a2 + 3a3 = 0 213 MATRICES LM1 3 5OP A = M1 1 3P MN0 3 3PQ ∴ R1 → R2 – R1 R3 → R3 + LM1 3 5 OP A ~ 0 −2 −2 MM0 3 3 PP N Q 3 R 2 2 LM1 3 5 OP ~ M0 −2 −2P MN0 0 0 PQ Here ρ(A) = 2 but n = 3 so let a3 = k. LM1 3 5 OP LM a OP L0O MM00 −02 −02PP MMaa PP = MM0PP Q N Q MN0PQ N 1 and 2 3 a1 + 3a2 + 5a3 = 0 ⇒ ...(i) a2 + a 3 = 0 a2 + k ⇒ a2 = – k ...(ii) From (i) a1 + 3 (–k) + 5k = 0 ⇒ a1 = 2k Since all a1, a2, a3 are not zero. Therefore, they are linearly dependent And the relation is ⇒ 2kX1 – kX2 + kX3 = O 2X1 – X2 + X3 = O. Example 5. Show that the vectors [2, 3, 1, –1], [2, 3, 1, – 2], [4, 6, 2, – 3] are linearly independent. Sol. Consider the relation a1X1 + a2X2 + a3X3 = O ⇒ a1[2, 3, 1, –1] + a2 [2, 3, 1, – 2] + a3 [4, 6, 2, –3] = O ⇒ 2a1 + 2a2 + 4a3 = 0 3a1 + 3a2 + 6a3 = 0 a1 + a2 + 2a3 = 0 a1 + 2a2 + 3a3 = 0 ∴ LM2 2 4OP 3 3 6P A = M MM1 1 3PP , R ↔ R , R ↔ R N1 2 2Q 1 4 2 3 LM1 2 2OP 1 1 3P ~ M MM3 3 6PP N2 2 4Q 214 A TEXTBOOK OF ENGINEERING MATHEMATICS—I R2 → R2 – R1, R3 → R3 – 3 R1, R4→ R4 – 2R1 LM1 2 2OP LM1 2 2 OP 0 −1 1P 0 −1 1 P ~ M , R → R – 3R , R → R – 2R ~ M MM0 −3 0PP MM0 0 −3PP N0 −2 0Q N0 0 −2Q 3 R4 → R4 − 2 R 3 3 3 2 4 4 2 LM1 2 2 OP 0 −1 1 P ~ M MM0 0 −3PP ⇒ ρ(A) = 3 and n = 3 N0 0 0 Q ∴ The system has a trivial solution Hence, a1 = a2 = a3 = 0 i.e., they are linearly independent. EXERCISE 3.5 1. Examine the following vectors for linear dependence and find the relation if it exists. X1 = (1, 2, 4), X2 = (2, –1, 3), X3 = (0, 1, 2), X4 = (–3, 7, 2). (U.P.T.U., 2002) [Ans. Linearly dependent, 9X1 – 12X2 + 5X3 – 5X4 = 0] 2. Show the vectors X1 = [1, 2, 1], X2 = [2, 1, 4], X3 = [4, 5, 6] and X4 = [1, 8, – 3] are linearly independent? 3. Show that the vectors [1, 2, 3], [3, –2, 1] , [1, – 6, – 5] are linearly dependent. 4. If the vectors (0, 1, a), (1, a, 1), (a, 1, 0) are linearly dependent, then find the value of a. [Ans. a = 0, 2 , – 2 ] 5. Examine for linear dependence [1, 0, 2, 1], [3, 1, 2, 1] [4, 6, 2, – 4], [–6. 0, – 3, –4] and find the relation between them, if possible. [Ans. Linear dependent and the relation is 2X1 – 6X2 + X3 – 2X4 = 0] 6. Show that the vectors X1 = [2, i, – i]. X2 = [2i, – 1, 1], X3 = [1, 2, 3] are linearly dependent. 7. If X1 = [1, 1, 2], X2 = [2, – 1, – 6], X3 = [13, 4, – 4] prove that 7X1 + 3X2 – X3 = 0. 8. Show that the vectors [3, 1, – 4], [2, 2, – 3] form a linearly independent set but [3, 1, – 4], [2, 2, – 3] and [0, – 4, 1] are linearly dependent. 3.17 EIGEN VALUES AND EIGEN VECTORS Introduction: At the start of 20th century, Hilbert studied the eigen values. He was the first to use the German word eigen to denote eigen value and eigen vectors in 1904. The word eigen mean— own characteristic or individual. More formally in a vector space, a vector function A (matrix) defined if each vector X of vector space, there corresponds a unique vector Y = AX. So here we consider a linear transform Y = AX 215 MATRICES transforms X into a scalar multiple of itself say λ so AX = λX which is called Eigen value equation. In this section, we study the problem AX = λX ...(i) _v >[ >l v where A is a n × n matrix, X is an unknown n × 1 vector and λ is an unknown scalar. From equation (i) it can be understand that AX is a scalar multiple of X say λX. Geometrically each vector on the line through the origin determined by X gets mapped back onto the same line under multiplication by A. Geometrical representation: The eigen value equation means that under the transformation, a eigen vector experiv ence only changes in magnitude and sign. The direction of lw AX is the same as that of X. Here A acts to stretch the vector X, not change its direction. So X is an given vector of A. w The eigen value determines the amount, the eigen vector is scaled under the linear transformation. For example, the eigen value λ = 2 means that the eigen vector is doubled in length and point in the same direction. The eigen value λ = 1, means that the eigen vector is unchanged, while an eigen value λ = – 1 means that the eigen vector is reversed in direction. v m v v lv Fig. 3.1 Thus, the eigen value λ is simply the amount of “stretches” or “shrinks” to which a vector is subjected when transformed by A. (U.P.T.U., 2007) 3.17.1 Characteristic Equation If we re-express (i) as AX = λIX (where I is an identity matrix). AX – λIX = O or ⇒ (A – λI)X = O ...(ii) Which is homogeneous system of n equations in the n variables x1, x2 ... xn. The system (ii) must have non-trivial solutions otherwise X = 0 (which is impossible). ∴ For non-trivial solution the coefficient matrix (A – λI) will be singular ⇒ |A – λI| = 0 ...(iii) |singular matrix |A| = 0 Expansion of the determinant gives an algebraic equation in λ, known as the ‘‘characteristic equation’’ of A. The determinant |A – λI| is called characteristic polynomial of A. 3.17.2 Characteristic Roots or Eigen Values [U.P.T.U. (C.O.), 2003, 2007] The roots of characteristic equation are called characteristic roots or eigen values. 3.17.3 Eigen Vectors [U.P.T.U. (C.O.), 2003, 2007] The corresponding non-zero vector X is called characteristic eigen vector. Notes: 1. If there is one linearly independent solution and two eigen values are same then there will be same eigen vectors of both eigen values. 2. If there is two linearly independent solution and two eigen values are same then there will be different eigen vectors of each eigen value. 216 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 3.17.4 Properties of Eigen Values and Eigen Vectors 1. If A is real, its eigen values are real or complex conjugate in pairs. 2. Determinant of A = product of eigen values of A. 3. A and AT has same eigen values. 4. A–1 exists iff 0 is not an eigen value of A, eigen values of A–1 are 1 , 1 λ1 λ 2 ,...., 1 λn . (U.P.T.U., 2008) 5. Eigen vector cannot correspond to two distinct characteristic values. 6. Eigen values of diagonal, upper triangular or lower triangular matrices are the principal diagonal elements. 7. KA (scalar multiples) has eigen values Kλi. 8. Am has eigen values λm. (U.P.T.U., 2008) 9. Two vectors X and Y are said to be orthogonal if XTY = YTX = 0. Theorem 1. The latent roots of a Hermitian matrix are all real Proof. We have AX = λX [U.P.T.U. (C.O.), 2003] ...(i) To prove that λ is a real number, we have to prove λ = λ From (i) X*(AX) = X* (λX) ⇒ X*AX = λX*X ⇒ (X* AX)* = (λX* X)* ⇒ X*A*X = λ X*X, (X** = X) ⇒ X* AX = λ X*X (A* = A) ⇒ X* λX = λ X*X ⇒ λX*X = λ X*X ⇒ — (λ – λ ) X*X = 0 ⇒ λ = λ. Proved. Theorem 2. The latent roots of a skew-Hermitian matrix are either zero or purely imaginary. [U.P.T.U. (C.O.), 2003] Proof. Let A be a skew-Hermitian matrix so that A* = – A. Now we are to prove that λ = 0 or purely imaginary number Here (iA)* = i A* = –iA* = –i(–A) = iA (As A* = –A) Hence, iA is a Hermitian matrix. Let λ be an eigen value relative to the eigen vector X of A, then ⇒ AX = λX iAX = iλX ⇒ (iA)X = (iλ)X ...(i) ⇒ iλ is an eigen root relative to the vector X of Hermitian matrix iA. ⇒ iλ is a real number, for eigen values of a Hermitian matrix are all real. ⇒ λ = 0 or purely imaginary number. Proved. Theorem 3. The characteristic roots of a unitary matrix are of unit modulus. [U.P.T.U. Special Exam., 2001] Proof. Let A be unitary matrix so that A*A = I. We have AX = λX ...(i) To prove |λ| = 1 217 MATRICES From (i), we have (AX)* = (λX)* ⇒ X*A* = λ X* ...(ii) Pre-multiplying (i) by (ii), (X*A*) (AX) = ( λ X*) (λX) ⇒ X* (A*A)X = λ λX*X 2 ⇒ X*IX = |λ| X*X ⇒ (1 – |λ|2) X*X = 0 2 2 ⇒ 1 – |λ| = 0 ⇒ |λ| = 1 or |λ| = 1. Proved. Theorem 4. Prove that the product of all eigen values of A is equal to the determinant (A). (U.P.T.U., 2004) Proof. We have AX = λIX ⇒ (A – λI) X = 0 For non-trivial solution |A – λI| = 0 a11 − λ ⇒ a12 ... a1n # an1 m = 0 an2 ... ann − λ r ⇒ (–1)n λn + b1 λn− 1 + b2 λn− 2 +...+ bn = 0 ⇒ (λ – λ1) (λ – λ2) ... (λ – λn) = 0 putting λ = 0, we get |A| = λ1, λ2, λ3 ... λn. Proved. Theorem 5. The latent roots of real symmetric matrix are all real. Proof. Let A be a real symmetric matrix so that A = A, A′ = A. The A* = ( A )′ = A′ = A or A* = A, meaning thereby, A is a Hermitian matrix. Hence, the latent roots of A are all real, by Theorem 1. Theorem 6. The characteristic roots of an idempotent matrix are either zero or unity. Proof. Let A be idempotent matrix so that A2 = A. Let AX = λX ...(i) premultiplying by A on equation (i), we get A(AX) = A(λX) = λ(AX) (As A2 = A) ⇒ (AA)X = λ(λX) ⇒ A2X = λ2X or AX = λ2X 2 ⇒ λX = λ X 2 ⇒ (λ – λ)X = 0 ⇒ λ2 – λ = 0 (As X ≠ 0) ⇒ λ = 0, 1. Proved. Example 1. Show that 0 is a characteristic root of a matrix if the matrix is singular. (U.P.T.U., 2008) Sol. The characteristic equation is |A – λI| = 0 or |A| – λ|I| = 0 As |A| = 0 (singular matrix) 218 A TEXTBOOK OF ENGINEERING MATHEMATICS—I ∴ 0 = λ |I| = 0 ⇒ λ = 0 Conversely, if λ = 0 then |A| = 0. Proved. LM 1 2 2OP Example 2. Find the characteristic equation of the matrix 0 2 1 . Also find the eigen MM−1 2 2PP Q N values and eigen vectors of this matrix. Sol. Let Now, (U.P.T.U., 2006) LM 1 2 2OP MM−01 22 12PP Q N LM 1 2 2OP LM1 0 0OP LM1 − λ 2 2 OP A – λI = 0 2 1 – λ M0 1 0P = M 0 MM−1 2 2PP M0 0 1P M −1 2 −2 λ 2 −1 λPP Q N Q Q N N A = ∴ Characteristic equation is |A – λI| = 0 ⇒ LM1 − λ 2 2 OP MM 0 2 − λ 1 PP = 0 ⇒ (1 – λ) m( 2 − λ) − 2r − 2{0 + 1} + 2{0 + ( 2 − λ)} = 0 N −1 2 2 − λQ 2 ⇒ (1 – λ) (λ2 – 4λ + 2) – 2 + 4 – 2λ = 0 ⇒ λ2 – 4λ + 2 – λ3 + 4λ2 – 2λ + 2 – 2λ = 0 ⇒ λ3 – 5λ2 + 8λ – 4 = 0 ⇒ λ2(λ – 1) – 4λ (λ – 1) + 4 (λ – 1) = 0 ⇒ (λ – 1) (λ2 – 4λ + 4) = 0 Hence, λ = 1, 2, 2. These are eigen values of A. Now, we consider the relation AX = λX ⇒ AX = λIX ⇒ (A – λI)X = 0 Taking λ = 1, from (1), we get LM1 − 1 2 2 OP LMx OP MM −01 2 2− 1 2 1− 1PP MMxx PP = 0 QN Q N LM 0 2 2OP LMx OP MM−01 12 11PP MMxx PP = 0 QN Q N 1 2 3 1 ⇒ 2 3 R1 ↔ R3 LM−1 2 1OP LMx OP ~ M 0 1 1P M x P = 0 MN 0 2 2PQ MNx PQ 1 2 3 ...(i) 219 MATRICES R3 → R3 – 2R2, R1 → (– 1) R1 LM1 −2 −1OP LMx OP ~ M0 1 1 P M x P = 0 MN0 0 0 PQ MNx PQ 1 2 3 ⇒ Let and x1 – 2x2 – x3 = 0 x2 + x3 = 0 x3 = k then x2 = – k x1 + 2k – k = 0 ⇒ x1 = – k LM 1OP LM 1OP LM−kOP X = M− kP = − k M 1P or M 1P MN−1PQ MN−1PQ MN kPQ ∴ 1 Taking λ = 2, from (1), we get LM−1 2 2OP LMx1 OP MM−01 02 10PP MMxx2 PP = 0 N Q N 3Q R2 ↔ R3 LM−1 2 2OP LM x OP ~ −1 2 0 MM 0 0 1PP MMxx PP = 0 QN Q N 1 2 3 R2 → R2 – R1, R1 → –R1 LM1 −2 −2OP LM x OP ~ 0 0 −2 MM0 0 1 PP MMxx PP = 0 QN Q N 1 2 3 R2 → – 1 R and then R3 → R3 – R2, we get 2 2 LM1 −2 −2OP LM x OP ~ M0 0 1 P M x P = 0 NM0 0 0 PQ MNx PQ 1 2 3 Here the solution is one variable linearly independent. ⇒ x1 – 2x2 – x3 = 0 x3 = 0 Let x2 = k then x1 – 2k – 0 = 0 ⇒ x 1 = 2k LM x OP LM2kOP LM2OP LM2OP X = M x P = M k P = k M1P or M1P MNx PQ MN 0 PQ MN0PQ MN0PQ 1 Hence, 2 2 3 220 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Thus only one eigen vector X2 corresponds to the repeated eigen value λ = 2. Hence the eigen values are 1, 2, 2 LM2OP LM 1 OP X = – k 1 , X = X = k M1P MM−1PP MN0PQ N Q LM2OP LM 1 OP X = 1 , X = X = M1P MM−1PP MN0PQ N Q eigen vectors 1 or 2 1 2 3 3 Example 3. Find the eigen values and eigen vectors of the matrix LM2 1 1OP A = M1 2 1P MN0 0 1PQ Sol. The characteristic equation is |A – λI| = 0 ⇒ 2− λ 1 1 1 2− λ 1 0 0 1− λ = (1 – λ) (λ – 1) (λ – 3) = 0 Thus λ = 1, 1, 3 are the eigen values of A. Now, we consider the relation (A – λI)X = 0 For λ = 3, LM−1 1 1 OP LM x OP MM 10 −01 −12PP MMxx PP = 0 QN Q N 1 2 3 LMx OP let X = Mx P MM PP Nx Q 1 2 3 Applying R2 → R2 + R1, then R1 → – R1 on coefficient matrix, we get LM1 −1 −1OP L x O ~ M0 0 2 P M x P = 0 M P NM0 0 −2PQ MNx PQ 1 2 3 Again R3 → R3 + R2, R2 → 1 R 2 2 LM1 −1 −1OP LM x OP ~ 0 0 1 MM0 0 0 PP MMxx PP = 0 Q N Q N 1 2 3 ⇒ x1 – x2 – x3 = 0 221 MATRICES x3 = 0 Suppose x2 = k, then x1 = k Here, For r = 2, n = 3 n− r = 1 LMk OP LM1OP LM1OP X = Mk P = k M1P or M1P MM PP MM PP MM PP N0Q N0Q N0Q 1 λ=1 LM1 MM1 MN0 R2 → R2 – R1 LM1 ~ M0 MM N0 0 OP LMx OP 1P Mx P = 0 PM P 0PQ MNx PQ 1 1 1 1 0 0 1 1 2 3 OP LMx OP 0P Mx P = 0 PM P 0PQ MNx PQ 1 2 3 Here r = 1, n = 3 ⇒ n – r = 3 – 1 = 2. ∴ There is two variables linearly independent solution Let ⇒ x2 = k1, x3 = k2 and x1 + x2 + x3 = 0 x1 + k1 + k2 = 0 ⇒ x1 = – (k1 + k2) ⇒ LM−bk + k gOP X = M MM k PPP . N k Q 1 2 1 2 2 Since the vectors are linearly independent so, we choose k1 and k2 as follows: (a) If we suppose x2 = k1 = 0, x3 = k2 (any arbitrary) LM− k OP LM 1 OP LM 1 OP X = M 0 P = − k M 0 P or M 0 P MM PP MM PP MM PP Nk Q N−1Q N−1Q 2 Then 2 2 2 (b) If we suppose x2 = k1 (any arbitrary) and x3 = k2 = 0 LM− k OP LM 1OP LM 1OP X = M k P = − k M−1P or M−1P MM PP MM PP MM PP 0 N Q N 0Q N 0Q 1 Then 3 1 1 222 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Hence the eigen values are λ = 1, 2, 2 Eigen vectors are or LM1OP LM 1OP LM 1OP X = k M1 P; X = − k M 0 P; X = – k M−1P MM PP MM PP MM PP N0Q N−1Q N 0Q LM1OP LM 1 OP LM 1 OP X = M1P; X2 = M 0 P; X3 = M−1P . MN0PQ MN−1PQ MN 0 PQ 2 1 2 3 1 1 Example 4. Find the eigen values and eigen-vectors of matrix LM3 1 4OP A = M0 2 6P . MN0 0 5PQ [U.P.T.U., 2004 (C.O.), 2002] Sol. The characteristic equation is |A –λI| = 0 3− λ 1 4 0 2− λ 6 = ( 3 − λ ) (2 − λ )(5 − λ ) − 0 + 0 = 0 0 0 5− λ l ⇒ q ∴ λ = 2, 3, 5 Now, we consider the relation (A – λI) X = 0 For λ = 2, LM3 − 2 1 4 OP L x O MM 00 2 −0 2 5 −6 2PP MMx PP = 0 Q MNx PQ N LM1 1 4OP LM x OP MM00 00 63PP MMxx PP = 0 QN Q N 1 2 3 ...(i) LM x OP X= x MMx PP N Q 1 2 3 1 ⇒ 2 3 R3 → R3 – 1 R on coefficient matrix 2 2 LM1 1 4OP LM x OP ~ M0 0 6P M x P = 0 MN0 0 0PQ MNx PQ 1 2 3 ⇒ Let ∴ x1 + x2 + 4x3 = 0 x2 = k and x3 = 0 then x1 + k + 0 = 0 ⇒ x1 = –k LM−kOP LM 1OP LM 1OP X = M k P = − k M−1P or M−1P MN 0 PQ MN 0PQ MN 0PQ 1 223 MATRICES For λ = 3, from (i), we get LM0 MM0 MN0 1 −1 0 OP LMx OP 6 P Mx P = 0 PM P 2PQ MNx PQ 4 1 2 3 R2 → R2 + R1 on coefficient matrix LM0 ~ M0 MM N0 1 0 0 1 R3 → R3 − R2 5 LM0 ~ M0 MM N0 1 0 0 OP LMx OP 10 P Mx P = 0 PM P 2 PQ MNx PQ 4 1 2 3 OP LMx OP 10 P Mx P = 0 PM P 0 PQ MNx PQ 4 1 2 3 x2 + 4x3 = 0 and 10x3 = 0 ⇒ x3 = 0 x2 + 0 = 0 ⇒ x2 = 0, let x1 = K ⇒ ⇒ LMk OP LM1OP LM1OP X = M0P = k M0P or M0P MM PP MM PP MM PP N0Q N0Q N0Q ∴ 2 Again, for λ = 5, we get ⇒ LM−2 MM 0 MN 0 1 −3 0 OP Lx O L0O 6 P Mx P = M0P PP MMNx PPQ MMN0PPQ 0Q 4 1 2 3 2x1 – x2 – 4x3 = 0 3x2 – 6x3 = 0 Let x 3 = k, then 3x2 – 6k = 0 ⇒ x2 = 2k and 2x1 – 2k – 4k = 0 ⇒ x1 = 3k ∴ LM3kOP LM3OP LM3OP X = M 2k P = k M 2P or M 2P MM PP MM PP MM PP N k Q N1Q N1Q 3 Hence the eigen values are λ = 2, 3, 5 And eigen vectors are LM 1 OP LM1OP LM3OP X = M−1P , X = M0 P , X = M 2P . MM PP MM PP MM PP N0Q N0 Q N1 Q 1 2 3 224 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 5. Determine the latent roots and the corresponding vector of the matrix, LM 6 −2 2 OP A = M−2 3 −1P . NM 2 −1 3 PQ Sol. The characteristic equation is |A – λI| = 0 6 − λ −2 2 −2 3 − λ −1 = 0 2 −1 3 − λ ⇒ R2 → R2 + R3, we get 6−λ 0 2 −2 2− λ −1 2 2− λ 3− λ 6−λ 0 = (2 – λ) 2 −2 1 −1 0 2 2− λ = ( 2 − λ ) ( 6 − λ ) ( 2 − λ + 2) − 8 = 0 2 1 3− λ =0 Now, C3 → C3 + C2 gives (2 – λ) ∴ 6−λ 0 2 −2 1 −1 l (2 – λ)(λ2 – 10λ + 16) = 0 ⇒ (2 – λ) (2 – λ) (λ – 8) = 0 ⇒ λ = 2, 2, 8 Now, consider the relation (A – λI)X = 0 For λ = 8, the equation (i), gives LM6 − 8 −2 2 OP LMx OP LM0OP MM −2 3 − 8 −1 PP MMx PP = MM0PP N 2 −1 3 − 8Q Nx Q N0Q LM−2 −2 2 OP LMx OP L0O MM−2 −5 −1PP MMx PP = MM0PP N 2 −1 −5Q Nx Q MN0PQ 1 2 3 1 ⇒ 2 3 R2 → R2 – R1, R3 → R3 + R1 on coefficient matrix LM−2 −2 2 OP LMx OP LM0OP MM 0 −3 −3PP MMx PP = MM0PP N 0 −3 −3Q Nx Q N0Q 1 2 3 −1 −1 R1 , R2 → R2 R3 → R3 – R2 and R1 → 2 3 1 1 −1 x1 0 0 1 1 x2 = 0 0 0 0 x3 0 ⇒ q LM MM N OP L O PP MM PP Q MN PQ LM OP MM PP NQ x1 + x2 – x3 = 0 x2 + x3 = 0 ...(i) 225 MATRICES Let x3 = k, then x2 = – k and x1 = 2k LMx OP LM 2k OP LM 2OP LM 2OP X = M x P = M− k P = k M−1P or M−1P MNx PQ MN k PQ MN 1PQ MN 1PQ 1 ∴ 2 1 For λ = 2, the equation (i) gives 3 LM6 − 2 −2 2 OP Lx O L0O MM −2 3 − 2 −1 PP MMx PP = MM0PP N 2 −1 3 − 2Q MNx PQ MN0PQ LM 4 −2 2OP LM x OP LM0OP MM−22 −11 −11PP MMxx PP = MM00PP N Q N Q NQ 1 2 3 1 ⇒ 2 3 R2 → R2 + 1 1 R1 , R3 → R3 – R1 , we get 2 2 LM4 −2 2OP Lx O L0O MM00 00 00PP MMx PP = MM0PP N Q MNx PQ MN0PQ 1 2 3 Here r = 1, n = 3 ⇒ n – r = 3 – 1 = 2 (Two linearly independent) Let and we have ⇒ x2 = k1 and x3 = k2 4x1 – 2x2 + 2x3 = 0 4x1 – 2k1 + 2k2 = 0 1 ( k1 − k 2 ) 2 k1 2 k1 1 k2 = 2k2 ∴ X2 = 2 1 ( k1 − k2 ) ( k1 − k 2 ) 2 Since the vectors are linearly independent so, we choose, k1 and k2 as follows: (i) Let x2 = k1 = 0 and x3 = k2 (arbitrary) ⇒ x1 = LM MM MN LM MM N OP PP Q LM OP MM PP N Q LM OP MM PP N Q OP PP PQ OP PP Q LM MM N 0 0 0 k2 1 2k 2 = 2 or 2 X2 = 2 2 −k 2 −1 −1 then (ii) Let x3 = k2 = 0 and x2 = k1 (arbitrary) 1 X3 = 2 then LM2k OP k LM2OP LM2OP MM 0 PP = 2 MM0PP or MM0PP N k Q N1Q N1Q 1 1 1 Hence λ = 8, 2, 2 LM 2OP LM 2OP L 0O LM 0OP L2O LM2OP k M P k M P 2 or M 2P, X = 0 or M0P . X = k M−1P or M−1P, X = 2 M P 2 M P MN 1PQ MN 1PQ MN−1PQ MN−1PQ MN1PQ MN1PQ 1 2 2 3 1 226 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 6. Find the eigen values of matrix A. 1 A = 3 LM1 2 2OP MM2 1 −2PP · N2 −2 1Q Sol. The characteristic equation of B is, where LM1 2 2 OP B = M2 1 −2P MN2 −2 1 PQ 1− λ 2 2 2 1 − λ −2 = λ3 – 3λ2 – 9λ + 27 = 0 2 −2 1 − λ (λ – 3)2 (λ + 3) = 0 ⇒ λ = 3, 3, – 3 or 1 B = 1, 1, – 1. 3 Example 7. Show that the matrix So the eigen values of A = LM 3 10 5 OP A = M−2 −3 −4P MN 3 5 7 PQ has less than three independent eigen vectors. Is it possible to obtain a similarity transformation that will diagonalise this matrix. (U.P.T.U., 2001) Sol. The characteristic equation is |A – λI| = 0 or 3−λ 10 5 −2 −3 − λ −4 3 5 7−λ or λ3 – 7λ2 + 16λ – 12 = 0 ⇒ = 0 (λ – 2)2 (λ – 3) = 0 ⇒ λ = 2, 2, 3 Now, consider the relation (A – λI)X = 0 For λ = 3, we have LM3− 3 10 5 OP LMx OP MM −2 −3− 3 −4 PP MMx PP = 0 N 3 5 7 − 3Q Nx Q LM 0 10 5OP LMx OP MM−2 −6 − 4PP MMx PP = 0 N 3 5 4Q Nx Q 1 2 3 1 ⇒ 2 3 ...(i) 227 MATRICES R1 → 1 1 R1 , R2 → – R2 , we have 5 2 LM0 2 1OP LMx OP MM1 3 2PP MMx PP = 0 N 3 5 4Q N x Q LM1 3 2OP LM x OP MM03 25 14PP MMxx PP = 0 N QN Q 1 2 3 R1 ↔ R2 1 2 3 R3 → R3 – 3R1 on coefficient matrix LM1 3 2 OP Lx O MM0 2 1 PP MMx PP = 0 N0 −4 −2Q MNx PQ 1 2 3 R3 → R3 + 2R2 LM1 3 2OP LMx OP L0O MM0 2 1PP MMx PP = MM0PP N0 0 0Q Nx Q MN0PQ 1 2 3 ⇒ x1 + 3x2 + 2x3 = 0 0x1 + 2x2 + x3 = 0 Solving these equations by cross-multiplication x1` x3 x2 = − = 1− 0 3−4 2−0 x1 x2 x3 ⇒ = = −1 −1 2 x1 x2 x3 ⇒ = = −2 1 1 ∴ The proportional values are x1 = 1, x2 = 1, x3 = – 2 LM 1OP X = M 1P MN−2PQ 1 For λ = 2, we have, from equation (i) LM3 − 2 10 5 OP LMx OP MM −2 −3 − 2 −4 PP MMx PP = 0 N 3 5 7 − 2Q N x Q LM 1 10 5OP LMx OP MM−2 −5 −4PP MMx PP = 0 N 3 5 5Q N x Q 1 2 3 1 2 3 228 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Applying R2 → R2 + 2R1, R3 → R3 – 3R1 on coefficient matrix LM1 10 5 OP Lx O MM00 −1525 −610PP MMx PP = 0 N Q MNx PQ 1 2 3 R2 1 R2 → , R3 → – R 3 5 3 LM1 10 5OP LMx OP MM0 5 2PP MMxx PP = 0 N0 5 2 Q N Q 1 2 3 R3 → R3 – R2 LM1 10 5OP LMx OP MM0 5 2PP MMx PP = 0 N0 0 0 Q N x Q 1 2 3 This shows that the solution is one linearly independent ⇒ x1 + 10x2 + 5x3 = 0 5x2 + 2x3 = 0 x x x1 = − 2 = 3 −5 2 5 or x1 x x = 2 = − 3 5 2 5 Taking proportional values, x1 = 5, x2 = 2, x3 = – 5 LM 5 OP X = M2P MN−5PQ ∴ 2 And the third eigen vector corresponding to repeated root λ3 = 2 = λ2 will be same because the solution is not two linearly independent. Hence the vectors X2 and X3 are not linearly independent. Similarity transformation is not possible. LM 1 OP LM 5 OP Therefore λ = 2, 2, 3 and X = M 1 P . X = X = M 2 P . MN−2PQ MN−5PQ 1 2 3 Example 8. Verify the statement that the sum of the elements in the diagonal of a matrix is the sum of the eigen values of the matrix. LM−2 2 −3OP 1 −6P · A = M2 NM−1 −2 0 PQ Sol. The characteristic equation is |A – λI| = 0 ⇒ −2 − λ 2 −3 2 1 − λ −6 −1 −2 − λ = 0 229 MATRICES ⇒ ⇒ ⇒ (– 2 –λ) [(1 – λ)(– λ) – 12] – 2(– 2λ – 6] – 3[– 4 + (1 – λ)] = 0 (– 2 – λ) (λ2 – λ – 12) + 4 (λ + 3) + 3 (λ + 3) = 0 – 2λ2 + 2λ + 24 – λ3 + λ2 + 12λ + 4λ + 12 + 3λ + 9 = 0 ⇒ – λ3 – λ2 + 21λ + 45 = 0 ⇒ λ3 + λ2 – 21λ – 45 = 0. This is cubic equation in λ and hence it has 3 roots i.e., there are three eigen values. FG coefficient of λ IJ = – 1. H coefficient of λ K 2 The sum of the eigen values = – 3 The sum of the elements on the diagonal of the matrix A = – 2 + 1 + 0 = – 1. Hence the result. Example 9. Find the eigen values and corresponding eigen vectors of the matrix, A = LM−5 2 OP N 2 −2Q (U.P.T.U., 2008) Sol. The characteristic equation is |A – λI| = 0 −5 − λ 2 = (5 + λ) (2 + λ) – 4 = 0 2 −2 − λ ⇒ ⇒ λ2 + 7λ + 6 = 0 ⇒ ∴ (λ + 6) (λ + 1) = 0 λ = – 1, – 6 Now, we consider the relation (A – λI) X = 0 For ...(i) λ = –1 LM−5 + 1 2 OP LMx OP = 0 N 2 −2 + 1Q Nx Q LM−4 2 OP LMx OP = 0 N 2 −1Q Nx Q 1 2 ⇒ 1 2 R2 → R2 + 2R1, on coefficient matrix LM−4 2OP LMx OP = 0 N 0 0Q Nx Q 1 2 ⇒ – 2x1 + x2 = 0 Let x2 = k ∴ From (i) x1 = X1 = For λ = – 6, from (i), we get k 2 LMk 2OP = k LM1OP or LM1OP N k Q 2 N2Q N2Q LM1 2OP LMx OP = 0 N2 4Q Nx Q 1 2 ...(ii) 230 A TEXTBOOK OF ENGINEERING MATHEMATICS—I R2 → R2 – 2R1, on coefficient matrix LM1 2OP LMx OP = 0 N0 0Q Nx Q 1 2 ⇒ Let x1 + 2x2 = 0 ...(iii) x2 = k ∴ x1 = – 2k X2 = LM−2kOP = k LM−2OP or LM−2OP N k Q N1Q N1Q Hence, the eigen values are λ = –1, – 6 LM1OP N2Q L−2O X = M P. N1Q And eigen vectors are X1 = 2 EXERCISE 3.6 Find the eigen values and eigen vectors of: LM1 2OP. N1 0Q L6 8 OP. 2. M N8 −6Q LM1 1 3OP 1 5 1 . 3. M MN3 1 1PPQ LM 8 −6 2 OP −6 7 −4P. 4. M MN 2 −4 3 PQ LM −2 1 1OP −11 4 5P. 5. M MN −1 1 0PQ LM2 1 0OP 0 2 1. 6. M MN0 0 2PPQ LM1 −1 −1OP 1 −1 0 P. 7. M MN1 0 −1PQ 1. LMAns. 2, − 1, L2O , L 1 OOP MN1PQ MN−1PQQ N LMAns. 10, - 10, L2O , L 1OOP MN1PQ MN−2PQQ N LM LM−2OP LM−1OP LM1OPOP MMAns. − 2, 3, 6, M 0 P , M 1 P, M2PPP MN 2 PQ MN−1PQ MN1PQQ N LM LM1OP LM 2 OP LM 2 OPOP MMAns. 0, 3, 15, M2P, M 1 P, M−2PPP MN2PQ MN−2PQ MN 1 PQQ N LM LM 0 OP LM1OP LM2OPOP Ans. − 1 , 1 , 2 , MM MM−11PP, MM12PP, MM13PPPP N Q N Q N QQ N LM LM1OP LM1OP LM1OPOP MMAns. 2, 2, 2, M0P, M0P, M0PPP MN0PQ MN0PQ MN0PQQ N LM LM 0 OP LM1 + iOP LM1 − iOPOP MMAns. − 1, ± i, M 1 P, M 1 P, M 1 PPP MN−1PQ MN 1 PQ MN 1 PQQ N 231 MATRICES LM3 1 4OP 0 2 0P. 8. M MN0 0 5PQ LM−2 2 −3OP 2 1 −6 . 9. M MN−1 −2 0 PPQ LM 3 −2 −5OP 10. M 4 −1 −5P. MN−2 −1 −3PQ LM1 1 3OP 11. M1 5 1P. MN3 1 1PQ LM 4 −20 −10OP 4 P. 12. M−2 10 MN 6 −30 −13PQ LM−1 0 2OP 13. M 0 1 2P. MN 2 2 0PQ LM11 −4 −7OP 14. M 7 −2 −5P. MN10 −4 −6PQ LM −9 4 4OP 15. M −8 3 4P. MN−16 8 7 PQ LM 2 4 6 OP 16. M 4 2 −6 P. NM−6 −6 −15QP LM1 2 3OP 17. M2 4 6P. MN3 6 9PQ LM 1 −4 −1 −4OP 2 0 5 −4P 18. M MM−1 1 −2 3 PP. N−1 4 −1 6 Q LM LM 0 OP LM1OP LM3OPOP Ans. 2 , 3 , 5 , MM MM−01PP, MM00PP, MM21 PPPP N Q N Q N QQ N LM LM−1OP LM−2OP LM3OPOP MMAns. 5, − 3, − 3, M−2P, M 1 P, M0 PPP MN 1 PQ MN 0 PQ MN1 PQQ N LM LM3OP LM 1 OP LM 1 OPOP MMAns. 5, 2, 2, M2P, M 3 P, M 3 PPP MN4PQ MN−1PQ MN−1PQQ N LM LM−1OP LM 1 OP LM1OPOP MMAns. − 2, 3, 6, M 0 P, M−1P, M2PPP MN 1 PQ MN 1 PQ MN1PQQ N LM LM5OP LM2OP LM 0 OPOP MMAns. 0, − 1, 2, M1P, M0P, M 1 PPP NM0QP NM1QP NM−2QPQ N LM LM 2 OP LM1OP LM 2 OPOP MMAns. 0, 3, − 3, M−2P, M2P, M 1 PPP MN 1 PQ MN2PQ MN−2PQQ N LM LM1OP LM 1 OP LM2OPOP Ans. 0 , 1 , 2 , MM MM11PP, MM−21PP, MM12PPPP N Q N Q N QQ N LM LM 0 OP LM1OP LM1OPOP MMAns. − 1, − 1, 3, M 1 P, M1P, M1PPP MN−1PQ MN1PQ MN2PQQ N LM LM 1 OP LM 2 OP LM1 OPOP MMAns. − 2, 9, − 18, M−1P, M 2 P, M1 PPP MN 0 PQ MN−1PQ MN4PQQ N LM LM−2OP LM 3 OP LM1 OPOP MMAns. 0, 0, 14, M 1 P, M 6 P, M2 PPP NM 0 QP NM−5QP NM3QPQ N LM 2 OP LM 3 OP 6 3 [Ans. λ – 5λ + 9λ – 7λ + 2 = 0, λ = 2, 1, 1, 1 M P, M P MM−2PP MM−4PP N−3Q N−5Q 4 3 2 232 A TEXTBOOK OF ENGINEERING MATHEMATICS—I LM LM0OP LM−4OP LM−4OPOP Ans. 2 , 2 , − 2 , MM MM11PP, MM 71 PP, MM 71 PPPP N Q N Q N QQ N LM2 −2 2 OP 19. M1 1 1 P. MN1 3 −1PQ 20. Find the eigen values of A5 when LM3 0 0OP 5 4 0P A= M MN3 6 1PQ . Ans. 3 5 , 4 5 , 1 5 21. If A and B are n × n matrices and B is a non-singular matrix prove that A and B–1 AB have same eigen values. [Hint : Characteristic polynomial of B–1 AB = |B–1AB – λI| = |B–1 AB – B–1 (λI)B| = |B–1 (A – λI)B| = |B–1 |A – λI| |B| = |B–1||B||A – λI| = |B–1B||A – λI| = |I| |A – λI| = |A – λI| = characteristic polynomial of A]. 22. Find the sum and product of the eigen values. LM 2 3 −2OP A = M−2 1 1 P . MN 1 0 2 PQ 3.18 Ans. Sum = 2 + 1 + 2 = 5, product = |A| = 21 CAYLEY-HAMILTON THEOREM STATEMENT Any square matrix A satisfies its own characteristic equation. (U.P.T.U., 2005) i.e., if C0 + C1λ + C2λ + .... + Cnλ is the characteristic polynomial of degree n in λ then C0I + C1A + C2A2 + .... + CnAn = 0. 2 n Proof: Let A be a square matrix of order n. Let |A – λI| = C0 + C1λ + C2λ2 + ... + Cnλn ...(i) be the characteristic polynomial of A. Now, the elements of adj (A – λI) are cofactors of the matrix (A – λI) and are polynomials in λ of degree not exceeding n – 1. Thus adj (A – λI) = B0 + B1λ + B2λ2 + ... + Bn – 1 λn – 1 ...(ii) where Bi are n-square matrices whose elements are the functions of the elements of A and independent of λ. We know that (A – λI). adj (A – λI) = |A – λI) I From (i) and (ii), we have (A – λI) (B0 + B1λ + B2λ2 + ... + Bn – 1λn–1) = (C0 + C1λ + C2λ2 + ... + Cnλn)I Equating the like powers of λ on both sides of (iii), we get AB0 = C0I AB1 – B0 = C1I ...(iii) 233 MATRICES ® ® ® ABn – 1 – Bn – 2 = Cn – iI –Bn – 1 = CnI Pre-multiplying the above equations by I, A, A2, ... An respectively and adding; we get C0I + C1A + C2A2 + ... + CnAn = 0 ...(iv) Since all the terms on the L.H.S. cancel each other. Thus A satisfies its own characteristic equation. Proved. 3.18.1 Inverse by Cayley-Hamilton Theorem We have C0I + C1A + C2A2 + ... + CnAn = 0 ...(i) –1 Multiplying equation (i) by A , we get C0A–1 + C1I + C2A + ... + CnAn – 1 = 0 ⇒ C0A–1 = –[C1I + C2A + ... + CnAn – 1] ⇒ A–1 = – 1 [C I + C2A + ... + CnAn – 1] C0 1 Note: A–1 exists only if C0 = |A|. Definition. An expression of the form C0 + C1λ + C2λ2 +, + Cnλn where C0, C1, ..., Cn are square matrices of the same order and Cn C 0 is called a ‘‘matrix polynomial of degrees n’’. LM1 2 3OP Example 1. Verify the Cayley-Hamilton theorem for the matrix M2 4 5P . Also find its MN3 5 6PQ inverse using this theorem. (U.P.T.U., 2006) Sol. The characteristic equation of A is |A – λI| = 1− λ 2 2 4−λ 3 5 3 5 =0 6−λ ⇒ (1 – λ) [(4 – λ) (6 – λ) – 25] – 2[2(6 – λ) – 15] + 3 [10 – 3 (4 – λ)] = 0 ⇒ λ3 – 11λ2 – 4λ + 1 = 0 Cayley-Hamilton theorem is verified if A satisfies the above characteristic equation i.e. A3 – 11A2 – 4A + 1 = 0 Now, LM1 2 3OP LM1 2 3OP LM14 25 31OP A = A·A = M2 4 5P M2 4 5P = M25 45 56P NM3 5 6PQ MN3 5 6PQ MN31 56 70PQ LM1 2 3OP LM14 25 31OP LM157 283 353OP A = A·A = M 2 4 5P M25 45 56P = M283 510 636P NM3 5 6QP NM31 56 70QP MN353 636 793PQ 2 3 2 ...(i) 234 A TEXTBOOK OF ENGINEERING MATHEMATICS—I LM157 283 353OP LM14 25 31OP LM1 2 3OP Verification, A – 11A – 4A + I = M283 510 636P − 11 M25 45 56P – M2 4 5P MN353 636 793PQ MN31 56 70PQ MN3 5 6PQ LM1 0 0OP L0 0 0O + M0 1 0P = M0 0 0P MN0 0 1PQ MMN0 0 0PPQ 3 2 ⇒ A3 – 11A2 – 4A + 1 = 0 Hence theorem verified. To find )1 : Multiplying equation by A–1, we get A2 – 11A – 4I + A–1 = 0 ⇒ A–1 = – A2 + 11A + 4I ⇒ LM14 25 31OP LM1 2 3OP LM1 0 0OP = – M25 45 56P + 11 M2 4 5P + 4 M0 1 0P NM31 56 70PQ MN3 5 6PQ MN0 0 1PQ LM 1 −3 2 OP = M−3 3 −1P . MN 2 −1 0 PQ –1 A Example 2. Express B = A8 – 11A7 – 4A6 + A5 + A4 – 11A3 – 3A2 + 2A + I as a quadratic polynomial in A, where A is square matrix given in Example 1. Find B as well as A4. Sol. Thus, B = A8 – 11A7 – 4A6 + A5 + A4 – 11A3 – 3A2 + 2A + I = A5 (A3 – 11A2 – 4A + I) + A (A3 – 11A2 – 4A + I) + A2 + A + I = A5 (0) + A(0) + A2 + A + I |As A3 – 11A2 – 4A + I = 0 in Example 1 = A2 + A + I LM14 25 31OP LM1 2 3OP LM1 0 0OP B = M25 45 56P + M2 4 5P + M0 1 0P NM31 56 70PQ MN3 5 6PQ MN0 0 1PQ L16 27 34O B = MM27 50 61PP . MN34 61 77 PQ To find )4 : We have A3 – 11A2 – 4A + I = 0 ⇒ A3 = 11A2 + 4A – I Multiplying both sides by A A4 = 11A3 + 4A2 – IA LM157 283 353OP LM14 25 31OP LM1 2 3OP A = 11 M 283 510 636P + 4 M25 45 56P – M2 4 5P MN353 636 793PQ MN31 56 70PQ MN3 5 6PQ LM1782 3211 4004OP = M 3211 5786 7215P . MN4004 7215 8997PQ 4 235 MATRICES LM 2 –1 1 OP PP MM Q N Example 3. Find the characteristic equation of the matrix –1 2 –1 and hence also find 1 –2 2 –1 A by Cayley-Hamilton theorem. (U.P.T.U., 2003, 2004, 2005) Sol. The characteristic equation of A is 2– λ |A – λ I| = –1 1 –1 1 2 – λ –1 = 0 –2 2 – λ ⇒ (2 – λ) [λ2 – 4λ + 4 – 2] + [– 2 + λ + 1] + [2 – 2 + λ] = 0 ⇒ λ3 – 6λ2 + 8λ – 3 = 0 By Cayley-Hamilton theorem A3 – 6A2 + 8A – 3I = 0. Multiplying by A–1 on both sides, we get A2 – 6A + 8I – 3A–1 = 0 1 (A2 – 6A + 8I) 3 2 –1 1 2 2 −1 A = A.A = −1 1 −2 2 ⇒ Now, ∴ Thus, A–1 = OP LM 2 −1 1OP LM 6 −6 6OP PP MM−1 2 −1PP = MM−5 7 −5PP Q N 1 −2 2Q N 6 −9 7Q LM 6 −6 6 OP LM 2 −1 1 OP LM8 0 0OP LM2 0 −1OP A – 6A + 8I = M−5 7 −5P − 6 M−1 2 −1P + M0 8 0P = M1 3 1 P MN 6 −9 7 PQ MN 1 −2 2 PQ MN0 0 8PQ MN0 3 3 PQ L2 0 −1OP 1M 1 3 1P. A = 3M MN0 3 3 PQ LM MM N 2 –1 Example 4. Evaluate A–1, A–2, A–3 if LM 4 6 6 OP 3 2 A = M1 MN−1 −4 −3PPQ Sol. Characteristic equation is 4− λ 6 6 1 3− λ 2 |A – λI| = =0 −1 −4 −3 − λ ⇒ λ3 – 4λ2 – λ + 4 = 0 By Cayley-Hamilton theorem, we have ⇒ A3 – 4A2 – A + 4I = 0 A2 – 4A – I + 4A–1 = 0 (multiplying byA–1) A–1 = 1 [–A2 + 4A + I) 4 ...(i) 236 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Now, ∴ Thus, LM16 18 18OP MM 5 7 6 PP N−5 −6 −5Q LM−16 −18 −18OP LM16 24 24 OP LM1 0 0OP 8 + 0 1 0 – A + 4A + I = M −5 −7 −6 P + M 4 12 MN 5 6 5 PQ MN−4 −16 −12PPQ MMN0 0 1PPQ LM 1 6 6 OP 2 = M−1 6 MN 1 −10 −6PPQ L 1 6 6 OP 1M −1 6 2 A = 4M MN 1 −10 −6PPQ A2 = 2 –1 Multiplying equation (1) by A–1, we get –2 A Similarly, LM MM N 14 1 1 −1 4 − + + A I IA − 54 = = 4 4 54 A–3 = −1 2 52 32 OP PP Q −9 2 −3 2 11 2 1 [– I + 4A–1 + A–2] 4 LM MM N OP PP Q 1 78 78 1 −21 90 26 . = 64 21 −154 −90 Example 5. Verify Cayley-Hamilton theorem for the matrix A = A–1. LM1 2 OP and hence find N2 −1Q (U.P.T.U., 2008) Sol. The characteristic equation of A is |A – λI| = or 1− λ 2 = 0 ⇒ – (1 – λ2) – 4 = 0 2 −1 − λ λ2 – 5 = 0 By Cayley-Hamilton theorem A2 – 5 = 0 Now A2 = LM1 2 OP LM1 2 OP = LM5 0OP = 5 LM1 0OP = 5I N2 −1Q N2 −1Q N0 5Q N0 1Q Putting value of A2 in equation (i), we get A2 – 5 = 5I – 5I = 0 Hence, the Cayley-Hamilton theorem is verified. ...(i) 237 MATRICES To find )1 : Multiplying equation (i) by A–1, we get A – 5A–1 = 0 ⇒ A–1 = LM N OP Q 1 1 1 2 . A = 5 5 2 −1 LM 0 c −bOP a satisfies its characteristic equation. Also Example 6. Show that the matrix A = M − c 0 MN b − a 0 PPQ find A–1. −λ c −b |A – λI| = − c − λ a = 0 b −a −λ Sol. i.e., – λ[λ2 + a2] – c[λc – ab] – b[ac + bλ] = 0 i.e., λ3 + λ(a2 + b2 + c2) = 0 By Cayley-Hamilton theorem, A3 + A (a2 + b2 + c2) = 0 ...(i) OP ac LM 0 c −bOP LM 0 c −bOP LM−ec + b j ab PP A = M− c 0 − ec + a j a −c 0 a = M ab bc MN b −a 0 PPQ MMN b −a 0 PPQ MMN ac −eb + a jP bc Q 2 2 2 2 Now, 2 2 2 OP ac LM 0 c −bOP LM−eb + c j ab PP A = A·A = M− c 0 −e c + a j a M ab bc MN b −a 0 PPQ MMN ac −eb + a jP bc Q 2 3 2 2 2 2 2 2 LM − c e a + b + c j b ea + b + c j O 0 LM 0 c −bOP PP M + + − + + = − + + c a b c 0 a a b c a b c or A = e jP e j M− c 0 a P MM−be a + b + c j a a + b + c MN b −a 0 PQ 0 PQ j e j N e 2 3 or 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 Using (ii) in equation (i), we get A3 + A (a2 + b2 + c2) = – (a2 + b2 + c2) A + A (a2 + b2 + c2) = 0 To find )1: Multiplying equation (i) A–2 on both sides, we get A + A–1 (a2 + b2 + c2) = 0 1 ⇒ A–1 = − Hence, LM 0 c −bOP A = − −c 0 a · ea + b + c j MMN b −a 0 PPQ ea + b + c j 2 2 1 –1 2 2 2 2 2 A3 = – (a2 + b2 + c2) A 2 2 2 ⋅A ...(ii) 238 A TEXTBOOK OF ENGINEERING MATHEMATICS—I EXERCISE 3.7 1. Find the characteristic equation of the matrix A LM4 3 1 OP A = M 2 1 −2P . Hence find A . (U.P.T.U., 2001) MN1 2 1 PQ LM L 5 −1 −7OPOP 1 M MMAns. Characteristic equation : λ − 6λ + 6λ − 11 = 0, A = 11 M−4 3 10 PPP MN 3 −5 −2PQQ N –1 3 −1 2 2. Find the characteristic equation of the matrix LM2 1 1OP A = M0 1 0P . Verify Cayley-Hamilton theorem and hence MN1 1 2PQ (U.P.T.U., 2002) evaluate the matrix equation A8 – 5A7 + 7A6 – 3A5 + A4 – 5A3 + 8A2 – 2A + I. LM LM8 5 5OPOP λ − λ + λ − = + + Ans. 5 7 3 0 , A A I , MM MM0 3 0PPPP N5 5 8 Q Q N 3 2 2 3. Verify Cayley-Hamilton theorem for A and hence find A–1 LM LM−2 0 1OPOP LM 2 −1 1 OP MMAns. A = M−5 1 5PPP (i) M−15 6 −5P MN 0 1 3PQQ MN 5 −2 2 PQ N LM LM 3 −1 −1OPOP LM1 1 1OP MMAns. A = M−1 1 0 PPP (ii) M1 2 1P MN−1 0 1 PQQ MN1 1 2PQ N LM L 0 −3 −3OPOP LM1 0 3 OP 1M MMAns. A = − 9 M−3 −2 7 PPP (iii) M 2 1 −1P MN−3 1 1 PQQ MN1 −1 1 PQ N LM L−3 −2 2 OPOP LM 7 2 −2OP 1M MMAns. A = 3 M 6 5 −2PPP 4. Find A if A = M−6 −1 2 P . MN−6 −2 5 PQQ N MN 6 2 −1PQ LM1 2 0 OP 5. Show that the matrix A = M 2 −1 0 P satisfies its own characteristic equation and hence MN0 0 −1PQ LM LM1/5 0 0OPOP MMAns. A = M 0 1/5 0PPP or otherwise obtain the value of A . MN 0 0 1PQQ N −1 −1 −1 −1 –1 −1 –1 239 MATRICES LM1 2 3 OP 6. Find A if A = M1 3 5 P . MN1 5 12PQ LM0 1 0 0OP 0 0 0 1 7. If A = M MM0 0 1 0PPP find A . N1 0 0 0Q LM L11 −9 1 OPOP 1M MMAns. A = 3 M−7 9 −2PPP MN 2 −3 1 PQQ N LM LM0 0 0 1OPOP MMAns. A = M1 0 0 0PPP MM0 0 1 0PPP MM N0 1 0 0QPQ N −1 –1 −1 –1 8. Find B = A6 – 4A5 + 8A4 – 12A3 + 14A2 if A = LM 1 2OP . N−1 3Q LMAns. λ − 4λ + 5 = 0, B = 5I − 4 A = L1 −8OOP MN 4 −7PQQ N LM L−1 7 −24OPOP 1 M MMAns. A = 11 M 2 −3 4 PPP MN 2 −3 15 PQQ N 2 LM3 3 4OP 9. Find A if A = M 2 −3 4P · MN0 −1 1PQ LM 2 −3 1 OP 3 P satisfies the equation A (A – I) (A + 2I) = 0. 10. Show that the matrix A = M 3 1 MN−5 2 −4PQ −1 –1 3.19 DIAGONALIZATION OF A MATRIX A square matrix ‘A’ is said to be diagonalisable if there exists another non-singular square matrix P such that P–1AP is a diagonal matrix. In other words, A is diagonalisable if A is similar to a diagonal matrix. Similar matrix: Two matrices A and B are said to be similar if there exists a non-singular matrix P such that B = P–1AP. This transformation of A to B is known as similarity transformation. Note: Similar matrices A and B have same eigen values. Further, if X is an eigen vector of A then Y = P–1X is an eigen vector of the matrix B. 3.19.1 Working Rule The matrix P is called a model matrix for A. There are following steps to obtain diagonal form of matrix A. Step 1: First of all obtain eigen values λ1, λ2, ..., λn of the matrix A. Step 2: If there exists a pair of distinct complex eigen values then A is not diagonalisable over the field R of real numbers. Step 3: If there does not exist a set of n linearly independent eigen vectors, then A is not diagonalisable. Step 4: If there exist n linearly independent eigen vectors. 240 A TEXTBOOK OF ENGINEERING MATHEMATICS—I LMx OP LMx OP LMx OP x x x X = M P, X = M P ...., X = M P MM M PP MM M PP MM M PP Nx Q Nx Q Nx Q LMx x L x OP L x x x P = M MM M M M PPP Nx x L x Q 11 12 1n and diagonal matrix D = P–1 AP. n2 22 n 2 1 then let, n1 21 2n nn 11 21 n1 12 22 n2 1n 2n nn 3.19.2 Powers of Matrix A Consider ⇒ D = P−1 AP D2 = (P−1 AP) (P−1 AP) = P−1 A (PP−1) AP = P−1 A. IAP = P−1 AAP = P−1 A2 P Similarly, D3 = P−1 A3 P, ...., Dn = P−1 An P We can obtain An pre-multiplying by P and post-multiplying by P–1 ⇒ P Dn P−1 = P (P−1 An P) P−1 = (PP−1) An (PP−1) = IAnI = An ∴ An = PDn P−1 LM1 6 1OP Example 1. Diagonalize the matrix M1 2 0P . MN0 0 3PQ Sol. The characteristic equation of A is 1− λ 6 1 1 2− λ 0 =0 |A – λ I| = 0 0 3− λ ⇒ (λ + 1) (λ – 3) (λ – 4) = 0 ⇒ λ = – 1, 3, 4 Consider the equation (A – λ I) X = 0 For, λ = – 1, LM2 6 1OP LMx OP MM1 3 0PP MMx PP = 0 N0 0 4Q Nx Q 1 2 3 R1 ↔ R2 on coefficient matrix, we get LM1 3 0OP LMx OP MM2 6 1PP MMx PP = 0 N0 0 4Q Nx Q 1 2 3 (U.P.T.U., 2006) 241 MATRICES R2 → R2 – 2R1 LM1 3 0OP LMx OP MM0 0 1PP MMx PP = 0 N0 0 4Q Nx Q 1 2 3 R1 → R1 – R2, R3 → R3 – 4R2 LM1 3 −1OP LMx OP MM0 0 1 PP MMx PP = 0 N0 0 0 Q Nx Q 1 2 3 ⇒ Let, x1 + 3x2 – x3 = 0 x3 = 0 x2 = k, then x1 = – 3k ∴ X1 = For, λ = 3 LM−3kOP LM−3OP LM−3OP MM k PP = k MM 1 PP = MM 1 PP N 0 Q N0Q N0Q LM−2 6 1OP LMx OP MM 1 −1 0PP MMx PP = 0 N 0 0 0 Q Nx Q 1 2 3 ⇒ – 2x1 + 6x2 + x3 = 0 x1 – x2 + 0·x3 = 0 ⇒ ∴ For, λ = 4 x1 –x2 x3 x x x = or 1 = 2 = 3 = −1 −4 1 1 −4 1 x1 = 1, x2 = 1, x3 = – 4 LM 1 OP X = M1P MN−4PQ 2 LM−3 6 1 OP LMx OP MM 1 −2 0 PP MMx PP = 0 N 0 0 −1Q Nx Q 1 2 3 R1 ↔ R2 LM 1 −2 0 OP LMx OP MM−3 6 1 PP MMx PP = 0 N 0 0 −1Q Nx Q 1 2 3 R2 → R2 + 3R1 LM1 −2 0 OP LMx OP MM0 0 1 PP MMx PP = 0 N0 0 −1Q Nx Q 1 2 3 242 A TEXTBOOK OF ENGINEERING MATHEMATICS—I R1 → R1 + R2, R3 → R3 + R2 LM1 −2 1OP LMx OP MM0 0 1PP MMx PP = 0 N0 0 0Q Nx Q 1 2 3 ⇒ x1 – 2x2 + x3 = 0 0x1 + 0x2 + x3 = 0 x x x1 = – 2 = 3 −2 1 0 ⇒ x1 x x = 2 = 3 ⇒ x1 = 2, x2 = 1, x3 = 0 2 1 0 or ∴ Thus, To obtain 21: matrix of cofactors ∴ LM2OP X = M1 P MN0PQ LM−3 1 2OP P = M 1 1 1P MN 0 −4 0PQ 3 |P| = – 3 (4) – 1 (0) + 2 (– 4) = – 20 LM 4 0 −4 OP B = M−8 0 −12P MN−1 5 −4 PQ LM 4 −8 −1OP 0 5P adj P = B′ = M 0 MN−4 −12 −4PQ L 4 −8 −1OP 1 LM−4 8 1 OP adj P 1 M =– 0 0 5 = 0 0 −5 P = |P| 20 M MN−4 −12 −4PPQ 20 MMN 4 12 4 PPQ –1 and Diagonalisation: D = P–1 AP LM−4 8 1 OP LM1 6 1OP LM−3 1 2OP MM 0 0 −5PP MM1 2 0PP MM 1 1 1PP N 4 12 4 Q N0 0 3Q N 0 −4 0Q L 4 −8 −1 OP LM−3 1 2OP 1 M 0 0 −15 1 1 1 = 20 M MN16 48 16 PPQ MMN 0 −4 0PPQ L−20 0 0 OP LM−1 0 0OP 1 M 0 60 0 = 0 3 0 . = 20 M MN 0 0 80PPQ MMN 0 0 4PPQ D = 1 20 243 MATRICES LM−1 2 −2OP MM 1 2 1 PP to the diagonal form. N−1 −1 0 Q Example 2. Reduce the matrix A = (U.P.T.U., 2001, 2004) |A – λI| = Sol. ⇒ −1 − λ 2 −2 1 2− λ 1 = 0 −1 −1 −λ λ3 – λ2 – 5λ + 5 = 0 λ = 1, + 5 , – 5 On solving, we get For, λ = 1 (A – λI) X = 0 LM−2 2 −2OP LMx OP MM 1 1 1 PP MMx PP = 0 N−1 −1 −1Q Nx Q 1 ⇒ 2 3 R1 ↔ R2 LM 1 1 1 OP LMx OP MM−2 2 −2PP MMx PP = 0 N−1 −1 −1Q Nx Q 1 2 3 R2 → R2 + 2R1, R3 → R3 + R1 LM1 1 1OP LMx OP MM0 4 0PP MMx PP = 0 N0 0 0Q Nx Q 1 2 3 ⇒ x1 + x2 + x3 = 0 0x1 + 4x2 + 0x3 = 0 x1 x x x x x = − 2 = 3 or 1 = 2 = 3 −1 −4 0 4 1 0 x1 = 1, x2 = 0, x3 = – 1 ⇒ LM 1 OP X = M0P MN−1PQ ∴ For 1 λ = LM−1 − 5 2 −2 OP Lx O MM 1 2 − 5 1 PP MMx PP = 0 N −1 −1 − 5 Q MNx PQ 1 2 3 R1 ↔ R2 LM 1 2 − 5 1 OP Lx O MM−1 − 5 2 −2 PP MMx PP = 0 N −1 −1 − 5 Q MNx PQ 1 2 3 5 244 A TEXTBOOK OF ENGINEERING MATHEMATICS—I R2 → R2 + (1 + 5 ) R1, R3 → R3 + R1 LM1 2 − 5 1 OP Lx O MM0 5 − 1 5 − 1PP MMx PP = 0 N0 1 − 5 1 − 5Q MNx PQ 1 2 3 R3 → R3 + R2, and R2 → 1 5 −1 R2 LM1 2 − 5 1OP Lx O MM0 1 1PP MMx PP = 0 N0 0 0Q MNx PQ 1 2 3 ⇒ x1 + (2 – 5 ) x2 + x3 = 0 0x1 + x2 + x3 = 0 Solving by cross-multiplication x1 = – 1− 5 x1 or = 5 −1 x2 x3 = 1 1 x2 x3 = ⇒ x1 = −1 1 5 − 1 , x2 = 1, x3 = – 1 LM 5 − 1OP X = M 1 P MN −1 PQ ∴ 2 Similarly, the eigen vector corresponding λ = – 5 LM 5 + 1OP X = M −1 P MN 1 PQ LM 1 5 − 1 1 P = M0 MN−1 −1 3 ∴ The modal matrix To obtain 21: R3 → R3 + R1 LM 1 MM 0 N−1 LM1 MM0 N0 5 −1 1 −1 5 −1 1 5−2 5 +1 −1 1 OP L1 0 0O PP ~ MM0 1 0PP P Q MN0 0 1PQ OP LM1 0 0OP PP ~ MM0 1 0PP P 5 + 2Q N1 0 1Q 5 +1 −1 5 +1 −1 1 OP PP Q 245 MATRICES e 5 − 1j R LM1 0 2 5 OP LM1 1 − 5 0OP 1 0P P −1 P ~ M0 MM0 1 M N0 5 − 2 5 + 2PQ N1 0 1PQ R → R – e 5 − 2j R LM1 0 2 5 OP LM1 1 − 5 0OP MM0 1 −1 PP ~ MM0 1 0PP P N0 0 2 5 Q N1 2 − 5 1Q R1 → R1 – 3 2 3 2 R1 → R1 – R3 LM1 0 0 OP LM0 −1 −1OP MM0 1 −1 PP ~ MM01 2 −1 5 01 PP P Q N0 0 2 5 Q N R3 → 1 2 5 R3 LM1 0 0 OP LM 0 MM0 1 −1PP ~ MM 01 N0 0 1 Q MMN 2 5 R2 → R2 + R3 −1 1 2− 5 2 5 OP P 0 PP 1 P 2 5 PQ −1 LM LM1 0 0OP M 01 MM0 1 0PP ~ MM 2 5 N0 0 1Q MM 1 MN 2 5 ∴ Now, OP −1 −1 P 2+ 5 1 P P 2 5 2 5P P 2− 5 1 P 2 5 2 5 PQ L0 −2 5 −2 5 OP 1 M P = M1 2 + 5 1 PP 2 5 M N1 2 − 5 1 Q LM0 −2 5 −2 5 OP L−1 2 −2O LM 1 M P 1 M 2 1P M0 1 2+ 5 1 P M1 D = P AP = 2 5 M MN1 2 − 5 1 PPQ MMN−1 −1 0 PPQ MMN−1 L2 5 0 0 OP L1 0 0 O 1 M M 5 0 PP . = 0 10 0 P = M0 M 2 5 M N 0 0 −10PQ MN0 0 − 5 PQ –1 –1 5 −1 1 −1 OP −1 P P 1 PQ 5 +1 246 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 3. Diagonalise the matrix A = LM11 −4 −7OP MM 7 −2 −5PP and hence find A . N10 −4 −6Q 5 Sol. The characteristic equation of A is |A – λI| = 0 ⇒ LM11 − λ −4 −7 OP MM 7 −2 − λ −5 PP = 0 N 10 −4 −6 − λQ On simplification, we get λ3 – 3λ2 + 2λ = 0 ⇒ λ (λ –1) (λ – 2) = 0 ∴ λ = 0, 1, 2 The eigen values are real and distinct and hence A is diagonalisable. Now, consider the relation (A – λI) X = For λ = 0 0 LM11 −4 −7OP LMx OP MM 7 −2 −5PP MMx PP = 0 N10 −4 −6Q Nx Q 1 2 3 ⇒ 11x1 – 4x2 – 7x3 = 0 | Take any two equations. 7x1 – 2x2 – 5x3 = 0 10x1 – 4x2 – 6x3 = 0 Taking the first-two equations, we get x1 x x x x x = 2 = 3 ⇒ 1 = 2 = 3 6 6 6 1 1 1 ∴ LM1OP X = M1P MN1PQ 1 For λ = 1, we have LM10 −4 −7OP LMx OP MM 7 −3 −5PP MMx PP = 0 N10 −4 −7Q Nx Q 1 2 3 ⇒ 10x1 – 4x2 – 7x3 = 0 7x1 – 3x2 – 5x3 = 0 10x1 – 4x2 – 7x3 = 0 Taking the first-two equations, we get x1 x x = 2 = 3 −1 2 1 247 MATRICES LM 1OP MM−1PP , N 2Q ∴ X2 = Similarly, LM−2OP λ = 2, X = M −1P MN−2PQ For 3 The three vectors are linearly independent, LM1 1 −2OP P = M1 −1 −1P MN1 2 −2PQ LM−4 2 3OP P = M−1 0 1P MN−3 1 2PQ LM−4 2 3OP LM11 −4 −7OP LM1 1 −2OP D = P AP = M−1 0 1P M 7 −2 −5P M1 −1 −1P MN−3 1 2PQ MN10 −4 −6PQ MN1 2 −2PQ LM0 0 0OP = M0 1 0P . MN0 0 2PQ Hence the modal matrix ∴ –1 –1 Now, A5 = PD5 P–1 Next LM0 0 0 OP LM0 0 0 OP 0 P = M0 1 0 P But, D = M0 1 MN0 0 2 PQ MN0 0 32PQ LM1 1 −2OP LM0 0 0 OP LM−4 2 3OP ∴ A = M1 −1 −1P M0 1 0 P M−1 0 1P MN1 2 −2PQ MN0 0 32PQ MN−3 1 2PQ LM191 −64 −127OP = M 97 −32 −65 P . MN190 −64 −126PQ LM 0 −2 −2OP 2 P is not diagonalisable. Example 4. Prove that the matrix A = M−1 1 MN−1 −1 2 PQ 5 5 5 5 Sol. The characteristic equation is |A – λI| = 0 ⇒ –λ −2 −2 −1 1 − λ 2 −1 −1 2 − λ = 0 248 A TEXTBOOK OF ENGINEERING MATHEMATICS—I ⇒ λ3 – 3λ2 + 4 = 0 ⇒ (λ + 1) (λ – 2)2 = 0 ∴ λ = –1, 2, 2. Here the eigen values are real. Let X be an eigen vector corresponding to λ = 2 ∴ (A – 2I) X = 0 LM−2 −2 −2OP LMx OP MM−1 −1 2 PP MMx PP = 0 N−1 −1 0 Q Nx Q 1 ⇒ 2 3 −1 R1 → R1 2 1 1 1 −1 −1 2 −1 −1 0 OP LMx OP PP MMx PP = 0 Q Nx Q LM MM N 1 2 3 R2 → R2 + R1, R3 → R3 + R1 LM1 1 1OP LMx OP MM0 0 3PP MMx PP = 0 N0 0 1Q Nx Q 1 2 3 R2 → 1 R and R3 → R3 – R2 3 2 1 1 1 x1 0 0 1 x2 = 0 0 0 0 x3 LM MM N OP LM OP PP MM PP QN Q ⇒ r = 2, n = 3 ⇒ n – r = 1 there exist only one linearly independent eigen vector corresponding λ = 2. Hence, there does not exist three linearly independent eigen vectors. Thus A is not diagonalisable. Example 5. Find a matrix P which diagonalizes the matrix A = where D is the diagonal matrix. Sol. The characteristic equation of A is |A – λI| = 0 ⇒ 4− λ 1 2 3− λ ⇒ λ2 – 7λ + 10 = 0 ⇒ (λ – 2) (λ – 5) = 0 ∴ = 0 ⇒ LM4 1OP , verify P AP = D N2 3Q (U.P.T.U., 2008) (4 – λ) (3 – λ) – 2 = 0 λ = 2, 5 Now, consider the relation (A – λI) X = 0 For λ = 2 LM2 1OP LMx OP = 0, R → R − 2R LM2 1OP LMx OP = 0 N2 1Q Nx Q N0 0Q Nx Q 1 2 ⇒ 2x1 + x2 = 0, 1 2 2 1 2 Here let x2 = k, so x1 = − k 2 –1 249 MATRICES ∴ X1 = LMx OP = LM −2k OP = − k LM 1 OP or LM 1 OP Nx Q MN k PQ 2 N−2Q N−2Q 1 2 For λ = 5 LM−1 1 OP LMx OP = 0, R → R + 2R LM−1 1OP LMx OP = 0 N 2 −2Q Nx Q N 0 0Q Nx Q 1 1 2 2 1 2 2 ⇒ – x1 + x2 = 0. Here again let x2 = k so x1 = k ∴ X2 = LMx OP = LMkOP = k LM1OP or LM1OP Nx Q NkQ N1Q N1Q 1 2 The modal matrix P = LM 1 1OP N−2 1Q Verification: |P| = 1 + 2 = 3 LM OP N Q 1 L1 −1O L 4 1O L 1 1O 1 L1 −1O L 2 5O P AP = = M M P M P M P PM P 3 N2 1 Q N 2 3Q N−2 1Q 3 N 2 1 Q N−4 5Q L2 0OP = D (diagonal matrix). 1 L6 0 O = M = M P 3 N0 15Q N0 5Q ∴ P–1 = 1 1 −1 3 2 1 –1 Thus P–1 AP = D verified. 3.20 APPLICATION OF MATRICES TO ENGINEERING PROBLEMS Matrices have various engineering applications, they can be used to characterize connections in electrical networks, in nets of roads, in production processes, etc., as follows: 3 3.20.1 Nodal Incidence Matrix The network in figure consists of 6 branches (connections) and 4 nodes (point where two or more branches together). One node is the reference node (grounded node, whose voltage is zero). We number the other nodes and number and direct the branches. This we do arbitrarily. The network can now we described by a matrix A = [aij], where aij = 2 1 2 1 6 4 (Reference node) +1 if branch j leaves node (i) –1 if branch j enters node (i) 0 if branch j does not touch node (i) 3 5 Fig. 3.2 250 A TEXTBOOK OF ENGINEERING MATHEMATICS—I A is called the nodal incidence matrix of the network. The matrix for the network in figure is given below: Branch 1 2 3 4 5 6 Node 1 Node 2 1 0 –1 1 –1 0 0 1 0 1 0 0 Node 3 0 0 1 0 –1 –1 3.20.2 Mesh Incidence Matrix A network can also be characterized by the mesh incidence matrix M = [mij], where mij = +1 if branch j is in mesh i and has the same orientation –1 if branch j is in mesh i and has the opposite orientation 0 if branch j is not in mesh i And a mesh is a loop with no branch in its interior (or in its exterior). Here, the meshes are numbered and directed (oriented) in an arbitrary fashion. Show that for the network in Figure. The matrix M has the given form, where row 1 corresponds to mesh 1, etc. 3 4 3 2 2 1 LM1 1 0 −1 0 0OP 0 0 0 1 −1 1P M = M MM0 −1 1 0 1 0PP N1 0 1 0 0 1Q 5 6 1 Fig. 3.3 Example 1. There is a circuit (electrical network) in figure given below. Find the currents i1, i2 and i3 respectively. 20 ohms 10 ohms Q i1 i3 80 10 ohms 90 volts i2 P 15 ohms Fig. 3.4 Sol. We label the currents as shown, choosing directions arbitrarily; if a current will come out negative, this will simply mean that the current flows against the direction of our arrow. The current entering each battery will be the same as the current leaving it. The equations for the currents results from Kirchhoff’s laws. Kirchhoffs Current Law (KCL): At any point of a circuit, the sum of the inflowing currents equals the sum of the out flowing currents. Kirchhoffs Voltage Law (KVL): In any closed loop, the sum of all voltage drops equals the impressed electromotive force. 251 MATRICES Node P gives the first equation, node Q the second, the right loop the third, and the left loop the fourth, as indicated in the figure. Node P : i1 – i2 + i3 = 0 (i) Node Q : Right loop : – i1 + i2 – i3 = 0 10 i2 + 25 i3 = 90 (ii) (iii) Left loop : 20 i1 + 10 i2 = 80 (iv) We solve these equations by matrix method LM 1 −1 1 M 0OP −1 1 −1 M 0 Augmented matrix [A : B] = M MM 0 10 25 M 90PPP N20 10 0 M 80Q Applying R2 → R2 + R1, R4 → R4 – 20 R1 LM1 −1 1 M 0 OP 0 0 0 M 0P ~ M MM0 10 25 M 90PP N0 30 −20 M 80Q R2 ↔ R4 LM1 −1 1 M 0 OP 0 30 −20 M 80 ~ M MM0 10 25 M 90PPP N0 0 0 M 0 Q R3 → 3R3 – R4 LM1 −1 1 M 0 OP 0 30 −20 M 80 P ~ M MM0 0 95 M 190PP N0 0 0 M 0 Q R2 → 1 1 R,R → R 5 2 3 19 3 LM1 −1 1 M 0 OP 0 6 −4 M 16 ~ M MM0 0 5 M 10PPP N0 0 0 M 0 Q LM1 −1 1 OP Li O LM 0 OP MM0 6 −4PP MMi PP MM16PP MN00 00 05 PQ MNi PQ MN100PQ 1 Now 2 3 252 A TEXTBOOK OF ENGINEERING MATHEMATICS—I ⇒ i1 – i2 + i3 = 0 5i3 = 10 On solving these equations i1 = 2 amperes x2 i2 = 4 amperes P di r inc r e ip ct al io n 6i2 – 4i3 = 16 i3 = 2 amperes. Example 2. An elastic membrane in the x1, x2-plane x1 with boundary circle x12 + x22 = 1, is stretched 80 that a point P : (x1, x2) goes over into the point Q : (y1, y2) given by y= P di r inc re ip c t al io n LMy OP = Ax = LM5 3OP LMx OP; Ny Q N3 5Q Nx Q 1 1 2 2 Fig. 3.5 in components y1 = 5x1 + 3 x2 ; y2 = 3 x1 + 5x2 . Find the principal directions, that is, the directions of the position vector x of P for which the direction of the position vector y of Q is the same or exactly opposite. What shape does the boundary circle take under this deformation? Sol. We are looking for vectors x such that y = λx. Since y = AX ⇒ AX = λX ∴ The above form is 5x1 + 3x2 = λx1 (5 – λ) x1 + 3x2 = 0 or 3x1 + 5x2 = λx2 3x1 + (5 – λ) x2 = 0 ∴ The characteristic equation is ⇒ 5− λ 3 = (5 – λ)2 – 9 = 0 3 5− λ λ1 = 8, λ2 = 2 for λ1 = 8, – 3x1 + 3x2 = 0 3x1 – 3x2 = 0 Sol. x2 = x1, x1 arbitrary. Let ⇒ x1 – x2 = 0 x1 – x2 = 0 x1 = x2 = 1 For λ2 = 2 our system becomes 3x1+ 3x2 = 0 ⇒ 3x1 + 3x2 = 0 x2 = – x1, x1 arbitrary Let x1 = 1, x2 = – 1 Thus the eigen vector, for instance are X1 = LM1OP and X = LM 1 OP N1Q N−1Q 2 These vectors make 45° and 135° angles with the positive x1-direction. They give the principal directions. 253 MATRICES The eigen values show that in the principal directions the membrane is stretched by factors 8 and 2, respectively. Accordingly if we choose the principal directions as directions of a new cartesian u1, u2-coordinats system, say with the positive u1-semiaxis in the first quadrant and the positive u2-semiaxis in the second quadrant of the x1 x2-system and if we get u1 = r cos φ, u2 = r sin φ then a boundary point of the unstretched circular membrane has coordinates cos φ, sin φ. Hence after the stretch we have Z1 = 8 cos φ, Z2 = 2 sin φ Since cos φ + sin φ = 1, this shows that the deformed boundary is an ellipse. 2 2 Z12 Z22 + 2 = 1 82 2 with principal semiaxes 8 and 2 in the principal directions. EXERCISE 3.8 LM 1 2 −2OP 2 1 P . Find the modal matrix P and the 1. A square matrix A is defined by A = M 1 MN−1 −1 0 PQ resulting diagonal matrix D of A. (U.P.T.U., 2000) LM LM–2 –2 2 OP LM1 0 0OPOP MMAns. P = M 1 1 1 P, D = M0 −1 0PPP MN 1 –1 –1PQ MN0 0 3PQQ N LM 3 −1 1 OP 2. Diagonalise the matrix A = M −1 5 −1P and hence find A . MN 1 −1 3 PQ LM LM 251 −405 235 OPOP MMAns. A = M−405 81 −405PPP MN 235 −405 251 PQQ N 3. Diagonalise the matrix A. LM LM2 0 0OPOP LM3 1 4OP A = M0 2 6P · MMAns. D = M0 3 0PPP MN0 0 5PQQ MN0 0 5PQ N 4 4 4. Verify whether the following matrices are diagonalisable or not. LM−2 2 −3OP LM−3 2 2OP (i) M 2 1 −6P , (ii) M−6 5 2P · MN−1 −2 0 PQ MN−7 4 4PQ LM 8 −6 2 OP 5. Diagonalise M−6 7 −4P · MN 2 −4 3 PQ [Ans. (i) Not diagonalisable (ii) diagonalisable] LM LM0 0 0 OPOP Ans. D = MM MM0 3 0 PPPP N0 0 15QQ N 254 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 6. Diagonalise the following matrices: LM1 0 0OP L0 0O L0 0OOP LM1 0 2OP L 2 −1O L1 1O LM MAns. (i) D = M0 –2 0P, (ii) MN0 6PQ, (iii) MN0 2PQPP (i) M0 1 2P , (ii) M , (iii) M −8 4 PQ 1 1PQ M MN0 0 3PQ N N MN1 2 0PQ N Q LM LM1 0 0OPOP LM2 2 1OP Ans. D = 0 1 0PP 7. Diagonalise the matrix A = M1 3 1P . MM M MN0 0 5PQPQ MN1 2 2PQ N LM4 1OP hence find A . LMAns. D = L1 0O, A = L2344 781OOP MN0 5PQ MN2343 782PQQ N3 2Q N L0 1OP is diagonalisable. 9. Verify the matrix M N0 0 Q Lk O [Ans. Not diagonalisable since only one eigen vector M P exists] N0 Q 8. Diagonalise the matrix 5 5 10. Using matrix equation determine the loop current in the following circuits: 8V i1 109 &9 Ans. i1 = 1.0826, i 2 = 1. 4004, i 3 = 1.2775 (i) $9 i2 i3 4V 6V 40 V 19 i1 49 i4 49 10 V 59 (ii) 30 V i2 29 69 i3 39 20 V Ans. i1 = 11. 43, i 2 = 10.55, i 3 = 8.04, i 4 = 5.84 255 MATRICES OBJECTIVE TYPE QUESTIONS A. Pick the correct answer of the choices given below: LM1 aOP , then the value of A is N0 1Q L1 a OP L4 4aOP (i) M (ii) M N0 4 Q NM0 1 QP L4 a OP L1 4aOP (iii) M (iv) M MN0 4 PQ N0 1 Q LM1 b 2OP 2. If the matrix M1 2 5P is not invertible, then the value of b is MN2 1 1PQ 1. If A = 4 4 4 (i) 2 (ii) 0 (iii) 1 (iv) –1 LM3 2OP , then (A ) is equal to N0 1 Q 1 L1 −26O 1 L−1 26O (i) (ii) M P M P 27 N0 27 Q 27 N 0 27Q 1 L1 −26O 1 L−1 −26O (iii) (iv) M P M P 27 N0 −27Q 27 N 0 −27Q LM0 0 1OP 4. If A = M0 1 0P , then A is MN1 0 0PQ LM0 1 0OP LM1 0 1OP (ii) M1 0 0P (i) M0 1 0P MN0 0 1PQ MN0 0 0PQ LM0 0 1OP (iii) M0 1 0P (iv) None of these MN1 0 0PQ 3. If A = –1 3 –1 5. If A and B are square matrices of order 3 such that |A| = –1 and |B| = 3, then |3AB| = (i) – 9 (ii) – 27 (iii) – 81 (iv) 81 6. The rank of the unit matrix of order n is (i) 0 (ii) 1 (iii) 2 (iv) 3 256 A TEXTBOOK OF ENGINEERING MATHEMATICS—I LM1 1 1OP 7. If A = M1 2 4P , then rank (A) is MN1 4 8PQ (i) 0 (ii) 1 (iii) 2 (iv) 3 8. If a matrix A has a non-zero minor of order r and all minors of higher orders zero, then (i) ρ(A) < r (ii) ρ(A) ≤ r (iii) ρ(A) > r (iv) ρ(A) = r 9. The system of equations 3x + 2y + z = 0, x + 4y + z = 0, 2x + y + 4z = 0. (i) is consistent (ii) has infinite solutions (iii) has only a trivial solution (iv) None of these 10. The system of equations – 2x + y + z = a, x – 2y + z = b, x + y – 2z = c is consistent, if (i) a + b – c = 0 (ii) a – b + c = 0 (iii) a + b + c ≠ 0 (iv) a + b + c = 0 11. If X1 = (2, 3, 1, –1); X2 = (2, 3, 1, – 2), X3 = (4, 6, 2, – 3) then (i) X1 + X3 = X2 (ii) X2 + X3 = X1 (iii) X1 + X2 = 2X3 (iv) X1 + X2 = X3 B. Fill in the blanks: 1. The vectors (1, –1, 1), (2, 1, 1), (3, a, 2) are linearly dependent if the value of a = .......... 2. If A is non-singular matrix of order 3 then ρ(A) = .......... 3. If λ is a non-zero characteristic root of a non-singular matrix A, then the characteristic roots of A–1 are .......... 4. If λ1, λ2, λ3 are the characteristic roots of a non-singular matrix A, then the characteristic roots of A–1 are .......... 5. The characteristic of a real skew symmetric are either ..... or ..... 6. The product of the characteristic roots of a square matrix is .......... 7. A system of n linear homogeneous equations in n unknowns has a non-trivial solution if the coefficient matrix is .......... 8. A system of n linear homogeneous equations in n unknowns is always .......... 9. For the set of vectors (X1, X2, ... Xn) to be linearly independent it is necessary that no Xi is ........... LM2 2 1OP 10. The eigen values of A = M1 3 1P are .......... MN1 2 2PQ C. Indicate True or False for the following statements: 1. (i) If |A| = 0, then at least one eigen value is zero. (ii) A–1 exists iff 0 is an eigen value of A. 257 MATRICES (iii) If |A| ≠ 0 then A is known as singular matrix. (iv) Two vectors X and Y are said to be orthogonal if, XTY = YTX ≠ 0. 2. (i) The eigen values of a real skew symmetric matrix are all real. (ii) If A is a square matrix, then latent roots of A and A′ are identical. (iii) If A is a unit matrix, then |A| = 0. (iv) If |A| ≠ 0, then |A. adj A| = |A|n–1, where A = (aij)n × n. D. Match the following: 1. (i) If A is an invertible matrix then (a) |A| = ± 1 (ii) If A is orthogonal then (b) |A| = 0 (iii) (A – λI) is a matrix (c) is 1 (iv) The rank of unit matrix of order n (d) Characteristic 2. (i) For non-Trivial (a) A–1 is a diagonal (ii) A is diagonal (b) A –1 –1 (iii) (A ) (c) |A| = 0 (iv) For trivial solution (d) ρ(A) < n 3. (i) Rank of skew symmetric matrix (a) Have same rank (ii) Two equivalent matrices (b) – A (iii) The rank of I4 is (c) 1 (iv) A* (d) 4 ANSWERS TO OBJECTIVE TYPE QUESTIONS A. Pick the correct answer: 1. (iv) 4. (iii) 7. (iii) 10. (iv) 2. (iii) 5. (iii) 8. (iv) 3. (i) 6. (iv) 9. (iii) 2. 3 3. adj A B. Fill in the blanks: 1. 0 4. λ , λ , λ –1 1 –1 2 –1 3 5. All zero; pure imaginary] 6. |A| 7. Singular 8. Consistent 9. Independent 10. 1, 1, 5 C. True or False: 1. (i) T (ii) F (iii) F (iv) F 2. (i) F (ii) T (iii) T (iv) F D. Match the following: 1. (i) → (b), (ii) → (a), (iii) → (d), (iv) → (c) 3. (i) → (c), (ii) → (a), (iii) → (d), (iv) → (b) 2. (i) → (d) (ii) → (a) (iii) → (b) (iv) → (c) GGG UNIT 18 Multiple Integrals 4.1 MULTIPLE INTEGRALS The process of integration for one variable can be extended to the functions of more than one variable. The generalization of definite integrals is known as multiple integral. 4.2 DOUBLE INTEGRALS Consider the region R in the x, y plane we assume that R is a closed*, bounded** region in the x , y plane, by the curve y = f1(x), y = f2(x) and the lines x = a, x = b. Let us lay down a rectangular grid on R consisting of a finite number of lines parallel to the coordinate axes. The N rectangles lying entirely within R (the shaded ones in Fig. 4.1). Let (xr, yr) be an arbitrarily selected point in the rth partition rectangle for each r = 1, 2, ..., N. Then denoting the area δxr · δyr = δSr Thus, the total sum of areas Y y = f2 (x) R dx r (x r, y r) dy r ∑ f bxr , y r g δSr N SN = y = f1 (x) r =1 Let the maximum linear dimensions of each portion of areas approach zero, and n increases indefinitely then the sum SN will approach a limit, “namely the double integral zz b g O x=a x=b Fig. 4.1 f x, y dS and the value of this limit is given by R * Boundary included i.e., part of the region. ** Can be enclosed within a sufficiently large circle. 258 X 259 MULTIPLE INTEGRALS zz b g LM a f O z MN za ffbx, yg dyPPQ dx b f2 x f x, y dS = a f1 x R Similarly if the region is bounded by y = c, y = d and by the curves x = f1(y), x = f2(y), then LM b g OP , f x y dx = b g f x , y dS b g z z zz MN b g PQ dy d f2 y c f1 y R 4.3 WORKING RULE (a) We integrate with respect to y, x is to be regarded as constant and evaluate the result between the limits y = f1(x) and y = f2(x). (b) Then we integrate the result of (a) with respect to x between the limits x = a and x = b. 4.4 DOUBLE INTEGRATION FOR POLAR CURVES Let OP and OQ are two radii vectors of the curve r = f (θ) and the coordinates of P and Q be (r, θ) and (r + δr, θ + δθ). The area of portion K1 L1 K2 L2 = 1 1 2 (r + δr)2 δθ – r δθ 2 2 1 δr2δθ 2 = rδrδθ |As δr2δθ is very small Therefore, the area of OPQ = rδrδθ + = B lim [Σrδrδθ] Q (r + dr, q + dq) δr→0 = lim [Σrδr] δθ K2 δr→0 z L a f rdrOP δθ = M N Q P (r, q) f θ Hence the area OAB z L a f rdrOP δθ = lim ∑ M N Q δθ → 0 f θ O 0 z LMNz a f OPQ θ =β fθ θ=α 0 L2 L 1 dq 0 = K1 rdr dθ where α and β are the vectorial angles of A and B. q Fig. 4.2 A 260 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 1. Evaluate z z log 8 log y 1 0 Sol. We have I = = ex + y dx dy. z LMNz e jOPQ log 8 log y 1 0 z z log 8 = = Example 2. Evaluate z z π x 0 I = = zz 0 Sol. We have b g log 8 dy y – e 1 log 8 1 (As eloge y = y) sin y dy dx . z LMNz z π x 0 0 π z z sin y dy OP dx Q x – cos y 0 dx, treating x as constant 0 = − 2 dy e y e log y − e 0 ey y−1 = − Example 3. Evaluate ey dy = elog 8 · (log 8 – 1) – 0 – (elog 8 – e) = 8 log 8 – 8 – 8 + e = 8 log 8 – 16 + e. 0 Sol. We have log 8 1 log y y e 0 ex 1 e x dx π 0 π 0 (cos x – cos 0) dx cos x dx + π z π 0 dx (As cos 0 = 1) π = − sin x 0 + x 0 = π. d 4− 2 y i y dx dy . − d 4− 2 y i 2 2 z z 2 I = = z z 2 2 0 1 2 4− 2 y 2 dx j OP PQ y dy c 4− 2 y h y dy − c 4− 2 y h 2 e 4 − 2 y j y dy 2 2 2 z 0 4 1 2 = – · 2 3 = j x 0 = – 4− 2 y 2 − 0 = LM e MN e t dt LMt OP MN PQ 1 8 ·8= . 3 3 by putting 4 – 2y2 = t, – 4y dy = dt 3 0 2 4 MULTIPLE INTEGRALS Example 4. Evaluate zz b g x+ y x2 y2 2 + a b2 2 261 2 dx dy over the area bounded by the ellipse [U.P.T.U. (C.O.), 2004] = 1 = 1 ⇒ y = ± b 1− ⇒ y = ± ∴ zz b g x+ y 2 dx dy = = b a = x = –a zz e FH IK z zFH IK Fig. 4.3 j x + y + 2xy dx dy 2 2 b a −b a a −a zz −a zz a 2 2 zz 2 2 2 FG b IJ a − x 2 2 x + y dy dx + 0 z z z −a ex + y + 2xyj dx dy a2 − x 2 2 2 H aK −a 0 a a2 − x 2 a e ba j a − x e ba j a − x 2 2 xy dy dx + x y dx dy + 2 e j −a e− b j a − x e −ab j a −x a As = X O a2 − x2 a = x2 a2 x=a x y 2 + a b2 Sol. We have Y 2 2 z a 2 e af f x dx = 2 –a 2 2 af a af f x dx when f ( x) even = 0 when f x odd F bI a −x 3 2 j 0 LM2F x y + y I OPGH JK MN GH 3 JK PQ 2 z 2 2 2 a dx 0 L b a − x + 1 b a − x OP dx = 2 Mx × e jP 3a MN a Q Lb O b a − x j P dx (Again by definite integral) = 4 M x a −x + e 3a MN a PQ a 2 2 3 2 2 3 3 2 2 −a a 2 2 2 3 3 2 3 2 2 0 = 4 z FGH π 2 0 [On putting x = a sin θ and dx = a cos θ dθ] IJ × a cos θ dθ K b ⋅ a 2 sin 2 θ. a cos θ + b 3 a 3 cos 3 θ a 3a 3 262 A TEXTBOOK OF ENGINEERING MATHEMATICS—I = 4 z π 2 0 = 4a3b = FG a b sin θ cos θ dθ + ab cos θIJ dθ 3 H K 3 2 3 2 4a3b · π 2 sin 0 θ cos 4 θ dθ 0 3 3 1 5 4 ab 3 2 2 2 2 + · 3 2+ 2+ 2 0 + 4+ 2 2 2 2 2 = 4a3b · = z z π 4ab 3 2 sin 2 θ cos 2 θ dθ + 0 3 1 1 1 1 . 4 ab 3 = 4a3b. 2 2 2 2 + . 3 2 3 ⇒ 4 z π As 2 sin m θ 0 m+1 n+1 2 2 cos θ dθ = m+n+2 2 2 n 1 3 1 1 . . . 2 2 2 2 2 3 a π 4ab 3 3π ⋅ + 16 3 16 As n = n − 1 b f n – 1 and 12 = π g πa 3 b πab 3 π + = ⋅ ab a 2 + b 2 . 4 4 4 Example 5. Evaluate zz xy ( x + y) dx dy over the area between y = x2 and y = x. Sol. The area is bounded by the curves y = f1(x) = x2, y = f2(x) = x. When f1 (x) = f2 (x), x2 = x, i.e., x (x – 1) = 0 or x = 0, x = 1 i.e., the area of integration is bounded by y = x2, y = x, x = 0, x = 1 ∴ = = = = = zz b g z LMNz b g OPQ z LMNz e j OPQ LM OP z MN PQ OP z LMMN PQ xy x + y dx dy A 1 y =x x=0 y=x 1 x 0 x2 zz D xy x + y dy x 2 y + xy 2 dy x 2 y 2 xy 3 + 0 2 3 0 4 A dx dx y=x 2 x=1 X O x dx x2 6 Fig. 4.4 7 5x x x − − 6 2 3 dx L x x − x OP L 1 − 1 − 1 O 3 . = M − N 6 14 24 Q = MN 6 14 24 PQ = 56 5 Example 6. Find 2 1 1 Y 7 8 1 0 x 2 dx dy where D is the region in the first quadrant bounded by the hyperbola xy = 16 and the lines y = x, y = 0 and x = 8. (U.P.T.U., 2002) 263 MULTIPLE INTEGRALS Sol. We have Y xy = 16 ...(i) y = x y = 0 ...(ii) ...(iii) y=x x = 8 ...(iv) From eqns. (i) and (ii), we get x = 4, y = 4 x=8 A i.e., intersection point of curve and the line y = x = (4, 4). (4, 4) xy = 16 Similarly intersection point of (i) and (iv) = (8, 2) To evaluate the given integral, we divide the area OABEO into two parts by AG as shown in the Figure 4.5. Then, zz D x 2 dx dy = = = z z x= 4 y=x x=0 y=0 x 2 dx dy + zz x=8 y= 16 x x= 4 y =0 z z z z B O z z zz G (4, 0) 4 X Fig. 4.5 z bg z bg 4 0 x 8 0 4 x y dx + 2 LM x OP 8x N4Q + 4 E x 2 dx dy 16 x 8 2 4 2 x dy = x dx dy x dx + 0 4 0 0 4 3 8 x dx + 16 x dx = 0 4 (8, 2) x 2 16 y 0x dx 2 8 0 4 = 64 + 8(64 – 16) = 64 + 384 = 448. Example 7. Evaluate A xy dx dy over the positive quadrant of the circle x2 + y2 = a2. Sol. Here the region of integration is positive quadrant of circle x2 + y2 = a2, where x varies ea − x j . 2 from 0 to a and y varies from 0 to Here, zz A xy dx dy = zz = z 0 z Y d a −x i xy dx dy 2 a 0 0 a 2 LM y OP N2Q 2 a2 − x 2 2 x dx dx 0 2 y= a –x Ly O = M P x dx N2Q 1 = z x e a − x j dx 2 1 La x x O − P = 18 a . = M 2N 2 4Q Fig. 4.6 Example 8. Find zz e x + y j dx dy where D is bounded by y = x and y = 4x. Sol. zz ex + y jdx dy = ex + y j dy dx LMx y + y OP dx = MN 3 PQ a a2 −x2 2 x=0 0 0 a 2 2 0 2 2 4 O a y=0 4 0 2 D 2 D 2 2 2 2 zz z 4 2 x 2 2 0 x 4 0 2 3 2 x x x=a X 264 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Y B (4, 4) y =2 x y=x X O (0, 0) z FGH 4 = Example 9. Find in the first quadrant. Sol. zz D 0 IJ K 8 4 2x 5 2 + x 3 2 − x 3 dx 3 3 768 . 35 = zz Fig. 4.7 x 3 y dx dy where D is the region enclosed by the ellipse D x 3 y dx dy = zz b 2 2 a −x a x 3 y dy dx 0 y=0 a z LM x y OP MN 2 PQ a = 3 0 = x2 y2 + =1 a2 b2 b a 2 b2 2a 2 ze b2 2 4 a 0 Y a2 −x2 (0, b) 0 a 2x 3 − x 5 j dx (a, 0) LM a x − x OP = b a . = 24 6 PQ 2 a MN 4 6 a 2 4 2 Example 10. Evaluate I = Sol. I = = = z z z z z LMN OPQ z 0 2π a 0 a sin θ 2π a 0 r = a sin θ 2π r2 2 0 a2 2 2π 0 2 I = a 4 = πa . 2 2 Fig. 4.8 rdrdθ . rdrdθ a dθ = a sin θ cos2 θ dθ = LMθ + sin 2θ OP N 2 Q 2π 0 ze j 1 2π 2 a − a 2 sin 2 θ dθ 2 0 z FGH IJ K a 2 2 π 1 + cos 2θ dθ 2 0 2 X 265 MULTIPLE INTEGRALS Example 11. Evaluate zz r 2 sin θ dθ dr over the area of cordioid r = a(1 + cos θ) above the A initial line. Sol. The region of integration A can be covered by radial strips whose ends are at r = 0, r = a(1 + cos θ). The strips lie between θ = 0 and θ = π Thus zz A r 2 sin θ dθ dr = = = z z z LMNz a f OPQ LM OP a f z NQ a π a 1+ cos θ 0 0 π 0 π 0 a 1+ cosθ sin θ 0 z dθ p q=– 2 r dr dθ sin θ zb z z 16 a 3 3 θ θ π cos7 sin dθ 0 2 2 16 3 = a 3 π 7 2 sin φ cos φ ⋅ 2dφ 0 1 + cos θ 8 Put π 2 3 0 a 2 r dr a 1−cosθ a r 2 drdθ . a = π 2 dθ 0 r3 3 = π 2 dθ 0 3 a 3 a 1 − cosθ − 3 3 = a3 3 π 2 1 − 1 − cosθ 3 0 = a3 3 π 2 1 − 1 − 3 cos θ + 3 cos 2 θ − cos 3 θ 0 = a3 3 π 2 0 a 1−cosθ 3 dθ 3 cos θ − 3 cos 2 θ + cos 3 θ dθ LM 1π 2 O P + 3 sin θ − 3 2 2 3 ⋅1P MN Q a L 3 − 3π + 2 O = M 4 3 PQ 3 N a3 = 3 π 2 0 3 = a3 [44 – 9π]. 36 A Fig. 4.9 LM − cos φ OP = 4 a . N 8 Q 3 a 1− cos θ q=0 q O z za f z a f z LMN OPQ a f F a f IJ G z H K z a f z e j ze π 2 0 P (r, q) X g3 sin θ dθ = π q=p dθ a3 3 0 Q dq a 1+cosθ = Example 12. Evaluate Sol. Y 2 r3 3 0 16 3 =2× a 3 π 2 0 f r 2 sin θ dθ dr j dθ θ = φ 2 266 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 13. Evaluate Sol. We have z z z z z LMNz OPQ z LMN OPQ z π aθ 3 0 0 π I = aθ 3 r dθ dr 0 0 = π aθ 3 0 0 r dr dθ aθ r4 dθ 4 0 π = Example 14. Evaluate r dθ dr 0 π 4 4 a θ dθ 0 = 1 4 = a4 4 LM θ OP = a π . N 5 Q 20 5 z zb z z π 4 5 0 π a 1+ cos θ 0 0 g r 3 sin θ cos θ d θ dr . z LM a f r drOP dθ N Q a f L r O sin θ cos θ M P dθ MN 4 PQ π 4 a 1 + cos θg sin θ cos θ dθ = b z 0 4 Sol. We have π I = 0 sin θ cos θ a 1+ cosθ 3 0 4 a 1+cosθ π 0 0 4 1 + cos θ = t and – sin θ dθ = dt Put = a4 4 = a4 4 z a fa f ze j 0 2 t 4 t − 1 − dt 2 0 t 5 − t 4 dt = 16 4 a. 15 EXERCISE 4.1 z z z z log y log 8 1. Evaluate 1 1 2. Evaluate 3. π 2 0 0 1+ x 2 4. 0 z z b g z z cos x + y a 0 a2 − y2 0 Ans. 8 log 8 – 16 + e 0 0 π 2 e x + y dx dy. dy dx 1+ x + y 2 2 . dy dx. a 2 − x 2 − y 2 dx dy. LMAns. π loge 2 + 1jOP N 4 Q Ans. – 2 LMAns. πa OP 6 PQ MN 3 267 MULTIPLE INTEGRALS z z 5. LMAns. 1 OP N 2Q x2 x y 1 e dx dy . 0 0 6. Evaluate 7. Evaluate zz zz e xy − y 2 dy dx where S is the triangle with vertices (0, 0), (10, 1) and (1, 1). S Ans. 6 j x 2 + y 2 dx dy, where S is the area enclosed by the curves, y = 4x, x + y = 3, S LMAns. 463 OP 48 Q N LMAns. π OP N 24 Q y = 0 and y = 2. 8. Evaluate 9. Evaluate 10. Evaluate zz zz e x 2 y 2 dx dy over the region x2 + y2 ≤ 1. x2 + y2 zz j dx dy over the region bounded by x = 0, y = 0, x + y = 1. LMAns. 1 OP N 6Q xy dx dy, where the region of integration is the positive quadrant of the 1 − y2 A LMAns. 1 OP N 6Q circle x2 + y2 = 1. 11. Evaluate zz 12. Evaluate z z e 13. 14. zz e zz b D 1 1+ x −1 2 −x LMAns. 1 OP N 24 Q LMAns. 63 OP N 32 Q j x 2 + y dy dx. j 4xy − y 2 dx dy, where D is the rectangle bounded by x = 1, x = 2, y = 0, y = 3. Ans. 18 g 1 + x + y dx dy, A is the region bounded by the lines y = – x, x = A 15. xy dx dy over the region in the positive quadrant for which x + y ≤ 1. Hint: Limits x : – y to y ; y : 0 to 2 . zz dx dy. 1 0 1+ x2 0 16. Evaluate e1 + x + y j 2 zz A 2 −1 LMAns. 44 2 + 13 OP N 15 3 Q LMAns. π log e1+ 2 j OP N 4 Q y dx dy, where A is bounded by the parabolas y2 = 4x and x2 = 4y. LM x − y OP LM x − y OP dx 17. Show that MN bx + yg dyPQ dx ≠ z MNz bx + yg PQ dy. zz 1 0 y , y = 2, y = 0. 1 0 1 2 0 1 0 2 LMAns. 48 OP 5Q N 268 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 18. Evaluate 19. zz e 1 y 0 2 y LMAns. 41 OP N 240 Q j 1 + xy 2 dx dy. j dx dy, where A is bounded by x + y = a and x + y = b , where b > a. LMAns. π eb 4 − a4 jOP N 2 Q LMAns. logFG 25IJ OP z bxdy+ dxyg ⋅ H 24 K Q N x2 + y2 A 20. zze z 4 2 2 2 2 2 2 2 1 3 Using polar coordinates evaluate the following double integral. zz zz zz zz π 21. π cos θ 0 0 π 23. a 0 0 24. 4 sin θ 3 2 sin θ 0 22. r dr dθ. ρ sin θ dρ dθ. r 3 sin θ cos θ dr dθ. LMAns. 45π OP N 2Q LMAns. 1 OP N 3Q Ans. 0 r 3 dr dθ, over the area bounded between the circles r = 2 cos θ and r = 4 cos θ. 25. Show that r = 2a cosθ. 26. 2 z z π 2 0 zz R LMAns. 45 π OP N 2 Q r 2 sin θ dr dθ = 2a 3 , where r is the region bounded by the semicircle 3 LMAns. a OP MN an + 1f PQ n+ 1 a n 0 n r sin θ cos θ d θ dr , for n + 1 > 0. 2 4.5 CHANGE OF THE ORDER OF INTEGRATION If the limits of x and y are constant then zz b g F x, y dx dy can be integrated in either order, but if the limits of y are functions of x, then the new limits of x in functions of y are to be determined. The best method is by geometrical consideration. Thus in several problems, the evaluation of double integral becomes easier with the change of order of integration, which of course, changes the limits of integration also. Remark. If y depends on x and we want that x depend on y i.e., x = f (y) then construct a strip parallel to x-axis. zz ∞ x Example 1. Evaluate the integral 0 0 F x I dx dy by changing the order of integration. GH y JK x exp − 2 (U.P.T.U., 2005) 269 MULTIPLE INTEGRALS Sol. Here y = 0 and y = x x = 0 and x = ∞ Here x start from x = y and goes to x → ∞ and y varies from y = 0 to y → ∞ ∴ 0 2 x exp − 0 z z z z FGGH ∞ = x F xI GH y JK dx dy x ∞ y= zz ∞ Y x2 y − xe x®¥ ⋅ dy dx y = 0 x= y ∞ = x2 ∞ y 2x − − − e y 2 y y 0 I JJ dy dx K O z z ∞ x2 − y 0 y ∞ − z X Fig. 4.10 LM y OP = MN− 2 e PQ dy LM y OP y e dy = dy = 0+ e 2 MN 2 PQ 1 Ly O = M e− e j − ee jP 2 N2 Q F 1I 1 = (0 – 0) + GH 0 + JK = . 2 2 ∞ x®¥ − Put e x2 y x2 2x − y =t ⇒ – e. dx = dt y ∞ y2 y 0 –y 0 −y −y ∞ (Integration by parts) 0 Example 2. Change the order of integration in Sol. Here the limits are x2 = ay and y = 2a – x i.e., x2 = ay and x + y = 2a also x = 0 and x = a Now zz b g zz b g z z b g zz zz zz 2 a− x f x, y x2 a a 0 ay a = 0 + a 0 2 a −x f x2 a bx, yg dx dy. (U.P.T.U., 2007) B (0, 2a) x + y = 2a 2 x = ay dx dy L A (a, a) f x, y dy dx 0 2a 2a − y a 0 Example 3. Evaluate zz O f x , y dy dx . 2 ex 0 1 x=a Fig. 4.11 dx dy changing the order of integration. (U.P.T.U., 2003) Sol. The given limits are x = 0, x = 2, y = 1 and y = ex. Here 2 ex 0 1 dx dy = e2 2 1 x= log y dy dx As x varies from x = log y to x = 2 270 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Y C x y = e Þ x = lo g y (0, 1) A = zz Example 4. y=1 B O = 2 (2, e ) X x=2 zb e 2 1 Fig. 4.12 g 2 − log y dy b2 y − y log y + yg = b3y − y log yg e2 e2 1 1 = (3e2 – 2e2) – 3 = e2 – 3. 1 3x +2 −2 x2 + 4 x dy dx . Sol. Here the curves y = x2 + 4x, and the straight lines y = 3x + 2, x = – 2 and x = 1 as shown 2 shaded in Figure 4.13. Here A(– 2, – 4) B (0, 0), E (1, 5), F (0, 2), G − , 0 . 3 Considering horizontal strip, x varies from the FG H 4 + y − 2 the lower curve upper curve x = x = FG y − 2IJ and then y varies from – 4 to 5. H 3K Y E (1, 5) Changing the order of integration to first x and later to y, the double integral becomes = = z z z z z LMN 1 3x + 2 –2 y = x 2 + 4x 5 −4 5 −4 y = 3x + 2 2 y = x + 4x dy dx F (0, 2) y+ 4 − 2 y − 2 dx dy x= 3 FG y − 2 IJ OP dy H 3 KQ L2 9 y 4 O − yP = . = M b y + 4g − 2 6 3 PQ MN 3 = IJ K x=1 x = –2 y+ 4 −2− 3 2 2 G B (0,0) A 5 −4 Fig. 4.13 X 271 MULTIPLE INTEGRALS Example 5. Change the order of integration in I = same. Sol. The limits are y = x2 and y = 2 – x x = 0 and x = 1 The point of intersection of the parabola y = x2 and the line y = 2 – x is B (1, 1). We have taken a strip parallel to x axis in the area OBC and second strip in the area ABC. z z 1 2− x 0 x2 xy dx dy and hence evaluate the (U.P.T.U., 2004) Y y 2 zz 2− x 0 x xydxdy = 2 z z z z z LMMN OPPQ z LMMN OPPQ z zb g 1 0 y dy y 0 xdx + x2 = y dy 0 2 y 1 0 2 1 2− y y dy –y x=0 and the limits of x in the area ABC are 0 and 2 – y. 1 x= x= y The limits of x in the area OBC are 0 and ∴ A (0, 2) 0 x2 + ydy 1 2 2 B (1, 1) C x dx 2−y x=0 X O x=1 X Fig. 4.14 0 LM y OP 1 y 2− y MN 3 PQ + 2 z e 4y − 4y + y j dy 1 1 L 32 4 1O 1 1 L y O 4 8− + 4− 2+ − P = + = + 2 y − y + M P M 6 2 M 3 4Q N 3 4 PQ 6 2 N 3 1 1 L 96 − 128 + 48 − 24 + 16 − 3 O = + PQ 12 6 2 MN 1 = 2 1 y dy + 0 2 1 2 2 1 2 4 3 3 1 1 dy = 2 2 0 2 2 3 1 2 1 1 5 + 6 24 9 3 = = . 24 8 = Example 6. Change the order of integration in Sol. Here the limits are y = a −x 2 2 z z b g a x +2 a 0 a −x φ x, y dx dy . 2 2 and y = x + 2a and x = 0, x = a. 2 2 2 x 2 − a 2 i.e., x + y = a is a circle and the straight line y = x + 2a and the limits of x are given by the straight lines x = 0 and x = a. Now, y = The integral extends to all points in the space bounded by the axis of y the circle with centre O, and the straight line x = a. We draw BN, KP perpendiculars to AM. Now the order of integration is to be changed. For this, we consider parallel strips. The integral is broken into three parts. Ist part is BAN, bounded by lines x = 0, x = a and the circle. 272 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 2nd part is KPNB, bounded by lines y = a, y = 2a and x = 0, x = a. (a, 3a) M x = y – 2a 3rd part is triangle KPM bounded by the lines y = 2a, y = x + 2a, x = a. z dz ib g z z b g zz b g x +2 a a Hence P φ x, y dx dy 0 = (0, 2a) K a2 − x 2 φ x, y dy dx + 0 =+ zzb g a a a −y 2 2 3a a 2a y−2a 2a a a 0 φ x, y dy dx φ x, y dy dx . B A O (2a, 0) N (a, a) Example 7. Change the order of integration and hence evaluate zz a 0 Sol. y 2 dx dy a ax x=0 y −a x 4 2 2 . x=a Fig. 4.15 The limits are y2 = ax, y = a and x = 0, x = a, y = 0 to y = a 2 y . a By changing the order of integration. Hence, the given integral, and x varies from x = 0 to x = z z a a x= 0 y = ax y 2 dx dy y −a x 4 2 2 z z zz F I y 2 dy dx a y2 a y =0 x =0 y −a x 1 a = a 0 y2 a 0 y 2 dy dx = z z 1 y = a 0 π = 2a z B A y x=— a x=0 2 F I OP dy GH JK PQ y2 a O 0 [sin–1 (1) – sin–1 (0)] a 2 y dy 0 y=a 2 2 2 1 2 −1 ax y sin y2 a0 a 2 Y 2 2 GH ya JK − x LM MN a = 4 F I GH JK 3 π y = 2a 3 dy y=0 X Fig. 4.16 a 0 π πa . (a3) = 6a 6 2 = Example 8. By changing the order of integration of z ∞ sin px 0 x dx = π · 2 zz ∞ ∞ e − xy sin px dx dy 0 0 show that (U.P.T.U., 2004, 2008) 273 MULTIPLE INTEGRALS Sol. We have zz ∞ 0 ∞ − xy e sin px dx dy 0 z RSTz UVW z LMMN OPPQ z OP z LMNz Q LM OP b g z MN PQ LM F I OP z MN GH JK PQ ∞ = 0 ∞ sin px 0 e − xy dy dx e − xy sin px 0 −x ∞ = Again zz ∞ sin px = ∞ ∞ − xy 0 0 e sin px dx dy = ∞ ∞ 0 0 = dx 0 ...(i) e − xy sin px dx dy 0 − e − xy p cos px + y sin px p2 + y 2 ∞ p 0 p +y ∞ = ∞ dx x 0 As the limits are constants. 2 2 dy = tan −1 y p ∞ = 0 ∞ dy 0 π 2 ...(ii) Hence from (i) and (ii) ∞ π sin px . Hence proved. dx = 0 2 x Example 9. Change the order of integration in the following integral and evaluate: z zz 4a 2 ax 0 x2 4a dydx ⋅ x2 Sol. The limits are y = and y = 2 ax and 4a x = 0, x = 4a. From the parabola y2 = 4ax i.e., x = parabola x2 = 4ay i.e., x = 2 to 2 ay . ∴ zz 4a 0 2 ax dy dx x2 4a Y A y2 and from 4a 2 y y2 ay i.e., x varies from 4a = zz z FGH 4a 2 ay 0 y2 4a 4a 0 2 dy dx 2 ay − 3 2 y2 4a 4a 3 0 3 2 3 (4a, 4a) x = 4ay O I dy JK LM O y y P − = M2 a . MMN 23 12a PPP Q 4 64a = a . a 4 af − 3 12a = =4 ax Fig. 4.17 X 274 A TEXTBOOK OF ENGINEERING MATHEMATICS—I = 32a 2 16 a 2 – 3 3 = 16 a 2 . 3 zz a f a y Example 10. Change the order of integration evaluate 0 y 2 y a−x a ax − y 2 dx dy . y2 i.e., y2 = ax, x = y and y = 0, y = a a intersection point of parabola and line y = x is A(a, a). Y Sol. Here the limits are x = Here x varies from 0 to a and y varies from x to ∴ zz a f zz a f a y y 2 a−x a ax 0 y a = = = = 0 x z a 0 ax − y 2 a−x y= Q dx dy P O y ax − y 2 y= ax A(a, a ) x X (0,0) dy dx ax − y 2 = t ax 12 −2 ydy = dt − ax − y 2 dx 12 x y dy = − ax − y 2 2 ax − y L a f NM e 1 a−x 2 ax j OQP z LMN e j OPQ za f z a f z e Fig. 4.18 j a 12 1 dx 0 + ax − x 2 0 a−x a 0 ax − x 2 dx = a−x a x 0 a−x dx = a−x Let x = a sin2 θ i.e., dx = 2a sin θ cos θ dθ = z x 0 a−x a . cos θ π 2 = 2a 0 LM N a a . sin θ . π 2 0 z z = a θ− 2a sin θ cos θ dθ sin 2 θdθ = 2a OP Q π 2 sin 2θ 2 0 dx = a z FGH π 2 0 1 − cos 2θ 2 LM π − 0OP = aπ . N2 Q 2 IJ = az π 2 b1 − cos 2θg dθ K 0 4.6 CHANGE OF VARIABLES IN A MULTIPLE INTEGRAL Sometimes it becomes easier to evaluate definite integrals by changing one system of variables to another system of variables, such as cartesian coordinates system to polar coordinates system. 275 MULTIPLE INTEGRALS Let us consider the transformation of = zz b g S F x , y dx dy ...(i) when the variables are changed from x and y to u and v by the relations x = φ (u, v), y = ψ (u, v) ...(ii) Let these relations transform the function F (x, y) to F (u, v) to express dx dy in terms of new variables u, v, we proceed as follows: First solve equations (ii) for u, v, to get u = F1 (x, y) and v = F2 (x, y) ..(iii) Then u = constant and v = constant form two systems Y v + dv of curves in the xy-plane. Divide the region S into elementary areas by the curve u = constant, u + δu Q¢ v = constant, v = constant and v + δv = constant. Let P be the intersection of u = constant and P¢ Q v = constant, and Q is the intersection of u + δu = constant and v = constant. u + du Thus, if P is the point (x, y), so that x = φ (u, v), P y = ψ (u, v) Then Q is the point u [( φ(u + δu, v), ψ (u + δu, v)] X Now using Taylor ’s theorem to the first order of O approximation, we obtain Fig. 4.19 δφ δu δu δψ ψ (u + δu, v) = ψ (u, v) + δu δu φ (u + δu, v) = φ (u, v) + and a f y = ψ au, vf FG x + δx δu, y + δy δuIJ H δu δu K δy I F δx Similarly, P′ is the point G x + δv , y + δvJ H δv δv K δy δy I F δx δx and Q′ is the point GH x + δu δu + δv δv , y + δu δu + δv δvJK As x = φ u, v Therefore, the coordinates of Q are Therefore, to the first order of approximation PQQ′ P′ will be a parallelogram and its area would be double that of the triangle PQP′. Hence the elementary area PQP′ Q′ = [2 ∆ PQP′] = 2× = 1 2 x δx δu δu δx δv δv 1 x y δy δx δu 1 δu y + x+ δu δu δy δx x + δv y + δv 1 δv δv 1 y δy δu 0 = δu δy δv 0 δv δx δu δx δv δy δu . δuδv δy δv 276 A TEXTBOOK OF ENGINEERING MATHEMATICS—I b g . δu δv = J.δu δv a f δ x, y = ..(iv) δ u, v Thus, if the whole region S be divided into elementary areas by the system of curves F1 = constant and F2 = constant then, we have lim Σ F (x, y) δS = lim ΣF (u, v) J . δu δv zz b g i.e., S F x , y dx dy = zz a f S F u, v J du dv δx δy b g = δu δu J = δx δy δau, vf δ x, y δv Hence, δy δu δy δv δx δu δx δv dx dy = ...(v) δv or δx δu δy δu δx δv δy δv du dv. Example 11. Transform to polar coordinates and integrates zz FGH 11 −+ xx −+ yy IJK dx dy 2 2 2 2 the integral being extended over all positive values of x and y subject to x2 + y2 ≤ 1. Sol. Put x = r cos θ, y = r sin θ δx δx δr δθ b g = δr δθ = cos θ r sin θ = r δy δy sin θ r cos θ δ a r , θf δ x, y then Therefore, dx dy = rdθ dr and limits of r are from 0 to 1 and those of θ from 0 to Hence zz F 1 − x − y I dx dy GH 1 + x + y JK F1 − r I = z z G H 1 + r JK r dθ dr F 1 − r I r dr π = 2z G H 1 + r JK I = 2 2 2 2 π 1 2 0 0 Now, 2 2 1 2 0 2 suppose r2 = cos φ, 2rdr = – sin φ dφ = π 2 z 0 1 π 2 2 FG 1 − cos φ IJ (– sin φ) dφ H 1 + cos φ K π , 2 277 MULTIPLE INTEGRALS g dφ = π π π π −1 . φ − sin φ 02 = 4 2 4 x+y R π 2 1 − cos φ 0 π 4 zz b g Example 12. Evaluate zb = 2 As FG H IJ K z af z af a 0 F x dx = – 0 a F x dx dx dy, where R is region bounded by the parallelogram x + y = 0, x + y = 2, 3x – 2y = 0, 3x – 2y = 3. (U.P.T.U., 2006) Sol. By changing the variables x, y to the new variables u, v, by the substitution x + y = u, 3x – 2y = v, the given parallelogram R reduces to a rectangle R* as shown in the Figure 4.20. a f = b g δ v, v δ x, y δu δy δv δy = 1 1 3 −2 =–5 v v=3 3x – 2y = 3x 3 – 2y = 0 Y δu δx δv δx R O x x + y= R* X + y = u O 2 0 u=2 Fig. 4.20 So required Jacobian J = b g =– 1 δau, vf 5 δ x, y Since u = x + y = 0 and u = x + y = 2, u varies from 0 to 2, while v varies from 0 to 3 since 3x – 2y = v = 0, 3x – 2y = v = 3. Thus, the given integral in terms of the new variables u, v is zz b g R x+ y 2 dx dy = = zz u2 R* 1 5 1 –5 zz z LMN OPQ du dv 3 2 2 0 0 u du dv 2 1 3 u3 8 8 3 dv = ⋅ v0 = . = 5 0 3 0 5 15 z Example 13. zz c h ∞ ∞ 0 0 e − x2 + y 2 dx dy by changing to polar coordinates hence, show that π . 2 Sol. Let x = r cos θ, y = r sin θ ⇒ x2 + y2 = r2 Here r varies from 0 to ∞ y and θ = tan–1 x ∞ − x2 0 e dx = (U.P.T.U., 2002) 278 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Now, θ = tan–1 0 = 0, θ = tan–1 ∞ = ⇒ Hence, Let π 2 θ varies from 0 to zz ∞ ∞ 0 0 e c − x2 +y2 = – 1 2 = – 1 2 I = I = z z ∞ 0 ∞ 0 h dx dy = zz z e −x 2 e−y 2 zz π ∞ − r2 2 e r dr dθ 0 0 a−2rfe ze a0 − 1f dθ = 12 z 1 dθ = π4 . π ∞ 2 0 0 π 2 0 π 2 −r 2 dr dθ = – 1 2 π 2 0 −r 2 π 2 0 dx ...(i) dy ...(ii) ∞ dθ 0 Y r ®¥ [Property of definite integrals] Multiplying (i) and (ii), we get ∞ ∞ −ex 2 + y 2 j π e I2 = dx dy = . 0 0 4 [As obtained above] zz I = π = 4 Example 14. Evaluate O π . Proved. 2 zz 1 x 0 0 X q=0 r=0 Fig. 4.21 x 2 + y 2 dx dy, the transformation is x = u, y = uv. Sol. Region of integration R is the triangle bounded by y = 0, x = 1 and y = x Put x = u, y = uv δx δx δu δv Y B (1, 1) b g = δu δv J = Jacobian = δau, vf δy δy δ x, y 1 = v y= 0 x R u = u In the given region R, x: varies from 0 to 1 while y varies from 0 to x. Since u = x, so u varies from 0 to 1. Similarly, since 0 ≤ y = uv ≤ x = u, so v varies from 0 to 1. Thus = = zz zz 1 x 0 0 1 1 0 0 O (0, 0) A (1, 0) Fig. 4.22 x 2 + y 2 dx dy u 1 + v 2 udu dv = 1 3 z 1 0 x=1 1 + v 2 dv X 279 MULTIPLE INTEGRALS L OP 1 M v e1 + v j 1 + sin h vP 2 2 3 MM PQ N 1 L 2 1 = M + 2 sin h 1OPPQ. 3 MN 2 bx + yg dx dy, where R is the parallelogram in the xy-plane with 1 2 −1 0 −1 Example 15. Evaluate zz 2 R vertices (1, 0), (3, 1), (2, 2), (0, 1) using the transformation u = x + y and v = x – 2y. (U.P.T.U., 2003) Sol. Since u = x + y and v = x – 2y v ∴ u = 1 + 0 = 1, v = 1 – 0 = 1 B (4, 1) A (1, 1) v=1 Similarly, u = 4 v = 1, u = 4, v = – 2 and u = 1, v = – 2 and 2 3 1 3 J = 1 3 −1 3 O u 1 = − 3 u=1 u=4 1 J du dv = du dv 3 ∴ dx dy = Thus, zz b g x+ y 2 dx dy = R zz z LMN OPQ = D (1, –2) 1 u . du dv –2 1 3 1 4 2 4 u3 dv = –2 3 3 1 1 1 V = –2 C (4, –2) Fig. 4.23 z 1 –2 7 dv = 21. Example 16. Evaluate the following by changing into polar coordinates zz a 0 0 a 2 − y2 y 2 x 2 + y 2 dy dx . Sol. Here y = 0, y = a and x = 0, x = (U.P.T.U., 2007) a2 − y2 Let x = r cos θ, y = r sin θ ∴ x2 + y2 = r2 Also, when x = 0, r cos θ = 0 ⇒ r = 0, or θ = π 2 when y = 0, r sin θ = 0 ⇒ r = 0 or θ = 0 when y = a, r sin θ = a ⇒ a sin θ = a ⇒ θ = Hence, we have zz a 0 0 a2 − y2 y 2 . x 2 + y 2 dy dx = z z π 2 a θ=0 r=0 r 2 sin 2 θ r. r dr d θ = or r = a z π 2 0 π 2 sin 2 θ dθ. z a 4 0 r . dr 280 A TEXTBOOK OF ENGINEERING MATHEMATICS—I z π2 = a5 5 = πa . 20 0 FG 1 − cos 2θ IJ dθ = a LMθ − sin 2θ OP H 2 K 10 N 2 Q 5 π2 0 5 Example 17. Transform z z π 2 π 2 0 0 sin φ dφ dθ by the transformation x = sin φ cos θ, sin θ y = sin φ. sin θ and show that its value is π. Sol. We have x = sin φ cos θ, y = sin φ sin θ ∴ B (0,1) x2 + y2 = sin2 φ and θ = tan–1 θ = 0, tan–1 when y x 2 O A (1,0) y =0 ⇒y=0 x when y π π , tan–1 = ⇒x=0 x 2 2 φ = 0, x2 + y2 = 0 ⇒ x = 0 or y = 0 when φ = π 2 , x + y2 = 1 2 J = ∂x ∂θ ∂y ∂θ Now ∴ Thus x +y =1 θ = when dφ dθ = zz π2 0 π 2 0 sin φ dφ dθ = sin θ = zz zz 1 1− y 2 1 cos φ sin φ sin φ z z FH zz e 1− y 2 1− y 2 1 dy dx ⋅ dy dx IK y dy dx j −1 OP x 1 − y PQ ⋅ 1 π 2y1 2 = π. 0 2 0 dy y = π 2 2 z Hence proved. dy dx e1 − y j − x 0 0 1− y 2 2 zz 1− y 2 1 = 1 − x2 + y2 ⋅ y 0 0 0 1 1 − sin 2 φ 0 0 LM = z MNsin = sin φ 1 ⋅ ⋅ dy dx sin θ sin φ cos φ 1− y 2 0 0 1 = ∂x − sin φ.cos θ cos φ cos θ ∂φ = = – sin φ cos φ ∂y sin φ cos θ cos φ sin θ ∂φ 0 0 1 = Fig. 4.24 1 1 dxdy = . dxdy J sin φ cos φ 1 1 −1 2 0 y 2 x=0 dy 2 y 281 MULTIPLE INTEGRALS EXERCISE 4.2 Change the order of integration and then evaluate the following double integrals: 1. 2. 3. 4. 5. 6. 7. 8. zz z zb g y 2 dx dy. LMAns. 160 OP N 21 Q x + y dx dy. Ans. 5 2 4 2y 0 y3 2 4 1 3 zz zz 2 –1 – x zz zz e 0 − 4− 2 y 2 a x a zz zz LMA ns . MN b g 4− 2 y 2 zz b g zz 1 –2 2 -y f x , y dx dy + –y 2 1 y dx dy . a a2 −x2 − 2−y b g f x , y dx dy OP PQ LMAns. 8 OP N 3Q 3 2 2 ay − y 2 0 2−y LMAns. a + a OP MN 28 20 PQ LMAns. πa OP 2 PQ MN LMAns. πa OP a PQ MN j x 2 + y 2 dy dx. 0 x 0 x3 f x, y dy dx. 2 2a 1 f x, y dx dy. 2– x 2 0 x a LMAns. z z f bx, yg dy dxOP N Q b g 1 y1 3 0 y2 dx dy . 3 a − x − y dy dx. 2 0 0 2 2 Change the following integrals in to polar coordinates and show that 9. 10. 11. zz z z e je j zz d i e j 1 a a dy dx = 0 y x 2 + y2 1 2x − x 2 x 2 + y 2 dx dy = 0 x a πa . 4 a2 − x 2 y 0 0 x 2 + y2 3π –1. 8 dx dy = πa 5 . 20 Evaluate the following by changing the order of integration: 12. zz x dy dx . 13. zz db − y i xy dx dy. a bx a 0 0 a b 0 0 b 2 2 LMAns. 1 a bOP N 3 Q LMAns. a b OP 8 PQ MN 2 2 2 282 14. A TEXTBOOK OF ENGINEERING MATHEMATICS—I z z 0 16. 17. 18. 4 xy dx dy. a − a2 − y 2 zz 4a 15. LMAns. = 2 a OP 3 N Q a+ a2 − y 2 a 2 dy dx . 2 x 4a 0 zz e 1 1 j x 2 + y 2 dx dy + 0 0 zz zz LMAns. = 16 a OP 3 PQ MN 2 ax 1 1 e 0 e x log y ∞ ∞ e−y 0 x y zz e 2 2− y 1 0 LMAns. = 5 OP 3 Q N j x 2 + y 2 dx dy . ⋅ dx dy Ans. = e − 1 dx dy. Ans. = 1 Change the order of integration in the following integrals: 19. zz b g 20. z ez j a lx 0 mx LM MNAns. V x, y dx dy. a2 ax f 2a 0 21. 22. 23. z z a e a −x j f bx, yg dx dy. 2 2 x tan a zz e j zz d i e 2 x2 y l LM MMAns. z z N z z f x , y dy dx + LMAns. MN z a sin a y cot a 0 0 OP PQ OP ze Vdyj dx + z z Vdy dxPP Q b g 2− x2 0 x x dx dy x2 + y2 j zz b g la a y V x, y dy dx. ma l 2a 2a 2a y2 2a a a2 − y2 z z a a sin a 0 x 2 + y 2 dx dy. 1 x 1 0 b g V x, y dy dx + a a− a2 − y2 a Vdy dx + 0 y2 0 2a a+ Vdx dy. 2 ax − x 2 a cos a zz ma y m . d a − x i f bx , y gdy dx OP 2 2 PQ LMAns. 9 61 OP N 105 Q LMAns. 1 − 2 OP 2 PQ MN 24. Using the transformation x + y = u, y = uv show that zz 1 1− x 0 0 25. zz b g e y / x + y dy dx = 1 (e – 1). 2 2π , integration being taken over the area of the triangle 105 bounded by the lines x = 0, y = 0, x + y = 1. b g 1 [ xy 1 − x − y ] 2 dx dy = 283 MULTIPLE INTEGRALS zz b g 26. D y − x dx dy, D: region in xy-plane bounded by the straight lines y = x + 4, 1 7 1 x+ ,y=– x + 5. 3 3 3 y = x – 3, y = – zz b g 27. R x+y De dx dy. R: parallelogram in the xy-plane with vertices (1, 0), (3, 1), (2, 2), (0, 1). Ans. 21 zz b 28. 2 Ans. 8 gb g x − y / x + y dx dy, D: triangle bounded by y = 0, x = 1 and y = x. Use x = u – uv, LM ee − 1j OP MNAns. 4 e PQ 2 y = uv to transfrom the double integrals. 29. Find zz π a 0 0 r 3 sin θ cos θ dr dθ by transforming it into cartesian coordinate. Ans. = 0 (U.P.T.U., 2007) 4.7 BETA AND GAMMA FUNCTIONS The first and second Eulerian Integrals which are also called “Beta and Gamma functions” respectively are defined as follows: z z β (m, n) = 0 n = and a f 1 x m −1 1 − x ∞ − x n −1 0 e x n −1 dx dx [U.P.T.U. (C.O.), 2003] β (m, n) is read as “Beta m, n” and n is read as “Gamma n”. Here the quantities m and n are positive numbers which may or may not be integrals. 4.7.1 Properties of Beta and Gamma Functions (a) The function β (m, n) is symmetrical w.r.t. m, n i.e., β (m, n) = β (n, m) we have putting β (m, n) = z 1 0 n −1 dx 1 – x = t ⇒ dx = – dt = − = ⇒ a f x m −1 1 − x za f 0 1 1−t m za f 1 0 1– x m−1 β (m, n) = β (n, m) . −1 ⋅ tn−1 dt = ⋅ x n−1 · dx = za f 1 0 z −1 1− t m ⋅ tn−1 dt a f 1 n−1 m −1 x 1–x 0 z z F By f atf dt = f axf dxI H K b a b a 284 A TEXTBOOK OF ENGINEERING MATHEMATICS—I (b) Evaluation of Beta Function β (m, n) m−1 n −1 am + n − 1f β (m, n) = We have β (m, n) = z 1 0 m n = (U.P.T.U., 2008) m+ n x m−1 (1 – x)n–1. dx Lx O an − 1f x = M a1 − xf P + (1 – x) · dx m Nm Q (Integrate by part) n − 1f a x (1 – x) . dx β (m, n) = m z 1 n−1 0 or z 1 1 m n –2 0 m n–2 0 m Again integrate by parts above, we get z an − 1f . an − 2f 1 x m+1 (1 – x)n – 3. dx 0 m m+1 Continuing the above process of integrating by parts β (m, n) = an − 1fan − 2f...2 ⋅ 1 m am + 1f... am + n − 2f β (m, n) = a β (m, n) = or f a n −1 z 1 0 x m + n− 2 . dx fa f m m + 1 ... m + n − 2 m + n − 1 n is a positive integer In case n is alone a positive integer. Similarly, if m is positive integer, then β (m, n) = β (n, m) m−1 a f a β (n, m) = ...(i) f ...(ii) n n + 1 ... n + m − 1 In case both m and n are positive integer, then multiplying (i) numerator and denominator by 1·2·3... (m – 1) or (ii) by 1·2·3... (n – 1), then, we get f a f a fa f 1 ⋅ 2 ⋅ 3... m − 1 ⋅ m m + 1 ... m + n − 1 m−1 n−1 β (m, n) = or a 1 ⋅ 2 ⋅ 3.... m − 1 n − 1 β (m, n) = m+ n−1 = m n m+n . (c) Evaluation of Gamma Function n = (n 1) z ∞ we know that n = n − 1 or n+1 = n n = x n−1 . e − x · dx 0 Integrating by parts keeping xn–1 as first function n = − e – x ⋅ x n −1 ∞ 0 z ∞ + (n – 1) 0 x n − 2 ⋅ e − x dx n 285 MULTIPLE INTEGRALS L = M– lim e ⋅ x N –x z OP Q n− 1 ∞ + 0 + (n – 1) x →∞ x n − 2 . e − x dx ...(i) 0 As lim e – x . x n−1 = 0 x→ 0 n−1 lim e–x · xn–1 = But x→∞ = lim · x→∞ x ex = lim 1 1 x→∞ 1 lim 1 + n−2 + ... n−1 2 ⋅ x n− 3 x x x→∞ ∴ From (i), we get z ∞ n x n −1 x x2 xn + +...+ +... 1+ n 1 2 = 0 + (n – 1) = x n − 2 ⋅ e − x dx = (n – 1) 0 ⇒ n za f ∞ x n−1 −1 ⋅ e − x dx 0 n−1 . = (n – 1) 1 =0 ∞ ...(ii) Replace n by n + 1 in equation (ii) then, we get (d) n+1 = n n n+1 = n from (i) . ...(iii) n = (n – 1) (n – 2) .... 3. 2. 1 n = n−1 n+1 = n replace n by (n + 1), then . 4.8 TRANSFORMATIONS OF GAMMA FUNCTION We have n = z n = z = λ (1) Put x = λy or dx = λ dy Then from (i), we get or z ∞ 0 y n−1 e–λy dy = ∞ 0 e − x x n −1 dx ∞ − λy e z 0 n ∞ 0 n λ n bλyg n −1 ...(i) λ dy e − λy y n−1 dy ...(ii) 286 A TEXTBOOK OF ENGINEERING MATHEMATICS—I (2) Put xn = ∴ From (i), we get z ∞ − z1 n e 0 (3) Put e–x = – e–x dx = Then z FG 1 IJ dz H nK dz = n n = an + 1f n = or z in (i) then nxn–1 dx = dz and x = z1/n ∞ − z1 n e 0 ...(iii) t in (i). 1 t dt and ex = zb g (–dt) L F 1I O dt = z MlogG J P N H tKQ ∴ From (i), we get n = 0 1 n−1 – log t n−1 1 0 z LMNlogFGH 1t IJK OPQ dt = n F 1I 1 in (iii), we get (4) Value of GH JK . Putting n = 2 2 1 ∴ n−1 ...(iv) 0 1 2 1 = 2 z ∞ − z2 e 0 dz = aπ f . 1 = 2 or 1 π 2 As z ∞ −x2 0 e dx = π 2 4.9 TRANSFORMATIONS OF BETA FUNCTION β (m, n) = We know that (1) Putting x= z 1 0 x m−1 (1 – x) n–1 dx ...(i) 1 1 or dx = – dy 2 1+ y 1+ y Also b g (1 – x) = 1– 1 = 1+ y y 1+ y b g Also, when x = 0 then y = ∞ and when x = 1 then y = 0 ∴ From (i), we get β (m, n) = = or β (m, n) = z zb g zb g 0 ∞ ∞ F 1 I GH 1 + y JK m−1 F y I ⋅G H 1 + y JK y n−1 dy n −1 R| −1 U| S| b1 + yg V| dy T W 2 m − 1 + n −1 + 2 0 1+ y ∞ y n−1 dy 0 1+ y m+ n ...(ii) 287 MULTIPLE INTEGRALS (2) Also, as β (m, n) = β (n, m), therefore interchanging m and n in (ii), we have zb g β (n, m) = ∞ y m −1 dy 0 1+ y . m+ n (U.P.T.U., 2008) 4.10 RELATION BETWEEN BETA AND GAMMA FUNCTIONS [U.P.T.U. (C.O.), 2002, U.P.T.U., 2003] z We know that ∞ n y n−1 e – xy dy = 0 or n = Also Multiplying both sides of (i) by x z z ∞ n n−1 − xy x y 0 m = m –1 [ § 4.8 eqn. (ii)] xn ∞ m−1 − x x 0 –x e e dy ...(i) dx ...(ii) e , we get n ⋅ x m−1 e − x = z x n+ m −1 y n −1 e b ∞ g dy − y +1 x 0 Integrating both sides with respect to x within limits x = 0 to x = ∞, we have n z ∞ 0 x m−1 e − x dx = z a f bg ∞ But 0 x n + m −1 e − y +1 x dx = z LMNz ∞ 0 b g ∞ n + m −1 − y +1 x x e dx y n−1 dy 0 an + mf b1 + yg OP Q ...(iii) m+ n [by putting λ = 1 + y and ‘n’ = m + n in 4.8 on (ii)] Using this result in (ii), we get m = n = = ∴ 4.11 β (m, n) = z a ∞ 0 f 1 +y y b g n+m ⋅ n−1 m +n dy bm + ng z b1y+ ygdy ∞ n −1 m+ n 0 bm + ng ⋅ βbm, ng m⋅ n m+n . SOME IMPORTANT DEDUCTIONS (1) To prove that We know that n π (1 − n) = sin n π β (m, n) = (U.P.T.U., 2008) m⋅ n am + nf . 288 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Putting m + n = 1 or m = (1 – n), we get a f = β (n, 1 – n) n⋅ 1− n 1 zb g zb g β (m, n) = But ∴ y n−1 dy 0 1+ y m+ n As 1+ y 0 π sin nπ n (1 − n) = (3) To prove that 1+ x π sin n π (1 + n) (1 − n) = z As 1+n = n n F m + 1I F n + 1I H 2 K H 2 K· = F m + n + 2I 2 H 2 K β (p, q) = z x a1 − xf dx π 2 sin m θ cos n θ dθ 0 1 We know that 0 dx = . nπ · sin nπ π We have n . (1 – n) = sin nπ Multiplying both sides by n, we get nπ n n (1 − n) = sin nπ nπ · (1 + n) (1 − n) = sin nπ (2) To prove that z ∞ x n−1 π ,n<1 sin n π = or ∞ ∞ y n− 1 dy β (n, 1 – n) = ∴ From (i), we have ...(i) p −1 q −1 ...(i) 0 = q⋅ q b p + qg ⋅ Putting x = sin2θ and dx = 2 sin θ cos θ dθ, then, we get from (i) β(p, q) = z e z π2 0 sin 2 θ π2 ∴ z j e1 − sin θj p −1 2 q −1 2 p −1 θ cos 2 q −1 θ dθ = 2 0 sin π2 0 sin 2 p −1 θ cos 2 q −1 θ dθ = = 1 β(p, q) 2 p⋅ q 2 bp + qg . 2 sin θ cos θ dθ 289 MULTIPLE INTEGRALS Putting 2p – 1 = m and 2q – 1 = n, FG m + 1 IJ and q = FG n + 1IJ , we get H 2 K H 2K F m + 1I F n + 1I z sin θ cos θ dθ = H2 2F mK+ nH + 22 I K . H 2 K or p = π2 m n 0 4.12 DUPLICATION FORMULA FG m + 1 IJ = aπf . 2m . H 2K 2 F 1I Hence show that β (m, m) = 2 β G m, J . H 2K Since z sin θ cos θ dθ = 2 amamf +annff To prove that m 2 m− 1 (U.P.T.U., 2001) 1–2m π2 2 m −1 2 n−1 0 Putting 2n – 1 = 0 or n = z π2 0 z π2 0 1 , we obtain 2 sin 2 m −1 θ dθ = sin 2 m −1 θ dθ = amf FH 12 IK F 1I 2 m+ H 2K m⋅ π F 1I 2 m+ H 2K Again putting n = m in (i), we get a mf z sin θ cos θ dθ = o2 a2mt f o amft 1 sin θ cos θ θ 2 d = b g z 2 a2mf 2 o amf t 1 sin 2θg 2 dθ = b z 2 a 2mf 2 2 π2 2 m −1 2 m −1 0 2 i.e., π2 2 m −1 2 m −1 0 2 i.e., π2 2m 0 2 m −1 ...(i) ...(ii) LM3 FG 1 IJ = π OP MN H 2K PQ 290 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Putting 2θ = φ and 2 dθ = dφ, we get 1 z z π 22m 0 2 or 2 2m sin π2 0 z or o amft φ dφ = 2 a2 mf o amft φ dφ = 2 a2 mf 2 { amf} θ dθ = 2 a2mf 2 2 m− 1 2 sin 2 m −1 2 m −1 π2 0 sin 2 m −1 As z af a 0 f x dx = 2 z af π2 0 2 ...(iii) F As f axf dx = f atf dtI GH z z JK b b Equating two values of z a π2 0 amf 2 a2mf m ⇒ Multiplying above by 2 = a mf ⋅ a π f F 1I 2 m+ H 2K 1 = π 2 2 m −1 m , we get m m 2m 1 2 2m ⋅ m 2 2m − 1 21 − 2m m = β (m, m) = Sol. We have As FG m + 1 IJ = aπf a2mf H 2K 2 F 1I m m GH m + JK = 2 Example 1. Compute: a sin 2 m −1 θ dθ from (ii) and (iii), we get 2 2 m−1 Hence, f x dx m+ 21−2m β As π = 1 2 1 2 1 2 FG m, 1 IJ H 2K a f As, β m, n = m n m+n FG − 1 IJ , FG − 3 IJ , FG − 5 IJ . H 2K H 2K H 2K n (1 − n) = π sin n π ...(i) 1 , in (i), we get 2 π 1 3 = –π = − ⋅ 1 2 2 sin − π 2 (1) Putting n = – FG IJ FG IJ H K HK F H I K · 291 MULTIPLE INTEGRALS FG − 1 IJ = − π = − π = – 2 π H 2K 3 1 F 1I 2 2 H 2K 1 FG − IJ = −2 π . H 2K or ⇒ 3 , in (i), we get 2 (2) Putting n = – FG − 3 IJ FG 5 IJ H 2K H 2K or ...(ii) = F H π I K = – 3 sin − π 2 F H π 3 sin π 2 I =π K FG − 3IJ = π = π = 4 π . H 2K F 5 I 3 1 3 ⋅ π H 2K 2 2 ...(iii) 5 , in (i), we get 2 (3) Putting n = – FG − 5 IJ FG 7 IJ H 2K H 2K = π π =– = –π 5 I 5 I F F sin − π H 2 K H sin 2 πK FG − 5 IJ = π = − π = − 8 π . H 2K F 7I 5 3 1 15 ⋅ ⋅ ⋅ π H 2K 2 2 2 or Example 2. Evaluate z ∞ 0 Sol. Putting dx · 1 + x4 x4 = y or dx = z ∞ dx 1 + x4 = = z 4x 3 , we have 3 0 Also, we have dy z b g 1 −4 y dy ∞ 4 0 1+ y FG 1 − 1IJ H K 1 ∞y 4 z 4 0 1+ y dy ⋅ dx = π , we get sin nπ dx 0 1 + x4 = 1 ⋅ 4 ∞x 0 z n− 1 1+ x ∞ π sin F πI H 4K = π 2 · 4 ...(iv) 292 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 3. Show that Sol. We have z π2 0 z π2 0 tan n θ dθ tan n θ dθ = z FG IJ H K 1 1 π sec nπ ⋅ 2 2 b g dθ RS 1 an + 1fUV ⋅ RS 1 a–n + 1fUV T2 W T2 W = U R1 2 S an − n + 2fV W T2 an + 1f ⋅ F1 − n + 1I H 2K 2 = = π2 0 sin n θ cos θ −n 2 1 = 1 π ⋅ 1 2 sin n + 1 π 2 As n a1 − nf = sinπnπ a f 1 U R1 π cosec S an + 1fπ V = 2 W T2 1 F π nπ I 1 F nπ I π cosec G + J = π sec G J . Hence proved. = H2 2 K 2 H 2 K 2 Example 4. Show that Sol. We have z π2 0 z a π2 0 tan θ dθ f tan θ dθ = = z a = 1 3 1 , β 2 4 4 1 2 FG 3 IJ FG 1 IJ = 2 z x dx · H 4K H 4K 1+ x ∞ 0 f acos θf dθ RS 1 F 1 + 1I UV RS 1 F − 1 + 1I UV T2 H 2 KW T2 H 2 KW = R 1 F 1 − 1 + 2I UV 2 S T2 H 2 2 K W F 3I F 1I H 4K H 4K = 2 a1f 1 F 3I F 1I G J G J = 2 H 4K H 4K π2 0 2 −1 2 sin θ 12 FG H IJ β (m, n) = m ⋅ n K am + nf 4 293 MULTIPLE INTEGRALS e 3 − 1j z b g 1 ∞ y 4 dy β(m, n) = = 2 0 1+ y = 1 2 z z e ∞ x −1 ⋅ 4 x 3 dx 0 1 + x4 j zb g ∞ y m −1 dy 0 1+ y . Putting y = x4, dy = 4x3 dx x 2 dx ⋅ Hence proved. 0 1 + x4 Example 5. Show that β(m, n) = β(m + 1, n) + β(m, n + 1). Sol. We have = 2 m+ n ∞ (U.P.T.U., 2008) am + 1f ⋅ n + m ⋅ an + 1f am + 1 + nf am + n + 1f R.H.S. = m m⋅ n m⋅n n am + nf am + nf am + nf m + n m⋅ n L m n O + = ( m + n) MN m + n m + n PQ = m⋅ n = z 1 Example 6. Show that 0 ∴ F 1I H nK . = F 1 + 1I (1 − x ) n H n 2K π dx n dx = z = β (m, n). Hence proved. ( m + n) Sol. Put xn = sin2 θ or x = sin2/n θ ∴ + dx (1 − x n ) F2 I G − 1J 2 sin H n K θ cos θ dθ n z FG H 2 −1 π/2 sin n IJ K θ cos θ dθ = F I H K = 2 π/2 GH n − 1JK θ cos0 θ dθ sin n 0 2 n z 0 b1 − sin θg 2 F2 I RS 1 F 2 − 1 + 1)I UV RS 1 ( 0 + 1) UV KW T2 W T2 H n = R 1 F 2 − 1 + 0 + 2 I UV 2 S KW T2 H n F 1I ⋅ π F 1 I ⋅ F 1I H n K · Hence proved. 1 H nK H 2K ⋅ = = n R1 F 2 IU ST 2 H n + 1K VW n FH n1 + 12 IK 2 n 294 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 4.13 EVALUATE THE INTEGRALS z z ∞ (i) e − ax cos bx ⋅ x m−1dx 0 ∞ (ii) [U.P.T.U. (C.O.), 2004] e − ax ⋅ sin bx ⋅ x m−1dx (U.P.T.U., 2003) 0 Proof: Both the above integrals are respectively the real and imaginary parts of z ∞ e − ax e ibx ⋅ x m −1 dx 0 z ∞ or e −( a − ib)x ⋅ x m−1 dx 0 z ∞ Now, we have z ∞ ∴ e − λx x n −1 dx = e − ( a − ib )x x m−1 dx = 0 0 n λn m ( a − ib) m ( a + ib) m ( a2 + b 2 ) m Let us put a = r cos θ, b = r sin θ in R.H.S. ∴ z ∞ = m⋅ e − ax (cos bx + i sin bx)x m−1dx = m⋅ r m (cos θ + i sin θ)m r 2m 0 z ∞ or e − ax cos bx ⋅ x m −1 + ie − ax sin bx ⋅ x m −1 dx 0 m (cos m θ + i sin mθ) rm Equating real and imaginary parts, we get = z z ∞ e − ax cos bx ⋅ x m −1 dx = 0 ∞ and e − ax sin bx ⋅ x m −1 dx = 0 where r = ( a 2 + b 2 ) and θ = tan–1 b · a m rm m rm cos mθ sin mθ |r2 = x2+ y2 295 MULTIPLE INTEGRALS z ∞ Example 7. Prove that a2 − b 2 , where a > 0. (a2 + b2 )2 xe − ax cos bx ⋅ dx = 0 Sol. We have z ∞ e − ax cos bx ⋅ x m−1 dx = 0 m rm cos mθ Put m – 1 = 1, i.e., m = 2 z ∞ xe − ax cos bx dx = 0 2 r cos 2θ = 2 1 1 − tan 2 θ ⋅ 2 ( a + b ) 1 + tan 2 θ 2 FG1 − b IJ H aK 1 ⋅ = a +b F GH1 + ba IJK 2 2 2 2 2 2 = a2 − b 2 ( a 2 + b 2 )2 . Hence proved. F 1 I F 2 I F 3 I ..... F n − 1I , where n is a positive integer. H nK H nK H nK H n K F 1 I F 2 I ... F n − 2 I F n − 1 I P = ...(i) H nK H nK H n K H n K F 1 I F 2 I .... F1 − 2 I F1 − 1 I P = H n K H n K H nK H nK Example 8. Find the value of Sol. Let or Writing in the reverse order, P = F1 − 1 I F1 − 2 I .... F 2 I ⋅ F 1 I H nK H nK H n K H nK ...(ii) Multiplying (i) and (ii), we get P2 = Now, we know that LM F 1 I F1 − 1 I OP LM F 2 I F1 − 2 I OP ... N H nK H nK Q N H n K H n K Q LM F n − 2 I F1 − n − 2 I OP LM F n − 1 I ⋅ F1 − n − 1I OP N H n K H n KQ N H n K H n KQ n (1 − n) = π sin nπ F 1 I ⋅ F1 − 1 I = π H n K H n K sin π or ...(iii) n ∴ 2 P = π π π π ... ⋅ π n−2 n−1 2π sin sin sin π sin π n n n n ...(iv) 296 A TEXTBOOK OF ENGINEERING MATHEMATICS—I From Trigonometry, we know that sin nθ = 2n–1 sin sin θ Putting θ = 0 Fθ + π I sin F θ + 2π I ... sin F θ + n − 2 πI sin F θ + n − 1 πI H nK H n K H n K H n K F H lim sin nθ = lim n ⋅ sin nθ ⋅ θ θ→0 θ→0 nθ sin θ sin θ = n×1×1= n ∴ n = 2n−1 sin π 2π n−2 n−1 π sin .... sin π sin n n n n ∴ P2 = πn–1 ∴ P = 2 n−1 n (2π) n −1/2 n F 1 I F 2 I ... F n − 1I = (2π) H n K H nK H n K n ( n − 1/2) Hence I K 1 = 2 Example 9. Prove that Sol. We have π· ( n) (1 − n) = Putting ⋅ n = π sin n π 1 , we get 2 F 1I F 1I = π H 2 K H 2 K sin π 2 LM F 1 I OP = π N H 2KQ F 1 I = π . Hence proved. H 2K 2 Example 10. Show that zz D xl −1 y m−1 dx dy = (l) (m) (l + m + 1) al + m where D is the domain x ≥ 0, y ≥ 0 and x + y ≤ a. Sol. Putting x = aX and y = aY in the given integral then, we get I = zz D ,( aX) l −1 ( aY)m−1 a 2dXdY Where D' is the domain X ≥ 0, Y ≥ 0 and X + Y ≤ 1 zz 1 1− X I = al+m 0 0 X l − 1Y m − 1 dY dX 297 MULTIPLE INTEGRALS z 1 = al+m X l −1 l+m 1 l −1 a l + m ( l ) ( m + 1) al + m ⋅ β(l , m + 1) = m ( l + m + 1) m = al + m z ( l) ( m) ( l + m + 1) x m −1 + x n −1 Sol. The given integral z 1 I = 0 · Hence proved. dx . (1 + x ) m + n 0 m 0 = 1 or m 1− X 0 0 Example 11. Evaluate LM Y OP dX = a X (1 − X) dX m z NmQ (1 + x ) z 1 x m −1 dx + m+ n 0 x n−1 (1 + x ) m + n I = I1 + I2 (say) dx ...(i) 1 Putting x = in I2, we have z F 1 I FG − 1 IJ dz H zK H z K I = z LM1 + F 1 I OP N H zKQ n −1 1 2 2 m+n ∞ z z ∞ = 1 z m −1 dz (1 + z ) m + n ∞ = ∴ 1 x m −1 dx (By definite integral) (1 + x ) m + n From equation (i) z 0 (1 + x) z ∞ = 0 = + m+n 1 x m −1 dx (1 + x ) m + n (m) ⋅ ( n) ( m + n) z ∞ x m−1 dx 1 I = x m −1 dx (1 + x ) m + n = β(m, n) · Example 12. Using Beta and Gamma functions, evaluate z 1 0 FG x IJ dx· H1− x K 1/2 3 3 (U.P.T.U., 2006) 298 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Sol. Let x3 = sin2 θ ⇒ 3x2dx = 2 sin θ cos θ dθ ∴ z z π/2 π/2 sin θ sin θ cos θ 2 sin 2− 4/3 θ dθ ×2 dθ = 4 / 3 3 θ cos θ sin ⋅ 3 0 0 5 1 2 2 ⋅ π 2 3 2 3 3 = 3 2/ 3 + 0 + 2 4 2 2 2 3 z π/2 2 2 = sin 2/3 θ cos 0 θ dθ = 3 3 0 2 π 2 3 2 ⋅ = 9 1 1 3 3 3 = 2 π 3 . 1 3 Example 13. Prove that: β (l, m) β (l + m , n) β (l + m + n , p) = l m n p l+m+n+p Sol. β (l, m) · β (l + m, n) · β (l + m + n, p) = l m l+m l+m n · l+m+n z ∞ Example 14. Find l+m+n p · = l+m+n+p l m n p l + m+ n+ p . Proved. 1 3 x 1/2 ⋅ e − x dx · 0 or x = t3 ⇒ dx = 3t2 dt Sol. Let x1/3 = t z ∞ ∴ 1 3 x 1/2 ⋅ e − x dx = z ∞ 0 0 = 3 z ∞ (a) z ∞ 9 −1 t 2 ⋅ e −t ⋅ dt e − ax ⋅ x n−1 cos bx dx = 0 Put a = 0, z x n−1 cos bx dx = 0 Put xn = z so that xn–1 dx = cos −∞ 0 ∞ z ∞ (b) Sol. (a) We know that z =3 9 315 = π· 2 16 Evaluate— cos x 2dx ∞ t 7/2 e − t dt 0 0 Example 15. z ∞ t 3/2 ⋅ e − t ⋅ 3t 2 dt = 3 π 2 x dx 2 (n)cos nθ 2 n 2 , where θ = tan –1 (a + b ) 2 ( n) bn nπ cos 2 dz and x = z 1/n n θ = tan −1 F bI H 0K = tan −1 ( ∞) = π 2 F bI H aK 299 MULTIPLE INTEGRALS z ∞ Then z cos bz1/n dz = n (n) b 0 ∞ b 0 1 2 z cos x 2 dx = 0 n z ∞ (b) I = cos nπ (By definite integral) 2 ...(i) z ∞ cos −∞ Putting b = nπ 2 F 3 I cos π = π ⋅ 1 = 1 π · H 2K 4 2 2 2 2 ∞ ∴ cos ( n + 1) cos ( bx1/n ) dx = Putting b = 1, n = n πx 2 πx 2 dx = 2 cos dx 2 2 ...(ii) |By definite integral 0 π 1 and n = in equation (i), we get 2 2 F 3I H 2 K cos π F I H K 4 F πI H 2K F 3I ∞ 2 z−∞ cos πx2 dx = 2 F πHI21/K2 cos π4 H 2K z ∞ 0 ∴ From (ii) π cos x 2 dx = 2 1/2 = 2· 1 2 1 π⋅ ⋅ = 1. 2 π 2 EXERCISE 4.3 z z 1 1. Prove that 0 2 3. Prove that 0 z z 1 5. Show that 0 1 · x (1 − x ) dx = 280 4 3 x2 dx = 64 2 · (2 − x) 15 F log 1 I H xK 2. Prove that 0 x( 8 − x 3 )1/3 dx = 16 π · 9 3 0 ∞ 4. Show that 0 e − st t dt = π , s > 0. s LM F 1 I OP 1 N H nK Q 6. Prove that (1 − x ) dx = ⋅ · n 2I F 2 H nK 2 1 n −1 dx = (n) , n > 0 . x 2 (1 − x ) 3 dx = 60. n 1/n 0 1 7. Show that z z z 2 z 2 8. Show that 0 ( 4 − x 2 ) 3/2 dx = 3π. 300 A TEXTBOOK OF ENGINEERING MATHEMATICS—I z z 3 9. Show that 0 ∞ 11. Prove that 1+ x 0 π/2 13. Prove that = π. ( 3 x − x)2 x 2 dx z 0 6 z z 1 dx 10. Show that 0 F1I H 3K · = F 5I (1 − x ) 3 H 6K 3 1 π = 3 3 12. Prove that · 0 z π x 2 dx x 2 dx (1 − x ) 3 = 2 · 3 π/2 dθ × sin θ . dθ = π . (sin θ) 0 14. Show that, if n > – 1, z ∞ 1 x n e − k x dx = 2 2 2k n+1 z 0 F n + 1I . H 2 K 0 Hence or otherwise evaluate e − k x dx. 2 2 −∞ F 1 I ⋅ F 5I H 3 K H 6 K = ( π) 2 15. Show that F 2I H 3K z 17. Prove that 19. Evaluate e x 3 2 dx = z 20. Prove that 4.14 ∞ −4 x 0 ∞ m− 1 x cos bx dx 0 z ∞ 0 and · aπ f · 3 128 x m−1 e − ax dx = 2 1/ 3 z ∞ 0 n 2a n LM F 1 I OP H 3KQ 16. Show that N F 1I H 6K 18. Prove that x m−1 sin bx dx . z ∞ 0 2 = x 2 e − x dx = 2 ( π) ⋅ 21/3 31/2 · aπ f · 4 LMAns. m cos mπ ; m sin mπ OP N b 2 b 2Q m m · APPLICATION TO AREA (DOUBLE INTEGRALS) 4.14.1 Area in Cartesian Coordinates (a) Let the area ABCD is enclosed by y = f(x), y = 0 and x = a, x = b. Let P(x, y) and Q (x + δx, y + δy) be two neighbouring points on the curve AD whose equation is y = f(x). Then the area of the element is δx δy. Consequently the area of the strip PNMQ. z f ( x) dx dy , where y = f (x) is the equation of AD. y=0 301 MULTIPLE INTEGRALS dy ) Y y=0 y dx , x=a . dx dy A (b) In a similar way we can prove that the area bounded by the curve x = f(y), the y axis and the abscissae at y = a and y = b is given by z z b f ( y) y=a x=0 M Q x=b dy dx dy dx O B N M C X Fig. 4.25 y=b C x = f(y) dx Q dy N A (x, y) P x=a Y D D + z z f (x) (x ABCD = b + ∴ The required area P B y=a X O Fig. 4.26 (c) If we are to find the area bounded by the two curves y = f1(x) and y = f2(x) and the ordinates x = a and x = b i.e., the area ABCD in the Figure 4.27 then the required area z z b f1 ( x ) x=a y = f 2 (x) dx dy Y D y = f1 (x) C dx x=b dy x=a B y = f2 (x) A O M N X Fig. 4.27 Example 1. Find the area lying between the parabola y = 4x – x2 and line y = x by the method of double integration. (U.P.T.U., 2007) 302 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Y Sol. OA is the line y = x and OBAD is the parabola y = 4x – x2 Solving y = 0 and Eqn. (ii), we find that the parabola Eqn. (ii) meets the x-axis at O (0, 0) and D (4, 0). Solving Eqns. (i) and (ii), we find x = 4x – x2 or x(x – 3) = 0 i.e., x = 0, 3. Also x = 0 gives y = 0 and x = 3 gives y = 3. O x¢ ∴ The line (i) meets the parabola (ii) in O (0, 0) (0, 0) and A (3, 3) we are to find the area OBAO. ∴ Required area OBAO = area OCABO – area of ∆ OCA. zz z 3 = x=0 y=0 3 = 4x−x2 0 z ( y ) 04 x − x 2 ...(i) B y= dx dy ...(ii) x A (3, 3) D (4, 0) X C Fig. 4.28 F 1 × OC × CAI H2 K F 1 × 3 × 3I dx − H2 K dx dy − 1 1 9 ( 4x − x 2 )dx − ( 9) = ( 2x 2 − x 3 ) 30 − 2 3 2 9 9 1 3 9 = 2(3)2 – ( 3) − = 18 – 9 – = . 2 2 3 2 = 3 0 Example 2. a2y2 = a2x2 – x4 find the whole area within the curve. Sol. First of all trace the curve as shown in the Figure 4.29. Y Whole area = 4 × area OAC zz a = 4 where f ( x) x=0 y=0 dx dy C y = f(x) i.e., y= P(x,y) A X¢ (a2 x 2 − x 4 ) a O a is the equation of the curve z a f (x) = 4 x=0 y 0 z a dx z Y¢ Fig. 4.29 4 ( a 2 x 2 − x 4 )dx a 0 4 a 4 π/2 x ( a 2 − x 2 )dx = a sin θ a cos θ a cos θ dθ = a 0 a 0 Putting x = a sin θ = 4 0 f ( x)dx = z = 4a2 z π/2 0 a sin θ cos 2 θ dθ = 1 π 4 a2 2 = = · 3 1 3 2⋅ ⋅ π 2 2 4a 2 1 ⋅ z 4a 2 1 ( 3 / 2) 2 (5 / 2) a (a, 0) X 303 MULTIPLE INTEGRALS Example 3. Determine the area of region bounded by the curves Y (0, 4) y=4 4y = x2 xy = 2, 4y = x2, y = 4. [U.P.T.U. (C.O.), 2003, 2008] Sol. Required area of shaded region zz z FGH 4 = y =1 x = 4 = 2 y 1 2 dx dy y IJ K (2, 1) (0, 1) 2 2 y − dy y xy = 2 F 2 y − log yI H3 K LF 16 I 2O = 2 MH − 2 log 2K − P 3Q N 3 O 4 Fig. 4.30 3/2 = 2 X (2, 0) 1 28 − 4 log 2 . 3 Example 4. By double integration, find the whole area of the curve a2x2 = y3(2a – y). Sol. Required area = 2 (area OAB) = = 2 y 3/2 2 a − y where x = f(y) = a z z 2a f ( y) y=0 x=0 dy dx ...(i) is the equation of the given curve. From eqn. (i), required area =2 z z 2a 0 2 = a 0 = 32a2 z 2 2 a 3/2 f ( y)dy = y 2 a − y dy a 0 π/ 2 z (2a sin θ) 2 π/2 0 3/2 A (0, 2a) dy B 2 a − 2 a sin θ ⋅ 4 a sin θ cos θ dθ (Put y = 2a sin2 θ) F 5I F 3 I 3 1 1 ⋅ π⋅ ⋅ π H 2K H 2K 2 = 32a · = 32a · 2 2 = πa . 2 2 ( 4) Y 2 sin 4 θ cos 2 θ dθ 2 (U.P.T.U., 2001) 2⋅ 3⋅ 2⋅1 O X Fig. 4.31 2 Example 5. Show that the larger of the two areas into which the circle x2 + y2 = 64a2 is divided by the parabola y2 = 12ax is 16 2 a (8π − 3 ) . 3 Sol. y2 = 12ax is a parabola, whose vertex is (0, 0) latus rectum = 12a, x2 + y2 = 64a2 is a circle, whose centre is (0, 0) and radius = 8a. 304 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Solving x2 + y2 = 64a2 and y2 = 12ax, we get 2 Y 2 x + 12ax – 64a = 0 or x = 4a. ∴ The x-coordinate of P the point of intersection of the two curves is 4a. Also the circle x2 + y2 = 64a2 meets x axis at x2 = 64a2 (Putting y = 0) i.e., at x = ± 8a. Hence the coordinates of A and A′ are (8a, 0) and (–8a, 0). P X¢ Also for the parabola y = (12ax ) and for the circle we have y= O A¢ (8a, 0) N (4a, 0) X P¢ ( 64 a 2 − x 2 ) Required area = shaded area in the Figure 4.32. = area of the circle – area OP' AOP = π(8a)2– 2 [area OAP] = 64 πa2 – 2 [area OPN + area PAN] z A (8a, 0) z Y¢ Fig. 4.32 LM (12ax)dx + (64a − x )dxOP N Q L 1 R1 F x I UV OP xdx + S x ( 64a − x ) + ⋅ 64a sin = 64πa – 2 M 2 ( 3 a) H 8 a K W PQ 2 MN T2 1 R2 U R F 1 I UV = 64πa – 4 ( 3a) S x V − 2S 32a sin (1) − ⋅ 4a ⋅ 48 a − 32a sin H 2KW 2 T3 W T = 64πa2 – 2 2 4a 0 z 8a 2 2 4a 4a 2 2 8a −1 2 0 2 4a 4a 2 3/2 −1 2 −1 2 0 64a 2 32 2 = 64πa2 – πa 3 − 32πa 2 + 16 a 2 3 + 3 3 128 πa 2 16 a 2 − 3 = 3 3 16 2 a (8π − 3 ) . = 3 4.14.2 Area of Curves in Polar Coordinate B + dr, Q (r q) q+d The area bounded by the curve r = f(θ), where f(θ) is single valued function of θ in the domain (α, β) and redii vectors θ = α and θ = β is z z β f ( θ) θ=α r=0 D P (r, q) E r dθ dr (α < β). Let O be the pole, OX the initial line and AB be the portion of the arc the curve r = f(θ) included between the radii vectors OA and OB i.e., θ = α and θ = β. The area of element CDEF = area of sector OCD–area of sector OFE O 1 1 2 2 = (r + δr ) δθ − r δθ 2 2 = r δθ δr, neglecting higher powers of δr and δθ. Hence the element of area in polar coordinates is r δθ δr. q=b F C A q=a dq q X Fig. 4.33 305 MULTIPLE INTEGRALS ∴ The area of the figure OPQO = ∴ z z β f ( θ) θ =α r = 0 r dθ dr, where r = f(θ) is the curve AB. z z The required area = β f ( θ) θ=α r=0 r dθ dr . Example 6. Find the area of the region enclosed by x+ y = Sol. We have x+ y = a and x + y = a. ...(i) a x+ y = a a− y ...(ii) ⇒ x= From (i) x = From (ii) e a − yj + y = a e a − yj 2 Y (0,a) B 2 x + y a + y – 2 ay + y = a 2y – 2 ay = 0 ⇒ a A (A,0) ay y= O Squaring on both sides, we get y2 = ay = ⇒ y(y – a) = 0 ⇒ y = 0, a X x+ y=a Using the value of y in equation (ii), we get, x = a, 0. Fig. 4.34 So, the intersection point of given curve are (a, 0) and (0, a). Required area of the shaded region. zz z RST z a b a− y g b a − yg . dy dx dy = x 0 x = e a − yj 0 e a − yj a = 2 2 z e a − y j UVWdy = e2 ay − 2yjdy L 2 a. y − 2y OP = 4 a − a = a . Ans. = M 2 PQ 3 3 MN 3 2 a = 0 2 a− y− 32 a 0 2 a 2 2 2 0 Example 7. Find the area of the region enclosed by the curves y = 3x and 4y = x2. x +2 2 3x y = 2 x +2 ...(i) 4y = x2 ...(ii) Sol. We have 2 From (i) and (ii) ⇒ y= x 4 = 3x x2 + 2 x4 + 2x2 – 12x = 0 3x 2 x +2 (y = O (0, 0) Fig. 4.35 (2, 3 4 A x 2 4 ) ) 306 A TEXTBOOK OF ENGINEERING MATHEMATICS—I x(x3 + 2x – 12) = 0 x(x – 2) (x2 + 2x + 6) = 0 ⇒ x = 0, 2 (real values only) Using the values of x in (i), we get 3 4 y = 0, So the intersection points of curves are (0, 0) and FG 2, 3 IJ . H 4K Hence, the required area of the shaded region. z z 2 = e 3x x 2 + 2 x = 0 y = x2 4 j dy dx = z 2 0 e 3x x 2 + 2 y x2 4 j dx = z F 3x − x I dx GH x + 2 4 JK 2 2 0 2 L3 3 x O 8 = M log ex + 2j − = blog 6 − log 2g − P 12 Q 2 12 N2 3 2 2 0 3 6 2 log − 2 2 3 = 2 3 log 3 − . 3 2 Example 8. Calculate the area which is inside the cardioid r = 2 (1 + cos θ) and outside the circle r = 2. Y Sol. r = 2 is a circle centred at origin and of radius 2. The shaded area in Figure 4.36 is the region R which is outside the given circle and Cardioid inside the cardioid. So r varies from the circle Circle r = 2 the cardioid r = 2(1 + cos θ). While θ varies from – π/2 to π/2. Since R is symmetric about x–axis, the required area A of the region R is O X given by = A = zz z z R = 2 dA = zz π/ 2 − π /2 r = 2 π/2 2(1+ cos θ) 0 z 2 π/ 2 = 40 2( 1 + cos θ ) r dr dθ = 2 z π/2 r 2 0 ( 2 cos θ + cos 2 θ) dθ LM N R rdrd θ OP Q π/ 2 2(1+ cos θ) 2 2 dθ Fig. 4.36 1 1 = 4 2 sin θ + θ + sin 2θ = π + 8. 2 4 0 Example 9. Find, by double integration, the area lying inside the cardioid r = a(1 + cos θ) and outside the circle r = a. Sol. Required area = area ABCDA = 2(area ABDA) = 2 z z π/2 a (1 + cos θ) 0 a r dr dθ 307 MULTIPLE INTEGRALS 2 π/2 s g q) 0 r=a co π/2 = a2 a (1 + cos θ) 2 − 1 dθ 0 q = p/2 + = a2 dθ (1 z zb 0 A a( 1+ cos θ ) a z FG r IJ H 2K r= = 2 π/ 2 cos 2 θ + 2 cos θ dθ q=p F 1 π I a (π + 8). = a H ⋅ + 2K = 2 2 4 O B q=0 D 2 2 Example 10. Find, by double integration the area lying inside the circle r = a sin θ and outside the cardioid C r = a (1 – cos θ). Sol. Eliminating r between the equations of two curves, sin θ = 1 – cos θ or sin θ + cos θ = 1 Squaring 1 + sin 2θ = 1 or sin 2θ = 0 Fig. 4.37 p q= — 2 π 2 For the required area, r varies from π a (1 – cos θ) to a sin θ and θ varies from 0 to . 2 ∴ 2θ = 0 or π ⇒ θ = 0 or ∴ Required area = = z z z LMN OPQ z π/2 a sin θ 0 a( 1 − cos θ ) π/ 2 r2 2 0 r = a sin q r dr dθ a sin θ a( 1− cos θ) π/2 2 1 = 2 0 r = a (1 – cos q) dθ O Fig. 4.38 a sin θ − (1 − cos θ) 2 2 z z = a 2 π/ 2 (sin 2 θ − 1 − cos 2 θ + 2 cos θ ) dθ 2 0 = a 2 π/ 2 π ( −2 cos 2 θ + 2 cos θ) dθ = a 2 1 − . 2 0 4 zz z FH IK π/2 r for cardioid θ =0 r for parabola π/2 F H = 2 = z 0 π/2 0 F B E q rd dr q=0 q O A D 1+ cos θ l I K q = p/2 r dθ dr C 1 2 r dθ 2 1/(1 + cos θ) q dθ (1 + cos θ) 2 − 1 / (1 + cos θ) 2 X dθ Example 11. Find by double integration the area lying inside the cardioid r = 1 + cos θ and outside the parabola r(1 + cos θ) = 1. Sol. The required area = area CAFBEDC = 2 (area DAFBED) = 2 q=0 Fig. 4.39 308 A TEXTBOOK OF ENGINEERING MATHEMATICS—I z Now, = π/2 0 (1 + cos θ)2 dθ = z π/ 2 (1 + cos θ) 2 dθ − z zF zH 0 π/2 0 z π 2 0 1 dθ (1 + cos θ) 2 ...(i) (1 + 2 cos θ + cos2 θ) dθ π/2 = = 1 [(1 + 2 cos θ + (1 + cos 2θ)] dθ 0 2 π/2 3 1 + 2 cos θ + cos 2θ dθ 0 2 2 I K 1 L3 O = M θ + 2 sin θ + sin 2θP 4 N2 Q 3F1 I 1 F I 1 π + 2 sin π + sin π − 0 = H K H2 K 4 2 2 π 2 0 = and z π 2 1 / 1 + cos θ 2 0 a f 3 1 π + 2 = ( 3π + 8) 4 4 z z ze j ze j ...(ii) FG IJ HK π θ 1 2 sec 4 dθ dθ = 0 4 2 = As 1 + cos θ = 2 cos 2 FG 1 θIJ H2 K π 1 1 4 θ= φ sec 4 φ dφ, Putting 0 2 2 π 1 4 1 + tan 2 φ sec 2 φ dφ , [ sec2 φ = 1 + tan2 φ ] 2 0 1 1 1 + t 2 dt , where t = tan φ = 2 0 1 1 1 3 1 1 2 1+ = t + t = = 2 3 2 3 3 0 ∴ From (i) with the help of (ii) and (iii), we get the required area 1 2 3π + 8 − = 4 3 3 2 3 4 π+2− = π+ . = 4 3 4 3 = IJ K FG H a IJ K FG H ...(iii) f Example 12. Find the area of the curve r2 = a2 cos 2θ. Sol. Since the curve is symmetric about origin so the required area of shaded region. q=p 4 Y dq r = a cos2q X O q =-p 4 Fig. 4.40 309 MULTIPLE INTEGRALS = 2 = 2 z z LM OP z NQ π 4 a cos 2θ −π 4 r=0 π 4 r2 2 0 −π 4 2 = 2a z π 4 0 r dθ dr a cos 2θ . dθ = z π 4 −π 4 a 2 cos 2θ dθ LM N OP Q 2a 2 π π 4 sin 2θ 0 = a 2 sin − sin 0 = a 2 . 2 2 cos 2θ dθ = Example 13. Find the area included between the curves r = a(secθ + cosθ) and its asymptote r = a secθ. Sol. The curve is symmetric about the line θ = 0, so the required area of shaded region. = = z z z b π2 a sec θ + cos θ −π 2 a sec θ π 2 −π 2 g r dθ dr LM r OP b MN 2 PQ 2 a sec θ + cos θ g dθ = a sec θ q=p 2 r= a z {b a2 2 π 2 −π 2 sec θ + cos θ g − sec θ}dθ 2 ) sq co + cq (se (a,o) (2 a,o) r = asecq q =-p 2 = a2 = a2 ze ze zb π2 0 π 2 0 π 2 Fig. 4.41 j z FGH g a2 sin 2θ 5θ + 2 2 0 2 + cos 2 θ dθ = a 2 = a2 2 = a 2 5π 5πa 2 −0 = . 2 2 4 0 LM N j sec 2 θ + 2 sec θ cos θ + cos 2 θ − sec 2 θ dθ 5 + cos 2θ dθ = OP Q π2 0 LM N 2+ IJ K 1 + cos 2θ dθ 2 OP Q π 2 2 310 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 14. Find the area included between the curve x2y2 = a2 (y2 – x2) and its asymptotes. z ze a ax a Required area = 4 × area OAB = 4 0 y dx = 4 0 Sol. a − x2 2 j dx B Y Q x=a P X¢ A X N M (a, 0) dx O Y¢ = 4a z Fig. 4.42 π 2 a sin θ ⋅ a cos θ dθ a cos θ 0 = 4 a2 z π2 0 , Putting x = a sin θ π2 sin θ dθ = 4 a 2 – cos θ 0 = 4 a 2 . Example 15. Show that the entire area between the curve = 2 = 2 z z a f z e j z 2a 0 y dx = 2 x3 2 0 2a − x 2 a sin 2 θ π2 32 π2 0 Q P sin 4 θ dθ = 16a 2 = 3πa2 . Hence proved. x = 2a dx X¢ 4 a sin θ cos θ 2 a ⋅ cos θ 0 = 16a 2 2a B Y y2 (2a – x) = x3 and its asymptote is 3πa2. Sol. Required area = 2 × area OAB N O dθ 3 1 π ⋅ ⋅ 4 2 2 Y¢ Fig. 4.43 M A (2a, 0) dx X 311 MULTIPLE INTEGRALS EXERCISE 4.4 LMAns. 8 a OP 3 PQ MN 2 1. Find the area bounded by the parabola y2 = 4ax and its latus rectum. 2. Find by double integration the area of the ellipse x2 y2 + Ans. πab = 1. a 2 b2 3. Show by double integration that the area between the parabolas y2 = 4ax and x2 = 4ay 16 2 a . is 3 4. Find the area of the region bounded by quadrant of x2 + y2 = a2 and x + y = a. LMAns. 1 aπ − 2fa OP N 4 Q 5. Find by the double integration the area of the region bounded by y = x and y = x. LMAns. 1 OP N 10 Q 2 2 3 LMAns. 8 a OP MN 15 3 PQ 2 6. Find the area of the curve 3ay2 = x (x – a)2. 7. Find the area included between the curve and its asymptotes. Ans. πa 2 8. Find the area between the curve y2 (2a – x) = x3 and its asymptote. Ans. 3 πa 2 9. Find the area of the portion bounded by the curve x(x2 + y2) = a (x2 – y2) and its asymptote. LMAns. 2 a FG π − 1IJ OP H 4 KQ N L 8 ab OP 10. Find the area included between the curves y = 4a (x + a) and y = 4b (b – x). MAns. 3 PQ MN LMAns. 1 πa OP 11. Find the whole area of the curve r = a cos 2θ. N 2 Q LMAns. 1 πa OP 12. Find the area of a loop of the curve r = a sin 2θ. N 8 Q 2 2 2 2 2 13. Find the area enclosed by the curve r = 3 + 2 cos θ. Ans. 11π 14. Show by double integration that the area lying inside the cardioid r = a (1 + cos θ) and 1 2 a (π + 8). outside the circle r = a is 4 15. Find the area outside the circle r = 2a cos θ and inside the cardioid r = a (1 + cos θ). LMAns. πa OP 2 PQ MN 2 16. Find the area included between the curve x = a (θ – sin θ), y = a (1 – cos θ) and its base. Ans. 3πa 2 312 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 17. Find the area bounded by the curve xy = 4, y-axis and the lines y = 1 and y = 4. [Ans. 8 log 2] 18. Use a double integral to find the area enclosed by one loop of the four-leaved rose r = cos 2θ. LMAns. π OP N 8Q 4.15 TRIPLE INTEGRALS Triple integral is a generalization of a double integral. Let a function f (x, y, z) defined at every point of a three dimensional region V; Divide the region V into n elementary volumes δV1, δV2, ..., δVn and let (xr, yr, zr) be any point inside the rth sub-division δVr. n b g Find the sum ∑ f xr , y r , z r δVr ⋅ r =1 zzz b g Lim n δVr → 0 r =1 b g n → ∞ ∑ f xr , yr , zr δVr ⋅ f x, y, z dV = V Then To extend definition of repeated integrals for triple integrals, consider a function F (x, y, z) and keep x and y constant and integrate with respect to z between limits in general depending upon x and y. This would reduce F(x, y, z) to a function of x and y only. Thus let φ (x, y) = z bb gg b z2 x , y g F x, y, z dz z1 x , y Then in φ (x, y) we can keep x constant and integrate with respect to y between limits in general depending upon x this leads to a function of x alone say ψ (x) = z aa ff b g y2 x y1 x φ x, y dy Finally ψ(x) is integrated with respect to x assuming that the limits for x are from a to b. Thus zzz b V zzz b V g = g = F x, y, z dV F x, y, z dV z z aa ff z bb gg b g z LMNz aa ff RSTz bb gg b g UVW OPQ zz z b g zz z b g LM OP z z MN b g PQ LM OP z z MN b g PQ b y2 x z2 x , y a y1 x z1 x , y b y2 x z 2 x, y a y1 x z1 x , y 1 1 − x 1− x − y Example 1. Evaluate the integral Sol. Let Ι = F x, y, z dx dy dz 0 0 0 1 1− x 1− x − y F x, y, z dz dy dx . dx dy dz x + y + z+1 3 ⋅ dx dy dz x+ y + z+1 0 0 0 1 1− x 1 1 − 2 x + y+ z+1 2 3 1 −x − y = ⇒ 0 0 = − dx dy 0 1 1 1 −x 1 1 − dx dy 0 0 2 4 x + y +1 2 313 MULTIPLE INTEGRALS ⇒ z z LMMN b g OPPQ OP z LMN Q z LMN a f OPQ IJ z FGH K = − 1 1 1 −x 1 1 − dx dy 2 0 0 4 x + y +1 2 = − 1 1 1 1 y+ dx x + y+1 0 2 0 4 = − 1 1 1 1 1 dx 1− x + − 2 0 4 2 x+1 = − 1 1 3 1 1 dx − x− 0 x+1 2 4 4 1− x LM a fOPQ N 1 L3 1 5O O 1L = − M − − log 2P = Mlog 2 − P. 2 N4 8 8Q Q 2N 1 3 1 x − x 2 − log x + 1 2 4 8 = − Example 2. Evaluate Sol. We have zz ze j zz 4z− x2 4 2 z 0 0 = = 4 2 z 4 2 z 0 0 z 0 2 x z LMN z e4 z − x j dz dx 2 z0 zz e 0 0 4 1 0 dz dx dy . 0 I = 1 j 4 z − x 2 dz dx e 4z − x j 2 1 + ⋅ 4 z sin −1 2 FG x IJ OP H 4z K Q 2 z dz 0 a 4z − 4zf + 12 ⋅ 4z sin −1 a1fOPQ dz L1 O = πzdz = π M z P = 8π. N2 Q Example 3. Evaluate zzz ex + y + z j dx dy dz where V is the volume of the cube bounded 4 = 1 ⋅2 z 0 2 4 4 2 0 2 2 0 2 V by the coordinate planes and the planes x = y = z = a. Sol. Here a column parallel to z-axis is bounded by the planes z = 0 and z = a. Here the region S above which the volume V stands is the region in the xy-plane bounded by the lines x = 0, x = a, y = 0, y = a. Z Hence, the given integral = = = = zzze z z FGH z z FGH z LMN j dx dy dz z I x z + y z + J dx dy 3K 1 I x a + y a + a J dx dy 3 K 1 1 O x ay + y a + a yP dx 3 3 Q a a a 0 0 0 a a x +y +z 2 2 2 a 3 2 0 0 a a z=a 2 0 2 2 O 3 S 0 0 a 0 2 a 3 3 0 X z=0 Y Fig. 4.44 314 A TEXTBOOK OF ENGINEERING MATHEMATICS—I = z FGH a 0 x 2a 2 + IJ K 1 4 1 4 a + a dx 3 3 LM 1 x a + 1 a x + 1 a xOP = a . 3 3 N3 Q Example 4. Evaluate zzz b 2x + yg dx dy dz, where V is a 3 2 = 4 4 5 0 Z V the closed region bounded by the cylinder z = 4 – x2 and the planes x = 0, y = 0, y = 2 and z = 0. Sol. Here a column parallel to z-axis is bounded by the plane z = 0 and the surface z = 4 – x2 of the clinder. This cylinder z = 4 − x2 meets the z-axis x = 0, y = 0 at (0, 0, 4) and the x-axis y = 0, z = 0 at (2, 0, 0) in the given region. Therefore, it is evident that the limits of integration for z are from 0 to 4 –x2, for y from 0 to 2 and for x from 0 to 2. Hence, the given integral = = = = = = z z z b g z z b g z z b ge j e j z z e j OPQ z LMN e j z 2 4− x 2 2 x=0 y=0 z=0 2 2 x=0 y=0 2 2 x=0 y=0 2 2 x=0 y=0 2 x=0 2 0 z = 4–x O z=0 2 (2, 0, 0) dxdy (0, 2, 0) Y 2x + y dx dy dz 2x + y (0, 0, 4) Fig. 4.45 4− x 2 z 0 dx dy 2x + y 4 − x 2 dx dy 8x − 2x 3 + 4 − x 2 y dx dy 8xy − 2x 3 y + 2 1 4 − x 2 y 2 dx 2 0 16 x − 4 x 3 + 2 4 − x 2 dx LM8x − x + 8x − 2 x OP 3 Q N 16 I 80 F . = G 32 − 16 + 16 − J = H 3K 3 2 = 2 4 3 0 EXERCISE 4.5 e x + y + z dx dy dz. LMAns. 8 log 2 − 19 OP 9Q N 3 S = [(x, y, z) : x2 + y2 + z2 ≤ 1, x ≥ 0, y ≥ 0, z ≥ 0]. LMAns. 1 OP N 48 Q 1. Evaluate the integral 2. Evaluate zzz S z zz log 2 x x + log y 0 0 0 xyz dx dy dz , where X 315 MULTIPLE INTEGRALS 3. Evaluate zzz x 2 + y 2 dx dy dz, where S is the solid bounded by the surfaces x2 + y2 = z2, π Ans. 6 LM N S z = 0, z = 1. 4. Evaluate 5. Evaluate z z zzz 1 y = 1− x 2 x=0 y=0 V z z= z=0 OP Q LMAns. –5 OP N 48 Q e1− x − y j xyz dz dy dx. 2 2 xy 2 dx dy dz, where V is the region bounded between the xy-plane and the LMAns. π OP N 24 Q sphere x2 + y2 + z2 = 1. 6. Evaluate zzz zzz V dx dy dz over the region V enclosed by the cylinder x2 + z2 = 9 and the planes Ans. 18π x = 0, y = 0, z = 0, y = 8. 7. Evaluate a x x+y x+y+z e 0 0 0 [Ans. 8. Evaluate 9. Evaluate 10. Evaluate 4.16 dx dy dz and state precisely what is the region of integration. e j 1 4a e − 6e 2 a + 8 e a − 3 . Region of integration is the volume enclosed by 8 the planes x = a, y = 0, y = x , z = 0 and z = x + y] LMAns. 3 log 3 − 19 OP 9Q N 8 LMAns. 2πa e 2 − 1jOP N 32 Q LMAns. 5a π OP 64 Q N z zz zzz z z z log 2 x x + log y x + y + z e dx dy dz . 0 0 0 2π 0 π a 2 4 r sin θ dr dθ dθ. 0 0 π 2 a sin θ 0 0 3 3 a 2 −r 2 a r dθ dr dz . 0 APPLICATION TO VOLUME (TRIPLE INTEGRALS) Volume of a solid contained in the domain V is given by the triple integral with f (x, y, z) = 1 Hence and V= zzz dx dy dz Volume in cylindrical coordinates, V = Volume in spherical polar coordinates, V = zzz zzz r dr dθ dz r 2 sin θ dr dθ dφ Example 1. Find the volume bounded by the elliptic paraboloids Sol. We have z = x2 + 9y2 and z = 18 – x2 – 9y2. z = x2 + 9y2 z = 18 – x2 – 9y2 (U.P.T.U., 2006) ...(i) ...(ii) 316 A TEXTBOOK OF ENGINEERING MATHEMATICS—I y2 =1 ...(iii) 1 3 The projection of this volume on to xy plane is the plane region D enclosed by ellipse (iii) as shown in the Figure 4.46. Here z varies from z1(x, y) = x2 + 9y2 to z2(x, y) = 18 – x2 – 9y2 From eqns. (i) and (ii), we get x2 + 9y2 = 9 ⇒ y varies from y1 (x) = − 9 − x2 to y2(x) = 9 x2 + 2 9 − x2 9 x varies from – 3 to 3. Thus, the volume V bounded by the elliptic paraboloids zz z zz zz j z e ze j a f z bx , yg V = –3 y axf z bx , yg dz dy dx 3 y ax f 2 2 2 2 = –3 y ax f e18 − x − 9 y j − ex + 9 y j dy dx 3 y ax f 2 2 = 2 –3 y a x f e9 − x − 9 y j dy dx 3 y2 x 2 1 1 Z 2 z = x + 9y 2 2 Z = 18 – x – 9y 1 2 –3 9 y − x y − 3y 3 9 −x 2 9 9− x 2 − 9 3 3 9 − x 2 2 dx –3 O dx 8 9 x = 3 cos θ so dx = – 3 sin θ dθ = Putting 2 = 72 z π z 0 = 144 2 = 2 3 2 1 D X (3, 0, 0) 2 2 Y (0,1,0) x + 9y = 9 Fig. 4.46 sin 4 θ dθ π 2 0 sin 4 θ cos 0 θ dθ 5 1 = 144 ⋅ 2 2 2 3 3 1 ⋅ ⋅π 2 2 ⋅ 144 = 2⋅ 2 = 27π. Example 2. Calculate the volume of the solid bounded by the surface x = 0, y = 0, x + y + z = 1 and z = 0. (U.P.T.U., 2004) Sol. Here x = 0, y = 0, z = 0 and x + y + z = 1 and z varies from 0 to 1– x – y y varies from 0 to 1 – x x varies from 0 to 1 317 MULTIPLE INTEGRALS zzz z z z FGH z LMN z FGH dx dy dz = V = 1 1− x 0 0 dx = 1 = 0 dx z z z z z 1 0 dx 1− x − y dy z 0 y2 y − xy − 2 I JK 1− x 1− x − y 0 0 dy 1 1− x 0 0 = dx dz b dy 1 − x − y g 1− x 0 a f 12 a1 − xf OPQ F1 x I 1 x I 1 − x − x + x − + x − J dx = z G − x + J dx = 2 2K H2 2 K L x x + x OP = 1 − 1 + 1 = 1 . = M − MN 2 2 6 PQ 2 2 6 6 1 = 0 2 dx 1 − x − x 1 − x − 1 2 2 2 1 0 0 3 1 2 0 Example 3. Find the volume of a solid bounded by the spherical surface x2 + y2 + z2 = 4a2 and the cylinder x2 + y2 – 2ay = 0. Sol. We have x2 + y2 + z2 = 4a2 ...(i) x2 + y2 – 2ay = 0 ...(ii) Changing it in polar coordinates x = r cos θ, y = r sin θ From (i) z2 = 4a2 – r2 From (ii) r2 = 2ar sin θ = 0 ⇒ r = 2a sin θ Hence, r varies from r = 0 to r = 2a sin θ z varies from z = 0 to z = 4a 2 − r 2 π θ varies from θ = 0 to θ = 2 and ∴ V= zzz z z π dx dy dz = 4 2 0 z z LM z MN e ze π = 4 2 dθ 0 2 a sin θ 0 dθ r dr z 2 a sin θ r =0 rdr 4a 2 − r 2 z 4a 2 − r 2 z z LM z MN e ze π = 4 2 dθ 0 0 Volume in 4 quadrant dz z= 0 2 a sin θ 0 r dr 4a 2 − r 2 O O j PP = 43 − 4a − 4a sin θj + 8a PP dθ Q Q 8 × 4a = 4 – 8a cos θ + 8 a j dθ = 1 − cos θj dθ 3 3 π 32a 3 3 I F 1 2 G 1 – cos 3θ − cos θJ dθ = K 3 z0 H 4 4 π = 4 2 dθ − 0 π 2 0 3 2 a sin θ 1 4 a2 − r 2 2 3 3 LM N 32 a L π 2 O = − . 3 MN 2 3 PQ 3 0 π 2 0 3 3 OP Q π π 2 0 LM N 2 2 3 32a 3 1 3 32a 3 π 1 3 2 θ − sin 3θ − sin θ = + − = 3 12 4 3 2 12 4 0 3 2 OP Q 3 2 3 318 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 4. Find the volume of the cylindrical column standing on the area common to the parabolas x = y2, y = x2 as base and cut off by the surface z = 12 + y – x2. (U.P.T.U., 2001) Sol. Volume = = = = z z z z ze z FGH z FGH x 1 dx 0 x dy 2 1 0 dx x 12 + y − x 2 0 x2 dx 12y + 1 j I JK y2 − x2 y 2 x 2 y =x x2 5 x x4 x + − x 2 − 12x 2 − + x4 12 0 Y 12 + y − x 2 dy 1 0 dz 2 2 I dx JK F x 2 x x I − x − 4x − + = G 8x + J 4 7 10 5K H 3 2 7 2 2 3 5 5 2 x =y 1 X O 0 Fig. 4.47 1 1 1 2 569 + = . = 8 + − − 4− 10 5 4 7 140 Example 5. Find the volume enclosed between the cylinders x2 + y2 = ax and z2 = ax. Sol. We have x2 + y2 = ax z2 = ax = a ax − x 2 0 – ax − x 2 dx a ax − x 2 0 – ax − x 2 dx = 2 = 2 z z 0 – ax − x 2 FH a z ax − ax af ax dy ax O 2 IK z a z z π 2 a sin 2 θ 0 = 8a 3 π 2 sin 3 θ cos2 θ dθ 0 = 8a 3 0 a − a sin 2 θ ⋅ 2a sin θ cos θ dθ 3 3 3 3 2 = 4a 2 = 16 a ⋅ 5 3 3 15 7 ⋅ 2 2 2 2 2 2 2 Y x a − x dx Put x = a sin2 θ so that dx = 2a sin θ cos θ dθ = 4 a X (a, 0) x + y = ax ax dx 2 ax − x 2 = 4 a 0 dz dy z − ax ax − x 2 ax −x 2 ax dx y − ax− x 2 0 dy a = 2 dx a Z dx dy dz ax = zzz z z z z z z z 2 = V= Fig. 4.48 319 MULTIPLE INTEGRALS Example 6. Find the volume bounded above by the sphere x2 + y2 + z2 = a2 and below by the cone x2 + y2 = z2. ...(i) Sol. We have x2 + y2 + z2 = a2 x2 + y2 = z2 ...(ii) Let x = r cos θ, y = r sin θ From eqns. (i) and (ii) z2 = a2 – r2 and z2 = r2 Here z varies from r to r varies from 0 to a2 − r2 a r 2 = a2 – r 2 ⇒ r = 2 θ varies from 0 to 2π ∴ Volume V = = zz z zz a 2 2π 0 0 2π a 0 0 a2 − r 2 r 2 b dz ⋅ r dr dθ g ( z)r a − r r dr dθ = 2 2 zz d 2π a 0 2 0 a 2 i a 2 − r 2 − r ⋅ r dr dθ OP LM O LM F a I a −r j e a P 1 1 = P M z MM −2 ⋅ FH 3IK − 3 r PP dθ = 3 z MMN−GH 2 JK + a − 2 2 PPQ dθ Q N 2 1 F 1 I 1 F 1 I θf = a G 1 − a G1 − a = J J ⋅ 2π 3 H 3 H 2K 2K 2 2π a 3 2 2 2 2π 3 0 2 3 2 3 3 0 0 2π 0 3 3 e2 − 2 j πa3 . 3 = Example 7. Find the volume bounded above by the sphere x2 + y2 + z2 = a2 and below by the cone x2 + y2 = z2. Sol. The equation of the sphere is x2 + y2 + z2 = a2 ...(i) 2 2 2 ...(ii) and that of the cone is x + y = z In spherical coordinates x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ From eqn. (i) (r sin θ cos φ)2 + (r sin θ sin φ)2 + (r cos θ)2 = a2 Z ⇒ r2 sin2 θ cos2 φ + r2 sin2 θ sin2 φ + r2 cos2 θ = a2 ⇒ r2 sin2θ (cos2 φ + sin2 φ) + r2 cos2 θ = a2 ⇒ r2 sin2 θ + r2 cos2 θ = a2 2 2 2 2 x +y +z =a ⇒ r2 (sin2 θ + cos2 θ) = a2 ⇒ r2 = a2 Y ⇒ r=a O From (ii) (r sin θ cos φ)2 + (r sin θ sin φ)2 = (r cos θ)2 2 2 2 ⇒ r2 sin2 θ (cos2 φ + sin2φ) = r2 cos2 θ x +y =z X ⇒ r2 sin2 θ = r2 cos2 θ ⇒ tan2 θ = 1 Fig. 4.49 320 A TEXTBOOK OF ENGINEERING MATHEMATICS—I ⇒ π 4 tan θ = 1 ⇒ θ = ± π 4 The volume in the first octant is one-fourth only. Limits in the first octant r varies 0 to a, θ π π from 0 to and φ from 0 to 4 2 r = a and θ = ± Thus V zzz z z LMMN r3 OPPQ L 1 + 1OP a 4a 4a φg M− = dφ − cos θ = b = 4 z dφ z sin θ dθ ⋅ z 3 3 3 N 2 Q 4a π 2 F 1 IJ . 2 − 1j = π a G 1 − e = 2 3 H 2K 3 2 π π π π a 2 2 4 2 dφ 4 sin θ dθ = 4 r sin θ dr d θ d φ 4 = 0 0 0 0 0 ∴ π 2 0 π 4 0 3 3 3 0 π 4 0 π 2 0 a 3 3 π 2 0 3 Spherical coordinates: Fig. 4.50 Example 8. Find by triple integration, the volume of the paraboloid of revolution x2 + y2 = 4z cut off by the plane z = 4. (U.P.T.U., 2005) Sol. By symmetry, the required volume is 4 times the volume in the positive octant. The section of the paraboloid by the plane z = 4 is the circle x2 + y2 = 16, z = 4 and its projection on the xy-plane is the circle x2 + y2 = 16, z = 0. The volume in the positive octant is bounded by z = x = 0, x = 4. z z zd i 4 Volume V = 4 0 0 16 − x 2 4 x +y 2 2 dz dy dx = 4 4 zz x2 + y2 , z = 4, y = 0, y = 4 16 − x 2 4 0 0 z 4 dy dx x2 + y 2 d i 4 F 4 − x + y I dy dx = 4 LMF 4 − x I y − 1 ⋅ y OP = 4z z GH 4 JK z MNGH 4 JK 4 3 PQ LF x I O 1 = 4 z MG 4 − J 16 − x − e16 − x j P dx 12 MNH 4 K PQ 4 2 16 − x 2 2 0 0 4 0 4 2 16 − x 2 3 dx 0 2 2 16 − x 2 and 3 2 2 0 321 MULTIPLE INTEGRALS LM 1 16 − x 16 − x − 1 16 − x OP dx e jP 12 MN 4 e j Q 1 2 = 4 16 − x j dx = e e16 − x j dx 6 3 z z za f z = 4 4 2 3 2 2 z 0 4 0 4 2 32 0 π = 3 2 2 2 2 2 32 ⋅ cos 3 θ ⋅ 4 cos θ dθ, where x = 4 sin θ 16 0 3 π 512 2 cos 4 θ dθ 3 0 512 3 1 π ⋅ ⋅ ⋅ = 3 4 2 2 = 32π. = Example 9. Find the volume of the solid under the surface az = x2 + y2 and whose base R is the circle x2 + y2 = a2. (U.P.T.U., 2008) Sol. We have az = x2 + y2, x2 + y2 = a2 Here x varies from 0 to a y varies from 0 to a2 − x2 2 x2 + y2 . a By symmetry, the required volume is 4 times the volume in positive octant. and z varies from 0 to z z z z LMMN OPPQ V= 4 a a2 −x2 x=0 y=0 y3 4 a 2 = x y+ 3 a 0 x2 +y2 a z=0 0 z zz e j 2 2 x +y =a 2 4 a a −x 2 x + y 2 dx dy a 0 0 dx dy dz = a 2 − x2 2 az=x +y LM MM N 2 2 e a2 − x2 4 a 2 2 dx = x a − x2 + 3 a 0 Putting x = a sinθ i.e., dx = a cosθ dθ z j OPdx PP Q 32 z Fig. 4.51 LM O cos θ dθP N Q 3. π O 4 La . 3 2 3 2 a . 5 2 1 2O 4a L 1 1 = ⋅ π+ + M P = M a NM 2 3 3. 2 3 QP 4 N 2 2 3. 2. 2 PQ L 2π O πa . = a M P = N4Q 2 = a4 4 4 π2 2 a sin θ cos 2 θ dθ + 0 a 3 4 3 4 π 2 4 0 3 3 Example 10. A triangular prism is formed by planes whose equations are ay = bx, y = 0 and x = a. Find the volume of the prism between the plane z = 0 and surface z = c + xy. [U.P.T.U. (C.O.), 2003] Sol. Here x varies from 0 to a bx y varies from 0 to a 322 A TEXTBOOK OF ENGINEERING MATHEMATICS—I z varies from 0 to c + xy Hence, the volume is zz z z LMMN OPPQ V = a bx a c + xy 0 0 0 xy 2 cy + 2 a = 0 bx a dx = 0 zz b g a bx a dx dy dz = z 0 0 a 0 c + xy dx dy F bcx + b x I dx GH a 2a JK 2 3 2 L bcx + b x OP = bca + b a = ab b 4c + abg. = M MN 2a 8a PQ 2a 8a 8 2 2 4 a 2 2 4 2 2 0 1. Find zzz V EXERCISE 4.6 x 2 yz dx dy dz , where V is the volume bounded by the surface x2 + y2 = 9, LMAns. 648 OP 5 Q N z = 0, z = 2. 2. Compute the volume of the solid enclosed between the two surfaces elliptic paraboloids z = 8 – x2 – y2 and z = x2 + 3y2. [ Hint: Projection of the volume on to xy - plane is the ellipse x 2 + 2 y 2 = 4, so limits are z ; x 2 + 3 y 2 to 8 – x 2 – y 2 ; y : ± . 4 − x2 , x: ± 2] 2 Ans. 8π 2 3. Compute the volume of the solid bounded by the plane 2x + 3y + 4z = 12, xy-plane and the cylinder x2 + y2 = 1. LMHint: Limits z: 0 to 1 b12 − 2x − 3yg, x:−1 to 1, y : ± 1 − x OP ⋅ 4 N Q 2 Ans. 3π 4. Find the volume of the solid common to two cylinders x2 + y2 = a2, x2 + z2 = a2. LMHint: z: a − x , y : a − x OP ⋅ PQ MN ± 2 ± 2 x : ±a 2 2 LMAns. 16 a OP 3 PQ MN 3 5. Find the volume of the cylindrical column standing on the area common to the parabolas y2 = x, x2 = y and cut off by the surface z = 12 + y – x2. (U.P.T.U., 2001) LMAns. 566 OP N 140 Q 6. A triangular prism is formed by planes whose equations are ay = bx, y = 0 and x = a. Find the volume of the prism between the planes z = 0 and surface z = c + xy. [U.P.T.U. (C.O.), 2003] LMAns. ab a4c + abfOP 8 N Q 323 MULTIPLE INTEGRALS Ans. π 7. Compute the volume bounded by xy = z, z = 0 and (x – 1)2 + (y – 1)2 = 1. 8. Evaluate zzz zzz 2 z yz 0 1 0 LMAns. 7 OP N 2Q LMAns. FG π − 1IJ OP N H 2 KQ xyz dx dy dz . π π xy z 2 2 cos dz dy dx . 0 x 0 x 9. 10. Compute the volume of the solid bounded by x2 + y2 = z, z = 2x. Ans. 2π 11. Find the volume of the region bounded by the paraboloid az = x2 + y2 and the cylinder LMAns. πR OP 2a Q N 4 x2 + y2 = R2. 12. Find the volume of the tetrahedron by the coordinate planes and the plane 4.17 x y z + + = 1. a b c abc Ans. 6 LM N OP Q DRITCHLET’S* THEOREM (U.P.T.U., 2005) If V is a region bounded by x ≥ 0, y ≥ 0 and x + y + z ≤ 1, then zzz V blg amf amf bl + m + n + 1g xl −1 y m −1z n−1dx dy dz = The given triple integral may be written as Ι = zz z 1 1− x 1− x − y 0 0 0 x l −1 y m −1 z n−1 dx dy dz Z C z = 1–x–y @y = @x @y @z O z=0 B Y Fig. 4.52 * Peter Gustav Lyeune (1805–1859), German Mathematician. x+y= 1 A X 324 A TEXTBOOK OF ENGINEERING MATHEMATICS—I LM z OP dx dy NnQ x y b1 − x − yg dx dy zz zz z z ma f r ma f a f r a f z a fa f z a fa f 1 1 − x l −1 m −1 = x 0 0 n 1 −x − y y 0 1 1 − x l −1 m −1 Put 1 n n 0 0 y = (1 – x) t, so that dy = (1 – x) dt Then I = = 1 1 1 l −1 m n x 1 − x t ⋅ 1 − x − 1 − x t 1 − x dx dt n 0 0 1 1 1 l −1 m + n−1 −1 n +1 −1 x 1−x dx tm −1 1 − t dt = 0 n 0 1 = β l, m + n + 1 ⋅ β m, n + 1 n a f a f alf ⋅ am + n + 1f ⋅ amf an + 1f al + m + n + 1 f am + n + 1 f 1 alf ⋅ amf an + 1f = n al + m + n + 1f 1 alf ⋅ amf ⋅ n anf = n al + m + n + 1f 1 = n alf ⋅ amf anf a l + m + n + 1f = zzz Hence, V x l − 1 y m − 1 z n − 1 dx dy dz = 3 an + 1f = n anf alf ⋅ amf anf al + m + n + 1f where V is the region given by x ≥ 0, y ≥ 0, z ≥ 0 and x + y + z ≤ 1. This integral is known as Dritchlets integral. This is an important integral useful in evaluating multiple integrals. Remark: It is to be noted that S is a region bounded by x ≥ 0, y ≥ 0 and x + y ≤ 1 then zz S x l −1 y m−1 dx dy = = l m l+m+1 . Note 1: The above integral can be generalised for more than three variables, i.e., zz z ... x1 1 − ⋅ x2 2 − ⋅ x3 3 − ... xn pn −1 dx1 dx2 dx3 ... dxn = p 1 p 1 1 p bp g bp g b p g − − − bp g 1 2 n 3 b p + p + p + − − − − + p + 1g 1 2 3 n where the region of integration is given by xi = 1, 2, ...., n; such that x1 + x2 + .... + xn ≤ 1. Note 2: If the region of integration be x ≥ 0, y ≥ 0, z ≥ 0 and x + y + z ≤ h, then zzz V x l −1 y m −1 z n−1 dx dy dz = blg amf anf bl + m + n + 1g h. MULTIPLE INTEGRALS Example 1. Find the value of zzz 325 x l −1 y m−1 z n −1 dx dy dz where x, y, z are always positive with FG x IJ + FG y IJ + FG z IJ ≤ 1. H aK H bK H cK FG x IJ = u, FG y IJ = v, FG z IJ = w, we have Sol. Putting H aK H bK H cK q p r p q (U.P.T.U., 2005) r x = au1/p, y = bv1/q, z = cw1/r, F a I uFGH IJK du, dy = F b I vFGH IJK dv, dz = FG c IJ w FGH IJK dw GH pJK GH q JK H rK F I F I F I F aIu ⋅ b v ⋅ c w ∴ The given integral = zzz GGH au JJK GGHbu JJK GH cw JK GH pJK q r ∴ 1 −1 p dx = 1 p al bm c n = pqr l −1 1 q m−1 1 −1 r n− 1 1 r 1 −1 p 1 −1 q 1 −1 r du dv dw F l − 1I m − 1 F n − 1I G p JK q GH r JK uH v w du dv dw , where u + v + w ≤ 1 zzz F l I F mI F nI GH pJK GH q JK H r K F l + m + n + 1I . GH p q r JK l m n = 1 −1 q ab c pqr x 2 y2 z2 + + = 1. a 2 b2 c 2 Sol. The volume in the positive octant will be Example 2. Find the volume of the ellipsoid V= zzz For points within positive octant, Put x2 a2 dx dy dz x 2 y2 z2 + + ≤ 1. a 2 b2 c 2 = u or x = a u , y = b v , z = c w , 1 ∴ 1 1 dx = a −2 b − c − du, dy = v 2 dv, dz = w 2 dw u 2 2 2 V= abc 8 zzz abc ⋅ = 8 FG 1 −1IJ FG 1 −1IJ FG 1 −1IJ K H2 K H2 K uH 2 v w F 1I ⋅ F 1I ⋅ F 1I H 2K H 2K H 2K F1 + 1 + 1 + 1 I H 2 2 2K e j 3 π π abc abc = = 3 1 8 6 ⋅ ⋅ π 2 2 du dv dw , where u + v + w ≤ 1 326 A TEXTBOOK OF ENGINEERING MATHEMATICS—I ∴ π abc 4 = π abc. 6 3 Total volume = 8 × x y z + + = 1 meets the axis in A, B and C. Apply Dritchlet’s integral a b c to find the volume of the tetrahedran OABC. Also find its mass if the density at any point is kxyz. (U.P.T.U., 2004) Example 3. The plane y x z = u, = v, = w, then u ≥ 0, v ≥ 0, w ≥ 0 and u + v + w ≤ 1 a c b Also, dx = a du, dy = b dv, dz = c dw. Sol. Let Volume OABC = = zzz zzz D abc du dv dw , where u + v + w < 1 zzz D = abc dx dy dz u1 −1 v 1−1 w 1 −1 du dr dw D a1f a1f a1f = abc = abc a1 + 1 + 1 + 1f 3! 6 Mass = zzz kxyz dx dy dz = zzz k a aufabv facw f abc du dv dw = abc D 2 2 2 = ka b c zzz 2 2 2 = ka b c D′ D′ 2 u 2−1 v 2−1 w 2 −1 du dv dw 2 2 a2 + 2 + 2 + 1f = ka 2 b 2 c 2 1 ! 1 ! 1 ! ka 2 b 2 c 2 = . 6! 720 x 2 y2 z2 Example 4. Find the mass of an octant of the ellipsoid 2 + 2 + 2 = 1, the density at any a b c point being ρ = kxyz. (U.P.T.U., 2002) Sol. Putting x2 y2 z2 = u, = v, = w and u + v + w = 1 a2 c2 b2 So that mass 2xdx a2 = du, ρdv = zzz = = = dv, 2zdz = dw, then c2 zzz b g zzz a fb ga f z z z FGH IJK FGH IJK FGH IJK zzz zzz b 2 kxyz dx dy dz = k Mass = k = 2 ydy a 2 du 2 b 2 dv 2 xdx ydy zdz ⋅ c 2 dw 2 ka 2 b 2 c 2 8 du dv dw , where u + v + w ≤ 1 ka 2 b 2 c 2 8 u1−1 v1 −1 w 1 −1 du dv dw = ka 2 b 2 c 2 ka 2 b 2 c 2 = . 8×6 48 ka 2 b 2 c 2 1 1 1 8 3+1 327 MULTIPLE INTEGRALS Example 5. Find the volume of the solid surrounded by the surface FG x IJ + FG y IJ + FG z IJ = 1. H aK H bK H cK 2 3 2 3 2 3 (U.P.T.U., 2007) 3a F xI Sol. Let u = G J ⇒ x = au ⇒ dx = u du H aK 2 F yI 3b v = G J ⇒ y = bv ⇒ dy = v dv H bK 2 3c F zI w = G J ⇒ z = cw ⇒ dz = w dw H cK 2 2 3 ∴ V= 8 3 2 1 2 2 3 3 2 1 2 2 3 3 2 1 2 zzz dx dy dz = 8 = 27 abc zzz 3 zzz 3 −1 1 1 1 27abc 2 2 2 u ⋅ v w du dv dw 8 3 −1 −1 u 2 , v 2 , w 2 du dv dw 3 3 3 ⋅ 2 2 2 = 27 abc ⋅ 3 3 3 + + +1 2 2 2 1 1 1 1 1 1 ⋅ ⋅ ⋅ ⋅ = 27 abc ⋅ 2 2 2 2 2 2 11 2 3 π2 = 27 abc ⋅ 9 7 5 3 1 8⋅ ⋅ ⋅ ⋅ ⋅ π 2 2 2 2 2 = 4πabc . 35 Example 6. Find the value of zzz b planes and the plane x + y + z = 1. Sol. Here x varies from 0 to 1 y varies from 0 to 1 – x z varies from 0 to 1 – x – y Thus zz z b 1 1 −x 1 −x − y 0 0 0 1 x + y+ z+1 g 3 dx dy dz 1 x + y+ z+1 g 3 dx dy dz the region bounded by coordinate 328 A TEXTBOOK OF ENGINEERING MATHEMATICS—I z LM OP 1 = − MN 2bx + y + z + 1g PQ dx dy OP 1 L 1 1 = − M − dx dy 2 M bx + y + 1 − x − y + 1g P + + x y 1 b g N Q LM 1 OP 1 1 − = − 2 MN 4 bx + y + 1g PQdx dy I dx = − 1 FG 1 − x + 1 − 1 IJ dx 1 Fy 1 = − + G 2 H 4 bx + y + 1g JK 2 H 4 2 1 + xK O 1 L1 1 L − a1 − x f x 1O = − M + − log a1 + xfP = M − log 2 + P 2 MN 8 2 2 2 8Q N PQ zz zz 1− x− y 1 1− x (0,0,1) 2 0 0 0 1 1− x 2 0 0 zz z 2 B O 1 1 −x 2 0 0 z 1− x 1 0 A (1,0,0) x 1 y (0,1,0) Fig. 4.53 0 0 1 2 0 = 1 5 . log 2 − 2 16 Example 7. Find the area and the mass contained m the first quadrant enclosed by the curve FG x IJ + FG y IJ = 1, where α > 0, β > 0 given that density at any point ρ (xy) is k xy . H aK H bK β α [U.P.T.U., 2008] Sol. The area of the plane region is A = zz dx dy S FG x IJ = u or x = au H aK FG y IJ = v or y = bv H bK 1 α Let 1/α so dx = a α −1 ⋅ du u α 1/β so dy = b β −1 ⋅ dv v β 1 β ∴ A = zz s A = 1 1 ab α − 1 β − 1 u dudv .v αβ ab 1 α 1 β ⋅ αβ 1 1 + +1 α β As zz x l −1 y m−1dx dy = s Now the total mass M contained in the plane region A is : M = zz b g ρ x , y dx dy = S zz k xy dx dy l m l +m+1 329 MULTIPLE INTEGRALS = k zz 1 2 au α⋅ S′ aabf = k 32 αβ zz αβ 3 u 2α ab α − 1 β −1 ⋅ ⋅v ⋅ du dv u αβ ⋅v 2β −1 1 3 −1 1 du dv S′ aabf ⋅ = k 32 bv 1 2β 3 3 2α 2β . 3 3 + +1 2α 2β Example 8. Find the mass of the region in the xy-plane bounded by x = 0, y = 0, x + y = 1 with density k xy . Sol. Mass M contained in the plane region is L y OP 2k M = kz z xy dx dy = k z x M dx = x . a1 − x f dx MN 3 2 PQ 3 z 2k = . a1 − xf x dx 3 2k F 3 5 I 2k 3 2 5 2 2k 1 2. π ⋅ 3 2 ⋅ 1 2. π kπ = ⋅ = ⋅ = βG , J = . 3 H 2 2K 3 3 3 ⋅ 2 ⋅1 24 4 1 1− x 1 1 2 0 0 0 3 2 1− x 1 1 2 3 2 0 0 z 1 3 −1 2 0 5 −1 2 EXERCISE 4.7 zzz x 2 y2 z2 + + ≤ 1. a 2 b2 c 2 2. Find the volume enclosed by the surface 1. Evaluate dx dy dz , where FG x IJ + FG y IJ + FG z IJ H aK H b K H cK 2n 2n 2n = 1. LMAns. πabc OP 6 Q N LMAns. 1 a b c OP N 6 Q 2 2 2 3. Find the volume of the solid bounded by coordinate planes and the surface FG x IJ + FG y IJ + FG z IJ H aK H bK H cK 1 2 1 2 1 2 = 1. LMAns. abc OP N 90 Q 4. Find the volume of the solid bounded by coordinate planes and the surface LM FG 1 IJ OP MMAns. abc ⋅ H 2n K PP MM 12 n 23n PP N Q 3 FG x IJ + FG y IJ + FG z IJ H aK H b K H cK 2n 2n 2n = 1, n being a positive integer. 2 330 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 5. Evaluate zzz x α −1 ⋅ y β −1 z ϒ−1 dx dy dz , where V is the region in the first octant bounded by V the sphere x2 + y2 + z2 = 1 and the coordinate planes. [U.P.T.U. (C.O.), 2003] LM α β γ O MMAns. α2 2β 2ϒ PPP MN 8 2 + 2 + 2 PQ FG x IJ 2m + FG y IJ 2n = 1 m, n being positive integers. H aK H bK LMAns. ab ⋅ 1 2m 1 2m OP NM 4mn 1 2 m + 1 2n + 1 QP LM 1 1 O P 4 10 ab y x ⋅ 4 10 P Ans. M 7. Find the area enclosed by the curve FG IJ + FG IJ = 1. 27 H aK H b K MM 40 P N 20 QP 6. Find the area enclosed by the curve OBJECTIVE TYPE QUESTIONS A. Pick the correct answer of the choices given below : 1. The volume of the solid under the surface az = x2 + y2 and whose R is the circle x2 + y2 = a2 is given as [UPTU. 2008] π 2a 4 3 πa (iii) 3 (ii) (i) 2. The value of triple integral zz z 1 1 πa 3 2 (iv) None of these 1− x 0 y2 0 x dx dy dz is 4 5 (ii) − 35 38 −17 35 (iii) (iv) 35 4 3. The volume bounded by the parabolic cylinder z = 4 – x2 and the planes x = 0, y = 0, y = 6 and z = 0 is (i) 32.5 (ii) 56 (i) (iii) 33 (iv) 32 4. The area which is inside the cardioid r = 2 (1 + cosθ) and the outside the circle r = 2 is : (i) π + 8 (ii) π – 8 (iii) 3π – 7 (iv) None of these 331 MULTIPLE INTEGRALS B. Fill in the blanks: 1. The volume of the solid bounded by the surfaces; z = 0, x2 + y2 = 1, x + y + z = 3 is .......... 2. In spherical coordinate system dx dy dz = .......... 3. In cylindrical coordinate system dx dy dz = .......... 4. The volume of region bounded by the parabolic cylinder x = y2 and the planes x = z, z = 0 and x = 1 is .......... 5. 6. 7. 8. zz z z zz π x 0 0 ∞ 0 y n−1 . e − λy dy = .......... π 2 0 sin y dy dx = .......... sin m θ. cosn θ dθ = .......... π aθ 0 0 r 3 dθ dr = .......... C. Indicate True and False for the following statements: 1. (i) Volume in cylindrical coordinates is (ii) Volume in spherical coordinates is zzz zzz rdrdθ dz . r 3 sin θ dr dθ dφ . (iii) The total volume of a solid bounded by the spherical surface x2 + y2 + z2 = 4a2 and the cylinder x2 + y2 – 2ay = 0 is multiplied by 4. (iv) In spherical coordinates F(x, y, z) = F(r sinθ cosφ, r sinθ sinφ, z). 2. (i) Parallelopiped, ellipsoid and tetrahedron are regular three dimensional domain. (ii) In the area of the cardioid r = a (1 + cos θ), r varies from 0 to a(1 + cos θ) and θ varies from – π to π. (iii) zz x l −1 . y m −1 dx dy = l m l +m+1 (iv) Triple integral is not generalization of double integral. D. Match the following: 1. (i) β (p, q) p q (a) FG 1 IJ H 2K zb g ∞ y p −1 (ii) p+q (b) (iii ) π (c) β (p, q) (iv) π sin pπ (d) 0 1+ y p 1− p p+ q ⋅ dy 332 2. (i) A TEXTBOOK OF ENGINEERING MATHEMATICS—I zz b g F x, y dxdy (a) s (ii) β(m, n) z 1 0 a f dx x m −1 1 − x n−1 (b) n n (iii) n+1 (c) nπ/sin nπ (iv) 1+n 1−n (d) zz a f F u, v J du dv s ANSWERS TO OBJECTIVE TYPE QUESTIONS A. Pick the correct answer: 1. (ii) 2. (i) 3. (iv) 2. r2 sin θ dr dθ dφ] 3. r dr dθ dz 6. 4. (i) B. Fill in the blanks: 1. 3π 4. 4 5 5. π 7. m+1 n+1 2 2 m+n+ 2 2 2 8. C. True and False: 1. (i) T 2. (i) T n λn a 4π 5 20 (ii) F (ii) T (iii) T (iii) T (iv) F (iv) F D. Match the following: 1. (i) → (b), (ii) → (c), (iii) → (a), (iv) → (d) 2. (i) → (d), (ii) → (a), (iii) → (b), (iv) → (c). GGG UNIT 8 Vector Calculus VECTOR DIFFERENTIAL CALCULUS The vector differential calculus extends the basic concepts of (ordinary) differential calculus to vector functions, by introducing derivative of a vector function and the new concepts of gradient, divergence and curl. 5.1 VECTOR FUNCTION If the vector r varies corresponding to the variation of a scalar variable t that is its length and direction be known and determine as soon as a value of t is given, then r is called a vector function of t and written as r = f (t) and read it as r equals a vector function of t. Any vector f (t) can be expressed in the component form f (t) = f1(t) i + f2 (t) j + f3 (t) k Where f1(t), f2 (t), f3 (t) are three scalar functions of t. r = 5t2 i + t j – t3 k For example, where f1(t) = 5t2, f2(t) = t, f3 (t) = – t3. 5.2 VECTOR DIFFERENTIATION Let r = f (t) be a single valued continuous vector point function of a scalar variable t. Let O be the origin of vectors. Let OP represents the vector r corresponding to a certain value t to the scalar variable t. Then r = f (t) ...(i) 333 334 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Let OQ represents the vector r + δ r corresponding to the value t + δt of the scalar variable t, where δt is infinitesimally small. Then, Q ...(iii) f ( t + δt ) − f ( t ) δr = δt δt Taking the limit of both side as δt → 0. lim δ r = lim f (t + δt) − f (t) t→0 δt→0 δt δt We obtain dr ® δ r = f (t + δt) – f (t) Dividing both sides by δt, we get ® dr — dt ® ...(i) r+ r + δ r = f (t + δt) Subracting (i) from (ii) P ® r O dr lim f (t + δt) − f (t) = d f (t) = δt→0 dt δt dt Fig. 5.1 . 5.3 SOME RESULTS ON DIFFERENTIATION H H d dr d s (r − s ) = − ⋅ dt dt dt H d( sr ) ds dr = r+s dt dt dt (a) (b) (c) H db da d( a ⋅ b) = a⋅ + ⋅b dt dt dt (d) d da db ( a × b) = ×b+a× dt dt dt (e) n s = dadt × (b × cH) + a × FGH dbdt × cIJK + a × FGH b × dcdt IJK ⋅ d a × (b × c) dt Velocity v = r = dr ⋅ dt dv d2 r ⋅ = dt dt 2 Example 1. A particle moves along the curve x = t3 + 1, y = t2, z = 2t + 5, where t is the time. Find the component of its velocity and acceleration at t = 1 in the direction i + j + 3k. Acceleration Sol. Velocity a = dr d = ( xi + y j + zk) dt dt d 3 (t + 1)i + t 2 j + (2t + 5) k = dt 2 = 3t i + 2t j + 2 k 335 VECTOR CALCULUS = 3i + 2 j + 2k , at t = 1. Again unit vector in the direction of i + j + 3 k is = i + j + 3k d 1 +1 + 3 i 2 2 2 = i + j + 3 k 11 Therefore, the component of velocity at t = 1 in the direction of i + j + 3 k is d3i + 2 j + 2ki ⋅ di + j + 3ki = 3 + 2 + 6 = 11 11 d2 r dt 2 Acceleration = 11 FG IJ = 6ti + 2 j = 6i + 2 j , at t = 1 H K d dr dt dt Therefore, the component of acceleration at t = 1 in the direction i + j + 3 k is d6i + 2 j i ⋅ di + j + 3k i = 6 + 2 = 8 · 11 11 11 Example 2. A particle moves along the curve x = 4 cos t, y = 4 sin t, z = 6t. Find the velocity and acceleration at time t = 0 and t = π/2. Find also the magnitudes of the velocity and acceleration at any time t. Sol. Let r = 4 cos t i + 4 sin t j + 6t k at t = 0, at t = π , 2 t = 0, π t = , 2 at at H v = 4 j + 6 k H v = −4i + 6 k |v| = 16 + 36 = 52 = 2 13 |v| = 16 + 36 = 52 = 2 13 . at d2 r = – 4 cos t i – 4 sin t j dt 2 a t = 0, = – 4 i ∴ at t = 0, Again, acceleration, a = π , 2 π t = , 2 at t = at |a| = a = |a| = (−4)2 = 4 – 4 j (−4)2 = 4. Example 3. If r = a ent + b e–nt, where a, b are constant vectors, then prove that d2 r dt 2 Sol. − n2 r = 0 nt −nt r = ae + be dr nt − nt = ae ⋅ n + be ⋅ ( −n) dt ...(i) 336 A TEXTBOOK OF ENGINEERING MATHEMATICS—I d2 r dt 2 = ae nt ⋅ n 2 + be − nt ⋅ ( − n)2 d2 r dt 2 = n 2 ae nt + be − nt = n2r [From (i)] d2 r – n 2 r = 0. Hence proved. dt 2 H H H Example 4. If a , b , c are constant vectors then show that r = at 2 + bt + c is the path of a point moving with constant acceleration. ⇒ 2 r = at + bt + c Sol. dr = 2at + b dt d2 r = 2a which is a constant vector. dt 2 Hence, acceleration of the moving point is a constant. Hence proved. EXERCISE 5.1 1. A particle moves along a curve whose parametric equations are x = e–t, y = 2 cos 3t, z = sin 3t. Find the velocity and acceleration at t = 0. [Hint: r = xi + yj + zk ] Ans. Vel. = 10 , acc. = 5 13 2. A particle moves along the curve x = 3t2, y = t2 – 2t, z = t2. Find the velocity and acceleration at t = 1. H Ans. v = 2 10 , a = 2 11 3. Find the angle between the directions of the velocity and acceleration vectors at time t of 2 Ans. arc cos t a particle with position vector r = (t 2 + 1)i − 2tj + (t 2 − 1)k . 2 2t + 1 LM MN OP PQ 4. A particle moves along the curve x = 2t2, y = t2 – 4t, z = 3t – 5 where t is the time. Find the components of its velocity and acceleration at time t = 1 in the direction i − 3 j + 2 k . LMAns. 8 14 ; − 14 OP 7 7 PQ MN 5. If r = (sec t) i + (tan t) j be the position vector of P. Find the velocity and acceleration of P at t = π . 6 LMAns. 2 i + 4 j, 2 (5i + 4 j)OP N 3 3 33 Q 337 VECTOR CALCULUS 5.4 SCALAR POINT FUNCTION If for each point P of a region R, there corresponds a scalar denoted by f(P), then f is called a ‘‘scalar point function’’ for the region R. Example 1. The temperature f(P) at any point P of a certain body occupying a certain region R is a scalar point function. Example 2. The distance of any point P(x, y, z) in space from a fixed point (x0, y0, z0) is a scalar function. (x − x0 ) 2 + ( y − y0 ) 2 + ( z − z 0 ) 2 ⋅ f(P) = Scalar field (U.P.T.U., 2001) Scalar field is a region in space such that for every point P in this region, the scalar function f associates a scalar f(P). 5.5 VECTOR POINT FUNCTION If for each point P of a region R, there corresponds a vector f (P) then f is called “vector point function” for the region R. Example. If the velocity of a particle at a point P, at any time t be f (P), then f is a vector point function for the region occupied by the particle at time t. If the coordinates of P be (x, y, z ) then f (P) = f1 (x, y, z) i + f2 (x, y, z) j + f3 (x, y, z) k. Vector field (U.P.T.U., 2001) Vector field is a region in space such that with every point P in the region, the vector function f associates a vector f (P). Del operator: The linear vector differential (Hamiltorian) operator ‘‘del’’ defined and ^ denoted as ∇ = i ∂ ^ ∂ ^ ∂ +k +j ∂x ∂y ∂z This operator also read nabla. It is not a vector but combines both differential and vectorial properties analogous to those of ordinary vectors. 5.6 GRADIENT OR SLOPE OF SCALAR POINT FUNCTION If f(x, y, z) be a scalar point function and continuously differentiable then the vector ∂f ∂f ∂f ∇ f = i ∂x + j ∂y + k ∂z is called the gradient of f and is written as grad f. Thus (U.P.T.U., 2006) ∂f ∂f ∂f grad f = i ∂x + j ∂y + k ∂z = ∇ f It should be noted that ∇f is a vector whose three components are is a scalar point function, then ∇f is a vector point function. ∂f ∂f ∂f , , ⋅ Thus, if f ∂x ∂y ∂z 338 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 5.7 GEOMETRICAL MEANING OF GRADIENT, NORMAL Consider any point P in a region throught which a scalar field (U.P.T.U., 2001) f(x, y, z) = c defined. Suppose that ∇f ≠ 0 at P and that there is a f = const. surface S through P and a tangent plane T; for instance, if f is a temperature field, then S is an isothermal surface (level surface). If n , at P, is choosen as any vector in the tangent plane T, then surely Normal N at P Z ^ N df must be zero. dS S P df Since = ∇f . n = 0 dS n^ for every n at P in the tangent plane, and both ∇f and n are non-zero, it follows that ∇f is normal to the tangent plane T and hence to the surface S at P. f = Constant If letting n be in the tangent plane, we learn that ∇f is normal to S, then to seek additional information about ∇f it seems logical to let n be along the normal line at P,. Then Y X Fig. 5.2 df df = , N = n then ds dN df = ∇f ⋅ N = ∇ f ⋅ 1 cos 0 = ∇ f ⋅ dN So that the magnitude of ∇f is the directional derivative of f along the normal line to S, in the direction of increasing f. Hence, ‘‘The gradient e∇ f j of scalar field f(x, y, z) at P is vector normal to the surface df in that direction. dN f = const. and has a magnitude is equal to the directional derivative 5.8 DIRECTIONAL DERIVATIVE ∂f ∂f ∂f are the derivatives (rates of change) of f , , ∂x ∂y ∂z in the direction of the coordinate axes OX, OY, OZ respectively. This concept can be extended to Let f = f(x, y, z) then the partial derivatives define a derivative of f in a "given" direction PQ . Let P be a point in space and b be a unit vector from P in the given direction. Let s be the are length measured from P to another point Q along the ray C in the direction of b . Now consider f(s) = f(x, y, z) = f{x(s), y(s), z(s)} Then ∂f dx ∂f dy ∂f dz df + + = ∂x ds ∂y ds ∂z ds ds s P ^ b Q C Fig. 5.3 ...(i) 339 VECTOR CALCULUS df is called directional derivative of f at P in the direction b which gives the rate of ds change of f in the direction of b. Here dx dy dz i+ j + k = b = unit vector ds ds ds Eqn. (i) can be rewritten as Since, FG i ∂f + j ∂f + k ∂f IJ ⋅ FG dx i + dy j + dz kIJ H ∂x ∂y ∂z K H ds ds ds K LF ∂ + y ∂ + k ∂ IJ f OP ⋅ b = ∇f ⋅ b df = MG i ds NH δx δy δz K Q ...(ii) df = ds ...(iii) Thus the directional derivative of f at P is the component (dot product) of ∇f in the direction of (with) unit vector b. Hence the directional derivative in the direction of any unit vector a is F I GH JK a df = ∇f · ds a Normal derivative df = ∇f ⋅ n , where n is the unit normal to the surface f = constant. dn 5.9 PROPERTIES OF GRADIENT Property I: ( a ⋅ ∇) f = a ⋅ (∇f ) Proof: L.H.S. = ( a ⋅ ∇) f RSa ⋅ FG i ∂ + j ∂ + k ∂ IJ UV f T H ∂x ∂y ∂z K W R ∂ ∂ ∂U = S( a ⋅ i ) ∂x + ( a ⋅ j ) ∂y + ( a ⋅ k ) ∂z V f T W = ∂f ∂f ∂f + ( a ⋅ k) = ( a ⋅ i) + ( a ⋅ j) ∂x ∂y ∂z ...(i) b g F ∂f ∂f ∂f I = a ⋅ G i ∂x + j ∂y + k ∂z J K H R.H.S. = a ⋅ ∇f ∂f ∂f ∂f = ( a ⋅ i ) ∂x + ( a ⋅ j ) ∂y + ( a ⋅ k ) ∂z From (i) and (ii), baH ⋅∇g f = aH ⋅ b∇ f g . ...(ii) 340 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Property II: Gradient of a Constant The necessary and sufficient condition that scalar point function φ is a constant is that ∇φ = 0. Proof: Let φ(x, y, z) = c ∂φ ∂φ ∂φ = = =0 ∂y ∂x ∂z Then, ∂φ ∂φ ∂φ +j + k ∂y ∂x ∂z = i.0 + j.0 + k.0 = 0. Hence, the condition is necessary. Conversely: Let ∇φ = 0. ∴ ∇φ = i ∂φ ∂φ ∂φ +j + k = 0.i + 0.j + 0.k. ∂y ∂x ∂z Equating the coefficients of i, j, k. ∂φ ∂φ ∂φ = 0. = 0, =0 On both sides, we get ∂y ∂x ∂z ⇒ φ is independent of x, y, z ⇒ φ is constant. Hence, the condition is sufficient. Then, or i Property III: Gradient of the Sum of Difference of Two Functions If f and g are any two scalar point functions, then ∇ (f ± g) = ∇f ± ∇g grad (f ± g) = grad f ± grad g Proof: ∇ (f ± g) = FG i ∂ + j ∂ + k ∂ IJ ( f ± g) H ∂x ∂y ∂z K ∂ ∂ ∂ = i ∂x ( f ± g ) + j ∂y ( f ± g) + k ∂z ( f ± g) FG ∂f ± ∂g IJ + j FG ∂f ± ∂g IJ + k FG ∂f ± ∂g IJ H ∂x ∂x K H ∂y ∂y K H ∂z ∂z K F ∂f + j ∂g + k ∂f IJ ± FG i ∂g + j ∂g + k ∂g IJ = Gi H ∂x ∂y ∂z K H ∂x ∂y ∂z K = i ∇(f ± g) = grad f ± grad g. Property IV: Gradient of the Product of Two Functions If f and g are two scalar point functions, then ∇(fg) = f∇g + g∇ f or grad (fg) = f (grad g) + g (grad f). Proof: ∇(fg) = FG i ∂ + j ∂ + k ∂ IJ ( fg) H ∂x ∂y ∂z K 341 VECTOR CALCULUS = i ∂ ∂ ∂ ( fg) + j ( fg) + k ( fg) ∂x ∂y ∂z FG ∂g + g ∂f IJ + j FG f ∂g + g ∂f IJ + k FG f ∂g + g ∂f IJ H ∂x ∂x K H ∂y ∂y K H ∂z ∂z K F ∂g + j ∂g + k ∂g IJ + g FG i ∂f + j ∂f + k ∂f IJ = f Gi H ∂x ∂y ∂z K H ∂x ∂y ∂z K = i f ∇(fg) = f∇g + g∇f. Property V: Gradient of the Quotient of Two Functions If f and g are two scalar point functions, then FG f IJ = g∇f − f∇g , g ≠ 0 H gK g F f I g(grad f ) − f (grad g) , ≠ 0. grad G g J = H K g F fI F ∂ ∂ ∂I FfI ∇ G g J = G i ∂x + j ∂y + k ∂z J G g J KH K H K H ∂ FfI ∂ FfI ∂ F fI +j +k = i J J G G G J ∂y H g K ∂z H g K ∂x H g K ∇ or 2 2 Proof: g = i ∂f ∂g ∂f ∂g ∂f ∂g g −f −f g −f y y ∂ ∂ ∂x ∂x + j + k ∂z 2 ∂z g2 g2 g LM ∂f + j ∂f + k ∂f OP − f LMi ∂g + j ∂g + k ∂g OP N ∂z ∂y ∂z Q N ∂x ∂y ∂z Q g i = ∇ FG f IJ = g∇f − f∇g . H gK g g2 2 Example 1. If r = xi + yj + zk then show that (i) ∇( a ⋅ r) = a , where a is a constant vector (ii) grad r = (iii) grad r r 1 r = – 3 r r grad rn = nrn – 2 r , where r = r . H Sol. (i) Let a = a1 i + a2 j + a 3 k, r = xi + yj + zk , (iv) then a⋅r = ( a1i + a2 j + a3 k) ⋅ ( xi + yj + zk) = a1x + a2y + a3z. (U.P.T.U., 2007) 342 A TEXTBOOK OF ENGINEERING MATHEMATICS—I ∴ FG i ∂ + j ∂ + k ∂ IJ (a x + a y + a z) H ∂x ∂y ∂z K ∇( a ⋅ r) = 1 2 3 H a1 i + a 2 j + a 3 k = a . Hence proved. = ∂ 2 ( x + y 2 + z 2 )1/2 ∂x x x =r = Σi 2 2 2 1/2 = Σi r (x + y + z ) grad r = ∆r = Σi (ii) Hence, xi + yj + zk r = = r . r r ∂ 1 1 −1 r r = 2 = ∇ = ∂r r r r r grad r = (iii) grad F 1I H rK FI HK FI HK r = − 3 . Proved. r (iv) Let r = xi + yj + zk. ∂ 2 2 2 n/2 grad rn = ∇rn = Σi ( x + y + z ) ∂x = n (x2 + y2 + z2)n/2–1 xi + yj + zk Now, d 2 2 2 (n–1)/2 = n(x + y + z ) = nr n−1 i exi + yj + zkj ex + y + z j 2 2 2 1/2 r r = nr n−2 r . Example 2. If f = 3x2y – y3z2, find grad f at the point (1, –2, –1). Sol. (U.P.T.U., 2006) FG i ∂ + j ∂ + k ∂ IJ ( 3x y − y z ) H ∂x ∂y ∂z K ∂ ∂ ∂ = i c 3x y − y z h + j ( 3x y − y z ) + k ( 3x y − y z ) ∂x ∂y ∂z H grad f = ∇f = 2 2 3 2 3 2 2 3 2 2 3 2 = i(6xy) + j(3x2 – 3y2z2) + k(–2y3z) grad φ at (1, –2, –1) = i(6)(1)(–2) + j [(3)(1) – 3(4)(1)] + k(–2)(–8)(–1) = –12i – 9j – 16k. Example 3. Find the directional derivative of Sol. Here f(x, y, z) = Now 1 = r 1 x2 + y2 + z2 = (x2 + y2 + z2)–1/2 FG ∂ + j ∂ + k ∂ IJ cx + y + z h H ∂x ∂y ∂z K ∇f = i 2 2 1 in the direction r where r = xi + yj + zk. r (U.P.T.U., 2002, 2005) 2 − 1/2 343 VECTOR CALCULUS = ∂ 2 2 2 −1/2 −1/2 ∂ 2 ∂ k x + y2 + z2 i + ( x 2 + y 2 + z 2 ) −1/2 j + ∂z ( x + y + z ) ∂x ∂y d i RS− 1 (x + y + z ) T 2 −b xi + yj + zk g = cx + y + z h = 2 2 −3/2 2 2 2 2 3/2 UV RS W T 1 2x i + − (x 2 + y 2 + z 2 ) −3/2 2 y 2 UV j + RS− 1 (x + y + z ) W T2 2 2 2 −3/2 UV W 2z k and a = unit vector in the direction of xi + yj + zk = xi + yj + zk As a = x2 + y2 + z2 ∴ Directional derivative = ∇f · a = − xi + yj + zk ⋅ r r xi + yj + zk cx + y + z h (xi + yj + zk ) F xi + yj + zk IJ ⋅ = −G = − (x + y + z ) Hx + y +z K (x + y + z ) 2 2 2 3/2 2 2 1/2 2 2 2 2 2 2 2 2 2 2 Example 4. Find the directional derivative of φ = x2yz + 4xz2 at (1, – 2, –1) in the direction 2i – j – 2k. In what direction the directional derivative will be maximum and what is its magnitude? Also find a unit normal to the surface x2yz + 4xz2 = 6 at the point (1, – 2, – 1). φ = x2yz + 4xz2 Sol. ∂φ = 2xyz + 4z2 ∂x ∂φ 2 ∂y = x z, ∴ ∂φ = x2y + 8xz ∂z ∂φ ∂φ ∂φ +j +k grad φ = i ∂x ∂y ∂z 2 = (2xyz + 4z ) i + (x2z) j + (x2y + 8xz)k = 8i – j – 10k at the point (1, – 2, – 1) Let a be the unit vector in the given direction. 2i − j − 2k Then a = ∴ Directional derivative dφ = ∇φ ⋅ a ds 4+1+ 4 = = (8i – j – 10k) · = 2i − j − 2 k 3 FG 2i − j − 2k IJ H 3 K 37 16 + 1 + 20 = · 3 3 344 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Again, we know that the directional derivative is maximum in the direction of normal which is the direction of grad φ. Hence, the directional derivative is maximum along grad φ = 8i – j – 10k. Further, maximum value of the directional derivative = |grad φ| = |8i – j – 10k| 64 + 1 + 100 = Again, a unit vector normal to the surface = = 165 . grad φ |grad φ| 8i − j − 10k = · 165 Example 5. What is the greatest rate of increase of u = xyz2 at the point (1, 0, 3)? Sol. u = xyz2 ∴ grad u = i ∂u ∂u ∂u +j +k ∂x ∂y ∂z = yz2 i + xz2 j + 2xyz k = 9j at (1, 0, 3) point. Hence, the greatest rate of increase of u at (1, 0, 3) = |grad u| at (1, 0, 3) point. = |9j| = 9. Example 6. Find the directional derivative of φ = (x2 + y2 + z2)–1/2 at the points (3, 1, 2) in the direction of the vector yz i + zx j + xy k. Sol. ∴ φ = (x2 + y2 + z2)–1/2 grad φ = i ∂ 2 ∂ ( x + y 2 + z 2 ) −1/2 + j ( x 2 + y 2 + z 2 ) −1/2 ∂x ∂y + k LM N OP LM Q N O ( 2 z )P Q ∂ 2 ( x + y 2 + z 2 ) −1/2 ∂z 1 2 1 2 2 −3/2 ( 2 x ) + j − ( x 2 + y 2 + z 2 ) −3/2 (2 y ) = i − (x + y + z ) 2 2 LM N 1 + k − ( x 2 + y 2 + z 2 ) −3/2 2 = − = − = − xi + yj + zk ( x + y 2 + z 2 ) 3/2 2 3i + j + 2k (9 + 1 + 4) 3/2 3i + j + 2k 14 14 at (3, 1, 2) at (3, 1, 2) OP Q 345 VECTOR CALCULUS Let a be the unit vector in the given direction, then yzi + zxj + xyk a = = Now, y z + z 2 x2 + x 2 y 2 2 2 2i + 6 j + 3 k at (3, 1, 2) 7 dφ = a . grad φ ds = FG 2i + 6 j + 3k IJ ⋅ FG − 3i + j + 2k IJ H 7 K H 14 14 K = − = − ( 2 )( 3) + ( 6 )(1) + ( 3 )(2 ) 7.14 14 18 7.14 14 = − 9 49 14 · Example 7. Find the directional derivative of the function φ = x2 – y2 + 2z2 at the point P(1, 2, 3) in the direction of the line PQ, where Q is the point (5, 0, 4). Sol. Here Position vector of P = i + 2j + 3k Position vector of Q = 5i + 0j + 4k ∴ PQ = Position vector of Q – Position vector of P = (5i + 0j + 4k) – (i + 2j + 3k) = 4i – 2j + k. Let a be the unit vector along PQ, then 4i − 2 j + k a = Also, 16 + 4 + 1 grad φ = i = 4i − 2 j + k 21 ∂φ ∂φ ∂φ +j +k ∂y ∂z ∂x = 2x i – 2y j + 4z k = 2i – 4j + 12k at (1, 2, 3) Hence, dφ = a . grad φ ds = = = FG 4i − 2 j + k IJ ⋅ (2i − 4 j + 12k) H 21 K ( 4)( 2) + ( −2)( −4) + (1)(12) 21 28 21 ⋅ Example 8. For the function φ = y x2 + y2 , 346 A TEXTBOOK OF ENGINEERING MATHEMATICS—I find the magnitude of the directional derivative making an angle 30° with the positive X-axis at the point (0, 1). Sol. Here ∴ φ = y x + y2 2 ∂φ 2 xy = – 2 ∂x (x + y 2 )2 ∂φ x2 − y2 ∂φ = =0 2 2 2 and ∂y (x + y ) ∂z ∴ grad φ = i = i ∂φ ∂φ ∂φ +j +k ∂x ∂y ∂z ∂φ ∂φ +j ∂x ∂y LM3 ∂φ = 0OP N ∂z Q −2xy x2 − y2 + i j ( x 2 + y 2 )2 (x 2 + y 2 )2 = – j at (0, 1). Let a be the unit vector along the line making an angle 30° with the positive X-axis at the point (0, 1), then a = cos 30° i + sin 30° j. Hence, the directional derivative is given by dφ = a . grad φ ds = (cos 30° i + sin 30° j) · (–j) = 1 · 2 Example 9. Find the values of the constants a, b, c so that the directional derivative of φ = axy2 + byz + cz2x3 at (1, 2, –1) has a maximum magnitude 64 in the direction parallel to Z-axis. Sol. φ = axy2 + byz + cz2x3 = – sin 30° = – ∴ grad φ = ∂φ ∂φ ∂φ i+ j+ k ∂x ∂y ∂z = (ay2 + 3cz2x2) i + (2axy + bz) j + (by + 2czx3) k = (4a + 3c) i + (4a – b) j + (2b – 2c) k. at (1, 2, –1). Now, we know that the directional derivative is maximum along the normal to the surface, i.e., along grad φ. But we are given that the directional derivative is maximum in the direction parallel to Z-axis, i.e., parallel to the vector k. Hence, the coefficients of i and j in grad φ should vanish and the coefficient of k should be positive. Thus 4a + 3c = 0 ...(i) 4a – b = 0 ...(ii) and 2b – 2c > 0 b > c ...(iii) Then grad φ = 2(b – c) k. 347 VECTOR CALCULUS Also, maximum value of the directional derivative =|grad φ| ⇒ 64 = |2 (b – c) k| = 2 (b – c) ⇒ b – c = 32. Solving equations (i), (ii) and (iv), we obtain [3 b > c] ...(iv) a = 6, b = 24, c = – 8. Example 10. If the directional derivative of φ = ax2y + by2z + cz2x, at the point (1, 1, 1) has maximum magnitude 15 in the direction parallel to the line of a, b and c. Sol. Given φ = ax2y + by2z + cz2x ∴ ∇φ = i y−3 x −1 z = = , find the values 2 1 −2 (U.P.T.U., 2001) FG ∂φ IJ + j FG ∂φ IJ + k FG ∂φ IJ H ∂x K H ∂y K H ∂z K = i(2axy + cz2) + j (ax2 + 2byz) + k (by2 + 2czx) ∇φ at the point (1, 1, 1) = i (2a + c) + j (a + 2b) + k (b + 2c). We know that the maximum value of the directional derivative is in the direction of ∇ f i.e., |∇φ| = 15 ⇒ (2a + c)2 + (2b + a)2 + (2c + b)2 = (15)2 But, the directional derivative is given to be maximum parallel to the line. y−3 z x −1 = = 1 2 −2 2a + c 2b + a 2c + b = = 2 −2 1 ⇒ 2a + c = – 2b – a ⇒ 3a + 2b + c = 0 and 2b + a = – 2(2c + b) ⇒ 2b + a = – 4c – 2b ⇒ a + 4b + 4c = 0 Solving (i) and (ii), we get ⇒ a b c = = = k (say) −11 4 10 ⇒ a = 4k, b = – 11k, and c = 10k. 2 Now, (2a + c) + (2b + a)2 + (2c + b)2 = (15)2 ⇒ (8k + 10k)2 + (– 22k + 4k)2 + (20k – 11k)2 = (15)2 5 9 20 55 50 a = ± ,b= ± and c = ± · 9 9 9 ⇒ k = ± ⇒ z af Example 11. Prove that ∇ f u du = f(u) ∇u. Sol. Let Then z af f u du = F(u), a function of u so that z af ∇ f u du = ∂F = f(u) ∂u FG i ∂ + j ∂ + k ∂ IJ F H ∂x ∂y ∂z K (i) (ii) 348 A TEXTBOOK OF ENGINEERING MATHEMATICS—I = i ∂F ∂F ∂F + j ∂y + k ∂x ∂z = i ∂F ∂u ∂F ∂u ∂F ∂u + j + k ∂u ∂x ∂u ∂z ∂u ∂y FG H IJ K ∂F i ∂u + j ∂u + k ∂u ∂x ∂y ∂z ∂u = f (u) ∇u. Hence proved. = Example 12. Find the angle between the surfaces x2 + y2 + z2 = 9 and z = x2 + y2 – 3 at the point (2, – 1, 2) (U.P.T.U., 2002) 2 2 2 Sol. Let φ1 = x + y + z – 9 φ2 = x2 + y2 – z – 3 ∴ ∇φ1 = FG i ∂ + j ∂ + k ∂ IJ (x + y + z – 9) = 2xi + 2yj + 2zk H ∂x ∂y ∂z K 2 2 2 ∇φ1 at the point (2, – 1, 2) = 4i – 2j + 4k H ∇φ 2 = FG i ∂ + j ∂ + k ∂ IJ (x + y – z – 3) = 2xi + 2yj – k H ∂x ∂y ∂z K 2 2 ∇φ2 at the point (2, – 1, 2) = 4i – 2j – k Let θ be the angle between normals (i) and (ii) (4i – 2j + 4k) · (4i – 2j – k) = 16 + 4 + 16 (i) (ii) 16 + 4 + 1 cos θ 16 + 4 – 4 = 6 21 cos θ ⇒ 16 = 6 21 cos θ ⇒ cos θ = 8 3 21 ⇒ θ = cos–1 8 3 21 · Example 13. If u = x + y + z, v = x2 + y2 + z2, w = yz + zx + xy, prove that grad u, grad v and grad w are coplanar vectors. (U.P.T.U., 2002) FG i ∂ + j ∂ + k ∂ IJ (x + y + z) = i + j + k H ∂x ∂y ∂z K F ∂ + j ∂ + k ∂ IJ (x + y + z ) = 2x i + 2y j + 2z k grad v = G i H ∂x ∂y ∂z K F ∂ + j ∂ + k ∂ IJ (yz + zx + xy) grad w = G i H ∂x ∂y ∂z K Sol. grad u = 2 2 = i ( z + y) + j ( z + x) + k ( y + x) Now, grad u (grad v × grad w) = 1 2x 1 2y 1 2z z+y z+x y+x 2 349 VECTOR CALCULUS 1 1 1 y z = 2 x z+y z+x y+x 1 = 2 x+z+ y 1 y+z+x z+x z+ y 1 z + y + x |Applying R2 → R2 + R3 y+x 1 1 1 =0 1 1 1 y+z z+x x+y Hence, grad u, grad v and grad w are coplanar vectors. = 2 (x + y + z) Example 14. Find the directional derivative of φ = 5x2y – 5y2z + 5 2 z x at the point 2 x −1 y − 3 z = = · (U.P.T.U., 2003) −2 2 1 5 2 2 Sol. ∇φ = 10 xy + z i + ( 5x − 10 yz) j + ( −5y 2 + 5zx)k 2 25 i − 5j ∇ φ at P(1, 1, 1) = 2 y−3 z x −1 Direction Ratio of the line = are 2, – 2, 1 = 2 −2 1 −2 1 2 , Direction cosines of the line are , 2 2 2 2 + −2 + 1 2 2 + −2 + 1 (2) 2 + ( −2)2 + (1) 2 2 −2 1 , i.e., , 3 3 3 Directional derivative in the direction of the line P(1, 1, 1) in the direction of the line LM N OP Q a f = F 25 i − 5 jI ⋅ F 2 i − 2 j + 1 kI H 2 K H3 3 3 K 25 10 + 3 3 35 = · 3 = RS f (r)r UV = 1 d (r f ). T r W r dr R f (r)r UV = R|S exi + yj + zkj U|V ∇⋅S T r W ∇ ⋅ |T f (r) r |W ∂ R f ( r )x U ∂ R f ( r ) y U ∂ R f ( r ) z U S V+ S V+ S V = ∂x T r W ∂y T r W ∂z T r W ∂ R f (r )x U d R f (r) U ∂r f (r ) S V = x S V + r ∂x T r W dr T r W ∂x Example 15. Prove that ∇· Sol. Now, 2 2 a f 350 A TEXTBOOK OF ENGINEERING MATHEMATICS—I = x = Similarly, RS1 df − f (r) UV x + f (r) as ∂r = x ∂x r T r dr r W r r 2 f (r ) x 2 df (r ) x 2 − 3 f (r) + 2 dr r r r RS UV = y df (r) − y f (r) + f (r) T W r dr r r ∂ R f (r )z U S V = zr dfdr(r) − zr f (r) + f (rr) ∂z T r W ∂ f (r ) y r ∂y and 2 2 2 3 2 2 2 3 Now using these results, we get ∇· LM f (r)r OP = df (r) + 2 f (r) N r Q dr r = 1 d 2 (r f )⋅ Hence proved. r 2 dr r and f(1) = 0. r5 Example 16. Find f(r) such that ∇ f = Sol. It is given that ∂f ∂f ∂f xi + yj + zk r i+ j+ k = ∇ f = 5 = ∂x ∂y ∂z r5 r x ∂f ∂f ∂f y z So, = 5, = 5 and = 5 ∂y r ∂x ∂z r r ∂f ∂f ∂f y x z We know that df = ∂x dx + ∂y dy + ∂z dz = 5 dx + 5 dy + 5 dz r r r df = xdx + ydy + zdz r 5 = rdr = r–4dr r5 −3 Integrating Since So, Thus, f(r) = r +c −3 0 = f(1) = − 1 +c 3 1 3 1 1 1 f(r) = – · 3 r3 3 c = EXERCISE 5.2 1. Find grad f where f = 2xz4 – x2y at (2, –2, –1). Ans. 10i − 4 j − 16 k 2. Find ∇ f when f = (x2 + y2 + z2) e − x + y + z · H Ans. 2 − r e − r r 2 2 2 a f 351 VECTOR CALCULUS 3. Find the unit normal to the surface x2y + 2xz = 4 at the point (2, –2, 3). LMAns. ± 1 bi − 2 j − 2kgOP 3 N Q 4. Find the directional derivative of f = x yz + 4xz at (1, –2, –1) in the direction LMAns. 37 OP 2i – j – 2k. 3Q N 5. Find the angle between the surfaces x + y + z = 9 and z = x + y – 3 at the point LMAns. θ = cos FG 8 21 IJ OP (2, –1, 2) . H 63 K PQ MN 6. Find the directional derivative of f = xy + yz + zx in the direction of vector i + 2j + 2k at LMAns. 10 OP the point (1, 2, 0). 3Q N 2 2 2 2 2 2 2 −1 Ans. f = x 2 yz 3 + 20 7. If ∇f = 2xyz3i + x2z3j + 3x2yz2k, find f (x, y, z) if f (1, – 2, 2) = 4. Ans. f = x 2 + 2 y 2 + 4z 2 8. Find f given ∇f = 2xi + 4yj + 8zk. 9. Find the directional derivative of φ = (x2 + y2 + z2)–1/2 at the point P(3, 1, 2) in the direction LMAns. − 9 OP 49 14 Q N of the vector, yzi + xzj + xyk. 10. Prove that ∇2f(r) = f ′′(r) + 11. Show that ∇2 2 f ′ ( r) · r FG x IJ = 0, where r is the magnitude of position vector r = xi + yj + zk. Hr K 3 [U.P.T.U. (C.O.), 2002] 12. Find the direction in which the directional derivative of f(x, y) = (x2 – y2)/xy at LMAns. 1 + i OP 2 Q N (1, 1) is zero. 13. Find the directional derivative of 1 in the direction of r , where r = xi + yj + zk. r 1 Ans. − 2 (U.P.T.U., 2003) r LM N 14. Show that ∇r–3 = – 3r–5 r . 15. If φ = log | r |, show that ∇φ = 5.10 r · r2 OP Q [U.P.T.U., 2008] DIVERGENCE OF A VECTOR POINT FUNCTION If f (x, y, z) is any given continuously differentiable vector point function then the divergence of f scalar function defined as (U.P.T.U., 2006) 352 A TEXTBOOK OF ENGINEERING MATHEMATICS—I ∇⋅ f = 5.11 FG i ∂ + j ∂ + k ∂ IJ ⋅ f = i ⋅ ∂ f + j ⋅ ∂ f + k ⋅ ∂ f = div f ∂x ∂y ∂z H ∂x ∂y ∂z K . PHYSICAL INTERPRETATION OF DIVERGENCE Let v = vxi + vy j + vz k be the velocity of the fluid at P(x, y, z). Here we consider the case of fluid flow along a rectangular parallelopiped of dimensions δx, δy, δz Mass in = v·x δyδz Mass out = vx(x + δx) δyδz = (along x-axis) Z FG v + ∂v δxIJ δyδz H ∂x K x S x |By Taylor's theorem Net amount of mass along x-axis FG H = vx δyδz – vx + Vx R IJ K ∂v x δx δyδz ∂x ∂v x δxδyδz ∂x |Minus sign shows decrease. Y Similar net amount of mass along y-axis ∂v y ∂y C @z P Q @x B X Fig. 5.4 δxδyδz and net amount of mass along z-axis = − ∂v z δxδyδz ∂z ∴ Total amount of fluid across parallelopiped per unit time = − Negative sign shows decrease of amount ⇒ A O = – = – D v x (x + @x) @y F ∂v + ∂v + ∂v I δxδyδz GH ∂x ∂y ∂z JK x F ∂v + ∂v + ∂v I δxδyδz GH ∂x ∂y ∂z JK x Decrease of amount of fluid per unit time = y z Hence the rate of loss of fluid per unit volume = F ∂v + ∂v + ∂v I GH ∂x ∂y ∂z JK x y z = ∇ ⋅ v = div v. Therefore, div v represents the rate of loss of fluid per unit volume. y z 353 VECTOR CALCULUS Solenoidal: For compressible fluid there is no gain no loss in the volume element ∴ div v = 0 then v is called Solenoidal vector function. 5.12 CURL OF A VECTOR If f is any given continuously differentiable vector point function then the curl of f (vector function) is defined as Curl f = ∇ × f = i × ∂f ∂f ∂f + j× +k× ∂x ∂y ∂z (U.P.T.U., 2006) f = fx i + fy j + fz k, then Let i j k ∇ × f = ∂ / ∂x ∂ / ∂y ∂ / ∂z · fx fy fz 5.13 PHYSICAL MEANING OF CURL Here we consider the relation v = w × r , w is the angular velocity r is position vector of a point on the rotating body (U.P.T.U., 2006) curl v = ∇ × v = ∇ × (w × r ) LMwHH = w i + w j + w kOP N r = xi + yj + zk Q 1 = ∇ × [(w1 i + w 2 j + w 3 k ) × (xi + yj + zk )] i = ∇ × w1 x j w2 k w3 y z 2 3 = ∇ × [( w2 z − w 3 y)i − (w1 z − w 3 x) j + ( w1 y − w2 x)k ] = FG i ∂ + j ∂ + k ∂ IJ × [(w z − w y)i − (w z − w x) j + (w y − w x)k] H ∂x ∂y ∂z K i ∂ = ∂x w2 z − w 3 y 2 3 1 3 1 2 j k ∂ ∂ ∂y ∂z w 3 x − w1 z w 1 y − w 2 x = (w1 + w1) i – (–w2 – w2) j + (w3 + w3) k = 2 (w1i + w2j + w3k) = 2w Curl v = 2w which shows that curl of a vector field is connected with rotational properties of the vector field and justifies the name rotation used for curl. 354 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Irrotational vector: If curl f = 0, then the vector f is said to be irrotational. Vice-versa, if f is irrotational then, curl f = 0. 5.14 VECTOR IDENTITIES Identity 1: Proof: grad uv = u grad v + v grad u grad (uv) = ∇ (uv) = FG i ∂ + j ∂ + k ∂ IJ (uv) H ∂x ∂y ∂z K ∂ ∂ ∂ = i (uv) + j (uv) + k (uv) ∂x ∂y ∂z FG IJ FG IJ FG H K H K H F ∂v ∂v ∂v I F ∂u ∂u ∂u I = uG i ∂x + j ∂y + k ∂z J + v G i ∂x + j ∂y + k ∂z J H K H K ∂v ∂v ∂u ∂u ∂v ∂u = i u ∂x + v ∂x + j u ∂y + v ∂y + k u ∂z + v ∂z or IJ K grad uv = u grad v + v grad u . H H H H H grad ( a · b ) = a × curl b + b × curl a + a ⋅ ∇ b + (b ⋅ ∇) a H H ∂ H H ∂a H H ∂b Proof: grad ( a · b ) = Σi a ⋅ b = Σi ⋅b + a ⋅ ∂x ∂x ∂x H H H ∂a H ∂b = Σi b ⋅ . + Σi a ⋅ ∂x ∂x H H H H ∂b H ∂b H ∂ b a × i× Now, = a⋅ i − ( a ⋅ i) ∂x ∂x dx H H H H H ∂b ∂b H ∂b ⇒ = + ⋅ a i × × a i a⋅ i ∂x ∂x ∂x H H H H H ∂b ∂b H ∂b ⇒ Σ a⋅ + ∑ a ⋅i i = ∑a × i× ∂x ∂x ∂x H H H ∂b H H ∂ H ∂b ⇒ Σ a⋅ + ∑ a ⋅i i = a × ∑ i× b ∂x ∂x ∂x H H H H ∂b H H ⇒ Σ a⋅ i = a × curl b + a ⋅ ∇ b · ∂x L L Interchanging a and a , we get H ∂aH H H H H Σ b⋅ i = b × curl a + b ⋅ ∇ a ∂x From equations (i), (ii) and (iii), we get Identity 2: FG H FG H FG H FG H FG H IJ K IJ K IJ K IJ K FG H IJ K IJ K F e j GH FG IJ FG IJ H K H K FG IJ H K FG IJ b g H K FG IJ b g H K FG IJ FG IJ H K H K b g IJ K ...(i) b g ...(ii) e j ...(iii) H H H H H H H H H H grad ( a · b ) = a × curl b + b × curl a + ( a · ∇) b + ( b · ∇) a . 355 VECTOR CALCULUS Identity 3: div (u a ) = u div a + a · grad u Proof: div (u a ) = ∇·(u a ) = (U.P.T.U., 2004) FG i ∂ + j ∂ + k ∂ IJ ⋅(uaH) H ∂x ∂y ∂z K ∂ H ∂ H ∂ H (ua ) + j ⋅ (ua ) + k ⋅ (ua ) ∂x ∂y ∂z H H H ∂u H ∂a ∂u H ∂a ∂u H ∂a i ⋅ a + u + j ⋅ a + u + k ⋅ a + u = ∂x ∂x ∂y ∂y ∂z ∂z H H H H ∂u ∂u ∂u ∂a ∂a ∂a +a⋅ i +j +k = u i⋅ + j⋅ + k⋅ ∂x ∂y ∂z ∂x ∂y ∂z = i⋅ RS T UV RS W T RS T or div (u a ) = u div a + a · grad u div ( a × b ) = b · curl b – a · curl b Proof: div ( a × b ) = ∇ · ( a × b ) H ∂ H = Σi ⋅ ( a × b ) ∂x H H ∂a H H ∂b = Σi ⋅ ×b+a× ∂x ∂x H H H H ∂b ∂a = Σi ⋅ × b + Σi ⋅ a × ∂x ∂x H H H ∂b H ∂a = Σ i× ⋅b − Σ i × ⋅a ∂x ∂x H H H = (curl a ) · b – (curl b) · a H H H H div ( a × b ) = b · curl a – a · curl b . H H H curl (u a ) = u curl a + (grad u) × a H H curl (u a ) = ∇ × (u a ) ∂ H = Σi × (ua ) ∂x H ∂u H ∂a = Σi × a+u ∂x ∂x H ∂u H ∂a = Σ i × a + uΣ i × ∂x ∂x H H = (grad u) × a + u curl a H H H curl (u a ) = u curl a + (grad u) × a . or Identity 5: Proof: FG H FG H IJ K FG H FG IJ H K IJ K FG H FG H IJ K FG H UV W UV W . Identity 4: FG H or UV RS W T UV RS W T [U.P.T.U. (C.O.), 2003] IJ K IJ K IJ K IJ K [U.P.T.U. (C.O.), 2003] 356 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Identity 6: Proof: H H H H H H H H H H curl( a × b ) = a div b – b div a + ( b ·∇) a – ( a · ∇) b H H H H curl ( a × b ) = ∇ × ( a × b ) ∂ H H = Σi × ( a × b ) ∂x H H ∂a H H ∂b = Σi × ×b+a× ∂x ∂x H H H H ∂b ∂a = Σi × × b + Σi × a × ∂x ∂x FG H FG H IJ K FG H FG H IJ K IJ K IJ K FG H IJ K = Σ ( i. b ) ∂ a − Σ i ⋅ ∂ a b + Σ i ⋅ ∂ b a − Σ ( i . a ) ∂ b ∂x ∂x ∂x ∂x H H H ∂ H H ∂ H ∂b H ∂a H = Σ (i . b ) a − Σ i ⋅ b + Σ i⋅ a − Σ a⋅i b ∂x ∂x ∂x ∂x H H H H H H H H = ( b · ∇) a – (div a ) b + (div b ) a – ( a .∇) b H H H H H H H H = ( b · ∇) a – ( a · ∇) b + a div b – b div a . H H H H H H H H H H curl ( a × b ) = a div b – b div a + ( b · ∇) a – ( a · ∇) b . or Identity 7: div grad f = ∇ · (∇f) = Proof: div grad f = ∇ · (∇f) ∂2 f ∂x 2 FG H + ∂2 f ∂y 2 IJ K + ∂2 f ∂z 2 FG H IJ K FG H IJ K = ∇2 f FG i ∂ + j ∂ + k ∂ IJ ⋅ FG i ∂f + j ∂f + k ∂f IJ H ∂x ∂y ∂z K H ∂x ∂y ∂z K ∂ F ∂f I ∂ F ∂f I ∂ F ∂f I G J+ G J+ G J = ∂x H ∂x K ∂y H ∂y K ∂z H ∂z K = = ∂2 f ∂2 f ∂2 f ∂x ∂y ∂z 2 + 2 div grad f = ∇2 f or Identity 8: Proof: + 2 . curl grad f = 0 curl grad f = ∇ × (∇f) = Σi FG H ∂f ∂f ∂f ∂ +k +j × i ∂z ∂y ∂x ∂x IJ K F ∂ f + j ∂ f +k ∂ f I GH ∂x ∂x∂y ∂x∂z JK F ∂ f −j ∂ f I = Σ Gk H ∂x∂y ∂x∂z JK F ∂ f − j ∂ f I + Fi ∂ f − k ∂ f I + F j ∂ f − i ∂ f I = Gk H ∂x∂y ∂x∂z JK GH ∂y∂z ∂y∂x JK GH ∂z∂x ∂z∂y JK 2 = Σi × i 2 2 2 or curl grad f = 0 . 2 2 2 2 2 2 2 2 357 VECTOR CALCULUS Identity 9: Proof: H div curl f = 0 H H div curl f = ∇ · (∇ × f ) U|V |W R|S |T H H H ∂f ∂f ∂f ∂ = Σi ⋅ i× + j× +k× ∂x ∂x ∂y ∂z H H H ∂2 f ∂2 f ∂2 f = Σi ⋅ i × 2 + j × +k× ∂x∂y ∂x∂z ∂x H H H ∂2 f ∂2 f ∂2 f = Σ ( i × i) ⋅ 2 + ( i × j ) ⋅ + (i × k ) ⋅ ∂x∂y ∂x∂z ∂x H H ∂2 f ∂2 f = Σ k⋅ − j⋅ ∂x∂y ∂x∂z H H H H ∂2 f ∂2 f ∂2 f ∂2 f = k⋅ + i⋅ + −k⋅ − j⋅ ∂y∂z ∂y∂x ∂x∂y ∂ x∂ z R|S |T R|S |T R|S |T R|S |T U|V |W U|V |W U|V |W U|V R|S |W |T H div curl f = 0 . U|V |W H H H H H ∂ 2 f ∂ 2 f ∂2 f Identity 10: grad div f = curl curl f + 2 + 2 + 2 ∂x ∂y ∂z H H Proof: Curl curl f = ∇ × (∇ × f ) H H H ∂f ∂f ∂f ∂ = Σi × i× + j× +k× ∂x ∂y ∂z ∂x H H H ∂2 f ∂2 f ∂2 f = Σi × i × 2 + j × +k× ∂x∂y ∂x∂z ∂x H H H H ∂2 f ∂2 f ∂2 f ∂2 f = Σ i ⋅ 2 i − (i ⋅ i ) ⋅ 2 + i ⋅ j − ( i ⋅ j) ∂x∂y ∂x∂y ∂x ∂x |RS |T LMR|F MNS|TGH |RS |T |UV |W |UV |W U|V R|F |W S|TGH IJ K I JK R|S j ⋅ ∂ Hf − i ⋅ ∂ Hf U|V |T ∂z∂x ∂z∂y |W 2 U| V| W H H R F ∂ f I ∂ f UO P + SG i ⋅ TH ∂x∂z JK k − (i ⋅ k) ∂x∂z VWPQ 2 H H H H L F F ∂ fI F ∂ f I ∂ f I O ∂ f = Σ MG i ⋅ MNH ∂x JK i + GH i ⋅ ∂x∂y JK j + GH i ⋅ ∂x∂z JK kPPQ − Σ ∂x H H H H H H H L F F F ∂ fI ∂ f I ∂ f I O ∂ f ∂ f ∂ f j + Gi ⋅ ⇒ Curl curl f + = Σ MG i ⋅ + + J kP · J i + i⋅ ∂x ∂y ∂z NMH H∂x K H GH ∂xH∂y JK H ∂x∂z K QP H RSi ⋅ ∂f + j ⋅ ∂f + k ⋅ ∂f |UV ∂ | Again, grad div f = Σi ∂x T | ∂x ∂y ∂z W| H H H R ∂ f ∂ f ∂ f U | | = Σi Si ⋅ |T ∂x + j ⋅ ∂x∂y + k ⋅ ∂x∂z V|W 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ...(i) 358 A TEXTBOOK OF ENGINEERING MATHEMATICS—I H H H L F ∂ f I F ∂ f I F ∂ f I O = Σ MG i ⋅ MNH ∂x JK i + GH j ⋅ ∂x∂y JK i + GH k ⋅ ∂x∂z JK i PPQ H H H L F ∂ f I F ∂ f I F ∂ f I O = Σ MG i ⋅ MNH ∂x JK i + GH i ⋅ ∂z∂x JK j + GH i ⋅ ∂y∂z JK kPPQ 2 2 2 2 2 2 2 ...(ii) 2 From eqns. (i) and (ii), we prove H H H grad div f = curl curl f + ∇2 f H H H Example 1. If f = xy2 i + 2x2yz j – 3yz2 k then find div f and curl f at the point (1, – 1, 1). H Sol. We have f = xy2 i +2x2yz j – 3yz2 k H ∂ ∂ ∂ (xy 2 ) + ( 2x 2 yz) + (–3yz 2 ) div f = ∂z ∂x ∂y = y2 + 2x2z – 6yz = (– 1)2 + 2 (1)2 (1) – 6 (– 1)(1) at (1, – 1, 1) = 1 + 2 + 6 = 9. H Again, curl f = curl [xy2 i + 2x2yz j – 3yz2 k] i ∂ = ∂x xy 2 = i j k ∂ ∂ ∂y ∂z 2 x 2 yz −3 yz 2 RS ∂ (−3 yz ) − ∂ (2x yz)UV + j RS ∂ (xy ) − ∂ (−3 yz )UV ∂z ∂x W T ∂y W T ∂z ∂ R∂ U + k S ( 2 x yz ) − ( xy )V ∂ x y ∂ T W 2 2 2 2 2 = i [–3z2 – 2x2y] + j [0 – 0] + k [4xyz – 2xy] = (–3z2 – 2x2y)i + (4xyz – 2xy) k = {–3 (1)2 – 2(1)2(–1)} i + {4(1)(–1)(1) – 2(1)(–1)} k at (1, –1, 1) = – i – 2k. Example 2. Prove that H H (i) div r = 3. (ii) curl r = 0. H H Sol. (i) div r = ∇ · r ∂ ∂ ∂ ⋅ ( xi + yj + zk ) +j +k = i ∂x ∂y ∂z FG H IJ K FG H IJ K ∂ ∂ ∂ ( x ) + ( y ) + ( z) ∂x ∂y ∂z = 1 + 1 + 1 = 3. H H curl r = ∇ × r ∂ ∂ ∂ = i ∂x + j ∂y + k ∂z × ( xi + yj + zk ) = (ii) = i ∂ ∂x x j ∂ ∂y y k ∂ ∂z z 2 359 VECTOR CALCULUS = i LM ∂ (z) − ∂ ( y)OP + j LM ∂ (x) − ∂ (z)OP + k LM ∂ (y) − ∂ (x)OP N ∂y ∂z Q N ∂z ∂x Q N ∂x ∂y Q = i (0) + j (0) + k (0) = 0 + 0 + 0 = 0. Example 3. Find the divergence and curl of the vector (x2 – y2) i + 2xy j + (y2 – xy) k. Sol. Let Then H f = (x2 – y2) i + 2xy j + (y2 – xy) k. H div f = ∂ 2 ∂ ∂ ( x − y 2 ) + (2 xy) + ( y 2 − xy) ∂x ∂y ∂z = 2x + 2x + 0 = 4x H curl f = and i ∂ ∂x x2 − y2 = i j k ∂ ∂ ∂y ∂z 2xy y 2 − xy LM ∂ ( y − xy) − ∂ (2 xy)OP + j LM ∂ (x − y ) − ∂ ( y − xy)OP ∂z ∂x Q N ∂y Q N ∂z ∂ L∂ O + k M ( 2 xy ) − ( x − y )P ∂ x y ∂ N Q 2 2 2 2 2 2 = i [2y – x) – 0] + j [0 – (–y)] + k [(2y) – (– 2y)] = (2y – x) i + y j + 4y k. Ans. H H Example 4. If f (x, y, z) = xz3 i – 2x2yz j + 2yz4 k find divergence and curl of f (x, y, z) (U.P.T.U., 2006) Sol. H div f = = FG i ∂ + j ∂ + k ∂ IJ ⋅ cxz i − 2x yz j + 2yz k h H ∂x ∂y ∂z K 3 2 4 ∂ ∂ ∂ ( xz 3 ) − (2x 2 yz) + (2 yz 4 ) ∂x ∂y ∂z = z3 – 2x2z + 8yz3 H curl f = i ∂ ∂x xz 3 = i j k ∂ ∂ ∂y ∂z −2 x 2 yz 2 yz 4 RS ∂ (2yz ) + ∂ (2x yz)UV − j RS ∂ (2yz ) − ∂ (xz )UV ∂z ∂z W T ∂y W T ∂x ∂ R∂ U + k S ( −2x yz) − ( xz )V ∂ x ∂ y T W 4 2 4 3 2 = i (2z4 + 2x2y) – j (0 – 3z2x) + k (– 4xyz – 0) = 2 (x2y + z4) i + 3z2xj – 4xyz·k. 3 360 A TEXTBOOK OF ENGINEERING MATHEMATICS—I H Example 5. Find the directional derivative of ∇. u at the point (4, 4, 2) in the direction of H the corresponding outer normal of the sphere x2 + y2 + z2 = 36 where u = x4 i + y4 j + z4 k. H Sol. ∇. u = ∇. (x4i + y4j + z4k) = 4(x3 + y3 + z3) = f (say) ∴ (∇f)(4, 4, 2) = 12 (x2i + y2j + z2k)(4, 4, 2) = 48(4i + 4j + k) Normal to the sphere g ≡ x2 + y2 + z2 = 36 is (∇g)(4, 4, 2) = 2 (xi + yj + zk)(4, 4, 2) = 4(2i + 2j + k) a = unit normal = b g 4 2i + 2 j + k ∇g = ∇g 64 + 64 + 16 2i + 2 j + k 3 The required directional derivative is = ∇f. a 2i + 2 j + k 3 = 16(8 + 8 + 1) = 272. = 48 (4i + 4j + k). Example 6. A fluid motion is given by v = (y + z)i + (z + x)j + (x + y)k show that the motion is irrotational and hence find the scalar potential. (U.P.T.U., 2003) Sol. Curl v = ∆× v = F i ∂ + y ∂ + k ∂ I × by + zgi + az + xf j + bx + ygk GH ∂x ∂y ∂z JK = i j k ∂ ∂ ∂ = i (1 – 1) – j (1 – 1) + k (1 – 1) = 0 ∂x ∂y ∂z y+z z+x x+ y Hence v is irrotational. Now dφ = ∂φ ∂φ ∂φ dx + dy + dz ∂x ∂y ∂z F i ∂φ + j ∂φ + k ∂φ I ⋅ bidx + jdy + kdzg GH ∂x ∂y ∂z JK F ∂ ∂ + k ∂ I φ ⋅ dr = ∇φ⋅ dr = v ⋅ dr = Gi + j H ∂x ∂y ∂z JK = v = ∇φ = [(y + z) i + (z + x) j + (x + y) k]. (idx + jdy + kdz) = (y + z) dx + (z + x) dy + (x + y) dz = ydx + zdx + zdy + xdy + xdz + ydz On integrating φ = = Thus, zb g zb g zb zb g zb g za f ydx + xdy + zdy + ydz + d xy + d yz + d zx φ = xy + yz + zx + c velocity potential = xy + yz + zx + c. zdx + xdz g 361 VECTOR CALCULUS b g b g b g H H H H H H H Example 7. Prove that a × ∇ × r = ∇ a ⋅ r − a ⋅ ∇ r where a is a constant vector and H (U.P.T.U., 2007) r = xi + yj + zk. H Sol. Let a = a1i + a2j + a3k H r = r1i + r2j + r3k ∴ ∇× r = e j Now a × ∇ × r = i ∂ ∂x r1 j ∂ ∂y r2 F GH k ∂ ∂z r3 i a1 j a2 ∂r3 ∂r2 − ∂y ∂x I FG JK H ∂r3 ∂r1 ∂r3 ∂r2 = i ∂y − ∂z – ∂x − ∂z IJ j + FG ∂r − ∂r IJ k K H ∂x ∂y K 2 1 k a3 ∂r1 ∂r3 − ∂z ∂x ∂r2 ∂r1 − ∂x ∂y |RSFG a ∂r − a ∂r IJ – FG a ∂r − a ∂r IJ |UVi |TH ∂x ∂y K H ∂z ∂x K |W |RF ∂r ∂r I F ∂r ∂r I |U – SG a ∂x − a ∂y J − G a ∂y − a ∂z J V j K H K |W |TH |RF ∂r ∂r I F ∂r ∂r I |U + SGH − a ∂x + a ∂z JK − G a ∂y − a ∂z J Vk H K |W |T F ∂ ∂ + k ∂ I ba r + a r + a r g – LMa ∂ + a ∂ + a ∂ OP br i + r j + r kg = Gi + j H ∂x ∂y ∂z JK N ∂x ∂y ∂z Q = ∇ nb a j + a j + a k g ⋅ br j + r j + r k gs R| F ∂ ∂ ∂ I U| – Sb a i + a j + a k g ⋅ G i + j + k J Vbr i + r j + r k g H ∂x ∂y ∂z K W| T| = ∇e a ⋅ rj − e a ⋅ ∇jr · Hence proved. = 2 2 2 1 3 1 3 3 1 1 1 2 3 1 1 2 2 3 3 1 2 3 1 2 1 2 1 1 3 3 1 1 2 2 3 3 3 3 3 2 2 2 1 2 3 1 2 3 Example 8. Find the directional derivative of Ö.(Öf) at the point (1, –2, 1) in the direction of the normal to the surface xy2z = 3x + z2 where f = 2x3y2z4. (U.P.T.U., 2008) Sol. F i ∂ + j ∂ + k ∂ I (2x y z ) = (6x y z )i + (4x yz )j + 8x y z )k GH ∂x ∂y ∂z JK F ∂ + j ∂ + k ∂ I . {(6x y z )i + (4x yz )j + (8x y z )k} ∇.(∇f) = G i H ∂x ∂y ∂z JK ∇f = 3 2 4 2 2 2 4 2 4 3 3 4 4 3 2 3 3 2 3 ∇.(∇f) = 12x y2z4 + 4x3 z4 + 24x3 y2z2 = F(x, y, z) (say) or Now ∇F = F i ∂ + j ∂ + k ∂ I (12x y z + 4x z + 24x y z ) GH ∂x ∂y ∂z JK 2 4 3 4 3 2 2 = (12y2 z4 + 12x2 z4 + 72x2 y2z2)i + (24xyz4 + 48x3 yz2) j + (48x y2z3 + 16x3z3 + 48x3y2z)k 362 A TEXTBOOK OF ENGINEERING MATHEMATICS—I ∴ Let (∇F)(1, –2, 1) = 348i – 144j + 400k g(x, y, z) = xy2z – 3x – z2 = 0 ∇g = F i ∂ + j ∂ + k ∂ I (xy z – 3x – z ) GH ∂x ∂y ∂z JK 2 2 = (y2z – 3) i + (2xyz) j + (xy2 – 2z) k (∇g)(1, –2, 1) = i – 4j + 2k a = unit normal = ∇g i − 4 j + 2k i − 4 j + 2k = = ∇g 1 + 16 + 4 21 Hence, the required directional derivative is b i − 4 j + 2k g ∇F . a = (348i – 144j + 400k) . = 348 + 576 + 800 21 = 21 1724 21 . Example 9. Determine the values of a and b so that the surface ax2 – byz = (a + 2)x will be orthogonal to the surface 4x2y + z3 = 4 at the point (1, –1, 2). Sol. Let f ≡ ax2 – byz – (a + 2)x = 0 ...(i) g ≡ 4x y + z – 4 = 0 ...(ii) 2 grad f = ∇f = 3 F i ∂ + j ∂ + k ∂ I {ax – byz – (a + 2)x} GH ∂x ∂y ∂z JK 2 = (2ax – a – 2)i + (–bz)j + (–by)k (∇f)(1, –1, 2) = (a – 2)i – 2bj + bk grad g = ∇g = F i ∂ + j ∂ + k ∂ I (4x y + z – 4) GH ∂x ∂y ∂z JK 2 ...(iii) 3 = (8xy)i + (4x2)j + (3z2)k (∇g)(1, –1, 2) = – 8i + 4j + 12k Since the surfaces are orthogonal so ...(iv) e∇ f j . e∇gj = 0 ⇒ [(a – 2)i – 2bj + bk] . [– 8i + 4j + 12k] = 0 – 8(a – 2) – 8b + 12b = 0 ⇒ – 2a + b + 4 = 0 But the point (1, –1, 2) lies on the surface (i), so a + 2b – (a + 2) = 0 ⇒ 2b − 2 = 0 ⇒ b = 1 Putting the value of b in (v), we get − 2a + 1 + 4 = 0 ⇒ a = Hence, a = 5 , b = 1. 2 5 2 ...(v) 363 VECTOR CALCULUS Example 10. Prove that A = (x2 – yz)i + (y2 – zx)j + (z2 – xy)k is irrotational and find the scalar potential f such that A = ∇f. i j k ∂ ∂ ∂ Sol. ∇ × A = = (–x + x)i – (–y + y)j + (–z + z) = 0 ∂x ∂y ∂z x 2 − yz y 2 − zx z 2 − xy Hence, A is irrotational. A = ∇f = i Now ∂f ∂f ∂f +j +k = (x2 – yz)i + (y2 – zx)j + (z2 – xy)k ∂x ∂y ∂z Comparing on both sides, we get ∂f ∂f ∂f = (x2 – yz), = (y2 – zx) and = (z2 – xy) ∂y ∂x ∂z ∴ df = ∂f ∂f ∂f dx + dy + dz = (x2 – yz)dx + (y2 – zx)dy + (z2 – xy)dz ∂x ∂y ∂z = (x2 dx + y2 dy + z2 dz) – (yzdx + zxdy + xydz) or df = 1 d(x3 + y3 + z3) – d(xyz) 3 f = 1 (x3 + y3 + z3) – xyz + c. 3 On integrating, we get EXERCISE 5.3 Ans. 2xz – 6 y 2 z 2 + xy 2 1. Find div A , when A = x2zi – 2y3z2j + xy2zk. 2. If V = xi + yj + zk x2 + y2 + z2 , find the value of div V . (U.P.T.U., 2000) LMAns. 2 division ex + y + z j OP N Q 3. Find the directional derivative of the divergence of f (x, y, z) = xyi + xy j + z k at the point L 13 OP (2, 1, 2) in the direction of the outer normal to the sphere, x + y + z = 9. MAns. 3Q N 2 2 2 2 2 2 2 2 H H 3 4. Show that the vector field f = r / r is irrotational as well as solenoidal. Find the scalar potential. LM MN (U.P.T.U., 2001, 2005) Ans. – OP x + y + z PQ 1 2 2 2 364 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 5. If r is the distance of a point (x, y, z) from the origin, prove that curl FG k. grad 1IJ = 0, where k is the unit vector in the direction OZ. H rK FG k × grad 1 IJ + grad H rK (U.P.T.U., 2000) 6. Prove that A = (6xy + z3)i + (3x2 – z)j + (3xz2 – y) k is irrotational. Find a scalar function Ans. f = 3 x 2 y + xz 3 – zy + c2 f (x, y, z) such that A = ∇f. 7. Find the curl of yzi + 3zxj + zk at (2, 3, 4). Ans. – 6i + 3 j + 8 k 8. If f = x2yz, g = xy – 3z2, calculate ∇.(∇f × ∇g). Ans. zero 9. Determine the constants a and b such that curl of (2xy + 3yz)i + (x2 + axz – 4z2)j + (3xy Ans. a = 3, b = −4 + 2byz) k = 0. 10. Find the value of constant b such that 3 2 2 A = (bxy – z ) i + (b – 2)x j + (1 – b) xz k has its curl identically equal to zero. Ans. b = 4 FG 1 IJ = a ⋅ r · H rK r 12. Prove that ∇ e a ⋅ uj = e a ⋅ ∇ju + a × curl u a is a constant vector. F xI 13. Prove that ∇ GH JK = 0. r H 11. Prove that a . ∇ 2 3 3 14. Prove that ∇2f(r) = f ″(r) + 2 f ′ (r). r 15. If u = x2 + y2 + z2 and v = xi + yj + zk , show that div euvj = 5u F a × r I = − a + 3rea ⋅ rj . GH r JK r r 17. Find the curl of v = e ei + j + kj at the point (1, 2, 3). 16. Prove the curl 3 3 5 b Ans. e 6 i − 21 j + 3k xyz 18. Prove that ∇ × ∇f = 0 for any f (x, y, z). Ans. 4 j 19. Find curl of A = x 2 yi – 2 xzj + 2 yzk at the point (1, 0, 2). H 20. Determine curl of xyz2i + yzx2j + zxy2 k at the point (1, 2, 3). a f b g g b g Ans. xy 2z – x i + yz 2x − y j + zx 2 y − z k ;10i + 3k 21. Find f(r) such that f(r) r is solenoidal. LMAns. c OP N rQ 3 22. Find a, b, c when f = (x + 2y + az)i + (bx – 3y – z)j + (4x + cy + 2z)k is irrotational. [Ans. a = 4, b = 2, c = –1] 23. Prove that (y2 – z2 + 3yz – 2x)i + (3xz + 2xy)i + (3xy – 2xz + 2z)k are both solenoidal and irrotational. [U.P.T.U., 2008] 365 VECTOR CALCULUS 5.15 VECTOR INTEGRATION Vector integral calculus extends the concepts of (ordinary) integral calculus to vector functions. It has applications in fluid flow design of under water transmission cables, heat flow in stars, study of satellites. Line integrals are useful in the calculation of work done by variable forces along paths in space and the rates at which fluids flow along curves (circulation) and across boundaries (flux). 5.16 LINE INTEGRAL ej Let F r be a continuous vector point function. Then ej z C F ⋅ dr , is known as the line integral of F r along the curve C. Let F = F1 i + F2 j + F3 k where F1, F2, F3 are the components of F along the coordinate axes and are the functions of x, y, z each. r = xi + yj + zk dr = dxi + dyj + dzk Now, ∴ ∴ z C F ⋅ dr = = ze zb C C je F1i + F2 j + F3 k ⋅ dxi + dyj + dzk g F1 dx + F2 dy + F3 dz . Again, let the parameteric equations of the curve C be x = x (t) y = y (t) z = z (t) z j z LMN a f z z af a f OPQ dy dx dz + F2 t + F3 t dt t1 C dt dt dt were t1 and t2 are the suitable limits so as to cover the arc of the curve C. H Note: work done = F ⋅ dr then we can write F ⋅ dr = t2 C Circulation: The line integral C F1 t H F ⋅ dr of a continuous vector point functional F along a closed curve C is called the circulation of F round the closed curve C. This fact can also be represented by the symbol Ü. Irrotational vector field: A single valued vector point Function F (Vector Field F ) is called irrotational in the region R, if its circulation round every closed curve C in that region is zero that is z z C or F ⋅ dr = 0 F ⋅ dr = 0. 366 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 5.17 SURFACE INTEGRAL Any integral which is to be evaluated over a surface is called a surface integral. H Let F r be a continuous vector point function. Let r = F (u, v) be a smooth surface such that F (u, v) possesses continuous first order partial derivatives. Then the normal surface integral bg ej of F r over S is denoted by z ej z ej F r ⋅ da = F r ⋅ ndS S S H where da is the vector area of an element dS and n is a unit vector normal to the surface dS. Let F1, F2, F3 which are the functions of x, y, z be the components of F along the coordinate axes, then Surface Integral = = = = z z zz e zz b S S F ⋅ ndS F ⋅ da S S je F1 i + F2 j + F3 k ⋅ dydzi + dzdxj + dxdyk g j F1 dy dz + F2 dz dx + F3 dx dy . 5.17.1 Important Form of Surface Integral e Let dS = dS cos αi + cos βj + cos γ k j ...(i) where α, β and γ are direction angles of dS. It shows that dS cos α, dS cos β, dS cos γ are orthogonal projections of the elementary area dS on yz. plane, zx-plane and xy-plane respectively. As the mode of sub-division of the surface S is arbitrary we have chosen a sub-division formed by planes parallel to coordinate planes that is yz-plane, zx-plans and xy plane. Clearly, projection on the coordinate planes will be rectangles with sides dy and dz on yz plane, dz and dx on xz plane and dx and dy on xy plane. ⋅ i i ⋅ dS = dS cos αi + cos βj + cos γ kj Hence { i ⋅ n dS } = dS cos α = dy dz dy dz dS = . i. n Hence Similarly, multiplying both sides of (i) scalarly by j and k respectively, we have dz dx dS = j ⋅ n dx dy and dS = k ⋅ n dy dz F ⋅ n F ⋅ ndS Hence = S1 i ⋅ n dz dx F ⋅ n = j ⋅ n S2 z zz zz zz dx dy S3 k ⋅ n where S1, S2, S3 are projections of S on yz, zx and xy plane respectively. = F ⋅ n ...(ii) ...(iii) ...(iv) ...(v) 367 VECTOR CALCULUS 5.18 VOLUME INTEGRAL ej Let F r is a continuous vector point function. Let volume V be enclosed by a surface S given by a f r = f u, v sub-dividing the region V into n elements say of cubes having volumes ∆V1, ∆V2, .... ∆Vn ...(i) ∆Vk = ∆xk ∆yk ∆zk Hence k = 1, 2, 3, ... n where (xk, yk, zk) is a point say P on the cube. Considering the sum P ∑ Fbxk , yk , zk g∆Vk n Fig. 5.5 k =1 taken over all possible cubes in the region. The limits of sum when n → ∞ in such a manner that the dimensions ∆Vk tends to zero, if it exists is denoted by the symbol z ej z V F r dV ⋅ or V FdV or is called volume integral or space integral. If F = F1 i + F2 j + F3 k , then F r dV = i z ej V zzz V F1 dx dy dz + j zzz zzz V V F dx dy dz F2 dx dy dz + k zzz V F3 dx dy dz where F1, F2, F3 which are function of x, y, z are the components of F along X, Y, Z axes respectively. Independence of path If in a conservative field F ÜC F ⋅ dr = 0 along any closed curve C. Which is the condition of the independence of path. z Example 1. Evaluate F ⋅ dr where F = x 2 y 2 i + yj and the curve C, is y2 = 4x in the C xy-plane from (0, 0) to (4, 4). Sol. We know that r = xi + yj ∴ dr = dxi + dyj ∴ F ⋅ dr ∴ z = ex y i + yjj ⋅ edxi + dyjj 2 2 = x2y2dx + ydy F ⋅ dr C = ze C x 2 y 2 dx + ydy j 368 A TEXTBOOK OF ENGINEERING MATHEMATICS—I z z z af z z z = C x 2 y 2 dx + C y dy ⋅ But for the curve C, x and y both vary from 0 to 4. H H 4 2 4 F ⋅ dr = x 4x dx + y dy ∴ z 0 C 4 4 0 0 3 = 4 x dx + y dy Fx I F y I = 4G J +G J H 4K H 2K 4 4 4 2 0 zb Example 2. Evaluate [ y2 = 4x] 0 0 = 256 + 8 = 264. g x dy – y dx around the circle x2 + y2 = 1. Sol. Let C denote the circle x2 + y2 = 1, i.e., x = cos t, y = sin t. In order to integrate around C, t varies from 0 to 2π. zb ∴ C x dy − y dy z ze g = 02π FGH x dydt − y dxdt IJK dt 2π = 0 z = 0 dt = (t)2π 0 = 2π. z Example 3. Evaluate 2π j cos 2 t + sin 2 t dt C H H H F ⋅ dr , where F = (x2 + y2) i – 2xy j, the curve C is the rectangle in Y the xy-plane bounded by y = 0, x = a, y = b, x = 0. Sol. z C H H F ⋅ dr = z {e z {e }m j r x 2 + y 2 i – 2xyj ⋅ dxi + dyj C j } y=b x 2 + y 2 dx − 2xy dy = C Now, C is the rectangle OACB. B (0, b) ...(i) On OA, y = 0 ⇒ dy = 0 C (a, b) x=a x=0 On AC, x = a ⇒ dx = 0 On CB, y = b ⇒ dy = 0 On BO, x = 0 ⇒ dx = 0 ∴ From (i), H H F ⋅ dr = z C = O (0, 0) y=0 ze j zb g ze j z z z ze j z x + 0 dx + 2 OA AC –2ay dy + x + b dx + 2 CB 2 0 0 a 2 b x dx − 2 a y dy + x 2 + b 2 dx + 0 dy 0 0 a b Fx I F y I Fx I = G J − 2aG J + G + b xJ + 0 H 3K H 2K H 3 K 3 a 0 2 b 0 3 0 2 a BO Fig. 5.6 0 dy A (a, 0) X 369 VECTOR CALCULUS a3 a3 − ab 2 − − ab 2 3 3 = – 2ab2. = Example 4. Evaluate z C H H H F ⋅ dr , where F = yz i + zx j + xy k and C is the portion of the curve π H . r = (a cos t) i + (b sin t) j + (ct) k from t = 0 to 2 H Sol. We have r = (a cos t) i + (b sin t) j + (ct) k. Hence, the parametric equations of the given curve are x = a cos t y = b sin t z = ct H dr Also, = (– a sin t) i + (b cos t) j + ck dt H H H drH F ⋅ dr = Now, F ⋅ dt C C dt z z = = = zb zb ze C C C gb g yzi + zxj + xyk ⋅ − a sin ti + b cos t j + ck dt gb g bct sin t i + act cos t j + ab sin t cos t k ⋅ − a sin ti + b cos t j + ck dt j − abc t sin 2 t + abc t cos 2 t + abc sin t cos t dt z e z FGH z FGH j sin 2 t I = abc t cos 2 t + J dt C 2 K π sin 2 t I = abc 2 t cos 2 t + J dt 0 2 K π sin 2t cos 2t cos 2t O 2 L = abc Mt N 2 + 4 − 4 PQ0 abc = bt sin 2tg0 = 0. 2 2 2 = abc C t cos t − sin t + sin t cos t dt π 2 Example 5. Evaluate z Y C (4,12) F. dr where F = xyi + (x2 + y2)j and C c is the x-axis from x = 2 to x = 4 and the line x = 4 from y = 0 to y = 12. Sol. Here the curve C consist the line AB and BC. Since so z C r = xi + yj (as z = 0) dr = dxi + dyj F . dr = z{ e j} m r xyi + x 2 + y 2 j . dxi + dyj ABC A B (2, 0) (4, 0) O Fig. 5.7 X 370 = z{ e = z 4 x=2 0 . dx + z j } {xy dx + ex + y jdy} xy dx + x 2 + y 2 dy + AB A TEXTBOOK OF ENGINEERING MATHEMATICS—I z 2 BC 12 y=0 e j ze 12 0 + 16 + y 2 dy = 0 L y OP = [192 + 576] = 768. = M16 y + 3 PQ MN 2 j 16 + y 2 dy 3 12 0 Example 6. If F = (–2x + y)i + (3x + 2y)j, compute the circulation of F about a circle C in the xy plane with centre at the origin and radius 1, if C is transversed in the positive direction. Sol. Here the equation of circle is x2 + y2 = 1 Let x = cos θ, y = sin θ |As r = 1 2 2 x + y =1 C F = (–2cos θ + sin θ)i + (3 cos θ + 2sin θ)j r = xi + yj = (cos θ)i + (sin θ)i So dr = {(– sin θ)i + (cos θ)j}dθ Thus, the circulation along circle C = = z nb ze ze 2π θ=0 2π = = 0 2π 0 O z F . dr Fig. 5.8 C g b g s nb− sin θgi + bcos θg jsdθ −2 cos θ + sin θ i + 3 cos θ + 2 sin θ j . j 2 sin θ cos θ − sin 2 θ + 3 cos 2 θ + 6 sin θ cos θ dθ j 8 sin θ cos θ + 4 cos 2 θ − 1 dθ = 2π 2π 2π = −2 cos 2θ 0 + sin 2θ 0 + θ 0 z b 2π 0 g 4 sin 2θ + 2 cos 2θ + 1 dθ = − 2 cos 4π − cos 0 + sin 4π − sin 0 + 2π = 2π. Example 7. Compute the work done in moving a particle in the force field F = 3x2 i + (2xz – y)j + zk along. (i) A straight line from P(0, 0, 0) to Q(2, 1, 3). (ii) Curve C : defined by x2 = 4y, 3x3 = 8z from x = 0 to x = 2. is Sol. (i) We know that the equation of straight line passing through (x1, y1, z1) and (x2, y2, z2) x − x1 x 2 − x1 ⇒ or = y − y1 z − z1 = y 2 − y1 z 2 − z1 x−0 y−0 z−0 = = 2−0 1−0 3−0 ⇒ y z x = = 1 3 2 y z x = = = t (say), so x = 2t, y = t, z = 3t 1 3 2 371 VECTOR CALCULUS ∴ r = xi + yj + zk = 2ti + tj + 3tk ⇒ dr = (2i + j + 3k)dt and The work done = z F = (12t2)i + (12t2 – t)j + (3t)k Q P F . dr = = z ze e12t ji + e12t − tj j + a3tfk . 2i + j + 3k dt from 0 to 1 1 0 1 0 2 As t varies 2 j 24t 2 + 12t 2 − t + 9t dt = ze 1 0 j 36t 2 + 8t dt L 36t + 8t OP = 12 + 4 = 16. = M N 3 2Q F 3x − x I j + 3x k = 3x i + F 3x − x I j + 3x k (ii) F = 3x i + G 2x. GH 4 4 JK 8 H 8 4 JK 8 2 1 3 0 3 2 2 3 4 2 2 2 3 3 x 3x r = xi + yj + zk = xi + j+ k 4 8 F i + x j + 9x kI dx GH 2 8 JK L F 3x − x I j + 3x kOP . Li + x j + 9x kOdx F . dr = M 3x i + G MN H 4 4 JK 8 PQ MN 2 8 PQ F 3x + 3x − x + 27x I dx = GH 8 8 64 JK L x − x + 27x OP = Mx + N 16 32 64 × 6 Q 2 dr = Work done = z 2 x=0 z 2 4 2 2 3 2 0 z 2 5 2 3 5 0 6 3 4 6 2 0 = 8+ 64 16 27 × 64 1 9 − + = 8+ 4− + 16 32 64 × 6 2 2 = 16. Example 8. If V is the region in the first octant bounded by y2 + z2 = 9 and the plane x = 2 and F = 2x2yi – y2j + 4xz2k. Then evaluate zzz e j V ∇ ⋅ F dV. ∇ ⋅ F = 4 xy − 2 y + 8xz The volume V of the solid region is covered by covering the plane region OAB while x varies from 0 to 2. Thus, Sol. zzz e j zzz b V = 2 ∇ ⋅ F dV 3 9− y 2 x =0 y =0 z=0 g 4xy − 2y + 8xz dz dy dx 372 A TEXTBOOK OF ENGINEERING MATHEMATICS—I = = = = = zz z z LMNb g LM z MMNa fFGH c z a f 2 3 0 0 2 3 0 0 2 0 2 0 4xyz − 2 yz + 4xz 2 e 3 3 2 2 3 0 9 4 x − 2 + 72x dx 18x 2 − 18x + 36x 2 zz e S z2 = a2 in the first octant. S dy dx jOPQ dy dx 1 I F y IO 4x − 2 − 9 − y hJ + 4xG 9 y − J P dx K H 3 K PPQ 3 Example 9. Evaluate zz e 0 4xy − 2y 9 − y 2 + 4 x 9 − y 2 = 180. Sol. 9− y2 2 0 Fig. 5.9 j yzi + zxj + xyk ⋅ dS , where S is the surface of the sphere x2 + y2 + H yzi + zxj + xyk ⋅ dS j = = = zz e zz b zz S S (U.P.T.U., 2005) je yzi + zxj + xyk ⋅ dy dz i + ⋅ dz dx j + dx dy k yz dy dz + zx dz dx + xy dx dy a a2 − z2 0 0 yz dy dz + zz a2 − x 2 a 0 0 g zx dz dx + zz j a2 − y 2 a 0 0 xy dx dy Fy I Fz I Fx I = z zG J dz + z xG J dx + z y G J dy H2K H 2K H 2K 1 1 1 ze a − z jdz + xea − x jdx + yea − y jdy = 2 2 2 y I 1Fa z z I 1Fa x x I 1Fa y = 2G 2 − 4 J + 2G 2 − 4 J +2G 2 − 4 J H K H K H K a2 − z 2 2 a 0 a2 − x 2 2 a 0 z 0 z 0 a 2 2 0 a 0 2 Example 10. Evaluate zz = 2 4 a a 2 2 4 a 0 4 4 2 2 4 a 0 1a 1a 1a 3a . + + = 8 2 4 2 4 2 4 F . n ds , where F = zi + xj – 3y2 zk and S is the surface of the S f ≡ x2 + y2 – 16 ∇f = 2 4 cylinder x2 + y2 = 16 included in the first octant between z = 0 and z = 5. Sol. Since surface S : x2 + y2 = 16 Let 2 0 0 4 z 0 0 2 2 a2 − y 2 2 a F i ∂ + j ∂ + k ∂ I (x + y – 16) GH ∂x ∂y ∂z JK = 2xi + 2yi 2 2 373 VECTOR CALCULUS unit normal n = n = ∇f = ∇f 2xi + 2 yj E 4x 2 + 4 y 2 b 2 xi + yj 2 2 x +y C 4 16 FG xi + yj IJ = 1 (zx + xy) H 4 K 4 Here the surface S is perpendicular to xy-plane so we will take the projection of S on zx-plane. Let R be that projection. ∴ ds = zz F . n ds = S z ze z z FGH R = X B 2 2 Y x +y =16 Fig. 5.10 j bxi +4 yjg . y4 dx dz I JK zx + xy dx dz y 16 − x 2 on S. x is also varies from 0 to 4. Since z varies from 0 to 5 and y = zz O zi + xj − 3 y 2 zk . R F zx + xy I dx dz = GH y JK A dx dz dx dz 4dx dz = = y n . j y 4 R = ∴ D g = xi + yj = xi + yi / 2 F . n = (zi + xj – 3y2 zk). Now Z I F xz GG + xJ dx dz H 16 − x JK LM− z 16 − x + x OP dz = b 4z + 8g dz 2Q N 5 zz z 4 2 z=0 x=0 5 0 2 2 4 z 5 0 0 = (2z2 + 8z)05 = 50 + 40 = 90. Example 11. Evaluate zzz φdV, where φ = 45x2 y and V is the closed region bounded by V the planes 4x + 2y + z = 8, x = 0, y = 0, z = 0. Sol. Putting y = 0, z = 0, we get 4x = 8 or x = 2 Here x varies from 0 to 2 y varies from 0 to 4 – 2x and z varies from 0 to 8 – 4x – 2y Thus zzz φdV = V = zzz z z z zz zz b 45x 2 y dx dy dz V 2 4 − 2x x=0 y=0 = 45 = 45 2 4− 2x 0 0 2 4 − 2x 0 0 8 − 4x − 2 y z=0 45x 2 y dx dy dz 8 − 4x − 2 y x2y z 0 dx dy x 2 y 8 − 4x − 2 y dx dy g 374 A TEXTBOOK OF ENGINEERING MATHEMATICS—I z z z ze 2 O LM y dx 3 PQ N 2 L O = 45 x M 4b 4 − 2xg − 2xb 4 − 2xg − b 4 − 2xg P dx 3 N Q 45 x b 4 − 2xg dx = 3 4 − 2x 2 = 45 x 2 4 y 2 − 2xy 2 − 3 0 2 0 2 2 2 3 0 2 3 2 0 j 2 = 15 x 2 64 − 8x 3 − 96x + 48x 2 dx 0 LM 64x − 8x − 96x + 48x OP N 3 6 4 5 Q L 512 − 256 − 384 + 1536 OP = 15M 5 Q N3 3 3 6 5 2 4 = 15 0 = 128. EXERCISE 5.4 1. Find the work done by a Force F = zi + xj + yk from t = 0 to 2π, where r = cos t i + sin tj + tk. LMHint: Work done = FH ⋅ drH OP. z PQ MN 2. Show that z C Ans. 3π C F ⋅ dr = –1, where F = (cos y) i – xj – (sin y) k and C is the curve y = 1 − x2 in xy-plane from (1, 0) to (0, 1). 3. Find the work done when a force F = (x2 – y2 + x) i – (2xy + y) j moves a particle from 2 Ans. origin to (1, 1) along a parabola y2 = x. 3 4. z C LM N OP Q xy 3 dS where C is the segment of the line y = 2x in the xy plane from A(– 1, – 2, 0) to LMAns. 16 OP 5Q N B(1, 2, 0). e j e j 5. F = 2xzi + x 2 − y j + 2z − x 2 k is conservative or not. e j 2 6. If F = 2x − 3z i − 2xyj − 4xk , then evaluate x = 0, y = 0, z = 0 and 2x + 2y + z = 4. zzz Ans. ∇ × F ≠ 0, so non - conservative H ∇FdV , where V is bounded by the plane V LMAns. 8 OP N 3Q 375 VECTOR CALCULUS zz 7. Show that F ⋅ ndS = S bounded by the planes; 3 , where F = 4xzi – y2j + yzk and S is the surface of the cube 2 LMAns. 3 OP N 2Q x = 0, x = 1, y = 0, y = 1, z = 0, z = 1. H 2 8. If F = 2yi − 3 j + x k and S is the surface of the parobolic cylinder y2 = 8x in the first octant bounded by the planes y = 4, and z = 6 then evaluate zz S . F ⋅ ndS Ans. 132 bx − ygi + bx + yg j evaluate Ü A ⋅ dr around the curve C consisting of y = x and LMAns. 2 OP y = x. N 3Q 9. If A = 2 2 10. Find the total work done in moving a particle in a force field A = 3xyi − 5zj + 10xk along the curve x = t2 + t, y =2t2, z = t3 from t = 1 to t = 2. Ans. 303 11. Find the surface integral over the parallelopiped x = 0, y = 0, z = 0, x = 1, y = 2, z = 3 H when A = 2xyi + yz2j + xzk. Ans. 33 12. Evaluate zzz ∇ × AdV , where A = (x + 2y) i – 3zj + x k and V is the closed region in the V LMAns. 8 b3i − j + 2kgOP N 3 Q first octant bounded by the plane 2x + 2y + z = 4. zzz 13. Evaluate V fdV where f = 45x2y and V denotes the closed region bounded by the planes 4x + 2y + z = 8, x = 0, y = 0, z = 0. e j Ans. 128 2 14. If A = 2x − 3z i − 2xyj − 4xk and V is the closed region bounded by the planes x = 0, y = 0, z = 0 and 2x + 2y + z = 4, evaluate e j 3 15. If A = x − yz i − 2 x yj + 2 k evaluate 3 zzz e zzz e j V V LMAns. 8 b j − kgOP N 3 Q j ∇ × A dV. ∇ ⋅ A dV over the volume of a cube of side b. LMAns. 1 b OP N 3 Q 3 16. Show that the integral zaa ffe 3, 4 1, 2 j e j xy 2 + y 3 dx + x 2 y + 3xy 2 dy is independent of the path joining the points (1, 2) and Ans. 254 (3, 4). Hence, evaluate the integral. 17. If F = ∇ φ show that the work done in moving a particle in the force field F from (x1, y1, z1) to B (x2, y2, z2) is independent of the path joining the two points. 18. If F = (y – 2x)i + (3x + 2y)j, find the circulation of F about a circle C in the xy-plane with centre at the origin and radius 2, if C is transversed in the positive direction. [Ans. 8π] 19. If F (2) = 2i – j + 2k, F (3) = 4i – 2j + 3k then evaluate z 3 2 F. dF dt dt . [Ans. 10] 376 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 2 0 . Prove that F = (4xy – 3x2z2)i + 2x2 j – 2x3 zk is a conservative field. 2 1 . Evaluate z Sz F . n ds where F = xyi – x j + (x + z)k, S is the portion of plane 2x + 2y + z 2 LMAns. 27 OP N 4Q = 6 included in the first octant. 2 2 . Find the volume enclosed between the two surfaces S1 : z = 8 – x2 – y2 and S2 : z = x2 + 3y2. 5.19 [ Ans. 8π 2 ] GREEN’S* THEOREM If C be a regular closed curve in the xy-plane bounding a region S and P(x, y) and Q (x, y) be continuously differentiable functions inside and on C then (U.P.T.U., 2007) zz b g Pdx + Qdy = C zz FGH S I JK ∂Q ∂P − dx dy ∂x ∂y Y y = f2(x) Proof: Let the equation of the curves AEB and AFB are y = f1(x) and y = f2 (x) respectively. Consider ∂P dx dy = S ∂y zz = = a f ∂P dy dx x = a y = f a x f ∂y b y = f axf P bx , y g dx y = f axf a b F d S f2 x C B zz z z b g b g zb g zb g zb g zb g zb g zb g zz z z b bg g z b g b g z b g zb g z b g zz A 1 c y = f1(x) E 2 1 b a = − P x, f 2 − P x, f1 dx a b = − P x, f 2 dx − P x, y dx − BFA = − O b a P x, f 1 dx a b X Fig. 5.11 P x, y dx AEB P x , y dx BFAEB ⇒ zz ∂P S ∂y dx dy = − C P x, y dx ...(i) Similarly, let the equations of the curve EAF and EBF be x = f1 (y) and x = f2 (y) respectively, ∂Q then S ∂x dx dy = = ⇒ ∂Q S ∂x dx dy = * George Green (1793–1841), English Mathematician. d f2 y y = c x = f1 y d c C d ∂Q dx dy = Q f2 , y − Q f1 , y dy c ∂x c Q f 2 , y dy + Q f1 , y dy d Q x, y dy ...(ii) 377 VECTOR CALCULUS Adding eqns. (i) and (ii), we get zb Pdx + Qdy C g= Vector form of Greens theorem: z F ⋅ dr = C Let F = Curl F = zz FGH S I JK ∂Q ∂P − dx dy ∂x ∂y zz e j ∇ × F ⋅ kdS S Pi + Qj, then we have i ∂ ∂x P j ∂ ∂y Q e∇ × Fj ⋅ k = FGH ∂∂Qx − ∂∂Py IJK ⇒ z Thus, F ⋅ dr = C k ∂ ∂z 0 = F ∂Q − ∂P I k GH ∂x ∂y JK zz e j ∇ × F ⋅ k dS S Corollary: Area of the plane region S bounded by closed curve C. Let Q = x, P = – y, then zb xdy − ydx C g= zz a f z z S 1 + 1 dxdy = 2 S dx dy = 2 A Thus, area A = 1 ( xdy − ydx ) 2C Example 1. Verify the Green’s theorem by evaluating ze j e x 3 − xy 3 dx + y 3 − 2 xy dy C j where C is the square having the vertices at the points (0, 0), (2, 0) (2, 2) and (0, 2). (U.P.T.U., 2007) ze Sol. We have C By Green’s theorem, we have zb C Here ∴ Pdx + Qdy g = j e x 3 − xy 3 dx + y 3 − 2 xy dy zz FGH I JK ∂Q ∂P − dx dy S ∂x ∂y P = (x3 – xy3), Q = (y3 – 2xy) ∂P ∂y = – 3xy2, ∂Q = – 2y ∂x j ...(A) 378 A TEXTBOOK OF ENGINEERING MATHEMATICS—I F ∂Q ∂P I So zz G ∂x − ∂y J dx dy = z z e–2y + 3xy jdx dy K H 2 z z 2 x =0 dx zz S z z 2 ydy + 3 y= 0 2 x= 0 xdx 2 y= 0 y 2 dy 2 2 2 2 3 2 0 0 0 Now, the line integral zb g Pdx + Qdy = C ze ze OA + O ...(i) j j e j x 3 − xy 3 dx + y 3 − 2xy dy + ze j e ze g Pdx + Qdy = AB j x 3 − xy 3 dx + y 3 − 2xy dy + C z ze 2 3 2 0 0 x dx + j y 3 − 4y dy + ze 0 2 j e ze j e x 3 − xy 3 dx + y 3 − 2xy dy CO j z 2 4 0 ⇒ zb g 2 2 4 0 0 2 0 4 0 2 2 2 = 4 – 4 + 12 – 4 Pdx + Qdy = 8 C j x 3 − 8x dx + y 3 dy O Lx Lx O L y O Ly O = M 4 P + M 4 − 2 y P + M 4 − 4x P + M 4 P PQ N N Q MN Q MN PQ 4 j x 3 − xy 3 dx + y 3 − 2xy dy along OA, y = 0 ⇒ dy = 0 and x = 0 to 2 along AB, x = 2 ⇒ dx = 0 and y = 0 to 2 along BC, y = 2 ⇒ dy = 0 and x = 2 to 0 along CO, x = 0 ⇒ dx = 0 and y = 2 to 0 zb X A (2, 0) (0, 0) Fig. 5.12 j e BC ∴ B (2, 2) x 3 − xy 3 dx + y 3 − 2xy dy C = But C (0, 2) LM y OP + 3LM x OP LM y OP = −8 + 16 MN 2 PQ MN 2 PQ MN 3 PQ F ∂Q − ∂P I dx dy = 8 GH ∂x ∂y JK = –2 x 02 ⇒ Y 2 x =0 y = 0 S = –2 2 ...(ii) Thus from eqns. (i) and (ii) relation (A) satisfies. Hence, the Green’s theorem is verified. Hence proved. Example 2. Verify Green’s theorem in plane for ze C boundary of the region defined by y2 = 8x and x = 2. Sol. By Green’s theorem z C b Pdx + Qdyg = zz FGH S I JK ∂Q ∂P − dx dy ∂x ∂y j e j x 2 − 2xy dx + x 2 y + 3 dy , where C is the 379 VECTOR CALCULUS i.e., Line Integral (LI) = Double Integral (DI) P = x2 – 2xy, Q = x2y + 3 Here, ∂P ∂y ∂Q = 2 xy ∂x So the R.H.S. of the Green’s theorem is the double integral given by DI = = = −2x, I JK zz FGH zz c ∂Q ∂P − dx dy ∂x ∂y S a fh 2 xy − −2 x dx dy S The region S is covered with y varying from –2 2 x of the lower branch of the parabola to its upper branch 2 2 x while x varies from 0 to 2. Thus DI = = zz b g z z ze j e j 2 8x x = 0 y= − 8x 2 0 xy 2 + 2xy 2xy + 2x dy dx 8x dx − 8x 3 128 5 The L.H.S. of the Green’s theorem result is the line integral 2 = 8 2 LI = 0 x 2 dx = x 2 − 2 xy dx + x 2 y + 3 dy ⋅ C Here C consists of the curves OA, ADB. BO. so LI = zz z z z = OA + ADB + BO C = + OA ADB + BO = LI 1 + LI 2 + LI 3 Y 2 Along OA: y = –2 2 x , so dy = – dx x LI1 = ze z j e 0 2 y = j 8x B (2, 4) x 2 − 2xy dx + x 2 y + 3 dy OA 2 j F 2I + x e−2 2 x j + 3 G − J dx H xK F I = z G 5x + 4 2 ⋅ x − 3 2 x J dx H K = x=2 e x 2 − 2x –2 2 x dx D (2, 0) O (0, 0) 2 2 0 2 32 − 1 2 A (2, –4) Fig. 5.13 X 380 A TEXTBOOK OF ENGINEERING MATHEMATICS—I L 5x + 4 2 2 x − 3 2 ⋅ 2 x OP = M 5 MN 3 PQ 40 64 + − 12 = 3 5 x = 2, dx = 0 Along ADB : LI2 = = Along BO : z e zb ADB 4 2 5 2 3 j e 0 j x 2 − 2xy dx + x 2 y + 3 dy g 4 y + 3 dy = 24 –4 y = 2 2 x , with x : 2 to 0. dy = LI3 = 2 dx x ze z FH BO j e 0 = 3 − 1 IK 5x 2 − 4 2 x 2 + 3 2 x 2 dx 2 = − j x 2 − 2xy dx + x 2 y + 3 dy 40 64 + − 12 3 5 LI = LI1 + LI2 + LI3 = ⇒ Hence the Green’s theorem is verified. Example 3. Apply Green’s theorem to evaluate FG 40 + 64 − 12IJ + a24f +FG − 40 + 64 − 12IJ = 128 H3 5 K H 3 5 K 5 ze C j e j 2x 2 − y 2 dx + x 2 + y 2 dy where C is the boundary of the area enclosed by the x-axis and the upper half of the circle x2 + y2 = a2. (U.P.T.U., 2005) Sol. By Green’s theorem 2 2 zz z z zz z bPdx + Qdyg = FGH ∂∂Qx − ∂∂Py IJK dx dy LM ∂ ex + y j − ∂ e2x − y jOPdx dy = ∂y N ∂x Q b2x + 2ygdx dy = LMN2xy + 2y2 OPQ dx. = = z 2x a − x + e a − x j dx = 0 + 2z e a − x jdx C y= a –x S a a2 −x2 2 2 2 2 x =− a y = 0 a –a z a 2 − x2 −a 0 a 2 2 2 O Fig. 5.14 a a 2 −x 2 2 a −a 0 a 2 −a S 2 2 0 [First integral vanishes as function is odd] LM N = 2 a 2x − OP Q a LM N x3 a3 = 2 a3 − 3 0 3 OP = 4 a . Q 3 3 382 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Let y = 3 sin θ in first and y = sin φ in second integral. z z π π sin 4 θ cos θ sin 4 φ cos φ dθ − 4 2 dφ 0 0 cos θ cos φ L.H.S. = 4 × 81 2 5 1 5 1 = 4 × 81 2 2 − 4 2 2 = 60 π 2 3 2 3 ⇒ L.H.S. = R.H.S. Hence Green’s theorem is verified. Example 5. Find the area of the loop of the folium of Descartes x3 + y3 = 3axy, a > 0. Sol. Let y = tx. ...(i) 3 3 3 ∴ x + t x = 3ax.tx giving ...(ii) 0 x dy − y dx 1 x2 ⋅ 2 C x2 = y 1 x 2d x 2 C = 1 2 x dt , as y = tx 2 C = 1 9a 2 t 2 dt using (ii) 2 C 1 + t3 2 1 3t 2 0 1 + t3 t=0 = = 2 = 3a O a z z FGH IJK z ze j ze j g + zb 1 x dy − y dx 2 C y required area = x t=1 x+ Hence, 3at x = 1 + t3 y= Fig. 5.16 dt, by summetry 2 LM 1 OP = 3 a . N 1+t Q 2 = 3a2 − 1 2 3 0 Example 6. Using Green’s theorem, find the area of the region in the first quadrant bounded 1 x by the curves y = x, y = ,y= . (U.P.T.U., 2008) x 4 Sol. By Green’s theorem the area of the region is given by A = = zb 1 xdy − ydx 2 C 1 2 g LM O xdy − ydx g + b xdy − ydxg + bxdy − ydxgP b MMC PP C C N Q z 1 z 2 z 3 ...(i) 383 VECTOR CALCULUS Y B( C3 y = x ) 1,1 C2 A (2,1) 2 C1 y=x 4 (0,0) O y=1 x X Fig. 5.17 Now along the curve C1 : y = zb xdy − ydx z g = FGH x dx − x dxIJK = 0 4 4 2 ...(ii) 0 C1 along, the curve C2 : y = zb zb 1 1 , dy = − 2 dx and x varies from 2 to 1. x x xdy − ydx g= C2 xdy − ydx or x dx or dy = and x varies from 0 to 2. 4 4 z RST FGH IJK 1 2 x. 1 1 dx − dx 2 x −x z UV = W 1 2 − 2 dx x g = − 2 log x = – 2[log 1 – log 2] = 2 log 3 1 ...(iii) 2 C2 along, the curve C3 : y = x, dy = dx and x varies from 1 to 0. zb xdy − ydx z g = bxdx − xdxg = 0 0 ...(iv) 1 C3 Using (ii), (iii) and (iv) in (i), we get the required area A = 1 [0 + 2 log 2 + 0] = log 2. 2 Example 7. Verify Green’s theorem in the xy-plane for ze C C is the boundary of the region enclosed by y = x2 and y2 = x. Sol. Here P(x, y) = 2xy – x2 j e Y 2 x=y Q(x, y) = x2 + y2 ∂Q = 2x, ∂x zb Pdx + Qdy C g = z z FGH S 2 I JK ∂Q ∂P − dx dy ∂x ∂y y=x A ∂P = 2x ∂y By Green’s theorem, we have j 2xy − x 2 dx + x 2 + y 2 dy , where C2 (1, 1) C1 O ...(i) Fig. 5.18 X 384 ∴ z zb ze R.H.S. = A TEXTBOOK OF ENGINEERING MATHEMATICS—I g 2x − 2x dx dy = 0 S and = ze ze j C j e j 2xy − x 2 dx + x 2 + y 2 dy + C1 e j 2xy − x 2 dx + x 2 + y 2 dy L.H.S. = ze j e j 2xy − x 2 dx + x 2 + y 2 dy C2 ...(ii) Along C1 : y = x2 i.e., dy = 2xdx and x varies from 0 to 1 C1 j e j 2xy − x 2 dx + x 2 + y 2 dy ze ze 1 = 0 1 = 0 C2 j e j 2xy − x 2 dx + x 2 + y 2 dy L x + x OP = 1 4x − x + 2x jdx = Mx − N 3 3Q 3 2x1 2 z LMNe 0 = 1 z FGH z FGH 0 = 1 2 5 and x varies from 1 to 0. j j 2xdx OPQ e 2x ⋅ x 1 2 − x 2 dx + x 2 + x . 2x 3 2 − x 2 + 1 1 2 IJ K 1 32 1 12 x + x dx 2 2 IJ K 5 Lx O = M P − LMN x3 OPQ + 12 LMN x3 2 OPQ 2N5 2Q 0 = 6 1 3 4 0 dx Along C2 : y2 = x, 2y dy = dx or dy = ze j 2x 3 − x 2 + 2x 3 + 2x 5 dx 5 32 1 x − x 2 + x1 2 dx 2 2 52 0 3 0 32 0 1 1 1 = −1+ 1 1 − = −1 3 3 Using the above values in (ii), we get L.H.S. = 1 – 1 = 0 Thus L.H.S. = R.H.S.; Hence, the Green’s theorem is verified. EXERCISE 5.5 1. Using Green’s theorem evaluate ze C j x 2 y dx + x 2 dy , where C is the boundary described counter clockwise of the triangle with vertices (0, 0), (1, 0), (1, 1). z {e j } (U.P.T.U., 2003) 5 Ans. 12 LM N OP Q xy + y 2 dx + x 2dy , where C is the closed curve of 1 Ans. – the region bounded by y = x2 and y = x. 20 2. Verify Green’s theorem in plane for C LM N OP Q 385 VECTOR CALCULUS 3. Evaluate z (cos x sin y − xy)dx + sin x ⋅ cos y dy by Green's theorem where C is the circle C x2 + y2 = 1. 4. Evaluate by Green's theorem zn s e − x sin y dx + e − x cos y dy , where C is the rectangle with C vertices (0, 0) (π, 0), [Ans. 0] F π, 1 πI F 0, 1 πI . H 2 K H 2 K [Ans. 2(e–π–1)] 5. Find the area of the ellipse by applying the Green's theorem that for a closed curve C in the xy-plane. [Hint: Parametric eqn. of ellipse x = a cos φ, y = a sin φ and φ vary from φ1 = 0 to φ2 = 2π] [Ans. π ab] 6. Verify the Green's theorem to evaluate the line integral z (2y 2 dx + 3x dy) , where C is the LMAns. 27 OP N 4Q C boundary of the closed region bounded by y = x and y = x2 . 7. Find the area bounded by the hypocycloid x2/3 + y2/3 = a2/3 with a > 0. O [Hint: x = a cos3 φ, y = a sin3 φ, φ varies from φ1 = 0 to φ1 = π/2 A] LMAns. 3πa OP N 8 Q 2 8. Verify Green's theorem z ( 3x + 4 y ) dx + (2 x − 3 y) dy with C; x2 + y2 = 4. C [Ans. Common value: – 8π] LMAns. 3a OP N 2Q 2 9. Find the area of the loop of the folium of descartes x3 + y3 = 3axy, a > 0. 10. Evaluate the integral z [Hint: Put y = tx, t : 0 to ∞] ( x 2 − cos hy ) dx + ( y + sin x ) dy, where C: 0 ≤ x ≤ π, 0 ≤ y ≤ l. C 11. Verify Green’s theorem in the xy-plane for ze [Ans. π (cos h 1 – 1)] j b g 3x 2 − 8 y 2 dx + 4 y − 6xy dy , where C is the C LMAns. 3 OP N 2Q 1 9π O L r dθ = 12. Find the area of a loop of the four - leafed rose r = 3 sin2θ. MHint : A = z 2 8 PQ N 13. Verify the Green’s theorem for e y dx + x dyj , where C is the boundary of the square – 2 x and y = x . region bounded by parabolas y = 1 ≤ x ≤ 1 and –1 ≤ y ≤ 1. z C π 2 2 0 2 2 386 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 14. Using Green’s theorem in the plane evaluate z b g e j 2 tan −1 y x dx + log x 2 + y 2 dy , where C C is the boundary of the circle (x – 1)2 + (y + 1)2 = 4. 5.20 STOKE’S THEOREM H If F is any continuously differentiable vector function and S is a surface enclosed by a curve C, then z F ⋅ dr = C zz . (∇ × F) ⋅ ndS S where n is the unit normal vector at any point of S. (U.P.T.U., 2006) Proof: Let S is surface such that its projection on the xy, yz and xz planes are regions bounded by simple closed curves. Let equation of surface f(x, y, z) = 0, can be written as z = f1 (x, y) y = f2 (x, z) x = f3 (y, z) H Let F = F1 i + F2 j + F3 k Then we have to show that zz S = ∇ × { F1 i + F2 j + F3 k } ⋅ ndS Considering integral zz z H H F ⋅ dr C , we have ∇ × ( F1 i ) ⋅ ndS S LMRSi ∂ + j ∂ + k ∂ UV × F iOP ⋅ ndS ...(i) NT ∂x ∂y ∂z W Q L ∂F j − ∂F kOP ⋅ ndS = M N ∂z ∂y Q L ∂F n ⋅ j − ∂F n ⋅ kOP dS = M ∂y N ∂z Q H [∇ × (F1 i)]· n dS = 1 1 1 1 Also, So, But So, Fig. 5.19 1 r = xi + yj + zk = xi + yj + f1(x, y)k H ∂r ∂f = j+ 1k ∂y ∂y [As z = f(x, y)] H ∂r is tangent to the surface S. Hence, it is perpendicular to n . ∂y n ⋅ ∂r ∂y = n ⋅ j + ...(ii) ∂f1 n ⋅ k = 0 ∂y 387 VECTOR CALCULUS n ·j = − Hence, Hence, (ii) becomes ∂f 1 ∂z n ⋅ k = − n ⋅ k ∂y ∂y LM ∂F ∂z + ∂F OPn ⋅ k dS N ∂z ∂y ∂y Q [∇ × (F1 i)]· n dS = – 1 But on surface S 1 ...(iii) F1 (x, y, z) = F1 [x, y, f1 (x, y)] ∂F1 ∂F1 ∂z + ⋅ ∂y ∂z ∂y ∴ = F(x, y) ...(iv) ∂F ∂y ...(v) = Hence, relation (iii) with the help of relation (v) gives ∂F ∂F ( n ·k) dS = – dx dy ∂y ∂y H ∂F − dx dy R ∂y [∇ × (F1 i)]· n dS = – zz S ( ∇ × F1 i) ⋅ n dS zz = ...(vi) where R is projection of S on xy-plane. Now, by Green’s theorem in plane, we have H H ∂F dx dy, Fdx = − R ∂y C1 zz z where C1 is the boundary of R. As at each point (x, y) of the curve C1 the value of F is same as the value of F1 at each point (x, y, z) on C and dx is same for both curves. Hence, we have H H F dx = F1 dx . z z C1 Hence, z C F1 dx = – C From eqns. (vi) and (vii), we have zz l zz l zz l S q ∇ × F1 i ⋅ n dS = z z z C zz ...(vii) H F1 dx ...(viii) H ∂F dx dy R ∂y Similarly, taking projection on other planes, we have H F2 dy . ∇ × F2 j ⋅ n dS = C S H F3 dz ∇ × F3 k ⋅ n dS = S q q C Adding eqns. (viii), (ix), (x), we get zz S ⇒ ∇ × { F1 i + F2 j + F3 k } ⋅ n dS = z H F ⋅ dr = C z C zz {F1 dx + F2 dy + F3 dz} S ( ∇ × F) ⋅ ndS . ...(ix) ...(x) 388 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 5.21 CARTESIAN REPRESENTATION OF STOKE'S THEOREM H F = F1i + F2 j + F3 k Let i ∂ ∂x F1 H curl F = = So the relation z H H F ⋅ dr = C is transformed into the form z C zz LMNFGH {F1 dx + F2 dy + F3 dz} = S j ∂ ∂y F2 k ∂ ∂z F3 RS ∂F − ∂F UV i + RS ∂F − ∂F UV j + RS ∂F − ∂F UV k T ∂y ∂z W T ∂z ∂x W T ∂x ∂y W 3 z S 2 1 3 2 1 H curl F ⋅ n dS, IJ K FG H IJ K FG H OP Q IJ K ∂F3 ∂F2 ∂F1 ∂F3 ∂F2 ∂F1 − dy dz + − − dz dx + dx dy . ∂y ∂z ∂x ∂z ∂x ∂y H Example 1. Verify Stoke's theorem for F = (x2 + y2) i – 2xy j taken round the rectangle bounded by x = ± a, y = 0, y = b. (U.P.T.U., 2002) H Sol. We have F ⋅ dr = {(x2 + y2) i – 2xy j}· {dx i + dy j} = (x2 + y2) dx – 2xy dy Y H H H H H H H H H H ∴ F ⋅ dr + F ⋅ dr + F ⋅ dr + F ⋅ dr F ⋅ dr = z z ∴ z C1 C C2 zn zn z C3 = I1 + I2 + I3 + I4 z C4 s LM 3 y = b OP = ( x + b ) dx − 0s N∴ dy = 0Q = FG x + b xIJ H3 K F 2 a + 2b aI = – H3 K n(x + y )dx − 2xy dys I = I1 = (–a, b) D A (a, b) C1 ( x 2 + y 2 )dx − 2 xy dy C1 −a 2 C2 C4 2 a 2 a 3 z 2 2 zn 0 b = 2a z s b y dy (–a, 0) O C3 B (a, 0) X Fig. 5.20 2 ( −a)2 + y 2 0 − 2(−a) y dy 0 X¢ 2 C2 = C −a 3 LM3 x = − aOP N∴ dx = 0Q 389 VECTOR CALCULUS Fy I = 2a G J = – ab H2K 2 z z z z I3 = 2 b C3 = ( x 2 + y 2 )dx − 2 xy dy +a −a I4 = C4 x 2 dx = FG x IJ = 2 a H3K 3 3 a 3 −a LM 3 x = 0 OP N∴ dx = 0Q −2ay dy = –2a = – ab ∴ LM 3 y = 0 OP N∴ dy = 0Q x 2 dx C3 = 0 z b 0 2 Fy I y dy = −2 aG J H2K b 2 z 0 H H F ⋅ dr = I1 + I 2 + I 3 + I 4 C FG 2a + 2b aIJ − ab + 2 a − ab H3 K 3 3 = – 2 2 = – 4ab2 H curl F = Again, 3 2 ...(i) i ∂ ∂x x2 + y2 j ∂ ∂y −2xy k ∂ ∂z 0 = – 4yk ∴ ∴ zz S n = k H n · curl F = k·(– 4yk) = – 4y H n ⋅ curl F dS = zz z FGH IJK a b −a 0 +a = −a − 4y dx dy −4 y2 2 b dx 0 = – 2b 2 ( x) −a a = – 4ab2. From eqns. (i) and (ii), we verify Stoke’s theorem. → ...(ii) Example 2. Verify Stoke's theorem when F = yi + zj + xk and surface S is the part of the sphere x2 + y2 + z2 = 1, above the xy-plane. Sol. Stoke’s theorem is H H H F ⋅ dr = (curl F ) ⋅ ndS z C zz S 390 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Here, C is unit circle x2 + y2 = 1, z = 0 H H Also, F ⋅ dr = (yi + zj + xk) · (dxi + dyj + dzk) = ydx + zdy + xdz H H ydx + zdy + xdz F ⋅ dr = ∴ z z z z z z z C C C C Again, on the unit circle C, z = 0 dz = 0 Let x = cos φ, ∴ dx = – sin φ.dφ and y = sin φ, ∴ dy = cos φ.dφ H H y dx F ⋅ dr = ∴ C C = 2π 0 = – sin φ ( − sin φ) dφ z = –π H curl F = Again, 2π 0 sin 2 φ dφ ...(i) i ∂ ∂x y j ∂ ∂y z k ∂ ∂z = – i – j – k x Using spherical polar coordinates n = sin θ cos φ i + sin θ sin φ j + cos θ k H ∴ curl F · n = – (sin θ cos φ + sin θ sin φ + cos θ). H π/ 2 2 π (sin θ cos φ + sin θ sin φ + cos θ) sin θ dθ dφ ( curl F) ⋅ n dS = – Hence, z z zz θ= 0 φ = 0 S = – z π/2 = – 2π = –π [sin θ sin φ − sin θ cos φ + φ cos θ]20 π sin θ dθ z z θ= 0 π/2 0 sin θ cos θ dθ π/ 2 0 sin 2θ dθ π (cos 2θ) π0 /2 2 = –π ...(ii) From eqns. (i) and (ii), we verify Stoke’s theorem. H Example 3. Verify Stoke’s theorem for F = xzi − yj + x2yk, where S is the surface of the region bounded by x = 0, y = 0, z = 0, 2x + y + 2z = 8 which is not included in the xz-plane. (U.P.T.U., 2006) Sol. Stoke’s theorem states that = z F ⋅ dr = C zz e j S ∇ × F ⋅ ndS 391 VECTOR CALCULUS Here C is curve consisting of the straight lines AO, OD and DA. L.H.S. = = z F ⋅ dr = z z z z z z C AO + OD Z D (0, 0, 4) AO + OD + DA + = LI 1 + LI 2 + LI 3 DA H On the straight line AO : y = 0, z = 0, F = 0, so LI1 = AO F ⋅ dr = 0 O On the straight line OD : x = 0, y = 0. F = 0, so LI2 = OD Y B (0, 8, 0) X F ⋅ dr = 0 A (4, 0, 0) Fig. 5.21 On the straight line DA: x + z = 4 and y = 0, so F = xzi = x (4 – x) i LI3 = z DA F ⋅ dr = z 4 0 x( 4 − x) i ⋅ dxi = z 4 0 x( 4 − x) dx = 32 3 32 32 = 3 3 Here the surface S consists of three surfaces (planes) S1 : OAB, S2 : OBD, S3 : ABD, so that LI = 0 + 0 + R.H.S. = = ∇× F = zz zz zz zz zz ^ S ( ∇ × F ) ⋅ n dS = S1 i ∂ ∂x xz + S2 j ∂ ∂y −y On the surface S1: Plane OAB : z = 0, (∇ × F ) ⋅ n = + S3 = SI 1 + SI 2 + SI 3 k ∂ = x2i + x(1 – 2y) j ∂z x2 y n = − k , so x 2 i + x (1 − 2 y ) j ⋅ ( − k ) = 0 zz =0 (∇ × F) ⋅ ndS zz =0 (∇× F) ⋅ ndS Unit normal n to the surface S3 = ∇(2x + y + 2 z) |∇(2x + y + 2 z)| SI1 = S1 + S2 + S3 S1 On surface S2: Plane OBD : Plane x = 0, n = – i, so ∇×F = 0 SI2 = S2 On surface S3: Plane ABD : 2x + y + 2z = 8. 392 A TEXTBOOK OF ENGINEERING MATHEMATICS—I n = 2i + j + 2k 4 +1+ 4 = 2i + j + 2 k 3 2 2 1 x + x (1 − 2 y ) 3 3 To evaluate the surface integral on the surface S3, project S3 on to say xz-plane i.e., projection of ABD on xz-plane is AOD (∇ × F ) ⋅ n = dS = Thus SI3 = = = dx dz dx dz = 3dx dz = n⋅ j 13 zz S3 (∇× F) ⋅ ndS zz LMN AOD zz 4 4− x x=0 z=0 OP Q 2 2 x x + (1 − 2 y) 3 dx dz 3 3 2x2 + x (1 − 2y) dz dx since the region AOD is covered by varying z from 0 to 4 – x, while x varies from 0 to 4. Using the equation of the surface S3, 2x + y + 2z = 8, eliminate y, then SI3 = = zz n zz 4 4 −x 4 4− x 0 0 s 2x 2 + x [1 − 2 (8 − 2x − 2z)] dz dx 0 0 (6x2 − 15x + 4xz) dz dx = z LMN6x z − 15xz + 4xz2 OPQ 4 2 2 0 = z 4 0 4− x dx 0 (23x 2 − 4 x 3 − 28 x) dx = 32 3 Thus L.H.S. = L.I. = R.H.S. = S.I. zz Hence Stoke’s theorem is verified. Example 4. Evaluate S over the surface of intersection of the cylinders x2 + y2 (∇× F) ⋅ ndS = a2, x2 + z2 = a2 which is included in the first octant, given that F = 2yzi – (x + 3y – 2) j + (x2 + z)k. Sol. By Stoke’s theorem the given surface integral can be converted to a line integral i.e., SI = zz S = ( ∇ × F ) ⋅ ndS z F ⋅ dr = LI C Here C is the curve consisting of the four curves C1: x2 + z2 = a2, y = 0; C2: x2 + y2 = a2, z = 0, C3: x = 0, y = a, 0 ≤ z ≤ a: C4: x = 0, z = a, 0 ≤ y ≤ a. 393 VECTOR CALCULUS Z 2 2 x +z =a 2 C4 C1 C3 Y 2 2 x +y =a X 2 C2 LI = Fig. 5.22 z F ⋅ dr = C z C1 + C2 + C 3 + C 4 z z z z = C1 + C2 + C3 + C4 = LI1 + LI2 + LI3 + LI4 On the curve C1: y = 0; x2 + z2 = a2 LI1 = = z C1 z 0 a z F ⋅ dr = C1 ( a 2 − z 2 ) + z dz = − On the curve C2: z = 0, x2 + y2 = a2 LI2 = z C2 z F ⋅ dr = ze a = – ( x 2 + z) dz 0 C2 2 3 a2 a − 3 2 − ( x + 3 y − 2) dy j a 2 − y 2 + 3 y − 2 dy πa 2 3 2 − a + 2a 4 2 On the curve C3: x = 0, y = a, 0 ≤ z ≤ a = – LI3 = On C4; x = 0, z = a, 0 ≤ y ≤ a LI4 = SI = z C3 z z zz e j F ⋅ dr = S z F ⋅ dr = a 0 zdz = a2 2 0 a ( 2 − 3 y) dy = −2a + = LI = ∇ × F ⋅ ndS 3a 2 2 FG −2a − a IJ H 3 2K F πa − 3a + 2aIJ + a + FG −2a + 3a IJ + G− H 4 2 K 2 H 2 K 3 2 2 SI = − a2 ( 3 π + 8 a). 12 2 2 2 394 A TEXTBOOK OF ENGINEERING MATHEMATICS—I z Example 5. Evaluate S F ⋅ dr by Stoke’s theorem, where F = y 2 i + x 2 j − ( x + z) k and C is (U.P.T.U., 2001) the boundary of the triangle with vertices at (0, 0, 0) (1, 0, 0) and (1, 1, 0) Sol. Since z-coordinates of each vertex of the triangle is zero, therefore, the triangle lies in the xy-plane and n = k i j k ∂ ∂ ∂ curl F = = j + 2 ( x − y) k ∂x ∂y ∂z Y y 2 x 2 −( x + z) ∴ curl F ⋅ n = j + 2 ( x − y ) k ⋅ k = 2 ( x − y ) B (1, 1) The equation of line OB is y = x. By Stoke’s theorem z C F ⋅ dr = = = y= zz zz z LMN S curl F ⋅ n dS 1 x 0 0 x S 2 ( x − y ) dy dx OP Q x O z FGH Fig. 5.23 IJ K A (1, 0) X 1 1 y2 x2 1 2 xy − dx = 2 x2 − dx = x 2 dx = . 0 0 0 2 0 2 3 1 z Example 6. Apply Stoke’s theorem to prove that z ( ydx + zdy + xdz) = −2 2 πa 2 , where C is the curve given by x2 + y2 + z2 – 2ax – 2ay = 0, x + y = 2a and begins of the point (2a, 0, 0). Sol. The given curve C is x2 + y2 + z2 – 2ax – 2ay = 0 x + y = 2a C ⇒ (x – a)2 + (y – a)2 + z2 = b a 2 g2 x + y = 2a which is the curve of intersection of the sphere (x – a)2 + (y – a)2 + z2 = ba 2 g 2 and the plane x + y = 2a . Clearly, the centre of the sphere is (a, a, 0) and radius is a 2 . Also, the plane passes through (a, a, 0). Hence, the circle C is a great circle. ∴ Radius of circle C = Radius of sphere = Now, z C ( ydx + zdy + xdz) = z C 2a ( yi + zj + xk ) ⋅ ( dxi + dyj + dzk ) Fig. 5.24 VECTOR CALCULUS = = But curl (yi + zj + xk) = z z 395 C C H ( yi + zj + xk ) ⋅ dr . curl ( yi + zj + xk ) ⋅ ndS i ∂ ∂x y j ∂ ∂y z [Using Stoke’s theorem] k ∂ = – i – j – k. ∂z x Since S is the surface of the plane x + y = 2a bounded by the circle C. Then n = = ∴ b g ∇bx + y − 2ag ∇ x + y − 2a i+ j 2 curl (yi + zj + xk) · n = (– i – j – k) · = − Hence, the given line integral = = – z Example 7. Evaluate 1+1 z 2 S FG i + j IJ H 2K =– 2. − 2 dS 2 (Area of the circle C). b 2 ag = −2 2πa . exy dx + xy dyj taken round the positively oriented square with = – 2 2π 2 2 vertices (1, 0), (0, 1), (– 1, 0) and (0, – 1) by using Stoke’s theorem and verify the theorem. Sol. We have z C Y ( xy dx + xy 2 dy) = = = z C z zz C B (0, 1) ( xy i + xy 2 j ) ⋅ ( dx i + dy j ) y–x=1 H ( xy i + xy 2 j ) ⋅ dr C S by Stoke’s theorem, where S is the area of the square ABCD. Now, 2 curl (xy i + xy j) = A (1, 0) X¢ (–1, 0) curl ( xy i + xy 2 j) ⋅ ndS i ∂ ∂x xy x+y=1 j ∂ ∂y xy 2 = (y2 – x) k k ∂ ∂z 0 O X x–y=1 x + y = –1 (0, –1) D Y¢ Fig. 5.25 396 A TEXTBOOK OF ENGINEERING MATHEMATICS—I ∴ ∴ curl (xy i + xy2j ) · n^ = (y2 – x) k · k = (y2 – x) zz S = curl (xy i + xy 2 j) ⋅ ndS = = zz zz zz S S S ( y 2 − x) dS ( y 2 − x) dx dy y 2 dx dy − zz = 4 = 4 = 4 = 4 1 1− x 0 0 1 1− x 0 0 1− x 0 0 1 y3 3 0 S x dx dy y 2 dx dy − Sx [By symmetry] y 2 dx dy − S ⋅ 0 [3 x = x-coordinate of zz zz FG IJ zH K z 1 zz the C.G. of ABCD = 0] y 2 dx dy 1− x dx 0 1 4 1 (1 − x) 3 dx = . 3 3 0 Verification of Stokes theorem: The given line integral = = z C ...(i) ( xy dx + xy 2 dy) , where C is the boundary of the square ABCD. Now C can be broken up into four parts namely: (i) the line AB whose equation is x + y = 1, (ii) the line BC whose equation is y – x = 1, (iii) the line CD whose equation is x + y = – 1, and (iv) the line DA whose equation is x – y = 1. Hence, the given line integral = = z AB z ( xy dx + xy 2 dy) + z z BC ( xy dx + xy 2 dy) + z z z CD ( xy dx + xy 2 dy) + z z DA ( xy dx + xy 2 dy) RS x (1 − x) dx + (1 − y) y dyUV + RS x (1 + x) dx + (y − 1) y dyUV W T W T + { x ( −x − 1) dx + − y (1 + y) dy} + R ST x ( x − 1) dx + = 2 1 0 z 1 0 1 2 −1 0 0 z z 0 −1 −1 x ( x − 1) dx + 2 x (1 + x) dx + 2 0 1 0 1 0 z 2 0 z 2 1 (1 − y) y 2 dy + 2 z 1 0 0 −1 y 2 (1 + y) dy z 0 −1 y 2 ( 1 + y ) dy UV W 397 VECTOR CALCULUS F x − x IJ + 2 FG x + x IJ + 2FG y − y IJ + 2FG y + y IJ = 2 G H3 2K H 3 2K H 3 4K H 3 4K F 1 1I F 1 1I F 1 1I F 1 1I = 2 H − K + 2 H− + K + 2 H − K + 2 H − K 3 2 3 2 3 4 3 4 F 1 1I = 4 H − K 3 4 2 2 1 3 −1 2 0 = 3 4 0 1 3 4 0 0 −1 1 · 3 ...(ii) From eqns. (i) and (ii), it is evident that H H H F ⋅ dr = curl F ⋅ n dS z z C C Hence, Stoke’s theorem is verified. Example 8. Verify Stoke’s theorem for the function H F = (x + 2y) dx + (y + 3x) dy where C is the unit circle in the xy-plane. H F = F1 i + F2 j + F3 k Sol. Let H F · dr = (F1 i + F2 j + F3 k) · (dx i + dy j + dz k) = F1dx + F2dy + F3dz Here, F1 = x + 2y, F2 = y + 3x, F3 = 0 or Unit circle in xy-plane is x2 + y2 = 1 x = cos φ, dx = – sin φ dφ Hence, z y = sin φ, dy = cos φ dφ. H H F ⋅ dr = ( x + 2 y ) dx + ( y + 3 x) dy C = = = z z 2π 0 [ −(cos φ + 2 sin φ) sin φ dφ + (sin φ + 3 cos φ) cos φ dφ] z m z 2π 0 2π 0 0 2π = ( 3 cos2 φ − 2 sin 2 φ) dφ zm zm 2π = r − sin φ cos φ − 2 sin 2 φ + sin φ cos φ + 3 cos 2 φ dφ 0 r ( 3 cos 2 φ − 2(1 − cos2 φ) dφ r 5 cos 2 φ − 2 dφ 398 A TEXTBOOK OF ENGINEERING MATHEMATICS—I z LMN z RST 2π = 0 2π = 0 UV W 1 5 + cos 2 φ dφ 2 2 LM N OP Q 2π = 1 5 φ + sin 2 φ 2 2 0 = 1 [2π + 0] = π. 2 H curl F = i ∂ ∂x x + 2y j ∂ ∂y y + 3x = i {0} – j {0} + k Hence, zz S k ∂ ∂z 0 RS ∂ ( y + 3 x) − ∂ (x + 2 y)UV ∂y T ∂x W = k (3 – 2) = k. H curl F ⋅ n dS = = = = z z zz z z ( k ⋅ k ) dS dS dx dy xdy 2π = 0 z cos2 φ dφ = 1 2π (1 + cos 2φ) dφ 2 0 = 1 sin 2 φ φ+ 2 2 0 = π. So Stoke’s theorem is verified. LM N OP Q 5(1 + cos 2 φ) − 2 dφ 2 OP Q 2π 399 VECTOR CALCULUS 1. Evaluate EXERCISE 5.6 zz e S j ∇ × A ⋅ ndS where S is the surface of the hemisphere x2 + y2 + z2 = 16 above the xy-plane and A = (x2 + y – 4) i + 3xyj + (2xz + z2) k. 2. If F = (y2 + z2 + x2) i + (z2 + x2 – y2) j + (x2 + y2 – z2) k evaluate the surface S = x2 + y2 – 2ax + az = 0, z ≥ 0. 3. Evaluate zz S b zz e S Ans. − 16 π j ∇ × F ⋅ n dS taken over Ans. 2πa 3 g ∇ × yi + zj + xk ⋅ n dS over the surface of the paraboloid z = 1 – x2 – y2, z ≥ 0. Ans. π 4. F = (2x – y) i – yz2j – y2zk, where S upper half surface of the sphere x2 + y2 + z2 = 1. Hint: Here C, x 2 + y 2 = 1, z = 0 Ans. π 5. Using Stoke’s theorem or otherwise, evaluate zb C g 2 x − y dx – yz 2 dy − y 2 z dz . where C is the circle x2 + y2 = 1, corresponding to the surface of sphere of unit radius. Ans. π 6. Use the Stoke’s theorem to evaluate zb C g a f b g x + 2 y dx + x − z dy + y − z dz . where C is the boundary of the triangle with vertices (2, 0, 0) (0, 3, 0) and (0, 0, 6) oriented Ans. 15 in the anti-clockwise direction. H 7. Verify Stoke’s theorem for the Function F = x2 i – xy j integrated round the square in the plane z = 0 and bounded by the lines x = 0, y = 0, x = a, y = a. LMAns. Common value – a OP 2 PQ MN 3 8. Verify Stoke’s theorem for F = (x2 + y – 4)i + 3xy j + (2xz + z2)k over the surface of hemisphere x2 + y2 + z2 = 16 above the xy plane. Ans. Common value – 16π 9. Verify Stoke’s theorem for the function F = zi + xj + yk, where C is the unit circle in xy plane bounding the hemisphere z = 10. Evaluate z C e1 − x − y j . 2 2 (U.P.T.U., 2002) Ans. Common value π H F ⋅ dr by Stoke’s theorem for F = yzi + zxj + xyk and C is the curve of inter- section of x2 + y2 = 1 and y = z2. Ans. 0 400 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 5.22 GAUSS’S DIVERGENCE THEOREM If F is a continuously differentiable vector point function in a region V and S is the closed surface enclosing the region V, then zz S F ⋅ n dS = zzz V diV F dV ...(i) where n is the unit outward drawn normal vector to the surface S. (U.PT.U., 2006) H Proof: Let i, j, k are unit vectors along X, Y, Z axes respectively. Then F = F1i + F2j + F3k, where F1, F2, F3, and their derivative in any direction are assumed to be uniform, finite and continuous. Let S is a closed surface which is such that any line parallel to the coordinate axes cuts S at the most on two points. Let z coordinates of these points be z = F1 (x, y) and z = F2 (x, y), we have assumed that the equations of lower and upper portions S2 and S1 of S are z = F2 (x, y) and z = F1 (x, y) respectively. The result of Gauss divergence theorem (i) incomponent form is zzz b S z z z FGH g = F1 i + F2 j + F3 k ⋅ nds V I JK ∂F1 ∂F2 ∂F3 + + dV ∂x ∂y ∂z Now, consider the integral I1 = = zzz zz LMNz ...(ii) Z ∂F3 dx dy dz V ∂z R F1 ∂F3 F2 ∂z r n1 1 K S1 S OP Q r2 dz dx dy n2 S2 –K where R is projection of S on xy-plane. I1 = = zz zz b zz b R R F1 x , y 2 g b g dx dy X F x , y , F g dx dy − zz F b x , y , F g dx dy F3 x , y , F1 − F3 x , y , F2 3 1 R 3 2 For the upper portion S1 of S, dx dy = k ⋅ n 1 ⋅ dS1 where n 1 is unit normal vector to surface dS1 in outward direction. For the lower portion S2 of S. dx dy = – k ⋅ n 1 ⋅ dS2 where n 2 is unit normal vector to surface dS2 in outward direction. Thus, we have zz b zz b R and R g F x, y, F g dx dy zz zz F3 x, y, F1 dx dy = 3 = – 2 S1 F3 k ⋅ n 1 dS1 S2 F3 k ⋅ n 2 dS 2 dS2 Y O b g dx dy F3 bx, y, zg F bx, yg R dS1 R dx dy Fig. 5.26 VECTOR CALCULUS So zz b R g F3 x , y , F1 dx dy − 401 zz b R g F3 x , y , F2 dx dy = = = Hence, I1 = = zz b g zz b g zz S1 F3 k ⋅ n 1 dS1 + S2 b F3 k ⋅ n 2 dS2 F3 k . n 1 dS1 + n 2 dS 2 g zz c h zzz zz c h zz c h zz c h zz n c h c h c hs S F3 k ⋅ n dS = n 1 S1 + n 2 S2 3 nS ∂F3 dx dy dz V ∂z S1 F3 k ⋅ n dS ...(iii) Similarly, projecting S on other coordinate planes, we have zzz zzz ∂F3 dx dy dz V ∂y ∂F1 V ∂x = dx dy dz = Adding eqns. (iii), (iv), (v) zzz RST V ⇒ ⇒ or 5.23 UV W ∂F1 ∂F2 ∂F3 + + dx dy dz = ∂x ∂y ∂z zzz RST V i UV W S S F2 j ⋅ n dS ...(iv) F1 i ⋅ n dS ...(v) S F1 i . n + F2 j . n + F3 k . n dS ∂ ∂ ∂ ⋅ {F1i + F2 j + F3k} dx dy dz +k +j ∂x ∂y ∂z zzz = V H div F dV zz S = = H F ⋅ n ds = zz m zz zz zzz S S S r F1 i + F2 j + F3 k ⋅ n dS H F ⋅ n dS H F ⋅ n dS V H div F dV . CARTESIAN REPRESENTATION OF GAUSS’S THEOREM H F = F1 i + F2 j + F3 k Let where F1, F2, F3 are functions of x, y, z. and dS = dS (cos α i + cos β j + cos γ k) where α, β, γ are direction angles of dS. Hence, dS cos α, dS cos β, dS cos γ are the orthogonal projections of the elementary area dS on yz-plane, zx-plane and xy-plane respectively. As the mode of sub-division of surface is arbitrary, we choose a sub-division formed by planes parallel to yz-plane, zx-plane and xy-plane. Clearly, its projection on coordinate planes will be rectangle with sides dy and dz on yz-plane, dz and dx on zx-plane, dx and dy on xy-plane. 402 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Hence, projected surface elements are dy dz on yz-plane, dz dx on zx-plane and dx dy on xy-plane. H F1 dy dz + F2 dz dx + F3 dx dy ∴ ...(i) F ⋅ n dS = zz zzz z zz S S By Gauss divergence theorem, we have H H div F dV. F ⋅ n dS = ...(ii) V S In cartesian coordinates, dV = dx dy dz. H H Also, div F = ∇ · F = Hence, z V H div F dV = ∂F1 ∂F2 ∂F3 + + ∂x ∂y ∂z zzz RST zz m zzz RST V UV W ∂F1 ∂F2 ∂F3 + + dx dy dz ∂x ∂y ∂z Hence cartesian form of Gauss theorem is, ...(iii) r F1 dy dz + F2 dz dx + F3 dx dy S UV W Example 1. Find zz F ⋅ n dS , where F = b2 x + 3 zg i – b xz + y g j + e y + 2 zj k and S is the = V ∂F1 ∂F2 ∂F3 + + dx dy dz . ∂x ∂y ∂z 2 S surface of the sphere having centre at (3, – 1, 2) and radius 3. (U.P.T.U., 2000, 2005) Sol. Let V be the volume enclosed by the surface S. Then by Gauss divergence theorem, we have zz S F ⋅ n dS = = = zzz zzz LMN b g b g e zzz a f zzz V V V div F dV jOPQ ∂ ∂ ∂ 2 y + 2z dV − xz − y + 2x + 3z + ∂x ∂y ∂z 2 − 1 + 2 dV = 3 V dV = 3V But V is the volume of a sphere of radius 3. ∴ Hence zz S V = F ⋅ n dS Example 2. Evaluate 2 2 2 zz e S af 4 3 π 3 = 36π. 3 = 3 × 36π = 108π . j y 2 z 2i + z 2x 2 j + z 2 y 2 k ⋅ n dS , where S is the part of the sphere x + y + z = 1 above the xy-plane and bounded by this plane. Sol. Let V be the volume enclosed by the surface S. Then by divergence theorem, we have zz e S j y 2 z 2 i + z 2 x 2 j + z 2 y 2 k ⋅ n dS = = zzz e zzz LMN e j V V j div y 2 z 2 i + z 2 x 2 j + z 2 y 2 k dV e j e ∂ 2 2 ∂ 2 2 ∂ 2 2 y z + z x + z y ∂x ∂y ∂z jOPQ dV VECTOR CALCULUS zzz = V 2zy 2 dV = 2 403 zzz zy 2 dV V Changing to spherical polar coordinates by putting x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ dV = r2 sin θ dr dθ dφ π and those of φ will be 2 To cover V, the limits of r will be 0 to 1, those of θ will be 0 to 0 to 2π. ∴ 2 zzz V zy 2 dV = 2 = 2 z z za 2π π 2 1 0 0 0 z z z 2π π2 1 0 0 0 2π π 2 0 0 fe j r cosθ r 2 sin 2 θ sin 2 φ r 2 sin θ dr dθ dφ r 5 sin3 θ cos θ sin 2 φ dr dθ dφ Lr O = 2 z z sin θ cos θ sin φ M P dθ dφ N6Q Example 3. Evaluate zz S z 3 2 z 6 1 0 π 1 1 sin 2 φ ⋅ dφ = sin 2 φ dφ = . 12 0 12 0 12 = 2π 2π F ⋅ n dS over the entire surface of the region above the xy-plane bounded by the cone z2 = x2 + y2 and the plane z = 4, if F = 4xzi + xyz 2 j + 3zk . Sol. If V is the volume enclosed by S, then V is bounded by the surfaces z = 0, z = 4, z2 = x2 + y2. By divergence theorem, we have = = zz S F ⋅ n dS = zzz LMN a f e j a fOPQ j zz z e zz z a f z zza f zza LM OP a f z MN PQ ze z a fLMN OPQ ∂ ∂ ∂ xyz 2 + 4xz + 3z dV = V ∂x ∂y ∂z z2 − y2 4 z 0 − z − z2 − y2 = 2 = 2 = 4 z2 − y2 4 z 0 –z 0 4 z 0 −z 4 0 V div F dV V j 4z + xz 2 + 3 dV 4 z + xz 2 + 3 dx dy dz 4z + 3 dx dy dz, since 4z + 3 z 2 − y 2 dy dz = 4 y z2 − y2 4z + 3 zzz zzz e 2 − z2 − y2 4 z 0 0 y z2 + sin −1 z 2 z2 − y2 x dx = 0 f z − y dy dz 4z + 3 2 z dz 0 = 4 4 0 4 3 = π z +z j π 4 3 z2 4z + 3z 2 dz sin −1 1 dz = 4 × 4 0 2 4z + 3 4 0 a f = π 256 + 64 = 320π . 2 404 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Example 4. By transforming to a triple integral evaluate I = zz e S x 3 dy dz + x 2 y dz dx + x 2 z dx dy j (U.P.T.U., 2006) where S is the closed surface bounded by the planes z = 0, z = b and the cylinder x2 + y2 = a2. Sol. By divergence theorem, the required surface integral I is equal to the volume integral zzz LMN e j V e jOPQ z e e j j e3x + x + x j dx dy dz e j ∂ 3 ∂ 2 ∂ 2 x + x y + x z dV ∂x ∂y ∂z = zz z z ze j zze j b a2 − y2 a z =0 y = – a x = – = 4×5 b 2 2 2 a2 − y2 a2 − y2 a z = 0 y= 0 x = 0 3 20 b a a2 − y2 2 = 3 z = 0 y= 0 LM x OP e x dx dy dz = 20 N3Q LM a − y zOP 20 dy dz = 3 MNe j PQ 2 z π 3 a z = 0 y= 0 z a a f 20 2 3 20 4 b a cos 3 t a cos t dt = a b 3 0 3 a2 − y2 j dy dz x= 0 3 2 2 2 y=0 Put y = a sin t so that dy = a cos t dt. ∴ I= zz b b dy = z=0 20 3 z a y=0 z e j 3 b a 2 − y 2 2 dy . π 20 4 3 π 5 4 2 cos 4 t dt = a b = πa b . 0 3 4.2 2 4 H Example 5. Verify divergence theorem for F = (x2 – yz) i + (y2 – zx) j + (z2 – xy) k taken over the rectangular parallelopiped 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c. [U.P.T.U. (C.O.), 2006] H H ∂ 2 ∂ 2 ∂ 2 x − yz + y − zx + z − xy = 2x + 2 y + 2z . Sol. We have div F = ∇ · F = ∂x ∂y ∂z H ∇ ⋅ F dV = 2 x + y + z dV ∴ Volume integral = = 2 zz zb z z LMN z LMN c b a z =0 y = 0 x = 0 e zzz V j e b x2 + yx + zx dy dz 2 x =0 zzz b g z z LMN z LMMN j V g x + y + z dx dy dz = 2 c z= 0 y = 0 e OP Q a OP dz OP = 2 PQ Q O a b ab = 2 + + abz P dz 2 2 Q L a b z + ab z + ab z OP = [a bc + ab c + abc ] = abc (a + b + c). = 2M 2Q N2 2 c b z = 0 y= 0 c y2 a2 a2 + ay + az dy dz = 2 + azy y+a z =0 2 2 2 2 c 2 z= 0 2 2 2 c 2 2 2 0 Surface integral: Now we shall calculate H F ⋅ n dS zz S Over the six faces of the rectangular parallelopiped. Over the face DEFG, n = i, x = a. b y =0 j 405 VECTOR CALCULUS Therefore, zz e zz e DEFG c = b = b z = 0 y= 0 z c Z j e C a 2 − yz i + y 2 − 2a j + z 2 − ay k ⋅ i dy dz c 2 z= 0 2 c 2 2 = zz zz 2 G 0 ABCO c zz b g z LMMN OPPQ b z =0 y = 0 A O Y 2 2 F X Over the face ABCO, n = – i, x = 0. Therefore = E y =0 2 F ⋅ n dS = B D 2 b 2 LM a b − zb OP dz = LM a bz − z b OP = a bc – c b 2 PQ 4 4 . MN MN PQ 2 z= 0 F ⋅ n dS j e j L y O a − yzj dy dz = z Ma y − z P dz 2 PQ MN z = 0 y= 0 c = zz Fig. 5.27 a f 0 − yz i + y j + z k ⋅ −i dy dz y 2z 2 c yz dy dz = z=0 2 2 b dz = y=0 z b2 b2c 2 zdz = z=0 2 4 c Over the face ABEF, n = j, y = b. Therefore H c a x 2 − bz i + b 2 − zx j + z 2 − bx k ⋅ j dx dz F ⋅ n dS = zz zze zz e = c a z = 0 x= 0 j e z =0 x = 0 ABEF j b 2 − zx dx dz = b 2 ca − j e j a 2c2 . 4 Over the face OGDC, n = – j, y = 0. Therefore zz H F ⋅ n dS = zz H F ⋅ n dS = OGDC z z c a z= 0 x =0 zx dx dz = c 2 a2 . 4 Over the face BCDE, n = k, z = c. Therefore BCDE zz e c 2 − xy dx dy = c 2 ab − z z xy dx dy = b a y = 0 x =0 j a 2b 2 · 4 Over the face AFGO, n = – k, z = 0. Therefore zz zz AFGO H F ⋅ n dS = b a y=0 x=0 a2 b 2 · 4 Adding the six surface integrals, we get S H F ⋅ n dS = F a bc − c b + c b I + F b ca − a c + a c I + F c ab − a b + a b I GH 4 4 JK GH 4 4 JK GH 4 4 JK 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 = abc (a + b + c). Hence, the theorem is verified. H Example 6. If F = 4xzi – y2j + yzk and S is the surface bounded by x = 0, y = 0, z = 0, H F ⋅ n dS . x = 1, y = 1, z = 1, evaluate zz S 406 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Sol. By Gauss divergence theorem, H H ∇ ⋅ F dV , where V is the volume enclosed by the surface S F ⋅ n dS = zz zzz zzz LMN a f e j b gOPQ zzz b zzz b g z z z b g zz z z b g z LMMN OPPQ z LMN OPQ z j ze V S = = V 1 = = ∂ ∂ ∂ − y2 + yz dV = 4xz + ∂z ∂x ∂y V 4z − y dx dy dz = 1 x =0 y = 0 x =0 Example 7. Evaluate S z=0 1 dx = y =0 1 1 0 2− g 4z − 2 y + y dV 4z − y dx dy dz x =0 y = 0 z= 0 dx dy = 1 y2 2y − 2 1 1 2z 2 − yz 1 V 1 1 x=0 y= 0 2 − y dx dy 1 3 1 3 dx = dx = . 2 2 0 2 y 2 z 2 i + z 2 x 2 j + z 2 y 2 k ⋅ n dS, where S is the part of the sphere x2 + y2 + z2 = 1, above the xy-plane and bounded by this plane. H Sol. We have F = y2z2 i + z2x2 j + z2y2 k H ∂ 2 2 ∂ 2 2 ∂ 2 2 y z + z x + z y div F = ∂x ∂y ∂z e j e j e j = 2zy2 ∴ zzz Given integral = V 2 zy 2 dV [By Gauss’s divergence theorem] where V is the volume enclosed by the surface S, i.e., it is the hemisphere x2 + y2 + z2 = 1, above the xy-plane. Z Y 2 z= 1–x –y 2 y= 1–x 2 dx dy dz dx dy Z=0 X¢ O X O X dx dy y=–1–x Y¢ Y (i) (ii) Fig. 5.28 From the above Figure 5.28 (i) and (ii), it is evident that limits of z are from 0 to 1 − x2 − y2 limits of y are from – 1 − x 2 to and limits of x are from –1 to + 1. 1 − x2 2 407 VECTOR CALCULUS ∴ Given integral = 2 = 2 z z 1 z 1− x 2 1 −x 2 − y 2 zy 2 dx dy dz x = –1 y = – 1 − x 2 z = 0 z z 1 Fz I GH 2 JK x = –1 y = – 1 − x 2 z z z LMMNe 1− x 2 − y 2 2 1− x 2 y 2 dx dy 0 e1 − x − y j y dx dy y y O = − P 1− x j dx 3 5 PQ R| 1 1 − x − 1 1 − x U| dx = 2z S| 3 e j 5 e j V| T W R| |V dx 1− x j U = 4z S b1 − x g − e |W |T 3 5 8 = e1 − x j dx 15 = 1 1− x 2 2 2 2 x = –1 y = – 1− x 2 1 3 2 5 x = –1 1− x 2 – 1− x 2 5 2 2 1 x =−1 2 5/2 2 5/2 1 5 2 2 0 z ze z 5 2 2 1 0 π = j 5 8 2 1 − sin 2 θ 2 cos θ dθ 15 0 [Putting x = sin θ] π 8 2 cos 2 θ dθ 15 0 8 5π π ⋅ = = · 15 32 12 H H Example 8. Evaluate F ⋅ n dS where F = (x + y2) i – 2x j + 2yz k where S is surface bounded = z S by coordinate planes and plane 2x + y + 2z = 6. Sol. We know from Gauss divergence theorem, H H F ⋅ n dS = div F dV S V H F = (x + y2) i – 2x j + 2yz k zz zzz F i ∂ + j ∂ + k ∂ I ⋅ {ex + y j i − 2x j + 2yz k} GH ∂x ∂y ∂z JK ∂ = ex + y j + ∂∂y a–2xf + ∂∂z b2yzg ∂x H div F = 2 2 Let zzz zzz zzz b g zzz b g = 1 + 2y H F ⋅ n dS = I = S = = V V 1 + 2 y dV 1 + 2 y dx dy dz H div F dV 408 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 6 − 2x − y 2 Limit of y is 0 to 6 – 2x Limit of x is 0 to 3 Limit of z is 0 to Hence, zzz b g b g zz b g k p zz b gb g t zz o UV z RST W z LMN b g b g b g UV z RST W 1 + 2 y dx dy dz I = 6 − 2 x − y /2 1 + 2y = 1 2 1 = 2 = = = = z 0 dx dy 1 + 2 y 6 − 2x − y dx dy 6 − 2x + 11y − 4xy − 2 y 2 dx dy 6 − 2x 1 2 6 y − 2xy + 1 2 6 6 − 2 x − 2 x 6 − 2x + 1 2 − LM N 11 2 2 y − 2xy 2 − y 3 2 3 0 dx b g b g OPQ 11 2 2 2 3 dx 6 − 2x 6 − 2 x − 2 x 6 − 2x − 2 3 8 3 x + 26x 2 − 84x + 90 dx 3 OP Q 3 1 2 26 3 − x4 + x − 42x 2 + 90x 2 3 3 0 = 1 − 54 + 234 − 378 + 270 2 1 = 72 = 36 . 2 = Example 9. Verify Gauss divergence theorem for zz {e S j } x 3 − yz dy dz − 2 x 2 y dz dx + z dx dy over the surface of cube bounded by coordinate planes and the planes x = y = z = a H Sol. Let F = F1 i + F2 j + F3 k. From Gauss divergence theorem, we know H H F ⋅ n dS = F1 dy dz + F2 dz dx + F3 dx dy = div F dV zz zz S Here, So, S V F1 = x3 – yz, F2 = – 2x2y, F3 = z H F = (x3 – yz) i – 2x2y j + z k H div F F i ∂ + j ∂ + k ∂ I ⋅ {ex − yzj i − 2x y j + z k} GH ∂x ∂y ∂z JK ∂ = ex − yzj + ∂∂y e– 2x yj + ∂∂z azf ∂x 3 = 3 Hence, zz S H F ⋅ n dS zzz 2 = 3x2 − 2x2 + 1 = x2 + 1 = zzz e V j x 2 + 1 dV 2 ...(i) 409 VECTOR CALCULUS = = zzze j z z e jk p zze j z e jm r ze j a a a x 2 + 1 dx dy dz 0 0 0 a a a a = a x 2 + 1 dx dy 0 0 a = a a = a2 a x2 + 1 0 a x 2 + 1 z 0 dx dy 0 0 y 0 dx x 2 + 1 dx 0 Ra U = a S + xV T3 W |R a |U a + a = a S + aV = |T 3 |W 3 a 3 2 0 3 2 5 3 ...(ii) Verification by direct integral: Outward drawn unit vector normal to face OEFG is – i and dS is dy dz. If I1 is integral along this face, H H F ⋅ n dS = F ⋅ − i dy dz I1 = z zz b g zz e j zz z R|S|T U|V|W z LMMN OPPQ S = = = S x − yz dy dz 3 S a a 0 0 a 0 [As x = 0 for this face] yz dy dz z2 2 y a dy 0 2 a2 y y dy = 0 2 2 a2 = 2 a a 0 = a4 4 For face ABCD, its equation is x = a and n dS = i dy dz , If I2 is integral along this face I2 = = = = = zz zz e zze z R|S|T z RST S H F ⋅ i dy dz S a 0 a 0 k j G x 3 − yz dy dz a a 0 0 Z j D a 3 − yz dy dz U|V dy |W a U a a − y V dy 2W a3 z − y 3 z2 z F –i C j a a –j a O 0 a E A 2 B X i –k Fig. 5.29 Y 410 A TEXTBOOK OF ENGINEERING MATHEMATICS—I L a y OP = Ma y − MN 2 2 PQ a 2 2 4 0 4 a 4 If I3 is integral along face OGDA whose equation is y = 0 = a5 − Hence, n dS = – j dxdz H I3 = F ⋅ − j dx dz zz b g zz S = – S – 2x 2 y dx dz = 0, as y = 0. If I4 is integral along face BEFC whose equation is y = a n dS = Then zz j dx dz I4 = – S 2 x 2 y dx dz zz z kp z a a x 2 dx dz = – 2a = – 2a x 2 z 0 dx = – 2a 2 x 2 dx 0 0 a a 0 a 0 Lx O 2 = – 2a M P = – a . N3Q 3 3 2 a 5 0 If I5 is integral along face OABE whose equation is z = 0 n dS = – k dx dy H F ⋅ – k dx dy I5 = zz b zz S = – S g z dx dy = 0 as z = 0. If I6 is integral along face CFGD whose equation is z = a n dS = k dx dy I6 = zz z = a Total surface S z dx dy = a 0 a zz a a y 0 dx = a 2 a dx dy z 0 0 a 0 dx = a 3 I = I1 + I2 + I3 + I4 + I5 + I6 411 VECTOR CALCULUS a4 a4 2 + a5 – + 0 – a5 + 0 + a3 4 4 3 = a5 + a3 3 which is equal to volume integral. Hence Gauss theorem is verified. = Example 10. Evaluate by Gauss divergence theorem zz { S e j e j zz S Here, Hence, zzz S V F1 = xz2 ; F2 = x2y – z3, F3 = 2xy + y2z. H F = xz2 i + (x2y – z3) j + (2xy – y2z) k H div F 2 zz zzz e 2 = z2 + x2 + y2. H F ⋅ n ds = I = S = Limit of z is 0 to H div F dV F ∂ i + ∂ j + ∂ kI GH ∂x ∂y ∂z JK {xz i + (x y – z ) j + (2xy + y z) k} ∂ = exz j + ∂∂y ex y – z j + ∂∂z e2xy + y zj ∂x = 2 Let } xz 2 dy dz + x 2 y − z 3 dz dx + 2 xy + y 2 z dx dy where S is surface bounded by z = 0 and z = a 2 − x 2 − y 2 . H F = F1 i + F2 j + F3 k. Sol. Let Cartesian form of Gauss divergence theorem is H F ⋅ n dS = F1 dy dz + F2 dz dx + F3 dx dy = zz ...(iii) 2 2 zzz V 2 3 3 2 2 H div F dV j x + y + z 2 dx dy dz a2 − x2 − y 2 − a 2 − x 2 to Limit of x is – a to a a2 − x2 Limit of y is I = = zzz e zz LMMNe j x 2 + y 2 + z 2 dx dy dz x +y 2 2 j z3 z+ 3 OP PQ a2 − x 2 − y2 dx dy 0 LM O a −x −y j P e PP dx dy = zz Mex + y j a − x − y + 3 MM PQ N a − x − y |U |R a − x − y Sx + y + = V| dx dy 3 |T W 2 zz 2 2 2 2 2 2 2 2 3 2 2 2 2 2 2 2 2 412 A TEXTBOOK OF ENGINEERING MATHEMATICS—I = = Let 1 3 1 3 zz zz zz zz o t a − x − y o2x + 2 y + a t dx dy a 2 − x 2 − y 2 3x 2 + 3y 2 + a 2 − x 2 − y 2 dx dy 2 2 = 1 a a2 −x2 3 − a − a2 −x2 = a2 − x 2 2 a 3 −a 0 2 2 2 2 {e2x + a j a − x − y + 2y a − x − y } dx dy RSe2x + a j a − x − y + 2y a − x − y UV dx dy T W y = a 2 − x 2 sin θ dy = a 2 − x 2 cosθ dθ I = 2 a 2 3 −a 0 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 z z LNMe je j e j LM OP z MMNe je j e j PPQ z LMNe je j e j OPQ z LMN e je j e j OPQ π 2x 2 + a 2 a 2 − x 2 cos 2 θ + 2 a 2 − x 2 2 OP Q sin 2 θ cos 2 θ dx dθ 3 1 3 3 2 2 + 2 a 2 − x 2 2 2 2 dx 2 2 2 3 = 2 a 2 2 3 − a 2x + a a −x = 2 a 2x 2 + a 2 3 −a a2 − x2 = 4 π a × 2 2x 2 + a 2 3 8 0 = a π 2 2x 2 a 2 − 2x 4 + a 4 − a 2x 2 + a 4 + x 4 − 2 2 x 2 dx 0 3×2 = ze ze z e 2 2 2 π π dx + 2 a2 − x2 4 16 a2 − x 2 + a2 − x 2 2 dx j j π a 4 3a − 3x 4 dx 0 6 π a 4 a − x 4 dx 2 0 j πL x O = a x− M P 2N 5Q = 5 4 a 0 = z zb π 4 5 2π 5 × a = a . 2 5 5 Example 11. Using the divergence theorem, evaluate the surface integral g yz dy dz + zx dz dx + xy dy dx , where S : x2 + y2 + z2 = 4. (U.P.T.U., 2008) S Sol. Let F = F1i + F2j + F3k From Gauss divergence theorem, we have zz S F . n dS = zz S F1 dy dz + F2 dz dx + F3 dx dy = zzz V div F dV ...(i) 413 VECTOR CALCULUS Comparing L.H.S. of (i) with given integral, we get F1 = yz, F2 = zx, F3 = xy So ⇒ F = F1i + F2 j + F3k F = (yz)i + (zx)j + (xy)k F i ∂ + j ∂ + k ∂ I .{(yz)i + (zx)j + (xy)k} GH ∂x ∂y ∂z JK ∂ ∂ ∂ = yzg + zxf + b xyg = 0 a b ∂x ∂y ∂z div F = z zb Thus g yz dy dz + zx dz dx + xy dy dx = S zzz 0. dV = 0. V EXERCISE 5.7 1. Use divergence theorem to evaluate zz H 3 3 3 F ⋅ dS where F = x i + y j + z k and S is the surface S LMAns. 12πa OP 5 Q N 5 of the sphere x2 + y2 + z2 = a2. 2. Use divergence theorem to show that surface enclosing volume V. zz e zz S j ∇ x 2 + y 2 + z 2 dS = 6V Where S is any closed 3. Apply divergence theorem to evaluate SFn dS , where F = 4x 3 i − x2 yj + x2 zk and S is the surface of the cylinder x2 + y2 = a2 bounded by the planes z = 0 and z = b. zz b 4. Use the divergence theorem to evaluate S Ans. 3ba 4 π g x dy dz + y dz dx + z dx dy , where S is the portion of the plane x + 2y + 3z = 6 which lies in the first octant. (U.P.T.U., 2003) Ans. 18 zz 5. The vector field F = x i + zj + yzk is defined over the volume of the cuboid given by 0 2 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c enclosing the surface S. Evaluate the surface integral 6. Evaluate zz e S LM N j b 2 IJ OP KQ yzi + zxj + xyk dS where S is the surface of the sphere x2 + y2 + z2 = a2 in the zz e S F ⋅ dS. (U.P.T.U., 2001) Ans. abc a + first octant. 7. Evaluate FG H S j (U.P.T.U., 2004) Ans. 0 e x dy dz – ye x dz dx + 3z dx dy , where S is the surface of the cylinder x2 + y2 = c2, 0 ≤ z ≤ h. Ans. 3π hc 2 414 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 8. Evaluate zz S F ⋅ n ds , where F = 2xyi + yz2j + xzk, and S is the surface of the region LMAns. 351 OP 2 Q N bounded by x = 0, y = 0, z = 0, y = 3 and x + 2z = 6. 9. F = 4xi – 2y2j + z2k taken over the region bounded by x2 + y2 = 4, z = 0 and z = 3. e Ans. Common value 8 π j 10. F = x 3 − yz i – 2x 2 yj + zk taken over the entire surface of the cube 0 ≤ x ≤ a, 0 ≤ y ≤ a, LMAns. Common value a + a OP 3 N Q 5 0 ≤ z ≤ a. 3 11. F = 2xyi + yz2j + xzk and S is the total surface of the rectangular parallelopiped bounded by the coordinate planes and x = 1, y = 2, z = 3. Ans. Common value 33 12. F = x 2 i + y 2 j + z 2 k taken over the surface of the ellipsoid x2 y2 z2 + + = 1. a 2 b2 c 2 Ans. Common value 0 13. F = xi + yj taken over the upper half on the unit sphere x2 + y2 + z2 = 1. 14. Prove that 15. Evaluate zzz zz dV zz V r 2 = S r. n r2 LMAns. Common value 4π OP 3 Q N ds . r. n ds , where S : surface of cube bounded by the planes x = –1, y = –1, S z = –1, x = 1, y = 1, z = 1. [Ans. 24] OBJECTIVE TYPE QUESTIONS A. Pick the correct answer of the choices given below: 1. If r = xi + yj + zk is position vector, then value of ∇(log r) is (i) r r (iii) − (ii) r r2 r (iv) None of these r3 2. The unit vector normal to the surface x2y + 2xz = 4 at (2, –2, 3) is 1 (i – 2j + 2k) 3 1 (iii) (i + 2j – 2k) 3 (i) (ii) 1 (i – 2j – 2k) 3 (iv) None of these [U.P.T.U., 2008] 415 VECTOR CALCULUS 3. If r is a position vector then the value of ∇rn is (i) nrn–2 r (ii) nrn–2 2 (iii) nr (iv) nrn–3 4. If f(x, y, z) = 3x2y – y3z2, then |∇f| at (1, –2, –1) is (i) 481 (ii) 381 (iii) (iv) 481 581 e j 5. If a is a constant vector, then grad r . a is equal to (i) r (ii) − a (iii) 0 (iv) a 6. The vector rn r is solenoidal if n equals (i) 3 (ii) – 3 (iii) 2 (iv) 0 7. If r is a position vector then div r is equal to (i) 3 (ii) 0 (iii) 5 (iv) –1 8. If r is a position vector then curl r is equal to (i) – 5 (ii) 0 (iii) 3 (iv) –1 9. If F = ∇φ, ∇2φ = – 4πρ where P is a constant, then the value of zz F. n dS is: S (i) 4π (ii) – 4πρ (iii) – 4πρV (iv) V 10. If n is the unit outward drawn normal to any closed surface S, the value of zzz div F dV V is (i) V (ii) S (iii) 0 (iv) 2S zz 11. If S is any closed surface enclosing a volume V and F = xi + 2yj + 3zk then the value of the integral F. n dS is S (i) 3V (iii) 2V 12. The integral (ii) 6V zz (iv) 6S r 5 n dS is equal S (i) 0 (ii) zzz V 5r 3 . r dV 416 A TEXTBOOK OF ENGINEERING MATHEMATICS—I (iii) zzz zzz 5r −3 . r dV (iv) None of these V 13. A vector F is always normal to a given closed surface S in closing V the value of the integral curl F dV is : V (i) 0 (ii) 0 (iii) V (iv) S B. Fill in the blanks: 1. If f = (bxy – z3)i + (b – 2)x2 j + (1 – b)xz2 k has its curl identically equal to zero then b = .......... 2. ∇ 2 FG 1IJ = .......... H rK 3. div grad f = .......... 4. curl grad f = .......... 5. grad r = .......... 6. grad 1 = .......... r 7. If A = 3x yz2 i + 2x y3 j – x2 yz k and f = 3x2 – yz then A . ∇f = .......... 8. If r = r, then ∇f(r) × r = .......... 9. If r = r, then a f = .......... ∇f r ∇r 10. The directional derivative of φ = xy + yz + zx in the direction of the vector i + 2j + k at (1, 2, 0) is = .......... 11. The value of zb g xdy − ydx around the circle x2 + y2 = 1 is .......... 12. If ∇2φ = 0, ∇2ψ = 0, then 13. zz z z FGH S IJ K ∂ψ ∂φ dS = .......... −ψ ∂n ∂n F . n dS is called the .......... dF over S. 14. If S is a closed surface, then 15. φ zzz V ∇. F dV = zz zz r . n dS = .......... S A dS , then A is equal to .......... S C. Indicate True or False for the following statements: 1. (i) If v is a solenoidal vector then div v = 0. 417 VECTOR CALCULUS (ii) div v represents the rate of loss of fluid per unit volume. (iii) If f is irrotational then curl f ≠ 0. (iv) The gradient of scalar field f(x, y, z) at any point P represents vector normal to the surface f = const. 2. (i) The gradient of a scalar is a scalar. (ii) Curl of a vector is a scalar. (iii) Divergence of a vector is a scalar. (iv) ∇f is a vector along the tangent to the surface f = 0. 3. (i) The directional derivative of f along a is f. a . (ii) The divergence of a constant vector is zero vector. (iii) The family of surfaces f(x, y, z) = c are called level surfaces. (iv) If a and b are irrotational then div ea × bj = 0. 4. (i) Any integral which is evaluated along a curve is called surface integral. (ii) Green’s theorem in a plane is a special case of stoke theorem. (iii) If the surface S has a unique normal at each of its points and the direction of this normal depends continuously on the points of S, then the surface is called smooth surface. (iv) The integral z z ze j F . dr is called circulation. S 5. (i) The formula curl F . n ds = S z F . dr is governed by Stoke’s theorem. C (ii) If the initial and terminal points of a curve coincide, the curve is called closed curve. (iii) If n is the unit outward drawn normal to any closed surface S, then (iv) The integral 1 2 zb V g xdy − ydx represents the area. C D. Match the Following: e j 1. (i) ∇. ∇ × a (ii) curl (φ grad φ) e j (a) dφ (b) 0 (iii) div a × r (c) 0 (iv) ∇φ . d r (d) r curl a 2. (i) ∇2 r2 (ii) df/ds zzz (a) a . (∇f) (b) grad f ± grad g ∇. n dv ≠ S . 418 A TEXTBOOK OF ENGINEERING MATHEMATICS—I (iii) a . ∇ f e j (c) ∇f . (iv) ∇(f ± g) (d) 6 3. (i) grad of φ along n (ii) d2r dt 2 F aI GH a JK (a) a (acceleration) ∂φ n ∂n (b) (iii) curl v (c) curl f = 0 (iv) irrotational (d) 2 ω ANSWERS TO OBJECTIVE TYPE QUESTIONS A. Pick the correct answer: 1. (ii) 5. (iv) 9. (iii) 2. (ii) 6. (ii) 10. (ii) 3. (i) 7. (i) 11. (ii) 4. (iv) 8. (ii) 12. (ii) 2. 0 3. ∇2 f 4. 0 7. –15 8. 0 11. 2π 12. 0 13. (i) B. Fill in the blanks: 1. 4 r r r 6. 9. f ′(r) 10. 13. flux 14. 3V 15. F . n 1. (i) T (ii) T (iii) F (iv) F 2. (i) F (ii) T (iii) T (iv) F 3. (i) F (ii) F (iii) T (iv) T 4. (i) F (ii) F (iii) T (iv) T 5. (i) T (ii) T (iii) F (iv) T 1. (i) ® (b) (ii) ® (c) (iii) ® (d) (iv) ® (a) 2. (i) ® (d) (ii) ® (c) (iii) ® (a) (iv) ® (b) 3. (i) ® (b) (ii) ® (a) (iii) ® (d) (iv) ® (c) 5. r3 10 3 C. True or False: C. Match the following: GGG UNSOLVED QUESTION PAPERS (2004–2009) B.Tech. (Only for the Candidates Admitted/Readmitted in the Session 200809) (SEM. I) Examination, 2008-09 MATHEMATICSI Time: 3 Hours] [Total Marks: 100 SECTION A All parts of this question are compulsory compulsory.. 1. (a) For which value of ‘b’ the rank of the martix. LM1 5 4 OP A = M0 3 2 P is 2, b = 4cm → MNb 13 10PQ 2 × 10 = 20 — ^ (b) Determine the constants a and b such that the curl of vector A = (2xy + 3yz)i + ^ ^ (x2 + axz – 4z2)j + (3xy + byz)k is zero, a = .........., b = .......... . (c) The nth derivative (yn) of the function y = x2 sin x at x = 0 is .......... . (d) With usual notations, match the items on right hand side with those on left hand side for properties of maximum and minimum: (i) Maximum (p) rt – s2 = 0 (ii) Minimum (q) rt – s2 < 0 (iii) Saddle point (r) rt – s2 > 0, r > 0 (iv) Failure case (s) rt – s2 > 0 and r < 0 (e) Match the items on the right hand side with those on left hand side for the following special functions : (Full marks is awarded if all matchings are correct). (i) β(p, q) (ii) p q p+q (p) b 1 2g z b gb g ∞ (q) 0 y p−1 dy 1+ y p+q (iii) π (r) β(p, q) (iv) π sin pπ (s) p 1− p Indicate T Trr ue or False for the following statements: (f) (i) If |A| = 0, then at least one eigen value is zero. (ii) A–1 exists iff 0 is an eigen value of A. 419 (True/False) (True/False) 420 A TEXTBOOK OF ENGINEERING MATHEMATICS—I (iii) If |A| ≠ 0, then A is known as singular matrix. (True/False) (iv) Two vectors X and Y is said to be orthogonal Y, XT Y = YT X ≠ 0. (True/False) (g) (i) The curve y2 = 4ax is symmetric about x-axis. 3 3 2 2 (True/False) (ii) The curve x + y = 3axy is symmetric about the line y = – x. (True/False) 2 (iii) The curve x + y = a is symmetric about both the axis x and y. (True/False) (iv) The curve x3 – y3 = 3axy is symmetric about the line y = x. Pick the correct answer of the choices given below: ^ ^ (True/False) ^ (h) If r = xi + yj + zk is position vector, then value of ∇ (log r) is : — (i) r⋅ r (iii) − (i) (j) r⋅ (ii) r⋅ r2 (iv) None of these r3 a f b g The Jacobian ∂ uv for the function u = ex sin y, v = (x + log sin y) is ∂ xy (i) 1 (ii) sin x sin y – xy cos x cos y ex . x The volume of the solid under the surface az = x2 + y2 and whose base R is the circle x2 + y2 = a2 is given as (iii) 0 (iv) (i) π|2a (ii) πa3|2 (iii) 4 3 πa 3 (iv) none of these. SECTION B 10 × 3 = 30 Attempt any three parts of the following: –1 2 2 2 2. (a) If y = (sin x) prove that yn(0) = 0 for n odd and yn(0) = 2, 2 , 4 , 62 ...(n – 2)2, n ≠ 2 for n is even. (b) Find the dimension of rectangular box of maximum capacity whose surface area is given when (a) box is open at the top (b) box is closed. (c) Find a matrix P which diagonalizes the matrix A = LM4 1OP , verify P . AP = D where N2 3Q –1 D is the diagonal matrix. (d) Find the area and the mass contained m the first quadrant enclosed by the curve FG x IJ + FG y IJ = 1 where α > 0, β > 0 given that density at any point p(xy) is k xy . H aK H bK α z zb β (e) Using the divergence theorem, evaluate the surface integral S g yz dy dz + zx dz dx + xy dy dx where S : x2 + y2 + z2 = 4. 421 UNSOLVED QUESTION PAPERS SECTION C Attempt any two parts from each question. All questions are compulsory. 3. (a) Trace the curve r2 = a2 cos 2θ F ex + y j I GG bx + yg JJ , prove that K H 2 (b) If u = log x 2 ∂u ∂u +y = 1. ∂x ∂y (c) If V = f(2x – 3y, 3y – 4z, 4z – 2x), compute the value of 6Vx + 4Vy + 3Vz. 4. (a) The temperature ‘T’ at any point (xyz) in space is T(xyz) = K xyz2 where K is constant. Find the highest temperature on the surface of the sphere x2 + y2 + z2 = a2. (b) Verify the chain rule for Jacobians if x = u, y = u tan v, z = w. (c) The time ‘T’ of a complete oscillation of a simple pendulum of length ‘L’ is governed ⋅L , g is constant, find the approximate error in the calculated g value of T corresponding to an error of 2% in the value of L. 5. (a) Determine ‘b’ such that the system of homogeneous equation 2x + y + 2z = 0; x + y + 3z = 0; 4x + 3y + bz = 0 has (i) Trivial solution, (ii) Non-trivial solution. Find the Non-trivial solution using matrix method. by the equation T = 2π (b) Verify Cayley-Hamilton theorem for the matrix A = FG 1 2 IJ and hence find A . H 2 −1K –1 (c) Find the eigen value and corresponding eigen vectors of the matrix. I= FG −5 2 IJ . H 2 −2K 6. (a) Find the directional derivative of ∇(∇f) at the point (1, –2, 1) in the direction of the normal to the surface xy2z = 3x + z2 where f = 2x3 y2z4. (b) Using Green’s theorem, find the area of the region in the first quadrant bounded by the curves 1 x ,y = . x 4 ^ ^ ^ (c) Prove that (y2 – z2 + 3yz)i + (3xz + 2xy) j + (3xy – 2xz + 2z) k is both solenoidal and irrotational. 7. (a) Changing the order of integration of y = x, y = zz ∞ ∞ 0 0 e − xy sin nx dx dy Show that z FGH ∞ 0 IJ K sin nx π dx = . x 2 (b) Determine the area bounded by the curves xy = 2, 4y = x2 and y = 4. (c) For a β function, show that β(p, q) = β(p + 1, q) + β(p, q + 1) GGG B.Tech., First Semester Examination, 2007-08 MATHEMATICSI Time: 3 Hours] [Total Marks: 100 Note: Attempt all the problems. Internal choices are mentioned in every problem. 1. Attempt any two parts of the following: (10 × 2 = 20) (a) Define the eigen values, eigen vectors and the characteristic equation of a square matrix. Find the characteristic equation/polynomial, eigen values and eigen vectors of the matrix: LM2 MM5 N7 5 3 0 7 1 2 OP PP Q (b) Check the consistency of the following system of linear non-homogeneous equations and find the solution, if exists: 7x1 + 2x2 + 3x3 = 16 2x1 + 11x2 + 5x3 = 25 x1 + 3x2 + 4x3 = 13 (c) Find the inverse of the matrix LM 1 1 1 OP MM 21 13 15 PP MM 3 5 7 PP MN 15 71 111 PQ 2. Attempt any two parts of the following: (10 × 2 = 20) (a) State Leibnitz theorem for nth differential coefficient of the product of two functions. 1 − 1 If y m + y m = 2x, prove that (x2 – 1)yn + 2 + (2n + 1)xyn + 1 + (n2 – m2)yn = 0 2 (b) Verify that (c) If u = x sin F GH ∂ u x2 + y2 ∂2 u , where u(x, y) = log e = ∂y ∂x ∂x ∂y xy −1 I JK F x I + y sin FG y IJ , find the value of x ∂ u + 2xy ∂ u + y ∂ u . GH y JK H xK ∂x ∂y ∂y ∂x −1 2 2 2 2 3. Attempt any four parts of the following: FG π IJ . H 4K (a) Expand ex cos y about the point 1, 422 2 2 2 (5 × 4 = 20) 423 UNSOLVED QUESTION PAPERS a b f g ∂ u, v, w of the following: ∂ x, y, z (b) Calculate the Jacobian u = x + 2y + z v = x + 2y + 3z w = 2x + 3y + 5z (c) Discuss the maxima and minima of the function: f(x, y) = cos x cos y cos(x + y) (d) Find a point on the ellipse 4x2 + y2 = 4 nearest to the point (1, 2). (e) Find the extreme value x2 + y2 + z2 subject to the condition xy + yz + zx = p. 1 (f) If f(x, y) = x 2 y 10 , compute the value of f when x = 1.99 and y = 3.01. 4. Attempt any four of the following: (a) Evaluate the following by changing into polar coordinates: zz a (5 × 4 = 20) a −y 2 0 0 2 y 2 x 2 + y 2 dx dx (b) Find the area enclosed between the parabola y = 4x – x2 and the line y = x. (c) Change the order of integration in zz a 0 2a−x x2 a b g f x , y dx dy F xI F yI F zI (d) Find the volume of the solid surrounded by the surface G J + G J + G J = 1. H aK H b K H cK 2 3 2 3 2 3 (e) Define Gamma and Beta functions. Prove that B(l, m) B(l + m, n) B(l + m + n, p) = l m n p l+m+n+ p z e j 10 1 5 x 1 − x 3 dx 0 (f) Show that = 5. Attempt any two of the following: 1 396 (10 × 2 = 20) (a) If r = xi + yj + zk then show that (i) e j ∇ a ⋅ r = a , where a is a constant vector. (ii) grad r = (iii) grad r r 1 r = – 3 , where a is a constant vector. r r (iv) grad rn = nr n−2 r → r when r = e j e j e j (b) Prove that a × ∇ × r = ∇ a ⋅ r − a ⋅ ∇ r , where a r = xi + yj + zk . (c) State the Green’s theorem. Verify it by evaluating zc c is a constant vector and h c h x 3 − xy 3 dx + y 2 − 2xy dy where C is the square having the vertices at the points (0, 0), (2, 0), (2, 2) and (0, 2). GGG B.Tech., First Semester Examination, 2006-07 MATHEMATICSI Time: 3 Hours] Note: [Total Marks: 100 (i) Attempt All questions. (ii) All questions carry equal marks. (iii) In case of numerical problems assume data wherever not provided. (iv) Be precise in your answer. 1. Attempt any four parts of the following: (a) If y = x log (1 + x), prove that yn = e a−1f a j n− 2 f a f a f n− 2 x+n / 1+x (5 × 4 = 20) n a f 2 (b) If x = tan y, prove that 1 + x y n + 1 + 2 nx − 1 y n + n n − 1 y n − 1 = 0 (c) If u(x, y, z) = log(tan x + tan y + tan z) prove that sin 2x ∂u ∂u ∂u + sin 2 y + sin 2 z = 2. ∂x ∂y ∂z (d) State and prove Euler’s theorem for partial differentiation of a homogeneous function f(x, y). ∂u ∂u y x + tan −1 , prove that x ∂x + y ∂ y = 0 y x (f) Trace the curve y2(a – x) = x3, a > 0 2. Attempt any two parts of the following: (a) If u3 + v3 = x + y, u2 + v2 = x3 + y3 (e) If u(x, y) = sin −1 show that (10 × 2 = 20) a f = y −x b g 2uvau − vf ∂ u, v ∂ x, y 2 2 2 2 (b) Find Taylor series expansion of function on f(x, y) = e–x –y cos xy about the point x0 = 0, y0 = 0 up to three terms. (c) Find the minimum distance from the point (1, 2, 0) to the cone z2 = x2 + y2. 3. Attempt any two parts of the following: (10 × 2 = 20) (a) Define the gradient, divergence and curl. (i) If f(x, y, z) = 3x2y – y3z2, find grad f at the point (1, – 2, – 1). H (ii) If F x, y, z = xz 3 i − 2x 2 yzj + 2 yz 4 k , find divergence and curl of F x, y, z . b g b g (b) State Gauss divergence theorem. Verify this theorem by evaluating the surface integral as a triple integral z ze S j x 3 dy dz + x 2 ydz dx + x 2 z dx dy , where S is the closed surface consisting of the cylinder x2 + y2 = a2, (0 ≤ z ≤ b) and the circular discs z = 0 and z = b (x2 + y2 ≤ a2). b g (c) State the Stoke’s theorem. Verify this theorem for F x, y, z = xzi − yj + x 2 yk , where the surface S is the surface of the region bounded by x = 0, y = 0, z = 0, 2x + y + 2z = 8 which is not included on xz-plane. 424 425 UNSOLVED QUESTION PAPERS 4. Attempt any four parts of the following: (a) Find the rank of matrix (5 × 4 = 20) LM 2 3 − 2 4 OP MM 3 − 2 1 2 PP MN − 32 24 30 45 PQ (b) Solve the system of equations 2x + 3x2 + x3 = 9 x + 2x2 + 3x3 = 6 3x1 + x2 + 2x3 = 8 by Gaussian elimination method. (c) Find the value of λ for which the vectors (1, –2, λ), (2, –1, 5) and (3, –5, 7λ) are linearly dependent. F 1 2 2I (d) Find the characteristic equation of the matrix G 0 2 1 J . Also, find the eigen GH − 1 2 2 JK values and eigen vectors of this matrix. F1 (e) Verify the Cayley-Hamilton theorem for the matrix G 2 GH 3 using this theorem. LM1 (f) Diagonalize the matrix M1 MN0 6 2 0 2 4 5 I JJ K 3 5 . Also, find its inverse 6 OP PP Q 1 0 . 3 Note: Following question no. 5 is for New Syllabus only (TAS–104/MA–101 (New)). 5. Attempt any two parts of the following: (a) Evaluate by changing the variable zz b g R (10 × 2 = 20) 2 x + y dx dy where R is the region bounded by the parallelogram x + y = 0, x + y = 3x – 2y = 0 and 3x – 2y = 3. (b) Find the volume bounded by the elliptical paraboloids z = x2 + 9y2 and z = 18 – x2 – 9y2. (c) Using Beta and Gamma functions, evaluate z FGH 1 −x x IJK dx 1 0 1 2 3 3 Note: Following question no. 6 is for Old Syllabus only (MA-101 (old)). 6. Attempt any two parts of the following: (10 × 2 = 20) (a) In a binomial distribution the sum and product of the mean and variance of the distribution are 25 50 and respectively. Find the distribution. 3 3 426 A TEXTBOOK OF ENGINEERING MATHEMATICS—I (b) From the following data which shows the ages X and systolic blood pressure Y of 12 women, find out whether the two variables ages X and blood pressure Y are correlated? Ages (X) : B.P. (Y) : 56 147 42 125 72 160 36 118 63 149 47 128 55 150 49 145 38 115 42 140 68 152 60 155 (c) (i) If θ is the acute angle between the two regression lines in case of two variables x and y, show that tan θ = 1 − r2 σ xσ y ⋅ 2 r σ x + σ 2y where r, σx and σy have their usual meanings. Explain the significance of the formula when r = 0 and r = ± 1. (ii) Two variables x and y are correlated by the equation ax + by + c = 0. Show that the correlation between them is –1 if signs of a and b are alike and +1, if they are different. GGG B.Tech., First Semester Examination, 2005-06 MATHEMATICSI Time: 3 Hours] Note: [Total Marks: 100 (i) Attempt All questions. (ii) All questions carry equal marks. (iii) Question no. 1–4 are common to all candidates. (iv) Be precise in your answer. 1. Attempt any four parts of the following: (5 × 4 = 20) (a) Use elementary transformation to reduce following matrix A to triangular form and hence the rank of A. LM 2 3 − 1 − 1 OP 1 −1 − 2 − 4 P A = M MM 3 1 3 − 2 PP N6 3 0 −7 Q (b) Define unitary matrix. Show that the matrix LMα + iγ −β + iδOP is a unitary matrix if α + β + γ + δ = 1. N β + iδ α − iγ Q 2 2 2 2 (c) Reduce the matrix A to diagonal form A = LM− 1 2 − 2 OP MM 1 2 1 PP N− 1 − 1 0 Q (d) Find the eigen values and eigen vectors of matrix A LM1 7 13OP 5 7P A = M2 MN3 11 5 PQ [Hint: λ3 – 11λ2 – 95λ – 116 = 0, which cannot be solve.] (e) Test the consistency of following system of linear equations and hence find the solution 4x1 – x2 = 12 – x1 + 5x2 – 2x3 = 0 – 2x2 + 4x3 = – 8 (f) State Cayley-Hamilton theorem. Using this theorem find the inverse of the matrix. A = LM 2 − 1 1 OP MM− 1 2 − 1 PP N 1 −1 2Q 427 428 A TEXTBOOK OF ENGINEERING MATHEMATICS—I 2. Attempt any four parts of the following: 1 in the direction of r , where r = x i + y j + z k . r2 (a) Find the directional derivative of (b) Find zz (5 × 4 = 20) F ⋅ n ds , where a f b g e j 2 F = 2x + 3z i − xz + y j + y + 2z k and s is the surface of sphere having centre (3, –1, 2) and radius 3. (c) Show that the vector field F = potential. r is irrotational as well as solenoidal. Find the scalar r3 e j= (d) If A is a vector function and φ is a scalar function, then show that ∇ ⋅ φA φ∇ ⋅ A + A ⋅ ∇φ. (e) Apply Green’s theorem to evaluate z C 2y 2 dx + 3x dy, where C is the boundary of closed region bounded between y = x and y = x2. b g (f) Suppose F x, y, z = x 3 i + yj + zk is the force field. Find the work done by F along the line from the (1, 2, 3) to (3, 5, 7). 3. Attempt any four parts of the following: (a) If y = (sin–1 x)2, prove that (1 – x2) yn + 2 – (2n + 1) xyn + 1– n2yn = 0 Hence find the value of yn at x = 0. (b) If u = f(r) and x = r cos θ, y = r sin θ, prove that af (5 × 4 = 20) af ∂ 2u ∂ 2u 1 + f″ r + f′ r . ∂x 2 ∂y 2 = r 2 2 2 (c) Trace the curve x 3 + y 3 = a 3 . (d) Expand tan–1 FG y IJ in the neighbourhood of (1, 1). H xK du . dx (f) State Euler’s theorem of differential calculus and verify the theorem for the function (e) If u = x log(xy), where x3 + y3 + 3xy = 1. Find u = log F x4 + y4 I . GH x + y JK 4. Attempt any two parts of the following: (10 × 2 = 20) (a) If J be the Jacobian of the system u, v with respect to x, y and J′ the Jacobian of the system x, y with respect to u, v then prove that JJ′ = 1. (b) A rectangular box open at top is to have capacity of 32 c.c. Find the dimensions of the box least material. (c) A balloon in the form of right circular cylinder of radius 1.5 m and length 4.0 m and is surmounted by hemispherical ends. If the radius is increased by 0.01 m and length by 0.05 m, find the percentage change in the volume of the balloon. 429 UNSOLVED QUESTION PAPERS For New Syllabus Only 5. Attempt any two parts of the following: (a) Evaluate the integral zz ∞ x 0 0 (10 × 2 = 20) F x I dy dx GH y JK x exp − 2 changing the order of integration. (b) Find the triple integration, the volume paraboloid of revolution x4 + y2 = 4z cut off plane z = 4. (c) State the Dirichlet’s theorem for three variables. Hence evaluate the integral zzz x l − 1 y m − 1 z n − 1 dx dy dz . FG x IJ + FG y IJ + FG z IJ ≤ 1. H aK H bK H cK q p where x, y, z are all positive but limited condition r For Old Syllabus Only 6. Attempt any two parts of the following: (10 × 2 = 20) (a) The following data regarding the heights (y) and weights (x) of 100 college students are given Σx = 15000, Σx2 = 2272500, Σy = 6800, Σy = 463025 and Σxy = 1022250. Find the correlation coefficient between height and weight and equation of regression line of height on weight. (b) Fit a Poisson distribution to the following data and calculate the theoretical frequencies: x 0 1 2 3 4 f 192 100 24 3 1 (c) Assume the mean height of soldiers to be 68.22 inches with a variance of 10.8 inches square. How many soldiers in a regiment of 10,000 would you expect to be over 6 feet tall, given that the area under the standard normal curve between x = 0 and x = 0.31 is 0.1368 and between x = 0 and x = 1.15 is 0.3746. GGG B.Tech., First Semester Examination, 2004-05 MATHEMATICSI Time: 3 Hours] Note: [Total Marks: 100 (1) There are five questions in all. (2) Each question has three parts. (3) Attempt any two parts in each question. 1. (a) Find the eigen values and eigen vectors of the matrix LM3 A = M0 MN0 1 2 0 4 6 5 OP PP Q (5 × 4 = 20) (b) Verify Cayley-Hamilton theorem for the matrix LM 2 A = M−1 MN 1 OP PP Q −1 2 2 −1 −1 2 Hence compute A–1. (c) Reduce the matrix A to its normal form when LM 1 2 A = M MM 1 N−1 OP PP PQ 2 −1 4 4 3 4 2 3 4 −2 6 −7 Hence find the rank of A. 2. (a) If y = sin (m sin–1x), prove that (10 × 2 = 20) (1 – x2) yn + 2 – (2n + 1) xyn + 1 + (m2 + n2)yn= 0, and hence find yn at x = 0. (b) If u = u F y − x , z − x I , show that GH xy xz JK ∂u ∂u 2 ∂u + y2 +z = 0. ∂x ∂y ∂z (c) Trace the curve y2 (2a – x) = x3. x2 x1 x2 x2 x3 x x , show that Jacobian of y1, y2, y3 with respect to , y2 = 3 1 and y3 = x3 x1 x2 x1, x2, x3 is 4. (10 × 2 = 20) (b) In estimating the cost of a pile of bricks measured as 6 m × 50 m × 4 m, the tape is stretched 1% beyond the standard length. If the count is 12 bricks in 1 m3 and bricks cost Rs. 100 per 1000, find the approximate error in the cost. (c) Find the extreme values of x3 + y3 – 3axy. 3. (a) If y1 = 430 431 UNSOLVED QUESTION PAPERS 4. (a) Calculate the volume of the solid bounded by the surface x = 0, y = 0, x + y + z = 1 and z = 0, where D is the domain x ≥ 0, y ≥ 0 and x + y ≤ h. (10 × 2 = 20) (b) Prove that z x l − 1 y m − 1 dx dy = D l m l+m+1 hl + m zz 1 (c) Change the order of integration in l = 0 2 −x x2 xy dx dy and hence evaluate the same. e j 1 5. (a) Prove div(grad rn) = n(n + 1)rn – 2, where r = x 2 + y 2 + z 2 2 . Hence show that ∇ (b) Apply Green’s theorem to evaluate ze C j e j 2 FG 1IJ = 0. H rK (10 × 2 = 20) 2x 2 – y 2 dx + x 2 + y 2 dy , where C is the boundary of the area enclosed by the x-axis and the upper half of circle x2 + y2 = a2. (c) Evaluate zzd i yz i + zx j + xy k ds , where S is the surface of the sphere x2 + y2 + z2 = a2 S in the first octant. GGG This page intentionally left blank Index A F Absolute error, 112 Absolute maximum, 121 Absolute minimum, 121 Acceleration, 334 Associative law, 156 Asymptotes, 65 Augmented matrix, 189 Functional dependence, 107 G Gamma function, 283 Geometrical representation, 215 H Hermitian matrix, 154 B I Beta, 283 Idempotent matrix, 154 Identity matrix, 152 Implicit functions, 102 Independence of path, 367 Intercepts, 64 Intersection point with x- and y-axis, 64 Involutory matrix, 155 Irrotational vector, 354 Irrotational vector field, 365 C Chain rule, 50, 96 Circulation, 365 Column matrix, 152 Conjugate of a matrix, 153 Consistent system, 189 Curve tracing, 63 D J Del operator, 337 Diagonal matrix, 152 Directional derivative, 339 Distributive law, 156 Dritchlet’s integral, 324 Jacobian for functional dependence functions, 107 Jacobians, 95 K Kirchhoff’s current law, 250 Kirchhoff’s voltage law (KVL), 250 E Eigen, 214 Eigen values of diagonal, 216 Equivalent matrices, 167 Expansion of function of several variables, 81 Exponential function, 2 Extremum value or point, 122 L Lagrange’s conditions for maximum or minimum, 123 Linear independence of vectors, 210 Linearly dependent vectors, 210 433 434 A TEXTBOOK OF ENGINEERING MATHEMATICS—I Logarithm case, 2 Lower triangular matrix, 153 Scalar matrix, 152 Similar matrix, 239 Skew hermitian matrix, 154 Skew symmetric matrix, 153 Solenoidal, 353 Square matrix, 152 Symmetric about origin, 64 Symmetric about the line, 64 Symmetric about x-axis, 64 Symmetric about y-axis, 64 Symmetry, 64 Symmetric matrix, 153 M Maclaurin’s series expansion, 83 Matrix polynomial, 233 Minor of a matrix, 176 Multiple integral, 258 N Nilpotent matrix, 154 Non-singular, 163 Non-trivial solution, 197 Normal derivative, 339 Null matrix, 152 O Oblique asymptotes (not parallel to x-axis and y-axis), 65 Oblique asymptotes (when curve is represented by implicit equation f ), 65 Origin and tangents at the origin, 64 Orthogonal matrix, 155 Own characteristic or individual or eigen, 214 P Partial differentiation, 20 Points of intersection, 65 Power function, 1 Product functions, 3 Properties of beta and gamma functions, 283 Proportional or relative error, 112 R Rectangular matrix, 152 Regions, 64 Row matrix, 152 S Saddle point, 122 Scalar field, 337 T Total differential, 49 Tranjugate or conjugate transpose of a matrix, 154 Transpose of a matrix, 153 Trigonometric functions cos (ax + b) or sin (ax + b), 2 Triangular matrix, 153 Trivial solution, 189 U Unitary matrix, 155 Upper triangular matrix, 153 V Vector differential calculus, 333 Vector field, 337 Vector form of green’s theorem, 377 Velocity, 334 Volume in cylindrical coordinates, 315 Volume in spherical polar coordinates, 315 Z Zero matrix, 152 GGG