Engineering Mathematics

SECOND EDITION

A Textbook of

ENGINEERING MATHEMATICS-I

H.S. Gangwar l Prabhakar Gupta

A Textbook of

ENGINEERING MATHEMATICS-I

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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 [email protected] ISBN (13) : 978-81-224-2847-6

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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


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

8. 9. 10. 11. 12. 13. 14. 15. 16. 17.

cos ix = cos hx

z z z z z z z

1 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

a 2 − x 2 dx =

x 2

a2 − x2 +

a2 x sin −1 a 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 =

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 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11

3.12 3.13 3.14 3.15 3.16 3.17 3.18 3.19 3.20

Introduction Definition of Matrix Types of Matrices Operations on Matrices Trace of Matrix Properties of Transpose Properties of Conjugate Matrices Singular and Non-Singular Matrices Adjoint of a Square Matrix Inverse of a Matrix (Reciprocal) Exercise 3.1 Elementary Row and Column Transformations Method of Finding Inverse of a Non-Singular Matrix by Elementary Transformations Exercise 3.2 Rank of a Matrix Exercise 3.3 System of Linear Equations (Non-Homogeneous) System of Homogeneous Equations Gaussian Elimination Method Exercise 3.4 Linear Dependence of Vectors Exercise 3.5 Eigen Values and Eigen Vectors Exercise 3.6 Cayley-Hamilton Theorem Exercise 3.7 Diagonalization of a Matrix 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 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 4.10 4.11 4.12 4.13 4.14 4.15 4.16 4.17

Multiple Integrals Double Integrals Working Rule Double Integration for Polar Curves Exercise 4.1 Change of the Order of Integration Change of Variables in a Multiple Integral Exercise 4.2 Beta and Gamma Functions Transformations of Gamma Function Transformations of Beta Function Relation between Beta and Gamma Functions Some Important Deductions Duplication Formula Evaluate the Integrals Exercise 4.3 Application to Area (Double Integrals) Exercise 4.4 Triple Integrals Exercise 4.5 Application to Volume (Triple Integrals) Exercise 4.6 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

n yn =

0

a n (ax + b)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 W

(−1)n n an

=

(ax + b)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

q

dn log( ax + b) dx n

=

( −1) n −1 ( n − 1) a n ( ax + b) 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 cos ax + b + H 2 K 3π I F cos ax + b + H 2 K

y1 = – a sin (ax + b) = a cos y2 = – a2 cos (ax + b) = a2 y3 = + a3 sin (ax + b) = a3

................................................................................. .................................................................................

3

DIFFERENTIAL CALCULUS-I

yn =

Similarly,

yn =

F I H K nπ I F sin ax + b + H 2K

nπ dn cos ( ax + b) = a n cos ax + b + n 2 dx dn dx

n

sin( ax + b) = a 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 =

–1 a 2 + b 2 and φ = tan (b/a)

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


L(−1) n 2 OP LM(−1) n 3 OP 2 M MN (2x − 1) PQ – 3 MN (3x − 1) PQ L 2 − 3 OP . ( −1) n M N (2x − 1) ( 3x − 1) Q n

=

or

yn =

n

n

n+1

n +1

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

(−1) n n a n dn −1 ( ax + b ) = dx n ( ax + b) n+1

n+1

l

or

As

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+

F 2x I H1−x K

2

⋅

(U.P.T.U., 2002)

FG H

d 2x dx 1 − x 2

2

2(1 + x 2 )

=

2

=

IJ K

=

e1 + x

(1 − x 2 ) 2 4

− 2x 2 + 4x 2

(1 + x 2 ) 2

y1 =

1 1 1 − , (by Partial fractions) i x−i x+i

LM N

OP Q

Differentiating both sides (n–1) times w.r. to ‘x’, we get yn =

=

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(1 − x 2 ) + 4x 2 (1 − x 2 ) 2

2 ( x + i)( x − i )

y1 =

(1 + x 2 )

j

⋅

( 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


=

( −1) n − 1 ( n − 1) r − n i

n − 1 r–n sin nθ, where r =

yn = 2(–1)n–1

x −1 . Show that x +1

LM x − n N (x − 1)

n−2

yn = (–1)n–2 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

n

−

x+n ( x + 1)

n

OP Q

(U.P.T.U., 2002)

x −1 = x [log(x – 1) – log (x + 1)] x +1

y = x log

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 dx n −1

log (x − 1) −

d n− 2

d n−1

log (x + 1) +

dx n −1

d n −1

d n −1

dx

dx 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

LM x − 1 N (x − 1) L x−n n−2M N (x − 1)

n−2 n−2 = ( −1)

n

−

n −2 = ( −1)

n

−

Example 5. Find yn (0) if y = Sol. We have

y =

+

x+1 ( x + 1)

n

x+n ( x + 1)

x3 x −1 2

n

( −1)n−1 (n − 1) n − 2

−

OP . Q

(x − 1)n ( n − 1) ( x − 1)

n

−

( n − 1) ( x + 1)

n

+

( −1)n − 1 (n − 1) n − 2

OP Q

.

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


or

y =

x2 + x + 1 1 + ( x + 1) ( x − 1)( x + 1)

y =

x2 − 1 + 1 1 +1+ x +1 x−1 x+1

y = or

y =

a fa f 1 1L 1 1 O + M − x+ x + 1 2 N x − 1 x + 1 PQ 1L 1 1 O + x+ M 2 N x − 1 x + 1 PQ (−1) n O 1 L ( −1) n P + 0+ M 2 NM ( x − 1) ( x + 1) QP ( −1) n L 1 1 MN (x − 1) + (x + 1) OPQ 2 ( −1) n L 1 1 O + M 2 N (−1) (1) PQ n

∴

yn =

n

n +1

n+1

n

or

yn =

n +1

n +1

n

At

x = 0, yn (0) =

When n is odd, yn(0) = When n is even, yn(0) =

n +1

( −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 MN

2 . Find the nth derivative of

x2 . ( x − a)( x − b)

3 . Find the nth derivative of tan–1

LM 1 + x OP . N1 − x Q

n

=

( −1) n n + 1 3( x − 1) n + 2

LMAns. MN

+

( −1) n n ( a − b)

5( −1) n n 9( x − 1) n +1

LM a N ( x − a) 2

n +1

+

−

OP PQ OPOP QPQ

4( −1) n n

9( x + 2) n +1

b2 (x − b)n +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

F xI . H aK

LMAns. N

F I H K Ans. a −1f n−1 a

F H

π π 3 1 − . 3n . sin 3x + n sin x + n 4 2 4 2 n− 1

−n

I OP KQ

sin n θ sin nθ

7


a f 2( −1) n O PP ( x + 1) Q

Ans. e x x + n

6 . Find yn, where y = ex.x.

LMAns. MN Ans. a −1f n − 1 ⋅ 2x n

1− x 7 . Find yn, when y = . 1+ x

n+1

n− 1

8 . Find nth derivative of log x2.

LMAns. N

9 . Find yn, y = ex sin2 x.

ex 1 − 5n/2 cos( 2x + n tan −1 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 N 4 MN

n

F H

cos 6x + n

−n

I K

F H

I K

F H

π π π + 4n cos 4x + n + 2n cos 2x + n 2 2 2

OP Q

I OPOP K QQ

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 Dm+1 (uv) =

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

c0 D m+1 ( u) ⋅ v + D m ( u) ⋅ D( v) + mc1 D m ( u) D( v) + Dm−1 ( u) D 2 ( v)

m

+ mc2

mD

m− 1

r

m

m

+ ..... + mcm D( u) ⋅ D ( v ) + uD But from Algebra we know that mcr + mcr+1 = ∴ Dm+1(uv) =

m +1

c 0 D m + 1 ( u) ⋅ v +

+ ... +

c

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

c

m

h

h

m+1

cr+1 and mc0 =

c 0 + m c 1 D m ( u ) ⋅ D( v ) +

c

m

m

m +1

m+1

c0 = 1

h

c1 + m c 2 D m − 1 u ⋅ D 2 v

c r + m c r +1 D m − r ( u) ⋅ D r +1 ( v ) +...+ m +1 c m +1 u ⋅ D m +1 ( v )

c As

m

cm = 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


⇒ 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

LM1 ± N 2

2x x2

m

OP = −1 Q

...(i)

m x ± x2 − 1 x2 − 1

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

2

m

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


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

1 n(n − 1) x 1 e − 2 = e · log x + ne · + x 2 x x

⇒

x

LM MN

Dn (ex log x) = ex log x +

IJ K

x

+ ... + e

OP PQ

xn

( −1) n −1 n − 1 n n( n − 1) − + + ... . x 2x 2 xn

Example 3. Find the nth derivative of x2 sin 3x. Sol. Let u = sin 3x and v = x2 ∴

( −1)n−1 n − 1

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)

F H

= 3n sin 3x +

nπ 2

I K

+

F H

= 3nx2 sin 3x +

FG 3x + n − 1π IJ · 2x H 2 K n(n − 1) F n − 2π IJ ⋅ 2 · 3 sin G 3x + H 2 2 K F n − 1π IJ sin G 3x + H 2 K F n − 2π IJ . + 3 n(n−1) · sin G 3x + H 2 K

· x2 + n3n–1 sin

I K

nπ + 2nx · 3n-1 2

n–2

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


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 1I H xK

– b sin (log x)

F 1I 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 N1 − x Q

= yα

LM x OP 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.

FG y IJ = log F x I H bK H mK FG y IJ = log F x I H bK H mK

Hence proved.

m

Example 7. If cos–1 Sol. We have cos–1

, prove that x2yn+2 + (2n + 1)xyn+1 + (n2 + m2) yn = 0. m

= m log

x m

11


⇒

y = b cos

F m log x I H mK

On differentiating, we have

F m log x I · m H mK x xI F – mb sin H m log K m

2

y1 = – b sin ⇒

xy1 =

Again differentiating w.r.t. x, we get

F H

xy2 + y1 = – mb cos m log

x (xy2 + y1) = – m2b cos

⋅

1 m

I K

1 1 x m· · x m m m

F m log x I H mK

= –m2y

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


Second part: Let y = (x2 – 1)n ∴ Now

Pn =

RS T

d d yn (1 − x 2 ) dx dx

UV = W

dn y = yn dx n

n

d (1 − x 2 )yn +1 dx

s

= (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. dx T d d (1 − x 2 ) Pn dx dx 2

or

[Using equation (i)]

n

n

n

Hence proved.

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 =

c1 yn-1 = (n – 1)yn–1 +

n−1

⇒

xyn + (n – 1)yn–1 = (n – 1)yn–1 +

n−1

⇒

xyn =

xyn +

n–1

At

x =

yn

FG 1 IJ H 2K

n − 1 i.e. y = n

d n −1 n−1 = ( n − 1)( n − 2)...2.1 = n − 1 x dx n −1

n−1 x

1 2

= 2

n−1 .

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 = –

1 2 −2 x(1 − x )

xy + 1 sin −1 x 1 = + 2 2 (1 − x 2 ) (1 − x ) (1 − x ) y1 (1–x2) = xy + 1 y1 =

⇒

1 (1 – x2)–3/2 (–2x) sin–1x + (1 – x2)–1/2· 2

1 1 − x2

13


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

a f

n

(ii) yn = nyn–1 + n − 1 . x Sol. (i) We have y = xn log x Differentiating w.r. to x, we get (i) yn+1 =

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 GH x JK ex .log xj + dxd . x

dn d n−1 x n .log x = n dx dx n −1

e

d n−1 dx n −1

= n

j

n

d n−1 dx n −1

= nyn–1 +

n

n−1

n−1

n −1

an − 1f .

n −1

n −1

Proved.

∴ yn −1

e

j

dn x n log x n dx d n −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 + MN N x

n

c1 ⋅

1 n 1 1 − c 2 ⋅ 2 + 2 n c 3 ⋅ 3 +...+( −1) n−1 n − 1 n c n x −n x x x

OP OP QQ

Ans. e x x 2 + 2nx + n(n − 1)

j

14


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 N n

2

F H

sin 3x +

I K

LM N

UVOP WQ + 12(n − 1)( 2x + 3) + 8n(n − 1)(n − 2)t OP Q

RS T

π 1 nπ + 2nx ⋅ 3 n −1 sin{ 3x + ( n − 1) π } + 3 n− 2 n( n − 1) sin 3x + ( n − 2) 2 2 2

o

Ans. e x ( 2x + 3) 2 + 6n( 2x + 3) 2

6 . ex (2x + 3)3.

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

LMF 1 I log yOP , prove that (x – 1)y + xy – m y = 0 and (x – 1)y NH m K Q 2

FG y IJ H bK

2

2

+ (n2 – m2)yn = 0. 1 1 . If cos–1

(U.P.T.U., 2004, 2002)

= log

F xI H nK

2

1

n+2

+ (2n + 1)xyn+1

n

, prove that x2yn+2 + (2n + 1)xyn+1 + 2n2yn = 0.

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

n

d

+ C2 x − x 2 − 1

i , prove that (x n

2

– 1)yn+2 + (2n + 1)xyn+1 = 0.

1 5 . If x = cos [log (y1/a)], then show that (1 – x2)yn+2 – (2n + 1) xyn+1 – (n2 + a2) yn = 0.

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

y = em.cos

–1x

Differentiating w.r.t. x, we get y 1 = em or

cos–1x

m

FG H

−1 1−x

2

IJ K

...(i) ⇒

2 2 2 2 1 − x 2 y1 = – my ⇒ (1 – x )y1 = m y

m 1 − x 2 . y1 = – me

cos–1x

15


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 At x = 0, y1(0) = −me

mπ 2

...(iv)

...(v) As cos −1 0 =

Using (v) in (iv), we get yn(0) = −

{an − 2f

2

+ m2

} {an − 4f

2

} e

j

+ m 2 .... 1 2 + m 2 me

mπ 2

π 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)

As y = e m cos

...(vii)

∴ y(0) = e m cos

−1

0

−1

x

= 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


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) = From (ii) y2 (0) = Using this value in eqn. (iv), we yn(0) =

...(iv)

yn(0) = 2.22.42 ... (n – 2)2, n ≠ 2 otherwise 0. Proved.

or Example 3. If

y =

x + 1 + x2

Sol. Given

y =

x + 1 + x2

= y1 1 + x 2

or Squaring

m

m x + 1 + x2 1+ x

, find yn 0).

m

2 y1 = m x + 1 + x

∴

m −1

m

LM1 + N

=

2

x 1+x

2

OP Q

...(i)

my 1 + x2

= my

y21(1+x2) = m2y2 2 1

or

(n – 2)2 (n – 4)2 ... 22· y2 (0) 2 get (n – 2)2 (n – 4)2 .... 22·2

...(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


(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) y1(0) y1(0) yn(0)

= = = =

– {(n – 2)2 – m2} {(n – 4)2 – m2} ... {12 – m2} y1 (0) my(0) m (As y (0) = 1) {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 MN

−2 1− x2

OP PQ

...(i)

–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


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 0

FG H

nπ = x2 . sin x + 2

FG nπ IJ H 2K F nπ IJ sin G x + H 2K

yn = (x2 – n2

FG H

+ 2nx . sin x +

IJ K

n−1 π 2

2

FG H

+ n(n – 1) sin x +

IJ K

n−2 π 2

FG nπ − π IJ + n(n – 1) sin FG x + nπ − πIJ H 2 K H 2 2K F nπ IJ – n(n – 1) sin FG x + nπ IJ – 2nx cos G x + H 2K H 2K F nπ IJ – 2nx cos FG x + nπ IJ + n) sin G x + H 2K H 2K

= x2 sin x + = x2

IJ K

1

+ 2nx sin x +

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


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

(1 – x2)yn+2 – 2nxyn+1 − 2n n − 1 2 or

yn – xyn+1 – nyn = – a2yn

(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

{an − 2f

yn

2

2

} {an − 4f + a } ... (3 + a ) (1 + a )a n is even, (0) = {an − 2f + a } {an − 4f + a } ...... (4 + a )(2 + a ) a . 2

2

+ a2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

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


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 fFGH

Ans. n is odd, yn (0) = –1

n−1 2

IJ K

{an − 2f

2

+ m2

} {an − 4f

2

}

+ m 2 .... (12 + m2) m · n is even, yn = (0) = 0.

LM { N

7 . If y = log x + 1 + x 2

}OPQ

2

, show that

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


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


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

= –

ex

4xy 2

j

+ y2

...(A)

2

Again differentiate (i) partially w.r.t. y, we obtain

∂u ∂y

=

ex

2y +y

2

2

j

1 y

–

Next, we differentiate above equation w.r.t. x.

∂ 2u 4xy = – 2 ∂x∂y x + y2

e

j

...(B)

2

Thus, from (A) and (B), we find ∂ 2u ∂ 2u = . ∂y∂x ∂x∂y

Hence proved.

FG y IJ , verify that ∂ f H xK ∂y∂x F yI f = tan G J H xK 2

Example 2. If f = tan–1 Sol. We have

=

∂2 f . ∂x∂y

–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


=

y2 − x2

ex

2

+ y2

j

...(iv)

2

Differentiating (iii) partially with respect to x, we get

F GH

∂2 f x ∂ = 2 ∂x∂y ∂x x + y 2 =

Example 3. If u(x + y) = x Sol. Given ∴

and ∴

or

ex

2

and

+ y2

j

...(v)

F ∂u ∂u I + y , prove that G − J H ∂x ∂y K

F1 − ∂u − ∂u I . GH ∂x ∂y JK

2

=4

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 bx + yg bx + yg 2

=

∂u ∂y

=

∂u ∂u + ∂x ∂y

=

b x + yg

=

1−

=

ex

2

2

2

2

2

2

=

2 2

2

2

2

2

2

2

4xy

2

4xy ( x + y) 2

=

(x − y) 2 (x + y) 2

j e bx + y g 2 ex − y j 2 bx − yg = bx + yg bx + y g 2

+ 2xy − y 2 − y 2 + 2xy − x 2 2

2

∴

3

∂2 f ∂2 f = . Hence proved. ∂y∂x ∂x∂y

∂u ∂x

∂u ∂u – ∂y ∂x

2

2

u =

∂u ∂u 1– – ∂x ∂y

2

y2 − x2

∴ From eqns. (iv) and (v), we get

2

I = ex + y ja1f − xa2xf JK ex + y j

...(i)

j

2

2

...(ii)

From (ii), we get

F ∂u − ∂u I GH ∂x ∂y JK

2

=

b g bx + yg

4 x− y

2

2

=4

F1 − ∂u − ∂u I , from (i). GH ∂x ∂y JK

Hence proved.

24


Example 4. If u = sin–1 Sol. Given

x

and from (i),

or

y

∂u ∂x

=

∂u ∂x

=

∂u ∂y

=

∂u ∂y

=

Adding (ii) and (iii), we get x Example 5. If f(x, y) = x2 tan–1

Sol.

or

∂u ∂u +y = 0. ∂y ∂x

–1

–1

∴

or

F x I + tan FG y IJ , show that x GH y JK H xK F xI F yI u = sin G J + tan G J H xK y H K –1

1

R| F x I S|1 − GH y JK T

ey

x 2

− x2

2

U| V| W

1

2

− x2

– y2 tan–1

FG y IJ H xK

+ x2 ·

∂f = 2x · tan–1 ∂x

FG y IJ H xK

–

1 x

...(iii)

x + y2 Hence proved.

F xI GH y JK

∂f = 2x · tan–1 ∂x

FG y IJ H xK

⋅

2

∂u ∂u +y = 0. ∂y ∂x

FG y IJ H xK

1+

2

xy

+

j

1

+

2

2

ey

...(ii)

x + y2

F– x I R| F x I U| GH y JK S|1 − GH y JK V| T W x

2

2

1

−

.

2

yx

−

j

FG − y IJ F yI H x K 1+G J H xK

1 + y

.

...(i)

then prove that

1

1+

FG y IJ H xK

yx 2 x2 + y2

–

2

×

FG − y IJ H xK 2

y3 x2 + y2

∂2 f ∂2 f = . ∂x∂y ∂y∂x – y2 ·

= 2x tan–1

F 1I GH y JK F xI 1+G J H yK FG y IJ – y H xK 1

2

·

Differentiating both sides with respect to y, we get

∂2 f = 2x · ∂y∂x

Again

or

∂f = x2 · ∂y

FG 1 IJ F yI H xK 1+G J H xK 1

2

1

1+

FG y IJ H xK

2

·

·

–1 =

2x 2 x2 − y2 –1 = x2 + y2 x2 + y2

1 – 2y tan–1 x

∂f x3 = 2 – 2y tan–1 ∂y x + y2

F xI GH y JK

+

F xI GH y JK

– y2 ·

xy 2 x2 + y2

...(i)

F− x I GH y JK F xI 1+G J H yK 1

2

·

2

25


e

x x2 + y2 =

x +y 2

2

j

– 2y tan–1

F xI GH y JK

= x – 2y tan–1

F xI GH y JK .

Differentiating both sides with respect to x, we get

∂2 f = 1 – 2y ∂x ∂y

F 1I F x I GH y JK 1+G J H yK 1

2

=1–

2y 2 x2 + y 2

=

x2 − y2

...(ii)

x2 + y2

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

= –

...(i)

1 (x2 + y2 + z2)–3/2 2x = – x (x2 + y2 + z2)–3/2 2

LM RS 3 ex NT 2

= – x –

2

+ y2 + z2

j

–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


Sol. Given

r2 = x2 + y2

...(i)

Differentiating both sides partially with respect to x, we have 2r

Similarly, Now,

∂r ∂x

= 2x or

∂r ∂y

=

∂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 ∴

∂ 2u ∂x 2

or

Similarly,

Adding,

LM f ′ arfx OP = r f ′ arf.1 + xf ′′arfb∂r / ∂xg – xf ′ arfb∂r / ∂xg r N r Q LMrf ′ arf + x f ′′arf − x f ′ arfOP r N Q , from (ii). LMrf ′ arf + y f ″ arf – y f ′ arfOP r MN PQ LM ex + y j f ′ arfOP 2rf ′ ar f + ex + y j f ″ arf − MN PQ r

=

∂ ∂x

∂ 2u ∂x 2

=

1 r2

∂ 2u ∂y 2

=

1 r2

=

1 r2

=

1 [2rf ′(r) + r2f ″(r) – rf ′(r)], from (i) r2

=

1 f ′(r) + f ″(r). Hence proved. r

∂ 2u ∂ 2u + ∂x 2 ∂y 2

2

2

2

2

2

2

2

2

2

Example 8. If u = log (x3 + y3 + z3 – 3xyz); show that

F ∂ + ∂ + ∂I GH ∂x ∂y ∂z JK Sol. Given

and

2

u=–

9

bx + y + zg

2

.

(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


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

=

bx + y + zgex

j

+ y + z − xy − yz − zx

2

2

2

a

j

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 GH ∂x ∂y ∂z JK H ∂x ∂y ∂z K H ∂x ∂y ∂z K 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 , 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

Now,

2

2

2

from (iv)

2

Hence proved.

Example 9. If u = tan–1

∂ 2u ∂x∂y Sol. Given

∴

e

xy

1 + x2 + y2

=

e1 + x

1 2

=

+ y2

j

3/2

⋅

xy

u = tan–1

∂u ∂y

j

, show that

e1 + x

2

+ y2

⋅

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

=

j

2

2

2 2

2

2

2 −1/2

2

2

2

2

2

2

2

2

2y

OP PP Q

28


∂u ∂y

or

=

x

1 + x2

e1 + x je1 + y j e1 + x 2

.

2

+y

2

2

x

=

j e1 + y j e1 + x 2

2

+ y2

j

Again differentiating partially w.r.to x ∂ 2u ∂x ∂y

=

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 ∂ ∂x

2

2

2

=

Example 10. If

2

2

2

2

2 −1/2

2

2

2

2

2

2 3/2

2

2 3/2

y2 x2 z2 + + = 1, show that a2 + u c2 + u b2 + u

FG ∂u IJ + FG ∂u IJ + FG ∂u IJ H ∂x K H ∂y K H ∂z K 2

2

2

2

2

=

2

2

2

=2

F x ∂u + y ∂u + z ∂u I . GH ∂x ∂y ∂z JK

(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 =

=

2

=

Adding with square

FG IJ + FG ∂u IJ + FG ∂u IJ H K H ∂y K H ∂z K ∂u ∂x

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

LM e N

=

2

2

2

2

2

2

2

2

e j + z / e c + uj OPQ LM∑ RSx / e a + uj UVOP WQ N T j

4 x2 / a2 + u

2

2

2

2

∂u ∂y

Similarly,

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 = 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 = ; ∑ LMNx / e a + uj OPQ ∂z ∑ LMNx / ea + uj OPQ 2x − a2 + u

2

∂u ∂x

or

...(i)

2

+ y 2 / b2 + u 2

2

2

2

2

2

2

2

2

2

29


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

From (ii) and (iii), we have 2

+

F ∂u I GH ∂y JK

FG ∂u IJ H ∂z K

2

+

∑ LMNx / e a + uj OPQ 2

=

FG ∂u IJ H ∂x K

4

2

= 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

2

2

OP Q

2

2

1

2

2

2

[1], from (i)

F x ∂u + y ∂u + z ∂u I GH ∂x ∂y ∂z JK .

...(iii)

Hence proved.

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 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

=0

...(ii)

...(iii)

30


∂2z 1 1 ∂x∂y = − x 1 + log x = – x log e e + log x = (As log

b

⇒

=–

g

1 x log ex

a f

b

m

a fr

= – x log ex

g

−1

e

e = 1)

. Hence proved.

Example 12. If u = log (x3 + y3 – x2y – xy2) then show that

∂ 2u 4 ∂ 2u ∂ 2u + 2 + = − 2 2 ∂x∂y ∂x ∂y x+y

b

Sol. We have

g

2

.

u = log (x3 + y3 – x2y – xy2)

∂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

∂ 2u ∂ 2u ∂2u Now, + 2 + 2 ∂x ∂x∂y ∂y 2

2 x− y

2

2

2

bx + yg ex + y – 2xyj 2 bx − yg 2 = = bx + ygbx − yg bx + yg F∂ ∂I = G + J u H ∂x ∂y K 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 =

∂u ∂u + ∂x ∂y

b g

j

j

2

... (iii)

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


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

u = exyz ∴

Sol. We have

= (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, Now, ∴ or ∴ or

Similarly,

∴

FG x − a IJ . H r K az − cf =

∂r ∂x

= 2 (x – a) or

∂r ∂y

=

∂u ∂x

=

1 ∂r 1 x−a = , from (ii) r r ∂x r

∂u ∂x

=

x−a r2

∂ 2u ∂x 2

=

∂ ∂x

∂ 2u ∂x 2

=

∂ 2u ∂y 2

=

b y − bg

∂r and ∂y

r u = log r.

∂ 2u ∂ 2u ∂ 2u + + ∂y 2 ∂x 2 ∂z 2

=

=

r2

∂r = ∂x

...(ii)

r

IJ K

FG H

FG x − a IJ = r a1f − ax − af 2rb∂r / ∂xg Hr K r − 2 ax − af , from (ii) 2

2

4

2

r4

b g

r2 − 2 y − b

2

r4

;

∂ 2u r 2 − 2 ( z − c) 2 = . ∂z 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


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 1 + b y / xg H 1

x2

= –

= –

2

x2 y x +y 2

–

2

e

y x2 + y2

ex

2

+y

2

j

j

⋅ −

y x

2

IJ K

y3 x +y 2

2

+ 2x tan −1

+ 2x tan–1

+ 2x tan–1

FG y IJ H xK

FG y IJ H xK

− y2 ⋅

1

b g

1 + x/ y

2

⋅

1 y

y x

FG y IJ H xK

= – y + 2x tan–1

Again differentiating partially with respect to y, we get ∂ 2u ∂y∂x

=

∂ ∂y

RS– y + 2x tan FG y IJ UV = – 1 + 2x 1 H xKW F yI T 1+G J H xK

= –1+

–1

2

⋅

1 x

x2 − y2 2x 2 = . Hence proved. x2 + y2 x2 + y2

Example 16. If z = f (x – by) + φ (x + by), prove that

∂2z b ∂x 2 z ∂z ∂x ∂2z ∂x 2 ∂z ∂y 2

Sol. Given ∴ and Again from (i),

∂ 2z

. ∂y 2 = f (x – by) + φ (x + by) =

...(i)

= f ′ (x – by) + φ′ (x + by) = f ″ (x – by) + φ″ (x + by).

...(ii)

= – bf ′ (x – by) + bφ′ (x + by)

∂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


∂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. N

(i) u = xxy.

(ii) u = log

3. If u = sin–1

FH x + x − y IK LMAns. ∂u = 1 ∂u = − y ex MN ∂x x − y ∂y F x − y I , show that ∂u = – y ∂u . GH x + y JK ∂x x ∂y 2

2

2

4. If u = ex (x cos y – y sin y), prove that

5. If u = a log (x2 + y2) + b tan–1

x2 x

y2 y

z2 z

FG y IJ , prove that H xK

1

1

1

, prove that

g

2

2

− y2

∂ 2u ∂ 2u + = 0. ∂x 2 ∂y 2

6. If z = tan (y – ax) + (y + ax)3/2, prove that

7. If u =

b

OP Q IK OP PQ

∂u ∂u = x xy y log x + y ; = x xy + 1 log x ∂x ∂y

∂ 2u ∂ 2u + = 0. ∂x 2 ∂y 2

2 ∂ 2z 2 ∂ z a – = 0. ∂x 2 ∂y 2

∂u ∂u ∂u + ∂y + = 0. ∂x ∂z

j FH x + −1 2

x2 − y2

−1

34


8. If u = 2(ax + by)2 – (x2 + y2) and a2 + b2 = 1, find the value of

∂ 2u ∂ 2u . 2 + ∂x ∂y 2

LM MMAns. ex N

∂ 2u . 9. If u = log (x2 + y2 + z2), find the value of ∂y∂z

e

10. If u = x 2 + y 2 + z 2

j

1 2

, then prove that

∂2u ∂x 2

11. If z = f (x + ay) + φ (x – ay), prove that

13. If u = log (x2 + y2), show that

∂2u ∂x 2

16. If u = r , where r =

x +y +z 2

17. If u = (x2 + y2 + z2)–1, prove that

18. If θ = t e

−

r2 4t

+ y2

= 0.

∂ 2u = 0. ∂y 2

+

∂2 f ∂x

2

= b2

∂2 f ∂t 2

.

∂ 2u ∂ 2u ∂ 2u . find 2 + ∂y 2 + ∂x ∂z 2

2

[Ans. m (m + 1)rm−2]

∂ 2u ∂ 2u ∂2u + + = 2 (x2 + y2 + z2)–2. ∂x 2 ∂z 2 ∂y 2

, find the value of n, when

1 ∂ r 2 ∂r

LMAns. n = − 3 OP 2Q N

FG r ∂θ IJ = ∂θ . H ∂r K ∂t 2

19. For n = 2 or – 3 show that u = rn (3 cos2 θ – 1) satisfies the differential equation

∂ ∂r

FG r ∂u IJ H ∂r K 2

FG sin θ ∂u IJ = 0. H ∂θ K F ∂ uI cos (a log r), show that G H ∂r JK +

∂ 1 sin θ ∂θ

2

20. If u = eaθ −

21. If e

2

z

ex − y j 2

2

= (x – y), show that y

+

1 1 ∂u + 2 r ∂ r r

∂ 2u ∂θ 2

= 0.

∂z ∂z +x = x2 – y2. ∂y ∂x

[Hint: Solve for z = (y2 – x2) 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

2

∂u ∂u ∂u + ∂y + = (x + y + z)2. ∂x ∂z

14. If u = x2y + y2z + z2x, show that

m

OP P +z j P Q

−4 yz

∂ 2u ∂2u 2 = . 2 + 2 ∂y u ∂z

bx − yg / bx + yg , prove that x ∂∂ux + y ∂∂uy

12. If u = cos–1

n

+

[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


∂ 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 = H ∂x K H ∂u K F ∂x I F ∂x I F ∂θI F ∂θ I If x = r cos θ, y = r sin θ, find G J , G J G J G J H ∂r K H ∂θ K H ∂xK H ∂y K y

25.

θ

v

r,

y,

F I FG ∂y IJ GH JK H ∂v K FG ∂y IJ . H ∂x K

1 ∂v 2 ∂y

24. If x2 = au + bv; y2 = au – bv, prove that

x

. 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 = xn

LMa MN

f (x, y) = xn F

or

0

+ a1

FG y IJ H xK

FG y IJ + a FG y IJ H xK H xK 2

FG y IJ OP H x K PQ n

2

+...+ a n

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 xI F xI x + a G J + .... + a G J S| y H yK H yK T 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 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 GH y JK

n

F(y/x)

n

n

U| V| W

36


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 ∴

f (x, y) = xn F

FG y IJ H xK

...(i)

Differentiating partially w.r.t. x and y, we get

FG y IJ + x H xK F yI F 1I F′ G J GH JK H xK x

∂f = nxn−1 F ∂x ∂f ∂y

= xn

n

F′

FG y IJ FG − y IJ H xK H x K

...(iii)

Multiplying (ii) by x and (iii) by y and adding, we have x

= nxn ⇒

x

FG y IJ H xK F yI F G J H xK

∂f ∂f +y = nxn F ∂x ∂y

∂f ∂f +y = nf ∂y ∂x

– xn−1 y F′

...(ii)

2

FG y IJ H xK

+ xn−1 y F′

FG y IJ H xK

(from (i)).

In general if f (x1, x2, ..., xn) be a homogeneous function in x1, x2, ..., xn then x1

∂f ∂f + x2 ∂x1 ∂x 2

∂f + ... + xn ∂x = nf. n 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)

x

∂f ∂2 f ∂2 f ∂f +x + y = n ∂x ∂x∂y ∂x ∂x 2 ∂f ∂2 f ∂2 f ∂f x + ∂y + y = n 2 ∂y∂x ∂y ∂y

and

...(ii) ...(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


⇒

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

...(i)

FG y IJ H xK F yI F yI F y I sin G J + x cos G J G − J H xK H xK H x K F yI F yI sin G J – x y cos G J H xK H xK F yI cos G J H xK

u = xn sin

We have ∴ or

x

∂u ∂x

= nxn–1

∂u ∂x

= nxn

∂u = yxn–1 ∂y Adding (ii) and (iii), we get

Similarly,

y

x

∂u ∂u + y ∂y ∂x

= nxn sin

∂u ∂u + y ∂y = nu ∂x which verifies Euler’s theorem. ⇒

x

n

n–1

FG y IJ H xK

2

...(ii) ...(iii)

38


Example 3. If u = sin–1

w xy

z { , show by Euler’s theorem that y|

x− y x+

y ∂u ∂u = – · ∂x x ∂y

w xy

Sol. We have

u = sin–1

Let

f = sin u =

z { y|

x− y x+

⇒ sin u =

x− y x+ y

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


⇒

F x ∂u + y ∂u I GH ∂x ∂y JK

⇒

x

eu = 3eu

∂u ∂u +y = 3. Hence proved. ∂x ∂y

Example 5. Verify Euler’s theorem for u = sin–1

F xI GH y JK

+ tan–1

FG y IJ . H xK

(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

F xI 1− G J H yK

2

x 2

y −x

⋅

–

2

FG − y IJ FG IJ H x K H K

1 + y

1 y 1+ x

2

−x 2

y −x

2

xy

...(i)

x + y2

.

+

2

2

F− x I GH y2 JK F xI 1−G J H yK 1

2

ex

+

xy 2

+ y2

FG 1IJ FG IJ H xK H K

1 y 1+ x

2

.

...(ii)

j

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

F xI GH y JK

+ y sin–1

∂ 2u ∂ 2u 2 + 2xy ∂x∂y ∂x

Sol. We have

+ y2

∂u = sin–1 ∂x

∂ 2u . ∂y2

(U.P.T.U., 2007)

F xI GH y JK + y sin F xI + x GH y JK x 1−

u = x sin–1 ∴

FG y IJ , find the value of H xK

–1

2

y2

FG y IJ H xK F 1I GH y JK

+y·

1 1−

y

2

x2

FG − y IJ H xK 2

40


⇒

x

∂u = x sin–1 ∂x ∂u = x· ∂y

and

∂u y ∂y = –

⇒

1 1−

2

x y2

x2 y2 − x2

Adding (i) and (ii), we get x

F xI GH y JK

∂u ∂u + y ∂y = x sin–1 ∂x

F xI GH y JK

+

x2 y2 − x2

F− x I GH y JK 2

+ y sin

–1

+ y sin–1

y2

–

+ sin–1

FG y IJ H xK FG y IJ H xK

...(i)

x2 − y2

FG y IJ H xK

+

+ y

1 1−

2

y x2

FG 1 IJ H xK

y2 x2 − y2

=u

...(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


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 y2

or

∂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 H 2 K

FG 4u − 2uIJ . H 2 K

. cos

∂ 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 H xK

+ xn g

+ 2xy

F xI GH y JK , prove that

∂ 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


∂ 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)


LM MN

u = sin–1

Sol. We have

⇒

sin u =

LM MN

x + 2y + 3z x8 + y8 + z8

x + 2y + 3z x8 + y8 + z8

OP PQ

OP PQ , let f = sin u

∴ 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

or

or

x

cos u = – 3 sin u

∂u ∂u ∂u + y ∂y + z + 3 tan u = 0. ∂x ∂z

Example 10. If V = loge sin

x

Sol. We have

Hence proved.

R| π 2x 2 + y 2 + zx 12 U| j |V , find the value of |S e || 2ex 2 + xy + 2yz + z 2 j 13 || T W

∂V ∂V ∂V + y ∂y + z , when x = 0, y = 1, z = 2. ∂x ∂z V =

R| π 2x 2 + y 2 + zx 12 U| j |V | e log sin S || 2ex 2 + xy + 2yz + z 2 j 13 || T W R π 2x + y + zx U| j V | e sin S || 2 ex + xy + 2yz + z j || T W R| π 2x 2 + y 2 + zx 12 U| j |V |S e || 2 ex 2 + xy + 2yz + z 2 j 13 || T W R| π 2x 2 + y 2 + zx 12 U| j |V |S e || 2 ex 2 + xy + 2yz + z 2 j 13 || T W 2

∴

eV =

2

sin–1 (eV) =

or

Let

f = sin–1 (ev) =

Since f is a homogeneous function ∴ n = 1 –

2 1 = 3 3

2

1 2

1 2 3

44


By Euler’s theorem, we get ∂f ∂f ∂f x +y +z = ∂x ∂z ∂y ⇒ x

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 GH ∂x ∂y ∂z JK H

⇒

1 f 3

⇒

x

I J 1 − e 2V K

1 × eV

=

1 sin–1 eV 3

∂V ∂V ∂V +y +z = ∂y ∂x ∂z

ee j V

Now,

esin

and

−1

eV

j

x = 0 y =1 z =2

x = 0 y =1 z =2

1 1 − e 2V × sin–1 eV V 3 e

RS π UV = T 2 × 2W

= sin =

...(i)

1 1 , e2V = 2 2

π

4

Putting all these values in equation (i), we get

⇒

x

∂V ∂V ∂V +y +z = ∂x ∂y ∂z

x

∂V ∂V ∂V +y +z = ∂x ∂y ∂z

Example 11. If u = x3y2 sin–1 x

1− 1 2 1 π × × = 1 3 4 2 π

12

12 π × 12 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.

=

x5

FG yIJ H xK

FG y IJ H xK

2

sin–1

FG y IJ H xK

=

x5

F yI F yI F yI F GH JK F G J = G J H xK H xK x

2

sin −1

y x

∴ 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

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


FG y IJ H xK

Example 12. If u = x f1

x2

+ f2

FG y IJ , prove that H xK

2 ∂ 2u ∂ 2u 2 ∂ u = 0. + 2 xy + y ∂x∂y ∂x 2 ∂y 2

u1 =

Sol. Let

FG y IJ H xK

x f1

and u2 = x0f2

FG y IJ , then u = u H xK

1

+ u2

Since u1 is a homogeneous function of degree one. So by Corollary 1 on page 36, we get

x2

∂2u1

+ 2xy

∂2u1 ∂ 2 u1 + y2 ∂x ∂y ∂y2

...(i)

= 1 (1 – 1) = 0 ∂x2 and u2 is also a homogeneous function of degree 0

∂ 2u2 ∂ 2 u2 ∂ 2 u2 2 + 2xy +y ∴ ∂y 2 ∂x∂y ∂x 2 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

g

F1 + y I GH x JK 3

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

6/5

b

g

=

6 sin u 5

∂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


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


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 GH

5.

b x + yg .

1 4

1

+ y4

I Fx JK GH

1 5

1

I JK

+ y5 .

xy

3. cos–1

6. cos–1

F xI GH y JK F xI GH y JK

+ cot–1

+ cot–1

FG y IJ. H xK FG y IJ . H xK

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 Nx + y Q

∂u ∂u + y ∂y = 0. ∂x

12. If u = log [(x2 + y2)/(x + y)], prove that x 13. If u = sin–1 {(x2 + y2)/(x + y)}, show that x 14. If u = sin–1 {(x + y)/( x +

∂u ∂u + y ∂y = 1. ∂x

(U.P.T.U., 2008)

∂u ∂u + y ∂y = tan u. ∂x

y )}. Show that x

∂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


FG y IJ , show that x ∂u + y H xK ∂x LM x + y OP , prove that x 1 8 . If u = sin M MN x + y PPQ 17. If u = f

–1

1 4

1 4

1 6

1 6

2

2 0 . Verify Euler’s theorem for f =

2 1 . If u = y2

∂ 2u 1 ∂ 2u ∂ 2u 2 2 + 2xy + y 2 = 144 tan u [tan u – 11]. 2 ∂ ∂ x y ∂y ∂x

5 ∂u ∂u +y = tan u if u = sin–1 2 ∂x ∂y

1 9 . Prove that x

y2

∂u = 0. ∂y

e y x + x2 tan–1

z x+ y

+

3

3

y x + . z+x y+ z

F xI GH y JK , show that x

∂u ∂ 2u ∂u ∂ 2u 2 + y ∂y = 2u and x + 2xy + ∂x∂y ∂x ∂x2

∂ 2u = 2u. ∂y2

j sin u = x + y , prove that x 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 22. If

F x +y I GH x + y JK .

e

x+ y

2

1 3

1 3

2

2 tanu ∂ 2u ∂ 2u 2 ∂ u = + 2xy + y 2 2 12 ∂x∂y ∂x ∂y

2

–1

2

∂f ∂f log x − log y 1 + , show that x + y ∂y + 2f (x, y) = 0. 2 2 ∂x xy x x +y ∂u ∂u 2 5 . If u = sec–1 {(x3 + y3) / (x + y)}, show that x + y ∂y = 2 cot u. ∂x

2 4 . If f (x, y) =

1.7 Let

1

2

+

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


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


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

...(ii)

∂z ∂z = 2xy, ∂y = x2 and we have ∂x x2 + xy + y2 = 1 Differentiating w.r.t. x, we get 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 JK

= 2xy –

b

x 2 2x + y

bx + 2 y g

g.

Example 2. If f (x, y) = 0, φ (y, z) = 0, show that ∂f ∂φ dz ∂y ∂z dx =

Sol. We have

∂f ∂φ . · ∂x ∂y

f (x, y) = 0 φ (y, z) = 0

From (i)

df dx

=

∂f ∂f dy dy + =0⇒ ⋅ ∂x ∂y dx dx

From (ii)

dφ = dy

∂φ ∂φ dz dz ⋅ + =0⇒ dy ∂y ∂z dy

...(i) ...(ii)

FG ∂f IJ H ∂x K = − F ∂f I GH ∂y JK F ∂φ I GH ∂y JK = − FG ∂φ IJ H ∂z K

51


Multiplying these two results, we get

dz = dx ∂f ∂φ dz ∂y ∂z dx

or

Example 3. If u = u x2

F ∂φ I GH ∂y JK FG ∂φ IJ H ∂z K

×

∂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 ∂f IJ H ∂x K F ∂f I GH ∂y JK

FG – 1 IJ H xK 2

+

∂u ∂t

FG − 1 IJ H xK 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


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

FG ∂uIJ = F cos α ∂ H ∂ξ K GH ∂x

+ sin α

∂ ∂y

I F cos α ∂u JK GH ∂x

+ sin α

...(i) ...(ii) ∂u ∂y

2

∂ u ∂ 2u ∂ 2u 2 ...(iii) 2 + 2 sin α cos α ∂x∂y + sin α ∂y 2 ∂x ∂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 JK

=

...(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


∂u ∂z

and

=

∂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

Sol. We have ⇒ ∴ and we have

∂x ∂x ∂θ ∂ 2θ ∂ 2θ ; =r and find the value of + · r∂θ ∂r ∂x ∂y2 ∂x2 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

FG IJ H K

x =

r cos θ ⇒

∂x = cos θ ∂r

...(ii)

From equations (i) and (ii), we get ∂r ∂x = . Hence proved. ∂x ∂r 1 ∂x ∂x Now, = – r sin θ ⇒ = – sin θ 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 ∂x r From equations (iii) and (iv), we obtain 1 ∂x ∂θ = r . Hence proved. ∂x r ∂θ

FG H

Since

and

∂θ ∂x

= –

∂θ ∂y

=

y 2

x +y 1

y2 x2 Adding equations (v) and (vi), we get

∂ 2θ ∂ 2θ + ∂y2 ∂x2

1+

= 0.

2

...(iii)

IJ K

⇒

∂ 2θ

y × 2x 2 = ∂x x2 + y2

FG 1IJ = H x K ex

e

x 2

+ y2

j

∴

j

2

⇒

∂ 2θ ∂x

2

...(iv)

=

2 xy

ex

2xy ∂ 2θ 2 = – 2 ∂y x + y2

e

2

j

+ y2 2

j2

...(v)

...(vi)

54


Example 7. If φ (x, y, z) = 0, show that

FG ∂y IJ FG ∂z IJ FG ∂x IJ H ∂z K H ∂xK y H ∂y K x

= – 1.

(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

y

FG ∂z IJ H ∂xK

y

⇒

= 0

= –

FG ∂φ IJ H ∂x K FG ∂φ IJ H ∂z K

...(i)

Next, keeping z as constant, differentiate partially w.r.t. y, we obtain

∂φ ∂x

⇒

Similarly,

F ∂x I GH ∂y JK

+ z

∂φ ∂y

F ∂x I GH ∂y JK

z

FG ∂yIJ H ∂z K

x

= 0

=

=

F ∂φ I GH ∂y JK − FG ∂φ IJ H ∂x K FG ∂φ IJ H ∂z K – F ∂φ I GH ∂y JK

...(ii)

...(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

Sol. We have

x+y x–y Adding 2x x and subtracting, we get y Let ∴

∂ 2V ∂ 2V ∂ 2V · 2 = 4xy 2 + ∂x ∂y ∂φ ∂θ = 2eθ cos φ = 2ieθ sin φ = 2 (eθ) (cos φ + i sin φ) = eθ + i φ = eθ – i φ

(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


∴

∂ ∂ 2V 2 = ∂θ ∂θ

⇒

∂ 2V = x2 ∂θ 2

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 +y + 2xy + G x ∂x + y ∂y J H K ∂y ∂x ∂y ∂x 2

2

2

2

2

...(iii)

2

Now

∂V ∂y ∂V ∂x ∂V ∂y ∂V ∂V ∂x = ix = = –iy = · + = (ix) + (–iy) As ∂φ ∂φ ∂x ∂y ∂φ ∂x ∂y ∂φ ∂φ

⇒

∂V = i ∂φ ∂ 2V ∂φ 2

Next

= =

F x ∂V − y ∂V I GH ∂x ∂y JK F ∂ ∂ I F ∂V ∂V I ∂ 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 F ∂ 2V ∂ 2V ∂ 2V + x ∂V + y ∂VI − G x2 2 + y2 2 − 2xy ∂x∂y ∂x ∂y JK H ∂x ∂y

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) rn – 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


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

j

= nrn–1 sin (θ – nθ)

∂y Now,

e

∂ 2 r n cos nθ

j

∂x∂y

...(v)

LM ∂ er n cos nθ j OP MN ∂y PQ ∂ nr sin a1 − nfθ , from (v) = n ∂x

∂ = ∂x =

n– 1

LM ∂ {r sina1 − nfθ} − sin θ ∂ {r sina1 − nfθ}OP r ∂θ N ∂r Q sin θ n LMcos θ an − 1f r sin a1 − nf θ – r a1 − nf cos a1 − nf θOP r N Q

= n cos θ =

a f

∂ n– 1 r sin 1 − n θ ∂x

n −1

n −1

n− 2

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 2

2

= 0 into polars and show that u = (Arn + Br–n) sin nθ satisfies

the above equation. Sol. We know x = r cos θ, y = r sin θ ∴

r2

=

x2

+

θ = tan–1

and From (ii), we get 2r or Similarly,

...(i)

y2

...(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),

∂θ = ∂x

...(iv) ...(v)

FG − y IJ FG IJ H x K H K

1 y 1+ x

2

·

2

=–

y 2

x + y2

57


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

= cos θ

or

LMcos θ ∂ 2u − sin θ ⋅ ∂ FG 1 ⋅ ∂u IJ OP ∂r H r ∂θ K PQ MN ∂r 2

LM MN

RS T

LMF – sin θ ∂u + cos θ ⋅ ∂ 2u I − 1 ∂ F sin θ ⋅ ∂u I OP G ∂θ JK P ∂θ ∂r JK r ∂θ H ∂r MNGH Q

∂ 2u 1 ∂ 2u 1 ∂u ∂ 2u cos sin θ ⋅ − θ ⋅ − = cos θ r ∂r ∂θ r 2 ∂θ ∂r 2 ∂x2 –

or

sin θ – r

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


If

u =

∂u ∂r and

∴

1 ∂ 2u 2 + r2 ∂r

(Arn

+

Br–n)

Or sin nθ, then

= n (Arn–1 – Br–n – 1) sin nθ;

∂u = (Arn + Br–n) n cos nθ ∂θ

∂ 2u ∂r 2

= n [A(n – 1)rn – 2 + B (n + 1) r–n –2] sin nθ;

∂ 2u ∂θ 2

=

FG ∂uIJ H ∂θ K

∂ ∂θ

= – (Arn + Br – n) n2 sin nθ

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

R|F ∂r I F ∂r I U| S|GH ∂x JK + GH ∂y JK V| · T W

∂ 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

2 y ∂ 2r x ∂r r 2 − x2 ∂ r r2 − y2 ; = ; = ; = , ∂y2 r ∂y r ∂x2 r3 r3

er

2

− x2 r

3

2

∂ 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

j + er

2

− y2 r

3

j

=

r

j

3

2

=

2r − r , ä x2 + y2 = r2 r3

=

1 r

=

1 r

2

e

2r 2 − x 2 + y 2

...(i)

R|FG x IJ + F y I U| 1 F x2 + y2 I S|H r K GH r JK V| = r G r2 J K T W H 1 Fr I G J,äx +y =r r Hr K 2

2

2

= =

2

2

2

2

∂ 2r 1 ∂ 2r = + , from (i) Hence proved. ∂y2 r ∂x2

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


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)

we have u = x log xy

1 ∂u = log xy + ⋅ y = log xy + 1 y ∂x

af

∂u x x x = = xy y ∂y

...(ii) ...(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


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 ∂2 f ∂x 2

Sol. We have

+

∂2 f ∂y 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


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

FG H

FG IJ IJ H K K F −2y ∂ φ + 2x ∂ φ I b−2yg + F −2y ∂ φ + 2x ∂ φ I a2xf GH ∂u ∂u∂v JK GH ∂u∂v ∂v JK ∂ ∂φ ∂φ ∂u ∂ ∂φ ∂φ ∂v + 2x −2 y + 2x + −2 y ∂u ∂u ∂v ∂y ∂v ∂u ∂ v ∂y 2

= ∂2 f

or

∂y

2

IJ K

2

2

2

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

e

j ∂∂uφ

e

j FGH ∂∂uφ + ∂∂vφ IJK .

= 4 x2 + y 2

= 4 x2 + y2

e

+ 4 x2 + y2

2

2

2

j ∂∂vφ 2

2

2

2

Example 16. If x2 + y2 + z2 – 2xyz = 1, show that

Hence Proved. 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

2

2

u h x2 + y2 + z2 – 2xyz – 1 = 0

...(ii) ...(iii)

62


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 2

2

Dividing by dx 1−x

2

+

2

2

e1 − x j e1 − y j 2

dy +

2

dz = 0

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 dy 1 . Find if xy + yx = c. dx

2 . If u = x log xy, where x3 + y3 + 3xy = 1, find

3 . If u = x2 y, where x2 + xy + y2 = 1, find

LM MN

LMA ns. MN

du A ns. . dx

du . dx

OPOP PQPQ du x L ( x + y) O O = b1 + log xyg + M− PP y MN ( y + x) QP PQ dx LMA ns. du = 2xy − x L b2x + yg OOP MM bx + 2yg PPP MN dx N QQ LM MN

dy yx y −1 + y x log y =− y dx x log x + xy x -1 2

2

2

4 . If V is a function of u, v where u = x – y and v = x – y, prove that x

∂ 2V ∂ 2V + y = (x + y) ∂x 2 ∂y 2

F∂ V GH ∂u 2

2

5 . Transform the Laplacian equation

+ xy

∂ 2V ∂v

2

I. JK

∂ 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 PQ MN dx −2r

2

2

2

2

2

2

63


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. N

dy y tan x + log sin y = dx log cos x − x cot y

OP 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. N

∂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 If z = z(u, v), u = x – 2xy – y and v = y, show that (x + y) + bx − yg = 0 is equivalent ∂x ∂y x

15.

OP Q

∂u ∂u = 8t – 4 = 8, ∂s ∂t

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


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


(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 H xK

and c = xlim → ∞ (y – mx).

(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


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

m

P

v> FQ J>NG

Q

FP J>R G>

Fig. 1.3

FG d y IJ H dx K 2

2

= 6x – 6a,

>

= – 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 –∞ 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


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 J Q 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 2

3m

=

I>

Fig. 1.5

φ3(m) = 1 + m3 = 0 ⇒

´>

a m

m = –1, c = – a. Asymptotes y = mx + c ⇒ y = –x–a or y + x + a = 0.

>

[>

N

v

68


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

I K

= – 1 i.e., the tangent at

F 3a , 3a I H 2 2K

making

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 5. Region: G J Ha K 2

2

∴ if

1 3

Fx I =1– G J Ha K

1 2 3

a F J>NG

2

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


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


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

Fig. 1.10

x2y2 – b2x2 – a2y2 = 0 or x2y2 = b2x2 + a2y2 1. Symmetry about both the axes.

w

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>[>

or

´>[>¡

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


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


3.

F x I + FG y IJ H aK H bK 2 3

2 3

4 . x6 + y6 = a2x2y2

=1 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


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


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


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


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 a+

F bI H 2K

π 2 a

2π 3

Fa − bI H 2K

b then r is positive for all values of θ from 0 to 2 3 2π = π or π. Here, r must vanish somewhere between θ = and 4 3 2π 3π and ). 3 4 For which r = 0 then θ = α is the straight line which is 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 negative for them. 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.

a b 3π 4

FG a − b IJ H 2K

π (a – b)

2π but r is negative when θ 3 3π . Let θ = α (lying between 4 q = 2 p/3 θ= α θ= π

q = p/2

q = p/3

O A

when θ = α = cos–1

Fig. 1.19

77


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


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


3.

dx dy = a (1 – cos t), = a sin t dt dt dy ⇒ = dx dy Corresponding values of x, y, dx π t = 0 2 π x = 0 a −1 2 y = 0 a

F H

dy dy/dt a sin t = = dx/dt dx a (1 − cos t) t 2a sin t/2.cos t/2 = cot 2 2 a(1 − 1 + 2 sin t/2) ∴

for different values of t are given below: π

I K

aπ 2a

3π 2 3π a +1 2 a

F H

2π

I K

2aπ 0

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

5π 4 a 2 2 −a 2 2

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

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


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¢

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

t)

Fig. 1.22

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


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 N

f(a + h, b + k) = f(a, b) + h +

+

LM MN 1 L Mh 3 MN

∂f ( a, b) ∂f ( a, b) +k ∂x ∂y

OP 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

OP PQ

OP PQ

3 3 3 ∂ 3 f ( a, b) 2 ∂ f ( a, b) 2 ∂ f ( a, b) 3 ∂ f ( a, b) +... 3 3 h k hk k + + + ∂y 3 ∂x 3 ∂x 2 ∂y ∂x∂y 2

LM N

f(a + h, b + k) = f(a, b) + h

Or

OP Q

LM N

∂ ∂ ∂ ∂ 1 +k f ( a, b ) + h +k ∂y ∂x ∂y 2 ∂x

LM N

OP Q

2

f(a, b)

∂ ∂ 1 + 3 h ∂x + k ∂y

OP Q

3

f ( a, b)+...

82


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, 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

f(x + h, y + k)

2

2

2

2

2

2

2

For any point (a, b) putting x = a, y = b in above equation then, we get

LM N

f (a + h, b + k) = f (a, b) + h

+

LM MN

∂f ( a, b) ∂f ( a, b) +k ∂x ∂y

OP Q

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) + Q

1 2

LM(x − a) ∂ + (y − b) ∂ OP ∂y Q N ∂x

* 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

2

f ( a, b) + ... ...(ii)

83


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 N

f (x, y) = f (0, 0) + x

OP Q

Example 1. Expand ex cos y about the point Sol. We have and

f (x, y) = ex cos y π π , f 1, a = 1, b = 4 4 ∂f = ex cos y ⇒ ∂x

∴ From (1)

F 1, π I H 4K

F I H K F πI ∂f 1, H 4K ∂x

F πI H 4K

= ex cos y ⇒

∂ 2 f 1,

F H

I K

e

= –

∂x∂y By Taylor's theorem, we have π f 1 + h, + k 4

F πI H 4K

=

2

,

∂2 f ∂y

2

∂x 2

e 2

= – ex cos y = –

2

,

e 2

π =– 4

e 2

∂2 f = – ex sin y ∂x∂y e 2

.

F ∂f F1, π I ∂f F1, π I I F π I + GG h H 4 K + k H 4 K JJ f 1, H 4 K H ∂x ∂y K

+ ...

...(ii)

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 =

=

f (0,0) + ...

e

π = 4

= – e sin

∂ 2 f 1,

2

...(i)

= e cos

F I H K

OP Q

(U.P.T.U., 2007)

π = e cos = 4

π ∂f 1, ∂f 4 x ∂y = – e sin y ⇒ ∂y

∂2 f ∂x 2

LM N

1 ∂ ∂ ∂ ∂ +y +y f (0,0) + x 2 ∂x ∂y ∂x ∂y

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 . 1 + (x − 1) − y − M H 4K 2 H 4 K H 4K Q 2N 2

+ 2 (x – 1)

⇒ f (x, y) =

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


∂2 f ey = ∂x∂y 1+ x ∂2 B ∂N 2 ∂2 f ∂y

2

ey (1 + x ) 2

⇒

= ey log (1 + x)

⇒

= −

∂2 f (0,0) ∂x∂y

⇒

Now, applying Taylor's theorem, we get f (0 + h, 0 + k) = f (h, k) = f (0, 0) +

∂ 2 f (0, 0) + k2 ∂x∂y

+ 2hk

∂2 f (0,0)

= – 1

∂x2 ∂2 f (0,0)

= 0

∂y2

FG h ∂f (0,0) + k ∂f (0,0) IJ + 1 FG h ∂y K 2H H ∂x ∂ f (0, 0) I JK + ... ∂y

2

∂ 2 f (0,0) ∂x 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) + ⇒

= 1

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 point x0 = 0, y0 = 0 up to three terms.

2

−y2

. cos xy about the (U.P.T.U., 2006)

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 f (0, 0) ∂ 2 f ( 0 ,0 ) ∂ 2 f (0, 0) − , 2 Similarly, = 0, = =–2 ∂x∂y ∂x 2 ∂y 2 2

∂3 f (0,0) ∂x ∂y 2

∂ f ( 0, 0 )

= 0,

2

∂ 3 f (0, 0) ∂x∂y

2

= 0,

∂ 3 f ( 0, 0 ) =0 ∂x 3

3

∂y 3

= 0

Applying Taylor's theorem f (h, k) = f (0, 0) +

FG h ∂ + k ∂ IJ f (0, 0) + 1 FG h ∂ + k ∂ IJ ∂y K 2 H ∂x H ∂x ∂y K ∂I 1 F ∂ h + k J f (0, 0) + ... G ∂y K 3 H ∂x 3

+

2

f (0, 0)

85


Putting h = x, k = y and all values then, we get f (x, y) =

e −x

2

− y2

· cos xy = 1 + (x × 0 + y × 0) 1 2 x ( −2) + 2 xy × 0 + y 2 ( −2) + 2 +

⇒

A− N

2

− O2

1 3 x × 0 + 3x 2 y × 0 + 3xy 2 × 0 + y 3 × 0 3

. 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 f (– 0.5, 2) = cot–1(–1) = Now

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 ∂x 2 ∂2 f ∂y

2

=

=

2 xy 3

∂ 2 f ( −0.5, 2) = 0 ∂x∂y ∂ 2 f ( −0.5, 2)

⇒

(1 + x 2 y 2 )2 2x 3 y

∂ 2 f ( −0.5, 2)

⇒

(1 + x y )

2 2 2

∂y

FG H

Let ∴

or

IJ K

1 8

FG H

∂ ∂ ∂ ∂ 1 ⋅ f ( −0.5, 2) + +k h +k ∂x ∂y ∂x ∂y 2

IJ K

2

f (–0.5, 2) +…

– 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

F H

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

= –

2

Now applying Taylor's series expansion, we get f (– 0.5 + h, 2 + k) = f (–0.5, 2) + h

= – 2

∂x 2

3π 0.2 1 – (0.1) + – (0.1)2 – (0.2)2 4 4 16 = 2.29369.

f (– 0.4, 2.2) =

86


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

Sol. Let f (x, y) = log

1 3

IK

1

+ y4 −1

f (1, 1) = log 1 = 0

∂f = ∂x

Now,

3

FH

1× x 1 x3

+

−

2 3

1 y4

IK

∂f (1,1) = ∂x

⇒

−1

1 3 Taking a = 1, b = 1

3 − 1× y 4

∂f ∂y =

F 4 Hx

1 3

+

1 y4

I − 1K

∂f (1,1) = ∂x

⇒

1 4

Now, applying Taylor's theorem

FG h ∂ + k ∂ IJ f (1,1) + ... H ∂x ∂y K L O log M(1 + h) + (1 + k ) − 1P N Q L O log M(1 + h) + (1 + k ) − 1P = 0 + h × N Q

f (1 + h, 1 + k) = f (1, 1) + But

f (1 + h, 1 + k) =

∴

f (1 + h, 1 + k) =

1 3

1 4

1 3

1 4

1 1 +k× 3 4

...(i)

Putting h = 0.03 and k = – 0.02 in equation (i) then, we get

LM N

1

OP Q

1

log (1.03 ) 3 + ( 0.98 ) 4 − 1

= 0.03 ×

1 1 + (– 0. 02) × 3 4

= 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) y Sol. Here f (x, y) = x f (1, 1) = 1 Now

∂f = yxy–1 ∂x

⇒

∂f (1, 1) = 1 ∂x

∂f y ∂y = x log x

⇒

∂f (1, 1) = 0 ∂y

⇒

∂ 2 f (1, 1) = 1 ∂x∂y

∂2 f = xy–1 + yxy–1 · log x ∂x∂y ∂2 f ∂x 2 ∂2 f ∂y 2

= y (y – 1) xy–2

⇒

= xy · (log x)2

⇒

∂2 f (1, 1) ∂x2 ∂ 2 f (1, 1) ∂y 2

= 0 = 0

87


∂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 H ∂x ∂y K ∂y K 2 H ∂x 1 F ∂ G h + k ∂∂y IJK · f (1, 1) + ... 3 H ∂x

f (1 + h, l + k) = f (1, 1) +

2

f (1, 1)

3

+ Let ⇒ ∴

1 + h = x and 1 + k = y h = x – 1 and k = y – 1 f (x, y)

∂ ∂O ∂ ∂O 1 L L = f (1, 1) + M( x − 1) ∂x + ( y − 1) ∂y P f + 2 M( x − 1) ∂x + ( y − 1) ∂y P N Q N Q ∂ ∂O 1 L + 3 M( x − 1) ∂x + ( y − 1) ∂y P f + ... N Q

2

f

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


∂f x ∂y = ( x 2 + y 2 ) ∂2 f ∂x 2 ∂2 f ∂y

2

= =

2 xy (x + y ) 2

2 2

−2 xy 2 2

2

∂ B y2 − x2 = ∂N∂O ( x 2 + y 2 )2 By Taylor’s theorem

∴

=

1 2

∂2 f ∂y 2

(1, 1) =

−

1 2

∂2 f = 0 ∂x∂y

LMh ∂ + k ∂ OP f (1, 1) + 1 LMh ∂ + k ∂ OP ∂y Q 2 N ∂x N ∂x ∂y Q

2

f (1,1) + ...

LM OP f + ... N Q 1 L F 1I 1 1 + (x – 1) F − I + (y – 1) F I + H 2K H 2 K 2 MN(x − 1) H 2 K + 2(x − 1)( y − 1) F 1IO × 0 + ( y − 1) − P H 2KQ LM N

f (x, y) = f (1, 1) + ( x − 1)

π 4

b g

∂2 f 1, 1 ∂x 2

l+h = x⇒ h= x–1 1+k = y⇒k =y–1

=

1 2

⇒

⇒

f (1 + h, 1 + k) = f (1, 1) + Let

=

∂f 1, 1 ∂y

⇒

(x + y ) 2

b g

⇒

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

⇒

∂ B (1, 0) = 0 ∂O

⇒

∂2 B 1 1, 0 = – 2 4.2 3/2 ∂N

∂2 f ∂x 2 ∂2 f ∂y

2

= – =

y 1+ x+ y

2

1 4 (1 + x + y 2 ) 3/2 1 1 + x + y2

–

y2 (1 + x + y )

2 3/2

⇒

b g

∂ 2 f (1, 0) ∂y

2

=

1 2

89


y ∂2 f = – 2(1 + x + y 2 ) 3/2 ∂x∂y By Taylor's theorem, we get

LM N

f (1 + h, 0 + k) = f (1, 0) + h Let ⇒ f (x, y) =

∂ 2 f (1, 0) = 0. ∂x∂y

⇒

∂ ∂ +k ∂x ∂y

OP f (1, 0) + Q

LM N

∂ ∂ 1 h +k ∂y 2 ∂x

OP Q

2

f (1, 0) + ...

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. N

f ( x , y) = 1 −

F H

I F K H

1 π2 π π π − (x − 1) 2 − (x − 1) y − y− 8 2 2 2 2

I OP + ... KQ 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) 2 T N

2

+ ( x − 1)( y − 1) +

UVOP WQ

( y − 1)2 + ... 2

6 . Expand cosx cosy in powers of x and y.

LMAns. N

f ( x, y) = 1 −

7 . Expand f (x, y) = e2x cos 3y up to second degree.

OP Q

1 2 1 4 (x + y 2 ) + ( x + 6 x 2 y 2 + y 4 ) + ... 2 24

[Ans. 1 + 2x + 2x2 –

9 2 y + ...] 2

8 . Obtain Taylor's series expansion of (x + h) (y + k) /(x + h + y + k)

LMAns. xy + hy MN x + y (x + y) 2

2

+

OP PQ

h2 y 2 2hkxy kx 2 k2x2 − + − + ... 2 3 3 (x + y) ( x + y) (x + y) ( x + y) 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 2 2 3 N 2

1 0 . Find Machaurin's expansion of ex log (1 + y).

2

2

3

OP Q

+ ...

90


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)

∂ u ∂ u +y is 2 ∂ y∂x ∂x 2

2

∂z ∂x

(ii) nz (iv) n

(iii) n(n – 1)z

∂z ∂y

∂x ∂y ∂z is ⋅ ⋅ ∂y ∂z ∂x

(i) –1 (iii) 0 7 . If u = x3 y2 sin–1 (y/x), then x

(iii) – 8u

x −y

(iv) 8xyzu

5 . If z = xn–1 y f(y/x) then x

(i) 0

x 2

(ii) 6xyzu

(iii) – 8xyzu

6 . If f(x, y, z) = 0 then

x x2 + y2

∂ 3u is ∂x∂y∂z

(i) 7xy

(i) n

∂u ∂y

∂z ∂z and are ∂y ∂x

y 2

∂u ∂y

(ii) 1 (iv) 2

∂u ∂u +y is: ∂x ∂y (ii) 6u (iv) 5u

2

,− ,

y x2 + y2 y

x − y2 2

91


8 . If u = xy f(y/x), then x (i) yu (iii) u

∂u ∂u +y is ∂x ∂y (ii) xu (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

(i) (iii)

FG 3a, 3a IJ H 2K FG 3a , 3a IJ H 2 2K

(ii) (iv)

FG 3a , − 3a IJ H 2 2K FG − 3a , − 3a IJ H 2 2K

1 3 . The equation of the curve x2y – y – x = 0 has maximum (i) One asymptote (iii) Three asymptotes

(ii) Four 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 (iii) both axis

(ii) y-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


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. 9. 10. 11.

The nth derivative (yn) of y = x2 sin x at x = 0 is .......... If y = tan–1 x and n is even, then yn (0) = .......... If y = cos(m sin–1 x) and n is odd then yn (0) = .......... 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 =

1 9 . If f(x, y) = 0, φ(y, z) = 0, then

∂x 2

+ 2xy

∂2 f ∂2 f + y 2 2 = .......... ∂x∂y ∂y

∂u ∂u ∂u +y +z = .......... ∂x ∂y ∂z ∂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

∂u ∂u x 2 y2 , then x = .......... +y ∂y ∂x x+y

1 8 . If u = (x2 + y2 + z2)1/2, then x

2 1 . If u =

(U.P.T.U., 2008)

x4 − y4 x +y 3

3

2

∂θ ∂θ = .......... and = .......... ∂y ∂x

, then x

, then 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


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 b g

∂ 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

=

a f b g b g a f

∂ x, y ∂ u, v × ∂ r, s ∂ r, s

∂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


(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

IJ K

FG H

en − n2 j sin n2π

1 1 . Zero

∂ 2u ∂y∂x 1 6 . n(n – 1)f 13.

∂y 2 1 5 . 2x + 4y

IJ K

8.

1 0 . Zero

∂ 2z

(iii) (i) (iii) (iii) (iii)

nπ nπ − n cos x + 2 2 5 . sin hx

7 . Try yourself

9 . Zero 12.

FG H

x sin x +

4 . Zero

f

3. 6. 9. 12. 15.

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 −

LMR ∂f O SMTaz − 1f ∂x + by ⋅ 2g ∂∂yf UVW + 12 RSTax − 1f ∂∂xf + by ⋅ 2g ∂∂yf UVW +......PP N Q 2

33.

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


2.1.2 Properties of Jacobians 1. If u =u (x, y) and v = v (x, y), then

a f ∂ b x , yg ∂ u, v

×

b g ∂ a u , vf ∂ x, y

where,

=

J =

1 or JJ′′ = 1

u = u (x, v = v (x, Differentiating partially equations (i) ∂u ∂u =1 = ∂u ∂x

a f b g

b g a f

a f b g

∂ x, y ∂ u, v and J′ = ∂ u, v ∂ x, y

Proof: Since

Now,

(U.P.T.U., 2005)

y) y) and (ii) w.r.t. u and v, we get ∂y ∂x ∂u · + × ∂u ∂y ∂u

∂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

=

∂u ∂x ∂v ∂x (By

∂u ∂x ∂y ∂ u ∂v × ∂x ∂y ∂v interchanging

=

∂u ∂x ∂u ∂y + ∂x ∂u ∂y ∂u ∂v ∂x ∂v ∂y + . ∂x ∂u ∂y ∂u

=

×

∂x ∂u ∂y ∂u

...(i) ...(ii)

...(iii) ...(iv) ...(v) ...(vi)

∂x ∂v ∂y ∂v

∂y ∂u ∂y ∂v rows and columns in II determinant)

∂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

∂ x, y ∂ u, v × ∂ x, y ∂ u, v

=

JJ′ =

or

1 0 1.

0 =1 1 Hence proved.

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 = · + ∂x ∂r ∂x ∂u ∂u ∂r = · + ∂y ∂r ∂y

∂v ∂x ∂v ∂y

Now,

a f a f

∂ u, v ∂ r, s

a f b g

∂ r, s · ∂ x, y

= =

∂u · ∂s ∂u · ∂s

∂s ∂x ∂s ∂y

...(i) ...(ii)

∂v ∂s ∂v ∂r · + · ∂s ∂x ∂r ∂x ∂v ∂r ∂v ∂s · + · r y ∂ ∂ ∂s ∂y

=

∂u ∂r ∂v ∂r

∂u ∂s ∂v ∂s

=

∂u ∂r ∂v ∂r

∂u ∂s ∂v ∂s

...(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 ∂v ∂r

∂r ∂u ∂s + ∂x ∂s ∂x ∂r ∂v ∂s + ∂x ∂s ∂x

∂u ∂r ∂v ∂r

∂r ∂u ∂s + ∂y ∂s ∂y ∂r ∂v ∂s + ∂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 , when u = 3x + 5y, v = 4x – 3y. ∂ bx, yg ∂ u, v ∂ x, y

Example 1. Find 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 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.

b g ∂ au, v, wf ∂ x, y, z

Example 3. Calculate Sol. Given

u =

if u =

3zx 2yz 4xy , v = , w = . y x z

3zx 2yz 4xy ,v= ,w= y x 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

= –

LM MN

∂u ∂y ∂v ∂y ∂w ∂y

∂u ∂z ∂v ∂z ∂w ∂z

2 2 yz 12x yz

x

2

2 2

y z

−

–

=

12x2 yz

2 yz 2

x 3z y 4y z

OP – PQ

2z x −3 zx y2 4x z

LM MN

∂ u, v, w = 0 + 48 + 48 = 96. ∂ x, y, z

2y x 3x y –4 xy z2

OP PQ

LM MN

2y 12xz 12xyz 2z −12xyz − 12xy + + 2 yz x yz x yz zy 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

OP PQ

99


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

= = = J(x, y, z) · J(u, v, w) =

∴

2 (x – y)[z (z – x) – y (z – x)] 2 (x – y) (z – y) (z – x) – 2 (x – y) (y – z) (z – x) 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 g

∂ x, y, z ∂ r, θ, φ

Consider


=

=

b g · ∂ au, v, wf ∂ b r , θ, φ g ∂ au, v, wf ∂ x, y, z ∂x ∂u ∂y ∂u ∂z ∂u

∂x ∂v ∂y ∂v ∂z ∂v

=

=

1 2

w u

1 2

v u

1 8

LM MN

∂x ∂w ∂y ∂w ∂z ∂w 1 2

0

...(i)

w v 0

1 2

u v

1 2

v w

1 2

u w 0

wv u v wu + v uw w u v

OP PQ

=

1 8

1+ 1 = 2 = 1

8

4

100


b a

Next

g f

∂ x, y, z

⇒

=

∂ u, v, w

a b

f g

∂ u, v, w ∂ r , θ, φ

...(ii)

∂u ∂r ∂v ∂r ∂w ∂r

=

∂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

⇒

b g ∂ br, θ, φg

∂ x, y, z

...(iii)

1 r 2 sin θ . × r2 sin θ = 4 4

=

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

= =

⇒

∂u ∂x

=

FG −1IJ (– 2r) e1 − r j . ∂r e j H 2K ∂x e1 − r j + rx e1 − r j . xr = 1 1− r + 1 − r2 2

−1 2

−3 2

2

+x

−1 2

2

−3 2

2

x2 3 2 2

e1 − r j

1 − r2 + x2

e

1 − r2

j

3 2

Differentiating partially u w.r.t. y, we get

⇒

and

∂u ∂y

= x

∂u ∂y

=

∂u ∂z

=

FG −1IJ e1 − r j H 2K 2

xy 3 2 2

e1 − r j xz

3 2 2

e1 − r j

−3 2

· (– 2r)

∂r = ∂y

xr 3 1 − r2 2

e

j

·

y r

101


∂v ∂x

Similarly,

∂w ∂x

Thus,

a b

f g

∂ u, v, w ∂ x, y, z

=

=

yz 1 − r 2 + y 2 ∂v ∂v , = 3 3 , ∂z = 3 ∂y 1 − r2 2 1 − r2 2 1 − r2 2 yx

e

e

j

zx 3 2 2

e1 − r j

e

3 1 − r2 2

e

e

j

1−r + y

j

2

e

3 2 2

1 9 1 − r2 2

e j 9 2 2 1 − r e j −

j

3 2

j

3 2

yz

j

e

1 − r2

1 − r2 + z2

zy

e1 − r j

=

e

1 − r2

3 1 − r2 2

zx

j

xz

2

3 1 − r2 2

e

j

xy

yx

=

1 − r + z2 ∂w , = 3 . 3 ∂z 2 2 2 2 1−r 1− r

3 1 − r2 2

j

j

2

zy

1 − r 2 + x2

e =

∂w = ∂y

,

e

j

3 2 2

3 2 2

e1 − r j

e1 − r j

1 − r2 + x2

xy

yx

2

xz

1−r + y

zx

2

yz 1 − r 2 + z2

zy

[(1 – r2 + x2) {(1 – r2 + y2)(1 – r2 + z2)– y2z2} – 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]

= = =

e1 − r j e1 − r j e1 − r j 2

−9 2

−9 2

2

2

−9 2

[(1 – r2)3 + (1 – r2)2 (x2 + y2 + z2)] [(1 – r2)3 + (1 – r2)2 r2]

e

. (1 – r2)2 [1 – r2 + r2] = 1 − r 2

j

−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

J =

⇒

b g ∂ bu, v, wg ∂ x, y, z

∂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 = tan v

0

0

0 2

u sec v 0

0 = u sec2 v 1

...(i)

102


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

J′ =

b g ∂ bx, y, zg

∂ u, v, w

1 =

−

0 x

y

2

x +y 0

2

2

x +y 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,

b g ∂ au, vf

∂ f1 , f2

a f b g

∂ u, v × ∂ x, y

=

=

∂ f1 ∂u

∂ f1 ∂v

∂ f2 ∂u

∂ 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


Using (iii), (iv), (v) and (vi) in above, we get ∂f ∂f − 1 − 1 ∂x ∂y = ∂ f2 ∂ f2 − − ∂x ∂y Thus,

b g ∂ au, vf

∂ f1 , f2

×

a f b g

∂ u, v ∂ x, y

a f b g

∂ u, v ∂ x, y

⇒

= (– 1)2

= (– 1)2

b g ∂ bx, yg ∂bf , f g ∂ bx, yg ∂bf , f g ∂ au, vf

a b

f g

∂ f1 ∂y

∂ f2 ∂x

∂ f2 ∂y

∂ f1 , f2

Similarly for three variables u, v, w

∂ u, v, w ∂ x, y, z

= (– 1)2

∂ f1 ∂x

b

1

2

1

2

∂ f1 , f 2 , f 3

b

∂ x , y, z

b

= (– 1)3

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

g

∂ u, v

Thus,

a f b g

∂ u, v ∂ x, y

=

∂ f1 ∂u

∂ f1 ∂v

∂ f2 ∂u

∂ f2 ∂v

b g b g ∂ b f1 , f2 g ∂ au, vf

3u2 = 2u

∂ f1 , f2

=

∂ x, y

=

(U.P.T.U., 2006)

–1 –3 y

3v 2 2v

2

= 3 (y2 – x2)

= 6 uv (u – v)

e

j = (y2 − x2 ) . 6uv au – vf 2uv (u − v)

3 y2 − x2

Example 9. If u, v, w are the roots of the equation in k,

that


=–

au − vfav − wfaw − uf . aa − bfab − cfac − af

x a+ k

Hence Proved.

+

y z + = 1, prove c+ k b+ k

104


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,

∴

a b

f g

b

∂ f1 , f2 , f3

b

g

∂ x, y , z ∂ u, v, w = (–1)3 ∂ f1 , f2 , f3 ∂ x, y, z ∂ u, v , w

b g =– ∂ au, v, wf ∂ x, y, z

b

g

aa − bf aa − cfab − cf au − vfau − wf av − wf

g a f au − vfav − wfau − wf . aa − bfab − cfaa − cf =–

Hence proved.

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


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 = − 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

= r2

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

∂ f1 , f2 , f3

b

∂ x, y , z

b

g

∂ f1 , f2 , f3

a

f

∂ u, v, w

g

g

b

g

2 x− y

–1 –3x2 –2x

=

0 3 x2 − z2

j e

a f

–1 –3y2 –2y

j

–1 –3z2 –2z

C2 → C2 − C1 , C3 → C3 − C1

2 x− z

2

2

= – 6[(x – y ) (x – z) – (x – z2) (x – y)] = – 6 (x – y) (x – z) [(x + y) – (x + z)] = 6 (x – y) (y – z) (z – x)

=

∂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


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

g

= – 6 (u – v) (v – w) (w – u)

∂ u, v, w

Hence

a b

b

∂ f1 , f2 , f3

f g

∂ u, v, w ∂ x, y, z

= (–1)3

b

∂ x, y, z

b

g

∂ f1 , f2 , f3

a

∂ u, v, w

=

f

g g

=+

b gb ga f a fa fa f

6 x− y y− z z− x 6 u− v v − w w− u

bx − ygby − zgaz − xf · Hence proved. au − vfav − wfaw − uf

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

b

∂ f1 , f 2 , f 3

b

∂ a, b, c

g

g

−1

−1

a 3 + b3 + c 3 3

−1

−1

= −2a −2b −2c = −2a − a 2 − b2 − c 2 − a2

a f a f FGH e j e j 0 2 a− b a 2 − b2

0 c → c2 − c1 2 a− c , 2 c3 → c3 − c1 a2 − c 2

IJ K

= – 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

1

1

1

1

0

= v+w u+w v+u = v+w u−v vw wu uv vw w u − v

a

FG c f va u − w f H c 0

u−w ,

2

→ c 2 − c1

3

→ c 3 − c1

IJ K

107


= (u – v) v(u – w) – (u – w) w(u – v) = – (u – v) (v – w) (w – u)

Thus

a a

f f

∂ u, v, w ∂ a, b, c

b

∂ f1 , f 2 , f 3

=

g

a−1f ∂b∂faa, ,fb,,cff g ∂au, v, wf 3

1

2

3

= −

a fa fa f a fa fa f

−2 a − b b − c c − a − u−v v−w w−u

= −2

aa − bfab − cf ac − af . au − vfav − wf aw − uf

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

∂ u, v ∂ x, y

=

0

a f b g

Hence, two functions u and v are “functionally dependent” if their Jacobian is equal to zero.

108


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, v, w ∂ x, y, z

Now,

=

∂u ∂x ∂v ∂x ∂w ∂x

∂u ∂y ∂v ∂y ∂w ∂y

∂u ∂z ∂v ∂z ∂w ∂z

=

1 1 2x

1 –1 2y − 2z

=

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 b g

∂ u, v ∂ x, y

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


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 b g

∂ u, v ∂ x, y

2ax + 2hy 2 Ax + 2Hy

=0

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

f g

∂ u, v, w ∂ x, y, z

=

∂u ∂x

∂u ∂y

∂u ∂z

2xe − y cos hz

−x 2 e − y cos hz

x 2 e − y sin hz

∂v ∂x

∂v ∂y

∂v ∂z = 2xe − y sin hz

− x 2 e − y sin hz

x 2 e − y cos hz

∂w ∂x

∂w ∂y

∂w ∂z

−6x 4 e −2 y

0

12x 3 e −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


3 . If x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ, show that

b g ∂ br, θ, φg

∂ x, y, z

= r2 sin θ.

4 . If u = x + y + z, uv = y + z, uvw = z, evaluate

e

b g b g

j

(U.P.T.U., 2000)

b g. ∂ au, v, wf ∂ x, y, z

2 (U.P.T.U., 2003) Ans. u v

LMAns. 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 ∂ bξ, ηg ∂ x, y

=

−

y 2x

OP Q

1 2 a (cos h 2ξ – cos 2η). 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 −2( x − y ) ( y − z) ( z − x) O au − vfav − wfaw − uf PQ

8 . If u, v, w are the roots of the equation (λ – x)3 + (λ – y)3 + (λ – z)3 = 0 in λ, find

LM N

(U.P.T.U., 2001) Ans.

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

a f = u v w. ∂ 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

LMAns. J = 2 y , J ′ = x OP x 2y Q N

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

∂ 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 1 3 . Find 1 4 . Find

a fL b g MN

∂ X, Y 2 2 2 2 2 2 . Ans. 6x y x + y x – y ∂ x, y

e

a f b g ∂ au, v, wf , if u = 3x + 2y – z, v = x – y + z, w = x + 2y – z. ∂ bx, y, zg

∂ u, v, w , if u = x2, v = sin y, w = e–3z. ∂ x, y, z

1 5 . Find J (u, v, w) if u = xyz, v = xy + yz + zx, w = x + y + z.

je

j OPQ 2

Ans. – 6 e −3 z x cos y

Ans. – 2

b

gb

ga f

Ans. x − y y − z z − x

111


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

17. u =

b g a b

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 =

2 2 . If u = x + y + z, uv = y + z, uvw = z, show that

b a

2

∂ u, v, w

2 3 . If x2 + y2 + u2 – v2 = 0 and uv + xy = 0 prove that 2 4 . Find Jacobian of u, v, w w.r.t. x, y, z when u =

g = u v. f ∂au, vf x −y = ∂bx, yg u +v

∂ x, y, z

2

2

2

2

.

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 ∂f ∂f δx + δy ∂x ∂y

...(ii)

...(iii)

112


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


⇒

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

IJ K

4

+ 100 ×

dy y

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 + ∴

or

= =

a f F I πr h + H 3K 3 × 2a h + 2r f ra 3h + 4r f

πr

dr +

⇒

3

a3h + 4rf

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%.

OLS

Fh+ I H 3K

dh =

Fig. 2.1

3 [2(h + 2r) dr +rdh] r 3h + 4r

a

3 [2(4 + 3)(0.01) + 1.5 (0.05)] 15 . 12 + 6

a

R>ª

f

δr = dr δh = dh

114


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


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 a a a

· db + c · dc – db · c cos A – b · dc cos A + bc sin A · dA · da – b · db – c · dc + db · c cos A + b · dc cos A · da – (b – c cos A) · db – (c – b cos A) dc 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, i.e., ⇒

dc will be minimum if sin2θ is maximum c sin 2θ = 1 = sin 90 2θ = 90 ⇒ θ = 45°. Hence proved.

Since sin θ lies between – 1 and 1

116


Example 8. Find approximate value of

a0.98f + a2.01f + a1.94f f (x, y, z) = e x + y + z j ∂f = x ex + y + z j ∂x 2

Sol. Suppose

2 12

2

2

Now

2 12

2

2

∴

.

2 −1 2

2

∂f 2 2 2 = z x +y +z ∂z

e

j

,

∂f 2 2 2 = y x +y +z ∂y

e

j

−1 2

,

−1 2

∂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

× ( −0.02) + 2 (9)

−1 2

a f a−0.06f

(.01) +2 9

−1 2

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


⇒

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 dD dt –2 D t dt 2 dD 100 × + – 2 100 × t D 3 −4 (1) – 2 (–1) = 3

log W = log K + 2 log S +

dW dS 2 = 2 + W S 3 dW dS ⇒ 100 × = 2 100 × S W 2 = 2 (–2) + 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 On differentiating

IJ K

FG H

FG H

IJ K

FG H

IJ K

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=

IJ LM KN

FG H

⇒ 2·

a

δh + tan α + sec α sec α. δα . h

FG 1 + 2 IJ · H 3 3K

2 3

· δα = 0

FG H

IJ K

FG H

IJ K

OP Q

118

⇒


1 100 3

+

FG 3IJ H 3K

δα= 0 ⇒ δα = – δα = –

or

=– =–

1 100 3 9 5π 3

×

1 100 3

radian.

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


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


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 x

2

, where r is the radius of the earth.

16. Calculate the error in R if RI = E and possible errors in E and I are 20% and 10% respectively.

Ans. 10%

121


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

18. Find 83.7

1 4

a f

a f

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


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

Fh GH

2

I JK

∂2 f

2 ∂2 f 2 ∂ f +.... ...(i) + hk + k 2 ∂x∂y ∂x 2 ∂y 2 ( a, b)

2

Neglecting higher order terms of h , 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 ⇒ f (a + h, b + k) – f (a, b) =

h

a f ∂f a a, bf ∂f a a, bf + k ∂x

∂y

a f

...(ii)

123


The necessary condition for a maximum or minimum value (L.H.S. of eqn (ii) negative or positive) is h

a f

∂f a, b ∂x

+ k

a f

∂f a, b = 0 ∂y

a f

∂f a, b

⇒

∂x

= 0,

a f

∂f a, b ∂y

= 0 h and k can take both + ve and – ve value ...(iii)

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

∂2 f ∂2 f ∂2 f 1 + k2 2 h 2 2 + 2hk ∂x∂y 2 ∂y ∂x

r,

( a, b )

∂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) =

OP PQ

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


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

(x – y) (x + y) – 4 (x – y) ⇒ (x – y) (x + y – 4) ⇒ x Putting x = y in (i), we get ∴ Again putting

...(i) ...(ii)

= 0 = 0 = y and x + y = 4 x2 + 4x – 21 = 0 ⇒ (x + 7) (x – 3) = 0

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


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. = a = y, we have r = 2a3a–3 = 2, s = 1, t = 2 = (2) (2) – 12 = 3 > 0 = 2 > 0. Hence, there is minima at x = a = y 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

At x ∴ rt – s2 Also r ∴ The minimum at

∂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


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


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


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 = 0, we get cos x – sin ∂x ∂u From ∂y = 0, we get cos y – sin Solving (i) and (ii), we get cos x = cos x = 2nπ ± y, where n is any integer. In

From

∂u ∂u = 0, =0 ∂x ∂y

(x + y) = 0

...(i)

(x + y) = 0.

...(ii)

y which gives particular x = y.

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 Q

< 0. (Note)

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


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 = –

x 2 + y 2 + ( 2x + 3 y − 12) 2 .

l =

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 or 2 H7K H7K

Solving (ii) and (iii), we get x = Also or

2x + 3y – z = 12

– z = 12

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


∴

∂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 =

F 3I H 4K

or x = ±

...(ii)

1 4 1 3 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 –

e2 3 j

2

F 3 I + 4F − 1 I − 2 = 6 , s = 4 FH 1 3 IK = 2 H 4K H 4K 2

> 0. Also r > 0

∴ There is a minimum when x = Again when x = –

3,t=6

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

2

> 0. Also r > 0.

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 = ⇒

b 3 + λ + 1 − 7µ g

2

+ ( 5 − 2λ + 1 + 6µ) 2 + (7 + λ + 1 − µ) 2

D2 = u(Say) = 6λ2 +86µ2 – 40λµ + 105.

131


∂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


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 = H K

a2 a2 4 2

Ka4 . 8

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 From From

∂z ∂z = 0, =0 ∂y ∂x

∂z = 0, we get cos 2x – cos (2x + 2y) = 0 ∂x ∂z = 0, we get cos 2y – cos (2x + 2y) = 0 ∂y

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

...(i) ...(ii)

133


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 = – ∴

rt – s2 =

2π 4π + sin , from (C) 3 3

3 – 2

3 =– 3 2

e− 3 j e− 3 j – FGH − 23 IJK

2

=3–

3 9 = 4 4

= 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 If we take x = – , then y = x = – 3 1 3,t= and r = 3, s= 2 9 ∴ rt – s2 = >0, r>0 4 π There is a minimum at x = – = y. 3 =

Hence, the minimum value = sin

F− πI H 3K

I K

.

π 3 3

sin

F− πI H 3K

sin

RSF − π I + F − π I UV TH 3 K H 3 K W

134


= – 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 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

2.

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


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 constraint g(x, y, z) = 0 Thus the problem is Extrema u = Subject to g (x, y, z) = For stationary point ∴

independent variables, where x, y, z are related by a known of f(x, y, z) 0

∂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)

...(i) ...(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


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


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 HpK

= 0, y –

FG bu IJ HpK

= 0, z –

x y u z = = = a p b c

or

FG cu IJ HpK

x y z = = a b c

or

...(vi)

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,

2

bp

y =

a +b +c These give the minimum value of u. Hence minimum value of u is

u = =

2

2

2

,

a2 p2 (a 2 + b2 + c 2 )2 ( a 2 + b 2 + c 2 )p 2 ( a2 + b2 + c 2 ) 2

2

2

z=

+ =

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

=0

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


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 Ha a K Hb b K K Hc c 1 2

4

2

1 2

4

2

1 2

4

2

dz = 0

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

FG x Ha

or or

or or

IJ K

FG H

IJ K

z2 x2 y2 z2 + + + λ1 + λ2 (lx + my + nz) = 0 4 b4 c 4 a2 b 2 c 2 u + λ1 (1) + λ2 (0) = 0, using (i), (ii) and (iii) u + λ1 = 0 or λ1 = – u ∴ From (vii), we have 2

+

y2

...(vii)

+

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

−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 =

or or or

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

or

a2b2c2 (l2a2 + m2b2 + n2c2)u2 – a2b2c2

or

(l2a2 + m2b2 + n2c2) u2 –

RSΣl a FG 1 + 1 IJ UV u + Σl a = 0 T H c b KW R|SΣl F a + a I U|V u + FG a l + b m + c n IJ |T GH c b JK |W H b c c a a b K 2

2 2

2

2 4

2

2

2

2 2

2

2

2 2

2

2

2 2

2 2

2 2

= 0,

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 du = 2x dx + 2y dy + 2z dz = 0

...(i) ...(ii) ...(iii)

139


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 −6 a 2

u

u

−6 a

u

or

= 0, which gives the max. and min. values of u.

u −6 a

u

2

−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


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 D =

F 1 − 1I H2 K

D2 =

5 ⇒ D= 2

2

or

2

+ (1 − 2)

2

F 5 IJ +G H2K

2

=

5 1 10 +1+ = 4 4 4

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


Multiplying equation (v) by x, y, z respectively, we get λx2 = – 4xyz, λy2 = – 4xyz, λz2 = – 4xyz From these, we get λx2 = ⇒ x2 = ⇒ x = Thus, the rectangular solid

λy2 = λz2 y2 = z2 y= z 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 b h V

∴

= = = =

2x 2y 2z 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 +

Lx 3xyz + λ M Na

2

y2

OP Q

z2 = 2 b2 c2 ⇒ 3xyz + λ = ⇒ λ = Putting the value of λ in any one of +

+

yz – 3xyz · ⇒ Similarly,

1–

x a2

3x2 a2

0 0 (using ii) – 3xyz equation (v), we get

= 0⇒ x=

y =

Hence, the largest volume V = 8 ·

FG H

3x 2 a2

= 0 ⇒ yz 1 −

b

abc 3 3

3 ⋅

, z=

a

,

3 c

3

IJ K

=0

...(v)

142


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

( x − 3)2 + ( y + 6)2 + ( z − 4) 2

D2 = (x – 3)2 + (y + 6)2 + (z – 4)2 u = (x – 3)2 + (y + 6)2 + (z – 4)2 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


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 = radius of circle = area of square = area of circle =

x , 4 y |As 2πr = y 2π x2 16

πy 2 4π 2

=

y2 4π

y2 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


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


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 (b) When box is closed n = 2

∴

x2 =

x = y=

S 6

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


4. Find the maxima and minima of x2 + y2 + z2 subject to the conditions : ax2 + by2 + cz2 = 1, lx + my + nz = 0.

LMAns. N

l2 m2 n2 + + =0 ( au − 1) (bu − 1) (cu − 1)

OP Q

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


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 = N

= 0 by the Lagrange’s method.

la + mb + nc l +m +n 2

2

2

OP Q

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

a f b g

∂ u, v ∂ x, y

2. The Jacobian

for the function u = 3x + 5y, v = 4x – 3y is

(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

(iv)

5 r

4. If u = x sin y, v = y sin x, then (i) sin x sin y (iii) cos x cos y – xy sin y

b a

∂ x, y, z

g f

∂ r , θ, Z

is

(ii) r2 – 5 (iv) r

a f b g

∂ u, v ∂ x, y

is

(ii) sin x sin y – xy cos x cos y (iv) 0

148


5. If u = 3x + 2y – z, v = x – y + z, w = x + 2y – z, then J (u, v, w) is (i) 5

(ii) 0 (iv) xy + 3x3

(iii) – 2

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)

(iv) (5, 6)

11. The extreme values of 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 = .......... ∂bx, yg

2. If x = r cos θ, y = r sin θ then 3. If u = x (1 – y), v = xy then 4.

a f b g b g a f

∂ x, y ∂ u, v × ∂ u, v ∂ x, y

= ..........

∂ x, y ∂ r, θ

a f b g

∂ u, v ∂ x, y

= ..........

149


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 ∂ar , θf

∂ x, y

2

– 2r + 1.

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 (ii)

Minm

(iii) Saddle point (iv) Failure case 2. (i) δx or dx (ii)

δx dx or x x

dx x (iv) f(x, y) ≤ f(a, b)

(iii) 100 ×

(a) rt – s2 = 0 (b) rt –

s2

<0

(c) rt –

s2

> 0, r > 0

(d) rt –

s2

> 0, r < 0

(a) Local maximum (b) Absolute error (c) Relative error (d) Percentage error

(U.P.T.U., 2008)

150


ANSWERS TO OBJECTIVE TYPE QUESTIONS A. Pick the correct answer: 1. (ii) 4. (ii) 7. (iii) 10. (i)

2. (iv) 5. (iii) 8. (iii)

3. (iv) 6. (v) 9. (iv)

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

A=

LM a MM a MNa M

11

a12

L

a1n

21

a22

L

a2n

M

M

M

am2

L

amn

m1

OP PP PQ

...(i)

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


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,

i.e., (m × 1) order is called column matrix, e.g.,

LM 1 OP MM 2 PP MN 05 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 N0

0 0

OP Q

0 . 0

(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

aij =

when i ≠ j when i = j

RS 0 TK

is a scalar matrix if

(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

A =

LM 2 MM1 N5

3

5

2 1

3 2

OP 0P , 1PQ 6

then A′ =

LM 2 MM53 MN6

1 2 3 0

OP P 2P P 1Q 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 N 3

OP Q

−4i , 1 − 2i

then A =

LM1 − 2i N 6

4i 1 + 2i

OP Q

154


(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 N 6

OP Q

−5i , 1 − 2i

then A* =

LM 2 − 3i N 5i

6 1 + 2i

OP Q

(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.

OP PP Q

LM 1 MM3 − i N −i

3+ i 2 −5i

Thus A is skew hermitian if

aii =

−a ji ∀ i, j

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

e.g.,

i 5i 0

(p) Skew hermitian matrix: Any square matrix A = [aij] is said to be a skew hermitian matrix if A* = – A

If

or

x = 0 Hence all the diagonal elements of a skew hermitian matrix are either zero or pure imaginary. e.g.,

LM i MM −3 + i 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 N1 LM0 0OP LM0 N1 0 Q N1

A2 = 0; e.g.,

As

A2 =

OP 0Q 0

0 0

is a nilpotent matrix

OP = LM0 0OP = 0 Q N0 0 Q

(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. A =

A2 =

LM1 N0 LM1 N0

OP Q 0O L1 0O 1 0 P1Q MN0 1PQ = LMN0 1OPQ = A 0 , 1

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 , N0 1 Q

A2 =

LM1 0OP LM1 0OP = L1 0O N0 1Q N0 1Q MN0 1PQ

= I

(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 Q

=

LM 3 N –3

OP Q

6 12

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 × e.g.,

Let A =

LM N

2 3

OP Q

5 1 ,B=

LM N

5 2

OP Q

–4 0 , then A + B =

LM N

7 5

OP Q

n

1 1

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 ×

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

n

156


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’. e.g.,

Let A =

AB =

=

LM a Na

21

LM a Na LM a Na

OP Q

a12 a 22

11

a13 a 23 , B =

a12 a 22

11 21

a13 a 23

OP Q

LMb MMb Nb

11 21 31

+ a12 b21 + a13b31 + b 21 11 a 22 b21 + a 23 b31 11 b11

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 MM 0 N –1

0 2 1

OP PP Q

5 3 , the trace of A = 1 + 2 + (– 2) = 1. 2

3.5 PROPERTIES OF TRANSPOSE (i) (ii) (iii) (iv)

(A′)′ = A (KA)′ = KA′, K being scalar (A + B)′ = A′ + B′ (AB)′ = B′A′.

3.6 PROPERTIES OF CONJUGATE MATRICES (i) (ii) (iii)

dKAi d A + Bi dA Bi

=

K A , K being a scalar,

=

A + B,

=

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


LM 1 a A + A *fOP N2 Q LM 1 a A – A *fOP N 2i Q

*

so that

P* = Q* =

= *

=

1 1 (A* + A∗∗) = (A* + A) = P, 2 2

FG 1 IJ H −2i K

A* – A** = –

1 (A* – A) = Q 2i

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. Example 4. If A =

Sol. Here

LM1 MM 2 N2

(A*A)′ = (I)′ A′ (A*)′ = I A′ (A′)* = I Proved.

2 1 2

2 2 show that A2 – 4A – 5I = 0. 1

A2

=

=

and

OP PP Q

a f

As A * * = A

4A =

LM1 2 2OP L1 2 2O MM 2 1 2PP MM 2 1 2PP N 2 2 1Q MN 2 2 1PQ LM1 + 4 + 4 2 + 2 + 4 2 + 4 + 2OP MM 2 + 2 + 4 4 + 1 + 4 4 + 2 + 2PP N 2 + 4 + 2 4 + 2 + 2 4 + 4 + 1Q LM1 2 2OP L 4 8 8O M 4 8PP 4 M 2 1 2P = M 8 MN 2 2 1PQ MN8 8 4PQ

=

LM 9 MM8 N8

8 9 8

8 8 9

OP PP Q

159

MATRICES

∴

A2 – 4A − 5I =

=

Example 5. If A = Sol. Let A = To prove

LM 3 N1

LM 3 N1

OP Q

–4 –1

OP Q

–4 –1

LM 9 8 MM8 9 N8 8 LM 9 – 9 MM8 – 8 N8 – 8

8 8 9

OP PP Q

–

8–8 9– 9 8–8

prove that AK =

LM 4 MM8 N8

OP PP Q

8 4 8

8–8 8–8 9– 9

LM1 + 2K N K

8 8 4 =

LM0 MM0 N0

LM5 MM0 N0

–

0 0 0

OP Q

0 0 0

OP PP Q

0 5 0

0 0 5

OP PP Q

= 0. Proved.

– 4K , K being any positive integer. 1 − 2K

and K be any positive integer AK =

A1 = A =

LM1 + 2K N K LM 3 −4OP N1 –1Q

OP Q

– 4K , we see that 1 − 2K =

LM1 + 2 ⋅1 N 1

OP Q

−4 ⋅ 1 1 – 2⋅1

This proves that result is true for K = 1 Now

OP PP Q

A2 =

=

LM 3 –4OP LM 3 –4OP N1 –1Q N1 –1Q LM1 + 2 ⋅ 2 –4⋅ 2 OP N 2 1 – 2⋅ 2Q

=

LM 9 – 4 N3 – 1

OP Q

–12 + 4 –4 + 1

=

LM5 N2

–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 An =

Now,

An + 1 = An A =

=

LM1 + 2n – 4n OP . N n 1 − 2n Q LM1 + 2n – 4n OP LM 3 – 4OP N n 1 − 2n Q N1 – 1Q LM 3 + 6n − 4n –4 − 8n + 4nOP N 3n + 1 − 2n –4n − 1 + 2nQ

=

LM1 + 2an + 1f N n+1

a f OP 1 − 2an + 1fQ –4 n + 1

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


Example 7. Show that the matrix A = α2 + β2 + γ2 + δ2 = 1. Sol. A =

LMα + iγ N β + iδ

OP Q

LMα + iγ N β + iδ

LM N

β + i δ α iγ β iδ ∴ A* = β i δ α + i γ α iγ

But for unitary matrix ∴

β + iδ α iγ 2

OP LM α iγ Q N β i δ

2

β iδ α + iγ

OP Q

⇒

LMα MN

2

+ β2 + γ 2 + δ2 0

0 2

α +β + γ

(U.P.T.U., 2005)

OP Q

LM1 N0 LM1 N0

2

OP + δ PQ 2

=

OP Q 0O 1PQ

OP PQ

0 1

α2 + β2 + γ2 + δ2 = 1. Proved.

Example 8. Prove that the matrix Sol. Let

is a unitary matrix if

αβ – iαδ + iβγ + γδ – αβ – iβγ + iαδ – δγ β2 + δ 2 + α 2 + γ 2

2

2

OP Q

LM1 0OP N 0 1Q

=

=

⇒

β + iδ α iγ

AA* = I

L α +γ +β +δ ⇒ M αβ βγ + αδ + γδ – αβ – iαδ + iβγ – δγ i i – MN 2

LMα + iγ Nβ + iδ

A =

A* =

A*·A =

=

LM 1 3 N1 – i

1

1+i –1

OP Q

is unitary.

(U.P.T.U., 2001)

LM 1 + iOP –1 Q N 1 + iO 1 L 1 M –1 PQ 3 N1 – i 1 + iO 1 L 1 1 + iO 1 L 1 × 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 ) 1 1 1 3 –i

Therefore, A is a unitary matrix. Example 9. Define a unitary matrix. If N =

LM 0 N –1 + 2i

1 + 2i 0

OP Q

is a matrix, then show that

(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 N0 1Q N–1 + 2i

OP LM Q N

1 + 2i 1 = 0 1 – 2i

(U.P.T.U., 2000)

–1 – 2i 1

OP Q

161

MATRICES

and

I+ N =

LM1 0OP + LM 0 N0 1Q N –1 + 2i

OP LM Q N

1 + 2i 1 = 0 –1 + 2i

|I + N| = 1 – (– 1 – 4) = 6 Let ∴

⇒ ⇒

A =

1 + 2i 1

B =

LM 1 N –1 – 2i

1 – 2i 1 or B′ = 1 1 – 2i

OP Q

OP Q

A11 = 1, A12 = – (– 1 + 2i) = 1 – 2i A21 = – (1 + 2i) = – 1 – 2i, A22 = 1

OP Q

Adj A = Adj (I + N) = B′ = (I + N)–1 =

and

LM 1 N–1 + 2i

1 + 2i 1

LM N

LM N

LM 1 N1 – 2i

Adj ( I + N) 1 1 = |I + N| 6 1 – 2i

–1 – 2i 1

–1 – 2i 1

–1 – 2i 1

OP Q

OP Q

OP Q

–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


LM OP Example 11. If A = [x y z], B = MM PP NQ 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 ABC = (AB) C = [ax + hy + gz hx + by + fz gx + fy + cz] M yP MN zPQ LM a MMh Ng

and

OP PP Q

h b f

g x f , C = y then find out ABC. c z

= [(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 LMx xOP LMx xOP = LM2x 2x OP = 2x LM1 1OP 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.

A2 =

=

2

2

2

2

2

Similarly,

A3 = 22x3 B, A4 = 23x4 B, ... etc.

∴

eA = 1 + A +

= I + xB +

=

=

=

=

2

1 2

A2 A3 + + ..... 2 3 2x 2 B 2 2 x 3 B + + ..........|As A = xB 2 3

F GH 2 I + ( 2x )B + ( 2 x2)

2

B+

I JK

( 2x) 3 B+. . . . . 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 MM a Na

21

a12 a 22

31

a 32

11

Let

A =

OP PP Q

a13 a 23 , then a 33

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 =

a 23 a 21 A12 = – a 33 a 31

a12 a 32

a13 a11 , A22 = a 33 a 31

a12 a 22

a13 a11 , A32 = – a 23 a 21

LM A MM A NA

A12 A22 A32

11

Let

B =

21 31

OP PP Q

a 23 a 21 , A13 = a 33 a 31

a 22 a 32

a13 a11 , A23 = – a 33 a 31 a13 a11 , A33 = a 23 a 21

LM MM N

A13 A11 A23 then Adj A = B′ = A12 A33 A13

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


(iii) Inverse of a product is the product of inverses in the reverse order i.e., (AB)–1 = B–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.

A–1,

Example 13. Find the inverse of matrix A, where

OP PP Q

LM–1 A= M 3 MN–1

1 2 –1 1 . 3 4

Sol. Here we find A11, A12......... cofactors of A as follow: A11 =

–1 3

A21 = – A31 =

Let B =

∴

A–1

1 3

1 –1

LM –7 MM 2 N3

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 PP Q

LM MM 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 A=

OP PP Q

LM1 MM3 N5

2. If A =

3. If A =

4. If A =

5. If A =

LM MM N

–3 2 4 ,B= 1 6 4

LM0 MM0 Np LM2 N0 LM1 MM2 N3 LM1 MM2 N4

1 0 q 3 1 –2 3 1 –3 1 –3

OP PP Q

LM MM N

–2 p –5 and C = r t 3

OP PP Q

q s so that A + B – C = 0. u [Ans. p = – 2, q = 0, r = 4, s = – 1, t = 9, u = 9]

OP PP Q 1O 1 2 –6 P5Q, B = LMN0 –1 3 OPQ , find 3A – 4B. 3O P –1P , show that 6A + 25A – 42I = 0. 2 PQ 2O LM1 4 1 0OP LM2 1 –1 P –3P, B = M2 1 1 1 , C = 3 –2 –1 MN1 –2 1 2PPQ MMN2 –5 –1 –1PQ

0 1 , then show that A3 = pI + qA + rA2. r

6. If A =

OP P prove that 0 P Q Lcos α I + A = (I – A) M Nsin α

– tan

A=

LM cos α N– sin α

27

1

OP PP Q

–2 –1 . 0

α 2

7. Find the product of the matrices

8. If Aα =

OPOP 3 QQ

1

2

Show that (i) AB = AC, (ii) (B + C) A = BA + CA.

LM 0 MMtan α N 2

LMAns. L2 N MN0

LM2 N1

OP Q

1 1

2 1

OP Q

– sin α · cos α

LM 2 1O 0 P1Q, B = MM–2 MN 1

–1 4 1 3

OP PP PQ

0 1 . 0 2

LMAns. L1 N MN1

1 1

OP OP 3Q Q 3

sin α , then show that cos α (Aα)n =

LM cos nα N– sin nα

OP Q

sin nα = Anα, cos 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


9. If A =

LM0 1OP , and B = LM 1 OP , find BA and AB if they exist. N2Q N1 2Q Ans. AB =

10. If A =

LM3 MM1 N2

OP L PP MN Q

4 2 1 , B= 1 0

1 2

OP Q

LM 2OP ; BA does not exist N 5Q

2 , find (AB)′. 4

Hence verify that (AB)′ = B′A′.

LM LM10 MMAns. ( AB)′ = M11 MN 22 N

3 3 6

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.

A=

19. Prove that

LM 1 MM 1 N–2

O cos αPQ

LMcos α Nsin α

sin α

1 3 –4 –1

OP PP Q

3 –3 . 4

= sec 2α

LM cos α N– sin α

– sin α cos α

20. If ω is one the imaginary cube roots of unity and if A=

LM1 MM1 N1

1 ω2 ω

OP PP Q

LM MM N

1 1 1 –1 ω , then show that A = 1 3 2 ω 1

1 ω2 ω

OP. Q

LM MMAns. A N

–1

=

LM MM N

12 1 –5 4 –1

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


(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.

1 1 2 3

2 2 2 3

2 3 . 3 3

OP PP PQ

LM MM MN

0 1 0 0

0 0 1 0

0 0 A 0 1

OP PP PQ

LM MM MN

1 0 0 0

0 0 1 0

0 0 A 0 1

Example 1. Find the inverse of A =

Sol. Let A = IA

⇒

R1 ↔ R2

OP PP PQ

LM0 MM1 MN22

LM0 MM12 MN2

1 1 2 3

2 2 2 3

1 2 0 3 = 0 3 0 3

LM1 MM0 MN22

1 1 2 3

2 2 2 3

0 3 1 2 = 0 3 0 3

OP PP PQ OP PP PQ

Applyirg R3 → R3 – 2R1, R4 → R4 – 2R1, we get

LM1 MM00 MN0

1 1 0 1

2 2 –2 –1

OP PP PQ

Applying R2 → R2 + R3, we have

LM1 MM00 MN0

1 1 0 1

2 0 –2 –1

LM MM MN

0 3 1 2 = 0 –3 0 –3

OP PP PQ

LM MM MN

3 0 –1 1 = –3 0 –3 0

1 0 –2 –2

1 –2 –2 –2

OP PP PQ

0 0 1 0

0 0 A 0 1

0 1 1 0

OP PP PQ

0 0 A 0 1

169

MATRICES

Applying R1 → R1 – R2, R4 → R4 – R2, we have

OP PP PQ

LM MM MN

3 –2 –2 0

OP PP PQ

LM MM MN

1 –2 –2 2

OP PP PQ

LM MM MN

3 –4 –8 2

LM MM MN

3 –4 4 –2

LM1 MM00 MN0

0 1 0 0

LM1 MM0 MN00

0 0 1 –1 1 0 1 1 = 0 2 3 0 0 0 1 –2

LM1 MM0 MN00

0 0 0 –3 1 0 0 3 = 0 2 0 6 0 0 1 –2

2 0 –2 –1

4 –1 –1 1 –3 = 0 –2 –1

OP PP PQ

–1 1 1 –1

0 0 A 0 1

Applying R1 → R1 + R3, R4 → 2R4 – R3, we get

OP PP PQ

0 1 1 –3

0 0 A 0 2

Applying R1 → R1 + R4, R2 → R2 – R4, R3 → R3 – 3R4

Applying R3 → –

–3 4 10 –3

1 R , R → (– 1) R4, we obtain 2 3 4

LM1 MM0 MN00

0 1 0 0

0 0 1 0

A–1

Hence

OP PP PQ

0 –3 0 3 0 = –3 1 2

LM–3 3 = M MM–3 N2

3 –4 4 –2

–3 4 –5 3

–3 4 –5 3

OP PP PQ

2 –2 A –6 2

OP PP PQ

OP PP PQ

2 –2 A 3 –2

2 –2 . 3 –2

Example 2. Find by the elementary row transformation inverse of the matrix.

LM0 MM1 N3

Now ∴

2 1

LM0 A = M1 MN3

Sol. Let

LM0 MM1 N3

OP 3P. 1PQ

1 2

OP PP Q

LM1 MM N

2

2 1

3 = 0 1 0

OP PP Q

1 2 1

2 3 1

0

0

1 0

0 A 1

A = IA

1

(U.P.T.U., 2000, 2003)

OP PP Q

170


Applying R1 ↔ R2, we get

R3 → R3 – 3R1

LM1 MM0 N3

LM1 MM0 N0

2 1 1 2 1 –5

R1 → R1 +

R3 →

1 R 2 3

1 R , R → R2 2 3 2 1 0 0

LM MM N

LM1 MM0 N0

LM MM N

1 0 0

OP PP Q

LM MM N

1 0 –3

OP PP Q

LM MM N

3 0 2 = 1 –8 0

R1 → R1 – 2R2, R3 → R3 + 5R2

LM1 MM0 N0

OP PP Q

3 0 2 = 1 1 0

0 1 0

–1 –2 2 = 1 2 5

OP PP Q

0 0 A 1

OP PP Q

0 0 A 1

OP PP Q

1 0 –3

0 0 A 1

– R3, we get

OP PP Q

LM MM N

OP PP Q

LM MM N LM1 2 = M –4 MN5 2

0 1 0

0 1 2 0 = –4 2 5

0 1 0

0 1 2 0 = –4 1 5 2

A–1

Therefore,

–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 MM2 N0

LM3 MM2 N0

Applying R1 → R1 – R2, we get

LM1 MM2 N0

4 4 · 1

0 0 A 1

A = IA

–3 –3 –1

0 –3 –1

OP PP Q

LM MM N

0 1 0

OP PP Q

LM MM N

–1 1 0

4 1 4 = 0 1 0

0 1 4 = 0 1 0

OP PP Q

–3 –3 –1

OP PP Q

OP PP Q

0 0 A 1

[U.P.T.U. (C.O.), 2002]

171

MATRICES

R2 → R2 – 2R1

LM1 MM0 N0

R3 → 3R3 – R2

R2 → R2 + 4R3

R2 → –

0 –3 –1

OP PP Q

LM MM N

–1 3 0

0 0 A 1

OP PP Q

LM MM N

–1 3 –3

0 0 A 3

OP PP Q

LM MM N

–1 –9 –3

0 12 A 3

–1 3 3

0 –4 A –3

–1 3 3

0 –4 . –3

0 1 4 = –2 1 0

LM1 MM0 N0

0 –3 0

0 1 4 = –2 –1 2

LM1 MM0 N0

0 –3 0

0 1 0 = 6 –1 2

0 1 0

OP PP Q

LM MM N LM 1 = M –2 MN –2

0 1 0 = –2 1 –2

A–1

Hence

Example 4. Find the inverse of the matrix

⇒

OP PP Q

OP PP Q

1 R , R → (– 1) R3, we obtain 3 2 3

LM1 MM0 N0

Sol. Let

OP PP Q

LM–1 M1 A= M 2 MN–1

LM–1 MM 12 MN–1

3 –1 2 0

OP PP Q

OP PP PQ

–3 3 –1 1 –1 0 . –5 2 –3 1 0 1

A = IA

–3 1 –5 1

OP PP Q

OP PP PQ

LM MM MN

1 –1 0 0 = 0 –3 0 1

0 1 0 0

0 0 1 0

OP PP PQ

0 0 A 0 1

Applying R2 → R2 + R1, R3 → R3 + 2R1, R4 → R4 – R1, we get

LM–1 MM 00 MN 0

–3 –2

3 2

–11 4

8 –3

OP LM 1 1 P = M MM 2 –5P P 2Q N−1

–1 –1

0 1 0 0

0 0 1 0

OP PP PQ

0 0 A 0 1

172


R2 → –

1 R 2 2

LM –1 MM 0 MN 00

–3 1 –11 4

OP PP PQ

LM MM MM N

OP PP PQ

LM MM MN

OP PP PQ

LM MM MM N

OP PP PQ

LM 1 M– 1 2 = M 1 MN –7

OP PP PQ

LM MM MN

1 –1 1 – 12 2 = –5 2 2 –1

3 –1 8 –3

R3 → R3 + 11R2, R4 → R4 – 4R2

R3 ↔ R4

R4 → 2R4

LM–1 MM 00 MN 0

–3 1 0 0

3 –1 –3 1

–1 1 12 –1 2 = 12 –7 2 0 1

LM–1 MM 00 MN 0

–3 1 0 0

3 –1 1 –3

1 –1 –1 2 12 = 1 0 7 – 12 2

LM –1 MM 0 MN 00

–3 1 0 0

3 –1 –1 1 2 1 0 1 –6

LM–1 MM 0 MN 00

–3 1 0 0

Applying R4 → R4 + 6R3, we get

R2 → R2 + R3

R2 → R2 –

LM–1 MM 0 MN 00

1 R 2 4

LM–1 MM 0 MN 00

3 –1 1 0

–3 1

3 0

0 0

1 0

–3 1 0 0

0

0

0

OP LM 1 PP = MM1 2 0 P MN –11 1 Q OP PP PQ

LM MM MN

–1 1 0 1 = 0 1 1 –1

0 1

OP 0P A 0P P 1PQ 0

0 –1 2 – 11 2 2

–1 1 12 –1 2 = 0 1 1 –1

–1 1/2

3 0 1 0

0 1 – 2 0

0 0 A 0 1

OP 0 P 1P A P 0P Q

0

0

–1 2

0

2 11 – 2

0 1

0 –1 2 2 –11

0 0 0 2

0 0 1 0

0 –1 2 2 1

0 0 0 2

0 0 A 1 6

0 32

0 0

2 1

0 2

0 1 2 1

OP PP PQ

0 0 1 0

0 –1 0 2

0

OP PP PQ OP PP PQ

OP PA 1P P 6Q

0 1

OP PP PQ

0 –2 A 1 6

173

MATRICES

Applying R1 → R1 + 3R2, we get

LM–1 MM 0 MN 00

0 1 0 0

LM–1 MM 00 MN 0

0 1 0 0

3 0 1 0

OP PP PQ

LM MM MN

3 1 2 1

OP PP PQ

LM MM MN

–3 1 2 1

–3 –1 0 2

–9 –2 A 1 6

OP PP PQ

LM MM MN

–2 1 2 1

–1 –1 0 2

–3 –2 A 1 6

OP PP PQ

LM MM MN LM 0 1 = M MM 1 N–1

–1 4 0 1 = 0 1 1 –1

Applying R1 → R1 – 3R3, we obtain

0 0 1 0

–1 1 0 1 = 0 1 1 –1

Applying R1 → R1 + R4

LM–1 MM 00 MN 0

0 1 0 0

LM1 MM00 MN0

0 1 0 0

0 0 1 0

0 0 0 1 = 0 1 1 –1

In the last, we apply R1 → (– 1) R1

0 0 1 0

0 0 0 1 = 0 1 1 –1

A–1

Therefore

–3 –1 0 2

2 1 2 1

1 –1 0 2

2 1 2 1

1 –1 0 2

OP PP PQ

–6 –2 A 1 6

OP PP PQ OP PP PQ

OP PP PQ 3O –2P P. 1P P 6Q

3 –2 A 1 6

Example 5. With the help of elementary operations, find the inverse of A = Sol. Since

OP PP Q

LM1 MM3 N1

2 1 2 3. 1 2

LM1 MM3 N1

A = IA

2 2 1

OP PP Q

LM MM N

OP PP Q

LM MM N

1 1 3 = 0 2 0

0 1 0

OP PP Q

0 0 A 1

Applying R2 → R2 – 3R1, R3 → R3 – R1, we get

LM1 MM0 N0

2 –4 –1

1 1 0 = –3 1 –1

0 1 0

OP PP Q

0 0 A 1

174


Applying R2 →

1 R2 , we get 4

LM1 MM0 N0

2 1 –1

OP PP Q

LM MM N

OP PP Q

LM MM N

OP PP Q

LM MM N LM– 1 4 = M34 MN– 1 4

1 1 0 = 3 4 1 –1

0 –1 4 0

OP PP Q

0 0 A 1

Applying R3 → R3 + R2, R1 → R1 – 2R2, we have

LM1 MM0 N0

0 1 0

LM1 MM0 N0

0 1 0

1 –1 2 0 = 3 4 1 –1 4

Now applying R1 → R1 – R3, we get

Hence

0 –1 4 0 = 3 4 1 –1 4

A–1

OP PP Q

12 –1 4 –1 4

0 0 A 1

OP PP Q –1O P 0 P. 1 PQ

3 4 –1 4 –1 4

–1 0 A 1

34 –1 4 –1 4

EXERCISE 3.2 1. Find the inverse of the following matrices:

2.

3.

4.

LM1 –1 0 2 OP MM02 11 12 –11 PP. MN 3 2 1 6 PQ LM 7 6 2OP MM –1 2 4PP. N 3 3 8Q LM 1 1 3 OP MM 1 3 –3PP. N–2 –4 –4Q LM2 4 3 OP MM 3 6 5 PP. MN24 55 142 PQ

LM L 2 –1 1 MMAns. MM –5 –3 1 MM MMN 23 –13 –10 N LM L 4 –42 1 M MMAns. 130 M 20 50 MN –9 –3 N LM 1 L12 4 MMAns. 4 MM–5 –1 MN–1 –1 N

LM L–23 MMAns. MM 10 MM MMN 12 N

29

– 64 5

–12

26 5

–2

65

–2

35

OPOP PPPP PQPPQ 20 O O PP –30P P 20 PQ PQ 6 OO PP –3P P –1PQ PQ – 18 5O O P 7 5 PP P 2 5 PP PP 1 5 Q PQ –1 1 0 1

175

MATRICES

5.

6.

7.

8.

9.

10.

11.

12.

LM 1 2 –1OP MM –1 1 2 PP. N 2 –1 1 Q LM4 –1 1 OP MM2 0 –1PP. N1 –1 3 Q LM i –1 2iOP MM 2 0 2 PP. N –1 0 1 Q LM1 2 3 1OP MM1 3 3 2PP. MN21 14 13 31PQ LM 2 –6 –2 –4 MM –15 –13 4 MN 0 1 01 LM0 1 3OP MM1 2 3PP. N 3 1 1Q LM2 1 2OP MM2 2 1PP. N1 2 2 Q LM 1 2 –2OP MM–1 3 0 PP. N 0 –2 1 Q

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

OP PP PQ

–3 –7 . 2 1

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


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 N1

OP Q

3 2

5 2 , then 3 1

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

LM 2 MM 3 MN –23

Sol. Let

OP PP PQ LM 2 MM 33 MN–2

3 –2 4 –2 1 2 . 2 3 4 4 0 5

A =

(U.P.T.U., 2006)

3 –2 2 4

–2 1 3 0

OP PP PQ

4 2 4 5

177

MATRICES

R1 → R1 + R4, R2 → R2 – R3, we get

A ~

7 –4 2 4

–2 –2 3 0

9 –2 4 5

~

LM 0 MM 01 MN–2

7 –4 6 4

–2 –2 3 0

9 –2 9 5

~

LM 1 MM 0 MN–20

6 –4 7 4

3 –2 –2 0

9 –2 9 5

R3 → R3 + R4

Q R1 ↔ R3

R4 → R4 + 2R1

A ~

Now,

OP PP PQ

LM 0 MM 03 MN–2

LM1 MM0 MN00

6 –4 7 16

–4 7 16

|A| =

OP PP PQ OP PP PQ

OP PP PQ

3 –2 –2 6

–2 –2 6

9 –2 9 23 –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.

A =

LM2 MM13 MN6

OP PP PQ

3 –1 –1 –1 –2 –4 . 1 3 –2 3 0 –7

Sol. Applying R1 ↔ R3, we get

A ~

LM1 MM2 MN63

–1 3 1 3

(U.P.T.U., 2005)

–2 –1 3 0

OP PP PQ

–4 –1 –2 −7

178


R2 → R2 – 2R1, R3 → R3 – 3R1, R4 → R4 – 6R1

A ~

Applying R3 → R3 –

LM1 MM0 MN00

–1 5 4 9

–2 3 9 12

OP PP PQ

–4 7 10 17

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

OP PP PQ

LM MM MN

In the last, we apply R4 → R4 – R3, we get

A ~

LM1 MM0 MN00

–1 5 0 0

OP PP PQ

–2 3 33 5 0

–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:

A =

LM 3 MM 45 MM10 MN15

OP PP PP PQ

4 5 6 7 5 6 7 8 6 7 8 9 . 11 12 13 14 16 17 18 16

Sol. Applying R1 → R1 – R2 and then again R1 → – R1

A ~

LM 1 MM 54 MM10 MN15

1

1

1

5 6

6 7

7 8

11 16

12 17

13 18

OP PP P 14P 16PQ 1

8 9

R2 → R2 – 4R1, R3 → R3 – 5R1, R4 → R4 – 10R1, R5 → R5 – 15R1

~

LM1 MM00 MM0 MN0

1 1

1

1 1 1 1

3 3 3 3

2 2 2 2

OP 4 P 4P P 4P 4PQ

1

179

MATRICES

Applying R3 → R3 – R2, R4 → R4 – R2, R5 → R5 – R2, we get

A ~

LM1 MM00 MM0 MN0

1 1 0 0 0

1 2 0 0 0

1 3 0 0 0

OP PP PP PQ

1 4 0 0 0

Here number of non-zero rows = 2 Therefore ρ(A) = 2. Example 4. Reduce the matrix A to its normal form, where

A =

LM0 MM13 MN1

1 0 1 1

OP PP PQ

–3 1 0 –2

–1 1 and hence find the rank of A. 2 0

Sol. Applying R1 ↔ R2, we have

OP PP PQ

LM1 MM0 MN13

0 1 1 1

1 –3 0 –2

1 –1 2 0

~

LM1 MM00 MN0

0 1 1 1

1 –3 –3 –3

1 –1 –1 –1

~

LM1 MM0 MN00

0 1 0 0

1 –3 0 0

1 –1 0 0

A ~

LM1 MM00 MN0

0 1 0 0

0 –3 0 0

0 –1 0 0

A ~

R3 → R3 – 3R1, R4 → R4 – R1

R3 → R3 – R2, R4 → R4 – R2

C3 → C3 – C1, C4 → C4 – C1

OP PP PQ OP PP PQ OP PP PQ

180


C3 → C3 + 3C2, C4 → C4 + C2, we get

A ~

Hence ρ (A) = 2.

LM1 MM00 MN0

0 1 0 0

OP PP LM PQ N

0 0 0 0

0 I2 0 = 0 0 0

OP Q

0 0

Example 5. Reduce the matrix A to its normal form, when

A =

LM 1 MM 21 MN–1

OP PP PQ

2 –1 4 4 3 4 . 2 3 4 –2 6 –7

Hence, find the rank of A.

Sol.

A =

LM 1 MM 12 MN–1

(U.P.T.U., 2001, 2004)

2 4 2 –2

OP PP PQ

–1 3 3 6

4 4 4 –7

Applying R2 → R2 – 2R1, R3 → R3 – R1, R4 → R4 + R1

OP PP PQ

LM1 MM00 MN0

2 0 0 0

~

LM1 MM00 MN0

0 0 0 0

0 5 4 5

0 –4 0 –3

~

LM1 MM0 MN00

0 5

0 0

0 –4

4 5

0 0

0 5

0 0

0 0

0 0

~

–1 5 4 5

4 –4 0 –3

Applying C2 → C2 – 2C1, C3 → C3 + C1, C4 → C4 – 4C1

C3 ↔ C2

R3 → R3 –

4 R , R → R4 – R2 5 2 4

~

LM1 MM0 MN00

OP PP PQ

OP P 0P P –3Q OP P 16 5P P 1 Q 0 –4

181

MATRICES

C3 ↔ C4

~

0 5 0 0

0 –4 16 5 1

0 0 0 0

~

LM1 MM0 MN00

0 5 0 0

0 0 16 5 1

0 0 0 0

R2 → R2 + 4R4,

16 1 R3, then R2 → 5 5 1 0 0 0 1 0 ~ 0 0 1 0 0 0

Applying R4 → R4 –

OP PP PQ

LM1 MM00 MN0

LM MM MN

OP PP PQ

R2 and R3 →

OP PP LM PQ N

0 0 I3 = 0 0 0

ρ (A) = 3.

5 R. 16 3

OP Q

0 0

Example 6. Find the non-singular matrices P and Q such that the normal form of A is PAQ where

LM1 MM1 N1

A =

3 4 5

6 5 4

–1 1 3

OP PP Q

. Hence, find its rank. 3×4

Sol. Here we consider A3 × 4 = I3 × 3· A3 × 4· I4 × 4

LM1 MM1 N1

3

6

4 5

5 4

OP 1P 3 PQ

–1

=

LM1 MM0 N0

OP LM10 0P A M M0 1PQ M N0

0

0

1 0

As Am ×

0 1 0 0

0 0 1 0

0 0 0 1

Applying R2 → R2 – R1, R3 → R3 – R1 (pre), we get

LM1 MM0 N0

3

6

1 2

–1 –2

OP 2P 4 PQ

=

LM 1 MM –1 N –1

–1 2 = 0

LM 1 MM–1 N1

–1

R3 → R3 – 2R1 (pre)

LM1 MM0 N0

3 1 0

6 –1 0

OP PP Q

0 1 0

0 1 –2

OP LM10 0P A M M0 1PQ M N0

0 1 0 0

0

OP PP Q

LM MM MN

1 0 0 0 A 0 1 0

0 1 0 0

0 0 1 0 0 0 1 0

OP PP PQ

0 0 0 1

OP PP PQ

0 0 0 1

OP PP PQ

n

= Im × m· Am × n· In ×

n

182


Applying C2 → C2 – 3C1, C3 → C3 – 6C1, C4 → C4 + C1 (post), we get

LM1 MM0 N0

0 1 0

0 –1 0

OP PP Q

0 2 = 0

LM 1 MM–1 N1

OP PP Q

LM MM MN

OP PP Q

LM MM MN

0 1 –2

1 0 0 0 ×A 0 1 0

0 1 –2

1 0 0 0 ×A 0 1 0

C3 → C3 + C2, C4 → C4 – 2C2 (post)

LM1 MM0 N0

0 1 0

0 0 0

Therefore,

OP PP Q

0 0 = 0

LM 1 MM–1 N1

I2 = N = PAQ, where

P =

and

LM 1 MM–1 N1

0 1 –2

OP PP Q

OP PP PQ

–3

–6

1 0 0

0 1 0

–3

–9

1 0 0

1 1 0

–3

–9

1 0 0

1 1 0

LM MM MN

1 0 0 0 , Q= 0 1 0

1 0 0 1

OP P 0P P 1Q

7 –2

OP P 0P P 1Q

7 –2

Rank of A = 2. Example 7. Find non-singular matrices P and Q such that PAQ is the normal form where

LM 1 MM 4 N2

A =

–1 2 2

2 –1 –2

–1 2 0

OP PP Q

. 3 × 4

Sol. Here we consider A3 × 4 = I3 × 3· A3 × 4· I4 × 4

LM1 MM4 N2

–1 2 2

2 –1 –2

OP PP Q

–1 2 0

=

LM1 MM0 N0

As Am

LM MM MN

OP PP Q

1 0 0 0 A 0 1 0

0 1 0

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

OP PP Q

2 –9 –6

–1 6 2

=

LM 1 MM – 4 N– 2

OP PP Q

LM MM MN

1 0 0 0 A 0 1 0

0 1 0

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 MM0 N0

0

0

6 4

–9 –6

OP 6P 2PQ 0

=

LM 1 MM–4 N–2

0 1 0

OP PP Q

LM MM MN

1 0 0 0 A 0 1 0

1 1 0 0

–2 0 1 0

×n

OP PP PQ

1 0 0 1

OP PP PQ

= Im

×m

Am × n In × n

183

MATRICES

R2 →

1 1 R2, R3 → R (pre) 3 2 3

LM1 MM0 N0

0 2 2

0 –3 –3

OP PP Q

LM 1 MM– 4 3 N –1

OP PP Q

=

LM 14 MM– 3 N1 3

=

LM 1 MM– 4 3 N13

0 2 = 1

1 0 0 0 A 0 12 0

1 1 0 0

–2

0 1 3 –1 3

OP LM10 0 P AM M0 1 2PQ M N0

1 1 0 0

–2

R3 → R3 – R2 (pre)

LM1 MM0 N0

0 2 0

0 –3 0

0 2 –1

C3 ↔ C4 (post)

OP PP Q

LM1 MM0 N0

0 0 0 2 2 3 0 1 0

C3 → C3 – C2, C4 → C4 +

LM1 MM0 N0 Applying R2 →

LM1 MM0 N0 ⇒

Therefore,

And

0 2 0

0 0 –1

LM MM MN

OP PP Q

0 13 0

0

LM MM MN

OP PP Q

0 1/3 –1 3

1 0 0 0 A 0 12 0

0 1 3 –1 3

1 0 0 0 A 0 1 2 0

3 C (post) 2 2

OP PP Q

0 0 = 0

LM 1 MM – 4 3 N13

0 0 1

OP PP Q

0 0 = 0

LM 1 MM– 2 3 N–1 3

P =

1 1 0 0

0 16 13

1 0 0 0 A 0 –1 2 0

1 1 0 0

0 –1

0 16 13

1 0 0 0 ,Q= 0 –1 2 0

ρ (A) = 3.

Note: In such problems other Ans. is possible.

OP PP Q

LM MM MN

LM MM MN

1 0 0 1

OP PP PQ

0 1 0

0 –1

OP PP Q

OP PP PQ

–1 2 3 2 1 0

0 1

OP PP PQ OP PP PQ

–1 2 32 1 0

0 1

1 1 0 0

OP PP PQ

–2

1 0 0 1

1 1 0 0

I3 = PAQ

LM 1 MM– 2 3 N– 1 3

0 1 0

LM MM MN

OP PP Q

1 R , R → (– 1) R3 (pre) 2 2 3 0 1 0

0 1 0

1 0 0 1

0 –1 0 1

–1 2 32 1 0

OP PP PQ

184


Example 8. Find the rank of

OP PP PQ

LM 6 16 A= M MM 5 N4

1 3 8 4 12 15 . 3 3 8 2 6 –1

Sol. Applying C1 ↔ C2, C3 →

1 C 3 3 6 16 5 4

~

LM1 MM4 MN32

4 8 –1 0

~

LM1 MM2 MN32

4 0 8 0 0 1 1 2 4 0 0 1

LM1 MM12 MN0

4 0 –1 0

A ~

1 4 1 2

By C2 → C2 – 2C1, C3 → C3 – C1, we have

By R2 → R2 – 2R1

0 0 –2 0

~

0 0 –2 0

⇒ |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

8 15 8 –1

OP PP PQ

8 15 4 –1

OP PP PQ

R3 → R3 – R2, R4 → R4 – R2, we get

∴

OP PP PQ

LM1 MM4 MN32

OP PP PQ

8 –1 5 0

= 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 4 A= M MM a N9

OP PP PQ

1 –1 0 4 –3 1 . 2 2 2 9 a 3

185

MATRICES

Sol.

A =

LM1 MM4 MN9a

1 4 2 9

OP PP PQ

–1 –3 2 a

0 1 2 3

Applying R2 → R2 – 4R1, R3 → R3 – 2R1, R4 → R4 – 9R1, we have

OP PP PQ

LM 1 MMa –0 2 MN 0

1 0 0 0

–1 1 4 a+ 9

0 1 2 3

~

LM 1 MM 0 MNa –0 2

1 0 0 0

–1 1 0 a+6

0 1 –2 0

~

LM 1 MM 0 MNa 0– 2

1 0 0 0

–1 1 a+6 0

0 1 0 –2

A ~

Again R3 → R3 – 4R2, R4 → R4 – 3R2

R4 ↔ R3

OP PP PQ OP PP PQ

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.

A =

LM1 MM0 Nb

5 3 13

OP P 10PQ 4 2

is 2.

(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


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 MN

3.

5.

7.

9.

OP PP Q

LM1 MM1 N2 LM 1 MM 43 MN 1 LM1 MM23 MN 4 LM1 MM23 MM 5 MN1

2 3 4 2 . 6 5

OP PP PQ

LM MM MM MN

OP PP PQ

LM9 4. M 5 MN6 LM0 M0 6. M0 MN0

[Ans. 2]

2 –1 3 1 2 1 . [Ans. 3] –1 1 2 2 0 1 (U.P.T.U. special exam., 2001) 2 3 4 5

3 4 5 6

4 5 6 7

2 1 2 1 –1 2 6 3 3 –1

OP PP PQ 0O 2 PP 5 P. P 2P –3PQ 5 6 . 7 8

[Ans. 3]

[Ans. 3]

Find the rank of the following matrix.

11.

13.

15.

LM1 2 MM2 3 MM 34 45 N LM 3 –2 MM 0 2 MN 10 –21 LM 1 2 MM 2 4 N–1 2 2

2

2

2

2

2

2

2

32 42 52 62 0 2 –3 2

OP P. 6 P P 7 PQ 42 52

[Ans. 3]

2 2

OP PP PQ

–1 –7 1 –5 . –2 1 1 –6

OP PP Q

–1 4 3 5 . 6 –7

[Ans. 4]

[Ans. 2]

10.

LM2 M0 12. M2 MN2

OP PP Q

7 3 6 –1 4 1 . 8 2 4 0 1 2 3

0 2 3 4

0 3 4 1

[Ans. 4]

[Ans. 3]

OP PP PQ

0 4 . 1 2

[Ans. 3]

OP PP Q

[Ans. 3]

1 2 2 3 . –1 –1

OP PP Q

[Ans. 2]

OP P. 5P P 6Q

[Ans. 3]

LM –1 8. M 1 MN –1 LM1 MM1 N0

OP PP PP PQ

9 1 –1 . 9 9

2 –2 2 1 . –1 0

1 3

3 4

3 7 5 11

4 1

LM1 2 3OP 14. M2 3 1P. MN3 1 2PQ LM 0 i –iOP 16. M – i 0 – i P. MN–3 1 0 PQ

[Ans. 3]

[Ans. 3]

187

MATRICES

17.

19.

LM 3 –1 MM –6 2 N3 1 LM1 2 3 MM2 4 6 N1 2 3

OP PP Q 1O P 2P. 2PQ

–2 –4 . –2

[Ans. 2]

[Ans. 2]

LM0 2 3OP 18. M0 4 6P. MN0 6 9PQ LM 4 2 3 OP 20. 8 MM 4 6 PP. N –2 –1 –15Q

[Ans. 1]

[Ans. 1]

Find the Echelon form of the following matrix and hence find the rank. 21.

23.

25.

27.

LM1 –2 3OP MM2 –1 2PP. N 3 1 2Q LM 3 4 5 6 7 OP MM 45 65 76 78 98 PP MM10 11 12 13 14PP. MN15 16 17 18 19PQ LM 5 6 7 8 OP MM116 127 138 149 PP. MN16 17 18 19PQ LM 5 3 14 4OP MM0 1 2 1PP. N1 –1 2 0Q

[Ans. 3]

[Ans. 2]

[Ans. 3]

[Ans. 3]

LM 1 22. M –4 MN 6

OP PP Q

2 –5 1 –6 . 3 –4

LM2 3 –1 –1OP M1 –1 –2 –4P 24. M 3 1 3 –2 P. MN6 3 0 –7 PQ LM–2 –1 –3 –1OP M 1 2 –3 –1P 26. M 1 0 1 1 P. MN 0 1 1 –1PQ LM0 1 3 –2OP M0 4 –1 3 P 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. A=

LM3 MM 5 N1

2

–1

1 4 –4 11

LM MM LM 0 MMAns. P = M 01 MM MM N2 MN

OP –2 P. –19PQ 5

0 13 –1 3

OP P 16P PQ

1 –5 3

LM1 MM0 Q=M MM0 MM0 N

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


30. A =

31.

LM2 MM 3 N1

LM LM 0 0 MMAns. P = M –1 0 MM –3 1 MM N 14 28 N LM1 1 2 OP A = M1 2 3 P. MN0 –1 –1PQ

32. A =

LM 1 MM 2 N3

OP PP Q

1 –3 –6 –3 1 2 . 1 1 2

OP PP Q

2 3 –2 –2 1 3 . 0 4 1

(U.P.T.U., 2002)

LM MM MN

OP PP PQ

–1

1 1 0 2 ,Q= 0 9 0 28

1 0 0

LM MMAns. P = LM–21 MM MM N–1 N

4 –5 1 0

0 1 –1

OP PP PP Q

OP PP PQ

0 0 and rank = 3 –2 1

LM LM 1 0 MMAns. P = M–1 1 MN–1 1 N LM1 1 3 0O P M 0 P , Q = M0 – 1 6 MM0 0 1PQ N0 0

OP LM1 0 P, Q = M0 MN0 1 PQ 0

–4/ 3 –5 6 1 0

1 3 76 0 1 –

–1 1 0

OPOP –1P P 1 PQ PQ

–1

OP OP PP PP rank = 2 PP PP QQ

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 MM a.... MNa 21

a12 a22

m1

.... am2

11

where

A =

X =

OP P .... P P a Q a1n a2n

.... ....

mn

LM x OP LM b OP MM xM PP and B = MM bM PP MNx PQ MNb PQ 1

1

2

2

n

n

189

MATRICES

Augmented matrix: The augmented matrix [A : B] or Ã of system (i) is obtained by augmenting A by the column B.

i.e.,

LM a MM a MNaL

Ã = [A : B] =

11

a12

L

a1n

M

21

a22

L

a2n

M

L

m1

L am2

L amn

M M

OP P LP P b Q b1

b2 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. Sol. Here,

The augmented matrix

A =

[A : B] =

Applying R1 ↔ R3, we get [A : B] ~ Again R2 → R2 – 2R1, R3 → R3 – 7R1 ~

LM7 MM2 N1 LM7 MM2 N1

2 11 3

OP PP Q

LM MM N 16O P 25P 13PQ

OP PP Q

LM MM N

OP PP Q

x1 3 16 5 , X = x2 , B = 25 4 x3 13

2 11 3

3 M 5 M 4 M

LM1 MM 2 N7

3 11 2

4 5 2

LM1 MM0 N0

3 5 –19

M M M

OP PP Q

13 25 16

4 M –3 M –25 M

OP PP Q

13 –1 –75

190


R3 → R3 +

19 R2 5

~

LM1 MM0 MM0 N

=

LM 13 OP MM –1 PP P MN –394 5 Q

3

M

4

OP –1 PP 394 P – 5 PQ 13

–3 M –182 0 M 5 ⇒ 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 M P –3 P Mx P –182 P M P PM P 5 Q Nx Q

3

4

5

2

0

LM x MM MM N

1

3

1

+ 3x 2 + 4x 3 5x 2 – 3x 3 –182 x3 5

OP PP PP Q

5

LM 1 3 OP MM – 1 PP MM – 39 4 PP N 5 Q

=

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 =

Sol. The augmented matrix [A : B] = Applying R1 ↔ R2, we get [A : B] ~

LM –1 MM 4 N0

LM 4 MM–1 N0

–1 0 M 12 5 –2 M 0 –2 4 M –8

5 –2 M 0 –1 0 M 12 –2 4 M –8

OP PP Q

OP PP Q

191

MATRICES

R2 → R2 + 4R1 ~

R1 → – R1, R3 → R3 +

2 R, 19 2

~

LM –1 MM 0 N0 LM1 MM0 MN0

5 –2 M 0 19 –8 M 12 –2 4 M –8

–5

2

M

19

–8 60 19

M

0

M

OP PP Q 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 P –8 P 60 P 19 PQ 2

LM x OP MMx PP MNx PQ 1

2 3

=

LM 0 OP MM 12 PP MN –128 P 19 Q

x1 – 5x2 + 2x3 = 0

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

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

...(i) ...(ii) ...(iii)

192


Example 3. Solve 2x1 – 2x2 + 4x3 + 3x4 x1 – x2 + 2x3 + 2x4 2x1 – 2x2 + x3 + 2x4 x1 – x2 + x4 Sol. The augmented matrix is

[A : B] =

LM 2 MM1 MN12

–2 –1 –2 –1

~

LM1 MM2 MN21

–2 –2

R1 ↔ R2

= = = =

4 2 1 0

–1 2

9 6 3 2

M M M M

3 2 2 1

OP PP PQ

9 6 3 2

OP PP P 2Q

2 M 6

4 3 M 9 1 2 M 3

–1 0

1 M

R2 → R2 – 2R1, R4 → R4 – R1, R3 → R3 – 2R1

~

LM1 MM0 MN00

2

2

0

–1 M –3

0 0

–3 –2 M –2 –1 M

R2 → (– 1) R2, R3 → (– 1) R3, R4 → (– 1) R4

~

LM1 MM0 MN00

~

LM1 MM0 MN00

R3 → R3 – R4

R3 ↔ R2

~

LM1 MM0 MN00

OP P –9 P P –4 Q

−1 0

M

6

OP P 9P P 4Q

–1 2

2 M 6

0

0

1 M 3

0 0

3 2

2 M 1 M

OP 3P 5P P 4Q

–1 2 2 M 6 0 0

0 1 M 1 1 M

0

2 1 M

OP 5P 3P P 4Q

–1 2 2 M 6 0 0

1 1 M 0 1 M

0

2 1 M

193

MATRICES

R4 ↔ R3

~

R3 → R3 – 2R2

~

LM1 MM0 MN00 LM1 MM0 MN00

OP 5P 4P P 3Q

–1 2 2 M 6 0 0

1 1 M 2 1 M

0

0 1 M

–1 0 0 0

2 2 1 1 0 −1 0 1

OP PP PQ

LM MM MN

M 6 1 –1 2 2 M M 5 0 0 1 1 M ~ M –6 0 0 0 1 M M 3 0 0 0 1 M

OP PP b PQ

6 5 R3 → – R3 6 3

g

R4 → R4 – R3 and then R4 → (– 1) R4

~

LM1 MM0 MN00

OP 5P 6P P 3Q

–1 2 2 M 6 0

1 1 M

0 0

0 1 M 0 0 M

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

OP PP Q

LM1 MM1 N1

1 1 M 6 2 3 M 10 2 λ M µ

~

LM1 MM0 N0

1 1 M 6 1 2 M 4 1 λ–1 M µ–6

~

LM1 MM0 N0

1 1 M 6 1 2 M 4 0 λ – 3 M µ – 10

[A : B] = R2 → R2 – R1, R3 → R3 – R1

R3 → R3 – R1

OP PP Q OP PP Q

(i) For no solution ρ (A) ≠ ρ [A; B] it is only possible when λ = 3.

(U.P.T.U., 2001)

194


(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

OP PP Q

LM1 MM1 N1

1 1 M 6 2 3 M 14 4 7 M 30

~

LM1 MM0 N0

1 1 M 6 1 2 M 8 3 6 M 24

~

LM1 MM0 N0

1 1 M 6 1 2 M 8 0 0 M 0

[A : B] = R2 → R2 – R1, R3 → R3 – R1

R3 → R3 – 3R2

OP PP Q

OP PP Q

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 MM0 N0

OP LMx OP PP MM y PP 0Q N z Q

1 1 1 2 0

=

LM6OP MM8PP 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

2 0 3 −1

: : : :

0 2 3 −1

: : : :

OP PP PQ 4O 1P P −2P P 5Q

0 2 3 −1

: : : :

4 −11 −2 −3

1 4 −2 5

R2 → R2 – 3R1, R4 → R4 – 2R1

~

R3 → R3 +

2 −3 10 −7

OP PP PQ

10 7 R , R → R4 – R2 3 2 4 3

~

R3 →

LM1 MM0 MN00

3 3 R3, R4 → R 29 17 4 ~

R4 → R4 + R3 ~

LM1 MM0 MM0 MM0 N


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


⇒

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 MM00 MN0

⇒

2 −3 0 0

OP LxO PP MMyPP PQ MNz PQ

0 2 1 0

=

LM 4OP MM−−114PP 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

[A : B] =

LM1 MM32 MN1

2 −1 −2 −1

−1

2 3 1

OP PP PQ

: : : :

3 1 2 −1

R2 → R2 – 3R1, R3 → R3 – 2R1, R4 → R4 – R1

~

R3 → R3 –

LM1 MM00 MN0

2 −7 −6 −3

−1

5 5 2

OP PP PQ

: : : :

3 −8 −4 −4

6 3 R2, R4 → R4 – R 7 7 2

~

LM1 MM0 MM0 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 MM0 MM0 N0

2 −1 −7

0 0

:

5 5 7 0

: :

OP −8P 20 P 7 PP 0Q 3

ρ (A) = ρ [A : B] = 3 = number of variables.

Hence, the system is consistent and has a unique solution. Now, AX = B

⇒

LM1 MM0 MM0 N0

⇒

OP x 5P LM OP 5 P M yP 7 PP MN z PQ 0Q

2 −1 −7

0 0

=

LM 3 OP MM −208 PP MM 7 PP MN 0 PQ

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 MM3 N7

2 4 10

OP PP Q

3 4 12

198


R3 → R3 – 2R2

OP PP Q

LM1 ~ M0 MN0

2 −2 −4

3 −8 −9

LM1 A ~ M0 MN0

2 −2 0

3 −5 1

OP PP Q

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 A = M MM4 N5

1 −2 1 −7

−2

1 −5

2

OP PP PQ

3 −1 8 −1

R2 → R2 – R1, R3 → R3 – 4R1, R5 → R5 – 5R1

R3 → R3 – R2, R4 → R4 – 4R2

LM1 0 ~ M MM0 N0

1 −3 −3 −12

LM1 0 ~ M MM0 N0

1 −3 0 0

−2

3 3 12 −2

3 0 0

OP PP PQ

3 −4 −4 −16

OP PP PQ

3 −4 0 0

⇒ ρ (A) = 2 < 4 (n = 4), so there exist a non-trivial solution. Now,

⇒

LM1 MM00 MN0

AX = B

1 −3 0 0

−2

3 0 0

OP LMx OP PP MMyz PP PQ MNwPQ

3 −4 0 0

LM 0 OP 0 = M P MM 0 PP N0Q

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 A = M4 MN 6

2 2λ 6

OP PP Q

−2 −1 λ

For unique solution |A| ≠ 0 ∴ ⇒

LMλ MM 4 N6

2 2λ 6

OP PP Q

−2 −1 λ

= λ3 + 11λ – 30 ≠ 0

(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


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:

OP PP Q

[A : B] = 1

LM2 MM N3

3 2 1

1 3 2

: : :

9 6 8

LM1 ~ M2 MN3

2 3 1

3 1 2

: : :

6 9 8

LM1 ~ M0 MN0

2 −1 −5

3 −5 −7

: : :

6 −3 −10

LM1 ~ M0 MN0

2 −1 0

3 −5 18

: : :

6 −3 5

R1 ↔ R2

OP PP Q

R2 → R2 – 2R1, R3 → R3 – 3R1

R3 → R3 – 5R2

OP PP Q

OP PP Q

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

[A : B]

LM 10 −6 = M MM 3 N5

−7

8 1 −9

OP PP PQ

3 −1 4 −2

5 −4 11 4

: : : :

6 5 2 7

3 4 −1 −2

5 11 −4 4

: : : :

6 2 5 7

R3 ↔ R2

LM 10 M3 ~ M−6 MN 5 R2 → R2 –

1 8 −9

OP PP PQ

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

−

−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


R3 → R3 – R2, R4 → R4 – R2

~

R3 → −

LM10 MM 0 MM MM 0 MM 0 N

−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

~

R4 → R4 – R3

~

LM10 MM 0 MM MM 0 MM 0 N LM10 MM 0 MM 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

LM OP MM PP NQ

A =

1 3 1 4 7 , B= 1 5 9 3

[A : B] =

LM2 MM24 N

1 3 : 1 4 7 : 1 5 9 : 3

~

LM2 MM00 N

1 3 : 1 2 1 : −1 4 6 : 2

~

LM2 MM00 N

1 3 : 1 2 1 : −1 0 4 : 4

Sol. Here ∴ Augmented matrix

R2 → R2 – 2R1, R3 → R3 – R1

R3 → R3 – 2R2

R1 →

OP PP Q

LM2 MM24 N

OP PP Q OP PP Q

OP PP Q

1 1 1 R , R → R2 , R3 → R3 2 1 2 2 4

~

R1 → R1 −

1 R 2 2

~

LM1 MM MM0 MN0

LM1 MM MM0 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


R1 → R1 –

5 1 R3 , R2 → R2 – R3 4 2 ~

LM1 MM0 MN0

OP PP 1 PQ

1 2 −1

0 0 : − 1 0 : 0 1 :

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.

[A : B] =

LM2 MM34 MN5

5 2 −3 6 5 2 5 14 14 10 8 4

OP PP PQ

: : : :

3 2 11 4

R2 → R2 – R1, R3 → R3 – R2, R4 → R4 – R3

~

R1 ↔ R4, R2 ↔ R3

~


5 2 −3 1 3 5 −1 9 12 5 −6 −10 5

−6 −10 :

−1

9 3 2

1 5

12 5 −3

OP PP PQ −7O 9 PP −1P P 3Q

: 3 : −1 : 9 : −7

: : :

R2 → R2 – R1, R3 → R3 – R1, R4 → R4 – 2R1

~

LM1 MM00 MN0

5 −6 −10 −6 15 22 −4 9 15 −5 14 17

: −7 : 16 : 6 : 17

OP PP PQ

R2 → R2 – R4, then R2 → – R2, R3 → R3 – 4R2, then again R3 → –R3 , R4 → R4 – 5R2 , then also again R4 → –R4

~

LM1 MM00 MN0

OP PP PQ

5 −6 −10 : −7 1 −1 −5 : 1 0 −5 5 : −10 0 9 −8 : 22

205

MATRICES

R3 → –

1 R , then R4 → R4 – 9R3 5 3

~

LM1 MM00 MN0

5 −6 −10 : −7 1 −1 −5 : 1 0 1 −1 : 2 0 0 1 : 4

R3 → R3 + R4, R2 → R2 + 5R4, R1 → R1 + 10R4

~

LM1 MM00 MN0

5 −6 1 −1 0 1 0 0

0 0 0 1

~

R1 → R1 – 5R2

~

LM1 MM00 MN0

5 0 0 : 1 0 0 : 0 1 0 : 0 0 1 : 0 1 0 0

0 0 1 0

0 0 0 1

OP PP PQ

: 33 : 21 : 6 : 4

R2 → R2 + R3, R1 → R1 + 6R3

LM1 MM0 MM0 N0

OP PP PQ

OP 27P P 26P 4 PQ

69

OP PP PQ

: −66 : 27 : 6 : 4

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 A =

LM 3 MM 4 N2λ

1 −λ −2 −3 4 λ

OP PP Q

for non-trivial solution A = 0 ∴

A =

LM 3 MM 4 N2λ

OP PP Q

1 −λ −2 −3 = 0 ⇒ 3 (–2λ + 12) – (4λ + 6λ) – λ (16 + 4λ) = 0 4 λ

⇒ – 6λ + 36 – 10λ – 16λ – 4λ2 = 0 ⇒ – 4λ2 – 32λ + 36 = 0 ⇒ λ2 + 8λ – 9 = 0

206


⇒ ⇒

λ2 + 9λ – λ – 9 = 0 ⇒ (λ – 1) (λ + 9) = 0 λ = 1, – 9.

Case I. If λ = 1,

A =

R2 → R2 –

LM 3 MM 4 NM 2

OP −3 P P 1 PQ

1 −1 −2

4

4 2 R , R → R3 – R1 3 1 3 3

~

LM3 MM0 MM0 N

~

LM3 MM0 MN0

R3 → R3 + R2

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

Case II. If λ = – 9 A =

k k –k=0⇒x= . Ans. 2 2

⇒ 3x –

3x + y – z = 0

LM 3 MM−418 N

OP PP Q

1 9 −2 −3 4 −9

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

10y + 3z = – 2 2x – 3y – z = 5 3.

4.

5.

6.

x1 + 2x2 − x3 3x1 – x2 + 2x3 2x1 – 2x2 + 3x3 x1 – x2 + x3

= = = =

3 1 2 – 1

[Ans. x = 2, y = 1 and z = – 4]

(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.

x + y + 3z x–y+z x – 2y x–y+z

= = = =

[Ans. x = y = z = 0]

0 0 0 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.


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. 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.

4x +

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 N

=

1

29.

OP Q

2x1 + x2 + 2x3 + x4 = 6 6x1 – 6x2 + 6x3 + 12x4 = 36 4x1 + 3x2 + 3x3 – 3x4 = –1 2x1 + 2x2 – x3 + x4 2x1 – 7x2 + 4x3 x1 + 9x2 – 6x3 – 3x1 + 8x2 + 5x3

= = = =

10 9 1 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.

117 81 148 , x2 = − , x3 = 71 71 71

[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 = x+z = 4x + 4y + 3z = 31. Test the consistency and solve 2x

– 2 0 – 3 [Ans. x = 1, y = 1, z = – 1] – 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 x – 4y – 3z 2x + 7y + 12z 5x – 5y + 3z

= = = =

16 – 16 48 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 N

1

= 2, x2 =

1 4 , x3 = 0, x4 = 5 5

OP Q

LMA ns. x = 7 , y = 3 , z = 0OP 11 11 N Q

210


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 A = R1 ↔ R2 ~ R2→ R2 – 3R1, R3 → R3 + 4R1 ~

R2 → –

1 1 R, R → R 4 2 3 5 3 ~

LM 3 MM 1 N− 4

LM 1 MM 3 N− 4

OP − 4P P 1Q

2

0

2 −3

OP 0P P 1Q

2 −4 2 −3

OP 12P P −15Q

LM1 MM0 N0

−4

LM1 MM00 N

2 −4 1 −3 1 −3

−4

2 5

OP PP Q LMx OP MMx PP MNx: PQ 1

* An ordered set of n numbers belonging to a field F and denoted by X = [x1, x2, ... xn] or an n-dimensional vector over F.

2

n

is called

211

MATRICES

R3 → R3 – R2 ~

and

⇒

ρ(A) = 2 and n = 3

Let

a3 = k

LM1 MM00 N

OP PP Q

2 −4 1 −3 0 0

a2 – 3a3 = 0 ⇒ a2 – 3k = 0 ⇒ a2 = 3k Again a1 + 2a2 – 4a3 = 0 ⇒ a1 + 6k – 4k = 0 ⇒ a1 = – 2k Therefore,

a1 = – 2k, a2 = 3k, a3 = k.

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 0 |W

a1 + 2a2 + 3a3 = 0 – 2a1 – a2 – 5a3 = 0

λa1 + 5a2 + 7λa3 = The system is homogeneous

...(i)

∴ 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


– 2a1 + a2 + a3 = 0 Now

LM 0 MM−12 N

A =

R1 ↔ R2

LM 1 MM−02 N

A ~ R3 → R3 + 2R1

LM1 MM00 N

~ R3 → R3 + R2

Now

OPLM OP 1P M a P 6PQ MNa PQ

−1 2 a1

1 0

2

=

3

⇒

OP 1P 1PQ

−1 2

1 1

OP 1P 5QP

−1 2

1 −1

LM1 −1 MM00 10 N LM0OP MM00PP NQ

~

LM1 MM00 N

OP PP Q

1 1 −1 2 1 1

OP PP Q

2 1 = ρ (A) = 3 6

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. Sol. Here X1 Now ⇒

LM1OP = M1P , X MN0PQ

2

⇒

3

LM5OP = M 3P MN3PQ

a1X1 + a2X2 + a3X3 = O

LM1OP LM3OP LM5OP a M1P + a M1P + a M 3P MN0PQ MN3PQ MN3PQ LMa + 3a + 5a OP MM 0a ++3aa ++ 33aa PP Q N 1

2

1

⇒

LM3OP = M 1P , X MN3PQ

1

3

2

=

3

2

3

2

3

=

LM0OP MM00PP NQ LM0OP MM00PP NQ

a1 + 3a2 + 5a3 = 0 a1 + a2 + 3a3 = 0 0.a1 + 3a2 + 3a3 = 0

213

MATRICES

∴

A =

3 5 1 3 3 3

A ~

LM1 MM00 N

3 5 −2 −2 3 3

~

LM1 MM00 N

−2 −2

R1 → R2 – R1

R3 → R3 +

OP PP Q

LM1 MM10 N

3 R 2 2

OP PP Q

3 0

OP P 0 PQ 5

Here ρ(A) = 2 but n = 3 so let a3 = k.

LM1 MM00 N

and

OP PP Q

3 5 −2 −2 0 0

LM a OP MMaa PP N Q 1 2

=

3

LM0OP MM00PP NQ

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

∴

A =

LM2 MM13 MN1

2 3 1 2

OP PP PQ

4 6 , R1 ↔ R4, R2 ↔ R3 3 2

LM1 1 ~ M MM3 N2

2 1 3 2

OP PP PQ

2 3 6 4

214


R2 → R2 – R1, R3 → R3 – 3 R1, R4→ R4 – 2R1

R4 → R4 −

~

LM1 MM00 MN0

2 −1 −3 −2

~

LM1 MM00 MN0

2 2 −1 1 ⇒ ρ(A) = 3 and n = 3 0 −3 0 0

2 R 3 3

OP PP PQ

LM MM MN

OP PP PQ

2 1 2 2 1 0 −1 1 , R3 → R3 – 3R2, R4 → R4 – 2R2 ~ 0 0 0 −3 0 0 0 −2

OP PP PQ

∴ 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. 5.

6. 7. 8.

3.17

[Ans. a = 0, 2 , – 2 ] 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] Show that the vectors X1 = [2, i, – i]. X2 = [2i, – 1, 1], X3 = [1, 2, 3] are linearly dependent. If X1 = [1, 1, 2], X2 = [2, – 1, – 6], X3 = [13, 4, – 4] prove that 7X1 + 3X2 – X3 = 0. 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.

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


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


∴ 0 = λ |I| = 0 ⇒ λ = 0 Conversely, if λ = 0 then |A| = 0. Proved.

OP PP Q

LM 1 Example 2. Find the characteristic equation of the matrix 0 MM−1 N

values and eigen vectors of this matrix. Sol. Let

A =

A – λI =

Now,

LM 1 MM−01 N LM 1 MM−01 N

OP PP Q L1 2O P1P – λ MM0 MN0 2PQ

2 2 2 1 . Also find the eigen 2 2 (U.P.T.U., 2006)

2 2 2 1 2 2 2 2 2

OP PP Q

LM MM N

0 0 1− λ 2 2 1 0 = 0 2− λ 1 0 1 −1 2 2− λ

OP PP Q

∴ Characteristic equation is |A – λI| = 0 ⇒

LM1 − λ MM 0 N −1

OP P 2 − λ PQ

2

2 1

2− λ 2

m

⇒ (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 MM −01 N ⇒ R1 ↔ R3

OP LMx OP PP MMxx PP QN Q 2O L x O 1PP MMx PP 1PQ MNx PQ

2 2 2−1 1 2 2−1

LM 0 MM−01 N LM−1 ~ M0 MN 0

2 1 2

1 2

= 0

3

1 2

= 0

3

OP LMx OP PP MMxx PP QN Q

2 1 1 1 2 2

1 2

3

r

= 0 ⇒ (1 – λ) ( 2 − λ ) 2 − 2 − 2{0 + 1} + 2{0 + ( 2 − λ )} = 0

= 0

...(i)

219

MATRICES

R3 → R3 – 2R2, R1 → (– 1) R1

LM1 ~ M0 MN0

⇒ Let and

OP LMx OP 1 P Mx P 0 PQ MNx PQ

−2 −1

1 0

1 2

= 0

3

x1 – 2x2 – x3 = 0 x2 + x3 = 0 x3 = k then x2 = – k x1 + 2k – k = 0 ⇒ x1 = – k

∴

X1 =

LM 1OP LM 1OP LM−kOP MM−kPP = −k MM 1PP or MM 1PP N−1Q N−1Q N kQ

Taking λ = 2, from (1), we get

R2 ↔ R3

LM−1 MM−01 N

2 2 0 1 2 0

L−1 ~ M−1 MM 0 N

2 2 2 0 0 1

R2 → R2 – R1, R1 → –R1

LM1 ~ 0 MM0 N

R2 → –

OP LMx1 OP PP MMxx2 PP Q N 3Q

= 0

OP PP Q

LM x OP MMxx PP N Q

= 0

OP P 1 PQ

LM x OP MMxx PP N Q

= 0

−2 −2 0 −2

0

1 2 3

1 2 3

1 R and then R3 → R3 – R2, we get 2 2

LM1 ~ M0 NM0

OP 1P 0 QP

−2 −2

0 0

LM x OP MMxx PP N Q 1 2

= 0

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 MMxx PP N Q 1

Hence,

X2 =

2 3

LM2kOP L2O L2O MP MP = M k P = k M1P or M1P MN 0 PQ MN0PQ MN0PQ

220


Thus only one eigen vector X2 corresponds to the repeated eigen value λ = 2. Hence the eigen values are 1, 2, 2

eigen vectors

X1

or

X1

LM2OP LM 1 OP = – k 1 , X = X = k M1P MM−1PP MN0PQ N Q LM2OP LM 1 OP = 1 , X = X = M1P MM−1PP MN0PQ N Q 2

2

3

3

Example 3. Find the eigen values and eigen vectors of the matrix

A =

LM2 MM10 N

OP PP Q

1 1 2 1 0 1

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 MM 10 N

OP PP Q

1 1 −1 1 0 −2

LM x OP MMxx PP N Q 2

LMx OP let X = Mx P MM PP Nx Q 1

1

= 0

3

2 3

Applying R2 → R2 + R1, then R1 → – R1 on coefficient matrix, we get

OP 2P −2QP

LM x OP MMxx PP N Q

R3 → R3 + R2, R2 →

1 R 2 2

LM1 ~ M0 NM0 Again

LM1 ~ 0 MM0 N ⇒

−1 −1

0 0

OP 1P 0 PQ

−1 −1

0 0

1 2

= 0

3

LM x OP MMxx PP N Q 1 2

= 0

3

x1 – x2 – x3 = 0

221

MATRICES

x3 = 0 Suppose x2 = k, then x1 = k

Here,

For

X1 =

r = 2, n = 3 n− r = 1

LMk OP LM1OP LM1OP MMk PP = k MM1PP or MM1PP MN0PQ MN0PQ MN0PQ

λ=1

LM1 MM1 MN0 R2 → R2 – R1

LM1 ~ M0 MM N0

= 0

0

OP LMx OP 1P Mx P PM P 0PQ MNx PQ

1

1

OP LMx OP 0P Mx P PM P 0PQ MNx PQ

= 0

1 1

0 0

1

1

2 3

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 MM k PP . MN k PQ 1

⇒

X2 =

2

1

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 MM 0 PP = − k MN k PQ 2

Then

X2 =

2

2

LM 1 OP LM 1 OP MM 0 PP or MM 0 PP MN−1PQ MN−1PQ

(b) If we suppose x2 = k1 (any arbitrary) and x3 = k2 = 0

LM− k OP MM k PP = − k MN 0 PQ 1

Then

X3 =

1

1

LM 1OP LM 1OP MM−1PP or MM−1PP MN 0PQ MN 0PQ

222


Hence the eigen values are λ = 1, 2, 2

Eigen vectors are

X1 =

or

X1 =

LM1OP LM 1OP LM 1OP 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 MM1PP; X2 = MM 0 PP; X3 = MM−1PP . N0Q N−1Q N0Q 2

2

3

1

Example 4. Find the eigen values and eigen-vectors of matrix

A =

LM3 MM00 N

OP PP Q

1 4 2 6 . 0 5

[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 MM 00 N ⇒

R3 → R3 –

1

∴

OP PP Q 4O 6PP 3PQ

2− 2 0 5− 2

LM1 MM00 N

1 0 0

LM x OP MMxx PP N Q LM x OP MMxx PP N Q 1 2

LM x OP X= x MMx PP N Q 1

= 0

3

2 3

1 2

= 0

3

1 R on coefficient matrix 2 2

LM1 ~ M0 MN0 ⇒ Let

4 6

...(i)

OP PP Q

1 4 0 6 0 0

LM x1 OP MMxx2 PP N 3Q

= 0

x1 + x2 + 4x3 = 0 x2 = k X1 =

and x3 = 0 then x1 + k + 0 = 0 ⇒ x1 = –k

LM−kOP LM 1OP LM 1OP MM k PP = −k MM−1PP or MM−1PP N 0 Q N 0Q N 0Q

223

MATRICES

For λ = 3, from (i), we get

LM0 MM0 MN0

1 −1 0

OP LMx OP 6 P Mx P PM P 2PQ MNx PQ 4

1

2

= 0

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 PM P 2 PQ MNx PQ

= 0

OP LMx OP 10 P Mx P PM P 0 PQ MNx PQ

= 0

4

1

2 3

4

1

2 3

x2 + 4x3 = 0 and 10x3 = 0 ⇒ x3 = 0 x2 + 0 = 0 ⇒ x2 = 0, let x1 = K

⇒ ⇒

∴

X2 =

Again, for λ = 5, we get

⇒

LMk OP LM1OP LM1OP MM0PP = k MM0PP or MM0PP MN0PQ MN0PQ MN0PQ 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 ∴

X3 =

Hence the eigen values are λ = 2, 3, 5 And eigen vectors are

X1 =

LM3kOP LM3OP LM3OP MM2kPP = k MM2PP or MM2PP MN k PQ MN1PQ MN1PQ LM 1 OP MM−1PP , X MN 0 PQ

2

LM1OP = M0 P , X MM PP N0 Q

3

LM3OP = M 2P . MM PP N1 Q

224


Example 5. Determine the latent roots and the corresponding vector of the matrix, A=

LM 6 MM−2 N2

OP PP Q

−2 2 3 −1 . −1 3

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

2 1 3− λ

=0

Now, C3 → C3 + C2 gives (2 – λ) ∴

6−λ 0 2

−2 1 −1

0 2 2− λ

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 MM −2 N2

⇒

OP PP Q 2O −1PP −5PQ

LMx OP MMx PP Nx Q LMx OP MMx PP Nx Q

OP PP Q

LMx OP MMx PP Nx Q

OP PP Q

LM MM N

−2 2 3 − 8 −1 −1 3 − 8

LM−2 MM−2 N2

−2 −5 −1

LM−2 MM 0 N0

−2 2 −3 −3 −3 −3

1 2

=

3 1 2

=

3

LM0OP MM0PP N0 Q LM0OP MM0PP N0 Q

R2 → R2 – R1, R3 → R3 + R1 on coefficient matrix

R3 → R3 – R2 and R1 →

⇒

q

= ( 2 − λ ) ( 6 − λ ) ( 2 − λ + 2) − 8 = 0

LM1 MM0 N0

1 1 0

1 2 3

=

LM0OP MM0PP N0 Q

−1 −1 R1 , R2 → R2 2 3 −1 x1 0 1 x2 = 0 0 x3 0

OP PP Q

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 MMx PP = Nx Q 1

∴

X1 =

3

For λ = 2, the equation (i) gives

⇒

LM6 − 2 −2 2 OP MM −2 3 − 2 −1 PP N 2 −1 3 − 2Q LM 4 −2 2OP MM−22 −11 −11PP N Q

R2 → R2 +

2

LMx OP MMx PP Nx Q LM x OP MMxx PP N Q 1 2

=

3 1 2

=

3

LM 2k OP LM 2OP LM 2OP MM− kPP = k MM−1PP or MM−1PP N k Q N 1 Q N 1Q

LM0OP MM0PP N0 Q LM0OP MM00PP NQ

1 1 R1 , R3 → R3 – R1 , we get 2 2

LM4 MM00 N

−2 2 0 0 0 0

OP PP Q

LMx OP MMx PP Nx Q

LM0OP MM0PP N0 Q

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

OP PP PQ

OP PP Q

LM MM N

OP PP Q

LM MM N

LM MM N

OP PP Q

LM MM N

OP PP Q

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 X1

LM 2OP LM 2OP = k M−1P or M−1P, X MN 1PQ MN 1PQ

2

k = 2 2

LM 0OP LM 0OP MM 2PP or MM 2PP, X N−1Q N−1Q

3

k = 1 2

LM2OP LM2OP MM0PP or MM0PP . N1Q N1Q

226


Example 6. Find the eigen values of matrix A.

1 A = 3

LM1 MM2 N2

Sol. The characteristic equation of B is, where

B =

LM1 MM2 N2

OP PP Q

2 1 −2

2 2 1 −2 −2 1

2 −2 · 1

OP PP Q

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 =

A =

LM 3 MM−32 N

10 5 −3 −4 5 7

OP PP Q

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 MM −2 −3− 3 N3 5 LM 0 10 MM−2 −6 N3 5

OP PP Q 5O − 4PP 4PQ

5 −4 7− 3

LMx OP MMx PP Nx Q LMx OP MMx PP Nx Q 1 2

= 0

3

1 2 3

= 0

...(i)

227

MATRICES

R1 →

1 1 R1 , R2 → – R2 , we have 5 2

LM0 MM1 N3 LM1 MM03 N

R1 ↔ R2

OP PP Q 2O 1P P 4PQ

LMx OP MMx PP Nx Q LM x OP MMxx PP N Q

OP PP Q

LMx OP MMx PP Nx Q

OP PP Q

LMx OP MMx PP Nx Q

2 1 3 2 5 4 3 2 5

1 2

= 0

3

1 2

= 0

3

R3 → R3 – 3R1 on coefficient matrix

R3 → R3 + 2R2

LM1 MM0 N0

3 2 2 1 −4 −2

LM1 MM0 N0

⇒

3 2 2 1 0 0

1 2

= 0

3

1 2

=

3

LM0OP MM0PP N0 Q

x1 + 3x2 + 2x3 = 0 0x1 + 2x2 + x3 = 0

Solving these equations by cross-multiplication

x1` 3−4 x1 ⇒ −1 x1 ⇒ 1 ∴ The proportional values are x1

x3 x2 = 1− 0 2−0 x2 x3 = = −1 2 x2 x3 = = −2 1 = −

= 1, x2 = 1, x3 = – 2

X1 = For λ = 2, we have, from equation (i)

LM3 − 2 MM −2 N3

OP PP Q 5O −4PP 5PQ

10 5 −3 − 2 −4 5 7−2

LM 1 MM−2 N3

10 −5 5

LMx OP MMx PP Nx Q LMx OP MMx PP Nx Q

LM 1OP MM−12PP N Q

1 2

= 0

3 1 2 3

= 0

228


Applying R2 → R2 + 2R1, R3 → R3 – 3R1 on coefficient matrix

LM1 MM00 N

10 5 15 6 −25 −10

R2 1 R2 → , R3 → – R 3 5 3

R3 → R3 – R2

OP PP Q

LMx OP MMx PP Nx Q

= 0

1 2 3

LM1 MM0 N0

10 5 5 2 5 2

OP PP Q

LMx OP MMx PP Nx Q

= 0

LM1 MM0 N0

10 5 5 2 0 0

OP PP Q

LMx OP MMx PP Nx Q

= 0

1 2 3

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 ∴

X2 =

LM 5 OP MM 2 PP N−5Q

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. Therefore λ = 2, 2, 3 and X1 =

LM 1 OP MM 1 PP . X = X = N−2Q

LM 5 OP MM 2 PP . N−5Q

LM−2 MM 2 N−1

OP PP Q

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. A =

2 −3 1 −6 · −2 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 N 2 −2 + 1Q Nx Q LM−4 2 OP LMx OP N 2 −1Q Nx Q 1

= 0

2

⇒

1

= 0

2

R2 → R2 + 2R1, on coefficient matrix

LM−4 2OP LMx OP N 0 0Q Nx Q 1

= 0

2

⇒

– 2x1 + x2 = 0

Let

x2 = k

∴ From (i)

x1 = X1 =

k 2

LMk 2OP = k LM1OP or LM1OP N k Q 2 N2Q N2Q

For λ = – 6, from (i), we get

LM1 2OP LMx OP N2 4Q Nx Q 1

2

= 0

...(ii)

230


R2 → R2 – 2R1, on coefficient matrix


= 0

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 And eigen vectors are X1 = X2 =

LM1OP N2Q LM−2OP . N1Q

EXERCISE 3.6 Find the eigen values and eigen vectors of: 1. 2.

3.

4.

5.

6.

7.

LM1 2OP. N1 0Q LM6 8 OP. N8 −6Q LM1 1 3OP MM1 5 1PP. N3 1 1 Q LM 8 −6 2 OP MM−6 7 −4PP. N 2 −4 3 Q LM −2 1 1OP MM−11 4 5PP. N −1 1 0Q LM2 1 0OP MM0 2 1PP. N0 0 2Q LM1 −1 −1OP MM1 −1 0 PP. N1 0 −1Q

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

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

LM3 1 4OP MM0 2 0PP. N0 0 5Q LM−2 2 −3OP MM 2 1 −6PP. N−1 −2 0 Q LM 3 −2 −5OP MM 4 −1 −5PP. N−2 −1 −3Q LM1 1 3OP MM1 5 1PP. N3 1 1 Q LM 4 −20 −10OP MM−2 10 4 PP. N 6 −30 −13Q LM−1 0 2OP MM 0 1 2PP. N 2 2 0Q LM11 −4 −7OP MM 7 −2 −5PP. N10 −4 −6Q LM −9 4 4OP MM −8 3 4PP. N−16 8 7 Q LM 2 4 6 OP MM 4 2 −6 PP. N−6 −6 −15Q LM1 2 3OP MM2 4 6PP. N3 6 9 Q LM 1 −4 −1 −4OP MM 2 0 5 −4PP. MN−−11 14 −−21 63 PQ

[Ans. λ4 – 5λ3 + 9λ2

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 – 7λ + 2 = 0, λ = 2, 1, 1, 1 M P, M P MM−2PP MM−4PP N−3Q N−5Q

232


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

OP PP Q

LM2 MM1 N1

−2 2 1 1 . 19. 3 −1 20. Find the eigen values of A5 when

A=

LM3 MM5 N3

OP PP Q

0 0 4 0 . 6 1

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. A=

3.18

LM 2 MM−2 N1

OP PP Q

3 −2 1 1 . 0 2

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’’. Example 1. Verify the Cayley-Hamilton theorem for the matrix

LM1 MM2 N3

OP PP Q

2 3 4 5 . Also find its 5 6

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,

2

A

3

A

=

=

LM1 A·A = M2 NM3 LM1 A·A = M 2 NM3 2

OP LM1 2 2 4 P M 5 6 QP MN3 5 2 3O L14 4 5PP MM25 5 6QP NM31

2 3 4 5

OP PP Q

LM MM N 31O 56PP = 70QP

14 25 31 3 5 = 25 45 56 31 56 70 6 25 45 56

LM157 MM283 N353

OP PP Q

283 353 510 636 636 793

...(i)

OP PP Q

234


Verification,

3

2

A – 11A

LM157 – 4A + I = M283 MN353

OP PP Q

LM MM N

283 353 14 25 31 510 636 − 11 25 45 56 636 793 31 56 70

⇒ 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 ⇒

–1

=

A

=

LM14 25 31OP LM1 – M25 45 56P + 11 M2 NM31 56 70PQ MN3 LM 1 −3 2 OP MM−3 3 −1PP . N 2 −1 0 Q

LM1 + M0 MN0

OP PP Q

OP PP Q

LM MM N

LM1 2 3OP – M2 4 5P MN3 5 6PQ 0 0O LM0 0 P 1 0P = 0 0 MM 0 1PQ N0 0

2 3 1 0 0 4 5 +4 0 1 0 5 6 0 0 1

0 0 0

OP PP Q

OP PP Q

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.

B = = = =

Thus,

B =

B =

A8 A5 A5 A2

– 11A7 – 4A6 + A5 + A4 – 11A3 – 3A2 + 2A + I (A3 – 11A2 – 4A + I) + A (A3 – 11A2 – 4A + I) + A2 + A + I (0) + A(0) + A2 + A + I |As A3 – 11A2 – 4A + I = 0 in Example 1 + A+ I

LM14 MM25 N31 LM16 MM27 N34

OP LM PP MM Q N 34O 61PP . 77 PQ

OP PP Q

LM MM N

25 31 1 2 3 1 0 0 45 56 + 2 4 5 + 0 1 0 56 70 3 5 6 0 0 1 27 50 61

To find ) : We have A3 – 11A2 – 4A + I = 0 ⇒ A3 = 11A2 + 4A – I Multiplying both sides by A A4 = 11A3 + 4A2 – IA 4

4

A =

=

LM157 283 11 M 283 510 MN353 636 LM1782 3211 MM3211 5786 N4004 7215

OP P 793PQ

OP PP Q

LM MM N31

OP P 70PQ

LM MM N3

OP P 6PQ

353 14 25 31 1 2 3 636 + 4 25 45 56 – 2 4 5

OP PP Q

4004 7215 . 8997

56

5

235

MATRICES

LM 2 MM N

–1

1

OP PP Q

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 3 2 2 A = A.A = −1 1

⇒

A–1 =

LM MM N

Now,

∴

A2 – 6A + 8I =

A–1 =

Thus,

LM 6 −6 MM−5 7 N 6 −9 L2 0 1M 1 3 3M MN0 3

+ 8I)

–1 2 −2

OP PP Q −1O P 1P. 3 PQ

OP LM 2 PP MM−1 QN1

1 −1 2

LM MM N

−1 2 −2

OP PP Q

OP PP Q

LM MM N

1 6 −1 = −5 2 6

LM MM N

OP PP Q

−6 7 −9

LM MM N

OP PP Q

6 −5 7

6 2 −1 1 8 0 0 2 0 −1 −5 − 6 −1 2 −1 + 0 8 0 = 1 3 1 7 1 −2 2 0 0 8 0 3 3

OP PP Q

Example 4. Evaluate A–1, A–2, A–3 if A =

LM 4 MM 1 N−1

6 6 3 2 −4 −3

Sol. Characteristic equation is

OP PP Q

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


A2 =

Now,

∴

2

– A + 4A + I =

=

Thus,

A–1 =

LM16 18 18OP MM 5 7 6 PP N−5 −6 −5Q LM−16 −18 −18OP LM16 MM −5 −7 −6 PP + MM 4 N 5 6 5 Q N−4 LM 1 6 6 OP MM−1 6 2 PP N 1 −10 −6Q L 1 6 6 OP 1M −1 6 2 4M MN 1 −10 −6PPQ

Multiplying equation (1) by A–1, we get –2

A

Similarly,

OP PP Q

LM MM N

14 1 1 −1 4 − + + A I IA − 54 = = 4 4 54

A–3 =

LM MM N

24 24 1 0 0 12 8 + 0 1 0 −16 −12 0 0 1

OP PP Q

OP PP Q

−1 2 52 32

−9 2 −3 2 11 2

1 [– I + 4A–1 + A–2] 4

OP PP Q

LM MM N

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 N2 −1Q

and hence find (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 N2

2 −1

OP LM1 Q N2

...(i)

OP LM Q N

2 5 = −1 0

Putting value of A2 in equation (i), we get A2 – 5 = 5I – 5I = 0 Hence, the Cayley-Hamilton theorem is verified.

OP Q

LM N

0 1 = 5 5 0

OP Q

0 = 5I 1

237

MATRICES

To find )1 : Multiplying equation (i) by A–1, we get A – 5A–1 = 0 ⇒

A–1 =

LM 0 Example 6. Show that the matrix A = M − c MN b

LM N

OP Q

1 1 1 2 . A = 5 5 2 −1

OP PP Q

c −b 0 a satisfies its characteristic equation. Also −a 0

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 PP Now, A = − ec + a j bc −eb + a jP bc Q 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 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 LM 0 MM− c Nb

2

OP LM 0 PP MM−c QNb

2

3

LM e MM MN

OP PP Q

− c 2 + b2 c −b 0 a = ab −a 0 ac

c −b 0 a −a 0

j

ab

2

2

2

3

2

2

or

2

2

2

2

2

2

2

2

2

2

2

2

To find )1: Multiplying equation (i) A–2 on both sides, we get A + A–1 (a2 + b2 + c2) = 0

Hence,

–1

A

= −

ea ea

1 + b2 + c2

1 2

+ b2

2

2

...(ii)

A3 + A (a2 + b2 + c2) = – (a2 + b2 + c2) A + A (a2 + b2 + c2) = 0

2

2

2

A3 = – (a2 + b2 + c2) A

A–1 = −

2

2

Using (ii) in equation (i), we get

⇒

2

2

2

2

2

2

2

2

ac

j

⋅A

LM 0 M− c +c jMb N 2

OP PP Q

c −b 0 a · −a 0

238


EXERCISE 3.7 1. Find the characteristic equation of the matrix A A=

LM4 MM2 N1

OP PP Q

3 1 1 −2 . Hence find A–1. 2 1

(U.P.T.U., 2001)

LM MMAns. Characteristic equation : λ N

3

− 6λ + 6λ − 11 = 0, A 2

LM MM N

5 −1 −7 1 = −4 3 10 11 3 −5 −2

−1

2. Find the characteristic equation of the matrix A=

LM2 MM0 N1

OP PP Q

1 1 1 0 . Verify Cayley-Hamilton theorem and hence 1 2

(U.P.T.U., 2002)

evaluate the matrix equation A8 – 5A7 + 7A6 – 3A5 + A4 – 5A3 + 8A2 – 2A + I.

LM MMAns. λ N

3

LM MM N

8 5 5 − 5λ2 + 7λ − 3 = 0, A 2 + A + I , 0 3 0 5 5 8

3. Verify Cayley-Hamilton theorem for A and hence find A–1

LM 2 −1 1 OP (i) M−15 6 −5P MN 5 −2 2 PQ LM1 1 1OP (ii) M1 2 1P MN1 1 2PQ LM1 0 3 OP (iii) M 2 1 −1P MN1 −1 1 PQ LM 7 4. Find A if A = M−6 MN 6

LM MMAns. A N LM MMAns. A N

LM−2 = M −5 MN 0 LM 3 = M−1 MN−1 LM L0 1M MMAns. A = − 9 M−3 MN−3 N LM L−3 1M MMAns. A = 3 M 6 MN−6 N −1

−1

−1

OP −1 2 P . 2 −1PQ LM1 2 5. Show that the matrix A = M 2 −1 MN0 0 2

−2

−1

–1

OP PP Q

OPOP PPPP QQ

0 1 1 −1 1 0 −3 −2 1 −2 5 −2

OPOP PPPP QQ

OPOP 5 PP 3PQPQ −1OO PP 0 PP 1 PQPQ −3OO PP 7 PP 1 PQPQ 2 OO PP −2PP 5 PQPQ 1

0 0 satisfies its own characteristic equation and hence −1 1/5 0 0 −1 1 /5 0 Ans. A = 0 or otherwise obtain the value of A–1. 0 0 1

LM MM N

LM MM N

OPOP PPPP QQ

239

MATRICES

6. Find A–1 if A =

7. If A =

LM0 MM00 MN1

1 0 0 0

0 0 1 0

LM1 MM1 N1

OP PP Q

2 3 3 5 . 5 12

OP PP PQ

−1

8. Find B = A6 – 4A5 + 8A4 – 12A3 + 14A2 if A =

LM 1 2OP . N−1 3Q

LMAns. λ N

2

9. Find A–1 if A =

10. Show that the matrix A =

3.19

OP PP Q

3 4 −3 4 · −1 1

LM 2 MM 3 N−5

−9

−1

0 1 find A–1. 0 0

LM3 MM2 N0

LM L11 1M MMAns. A = 3 M−7 MN 2 N LM LM0 MMAns. A = M1 MM0 MM N0 N

OP PP Q

9 −3 0

0

0

0

0

1

1

0

− 4λ + 5 = 0 , B = 5I − 4 A =

LM MMAns. A N

−1

=

LM MM N

−1 1 2 11 2

LM1 N4

7 −3 −3

OPOP −2PP 1 PQPQ 1OO P 0PP P0PP PP 0QPQ 1

OPOP −7 Q Q −24OO PP 4 PP 15 PQPQ −8

−3 1 1 3 satisfies the equation A (A – I) (A + 2I) = 0. 2 −4

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


LMx OP MMx PP, X MNxM PQ LMx x MMxM xM MNx x 11 12

X1 =

2

1n

then let,

P =

and diagonal matrix D = P–1 AP.

11

21

12

22

1n

2n

LMx OP x = M P ...., X MM M PP Nx Q L x O L x P PP M P L x Q

n2

22

2n

LMx OP x =M P MM M PP Nx Q n1

21

n

nn

n1

n2

nn

3.19.2 Powers of Matrix A D = P−1 AP

Consider ⇒

D2 = (P−1 AP) (P−1 AP) = P−1 A (PP−1) AP = P−1 A. IAP = P−1 AAP = P−1 A2 P D3 = P−1 A3 P, ...., Dn = P−1 An P

Similarly,

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

Example 1. Diagonalize the matrix

LM1 MM1 N0

OP PP Q

6 1 2 0 . 0 3

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,

OP LMx OP PP MMx PP = 0 Q Nx Q

LM2 MM1 N0

6 1 3 0 0 4

LM1 MM2 N0

3 0 6 1 0 4

1

2 3

R1 ↔ R2 on coefficient matrix, we get

OP LMx OP PP MMx PP = 0 Q Nx Q 1

2 3

(U.P.T.U., 2006)

241

MATRICES

R2 → R2 – 2R1

LM1 MM0 N0


3 0 0 1 0 4

R1 → R1 – R2, R3 → R3 – 4R2

LM1 MM0 N0

⇒

3 −1 0 1 0 0

1

2 3


2 3

Let,

x1 + 3x2 – x3 = 0 x3 = 0 x2 = k, then x1 = – 3k

∴

X1 =

For, λ = 3

⇒

LM−2 MM 1 N0

6 1 −1 0 0 0

OP LMx OP PP MMx PP Q Nx Q

LM−3kOP LM−3OP LM−3OP MM k PP = k MM 1 PP = MM 1 PP N 0 Q N0Q N0Q

1

2

= 0

3

– 2x1 + 6x2 + x3 = 0 x1 – x2 + 0·x3 = 0

⇒

x1 –x2 x3 x x x = or 1 = 2 = 3 = −1 −4 1 1 −4 1 x1 = 1, x2 = 1, x3 = – 4

∴

X2 =

For, λ = 4

R1 ↔ R2

R2 → R2 + 3R1

LM−3 MM 1 N0

6 −2 0



1 0 −1

1

2 3


LM 1 MM−3 N0

−2 0 6 1 0 −1

LM1 MM0 N0

−2 0 0 1 0 −1

1

2 3


2 3

242


R1 → R1 + R2, R3 → R3 + R2

⇒

LM1 MM0 N0

−2 1 0 1 0 0


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

∴

X3 =

P =

1 2 1 1 −4 0

OP PP Q

|P| = – 3 (4) – 1 (0) + 2 (– 4) = – 20 B =

LM 4 MM−8 N−1

adj P = B′ =

–1

and

LM2OP MM1PP N0 Q LM−3 MM 1 N0

P

0 −4 0 −12 5 −4

LM 4 MM 0 N−4

D = P–1 AP D =

=

=

OP PP Q

−8 −1 0 5 −12 −4

adj P 1 =– = |P| 20

Diagonalisation:

OP PP Q

LM 4 MM 0 N−4

OP PP Q

LM−4 8 1 OP LM1 6 1OP LM−3 MM 0 0 −5PP MM1 2 0PP MM 1 N 4 12 4 Q N0 0 3Q N 0 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 1 20

LM MM N

−8 −1 −4 8 1 0 5 = 0 0 20 −12 −4 4 12

OP PP Q

1 2 1 1 −4 0

1 −5 4

OP PP Q

243

MATRICES

Example 2. Reduce the matrix A =

|A – λI| =

Sol. ⇒

OP PP Q

LM−1 MM 1 N−1

2 −2 2 1 to the diagonal form. −1 0 (U.P.T.U., 2001, 2004)

−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 ⇒ R1 ↔ R2

(A – λI) X = 0

LM−2 MM 1 N−1

2 −2 1 1 −1 −1


= 0

LM 1 MM−2 N−1

1 1 2 −2 −1 −1


= 0


= 0

R2 → R2 + 2R1, R3 → R3 + R1

LM1 MM0 N0

⇒

1 1 4 0 0 0

1

2 3

1

2 3

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

∴

X1 =

For

λ =


LM−1 − 5 MM 1 N −1

2 2− 5 −1

−2 1 − 5

OP Lx O PP MMx PP Q MNx PQ

= 0

LM 1 MM−1 − 5 N −1

2− 5 2 −1

1 −2 − 5


= 0

R1 ↔ R2

1

2 3

1

2 3

5

244


R2 → R2 + (1 +

LM1 MM0 N0

5 ) R1, R3 → R3 + R1


2− 5 5 −1 1− 5

1 5 −1 1− 5

R3 → R3 + R2, and R2 →

LM1 MM0 N0 ⇒

2− 5 1 0

x1 + (2 –

1

2

= 0

3

1 5 −1

R2


1 1 0

1

2

= 0

3

5 ) x2 + x3 = 0 0x1 + x2 + x3 = 0

Solving by cross-multiplication x1 1− 5 x1

or

5 −1

∴

= – =

X2 =

x2 x3 = 1 1

x2 x3 = ⇒ x1 = −1 1

LM MM N

OP PP Q

5 −1 1 −1

Similarly, the eigen vector corresponding λ = –

X3 =

∴ The modal matrix To obtain 21:

R3 → R3 + R1

LM 1 MM 0 N−1 LM1 MM0 N0

P =

5 −1 1 −1 5 −1 1 5−2

5 +1 −1 1

OP PP Q

OP P 5 + 2PQ 5 +1 −1

5 − 1 , x2 = 1, x3 = – 1

5

LM 5 + 1OP MM −1 PP N 1 Q LM 1 5 − 1 MM 0 1 N−1 −1 OP PP Q

~

LM1 MM0 N0

0 0 1 0 P 0 1

~

LM1 MM0 N1

0 0 1 0 P 0 1

OP PP Q

5 +1 −1 1

OP PP Q

245

MATRICES

e

R1 → R1 –

LM1 MM0 N0

e

R3 → R3 –

LM1 MM0 N0

0 1 0

LM1 MM0 N0

0 1 0 1

2 5

LM1 MM0 N0

0 1 5−2

j 2 5O P −1 P 2 5 PQ

5 − 2 R2

R1 → R1 – R3

R3 →

j

5 − 1 R2

0 −1 2 5

OP PP Q

∴

Now,

OP PP Q

LM1 MM0 N1

~

~

LM1 MM0 N1

1− 5 1 2− 5

0 0 P 1

~

LM0 MM0 N1

−1 1 2− 5

−1 0 P 1

OP PP Q

1− 5 0 1 0 P 0 1

OP PP Q

OP PP Q

R3

OP PP Q

LM 0 MM 0 MM 1 N2 5

OP PP Q

LM MM 01 MM 2 5 MM 1 N2 5

0 0 1 −1 ~ 0 1

R2 → R2 + R3

LM1 MM0 N0

2 5 −1 5+2

0 0 1 0 0 1

~

P–1 =

1 2

LM0 M1 5 M N1

D = P–1 AP =

=

1 2

−1 1 2− 5 2 5

−1 2+ 5 2 5 2− 5 2 5 −2 5 2+ 5 2− 5

LM0 1 M 1 2 5 M MN1

LM2 5 M0 5 M N0

0 10 0

OP P 0 PP 1 P 2 5 PQ −1

OP −1 P 1 P P 2 5P P 1 P 2 5 PQ −2 5 O P 1 P 1 PQ −2 5

−2 5

2+ 5

1

2− 5

1

OP PP Q

LM MM N

0 1 0 = 0 −10 0

0 5 0

OP L−1 PP MM 1 PQ MMN−1 0 0 −

2 2 −1

OP PP . 5Q

OP LM 1 1P M0 PM 0 PQ MN−1

−2

5 −1

5 +1

1

−1

−1

1

OP PP PQ

246


Example 3. Diagonalise the matrix A =

LM11 MM 7 N10

−4 −7 −2 −5 −4 −6

OP PP Q

and hence find A5.

Sol. The characteristic equation of A is |A – λI| = 0 ⇒

LM11 − λ MM 7 N 10

−4 −7 −2 − λ −5 −4 −6 − λ

OP PP Q

= 0

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

⇒

LM11 MM 7 N10

−4 −7 −2 −5 −4 −6


0

1

2

= 0

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 ∴

X1 =

For λ = 1, we have

⇒

LM10 MM 7 N10

−4 −7 −3 −5 −4 −7


LM1OP MM1PP N1Q

1

2

= 0

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

∴

X2 =

Similarly,

LM 1OP MM−1PP , N 2Q

λ = 2, X3 =

For

The three vectors are linearly independent, Hence the modal matrix

P =

∴

–1

P

=

LM−2OP MM−1PP N−2Q

LM1 1 MM1 −1 N1 2 LM−4 2 MM−1 0 N−3 1

−2 −1 −2

Now,

D = P AP =

=

LM0 MM0 N0

OP PP Q LM−4 MM−1 N−3

3 1 2

–1

OP PP Q

OP LM11 PP MM 7 Q N10

2 3 0 1 1 2

−4 −7 −2 −5 −4 −6

OP LM1 PP MM1 Q N1

0 0 1 0 . 0 2

A5 = PD5 P–1

Next

OP PP Q

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


–λ −2 −2 −1 1 − λ 2 −1 −1 2 − λ

= 0

OP PP Q

1 −2 −1 −1 2 −2

248


⇒

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 MM−1 N−1


= 0


= 0


= 0

−2 −2 −1 2 −1 0

−1 R1 → R1 2 1 1 1 −1 −1 2 −1 −1 0

LM MM N

1

2 3

1

2 3

R2 → R2 + R1, R3 → R3 + R1

LM1 MM0 N0

R2 →

1 1 0 3 0 1

1 R and R3 → 3 2 1 1 1 0 0 1 0 0 0

LM MM N

1

2 3

R3 – R2

OP LMx OP PP MMx PP Q Nx Q 1

2

= 0

3

⇒ 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

⇒

= 0 ⇒

(λ – 2) (λ – 5) = 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


2

⇒

= 0, R2 → R2 − 2R1

2x1 + x2 = 0,

LM2 1OP LMx OP = 0 N0 0Q Nx Q 1

2

Here let x2 = k, so x1 = −

k 2

–1

249

MATRICES

∴

X1 =

LMx OP = Nx Q 1

2

LM − k OP k L 1 O L 1 O MN k2 PQ = − 2 MN−2PQ or MN−2PQ

For λ = 5

LM−1 1 OP LMx OP N 2 −2Q Nx Q 1

= 0, R2 → R2 + 2R1

2

1

2

⇒ – x1 + x2 = 0. Here again let x2 = k so x1 = k ∴

LM−1 1OP LMx OP = 0 N 0 0Q Nx Q

X2 =

LMx OP = LMkOP = k LM1OP or LM1OP Nx Q NkQ N1Q N1Q 1

2

The modal matrix P = Verification: ∴

LM 1 1OP N−2 1Q

|P| = 1 + 2 = 3 P–1 =

P–1 AP =

=

LM N 1 L1 M 3 N2 1 L6 M 3 N0

OP Q −1O L 4 1O L 1 1O 1 L1 −1O L 2 5O = M P M P M P PM P 1 Q N 2 3Q N−2 1Q 3 N 2 1 Q N−4 5Q 0O L2 0OP = D (diagonal matrix). = M P 15Q N0 5Q

1 1 −1 3 2 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 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.

M =

LM1 MM0 MN01

1 0 −1 0

0 0 1 1

−1 1 0 0

0 −1 1 0

0 1 0 1

3 4 3

2

2

1

OP PP PQ

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

Augmented matrix [A : B] =

LM 1 MM−1 MN200

−1 1 1 −1 10 25 10 0

Applying R2 → R2 + R1, R4 → R4 – 20 R1

~

LM1 MM0 MN00

−1 0

1

M

0

M

~

OP P 90P P 80Q 0

10 25 M 30 −20 M

OP P 90P P 0Q

−1 1 M 0 30 −20 M 80 10 0

25 0

M M

R3 → 3R3 – R4

~

LM1 MM0 MN00

−1 1 30 −20 0 95 0 0

R2 →

~

Now


OP PP PQ

0

R2 ↔ R4

LM1 MM0 MN00

M 0 M 0 M 90 M 80

M 0 M 80 M 190 M 0

OP PP PQ

1 1 R,R → R 5 2 3 19 3

OP P 5 M 10 P P 0 M 0Q 1O L0O i O M P L P −4P M P M16P i 5 P M P M10P P Mi P M P 0Q N Q N0Q

−1 1 M 0 6 −4 M 16 0 0

−1 6 0 0

1 2

3

252


⇒

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 N1Q

2

=

LM 1 OP N−1Q

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 1. A square matrix A is defined by A = resulting diagonal matrix D of A.

2. Diagonalise the matrix A =

3. Diagonalise the matrix A. A=

OP PP Q

LM 3 MM−1 N1 LM3 MM0 N0

LM 1 MM 1 N−1

OP PP Q

2 −2 2 1 . Find the modal matrix P and the −1 0 (U.P.T.U., 2000)

LM LM–2 MMAns. P = M 1 MN 1 N

OP PP Q

LM MMAns. A N

1 4 2 6 · 0 5

OP PP Q

LM MM N

OP PP Q

−3 −3 2 2 1 −6 , (ii) −6 5 2 · −2 0 −7 4 4 2

5. Diagonalise

LM 8 MM−6 N2

−6 7 −4

OP −4P · 3 PQ 2

1 –1

−1 1 5 −1 and hence find A4. −1 3

4. Verify whether the following matrices are diagonalisable or not.

LM−2 (i) M 2 MN−1

–2

4

OP LM1 1 P, D = M0 MN0 –1PQ 2

LM 251 = M−405 MN 235

−405 81 −405

LM LM2 Ans. D = MM MM0 N0 N

0 −1 0

OPOP 0PP 3PQ PQ

0

OPOP −405PP 251 PQPQ 235

OPOP 0PP 5PQ PQ

0 0 3 0

[Ans. (i) Not diagonalisable (ii) diagonalisable]

LM LM0 Ans. D = MM MM0 N0 N

0 3 0

OPOP PP 15PQPQ 0

0

254


6. Diagonalise the following matrices:

L O L1 0 0O LM OP MMAns. (i) D = MM0 –2 0PP, (ii) LM00 06OP, (iii) LM00 02OPPP MN0 0 3PQ N Q N QPQ N Q MN 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 L4 1OP hence find A . LMAns. D = L1 0O, A = L2344 781OOP 8. Diagonalise the matrix M 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

LM1 (i) M0 MN1

OP PP Q

LM N

OP Q

0 2 2 1 2 , (ii) −8 2 0

−1 1 1 , (iii) 4 1 1

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 (i) (ii) M P 27 N0 27 Q 1 L1 −26O (iii) (iv) M P 27 N0 −27Q LM0 0 1OP 4. If A = M0 1 0P , then A is MN1 0 0PQ LM1 0 1OP (ii) (i) M0 1 0P MN0 0 0PQ LM0 0 1OP (iii) M0 1 0P (iv) MN1 0 0PQ 3. If A =

–1 3

LM OP N Q 1 L−1 −26O M P 27 N 0 −27Q 1 −1 26 27 0 27

–1

LM0 MM1 N0

1 0 0 0 0 1

OP PP Q

None of these

5. If A and B are square matrices of order 3 such that |A| = –1 and |B| = 3, then |3AB| = (i) – 9 (iii) – 81

(ii) – 27 (iv) 81

6. The rank of the unit matrix of order n is (i) 0

(ii) 1

(iii) 2

(iv) 3

256


7. If A =

LM1 MM1 N1

OP PP Q

1 1 2 4 , then rank (A) is 4 8

(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 ........... 10. The eigen values of A =

LM2 MM1 N1

OP PP Q

2 1 3 1 are .......... 2 2

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

7. Singular 9. Independent

5. All zero; pure imaginary] 6. |A| 8. Consistent 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

∑ f bxr , y r g

Y

y = f2 (x)

R

dx r (x r, y r)

dy r

N

SN =

r =1

δSr

y = f1 (x)

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

LM a f O z MN za ffbx, yg dyPPQ b f2 x

g

f x, y dS =

dx

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

zz b

LM b g O z MN zb fgbx, yg dxPPQ dy d f2 y

g

f x, y dS =

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 δr→0

=

LM N

z

af

f θ

0

Hence the area OAB = =

O rdrP Q

L lim ∑ M N

δθ → 0 θ =β

δθ

z

af

f θ

0

fθ

0

K1 P (r, q)

z LMNz a f θ=α

K2

L2 L 1

dq

O rdr P δθ Q

OP Q

O

rdr dθ

where α and β are the vectorial angles of A and B.

q

Fig. 4.2

A

260


Example 1. Evaluate

z z log 8

log y

1

0

Sol. We have

I = =

ex +

z LMNz

=

Example 2. Evaluate

z z

log y

1

0

z z

log 8

1

π

x

0

I = =

zz 0

Sol. We have

b g

log 8

dy dy

y – e

1

dy

log 8 1

(As eloge y = y)

sin y dy dx .

z LMNz z π

x

0

0

π

z z

sin y dy

– cos y

0

= −

2

y

e y e log y − e 0

ey y−1

= −

Example 3. Evaluate

x

log y y e 0

ex

log 8

ee dxjOPQ e

= elog 8 · (log 8 – 1) – 0 – (elog 8 – e) = 8 log 8 – 8 – 8 + e = 8 log 8 – 16 + e.

0

Sol. We have

dx dy.

log 8

1

=

y

π

0

π

0

x 0

OP Q

dx

dx, treating x as constant

(cos x – cos 0) dx

cos x dx + π

z

π

dx

0

(As cos 0 = 1)

π

= − sin x 0 + x 0 = π.

d 4− 2 y i − d 4− 2 y i 2

2

z z 2

I =

0

= =

y dx dy .

z z

2

LM MN

e 4− 2 y j − e 4− 2 y j

0

= –

1 2

2

c 4− 2 y h y − c 4− 2 y h

2

e4 − 2y j

OP PQ y dy

2

2

2

z

0

4

1 2 = – · 2 3 =

dx

x

0 2

2

t dt

LMt MN

1 8 ·8= . 3 3

OP PQ

dy y dy by putting 4 – 2y2 = t, – 4y dy = dt

3 0 2 4

MULTIPLE INTEGRALS

Example 4. Evaluate

261

zz b

x+ y

= 1

x y 2 + a b2

= 1

⇒

y = ±

zz b

x+ y

g

dx dy =

=

b a

zz e z

=

zz −a

2

−a

z z z

2

2

2

z

a2 − x 2

a

ex

a2 − x 2

2

ex

ex 2

2

af

f x dx = 2

–a

= 2

−a

a

= 4

0

+y

2

j dx dy + 2

zz a

−a

e ba j a − x e− ba j a − x 2

2

2

2

xy dy dx

j

z

af

a

F bI a

af

f x dx when f ( x) even = 0 when f x odd

a 2 − x2

dx

0

LMx × b a − x + 1 b a − x OP dx e jP 3a MN a Q LM b x a − x + b a − x OP dx (Again by definite integral) e jP 3a MN a Q

z FGH π 2

0

+ y 2 + 2 xy dx dy

0

3

2

j

2

+ y 2 dy dx + 0

LM2F x y + y I OPGH JK MN GH 3 JK PQ

a

= 4

a2 − x 2

a

FG b IJ H aK 0

zz

−a

FH b IK a FH − b IK

e ba j a − x e −ab j a −x

As

=

j

2

z

−a

a

Fig. 4.3

x + y + 2xy dx dy 2

a

a

X

O

a2 − x2

a

=

x2 a2

1−

x=a

y = ± b

∴

Y

2

⇒

2

dx dy over the area bounded by the ellipse

x = –a

Sol. We have

2

[U.P.T.U. (C.O.), 2004]

x2 y2 2 + a b2 2

g

2

2

3

2

2

3

2

2

2

3

3

2

3 2 2

3 2 2

[On putting x = a sin θ and dx = a cos θ dθ]

IJ K

b ⋅ a 2 sin 2 θ. a cos θ + b 3 a 3 cos 3 θ a 3a 3

× a cos θ dθ

262


= 4

z

π 2

0

= 4a3b

=

4a3b

FG a b sin H 3

2

z

π 2 sin 2 θ cos 2 θ 0

z

4ab 3 3

dθ +

sin 0 θ cos 4 θ dθ

3 3 1 5 4 ab 3 2 2 2 2 · + · 3 2+ 2+ 2 0 + 4+ 2 2 2 2 2

= 4a3b · =

π 2

dθ

0

1 1 1 1 . 4 ab 3 = 4a3b. 2 2 2 2 + . 3 2 3

⇒

IJ K

3 θ cos 2 θ dθ + ab cos 4 θ 3

As

z

π 2 sin m 0

m+1 n+1 2 2 θ 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

1 = 2

n – 1 and

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.

= = = is =

f2 (x), x, i.e., x (x – 1) = 0 0, x = 1 bounded by x2, y = x, x = 0, x = 1

Sol. The area is bounded by the curves y = f1(x) = x2, y = f2(x) = x. When

f1 (x) x2 or x i.e., the area of integration y ∴

= = = =

=

zz b z LMNz z LMNz e LM z MN z LMMN

Example 6. Find

zz

D

g

xy x + y dx dy

A

b g OPQ dx x=0 y = x x y + xy j dy OP dx Q x y xy O + P dx 2 3 PQ y =x

1

1

x

0

x2

2

2

0

y=x

3 x

6

5x x − 6 2

Fig. 4.4

x O − P dx 3 PQ 7

LM x − x − x OP N 6 14 24 Q 7

X

O

x2

4

2

x=1

2

0

1

A

xy x + y dy

2 2

1

5

=

Y

8 1

= 0

LM 1 − 1 − 1 OP = N 6 14 24 Q

3 . 56

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

x=0

y=x

y=0

x 2 dx dy +

zz x=8

z z zz

4 3 8 x dx + 0 4

16 x

16 8 2 x dy x dx 4 0

+

B O

LM x OP N4Q 4

16 x dx =

G (4, 0)

E

= 4

z

4

0

bg

x

x y dx + 2

2 + 8x 0

0

X

Fig. 4.5

x 2 dx dy

x= 4 y =0

z z z z x 4 2 x dx 0 dy 0

y=

(8, 2)

z

8

4

2

bg

x y

16 x dx 0

8 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

from 0 to a and y varies from 0 to Here,

zz

A

xy dx dy =

zz

=

z 0

z

j

− x2 . 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

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 0 y=0 a

=

z

3

0

= = Example 10. Evaluate I = I = =

a

0

2 4

2a2

z z z z z LMN OPQ z 2π

a

0

a sin θ

= I =

a

r = a sin θ

2π

r2 2

a 2

2π

0

OP PQ

πa . 2

j dx

a

= 0

(a, 0)

b2 a 4 . 24 Fig. 4.8

rdrdθ .

rdrdθ

a

dθ = a sin θ

cos2 θ dθ =

LMθ + sin 2θ OP N 2 Q

a2 4 2

=

x6 − 6

0

2

(0, b)

0

2π

0

a2 −x2

a 2x 3 − x 5

LM a x MN 4

b2

Y b a

2

ze

b2 2a 2

x2 y2 + =1 a2 b2

x 3 y dy dx

LM x y OP MN 2 PQ

a

Sol.

IJ K

8 4 2x 5 2 + x 3 2 − x 3 dx 3 3

768 . 35

=

zz

Fig. 4.7

2π 0

1 2

z

2π

0

z

ea − a sin θj dθ FG 1 + cos 2θ IJ dθ H 2 K

a 2 2π 2 0

2

2

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 LM OP a z NQ a

π

a 1+ cos θ

0

0

π

0

π

0

3

a 1+ cosθ

sin θ

π

=

16 a 3 3

0

Sol.

z

π 2 0

dθ

za

a 2

r dr

a 1−cosθ

f

= =

= = =

f

fr

OP Qa

2

Put 4 3 a. 3

= 0

drdθ .

a

r3 3

π 2 dθ 0

a 3 a 1 − cosθ − 3 3

a 1−cosθ

a

f

f IJ K 3

a3 3

π 2 0

1 − 1 − cosθ

a3 3

π 2 0

1 − 1 − 3 cos θ + 3 cos 2 θ − cos 3 θ

a3 3

π 2 0

3 cos θ − 3 cos 2 θ + cos 3 θ dθ

a3 = 3

a

f

3

dθ

e

LM 1π 2 O P + 3 sin θ − 3 2 2 3 ⋅1P MN Q LM3 − 3π + 2 OP N 4 3Q π 2 0

=

a3 3

=

a3 [44 – 9π]. 36

A

Fig. 4.9

π 2

π 2 dθ 0

3

q=0

q O

φ ⋅ 2dφ

8

a

P (r, q)

dq q=p

θ θ sin dθ 2 2

LM − cos φ OP N 8 Q

a 1− cos θ

Q

X

g3 sin θ dθ

π 7 2 sin φ cos 0

z za LM z N F z GH z z ze π 2 0


p q=– 2

dθ

1 + cos θ

16 3 =2× a 3

O r dr P dθ Q

0

π cos7 0

16 3 = a 3

Y

2

a 1+cosθ

r3 3

zb z z

a 3

f

0

sin θ

=

f r 2 sin θ dθ dr

j

j

dθ

θ = φ 2

266


Example 13. Evaluate Sol. We have

z z z z z LMNz z LMN OPQ z π

aθ 3

0

0

π

I =

π

aθ 3

0

0

r4 4

π

=

Sol. We have

aθ 3 r dθ 0

0

=


r dθ dr

0

r dr

aθ

OP Q

dθ

dθ 0

π 4 4 a θ dθ 0

=

1 4

=

a4 4

LM θ OP N5Q 5

z zb z z π

a 1+ cos θ

0

0

I =

dr

π

a4π 5 . 20

= 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 π

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 ze 0

2

2

0

fa f

t 4 t − 1 − dt

j

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

π 2

0

0

z z z z

4.

0

1+ x 2

0

dy dx 1+ x + y

0

b

g dy dx.

2

0

a2 − y2

Ans. 8 log 8 – 16 + e

0

cos x + y

a

e x + y dx dy.

2

.

a 2 − x 2 − y 2 dx dy.

LMAns. N

e

π log 4

jOPQ

2 +1

Ans. – 2

LMAns. MN

πa 3 6

OP PQ

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

13.

14.

zz e zz b D

xy dx dy over the region in the positive quadrant for which x + y ≤ 1.

1

1+ x

−1 2

−x

ex

2

j

+ 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

Hint: Limits x : – y to

15.

LMAns. 1 OP N 24 Q LMAns. 63 OP N 32 Q

zz 1

0

1+ x2

0

16. Evaluate

e1 + x

zz

A

0

+ y2

j

−1

LMAns. N

y ; y : 0 to 2 .

LMAns. N

dx dy.

OP Q π O log e1+ 2 j P 4 Q 44 13 2+ 15 3

y dx dy, where A is bounded by the parabolas y2 = 4x and x2 = 4y.

LM x − y OP MN bx + yg dyPQ dx ≠ z

zz 1

17. Show that

2

y , y = 2, y = 0.

1

0

2

1

0

LM x − y MNz bx + yg 1

0

2

O dx P PQ

dy.

LMAns. 48 OP 5Q N

268


18. Evaluate 19.

zz e

zze 1

0

x2 + y2

A

20.

z zb 4

2

3

1

zz zz zz zz π

0 a

0

0

24.

j dx dy, where A is bounded by x

+ y2 = a2 and x2 + y2 = b2, where b > a.

g

r dr dθ.

ρ sin θ dρ dθ.

r 3 sin θ cos θ dr dθ.

LMAns. π eb 4 − a4 jOP N 2 Q LMAns. logFG 25IJ OP H 24 K Q N 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

4 sin θ 3

cos θ

0 π

23.

LMAns. 41 OP N 240 Q

j

1 + xy 2 dx dy.

dy dx

2 sin θ

0

22.

y

2

⋅ 2 x+y Using polar coordinates evaluate the following double integral. π

21.

y

z z π 2 0

zz

R

LMAns. 45 π OP N 2 Q

r 2 sin θ dr dθ =

a n r sin n θ cos θ 0

2a 3 , where r is the region bounded by the semicircle 3

LMAns. a MN an + 1f n+ 1

d θ dr , for n + 1 > 0.

2

OP PQ

4.5 CHANGE OF THE ORDER OF INTEGRATION If the limits of x and y are constant then

zz b

g

F GH

x2 y

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

x exp −

I dx dy by changing the order of integration. JK (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

x exp −

0

x2 y

I JK

x

F GH

x

y=

zz ∞

dx dy

z z z z FGGH ∞

=

∞

x2 y

−

xe

x®¥

⋅ dy dx

y = 0 x= y

∞

=

∞

y 2x − − − e 2 y y

0

Y

x2 y

I JJ dy dx K

LM y OP = MN− 2 e PQ dy LM y OP = MN0 + 2 e PQ dy = 1 Ly O = M e− e j − ee jP 2 N2 Q F 1I 1 = (0 – 0) + GH 0 + JK = . 2 2

z z

∞

x®¥

0

y

−

y2 y

0

−y

−y

∞

−

Put e

z ∞

0

X

Fig. 4.10

∞

x2 − y

∞

O

x2 y

x2

2x − y =t ⇒ – e. dx = dt y

y –y e dy 2 (Integration by parts)

0

Example 2. Change the order of integration in Sol. Here the x2 = i.e., x2 = also x = Now

zz a

0

2 a −x x2 a

limits are ay and y = 2a – x ay and x + y = 2a 0 and x = a

zz b g zz b g z z b g zz zz zz 2 a− x x2 a

a

0

0

+

(U.P.T.U., 2007)

B (0, 2a) x + y = 2a 2

x = ay

f x , y dx dy ay

a

=

b g

f x, y dx dy.

0

2a

2a − y

a

0

Example 3. Evaluate

L

A (a, a)

f x, y dy dx 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

Þ

x

y

=

e

=

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

e2

1

b

= 3 y − y log y

g

e2 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 H 3K

Y

and then y varies from – 4 to 5.

E (1, 5)

Changing the order of integration to first x and later to y, the double integral becomes = = =

=

IJ K

z z z z z LMN 1

3x + 2

–2

y = x 2 + 4x

5

−4

−4

2

y = x + 4x

dy dx F (0, 2)

y+ 4 − 2 y − 2 dx x= 3

dy

FG y − 2 IJ OP dy H 3 KQ LM 2 b y + 4g − y − 4 yOP = 9. 2 6 3 PQ MN 3 5

y = 3x + 2

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 z zb g 1

0

y dy

y

0

xdx +

x2 = y dy 0 2

y

1

1 = 2 =

0

1 y dy + 0 2 1 2

1 1 + 6 2

2

1

y dy

2− y

0

x2 + ydy 1 2 2

2

1

y 2− y

2

4

3

x=0

OP2−y PQ0

X

3 1

0

2

Example 6. Change the order of integration in Sol. Here the limits are y =

a −x 2

2

x=1

X

LM y OP 1 MN 3 PQ + 2 z e 4y − 4y + y j dy 32 4 1O 1 L 8− + 4− 2+ − P M 3 3 4Q 2 N

1 dy = 2

1 1 + 6 2 1 5 = + 6 24 9 3 = = . 24 8

O

Fig. 4.14

1

=

B (1, 1)

C

x dx

LM2y − 4 y + y OP = 1 + MN 3 4 PQ 6 LM 96 − 128 + 48 − 24 + 16 − 3 OP 12 N Q 2

–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)

z z a

x +2 a

0

a −x 2

2

2

3

1

b g

φ x, y dx dy . 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


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

=+

zz

a

a

a −y 2

2

3a

a

2a

y−2a

2a

a

a

0

b g

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. The limits are

y 2 dx dy

a

y2

ax

x=0

y −a x 4

2 2

.

x=a

Fig. 4.15

= ax, y = a and x = 0, x = a, y = 0 to y = a

y2 . 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 a

y2 a

y =0

x =0

y2 a 0

1 a = a 0

z z

4

1 = a

π = 2a

z

B

y 2 dy dx

Fy I GH a JK

2 2

F GH

1 2 −1 ax y sin y2 a0 a 2 y 0

Y

2 2

y −a x

LM MN

a

=

y 2 dy dx

[sin–1

a 2 y dy 0

− x2

(1) –

A

2

x=0

I OP JK PQ

y=a

y x=— a

y2 a

dy O

0

sin–1

F I GH JK

3 π y = 2a 3

(0)] 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

∞

0

sin px π dx = · x 2

zz

∞ ∞ e − xy 0 0

sin px dx dy show that

(U.P.T.U., 2004, 2008)

273

MULTIPLE INTEGRALS

Sol. We have

zz ∞

0

∞ − xy e 0

z RSTz z LMMN z z LMNz LM z MN z ∞

sin px dx dy =

0

sin px

∞

=

Again

zz

0

∞

=

∞

∞ − xy

0

0

e

0

sin px dx dy =

=

0

OP PQ

∞

∞

∞

0

As the limits are constants.

dx 0

sin px dx x

0

UV W

e − xy dy dx

e − xy sin px −x

...(i)

OP Q

e − xy sin px dx dy

b

0

− e − xy p cos px + y sin px p2 + y 2

∞

p

∞

=

∞

0

p +y 2

2

dy =

LMtan F y I OP MN GH p JK QP

∞

−1

gOPP Q

= 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 parabola to 2 ay . ∴

x2

y2

= 4ay i.e., x = 2

zz 4a

0

2 ax x2 4a

dy dx

Y A

y2 = 4ax i.e., x = and from 4a

2

y

y2 ay i.e., x varies from 4a

=

=

=

zz z FGH 4a

0

4a

0

LM MM2 MN

4 = 3

=4

2

x = 4ay O

2 ay

y2 4a

dy dx

I dy JK O y y P a. − 3 12a PP PQ 2 64a a . a 4 af − 12a 2 ay − 3 2

y2 4a

4a

3

0

3 2

3

(4a, 4a)

ax

Fig. 4.17

X

274


=

32a 2 16 a 2 – 3 3

=

16 a 2 . 3

zz

a y

Example 10. Change the order of integration evaluate

0 y

2

aa − xf

a

y 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

0 y a

=

=

=

=

a

0 x

z

a

0

z z

a

0 a

0

a−x

P O

y

L O a f NM− eax − y j QP 2 12

LM e N

1 0 + ax − x 2 a−x ax − x 2 dx = a−x

z

y=

A(a, a )

x

X

(0,0)

dy dx

ax − y 2

1 a−x

a f

2 ax y=

Q

dx dy

ax − y 2

ax

ax

ax − y 2 = t ax −2 ydy = dt dx x y dy = − ax − y 2 2 ax − y

z

j OPQ dx

e

Fig. 4.18

j

12

12

a

0

x

a−x dx = a−x

a f

Let x = a sin2 θ i.e., dx = 2a sin θ cos θ dθ =

z

π 2

0

z

a

x

0

a−x

dx

a . sin θ . 2a sin θ cos θ dθ a . cos θ

FG 1 − cos 2θ IJ H 2 K L sin 2θ OP = a LM π − 0OP = aπ . a Mθ − N2 Q 2 N 2 Q

z

π 2

= 2a 0

sin 2 θdθ = 2a

z

π 2

0

= a

z

π 2

0

b1 − cos 2θg dθ

π 2

=

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

S

b g

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


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

i.e.,

S

b g

F x , y dx dy

=

zz

S

J du dv

δx δu δx δv

b g= δau, vf δ x, y

J =

Hence,

a f

F u, v

δy δu δy δv

δx δu δx δv

dx dy =

...(v)

δy δu δy δv

or

δx δu δy δu

δx δv δy δv

du dv.

Example 11. Transform to polar coordinates and integrates

zz

F1−x GH 1 + x

2

− y2

2

+y

2

I JK

dx dy

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

b g δ a r , θf

δ x, y

then

=

δx δr δy δr

δx δθ = δy δθ

cos θ r sin θ sin θ r cos θ

= r

Therefore, dx dy = rdθ dr and limits of r are from 0 to 1 and those of θ from 0 to

Hence

I =

=

= Now,

zz

F 1 − x − y I dx dy GH 1 + x + y JK z z FGH 11 −+ rr IJK r dθ dr π z2 FGH 11 −+ rr IJK r dr 2

2

2

2

π 1 2 0 0

2 2

1

2

0

2

suppose r2 = cos φ, 2rdr = – sin φ dφ =

π 2

z

0 π 2

1 2

FG 1 − cos φ IJ H 1 + cos φ K

(– sin φ) dφ

π , 2

277

MULTIPLE INTEGRALS

zz b


π 2 0

b1 − cos φ g dφ

π 4

=

π φ − sin φ 4

x+y

R

z

=

g

2

π 2 0

=

As

FG H

IJ K

π π −1 . 4 2

z

a

0

af

F x dx = –

z

0

a

af

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 δx δv δx

=

=

1 1 3 −2

=–5 v

v=3

3x

–

2y

=

3x

3

–

2y

=

0

Y

δu δy δv δy

R

O

x x

+

y=

R*

X +

y

=

u

O

2

0

u=2

Fig. 4.20

b g δau, vf δ x, y

1 5 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 So required Jacobian

zz b R

x+ y

g

2

J =

dx dy = =

z

Example 13.

zz

∞ ∞

0

0

e

c

− x2 + y 2

zz

u2

R*

1 5

zz z LMN

e

dx =

1 –5

du dv

3 2 2

0 0

u du dv

OP2 Q

1 3 u3 8 8 3 dv = ⋅ v0 = . = 5 0 3 0 5 15

h 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

=–

(U.P.T.U., 2002)

278


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

π ∞ 2 0 0 π 2 0

e −x

2

e−y

2

π 2

zz

π ∞ − r2 2 e r 0 0

a−2rfe a0 − 1f dθ = −r 2

dr dθ

dr dθ = –

1 2

z

π 2 0

1 2

z

π 2 0

e −r

2

π 1 dθ = 4 .

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

O

π . Proved. 2


zz

1 x

0 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

b g= δau, vf δ x, y

J = Jacobian =

1 = v

X

q=0

r=0

δx δu δy δu

Y

B (1, 1)

δx δv δy δv

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

LM v e1 + v j 1 MM 2 + 2 sin h N LM 2 + 1 sin h 1OP. PQ MN 2 2 2

1 3 1 3

= Example 15. Evaluate

zz b R

g

x+ y

−1

OP vP PQ

1

0

−1

2

dx dy, where R is the parallelogram in the xy-plane with

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

x+ y

R

g

2

zz z LMN 1

dx dy =

–2 1

D (1, –2)

1 u . du dv 3 2

1 u3 –2 3 3 1

=

4

OP Q

4

dv = 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


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

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

A (1,0)

O

θ =

when

dφ dθ =

zz π2

0

π 2

0

sin φ dφ dθ = sin θ =

zz zz 1

=

=

sin φ 1 ⋅ ⋅ dy dx sin θ sin φ cos φ

1− y 2

1− y 2

zz zz

1− y 2

0 0

1

0

FH

1 1 − sin 2 φ

1− y 2

e

LM MNsin

dy dx

IK

⋅ dy dx

dy dx y

−1

x 1−y

2

1 π 2y1 2 = π. 0 2

OP PQ

zz

j

0

dy y

=

π 2

2

z

Hence proved.

dy dx

e1 − y j − x

0 0

1− y 2

⋅

1− y 2

1

=

1 − x2 + y2 ⋅ y

0 0

z

1 cos φ sin φ sin φ

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

dy

2

x +y =1

x=0

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

2 4 2y

2

4

1

3

g

x + y dx dy.

zz 2

–1 – x

zz zz 0

− 4− 2 y 2 x a

0 x a

zz zz

ex

2

0

0

a

a2 −x2

zz 1

–2

2 -y

–y

b g

f x , y dx dy +

zz 2

1

y dx dy .

x

2−y

− 2−y

b g

f x , y dx dy

OP PQ

LMAns. 8 OP N 3Q

LMAns. a + a OP MN 28 20 PQ LMAns. πa OP 2 PQ MN LMAns. πa OP a PQ MN 3

j

+ y 2 dy dx.

2 ay − y 2

2a

LMA ns . MN

b g

4− 2 y 2

a

1

0 x3

f x, y dy dx.

2

OP Q

LMAns. z z f bx, yg dy dxOP N Q

f x, y dx dy.

2– x 2

160 21

Ans. 5

b g

1 y1 3

0 y2

zz

LMAns. N

y 2 dx dy.

0 y3

2

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 zz e zz d a a

0 y

1

1 πa . dy dx = x 2 + y2 4 2x − x 2

j

a2 − x 2

iy

0 x

a

0 0

ex

2

j

+ y 2 dx dy =

ex

2

+ y2

j

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.


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

dy dx .

2

x 4a

0

zz e 1 1

j

x 2 + y 2 dx dy +

0 0

zz zz

LMAns. = 16 a 3 MN

2 ax

zz

2 2− y

1 0

ex

2

e

ex

∞

∞

0

x

Ans. = e − 1

e−y dx dy. y

Ans. = 1

Change the order of integration in the following integrals: 19.

zz

20.

z ez

a lx

0 mx

b g

22.

23.

z z a cos a

a

ea

j 2

−x 2

x tan a

zz e zz d 2 x2

1 x

0 x

y l

j f bx, yg

dx dy.

LMAns. MN z

z

a sin a y cot a

0

0

b g

V x, y dy dx +

Vdy dx +

b g

z ez 0

x dx dy

e

x2 + y2

j

a yV ma l

a+

f x , y dy dx +

Vdy dx +

a2 − y2

z z

j

a

a sin a 0

j

i

zz la

2a

a

x 2 + y 2 dx dy.

2− x2

1

0

a a− a2 − y2 0 y2 2a

Vdx dy.

0

zz

ma y m

LM MMAns. z z N

a2 ax f

2 ax − x 2

21.

LM MNAns.

V x, y dx dy.

2a

OP PQ

LMAns. = 5 OP 3 Q N

j

+ y 2 dx dy .

1 ⋅ dx dy log y

1

0

2

.

z

O bx, ygdy dx.PP Q OP z Vdy dxPP Q

2a 2a y2 2a

a

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

26.

D

g

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

27.

R

x+y

g

2

Ans. 8

dx dy. R: parallelogram in the xy-plane with vertices (1, 0), (3, 1), (2, 2), (0, 1).

Ans. 21

zz

bx − yg / bx + yg dx dy, D: triangle bounded by y = 0, x = 1 and y = x. Use x = u – uv, LM ee − 1j OP y = uv to transfrom the double integrals. MNAns. 4 e PQ

28.

De

2

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 0

za 1

0

1

f

za

f a1 – xf

1 −1 −1 1 − t m ⋅ tn−1 dt = 1 − t m ⋅ tn−1 dt

1– x

fm−1 ⋅ xn−1 · dx =

β (m, n) = β (n, m) .

0

z

1 n−1 x 0

m −1

z

z

F By f atf dt = f axf dxI H K b

a

b

a

284


(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

(U.P.T.U., 2008)

m+ n

x m−1 (1 – x)n–1. dx

Lx O = M a1 − xf P Nm Q an − 1f x β (m, n) = m

or

m n

=

n−1

z

m Again integrate by parts above, we get

1

1

+

z

an − 1f m

0

1

0

x m (1 – x)n – 2 · dx (Integrate by part)

m

(1 – x)n – 2. dx

0

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

=

LM– lim e N

z

OP Q

∞

⋅ x n−1 + 0 + (n – 1)

–x

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→∞

=

x ex

lim ·

x→∞

= lim

x→∞

1

lim

1

x→∞

x n−1

+

∴ From (i), we get

1 xn−2

z

∞

n

x n −1 x x2 xn + +...+ +... 1+ n 1 2

= 0 + (n – 1)

1 + ... 2 ⋅ x n− 3

=

x n − 2 ⋅ e − x dx = (n – 1)

0

⇒

n

za

∞

x

f

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


(2) Put

xn =

∴ From (i), we get

z

∞ − z1 n e 0

or (3) Put

z in (i) then nxn–1 dx = dz and x = z1/n

n =

z

dz =

n n =

e–x =

t in (i).

– e–x dx =

Then ∴ From (i), we get

FG 1 IJ dz H nK a n + 1f

∞ − z1 n e 0

1 t

dt and ex =

zb

g LMlogFG 1IJ OP N H tKQ

0

n =

1

z

=

1

0

n−1

– log t

...(iii)

(–dt)

n−1

dt

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

z

∞ − z2 e 0

1 = 2

aπ f

1 = 2

or

1 π 2

dz =

As

z

∞ −x2

0

e

dx =

π 2

.

4.9 TRANSFORMATIONS OF BETA FUNCTION

(1) Putting

z

β (m, n) =

We know that x=

1 1 or dx = – 1+ y 1+ y

Also

b g

(1 – x) =

1

0

2

x m−1 (1 – x) n–1 dx

...(i)

dy

1 = 1+ y

1–

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 zb 0

∞

∞

0

∞

0

F 1 I GH 1 + y JK

m−1

F y I ⋅G H 1 + y JK

y n−1 dy

1+ y

g

g

R| −1 S| b1 + yg T

2

U| V| dy W

m − 1 + n −1 + 2

y n−1 dy

1+ y

n −1

m+ n

...(ii)

287

MULTIPLE INTEGRALS

(2) Also, as β (m, n) = β (n, m), therefore interchanging m and n in (ii), we have

z

β (n, m) =

y m −1 dy

∞

b1 + yg

0

.

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

dx

e , we get

n ⋅ x m−1 e − x =

z

∞

x n+ m −1 y n −1 e

0

...(i) ...(ii)

b g

− y +1 x

dy

Integrating both sides with respect to x within limits x = 0 to x = ∞, we have

n

z

But

∞

0

z

∞

0

x m−1 e − x dx =

x a n + m f −1 e

b g

− y +1 x

dx =

z LMNz ∞

0

b g

∞ n + m −1 − y +1 x x e dx 0

y n−1 dy

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

an + mf ⋅ 1 +y y b g bm + ng z b1y+ ygdy n−1

∞

m +n

0

∞

dy

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


Putting m + n = 1 or m = (1 – n), we get

a f

n⋅ 1− n

=

1

β (n, 1 – n)

zb zb ∞

β (m, n) =

But

∴

0

g

As

π sin nπ

n (1 − n) =

(3) To prove that

0

π x n−1 dx = 1+ x sin n π

(1 + n) (1 − n) =

z

π 2 sin m θ cos n θ 0

dθ =

β (p, q) =

We know that

z

∞

.

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

or

m+ n

π ,n<1 sin n π

= ∴ From (i), we have

g

y n−1 dy 1+ y

∞

β (n, 1 – n) =

y n−1 dy

1+ y

0

...(i)

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 z x a1 − xf dx 1

q −1

p −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

z

0

sin 2 p −1 θ cos 2 q −1 θ dθ =

=

z

0

= 2 π2

π2

esin θj e1 − sin θj

π2

0

2

2

q −1

sin 2 p −1θ cos 2 q −1 θ dθ

1 β(p, q) 2 p⋅ q 2

p −1

bp + qg

. 2 sin θ cos θ dθ

289

MULTIPLE INTEGRALS

Putting 2p – 1 = m and 2q – 1 = n, or

FG m + 1 IJ and q = FG n + 1IJ , we get H 2 K H 2K F m + 1I F n + 1I H 2 K H 2K . F m + n + 2I 2 H 2 K

p =

z

π2

0

sin m θ cos n θ dθ =

4.12 DUPLICATION FORMULA To prove that

FG m + 1 IJ = aπf H 2K 2

m

2 m− 1

.

Hence show that β (m, m) = 21–2m β

z

Since

π2

0

sin 2 m −1 θ cos 2 n−1 θ dθ

Putting 2n – 1 = 0 or n =

z

π2

0

z

π2

0

2m .

(U.P.T.U., 2001)

FG m, 1 IJ . H 2K a mf a n f = 2 am + nf

1 , we obtain 2

sin 2 m −1 θ dθ =

amf FH 12 IK F 1I 2 m+ H 2K m⋅ π


2

Fm + 1I H 2K

Again putting n = m in (i), we get

z

0

i.e.,

i.e.,

1 2 2 m −1

o amft 2 a2mf o amft 2 a2mf o amf t 2 a 2mf 2

π2

z

sin 2 m −1 θ cos 2 m −1 θ dθ =

π2

0

1 2 2m

b2 sin θ cos θg

z

π2

0

bsin 2θg

2

2 m −1

dθ =

2

2 m −1

2 dθ =

...(i)

...(ii)

LM3 FG 1 IJ MN H 2K

= π

OP PQ

290


Putting 2θ = φ and 2 dθ = dφ, we get 1

z z

π

22m 0

2

or

2

2m

sin

φ dφ =

2

π2

sin

0

z

or

o amft 2 a2 mf o amft 2 a2 mf 2 { amf} 2 a2mf 2

2 m− 1

2 m −1

φ dφ =

2 m −1

π2

0


As

z

a

0

af

f x dx = 2

z

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

2 2 m−1 2

m

Hence, ⇒ Multiplying above by

m

amf a2mf

2

=

FG m + 1 IJ H 2K

=

a mf ⋅ a π f F 1I 2 m+ H 2K

As

1 = π 2

aπf a2mf 2 2 m −1

m , we get

FG m + 1 IJ H 2K

m

1 2

=

2m

2m ⋅ m 2 2m − 1

21 − 2m m =

β (m, m) =

Sol. We have

a

sin 2 m −1 θ dθ from (ii) and (iii), we get

m m

Example 1. Compute:

af

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 = − H 2K

or

π 3 2

FG − 1 IJ H 2K

⇒

FG − 3 IJ FG 5 IJ H 2K H 2K FG − 3IJ H 2K

π

=

F 1I = – 2 H 2K

π

...(ii)

= −2 π .

F 5I H 2K

=

F H

π

I K

= –

3 sin − π 2

F H

π 3 sin π 2

I K

π 4 π = . 3 1 3 ⋅ π 2 2

=

= π

...(iii)

5 , in (i), we get 2

(3) Putting n = –

or

π 1 2

3 , in (i), we get 2

(2) Putting n = –

or

= −

FG − 5 IJ FG 7 IJ H 2K H 2K

=

FG − 5 IJ = π H 2K F 7I H 2K

=– = –π 5 I 5 I F F sin − π H 2 K H sin 2 πK

=

−

Example 2. Evaluate Sol. Putting

∞

0

π

π 8 π =− . 5 3 1 15 ⋅ ⋅ ⋅ π 2 2 2

dx · 1 + x4 x4 = y or dx =

z

∞

0

Also, we have

z

π

z

∞

0

z

dx 1 + x4

x n−1 dx 1+ x ∞

0

dx 1 + x4

=

z

dy 4x 3

, we have

3

1 −4 y dy 4 1+ y

∞

b g

0

z

FG 1 − 1IJ H K ∞ y 4

=

1 4

=

π , we get sin nπ

=

1 ⋅ 4

0

1+ y

π sin

F πI H 4K

=

dy ⋅

π 2 · 4

...(iv)

292


Example 3. Show that Sol. We have

z

π2

0

z

π2

0

tan n θ dθ

tan n θ dθ =

=

z

=

=

=

= Example 4. Show that Sol. We have

z

π2

0

z

π2

0

tan θ dθ

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

=

FG IJ H K

1 1 π sec nπ ⋅ 2 2

sin n θ cos θ

2 1 1 π ⋅ 1 2 sin n + 1 π 2

a

=

=

= =

As n

a1 − nf =

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 f

1 2

tan θ dθ =

=

−n

z

FG 3 IJ FG 1 IJ H 4K H 4K

=2

z

∞

0

x 2 dx 1 + x4

asin θf 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

12

−1 2

0

FG H

1 3 1 , β 2 4 4

IJ K

β (m, n) =

m⋅ n

am + nf

·

π sin nπ

293

MULTIPLE INTEGRALS

1 = 2 =

z

e 3 − 1j

y 4 dy β(m, n) = 1+ y

∞

b g

0

1 2

z

∞

x −1 ⋅ 4 x 3 dx

e

1 + x4

0

z

j

z

∞

0

y m −1 dy

b1 + yg

. 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

R.H.S. = =

=

=

z 1

Example 6. Show that

0

∞

m m⋅ n

m⋅ n

∴

z

(1 − x n )

+

= β (m, n). Hence proved.

( m + n)

F 1I H nK . F 1 + 1I n H n 2K π

=

(1 − x n )

dx

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

dx

dx =

(U.P.T.U., 2008)

am + 1f ⋅ n + m ⋅ an + 1f am + 1 + nf am + n + 1f

Sol. Put xn = sin2 θ or x = sin2/n θ ∴

m+ n

F2 I

G − 1J 2 sin H n K θ cos θ dθ n

=

F I H K

=

2 n

=

2 n

=

1 ⋅ n

2 n

z

z

0

π/ 2

0

FG H

2 −1 π/2 sin n

IJ K θ cos θ dθ

b1 − sin θg

FG 2 − 1IJ sin H n K θ

2

cos0 θ dθ

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 F 1 I ⋅ F 1I H n K H 2 K = H n K ⋅ π · Hence proved. RS 1 F 2 + 1I UV n F 1 + 1 I H n 2K T2 H n KW

294


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

xe − ax cos bx ⋅ dx =

0

a2 − b 2 , where a > 0. (a2 + b2 )2

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

=

=

1 1 − tan 2 θ ⋅ = cos θ 2 2 2 ( a + b ) 1 + tan 2 θ r2 2

1 a2 + b2

a2 − b 2 ( a 2 + b 2 )2

P =

or

P =

.

F 1 I F 2 I F 3 I ..... H nK H nK H nK F 1 I F 2 I ... F n − 2 I H nK H nK H n K F 1 I F 2 I .... F1 − 2 I H n K H n K H nK

Example 8. Find the value of Sol. Let

FG1 − b H a ⋅ FG1 + b H a

2 2 2 2

IJ K IJ K

Hence proved.

F n − 1I , where n is a positive integer. H nK F n − 1I ...(i) H nK F1 − 1 I H nK

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) =

F 1 I ⋅ F1 − 1 I H nK H nK

or

∴

=

P2 =

π sin nπ π sin π

π n

π π π ... ⋅ π n−2 n−1 2π sin sin sin π sin π n n n n

...(iii)

...(iv)

296


From Trigonometry, we know that

sin nθ = 2n–1 sin sin θ Putting θ = 0

Fθ + π I H nK

sin

F θ + 2π I H nK

... sin

F θ + n − 2 πI 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

I K

π 2π n−2 n−1 π sin .... sin π sin n n n n

∴

P2 = πn–1

∴

P =

Hence

sin

F 1 I F 2 I ... F n − 1I H n K H nK H n K

=

1 = 2

Example 9. Prove that Sol. We have

(2π) n −1/2 n ( 2π)( n−1/2) n

⋅

π·

( n) (1 − n) = Putting

2 n−1 n

n =

F 1I F 1I H 2K H 2K

π sin n π 1 , we get 2 π

=

sin

LM F 1 I OP N H 2KQ F 1I H 2K

π 2

2

Example 10. Show that

zz

D

= π π . Hence proved.

=

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

F θ + n − 1 πI H n K

297

MULTIPLE INTEGRALS

z 1

= al+m

X l −1

z

al + m X l −1 (1 − X )m dX m 0 1

=

a l + m ( l ) ( m + 1) al + m ⋅ β(l , m + 1) = m ( l + m + 1) m

=

al + m

z 1

( l) ( m) ( l + m + 1)

x m −1 + x n −1 (1 + x ) m + n

0

Sol. The given integral

z 1

I =

0

or

m 1− X 0

0


LM Y OP dX = NmQ

· Hence proved.

dx .

x m −1 (1 + x ) m + n

z 1

dx +

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 z H zL1K + F H1I Oz K MN H z K PQ n −1

1

I2 =

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) m+ n

z

∞

=

0

=

x m −1 dx (1 + x ) m + n

(m) ⋅ ( n) ( m + n)

z

∞

x m−1 dx

1

I =

+

1

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 H1− x K 3

3

1/2

dx ·

(U.P.T.U., 2006)

298


Sol. Let x3 = sin2 θ ⇒ 3x2dx = 2 sin θ cos θ dθ ∴

z

π/2 0

z

π/2

sin θ sin θ cos θ 2 sin 2− 4/3 θ dθ ×2 dθ = 4 / 3 3 0 cos θ θ 3 ⋅ sin 5 1 2 2 ⋅ π π/2 2 2 2 3 2 3 3 = = sin 2/3 θ cos 0 θ dθ = 3 3 2/ 3 + 0 + 2 4 3 0 2 2 2 3

z

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 t2

⋅ e −t ⋅ dt = 3

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

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 (a + b ) 2

( n) bn

2 n2

nπ cos 2

dz and x = z 1/n n

, where θ = tan–1

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

=

π 1 1 ⋅ = 2 2 2

...(i)

π · 2

z

∞

cos

−∞

Putting b =

nπ 2

F 3 I cos π H 2K 4

∞

∴

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

z

F H

∞

I K

π cos x 2 dx = 2

0

z

∞

∴ From (ii)

cos

−∞

πx 2 dx = 2

F 3I H 2 K cos π 4 F πI H 2K F 3I H 2 K cos 2 F π I 1/2 H 2K 1/2

= 2·

π 4

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

1

n −1

dx =

(n) , n > 0 .

6. Prove that

0

1

7. Show that

z z z 2

x 2 (1 − x ) 3 dx = 60.

z

2

8. Show that

0

(1 − x n ) 1/n dx =

LM F 1 I OP 1 N H nK Q ⋅ n F 2I 2 H nK

( 4 − x 2 ) 3/2 dx = 3π.

2

·

300


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

6

=

z

3 3

xne−k

12. Prove that

·

0

z

F1I H 3K · F 5I H 6K

π

x 2 dx

=

(1 − x 3 )

1

π

3

x 2 dx (1 − x ) 3

=

2 · 3

π/2

14. Show that, if n > – 1, ∞

10. Show that

0

dθ × (sin θ)

0

z z 1

dx

sin θ . dθ = π .

0

2 2

x

1

dx =

2k n+1

z

0

F n + 1I . H 2 K

0

Hence or otherwise evaluate

15. Show that

z

17. Prove that 19. Evaluate

z

20. Prove that

4.14

∞ −4 x

e

∞ m− 1 x 0

z

∞

0

2 2

x

dx.

−∞

F 1 I ⋅ F 5I H 3K H 6 K F 2I H 3K

0

e −k

=

x 3 2 dx =

( π) 2 1/ 3 ·

cos bx dx and

x m−1 e − ax dx = 2

aπ f ·

3 128

z

∞

0

n 2a n

16. Show that

18. Prove that

x m−1 sin bx dx .

LM F 1 I OP N H 3KQ F 1I H 6K

z

∞

0

2

( π) ⋅ 21/3

=

31/2

x 2 e − x dx =

LMAns. N

2

m bm

cos

·

aπ f · 4

mπ m mπ ; m sin 2 b 2

OP Q

·

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

y=a

f ( y)

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

x=a

f1 ( x )

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


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

x

...(ii)

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. Whole area = 4 × area OAC

zz a

= 4 where

f ( x)

x=0 y=0

Y

dx dy

C

y = f(x) i.e.,

X¢

(a2 x 2 − x 4 )

y=

a

O

a

is the equation of the curve = 4 = 4 =

z

z

a

x=0

a

0

y

f (x) dx 0

f ( x)dx =

z

4 a

z

Y¢ a

0

( a 2 x 2 − x 4 )dx

z

Fig. 4.29

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

= 4a2

z

π/2

0

sin θ cos 2 θ dθ =

1 π 4 a2 2 = = · 3 1 3 2⋅ ⋅ π 2 2 4a 2 1 ⋅

4a 2 1 ( 3 / 2) 2 (5 / 2)

a

P(x,y) 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 y

dx dy

IJ K

xy = 2

F 2 y − log yI H3 K LF 16 − 2 log 2I − 2 OP 2M K 3Q NH 3 4

3/2

= 2 =

(2, 1)

(0, 1)

2 2 y − dy y

O

X

(2, 0)

Fig. 4.30

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

y=0

f ( y)

x=0

dy dx

is the equation of the given

curve. From eqn. (i), required area =2

2 = a

z z

2a

0

2 f ( y)dy = a

π/ 2

0

= 32a2

= 32a2·

z

(2a sin θ) 2

π/2

0

z

2a

0

3/2

y

3/2

2 ( 4)

...(i) Y

A

(0, 2a)

dy

B

2 a − y dy

2 a − 2 a sin θ ⋅ 4 a sin θ cos θ dθ


F 5I F 3 I H 2K H 2K

(U.P.T.U., 2001)

2

(Put y = 2a sin2 θ)

3 1 1 ⋅ π⋅ ⋅ π 2 = 32a2· 2 2 = πa2. 2⋅ 3⋅ 2⋅1

O

X

Fig. 4.31

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


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= Required area = = = =

O

A¢ (8a, 0)

N (4a, 0)

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]

2

= 64πa – 2

z

LM N LM MN2

X

P¢

( 64 a 2 − x 2 )

= 64πa2 – 2

A (8a, 0)

4a

0

(12 ax ) dx +

( 3 a)

= 64πa2 – 4 ( 3a)

z

RS 2 x T3

4a

0

3/2

z

8a

4a

Y¢

Fig. 4.32

( 64 a 2 − x 2 ) dx

OP Q

RS 1 x (64a 2 − x 2 ) + 1 ⋅ 64a 2 sin −1 F x I UV 8a OP H 8 a K W 4a PQ 2 T2 UV − 2RS32a sin (1) − 1 ⋅ 4a ⋅ 48 a − 32a sin F 1 I UV H 2KW 2 W T xdx +

4a

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

θ=α

f ( θ)

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 β

The required area =

θ=α

f ( θ)

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 From (i)

x =

a− y

From (ii)

e

j

a− y

2

...(ii)

⇒ x=

e

a− y

j

2

Y (0,a)

B

+y = a

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 a

=

0

a

=

=

0

LM 2 MN

b a− y g x = e a − yj a− y−

e

2

dx dy =

a− y

a . y 3 2 2y 2 − 32 2

OP PQ

z

a

0

x

b a − yg e a − yj

z

j UVWdy = e2 2

a

0

a

= 0

2

. dy

j

ay − 2 y dy

4 2 a2 . a − a2 = 3 3

Ans.

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

Sol. We have

4y = x2 2

From (i) and (ii) ⇒

x 4

=

3x x2 + 2

x4 + 2x2 – 12x = 0

y=

...(i) ...(ii)

3x 2 x +2

(y = O (0, 0)

Fig. 4.35

(2,

3 4

A

x

2

4

)

)

306


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 =

LM 3 log ex + 2j − x OP 12 Q N2 3

2

=

z

2

0

2

= 0

y

e

3x x 2 + 2 2

x 4

b

j dx

=

z

F 3x − x I dx GH x + 2 4 JK 2

2

0

2

g

3 8 log 6 − log 2 − 2 12

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

= 4

π/ 2

− π /2 r = 2

π/2 2(1+ cos θ)

0

z

2

π/ 2

0

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

2(1+ cos θ)

dθ 2

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

a

(1 + cos θ) 2 − 1 dθ

r=a

s

π/2

0

bcos

2

g

q)

= a2

θ + 2 cos θ dθ q=p

F 1 ⋅ π + 2I = a (π + 8). H2 2 K 4

= a2

q = p/2

co

0

dθ

+

= a2

π/2

A

a( 1+ cos θ )

(1

z z

0

2

a

z

FG r IJ H 2K

r=

= 2

π/ 2

O

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

z z z LMN z

π/2 a sin θ

∴ Required area =

=

0

a( 1 − cos θ )

π/ 2

r2 2

0

1 = 2 =

a2 2

=

a2 2

OP Q

z z

r dr dθ

a( 1− cos θ)

dθ

O

Fig. 4.38

a sin θ − (1 − cos θ)

π/ 2

0 π/ 2

0

r = a sin q

a sin θ

π/2 2

0

r = a (1 – cos q)

2

2

zz z FH

π/2 r for cardioid

θ =0 r for parabola π/2

= 2 =

z

0

π/2

0

1 2 r 2

I K

X

(sin 2 θ − 1 − cos 2 θ + 2 cos θ ) dθ

F H

( −2 cos 2 θ + 2 cos θ) dθ = a 2 1 −

1+ cos θ

I K

π . 4 q = p/2 F

B E

q rd

dr

q=0

q

O

A

D

r dθ dr C

1/(1 + cos θ)

l

q=0

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

B q=0

D

2

dθ

q

(1 + cos θ) 2 − 1 / (1 + cos θ)

2

dθ

Fig. 4.39

308


z

Now,

= π/2

0

(1 + cos θ)2 dθ = = = =

z

z z z

π/ 2

0 π/2

0

(1 + cos θ) 2 dθ −

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

F H

I K

LM 3 θ + 2 sin θ + 1 sin 2θOP 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 0

a

1 / 1 + cos θ

f

2

3 1 π + 2 = ( 3π + 8) 4 4

1 dθ = 4 =

1 2

z z ze ze π 2 0

sec 4

π 4 sec 4 0

...(ii)

FG θ IJ dθ H 2K

As 1 + cos θ = 2 cos 2

φ dφ, Putting

π

1 θ= φ 2

j

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 =

j IJ K

FG H

a

FG 1 θIJ H2 K

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

−π 4

2 = 2a

z

π 4

0

r dθ dr a cos 2θ

. dθ = 0

z

π 4

−π 4

a 2 cos 2θ dθ

2a 2 sin 2θ 2

cos 2θ dθ =

π 4 0

LM N

= a 2 sin

OP Q

π − sin 0 = a 2 . 2

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

a2 2

π 2 −π 2

{bsec θ + cos θg

2

) sq co + cq (se

(a,o) (2 a,o) r = asecq q =-p 2

= a2 = a2

=

=

a2 2

ze ze zb π2

0

π 2

0

π 2

0

LM N

Fig. 4.41

j

sec 2 θ + 2 sec θ cos θ + cos 2 θ − sec 2 θ dθ

j

g

5 + cos 2θ dθ =

OP Q

a 2 5π 5πa 2 −0 = . 2 2 4

z

FG 2 + 1 + cos 2θ IJ dθ H K 2 a L sin 2θ O 5θ + M 2 N 2 PQ

2 + cos 2 θ dθ = a 2

π2

0

2

π 2

0

}

− sec 2 θ dθ

310


Example 14. Find the area included between the curve x2y2 = a2 (y2 – x2) and its asymptotes. Required area = 4 × area OAB = 4

Sol.

z

a

0

y dx = 4

z

ax

a

ea

0

2

− x2

j

dx

B

Y

Q x=a

P

X¢

A X N M (a, 0) dx

O

Y¢

= 4a

z

Fig. 4.42

a sin θ ⋅ a cos θ dθ , a cos θ

π2

0

= 4 a2

z

π2

0

Putting x = a sin θ

sin θ dθ = 4 a 2 – cos θ

π2 0

= 4 a2 .

Example 15. Show that the entire area between the curve

= 2

= 2

z z

2a

0

y dx = 2

= 16a 2

Q 2a

0

e2a sin θj 2

π2

0

z

z

x3 2

0

P

a2 a − xf dx 32

4 a sin θ cos θ

2 a ⋅ cos θ

π2

B

Y

y2 (2a – x) = x3 and its asymptote is 3πa2. Sol. Required area = 2 × area OAB

sin 4 θ dθ = 16a 2

= 3πa2 . Hence proved.

X¢

N

O

dθ

3 1 π ⋅ ⋅ 4 2 2

Y¢

Fig. 4.43

x = 2a

M A (2a, 0) dx

X

311

MULTIPLE INTEGRALS

EXERCISE 4.4

LMAns. MN

1. Find the area bounded by the parabola y2 = 4ax and its latus rectum. 2. Find by double integration the area of the ellipse

8 a2 3

OP PQ

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. MN

6. Find the area of the curve 3ay2 = x (x – a)2. 7. Find the area included between the curve and its asymptotes. 8. Find the area between the curve y2 (2a – x) = x3 and its asymptote.

OP 3 PQ

8 a2 15

Ans. πa 2 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 Find the area included between the curves y = 4a (x + a) and y = 4b (b – x). MAns. 3 PQ MN LMAns. 1 πa OP Find the whole area of the curve r = a cos 2θ. N 2 Q LMAns. 1 πa OP Find the area of a loop of the curve r = a sin 2θ. N 8 Q 2

10. 11. 12.

2

13. Find the area enclosed by the curve r = 3 + 2 cos θ.

2

2

2

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. MN

πa 2 2

OP PQ

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


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.

∑ f bxr , y r , zr gδVr ⋅ n

Find the sum

r =1

Then

zzz

V

b

g

Lim

n

δVr → 0

r =1

b

g

n → ∞ ∑ f xr , yr , zr δVr ⋅

f x, y, z dV =

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 y2 x

y1 x

g

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 zzz

V

V

b

g

=

b

g

=

F x, y, z dV F x, y, z dV

z z aa ff z bb gg b z LMNz aa ff RSTz bb gg b zz z b zz z b LM z z MN b LM z z MN b 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 =

=

⇒

0 0

0

1 1− x 1− x − y

0 0

0 0

= −

1 2

g UVW OPQ

F x, y, z dz dy dx . dx dy dz

g

x + y + z+1

3

⋅

dx dy dz

g OP 1 1 − dx dy 2 x + y + z + 1g P Q OP 1 1 − dx dy 4 x + y + 1g P Q

0

1 1− x

g

F x, y, z dx dy dz

x+ y + z+1

1 −x − y

2

0

1 1 −x

0 0

3

2

313

MULTIPLE INTEGRALS

⇒

= −

1 2

= −

1 2

z z LMMN z LMN z LMN a z FGH 1 1 −x

0 0

1

0

1 2

= −

1

0

OP dx dy g PQ

1 1 − 4 x + y +1

b

1 1 y+ x + y+1 4

f

OP Q

2

1− x

dx 0

OP Q

1 1 1 dx 1− x + − 4 2 x+1

IJ K 1 L3 1 O − M x − x − log ax + 1fP 2 N4 8 Q 1 L3 1 5O O 1L − M − − log 2P = Mlog 2 − P. 2 N4 8 8Q Q 2N 1 2

= −

1

0

3 1 1 dx − x− x+1 4 4

1

=

2

0

= Example 2. Evaluate Sol. We have

zz ze 4 2 z

0 0

0

I = = =

j dz dx dy .

4z− x2

zz

4 2 z

zz

0 0

4 2 z

0 0

z

4

0

1 x 2

z LMN z

e4 z − x j dz dx 2

z

0

e4z − x j dz dx 2

e 4z − x j 2

1 + ⋅ 4 z sin −1 2

FG H

IJ OP 4z K Q

x

2 z

dz 0

a 4z − 4zf + 12 ⋅ 4z sin −1 a1fOPQ dz L1 O πzdz = π M z P = 8π. N2 Q + z j dx dy dz where V is the volume of the cube bounded

4

=

zzz e

=

1 ⋅2 z 0 2 4

4

2

0

0

x +y Example 3. Evaluate 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 = =

= =

2

2

2

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


=

z FGH a

0

IJ K

1 4 1 4 a + a dx 3 3

x 2a 2 +

LM 1 x a + 1 a x + 1 a xOP = a . 3 3 N3 Q zzz b2x + yg dx dy dz, where V is a

3 2

=

4

4

5

0

Example 4. Evaluate

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 z z z

Z

V

z z z b z b z

b2x + yg dx dy dz 2x + yg z dx dy 2x + yge 4 − x j dx dy 8x − 2x + e 4 − x j y dx dy LM8xy − 2x y + 1 e4 − x jy OP dx 2 N Q 16 x − 4 x + 2e4 − x j dx LM8x − x + 8x − 2 x OP 3 Q N FG 32 − 16 + 16 − 16 IJ = 80 . H 3K 3 2

2

4− x 2

x=0 y=0 z=0 2

2

4− x 2 0

x=0 y=0

2

2

(0, 0, 4) z = 4–x

2

(2, 0, 0)

O

dxdy

z=0

(0, 2, 0) Y

Fig. 4.45

2

x=0 y=0 2

2

3

2

x=0 y=0

2

2

3

2

x=0

2

0

2

3

2

0

2

2

4

3

0

=

EXERCISE 4.5 1. Evaluate the integral 2. Evaluate

zzz

S

z zz

log 2 x x + log y

0

0 0

e x + y + z dx dy dz.

LMAns. N

8 19 log 2 − 3 9

OP Q

xyz dx dy dz , where

S = [(x, y, z) : x2 + y2 + z2 ≤ 1, x ≥ 0, y ≥ 0, z ≥ 0].

LMAns. 1 OP N 48 Q

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

LMAns. N

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]

z zz zzz z z z

log 2 x x + log y x + y + z e dx 0 0 0 2π

0

OP Q –5 O 48 PQ

π a 2 4 r 0 0

π 2

a sin θ

0

0

LMAns. N

dy dz .

LMAns. N

sin θ dr dθ dθ.

a 2 −r 2 a 0

OP Q O 2πa 2 − 1jP e 3 2 Q LMAns. 5a π OP 64 Q N 3 19 log 3 − 8 9 3

3

r dθ dr dz .

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


y2 =1 ...(iii) 1 32 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

+

9 − x2 9

x varies from – 3 to 3. Thus, the volume V bounded by the elliptic paraboloids V= =

y2 x 1

Z

2

1

2

2

z = x + 9y

dy dx

1

=

2

–3

9 y − x y − 3y

j

9 −x 2 9 9− x 2 − 9

3 3 9 − x2 2 –3

j

O

dx

8 dx 9 x = 3 cos θ so dx = – 3 sin θ dθ =

Putting

3

= 72

z

π

z

0

= 144

2

= 2

2

2

1

3

2

Z = 18 – x – 9y

zz z zz zz z e ze

a f z bx , yg dz dy dx –3 y a x f z b x , y g 3 y ax f e18 − x 2 − 9y 2 j − ex 2 + 9y 2 j –3 y a x f 3 y ax f 2 e9 − x 2 − 9 y 2 j dy dx –3 y a x f 3

D X (3, 0, 0)

2

2

Y (0,1,0)

x + 9y = 9

Fig. 4.46

sin 4 θ dθ π 2

0


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

dy z

z z z z z 1

0

dx

1− x − y 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 LM x − x + x OP = 1 − 1 + 1 = 1 . MN 2 2 6 PQ 2 2 6 6 1

=

0

1

=

2

dx 1 − x − x 1 − x −

2

2

0

0

3 1

2

=

2

1

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

dx dy dz = 4

= 4

z z LM z MN ze π 2 0

dθ

z z π 2 0

2 a sin θ

0

dθ

r dr z

2 a sin θ

r =0

rdr

4a 2 − r 2 0

z

4a 2 − r 2

z= 0

=4

z z LM z MN e ze π 2 0

dθ

O e j PP = 43 Q 8 × 4a = 4 – 8a cos θ + 8 a j dθ = 3 3 3 π 32a 3 I F 1 2 G 1 – cos 3θ − cos θJ dθ = K 3 z0 H 4 4 = 4

π 2 0

1 dθ − 4 a 2 − r 2 3

π 2 0

3

LM N LM π − 2 OP. N 2 3Q

3

3 2 a sin θ 2 0

3

32a 3 1 3 θ − sin 3θ − sin θ = 3 12 4 =

32 a 3

3

OP Q

π 2

0

Volume in 4 quadrant

dz

2 a sin θ

0

π 2 0

=

r dr 4a 2 − r 2

− 4 a − 4a sin θ

3

π 2 0

LM N

2

2

2

j

1 − cos 3 θ dθ

32a 3 π 1 3 + − 3 2 12 4

OP Q

j

3 2

OP PQ

+ 8 a 3 dθ

318


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

x2

1

0

dx

dy

x

12 + y − x 2

0

x2

dx 12y +

1

12 x +

0

F = G 8x H

3 2

Y

j

12 + y − x 2 dy

1

0

dz

I JK

y2 − x2 y 2

x 2

5 − x2

x 2

y =x x2

− 12x 2 −

I JK

x4 + x 4 dx 2

7

x2 2 2 x5 x5 + − x − 4x 3 − + 4 7 10 5

I JK

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

=

0

– ax − x 2

a

ax − x 2

0

– ax − x 2

dx

= 2 = 2

z z

ax dx y

0

FH

a

dy

ax − x 2

0

– ax − x 2

ax

− ax

af

dy z

a

ax −x 2 − ax − x 2

z

ax − ax

dy ax

O 2

IK

z

a

z z

= 8a 3

= 8a

3

π 2 0

0

a sin 2 θ a − a sin 2 θ ⋅ 2a sin θ cos θ dθ

π 2 sin 3 0

θ cos2 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 ax

ax − x 2

dx

=

a

= 2 dx a

Z

dx dy dz

2

=

zzz z z z z z z

z

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

( z) r a

2

b

dz ⋅ r dr dθ

−r 2

g

r dr dθ =

zz

2π a

0

d

2

0

2π

a

3 2 2

i

2

2π

3

0

2

3 2

0

0

= =

2π 0

3

e2 − 2 j πa3

2

a 2 − r 2 − r ⋅ r dr dθ

O LM eM a − r j − 1 r PP dθ = 1 LM−F a I z MM −2 ⋅ FH 3IK 3 PP 3 z MMN GH 2 JK 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

a

+a − 3

OP P dθ 2P Q

a3 2

3

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


⇒

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 = 4

∴

= 4

zzz z z

π π a 2 2 4 r 0 0 0

π π 2 dφ 4 sin θ 0 0

e

z z z

sin θ dr dθ dφ = 4 a3 4a3 dθ ⋅ = 3 3

FG H

j

π 2 0

dφ

π 2 dφ 0

π 4 sin θ 0

− cos θ

=

IJ K

a

3

0

π 4 0

4a π 2 3 1 2 −1 = πa 1 − . 2 3 3 2 2 Spherical coordinates: 3

Lr O dθ M P MN 3 PQ L 4a φg M− = b 3 N 3

π 2 0

1 2

OP Q

+1

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. Volume V = 4

z z zd 4

16 − x 2

4

0 0

x +y 2

2

i

dz dy dx = 4

4

zz

x2 + y2 , z = 4, y = 0, y = 4

16 − x 2

4

0 0

z

4 x2 + y 2

d

i dy dx

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

= 4 =

LM 1 16 − x 16 − x − 1 16 − x OP dx e jP 12 MN 4 e j Q 1 2 16 − x j dx = e e16 − x j dx 6 3

z z za z

= 4

2 3

4

0

4

0

π 2 0

3 2 2

2

2

3 2 2

z

4

2 32

0

f

16

32

⋅ cos 3 θ ⋅ 4 cos θ dθ, where x = 4 sin θ

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

4 = a

a2 −x2

a

x=0 y=0 a

0

y3 x y+ 3

x2 +y2 a

z=0

0

4 dx = a

z

a

0

zz a

4 a

dx dy dz =

a 2 − x2

2

2

az=x +y

LM MMx N

a2 −x2

0 0

2

a −x 2

ex

2

2

j

z

=

2

ea +

2

− x2 3

j OPdx PP Q 32

z

Fig. 4.51

LM sin θ cos θ dθ + a cos θ dθOP 3 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 MN 2 3 3. 2 3 PQ 4 N 2 2 3. 2. 2 PQ L 2π O πa . a M P = N4Q 2 4 4 a a

π 2

2

3

4

2

0

π 2

4

0

4

=

2

+ y 2 dx dy

Putting x = a sinθ i.e., dx = a cosθ dθ =

2

x +y =a

4

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


z varies from 0 to c + xy Hence, the volume is V =

=

zz z z LMMN a bx a

c + xy

0 0

0

a

0

xy 2 cy + 2

LM bcx MN 2a

2

=

1. Find

zzz

V

+

OP PQ

b 2x 4 8a 2

bx a

dx = 0

OP PQ

a

= 0

zz

bc + xygdx dy F bcx + b x I dx GH a 2a JK

a bx a

dx dy dz =

z

0 0

a

0

2 3 2

b

g

bca 2 b 2 a 4 ab + = 4c + ab . 2 2a 8 8a

EXERCISE 4.6 x 2 yz dx dy dz , where V is the volume bounded by the surface x2 + y2 = 9,

LMAns. N

OP Q

648 5 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 . 4 − x2 x 2 + 2 y 2 = 4, so limits are z ; x 2 + 3 y 2 to 8 – x 2 – y 2 ; y : ± , x: ± 2] 2 z = 0, z = 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 : ± 4 N

OP Q

1 − x2 ⋅

Ans. 3π

4. Find the volume of the solid common to two cylinders x2 + y2 = a2, x2 + z2 = a2.

LMHint: z: MN

±

a2 − x2 ,

y :

x : ±a

±

OP PQ

a2 − x 2 ⋅

LMAns. MN

16 a 3 3

OP PQ

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. N

566 140

OP 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

π π xy z 2 2 cos dz dy 0 x 0 x

9.

LMAns. 7 OP N 2Q LMAns. FG π − 1IJ OP N H 2 KQ

xyz dx dy dz .

dx .

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. N

x2 + y2 = R2. 12. Find the volume of the tetrahedron by the coordinate planes and the plane

4.17

πR 4 2a

OP Q

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.

1 x+y=

A

X

324


zz zz zz z

1 1 − x l −1 m −1

=

x

0 0

y

LM z OP dx dy NnQ b1 − x − yg dx dy n 1 −x − y 0

1 1 − x l −1 m −1

Put

1 n x y n 0 0 y = (1 – x) t, so that dy = (1 – x) dt

Then

I =

=

ma f r ma f a f r a f a f z a f a f a f

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 l −1 am+ n−1f−1 dx 1tm−1 1 − t an +1f−1 dt x 1−x = 0 n 0 1 = β l, m + n + 1 ⋅ β m, n + 1 n 1

= n =

1 n

=

1 n

alf ⋅ amf anf a l + m + n + 1f

=

zzz

Hence,

alf ⋅ am + n + 1f ⋅ amf an + 1f al + m + n + 1 f am + n + 1 f alf ⋅ amf an + 1f al + m + n + 1f alf ⋅ amf ⋅ n anf al + m + n + 1f

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

bp

1

1

2

3

n

+ p2 + p3 + − − − − + pn + 1

g

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

FG x IJ + FG y IJ + FG z IJ H aK H bK H cK q

p

Sol. Putting

r

zzz

325

x l −1 y m−1 z n −1 dx dy dz where x, y, z are always positive with

≤ 1.

(U.P.T.U., 2005)

FG x IJ H aK

FG y IJ H bK

p

= u,

FG z IJ H cK

q

= v,

r

= w, we have

x = au1/p, y = bv1/q, z = cw1/r,

F a I uFGH GH pJK F ∴ The given integral = zzz GGH au ∴

I JK F b I vFGH 1q − 1IJK dv, dz = FG c IJ w FGH 1r − 1IJK dw du, dy =

1 −1 p

dx =

al bm c n = pqr

1 p

I JJ K

1 q

m−1

1 r

n− 1

1 −1 p

1 −1 q

F l − 1I m − 1 F n − 1I G p JK q GH r JK uH v w du dv dw , where

zzz

1

−1 c ⋅ w r du dv dw r

u+ v+ w ≤1

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

=

l −1

GH q JK H rK F I F I F aIu ⋅ b v GG bu JJ GH cw JK GH pJK q H K

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 v H 2 K wH 2 K du dv dw , where

uH 2

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

u+ v+ w≤1

326


∴

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

zzz

u1 −1 v 1−1 w 1 −1 du dr dw

D

= abc Mass =

abc du dv dw , where u + v + w < 1

zzz

D

= abc

dx dy dz

D

a1f a1f a1f = abc = abc a1 + 1 + 1 + 1f 3! 6 kxyz dx dy dz = zzz k a aufabv facw f abc du dv dw

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 2xdx a2

So that mass

zzz

ρdv

= du, =

=

=

= dv,

2zdz = dw, then c2

zzz b g zzz a fb 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

ga f

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 H aK H bK H cK 2 3

2 3

Sol. Let

u=

v=

w= ∴

FG x IJ H aK FG y IJ H bK FG z IJ H cK

V= 8

2 3

2 3

2 3

zzz

= 1.

(U.P.T.U., 2007) 1

3

3a 2 u du 2

⇒ x = au 2 ⇒ dx = 3

1

3b 2 ⇒ y = bv 2 ⇒ dy = v dv 2 ⇒z=

3 cw 2

dx dy dz = 8

= 27 abc

2 3

zzz

3

zzz 3

−1

1

⇒ dz =

3c 2 w dw 2 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 1 1 −x

1 −x − y

0 0

0

1

bx + y + z + 1g

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

0

1 1− x

2

0 0

zz z

2

B

O

2

0 0

z

1− x

1

A (1,0,0)

x

1

y

(0,1,0)

1 1 −x

0

Fig. 4.53

0

0

1

2

=

(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 H aK H bK α

β

= 1, where α > 0, β > 0 given that density at any point ρ (xy) is k xy . [U.P.T.U., 2008]

Sol. The area of the plane region is A =

zz

dx dy

S

FG x IJ H aK FG y IJ H bK

1

α

Let

∴

or x = au1/α so dx =

= u

a α −1 ⋅ du u α 1

β

= v or

A =

zz s

A =

y = bv1/β 1

so dy =

b β −1 ⋅ dv v β

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

zz

a

aabf k

32

= k

S′

=

αβ

aabf k

=

32

αβ

1 2 u α

zz

⋅ bv 3

u 2α

1 2β

ab α − 1 β −1 ⋅ ⋅v ⋅ du dv u αβ

⋅v

2β

−1

1

1

3

−1

du dv

S′

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 M = k

zz

1 1− x

xy dx dy = k

0 0

z

a1 − xf 2k F 3 5 I 2k ⋅ βG , J = 3 H 2 2K 3 2k 3

= =

1 3 −1 x2 . 0

z

1 1 2 x 0

LM y OP MN 3 2 PQ 3 2

1− x

dx = 0

2k 3

z

1 1 2

0

x

a f

. 1−x

3 2

dx

5 −1 2 dx

32

52 4

=

2k 1 2. π ⋅ 3 2 ⋅ 1 2. π kπ ⋅ = . 3 3 ⋅ 2 ⋅1 24

EXERCISE 4.7

zzz

LMAns. N

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

LMAns. N

2n

= 1.

πabc 6

OP Q

1 2 2 2 a b c 6

OP Q

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

LMAns. N

= 1.

4. 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 b K H cK 2n

2n

2n

= 1, n being a positive integer.

LM MMAns. MM N

abc 90

FG 1 IJ OP H 2n K P ⋅ 3 P P 2n P Q 3

abc 12 n 2

OP Q

330


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.

6. Find the area enclosed by the curve

7. Find the area enclosed by the curve

[U.P.T.U. (C.O.), 2003]

LM MMAns. MN

α β γ 2 2 2 α β ϒ + + 8 2 2 2

OP PP 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 FG x IJ + FG y IJ = 1 . ⋅ 4 10 P Ans. M 27 H aK H b K MM 40 P N 20 QP

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 (iii) 3π – 7

(ii) π – 8 (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

π

0

sin y dy dx = ..........

sin m θ. cosn θ dθ = .......... aθ

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)

(ii) (iii ) (iv)

p q

(a)

FG 1 IJ H 2K

z

∞

y p −1

b1 + yg

p+q

(b)

π

(c) β (p, q)

π sin pπ

(d)

0

p 1− p

p+ q

⋅ dy

332

2. (i)

zz b


g

F x, y dxdy

(a)

s

(ii) β(m, n)

z

1

0

a f

x m −1 1 − x

n−1

dx

(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

5. π

6.

4. (i) B. Fill in the blanks: 1. 3π 4.

7.

4 5 m+1 n+1 2 2 m+n+ 2 2 2

C. True and False: 1. (i) T 2. (i) T

8.

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


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

r + δ r = f (t + δt) Subracting (i) from (ii)

...(i)

®

r+

...(iii)

f ( t + δt ) − f ( t ) δr = δt δt Taking the limit of both side as δt → 0.

P ®

r

O

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

Fig. 5.1

dr lim f (t + δt) − f (t) = d f (t) = δt→0 dt δt dt

.

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

d a × (b × c) dt

Velocity

=

FG H

IJ K

FG H

IJ K

H da db dc ⋅ × (b × c ) + a × ×c +a× b× dt dt dt

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

d1

2

+ 12 + 32

i

=

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 11 d2 r dt 2

Acceleration

= =

3+2+6 11

FG IJ H K

d dr dt dt

=

11

= 6ti + 2 j = 6i + 2 j , at t = 1

Therefore, the component of acceleration at t = 1 in the direction i + j + 3 k is

d6i + 2 j i ⋅ di + j + 3k i 11

=

6+2 11

=

8 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


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. MN

8 14 14 ;− 7 7

OP PQ

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. N

2 4 2 i + j, (5i + 4 j ) 3 3 3 3

OP 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


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,

df = ds

...(ii)

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 LMFG i ∂ + y ∂ + k ∂ IJ f OP ⋅ b = ∇f ⋅ b NH δx δy δz K Q

df = ...(iii) ds 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 RS(a ⋅ i) ∂ + (a ⋅ j) ∂ + ( a ⋅ k) ∂ UV f ∂y ∂z W T ∂x

∂f ∂f ∂f + ( a ⋅ k) = ( a ⋅ i) + ( a ⋅ j) ∂x ∂y ∂z

...(i)

b g F ∂f + j ∂f + k ∂f IJ a ⋅Gi H ∂x ∂y ∂z K

R.H.S. = a ⋅ ∇f =

∂f ∂f ∂f = ( a ⋅ i ) ∂x + ( a ⋅ j ) ∂y + ( a ⋅ k ) ∂z From (i) and (ii),

baH ⋅∇g f

b g

H = a⋅ ∇ f .

...(ii)

340


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) = i

=

FG ∂f H ∂x

FG i ∂f H ∂x

±

∂g ∂x

+ j

IJ K

∂g ∂y

FG ∂f ± ∂g IJ + k FG ∂f H ∂y ∂y K H ∂z ∂f I F ∂g + j ∂g + k J ± Gi ∂z K H ∂x ∂y

+ j

∇(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

±

∂g ∂z

+ k

IJ K

∂g ∂z

IJ 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 H gK F fI grad G g J H K F fI ∇ G gJ H K ∇

or Proof:

=

=

= =

g∇f − f∇g g2

, g≠0

g(grad f ) − f (grad g) g2

, ≠ 0.

FG i ∂ + j ∂ + k ∂ IJ FG f IJ H ∂x ∂y ∂z K H g K ∂ 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 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 H gK

=

g2

g∇f − f∇g g2

.

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 ⋅ r) =

FG i ∂ + j ∂ + k ∂ IJ H ∂x ∂y ∂z K

(a1x + a2y + a3z)

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

= −

r r3

FI HK

. Proved.

(iv) Let r = xi + yj + zk.

∂ 2 ( x + y 2 + z 2 )n/2 ∂x = n (x2 + y2 + z2)n/2–1 xi + yj + zk

grad rn = ∇rn = Σi

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.

H grad f = ∇f = =

(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 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

FG H

∇f = i

1 = r

1 x2 + y2 + z2

IJ c K

1 in the direction r where r = xi + yj + zk. r (U.P.T.U., 2002, 2005)

= (x2 + y2 + z2)–1/2

∂ ∂ ∂ +j x2 + y2 + z2 +k ∂x ∂y ∂z

h

− 1/2

343

VECTOR CALCULUS

=

d

∂ 2 x + y2 + z2 ∂x

i

−1/2

RS− 1 (x + y + z ) T 2 −b xi + yj + zk g = cx + y + z h =

2

i+

2 −3/2

2

2

2

2 3/2

∂ 2 2 2 −1/2 ∂ 2 k ( x + y 2 + z 2 ) −1/2 j + ∂z ( x + y + z ) ∂y

UV RS W T

1 2x i + − (x 2 + y 2 + z 2 ) −3/2 2 y 2

UV W

RS T

UV W

1 j + − ( x 2 + y 2 + z 2 ) −3/2 2 z k 2

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 (x + y + z ) 2

2

(xi + yj + zk ) 2 (x 2 + y 2 + z 2 ) 2

⋅

xi + yj + zk

cx + y + z h F xi + yj + zk IJ = −G Hx + y +z K

2 3/2

r r

2

2 1/2

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. Then ∴ Directional derivative

a =

2i − j − 2k 4+1+ 4

=

dφ = ∇φ ⋅ a ds = (8i – j – 10k) · =

2i − j − 2 k 3

FG 2i − j − 2k IJ H 3 K

37 16 + 1 + 20 = · 3 3

344


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


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 −2xy

i+

LM3 ∂φ = 0OP N ∂z Q

x2 − y2

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

FG ∂φ IJ H ∂x K

+ j

FG ∂φ IJ H ∂y K

+ k

y−3 x −1 z = = , find the values 2 1 −2 (U.P.T.U., 2001)

FG ∂φ IJ 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 2 ⇒ 2a + c and 2b + a ⇒ 2b + a Solving (i) and (ii), we get ⇒

2b + a 2c + b = −2 1 = – 2b – a ⇒ 3a + 2b + c = 0 = – 2(2c + b) = – 4c – 2b ⇒ a + 4b + 4c = 0 =

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


= 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 H ∂x ∂y ∂z K

(x2 + y2 + z2 – 9) = 2xi + 2yj + 2zk

∇φ1 at the point (2, – 1, 2) = 4i – 2j + 4k

H ∇φ 2 =

FG i ∂ + j ∂ + k ∂ IJ (x H ∂x ∂y ∂z K

(i) 2

+ y2 – z – 3) = 2xi + 2yj – k

∇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

(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) Sol.

grad u =

grad v = grad w =

FG i ∂ + j ∂ + k ∂ IJ (x + y + z) = i + j + k H ∂x ∂y ∂z K FG i ∂ + j ∂ + k ∂ IJ (x + y + z ) = 2x i + 2y j + 2z k H ∂x ∂y ∂z K FG i ∂ + j ∂ + k ∂ IJ (yz + zx + xy) H ∂x ∂y ∂z K 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


= x

Similarly,

RS UV T W ∂ R f (r )z U S V ∂z T r W

∂ f (r ) y r ∂y

and

RS1 df − f (r) UV x + f (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

=

y 2 df (r ) y 2 f (r ) − 3 f (r ) + 2 dr r r r

=

f ( r) z 2 df (r) z 2 − 3 f (r ) + r r r 2 dr

as

∂r x = ∂x r

Now using these results, we get ∇·

LM f (r)r OP N r Q

=

df ( r ) 2 + f (r ) dr r

=

1 d 2 (r f )⋅ Hence proved. r 2 dr

Example 16. Find f(r) such that ∇ f = Sol. It is given that

r and f(1) = 0. r5

∂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). 2. Find ∇ f when f = (x2 + y2 + z2) e −

x2 + y2 + z2

·

Ans. 10i − 4 j − 16 k

a f

H Ans. 2 − r e − r r

351

VECTOR CALCULUS

3. Find the unit normal to the surface x2y + 2xz = 4 at the point (2, –2, 3).

LMAns. N

gOPQ

b

1 i − 2 j − 2k 3 4. Find the directional derivative of f = x2yz + 4xz 2 at (1, –2, –1) in the direction 37 Ans. 2i – j – 2k. 3 5. Find the angle between the surfaces x2 + y2 + z2 = 9 and z = x2 + y2 – 3 at the point ±

LM N

OP Q 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 −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. N

of the vector, yzi + xzj + xyk. 10. Prove that ∇2f(r) = f ′′(r) + 11. Show that ∇2

FG x IJ Hr K 3

−

9 49

OP 14 Q

2 f ′ ( r) · r

= 0, where r is the magnitude of position vector r = xi + yj + zk.

[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. 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

OP 2 Q

1+i

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


∇⋅ 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

FG v H

(along x-axis) Z

IJ K

∂vx δx δyδz ∂x |By Taylor's theorem Net amount of mass along x-axis =

x

+

FG H

= vx δyδz – vx +

S

Vx R

IJ K

∂v y ∂y

@x

B

O

X

Y

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

P

Q

= –

= –

C

@z

∂v x δx δyδz ∂x

∂v x δxδyδz ∂x |Minus sign shows decrease. Similar net amount of mass along y-axis

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


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 + j ∂v + k ∂v IJ + vFG i ∂u + j ∂u + k ∂u IJ uG i H ∂x ∂y ∂z K H ∂x ∂y ∂z 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 grad ( a · b ) = a × curl b + b × curl 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 ∂ ∂b ⇒ Σ a⋅ + ∑ a ⋅i i = a × ∑ i× ∂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

FG H

FG H

F e j GH IJ FG K H IJ K FG IJ b H K FG IJ H K FG IJ H K

IJ K

H

H

b aH ⋅ ∇g b + (b ⋅ ∇) aH IJ K ...(i)

g

b g FG IJ bH H K

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


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

FG H

FG H

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) = = =

Identity 8: Proof:

FG H

∂2 f ∂x

+

2

∂2 f ∂y

2

IJ K

+

∂2 f ∂z

∂2 f

+

∂x 2

∂2 f ∂y 2

+

IJ K

FG H

IJ K

= ∇2 f

∂2 f ∂z 2

.

curl grad f = 0 curl grad f = ∇ × (∇f) = Σi

FG H

∂f ∂f ∂f ∂ +k +j × i ∂z ∂y ∂x ∂x

F GH

= Σi × i

F GH

= Σ k

∂2 f ∂x

2

+j

IJ K

∂2 f ∂2 f +k ∂x∂y ∂x∂z

∂2 f ∂2 f −j ∂ x∂ y ∂x∂z

I JK

I JK

Fk ∂ f − j ∂ f I + Fi ∂ f − k ∂ f I + F j ∂ f − i ∂ f I GH ∂x∂y ∂x∂z JK GH ∂y∂z ∂y∂x JK GH ∂z∂x ∂z∂y JK 2

=

2

FG H

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

div grad f = ∇2 f

or

or

IJ K

IJ K

curl grad f = 0 .

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

H div curl f = 0 .

U|V |W

U|V |W

U|V |W

U|V |W

R|S |T

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

U|V |W

IJ K

|UV |W

R|F S|GH T

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 L F F F ∂ fI ∂ f I ∂ f I O ∂ f ∂ f ∂ f j + Gi ⋅ = Σ MG i ⋅ + + J kP · J i + i⋅ ∂x ∂y ∂z NMH H∂x K H GH ∂xH∂y JK H ∂x∂z K PQ H RSi ⋅ ∂f + j ⋅ ∂f + k ⋅ ∂f |UV ∂ | 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

Again,

2

2

2

2

2

2

2

2

2

H ⇒ Curl curl f +

2

2

2

2

2

2

2

2

2

2

...(i)

358


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

LM ∂ ( y N ∂y

j k ∂ ∂ ∂y ∂z 2xy y 2 − xy 2

− xy ) −

OP LM Q N

∂ 2 ∂ ∂ 2 ( 2 xy ) + j (x − y 2 ) − ( y − xy ) ∂z ∂z ∂x

+ k = i [2y – x) – 0] + j [0 – (–y)] + k [(2y) – (– 2y)] = (2y – x) i + y j + 4y k. Ans.

OP Q

LM ∂ (2xy) − ∂ (x ∂y N ∂x

2

− y2 )

OP Q

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


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 i ∂ + j ∂ + k ∂ I φ ⋅ dr = ∇φ⋅ dr = v ⋅ dr GH ∂x ∂y ∂z JK

= [(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 zb g zb g za ydx + xdy +

d xy + d yz + d zx

φ = xy + yz + zx + c velocity potential = xy + yz + zx + c.

g z bzdx + xdzg

zdy + ydz +

f

v = ∇φ

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 i a1

I FG JK H

∂r3 ∂r1 ∂r3 ∂r2 = i ∂y − ∂z – ∂x − ∂z

j a2

∂r3 ∂r2 − ∂y ∂x

IJ j + FG ∂r − ∂r IJ k K H ∂x ∂y K 2

1

k a3

∂r1 ∂r3 − ∂z ∂x

∂r2 ∂r1 − ∂x ∂y

IJ |UVi 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. =

|RSFG a |TH

F GH

k ∂ ∂z r3

2

I FG JK H

∂r2 ∂r ∂r ∂r − a 2 1 – a3 1 − a 3 3 ∂x ∂y ∂z ∂x

1

1

1 1

1

2

3

1

2 2

2

3 3

1

2

1

1

3

3

1

1

2

2

3

3

3

3

2

2

2

1

2

3

1

2

3

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 = ∇.(∇f) =

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 i ∂ + j ∂ + k ∂ I . {(6x y z )i + (4x yz )j + (8x y z )k} GH ∂x ∂y ∂z JK 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 =

Fi ∂ + j ∂ + k ∂ I GH ∂x ∂y ∂z JK

(12x y2z4 + 4x3 z4 + 24x3 y2z2)

= (12y2 z4 + 12x2 z4 + 72x2 y2z2)i + (24xyz4 + 48x3 yz2) j + (48x y2z3 + 16x3z3 + 48x3y2z)k

362


∴ Let

(∇F)(1, –2, 1) = 348i – 144j + 400k g(x, y, z) = xy2z – 3x – z2 = 0 ∇g =

(∇g)(1, –2, 1)

Fi ∂ + j ∂ + k ∂ I GH ∂x ∂y ∂z JK

(xy2z – 3x – z2)

= (y2z – 3) i + (2xyz) j + (xy2 – 2z) k = 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

Fi ∂ + j ∂ + k ∂ I GH ∂x ∂y ∂z JK

{ax2 – byz – (a + 2)x}

= (2ax – a – 2)i + (–bz)j + (–by)k (∇f)(1, –1, 2) = (a – 2)i – 2bj + bk grad g = ∇g =

Fi ∂ + j ∂ + k ∂ I GH ∂x ∂y ∂z JK

...(iii) (4x2y + z3 – 4)

= (8xy)i + (4x2)j + (3z2)k (∇g)(1, –1, 2) = – 8i + 4j + 12k Since the surfaces are orthogonal so

e∇ f j . e∇gj

...(iv)

= 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 1. Find div A , when A = x2zi – 2y3z2j + xy2zk. 2. If V =

xi + yj + zk x2 + y2 + z2

Ans. 2xz – 6 y 2 z 2 + xy 2

, find the value of div V .

(U.P.T.U., 2000)

LMAns. 2 division ex N

2

+ y2 + z2

j OPQ

3. Find the directional derivative of the divergence of f (x, y, z) = xyi + xy2j + z2k at the point 13 (2, 1, 2) in the direction of the outer normal to the sphere, x2 + y2 + z2 = 9. Ans. 3 H H 3 4. Show that the vector field f = r / r is irrotational as well as solenoidal. Find the scalar

LM N

potential.

LM MN

(U.P.T.U., 2001, 2005) Ans. –

OP Q

1 x 2 + y2 + z2

OP PQ

364


5. If r is the distance of a point (x, y, z) from the origin, prove that curl

FG k. grad 1IJ H rK

FG k × grad 1 IJ H rK

= 0, where k is the unit vector in the direction OZ.

+ grad

(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 F xI 13. Prove that ∇ GH JK = 0. r H 11. Prove that a . ∇

2

3

u a is a constant vector.

3

2 f ′ (r). r

14. Prove that ∇2f(r) = f ″(r) +

15. If u = x2 + y2 + z2 and v = xi + yj + zk , show that div 16. Prove the curl

F a × rI = GH r JK 3

e j

−

3r a ⋅ r a + . r3 r5

e

j

euvj = 5u

b

17. Find the curl of v = exyz i + j + k at the point (1, 2, 3). 18. Prove that ∇ × ∇f = 0 for any f (x, y, z).

Ans. e 6 i − 21 j + 3k

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


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

V

ej

F r dV ⋅ or

is called volume integral or space integral. If

z

V

FdV or

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

∴

dr

∴

F ⋅ dr

∴

z

= xi + yj = dxi + dyj =

ex y i + yjj ⋅ edxi + dyjj 2 2

= x2y2dx + ydy F ⋅ dr

C

=

ze C

x 2 y 2 dx + ydy

j

368

z z

=

C

x 2 y 2 dx +

z


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

a f

0

C

z

4

3 = 4 x dx + 0

z

z

4

0

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

g

z FGH ze 2π

=

0

2π

=

0

z

=

0

IJ K

dy dx −y dt dt dt

j

cos 2 t + sin 2 t dt

dt

= (t)2π 0 = 2π.

z

Example 3. Evaluate

2π

x

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 = C

z

z z

O (0, 0)

y=0

ex + 0jdx + z b–2aygdy + z ex + b jdx + z = x dx − 2 az y dy + z ex + b jdx + z 0 dy Fx I F y I Fx I = G J − 2aG J + G + b xJ + 0 H 3K H 2K H 3 K 2

2

OA

AC

a 2 0 3

0

a

0

CB

0

b

2

2

0

BO

0

2

a

b

2

b

3

0

2

a

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 x = y = z = H dr Also, = dt H H F ⋅ dr = Now,

z

C

= = =

equations of the given curve are a cos t b sin t ct (– a sin t) i + (b cos t) j + ck H drH F⋅ dt C dt

z

zb zb ze C C C

=

=

g

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

= abc =

gb

yzi + zxj + xyk ⋅ − a sin ti + b cos t j + ck dt

C

t cos 2 t − sin 2 t + sin t cos t dt

π

= Example 5. Evaluate

z

2

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) A

dr = dxi + dyj F . dr =

z{

O

e

j} m

r

xyi + x 2 + y 2 j . dxi + dyj

ABC

(2, 0)

Fig. 5.7

B (4, 0)

X

370

=

z{

e

=

z

j } {xy dx + ex

xy dx + x 2 + y 2 dy +

AB

z

4

x=2

0 . dx +

L y OP = M16 y + 3 PQ MN


z

BC

12 y=0

e

z

j

0 + 16 + y 2 dy =

2

12

0

j }

+ y 2 dy

e16 + y jdy 2

3 12

= [192 + 576] = 768. 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

O

dr = {(– sin θ)i + (cos θ)j}dθ

z

Thus, the circulation along circle C = = = =

z z z

F . dr Fig. 5.8

C

2π

θ=0 2π

0 2π

0

nb−2 cos θ + sin θgi + b3 cos θ + 2 sin θg js . nb− sin θgi + bcos θg jsdθ

e 2 sin θ cos θ − sin

2

e8 sin θ cos θ + 4 cos 2π

j

θ + 3 cos 2 θ + 6 sin θ cos θ dθ 2

2π

j

θ − 1 dθ = 2π

= −2 cos 2θ 0 + sin 2θ 0 + θ 0 = 2π.

z

2π

0

b 4 sin 2θ + 2 cos 2θ + 1gdθ

= − 2 cos 4π − cos 0 + sin 4π − sin 0 + 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 =

=

=

F GH

(ii) F = 3x 2 i + 2x.

I JK

z ze

e12t ji + e12t − tj j + a3tfk . 2i + j + 3k dt

1

0

1

0

2

2

j

24t 2 + 12t 2 − t + 9t dt =

LM 36t N 3

3

8t 2 + 2

OP Q

ze 1

0

As t varies from 0 to 1

j

36t 2 + 8t dt

1

= 12 + 4 = 16. 0

F GH

I JK

3x 3 x 2 3x 3 3x 4 x 2 3x 3 − − j+ k = 3x 2 i + j+ k 8 4 8 4 4 8 x2 3x 3 r = xi + yj + zk = xi + j+ k 4 8

F i + x j + 9x kI dx GH 2 8 JK LM3x i + F 3x − x I j + 3x kOP . Li + x j + 9x kOdx MN GH 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

F . dr =

x=0

z

2

4

2

2

3

2

0

z

2

5

2

3

5

0

3

6

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 V

j

∇ ⋅ 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 V

=

2

∇ ⋅ F dV 3

9− y 2

x =0 y =0 z=0

b4xy − 2y + 8xzgdz dy dx

372


= =

=

= =

zz z z LMNb LM z MMNa z a 2 3

0 0

0

2

0

g

=

18x 2 − 18x + 36x 2

zz e S

in the first octant.

Sol.

zz e S

dy dx

e

3 2

2

Example 9. Evaluate a2

0

4xy − 2y 9 − y 2 + 4 x 9 − y 2

= 180.

z2

9− y2

jOPQ dy dx I F y IO F 1 4x − 2fG − c 9 − y hJ + 4xG 9 y − J P K H 3 K PPQ H 3 9 4 x − 2f + 72x dx

2 3

0 0

2

4xyz − 2 yz + 4xz 2

3

3

dx 0

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

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

yz dy dz +

0 0

zz

a2 − x 2

a

0 0

a2 − z 2

2

zz

0

0

z

a

2

z

2

0

z

0

a

2

2

a

4

a

2

0

2 2

4

a

0

4

xy dx dy

2 2

j y I − J 4K

− y 2 dy 4

a

0

4

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 =

0 0

0

0

4

a2 − y 2

a

0

0

cylinder x2 + y2 = 16 included in the first octant between z = 0 and z = 5. Sol. Since surface S : x2 + y2 = 16 Let

zz

j

a2 − y 2

2

a

0


zx dz dx +

a2 − x 2

2

a

4

=

g

z zFGH y2 IJK dz + z xFGH z2 IJK dx + z y FGH x2 IJK dy 1 1 1 ze a − z jdz + xea − x jdx + ye a 2 2 2 1Fa z z I 1Fa x x I 1Fa y − J + G − J + G G 2H 2 4K 2H 2 4K 2H 2 a

2 2

=

(U.P.T.U., 2005)

Fi ∂ + j ∂ + k ∂ I GH ∂x ∂y ∂z JK

= 2xi + 2yi

(x2 + y2 – 16)

373

VECTOR CALCULUS

unit normal

n =

n =

∇f = ∇f

2xi + 2 yj

E

4x 2 + 4 y 2

2 xi + yj

b

g

2

2

2 x +y

C

xi + yj

=

16

F . n = (zi + xj – 3y2 zk).

Now

Z

xi + yi / 4

=

FG xi + yj IJ H 4 K

1 (zx + xy) 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

R

zz

R

F zx + xy I dx dz GH y JK

=

=

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 = ∴

O

zi + xj − 3 y 2 zk .

R

=

A

dx dz dx dz 4dx dz = = y n . j y 4

z ze z z FGH

D

5

zz z LMN

4

z=0 x=0 5

0

F GG H

xz 16 − x 2

− z 16 − x 2 +

x2 2

I JJ K

+ x dx dz 4

OP dz = b 4z + 8g dz Q

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

x2y z

45x 2 y dx dy dz

8 − 4x − 2 y dx 0

dy

x 2 y 8 − 4x − 2 y dx dy

g

374


z z z ze

2 O LM y dx 3 PQ N 2 L O x M 4b 4 − 2xg − 2xb 4 − 2xg − b 4 − 2xg P dx 3 N Q x b 4 − 2xg dx 4 − 2x

2

= 45 x 2 4 y 2 − 2xy 2 − 0

2

= 45

0

2

45 3

=

3

0

2

2

2

3

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. N

B(1, 2, 0).

e

j e

j

OP 5Q

16

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

bx − ygi + bx + yg j evaluate Ü

9. If A =

zz

S

. F ⋅ ndS

Ans. 132

A ⋅ dr around the curve C consisting of y = x2 and

LMAns. 2 OP N 3Q

y2 = x.

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

Ans. 128

j

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 V

zzz e V

LMAns. 8 b j − kgOP N 3 Q

j

∇ × A dV.

j

∇ ⋅ A dV over the volume of a cube of side b.

LMAns. N

16. Show that the integral

zaa ffe 3, 4

1, 2

j e

1 3 b 3

OP Q

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


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 – x2 j + (x + z)k, S is the portion of plane 2x + 2y + z

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 C

g

Pdx + Qdy =

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

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 z b g zb g z b g zz

C

B

A

y = f1(x)

1

c

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

= −

a

a

O

b

P x, f 1 dx

b

X

Fig. 5.11

P x, y dx

AEB

P x , y dx

BFAEB

⇒

zz

S

∂P dx dy = − ∂y

C

...(i)

P x, y dx

Similarly, let the equations of the curve EAF and EBF be x = f1 (y) and x = f2 (y) respectively, then

S

∂Q dx dy = ∂x =

⇒

S

∂Q dx dy = ∂x

* George Green (1793–1841), English Mathematician.

d

f2 y

y = c x = f1 y d

c

C

∂Q dx dy = ∂x

d

c

b g b g

Q f2 , y − Q f1 , y dy

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

=

zz FGH

F ⋅ dr =

zz e

Pdx + Qdy

C

g

Vector form of Greens theorem:

z

C

Let

F =

Curl F =

e∇ × Fj ⋅ k

⇒

z

Thus,

=

F ⋅ dr =

C

S

I JK


j

∇ × F ⋅ kdS

S

Pi + Qj, then we have

i ∂ ∂x P

j ∂ ∂y Q

k ∂ ∂z 0

=

F ∂Q − ∂P I k GH ∂x ∂y JK

F ∂Q − ∂P I GH ∂x ∂y JK e∇ × Fj ⋅ k dS

zz

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 z z S

f

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 ∴

j e

x 3 − xy 3 dx + y 3 − 2 xy dy

Pdx + Qdy

g

=

zz FGH S

I JK


P = (x3 – xy3), Q = (y3 – 2xy)

∂P ∂y

= – 3xy2,

∂Q = – 2y ∂x

j ...(A)

378


So

zz FGH ∂∂Qx − ∂∂Py IJK dx dy = z z z 2

x =0

= –2 x

zz FGH S

dx

2 0

2

2

ydy + 3

y= 0

LM y OP MN 2 PQ

2 2 0

+3

I JK

2

x= 0

xdx

LM x OP LM y OP MN 2 PQ MN 3 PQ

e–2y + 3xy jdx dy

2

y= 0

2 2

3 2

0

0

y 2 dy C (0, 2)

zb

g

Pdx + Qdy =

C

O

...(i)

ze ze

j e

+

j e

∴

zb

g

=0 =2 =2 =0

Pdx + Qdy =

C

j

x 3 − xy 3 dx + y 3 − 2xy dy +

ze

x 3 − xy 3 dx + y 3 − 2xy dy +

dy dx dy dx

= = = =

j e

⇒ ⇒ ⇒ ⇒

and and and and

2 3

2

0

0

y 3 − 4y dy +

z ze x dx +

x y x y

= = = =

0 0 2 2

j

AB

j

0 0 0 0

to to to to

0

2

ze

j e

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

CO

2 2 0 0

ze

j e

x 3 − xy 3 dx + y 3 − 2xy dy

ze

BC

y x y x

j

x 3 − xy 3 dx + y 3 − 2xy dy

OA

along OA, along AB, along BC, along CO,

X

A (2, 0)

(0, 0)

Fig. 5.12

C

=

B (2, 2)

= −8 + 16

∂Q ∂P − dx dy = 8 ∂x ∂y

Now, the line integral

But

Y

2

x =0 y = 0

S

= –2

⇒

z z z 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


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


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 z z ze 2

8x

x = 0 y= − 8x

2

0

xy 2 + 2xy

b2xy + 2xgdy dx 8x − 8x

dx

3

128 5 The L.H.S. of the Green’s theorem result is the line integral = 8 2

LI =

2

0

x 2 dx =

j e

j

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 = =

z z

OA 2

ex

2

j e

2 8x y =

j

B (2, 4)

− 2xy dx + x 2 y + 3 dy

j F 2I + x e−2 2 x j + 3 G − J dx H xK z FGH 5x + 4 2 ⋅ x − 3 2x IJK dx 0

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


=

LM 5x MN 3

3

5

2 + 4 2 x2 − 3 2 ⋅ 2 x 5

40 64 + − 12 = 3 5 x = 2, dx = 0

Along ADB :

LI2 = = Along BO :

z z

ADB 4 –4

ex

2

j e

OP PQ

2

0

j

− 2xy dx + x 2 y + 3 dy

b 4 y + 3 gd y = 2 4

y = 2 2 x , with x : 2 to 0. dy = LI3 =

2 dx x

ze z FH BO

j e

0

=

3

5x 2 − 4 2 x 2 + 3 2 x

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

−

1 2

IK dx

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

zb C

g

Pdx + Qdy =

=

zz z z zz z S

a

2

2

2

z

a 2 − x2

−a

LM N

= 2 a 2x −

2

2

O

Fig. 5.14

a

0

a

2

S

a 2 −x 2

2

a

−a

2

y= a –x

2

–a

−a 0 a

=

a2 −x2

x =− a y = 0

a

=

F ∂Q − ∂P I dx dy GH ∂x ∂y JK LM ∂ ex + y j − ∂ e2x − y jOPdx dy ∂y N ∂x Q b2x + 2ygdx dy = LMN2xy + 2y2 OPQ dx. 2x a − x + e a − x j dx = 0 + 2z e a − x jdx 2

2

0

[First integral vanishes as function is odd]

x3 3

OP Q

a

LM N

= 2 a3 − 0

a3 3

OP = Q

4 3 a . 3

382


Let

y = 3 sin θ in first and y = sin φ in second integral. L.H.S. = 4 × 81

z

π 2 0

sin 4 θ cos θ dθ − 4 cos θ

z

π 2 0

sin 4 φ cos φ dφ cos φ

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)

z bx dy − y dxg

1 2

=

1 2

=

1 2

=

1 2 x dt , as y = tx 2 C

=

1 2

2 = 3a

x2

y x

LM N

1 + t3

2

1

3t 2

0

1 + t3

= 3a2 −

0

x dy − y dx

9a 2 t 2

C

=

C

x 2d

a

x2 ⋅

t=0

+

C

z z FGH IJK z ze j ze j

O

y

required area =

C

x

t=1

x+

Hence,

3at x = 1 + t3

y=

1 1 + t3

OP Q

Fig. 5.16

dt using (ii)

dt, by summetry

2

1

=

0

3 2 a . 2

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

g

z FGH 2

=

C1

0

IJ K

x x dx − dx 4 4

= 0

...(ii)

1 1 , dy = − 2 dx and x varies from 2 to 1. x x

along, the curve C2 : y =

zb zb

xdy − ydx xdy − ydx

z RST FGH

g

=

g

= − 2 log x

1

C2

or

x dx or dy = and x varies from 0 to 2. 4 4

x.

2

C2

IJ K

1 1 dx − dx 2 x −x 1 2

z

UV = W

1

2

−

2 dx x

= – 2[log 1 – log 2] = 2 log 3

...(iii)

along, the curve C3 : y = x, dy = dx and x varies from 1 to 0.

zb

xdy − ydx

g

zb 0

=

C3

1

g

xdx − xdx = 0

...(iv)

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


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. =


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

2x1 2

z LMNe 0

=

1

z FGH z FGH 0

=

j

4x − x + 2x dx = 3

dx

Along C2 : y2 = x, 2y dy = dx or dy =

ze

j

2x 3 − x 2 + 2x 3 + 2x 5 dx

1

2

5

LMx N

4

1

=

OP Q

1

= 1 0

and x varies from 1 to 0.

2x ⋅ x 1

2

j

j 2xdx OPQ

e

− x 2 dx + x 2 + x .

2x 3 2 − x 2 +

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

=

x 3 x6 − + 3 3

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

region bounded by parabolas y =

2 x and y = x .

LM N

12. Find the area of a loop of the four - leafed rose r = 3 sin2θ. Hint : A = 13. Verify the Green’s theorem for 1 ≤ x ≤ 1 and –1 ≤ y ≤ 1.

ze C

j

LMAns. 3 OP N 2Q 1 z r dθ = 98π OPQ 2 π 2 2 0

y 2 dx + x 2 dy , where C is the boundary of the square –

386


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

[∇ × (F1 i)]· n dS = =

=

Also, So, But So,

LMRSi ∂ + j ∂ + k ∂ UV × F iOP ⋅ ndS ...(i) NT ∂x ∂y ∂z W Q LM ∂F j − ∂F kOP ⋅ ndS N ∂z ∂y Q LM ∂F n ⋅ j − ∂F n ⋅ kOP dS ∂y N ∂z Q 1

1

1

1

Fig. 5.19

1

H 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


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 =

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

I2

3

z

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 H2K 2

=

z z z z

I3 =

C3

=

I4 =

C4

x 2 dx =

FG x IJ H3K 3

a

=

−a

2a3 3

LM 3 x = 0 OP N∴ dx = 0Q

−2ay dy

= –2a = – ab ∴

LM 3 y = 0 OP N∴ dy = 0Q

x 2 dx

+a

−a

= – ab2 b

( x 2 + y 2 )dx − 2 xy dy

C3

=

0

z

b

0

2

Fy I y dy = −2 aG J H2K 2

z

b 0

H H F ⋅ dr = I1 + I 2 + I 3 + I 4 C

FG 2a H3

= –

3

IJ K

+ 2b 2 a − ab 2 +

= – 4ab2 H curl F =

Again,

i ∂ ∂x x2 + y2

2 3 a − ab 2 3 ...(i)

j ∂ ∂y −2xy

k ∂ ∂z 0

= – 4yk ∴ ∴

zz

S

n = k H n · curl F = k·(– 4yk) = – 4y a b H − 4y dx dy n ⋅ curl F dS =

zz z FGH −a 0 +a

=

−a

−4

y2 2

IJ K

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


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 C

C

Again, on the unit circle C, z dz = Let x = and y = H H F ⋅ dr = ∴

z

C

C

=

z z

y dx

C

2π

0

= –

sin φ ( − sin φ) dφ

z

= –π H curl F =

Again,

C

=0 0 cos φ, ∴ dx = – sin φ.dφ sin φ, ∴ dy = cos φ.dφ

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 S

j

∇ × 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 = SI1 =

+

S3

S1 + S2 + 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

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. Unit normal n to the surface S3 =

∇(2x + y + 2 z) |∇(2x + y + 2 z)|

392


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

LM 2 x N3

AOD

zz 4

4− x

x=0 z=0

2

+

OP Q

x (1 − 2 y) 3 dx dz 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 zz

4 4 −x

0 0 4 4− x

0 0

z

4

0

z

4

0

n2x

2

s

+ x [1 − 2 (8 − 2x − 2z)] dz dx

(6x2 − 15x + 4xz) dz dx

LM6x z − 15xz + 4xz OP 2 Q N 2

2

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 + G− H 4 2 3

2

2

SI =

− a2 ( 3 π + 8 a). 12

2

IJ K

+ 2a +

FG H

a2 3 a2 + −2 a + 2 2

IJ K

394


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

x

curl F ⋅ n dS

1 x

0 0

1

0

S

2 ( x − y ) dy dx

y2 2 xy − 2

OP Q

x

dx = 2

0

O

z

1

0

FG x H

2

−

Fig. 5.23

IJ K

x2 dx = 2

1 2 x dx 0

z

A (1, 0)

X

1 = . 3

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 = = –

ze

Example 7. Evaluate

1+1

z

= –

=– 2.

2

S

FG i + j IJ H 2K

− 2 dS

2 (Area of the circle C).

b 2 ag = −2 2πa . dyj taken round the positively oriented square with 2

2π

xy dx + xy 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 ) H ( xy i + xy 2 j ) ⋅ dr

S

Now,

curl (xy i + xy j) =

A (1, 0)

X¢ (–1, 0)

by Stoke’s theorem, where S is the area of the square ABCD.

2

x+y=1

C

curl ( xy i + xy 2 j) ⋅ ndS

i ∂ ∂x xy

y–x=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


∴

∴

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

zz zz FG zH z 1

1− x

0

0

S

x dx dy

y 2 dx dy − Sx

[By symmetry]

y 2 dx dy − S ⋅ 0

[3 x = x-coordinate of the C.G. of ABCD = 0]

1

1− x

0

0

1

y3 3

0

zz

y 2 dx dy

IJ K

1− x

dx 0

1 4 1 (1 − x) 3 dx = . 3 3 0 Verification of Stokes theorem: The given line integral =

= where C is (i) (ii) (iii)

the the the the

z

C

...(i)

( xy dx + xy 2 dy) ,

boundary of the square ABCD. Now C can be broken up into four parts namely: line AB whose equation is x + y = 1, line BC whose equation is y – x = 1, 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

RS T

z

1

0

( xy dx + xy 2 dy) + x (1 − x ) dx +

+

= 2

z

1

0

z

z

1

0

BC

( xy dx + xy 2 dy) +

(1 − y ) y 2 dy

{z

z

0

−1

−1

x ( −x − 1) dx +

x ( x − 1) dx + 2 x (1 + x) dx + 2 0

z z

UV + RS W T

−1

0

1

0

z

z

CD

( xy dx + xy 2 dy) +

x (1 + x) dx +

z

0

1

0

DA

( xy dx + xy 2 dy)

UV W

(y − 1) y 2 dy

} RST

− y 2 (1 + y) dy +

1

z

(1 − y) y 2 dy + 2

z

0

−1

z

1

0

x ( x − 1 ) dx +

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,

y = sin φ, dy = cos φ dφ. H H F ⋅ dr = ( x + 2 y ) dx + ( y + 3 x) dy C

z

= = = = =

z z

2π

0

z z

[ −(cos φ + 2 sin φ) sin φ dφ + (sin φ + 3 cos φ) cos φ dφ]

2π

0 2π

0

z z

2π

0 2π

0

m− sin φ cos φ − 2 sin

2

r

φ + sin φ cos φ + 3 cos 2 φ dφ

( 3 cos2 φ − 2 sin 2 φ) dφ

m(3 cos m5 cos

2

2

r

φ − 2(1 − cos2 φ) dφ

r

φ − 2 dφ

398


=

=

z z

LM 5(1 + cos 2φ) − 2OP dφ N 2 Q RS 1 + 5 cos 2φUV dφ T2 2 W 1L 5 O φ + sin 2 φ P Q 2 NM 2 2π

0

2π

0

2π

=

=

H curl F =

0

1 [2π + 0] = π. 2 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

2π

=

1 2

=

1 sin 2 φ φ+ 2 2

= π. So Stoke’s theorem is verified.

0

LM N

(1 + cos 2φ) dφ

OP Q

2π 0

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 MN

–a3 2

OP PQ

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

2

j

− y2 .

(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


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

...(ii) Z

Now, consider the integral I1 = =

zzz zz LMNz

∂F3 dx dy dz V ∂z

R

F1 ∂F3

F2

∂z

n1

r1

K

S1

S

OP Q

r2

dz dx dy

n2

S2 –K

where R is projection of S on xy-plane. I1 = =

zz zz zz

R

R

R

F1 x , y 2

b g b g dx dy F b 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

R dx dy X

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 zz

R

and

R

b g F bx, y, F g dx dy F3 x, y, F1 dx dy 3

2

=

zz zz

= –

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

dS1

Fig. 5.26

VECTOR CALCULUS

So

zz

R

b

g

F3 x , y , F1 dx dy −

zz

R

401

b

g

F3 x , y , F2 dx dy

= = = Hence,

I1 = =

zz zz

S1

b

g

F3 k ⋅ n 1 dS1 +

b

zz

S2

F3 k . n 1 dS1 + n 2 dS 2

zz c h zzz zz c h zz c h zz c h zz n c h S

b

g

F3 k ⋅ n 2 dS2

g

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 = ∂y

V

∂F1 dx dy dz = ∂x Adding eqns. (iii), (iv), (v) 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

c h

c hs

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


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,

zz

r

F1 dy dz + F2 dz dx + F3 dx dy

S

=

...(iii)

UV W b2 x + 3zg i – b xz + yg j + ey

∂F1 ∂F2 ∂F3 + + dx dy dz . ∂x ∂y ∂z

V

j

2 F ⋅ n dS , where F = + 2 z k and S is the Example 1. Find 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 zzz a V

V

div F dV

b

g

b

V

f

2 − 1 + 2 dV = 3

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

g

e

jOPQ

∂ ∂ ∂ 2 y + 2z dV − xz − y + 2x + 3z + ∂x ∂y ∂z

af

4 π3 3

3

zzz

V

dV = 3V

= 36π.

= 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 zzz LMN V

V

e

j

div y 2 z 2 i + z 2 x 2 j + z 2 y 2 k dV

e

j

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

V

zy 2 dV

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 = 2

zz

Example 3. Evaluate

S

z z za 2π

π 2 1

0

0

z z z 2π

π2

1

0

0

0

z z 2π

π 2

0

0

1 12

=

0

z

2π

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

sin θ cos θ sin 3

sin 2 φ ⋅ dφ =

1 12

z

2

2π

0

Lr O φM P N6Q

6 1

dθ dφ

0

sin 2 φ dφ =

π . 12

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 = =

zzz LMN a f zz z e zz z a zza f LM a f z MN z a fLMN z2 − y2

4 z

0 − z − z2 − y2

= 2

= 4

= 4

z2 − y2

4 z

0 –z 0

4

0

4

0

4

0

S

F ⋅ n dS =

a fOP Q

∂ ∂ ∂ xyz 2 + 4xz + 3z dV = ∂x ∂y ∂z

V

= 2

e j

zz

z

−z

j

4 3 = π z +z

2

OP Q

4

0

y z2 + sin −1 z 2

0

a

f

= π 256 + 64 = 320π .

z

OP PQ

z2 − y2

− z2 − y2

zza

z 2 − y 2 dy dz = 4

π z2 sin −1 1 dz = 4 × 4 2

4

z

f

4z + 3 dx dy dz, since

y z2 − y2

4z + 3

V

div F dV

V

j

4z + xz 2 + 3 dV

4 z + xz 2 + 3 dx dy dz

4z + 3

4z + 3

zzz zzz e

0

x dx = 0

f

4z + 3

z 2 − y 2 dy dz

z

dz 0

ze 4

0

j

4z 3 + 3z 2 dz

404


Example 4. By transforming to a triple integral evaluate

zz e

I =

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 V

e j

e j

e jOPQ j 3x e je

∂ 3 ∂ 2 ∂ 2 x + x y + x z dV ∂x ∂y ∂z =

zz z z z ze zze b

ea − y 2

a

z =0 y = – a x = –

b

= 4×5 20 = 3

2

a

a2 − y2

z = 0 y= 0 x = 0

b

a

a − 2

z = 0 y= 0

2

a2 − y2

j

+ x 2 + x 2 dx dy dz

j x 2 dx dy dz = 20

3 y 2 2 dy

j

20 dz = 3

z

∴ I=

20 b 3

z

π 2 0

a

f

a 3 cos 3 t a cos t dt =

zz

20 4 a b 3

3

a

z = 0 y= 0

a

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.

LM x OP e N3Q LM a − y zOP MNe j PQ b

b

dy =

z=0

z

π 2 0

cos 4 t dt =

20 3

z

a

y=0

e

b a2 − y2

j

3 2

dy .

20 4 3 π 5 4 a b = πa b . 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

zzz b zz z LMMN V

g

x + y + z dx dy dz = 2

e

j

e

g

LM x + yx + zxOP N2 Q O y a + azyP y+a 2 2 PQ

c

j

b

2

z= 0 y = 0

a

dy dz x =0

OP = 2 dz 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

a2 + ay + az dy dz = 2 2

2

c

c

z =0

2

2

y =0

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

405

VECTOR CALCULUS

Therefore,

zz zz c

=

=

DEFG

b

z = 0 y= 0 c

=

zz

b

z = 0 y= 0

z

c z= 0

LM a MN

2

ea e

2

F ⋅ n dS

Z

j e

j

a 2 − yz dy dz =

b−

j e j LMa y − z y OP 2 PQ MN

zb 2

2

OP dz = LM a PQ MN

2

z

c

z= 0

bz −

2

z b2 4

OP PQ

=

zz zz

ABCO

c

zz b z LMMN

F ⋅ n dS =

b

z =0 y = 0

g y zO P 2 PQ

B

D

2 b

2

E

dz y =0

A

O

c

= a 2 bc – 0

Y

2 2

c b 4

.

G

F

X

Over the face ABCO, n = – i, x = 0. Therefore =

C

− yz i + y 2 − 2a j + z 2 − ay k ⋅ i dy dz

Fig. 5.27

a f

0 − yz i + y j + z k ⋅ −i dy dz

c

yz dy dz =

z=0

2

2

b

2

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 zz

zz

=

c

e

z =0 x = 0

ABEF a

z = 0 x= 0

eb

2

j

j e

− 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 b

a

y = 0 x =0

ec

2

j

− xy dx dy = c 2 ab −

a 2b 2 · 4

Over the face AFGO, n = – k, z = 0. Therefore

zz zz

AFGO

H F ⋅ n dS =

z z b

a

y=0 x=0

xy dx dy =

Adding the six surface integrals, we get S

H F ⋅ n dS =

F a bc − c b GH 4

2 2

2

+

I F JK GH

a2 b 2 · 4

I F JK GH

c 2b 2 a2 c2 a2 c2 a 2b 2 a 2b 2 + b 2 ca − + + c 2 ab − + 4 4 4 4 4

I JK

= 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


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 zzz b zz z LMMN V

S

zzz

b gOP b 4z − 2y + yg dV Q 4z − yg dx dy dz = b 4z − yg dx dy dz = 2z − yz dx dy = b2 − yg dx dy = y O 3 L 1O 3 2y − dx = z M2 − P dx = z dx = . = P 2 PQ 2 N 2Q 2 Example 7. Evaluate z e y z i + z x j + z y k j ⋅ n dS, where S is the part of the sphere =

V

a f

e j

∂ ∂ ∂ − y2 + yz dV = 4xz + ∂z ∂x ∂y

zzz z z 1

1

x =0 y = 0 z= 0

V

1

1

V

1

1

2

z=0

x =0 y = 0

2 1

1

x =0

y =0

2 2

1

1

x=0

y= 0

1

1

0

0

2 2

2 2

S

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 z 1

Fz I GH 2 JK

zy 2 dx dy dz

1

1− x 2

x = –1 y = – 1− x 2

2

x = –1

1

0

8 15

=

8 15

2

j

− y 2 y 2 dx dy

y 3 y5 − 3 5

1− x 2

– 1− x 2

5 2 2

2 5/2

2 5/2

1

=

0

5 2 2

x =−1

z

e1 − x

y 2 dx dy

OP dx 1− x j PQ R| 1 1 − x − 1 1 − x U| dx S| 3 e j 5 e j V| T W R| b1 − x g e1 − x j U| dx V S| T 3 − 5 |W e1 − x j dx

1

z

1− x 2 − y 2

2

1− x 2

x = –1 y = – 1 − x 2

= 4

1 −x 2 − y 2

x = –1 y = – 1 − x 2 z = 0

z z z LMMNe

= 2

z

1− x 2

z ze z

5 2 2

1

0

π 2 0

1 − sin 2 θ

j

5 2

cos θ dθ

[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 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 =

z

zz

zzz

H div F = =

Let

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 2

2

zzz zzz b zzz b

= 1 + 2y H F ⋅ n dS = I = S

= =

V

g

zzz

V

1 + 2 y dV

g

1 + 2 y dx dy dz

H div F dV

408


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 zz b zz b zz o z RST z LMN b z RST

g g dx dy 1 + 2 y g kzpb 1 + 2 ygb 6 − 2x − yg dx dy 1 + 2 y dx dy dz

I = =

1 2 1 = 2 =

= = =

6 − 2 x − y /2 0

t

6 − 2x + 11y − 4xy − 2 y 2 dx dy

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

g

b

g

6 − 2x

dx 0

11 6 − 2x 2

UV W

g

2

b

− 2 x 6 − 2x

g

2

−

b

2 6 − 2x 3

g OPQ dx 3

8 3 x + 26x 2 − 84x + 90 dx 3

1 2 26 3 − x4 + x − 42x 2 + 90x 2 3 3

=

b

UV W

OP Q

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

= =

Hence,

zz

S

V

F1 = x3 – yz, F2 = – 2x2y, F3 = z H F = (x3 – yz) i – 2x2y j + z k H div F

H F ⋅ n dS

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

2

= 3x2 − 2x2 + 1 = x2 + 1 =

zzz

zzz e V

j

x 2 + 1 dV

2

...(i)

409

VECTOR CALCULUS

= = = = = =

zzze zze zze ze ze

j x + 1j kzp dx dy a x + 1j dx dy a x + 1j myr dx a x + 1j dx Ra U a S + xV T3 W |R a + a|UV = a + a a S |T 3 |W 3 a a a

0 0 0 a a

x 2 + 1 dx dy dz a 0

2

0 0

a a

0 0 a

2

a

2

0

0

a

2

2

0

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


=

LMa y − a y OP MN 2 2 PQ 2

2

4

a

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 zz

b g

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

=

– 2a 2 x 2 dx

0 0 a

0

a 0

dx

a

0

=

– 2a

LM x OP N3Q 3

2

a

= – 0

2 5 a . 3

If I5 is integral along face OABE whose equation is z = 0 n dS = – k dx dy H F ⋅ – k dx dy I5 =

zz

zz

b

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

y

a 0

zz

a a

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

= =

Let

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

2

zz zzz e

2

= z2 + x2 + y2. H F ⋅ n ds = I = S =

}

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

LM zz MMMex N

zz

2

2

2

+ y2

j

j

z3 z+ 3

OP PQ

a2 − x 2 − y2

dx dy 0

a2 − x 2 − y 2

|RS |T

e +

a2 − x 2 − y 2 x 2 + y 2 +

O j PP dx dy PP Q U| dx dy V| W

a2 − x 2 − y 2 3

a2 − x 2 − y 2 3

3 2

412


= =

Let

1 3 1 3

=

1 3

=

2 3

zz zz zz zz

a2 − x2 − y 2 a2 −x2

a

− a − a2 −x2

a2 − x 2

a

−a 0

I =

2 3

=

2 3

z z LNMe LM z MMNe z LMNe z LMN e π 2 −a 0 a

= = =

2

+ a2

j

a2 − x 2 − y 2 + 2y 2

2

+ a2

j

a2 − x 2 − y 2 + 2 y 2 a 2 − x 2

2x 2 + a 2

a

} −y U VW dx dy

a 2 − x 2 − y 2 dx dy

jea

jea

2

j

e

− x 2 cos 2 θ + 2 a 2 − x 2

2

−x

2

j

3 1 2 2 + 2 a2 − x2 2 2

2

2

2

2

2

2

2

2

2

e

j

2

2

2

OP Q

sin 2 θ cos 2 θ dx dθ

OP PP Q

3 3 2 2 dx 2 3

j π πO 2 dx 2x + a je a − x j + 2e a − x j 3 4 16 PQ 4 π O × 2 2x + a je a − x j + e a − x j P dx 3 8 Q π 2 e2x a − 2x + a − a x j + a + x − 2 x 3×2 2x + a

−a

a

=

{e2x RSe2x T

t

+ 2 y 2 + a 2 dx dy

2

a 2 − x 2 cosθ dθ

dy =

=

t

a 2 − x 2 sin θ

y =

=

o o2x

a 2 − x 2 − y 2 3x 2 + 3y 2 + a 2 − x 2 − y 2 dx dy

2 2

2

−a

a

π 6 π 2

ze ze a

0

z

2

2

2 2

0

a

2 2

4

4

2 2

4

4

2 2

0

dx

j

3a 4 − 3x 4 dx

j πL x O a x− M P 2N 5Q a

0

a 4 − x 4 dx 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

div F = =

z zb

⇒

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

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

of the sphere x2 + y2 + z2 = a2. 2. Use divergence theorem to show that surface enclosing volume V.

zz zz

S

e

LMAns. N

j

12πa 5 5

OP Q

∇ 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


8. Evaluate

zz

S

F ⋅ n ds , where F = 2xyi + yz2j + xzk, and S is the surface of the region

LMAns. N

bounded by x = 0, y = 0, z = 0, y = 3 and x + 2z = 6.

351 2

OP Q

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 (iii)

581

(ii)

381

(iv)

481

e j

5. If a is a constant vector, then grad r . a is equal to (ii) − a

(i) r (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


(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

af

∇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

F . dr is called circulation.

S

5. (i) The formula

S

j

curl F . n ds =

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


e j

(iii) a . ∇ f

(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)

13. (i) B. Fill in the blanks: 1. 4 5.

r r

9. f ′(r)

2. 0 6. 10.

r r3 10 3

3. ∇2 f

4. 0

7. –15

8. 0

11. 2π

12. 0

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)

13. flux 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 A = M0 MNb

OP PP Q

2 × 10 = 20

5 4 3 2 is 2, b = 4cm → 13 10 —

^

(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) (iii) (iv)

p q p+q π

π sin pπ

(p) (q)

z

b 1 2g ∞

0

y p−1 dy 1+ y p+q

b gb g

(r) β(p, q) (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


(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

(True/False)

3

(ii) The curve x + y = 3axy is symmetric about the line y = – x. 2

2

(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: ^ ^ ^ — (h) If r = xi + yj + zk is position vector, then value of ∇ (log r) is : (i)

r⋅ r

(iii) − (i)

r2

(iv) None of these

r3

The Jacobian (i) 1

a f b g

∂ uv for the function u = ex sin y, v = (x + log sin y) is ∂ xy (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

(j)

r⋅

(ii) r⋅

(True/False)

(iv)

(i) π|2a (iii)

(ii) πa3|2

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 H aK H bK α

z zb

β

= 1 where α > 0, β > 0 given that density at any point p(xy) is k xy .

(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 H 2 −1K

and hence find A–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 MM 21 MM 3 MN 15

1 3 1 5 1 7

1 5 1 7 1 11

OP PP PP 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

F GH

∂ u x2 + y2 ∂2 u , where u(x, y) = log e = ∂y ∂x ∂x ∂y xy

−1 (c) If u = x sin

F x I + y sin FG y IJ , find the value of GH y JK H xK −1

3. Attempt any four parts of the following:

FG π IJ . H 4K

(a) Expand ex cos y about the point 1,

422

x2

I JK

∂2u ∂x 2

+ 2 xy

∂2u ∂2u + y2 2 . ∂x ∂y ∂y (5 × 4 = 20)

423


(b) Calculate the Jacobian

a b

f g

∂ u, v, w of the following: ∂ x, y, z

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: (5 × 4 = 20) (a) Evaluate the following by changing into polar coordinates:

zz a

a2 − y 2

0 0

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 H aK H b K H cK 2 3

2 3

2 3

= 1.

(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

1 5 x 0

(f) Show that

e1 − x j

3 10

dx =

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:

(i) (ii) (iii) (iv)

[Total Marks: 100

Attempt All questions. All questions carry equal marks. In case of numerical problems assume data wherever not provided. 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

a f b g

∂ u, v ∂ x, y

=

(10 × 2 = 20)

y2 − x2 2uv u − v

a f 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


4. Attempt any four parts of the following: (a) Find the rank of matrix

LM 2 MM 3 MN − 32

3 −2 2 4

−2

(5 × 4 = 20)

4 2 4 5

1 3 0

OP PP 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. (d) Find the characteristic equation of the matrix

F 1 GG 0 H −1

I JJ K

2 2 2

2 1 . Also, find the eigen 2

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 R

g

(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 1

0

3

3

1 2

dx

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


(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

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. tan θ

=

GGG


(i) (ii) (iii) (iv)

[Total Marks: 100

Attempt All questions. All questions carry equal marks. Question no. 1–4 are common to all candidates. 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 N β + iδ α − iγ Q 2

2

+ γ2 + δ2 = 1.

(c) Reduce the matrix A to diagonal form A =

LM− 1 MM 1 N− 1

2 2 −1

−2 1 0

OP PP Q

(d) Find the eigen values and eigen vectors of matrix A A =

LM1 MM2 N3

7 5 11

OP PP Q

13 7 5

[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 MM− 1 N1

−1 2 −1

1 −1 2

OP PP Q

427

428


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 H xK

in the neighbourhood of (1, 1).

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


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 H aK H bK H cK q

p

where x, y, z are all positive but limited condition

r

≤ 1.

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


[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 A =

LM3 MM0 N0

1 2 0

4 6 5

OP PP Q

(5 × 4 = 20)

(b) Verify Cayley-Hamilton theorem for the matrix

LM 2 MM−1 N1

OP PP Q

−1 2 2 −1 A = −1 2 Hence compute A–1. (c) Reduce the matrix A to its normal form when

LM 1 MM 2 MN−11

OP PP PQ

2 −1 4 4 3 4 A = 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


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

5. (a) Prove div(grad rn) = n(n + 1)rn – 2, where r = x 2 + y 2 + z 2

(b) Apply Green’s theorem to evaluate

ze C

j

j

1 2

e

2 . Hence show that ∇

j

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


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


Logarithm case, 2 Lower triangular 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

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

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

Engineering Mathematics

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