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Engineering Mathematics 1

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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
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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
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Some Useful Formulae
1.
2.
sin ix = i sin hx
3.
sin x =
e ix − e − ix
2i
4.
cos x =
e ix + e − ix
2
5.
Sin h2x =
1
(cosh 2x – 1)
2
6.
cos h2x =
1
(cosh 2x + 1)
2
7.
z
a x dx =
ax
a ≠ 1, a > 0
log a
z
z
z
sin haxdx =
1
cos hax
a
cos haxdx =
1
sin hax
a
tan haxdx =
1
log cos hax
a
z
z
z
z
z
z
z
1
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
cos ix = cos hx
a2 − x2
1
2
x −a
1
2
x +a
2
2
dx = sin −1
x
x
= arc sin
a
a
dx = log x + x 2 − a 2
dx =
1
x
x
tan −1
= arc tan
a
a
a
a2
x
sin −1
a
2
a 2 − x 2 dx =
x
2
e ax sin bx dx =
e ax
(a sin bx – b cos bx)
a2 + b2
e ax cos bx dx =
e ax
(a cos bx + b sin bx)
a 2 + b2
sec ax dx =
a2 − x2 +
1
log |sec ax + tan ax|
a
1
log |cosec ax – cot ax|
a
18.
z
19.
sin x = x −
x 3 x5
+
......
3
5
20.
cos x = 1 −
x2 x 4
+
........
2
4
21.
tan x = x +
x3
2 5 17 7
+
x −
x + ........
3 15
315
22.
log (1 + x) = x −
23.
log (1 – x) = − x −
24.
sin hx = x +
25.
d x
a = ax loge a
dx
26.
1
d
cos −1 x = −
dx
1 − x2
27.
1
d
cot −1 x = −
dx
1 + x2
28.
29.
30.
cosec ax dx =
d
dx
d
dx
x2 x3
−
− .........
2
3
x 3 x5
+
+ .........
3
5
cosec −1 x = −
log a x =
x2 x3 x 4
+
−
+ ........
2
3
4
1
x
1
x x2 − 1
log a e
d
x
a
tan −1
.
= 2
dx
a
a + x2
Contents
PREFACE TO THE SECOND REVISED EDITION
SOME USEFUL FORMULAE
U NIT I. Differential Calculus-I
1.0 Introduction
1.1 nth Derivative of Some Elementary Functions
Exercise 1.1
1.2 Leibnitz’s Theorem
Exercise 1.2
Exercise 1.3
Partial Differentiation
1.3 Function of Two Variables
1.4 Partial Differential Coefficients
Exercise 1.4
1.5 Homogeneous Function
1.6 Euler’s Theorem on Homogeneous Functions
Exercise 1.5
1.7 Total Differential Coefficient
Exercise 1.6
Curve Tracing
1.8 Procedure for Tracing Curves in Cartesian Form
Exercise 1.7
1.9 Polar Curves
Exercise 1.8
1.10 Parametric Curves
Exercise 1.9
Expansion of Function of Several Variables
1.11 Taylor’s Theorem for Functions of Two Variables
Exercise 1.10
Objective Type Questions
Answers to Objective Type Questions
U NIT II. Differential Calculus-II
2.1 Jacobian
Exercise 2.1
1–94
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
95–150
95
109
2.2 Approximation of Errors
Exercise 2.2
2.3 Extrema of Function of Several Variables
Exercise 2.3
2.4 Lagrange’s Method of Undetermined Multipliers
Exercise 2.4
Objective Type Questions
Answers to Objective Type Questions
U NIT III. Matrices
3.0 Introduction
3.1 Definition of Matrix
3.2 Types of Matrices
3.3 Operations on Matrices
3.4 Trace of Matrix
3.5 Properties of Transpose
3.6 Properties of Conjugate Matrices
3.7 Singular and Non-Singular Matrices
3.8 Adjoint of a Square Matrix
3.9 Inverse of a Matrix (Reciprocal)
Exercise 3.1
3.10 Elementary Row and Column Transformations
3.11 Method of Finding Inverse of a Non-Singular Matrix by Elementary
Transformations
Exercise 3.2
3.12 Rank of a Matrix
Exercise 3.3
3.13 System of Linear Equations (Non-Homogeneous)
3.14 System of Homogeneous Equations
3.15 Gaussian Elimination Method
Exercise 3.4
3.16 Linear Dependence of Vectors
Exercise 3.5
3.17 Eigen Values and Eigen Vectors
Exercise 3.6
3.18 Cayley-Hamilton Theorem
Exercise 3.7
3.19 Diagonalization of a Matrix
3.20 Application of Matrices to Engineering Problems
Exercise 3.8
Objective Type Questions
Answers to Objective Type Questions
111
119
121
134
135
145
147
150
151–257
151
151
152
155
156
156
156
163
163
163
165
167
168
174
175
186
188
197
200
206
210
214
214
230
232
238
239
249
253
255
257
U NIT IV. Multiple Integrals
4.1 Multiple Integrals
4.2 Double Integrals
4.3 Working Rule
4.4 Double Integration for Polar Curves
Exercise 4.1
4.5 Change of the Order of Integration
4.6 Change of Variables in a Multiple Integral
Exercise 4.2
4.7 Beta and Gamma Functions
4.8 Transformations of Gamma Function
4.9 Transformations of Beta Function
4.10 Relation between Beta and Gamma Functions
4.11 Some Important Deductions
4.12 Duplication Formula
4.13 Evaluate the Integrals
Exercise 4.3
4.14 Application to Area (Double Integrals)
Exercise 4.4
4.15 Triple Integrals
Exercise 4.5
4.16 Application to Volume (Triple Integrals)
Exercise 4.6
4.17 Dritchlet’s Theorem
Exercise 4.7
Objective Type Questions
Answers to Objective Type Questions
U NIT V. Vector Calculus
Vector Differential Calculus
5.1 Vector Function
5.2 Vector Differentiation
5.3 Some Results on Differentiation
Exercise 5.1
5.4 Scalar Point Function
5.5 Vector Point Function
5.6 Gradient or Slope of Scalar Point Function
5.7 Geometrical Meaning of Gradient, Normal
5.8 Directional Derivative
5.9 Properties of Gradient
Exercise 5.2
5.10 Divergence of a Vector Point Function
5.11 Physical Interpretation of Divergence
5.12 Curl of a Vector
258–332
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
333–418
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 (2004–2009)
419–431
Index
433–434
UNIT
1
Differential Calculus-I
1.0
INTRODUCTION
Calculus is one of the most beautiful intellectual achievements of human being. The mathematical
study of change motion, growth or decay is calculus. One of the most important idea of differential
calculus is derivative which measures the rate of change of a given function. Concept of derivative
is very useful in engineering, science, economics, medicine and computer science.
dy
d2 y
, second order derivative, denoted by
dx
dx 2
d3 y
third order derivative by
and so on. Thus by differentiating a function y = f(x), n times,
dx 3
dny
or Dny or yn(x). Thus, the process
successively, we get the nth order derivative of y denoted by
dx n
of finding the differential co-efficient of a function again and again is called Successive
Differentiation.
The first order derivative of y denoted by
1.1 nth DERIVATIVE OF SOME ELEMENTARY FUNCTIONS
1. Power Function (ax + b)m
Let
y = (ax + b)m
y1 = ma (ax + b)m–1
y2 = m (m–1)a2 (ax + b)m–2
..... ...... ...................................
..... ...... ...................................
yn = m(m–1) (m–2) ... (m – n – 1) an (ax + b)m–n
Case I.
⇒
When m is positive integer, then
m (m – 1)...(m – n + 1)(m – n)...3 ⋅ 2 ⋅ 1 n
yn =
a (ax + b)m–n
(m – n)...3 ⋅ 2 ⋅ 1
yn =
m
dn
m
(
+
)
=
ax
b
a n ( ax + b) m− n
m−n
dx n
1
2
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Case II. When m = n = +ve integer
yn =
n n
a (ax + b)0 =
0
n an ⇒
dn
( ax + b)n = n a n
n
dx
Case III. When m = –1, then
y = (ax + b)–1 =
∴
1
( ax + b)
yn = (–1) (–2) (–3) ... (–n) an (ax + b)–1–n
⇒
RS
T
dn
1
n
dx ax + b
UV = (−1) n a
W (ax + b)
n
n
n+1
Case IV. Logarithm case: When y = log (ax + b), then
a
ax + b
Differentiating (n–1) times, we get
y1 =
yn = an
d n −1
( ax + b) −1
dx n−1
Using case III, we obtain
⇒
l
n −1
q = (−1) ( ax +(nb)− 1) a
dn
log( ax + b)
dx n
n
n
2. Exponential Function
(i) Consider
y = amx
y1 = mamx. loge a
y2 = m2amx (loge a)2
...................................
...................................
yn = mn amx (loge a)n
(ii) Consider
Putting
y = emx
a = e in above
yn = mnemx
3. Trigonometric Functions cos (ax + b) or sin (ax + b)
Let
y = cos (ax + b), then
F ax + b + π I
H
2K
2π I
F
y = – a cos (ax + b) = a cos ax + b +
H
2 K
3π I
F
y = + a sin (ax + b) = a cos ax + b +
H
2 K
y1 = – a sin (ax + b) = a cos
2
2
3
3
2
3
.................................................................................
.................................................................................
3
DIFFERENTIAL CALCULUS-I
F
I
H
K
d
nπ I
F
y =
sin( ax + b) = a sin ax + b +
H
2K
dx
yn =
nπ
dn
cos ( ax + b) = a n cos ax + b +
n
2
dx
n
Similarly,
n
n
n
4. Product Functions eax sin (bx + c ) or e ax cos (bx + c )
Consider the function y = eax sin (bx + c)
y1 = eax·b cos (bx + c) + aeax sin (bx + c)
= eax [b cos (bx + c) + a sin (bx + c)]
To rewrite this in the form of sin, put
a = r cos φ, b = r sin φ, we get
y1 = eax [r sin φ cos (bx + c) + r cos φ sin (bx + c)]
y1 = reax sin (bx + c + φ)
Here,
r =
a 2 + b 2 and φ = tan (b/a)
–1
Differentiating again w.r.t. x, we get
y2 = raeax sin (bx + c + φ) + rbeax cos (bx + c + φ)
Substituting for a and b, we get
y2 = reax. r cos φ sin (bx + c + φ) + reax r sin φ cos (bx + c+ φ)
y2 = r2eax [cos φ sin (bx + c + φ) + sin φ cos (bx + c + φ)]
∴
= r2 eax sin (bx + c + φ + φ)
y2 = r2 eax sin (bx + c+ 2φ)
Similarly,
y3 = r3eax sin (bx + c + 3φ)
......................................................
yn =
dn
dx
n
e ax sin( bx + c ) = r n e ax sin (bx + c + nφ)
In similar way, we obtain
d n ax
e cos(bx + c) = r ne ax cos (bx + c + nφ)
yn =
dx n
1
1 − 5x + 6x 2
1
=
( 2 x − 1)( 3 x − 1)
Example 1. Find the nth derivative of
Sol. Let
or
∴
1
1 − 5x + 6x 2
2
3
−
y =
2x − 1 3x − 1
y =
yn = 2
(By Partial fraction)
dn
dn
–1
(2x
–
1)
–
3
(3x – 1)–1
dx n
dx n
4
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
L(−1) n 2 OP LM(−1) n 3 OP
(−1) n a
d
As
( ax + b) =
= 2 M
–3
MN (2x − 1) PQ MN (3x − 1) PQ
dx
( ax + b)
L 2 − 3 OP .
y = ( −1) n M
N (2x − 1) ( 3x − 1) Q
n
n
n
n+1
n+1
n
or
n
n+1
n
n+1
Example 2. Find the nth derivative of eax cos2 x sin x.
(1 + cos 2 x)
sin x
Sol. Let
y = eax cos2 x sin x = eax
2
1 ax
1 ax
e sin x +
e ( 2 cos 2x sin x)
=
2
2× 2
1 ax
1
e sin x + e ax sin( 3x) − sin x
=
2
4
1 ax
1 ax
e sin x + e sin 3x
y =
4
4
1 n ax
1
∴
yn =
r e sin (x + nφ) + r1 n e ax sin ( 3x + nθ) .
4
4
where
and
n
n +1
l
or
n
−1
n+1
n +1
n
n
r =
q
a 2 + 1 ; tan φ = 1/a
a 2 + 9 ; tan θ = 3/a.
2x
, find yn.
Example 3. If y = tan–1
1 − x2
2x
Sol. We have
y = tan −1
1 − x2
Differentiating y w.r.t. x, we get
r1 =
y1 =
1
1+
⋅
2
F 2x I
H1−x K
(U.P.T.U., 2002)
FG
H
d
2x
dx 1 − x 2
2
2(1 + x 2 )
=
2
=
(1 − x )
2(1 − x ) + 4x
IJ =
K e1 + x − 2x + 4x j ⋅ (1 − x )
2 2
4
(1 + x 2 ) 2
y1 =
1 1
1
−
, (by Partial fractions)
i x−i x+i
LM
N
=
LM
MN
n −1
n −1
1 ( −1) ( n − 1) ( −1) ( n − 1)
−
i
( x − i) n
( x + i) n
( −1)n−1 ( n − 1)
i
=
2 2
OP
Q
Differentiating both sides (n–1) times w.r. to ‘x’, we get
yn =
2
2
2
( x + i)( x − i )
y1 =
(1 + x 2 )
2
2
( x − i ) − n − ( x + i ) −n
( −1)n−1 ( n − 1)
i
OP
PQ
r − n (cos θ − i sin θ) − n − r − n (cos θ + i sin θ) − n
(where x = r cos θ, 1 = r sin θ)
5
DIFFERENTIAL CALCULUS-I
=
( −1) n − 1 ( n − 1) r − n
i
n − 1 r–n sin nθ, where r =
yn = 2(–1)n–1
Sol. We have
x2 + 1
F 1I.
H xK
θ = tan–1
Example 4. If y = x log
cos nθ + i sin nθ − cos nθ + i sin nθ
x −1
. Show that
x +1
LM x − n − x + n OP
N (x − 1) (x + 1) Q
yn = (–1)n–2
n−2
y = x log
x −1
= x [log(x – 1) – log (x + 1)]
x +1
n
(U.P.T.U., 2002)
n
Differentiating w.r. to ‘x’, we get
LM 1 − 1 OP
N x − 1 x + 1Q
F 1 I + F−1 + 1 I
= log (x – 1) – log (x + 1) + 1 +
H x − 1K H x + 1 K
y1 = log (x – 1) – log (x + 1) + x
or
y1 = log (x –1) – log (x + 1) +
1
1
+
x −1 x +1
Differentiating (n–1) times with respect to N, we get
yn =
=
=
=
d n−1
d n−1
d n −1
d n −1
dx
dx
dx
dx n −1
log (x − 1) −
n −1
d n− 2
log (x + 1) +
n −1
(x − 1) −1 +
n −1
RS d log(x − 1)UV − d RS d log(x + 1)UV + (−1) n − 1 + (−1) n − 1
Tdx
W dx Tdx
W (x − 1)
(x + 1)
n− 2
dxn−2
(x + 1) −1
n −1
n− 2
F
H
I
K
F
H
n− 1
n
n
I
K
( −1)n −1 n − 1 ( −1) n−1 n − 1
d n− 2
d n− 2
1
1
−
+
+
(x − 1) n
( x + 1) n
dx n − 2 x − 1
dx n − 2 x + 1
( −1)n− 2 n − 2
(x − 1) n−1
−
( −1) n− 2 n − 2
(x + 1)n−1
+
(x − 1)n
+
( −1)n − 1 (n − 1) n − 2
LM x − 1 − x + 1 − (n − 1) − (n − 1) OP
N (x − 1) (x + 1) (x − 1) (x + 1) Q
L x − n − x + n OP .
n−2M
N (x − 1) (x + 1) Q
n−2
n−2
= ( −1)
n
n
n −2
= ( −1)
n
n
Example 5. Find yn (0) if y =
Sol. We have
( −1)n−1 (n − 1) n − 2
y =
x3
x −1
2
n
n
.
x3
x 3 − 1 + 1 ( x − 1)( x 2 + x + 1)
1
=
+ 2
=
2
2
( x − 1)( x + 1)
x −1
x −1
x −1
(x + 1)n
6
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
or
y =
x2 − 1 + 1
1
+1+
x +1
x−1 x+1
a fa f
1
1L 1
1 O
+ M
−
y = x+
x + 1 2 N x − 1 x + 1 PQ
1L 1
1 O
+
y = x+ M
2 N x − 1 x + 1 PQ
(−1) n O
1 L ( −1) n
P
+
y = 0+ M
2 NM ( x − 1)
( x + 1) QP
( −1) n L
1
1
y =
MN (x − 1) + (x + 1) OPQ
2
( −1) n L 1
1 O
+
x = 0, y (0) =
M
2
N (−1) (1) PQ
y =
or
x2 + x + 1
1
+
( x + 1)
( x − 1)( x + 1)
n
∴
n
n +1
n
n+1
n
or
n +1
n
n +1
n
At
n +1
n
When n is odd, yn(0) =
When n is even, yn(0) =
( −1) n n
2
( −1) n n
2
n +1
1+1 = − n
−1 + 1 = 0 .
EXERCISE 1.1
1 . If y =
x2
, find nth derivative of y.
( x − 1)2 ( x + 2)
(U.P.T.U., 2002)
LMAns. y = (−1) n + 1 + 5(−1) n + 4(−1) n OP
3( x − 1)
9( x − 1)
9( x + 2) PQ
MN
LMAns. (−1) n L a − b OOP
x
2 . Find the nth derivative of
.
M
P
( a − b) N (x − a)
( x − a)( x − b)
(x − b) Q PQ
MN
L 1 + x OP .
3 . Find the nth derivative of tan M
N1 − x Q
n
n
2
n
n+2
n
n +1
n
n +1
2
2
n +1
n +1
–1
[Ans. ( −1)n −1 n − 1 sin n θ sin nθ where θ = cot–1x ]
4 . If y = sin3 x, find yn.
5 . Find nth derivative of tan–1
LMAns. 3 sinF x + n π I − 1 .3 . sinF 3x + n π I OP
H 2KQ
N 4 H 2K 4
Ans. a −1f
n − 1 a sin θ sin nθ
n
F xI .
H aK
n− 1
−n
n
7
DIFFERENTIAL CALCULUS-I
a f
LMAns. 2(−1) n OP
MN (x + 1) PQ
Ans. a −1f
n − 1 ⋅ 2x
LMAns. e 1 − 5 cos(2x + n tan ) OP
2
N
Q
Ans. e x x + n
6 . Find yn, where y = ex.x.
n
1− x
7 . Find yn, when y =
.
1+ x
n+1
n− 1
8 . Find nth derivative of log x2.
x
9 . Find yn, y = ex sin2 x.
−n
−1
n/2
2
1 0 . If y = cos x · cos 2x · cos 3x find yn.
1
(cos 6 x + cos 4x + cos 2x + 1)
[Hint: cos x · cos 2x · cos 3x =
4
LMAns. 1 L6 cosF 6x + n π I + 4 cosF 4x + n π I + 2 cosF 2x + n π I OOP
H 2K
H 2 K PQQ
N 4 MN H 2 K
n
n
n
1.2 LEIBNITZ'S* THEOREM
Statement. If u and v be any two functions of x, then
Dn (u.v) = nc0 Dn (u).v + nc1Dn–1(u). D(v) + nc2 Dn–2(u).D2 (v) + ...
+ ncr Dn–r (u).Dr(v) + ... + ncn u. Dn v ...(i)
(U.P.T.U., 2007)
Proof. This theorem will be proved by Mathematical induction.
D (u.v) = D (u).v + u.D(v) = 1c0 D (u).v + 1c1 u.D(v)
Now,
...(ii)
This shows that the theorem is true for n = 1.
Next, let us suppose that the theorem is true for, n = m from (i), we have
Dm (u.v) =
m
Differentiating w.r. to x, we have
m
c0 Dm(u).v + mc1 Dm–1 (u) D (v) + mc2 Dm–2(u) D2 (v) + ... + mcr
Dm–r(u) Dr (v) + ... + mcm u Dm(v)
r
m
r
Dm+1 (uv) = mc0 D m+1 ( u) ⋅ v + D m ( u) ⋅ D( v) + mc1 D m ( u) D( v) + Dm−1 ( u) D 2 ( v)
+ mc2
mD
m− 1
r
m
r
( v )r
( u) D 2 ( v ) + D m − 2 ( u). D 3 ( v) +...+ m c r Dm − r +1 ( u) Dr v + D m − r ( u) Dr +1 ( v)
m
+ ..... + mcm D( u) ⋅ D ( v ) + uD
m
m +1
But from Algebra we know that mcr + mcr+1 = m+1cr+1 and mc0 = m+1c0 = 1
c
h
c
h
∴ Dm+1(uv) = m +1 c 0 D m +1 ( u) ⋅ v + m c 0 + m c1 D m ( u) ⋅ D( v) + m c1 + m c 2 D m −1 u ⋅ D 2 v
+ ... +
c c + c hD
m
r
m
r +1
m−r
( u) ⋅ D r +1 ( v ) +...+ m +1 c m +1 u ⋅ D m +1 ( v )
c As c =
m
m
m +1
cm +1 = 1
h
* Gottfried William Leibnitz (1646–1716) was born Leipzig (Germany). He was Newton’s rival in
the invention of calculus. He spent his life in diplomatic service. He exhibited his calculating machine in
1673 to the Royal society. He was linguist and won fame as Sanskrit scholar. The theory of determinants
is said to have originated with him in 1683. The generalization of Binomial theorem into multinomial
theorem is also due to him. His works mostly appeared in the journal ‘Acta eruditorum’ of which he
was editor-in-chief.
8
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
⇒ Dm+1(uv) =
m+1
c0 D m+1 (u) ⋅ v + m+1 c1D m (u) ⋅ D(v) + m +1 c2 D m−1 (u) ⋅ D 2 ( v)+...
+ m +1 c r +1 D m− r (u) ⋅ D r +1 (v)+...+ m+1 cm+1u ⋅ D m +1 (v)
...(iii)
Therefore, the equation (iii) shows that the theorem is true for n = m + 1 also. But from (2)
that the theorem is true for n = 1, therefore, the theorem is true for (n = 1 + 1) i.e., n = 2, and so
for n = 2 + 1 = 3, and so on. Hence, the theorem is true for all positive integral value of n.
Example 1. If y1/m + y–1/m = 2x, prove that
(x2 – 1) yn+2 + (2n + 1) xyn+1 + (n2 – m2) yn = 0.
Sol. Given
⇒
y1/m +
1
y 1/ m
(U.P.T.U., 2007)
= 2x
y2/m – 2xy1/m + 1 = 0
(y ) – 2x(y1/m) + 1 = 0
(y1/m = z)
z2 – 2xz + 1
1/m 2
or
⇒
∴
z =
2x ± 4x 2 − 4
=x ±
2
x2 − 1
x ± x2 − 1 ⇒ y = x ±
Differentiating equation (i) w.r.t. x, we get
⇒
y1 =
⇒
or
x2 − 1
y1/m =
m x ± x −1
y1 =
2
my
m −1
⇒ y1
m
LM1 ± 2x OP = m x ± x − 1
x −1
N 2 x −1 Q
2
2
⇒
2
m
2
x 2 − 1 = my
x −1
y (x – 1) = m y
Differentiating both sides equation (ii) w.r.t. x, we obtain
2y1y2(x2 – 1) + 2xy21 = 2m2 yy1
2
1
2
...(i)
2 2
...(ii)
y2 (x2 – 1) + xy1 – m2y = 0
Differentiating n times by Leibnitz's theorem w.r.t. x, we get
Dn (y2) · (x2 – 1) + nc1 Dn–1y2·D2(x2 – 1) + nc2 Dn−2 y2 D2 (x2 − 1) + Dn (y1)x + nc1 Dn–1 (y1) Dx–m2yn
=0
⇒ yn+2 (x2 – 1) + nyn+1· 2x +
n(n − 1)
yn · 2 + yn+1 · x + nyn – m2yn = 0
2
⇒ (x2 – 1)yn+2 + (2n + 1) xyn+1 + (n2 – n + n – m2)yn = 0
⇒ (x2 – 1) yn+2 + (2n + 1) xyn+1 (n2 – m2) yn = 0. Hence proved.
Example 2. Find the nth derivative of ex log x.
Sol. Let u = ex and v = log x
n
x
n
Then D (u) = e and D (v) =
( −1) n−1 n − 1
xn
D n ( ax + b) −1 =
(−1) n n
( ax + b)n+1
9
DIFFERENTIAL CALCULUS-I
By Leibnitz’s theorem, we have
Dn (ex log x) = Dnex log x + nc1 Dn–1 (ex) D(log x) + nc2 Dn–2 (ex)D2 (log x)
+ ... + ex Dn (log x)
FG
H
IJ + ... + e (−1) n − 1
K
x
L
( −1)
n − 1O
n n( n − 1)
PP .
+...+
D (e log x) = e M log x + −
MN
x
2x
x
Q
n−1
1 n(n − 1) x
1
e − 2
= e · log x + ne · +
x
2
x
x
x
x
n
n −1
⇒
n
x
x
2
n
Example 3. Find the nth derivative of x2 sin 3x.
Sol. Let u = sin 3x and v = x2
∴
F 3x + n π I
H 2K
Dn(u) = Dn (sin 3x) = 3n sin
D(u) = 2x, D2 (v) = 2, D3 (v) = 0
By Leibnitz’s theorem, we have
Dn (x2 sin 3x) = Dn (sin 3x)x2 + nc1 Dn–1 (sin 3x) · D (x2) + nc2 Dn–2(sin 3x) · D2(x2)
I · x + n3 sin FG 3x + n − 1π IJ · 2x
H
K
2 K
n(n − 1)
F n − 2π IJ ⋅ 2
+
· 3 sin G 3x +
H
2
2 K
F nπ I + 2nx · 3 sin FG 3x + n − 1π IJ
= 3 x sin 3x +
H
H 2K
2 K
F n − 2π IJ .
+ 3 n(n−1) · sin G 3x +
H
2 K
F
H
= 3n sin 3x +
nπ
2
2
n–1
n–2
n 2
n-1
n−2
Example 4. If y = x log (1 + x), prove that
yn =
( −1) n − 2 n − 2 (x + n)
(x + 1) n
.
Sol. Let u = log (1 + x), v = x
Dn (u) =
=
⇒
dn
d n −1
log
(
1
+
)
x
=
dx n
dx n−1
d n −1
dx
⋅
n −1
(U.P.T.U., 2006)
F d log (1 + x)I
H dx
K
d n −1
1
( x + 1) −1
=
x+1
dx n −1
( −1) n−1 n − 1
n
D (u) =
( x + 1) n
and D(v) = 1, D2(v) = 0
By Leibnitz’s theorem, we have
yn = Dn (x log (1 + x) = Dn (log (1 + x)) x + nc1 Dn–1 (log (1 + x)) Dx
= x
( −1) n −1 n − 1
( x + 1) n
+
n( −1) n −2 n − 2
( x + 1) n−1
10
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
LM − x(n − 1) + n(x + 1) OP
N (x + 1) (x + 1) Q
L − xn + x + xn + n OP
n−2 M
N (x + 1) Q
L x + n OP . Hence proved.
n−2 M
N (x + 1) Q
yn = (–1)n–2 n − 2
⇒
= (–1)n–2
= (–1)n–2
n
n
n
n
Example 5. If y = a cos (log x) + b sin (log x). Show that
x2y2 + xy1 + y = 0
x2yn+2 + (2n +1) xyn+1 + (n2 + 1) yn = 0.
y = a cos (log x) + b sin (log x)
and
Sol.
Given
∴
or
y1 = – a sin (log x)
(U.P.T.U., 2003)
F 1 I + b cos (log x) F 1 I
H xK
H xK
xy1 = – a sin (log x) + b cos (log x)
Again differentiating w.r.t. x, we get
xy2 + y1 = – a cos (log x)
⇒
⇒
F 1 I – b sin (log x) F 1 I
H xK
H xK
x2y2 + xy1 = – {a cos (log x) + b sin (log x)} = – y
x2y2 + xy1 + y = 0. Hence proved.
...(i)
Differentiating (i) n times, by Leibnitz’s theorem, we have
yn+2 · x2 + n yn+1 · 2x +
n(n − 1)
yn · 2 + yn+1 · x + nyn + yn = 0
2
⇒
x2yn+2 + (2n + 1) xyn+1 + (n2 – n + n + 1) yn = 0
⇒
x2yn+2 + (2n + 1) xyn+1 + (n2 + 1) yn = 0.
Hence proved.
Example 6. If y = (1 – x)–α e–αx, show that
(1 – x)yn+1 – (n + αx) yn – nαyn–1 = 0.
Sol. Given y = (1 – x)–α. e–αx
Differentiating w.r.t. x, we get
y1 = α (1 – x)–α–1 e–αx – (1 – x)–α e–αx·α
y1 = (1 – x)–α e–αx ·α
=
y1 (1 – x) = αxy
LM 1 − 1OP = yα LM x OP
N1 − x Q
N1 − x Q
Differentiating n times w.r.t. x, by Leibnitz’s theorem, we get
yn+1 (1 – x) – nyn = αyn· x + nαyn–1
⇒ (1 – x)yn+1 – (n + αx) yn – nαyn–1 = 0.
Hence proved.
FG y IJ = log F x I , prove that x y + (2n + 1)xy + (n + m ) y = 0.
H bK
H mK
F yI
F x I = m log x
Sol. We have cos GH JK = log
H mK
b
m
m
Example 7. If cos–1
2
m
–1
2
n+2
n+1
2
n
11
DIFFERENTIAL CALCULUS-I
⇒
y = b cos
F m log x I
H mK
On differentiating, we have
F m log x I · m ⋅ 1
H mK x m
xI
F
xy = – mb sin H m log K
m
2
y1 = – b sin
⇒
1
Again differentiating w.r.t. x, we get
F
H
xy2 + y1 = – mb cos m log
x (xy2 + y1) = – m2b cos
I
K
1
1
x
m·
·
x
m
m
m
F m log x I = –m y
H mK
2
x2y2 + xy1 + m2y = 0
or
Differentiating n times with respect to x, by Leibnitz’s theorem, we get
yn+2 · x2 + nyn+1 · 2x + n(n − 1) · 2yn + xyn+1 + nyn + m2yn = 0
2
⇒ x2yn+2 + (2n + 1) xyn+1 + (n2 – n + n + m2)yn = 0
⇒ x2yn+2 + (2n + 1) xyn+1 + (n2 + m2) yn = 0.
Hence proved.
Example 8. If y = (x2 – 1)n, prove that
(x2 – 1)yn+2 + 2xyn+1 – n (n+1)yn = 0
(U.P.T.U., 2000, 2002)
UV
W
RS
T
dP
d
dn
(1 – x 2 ) n + n(n + 1) Pn = 0.
(x2 – 1)n, show that
n
dx
dx
dx
2
n
Sol. Given
y = (x – 1)
Differentiating w.r. to x, we get
Hence, if Pn =
y1 = n(x2 – 1)n–1. 2x =
⇒
2nx (x 2 − 1)n
(x 2 − 1)
(x2 – 1) y1 = 2nxy
Again differentiating, w.r.t. x, we obtain
(x2 – 1) y2 + 2xy1 = 2nxy1 + 2ny
Now, differentiating n times, w.r.t. x by Leibnitz's theorem
(x2 – 1)yn+2 + 2nxyn+1 +
or
or
2n(n − 1)
yn + 2xyn+1 + 2nyn = 2nxyn+1 + 2n2yn + 2nyn
2
(x2 – 1)yn+2 + 2xyn+1 (n+1 – n) + (n2 – n + 2n – 2n2 – 2n)yn = 0
(x2 – 1) yn+2 + 2xyn+1 – (n2 + n)yn = 0
⇒ (x2 – 1) yn + 2 + 2xyn+1 − n(n + 1) yn = 0. Hence proved.
...(i)
12
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Second part: Let y = (x2 – 1)n
∴
Now
Pn =
RS
T
d
d
yn
(1 − x 2 )
dx
dx
dn
y = yn
dx n
UV = d n(1 − x )y s
W dx
2
n +1
= (1 – x2)yn+2 – 2xyn+1 = – (x 2 − 1) y n+ 2 + 2xyn +1
⇒
RS
UV = – n(n + 1)y
T
W
d R
S(1 − x ) dPdx UVW + n (n + 1)y = 0. Hence proved.
dx T
d
d
(1 − x 2 )
Pn
dx
dx
2
or
[Using equation (i)]
n
n
n
Example 9. Find the nth derivative of y = xn–1 log x at x =
1
.
2
Sol. Differentiating
y1 = (n – 1) xn–1–1 log x + xn–1
or
1
x
(n − 1)x n −1 ⋅ log x x n−1
+
⇒ xy1 = (n – 1)y + xn–1
x
x
Differentiating (n–1) times by Leibnitz's theorem, we get
y1 =
xyn + n–1c1 yn-1 = (n – 1)yn–1 +
n−1
⇒
xyn + (n – 1)yn–1 = (n – 1)yn–1 +
n−1
⇒
xyn =
At
x =
yn
n − 1 i.e. y =
d n −1 n−1
= ( n − 1)( n − 2)...2.1 = n − 1
x
dx n −1
n−1
n
x
1
2
FG 1 IJ = 2 n − 1 .
H 2K
Example 10. If y = (1 – x2)–1/2 sin–1x, when –1 < x < 1 and −
π
π
< sin–1x < , then show
2
2
that (1 – x2)yn+1 – (2n + 1) xyn – n2 yn–1 = 0.
Sol. Given y = (1 – x2)–1/2 sin–1x
Differentiating
y1 = –
y1 =
⇒
1
(1 – x2)–3/2 (–2x) sin–1x + (1 – x2)–1/2·
2
1
2 −2
x(1 − x ) sin −1 x
(1 − x )
y1 (1–x ) = xy + 1
2
2
+
xy + 1
1
=
2
(1 − x 2 )
(1 − x )
1
1 − x2
13
DIFFERENTIAL CALCULUS-I
Differentiating n times w.r.t. x, by Leibnitz's theorem, we get
n(n − 1)
yn+1 (1 – x2) + nyn (–2x) +
yn–1 · (–2) = xyn + nyn–1
2
(1 – x2)yn+1 – (2n + 1)xyn – (n2 – n + n)yn–1 = 0
⇒
Hence proved.
(1 – x2)yn+1 – (2n + 1)xyn – n2yn–1 = 0.
Example 11. If y = xn log x, then prove that
(i) yn+1 =
n
(ii) yn = nyn–1 +
x
Sol. (i) We have y = xn log x
an − 1f .
Differentiating w.r. to x, we get
y1 = nxn–1 · log x +
⇒
xn
x
xy1 = nxn . log x + xn
xy1 = ny + xn
...(i)
Differentiating equation (i) n times, we get
n
xyn+1 + nyn = nyn +
⇒
yn+1 =
(ii)
yn =
n
Proved.
x
FG d x . log xIJ
H dx
K
F x + nx . log xI
d
=
GH x
JK
dx
d
d
= n
x .log xj +
.x
e
dx
dx
dn
d n−1
x n .log x =
n
dx
dx n −1
e
j
n−1
n
n−1
n −1
n−1
n−1
n −1
n −1
n
n −1
n −1
an − 1f . Proved.
= nyn–1 +
e
j
dn
x n log x
n
dx
d n −1
∴ yn −1 = n−1 x n−1 log x
dx
As yn =
e
EXERCISE 1.2
Find the nth derivative of the following:
1 . ex log x.
2 . x2 ex.
LMAns. e Llog x + c ⋅ 1 − c ⋅ 1 + 2 c ⋅ 1 +...+(−1)
MN
x
x
x
N
x
n
1
n
2
n
2
3
3
n−1
n − 1 n c n x −n
OP OP
QQ
Ans. e x x 2 + 2nx + n(n − 1)
j
14
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
LMAns. 6(−1) n − 4 OP
x
MN
PQ
LMAns. 2 (−1) n OP
MN (1 + x) PQ
n
3 . x log x.
3
4.
n− 3
1− x
.
1+ x
n
n+1
5 . x2 sin 3x.
LMAns. 3 x sinF 3x + nπ I + 2nx ⋅ 3 Lsin{3x + 1 (n − 1)π}O + 3 n(n − 1) sin RS3x + π (n − 2)UVOP
MN
PQ
H 2K
2
T 2 WQ
N
6 . e (2x + 3) .
Ans. e o( 2x + 3) + 6n( 2x + 3) + 12(n − 1)( 2x + 3) + 8n(n − 1)(n − 2)t
n −1
2
n
x
n− 2
x
3
2
2
7 . If x = tan y, prove that
(1 + x2) yn+1 + 2nxyn + n(n–1)yn–1 = 0.
(U.P.T.U., 2006)
8 . If y = ex sin x, prove that y′′–2y′ + 2y = 0.
9 . If y = sin (m sin–1x), prove that
(1 – x2)yn+2 – (2n + 1)x · yn+1 + (m2 – n2)yn = 0.
1 0 . If x = cos h
2
2
(U.P.T.U., 2004, 2002)
LMF 1 I log yOP , prove that (x – 1)y + xy – m y = 0 and (x – 1)y + (2n + 1)xy
NH m K Q
2
2
2
2
1
n+2
n+1
+ (n – m )yn = 0.
FG y IJ = log F x I , prove that x y + (2n + 1)xy + 2n y = 0.
H bK
H nK
n
1 1 . If cos–1
2
2
n+2
n+1
n
1 2 . If y = etan–1x, prove that (1 + x2)yn+2 + {2(n + 1) x –1}yn+1 + n(n+1)yn = 0.
1 3 . If sin–1y = 2 log (x + 1), show that
(x + 1)2yn+2 + (2n + 1)(x + 1)yn+1 + (n2 + 4)yn = 0.
d
1 4 . If y = C1 x + x 2 − 1
i + C dx − x − 1 i , prove that (x – 1)y + (2n + 1)xy = 0.
n
2
2
n
2
n+2
n+1
1 5 . If x = cos [log (y )], then show that (1 – x )yn+2 – (2n + 1) xyn+1 – (n + a ) yn = 0.
1/a
2
2
2
1.2.1 To Find (yn)0 i.e., nth Differential Coefficient of y, When x = 0
Sometimes we may not be able to find out the nth derivative of a given function in a compact form
for general value of x but we can find the nth derivative for some special value of x generally
x = 0. The method of procedure will be clear from the following examples:
–1
Example 1. Determine yn(0) where y = em.cos x.
Sol. We have
–1
y = em.cos x
Differentiating w.r.t. x, we get
–1
y 1 = em cos x m
or
FG
IJ ⇒
H 1 − x K 1 − x . y = – me
−1
2
2
2
2 2
1 − x 2 y1 = – my ⇒ (1 – x )y1 = m y
2
1
...(i)
m cos–1x
15
DIFFERENTIAL CALCULUS-I
Differentiating again
(1 – x2) 2y1 y2 – 2xy21 = 2m2yy1
⇒
(1 – x2)y2 – xy1 = m2y
...(ii)
Using Leibnitz's rule differentiating n times w.r.t. x
2n(n − 1)
yn – xyn+1 – nyn = m2yn
(1 – x2)yn+2 – 2nxyn+1 –
2
(1 – x2)yn+2 – (2n +1)xyn+1 – (n2 + m2)yn = 0
or
Putting x = 0
yn+2 (0) – (n2 + m2)yn (0) = 0
⇒
replace n by (n – 2)
yn+2 (0) = (n2 + m2)yn(0)
...(iii)
yn (0) = {(n – 2)2 + m2} yn–2 (0)
replace n by (n – 4) in equation (iii), we get
yn–2 (0) = {(n – 4 )2 + m2} yn–4 (0)
∴
yn(0) = {(n – 2)2 + m2} {(n – 4)2 + m2} yn–4 (0)
Case I. When n is odd:
yn(0) = {(n – 2)2 + m2} {(n – 4)2 + m2} .... (12 + m2)y1 (0)
[The last term obtain putting n = 1 in eqn. (iii)]
1
m cos −1 x
m⋅
Now we have
y1 = − e
1 − x2
mπ
...(iv)
...(v) As cos −1 0 =
At x = 0, y1(0) = −me 2
Using (v) in (iv), we get
yn(0) = −
π
2
mπ
2
{an − 2f + m } {an − 4f + m } .... e1 + m j me .
2
2
2
2
2
2
Case II. When n is even:
yn(0) = {(n – 2)2 + m2} {(n – 4)2 + m2} .... (22 + m2)y2 (0)
[The last term obtain by putting n = 2 in (iii)]
From (ii),
y2(0) = m2(y)0
∴
y2(0) = m2 emπ/2
...(vi)
−1
As y = e m cos x
...(vii)
−1
∴ y(0) = e m cos 0 = emπ/2
From eqns. (vi) and (vii), we get
yn(0) = {(n – 2)2 + m2} {(n – 4)2 + m2} .... (22 + m2) m2emπ/2.
Example 2. If y = (sin–1x)2. Prove that yn(0) = 0 for n odd and yn(0) = 2.22.42....(n – 2)2,
n ≠ 2 for n even.
(U.P.T.U., 2005, 2008)
–1
2
Sol. We have
y = (sin x)
...(i)
1
−1
On differentiating
y 1 = 2 sin–1 x ·
⇒ y1 1 − x 2 = 2 y , As y = sin x
2
1− x
Squaring on both sides,
y21 (1 – x2) = 4y
e
j
16
or
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Again differentiating
2(1 – x2) y1 y2 – 2xy12 = 4y1
(1 – x2) y2 – xy1 = 2
...(ii)
Differentiating n times by Leibnitz’s theorem
2n n − 1
(1 – x2) yn+2 – 2nxyn+1 –
yn – xyn+1 – nyn = 0
2
a f
(1 – x2) yn+2 – (2n + 1) xyn+1 – n2yn = 0
Putting x = 0 in above equation
yn+2 (0) – n2 yn (0) = 0
⇒
yn+2 (0) = n2 yn(0)
replace n by (n – 2)
yn(0) = (n – 2)2 yn –2 (0)
or
...(iii)
Again replace n by (n – 4) in (iii) and putting the value of yn–2 (0) in above equation
yn (0) = (n – 2)2 (n – 4)2 yn–4 (0)
Case I. If n is odd, then
yn(0) = (n – 2)2 (n – 4)2 (n – 6)2 ... 12·y1 (0)
1
But
y1(0) = 2 sin–1 0·
=0
1–0
∴
yn(0) = 0. Hence proved.
Case II. If n is even, then
yn(0) = (n – 2)2 (n – 4)2 ... 22· y2 (0)
From (ii)
y2 (0) = 2
Using this value in eqn. (iv), we get
yn(0) = (n – 2)2 (n – 4)2 .... 22·2
yn(0) = 2.22.42 ... (n – 2)2, n ≠ 2 otherwise 0. Proved.
or
m
Example 3. If
y =
x + 1 + x2
Sol. Given
y =
x + 1 + x2
∴
2
y1 = m x + 1 + x
=
y1 1 + x 2
or
Squaring
m
m x + 1 + x2
1+ x
, find yn 0).
m −1
LM1 + x OP
N 1+x Q
...(i)
2
m
=
2
my
1 + x2
= my
y21(1+x2) = m2y2
2
1
or
...(iv)
...(ii)
2
2
Again differentiating, y (2x) + (1 + x )2y1y2 = m ·2yy1
y2 (1 + x2) + xy1 – m2y = 0
Differentiating n times by Leibnitz’s theorem
2n(n − 1)
(1 + x2)yn+2+ 2nxyn+1 +
yn + xyn+1 + nyn – m2yn = 0
2
...(iii)
17
DIFFERENTIAL CALCULUS-I
(1 + x2)yn+2 + (2n + 1)xyn+1 + (n2 – m2)yn = 0
Putting x = 0, we get
yn+2 (0) + (n2 – m2) yn(0) = 0
⇒
yn+2 (0) = – (n2 – m2) yn (0)
replace n by n – 2
yn (0) = – { (n – 2)2 – m2} yn–2 (0)
Again replace n by (n – 4) in (iv) and putting yn–2 (0) in above equation
or
...(iv)
yn(0) = (–1)2 {(n – 2)2 – m2} {(n – 4}2 – m2} yn–4 (0)
Case I. If n is odd
But
or
⇒
Case II. If n is even
yn(0) = – {(n – 2)2 – m2} {(n – 4)2 – m2} ... {12 – m2} y1 (0)
y1(0) = my(0)
y1(0) = m (As y (0) = 1)
yn(0) = {m2 – (n – 2)2}{m2 – (n – 4)2} ... (m2 – 12)·m.
yn (0) = {m2 – (n – 2)2} {m2 – (n – 4)2}... (m2 – 22) y2 (0)
⇒
yn (0) = {m2 – (n – 2)2} {m2 – (n – 4)2} ... (m2 – 22)·m2.
(As y2 (0) = m2).
–1
Example 4. Find the nth differential coefficient of the function on cos (2 cos x) at the point
x = 0.
or
Sol. Let
y = cos (2 cos–1 x)
On diffentiating,
y1 = – sin (2 cos–1 x)
LM −2 OP
MN 1 − x PQ
...(i)
2
–1
1 − x 2 = 2 sin (2 cos x)
Squaring on both sides, we get
y1
e
−1
= 4 sin2 2 cos x
y21 (1 – x2)
{
e
j
j}
2
−1
= 4 1 − cos 2 cos x
or
2
1
2
2
y (1 – x )
= 4 (1 – y )
Again differentiating w.r.t. x, we get
2y1y2 (1 – x2) – 2xy21 = – 8yy1
or
y2 (1 – x2) – xy1 + 4y = 0
...(ii)
Differentiating n times by Liebnitz’s theorem
2n n − 1
(1 – x2) yn+2 – 2nxyn+1 –
yn – xyn+1 – nyn + 4yn = 0
2
a f
(1 – x2) yn+2 – (2n + 1) xyn+1 – (n2– 4) yn = 0
Putting x = 0 in above equation, we get
yn+2 (0) – (n2 – 4) yn (0) = 0
or
yn+2 (0) = (n2 – 4) yn(0)
Replace n by n – 2, we get
yn (0) = {(n – 2)2 – 4} yn–2 (0)
...(iii)
18
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Again replace n by (n – 4) in (iii) and putting yn –2 (0) in above then, we get
yn (0) = {(n – 2)2 – 4} {(n – 4)2 – 4} yn – 4 (0)
Case I. If n is odd
yn (0) = {(n – 2)2 – 4} {(n – 4)2 – 4} .......... (12 – 4) y1 (0)
But
y1 (0) = 2 sin (2 cos–1 0) = 2 sin (π) = 0
∴
yn (0) = 0.
Case II. If n is even
yn (0) = {(n – 2)2 – 4}{(n – 4)2 – 4} .......... {22 – 4} y2 (0)
yn (0) = 0
Hence for all values of n, even or odd,
yn (0) = 0.
Example 5. Find the nth derivative of y = x2 sin x at x = 0.
Sol. We have
(U.P.T.U., 2008)
y = x2 sin x = sin x. x2
...(i)
Differentiate n times by Leibnitz’s theorem, we get
yn = nC Dn (sin x). x2 + nC Dn–1 (sin x) D(x2) + nC Dn–2 (sin x) D2 (x2) + 0
FG
H
IJ + 2nx . sin FG x + n − 1 πIJ + n(n – 1) sin FG x + n − 2 πIJ
K
H 2 K
H 2 K
F nπ IJ + 2nx sin FG x + nπ − π IJ + n(n – 1) sin FG x + nπ − πIJ
= x sin G x +
H 2K
H 2 K
H 2 2K
F nπ IJ – 2nx cos FG x + nπ IJ – n(n – 1) sin FG x + nπ IJ
= x sin G x +
H 2K
H 2K
H 2K
F nπ IJ – 2nx cos FG x + nπ IJ
y = (x – n + n) sin G x +
H 2K
H 2K
0
nπ
= x2 . sin x +
2
1
2
2
2
2
2
n
Putting x = 0, we obtain
nπ
.
2
Example 6. If y = sin (a sin–1 x), prove that
yn (0) = (n – n2) sin
(1 – x2)yn+2 – (2n + 1)xyn+1 – (n2 – a2)yn = 0
Also find nth derivative of y at x = 0.
Sol. We have
[U.P.T.U. (C.O.), 2007]
y = sin(a sin–1 x)
Differentiating w.r.t. x, we get
y1 = cos(a sin–1 x).
or
or
a
1 − x2
...(i)
y1 1 − x 2 = a cos(a sin–1 x)
Squaring on both sides, we obtain
y12 (1 – x2) = a2 cos2 (a sin–1 x) = a2 [1 – sin2 (a sin–1 x)]
y12 (1 – x2) = a2 (1 – y2)
...(ii)
19
DIFFERENTIAL CALCULUS-I
Differentiating again, we get
2y1 y2 (1 – x2) – 2xy12 = – 2 a2yy1
y2 (1 – x2) – xy1 = – a2y
or
...(iii)
Differentiating n times by Leibnitz’s theorem
a f y – xy – ny = – a y
(1 – x2)yn+2 – 2nxyn+1 − 2n n − 1
2
or
2
n
n+1
n
n
(1 – x2)yn+2 – (2n + 1)xyn+1 – (n2 – a2)yn = 0
...(iv)
Putting x = 0 in relation (iv), we get
yn+2 (0) – (n2 – a2)yn (0) = 0
yn+2 (0) = (n2 – a2)yn (0)
or
...(v)
Replace n by (n – 2) in relation (v), we get
yn (0) = {(n – 2)2 – a2}yn–2 (0)
Again replace n by (n – 4) in equation (v) and putting yn–2 (0) in above relation, we get
yn (0) = {(n – 2)2 – a2} {(n – 4)2 – a2}yn–4 (0)
Case I. When n is odd:
yn (0) = {(n – 2)2 – a2} {(n – 4)2 – a2} ..... {12 – a2}y1 (0)
...(vi)
[The last term in (vi) obtain by putting n = 1 in equation (v)]
Putting x = 0, in equation (i), we get
y1 (0) = cos(a sin–1 0). a = cos 0 . a ⇒ y1 (0) = a
yn (0) = {(n – 2)2 – a2} {(n – 4)2 – a2} ..... {12 – a2}. a
Hence,
Case II. When n is even:
yn (0) = {(n – 2)2 – a2} {(n – 4)2 – a2} ..... {22 – a2}y2 (0).
Putting x = 0 in (iii), we get
[The last term obtain by putting n = 2 in equation (v)]
y2 (0) = – a2 y(0) = – a2 x0 = 0
Hence,
(As y(0) = 0)
yn (0) = 0.
EXERCISE 1.3
1 . If y = tan–1 x, find the value of y7 (0) and y8 (0).
a sin–1 x
[Ans.
6
and 0.]
2 . If y = e
, prove that (1 – x ) yn + 2 – (2n + 1)xyn + 1 – (n + a ) yn = 0 and hence find the
value of yn when x = 0.
Ans. n is odd, yn (0) =
2
2
2
{an − 2f + a } {an − 4f + a } ... (3 + a ) (1 + a )a n is even,
y (0) = {an − 2f + a } {an − 4f + a } ...... (4 + a )(2 + a ) a .
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
n
3 . If log y = tan–1x, show that (1 + x2) yn + 2 + {2(n + 1)x – 1} yn + 1 + n (n + 1) yn = 0 and hence
find y3, y4 and y5 at x = 0.
[U.P.T.U. (C.O.), 2003]
Ans. y3 (0) = – 1, y4 (0) = – 1, y5 (0) = 5
20
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
4 . If f (x) = tan x, then prove that
fn (0) – nc2 fn – 2 (0) + nc4 fn – 4 (0) – ....... = Sin
5 . If y = sin–1x, find yn (0).
FG nπ IJ .
H 2K
Ans. n is odd, yn (0) = (n – 2)2 (n – 4)2 .... 52 . 32 . 1 n is even, yn (0) = 0.
6 . Find yn (0) when y = sin (m sin h–1 x).
a fFG IJ {an − 2f + m } {an − 4f + m } .... (1 + m ) m · n
n−1
Ans. n is odd, yn (0) = –1 H 2 K
2
2
2
2
2
2
is even, yn = (0) = 0.
LM {
N
7 . If y = log x + 1 + x 2
}OPQ , show that
2
yn + 2 (0) = – n2 yn (0) hence find yn (0).
a f
Ans. n is odd, yn (0), = 0 n is even yn(0) = −1
n− 2
2
(n – 2)2 (n – 4)2 .... 42 22.2.
PARTIAL DIFFERENTIATION
Introduction
Real world can be described in mathematical terms using parametric equations and functions such
as trigonometric functions which describe cyclic, repetitive activity; exponential, logrithmic and
logistic functions which describe growth and decay and polynomial functions which approximate
these and most other functions.
The problems in computer science, statistics, fluid dynamics, economics etc., deal with functions of two or more independent variables.
1.3
FUNCTION OF TWO VARIABLES
If f (x, y) is a unique value for every x and y, then f is said to be a function of the two independent
variables x and y and is denoted by
z = f (x, y)
Geometrically the function z = f (x, y) represents a surface.
The graphical representation of function of two variables is shown in Figure 1.1.
y
¦
(x,y)
z
O
z = ¦ (x,y)
Fig. 1.1
x
21
DIFFERENTIAL CALCULUS-I
1.4
PARTIAL DIFFERENTIAL COEFFICIENTS
The partial derivative of a function of several variables is the ordinary derivative with respect to
any one of the variables whenever, all the remaining variables are held constant. The difference
between partial and ordinary differentiation is that while differentiating (partially) with respect to
one variable, all other variables are treated (temporarily) as constants and in ordinary differentiation no variable taken as constant,
Definition: Let
z = f (x, y)
Keeping y constant and varying only x, the partial derivative of z w.r.t. ‘x’ is denoted by
and is defined as the limit
∂z
∂x
= Lim
b
∂z
∂x
g b g
f x + δx, y − f x, y
δx→0
δx
∂z
Partial derivative of z, w.r.t. y is denoted by ∂y and is defined as
∂z
∂y
= Lim
δy→0
Notation: The partial derivative
b
g b g.
f x, y + δy − f x, y
δy
∂z
∂z
∂f
is also denoted by
or fx similarly ∂y is denoted
∂x
∂x
∂f
by ∂y or fy. The partial derivatives for higher order are calculated by successive differentiation.
Thus,
∂2 f
∂ 2z
=
∂x 2
∂x 2
= fxx,
∂2z
∂2 f
=
= fyy
∂y 2
∂y 2
∂2z
∂2 f
=
∂x∂y
∂x∂y
= fxy,
∂2 f
∂2z
=
= fyx and so on.
∂y∂x
∂y∂x
∂z
∂z
and
:
∂
y
∂x
Let z = f(x, y) represents the equation of a surface in xyzcoordinate system. Suppose APB is the curve which a plane
through any point P on the surface O to the xz-plane, cuts. As
point P moves along this curve APB, its coordinates z and x
vary while y remains constant. The slope of the tangent line at
P to APB represents the rate at which z-changes w.r.t. x.
Geometrical interpretation of
Hence,
and
Z
A
P
O
∂z
= tan θ (slope of the curve APB at the point P)
∂x
∂z
= tan φ (slope of the curve CPD at point P)
∂y
Y
f
q
B
D
X
Fig. 1.2
22
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 1. Verify that
∂ 2u
∂ 2u
=
where u(x, y) = loge
∂x∂y
∂y∂x
Fx + y I
GH xy JK
2
Fx + y I
GH xy JK
2
2
(U.P.T.U., 2007)
2
Sol. We have
u(x, y) = loge
⇒
u(x, y) = log (x2 + y2) – log x – log y
...(i)
Differentiating partially w.r.t. x, we get
∂u
∂x
=
2x
1
–
x2 + y2
x
Now differentiating partially w.r.t. y.
∂ 2u
∂y∂x
= –
4xy
...(A)
ex + y j
2 2
2
Again differentiate (i) partially w.r.t. y, we obtain
∂u
∂y
=
2y
ex + y j
2
2
1
y
–
Next, we differentiate above equation w.r.t. x.
∂ 2u
4xy
= –
2
2
∂x∂y
x + y2
e
...(B)
j
Thus, from (A) and (B), we find
∂ 2u
∂ 2u
=
.
∂y∂x
∂x∂y
Hence proved.
FG y IJ , verify that ∂ f = ∂ f .
H xK
∂y∂x
∂x∂y
F yI
f = tan G J
H xK
2
Example 2. If f = tan–1
Sol. We have
2
–1
...(i)
Differentiating (i) partially with respect to x, we get
∂f
=
∂x
FG –y IJ = F –y I
F y I H x K GH x + y JK
1+G J
H xK
1
2
2
2
...(ii)
2
Differentiating (i) partially with respect to y, we get
∂f
1
1
x
=
= 2
2
x
∂y
x + y2
y
1+
x
Differentiating (ii) partially with respect to y, we get
FG IJ
H K
F
GH
∂2 f
−y
∂
=
∂y∂x
∂y x 2 + y 2
...(iii)
I = ex + y ja–1f − b– ygb2yg
JK
ex + y j
2
2
2
2 2
23
DIFFERENTIAL CALCULUS-I
=
y2 − x2
...(iv)
ex + y j
2 2
2
Differentiating (iii) partially with respect to x, we get
F
GH
∂2 f
x
∂
=
2
∂x∂y
∂x x + y 2
=
I = ex + y ja1f − xa2xf
JK
ex + y j
2
2
3
2 2
y2 − x2
...(v)
ex + y j
2 2
2
∂2 f
∂2 f
=
. Hence proved.
∂y∂x
∂x∂y
∴ From eqns. (iv) and (v), we get
F ∂u ∂u I
F ∂u ∂u I
Example 3. If u(x + y) = x + y , prove that G − J = 4 G 1 − ∂x − ∂y J .
H
K
H ∂x ∂y K
2
2
Sol. Given
∴
2
u =
x2 + y2
x+y
bx + yga2xf − ex + y ja1f = x + 2xy − y
b x + yg
bx + y g
bx + ygb2yg − ex + y ja1f = y + 2xy − x
∂u
=
∂y
bx + yg
bx + yg
∂u
∂x
2
=
2
2
2
2
2
and
2
2
2
2
2
∴
∂u
∂u
+
∂x
∂y
∂u
∂u
1–
–
∂x
∂y
or
∂u
∂u
–
∂y
∂x
and
4xy
=
b x + yg
=
1−
2
4xy
( x + y)
=
2
(x − y) 2
...(i)
(x + y) 2
ex + 2xy − y j − e y + 2xy − x j
=
bx + y g
2 ex − y j
2 bx − yg
=
=
bx + yg
bx + y g
2
2
2
2
2
2
2
2
∴
2
...(ii)
From (ii), we get
F ∂u − ∂u I
GH ∂x ∂y JK
2
=
b g = 4 F1 − ∂u − ∂u I , from (i). Hence proved.
GH ∂x ∂y JK
bx + yg
4 x− y
2
2
24
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 4. If u = sin–1
Sol. Given
and from (i),
y
∂u
∂x
=
∂u
∂y
=
∂u
∂y
=
x
Example 5. If f(x, y) = x2 tan–1
2
yx
...(ii)
x + y2
2
F– x I + 1 ⋅ 1
R| F x I U| GH y JK 1 + FG y IJ x
S|1 − GH y JK V|
H xK
T
W
1
2
2
x
−
+
ey − x j
2
2
2
xy
...(iii)
x + y2
2
∂u
∂u
+y
= 0.
∂y
∂x
Hence proved.
FG y IJ – y tan FG x IJ then prove that ∂ f = ∂ f .
H xK
H yK
∂x∂y
∂y∂x
2
2
2
–1
FG y IJ + x · 1 × FG − y IJ – y · 1 · FG 1 IJ
H xK
F yI H x K
F xI H yK
1+G J
1+G J
H xK
H yK
∂f
F y I yx – y = 2x tan FG y IJ – y
= 2x · tan G J –
H xK
∂x
H xK x + y x + y
∂f
= 2x · tan–1
∂x
2
2
2
2
2
2
3
–1
or
...(i)
2
2
−
ey − x j
2
Adding (ii) and (iii), we get x
Sol.
–1
2
x
or
–1
–1
∴
or
F x I + tan FG y IJ , show that x ∂u + y ∂u = 0.
GH y JK
∂y
H xK
∂x
F xI
F yI
u = sin G J + tan G J
H xK
y
H K
∂u
F yI
1
1
1
=
.
+
. G− J
∂x
R| F x I U| y 1 + FG y IJ H x K
S|1 − GH y JK V|
H xK
W
T
–1
2
2
2
2
Differentiating both sides with respect to y, we get
FG 1 IJ –1 = 2x –1 = x − y
...(i)
H xK
yI
x +y
x +y
F
1+G J
H xK
F xI
F xI
∂f
1
1
1
= x ·
·
– 2y tan G J – y ·
· G− J
∂y
H yK
F yI x
F xI H y K
1+G J
1+G J
H xK
H yK
F x I xy
∂f
x
=
– 2y tan G y J +
H K x +y
∂y
x +y
∂2 f
= 2x ·
∂y∂x
Again
2
1
2
2
·
2
2
–1
2
2
2
2
2
2
–1
2
2
2
3
or
2
2
2
2
25
DIFFERENTIAL CALCULUS-I
j – 2y tan FG x IJ = x – 2y tan FG x IJ .
H yK
H yK
x +y
e
x x2 + y2
=
2
–1
2
–1
Differentiating both sides with respect to x, we get
∂2 f
= 1 – 2y
∂x ∂y
F 1I
2y
x −y
G
J
=1–
=
y
H
K
x +y
x
+
y
F xI
1+G J
H yK
2
1
2
2
2
2
2
2
2
...(ii)
Thus, from (i) and (ii), we get
∂2 f
∂2 f
=
. Hence proved.
∂x∂y
∂y∂x
Example 6. If V = (x2 + y2 + z2)–1/2, show that
∂ 2V
∂ 2V
∂ 2V
+
+
= 0.
2
∂y
∂x 2
∂z 2
Sol. Given
V = (x2 + y2 + z2)–1/2.
∴
∂V
∂x
= –
∴
∂ 2V
∂x 2
= – x –
...(i)
1
(x2 + y2 + z2)–3/2 2x = – x (x2 + y2 + z2)–3/2
2
LM RS 3 ex + y + z j
NT 2
2
2
2 –5/2
UV e
W
⋅ 2x + x 2 + y 2 + z 2
= 3x2 (x2 + y2 + z2)–5/2 – (x2 + y2 + z2)–3/2
= (x2 + y2 + z2)–5/2 [3x2 – (x2 + y2 + z2)]
or
and
∂ 2V
= (x2 + y2 + z2)–5/2 (2x2 – y2 – z2)
∂x 2
Similarly from (i), we can find
j
–3/2.
⋅1
OP
Q
...(ii)
∂ 2V
∂y 2
= (x2 + y2 + z2)–5/2 (2y2 – x2 – z2)
...(iii)
∂ 2V
∂z 2
= ( x2 + y2 + z2)–5/2 (2z2 – x2 – y2)
...(iv)
Adding (ii), (iii) and (iv), we get
∂ 2V
∂ 2V
∂ 2V
+
+
2
∂y
∂x 2
∂z 2
= (x2 + y2 + z2)–5/2 [(2x2 – y2 – z2) + (2y2 – x2 – z2) + (2z2 – x2 – y2)]
= (x2 + y2 + z2)–5/2 [0] = 0. Hence proved.
Example 7. If u = f (r), where r2 = x2 + y2, show that
∂ 2u
∂ 2u
1
f ′ (r).
2 + ∂y 2 = f ′′ (r) +
r
∂x
(U.P.T.U., 2001, 2005)
26
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Sol. Given
r2 = x2 + y2
...(i)
Differentiating both sides partially with respect to x, we have
2r
Similarly,
∂r
∂x
= 2x or
∂r
∂y
=
Now,
∂r
x
=
∂x
r
...(ii)
y
r
...(iii)
u = f(r)
∂u
∂r
x
= f ′ (r).
= f ′ (r) · , from (ii)
∂x
∂x
r
Again differentiating partially w.r.to x, we get
∴
LM f ′ arfx OP = r f ′ arf.1 + xf ′′arfb∂r / ∂xg – xf ′ arfb∂r / ∂xg
r
N r Q
O
x
1 L
∂ u
rf ′ ar f + x f ′′ar f −
f ′ arfP , from (ii).
M
=
r
r N
∂x
Q
∂ u
O
1 L
y
rf ′ ar f + y f ″ ar f –
f ′ arfP
Similarly,
=
M
∂y
r MN
r
PQ
L
ex + y j f ′ arfOP
∂ u
∂ u
1 M
2rf ′ ar f + ex + y j f ″ arf −
Adding,
+
=
PQ
r
∂x
∂y
r M
N
∂ 2u
∂x 2
=
∂
∂x
2
2
or
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
=
1
[2rf ′(r) + r2f ″(r) – rf ′(r)], from (i)
r2
=
1
f ′(r) + f ″(r). Hence proved.
r
Example 8. If u = log (x3 + y3 + z3 – 3xyz); show that
F ∂ + ∂ + ∂I u=– 9
.
GH ∂x ∂y ∂z JK
bx + y + zg
2
2
Sol. Given
and
(U.P.T.U., 2003)
u = log (x3 + y3 + z3 – 3xyz).
∴
∂u
∂x
=
x 3 + y 3 + z 3 − 3xyz
Similarly,
∂u
∂y
=
x + y 3 + z 3 – 3xyz
∂u
∂z
=
3x 2 − 3 yz
...(i)
3 y 2 − 3xz
...(ii)
3
3z 2 − 3xy
x 3 + y 3 + z 3 – 3xyz
.
...(iii)
27
DIFFERENTIAL CALCULUS-I
Adding eqns. (i), (ii) and (iii), we get
∂u
∂u
∂u
+ ∂y +
∂x
∂z
=
e
3 x 2 + y 2 + z 2 – xy − yz − zx
x + y + z − 3xyz
3
3
3
j
e
3 x 2 + y 2 + z 2 – xy − yz − zx
=
j
bx + y + zgex + y + z − xy − yz − zxj
2
2
2
a
f
As a 3 + b 3 + c 3 – 3abc = a + b + c ( a 2 + b 2 + c 2 − ab − bc − ca)
∂u
∂u
∂u
+
+
∂x
∂y
∂z
or
=
3
.
x+y+z
...(iv)
F ∂ ∂ + ∂ I u = FG ∂ + ∂ + ∂ IJ FG ∂ + ∂ + ∂ IJ u
Now, G +
H ∂x ∂y ∂z K H ∂x ∂y ∂z K
H ∂x ∂y ∂z JK
F ∂ ∂ ∂ I F ∂u ∂u ∂u I F ∂ ∂ ∂ I F 3 I
= G ∂x + ∂y + ∂z J G ∂x + ∂y + ∂z J = G ∂x + ∂y + ∂z J G x + y + z J , from (iv)
H
KH
K H
KH
K
L ∂ F 1 I ∂ F 1 I ∂ F 1 IO
= 3 M ∂x G x + y + z J + ∂y G x + y + z J + ∂z G x + y + z J P
K H
K H
K PQ
MN H
L 1
OP
1
1
−9
−
−
= 3 M−
MN bx + y + zg bx + y + zg bx + y + zg PQ = bx + y + zg .
2
2
2
2
2
Hence proved.
Example 9. If u = tan–1
∂ 2u
∂x∂y
Sol. Given
∴
e
xy
1 + x2 + y2
=
1
e1 + x + y j
2 3/2
2
⋅
xy
u = tan–1
∂u
∂y
j
, show that
e1 + x + y j
2
2
⋅
1
LM1 + {x y /e1 + x + y j}OP
N
Q
LM √ e1 + x + y j.1 − y 1 e1 + x + y j
2
×x M
+
1
x
e +y j
MN
e1 + x + y j − y
x
=
.
√ e1 + x + y j
1+ x + y + x y
=
2 2
2
2
2
2
2 2
2
2
2 −1/2
2
2
2
2
2
2
2
2
2y
OP
PP
Q
28
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
∂u
∂y
or
=
x
1 + x2
x
=
e1 + x je1 + y j e1 + x + y j e1 + y j e1 + x + y j
.
2
2
2
2
2
2
2
Again differentiating partially w.r.to x
LM
OP
L x OP
∂ M
x
1
=
MMe1 + y j e1 + x + y j PP = e1 + y j ∂x MM e1 + x + y j PP
N
Q
N
Q
LM e1 + x + y j1 − x 1 e1 + x + y j 2x OP
1
2
=
M
PP
e1 + y j MN
e1 + x + y j
Q
e1 + x + y j – x =
1
1
.
=
. Hence proved.
e1 + y j e1 + x + y j
e1 + x + y j
∂ 2u
∂x ∂y
∂
∂x
2
2
2
2
2
2
2
2
2
2
2
2 3/2
2
Example 10. If
2
2
2 −1/2
2
2 3/2
2
2
y2
x2
z2
+
+
= 1, show that
a2 + u
c2 + u
b2 + u
FG ∂u IJ + FG ∂u IJ + FG ∂u IJ = 2 FG x ∂u + y ∂u + z ∂u IJ .
H ∂x K H ∂y K H ∂z K
H ∂x ∂y ∂z K
2
2
2
2
(U.P.T.U., 2002)
Sol. We have
y2
x2
z2
+
+
= 1
a2 + u b2 + u c2 + u
where u is a function of x, y and z
Differentiating (i) partially with respect to x, we get
...(i)
LM x
OP ∂u
y
z
+
+
MMe a + uj eb + uj e c + uj PP ∂x = 0
N
Q
2x / e a + uj
2x / e a + uj
∂u
=
=
∂x
LMx / ea + uj + y / eb + uj + z / e c + uj OP ∑ LMx / e a + uj OP
Q
N
N
Q
2 y / e b + uj
2z / e c + uj
∂u
∂u
=
;
=
∂y
∑ LMNx / e a + uj OPQ ∂z ∑ LMNx / ea + uj OPQ
=
2
2
2x
−
a2 + u
2
2
2
2
2
2
2
2
2
or
2
2
2
2
2
2
2
Similarly,
2
2
2
2
2
2
2
2
2
2
2
Adding with square
L
O
FG IJ + FG ∂u IJ + FG ∂u IJ = 4 MNx / e a + uj + y / eb + uj + z / e c + uj PQ
H K H ∂y K H ∂z K
LM∑ RSx / e a + uj UVOP
WQ
N T
∂u
∂x
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
29
DIFFERENTIAL CALCULUS-I
FG ∂u IJ + FG ∂u IJ + FG ∂u IJ =
H ∂x K H ∂y K H ∂z K
2
2
or
Also, x
2
∂u
∂u
∂u
+y
+z
=
∂x
∂z
∂y
=
4
∑ LMNx / e a + uj OPQ
2
...(ii)
2
2
LM 2x
O
2y
2z P
+
+
∑ LMNx / e a + uj OPQ MN e a + uj eb + uj e c + uj PQ
2
2
2
2
∑ LMNx 2 / e a 2 + uj OPQ
2
2
2
1
2
2
2
2
[1], from (i)
...(iii)
From (ii) and (iii), we have
FG ∂u IJ + FG ∂u IJ + FG ∂u IJ = 2 FG x ∂u + y ∂u + z ∂u IJ . Hence proved.
H ∂x K H ∂y K H ∂z K H ∂x ∂y ∂z K
2
2
2
Example 11. If xx yy zz = c, show that at x = y = z,
∂ 2z
= – (x log ex)–1. Where z is a function of x and y.
∂x ∂y
Sol. Given
xx.yy.zz = c, where z is a function of x and y.
Taking logarithms, x log x + y log y + z log z = log c.
...(i)
Differentiating (i) partially with respect to x, we get
LMx FG 1 IJ + blog xg1OP + LMz FG 1 IJ + blog zg1OP ∂z = 0
N H xK
Q N H zK
Q ∂x
b1 + log xg
∂z
= −
b1 + log zg
∂x
∂z
b1 + log yg
Similarly, from (i), we have ∂y = –
b1 + log zg
∂ F ∂z I
∂ z
∂ L F 1 + log y I O
∴
= ∂x G ∂y J =
∂x∂y
H K ∂x MMN−GH 1 + log z JK PPQ , from (iii)
=
or
2
∂2z
∂
–1
∂x∂y = – (1 + log y) · ∂x [(1 + log z) ]
or
LM– b1 + log zg ⋅ 1 ⋅ ∂z OP
z ∂x Q
N
∂ z
b1 + log yg · R|S−F 1 + logxIU|V , from (iii)
=
∂x∂y
zb1 + log zg
T| GH 1 + log zJKW|
∂ z
b1 + log xg
At x = y = z, we have
=–
∂x∂y
x b1 + log xg
−2
= – (1 + log y) ·
2
or
2
2
∴
2
3
...(ii)
...(iii)
30
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
∂2z
1
1
∂x∂y = − x 1 + log x = – x log e e + log x = (As log e e = 1)
b
⇒
g
b
g
a f = – mx log aexfr . Hence proved.
1
x log ex
=–
−1
Example 12. If u = log (x3 + y3 – x2y – xy2) then show that
∂ 2u
4
∂ 2u
∂ 2u
.
+
2
+
= −
2
2
2
∂x∂y
∂x
∂y
x+y
b
Sol. We have
3
3
2
g
2
u = log (x + y – x y – xy )
∂u
3x 2 − 2xy − y 2
= 3
∂x
x + y 3 − x 2 y − xy 2
e
3 y − x − 2xy
∂u
= 3
∂y
x + y 3 − x 2 y − xy 2
2
e
...(i)
j
...(ii)
2
e
Adding (i) and (ii), we get
j
j e
3x 2 − 2xy − y 2 + 3 y 2 − x 2 − 2xy
∂u
∂u
+
=
∂y
∂x
x 3 + y 3 − x 2 y − xy 2
e
b g
j
2 x− y
2
2
2
j
bx + yg ex + y – 2xyj
2 bx − yg
∂u
∂u
2
+
=
=
∂x
∂y
bx + ygbx − yg bx + yg
F∂ ∂I
∂ u
∂ u
∂ u
Now,
+2
+
= G + J u
H ∂x ∂y K
∂x
∂x∂y
∂y
F∂ ∂I F∂ ∂I
= G ∂x + ∂y J G ∂x + ∂y J u
H
KH
K
F ∂ ∂ I 2 (from iii)
= G + J .
H ∂x ∂y K x + y
∂ F 1 I
∂ F 1 I
G
J
+
2
=2
∂y GH x + y JK
∂x H x + y K
=
2
... (iii)
2
2
2
2
2
2
2
=–
2
b x + yg
2
Example 13. If u = exyz, show that
∂ 3u
= (1 + 3xyz + x2y2z2)exyz.
∂x∂y∂z
–
2
b x + yg
2
=–
4
bx + yg
2
. Hence proved.
31
DIFFERENTIAL CALCULUS-I
u = exyz ∴
Sol. We have
or
∂
xyz
xyz 2
xyz
∂y (e ·xy) = e x yz + e .x
∂ 2u
∂y∂z
=
∂ 2u
∂y∂z
= (x2yz + x) exyz
∂ 3u
∂x∂y∂z
Hence
∂u
= exyz.xy
∂z
= (2xyz + 1) exyz + (x2yz + x) exyz·yz
= (1 + 3xyz + x2y2z2) exyz. Hence proved.
Example 14. If u = log r, where r2 = (x – a)2 + (y – b)2 + (z – c)2, show that
∂ 2u
∂ 2u
∂ 2u
1
+
+
= 2.
2
2
2
∂x
∂z
∂y
r
Sol. Given
r2 = (x – a)2 + (y – b)2 + (z – c)2,
...(i)
Differentiating partially with respect to x, we get
2r
Similarly,
FG x − a IJ .
H r K
∂r
b y − bg and ∂r = az − cf
=
∂r
∂x
∂y
or
...(ii)
r
IJ
K
FG
H
∂u
∂x
=
1 ∂r
1 x−a
=
, from (ii)
r
r ∂x
r
∂u
∂x
=
x−a
r2
∂ 2u
∂x 2
=
FG x − a IJ = r a1f − ax − af 2rb∂r / ∂xg
Hr K
r
∂ u
r − 2 ax − af
=
, from (ii)
∴
∂x 2
Similarly,
∂ 2u
∂y 2
∂ 2u
∂ 2u
∂ 2u
+
+
∂y 2
∂x 2
∂z 2
∂
∂x
2
2
4
2
2
2
or
∴
∂y
r
u = log r.
Now,
∴
∂r
=
∂x
= 2 (x – a) or
r4
=
=
=
b g ; ∂ u = r − 2 ( z − c) .
r2 − 2 y − b
2
2
2
∂z 2
r4
2
r4
{a f + b y − bg + az − cf }
3r 2 − 2 x − a
2
2
2
r4
1
3r 2 − 2r 2
, from (i) = 2 . Hence proved.
4
r
r
32
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 15. If u = x2 tan–1 (y/x) – y2 tan–1 (x/y), prove that
∂ 2u
x2 − y2
= 2
.
∂y∂x
x + y2
Sol. Given
u = x2 tan–1 (y/x) – y2 tan–1 (x/y).
Differentiating partially with respect to x, we get
∂u
∂x
=
FG y IJ + 2x tan FG y IJ − y ⋅ 1 ⋅ 1
H xK
1 + b y / xg H x K
1 + bx / y g y
1
x2
= –
= –
2
x2 y
x +y
2
–
2
⋅ −
−1
y3
x +y
2
2
2
2
2
+ 2x tan–1
y
x
j + 2x tan FG y IJ = – y + 2x tan FG y IJ
H xK
H xK
ex + y j
e
y x2 + y2
2
–1
–1
2
Again differentiating partially with respect to y, we get
∂ 2u
∂y∂x
=
∂
∂y
RS– y + 2x tan FG y IJ UV = – 1 + 2x 1 ⋅ 1
H xKW
F yI x
T
1+G J
H xK
= –1+
–1
2
x2 − y2
2x 2
=
. Hence proved.
x2 + y2
x2 + y2
Example 16. If z = f (x – by) + φ (x + by), prove that
∂ 2z
∂2z
.
b
=
∂y 2
∂x 2
z = f (x – by) + φ (x + by)
∂z
= f ′ (x – by) + φ′ (x + by)
∂x
∂2z
= f ″ (x – by) + φ″ (x + by).
∂x 2
∂z
∂y = – bf ′ (x – by) + bφ′ (x + by)
2
Sol. Given
∴
and
Again from (i),
...(i)
...(ii)
∂2z
∂ 2z
2
2
2
from (ii). Hence proved.
2 = b f ″ (x – by) + b φ′′ (x + by) = b
∂y
∂x 2 ′
and
Example 17. If u (x, y, z) = log (tan x + tan y + tan z). Prove that
sin 2x
∂u
∂u
∂u
+ sin 2y ∂y + sin 2z
= 2.
∂x
∂z
Sol.
∂u
∂x
=
sec 2 x
tan x + tan y + tan z
(U.P.T.U., 2006)
33
DIFFERENTIAL CALCULUS-I
∂u
∂y
=
sec 2 y
tan x + tan y + tan z
sec 2 z
∂u
=
tan x + tan y + tan z
∂z
∂u
∂u
∂u
∴ sin 2x
+ sin 2y ∂y + sin 2z
∂x
∂z
=
=
sin 2x sec 2 x + sin 2y sec 2 y + sin 2z sec 2 z
tan x + tan y + tan z
b
2 tan x + tan y + tan z
btan x + tan y + tan zg
g
= 2. Hence proved.
EXERCISE 1.4
∂ 3u
2 2 2
1. Find ∂x∂y∂z if u = ex +y +z .
Ans. 8xyzu
2. Find the first order derivatives of
LMAns. ∂u = x b y log x + yg; ∂u = x log xOP
∂y
N ∂x
Q
F
I LMAns. ∂u = 1 ∂u = − y ex − y j FH x + x − y IK OP
(ii) u = log H x + x − y K
MN ∂x x − y ∂y
PQ
F x − y I , show that ∂u = – y ∂u .
3. If u = sin G
∂x
x ∂y
H x + y JK
xy + 1
xy
(i) u = xxy.
2
2
2
2
2
–1
4. If u = ex (x cos y – y sin y), prove that
∂ 2u
∂ 2u
+
= 0.
∂x 2
∂y 2
FG y IJ , prove that ∂ u + ∂ u = 0.
H xK
∂x
∂y
2
5. If u = a log (x2 + y2) + b tan–1
2
6. If z = tan (y – ax) + (y + ax)3/2, prove that
7. If u =
x2
x
y2
y
z2
z
1
1
1
, prove that
2
2
2
∂ 2z
2 ∂ z
a
–
= 0.
∂x 2
∂y 2
∂u
∂u
∂u
+ ∂y +
= 0.
∂x
∂z
2
−1
2
2
2
−1
34
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
8. If u = 2(ax + by)2 – (x2 + y2) and a2 + b2 = 1, find the value of
∂ 2u
∂ 2u
.
2 +
∂x
∂y 2
LM
OP
−4 yz
MMAns. ex + y + z j PP
N
Q
∂ 2u
.
9. If u = log (x2 + y2 + z2), find the value of
∂y∂z
e
j
1
10. If u = x 2 + y 2 + z 2 2 , then prove that
∂2u
∂x 2
11. If z = f (x + ay) + φ (x – ay), prove that
12. If u = cos–1
+
∂2u
∂x 2
16. If u = r , where r =
x +y +z
2
2
17. If u = (x2 + y2 + z2)–1, prove that
18. If θ = t e
∂ 2u
∂2u
2
= .
2 +
2
∂y
u
∂z
∂ 2u
= 0.
∂y 2
+
2
∂2 f
∂x
2
= b2
∂2 f
∂t 2
.
∂ 2u
∂ 2u
∂ 2u
.
find
2 + ∂y 2 +
∂x
∂z 2
[Ans. m (m + 1)rm−2]
∂ 2u
∂ 2u
∂2u
+
+
= 2 (x2 + y2 + z2)–2.
∂x 2
∂z 2
∂y 2
r2
4t , find the value of n, when
1 ∂
r 2 ∂r
FG r ∂θ IJ = ∂θ .
H ∂r K ∂t
2
LMAns. n = − 3 OP
2Q
N
19. For n = 2 or – 3 show that u = rn (3 cos2 θ – 1) satisfies the differential equation
FG r ∂u IJ + 1 ∂ FG sin θ ∂u IJ = 0.
H ∂r K sin θ ∂θ H ∂θ K
F ∂ u I + 1 ∂u + 1 ∂ u = 0.
20. If u = e cos (a log r), show that G
H ∂r JK r ∂r r ∂θ
∂
∂r
2
2
2
aθ
−
2
2
2
z
∂z
x −y j
∂z
21. If e e
= (x – y), show that y
+x
= x2 – y2.
2
2
∂x
2
∂y
2
[Hint: Solve for z = (y – x ) log (x – y)].
22. If u =
2 2
∂2z
∂2z .
2
=
a
∂x 2
∂y 2
15. Prove that f (x, t) = a sin bx. cos bt satisfies
−
2
∂u
∂u
∂u
+ ∂y +
= (x + y + z)2.
∂x
∂z
14. If u = x2y + y2z + z2x, show that
n
2
bx − yg / bx + yg , prove that x ∂∂ux + y ∂∂uy = 0.
13. If u = log (x2 + y2), show that
m
[Ans. 0]
∂ 2u
1
∂ 2u
∂2u
and r2 = (x – a)2 + (y – b)2 + (z – c)2, prove that
+
+
= 0.
∂y 2
r
∂x 2
∂z 2
35
DIFFERENTIAL CALCULUS-I
∂ 2u
∂ 2u
∂2u
23. If u (x, y, z) = cos 3x cos 4y sin h 5z, prove that
= 0.
2 + ∂y 2 +
∂x
∂z 2
FG ∂u IJ FG ∂x IJ = 1 FG ∂v IJ FG ∂y IJ .
H ∂x K H ∂u K 2 H ∂y K H ∂v K
F ∂x I F ∂x I F ∂θI F ∂θ I F ∂y I
25. If x = r cos θ, y = r sin θ, find G J , G J G J G J G J .
H ∂r K H ∂θ K H ∂xK H ∂y K H ∂x K
24. If x2 = au + bv; y2 = au – bv, prove that
y
θ
v
r,
x
y,
u
r
x,
[Ans. cos θ, – r sin θ, –r –1 sin θ, r –1 cos θ, – cot θ.]
1.5 HOMOGENEOUS FUNCTION
A polynomial in x and y i.e., a function f (x, y) is said to be homogeneous if all its terms are of
the same degree. Consider a homogeneous polynomial in x and y
f (x, y) = a0xn + a1 xn−1 y + a2 xn-2y2 + ..... + an yn
LMa + a F y I + a F y I +...+ a F y I OP
GH x JK P
MN GH x JK GH x JK
Q
F yI
f (x, y) = x F G J
H xK
n
2
= xn
0
1
n
2
n
or
Hence every homogeneous function of x and y of degree n can be written in above form.
NOTE: Degree of Homogeneous function = degree of numerator – degree of denominator.
Remark 1: If
f (x, y) = a0xn + a1xn+1 . y–1 + a2xn+2 . y–2 + ...... + anxn+n y –n
R|
F
F
x
xI
xI U
+
+
+
a
a
....
S|
GH y JK
GH y JK |V|
y
T
W
F xI
f (x, y) = x F G J ; degree = n
H yK
f (x, y) = a y + a y . x + ...... + a y . x
R| F x I
F x I U|
= y Sa + a G J + .... + a G J V
H y K |W
|T H y K
F xI
f (x, y) = y F G J ; degree = –n
H yK
n
2
= x n a0 + a1
⇒
Remark 2: If
n
2
n
–n
–n–1
0
–n–n
1
n
n
n
⇒
0
n
1
–n
Another forms are also possible i.e.,
f (x, y) = yn F
F x I ; f (x, y) = y F(y/x)
GH y JK
n
n
36
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
1.6 EULER’S THEOREM ON HOMOGENEOUS FUNCTIONS
Statement: If f is a homogeneous function of x, y of degree n then
x
∂f
∂f
+ y ∂y = nf.
∂x
(U.P.T.U., 2006)
Proof. Since f is a homogeneous function
∴
FG y IJ
H xK
f (x, y) = xn F
...(i)
Differentiating partially w.r.t. x and y, we get
FG y IJ + x F′ FG y IJ FG − y IJ
H xK
H xK H x K
∂f
FG y IJ FG 1 IJ
=
x
F′
∂y
H xK H xK
∂f
= nxn−1 F
∂x
n
2
n
...(ii)
...(iii)
Multiplying (ii) by x and (iii) by y and adding, we have
x
FG y IJ – x y F′ FG y IJ + x y F′ FG y IJ
H xK
H xK
H xK
F yI
= nx F G J
H xK
∂f
∂f
+y
= nxn F
∂x
∂y
n−1
n−1
n
⇒
x
∂f
∂f
+y
= nf
∂y
∂x
(from (i)).
In general if f (x1, x2, ..., xn) be a homogeneous function in x1, x2, ..., xn then x1
∂f
+ ... + xn ∂x
n
∂f
∂f
+ x2
∂x1
∂x 2
= nf.
Corollary 1. If f is a homogeneous function of degree n, then
2
x
∂2 f
∂x 2
∂2 f
∂2 f
2
+ 2xy
+y
= n (n –1)f.
∂x∂y
∂y 2
Proof. We have
∂f
∂f
+ y ∂y = n f
∂x
Differentiating (i) w.r.t. x and y respectively, we get
...(i)
∂f
∂2 f
∂2 f
∂f
+x
+
y
= n
∂x
∂x∂y
∂x
∂x 2
...(ii)
x
and
x
∂f
∂2 f
∂2 f
+ ∂y + y
∂y∂x
∂y 2
= n
∂f
∂y
...(iii)
Multiplying (ii) by x and (iii) by y and adding, we have
x2
∂2 f
∂x
2
+ 2xy
∂2 f
∂2 f
∂f
∂f
+ y2
+y
=n
2 + x
∂x∂y
xy
∂y
∂x
F x ∂f + y ∂t I
GH ∂x ∂y JK
37
DIFFERENTIAL CALCULUS-I
⇒
x
2
∂2 f
∂x 2
∂2 f
∂2 f
2
+ 2xy
+y
∂x∂y
∂y 2
= n2f − nf = n (n – 1) f
Example 1. Verify Euler’s theorem for the function
f (x, y) = ax2 + 2hxy + by2.
Sol. Here the given function f (x, y) is homogeneous of degree n = 2. Hence the Euler’s
theorem is
∂f
∂f
x
+y
= 2f
...(i)
∂y
∂x
Now, we are to prove equation (i) as follows:
∂f
∂f
= 2 ax + 2 hy, ∂y = 2hx + 2by
∂x
∴
x
∂f
∂f
+ y ∂y
∂x
= 2ax2 + 2hxy + 2hxy + 2by2
= 2(ax2 + 2 hxy + by2) = 2f
⇒
x
∂f
∂f
+y
∂x
∂y
= 2f, which proves equation (i).
Example 2. Verify Euler’s theorem for the function u = xn sin
FG y IJ .
H xK
Sol. Since u is homogeneous function in x and y of degree n, hence we are to prove that
x
∂u
∂u
+ y ∂y
∂x
= nu
FG y IJ
H xK
∂u
F yI
F yI F y I
= nx sin G J + x cos G J G − J
H xK
H xK H x K
∂x
F yI
F yI
∂u
x
= nx sin G J – x y cos G J
H
H xK
K
x
∂x
F yI
∂u
= yx cos G J
y
H xK
∂y
u = xn sin
We have
∴
or
n–1
n
n
n–1
n–1
Similarly,
Adding (ii) and (iii), we get
x
∂u
∂u
+ y ∂y
∂x
= nxn sin
∂u
∂u
+ y ∂y = nu
∂x
which verifies Euler’s theorem.
⇒
...(i)
x
FG y IJ
H xK
2
...(ii)
...(iii)
38
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 3. If u = sin–1
w x − yz
xy x + y {| , show by Euler’s theorem that
y ∂u
∂u
= –
·
∂x
x ∂y
w x − y z ⇒ sin u = x − y
xy x + y {|
x+ y
Sol. We have
u = sin–1
Let
f = sin u =
x− y
x+ y
Here, f is a homogeneous function in x and y
where,
degree n =
1
1
–
=0
2
2
∴ By Euler’s theorem, we have
x
or
or
∂f
∂f
+y
= 0. f = 0
∂x
∂y
∂
∂
(sin u) + y ∂y (sin u) = 0
∂x
∂u
∂u
x cos u ·
+ y cos u ∂y = 0
∂x
x
⇒
x
As f = sin u
∂u
∂u
+y
= 0
∂x
∂y
y ∂u
∂u
= −
.
∂x
x ∂y
⇒
Hence proved.
Example 4. If u = log [(x4 + y4)/ (x + y)], show that
x
∂u
∂u
+y
= 3.
∂x
∂y
Sol. We have
x4 + y4
x4 + y4
u
u = loge
⇒e =
x+ y
x+y
Let
f = eu =
(U.P.T.U., 2000)
x4 + y4
x+y
Here the function f is a homogeneous function in x and y of degree, n = 4 – 1 = 3
∴
By Euler’s theorem
x
⇒
x
∂f
∂f
+ y ∂y = nf = 3f
∂x
∂
∂
(eu) + y ∂y (eu) = 3f
∂x
DIFFERENTIAL CALCULUS-I
⇒
F x ∂u + y ∂u I
GH ∂x ∂y JK e = 3e
⇒
u
x
u
∂u
∂u
+y
= 3. Hence proved.
∂x
∂y
Example 5. Verify Euler’s theorem for
u = sin–1
F x I + tan FG y IJ .
GH y JK
H xK
–1
(U.P.T.U., 2006)
Sol. Here u is a homogeneous function of degree,
n = 1 – 1 = 0; hence by Euler’s theorem
x
∂u
∂u
+y
= 0
∂x
∂y
∂u
=
∂x
Now,
or
x
∂u
=
∂x
∂u
∂y =
and
or
y
∂u
=
∂y
1
1−
FG x IJ
H yK
2
x
2
y −x
⋅
–
2
FG
FG IJ H
H K
y
1
2 − 2
x
y
1+
x
1
+
y
IJ
K
xy
...(i)
2
x + y2
F − x I + 1 . FG 1IJ
GH y2 JK
yI 2 H xK
F
F
xI
+
1
GH x JK
1−G J
H yK
1
2
−x
2
y −x
2
.
+
xy
...(ii)
ex + y 2 j
2
Adding (i) and (ii), we get
x
∂u
∂u
+y
= 0, hence Euler’s theorem is verified.
∂x
∂y
Example 6. If u = x sin–1
x2
Sol. We have
F xI
GH y JK + y sin FGH xy IJK , find the value of
–1
∂ 2u
∂ 2u
2 + 2xy ∂x∂y
∂x
∂ 2u
.
∂y2
(U.P.T.U., 2007)
F xI
GH y JK + y sin FGH xy IJK
F xI
F 1 I + y · 1 FG − y IJ
∂u
x
= sin G J +
H xK
∂x
H y K 1 − x GH y JK
y
1−
u = x sin–1
∴
+ y2
–1
–1
2
2
y2
x2
2
40
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
⇒
x
2
F xI
GH y JK + 2x 2 – x y− y
y −x
F − x I + sin FG y IJ + y 1 FG 1 IJ
1
∂u
= x·
GH y2 JK
H xK
∂y
y H xK
x
1−
1−
2
∂u
= x sin–1
∂x
2
2
...(i)
–1
and
∂u
y ∂y = –
⇒
2
y2
x2
y
F yI
+ y sin G J +
H xK x − y
y2 − x2
2
x2
–1
2
Adding (i) and (ii), we get
x
2
∂u
∂u
+ y ∂y = x sin–1
∂x
2
F x I + y sin FG y IJ = u
GH y JK
H xK
–1
...(ii)
...(iii)
Differentiating (iii) partially w.r.t. x and y respectively.
and
∂u
∂u
∂ 2u
∂ 2u
∂ 2u
∂ 2u
+x
+
y
=
⇒
x
+
y
=0
∂x
∂x
∂x2
∂x2
∂x∂y
∂x∂y
...(iv)
∂ 2u
∂ 2u
∂ 2u
∂u
∂u
∂ 2u
+
+y
=
⇒
y
+
x
=0
∂x∂y
∂y∂x
∂y
∂y
∂y 2
∂y2
...(v)
x
Multiplying equation (iv) by x and (v) by y and adding, we get
x2
Example 7. If u = tan–1
x2
∂ 2u
∂ 2u
∂ 2u
2
+
2xy
+
y
= 0.
∂y2
∂x∂y
∂x2
x 3 + y3
∂u
∂u
, prove that x
+y
= sin 2u and evaluate
x− y
∂x
∂y
∂ 2u
∂ 2u
+
2xy
+ y2
∂x∂y
∂x2
∂ 2u .
∂y2
Sol. We have
u = tan
–1
x 3 + y3
x 3 + y3
⇒ tan u =
x− y
x− y
x 3 + y3
Let f = tan u =
x− y
∴
Since f (x, y) is a homogeneous function of degree
n = 3–1=2
By Euler’s theorem, we have
x
⇒ x
∂f
∂f
∂
∂
+y
= nf ⇒ x
(tan u) + y
(tan u) = 2 tan u
∂x
∂x
∂y
∂y
∂u
∂u
· sec2 u + y
· sec2 u = 2 tan u
∂y
∂x
DIFFERENTIAL CALCULUS-I
or
x
∂u
tan u
∂u
+y
= 2
= sin 2u. Proved.
∂x
sec 2 u
∂y
...(i)
Differentiating (i) partially w.r.t. x, we get
x
or
∂u
∂ 2u
∂u
∂ 2u
+
+
y
= 2 cos 2u .
2
∂x∂y
∂x
∂x
∂x
∂ 2u
∂u
∂ 2u
+
y
= (2 cos 2u –1)
2
∂
∂
x
y
∂x
∂x
Multiplying by x, we obtain
x
∂ 2u
∂u
∂ 2u
= x (2 cos 2u –1)
2 + xy
∂x
∂x∂y
∂x
Again differentiating equation (i) partially w.r.t. y, we get
x2
x
...(ii)
∂ 2u
∂u
∂ 2u
∂u
+y
+
= 2 cos 2u
2
∂y∂x
∂y
∂y
∂y
or
y
or
y2
∂u
∂ 2u
∂ 2u
+
x
= (2 cos 2u – 1)
2
∂y
∂
∂
y
x
∂y
∂u
∂ 2u
∂ 2u
2 + xy ∂x∂y = y ( 2 cos 2u – 1) ∂y (multiply by y)
∂y
...(iii)
Adding (ii) and (iii), we get
x2
∂ 2u
∂ 2u
∂ 2u
2
= ( 2cos 2u – 1)
2 + 2xy ∂x∂y + y
∂x
∂y2
F x ∂u + y ∂u I
GH ∂x ∂y JK
= (2cos 2u – 1) sin 2u, (from (i))
= (2sin 2u cos 2u – sin 2u)
= sin 4u – sin 2u
= 2 cos
FG 4u + 2uIJ . cos FG 4u − 2uIJ .
H 2 K
H 2 K
∂ 2u
∂ 2u
2
Hence, x
+ 2xy
+y
∂y2 = 2 cos 3u · cos u.
∂x∂y
∂x2
2
∂ 2u
Example 8. If z = xm f
x2
Sol. Let
then
∂ 2z
∂x
2
FG y IJ + x g FG x IJ , prove that
H xK
H yK
n
+ 2xy
∂ 2z
∂ x∂ y
+ y2
u = xm f
∂ 2z
∂y
2
+ mnz = (m + n – 1)
F x ∂z + y ∂z I
GH ∂x ∂y JK ·
FG y IJ , v = x g FG x IJ
H xK
H yK
n
z = u+v
...(i)
Now, u is homogeneous function of degree m. Therefore with the help of (Corollary 1, on
page 36), we have
42
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
∂ 2u
∂ 2u
∂ 2u
2
x
+ 2xy
+y
= m (m – 1) u
∂y2
∂x∂y
∂x2
2
F x I , we have
GH y JK
Similarly for v = xn g
x
∂ 2v
∂ 2v
2
+ 2xy
+y
∂ x∂ y
∂y 2
∂x 2
∂ 2v
2
...(ii)
= n (n – 1) v
...(iii)
Adding (ii) and (iii), we get
∂
∂2
∂2
2
x
2 (u + v) + 2xy ∂x∂y (u + v) + y ∂y 2 (u + v) = m (m – 1) u + n (n – 1) v
∂x
2
⇒
x2
∂ 2z
∂2z
∂ 2z
2
+
2xy
+
y
∂y2 = m (m – 1) u + n (n – 1) v (As z = u + v).
∂x ∂y
∂x2
...(iv)
Again from Euler’s theorem, we get
x
Adding x
∂u
∂v
∂u
∂v
+ y ∂y = m u and x
+ y ∂ y = nv
∂x
∂x
∂
∂
(u + v) + y ∂y (u + v) = mu + nv
∂x
∂z
∂z
+ y ∂y = mu + nv
...(v)
∂x
m (m –1) u + n (n – 1) v = (m2 u + n2v) – (mu + nv)
= m (m + n) u + n (m + n) v – mn (u + v) – (mu + nv)
= (mu + nv) (m + n) − (mu + nv) − mnz
= (mu + nv) (m + n – 1) – mnz
⇒
x
Now,
= (m + n –1)
F x ∂z + y ∂z I
GH ∂x ∂y JK – mnz, from (v)
Putting this value in equation (iv), we get
F x ∂z + y ∂z I
GH ∂x ∂y JK – mnz
F x ∂z + y ∂z I
∂ 2z
∂ z
∂ 2z
⇒ x
2 + 2xy ∂x∂y + y ∂y2 + mnz = (m + n –1) GH ∂x
∂y JK . Hence proved.
∂x
L x + 2y + 3z OP
Example 9. If u = sin M
MN x8 + y8 + z8 PQ , show that
x2
∂ 2z
∂2z
∂ 2z
2
+
2xy
+
y
= (m + n –1)
∂x ∂y
∂y2
∂x2
2
2
2
–1
x
∂u
∂u
∂u
+ y ∂y + z
+ 3 tan u = 0.
∂x
∂z
(U.P.T.U., 2003)
DIFFERENTIAL CALCULUS-I
LM x + 2y + 3z OP
MN x + y + z PQ
L x + 2y + 3z OP
sin u = M
MN x + y + z PQ , let f = sin u
u = sin–1
Sol. We have
⇒
8
8
8
8
8
8
∴ Degree of homogeneous function f, n = 1 – 4 = – 3, from Euler’s theorem, we have
x
⇒
x
∂f
∂f
∂f
+ y ∂y + z
= – 3 sin u
∂x
∂z
∂
∂
∂
(sin u) + y ∂y (sin u) + z
(sin u) = – 3 sin u
∂x
∂z
F x ∂u + y ∂u + z ∂u I
GH ∂x ∂y ∂z JK cos u = – 3 sin u
or
or
x
∂u
∂u
∂u
+ y ∂y + z
+ 3 tan u = 0.
∂x
∂z
Hence proved.
R| π 2x 2 + y 2 + zx 12 U|
j |V , find the value of
| e
Example 10. If V = log sin S
|| 2ex 2 + xy + 2yz + z 2 j 13 ||
T
W
e
x
Sol. We have
∂V
∂V
∂V
+ y ∂y + z
, when x = 0, y = 1, z = 2.
∂x
∂z
R| π 2x 2 + y 2 + zx 12 U|
j |V
| e
V = log sin S
|| 2ex 2 + xy + 2yz + z 2 j 13 ||
T
W
R π 2x + y + zx U
j |V
| e
e = sin S
|| 2 ex + xy + 2yz + z j ||
T
W
R| π 2x 2 + y 2 + zx 12 U|
j |V
| e
sin (e ) = S
|| 2 ex 2 + xy + 2yz + z 2 j 13 ||
T
W
R| π 2x 2 + y 2 + zx 12 U|
j |V
| e
f = sin (e ) = S
|| 2 ex 2 + xy + 2yz + z 2 j 13 ||
T
W
2
∴
V
2
or
Let
–1
V
–1
v
Since f is a homogeneous function ∴ n = 1 –
2
1
=
3
3
2
1
2
1
2 3
44
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
By Euler’s theorem, we get
∂f
∂f
∂f
x
+y
+z
=
∂x
∂z
∂y
⇒ x
1
f
3
1
∂
∂
∂
(sin–1 eV) + y
(sin–1 eV) + z
(sin–1 eV) =
sin–1 eV
3
∂x
∂z
∂y
Fx ∂V + y ∂V + z ∂V I FG 1 × eV IJ 1
GH ∂x ∂y ∂z JK H 1 − e 2V K = 3 sin e
⇒
–1
⇒
x
∂V
∂V
∂V
+y
+z
=
∂y
∂x
∂z
ee j
V
Now,
esin e j
−1
and
V
1
1 − e 2V
×
sin–1 eV
V
3
e
=
x = 0
y =1
z =2
...(i)
RS π UV = 1 , e = 1
T 2 × 2W 2
2
2V
= sin
x = 0
y =1
z =2
V
π
4
Putting all these values in equation (i), we get
⇒
x
∂V
∂V
∂V
+y
+z
=
∂x
∂y
∂z
1− 1 2
1
π
×
×
=
1
3
4
2
x
∂V
∂V
∂V
+y
+z
=
∂x
∂y
∂z
π
Example 11. If u = x3y2 sin–1
x
12
.
FG y IJ , show that
H xK
∂u
∂ 2u
∂u
∂ 2u
∂ 2u
2
+ y ∂y = 5u and x2
+
2xy
+
y
= 20u.
∂y2
∂x
∂x2
∂x∂y
u = x3y2 sin–1
Sol.
FG y IJ
H xK
FG yIJ sin FG y IJ = x F FG y IJ F FG y IJ = FG y IJ sin y
H xK
H xK H xK H xK
H xK
x
2
=
x5
2
–1
5
−1
∴ u is a homogeneous function of degree 5 i.e., n = 5
By Euler’s theorem, we get x
∂u
∂u
+ y ∂y = 5u. Proved.
∂x
Next, we know that (from Corollary 1, on page 36)
or
12
π
×
12
12
x2
∂ 2u
∂ 2u
∂ 2u
2
+ 2xy
+y
∂y2 = n (n – 1) u = 5 (5 – 1) u
∂x∂y
∂x2
x2
∂ 2u
∂ 2u
∂ 2u
2
+
2xy
+
y
= 20 u. Hence proved.
∂x∂y
∂y2
∂x2
45
DIFFERENTIAL CALCULUS-I
FG y IJ + f FG y IJ , prove that
H xK
H xK
Example 12. If u = x f1
x2
2
2
∂ 2u
∂ 2u
2 ∂ u = 0.
+
2
xy
+
y
∂x∂y
∂x 2
∂y 2
u1 =
Sol. Let
FG y IJ and u = x f FG y IJ , then u = u + u
H xK
H xK
x f1
0
2
2
1
2
Since u1 is a homogeneous function of degree one. So by Corollary 1 on page 36, we get
x2
∂2u1
2
+ 2xy
∂ u1
∂ 2 u1
+ y2
∂x ∂y
∂y2
...(i)
= 1 (1 – 1) = 0
∂x2
and u2 is also a homogeneous function of degree 0
2
∂ u2
∂ 2 u2
∂ 2 u2
+ 2xy
+ y2
∴
2
∂y 2
∂x∂y
∂x
Adding (i) and (ii), we get
x2
x2
∂2
∂x 2
(u1 + u2) + 2xy
⇒
x2
...(ii)
= 0
∂2
∂2
(u1 + u2) + y2
(u1 + u2) = 0
∂x∂y
∂y 2
∂ 2u
∂ 2u
∂ 2u
2
+ 2xy
+y
∂x∂y
∂x 2
∂y 2
= 0. Hence proved.
Example 13. If u = sin–1 (x3 + y3)2/5, evaluate
2
∂ 2u
x 2∂ 2u
2 ∂ u
.
+
+
xy
y
2
∂x∂y
∂x 2
∂y 2
u = sin–1 (x3 + y3)2/5
Sol. Given
F1 + y I
GH x JK
3
sin u = (x + y )
3
Let
f = sin u
∴
f = x
6/5
which is homogeneous of degree n =
x
⇒
or
3 2/5
x
=x
F1 + y I
GH x JK
3
6/5
25
3
25
3
...(i)
6
. By Euler’s theorem
5
∂f
∂f
6
+y
=
f
∂x
∂y
5
∂
∂
sin u + y
sin u
∂x
∂y
b
x
g
b
g = 65 sin u
∂u
∂u
cos u + y
cos u = 6 sin u
∂x
∂y
5
x
6
∂u
∂u
tan u
=
+y
5
∂x
∂y
...(ii)
46
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Differentiating (ii) w.r. to ’x‘, we get
6
∂u
∂u
∂ 2u
∂ 2u
sec2 u.
+x 2 +y
=
5
∂x
∂x
∂x∂y
∂x
Multiplying by x
x
∂u
∂ 2u
∂ 2u
6
∂u
+ x 2 2 + xy
=
sec2 u. x
∂x
∂x∂y
∂x
5
∂x
...(iii)
Differentiating (ii) w.r. to ‘y’, we get
x
6
∂ 2u ∂u
∂ 2u
∂u
sec2 u
+
+y 2 =
∂y
5
∂y∂x ∂y
∂y
Multiplying by y
y
∂u
∂ 2u
∂ 2u
6
∂u
+ y 2 2 + xy
sec2 u. y
=
x
y
∂y
∂
∂
5
∂y
∂y
...(iv)
Adding (iii) and (iv), we get
F x ∂u + y ∂u I FG 6 sec u − 1IJ
GH ∂x ∂y JK H 5
K
∂u
∂u 6
∂ 2u
∂ 2u
∂ 2u
6
6
As x
+y
= tan u
x 2 2 + 2xy
+ y2 2 =
tan u FG sec2 u − 1IJ
K
H
∂
x
∂y 5
∂x∂y
5
5
∂x
∂y
F xI
F yI
Example 14. If u = 3x cot G J + 16y cos G J then prove that
H xK
H yK
x2
or
2
∂ 2u
∂ 2u
2∂ u
=
2 + 2xy ∂x∂y + y
∂x
∂y 2
4
x
–1
2
4
–1
2
∂ 2u
∂ 2u
2∂ u
xy
2
+
+
y
= 12u.
∂x∂y
∂x2
∂y 2
Sol. The given function is homogeneous function of degree 4. By Euler’s theorem, we have
x
∂u
∂u
+y
= 4u
∂x
∂y
...(i)
Differentiating (i) w.r. to x, we get
∂u
∂u
∂ 2u
∂ 2u
+x 2 +y 2 = 4
∂x
∂x
∂x
∂y
or
x
∂u
∂u
∂ 2u
∂ 2u
+ x2 2 + xy 2 = 4x
∂x
∂x
∂x
∂y
...(ii)
Differentiating (i) w.r. to y, we get
x
or
xy
∂u
∂ 2u ∂u
∂ 2u
+
+y 2 = 4
∂y
∂x∂y ∂y
∂y
∂u
∂ 2u
∂u
∂ 2u
+y
+ y 2 2 = 4y
∂y
∂x∂y
∂y
∂y
...(iii)
47
DIFFERENTIAL CALCULUS-I
Adding (ii) and (iii), we get
or
F
GH
I
JK
x2
2
∂u
∂u
∂ 2u
∂ 2u
2∂ u
+y
= 3 x
xy
y
2
+
+
2
2
∂x
∂y
∂x∂y
∂x
∂y
x2
2
∂ 2u
∂ 2u
2∂ u
= 3 × 4u = 12u. Hence proved.
xy
y
2
+
+
∂x∂y
∂x 2
∂y 2
EXERCISE 1.5
Verify Euler’s theorem
1.
2
2
x −y .
1
4
4. x 3 y − 3 tan–1
FG y IJ.
H xK
2.
Fx + y I Fx + y I.
GH
JK GH
JK
5.
b x + yg .
1
4
1
4
1
5
1
5
xy
F xI
GH y JK + cot FGH yx IJK.
F xI
F yI
6. cos G y J + cot GH JK .
H K
x
3. cos–1
–1
–1
–1
F x + y I , show that x ∂u + y ∂u = 3.
(U.P.T.U., 2000)
GH x + y JK
∂y
∂x
∂u
9
∂u
I
F
I F
8. If u = H x + y K H x + y K , show that x
+ y ∂y =
u
[U.P.T.U. (AG), 2005]
20
∂x
F xI
F xI
∂z
∂z
9. If z = x y sin G y J + log x – log y, show that x
+y
= 6x y sin G y J .
H K
H K
∂x
∂y
4
4
7. If u = loge
1
4
1
4
4 2
–1
1
5
1
5
4 2
–1
[U.P.T.U. (C.O.)., 2003]
∂
∂
∂
u
u
u
10. If u = x3 + y3 + z3 + 3xyz; show that x
+y
+z
= 3u.
∂x
∂y
∂z
11. If u = cos–1
LM x − y OP , prove that x ∂u + y ∂u = 0.
∂y
∂x
Nx + y Q
12. If u = log [(x2 + y2)/(x + y)], prove that x
∂u
∂u
+ y ∂y = 1.
∂x
13. If u = sin–1 {(x2 + y2)/(x + y)}, show that x
∂u
∂u
+ y ∂y = tan u.
∂x
14. If u = sin–1 {(x + y)/( x +
y )}. Show that x
(U.P.T.U., 2008)
∂u
1
∂u
+ y ∂y =
tan u.
2
∂x
∂u
1
∂u
+ y ∂y =
sin 2u.
2
∂x
16. If u is a homogeneous function of degree n, show that
15. If u = tan–1 [(x2 + y2)/(x + y)], then prove that x
∂ 2u
∂ 2u
∂u
.
(a) x
2 + y ∂x∂y = (n – 1) ∂x
∂x
∂ 2u
∂u
∂ 2u
.
(b) y
2 + x ∂x∂y = (n – 1) ∂y
∂y
48
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
FG y IJ , show that x ∂u + y ∂u = 0.
H xK
∂x
∂y
LM x + y OP
∂ u
1
∂ 2u
∂ 2u
, prove that x
+
2xy
+
y
1 8 . If u = sin M
P
2 = 144 tan u [tan u – 11].
2
∂
∂
x
y
∂y
∂x
MN x + y PQ
F x +y I
5
∂u
∂u
1 9 . Prove that x
+y
=
tan u if u = sin G
2
∂x
∂y
H x + y JK .
17. If u = f
–1
1
4
1
4
1
6
1
6
2
2
2
3
2
3
–1
2 0 . Verify Euler’s theorem for f =
z
x+ y
+
y
x
+
.
z+x
y+ z
2
F xI
GH y JK , show that x ∂∂ux + y ∂∂uy = 2u and x ∂∂xu2 + 2xy ∂∂x∂uy +
2
21.
If u = y2
y2
e y x + x2 tan–1
2
∂ 2u
= 2u.
∂y2
2
2
2
e x + y j sin u = x + y , prove that x ∂∂xu2 + 2xy ∂∂x∂uy + y ∂∂yu2 = tanu
12
F 13 + tan u I
GH 12 12 JK .
F y2 I
∂ 2u
2 3 . If u = tan G x J , show that x
H K
∂x2
2
22. If
1
3
1
3
2
2
2
–1
2 4 . If f (x, y) =
2
1
x
2
+
∂f
∂f
log x − log y
1
+
, show that x
+ y ∂y + 2f (x, y) = 0.
2
2
∂x
xy
x +y
2 5 . If u = sec–1 {(x3 + y3) / (x + y)}, show that x
1.7
Let
∂u
∂u
+ y ∂y = 2 cot u.
∂x
TOTAL DIFFERENTIAL COEFFICIENT
z = f (x, y)
...(i)
where x = φ (t) and y = ψ (t), then z can be expressed as a function of t alone by substituting the
values of x and y in terms of t from the last two equations in equation (i).
And we can find the ordinary differential coefficient
dz
, which is called total differential
dt
coefficient of z with respect to t. Since it is very difficult sometimes to express z in terms of t alone
by eliminating x and y. So we are now to find
and y in terms of t in z = f (x, y).
dz
without actually substituting the values of x
dt
49
DIFFERENTIAL CALCULUS-I
Let δx, δy and δz be the increaments in x, y and z corresponding to a small increament δt in
the value of t.
then
z + δz = f (x + δx, y + δy)
...(ii)
where
x + δx = φ (t + δt), y + δy = ψ (t + δt)
Now,
b
g
f x + δx, y + δy − f (x, y)
δz
dz
= Lim
= Lim
(from ii)
δt→0 δt
δt→0
dt
δt
= Lim
δt→0
b
g
f x + δx, y + δy − f (x + δx, y) + f (x + δx, y) − f (x, y)
δt
b
{Adding and subtracting f (x + δx, y)}
b
g
g
f x + δx, y + δy − f (x + δx, y)
Lim f x + δx, y − f (x, y)
= Lim
+
0
δt→
δt→0
δt
δt
LM f bx + δx, y + δyg − f bx + δx, yg ⋅ δy OP Lim LM f bx + δx, yg − f bx, yg ⋅ δx OP
δy
δt PQ + t 0 MN
δx
δt PQ
MN
Also, as δt → 0, δx → 0, δy → 0
L ∂f bx, yg δy OP + Lim LM ∂f bx, yg δx OP
dz
∴
= Lim M
t 0 M ∂y
t 0 M ∂x
δt PQ
δt PQ
dt
N
N
∂f bx, yg dx
∂f bx, yg dy
=
⋅
+
= Lim
δt→0
δ→
δ→
∂y
∴
dz
dt
=
δ→
dt
∂x
∂z dx
∂z dy
+
∂y dt
∂x dt
dt
...(iii) (As z = f (x, y))
∂z dx1
∂z dx2
∂z dxn
+
+....+
∂x1 dt
∂x2 dt
∂xn dt
The above relation can be also written as
In general
dz
dt
=
dz =
∂z
∂z
dx + ∂y dy , which is called total differential of z.
∂x
Corollary: If z = f (x, y) and suppose y is the function of x, then f is a function of one
independent variable x. Here y is intermediate variable. Identifying t with x in (iii), we get
dz
dx
=
∂z dy
∂z dx
∂z
∂z dy
dz
+ ∂y
⇒
=
+
.
∂x dx
dx
∂x
∂y dx
dx
1.7.1 Change of Variables
Let z = f (x, y) where x = φ (s, t) and y = ψ (s, t) then z is considered as function of s and t.
Now the derivative of z with respect s is partial but not total. Keeping t constant the
equation (iii) modified as
∂z
∂s
=
∂f
∂f
∂y
∂x
·
+
·
∂x
∂y
∂s
∂s
...(A)
50
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
In a similar way, we get
∂z
∂t
=
∂f
∂f ∂y
∂x
·
+
·
∂x
∂y ∂t
∂t
...(B)
The equations (A) and (B) are known as chain rule for partial differentiation.
Example 1. Find the total differential coefficient of x2 y w.r.t. x when x, y are connected by
x2 + xy + y2 = 1.
Sol. Let
z = x2 y
...(i)
Then the total differential coefficient of z
dz
=
dx
∂z dy
∂z dx
∂z dy
∂z
·
+
·
=
+ ∂y ·
∂x dx
∂x dx
∂x
dx
∂z
∂z
= 2xy, ∂y = x2 and we have
∂x
x2 + xy + y2 = 1
Differentiating w.r.t. x, we get
...(ii)
From (i)
2x +
dy
dy
x + y + 2y
dx
dx
= 0
dy
dx
= 0
dy
dx
= –
(2x + y) + (x + 2y)
⇒
...(iii)
2x + y
x + 2y
Putting these values in equation (ii), we get
F
GH
2x + y
dz
= 2xy + x2 −
x + 2y
dx
I = 2xy – x b2x + yg .
JK
bx + 2 y g
2
Example 2. If f (x, y) = 0, φ (y, z) = 0, show that
∂f ∂φ dz
∂y ∂z dx =
Sol. We have
From (i)
From (ii)
∂f ∂φ
.
·
∂x ∂y
f (x, y) = 0
φ (y, z) = 0
FG ∂f IJ
H ∂x K
∂f
∂f
dy
dy
=
+
=0⇒
= −
⋅
F ∂f I
∂x
∂y dx
dx
GH ∂y JK
F ∂φ I
GH ∂y JK
∂φ
∂φ dz
dφ
dz
⋅
=
+
=0⇒
= −
dy
dy
∂y
∂z dy
FG ∂φ IJ
H ∂z K
df
dx
...(i)
...(ii)
51
DIFFERENTIAL CALCULUS-I
Multiplying these two results, we get
FG ∂f IJ FG ∂φ IJ
H ∂x K × H ∂y K
dz
=
dx
F ∂f I FG ∂φ IJ
GH ∂y JK H ∂z K
∂f ∂φ dz
∂y ∂z dx
or
Example 3. If u = u
x2
∂f
∂φ
·
⋅
∂x ∂y
Hence proved.
F y − x , z − x I , show that
GH xy xz JK
∂u
∂u
∂u
+ y2
+ z2
= 0.
y
∂
∂x
∂z
Sol. Let
s =
∂s
∂x
So
(U.P.T.U., 2005)
y−x
1
1
1
1
z− x
=
–
and t =
=
–
xy
x
y
x
z
zx
1
= –
,
x2
∂s
∂t
1
∂t
1 ∂t
1
= 2 ,
= − 2,
= 2,
=0
∂y
∂x
y
y
∂
∂
z
x
z
∂s
= 0
∂z
u = u(s, t)
Since
∴
∂u
∂x
=
∂u ∂s
∂u ∂t
⋅
⋅
+
∂s ∂x
∂t ∂x
⇒
∂u
∂x
=
∂u
∂s
x2
or
Next,
∂u
∂x
∂u
∂y
∂u
∂y
or
and
=
FG – 1 IJ + ∂u FG − 1 IJ
H x K ∂t H x K
2
2
∂u
∂u
–
∂s
∂t
∂u ∂s
∂u ∂t
+
∂s ∂y
∂t ∂y
= –
=
∂u
∂u
∂u
2
+
0
⇒
y
y
∂y = ∂s
∂s
1 ∂u
∂u ∂s
∂u ∂t
+
=0+ 2
z
∂s ∂z
∂t ∂z
∂t
∂u
∂t
1
=
2
∂u
=
∂z
∂u
⇒
z2
=
∂z
Adding (i), (ii) and (iii), we get
∂u
∂u
∂u
∂u
∂u
∂u
∂u
x2
+ y2 ∂y + z2
= –
–
+
+
= 0. Hence proved.
∂x
∂z
∂s
∂t
∂s
∂t
Example 4. Prove that
...(i)
∂ 2u
∂ 2u
∂ 2u
∂ 2u
+
=
+
, where
∂ξ 2
∂η2
∂y2
∂x2
x = ξ cos α – η sin α, y = ξ sin α + η cos α.
...(ii)
...(iii)
52
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Sol. We have
∴
Now, Let
⇒
x = ξ cos α – η sin α, y = ξ sin α + η cos α.
∂y
∂y
∂x
∂x
= cos α,
= – sin α and
= sin α,
= cos α
∂ξ
∂η
∂ξ
∂η
u = u (x, y)
∂u
∂u ∂x
∂u ∂y
∂u
∂u
=
·
+
·
= cos α
+ sin α
∂ξ
∂x ∂ξ
∂y ∂ξ
∂x
∂y
∂u
∂η
=
∂u ∂x
∂u
∂u ∂y
∂u
·
+
·
= – sin α
+ cos α
∂η
∂x
∂x
∂y ∂η
∂y
Again
∂ 2u
∂ξ 2
=
∂
∂ξ
⇒
∂ 2u
∂ξ 2
=
cos2α
∂ 2u
∂η2
=
∂
∂η
and
and
=
...(ii)
FG ∂uIJ = F cos α ∂ + sin α ∂ I F cos α ∂u + sin α ∂u I
H ∂ξ K GH ∂x
∂y JK GH
∂x
∂y JK
∂ 2u
∂ 2u
∂ 2u
2
+ 2 sin α cos α
+ sin α
...(iii)
∂y2
∂x∂y
∂x 2
∂u
∂
∂
∂u
∂u
+ cos α
– sin α
+ cos α
= – sin α
∂η
∂x
∂y
∂x
∂y
FG IJ F
H K GH
IF
JK GH
I
JK
∂ 2u
∂ 2u
∂ 2u
∂ 2u
2α
= sin2α
–
2
sin
α
cos
α
+
cos
2
∂x∂y
∂y2
∂η
∂x2
Adding (iii) and (iv), we get
∂ 2u
∂ 2u
+
∂ξ 2
∂η2
...(i)
...(iv)
∂ 2u
∂ 2u
2 α + sin2α) + (cos2α + sin2α)
·
(cos
∂y2
∂x2
∂ 2u
∂ 2u
. Hence proved.
2 +
∂x
∂y2
Example 5. If u = f (y – z, z – x, x – y), show that
∂u
∂u
∂u
+
+
= 0.
∂
∂z
y
∂x
Sol. Let
r = y – z, s = z – x, t = x – y
Then
u = f (r, s, t)
=
(U.P.T.U., 2003)
∴
∂u
∂x
=
∂f ∂t
∂f ∂r
∂f ∂s
·
+
·
+
·
∂t ∂x
∂r ∂x
∂s ∂x
⇒
∂u
∂x
=
∂r
∂s
∂t
∂f
∂f
∂f
= 0,
= −1,
=1
×0+
(–1) +
(1) As
∂x
∂x
∂x
∂r
∂s
∂t
⇒
⇒
∂u
∂f
∂f
= –
+
∂s
∂t
∂x
∂u
∂f ∂r
∂f ∂s
∂f ∂t
∂y = ∂r · ∂y + ∂s · ∂y + ∂t · ∂y
∂u
∂y
=
∂f
∂f
∂f
∂r
∂s
∂t
= 1,
= 0,
= −1
(1) +
(0) +
(–1) As
∂y
∂y
∂y
∂r
∂s
∂t
=
∂f
∂f
–
∂r
∂t
...(i)
...(ii)
53
DIFFERENTIAL CALCULUS-I
and
∂u
∂z
=
∂f ∂r
∂f ∂s
∂f ∂t
·
+
·
+
·
∂r ∂z
∂s ∂z
∂t ∂z
∂r
∂s
∂t
∂f
∂f
∂f
= −1, = 1, = 0
(–1) +
(1) +
(0) As
∂z
∂z
∂z
∂r
∂s
∂t
∂u
∂f
∂f
⇒
= –
+
...(iii)
z
∂
∂s
∂r
Adding equations (i), (ii) and (iii), we get
∂u
∂u
∂u
+ ∂y +
= 0. Hence proved.
∂x
∂z
=
Example 6. If x = r cos θ, y = r sin θ, show that
∂r
=
∂x
∂x
∂x
∂θ
∂ 2θ
∂ 2θ
;
=r
and find the value of
+
·
r∂θ
∂r
∂x
∂y2
∂x2
Sol. We have
x = r cos θ, y = r sin θ
y
⇒
r2 = x2 + y2 and θ = tan–1
x
∂r
∂r
x = r cos θ = cos θ
∴
2r ·
= 2x ⇒
=
...(i)
∂
x
r
∂x
r
∂x
and we have
x = r cos θ ⇒
= cos θ
...(ii)
∂r
From equations (i) and (ii), we get
∂r
∂x
=
. Hence proved.
∂x
∂r
1 ∂x
∂x
Now,
= – r sin θ ⇒
= – sin θ
...(iii)
r ∂θ
∂θ
y
∂θ
1
y
r sin θ
− 2 =–
and
=
=–
2
2
2
x
∂x
y
r2
x +y
1+ 2
x
1
∂θ
= –
sin θ ⇒ r
= – sin θ
...(iv)
∂x
r
From equations (iii) and (iv), we obtain
1 ∂x
∂θ
= r
. Hence proved.
∂x
r ∂θ
2 xy
∂θ
∂ 2θ
∂ 2θ
y × 2x
y
Since
= – 2
⇒
=
⇒
2 ...(v)
2
2 =
2
2
2
∂x
∂x
∂x
x +y
x + y2
x2 + y2
FG IJ
H K
FG
H
IJ
K
e
and
∂θ
∂y
1
=
y2
x2
Adding equations (v) and (vi), we get
∂ 2θ
∂ 2θ
+
∂y2
∂x2
1+
= 0.
e
j
FG 1IJ = x ∴ ∂ θ = – 2xy
H xK ex + y j ∂y
ex + y j
j
2
2
2
2
2
2 2
...(vi)
54
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 7. If φ (x, y, z) = 0, show that
FG ∂y IJ FG ∂z IJ FG ∂x IJ = – 1.
H ∂z K H ∂xK y H ∂y K
x
(U.P.T.U., 2004)
z
Sol. We have φ (x, y, z) = 0
Keeping y as constant, differentiate partially w.r.t. x, we get
FG ∂φ IJ + FG ∂φ IJ · FG ∂z IJ
H ∂xK H ∂z K H ∂xK
FG ∂z IJ
H ∂xK
⇒
= 0
y
FG ∂φ IJ
H ∂x K
= –
FG ∂φ IJ
y
H ∂z K
...(i)
Next, keeping z as constant, differentiate partially w.r.t. y, we obtain
∂φ
∂x
⇒
F ∂x I + ∂φ = 0
GH ∂y JK ∂y
z
F ∂φ I
F ∂x I = − GH ∂y JK
GH ∂y JK
FG ∂φ IJ
H ∂x K
FG ∂φ IJ
∂
y
FG IJ = – H ∂z K
H ∂z K x
F ∂φ I
GH ∂y JK
...(ii)
z
Similarly,
...(iii)
Multiplying equations (i), (ii) and (iii), we get
FG ∂y IJ FG ∂z IJ FG ∂x IJ
H ∂z K H ∂xK y H ∂y K
x
= –1. Hence proved.
z
Example 8. If x + y = 2eθ cos φ and x – y = 2i eθ sin φ, show that
∂ 2V
∂ 2V
∂ 2V
·
2 = 4xy
2 +
∂x ∂y
∂φ
∂θ
Sol. We have x + y = 2eθ cos φ
x – y = 2ieθ sin φ
Adding
2x = 2 (eθ) (cos φ + i sin φ)
x = eθ + i φ
and subtracting, we get y = eθ – i φ
Let
∴
(U.P.T.U., 2001)
...(i)
...(ii)
V = V (x, y)
∂V
∂V ∂x
∂V ∂y
=
·
+
·
∂θ
∂x
∂θ
∂y
∂θ
∂V
∂V
∂V
= x
+y
∂θ
∂y
∂x
As
∂y
∂x
= e θ+iφ = x,
= e θ − iφ = y
∂θ
∂θ
55
DIFFERENTIAL CALCULUS-I
FG ∂VIJ = FG x ∂ + y ∂ IJ FG x ∂V + y ∂V IJ
H ∂θ K H ∂x ∂y K H ∂x ∂y K
F ∂V ∂V I
∂ V
∂ V
∂ V
∂ V
= x
+y
+ 2xy
+ G x ∂x + y ∂y J
H
K
∂θ
∂y
∂x ∂y
∂x
∂
∂ 2V
2 = ∂θ
∂θ
∴
2
⇒
2
2
2
2
2
2
2
...(iii)
2
∂V
∂y
∂V ∂x
∂V ∂y
∂V
∂V
∂x
= ix = = –iy
=
·
+
=
(ix) +
(–iy) As
∂φ
∂φ
∂x
∂y ∂φ
∂x
∂y
∂φ
∂φ
Now
F x ∂V − y ∂V I
GH ∂x ∂y JK
F ∂ ∂ I F ∂V ∂V I
∂ V
∂ F ∂V I
G
J
=
= i G x ∂x − y ∂y J i G x ∂x − y ∂y J
∂φ H ∂φ K
H
K H
K
∂φ
F ∂ 2V ∂ 2V ∂ 2V + x ∂V + y ∂VI
= − G x2 2 + y2 2 − 2xy
∂x∂y
∂x
∂y JK
H ∂x ∂y
∂V
= i
∂φ
⇒
2
Next
2
2
∂ 2v
∂ 2v
∂v
∂v
∂ 2V
2 ∂ v
2
y
2
xy
−x
−y
−
+
= –x
2
2
2
∂x ∂y
∂x
∂y
∂x
∂y
∂φ
Adding equations (iii) and (iv), we get
∂ 2V
∂ 2V
+
∂φ 2
∂θ 2
= 4xy
...(iv)
∂ 2V
. Hence proved.
∂x ∂y
Example 9. If x = r cos θ, y = r sin θ, z = f (x, y), prove that
1 ∂z
1 ∂z
∂z
∂z
∂z
∂z
=
cos θ –
sin θ ;
=
sin θ +
cos θ.
r ∂θ
r ∂θ
∂x
∂r
∂y
∂r
e
∂ 2 r n .cos nθ
j = – n (n – 1) r
n – 2 · sin (n – 2) θ.
∂x ∂y
Sol. Here z is a function of x and y where x and y are functions of r and θ.
Prove also that
∴ We have
and
Now,
∂z
∂z ∂r
∂z ∂θ
=
·
+
·
∂x
∂r ∂x
∂θ ∂x
...(i)
∂z
∂z ∂r
∂z ∂θ
=
·
+
·
∂y
∂r ∂y
∂θ ∂y
...(ii)
x = r cos θ, y = r sin θ, so r2 = x2 + y2 and θ = tan–1
FG y IJ .
H xK
∂r
∂θ
sin θ
cos θ
∂r
∂θ
= cos θ, ∂y = sin θ,
=–
and ∂y =
r
r
∂x
∂x
Substituting these values in equations, (i) and (ii), we have
Then,
and
1
∂z
∂z
∂z
. Hence proved.
= cos θ
–
sin θ
r
∂x
∂r
∂θ
...(iii)
∂z
∂y
...(iv)
= sin θ
1
∂z
∂z
. Hence proved.
+
cos θ
r
∂r
∂θ
56
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Again substituting rn cos nθ for z in (iv), we get
∂
cos θ ∂ n
∂
n
n
(r cos nθ)
∂y (r cos nθ) = sin θ · ∂r (r cos nθ) + r
∂θ
cos θ
= sin θ (nrn – 1 cos nθ) +
(– rnn sin nθ)
r
= nrn−1(cos nθ sin θ − cos θ sin nθ)
e
∂ r n cos nθ
or
∂y
Now,
e
j = nr
∂ 2 r n cos nθ
∂x∂y
n–1 sin (θ – nθ)
...(v)
j = ∂ LM ∂ er n cos nθ j OP
∂x M
N ∂y PQ
∂
∂
nr sin a1 − nfθ , from (v) = n
r sin a1 − nfθ
=
∂x
∂x
sin θ ∂ n−1
L ∂
O
r sina1 − nfθ}P
= n Mcos θ {rn−1 sina1 − nfθ} −
r ∂θ {
N ∂r
Q
sin θ n−1
L
r a1 − nf cos a1 − nf θOP
= n Mcos θ an − 1f rn − 2 sin a1 − nf θ –
r
N
Q
n– 1
n– 1
= – n (n –1) rn–2 [sin (n – 1)θ cos θ – cos (n – 1) θ sin θ]
= – n (n – 1) rn-2 sin (n – 2) θ. Hence proved.
Example 10. If u = f (x, y) and x = r cos θ, y = r sin θ, prove that
∂ 2u
∂ 2u
∂ 2u
1
2 =
2 +
2 + 2
∂y
∂x
∂r
r
1 ∂u
∂ 2u
+
·
2
r ∂r
∂θ
Or
F ∂ u I F ∂ 2uI
Transform G
H ∂x JK + GH ∂y2 JK = 0 into polars and show that u = (Ar + Br ) sin nθ satisfies
2
n
2
the above equation.
Sol. We know x = r cos θ, y = r sin θ
∴
r2
=
From (ii), we get 2r
or
Similarly,
...(ii)
FG y IJ
H xK
...(iii)
∂r
∂r
x
r cos θ
= 2x or
=
=
, from (i)
r
r
∂x
∂x
∂r
= cos θ
∂x
∂r
r sin θ
y
=
=
= sin θ
r
∂y
r
Also from (iii),
...(i)
x2 + y2
θ = tan–1
and
–n
∂θ
=
∂x
FG
FG IJ H
H K
y
1
− 2
2 ·
x
y
1+
x
...(iv)
...(v)
IJ = – y
K x +y
2
2
57
DIFFERENTIAL CALCULUS-I
r sin θ
sin θ
∂θ
= –
=–
2
r
r
∂x
∂θ
1
x
r cosθ
1
cos θ
=
2
2 · x = x2 + y2 =
∂y =
r
r
y
1+
or
FG IJ
FG IJ H K
H xK
and
Now, we know
∂u
∂u ∂r
∂u ∂θ
=
·
+
·
∂x
∂r ∂x
∂θ ∂x
...(vii)
IJ
K
FG
H
sin θ
∂u
∂u
−
(cos θ) +
, from (iv), (vi)
r
∂r
∂θ
sin θ ∂
∂
∂
(u) = cos θ
(u) –
(u)
r ∂θ
∂x
∂r
∂u
Replacing u by
∂x
2
∂
u
sin θ ∂ ∂u
∂ u
∂
∂
∂u
= cos θ.
–
,
2 = ∂x
∂x
r
∂r ∂x
∂θ ∂x
∂x
=
or
...(vi)
...(viii)
FG IJ
H K
FG IJ
FG IJ
H K
H K
∂ L
∂u sin θ ∂u O sin θ ∂ L
∂u sin θ ∂u O
cos θ −
cos θ −
−
= cos θ ⋅
M
P
M
r ∂θ Q
r ∂θ N
r ∂θ PQ
∂r N
∂r
∂r
2
sin θ LMF – sin θ ∂u + cos θ ⋅ ∂ u I − 1 ∂ FG sin θ ⋅ ∂u IJ OP
L
∂ 2u
∂ F 1 ∂u I O
G
= cos θ Mcos θ 2 − sin θ ⋅ G ⋅ J P –
∂θ ∂r JK r ∂θ H
∂θ K PQ
∂r
r MNH
∂r H r ∂θ K PQ
MN ∂r
or
LM
MN
RS
T
∂ 2u
1 ∂ 2u 1 ∂u
∂ 2u
cos
sin
θ
⋅
−
θ
⋅
−
=
cos
θ
r ∂r ∂θ r 2 ∂θ
∂r 2
∂x2
–
or
LM
MN
UVOP
WPQ
F
GH
2
2
sin θ – sin θ ∂u + cos θ ⋅ ∂ u − 1 sin θ ∂ u + cos θ ⋅ ∂u
∂θ
∂r ∂θ r
∂r
∂θ 2
r
I OP
JK PQ
2sin θ cos θ ∂ 2u
sin 2 θ ∂ 2u
∂ 2u
∂ 2u
sin 2 θ ∂u
2 θ
=
cos
–
+
+
2
2
2
2
∂r ∂θ
r
∂r
∂x
∂r
∂θ
r
r
+
Similarly,
2 cos θ sin θ ∂u
...(ix)
r2
∂θ
∂ 2u
2sin θ cos θ ∂ 2u
cos2 θ ∂ 2u
∂ 2u
+
+
2 = sin2 θ
∂y
r
r2
∂r 2
∂r ∂θ
∂θ 2
2 cos θ sin θ
cos 2 θ ∂u
–
r2
∂r
r
Adding equations (ix) and (x), we get
+
∂u
∂θ
...(x)
∂ 2u
1
∂ 2u
∂ 2u
∂ 2u
2 θ + sin2 θ)
2 θ + cos2 θ)
+
=
(cos
+
(sin
2
∂y
r2
∂x2
∂r 2
∂θ 2
+
1 ∂ 2u
1
1 ∂u
∂u
∂ 2u
(sin2 θ + cos2 θ)
=
+ 2
2
2 + r ∂r .
r
r
∂r
∂r
∂θ
Hence proved.
58
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Or
If
u =
and
∴
1
∂ 2u
2 + r2
∂r
(Arn + Br–n) sin nθ, then
∂u
∂r
= n (Arn–1 – Br–n – 1) sin nθ;
∂ 2u
∂r 2
= n [A(n – 1)rn – 2 + B (n + 1) r–n –2] sin nθ;
∂ 2u
∂θ 2
=
∂
∂θ
∂u
= (Arn + Br–n) n cos nθ
∂θ
FG ∂uIJ = – (Ar + Br ) n sin nθ
H ∂θ K
n
–n
2
1 ∂u
∂ 2u
n–2 + B (n + 1) r–n–2] sin nθ
2 + r ∂r = n[A (n – 1) r
∂θ
1
1
(Arn + Br–n)n2 sin n θ +
n(Arn–1 – Br–n–1) sin nθ
r
r2
= [A{n2–n–n2 + n}rn–2 + B(n2 + n – n2 –n)r–n – 2] sin nθ
–
= 0. Hence proved.
Example 11. If x = r cos θ, y = r sin θ or r2 = x2 + y2, prove that
∂ 2r
1
∂ 2r
+
2 =
2
∂y
r
∂x
Sol. Since x = r cos θ and y = r sin θ
∂r
∂x
=
∂ 2r
∂ 2r
+
∂y2
∂x2
=
∴
Adding
∂ 2r
∂ 2r
+
∂y2
∂x2
or
Also,
1
r
R|F ∂r I F ∂r I U|
S|GH ∂x JK + GH ∂y JK V| ·
T
W
2
2
2
y ∂ 2r
x ∂r
r 2 − x2 ∂ r
r2 − y2
;
= ;
=
;
=
,
∂y2
r ∂y
r ∂x2
r3
r3
er − x j + er − y j = 2r − ex + y j
2
2
2
r3
2
2
r3
2
2
r3
=
2r 2 − r 2
, ä x2 + y2 = r2
r3
=
1
r
...(i)
R|F ∂r I F ∂r I U| 1 R|F x I F y I U| 1 F x2 + y2 I
S|GH ∂x JK + GH ∂y JK V| = r S|GH r JK + GH r JK V| = r G r2 J
K
T
W H
T
W
1 Fr I
=
G J,äx +y =r
r Hr K
2
2
2
2
2
2
2
2
2
2
=
∂ r
1
∂ 2r
=
+
, from (i) Hence proved.
2
∂y2
r
∂x
Example 12. If V = f (2x – 3y, 3y – 4z, 4z – 2x), compute the value of 6Vx + 4Vy + 3Vz.
(U.P.T.U., 2008)
Sol. Let
r = 2x – 3y, s = 3y – 4z, t = 4z – 2x
∴
V = f(r, s, t)
59
DIFFERENTIAL CALCULUS-I
Vx =
=
⇒
∂f ∂r ∂f ∂s ∂f ∂t
∂V
=
⋅
+
⋅
+
⋅
∂x
∂r ∂x ∂s ∂x ∂t ∂x
a f
∂f
∂f
∂f
⋅2 +
⋅0 +
−2
∂r
∂s
∂t
As =
∂r
∂s
∂t
= 2,
= 0,
= −2
∂x
∂x
∂x
Vx = 2fr – 2ft
Vy =
=
⇒
...(i)
∂f ∂r ∂f ∂s ∂f ∂t
∂V
=
⋅
+
⋅
+
⋅
∂y
∂r ∂y ∂s ∂y ∂t ∂y
a f
af
∂f
∂f
∂f
−3 +
3 +
⋅0
∂r
∂s
∂t
As
∂r
∂s
∂t
= − 3,
= 3,
= 0
∂y
∂y
∂y
Vy = – 3fr + 3fs
...(ii)
Vz = 4fr – 4fs
...(iii)
Similarly
Multiplying (i), (ii) and (iii) by 6, 4, 3 respectively and adding, we get
6Vx + 4Vy + 3Vz = 12fr – 12ft – 12fr + 12fs + 12fr – 12fs
⇒
6Vx + 4Vy + 3Vz = 0.
Hence Proved.
Example 13. If u = x log xy, where x3 + y3 + 3xy = 1, find
du
.
dx
[U.P.T.U. (C.O.), 2005]
Sol. By total differentiation, we know that
∂u dx ∂u dy
∂u ∂u dy
du
+
⋅
⋅
+
⋅
=
=
∂x dx ∂y dx
∂x ∂y dx
dx
...(i)
1
∂u
= log xy + ⋅ y = log xy + 1
y
∂x
...(ii)
we have u = x log xy
af
∂u
x
x
x =
=
xy
y
∂y
...(iii)
Also, given that
x3 + y3 + 3xy = 1
Differentiating w.r.t. ‘x’, we get
3x2 + 3y2
⇒
or
dy
dy
+ 3y + 3x
= 0
dx
dx
(x2 + y) + (x + y2)
dy
= 0
dx
e
e
x2 + y
dy
= −
dx
x + y2
Using (ii), (iii) and (iv) in (i), we get
j
j
...(iv)
F
GH
x x2 + y
du
−
= (1 + log xy)
y x + y2
dx
I
JK
60
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 14. If x = u + v + w, y = vw + wu + uv, z = uvw and f is a function of x, y, z, show
∂f
∂f
∂f
∂f
∂f
∂f
that u
.
+v
+w
= x
+ 2y
+ 3z
∂u
∂v
∂w
∂x
∂y
∂z
Sol. Let
f = f(x, y, z)
∂f
∂f ∂x ∂f ∂y ∂f ∂z
⋅
+
⋅
+
=
∂u
∂x ∂u ∂y ∂u ∂z ∂u
=
or
a
f
∂f
∂f
∂f
+ w+v
+ vw
∂x
∂y
∂z
a
As
a
f
∂y
∂x
∂z
= 1,
= w+v ,
= vw
∂u
∂u
∂u
f
∂f
∂f
∂f
∂f
+ u w+v
+ uvw
= u
∂x
∂y
∂z
∂u
u
...(i)
∂f
∂f ∂x ∂f ∂y ∂f ∂z
⋅
+
⋅
+
⋅
=
∂v
∂x ∂v ∂y ∂v ∂z ∂v
=
or
v
a
a
f
∂y
∂f
∂f
∂f
∂x
∂z
As
= 1,
= u+w ,
= uw
+ u+w
+ uw
∂v
∂v
∂v
∂z
∂x
∂y
f
a
∂f
∂f
∂f
∂f
= v
+ v u+w
+ uvw
∂
∂
∂z
x
y
∂v
f
...(ii)
a
f
...(iii)
Similarly
w
∂f
∂f
∂f
∂f
+ w u+v
+ uvw
= w
∂x
∂y
∂z
∂w
Adding (i), (ii) and (iii), we get
or
au + v + wf ∂∂xf + 2avw + wu + uvf ∂∂yf + 3uvw ∂∂fz
u
∂f
∂f
∂f
+v
+w
=
∂u
∂v
∂w
u
∂f
∂f
∂f
∂f
∂f
∂f
+v
+w
= x
+ 2y
+ 3z . Proved.
∂u
∂v
∂w
∂x
∂y
∂z
Example 15. If by the substitution u = x2 – y2, v = 2xy, f(x, y) = φ(u, v) show that
Sol. We have
∂2 f
∂2 f
∂x
∂y 2
+
2
F ∂ φ + ∂ φI .
GH ∂u ∂v JK
2
= 4(x2 + y2)
2
2
2
f(x, y) = φ(u, v)
Differentiating partialy w.r.t. ‘x’.
∂φ
∂φ
∂f
∂φ ∂u ∂φ ∂v
=
⋅
+
⋅
= 2x
+ 2y
∂u ∂x ∂v ∂x
∂u
∂v
∂x
∂2 f
∂x
2
As
∂u
∂v
= 2x,
= 2y
∂x
∂x
FG IJ = ∂ FG 2x ∂φ + 2y ∂φ IJ
H K ∂x H ∂u ∂v K
∂ F ∂φ
∂φ I ∂u ∂ F ∂φ
∂φ I ∂v
2x + 2y J
2x + 2y J ⋅
+
=
G
G
H
K
H
∂u
∂u
∂v ∂x ∂v
∂u
∂v K ∂x
=
∂ ∂f
∂x ∂x
61
DIFFERENTIAL CALCULUS-I
F 2x ∂ φ + 2y ∂ φ I .2x + F 2x ∂ φ + 2y ∂ φ I .2y
GH ∂u ∂u∂v JK GH ∂v∂u ∂v JK
2
=
or
2
2
2
2
2
2
∂ 2φ
∂ 2φ
2∂ φ
8
4
xy
y
+
+
∂u∂v
∂x 2
∂u2
∂v2
Again differentiating f(x, y) partially w.r. to y
∂2 f
= 4x2
...(i)
∂f
∂u
∂v
∂φ
∂φ
∂φ ∂u ∂φ ∂v
As
= − 2 y,
= 2x
=
⋅
+
= − 2y
+ 2x
∂y
∂y
∂y
∂u ∂y ∂v ∂y
∂u
∂v
∂2 f
∂y
2
=
FG
H
∂
∂φ
∂φ
−2 y
+ 2x
∂y
∂u
∂v
IJ
K
FG
IJ
IJ
H
K
K
F ∂ φ + 2x ∂ φ I b−2yg + F −2y ∂ φ + 2x ∂ φ I a2xf
= G −2 y
GH ∂u∂v ∂v JK
H ∂u ∂u∂v JK
=
FG
H
∂
∂φ
∂φ ∂u
∂
∂φ
∂φ ∂v
+ 2x
−2 y
+ 2x
+
−2 y
∂u
∂u
∂v ∂y ∂v
∂u
∂ v ∂y
2
2
2
2
2
∂2 f
or
∂y
2
= 4y2
2
2
∂ 2φ
∂ 2φ
2∂ φ
8
4
xy
x
−
+
∂u∂v
∂u2
∂v2
...(ii)
Adding (i) and (ii), we get
∂2 f
∂x
2
+
∂2 f
∂y
2
2
2
e
j ∂∂uφ + 4 ex + y j ∂∂vφ
e
j FGH ∂∂uφ + ∂∂vφ IJK . Hence Proved.
= 4 x2 + y 2
= 4 x2 + y2
2
2
2
2
2
2
2
2
Example 16. If x2 + y2 + z2 – 2xyz = 1, show that
dx
1−x
2
+
dy
1− y
2
+
dz
1 − z2
= 0.
Sol. We have
x2 + y2 + z2 – 2xyz = 1
x2 – 2xyz + y2 z2 = 1 – y2 – z2 + y2 z2
or
(x – yz)2 = (1 – y2) (1 – z2)
or
(x – yz) =
Again
e1 − y j e1 − z j
2
(y – zx)2 = (1 – x2) (1 – z2)
or
(y – zx) =
Let
...(i)
y2 – 2xyz + z2 x2 = 1 – x2 – z2 + z2 x2
or
Similarly
2
(z – xy) =
e1 − x j e1 − z j
e1 − x j ⋅ e1 − y j
2
2
...(ii)
2
2
...(iii)
u h x2 + y2 + z2 – 2xyz – 1 = 0
62
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
By total differentiation, we get
∂u
∂u
∂u
dx + dy + dz = 0
∂x
∂y
∂z
du =
As
or (2x – 2yz)dx + (2y – 2zx)dy + (2z – 2xy)dz = 0
∂u
= 2x − 2 yz
∂x
∂u
= 2 y − 2zx
∂y
∂u
= 2z − 2xy
∂z
or (x – yz)dx + (y – zx)dy + (z – xy)dz = 0
...(iv)
Putting (i), (ii) and (iii) in equation (iv), we get
e1 − y j e1 − z j . dx + e1 − x j e1 − z j dy + e1 − x j e1 − y j dz = 0
2
2
Dividing by
dx
1−x
2
+
2
2
2
2
e1 − x j e1 − y j e1 − z j , we get
2
dy
1− y
2
+
2
dz
1 − z2
2
= 0. Hence proved.
EXERCISE 1.6
LMA ns. dy = − L yx + y log y OOP
MN dx MMN x log x + xy PPQPQ
du LA ns. du = 1 + log xy + x L− (x + y) O O
b
g y MM ( y + x) PPPP
2 . If u = x log xy, where x + y + 3xy = 1, find
.M
dx
dx MN
N
QQ
LMA ns. du = 2xy − x L b2x + yg OOP
du
MM bx + 2yg PPP
3 . If u = x y, where x + xy + y = 1, find
.
MN dx
N
QQ
dx
4 . If V is a function of u, v where u = x – y and v = x – y, prove that
F ∂ V + xy ∂ V I .
∂ 2V
∂ 2V
x
+
y
J
2
2 = (x + y) G
∂x
∂y
∂v K
H ∂u
y −1
dy
1 . Find
if xy + yx = c.
dx
x
x -1
y
2
3
2
2
3
2
2
2
2
2
2
5 . Transform the Laplacian equation
2
∂ 2u
∂ 2u
+
∂y2 = 0 by change of variables from x, y to r, θ
∂x2
when x = er cos θ, y = er sin θ.
6 . Find
du
, if u = x log xy where x3 + y3 + 3xy = 1.
dx
LMA ns. e FG ∂ u + ∂ uIJ = 0OP
H ∂r ∂θ K PQ
MN
LMA ns. du = 1 + log xy − x . x + y OP
y x + y QP
NM dx
−2r
2
2
2
2
2
2
63
DIFFERENTIAL CALCULUS-I
7 . If the curves f (x, y) = 0 and φ (x, y) = 0 touch, show that at the point of contact
∂f ∂φ
∂f ∂φ
·
=
·
.
∂
∂x y
∂y ∂x
dy
, when (cos x)y = (sin y)x.
dx
9 . If x = r cos θ, y = r sin θ, prove that
8 . Find
F
GH
I
JK
LMA ns. dy = y tan x + log sin y OP
N dx log cos x − x cot y Q
2
2
∂ 2r
∂ 2r ∂ r
·
.
2 =
∂y∂x
∂x2 ∂y
1 0 . If z is a function of x and y and x = eu + e–v, y= e–u –ev prove that
∂z
∂z
∂z
∂z
.
–
=x
– y
∂v
∂y
∂u
∂x
1 1 . If u = log (tan x + tan y + tan z) prove that
∂u
∂u
∂u
+ (sin 2y) ∂y + (sin 2z)
= 2.
∂x
∂z
1 2 . If u = 3 (lx + my + nz)2 – (x2 + y2 + z2) and l2 + m2 + n2 = 1, show that
(sin 2x)
∂ 2u
∂ 2u
∂ 2u
+
+
= 0.
∂x2
∂y2
∂z2
1 3 . If u = x2 + 2xy – y log z, where x = s + t2, y = s – t2, z = 2t, find
LMA ns. ∂u = 8, ∂u = 8t – 4OP
N ∂s ∂t
Q
∂u ∂u
,
at (1, 2, 1).
∂s ∂t
1 4 . If u = x2 – y2 + sin yz, where y = ex and z = log x, find
du
.
dx
LMA ns. 2ex – e j + e cosee log xjFG log x + 1 IJ OP
H
xK Q
N
∂z
∂z
1 5 . If z = z(u, v), u = x – 2xy – y and v = y, show that (x + y)
+ bx − yg
= 0 is equivalent
∂x
∂y
x
2
to
x
x
2
∂z
= 0.
∂v
CURVE TRACING
Introduction
It is analytical method in which we draw approximate shape of any curve with the help of
symmetry, intercepts, asymptotes, tangents, multiple points, region of existence, sign of the first
and second derivatives. In this section, we study tracing of standard and other curves in the
cartesian, polar and parametric form.
64
1.8
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
PROCEDURE FOR TRACING CURVES IN CARTESIAN FORM
The following points should be remembered for tracing of cartesian curves:
1.8.1 Symmetry
(a) Symmetric about x-axis: If all the powers of y occurring in the equation are even then the
curve is symmetrical about x-axis.
Example.
y2
x2
+
= 1, y2 = 4ax.
a2
b2
(U.P.T.U., 2008)
(b) Symmetric about y-axis: If all the powers of x occurring in the equation are even then
the curve is symmetrical about y-axis.
Example.
x2 = 4ay, x4 + y4 = 4x2y2.
(c) Symmetric about both x- and y-axis: If only even powers of x and y appear in equation
then the curve is symmetrical about both axis.
Example.
x2 + y2 = a2.
(U.P.T.U., 2008)
(d) Symmetric about origin: If equation remains unchanged when x and y are replaced by
– x and – y.
Example.
x5 + y5 = 5a2x2y.
Remark: Symmetry about both axis is also symmetry about origin but not the converse (due
to odd powers).
(e) Symmetric about the line y = x : A curve is symmetrical about the line y = x, if on
interchanging x and y its equation does not change.
Example.
x3 + y3 = 3axy.
(U.P.T.U., 2008)
(f) Symmetric about y = – x : A curve is symmetrical about the line y = – x, if the equation
of curve remains unchanged by putting x = – y and y = – x in equation.
Example.
x3 – y3 = 3axy.
(U.P.T.U., 2008)
1.8.2 Regions
(a) Region where the curve exists: It is obtained by solving y in terms of x or vice versa. Real
horizontal region is defined by values of x for which y is defined. Real vertical region is defined
by values of y for which x is defined.
(b) Region where the curve does not exist: This region is also called imaginary region, in
this region y becomes imaginary for values of x or vice versa.
1.8.3 Origin and Tangents at the Origin
If there is no constant term in the equation then the curve passes through the origin otherwise not.
If the curve passes through the origin, then the tangents to the curve at the origin are
obtained by equating to zero the lowest degree terms.
Example. The curve a2y2 = a2x2 – x4, lowest degree term (y2 – x2) equating to zero gives
y = ± x as the two tangents at the origin.
1.8.4 Intercepts
(a) Intersection point with x- and y-axis: Putting y = 0 in the equation we can find points where
the curve meets the x–axis. Similarly, putting x = 0 in the equation we can find the points where
the curve meets y-axis.
65
DIFFERENTIAL CALCULUS-I
(b) Points of intersection: When curve is symmetric about the line y = ± x, the points of
intersection are obtained by putting y = ± x in given equation of curve.
(c) Tangents at other points say (h, k) can be obtained by shifting the origin to these points
(h, k) by the substitution x = x + h, y = y + k and calculating the tangents at origin in the new
xy plane.
Remark. The point where dy/dx = 0, the tangent is parallel to x-axis. And the point where
dy/dx = ∞, the tangent is vertical i.e., parallel to y-axis.
1.8.5 Asymptotes
If there is any asymptotes then find it.
(a) Parallel to x-axis: Equate the coefficient of the highest degree term of x to zero, if it is
not constant.
(b) Parallel to y-axis: Equate the coefficient of the highest degree term of y to zero, if it is
not constant.
Example. x2y – y – x = 0
highest power coefficient of x i.e., x2 = y
Thus asymptote parallel to x-axis is y = 0
Similarly asymptote parallel to y-axis are x2 – 1 = 0 ⇒ x = ± 1.
(c) Oblique asymptotes (not parallel to x-axis and y-axis): The asymptotes are given by
y = mx + c, where m = lim
x→∞
FG y IJ and c = lim (y – mx).
H xK
x→∞
(d) Oblique asymptotes (when curve is represented by implicit equation f (x, y) = 0):
The asymptotes are given by y = mx + c where m is solution of φn(m) = 0 and c is the solution of
cφ′n(m) + φn–1(m) = 0 or c =
af
af
− φ n−1 m
. Here φn(m) and φn–1(m) are obtained by putting x = 1 and
φ ′n m
y = m in the collection of highest degree terms of degree n and in the collection of the next highest
degree terms of degree (n – 1).
1.8.6 Sign of First Derivatives dy/dx (a ≤ x ≤ b)
dy
> 0 , then curve is increasing in [a, b].
dx
dy
< 0 , then curve is decreasing in [a, b].
(b)
dx
dy
= 0 , then the point is stationary point where maxima and minima can occur.
(c) If
dx
(a)
1.8.7 Sign of Second Derivative
(a)
(b)
d2 y
dx 2
d2 y
dx 2
£ P´
(a ≤ x ≤ b)
£³P
> 0 , then curve is convex or concave upward (holds water).
< 0 , then the curve is concave or concave downward (spills water).
66
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
1.8.8 Point of Inflexion
A point where d2y/dx2 = 0 is called an inflexion point where the curve changes the direction of
concavity from downward to upward or vice versa.
w
Example 1. Trace the curve y = x3 – 3ax2.
Sol. 1. Symmetry: Here the equation of curve do not hold
any condition of symmetry. So there is no symmetry.
2. Origin: Since there is no constant add in equation so the
curve passes through the origin. The equation of tangent at origin
is y = 0 i.e., x-axis (lowest degree term).
3. Intercepts: Putting y = 0 in given equation, we get
x3 – 3ax2 = 0 ⇒ x = 0, 3a
Thus, the curve cross x-axis at (0, 0) and (3a, 0) .
4. There is no asymptotes.
dy
5.
= 3x2 – 6ax.
dx
dy
For stationary point
= 0 = 3x2 – 6ax = 0 ⇒ x = 0, 2a.
dx
6.
d2y
FG d y IJ
H dx K
P
v>
FQ J>NG
Q
FP J>R G>
Fig. 1.3
FG d y IJ
H dx K
2
2
= 6x – 6a,
>
m
= – 6a < 0 (concave) and ymax = 0 and
2
dx 2
x=0
= 6a > 0 (convax) and ymin. = – 4a3.
2
x=2a
= 12a – 6a
d2y
= 0 ⇒ 6x – 6a = 0 ⇒ x = 0, a.
dx 2
8. Region: – ∞ < x <∞ since y is defined for all x.
7. Inflexion point:
9. Sign of derivative
Interval
Sign of y
Quadrant
Sign of y′
Nature of curve
–∞<x<0
y<0
III
y′ > 0
increasing
0 < x < 2a
y<0
IV
y′ < 0
decreasing
2a < x < 3a
y<0
IV
y′ > 0
increasing
3a < x < ∞
y>0
I
y′ > 0
increasing
Using the above calculations. We draw the graph in Figure 1.3.
Example 2. Trace the curve y2 (a – x) = x3, a > 0.
Sol. 1. Symmetry: Since y has even power so the curve
is symmetric about x-axis.
2. Origin: The curve passes through the origin.
3. Tangent at origin: The coefficient of lowest degree
term is y2 = 0 or y = 0 and y = 0 i.e., there is a cusp at the
origin.
4. Intercepts: Putting x = 0, then y = 0 ⇒ origin is the
only point where the curve meets the co-ordinate axes.
5. Asymptotes: Asymptotes parallel to y-axis obtained
by equating to zero the highest degree term of y i.e., (x – a) =
0 ⇒ x = a.
(U.P.T.U., 2006)
w
³>[>
b¬°¡©¤>>>¯ «¦¤«¯>
m
F J>NG
Fig. 1.4
v>
67
DIFFERENTIAL CALCULUS-I
x3
this shows that the values of x > a, y is imaginary. So the curve does
a−x
not exist x > a. Similarly, the curve does not exist when x < 0.
Here the curve only for 0 ≤ x < a.
7. Sign of derivation:
6 . Region: y2 =
1
F
H
I
K
3a
−x
dy
dy
2
=±
, in the first quadrant for 0 ñ x < a,
> 0, curve increasing in the first
dx
dx
( a − x) a − x
quadrant.
The shape of figure is shown in the (Fig. 1.4).
x2
Example 3. Trace the following curve and write its asymptotes.
x3 + y3 = 3axy.
(U.P.T.U., 2003)
Sol. 1. Symmetry: Interchage x and y. Then equation of curve remain unchanged.
∴ The curve is symmetric about the line y = x.
2. Origin: The curve passes through origin.
3. Tangent at origin: The coefficient of lowest degree term is x = 0 or y = 0
∴
x = 0 and y = 0 are tangents at origin.
4. Intercepts: Putting y = x, we get
w
2x3 = 3ax2 ⇒ 2x3 – 3ax2 = 0
>³
´> [
⇒ x2 (2x – 3a) = 0
3a
2
Q JQ
Å Å
P P
⇒
x = 0 and x =
At
x = 0, y = 0 and at x =
3a
3a
,y=
2
2
Thus, points of intersection are (0, 0) and
F 3a , 3a I
H 2 2K
along the line y = x.
5. Asymptotes: Since the coefficients of highest
powers of x and y are constants, there are no asymptotes
parallel to x-and y-axis.
Putting x = 1 and y = m in highest degree term
(x3 + y3), we get
OQSÝ
>
m>FNJ>NG
³>
I>
m = –1 (Real solution)
Again putting
x = 1 and y = m
in next highest degree term (– 3axy), we get
c =
At
a−3amf = a
3m2
I>
Fig. 1.5
φ3(m) = 1 + m3 = 0
⇒
´>
m
m = –1, c = – a.
Asymptotes y = mx + c ⇒ y = –x–a or y + x + a = 0.
>[
>N
v
68
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
dy
ay − x 2
= 2
⇒
dx
y − ax
an angle 135° with x-axis.
6 . Derivative:
FG dy IJ
H dx K FH
3a 3a
,
2 2
F 3a 3a I
I = – 1 i.e., the tangent at H 2 , 2 K making
K
7 . Region: When both x and y are negative simultaneously equation of curve is not satisfied
(–x)3 + (–y)3 = 3a (–x) (–y) ⇒ – (x3 + y3) = 3axy
⇒ There is no part of the curve exists in 3rd quadrant
The shape of the curve is shown in Fig. 1.5.
2
2
2
Example 4. Trace the curve x 3 + y 3 = a 3 (Astroid).
2
2
2
F x I + FG y IJ = 1
H aK H aK
FG x2 IJ 1 3 + FG y2 IJ 1 3 = 1
H a2 K H a 2 K
Sol. We have x 3 + y 3 = a 3 ⇒
or
23
23
...(i)
1 . Symmetry: Since there are even powers of x and y so the curve is symmetric about both
axis.
2. Origin: Since there is constant term so the curve does not passes through the origin.
3. Intercept: Putting y = 0, we get x = ± a.
⇒ The curve cross x-axis at (a, 0) and (–a, 0)
Similarly, putting x = 0 then y = ± a i.e., the curve cross y-axis at (0, a) and (0, –a).
FG yIJ ⇒ FGH dyIJK = 0 i.e., tangent
dx
H xK
F dyI = – ∞ = tan F − π I i.e.,
at (a, 0) is along x-axis and GH dx JK
H 2K
dy
4. Derivative:
=–
dx
1
3
w
( a, 0)
`> FNJ G
(0, a)
tangent at (0, a) is y-axis.
Fy I
F x2 I 3
5. Region: G J = 1 – G 2 J
Ha K
Ha K
2
1
3
1
a
F J>NG
2
∴ if
y2
x
> 1 i.e., x > a, we have 2 < 0 i.e., y2 < 0 = y is
a
a
_>
m
F J>NG
v
b
FNJ> G
Fig. 1.6
imaginary, so the curve does not exist for x > a. Similarly, the curve does not exist for x < – a and
y > a, y < – a.
6. Asymptotes: No asymptotes.
The shape of the figure is shown in the Fig. 1.6.
Example 5. Trace the curve x2y2 = a2 (y2 – x2).
Sol. 1. Symmetry: The curve is symmetric about both axis.
2. Origin: The curve passes through origin.
69
DIFFERENTIAL CALCULUS-I
3 . Tangents: The tangents at the origin y2 – x2 = 0. i.e.,
y = ± x.
4. Intercepts: Cross the line at origin i.e., (0, 0).
5. Region: y2 = a2x2/(a2 – x2) this shows that the curve
does not exist for x2 > a2 i.e., for x > a and x < – a.
6. As x → a, y2 → ∞ and as x → – a, y2 → ∞
The shape of the curve is shown in Fig. 1.7.
w
³>[>
m
Example 6. Trace the curve y2 (a + x) = x 2 (b – x),
(Strophoid).
Sol. 1. Symmetry: There is only even power of y so
the curve is symmetric about x-axis.
2. Origin: The curve passes through the origin.
3. Tangents at origin: The lowest degree term is
ay2 – bx2
b
∴ tangents are ay2 – ax2 = 0 ⇒ y = ±
x.
a
4. Intercepts: Putting
y = 0, so x2 (b – x) = 0
⇒
x = 0, b
The curve meets x-axis at (0, 0) and (b, 0). y-intercept:
put x = 0 then y = 0. So (0, 0) is the y-intercept.
5. Asymptotes: Asymptotes parallel to y-axis is
x + a = 0 ⇒ x = – a.
³>[>
w¢
Fig. 1.7
w
³>[>¡
m
³>[>>
F¡J>NG
b−x
, y becomes imaginary
a+x
when x > b and x < – a. Thus curve exist only in the region
– a < x < b.
6. Region: y = ± x
7. Derivative:
( −2 x 2 − 3 ax + bx + 2 ab)
dy
=
⇒
2( a + x) 3/2 (b − x )
dx
FG dyIJ
H dxK
v
Fig. 1.8
(b, 0)
= ∞ = tan
π
2
Thus, the tangent at (b, 0) is parallel to y-axis.
Therefore from above calculations, we draw the curve (Fig. 1.8) of given equation.
Example 7. Trace the curve y2 (x2 + y2) + a2 (x2 – y2) = 0.
Sol. 1. Symmetry: The curve is symmetric about both axis.
2. Origin: Curve passes through origin.
3. Tangents at origin: The tangents at the origin are x2 – y2 = 0 or y = ± x.
4. Intercepts: At x-axis = (0, 0)
At y-axis ⇒ y4 – a2y2 = 0
(put x = 0)
y2 (y2 – a2) = 0
⇒
y = ± a, 0
Thus, the curve cross y-axis at (0, a) and (0, – a) and (0, 0).
v
70
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
5 . Tangent at new point (0, – a) and (0, a)
Putting
y = y + a, we get
(y + a)2 x 2 + ( y + a) 2 + a 2 x 2 − ( y + a) 2
FNJ> G
w
´>[>³
´>[>³
=0
³>[>>
³>[>
∴ The tangent at the new points are y = 0, y = 0
v
i.e., parallel to x-axis.
m
2
2
2
2
2
2
6 . Region: x = y (a – y )/(a + y )
The curve does not exist when y2 > a2
or y > a and y < – a
FNJ> G
∴ The curve exist in the region when – a < y < a.
7 . No asymptotes: The shape of the curve shown in Fig. 1.9.
Fig. 1.9
Example 8. Trace the curve x2y2 = a2 (x2 + y2).
Sol. 1. Symmetry: Curve is symmetric about both axis.
2. Origin: Curve passes through the origin. Tangnt at (0, 0) are x2 + y2 = 0 ⇒ y = ± ix, which
give imaginary tangents. So (0, 0) is a conjugate point.
w
3. The curve does not cross the axis.
4. Asymptotes: Asymptotes are x = ± a and y = ± a.
5. Region: y2 = a2x2/(x2 – a2). If x2 < a2 i.e., x < ± a
´>[>
then y is imaginary i.e., the curve does not exist when
x < a and x < – a.
v¢
v
m
Similarly does not exist when y < a and y < – a
´>[>>
´>[>>
6. y2 → ∞ as x → a and x2 → ∞ as y → a
∴ The shape of the curve is shown in Fig. 1.10.
Example 9. Trace the curve (x2 – a2) (y2 – b2) = a2b2.
w¢
Sol. The given curve is (x2 – a2) (y2 – b2) = a2b2
or
Fig. 1.10
x2y2 – b2x2 – a2y2 = 0 or x2y2 = b2x2 + a2y2
1. Symmetry about both the axes.
w
³>[>
2. (0, 0) satisfies the equation of the curve.
Tangents at the origin are a2y2 + b2x2 = 0, which give
imaginary tangents. So (0, 0) is a conjugate point.
´>[>¡
3. The curve does not cross the axes.
4. Equating the coefficients of highest powers of
x and y we find that x = ± a and y = ± b are the asymptotes.
v¢
v
m
´>[>>¡
5. Solving for y, we get y2 =
b x
.
x 2 − a2
∴ If x2 < a2 i.e., x is numerically less than a, y2 is
negative i.e., y is imaginary i.e., the curve does not exist
between the lines x = − a and x = a.
³>[>>
2 2
w¢
Fig. 1.11
Similarly arguing we find that the curve does not exist between the lines y = – b and y = b.
6. y2 → ∞ as x → a and x2 → ∞ as y → a.
With the above data, the shape of the curve is as shown in Fig. 1.11.
71
DIFFERENTIAL CALCULUS-I
Example 10. Trace the curve y2 (a – x) = x2 (a + x).
Sol. 1. Symmetry about x-axis.
2. Passes through (0, 0), the tangents are y2 = x2 or y = ±x. Tangents being real and distinct,
node is expected at the origin.
3. Curve crosses the x-axis at (– a, 0) and (0, 0). Shifting the origin to (–a, 0) and equating
the lowest degree terms to zero, we get new y-axis as the tangent at the new origin.
4. x = a is the asymptote.
5. For x < – a, the curve does not exist. Similarly for x > a, the curve does not exist.
6. As x → a, y2 → ∞.
7. No point of inflexion.
∴ Shape of the curve is as shown in Fig. 1.12.
x=a
Y
(–a, 0)
(a, 0)
O
X¢
X
Y¢
Fig. 1.12
EXERCISE 1.7
1
1
1
2. x 2 + y 2 = a 2
1 . Trace the curve y2 (2a – x) = x3
(U.P.T.U. (C.O.), 2003)
(U.P.T.U., 2004)
Y
Y
O
X
Asymptote
Ans.
=
(0, a)
y
Ans.
x
Tangent
a/4
O a/4
(a, 0)
X
72
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
F xI F yI
3 . H K + GH JK = 1
a
b
2
3
2
3
4 . x6 + y6 = a2x2y2
Y
Y
y = –x
(0, b)
y=x
b
Ans.
–a
a
Ans.
X
O
–(a, 0)
x
(a, 0)
–b
(0, –b)
5 . a2y2 = x3 (2a – x)
6 . 9ay2 = x(x – 3a)2
Y
Y
O
Ans.
(2a, 0)
7 . a4y2 = a2 x4 – x6
O
Ans.
X
X
8 . xy2 = 4a2 (2a – x)
Y
Ans.
3a
Y
(–a, 0)
–a
O
a
Ans.
X
2a
O
(a, 0)
9 . y3 = x(a2 – x2)
X
(2a, 0)
1 0 . y = 8a3/(x2 + 4a2)
Y
Y
A
Ans.
(a, 0)
(–a, 0)
X
O
(0 ,
2a)
y = 2a
Ans.
x1
x2
O Asymptote
X
73
DIFFERENTIAL CALCULUS-I
1.9
POLAR CURVES
The general form (explicit) of polar curve is r = f(θ) or θ = f(r) and the implicit form is F(r, θ) = 0.
Procedure:
1. Symmetry: (a) If we replace θ by −θ, the equation of the curve remains unchanged then
there is symmetry about initial line θ = 0 (usually the positive x-axis in cartesian form).
Example. r = a (1 ± cos θ).
(b) If we replace θ by π − θ, the equation of the curve remains unchanged, then there is a
π
symmetry about the line θ =
(passing through the pole and z to the initial line) which is usually
2
the positive y-axis in cartesian).
Example. r = a sin 3θ.
(c) There is a symmetry about the pole (origin) if the equation of the curve remains unchanged
by replacing r into – r.
Example. r2 = a cos 2θ.
(d) Curve is symmetric about pole if f(r, θ) = f(r, θ + π)
Example.
r = 4 tan θ.
(e) Symmetric about θ =
F
H
I
K
π
π
i.e., (y = x), if f(r, θ) = f r, − θ
4
2
F
H
I
K
3π
3π
i.e., (y = – x), if f (r, θ) = f r,
−θ
4
2
2 . Pole (origin): If r = f(θ1) = 0 for some θ = θ1 = constant then curve passes through the
pole (origin) and the tangent at the pole (origin) is θ = θ1.
(f) Symmetric about θ =
Example. r = a (1 + cos θ) = 0, at θ = π.
3 . Point of intersection: Points of intersection of the curve with initial line and line
θ=
π
π
are obtained by putting θ = 0 and θ =
.
2
2
4 . Region: If r2 is negative i.e., imaginary for certain values of θ then the curve does not
exist for those values of θ.
5 . Asymptote: If lim r = ∞ then an asymptote to the curve exists and is given by equation
θ→ α
r sin (θ – α) = f ′(α)
where α is the solution of
1
= 0.
f(θ)
6 . Tangent at any point (r, θ ): Tangent at this is obtained from tan φ =
rdθ
, where
dr
φ is the angle between radius vector and the tangent.
7. Plotting of points: Solve the equation for r and consider how r varies as θ varies from
0 to ∞ or 0 to – ∞. The corresponding values of r and θ give a number of points. Plot these points.
This is sufficient for tracing of the curve. (Here we should observe those values of θ for which r
is zero or attains a minimum or maximum value).
74
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 1. Trace the curve r2 = a2 cos 2θ
(U.P.T.U., 2000, 2008)
q = 3 p/4
Y
q=
π
Hence, the straight lines θ = ±
are the tangents
4
at origin to the curve.
3 . Intersection: Putting θ = 0
∴ r2 = a2 ⇒ r = ± a the curve meets initial line
to the points (a, 0) and (–a, π).
4 . As θ varies from 0 to π, r varies as given below:
θ = 0
30
45
90
135
150
180
2
2
2
2
2
r = a
a /2
0
– a
0
a /2
a2
←imaginary→
p/4
π
π
or θ = ±
2
4
q=
i.e., cos 2θ = 0 ⇒ 2θ = ±
p/2
Sol. 1. Symmetry: Since there is no change in the curve when θ replace by – θ. So the curve
is symmetric about initial line.
2 . Pole: Curve passes through the pole when r2 = a2 cos 2θ = 0
(–a, p)
X
(a, 0)
–3
p/4
O
q=
q = – p/4
Fig. 1.13
5 . Region: The above data shows that curve does
not exist for values of θ which lying between 45º and 135º.
Example 2. Trace the curve r = a sin 3θ
(U.P.T.U., 2002)
Sol. 1. Symmetry: The curve is not symmetric about the initial line.
2 . Origin: Curve passes through the origin
when r = 0
⇒
⇒
a sin 3θ = 0
3θ = 0, π, 2π, 3π, 4π, 5π
⇒
θ = 0,
π 2π
4 π 5π
,
, π,
,
3
3
3
3
are the tangents at the pole.
3 . Asymptote: No asymptote since r is finite for any value of θ.
4 . Region: Since the maximum value of sin 3θ is 1.
π 3 π 5π
,
,
, etc.
2
2
2
π 3 π 5π
or
θ=
,
,
, etc.
6
6
6
for which r = a (maximum value).
∴ The curve exist in all quadrant about the
lines θ =
q
q = p/2
q = 2p/3
=
5p
/
q = p/3
p/6
3θ =
6
q=
So,
a
a
(q = p )
π 3π 5 π
,
,
at distance r = a.
q = 7p/6
6
6
6
The shape of the curve is given in the (Fig. 1.14).
q = 4 p/3
q=0
O
q = 11p/4
q = 3 p/2
q = 5 p/3
Fig. 1.14
75
DIFFERENTIAL CALCULUS-I
Example 3. Trace the curve r = aeθ
Sol. 1. No symmetry about the initial line.
2 . As θ → ∞ , r → ∞ and r always positive.
3. Corresponding values of θ and r are given below:
θ =
r =
π/2
aeπa/2
0
a
π
aeπa
3π/2
ae3π/2
Y
q = p/2
q=0
X
q = p/2
2π
ae2πa
O
with the above data the shape of the curve is shown below:
Example 4. Trace the curve r2 cos 2θ = a2 (Hyperbola)
1
Sol. 1. Symmetry about pole and about the line θ = π .
2
2 . Changing to cartesian the equation becomes
q = 3 p/2
Fig. 1.15
q = p/2
Y
x2 – y2 = a2.
∴ The equation of the asymptotes are y = ± x or
1
π are its polar asymptotes.
4
3 . When θ = 0, r2 = a2 or r = ±a i.e., the points (a, 0)
and (– a, 0) lie on the curve. (Here co-ordinates of the points
are polar coordinates).
4 . Solving for r we get r2 = a2/cos 2θ. This shows
1
that as θ increases from 0 to π , r increases from a to ∞.
4
1
3
5 . For values of θ lying between π and π , r2 is
4
4
negative i.e., r is imaginary. So the curve does not exist for
1
3
π< θ < π.
4
4
q = p/4
θ=±
(a = p)
Fig. 1.16
p/4
q = p/2
q=
q = 3p/4
a
a
π
i.e., y-axis.
2
a
q=0
q=p
30°
45°
60°
90°
120°
q=
5p
/4
a
1
π or
2 . Putting r = 0, cos 2θ = 0 or 2θ = ±
2
1
1
θ = ± π , i.e., the straight lines θ = ± π are the tangents
4
4
to the curve at the pole.
3 . Corresponding values of θ and r are given below:
θ = 0°
X
q = – p/4
q = 3p/2
Example 5. Trace the curve r = a cos 2θ.
Sol. 1. Symmetry about the initial line and the line
θ=
(a = 0)
O
q=p
q = 7 p/4
q = 3p/2
Fig. 1.17
135°
150°
180°
1
1
1
1
− a
− a
a
a
0
– a
0
a
2
2
2
2
Plot these points and due to symmetry about the initial line the other portion can be
traced.
r =a
76
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 6. Trace the curve r = a (1 – cos θ) (cardoid).
Sol. 1. Symmetry: No change in equation when we replace θ by – θ. So the equation of
curve is symmetric about initial line.
2 . Pole (region): If the curve passes through origin then
r = 0 ⇒ a (1 – cos θ) = 0
cos θ = 1 = cos 0 ⇒ θ = 0.
⇒
Here, the straight line θ = 0 is tangent at origin.
3 . Intersection: Putting θ = 0 then r = 0
and putting
θ = π, then r = 2a
∴ Intersection points on initial line = (0, 0) and (2a, π).
4 . Region: It exists in all quadrant.
5 . Asymptotes: No. asymptotes.
6 . As θ increases from 0 to π, r also increases from 0 to 2a.
The corresponding values of r and θ given below:
θ= 0
60°
90°
120°
0 = π/2
(2a, π)
(0, 0)
θ=0
180°
a
3a
Fig. 1.18
r =a
a
2a
2
2
With the above data the shape of the curve is given in (Fig. 1.18).
Example 7. Trace the curve r = a + b cos θ, when a < b.
Sol. 1. Symmetry: The curve is symmetric about initial line.
2 . Origin: For origin r = 0 ⇒ a + b cos θ = 0 ⇒ cos θ = −
3 . Corresponding values of θ and r are given below:
π
4
θ= 0
r =a+b
a+
FG b IJ
H 2K
π
3
π
2
a+
a
F bI
H 2K
2π
3
Fa − bI
H 2K
a
b
3π
4
FG a − b IJ
H 2K
π
(a – b)
b
2π
then r is positive for all values of θ from 0 to
but r is negative when θ
2
3
3
2π
3π
= π or π. Here, r must vanish somewhere between θ =
and
. Let θ = α (lying between
4
3
4
2π
3π
and
).
3
4
q = 2 p/3
q = p/2
q = p/3
For which r = 0 then θ = α is the straight line which is
θ=
α
tangent to the curve at the pole and for values of θ lying between
α and π, r is negative and points corresponding to such values of
θ will be marked in the opposite direction on these lines as r is
θ= π O
negative for them.
A
Thus in this case a < b, the curve passes through the origin
Let a >
RS−F a I UV and form two loops, one inside the
T H bK W
other as shown in the Fig. 1.19.
when θ = α = cos–1
Fig. 1.19
77
DIFFERENTIAL CALCULUS-I
EXERCISE 1.8
1 . Trace the curve
r = a (1 + cos θ)
2 . Trace the curve
r = a sin 2θ
Y
(a, p/2)
q = 3 p/4
(a, p)
Ans.
X
O
(2a, 0)
q = p/2
q = p/4
q=p
Ans.
q= 0
0
q = 5 p/4
3 . Trace the curve
q = 7 p/4
4 . Trace the curve
r = 2a cos θ
r=
a sin 2 θ
cos θ
Y
Y
P (r, q)
x=0
r
Ans.
O
q
a
a
C (a, 0)
A
X
Ans.
A (a, 0)
X
O
[Hint: Change it in cartesian coordinates,
y2 (x – a) = – x3]
5 . Trace the curve
r = 2 (1 – 2 sin θ)
6 . Trace the curve
r = a cos 2θ
q = 3 p/4
q = p/2
q=p
O
q = p/2
q = p/4
X
Ans.
Ans.
(U.P.T.U., 2003)
q=p
q=0
q = 3p/2
q = 5 p/4
q = 3p/2
x
78
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
8 . Trace the curve
r2 cos 2θ = a2
7 . Trace the curve
r = a (1 + sin θ)
q = p/2
q = p/2
Y
q = p/4
(0, 2a)
Ans.
Ans.
O
(–a, 0)
q=0
X
(a, 0)
q=p
[Hint: Change in cartesian form x2 – y2 = a2]
1.10
O
(a, 0)
(a, p)
q = 3p/2
q = – p/4
PARAMETRIC CURVES
Let x = f1(t) and y = f2(t) be the parametric equations of a curve where t is a parameter.
Method I: Eliminate the parameter ‘t’ if possible and we shall get the cartesian equation
of curve which can easily traced.
Example. x = a cos t, y = a sin t ⇒ x2 + y2 = a2.
Method II: When the parameter ‘t’ cannot be eliminated.
1 . Symmetry: If x = f1(t) is even and y = f2(t) is odd then curve is symmetric about x-axis:
Similarly, if x = f1(t) is odd and y = f2 (t) is even then curve is symmetric about y-axis.
2 . Origin: Find ‘t’ for which x = 0 and y = 0.
3 . Intercept: x-intercept obtained for values of t for which y = 0, y-intercept for values
of t for which x = 0.
4 . Determine least and greatest values of x and y.
lim
lim
5 . Asymptotes: t → t x(t) = ∞, t → t1 y(t) = ∞, then t = t1 is asymptote.
1
6 . Tangents:
dy
dy
= ∞ (vertical tangent) and
= 0 (horizontal tangent).
dx
dx
Remark. If the given equations of the curves are periodic functions of t having a common
period, then it is enough to trace the curve for one period.
Example 1. Trace the curve
x = a (t – sin t), y = a (1 – cos t)
(Cycloid)
Sol. 1. Symmetry: Since x is odd and y is even so the curve is symmetric about y-axis.
2 . Origin: Putting x = 0 and y = 0, we get
and
0 = a (t – sin t) ⇒ t = 0
0 = a (1 – cos t) ⇒ cos t = 1 ⇒ t = 0, 2π, 4π, etc.
79
DIFFERENTIAL CALCULUS-I
3.
dx
dy
dy/dt
dy
a sin t
= a (1 – cos t),
= a sin t ∴
=
=
dt
dx/dt
dt
dx
a (1 − cos t)
dy
t
2a sin t/2.cos t/2
⇒
=
= cot
2
2
dx
a(1 − 1 + 2 sin t/2)
dy
Corresponding values of x, y,
for different values of t are given below:
dx
3π
π
π
2π
t = 0
2
2
π
3π
x = 0
a
aπ
a
2aπ
−1
+1
2
2
y = 0
a
2a
a
0
F
H
I
K
F
H
I
K
dy
= ∞
1
0
– 1
–∞
dx
4 . Tangents at y = 0 are vertical and at y = 2a is horizontal. Curve is periodic for period
2π in the interval [0, 2π]. Curve repeats over intervals of [0, 2 aπ] refer Fig. 1.20.
Y
2a
– 4ap
– 2ap
2ap
O
4ap
Fig. 1.20
Y
t = p/2 B (0, a)
Example 2. Trace the curve x = a cos3 t, y = a sin3 t. (Astroid)
Sol. 1. dy/dt = 3a sin2 t cos t.
dx/dt = –3a cos2 t sin t
2
A t=0
X¢ (–a, 0)
dy
dy dt
3 a sin t cos t
=
=
dx dt
dx
−3 a cos2 t sin t
∴
a
A¢
x= a
y= 0
π
4
a
2 2
a
2 2
π
2
0
a
3π
2
−a
2 2
a
2 2
π
– a
0
a
(a, 0)
X
t = 3 p/2
B¢ (0, –a)
Y¢
dy
or
= – tan t.
dx
2 . Corresponding values of x, y and dy/dx for different
values of t are given below:
t = 0
O
5π
4
a
2 2
−a
2 2
Fig. 1.21
3π
2
0
– a
7π
4
a
2 2
−a
2 2
2π
a
0
dy
= 0 – 1
–∞
1
0
–∞
1
1
0
dx
Plotting the above points and observing the inclinations of the tangents at these points
the shape of the curve is as shown in Fig. 1.21.
80
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 3.
Sol.
1.
x = a (t + sin t),
Y
y = a (1 – cos t)
dy/dt = a (sin t);
B t = –p
ap
dx/dt = a (1 + cos t)
∴
t = –p/2 t = p/2
2
a(2 cos
2 1
2
X
X¢
1
O t=0
2
Y¢
t)
Fig. 1.22
2 a sin t cos t
=
t=p
2a
dy
a(sin t )
dy dt
=
=
dx
a(1 + cos t )
dx dt
1
C
ap
1
t.
2
2 . Corresponding values of x, y and dy/dx for different values of t are given below:
or
t = – π
x = – πa
y = 2a
dy/dx = – ∞
dy|dx = tan
1
− π
2
1
– a ( π – 1)
2
a
– 1
1
π
2
0
0
1
π + 1)
2
a
1
a(
0
0
EXERCISE 1.9
Trace the following curves:
1 . x = a sin 2t (1 + cos 2t), y = a cos 2t (1 – cos 2t)
Y
C
A
O
Ans.
2 . x = a (t – sin t), y = a (1 + cos t)
X
B
Y
C
B
ap
ap
2a
t = p/2 t = 3p/2
Ans.
X¢
A t=p
O
Y¢
t = 2p
X
π
aπ
2a
∞
81
DIFFERENTIAL CALCULUS-I
3 . x = a cos t +
1
a log tan2
2
F t I , y = a sin t (Tractrix)
H 2K
Y
B (0, a)
t = p/2
a
t=0
X¢
Ans.
t = –p
X
O
a
(0, –a)
B¢
4. x =
a (t + t 3 )
1 + t4
,y=
a (t − t 3 )
1 + t4
Y
y=x
Ans.
X
(–a, 0)
(a, 0)
O
y = –x
EXPANSION OF FUNCTION OF SEVERAL VARIABLES
1.11
TAYLOR'S THEOREM FOR FUNCTIONS OF TWO VARIABLES
Let f(x, y) be a function of two independent variables x and y. If the function f(x, y) and its partial
derivatives up to nth order are continuous throughout the domain centred at a point (x, y). Then
LM ∂f (a, b) + k ∂f (a, b) OP
∂y Q
N ∂x
f(a + h, b + k) = f(a, b) + h
LM
OP
MN
PQ
∂ f ( a, b)
∂ f ( a, b)
∂ f ( a, b) O
1 L ∂ f ( a, b)
+ 3h k
+ 3hk
+k
PP+...
+ 3 Mh
∂y
∂x ∂y
∂x∂y
MN ∂x
Q
+
2
∂ 2 f ( a, b)
∂ 2 f ( a, b)
1
2 ∂ f ( a, b)
h2
2
hk
+
k
+
2
∂x∂y
∂y 2
∂x 2
3
3
2
3
3
2
2
3
3
2
3
3
Or
LM ∂ + k ∂ OP f (a, b) + 1 LMh ∂ + k ∂ OP f(a, b)
∂y Q
2 N ∂x
N ∂x ∂y Q
∂O
1 L ∂
+ 3 Mh ∂x + k ∂y P f ( a, b)+...
N
Q
2
f(a + h, b + k) = f(a, b) + h
3
82
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Proof. Suppose P(x, y) and Q (x + h, y + k) be two neighbouring points. Then f (x + h, y + k),
the value of f at Q can be expressed in terms of f and its derivatives at P.
Here, we treat f(x + h, y + k) as a function of single variable x and keeping y as a constant.
Expanded as follows using Taylor's theorem for single variable.*
∂f ( x , y + k ) h 2 ∂ 2 f ( x , y + k )
+
+...
∂x
2
∂x 2
f(x + h, y + k) = f(x, y + k) + h
...(i)
Now expanding all the terms on the R.H.S. of (i) as function of y, keeping x as constant.
LM f (x, y) + k ∂f (x, y) + k ∂ f (x, y) +...OP
2 ∂y
∂y
MN
PQ
O
∂ f ( x , y ) k ∂ f ( x , y)
∂ L
+
+ …P
+ h M f ( x , y) + k
2 ∂y
∂x M
∂y
PQ
N
O
∂f ( x, y ) k ∂ f ( x, y )
h ∂ L
f ( x, y ) + k
+
+...P+...
+
M
∂y
2 ∂x M
2 ∂y
PQ
N
L ∂f (x, y) + k ∂f (x, y) OP
f(x + h, y + k) = f(x, y) + M h
∂y Q
N ∂x
∂ f ( x , y) O
∂ f ( x , y)
1 L ∂ f ( x , y)
h
+k
+ 2hk
+
M
PP + ...
2 MN
∂x∂y
∂y
∂x
Q
2
2
f(x + h, y + k) =
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
For any point (a, b) putting x = a, y = b in above equation then, we get
LM ∂f (a, b) + k ∂f (a, b) OP
∂y Q
N ∂x
f (a + h, b + k) = f (a, b) + h
+
LM
MN
OP
PQ
2
∂ 2 f ( a, b)
1 2 ∂ 2 f ( a, b)
2 ∂ f ( a, b)
+ ...
h
2
hk
k
+
+
2
∂x 2
∂y 2
∂y 2
Or
LM ∂ + k ∂ OP f (a, b) + 1 LMh ∂ + k ∂ OP f (a, b)
∂y Q
2 N ∂x
N ∂x ∂y Q
∂O
1 L ∂
h + k P f ( a, b) + ...
+
Hence proved.
M
∂y Q
3 N ∂x
2
f (a + h, b + k) = f (a, b) + h
3
Alternative form:
Putting
a+h = x⇒h=x–a
b+k = y⇒k=y–b
LM
N
then f (x, y) = f (a, b) + ( x − a)
∂
∂
+ ( y − b)
∂x
∂y
OP f (a, b) + 1 L(x − a) ∂ + (y − b) ∂ O f (a, b) + ... ...(ii)
2 MN
∂x
∂y PQ
Q
2
* Taylor's theorem for single variable
f (x + h) = f (x) + h
∂f h 2 ∂2 f h 3 ∂ 3 f
+ ...
+
+
∂x
2 ∂x 2
3 ∂x 3
83
DIFFERENTIAL CALCULUS-I
1.11.1 Maclaurin's Series Expansion
It is a special case of Taylor's series when the expansion is about the origin (0, 0).
So, putting a = 0 and b = 0 in equation (2), we get
LM ∂ + y ∂ OP f (0,0) + 1 LMx ∂ + y ∂ OP f (0,0) + ...
2 N ∂x
∂y Q
N ∂x ∂y Q
π
Example 1. Expand e cos y about the point F 1, I
(U.P.T.U., 2007)
H 4K
Sol. We have
f (x, y) = e cos y
...(i)
π
π
e
π
F I = e cos =
, f 1,
and
a = 1, b =
H 4K
4
4
2
F πI
∂f 1,
∂f
H 4 K = e cos π = e
∴ From (1)
= e cos y ⇒
2
f (x, y) = f (0, 0) + x
x
x
x
∂x
4
∂x
2
F I
H K = – e sin π = – e
π
∂f 1,
∂f
4
x
∂y = – e sin y ⇒
∂y
∂2 f
∂x 2
4
2
F πI
H 4 K = e , ∂ f = – e sin y
∂ 2 f 1,
= ex cos y ⇒
2
∂x 2
2
x
∂x∂y
F πI
H 4 K = – e , ∂ f = – e cos y = – e .
∂ 2 f 1,
2
∂x∂y
By Taylor's theorem, we have
F
H
π
f 1 + h, + k
4
2
x
∂y 2
2
F ∂f F1, π I ∂f F1, π I I
I = f F1, π I + GG h H 4 K + k H 4 K JJ + ...
K
H 4 K H ∂x
∂y K
π
π
+k=y⇒k=y–
, equation (2) reduce in the form
4
4
e
e
e
1
π
e
−
( x − 1) 2 ⋅
+ (x – 1) ·
+ O−
+
2
2
4
2
2
2
Let 1 + h = x ⇒ h = x – 1 and
f ((x, y) = ex cos y =
...(ii)
F
H
I FG
K H
IJ
K
LM
N
F y − π I FG − e IJ + F y − π I FG − e IJ OP + ...
H 4 K H 2 K H 4 K H 2 K PQ
e L
F π I + (x − 1) − (x − 1) F y − π I − F y − π I + ...OP .
⇒ f (x, y) =
1 + (x − 1) − y −
M
H 4K 2
H 4 K H 4K Q
2N
2
+ 2 (x – 1)
2
2
Example 2. Expand f (x, y) = ey log (1 + x) in powers of x and y about (0, 0)
Sol. We have
f (x, y) = ey log (1 + x)
Here,
a = 0 and b = 0, then f (0, 0) = e0 log 1 = 0
Now,
∂f
ey
=
∂x
1+ x
∂f
y
∂y = e log (1 + x)
⇒
∂f
( 0, 0 ) = 1
∂x
⇒
∂f (0,0)
= 0
∂x
84
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
∂2 f
ey
=
∂x∂y
1+ x
⇒
∂2 B
∂N 2
ey
(1 + x ) 2
⇒
= ey log (1 + x)
⇒
∂2 f
∂y
2
= −
∂2 f (0,0)
∂x∂y
= 1
∂2 f (0,0)
= – 1
∂x2
∂2 f (0,0)
= 0
∂y2
Now, applying Taylor's theorem, we get
FG h ∂f (0,0) + k ∂f (0,0) IJ + 1 FG h ∂ f (0,0)
∂y K
2H
∂x
H ∂x
∂ f (0, 0)
∂ f (0, 0) I
+k
+ 2hk
JK + ...
∂x∂y
∂y
2
2
2
2
f (0 + h, 0 + k) = f (h, k) = f (0, 0) +
2
2
2
Let h = x, k = y, then, we get
f (x, y) = ey log (1 + x) = f (0,0) +
= 0 + (x × 1 + y × 0) +
⇒
FG x ∂f (0, 0) + y ∂f (0, 0) IJ + ...
∂y K
H ∂x
1
[x2 (–1) + 2xy × 1 + y2 × 0] + ...
2
x2
+ xy + ... .
2
ey log (1 + x) = x –
Example 3. Find Taylor’s series expansion of function f (x, y) = e − x − y . cos xy about the
point x0 = 0, y0 = 0 up to three terms.
(U.P.T.U., 2006)
2
2
Sol. We have
f (x, y) = e − x − y cos xy.
Now, we get the following terms f (0, 0) = 1.
∂f
∂ f ( 0, 0 )
−x 2 − y 2
= – e
(2x cos xy + y sin xy) ⇒
=0
∂x
∂x
∂f
∂ f ( 0, 0 )
−x 2 − y 2
(2y cos xy + x sin xy) ⇒
=0
∂y = – e
∂y
2
Similarly,
∂ 2 f (0, 0)
= 0,
∂x∂y
∂3 f (0,0)
∂x ∂y
2
∂ f ( 0, 0 )
= 0,
2
∂ 2 f ( 0 ,0 )
∂x 2
3
∂ f (0, 0)
∂x∂y
2
= −2,
= 0,
∂ 2 f (0, 0)
∂y 2
=–2
∂ 3 f ( 0, 0 )
=0
∂x 3
3
∂y 3
= 0
Applying Taylor's theorem
FG h ∂ + k ∂ IJ f (0, 0) + 1 FG h ∂ + k ∂ IJ f (0, 0)
∂y K
2 H ∂x
H ∂x ∂y K
∂I
1 F ∂
h + k J f (0, 0) + ...
+
G
∂y K
3 H ∂x
2
f (h, k) = f (0, 0) +
3
85
DIFFERENTIAL CALCULUS-I
Putting h = x, k = y and all values then, we get
f (x, y) =
e − x − y · cos xy = 1 + (x × 0 + y × 0)
1 2
x ( −2) + 2 xy × 0 + y 2 ( −2)
+
2
2
2
+
⇒
2
1 3
x × 0 + 3x 2 y × 0 + 3xy 2 × 0 + y 3 × 0
3
2
A − N − O . cos xy = 1 – x2 – y2 .... .
Example 4. Find Taylor's expansion of f (x, y) = cot–1 xy in powers of (x + 0. 5) and
(y – 2) up to second degree terms. Hence compute f (– 0.4, 2.2) approximately.
Sol. Here f (x, y) = cot–1 xy
Now
f (– 0.5, 2) = cot–1(–1) =
3π
4
−y
∂f
=
∂x
1 + x2y2
⇒
∂f ( −0.5, 2 )
= – 1
∂x
−x
∂f
=
1 + x2y2
∂y
⇒
∂f ( −0.5, 2)
= 1/4
∂y
∂2 f
( x 2 y 2 − 1)
=
∂x∂y
(1 + x 2 y 2 )2
⇒
∂ 2 f ( −0.5, 2)
= 0
∂x∂y
∂2 f
∂x 2
⇒
∂2 f
∂y
2
=
=
2 xy 3
(1 + x 2 y 2 )2
2x 3 y
⇒
(1 + x y )
2 2 2
∂ 2 f ( −0.5, 2)
∂x 2
∂ 2 f ( −0.5, 2)
∂y
2
= – 2
= –
1
8
Now applying Taylor's series expansion, we get
FG ∂ + k ∂ IJ ⋅ f (−0.5, 2) + 1 FG h ∂ + k ∂ IJ f (–0.5, 2) +…
∂y K
H ∂x ∂y K
2 H ∂x
2
f (– 0.5 + h, 2 + k) = f (–0.5, 2) + h
Let
∴
or
– 0.5 + h = x ⇒ h = x + 0.5
2+k = y⇒k=y–2
|As a = – 0.5, b = 2
π
3
1
f (x, y) = cot–1 xy =
+ (x + 0.5) (–1) + (y – 2) ×
4
4
1
1
(x + 0.5) 2 ( −2) + 2 (x + 0.5)( y − 2) × 0 + ( y − 2) 2 −
.
+
2
8
LM
N
I OP
KQ
3π
1
1
– (x + 0.5) +
(y – 2) – (x + 0.5)2 –
(y – 2)2 + ....
4
4
16
x = – 0.4 and y = 2.2
f (x, y) =
Putting
F
H
3π
0.2
1
– (0.1) +
– (0.1)2 –
(0.2)2
4
4
16
= 2.29369.
f (– 0.4, 2.2) =
86
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 5. Calculate log (1.03)1/3 + ( 0.98 ) 1/ 4 − 1 approximately by using Taylor's expansion up to first order terms.
FH x + y − 1IK
1
3
Sol. Let f (x, y) = log
1
4
f (1, 1) = log 1 = 0
∂f
=
∂x
Now,
−
3
FH
2
1× x 3
IK
1
1
x3 + y4 −1
∂f
(1,1) =
∂x
⇒
1
3
Taking a = 1, b = 1
∂f
∂y =
3
−
1× y 4
F
I
4 H x + y − 1K
1
3
1
4
∂f
(1,1) =
∂x
⇒
1
4
Now, applying Taylor's theorem
FG h ∂ + k ∂ IJ f (1,1) + ...
H ∂x ∂y K
L
O
f (1 + h, 1 + k) = log M(1 + h) + (1 + k ) − 1P
N
Q
1
1
L
O
f (1 + h, 1 + k) = log M(1 + h) + (1 + k ) − 1P = 0 + h ×
+k×
N
Q
3
4
f (1 + h, 1 + k) = f (1, 1) +
But
∴
1
3
1
4
1
3
1
4
...(i)
Putting h = 0.03 and k = – 0.02 in equation (i) then, we get
LM
N
1
1
OP = 0.03 × 1 + (– 0. 02) × 1
Q
3
4
log (1.03 ) 3 + ( 0.98 ) 4 − 1
= 0.005.
Example 6. Expand xy in powers of (x – 1) and (y – 1) up to the third degree terms.
(U.P.T.U., 2003)
Sol.
Here f (x, y) = xy
Now
∂f
= yxy–1
∂x
⇒
∂f (1, 1)
= 1
∂x
∂f
y
∂y = x log x
⇒
∂f (1, 1)
= 0
∂y
∂2 f
= xy–1 + yxy–1 · log x
∂x∂y
⇒
∂ 2 f (1, 1)
= 1
∂x∂y
∂2 f
∂x 2
= y (y – 1) xy–2
⇒
= xy · (log x)2
⇒
∂2 f
∂y 2
f (1, 1) = 1
∂2 f (1, 1)
∂x2
∂ 2 f (1, 1)
∂y 2
= 0
= 0
87
DIFFERENTIAL CALCULUS-I
∂3 f
∂x∂y
2
∂3 f
∂x 2 ∂y
∂3 f
∂x 3
∂3 f
∂y 3
= yxy–1(log x)2 + 2xy–1 · log x
⇒
= (y – 1)xy–2 + y(y – 1) xy–2 · log x + yxy–2
⇒
= y(y – 1) (y – 2)xy–3
⇒
= xy(log x)3
⇒
∂ 3 f (1, 1)
∂x∂y 2
∂3 f (1, 1)
∂x 2∂y
∂3 f (1,1)
∂x3
∂3 f (1, 1)
∂y3
= 0
= 1
= 0
= 0
Now applying Taylor's theorem, we get
FG h ∂ + k ∂ IJ f (1, 1) + 1 FG h ∂ + k ∂ IJ f (1, 1)
H ∂x ∂y K
∂y K
2 H ∂x
1 F ∂
+
G h + k ∂∂y IJK · f (1, 1) + ...
3 H ∂x
2
f (1 + h, l + k) = f (1, 1) +
3
Let
⇒
1 + h = x and 1 + k = y
h = x – 1 and k = y – 1
∂
∂O
∂
∂O
1 L
L
f (x, y) = f (1, 1) + M( x − 1) ∂x + ( y − 1) ∂y P f + 2 M( x − 1) ∂x + ( y − 1) ∂y P f
N
Q
N
Q
∂
∂O
1 L
+ 3 M( x − 1) ∂x + ( y − 1) ∂y P f + ...
N
Q
2
∴
3
Using all values in above equation, we get
f (x, y) = xy = 1 + (x – 1) + 0 +
+
1
0 + 2 ( x − 1)( y − 1) + 0
2
1
0 + 3 ( x − 1)2 ( y − 1) + 0 + 0
3
= 1 + (x – 1) + (x – 1) (y – 1) +
Example 7. Obtain Taylor's expansion of tan–1
second degree terms. Hence compute f (1.1, 0.9).
Sol. Here
f (x, y) = tan–1
Now,
y
x
∂f
y
= –
2
∂x
(x + y 2 )
1
( x − 1)2 ( y − 1) .
2
y
about (1, 1) up to and including the
x
(U.P.T.U., 2002, 2005)
∴
f (1, 1) =
π
4
⇒
∂f (1, 1)
=
∂x
−
1
2
88
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
b g = 12
∂f
x
∂y = ( x 2 + y 2 )
⇒
∂f
1, 1
∂y
∂2 f
∂x 2
⇒
∂2 f
1, 1
∂x 2
∂2 f
∂y
2
=
=
2 xy
(x + y )
2
2 2
−2 xy
∂2 f
⇒
(x + y )
2
b g = 12
2 2
2
∂ B
y2 − x2
=
∂N∂O
( x 2 + y 2 )2
By Taylor’s theorem
∂y 2
(1, 1) =
−
1
2
∂2 f
= 0
∂x∂y
⇒
LMh ∂ + k ∂ OP f (1, 1) + 1 LMh ∂ + k ∂ OP f (1,1) + ...
∂y Q
2 N ∂x
N ∂x ∂y Q
2
f (1 + h, 1 + k) = f (1, 1) +
Let
l+h = x⇒ h= x–1
1+k = y⇒k =y–1
∴
f (x, y) = f (1, 1) + ( x − 1)
LM
OP f + ...
N
Q
π
1 L
F 1 I + 2(x − 1)( y − 1)
1
1
( x − 1)
=
+ (x – 1) F − I + (y – 1) F I +
M
H 2K
H
K
H
K
2
4
N
2
2
F 1IO
× 0 + ( y − 1) − P
H 2KQ
LM
N
OP
Q
∂
∂
1
∂
∂
( x − 1)
+ ( y − 1)
+ ( y − 1)
f+
∂x
∂y
2
∂x
∂y
2
2
2
or
1
π
1
1
1
–
(x – 1) +
(y – 1) +
(x – 1)2 –
(y – 1)2 + … .
4
4
2
2
4
Putting x = 1.1, y = 0.9, we get
f (x, y) =
π 1
1
1
1
–
· (1.1 – 1) +
(0.9 – 1) +
(1.1 – 1)2 –
(0.9 – 1)2
4 2
2
4
4
= 0.785 – 0.05 – 0.05 + 0.0025 – 0.0025
= 0.685.
f (1.1, 0. 9) =
Example 8. Find Taylor's expansion of
Sol. Here
∴
f (x, y) =
1 + x + y 2 in power of (x – 1) and (y – 0).
1 + x + y2
f (1, 0) =
2
∂f
1
=
∂x
2 1 + x + y2
⇒
∂
B (1, 0) = 1/2 2
∂N
∂f
=
∂y
1+ x+ y
⇒
∂
B (1, 0) = 0
∂O
∂2 f
∂x 2
1
4 (1 + x + y 2 ) 3/2
⇒
∂2 B
1
1, 0 = –
2
4.2 3/2
∂N
∂2 f
∂y
2
= –
=
y
2
1
1 + x + y2
–
y2
(1 + x + y )
2 3/2
⇒
b g
∂ 2 f (1, 0)
∂y
2
=
1
2
89
DIFFERENTIAL CALCULUS-I
y
∂2 f
= – 2(1 + x + y 2 ) 3/2
∂x∂y
By Taylor's theorem, we get
∂ 2 f (1, 0)
= 0.
∂x∂y
⇒
LM ∂ + k ∂ OP f (1, 0) + 1 Lh ∂ + k ∂ O f (1, 0) + ...
∂y PQ
2 MN ∂x
N ∂x ∂y Q
2
f (1 + h, 0 + k) = f (1, 0) + h
Let
⇒ f (x, y) =
1 + h = x ⇒ h = x – 1, k = y
1
1
1
1
(x − 1)2 −
+ ( x − 1)y × 0 + y2.
+ ...
2 + ( x − 1) ⋅
+ y×0 +
3/2
2
4.2
2 2
2 2
=
LM
N
2 1+
FG
H
IJ
K
OP
Q
x − 1 ( x − 1) 2 y 2
−
+
+ ... .
4
32
4
EXERCISE 1.10
1 . Expand f (x, y) = x2 + xy + y2 in powers of (x – 1) and (y – 2).
[Ans. f (x, y) = 7 + 4 (x – 1) + 5 (y – 2) + (x – 1)2 + (x – 1) (y – 2) + (y – 2)2 + ...]
2 . Evaluate tan–1
F 0.9 I .
H 1.1 K
[Ans. 0.6904]
3 . Expand f (x, y) = sin (xy) about the point (1, π/2) up to and second degree term.
LMAns. f (x, y) = 1 − π (x − 1) − π (x − 1)F y − π I − 1 F y − π I OP + ...
H 2K 2 H 2K Q
8
2
N
2
2
2
4 . Obtain Taylor's expansion of x2y + 3y – 2 in powers of (x – 1) and (y + 2).
[Ans. f (x, y) = – 10 – 4 (x – 1) + 4 (y + 2) – 2(x – 1)2 + ...]
5 . Expand exy in powers of (x – 1) and (y – 1).
LMAns. eRS1 + (x − 1) + ( y − 1) + (x − 1) + (x − 1)(y − 1) + (y − 1) + ...UVOP
2
2
T
WQ
N
2
2
6 . Expand cosx cosy in powers of x and y.
LMAns. f (x, y) = 1 − 1 (x + y ) + 1 (x + 6 x y + y ) + ...OP
2
24
N
Q
2
7 . Expand f (x, y) = e2x cos 3y up to second degree.
2
4
2 2
[Ans. 1 + 2x + 2x2 –
4
9 2
y + ...]
2
8 . Obtain Taylor's series expansion of (x + h) (y + k) /(x + h + y + k)
LMAns. xy + hy + kx − h y + 2hkxy − k x + ...OP
MN x + y (x + y) (x + y) (x + y) (x + y) (x + y) PQ
2
2 2
2
2
2
2 2
3
3
3
9 . Obtain Taylor's expansion of (1 + x – y)–1 in powers of (x – 1) and (y – 1).
[Ans. 1 – x + y + x2 – 2xy + y2 + ...]
LMAns. y + xy − y + (x y − xy ) + y + ...OP
2
2
3
N
Q
2
1 0 . Find Machaurin's expansion of ex log (1 + y).
2
2
3
90
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
OBJECTIVE TYPE QUESTIONS
A. Pick the correct answer of the choices given below:
∂ 3u
is
∂x∂y∂z
1 . If u = exyz, then
(i) (x2yz + x)
(ii) (x2yz – x)exyz
(iii) exyz.xy
(iv) (1 + 3xyz + x2y2z2)exyz
x− y
2 . If u = sin–1
x+ y
, then
∂u
is equal to
∂x
(i) −
x ∂u
y ∂y
(ii)
(iii) −
y ∂u
x ∂y
(iv) xy
3 . If z = log
x
(i) −
(iii)
x 2 + y 2 then
x +y
2
2
,
x +y
(ii)
2
y
x
,
x2 + y2 x2 + y2
4 . If u = ex2 + y2 + z2, then
(iv)
x
x2 + y2
x
x −y
2
∂ 3u
is
∂x∂y∂z
(i) 7xy
(ii) 6xyzu
(iii) – 8xyzu
(iv) 8xyzu
5 . If z = xn–1 y f(y/x) then x
∂ u
∂ u
+y
is
2
∂
y∂x
∂x
2
2
∂z
∂x
(ii) nz
(iii) n(n – 1)z
(iv) n
(i) n
6 . If f(x, y, z) = 0 then
∂u
∂y
∂z
∂z
and
are
∂y
∂x
y
2
∂u
∂y
∂z
∂y
∂x ∂y ∂z
is
⋅ ⋅
∂y ∂z ∂x
(i) –1
(ii) 1
(iii) 0
(iv) 2
7 . If u = x3 y2 sin–1 (y/x), then x
∂u
∂u
+y
is:
∂x
∂y
(i) 0
(ii) 6u
(iii) – 8u
(iv) 5u
2
,−
,
y
x2 + y2
y
x − y2
2
91
DIFFERENTIAL CALCULUS-I
8 . If u = xy f(y/x), then x
∂u
∂u
+y
is
∂x
∂y
(i) yu
(ii) xu
(iii) u
(iv) 2u
9 . The degree of u = [z2/(x4 + y4)]1/3 is
2
1
(i)
(ii)
3
3
−2
(iii)
(iv) 3
3
(i) t22
dz
is
dt
(ii) 0
(iii) 23t17
(iv) 23t22
1 0 . If z = x4 y5 where x = t2 and y = t3 then
∂y
∂z
and
at (1, –1, 2) are
∂x
∂x
1
1
(i) − , 1
(ii) − , − 1
2
2
1
1
, −1
,1
(iii)
(iv)
2
2
1 2 . The equation of curve x3 + y3 = 3axy inersects the line y = x at the point
1 1 . If x2 + y2 + z2 = a2 then
FG 3a, 3a IJ
H 2K
F 3a 3a I
(iii) G , J
H 2 2K
(i)
FG 3a , − 3a IJ
H 2 2K
F 3a , − 3a IJ
(iv) G −
H 2 2K
(ii)
1 3 . The equation of the curve x2y – y – x = 0 has maximum
(i) One asymptote
(ii) Four asymptotes
(iii) Three asymptotes
(iv) None of these
1 4 . The curve r2 cos 2θ = a2 is symmetric about the line
π
π
(i) θ =
(ii) θ = −
2
2
3π
(iii) θ = −
(iv) θ = π
2
1 5 . The parametric form of the curve x = a cost, y = a sin t is symmetric about
(i) x-axis
(ii) y-axis
(iii) both axis
(iv) about y = x
B. Fill in the blanks:
1 . The nth derivative of y = xn–1 log x at x =
2 . The nth derivative of y = x sin x is ..........
3 . If y = ex sin x, then y″ – 2y′ + 2y = ..........
4 . If y = sin 2x then y6 (0) = ...........
1
is ..........
2
92
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
5 . If y = sin hx, then y2n (x) = ..........
6 . If y = ex.x, then yn(x) = ...........
7 . If y = sin3 x, then yn (x) = ..........
8 . The nth derivative (yn) of y = x2 sin x at x = 0 is ..........
9 . If y = tan–1 x and n is even, then yn (0) = ..........
1 0 . If y = cos(m sin–1 x) and n is odd then yn (0) = ..........
1 1 . If y = (sin–1 x)2 and n is odd then yn (0) = ..........
1 2 . If z = f(x – by) + φ(x + by), then b 2
1 3 . If u = log(2x + 3y), then
∂2z
= ..........
∂x 2
∂ 2u
= ...........
∂x∂y
1 4 . If u = eax + by f(ax – by), then b
∂u
∂u
+a
= ..........
∂x
∂y
1 5 . If u = x2 + y2, x = s + 3t, y = 2s – t, then
∂u
= ..........
∂s
1 6 . If f(x, y) be a homogeneous function of degree n, then x 2
1 7 . If log u =
∂x
+ 2xy
2
∂2 f
∂2 f
+ y 2 2 = ..........
∂x∂y
∂y
∂u
∂u
x y
, then x
= ..........
+y
∂y
∂x
x+y
∂u
∂u
∂u
+y
+z
= ..........
∂x
∂y
∂z
1 9 . If f(x, y) = 0, φ(y, z) = 0, then
∂f ∂φ dz
= ..........
∂y ∂z dx
2 0 . If x = er cos θ, y = er sin θ then
x 3 − x 2 y + xy 2 + y 3
x − xy − y
2
2 2 . If u = log
∂2 f
2 2
1 8 . If u = (x2 + y2 + z2)1/2, then x
2 1 . If u =
(U.P.T.U., 2008)
2
x4 − y4
x +y
3
3 , then
∂θ
∂θ
= .......... and
= ..........
∂y
∂x
, then x
x
∂u
∂u
+y
= ..........
∂x
∂y
∂u
∂u
+y
= ..........
∂x
∂y
2 3 . If tlim
x(t) = ∞, tlim
y(t) = ∞, then t = .......... is asymptote.
→t
→t
1
1
2 4 . The curve r = a (1 – cos θ) is symmetric about the ..........
dy
= ..........
dx
2 6 . The curve x2y2 = a2 (y2 – x2) has tangents at origin ..........
2 5 . The tangent is parallel to x-axis if
2 7 . The curve y2 (a – x) = x2(a + x) exists if ..... x .....
2 8 . The curve y2 (x2 + y2) + a2 (x2 – y2) = 0, cross y-axis at ..... and .....
2 9 . If f(x, y) = ey log(1 + x), then expansion of this function about (0, 0) up to second degree
term is ..........
3 0 . tan–1 {(0.9) (–1.2)} = ..........
93
DIFFERENTIAL CALCULUS-I
3 1 . If f(x, y) = ex . sin y, then
b g = ..........
∂ 3 f 0, 0
∂x 3
3 2 . Expansion of exy up to first order tem is ..........
3 3 . f(x, y) = f(1, 2) + .........
C. Indicate True or False for the following statements:
1 . If u, v are functions of r, s are themselves functions of x, y then
a f = ∂au, vf × ∂bx, yg
b g ∂b r , s g ∂a r , s f
∂ u, v
∂ x, y
2 . Geometrically the function z = f(x, y) represents a surface in space.
3 . If f(x, y) = ax2 + 2hxy + by2 then x
∂f
∂f
+y
= 2f.
∂x
∂y
4 . If z is a function of two variables then dz is defined as dz =
∂z
∂z
dx +
dy .
∂x
∂y
5 . If f(x 1, x 2, ..., x n ) be a homogeneous function then x1
∂f
∂f
∂f
+ x2
+ ... + xn
∂x1
∂x 2
∂xn
= n (n – 1) f.
6 . If u =
x2 + y2
x −y
2
2
+ 4 , then x
∂u
∂u
+y
= 4.
∂x
∂y
7 . When the function z = f(x, y) differentiating (partially) with respect to one variable, other
variable is treated (temporarily) as constant.
8 . To satisfied Euler’s theorem the function f(x, y) should not be homogeneous.
∂z
∂z
9 . The partial derivatives
and
are interpreted geometrically as the slopes of the
∂y
∂x
tangent lines at any point.
10.
(i) The curve y2 = 4ax is symmetric about x-axis.
(ii) The curve x3 + y3 = 3axy is symmetric about the line y = – x.
(iii) The curve x2 + y2 = a2 is symmetric about both the axis x and y.
(iv) The curve x3 – y3 = 3axy is symmetric about the line y = x.
11.
(i) If there is no constant term in the equation then the curve passes through the origin
otherwise not.
d2 y
≠ 0 is called an inflexion point.
dx 2
(iii) If r2 is negative i.e., imaginary for certain values of θ then the curve does not exist for
those values of θ.
(ii) A point where
π
.
2
(i) Maclaurin’s series expansion is a special case of Taylor’s series when the expansion
is about the origin (0, 0).
(ii) Taylor’s theorem is important tool which provide polynomial approximations of real
valued functions.
(iv) The curve r = aeθ is symmetric about the line θ =
12.
94
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
(iii) Taylor’s theorem fail to expand f(x, y) in an infinite series if any of the functions
fx(x, y), fxx(x, y), fxy(x, y) etc., becomes infinite or does not exist for any value of x, y
in the given interval.
(iv) f(x, y) = f(a, b) +
LMax − af ∂ − ax − bf ∂ OP f(a, b) + ..... .
∂y Q
N ∂x
ANSWERS TO OBJECTIVE TYPE QUESTIONS
A. Pick the correct answer:
1 . (iv)
2 . (iii)
4 . (iv)
5 . (i)
7 . (iv)
8 . (iv)
1 0 . (iv)
1 1 . (i)
1 3 . (iii)
1 4 . (i)
B. Fill in the blanks:
1.
2 n −1
2.
3 . Zero
a
6 . ex x + n
9 . Zero
12.
∂ 2z
FG
H
x sin x +
4 . Zero
f
3 . (iii)
6 . (i)
9 . (iii)
1 2 . (iii)
1 5 . (iii)
IJ
K
FG
H
en − n2 j sin n2π
7 . Try yourself
8.
1 0 . Zero
1 1 . Zero
∂ 2u
∂y∂x
1 6 . n(n – 1)f
13.
∂y 2
1 5 . 2x + 4y
IJ
K
nπ
nπ
− n cos x +
2
2
5 . sin hx
1 4 . 2 abu
1 7 . 3u log u
18. u
∂f ∂φ
1 9 . ∂x ⋅ ∂y
20. −
21. u
2 4 . initial line
22. 1
25. 0
2 3 . t1
26. y = ± x
27. – a < x < a
2 8 . (0, a) and (0, – a)
3 0 . – 0.823
31. 0
sin 2 θ cos 2 θ
,
y
x
x2
+ xy
2
3 2 . e{1 + (x – 1) + (y – 1)}
29. x −
L R ∂f
O
∂f U 1 R
∂f U
∂f
x − 1f + b y ⋅ 2g V +......P
a
S
3 3 . MSaz − 1f + b y ⋅ 2g V +
∂y W 2 T
∂y W
∂x
MNT ∂x
PQ
2
C. True or False:
1. F
5. F
9. T
1 0 . (i) T
1 1 . (i) T
1 2 . (i) T
2. T
6. F
3. T
7. T
4. T
8. F
(ii) F
(iii) T
(iv) F
(ii) F
(ii) T
(iii) T
(iii) T
(iv) F
(iv) F
GGG
UNIT
11
Differential Calculus-II
2.1 JACOBIAN
The Jacobians* themselves are of great importance in solving for the reverse (inverse functions)
derivatives, transformation of variables from one coordinate system to another coordinate system
(cartesian to polar etc.). They are also useful in area and volume elements for surface and volume
integrals.
2.1.1 Definition
If u = u (x, y) and v = v (x, y) where x and y are independent, then the determinant
∂u
∂x
∂v
∂x
∂u
∂y
∂v
∂y
is known as the Jacobian of u, v with respect to x, y and is denoted by
a f
b g
∂ u, v
or J (u, v)
∂ x, y
as
Similarly, the Jacobian of three functions u = u (x, y, z), v = v (x, y, z), w = w (x, y, z) is defined
J (u, v, w) =
a
b
f
g
∂ u, v, w
=
∂ x, y , z
∂u
∂x
∂v
∂x
∂w
∂x
* Carl Gustav Jacob Jacobi (1804–1851), German mathematician.
95
∂u
∂y
∂v
∂y
∂w
∂y
∂u
∂z
∂v
∂z
∂w
∂z
96
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
2.1.2 Properties of Jacobians
1. If u =u (x, y) and v = v (x, y), then
a f × ∂ bx, yg = 1 or JJ′′ = 1
∂ a u , vf
∂ b x , yg
∂ bx, yg
∂ au, vf
and J′ =
J =
∂ au, vf
∂ bx, yg
∂ u, v
where,
Proof: Since
u = u (x, y)
v = v (x, y)
Differentiating partially equations (i) and (ii) w.r.t. u and v, we get
∂y
∂u
∂u
∂x
∂u
=1 =
·
+
×
∂u
∂x
∂u
∂y
∂u
Now,
a f
b g
∂u
=0 =
∂v
∂y
∂u ∂x
∂u
·
+
×
∂v
∂x
∂v
∂y
∂v
=0 =
∂u
∂v
=1 =
∂v
∂v
∂v
∂x
∂y
·
+ ∂y ×
∂x
∂u
∂u
∂v
∂x
∂
y
∂v
·
+
×
∂x
∂v
∂v
∂y
b g =
a f
∂ x, y
∂ u, v
×
∂ x, y
∂ u, v
∂u
∂x
∂v
∂x
∂u
∂y
∂v
∂y
×
∂x
∂u
∂y
∂u
(U.P.T.U., 2005)
...(i)
...(ii)
...(iii)
...(iv)
...(v)
...(vi)
∂x
∂v
∂y
∂v
=
∂u
∂u
∂y
∂x
∂x
∂y
∂
∂
u
u
∂v
∂v × ∂x
∂y
∂x
∂y
∂v
∂v
(By interchanging rows and columns in II determinant)
=
∂u ∂x ∂u ∂y
+
∂x ∂u ∂y ∂u
∂v ∂x ∂v ∂y
+ .
∂x ∂u ∂y ∂u
∂u ∂x ∂u ∂y
+
∂x ∂v ∂y ∂v
∂v ∂x ∂v ∂y
. +
∂x ∂v ∂y ∂v
(multiplying row-wise)
Putting equations (iii), (iv), (v) and (vi) in above, we get
a f
b g
b g =
a f
1
0
0
=1
1
JJ′ =
1.
Hence proved.
∂ x, y
∂ u, v
×
∂ x, y
∂ u, v
or
2. Chain rule: If u, v, are function of r, s and r, s are themselves functions of x, y i.e.,
u = u (r, s), v = v (r, s) and r = r (x, y), s = s (x, y)
a f =
b g
∂ u, v
∂ x, y
then
Proof: Here
and
u =
r =
a f
a f
a f
b g
∂ u, v
∂ r, s
⋅
∂ r, s
∂ x, y
u (r, s), v = v (r, s)
r (x, y), s = s (x, y)
97
DIFFERENTIAL CALCULUS-II
Differentiating u, v partially w.r.t. x and y
∂u
∂u ∂r
∂u ∂s
=
·
+
·
∂x
∂r ∂x
∂s ∂x
∂u
∂u ∂r
∂u ∂s
=
·
+
·
∂y
∂r ∂y
∂s ∂y
∂v
∂x
∂v
∂y
Now,
=
=
...(ii)
∂v ∂s
∂v ∂r
·
+
·
∂s ∂x
∂r ∂x
∂v ∂r
∂v ∂s
·
+
·
r
y
∂ ∂
∂s ∂y
a f · ∂ar, sf =
a f ∂bx, yg
∂u
∂r
∂v
∂r
∂u
∂s
∂v
∂s
=
∂u
∂r
∂v
∂r
∂u
∂s
∂v
∂s
∂ u, v
∂ r, s
...(i)
...(iii)
...(iv)
.
∂r
∂x
∂s
∂x
∂r
∂y
∂s
∂y
.
∂r
∂x
∂r
∂y
∂s
∂x
∂s
∂y
By interchanging the rows and columns in second determinant
=
∂u ∂r ∂u ∂s
+
∂r ∂x ∂s ∂x
∂v ∂r ∂v ∂s
+
∂r ∂x ∂s ∂x
∂u ∂r ∂u ∂s
+
∂r ∂y ∂s ∂y
∂v ∂r ∂v ∂s
+
∂r ∂y ∂s ∂y
|multiplying row-wise
Using equations (i), (ii), (iii) and (iv) in above, we get
a f
a f
a f =
b g
∂ u, v
∂ r, s
.
∂ r, s
∂ x, y
∂u
∂x
∂v
∂x
∂u
∂y
∂v
∂y
=
a f
b g
∂ u, v
.
∂ x, y
Or
a f = ∂ au, vf ⋅ ∂ ar, sf . Hence proved.
∂ ar, sf ∂ b x, yg
b g
∂ au, vf
Example 1. Find
, when u = 3x + 5y, v = 4x – 3y.
∂ bx, yg
∂ u, v
∂ x, y
Sol. We have
∴
Thus,
u = 3x + 5y
v = 4x – 3y
∂u
∂u
∂v
∂v
= 3,
= 5,
= 4 and
=–3
∂x
∂y
∂x
∂y
a f =
b g
∂ u, v
∂ x, y
∂u
∂x
∂v
∂x
∂u
∂y
∂v
∂y
=
3
4
5
= – 9 – 20 = – 29.
−3
98
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
a
b
Sol. We have
∴
f
g
∂ u, v, w
of the following:
∂ x, y, z
Example 2. Calculate the Jacobian
u = x + 2y + z, v = x + 2y + 3z, w = 2x + 3y + 5z.
u = x + 2y + z
v = x + 2y + 3z
w = 2x + 3y + 5z
∂u
∂u
∂u
∂v
∂v
∂v
= 1,
= 2,
= 1,
= 1,
= 2,
= 3,
∂x
∂y
∂z
∂x
∂y
∂z
(U.P.T.U., 2007)
∂w
∂w
∂w
= 2, ∂y = 3 and
= 5.
∂x
∂z
Now,
a
b
∂u
∂x
∂v
∂x
∂w
∂x
f
g
∂ u, v, w
=
∂ x, y, z
∂u
∂y
∂v
∂y
∂w
∂y
∂u
∂z
∂v
∂z
∂w
∂z
1
= 1
2
2
2
3
1
3
5
= 1 (10 – 9) – 2 (5 – 6) + 1 (3 – 4) = 2.
Example 3. Calculate
b g if u = 2yz , v = 3zx , w = 4xy .
y
x
z
∂ au, v, wf
Sol. Given
3zx
2yz
4xy
,v=
,w=
y
x
z
u =
∂ x, y, z
3zx ∂v 3x
−2 yz ∂u
2z ∂u
3z ∂v
2y ∂v
∂u
= ,
=
=
,
=
,
=
,
=− 2 ,
2 ,
∂z y
x
y
y
z
x
y
y
∂x
∂
∂
∂
∂
x
x
∴
∂w
4xy
4x
4y ∂w
∂w
=
,
=
and
=– 2
∂x
z
∂y
∂z
z
z
Now,
a
b
f
g
∂ u, v, w
=
∂ x, y, z
= –
∂u
∂y
∂v
∂y
∂w
∂y
∂u
∂z
∂v
∂z
∂w
∂z
–
=
2 yz
2
x
3z
y
4y
z
2z
x
−3 zx
y2
4x
z
2y
x
3x
y
–4 xy
z2
LM
2y L 12xz 12xyz O
2z L −12xyz − 12xy O
12x2 O
−
M
P
–
+
P
M + zy2 PP
2
2 2
yz PQ
x MN yz
yz PQ
x MN yz
x MN y z
Q
2
2 yz 12x yz
2
f
g
∂ bx, y, zg
∂ au, v, wf
But, we have
×
= 1 (Property 1)
∂ au, v, wf
∂ bx, y, zg
∂ bx, y, zg
1
.
∴
=
96
∂ au, v, wf
⇒
a
b
∂u
∂x
∂v
∂x
∂w
∂x
∂ u, v, w
= 0 + 48 + 48 = 96.
∂ x, y, z
99
DIFFERENTIAL CALCULUS-II
Example 4. If u = xyz, v = x2 + y2 + z2, w = x + y + z find J (x, y, z).
Sol. Here we calculate J(u, v, w) as follows:
J(u, v, w) =
∂u
∂x
∂v
∂x
∂w
∂x
∂u
∂y
∂v
∂y
∂w
∂y
∂u
∂z
∂v
∂z
∂w
∂z
yz
= 2x
1
zx
2y
1
(U.P.T.U., 2002)
xy
2z
1
= yz (2y – 2z) – zx (2x – 2z) + xy (2x – 2y)
= 2 [y2z – yz2 – zx2 + z2x + xy (x – y)]
= 2 [– z (x2 – y2) + z2 (x – y) + xy (x – y)]
As x 2 – y 2 = ( x – y )( x + y)
= 2 (x – y)[– zx – zy + z2 + xy]
But
= 2 (x – y)[z (z – x) – y (z – x)]
= 2 (x – y) (z – y) (z – x)
= – 2 (x – y) (y – z) (z – x)
J(x, y, z) · J(u, v, w) = 1
∴
J (x, y, z) =
Example 5. If x =
vw , y =
uv and u = r sin θ cos φ,
wu , z =
v = r sin θ sin φ, w = r cos θ, calculate
b g.
∂ br, θ, φg
∂ x, y, z
Sol. Here x, y, z are functions of u, v, w and u, v, w are functions of r, θ, φ so we apply IInd
property.
b
b
g ∂ bx, y, zg ∂ au, v, wf
g = ∂ au, v, wf · ∂ br, θ, φg
∂ x, y, z
...(i)
∂ r, θ, φ
Consider
b g =
∂ au, v, wf
∂ x, y, z
∂x
∂u
∂y
∂u
∂z
∂u
∂x
∂v
∂y
∂v
∂z
∂v
1
2
0
=
=
1
2
w
u
1
2
v
u
1
8
∂x
∂w
∂y
∂w
∂z
∂w
w
v
0
1
2
u
v
1
2
v
w
1
2
u
w
0
LM w v u + v w u OP 1
MN v u w w u v PQ = 8
1+ 1 = 2 = 1
8
4
100
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
b
a
g =
f
f =
g
∂u
∂r
∂v
∂r
∂w
∂r
∂ x, y, z
⇒
Next
...(ii)
∂ u, v, w
a
b
∂ u, v, w
∂ r , θ, φ
∂u
∂θ
∂v
∂θ
∂w
∂θ
∂u
sin θ cos φ
∂φ
∂v
= sin θ sin φ
∂φ
∂w
cos θ
∂φ
r cos θ cos φ
−r sin θ sin φ
r cos θ sin φ
r sin θ cos φ
−r sin θ
0
= sinθ cos φ (r2 sin2 θ cos φ) – r cos θ cos φ (–r2 sin θ cos θ cos φ)
+ r2 sin θ sin φ (sin2 θ sin φ + cos2 θ sin φ)
= r2 sin θ cos2 φ (sin2 θ + cos2 θ) + r2 sin θ sin2 φ
a
b
f
g
∂ u, v, w
= r2 sin θ cos2 φ + r2 sin θ sin2 φ = r2 sin θ
∂ r, θ, φ
Using (ii) and (iii) in equation (i), we get
⇒
...(iii)
b g = 1 × r sin θ = r sin θ .
4
∂ br, θ, φg
4
∂ x, y, z
2
2
Example 6. If u = x (1 – r2)−1/2, v = y (1 – r2)−1/2, w = z (1 – r2)−1/2
where
Sol. Since
a
b
f
g
∂ u, v, w
x 2 + y 2 + z 2 , then find ∂ x, y, z .
r =
r2 = x2 + y2 + z2
∂r
x ∂r
y ∂r
z
=
,
= ,
=
r ∂y
r
∂x
r ∂z
2 –1/2
Differentiating partially u = x (1 – r )
w.r.t. x, we get
∴
∂u
∂x
FG −1IJ (– 2r) e1 − r j . ∂r
e j
H 2K
∂x
1
x
= e1 − r j
+ rx e1 − r j .
=
+
r
1−r
=
1 − r2
2
⇒
∂u
∂x
=
−1
2
2
+x
−1
2
2
−3
2
−3
2
2
e
j
3
1 − r2 2
and
∂u
∂y
= x
∂u
∂y
=
∂u
∂z
=
FG −1IJ e1 − r j · (– 2r) ∂r = xr · y
H 2K
∂y
e1 − r j r
2
xy
3
2 2
e1 − r j
xz
3
2 2
e1 − r j
−3
2
3
2 2
3
2 2
e1 − r j
1 − r2 + x2
Differentiating partially u w.r.t. y, we get
⇒
x2
101
DIFFERENTIAL CALCULUS-II
∂v
∂x
Similarly,
∂w
∂x
Thus,
a
b
=
yx
e1 − r j
yz
1 − r 2 + y 2 ∂v
∂v
=
3
3 , ∂z =
∂y
1 − r2 2
1 − r2 2
3 ,
e
2 2
=
zx
e1 − r j
∂w
=
∂y
3 ,
2 2
e1 − r j
xy
3
1 − r2 2
j
yx
f
g = e1 − r j
∂ u, v, w
∂ x, y, z
3
2 2
e
j
1−r + y
e
2
3
2 2
1
9
1 − r2 2
j
j
3
j
3
yz
j
e
1 − r2 2
1 − r2 + z2
zy
e1 − r j
e
1 − r2 2
3
1 − r2 2
e
2
1−r + z
∂w
=
3 .
∂z
2 2
1−r
xz
2
zx
=
3 ,
2 2
3
1 − r2 2
j
2
zy
1 − r 2 + x2
e
e
j
3
2 2
3
2 2
e1 − r j
e1 − r j
1 − r2 + x2
xy
yx
2
xz
1−r + y
2
yz
1−r + z
zx
zy
e j
9
2 2 [(1 – r + x ) {(1 – r + y )(1 – r + z )– y z }
1
−
r
e j
2
−
=
2
2
2
2
2
2
2
2 2
– xy {xy (1 – r2 + z2) – xyz2} + xz{xy2z – zx(1 – r2 + y2)}]
=
−9
1 − r2 2
e
j
[(1 – r2 + x2) (1 – r2 + y2) (1 – r2 + z2)
– (1 – r2) (y2z2 + x2y2 + x2z2) – x2y2z2]
−9
2
e1 − r j [(1 – r ) + (1 – r ) (x + y + z )]
= e1 − r j
[(1 – r ) + (1 – r ) r ]
= e1 − r j . (1 – r ) [1 – r + r ] = e1 − r j .
=
2
2
2
2 3
−9
2
2 3
−9
2
2 2
2 2
2
2 2
2
2
2
2
2
2
−5
2
Example 7. Verify the chain rule for Jacobians if x = u, y = u tan v, z = w. (U.P.T.U., 2008)
Sol. We have
x = u
⇒
y = u tan v ⇒
z = w
⇒
∂x
∂x
∂x
= 1,
=
= 0
∂u
∂v
∂w
∂y
∂y
∂y
= tan v,
= u sec2 v,
=0
∂u
∂v
∂w
∂z
∂z
∂z
=
=1
= 0,
∂u
∂v
∂w
1
0
0
b g = tan v u sec2 v 0 = u sec v
J =
∂ bu, v, wg
0
0
1
∂ x, y, z
2
...(i)
102
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Solving for u, v, w in terms of x, y, z, we have
u = x
v = tan–1
y
y
= tan −1
u
x
w = z
y
∂u
∂u
∂u
∂v
∂v
x
∂v
∂w
∂w
∴
,
,
= 1,
=
= 0,
= − 2
= 2
= 0,
=
= 0 and
2
2
∂x
∂y
∂z
∂x
∂y
∂z
∂x
∂y
x +y
x +y
∂w
= 1
∂z
1
0
x
b g= − y
J′ =
∂ bx, y, zg
x2 + y2
∂ u, v, w
2
x +y
0
0
0
0 =
2
x
2
x +y
2
=
1
1
F yI
xG 1 + 2 J
H xK
2
=
1
u sec2 v
...(ii)
Hence from (i) and (ii), we get
J.J′ = u sec2 v.
1
u sec 2 v
= 1.
2.1.3 Jacobian of Implicit Functions
If the variables u, v and x, y be connected by the equations
f1 (u, v, x, y) = 0
f2 (u, v, x, y) = 0
i.e., u, v are implicit functions of x, y.
Differentiating partially (i) and (ii) w.r.t. x and y, we get
...(i)
...(ii)
∂f1
∂f1
∂f1
∂u
∂v
+
·
+
·
= 0
∂x
∂x
∂x
∂u
∂v
...(iii)
∂f1
∂f1
∂f1
∂u
∂u
∂ y + ∂u · ∂ y + ∂v · ∂ y
= 0
...(iv)
∂f 2
∂f 2
∂f 2
∂u
∂v
+
·
+
·
= 0
∂x
∂x
∂x
∂u
∂v
...(v)
∂f 2
∂f
∂u
∂f
∂v
+ 2 ·
+ 2 ·
= 0
∂y
∂y
∂y
∂u
∂v
...(vi)
Now,
∂ f1
b g × ∂ au, vf = ∂ u
∂f
∂ bx, yg
∂ au, vf
∂ f1 , f2
2
∂u
=
∂ f1
∂v
∂ f2
∂v
∂u
∂x
∂v
∂x
∂u
∂y
∂v
∂y
∂ f1 ∂ u ∂ f1 ∂v
+
∂u ∂x ∂v ∂x
∂f1 ∂u ∂ f1 ∂ v
+
∂ u ∂ y ∂v ∂ y
∂ f2 ∂ u ∂ f 2 ∂ v
+
∂u ∂ x ∂ v ∂ x
∂f2 ∂u ∂ f2 ∂ v
+
∂u ∂y ∂v ∂y
103
DIFFERENTIAL CALCULUS-II
Using (iii), (iv), (v) and (vi) in above, we get
∂f
∂f
− 1 − 1
∂x
∂y
=
∂ f2
∂ f2
−
−
∂x
∂y
Thus,
⇒
= (– 1)2
b g × ∂ au, vf = (– 1) ∂ b f , f g
∂ bx, yg
∂ bx, yg
∂ au, vf
∂bf , f g
∂ bx, yg
∂ au, vf
= (– 1)
∂ bx, yg
∂bf , f g
∂ au, vf
∂ f1 , f2
2
1
∂ f1
∂x
∂ f1
∂y
∂ f2
∂x
∂ f2
∂y
2
1
2
1
2
2
Similarly for three variables u, v, w
a
b
f = (– 1)
g
∂ u, v, w
∂ x, y, z
b
∂ f1 , f 2 , f 3
3
b
∂ x , y, z
b
g
∂ f1 , f 2 , f 3
a
∂ u, v, w
f
g
g
and so on.
Example 8. If u3 + v3 = x + y, u2 + v2 = x3 + y3, show that
a f
b g
y2 − x2
∂ u, v
.
=
2uv u − v
∂ x, y
Sol. Let
f 1 ≡ u3 + v3 − x − y = 0
f 2 ≡ u2 + v2 − x3 − y3 = 0
∂ f1 ∂ f1
–1
∂x
∂y
∂ f1 , f2
Now,
=
=
∂
∂
f
f
∂ x, y
2
2
–3x 2
∂x
∂y
a
f
b g
b g
b
∂ f1 , f2
and
a f
∂ u, v
g=
∂ f1
∂u
∂ f1
∂v
∂ f2
∂u
∂ f2
∂v
3u2
=
2u
–1
–3 y
3v 2
2v
2
= 3 (y2 – x2)
= 6 uv (u – v)
b g
a f = ∂ bx, yg = 3 ey − x j = ( y − x ) . Hence Proved.
b g ∂ b f , f g 6uv au – vf 2uv (u − v)
∂ au, vf
∂ f1 , f2
Thus,
(U.P.T.U., 2006)
∂ u, v
∂ x, y
1
2
2
2
2
Example 9. If u, v, w are the roots of the equation in k,
b g = – au − vfav − wfaw − uf .
that
aa − bfab − cfac − af
∂ au, v, wf
∂ x, y, z
2
x
a+ k
+
y
z
+
= 1, prove
c+ k
b+ k
104
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
x
y
z
+
=1
a+ k
c+ k
b+ k
x (b + k) (c + k) + y (a + k) (c + k) + z (a + k) (c + k) = (a + k) (b + k) (c + k)
k3 + k2 (a + b + c – x – y – z) + k{bc + ca + ab – (b + c) x – (c + a)
y – (a + b)z} + (abc – bcx – cay – abz) = 0
Since its roots are given to be u, v, w, so we have
u + v + w = – (a + b + c – x – y – z)
uv + vw + wu = bc + ca + ab – (b + c) x – (c + a)y – (a + b)z
uvw = – (abc – bcx – cay – abz)
Let
f1 ≡ u + v + w + a + b + c – x – y – z = 0
f 2 ≡ uv + vw + wu – bc – ca – ab + (b + c) x + (c + a) y + (a + b)z = 0
f 3 ≡ uvw + abc – bcx – cay – abz = 0
Sol. We have
or
or
Now,
b
+
g=
∂ f1 , f2 , f3
b
∂ x, y , z
g
∂ f1
∂x
∂ f2
∂x
∂ f3
∂x
∂ f1
∂y
∂ f2
∂y
∂ f3
∂y
∂ f1
∂z
∂ f2
∂z
∂ f3
∂z
1
0
b
+
c
a
−
b
=
bc c a − b
–1
b+ c
–bc
b
∂ f1 , f 2 , f 3
and
a
f
∂ u, v, w
g=
∂ f1
∂v
∂ f2
∂v
∂ f3
∂v
0
a−c
b a−c
∂ f1
∂w
∂ f2
∂w
∂ f3
∂w
1
= v+w
vw
=
= (a – b)(a – c)(b – c)
1
1
1
vw
wu
uv
a v + w f a w + uf au + v f
0
0
u−v
u−w
a
w u− v
–1
a+ b
–ab
a f a f a f
=
a f a f
∂ f1
∂u
∂ f2
∂u
∂ f3
∂u
–1
c+ a
–ca
f va u − w f
(C2 → C2 – C1, C3 → C3 – C1)
= (u – v)(u – w)(v – w)
Thus,
∴
b
∂ f1 , f2 , f3
g = – aa − bf aa − cfab − cf
b
g au − vfau − wf av − wf
a f
∂ bx, y, zg
au − vfav − wfau − wf . Hence proved.
=–
∂ au, v, wf
aa − bfab − cfaa − cf
a
b
f
g
b
g
∂ x, y , z
∂ u, v, w
= (–1)3
∂ f1 , f2 , f3
∂ x, y, z
∂ u, v , w
As JJ ′ = 1.
Example 10. If u = 2axy, v = a (x2 – y2) where x = r cosθ, y = r sin θ, then prove that
a f
a f
∂ u, v
= – 4a2r3.
∂ r, θ
105
DIFFERENTIAL CALCULUS-II
Sol. We have
u = 2axy, v = a (x2 – y2)
∂u
∂u
2ay
∂ u, v
∂x
∂y
= ∂v
=
∂v
2ax
∂ x, y
∂x
∂y
2ax
a f
= – 4a (x + y )
−2ay
b g
∂ au, vf
As x + y = r
= − 4a r
∂ bx, yg
∂x
∂x
cos θ –r sin θ
∂ bx, yg
∂r
∂θ
= ∂y
=
=r
y
∂
sin θ
r cos θ
∂ ar, θf
∂r
∂θ
∂ au, vf
∂ bx, yg
∂ au, vf
=
·
= (− 4a r ).r = − 4a r . Hence proved.
∂ ar, θf
∂ bx, yg
∂ ar, θf
Now,
2
2
2
2
2 2
or
and
2 2
Hence
2
2
2 3
Example 11. If u3 + v3 + w3 = x + y + z, u2 + v2 + w2 = x3 + y3 + z3, u + v + w = x2 + y2 + z2,
then show that
f b gb ga f
g a fa fa f
a
b
x− y y− z z− x
∂ u, v, w
⋅
=
∂ x, y, z
u− v v− w w− u
f 1 ≡ u3 + v3 + w3 – x – y – z = 0
f 2 ≡ u2 + v2 + w2 – x3 – y3 – z3 = 0
f 3 ≡ u + v + w – x2 – y2 – z2 = 0
Sol. Let
Now,
b
∂ f1 , f2 , f3
b
∂ x, y , z
g
g =
=
∂f 1
∂f 1
∂f 1
∂x
∂f 2
∂y
∂f 2
∂z
∂f 2
∂x
∂f 3
∂y
∂f 3
∂z
∂f 3
∂x
∂y
∂z
–1
0
–3x2 3 x2 − y2
e
–2x
2
⇒
and
b
g
=
0
3 x2 − z2
j e
a f
–1
–3y2
–2y
j
–1
–3z2
–2z
C2 → C2 − C1 , C3 → C3 − C1
2 x− z
= – 6[(x – y ) (x – z) – (x – z2) (x – y)]
= – 6 (x – y) (x – z) [(x + y) – (x + z)]
∂ f1 , f2 , f3
b
∂ x, y , z
b
b
2 x− y
–1
–3x2
–2x
g
∂ f1 , f2 , f3
a
f
∂ u, v, w
2
2
g = 6 (x – y) (y – z) (z – x)
g =
∂f1
∂u
∂f 2
∂u
∂f 3
∂u
∂f 1
∂v
∂f 2
∂v
∂f 3
∂v
∂f1
∂w
∂f 2
∂w
∂f 3
∂w
=
3u 2
3v 2
3w 2
2u
2v
2w
1
1
1
106
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
b
g 3 bw 2 − u2 g
2a v − u f
2a w − u f
3u 2
=
3 v2 − u2
2u
1
0
C2 → C2 − C1 , C3 → C3 − C1
0
Expand it with respect to third row, we get
= 6[(v2 – u2) (w – u) – (w2 – u2) (v – u)]
= 6 (v – u) (w – u) [(v + u) – (w + u)]
b
∂ f1 , f2 , f3
⇒
a
f
∂ u, v, w
Hence
a
b
g = – 6 (u – v) (v – w) (w – u)
b
∂ f1 , f2 , f3
g
f = (–1) ∂ bx, y, zg = + 6 bx − ygby − zgaz − xf
∂ bf , f , f g
6 au − vfav − wfaw − uf
g
∂ au, v, w f
bx − ygby − zgaz − xf · Hence proved.
=
au − vfav − wfaw − uf
∂ u, v, w
∂ x, y, z
3
1
2
3
Example 12. u, v, w are the roots of the equation
(x – a)3 + (x – b)3 + (x – c)3 = 0, find
b g.
∂ ba, b, cg
∂ u, v, w
Sol. We have (x – a)3 + (x – b)3 + (x – c)3 = 0
x3 – a3 – 3xa (x – a) + x3 – b3 – 3xb (x – b) + x3 – c3 – 3xc (x – c) = 0
or
3x3 – 3x2 (a + b + c) + 3x (a2 + b2 + c2) – (a3 + b3 + c3) = 0
Since u, v, w are the roots of this equation, we have
u+v+w = a+b+c
uv + vw + wu = a2 + b2 + c2
α +β+ γ = −b a
As
αβ + βγ + γα = c a
a 3 + b3 + c 3
3
f1 ≡ u + v + w – a – b – c = 0
αβγ = − d a
uvw =
Let
f 2 ≡ uv + vw + wu – a2 – b2 – c2 = 0
f 3 = uvw –
Now
a 3 + b3 + c 3
3
g = −−21a −−21b −−21c = −−21a 2aa0− bf 2aa0− cf , FG c2 → c2 − c1IJ
H c3 → c3 − c1K
∂b a, b, cg
− a 2 − b2 − c 2
− a 2 e a 2 − b2 j e a 2 − c 2 j
b
∂ f1 , f 2 , f 3
= – 2{(a – b) (a2 – c2) – (a – c) (a2 – b2)} = – 2(a – b) (b – c) (c – a)
and
b
∂ f1 , f 2 , f 3
b
∂ u, v , w
g
g = v +1 w u +1 w v +1 u = v +1 w
vw
wu
uv
FG c → c − c IJ
Hc → c − c K
vw wau − vf vau − wf
0
0
u−v
u−w ,
2
2
1
3
3
1
107
DIFFERENTIAL CALCULUS-II
= (u – v) v(u – w) – (u – w) w(u – v)
= – (u – v) (v – w) (w – u)
Thus
b
∂ f1 , f 2 , f 3
a
a
g
f = a−1f ∂aa, b, cf = − −2aa − bf ab − cf ac − af
f
∂b f , f , f g
− au − vfa v − wfaw − uf
∂au, v, wf
aa − bfab − cf ac − af .
= −2
au − vfav − wf aw − uf
∂ u, v, w
∂ a, b, c
3
1
2
3
2.1.4 Functional Dependence
Let u = f1 (x, y), v = f2 (x, y) be two functions. Suppose u and v are connected by the relation
f (u, v) = 0, where f is differentiable. Then u and v are called functionally dependent on one another
∂v
∂u ∂u ∂v
(i.e., one function say u is a function of the second function v) if the
,
,
and ∂y are not
∂x ∂y ∂x
all zero simultaneously.
Necessary and sufficient condition for functional dependence (Jacobian for
functional dependence functions):
Let u and v are functionally dependent then
f (u, v) =
0
...(i)
Differentiate partially equation (i) w.r.t. x and y, we get
∂f ∂v
∂f ∂u
·
+
∂v ∂x
∂u ∂x
∂f ∂v
∂f ∂u
·
+
·
∂v ∂y
∂u ∂y
=
0
...(ii)
=
0
...(iii)
There must be a non-trivial solution for
Thus,
or
or
∂f
∂f
≠ 0,
≠ 0 to this system exists.
∂u
∂v
∂u
∂x
∂u
∂y
∂v
∂x
∂v
∂y
=
0
For non-trivial solution xAx = 0
∂u
∂x
∂v
∂x
∂u
∂y
∂v
∂y
=
0
Changing all rows in columns
a f = 0
b g
∂ u, v
∂ x, y
Hence, two functions u and v are “functionally dependent” if their Jacobian is equal to
zero.
108
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Note: The functions u and v are said to be “functionally independent” if their Jacobian is not equal
to zero i.e., J(u, v) ≠ 0
Similarly for three functionally dependent functions say u, v and w.
J(u, v, w) =
a f = 0.
∂ bx, y, zg
∂ u, v, w
Example 13. Show that the functions u = x + y – z, v = x – y + z, w = x2 + y2 + z2 – 2yz
are not independent of one another. Also find the relation between them.
Sol. Here
u = x + y – z, v = x – y + z and w = x2 + y2 + z2 – 2yz
a
b
f =
g
∂u
∂x
∂v
∂x
∂w
∂x
∂u
∂y
∂v
∂y
∂w
∂y
=
1
1
2x
1
–1
2y − 2z
∂ u, v, w
∂ x, y, z
Now,
∂u
∂z
∂v
∂z
∂w
∂z
=
1
1
2x
1
–1
2y − 2z
–1
1
2z − 2y
0
0 (C3 → C3 + C2)
0
= 0. Hence u, v, w are not independent.
Again
or
u + v = x + y – z + x – y + z = 2x
u – v = x + y – z – x + y – z = 2 (y – z)
∴ (u + v)2 + (u – v)2 = 4x2 + 4 (y – z)2
= 4(x2 + y2 + z2 – 2yz) = 4w
2
2
⇒ (u + v) + (u – v) = 4w
2(u2 + v2) = 4w or u2 + v2 = 2w.
Example 14. Find Jacobian of u = sin–1 x + sin–1 y and v = x 1 − y 2 + y 1 − x 2 . Also find
relation between u and v.
u = sin–1 x + sin–1 y, v = x 1 − y 2 + y 1 − x 2
Sol. We have
a f =
∂bx, yg
∂ u, v
Now,
= −
Next,
xy
1−x
2
1− y
2
∂u
∂x
∂v
∂x
+1−1+
∂u
∂y
∂v
∂y
1
1− x
=
1 − y2 −
xy
1−x
2
1
1 − y2
1 − y2
2
xy
1− x
2
−
xy
1− y
2
+ 1 − x2
= 0. Hence u and v are dependent.
{
2
2
u = sin–1 x + sin–1 y ⇒ u = sin–1 x 1 − y + y 1 − x
{
}
As sin −1 A + sin −1 B = sin −1 A 1 − B 2 + B 1 − A 2
⇒
or
x 1 − y 2 + y 1 − x2 = v
v = sin u.
sin u =
}
109
DIFFERENTIAL CALCULUS-II
Example 15. Show that ax2 + 2hxy + by2 and Ax2 + 2Hxy + By2 are independent unless
a
h
b
=
= .
A
H
B
Sol. Let
u = ax2 + 2hxy + by2
v = Ax2 + 2Hxy + By2
If u and v are not independent, then
∂u
∂x
∂v
∂x
or
∂u
∂y
∂v
∂y
=
a f =0
b g
∂ u, v
∂ x, y
2ax + 2hy
2 Ax + 2Hy
2hx + 2by
2Hx + 2By
=0
⇒ (ax + hy) (Hx + By) – (hx + by) (Ax + Hy) = 0
⇒ (aH – hA) x2 + (aB – bA) xy + (hB – bH) y2 = 0
But variable x and y are independent so the coefficients of x2 and y2 must separately vanish
and therefore, we have
aH – hA = 0 and hB – bH = 0 i.e.,
a
A
i.e,
=
h
b
a
h
=
and
=
H
B
A
H
h
b
=
· Hence proved.
H
B
Example 16. If u = x2 e–y cos hz, v = x2 e–y sin hz and w = 3x4 e–2y then prove that u, v, w
are functionally dependent. Hence establish the relation between them.
Sol. We have u = x2 e–y cos hz, v = x2 e–y sin hz, w = 3x4 e–2y
a
b
∂u ∂x
∂u ∂y
∂u ∂z
2xe − y cos hz
−x 2 e − y cos hz
x 2 e − y sin hz
f = ∂v ∂x ∂v ∂y ∂v ∂z = 2xe sin hz −x e sin hz x e cos hz
g ∂w ∂x ∂w ∂y ∂w ∂z 12x e
−6x e
0
∂ u, v, w
∂ x, y, z
−y
3 −2 y
2 −y
2 −y
4 −2 y
= 2x e–y cos hz{0 + 6x6 e–3y cos hz} + x2 e–y cos hz{0 – 12x5 e–3y cos hz}
+ x2 e–y sin hz{–12x5 e–3y sin hz + 12x5 e–3y sin hz}
= 12x7 e–4y cos h2 z – 12x7 e–4y cos h2 z = 0
Thus u, v and w are functionally dependent.
Next,
3u2 – 3v2 = 3(x4 e–2y cos h2 z – x4 e–2y sin h2 z) = 3x4 e–2y (cos h2 z – sin h2 z)
= 3x4 e–2y
⇒
3u2 – 3v2 = w.
EXERCISE 2.1
1 . If x = r cos θ, y = r sin θ find
b g.
∂ ar, θf
∂ x, y
LM Ans. 1 OP
rQ
N
x 2x 3
x x
xx
, y2 = 3 1 , y3 = 1 2 show that the Jacobian of y1, y2, y3 with respect to x1,
x1
x2
x3
x2, x3 is 4.
(U.P.T.U., 2004(CO), 2002)
2 . If y1 =
110
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
3 . If x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ, show that
b g = r sin θ.
∂ br, θ, φg
∂ x, y, z
2
4 . If u = x + y + z, uv = y + z, uvw = z, evaluate
e
j
b g
b g
(U.P.T.U., 2000)
b g.
∂ au, v, wf
∂ x, y, z
2
(U.P.T.U., 2003) Ans. u v
LMAns. − y OP
2x Q
N
∂ u, v
x2 + y2
y2
5 . If u =
,v=
, find ∂ x , y .
2x
2x
6 . If x = a cos h ξ cos η, y = a sin h ξ sin η, show that
b g = 1 a (cos h 2ξ – cos 2η).
2
∂ bξ, ηg
∂ x, y
2
7 . If u3 + v + w = x + y2 + z2, u + v3 + w = x2 + y + z2, u + v + w3 = x2 + y2 + z, then evaluate
a
b
LM
b1 − 4xy − 4yz − 4zx + 6xyzg OP
Ans.
MM 27u2v2w2 + 2 − 3 eu2 + v2 + w2 j PP
Q
N
f
g
∂ u, v, w
.
∂ x, y, z
f
g
L −2(x − y) ( y − z) ( z − x) O
(U.P.T.U., 2001) MAns. au − vfav − wfaw − uf P
N
Q
8 . If u, v, w are the roots of the equation (λ – x)3 + (λ – y)3 + (λ – z)3 = 0 in λ, find
a
b
∂ u, v, w
.
∂ x, y, z
9 . If u = x1 + x2 + x3 + x4, uv = x2 + x3 + x4, uvw = x3 + x4 and uvwt = x4, show that
b
∂ x1 , x2 , x3 , x4
g = u v w.
a
f
∂ bx, yg
∂ au, vf
1 0 . Calculate J =
and J′ =
. Verify that JJ′ = 1 given
∂ bx, yg
∂ au, vf
∂ u, v, w, t
(i) u = x +
3 2
y2
y2
.
,v=
x
x
Ans. J = e 2 u , J ′ = e −2 u .
(ii) x = eu cos v, y = eu sin v.
1 1 . Show that
a f
a f
LMAns. J = 2 y , J ′ = x OP
x
2y Q
N
∂ u, v
= 6r3 sin 2θ given u = x2 – 2y2, v = 2x2 – y2 and x = r cos θ, y = r sin θ.
∂ r, θ
1 2 . If X = u2v, Y = uv2 and u = x2 – y2, v = xy, find
a fL
b g MN
j OPQ
∂ X, Y
2 2 2
2
2
2 2
. Ans. 6x y x + y x – y
∂ x, y
e
je
a f
Ans. – 6 e x cos y
b g
∂ au, v, wf
Ans. – 2
1 4 . Find
, if u = 3x + 2y – z, v = x – y + z, w = x + 2y – z.
∂ bx, y, zg
Ans. bx − ygb y − zga z − xf
1 5 . Find J (u, v, w) if u = xyz, v = xy + yz + zx, w = x + y + z.
1 3 . Find
∂ u, v, w
, if u = x2, v = sin y, w = e–3z.
∂ x, y, z
−3 z
111
DIFFERENTIAL CALCULUS-II
1 6 . Prove that u, v, w are dependent and find relation between them if u = xey sin z, v = xey
Ans. dependent, u 2 + v 2 = w
cos z, w = x2e2y.
b g
b g
f
g
2 y+z
x− y
3x 2
Ans. dependent, uvw 2 = 1
,v=
.
2 , w =
x
2 y+z
3 x−y
1 8 . If X = x + y + z + u, Y = x + y – z – u, Z = xy – zu and U = x2 + y2 – z2 – u2, then show
∂ X, Y, Z, U
that J =
= 0 and hence find a relation between X, Y, Z and U.
∂ x, y, z, u
b g
a
b
17. u =
Ans. XY = U + 2Z
1 9 . If u =
x
y− z
,v=
y
z
,w=
, then prove that u, v, w are not independent and also
x− y
z−x
Ans. uv + vw + wu + 1 = 0
find the relation between them.
2 0 . If u = x + 2y + z, v = x – 2y + 3z, w = 2xy – xz + 4yz – 2z2, show that they are not
Ans. 4w = u 2 − v 2
independent. Find the relation between u, v and w.
a f
b g
∂ u, v
x+y
and v = tan–1 x + tan–1 y, find
. Are u and v functionally related? If
∂ x, y
1 − xy
yes find the relationship.
[Ans. yes, u = tan v]
2 1 . If u =
g = u v.
f
∂au, vf
x −y
2 3 . If x + y + u – v = 0 and uv + xy = 0 prove that
.
=
∂bx, yg
u +v
2 2 . If u = x + y + z, uv = y + z, uvw = z, show that
2
2
2
b
a
∂ x, y, z
2
∂ u, v, w
2
2
2
2
2
2 4 . Find Jacobian of u, v, w w.r.t. x, y, z when u =
yz
xy
zx
.
,v =
,w =
z
x
y
[Ans. 4]
2.2 APPROXIMATION OF ERRORS
Let u = f (x, y) then the total differential of u, denoted by du, is given by
du =
∂f
∂f
dx + ∂y dy
∂x
...(i)
If δx and δy are increments in x and y respectively then the total increment δu in u
is given by
or
But
δu =
f (x + δx, y + δy) =
δu ≈
∴ From (i)
δu ≈
f (x + δx, y + δy) – f (x, y)
f (x, y) + δu
du, δx ≈ dx and δy ≈ dy
...(ii)
∂f
∂f
δx +
δy
∂x
∂y
...(iii)
112
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Using (iii) in (ii), we get the approximate formula
f (x + δx, y + δy) ≈
f (x, y) +
∂f
∂f
δx + ∂y δy
∂x
...(iv)
Thus, the approximate value of the function can be obtained by equation (iv). Hence
δx or dx = Absolute error
δx
x
and
100 ×
or
dx
=
x
Proportional or Relative error
dx
dx
or 100 ×
=
x
x
Percentage error in x.
1
2
Example 1. If f (x, y) = x y 10 , compute the value of f when x = 1.99 and y = 3.01.
(U.P.T.U., 2007)
1
Sol. We have f (x, y) =
∴
Let
∂f
∂x
x 2 y 10
9
∂f
1 2 − 10
=
x y
∂y
10
1
=
2xy 10 ,
As
x = 2, δx = – 0.01
a f
a f
x + δx = 2 + –0.01 = 1.99
y + δy = 1 + 2.01 = 3.01
y = 1, δy = 2.01
Now, f (x + δx, y + δy) = f (x, y) +
∂f
∂f
δx + ∂y δy
∂x
af
1
⇒ f{2 + (– 0.01), 1 + 2.01} = f (2, 1) + 2 × 2 1 10 × (– 0.01) +
af
9
1
−
(2)2. 1 10 × (2.01)
10
1
⇒
f (1.99, 3.01) ≈ 2 × 1 10 + (– 0.04) + 0.804
≈ 4 – 0.04 + 0.804 = 4.764.
2
Example 2. The diameter and height of a right circular cylinder are measured to be 5 and
8 cm. respectively. If each of these dimensions may be in error by ± 0.1 cm, find the relative
percentage error in volume of the cylinder.
Sol. Let diameter of cylinder = x cm.
height of cylinder = y cm.
then
V =
πx2 y
4
(radius =
∴
πxy ∂V
∂V
πx2
=
,
=
∂x
2
4
∂y
⇒
dV =
∂V
∂V
dx +
dy
∂x
∂y
x
)
2
113
DIFFERENTIAL CALCULUS-II
⇒
dV =
1
1
π xy. dx +
πx2. dy
2
4
1
1
2
πxy. dx
πx dy
dV
dy
dx
2
+ 4 2
=
= 2.
+
2
V
x
y
x
y
π
π
x
y
or
or
100 ×
FG
H
4
dV
dx
= 2 100 ×
V
x
4
IJ + 100 × dy
y
K
Given x = 5 cm., y = 8 cm. and error dx = dy = ± 0.1. So 100 ×
IJ
K
FG
H
01
. 01
.
dV
+
= ± 100 2 ×
5
8
V
= ± 5.25.
Thus, the percentage error in volume = ± 5.25.
Example 3. A balloon is in the form of right circular cylinder of radius 1.5 m and length
4 m and is surmounted by hemispherical ends. If the radius is increased by 0.01 m and the length
by 0.05 m, find the percentage change in the volume of the balloon. [U.P.T.U., 2005 (Comp.), 2002]
Sol. Let radius = r = 1.5 m, δr = 0.01 m
height = h = 4 m, δh = 0.05 m
OLS
2 3
2 3
4 3
πr +
πr = πr2h +
πr
3
3
3
∂V
∂V
= 2πrh + 4πr2,
= πr2
∂r
∂h
∂V
∂V
dV =
dr +
dh = (2πrh + 4πr2) dr + πr2dh
∂r
∂h
dV
2πr h + 2r
πr 2
=
dr +
dh
V
4r
4r
2
2
OLS>ª
volume (V) = πr2 h +
∴
R>ª
OLS
a f
F I
F I
Fig. 2.1
πr h +
πr h +
H 3K
H 3K
3 × 2a h + 2r f
3
3
=
dr +
dh =
[2(h + 2r) dr +rdh]
ra 3h + 4r f
ra 3h + 4rf
a3h + 4rf
or
=
3
[2(4 + 3)(0.01) + 1.5 (0.05)]
15
. 12 + 6
a
f
1
0.215
[0.14 + 0.075] =
9
9
0.215
dV
100 ×
= 100 ×
= 2.389%
9
V
=
⇒
Thus, change in the volume = 2.389%.
δr = dr
δh = dh
114
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 4. Calculate the percentage increase in the pressure p corresponding to a reduction
1
of % in the volume V, if the p and V are related by pV1.4 = C, where C is a constant.
2
Sol. We have
pV1.4 = C
Taking log of equation (i)
log p + 1.4 log V = log C
Differentiating
...(i)
dp
1
1.4
dV
· dp +
· dV = 0 ⇒
= – 1.4
p
p
V
V
or
100 ×
FG
H
dp
dV
= −1.4 100 ×
V
p
= −1.4 ×
IJ
K
FG −1IJ
H 2K
1
dV 1
= %=
2 × 100
V 2
= 0.7
Hence increase in the pressure p = 0.7%.
Example 5. In estimating the cost of a pile of bricks measured as 6′ × 50′ × 4′, the tape is
stretched 1% beyond the standard length. If the count is 12 bricks to one ft3, and bricks cost Rs.
100 per 1000, find the approximate error in the cost.
(U.P.T.U., 2004)
Sol. Let
length (l) = x
...(i)
breadth (b) = y
height (h) = z
∴
V = lbh = xyz
or
log V = log x + log y + log z
On differentiating
or
1
dV =
V
100 ×
⇒
100 ×
1
1
1
dx + y y +
dz
x
z
dy
dV
dx
dz
= 100 ×
+ 100 ×
+ 100 ×
y
V
x
z
dV
= 1+1+1=3
V
3V
3 6 × 50 × 4
dV =
=
= 36 cube fit
100
100
a
∴
f
Number of bricks in dV = 36 × 12 = 432
Hence,
the error in cost = 432 ×
.100
= Rs. 43.20.
1000
Example 6. The angles of a triangle are calculated from the sides a, b, c of small changes
δa, δb, δc are made in the sides, show that approximately δA =
a
[δa – δb cos C – δc. cos B]
2∆
where ∆ is the area of the triangle and A, B, C are the angles opposite to a, b, c respectively. Verify
that δA + δB + δC = 0.
(U.P.T.U., 2001)
115
DIFFERENTIAL CALCULUS-II
Sol. From trigonometry, we have
b 2 + c2 − a2
⇒ b2 + c2 – a2 = 2bc cos A
2bc
= b2 + c2 – 2bc cos A
cos A =
⇒
a2
On differentiating, we get
a · da
⇒
bc sin A · dA
⇒
2∆ · dA
⇒
2 ∆ · dA
=
=
=
=
b · db + c · dc – db · c cos A – b · dc cos A + bc sin A · dA
a · da – b · db – c · dc + db · c cos A + b · dc cos A
a · da – (b – c cos A) · db – (c – b cos A) dc
a da – (a cos C + c cos A – c cos A) db – (a cos B + b cos A
– b cos A) dc
1
bc sin A
2
a = b cos C + c cos B
As ∆ =
2 ∆ · dA = a da – db.a cos C – dc.a cos B
or
⇒
δA =
a
[δa – δb. cos C – δc. cos B]
2∆
As δA ≈ dA
δa ≈ da, δb ≈ db
δc ≈ dc
Hence proved.
Similarly,
δB =
δC =
and
Adding δA, δB and δC, we get
δA + δB + δC =
b
[δb – δc. cos A – δa. cos C]
2∆
c
[δc – δa. cos B – δb. cos A]
2∆
1
[(a – b cos C – c cos B) δa + (b – a cos C – c cos A) δb
2∆
+ (c – a cos B – b cos A) δc]
=
⇒
1
[(a – a) δa + (b – b) δb + (c – c) δ c]
2∆
δA + δB + δC = 0. Verified.
Example 7. Show that the relative error in c due to a given error in θ is minimum when
θ = 45° if c = k tan θ.
Sol. We have
c = k tanθ
...(i)
On differentiating, we get
dc = k sec2 θ dθ
...(ii)
2
2
d
θ
dc sec θdθ
From (i) and (ii), we get
=
=
c
sin 2θ
tan θ
Thus,
dc
will be minimum if sin2θ is maximum
c
i.e.,
sin 2θ = 1 = sin 90
⇒
2θ = 90 ⇒ θ = 45°. Hence proved.
Since sin θ lies
between – 1 and 1
116
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 8. Find approximate value of
a0.98f + a2.01f + a1.94f .
Sol. Suppose f (x, y, z) = e x + y + z j
∂f
∂f
∴
= x ex + y + z j
,
= y ex + y + z j
∂y
∂x
2
2 12
2
2
2
Now
2 12
2
2 −1 2
2
2
2
2 −1 2
,
∂f
2
2
2 −1 2
= z x +y +z
∂z
e
j
∂f
∂f
∂f
δx + ∂y δy +
δz
∂x
∂z
x = 1, y = 2, z = 2, δx = – .02, δy = .01, δz = –.06.
f (x + δx, y + δy,z + δz) = f (x, y, z) +
Let
⇒
e
f (0.98, 2.01, 1.94) = 12 + 2 2 + 2 2
j
12
+ (9)
−1 2
af a
−1 2
(.01) +2 9 −1 2 −0.06
× ( −0.02) + 2 (9)
f
1
(0.02 – 0.02 + 0.12)
3
= 3 – 0.04 = 2.96.
= 3–
Example 9. Prove that the relative error of a quotient does not exceed the sum of the relative
errors of dividend and the divisor.
Sol. Let
x = dividend
y = divisor
z = quotient
x
Then
= z
y
Taking log on both sides
log x – log y = log z
dy
dz
dx
–
=
z
y
x
⇒ the relative error in quotient is equal to difference of the relative errors of dividend and
divisor.
dz
Hence
(relative error) in quotient does not exceed the sum of relative errors of dividend
z
and the divisor
dz
dx
dy
i.e.,
<
+
. Hence proved.
z
x
y
Differentiating
Example 10. The work that must be done to propel a ship of displacement D for a distance
2
S2 D 3
. Find approximately the increase of work necessary when the
‘S in time ‘t’ proportional to
t2
displacement is increased by 1%, the time diminished by 1% and the distance diminished by 2%.
Sol. Let
the work = W
2
then
S2D 3
W ∝
t2
117
DIFFERENTIAL CALCULUS-II
⇒
W = K·
S 2 D −2 3
, where K is proportional constant.
t2
Taking log on both sides
2
log D – 2 log t
3
dW
dS
2 dD
dt
On differentiating
= 2
+
–2
W
S
3 D
t
dt
dW
dS
2
dD
100 ×
⇒
100 ×
= 2 100 ×
+
– 2 100 ×
t
S
D
W
3
−4
2
= 2 (–2) +
(1) – 2 (–1) =
3
3
−4
∴ Approximate increase of work =
%.
3
Example 11. The height h and semi-vertical angle α, of a cone are measured from there A,
the total area of the cone, including the base, is calculated. If h and α are in error by small
quantities δh and δα respectively, find the corresponding error in the area. Show further that, if
π
, an error of + 1 per cent in h will be approximately compensated by an error of –19.8
α=
6
minutes in α.
Sol. Total area of the cone
A = πr2 + πrl
log W = log K + 2 log S +
IJ
K
FG
H
IJ
K
FG
H
IJ
K
FG
H
A = πh2 tan2 α + π (h tan α) (h sec α)
or
= πh2 (tan2α + tan α sec α)
r
= tan α ⇒ r = h tan α
h
l
= sec α ⇒ l = h sec α
h
δA = 2πh δh (tan2 α + tan α sec α)
Differentiating
Fig. 2.2
+ πh2 (2tan α sec2 α. δα + sec3α δα + tan2 α · sec α · δα)
δA = 2πh tan α.δh (tan α + sec α) + πh2 (sec2 α + 2.tan α sec α
or
+ tan2 α) sec α . δα
= 2πh tan α. δh (tan α + sec α) + πh2 (sec α + tan α)2 sec α.δα
LM
N
δA = πh2 (tan α + sec α) 2tan α.
Now, putting
1
3
.
1
+
100
f
OP
Q
π δh
,
× 100 = 1 and δA = 0 in above.
6 h
π
π
π
δh 1
π
π
π
+ tan + sec . sec . δα
2 tan
100 ×
.
0 = πh2 tan + sec
6
h 100
6
6
6
6
6
α=
FG
H
⇒ 2·
a
δh
+ tan α + sec α sec α. δα .
h
IJ LM
KN
FG 1 + 2 IJ · 2 · δα = 0
H 3 3K 3
FG
H
IJ
K
FG
H
IJ
K
OP
Q
118
⇒
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
1
100 3
+
FG 3IJ δα= 0 ⇒ δα = – 1 radian.
H 3K
100 3
δα = –
or
=–
=–
1
100 3
9
5π 3
×
180
degree
π
× 60 minutes (1° = 60 minutes)
9 × 60
= – 19.858 minutes. Hence proved.
5 × 3.14 × 1.732
Example 12. If the sides and angles of a plane triangle vary in such a way that its circum
da
db
dc
+
+
= 0 where da, db, dc are small
cos A
cos B
cos C
increaments in the sides a, b, c respectively.
C
Sol. Let R be the circum radius.
a
We know that R =
R
2sin A
∂R
1
a × cos A
∂R
∴
=
,
=–
O
∂
A
2sin
A
2 sin 2 A
∂a
∂R
∂R
⇒
dR =
da +
dA
∂a
∂A
B
A
a cos A
1
da –
. dA
=
2 sin 2 A
2sin A
Fig. 2.3
1
a cos A
or
0 =
da −
. dA As R = constant
2sin A
2 sin A
radius remains constant, prove that
RS
T
⇒
da −
UV
W
a cosA
da
a
dA = 0 ⇒
=
. dA
cosA
sinA
sinA
da
= 2R dA
cosA
db
Similarly,
= 2R dB
cosB
dc
= 2R dC
cosC
Adding these equations, we get
da
db
dc
+
+
= 2R (dA + dB + dC)
cosA
cosB
cosC
Œ
A+ B+ C = π
∴
dA + dB + dC = 0
From (i) and (ii), we get
da
db
dc
+
+
= 0. Hence proved.
cosA
cosB
cosC
⇒
...(i)
...(ii)
119
DIFFERENTIAL CALCULUS-II
Example 13. Considering the area of a circular ring as an increment of area of a circle, find
approximately the area of the ring whose inner and outer radii are 3 cm. and 3.02 cm.
respectively.
Sol. Let
area of a circle = πr2
or
A = πr2
⇒
dA = 2πr dr
...(i)
Now area of circular ring δA is the difference between A1 of outer ring and A2 of inner
sphere
choose r = 3 cm. and δr = 0.02, then
⇒
⇒
A1 – A2 = A (r + δr) – A (r) = δA ≈ dA
A1 – A2 = dA = 2πr.dr (from i)
A1 – A2 = 2 π × 3 × .02
= .12 π cm2.
O
Example 14. Calculate (2.98)3
Sol. Let
f (x, y) = yx
⇒
Now,
Let
⇒
∂f
∂x
3.02
3 cm
0.02
∂f
= yx. log y, ∂y = xyx–1
∂f
∂f
f (x + δx, y + δy) = f (x, y) +
δx + ∂y δy
∂x
Fig. 2.4
x = 3, δx = – 0.02, y = 3, δy = 0
f (2.98, 3) = f (3, 3) + 33. log 3 × (– 0.02) + 0
= 33 – 33 × (.4771213) × 0.02
log 3 = .4771213
= 27 (1 – 0.00954) = 26.74.
EXERCISE 2.2
1. The period T of a simple pendulum is T = 2π
possible errors up to 1% in l and 2.5 % in g.
l
. Find the maximum error in T due to
g
(U.P.T.U., 2003) Ans. 1.75%
2. In the estimating the number of bricks in a pile which is measured to be (5 m × 10 m ×
5 m), count of bricks is taken as 100 bricks per m3. Find the error in the cost when the tape
is stretched 2% beyond its standard length. The cost of bricks is Rs. 2,000 per thousand
bricks.
(U.P.T.U., 2000) Ans. Rs. 3000
3. Find the approximately value of f(0.999) where f(x) = 2x4 + 7x3 – 8x2 + 3x + 1.
LMHint: f a0.999f = f bx + δxg = f a x f + ∂f . δx = f a1f + f ′ a1fa –0.001fOP
∂x
N
Q
Ans. 4.984
120
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
1
mv2e find approximate change in T as the mass
2
m changes from 49 to 49.5 and the velocity v changes from 1600 to 1590.
4. If the kinetic energy T is given by T =
Ans. 144000 units
5. Find the approximate value of (1.04)3.01.
Ans. 1.12
6. If ∆ be the area of a triangle, prove that the error in ∆ resulting from a small error in c
is given by
1
1
1
∆ 1
–
+
+
δ∆ =
δc
4 s s−a s−b s− c
7. Considering the volume of a spherical shell as an increment of volume of a sphere,
calculate approximately the volume of a spherical shell whose inner diameter is 8 inches
1
inch.
and whose thickness is
Ans. 4π cubic inches
6
8. A diameter and altitude of a can in the form of right circular cylinder are measured as
4 cm. and 6 cm. respectively. The possible error in each measurement is 0.1 cm. Find
approximately the maximum possible error in the value computed for the volume and
LM
N
OP
Q
Ans. 5.0336 cm 3 , 3.146 cm 2
lateral surface.
x2
y2
9. Find the percentage error in calculating the area of ellipse 2 + 2 = 1, when error of
a
b
Ans. 2%
+ 1 % is made in measuring the major and minor axis.
5
10. The quantity Q of water flowing over a v-notch is given by the formula Q = cH 2 where
H is the head of water and c is a constant. Find the error in Q if the error in H is 1.5%.
Ans. 3.75%
11. Find the percentage error in calculated value of volume of a right circular cone whose
altitude is same as the base radius and is measured as 5 cm. with a possible error of
0.02 cm.
Ans. 1.2%
12. Find possible percentage error in computing the parallel resistance r of three resistance
1
1
1
1
r1, r2, r3 from the formula
=
+
+
if r1, r2, r3 are each in error by plus 1.2%.
2
r
r
r
r
1
3
Ans. 12%
13. The diameter and the height of a right circular cylinder are measured as 4 cm. and 6 cm.
respectively, with a possible error of 0.1 cm. Find approximately the maximum possible
error in the computed value of the volume and surface area. Ans. 1.6 π cu. cm, π sq. cm
14. Find
a3.82f + 2a2.1f
2
1
3 5
Ans. = 2.012
.
15. Show that the acceleration due to gravity is reduced nearly by 1% at an altitude equal to
0.5 % of earth’s radius. Given that an external point x kilometers from the earth’s centre,
F rI
such an acceleration is given by g GH JK , where r is the radius of the earth.
x
2
16. Calculate the error in R if RI = E and possible errors in E and I are 20% and 10%
respectively.
Ans. 10%
121
DIFFERENTIAL CALCULUS-II
17. In the manufacture of closed cylindrical boxes with specified sides a, b, c (a ≠ b ≠ c) small
changes of A %, B %, C % occurred in a, b, c respectively from box to box from the
specified dimension. However, the volume and surface area of all boxes were according
to specification, show that
A
B
C
=
=
a b−c
b c−a
c a−b
a f
a f
a f
a f
1
18. Find 83.7 4 .
Ans. 3.025
19. Find y (1.997) where y(x) = x4 – 2x3 + 9x + 7.
Ans. 24.949
20. The time T of a complete oscillation of a simple pendulum of length l is governed by T
l
= 2π g where g is a constant.
(a) Find approximate error in the calculated value of T corresponding to an error of 2%
in the value of L.
(U.P.T.U., 2008) Ans. 1%
(b) By what percentage should the length be changed in order to correct a loss of 2
Ans. – 0.278%
minutes per day?
2.3 EXTREMA OF FUNCTION OF SEVERAL VARIABLES
Introduction
In some practical and theoretical problems, it is required to find the largest and smallest values
of a function of two variables where the variables are connected by some given relation or
condition known as a constraint. For example, if we plot the function z = f(x, y) to look like a
mountain range, then the mountain tops or the high points are called local maxima of f(x, y) and
valley bottoms or the low points are called local minima of f(x, y). The highest mountain and
lowest valley in the entire range are said to be absolute maximum and absolute minimum. The
graphical representation is as follows.
Absolute maximum
Z
Local maximum
Y
O
X
Absolute minimum
Local minimum
Fig. 2.5
Definition
Let f (x, y) be a function of two independent variables x, y such that it is continuous and finite
for all values of x and y in the neighbourhood of their values a and b (say) respectively.
122
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Maximum value: f(a, b) is called maximum value of f(x, y) if f (a, b) > f(a + h, b + k). For small
positive or negative values of h and k i.e., f(a, b) is greater than the value of function f (x, y,) at
all points in some small nbd of (a, b).
Minimum value: f(a, b) is called minimum value of f(x, y) if f (a, b) < f(a + h, b + k).
Note:
f(a + h, b + k) – f (a, b) = positive, for Minimum value.
f(a + h, b + k) – f (a, b) = negative, for Maximum value.
Extremum: The maximum or minimum value of the function f(x, y) at any point x = a and
y = b is called the extremum value and the point is called “extremum point”.
Geometrical representation of maxima and minima: The function f(x, y) represents a
surface. The maximum is a point on the surface (hill top). The minimum is a point on the surface
(bottom) from which the surface ascends (climbs up) in every direction.
Z = f (x, y)
Z = f (x, y)
P (maximum)
Q (minimum)
Y
Y
X
X
(a)
(b)
Fig. 2.6
Saddle point: It is a point where function is neither maximum nor minimum. At such point
f is maximum in one direction while minimum in another direction.
Example: z = xy, hyperbolic paraboloid has a saddle point at the origin.
Remark 1 :
If f(x, y) ≤ f(a, b) where (x, y) is a neighbourhood of (a, b). The number f(a, b)
is called local maximum value of f(x, y).
Remark 2 :
If f(x, y) ≥ f(a, b) where (x, y) is a neighbourhood of f(a, b). The number f(a, b)
is called local minimum value of f(x, y).
2.3.1 Condition for the Existence of Maxima and Minima (Extrema)
By Taylor’s theorem
f (a + h, b + k) = f (a, b) +
F h ∂f + k ∂f I
GH ∂x ∂y JK
+
( a, b )
1
2
F h ∂ f + 2 hk ∂ f + k ∂ f I
GH ∂x
J
∂x∂y
∂y K
2
2
2
2
2
2
+.... ...(i)
2
( a, b )
Neglecting higher order terms of h2, hk, k2, etc. Since h, k are small, the above expansion
reduce to
∂f a, b
∂f a, b
f (a + h, b + k) = f (a, b) + h
+ k
∂y
∂x
a f
∂f a a, bf
∂f a a, bf
⇒ f (a + h, b + k) – f (a, b) = h
+ k
∂x
∂y
a f
...(ii)
123
DIFFERENTIAL CALCULUS-II
The necessary condition for a maximum or minimum value (L.H.S. of eqn (ii) negative or
positive) is
h
a f + k ∂f a a, bf = 0
∂f a, b
∂x
∂y
a f = 0, ∂f a a, bf = 0 h and k can take both + ve and – ve value ...(iii)
∂f a, b
⇒
∂y
∂x
The conditions (iii) are necessary conditions for a maxmium or a minimum value of f(x, y).
Note: The conditions given by (iii) are not sufficient for existence of a maximum or a minimum value
of f(x, y).
2.3.2 Lagrange’s Conditions for Maximum or Minimum (Extrema)
Using the conditions (iii) in equation (i) (2.3.1) and neglecting the higher order term h3, k3, h2 k etc.
we get
f (a + h, b + k) – f (a, b) =
Putting
∂2 f
∂x 2
=
f (a + h, b + k) – f (a, b) =
=
LM
MN
OP
PQ
∂2 f
∂2 f
∂2 f
1
+ k2 2
h 2 2 + 2hk
∂x∂y
2
∂y ( a, b )
∂x
r,
∂2 f
∂2 f
= s,
= t, then
∂x∂y
∂y 2
1
[h2r + 2hks + k2 t]
2
LM h r + 2hkrs + k tr OP
r
N
Q
L
O
1 M a hr + ksf + k ert − s j P
PQ
r
2M
N
1
2
2 2
2
2
⇒ f (a + h, b + k) – f (a, b) =
2
2
...(iv)
If rt – s2 > 0 then the numerator in R.H.S. of (iv) is positive. Here sign of L.H.S. = sign of r.
Thus,
if rt –s2 > 0 and r < 0, then f (a + h, b + k) – f (a, b) < 0
if rt – s2 > 0 and r > 0, then f (a + h, b + k) – f (a, b) > 0.
Therefore, the Lagrange’s conditions for maximum or minimum are: (U.P.T.U., 2008)
1. If rt –s2 > 0 and r < 0, then f (x, y) has maximum value at (a, b).
2. If rt –s2 > 0 and r > 0, then f (x, y) has minimum value at (a, b).
3. If rt –s2 < 0, then f (x, y) has neither a maximum nor minimum i.e., (a, b) is saddle point.
4. If rt –s2 = 0, then case fail and here again investigate more for the nature of function.
2.3.3 Method of Finding Maxima or Minima
1. Solve
∂f
∂f
= 0 and ∂y = 0, for the values of x and y. Let x = a, y = b.
∂x
The point P (a, b) is called critical or stationary point.
124
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
2. Find r, s and t at x = a, y = b.
3. Now check the following conditions:
(i) If rt – s2 > 0 and r < 0, f (x, y) has maximum at x = a, y = b.
(ii) If rt – s2 > 0 and r > 0, f (x, y) has minimum at x = a, y = b.
(iii) If rt – s2 < 0, f(x, y) has neither maximum nor minimum.
(iv) If rt – s2 = 0, case fail.
Example 1. Find the maximum and minimum of u = x3 + y3 – 63(x + y) + 12xy.
Sol.
∂u
∂u
= 3x2 – 63 + 12y; ∂y = 3y2 – 63 + 12x
∂x
r =
∂ 2u
∂ 2u
∂ 2u
= 12, t =
= 6y
2 = 6x, s =
∂x
∂y 2
∂x∂y
Now for maximum and minimum value
⇒
and
∂u
∂u
= 0 and ∂y = 0
∂x
2
3x + 12y – 63 = 0 ⇒ x2 + 4y – 21 = 0
3y + 12x – 63 = 0 ⇒ y + 4x – 21 = 0
Subtracting (i) and (ii), we get
2
2
...(i)
...(ii)
(x – y) (x + y) – 4 (x – y) = 0
⇒
(x – y) (x + y – 4) = 0
⇒
x = y and x + y = 4
Putting x = y in (i), we get x2 + 4x – 21 = 0 ⇒ (x + 7) (x – 3) = 0
∴
Again putting
x = 3, x = – 7
y = 3, y = – 7
y = 4 – x in eqn (i), we get
∴
x2 + 4 (4 – x) – 21 = 0 ⇒ x2 – 4x – 5 = 0
⇒
(x – 5) (x + 1) = 0 ⇒ x = 5, x = – 1
∴
y = 4 – 5 = – 1, y = 4 – (–1) = 5.
Hence (3, 3), (5, –1), (–7, –7) and (–1, 5) may be possible extremum.
At x = 3, y = 3, we have r = 18; s = 12, t = 18
∴
rt – s2 = 18 × 18 – (12)2 > 0 and r = 18 > 0.
So there is minima at x = 3, y = 3, and the minimum value of u is
(3)3 + (3)3 – 63 (3 + 3) + 12 (3) (3) = – 216.
At x = 5, y = – 1, we have r = 30; s = 12, t = – 6
∴ rt – s2 = 30 (– 6) – (12)2 < 0, so there is neither maxima nor minima at x = 5, y = – 1.
At x = – 7, y = – 7, we have r = – 42; s = 12, t = – 42
∴ rt – s2 = (– 42) (– 42) – (12)2 > 0 and r < 0, so there is maxima at x = – 7, y = – 7 and its
maximum value is
(– 7)3 + (– 7)3 – 63 (– 7 – 7) + 12 (– 7) (– 7) = 784.
At x = – 1, y = 5, we have r = – 6 ; s = 12, t = 30
∴rt – s2 = (– 6) (30) – (12)2 < 0, so there is neither maxima nor minima at x = – 1, y = – 5.
125
DIFFERENTIAL CALCULUS-II
Example 2. Show that minimum value of u = xy +
Sol.
a3 a3
+
is 3a2.
x
y
∂u
∂u
∂2u
= y − a 3 x −2 ;
= x − a 3 y −2 , r = 2 = 2 a 3 x −3 ;
∂x
∂y
∂x
s =
∂2u
∂2u
= 1; t =
= 2 a 3 y −3 .
∂ x∂ y
∂y 2
Now for maximum or minimum we must have
So from
∂u
∂u
=0,
=0
∂x
∂y
∂u
= 0, we get y – a3x–2 = 0 or x2y = a3
∂x
∂u
= 0, we get x – a3y−2 = 0 or xy2 = a3
∂y
Solving (i) and (ii), we get x2y = xy2 or xy (x – y) = 0
and from
or
...(i)
...(ii)
x = 0, y = 0 and x = y.
From (i) and (ii), we find that x = 0 and y = 0 do not hold as it gives a = 0, which is against
hypothesis.
∴ We have x = y and from (i) we get x3 = a3 or x = a and therefore, we have x = a = y.
This satisfies (ii) also. Hence it is a solution.
at
At
x = a = y, we have r = 2a3a–3 = 2, s = 1, t = 2
∴
rt – s2 = (2) (2) – 12 = 3 > 0
Also
r = 2 > 0. Hence, there is minima at x = a = y
∴ The minimum value of u
= xy + (a3/x) + (a3/y)
x = a=y
= a.a + (a3/a) + (a3/a) = a2 + a2 + a2 = 3a2. Hence proved.
Example 3. Discuss the maximum or minimum values of u when u = x3 + y3 – 3axy.
(U.P.T.U., 2004)
2
∂
u
∂u
∂u
= 3x2 – 3ay; ∂y = 3y2 – 3ax; r =
= 6x;
Sol.
∂x
∂x2
∂2u
∂2u
s = ∂x∂y = −3a, t = 2 = 6 y.
∂y
∂u
∂u
Now for maximum or minimum, we must have ∂x = 0, ∂y = 0
∂u
So from
= 0, we get x2 – ay = 0
∂y
∂u
and from
= 0, we get y2 – ax = 0
∂y
Solving (i) and (ii), we get (y2/a)2 – ay = 0
or
y4 – a3y = 0 or y (y3 – a3) = 0 or y = 0, a.
Now from (i), we have when y = 0, x = 0, and when y = a, x = ± a.
...(i)
...(ii)
126
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
But x = – a, y = a, do not satisfy (ii), here are not solutions.
Hence the solutions are x = 0, y = 0; x = a, y = a;
At x = 0, y = 0, we have r = 0, s = – 3a, t = 0.
∴ rt – s2 = 0 – (–3a)2 = negative and there is neither maximum nor minimum at
x = 0, y = 0.
At x = a, y = a, we get r = 6a, s = – 3a, t = 6a
∴ rt – s2 = (6a)(6a) – (–3a)2 = 36a2 – 9a2 > 0
Also r = 6a > 0 if a > 0 and r < 0 if a < 0.
Hence there is maximum or minimum according as a < 0
or a > 0. The maximum or minimum value of u = – a3 according
as a < 0 or a > 0.
Example 4. Determine the point where the function
u = x2 + y2 + 6x + 12 has a maxima or minima.
∂u
∂x
Sol.
= 2x + 6;
r ≡
∂u
= 2y
∂y
∂2 u
∂2u
∂2u
=
2;
s
≡
=
0;
t
≡
=2
∂x∂y
∂x 2
∂y 2
∂u
∂u
= 0,
= 0.
∂x
∂y
Now for maxima or minima we must have
From
∂u
= 0, we get 2x + 6 = 0
∂x
or x = – 3
∂u
From ∂y = 0, we get 2y = 0 or y = 0
Also at x = – 3, y = 0, r = 2, s = 0, t = 2
∴ rt – s2 = 2 (2) – (0)2 = 4 > 0 and r = 2 > 0
Hence, there is minima at x = –3, y = 0.
Example 5. A rectangular box, open at the top, is to have a volume of 32 c.c. Find the
dimensions of the box requiring least material for its construction.
(U.P.T.U., 2005)
Sol.
V = 32 c.c.
Let length = l, breadth = b and height = h
Total surface area
S = 2lh + 2bh + lb
...(i)
S = 2(l + b)h + lb
Now volume
V = lbh = 32 ⇒ b =
32
lh
Putting the value of ‘b’ in equation (i)
S = 2
F l + 32 I h + l F 32 I
H lh K
H lh K
64 32
+
l
h
64 ∂S
32
= 2l − 2
= 2h – 2 ,
∂h
l
h
S = 2lh +
∴
∂S
∂l
...(ii)
h
b
l
Fig. 2.7
...(iii)
127
DIFFERENTIAL CALCULUS-II
For minimum S, we get
∂S
∂l
64
32
=0 ⇒h= 2
2
l
l
∂S
32
16
= 0 ⇒ 2l – 2 = 0 ⇒ l = 2
and
∂h
h
h
From (iv) and (v), we get
= 0 ⇒ 2h –
Putting h = 2, in equation (v), we get l =
Now,
and
∂l 2
∂ 2S
= 2 ⇒ s = 2 and
∂ l ∂h
∴
rt – s2
16
=4
4
32
=4
4×2
b =
∂2S
...(v)
32 × h 4
⇒ h3 = 8 ⇒ h = 2
256
h =
From (ii)
...(iv)
128
128
=
= 2 ⇒r=2>0
64
l3
=
∂2S
64 64
=
=8 ⇒t = 8
8
h3
∂h
= 2 × 8 – 4 = 12 > 0
=
2
⇒ rt – s2 > 0 and r > 0
Hence, S is minimum, for least material
l = 4, b = 4, h = 2.
Example 6. Examine the following surface for high and low points z = x2 + xy + 3x + 2y + 5.
Sol.
∂z
∂x
= 2x + y + 3;
r =
∂z
= x+2
∂y
∂2 z
∂2 z
∂2 z
=
=
=
=
=0
s
t
2
;
1
;
∂x∂y
∂x 2
∂y 2
For the maximum or minimum we must have
∂z
∂x
From
From
∂z
∂x
∂z
∂y
= 0,
∂z
= 0.
∂y
= 0, we get 2x + y + 3 = 0
...(i)
= 0, we get x + 2 = 0.
...(ii)
From (ii), we get x = – 2 and ∴ from (i) y = – 3 + 4 = 1.
Hence, the solution is x = – 2, y = 1 and for these values we have r = 2, s = 1, t = 0.
∴ rt – s2 = (2) (0) – (1)2 = – 1 < 0.
∴ There is neither maximum nor minimum at x = – 2, y = 1.
128
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 7. Discuss the maximum and minimum values of 2 sin
+ cos (x + y).
Sol. Let
∴
1
1
(x + y) cos
(x – y)
2
2
1
1
(x + y) cos
(x – y) + cos (x + y)
2
2
= sin x + sin y + cos (x + y)
∂u
= cos x – sin (x + y);
= cos y – sin (x + y)
∂y
u = 2 sin
∂u
∂x
r =
∂2 u
∂2u
=
–
sin
x
–
cos
(x
+
y);
s
=
= – cos (x + y);
∂x 2
∂x∂y
t =
∂2u
= – sin y – cos (x + y)
∂y 2
For maximum or minimum, we must have
∂u
∂u
= 0,
=0
∂x
∂y
∂u
= 0, we get
cos x – sin (x + y) = 0
∂x
∂u
From ∂y = 0, we get
cos y – sin (x + y) = 0.
Solving (i) and (ii), we get cos x = cos y which gives
x = 2nπ ± y, where n is any integer. In particular x = y.
From
...(i)
...(ii)
When x = y, from (i), we get cos x – sin 2x = 0
or cos x (1 – 2 sin x) = 0 which gives cos x = 0, sin x =
1
1
, then x = nπ + (–1)n. π = y,
2
6
and for these value of x and y, we have
If sin x =
1
2
Œx= y
LM
N
1
1
– cos 2 nπ + ( −1) n . π
3
2
Similarly, t < 0 and s < 0 and r > s, t > s.
∴ rt – s2 > 0. Also r < 0.
r = –
Hence, there is a maximum when x = nπ + (–1)n.
OP < 0. (Note)
Q
1
π = y.
6
1
π = y, Œ x = y
2
1
From here, we get x = y = ± π , (3/2)π, (5/2)π etc.
2
1
If x = π = y, then r = – 1 + 1 = 0, s = 1, t = 0
2
If cos x = 0, then x = 2nπ +
∴ rt – s2 < 0. Hence, there is neither maximum nor minimum at x =
1
π = y.
2
129
DIFFERENTIAL CALCULUS-II
1
If x = − π = y, then r = 1 + 1 = 2 = t, s = 1
2
∴ rt – s2 = (2 × 2) –1 = 3 > 0. Also r > 0
1
π = y. In a similar way we can discuss other values too.
2
Example 8. Show that the distance of any point (x, y, z) on the plane 2x + 3y – z = 12, from
the origin is given by
Hence, there is minimum at x = –
l =
x 2 + y 2 + ( 2x + 3 y − 12) 2 .
Hence find a point on the plane that is nearest to the origin.
Sol.
l = distance between (x, y, z) and (0, 0, 0)
=
( x − 0 ) 2 + ( y − 0) 2 + ( z − 0 ) 2
=
x 2 + y 2 + (2x + 3 y − 12) 2 , Œ z = 2x + 3y – 12
x2 + y 2 + z2
=
l2 = x2 + y2 + (2x + 3y – 12)2 = u (say)
l2 = u = 5x2 + 10y2 + 12xy – 48x – 72y + 144
or
or
∴
∂u
∂x
= 10x + 12y – 48;
...(i)
∂u
= 20y + 12x – 72
∂y
∂2 u
∂ 2u
∂2u
=
10;
s
=
=
12
;
t
=
= 20
∂ y∂ x
∂x 2
∂y 2
= (10) (20) – (12)2 = 56 > 0 and r > 0, so there is a minimum
value of l.
r =
∴
Also
and
rt – s2
∂u
∂x
∂u
∂y
= 0 ⇒ 10x + 12y – 48 = 0 or 5x + 6y = 24
...(ii)
= 0 ⇒ 20y + 12x – 72 = 0 or 5y + 3x = 18
...(iii)
F 12 I , y = F 18 I
H7K
H7K
F 12 I + 3 F 18 I – z = 12
2x + 3y – z = 12 or 2
H7K H7K
Solving (ii) and (iii), we get x =
Also
or
24 + 54 – 7z = 84 or 7z = 24 + 54 – 84 = – 6
∴ The required point is
or z = −
F 12 , 18 , − 6 I .
H 7 7 7K
Example 9. Discuss the maximum and minimum values of
Sol. Let
Then
x4 + 2x2y – x2 + 3y2.
u = x4 + 2x2y – x2 + 3y2
∂u
∂x
= 4x3 + 4xy – 2x ;
∂u
= 2x2 + 6y;
∂y
6
7
130
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
∴
∂2 u
∂2u
∂ 2u
= 12x2 + 4y – 2; s =
= 4x; t =
=6
2
∂x
∂y 2
∂ x∂ y
r =
∂u
∂u
= 0,
= 0.
∂y
∂x
For maximum and minimum, we must have
From
∂u
= 0, we get 2x(2x2 + 2y – 1) = 0
∂x
or 2x2 + 2y – 1 = 0
...(i)
∂u
From ∂y = 0, we get 2x2 + 6y = 0 or x2 + 3y = 0
Solving (i) and (ii), we get 4y + 1 = 0 or y = –
∴ From (ii), we get x2 = – 3y =
∴ The solutions are x =
When x =
...(ii)
1
4
F 3 I or x = ± 1 3
H 4K
2
1
1
1
1
3,y=–
3,y=–
and x = –
2
4
2
4
1
1
3,y=–
, we get
2
4
r = 12
∴ rt – s2 = 6 × 6 –
F 3 I + 4F − 1 I − 2 = 6 , s = 4 FH 1 3 IK = 2 3 , t = 6
H 4K H 4K
2
e2 3 j > 0. Also r > 0
2
∴ There is a minimum when x =
Again when x = –
1
1
3, y =–
2
4
1
1
3, y=–
, we have r = 6, s = – 2 3 , t = 6
2
4
∴ rt – s2 = (6) (6) –
e−2 3 j > 0. Also r > 0.
2
Hence as before there is a minimum when
1
1
3,y=–
x =
.
2
4
Example 10. Find the shortest distance between the lines
x − 3 y − 5 z −7
x +1 y +1 z+1
=
=
.
=
=
and
1
−2
1
7
−6
1
x − 3 y − 5 z −7
=
=
Sol. Let
= λ ⇒ x = λ + 3, y = 5 – 2λ, z = 7 + λ
1
−2
1
Thus any point P on the line is (3 + λ, 5 – 2λ, 7 + λ)
x +1
y +1 z+1
=
= µ ⇒ x = – 1 + 7µ, y = – 1 – 6µ , z = – 1 + µ
=
7
−6
1
The point Q is (– 1 + 7 µ, – 1 – 6µ, – 1 + µ)
∴ Distance between these two lines is
and let
D =
⇒
b3 + λ + 1 − 7µg + ( 5 − 2λ + 1 + 6µ) + (7 + λ + 1 − µ)
2
D2 = u(Say) = 6λ2 +86µ2 – 40λµ + 105.
2
2
131
DIFFERENTIAL CALCULUS-II
∂u
∂u
= 12λ – 40µ,
= 172µ – 40λ.
∂λ
∂µ
∂u
∂u
= 0 and
= 0 ⇒ 12λ – 40µ = 0, 172µ – 40λ = 0.
But
∂λ
∂µ
Solving these equations, we get λ = 0, µ = 0
∴
∂2u
∂2 u
∂2u
=
172
=
12,
t
=
,
= s = – 40
∂λ∂µ
∂λ2
∂µ 2
Now,
rt – s2 = (12)·(172) – (– 40)2 = 464 > 0
⇒
rt – s2 > 0 and r > 0
Hence u occurs minimum value at λ = 0 and µ = 0.
The shortest distance is given by
r =
D =
4 2 + 62 + 82 =
116 = 2 29 .
Example 11. The temperature T at any point (x, y, z) in space is T (x, y, z) = Kxyz2 where
K is a constant. Find the highest temperature on the surface of the sphere x2 + y2 + z2 = a2.
(U.P.T.U., 2008)
Sol.
T = Kxyz2
...(i)
2
2
2
2
2
2
2
2
x + y + z = a ⇒z = a – x – y
From (i)
T = Kxy (a2 – x2 – y2)
∂T
∂x
∴
= Ky (a2 – x2 – y2) – 2 Kx2y = Ky(a2 – 3x2 – y2)
∂T
2
2
2
∂y = Kx(a – x – 3y )
for maximum and minimum value
∂T
∂T
= 0 and
= 0 ⇒ x = 0 and y = 0
∂x
∂y
or
3x2 + y2 = a2
x2 + 3y2 = a2
Similarly,
Solving
x=y=±
r =
a
2
∂ 2T
∂2T
=
–
6
Kxy,
s
=
= K (a2 – 3x2 – 3y2)
∂x 2
∂ x∂ y
∂ 2T
t =
∂y 2 = – 6Kxy
and
At (0, 0)
r = 0, s = Ka2 and t = 0
∴
rt – s2 = 0.0 – Ka2 = – Ka2 < 0
So, there is neither maximum nor minimum at x = 0 and y = 0
At
a
a
a
a
,y=
and x = –
,y=–
2
2
2
2
3
a 2K
6 2
3 2
r = – Ka = – Ka < 0, t = − Kxy, s = −
2
2
4
2
x =
132
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
9 2 4 a4K2
K a –
= 2K2a4 > 0
4
4
∴ rt – s2 > 0 and r < 0
Hence, T has a maximum value at x = + a/2 and y = ± a/2
rt – s2 =
The maximum value of T = K ·
FG IJ = Ka .
H K 8
4
a2 a2
4 2
Example 12. Find the maximum and minimum values of the function
z = sin x sin y sin (x + y).
Sol. Given z = sin x sin y sin (x + y)
or
=
1
[2 sin x sin y] sin (x + y)
2
=
1
[cos (x – y) – cos (x + y)] sin (x + y)
2
=
1
[2 sin (x + y) cos (x – y) – 2 sin (x + y) cos (x + y)]
4
z =
∴
1
[sin 2x + sin 2y – sin(2x + 2y)]
4
1
∂z
=
[cos 2x – cos (2x + 2y)]
2
∂x
∂z
1
=
[cos 2y – cos (2x + 2y)]
∂y
2
r =
∂2 z
= – sin 2x + sin (2x + 2y)
∂x 2
...(A)
s =
∂2z
= sin (2x + 2y)
∂x∂y
...(B)
t =
∂2 z
= – sin 2y + sin (2x + 2y)
∂y 2
...(C)
For maximum or minimum, we must have
∂z
∂z
= 0,
=0
∂y
∂x
From
∂z
= 0, we get cos 2x – cos (2x + 2y) = 0
∂x
...(i)
From
∂z
= 0, we get cos 2y – cos (2x + 2y) = 0
∂y
...(ii)
Solving (i) and (ii), we get cos 2x = cos 2y which gives
2x = 2nπ ± 2y. In particular 2x = 2y or x = y
When x = y, from (i), we get cos 2x – cos 4x = 0
or
cos 2x –(2 cos2 2x – 1) = 0
or
2 cos2 2x – cos 2x – 1 = 0
Œ cos 2θ = 2 cos2 θ – 1
133
DIFFERENTIAL CALCULUS-II
1 ± (1 + 8 )
or
cos 2x =
or
2x = 2nπ ± 0, 2mπ ±
or
x = nπ, mπ ±
4
π
3
In particular x =
=
1± 3
1
= 1, –
4
2
2π
, where m, n are zero or any integers
3
π
3
π
π
, we have y = x =
3
3
2π
4π
and then r = – sin
+ sin
, from (A)
3
3
When x =
3
3
–
=– 3;
2
2
4π
s = sin
, from (B)
3
= –
3
2
or
s = –
and
t = – sin
= –
2π
4π
+ sin
, from (C)
3
3
3
–
2
3
=– 3
2
e− 3 j e− 3 j – FGH − 23 IJK = 3 – 34 = 94
2
∴
rt – s2 =
= positive.
π
π
= y, rt – s2 > 0 , r < 0, so there is a maximum at x =
= y.
Thus at x =
3
3
π π
π
π
+
Hence, maximum value = sin
. sin
. sin
3 3
3
3
F
H
3
3
3
3 3
⋅
⋅
=
.
2
2
2
8
π
π
If we take
x = –
, then y = x = –
3
3
1
3,t= 3
and
r = 3, s=
2
9
∴
rt – s2 =
>0, r>0
4
π
There is a minimum at x = –
= y.
3
I
K
=
Hence, the minimum value = sin
F − π I sin F − π I sin RSF − π I + F − π I UV
H 3 K H 3 K TH 3 K H 3 K W
134
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
= – sin
= −
π
π
2π
· sin
sin
3
3
3
3
3
3
3 3
⋅
⋅
= −
·
2
2
2
8
Example 13. Find the local minima or local maxima of the function
f(x, y) = 4x2 + y2 – 4x + 6y – 15.
Sol. We have
f(x, y) = 4x2 + y2 – 4x + 6y – 15
∂f
= 8x – 4
∂x
For stationary point
and
∂f
= 2y + 6.
∂y
∂f
∂f
= 0
= 0 and
∂y
∂x
1
and 2y + 6 = 0 ⇒ y = – 3
2
Now, the given function can be written as
f(x, y) = (2x – 1)2 + (y + 3)2 – 25
...(i)
2
2
Since (2x – 1) ≥ 0 and (y + 3) ≥ 0.
From (i), we get
f(x, y) ≥ – 25 for all values of x and y. Hence f(1/2, – 3) = – 25 is a local minimum.
∴
8x – 4 = 0 ⇒ x =
Example 14. Find the local minima or local maxima of the function
f(x, y) = – x2 – y2 + 2x + 2y + 16.
∂f
= – 2x + 2 = 0 ⇒ x = 1
∂x
Sol.
∂f
= – 2y + 2 = 0 ⇒ y = 1
∂y
f(x, y) = 18 – {(x – 1)2 + (y – 1)2}
Now,
...(i)
Since (x – 1) ≥ 0 and (y – 1) ≥ 0, from (i) we get f(x, y) ≤ 18 for all values of x and y.
2
2
Hence f(1, 1) = 18 is a local maxima.
EXERCISE 2.3
1. Discuss the maximum values of u, where
LMAns. x = a , y = a OP
2
2Q
N
2. Find the points (x, y) where the function u = xy (1 – x – y) is maximum or minimum.
LMAns. Maxima at x = 1 , y = 1 OP
3
3Q
N
u = 2a2xy – 3ax2y – ay3 + x3y + xy3
3. Find the extrema of f (x, y) = (x2 + y2) e(6x + 2x2).
[Ans. minima at (0, 0) minimum value = 0 and at (–1, 0) min. value = e–4,
1
saddle point (– , 0)].
2
135
DIFFERENTIAL CALCULUS-II
4. If the perimeter of a triangle is constant prove that the area of this triangle is maximum
when the triangle is equilateral.
[Hint: 2s = a + b + c, ∆ =
s( s − a)( s − b)( s − c ) ]
2s
]
3
5. Show that the rectangular solid of maximum volume that can be inscribed in a sphere is
a cube.
[Ans. Maximum when a = b = c =
[Hint: V = xyz, diagonal of cubic =
V = xy
x2 + y 2 + z2 = d ⇒ z =
d2 − x2 − y2
OP
Q
d 2 − x 2 − y 2 so
6. Find the shortest distance from origin to the surface xyz2 = 2.
[Ans. 2]
7. In a plane triangle ABC find the maximum value of cos A cos B cos C.
[Ans. 1/8]
8. Discuss the maximum or minimum values of u given by u = x3y2 (1 – x – y).
[Ans. Maximum at x = 0, y = 1/3]
9. Find the maximum and minimum values of u = 6xy + (47 – x – y) (4x + 3y).
[Ans. Max. value of u = 3384]
10. Discuss the maxima and minima of the function
f (x, y) = cos x, cos y cos (x + y)
(U.P.T.U., 2007)
11. Divide 24 into three parts such that the continued product of the first, square of the
second and the cube of the third may be maximum.
[Ans. 4, 8, 12]
12. Examine for extreme values: u (x, y) = x2 + y2 + 6x + 12.
[Ans. Min. value = 3]
2.4 LAGRANGE'S* METHOD OF UNDETERMINED MULTIPLIERS
Let φ (x, y, z) is a function of three independent variables, where x, y, z are related by a known
constraint g(x, y, z) = 0
Thus the problem is Extrema of
u = f(x, y, z)
...(i)
Subject to
g (x, y, z) = 0
...(ii)
For stationary point
∂f
∂f
∂f
=
=
=0
∂x
∂z
∂y
∴
df =
∂f
∂f
∂f
dx +
dy +
dz = 0
∂x
∂z
∂y
∂g
∂g
∂g
dx +
dy +
dz = 0
∂z
∂y
∂x
Multiplying eqn. (iv) by λ and adding to (iii), we obtain
From (ii)
dg =
FG ∂f + λ ∂g IJ dx + FG ∂f + λ ∂g IJ dy + FG ∂f + λ ∂g IJ dz = 0
H ∂x ∂x K H ∂y ∂y K H ∂z ∂z K
*Joseph Louis Lagrange (1736–1813).
...(iii)
...(iv)
...(v)
136
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Since x, y, z are independent variables
∴
∂g
∂f
+λ
= 0
∂x
∂x
∂f
∂g
+λ
∂y
∂y = 0
...(vi)
...(vii)
∂g
∂f
+λ
= 0
...(viii)
∂z
∂z
On solving (ii), (vi), (vii) and (viii), we can find x, y, z and λ for which f (x, y, z) has maximum
or minimum.
Notes: 1. The Lagrange's method of undetermined multiplier's introduces an additional unknown
constant λ known as Lagrange's multiplier.
2. Nature of stationary points cannot be determined by Lagranges method.
Example 1. Determine the maxima and minima of x2 + y2 + z2 when ax2 + by2 + cz2 = 1.
Sol. Let
f (x, y, z) = x2 + y2 + z2
...(i)
and
g(x, y, z) ≡ ax2 + by2 + cz2 – 1 = 0
...(ii)
∂f
∂f
∂f
= 2x, ∂y = 2y,
= 2z
∂x
∂z
∂g
∂g
∂g
From (ii)
= 2ax, ∂y = 2by,
= 2cz.
∂x
∂z
Now from Lagrange's equations, we get
From (i)
∂f
∂g
= 2x + λ . 2ax = 0 ⇒ 2x (1 + λa) = 0 ⇒ x (1 + λa) = 0
...(iii)
+λ
∂x
∂x
∂f
∂g
= 0 ⇒ 2y + λ · 2by = 0 ⇒ 2y (1 + λb) = 0 ⇒ y ( 1 + λb) = 0 ...(iv)
+ λ
∂y
∂y
and
or
∂f
∂g
= 0 ⇒ 2z + λ · 2cz = 0 ⇒ 2z (1 + λc) = 0 ⇒ z ( 1 + λb) = 0 ...(v)
+ λ
∂z
∂z
Multiplying these equations by x, y, z respectively and adding, we get
x2 (1 + λa) + y2 (1 + λb) + z2 (1 + λc) = 0
(x2 + y2 + z2) + λ (ax2 + by2 + cz2) = 0
...(vi)
Using (i) and (ii) in above equation, we get
f+ λ = 0⇒λ=– f
Putting λ = – f in equations (iii), (iv) and (v), we get
⇒
i.e.,
x (1 – fa) = 0, y (1 – fb) = 0, z (1 – fc) = 0
1 – fa = 0, 1 – fb = 0, 1 – fc = 0
f =
1 1 1
, , . These give the max. and min. values of f.
a b c
Example 2. Find the extreme value of x2 + y2 + z2, given that ax + by + cz = p.
(U.P.T.U., 2007)
Sol. Let
u = x2 + y2 + z2
...(i)
Given
ax + by + cz = p.
...(ii)
137
DIFFERENTIAL CALCULUS-II
For max. or min. from (i), we have
du = 2x dx + 2y dy + 2z dz = 0.
...(iii)
Also from (ii),
a dx + b dy + c dz = 0.
...(iv)
Multiplying (iv) by λ and adding in (iii), we get (x dx + y dy + z dz) + λ (a dx + b dy + c dz)
= 0.
Equating the coefficients of dx, dy and dz to zero, we get
x + λa = 0, y + λb = 0, z + λc = 0
These are Lagrange's equations.
Multiplying these by x, y, z respectively and adding, we get
...(v)
x (x + λa) + y (y + λb) + z (z + λc) = 0
or
or
(x2 + y2 + z2) + λ (ax + by + cz) = 0
u + λp = 0 or λ = – u/p.
∴ From (v), we get
x–
FG au IJ = 0, y – FG bu IJ = 0, z – FG cu IJ = 0
HpK
HpK
HpK
x
y
u
z
=
=
=
a
p
b
c
or
x y z
= =
a b c
or
F x I + b FG y IJ + c F z I = p
H aK H bK H cK
F xI F xI F xI
a H K + b H K + c H K = p, from (vi)
a
a
a
F x I = p or x = ap
(a + b + c )
H aK
a +b +c
From (ii), we get a2
2
2
2
or
2
or
2
2
2
2
Similarly,
bp
y =
a +b +c
These give the minimum value of u.
Hence minimum value of u is
u =
=
2
2
2
2
2
,
z=
a2 p2
(a 2 + b2 + c 2 )2
( a 2 + b 2 + c 2 )p 2
( a2 + b2 + c 2 ) 2
+
=
2
cp
a + b2 + c 2
2
b2 p2
= 1 and lx + my + nz = 0.
Sol. Let
Given
u =
x 2 y2 z2
+
+
a4 b4 c 4
x2 y2 z2
+
+
= 1
a2 b2 c2
lx + my + nz = 0
+
( a2 + b2 + c 2 )2
p2
(a2 + b2 + c 2 )
Example 3. Find the maximum and minimum values of
and
...(vi)
c 2 p2
(a 2 + b2 + c 2 )2
·
x 2 y2 z2
x2 y2 z2
+
+
+
+
where
a4 b4 c 4
a2 b2 c2
...(i)
...(ii)
...(iii)
138
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
From (i), (ii) and (iii), we get
FG x IJ dx + FG y IJ dy + FG z IJ dz = 0
Ha K Hb K Hc K
FG x IJ dx + FG y IJ dy + FG z IJ dz = 0
Ha K Hb K Hc K
and
4
4
4
...(iv)
2
2
2
...(v)
l dx + m dy + n dz = 0
Multiplying (v) and (vi) by λ1, λ2 and adding, we get
...(vi)
FG x + λ x + λ lIJ dx + FG y + λ y + λ mIJ dy + FG z + λ z + λ nIJ dz = 0
Ha a
K Hb b
K
K Hc c
1
2
4
2
1
2
4
2
1
2
4
2
Equating to zero the coefficients of dx, dy and dz, we get
x λ1x
y λ y
z λ z
+ 2 + λ 2 l = 0 , 4 + 12 + λ 2 m = 0 , 4 + 12 + λ 2 n = 0
a4
a
c
c
b
b
These are Lagrange’s equations.
Multiplying these by x, y, z and adding, we get
...(vii)
FG x + y + z IJ + λ FG x + y + z IJ + λ (lx + my + nz) = 0
Ha b c K Ha b c K
or
or
2
2
2
4
4
4
or
or
or
or
or
2
2
2
2
2
2
u + λ1 (1) + λ2 (0) = 0, using (i), (ii) and (iii)
u + λ1 = 0 or λ1 = – u
∴ From (vii), we have
y uy
x ux
z uz
− 2 + λ 2 l = 0,
−
+ λ 2n = 0
− 2 + λ 2 m = 0,
4
4
a
a
c4 c2
b
b
x (1 – a2u) = – la4 λ2, y (1 – b2u) = – mb4 λ2, z (1 – c2u) = – nc4λ2
or
or
1
2
−la 4 λ 2
− mb 4 λ 2
− nc 4 λ 2
, z=
2 , y =
2
1− a u
1−b u
1 − c2u
Substituting these values in (iii), we get
x =
l2 a 4
m2 b 4
n2 c 4
+
+
=0
1 − a2 u 1 − b 2 u 1 − c 2 u
Σl2 a4 (1 – b2u) (1 – c2u) = 0
Σl2 a4 {b2c2u2 – (b2 + c2) u + 1} = 0
u2 (Σl2a4b2c2) – u {Σl2a4 (b2 + c2)} + Σl2a4 = 0
RSΣl a FG 1 + 1 IJ UV u + Σl a = 0
T H c b KW
R| F a a I U| F a l + b m + c n IJ = 0,
(l a + m b + n c ) u – SΣl G + J V u + GH
b c
c a
a b K
|T H c b K |W
2 2
a2b2c2 (l2a2 + m2b2 + n2c2)u2 – a2b2c2
2 2
2 2
2 2
2
2
2
2 4
2
2
2
2 2
2
2
2 2
2
2
2 2
2 2
2 2
which gives the max. and min. values of u.
Example 4. Find the minimum value of x2 + y2 + z2 when yz + zx + xy = 3a2.
Sol. Let u = x2 + y2 + z2
Given
yz + zx + xy = 3a2.
For max. or min. from (i), we have
...(i)
...(ii)
du = 2x dx + 2y dy + 2z dz = 0
...(iii)
139
DIFFERENTIAL CALCULUS-II
Also from (ii), (y dz + z dy) + (z dx + x dz) + (x dy + y dx) = 0
(z + y) dx + (x + z) dy + (y + x) dz = 0.
...(iv)
Multiplying equation (iv), by λ and adding in (iii), we get
[x + λ (z + y)] dx + [y + λ (x + z)] dy + [z + λ (y + x)] dz = 0
Equating the coefficients of dx, dy, dz to zero, we get
x + λ (z + y) = 0, y + λ (x + z) = 0, z + λ (x + y) = 0
...(v)
These are Lagrange’s equations
Multiplying these by x, y, z respectively and adding, we get
[x2 + λx (z + y)] + [y2 + λy (x + z)] + [z2 + λz (y + x)] = 0
(x2 + y2 + z2) + 2λ (xy + yz + zx) = 0
or
u + 2λ (3a2) = 0 or λ = −
or
or
6a2
u ( z + y)
∴ From (v), we get x =
or
u
6a
2
.
,y=
u ( x + z)
6a
2
,z=
u ( y + x)
6 a2
x
z
u
y
= 2
=
=
z+y
x + y 6a
x+z
2
– 6a x + uy + uz = 0, ux – 6a2y + uz = 0, ux + uy – 6a2z = 0
Eliminating x, y, z, we get
or
−6 a 2
u
u
−6 a
= 0, which gives the max. and min. values of u.
u
−6 a
2
u
u
−6 a 2
u + 6a 2
u + 6 a2
u
−6 a 2 − u
0
0
−6 a − u
−6 a 2
1
1
u
−1
0
0
−1
u
(u + 6a2)2
or
u
2
2
u
2 2
= 0, C2 → C2 – C1 C3 → C3 – C1
= 0 or (u + 6a2)2
−6 a 2
1
1
u
−1
0
0
1
−1
= 0, R3 → R3 – R2
2
or
(u + 6a ) [– 6a – u(–1–1)] = 0
|Expand with respect to first column
2 2
2
2
2
or
(u + 6a ) [– 6a + 2u] = 0 or u = – 6a , 3a . But u cannot be equal to – 6a2, since sum
of three squares (viz. x2, y2, z2) from (i), cannot be negative. Hence u = 3a2 gives max. or min. value
of u.
Example 5. Find the minimum distance from the point (1, 2, 0) to the cone z2 = x2 + y2.
(U.P.T.U., 2006)
Sol. Let (x, y, z) be any point on the cone then distance from the point (1, 2, 0) is
D2 = (x – 1)2 + (y – 2)2 + (z – 0)2
u = (x – 1)2 + (y – 2)2 + z2
Let
Subject to
2
2
2
x + y – z = 0
...(i)
...(ii)
140
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
for minimum, from (i) and (ii), we get
du = (x – 1)dx + (y – 2)dy + zdz = 0
xdx + ydy – zdz = 0
...(iii)
...(iv)
Multiplying equation (iv) by λ and adding in (iii), we get
(x – 1)dx + (y – 2)dy + zdz + λ (xdx + ydy – zdz) = 0
⇒
⇒
{x (1 + λ) –1} dx + {y (1 + λ) – 2} dy + {z (1 – λ)} dz = 0
x (1 + λ) –1 = 0, y ( 1 + λ) –2 = 0, z (1 – λ) = 0
⇒
x =
1
,
1+λ
y=
2
, λ=1
1+λ
...(v)
1
1
2
=
, y=
=1
2
1+1
1+1
Putting the value of x and y in equation (ii), we get
∴
x =
1
5
5
+ 1 – z2 = 0 ⇒ z2 =
⇒z=±
4
4
2
Hence, the minimum distance from the point (1, 2, 0) is
F 1 − 1I + (1 − 2) + FG 5 IJ = 1 + 1 + 5 = 10
D =
H2 K
H2K 4 4 4
2
2
2
2
D2 =
or
5
⇒ D=
2
5
.
2
Example 6. Show that the rectangular solid of maximum volume that can be inscribed in
a sphere is a cube.
(U.P.T.U., 2003)
Sol. Let the length, breadth and height of solid are
Z
l = 2x
b = 2y
h = 2z
∴ Volume of the solid V = lbh = 2x·2y·2z
⇒
V = 8xyz
Equation of the sphere
⇒
x2 + y2 + z2 = R2
x + y2 + z2 – R2 = 0
2
...(i)
...(ii)
For maximum differentiating (i), (ii), we get
dV = 8yzdx + 8xzdy + 8xydz = 0
and
2xdx + 2ydy + 2zdz = 0
Multiplying (iv) by λ and adding in (iii), we get
Y
X
Fig. 2.8
...(iii)
...(iv)
8yzdx + 8xzdy + 8xydz + λ (2xdx + 2ydy + 2zdz) = 0
⇒
(2λx + 8yz)dx + (2λy + 8xz)dy + (2λz + 8xy)dz = 0
Equating the coefficient of dx, dy and dz to zero, we get
⇒ λx = – 4yz , λy = – 4xz, λz = – 4xy
These are Lagrange’s equations
...(v)
141
DIFFERENTIAL CALCULUS-II
Multiplying equation (v) by x, y, z respectively, we get
λx2 = – 4xyz, λy2 = – 4xyz, λz2 = – 4xyz
From these, we get
λx2 = λy2 = λz2
⇒
x2 = y2 = z2
⇒
x = y= z
Thus, the rectangular solid is a cube. Proved.
Example 7. Use the method of Lagrange's multiplier to find the volume of the largest
rectangular parallelopiped that can be inscribed in the ellipsoid
Sol. Let
l = 2x
b = 2y
h = 2z
V = 8xyz
∴
...(i)
2
2
and
x 2 y2 z2
+
+
=1.
a2 b2 c2
(U.P.T.U., 2002, 2000)
2
y
x
z
+ 2 + 2 −1= 0
2
a
b
c
For largest volume, from (i) and (ii), we get
...(ii)
dV = yz·dx + xz·dy + xy·dz = 0
and
...(iii)
y
x
z
dx + 2 dy + 2 dz = 0
2
a
b
c
Now, equation (iii) + λ × equation (iv), we get
...(iv)
λ
λ
λ
z) dz = 0
2 x) dx + (xz +
2 y) dy + (xy +
a
b
c2
λ
λ
λ
⇒
yz + 2 x = 0, xz + 2 y = 0, xy + 2 z = 0
a
b
c
Multiplying (v) with x, y, z respectively and adding then, we get
(yz +
L x y + z OP = 0
3xyz + λ M +
Na b c Q
2
2
2
2
2
2
⇒
3xyz + λ = 0 (using ii)
⇒
λ = – 3xyz
Putting the value of λ in any one of equation (v), we get
⇒
Similarly,
1–
FG
H
x
a2
= 0 ⇒ yz 1 −
3x2
a2
= 0⇒ x=
yz – 3xyz ·
y =
Hence, the largest volume V = 8 ·
b
abc
3 3
3
⋅
, z=
3x 2
a2
a
,
3
c
3
IJ = 0
K
...(v)
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 8. Determine the point on the paraboloid z = x2 + y2 which is closest to the point
(3, – 6, 4).
Sol. Let (x, y, z) be any point on paraboloid nearest to the point (3, – 6, 4)
∴
D =
⇒
Let
D2 = (x – 3)2 + (y + 6)2 + (z – 4)2
u = (x – 3)2 + (y + 6)2 + (z – 4)2
2
( x − 3)2 + ( y + 6)2 + ( z − 4) 2
...(i)
2
Subject to
x + y – z = 0
For minimum distance, differentiating (i) and (ii), we get
du = 2(x – 3) dx + 2(y + 6) dy + 2(z – 4) dz = 0
and
2xdx + 2ydy – dz = 0
...(ii)
...(iii)
...(iv)
Now, equation (iii) + λ × (iv), we get
{2x (1 + λ) – 6} dx + {2y (1 + λ) + 12} dy + {2z – λ – 8) dz = 0
⇒
x(1 + λ) – 3 = 0, y (1 + λ) + 6 = 0, 2z – (λ + 8) = 0
3
6
(λ + 8)
, y=–
,z=
1+λ
1+λ
2
Putting the values of x, y, z in equation (ii), we get
⇒
x=
...(v)
9
36
(λ + 8)
–
= 0
2 +
(1 + λ )
(1 + λ ) 2
2
45
(λ + 8)
–
= 0
(1 + λ )2
2
or
⇒
90 – (λ + 1)2 (λ + 8) = 0
⇒
⇒
λ3 + 10λ2 + 17λ – 82 = 0
λ = 2
Hence
x = 1, y = – 2, z = 5.
Example 9. Find the maximum and minimum distances from the origin to the curve
3x2 + 4xy + 6y2 = 140.
Sol. Let (x, y) be any point on the curve
∴ distance from (0, 0) is given by
D2 = x2 + y2 = u (say)
Subject to
3x2 + 4xy + 6y2 – 140 = 0
...(i)
...(ii)
For maximum and minimum from (i) and (ii), we get
du = 2xdx + 2ydy = 0
6xdx + 4dx y + 4xdy+ 12ydy = 0
...(iii)
...(iv)
Now, equation (iii) + λ × (iv), we get
{x (2 + 6λ) + 4yλ} dx + {y (2 + 12λ) + 4xλ}dy = 0
⇒
2x + λ (6x + 4y) = 0
...(v)
2y + λ (12y + 4x) = 0
...(vi)
143
DIFFERENTIAL CALCULUS-II
Multiplying the above equations by x, y respectively and adding, we get
2 (x2 + y2) + 2λ (3x2 + 4xy + 6y2) = 0
⇒
2u + 2λ (140) = 0
(using (i) and (ii))
u
140
Putting the value of λ in eqns. (v) and (vi), we get
∴
λ = –
u
(6x + 4y) = 0 ⇒ (140 – 3u)x – 2uy = 0
140
u
(12y + 4x) = 0 ⇒ – 2ux + (140 – 6u)y = 0
and
2y –
140
This system has non-trivial solution if
2x –
140 − 3u
−2u
−2u
(140 − 6u) = 0
⇒
(140 – 3u) (140 – 6u) – 4u2 = 0
⇒
14u2 – 1260u + (140)2 = 0
u2 – 90u – 1400 = 0
⇒
(u – 70) (u – 20) = 0
u = 70, 20
Thus, the maximum and minimum distances are
70 ,
20 .
| As D2 = u
Example 10. A wire of length b is cut into two parts which are bent in the form of a square
and circle respectively. Find the least value of the sum of the areas so found.
Sol. Let part of square = x
and
part of circle = y ⇒ x + y = b
∴
side of square =
x
,
4
radius of circle =
y
|As 2πr = y
2π
area of square =
x2
16
area of circle =
πy 2
4π 2
=
y2
4π
2
y
x2
+
16
4π
Subject to b =x + y ⇒ x + y – b = 0
For minimum from (i) and (ii), we get
Here, let
u = sum of areas =
y
x
dx +
dy = 0
8
2π
dx + dy = 0
du =
and
...(i)
...(ii)
...(iii)
...(iv)
144
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Now, (iii) + λ × (iv), we get
xdx ydy
+
+ λ (dx + dy) = 0
8
2π
y
x
+ λ dy = 0
+ λ dx +
⇒
8
2π
y
x
+ λ = 0,
+λ = 0
⇒
8
2π
⇒
x = – 8λ, y = – 2πλ
Putting x and y in equation (ii), we get
b
– 8λ – 2πλ = b ⇒ λ = –
8 + 2π
8b
2πb
Thus
x = – 8λ =
, y = – 2πλ =
8 + 2π
8 + 2π
∴ The least value of areas is, from (i)
F
H
I
K
IJ
K
FG
H
...(v)
x2 y 2
+
16 4π
64b 2
4π 2 b 2
+
=
16(8 + 2π) 2 4π (8 + 2π) 2
u =
=
b 2 ( π + 4)
4( π + 4) 2
=
b2
.
4( π + 4)
Example 11. Find the dimension of rectangular box of maximum capacity whose surface
area is given when (a) box is open at the top (b) box is closed.
(U.P.T.U., 2008)
Sol. Let the length, breadth and height of box are x, y, z respectively.
So volume
V = xyz
...(i)
There will be two surface area one for open and one for closed box
∴
or
Here
nxy + 2yz + 2zx = S (say)
...(ii)
g(x, y, z) ≡ nxy + 2yz + 2zx – S = 0
n = 1,
when the box is open on the top
n = 2,
when the box is closed.
...(iii)
The Lagrange’s equations are
∂g
∂V
= yz + λ(ny + 2z) = 0
+λ
∂x
∂x
...(iv)
∂g
∂V
= xz + λ(nx + 2z) = 0
+λ
∂y
∂y
...(v)
∂g
∂V
= xy + λ(2y + 2x) = 0
+λ
∂z
∂z
Multiplying (iv), (v), (vi) by x, y, z respectively and adding, we get
...(vi)
3xyz + λ [2(nxy + 2yz + 2zx)]= 0
or
3V + λ[2S] = 0 ⇒
λ= −
3V
2S
...(vii)
145
DIFFERENTIAL CALCULUS-II
Putting value of λ from (vii) in (iv), (v) and (vi), we get
3xyz
3V
(ny + 2z) = 0 ⇒ yz −
(ny + 2z) = 0
2S
2S
2S
or
nxy + 2xz =
3
2S
Similarly
nxy + 2yz =
3
2S
2yz + 2zx =
3
From (viii) and (ix), we get
x = y
and from (ix), (x), we get
yz −
ny
nx
=
2
2
Putting (xi) and (xii) in equation (ii), we have
ny = 2z ⇒ z =
...(viii)
...(ix)
...(x)
...(xi)
...(xii)
nx
nx
+2·
· x = S ⇒ 3nx2 = S
2
2
S
or
x2 =
3n
(a) When box is open n = 1
nx · x + 2 · x ·
∴
x2 =
S
3
⇒
x=
S
3
Hence, the dimensions of the open box are x = y =
1 S
S
and z =
2 3
3
S
6
S
⇒ x=
6
Hence, the dimensions of the closed box are
∴
x2 =
x = y=
S
6
(b) When box is closed n = 2
and
z=
S
·
6
EXERCISE 2.4
1. Find the maximum and minimum distances of the point (3, 4, 12) from the sphere,
x2 + y2 + z2 = 1.
(U.P.T.U., 2001) Ans. Dmin = 12, Dmax = 14
2. Find the maximum and minimum distances from the origin to the curve
x2 + 4xy + 6y2 = 140.
[U.P.T.U. (C.O.), 2003] Ans. Dmin = 4.5706, Dmax. = 21.6589
3. The temperature T at any point (x, y, z) in space is T = 400 xyz2. Find the highest temperature at the surface of a sphere x2 + y2 + z2 = 1.
Ans. T = 50
146
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
4. Find the maxima and minima of x2 + y2 + z2 subject to the conditions : ax2 + by2 + cz2
LMAns. l + m + n = 0OP
N ( au − 1) (bu − 1) (cu − 1) Q
2
= 1, lx + my + nz = 0.
2
2
5. Find the maximum value of u = xpyqzr when the variables x, y, z are subject to the
LMAns. u = FG p IJ FG q IJ F r I OP
H a K H bK H cK Q
N
p
condition ax + by + cz = p + q + r.
q
r
6. A rectangular box, which is open at the top has a capacity of 256 cubic feet. Determine
the dimensions of the box such that the least material is required for the construction of
the box. Use Lagrange's method of multipliers to obtain the solution.
Ans. length = breadth = 8′, height = 4′
7. Find the minimum value of x2 + y2 + z2 subject to condition
1 1 1
+ + = 0.
x y z
Ans. Minimum value = 27
8. Determine the point in the plane 3x – 4y + 5z = 50 nearest to the origin.
Ans. (3, – 4, 5)
9. Find the length and breadth of a rectangle of maximum area that can be inscribed in the
3 2
, b = 2 , area = 12
2
10. Divide 24 into three parts such that the continued product of the first square of the
second and the cube of the third may be maximum.
ellipse 4x2 + 9y2 = 36.
Ans. l =
Ans. 4, 8, 12, maximum value = 4·82·123
11. Find the volume of the largest rectangular parallelopiped that can be inscribed in the
ellipsoid of revolution 4x2 + 4y2 + 9z2 = 36.
Ans. Maximum volume = 16 3
12. Using the Lagrange's method of multipliers, find the largest product of the numbers x,
y and z when x2 + y2 + z2 = 9.
Ans. 3 3
13. A torpedo has the shape of a cylinder with conical ends. For given surface area, show
2
that the dimensions which give maximum value are l = h =
r, where l is the length
5
of the cylinder, r is the radius and h is the altitude of cone.
14. Find the dimensions of a rectangular box, with open top of given capacity (volume) such
that the sheet metal (surface area) required is least.
Ans. x = y = 2z = (2V)1/3, V = volume
15. Find the maximum and minimum values of
x 2 + y 2 when 13x2 – 10xy + 13y2 = 72.
Ans. Maximum = 3, minimum = 2
147
DIFFERENTIAL CALCULUS-II
16. A tent of given volume has a square base of side 2a and has its four sides of height b
vertical and is surmounted by a pyramid of height h. Find the values of a and b in terms
of h so that the canvas required for its construction be minimum.
1
h
5
h, b =
(4a2)h, S = 8ab + 4a a 2 + h 2 ; a =
].
3
2
2
17. If x and y satisfy the relation ax2 + by2 = ab, prove that the extreme values of function
u = x2 + xy + y2 are given by the roots of the equation 4(u – a) (u – b) = ab.
18. Find the maximum value of xmynzp when x + y + z = a.
[Hint: V = 4a2b +
Ans. am+n+b· mm·nn·pp/(m + n + p)m + n + p
19. Find minimum distance from the point (1, 2, 2) to the sphere x2 + y2 + z2 = 30. Ans. 3
20. Determine the perpendicular distance of the point (a, b, c) from the plane lx + my + nz
LMAns. Minimum distance = la + mb + nc OP
l +m +n Q
N
= 0 by the Lagrange’s method.
2
2
2
OBJECTIVE TYPE QUESTIONS
A. Pick the correct answer of the choices given below:
b g for the function u = e sin y, v = (x + log sin y) is (U.P.T.U., 2008)
∂b x , y g
∂ u, v
1. The Jacobian
x
(i) 1
(ii) 0
(iii) sin x sin y – xy cos x cos y
(iv)
a f for the function u = 3x + 5y, v = 4x – 3y is
b g
∂ u, v
∂ x, y
2. The Jacobian
(i) 29
2
(ii) xy
2
3
(iii) x y – y
(iv) – 29
3. If x = r cos θ, y = r sin θ, z = Z, then
(i) 2r
(iii)
ex
x
5
r
4. If u = x sin y, v = y sin x, then
b
a
∂ x, y, z
g is
f
∂ r , θ, Z
(ii) r2 – 5
(iv) r
a f is
b g
∂ u, v
∂ x, y
(i) sin x sin y
(ii) sin x sin y – xy cos x cos y
(iii) cos x cos y – xy sin y
(iv) 0
148
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
5. If u = 3x + 2y – z, v = x – y + z, w = x + 2y – z, then J (u, v, w) is
(i) 5
(ii) 0
(iii) – 2
(iv) xy + 3x3
6. If in the area of an ellipse if a two per cent error is made in measuring the major and
minor axis then the percentage error in area is
(i) 2%
(ii) 3%
(iii) 5%
(iv) 4%
7. The value of (1.05)3.02 is
(i) 2.35
(ii) 1.25
(iii) 1.15
(iv) 0.57
8. If an error of 2% is made in measuring the sides of a rectangle, then what is the
percentage of error in calculating its area?
(i) 1%
(ii) 2%
(iii) 4%
(iv) 8%
9. The relation among relative error of quotient, relative errors of dividend and the divisor (Take x = dividend, y = divisor, z = quotient) is
(i)
dz
dx dy
=
+
z
x
y
(ii)
dy
dz dx
<
+2
z
x
y
(iii)
dz dx dy
<
−
z
x
y
(iv)
dz
dx dy
=
−
z
x
y
10. If u = x2 + y2 + 6x + 12 then the stationary point is
(i) (– 3, 0)
(ii) (– 3, 5)
(iii) (– 2, 6)
11. The extreme values of
(iv) (5, 6)
f(x, y) = x2 + 2y2 on the circle
x2 + y2 = 1 are
(i) fmax = 2, fmin = 1
(ii) fmax = 0, fmin = – 2
(iii) fmax = 7, fmin = – 5
(iv) None of these
12. If u = x4 + 2x2y – x2 + 3y2 then rt – s2 is equal to
(i) 24
(ii) 36
(iii) – 58
(iv) 14
B. Fill in the blanks:
1. If u = u (r, s), v = v (r, s) and r = r (x, y), s = s (x, y) then
b g is ..........
a f
∂au, vf
3. If u = x (1 – y), v = xy then
= ..........
∂bx, yg
∂au, vf ∂bx, yg
×
4.
= ..........
∂bx, yg ∂au, vf
2. If x = r cos θ, y = r sin θ then
∂ x, y
∂ r, θ
a f = ..........
b g
∂ u, v
∂ x, y
149
DIFFERENTIAL CALCULUS-II
C. Indicate True or False for the following statements:
1. (i) The functions u and v are said to be functionally independent if their Jacobian is not
equal to zero.
(ii) If f1 (u, v, x, y) = 0 and f2 (u, v, x, y) = 0 then u, v are said to be implicit functions.
(iii) m functions of n variables are always functionally dependent when m > n.
(iv) If x = r cos θ, y = r sin θ then
2. (i)
b g = r – 2r + 1.
∂ar , θf
∂ x, y
2
dx
represents relative error.
x
(ii) If f(a, b) < f(a + h, b + k) then f(a, b) is a maximum value.
(iii) If f(a, b) > f(a + h, b + k) then f(a, b) is a minimum value.
(iv) A point where function is neither maximum nor minimum is called saddle point.
3. (i) Nature of stationary points cannot be determined by Lagrange’s method.
(ii) Solving fx = 0 and fy = 0 for stationary point.
(iii) Extremum is a point which is either a maximum or minimum.
(iv) Extrema occur only at stationary points. However, stationary points need not be
extrema.
D. Match the Following:
1. (i) Maxm
(a) rt – s2 = 0
(ii) Minm
(b) rt – s2 < 0
(iii) Saddle point
(c) rt – s2 > 0, r > 0
(iv) Failure case
(d) rt – s2 > 0, r < 0
2. (i) δx or dx
(ii)
δx
dx
or
x
x
dx
x
(iv) f(x, y) ≤ f(a, b)
(iii) 100 ×
(a) Local maximum
(b) Absolute error
(c) Relative error
(d) Percentage error
(U.P.T.U., 2008)
150
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
ANSWERS TO OBJECTIVE TYPE QUESTIONS
A. Pick the correct answer:
1. (ii)
4. (ii)
7. (iii)
2. (iv)
5. (iii)
8. (iii)
3. (iv)
6. (v)
9. (iv)
10. (i)
11. (i)
12. (i)
2. r
3. x
1. (i) T
(ii) T
(iii) T
(iv) F
2. (i) T
(ii) F
(iii) F
(iv) T
3. (i) T
(ii) T
(iii) T
(iv) T
B. Fill in the blanks:
1.
4. 1
a f a f
a f b g
∂ u, v ∂ r , s
⋅
∂ r , s ∂ x, y
C. True or False:
D. Match the following:
1. (i) → (d) (ii) → (c) (iii) → (b) (iv) → (a)
2. (i) → (b) (ii) → (c) (iii) → (d) (iv) → (a)
GGG
UNIT
III
Matrices
3.0 INTRODUCTION
The term matrix was apparently coined by sylvester about 1850, but introduced first by Cayley
in 1860. In this unit, we focus on matrix theory itself which theory will enable us to obtain
additional important results regarding the solution of systems of linear algebraic equations.
One way to view matrix theory is to think in terms of a parallel with function theory. In
mathematics, we first study numbers—the points on a real number axis. Then we study functions,
which are mappings or transformations, from one real axis to another. For example, f (x) = x2
maps the point x = 2, say on x-axis to the point f = 4 on f or y-axis. Just as functions act upon
numbers, we shall see that matrices act upon vectors and are mappings from one vector space to
another.
3.1 DEFINITION OF MATRIX
A matrix is a collection of numbers arranged in the form of a rectangular array. These numbers
known as elements or entries are enclosed in brackets [ ] or ( ) or O O.
Therefore a matrix A may be expressed as
LM a
a
A = MM
MNa M
OP
P
M
M P
P
L a Q
11
a12
L
a1n
21
a22
L
a2n
m1
M
am2
...(i)
mn
The horizontal lines are called rows and vertical lines are called columns. The order of
matrix A is m × n and is said to be a rectangular matrix.
3.1.1 Notation
Elements of matrix are located by the double subscript ij where i denotes the row and j the
column. In view of subscript notation in (1), one also writes
A = [aij], where i = 1, 2, ...,m and j = 1, 2, ... n
151
152
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
3.2 TYPES OF MATRICES
There are following types of matrices:
(a) Rectangular matrix: A matrix in which the number of rows and columns are not equal,
i.e., (m ≠ n) is called a rectangular matrix, e.g.,
LM1
N3
OP
Q
2
2
5
.
1 2×3
(b) Square matrix: A matrix in which the number of rows and columns are equal, i.e.,
(m = n) is called a square matrix, e.g.,
LM 5
MM 3
N0
OP
P
2PQ
2
1
1
0
1
.
3× 3
(c) Row matrix: A matrix which has only single row and any number of columns is called
a row matrix, e.g., [1 2 0 5]1 × 4
(d) Column matrix: A matrix which has only single column and any number of rows,
LM 1 OP
2
i.e., (m × 1) order is called column matrix, e.g., MM PP
0
MN 5 PQ
.
4×1
(e) Null matrix or zero matrix: Any m × n matrix is called a null matrix if each of its
elements is zero and is denoted by Om × n or simply by O simply, e.g.,
LM0 0 0OP .
N0 0 0Q
(f) Diagonal matrix: A square matrix A = [aij] is called diagonal matrix, if all the elements
except principal diagonal are zero. Thus, for diagonal matrix aij ≠ 0, i = j and aij = 0, i ≠ j, e.g.,
LM1
MM0
N0
0
2
0
0
0
5
OP
PP
Q
(g) Scalar matrix: Any diagonal matrix in which all its diagonal elements are equal to a
scalar, say (K) is called a scalar matrix
LM 5
MM0
N0
Thus,
i.e.,
0
5
0
OP
P
5PQ
0
0
A = [aij]n × n is a scalar matrix if
aij =
RS 0
TK
when i ≠ j
when i = j
(h) Identity matrix (or unit matrix): Any diagonal matrix is called an identity matrix,
if each of its diagonal elements is unity. Thus a matrix A = {aij}n × n is called identity matrix iff
aij =
RS0,
T1,
i≠ j
. An identity matrix of order n is denoted by I or In.
i =j
153
MATRICES
Thus,
I4 =
LM1
MM0
N0
0
1
0
0
0
0
0
0
1
OP
PP
Q
(i) Symmetric matrix: A square matrix A = [aij]n × n is said to be symmetric matrix if its
(i, j)th element is equal to (j, i)th element. Thus matrix A is said to be symmetric matrix if
aij = aji ∀ i, j.
e.g.,
LM 1
MM−3
N−4
−3
2
−5
LM 0
MM−2
N−5
2
0
−3
−4
−5
3
OP
PP
Q
(j) Skew symmetric matrix: A sqaure matrix A = [aij] is said to be skew symmetric if
aij = – aji ∀ i, j.
But for diagonal elements aii = – aii ⇒ 2aii = 0 ⇒ aii = 0. This proves that every leading
diagonal element of a skew symmetric matrix is zero.
Thus,
OP
PP
Q
5
3 is a skew symmetric matrix.
0
(k) Triangular matrix: If every element above or below the leading diagonal of a square
matrix is zero, the matrix is called a triangular matrix. It has the following two forms:
(i) Upper triangular matrix: A square matrix in which all the elements below the leading
diagonal are zero i.e., aij = 0; i > j is called an upper triangular matrix.
LM 2
MM0
N0
e.g.,
3
5
0
1
2
4
OP
PP
Q
(ii) Lower triangular matrix: A square matrix in which all the elements above the leading
diagonal are zero i.e., aij = 0; i < j is called a lower triangular matrix.
LM5
MM 3
N6
e.g.,
0
2
3
0
0
1
OP
PP
Q
(l) Transpose of a matrix: The matrix is obtained by interchange the rows and columns of
a given matrix A, is called the transpose of A and is denoted by A′ or AT e.g.,
If
LM 2 3
A = M1 2
MN5 1
LM 2
OP
3
3 0P , then A′ = M
MM5
2 1PQ
N6
5
6
OP
P
3 2P
P
0 1Q
1
2
5
1
(m) Conjugate of a matrix: The matrix obtained by replacing each element by its conjugate
complex number of a given matrix A, is called the conjugate of A and is denoted by A .
Thus, if
A =
LM1 + 2i −4i OP ,
N 3 1 − 2iQ
then A =
LM1 − 2i 4i OP
N 6 1 + 2iQ
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
(n) Tranjugate or conjugate transpose of a matrix: The conjugate of a transposed matrix
A or transpose of a conjugate matrix A is called a tranjugate matrix of A and is denoted by A*.
Thus, if
A =
LM 2 + 3i −5i OP ,
N 6 1 − 2iQ
then A* =
LM 2 − 3i 6 OP
N 5i 1 + 2iQ
(o) Hermitian matrix: A square matrix A = [aij] is said to be “Hermitian matrix” if its
conjugate transpose matrix A* is equal to itself i.e., A* = A.
Thus A = [aij] is hermitian matrix if aij = a ji ∀ i and j. Hence for diagonal elements, aii = a ii
i.e., every leading diagonal element in a hermitian matrix is wholly real.
LM 1 3 + i i OP
MM3 − i 2 5iPP
N −i −5i 0 Q
e.g.,
(p) Skew hermitian matrix: Any square matrix A = [aij] is said to be a skew hermitian
matrix if A* = – A
−a ji ∀ i, j
Thus A is skew hermitian if
aii =
For diagonal elements
and
aii = – a ii ⇒ aii + a ii = 0
aii = x + iy
a ii = x – iy
then
a ii + a ii = x + iy + x – iy = 2x
If
or
x = 0
Hence all the diagonal elements of a skew hermitian matrix are either zero or pure imaginary.
LM i
e.g., −3 + i
MM
N−4 + 5i
3+i
0
2i
4 + 5i
2i
0
OP
PP
Q
(q) Nilpotent matrix: A square matrix A is said to be nilpotent of index p if p is the least
positive integer such that Ap = 0. Thus, a square matrix A is said to be nilpotent of index 2, if
LM0 0OP is a nilpotent matrix
N1 0 Q
L0 0 O L0 0 O L0 0 O
A = M1 0 P M1 0 P = M0 0 P = 0
N Q N Q N Q
A2 = 0; e.g.,
As
2
(r) Idempotent matrix: A square matrix A is said to be periodic of period p if p is the least
positive integer such that Ap+1 = A. If p = 1 so that A2 = A, then A is called idempotent. Thus a
square matrix A is said to be “idempotent” (or of period 1)
if
e.g.,
A2 = A.
LM1 0OP ,
N0 1 Q
L1 0OP LM1 0OP = LM1 0OP = A
A = M
N0 1Q N0 1Q N0 1Q
A =
2
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MATRICES
(s) Involutory matrix: A square matrix A is said to be involutory if A2 = I, I being the
identity matrix.
e.g.,
A =
LM1 0OP , A = LM1 0OP LM1 0OP = LM1 0OP = I
N0 1 Q N0 1 Q N0 1 Q
N0 1 Q
2
(t) Orthogonal matrix: A square matrix A is said to be an orthogonal matrix, if
A′A = AA′ = I
(u) Unitary matrix: A square matrix A is said to be unitary matrix if AA* = A*A = I.
(U.P.T.U., 2001, 2005)
3.3 OPERATIONS ON MATRICES
3.3.1 Scalar Multiple of a Matrix
Consider a matrix A = [aij]m × n. Let k be any scalar belonging to a field over which A is defined.
The scalar multiple of k and A, denoted by kA, is defined as
kA = [kaij]m × n
i.e., each element of A is multiplied by k.
If
k = –1, then (–1) A = [–aij]
(–1) A is denoted by –A and is called the negative of matrix A.
–A is also called “additive inverse of A”.
Thus, if
A =
LM 1 2OP , then 3A = L 1 ⋅ 3
MN –1 ⋅ 3
N –1 4Q
2⋅ 3
4⋅ 3
OP = LM 3 6 OP
Q N –3 12Q
3.3.2 Addition of Matrices
Any two matrices can be added if they are of the same order.
If
A = [aij]m × n, B = [bij]m × n then A + B = [aij + bij]m × n
e.g.,
Let A =
LM 2 5OP , B = LM5
N 3 1Q
N2
OP
Q
–4
0 , then A + B =
LM7 1OP
N5 1Q
3.3.3 Subtraction of Matrices
Any two matrices can be subtracted if they are of the same order.
If
A = [aij]m × n, B = [bij]m × n; then A – B = [aij – bij]m × n
3.3.4 Properties of Addition of Matrices
(i) Commulative law
A + B =B+ A
(ii) Associative law
(A + B) + C = A + (B + C)
(iii) Each matrix has an additive inverse
(iv) Cancellation law
A + B =A+ C⇒ B= C
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
3.3.5 Multiplication of Matrices
The product AB of two matrices A and B is only possible if the number of columns of A = number
of rows of B. In the product AB, A is called the ‘prefactor’ and B is called the ‘post factor’.
La
Let A = M a
N
11
e.g.,
21
a12
a 22
OP
Q
a13
a 23 , B =
Lb
a
a
a O M
L
AB = M a
N a a PQ MMNbb
La b + a b + a b
= Ma b + a b + a b
N
11
12
13
21
22
23
11
21
31
11 11
12 21
13 31
21 11
22 21
23 31
LMb
MMb
Nb
11
21
31
b12
b22
b32
OP
PP
Q
b12
b22
b32
OP
PP
Q
a11b12 + a12 b22 + a13 b32
a 21b12 + a 22 b22 + a 23 b32
OP
Q
3.3.6 Properties of Multiplication of Matrices
(i) Associative law: (AB) C = A (BC)
(ii) Distributive law: A(B + C) = AB + AC.
3.4 TRACE OF MATRIX
If A = [aij]n × n be a square matrix, then the sum of its diagonal elements is defined as the trace
of the matrix, hence
∞
trace of A =
∑ aii
i= 1
e.g.,
A=
LM 1 0 5OP
MM 0 2 3PP , the trace of A = 1 + 2 + (– 2) = 1.
N –1 1 2Q
3.5 PROPERTIES OF TRANSPOSE
(i)
(A′)′ = A
(ii)
(KA)′ = KA′, K being scalar
(iii)
(iv)
(A + B)′ = A′ + B′
(AB)′ = B′A′.
3.6 PROPERTIES OF CONJUGATE MATRICES
(i)
(ii)
(iii)
dKAi = K A , K being a scalar,
d A + Bi = A + B,
d A B i = A B.
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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)
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
L1
O 1 (A* + A ) = 1 (A* + A) = P,
P* = M a A + A *fP =
2
2
N
Q 2
L1
O F 1 IJ A* – A** = – 1 (A* – A) = Q
Q* = M a A – A *fP = G
2i
2i
N
Q H −2i K
*
∗∗
so that
*
Thus
P* = P, Q* = Q ⇒ P and Q are hermitian.
In this event (i) proves that A is expressible as P + iQ where P and Q are hermitian.
Deduction: To prove that this representation is unique, let if possible. A = R + iS be another
representation where R and S are hermitian so that R* = R, S* = S
A* = (R + iS)* = R* + i S* = R + (–i)S = R – iS
or
A + A* = (R + iS) + (R – iS) = 2R
1
R =
(A + A*) = P
2
A – A* = (R + iS) – (R – iS) = 2iS
1
(A + A*) = Q
⇒
S =
2i
Finally,
R = P, S = Q.
Hence the representation is unique.
Example 3. If A is unitary matrix, show that A′ is also unitary.
Sol.
⇒
AA* = A*A = I
(AA*)* = (A*A)* = I* = I (I* = I)
(A*)* A* = A* (A*)* = I
AA* = A* A = I
(AA*)′ =
(A*)′ A′ =
⇒
(A′)* · A′ =
Hence, A′ is a unitary matrix.
LM1
Example 4. If A = M 2
MN 2
Sol. Here
and
2
1
2
OP
PP
Q
a f
As A * * = A
(A*A)′ = (I)′
A′ (A*)′ = I
A′ (A′)* = I
Proved.
2
2 show that A2 – 4A – 5I = 0.
1
LM1 2 2OP L1 2 2O
= M 2 1 2P M 2 1 2P
MN 2 2 1PQ MMN 2 2 1PPQ
LM1 + 4 + 4 2 + 2 + 4 2 + 4 + 2OP L 9
M
= M 2 + 2 + 4 4 + 1 + 4 4 + 2 + 2P = M 8
MN 2 + 4 + 2 4 + 2 + 2 4 + 4 + 1PQ MN8
LM1 2 2OP L 4 8 8O
M 4 8PP
4A = 4 M 2 1 2P = M 8
MN 2 2 1PQ MN8 8 4PQ
A2
8
9
8
8
8
9
OP
PP
Q
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MATRICES
LM 9 8 8OP LM 4 8 8OP LM5 0 0OP
4 8 P – M 0 5 0P
∴
A – 4A − 5I = M 8 9 8P – M 8
MN8 8 9PQ MN8 8 4PQ MN0 0 5PQ
LM 9 – 9 8 – 8 8 – 8OP LM0 0 0OP
= M 8 – 8 9 – 9 8 – 8P = M 0 0 0P = 0. Proved.
MN8 – 8 8 – 8 9 – 9PQ MN0 0 0PQ
L 3 –4OP prove that A = L1 + 2K – 4K O , K being any positive integer.
Example 5. If A = M
MN K 1 − 2KPQ
N1 –1Q
L 3 –4OP and K be any positive integer
Sol. Let A = M
N1 –1Q
L1 + 2K – 4K OP , we see that
To prove
A = M
N K 1 − 2KQ
L 3 −4OP = LM1 + 2 ⋅1 −4⋅ 1 OP
A = A = M
N1 –1Q N 1 1 – 2⋅1Q
2
K
K
1
This proves that result is true for K = 1
LM 3 –4OP LM 3 –4OP = LM 9 – 4
N1 –1Q N1 –1Q N 3 – 1
L1 + 2 ⋅ 2 –4⋅ 2 OP
= M 2
1 – 2 ⋅ 2Q
N
A2 =
Now
OP = LM5
Q N2
–12 + 4
–4 + 1
–8
–3
OP
Q
This proves that the result is true for K = 2.
Let us suppose that the result is true for K = n, so that
LM1 + 2n – 4n OP .
N n 1 − 2n Q
L1 + 2n – 4n OP LM 3 – 4OP
= A A = M
N n 1 − 2n Q N1 – 1Q
L 3 + 6n − 4n –4 − 8n + 4nOP = LM1 + 2an + 1f –4an + 1f OP
= M 3n + 1 − 2n
–4n − 1 + 2n Q
N n + 1 1 − 2an + 1fQ
N
An =
Now,
An + 1
n
This proves that the result is true for K = n + 1, if it is true for K = n.
Also, we have shown that the result is true for K = 1, 2. Hence by mathematical induction
the required result follows.
Example 6. Find the nature of the following matrices
A + A*, AA* and A – A*.
Sol.
and
(U.P.T.U., 2001)
z
A* = A → Hermitian matrix, (A + A*)* = A* + A { Hermitian
(AA*)* = (A*)*·A* = A·A* ⇒ Hermitian matrix.
(A – A*)* = A* – (A*)* = A* – A = – (A – A*) ⇒ skew symmetric matrix.
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 7. Show that the matrix A =
LMα + iγ – β + iδ OP is a unitary matrix if
Nβ + iδ α – iγ Q
α2 + β2 + γ2 + δ2 = 1.
Sol. A =
(U.P.T.U., 2005)
LMα + iγ –β + iδOP ∴ A = LM α – iγ β – iδ OP
N β + iδ α – iγ Q
N –β – i δ α + i γ Q
*
But for unitary matrix
AA* = I
LMα + iγ –β + iδOP LM α – iγ β – iδ OP = LM1 0OP
N β + i δ α – i γ Q N –β – i δ α + i γ Q N 0 1 Q
∴
L
α +γ +β +δ
⇒ M
αβ
βγ
+
αδ + γδ – αβ – iαδ + iβγ – δγ
i
i
–
MN
2
2
2
αβ – iαδ + iβγ + γδ – αβ – iβγ + iαδ – δγ
β2 + δ 2 + α 2 + γ 2
2
LM1 0OP
N 0 1Q
OP L1 0O
0
= M
α + β + γ + δ PQ
N0 1PQ
OP
PQ
=
LMα + β + γ + δ
0
MN
2
⇒
⇒
2
2
2
2
2
2
α2 + β2 + γ2 + δ2 = 1. Proved.
LM 1 1 + iOP is unitary.
(U.P.T.U., 2001)
–1 Q
3 N1 – i
1 + iO
1 L 1
A =
M
1
–
i
–1 PQ
3 N
1 + iO
1 L 1
A* =
M
–1 PQ
3 N1 – i
1 + iO 1 L 1
1 + iO
1 L 1
×
A*·A =
M
P
M
–1 Q
–1 PQ
3 N1 – i
3 N1 – i
(1 + i ) – (1 + i)O 1 L 3 0O L1 0O
1 L 1 + (1 + 1)
=
=
=I
=
M
(1 + 1) + 1 PQ 3 MN0 3PQ MN0 1PQ
3 N(1 – i ) – (1 – i )
Example 8. Prove that the matrix
Sol. Let
2
1
Therefore, A is a unitary matrix.
Example 9. Define a unitary matrix. If N =
LM 0 1 + 2iOP is a matrix, then show that
N –1 + 2i 0 Q
(I – N) (I + N)–1 is a unitary matrix where I is an identity matrix.
Sol. A square matrix A is said to be unitary if A*A = I.
Now
I– N =
LM1 0OP – LM 0 1 + 2iOP = LM 1
N0 1Q N–1 + 2i 0 Q N1 – 2i
(U.P.T.U., 2000)
OP
1 Q
–1 – 2i
161
MATRICES
and
I+ N =
LM1 0OP + LM 0 1 + 2iOP = LM 1 1 + 2iOP
N0 1Q N –1 + 2i 0 Q N–1 + 2i 1 Q
|I + N| = 1 – (– 1 – 4) = 6
Let
∴
⇒
⇒
A =
A11 = 1, A12 = – (– 1 + 2i) = 1 – 2i
A21 = – (1 + 2i) = – 1 – 2i, A22 = 1
B =
LM 1
N –1 – 2i
OP
Q
LM
N
1 – 2i
1
or B′ =
1
1 – 2i
Adj A = Adj (I + N) = B′ =
(I + N)–1 =
and
LM 1 1 + 2iOP
N–1 + 2i 1 Q
LM
N
LM 1
N1 – 2i
Adj ( I + N) 1 1
=
|I + N|
6 1 – 2i
–1 – 2i
1
–1 – 2i
1
OP
1 Q
OP
Q
OP
Q
–1 – 2i
–1 – 2iO L 1
–1 – 2iO
LM
P
M
1 Q N1 – 2i
1 PQ
N
–2 – 4i O
1 L –4
=
= C (say)
M
–4 PQ
6 N 2 – 4i
2 + 4i O
1L –4
∴
C* =
M
– 4 PQ
6 N –2 + 4i
2 + 4i O L – 4
–2 – 4i O 1 L 36
0O
1 L –4
⇒
C*C =
=
M
P
M
P
M
PQ
i
i
4
4
2
4
4
–2
+
–
–
–
0
36
36 N
QN
Q 36 N
L1 0OP = I . Hence Proved.
= M
N0 1Q
LM 1 2 3 OP
2
3 is nilpotent of index 2.
Example 10. Show that A = M 1
MN–1 –2 –3 PPQ
LM 1 2 3OP LM 1 2 3OP
2
3
1
2 3
Sol. Here
A = M 1
MN–1 –2 –3PPQ MMN–1 –2 –3PPQ
LM 1 + 2 – 3 2 + 4 – 6 3 + 6 – 9OP LM0 0 0OP
2+ 4– 6
3 + 6 – 9P = M0 0 0P = 0
= M 1+2– 3
MN–1 – 2 + 3 –2 – 4 + 6 –3 – 6 + 9PQ MN0 0 0PQ
Now
(I – N) (I + N)–1 =
2
Here A is nilpotent of index 2.
1
1
6 1 – 2i
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
LM a h gOP LMxOP
f P, C = M yP then find out ABC.
Example 11. If A = [x y z], B = M h b
MNg f c PQ MNzPQ
LMa h gOP
Sol.
AB = [x y z] M h b f P = [ax + hy + gz hx + by + fz gx + fy + cz]
MNg f cPQ
LM xOP
and
ABC = (AB) C = [ax + hy + gz hx + by + fz gx + fy + cz] M yP
MN zPQ
= [(ax + hy + gz)x + (hx + by + fz)y + z(gx + fy + cz)]
= [ax2 + by2 + cz2 + 2fy + 2zx + 2hxy].
A2 A3
+
+..., show that
2
3
Example 12. If eA is defined as I + A +
LMcos hx sin hx OP, where A = LMx xOP.
N sin hx cos hxQ
N x xQ
Lx xOP LMx xOP = LM2x 2x OP = 2x LM1 1OP
A = M
Nx xQ Nx xQ MN2x 2x PQ N1 1Q
F
L1 1OI
= 2x B G say B = M1 1PJ
H
N QK
eA = ex
Sol.
2
2
2
2
2
2
2
Similarly,
A3 = 22x3 B, A4 = 23x4 B, ... etc.
∴
eA = 1 + A +
= I + xB +
=
1
2
A2 A3
+
+ .....
2
3
2x 2 B 2 2 x 3 B
+
+ ..........|As A = xB
2
3
F
I
GH 2 I + ( 2x )B + ( 2 x2) B + ( 2 x3) B + . . . . .JK
2
3
LM 2 + 2x + (2x) + (2x) +... 0 + 2x + ( 2x) + ( 2x) +...OP
2
3
2
3
1 M
PP
=
( 2x )
( 2x )
( 2x)
( 2x )
2M
MM0 + 2x + 2 + 3 +... 2 + 2x + 2 + 3 +...PP
N
Q
LM 1 ee + e j 1 ee – e jOP
1 L e + 1 e – 1O
2
=
M
P = e M 12
1
2 MN e – 1 e + 1PQ
MN 2 ee – e j 2 ee + e jPPQ
Lcos hx sin hxOP . Hence Proved.
= e M
N sin hx cos hxQ
x
2
3
2
3
2
3
2
3
2x
2x
2x
2x
x
–x
x
–x
x
–x
x
–x
x
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MATRICES
3.7 SINGULAR AND NON-SINGULAR MATRICES
A square matrix A is said to be singular, if |A| = 0. If |A|≠ 0, then A is called a non-singular
matrix or a regular matrix.
(U.P.T.U., 2008)
3.8 ADJOINT OF A SQUARE MATRIX
Adjoint of A is obtained by first replacing each element of A by its cofactor in |A| and then
taking transpose of the new matrix or by first taking transpose of A and then replacing each
element by its cofactor in the determinant of A.
LM a
A = Ma
MN a
Let
OP
P
a PQ
21
a12
a 22
a13
a 23 , then
31
a 32
33
11
a11
A = a 21
a12
a 22
a13
a 23
a 31
a 32
a 33
The cofactors of A in |A| are as follows:
A11 =
a 22
a 32
A21 = –
A31 =
a12
a 32
a12
a 22
LM A
B = MA
MN A
a 23
a 21
, A13 =
a 33
a 31
a13
a11
, A22 =
a 33
a 31
a13
a11
, A23 = –
a 33
a 31
a13
a11
, A32 = –
a 23
a 21
a13
a11
, A33 =
a 23
a 21
A12
A22
A32
11
Let
a 23
a 21
A12 = –
a 33
a 31
21
31
OP
PP
Q
LM
MM
N
A13
A11
A23 then Adj A = B′ = A12
A33
A13
a 22
a 32
a12
a 32
a12
a 22
A21
A22
A23
A31
A32
A33
OP
PP
Q
3.8.1 Properties of Adjoint
If A = [aij] is a square matrix of order n then
(i) adj A′ = (adj A)′ (ii) adj A* = (adj A)* (iii) adjoint of a symmetric (Hermitian) matrix is
symmetric (Hermitian).
3.9 INVERSE OF A MATRIX (RECIPROCAL)
Consider only square matrices.
Inverse of a n-square matrix A is denoted by A–1 and is defined as follows:
where I is n × n unit matrix
AA–1 = A–1A = I
adj A
or
A–1 =
·
A
3.9.1 Properties of Inverse
(i) Inverse of A exists only if |A| ≠ 0 i.e., A is non-singular.
(ii) The inverse of a matrix is unique. If B and C are two inverses of the same matrix A
then (CA) B = C (AB), IB = CI i.e., B = C, so inverse is unique.
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
(iii) Inverse of a product is the product of inverses in the reverse order i.e., (AB)–1 = B–1
A–1, since (AB) (B–1 A–1) = A (BB–1) A–1 = AIA–1 = AA–1 = I.
(iv) For a diagonal matrix D with dii as diagonal elements, D–1 is a diagonal matrix with
1
reciprocal d as the diagonal elements.
ii
(v) Transportation and inverse are commutative i.e.,
(A–1)T = (AT)–1, taking transpose of AA–1 = A–1A = In
(A–1)T AT = AT (A–1)T = IT = I i.e.,
(A–1)T is the inverse of AT or (A–1)T = (AT)–1.
(vi) (A–1)–1 = A
Taking inverse of (AA–1) = I,
(AA–1)–1 = (A–1)–1 A–1 = I–1 = I = AA–1. Thus, A = (A–1)–1.
Example 13. Find the inverse of matrix A, where
LM–1 1 2OP
A = M 3 –1 1P.
MN–1 3 4PQ
Sol. Here we find A11, A12......... cofactors of A as follow:
A11 =
–1
3
A21 = –
A31 =
1
3
1
–1
LM –7
Let B = M 2
MN 3
∴
A–1 =
1
= – 7, A12 = –
4
3
–1
3
1
= – (12 + 1) = – 13, A13 =
–1
4
–1
2
= – (4 – 6) = 2, A22 =
–1
4
2
–1
= 1 + 2 = 3, A32 = –
1
3
–13
–2
7
LM
MM
N
OP
PP
Q
LM
MM
N
–7
adj A 1
=
–13
A
10
10
2
–2
2
OP LM
PP MM
Q N
2
–1
= – 4 + 2 = – 2, A23 = –
4
–1
2
–1
= – (– 1 – 6) = 7, A33 =
1
3
10
–7
T
2 ∴ adj A = (B) = –13
–2
10
3
–0 ⋅ 7
7 = –1 ⋅ 3
–2
1
–1
= 10
3
2
–2
2
0⋅2
–0 ⋅ 2
0⋅2
OP
PP
Q
1
=2
3
1
= 1– 3 = –2
–1
3
7 , A = 10
–2
OP
PP
Q
0⋅3
0⋅7 .
–0 ⋅ 2
Example 14. If A and B are n-rowed orthogonal (unitary) matrices, then AB and BA are also
orthogonal (unitary).
Sol. (i) Let A and B be n-rowed orthogonal matrices
then
A′A = AA′ = I and B′B = BB′ = I
To prove AB and BA are orthogonal we have
(AB)′ (AB) = (B′A′) (AB) = B′ (A′A) B = B′IB = B′B = I
(BA)′ (BA) = (A′B′) (BA) = A′ (B′B) A = A′IA = A′A = I
Thus
(AB)′ (AB) = I and (BA)′ (BA) = I
This ⇒ AB and BA are orthogonal.
(ii) Let A and B be unitary, then A*A = AA* = I and B*B = BB* = I.
To prove AB and BA are unitary complete the proof yourself.
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MATRICES
EXERCISE 3.1
1. Find the values of p, q, r, s, t and u if
LM1 2OP LM–3 –2OP
LM p qOP
=
=
,
B
–5
and
C
r
s P so that A + B – C = 0.
3
4
1
A= M
P
P
M
M
MN5 6PQ MN 4 3PQ
MN t uPQ
[Ans. p = – 2, q = 0, r = 4, s = – 1, t = 9, u = 9]
LM0 1 0OP
2. If A = M0 0 1P , then show that A = pI + qA + rA .
MNp q rPQ
L2 3 1OP, B = LM1 2 –6OP , find 3A – 4B.
3. If A = M
N0 1 5Q N0 –1 3 Q
LM1 –2 3 OP
4. If A = M2 3 –1P , show that 6A + 25A – 42I = 0.
MN3 1 2 PQ
LM1 –3 2OP LM1 4 1 0OP LM2 1 –1 –2OP
1 –3 , B = 2
1 1 1 , C = 3 –2 –1 –1 .
5. If A = M 2
MN4 –3 –1PPQ MMN1 –2 1 2PPQ MMN2 –5 –1 0PPQ
3
2
LMAns. L2
N MN0
OPOP
3 QQ
1
27
1
2
Show that (i) AB = AC, (ii) (B + C) A = BA + CA.
LM 0
6. If A = M
MNtan α2
OP
P prove that
0 P
Q
Lcos α – sin αOP ·
I + A = (I – A) M
Nsin α cos α Q
– tan
α
2
7. Find the product of the matrices
LM 2 –1 0OP
L2 1 2 1OP, B = M 0 4 1P .
A= M
N1 1 1 1Q MM–2 1 0PP
N 1 3 2Q
8. If Aα =
LMAns. L1
N MN1
1
1
OP OP
3Q Q
3
LM cos α sin αOP , then show that
N– sin α cos αQ
L cos nα sin nα OP = A ,
(A ) = M
N– sin nα cos nαQ
α
n
nα
where n is any positive integer. Also prove that Aα and Aβ commute and that Aα Aβ
= Aα + β. Also prove that Aα A–α = I.
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
9. If A =
LM0 1OP , and B = LM 1 OP , find BA and AB if they exist.
N2Q
N1 2Q
Ans. AB =
LM3 4OP L2 1 2O
10. If A = M1 1P, B = M
P , find (AB)′.
MN2 0PQ N1 2 4Q
Hence verify that (AB)′ = B′A′.
LM 2OP ; BA does not exist
N 5Q
LM
LM10 3
MMAns. ( AB)′ = M11 3
MN 22 6
N
4
2
4
OPOP
PPPP
QQ
LM 2 –2 –4OP
MM–1 3 4 PP is idempotent.
N 1 –2 3 Q
L 1 2 2 OP
1 M
2 1 –2 is orthogonal.
12. Show that the matrix A =
3 M
MN–2 2 1 PPQ
1 L1 + i i – 1O
13. Prove that the matrix A =
M
P is unitary.
2 N1 + i 1 – i Q
LM–2 + 3i 1 – i 2 + i OP
4 – 5i
5 P as sum of hermitian and skew hermitian matrices.
14. Express M 3
MN 1 1 + i –2 + 2iPQ
LM 1 L –4 4 – i 4 O 1 L 6i –2 – i 2iO OP
MMAns. 2 MM 4 + i 8 10PP + 2 MM 2 – i –10i 0 PP PP
MN 3 – i 2 –4PQ MN–1 + i 2i 4iPQ Q
N
LM 1 2 3OP
2
3P is nilpotent.
15. Show that M 1
MN–1 –2 –3PQ
16. Express given matrix A as sum of a symmetric and skew symmetric matrices.
LM L6 6 7O L 0 2 –2OOP
LM6 8 5OP
MMAns. MM6 2 5PP + MM –2 0 –2PPPP
A = M4 2 3P .
N MN7 5 1PQ MN 2 2 0PQQ
MN1 7 1PQ
LM1 0 0 0OP
1 –1 0
0
17. Show that A = M
MM1 –2 1 0PPP is involutory.
N1 –3 3 –1Q
11. Show that the matrix A =
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MATRICES
18. Find the inverse of the matrix A.
LM 1 1 3 OP
3 –3 .
A = M1
MN–2 –4 4 PPQ
Lcos α sin α OP = sec 2α LM cos α
19. Prove that M
Nsin α cos αQ
N– sin α
–1
–1
– sin α
cos α
20. If ω is one the imaginary cube roots of unity and if
LM1 1
A = M1 ω
MN1 ω
2
OP
PP
Q
LM
L12
1M
=
Ans.
A
–5
MM
4M
MN –1
N
LM
MM
N
1
1
1
–1
ω , then show that A =
1
3
2
ω
1
1
ω2
ω
OP.
Q
4
–1
–1
OPOP
PP
–1PQ PQ
6
–3
OP
PP
Q
1
ω .
ω2
3.10 ELEMENTARY ROW AND COLUMN TRANSFORMATIONS
The following transformations on a given matrix are defined as elementary transformations:
(i) Inter-change of any two rows (columns).
(ii) Multiplication of any row (column) by any non-zero scalar k.
(iii) Addition to one row (column) of another row (column) multiplied by any non-zero
scalar.
We will use the following notations to represent the elementary row (column)
operations:
(a) Rij (Cij) or Ri ↔ Rj (Ci ↔ Cj) is used for the inter-change of ith and jth rows (columns).
(b) Ri (K) [Ci (K)] or Ri → KRi (Ci → KCi) will denote the multiplication of the elements of
the ith row (column) by a non-zero scalar K.
(c) Rij (K) [Cij (K)] or Ri → Ri + KRj (Ci → Ci + KCj) is used for addition to the elements of
ith row (column) the elements of jth row (column) multiplied by the constant K.
3.10.1 Elementary Matrices
The square matrices obtained from an identity or unit matrix by any single elementary
transformation (i), (ii) or (iii) are called “elementary matrices”.
3.10.2 Properties of Elementary Transformations
(i) Every elementary row (column) transformation on a matrix can be effected by pre-post
multiplication by the corresponding elementary matrix of an appropriate order.
(ii) The inverse of an elementary matrix is an elementary matrix.
3.10.3 Equivalent Matrices
Two matrices A and B are said to be equivalent, denoted by A ~ B, if one matrix say A can be
obtained from B by the application of elementary transformations.
3.10.4 Properties of Equivalent Matrices
(i) If A and B are equivalent matrices, then there exist non-singular matrices R and C such
that B = RAC.
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
(ii) Every non-singular square matrix can be expressed as the product of elementary
matrices.
(iii) If there exist a finite system of elementary matrices R1, R2,....., Rm such that (Rm... R2
R1) A = I and A is non-singular, than A−1 = (Rm ... R2 R1) I.
3.11
METHOD OF FINDING INVERSE OF A NON-SINGULAR MATRIX
BY ELEMENTARY TRANSFORMATIONS
The property III gives a method of finding the inverse of a non-singular matrix A.
In this method, we write A = IA. Apply row transformations successively till A of L.H.S.
becomes identity matrix I. Therefore, A reduces to I, I reduces to A–1.
LM0 1 2 2OP
1 1 2 3P
Example 1. Find the inverse of A = M
MM2 2 2 3PP .
N2 3 3 3Q
Sol. Let A = IA
⇒
R1 ↔ R2
LM0 1 2 2OP LM1 0 0 0OP
MM12 12 22 33PP = MM00 10 01 00PP A
MN2 3 3 3PQ MN0 0 0 1PQ
LM1 1 2 3OP LM0 1 0 0OP
MM0 1 2 2PP = MM10 00 01 00PP A
MN22 23 23 33PQ MN0 0 0 1PQ
Applyirg R3 → R3 – 2R1, R4 → R4 – 2R1, we get
LM1 1 2 3 OP LM0 1 0 0OP
MM00 10 –22 –32 PP = MM10 –20 10 00PP A
MN0 1 –1 –3PQ MN0 –2 0 1PQ
Applying R2 → R2 + R3, we have
LM1 1 2 3 OP L0
MM00 10 –20 –1–3PP = MM1
MN0 1 –1 –3PQ MMN00
1
–2
–2
–2
0
1
1
0
OP
PP
PQ
0
0
A
0
1
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MATRICES
Applying R1 → R1 – R2, R4 → R4 – R2, we have
LM1 0 2 4OP LM–1 3 –1 0OP
MM00 01 –20 –1–3PP = MM 01 –2–2 11 00PP A
MN0 0 –1 –2PQ MN–1 0 –1 1PQ
Applying R1 → R1 + R3, R4 → 2R4 – R3, we get
LM1 0 0 1 OP L–1 1 0 0O
MM0 1 0 –1PP = MM 1 –2 1 0PP A
P MMN–20 –22 –31 02PPQ
MN00 00 –20 –3
–1Q
Applying R1 → R1 + R4, R2 → R2 – R4, R3 → R3 – 3R4
LM1 0 0 0 OP L–3 3 –3 2 O
MM0 1 0 0 PP = MM 3 –4 4 –2PP A
MN00 00 –20 –10 PQ MM–26 –82 10–3 –62 PP
Q
N
Applying R3 → –
Hence
1
R , R → (– 1) R4, we obtain
2 3 4
LM1 0 0 0OP L–3 3 –3 2 O
MM0 1 0 0PP = MM 3 –4 4 –2PP A
MN00 00 10 10PQ MM–32 –24 –53 –23 PP
Q
N
LM–3 3 –3 2OP
3 –4
4 –2
A = M
MM–3 4 –5 3PPP .
N 2 –2 3 –2Q
–1
Example 2. Find by the elementary row transformation inverse of the matrix.
Sol. Let
Now
∴
LM0 1 2OP
MM1 2 3PP.
N 3 1 1Q
LM0 1 2OP
A = M1 2 3P
MN3 1 1PQ
A = IA
LM0 1 2OP LM1 0 0OP
MM1 2 3PP = MM0 1 0PP A
N3 1 1Q N0 0 1Q
(U.P.T.U., 2000, 2003)
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Applying R1 ↔ R2, we get
R3 → R3 – 3R1
LM1 2 3OP LM0 1 0OP
MM0 1 2PP = MM1 0 0PP A
N3 1 1Q N0 0 1Q
LM1 2 3OP LM0 1 0OP
MM0 1 2PP = MM1 0 0PP A
N0 –5 –8Q N0 –3 1Q
R1 → R1 – 2R2, R3 → R3 + 5R2
LM1 0 –1OP LM–2 1 0OP
MM0 1 2 PP = MM 1 0 0PP A
N0 0 2 Q N 5 –3 1Q
R1 → R1 +
R3 →
1
R
2 3
Therefore,
1
R , R → R2 – R3, we get
2 3 2
1 0 0
1 2
0 1 0 = –4
0 0 2
5
LM
MM
N
OP
PP
Q
LM
MM
N
LM1 0 0OP LM1 2
MM0 1 0PP = MM –4
N0 0 1Q N5 2
LM1 2
A = M –4
MN5 2
–1
–1 2
3
–3
–1 2
3
–3 2
–1 2
3
–3 2
OP
PP
Q
1 2
–1 A
1
OP
PP
Q
1 2O
P
–1 P .
1 2PQ
1 2
–1 A
1 2
Example 3. Find the inverse of the following matrix employing elementary transformations:
Sol. Let
∴
LM3 –3 4OP
MM2 –3 4PP ·
N0 –1 1Q
A = IA
LM3 –3 4OP LM1 0 0OP
MM2 –3 4PP = MM0 1 0PP A
N0 –1 1Q N0 0 1Q
Applying R1 → R1 – R2, we get
LM1 0 0OP LM1 –1 0OP
MM2 –3 4PP = MM0 1 0PP A
N0 –1 1Q N0 0 1Q
[U.P.T.U. (C.O.), 2002]
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MATRICES
R2 → R2 – 2R1
LM1 0 0OP LM 1 –1 0OP
MM0 –3 4PP = MM–2 3 0PP A
N0 –1 1Q N 0 0 1Q
R3 → 3R3 – R2
R2 → R2 + 4R3
R2 → –
Hence
LM1 0 0 OP LM 1 –1 0OP
MM0 –3 4 PP = MM–2 3 0PP A
N0 0 –1Q N 2 –3 3Q
LM1 0 0 OP LM1
MM0 –3 0 PP = MM6
N0 0 –1Q N 2
–1
–9
–3
OP
PP
Q
0
12 A
3
1
R , R → (– 1) R3, we obtain
3 2 3
LM1 0 0OP LM 1 –1 0 OP
MM0 1 0PP = MM–2 3 –4PP A
N0 0 1Q N–2 3 –3Q
LM 1 –1 0OP
3 –4 .
A = M –2
MN –2 3 – 3PPQ
–1
Example 4. Find the inverse of the matrix
Sol. Let
⇒
LM–1 –3 3 –1OP
M 1 1 –1 0P
A = M 2 –5 2 –3 P.
MN–1 1 0 1PQ
A = IA
LM–1 –3 3 –1OP LM1 0 0 0OP
MM 12 –51 –12 –30PP = MM00 10 01 00PP A
MN–1 1 0 1PQ MN0 0 0 1PQ
Applying R2 → R2 + R1, R3 → R3 + 2R1, R4 → R4 – R1, we get
LM–1 –3 3 –1OP LM 1 0
1
1
–2
2 –1
P
MM 00 –11
= M
M2 0
MN 0 4 –38 –52 PPQ MN−1 0
0
0
1
0
OP
PP
PQ
0
0
A
0
1
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
R2 → –
1
R
2 2
LM –1
MM 0
MN 00
–3
1
–11
4
3
–1
8
–3
OP
PP
PQ
LM
MM
MM
N
1
–1
1
–
12
2
=
–5
2
2
–1
R3 → R3 + 11R2, R4 → R4 – 4R2
LM–1 –3 3 –1 OP L 1
MM 00 10 –1–3 11 22PP = MM – 1 2
MN 0 0 1 0 PQ MM – 71 2
N
R3 ↔ R4
OP
0 0P
A
1 0P
P
0 0 1PQ
0
1
–
2
0
0
0
–1 2
– 11 2
2
0
0
0
1
0
LM–1 –3 3 –1 OP LM 1 0 0 0OP
MM 00 10 –11 102PP = MM– 11 2 – 12 2 00 01PP A
MN 0 0 –3 1 2PQ MMN – 72 – 112 1 0PPQ
R4 → 2R4
LM –1 –3 3 –1 OP LM 1 0 0 0OP
MM 0 1 –1 1 2PP = MM– 1 2 – 1 2 0 0PP
MN 00 00 –61 10 PQ MN –71 –112 02 01PQ
Applying R4 → R4 + 6R3, we get
R2 → R2 + R3
R2 → R2 –
OP
PP
PQ
0
0
A
0
1
LM–1 –3 3 –1 OP LM 1 0 0 0OP
MM 0 1 –1 1 2PP = MM– 1 2 – 1 2 0 0PP A
MN 00 00 10 01 PQ MN –11 12 02 61PQ
LM–1 –3 3 –1 OP LM 1 0 0 0OP
MM 0 1 0 1/2PP = MM1 2 3 2 0 1PP A
MN 00 00 10 01 PQ MN –11 12 02 61PQ
1
R
2 4
LM–1 –3 3 –1OP L 1 0 0 0 O
MM 0 1 0 0 PP = MM 1 1 –1 –2PP A
MN 00 00 10 01 PQ MM–11 12 02 61 PPQ
N
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MATRICES
Applying R1 → R1 + 3R2, we get
LM–1 0 3 –1OP LM 4 3 –3 –6OP
MM 0 1 0 0 PP = MM 1 1 –1 –2PP A
MN 00 00 10 01 PQ MN–11 12 02 16 PQ
Applying R1 → R1 – 3R3, we obtain
LM–1 0 0 –1OP LM 1 –3 –3 –9OP
MM 00 10 01 00 PP = MM 11 12 –10 –21 PP A
MN 0 0 0 1 PQ MN–1 1 2 6 PQ
Applying R1 → R1 + R4
LM–1 0 0 0OP LM 0 –2 –1 –3OP
MM 00 10 01 00PP = MM 11 12 –10 –21 PP A
MN 0 0 0 1PQ MN–1 1 2 6 PQ
In the last, we apply R1 → (– 1) R1
Therefore
LM1 0 0 0OP LM 0 2 1 3 OP
MM00 10 01 00PP = MM 11 12 –10 –21 PP A
MN0 0 0 1PQ MN–1 1 2 6 PQ
LM 0 2 1 3 OP
1 1 –1 –2
A = M
MM 1 2 0 1 PPP .
N–1 1 2 6 Q
–1
Example 5. With the help of elementary operations, find the inverse of
Sol. Since
LM1 2 1OP
A = M 3 2 3P.
MN1 1 2PQ
A = IA
LM1 2 1OP LM1 0 0OP
MM3 2 3PP = MM0 1 0PP A
N1 1 2Q N0 0 1Q
Applying R2 → R2 – 3R1, R3 → R3 – R1, we get
LM1 2 1OP LM 1 0 0OP
MM0 –4 0PP = MM–3 1 0PP A
N0 –1 1Q N–1 0 1Q
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Applying R2 → –
1
R2 , we get
4
LM1 2 1OP LM 1
MM0 1 0PP = MM 3 4
N0 –1 1Q N –1
0
–1 4
0
LM1 0 1OP LM – 1 2
MM0 1 0PP = MM 3 4
N0 0 1Q N – 1 4
12
–1 4
–1 4
OP
PP
Q
0
0 A
1
Applying R3 → R3 + R2, R1 → R1 – 2R2, we have
OP
PP
Q
0
0 A
1
Now applying R1 → R1 – R3, we get
Hence
LM1 0 0OP LM – 1 4 3 4 –1OP
MM0 1 0PP = MM 3 4 – 1 4 0 PP A
N0 0 1Q N – 1 4 – 1 4 1 Q
LM– 1 4 3 4 –1OP
–1 4
0 P.
A = M34
MN– 1 4 – 1 4 1 PQ
–1
EXERCISE 3.2
1. Find the inverse of the following matrices:
LM1 –1 0 2 OP
MM02 11 12 –11 PP.
MN 3 2 1 6 PQ
LM 7 6 2OP
2. M –1 2 4P.
MN 3 3 8PQ
LM 1 1 3 OP
3 –3 .
3. M 1
MN–2 –4 –4PPQ
LM2 4 3 OP
3 6 5P
4. M
MM2 5 2 PP.
N 4 5 14Q
LM L 2 –1 1 –1OOP
MMAns. MM –5 –3 1 1 PPPP
MM MMN 23 –13 –10 10 PPQPP
N
Q
LM
L 4 –42 20 OP OP
1 M
MMAns. 130 M 20 50 –30P PP
MN –9 –3 20 PQ Q
N
LM 1 L12 4 6 O OP
MMAns. 4 MM–5 –1 –3PP PP
MN–1 –1 –1PQ Q
N
LM L–23 29 – 64 5 – 18 5O OP
MMAns. MM 10 –12 26 5 7 5 PP PP
65
25 P
MM MMN 12 –2
PP
3
5
1
5 Q PQ
–2
N
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MATRICES
LM
L 3 –1 5 OP OP
1 M
MMAns. 14 M 5 3 –1P PP
MN –1 5 3 PQ Q
N
LM L –1 2 1OOP
MMAns. MM –7 11 6PPPP
N MN –2 3 2PQQ
LM L 0 1 4 – 1 2 OOP
MMAns. MM –1 b3 4g i b1 2g iPPPP
N MN 0 1 4 1 2 PQQ
LM L 1 –2 1 0 OOP
MMAns. MM 1 –2 2 –3PPPP
MM MMN –20 13 –2–1 13 PPQPP
N
Q
LM L–2 1 0 1 OOP
MMAns. MM 1 0 2 –1PPPP
1 –3
1P
MM MMN–4
P
N –1 0 –2 2 PQPQ
LM 1 L 1 –1 1 O OP
MMAns. 2 MM–8 6 –2PP PP
MN 5 –3 1 PQ Q
N
LM 1 L 2 2 –3OOP
MMAns. 5 MM–3 2 2 PPPP
MN 3 –3 2 PQQ
N
LM L 3 2 6O OP
MMAns. MM1 1 2PP PP
N MN 2 2 5PQ Q
LM 1 2 –1OP
5. M –1 1 2 P.
MN 2 –1 1 PQ
LM4 –1 1 OP
6. M 2 0 –1P.
MN1 –1 3 PQ
LM i –1 2iOP
7. M 2 0 2 P.
MN –1 0 1 PQ
LM1 2 3 1OP
M1 3 3 2 P
8. M2 4 3 3P.
MN1 1 1 1PQ
LM 2 –6 –2 –3 OP
M 5 –13 –41 –72 PP.
9. M –1 4
MN 0 1 0 1 PQ
LM0 1 3OP
10. M 1 2 3P.
MN3 1 1PQ
LM2 1 2OP
11. M2 2 1P.
MN1 2 2PQ
LM 1 2 –2OP
12. M –1 3 0 P.
MN 0 –2 1 PQ
3.12 RANK OF A MATRIX
A positive number r is said to be the rank of matrix A if matrix A satisfies the following conditions.
(i) There exists at least one non-zero minor of order r.
(ii) Every minor of order (r + 1) and higher, if any, vanishes.
The rank of matrix A is denoted by ρ(A) or r (A).
Or
The rank of a matrix or a linear map is the dimension of the image of the matrix or the linear
map corresponding to the number of linearly independent rows or columns of the matrix or to
the number of non-zero singular values of the map.
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Result: (a) Rank of A and A′ is same.
(b) For a rectangular matrix A of order m × n, rank of A ≤ min (m, n) i.e., rank cannot be
exceed the smaller of m and n.
Minor of ) matrix. The determinant corresponding to any r × r submatrix of m × n matrix.
A is called a minor of order r of the matrix A of order m × n.
Example. If A =
LM2 3 5OP , then 2
1
N1 2 3Q
3
,
2
3
2
5
2
,
3
1
5
are all minors of order 2 of A.
3
3.12.1 Normal Form
The normal form of matrix A of rank r is one of the forms
Ir,
LMI 0OP, I
N 0 0Q
r
r
0,
LMI OP
N 0Q
r
where Ir is an identity matrix of order r. This form can be obtained by the application of both
elementary row and column operations on any given matrix A.
3.12.2 Procedure to Obtain Normal Form
Consider Am × n = Im × m· Am × n· In × n
Apply elementary row operations on A and on the prefactor Im × m and apply elementary
column operations on A and on the postfactor In × n, such that A on the L.H.S. reduces to normal
form. Then Im × m reduces to Pm × m and In × n reduces to Qn × n; resulting in N = PAQ.
Here P and Q are non-singular matrices. Thus for any matrix of rank r, there exist nonsingular matrices P and Q such that
PAQ = N =
LM I 0OP .
N 0 0Q
r
3.12.3 Echelon Form
A matrix A = [aij] is an echelon matrix or is said to be in echelon form, if the number of zeros
preceding the first non-zero entry (known as distinguished elements) of a row increases row by
row until only zero rows remain.
In row reduced echelon matrix, the distinguished elements are unity and are the only
non-zero entry in their respective columns.
“The number of non-zero rows in an Echelon form is the rank”.
Example 1. Find the rank of matrix
Sol. Let
LM 2 3 –2 4OP
MM 3 –2 1 2PP.
MN –23 42 03 45PQ
LM 2 3 –2 4OP
3 –2 1 2
A = M
MM 3 2 3 4PPP
N–2 4 0 5Q
(U.P.T.U., 2006)
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MATRICES
R1 → R1 + R4, R2 → R2 – R3, we get
LM 0 7 –2 9 OP
0 –4 –2 –2
A ~ M
MM 3 2 3 4 PPP
N–2 4 0 5 Q
R3 → R3 + R4
LM 0 7 –2 9 OP
0 –4 –2 –2
~ M
MM 1 6 3 9 PPP
N–2 4 0 5 Q
Q
R1 ↔ R3
R4 → R4 + 2R1
Now,
LM 1 6 3 9 OP
0 –4 –2 –2P
~ M
MM 0 7 –2 9 PP
N–2 4 0 5 Q
LM1 6 3 9 OP
0 –4 –2 –2
A ~ M
MM0 7 –2 9 PPP
N0 16 6 23Q
|A| =
–4
7
16
–2
–2
6
–2
–9
23
(Expanded w.r. to first column)
= – 4 (–46 –54) + 2 (161 – 144) –2 (42 + 32)
= 400 + 34 – 148 = 286
⇒
|A| ≠ 0
Thus there is a non-singular minor of order 4.
Hence
ρ(A) = 4.
Example 2. Find the rank of matrix A by echelon form.
LM2 3 –1 –1OP
1 –1 –2 –4
A = M
MM 3 1 3 –2PPP.
N6 3 0 –7 Q
Sol. Applying R1 ↔ R3, we get
LM1 –1 –2 –4OP
M2 3 –1 –1PP
A ~ M3 1
MN6 3 03 −–27PQ
(U.P.T.U., 2005)
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
R2 → R2 – 2R1, R3 → R3 – 3R1, R4 → R4 – 6R1
LM1 –1 –2 –4OP
M0 5 3 7 PP
A ~ M0 4
MN0 9 129 1710PQ
Applying R3 → R3 –
4
9
R , R → R4 –
R , we get
5 2 4
5 2
1 –1
–2
–4
0
5
3
7
~
0
0
33 5 22 5
0
0
33 5 22 5
LM
MM
MN
In the last, we apply R4 → R4 – R3, we get
LM1
0
A ~ M
MM0
N0
–1
5
0
0
–2
3
33 5
0
OP
PP
PQ
OP
PP
PQ
–4
7
22 5
0
Here the number of non-zero rows = 3
Therefore
ρ(A) = 3
Example 3. Find the rank of following matrix:
LM 3 4 5 6 7 OP
MM 45 65 76 78 89 PP
A =
MM10 11 12 13 14PP.
MN15 16 17 18 16PQ
Sol. Applying R1 → R1 – R2 and then again R1 → – R1
LM 1 1 1 1 1 OP
MM 54 65 67 78 89 PP
A ~
MM10 11 12 13 14PP
MN15 16 17 18 16PQ
R2 → R2 – 4R1, R3 → R3 – 5R1, R4 → R4 – 10R1, R5 → R5 – 15R1
LM1 1 1 1 1OP
MM00 11 22 33 44PP
~
MM0 1 2 3 4PP
MN0 1 2 3 4PQ
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MATRICES
Applying R3 → R3 – R2, R4 → R4 – R2, R5 → R5 – R2, we get
LM1 1 1 1 1OP
MM00 10 02 03 04PP
A ~
MM0 0 0 0 0PP
MN0 0 0 0 0PQ
Here number of non-zero rows = 2
Therefore
ρ(A) = 2.
Example 4. Reduce the matrix A to its normal form, where
LM0 1 –3 –1OP
1 0 1
1
A = M
MM3 1 0 2 PPP and hence find the rank of A.
N1 1 –2 0 Q
Sol. Applying R1 ↔ R2, we have
LM1 0 1 1 OP
0 1 –3 –1P
A ~ M
MM3 1 0 2 PP
N1 1 –2 0 Q
R3 → R3 – 3R1, R4 → R4 – R1
LM1 0 1 1 OP
0 1 –3 –1
~ M
MM0 1 –3 –1PPP
N0 1 –3 –1Q
R3 → R3 – R2, R4 → R4 – R2
C3 → C3 – C1, C4 → C4 – C1
LM1 0 1 1 OP
0 1 –3 –1P
~ M
MM0 0 0 0 PP
N0 0 0 0 Q
LM1 0 0 0 OP
0 1 –3 –1
A ~ M
MM0 0 0 0 PPP
N0 0 0 0 Q
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
C3 → C3 + 3C2, C4 → C4 + C2, we get
LM1 0 0 0OP
0 1 0 0 LI
0O
P
A ~ M
=M
MM0 0 0 0PP N 0 0PQ
N0 0 0 0Q
2
Hence ρ (A) = 2.
Example 5. Reduce the matrix A to its normal form, when
LM 1 2 –1 4 OP
2 4 3
4
A = M
MM 1 2 3 4 PPP.
N–1 –2 6 –7 Q
Hence, find the rank of A.
Sol.
LM 1 2 –1 4 OP
2
4
3
4
A = M
MM 1 2 3 4 PPP
N–1 –2 6 –7Q
Applying R2 → R2 – 2R1, R3 → R3 – R1, R4 → R4 + R1
LM1 2 –1 4 OP
0 0 5 –4
~ M
MM0 0 4 0 PPP
N0 0 5 –3Q
Applying C2 → C2 – 2C1, C3 → C3 + C1, C4 → C4 – 4C1
C3 ↔ C2
R3 → R3 –
LM1 0 0 0 OP
0 0 5 –4
~ M
MM0 0 4 0 PPP
N0 0 5 –3Q
LM1 0 0 0 OP
0 5 0 –4
~ M
MM0 4 0 0 PPP
N0 5 0 –3Q
4
R , R → R4 – R2
5 2 4
LM1 0 0 0 OP
0 5 0
–4
~ M
MM0 0 0 16 5PPP
N0 0 0 1 Q
(U.P.T.U., 2001, 2004)
181
MATRICES
C3 ↔ C4
R2 → R2 + 4R4,
Applying R4 → R4 –
LM1 0 0 0OP
0 5
–4
0
~ M
MM0 0 16 5 0PPP
N0 0 1 0Q
LM1 0 0 0OP
0 5
0
0
~ M
MM0 0 16 5 0PPP
N0 0 1 0Q
16
1
5
R3, then R2 →
R and R3 →
R.
5
5 2
16 3
1 0 0 0
0 1 0 0
I3 0
~
=
0 0 1 0
0 0
0 0 0 0
OP
PP LM
PQ N
LM
MM
MN
ρ (A) = 3.
OP
Q
Example 6. Find the non-singular matrices P and Q such that the normal form of A is PAQ
where
LM1 3 6 –1OP
A = M1 4 5 1 P
MN1 5 4 3 PQ
. Hence, find its rank.
3×4
Sol. Here we consider
A3 × 4 = I3 × 3· A3 × 4· I4 × 4
LM1 3 6 –1OP L1
MM1 4 5 1 PP = MM0
N1 5 4 3 Q MN0
0
1
0
As Am × n = Im × m· Am × n· In × n
OP LM10
0P A M
M0
1PQ M
N0
0
1
0
0
OP LM10
0P A M
M0
1PQ M
N0
0
1
0
0
0
0
0
0
1
0
0
1
0
0
0
0
1
Applying R2 → R2 – R1, R3 → R3 – R1 (pre), we get
LM1 3 6 –1OP LM 1 0
MM0 1 –1 2 PP = MM –1 1
N0 2 –2 4 Q N –1 0
R3 → R3 – 2R1 (pre)
0
OP
PP
PQ
0
0
1
0
LM1 3 6 –1OP LM 1 0 0OP LM10 01 00 00OP
MM0 1 –1 2 PP = MM–11 –21 01PP A MM0 0 1 0PP
Q MN0 0 0 1PQ
N0 0 0 0 Q N
OP
PP
PQ
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Applying C2 → C2 – 3C1, C3 → C3 – 6C1, C4 → C4 + C1 (post), we get
LM1 0 0 0OP LM 1 0 0OP LM10 –31 –60 01OP
MM0 1 –1 2PP = MM–11 –21 10PP × A MM0 0 1 0PP
Q MN0 0 0 1PQ
N0 0 0 0Q N
C3 → C3 + C2, C4 → C4 – 2C2 (post)
LM1 0 0 0OP LM 1 0 0OP LM10 –31 –91 –27 OP
MM0 1 0 0PP = MM–11 –21 10PP × A MM0 0 1 0 PP
Q MN0 0 0 1 PQ
N0 0 0 0Q N
Therefore,
I2 = N = PAQ, where
LM 1 0 0OP LM10 –31 –91 –27 OP
P
P = M–1 1 0P, Q = M
M
0 0
1
0P
MN 1 –2 1PQ M0 0 0 1 P
Q
N
and
Rank of A = 2.
Example 7. Find non-singular matrices P and Q such that PAQ is the normal form where
LM 1 –1 2 –1 OP
A = M 4 2 –1 2 P
MN 2 2 –2 0 PQ
.
3 × 4
Sol. Here we consider
A3 × 4 = I3 × 3· A3 × 4· I4 × 4
LM1 –1 2 –1OP L1
MM4 2 –1 2 PP = MM0
N2 2 –2 0 Q MN0
0
1
0
OP LM
PP MM
Q MN
1
0
0
0 A
0
1
0
As Am × n = Im × m Am × n In × n
0
1
0
0
0
0
1
0
0
0
0
1
Applying R2 → R2 – 4R1, R2 → R2 – 2R1 (pre), we get
LM1
MM0
N0
–1
6
4
2
–9
–6
OP LM 1 0 0OP LM10
PP = MM – 4 1 0PP A MM0
Q N – 2 0 1Q MN0
–1
6
2
0
1
0
0
0
0
1
0
OP
PP
PQ
0
0
0
1
Applying C2 → C2 + C1, C3 → C3 – 2C1, C4 → C4 + C1 (post)
LM1 0 0 0OP LM 1 0 0OP LM10 11 –20 01OP
MM0 6 –9 6PP = MM–4–2 10 01PP A MM0 0 1 0PP
Q MN0 0 0 1PQ
N0 4 –6 2Q N
OP
PP
PQ
183
MATRICES
R2 →
1
1
R2, R3 →
R (pre)
3
2 3
LM1 0 0 0OP LM 1 0 0 OP LM10 11 –20 10OP
MM0 2 –3 2PP = MM– 4 3 1 3 0 PP A MM0 0 1 0PP
N0 2 –3 1Q N –1 0 1 2Q MN0 0 0 1PQ
R3 → R3 – R2 (pre)
–2
LM1 0 0 0 OP LM 14 10 0 OP LMM10 11 0 10OPP
MM0 2 –3 2 PP = MM– 3 3 0 PP A MM0 0 1 0PP
N0 0 0 –1Q N1 3 – 1 3 1 2Q N0 0 0 1Q
C3 ↔ C4 (post)
LM1 0 0 0 OP LM 1 0 0 OP LM10 11 10 –20 OP
MM0 2 2 –3PP = MM– 4 3 1/ 3 0 PP A MM0 0 0 1 PP
N0 0 –1 0 Q N 1 3 – 1 3 1 2Q MN0 0 1 0 PQ
C3 → C3 – C2, C4 → C4 +
3
C (post)
2 2
LM1 0 0 0OP LM 1 0 0 OP LM 10 11 –10 –31 22 OP
MM0 2 0 0PP = MM – 4 3 1 3 0 PP A MM 0 0 0 1 PP
N0 0 –1 0Q N 1 3 – 1 3 1 2 Q MN 0 0 1 0 PQ
Applying R2 →
1
R , R → (– 1) R3 (pre)
2 2 3
LM1 0 0 0OP LM 1 0
MM0 1 0 0PP = MM– 2 3 1 6
N0 0 1 0Q N – 1 3 1 3
⇒
Therefore,
And
OP LM
PP MM
Q MN
1
0
0
0 A
0
–1 2
0
1
1
0
0
0
–1
0
1
OP
32
P
1 P
P
0 Q
–1 2
I3 = PAQ
LM 1 0 0 OP LM10 11 –10
0 P, Q = M
P = M– 2 3 1 6
MN – 1 3 1 3 – 1 2PQ MM00 00 01
N
ρ (A) = 3.
Note: In such problems other Ans. is possible.
–1 2
32
1
0
OP
PP
PQ
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 8. Find the rank of
LM 6 1 3 8 OP
16 4 12 15 P
A= M
MM 5 3 3 8 PP.
N 4 2 6 –1Q
Sol. Applying C1 ↔ C2, C3 →
1
C
3 3
LM1 6 1 8 OP
4 16 4 15P
A ~ M
MM3 5 1 8 PP
N2 4 2 –1Q
By C2 → C2 – 2C1, C3 → C3 – C1, we have
LM1 4 0 8 OP
M4 8 0 15P
~ M3 –1 –2 4 P
MN2 0 0 –1PQ
By R2 → R2 – 2R1
LM1 4 0 8 OP
2 0 0 –1P
~ M
MM3 –1 –2 4 PP
N2 0 0 –1Q
R3 → R3 – R2, R4 → R4 – R2, we get
LM1 4 0 8 OP
2 0
0 –1
~ M
MM1 –1 –2 5 PPP
N0 0 0 0 Q
⇒ |A| = 0 i.e., minor of order 4 = 0
Next, we consider a minor of order 3
∴
∴
1
2
1
4
0
–1
0
0
–2
= 1 (0 – 0) – 4 (– 4 – 0) + 0 = 16 ≠ 0
ρ (A) = 3.
Example 9. Find the value of a such that the rank of A is 3, where
LM1 1 –1 0OP
4 4 –3 1 P
A= M
MM a 2 2 2PP.
N 9 9 a 3Q
185
MATRICES
LM1 1 –1 0OP
4 4 –3 1P
A = M
MMa 2 2 2PP
N9 9 a 3Q
Sol.
Applying R2 → R2 – 4R1, R3 → R3 – 2R1, R4 → R4 – 9R1, we have
LM 1 1 –1 0OP
0
0
1
1
A ~ M
MMa – 2 0 4 2PPP
N 0 0 a + 9 3Q
Again R3 → R3 – 4R2, R4 → R4 – 3R2
LM 1 1 –1 0 OP
0
0
1
1P
~ M
MMa – 2 0 0 –2PP
N 0 0 a+6 0 Q
R4 ↔ R3
LM 1 1 –1 0 OP
0
0
1
1
~ M
MM 0 0 a + 6 0 PPP
Na – 2 0 0 –2Q
Cases: (i) If a = 2, |A| = 1·0·8. (– 2) = 0, rank of A = 3.
(ii) If a = – 6, no. of non-zero rows is 3, rank of A = 3.
Example 10. For which value of ‘b’ the rank of the matrix.
LM1 5 4 OP
A = M0 3 2 P is 2.
MN b 13 10PQ
(U.P.T.U., 2008)
Sol. Since the rank of matrix A is 2 so the minor of 3rd order must be zero i.e., |A| = 0.
Thus
1
5
4
0
3
2
b
13
10
= 0
(30 – 26) – 5 (0 – 2b) + 4 (0 – 3b) = 0
⇒
Hence
4 + 10b – 12b = 0 ⇒
b = 2
4 – 2b = 0
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
EXERCISE 3.3
Find the rank of the following matrix by reducing normal form:
1 2 –3
1 2 –1 3
1 0 –1
4 1 2 1
.
1.
(U.P.T.U., 2001)
2. 3 –1 1
3 –1 1 2
–1 1
0
1 2 0 1
3 1
0
[Ans. ρ(A) = 3]
LM
MM
MM
MN
OP
PP
PQ
LM
MM
MN
OP
PP
PP
PQ
9
1
–1 .
9
9
LM1 2 3OP
LM9 7 3 6OP
3. M1 4 2P.
[Ans. 2]
4. M 5 –1 4 1 P.
MN2 6 5PQ
MN6 8 2 4PQ
LM0 0 0 0 0OP
LM 1 2 –1 3OP
M0 1 2 3 4 P
M4 1 2 1P
5. M 3 –1 1 2 P.
[Ans. 3]
6. M0 2 3 4 1 P.
MN0 3 4 1 2PQ
MN 1 2 0 1PQ
(U.P.T.U. special exam., 2001)
LM1 2 3 4 5OP
LM –1 2 –2OP
2 3 4 5 6
P
M
7. M 3 4 5 6 7 P.
[Ans. 3]
8. M 1 2 1 P.
MN –1 –1 0 PQ
MN 4 5 6 7 8PQ
LM1 2 1 0 OP
LM1 1 2 OP
MM23 –12 21 25 PP
.
[Ans. 3] 10. M1 2 3 P.
9. M
P
5
6
3
2
MN0 –1 –1PQ
PP
MM
N1 3 –1 –3Q
[Ans. 4]
[Ans. 3]
[Ans. 3]
[Ans. 3]
[Ans. 2]
Find the rank of the following matrix.
LM1 2 3 4 OP
2
3 4
5
11. M
MM 3 4 5 6 PPP.
MN 4 5 6 7 PQ
LM 3 –2 0 –1 –7 OP
M 0 2 2 1 –5P
13. M 1 –2 –3 –2 1 P.
MN 0 1 2 1 –6PQ
LM 1 2 –1 4 OP
5 .
15. M 2 4 3
MN–1 2 6 –7PPQ
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
LM2 1 3 4OP
M0 3 4 1 P
[Ans. 3] 12. M2 3 7 5P.
MN2 5 11 6PQ
LM1 2 3OP
[Ans. 4] 14. M2 3 1P.
MN3 1 2PQ
LM 0 i –iOP
[Ans. 2] 16. M – i 0 – i P.
MN–3 1 0 PQ
[Ans. 3]
[Ans. 3]
[Ans. 3]
187
MATRICES
LM 3 –1 –2OP
17. M –6 2 –4 P.
MN 3 1 –2PQ
LM1 2 3 1OP
19. M2 4 6 2P.
MN1 2 3 2PQ
LM0 2 3OP
[Ans. 2] 18. M0 4 6P.
MN0 6 9PQ
LM 4 2 3 OP
[Ans. 2] 20. 8
MM 4 6 PP.
N –2 –1 –15Q
LM1 –2 3OP
21. M 2 –1 2P.
MN3 1 2PQ
LM 3 4 5 6 7 OP
MM 45 65 76 78 98 PP
23.
MM10 11 12 13 14PP.
MN15 16 17 18 19PQ
LM 5 6 7 8 OP
M6 7 8 9P
25. M11 12 13 14 P.
MN16 17 18 19PQ
LM 5 3 14 4OP
27. M 0 1 2 1 P.
MN1 –1 2 0PQ
LM 1 2 –5OP
[Ans. 3] 22. M –4 1 –6P.
MN 6 3 –4PQ
[Ans. 1]
[Ans. 1]
Find the Echelon form of the following matrix and hence find the rank.
LM2 3 –1 –1OP
M1 –1 –2 –4P
[Ans. 2] 24. M 3 1 3 –2 P.
MN6 3 0 –7 PQ
LM–2 –1 –3 –1OP
M 1 2 –3 –1P
[Ans. 3] 26. M 1 0 1 1 P.
MN 0 1 1 –1PQ
LM0 1 3 –2OP
M0 4 –1 3 P
[Ans. 3] 28. M 0 0 2 1 P.
MN0 5 –3 4 PQ
[Ans. 2]
[Ans. 3]
[Ans. 3]
[Ans. 2]
29. Determine the non-singular matrices P and Q such that PAQ is in the normal form for
A. Hence find the rank of A.
LM3 2 –1 5 OP
–2 P.
A = M5 1 4
MN1 –4 11 –19PQ
LM
LM1
MM
LM 0 0 1 OP MM0
MMAns. P = M 01 1 3 – 5 3P Q = M
MM – 1 3 1 6 PP MM0
MM
N2
Q M
MN
MN0
4 17
17
0
0
9
119
–1
7
–1
17
0
OP OP
PP PP
PP PP rank = 2
PP
1 PP
PP
31 Q Q
9
217
–1
7
0
(other forms are also possible)
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
LM2 1 –3 –6OP
30. A = M 3 –3 1 2 P.
(U.P.T.U., 2002)
MN1 1 1 2 PQ
LM
OP
LM 0 0 1 OP LM1 –1 4 0 OP
–5
0
1
0
MMAns. P = M –1 0 2 P, Q = M
PP and rank = 3PP
M
–2
0
0
1
MM –3 1 9 PP M
MM
PP
N 14 28 28 Q N0 0 0 1 PQ
N
Q
LM
LM 1 0 0OP LM1 –1 –1OPOP
LM1 1 2 OP
MMAns. P = M–1 1 0P, Q = M0 1 –1PPP
31. A = M1 2 3 P.
MN–1 1 1PQ MN0 0 1 PQQ
N
MN0 –1 –1PQ
LM
LM1 1 3 –4/ 3 – 1 OP OP
1
0
0
L
O
3
LM 1 2 3 –2OP M
M
M
P
0 –1 6
–5 6
7 6P P rank = 2
–2
1
0
,
=
Ans.
P
=
Q
M
PP PP
32. A = M 2 –2 1 3 P.
MM
PP MM0 0
M
1
0
MN 3 0 4 1 PQ M
N–1 –1 1Q M0 0
P
0
1 PQ Q
N
N
3.13 SYSTEM OF LINEAR EQUATIONS (NON-HOMOGENEOUS)
Let us consider the following system of m linear equations in n unknowns x1, x2, ....., xn:
a11 x1 + a12 x 2 + .... + a1n xn = b1
a 21 x1 + a 22 x 2 + .... + a 2n xn = b2
= ...
....
....
....
a m1 x1 + am 2 x 2 + .... + amn x n = bn
U|
|V
||
W
...(i)
In matrix notation these equations can be put in the form
AX = B
...(ii)
LM a
a
A = M
MM ....
Na
21
a12
a22
m1
....
am2
11
where
OP
P
.... P
P
a Q
a1n
a2n
....
....
mn
LM x OP
LM b OP
b
x
X = M P and B = M P
MM M PP
MM M PP
Nx Q
Nb Q
1
1
2
2
n
n
189
MATRICES
Augmented matrix: The augmented matrix [A : B] or à of system (i) is obtained by
augmenting A by the column B.
LM a
a
à = [A : B] = M
MM L
Na
11
21
i.e.,
m1
OP
P
L L L M LP
P
M b Q
a
a
a12
L
a1n
M
b1
a22
L
a2n
M
b2
m2
mn
n
[U.P.T.U., (C.O.), 2003]
3.13.1 Conditions for Solution of Linear Equations
Consistent: If the ranks of A and augmented matrix [A : B] are equal, then the system is said to
be consistent otherwise inconsistent. There are following conditions for exist the solution of any
system of linear equations :
(i) If ρ(A) = ρ[A : B] = r = n (where n is the number of variables) then the system has a
unique solution.
(ii) If ρ(A) = ρ[A : B] = r < n.
then the system has infinitely many solutions in terms of remaining n – r unknowns which are
arbitrary.
If n – r = 1 (then solution is one variable independent solution and let equal to K).
n – r = 2 (then solution is two variable independent solution and let variables equal to K1, and
K2) and so on.
Trivial solution: It is a solution where all xi are zero i.e., x1 = x2 ... = xn = 0.
Example 1. Check the consistency of the following system of linear nonhomogeneous
equations and find the solution, if exists:
(U.P.T.U., 2007)
7x1 + 2x2 + 3x3 = 16
2x1 + 11x2 + 5x3 = 25
x1 + 3x2 + 4x3 = 13.
LM7 2 3OP LMx OP LM16OP
A = M 2 11 5P , X = Mx P , B = M 25P
MN1 3 4PQ MNx PQ MN13PQ
LM7 2 3 M 16OP
[A : B] = M2 11 5 M 25P
MN1 3 4 M 13PQ
1
Sol. Here,
2
3
The augmented matrix
Applying R1 ↔ R3, we get
[A : B] ~
Again R2 → R2 – 2R1, R3 → R3 – 7R1
LM1 3 4 M 13OP
MM 2 11 5 M 25PP
N7 2 2 M 16Q
LM1 3 4 M 13 OP
~ 0
MM 5 –3 M –1 PP
N0 –19 –25 M –75Q
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
R3 → R3 +
19
R2
5
LM1 3 4 M 13 OP
M
–3
M
–1 PP
~ M0 5
MM0 0 –182 M – 394 PP
N
5
5 Q
⇒
ρ (A) = ρ [A : B] = 3
⇒
r = n = 3. ∴ The system is consistent.
The given system has a unique solution.
Now from
AX = B
LM1
MM0
MN0
OP Lx O LM 13 OP
M P
–3 P Mx P = M –1 P
MM –394 PP
–182 P M P
P
P
M
x
5 Q N Q
N 5 Q
3
4
5
2
0
3
LM x + 3x + 4x OP L 1 3 O
MM 5x – 3x PP = MMM – 1 PPP
MM –182 x PP MM – 39 4 PP
Q N 5 Q
N 5
1
⇒
1
2
2
3
3
3
⇒
x1 + 3x2 + 4x3 = 13
...(i)
5x2 – 3x3 = –1
...(ii)
182
394
x3 =
5
5
On solving these equations, we get the final solution
...(iii)
95
100
197
, x2 =
, x3 =
.
91
91
91
Example 2. Test the consistency of following system of linear equations and hence find the
solution.
(U.P.T.U., 2005)
4x1 – x2 = 12
– x1 + 5x2 – 2x3 = 0
– 2x2 + 4x3 = – 8
x1 =
LM 4 –1 0 M 12OP
Sol. The augmented matrix [A : B] = M –1 5 –2 M 0 P
MN 0 –2 4 M –8PQ
Applying R1 ↔ R2, we get
LM –1 5 –2 M 0 OP
[A : B] ~ M 4 –1 0 M 12 P
MN 0 –2 4 M –8PQ
191
MATRICES
R2 → R2 + 4R1
LM –1 5 –2 M 0 OP
~ 0 19 –8 M 12
P
MM
N 0 –2 4 M –8PQ
R1 → – R1, R3 → R3 +
2
R,
19 2
LM1
~ M0
MM
N0
–5
2
M
19
–8
60
19
M
0
M
OP
12 P
–128 P
P
19 Q
0
∴
ρ(A) = ρ [A : B] = 3
i.e.,
r = n = 3 (The system is consistent).
Hence, there is a unique solution
⇒
⇒
LM –1
MM 0
MM 0
N
–5
19
0
OP L x O L 0 O
PM P M P
–8 P M x P = M 12 P
MM –128 PP
60 P MM PP
x
N 19 Q
19 PQ N Q
2
1
2
3
x1 – 5x2 + 2x3 = 0
...(i)
19x2 – 8x3 = 12
60 x 3
–128
=
19
19
32
⇒
x3 = –
15
putting the value of x3 in equation (ii), we get
32
19x2 – 8 –
= 12
15
256
76
=–
⇒
19x2 = 12 –
15
15
–76
4
=–
⇒
x2 =
15 × 19
15
and putting the values of x1, x2 in equation (i), we get
4
32
+2 −
x1 – 5 −
= 0
15
15
...(ii)
FG
H
IJ
K
FG IJ FG IJ
H K H K
x1 +
⇒
⇒
Hence,
20 64
–
= 0
15 15
44
x1 –
= 0
15
44
x1 =
15
4
32
44
x1 =
, x2 = –
and x3 = – .
15
15
15
...(iii)
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 3. Solve
2x1 – 2x2 + 4x3 + 3x4 = 9
x1 – x2 + 2x3 + 2x4 = 6
2x1 – 2x2 + x3 + 2x4 = 3
x1 – x2 + x4 = 2
Sol. The augmented matrix is
LM 2 –2 4 3 M 9OP
1 –1 2 2 M 6
[A : B] = M
MM 2 –2 1 2 M 3PPP
N1 –1 0 1 M 2Q
R1 ↔ R2
LM1 –1 2 2 M 6OP
2 –2 4 3 M 9 P
~ MM
MN21 –2–1 10 21 MM 32PPQ
R2 → R2 – 2R1, R4 → R4 – R1, R3 → R3 – 2R1
LM1 −1 2 2 M 6 OP
M0 0 0 –1 M –3PP
~ M
MN00 00 –3–2 –2–1 MM –4–9PQ
R2 → (– 1) R2, R3 → (– 1) R3, R4 → (– 1) R4
LM1 –1 2 2 M 6 OP
M0 0 0 1 M 3 PP
~ M
MN00 00 23 21 MM 94 PQ
R3 → R3 – R4
R3 ↔ R2
LM1 –1 2 2 M 6 OP
0 0 0 1 M 3P
~ MM
MN00 00 12 11 MM 45PPQ
LM1 –1 2 2 M 6 OP
0 0 1 1 M 5P
~ MM
MN00 00 02 11 MM 34PPQ
193
MATRICES
R4 ↔ R3
LM1 –1 2 2 M 6 OP
0 0 1 1 M 5
~ M
MM0 0 2 1 M 4PPP
N0 0 0 1 M 3 Q
R3 → R3 – 2R2
LM1 –1 2 2 M 6 OP LM1 –1 2 2 M 6OP
0 0 1 1 M 5 P M0 0 1 1 M 5 P
~ M
MM0 0 0 −1 M –6PP ~ MM0 0 0 1 M 6PP bR → – R g
N 0 0 0 1 M 3 Q N 0 0 0 1 M 3Q
3
3
R4 → R4 – R3 and then R4 → (– 1) R4
LM1 –1 2 2 M 6 OP
M0 0 1 1 M 5PP
~ M
MN00 00 00 10 MM 63 PQ
Hence,
ρ (A) = 3 and ρ [A : B] = 4 ⇒ ρ (A) ≠ ρ [ A : B].
So the given system is inconsistent and therefore it has no solution.
Example 4. Investigate for what values of λ, µ the equations
x + y + z = 6, x + 2y + 3z = 10, x + 2y + λz = µ
have (i) no solution (ii) a unique solution (iii) an infinity of solutions.
Sol. The augmented matrix
LM1 1 1 M 6 OP
[A : B] = M1 2 3 M 10 P
MN1 2 λ M µ PQ
R2 → R2 – R1, R3 → R3 – R1
R3 → R3 – R1
LM1 1 1 M 6 OP
2
M
4 P
~ M0 1
MN0 1 λ – 1 M µ – 6PQ
LM1 1 1 M 6 OP
2
M
4
~ M0 1
MN0 0 λ – 3 M µ – 10PPQ
(i) For no solution ρ (A) ≠ ρ [A; B] it is only possible when λ = 3.
(U.P.T.U., 2001)
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
(ii) For unique solution ρ (A) = ρ [A ® B] it is only possible when λ – 3 ≠ 0 i.e., λ ≠ 3 and
µ ≠ 10.
(iii) For infinite number of solutions ρ (A) = ρ [A : B] = r < n it is only possible when
λ = 3 and µ = 10.
Example 5. Show that the equations
x+ y+ z=6
x + 2y + 3z = 14
x + 4y + 7z = 30
are consistent and solve them.
Sol. The augmented matrix is
LM1 1 1 M 6 OP
[A : B] = M1 2 3 M 14 P
MN1 4 7 M 30PQ
R2 → R2 – R1, R3 → R3 – R1
R3 → R3 – 3R2
LM1 1 1 M 6 OP
~ M0 1 2 M 8 P
MN0 3 6 M 24PQ
LM1 1 1 M 6OP
~ M0 1 2 M 8 P
MN0 0 0 M 0PQ
Hence,
i.e.,
ρ (A) = ρ [A : B] = 2
r = 2 < 3 (n = 3)
∴
n – r = 3 – 2 = 1 (one variable independent solution).
The system is consistent and have infinitely solutions.
Now
∴
AX = B
LM1 1 1OP LMx OP LM6OP
MM0 1 2PP MM y PP = MM8PP
N0 0 0 Q N z Q N0Q
x+y+z = 6
...(i)
y + 2z = 8
...(ii)
Let
z = k
Putting z = k in (ii), we get
y + 2k = 8 ⇒ y = 8 – 2k
From (i)
x + 8 – 2k + k = 6 ⇒ x = k – 2
Therefore, x = k – 2, y = 8 – 2k and z = k.
Ans.
195
MATRICES
Example 6. Solve
3x + 3y + 2z = 1
x + 2y = 4
10y + 3z = – 2
2 x – 3y – z = 5
Sol. The augmented matrix is
3
3
1
2
[A : B] =
0 10
2 −3
R1 ↔ R3
1
2
3
3
~
0 10
2 −3
LM
MM
MN
LM
MM
MN
OP
PP
PQ
0 :
4O
2 :
1P
P
3 : −2P
P
−1 :
5Q
2
0
3
−1
:
:
:
:
1
4
−2
5
0
2
3
−1
:
:
:
:
4
−11
−2
−3
R2 → R2 – 3R1, R4 → R4 – 2R1
LM1
0
~ M
MM0
N0
R3 → R3 +
OP
PP
PQ
10
7
R , R → R4 – R2
3 2 4
3
LM1
M0
~ M
MM0
MM0
N
R3 →
2
−3
10
−7
3
3
R3, R4 →
R
29
17 4
R4 → R4 + R3
LM1
0
~ M
MM0
N0
LM1
0
~ M
MM0
N0
2
0
:
−3
2
29
3
−17
3
:
0
0
2
−3
0
0
0
2
1
−1
2
−3
0
0
0
2
1
0
:
:
:
:
:
:
:
:
:
:
OP
−11 P
−116 P
P
3 P
68 P
3 PQ
4
OP
PP
PQ
4
−11
−4
4
OP
PP
PQ
4
−11
−4
0
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
⇒
i.e.,
ρ (A) = ρ [A : B] = 3
r = 3 = n = number of variables.
Hence, the system is consistent and has unique solution.
Now,
AX = B
⇒
LM1 2 0OP LxO LM 4OP
MM00 −30 21PP MMyPP = MM−−114PP
MN0 0 0PQ MNz PQ MN 0PQ
⇒
x + 2y = 4
...(i)
– 3y + 2z = – 11
z = –4
...(ii)
...(iii)
On solving (i) and (ii), we get x = 2, y = 1. Hence, x = 2, y = 1 and z = – 4.
Example 7. Apply the matrix method to solve the system of equations
x + 2y – z = 3
3x – y + 2z = 1
2x – 2y + 3z = 2
x– y +z = – 1
[U.P.T.U., 2003; U.P.T.U. (C.O.), 2003]
Sol. The augmented matrix is
LM1 2 −1 : 3OP
3 −1
2 :
1P
[A : B] = M
MM2 −2 3 : 2PP
N1 −1 1 : −1Q
R2 → R2 – 3R1, R3 → R3 – 2R1, R4 → R4 – R1
LM1 2
0 −7
~ M
MM0 −6
N0 −3
R3 → R3 –
−1
5
5
2
OP
PP
PQ
:
:
:
:
3
−8
−4
−4
6
3
R2, R4 → R4 –
R
7
7 2
LM1
MM0
M0
~ M
MM0
N
2
−1
:
−7
5
5
7
1
−
7
:
0
0
:
:
OP
−8 P
20 P
P
7 P
4
− PP
7Q
3
197
MATRICES
R4 → R4 +
⇒
1
R
5 3
LM1 2 −1 : 3OP
M0 −7 55 : 20−8PP
~ M
MM0 0 7 : 7 PP
0Q
N0 0 0
ρ (A) = ρ [A : B] = 3 = number of variables.
Hence, the system is consistent and has a unique solution.
Now,
AX = B
⇒
LM 3 OP
LM1 2 −1OP
MM0 −7 55PP LMxyOP = MM −208 PP
MM0 0 7 PP MMNz PPQ MM 7 PP
MN 0 PQ
N0 0 0Q
⇒
x + 2y – z = 3
– 7y + 5z = – 8
...(i)
...(ii)
5
20
z =
⇒z=4
7
7
From (i) and (ii), we get, x = – 1, y = 4
⇒
x = – 1, y = 4, z = 4.
3.14
SYSTEM OF HOMOGENEOUS EQUATIONS
If in the set of equations (1) of (3.13), b1 = b1 = ... = bn = 0, the set of equation is said to be
homogeneous.
Result 1: If r = n, i.e., the rank of coefficient matrix is equal to the number of variables, then
there is always a trivial solution (x1 = x2 = ... = xn = 0).
Result 2: If r < n, i.e., the rank of coefficient matrix is smaller than the number of variables,
then there exist a non-trivial solution.
Result 3: For non-trivial solution always |A| = 0.
Example 8. Solve the following system of homogeneous equations:
x + 2y + 3z = 0
3x + 4y + 4z = 0
7x + 10y + 12z = 0
Sol. Here,
A =
LM1 2 3OP
MM3 4 4PP
N7 10 12Q
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
LM1 2 3OP
~ M0 −2 −8P
MN0 −4 −9PQ
R3 → R3 – 2R2
LM1 2 3OP
A ~ M0 −2 −5P
MN0 0 1PQ
This shows rank (A) = 3 = number of unknowns. Hence, the given system has a trivial
solution i.e., x = y = z = 0.
Example 9. Solve
x + y – 2z + 3w = 0
x – 2y + z – w = 0
4x + y – 5z + 8w = 0
5x – 7y + 2z – w = 0
Sol. The coefficient matrix A is
LM1 1 −2 3OP
1 −2
1 −1
A = M
MM4 1 −5 8PPP
N5 −7 2 −1Q
R2 → R2 – R1, R3 → R3 – 4R1, R5 → R5 – 5R1
R3 → R3 – R2, R4 → R4 – 4R2
LM1 1 −2
−3
0
3
~ M
MM0 −3 3
N0 −12 12
3
−4
−4
−16
OP
PP
PQ
LM1 1 −2
0 −3
3
~ M
MM0 0 0
N0 0 0
3
−4
0
0
OP
PP
PQ
⇒ ρ (A) = 2 < 4 (n = 4), so there exist a non-trivial solution.
Now,
LM1 1 −2
0 −3
3
⇒ M
MM0 0 0
N0 0 0
AX = B
OP LMx OP LM 0 OP
PP MMyz PP = MM 00 PP
PQ MNwPQ MN 0 PQ
3
−4
0
0
199
MATRICES
⇒
x + y – 2z + 3w = 0
– 3y + 3z – 4w = 0
...(i)
...(ii)
Choose z = k1, w = k2, then from (ii) and (i), we get
4
k
3 2
– 3y + 3k1 – 4k2 = 0 ⇒ y = k1 –
and
x + k1 –
4
5
k2 – 2k1 + 3k2 = 0 ⇒ x = k1 –
k.
3
3 2
where k1 and k2 are arbitrary constants.
Example 10. Find the value of λ such that the following equations have unique solution.
λx + 2y – 2z – 1 = 0, 4x + 2λy – z – 2 = 0, 6x + 6y + λz – 3 = 0
(U.P.T.U., 2003)
Sol. We have
λx + 2y – 2z = 1
4x + 2λy – z = 2
6x + 6y + λz = 3
The coefficient matrix A is
LMλ 2
A = M 4 2λ
MN 6 6
OP
PP
Q
−2
−1
λ
For unique solution |A| ≠ 0
∴
⇒
LMλ 2 −2OP
MM 4 2λ −1PP = λ + 11λ – 30 ≠ 0
N 6 6 λQ
3
(λ – 2) (λ2 + 2λ + 15) ≠ 0 ⇒ λ ≠ 2.
Example 11. Determine b such that the system of homogeneous equations (U.P.T.U., 2008)
2x + y + 2z = 0
x + y + 3z = 0
4x + 3y + bz = 0
has (i) trivial solution and (ii) non-trivial solution.
Sol. (i) For trivial solution, |A| ≠ 0
∴
⇒
⇒
|A| =
2
1
4
1
1
3
2
3 ≠0
b
2 (b – 9) – (b – 12) + 2 (3 – 4) ≠ 0
2b – 18 – b + 12 – 2 ≠ 0 ⇒ b – 8 ≠ 0 ⇒ b ≠ 8.
(ii) For non-trivial solution, |A| = 0
⇒
b – 8 = 0 ⇒ b = 8.
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
3.15 GAUSSIAN ELIMINATION METHOD
Gaussian elimination method is an exact method which solves a given system of equations in n
unknowns by transforming the coefficient matrix into an upper triangular matrix and then solve
for the unknowns by back substitution.
Example 12. Solve the system of equations:
2x1 + 3x2 + x3 = 9
x1 + 2x2 + 3x3 = 6
3x1 + x2 + 2x3 = 8
(U.P.T.U., 2006)
by Gaussian elimination method.
Sol. The augmented matrix is:
LM2
MM
N3
[A : B] = 1
R1 ↔ R2
3
2
1
1
3
2
:
:
:
OP
PP
Q
9
6
8
LM1 2 3 : 6OP
~ M 2 3 1 : 9P
MN3 1 2 : 8PQ
R2 → R2 – 2R1, R3 → R3 – 3R1
LM1 2 3 : 6OP
~ M0 −1 −5 :
−3
MN0 −5 −7 : −10PPQ
R3 → R3 – 5R2
LM1 2 3 : 6OP
~ M0 −1 −5 : −3P
MN0 0 18 : 5PQ
which is upper triangular form
∴
and
18x 3 = 5 ⇒ x3 =
5
18
x1 + 2x2 + 3x3 = 6
– x2 – 5x3 = – 3
From (ii)
x2 +
25
25
29
= 3 ⇒ x2 = 3 –
=
18
18
18
again from (i), we have
x1 + 2 ×
⇒
Hence,
29
5
73
+3×
= 6 ⇒ x1 +
=6
18
18
18
73
35
x1 = 6 –
=
18
18
35
29
5 .
x1 =
, x2 =
and x3 =
18
18
18
...(i)
...(ii)
201
MATRICES
Example 13. Solve by Gaussian elimination method
10x1 – 7x2 + 3x3 + 5x4 = 6
– 6x1 + 8x2 – x3 – 4x4 = 5
3x1 + x2 + 4x3 + 11x4 = 2
5x1 – 9x2 – 2x3 + 4x4 = 7
Sol. The augmented matrix is
LM 10 −7 3 5 : 6OP
−6
8 −1 −4 : 5P
[A : B] = M
MM 3 1 4 11 : 2PP
N 5 −9 −2 4 : 7Q
R3 ↔ R2
LM 10 −7 3 5 : 6OP
M 3 81 −41 −114 :: 25PP
~ M−6
MN 5 −9 −2 4 : 7PQ
R2 → R2 –
3
6
1
R , R → R3 +
R , R → R4 –
R
10 1 3
10 1 4
2 1
LM10
MM 0
M
~ M
MM 0
MM
MN 0
R2 →
−
−7
3
5
:
31
10
31
10
19
2
:
19
5
4
5
−1
:
7
2
3
2
:
11
2
−
OP
1P
P
5P
43 PP
5P
P
4P
PQ
6
10
5
2
R2, R3 →
R2, R3 → − R3
31
19
11
LM10
MM 0
M
~ M
MM 0
MM
MN 0
−7
3
5
:
1
1
95
31
:
1
4
19
−
5
19
:
1
7
11
−
3
11
:
OP
2P
P
31 P
43 PP
19 P
P
−8 P
11 PQ
6
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
R3 → R3 – R2, R4 → R4 – R2
LM10
MM 0
~ M
MM 0
MM
MN 0
R3 → −
−7
3
1
1
0
0
5
:
95
31
1960
−
589
1138
−
341
15
19
4
−
11
−
:
:
:
OP
2 P
P
31 P
1295 P
P
589 P
270 P
−
341 PQ
6
19
11
R3, R4 → − R4
15
4
LM10
MM 0
~ M
MM 0
MM
MN 0
R4 → R4 – R3
LM10
MM 0
M
~ M
MM 0
MM 0
N
−7
3
1
1
0
1
0
1
−7
3
1
1
0
1
0
0
5
95
31
392
93
569
62
5
95
31
392
93
923
186
:
:
:
:
:
:
:
:
OP
2 P
P
31 P
259 P
−
P
93 P
135 P
62 PQ
6
OP
2 P
P
31 P
259 P
−
P
93 P
923 P
186 PQ
6
Hence, the coefficient matrix is an upper triangular form
∴
or
and
⇒
923
923
x4 =
⇒ x4 = 1
186
186
392
259
392
259
x3 +
x = −
⇒ x3 +
= −
93 4
93
93
93
259
392
651
−
x3 = −
= −
=–7
93
93
93
95
2
95
2
x2 + x3 +
x =
⇒ x2 – 7 +
=
31 4
31
31
31
x2 –
122
2
2 122 124
+
=
=
⇒ x2 =
=4
31
31
31 31
31
Again 10x1 – 7x2 + 3x3 + 5x4 = 6
⇒ 10x1 – 7 × 4 + 3 × (–7) + 5 × 1 = 6 ⇒ 10x1 – 28 – 21 + 5 = 6
203
MATRICES
⇒ 10x1 – 44 = 6 ⇒ 10x1 = 50 ⇒ x1 = 5
Therefore, x1 = 5, x2 = 4, x3 = – 7 and x4 = 1.
3.15.1 Gauss-Jordan Elimination Method
Apply elementary row operations on both A and B such that A reduces to the normal form. Then
the solution is obtained.
Example 14. Solve by Gauss-Jordan elimination method:
2x1 + x2 + 3x3 = 1
4x1 + 4x2 + 7x3 = 1
2x1 + 5x2 + 9x3 = 3
LM2 1 3OP
LM1OP
A = M 4 4 7P , B = M1P
MN2 5 9PQ
MN3PQ
Sol. Here
∴ Augmented matrix
LM2 1 3 : 1OP
[A : B] = M4 4 7 : 1P
MN2 5 9 : 3PQ
R2 → R2 – 2R1, R3 → R3 – R1
R3 → R3 – 2R2
R1 →
LM2 1 3 : 1 OP
~ M0 2 1 : −1P
MN0 4 6 : 2 PQ
LM2 1 3 : 1 OP
~ 0 2 1 : −1
MM0 0 4 : 4 PP
N
Q
1
1
1
R , R → R2 , R3 → R3
2 1 2
2
4
LM1
M
~ M0
MM
MN0
R1 → R1 −
1
R
2 2
LM1
M
~ M0
MM
MN0
1
2
1
0
0
1
0
3
2
1
2
1
5
4
1
2
1
:
:
:
:
:
:
OP
PP
PP
1 PQ
1
2
1
−
2
OP
PP
PP
1 PQ
3
4
1
−
2
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
R1 → R1 –
5
1
R3 , R2 → R2 – R3
4
2
LM1 0 0 : − 1 OP
2
~ M0 1 0 : − 1 P
MM
PP
N0 0 1 : 1 Q
Hence, the matrix A is in normal form
∴
x1 = –
1
, x = – 1, x3 = 1.
2 2
Example 15. Solve by Gauss-Jordan elimination method:
2x1 + 5x2 + 2x3 – 3x4 = 3
3x1 + 6x2 + 5x3 + 2x4 = 2
4x1 + 5x2 + 14x3 + 14x4 = 11
5x1 + 10x2 + 8x3 + 4x4 = 4
Sol.
LM2 5 2 −3 : 3 OP
3 6 5 2 : 2P
[A : B] = M
MM4 5 14 14 : 11PP
N5 10 8 4 : 4 Q
R2 → R2 – R1, R3 → R3 – R2, R4 → R4 – R3
R1 ↔ R4, R2 ↔ R3
LM2 5 2 −3 : 3 OP
1 1 3
5 : −1
~ M
MM1 −1 9 12 : 9 PPP
N1 5 −6 −10 : −7Q
LM1 5 −6 −10 : −7OP
1 −1 9 12 : 9 P
~ M
MM1 1 3 5 : −1PP
N2 5 2 −3 : 3 Q
R2 → R2 – R1, R3 → R3 – R1, R4 → R4 – 2R1
LM1 5 −6 −10 : −7OP
0 −6 15 22 : 16 P
~ M
MM0 −4 9 15 : 6 PP
N0 −5 14 17 : 17Q
R2 → R2 – R4, then R2 → – R2, R3 → R3 – 4R2, then again R3 → –R3 , R4 → R4 – 5R2 , then also
again R4 → –R4
LM1 5 −6 −10 : −7 OP
0 1 −1 −5 : 1 P
~ M
MM0 0 −5 5 : −10PP
N0 0 9 −8 : 22 Q
205
MATRICES
R3 → –
1
R , then R4 → R4 – 9R3
5 3
LM1 5 −6 −10 : −7OP
0 1 −1 −5 : 1
~ M
MM0 0 1 −1 : 2 PPP
N0 0 0 1 : 4 Q
R3 → R3 + R4, R2 → R2 + 5R4, R1 → R1 + 10R4
LM1 5 −6 0 : 33OP
0 1 −1 0 : 21P
~ M
MM0 0 1 0 : 6 PP
N0 0 0 1 : 4 Q
R2 → R2 + R3, R1 → R1 + 6R3
LM1 5 0 0 : 69OP
M0 1 0 0 : 27PP
~ M
MM0 0 1 0 : 26PP
N0 0 0 1 : 4 Q
R1 → R1 – 5R2
LM1 0 0 0 : −66OP
0 1 0 0 : 27 P
~ M
MM0 0 1 0 : 6 PP
N0 0 0 1 : 4 Q
Hence, the coefficient matrix is in normal form
∴
x1 = – 66, x2 = 27, x3 = 6 and x4 = 4.
Example 16. Find the values of λ for which the equations
3x + y – λz = 0; 4x – 2y – 3z = 0; 2 λx + 4y + λz = 0
have a non-trivial solution. Obtain the most general solutions in each case.
Sol. The coefficient matrix is
LM 3 1 −λOP
A = M 4 −2 −3 P
MN2λ 4 λ PQ
for non-trivial solution A = 0
∴
LM 3 1 −λOP
A = M 4 −2 −3 P = 0 ⇒ 3 (–2λ + 12) – (4λ + 6λ) – λ (16 + 4λ) = 0
NM2λ 4 λ PQ
⇒ – 6λ + 36 – 10λ – 16λ – 4λ2 = 0
⇒
– 4λ2 – 32λ + 36 = 0 ⇒ λ2 + 8λ – 9 = 0
206
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
⇒
⇒
λ2 + 9λ – λ – 9 = 0 ⇒ (λ – 1) (λ + 9) = 0
λ = 1, – 9.
Case I. If λ = 1,
R2 → R2 –
LM 3 1 −1 OP
A = M 4 −2 −3 P
MM 2 4 1 PP
N
Q
4
2
R , R → R3 – R1
3 1 3
3
LM3
~ M0
MM
MN0
R3 → R3 + R2
LM3
~ M0
MM
N0
1
−1
−10
3
10
3
5
−
3
5
3
1
−
10
3
0
OP
PP
PP
Q
OP
5
− P
3P
0 PQ
−1
This shows ρ (A) = 2 < 3 (n = 3)
∴ Let z = k
10 y 5z
k
−
−
= 0
⇒ 2y + k = 0 ⇒ y = –
2
3
3
and
3x + y – z = 0
Case II. If λ = – 9
⇒ 3x –
k
k
–k=0⇒x=
. Ans.
2
2
LM 3 1 9 OP
A = M 4 −2 −3P
MN−18 4 −9PQ
Solve itself like case I.
x = −
3k
9k
,y= −
, z = k.
2
2
EXERCISE 3.4
Examine whether the following systems of equations are consistent. If consistent solve.
1.
x+y+z = 6
2x + 3y – 2z = 2
5x + y + 2z = 13
(U.P.T.U., 2000) [Ans. x = 1, y = 2, z = 3]
2.
3x + 3y + 2z = 1
x + 2y = 4
207
MATRICES
3.
4.
5.
6.
10y + 3z = – 2
2x – 3y – z = 5
[Ans. x = 2, y = 1 and z = – 4]
x1 + 2x2 − x3 = 3
3x1 – x2 + 2x3 = 1
2x1 – 2x2 + 3x3 = 2
x1 – x2 + x3 = – 1
(U.P.T.U., 2002) [Ans. x1 = –1, x2 = 4, x3 = 4]
x+y+z = 7
x + 2y + 3z = 8
y + 2z = 6
[Ans. inconsistent; no solution]
5x + 3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5
[Ans. x = (7 – 6k)/11, y = (3 + k)/11, z = k]
– x1 + x2 + 2x3 = 2
3x1 – x2 + x3 = 6
– x1 + 3x2 + 4x3 = 4
[Ans. x1 = 1, x2 = – 1, x3 = 2]
7. Find the values of a and b for which the system has (i) no solution, (ii) unique solution
and (iii) infinitely many solution.
2x + 3y + 5z = 9
(i)
a = 5, b ≠ 9
7x + 3y – 2z = 8
Ans. (ii) a ≠ 5, b any value
2x + 3y + az = b
(iii)
a = 5, b = 9
8. Discuss the solutions of the system of equations for all values of λ.
x + y + z = 2, 2x + y – 2z = 2, λx + y + 4z = 2.
[Ans. Unique solution if λ ≠ 0; infinite number of solutions if λ = 0]
9.
10.
x+y+z = 6
x + 2y + 3z = 14
x + 4y + 7z = 30
[Ans. x = k – 2, y = 8 – 2k, z = k]
2x – y + z = 7
3x + y – 5z = 13
x+y+z = 5
[Ans. x = 4, y = 1, z = 0]
Solve the following homogeneous equations:
11.
x + 2y + 3z = 0
2x + y + 3z = 0
3x + 2y + z = 0
12.
13.
[Ans. x = y = z = 0]
x + y + 3z = 0
x–y+z = 0
x – 2y = 0
x–y+z = 0
[Ans. x = – 2k, y = –k, z = k]
x + 2y + 3z = 0
3x + 4y + 4z = 0
7x + 10y + 12z = 0
[Ans. x = y = z = 0]
208
14.
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
4x + 2y + z + 3w = 0
6x + 3y + 4z + 7w = 0
2x + y + w = 0
[Ans. x = k1, y = – 2k1 – k2, z = – k2, w = k2]
15. For what values of λ the given equations will have a non-trivial solution.
x + 2y + 3z = λx
2x + 3y + z = λx
3x + y + 2z = λy
[Ans. λ = 6]
16. Find the value of ‘a’ so that the following system of homogeneous equations have exactly
2 linearly independent solutions.
ax1 – x2 – x3 = 0
– x1 + ax2 – x3 = 0
[Ans. a = – 1]
– x1 – x2 + ax3 = 0
17. Apply the test of rank to examine if the following equations are consistent:
2x – y + 3z = 8
– x + 2y + z = 4
3x + y – 4z = 0
and if consistent, find the complete solution.
[Ans. x = y = z = 2]
18. Show that the equations
– 2x + y + z = a
x – 2y + z = b
x + y – 2z = c
have no solutions unless a + b + c = 0, in which case they have infinitely many solutions.
Find their solutions a = 1, b = 1 and c = – 2.
[Ans. x = k – 1, y = k – 1, z = k]
19. Show that the system of equations x + 2y – 2w = 0, 2x – y – w = 0, x + 2y – w = 0.
– y + 3z – w = 0 do not have a non-trivial solution.
4x
20. Show that the homogeneous system of equations x + y cos γ + z cos β = 0, x cos γ + y
z cos α = 0, x cos β + y cos γ + z = 0, has non-trivial solution if α + β + γ = 0.
+
Solve the following system of equations by Gaussian elimination method:
21.
x1 + 2x2 – x3 = 3
2x1 – 2x2 + 3x3 = 2
3x1 – x2 + 2x3 = 1
x1 – x2 + x3 = – 1
[Ans. x1 = 1, x2 = 4, x3 = 4]
22.
23.
2x1 + x2 + x3 = 10
3x1 + 2x2 + 3x3 = 18
x1 + 4x2 + 9x3 = 16
[Ans. x1 = 7, x2 = – 9, x3 = 5]
2x1 + x2 + 4x3 = 12
8x1 – 3x2 + 2x3 = 20
4x1 + 11x2 – x3 = 33
[Ans. x1 = 3, x2 = 2, x3 = 1]
209
MATRICES
24.
25.
26.
x1 + 4x2 – x3 = –5
x1 + x2 – 6x3 = – 12
3x1 – x2 – x3 = 4
LMA ns. x = 117 , x = − 81 , x = 148 OP
71
71
71 Q
N
1
2
2x1 + x2 + 2x3 + x4 = 6
6x1 – 6x2 + 6x3 + 12x4 = 36
4x1 + 3x2 + 3x3 – 3x4 = –1
2x1 + 2x2 – x3 + x4 = 10
2x1 – 7x2 + 4x3 = 9
x1 + 9x2 – 6x3 = 1
– 3x1 + 8x2 + 5x3 = 6
[Ans. x1 = 2, x2 = 1, x3 = – 1, x4 = 3]
[Ans. x1 = 4, x2 = 1, x3 = 2]
Solve by Gauss-Jordan method:
27.
x – 3y – 8z = – 10
3x + y = 4
2x + 5y + 6z = 13
28.
29.
3
[Ans. x = y = z = 1]
x+y+z = 6
2x + 3y – 2z = 2
5x + y + 2z = 13
[Ans. x = 1, y = 2, z = 3]
3x + y + 2z = 3
2x – 3y – z = – 3
x + 2y + z = 4
[Ans. x = 1, y = 2, z = – 1]
4x + 3y + 3z = – 2
x+z = 0
4x + 4y + 3z = – 3
[Ans. x = 1, y = 1, z = – 1]
31. Test the consistency and solve 2x – 3y + 7z = 5, 3x + y – 3z = 13, 2x + 19y – 47z = 32.
[Ans. Inconsistent]
Solve the following system by any method:
32.
2x + 6y + 7z = 0
6x + 20y – 6z = – 3
6y – 18z = – 1
[Ans. Inconsistent]
30.
33.
34.
4x – y + 6z = 16
x – 4y – 3z = – 16
2x + 7y + 12z = 48
5x – 5y + 3z = 0
2x1 + x2 + 5x3 + x4 = 5
x1 + x2 – 3x3 – 4x4 = – 1
3x1 + 6x2 – 2x3 + x4 = 8
2x1 + 2x2 + 2x3 – 3x4 = 2
35.
5x + 3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 11z = 5
LMA ns. x = −9k + 16 , y = −6k + 16 , z = kOP
5
3
5
3
N
Q
LMA ns. x = 2, x = 1 , x = 0, x = 4OP
5
5Q
N
1
2
3
4
LMA ns. x = 7 , y = 3 , z = 0OP
11
11
N
Q
210
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
3.16
LINEAR DEPENDENCE OF VECTORS
The set of vectors* (row or column matrices) X1, X2, ... Xn is said to be linearly dependent if there
exist scalars a1, a2,... an not all zero such that
a1X1 + a2X2 + ... + anXn = O
[O is null matrix]
3.16.1 Linear Independence of Vectors
If the set of vectors is not linearly dependent then it is said to be linearly independent.
i.e., if every relation of the type
a1X1 + a2X2 + ... + anXn = O
⇒
a1 = a2 = ... = an = 0.
Example 1. Show that the vectors (3, 1, – 4), (2, 2, – 3) and (0, – 4, 1) are linearly dependent.
Sol. Let X1 = (3, 1, – 4), X2 = (2, 2, – 3), X3 = (0, – 4, 1)
Now,
a1X1 + a2X2 + a3X3 = O
|linear dependence
a1(3, 1, – 4) + a2 (2, 2, – 3) + a3 (0, – 4, 1) = (0, 0, 0)
⇒
(3a1 + 2a2, a1 + 2a2 – 4a3, – 4a1 – 3a2 + a3) = (0, 0, 0)
⇒
3a1 + 2a2 = 0, a1 + 2a2 – 4a3 = 0, – 4a1 – 3a2 + a3 = 0
The system of equations is homogeneous.
Now the coefficient matrix is
R1 ↔ R2
LM 3 2 0OP
2 −4
A = M 1
MN− 4 −3 1PPQ
R2→ R2 – 3R1, R3 → R3 + 4R1
R2 → –
1
1
R, R →
R
4 2 3
5 3
LM 1 2 − 4OP
~ M 3
2
0
MN− 4 −3 1PPQ
LM1 2 − 4OP
~ M0 − 4
12
MN0 5 −15PPQ
LM1 2 −4OP
~ M0 1 −3P
MN0 1 −3PQ
LMx OP
x
* An ordered set of n numbers belonging to a field F and denoted by X = [x , x , ... x ] or M P is called
MM : PP
an n-dimensional vector over F.
Nx Q
1
2
1
2
n
n
211
MATRICES
LM1 2 −4OP
~ M0 1 −3P
MN0 0 0 PQ
R3 → R3 – R2
and
⇒
ρ(A) = 2 and n = 3
Let
a3 = k
a2 – 3a3 = 0 ⇒ a2 – 3k = 0 ⇒ a2 = 3k
Again a1 + 2a2 – 4a3 = 0 ⇒ a1 + 6k – 4k = 0 ⇒ a1 = – 2k
a1 = – 2k, a2 = 3k, a3 = k.
Therefore,
Hence, a1, a2 and a3 cannot be zero otherwise there is a trivial solution which is impossible.
So the vectors X1, X2 and X3 are linearly dependent and the relation is
– 2kX1 + 3kX2 + kX3 = O
⇒
2X1 – 3X2 – X3 = O.
Example 2. Find the value of λ for which the vectors (1, – 2, λ), (2, – 1, 5) and (3, – 5, 7λ) are
linearly dependent.
(U.P.T.U., 2006)
Sol. Let X1 = (1, – 2, λ), X2 = (2, – 1, 5), X3 = (3, – 5, 7λ)
Now
a1X1 + a2X2 + a3X3 = 0
|For linear dependence
⇒ a1 (1, – 2, λ) + a2 (2, – 1, 5) + a3 (3, – 5, 7λ) = (0, 0, 0)
⇒
U|
V
λa + 5a + 7λa = 0 |W
a1 + 2a2 + 3a3 = 0
– 2a1 – a2 – 5a3 = 0
1
2
...(i)
3
The system is homogeneous
∴ For non-trivial solution* |A| = 0
⇒
|A| =
1
2
3
−2
−1
−5 = 0
λ
5
7λ
= (– 7λ + 25) – 2 (– 14λ + 5λ) + 3 ( – 10 + λ) = 0
= – 7λ + 25 + 18 λ – 30 + 3 λ = 0
= 14λ – 5 = 0 ⇒ λ =
5
.
14
Example 3. Show that the vectors [0, 1, – 2], [1, – 1, 1] [1, 2, 1] form a linearly independent
set.
Sol. Let X1 = [0, 1, – 2], X2 = [1, – 1, 1] X3 = [1, 2, 1]
Also suppose a1X1 + a2X2 + a3X3 = O
⇒ a1 [0, 1, –2] + a2 [1, – 1, 1] + a3 [1, 2, 1] = O
⇒ {a2 + a3, a1 – a2 + 2a3, – 2a1 + a2 + a3] = [0, 0, 0]
⇒
a2 + a3 = 0
a1 – a2 + 2a3 = 0
* For linear dependence the solution must be non-trivial otherwise it will be linear independence.
212
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
– 2a1 + a2 + a3 = 0
LM 0 1 1OP
A = M 1 −1 2P
NM−2 1 1PQ
Now
R1 ↔ R2
LM 1 −1 2OP
0 1 1
A ~ M
MN−2 1 1PPQ
R3 → R3 + 2R1
R3 → R3 + R2
LM1 −1 2OP
~ M0 1 1P
NM0 −1 5PQ
LM1 −1 2OP
~ M0 1 1P = ρ (A) = 3
MN0 0 6PQ
LM1 −1 2OPLM a OP LM0OP
MM00 10 16PPMMaa PP = MM0PP
QN Q N0Q
N
1
Now
2
3
⇒
a1 – a2 + 2a3 = 0
a2 + a3 = 0
a3 = 0
⇒
a1 = a2 = a3 = 0
All a1, a2 and a3 are zero. Therefore, they are linearly independent.
Example 4. Examine the vectors
X1 = [1, 1, 0]T, X2 = [3, 1, 3]T, X3 = [5, 3, 3]T
are linearly dependent.
LM1OP
LM3OP
LM5OP
Sol. Here X = M1P , X = M1P , X = M 3P
MN0PQ
MN3PQ
MN3PQ
1
Now
⇒
⇒
⇒
2
3
a1X1 + a2X2 + a3X3 = O
LM1OP LM3OP LM5OP L0O
a M1P + a M1P + a M 3P = M0P
MN0PQ MN3PQ MN3PQ MMN0PPQ
LMa + 3a + 5a OP LM0OP
MM 0a ++3aa ++ 33aa PP = MM00PP
Q NQ
N
1
2
3
1
2
3
1
2
3
2
3
a1 + 3a2 + 5a3 = 0
a1 + a2 + 3a3 = 0
0.a1 + 3a2 + 3a3 = 0
213
MATRICES
LM1 3 5OP
A = M1 1 3P
MN0 3 3PQ
∴
R1 → R2 – R1
R3 → R3 +
LM1 3 5 OP
A ~ 0 −2 −2
MM0 3 3 PP
N
Q
3
R
2 2
LM1 3 5 OP
~ M0 −2 −2P
MN0 0 0 PQ
Here ρ(A) = 2 but n = 3 so let a3 = k.
LM1 3 5 OP LM a OP L0O
MM00 −02 −02PP MMaa PP = MM0PP
Q N Q MN0PQ
N
1
and
2
3
a1 + 3a2 + 5a3 = 0
⇒
...(i)
a2 + a 3 = 0
a2 + k ⇒ a2 = – k
...(ii)
From (i) a1 + 3 (–k) + 5k = 0 ⇒ a1 = 2k
Since all a1, a2, a3 are not zero.
Therefore, they are linearly dependent
And the relation is
⇒
2kX1 – kX2 + kX3 = O
2X1 – X2 + X3 = O.
Example 5. Show that the vectors [2, 3, 1, –1], [2, 3, 1, – 2], [4, 6, 2, – 3] are linearly
independent.
Sol. Consider the relation a1X1 + a2X2 + a3X3 = O
⇒ a1[2, 3, 1, –1] + a2 [2, 3, 1, – 2] + a3 [4, 6, 2, –3] = O
⇒
2a1 + 2a2 + 4a3 = 0
3a1 + 3a2 + 6a3 = 0
a1 + a2 + 2a3 = 0
a1 + 2a2 + 3a3 = 0
∴
LM2 2 4OP
3 3 6P
A = M
MM1 1 3PP , R ↔ R , R ↔ R
N1 2 2Q
1
4
2
3
LM1 2 2OP
1 1 3P
~ M
MM3 3 6PP
N2 2 4Q
214
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
R2 → R2 – R1, R3 → R3 – 3 R1, R4→ R4 – 2R1
LM1 2 2OP
LM1 2 2 OP
0 −1 1P
0 −1 1 P
~ M
, R → R – 3R , R → R – 2R ~ M
MM0 −3 0PP
MM0 0 −3PP
N0 −2 0Q
N0 0 −2Q
3
R4 → R4 −
2
R
3 3
3
2
4
4
2
LM1 2 2 OP
0 −1 1 P
~ M
MM0 0 −3PP ⇒ ρ(A) = 3 and n = 3
N0 0 0 Q
∴ The system has a trivial solution
Hence,
a1 = a2 = a3 = 0
i.e., they are linearly independent.
EXERCISE 3.5
1. Examine the following vectors for linear dependence and find the relation if it exists.
X1 = (1, 2, 4), X2 = (2, –1, 3), X3 = (0, 1, 2), X4 = (–3, 7, 2).
(U.P.T.U., 2002)
[Ans. Linearly dependent, 9X1 – 12X2 + 5X3 – 5X4 = 0]
2. Show the vectors X1 = [1, 2, 1], X2 = [2, 1, 4], X3 = [4, 5, 6] and X4 = [1, 8, – 3] are linearly
independent?
3. Show that the vectors [1, 2, 3], [3, –2, 1] , [1, – 6, – 5] are linearly dependent.
4. If the vectors (0, 1, a), (1, a, 1), (a, 1, 0) are linearly dependent, then find the value of a.
[Ans. a = 0, 2 , – 2 ]
5. Examine for linear dependence [1, 0, 2, 1], [3, 1, 2, 1] [4, 6, 2, – 4], [–6. 0, – 3, –4] and find
the relation between them, if possible.
[Ans. Linear dependent and the relation is 2X1 – 6X2 + X3 – 2X4 = 0]
6. Show that the vectors X1 = [2, i, – i]. X2 = [2i, – 1, 1], X3 = [1, 2, 3] are linearly dependent.
7. If X1 = [1, 1, 2], X2 = [2, – 1, – 6], X3 = [13, 4, – 4] prove that 7X1 + 3X2 – X3 = 0.
8. Show that the vectors [3, 1, – 4], [2, 2, – 3] form a linearly independent set but [3, 1, – 4],
[2, 2, – 3] and [0, – 4, 1] are linearly dependent.
3.17
EIGEN VALUES AND EIGEN VECTORS
Introduction: At the start of 20th century, Hilbert studied the eigen values. He was the first to use
the German word eigen to denote eigen value and eigen vectors in 1904. The word eigen mean—
own characteristic or individual.
More formally in a vector space, a vector function A (matrix) defined if each vector X of vector
space, there corresponds a unique vector Y = AX. So here we consider a linear transform Y = AX
215
MATRICES
transforms X into a scalar multiple of itself say λ so AX = λX which is called Eigen value equation.
In this section, we study the problem
AX = λX
...(i)
_v
>[
>l
v
where A is a n × n matrix, X is an unknown n × 1 vector and λ is an unknown scalar. From equation
(i) it can be understand that AX is a scalar multiple of X say λX. Geometrically each vector on the
line through the origin determined by X gets mapped back onto the same line under multiplication
by A.
Geometrical representation: The eigen value equation
means that under the transformation, a eigen vector experiv
ence only changes in magnitude and sign. The direction of
lw
AX is the same as that of X.
Here A acts to stretch the vector X, not change its
direction. So X is an given vector of A.
w
The eigen value determines the amount, the eigen vector
is scaled under the linear transformation. For example, the
eigen value λ = 2 means that the eigen vector is doubled in
length and point in the same direction. The eigen value λ =
1, means that the eigen vector is unchanged, while an eigen
value λ = – 1 means that the eigen vector is reversed in
direction.
v
m
v
v
lv
Fig. 3.1
Thus, the eigen value λ is simply the amount of “stretches” or “shrinks” to which a vector
is subjected when transformed by A.
(U.P.T.U., 2007)
3.17.1 Characteristic Equation
If we re-express (i) as AX = λIX (where I is an identity matrix).
AX – λIX = O
or
⇒
(A – λI)X = O
...(ii)
Which is homogeneous system of n equations in the n variables x1, x2 ... xn. The system (ii)
must have non-trivial solutions otherwise X = 0 (which is impossible).
∴ For non-trivial solution the coefficient matrix (A – λI) will be singular
⇒
|A – λI| = 0
...(iii) |singular matrix |A| = 0
Expansion of the determinant gives an algebraic equation in λ, known as the ‘‘characteristic
equation’’ of A. The determinant |A – λI| is called characteristic polynomial of A.
3.17.2 Characteristic Roots or Eigen Values
[U.P.T.U. (C.O.), 2003, 2007]
The roots of characteristic equation are called characteristic roots or eigen values.
3.17.3 Eigen Vectors
[U.P.T.U. (C.O.), 2003, 2007]
The corresponding non-zero vector X is called characteristic eigen vector.
Notes: 1. If there is one linearly independent solution and two eigen values are same then there
will be same eigen vectors of both eigen values.
2. If there is two linearly independent solution and two eigen values are same then there
will be different eigen vectors of each eigen value.
216
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
3.17.4 Properties of Eigen Values and Eigen Vectors
1. If A is real, its eigen values are real or complex conjugate in pairs.
2. Determinant of A = product of eigen values of A.
3. A and AT has same eigen values.
4. A–1 exists iff 0 is not an eigen value of A, eigen values of A–1 are
1
,
1
λ1 λ 2
,....,
1
λn
.
(U.P.T.U., 2008)
5. Eigen vector cannot correspond to two distinct characteristic values.
6. Eigen values of diagonal, upper triangular or lower triangular matrices are the principal
diagonal elements.
7. KA (scalar multiples) has eigen values Kλi.
8. Am has eigen values λm.
(U.P.T.U., 2008)
9. Two vectors X and Y are said to be orthogonal if XTY = YTX = 0.
Theorem 1. The latent roots of a Hermitian matrix are all real
Proof. We have
AX = λX
[U.P.T.U. (C.O.), 2003]
...(i)
To prove that λ is a real number, we have to prove λ = λ
From (i)
X*(AX) = X* (λX) ⇒ X*AX = λX*X
⇒
(X* AX)* = (λX* X)* ⇒ X*A*X = λ X*X, (X** = X)
⇒
X* AX = λ X*X (A* = A)
⇒
X* λX = λ X*X ⇒ λX*X = λ X*X
⇒
—
(λ – λ ) X*X = 0 ⇒ λ = λ. Proved.
Theorem 2. The latent roots of a skew-Hermitian matrix are either zero or purely imaginary.
[U.P.T.U. (C.O.), 2003]
Proof. Let A be a skew-Hermitian matrix so that A* = – A.
Now we are to prove that λ = 0 or purely imaginary number
Here (iA)* = i A* = –iA* = –i(–A) = iA (As A* = –A)
Hence, iA is a Hermitian matrix.
Let λ be an eigen value relative to the eigen vector X of A, then
⇒
AX = λX
iAX = iλX ⇒ (iA)X = (iλ)X
...(i)
⇒ iλ is an eigen root relative to the vector X of Hermitian matrix iA.
⇒ iλ is a real number, for eigen values of a Hermitian matrix are all real.
⇒ λ = 0 or purely imaginary number. Proved.
Theorem 3. The characteristic roots of a unitary matrix are of unit modulus.
[U.P.T.U. Special Exam., 2001]
Proof. Let A be unitary matrix so that A*A = I.
We have
AX = λX
...(i)
To prove
|λ| = 1
217
MATRICES
From (i), we have
(AX)* = (λX)* ⇒ X*A* = λ X*
...(ii)
Pre-multiplying (i) by (ii),
(X*A*) (AX) = ( λ X*) (λX)
⇒
X* (A*A)X = λ λX*X
2
⇒
X*IX = |λ| X*X ⇒ (1 – |λ|2) X*X = 0
2
2
⇒ 1 – |λ| = 0 ⇒ |λ| = 1 or |λ| = 1. Proved.
Theorem 4. Prove that the product of all eigen values of A is equal to the determinant (A).
(U.P.T.U., 2004)
Proof. We have AX = λIX ⇒ (A – λI) X = 0
For non-trivial solution |A – λI| = 0
a11 − λ
⇒
a12
... a1n
#
an1
m
= 0
an2
... ann − λ
r
⇒ (–1)n λn + b1 λn− 1 + b2 λn− 2 +...+ bn = 0
⇒ (λ – λ1) (λ – λ2) ... (λ – λn) = 0
putting λ = 0, we get
|A| = λ1, λ2, λ3 ... λn. Proved.
Theorem 5. The latent roots of real symmetric matrix are all real.
Proof. Let A be a real symmetric matrix so that
A = A, A′ = A.
The A* = ( A )′ = A′ = A or A* = A, meaning thereby, A is a Hermitian matrix. Hence, the
latent roots of A are all real, by Theorem 1.
Theorem 6. The characteristic roots of an idempotent matrix are either zero or unity.
Proof. Let A be idempotent matrix so that A2 = A.
Let
AX = λX
...(i)
premultiplying by A on equation (i), we get
A(AX) = A(λX) = λ(AX)
(As A2 = A)
⇒
(AA)X = λ(λX) ⇒ A2X = λ2X or AX = λ2X
2
⇒
λX = λ X
2
⇒
(λ – λ)X = 0 ⇒ λ2 – λ = 0
(As X ≠ 0)
⇒
λ = 0, 1. Proved.
Example 1. Show that 0 is a characteristic root of a matrix if the matrix is singular.
(U.P.T.U., 2008)
Sol. The characteristic equation is
|A – λI| = 0
or
|A| – λ|I| = 0
As
|A| = 0 (singular matrix)
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
∴
0 = λ |I| = 0 ⇒ λ = 0
Conversely, if λ = 0 then |A| = 0. Proved.
LM 1 2 2OP
Example 2. Find the characteristic equation of the matrix 0 2 1 . Also find the eigen
MM−1 2 2PP
Q
N
values and eigen vectors of this matrix.
Sol. Let
Now,
(U.P.T.U., 2006)
LM 1 2 2OP
MM−01 22 12PP
Q
N
LM 1 2 2OP LM1 0 0OP LM1 − λ 2 2 OP
A – λI = 0 2 1 – λ M0 1 0P = M 0
MM−1 2 2PP M0 0 1P M −1 2 −2 λ 2 −1 λPP
Q N
Q
Q N
N
A =
∴ Characteristic equation is |A – λI| = 0
⇒
LM1 − λ 2 2 OP
MM 0 2 − λ 1 PP = 0 ⇒ (1 – λ) m( 2 − λ) − 2r − 2{0 + 1} + 2{0 + ( 2 − λ)} = 0
N −1 2 2 − λQ
2
⇒
(1 – λ) (λ2 – 4λ + 2) – 2 + 4 – 2λ = 0
⇒
λ2 – 4λ + 2 – λ3 + 4λ2 – 2λ + 2 – 2λ = 0
⇒
λ3 – 5λ2 + 8λ – 4 = 0
⇒
λ2(λ – 1) – 4λ (λ – 1) + 4 (λ – 1) = 0
⇒
(λ – 1) (λ2 – 4λ + 4) = 0
Hence, λ = 1, 2, 2. These are eigen values of A.
Now, we consider the relation
AX = λX ⇒ AX = λIX
⇒
(A – λI)X = 0
Taking λ = 1, from (1), we get
LM1 − 1 2 2 OP LMx OP
MM −01 2 2− 1 2 1− 1PP MMxx PP = 0
QN Q
N
LM 0 2 2OP LMx OP
MM−01 12 11PP MMxx PP = 0
QN Q
N
1
2
3
1
⇒
2
3
R1 ↔ R3
LM−1 2 1OP LMx OP
~ M 0 1 1P M x P = 0
MN 0 2 2PQ MNx PQ
1
2
3
...(i)
219
MATRICES
R3 → R3 – 2R2, R1 → (– 1) R1
LM1 −2 −1OP LMx OP
~ M0 1 1 P M x P = 0
MN0 0 0 PQ MNx PQ
1
2
3
⇒
Let
and
x1 – 2x2 – x3 = 0
x2 + x3 = 0
x3 = k then x2 = – k
x1 + 2k – k = 0 ⇒ x1 = – k
LM 1OP LM 1OP
LM−kOP
X = M− kP = − k M 1P or M 1P
MN−1PQ MN−1PQ
MN kPQ
∴
1
Taking λ = 2, from (1), we get
LM−1 2 2OP LMx1 OP
MM−01 02 10PP MMxx2 PP = 0
N
Q N 3Q
R2 ↔ R3
LM−1 2 2OP LM x OP
~ −1 2 0
MM 0 0 1PP MMxx PP = 0
QN Q
N
1
2
3
R2 → R2 – R1, R1 → –R1
LM1 −2 −2OP LM x OP
~ 0 0 −2
MM0 0 1 PP MMxx PP = 0
QN Q
N
1
2
3
R2 → –
1
R and then R3 → R3 – R2, we get
2 2
LM1 −2 −2OP LM x OP
~ M0 0 1 P M x P = 0
NM0 0 0 PQ MNx PQ
1
2
3
Here the solution is one variable linearly independent.
⇒
x1 – 2x2 – x3 = 0
x3 = 0
Let x2 = k then x1 – 2k – 0 = 0
⇒
x 1 = 2k
LM x OP LM2kOP LM2OP LM2OP
X = M x P = M k P = k M1P or M1P
MNx PQ MN 0 PQ MN0PQ MN0PQ
1
Hence,
2
2
3
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Thus only one eigen vector X2 corresponds to the repeated eigen value λ = 2.
Hence the eigen values are 1, 2, 2
LM2OP
LM 1 OP
X = – k 1 , X = X = k M1P
MM−1PP
MN0PQ
N Q
LM2OP
LM 1 OP
X = 1 , X = X = M1P
MM−1PP
MN0PQ
N Q
eigen vectors
1
or
2
1
2
3
3
Example 3. Find the eigen values and eigen vectors of the matrix
LM2 1 1OP
A = M1 2 1P
MN0 0 1PQ
Sol. The characteristic equation is |A – λI| = 0
⇒
2− λ
1
1
1 2− λ
1
0
0 1− λ
= (1 – λ) (λ – 1) (λ – 3) = 0
Thus λ = 1, 1, 3 are the eigen values of A.
Now, we consider the relation (A – λI)X = 0
For λ = 3,
LM−1 1 1 OP LM x OP
MM 10 −01 −12PP MMxx PP = 0
QN Q
N
1
2
3
LMx OP
let X = Mx P
MM PP
Nx Q
1
2
3
Applying R2 → R2 + R1, then R1 → – R1 on coefficient matrix, we get
LM1 −1 −1OP L x O
~ M0 0 2 P M x P = 0
M P
NM0 0 −2PQ MNx PQ
1
2
3
Again
R3 → R3 + R2, R2 →
1
R
2 2
LM1 −1 −1OP LM x OP
~ 0 0 1
MM0 0 0 PP MMxx PP = 0
Q N Q
N
1
2
3
⇒
x1 – x2 – x3 = 0
221
MATRICES
x3 = 0
Suppose x2 = k, then x1 = k
Here,
For
r = 2, n = 3
n− r = 1
LMk OP LM1OP LM1OP
X = Mk P = k M1P or M1P
MM PP MM PP MM PP
N0Q N0Q N0Q
1
λ=1
LM1
MM1
MN0
R2 → R2 – R1
LM1
~ M0
MM
N0
0
OP LMx OP
1P Mx P = 0
PM P
0PQ MNx PQ
1
1
1
1
0
0
1
1
2
3
OP LMx OP
0P Mx P = 0
PM P
0PQ MNx PQ
1
2
3
Here r = 1, n = 3 ⇒ n – r = 3 – 1 = 2.
∴ There is two variables linearly independent solution
Let
⇒
x2 = k1, x3 = k2 and x1 + x2 + x3 = 0
x1 + k1 + k2 = 0 ⇒ x1 = – (k1 + k2)
⇒
LM−bk + k gOP
X = M
MM k PPP .
N k Q
1
2
1
2
2
Since the vectors are linearly independent so, we choose k1 and k2 as follows:
(a) If we suppose x2 = k1 = 0, x3 = k2 (any arbitrary)
LM− k OP
LM 1 OP LM 1 OP
X = M 0 P = − k M 0 P or M 0 P
MM PP
MM PP MM PP
Nk Q
N−1Q N−1Q
2
Then
2
2
2
(b) If we suppose x2 = k1 (any arbitrary) and x3 = k2 = 0
LM− k OP
LM 1OP LM 1OP
X = M k P = − k M−1P or M−1P
MM PP
MM PP MM PP
0
N Q
N 0Q N 0Q
1
Then
3
1
1
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Hence the eigen values are λ = 1, 2, 2
Eigen vectors are
or
LM1OP
LM 1OP
LM 1OP
X = k M1 P; X = − k M 0 P; X = – k M−1P
MM PP
MM PP
MM PP
N0Q
N−1Q
N 0Q
LM1OP
LM 1 OP
LM 1 OP
X = M1P; X2 = M 0 P; X3 = M−1P .
MN0PQ
MN−1PQ
MN 0 PQ
2
1
2
3
1
1
Example 4. Find the eigen values and eigen-vectors of matrix
LM3 1 4OP
A = M0 2 6P .
MN0 0 5PQ
[U.P.T.U., 2004 (C.O.), 2002]
Sol. The characteristic equation is |A –λI| = 0
3− λ
1
4
0
2− λ
6
= ( 3 − λ ) (2 − λ )(5 − λ ) − 0 + 0 = 0
0
0
5− λ
l
⇒
q
∴
λ = 2, 3, 5
Now, we consider the relation (A – λI) X = 0
For λ = 2,
LM3 − 2 1 4 OP L x O
MM 00 2 −0 2 5 −6 2PP MMx PP = 0
Q MNx PQ
N
LM1 1 4OP LM x OP
MM00 00 63PP MMxx PP = 0
QN Q
N
1
2
3
...(i)
LM x OP
X= x
MMx PP
N Q
1
2
3
1
⇒
2
3
R3 → R3 –
1
R on coefficient matrix
2 2
LM1 1 4OP LM x OP
~ M0 0 6P M x P = 0
MN0 0 0PQ MNx PQ
1
2
3
⇒
Let
∴
x1 + x2 + 4x3 = 0
x2 = k
and x3 = 0
then x1 + k + 0 = 0 ⇒ x1 = –k
LM−kOP LM 1OP LM 1OP
X = M k P = − k M−1P or M−1P
MN 0 PQ MN 0PQ MN 0PQ
1
223
MATRICES
For λ = 3, from (i), we get
LM0
MM0
MN0
1
−1
0
OP LMx OP
6 P Mx P = 0
PM P
2PQ MNx PQ
4
1
2
3
R2 → R2 + R1 on coefficient matrix
LM0
~ M0
MM
N0
1
0
0
1
R3 → R3 − R2
5
LM0
~ M0
MM
N0
1
0
0
OP LMx OP
10 P Mx P = 0
PM P
2 PQ MNx PQ
4
1
2
3
OP LMx OP
10 P Mx P = 0
PM P
0 PQ MNx PQ
4
1
2
3
x2 + 4x3 = 0 and 10x3 = 0 ⇒ x3 = 0
x2 + 0 = 0 ⇒ x2 = 0, let x1 = K
⇒
⇒
LMk OP LM1OP LM1OP
X = M0P = k M0P or M0P
MM PP MM PP MM PP
N0Q N0Q N0Q
∴
2
Again, for λ = 5, we get
⇒
LM−2
MM 0
MN 0
1
−3
0
OP Lx O L0O
6 P Mx P = M0P
PP MMNx PPQ MMN0PPQ
0Q
4
1
2
3
2x1 – x2 – 4x3 = 0
3x2 – 6x3 = 0
Let
x 3 = k, then 3x2 – 6k = 0
⇒ x2 = 2k and 2x1 – 2k – 4k = 0 ⇒ x1 = 3k
∴
LM3kOP LM3OP LM3OP
X = M 2k P = k M 2P or M 2P
MM PP MM PP MM PP
N k Q N1Q N1Q
3
Hence the eigen values are λ = 2, 3, 5
And eigen vectors are
LM 1 OP
LM1OP
LM3OP
X = M−1P , X = M0 P , X = M 2P .
MM PP
MM PP
MM PP
N0Q
N0 Q
N1 Q
1
2
3
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 5. Determine the latent roots and the corresponding vector of the matrix,
LM 6 −2 2 OP
A = M−2 3 −1P .
NM 2 −1 3 PQ
Sol. The characteristic equation is |A – λI| = 0
6 − λ −2
2
−2 3 − λ −1 = 0
2
−1 3 − λ
⇒
R2 → R2 + R3, we get
6−λ
0
2
−2
2− λ
−1
2
2− λ
3− λ
6−λ
0
= (2 – λ)
2
−2
1
−1
0
2
2− λ
= ( 2 − λ ) ( 6 − λ ) ( 2 − λ + 2) − 8 = 0
2
1
3− λ
=0
Now, C3 → C3 + C2 gives
(2 – λ)
∴
6−λ
0
2
−2
1
−1
l
(2 – λ)(λ2 – 10λ + 16) = 0 ⇒ (2 – λ) (2 – λ) (λ – 8) = 0
⇒
λ = 2, 2, 8
Now, consider the relation
(A – λI)X = 0
For λ = 8, the equation (i), gives
LM6 − 8 −2 2 OP LMx OP LM0OP
MM −2 3 − 8 −1 PP MMx PP = MM0PP
N 2 −1 3 − 8Q Nx Q N0Q
LM−2 −2 2 OP LMx OP L0O
MM−2 −5 −1PP MMx PP = MM0PP
N 2 −1 −5Q Nx Q MN0PQ
1
2
3
1
⇒
2
3
R2 → R2 – R1, R3 → R3 + R1 on coefficient matrix
LM−2 −2 2 OP LMx OP LM0OP
MM 0 −3 −3PP MMx PP = MM0PP
N 0 −3 −3Q Nx Q N0Q
1
2
3
−1
−1
R1 , R2 →
R2
R3 → R3 – R2 and R1 →
2
3
1 1 −1
x1
0
0 1 1
x2 = 0
0 0 0
x3
0
⇒
q
LM
MM
N
OP L O
PP MM PP
Q MN PQ
LM OP
MM PP
NQ
x1 + x2 – x3 = 0
x2 + x3 = 0
...(i)
225
MATRICES
Let x3 = k, then x2 = – k and x1 = 2k
LMx OP LM 2k OP LM 2OP LM 2OP
X = M x P = M− k P = k M−1P or M−1P
MNx PQ MN k PQ MN 1PQ MN 1PQ
1
∴
2
1
For λ = 2, the equation (i) gives
3
LM6 − 2 −2 2 OP Lx O L0O
MM −2 3 − 2 −1 PP MMx PP = MM0PP
N 2 −1 3 − 2Q MNx PQ MN0PQ
LM 4 −2 2OP LM x OP LM0OP
MM−22 −11 −11PP MMxx PP = MM00PP
N
Q N Q NQ
1
2
3
1
⇒
2
3
R2 → R2 +
1
1
R1 , R3 → R3 – R1 , we get
2
2
LM4 −2 2OP Lx O L0O
MM00 00 00PP MMx PP = MM0PP
N
Q MNx PQ MN0PQ
1
2
3
Here r = 1, n = 3 ⇒ n – r = 3 – 1 = 2 (Two linearly independent)
Let
and we have
⇒
x2 = k1 and x3 = k2
4x1 – 2x2 + 2x3 = 0
4x1 – 2k1 + 2k2 = 0
1
( k1 − k 2 )
2
k1
2 k1
1
k2
=
2k2
∴
X2 =
2
1
( k1 − k2 )
( k1 − k 2 )
2
Since the vectors are linearly independent so, we choose, k1 and k2 as follows:
(i) Let x2 = k1 = 0 and x3 = k2 (arbitrary)
⇒
x1 =
LM
MM
MN
LM
MM
N
OP
PP
Q
LM OP
MM PP
N Q
LM OP
MM PP
N Q
OP
PP
PQ
OP
PP
Q
LM
MM
N
0
0
0
k2
1
2k 2 =
2 or 2
X2 =
2
2
−k 2
−1
−1
then
(ii) Let x3 = k2 = 0 and x2 = k1 (arbitrary)
1
X3 =
2
then
LM2k OP k LM2OP LM2OP
MM 0 PP = 2 MM0PP or MM0PP
N k Q N1Q N1Q
1
1
1
Hence λ = 8, 2, 2
LM 2OP LM 2OP
L 0O LM 0OP
L2O LM2OP
k M P
k M P
2 or M 2P, X =
0 or M0P .
X = k M−1P or M−1P, X =
2 M P
2 M P
MN 1PQ MN 1PQ
MN−1PQ MN−1PQ
MN1PQ MN1PQ
1
2
2
3
1
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 6. Find the eigen values of matrix A.
1
A =
3
LM1 2 2OP
MM2 1 −2PP ·
N2 −2 1Q
Sol. The characteristic equation of B is, where
LM1 2 2 OP
B = M2 1 −2P
MN2 −2 1 PQ
1− λ
2
2
2
1 − λ −2
= λ3 – 3λ2 – 9λ + 27 = 0
2
−2 1 − λ
(λ – 3)2 (λ + 3) = 0 ⇒ λ = 3, 3, – 3
or
1
B = 1, 1, – 1.
3
Example 7. Show that the matrix
So the eigen values of A =
LM 3 10 5 OP
A = M−2 −3 −4P
MN 3 5 7 PQ
has less than three independent eigen vectors. Is it possible to obtain a similarity transformation
that will diagonalise this matrix.
(U.P.T.U., 2001)
Sol. The characteristic equation is
|A – λI| = 0
or
3−λ
10
5
−2 −3 − λ −4
3
5
7−λ
or
λ3 – 7λ2 + 16λ – 12 = 0
⇒
= 0
(λ – 2)2 (λ – 3) = 0 ⇒ λ = 2, 2, 3
Now, consider the relation
(A – λI)X = 0
For λ = 3, we have
LM3− 3 10 5 OP LMx OP
MM −2 −3− 3 −4 PP MMx PP = 0
N 3 5 7 − 3Q Nx Q
LM 0 10 5OP LMx OP
MM−2 −6 − 4PP MMx PP = 0
N 3 5 4Q Nx Q
1
2
3
1
⇒
2
3
...(i)
227
MATRICES
R1 →
1
1
R1 , R2 → – R2 , we have
5
2
LM0 2 1OP LMx OP
MM1 3 2PP MMx PP = 0
N 3 5 4Q N x Q
LM1 3 2OP LM x OP
MM03 25 14PP MMxx PP = 0
N
QN Q
1
2
3
R1 ↔ R2
1
2
3
R3 → R3 – 3R1 on coefficient matrix
LM1 3 2 OP Lx O
MM0 2 1 PP MMx PP = 0
N0 −4 −2Q MNx PQ
1
2
3
R3 → R3 + 2R2
LM1 3 2OP LMx OP L0O
MM0 2 1PP MMx PP = MM0PP
N0 0 0Q Nx Q MN0PQ
1
2
3
⇒
x1 + 3x2 + 2x3 = 0
0x1 + 2x2 + x3 = 0
Solving these equations by cross-multiplication
x1`
x3
x2
= −
=
1− 0
3−4
2−0
x1
x2
x3
⇒
=
=
−1
−1
2
x1
x2
x3
⇒
=
=
−2
1
1
∴ The proportional values are
x1 = 1, x2 = 1, x3 = – 2
LM 1OP
X = M 1P
MN−2PQ
1
For λ = 2, we have, from equation (i)
LM3 − 2 10 5 OP LMx OP
MM −2 −3 − 2 −4 PP MMx PP = 0
N 3 5 7 − 2Q N x Q
LM 1 10 5OP LMx OP
MM−2 −5 −4PP MMx PP = 0
N 3 5 5Q N x Q
1
2
3
1
2
3
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Applying R2 → R2 + 2R1, R3 → R3 – 3R1 on coefficient matrix
LM1 10 5 OP Lx O
MM00 −1525 −610PP MMx PP = 0
N
Q MNx PQ
1
2
3
R2
1
R2 →
, R3 → – R 3
5
3
LM1 10 5OP LMx OP
MM0 5 2PP MMxx PP = 0
N0 5 2 Q N Q
1
2
3
R3 → R3 – R2
LM1 10 5OP LMx OP
MM0 5 2PP MMx PP = 0
N0 0 0 Q N x Q
1
2
3
This shows that the solution is one linearly independent
⇒
x1 + 10x2 + 5x3 = 0
5x2 + 2x3 = 0
x
x
x1
= − 2 = 3
−5
2
5
or
x1
x
x
= 2 = − 3
5
2
5
Taking proportional values, x1 = 5, x2 = 2, x3 = – 5
LM 5 OP
X = M2P
MN−5PQ
∴
2
And the third eigen vector corresponding to repeated root λ3 = 2 = λ2 will be same
because the solution is not two linearly independent. Hence the vectors X2 and X3 are not linearly
independent. Similarity transformation is not possible.
LM 1 OP
LM 5 OP
Therefore λ = 2, 2, 3 and X = M 1 P . X = X = M 2 P .
MN−2PQ
MN−5PQ
1
2
3
Example 8. Verify the statement that the sum of the elements in the diagonal of a matrix
is the sum of the eigen values of the matrix.
LM−2 2 −3OP
1 −6P ·
A = M2
NM−1 −2 0 PQ
Sol. The characteristic equation is |A – λI| = 0
⇒
−2 − λ
2
−3
2
1 − λ −6
−1
−2 − λ
= 0
229
MATRICES
⇒
⇒
⇒
(– 2 –λ) [(1 – λ)(– λ) – 12] – 2(– 2λ – 6] – 3[– 4 + (1 – λ)] = 0
(– 2 – λ) (λ2 – λ – 12) + 4 (λ + 3) + 3 (λ + 3) = 0
– 2λ2 + 2λ + 24 – λ3 + λ2 + 12λ + 4λ + 12 + 3λ + 9 = 0
⇒
– λ3 – λ2 + 21λ + 45 = 0
⇒
λ3 + λ2 – 21λ – 45 = 0.
This is cubic equation in λ and hence it has 3 roots i.e., there are three eigen values.
FG coefficient of λ IJ = – 1.
H coefficient of λ K
2
The sum of the eigen values = –
3
The sum of the elements on the diagonal of the matrix A
= – 2 + 1 + 0 = – 1. Hence the result.
Example 9. Find the eigen values and corresponding eigen vectors of the matrix,
A =
LM−5 2 OP
N 2 −2Q
(U.P.T.U., 2008)
Sol. The characteristic equation is |A – λI| = 0
−5 − λ
2
= (5 + λ) (2 + λ) – 4 = 0
2
−2 − λ
⇒
⇒
λ2 + 7λ + 6 = 0 ⇒
∴
(λ + 6) (λ + 1) = 0
λ = – 1, – 6
Now, we consider the relation (A – λI) X = 0
For
...(i)
λ = –1
LM−5 + 1 2 OP LMx OP = 0
N 2 −2 + 1Q Nx Q
LM−4 2 OP LMx OP = 0
N 2 −1Q Nx Q
1
2
⇒
1
2
R2 → R2 + 2R1, on coefficient matrix
LM−4 2OP LMx OP = 0
N 0 0Q Nx Q
1
2
⇒
– 2x1 + x2 = 0
Let
x2 = k
∴ From (i)
x1 =
X1 =
For λ = – 6, from (i), we get
k
2
LMk 2OP = k LM1OP or LM1OP
N k Q 2 N2Q N2Q
LM1 2OP LMx OP = 0
N2 4Q Nx Q
1
2
...(ii)
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
R2 → R2 – 2R1, on coefficient matrix
LM1 2OP LMx OP = 0
N0 0Q Nx Q
1
2
⇒
Let
x1 + 2x2 = 0
...(iii)
x2 = k ∴ x1 = – 2k
X2 =
LM−2kOP = k LM−2OP or LM−2OP
N k Q N1Q N1Q
Hence, the eigen values are λ = –1, – 6
LM1OP
N2Q
L−2O
X = M P.
N1Q
And eigen vectors are X1 =
2
EXERCISE 3.6
Find the eigen values and eigen vectors of:
LM1 2OP.
N1 0Q
L6 8 OP.
2. M
N8 −6Q
LM1 1 3OP
1 5 1 .
3. M
MN3 1 1PPQ
LM 8 −6 2 OP
−6 7 −4P.
4. M
MN 2 −4 3 PQ
LM −2 1 1OP
−11 4 5P.
5. M
MN −1 1 0PQ
LM2 1 0OP
0 2 1.
6. M
MN0 0 2PPQ
LM1 −1 −1OP
1 −1 0 P.
7. M
MN1 0 −1PQ
1.
LMAns. 2, − 1, L2O , L 1 OOP
MN1PQ MN−1PQQ
N
LMAns. 10, - 10, L2O , L 1OOP
MN1PQ MN−2PQQ
N
LM
LM−2OP LM−1OP LM1OPOP
MMAns. − 2, 3, 6, M 0 P , M 1 P, M2PPP
MN 2 PQ MN−1PQ MN1PQQ
N
LM
LM1OP LM 2 OP LM 2 OPOP
MMAns. 0, 3, 15, M2P, M 1 P, M−2PPP
MN2PQ MN−2PQ MN 1 PQQ
N
LM
LM 0 OP LM1OP LM2OPOP
Ans.
−
1
,
1
,
2
,
MM
MM−11PP, MM12PP, MM13PPPP
N Q N Q N QQ
N
LM
LM1OP LM1OP LM1OPOP
MMAns. 2, 2, 2, M0P, M0P, M0PPP
MN0PQ MN0PQ MN0PQQ
N
LM
LM 0 OP LM1 + iOP LM1 − iOPOP
MMAns. − 1, ± i, M 1 P, M 1 P, M 1 PPP
MN−1PQ MN 1 PQ MN 1 PQQ
N
231
MATRICES
LM3 1 4OP
0 2 0P.
8. M
MN0 0 5PQ
LM−2 2 −3OP
2 1 −6 .
9. M
MN−1 −2 0 PPQ
LM 3 −2 −5OP
10. M 4 −1 −5P.
MN−2 −1 −3PQ
LM1 1 3OP
11. M1 5 1P.
MN3 1 1PQ
LM 4 −20 −10OP
4 P.
12. M−2 10
MN 6 −30 −13PQ
LM−1 0 2OP
13. M 0 1 2P.
MN 2 2 0PQ
LM11 −4 −7OP
14. M 7 −2 −5P.
MN10 −4 −6PQ
LM −9 4 4OP
15. M −8 3 4P.
MN−16 8 7 PQ
LM 2 4 6 OP
16. M 4 2 −6 P.
NM−6 −6 −15QP
LM1 2 3OP
17. M2 4 6P.
MN3 6 9PQ
LM 1 −4 −1 −4OP
2 0 5 −4P
18. M
MM−1 1 −2 3 PP.
N−1 4 −1 6 Q
LM
LM 0 OP LM1OP LM3OPOP
Ans.
2
,
3
,
5
,
MM
MM−01PP, MM00PP, MM21 PPPP
N Q N Q N QQ
N
LM
LM−1OP LM−2OP LM3OPOP
MMAns. 5, − 3, − 3, M−2P, M 1 P, M0 PPP
MN 1 PQ MN 0 PQ MN1 PQQ
N
LM
LM3OP LM 1 OP LM 1 OPOP
MMAns. 5, 2, 2, M2P, M 3 P, M 3 PPP
MN4PQ MN−1PQ MN−1PQQ
N
LM
LM−1OP LM 1 OP LM1OPOP
MMAns. − 2, 3, 6, M 0 P, M−1P, M2PPP
MN 1 PQ MN 1 PQ MN1PQQ
N
LM
LM5OP LM2OP LM 0 OPOP
MMAns. 0, − 1, 2, M1P, M0P, M 1 PPP
NM0QP NM1QP NM−2QPQ
N
LM
LM 2 OP LM1OP LM 2 OPOP
MMAns. 0, 3, − 3, M−2P, M2P, M 1 PPP
MN 1 PQ MN2PQ MN−2PQQ
N
LM
LM1OP LM 1 OP LM2OPOP
Ans.
0
,
1
,
2
,
MM
MM11PP, MM−21PP, MM12PPPP
N Q N Q N QQ
N
LM
LM 0 OP LM1OP LM1OPOP
MMAns. − 1, − 1, 3, M 1 P, M1P, M1PPP
MN−1PQ MN1PQ MN2PQQ
N
LM
LM 1 OP LM 2 OP LM1 OPOP
MMAns. − 2, 9, − 18, M−1P, M 2 P, M1 PPP
MN 0 PQ MN−1PQ MN4PQQ
N
LM
LM−2OP LM 3 OP LM1 OPOP
MMAns. 0, 0, 14, M 1 P, M 6 P, M2 PPP
NM 0 QP NM−5QP NM3QPQ
N
LM 2 OP LM 3 OP
6
3
[Ans. λ – 5λ + 9λ – 7λ + 2 = 0, λ = 2, 1, 1, 1 M P, M P
MM−2PP MM−4PP
N−3Q N−5Q
4
3
2
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
LM
LM0OP LM−4OP LM−4OPOP
Ans.
2
,
2
,
−
2
,
MM
MM11PP, MM 71 PP, MM 71 PPPP
N Q N Q N QQ
N
LM2 −2 2 OP
19. M1 1 1 P.
MN1 3 −1PQ
20. Find the eigen values of A5 when
LM3 0 0OP
5 4 0P
A= M
MN3 6 1PQ .
Ans. 3 5 , 4 5 , 1 5
21. If A and B are n × n matrices and B is a non-singular matrix prove that A and B–1 AB
have same eigen values.
[Hint : Characteristic polynomial of B–1 AB = |B–1AB – λI| = |B–1 AB – B–1 (λI)B|
= |B–1 (A – λI)B| = |B–1 |A – λI| |B| = |B–1||B||A – λI| = |B–1B||A – λI|
= |I| |A – λI| = |A – λI| = characteristic polynomial of A].
22. Find the sum and product of the eigen values.
LM 2 3 −2OP
A = M−2 1 1 P .
MN 1 0 2 PQ
3.18
Ans. Sum = 2 + 1 + 2 = 5, product = |A| = 21
CAYLEY-HAMILTON THEOREM
STATEMENT
Any square matrix A satisfies its own characteristic equation.
(U.P.T.U., 2005)
i.e., if C0 + C1λ + C2λ + .... + Cnλ is the characteristic polynomial of degree n in λ then
C0I + C1A + C2A2 + .... + CnAn = 0.
2
n
Proof: Let A be a square matrix of order n.
Let |A – λI| = C0 + C1λ + C2λ2 + ... + Cnλn
...(i)
be the characteristic polynomial of A.
Now, the elements of adj (A – λI) are cofactors of the matrix (A – λI) and are polynomials
in λ of degree not exceeding n – 1.
Thus adj (A – λI) = B0 + B1λ + B2λ2 + ... + Bn – 1 λn – 1
...(ii)
where Bi are n-square matrices whose elements are the functions of the elements of A and
independent of λ.
We know that (A – λI). adj (A – λI) = |A – λI) I
From (i) and (ii), we have
(A – λI) (B0 + B1λ + B2λ2 + ... + Bn – 1λn–1) = (C0 + C1λ + C2λ2 + ... + Cnλn)I
Equating the like powers of λ on both sides of (iii), we get
AB0 = C0I
AB1 – B0 = C1I
...(iii)
233
MATRICES
®
®
®
ABn – 1 – Bn – 2 = Cn – iI
–Bn – 1 = CnI
Pre-multiplying the above equations by I, A, A2, ... An respectively and adding; we get
C0I + C1A + C2A2 + ... + CnAn = 0
...(iv)
Since all the terms on the L.H.S. cancel each other. Thus A satisfies its own characteristic
equation. Proved.
3.18.1 Inverse by Cayley-Hamilton Theorem
We have C0I + C1A + C2A2 + ... + CnAn = 0
...(i)
–1
Multiplying equation (i) by A , we get
C0A–1 + C1I + C2A + ... + CnAn – 1 = 0
⇒
C0A–1 = –[C1I + C2A + ... + CnAn – 1]
⇒
A–1 = –
1
[C I + C2A + ... + CnAn – 1]
C0 1
Note: A–1 exists only if C0 = |A|.
Definition. An expression of the form C0 + C1λ + C2λ2 +, + Cnλn where C0, C1, ..., Cn are
square matrices of the same order and Cn C 0 is called a ‘‘matrix polynomial of degrees n’’.
LM1 2 3OP
Example 1. Verify the Cayley-Hamilton theorem for the matrix M2 4 5P . Also find its
MN3 5 6PQ
inverse using this theorem.
(U.P.T.U., 2006)
Sol. The characteristic equation of A is
|A – λI| =
1− λ
2
2
4−λ
3
5
3
5
=0
6−λ
⇒ (1 – λ) [(4 – λ) (6 – λ) – 25] – 2[2(6 – λ) – 15] + 3 [10 – 3 (4 – λ)] = 0
⇒
λ3 – 11λ2 – 4λ + 1 = 0
Cayley-Hamilton theorem is verified if A satisfies the above characteristic equation i.e.
A3 – 11A2 – 4A + 1 = 0
Now,
LM1 2 3OP LM1 2 3OP LM14 25 31OP
A = A·A = M2 4 5P M2 4 5P = M25 45 56P
NM3 5 6PQ MN3 5 6PQ MN31 56 70PQ
LM1 2 3OP LM14 25 31OP LM157 283 353OP
A = A·A = M 2 4 5P M25 45 56P = M283 510 636P
NM3 5 6QP NM31 56 70QP MN353 636 793PQ
2
3
2
...(i)
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
LM157 283 353OP LM14 25 31OP LM1 2 3OP
Verification, A – 11A – 4A + I = M283 510 636P − 11 M25 45 56P – M2 4 5P
MN353 636 793PQ MN31 56 70PQ MN3 5 6PQ
LM1 0 0OP L0 0 0O
+ M0 1 0P = M0 0 0P
MN0 0 1PQ MMN0 0 0PPQ
3
2
⇒
A3 – 11A2 – 4A + 1 = 0
Hence theorem verified.
To find )–1 : Multiplying equation by A–1, we get
A2 – 11A – 4I + A–1 = 0
⇒
A–1 = – A2 + 11A + 4I
⇒
LM14 25 31OP LM1 2 3OP LM1 0 0OP
= – M25 45 56P + 11 M2 4 5P + 4 M0 1 0P
NM31 56 70PQ MN3 5 6PQ MN0 0 1PQ
LM 1 −3 2 OP
= M−3 3 −1P .
MN 2 −1 0 PQ
–1
A
Example 2. Express B = A8 – 11A7 – 4A6 + A5 + A4 – 11A3 – 3A2 + 2A + I as a quadratic
polynomial in A, where A is square matrix given in Example 1. Find B as well as A4.
Sol.
Thus,
B = A8 – 11A7 – 4A6 + A5 + A4 – 11A3 – 3A2 + 2A + I
= A5 (A3 – 11A2 – 4A + I) + A (A3 – 11A2 – 4A + I) + A2 + A + I
= A5 (0) + A(0) + A2 + A + I |As A3 – 11A2 – 4A + I = 0 in Example 1
= A2 + A + I
LM14 25 31OP LM1 2 3OP LM1 0 0OP
B = M25 45 56P + M2 4 5P + M0 1 0P
NM31 56 70PQ MN3 5 6PQ MN0 0 1PQ
L16 27 34O
B = MM27 50 61PP .
MN34 61 77 PQ
To find )4 : We have
A3 – 11A2 – 4A + I = 0
⇒
A3 = 11A2 + 4A – I
Multiplying both sides by A
A4 = 11A3 + 4A2 – IA
LM157 283 353OP LM14 25 31OP LM1 2 3OP
A = 11 M 283 510 636P + 4 M25 45 56P – M2 4 5P
MN353 636 793PQ MN31 56 70PQ MN3 5 6PQ
LM1782 3211 4004OP
= M 3211 5786 7215P .
MN4004 7215 8997PQ
4
235
MATRICES
LM 2 –1 1 OP
PP
MM
Q
N
Example 3. Find the characteristic equation of the matrix –1 2 –1 and hence also find
1 –2 2
–1
A by Cayley-Hamilton theorem.
(U.P.T.U., 2003, 2004, 2005)
Sol. The characteristic equation of A is
2– λ
|A – λ I| = –1
1
–1
1
2 – λ –1 = 0
–2 2 – λ
⇒ (2 – λ) [λ2 – 4λ + 4 – 2] + [– 2 + λ + 1] + [2 – 2 + λ] = 0
⇒
λ3 – 6λ2 + 8λ – 3 = 0
By Cayley-Hamilton theorem A3 – 6A2 + 8A – 3I = 0.
Multiplying by A–1 on both sides, we get
A2 – 6A + 8I – 3A–1 = 0
1
(A2 – 6A + 8I)
3
2 –1
1
2
2 −1
A = A.A = −1
1 −2
2
⇒
Now,
∴
Thus,
A–1 =
OP LM 2 −1 1OP LM 6 −6 6OP
PP MM−1 2 −1PP = MM−5 7 −5PP
Q N 1 −2 2Q N 6 −9 7Q
LM 6 −6 6 OP LM 2 −1 1 OP LM8 0 0OP LM2 0 −1OP
A – 6A + 8I = M−5 7 −5P − 6 M−1 2 −1P + M0 8 0P = M1 3 1 P
MN 6 −9 7 PQ MN 1 −2 2 PQ MN0 0 8PQ MN0 3 3 PQ
L2 0 −1OP
1M
1 3 1P.
A =
3M
MN0 3 3 PQ
LM
MM
N
2
–1
Example 4. Evaluate A–1, A–2, A–3 if
LM 4 6 6 OP
3 2
A = M1
MN−1 −4 −3PPQ
Sol. Characteristic equation is
4− λ
6
6
1
3− λ
2
|A – λI| =
=0
−1
−4 −3 − λ
⇒
λ3 – 4λ2 – λ + 4 = 0
By Cayley-Hamilton theorem, we have
⇒
A3 – 4A2 – A + 4I = 0
A2 – 4A – I + 4A–1 = 0 (multiplying byA–1)
A–1 =
1
[–A2 + 4A + I)
4
...(i)
236
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Now,
∴
Thus,
LM16 18 18OP
MM 5 7 6 PP
N−5 −6 −5Q
LM−16 −18 −18OP LM16 24 24 OP LM1 0 0OP
8 + 0 1 0
– A + 4A + I = M −5 −7 −6 P + M 4 12
MN 5 6 5 PQ MN−4 −16 −12PPQ MMN0 0 1PPQ
LM 1 6 6 OP
2
= M−1 6
MN 1 −10 −6PPQ
L 1 6 6 OP
1M
−1 6
2
A =
4M
MN 1 −10 −6PPQ
A2 =
2
–1
Multiplying equation (1) by A–1, we get
–2
A
Similarly,
LM
MM
N
14
1
1
−1
4
−
+
+
A
I
IA
−
54
=
=
4
4
54
A–3 =
−1 2
52
32
OP
PP
Q
−9 2
−3 2
11 2
1
[– I + 4A–1 + A–2]
4
LM
MM
N
OP
PP
Q
1
78
78
1
−21 90
26 .
=
64
21 −154 −90
Example 5. Verify Cayley-Hamilton theorem for the matrix A =
A–1.
LM1 2 OP and hence find
N2 −1Q
(U.P.T.U., 2008)
Sol. The characteristic equation of A is
|A – λI| =
or
1− λ
2
= 0 ⇒ – (1 – λ2) – 4 = 0
2
−1 − λ
λ2 – 5 = 0
By Cayley-Hamilton theorem A2 – 5 = 0
Now
A2 =
LM1 2 OP LM1 2 OP = LM5 0OP = 5 LM1 0OP = 5I
N2 −1Q N2 −1Q N0 5Q N0 1Q
Putting value of A2 in equation (i), we get
A2 – 5 = 5I – 5I = 0
Hence, the Cayley-Hamilton theorem is verified.
...(i)
237
MATRICES
To find )–1 : Multiplying equation (i) by A–1, we get
A – 5A–1 = 0 ⇒
A–1 =
LM
N
OP
Q
1
1 1 2
.
A =
5
5 2 −1
LM 0 c −bOP
a satisfies its characteristic equation. Also
Example 6. Show that the matrix A = M − c 0
MN b − a 0 PPQ
find A–1.
−λ c −b
|A – λI| = − c − λ a = 0
b −a −λ
Sol.
i.e.,
– λ[λ2 + a2] – c[λc – ab] – b[ac + bλ] = 0
i.e.,
λ3 + λ(a2 + b2 + c2) = 0
By Cayley-Hamilton theorem, A3 + A (a2 + b2 + c2) = 0
...(i)
OP
ac
LM 0 c −bOP LM 0 c −bOP LM−ec + b j ab
PP
A = M− c 0
− ec + a j
a −c 0
a = M
ab
bc
MN b −a 0 PPQ MMN b −a 0 PPQ MMN ac
−eb + a jP
bc
Q
2
2
2
2
Now,
2
2
2
OP
ac
LM 0 c −bOP LM−eb + c j ab
PP
A = A·A = M− c 0
−e c + a j
a M
ab
bc
MN b −a 0 PPQ MMN ac
−eb + a jP
bc
Q
2
3
2
2
2
2
2
2
LM
− c e a + b + c j b ea + b + c j O
0
LM 0 c −bOP
PP
M
+
+
−
+
+
=
−
+
+
c
a
b
c
0
a
a
b
c
a
b
c
or A =
e
jP e
j M− c 0 a P
MM−be a + b + c j a a + b + c
MN b −a 0 PQ
0
PQ
j e
j
N e
2
3
or
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
Using (ii) in equation (i), we get
A3 + A (a2 + b2 + c2) = – (a2 + b2 + c2) A + A (a2 + b2 + c2) = 0
To find )–1: Multiplying equation (i) A–2 on both sides, we get
A + A–1 (a2 + b2 + c2) = 0
1
⇒
A–1 = −
Hence,
LM 0 c −bOP
A = −
−c 0 a ·
ea + b + c j MMN b −a 0 PPQ
ea + b + c j
2
2
1
–1
2
2
2
2
2
A3 = – (a2 + b2 + c2) A
2
2
2
⋅A
...(ii)
238
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
EXERCISE 3.7
1. Find the characteristic equation of the matrix A
LM4 3 1 OP
A = M 2 1 −2P . Hence find A .
(U.P.T.U., 2001)
MN1 2 1 PQ
LM
L 5 −1 −7OPOP
1 M
MMAns. Characteristic equation : λ − 6λ + 6λ − 11 = 0, A = 11 M−4 3 10 PPP
MN 3 −5 −2PQQ
N
–1
3
−1
2
2. Find the characteristic equation of the matrix
LM2 1 1OP
A = M0 1 0P . Verify Cayley-Hamilton theorem and hence
MN1 1 2PQ
(U.P.T.U., 2002)
evaluate the matrix equation A8 – 5A7 + 7A6 – 3A5 + A4 – 5A3 + 8A2 – 2A + I.
LM
LM8 5 5OPOP
λ
−
λ
+
λ
−
=
+
+
Ans.
5
7
3
0
,
A
A
I
,
MM
MM0 3 0PPPP
N5 5 8 Q Q
N
3
2
2
3. Verify Cayley-Hamilton theorem for A and hence find A–1
LM
LM−2 0 1OPOP
LM 2 −1 1 OP
MMAns. A = M−5 1 5PPP
(i) M−15 6 −5P
MN 0 1 3PQQ
MN 5 −2 2 PQ
N
LM
LM 3 −1 −1OPOP
LM1 1 1OP
MMAns. A = M−1 1 0 PPP
(ii) M1 2 1P
MN−1 0 1 PQQ
MN1 1 2PQ
N
LM
L 0 −3 −3OPOP
LM1 0 3 OP
1M
MMAns. A = − 9 M−3 −2 7 PPP
(iii) M 2 1 −1P
MN−3 1 1 PQQ
MN1 −1 1 PQ
N
LM
L−3 −2 2 OPOP
LM 7 2 −2OP
1M
MMAns. A = 3 M 6 5 −2PPP
4. Find A if A = M−6 −1 2 P .
MN−6 −2 5 PQQ
N
MN 6 2 −1PQ
LM1 2 0 OP
5. Show that the matrix A = M 2 −1 0 P satisfies its own characteristic equation and hence
MN0 0 −1PQ
LM
LM1/5 0 0OPOP
MMAns. A = M 0 1/5 0PPP
or otherwise obtain the value of A .
MN 0 0 1PQQ
N
−1
−1
−1
−1
–1
−1
–1
239
MATRICES
LM1 2 3 OP
6. Find A if A = M1 3 5 P .
MN1 5 12PQ
LM0 1 0 0OP
0 0 0 1
7. If A = M
MM0 0 1 0PPP find A .
N1 0 0 0Q
LM
L11 −9 1 OPOP
1M
MMAns. A = 3 M−7 9 −2PPP
MN 2 −3 1 PQQ
N
LM
LM0 0 0 1OPOP
MMAns. A = M1 0 0 0PPP
MM0 0 1 0PPP
MM
N0 1 0 0QPQ
N
−1
–1
−1
–1
8. Find B = A6 – 4A5 + 8A4 – 12A3 + 14A2 if A =
LM 1 2OP .
N−1 3Q
LMAns. λ − 4λ + 5 = 0, B = 5I − 4 A = L1 −8OOP
MN 4 −7PQQ
N
LM
L−1 7 −24OPOP
1 M
MMAns. A = 11 M 2 −3 4 PPP
MN 2 −3 15 PQQ
N
2
LM3 3 4OP
9. Find A if A = M 2 −3 4P ·
MN0 −1 1PQ
LM 2 −3 1 OP
3 P satisfies the equation A (A – I) (A + 2I) = 0.
10. Show that the matrix A = M 3 1
MN−5 2 −4PQ
−1
–1
3.19
DIAGONALIZATION OF A MATRIX
A square matrix ‘A’ is said to be diagonalisable if there exists another non-singular square matrix
P such that P–1AP is a diagonal matrix. In other words, A is diagonalisable if A is similar to a
diagonal matrix.
Similar matrix: Two matrices A and B are said to be similar if there exists a non-singular
matrix P such that B = P–1AP.
This transformation of A to B is known as similarity transformation.
Note: Similar matrices A and B have same eigen values. Further, if X is an eigen vector of A then
Y = P–1X is an eigen vector of the matrix B.
3.19.1 Working Rule
The matrix P is called a model matrix for A. There are following steps to obtain diagonal form
of matrix A.
Step 1: First of all obtain eigen values λ1, λ2, ..., λn of the matrix A.
Step 2: If there exists a pair of distinct complex eigen values then A is not diagonalisable
over the field R of real numbers.
Step 3: If there does not exist a set of n linearly independent eigen vectors, then A is not
diagonalisable.
Step 4: If there exist n linearly independent eigen vectors.
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
LMx OP LMx OP
LMx OP
x
x
x
X = M P, X = M P ...., X = M P
MM M PP MM M PP
MM M PP
Nx Q Nx Q
Nx Q
LMx x L x OP
L x
x
x
P = M
MM M M M PPP
Nx x L x Q
11
12
1n
and diagonal matrix D = P–1 AP.
n2
22
n
2
1
then let,
n1
21
2n
nn
11
21
n1
12
22
n2
1n
2n
nn
3.19.2 Powers of Matrix A
Consider
⇒
D = P−1 AP
D2 = (P−1 AP) (P−1 AP) = P−1 A (PP−1) AP
= P−1 A. IAP = P−1 AAP = P−1 A2 P
Similarly,
D3 = P−1 A3 P, ...., Dn = P−1 An P
We can obtain An pre-multiplying by P and post-multiplying by P–1
⇒
P Dn P−1 = P (P−1 An P) P−1 = (PP−1) An (PP−1)
= IAnI = An
∴
An = PDn P−1
LM1 6 1OP
Example 1. Diagonalize the matrix M1 2 0P .
MN0 0 3PQ
Sol. The characteristic equation of A is
1− λ
6
1
1
2− λ
0 =0
|A – λ I| =
0
0
3− λ
⇒
(λ + 1) (λ – 3) (λ – 4) = 0 ⇒ λ = – 1, 3, 4
Consider the equation
(A – λ I) X = 0
For, λ = – 1,
LM2 6 1OP LMx OP
MM1 3 0PP MMx PP = 0
N0 0 4Q Nx Q
1
2
3
R1 ↔ R2 on coefficient matrix, we get
LM1 3 0OP LMx OP
MM2 6 1PP MMx PP = 0
N0 0 4Q Nx Q
1
2
3
(U.P.T.U., 2006)
241
MATRICES
R2 → R2 – 2R1
LM1 3 0OP LMx OP
MM0 0 1PP MMx PP = 0
N0 0 4Q Nx Q
1
2
3
R1 → R1 – R2, R3 → R3 – 4R2
LM1 3 −1OP LMx OP
MM0 0 1 PP MMx PP = 0
N0 0 0 Q Nx Q
1
2
3
⇒
Let,
x1 + 3x2 – x3 = 0
x3 = 0
x2 = k, then x1 = – 3k
∴
X1 =
For, λ = 3
LM−3kOP LM−3OP LM−3OP
MM k PP = k MM 1 PP = MM 1 PP
N 0 Q N0Q N0Q
LM−2 6 1OP LMx OP
MM 1 −1 0PP MMx PP = 0
N 0 0 0 Q Nx Q
1
2
3
⇒
– 2x1 + 6x2 + x3 = 0
x1 – x2 + 0·x3 = 0
⇒
∴
For, λ = 4
x1
–x2 x3
x
x
x
=
or 1 = 2 = 3
=
−1
−4
1
1
−4
1
x1 = 1, x2 = 1, x3 = – 4
LM 1 OP
X = M1P
MN−4PQ
2
LM−3 6 1 OP LMx OP
MM 1 −2 0 PP MMx PP = 0
N 0 0 −1Q Nx Q
1
2
3
R1 ↔ R2
LM 1 −2 0 OP LMx OP
MM−3 6 1 PP MMx PP = 0
N 0 0 −1Q Nx Q
1
2
3
R2 → R2 + 3R1
LM1 −2 0 OP LMx OP
MM0 0 1 PP MMx PP = 0
N0 0 −1Q Nx Q
1
2
3
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
R1 → R1 + R2, R3 → R3 + R2
LM1 −2 1OP LMx OP
MM0 0 1PP MMx PP = 0
N0 0 0Q Nx Q
1
2
3
⇒
x1 – 2x2 + x3 = 0
0x1 + 0x2 + x3 = 0
x
x
x1
= – 2 = 3
−2
1
0
⇒
x1
x
x
= 2 = 3 ⇒ x1 = 2, x2 = 1, x3 = 0
2
1
0
or
∴
Thus,
To obtain 2–1:
matrix of cofactors
∴
LM2OP
X = M1 P
MN0PQ
LM−3 1 2OP
P = M 1 1 1P
MN 0 −4 0PQ
3
|P| = – 3 (4) – 1 (0) + 2 (– 4) = – 20
LM 4 0 −4 OP
B = M−8 0 −12P
MN−1 5 −4 PQ
LM 4 −8 −1OP
0
5P
adj P = B′ = M 0
MN−4 −12 −4PQ
L 4 −8 −1OP 1 LM−4 8 1 OP
adj P
1 M
=–
0
0
5 =
0
0 −5
P =
|P|
20 M
MN−4 −12 −4PPQ 20 MMN 4 12 4 PPQ
–1
and
Diagonalisation:
D = P–1 AP
LM−4 8 1 OP LM1 6 1OP LM−3 1 2OP
MM 0 0 −5PP MM1 2 0PP MM 1 1 1PP
N 4 12 4 Q N0 0 3Q N 0 −4 0Q
L 4 −8 −1 OP LM−3 1 2OP
1 M
0 0 −15
1 1 1
=
20 M
MN16 48 16 PPQ MMN 0 −4 0PPQ
L−20 0 0 OP LM−1 0 0OP
1 M
0 60 0 = 0 3 0 .
=
20 M
MN 0 0 80PPQ MMN 0 0 4PPQ
D =
1
20
243
MATRICES
LM−1 2 −2OP
MM 1 2 1 PP to the diagonal form.
N−1 −1 0 Q
Example 2. Reduce the matrix A =
(U.P.T.U., 2001, 2004)
|A – λI| =
Sol.
⇒
−1 − λ
2
−2
1
2− λ 1 = 0
−1
−1 −λ
λ3 – λ2 – 5λ + 5 = 0
λ = 1, + 5 , – 5
On solving, we get
For, λ = 1
(A – λI) X = 0
LM−2 2 −2OP LMx OP
MM 1 1 1 PP MMx PP = 0
N−1 −1 −1Q Nx Q
1
⇒
2
3
R1 ↔ R2
LM 1 1 1 OP LMx OP
MM−2 2 −2PP MMx PP = 0
N−1 −1 −1Q Nx Q
1
2
3
R2 → R2 + 2R1, R3 → R3 + R1
LM1 1 1OP LMx OP
MM0 4 0PP MMx PP = 0
N0 0 0Q Nx Q
1
2
3
⇒
x1 + x2 + x3 = 0
0x1 + 4x2 + 0x3 = 0
x1
x
x
x
x
x
= − 2 = 3 or 1 = 2 = 3
−1
−4
0
4
1
0
x1 = 1, x2 = 0, x3 = – 1
⇒
LM 1 OP
X = M0P
MN−1PQ
∴
For
1
λ =
LM−1 − 5 2 −2 OP Lx O
MM 1 2 − 5 1 PP MMx PP = 0
N −1 −1 − 5 Q MNx PQ
1
2
3
R1 ↔ R2
LM 1 2 − 5 1 OP Lx O
MM−1 − 5 2 −2 PP MMx PP = 0
N −1 −1 − 5 Q MNx PQ
1
2
3
5
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
R2 → R2 + (1 +
5 ) R1, R3 → R3 + R1
LM1 2 − 5 1 OP Lx O
MM0 5 − 1 5 − 1PP MMx PP = 0
N0 1 − 5 1 − 5Q MNx PQ
1
2
3
R3 → R3 + R2, and R2 →
1
5 −1
R2
LM1 2 − 5 1OP Lx O
MM0 1 1PP MMx PP = 0
N0 0 0Q MNx PQ
1
2
3
⇒
x1 + (2 –
5 ) x2 + x3 = 0
0x1 + x2 + x3 = 0
Solving by cross-multiplication
x1
= –
1− 5
x1
or
=
5 −1
x2 x3
=
1
1
x2 x3
=
⇒ x1 =
−1
1
5 − 1 , x2 = 1, x3 = – 1
LM 5 − 1OP
X = M 1 P
MN −1 PQ
∴
2
Similarly, the eigen vector corresponding λ = –
5
LM 5 + 1OP
X = M −1 P
MN 1 PQ
LM 1 5 − 1
1
P = M0
MN−1 −1
3
∴ The modal matrix
To obtain 2–1:
R3 → R3 + R1
LM 1
MM 0
N−1
LM1
MM0
N0
5 −1
1
−1
5 −1
1
5−2
5 +1
−1
1
OP L1 0 0O
PP ~ MM0 1 0PP P
Q MN0 0 1PQ
OP LM1 0 0OP
PP ~ MM0 1 0PP P
5 + 2Q
N1 0 1Q
5 +1
−1
5 +1
−1
1
OP
PP
Q
245
MATRICES
e 5 − 1j R
LM1 0 2 5 OP LM1 1 − 5 0OP
1
0P P
−1 P ~ M0
MM0 1
M
N0 5 − 2 5 + 2PQ N1 0 1PQ
R → R – e 5 − 2j R
LM1 0 2 5 OP LM1 1 − 5 0OP
MM0 1 −1 PP ~ MM0 1 0PP P
N0 0 2 5 Q N1 2 − 5 1Q
R1 → R1 –
3
2
3
2
R1 → R1 – R3
LM1 0 0 OP LM0 −1 −1OP
MM0 1 −1 PP ~ MM01 2 −1 5 01 PP P
Q
N0 0 2 5 Q N
R3 →
1
2 5
R3
LM1 0 0 OP LM 0
MM0 1 −1PP ~ MM 01
N0 0 1 Q MMN 2 5
R2 → R2 + R3
−1
1
2− 5
2 5
OP
P
0 PP
1 P
2 5 PQ
−1
LM
LM1 0 0OP M 01
MM0 1 0PP ~ MM 2 5
N0 0 1Q MM 1
MN 2 5
∴
Now,
OP
−1
−1 P
2+ 5
1 P
P
2 5
2 5P
P
2− 5
1 P
2 5
2 5 PQ
L0 −2 5 −2 5 OP
1 M
P =
M1 2 + 5 1 PP
2 5 M
N1 2 − 5 1 Q
LM0 −2 5 −2 5 OP L−1 2 −2O LM 1
M
P
1 M
2 1P M0
1 2+ 5
1 P M1
D = P AP =
2 5 M
MN1 2 − 5 1 PPQ MMN−1 −1 0 PPQ MMN−1
L2 5 0 0 OP L1 0 0 O
1 M
M 5 0 PP .
=
0
10
0 P = M0
M
2 5 M
N 0 0 −10PQ MN0 0 − 5 PQ
–1
–1
5 −1
1
−1
OP
−1 P
P
1 PQ
5 +1
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 3. Diagonalise the matrix A =
LM11 −4 −7OP
MM 7 −2 −5PP and hence find A .
N10 −4 −6Q
5
Sol. The characteristic equation of A is |A – λI| = 0
⇒
LM11 − λ −4 −7 OP
MM 7 −2 − λ −5 PP = 0
N 10 −4 −6 − λQ
On simplification, we get
λ3 – 3λ2 + 2λ = 0
⇒
λ (λ –1) (λ – 2) = 0
∴
λ = 0, 1, 2
The eigen values are real and distinct and hence A is diagonalisable.
Now, consider the relation (A – λI) X =
For λ = 0
0
LM11 −4 −7OP LMx OP
MM 7 −2 −5PP MMx PP = 0
N10 −4 −6Q Nx Q
1
2
3
⇒
11x1 – 4x2 – 7x3 = 0
| Take any two equations.
7x1 – 2x2 – 5x3 = 0
10x1 – 4x2 – 6x3 = 0
Taking the first-two equations, we get
x1
x
x
x
x
x
= 2 = 3 ⇒ 1 = 2 = 3
6
6
6
1
1
1
∴
LM1OP
X = M1P
MN1PQ
1
For λ = 1, we have
LM10 −4 −7OP LMx OP
MM 7 −3 −5PP MMx PP = 0
N10 −4 −7Q Nx Q
1
2
3
⇒
10x1 – 4x2 – 7x3 = 0
7x1 – 3x2 – 5x3 = 0
10x1 – 4x2 – 7x3 = 0
Taking the first-two equations, we get
x1
x
x
= 2 = 3
−1 2
1
247
MATRICES
LM 1OP
MM−1PP ,
N 2Q
∴
X2 =
Similarly,
LM−2OP
λ = 2, X = M −1P
MN−2PQ
For
3
The three vectors are linearly independent,
LM1 1 −2OP
P = M1 −1 −1P
MN1 2 −2PQ
LM−4 2 3OP
P = M−1 0 1P
MN−3 1 2PQ
LM−4 2 3OP LM11 −4 −7OP LM1 1 −2OP
D = P AP = M−1 0 1P M 7 −2 −5P M1 −1 −1P
MN−3 1 2PQ MN10 −4 −6PQ MN1 2 −2PQ
LM0 0 0OP
= M0 1 0P .
MN0 0 2PQ
Hence the modal matrix
∴
–1
–1
Now,
A5 = PD5 P–1
Next
LM0 0 0 OP LM0 0 0 OP
0 P = M0 1 0 P
But,
D = M0 1
MN0 0 2 PQ MN0 0 32PQ
LM1 1 −2OP LM0 0 0 OP LM−4 2 3OP
∴
A = M1 −1 −1P M0 1 0 P M−1 0 1P
MN1 2 −2PQ MN0 0 32PQ MN−3 1 2PQ
LM191 −64 −127OP
= M 97 −32 −65 P .
MN190 −64 −126PQ
LM 0 −2 −2OP
2 P is not diagonalisable.
Example 4. Prove that the matrix A = M−1 1
MN−1 −1 2 PQ
5
5
5
5
Sol. The characteristic equation is |A – λI| = 0
⇒
–λ −2
−2
−1 1 − λ
2
−1 −1 2 − λ
= 0
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
⇒
λ3 – 3λ2 + 4 = 0 ⇒ (λ + 1) (λ – 2)2 = 0
∴
λ = –1, 2, 2. Here the eigen values are real.
Let X be an eigen vector corresponding to λ = 2
∴
(A – 2I) X = 0
LM−2 −2 −2OP LMx OP
MM−1 −1 2 PP MMx PP = 0
N−1 −1 0 Q Nx Q
1
⇒
2
3
−1
R1 →
R1
2
1 1 1
−1 −1 2
−1 −1 0
OP LMx OP
PP MMx PP = 0
Q Nx Q
LM
MM
N
1
2
3
R2 → R2 + R1, R3 → R3 + R1
LM1 1 1OP LMx OP
MM0 0 3PP MMx PP = 0
N0 0 1Q Nx Q
1
2
3
R2 →
1
R and R3 → R3 – R2
3 2
1 1 1 x1
0 0 1 x2 = 0
0 0 0 x3
LM
MM
N
OP LM OP
PP MM PP
QN Q
⇒
r = 2, n = 3 ⇒ n – r = 1
there exist only one linearly independent eigen vector corresponding λ = 2.
Hence, there does not exist three linearly independent eigen vectors. Thus A is not
diagonalisable.
Example 5. Find a matrix P which diagonalizes the matrix A =
where D is the diagonal matrix.
Sol. The characteristic equation of A is |A – λI| = 0
⇒
4− λ
1
2
3− λ
⇒
λ2 – 7λ + 10 = 0
⇒
(λ – 2) (λ – 5) = 0 ∴
= 0 ⇒
LM4 1OP , verify P AP = D
N2 3Q (U.P.T.U., 2008)
(4 – λ) (3 – λ) – 2 = 0
λ = 2, 5
Now, consider the relation (A – λI) X = 0
For λ = 2
LM2 1OP LMx OP = 0, R → R − 2R LM2 1OP LMx OP = 0
N2 1Q Nx Q
N0 0Q Nx Q
1
2
⇒
2x1 + x2 = 0,
1
2
2
1
2
Here let x2 = k, so x1 = −
k
2
–1
249
MATRICES
∴
X1 =
LMx OP = LM −2k OP = − k LM 1 OP or LM 1 OP
Nx Q MN k PQ 2 N−2Q N−2Q
1
2
For λ = 5
LM−1 1 OP LMx OP = 0, R → R + 2R LM−1 1OP LMx OP = 0
N 2 −2Q Nx Q
N 0 0Q Nx Q
1
1
2
2
1
2
2
⇒ – x1 + x2 = 0. Here again let x2 = k so x1 = k
∴
X2 =
LMx OP = LMkOP = k LM1OP or LM1OP
Nx Q NkQ N1Q N1Q
1
2
The modal matrix P =
LM 1 1OP
N−2 1Q
Verification:
|P| = 1 + 2 = 3
LM OP
N Q
1 L1 −1O L 4 1O L 1 1O
1 L1 −1O L 2 5O
P AP =
= M
M
P
M
P
M
P
PM P
3 N2 1 Q N 2 3Q N−2 1Q
3 N 2 1 Q N−4 5Q
L2 0OP = D (diagonal matrix).
1 L6 0 O
= M
=
M
P
3 N0 15Q
N0 5Q
∴
P–1 =
1 1 −1
3 2 1
–1
Thus P–1 AP = D verified.
3.20
APPLICATION OF MATRICES TO ENGINEERING PROBLEMS
Matrices have various engineering applications, they can be used to characterize connections in
electrical networks, in nets of roads, in production processes, etc., as follows:
3
3.20.1 Nodal Incidence Matrix
The network in figure consists of 6 branches (connections) and 4
nodes (point where two or more branches together). One node is
the reference node (grounded node, whose voltage is zero). We
number the other nodes and number and direct the branches.
This we do arbitrarily. The network can now we described by a
matrix A = [aij], where
aij =
2
1
2
1
6
4
(Reference
node)
+1 if branch j leaves node (i)
–1 if branch j enters node (i)
0 if branch j does not touch node (i)
3
5
Fig. 3.2
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
A is called the nodal incidence matrix of the network. The matrix for the network in figure
is given below:
Branch
1
2
3
4
5
6
Node 1
Node 2
1
0
–1
1
–1
0
0
1
0
1
0
0
Node 3
0
0
1
0
–1
–1
3.20.2 Mesh Incidence Matrix
A network can also be characterized by the mesh incidence matrix M = [mij], where
mij =
+1 if branch j is in mesh
i
and has the same orientation
–1 if branch j is in mesh
i
and has the opposite orientation
0 if branch j is not in mesh
i
And a mesh is a loop with no branch in its interior (or in
its exterior). Here, the meshes are numbered and directed
(oriented) in an arbitrary fashion. Show that for the network
in Figure. The matrix M has the given form, where row 1
corresponds to mesh 1, etc.
3
4
3
2
2
1
LM1 1 0 −1 0 0OP
0 0 0 1 −1 1P
M = M
MM0 −1 1 0 1 0PP
N1 0 1 0 0 1Q
5
6
1
Fig. 3.3
Example 1. There is a circuit (electrical network) in figure given below. Find the currents
i1, i2 and i3 respectively.
20 ohms
10 ohms
Q
i1
i3
80
10 ohms
90 volts
i2
P
15 ohms
Fig. 3.4
Sol. We label the currents as shown, choosing directions arbitrarily; if a current will come
out negative, this will simply mean that the current flows against the direction of our arrow. The
current entering each battery will be the same as the current leaving it. The equations for the
currents results from Kirchhoff’s laws.
Kirchhoff’s Current Law (KCL): At any point of a circuit, the sum of the inflowing currents
equals the sum of the out flowing currents.
Kirchhoff’s Voltage Law (KVL): In any closed loop, the sum of all voltage drops equals the
impressed electromotive force.
251
MATRICES
Node P gives the first equation, node Q the second, the right loop the third, and the left loop
the fourth, as indicated in the figure.
Node P :
i1 – i2 + i3 = 0
(i)
Node Q :
Right loop :
– i1 + i2 – i3 = 0
10 i2 + 25 i3 = 90
(ii)
(iii)
Left loop :
20 i1 + 10 i2 = 80
(iv)
We solve these equations by matrix method
LM 1 −1 1 M 0OP
−1 1 −1 M 0
Augmented matrix [A : B] = M
MM 0 10 25 M 90PPP
N20 10 0 M 80Q
Applying R2 → R2 + R1, R4 → R4 – 20 R1
LM1 −1 1 M 0 OP
0 0
0 M 0P
~ M
MM0 10 25 M 90PP
N0 30 −20 M 80Q
R2 ↔ R4
LM1 −1 1 M 0 OP
0 30 −20 M 80
~ M
MM0 10 25 M 90PPP
N0 0 0 M 0 Q
R3 → 3R3 – R4
LM1 −1 1 M 0 OP
0 30 −20 M 80 P
~ M
MM0 0 95 M 190PP
N0 0 0 M 0 Q
R2 →
1
1
R,R →
R
5 2 3
19 3
LM1 −1 1 M 0 OP
0 6 −4 M 16
~ M
MM0 0 5 M 10PPP
N0 0 0 M 0 Q
LM1 −1 1 OP Li O LM 0 OP
MM0 6 −4PP MMi PP MM16PP
MN00 00 05 PQ MNi PQ MN100PQ
1
Now
2
3
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
⇒
i1 – i2 + i3 = 0
5i3 = 10
On solving these equations
i1 = 2 amperes
x2
i2 = 4 amperes
P
di r inc
r e ip
ct al
io
n
6i2 – 4i3 = 16
i3 = 2 amperes.
Example 2. An elastic membrane in the x1, x2-plane
x1
with boundary circle x12 + x22 = 1, is stretched 80 that a
point P : (x1, x2) goes over into the point Q : (y1, y2) given by
y=
P
di r inc
re ip
c t al
io
n
LMy OP = Ax = LM5 3OP LMx OP;
Ny Q
N3 5Q Nx Q
1
1
2
2
Fig. 3.5
in components y1 = 5x1 + 3 x2 ; y2 = 3 x1 + 5x2 .
Find the principal directions, that is, the directions of the position vector x of P for which
the direction of the position vector y of Q is the same or exactly opposite. What shape does the
boundary circle take under this deformation?
Sol. We are looking for vectors x such that y = λx.
Since y = AX ⇒ AX = λX
∴ The above form is
5x1 + 3x2 = λx1
(5 – λ) x1 + 3x2 = 0
or
3x1 + 5x2 = λx2
3x1 + (5 – λ) x2 = 0
∴ The characteristic equation is
⇒
5− λ
3
= (5 – λ)2 – 9 = 0
3
5− λ
λ1 = 8, λ2 = 2
for λ1 = 8,
– 3x1 + 3x2 = 0
3x1 – 3x2 = 0
Sol. x2 = x1, x1 arbitrary.
Let
⇒
x1 – x2 = 0
x1 – x2 = 0
x1 = x2 = 1
For λ2 = 2 our system becomes
3x1+ 3x2 = 0
⇒
3x1 + 3x2 = 0
x2 = – x1, x1 arbitrary
Let
x1 = 1, x2 = – 1
Thus the eigen vector, for instance are
X1 =
LM1OP and X = LM 1 OP
N1Q
N−1Q
2
These vectors make 45° and 135° angles with the positive x1-direction. They give the principal
directions.
253
MATRICES
The eigen values show that in the principal directions the membrane is stretched by factors
8 and 2, respectively. Accordingly if we choose the principal directions as directions of a new
cartesian u1, u2-coordinats system, say with the positive u1-semiaxis in the first quadrant and the
positive u2-semiaxis in the second quadrant of the x1 x2-system and if we get
u1 = r cos φ, u2 = r sin φ
then a boundary point of the unstretched circular membrane has coordinates cos φ, sin φ. Hence
after the stretch we have
Z1 = 8 cos φ, Z2 = 2 sin φ
Since cos φ + sin φ = 1, this shows that the deformed boundary is an ellipse.
2
2
Z12 Z22
+ 2 = 1
82
2
with principal semiaxes 8 and 2 in the principal directions.
EXERCISE 3.8
LM 1 2 −2OP
2 1 P . Find the modal matrix P and the
1. A square matrix A is defined by A = M 1
MN−1 −1 0 PQ
resulting diagonal matrix D of A.
(U.P.T.U., 2000)
LM
LM–2 –2 2 OP LM1 0 0OPOP
MMAns. P = M 1 1 1 P, D = M0 −1 0PPP
MN 1 –1 –1PQ MN0 0 3PQQ
N
LM 3 −1 1 OP
2. Diagonalise the matrix A = M −1 5 −1P and hence find A .
MN 1 −1 3 PQ
LM
LM 251 −405 235 OPOP
MMAns. A = M−405 81 −405PPP
MN 235 −405 251 PQQ
N
3. Diagonalise the matrix A.
LM
LM2 0 0OPOP
LM3 1 4OP
A = M0 2 6P ·
MMAns. D = M0 3 0PPP
MN0 0 5PQQ
MN0 0 5PQ
N
4
4
4. Verify whether the following matrices are diagonalisable or not.
LM−2 2 −3OP LM−3 2 2OP
(i) M 2 1 −6P , (ii) M−6 5 2P ·
MN−1 −2 0 PQ MN−7 4 4PQ
LM 8 −6 2 OP
5. Diagonalise M−6 7 −4P ·
MN 2 −4 3 PQ
[Ans. (i) Not diagonalisable (ii) diagonalisable]
LM
LM0 0 0 OPOP
Ans.
D
=
MM
MM0 3 0 PPPP
N0 0 15QQ
N
254
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
6. Diagonalise the following matrices:
LM1 0 0OP L0 0O L0 0OOP
LM1 0 2OP L 2 −1O L1 1O LM
MAns. (i) D = M0 –2 0P, (ii) MN0 6PQ, (iii) MN0 2PQPP
(i) M0 1 2P , (ii) M
, (iii) M
−8 4 PQ
1 1PQ M
MN0 0 3PQ
N
N
MN1 2 0PQ
N
Q
LM
LM1 0 0OPOP
LM2 2 1OP
Ans.
D
=
0 1 0PP
7. Diagonalise the matrix A = M1 3 1P .
MM
M
MN0 0 5PQPQ
MN1 2 2PQ
N
LM4 1OP hence find A . LMAns. D = L1 0O, A = L2344 781OOP
MN0 5PQ MN2343 782PQQ
N3 2Q
N
L0 1OP is diagonalisable.
9. Verify the matrix M
N0 0 Q
Lk O
[Ans. Not diagonalisable since only one eigen vector M P exists]
N0 Q
8. Diagonalise the matrix
5
5
10. Using matrix equation determine the loop current in the following circuits:
8V
i1
109
&9
Ans. i1 = 1.0826, i 2 = 1. 4004, i 3 = 1.2775
(i)
$9
i2
i3
4V
6V
40 V
19
i1
49
i4
49
10 V
59
(ii)
30 V
i2
29
69
i3
39
20 V
Ans. i1 = 11. 43, i 2 = 10.55, i 3 = 8.04, i 4 = 5.84
255
MATRICES
OBJECTIVE TYPE QUESTIONS
A. Pick the correct answer of the choices given below:
LM1 aOP , then the value of A is
N0 1Q
L1 a OP
L4 4aOP
(i) M
(ii) M
N0 4 Q
NM0 1 QP
L4 a OP
L1 4aOP
(iii) M
(iv) M
MN0 4 PQ
N0 1 Q
LM1 b 2OP
2. If the matrix M1 2 5P is not invertible, then the value of b is
MN2 1 1PQ
1. If A =
4
4
4
(i) 2
(ii) 0
(iii) 1
(iv) –1
LM3 2OP , then (A ) is equal to
N0 1 Q
1 L1 −26O
1 L−1 26O
(i)
(ii)
M
P
M
P
27 N0 27 Q
27 N 0 27Q
1 L1 −26O
1 L−1 −26O
(iii)
(iv)
M
P
M
P
27 N0 −27Q
27 N 0 −27Q
LM0 0 1OP
4. If A = M0 1 0P , then A is
MN1 0 0PQ
LM0 1 0OP
LM1 0 1OP
(ii) M1 0 0P
(i) M0 1 0P
MN0 0 1PQ
MN0 0 0PQ
LM0 0 1OP
(iii) M0 1 0P
(iv) None of these
MN1 0 0PQ
3. If A =
–1 3
–1
5. If A and B are square matrices of order 3 such that |A| = –1 and |B| = 3, then |3AB| =
(i) – 9
(ii) – 27
(iii) – 81
(iv) 81
6. The rank of the unit matrix of order n is
(i) 0
(ii) 1
(iii) 2
(iv) 3
256
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
LM1 1 1OP
7. If A = M1 2 4P , then rank (A) is
MN1 4 8PQ
(i) 0
(ii) 1
(iii) 2
(iv) 3
8. If a matrix A has a non-zero minor of order r and all minors of higher orders zero, then
(i) ρ(A) < r
(ii) ρ(A) ≤ r
(iii) ρ(A) > r
(iv) ρ(A) = r
9. The system of equations 3x + 2y + z = 0, x + 4y + z = 0, 2x + y + 4z = 0.
(i) is consistent
(ii) has infinite solutions
(iii) has only a trivial solution
(iv) None of these
10. The system of equations – 2x + y + z = a, x – 2y + z = b, x + y – 2z = c is consistent,
if
(i) a + b – c = 0
(ii) a – b + c = 0
(iii) a + b + c ≠ 0
(iv) a + b + c = 0
11. If X1 = (2, 3, 1, –1); X2 = (2, 3, 1, – 2), X3 = (4, 6, 2, – 3) then
(i) X1 + X3 = X2
(ii) X2 + X3 = X1
(iii) X1 + X2 = 2X3
(iv) X1 + X2 = X3
B. Fill in the blanks:
1. The vectors (1, –1, 1), (2, 1, 1), (3, a, 2) are linearly dependent if the value of a = ..........
2. If A is non-singular matrix of order 3 then ρ(A) = ..........
3. If λ is a non-zero characteristic root of a non-singular matrix A, then the characteristic
roots of A–1 are ..........
4. If λ1, λ2, λ3 are the characteristic roots of a non-singular matrix A, then the characteristic
roots of A–1 are ..........
5. The characteristic of a real skew symmetric are either ..... or .....
6. The product of the characteristic roots of a square matrix is ..........
7. A system of n linear homogeneous equations in n unknowns has a non-trivial solution
if the coefficient matrix is ..........
8. A system of n linear homogeneous equations in n unknowns is always ..........
9. For the set of vectors (X1, X2, ... Xn) to be linearly independent it is necessary that no
Xi is ...........
LM2 2 1OP
10. The eigen values of A = M1 3 1P are ..........
MN1 2 2PQ
C. Indicate True or False for the following statements:
1. (i) If |A| = 0, then at least one eigen value is zero.
(ii) A–1 exists iff 0 is an eigen value of A.
257
MATRICES
(iii) If |A| ≠ 0 then A is known as singular matrix.
(iv) Two vectors X and Y are said to be orthogonal if, XTY = YTX ≠ 0.
2. (i) The eigen values of a real skew symmetric matrix are all real.
(ii) If A is a square matrix, then latent roots of A and A′ are identical.
(iii) If A is a unit matrix, then |A| = 0.
(iv) If |A| ≠ 0, then |A. adj A| = |A|n–1, where A = (aij)n × n.
D. Match the following:
1. (i) If A is an invertible matrix then
(a) |A| = ± 1
(ii) If A is orthogonal then
(b) |A| = 0
(iii) (A – λI) is a matrix
(c) is 1
(iv) The rank of unit matrix of order n
(d) Characteristic
2. (i) For non-Trivial
(a) A–1 is a diagonal
(ii) A is diagonal
(b) A
–1 –1
(iii) (A )
(c) |A| = 0
(iv) For trivial solution
(d) ρ(A) < n
3. (i) Rank of skew symmetric matrix
(a) Have same rank
(ii) Two equivalent matrices
(b) – A
(iii) The rank of I4 is
(c) 1
(iv) A*
(d) 4
ANSWERS TO OBJECTIVE TYPE QUESTIONS
A. Pick the correct answer:
1. (iv)
4. (iii)
7. (iii)
10. (iv)
2. (iii)
5. (iii)
8. (iv)
3. (i)
6. (iv)
9. (iii)
2. 3
3. adj A
B. Fill in the blanks:
1. 0
4. λ , λ , λ
–1
1
–1
2
–1
3
5. All zero; pure imaginary] 6. |A|
7. Singular
8. Consistent
9. Independent
10. 1, 1, 5
C. True or False:
1. (i) T
(ii) F
(iii) F
(iv) F
2. (i) F
(ii) T
(iii) T
(iv) F
D. Match the following:
1. (i) → (b), (ii) → (a), (iii) → (d), (iv) → (c)
3. (i) → (c), (ii) → (a), (iii) → (d), (iv) → (b)
2. (i) → (d) (ii) → (a) (iii) → (b) (iv) → (c)
GGG
UNIT
18
Multiple Integrals
4.1
MULTIPLE INTEGRALS
The process of integration for one variable can be extended to the functions of more than one
variable. The generalization of definite integrals is known as “multiple integral”.
4.2 DOUBLE INTEGRALS
Consider the region R in the x, y plane we assume
that R is a closed*, bounded** region in the x , y
plane, by the curve y = f1(x), y = f2(x) and the
lines x = a, x = b. Let us lay down a rectangular
grid on R consisting of a finite number of lines
parallel to the coordinate axes. The N rectangles
lying entirely within R (the shaded ones in Fig.
4.1). Let (xr, yr) be an arbitrarily selected point in
the rth partition rectangle for each r = 1, 2, ..., N.
Then denoting the area δxr · δyr = δSr
Thus, the total sum of areas
Y
y = f2 (x)
R
dx r
(x r, y r)
dy r
∑ f bxr , y r g δSr
N
SN =
y = f1 (x)
r =1
Let the maximum linear dimensions of each
portion of areas approach zero, and n increases
indefinitely then the sum SN will approach a
limit, “namely the double integral
zz b g
O
x=a
x=b
Fig. 4.1
f x, y dS and the value of this limit is given by
R
* Boundary included i.e., part of the region.
** Can be enclosed within a sufficiently large circle.
258
X
259
MULTIPLE INTEGRALS
zz b g
LM a f
O
z MN za ffbx, yg dyPPQ dx
b f2 x
f x, y dS =
a f1 x
R
Similarly if the region is bounded by y = c, y = d and by the curves x = f1(y), x = f2(y), then
LM b g
OP
,
f
x
y
dx
=
b
g
f
x
,
y
dS
b
g
z
z
zz
MN b g
PQ dy
d f2 y
c f1 y
R
4.3 WORKING RULE
(a) We integrate with respect to y, x is to be regarded as constant and evaluate the result between
the limits y = f1(x) and y = f2(x).
(b) Then we integrate the result of (a) with respect to x between the limits x = a and x = b.
4.4 DOUBLE INTEGRATION FOR POLAR CURVES
Let OP and OQ are two radii vectors of the curve r = f (θ) and the coordinates of P and Q be
(r, θ) and (r + δr, θ + δθ).
The area of portion
K1 L1 K2 L2 =
1
1 2
(r + δr)2 δθ –
r δθ
2
2
1
δr2δθ
2
= rδrδθ |As δr2δθ is very small
Therefore, the area of OPQ
= rδrδθ +
=
B
lim [Σrδrδθ]
Q (r + dr, q + dq)
δr→0
= lim [Σrδr] δθ
K2
δr→0
z
L a f rdrOP δθ
= M
N
Q
P (r, q)
f θ
Hence the area OAB
z
L a f rdrOP δθ
= lim ∑ M
N
Q
δθ → 0
f θ
O
0
z LMNz a f OPQ
θ =β
fθ
θ=α
0
L2 L 1
dq
0
=
K1
rdr dθ
where α and β are the vectorial angles of A and B.
q
Fig. 4.2
A
260
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 1. Evaluate
z z
log 8
log y
1
0
Sol. We have
I =
=
ex + y dx dy.
z LMNz e jOPQ
log 8
log y
1
0
z
z
log 8
=
=
Example 2. Evaluate
z z
π
x
0
I =
=
zz
0
Sol. We have
b g
log 8
dy
y
– e
1
log 8
1
(As eloge y = y)
sin y dy dx .
z LMNz
z
π
x
0
0
π
z
z
sin y dy
OP dx
Q
x
– cos y 0 dx, treating x as constant
0
= −
2
dy
e y e log y − e 0
ey y−1
= −
Example 3. Evaluate
ey dy
= elog 8 · (log 8 – 1) – 0 – (elog 8 – e)
= 8 log 8 – 8 – 8 + e
= 8 log 8 – 16 + e.
0
Sol. We have
log 8
1
log y y
e
0
ex
1
e x dx
π
0
π
0
(cos x – cos 0) dx
cos x dx +
π
z
π
0
dx
(As cos 0 = 1)
π
= − sin x 0 + x 0 = π.
d 4− 2 y i
y dx dy .
− d 4− 2 y i
2
2
z z
2
I =
=
z
z
2
2
0
1
2
4− 2 y 2
dx
j
OP
PQ y dy
c 4− 2 y h y dy
− c 4− 2 y h
2
e 4 − 2 y j y dy
2
2
2
z
0
4
1 2
= – ·
2 3
=
j
x
0
= –
4− 2 y 2
−
0
=
LM e
MN e
t dt
LMt OP
MN PQ
1
8
·8=
.
3
3
by putting 4 – 2y2 = t, – 4y dy = dt
3 0
2
4
MULTIPLE INTEGRALS
Example 4. Evaluate
zz b g
x+ y
x2
y2
2 +
a
b2
2
261
2
dx dy over the area bounded by the ellipse
[U.P.T.U. (C.O.), 2004]
= 1
= 1
⇒
y
= ±
b
1−
⇒
y = ±
∴
zz b g
x+ y
2
dx dy =
=
b
a
=
x = –a
zz e
FH IK
z zFH IK
Fig. 4.3
j
x + y + 2xy dx dy
2
2
b
a
−b
a
a
−a
zz
−a
zz
a
2
2
zz
2
2
2
FG b IJ a − x 2 2
x + y dy dx + 0
z
z
z
−a
ex + y + 2xyj dx dy
a2 − x 2
2
2 H aK
−a 0
a
a2 − x 2
a
e ba j a − x
e ba j a − x 2 2
xy dy dx
+
x
y
dx
dy
+
2
e
j
−a e− b j a − x
e −ab j a −x
a
As
=
X
O
a2 − x2
a
=
x2
a2
x=a
x
y
2 +
a
b2
Sol. We have
Y
2
2
z
a
2
e
af
f x dx = 2
–a
2
2
af
a
af
f x dx when f ( x) even = 0 when f x odd
F bI a −x
3
2
j
0
LM2F x y + y I OPGH JK
MN GH 3 JK PQ
2
z
2
2
2
a
dx
0
L b a − x + 1 b a − x OP dx
= 2 Mx ×
e jP
3a
MN a
Q
Lb
O
b
a − x j P dx (Again by definite integral)
= 4 M x a −x +
e
3a
MN a
PQ
a
2
2
3
2
2
3
3
2 2
−a
a
2
2
2
3
3
2
3
2 2
0
= 4
z FGH
π
2
0
[On putting x = a sin θ and dx = a cos θ dθ]
IJ × a cos θ dθ
K
b ⋅ a 2 sin 2 θ. a cos θ + b 3 a 3 cos 3 θ
a
3a 3
262
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
= 4
z
π
2
0
= 4a3b
=
FG a b sin θ cos θ dθ + ab cos θIJ dθ
3
H
K
3
2
3
2
4a3b ·
π
2
sin 0 θ cos 4 θ dθ
0
3 3
1 5
4 ab 3
2 2
2 2
+
·
3
2+ 2+ 2
0 + 4+ 2
2
2
2
2
= 4a3b ·
=
z
z
π
4ab 3
2 sin 2 θ cos 2 θ dθ +
0
3
1 1 1 1
.
4 ab 3
= 4a3b. 2 2 2 2 +
.
3
2 3
⇒
4
z
π
As 2 sin m θ
0
m+1 n+1
2
2
cos θ dθ =
m+n+2
2
2
n
1 3 1 1
. . .
2 2 2 2
2 3
a
π
4ab 3 3π
⋅
+
16
3 16
As n = n − 1
b
f n – 1 and 12 = π
g
πa 3 b πab 3 π
+
= ⋅ ab a 2 + b 2 .
4
4
4
Example 5. Evaluate
zz
xy ( x + y) dx dy over the area between y = x2 and y = x.
Sol. The area is bounded by the curves y = f1(x) = x2, y = f2(x) = x.
When
f1 (x) = f2 (x),
x2 = x, i.e., x (x – 1) = 0
or
x = 0, x = 1
i.e., the area of integration is bounded by
y = x2, y = x, x = 0, x = 1
∴
=
=
=
=
=
zz b g
z LMNz b g OPQ
z LMNz e j OPQ
LM
OP
z MN
PQ
OP
z LMMN
PQ
xy x + y dx dy
A
1
y =x
x=0
y=x
1
x
0
x2
zz
D
xy x + y dy
x 2 y + xy 2 dy
x 2 y 2 xy 3
+
0
2
3
0
4
A
dx
dx
y=x
2
x=1
X
O
x
dx
x2
6
Fig. 4.4
7
5x
x
x
−
−
6
2
3
dx
L x x − x OP L 1 − 1 − 1 O 3 .
= M −
N 6 14 24 Q = MN 6 14 24 PQ = 56
5
Example 6. Find
2
1
1
Y
7
8 1
0
x 2 dx dy where D is the region in the first quadrant bounded by the
hyperbola xy = 16 and the lines y = x, y = 0 and x = 8.
(U.P.T.U., 2002)
263
MULTIPLE INTEGRALS
Sol. We have
Y
xy = 16
...(i)
y = x
y = 0
...(ii)
...(iii)
y=x
x = 8
...(iv)
From eqns. (i) and (ii), we get x = 4, y = 4
x=8
A
i.e., intersection point of curve and the line
y = x = (4, 4).
(4, 4)
xy = 16
Similarly intersection point of (i) and (iv) = (8, 2)
To evaluate the given integral, we divide the area
OABEO into two parts by AG as shown in the Figure 4.5.
Then,
zz
D
x 2 dx dy =
=
=
z z
x= 4
y=x
x=0
y=0
x 2 dx dy +
zz
x=8
y=
16
x
x= 4 y =0
z z z z
B
O
z z
zz
G
(4, 0)
4
X
Fig. 4.5
z bg z bg
4
0
x
8
0
4
x y dx +
2
LM x OP 8x
N4Q +
4
E
x 2 dx dy
16
x
8 2
4 2
x dy =
x
dx
dy
x
dx
+
0
4
0
0
4 3
8
x dx + 16 x dx =
0
4
(8, 2)
x
2
16
y 0x dx
2 8
0
4
= 64 + 8(64 – 16) = 64 + 384 = 448.
Example 7. Evaluate
A
xy dx dy over the positive quadrant of the circle x2 + y2 = a2.
Sol. Here the region of integration is positive quadrant of circle x2 + y2 = a2, where x varies
ea − x j .
2
from 0 to a and y varies from 0 to
Here,
zz
A
xy dx dy =
zz
=
z
0
z
Y
d a −x i xy dx dy
2
a
0 0
a
2
LM y OP
N2Q
2
a2 − x 2
2
x dx
dx
0
2
y= a –x
Ly O
= M P x dx
N2Q
1
= z x e a − x j dx
2
1 La x
x O
−
P = 18 a .
= M
2N 2
4Q
Fig. 4.6
Example 8. Find zz e x + y j dx dy where D is bounded by y = x and y = 4x.
Sol.
zz ex + y jdx dy = ex + y j dy dx
LMx y + y OP dx
=
MN 3 PQ
a
a2 −x2
2
x=0
0
0
a
2
2
0
2 2
4
O
a
y=0
4
0
2
D
2
D
2
2
2
2
zz
z
4 2 x
2
2
0 x
4
0
2
3 2 x
x
x=a
X
264
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Y
B
(4, 4)
y
=2
x
y=x
X
O (0, 0)
z FGH
4
=
Example 9. Find
in the first quadrant.
Sol.
zz
D
0
IJ
K
8
4
2x 5 2 + x 3 2 − x 3 dx
3
3
768
.
35
=
zz
Fig. 4.7
x 3 y dx dy where D is the region enclosed by the ellipse
D
x 3 y dx dy
=
zz
b 2 2
a −x
a
x 3 y dy dx
0 y=0
a
z
LM x y OP
MN 2 PQ
a
=
3
0
=
x2
y2
+
=1
a2
b2
b
a
2
b2
2a 2
ze
b2
2 4
a
0
Y
a2 −x2
(0, b)
0
a 2x 3 − x 5
j dx
(a, 0)
LM a x − x OP = b a .
=
24
6 PQ
2 a MN 4
6 a
2 4
2
Example 10. Evaluate I =
Sol.
I =
=
=
z z
z z
z LMN OPQ
z
0
2π
a
0
a sin θ
2π
a
0
r = a sin θ
2π
r2
2
0
a2
2
2π
0
2
I =
a
4
=
πa
.
2
2
Fig. 4.8
rdrdθ .
rdrdθ
a
dθ =
a sin θ
cos2 θ dθ =
LMθ + sin 2θ OP
N 2 Q
2π
0
ze
j
1 2π 2
a − a 2 sin 2 θ dθ
2 0
z FGH
IJ
K
a 2 2 π 1 + cos 2θ
dθ
2 0
2
X
265
MULTIPLE INTEGRALS
Example 11. Evaluate
zz
r 2 sin θ dθ dr over the area of cordioid r = a(1 + cos θ) above the
A
initial line.
Sol. The region of integration A can be covered by radial strips whose ends are at r = 0,
r = a(1 + cos θ).
The strips lie between θ = 0 and θ = π
Thus
zz
A
r 2 sin θ dθ dr =
=
=
z z
z LMNz a f OPQ
LM OP a f
z NQ
a
π
a 1+ cos θ
0
0
π
0
π
0
a 1+ cosθ
sin θ
0
z
dθ
p
q=–
2
r dr dθ
sin θ
zb
z
z
16 a 3
3
θ
θ
π
cos7 sin dθ
0
2
2
16 3
=
a
3
π
7
2 sin φ cos φ ⋅ 2dφ
0
1 + cos θ
8
Put
π
2
3
0
a 2
r dr
a 1−cosθ
a
r 2 drdθ .
a
=
π
2 dθ
0
r3
3
=
π
2 dθ
0
3
a 3 a 1 − cosθ
−
3
3
=
a3
3
π
2 1 − 1 − cosθ 3
0
=
a3
3
π
2 1 − 1 − 3 cos θ + 3 cos 2 θ − cos 3 θ
0
=
a3
3
π
2
0
a 1−cosθ
3
dθ
3 cos θ − 3 cos 2 θ + cos 3 θ dθ
LM
1π
2 O
P
+
3 sin θ − 3
2 2 3 ⋅1P
MN
Q
a L 3 − 3π + 2 O
=
M 4 3 PQ
3 N
a3
=
3
π
2
0
3
=
a3
[44 – 9π].
36
A
Fig. 4.9
LM − cos φ OP = 4 a .
N 8 Q 3
a 1− cos θ
q=0
q
O
z za f
z a f z LMN OPQ a f
F
a
f IJ
G
z H
K
z a f
z e
j
ze
π
2
0
P (r, q)
X
g3 sin θ dθ
=
π
q=p
dθ
a3
3
0
Q
dq
a 1+cosθ
=
Example 12. Evaluate
Sol.
Y
2
r3
3 0
16 3
=2×
a
3
π
2
0
f r 2 sin θ dθ dr
j dθ
θ
= φ
2
266
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 13. Evaluate
Sol. We have
z z
z z
z LMNz OPQ
z LMN OPQ
z
π
aθ 3
0
0
π
I =
aθ 3
r dθ dr
0
0
=
π
aθ 3
0
0
r dr
dθ
aθ
r4
dθ
4 0
π
=
Example 14. Evaluate
r dθ dr
0
π 4 4
a θ dθ
0
=
1
4
=
a4
4
LM θ OP = a π .
N 5 Q 20
5
z zb
z
z
π
4 5
0
π
a 1+ cos θ
0
0
g r 3 sin θ cos θ d θ dr .
z
LM a f r drOP dθ
N
Q
a
f
L
r O
sin θ cos θ M P dθ
MN 4 PQ
π
4
a
1 + cos θg sin θ cos θ dθ
=
b
z
0
4
Sol. We have
π
I =
0
sin θ cos θ
a 1+ cosθ
3
0
4 a 1+cosθ
π
0
0
4
1 + cos θ = t and – sin θ dθ = dt
Put
=
a4
4
=
a4
4
z a fa f
ze j
0
2
t 4 t − 1 − dt
2
0
t 5 − t 4 dt =
16 4
a.
15
EXERCISE 4.1
z z
z z
log y
log 8
1. Evaluate
1
1
2. Evaluate
3.
π
2
0
0
1+ x 2
4.
0
z z b g
z z
cos x + y
a
0
a2 − y2
0
Ans. 8 log 8 – 16 + e
0
0
π
2
e x + y dx dy.
dy dx
1+ x + y
2
2
.
dy dx.
a 2 − x 2 − y 2 dx dy.
LMAns. π loge 2 + 1jOP
N 4
Q
Ans. – 2
LMAns. πa OP
6 PQ
MN
3
267
MULTIPLE INTEGRALS
z z
5.
LMAns. 1 OP
N 2Q
x2 x
y
1
e dx dy .
0
0
6. Evaluate
7. Evaluate
zz
zz e
xy − y 2 dy dx where S is the triangle with vertices (0, 0), (10, 1) and (1, 1).
S
Ans. 6
j
x 2 + y 2 dx dy, where S is the area enclosed by the curves, y = 4x, x + y = 3,
S
LMAns. 463 OP
48 Q
N
LMAns. π OP
N 24 Q
y = 0 and y = 2.
8. Evaluate
9. Evaluate
10. Evaluate
zz
zz e
x 2 y 2 dx dy over the region x2 + y2 ≤ 1.
x2 + y2
zz
j dx dy over the region bounded by x = 0, y = 0, x + y = 1.
LMAns. 1 OP
N 6Q
xy
dx dy, where the region of integration is the positive quadrant of the
1 − y2
A
LMAns. 1 OP
N 6Q
circle x2 + y2 = 1.
11. Evaluate
zz
12. Evaluate
z z e
13.
14.
zz e
zz b
D
1
1+ x
−1 2
−x
LMAns. 1 OP
N 24 Q
LMAns. 63 OP
N 32 Q
j
x 2 + y dy dx.
j
4xy − y 2 dx dy, where D is the rectangle bounded by x = 1, x = 2, y = 0, y = 3.
Ans. 18
g
1 + x + y dx dy, A is the region bounded by the lines y = – x, x =
A
15.
xy dx dy over the region in the positive quadrant for which x + y ≤ 1.
Hint: Limits x : – y to
y ; y : 0 to 2 .
zz
dx dy.
1
0
1+ x2
0
16. Evaluate
e1 + x + y j
2
zz
A
2 −1
LMAns. 44 2 + 13 OP
N 15 3 Q
LMAns. π log e1+ 2 j OP
N 4
Q
y dx dy, where A is bounded by the parabolas y2 = 4x and x2 = 4y.
LM x − y OP
LM x − y OP
dx
17. Show that
MN bx + yg dyPQ dx ≠ z MNz bx + yg PQ dy.
zz
1
0
y , y = 2, y = 0.
1
0
1
2
0
1
0
2
LMAns. 48 OP
5Q
N
268
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
18. Evaluate
19.
zz e
1
y
0
2
y
LMAns. 41 OP
N 240 Q
j
1 + xy 2 dx dy.
j dx dy, where A is bounded by x + y = a and x + y = b , where b > a.
LMAns. π eb 4 − a4 jOP
N 2
Q
LMAns. logFG 25IJ OP
z bxdy+ dxyg ⋅
H 24 K Q
N
x2 + y2
A
20.
zze
z
4
2
2
2
2
2
2
2
1
3
Using polar coordinates evaluate the following double integral.
zz
zz
zz
zz
π
21.
π
cos θ
0
0
π
23.
a
0
0
24.
4 sin θ 3
2 sin θ
0
22.
r dr dθ.
ρ sin θ dρ dθ.
r 3 sin θ cos θ dr dθ.
LMAns. 45π OP
N 2Q
LMAns. 1 OP
N 3Q
Ans. 0
r 3 dr dθ, over the area bounded between the circles r = 2 cos θ and r = 4 cos θ.
25. Show that
r = 2a cosθ.
26.
2
z z
π
2
0
zz
R
LMAns. 45 π OP
N 2 Q
r 2 sin θ dr dθ =
2a 3
, where r is the region bounded by the semicircle
3
LMAns. a OP
MN an + 1f PQ
n+ 1
a n
0
n
r sin θ cos θ d θ dr , for n + 1 > 0.
2
4.5 CHANGE OF THE ORDER OF INTEGRATION
If the limits of x and y are constant then
zz b g
F x, y dx dy can be integrated in either order, but
if the limits of y are functions of x, then the new limits of x in functions of y are to be determined.
The best method is by geometrical consideration.
Thus in several problems, the evaluation of double integral becomes easier with the change
of order of integration, which of course, changes the limits of integration also.
Remark. If y depends on x and we want that x depend on y i.e., x = f (y) then construct a
strip parallel to x-axis.
zz
∞ x
Example 1. Evaluate the integral
0 0
F x I dx dy by changing the order of integration.
GH y JK
x exp −
2
(U.P.T.U., 2005)
269
MULTIPLE INTEGRALS
Sol. Here
y = 0 and y = x
x = 0 and x = ∞
Here x start from x = y and goes to x → ∞ and
y varies from y = 0 to y → ∞
∴
0
2
x exp −
0
z z
z z FGGH
∞
=
x
F xI
GH y JK dx dy
x
∞
y=
zz
∞
Y
x2
y
−
xe
x®¥
⋅ dy dx
y = 0 x= y
∞
=
x2
∞
y
2x −
− − e y
2
y
y
0
I
JJ dy dx
K
O
z
z
∞
x2
−
y
0
y
∞
−
z
X
Fig. 4.10
LM y OP
=
MN− 2 e PQ dy
LM y OP
y
e dy
=
dy =
0+ e
2
MN 2 PQ
1
Ly
O
= M e− e j − ee jP
2
N2
Q
F 1I 1
= (0 – 0) + GH 0 + JK = .
2
2
∞
x®¥
−
Put e
x2
y
x2
2x − y
=t ⇒ –
e.
dx = dt
y
∞
y2
y
0
–y
0
−y
−y
∞
(Integration by parts)
0
Example 2. Change the order of integration in
Sol. Here the limits are
x2 = ay and y = 2a – x
i.e.,
x2 = ay and x + y = 2a
also
x = 0 and x = a
Now
zz b g
zz b g
z z b g
zz
zz
zz
2 a− x
f x, y
x2
a
a
0
ay
a
=
0
+
a
0
2 a −x
f
x2
a
bx, yg dx dy.
(U.P.T.U., 2007)
B (0, 2a)
x + y = 2a
2
x = ay
dx dy
L
A (a, a)
f x, y dy dx
0
2a
2a − y
a
0
Example 3. Evaluate
zz
O
f x , y dy dx .
2
ex
0
1
x=a
Fig. 4.11
dx dy changing the order of integration.
(U.P.T.U., 2003)
Sol. The given limits are x = 0, x = 2, y = 1 and y = ex.
Here
2
ex
0
1
dx dy
=
e2 2
1
x= log y
dy dx
As x varies from x = log y to x = 2
270
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Y
C
x
y
=
e
Þ
x
=
lo
g
y
(0, 1)
A
=
zz
Example 4.
y=1
B
O
=
2
(2, e )
X
x=2
zb
e
2
1
Fig. 4.12
g
2 − log y dy
b2 y − y log y + yg = b3y − y log yg
e2
e2
1
1
= (3e2 – 2e2) – 3
= e2 – 3.
1
3x +2
−2
x2 + 4 x
dy dx .
Sol. Here the curves y = x2 + 4x, and the straight lines y = 3x + 2, x = – 2 and x = 1 as shown
2
shaded in Figure 4.13. Here A(– 2, – 4) B (0, 0), E (1, 5), F (0, 2), G − , 0 .
3
Considering horizontal strip, x varies from the
FG
H
4 + y − 2 the lower curve
upper curve x =
x =
FG y − 2IJ and then y varies from – 4 to 5.
H 3K
Y
E (1, 5)
Changing the order of integration to first x and later
to y, the double integral becomes
=
=
z z
z z
z LMN
1
3x + 2
–2
y = x 2 + 4x
5
−4
5
−4
y = 3x + 2
2
y = x + 4x
dy dx
F (0, 2)
y+ 4 − 2
y − 2 dx dy
x=
3
FG y − 2 IJ OP dy
H 3 KQ
L2
9
y
4 O
− yP = .
= M b y + 4g −
2
6 3 PQ
MN 3
=
IJ
K
x=1
x = –2
y+ 4 −2−
3
2
2
G
B (0,0)
A
5
−4
Fig. 4.13
X
271
MULTIPLE INTEGRALS
Example 5. Change the order of integration in I =
same.
Sol. The limits are y = x2 and y = 2 – x
x = 0 and x = 1
The point of intersection of the parabola
y = x2 and the line y = 2 – x is B (1, 1).
We have taken a strip parallel to x axis in
the area OBC and second strip in the area ABC.
z z
1
2− x
0
x2
xy dx dy and hence evaluate the
(U.P.T.U., 2004)
Y
y
2
zz
2− x
0
x
xydxdy =
2
z z z z
z LMMN OPPQ z LMMN OPPQ
z zb g
1
0
y dy
y
0
xdx +
x2
=
y dy
0
2
y
1
0
2
1
2− y
y dy
–y
x=0
and the limits of x in the area ABC are 0 and
2 – y.
1
x=
x=
y
The limits of x in the area OBC are 0 and
∴
A
(0, 2)
0
x2
+
ydy
1
2
2
B (1, 1)
C
x dx
2−y
x=0
X
O
x=1
X
Fig. 4.14
0
LM y OP 1
y 2− y
MN 3 PQ + 2 z e 4y − 4y + y j dy
1
1 L
32
4 1O
1
1 L
y O
4
8−
+ 4− 2+ − P
=
+
=
+
2
y
−
y
+
M
P
M
6
2 M
3 4Q
N 3 4 PQ 6 2 N 3
1
1 L 96 − 128 + 48 − 24 + 16 − 3 O
=
+
PQ
12
6
2 MN
1
=
2
1
y dy +
0
2
1 2
2
1
2
4
3
3 1
1
dy =
2
2
0
2
2
3
1
2
1
1
5
+
6
24
9
3
=
= .
24
8
=
Example 6. Change the order of integration in
Sol. Here the limits are y =
a −x
2
2
z z b g
a
x +2 a
0
a −x
φ x, y dx dy .
2
2
and y = x + 2a and x = 0, x = a.
2
2
2
x 2 − a 2 i.e., x + y = a is a circle and the straight line y = x + 2a and the limits
of x are given by the straight lines x = 0 and x = a.
Now, y =
The integral extends to all points in the space bounded by the axis of y the circle with centre
O, and the straight line x = a.
We draw BN, KP perpendiculars to AM.
Now the order of integration is to be changed. For this, we consider parallel strips.
The integral is broken into three parts.
Ist part is BAN, bounded by lines x = 0, x = a and the circle.
272
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
2nd part is KPNB, bounded by lines y = a, y = 2a and
x = 0, x = a.
(a, 3a)
M
x = y – 2a
3rd part is triangle KPM bounded by the lines y = 2a,
y = x + 2a, x = a.
z dz ib g
z z b g
zz b g
x +2 a
a
Hence
P
φ x, y dx dy
0
=
(0, 2a)
K
a2 − x 2
φ x, y dy dx +
0
=+
zzb g
a
a
a −y
2
2
3a
a
2a
y−2a
2a
a
a
0
φ x, y dy dx
φ x, y dy dx .
B
A
O
(2a, 0)
N (a, a)
Example 7. Change the order of integration and hence
evaluate
zz
a
0
Sol.
y 2 dx dy
a
ax
x=0
y −a x
4
2 2
.
x=a
Fig. 4.15
The limits are y2 = ax, y = a and x = 0,
x = a, y = 0 to y = a
2
y
.
a
By changing the order of integration. Hence, the given integral,
and x varies from x = 0 to x =
z z
a
a
x= 0
y = ax
y 2 dx dy
y −a x
4
2 2
z z
zz F I
y 2 dy dx
a
y2 a
y =0
x =0
y −a x
1 a
=
a 0
y2
a
0
y 2 dy dx
=
z
z
1
y
=
a 0
π
=
2a
z
B
A
y
x=—
a
x=0
2
F I OP dy
GH JK PQ
y2
a
O
0
[sin–1 (1) – sin–1 (0)]
a 2
y dy
0
y=a
2
2 2
1 2
−1 ax
y sin
y2
a0
a 2
Y
2 2
GH ya JK − x
LM
MN
a
=
4
F I
GH JK
3
π y
=
2a 3
dy
y=0
X
Fig. 4.16
a
0
π
πa
.
(a3) =
6a
6
2
=
Example 8. By changing the order of integration of
z
∞ sin px
0
x
dx =
π
·
2
zz
∞ ∞
e − xy sin px dx dy
0 0
show that
(U.P.T.U., 2004, 2008)
273
MULTIPLE INTEGRALS
Sol. We have
zz
∞
0
∞ − xy
e sin px dx dy
0
z RSTz UVW
z LMMN OPPQ
z
OP
z LMNz
Q
LM
OP
b
g
z MN
PQ
LM F I OP
z
MN GH JK PQ
∞
=
0
∞
sin px
0
e − xy dy dx
e − xy
sin px
0
−x
∞
=
Again
zz
∞ sin px
=
∞
∞ − xy
0
0
e
sin px dx dy =
∞
∞
0
0
=
dx
0
...(i)
e − xy sin px dx dy
0
− e − xy
p cos px + y sin px
p2 + y 2
∞
p
0
p +y
∞
=
∞
dx
x
0
As the limits are constants.
2
2
dy =
tan −1
y
p
∞
=
0
∞
dy
0
π
2
...(ii)
Hence from (i) and (ii)
∞
π
sin px
. Hence proved.
dx =
0
2
x
Example 9. Change the order of integration in the following integral and evaluate:
z
zz
4a 2 ax
0
x2
4a
dydx ⋅
x2
Sol. The limits are y =
and y = 2 ax and
4a
x = 0, x = 4a.
From the parabola y2 = 4ax i.e., x =
parabola x2 = 4ay i.e., x = 2
to 2 ay .
∴
zz
4a
0
2 ax
dy dx
x2
4a
Y
A
y2
and from
4a
2
y
y2
ay i.e., x varies from
4a
=
zz
z FGH
4a
2 ay
0
y2
4a
4a
0
2
dy dx
2 ay −
3
2
y2
4a
4a
3
0
3
2
3
(4a, 4a)
x = 4ay
O
I dy
JK
LM
O
y
y P
−
= M2 a .
MMN 23 12a PPP
Q
4
64a
=
a . a 4 af −
3
12a
=
=4
ax
Fig. 4.17
X
274
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
=
32a 2
16 a 2
–
3
3
=
16 a 2
.
3
zz a f
a y
Example 10. Change the order of integration evaluate
0 y
2
y
a−x
a
ax − y 2
dx dy .
y2
i.e., y2 = ax, x = y and y = 0, y = a
a
intersection point of parabola and line y = x is A(a, a).
Y
Sol. Here the limits are x =
Here x varies from 0 to a and y varies from x to
∴
zz a f
zz a f
a y
y
2
a−x
a
ax
0 y a
=
=
=
=
0 x
z
a
0
ax − y 2
a−x
y=
Q
dx dy
P
O
y
ax − y 2
y=
ax
A(a, a )
x
X
(0,0)
dy dx
ax − y 2 = t
ax
12
−2 ydy = dt
− ax − y 2
dx
12
x
y
dy = − ax − y 2
2
ax − y
L
a f NM e
1
a−x
2
ax
j OQP
z LMN e j OPQ
za f z a f
z
e
Fig. 4.18
j
a
12
1
dx
0 + ax − x 2
0 a−x
a
0
ax − x 2
dx =
a−x
a
x
0
a−x
dx =
a−x
Let x = a sin2 θ i.e., dx = 2a sin θ cos θ dθ
=
z
x
0
a−x
a . cos θ
π 2
= 2a 0
LM
N
a
a . sin θ .
π 2
0
z
z
= a θ−
2a sin θ cos θ dθ
sin 2 θdθ = 2a
OP
Q
π 2
sin 2θ
2
0
dx
= a
z FGH
π 2
0
1 − cos 2θ
2
LM π − 0OP = aπ .
N2 Q 2
IJ = az π 2 b1 − cos 2θg dθ
K 0
4.6 CHANGE OF VARIABLES IN A MULTIPLE INTEGRAL
Sometimes it becomes easier to evaluate definite integrals by changing one system of variables to
another system of variables, such as cartesian coordinates system to polar coordinates system.
275
MULTIPLE INTEGRALS
Let us consider the transformation of
=
zz b g
S
F x , y dx dy
...(i)
when the variables are changed from x and y to u and v by the relations
x = φ (u, v), y = ψ (u, v)
...(ii)
Let these relations transform the function F (x, y) to F (u, v) to express dx dy in terms of new
variables u, v, we proceed as follows:
First solve equations (ii) for u, v, to get
u = F1 (x, y) and v = F2 (x, y)
..(iii)
Then u = constant and v = constant form two systems Y
v + dv
of curves in the xy-plane. Divide the region S into
elementary areas by the curve u = constant, u + δu
Q¢
v
= constant, v = constant and v + δv = constant.
Let P be the intersection of u = constant and
P¢
Q
v = constant, and Q is the intersection of u + δu = constant
and v = constant.
u + du
Thus, if P is the point (x, y), so that x = φ (u, v),
P
y = ψ (u, v)
Then Q is the point
u
[( φ(u + δu, v), ψ (u + δu, v)]
X
Now using Taylor ’s theorem to the first order of O
approximation, we obtain
Fig. 4.19
δφ
δu
δu
δψ
ψ (u + δu, v) = ψ (u, v) +
δu
δu
φ (u + δu, v) = φ (u, v) +
and
a f
y = ψ au, vf
FG x + δx δu, y + δy δuIJ
H δu
δu K
δy I
F δx
Similarly, P′ is the point G x + δv , y + δvJ
H δv
δv K
δy
δy I
F δx δx
and Q′ is the point GH x + δu δu + δv δv , y + δu δu + δv δvJK
As x = φ u, v
Therefore, the coordinates of Q are
Therefore, to the first order of approximation PQQ′ P′ will be a parallelogram and its area
would be double that of the triangle PQP′.
Hence the elementary area PQP′ Q′ = [2 ∆ PQP′]
= 2×
=
1
2
x
δx
δu
δu
δx
δv
δv
1
x
y
δy
δx
δu 1
δu y +
x+
δu
δu
δy
δx
x + δv y + δv 1
δv
δv
1
y
δy
δu 0
=
δu
δy
δv 0
δv
δx
δu
δx
δv
δy
δu . δuδv
δy
δv
276
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
b g . δu δv = J.δu δv
a f
δ x, y
=
..(iv)
δ u, v
Thus, if the whole region S be divided into elementary areas by the system of curves
F1 = constant and F2 = constant then, we have
lim Σ F (x, y) δS = lim ΣF (u, v) J . δu δv
zz b g
i.e.,
S
F x , y dx dy
=
zz a f
S
F u, v
J du dv
δx
δy
b g = δu δu
J =
δx δy
δau, vf
δ x, y
δv
Hence,
δy
δu
δy
δv
δx
δu
δx
δv
dx dy =
...(v)
δv
or
δx
δu
δy
δu
δx
δv
δy
δv
du dv.
Example 11. Transform to polar coordinates and integrates
zz FGH 11 −+ xx −+ yy IJK dx dy
2
2
2
2
the integral being extended over all positive values of x and y subject to x2 + y2 ≤ 1.
Sol. Put
x = r cos θ, y = r sin θ
δx
δx
δr
δθ
b g = δr δθ = cos θ r sin θ = r
δy δy
sin θ r cos θ
δ a r , θf
δ x, y
then
Therefore, dx dy = rdθ dr and limits of r are from 0 to 1 and those of θ from 0 to
Hence
zz
F 1 − x − y I dx dy
GH 1 + x + y JK
F1 − r I
= z z G
H 1 + r JK r dθ dr
F 1 − r I r dr
π
= 2z G
H 1 + r JK
I =
2
2
2
2
π
1
2
0 0
Now,
2
2
1
2
0
2
suppose r2 = cos φ, 2rdr = – sin φ dφ
=
π
2
z
0 1
π
2
2
FG 1 − cos φ IJ (– sin φ) dφ
H 1 + cos φ K
π
,
2
277
MULTIPLE INTEGRALS
g dφ
=
π
π π
π
−1 .
φ − sin φ 02 =
4 2
4
x+y
R
π
2 1 − cos φ
0
π
4
zz b g
Example 12. Evaluate
zb
=
2
As
FG
H
IJ
K
z af z af
a
0
F x dx = –
0
a
F x dx
dx dy, where R is region bounded by the parallelogram
x + y = 0, x + y = 2, 3x – 2y = 0, 3x – 2y = 3.
(U.P.T.U., 2006)
Sol. By changing the variables x, y to the new variables u, v, by the substitution x + y = u,
3x – 2y = v, the given parallelogram R reduces to a rectangle R* as shown in the Figure 4.20.
a f =
b g
δ v, v
δ x, y
δu
δy
δv
δy
=
1 1
3 −2
=–5
v
v=3
3x
–
2y
=
3x
3
–
2y
=
0
Y
δu
δx
δv
δx
R
O
x
x
+
y=
R*
X
+
y
=
u
O
2
0
u=2
Fig. 4.20
So required Jacobian
J =
b g =– 1
δau, vf
5
δ x, y
Since u = x + y = 0 and u = x + y = 2, u varies from 0 to 2, while v varies from 0 to 3 since
3x – 2y = v = 0, 3x – 2y = v = 3.
Thus, the given integral in terms of the new variables u, v is
zz b g
R
x+ y
2
dx dy =
=
zz
u2
R*
1
5
1
–5
zz
z LMN OPQ
du dv
3 2 2
0 0
u du dv
2
1 3 u3
8
8
3
dv =
⋅ v0 = .
=
5 0 3 0
5
15
z
Example 13.
zz c h
∞ ∞
0
0
e
− x2 + y 2
dx dy by changing to polar coordinates hence, show that
π
.
2
Sol. Let x = r cos θ, y = r sin θ
⇒ x2 + y2 = r2
Here r varies from 0 to ∞
y
and θ = tan–1
x
∞ − x2
0
e
dx =
(U.P.T.U., 2002)
278
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Now, θ = tan–1 0 = 0, θ = tan–1 ∞ =
⇒
Hence,
Let
π
2
θ varies from 0 to
zz
∞ ∞
0
0
e
c
− x2 +y2
= –
1
2
= –
1
2
I =
I =
z
z
∞
0
∞
0
h dx dy =
zz
z
e −x
2
e−y
2
zz
π
∞ − r2
2
e r dr dθ
0 0
a−2rfe
ze
a0 − 1f dθ = 12 z 1 dθ = π4 .
π
∞
2
0 0
π
2
0
π
2
−r 2
dr dθ = –
1
2
π
2
0
−r 2
π
2
0
dx
...(i)
dy
...(ii)
∞
dθ
0
Y
r ®¥
[Property of definite integrals]
Multiplying (i) and (ii), we get
∞ ∞ −ex 2 + y 2 j
π
e
I2 =
dx dy =
.
0 0
4
[As obtained above]
zz
I =
π
=
4
Example 14. Evaluate
O
π
. Proved.
2
zz
1 x
0 0
X
q=0
r=0
Fig. 4.21
x 2 + y 2 dx dy, the transformation is x = u, y = uv.
Sol. Region of integration R is the triangle bounded
by y = 0, x = 1 and y = x
Put x = u, y = uv
δx
δx
δu
δv
Y
B (1, 1)
b g = δu δv
J = Jacobian =
δau, vf
δy δy
δ x, y
1
= v
y=
0
x
R
u = u
In the given region R, x: varies from 0 to 1 while y
varies from 0 to x. Since u = x, so u varies from 0 to 1.
Similarly, since 0 ≤ y = uv ≤ x = u, so v varies from
0 to 1. Thus
=
=
zz
zz
1 x
0 0
1 1
0 0
O (0, 0)
A (1, 0)
Fig. 4.22
x 2 + y 2 dx dy
u 1 + v 2 udu dv =
1
3
z
1
0
x=1
1 + v 2 dv
X
279
MULTIPLE INTEGRALS
L
OP
1 M v e1 + v j 1
+ sin h vP
2
2
3 MM
PQ
N
1 L 2 1
=
M + 2 sin h 1OPPQ.
3 MN 2
bx + yg dx dy, where R is the parallelogram in the xy-plane with
1
2
−1
0
−1
Example 15. Evaluate
zz
2
R
vertices (1, 0), (3, 1), (2, 2), (0, 1) using the transformation u = x + y and v = x – 2y.
(U.P.T.U., 2003)
Sol. Since u = x + y and v = x – 2y
v
∴ u = 1 + 0 = 1, v = 1 – 0 = 1
B (4, 1)
A (1, 1)
v=1
Similarly, u = 4 v = 1, u = 4, v = – 2
and
u = 1, v = – 2
and
2
3
1
3
J =
1
3
−1
3
O
u
1
= −
3
u=1
u=4
1
J du dv =
du dv
3
∴ dx dy =
Thus,
zz b g
x+ y
2
dx dy =
R
zz
z LMN OPQ
=
D
(1, –2)
1
u . du dv
–2 1
3
1
4
2
4
u3
dv =
–2 3
3 1
1 1
V = –2
C (4, –2)
Fig. 4.23
z
1
–2
7 dv = 21.
Example 16. Evaluate the following by changing into polar coordinates
zz
a
0 0
a 2 − y2
y 2 x 2 + y 2 dy dx .
Sol. Here y = 0, y = a and x = 0, x =
(U.P.T.U., 2007)
a2 − y2
Let
x = r cos θ, y = r sin θ ∴ x2 + y2 = r2
Also, when
x = 0, r cos θ = 0 ⇒ r = 0, or θ =
π
2
when
y = 0, r sin θ = 0 ⇒ r = 0 or θ = 0
when
y = a, r sin θ = a ⇒ a sin θ = a ⇒ θ =
Hence, we have
zz
a
0 0
a2 − y2
y 2 . x 2 + y 2 dy dx =
z z
π 2
a
θ=0
r=0
r 2 sin 2 θ r. r dr d θ =
or r = a
z
π 2
0
π
2
sin 2 θ dθ.
z
a 4
0
r . dr
280
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
z
π2
=
a5
5
=
πa
.
20
0
FG 1 − cos 2θ IJ dθ = a LMθ − sin 2θ OP
H 2 K 10 N 2 Q
5
π2
0
5
Example 17. Transform
z z
π 2 π 2
0
0
sin φ
dφ dθ by the transformation x = sin φ cos θ,
sin θ
y = sin φ. sin θ and show that its value is π.
Sol. We have x = sin φ cos θ, y = sin φ sin θ
∴
B (0,1)
x2 + y2 = sin2 φ and θ = tan–1
θ = 0, tan–1
when
y
x
2
O
A
(1,0)
y
=0 ⇒y=0
x
when
y
π
π
, tan–1
=
⇒x=0
x
2
2
φ = 0, x2 + y2 = 0 ⇒ x = 0 or y = 0
when
φ =
π 2
, x + y2 = 1
2
J =
∂x
∂θ
∂y
∂θ
Now
∴
Thus
x +y =1
θ =
when
dφ dθ =
zz
π2
0
π 2
0
sin φ
dφ dθ =
sin θ
=
zz
zz
1
1− y 2
1
cos φ sin φ sin φ
z z FH
zz e
1− y 2
1− y 2
1
dy dx
⋅ dy dx
IK y dy dx
j
−1
OP
x
1 − y PQ
⋅
1
π
2y1 2 = π.
0
2
0
dy
y
=
π
2
2
z
Hence proved.
dy dx
e1 − y j − x
0 0
1− y 2
2
zz
1− y 2
1
=
1 − x2 + y2 ⋅ y
0 0
0
1
1 − sin 2 φ
0 0
LM
=
z MNsin
=
sin φ
1
⋅
⋅ dy dx
sin θ sin φ cos φ
1− y 2
0 0
1
=
∂x
− sin φ.cos θ cos φ cos θ
∂φ
=
= – sin φ cos φ
∂y
sin φ cos θ cos φ sin θ
∂φ
0 0
1
=
Fig. 4.24
1
1
dxdy =
. dxdy
J
sin φ cos φ
1
1 −1 2
0
y
2
x=0
dy
2
y
281
MULTIPLE INTEGRALS
EXERCISE 4.2
Change the order of integration and then evaluate the following double integrals:
1.
2.
3.
4.
5.
6.
7.
8.
zz
z zb g
y 2 dx dy.
LMAns. 160 OP
N 21 Q
x + y dx dy.
Ans. 5
2 4 2y
0 y3
2
4
1
3
zz
zz
2
–1 – x
zz
zz e
0
− 4− 2 y 2
a
x
a
zz
zz
LMA ns .
MN
b g
4− 2 y 2
zz b g zz
1
–2
2 -y
f x , y dx dy +
–y
2
1
y dx dy .
a
a2 −x2
− 2−y
b g
f x , y dx dy
OP
PQ
LMAns. 8 OP
N 3Q
3
2
2 ay − y 2
0
2−y
LMAns. a + a OP
MN 28 20 PQ
LMAns. πa OP
2 PQ
MN
LMAns. πa OP
a PQ
MN
j
x 2 + y 2 dy dx.
0
x
0 x3
f x, y dy dx.
2
2a
1
f x, y dx dy.
2– x 2
0 x
a
LMAns. z z f bx, yg dy dxOP
N
Q
b g
1 y1 3
0 y2
dx dy .
3
a − x − y dy dx.
2
0 0
2
2
Change the following integrals in to polar coordinates and show that
9.
10.
11.
zz
z z e je j
zz d i e j
1
a a
dy dx =
0 y x 2 + y2
1
2x − x 2
x 2 + y 2 dx dy =
0 x
a
πa
.
4
a2 − x 2
y
0 0
x 2 + y2
3π
–1.
8
dx dy =
πa 5
.
20
Evaluate the following by changing the order of integration:
12.
zz
x dy dx .
13.
zz
db − y i xy dx dy.
a bx a
0 0
a
b
0 0
b
2
2
LMAns. 1 a bOP
N 3 Q
LMAns. a b OP
8 PQ
MN
2
2 2
282
14.
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
z z
0
16.
17.
18.
4
xy dx dy.
a − a2 − y 2
zz
4a
15.
LMAns. = 2 a OP
3
N
Q
a+ a2 − y 2
a
2
dy dx .
2
x
4a
0
zz e
1 1
j
x 2 + y 2 dx dy +
0 0
zz
zz
LMAns. = 16 a OP
3 PQ
MN
2 ax
1
1
e
0
e x log y
∞
∞ e−y
0
x
y
zz e
2 2− y
1 0
LMAns. = 5 OP
3 Q
N
j
x 2 + y 2 dx dy .
⋅ dx dy
Ans. = e − 1
dx dy.
Ans. = 1
Change the order of integration in the following integrals:
19.
zz b g
20.
z ez j
a lx
0 mx
LM
MNAns.
V x, y dx dy.
a2 ax f
2a
0
21.
22.
23.
z z
a
e a −x j f bx, yg dx dy.
2
2
x tan a
zz e j
zz d i e
2 x2
y l
LM
MMAns. z z
N
z
z
f x , y dy dx +
LMAns.
MN z
a sin a y cot a
0
0
OP
PQ
OP
ze Vdyj dx + z z Vdy dxPP
Q
b g
2− x2
0 x
x dx dy
x2 + y2
j
zz
b g
la
a
y V x, y dy dx.
ma
l
2a
2a 2a
y2
2a
a
a2 − y2
z z
a
a sin a 0
x 2 + y 2 dx dy.
1 x
1
0
b g
V x, y dy dx +
a a− a2 − y2
a
Vdy dx +
0 y2
0
2a
a+
Vdx dy.
2 ax − x 2
a cos a
zz
ma y m
.
d a − x i f bx , y gdy dx OP
2
2
PQ
LMAns. 9 61 OP
N 105 Q
LMAns. 1 − 2 OP
2 PQ
MN
24. Using the transformation x + y = u, y = uv show that
zz
1 1− x
0 0
25.
zz
b g
e y / x + y dy dx =
1
(e – 1).
2
2π
, integration being taken over the area of the triangle
105
bounded by the lines x = 0, y = 0, x + y = 1.
b
g
1
[ xy 1 − x − y ] 2 dx dy =
283
MULTIPLE INTEGRALS
zz b g
26.
D
y − x dx dy, D: region in xy-plane bounded by the straight lines y = x + 4,
1
7
1
x+
,y=–
x + 5.
3
3
3
y = x – 3, y = –
zz b g
27.
R
x+y
De
dx dy. R: parallelogram in the xy-plane with vertices (1, 0), (3, 1), (2, 2), (0, 1).
Ans. 21
zz b
28.
2
Ans. 8
gb
g
x − y / x + y dx dy, D: triangle bounded by y = 0, x = 1 and y = x. Use x = u – uv,
LM ee − 1j OP
MNAns. 4 e PQ
2
y = uv to transfrom the double integrals.
29. Find
zz
π
a
0
0
r 3 sin θ cos θ dr dθ by transforming it into cartesian coordinate.
Ans. = 0
(U.P.T.U., 2007)
4.7 BETA AND GAMMA FUNCTIONS
The first and second Eulerian Integrals which are also called “Beta and Gamma functions”
respectively are defined as follows:
z
z
β (m, n) =
0
n =
and
a f
1
x m −1 1 − x
∞ − x n −1
0
e x
n −1
dx
dx
[U.P.T.U. (C.O.), 2003]
β (m, n) is read as “Beta m, n” and n is read as “Gamma n”. Here the quantities m and
n are positive numbers which may or may not be integrals.
4.7.1 Properties of Beta and Gamma Functions
(a) The function β (m, n) is symmetrical w.r.t. m, n i.e.,
β (m, n) = β (n, m)
we have
putting
β (m, n) =
z
1
0
n −1
dx
1 – x = t ⇒ dx = – dt
= −
=
⇒
a f
x m −1 1 − x
za f
0
1
1−t m
za f
1
0
1– x
m−1
β (m, n) = β (n, m) .
−1
⋅ tn−1 dt =
⋅ x n−1 · dx =
za f
1
0
z
−1
1− t m
⋅ tn−1 dt
a f
1 n−1
m −1
x
1–x
0
z
z
F By f atf dt = f axf dxI
H
K
b
a
b
a
284
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
(b) Evaluation of Beta Function β (m, n)
m−1 n −1
am + n − 1f
β (m, n) =
We have
β (m, n) =
z
1
0
m n
=
(U.P.T.U., 2008)
m+ n
x m−1 (1 – x)n–1. dx
Lx
O an − 1f x
= M a1 − xf P +
(1 – x)
· dx
m
Nm
Q
(Integrate by part)
n − 1f
a
x (1 – x) . dx
β (m, n) =
m
z
1
n−1
0
or
z
1
1
m
n –2
0
m
n–2
0
m
Again integrate by parts above, we get
z
an − 1f . an − 2f
1
x m+1 (1 – x)n – 3. dx
0
m
m+1
Continuing the above process of integrating by parts
β (m, n) =
an − 1fan − 2f...2 ⋅ 1
m am + 1f... am + n − 2f
β (m, n) =
a
β (m, n) =
or
f a
n −1
z
1
0
x m + n− 2 . dx
fa
f
m m + 1 ... m + n − 2 m + n − 1
n is a positive integer
In case n is alone a positive integer.
Similarly, if m is positive integer, then β (m, n) = β (n, m)
m−1
a f a
β (n, m) =
...(i)
f
...(ii)
n n + 1 ... n + m − 1
In case both m and n are positive integer, then multiplying (i) numerator and denominator
by 1·2·3... (m – 1) or (ii) by 1·2·3... (n – 1), then, we get
f
a
f a
fa
f
1 ⋅ 2 ⋅ 3... m − 1 ⋅ m m + 1 ... m + n − 1
m−1 n−1
β (m, n) =
or
a
1 ⋅ 2 ⋅ 3.... m − 1 n − 1
β (m, n) =
m+ n−1
=
m n
m+n
.
(c) Evaluation of Gamma Function
n
= (n – 1)
z
∞
we know that
n
=
n − 1 or
n+1 = n n =
x n−1 . e − x · dx
0
Integrating by parts keeping xn–1 as first function
n
=
− e – x ⋅ x n −1
∞
0
z
∞
+ (n – 1)
0
x n − 2 ⋅ e − x dx
n
285
MULTIPLE INTEGRALS
L
= M– lim e ⋅ x
N
–x
z
OP
Q
n− 1
∞
+ 0 + (n – 1)
x →∞
x n − 2 . e − x dx
...(i)
0
As lim e – x . x n−1 = 0
x→ 0
n−1
lim e–x · xn–1 =
But
x→∞
=
lim ·
x→∞
x
ex
= lim
1
1
x→∞
1
lim
1
+ n−2 +
...
n−1
2 ⋅ x n− 3
x
x
x→∞
∴ From (i), we get
z
∞
n
x n −1
x
x2
xn
+
+...+
+...
1+
n
1
2
= 0 + (n – 1)
=
x n − 2 ⋅ e − x dx = (n – 1)
0
⇒
n
za f
∞
x n−1 −1 ⋅ e − x dx
0
n−1 .
= (n – 1)
1
=0
∞
...(ii)
Replace n by n + 1 in equation (ii) then, we get
(d)
n+1
=
n n
n+1
=
n
from (i)
.
...(iii)
n
= (n – 1) (n – 2) .... 3. 2. 1
n
=
n−1
n+1
=
n
replace n by (n + 1), then
.
4.8 TRANSFORMATIONS OF GAMMA FUNCTION
We have
n =
z
n =
z
=
λ
(1) Put x = λy or dx = λ dy
Then from (i), we get
or
z
∞
0
y n−1 e–λy dy =
∞
0
e − x x n −1 dx
∞ − λy
e
z
0
n
∞
0
n
λ
n
bλyg
n −1
...(i)
λ dy
e − λy y n−1 dy
...(ii)
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
(2) Put
xn =
∴ From (i), we get
z
∞ − z1 n
e
0
(3) Put
e–x =
– e–x dx =
Then
z
FG 1 IJ dz
H nK
dz = n n = an + 1f
n =
or
z in (i) then nxn–1 dx = dz and x = z1/n
∞ − z1 n
e
0
...(iii)
t in (i).
1
t
dt and ex =
zb
g (–dt)
L F 1I O dt
= z MlogG J P
N H tKQ
∴ From (i), we get
n =
0
1
n−1
– log t
n−1
1
0
z LMNlogFGH 1t IJK OPQ dt = n
F 1I
1
in (iii), we get
(4) Value of  GH JK . Putting n =
2
2
1
∴
n−1
...(iv)
0
1
2
1
=
2
z
∞ − z2
e
0
dz =
aπ f
.
1
=
2
or
1
π
2
As
z
∞ −x2
0
e
dx =
π
2
4.9 TRANSFORMATIONS OF BETA FUNCTION
β (m, n) =
We know that
(1) Putting
x=
z
1
0
x m−1 (1 – x) n–1 dx
...(i)
1
1
or dx = –
dy
2
1+ y
1+ y
Also
b g
(1 – x) =
1–
1
=
1+ y
y
1+ y
b g
Also, when x = 0 then y = ∞ and when x = 1 then y = 0
∴ From (i), we get
β (m, n) =
=
or
β (m, n) =
z
zb g
zb g
0
∞
∞
F 1 I
GH 1 + y JK
m−1
F y I
⋅G
H 1 + y JK
y n−1 dy
n −1
R| −1 U|
S| b1 + yg V| dy
T
W
2
m − 1 + n −1 + 2
0
1+ y
∞
y n−1 dy
0
1+ y
m+ n
...(ii)
287
MULTIPLE INTEGRALS
(2) Also, as
β (m, n) = β (n, m),
therefore interchanging m and n in (ii), we have
zb g
β (n, m) =
∞
y m −1 dy
0
1+ y
.
m+ n
(U.P.T.U., 2008)
4.10 RELATION BETWEEN BETA AND GAMMA FUNCTIONS
[U.P.T.U. (C.O.), 2002, U.P.T.U., 2003]
z
We know that
∞
n
y n−1 e – xy dy =
0
or
n =
Also
Multiplying both sides of (i) by x
z
z
∞ n n−1 − xy
x y
0
m =
m –1
[ § 4.8 eqn. (ii)]
xn
∞ m−1 − x
x
0
–x
e
e
dy
...(i)
dx
...(ii)
e , we get
n ⋅ x m−1 e − x =
z
x n+ m −1 y n −1 e b
∞
g dy
− y +1 x
0
Integrating both sides with respect to x within limits x = 0 to x = ∞, we have
n
z
∞
0
x m−1 e − x dx =
z a f bg
∞
But
0
x n + m −1 e
− y +1 x
dx =
z LMNz
∞
0
b g
∞ n + m −1 − y +1 x
x
e
dx y n−1 dy
0
an + mf
b1 + yg
OP
Q
...(iii)
m+ n
[by putting λ = 1 + y and ‘n’ = m + n in 4.8 on (ii)]
Using this result in (ii), we get
m =
n
=
=
∴
4.11
β (m, n) =
z a
∞
0
f 1 +y y
b g
n+m ⋅
n−1
m +n
dy
bm + ng z b1y+ ygdy
∞
n −1
m+ n
0
bm + ng ⋅ βbm, ng
m⋅ n
m+n
.
SOME IMPORTANT DEDUCTIONS
(1) To prove that
We know that
n
π
(1 − n) = sin n π
β (m, n) =
(U.P.T.U., 2008)
m⋅ n
am + nf .
288
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Putting m + n = 1 or m = (1 – n), we get
a f = β (n, 1 – n)
n⋅ 1− n
1
zb g
zb g
β (m, n) =
But
∴
y n−1 dy
0
1+ y
m+ n
As
1+ y
0
π
sin nπ
n (1 − n) =
(3) To prove that
1+ x
π
sin n π
(1 + n) (1 − n) =
z
As
1+n = n n
F m + 1I F n + 1I
H 2 K H 2 K·
=
F m + n + 2I
2
H 2 K
β (p, q) = z x a1 − xf dx
π
2 sin m θ cos n θ dθ
0
1
We know that
0
dx =
.
nπ
·
sin nπ
π
We have
n . (1 – n) =
sin nπ
Multiplying both sides by n, we get
nπ
n n (1 − n) =
sin nπ
nπ
·
(1 + n) (1 − n) =
sin nπ
(2) To prove that
z
∞ x n−1
π
,n<1
sin n π
=
or
∞
∞ y n− 1 dy
β (n, 1 – n) =
∴ From (i), we have
...(i)
p −1
q −1
...(i)
0
=
q⋅ q
b p + qg ⋅
Putting x = sin2θ and dx = 2 sin θ cos θ dθ, then, we get from (i)
β(p, q) =
z e
z
π2
0
sin 2 θ
π2
∴
z
j e1 − sin θj
p −1
2
q −1
2 p −1
θ cos 2 q −1 θ dθ
= 2 0 sin
π2
0
sin 2 p −1 θ cos 2 q −1 θ dθ =
=
1
β(p, q)
2
p⋅ q
2
bp + qg
. 2 sin θ cos θ dθ
289
MULTIPLE INTEGRALS
Putting 2p – 1 = m and 2q – 1 = n,
FG m + 1 IJ and q = FG n + 1IJ , we get
H 2 K
H 2K
F m + 1I F n + 1I
z sin θ cos θ dθ = H2 2F mK+ nH + 22 I K .
H 2 K
or
p =
π2
m
n
0
4.12 DUPLICATION FORMULA
FG m + 1 IJ = aπf . 2m .
H 2K 2
F 1I
Hence show that β (m, m) = 2
β G m, J .
H 2K
Since
z sin θ cos θ dθ = 2 amamf +annff
To prove that
m
2 m− 1
(U.P.T.U., 2001)
1–2m
π2
2 m −1
2 n−1
0
Putting 2n – 1 = 0 or n =
z
π2
0
z
π2
0
1
, we obtain
2
sin 2 m −1 θ dθ =
sin 2 m −1 θ dθ =
amf FH 12 IK
F 1I
2 m+
H 2K
m⋅ π
F 1I
2 m+
H 2K
Again putting n = m in (i), we get
a mf
z sin θ cos θ dθ = o2 a2mt f
o amft
1
sin
θ
cos
θ
θ
2
d
=
b
g
z
2 a2mf
2
o amf t
1
sin 2θg
2 dθ =
b
z
2 a 2mf
2
2
π2
2 m −1
2 m −1
0
2
i.e.,
π2
2 m −1
2 m −1
0
2
i.e.,
π2
2m
0
2 m −1
...(i)
...(ii)
LM3 FG 1 IJ = π OP
MN H 2K PQ
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Putting 2θ = φ and 2 dθ = dφ, we get
1
z
z
π
22m 0
2
or
2
2m
sin
π2
0
z
or
o amft
φ dφ =
2 a2 mf
o amft
φ dφ =
2 a2 mf
2
{ amf}
θ dθ =
2 a2mf
2
2 m− 1
2
sin
2 m −1
2 m −1
π2
0
sin 2 m −1
As
z af
a
0
f x dx = 2
z af
π2
0
2
...(iii)
F As f axf dx = f atf dtI
GH z
z JK
b
b
Equating two values of
z
a
π2
0
amf
2 a2mf
m
⇒ Multiplying above by
2
=
a mf ⋅ a π f
F 1I
2 m+
H 2K
1
= π
2
2 m −1
m , we get
m m
2m
1
2
2m ⋅ m
2 2m − 1
21 − 2m m
=
β (m, m) =
Sol. We have
As
FG m + 1 IJ = aπf a2mf
H 2K 2
F 1I
m m GH m + JK =
2
Example 1. Compute:
a
sin 2 m −1 θ dθ from (ii) and (iii), we get
2 2 m−1
Hence,
f x dx
m+
21−2m β
As π =
1
2
1
2
1
2
FG m, 1 IJ
H 2K
a f
As, β m, n =
m n
m+n
FG − 1 IJ , FG − 3 IJ , FG − 5 IJ .
H 2K H 2K H 2K
n (1 − n) =
π
sin n π
...(i)
1
, in (i), we get
2
π
1
3
= –π
=
−
⋅
1
2
2
sin − π
2
(1) Putting n = –
FG IJ FG IJ
H K HK
F
H
I
K
·
291
MULTIPLE INTEGRALS
FG − 1 IJ = − π = − π = – 2 π
H 2K 3
1 F 1I
2
2 H 2K
1
FG − IJ = −2 π .
H 2K
or
⇒
3
, in (i), we get
2
(2) Putting n = –
FG − 3 IJ FG 5 IJ
H 2K H 2K
or
...(ii)
=
F
H
π
I
K
= –
3
sin − π
2
F
H
π
3
sin π
2
I =π
K
FG − 3IJ = π = π = 4 π .
H 2K F 5 I
3 1
3
⋅
π
H 2K
2 2
...(iii)
5
, in (i), we get
2
(3) Putting n = –
FG − 5 IJ FG 7 IJ
H 2K H 2K
=
π
π
=–
= –π
5 I
5 I
F
F
sin − π
H 2 K H sin 2 πK
FG − 5 IJ = π = − π = − 8 π .
H 2K F 7I
5 3 1
15
⋅ ⋅ ⋅ π
H 2K
2 2 2
or
Example 2. Evaluate
z
∞
0
Sol. Putting
dx
·
1 + x4
x4 = y or dx =
z
∞
dx
1 + x4
=
=
z
4x 3
, we have
3
0
Also, we have
dy
z b g
1 −4
y dy
∞
4
0
1+ y
FG 1 − 1IJ
H K
1 ∞y 4
z
4 0
1+ y
dy ⋅
dx
=
π
, we get
sin nπ
dx
0 1 + x4
=
1
⋅
4
∞x
0
z
n− 1
1+ x
∞
π
sin
F πI
H 4K
=
π 2
·
4
...(iv)
292
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 3. Show that
Sol. We have
z
π2
0
z
π2
0
tan n θ dθ
tan n θ dθ =
z
FG IJ
H K
1
1
π sec nπ ⋅
2
2
b g dθ
RS 1 an + 1fUV ⋅ RS 1 a–n + 1fUV
T2 W T2
W
=
U
R1
2 S an − n + 2fV
W
T2
an + 1f ⋅ F1 − n + 1I
H 2K
2
=
=
π2
0
sin n θ cos θ
−n
2 1
=
1
π
⋅
1
2 sin n + 1 π
2
As n
a1 − nf = sinπnπ
a f
1
U
R1
π cosec S an + 1fπ V
=
2
W
T2
1
F π nπ I 1 F nπ I
π cosec G + J = π sec G J . Hence proved.
=
H2 2 K 2 H 2 K
2
Example 4. Show that
Sol. We have
z
π2
0
z a
π2
0
tan θ dθ
f
tan θ dθ =
=
z a
=
1
3 1
,
β
2
4 4
1
2
FG 3 IJ FG 1 IJ = 2 z x dx ·
H 4K H 4K
1+ x
∞
0
f acos θf dθ
RS 1 F 1 + 1I UV RS 1 F − 1 + 1I UV
T2 H 2 KW T2 H 2 KW
=
R 1 F 1 − 1 + 2I UV
2 S
T2 H 2 2 K W
F 3I F 1I
H 4K H 4K
=
2 a1f
1 F 3I F 1I
G J G J
=
2 H 4K H 4K
π2
0
2
−1 2
sin θ
12
FG
H
IJ Œ β (m, n) = m ⋅ n
K
am + nf
4
293
MULTIPLE INTEGRALS
e 3 − 1j
z b g
1 ∞ y 4
dy Œ β(m, n) =
=
2 0 1+ y
=
1
2
z
z e
∞
x −1 ⋅ 4 x 3 dx
0
1 + x4
j
zb g
∞
y m −1 dy
0
1+ y
. Putting y = x4, dy = 4x3 dx
x 2 dx
⋅ Hence proved.
0 1 + x4
Example 5. Show that β(m, n) = β(m + 1, n) + β(m, n + 1).
Sol. We have
= 2
m+ n
∞
(U.P.T.U., 2008)
am + 1f ⋅ n + m ⋅ an + 1f
am + 1 + nf am + n + 1f
R.H.S. =
m m⋅ n
m⋅n n
am + nf am + nf am + nf m + n
m⋅ n L m
n O
+
=
( m + n) MN m + n m + n PQ
=
m⋅ n
=
z
1
Example 6. Show that
0
∴
F 1I
H nK .
=
F 1 + 1I
(1 − x )
n
H n 2K
π
dx
n
dx =
z
= β (m, n). Hence proved.
( m + n)
Sol. Put xn = sin2 θ or x = sin2/n θ
∴
+
dx
(1 − x n )
F2 I
G − 1J
2
sin H n K θ cos θ dθ
n
z
FG
H
2
−1
π/2 sin n
IJ
K θ cos θ dθ
=
F I
H K
=
2 π/2 GH n − 1JK
θ cos0 θ dθ
sin
n 0
2
n
z
0
b1 − sin θg
2
F2 I
RS 1 F 2 − 1 + 1)I UV RS 1 ( 0 + 1) UV
KW T2
W
T2 H n
=
R 1 F 2 − 1 + 0 + 2 I UV
2 S
KW
T2 H n
F 1I ⋅ π
F 1 I ⋅ F 1I
H n K · Hence proved.
1 H nK H 2K
⋅
=
=
n R1 F 2 IU
ST 2 H n + 1K VW n FH n1 + 12 IK
2
n
294
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
4.13
EVALUATE THE INTEGRALS
z
z
∞
(i)
e − ax cos bx ⋅ x m−1dx
0
∞
(ii)
[U.P.T.U. (C.O.), 2004]
e − ax ⋅ sin bx ⋅ x m−1dx
(U.P.T.U., 2003)
0
Proof: Both the above integrals are respectively the real and imaginary parts of
z
∞
e − ax e ibx ⋅ x m −1 dx
0
z
∞
or
e −( a − ib)x ⋅ x m−1 dx
0
z
∞
Now, we have
z
∞
∴
e − λx x n −1 dx
=
e − ( a − ib )x x m−1 dx
=
0
0
n
λn
m
( a − ib) m
( a + ib) m
( a2 + b 2 ) m
Let us put a = r cos θ, b = r sin θ in R.H.S.
∴
z
∞
=
m⋅
e − ax (cos bx + i sin bx)x m−1dx =
m⋅
r m (cos θ + i sin θ)m
r 2m
0
z
∞
or
e − ax cos bx ⋅ x m −1 + ie − ax sin bx ⋅ x m −1 dx
0
m
(cos m θ + i sin mθ)
rm
Equating real and imaginary parts, we get
=
z
z
∞
e − ax cos bx ⋅ x m −1 dx =
0
∞
and
e − ax sin bx ⋅ x m −1 dx =
0
where r =
( a 2 + b 2 ) and θ = tan–1
b
·
a
m
rm
m
rm
cos mθ
sin mθ
|r2 = x2+ y2
295
MULTIPLE INTEGRALS
z
∞
Example 7. Prove that
a2 − b 2
, where a > 0.
(a2 + b2 )2
xe − ax cos bx ⋅ dx =
0
Sol. We have
z
∞
e − ax cos bx ⋅ x m−1 dx =
0
m
rm
cos mθ
Put m – 1 = 1, i.e., m = 2
z
∞
xe − ax cos bx dx =
0
2
r
cos 2θ =
2
1
1 − tan 2 θ
⋅
2
( a + b ) 1 + tan 2 θ
2
FG1 − b IJ
H aK
1
⋅
=
a +b F
GH1 + ba IJK
2
2
2
2
2
2
=
a2 − b 2
( a 2 + b 2 )2
.
Hence proved.
F 1 I F 2 I F 3 I ..... F n − 1I , where n is a positive integer.
H nK H nK H nK H n K
F 1 I F 2 I ... F n − 2 I F n − 1 I
P =
...(i)
H nK H nK H n K H n K
F 1 I F 2 I .... F1 − 2 I F1 − 1 I
P =
H n K H n K H nK H nK
Example 8. Find the value of
Sol. Let
or
Writing in the reverse order,
P =
F1 − 1 I F1 − 2 I .... F 2 I ⋅ F 1 I
H nK H nK H n K H nK
...(ii)
Multiplying (i) and (ii), we get
P2 =
Now, we know that
LM F 1 I F1 − 1 I OP LM F 2 I F1 − 2 I OP ...
N H nK H nK Q N H n K H n K Q
LM F n − 2 I F1 − n − 2 I OP LM F n − 1 I ⋅ F1 − n − 1I OP
N H n K H n KQ N H n K H n KQ
n (1 − n) =
π
sin nπ
F 1 I ⋅ F1 − 1 I = π
H n K H n K sin π
or
...(iii)
n
∴
2
P =
π
π
π
π
...
⋅
π
n−2
n−1
2π
sin
sin
sin
π sin
π
n
n
n
n
...(iv)
296
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
From Trigonometry, we know that
sin nθ
= 2n–1 sin
sin θ
Putting θ = 0
Fθ + π I sin F θ + 2π I ... sin F θ + n − 2 πI sin F θ + n − 1 πI
H nK H n K
H n K H n K
F
H
lim sin nθ = lim n ⋅ sin nθ ⋅ θ
θ→0
θ→0
nθ sin θ
sin θ
= n×1×1= n
∴ n = 2n−1 sin
π
2π
n−2
n−1
π
sin
.... sin
π sin
n
n
n
n
∴
P2 = πn–1
∴
P =
2 n−1
n
(2π) n −1/2
n
F 1 I F 2 I ... F n − 1I = (2π)
H n K H nK H n K
n
( n − 1/2)
Hence
I
K
1
=
2
Example 9. Prove that
Sol. We have
π·
( n) (1 − n) =
Putting
⋅
n =
π
sin n π
1
, we get
2
F 1I F 1I = π
H 2 K H 2 K sin π
2
LM F 1 I OP = π
N H 2KQ
F 1 I = π . Hence proved.
H 2K
2
Example 10. Show that
zz
D
xl −1 y m−1 dx dy =
(l) (m)
(l + m + 1)
al + m
where D is the domain x ≥ 0, y ≥ 0 and x + y ≤ a.
Sol. Putting x = aX and y = aY in the given integral then, we get
I =
zz
D
,( aX) l −1 ( aY)m−1 a 2dXdY
Where D' is the domain X ≥ 0, Y ≥ 0 and X + Y ≤ 1
zz
1 1− X
I = al+m
0
0
X l − 1Y m − 1 dY dX
297
MULTIPLE INTEGRALS
z
1
= al+m
X l −1
l+m 1
l −1
a l + m ( l ) ( m + 1)
al + m
⋅
β(l , m + 1) =
m
( l + m + 1)
m
=
al + m
z
( l) ( m)
( l + m + 1)
x m −1 + x n −1
Sol. The given integral
z
1
I =
0
· Hence proved.
dx .
(1 + x ) m + n
0
m
0
=
1
or
m 1− X
0
0
Example 11. Evaluate
LM Y OP dX = a X (1 − X) dX
m z
NmQ
(1 + x )
z
1
x m −1
dx +
m+ n
0
x n−1
(1 + x ) m + n
I = I1 + I2 (say)
dx
...(i)
1
Putting x =
in I2, we have
z
F 1 I FG − 1 IJ dz
H zK H z K
I = z
LM1 + F 1 I OP
N H zKQ
n −1
1
2
2
m+n
∞
z
z
∞
=
1
z m −1 dz
(1 + z ) m + n
∞
=
∴
1
x m −1 dx
(By definite integral)
(1 + x ) m + n
From equation (i)
z
0
(1 + x)
z
∞
=
0
=
+
m+n
1
x m −1 dx
(1 + x ) m + n
(m) ⋅ ( n)
( m + n)
z
∞
x m−1 dx
1
I =
x m −1 dx
(1 + x ) m + n
= β(m, n)
·
Example 12. Using Beta and Gamma functions, evaluate
z
1
0
FG x IJ dx·
H1− x K
1/2
3
3
(U.P.T.U., 2006)
298
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Sol. Let x3 = sin2 θ ⇒ 3x2dx = 2 sin θ cos θ dθ
∴
z
z
π/2
π/2
sin θ
sin θ cos θ
2
sin 2− 4/3 θ dθ
×2
dθ =
4
/
3
3
θ
cos
θ
sin
⋅
3
0
0
5 1
2 2
⋅ π
2
3 2
3 3
=
3
2/ 3 + 0 + 2
4
2
2
2
3
z
π/2
2
2
=
sin 2/3 θ cos 0 θ dθ =
3
3 0
2
π
2 3
2
⋅
=
9 1 1
3
3 3
=
2
π
3
.
1
3
Example 13. Prove that: β (l, m) β (l + m , n) β (l + m + n , p) =
l m n p
l+m+n+p
Sol. β (l, m) · β (l + m, n) · β (l + m + n, p)
=
l m
l+m
l+m n
·
l+m+n
z
∞
Example 14. Find
l+m+n p
·
=
l+m+n+p
l m n p
l + m+ n+ p
. Proved.
1
3
x 1/2 ⋅ e − x dx ·
0
or x = t3 ⇒ dx = 3t2 dt
Sol. Let x1/3 = t
z
∞
∴
1
3
x 1/2 ⋅ e − x dx =
z
∞
0
0
= 3
z
∞
(a)
z
∞ 9
−1
t 2 ⋅ e −t ⋅ dt
e − ax ⋅ x n−1 cos bx dx =
0
Put a = 0,
z
x
n−1
cos bx dx =
0
Put xn = z so that xn–1 dx =
cos
−∞
0
∞
z
∞
(b)
Sol. (a) We know that
z
=3
9
315
=
π·
2
16
Evaluate—
cos x 2dx
∞
t 7/2 e − t dt
0
0
Example 15.
z
∞
t 3/2 ⋅ e − t ⋅ 3t 2 dt = 3
π 2
x dx
2
(n)cos nθ
2 n 2 , where θ = tan
–1
(a + b )
2
( n)
bn
nπ
cos
2
dz
and x = z 1/n
n
θ = tan −1
F bI
H 0K
= tan −1 ( ∞) =
π
2
F bI
H aK
299
MULTIPLE INTEGRALS
z
∞
Then
z
cos bz1/n dz =
n (n)
b
0
∞
b
0
1
2
z
cos x 2 dx =
0
n
z
∞
(b)
I =
cos
nπ
(By definite integral)
2
...(i)
z
∞
cos
−∞
Putting b =
nπ
2
F 3 I cos π = π ⋅ 1 = 1 π ·
H 2K 4 2 2 2 2
∞
∴
cos
( n + 1)
cos ( bx1/n ) dx =
Putting b = 1, n =
n
πx 2
πx 2
dx = 2 cos
dx
2
2
...(ii) |By definite integral
0
π
1
and n =
in equation (i), we get
2
2
F 3I
H 2 K cos π
F I
H K
4
F πI
H 2K
F 3I
∞
2
z−∞ cos πx2 dx = 2 F πHI21/K2 cos π4
H 2K
z
∞
0
∴ From (ii)
π
cos x 2 dx =
2
1/2
= 2·
1
2 1
π⋅
⋅
= 1.
2
π
2
EXERCISE 4.3
z
z
1
1. Prove that
0
2
3. Prove that
0
z
z
1
5. Show that
0
1
·
x (1 − x ) dx =
280
4
3
x2
dx = 64 2 ·
(2 − x)
15
F log 1 I
H xK
2. Prove that
0
x( 8 − x 3 )1/3 dx = 16 π ·
9 3
0
∞
4. Show that
0
e − st
t
dt =
π
, s > 0.
s
LM F 1 I OP
1 N H nK Q
6. Prove that (1 − x ) dx = ⋅
·
n
2I
F
2
H nK
2
1
n −1
dx =
(n) , n > 0 .
x 2 (1 − x ) 3 dx = 60.
n 1/n
0
1
7. Show that
z
z
z
2
z
2
8. Show that
0
( 4 − x 2 ) 3/2 dx = 3π.
300
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
z
z
3
9. Show that
0
∞
11. Prove that
1+ x
0
π/2
13. Prove that
= π.
( 3 x − x)2
x 2 dx
z
0
6
z
z
1
dx
10. Show that
0
F1I
H 3K ·
=
F 5I
(1 − x )
3
H 6K
3
1
π
=
3 3
12. Prove that
·
0
z
π
x 2 dx
x 2 dx
(1 − x )
3
=
2
·
3
π/2
dθ
×
sin θ . dθ = π .
(sin θ) 0
14. Show that, if n > – 1,
z
∞
1
x n e − k x dx =
2 2
2k n+1
z
0
F n + 1I .
H 2 K
0
Hence or otherwise evaluate
e − k x dx.
2 2
−∞
F 1 I ⋅ F 5I
H 3 K H 6 K = ( π) 2
15. Show that
F 2I
H 3K
z
17. Prove that
19. Evaluate
e
x 3 2 dx =
z
20. Prove that
4.14
∞ −4 x
0
∞ m− 1
x
cos bx dx
0
z
∞
0
and
·
aπ f ·
3
128
x m−1 e − ax dx =
2
1/ 3
z
∞
0
n
2a n
LM F 1 I OP
H 3KQ
16. Show that N
F 1I
H 6K
18. Prove that
x m−1 sin bx dx .
z
∞
0
2
=
x 2 e − x dx =
2
( π) ⋅ 21/3
31/2
·
aπ f ·
4
LMAns. m cos mπ ; m sin mπ OP
N b 2 b 2Q
m
m
·
APPLICATION TO AREA (DOUBLE INTEGRALS)
4.14.1 Area in Cartesian Coordinates
(a) Let the area ABCD is enclosed by y = f(x), y = 0 and x = a, x = b.
Let P(x, y) and Q (x + δx, y + δy) be two neighbouring points on the curve AD whose
equation is y = f(x).
Then the area of the element is δx δy.
Consequently the area of the strip PNMQ.
z
f ( x)
dx dy , where y = f (x) is the equation of AD.
y=0
301
MULTIPLE INTEGRALS
dy
)
Y
y=0
y
dx
,
x=a
.
dx dy
A
(b) In a similar way we can prove that the area
bounded by the curve x = f(y), the y axis and the
abscissae at y = a and y = b is given by
z z
b
f ( y)
y=a
x=0
M
Q
x=b
dy
dx
dy dx
O
B
N
M
C
X
Fig. 4.25
y=b
C
x = f(y)
dx
Q
dy
N
A
(x, y) P
x=a
Y
D
D
+
z z
f (x)
(x
ABCD =
b
+
∴ The required area
P
B
y=a
X
O
Fig. 4.26
(c) If we are to find the area bounded by the two curves y = f1(x) and y = f2(x) and the
ordinates x = a and x = b i.e., the area ABCD in the Figure 4.27 then the required area
z z
b
f1 ( x )
x=a
y = f 2 (x)
dx dy
Y
D
y = f1 (x)
C
dx
x=b
dy
x=a
B
y = f2 (x)
A
O
M
N
X
Fig. 4.27
Example 1. Find the area lying between the parabola y = 4x – x2 and line y = x by the
method of double integration.
(U.P.T.U., 2007)
302
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Y
Sol. OA is the line y = x
and OBAD is the parabola
y = 4x – x2
Solving y = 0 and Eqn. (ii), we find that the
parabola Eqn. (ii) meets the x-axis at O (0, 0) and
D (4, 0).
Solving Eqns. (i) and (ii), we find x = 4x – x2
or x(x – 3) = 0
i.e., x = 0, 3.
Also x = 0 gives y = 0 and x = 3 gives y = 3.
O
x¢
∴ The line (i) meets the parabola (ii) in O (0, 0)
(0, 0)
and A (3, 3) we are to find the area OBAO.
∴ Required area OBAO = area OCABO – area of ∆ OCA.
zz
z
3
=
x=0 y=0
3
=
4x−x2
0
z
( y ) 04 x − x
2
...(i)
B
y=
dx dy
...(ii)
x
A (3, 3)
D (4, 0)
X
C
Fig. 4.28
F 1 × OC × CAI
H2
K
F 1 × 3 × 3I
dx −
H2 K
dx dy −
1
1
9
( 4x − x 2 )dx − ( 9) = ( 2x 2 − x 3 ) 30 −
2
3
2
9
9
1 3 9
= 2(3)2 – ( 3) − = 18 – 9 –
= .
2
2
3
2
=
3
0
Example 2. a2y2 = a2x2 – x4 find the whole area within the curve.
Sol. First of all trace the curve as shown in the Figure 4.29.
Y
Whole area = 4 × area OAC
zz
a
= 4
where
f ( x)
x=0 y=0
dx dy
C
y = f(x) i.e.,
y=
P(x,y) A
X¢
(a2 x 2 − x 4 )
a
O
a
is the equation of the curve
z
a
f (x)
= 4 x=0 y 0
z
a
dx
z
Y¢
Fig. 4.29
4
( a 2 x 2 − x 4 )dx
a 0
4 a
4 π/2
x ( a 2 − x 2 )dx =
a sin θ a cos θ a cos θ dθ
=
a 0
a 0
Putting x = a sin θ
= 4
0
f ( x)dx =
z
= 4a2
z
π/2
0
a
sin θ cos 2 θ dθ =
1
π
4 a2
2
=
=
·
3 1
3
2⋅ ⋅
π
2 2
4a 2 1 ⋅
z
4a 2 1 ( 3 / 2)
2 (5 / 2)
a
(a, 0)
X
303
MULTIPLE INTEGRALS
Example 3. Determine the area of region bounded
by the curves
Y
(0, 4)
y=4
4y =
x2
xy = 2, 4y = x2, y = 4.
[U.P.T.U. (C.O.), 2003, 2008]
Sol. Required area of shaded region
zz
z FGH
4
=
y =1 x =
4
=
2 y
1
2 dx dy
y
IJ
K
(2, 1)
(0, 1)
2
2 y − dy
y
xy = 2
F 2 y − log yI
H3
K
LF 16
I 2O
= 2 MH − 2 log 2K − P
3Q
N 3
O
4
Fig. 4.30
3/2
= 2
X
(2, 0)
1
28
− 4 log 2 .
3
Example 4. By double integration, find the whole area of the curve
a2x2 = y3(2a – y).
Sol. Required area = 2 (area OAB)
=
= 2
y 3/2 2 a − y
where x = f(y) =
a
z z
2a
f ( y)
y=0
x=0
dy dx
...(i)
is the equation of the given
curve.
From eqn. (i), required area
=2
z
z
2a
0
2
=
a 0
= 32a2
z
2 2 a 3/2
f ( y)dy =
y
2 a − y dy
a 0
π/ 2
z
(2a sin θ)
2
π/2
0
3/2
A
(0, 2a)
dy
B
2 a − 2 a sin θ ⋅ 4 a sin θ cos θ dθ
(Put y = 2a sin2 θ)
F 5I F 3 I
3 1
1
⋅
π⋅ ⋅ π
H
2K H 2K
2
= 32a ·
= 32a · 2 2
= πa .
2
2 ( 4)
Y
2
sin 4 θ cos 2 θ dθ
2
(U.P.T.U., 2001)
2⋅ 3⋅ 2⋅1
O
X
Fig. 4.31
2
Example 5. Show that the larger of the two areas into which the circle x2 + y2 = 64a2 is
divided by the parabola y2 = 12ax is
16 2
a (8π − 3 ) .
3
Sol. y2 = 12ax is a parabola, whose vertex is (0, 0) latus rectum = 12a, x2 + y2 = 64a2 is a circle,
whose centre is (0, 0) and radius = 8a.
304
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Solving x2 + y2 = 64a2 and y2 = 12ax, we get
2
Y
2
x + 12ax – 64a = 0 or x = 4a.
∴ The x-coordinate of P the point of
intersection of the two curves is 4a. Also the circle
x2 + y2 = 64a2 meets x axis at x2 = 64a2 (Putting
y = 0) i.e., at x = ± 8a. Hence the coordinates of
A and A′ are (8a, 0) and (–8a, 0).
P
X¢
Also for the parabola y = (12ax ) and for
the circle we have
y=
O
A¢ (8a, 0)
N
(4a, 0)
X
P¢
( 64 a 2 − x 2 )
Required area = shaded area in the Figure 4.32.
= area of the circle – area OP' AOP
= π(8a)2– 2 [area OAP]
= 64 πa2 – 2 [area OPN + area PAN]
z
A (8a, 0)
z
Y¢
Fig. 4.32
LM (12ax)dx + (64a − x )dxOP
N
Q
L
1
R1
F x I UV OP
xdx + S x ( 64a − x ) + ⋅ 64a sin
= 64πa – 2 M 2 ( 3 a)
H 8 a K W PQ
2
MN
T2
1
R2 U R
F 1 I UV
= 64πa – 4 ( 3a) S x V − 2S 32a sin (1) − ⋅ 4a ⋅ 48 a − 32a sin
H 2KW
2
T3 W T
= 64πa2 – 2
2
4a
0
z
8a
2
2
4a
4a
2
2
8a
−1
2
0
2
4a
4a
2
3/2
−1
2
−1
2
0
64a 2
32 2
= 64πa2 –
πa
3 − 32πa 2 + 16 a 2 3 +
3
3
128 πa 2 16 a 2
−
3
=
3
3
16 2
a (8π − 3 ) .
=
3
4.14.2 Area of Curves in Polar Coordinate
B
+ dr,
Q (r
q)
q+d
The area bounded by the curve r = f(θ), where
f(θ) is single valued function of θ in the domain
(α, β) and redii vectors θ = α and θ = β is
z z
β
f ( θ)
θ=α
r=0
D
P (r, q)
E
r dθ dr (α < β).
Let O be the pole, OX the initial line and
AB be the portion of the arc the curve r = f(θ)
included between the radii vectors OA and OB
i.e., θ = α and θ = β.
The area of element CDEF = area of sector
OCD–area of sector OFE
O
1
1 2
2
= (r + δr ) δθ − r δθ
2
2
= r δθ δr, neglecting higher powers of δr and δθ.
Hence the element of area in polar coordinates is r δθ δr.
q=b
F
C
A
q=a
dq
q
X
Fig. 4.33
305
MULTIPLE INTEGRALS
∴ The area of the figure OPQO
=
∴
z z
β
f ( θ)
θ =α r = 0
r dθ dr, where r = f(θ) is the curve AB.
z z
The required area =
β
f ( θ)
θ=α
r=0
r dθ dr
.
Example 6. Find the area of the region enclosed by
x+ y =
Sol. We have
x+ y =
a and x + y = a.
...(i)
a
x+ y = a
a− y
...(ii)
⇒ x=
From (i)
x =
From (ii)
e a − yj + y = a
e a − yj
2
Y
(0,a)
B
2
x
+
y
a + y – 2 ay + y = a
2y – 2 ay = 0 ⇒
a
A
(A,0)
ay
y=
O
Squaring on both sides, we get
y2 = ay
=
⇒ y(y – a) = 0 ⇒ y = 0, a
X
x+ y=a
Using the value of y in equation (ii), we get,
x = a, 0.
Fig. 4.34
So, the intersection point of given curve are (a, 0)
and (0, a).
Required area of the shaded region.
zz
z RST
z
a
b a− y g
b a − yg . dy
dx dy =
x
0 x = e a − yj
0
e a − yj
a
=
2
2
z
e a − y j UVWdy = e2 ay − 2yjdy
L 2 a. y − 2y OP = 4 a − a = a . Ans.
= M
2 PQ
3
3
MN 3 2
a
=
0
2
a− y−
32
a
0
2
a
2
2
2
0
Example 7. Find the area of the region enclosed by
the curves y =
3x
and 4y = x2.
x +2
2
3x
y = 2
x +2
...(i)
4y = x2
...(ii)
Sol. We have
2
From (i) and (ii)
⇒
y=
x
4
=
3x
x2 + 2
x4 + 2x2 – 12x = 0
3x
2
x +2
(y =
O (0, 0)
Fig. 4.35
(2,
3
4
A
x
2
4
)
)
306
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
x(x3 + 2x – 12) = 0
x(x – 2) (x2 + 2x + 6) = 0
⇒
x = 0, 2 (real values only)
Using the values of x in (i), we get
3
4
y = 0,
So the intersection points of curves are (0, 0) and
FG 2, 3 IJ .
H 4K
Hence, the required area of the shaded region.
z z
2
=
e
3x x 2 + 2
x = 0 y = x2 4
j
dy dx =
z
2
0
e
3x x 2 + 2
y x2 4
j dx =
z
F 3x − x I dx
GH x + 2 4 JK
2
2
0
2
L3
3
x O
8
= M log ex + 2j −
= blog 6 − log 2g −
P
12 Q
2
12
N2
3
2
2
0
3
6 2
log −
2
2 3
=
2
3
log 3 − .
3
2
Example 8. Calculate the area which is inside the cardioid r = 2 (1 + cos θ) and outside the
circle r = 2.
Y
Sol. r = 2 is a circle centred at origin and of
radius 2. The shaded area in Figure 4.36 is the
region R which is outside the given circle and
Cardioid
inside the cardioid. So r varies from the circle
Circle
r = 2 the cardioid r = 2(1 + cos θ). While θ varies
from – π/2 to π/2. Since R is symmetric about
x–axis, the required area A of the region R is
O
X
given by
=
A =
zz z z
R
= 2
dA =
zz
π/ 2
− π /2 r = 2
π/2 2(1+ cos θ)
0
z
2
π/ 2
= 40
2( 1 + cos θ )
r dr dθ = 2
z
π/2 r 2
0
( 2 cos θ + cos 2 θ) dθ
LM
N
R
rdrd θ
OP
Q
π/ 2
2(1+ cos θ)
2 2
dθ
Fig. 4.36
1
1
= 4 2 sin θ + θ + sin 2θ
= π + 8.
2
4
0
Example 9. Find, by double integration, the area lying inside the cardioid r = a(1 + cos θ)
and outside the circle r = a.
Sol. Required area
= area ABCDA = 2(area ABDA)
= 2
z z
π/2 a (1 + cos θ)
0
a
r dr dθ
307
MULTIPLE INTEGRALS
2
π/2
s
g
q)
0
r=a
co
π/2
= a2
a
(1 + cos θ) 2 − 1 dθ
0
q = p/2
+
= a2
dθ
(1
z
zb
0
A
a( 1+ cos θ )
a
z
FG r IJ
H 2K
r=
= 2
π/ 2
cos 2 θ + 2 cos θ dθ
q=p
F 1 π I a (π + 8).
= a H ⋅ + 2K =
2 2
4
O
B q=0
D
2
2
Example 10. Find, by double integration the area
lying inside the circle r = a sin θ and outside the cardioid
C
r = a (1 – cos θ).
Sol. Eliminating r between the equations of two curves,
sin θ = 1 – cos θ or sin θ + cos θ = 1
Squaring 1 + sin 2θ = 1 or sin 2θ = 0
Fig. 4.37
p
q= —
2
π
2
For the required area, r varies from
π
a (1 – cos θ) to a sin θ and θ varies from 0 to
.
2
∴
2θ = 0 or π ⇒ θ = 0 or
∴ Required area =
=
z z
z LMN OPQ
z
π/2 a sin θ
0
a( 1 − cos θ )
π/ 2
r2
2
0
r = a sin q
r dr dθ
a sin θ
a( 1− cos θ)
π/2 2
1
=
2 0
r = a (1 – cos q)
dθ
O
Fig. 4.38
a sin θ − (1 − cos θ)
2
2
z
z
=
a 2 π/ 2
(sin 2 θ − 1 − cos 2 θ + 2 cos θ ) dθ
2 0
=
a 2 π/ 2
π
( −2 cos 2 θ + 2 cos θ) dθ = a 2 1 −
.
2 0
4
zz
z FH IK
π/2 r for cardioid
θ =0 r for parabola
π/2
F
H
= 2
=
z
0
π/2
0
F
B
E
q
rd
dr
q=0
q
O
A
D
1+ cos θ
l
I
K
q = p/2
r dθ dr
C
1 2
r
dθ
2
1/(1 + cos θ)
q dθ
(1 + cos θ) 2 − 1 / (1 + cos θ)
2
X
dθ
Example 11. Find by double integration the area
lying inside the cardioid r = 1 + cos θ and outside the
parabola
r(1 + cos θ) = 1.
Sol. The required area = area CAFBEDC
= 2 (area DAFBED)
= 2
q=0
Fig. 4.39
308
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
z
Now,
=
π/2
0
(1 + cos θ)2 dθ =
z
π/ 2
(1 + cos θ) 2 dθ −
z
zF
zH
0
π/2
0
z
π 2
0
1
dθ
(1 + cos θ) 2
...(i)
(1 + 2 cos θ + cos2 θ) dθ
π/2
=
=
1
[(1 + 2 cos θ + (1 + cos 2θ)] dθ
0
2
π/2 3
1
+ 2 cos θ + cos 2θ dθ
0
2
2
I
K
1
L3
O
= M θ + 2 sin θ + sin 2θP
4
N2
Q
3F1 I
1
F I 1
π + 2 sin π + sin π − 0
=
H
K
H2 K 4
2 2
π
2
0
=
and
z
π
2 1 / 1 + cos θ 2
0
a
f
3
1
π + 2 = ( 3π + 8)
4
4
z
z
ze j
ze j
...(ii)
FG IJ
HK
π
θ
1 2
sec 4
dθ
dθ =
0
4
2
=
As 1 + cos θ = 2 cos 2
FG 1 θIJ
H2 K
π
1
1 4
θ= φ
sec 4 φ dφ, Putting
0
2
2
π
1 4
1 + tan 2 φ sec 2 φ dφ , [Œ sec2 φ = 1 + tan2 φ ]
2 0
1 1
1 + t 2 dt , where t = tan φ
=
2 0
1
1
1 3
1
1
2
1+
=
t
+
t
=
=
2
3
2
3
3
0
∴ From (i) with the help of (ii) and (iii), we get the required area
1
2
3π + 8 −
=
4
3
3
2 3
4
π+2− = π+ .
=
4
3 4
3
=
IJ
K
FG
H
a
IJ
K
FG
H
...(iii)
f
Example 12. Find the area of the curve r2 = a2 cos 2θ.
Sol. Since the curve is symmetric about origin so the required area of shaded region.
q=p
4
Y
dq
r = a cos2q
X
O
q =-p
4
Fig. 4.40
309
MULTIPLE INTEGRALS
= 2
= 2
z z
LM OP
z NQ
π 4
a cos 2θ
−π 4
r=0
π 4
r2
2 0
−π 4
2
= 2a
z
π 4
0
r dθ dr
a cos 2θ
. dθ =
z
π 4
−π 4
a 2 cos 2θ dθ
LM
N
OP
Q
2a 2
π
π 4
sin 2θ 0 = a 2 sin − sin 0 = a 2 .
2
2
cos 2θ dθ =
Example 13. Find the area included between the curves r = a(secθ + cosθ) and its asymptote
r = a secθ.
Sol. The curve is symmetric about the line θ = 0, so the required area of shaded region.
=
=
z z
z
b
π2
a sec θ + cos θ
−π 2
a sec θ
π 2
−π 2
g
r dθ dr
LM r OP b
MN 2 PQ
2 a sec θ + cos θ
g
dθ =
a sec θ
q=p
2
r=
a
z {b
a2
2
π 2
−π 2
sec θ + cos θ
g − sec θ}dθ
2
)
sq
co
+
cq
(se
(a,o)
(2 a,o)
r = asecq
q =-p
2
= a2
= a2
ze
ze
zb
π2
0
π 2
0
π 2
Fig. 4.41
j
z FGH
g
a2
sin 2θ
5θ +
2
2 0
2 + cos 2 θ dθ = a 2
=
a2
2
=
a 2 5π
5πa 2
−0 =
.
2 2
4
0
LM
N
j
sec 2 θ + 2 sec θ cos θ + cos 2 θ − sec 2 θ dθ
5 + cos 2θ dθ =
OP
Q
π2
0
LM
N
2+
IJ
K
1 + cos 2θ
dθ
2
OP
Q
π 2
2
310
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 14. Find the area included between the curve
x2y2 = a2 (y2 – x2) and its asymptotes.
z
ze
a
ax
a
Required area = 4 × area OAB = 4 0 y dx = 4 0
Sol.
a − x2
2
j
dx
B
Y
Q
x=a
P
X¢
A X
N M (a, 0)
dx
O
Y¢
= 4a
z
Fig. 4.42
π 2 a sin θ ⋅ a cos θ dθ
a cos θ
0
= 4 a2
z
π2
0
,
Putting x = a sin θ
π2
sin θ dθ = 4 a 2 – cos θ 0 = 4 a 2 .
Example 15. Show that the entire area between the
curve
= 2
= 2
z z a f
z e j
z
2a
0
y dx = 2
x3 2
0
2a − x
2 a sin 2 θ
π2
32
π2
0
Q
P
sin 4 θ dθ = 16a 2
= 3πa2 . Hence proved.
x = 2a
dx
X¢
4 a sin θ cos θ
2 a ⋅ cos θ
0
= 16a 2
2a
B
Y
y2 (2a – x) = x3 and its asymptote is 3πa2.
Sol. Required area = 2 × area OAB
N
O
dθ
3 1 π
⋅ ⋅
4 2 2
Y¢
Fig. 4.43
M A (2a, 0)
dx
X
311
MULTIPLE INTEGRALS
EXERCISE 4.4
LMAns. 8 a OP
3 PQ
MN
2
1. Find the area bounded by the parabola y2 = 4ax and its latus rectum.
2. Find by double integration the area of the ellipse
x2 y2
+
Ans. πab
= 1.
a 2 b2
3. Show by double integration that the area between the parabolas y2 = 4ax and x2 = 4ay
16 2
a .
is
3
4. Find the area of the region bounded by quadrant of x2 + y2 = a2 and x + y = a.
LMAns. 1 aπ − 2fa OP
N 4
Q
5. Find by the double integration the area of the region bounded by y = x and y = x.
LMAns. 1 OP
N 10 Q
2
2
3
LMAns. 8 a OP
MN 15 3 PQ
2
6. Find the area of the curve 3ay2 = x (x – a)2.
7. Find the area included between the curve and its asymptotes.
Ans. πa 2
8. Find the area between the curve y2 (2a – x) = x3 and its asymptote.
Ans. 3 πa 2
9. Find the area of the portion bounded by the curve x(x2 + y2) = a (x2 – y2) and its asymptote.
LMAns. 2 a FG π − 1IJ OP
H 4 KQ
N
L 8 ab OP
10. Find the area included between the curves y = 4a (x + a) and y = 4b (b – x). MAns.
3 PQ
MN
LMAns. 1 πa OP
11. Find the whole area of the curve r = a cos 2θ.
N 2 Q
LMAns. 1 πa OP
12. Find the area of a loop of the curve r = a sin 2θ.
N 8 Q
2
2
2
2
2
13. Find the area enclosed by the curve r = 3 + 2 cos θ.
Ans. 11π
14. Show by double integration that the area lying inside the cardioid r = a (1 + cos θ) and
1 2
a (π + 8).
outside the circle r = a is
4
15. Find the area outside the circle r = 2a cos θ and inside the cardioid r = a (1 + cos θ).
LMAns. πa OP
2 PQ
MN
2
16. Find the area included between the curve x = a (θ – sin θ), y = a (1 – cos θ) and its base.
Ans. 3πa 2
312
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
17. Find the area bounded by the curve xy = 4, y-axis and the lines y = 1 and y = 4.
[Ans. 8 log 2]
18. Use a double integral to find the area enclosed by one loop of the four-leaved rose
r = cos 2θ.
LMAns. π OP
N 8Q
4.15
TRIPLE INTEGRALS
Triple integral is a generalization of a double integral. Let a function f (x, y, z) defined at every
point of a three dimensional region V; Divide the region V into n elementary volumes δV1,
δV2, ..., δVn and let (xr, yr, zr) be any point inside the rth sub-division δVr.
n
b
g
Find the sum ∑ f xr , y r , z r δVr ⋅
r =1
zzz b
g
Lim
n
δVr → 0
r =1
b
g
n → ∞ ∑ f xr , yr , zr δVr ⋅
f x, y, z dV =
V
Then
To extend definition of repeated integrals for triple integrals, consider a function F (x, y, z)
and keep x and y constant and integrate with respect to z between limits in general depending
upon x and y. This would reduce F(x, y, z) to a function of x and y only. Thus let
φ (x, y) =
z bb gg b
z2 x , y
g
F x, y, z dz
z1 x , y
Then in φ (x, y) we can keep x constant and integrate with respect to y between limits in
general depending upon x this leads to a function of x alone say
ψ (x) =
z aa ff b g
y2 x
y1 x
φ x, y dy
Finally ψ(x) is integrated with respect to x assuming that the limits for x are from a to b.
Thus
zzz b
V
zzz b
V
g
=
g
=
F x, y, z dV
F x, y, z dV
z z aa ff z bb gg b g
z LMNz aa ff RSTz bb gg b g UVW OPQ
zz z b g
zz z b g
LM
OP
z z MN b g PQ
LM
OP
z z MN b g PQ
b y2 x
z2 x , y
a y1 x
z1 x , y
b
y2 x
z 2 x, y
a
y1 x
z1 x , y
1 1 − x 1− x − y
Example 1. Evaluate the integral
Sol. Let
Ι =
F x, y, z dx dy dz
0 0
0
1 1− x 1− x − y
F x, y, z dz dy dx .
dx dy dz
x + y + z+1
3
⋅
dx dy dz
x+ y + z+1
0 0
0
1 1− x
1
1
−
2 x + y+ z+1 2
3
1 −x − y
=
⇒
0 0
= −
dx dy
0
1 1 1 −x 1
1
−
dx dy
0
0
2
4 x + y +1 2
313
MULTIPLE INTEGRALS
⇒
z z LMMN b g OPPQ
OP
z LMN
Q
z LMN a f OPQ
IJ
z FGH
K
= −
1 1 1 −x 1
1
−
dx dy
2 0 0
4 x + y +1 2
= −
1 1 1
1
y+
dx
x + y+1 0
2 0 4
= −
1 1 1
1
1
dx
1− x + −
2 0 4
2 x+1
= −
1 1 3 1
1
dx
− x−
0
x+1
2
4 4
1− x
LM
a fOPQ
N
1 L3 1
5O
O 1L
= − M − − log 2P = Mlog 2 − P.
2 N4 8
8Q
Q 2N
1 3
1
x − x 2 − log x + 1
2 4
8
= −
Example 2. Evaluate
Sol. We have
zz ze j
zz
4z− x2
4 2 z
0 0
=
=
4 2 z
4 2 z
0 0
z
0
2
x
z LMN
z
e4 z − x j dz dx
2
z0
zz e
0 0
4 1
0
dz dx dy .
0
I =
1
j
4 z − x 2 dz dx
e 4z − x j
2
1
+ ⋅ 4 z sin −1
2
FG x IJ OP
H 4z K Q
2 z
dz
0
a 4z − 4zf + 12 ⋅ 4z sin −1 a1fOPQ dz
L1 O
=
πzdz = π M z P = 8π.
N2 Q
Example 3. Evaluate zzz ex + y + z j dx dy dz where V is the volume of the cube bounded
4
=
1
⋅2 z
0 2
4
4
2
0
2
2
0
2
V
by the coordinate planes and the planes x = y = z = a.
Sol. Here a column parallel to z-axis is bounded by the planes z = 0 and z = a.
Here the region S above which the volume V stands is the region in the xy-plane bounded
by the lines x = 0, x = a, y = 0, y = a.
Z
Hence, the given integral =
=
=
=
zzze
z z FGH
z z FGH
z LMN
j dx dy dz
z I
x z + y z + J dx dy
3K
1 I
x a + y a + a J dx dy
3 K
1
1
O
x ay + y a + a yP dx
3
3
Q
a a a
0 0 0
a a
x +y +z
2
2
2
a
3
2
0 0
a a
z=a
2
0
2
2
O
3
S
0 0
a
0
2
a
3
3
0
X
z=0
Y
Fig. 4.44
314
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
=
z FGH
a
0
x 2a 2 +
IJ
K
1 4 1 4
a + a dx
3
3
LM 1 x a + 1 a x + 1 a xOP = a .
3
3
N3
Q
Example 4. Evaluate zzz b 2x + yg dx dy dz, where V is
a
3 2
=
4
4
5
0
Z
V
the closed region bounded by the cylinder z = 4 – x2 and the
planes x = 0, y = 0, y = 2 and z = 0.
Sol. Here a column parallel to z-axis is bounded by
the plane z = 0 and the surface z = 4 – x2 of the clinder. This
cylinder z = 4 − x2 meets the z-axis x = 0, y = 0 at (0, 0, 4)
and the x-axis y = 0, z = 0 at (2, 0, 0) in the given region.
Therefore, it is evident that the limits of integration for
z are from 0 to 4 –x2, for y from 0 to 2 and for x from 0 to
2.
Hence, the given integral
=
=
=
=
=
=
z z z b g
z z b g
z z b ge j
e j
z z
e j OPQ
z LMN
e j
z
2
4− x 2
2
x=0 y=0 z=0
2
2
x=0 y=0
2
2
x=0 y=0
2
2
x=0 y=0
2
x=0
2
0
z = 4–x
O
z=0
2
(2, 0, 0)
dxdy
(0, 2, 0)
Y
2x + y dx dy dz
2x + y
(0, 0, 4)
Fig. 4.45
4− x 2
z 0 dx dy
2x + y 4 − x 2 dx dy
8x − 2x 3 + 4 − x 2 y dx dy
8xy − 2x 3 y +
2
1
4 − x 2 y 2 dx
2
0
16 x − 4 x 3 + 2 4 − x 2
dx
LM8x − x + 8x − 2 x OP
3 Q
N
16 I 80
F
.
= G 32 − 16 + 16 − J =
H
3K
3
2
=
2
4
3
0
EXERCISE 4.5
e x + y + z dx dy dz.
LMAns. 8 log 2 − 19 OP
9Q
N 3
S = [(x, y, z) : x2 + y2 + z2 ≤ 1, x ≥ 0, y ≥ 0, z ≥ 0].
LMAns. 1 OP
N 48 Q
1. Evaluate the integral
2. Evaluate
zzz
S
z zz
log 2 x x + log y
0
0 0
xyz dx dy dz , where
X
315
MULTIPLE INTEGRALS
3. Evaluate
zzz
x 2 + y 2 dx dy dz, where S is the solid bounded by the surfaces x2 + y2 = z2,
π
Ans.
6
LM
N
S
z = 0, z = 1.
4. Evaluate
5. Evaluate
z z
zzz
1
y = 1− x 2
x=0 y=0
V
z
z=
z=0
OP
Q
LMAns. –5 OP
N 48 Q
e1− x − y j xyz dz dy dx.
2
2
xy 2 dx dy dz, where V is the region bounded between the xy-plane and the
LMAns. π OP
N 24 Q
sphere x2 + y2 + z2 = 1.
6. Evaluate
zzz
zzz
V
dx dy dz over the region V enclosed by the cylinder x2 + z2 = 9 and the planes
Ans. 18π
x = 0, y = 0, z = 0, y = 8.
7. Evaluate
a x x+y x+y+z
e
0 0 0
[Ans.
8. Evaluate
9. Evaluate
10. Evaluate
4.16
dx dy dz and state precisely what is the region of integration.
e
j
1 4a
e − 6e 2 a + 8 e a − 3 . Region of integration is the volume enclosed by
8
the planes x = a, y = 0, y = x , z = 0 and z = x + y]
LMAns. 3 log 3 − 19 OP
9Q
N 8
LMAns. 2πa e 2 − 1jOP
N 32
Q
LMAns. 5a π OP
64 Q
N
z zz
zzz
z z z
log 2 x x + log y x + y + z
e
dx dy dz .
0
0 0
2π
0
π
a 2
4
r sin θ dr dθ dθ.
0 0
π 2
a sin θ
0
0
3
3
a 2 −r 2
a r dθ dr dz .
0
APPLICATION TO VOLUME (TRIPLE INTEGRALS)
Volume of a solid contained in the domain V is given by the triple integral with f (x, y, z) = 1
Hence
and
V=
zzz
dx dy dz
Volume in cylindrical coordinates, V =
Volume in spherical polar coordinates, V =
zzz
zzz
r dr dθ dz
r 2 sin θ dr dθ dφ
Example 1. Find the volume bounded by the elliptic paraboloids
Sol. We have
z = x2 + 9y2 and z = 18 – x2 – 9y2.
z = x2 + 9y2
z = 18 – x2 – 9y2
(U.P.T.U., 2006)
...(i)
...(ii)
316
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
y2
=1
...(iii)
1
3
The projection of this volume on to xy plane is the plane region D enclosed by ellipse (iii)
as shown in the Figure 4.46.
Here
z varies from z1(x, y) = x2 + 9y2 to z2(x, y) = 18 – x2 – 9y2
From eqns. (i) and (ii), we get x2 + 9y2 = 9 ⇒
y varies from y1 (x) = −
9 − x2
to y2(x) =
9
x2
+
2
9 − x2
9
x varies from – 3 to 3.
Thus, the volume V bounded by the elliptic paraboloids
zz z
zz
zz
j
z e
ze j
a f z bx , yg
V = –3 y axf z bx , yg dz dy dx
3
y ax f
2
2
2
2
= –3 y ax f e18 − x − 9 y j − ex + 9 y j dy dx
3
y ax f
2
2
= 2 –3 y a x f e9 − x − 9 y j dy dx
3
y2 x
2
1
1
Z
2
z = x + 9y
2
2
Z = 18 – x – 9y
1
2
–3
9 y − x y − 3y
3
9 −x 2
9
9− x 2
−
9
3
3
9 − x 2 2 dx
–3
O
dx
8
9
x = 3 cos θ so dx = – 3 sin θ dθ
=
Putting
2
= 72
z
π
z
0
= 144
2
= 2
3
2
1
D
X
(3, 0, 0)
2
2
Y
(0,1,0)
x + 9y = 9
Fig. 4.46
sin 4 θ dθ
π 2
0
sin 4 θ cos 0 θ dθ
5 1
= 144 ⋅ 2 2
2 3
3 1
⋅ ⋅π
2
2
⋅
144
=
2⋅ 2
= 27π.
Example 2. Calculate the volume of the solid bounded by the surface x = 0, y = 0,
x + y + z = 1 and z = 0.
(U.P.T.U., 2004)
Sol. Here x = 0, y = 0, z = 0 and x + y + z = 1
and
z varies from 0 to 1– x – y
y varies from 0 to 1 – x
x varies from 0 to 1
317
MULTIPLE INTEGRALS
zzz
z z
z FGH
z LMN
z FGH
dx dy dz =
V =
1
1− x
0
0
dx
=
1
=
0
dx
z z z
z z
1
0
dx
1− x − y
dy z 0
y2
y − xy −
2
I
JK
1− x
1− x − y
0
0
dy
1
1− x
0
0
= dx
dz
b
dy 1 − x − y
g
1− x
0
a f 12 a1 − xf OPQ
F1 x I
1
x I
1 − x − x + x − + x − J dx = z G − x + J dx
=
2
2K
H2 2 K
L x x + x OP = 1 − 1 + 1 = 1 .
= M −
MN 2 2 6 PQ 2 2 6 6
1
=
0
2
dx 1 − x − x 1 − x −
1
2
2
2
1
0
0
3 1
2
0
Example 3. Find the volume of a solid bounded by the spherical surface x2 + y2 + z2 = 4a2
and the cylinder x2 + y2 – 2ay = 0.
Sol. We have
x2 + y2 + z2 = 4a2
...(i)
x2 + y2 – 2ay = 0
...(ii)
Changing it in polar coordinates
x = r cos θ, y = r sin θ
From (i)
z2 = 4a2 – r2
From (ii)
r2 = 2ar sin θ = 0 ⇒ r = 2a sin θ
Hence, r varies from r = 0 to r = 2a sin θ
z varies from z = 0 to z =
4a 2 − r 2
π
θ varies from θ = 0 to θ =
2
and
∴
V=
zzz
z z
π
dx dy dz = 4 2
0
z z
LM
z MN e
ze
π
= 4 2 dθ
0
2 a sin θ
0
dθ
r dr z
2 a sin θ
r =0
rdr
4a 2 − r 2
z
4a 2 − r 2
z z
LM
z MN e
ze
π
= 4 2 dθ
0
0
Volume in 4 quadrant
dz
z= 0
2 a sin θ
0
r dr 4a 2 − r 2
O
O
j PP = 43 − 4a − 4a sin θj + 8a PP dθ
Q
Q
8 × 4a
= 4
– 8a cos θ + 8 a j dθ =
1 − cos θj dθ
3
3
π
32a 3
3
I
F 1
2 G 1 – cos 3θ − cos θJ dθ
=
K
3 z0 H
4
4
π
= 4 2 dθ −
0
π
2
0
3 2 a sin θ
1
4 a2 − r 2 2
3
3
LM
N
32 a L π 2 O
=
− .
3 MN 2 3 PQ
3
0
π
2
0
3
3
OP
Q
π
π
2
0
LM
N
2
2
3
32a 3
1
3
32a 3 π 1 3
2
θ − sin 3θ − sin θ =
+ −
=
3
12
4
3
2 12 4
0
3
2
OP
Q
3
2
3
318
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 4. Find the volume of the cylindrical column standing on the area common to the
parabolas x = y2, y = x2 as base and cut off by the surface z = 12 + y – x2.
(U.P.T.U., 2001)
Sol. Volume =
=
=
=
z z z
z ze
z FGH
z FGH
x
1
dx
0
x
dy
2
1
0
dx
x
12 + y − x 2
0
x2
dx 12y +
1
j
I
JK
y2
− x2 y
2
x
2
y =x
x2
5
x
x4
x + − x 2 − 12x 2 −
+ x4
12
0
Y
12 + y − x 2 dy
1
0
dz
2
2
I dx
JK
F
x
2
x
x I
− x − 4x −
+
= G 8x +
J
4 7
10
5K
H
3
2
7
2
2
3
5
5
2
x =y
1
X
O
0
Fig. 4.47
1 1
1 2
569
+ =
.
= 8 + − − 4−
10 5
4 7
140
Example 5. Find the volume enclosed between the cylinders x2 + y2 = ax and z2 = ax.
Sol. We have x2 + y2 = ax
z2 = ax
=
a
ax − x 2
0
– ax − x 2
dx
a
ax − x 2
0
– ax − x 2
dx
= 2
= 2
z
z
0
– ax − x 2
FH
a
z
ax
− ax
af
ax
dy ax
O
2
IK
z
a
z
z
π
2 a sin 2 θ
0
= 8a 3
π
2 sin 3 θ cos2 θ dθ
0
= 8a
3
0
a − a sin 2 θ ⋅ 2a sin θ cos θ dθ
3
3
3
3
2 = 4a
2 = 16 a ⋅
5 3 3
15
7
⋅
2
2 2 2
2
2
2
Y
x a − x dx
Put x = a sin2 θ so that dx = 2a sin θ cos θ dθ
= 4 a
X
(a, 0)
x + y = ax
ax dx 2 ax − x 2 = 4 a
0
dz
dy z − ax
ax − x 2
ax −x 2
ax dx y − ax− x 2
0
dy
a
= 2 dx
a
Z
dx dy dz
ax
=
zzz
z z
z z
z z
z 2
=
V=
Fig. 4.48
319
MULTIPLE INTEGRALS
Example 6. Find the volume bounded above by the sphere x2 + y2 + z2 = a2 and below by
the cone x2 + y2 = z2.
...(i)
Sol. We have x2 + y2 + z2 = a2
x2 + y2 = z2
...(ii)
Let
x = r cos θ, y = r sin θ
From eqns. (i) and (ii) z2 = a2 – r2
and
z2 = r2
Here
z varies from r to
r varies from 0 to
a2 − r2
a
r 2 = a2 – r 2 ⇒ r =
2
θ varies from 0 to 2π
∴
Volume V =
=
zz z
zz
a
2
2π
0
0
2π
a
0
0
a2 − r 2
r
2
b
dz ⋅ r dr dθ
g
( z)r a − r r dr dθ =
2
2
zz d
2π a
0
2
0
a
2
i
a 2 − r 2 − r ⋅ r dr dθ
OP
LM
O
LM F a I
a −r j
e
a P
1
1
=
P
M
z MM −2 ⋅ FH 3IK − 3 r PP dθ = 3 z MMN−GH 2 JK + a − 2 2 PPQ dθ
Q
N 2
1 F
1 I
1 F
1 I
θf = a G 1 −
a G1 −
a
=
J
J ⋅ 2π
3 H
3 H
2K
2K
2
2π
a
3
2 2
2
2π
3
0
2
3
2
3
3
0
0
2π
0
3
3
e2 − 2 j πa3 .
3
=
Example 7. Find the volume bounded above by the sphere x2 + y2 + z2 = a2 and below by
the cone x2 + y2 = z2.
Sol. The equation of the sphere is x2 + y2 + z2 = a2
...(i)
2
2
2
...(ii)
and that of the cone is x + y = z
In spherical coordinates x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ
From eqn. (i) (r sin θ cos φ)2 + (r sin θ sin φ)2 + (r cos θ)2 = a2
Z
⇒ r2 sin2 θ cos2 φ + r2 sin2 θ sin2 φ + r2 cos2 θ = a2
⇒ r2 sin2θ (cos2 φ + sin2 φ) + r2 cos2 θ = a2
⇒
r2 sin2 θ + r2 cos2 θ = a2
2
2
2
2
x +y +z =a
⇒
r2 (sin2 θ + cos2 θ) = a2
⇒
r2 = a2
Y
⇒
r=a
O
From (ii) (r sin θ cos φ)2 + (r sin θ sin φ)2
= (r cos θ)2
2
2
2
⇒
r2 sin2 θ (cos2 φ + sin2φ) = r2 cos2 θ
x +y =z
X
⇒
r2 sin2 θ = r2 cos2 θ
⇒
tan2 θ = 1
Fig. 4.49
320
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
⇒
π
4
tan θ = 1 ⇒ θ = ±
π
4
The volume in the first octant is one-fourth only. Limits in the first octant r varies 0 to a, θ
π
π
from 0 to
and φ from 0 to
4
2
r = a and θ = ±
Thus
V
zzz
z z LMMN r3 OPPQ
L 1 + 1OP
a
4a
4a
φg M−
=
dφ − cos θ =
b
= 4 z dφ z sin θ dθ ⋅
z
3
3
3
N 2 Q
4a
π 2
F 1 IJ .
2 − 1j = π a G 1 −
e
=
2 3
H 2K
3 2
π π
π
π
a 2
2 4
2 dφ 4 sin θ dθ
=
4
r
sin
θ
dr
d
θ
d
φ
4
= 0 0 0
0
0
∴
π
2
0
π
4
0
3
3
3
0
π
4
0
π
2
0
a
3
3
π
2
0
3
Spherical coordinates:
Fig. 4.50
Example 8. Find by triple integration, the volume of the paraboloid of revolution
x2 + y2 = 4z cut off by the plane z = 4.
(U.P.T.U., 2005)
Sol. By symmetry, the required volume is 4 times the volume in the positive octant.
The section of the paraboloid by the plane z = 4 is the circle x2 + y2 = 16, z = 4 and its
projection on the xy-plane is the circle x2 + y2 = 16, z = 0.
The volume in the positive octant is bounded by z =
x = 0, x = 4.
z z zd i
4
Volume V = 4 0 0
16 − x 2
4
x +y
2
2
dz dy dx = 4
4
zz
x2 + y2
, z = 4, y = 0, y =
4
16 − x 2
4
0 0
z
4
dy dx
x2 + y 2
d
i
4
F 4 − x + y I dy dx = 4 LMF 4 − x I y − 1 ⋅ y OP
= 4z z
GH 4 JK
z MNGH 4 JK 4 3 PQ
LF x I
O
1
= 4 z MG 4 − J 16 − x − e16 − x j P dx
12
MNH 4 K
PQ
4
2
16 − x 2
2
0 0
4
0
4
2
16 − x 2
3
dx
0
2
2
16 − x 2 and
3
2 2
0
321
MULTIPLE INTEGRALS
LM 1 16 − x 16 − x − 1 16 − x OP dx
e jP
12
MN 4 e j
Q
1
2
= 4
16 − x j dx =
e
e16 − x j dx
6
3
z
z
za f
z
= 4
4
2
3
2 2
z
0
4
0
4
2 32
0
π
=
3
2 2
2
2 2
32
⋅ cos 3 θ ⋅ 4 cos θ dθ, where x = 4 sin θ
16
0
3
π
512 2
cos 4 θ dθ
3 0
512 3 1 π
⋅ ⋅ ⋅
=
3 4 2 2
= 32π.
=
Example 9. Find the volume of the solid under the surface az = x2 + y2 and whose base R
is the circle x2 + y2 = a2.
(U.P.T.U., 2008)
Sol. We have az = x2 + y2, x2 + y2 = a2
Here x varies from 0 to a
y varies from 0 to
a2 − x2
2
x2 + y2
.
a
By symmetry, the required volume is 4 times the volume in
positive octant.
and z varies from 0 to
z z z
z LMMN OPPQ
V= 4
a
a2 −x2
x=0 y=0
y3
4 a 2
=
x y+
3
a 0
x2 +y2
a
z=0
0
z
zz
e
j
2
2
x +y =a
2
4 a a −x 2
x + y 2 dx dy
a 0 0
dx dy dz =
a 2 − x2
2
az=x +y
LM
MM
N
2
2
e
a2 − x2
4 a 2 2
dx =
x a − x2 +
3
a 0
Putting x = a sinθ i.e., dx = a cosθ dθ
z
j OPdx
PP
Q
32
z
Fig. 4.51
LM
O
cos θ dθP
N
Q
3. π O
4 La . 3 2 3 2 a . 5 2 1 2O
4a L 1 1
=
⋅ π+
+
M
P
=
M
a NM
2 3
3. 2 3
QP 4 N 2 2 3. 2. 2 PQ
L 2π O πa .
= a M P =
N4Q 2
=
a4
4 4 π2 2
a
sin θ cos 2 θ dθ +
0
a
3
4
3
4
π 2
4
0
3
3
Example 10. A triangular prism is formed by planes whose equations are ay = bx, y = 0 and
x = a. Find the volume of the prism between the plane z = 0 and surface z = c + xy.
[U.P.T.U. (C.O.), 2003]
Sol. Here x varies from 0 to a
bx
y varies from 0 to
a
322
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
z varies from 0 to c + xy
Hence, the volume is
zz z
z LMMN OPPQ
V =
a bx a
c + xy
0 0
0
xy 2
cy +
2
a
=
0
bx a
dx =
0
zz b g
a bx a
dx dy dz =
z
0 0
a
0
c + xy dx dy
F bcx + b x I dx
GH a 2a JK
2 3
2
L bcx + b x OP = bca + b a = ab b 4c + abg.
= M
MN 2a 8a PQ 2a 8a 8
2
2 4 a
2
2 4
2
2
0
1. Find
zzz
V
EXERCISE 4.6
x 2 yz dx dy dz , where V is the volume bounded by the surface x2 + y2 = 9,
LMAns. 648 OP
5 Q
N
z = 0, z = 2.
2. Compute the volume of the solid enclosed between the two surfaces elliptic paraboloids
z = 8 – x2 – y2 and z = x2 + 3y2.
[ Hint: Projection of the volume on to xy - plane is the ellipse
x 2 + 2 y 2 = 4, so limits are z ; x 2 + 3 y 2 to 8 – x 2 – y 2 ; y : ±
.
4 − x2
, x: ± 2]
2
Ans. 8π 2
3. Compute the volume of the solid bounded by the plane 2x + 3y + 4z = 12, xy-plane and
the cylinder x2 + y2 = 1.
LMHint: Limits z: 0 to 1 b12 − 2x − 3yg, x:−1 to 1, y : ± 1 − x OP ⋅
4
N
Q
2
Ans. 3π
4. Find the volume of the solid common to two cylinders x2 + y2 = a2, x2 + z2 = a2.
LMHint: z: a − x , y : a − x OP ⋅
PQ
MN
±
2
±
2
x : ±a
2
2
LMAns. 16 a OP
3 PQ
MN
3
5. Find the volume of the cylindrical column standing on the area common to the parabolas
y2 = x, x2 = y and cut off by the surface z = 12 + y – x2.
(U.P.T.U., 2001)
LMAns. 566 OP
N 140 Q
6. A triangular prism is formed by planes whose equations are ay = bx, y = 0 and x = a. Find
the volume of the prism between the planes z = 0 and surface z = c + xy.
[U.P.T.U. (C.O.), 2003]
LMAns. ab a4c + abfOP
8
N
Q
323
MULTIPLE INTEGRALS
Ans. π
7. Compute the volume bounded by xy = z, z = 0 and (x – 1)2 + (y – 1)2 = 1.
8. Evaluate
zzz
zzz
2 z yz
0 1 0
LMAns. 7 OP
N 2Q
LMAns. FG π − 1IJ OP
N H 2 KQ
xyz dx dy dz .
π π
xy
z
2 2
cos dz dy dx .
0 x 0
x
9.
10. Compute the volume of the solid bounded by x2 + y2 = z, z = 2x.
Ans. 2π
11. Find the volume of the region bounded by the paraboloid az = x2 + y2 and the cylinder
LMAns. πR OP
2a Q
N
4
x2 + y2 = R2.
12. Find the volume of the tetrahedron by the coordinate planes and the plane
4.17
x y z
+ + = 1.
a b c
abc
Ans.
6
LM
N
OP
Q
DRITCHLET’S* THEOREM
(U.P.T.U., 2005)
If V is a region bounded by x ≥ 0, y ≥ 0 and x + y + z ≤ 1, then
zzz
V
blg amf amf
bl + m + n + 1g
xl −1 y m −1z n−1dx dy dz =
The given triple integral may be written as
Ι =
zz z
1 1− x 1− x − y
0 0
0
x l −1 y m −1 z n−1 dx dy dz
Z
C
z = 1–x–y
@y = @x @y @z
O
z=0
B
Y
Fig. 4.52
* Peter Gustav Lyeune (1805–1859), German Mathematician.
x+y=
1
A
X
324
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
LM z OP dx dy
NnQ
x y
b1 − x − yg dx dy
zz
zz
z z ma f r ma f a f r a f
z a fa f z a fa f
1 1 − x l −1 m −1
=
x
0 0
n 1 −x − y
y
0
1 1 − x l −1 m −1
Put
1
n
n 0 0
y = (1 – x) t, so that dy = (1 – x) dt
Then
I =
=
1 1 1 l −1
m
n
x
1 − x t ⋅ 1 − x − 1 − x t 1 − x dx dt
n 0 0
1
1 1 l −1
m + n−1 −1
n +1 −1
x 1−x
dx tm −1 1 − t
dt
=
0
n 0
1
= β l, m + n + 1 ⋅ β m, n + 1
n
a
f a
f
alf ⋅ am + n + 1f ⋅ amf an + 1f
al + m + n + 1 f am + n + 1 f
1 alf ⋅ amf an + 1f
=
n al + m + n + 1f
1 alf ⋅ amf ⋅ n anf
=
n al + m + n + 1f
1
= n
alf ⋅ amf anf
a l + m + n + 1f
=
zzz
Hence,
V
x l − 1 y m − 1 z n − 1 dx dy dz =
3
an + 1f = n anf
alf ⋅ amf anf
al + m + n + 1f
where V is the region given by x ≥ 0, y ≥ 0, z ≥ 0 and x + y + z ≤ 1.
This integral is known as Dritchlet’s integral. This is an important integral useful in
evaluating multiple integrals.
Remark: It is to be noted that S is a region bounded by x ≥ 0, y ≥ 0 and x + y ≤ 1 then
zz
S
x l −1 y m−1 dx dy = =
l m
l+m+1
.
Note 1: The above integral can be generalised for more than three variables, i.e.,
zz z
... x1 1 − ⋅ x2 2 − ⋅ x3 3 − ... xn pn −1 dx1 dx2 dx3 ... dxn =
p
1
p
1
1
p
bp g bp g b p g − − − bp g
1
2
n
3
b p + p + p + − − − − + p + 1g
1
2
3
n
where the region of integration is given by xi = 1, 2, ...., n; such that x1 + x2 + .... + xn ≤ 1.
Note 2: If the region of integration be x ≥ 0, y ≥ 0, z ≥ 0 and x + y + z ≤ h, then
zzz
V
x l −1 y m −1 z n−1 dx dy dz =
blg amf anf
bl + m + n + 1g h.
MULTIPLE INTEGRALS
Example 1. Find the value of
zzz
325
x l −1 y m−1 z n −1 dx dy dz where x, y, z are always positive with
FG x IJ + FG y IJ + FG z IJ ≤ 1.
H aK H bK H cK
FG x IJ = u, FG y IJ = v, FG z IJ = w, we have
Sol. Putting
H aK
H bK H cK
q
p
r
p
q
(U.P.T.U., 2005)
r
x = au1/p, y = bv1/q, z = cw1/r,
F a I uFGH IJK du, dy = F b I vFGH IJK dv, dz = FG c IJ w FGH IJK dw
GH pJK
GH q JK
H rK
F I F I F I F aIu ⋅ b v ⋅ c w
∴ The given integral =
zzz GGH au JJK GGHbu JJK GH cw JK GH pJK q r
∴
1
−1
p
dx =
1
p
al bm c n
=
pqr
l −1
1
q
m−1
1
−1
r
n− 1
1
r
1
−1
p
1
−1
q
1
−1
r
du dv dw
F l − 1I m − 1 F n − 1I
G p JK q GH r JK
uH
v
w
du dv dw , where u + v + w ≤ 1
zzz
F l I F mI F nI
GH pJK GH q JK H r K
F l + m + n + 1I .
GH p q r JK
l m n
=
1
−1
q
ab c
pqr
x 2 y2 z2
+
+
= 1.
a 2 b2 c 2
Sol. The volume in the positive octant will be
Example 2. Find the volume of the ellipsoid
V=
zzz
For points within positive octant,
Put
x2
a2
dx dy dz
x 2 y2 z2
+
+
≤ 1.
a 2 b2 c 2
= u or x = a u , y = b v , z = c w ,
1
∴
1
1
dx =
a −2
b −
c −
du, dy = v 2 dv, dz = w 2 dw
u
2
2
2
V=
abc
8
zzz
abc
⋅
=
8
FG 1 −1IJ FG 1 −1IJ FG 1 −1IJ
K H2 K H2 K
uH 2
v
w
F 1I ⋅ F 1I ⋅ F 1I
H 2K H 2K H 2K
F1 + 1 + 1 + 1 I
H 2 2 2K
e j
3
π
π abc
abc
=
=
3
1
8
6
⋅ ⋅ π
2 2
du dv dw , where u + v + w ≤ 1
326
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
∴
π abc 4
= π abc.
6
3
Total volume = 8 ×
x y z
+ + = 1 meets the axis in A, B and C. Apply Dritchlet’s integral
a b c
to find the volume of the tetrahedran OABC. Also find its mass if the density at any point is kxyz.
(U.P.T.U., 2004)
Example 3. The plane
y
x
z
= u,
= v,
= w, then u ≥ 0, v ≥ 0, w ≥ 0 and u + v + w ≤ 1
a
c
b
Also, dx = a du, dy = b dv, dz = c dw.
Sol. Let
Volume OABC =
=
zzz
zzz
D
abc du dv dw , where u + v + w < 1
zzz
D
= abc
dx dy dz
u1 −1 v 1−1 w 1 −1 du dr dw
D
a1f a1f a1f = abc = abc
a1 + 1 + 1 + 1f 3! 6
Mass = zzz kxyz dx dy dz = zzz k a aufabv facw f abc du dv dw
= abc
D
2 2 2
= ka b c
zzz
2 2 2
= ka b c
D′
D′
2
u 2−1 v 2−1 w 2 −1 du dv dw
2 2
a2 + 2 + 2 + 1f
= ka 2 b 2 c 2
1 ! 1 ! 1 ! ka 2 b 2 c 2
=
.
6!
720
x 2 y2 z2
Example 4. Find the mass of an octant of the ellipsoid 2 + 2 + 2 = 1, the density at any
a
b
c
point being ρ = kxyz.
(U.P.T.U., 2002)
Sol. Putting
x2
y2
z2
=
u,
=
v,
= w and u + v + w = 1
a2
c2
b2
So that
mass
2xdx
a2
= du,
ρdv
=
zzz
=
=
= dv,
2zdz
= dw, then
c2
zzz b g
zzz a fb ga f
z z z FGH IJK FGH IJK FGH IJK
zzz
zzz
b
2
kxyz dx dy dz = k
Mass = k
=
2 ydy
a 2 du
2
b 2 dv
2
xdx ydy zdz ⋅
c 2 dw
2
ka 2 b 2 c 2
8
du dv dw , where u + v + w ≤ 1
ka 2 b 2 c 2
8
u1−1 v1 −1 w 1 −1 du dv dw =
ka 2 b 2 c 2 ka 2 b 2 c 2
=
.
8×6
48
ka 2 b 2 c 2 1 1 1
8
3+1
327
MULTIPLE INTEGRALS
Example 5. Find the volume of the solid surrounded by the surface
FG x IJ + FG y IJ + FG z IJ = 1.
H aK H bK H cK
2
3
2
3
2
3
(U.P.T.U., 2007)
3a
F xI
Sol. Let
u = G J ⇒ x = au ⇒ dx =
u du
H aK
2
F yI
3b
v = G J ⇒ y = bv ⇒ dy =
v dv
H bK
2
3c
F zI
w = G J ⇒ z = cw ⇒ dz =
w dw
H cK
2
2
3
∴
V= 8
3
2
1
2
2
3
3
2
1
2
2
3
3
2
1
2
zzz
dx dy dz = 8
= 27 abc
zzz
3
zzz
3
−1
1
1
1
27abc 2 2 2
u ⋅ v w du dv dw
8
3
−1
−1
u 2 , v 2 , w 2 du dv dw
3 3 3
⋅
2 2 2
= 27 abc ⋅
3 3 3
+ + +1
2 2 2
1 1 1 1 1 1
⋅ ⋅ ⋅
⋅
= 27 abc ⋅ 2 2 2 2 2 2
11
2
3
π2
= 27 abc ⋅
9 7 5 3 1
8⋅ ⋅ ⋅ ⋅ ⋅
π
2 2 2 2 2
=
4πabc
.
35
Example 6. Find the value of
zzz b
planes and the plane x + y + z = 1.
Sol. Here x varies from 0 to 1
y varies from 0 to 1 – x
z varies from 0 to 1 – x – y
Thus
zz z b
1 1 −x
1 −x − y
0 0
0
1
x + y+ z+1
g
3
dx dy dz
1
x + y+ z+1
g
3
dx dy dz the region bounded by coordinate
328
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
z
LM
OP
1
=
−
MN 2bx + y + z + 1g PQ dx dy
OP
1 L
1
1
=
− M
−
dx dy
2 M bx + y + 1 − x − y + 1g
P
+
+
x
y
1
b
g
N
Q
LM 1
OP
1
1
−
= −
2
MN 4 bx + y + 1g PQdx dy
I dx = − 1 FG 1 − x + 1 − 1 IJ dx
1 Fy
1
= −
+
G
2 H 4 bx + y + 1g JK
2 H 4
2 1 + xK
O 1 L1
1 L − a1 − x f
x
1O
= − M
+ − log a1 + xfP = M − log 2 + P
2 MN
8
2
2
2
8Q
N
PQ
zz
zz
1− x− y
1 1− x
(0,0,1)
2
0 0
0
1 1− x
2
0 0
zz
z
2
B
O
1 1 −x
2
0 0
z
1− x
1
0
A
(1,0,0)
x
1
y
(0,1,0)
Fig. 4.53
0
0
1
2
0
=
1
5
.
log 2 −
2
16
Example 7. Find the area and the mass contained m the first quadrant enclosed by the curve
FG x IJ + FG y IJ = 1, where α > 0, β > 0 given that density at any point ρ (xy) is k xy .
H aK H bK
β
α
[U.P.T.U., 2008]
Sol. The area of the plane region is
A =
zz
dx dy
S
FG x IJ = u or x = au
H aK
FG y IJ = v or y = bv
H bK
1
α
Let
1/α
so dx =
a α −1
⋅ du
u
α
1/β
so dy =
b β −1
⋅ dv
v
β
1
β
∴
A =
zz
s
A =
1
1
ab α − 1 β − 1
u
dudv
.v
αβ
ab 1 α 1 β
⋅
αβ 1 1
+ +1
α β
As
zz
x l −1 y m−1dx dy =
s
Now the total mass M contained in the plane region A is :
M =
zz b g
ρ x , y dx dy =
S
zz
k xy dx dy
l
m
l +m+1
329
MULTIPLE INTEGRALS
= k
zz
1
2
au α⋅
S′
aabf
= k
32
αβ
zz
αβ
3
u 2α
ab α − 1 β −1
⋅
⋅v
⋅ du dv
u
αβ
⋅v
2β
−1
1
3
−1
1
du dv
S′
aabf ⋅
= k
32
bv
1
2β
3
3
2α 2β
.
3
3
+
+1
2α 2β
Example 8. Find the mass of the region in the xy-plane bounded by x = 0, y = 0, x + y = 1
with density k xy .
Sol. Mass M contained in the plane region is
L y OP
2k
M = kz z
xy dx dy = k z x M
dx =
x . a1 − x f dx
MN 3 2 PQ
3 z
2k
=
. a1 − xf
x
dx
3
2k F 3 5 I
2k 3 2 5 2
2k 1 2. π ⋅ 3 2 ⋅ 1 2. π
kπ
=
⋅
=
⋅
=
βG , J =
.
3 H 2 2K
3
3
3 ⋅ 2 ⋅1
24
4
1 1− x
1 1 2
0 0
0
3
2
1− x
1 1 2
3 2
0
0
z
1 3 −1
2
0
5
−1
2
EXERCISE 4.7
zzz
x 2 y2 z2
+
+
≤ 1.
a 2 b2 c 2
2. Find the volume enclosed by the surface
1. Evaluate
dx dy dz , where
FG x IJ + FG y IJ + FG z IJ
H aK H b K H cK
2n
2n
2n
= 1.
LMAns. πabc OP
6 Q
N
LMAns. 1 a b c OP
N 6
Q
2 2 2
3. Find the volume of the solid bounded by coordinate planes and the surface
FG x IJ + FG y IJ + FG z IJ
H aK H bK H cK
1
2
1
2
1
2
= 1.
LMAns. abc OP
N 90 Q
4. Find the volume of the solid bounded by coordinate planes and the surface
LM
FG 1 IJ OP
MMAns. abc ⋅ H 2n K PP
MM 12 n 23n PP
N
Q
3
FG x IJ + FG y IJ + FG z IJ
H aK H b K H cK
2n
2n
2n
= 1, n being a positive integer.
2
330
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
5. Evaluate
zzz
x α −1 ⋅ y β −1 z ϒ−1 dx dy dz , where V is the region in the first octant bounded by
V
the sphere x2 + y2 + z2 = 1 and the coordinate planes.
[U.P.T.U. (C.O.), 2003]
LM
α β γ O
MMAns. α2 2β 2ϒ PPP
MN 8 2 + 2 + 2 PQ
FG x IJ 2m + FG y IJ 2n = 1 m, n being positive integers.
H aK H bK
LMAns. ab ⋅ 1 2m 1 2m OP
NM 4mn 1 2 m + 1 2n + 1 QP
LM
1
1 O
P
4
10
ab
y
x
⋅ 4 10 P
Ans.
M
7. Find the area enclosed by the curve FG IJ + FG IJ
= 1.
27
H aK H b K
MM 40
P
N
20 QP
6. Find the area enclosed by the curve
OBJECTIVE TYPE QUESTIONS
A. Pick the correct answer of the choices given below :
1. The volume of the solid under the surface az = x2 + y2 and whose R is the circle
x2 + y2 = a2 is given as
[UPTU. 2008]
π
2a
4 3
πa
(iii)
3
(ii)
(i)
2. The value of triple integral
zz z
1 1
πa 3
2
(iv) None of these
1− x
0 y2 0
x dx dy dz is
4
5
(ii) −
35
38
−17
35
(iii)
(iv)
35
4
3. The volume bounded by the parabolic cylinder z = 4 – x2 and the planes x = 0, y = 0,
y = 6 and z = 0 is
(i) 32.5
(ii) 56
(i)
(iii) 33
(iv) 32
4. The area which is inside the cardioid r = 2 (1 + cosθ) and the outside the circle r = 2
is :
(i) π + 8
(ii) π – 8
(iii) 3π – 7
(iv) None of these
331
MULTIPLE INTEGRALS
B. Fill in the blanks:
1. The volume of the solid bounded by the surfaces; z = 0, x2 + y2 = 1, x + y + z = 3 is ..........
2. In spherical coordinate system dx dy dz = ..........
3. In cylindrical coordinate system dx dy dz = ..........
4. The volume of region bounded by the parabolic cylinder x = y2 and the planes x = z,
z = 0 and x = 1 is ..........
5.
6.
7.
8.
zz
z
z
zz
π
x
0
0
∞
0
y n−1 . e − λy dy = ..........
π 2
0
sin y dy dx = ..........
sin m θ. cosn θ dθ = ..........
π
aθ
0
0
r 3 dθ dr = ..........
C. Indicate True and False for the following statements:
1. (i) Volume in cylindrical coordinates is
(ii) Volume in spherical coordinates is
zzz
zzz
rdrdθ dz .
r 3 sin θ dr dθ dφ .
(iii) The total volume of a solid bounded by the spherical surface x2 + y2 + z2 = 4a2 and
the cylinder x2 + y2 – 2ay = 0 is multiplied by 4.
(iv) In spherical coordinates F(x, y, z) = F(r sinθ cosφ, r sinθ sinφ, z).
2. (i) Parallelopiped, ellipsoid and tetrahedron are regular three dimensional domain.
(ii) In the area of the cardioid r = a (1 + cos θ), r varies from 0 to a(1 + cos θ) and θ
varies from – π to π.
(iii)
zz
x l −1 . y m −1 dx dy =
l
m
l +m+1
(iv) Triple integral is not generalization of double integral.
D. Match the following:
1. (i) β (p, q)
p q
(a)
FG 1 IJ
H 2K
zb g
∞
y p −1
(ii)
p+q
(b)
(iii )
π
(c) β (p, q)
(iv)
π
sin pπ
(d)
0
1+ y
p 1− p
p+ q
⋅ dy
332
2. (i)
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
zz b g
F x, y dxdy
(a)
s
(ii) β(m, n)
z
1
0
a f dx
x m −1 1 − x
n−1
(b) n n
(iii)
n+1
(c) nπ/sin nπ
(iv)
1+n 1−n
(d)
zz a f
F u, v J du dv
s
ANSWERS TO OBJECTIVE TYPE QUESTIONS
A. Pick the correct answer:
1. (ii)
2. (i)
3. (iv)
2. r2 sin θ dr dθ dφ]
3. r dr dθ dz
6.
4. (i)
B. Fill in the blanks:
1. 3π
4.
4
5
5. π
7.
m+1 n+1
2
2
m+n+ 2
2
2
8.
C. True and False:
1. (i) T
2. (i) T
n
λn
a 4π 5
20
(ii) F
(ii) T
(iii) T
(iii) T
(iv) F
(iv) F
D. Match the following:
1. (i) → (b), (ii) → (c), (iii) → (a), (iv) → (d)
2. (i) → (d), (ii) → (a), (iii) → (b), (iv) → (c).
GGG
UNIT
8
Vector Calculus
VECTOR DIFFERENTIAL CALCULUS
The vector differential calculus extends the basic concepts of (ordinary) differential calculus to
vector functions, by introducing derivative of a vector function and the new concepts of gradient,
divergence and curl.
5.1 VECTOR FUNCTION
If the vector r varies corresponding to the variation of a scalar variable t that is its length and
direction be known and determine as soon as a value of t is given, then r is called a vector
function of t and written as
r = f (t)
and read it as r equals a vector function of t.
Any vector f (t) can be expressed in the component form
f (t) = f1(t) i + f2 (t) j + f3 (t) k
Where f1(t), f2 (t), f3 (t) are three scalar functions of t.
r = 5t2 i + t j – t3 k
For example,
where
f1(t) = 5t2, f2(t) = t, f3 (t) = – t3.
5.2 VECTOR DIFFERENTIATION
Let r = f (t) be a single valued continuous vector point function of a scalar variable t. Let O be
the origin of vectors. Let OP represents the vector r corresponding to a certain value t to the
scalar variable t. Then
r = f (t)
...(i)
333
334
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Let OQ represents the vector r + δ r corresponding to the value t + δt of the scalar variable
t, where δt is infinitesimally small.
Then,
Q
...(iii)
f ( t + δt ) − f ( t )
δr
=
δt
δt
Taking the limit of both side as δt → 0.
lim δ r = lim f (t + δt) − f (t)
t→0
δt→0
δt
δt
We obtain
dr ®
δ r = f (t + δt) – f (t)
Dividing both sides by δt, we get
®
dr
—
dt
®
...(i)
r+
r + δ r = f (t + δt)
Subracting (i) from (ii)
P
®
r
O
dr
lim f (t + δt) − f (t) = d f (t)
= δt→0
dt
δt
dt
Fig. 5.1
.
5.3 SOME RESULTS ON DIFFERENTIATION
H
H
d
dr d s
(r − s ) =
− ⋅
dt
dt dt
H
d( sr )
ds
dr
=
r+s
dt
dt
dt
(a)
(b)
(c)
H db da
d( a ⋅ b)
= a⋅
+
⋅b
dt
dt dt
(d)
d
da
db
( a × b) =
×b+a×
dt
dt
dt
(e)
n
s = dadt × (b × cH) + a × FGH dbdt × cIJK + a × FGH b × dcdt IJK ⋅
d
a × (b × c)
dt
Velocity
v = r =
dr
⋅
dt
dv
d2 r
⋅
=
dt
dt 2
Example 1. A particle moves along the curve x = t3 + 1, y = t2, z = 2t + 5, where t is the time.
Find the component of its velocity and acceleration at t = 1 in the direction i + j + 3k.
Acceleration
Sol. Velocity
a =
dr
d =
( xi + y j + zk)
dt
dt
d 3
(t + 1)i + t 2 j + (2t + 5) k
=
dt
2
= 3t i + 2t j + 2 k
335
VECTOR CALCULUS
= 3i + 2 j + 2k , at t = 1.
Again unit vector in the direction of i + j + 3 k is
=
i + j + 3k
d 1 +1 + 3 i
2
2
2
=
i + j + 3 k
11
Therefore, the component of velocity at t = 1 in the direction of i + j + 3 k is
d3i + 2 j + 2ki ⋅ di + j + 3ki = 3 + 2 + 6 =
11
11
d2 r
dt 2
Acceleration
=
11
FG IJ = 6ti + 2 j = 6i + 2 j , at t = 1
H K
d dr
dt dt
Therefore, the component of acceleration at t = 1 in the direction i + j + 3 k is
d6i + 2 j i ⋅ di + j + 3k i = 6 + 2 = 8 ·
11
11
11
Example 2. A particle moves along the curve x = 4 cos t, y = 4 sin t, z = 6t. Find the velocity
and acceleration at time t = 0 and t = π/2. Find also the magnitudes of the velocity and acceleration at any time t.
Sol. Let r = 4 cos t i + 4 sin t j + 6t k
at
t = 0,
at
t =
π
,
2
t = 0,
π
t =
,
2
at
at
H
v
=
4 j + 6 k
H
v
=
−4i + 6 k
|v|
=
16 + 36 =
52 = 2 13
|v|
=
16 + 36 =
52 = 2 13 .
at
d2 r
= – 4 cos t i – 4 sin t j
dt 2
a
t = 0,
= – 4 i
∴ at
t = 0,
Again, acceleration,
a =
π
,
2
π
t =
,
2
at
t =
at
|a|
=
a
=
|a|
=
(−4)2 = 4
– 4 j
(−4)2 = 4.
Example 3. If r = a ent + b e–nt, where a, b are constant vectors, then prove that
d2 r
dt 2
Sol.
− n2 r = 0
nt
−nt
r = ae + be
dr
nt
− nt
= ae ⋅ n + be ⋅ ( −n)
dt
...(i)
336
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
d2 r
dt 2
= ae nt ⋅ n 2 + be − nt ⋅ ( − n)2
d2 r
dt 2
= n 2 ae nt + be − nt = n2r
[From (i)]
d2 r
– n 2 r = 0.
Hence proved.
dt 2
H H H
Example 4. If a , b , c are constant vectors then show that r = at 2 + bt + c is the path of a
point moving with constant acceleration.
⇒
2
r = at + bt + c
Sol.
dr
= 2at + b
dt
d2 r
= 2a which is a constant vector.
dt 2
Hence, acceleration of the moving point is a constant. Hence proved.
EXERCISE 5.1
1. A particle moves along a curve whose parametric equations are x = e–t, y = 2 cos 3t,
z = sin 3t.
Find the velocity and acceleration at t = 0.
[Hint: r = xi + yj + zk ]
Ans. Vel. =
10 , acc. = 5 13
2. A particle moves along the curve
x = 3t2, y = t2 – 2t, z = t2.
Find the velocity and acceleration at t = 1.
H
Ans. v = 2 10 , a = 2 11
3. Find the angle between the directions of the velocity and acceleration vectors at time t of
2
Ans. arc cos t
a particle with position vector r = (t 2 + 1)i − 2tj + (t 2 − 1)k .
2
2t + 1
LM
MN
OP
PQ
4. A particle moves along the curve x = 2t2, y = t2 – 4t, z = 3t – 5 where t is the time. Find
the components of its velocity and acceleration at time t = 1 in the direction i − 3 j + 2 k .
LMAns. 8 14 ; − 14 OP
7
7 PQ
MN
5. If r = (sec t) i + (tan t) j be the position vector of P. Find the velocity and acceleration
of P at t =
π
.
6
LMAns. 2 i + 4 j, 2 (5i + 4 j)OP
N 3 3 33
Q
337
VECTOR CALCULUS
5.4 SCALAR POINT FUNCTION
If for each point P of a region R, there corresponds a scalar denoted by f(P), then f is called a
‘‘scalar point function’’ for the region R.
Example 1. The temperature f(P) at any point P of a certain body occupying a certain region
R is a scalar point function.
Example 2. The distance of any point P(x, y, z) in space from a fixed point (x0, y0, z0) is a
scalar function.
(x − x0 ) 2 + ( y − y0 ) 2 + ( z − z 0 ) 2 ⋅
f(P) =
Scalar field
(U.P.T.U., 2001)
Scalar field is a region in space such that for every point P in this region, the scalar function
f associates a scalar f(P).
5.5 VECTOR POINT FUNCTION
If for each point P of a region R, there corresponds a vector f (P) then f is called “vector point
function” for the region R.
Example. If the velocity of a particle at a point P, at any time t be f (P), then f is a vector
point function for the region occupied by the particle at time t.
If the coordinates of P be (x, y, z ) then
f (P) = f1 (x, y, z) i + f2 (x, y, z) j + f3 (x, y, z) k.
Vector field
(U.P.T.U., 2001)
Vector field is a region in space such that with every point P in the region, the vector
function f associates a vector f (P).
Del operator: The linear vector differential (Hamiltorian) operator ‘‘del’’ defined and
^
denoted as
∇ = i
∂ ^ ∂ ^ ∂
+k
+j
∂x
∂y
∂z
This operator also read nabla. It is not a vector but combines both differential and vectorial
properties analogous to those of ordinary vectors.
5.6 GRADIENT OR SLOPE OF SCALAR POINT FUNCTION
If f(x, y, z) be a scalar point function and continuously differentiable then the vector
∂f ∂f ∂f
∇ f = i ∂x + j ∂y + k ∂z
is called the gradient of f and is written as grad f.
Thus
(U.P.T.U., 2006)
∂f
∂f
∂f
grad f = i ∂x + j ∂y + k ∂z = ∇ f
It should be noted that ∇f is a vector whose three components are
is a scalar point function, then ∇f is a vector point function.
∂f ∂f ∂f
,
,
⋅ Thus, if f
∂x ∂y ∂z
338
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
5.7 GEOMETRICAL MEANING OF GRADIENT, NORMAL
Consider any point P in a region throught which a scalar field
(U.P.T.U., 2001)
f(x, y, z) = c defined. Suppose that ∇f ≠ 0 at P and that there is
a f = const. surface S through P and a tangent plane T; for
instance, if f is a temperature field, then S is an isothermal surface (level surface). If n , at P, is choosen as any vector in the
tangent plane T, then surely
Normal N at P
Z
^
N
df
must be zero.
dS
S
P
df
Since
= ∇f . n = 0
dS
n^
for every n at P in the tangent plane, and both ∇f and n are
non-zero, it follows that ∇f is normal to the tangent plane T
and hence to the surface S at P.
f = Constant
If letting n be in the tangent plane, we learn that ∇f is normal
to S, then to seek additional information about ∇f it seems
logical to let n be along the normal line at P,.
Then
Y
X
Fig. 5.2
df
df
=
, N = n then
ds
dN
df
= ∇f ⋅ N = ∇ f ⋅ 1 cos 0 = ∇ f ⋅
dN
So that the magnitude of ∇f is the directional derivative of f along the normal line to S, in
the direction of increasing f.
Hence, ‘‘The gradient
e∇ f j of scalar field f(x, y, z) at P is vector normal to the surface
df
in that direction.
dN
f = const. and has a magnitude is equal to the directional derivative
5.8 DIRECTIONAL DERIVATIVE
∂f ∂f ∂f
are the derivatives (rates of change) of f
,
,
∂x ∂y ∂z
in the direction of the coordinate axes OX, OY, OZ respectively. This concept can be extended to
Let f = f(x, y, z) then the partial derivatives
define a derivative of f in a "given" direction PQ .
Let P be a point in space and b be a unit vector from
P in the given direction. Let s be the are length measured
from P to another point Q along the ray C in the direction
of b . Now consider
f(s) = f(x, y, z) = f{x(s), y(s), z(s)}
Then
∂f dx
∂f dy
∂f dz
df
+
+
=
∂x ds
∂y ds
∂z ds
ds
s
P
^
b
Q
C
Fig. 5.3
...(i)
339
VECTOR CALCULUS
df
is called directional derivative of f at P in the direction b which gives the rate of
ds
change of f in the direction of b.
Here
dx dy dz i+
j + k = b = unit vector
ds
ds
ds
Eqn. (i) can be rewritten as
Since,
FG i ∂f + j ∂f + k ∂f IJ ⋅ FG dx i + dy j + dz kIJ
H ∂x ∂y ∂z K H ds ds ds K
LF ∂ + y ∂ + k ∂ IJ f OP ⋅ b = ∇f ⋅ b
df
= MG i
ds
NH δx δy δz K Q
...(ii)
df
=
ds
...(iii)
Thus the directional derivative of f at P is the component (dot product) of ∇f in the direction
of (with) unit vector b.
Hence the directional derivative in the direction of any unit vector a is
F I
GH JK
a
df
= ∇f ·
ds
a
Normal derivative
df
= ∇f ⋅ n , where n is the unit normal to the surface f = constant.
dn
5.9 PROPERTIES OF GRADIENT
Property I: ( a ⋅ ∇) f = a ⋅ (∇f )
Proof:
L.H.S. = ( a ⋅ ∇) f
RSa ⋅ FG i ∂ + j ∂ + k ∂ IJ UV f
T H ∂x ∂y ∂z K W
R ∂ ∂ ∂U
= S( a ⋅ i ) ∂x + ( a ⋅ j ) ∂y + ( a ⋅ k ) ∂z V f
T
W
=
∂f
∂f
∂f
+ ( a ⋅ k)
= ( a ⋅ i) + ( a ⋅ j)
∂x
∂y
∂z
...(i)
b g
F ∂f ∂f ∂f I
= a ⋅ G i ∂x + j ∂y + k ∂z J
K
H
R.H.S. = a ⋅ ∇f
∂f
∂f
∂f
= ( a ⋅ i ) ∂x + ( a ⋅ j ) ∂y + ( a ⋅ k ) ∂z
From (i) and (ii),
baH ⋅∇g f = aH ⋅ b∇ f g .
...(ii)
340
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Property II: Gradient of a Constant
The necessary and sufficient condition that scalar point function φ is a constant is that
∇φ = 0.
Proof: Let
φ(x, y, z) = c
∂φ
∂φ
∂φ
=
=
=0
∂y
∂x
∂z
Then,
∂φ
∂φ
∂φ
+j
+ k
∂y
∂x
∂z
= i.0 + j.0 + k.0
= 0.
Hence, the condition is necessary.
Conversely: Let
∇φ = 0.
∴
∇φ = i
∂φ
∂φ
∂φ
+j
+ k
= 0.i + 0.j + 0.k.
∂y
∂x
∂z
Equating the coefficients of i, j, k.
∂φ
∂φ
∂φ
= 0.
= 0,
=0
On both sides, we get
∂y
∂x
∂z
⇒ φ is independent of x, y, z
⇒ φ is constant.
Hence, the condition is sufficient.
Then,
or
i
Property III: Gradient of the Sum of Difference of Two Functions
If f and g are any two scalar point functions, then
∇ (f ± g) = ∇f ± ∇g
grad (f ± g) = grad f ± grad g
Proof:
∇ (f ± g) =
FG i ∂ + j ∂ + k ∂ IJ ( f ± g)
H ∂x ∂y ∂z K
∂
∂
∂
= i ∂x ( f ± g ) + j ∂y ( f ± g) + k ∂z ( f ± g)
FG ∂f ± ∂g IJ + j FG ∂f ± ∂g IJ + k FG ∂f ± ∂g IJ
H ∂x ∂x K H ∂y ∂y K H ∂z ∂z K
F ∂f + j ∂g + k ∂f IJ ± FG i ∂g + j ∂g + k ∂g IJ
= Gi
H ∂x ∂y ∂z K H ∂x ∂y ∂z K
= i
∇(f ± g) = grad f ± grad g.
Property IV: Gradient of the Product of Two Functions
If f and g are two scalar point functions, then
∇(fg) = f∇g + g∇ f
or
grad (fg) = f (grad g) + g (grad f).
Proof:
∇(fg) =
FG i ∂ + j ∂ + k ∂ IJ ( fg)
H ∂x ∂y ∂z K
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VECTOR CALCULUS
= i
∂
∂
∂
( fg) + j
( fg) + k
( fg)
∂x
∂y
∂z
FG ∂g + g ∂f IJ + j FG f ∂g + g ∂f IJ + k FG f ∂g + g ∂f IJ
H ∂x ∂x K H ∂y ∂y K H ∂z ∂z K
F ∂g + j ∂g + k ∂g IJ + g FG i ∂f + j ∂f + k ∂f IJ
= f Gi
H ∂x ∂y ∂z K H ∂x ∂y ∂z K
= i f
∇(fg) = f∇g + g∇f.
Property V: Gradient of the Quotient of Two Functions
If f and g are two scalar point functions, then
FG f IJ = g∇f − f∇g , g ≠ 0
H gK
g
F f I g(grad f ) − f (grad g) , ≠ 0.
grad G g J =
H K
g
F fI F ∂ ∂ ∂I FfI
∇ G g J = G i ∂x + j ∂y + k ∂z J G g J
KH K
H K H
∂ FfI
∂ FfI
∂ F fI
+j
+k
= i
J
J
G
G
G J
∂y H g K
∂z H g K
∂x H g K
∇
or
2
2
Proof:
g
= i
∂f
∂g
∂f
∂g
∂f
∂g
g
−f
−f
g −f
y
y
∂
∂
∂x
∂x + j
+ k ∂z 2 ∂z
g2
g2
g
LM ∂f + j ∂f + k ∂f OP − f LMi ∂g + j ∂g + k ∂g OP
N ∂z ∂y ∂z Q N ∂x ∂y ∂z Q
g i
=
∇
FG f IJ = g∇f − f∇g .
H gK
g
g2
2
Example 1. If r = xi + yj + zk then show that
(i)
∇( a ⋅ r) = a , where a is a constant vector
(ii)
grad r =
(iii) grad
r
r
1
r
= – 3
r
r
grad rn = nrn – 2 r , where r = r .
H
Sol. (i) Let
a = a1 i + a2 j + a 3 k, r = xi + yj + zk ,
(iv)
then
a⋅r
=
( a1i + a2 j + a3 k) ⋅ ( xi + yj + zk) = a1x + a2y + a3z.
(U.P.T.U., 2007)
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
∴
FG i ∂ + j ∂ + k ∂ IJ (a x + a y + a z)
H ∂x ∂y ∂z K
∇( a ⋅ r) =
1
2
3
H
a1 i + a 2 j + a 3 k = a . Hence proved.
=
∂ 2
( x + y 2 + z 2 )1/2
∂x
x
x =r
= Σi 2
2
2 1/2 = Σi
r
(x + y + z )
grad r = ∆r = Σi
(ii)
Hence,
xi + yj + zk
r
= = r .
r
r
∂ 1
1
−1 r
r = 2
= ∇
=
∂r r
r
r r
grad r =
(iii)
grad
F 1I
H rK
FI
HK
FI
HK
r
= − 3 . Proved.
r
(iv) Let r = xi + yj + zk.
∂ 2
2
2 n/2
grad rn = ∇rn = Σi ( x + y + z )
∂x
= n (x2 + y2 + z2)n/2–1 xi + yj + zk
Now,
d
2
2
2 (n–1)/2
= n(x + y + z )
= nr n−1
i
exi + yj + zkj
ex + y + z j
2
2
2 1/2
r
r
= nr n−2 r .
Example 2. If f = 3x2y – y3z2, find grad f at the point (1, –2, –1).
Sol.
(U.P.T.U., 2006)
FG i ∂ + j ∂ + k ∂ IJ ( 3x y − y z )
H ∂x ∂y ∂z K
∂
∂
∂
= i c 3x y − y z h + j ( 3x y − y z ) + k ( 3x y − y z )
∂x
∂y
∂z
H
grad f = ∇f =
2
2
3 2
3 2
2
3 2
2
3 2
= i(6xy) + j(3x2 – 3y2z2) + k(–2y3z)
grad φ at
(1, –2, –1) = i(6)(1)(–2) + j [(3)(1) – 3(4)(1)] + k(–2)(–8)(–1)
= –12i – 9j – 16k.
Example 3. Find the directional derivative of
Sol. Here f(x, y, z) =
Now
1
=
r
1
x2 + y2 + z2
= (x2 + y2 + z2)–1/2
FG ∂ + j ∂ + k ∂ IJ cx + y + z h
H ∂x ∂y ∂z K
∇f = i
2
2
1
in the direction r where r = xi + yj + zk.
r
(U.P.T.U., 2002, 2005)
2 − 1/2
343
VECTOR CALCULUS
=
∂ 2
2
2 −1/2
−1/2
∂ 2
∂
k
x + y2 + z2
i + ( x 2 + y 2 + z 2 ) −1/2 j + ∂z ( x + y + z )
∂x
∂y
d
i
RS− 1 (x + y + z )
T 2
−b xi + yj + zk g
=
cx + y + z h
=
2
2 −3/2
2
2
2
2 3/2
UV RS
W T
1
2x i + − (x 2 + y 2 + z 2 ) −3/2 2 y
2
UV j + RS− 1 (x + y + z )
W T2
2
2
2 −3/2
UV
W
2z k
and a = unit vector in the direction of xi + yj + zk
=
xi + yj + zk
As a =
x2 + y2 + z2
∴ Directional derivative = ∇f · a = −
xi + yj + zk
⋅
r
r
xi + yj + zk
cx + y + z h
(xi + yj + zk )
F xi + yj + zk IJ ⋅
= −G
= −
(x + y + z )
Hx + y +z K
(x + y + z )
2
2
2 3/2
2
2 1/2
2
2
2
2
2
2 2
2
2
2
Example 4. Find the directional derivative of φ = x2yz + 4xz2 at (1, – 2, –1) in the direction
2i – j – 2k. In what direction the directional derivative will be maximum and what is its magnitude? Also find a unit normal to the surface x2yz + 4xz2 = 6 at the point (1, – 2, – 1).
φ = x2yz + 4xz2
Sol.
∂φ
= 2xyz + 4z2
∂x
∂φ
2
∂y = x z,
∴
∂φ
= x2y + 8xz
∂z
∂φ
∂φ
∂φ
+j
+k
grad φ = i
∂x
∂y
∂z
2
= (2xyz + 4z ) i + (x2z) j + (x2y + 8xz)k
= 8i – j – 10k at the point (1, – 2, – 1)
Let a be the unit vector in the given direction.
2i − j − 2k
Then
a =
∴ Directional derivative
dφ
= ∇φ ⋅ a
ds
4+1+ 4
=
= (8i – j – 10k) ·
=
2i − j − 2 k
3
FG 2i − j − 2k IJ
H 3 K
37
16 + 1 + 20
=
·
3
3
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Again, we know that the directional derivative is maximum in the direction of normal
which is the direction of grad φ. Hence, the directional derivative is maximum along grad
φ = 8i – j – 10k.
Further, maximum value of the directional derivative
= |grad φ|
= |8i – j – 10k|
64 + 1 + 100 =
Again, a unit vector normal to the surface
=
=
165 .
grad φ
|grad φ|
8i − j − 10k
=
·
165
Example 5. What is the greatest rate of increase of u = xyz2 at the point (1, 0, 3)?
Sol.
u = xyz2
∴
grad u = i
∂u
∂u
∂u
+j
+k
∂x
∂y
∂z
= yz2 i + xz2 j + 2xyz k
= 9j at (1, 0, 3) point.
Hence, the greatest rate of increase of u at (1, 0, 3)
= |grad u| at (1, 0, 3) point.
= |9j| = 9.
Example 6. Find the directional derivative of
φ = (x2 + y2 + z2)–1/2
at the points (3, 1, 2) in the direction of the vector yz i + zx j + xy k.
Sol.
∴
φ = (x2 + y2 + z2)–1/2
grad φ = i
∂ 2
∂
( x + y 2 + z 2 ) −1/2 + j ( x 2 + y 2 + z 2 ) −1/2
∂x
∂y
+ k
LM
N
OP LM
Q N
O
( 2 z )P
Q
∂ 2
( x + y 2 + z 2 ) −1/2
∂z
1 2
1
2
2 −3/2
( 2 x ) + j − ( x 2 + y 2 + z 2 ) −3/2 (2 y )
= i − (x + y + z )
2
2
LM
N
1
+ k − ( x 2 + y 2 + z 2 ) −3/2
2
= −
= −
= −
xi + yj + zk
( x + y 2 + z 2 ) 3/2
2
3i + j + 2k
(9 + 1 + 4) 3/2
3i + j + 2k
14 14
at (3, 1, 2)
at (3, 1, 2)
OP
Q
345
VECTOR CALCULUS
Let a be the unit vector in the given direction, then
yzi + zxj + xyk
a =
=
Now,
y z + z 2 x2 + x 2 y 2
2 2
2i + 6 j + 3 k
at (3, 1, 2)
7
dφ
= a . grad φ
ds
=
FG 2i + 6 j + 3k IJ ⋅ FG − 3i + j + 2k IJ
H 7 K H 14 14 K
= −
= −
( 2 )( 3) + ( 6 )(1) + ( 3 )(2 )
7.14 14
18
7.14 14
= −
9
49 14
·
Example 7. Find the directional derivative of the function φ = x2 – y2 + 2z2 at the point
P(1, 2, 3) in the direction of the line PQ, where Q is the point (5, 0, 4).
Sol. Here
Position vector of P = i + 2j + 3k
Position vector of Q = 5i + 0j + 4k
∴
PQ = Position vector of Q – Position vector of P
= (5i + 0j + 4k) – (i + 2j + 3k)
= 4i – 2j + k.
Let a be the unit vector along PQ, then
4i − 2 j + k
a =
Also,
16 + 4 + 1
grad φ = i
=
4i − 2 j + k
21
∂φ
∂φ
∂φ
+j
+k
∂y
∂z
∂x
= 2x i – 2y j + 4z k
= 2i – 4j + 12k at (1, 2, 3)
Hence,
dφ
= a . grad φ
ds
=
=
=
FG 4i − 2 j + k IJ ⋅ (2i − 4 j + 12k)
H 21 K
( 4)( 2) + ( −2)( −4) + (1)(12)
21
28
21
⋅
Example 8. For the function
φ =
y
x2 + y2
,
346
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
find the magnitude of the directional derivative making an angle 30° with the positive X-axis at
the point (0, 1).
Sol. Here
∴
φ =
y
x + y2
2
∂φ
2 xy
= –
2
∂x
(x + y 2 )2
∂φ
x2 − y2
∂φ
=
=0
2
2 2 and
∂y
(x + y )
∂z
∴
grad φ = i
= i
∂φ
∂φ
∂φ
+j
+k
∂x
∂y
∂z
∂φ
∂φ
+j
∂x
∂y
LM3 ∂φ = 0OP
N ∂z Q
−2xy
x2 − y2
+
i
j
( x 2 + y 2 )2
(x 2 + y 2 )2
= – j at (0, 1).
Let a be the unit vector along the line making an angle 30° with the positive X-axis at the
point (0, 1), then
a = cos 30° i + sin 30° j.
Hence, the directional derivative is given by
dφ
= a . grad φ
ds
= (cos 30° i + sin 30° j) · (–j)
=
1
·
2
Example 9. Find the values of the constants a, b, c so that the directional derivative of
φ = axy2 + byz + cz2x3 at (1, 2, –1) has a maximum magnitude 64 in the direction parallel to Z-axis.
Sol.
φ = axy2 + byz + cz2x3
= – sin 30° = –
∴
grad φ =
∂φ
∂φ
∂φ
i+
j+
k
∂x
∂y
∂z
= (ay2 + 3cz2x2) i + (2axy + bz) j + (by + 2czx3) k
= (4a + 3c) i + (4a – b) j + (2b – 2c) k. at (1, 2, –1).
Now, we know that the directional derivative is maximum along the normal to the surface,
i.e., along grad φ. But we are given that the directional derivative is maximum in the direction
parallel to Z-axis, i.e., parallel to the vector k.
Hence, the coefficients of i and j in grad φ should vanish and the coefficient of k should be
positive. Thus
4a + 3c = 0
...(i)
4a – b = 0
...(ii)
and
2b – 2c > 0
b > c
...(iii)
Then
grad φ = 2(b – c) k.
347
VECTOR CALCULUS
Also, maximum value of the directional derivative =|grad φ|
⇒
64 = |2 (b – c) k| = 2 (b – c)
⇒
b – c = 32.
Solving equations (i), (ii) and (iv), we obtain
[3 b > c]
...(iv)
a = 6, b = 24, c = – 8.
Example 10. If the directional derivative of φ = ax2y + by2z + cz2x, at the point (1, 1, 1) has
maximum magnitude 15 in the direction parallel to the line
of a, b and c.
Sol. Given
φ = ax2y + by2z + cz2x
∴
∇φ = i
y−3
x −1
z
=
= , find the values
2
1
−2
(U.P.T.U., 2001)
FG ∂φ IJ + j FG ∂φ IJ + k FG ∂φ IJ
H ∂x K H ∂y K H ∂z K
= i(2axy + cz2) + j (ax2 + 2byz) + k (by2 + 2czx)
∇φ at the point (1, 1, 1) = i (2a + c) + j (a + 2b) + k (b + 2c).
We know that the maximum value of the directional derivative is in the direction of ∇ f
i.e.,
|∇φ| = 15 ⇒ (2a + c)2 + (2b + a)2 + (2c + b)2 = (15)2
But, the directional derivative is given to be maximum parallel to the line.
y−3
z
x −1
=
=
1
2
−2
2a + c
2b + a
2c + b
=
=
2
−2
1
⇒
2a + c = – 2b – a ⇒ 3a + 2b + c = 0
and
2b + a = – 2(2c + b)
⇒
2b + a = – 4c – 2b ⇒ a + 4b + 4c = 0
Solving (i) and (ii), we get
⇒
a
b
c
=
=
= k (say)
−11
4
10
⇒
a = 4k, b = – 11k, and c = 10k.
2
Now,
(2a + c) + (2b + a)2 + (2c + b)2 = (15)2
⇒ (8k + 10k)2 + (– 22k + 4k)2 + (20k – 11k)2 = (15)2
5
9
20
55
50
a = ±
,b= ±
and c = ±
·
9
9
9
⇒
k = ±
⇒
z af
Example 11. Prove that ∇ f u du = f(u) ∇u.
Sol. Let
Then
z af
f u du = F(u), a function of u so that
z af
∇ f u du =
∂F
= f(u)
∂u
FG i ∂ + j ∂ + k ∂ IJ F
H ∂x ∂y ∂z K
(i)
(ii)
348
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
= i
∂F
∂F
∂F
+ j ∂y + k
∂x
∂z
= i
∂F ∂u
∂F ∂u
∂F ∂u
+ j
+ k
∂u ∂x
∂u ∂z
∂u ∂y
FG
H
IJ
K
∂F i ∂u + j ∂u + k ∂u
∂x
∂y
∂z
∂u
= f (u) ∇u. Hence proved.
=
Example 12. Find the angle between the surfaces x2 + y2 + z2 = 9 and z = x2 + y2 – 3 at the
point (2, – 1, 2)
(U.P.T.U., 2002)
2
2
2
Sol. Let
φ1 = x + y + z – 9
φ2 = x2 + y2 – z – 3
∴
∇φ1 =
FG i ∂ + j ∂ + k ∂ IJ (x + y + z – 9) = 2xi + 2yj + 2zk
H ∂x ∂y ∂z K
2
2
2
∇φ1 at the point (2, – 1, 2) = 4i – 2j + 4k
H
∇φ 2 =
FG i ∂ + j ∂ + k ∂ IJ (x + y – z – 3) = 2xi + 2yj – k
H ∂x ∂y ∂z K
2
2
∇φ2 at the point (2, – 1, 2) = 4i – 2j – k
Let θ be the angle between normals (i) and (ii)
(4i – 2j + 4k) · (4i – 2j – k) =
16 + 4 + 16
(i)
(ii)
16 + 4 + 1 cos θ
16 + 4 – 4 = 6 21 cos θ ⇒ 16 = 6 21 cos θ
⇒
cos θ =
8
3 21
⇒ θ = cos–1
8
3 21
·
Example 13. If u = x + y + z, v = x2 + y2 + z2, w = yz + zx + xy, prove that grad u, grad v
and grad w are coplanar vectors.
(U.P.T.U., 2002)
FG i ∂ + j ∂ + k ∂ IJ (x + y + z) = i + j + k
H ∂x ∂y ∂z K
F ∂ + j ∂ + k ∂ IJ (x + y + z ) = 2x i + 2y j + 2z k
grad v = G i
H ∂x ∂y ∂z K
F ∂ + j ∂ + k ∂ IJ (yz + zx + xy)
grad w = G i
H ∂x ∂y ∂z K
Sol.
grad u =
2
2
= i ( z + y) + j ( z + x) + k ( y + x)
Now,
grad u (grad v × grad w) =
1
2x
1
2y
1
2z
z+y z+x y+x
2
349
VECTOR CALCULUS
1
1
1
y
z
= 2 x
z+y z+x y+x
1
= 2 x+z+ y
1
y+z+x
z+x
z+ y
1
z + y + x |Applying R2 → R2 + R3
y+x
1
1
1
=0
1
1
1
y+z z+x x+y
Hence, grad u, grad v and grad w are coplanar vectors.
= 2 (x + y + z)
Example 14. Find the directional derivative of φ = 5x2y – 5y2z +
5 2
z x at the point
2
x −1 y − 3 z
=
= ·
(U.P.T.U., 2003)
−2
2
1
5 2
2
Sol.
∇φ = 10 xy + z i + ( 5x − 10 yz) j + ( −5y 2 + 5zx)k
2
25
i − 5j
∇ φ at P(1, 1, 1) =
2
y−3 z
x −1
Direction Ratio of the line
=
are 2, – 2, 1
=
2
−2
1
−2
1
2
,
Direction cosines of the line are
,
2
2
2 2 + −2 + 1
2 2 + −2 + 1
(2) 2 + ( −2)2 + (1) 2
2 −2 1
,
i.e., ,
3 3 3
Directional derivative in the direction of the line
P(1, 1, 1) in the direction of the line
LM
N
OP
Q
a f
=
F 25 i − 5 jI ⋅ F 2 i − 2 j + 1 kI
H 2 K H3 3 3 K
25 10
+
3
3
35
=
·
3
=
RS f (r)r UV = 1 d (r f ).
T r W r dr
R f (r)r UV = R|S exi + yj + zkj U|V
∇⋅S
T r W ∇ ⋅ |T f (r) r |W
∂ R f ( r )x U ∂ R f ( r ) y U ∂ R f ( r ) z U
S V+ S V+ S V
=
∂x T r W ∂y T r W ∂z T r W
∂ R f (r )x U
d R f (r) U ∂r f (r )
S
V
= x S
V + r
∂x T r W
dr T r W ∂x
Example 15. Prove that ∇·
Sol.
Now,
2
2
a f
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A TEXTBOOK OF ENGINEERING MATHEMATICS—I
= x
=
Similarly,
RS1 df − f (r) UV x + f (r) as ∂r = x
∂x r
T r dr r W r r
2
f (r )
x 2 df (r ) x 2
− 3 f (r) +
2
dr
r
r
r
RS UV = y df (r) − y f (r) + f (r)
T W r dr r
r
∂ R f (r )z U
S V = zr dfdr(r) − zr f (r) + f (rr)
∂z T r W
∂ f (r ) y
r
∂y
and
2
2
2
3
2
2
2
3
Now using these results, we get
∇·
LM f (r)r OP = df (r) + 2 f (r)
N r Q dr r
=
1 d 2
(r f )⋅ Hence proved.
r 2 dr
r
and f(1) = 0.
r5
Example 16. Find f(r) such that ∇ f =
Sol. It is given that
∂f
∂f
∂f
xi + yj + zk
r
i+
j+ k = ∇ f =
5 =
∂x
∂y
∂z
r5
r
x ∂f
∂f
∂f
y
z
So,
= 5,
= 5 and
= 5
∂y
r
∂x
∂z
r
r
∂f
∂f
∂f
y
x
z
We know that
df = ∂x dx + ∂y dy + ∂z dz = 5 dx + 5 dy + 5 dz
r
r
r
df =
xdx + ydy + zdz
r
5
=
rdr
= r–4dr
r5
−3
Integrating
Since
So,
Thus,
f(r) =
r
+c
−3
0 = f(1) = −
1
+c
3
1
3
1 1
1
f(r) =
–
·
3 r3
3
c =
EXERCISE 5.2
1. Find grad f where f = 2xz4 – x2y at (2, –2, –1).
Ans. 10i − 4 j − 16 k
2. Find ∇ f when f = (x2 + y2 + z2) e − x + y + z ·
H
Ans. 2 − r e − r r
2
2
2
a f
351
VECTOR CALCULUS
3. Find the unit normal to the surface x2y + 2xz = 4 at the point (2, –2, 3).
LMAns. ± 1 bi − 2 j − 2kgOP
3
N
Q
4. Find the directional derivative of f = x yz + 4xz at (1, –2, –1) in the direction
LMAns. 37 OP
2i – j – 2k.
3Q
N
5. Find the angle between the surfaces x + y + z = 9 and z = x + y – 3 at the point
LMAns. θ = cos FG 8 21 IJ OP
(2, –1, 2) .
H 63 K PQ
MN
6. Find the directional derivative of f = xy + yz + zx in the direction of vector i + 2j + 2k at
LMAns. 10 OP
the point (1, 2, 0).
3Q
N
2
2
2
2
2
2
2
−1
Ans. f = x 2 yz 3 + 20
7. If ∇f = 2xyz3i + x2z3j + 3x2yz2k, find f (x, y, z) if f (1, – 2, 2) = 4.
Ans. f = x 2 + 2 y 2 + 4z 2
8. Find f given ∇f = 2xi + 4yj + 8zk.
9. Find the directional derivative of φ = (x2 + y2 + z2)–1/2 at the point P(3, 1, 2) in the direction
LMAns. − 9 OP
49 14 Q
N
of the vector, yzi + xzj + xyk.
10. Prove that ∇2f(r) = f ′′(r) +
11. Show that ∇2
2
f ′ ( r) ·
r
FG x IJ = 0, where r is the magnitude of position vector r = xi + yj + zk.
Hr K
3
[U.P.T.U. (C.O.), 2002]
12. Find the direction in which the directional derivative of f(x, y) = (x2 – y2)/xy at
LMAns. 1 + i OP
2 Q
N
(1, 1) is zero.
13. Find the directional derivative of
1
in the direction of r , where r = xi + yj + zk.
r
1
Ans. − 2 (U.P.T.U., 2003)
r
LM
N
14. Show that ∇r–3 = – 3r–5 r .
15. If φ = log | r |, show that ∇φ =
5.10
r
·
r2
OP
Q
[U.P.T.U., 2008]
DIVERGENCE OF A VECTOR POINT FUNCTION
If f (x, y, z) is any given continuously differentiable vector point function then the divergence of
f scalar function defined as
(U.P.T.U., 2006)
352
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
∇⋅ f =
5.11
FG i ∂ + j ∂ + k ∂ IJ ⋅ f = i ⋅ ∂ f + j ⋅ ∂ f + k ⋅ ∂ f = div f
∂x
∂y
∂z
H ∂x ∂y ∂z K
.
PHYSICAL INTERPRETATION OF DIVERGENCE
Let v = vxi + vy j + vz k be the velocity of the fluid at P(x, y, z).
Here we consider the case of fluid flow along a rectangular parallelopiped of dimensions
δx, δy, δz
Mass in = v·x δyδz
Mass out = vx(x + δx) δyδz
=
(along x-axis)
Z
FG v + ∂v δxIJ δyδz
H ∂x K
x
S
x
|By Taylor's theorem
Net amount of mass along x-axis
FG
H
= vx δyδz – vx +
Vx R
IJ
K
∂v x
δx δyδz
∂x
∂v x
δxδyδz
∂x
|Minus sign shows decrease.
Y
Similar net amount of mass along y-axis
∂v y
∂y
C
@z
P
Q
@x
B
X
Fig. 5.4
δxδyδz
and net amount of mass along z-axis = −
∂v z
δxδyδz
∂z
∴ Total amount of fluid across parallelopiped per unit time = −
Negative sign shows decrease of amount
⇒
A
O
= –
= –
D
v x (x + @x)
@y
F ∂v + ∂v + ∂v I δxδyδz
GH ∂x ∂y ∂z JK
x
F ∂v + ∂v + ∂v I δxδyδz
GH ∂x ∂y ∂z JK
x
Decrease of amount of fluid per unit time =
y
z
Hence the rate of loss of fluid per unit volume
=
F ∂v + ∂v + ∂v I
GH ∂x ∂y ∂z JK
x
y
z
= ∇ ⋅ v = div v.
Therefore, div v represents the rate of loss of fluid per unit volume.
y
z
353
VECTOR CALCULUS
Solenoidal: For compressible fluid there is no gain no loss in the volume element
∴ div v = 0
then v is called Solenoidal vector function.
5.12
CURL OF A VECTOR
If f is any given continuously differentiable vector point function then the curl of f (vector
function) is defined as
Curl f = ∇ × f = i ×
∂f
∂f
∂f
+ j×
+k×
∂x
∂y
∂z
(U.P.T.U., 2006)
f = fx i + fy j + fz k, then
Let
i
j
k
∇ × f = ∂ / ∂x ∂ / ∂y ∂ / ∂z ·
fx
fy
fz
5.13
PHYSICAL MEANING OF CURL
Here we consider the relation v = w × r , w is the angular velocity r is position vector of a point
on the rotating body
(U.P.T.U., 2006)
curl v = ∇ × v
= ∇ × (w × r )
LMwHH = w i + w j + w kOP
N r = xi + yj + zk Q
1
= ∇ × [(w1 i + w 2 j + w 3 k ) × (xi + yj + zk )]
i
= ∇ × w1
x
j
w2
k
w3
y
z
2
3
= ∇ × [( w2 z − w 3 y)i − (w1 z − w 3 x) j + ( w1 y − w2 x)k ]
=
FG i ∂ + j ∂ + k ∂ IJ × [(w z − w y)i − (w z − w x) j + (w y − w x)k]
H ∂x ∂y ∂z K
i
∂
=
∂x
w2 z − w 3 y
2
3
1
3
1
2
j
k
∂
∂
∂y
∂z
w 3 x − w1 z w 1 y − w 2 x
= (w1 + w1) i – (–w2 – w2) j + (w3 + w3) k
= 2 (w1i + w2j + w3k) = 2w
Curl v = 2w which shows that curl of a vector field is connected with rotational properties
of the vector field and justifies the name rotation used for curl.
354
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Irrotational vector: If curl f = 0, then the vector f is said to be irrotational. Vice-versa, if
f is irrotational then, curl f = 0.
5.14 VECTOR IDENTITIES
Identity 1:
Proof:
grad uv = u grad v + v grad u
grad (uv) = ∇ (uv)
=
FG i ∂ + j ∂ + k ∂ IJ (uv)
H ∂x ∂y ∂z K
∂
∂
∂
= i (uv) + j (uv) + k (uv)
∂x
∂y
∂z
FG
IJ FG
IJ FG
H
K H
K H
F ∂v ∂v ∂v I F ∂u ∂u ∂u I
= uG i ∂x + j ∂y + k ∂z J + v G i ∂x + j ∂y + k ∂z J
H
K H
K
∂v
∂v
∂u
∂u
∂v
∂u
= i u ∂x + v ∂x + j u ∂y + v ∂y + k u ∂z + v ∂z
or
IJ
K
grad uv = u grad v + v grad u .
H
H
H
H
H
grad ( a · b ) = a × curl b + b × curl a + a ⋅ ∇ b + (b ⋅ ∇) a
H
H
∂ H H
∂a H H ∂b
Proof:
grad ( a · b ) = Σi
a ⋅ b = Σi
⋅b + a ⋅
∂x
∂x
∂x
H
H
H ∂a
H ∂b
= Σi b ⋅
.
+ Σi a ⋅
∂x
∂x
H
H
H
H
∂b
H ∂b
H
∂
b
a × i×
Now,
= a⋅
i − ( a ⋅ i)
∂x
∂x
dx
H
H
H
H
H
∂b
∂b
H ∂b
⇒
=
+
⋅
a
i
×
×
a
i
a⋅
i
∂x
∂x
∂x
H
H
H
H
H
∂b
∂b
H ∂b
⇒
Σ a⋅
+ ∑ a ⋅i
i = ∑a × i×
∂x
∂x
∂x
H
H
H ∂b
H
H ∂ H
∂b
⇒
Σ a⋅
+ ∑ a ⋅i
i = a × ∑ i×
b
∂x
∂x
∂x
H
H
H
H ∂b
H
H
⇒
Σ a⋅
i = a × curl b + a ⋅ ∇ b ·
∂x
L
L
Interchanging a and a , we get
H ∂aH
H
H
H
H
Σ b⋅
i = b × curl a + b ⋅ ∇ a
∂x
From equations (i), (ii) and (iii), we get
Identity 2:
FG
H
FG
H
FG
H
FG
H
FG
H
IJ
K
IJ
K
IJ
K
IJ
K
FG
H
IJ
K
IJ
K
F
e j GH
FG IJ FG IJ
H K H K
FG IJ
H K
FG IJ b g
H K
FG IJ b g
H K
FG IJ FG IJ
H K H K
b g
IJ
K
...(i)
b g
...(ii)
e j
...(iii)
H
H
H
H
H
H
H
H
H
H
grad ( a · b ) = a × curl b + b × curl a + ( a · ∇) b + ( b · ∇) a
.
355
VECTOR CALCULUS
Identity 3:
div (u a ) = u div a + a · grad u
Proof:
div (u a ) = ∇·(u a )
=
(U.P.T.U., 2004)
FG i ∂ + j ∂ + k ∂ IJ ⋅(uaH)
H ∂x ∂y ∂z K
∂ H
∂ H
∂ H
(ua ) + j ⋅ (ua ) + k ⋅ (ua )
∂x
∂y
∂z
H
H
H
∂u H
∂a
∂u H
∂a
∂u H
∂a
i
⋅
a
+
u
+
j
⋅
a
+
u
+
k
⋅
a
+
u
=
∂x
∂x
∂y
∂y
∂z
∂z
H
H
H
H
∂u
∂u
∂u
∂a
∂a
∂a
+a⋅ i
+j
+k
= u i⋅ + j⋅ + k⋅
∂x
∂y
∂z
∂x
∂y
∂z
= i⋅
RS
T
UV RS
W T
RS
T
or
div (u a ) = u div a + a · grad u
div ( a × b ) = b · curl b – a · curl b
Proof:
div ( a × b ) = ∇ · ( a × b )
H
∂ H
= Σi ⋅ ( a × b )
∂x
H
H
∂a H H ∂b
= Σi ⋅
×b+a×
∂x
∂x
H
H H
H ∂b
∂a
= Σi ⋅
× b + Σi ⋅ a ×
∂x
∂x
H
H H
∂b H
∂a
= Σ i×
⋅b − Σ i ×
⋅a
∂x
∂x
H
H
H
= (curl a ) · b – (curl b) · a
H
H
H
H
div ( a × b ) = b · curl a – a · curl b .
H
H
H
curl (u a ) = u curl a + (grad u) × a
H
H
curl (u a ) = ∇ × (u a )
∂ H
= Σi × (ua )
∂x
H
∂u H
∂a
= Σi ×
a+u
∂x
∂x
H
∂u H
∂a
= Σ i
× a + uΣ i ×
∂x
∂x
H
H
= (grad u) × a + u curl a
H
H
H
curl (u a ) = u curl a + (grad u) × a .
or
Identity 5:
Proof:
FG
H
FG
H
IJ
K
FG
H
FG IJ
H K
IJ
K
FG
H
FG
H
IJ
K
FG
H
UV
W
UV
W
.
Identity 4:
FG
H
or
UV RS
W T
UV RS
W T
[U.P.T.U. (C.O.), 2003]
IJ
K
IJ
K
IJ
K
IJ
K
[U.P.T.U. (C.O.), 2003]
356
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Identity 6:
Proof:
H
H
H
H
H
H
H
H
H
H
curl( a × b ) = a div b – b div a + ( b ·∇) a – ( a · ∇) b
H
H
H
H
curl ( a × b ) = ∇ × ( a × b )
∂ H H
= Σi × ( a × b )
∂x
H
H
∂a H H ∂b
= Σi ×
×b+a×
∂x
∂x
H
H H
H ∂b
∂a
= Σi ×
× b + Σi × a ×
∂x
∂x
FG
H
FG
H
IJ
K
FG
H
FG
H
IJ
K
IJ
K
IJ
K
FG
H
IJ
K
= Σ ( i. b ) ∂ a − Σ i ⋅ ∂ a b + Σ i ⋅ ∂ b a − Σ ( i . a ) ∂ b
∂x
∂x
∂x
∂x
H
H
H ∂
H H ∂ H
∂b H
∂a H
= Σ (i . b ) a − Σ i ⋅
b + Σ i⋅
a − Σ a⋅i
b
∂x
∂x
∂x
∂x
H
H H
H
H
H H
H
= ( b · ∇) a – (div a ) b + (div b ) a – ( a .∇) b
H
H
H
H
H
H
H
H
= ( b · ∇) a – ( a · ∇) b + a div b – b div a .
H
H
H
H
H
H
H
H
H
H
curl ( a × b ) = a div b – b div a + ( b · ∇) a – ( a · ∇) b .
or
Identity 7:
div grad f = ∇ · (∇f) =
Proof:
div grad f = ∇ · (∇f)
∂2 f
∂x
2
FG
H
+
∂2 f
∂y
2
IJ
K
+
∂2 f
∂z
2
FG
H
IJ
K
FG
H
IJ
K
= ∇2 f
FG i ∂ + j ∂ + k ∂ IJ ⋅ FG i ∂f + j ∂f + k ∂f IJ
H ∂x ∂y ∂z K H ∂x ∂y ∂z K
∂ F ∂f I ∂ F ∂f I ∂ F ∂f I
G J+ G J+ G J
=
∂x H ∂x K ∂y H ∂y K ∂z H ∂z K
=
=
∂2 f
∂2 f
∂2 f
∂x
∂y
∂z 2
+
2
div grad f = ∇2 f
or
Identity 8:
Proof:
+
2
.
curl grad f = 0
curl grad f = ∇ × (∇f)
= Σi
FG
H
∂f
∂f
∂f
∂
+k
+j
× i
∂z
∂y
∂x
∂x
IJ
K
F ∂ f + j ∂ f +k ∂ f I
GH ∂x ∂x∂y ∂x∂z JK
F ∂ f −j ∂ f I
= Σ Gk
H ∂x∂y ∂x∂z JK
F ∂ f − j ∂ f I + Fi ∂ f − k ∂ f I + F j ∂ f − i ∂ f I
= Gk
H ∂x∂y ∂x∂z JK GH ∂y∂z ∂y∂x JK GH ∂z∂x ∂z∂y JK
2
= Σi × i
2
2
2
or
curl grad f = 0 .
2
2
2
2
2
2
2
2
357
VECTOR CALCULUS
Identity 9:
Proof:
H
div curl f = 0
H
H
div curl f = ∇ · (∇ × f )
U|V
|W
R|S
|T
H
H
H
∂f
∂f
∂f
∂
= Σi ⋅
i×
+ j×
+k×
∂x
∂x
∂y
∂z
H
H
H
∂2 f
∂2 f
∂2 f
= Σi ⋅ i × 2 + j ×
+k×
∂x∂y
∂x∂z
∂x
H
H
H
∂2 f
∂2 f
∂2 f
= Σ ( i × i) ⋅ 2 + ( i × j ) ⋅
+ (i × k ) ⋅
∂x∂y
∂x∂z
∂x
H
H
∂2 f
∂2 f
= Σ k⋅
− j⋅
∂x∂y
∂x∂z
H
H
H
H
∂2 f
∂2 f
∂2 f
∂2 f
= k⋅
+ i⋅
+
−k⋅
− j⋅
∂y∂z
∂y∂x
∂x∂y
∂ x∂ z
R|S
|T
R|S
|T
R|S
|T
R|S
|T
U|V
|W
U|V
|W
U|V
|W
U|V R|S
|W |T
H
div curl f = 0 .
U|V
|W
H
H
H
H
H ∂ 2 f ∂ 2 f ∂2 f
Identity 10: grad div f = curl curl f + 2 + 2 + 2
∂x
∂y
∂z
H
H
Proof: Curl curl f = ∇ × (∇ × f )
H
H
H
∂f
∂f
∂f
∂
= Σi
× i×
+ j×
+k×
∂x
∂y
∂z
∂x
H
H
H
∂2 f
∂2 f
∂2 f
= Σi × i × 2 + j ×
+k×
∂x∂y
∂x∂z
∂x
H
H
H
H
∂2 f
∂2 f
∂2 f
∂2 f
= Σ
i ⋅ 2 i − (i ⋅ i ) ⋅ 2 + i ⋅
j − ( i ⋅ j)
∂x∂y
∂x∂y
∂x
∂x
|RS
|T
LMR|F
MNS|TGH
|RS
|T
|UV
|W
|UV
|W
U|V R|F
|W S|TGH
IJ
K
I
JK
R|S j ⋅ ∂ Hf − i ⋅ ∂ Hf U|V
|T ∂z∂x ∂z∂y |W
2
U|
V|
W
H
H
R
F
∂ f I
∂ f UO
P
+ SG i ⋅
TH ∂x∂z JK k − (i ⋅ k) ∂x∂z VWPQ
2
H
H
H
H
L
F
F
∂ fI F ∂ f I
∂ f I O
∂ f
= Σ MG i ⋅
MNH ∂x JK i + GH i ⋅ ∂x∂y JK j + GH i ⋅ ∂x∂z JK kPPQ − Σ ∂x
H
H
H
H
H
H
H
L
F
F
F
∂ fI
∂ f I
∂ f I O
∂ f ∂ f ∂ f
j + Gi ⋅
⇒ Curl curl f +
= Σ MG i ⋅
+
+
J kP ·
J i + i⋅
∂x
∂y
∂z
NMH H∂x K H GH ∂xH∂y JK H ∂x∂z K QP
H
RSi ⋅ ∂f + j ⋅ ∂f + k ⋅ ∂f |UV
∂ |
Again,
grad div f = Σi
∂x T
| ∂x ∂y ∂z W|
H
H
H
R
∂ f
∂ f
∂ f U
|
|
= Σi Si ⋅
|T ∂x + j ⋅ ∂x∂y + k ⋅ ∂x∂z V|W
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
...(i)
358
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
H
H
H
L
F
∂ f I F ∂ f I F ∂ f I O
= Σ MG i ⋅
MNH ∂x JK i + GH j ⋅ ∂x∂y JK i + GH k ⋅ ∂x∂z JK i PPQ
H
H
H
L
F
∂ f I F ∂ f I F ∂ f I O
= Σ MG i ⋅
MNH ∂x JK i + GH i ⋅ ∂z∂x JK j + GH i ⋅ ∂y∂z JK kPPQ
2
2
2
2
2
2
2
...(ii)
2
From eqns. (i) and (ii), we prove
H
H
H
grad div f = curl curl f + ∇2 f
H
H
H
Example 1. If f = xy2 i + 2x2yz j – 3yz2 k then find div f and curl f at the point (1, – 1, 1).
H
Sol. We have f = xy2 i +2x2yz j – 3yz2 k
H
∂
∂
∂
(xy 2 ) + ( 2x 2 yz) + (–3yz 2 )
div f =
∂z
∂x
∂y
= y2 + 2x2z – 6yz
= (– 1)2 + 2 (1)2 (1) – 6 (– 1)(1) at (1, – 1, 1)
= 1 + 2 + 6 = 9.
H
Again, curl f = curl [xy2 i + 2x2yz j – 3yz2 k]
i
∂
= ∂x
xy 2
= i
j
k
∂
∂
∂y
∂z
2 x 2 yz −3 yz 2
RS ∂ (−3 yz ) − ∂ (2x yz)UV + j RS ∂ (xy ) − ∂ (−3 yz )UV
∂z
∂x
W
T ∂y
W T ∂z
∂
R∂
U
+ k S ( 2 x yz ) − ( xy )V
∂
x
y
∂
T
W
2
2
2
2
2
= i [–3z2 – 2x2y] + j [0 – 0] + k [4xyz – 2xy]
= (–3z2 – 2x2y)i + (4xyz – 2xy) k
= {–3 (1)2 – 2(1)2(–1)} i + {4(1)(–1)(1) – 2(1)(–1)} k at (1, –1, 1)
= – i – 2k.
Example 2. Prove that
H
H
(i) div r = 3. (ii) curl r = 0.
H
H
Sol. (i) div r = ∇ · r
∂
∂
∂
⋅ ( xi + yj + zk )
+j
+k
= i
∂x
∂y
∂z
FG
H
IJ
K
FG
H
IJ
K
∂
∂
∂
( x ) + ( y ) + ( z)
∂x
∂y
∂z
= 1 + 1 + 1 = 3.
H
H
curl r = ∇ × r
∂
∂
∂
= i ∂x + j ∂y + k ∂z × ( xi + yj + zk )
=
(ii)
=
i
∂
∂x
x
j
∂
∂y
y
k
∂
∂z
z
2
359
VECTOR CALCULUS
= i
LM ∂ (z) − ∂ ( y)OP + j LM ∂ (x) − ∂ (z)OP + k LM ∂ (y) − ∂ (x)OP
N ∂y ∂z Q N ∂z ∂x Q N ∂x ∂y Q
= i (0) + j (0) + k (0)
= 0 + 0 + 0 = 0.
Example 3. Find the divergence and curl of the vector
(x2 – y2) i + 2xy j + (y2 – xy) k.
Sol. Let
Then
H
f = (x2 – y2) i + 2xy j + (y2 – xy) k.
H
div f =
∂ 2
∂
∂
( x − y 2 ) + (2 xy) + ( y 2 − xy)
∂x
∂y
∂z
= 2x + 2x + 0 = 4x
H
curl f =
and
i
∂
∂x
x2 − y2
= i
j
k
∂
∂
∂y
∂z
2xy y 2 − xy
LM ∂ ( y − xy) − ∂ (2 xy)OP + j LM ∂ (x − y ) − ∂ ( y − xy)OP
∂z
∂x
Q
N ∂y
Q N ∂z
∂
L∂
O
+ k M ( 2 xy ) − ( x − y )P
∂
x
y
∂
N
Q
2
2
2
2
2
2
= i [2y – x) – 0] + j [0 – (–y)] + k [(2y) – (– 2y)]
= (2y – x) i + y j + 4y k. Ans.
H
H
Example 4. If f (x, y, z) = xz3 i – 2x2yz j + 2yz4 k find divergence and curl of f (x, y, z)
(U.P.T.U., 2006)
Sol.
H
div f =
=
FG i ∂ + j ∂ + k ∂ IJ ⋅ cxz i − 2x yz j + 2yz k h
H ∂x ∂y ∂z K
3
2
4
∂
∂
∂
( xz 3 ) − (2x 2 yz) + (2 yz 4 )
∂x
∂y
∂z
= z3 – 2x2z + 8yz3
H
curl f =
i
∂
∂x
xz 3
= i
j
k
∂
∂
∂y
∂z
−2 x 2 yz 2 yz 4
RS ∂ (2yz ) + ∂ (2x yz)UV − j RS ∂ (2yz ) − ∂ (xz )UV
∂z
∂z
W
T ∂y
W T ∂x
∂
R∂
U
+ k S ( −2x yz) − ( xz )V
∂
x
∂
y
T
W
4
2
4
3
2
= i (2z4 + 2x2y) – j (0 – 3z2x) + k (– 4xyz – 0)
= 2 (x2y + z4) i + 3z2xj – 4xyz·k.
3
360
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
H
Example 5. Find the directional derivative of ∇. u at the point (4, 4, 2) in the direction of
H
the corresponding outer normal of the sphere x2 + y2 + z2 = 36 where u = x4 i + y4 j + z4 k.
H
Sol.
∇. u = ∇. (x4i + y4j + z4k) = 4(x3 + y3 + z3) = f (say)
∴
(∇f)(4, 4, 2) = 12 (x2i + y2j + z2k)(4, 4, 2) = 48(4i + 4j + k)
Normal to the sphere g ≡ x2 + y2 + z2 = 36 is
(∇g)(4, 4, 2) = 2 (xi + yj + zk)(4, 4, 2) = 4(2i + 2j + k)
a
= unit normal =
b
g
4 2i + 2 j + k
∇g
=
∇g
64 + 64 + 16
2i + 2 j + k
3
The required directional derivative is
=
∇f. a
2i + 2 j + k
3
= 16(8 + 8 + 1) = 272.
= 48 (4i + 4j + k).
Example 6. A fluid motion is given by v = (y + z)i + (z + x)j + (x + y)k show that the motion
is irrotational and hence find the scalar potential.
(U.P.T.U., 2003)
Sol.
Curl v
= ∆× v
=
F i ∂ + y ∂ + k ∂ I × by + zgi + az + xf j + bx + ygk
GH ∂x ∂y ∂z JK
=
i
j
k
∂
∂
∂
= i (1 – 1) – j (1 – 1) + k (1 – 1) = 0
∂x
∂y
∂z
y+z z+x x+ y
Hence v is irrotational.
Now
dφ =
∂φ
∂φ
∂φ
dx +
dy + dz
∂x
∂y
∂z
F i ∂φ + j ∂φ + k ∂φ I ⋅ bidx + jdy + kdzg
GH ∂x ∂y ∂z JK
F ∂ ∂ + k ∂ I φ ⋅ dr = ∇φ⋅ dr = v ⋅ dr
= Gi + j
H ∂x ∂y ∂z JK
=
v = ∇φ
= [(y + z) i + (z + x) j + (x + y) k]. (idx + jdy + kdz)
= (y + z) dx + (z + x) dy + (x + y) dz
= ydx + zdx + zdy + xdy + xdz + ydz
On integrating
φ =
=
Thus,
zb g zb g zb
zb g zb g za f
ydx + xdy +
zdy + ydz +
d xy + d yz + d zx
φ = xy + yz + zx + c
velocity potential = xy + yz + zx + c.
zdx + xdz
g
361
VECTOR CALCULUS
b
g
b g b g
H H
H
H
H
H
H
Example 7. Prove that a × ∇ × r = ∇ a ⋅ r − a ⋅ ∇ r where a is a constant vector and
H
(U.P.T.U., 2007)
r = xi + yj + zk.
H
Sol. Let
a = a1i + a2j + a3k
H
r = r1i + r2j + r3k
∴
∇× r =
e
j
Now a × ∇ × r =
i
∂
∂x
r1
j
∂
∂y
r2
F
GH
k
∂
∂z
r3
i
a1
j
a2
∂r3 ∂r2
−
∂y ∂x
I FG
JK H
∂r3 ∂r1
∂r3 ∂r2
= i ∂y − ∂z – ∂x − ∂z
IJ j + FG ∂r − ∂r IJ k
K H ∂x ∂y K
2
1
k
a3
∂r1 ∂r3
−
∂z ∂x
∂r2 ∂r1
−
∂x ∂y
|RSFG a ∂r − a ∂r IJ – FG a ∂r − a ∂r IJ |UVi
|TH ∂x ∂y K H ∂z ∂x K |W
|RF ∂r ∂r I F ∂r ∂r I |U
– SG a ∂x − a ∂y J − G a ∂y − a ∂z J V j
K H
K |W
|TH
|RF ∂r ∂r I F ∂r ∂r I |U
+ SGH − a ∂x + a ∂z JK − G a ∂y − a ∂z J Vk
H
K |W
|T
F ∂ ∂ + k ∂ I ba r + a r + a r g – LMa ∂ + a ∂ + a ∂ OP br i + r j + r kg
= Gi + j
H ∂x ∂y ∂z JK
N ∂x ∂y ∂z Q
= ∇ nb a j + a j + a k g ⋅ br j + r j + r k gs
R|
F ∂ ∂ ∂ I U|
– Sb a i + a j + a k g ⋅ G i + j + k J Vbr i + r j + r k g
H ∂x ∂y ∂z K W|
T|
= ∇e a ⋅ rj − e a ⋅ ∇jr · Hence proved.
=
2
2
2
1
3
1
3
3
1
1
1
2
3
1 1
2 2
3 3
1
2
3
1
2
1
2
1
1
3
3
1
1
2
2
3
3
3
3
3
2
2
2
1
2
3
1
2
3
Example 8. Find the directional derivative of Ö.(Öf) at the point (1, –2, 1) in the direction of
the normal to the surface xy2z = 3x + z2 where f = 2x3y2z4.
(U.P.T.U., 2008)
Sol.
F i ∂ + j ∂ + k ∂ I (2x y z ) = (6x y z )i + (4x yz )j + 8x y z )k
GH ∂x ∂y ∂z JK
F ∂ + j ∂ + k ∂ I . {(6x y z )i + (4x yz )j + (8x y z )k}
∇.(∇f) = G i
H ∂x ∂y ∂z JK
∇f =
3 2 4
2
2 2 4
2 4
3
3
4
4
3 2 3
3
2 3
∇.(∇f) = 12x y2z4 + 4x3 z4 + 24x3 y2z2 = F(x, y, z) (say)
or
Now
∇F =
F i ∂ + j ∂ + k ∂ I (12x y z + 4x z + 24x y z )
GH ∂x ∂y ∂z JK
2 4
3
4
3
2 2
= (12y2 z4 + 12x2 z4 + 72x2 y2z2)i + (24xyz4 + 48x3 yz2) j
+ (48x y2z3 + 16x3z3 + 48x3y2z)k
362
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
∴
Let
(∇F)(1, –2, 1) = 348i – 144j + 400k
g(x, y, z) = xy2z – 3x – z2 = 0
∇g =
F i ∂ + j ∂ + k ∂ I (xy z – 3x – z )
GH ∂x ∂y ∂z JK
2
2
= (y2z – 3) i + (2xyz) j + (xy2 – 2z) k
(∇g)(1, –2, 1) = i – 4j + 2k
a = unit normal =
∇g
i − 4 j + 2k
i − 4 j + 2k
=
=
∇g
1 + 16 + 4
21
Hence, the required directional derivative is
b i − 4 j + 2k g
∇F . a = (348i – 144j + 400k) .
=
348 + 576 + 800
21
=
21
1724
21
.
Example 9. Determine the values of a and b so that the surface ax2 – byz = (a + 2)x will be
orthogonal to the surface 4x2y + z3 = 4 at the point (1, –1, 2).
Sol. Let
f ≡ ax2 – byz – (a + 2)x = 0
...(i)
g ≡ 4x y + z – 4 = 0
...(ii)
2
grad f = ∇f =
3
F i ∂ + j ∂ + k ∂ I {ax – byz – (a + 2)x}
GH ∂x ∂y ∂z JK
2
= (2ax – a – 2)i + (–bz)j + (–by)k
(∇f)(1, –1, 2) = (a – 2)i – 2bj + bk
grad g = ∇g =
F i ∂ + j ∂ + k ∂ I (4x y + z – 4)
GH ∂x ∂y ∂z JK
2
...(iii)
3
= (8xy)i + (4x2)j + (3z2)k
(∇g)(1, –1, 2) = – 8i + 4j + 12k
Since the surfaces are orthogonal so
...(iv)
e∇ f j . e∇gj = 0
⇒ [(a – 2)i – 2bj + bk] . [– 8i + 4j + 12k] = 0
– 8(a – 2) – 8b + 12b = 0 ⇒
– 2a + b + 4 = 0
But the point (1, –1, 2) lies on the surface (i), so
a + 2b – (a + 2) = 0 ⇒ 2b − 2 = 0 ⇒ b = 1
Putting the value of b in (v), we get
− 2a + 1 + 4 = 0 ⇒ a =
Hence, a =
5
, b = 1.
2
5
2
...(v)
363
VECTOR CALCULUS
Example 10. Prove that A = (x2 – yz)i + (y2 – zx)j + (z2 – xy)k is irrotational and find the
scalar potential f such that A = ∇f.
i
j
k
∂
∂
∂
Sol. ∇ × A =
= (–x + x)i – (–y + y)j + (–z + z) = 0
∂x
∂y
∂z
x 2 − yz y 2 − zx z 2 − xy
Hence, A is irrotational.
A = ∇f = i
Now
∂f
∂f
∂f
+j
+k
= (x2 – yz)i + (y2 – zx)j + (z2 – xy)k
∂x
∂y
∂z
Comparing on both sides, we get
∂f
∂f
∂f
= (x2 – yz),
= (y2 – zx) and
= (z2 – xy)
∂y
∂x
∂z
∴
df =
∂f
∂f
∂f
dx +
dy + dz = (x2 – yz)dx + (y2 – zx)dy + (z2 – xy)dz
∂x
∂y
∂z
= (x2 dx + y2 dy + z2 dz) – (yzdx + zxdy + xydz)
or
df =
1
d(x3 + y3 + z3) – d(xyz)
3
f =
1
(x3 + y3 + z3) – xyz + c.
3
On integrating, we get
EXERCISE 5.3
Ans. 2xz – 6 y 2 z 2 + xy 2
1. Find div A , when A = x2zi – 2y3z2j + xy2zk.
2. If V =
xi + yj + zk
x2 + y2 + z2
, find the value of div V .
(U.P.T.U., 2000)
LMAns. 2 division ex + y + z j OP
N
Q
3. Find the directional derivative of the divergence of f (x, y, z) = xyi + xy j + z k at the point
L 13 OP
(2, 1, 2) in the direction of the outer normal to the sphere, x + y + z = 9. MAns.
3Q
N
2
2
2
2
2
2
2
2
H H 3
4. Show that the vector field f = r / r is irrotational as well as solenoidal. Find the scalar
potential.
LM
MN
(U.P.T.U., 2001, 2005) Ans. –
OP
x + y + z PQ
1
2
2
2
364
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
5. If r is the distance of a point (x, y, z) from the origin, prove that curl
FG k. grad 1IJ = 0, where k is the unit vector in the direction OZ.
H
rK
FG k × grad 1 IJ + grad
H
rK
(U.P.T.U., 2000)
6. Prove that A = (6xy + z3)i + (3x2 – z)j + (3xz2 – y) k is irrotational. Find a scalar function
Ans. f = 3 x 2 y + xz 3 – zy + c2
f (x, y, z) such that A = ∇f.
7. Find the curl of yzi + 3zxj + zk at (2, 3, 4).
Ans. – 6i + 3 j + 8 k
8. If f = x2yz, g = xy – 3z2, calculate ∇.(∇f × ∇g).
Ans. zero
9. Determine the constants a and b such that curl of (2xy + 3yz)i + (x2 + axz – 4z2)j + (3xy
Ans. a = 3, b = −4
+ 2byz) k = 0.
10. Find the value of constant b such that
3
2
2
A = (bxy – z ) i + (b – 2)x j + (1 – b) xz k has its curl identically equal to zero.
Ans. b = 4
FG 1 IJ = a ⋅ r ·
H rK r
12. Prove that ∇ e a ⋅ uj = e a ⋅ ∇ju + a × curl u a is a constant vector.
F xI
13. Prove that ∇ GH JK = 0.
r
H
11. Prove that a . ∇
2
3
3
14. Prove that ∇2f(r) = f ″(r) +
2
f ′ (r).
r
15. If u = x2 + y2 + z2 and v = xi + yj + zk , show that div
euvj = 5u
F a × r I = − a + 3rea ⋅ rj .
GH r JK r r
17. Find the curl of v = e
ei + j + kj at the point (1, 2, 3).
16. Prove the curl
3
3
5
b
Ans. e 6 i − 21 j + 3k
xyz
18. Prove that ∇ × ∇f = 0 for any f (x, y, z).
Ans. 4 j
19. Find curl of A = x 2 yi – 2 xzj + 2 yzk at the point (1, 0, 2).
H
20. Determine curl of xyz2i + yzx2j + zxy2 k at the point (1, 2, 3).
a
f
b
g
g
b
g
Ans. xy 2z – x i + yz 2x − y j + zx 2 y − z k ;10i + 3k
21. Find f(r) such that f(r) r is solenoidal.
LMAns. c OP
N rQ
3
22. Find a, b, c when f = (x + 2y + az)i + (bx – 3y – z)j + (4x + cy + 2z)k is irrotational.
[Ans. a = 4, b = 2, c = –1]
23. Prove that (y2 – z2 + 3yz – 2x)i + (3xz + 2xy)i + (3xy – 2xz + 2z)k are both solenoidal and
irrotational.
[U.P.T.U., 2008]
365
VECTOR CALCULUS
5.15 VECTOR INTEGRATION
Vector integral calculus extends the concepts of (ordinary) integral calculus to vector functions. It
has applications in fluid flow design of under water transmission cables, heat flow in stars, study
of satellites. Line integrals are useful in the calculation of work done by variable forces along
paths in space and the rates at which fluids flow along curves (circulation) and across boundaries
(flux).
5.16
LINE INTEGRAL
ej
Let F r be a continuous vector point function. Then
ej
z
C
F ⋅ dr , is known as the line integral of
F r along the curve C.
Let F = F1 i + F2 j + F3 k where F1, F2, F3 are the components of F along the coordinate axes
and are the functions of x, y, z each.
r = xi + yj + zk
dr = dxi + dyj + dzk
Now,
∴
∴
z
C
F ⋅ dr =
=
ze
zb
C
C
je
F1i + F2 j + F3 k ⋅ dxi + dyj + dzk
g
F1 dx + F2 dy + F3 dz .
Again, let the parameteric equations of the curve C be
x = x (t)
y = y (t)
z = z (t)
z
j
z LMN a f
z
z
af
a f OPQ
dy
dx
dz
+ F2 t
+ F3 t
dt
t1
C
dt
dt
dt
were t1 and t2 are the suitable limits so as to cover the arc of the curve C.
H
Note:
work done =
F ⋅ dr
then we can write
F ⋅ dr =
t2
C
Circulation: The line integral
C
F1 t
H
F ⋅ dr of a continuous vector point functional F along a
closed curve C is called the circulation of F round the closed curve C.
This fact can also be represented by the symbol Ü.
Irrotational vector field: A single valued vector point Function F (Vector Field F ) is called
irrotational in the region R, if its circulation round every closed curve C in that region is zero
that is
z
z
C
or
F ⋅ dr = 0
F ⋅ dr = 0.
366
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
5.17
SURFACE INTEGRAL
Any integral which is to be evaluated over a surface is called a surface integral.
H
Let F r be a continuous vector point function. Let r = F (u, v) be a smooth surface such
that F (u, v) possesses continuous first order partial derivatives. Then the normal surface integral
bg
ej
of F r over S is denoted by
z ej
z ej
F r ⋅ da = F r ⋅ ndS
S
S
H
where da is the vector area of an element dS and n is a unit vector normal to the surface dS.
Let F1, F2, F3 which are the functions of x, y, z be the components of F along the coordinate
axes, then
Surface Integral =
=
=
=
z
z
zz e
zz b
S
S
F ⋅ ndS
F ⋅ da
S
S
je
F1 i + F2 j + F3 k ⋅ dydzi + dzdxj + dxdyk
g
j
F1 dy dz + F2 dz dx + F3 dx dy .
5.17.1 Important Form of Surface Integral
e
Let dS = dS cos αi + cos βj + cos γ k
j
...(i)
where α, β and γ are direction angles of dS. It shows that dS cos α, dS cos β, dS cos γ are
orthogonal projections of the elementary area dS on yz. plane, zx-plane and xy-plane respectively.
As the mode of sub-division of the surface S is arbitrary we have chosen a sub-division formed
by planes parallel to coordinate planes that is yz-plane, zx-plans and xy plane.
Clearly, projection on the coordinate planes will be rectangles with sides dy and dz on yz
plane, dz and dx on xz plane and dx and dy on xy plane.
⋅ i
i ⋅ dS = dS cos αi + cos βj + cos γ kj
Hence
{
i ⋅ n dS
}
= dS cos α = dy dz
dy dz
dS =
.
i. n
Hence
Similarly, multiplying both sides of (i) scalarly by j and k respectively, we have
dz dx
dS =
j ⋅ n
dx dy
and
dS =
k ⋅ n
dy dz
F ⋅ n
F ⋅ ndS
Hence
=
S1
i ⋅ n
dz dx
F ⋅ n
=
j ⋅ n
S2
z
zz
zz
zz
dx dy
S3
k ⋅ n
where S1, S2, S3 are projections of S on yz, zx and xy plane respectively.
=
F ⋅ n
...(ii)
...(iii)
...(iv)
...(v)
367
VECTOR CALCULUS
5.18 VOLUME INTEGRAL
ej
Let F r is a continuous vector point function. Let volume V be enclosed by a surface S given by
a f
r = f u, v
sub-dividing the region V into n elements say of cubes having
volumes
∆V1, ∆V2, .... ∆Vn
...(i)
∆Vk = ∆xk ∆yk ∆zk
Hence
k = 1, 2, 3, ... n
where (xk, yk, zk) is a point say P on the cube. Considering the
sum
P
∑ Fbxk , yk , zk g∆Vk
n
Fig. 5.5
k =1
taken over all possible cubes in the region. The limits of sum when n → ∞ in such a manner that
the dimensions ∆Vk tends to zero, if it exists is denoted by the symbol
z ej z
V
F r dV ⋅ or
V
FdV or
is called volume integral or space integral.
If
F
=
F1 i + F2 j + F3 k , then
F r dV
=
i
z ej
V
zzz
V
F1 dx dy dz + j
zzz
zzz
V
V
F dx dy dz
F2 dx dy dz + k
zzz
V
F3 dx dy dz
where F1, F2, F3 which are function of x, y, z are the components of F along X, Y, Z axes
respectively.
Independence of path
If in a conservative field F
ÜC F ⋅ dr
= 0
along any closed curve C.
Which is the condition of the independence of path.
z
Example 1. Evaluate
F ⋅ dr where F = x 2 y 2 i + yj and the curve C, is y2 = 4x in the
C
xy-plane from (0, 0) to (4, 4).
Sol. We know that
r
= xi + yj
∴
dr
= dxi + dyj
∴
F ⋅ dr
∴
z
=
ex y i + yjj ⋅ edxi + dyjj
2 2
= x2y2dx + ydy
F ⋅ dr
C
=
ze
C
x 2 y 2 dx + ydy
j
368
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
z z
z af z
z z
=
C
x 2 y 2 dx +
C
y dy ⋅
But for the curve C, x and y both vary from 0 to 4.
H H
4 2
4
F ⋅ dr =
x 4x dx + y dy
∴
z
0
C
4
4
0
0
3
= 4 x dx +
y dy
Fx I F y I
= 4G J +G J
H 4K H 2K
4
4
4
2
0
zb
Example 2. Evaluate
[Œ y2 = 4x]
0
0
= 256 + 8 = 264.
g
x dy – y dx around the circle x2 + y2 = 1.
Sol. Let C denote the circle x2 + y2 = 1, i.e., x = cos t, y = sin t. In order to integrate around
C, t varies from 0 to 2π.
zb
∴
C
x dy − y dy
z
ze
g = 02π FGH x dydt − y dxdt IJK dt
2π
=
0
z
=
0
dt
= (t)2π
0
= 2π.
z
Example 3. Evaluate
2π
j
cos 2 t + sin 2 t dt
C
H
H H
F ⋅ dr , where F = (x2 + y2) i – 2xy j, the curve C is the rectangle in
Y
the xy-plane bounded by y = 0, x = a, y = b, x = 0.
Sol.
z
C
H H
F ⋅ dr =
z {e
z {e
}m
j
r
x 2 + y 2 i – 2xyj ⋅ dxi + dyj
C
j
}
y=b
x 2 + y 2 dx − 2xy dy
=
C
Now, C is the rectangle
OACB.
B
(0, b)
...(i)
On OA, y = 0 ⇒ dy = 0
C (a, b)
x=a
x=0
On AC, x = a ⇒ dx = 0
On CB, y = b ⇒ dy = 0
On BO, x = 0 ⇒ dx = 0
∴ From (i),
H H
F ⋅ dr =
z
C
=
O
(0, 0)
y=0
ze j zb g ze j z
z z ze j z
x + 0 dx +
2
OA
AC
–2ay dy +
x + b dx +
2
CB
2
0
0
a 2
b
x dx − 2 a y dy + x 2 + b 2 dx + 0 dy
0
0
a
b
Fx I F y I Fx I
= G J − 2aG J + G + b xJ + 0
H 3K H 2K H 3 K
3
a
0
2
b
0
3
0
2
a
BO
Fig. 5.6
0 dy
A
(a, 0)
X
369
VECTOR CALCULUS
a3
a3
− ab 2 −
− ab 2
3
3
= – 2ab2.
=
Example 4. Evaluate
z
C
H H
H
F ⋅ dr , where F = yz i + zx j + xy k and C is the portion of the curve
π
H
.
r = (a cos t) i + (b sin t) j + (ct) k from t = 0 to
2
H
Sol. We have
r = (a cos t) i + (b sin t) j + (ct) k.
Hence, the parametric equations of the given curve are
x = a cos t
y = b sin t
z = ct
H
dr
Also,
= (– a sin t) i + (b cos t) j + ck
dt
H H
H drH
F ⋅ dr =
Now,
F
⋅
dt
C
C
dt
z
z
=
=
=
zb
zb
ze
C
C
C
gb
g
yzi + zxj + xyk ⋅ − a sin ti + b cos t j + ck dt
gb
g
bct sin t i + act cos t j + ab sin t cos t k ⋅ − a sin ti + b cos t j + ck dt
j
− abc t sin 2 t + abc t cos 2 t + abc sin t cos t dt
z e
z FGH
z FGH
j
sin 2 t I
= abc
t cos 2 t +
J dt
C
2 K
π
sin 2 t I
= abc 2 t cos 2 t +
J dt
0
2 K
π
sin 2t cos 2t cos 2t O 2
L
= abc Mt
N 2 + 4 − 4 PQ0
abc
=
bt sin 2tg0 = 0.
2
2
2
= abc C t cos t − sin t + sin t cos t dt
π
2
Example 5. Evaluate
z
Y
C
(4,12)
F. dr where F = xyi + (x2 + y2)j and C
c
is the x-axis from x = 2 to x = 4 and the line x = 4 from y = 0 to
y = 12.
Sol. Here the curve C consist the line AB and BC.
Since
so
z
C
r = xi + yj (as z = 0)
dr = dxi + dyj
F . dr =
z{
e
j} m
r
xyi + x 2 + y 2 j . dxi + dyj
ABC
A
B
(2, 0)
(4, 0)
O
Fig. 5.7
X
370
=
z{
e
=
z
4
x=2
0 . dx +
z
j } {xy dx + ex + y jdy}
xy dx + x 2 + y 2 dy +
AB
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
z
2
BC
12
y=0
e
j
ze
12
0 + 16 + y 2 dy =
0
L y OP = [192 + 576] = 768.
= M16 y +
3 PQ
MN
2
j
16 + y 2 dy
3 12
0
Example 6. If F = (–2x + y)i + (3x + 2y)j, compute the circulation of F about a circle C in
the xy plane with centre at the origin and radius 1, if C is transversed in the positive direction.
Sol. Here the equation of circle is x2 + y2 = 1
Let
x = cos θ, y = sin θ |As r = 1
2
2
x + y =1
C
F = (–2cos θ + sin θ)i + (3 cos θ + 2sin θ)j
r = xi + yj = (cos θ)i + (sin θ)i
So
dr = {(– sin θ)i + (cos θ)j}dθ
Thus, the circulation along circle C =
=
z nb
ze
ze
2π
θ=0
2π
=
=
0
2π
0
O
z
F . dr
Fig. 5.8
C
g b
g s nb− sin θgi + bcos θg jsdθ
−2 cos θ + sin θ i + 3 cos θ + 2 sin θ j .
j
2 sin θ cos θ − sin 2 θ + 3 cos 2 θ + 6 sin θ cos θ dθ
j
8 sin θ cos θ + 4 cos 2 θ − 1 dθ =
2π
2π
2π
= −2 cos 2θ 0 + sin 2θ 0 + θ 0
z b
2π
0
g
4 sin 2θ + 2 cos 2θ + 1 dθ
= − 2 cos 4π − cos 0 + sin 4π − sin 0 + 2π
= 2π.
Example 7. Compute the work done in moving a particle in the force field F = 3x2 i +
(2xz – y)j + zk along.
(i) A straight line from P(0, 0, 0) to Q(2, 1, 3).
(ii) Curve C : defined by x2 = 4y, 3x3 = 8z from x = 0 to x = 2.
is
Sol. (i) We know that the equation of straight line passing through (x1, y1, z1) and (x2, y2, z2)
x − x1
x 2 − x1
⇒
or
=
y − y1
z − z1
=
y 2 − y1
z 2 − z1
x−0
y−0
z−0
=
=
2−0
1−0
3−0
⇒
y
z
x
=
=
1
3
2
y
z
x
=
=
= t (say), so x = 2t, y = t, z = 3t
1
3
2
371
VECTOR CALCULUS
∴
r = xi + yj + zk = 2ti + tj + 3tk
⇒
dr = (2i + j + 3k)dt
and
The work done =
z
F = (12t2)i + (12t2 – t)j + (3t)k
Q
P
F . dr =
=
z
ze
e12t ji + e12t − tj j + a3tfk . 2i + j + 3k dt from 0 to 1
1
0
1
0
2
As t varies
2
j
24t 2 + 12t 2 − t + 9t dt =
ze
1
0
j
36t 2 + 8t dt
L 36t + 8t OP = 12 + 4 = 16.
= M
N 3 2Q
F 3x − x I j + 3x k = 3x i + F 3x − x I j + 3x k
(ii) F = 3x i + G 2x.
GH 4 4 JK 8
H 8 4 JK 8
2 1
3
0
3
2
2
3
4
2
2
2
3
3
x
3x
r = xi + yj + zk = xi +
j+
k
4
8
F i + x j + 9x kI dx
GH 2 8 JK
L F 3x − x I j + 3x kOP . Li + x j + 9x kOdx
F . dr = M 3x i + G
MN H 4 4 JK 8 PQ MN 2 8 PQ
F 3x + 3x − x + 27x I dx
=
GH
8
8
64 JK
L x − x + 27x OP
= Mx +
N 16 32 64 × 6 Q
2
dr =
Work done =
z
2
x=0
z
2
4
2
2
3
2
0
z
2
5
2
3
5
0
6
3
4
6
2
0
= 8+
64 16 27 × 64
1 9
−
+
= 8+ 4− +
16 32 64 × 6
2 2
= 16.
Example 8. If V is the region in the first octant bounded by y2 + z2 = 9 and the plane
x = 2 and F = 2x2yi – y2j + 4xz2k. Then evaluate
zzz e j
V
∇ ⋅ F dV.
∇ ⋅ F = 4 xy − 2 y + 8xz
The volume V of the solid region is covered by covering the plane region OAB while x varies
from 0 to 2. Thus,
Sol.
zzz e j
zzz b
V
=
2
∇ ⋅ F dV
3
9− y 2
x =0 y =0 z=0
g
4xy − 2y + 8xz dz dy dx
372
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
=
=
=
=
=
zz
z z LMNb g
LM
z MMNa fFGH c
z a f
2 3
0 0
2 3
0 0
2
0
2
0
4xyz − 2 yz + 4xz 2
e
3
3
2
2
3
0
9 4 x − 2 + 72x dx
18x 2 − 18x + 36x 2
zz e
S
z2 = a2 in the first octant.
S
dy dx
jOPQ dy dx
1
I F y IO
4x − 2 − 9 − y hJ + 4xG 9 y − J P dx
K H 3 K PPQ
3
Example 9. Evaluate
zz e
0
4xy − 2y 9 − y 2 + 4 x 9 − y 2
= 180.
Sol.
9− y2
2
0
Fig. 5.9
j
yzi + zxj + xyk ⋅ dS , where S is the surface of the sphere x2 + y2 +
H
yzi + zxj + xyk ⋅ dS
j
=
=
=
zz e
zz b
zz
S
S
(U.P.T.U., 2005)
je
yzi + zxj + xyk ⋅ dy dz i + ⋅ dz dx j + dx dy k
yz dy dz + zx dz dx + xy dx dy
a
a2 − z2
0 0
yz dy dz +
zz
a2 − x 2
a
0 0
g
zx dz dx +
zz
j
a2 − y 2
a
0 0
xy dx dy
Fy I
Fz I
Fx I
= z zG J dz + z xG J dx + z y G J dy
H2K
H 2K
H 2K
1
1
1
ze a − z jdz +
xea − x jdx +
yea − y jdy
=
2
2
2
y I
1Fa z
z I
1Fa x
x I
1Fa y
= 2G 2 − 4 J + 2G 2 − 4 J +2G 2 − 4 J
H
K H
K H
K
a2 − z 2
2
a
0
a2 − x 2
2
a
0
z
0
z
0
a
2
2
0
a
0
2
Example 10. Evaluate
zz
=
2
4
a
a
2 2
4
a
0
4
4
2 2
4
a
0
1a
1a
1a
3a
.
+
+
=
8
2 4 2 4 2 4
F . n ds , where F = zi + xj – 3y2 zk and S is the surface of the
S
f ≡ x2 + y2 – 16
∇f =
2
4
cylinder x2 + y2 = 16 included in the first octant between z = 0 and z = 5.
Sol. Since surface S : x2 + y2 = 16
Let
2
0
0
4
z
0
0
2 2
a2 − y 2
2
a
F i ∂ + j ∂ + k ∂ I (x + y – 16)
GH ∂x ∂y ∂z JK
= 2xi + 2yi
2
2
373
VECTOR CALCULUS
unit normal
n =
n =
∇f
=
∇f
2xi + 2 yj
E
4x 2 + 4 y 2
b
2 xi + yj
2
2 x +y
C
4
16
FG xi + yj IJ = 1 (zx + xy)
H 4 K 4
Here the surface S is perpendicular to xy-plane so we will
take the projection of S on zx-plane. Let R be that projection.
∴
ds =
zz
F . n ds =
S
z ze
z z FGH
R
=
X
B
2
2
Y
x +y =16
Fig. 5.10
j bxi +4 yjg . y4 dx dz
I
JK
zx + xy
dx dz
y
16 − x 2 on S. x is also varies from 0 to 4.
Since z varies from 0 to 5 and y =
zz
O
zi + xj − 3 y 2 zk .
R
F zx + xy I dx dz =
GH y JK
A
dx dz
dx dz
4dx dz
=
=
y
n . j
y
4
R
=
∴
D
g = xi + yj = xi + yi /
2
F . n = (zi + xj – 3y2 zk).
Now
Z
I
F xz
GG
+ xJ dx dz
H 16 − x JK
LM− z 16 − x + x OP dz = b 4z + 8g dz
2Q
N
5
zz
z
4
2
z=0 x=0
5
0
2
2
4
z
5
0
0
= (2z2 + 8z)05 = 50 + 40 = 90.
Example 11. Evaluate
zzz
φdV, where φ = 45x2 y and V is the closed region bounded by
V
the planes 4x + 2y + z = 8, x = 0, y = 0, z = 0.
Sol. Putting y = 0, z = 0, we get 4x = 8 or x = 2
Here x varies from 0 to 2
y varies from 0 to 4 – 2x
and z varies from 0 to 8 – 4x – 2y
Thus
zzz
φdV =
V
=
zzz
z z z
zz
zz b
45x 2 y dx dy dz
V
2
4 − 2x
x=0 y=0
= 45
= 45
2 4− 2x
0 0
2 4 − 2x
0 0
8 − 4x − 2 y
z=0
45x 2 y dx dy dz
8 − 4x − 2 y
x2y z 0
dx dy
x 2 y 8 − 4x − 2 y dx dy
g
374
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
z
z
z
ze
2 O
LM
y
dx
3 PQ
N
2
L
O
= 45 x M 4b 4 − 2xg − 2xb 4 − 2xg − b 4 − 2xg P dx
3
N
Q
45
x b 4 − 2xg dx
=
3
4 − 2x
2
= 45 x 2 4 y 2 − 2xy 2 −
3
0
2
0
2
2
2
3
0
2
3
2
0
j
2
= 15 x 2 64 − 8x 3 − 96x + 48x 2 dx
0
LM 64x − 8x − 96x + 48x OP
N 3 6 4 5 Q
L 512 − 256 − 384 + 1536 OP
= 15M
5 Q
N3 3
3
6
5 2
4
= 15
0
= 128.
EXERCISE 5.4
1. Find the work done by a Force F = zi + xj + yk from t = 0 to 2π, where r = cos t i + sin
tj + tk.
LMHint: Work done = FH ⋅ drH OP.
z PQ
MN
2. Show that
z
C
Ans. 3π
C
F ⋅ dr = –1, where F = (cos y) i – xj – (sin y) k and C is the curve y =
1 − x2
in xy-plane from (1, 0) to (0, 1).
3. Find the work done when a force F = (x2 – y2 + x) i – (2xy + y) j moves a particle from
2
Ans.
origin to (1, 1) along a parabola y2 = x.
3
4.
z
C
LM
N
OP
Q
xy 3 dS where C is the segment of the line y = 2x in the xy plane from A(– 1, – 2, 0) to
LMAns. 16 OP
5Q
N
B(1, 2, 0).
e
j e
j
5. F = 2xzi + x 2 − y j + 2z − x 2 k is conservative or not.
e
j
2
6. If F = 2x − 3z i − 2xyj − 4xk , then evaluate
x = 0, y = 0, z = 0 and 2x + 2y + z = 4.
zzz
Ans. ∇ × F ≠ 0, so non - conservative
H
∇FdV , where V is bounded by the plane
V
LMAns. 8 OP
N 3Q
375
VECTOR CALCULUS
zz
7. Show that
F ⋅ ndS
=
S
bounded by the planes;
3
, where F = 4xzi – y2j + yzk and S is the surface of the cube
2
LMAns. 3 OP
N 2Q
x = 0, x = 1, y = 0, y = 1, z = 0, z = 1.
H
2
8. If F = 2yi − 3 j + x k and S is the surface of the parobolic cylinder y2 = 8x in the first
octant bounded by the planes y = 4, and z = 6 then evaluate
zz
S
.
F ⋅ ndS
Ans. 132
bx − ygi + bx + yg j evaluate Ü A ⋅ dr around the curve C consisting of y = x and
LMAns. 2 OP
y = x.
N 3Q
9. If A =
2
2
10. Find the total work done in moving a particle in a force field A = 3xyi − 5zj + 10xk along
the curve x = t2 + t, y =2t2, z = t3 from t = 1 to t = 2.
Ans. 303
11. Find the surface integral over the parallelopiped x = 0, y = 0, z = 0, x = 1, y = 2, z = 3
H
when A = 2xyi + yz2j + xzk.
Ans. 33
12. Evaluate
zzz
∇ × AdV , where A = (x + 2y) i – 3zj + x k and V is the closed region in the
V
LMAns. 8 b3i − j + 2kgOP
N 3
Q
first octant bounded by the plane 2x + 2y + z = 4.
zzz
13. Evaluate
V
fdV where f = 45x2y and V denotes the closed region bounded by the
planes 4x + 2y + z = 8, x = 0, y = 0, z = 0.
e
j
Ans. 128
2
14. If A = 2x − 3z i − 2xyj − 4xk and V is the closed region bounded by the planes x = 0,
y = 0, z = 0 and 2x + 2y + z = 4, evaluate
e
j
3
15. If A = x − yz i − 2 x yj + 2 k evaluate
3
zzz e
zzz e j
V
V
LMAns. 8 b j − kgOP
N 3 Q
j
∇ × A dV.
∇ ⋅ A dV over the volume of a cube of side b.
LMAns. 1 b OP
N 3 Q
3
16. Show that the integral
zaa ffe
3, 4
1, 2
j e
j
xy 2 + y 3 dx + x 2 y + 3xy 2 dy is independent of the path joining the points (1, 2) and
Ans. 254
(3, 4). Hence, evaluate the integral.
17. If F = ∇ φ show that the work done in moving a particle in the force field F from (x1, y1,
z1) to B (x2, y2, z2) is independent of the path joining the two points.
18. If F = (y – 2x)i + (3x + 2y)j, find the circulation of F about a circle C in the xy-plane with
centre at the origin and radius 2, if C is transversed in the positive direction. [Ans. 8π]
19. If F (2) = 2i – j + 2k, F (3) = 4i – 2j + 3k then evaluate
z
3
2
F.
dF
dt
dt .
[Ans. 10]
376
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
2 0 . Prove that F = (4xy – 3x2z2)i + 2x2 j – 2x3 zk is a conservative field.
2 1 . Evaluate
z Sz F . n ds where F = xyi – x j + (x + z)k, S is the portion of plane 2x + 2y + z
2
LMAns. 27 OP
N 4Q
= 6 included in the first octant.
2 2 . Find the volume enclosed between the two surfaces S1 : z = 8 – x2 – y2 and S2 : z
= x2 + 3y2.
5.19
[ Ans. 8π 2 ]
GREEN’S* THEOREM
If C be a regular closed curve in the xy-plane bounding a region S and P(x, y) and Q (x, y) be
continuously differentiable functions inside and on C then
(U.P.T.U., 2007)
zz b
g
Pdx + Qdy =
C
zz FGH
S
I
JK
∂Q ∂P
−
dx dy
∂x ∂y
Y
y = f2(x)
Proof: Let the equation of the curves AEB and AFB
are y = f1(x) and y = f2 (x) respectively.
Consider
∂P
dx dy =
S ∂y
zz
=
=
a f ∂P
dy dx
x = a y = f a x f ∂y
b
y = f axf
P bx , y g
dx
y = f axf
a
b
F
d
S
f2 x
C
B
zz
z
z b g b g
zb g zb g
zb g zb g
zb g
zb g
zz
z z b bg g
z b g b g
z b g zb g
z b g
zz
A
1
c
y = f1(x)
E
2
1
b
a
= −
P x, f 2 − P x, f1 dx
a
b
= −
P x, f 2 dx −
P x, y dx −
BFA
= −
O
b
a
P x, f 1 dx
a
b
X
Fig. 5.11
P x, y dx
AEB
P x , y dx
BFAEB
⇒
zz
∂P
S ∂y
dx dy = −
C
P x, y dx
...(i)
Similarly, let the equations of the curve EAF and EBF be x = f1 (y) and x = f2 (y) respectively,
∂Q
then
S ∂x
dx dy =
=
⇒
∂Q
S ∂x
dx dy =
* George Green (1793–1841), English Mathematician.
d
f2 y
y = c x = f1 y
d
c
C
d
∂Q
dx dy =
Q f2 , y − Q f1 , y dy
c
∂x
c
Q f 2 , y dy + Q f1 , y dy
d
Q x, y dy
...(ii)
377
VECTOR CALCULUS
Adding eqns. (i) and (ii), we get
zb
Pdx + Qdy
C
g=
Vector form of Green’s theorem:
z
F ⋅ dr =
C
Let
F =
Curl F =
zz FGH
S
I
JK
∂Q ∂P
−
dx dy
∂x ∂y
zz e j
∇ × F ⋅ kdS
S
Pi + Qj, then we have
i
∂
∂x
P
j
∂
∂y
Q
e∇ × Fj ⋅ k = FGH ∂∂Qx − ∂∂Py IJK
⇒
z
Thus,
F ⋅ dr =
C
k
∂
∂z
0
=
F ∂Q − ∂P I k
GH ∂x ∂y JK
zz e j
∇ × F ⋅ k dS
S
Corollary: Area of the plane region S bounded by closed curve C.
Let
Q = x, P = – y, then
zb
xdy − ydx
C
g=
zz a f
z
z
S
1 + 1 dxdy
= 2 S dx dy = 2 A
Thus,
area A =
1
( xdy − ydx )
2C
Example 1. Verify the Green’s theorem by evaluating
ze
j e
x 3 − xy 3 dx + y 3 − 2 xy dy
C
j where
C is the square having the vertices at the points (0, 0), (2, 0) (2, 2) and (0, 2). (U.P.T.U., 2007)
ze
Sol. We have
C
By Green’s theorem, we have
zb
C
Here
∴
Pdx + Qdy
g =
j e
x 3 − xy 3 dx + y 3 − 2 xy dy
zz FGH
I
JK
∂Q ∂P
−
dx dy
S ∂x
∂y
P = (x3 – xy3), Q = (y3 – 2xy)
∂P
∂y
= – 3xy2,
∂Q
= – 2y
∂x
j
...(A)
378
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
F ∂Q ∂P I
So zz G ∂x − ∂y J dx dy = z z e–2y + 3xy jdx dy
K
H
2
z z
2
x =0
dx
zz
S
z z
2
ydy + 3
y= 0
2
x= 0
xdx
2
y= 0
y 2 dy
2 2
2 2
3 2
0
0
0
Now, the line integral
zb
g
Pdx + Qdy =
C
ze
ze
OA
+
O
...(i)
j
j e
j
x 3 − xy 3 dx + y 3 − 2xy dy +
ze
j e
ze
g
Pdx + Qdy =
AB
j
x 3 − xy 3 dx + y 3 − 2xy dy +
C
z ze
2 3
2
0
0
x dx +
j
y 3 − 4y dy +
ze
0
2
j e
ze
j e
x 3 − xy 3 dx + y 3 − 2xy dy
CO
j
z
2
4
0
⇒
zb
g
2
2
4
0
0
2
0
4 0
2
2
2
= 4 – 4 + 12 – 4
Pdx + Qdy = 8
C
j
x 3 − 8x dx + y 3 dy
O Lx
Lx O L y
O Ly O
= M 4 P + M 4 − 2 y P + M 4 − 4x P + M 4 P
PQ N
N Q MN
Q MN PQ
4
j
x 3 − xy 3 dx + y 3 − 2xy dy
along OA, y = 0 ⇒ dy = 0 and x = 0 to 2
along AB, x = 2 ⇒ dx = 0 and y = 0 to 2
along BC, y = 2 ⇒ dy = 0 and x = 2 to 0
along CO, x = 0 ⇒ dx = 0 and y = 2 to 0
zb
X
A
(2, 0)
(0, 0)
Fig. 5.12
j e
BC
∴
B
(2, 2)
x 3 − xy 3 dx + y 3 − 2xy dy
C
=
But
C
(0, 2)
LM y OP + 3LM x OP LM y OP = −8 + 16
MN 2 PQ MN 2 PQ MN 3 PQ
F ∂Q − ∂P I dx dy = 8
GH ∂x ∂y JK
= –2 x 02
⇒
Y
2
x =0 y = 0
S
= –2
2
...(ii)
Thus from eqns. (i) and (ii) relation (A) satisfies. Hence, the Green’s theorem is verified.
Hence proved.
Example 2. Verify Green’s theorem in plane for
ze
C
boundary of the region defined by y2 = 8x and x = 2.
Sol. By Green’s theorem
z
C
b Pdx + Qdyg =
zz FGH
S
I
JK
∂Q ∂P
−
dx dy
∂x ∂y
j e
j
x 2 − 2xy dx + x 2 y + 3 dy , where C is the
379
VECTOR CALCULUS
i.e.,
Line Integral (LI) = Double Integral (DI)
P = x2 – 2xy, Q = x2y + 3
Here,
∂P
∂y
∂Q
= 2 xy
∂x
So the R.H.S. of the Green’s theorem is the double integral given by
DI =
=
= −2x,
I
JK
zz FGH
zz c
∂Q ∂P
−
dx dy
∂x ∂y
S
a fh
2 xy − −2 x dx dy
S
The region S is covered with y varying from –2 2 x of the lower branch of the parabola
to its upper branch 2 2 x while x varies from 0 to 2. Thus
DI =
=
zz b g
z
z
ze j e j
2
8x
x = 0 y= − 8x
2
0
xy 2 + 2xy
2xy + 2x dy dx
8x
dx
− 8x
3
128
5
The L.H.S. of the Green’s theorem result is the line integral
2
= 8 2
LI =
0
x 2 dx =
x 2 − 2 xy dx + x 2 y + 3 dy ⋅
C
Here C consists of the curves OA, ADB. BO. so
LI =
zz
z z z
=
OA + ADB + BO
C
=
+
OA
ADB
+
BO
= LI 1 + LI 2 + LI 3
Y
2
Along OA: y = –2 2 x , so dy = – dx
x
LI1 =
ze
z
j e
0
2
y =
j
8x
B (2, 4)
x 2 − 2xy dx + x 2 y + 3 dy
OA
2
j
F 2I
+ x e−2 2 x j + 3 G − J dx
H xK
F
I
= z G 5x + 4 2 ⋅ x − 3 2 x J dx
H
K
=
x=2
e
x 2 − 2x –2 2 x dx
D (2, 0)
O (0, 0)
2
2
0
2
32
−
1
2
A (2, –4)
Fig. 5.13
X
380
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
L 5x + 4 2 2 x − 3 2 ⋅ 2 x OP
= M
5
MN 3
PQ
40 64
+
− 12
=
3
5
x = 2, dx = 0
Along ADB :
LI2 =
=
Along BO :
z e
zb
ADB
4
2
5
2
3
j e
0
j
x 2 − 2xy dx + x 2 y + 3 dy
g
4 y + 3 dy = 24
–4
y = 2 2 x , with x : 2 to 0.
dy =
LI3 =
2
dx
x
ze
z FH
BO
j e
0
=
3
−
1
IK
5x 2 − 4 2 x 2 + 3 2 x 2 dx
2
= −
j
x 2 − 2xy dx + x 2 y + 3 dy
40 64
+
− 12
3
5
LI = LI1 + LI2 + LI3 =
⇒ Hence the Green’s theorem is verified.
Example 3. Apply Green’s theorem to evaluate
FG 40 + 64 − 12IJ + a24f +FG − 40 + 64 − 12IJ = 128
H3 5 K
H 3 5 K 5
ze
C
j e
j
2x 2 − y 2 dx + x 2 + y 2 dy where C is the
boundary of the area enclosed by the x-axis and the upper half of the circle x2 + y2 = a2.
(U.P.T.U., 2005)
Sol. By Green’s theorem
2
2
zz
z z
zz
z bPdx + Qdyg = FGH ∂∂Qx − ∂∂Py IJK dx dy
LM ∂ ex + y j − ∂ e2x − y jOPdx dy
=
∂y
N ∂x
Q
b2x + 2ygdx dy = LMN2xy + 2y2 OPQ dx.
=
= z 2x a − x + e a − x j dx = 0 + 2z e a − x jdx
C
y= a –x
S
a
a2 −x2
2
2
2
2
x =− a y = 0
a
–a
z
a 2 − x2
−a 0
a
2
2
2
O
Fig. 5.14
a
a 2 −x 2
2
a
−a
0
a
2
−a
S
2
2
0
[First integral vanishes as function is odd]
LM
N
= 2 a 2x −
OP
Q
a
LM
N
x3
a3
= 2 a3 −
3 0
3
OP = 4 a .
Q 3
3
382
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Let
y = 3 sin θ in first and y = sin φ in second integral.
z
z
π
π
sin 4 θ cos θ
sin 4 φ cos φ
dθ − 4 2
dφ
0
0
cos θ
cos φ
L.H.S. = 4 × 81 2
5 1
5 1
= 4 × 81 2 2 − 4 2 2 = 60 π
2 3
2 3
⇒
L.H.S. = R.H.S.
Hence Green’s theorem is verified.
Example 5. Find the area of the loop of the folium of Descartes
x3 + y3 = 3axy, a > 0.
Sol. Let
y = tx.
...(i)
3
3
3
∴
x + t x = 3ax.tx
giving
...(ii)
0
x dy − y dx
1
x2 ⋅
2 C
x2
=
y
1
x 2d
x
2 C
=
1 2
x dt , as y = tx
2 C
=
1
9a 2 t 2
dt using (ii)
2 C 1 + t3 2
1
3t 2
0
1 + t3
t=0
=
=
2
= 3a
O
a
z
z FGH IJK
z
ze j
ze j
g
+
zb
1
x dy − y dx
2 C
y
required area =
x
t=1
x+
Hence,
3at
x =
1 + t3
y=
Fig. 5.16
dt, by summetry
2
LM 1 OP = 3 a .
N 1+t Q 2
= 3a2 −
1
2
3
0
Example 6. Using Green’s theorem, find the area of the region in the first quadrant bounded
1
x
by the curves y = x, y =
,y=
.
(U.P.T.U., 2008)
x
4
Sol. By Green’s theorem the area of the region is given by
A =
=
zb
1
xdy − ydx
2 C
1
2
g
LM
O
xdy − ydx g + b xdy − ydxg + bxdy − ydxgP
b
MMC
PP
C
C
N
Q
z
1
z
2
z
3
...(i)
383
VECTOR CALCULUS
Y
B(
C3
y
=
x
)
1,1
C2
A (2,1)
2
C1
y=x
4
(0,0)
O
y=1
x
X
Fig. 5.17
Now along the curve C1 : y =
zb
xdy − ydx
z
g = FGH x dx − x dxIJK = 0
4
4
2
...(ii)
0
C1
along, the curve C2 : y =
zb
zb
1
1
, dy = − 2 dx and x varies from 2 to 1.
x
x
xdy − ydx
g=
C2
xdy − ydx
or
x
dx
or dy =
and x varies from 0 to 2.
4
4
z RST FGH IJK
1
2
x.
1
1
dx − dx
2
x
−x
z
UV =
W
1
2
−
2
dx
x
g = − 2 log x = – 2[log 1 – log 2] = 2 log 3
1
...(iii)
2
C2
along, the curve C3 : y = x, dy = dx and x varies from 1 to 0.
zb
xdy − ydx
z
g = bxdx − xdxg = 0
0
...(iv)
1
C3
Using (ii), (iii) and (iv) in (i), we get the required area
A =
1
[0 + 2 log 2 + 0] = log 2.
2
Example 7. Verify Green’s theorem in the xy-plane for
ze
C
C is the boundary of the region enclosed by y = x2 and y2 = x.
Sol. Here P(x, y) = 2xy – x2
j
e
Y
2
x=y
Q(x, y) = x2 + y2
∂Q
= 2x,
∂x
zb
Pdx + Qdy
C
g =
z z FGH
S
2
I
JK
∂Q ∂P
−
dx dy
∂x ∂y
y=x
A
∂P
= 2x
∂y
By Green’s theorem, we have
j
2xy − x 2 dx + x 2 + y 2 dy , where
C2
(1, 1)
C1
O
...(i)
Fig. 5.18
X
384
∴
z zb
ze
R.H.S. =
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
g
2x − 2x dx dy = 0
S
and
=
ze
ze
j
C
j
e
j
2xy − x 2 dx + x 2 + y 2 dy +
C1
e
j
2xy − x 2 dx + x 2 + y 2 dy
L.H.S. =
ze
j
e
j
2xy − x 2 dx + x 2 + y 2 dy
C2
...(ii)
Along C1 : y = x2 i.e., dy = 2xdx and x varies from 0 to 1
C1
j
e
j
2xy − x 2 dx + x 2 + y 2 dy
ze
ze
1
=
0
1
=
0
C2
j
e
j
2xy − x 2 dx + x 2 + y 2 dy
L x + x OP = 1
4x − x + 2x jdx = Mx −
N 3 3Q
3
2x1 2
z LMNe
0
=
1
z FGH
z FGH
0
=
1
2
5
and x varies from 1 to 0.
j
j 2xdx OPQ
e
2x ⋅ x 1 2 − x 2 dx + x 2 + x .
2x 3 2 − x 2 +
1
1 2
IJ
K
1 32 1 12
x + x
dx
2
2
IJ
K
5 Lx O
=
M P − LMN x3 OPQ + 12 LMN x3 2 OPQ
2N5 2Q
0
=
6 1
3
4
0
dx
Along C2 : y2 = x, 2y dy = dx or dy =
ze
j
2x 3 − x 2 + 2x 3 + 2x 5 dx
5 32
1
x − x 2 + x1 2 dx
2
2
52 0
3 0
32 0
1
1
1
= −1+
1 1
− = −1
3 3
Using the above values in (ii), we get
L.H.S. = 1 – 1 = 0
Thus L.H.S. = R.H.S.; Hence, the Green’s theorem is verified.
EXERCISE 5.5
1. Using Green’s theorem evaluate
ze
C
j
x 2 y dx + x 2 dy , where C is the boundary described
counter clockwise of the triangle with vertices (0, 0), (1, 0), (1, 1).
z {e
j
}
(U.P.T.U., 2003)
5
Ans.
12
LM
N
OP
Q
xy + y 2 dx + x 2dy , where C is the closed curve of
1
Ans. –
the region bounded by y = x2 and y = x.
20
2. Verify Green’s theorem in plane for
C
LM
N
OP
Q
385
VECTOR CALCULUS
3. Evaluate
z
(cos x sin y − xy)dx + sin x ⋅ cos y dy
by Green's theorem where C is the circle
C
x2 + y2 = 1.
4. Evaluate by Green's theorem
zn
s
e − x sin y dx + e − x cos y dy , where C is the rectangle with
C
vertices (0, 0) (π, 0),
[Ans. 0]
F π, 1 πI F 0, 1 πI .
H 2 K H 2 K
[Ans. 2(e–π–1)]
5. Find the area of the ellipse by applying the Green's theorem that for a closed curve C in
the xy-plane.
[Hint: Parametric eqn. of ellipse x = a cos φ, y = a sin φ and φ vary from φ1 = 0 to φ2 = 2π]
[Ans. π ab]
6. Verify the Green's theorem to evaluate the line integral
z
(2y 2 dx + 3x dy) , where C is the
LMAns. 27 OP
N 4Q
C
boundary of the closed region bounded by y = x and y = x2 .
7. Find the area bounded by the hypocycloid x2/3 + y2/3 = a2/3 with a > 0.
O
[Hint: x = a cos3 φ, y = a sin3 φ, φ varies from φ1 = 0 to φ1 = π/2
A]
LMAns. 3πa OP
N
8 Q
2
8. Verify Green's theorem
z
( 3x + 4 y ) dx + (2 x − 3 y) dy with C; x2 + y2 = 4.
C
[Ans. Common value: – 8π]
LMAns. 3a OP
N 2Q
2
9. Find the area of the loop of the folium of descartes x3 + y3 = 3axy, a > 0.
10. Evaluate the integral
z
[Hint: Put y = tx, t : 0 to ∞]
( x 2 − cos hy ) dx + ( y + sin x ) dy, where C: 0 ≤ x ≤ π, 0 ≤ y ≤ l.
C
11. Verify Green’s theorem in the xy-plane for
ze
[Ans. π (cos h 1 – 1)]
j b
g
3x 2 − 8 y 2 dx + 4 y − 6xy dy , where C is the
C
LMAns. 3 OP
N 2Q
1
9π O
L
r dθ =
12. Find the area of a loop of the four - leafed rose r = 3 sin2θ. MHint : A =
z
2
8 PQ
N
13. Verify the Green’s theorem for e y dx + x dyj , where C is the boundary of the square –
2
x and y = x .
region bounded by parabolas y =
1 ≤ x ≤ 1 and –1 ≤ y ≤ 1.
z
C
π 2 2
0
2
2
386
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
14. Using Green’s theorem in the plane evaluate
z
b g
e
j
2 tan −1 y x dx + log x 2 + y 2 dy , where C
C
is the boundary of the circle (x – 1)2 + (y + 1)2 = 4.
5.20
STOKE’S THEOREM
H
If F is any continuously differentiable vector function and S is a surface enclosed by a curve C,
then
z
F ⋅ dr =
C
zz
.
(∇ × F) ⋅ ndS
S
where n is the unit normal vector at any point of S.
(U.P.T.U., 2006)
Proof: Let S is surface such that its projection on the xy, yz and xz planes are regions
bounded by simple closed curves. Let equation of surface f(x, y, z) = 0, can be written as
z = f1 (x, y)
y = f2 (x, z)
x = f3 (y, z)
H
Let
F = F1 i + F2 j + F3 k
Then we have to show that
zz
S
=
∇ × { F1 i + F2 j + F3 k } ⋅ ndS
Considering integral
zz
z
H H
F ⋅ dr
C
, we have
∇ × ( F1 i ) ⋅ ndS
S
LMRSi ∂ + j ∂ + k ∂ UV × F iOP ⋅ ndS
...(i)
NT ∂x ∂y ∂z W Q
L ∂F j − ∂F kOP ⋅ ndS
= M
N ∂z ∂y Q
L ∂F n ⋅ j − ∂F n ⋅ kOP dS
= M
∂y
N ∂z
Q
H
[∇ × (F1 i)]· n dS =
1
1
1
1
Also,
So,
But
So,
Fig. 5.19
1
r = xi + yj + zk
= xi + yj + f1(x, y)k
H
∂r
∂f
= j+ 1k
∂y
∂y
[As z = f(x, y)]
H
∂r
is tangent to the surface S. Hence, it is perpendicular to n .
∂y
n ⋅
∂r
∂y
= n ⋅ j +
...(ii)
∂f1
n ⋅ k = 0
∂y
387
VECTOR CALCULUS
n ·j = −
Hence,
Hence, (ii) becomes
∂f 1
∂z
n ⋅ k = −
n ⋅ k
∂y
∂y
LM ∂F ∂z + ∂F OPn ⋅ k dS
N ∂z ∂y ∂y Q
[∇ × (F1 i)]· n dS = –
1
But on surface S
1
...(iii)
F1 (x, y, z) = F1 [x, y, f1 (x, y)]
∂F1 ∂F1 ∂z
+
⋅
∂y
∂z ∂y
∴
= F(x, y)
...(iv)
∂F
∂y
...(v)
=
Hence, relation (iii) with the help of relation (v) gives
∂F
∂F
( n ·k) dS = –
dx dy
∂y
∂y
H
∂F
−
dx dy
R ∂y
[∇ × (F1 i)]· n dS = –
zz
S
( ∇ × F1 i) ⋅ n dS
zz
=
...(vi)
where R is projection of S on xy-plane.
Now, by Green’s theorem in plane, we have
H
H
∂F
dx dy,
Fdx = −
R ∂y
C1
zz
z
where C1 is the boundary of R.
As at each point (x, y) of the curve C1 the value of F is same as the value of F1 at each point
(x, y, z) on C and dx is same for both curves. Hence, we have
H
H
F dx = F1 dx .
z
z
C1
Hence,
z
C
F1 dx = –
C
From eqns. (vi) and (vii), we have
zz l
zz l
zz l
S
q
∇ × F1 i ⋅ n dS =
z
z
z
C
zz
...(vii)
H
F1 dx
...(viii)
H
∂F
dx dy
R ∂y
Similarly, taking projection on other planes, we have
H
F2 dy .
∇ × F2 j ⋅ n dS =
C
S
H
F3 dz
∇ × F3 k ⋅ n dS =
S
q
q
C
Adding eqns. (viii), (ix), (x), we get
zz
S
⇒
∇ × { F1 i + F2 j + F3 k } ⋅ n dS =
z
H
F ⋅ dr =
C
z
C
zz
{F1 dx + F2 dy + F3 dz}
S
( ∇ × F) ⋅ ndS
.
...(ix)
...(x)
388
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
5.21
CARTESIAN REPRESENTATION OF STOKE'S THEOREM
H
F = F1i + F2 j + F3 k
Let
i
∂
∂x
F1
H
curl F =
=
So the relation
z
H H
F ⋅ dr =
C
is transformed into the form
z
C
zz LMNFGH
{F1 dx + F2 dy + F3 dz} =
S
j
∂
∂y
F2
k
∂
∂z
F3
RS ∂F − ∂F UV i + RS ∂F − ∂F UV j + RS ∂F − ∂F UV k
T ∂y ∂z W T ∂z ∂x W T ∂x ∂y W
3
z
S
2
1
3
2
1
H
curl F ⋅ n dS,
IJ
K
FG
H
IJ
K
FG
H
OP
Q
IJ
K
∂F3 ∂F2
∂F1 ∂F3
∂F2 ∂F1
−
dy dz +
−
−
dz dx +
dx dy .
∂y
∂z
∂x
∂z
∂x
∂y
H
Example 1. Verify Stoke's theorem for F = (x2 + y2) i – 2xy j taken round the rectangle
bounded by x = ± a, y = 0, y = b.
(U.P.T.U., 2002)
H
Sol. We have F ⋅ dr = {(x2 + y2) i – 2xy j}· {dx i + dy j}
= (x2 + y2) dx – 2xy dy
Y
H
H H
H H
H H
H H
H
∴
F ⋅ dr + F ⋅ dr +
F ⋅ dr + F ⋅ dr
F ⋅ dr =
z
z
∴
z
C1
C
C2
zn
zn
z
C3
= I1 + I2 + I3 + I4
z
C4
s
LM 3 y = b OP
=
( x + b ) dx − 0s
N∴ dy = 0Q
= FG x + b xIJ
H3 K
F 2 a + 2b aI
= –
H3
K
n(x + y )dx − 2xy dys
I =
I1 =
(–a, b) D
A (a, b)
C1
( x 2 + y 2 )dx − 2 xy dy
C1
−a
2
C2
C4
2
a
2
a
3
z
2
2
zn
0
b
= 2a
z
s
b
y dy
(–a, 0)
O
C3
B
(a, 0)
X
Fig. 5.20
2
( −a)2 + y 2 0 − 2(−a) y dy
0
X¢
2
C2
=
C
−a
3
LM3 x = − aOP
N∴ dx = 0Q
389
VECTOR CALCULUS
Fy I
= 2a G J = – ab
H2K
2
z
z
z
z
I3 =
2
b
C3
=
( x 2 + y 2 )dx − 2 xy dy
+a
−a
I4 =
C4
x 2 dx =
FG x IJ = 2 a
H3K 3
3
a
3
−a
LM 3 x = 0 OP
N∴ dx = 0Q
−2ay dy
= –2a
= – ab
∴
LM 3 y = 0 OP
N∴ dy = 0Q
x 2 dx
C3
=
0
z
b
0
2
Fy I
y dy = −2 aG J
H2K
b
2
z
0
H H
F ⋅ dr = I1 + I 2 + I 3 + I 4
C
FG 2a + 2b aIJ − ab + 2 a − ab
H3
K
3
3
= –
2
2
= – 4ab2
H
curl F =
Again,
3
2
...(i)
i
∂
∂x
x2 + y2
j
∂
∂y
−2xy
k
∂
∂z
0
= – 4yk
∴
∴
zz
S
n = k
H
n · curl F = k·(– 4yk) = – 4y
H
n ⋅ curl F dS =
zz
z FGH IJK
a
b
−a 0
+a
=
−a
− 4y dx dy
−4
y2
2
b
dx
0
= – 2b 2 ( x) −a a
= – 4ab2.
From eqns. (i) and (ii), we verify Stoke’s theorem.
→
...(ii)
Example 2. Verify Stoke's theorem when F = yi + zj + xk and surface S is the part of the
sphere x2 + y2 + z2 = 1, above the xy-plane.
Sol. Stoke’s theorem is
H H
H
F ⋅ dr =
(curl F ) ⋅ ndS
z
C
zz
S
390
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Here, C is unit circle x2 + y2 = 1, z = 0
H H
Also,
F ⋅ dr = (yi + zj + xk) · (dxi + dyj + dzk)
= ydx + zdy + xdz
H H
ydx + zdy + xdz
F ⋅ dr =
∴
z
z z z
z
z
z
C
C
C
C
Again, on the unit circle C, z = 0
dz = 0
Let
x = cos φ, ∴ dx = – sin φ.dφ
and
y = sin φ, ∴ dy = cos φ.dφ
H H
y dx
F ⋅ dr =
∴
C
C
=
2π
0
= –
sin φ ( − sin φ) dφ
z
= –π
H
curl F =
Again,
2π
0
sin 2 φ dφ
...(i)
i
∂
∂x
y
j
∂
∂y
z
k
∂
∂z = – i – j – k
x
Using spherical polar coordinates
n = sin θ cos φ i + sin θ sin φ j + cos θ k
H
∴
curl F · n = – (sin θ cos φ + sin θ sin φ + cos θ).
H
π/ 2 2 π
(sin θ cos φ + sin θ sin φ + cos θ) sin θ dθ dφ
( curl F) ⋅ n dS = –
Hence,
z z
zz
θ= 0 φ = 0
S
= –
z
π/2
= – 2π
= –π
[sin θ sin φ − sin θ cos φ + φ cos θ]20 π sin θ dθ
z
z
θ= 0
π/2
0
sin θ cos θ dθ
π/ 2
0
sin 2θ dθ
π
(cos 2θ) π0 /2
2
= –π
...(ii)
From eqns. (i) and (ii), we verify Stoke’s theorem.
H
Example 3. Verify Stoke’s theorem for F = xzi − yj + x2yk, where S is the surface of the
region bounded by x = 0, y = 0, z = 0, 2x + y + 2z = 8 which is not included in the xz-plane.
(U.P.T.U., 2006)
Sol. Stoke’s theorem states that
=
z
F ⋅ dr =
C
zz e j
S
∇ × F ⋅ ndS
391
VECTOR CALCULUS
Here C is curve consisting of the straight lines
AO, OD and DA.
L.H.S. =
=
z
F ⋅ dr =
z
z z z
z
z
C
AO
+
OD
Z
D (0, 0, 4)
AO + OD + DA
+
= LI 1 + LI 2 + LI 3
DA
H
On the straight line AO : y = 0, z = 0, F = 0, so
LI1 =
AO
F ⋅ dr = 0
O
On the straight line OD : x = 0, y = 0. F = 0, so
LI2 =
OD
Y
B (0, 8, 0)
X
F ⋅ dr = 0
A (4, 0, 0)
Fig. 5.21
On the straight line DA: x + z = 4 and y = 0, so
F = xzi = x (4 – x) i
LI3 =
z
DA
F ⋅ dr =
z
4
0
x( 4 − x) i ⋅ dxi =
z
4
0
x( 4 − x) dx =
32
3
32
32
=
3
3
Here the surface S consists of three surfaces (planes) S1 : OAB, S2 : OBD, S3 : ABD, so that
LI = 0 + 0 +
R.H.S. =
=
∇× F =
zz
zz
zz zz zz
^
S
( ∇ × F ) ⋅ n dS =
S1
i
∂
∂x
xz
+
S2
j
∂
∂y
−y
On the surface S1: Plane OAB : z = 0,
(∇ × F ) ⋅ n =
+
S3
= SI 1 + SI 2 + SI 3
k
∂
= x2i + x(1 – 2y) j
∂z
x2 y
n = − k , so
x 2 i + x (1 − 2 y ) j ⋅ ( − k ) = 0
zz
=0
(∇ × F) ⋅ ndS
zz
=0
(∇× F) ⋅ ndS
Unit normal n to the surface S3 =
∇(2x + y + 2 z)
|∇(2x + y + 2 z)|
SI1 =
S1 + S2 + S3
S1
On surface S2: Plane OBD : Plane x = 0, n = – i, so
∇×F = 0
SI2 =
S2
On surface S3: Plane ABD : 2x + y + 2z = 8.
392
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
n =
2i + j + 2k
4 +1+ 4
=
2i + j + 2 k
3
2 2 1
x + x (1 − 2 y )
3
3
To evaluate the surface integral on the surface S3, project S3 on to say xz-plane i.e., projection
of ABD on xz-plane is AOD
(∇ × F ) ⋅ n =
dS =
Thus
SI3 =
=
=
dx dz dx dz
= 3dx dz
=
n⋅ j
13
zz
S3
(∇× F) ⋅ ndS
zz LMN
AOD
zz
4
4− x
x=0 z=0
OP
Q
2 2 x
x + (1 − 2 y) 3 dx dz
3
3
2x2 + x (1 − 2y) dz dx
since the region AOD is covered by varying z from 0 to 4 – x, while x varies from 0 to 4. Using
the equation of the surface S3, 2x + y + 2z = 8, eliminate y, then
SI3 =
=
zz n
zz
4 4 −x
4 4− x
0 0
s
2x 2 + x [1 − 2 (8 − 2x − 2z)] dz dx
0 0
(6x2 − 15x + 4xz) dz dx
=
z LMN6x z − 15xz + 4xz2 OPQ
4
2
2
0
=
z
4
0
4− x
dx
0
(23x 2 − 4 x 3 − 28 x) dx =
32
3
Thus L.H.S. = L.I. = R.H.S. = S.I.
zz
Hence Stoke’s theorem is verified.
Example 4. Evaluate
S
over the surface of intersection of the cylinders x2 + y2
(∇× F) ⋅ ndS
= a2, x2 + z2 = a2 which is included in the first octant, given that F = 2yzi – (x + 3y – 2) j
+ (x2 + z)k.
Sol. By Stoke’s theorem the given surface integral can be converted to a line integral i.e.,
SI =
zz
S
=
( ∇ × F ) ⋅ ndS
z
F ⋅ dr = LI
C
Here C is the curve consisting of the four curves C1: x2 + z2 = a2, y = 0; C2: x2 + y2 = a2, z = 0,
C3: x = 0, y = a, 0 ≤ z ≤ a: C4: x = 0, z = a, 0 ≤ y ≤ a.
393
VECTOR CALCULUS
Z
2
2
x +z =a
2
C4
C1
C3
Y
2
2
x +y =a
X
2
C2
LI =
Fig. 5.22
z
F ⋅ dr =
C
z
C1 + C2 + C 3 + C 4
z z z z
=
C1
+
C2
+
C3
+
C4
= LI1 + LI2 + LI3 + LI4
On the curve C1: y = 0; x2 + z2 = a2
LI1 =
=
z
C1
z
0
a
z
F ⋅ dr =
C1
( a 2 − z 2 ) + z dz = −
On the curve C2: z = 0, x2 + y2 = a2
LI2 =
z
C2
z
F ⋅ dr =
ze
a
= –
( x 2 + z) dz
0
C2
2 3 a2
a −
3
2
− ( x + 3 y − 2) dy
j
a 2 − y 2 + 3 y − 2 dy
πa 2 3 2
− a + 2a
4
2
On the curve C3: x = 0, y = a, 0 ≤ z ≤ a
= –
LI3 =
On C4; x = 0, z = a, 0 ≤ y ≤ a
LI4 =
SI =
z
C3
z z
zz e j
F ⋅ dr =
S
z
F ⋅ dr =
a
0
zdz =
a2
2
0
a
( 2 − 3 y) dy = −2a +
= LI =
∇ × F ⋅ ndS
3a 2
2
FG −2a − a IJ
H 3 2K
F πa − 3a + 2aIJ + a + FG −2a + 3a IJ
+ G−
H 4 2 K 2 H
2 K
3
2
2
SI =
− a2
( 3 π + 8 a).
12
2
2
2
394
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
z
Example 5. Evaluate
S
F ⋅ dr by Stoke’s theorem, where F = y 2 i + x 2 j − ( x + z) k and C is
(U.P.T.U., 2001)
the boundary of the triangle with vertices at (0, 0, 0) (1, 0, 0) and (1, 1, 0)
Sol. Since z-coordinates of each vertex of the triangle is zero, therefore, the triangle lies in
the xy-plane and n = k
i
j
k
∂ ∂
∂
curl F =
= j + 2 ( x − y) k
∂x ∂y
∂z
Y
y 2 x 2 −( x + z)
∴
curl F ⋅ n =
j + 2 ( x − y ) k ⋅ k = 2 ( x − y )
B (1, 1)
The equation of line OB is y = x.
By Stoke’s theorem
z
C
F ⋅ dr =
=
=
y=
zz
zz
z LMN
S
curl F ⋅ n dS
1 x
0 0
x
S
2 ( x − y ) dy dx
OP
Q
x
O
z FGH
Fig. 5.23
IJ
K
A (1, 0)
X
1
1
y2
x2
1
2 xy −
dx = 2
x2 −
dx = x 2 dx = .
0
0
0
2 0
2
3
1
z
Example 6. Apply Stoke’s theorem to prove that
z
( ydx + zdy + xdz) = −2 2 πa 2 , where C is the curve given by
x2 + y2 + z2 – 2ax – 2ay = 0, x + y = 2a and begins of the point (2a, 0, 0).
Sol. The given curve C is
x2 + y2 + z2 – 2ax – 2ay = 0
x + y = 2a
C
⇒
(x – a)2 + (y – a)2 + z2 =
b a 2 g2
x + y = 2a
which is the curve of intersection of the sphere
(x – a)2 + (y – a)2 + z2 =
ba 2 g
2
and the plane
x + y = 2a .
Clearly, the centre of the sphere is (a, a, 0) and radius
is a 2 .
Also, the plane passes through (a, a, 0).
Hence, the circle C is a great circle.
∴ Radius of circle C = Radius of sphere =
Now,
z
C
( ydx + zdy + xdz)
=
z
C
2a
( yi + zj + xk ) ⋅ ( dxi + dyj + dzk )
Fig. 5.24
VECTOR CALCULUS
=
=
But
curl (yi + zj + xk) =
z
z
395
C
C
H
( yi + zj + xk ) ⋅ dr
.
curl ( yi + zj + xk ) ⋅ ndS
i
∂
∂x
y
j
∂
∂y
z
[Using Stoke’s theorem]
k
∂
= – i – j – k.
∂z
x
Since S is the surface of the plane x + y = 2a bounded by the circle C. Then
n =
=
∴
b
g
∇bx + y − 2ag
∇ x + y − 2a
i+ j
2
curl (yi + zj + xk) · n = (– i – j – k) ·
= −
Hence, the given line integral =
= –
z
Example 7. Evaluate
1+1
z
2
S
FG i + j IJ
H 2K
=– 2.
− 2 dS
2 (Area of the circle C).
b 2 ag = −2 2πa .
exy dx + xy dyj taken round the positively oriented square with
= –
2
2π
2
2
vertices (1, 0), (0, 1), (– 1, 0) and (0, – 1) by using Stoke’s theorem and verify the theorem.
Sol. We have
z
C
Y
( xy dx + xy 2 dy)
=
=
=
z
C
z
zz
C
B (0, 1)
( xy i + xy 2 j ) ⋅ ( dx i + dy j )
y–x=1
H
( xy i + xy 2 j ) ⋅ dr
C
S
by Stoke’s theorem, where S is the area of the square ABCD.
Now,
2
curl (xy i + xy j) =
A (1, 0)
X¢ (–1, 0)
curl ( xy i + xy 2 j) ⋅ ndS
i
∂
∂x
xy
x+y=1
j
∂
∂y
xy 2
= (y2 – x) k
k
∂
∂z
0
O
X
x–y=1
x + y = –1
(0, –1) D
Y¢
Fig. 5.25
396
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
∴
∴
curl (xy i + xy2j ) · n^ = (y2 – x) k · k
= (y2 – x)
zz
S
=
curl (xy i + xy 2 j) ⋅ ndS
=
=
zz
zz
zz
S
S
S
( y 2 − x) dS
( y 2 − x) dx dy
y 2 dx dy −
zz
= 4
= 4
= 4
= 4
1
1− x
0
0
1
1− x
0
0
1− x
0
0
1
y3
3
0
S
x dx dy
y 2 dx dy − Sx
[By symmetry]
y 2 dx dy − S ⋅ 0
[3 x = x-coordinate of
zz
zz
FG IJ
zH K
z
1
zz
the C.G. of ABCD = 0]
y 2 dx dy
1− x
dx
0
1
4 1
(1 − x) 3 dx = .
3
3 0
Verification of Stoke’s theorem: The given line integral
=
=
z
C
...(i)
( xy dx + xy 2 dy) ,
where C is the boundary of the square ABCD. Now C can be broken up into four parts namely:
(i) the line AB whose equation is x + y = 1,
(ii) the line BC whose equation is y – x = 1,
(iii) the line CD whose equation is x + y = – 1, and
(iv) the line DA whose equation is x – y = 1.
Hence, the given line integral
=
=
z
AB
z
( xy dx + xy 2 dy) +
z
z
BC
( xy dx + xy 2 dy) +
z
z
z
CD
( xy dx + xy 2 dy) +
z
z
DA
( xy dx + xy 2 dy)
RS x (1 − x) dx + (1 − y) y dyUV + RS x (1 + x) dx + (y − 1) y dyUV
W
T
W T
+ { x ( −x − 1) dx + − y (1 + y) dy} + R
ST x ( x − 1) dx +
= 2
1
0
z
1
0
1
2
−1
0
0
z
z
0
−1
−1
x ( x − 1) dx + 2 x (1 + x) dx + 2
0
1
0
1
0
z
2
0
z
2
1
(1 − y) y 2 dy + 2
z
1
0
0
−1
y 2 (1 + y) dy
z
0
−1
y 2 ( 1 + y ) dy
UV
W
397
VECTOR CALCULUS
F x − x IJ + 2 FG x + x IJ + 2FG y − y IJ + 2FG y + y IJ
= 2 G
H3 2K H 3 2K H 3 4K H 3 4K
F 1 1I F 1 1I F 1 1I F 1 1I
= 2 H − K + 2 H− + K + 2 H − K + 2 H − K
3 2
3 2
3 4
3 4
F 1 1I
= 4 H − K
3 4
2
2
1
3
−1
2
0
=
3
4
0
1
3
4
0
0
−1
1
·
3
...(ii)
From eqns. (i) and (ii), it is evident that
H H
H
F ⋅ dr =
curl F ⋅ n dS
z
z
C
C
Hence, Stoke’s theorem is verified.
Example 8. Verify Stoke’s theorem for the function
H
F = (x + 2y) dx + (y + 3x) dy
where C is the unit circle in the xy-plane.
H
F = F1 i + F2 j + F3 k
Sol. Let
H
F · dr = (F1 i + F2 j + F3 k) · (dx i + dy j + dz k)
= F1dx + F2dy + F3dz
Here, F1 = x + 2y, F2 = y + 3x, F3 = 0
or
Unit circle in xy-plane is x2 + y2 = 1
x = cos φ, dx = – sin φ dφ
Hence,
z
y = sin φ, dy = cos φ dφ.
H H
F ⋅ dr = ( x + 2 y ) dx + ( y + 3 x) dy
C
=
=
=
z
z
2π
0
[ −(cos φ + 2 sin φ) sin φ dφ + (sin φ + 3 cos φ) cos φ dφ]
z m
z
2π
0
2π
0
0
2π
=
( 3 cos2 φ − 2 sin 2 φ) dφ
zm
zm
2π
=
r
− sin φ cos φ − 2 sin 2 φ + sin φ cos φ + 3 cos 2 φ dφ
0
r
( 3 cos 2 φ − 2(1 − cos2 φ) dφ
r
5 cos 2 φ − 2 dφ
398
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
z LMN
z RST
2π
=
0
2π
=
0
UV
W
1 5
+ cos 2 φ dφ
2 2
LM
N
OP
Q
2π
=
1
5
φ + sin 2 φ
2
2
0
=
1
[2π + 0] = π.
2
H
curl F =
i
∂
∂x
x + 2y
j
∂
∂y
y + 3x
= i {0} – j {0} + k
Hence,
zz
S
k
∂
∂z
0
RS ∂ ( y + 3 x) − ∂ (x + 2 y)UV
∂y
T ∂x
W
= k (3 – 2) = k.
H
curl F ⋅ n dS =
=
=
=
z
z
zz
z
z
( k ⋅ k ) dS
dS
dx dy
xdy
2π
=
0
z
cos2 φ dφ
=
1 2π
(1 + cos 2φ) dφ
2 0
=
1
sin 2 φ
φ+
2
2
0
= π.
So Stoke’s theorem is verified.
LM
N
OP
Q
5(1 + cos 2 φ)
− 2 dφ
2
OP
Q
2π
399
VECTOR CALCULUS
1. Evaluate
EXERCISE 5.6
zz e
S
j
∇ × A ⋅ ndS where S is the surface of the hemisphere x2 + y2 + z2 = 16 above
the xy-plane and A = (x2 + y – 4) i + 3xyj + (2xz + z2) k.
2. If F = (y2 + z2 + x2) i + (z2 + x2 – y2) j + (x2 + y2 – z2) k evaluate
the surface S = x2 + y2 – 2ax + az = 0, z ≥ 0.
3. Evaluate
zz
S
b
zz e
S
Ans. − 16 π
j
∇ × F ⋅ n dS taken over
Ans. 2πa 3
g
∇ × yi + zj + xk ⋅ n dS over the surface of the paraboloid z = 1 – x2 – y2, z ≥ 0.
Ans. π
4. F = (2x – y) i – yz2j – y2zk, where S upper half surface of the sphere x2 + y2 + z2 = 1.
Hint: Here C, x 2 + y 2 = 1, z = 0
Ans. π
5. Using Stoke’s theorem or otherwise, evaluate
zb
C
g
2 x − y dx – yz 2 dy − y 2 z dz .
where C is the circle x2 + y2 = 1, corresponding to the surface of sphere of unit radius.
Ans. π
6. Use the Stoke’s theorem to evaluate
zb
C
g a f
b g
x + 2 y dx + x − z dy + y − z dz .
where C is the boundary of the triangle with vertices (2, 0, 0) (0, 3, 0) and (0, 0, 6) oriented
Ans. 15
in the anti-clockwise direction.
H
7. Verify Stoke’s theorem for the Function F = x2 i – xy j integrated round the square in the
plane z = 0 and bounded by the lines x = 0, y = 0, x = a, y = a.
LMAns. Common value – a OP
2 PQ
MN
3
8. Verify Stoke’s theorem for F = (x2 + y – 4)i + 3xy j + (2xz + z2)k over the surface of
hemisphere x2 + y2 + z2 = 16 above the xy plane.
Ans. Common value – 16π
9. Verify Stoke’s theorem for the function F = zi + xj + yk, where C is the unit circle in xy
plane bounding the hemisphere z =
10. Evaluate
z
C
e1 − x − y j .
2
2
(U.P.T.U., 2002)
Ans. Common value π
H
F ⋅ dr by Stoke’s theorem for F = yzi + zxj + xyk and C is the curve of inter-
section of x2 + y2 = 1 and y = z2.
Ans. 0
400
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
5.22
GAUSS’S DIVERGENCE THEOREM
If F is a continuously differentiable vector point function in a region V and S is the closed surface
enclosing the region V, then
zz
S
F ⋅ n dS =
zzz
V
diV F dV
...(i)
where n is the unit outward drawn normal vector to the surface S.
(U.PT.U., 2006)
H
Proof: Let i, j, k are unit vectors along X, Y, Z axes respectively. Then F = F1i + F2j + F3k,
where F1, F2, F3, and their derivative in any direction are assumed to be uniform, finite and
continuous. Let S is a closed surface which is such that any line parallel to the coordinate axes
cuts S at the most on two points. Let z coordinates of these points be z = F1 (x, y) and z = F2
(x, y), we have assumed that the equations of lower and upper portions S2 and S1 of S are z = F2
(x, y) and z = F1 (x, y) respectively.
The result of Gauss divergence theorem (i) incomponent form is
zzz b
S
z z z FGH
g
=
F1 i + F2 j + F3 k ⋅ nds
V
I
JK
∂F1 ∂F2 ∂F3
+
+
dV
∂x ∂y
∂z
Now, consider the integral
I1 =
=
zzz
zz LMNz
...(ii)
Z
∂F3
dx dy dz
V ∂z
R
F1 ∂F3
F2
∂z
r
n1 1
K
S1
S
OP
Q
r2
dz dx dy
n2
S2
–K
where R is projection of S on xy-plane.
I1 =
=
zz
zz b
zz b
R
R
F1 x , y
2
g b
g dx dy
X
F x , y , F g dx dy − zz F b x , y , F g dx dy
F3 x , y , F1 − F3 x , y , F2
3
1
R
3
2
For the upper portion S1 of S,
dx dy = k ⋅ n 1 ⋅ dS1
where n 1 is unit normal vector to surface dS1 in outward direction.
For the lower portion S2 of S.
dx dy = – k ⋅ n 1 ⋅ dS2
where n 2 is unit normal vector to surface dS2 in outward direction.
Thus, we have
zz b
zz b
R
and
R
g
F x, y, F g dx dy
zz
zz
F3 x, y, F1 dx dy
=
3
= –
2
S1
F3 k ⋅ n 1 dS1
S2
F3 k ⋅ n 2 dS 2
dS2
Y
O
b g dx dy
F3 bx, y, zg
F bx, yg
R
dS1
R dx dy
Fig. 5.26
VECTOR CALCULUS
So
zz b
R
g
F3 x , y , F1 dx dy −
401
zz b
R
g
F3 x , y , F2 dx dy
=
=
=
Hence,
I1 =
=
zz b g zz b g
zz
S1
F3 k ⋅ n 1 dS1 +
S2
b
F3 k ⋅ n 2 dS2
F3 k . n 1 dS1 + n 2 dS 2
g
zz c h
zzz
zz c h
zz c h
zz c h
zz n c h c h c hs
S
F3 k ⋅ n dS
= n 1 S1 + n 2 S2
3 nS
∂F3
dx dy dz
V ∂z
S1
F3 k ⋅ n dS
...(iii)
Similarly, projecting S on other coordinate planes, we have
zzz
zzz
∂F3
dx dy dz
V ∂y
∂F1
V ∂x
=
dx dy dz =
Adding eqns. (iii), (iv), (v)
zzz RST
V
⇒
⇒
or
5.23
UV
W
∂F1 ∂F2 ∂F3
+
+
dx dy dz =
∂x
∂y
∂z
zzz RST
V
i
UV
W
S
S
F2 j ⋅ n dS
...(iv)
F1 i ⋅ n dS
...(v)
S
F1 i . n + F2 j . n + F3 k . n dS
∂
∂
∂
⋅ {F1i + F2 j + F3k} dx dy dz
+k
+j
∂x
∂y
∂z
zzz
=
V
H
div F dV
zz
S
=
=
H
F ⋅ n ds
=
zz m
zz
zz
zzz
S
S
S
r
F1 i + F2 j + F3 k ⋅ n dS
H
F ⋅ n dS
H
F ⋅ n dS
V
H
div F dV .
CARTESIAN REPRESENTATION OF GAUSS’S THEOREM
H
F = F1 i + F2 j + F3 k
Let
where F1, F2, F3 are functions of x, y, z.
and
dS = dS (cos α i + cos β j + cos γ k)
where α, β, γ are direction angles of dS. Hence, dS cos α, dS cos β, dS cos γ are the orthogonal
projections of the elementary area dS on yz-plane, zx-plane and xy-plane respectively. As the
mode of sub-division of surface is arbitrary, we choose a sub-division formed by planes parallel
to yz-plane, zx-plane and xy-plane. Clearly, its projection on coordinate planes will be rectangle
with sides dy and dz on yz-plane, dz and dx on zx-plane, dx and dy on xy-plane.
402
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Hence, projected surface elements are dy dz on yz-plane, dz dx on zx-plane and dx dy on
xy-plane.
H
F1 dy dz + F2 dz dx + F3 dx dy
∴
...(i)
F ⋅ n dS =
zz
zzz
z
zz
S
S
By Gauss divergence theorem, we have
H
H
div F dV.
F ⋅ n dS =
...(ii)
V
S
In cartesian coordinates,
dV = dx dy dz.
H
H
Also,
div F = ∇ · F
=
Hence,
z
V
H
div F dV
=
∂F1 ∂F2 ∂F3
+
+
∂x
∂y
∂z
zzz RST
zz m
zzz RST
V
UV
W
∂F1 ∂F2 ∂F3
+
+
dx dy dz
∂x
∂y
∂z
Hence cartesian form of Gauss theorem is,
...(iii)
r
F1 dy dz + F2 dz dx + F3 dx dy
S
UV
W
Example 1. Find zz F ⋅ n dS , where F = b2 x + 3 zg i – b xz + y g j + e y + 2 zj k and S is the
=
V
∂F1 ∂F2 ∂F3
+
+
dx dy dz .
∂x
∂y
∂z
2
S
surface of the sphere having centre at (3, – 1, 2) and radius 3.
(U.P.T.U., 2000, 2005)
Sol. Let V be the volume enclosed by the surface S. Then by Gauss divergence theorem, we
have
zz
S
F ⋅ n dS
=
=
=
zzz
zzz LMN b g b g e
zzz a f zzz
V
V
V
div F dV
jOPQ
∂
∂
∂ 2
y + 2z dV
− xz − y +
2x + 3z +
∂x
∂y
∂z
2 − 1 + 2 dV = 3
V
dV = 3V
But V is the volume of a sphere of radius 3.
∴
Hence
zz
S
V =
F ⋅ n dS
Example 2. Evaluate
2
2
2
zz e
S
af
4
3
π 3 = 36π.
3
= 3 × 36π = 108π .
j
y 2 z 2i + z 2x 2 j + z 2 y 2 k ⋅ n dS , where S is the part of the sphere
x + y + z = 1 above the xy-plane and bounded by this plane.
Sol. Let V be the volume enclosed by the surface S. Then by divergence theorem, we have
zz e
S
j
y 2 z 2 i + z 2 x 2 j + z 2 y 2 k ⋅ n dS =
=
zzz e
zzz LMN e j
V
V
j
div y 2 z 2 i + z 2 x 2 j + z 2 y 2 k dV
e j
e
∂ 2 2
∂ 2 2
∂ 2 2
y z +
z x +
z y
∂x
∂y
∂z
jOPQ dV
VECTOR CALCULUS
zzz
=
V
2zy 2 dV = 2
403
zzz
zy 2 dV
V
Changing to spherical polar coordinates by putting
x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ
dV = r2 sin θ dr dθ dφ
π
and those of φ will be
2
To cover V, the limits of r will be 0 to 1, those of θ will be 0 to
0 to 2π.
∴
2
zzz
V
zy 2 dV
= 2
= 2
z z za
2π
π 2 1
0
0
0
z z z
2π
π2
1
0
0
0
2π
π 2
0
0
fe
j
r cosθ r 2 sin 2 θ sin 2 φ r 2 sin θ dr dθ dφ
r 5 sin3 θ cos θ sin 2 φ dr dθ dφ
Lr O
= 2 z z sin θ cos θ sin φ M P dθ dφ
N6Q
Example 3. Evaluate
zz
S
z
3
2
z
6 1
0
π
1
1
sin 2 φ ⋅ dφ =
sin 2 φ dφ = .
12 0
12 0
12
=
2π
2π
F ⋅ n dS over the entire surface of the region above the xy-plane
bounded by the cone z2 = x2 + y2 and the plane z = 4, if F = 4xzi + xyz 2 j + 3zk .
Sol. If V is the volume enclosed by S, then V is bounded by the surfaces z = 0, z = 4, z2 =
x2 + y2.
By divergence theorem, we have
=
=
zz
S
F ⋅ n dS =
zzz LMN a f e j a fOPQ
j
zz z e
zz z a f
z
zza f
zza
LM
OP
a
f
z MN
PQ
ze
z a fLMN OPQ
∂
∂
∂
xyz 2 +
4xz +
3z dV =
V ∂x
∂y
∂z
z2 − y2
4 z
0 − z − z2 − y2
= 2
= 2
= 4
z2 − y2
4 z
0 –z 0
4
z
0
−z
4
0
V
div F dV
V
j
4z + xz 2 + 3 dV
4 z + xz 2 + 3 dx dy dz
4z + 3 dx dy dz, since
4z + 3
z 2 − y 2 dy dz = 4
y z2 − y2
4z + 3
zzz
zzz e
2
− z2 − y2
4
z
0
0
y
z2
+ sin −1
z
2
z2 − y2
x dx = 0
f z − y dy dz
4z + 3
2
z
dz
0
= 4
4
0
4
3
= π z +z
j
π 4 3
z2
4z + 3z 2 dz
sin −1 1 dz = 4 ×
4 0
2
4z + 3
4
0
a
f
= π 256 + 64 = 320π .
2
404
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 4. By transforming to a triple integral evaluate
I =
zz e
S
x 3 dy dz + x 2 y dz dx + x 2 z dx dy
j
(U.P.T.U., 2006)
where S is the closed surface bounded by the planes z = 0, z = b and the cylinder x2 + y2 = a2.
Sol. By divergence theorem, the required surface integral I is equal to the volume integral
zzz LMN e j
V
e jOPQ
z e e j j e3x + x + x j dx dy dz
e j
∂ 3
∂ 2
∂ 2
x +
x y +
x z dV
∂x
∂y
∂z
=
zz
z z ze j
zze j
b
a2 − y2
a
z =0 y = – a x = –
= 4×5
b
2
2
2
a2 − y2
a2 − y2
a
z = 0 y= 0 x = 0
3
20 b a
a2 − y2 2
=
3 z = 0 y= 0
LM x OP e
x dx dy dz = 20
N3Q
LM a − y zOP
20
dy dz =
3
MNe j PQ
2
z
π
3
a
z = 0 y= 0
z
a
a
f
20 2 3
20 4
b a cos 3 t a cos t dt =
a b
3 0
3
a2 − y2
j
dy dz
x= 0
3
2 2
2
y=0
Put y = a sin t so that dy = a cos t dt.
∴ I=
zz
b
b
dy =
z=0
20
3
z
a
y=0
z
e
j
3
b a 2 − y 2 2 dy .
π
20 4 3 π 5 4
2 cos 4 t dt =
a b
= πa b .
0
3
4.2 2 4
H
Example 5. Verify divergence theorem for F = (x2 – yz) i + (y2 – zx) j + (z2 – xy) k taken over
the rectangular parallelopiped 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c.
[U.P.T.U. (C.O.), 2006]
H
H
∂ 2
∂ 2
∂ 2
x − yz +
y − zx +
z − xy = 2x + 2 y + 2z .
Sol. We have div F = ∇ · F =
∂x
∂y
∂z
H
∇ ⋅ F dV =
2 x + y + z dV
∴ Volume integral =
= 2
zz zb
z z LMN
z LMN
c
b
a
z =0 y = 0 x = 0
e
zzz
V
j
e
b
x2
+ yx + zx
dy dz
2
x =0
zzz b g
z z LMN
z LMMN
j
V
g
x + y + z dx dy dz = 2
c
z= 0 y = 0
e
OP
Q
a
OP dz
OP
= 2
PQ
Q
O
a b ab
= 2
+
+ abz P dz
2
2
Q
L a b z + ab z + ab z OP = [a bc + ab c + abc ] = abc (a + b + c).
= 2M
2Q
N2 2
c
b
z = 0 y= 0
c
y2
a2
a2
+ ay + az dy dz = 2
+ azy
y+a
z =0
2
2
2
2
c
2
z= 0
2
2
2
c
2
2
2
0
Surface integral: Now we shall calculate
H
F ⋅ n dS
zz
S
Over the six faces of the rectangular parallelopiped.
Over the face DEFG,
n
= i, x = a.
b
y =0
j
405
VECTOR CALCULUS
Therefore,
zz e
zz e
DEFG
c
=
b
=
b
z = 0 y= 0
z
c
Z
j e
C
a 2 − yz i + y 2 − 2a j + z 2 − ay k ⋅ i dy dz
c
2
z= 0
2
c
2
2
=
zz
zz
2
G
0
ABCO
c
zz b g
z LMMN OPPQ
b
z =0 y = 0
A
O
Y
2 2
F
X
Over the face ABCO, n = – i, x = 0. Therefore
=
E
y =0
2
F ⋅ n dS =
B
D
2 b
2
LM a b − zb OP dz = LM a bz − z b OP = a bc – c b
2 PQ
4
4 .
MN
MN
PQ
2
z= 0
F ⋅ n dS
j e j
L
y O
a − yzj dy dz = z Ma y − z P dz
2 PQ
MN
z = 0 y= 0
c
=
zz
Fig. 5.27
a f
0 − yz i + y j + z k ⋅ −i dy dz
y 2z
2
c
yz dy dz =
z=0
2
2
b
dz =
y=0
z
b2
b2c 2
zdz =
z=0 2
4
c
Over the face ABEF, n = j, y = b. Therefore
H
c
a
x 2 − bz i + b 2 − zx j + z 2 − bx k ⋅ j dx dz
F ⋅ n dS =
zz
zze
zz e
=
c
a
z = 0 x= 0
j e
z =0 x = 0
ABEF
j
b 2 − zx dx dz = b 2 ca −
j e
j
a 2c2
.
4
Over the face OGDC, n = – j, y = 0. Therefore
zz
H
F ⋅ n dS =
zz
H
F ⋅ n dS =
OGDC
z z
c
a
z= 0 x =0
zx dx dz =
c 2 a2
.
4
Over the face BCDE, n = k, z = c. Therefore
BCDE
zz e
c 2 − xy dx dy = c 2 ab −
z z
xy dx dy =
b
a
y = 0 x =0
j
a 2b 2
·
4
Over the face AFGO, n = – k, z = 0. Therefore
zz
zz
AFGO
H
F ⋅ n dS =
b
a
y=0 x=0
a2 b 2
·
4
Adding the six surface integrals, we get
S
H
F ⋅ n dS =
F a bc − c b + c b I + F b ca − a c + a c I + F c ab − a b + a b I
GH
4
4 JK GH
4
4 JK GH
4
4 JK
2 2
2
2 2
2
2 2
2 2
2
2 2
2 2
= abc (a + b + c).
Hence, the theorem is verified.
H
Example 6. If F = 4xzi – y2j + yzk and S is the surface bounded by x = 0, y = 0, z = 0,
H
F ⋅ n dS .
x = 1, y = 1, z = 1, evaluate
zz
S
406
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Sol. By Gauss divergence theorem,
H
H
∇ ⋅ F dV , where V is the volume enclosed by the surface S
F ⋅ n dS =
zz
zzz
zzz LMN a f e j b gOPQ zzz b
zzz b g z z z b g
zz
z z b g
z LMMN OPPQ z LMN OPQ z
j
ze
V
S
=
=
V
1
=
=
∂
∂
∂
− y2 +
yz dV =
4xz +
∂z
∂x
∂y
V
4z − y dx dy dz =
1
x =0 y = 0
x =0
Example 7. Evaluate
S
z=0
1
dx =
y =0
1
1
0
2−
g
4z − 2 y + y dV
4z − y dx dy dz
x =0 y = 0 z= 0
dx dy =
1
y2
2y −
2
1
1
2z 2 − yz
1
V
1
1
x=0
y= 0
2 − y dx dy
1
3 1
3
dx =
dx = .
2
2 0
2
y 2 z 2 i + z 2 x 2 j + z 2 y 2 k ⋅ n dS, where S is the part of the sphere
x2 + y2 + z2 = 1, above the xy-plane and bounded by this plane.
H
Sol. We have
F = y2z2 i + z2x2 j + z2y2 k
H
∂ 2 2
∂ 2 2
∂ 2 2
y z +
z x +
z y
div F =
∂x
∂y
∂z
e
j
e j
e
j
= 2zy2
∴
zzz
Given integral =
V
2 zy 2 dV
[By Gauss’s divergence theorem]
where V is the volume enclosed by the surface S, i.e., it is the hemisphere x2 + y2 + z2 = 1, above
the xy-plane.
Z
Y
2
z= 1–x –y
2
y= 1–x
2
dx dy dz
dx dy
Z=0
X¢
O
X
O
X
dx dy
y=–1–x
Y¢
Y
(i)
(ii)
Fig. 5.28
From the above Figure 5.28 (i) and (ii), it is evident that
limits of z are from 0 to
1 − x2 − y2
limits of y are from – 1 − x 2 to
and limits of x are from –1 to + 1.
1 − x2
2
407
VECTOR CALCULUS
∴ Given integral
= 2
= 2
z z
1
z
1− x 2
1 −x 2 − y 2
zy 2 dx dy dz
x = –1 y = – 1 − x 2 z = 0
z z
1
Fz I
GH 2 JK
x = –1 y = – 1 − x 2
z z
z LMMNe
1− x 2 − y 2
2
1− x 2
y 2 dx dy
0
e1 − x − y j y dx dy
y
y O
=
− P
1− x j
dx
3
5 PQ
R| 1 1 − x − 1 1 − x U| dx
= 2z
S| 3 e j 5 e j V|
T
W
R|
|V dx
1− x j U
= 4z S b1 − x g − e
|W
|T 3
5
8
=
e1 − x j dx
15
=
1
1− x 2
2
2
2
x = –1 y = – 1− x 2
1
3
2
5
x = –1
1− x 2
– 1− x 2
5
2 2
1
x =−1
2 5/2
2 5/2
1
5
2 2
0
z
ze
z
5
2 2
1
0
π
=
j
5
8 2
1 − sin 2 θ 2 cos θ dθ
15 0
[Putting x = sin θ]
π
8 2
cos 2 θ dθ
15 0
8 5π π
⋅
=
=
·
15 32 12
H
H
Example 8. Evaluate F ⋅ n dS where F = (x + y2) i – 2x j + 2yz k where S is surface bounded
=
z
S
by coordinate planes and plane 2x + y + 2z = 6.
Sol. We know from Gauss divergence theorem,
H
H
F ⋅ n dS =
div F dV
S
V
H
F = (x + y2) i – 2x j + 2yz k
zz
zzz
F i ∂ + j ∂ + k ∂ I ⋅ {ex + y j i − 2x j + 2yz k}
GH ∂x ∂y ∂z JK
∂
=
ex + y j + ∂∂y a–2xf + ∂∂z b2yzg
∂x
H
div F =
2
2
Let
zzz zzz
zzz b g
zzz b g
= 1 + 2y
H
F ⋅ n dS =
I =
S
=
=
V
V
1 + 2 y dV
1 + 2 y dx dy dz
H
div F dV
408
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
6 − 2x − y
2
Limit of y is 0 to 6 – 2x
Limit of x is 0 to 3
Limit of z is 0 to
Hence,
zzz b g b g
zz b g k p
zz b gb g
t
zz o
UV
z RST
W
z LMN b g b g b g
UV
z RST
W
1 + 2 y dx dy dz
I =
6 − 2 x − y /2
1 + 2y
=
1
2
1
=
2
=
=
=
=
z 0
dx dy
1 + 2 y 6 − 2x − y dx dy
6 − 2x + 11y − 4xy − 2 y 2 dx dy
6 − 2x
1
2
6 y − 2xy +
1
2
6 6 − 2 x − 2 x 6 − 2x +
1
2
−
LM
N
11 2
2
y − 2xy 2 − y 3
2
3
0
dx
b
g
b
g OPQ
11
2
2
2
3
dx
6 − 2x
6 − 2 x − 2 x 6 − 2x −
2
3
8 3
x + 26x 2 − 84x + 90 dx
3
OP
Q
3
1
2
26 3
− x4 +
x − 42x 2 + 90x
2
3
3
0
=
1
− 54 + 234 − 378 + 270
2
1
=
72 = 36 .
2
=
Example 9. Verify Gauss divergence theorem for
zz {e
S
j
}
x 3 − yz dy dz − 2 x 2 y dz dx + z dx dy
over the surface of cube bounded by coordinate planes and the planes x = y = z = a
H
Sol. Let
F = F1 i + F2 j + F3 k.
From Gauss divergence theorem, we know
H
H
F ⋅ n dS =
F1 dy dz + F2 dz dx + F3 dx dy =
div F dV
zz
zz
S
Here,
So,
S
V
F1 = x3 – yz, F2 = – 2x2y, F3 = z
H
F = (x3 – yz) i – 2x2y j + z k
H
div F
F i ∂ + j ∂ + k ∂ I ⋅ {ex − yzj i − 2x y j + z k}
GH ∂x ∂y ∂z JK
∂
=
ex − yzj + ∂∂y e– 2x yj + ∂∂z azf
∂x
3
=
3
Hence,
zz
S
H
F ⋅ n dS
zzz
2
= 3x2 − 2x2 + 1 = x2 + 1
=
zzz e
V
j
x 2 + 1 dV
2
...(i)
409
VECTOR CALCULUS
=
=
zzze j
z z e jk p
zze j
z e jm r
ze j
a a a
x 2 + 1 dx dy dz
0 0 0
a a
a a
= a
x 2 + 1 dx dy
0 0
a
= a
a
= a2
a
x2 + 1
0
a
x 2 + 1 z 0 dx dy
0 0
y 0 dx
x 2 + 1 dx
0
Ra U
= a S + xV
T3 W
|R a |U a + a
= a S + aV =
|T 3 |W 3
a
3
2
0
3
2
5
3
...(ii)
Verification by direct integral: Outward drawn unit vector normal to face OEFG is – i and
dS is dy dz.
If I1 is integral along this face,
H
H
F ⋅ n dS =
F ⋅ − i dy dz
I1 =
z zz b g
zz e j
zz
z R|S|T U|V|W
z LMMN OPPQ
S
=
=
=
S
x − yz dy dz
3
S
a a
0 0
a
0
[As x = 0 for this face]
yz dy dz
z2
2
y
a
dy
0
2
a2 y
y dy =
0
2 2
a2
=
2
a
a
0
=
a4
4
For face ABCD, its equation is x = a and n dS = i dy dz ,
If I2 is integral along this face
I2 =
=
=
=
=
zz
zz e
zze
z R|S|T
z RST
S
H
F ⋅ i dy dz
S
a
0
a
0
k
j
G
x 3 − yz dy dz
a a
0 0
Z
j
D
a 3 − yz dy dz
U|V dy
|W
a U
a a − y V dy
2W
a3 z − y
3
z2
z
F
–i
C
j
a
a
–j
a O
0
a
E
A
2
B
X
i
–k
Fig. 5.29
Y
410
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
L a y OP
= Ma y −
MN 2 2 PQ
a
2
2
4
0
4
a
4
If I3 is integral along face OGDA whose equation is
y = 0
= a5 −
Hence,
n dS = – j dxdz
H
I3 =
F ⋅ − j dx dz
zz b g
zz
S
= –
S
– 2x 2 y dx dz
= 0, as y = 0.
If I4 is integral along face BEFC whose equation is
y = a
n dS =
Then
zz
j dx dz
I4 = –
S
2 x 2 y dx dz
zz
z kp
z
a a
x 2 dx dz
=
– 2a
=
– 2a x 2 z 0 dx
=
– 2a 2 x 2 dx
0 0
a
a
0
a
0
Lx O 2
= – 2a M P = – a .
N3Q 3
3
2
a
5
0
If I5 is integral along face OABE whose equation is
z = 0
n dS = – k dx dy
H
F ⋅ – k dx dy
I5 =
zz b
zz
S
=
–
S
g
z dx dy = 0 as z = 0.
If I6 is integral along face CFGD whose equation is
z = a
n dS = k dx dy
I6 =
zz
z
= a
Total surface
S
z dx dy =
a
0
a
zz
a a
y 0 dx = a 2
a dx dy
z
0 0
a
0
dx = a 3
I = I1 + I2 + I3 + I4 + I5 + I6
411
VECTOR CALCULUS
a4
a4
2
+ a5 –
+ 0 – a5 + 0 + a3
4
4
3
=
a5
+ a3
3
which is equal to volume integral. Hence Gauss theorem is verified.
=
Example 10. Evaluate by Gauss divergence theorem
zz {
S
e
j
e
j
zz
S
Here,
Hence,
zzz
S
V
F1 = xz2 ; F2 = x2y – z3, F3 = 2xy + y2z.
H
F = xz2 i + (x2y – z3) j + (2xy – y2z) k
H
div F
2
zz
zzz e
2
= z2 + x2 + y2.
H
F ⋅ n ds =
I =
S
=
Limit of z is 0 to
H
div F dV
F ∂ i + ∂ j + ∂ kI
GH ∂x ∂y ∂z JK {xz i + (x y – z ) j + (2xy + y z) k}
∂
=
exz j + ∂∂y ex y – z j + ∂∂z e2xy + y zj
∂x
=
2
Let
}
xz 2 dy dz + x 2 y − z 3 dz dx + 2 xy + y 2 z dx dy
where S is surface bounded by z = 0 and z = a 2 − x 2 − y 2 .
H
F = F1 i + F2 j + F3 k.
Sol. Let
Cartesian form of Gauss divergence theorem is
H
F ⋅ n dS =
F1 dy dz + F2 dz dx + F3 dx dy =
zz
...(iii)
2
2
zzz
V
2
3
3
2
2
H
div F dV
j
x + y + z 2 dx dy dz
a2 − x2 − y 2
− a 2 − x 2 to
Limit of x is – a to a
a2 − x2
Limit of y is
I =
=
zzz e
zz LMMNe
j
x 2 + y 2 + z 2 dx dy dz
x +y
2
2
j
z3
z+
3
OP
PQ
a2 − x 2 − y2
dx dy
0
LM
O
a −x −y j P
e
PP dx dy
= zz Mex + y j a − x − y +
3
MM
PQ
N
a − x − y |U
|R
a − x − y Sx + y +
=
V| dx dy
3
|T
W
2
zz
2
2
2
2
2
2
2
2
3
2 2
2
2
2
2
2
2
412
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
=
=
Let
1
3
1
3
zz
zz
zz
zz
o
t
a − x − y o2x + 2 y + a t dx dy
a 2 − x 2 − y 2 3x 2 + 3y 2 + a 2 − x 2 − y 2 dx dy
2
2
=
1 a a2 −x2
3 − a − a2 −x2
=
a2 − x 2
2 a
3 −a 0
2
2
2
2
{e2x + a j a − x − y + 2y a − x − y } dx dy
RSe2x + a j a − x − y + 2y a − x − y UV dx dy
T
W
y =
a 2 − x 2 sin θ
dy =
a 2 − x 2 cosθ dθ
I =
2 a 2
3 −a 0
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
z z LNMe je j e j
LM
OP
z MMNe je j e j PPQ
z LMNe je j e j OPQ
z LMN e je j e j OPQ
π
2x 2 + a 2
a 2 − x 2 cos 2 θ + 2 a 2 − x 2
2
OP
Q
sin 2 θ cos 2 θ dx dθ
3 1
3 3
2 2 + 2 a 2 − x 2 2 2 2 dx
2 2
2 3
=
2 a
2
2
3 − a 2x + a
a −x
=
2 a
2x 2 + a 2
3 −a
a2 − x2
=
4 π a
×
2 2x 2 + a 2
3 8 0
=
a
π
2 2x 2 a 2 − 2x 4 + a 4 − a 2x 2 + a 4 + x 4 − 2 2 x 2 dx
0
3×2
=
ze
ze
z e
2
2
2 π
π
dx
+ 2 a2 − x2
4
16
a2 − x 2 + a2 − x 2
2
dx
j
j
π a 4
3a − 3x 4 dx
0
6
π a 4
a − x 4 dx
2 0
j
πL
x O
=
a x−
M
P
2N
5Q
=
5
4
a
0
=
z zb
π 4 5 2π 5
× a =
a .
2 5
5
Example 11. Using the divergence theorem, evaluate the surface integral
g
yz dy dz + zx dz dx + xy dy dx , where S : x2 + y2 + z2 = 4.
(U.P.T.U., 2008)
S
Sol. Let
F = F1i + F2j + F3k
From Gauss divergence theorem, we have
zz
S
F . n dS =
zz
S
F1 dy dz + F2 dz dx + F3 dx dy =
zzz
V
div F dV
...(i)
413
VECTOR CALCULUS
Comparing L.H.S. of (i) with given integral, we get
F1 = yz, F2 = zx, F3 = xy
So
⇒
F = F1i + F2 j + F3k
F = (yz)i + (zx)j + (xy)k
F i ∂ + j ∂ + k ∂ I .{(yz)i + (zx)j + (xy)k}
GH ∂x ∂y ∂z JK
∂
∂
∂
=
yzg +
zxf + b xyg = 0
a
b
∂x
∂y
∂z
div F =
z zb
Thus
g
yz dy dz + zx dz dx + xy dy dx =
S
zzz
0. dV = 0.
V
EXERCISE 5.7
1. Use divergence theorem to evaluate
zz
H
3
3
3
F ⋅ dS where F = x i + y j + z k and S is the surface
S
LMAns. 12πa OP
5 Q
N
5
of the sphere x2 + y2 + z2 = a2.
2. Use divergence theorem to show that
surface enclosing volume V.
zz e
zz
S
j
∇ x 2 + y 2 + z 2 dS = 6V Where S is any closed
3. Apply divergence theorem to evaluate SFn dS , where F = 4x 3 i − x2 yj + x2 zk and S is
the surface of the cylinder x2 + y2 = a2 bounded by the planes z = 0 and z = b.
zz b
4. Use the divergence theorem to evaluate
S
Ans. 3ba 4 π
g
x dy dz + y dz dx + z dx dy , where S is the
portion of the plane x + 2y + 3z = 6 which lies in the first octant.
(U.P.T.U., 2003)
Ans. 18
zz
5. The vector field F = x i + zj + yzk is defined over the volume of the cuboid given by 0
2
≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c enclosing the surface S. Evaluate the surface integral
6. Evaluate
zz e
S
LM
N
j
b
2
IJ OP
KQ
yzi + zxj + xyk dS where S is the surface of the sphere x2 + y2 + z2 = a2 in the
zz e
S
F ⋅ dS.
(U.P.T.U., 2001) Ans. abc a +
first octant.
7. Evaluate
FG
H
S
j
(U.P.T.U., 2004) Ans. 0
e x dy dz – ye x dz dx + 3z dx dy , where S is the surface of the cylinder x2 + y2
= c2, 0 ≤ z ≤ h.
Ans. 3π hc 2
414
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
8. Evaluate
zz
S
F ⋅ n ds , where F = 2xyi + yz2j + xzk, and S is the surface of the region
LMAns. 351 OP
2 Q
N
bounded by x = 0, y = 0, z = 0, y = 3 and x + 2z = 6.
9. F = 4xi – 2y2j + z2k taken over the region bounded by x2 + y2 = 4, z = 0 and z = 3.
e
Ans. Common value 8 π
j
10. F = x 3 − yz i – 2x 2 yj + zk taken over the entire surface of the cube 0 ≤ x ≤ a, 0 ≤ y ≤ a,
LMAns. Common value a + a OP
3
N
Q
5
0 ≤ z ≤ a.
3
11. F = 2xyi + yz2j + xzk and S is the total surface of the rectangular parallelopiped bounded
by the coordinate planes and x = 1, y = 2, z = 3.
Ans. Common value 33
12. F = x 2 i + y 2 j + z 2 k taken over the surface of the ellipsoid
x2 y2 z2
+
+
= 1.
a 2 b2 c 2
Ans. Common value 0
13. F = xi + yj taken over the upper half on the unit sphere
x2 + y2 + z2 = 1.
14. Prove that
15. Evaluate
zzz zz
dV
zz
V
r
2
=
S
r. n
r2
LMAns. Common value 4π OP
3 Q
N
ds .
r. n ds , where S : surface of cube bounded by the planes x = –1, y = –1,
S
z = –1, x = 1, y = 1, z = 1.
[Ans. 24]
OBJECTIVE TYPE QUESTIONS
A. Pick the correct answer of the choices given below:
1. If r = xi + yj + zk is position vector, then value of ∇(log r) is
(i)
r
r
(iii) −
(ii)
r
r2
r
(iv) None of these
r3
2. The unit vector normal to the surface x2y + 2xz = 4 at (2, –2, 3) is
1
(i – 2j + 2k)
3
1
(iii)
(i + 2j – 2k)
3
(i)
(ii)
1
(i – 2j – 2k)
3
(iv) None of these
[U.P.T.U., 2008]
415
VECTOR CALCULUS
3. If r is a position vector then the value of ∇rn is
(i) nrn–2 r
(ii) nrn–2
2
(iii) nr
(iv) nrn–3
4. If f(x, y, z) = 3x2y – y3z2, then |∇f| at (1, –2, –1) is
(i) 481
(ii)
381
(iii)
(iv)
481
581
e j
5. If a is a constant vector, then grad r . a is equal to
(i) r
(ii) − a
(iii) 0
(iv) a
6. The vector rn r is solenoidal if n equals
(i) 3
(ii) – 3
(iii) 2
(iv) 0
7. If r is a position vector then div r is equal to
(i) 3
(ii) 0
(iii) 5
(iv) –1
8. If r is a position vector then curl r is equal to
(i) – 5
(ii) 0
(iii) 3
(iv) –1
9. If F = ∇φ, ∇2φ = – 4πρ where P is a constant, then the value of
zz
F. n dS is:
S
(i) 4π
(ii) – 4πρ
(iii) – 4πρV
(iv) V
10. If n is the unit outward drawn normal to any closed surface S, the value of
zzz
div F dV
V
is
(i) V
(ii) S
(iii) 0
(iv) 2S
zz
11. If S is any closed surface enclosing a volume V and F = xi + 2yj + 3zk then the value
of the integral
F. n dS is
S
(i) 3V
(iii) 2V
12. The integral
(ii) 6V
zz
(iv) 6S
r 5 n dS is equal
S
(i) 0
(ii)
zzz
V
5r 3 . r dV
416
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
(iii)
zzz
zzz
5r −3 . r dV
(iv) None of these
V
13. A vector F is always normal to a given closed surface S in closing V the value of the
integral
curl F dV is :
V
(i) 0
(ii) 0
(iii) V
(iv) S
B. Fill in the blanks:
1. If f = (bxy – z3)i + (b – 2)x2 j + (1 – b)xz2 k has its curl identically equal to zero then
b = ..........
2. ∇ 2
FG 1IJ = ..........
H rK
3. div grad f = ..........
4. curl grad f = ..........
5. grad r = ..........
6. grad
1
= ..........
r
7. If A = 3x yz2 i + 2x y3 j – x2 yz k and f = 3x2 – yz then A . ∇f = ..........
8. If r = r, then ∇f(r) × r = ..........
9. If r = r, then
a f = ..........
∇f r
∇r
10. The directional derivative of φ = xy + yz + zx in the direction of the vector i + 2j + k
at (1, 2, 0) is = ..........
11. The value of
zb
g
xdy − ydx around the circle x2 + y2 = 1 is ..........
12. If ∇2φ = 0, ∇2ψ = 0, then
13.
zz
z z FGH
S
IJ
K
∂ψ
∂φ
dS = ..........
−ψ
∂n
∂n
F . n dS is called the .......... dF over S.
14. If S is a closed surface, then
15.
φ
zzz
V
∇. F dV =
zz
zz
r . n dS = ..........
S
A dS , then A is equal to ..........
S
C. Indicate True or False for the following statements:
1. (i) If v is a solenoidal vector then div v = 0.
417
VECTOR CALCULUS
(ii) div v represents the rate of loss of fluid per unit volume.
(iii) If f is irrotational then curl f ≠ 0.
(iv) The gradient of scalar field f(x, y, z) at any point P represents vector normal to the
surface f = const.
2. (i) The gradient of a scalar is a scalar.
(ii) Curl of a vector is a scalar.
(iii) Divergence of a vector is a scalar.
(iv) ∇f is a vector along the tangent to the surface f = 0.
3. (i) The directional derivative of f along a is f. a .
(ii) The divergence of a constant vector is zero vector.
(iii) The family of surfaces f(x, y, z) = c are called level surfaces.
(iv) If a and b are irrotational then div
ea × bj = 0.
4. (i) Any integral which is evaluated along a curve is called surface integral.
(ii) Green’s theorem in a plane is a special case of stoke theorem.
(iii) If the surface S has a unique normal at each of its points and the direction of this
normal depends continuously on the points of S, then the surface is called smooth
surface.
(iv) The integral
z
z ze j
F . dr is called circulation.
S
5. (i) The formula
curl F . n ds =
S
z
F . dr is governed by Stoke’s theorem.
C
(ii) If the initial and terminal points of a curve coincide, the curve is called closed
curve.
(iii) If n is the unit outward drawn normal to any closed surface S, then
(iv) The integral
1
2
zb
V
g
xdy − ydx represents the area.
C
D. Match the Following:
e
j
1. (i) ∇. ∇ × a
(ii) curl (φ grad φ)
e j
(a) dφ
(b) 0
(iii) div a × r
(c) 0
(iv) ∇φ . d r
(d) r curl a
2. (i) ∇2 r2
(ii) df/ds
zzz
(a) a . (∇f)
(b) grad f ± grad g
∇. n dv ≠ S .
418
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
(iii) a . ∇ f
e j
(c) ∇f .
(iv) ∇(f ± g)
(d) 6
3. (i) grad of φ along n
(ii)
d2r
dt 2
F aI
GH a JK
(a) a (acceleration)
∂φ
n
∂n
(b)
(iii) curl v
(c) curl f = 0
(iv) irrotational
(d) 2 ω
ANSWERS TO OBJECTIVE TYPE QUESTIONS
A. Pick the correct answer:
1. (ii)
5. (iv)
9. (iii)
2. (ii)
6. (ii)
10. (ii)
3. (i)
7. (i)
11. (ii)
4. (iv)
8. (ii)
12. (ii)
2. 0
3. ∇2 f
4. 0
7. –15
8. 0
11. 2π
12. 0
13. (i)
B. Fill in the blanks:
1. 4
r
r
r
6.
9. f ′(r)
10.
13. flux
14. 3V
15. F . n
1. (i) T
(ii) T
(iii) F
(iv) F
2. (i) F
(ii) T
(iii) T
(iv) F
3. (i) F
(ii) F
(iii) T
(iv) T
4. (i) F
(ii) F
(iii) T
(iv) T
5. (i) T
(ii) T
(iii) F
(iv) T
1. (i) ® (b)
(ii) ® (c)
(iii) ® (d)
(iv) ® (a)
2. (i) ® (d)
(ii) ® (c)
(iii) ® (a)
(iv) ® (b)
3. (i) ® (b)
(ii) ® (a)
(iii) ® (d)
(iv) ® (c)
5.
r3
10
3
C. True or False:
C. Match the following:
GGG
UNSOLVED QUESTION PAPERS (2004–2009)
B.Tech.
(Only for the Candidates Admitted/Readmitted in the Session 2008–09)
(SEM. I) Examination, 2008-09
MATHEMATICS–I
Time: 3 Hours]
[Total Marks: 100
SECTION A
All parts of this question are compulsory
compulsory..
1. (a) For which value of ‘b’ the rank of the martix.
LM1 5 4 OP
A = M0 3 2 P is 2, b = 4cm
→
MNb 13 10PQ
2 × 10 = 20
—
^
(b) Determine the constants a and b such that the curl of vector A = (2xy + 3yz)i +
^
^
(x2 + axz – 4z2)j + (3xy + byz)k is zero, a = .........., b = .......... .
(c) The nth derivative (yn) of the function y = x2 sin x at x = 0 is .......... .
(d) With usual notations, match the items on right hand side with those on left hand
side for properties of maximum and minimum:
(i) Maximum
(p) rt – s2 = 0
(ii) Minimum
(q) rt – s2 < 0
(iii) Saddle point
(r) rt – s2 > 0, r > 0
(iv) Failure case
(s) rt – s2 > 0 and r < 0
(e) Match the items on the right hand side with those on left hand side for the following
special functions : (Full marks is awarded if all matchings are correct).
(i) β(p, q)
(ii)
p q
p+q
(p)
b 1 2g
z b gb g
∞
(q)
0
y p−1
dy
1+ y p+q
(iii)
π
(r) β(p, q)
(iv)
π
sin pπ
(s)
p 1− p
Indicate T
Trr ue or False for the following statements:
(f)
(i) If |A| = 0, then at least one eigen value is zero.
(ii) A–1 exists iff 0 is an eigen value of A.
419
(True/False)
(True/False)
420
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
(iii) If |A| ≠ 0, then A is known as singular matrix.
(True/False)
(iv) Two vectors X and Y is said to be orthogonal Y, XT Y = YT X ≠ 0. (True/False)
(g)
(i) The curve y2 = 4ax is symmetric about x-axis.
3
3
2
2
(True/False)
(ii) The curve x + y = 3axy is symmetric about the line y = – x.
(True/False)
2
(iii) The curve x + y = a is symmetric about both the axis x and y. (True/False)
(iv) The curve x3 – y3 = 3axy is symmetric about the line y = x.
Pick the correct answer of the choices given below:
^
^
(True/False)
^
(h) If r = xi + yj + zk is position vector, then value of ∇ (log r) is :
—
(i)
r⋅
r
(iii) −
(i)
(j)
r⋅
(ii)
r⋅
r2
(iv) None of these
r3
a f
b g
The Jacobian
∂ uv
for the function u = ex sin y, v = (x + log sin y) is
∂ xy
(i) 1
(ii) sin x sin y – xy cos x cos y
ex
.
x
The volume of the solid under the surface az = x2 + y2 and whose base R is the circle
x2 + y2 = a2 is given as
(iii) 0
(iv)
(i) π|2a
(ii) πa3|2
(iii)
4 3
πa
3
(iv) none of these.
SECTION B
10 × 3 = 30
Attempt any three parts of the following:
–1
2
2
2
2. (a) If y = (sin x) prove that yn(0) = 0 for n odd and yn(0) = 2, 2 , 4 , 62 ...(n – 2)2,
n ≠ 2 for n is even.
(b) Find the dimension of rectangular box of maximum capacity whose surface area is
given when (a) box is open at the top (b) box is closed.
(c) Find a matrix P which diagonalizes the matrix A =
LM4 1OP , verify P . AP = D where
N2 3Q
–1
D is the diagonal matrix.
(d) Find the area and the mass contained m the first quadrant enclosed by the curve
FG x IJ + FG y IJ = 1 where α > 0, β > 0 given that density at any point p(xy) is k xy .
H aK H bK
α
z zb
β
(e) Using the divergence theorem, evaluate the surface integral
S
g
yz dy dz + zx dz dx + xy dy dx where S : x2 + y2 + z2 = 4.
421
UNSOLVED QUESTION PAPERS
SECTION C
Attempt any two parts from each question. All questions are compulsory.
3. (a) Trace the curve r2 = a2 cos 2θ
F ex + y j I
GG bx + yg JJ , prove that
K
H
2
(b) If u = log
x
2
∂u
∂u
+y
= 1.
∂x
∂y
(c) If V = f(2x – 3y, 3y – 4z, 4z – 2x), compute the value of 6Vx + 4Vy + 3Vz.
4. (a) The temperature ‘T’ at any point (xyz) in space is T(xyz) = K xyz2 where K is constant. Find the highest temperature on the surface of the sphere x2 + y2 + z2 = a2.
(b) Verify the chain rule for Jacobians if x = u, y = u tan v, z = w.
(c) The time ‘T’ of a complete oscillation of a simple pendulum of length ‘L’ is governed
⋅L
, g is constant, find the approximate error in the calculated
g
value of T corresponding to an error of 2% in the value of L.
5. (a) Determine ‘b’ such that the system of homogeneous equation 2x + y + 2z = 0;
x + y + 3z = 0; 4x + 3y + bz = 0 has (i) Trivial solution, (ii) Non-trivial solution. Find
the Non-trivial solution using matrix method.
by the equation T = 2π
(b) Verify Cayley-Hamilton theorem for the matrix A =
FG 1 2 IJ and hence find A .
H 2 −1K
–1
(c) Find the eigen value and corresponding eigen vectors of the matrix.
I=
FG −5 2 IJ .
H 2 −2K
6. (a) Find the directional derivative of ∇(∇f) at the point (1, –2, 1) in the direction of the
normal to the surface xy2z = 3x + z2 where f = 2x3 y2z4.
(b) Using Green’s theorem, find the area of the region in the first quadrant bounded by
the curves
1
x
,y = .
x
4
^
^
^
(c) Prove that (y2 – z2 + 3yz)i + (3xz + 2xy) j + (3xy – 2xz + 2z) k is both solenoidal and
irrotational.
7. (a) Changing the order of integration of
y = x, y =
zz
∞
∞
0
0
e − xy sin nx dx dy
Show that
z FGH
∞
0
IJ
K
sin nx
π
dx = .
x
2
(b) Determine the area bounded by the curves xy = 2, 4y = x2 and y = 4.
(c) For a β function, show that
β(p, q) = β(p + 1, q) + β(p, q + 1)
GGG
B.Tech., First Semester Examination, 2007-08
MATHEMATICS–I
Time: 3 Hours]
[Total Marks: 100
Note: Attempt all the problems. Internal choices are mentioned in every problem.
1. Attempt any two parts of the following:
(10 × 2 = 20)
(a) Define the eigen values, eigen vectors and the characteristic equation of a square
matrix. Find the characteristic equation/polynomial, eigen values and eigen vectors of
the matrix:
LM2
MM5
N7
5
3
0
7
1
2
OP
PP
Q
(b) Check the consistency of the following system of linear non-homogeneous equations
and find the solution, if exists:
7x1 + 2x2 + 3x3 = 16
2x1 + 11x2 + 5x3 = 25
x1 + 3x2 + 4x3 = 13
(c) Find the inverse of the matrix
LM 1 1 1 OP
MM 21 13 15 PP
MM 3 5 7 PP
MN 15 71 111 PQ
2. Attempt any two parts of the following:
(10 × 2 = 20)
(a) State Leibnitz theorem for nth differential coefficient of the product of two functions.
1
−
1
If y m + y m
= 2x, prove that
(x2 – 1)yn + 2 + (2n + 1)xyn + 1 + (n2 – m2)yn = 0
2
(b) Verify that
(c) If u = x sin
F
GH
∂ u
x2 + y2
∂2 u
, where u(x, y) = log e
=
∂y ∂x
∂x ∂y
xy
−1
I
JK
F x I + y sin FG y IJ , find the value of x ∂ u + 2xy ∂ u + y ∂ u .
GH y JK
H xK
∂x ∂y
∂y
∂x
−1
2
2
2
2
3. Attempt any four parts of the following:
FG π IJ .
H 4K
(a) Expand ex cos y about the point 1,
422
2
2
2
(5 × 4 = 20)
423
UNSOLVED QUESTION PAPERS
a
b
f
g
∂ u, v, w
of the following:
∂ x, y, z
(b) Calculate the Jacobian
u = x + 2y + z
v = x + 2y + 3z
w = 2x + 3y + 5z
(c) Discuss the maxima and minima of the function:
f(x, y) = cos x cos y cos(x + y)
(d) Find a point on the ellipse 4x2 + y2 = 4 nearest to the point (1, 2).
(e) Find the extreme value x2 + y2 + z2 subject to the condition
xy + yz + zx = p.
1
(f) If f(x, y) = x 2 y 10 , compute the value of f when x = 1.99 and y = 3.01.
4. Attempt any four of the following:
(a) Evaluate the following by changing into polar coordinates:
zz
a
(5 × 4 = 20)
a −y
2
0 0
2
y 2 x 2 + y 2 dx dx
(b) Find the area enclosed between the parabola y = 4x – x2 and the line y = x.
(c) Change the order of integration in
zz
a
0
2a−x
x2
a
b g
f x , y dx dy
F xI F yI F zI
(d) Find the volume of the solid surrounded by the surface G J + G J + G J = 1.
H aK H b K H cK
2
3
2
3
2
3
(e) Define Gamma and Beta functions. Prove that B(l, m) B(l + m, n) B(l + m + n, p)
=
l m n p
l+m+n+ p
z e
j
10
1 5
x 1 − x 3 dx
0
(f) Show that
=
5. Attempt any two of the following:
1
396
(10 × 2 = 20)
(a) If r = xi + yj + zk then show that
(i)
e j
∇ a ⋅ r = a , where a is a constant vector.
(ii) grad r =
(iii) grad
r
r
1
r
= – 3 , where a is a constant vector.
r
r
(iv) grad rn = nr n−2 r
→
r
when r =
e
j
e j e j
(b) Prove that a × ∇ × r = ∇ a ⋅ r − a ⋅ ∇ r , where a
r = xi + yj + zk .
(c) State the Green’s theorem. Verify it by evaluating
zc
c
is a constant vector and
h
c
h
x 3 − xy 3 dx + y 2 − 2xy dy where
C is the square having the vertices at the points (0, 0), (2, 0), (2, 2) and (0, 2).
GGG
B.Tech., First Semester Examination, 2006-07
MATHEMATICS–I
Time: 3 Hours]
Note:
[Total Marks: 100
(i) Attempt All questions.
(ii) All questions carry equal marks.
(iii) In case of numerical problems assume data wherever not provided.
(iv) Be precise in your answer.
1. Attempt any four parts of the following:
(a) If y = x log (1 + x), prove that yn =
e
a−1f
a
j
n− 2
f
a f a f
n− 2 x+n / 1+x
(5 × 4 = 20)
n
a f
2
(b) If x = tan y, prove that 1 + x y n + 1 + 2 nx − 1 y n + n n − 1 y n − 1 = 0
(c) If u(x, y, z) = log(tan x + tan y + tan z) prove that sin 2x
∂u
∂u
∂u
+ sin 2 y
+ sin 2 z
= 2.
∂x
∂y
∂z
(d) State and prove Euler’s theorem for partial differentiation of a homogeneous function
f(x, y).
∂u
∂u
y
x
+ tan −1 , prove that x ∂x + y ∂ y = 0
y
x
(f) Trace the curve y2(a – x) = x3, a > 0
2. Attempt any two parts of the following:
(a) If u3 + v3 = x + y, u2 + v2 = x3 + y3
(e) If u(x, y) = sin −1
show that
(10 × 2 = 20)
a f = y −x
b g 2uvau − vf
∂ u, v
∂ x, y
2
2
2
2
(b) Find Taylor series expansion of function on f(x, y) = e–x –y cos xy about the point x0 = 0,
y0 = 0 up to three terms.
(c) Find the minimum distance from the point (1, 2, 0) to the cone z2 = x2 + y2.
3. Attempt any two parts of the following:
(10 × 2 = 20)
(a) Define the gradient, divergence and curl.
(i) If f(x, y, z) = 3x2y – y3z2, find grad f at the point (1, – 2, – 1).
H
(ii) If F x, y, z = xz 3 i − 2x 2 yzj + 2 yz 4 k , find divergence and curl of F x, y, z .
b
g
b
g
(b) State Gauss divergence theorem. Verify this theorem by evaluating the surface integral
as a triple integral
z ze
S
j
x 3 dy dz + x 2 ydz dx + x 2 z dx dy , where S is the closed surface consisting of the
cylinder x2 + y2 = a2, (0 ≤ z ≤ b) and the circular discs z = 0 and z = b (x2 + y2 ≤ a2).
b
g
(c) State the Stoke’s theorem. Verify this theorem for F x, y, z = xzi − yj + x 2 yk , where the
surface S is the surface of the region bounded by x = 0, y = 0, z = 0, 2x + y + 2z = 8
which is not included on xz-plane.
424
425
UNSOLVED QUESTION PAPERS
4. Attempt any four parts of the following:
(a) Find the rank of matrix
(5 × 4 = 20)
LM 2 3 − 2 4 OP
MM 3 − 2 1 2 PP
MN − 32 24 30 45 PQ
(b) Solve the system of equations
2x + 3x2 + x3 = 9
x + 2x2 + 3x3 = 6
3x1 + x2 + 2x3 = 8
by Gaussian elimination method.
(c) Find the value of λ for which the vectors (1, –2, λ), (2, –1, 5) and (3, –5, 7λ) are linearly
dependent.
F 1 2 2I
(d) Find the characteristic equation of the matrix G 0 2 1 J . Also, find the eigen
GH − 1 2 2 JK
values and eigen vectors of this matrix.
F1
(e) Verify the Cayley-Hamilton theorem for the matrix G 2
GH 3
using this theorem.
LM1
(f) Diagonalize the matrix M1
MN0
6
2
0
2
4
5
I
JJ
K
3
5 . Also, find its inverse
6
OP
PP
Q
1
0 .
3
Note: Following question no. 5 is for New Syllabus only (TAS–104/MA–101 (New)).
5. Attempt any two parts of the following:
(a) Evaluate by changing the variable
zz b g
R
(10 × 2 = 20)
2
x + y dx dy where R is the region bounded by
the parallelogram x + y = 0, x + y = 3x – 2y = 0 and 3x – 2y = 3.
(b) Find the volume bounded by the elliptical paraboloids z = x2 + 9y2 and z = 18 – x2 – 9y2.
(c) Using Beta and Gamma functions, evaluate
z FGH 1 −x x IJK dx
1
0
1
2
3
3
Note: Following question no. 6 is for Old Syllabus only (MA-101 (old)).
6. Attempt any two parts of the following:
(10 × 2 = 20)
(a) In a binomial distribution the sum and product of the mean and variance of the
distribution are
25
50
and
respectively. Find the distribution.
3
3
426
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
(b) From the following data which shows the ages X and systolic blood pressure Y of 12
women, find out whether the two variables ages X and blood pressure Y are correlated?
Ages (X) :
B.P. (Y)
:
56
147
42
125
72
160
36
118
63
149
47
128
55
150
49
145
38
115
42
140
68
152
60
155
(c) (i) If θ is the acute angle between the two regression lines in case of two variables x
and y, show that
tan θ
=
1 − r2 σ xσ y
⋅ 2
r
σ x + σ 2y
where r, σx and σy have their usual meanings. Explain the significance of the
formula when r = 0 and r = ± 1.
(ii) Two variables x and y are correlated by the equation ax + by + c = 0. Show that the
correlation between them is –1 if signs of a and b are alike and +1, if they are
different.
GGG
B.Tech., First Semester Examination, 2005-06
MATHEMATICS–I
Time: 3 Hours]
Note:
[Total Marks: 100
(i) Attempt All questions.
(ii) All questions carry equal marks.
(iii) Question no. 1–4 are common to all candidates.
(iv) Be precise in your answer.
1. Attempt any four parts of the following:
(5 × 4 = 20)
(a) Use elementary transformation to reduce following matrix A to triangular form and
hence the rank of A.
LM 2 3 − 1 − 1 OP
1 −1 − 2 − 4 P
A = M
MM 3 1 3 − 2 PP
N6 3 0 −7 Q
(b) Define unitary matrix. Show that the matrix
LMα + iγ −β + iδOP is a unitary matrix if α + β + γ + δ = 1.
N β + iδ α − iγ Q
2
2
2
2
(c) Reduce the matrix A to diagonal form
A =
LM− 1 2 − 2 OP
MM 1 2 1 PP
N− 1 − 1 0 Q
(d) Find the eigen values and eigen vectors of matrix A
LM1 7 13OP
5
7P
A = M2
MN3 11 5 PQ
[Hint: λ3 – 11λ2 – 95λ – 116 = 0, which cannot be solve.]
(e) Test the consistency of following system of linear equations and hence find the
solution
4x1 – x2 = 12
– x1 + 5x2 – 2x3 = 0
– 2x2 + 4x3 = – 8
(f) State Cayley-Hamilton theorem. Using this theorem find the inverse of the matrix.
A =
LM 2 − 1 1 OP
MM− 1 2 − 1 PP
N 1 −1 2Q
427
428
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
2. Attempt any four parts of the following:
1
in the direction of r , where r = x i + y j + z k .
r2
(a) Find the directional derivative of
(b) Find
zz
(5 × 4 = 20)
F ⋅ n ds , where
a
f b
g e
j
2
F = 2x + 3z i − xz + y j + y + 2z k and s is the surface of sphere having centre
(3, –1, 2) and radius 3.
(c) Show that the vector field F =
potential.
r
is irrotational as well as solenoidal. Find the scalar
r3
e j=
(d) If A is a vector function and φ is a scalar function, then show that ∇ ⋅ φA
φ∇ ⋅ A + A ⋅ ∇φ.
(e) Apply Green’s theorem to evaluate
z
C
2y 2 dx + 3x dy, where C is the boundary of closed
region bounded between y = x and y = x2.
b
g
(f) Suppose F x, y, z = x 3 i + yj + zk is the force field. Find the work done by F along the
line from the (1, 2, 3) to (3, 5, 7).
3. Attempt any four parts of the following:
(a) If y = (sin–1 x)2, prove that
(1 – x2) yn + 2 – (2n + 1) xyn + 1– n2yn = 0
Hence find the value of yn at x = 0.
(b) If u = f(r) and x = r cos θ, y = r sin θ, prove that
af
(5 × 4 = 20)
af
∂ 2u ∂ 2u
1
+
f″ r + f′ r .
∂x 2 ∂y 2 =
r
2
2
2
(c) Trace the curve x 3 + y 3 = a 3 .
(d) Expand tan–1
FG y IJ in the neighbourhood of (1, 1).
H xK
du
.
dx
(f) State Euler’s theorem of differential calculus and verify the theorem for the function
(e) If u = x log(xy), where x3 + y3 + 3xy = 1. Find
u = log
F x4 + y4 I .
GH x + y JK
4. Attempt any two parts of the following:
(10 × 2 = 20)
(a) If J be the Jacobian of the system u, v with respect to x, y and J′ the Jacobian of the
system x, y with respect to u, v then prove that JJ′ = 1.
(b) A rectangular box open at top is to have capacity of 32 c.c. Find the dimensions of the
box least material.
(c) A balloon in the form of right circular cylinder of radius 1.5 m and length 4.0 m and
is surmounted by hemispherical ends. If the radius is increased by 0.01 m and length
by 0.05 m, find the percentage change in the volume of the balloon.
429
UNSOLVED QUESTION PAPERS
For New Syllabus Only
5. Attempt any two parts of the following:
(a) Evaluate the integral
zz
∞ x
0
0
(10 × 2 = 20)
F x I dy dx
GH y JK
x exp −
2
changing the order of integration.
(b) Find the triple integration, the volume paraboloid of revolution x4 + y2 = 4z cut off
plane z = 4.
(c) State the Dirichlet’s theorem for three variables. Hence evaluate the integral
zzz
x l − 1 y m − 1 z n − 1 dx dy dz .
FG x IJ + FG y IJ + FG z IJ ≤ 1.
H aK H bK H cK
q
p
where x, y, z are all positive but limited condition
r
For Old Syllabus Only
6. Attempt any two parts of the following:
(10 × 2 = 20)
(a) The following data regarding the heights (y) and weights (x) of 100 college students
are given Σx = 15000, Σx2 = 2272500, Σy = 6800, Σy = 463025 and Σxy = 1022250.
Find the correlation coefficient between height and weight and equation of regression
line of height on weight.
(b) Fit a Poisson distribution to the following data and calculate the theoretical
frequencies:
x
0
1
2
3
4
f
192
100
24
3
1
(c) Assume the mean height of soldiers to be 68.22 inches with a variance of 10.8 inches
square. How many soldiers in a regiment of 10,000 would you expect to be over 6 feet
tall, given that the area under the standard normal curve between x = 0 and x = 0.31
is 0.1368 and between x = 0 and x = 1.15 is 0.3746.
GGG
B.Tech., First Semester Examination, 2004-05
MATHEMATICS–I
Time: 3 Hours]
Note:
[Total Marks: 100
(1) There are five questions in all.
(2) Each question has three parts.
(3) Attempt any two parts in each question.
1. (a) Find the eigen values and eigen vectors of the matrix
LM3
A = M0
MN0
1
2
0
4
6
5
OP
PP
Q
(5 × 4 = 20)
(b) Verify Cayley-Hamilton theorem for the matrix
LM 2
A = M−1
MN 1
OP
PP
Q
−1
2
2 −1
−1
2
Hence compute A–1.
(c) Reduce the matrix A to its normal form when
LM 1
2
A = M
MM 1
N−1
OP
PP
PQ
2 −1
4
4
3
4
2
3
4
−2
6 −7
Hence find the rank of A.
2. (a) If y = sin (m sin–1x), prove that
(10 × 2 = 20)
(1 – x2) yn + 2 – (2n + 1) xyn + 1 + (m2 + n2)yn= 0, and hence find yn at x = 0.
(b) If u = u
F y − x , z − x I , show that
GH xy xz JK
∂u
∂u 2 ∂u
+ y2
+z
= 0.
∂x
∂y
∂z
(c) Trace the curve y2 (2a – x) = x3.
x2
x1 x2
x2 x3
x x
, show that Jacobian of y1, y2, y3 with respect to
, y2 = 3 1 and y3 =
x3
x1
x2
x1, x2, x3 is 4.
(10 × 2 = 20)
(b) In estimating the cost of a pile of bricks measured as 6 m × 50 m × 4 m, the tape is
stretched 1% beyond the standard length. If the count is 12 bricks in 1 m3 and bricks
cost Rs. 100 per 1000, find the approximate error in the cost.
(c) Find the extreme values of x3 + y3 – 3axy.
3. (a) If y1 =
430
431
UNSOLVED QUESTION PAPERS
4. (a) Calculate the volume of the solid bounded by the surface x = 0, y = 0, x + y + z = 1
and z = 0, where D is the domain x ≥ 0, y ≥ 0 and x + y ≤ h.
(10 × 2 = 20)
(b) Prove that
z
x l − 1 y m − 1 dx dy =
D
l m
l+m+1
hl + m
zz
1
(c) Change the order of integration in l =
0
2 −x
x2
xy dx dy and hence evaluate the same.
e
j
1
5. (a) Prove div(grad rn) = n(n + 1)rn – 2, where r = x 2 + y 2 + z 2 2 . Hence show that ∇
(b) Apply Green’s theorem to evaluate
ze
C
j
e
j
2
FG 1IJ = 0.
H rK
(10 × 2 = 20)
2x 2 – y 2 dx + x 2 + y 2 dy , where C is the
boundary of the area enclosed by the x-axis and the upper half of circle x2 + y2 = a2.
(c) Evaluate
zzd
i
yz i + zx j + xy k ds , where S is the surface of the sphere x2 + y2 + z2 = a2
S
in the first octant.
GGG
This page
intentionally left
blank
Index
A
F
Absolute error, 112
Absolute maximum, 121
Absolute minimum, 121
Acceleration, 334
Associative law, 156
Asymptotes, 65
Augmented matrix, 189
Functional dependence, 107
G
Gamma function, 283
Geometrical representation, 215
H
Hermitian matrix, 154
B
I
Beta, 283
Idempotent matrix, 154
Identity matrix, 152
Implicit functions, 102
Independence of path, 367
Intercepts, 64
Intersection point with x- and y-axis, 64
Involutory matrix, 155
Irrotational vector, 354
Irrotational vector field, 365
C
Chain rule, 50, 96
Circulation, 365
Column matrix, 152
Conjugate of a matrix, 153
Consistent system, 189
Curve tracing, 63
D
J
Del operator, 337
Diagonal matrix, 152
Directional derivative, 339
Distributive law, 156
Dritchlet’s integral, 324
Jacobian for functional dependence
functions, 107
Jacobians, 95
K
Kirchhoff’s current law, 250
Kirchhoff’s voltage law (KVL), 250
E
Eigen, 214
Eigen values of diagonal, 216
Equivalent matrices, 167
Expansion of function of several variables, 81
Exponential function, 2
Extremum value or point, 122
L
Lagrange’s conditions for maximum or
minimum, 123
Linear independence of vectors, 210
Linearly dependent vectors, 210
433
434
A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Logarithm case, 2
Lower triangular matrix, 153
Scalar matrix, 152
Similar matrix, 239
Skew hermitian matrix, 154
Skew symmetric matrix, 153
Solenoidal, 353
Square matrix, 152
Symmetric about origin, 64
Symmetric about the line, 64
Symmetric about x-axis, 64
Symmetric about y-axis, 64
Symmetry, 64
Symmetric matrix, 153
M
Maclaurin’s series expansion, 83
Matrix polynomial, 233
Minor of a matrix, 176
Multiple integral, 258
N
Nilpotent matrix, 154
Non-singular, 163
Non-trivial solution, 197
Normal derivative, 339
Null matrix, 152
O
Oblique asymptotes (not parallel to x-axis and
y-axis), 65
Oblique asymptotes (when curve is represented
by implicit equation f ), 65
Origin and tangents at the origin, 64
Orthogonal matrix, 155
Own characteristic or individual or eigen, 214
P
Partial differentiation, 20
Points of intersection, 65
Power function, 1
Product functions, 3
Properties of beta and gamma functions, 283
Proportional or relative error, 112
R
Rectangular matrix, 152
Regions, 64
Row matrix, 152
S
Saddle point, 122
Scalar field, 337
T
Total differential, 49
Tranjugate or conjugate transpose of a matrix,
154
Transpose of a matrix, 153
Trigonometric functions cos (ax + b) or
sin (ax + b), 2
Triangular matrix, 153
Trivial solution, 189
U
Unitary matrix, 155
Upper triangular matrix, 153
V
Vector differential calculus, 333
Vector field, 337
Vector form of green’s theorem, 377
Velocity, 334
Volume in cylindrical coordinates, 315
Volume in spherical polar coordinates, 315
Z
Zero matrix, 152
GGG
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