Engineering Mathematics
A Tutorial Approach
About the Authors
Ravish R Singh is presently Vice-Principal and Head, Department of Electronics and Telecommunication Engineering at
Thakur College of Engineering and Technology, Mumbai. He
obtained his BE degree from University of Mumbai, in 1991
and MTech from IIT Bombay, in 2001. He is pursuing PhD
from Faculty of Technology, University of Mumbai. He has
published two books, namely, Electrical Networks, and Basic
Electrical and Electronics Engineering with Tata McGraw Hill
Education Private Limited. He is a member of IEEE, ISTE,
IETE and CSI, and has published research papers in national journals. His fields of
interest include Circuits, Signals & Systems and Engineering Mathematics.
Mukul Bhatt is presently Senior Lecturer, Department of
Humanities and Sciences at Thakur College of Engineering and
Technology, Mumbai. She obtained her MSc (Mathematics)
degree from H N B Garhwal University, in 1992. She has
fifteen years of teaching experience at various levels in
engineering colleges of Mumbai. Her fields of interest include
Integral Calculus, Complex Analysis and Operation Research.
She is a member of ISTE.
Engineering Mathematics
A Tutorial Approach
Ravish R Singh
Vice Principal and Head
Department of Electronics and Telecommunication, Engineering
Thakur College of Engineering and Technology, Mumbai
Mukul Bhatt
Senior Lecturer
Department of Humanities and Sciences
Thakur College of Engineering and Technology, Mumbai
Tata McGraw Hill Education Private Limited
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RYZYYRYZDARCL
Dedicated
To Our Parents
Late Shri Ramsagar Singh
and
Shrimati Premsheela Singh
Ravish R Singh
Shri Ved Prakash Sharma
and
Late Shrimati Vidyavati Hemdan
Mukul Bhatt
Preface
O
Engineering Mathematics is a key area in the study of an engineering course. It is
the study of numbers, structures, and associated relationships using rigorously defined
literal, numerical and operational symbols. A sound knowledge of the subject develops
analytical skills, thus enabling engineering graduates to solve numerical problems
encountered in daily life, as well as apply mathematical principles to physical problems,
particularly in the area of engineering.
Rationale
We have observed that many students who opt for engineering find it difficult to
conceptualise the subject since very few available texts have syllabus compatibility
and right pedagogy. Feedback received from students and teachers have highlighted
the need for a comprehensive textbook on Mathematics that covers all topics of first
year engineering along with suitable solved problems. This book—an outcome of our
vast experience of teaching the undergraduate students of engineering—provides a
solid foundation in mathematical principles, enabling students to solve mathematical,
scientific and associated engineering principles.
Users
This book on Engineering Mathematics, meant for first year engineering students,
covers both Mathematics-I and Mathematics-II papers (first year engineering
mathematics course) in a single volume. The structuring of the book takes into account
the commonly featuring topics in the syllabi of major Indian universities.
Intent
An easy-to-understand and student-friendly text, it presents concepts in adequate
depth using step-by-step problem solving approach. The text is well supported with
plethora of solved examples at varied difficulty levels, practice problems and engineering applications. It is intended that students will gain logical understanding from
solved problems and then through solving similar problems themselves.
Features
Each topic has been thoroughly covered from the examination point of view. The
theory part of the text is explained in a lucid manner. For each topic, problems of all
viii
Preface
possible combinations have been worked out. This is followed by an exercise with
answers. Objective type questions provided in each chapter help students in mastering
concepts. Salient features of the book are summarised below:
Multiple Choice Questions (350).
maxima and minima under Partial Differential Equation) have been provided.
the text.
Organisation
The contents of this book are divided into 15 chapters, keeping in mind the syllabus
structure in major Indian universities.
first chapter on Complex Numbers covers De Moivre’s theorem, hyperbolic
functions and logarithm of complex number.
Chapter 2 on Differential Calculus I offers a detailed exposition of successive
differentiation, mean value theorems, expansion of functions and indeterminate
forms.
Chapter 3 on Differential Calculus II are tangents
and normals, radius of curvature, evolutes, envelopes and curve tracing.
Chapter 4 on Partial Differentiation elucidates composite function, homogemaxima and minima and Lagrange’s multipliers.
Chapter 5 on Infinite Series deals with various tests to check the convergence of
the series.
Chapter 6 on Integral Calculus explains reduction formulae, rectification of
curves, area under the curves, volume and surface area of solid of revolution.
Chapter 7 gives a clear understanding of Gamma and Beta functions and their
properties.
Chapter 8 on Multiple Integrals includes double and triple integrals and their
applications.
Chapter 9 on Vector Calculus provides comprehensive coverage of vector
differentiation and integration.
Chapter 10 on Differential Equations explains first order differential equations,
linear differential equations of higher order, homogeneous differential equations
and applications of differential equations.
Chapter 11 on Matrices covers inverse, rank, normal form, solution of
homogeneous and non homogeneous equations, eigen values, eigen vectors and
quadratic forms.
Chapter 12 on Laplace Transform explains properties of Laplace transform,
inverse Laplace transform and its applications.
Preface
ix
Chapter 13 on Fourier Series gives a detailed account of orthogonal functions,
trigonometric and exponential Fourier series and half range Fourier series.
Chapter 14 on Fourier Transform covers Fourier integral theorem, Fourier sine
and cosine transforms, and finite Fourier transforms.
Chapter 15 on Z-Transform deals with properties of Z-Transform, inverse
Z-Transform and applications of Z-Transform.
Exhaustive OLC Supplements
The website accompanying the book http://www.mhhe.com/ravish/mukul/em provides
valuable resources such as additional solved examples. Instructors can access a
solution manual, chapter wise PowerPoint slides with diagrams and notes for effective
lecture presentations, and a test bank. Students can avail a sample chapter and link to
reference material.
Acknowledgements
We would like to express our gratitude to our colleagues in Thakur college of Engineering and Technology for their support and suggestions. We extend our appreciation
with us during the editorial, copyediting and production stages of the book. We would
also like to thank our family members for encouraging, inspiring and supporting us
while the making of the book was in progress. A note of acknowledgement is due to
the following reviewers for their valuable suggestions.
S B Singh, G. B. Pant University of
Agriculture & Technology, Pantnagar
Vinai K Singh, R. D. Engineering College
(UPTU), Ghaziabad
K H Patil, University of Pune, Pune
S Jha, National Institute of Technology,
Jamshedpur
Debdas Mishra, C V Raman College of
Engineering, Bhubaneswar
G Prema, Amrita Vishwa Vidyapeetham
(Deemed University), Coimbatore
BV Appa Rao, Koneru Lakshmaiah College
of Engineering, Guntur
Y P Anand, Kakinada Institute of
Engineering and Technology, Kakinada
Ravish R Singh
Mukul Bhatt
Publisher’s Note:
Tata McGraw Hill Education looks forward to receiving from teachers and
students their valuable views, comments and suggestions for improvements, all
of which may be sent to tmh.corefeedback@gmail.com (mentioning the title and
author’s name). Also, please inform any observations on piracy related issues.
Contents
O
Preface
vii
1. COMPLEX NUMBERS
1.1
1.1 Introduction 1.1
1.2 Complex Numbers 1.1
1.3 Geometrical Representation of
Complex Numbers (Argand’s Diagram) 1.2
1.4 Algebra of Complex Numbers 1.2
1.5 Different Forms of Complex Numbers 1.2
1.6 Modulus and Argument (or Amplitude) of Complex Numbers 1.3
1.7 Properties of Complex Numbers 1.3
1.21
1.9 Applications of De Moivre’s Theorem 1.37
1.10 Circular and Hyperbolic Functions 1.71
1.11 Inverse Hyperbolic Functions 1.74
1.12 Separation Into Real and Imaginary Parts 1.86
1.13 Logarithm of a Complex Number 1.108
Formulae 1.122
Multiple Choice Questions
1.123
2. DIFFERENTIAL CALCULUS I
2.1
2.2
2.3
2.4
2.5
2.6
2.7
Introduction 2.1
Successive Differentiation 2.1
Leibnitz’s Theorem 2.22
Mean Value Theorem 2.42
Rolle’s Theorem 2.42
Lagrange’s Mean Value Theorem (L.M.V.T.) 2.53
Cauchy’s Mean Value Theorem (C.M.V.T.) 2.71
78
2.9 Maclaurin’s Series 2.90
2.1
xii
Contents
2.10 Indeterminate Forms 2.121
Formulae 2.161
Multiple Choice Questions
2.162
3. DIFFERENTIAL CALCULUS II
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.1
Introduction 3.1
Tangent and Normal 3.1
Length of an Arc and its Derivative 3.28
Curvature 3.31
Centre and Circle of Curvature 3.50
Evolute 3.51
Envelopes 3.64
3.76
Formulae 3.109
Multiple Choice Questions
3.110
4. PARTIAL DIFFERENTIATION
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.1
Introduction 4.1
Partial Derivative 4.1
Higher Order Partial Derivatives 4.2
Variables to be Treated as Constants 4.33
Composite Function 4.40
Implicit Functions 4.63
Homogeneous Functions and Euler’s Theorem 4.68
4.98
Formulae 4.154
Multiple Choice Questions
4.155
5. INFINITE SERIES
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.1
Introduction 5.1
Sequence 5.1
Infinite Series 5.2
Geometric Series 5.4
Standard Limits 5.5
Comparison Test 5.5
D’Alembert’s Ratio Test 5.11
5.18
5.9
5.10
5.11
5.12
Logarithmic Test 5.23
Cauchy’s Root Test 5.27
Cauchy’s Integral Test 5.31
Alternating Series 5.34
xiii
Contents
5.13 Absolute Convergence of a Series 5.35
5.14 Uniform Convergence of a Series 5.38
Formulae 5.41
Multiple Choice Questions
5.42
6. INTEGRAL CALCULUS
6.1
6.2
6.3
6.4
6.5
6.6
6.1
Introduction 6.1
Reduction Formulae 6.1
Rectification of Curves 6.16
Areas of Plane Curves (Quadrature) 6.47
Volume of Solid of Revolution 6.68
Surface of Solid of Revolution 6.90
Formulae 6.106
Multiple Choice Questions
6.108
7. GAMMA AND BETA FUNCTIONS
7.1
7.2
7.3
7.4
7.5
7.6
7.1
Introduction 7.1
Gamma Function 7.1
Properties of Gamma Function 7.2
Beta Function 7.10
Properties of Beta Function 7.11
Beta Function as Improper Integral 7.25
Formulae 7.32
Multiple Choice Questions
7.33
8. MULTIPLE INTEGRAL
8.1
8.1
8.1
Order of Integration 8.21
8.39
Variables of Integration 8.57
8.67
Multiple Integrals 8.85
Multiple Choice Questions
133
9. VECTOR CALCULUS
9.1
9.2
9.3
9.4
Introduction 9.1
Unit Vector 9.1
Components of a Vector 9.1
Triple Product 9.2
9.1
xiv
Contents
9.5 Product of Four Vectors 9.8
9.6 Vector Function of a Single Scalar Variable 9.12
9.7 Velocity and Acceleration 9.13
9.13
9.9 Tangent Vector to a Curve at a Point 9.14
9.10 Scalar and Vector Point Function 9.24
9.11 Gradient 9.25
9.12 Divergence 9.46
9.13 Curl 9.48
9.14 Properties of Gradient, Divergence and Curl 9.60
9.15 Second Order Differential Operator 9.64
9.16 Line Integrals 9.81
9.17 Green’s Theorem in the Plane 9.98
9.115
9.19 Volume Integral 9.121
9.20 Stoke’s Theorem 9.124
9.21 Gauss Divergence Theorem 9.147
Formulae 9.165
Multiple Choice Questions
9.165
10. DIFFERENTIAL EQUATIONS
10.1
10.1 Introduction 10.1
10.2 Differential Equation 10.1
10.3 Ordinary Differential Equations of
First Order and First Degree 10.2
10.4 Homogeneous Linear Differential Equations of
Higher Order with Constant Coefficients 10.77
10.5 Non-Homogeneous Linear Differential Equations of
Higher Order with Constant Coefficients 10.85
10.6 Higher Order Linear Differential Equations with
Variable Coefficients 10.111
10.7 Method of Variation of Parameters 10.125
10.131
10.9 Simultaneous Linear Differential Equations with
Constant Coefficients 10.142
10.10 Applications of Ordinary Differential Equations of
First Order and First Degree 10.150
10.11 Applications of Higher Order Linear Differential Equations 10.175
Formulae 10.197
Multiple Choice Questions
10.200
Contents
11. MATRICES
11.1
11.2
11.3
11.4
11.5
11.6
11.7
11.9
11.10
11.11
11.12
11.13
11.14
11.15
11.16
xv
11.1
Introduction 11.1
Matrix 11.1
Some Definitions Associated with Matrices 11.1
Adjoint of a Square Matrix 11.19
Inverse or Reciprocal of a Matrix 11.23
Elementary Transformations 11.38
Rank of a Matrix 11.45
11.63
Homogeneous Linear Equations 11.73
Linear Dependence and Independence of Vectors 11.81
Eigen Values and Eigen Vectors 11.90
Cayley–Hamilton Theorem 11.112
Minimal Polynomial and Minimal Equation of a Matrix 11.120
Function of Square Matrix 11.124
Similarity of Matrices 11.131
Quadratic Form 11.152
Multiple Choice Questions
11.173
12. LAPLACE TRANSFORM
12.1
12.1
12.2
12.3
12.4
12.5
12.6
12.7
Introduction 12.1
Laplace Transform 12.1
Laplace Transform of Some Standard Functions 12.2
Properties of Laplace Transform 12.6
Evaluation of an Integral using Laplace Transform 12.37
Heaviside’s Unit-step Function 12.44
Dirac Delta or Unit Impulse Function 12.50
Transform of Periodic Functions 12.53
12.9 Inverse Laplace Transform 12.58
12.10 Application of Laplace Transform to
Differential Equations with Constant Coefficients 12.87
12.11 Application of Laplace Transform to a System of
Simultaneous Differential Equations 12.100
Formulae 12.108
Multiple Choice Questions
12.109
13. FOURIER SERIES
13.1 Introduction 13.1
13.2 Orthogonality of Functions 13.1
13.3 Fourier Series 13.10
13.1
xvi
Contents
13.4
13.5
13.6
13.7
Parseval’s Identity 13.14
Fourier Series of Even and Odd Functions 13.37
Half-range Fourier Series 13.52
Complex Form of Fourier Series 13.62
Formulae 13.70
Multiple Choice Questions
13.71
14. FOURIER TRANSFORM
14.1
14.2
14.3
14.4
14.5
14.1
Introduction 14.1
Fourier Integral Theorem 14.1
Fourier Transform 14.9
Properties of the Fourier Transform 14.11
Finite Fourier Transforms 14.29
Formulae 14.35
Multiple Choice Questions
14.35
15. Z-TRANSFORM
15.1
15.2
15.3
15.4
15.5
15.6
15.1
Introduction 15.1
Sequence 15.1
Z-transform 15.6
Properties of Z-transform 15.6
Inverse Z-transform 15.18
Application of Z-transform to Difference Equations 15.32
Formulae 15.138
Multiple Choice Questions
Appendix A Differential Formulae
Appendix B Integral Formulae
Appendix C Standard Curves
Index
15.140
A.1.1
A.2.1
A.3.1
I.1
Visual Guide
O
4.2.1 Geometrical Interpretation
The function u = f (x, y) represents a surface. The point
P [x1, y1, f (x1, y1)] on the surface corresponds to the values
x1, y1 of the independent variables x, y. The intersection
of the plane y = y1 (parallel to the zox–plane) and the surface u = f (x, y) is the curve shown by the dotted line in the
Figure. On this curve, x and u vary according to the relation
u = f (x, y1). The ordinary derivative of f (x, y1) w.r.t. x at x1
Lucid Text
Fig. 4.1
⎛ ∂u ⎞
⎛ ∂u ⎞
is ⎜ ⎟
. Hence, ⎜ ⎟
is the slope of the tangent to
⎝ ∂x ⎠ ( x, y1 )
⎝ ∂x ⎠ ( x1 , y1 )
the curve of the intersection of the surface u = f (x, y) with the plane y = y1 at the point
P[x1, y1, f (x1, y1)].
⎛ ∂u ⎞
Similarly, ⎜ ⎟
is the slope of the tangent to the curve of the intersection of the
⎝ ∂y ⎠ ( x , y )
1 1
surface u = f (x, y) with the plane x = x1 at the point P[x1, y1, f (x1, y1)].
4.3 HIGHER ORDER PARTIAL DERIVATIVES
Partial derivatives of higher order, of a function u = f (x, y), are obtained by partial differentiation of first order partial derivative. Thus, if u = f (x, y), then
∂ 2 u ∂ ⎛ ∂u ⎞
=
⎜ ⎟
∂x 2 ∂x ⎝ ∂x ⎠
∂2 u
∂ ⎛ ∂u ⎞
=
⎜ ⎟
∂y ∂x ∂y ⎝ ∂x ⎠
O
Chapter
5
Organised Sections
In this chapter, we will learn about the convergence and divergence of an infinite series. There are various methods to test the convergence and divergence of an infinite
series. In this chapter, we will study Comparision Test, D’Alembert’s ratio test, Raabe’s
test, Logarithmic test, Cauchy’s root test and Cauchy’s integral test. We will also study
alternating series, absolute and uniform convergence of the series.
An ordered set of real numbers as u1, u2, u3, ……..un, …… is called a sequence and is
denoted by {un}. If the number of terms in a sequence is infinite, it is said to be infinite
sequence, otherwise it is a finite sequence and un is called the nth term of the sequence.
A sequence is said to be monotonically increasing if un 1 un for each value of n
and is monotonically decreasing if un 1 un for each value of n, whereas the sequence
is called alternating sequence if the terms are alternate positive and negative.
e.g. (i) 1, 2, 3, 4, … is a monotonically increasing sequence.
1 1 1
(ii) 1, , , , … is a monotonically decreasing sequence.
2 3 4
(iii) 1, –2, 3, – 4, … is an alternating sequence.
Example 18: If ` = i + 1, a = 1 - i and tanφ =
Solved Examples
( x + ` )n - ( x + a )n
= sin ne cosec ne .
` -a
Solution: a
1, b
i
i, tan =
1
cot f
n
x
1, x
1
x +1
cot f
1
, then prove that
x +1
1
n
(x + α ) − (x + β )
(cot φ − 1 + i + 1) n − (cot φ − 1 + 1 − i ) n
=
α −β
i +1−1+ i
n
n
⎛ cos φ ⎞ ⎛ cos φ ⎞
+i⎟ −⎜
−i⎟
⎜
⎝ sin φ
⎠ ⎝ sin φ
⎠
2i
n
(cos f + i sin f ) − (cos f − i sin f ) n
=
2i sin n f
=
=
(e if ) n − (e − if ) n e inf − e − inf 2i sin nf
=
=
2i sin n f
2i sin n f
2i sin n f
= sin nf cosec n f
Example 19: If (1 + cos p + i sin p ) (1 + cos 2p + i sin 2p ) = u + iv, prove that
θ
v
3
(ii)
.
(i) u2 + v 2 = 16 cos 2 cos 2θ
= tan
2
u
2
Solution: u iv (1 cos q i sin q ) (1 cos 2q i sin 2q )
q
q
q
⎛
⎞
= ⎜ 2 cos 2 + i 2 sin cos ⎟ (2 cos 2 q + i 2 sin q cos q )
⎝
2
2
2⎠
= 2 cos
q⎛
q
q⎞
⎜ cos + i sin ⎟⎠ 2 cos q (cosq + i sin q )
2⎝
2
2
xviii
Visual Guide
4.8 APPLICATIONS OF PARTIAL
DIFFERENTIATION
4.8.1 Jacobians
If u and v are continuous and differentiable functions of two independent variables x
and y, i.e., u
f1(x, y) and v
f2(x, y), then the determinant
u
x
u
y
v
x
v
y
Application Focus
is called the
Jacobian of u, v with respect to x, y and is denoted as J = (u, v) .
( x, y )
Similarly, if u, v and w are continuous and differentiable functions of three independent variables x, y, z, then the Jacobian of u, v, w with respect to x, y, z is
u
x
(u, v, w)
=
( x, y , z )
u
y
u
z
v v v
x y z
w w w
x y z
Jacobian is useful in transformation of variables from cartesian to polar, cylindrical
and spherical coordinates in multiple integrals.
Exercise 2.2
1. Find the nth order derivative w.r.t. x
(i) xex
(ii) x2e2x
(iii) x log (x 1)
(iv) x3 sin 2x
(v) y x2 sin x
⎡ Ans. : (i) e x ( x + n)
⎢
2x
n 2
n
n −1
⎢ (ii) e [2 x + 2 nx + n ( n − 1) 2 ]
⎢
( −1) n − 2 ( n − 2)!( x + n)
⎢(iii)
( x + 1) n
⎢
⎢
⎢(iv) 2n x 3 sin ⎛⎜ 2 x + np ⎞⎟ +
⎝
⎢
2 ⎠
⎢
p⎤
⎡
⎢
3n x 2 2n −1 sin ⎢ 2 x + ( n − 1) ⎥
⎢
2⎦
⎣
⎢
p⎤
⎡
⎢
+ 3n ( n − 1) x 2n − 2 sin ⎢ 2 x + ( n − 2) ⎥
⎢
2⎦
⎣
⎢
+ n ( n − 1) ( n − 2) 2n − 3
⎢
⎢
n
p
⎡
⎤
⎢
sin ⎢ 2 x + ( n − 3) ⎥
2 ⎦
⎢
⎣
⎢
⎢( v) x 2 sin ⎛ x + np ⎞
⎜
⎟
⎝
⎢
2 ⎠
⎢
p⎤
⎡
⎢
+ 2nx sin ⎢ x + ( n − 1) ⎥
⎢
2⎦
⎣
⎢
p
⎡
⎤
⎢
+ ( n2 − n) sin ⎢ x + ( n − 2) ⎥
2⎦
⎣
⎣⎢
Exercises
3. If y e ax [a2 x2 - 2nax n (n
prove that yn an 2 x2 eax.
4. If y
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦
1)],
x2 sin x, prove that
np ⎞
⎛
yn = ( x 2 − n2 + n) sin ⎜ x +
⎟
⎝
2 ⎠
np ⎞
⎛
− 2nx cos ⎜ x +
⎟
⎝
2 ⎠
5. If x
tan log y, prove that
(1 + x 2 ) yn +1 + (2nx − 1) yn
+ n (n − 1) yn −1 = 0
⎡ Hint : log y = tan −1 x, y = e tan
⎣
−1
x
⎤
⎦
6. If y cos (m sin−1 x), prove that
(1 - x2) yn 2 - (2n 1) xyn 1
(m2 - n2) yn 0
Hence, obtain yn (0).
⎡ Ans. : yn (0) = ( n2 − m 2 )........... ⎤
⎥
⎢
( 4 2 − m 2 )( 22 − m 2 )( − m 2 ) ⎥⎦
⎢⎣
7. If x sin q, y sin 2q, prove that
1) xyn 1 (1 - x2) yn 2 - (2n
(n2 - 4) yn 0
⎡ Hint : y = 2 sin q cos q = 2 x 1 − x 2 ⎤
⎣
⎦
1.8 DE MOIVRE’S THEOREM
Statement: For any real number n, one of the values of (cos q
cos nq i sin nq.
Hence, (cos q
i sin q )n
cos nq
i sin q )n is
i sin nq
Proof: Case I: If n is a positive integer
Let z1 r1 (cos q1
i sin q1), z2 r2 (cos q2
z1 z2 r1 (cos q1
i sin q1) r2 (cos q2
r1 r2 [(cos q1 cos q2
r1 r2 [cos (q1
Similarly,
z1 z2……. zn
r1 (cos q1
q2)
i sin q2) , …… , zn rn (cos qn
sin q1 sin q2)
i sin (q1
i sin qn).
i sin q2)
i(sin q1 cos q2
cos q1 sin q2)]
q2)]
i sin q1) r2 (cos q2
i sin q2)……. rn (cos qn
i sin qn)
(r1 r2…….. rn) (cos q1 i sin q1) (cos q2 i sin q2)….. (cos qn i sin qn)
(r1 r2…….. rn)[cos (q1 q2 …… qn)
If z1
z2
…….
zn rn (cos q
(cos q
zn z
i sin q )n
i sin q )n
r (cos q
(cos nq
i sin (q1
q2…… qn)] … (1)
i sin q ) , then Eq. (1) reduces to
rn (cos nq
i sin nq )
i sin nq ), where n is a positive integer.
Theorems and Derivations
xix
Visual Guide
FORMULAE
Important Formulae
Tangent and Normal
Equation of the tangent at any point
(x, y): Y – y = fÄ (x) (X – x)
Equation of the normal at any point
1
(x, y): Y – y = −
(X – x)
f ′( x )
Length of polar normal
Angle of Intersection of Curves
Derivative of Length of an arc
= tan −1
m2 − m1
1 + m2 m1
=
dr
d
2
(i)
⎛ dx ⎞
Length of tangent = y 1 + ⎜ ⎟
⎝ dy ⎠
dy
dx
ds
⎛ dy ⎞
= 1 + ⎜ ⎟ Cartesian form
⎝ dx ⎠
dx
⎛ dx ⎞
ds
= 1+ ⎜ ⎟
dy
⎝ dy ⎠
2
dx
dy
⎛ dy ⎞
Length of normal = y 1 + ⎝⎜ ⎠⎟
dx
Length of sub-normal = y
2
Length of polar sub-normal =
Length of Tangent, Sub-tangent, Normal
and Sub-normal
Length of sub-tangent = y
⎛ dr ⎞
r2 + ⎜ ⎟
⎝d ⎠
2
2
2
(ii)
ds
⎛ dx ⎞ ⎛ dy ⎞
= ⎜ ⎟ + ⎜ ⎟ Parametric
⎝ dt ⎠ ⎝ dt ⎠ form
dt
(iii)
ds
⎛ dr ⎞
= r 2 + ⎜ ⎟ Polar form
⎝d ⎠
d
2
2
ds
⎛d ⎞
= 1 + r2 ⎜ ⎟
⎝ dr ⎠
dr
2
MULTIPLE CHOICE QUESTIONS
Choose the correct alternative in each of the following:
1. The equation of the tangent to the
curve y = 2 sin x + sin 2x at x = p is
3
equal to
(a) 2y = 3 3
(b) y = 3 3
(c) 2y + 3 3 = 0
(d) y + 3 3 = 0
2. The sum of the squares of the
intercept made on the co-ordinate
axis by the tangents to the curve
2
2
2
x 3 + y 3 = a 3 is
(a) a2
(b) 2a2
(c) 3a2
(d) 4a2
3. The equation of the normal to the
curve y = x (2 – x) at the point (2, 0) is
(a) x – 2y = 2
(b) 2x + y = 4
(c) x – 2y + 2 = 0
(d) none of these
4. The length of the normal at t on the
curve x = a (t + sin t), y = a(1 – cos t)
is
(b) 2a sin3 t sec t
2
2
(c) 2a sin t tan t
2
2
t
(d) 2a sin
2
5. The length of the sub-tangent to the
curve x2 + xy + y2 = 7 at (1, –3) is
(a) 3
(b) 5
(c) 15
(d) 3
5
6. The angle of intersection of the
curves y = 4 – x2 and y = x2 is
4
p
(b) tan–1
(a)
3
2
(d) none of these
(c) tan–1 4 2
7
7. The length of the sub-normal to the
parabola y2 = 4ax at any point is
equal to
()
2a
(b) 2 2a
a
(d) 2a
2
8. If x = a (q + sin q) and y =
dy
will be equal to
a (1 – cos q), then
dx
(a)
(c)
(a) a sin t
Exhaustive Online
Learning Center
Multiple Choice Questions
Complex Numbers
Complex
Numbers
Chapter
1.1
1
1.1 INTRODUCTION
The complex numbers are an extension of the real numbers obtained by introducing
an imaginary unit i, where i =
1 . The operations of addition, subtraction,
multiplication and division are applicable on complex numbers. A negative real
number can be obtained by squaring a complex number. With a complex number,
it is always possible to find solutions to polynomial equations of degree more than
one. Complex numbers are used in many applications, such as control theory, signal
analysis, quantum mechanics, relativity, etc.
1.2 COMPLEX NUMBERS
A complex number z is an ordered pair (x, y) of real numbers x and y. It is written as
z = (x, y) orz = x + iy, where i =
1 is known as the imaginary unit. Here, x is called
the real part of z and is written as “Re (z)” and y is called the imaginary part of z and
is written as “Im (z)”.
If x = 0 and y 0, then z = 0 + iy = iy which is purely imaginary.
If x 0 and y = 0, then z = x + i 0 = x which is real.
Hence, z is purely imaginary, if its real part is zero and is real, if its imaginary
part is zero.
This shows that every real number can be written in the form of a complex number
by taking its imaginary part as zero. Hence, the set of real numbers is contained in the
set of complex numbers.
The even power of i is either 1 or 1 and odd power of i is either i or i.
i2 = i.i = 1,
i3 = i2.i = i,
i4 = (i2)2 = ( 1)2 = 1,
i5 = i. i4 = i, etc.
Two complex numbers are equal if and only if their corresponding real and imaginary parts are equal.
If z = x + iy
as z = x iy.
1.2
Engineering Mathematics
1.3 GEOMETRICAL REPRESENTATION OF
COMPLEX NUMBERS (ARGAND’S DIAGRAM)
Any complex number z = x + iy can be represented
as a point P(x, y) in the xy-plane with reference to
the rectangular x and y axes.
The plot of a given complex number z = x +
iy, as the point P(x, y) in the xy-plane is known as
Argand’s diagram. The x-axis is called the real
axis, y-axis is called the imaginary axis and the
xy-plane is called the complex plane.
y
P(x, y)
x'
x
O
y'
Fig. 1.1
1.4 ALGEBRA OF COMPLEX NUMBERS
Let z1 = x1 + iy1 and z2 = x2 + iy2 be two complex numbers.
(a) Addition: z1 + z2 = (x1 + iy1) + (x2 + iy2)
= (x1 + x2) + i (y1 + y2)
(b) Subtraction: z1 z2 = (x1 + iy1) (x2 + iy2)
= (x1 x2) + i (y1 y2)
(c) Multiplication: z1 z2 = (x1 + iy1) (x2 + iy2)
= (x1 x2 y1 y2) + i (x2 y1 + y2 x1)
(d) Division:
[∵ i2 = –1]
z1
x + iy1
= 1
z2 x2 + iy2
=
=
( x1 + iy1 ) ( x2 − iy2 )
⋅
( x2 + iy2 ) ( x2 − iy2 )
x1 x2 + y1 y2
x22 + y22
+i
( y1 x2 − x1 y2 )
( x22 + y22 )
1.5 DIFFERENT FORMS OF COMPLEX NUMBERS
1.5.1 Cartesian or Rectangular Form
If x and y are real numbers, then z = x + iy is called the Cartesian form of the complex
number.
1.5.2 Polar Form
The complex number z = x + iy can be represented by the point P whose cartesian coordinates are (x, y). We know that if polar coordinates of the same point P are (r, q ),
then x = r cos q and y = r sin q.
Complex Numbers
1.3
y
Hence, polar form of z is
z = r cos q + ir sin q
= r (cos q + i sin q )
Polar form can also be written as r q.
P(r, )
r
x'
We know that eiq = cos q + i sin q
Using polar form, z = r (cos q + i sin q) = reiq
This is called the exponential form or Euler’s
form of a complex number z.
Note: eiq = cos q + i sin q, e iq = cos q i sin q.
1
1
Hence, cos = (ei + e −i ) and sin = (ei
2i
2
x
O
1.5.3 Exponential Form
y'
Fig. 1.2
e
i
)
1.6 MODULUS AND ARGUMENT (OR AMPLITUDE)
OF COMPLEX NUMBER
Let z be a complex number such that z = x + iy = r (cos q + i sin q )
where,
x = r cos q, y = r sin q
y
r = x 2 + y 2 and tan =
or
then
x
⎛ y⎞
= tan −1 ⎜ ⎟
⎝x⎠
Here ‘r’ is called the modulus or absolute value of z and is denoted by |z| or mod (z)
and q is called argument or amplitude of z and is denoted by arg (z) or amp (z).
Hence,
z = r = x2 + y 2
y
x
Note: The value of q which satisfies both the equations x = r cos q and y = r sin q, gives
the argument of z.
Argument q has infinite number of values. The value of q lying between p and p
is called the principal value of argument.
arg (z) =
= tan −1
1.7 PROPERTIES OF COMPLEX NUMBER
Let z = x + iy and z = x iy.
1
(a) Re (z) = x = ( z + z )
2
1
(b) Im (z) = y =
(z z )
2i
(c) ( z1 + z2 ) = z1 + z2
1.4
Engineering Mathematics
(d) ( z1 z2 ) = z1 z2
⎛ z1 ⎞
z1
⎟ =
z
z
2
⎝ 2⎠
(e) ⎜
(f ) z z = |z|2 = | z |2
[∵ z = | z | = x 2 + y 2 ]
(g) |z1z2|= |z1| |z2|
and arg (z1z2) = arg (z1) + arg (z2)
Proof:
Let z1 = r1 eiq1 , z2 = r2 eiq 2
z1z2 = r1 eiq1 r2 eiq 2
= (r1 r2) ei(q1 +q 2 )
Comparing with exponential form,
|z1 z2| = r1 r2 = |z1| |z2|
and
arg (z1 z2) = q1 + q2 = arg (z1) + arg (z2)
(h)
and
z1
z1
=
z2
z2
⎛z ⎞
arg ⎜ 1 ⎟ = arg (z1)
⎝ z2 ⎠
arg (z2)
z1 r1ei 1 ⎛ r1 ⎞ i ( 1 − 2 )
=
= ⎜ ⎟e
z2 r2 ei 2 ⎝ r2 ⎠
Comparing with exponential form,
z1
z1
r
= 1 =
z2
r2
z2
Proof:
⎛z ⎞
arg ⎜ 1 ⎟ = q1 q2 = arg (z1) arg (z2)
⎝ z2 ⎠
Example 1: Find the modulus and principal value of argument.
and
(ii) (4 + 2i ) ( -3 + 2 i )
(i) -1 + i 3
⎛ 4 - 5i ⎞ ⎛ 3 + 2i ⎞
(iii) ⎜
.
⎝ 2 + 3i ⎟⎠ ⎜⎝ 7 + i ⎟⎠
Solution: (i)
z= 1+ i 3
Re (z) = x = 1, Im (z) = y =
r = |z| =
3
( −1) 2 + ( 3 ) = 2
2
Complex Numbers
q = arg (z) = tan
1
y
= tan
x
1
⎛ 3⎞
⎜⎜
⎟⎟ = tan
⎝ −1 ⎠
1.5
1
( − 3 ) = 2p
3
⎡⎣∵ Point ( −1, 3 ) lies in the second quadrant ⎤⎦
(
)
(ii) z = (4 + 2i ) −3 + 2i = z1 z2
r = z = z1 z2 = z1 z2 = 4 + 2i −3 + 2i = ( 16 + 4 ) ( 9 + 2 ) = 220 = 2 55
= arg ( z ) = arg ( z1 z2 )
= arg ( z1 ) + arg ( z2 )
= arg (4 + 2i ) + arg ( −3 + 2i )
⎛ 2⎞
⎛2⎞
= tan −1 ⎜ ⎟ + tan −1 ⎜
⎜ −3 ⎟⎟
⎝4⎠
⎝
⎠
⎛ 2⎞
⎛1⎞
= tan −1 ⎜ ⎟ − tan −1 ⎜
⎜ 3 ⎟⎟
2
⎝ ⎠
⎝
⎠
⎛ 1
2
−
⎜
2
3
= tan
⎜
1 2
⎜ 1+ ⋅
2 3
⎝
−1 ⎜
⎞
⎟
⎟
⎟
⎟
⎠
⎛ 3− 2 2 ⎞
= tan −1 ⎜
⎜ 6 + 2 ⎟⎟
⎝
⎠
(iii) z =
(4 − 5i )(3 + 2i ) z1 z2
=
z3 z4
(2 + 3i )(7 + i )
r= z =
z1 z2
4 − 5i 3 + 2i
z1 z2
=
=
=
2 + 3i 7 + i
z3 z4
z3 z4
q = arg ( z ) = arg
( 16 + 25 ) ( 9 + 4 )
=
( 4 + 9 ) ( 49 + 1 )
z1 z2
z3 z4
= arg ( z1 z2 ) − arg ( z3 z4 )
= arg ( z1 ) + arg ( z2 ) − [ arg ( z3 ) + arg ( z4 ) ]
= arg ( 4 − 5i ) + arg (3 + 2i ) − arg ( 2 + 3i ) − arg (7 + i )
⎛ −5 ⎞
⎛2⎞
⎛3⎞
⎛1⎞
= tan −1 ⎜ ⎟ + tan −1 ⎜ ⎟ − tan −1 ⎜ ⎟ − tan −1 ⎜ ⎟
4
3
2
⎝ ⎠
⎝ ⎠
⎝ ⎠
⎝7⎠
5
1⎞ ⎛
2
3⎞
⎛
= − ⎜ tan −1 + tan −1 ⎟ + ⎜ tan −1 − tan −1 ⎟
4
7⎠ ⎝
3
2⎠
⎝
41
50
1.6
Engineering Mathematics
⎛ 5 1 ⎞
⎛ 2 3 ⎞
−
+
⎜
⎟
⎜
⎟
⎛ 39 ⎞
⎛ 5⎞
= − tan −1 ⎜ 4 7 ⎟ + tan −1 ⎜ 3 2 ⎟ = − tan −1 ⎜ ⎟ + tan −1 ⎜ − ⎟
5.1
2.3
⎝ 23 ⎠
⎝ 12 ⎠
⎜1−
⎟
⎜ 1+
⎟
⎝
⎠
⎝
⎠
4 7
3 2
⎛ 39 5 ⎞
+
⎜
⎟
= − tan −1 ⎜ 23 12 ⎟ = − tan −1 (7.19)
39 . 5
⎜ 1−
⎟
⎝
23 12 ⎠
Example 2: Express in polar form
⎛ 2+ i ⎞
(i) ⎜
⎟
⎝ 3−i ⎠
2
(ii) 1 + sin ` + i cos `
2
Solution: (i)
4 + i 2 + 4i 3 + 4i
⎛ 2 + i⎞
=
z=⎜
=
⎟
⎝ 3−i⎠
9 + i 2 − 6i 8 − 6i
=
3 + 4i 8 + 6i 1
⋅
= i
8 − 6i 8 + 6i 2
Comparing with polar form,
2
1
⎛1⎞
r = z = 02 + ⎜ ⎟ =
2
⎝2⎠
and
⎛1⎞
⎜⎝ ⎟⎠
p
2
= tan −1 ∞ =
q = tan −1
0
2
2
Hence,
(ii)
1⎛
⎛ 2+i ⎞
⎞
⎜ 3 − i ⎟ = 2 ⎜ cos 2 + i sin 2 ⎟
⎝
⎠
⎝
⎠
z = 1 + sin a + i cos a
⎛p
⎞
⎛p
⎞
= 1 + cos ⎜ − a ⎟ + i sin ⎜ − a ⎟
⎝2
⎠
⎝2
⎠
⎛p a ⎞
⎛p a ⎞
⎛p a ⎞
= 2 cos 2 ⎜ − ⎟ + 2i sin ⎜ − ⎟ cos ⎜ − ⎟
⎝ 4 2⎠
⎝ 4 2⎠
⎝ 4 2⎠
q
q⎤
⎡
2q
⎢∵1 + cos q = 2 cos 2 , sin q = 2 sin 2 cos 2 ⎥
⎦
⎣
⎛π α ⎞⎡ ⎛π α ⎞
⎛ π α ⎞⎤
z = 2 cos ⎜ − ⎟ ⎢cos ⎜ − ⎟ + i sin ⎜ − ⎟ ⎥
⎝ 4 2 ⎠⎣ ⎝ 4 2 ⎠
⎝ 4 2 ⎠⎦
Complex Numbers
1.7
Comparing with polar form,
⎛π α ⎞
r = 2 cos ⎜ − ⎟
⎝4 2⎠
π α
θ= −
4 2
⎛π α ⎞⎡ ⎛π α ⎞
⎛ π α ⎞⎤
Hence, 1 + sin a + i cos a = 2 cos ⎜ − ⎟ ⎢cos ⎜ − ⎟ + i sin ⎜ − ⎟ ⎥
⎝ 4 2 ⎠⎣ ⎝ 4 2 ⎠
⎝ 4 2 ⎠⎦
Example 3: Find the value of
Solution: Let x + iy =
- 5 + 12i .
− 5 + 12i
(x + iy)2 = 5 + 12i
(x2 y2) + i (2xy) = 5 + 12i
Comparing real and imaginary parts on both the sides,
x2 y2 = 5,
2xy = 12, xy = 6
6
Putting y = in Eq. (1),
x
36
x 2 − 2 = −5
x
4
2
x + 5x 36 = 0
(x2 + 9) (x2
4) = 0
x2 = 9, x2 = 4
Since x is real,
x=±2
When
x = 2, y =
When
x = 2, y =
Hence,
5 12i = 2 + 3i or 2
6
=3
2
6
= −3
−2
3i
Example 4: If x and y are real, solve the equation
Solution:
... (1)
iy
3 y + 4i
−
= 0.
ix + 1 3 x + y
iy
3 y + 4i
−
=0
ix + 1 3x + y
iy (3 x + y ) − (3 y + 4i )(ix + 1)
=0
(ix + 1)(3x + y )
(−3 y + 4 x) + i (3 xy + y 2 − 3 xy − 4)
= 0 + i0
(ix + 1)(3 x + y )
1.8
Engineering Mathematics
Comparing real and imaginary parts on both the sides,
3y + 4x = 0 and y2 4 = 0, y = ± 2
3
x=±
2
3
Hence,
x = ± , y = ± 2.
2
Example 5: Prove that Re (z) > 0 and |z - 1| < |z + 1| are equivalent, where
z = x + iy.
z = x + iy
Solution:
Now,
Re (z) > 0
x>0
|z 1| < |z + 1|
|x + iy 1| < |x + iy + 1|
... (1)
( x − 1) 2 + y 2 < ( x + 1) 2 + y 2
x2 + 1
2x + y2 < x2 + 1 + 2x + y2
2x < 2x
0 < 4x
0 < x or x > 0
From Eqs. (1) and (2),
Re (z) > 0 and |z
... (2)
1| < |z + 1| are equivalent.
a + ib 1 + iz
, then prove that a2 + b2 + c2 = 1,
=
1 + c 1 − iz
where a, b and c are real numbers and z is a complex number.
Example 6: If b + ic = (1 + a) z and
Solution: We have b + ic = (1 + a) z
b + ic
z=
1+ a
a + ib 1 + iz
and
=
1 + c 1 − iz
Substituting z in the above equation,
⎛ b + ic ⎞
1+ i ⎜
⎟
a + ib
⎝ 1+ a ⎠
=
1+ c
⎛ b + ic ⎞
1− i ⎜
⎟
⎝ 1+ a ⎠
=
1 + a + ib + i 2 c
1 + a − ib − i 2 c
(1 + a − c) + ib
=
(1 + a + c) − ib
[∵i 2 = −1]
Complex Numbers
1.9
(a + ib) [(1 + a + c) ib] = (1 + c) [(1 + a c) + ib]
a (1 + a + c) i ab + ib (1 + a + c) i2 b2 = 1 + a c + c + ac c2 + ib
(a + a2 + ac + b2) + i (b + bc) = (1 + a + ac c2) + ib
Comparing real parts on both the sides,
a + a2 + ac + b2 = 1 + a + ac c2
a2 + b2 + c2 = 1
π
2π
Example 7: Find z if arg ( z + 1) =
and arg ( z − 1) =
.
6
3
Solution: Let
z = x + iy
arg (z + 1) =
arg (x +iy + 1) =
6
6
arg[( x + 1) + iy ] =
tan −1
6
y
=
x +1 6
y
1
= tan =
x +1
6
3
x − y 3 = −1
Also,
2
3
2
arg ( x + iy − 1) =
3
2
arg [( x − 1) + iy ] =
3
y
2
tan −1
=
x −1 3
y
2
= tan
=− 3
x −1
3
arg ( z − 1) =
x 3+y= 3
Solving Eqs. (1) and (2),
x=
1
,
2
y=
3
2
⎛z+i⎞ π
Example 8: Find z if |z + i| = |z| and arg ⎜
⎟= .
⎝ z ⎠ 4
Solution: We have |z + i| = |z|
z+i
=1
z
1.10
Engineering Mathematics
⎡ z1
z1 ⎤
=
⎢∵
⎥
z2 ⎥⎦
⎢⎣ z2
⎛ z +i ⎞
arg ⎜
⎟=
⎝ z ⎠ 4
z +i
=1
z
Also,
Let
z +i
= rei
z
z +i
=1
z
where,
r=
and
⎛ z +i ⎞ π
θ = arg ⎜
⎟=
⎝ z ⎠ 4
iπ
⎛ z +i ⎞
iθ
4
⎜ z ⎟ = re = 1.e
⎝
⎠
Hence,
iπ
z + i = ze 4
iπ
⎛
z ⎜1 − e 4
⎜
⎝
−i
z=
1− e
ip
4
⋅
1− e
1− e
⎞
⎟ = −i
⎟
⎠
ip
−
4
−
ip
4
ip
− ⎞
⎛
−i ⎜1 − e 4 ⎟
⎝
⎠
=
ip
1− e 4 − e
−
ip
4
i
⎛
−
−i ⎜1 − e 4
=
⎜
⎝
2 − 2 cos
=
⎞
⎟
⎟
⎠
+1
[∵ ei + e −i = 2 cos ]
4
⎛
⎞
−i ⎜ 2 sin 2 + i 2 sin cos ⎟
8
8
8
⎝
⎠
⎛
⎞
−i ⎜1 − cos + i sin ⎟
4
4
⎝
⎠
=
⎛
⎞
⎛
⎞
2 ⎜ 2 sin 2 ⎟
2 ⎜1 − cos ⎟
8⎠
4⎠
⎝
⎝
1⎛
1
⎞
⎛
⎞
= ⎜ −i − i 2 cot ⎟ = ⎜ −i + cot ⎟
2⎝
8 ⎠ 2⎝
8⎠
Example 9: Determine the locus of z if |z - 3| - |z + 3| = 4.
Solution: Let z = x + iy
|z
3|
|z + 3| = 4
|z
|x + iy
3| = 4 + |z + 3|
3| = 4 + |x + iy + 3|
Complex Numbers
|(x
3) + iy| = 4 + |(x + 3) + iy|
( x − 3) 2 + y 2 = 4 + ( x + 3) 2 + y 2
Squaring both the sides,
(x
x2 + 9
3)2 + y2 = 16 + (x + 3)2 + y2 + 8 ( x + 3) 2 + y 2
6x + y2 = 16 + x2 + 9 + 6x + y2 + 8 ( x + 3) 2 + y 2
12x = 8 ( x + 3) 2 + y 2
16
(4 + 3x) = 2 ( x + 3) 2 + y 2
Squaring again both the sides,
16 + 9x2 + 24x = 4 (x2 + 9 + 6x + y2)
5x2 4y2 = 20
x2 y 2
−
=1
4
5
x2 y 2
Hence, locus of z is
−
= 1, which represents a hyperbola.
4
5
z+i
Example 10: If u =
and z = x + iy, then show that
z+2
(i) locus of (x, y) is a straight line, if u is real.
(ii) locus of (x, y) is a circle, if u is purely imaginary.
Find the centre and radius of the circle.
Solution:
u=
u=
=
Re (u ) =
Im (u ) =
z +i
and z = x + iy
z+2
x + iy + i x + i ( y + 1) ( x + 2 − iy )
=
⋅
x + iy + 2 ( x + 2) + iy ( x + 2) − iy
[ x ( x + 2) + y ( y + 1)] + i [( y + 1)( x + 2) − xy ]
( x + 2) 2 + y 2
x ( x + 2) + y ( y + 1)
( x + 2) 2 + y 2
( y + 1)( x + 2) − xy
2
( x + 2) + y
2
=
x + 2y + 2
( x + 2) 2 + y 2
(i) If u is real, then
Im (u) = 0
x + 2y + 2
=0
( x + 2) 2 + y 2
x + 2y + 2 = 0
Hence, locus of (x, y) is x + 2y + 2 = 0, which represents a straight line.
1.11
1.12
Engineering Mathematics
(ii) If u is purely imaginary, then
Re (u) = 0
x ( x + 2) + y ( y + 1)
=0
( x + 2) 2 + y 2
x2 + y2 + 2x + y = 0
Hence, locus of (x, y) is x2 + y2 + 2x + y = 0, which represents a circle with centre
5
1⎞
⎛
at ⎜ −1, − ⎟ and radius
unit.
2
2⎠
⎝
Example 11: If sum and product of two numbers are real, show that the two
numbers must be either real or conjugate.
Solution: Let z1 = x1 + iy1 and z2 = x2 + iy2 are two complex numbers.
Let z1 + z2 = a, where a is real
(x1 + iy1) + (x2 + iy2) = a + i · 0
(x1 + x2) + i (y1 + y2) = a + i · 0
Comparing real and imaginary parts on both the sides,
x1 + x2 = a
y1 + y2 = 0
… (1)
… (2)
Let z1 z2 = b, where b is real
(x1x2
(x1 + iy1) (x2 + iy2) = b + i · 0
y1y2) + i (x2y1 + x1y2) = b + i · 0
Comparing real and imaginary parts on both the sides,
x1x2 y1y2 = b
x2y1 + x1y2 = 0
… (3)
… (4)
Substituting y2 =
y1 from Eq. (2) in Eq. (4),
x2y1 x1y1 = 0
y1(x2 x1) = 0
y1 = 0 or x2
If y1 = 0, then y2 = 0
Hence,
z1 = x1 and z2 = x2
If x1 = x2, then z1 = x1 + iy1 and z2 = x1 iy1
x1 = 0, x1 = x2
Hence, z1 and z2 both are either real or conjugate.
Example 12: If z1 and z2 are two complex numbers such that |z1 + z2| = |z1 - z2|,
prove that the difference of their amplitude is .
2
Solution: Let z1 = x1 + iy1 and z2 = x2 + iy2 are two complex numbers.
|z1 + z2| = |z1
z2|
|x1 + iy1 + x2 + iy2| = |x1 + iy1
x2
iy2|
Complex Numbers
|(x1 + x2) + i (y1 + y2)| = |(x1
1.13
x2) + i (y1
y2)|
( x1 + x2 ) 2 + ( y1 + y2 ) 2 = ( x1 − x2 ) 2 + ( y1 − y2 ) 2
Squaring both the sides,
x12 + x22 + 2 x1 x2 + y12 + y22 + 2 y1 y2 = x12 + x22 − 2 x1 x2 + y12 + y22 − 2 y1 y2
4x1x2 + 4y1y2 = 0
x1x2 + y1y2 = 0
Now, amp (z1)
amp (z2) = amp (x1 + iy1)
… (1)
amp (x2 + iy2)
⎛y ⎞
⎛y ⎞
= tan −1 ⎜ 1 ⎟ − tan −1 ⎜ 2 ⎟
⎝ x1 ⎠
⎝ x2 ⎠
⎛ y1 y2
−
⎜ x x
−1
1
2
= tan ⎜
y
y
1
⎜1 + ⋅ 2
⎜⎝
x1 x2
⎞
⎟
⎛x y −x y ⎞
⎟ = tan −1 ⎜ 2 1 1 2 ⎟
⎝ x1 x2 + y1 y2 ⎠
⎟
⎟⎠
⎛x y −x y ⎞
= tan −1 ⎜ 2 1 1 2 ⎟
⎝
⎠
0
[Using Eq. (1)]
p
2
Hence, the difference of amplitude of z1 and z2 is
= tan −1 ( ∞) =
Example 13: Show that
2
.
z
− 1 ≤ arg ( z ) .
z
Solution: Let z = reiq, where |z| = r and arg (z) = q
z
rei
−1 =
− 1 = |eiq
z
r
= |cos q + i sin q
= −2 sin 2
= 2 sin
≤2
q
2
q
2
arg( z )
1|
1| = |cos q
1+ i sin q |
q
q
q
q
q
q
+ i 2 sin cos = 2 sin − sin + i cos
2
2
2
2
2
2
sin 2
q
q
q
+ cos 2 = 2 sin
2
2
2
⎡ sin
⎢⎣∵
⎤
≤ 1⎥
⎦
1.14
Engineering Mathematics
π α
Example 14: If sin ` = i tan p, prove that cos p + i sin p = tan ⎛⎜ + ⎞⎟ .
⎝4 2⎠
Solution:
i tan q = sin
i sin θ sin α
=
1
cos θ
Applying componendo—dividendo,
cos q + i sin q 1 + sin a
=
cos q − i sin q 1 − sin a
iq
e
=
e − iq
e 2iq =
⎛p
⎞
1 + cos ⎜ − a ⎟
⎝2
⎠
⎛p
⎞
1 − cos ⎜ − a ⎟
⎝2
⎠
a⎞
⎛
2 cos 2 ⎜ p − ⎟
⎝4 2⎠
a⎞
⎛
2 sin 2 ⎜ p − ⎟
⎝4 2⎠
⎡ ⎛ p a ⎞⎤
(e iq ) 2 = ⎢cot ⎜ − ⎟ ⎥
⎣ ⎝ 4 2 ⎠⎦
2
⎡p ⎛ p a ⎞⎤
⎛p a ⎞
e iq = cot ⎜ − ⎟ = tan ⎢ − ⎜ − ⎟ ⎥
⎝4 2⎠
⎣ 2 ⎝ 4 2 ⎠⎦
⎛p a ⎞
= tan ⎜ − ⎟
⎝4 2⎠
⎛p a ⎞
cos q + i sin q = tan ⎜ − ⎟
⎝4 2⎠
-
ip
Example 15: Prove that (1 - e )
i
Solution: (1 − e )
−
1
2
+ (1 − e
−i
)
−
1
2
1
2
+ (1 - e
- ip
-
)
q
q
q
⎛
= ⎜ 2 sin 2 − i 2 sin cos ⎟
⎝
2
2
2⎠
q⎞
⎛
= ⎜ 2 sin ⎟
⎝
2⎠
−
1
2
−
1
2
1
p ⎞2
⎛
= ⎜ 1 + cosec ⎟ .
2⎠
⎝
= (1 − cos − i sin )
1
−
⎞ 2
q⎞
⎛
= ⎜ 2 sin ⎟
⎝
2⎠
1
2
−
1
2
+ (1 − cos + i sin )
−
q
q
q⎞
⎛
+ ⎜ 2 sin 2 + i 2 sin cos ⎟
⎝
2
2
2⎠
1
2
−
1
2
1
1
⎡
−
− ⎤
2
2
q
q
q
q
⎛
⎞
⎛
⎞
⎢ sin − i cos
+ ⎜ sin + i cos ⎟ ⎥
⎟
⎢⎣ ⎜⎝
⎥⎦
⎠
⎝
⎠
2
2
2
2
1
1
− ⎤
−
⎡
2
2
⎢⎧cos ⎛ p − q ⎞ − i sin ⎛ p − q ⎞ ⎫ + ⎧cos ⎛ p − q ⎞ + i sin ⎛ p − q ⎞ ⎫ ⎥
⎨
⎬
⎨
⎬
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎢⎩ ⎝ 2 2 ⎠
⎝ 2 2 ⎠⎭
⎝ 2 2 ⎠⎭ ⎥
⎩ ⎝ 2 2⎠
⎢⎣
⎥⎦
Complex Numbers
q⎞
⎛
= ⎜ 2 sin ⎟
⎝
2⎠
−
1
2
q⎞
⎛
= ⎜ 2 sin ⎟
⎝
2⎠
−
1
2
q⎞
⎛
= ⎜ 2 sin ⎟
⎝
2⎠
−
1
2
1.15
1
1
1
− ⎤
−
⎡
−
⎛p q ⎞
⎛p q ⎞
⎛p q ⎞
⎛p q ⎞
2
2
⎧
⎫
⎫
⎧
−
−
i
i
−i ⎜ − ⎟ ⎤
−
⎜
⎟
⎜
⎟
q ⎞ 2 ⎡ i ⎜⎝ 4 − 4 ⎟⎠
⎪ ⎝ 2 2⎠ ⎪ ⎥ ⎛
⎢⎪ ⎝ 2 2 ⎠ ⎪
⎝ 4 4⎠
e
+
e
2
sin
e
e
=
+
⎢
⎥
⎬
⎨
⎬ ⎥ ⎜
⎟⎠
⎢⎨
⎝
2
⎢
⎥⎦
⎪
⎪
⎪
⎪
⎩
⎭
⎩
⎭
⎣
⎢⎣
⎥⎦
⎡
q⎞
⎛ p q ⎞⎤ ⎛
⎢ 2 cos ⎜⎝ 4 − 4 ⎟⎠ ⎥ = ⎜⎝ 2 sin 2 ⎟⎠
⎦
⎣
−
1
2
1
⎡
q ⎞⎤ 2
2 ⎛p
⎢ 4 cos ⎜⎝ 4 − 4 ⎟⎠ ⎥
⎦
⎣
1
⎡
2 ⎤
⎢∵1 + cos = 2 cos 2 ⎥
⎣
⎦
⎡ ⎧
⎛ p q ⎞ ⎫⎤ 2
⎢ 2 ⎨1 + cos ⎜⎝ − ⎟⎠ ⎬⎥
2 2 ⎭⎦
⎣ ⎩
1
q ⎤2
⎡
1
⎢1 + sin 2 ⎥
q ⎤2
⎡
=⎢
⎥ = ⎢cosec + 1⎥
2 ⎦
⎣
⎢ sin q ⎥
⎣
2 ⎦
Example 16: If a = cos ` + i sin ` and b = cos a + i sin a, then show that
(a + b )(ab − 1) sinα + sinβ
.
=
(a − b )(ab + 1) sinα − sinβ
a = cos a + i sin a = eia, b = cos b + i sin b = eib
Solution:
( a + b)( ab − 1) (e ia + e ib )(e ia e ib − 1)
=
( a − b)( ab + 1) (e ia − e ib )(e ia e ib + 1)
=
(e 2ia e ib + e 2ib e ia − e ia − e ib ) e − i ( b +a )
⋅
(e 2ia e ib − e 2ib e ia + e ia − e ib ) e − i ( b +a )
e ia + e ib − e − ib − e − ia (e ia − e − ia ) + (e ib − e − ib )
=
e ia − e ib + e − ib − e − ia (e ia − e − ia ) − (e ib − e − ib )
2i sin a + 2i sin b sin a + sin b
=
=
2i sin a − 2ii sin b sin a − sin b
=
Example 17: If a = cos ` + i sin `, b = cos a + i sin a, c = cos f + i sin f , then
prove that
(b + c )(c + a )(a + b )
⎛ β −γ
= 8cos ⎜
abc
⎝ 2
⎞
⎛ γ −α ⎞
⎛α − β ⎞
⎟ cos ⎜ 2 ⎟ cos ⎜ 2 ⎟ .
⎠
⎝
⎠
⎝
⎠
Solution: a = cos a + i sin a = eia, b = cos b + i sin b = e ib, c = cos g + i sin g = e ig,
(b + c)(c + a)( a + b) (e ib + e ig )(e ig + e ia )(e ia + e ib )
=
abc
e ia e ib e ig
=
e ib + e ig eig + e ia eia + eib
⋅ ig ia ⋅ ia ib
i b ig
e2e2
e2e2
e2e2
ia ia
⎡ ia
⎤
2
2
⎢∵ e = e e etc.⎥
⎣
⎦
1.16
Engineering Mathematics
− i ( b −g )
− i (g −a )
− i (a − b )
⎡ i ( b −g )
⎤ ⎡ i (g −a )
⎤ ⎡ i (a − b )
⎤
= ⎣e 2 + e 2 ⎦ ⎣e 2 + e 2 ⎦ ⎣e 2 + e 2 ⎦
⎛ b −g ⎞
⎛g − a ⎞
⎛a − b ⎞
2 cos ⎜
2 cos ⎜
= 2 cos ⎜
⎝ 2 ⎟⎠
⎝ 2 ⎟⎠
⎝ 2 ⎟⎠
⎛ b −g ⎞
⎛g − a
cos ⎜
= 8 cos ⎜
⎝ 2 ⎟⎠
⎝ 2
⎞
⎛a − b ⎞
⎟⎠ cos ⎜⎝
⎟
2 ⎠
Example 18: If ` = i + 1, a = 1 - i and tanφ =
( x + ` )n - ( x + a )n
= sin ne cosec ne .
` -a
1
x +1
cot f = x + 1, x = cot f
Solution: a = i + 1, b = 1
1
, then prove that
x +1
i, tan =
1
(x + α ) − (x + β )
(cot φ − 1 + i + 1) − (cot φ − 1 + 1 − i ) n
=
α −β
i +1−1+ i
n
n
n
n
n
⎛ cos φ ⎞ ⎛ cos φ ⎞
+i⎟ −⎜
−i⎟
⎜
⎝ sin φ
⎠ ⎝ sin φ
⎠
=
2i
(cos f + i sin f ) n − (cos f − i sin f ) n
=
2i sin n f
=
(e if ) n − (e − if ) n e inf − e − inf 2i sin nf
=
=
2i sin n f
2i sin n f
2i sin n f
= sin nf cosec n f
Example 19: If (1 + cos p + i sin p ) (1 + cos 2p + i sin 2p ) = u + iv, prove that
θ
v
3
(i) u2 + v 2 = 16 cos 2 cos 2θ
(ii)
.
= tan
2
u
2
Solution: u + iv = (1 + cos q + i sin q ) (1 + cos 2q + i sin 2q )
q
q
q⎞
⎛
= ⎜ 2 cos 2 + i 2 sin cos ⎟ (2 cos 2 q + i 2 sin q cos q )
2
2
2⎠
⎝
= 2 cos
q⎛
q
q⎞
cos + i sin ⎟ 2 cos q (cosq + i sin q )
2 ⎜⎝
2
2⎠
= 4 cos
i
q
cos q ⋅ e 2 ⋅ eiq
2
= 4 cos
q
cosq ⋅ e
2
q
= reif
i 3q
2
Complex Numbers
r = u + iv = 4 cos
where,
u 2 + v 2 = 4 cos
2
u 2 + v 2 = 16 cos 2
tan −1
cos
cos
2
cos 2
φ = arg(u + iv) =
and
2
1.17
3θ
2
v 3
=
u
2
v
3
= tan
u
2
Example 20: If (a1 + ib1)(a2 + ib2) . . . . . (an + ibn) = A + iB, prove that
(i) (a12 + b12 )(a22 + b22 )…… (an2 + bn2 ) = A2 + B 2
⎛b
(ii) tan −1 ⎜ 1
⎝ a1
⎞
−1 ⎛ b2
⎟ + tan ⎜
⎠
⎝ a2
⎞
−1 ⎛ bn ⎞
−1 ⎛ B ⎞
⎟ + ...... + tan ⎜ ⎟ = tan ⎜ ⎟ .
⎝ A⎠
⎠
⎝ an ⎠
Solution:
(a1 + ib1) (a2 + ib2) . . . . . (an + ibn) = A + iB
(i) Taking modulus of Eq. (1) on both the sides,
|(a1 + ib1)(a2 + ib2) . . . . . (an + ibn)| = |A + iB |
|a1 + ib1| |a2 + ib2| . . . . . |an + ibn| = | A + iB |
a12 + b12 a22 + b22
an2 + bn2 =
… (1)
A2 + B 2
Squaring both the sides,
(a12 + b12 )(a22 + b22 )
(an2 + bn2 ) = A2 + B 2
(ii) Taking argument of Eq. (1) on both the sides,
Arg [(a1 + ib1) (a2 + ib2) . . . . . . (an + ibn)] = Arg (A + iB)
Arg (a1 + ib1) + Arg (a2 + ib2) + . . . . . . + Arg (an + ibn) = Arg (A + iB)
⎛b ⎞
⎛b
tan −1 ⎜ 1 ⎟ + tan −1 ⎜ 2
⎝ a1 ⎠
⎝ a2
Example 21: If
⎞
−1 ⎛ bn
⎟ + .... + tan ⎜
⎠
⎝ an
⎞
−1 ⎛ B ⎞
⎟ = tan ⎜ ⎟
⎝ A⎠
⎠
1
1
+
= 1, where ` , a , a and b are real, express b in
` + i a a + ib
terms of ` and a.
Solution:
1
1
+
=1
a + i b a + ib
1
1
a + ib − 1
= 1−
=
a + ib
a + ib
a + ib
1.18
Engineering Mathematics
a + ib
a + ib
(a − 1) − i b
=
⋅
(a − 1) + i b (a − 1) + i b (a − 1) − i b
a + ib =
=
a (a − 1) − i 2 b 2 + i b (a − 1) − iab
(a − 1) 2 + b 2
=
a (a − 1) + b 2
b
−i
2
2
(a − 1) + b
(a − 1) 2 + b 2
Comparing the imaginary part on both the sides,
b=
Example 22: If x
Solution:
iy = 3 a
−β
(α − 1) 2 + β 2
ib , prove that
a b
+ = 4( x 2 − y 2 ).
x y
x + iy = 3 a + ib ,
1
(a + ib) 3 = x + iy
a + ib = ( x + iy )3
= x3 + i 3 y 3 + 3 x 2 iy + 3 xi 2 y 2
[ i 3 = −i ]
= ( x3 − 3 xy 2 ) + i (3 x 2 y − y 3 )
Comparing real and imaginary parts on both the sides,
Hence,
a = x3 3xy2 and b = 3x2y y3
a
b
= x 2 − 3 y 2 and = 3 x 2 − y 2
x
y
a b
+ = 4( x 2 − y 2 )
x y
⎛π
Example 23: If xr = cos ⎜ r
⎝2
⎛π
⎞
⎟ + i sin ⎜ r
⎝2
⎠
⎞
x1 x2 x3 ......xn = -1.
⎟ , show that nlim
ãÇ
⎠
i
xr = cos
Solution:
2r
i
lim x1 ⋅ x2 ⋅ x3
n →∞
+ i sin
2r
i
i
2
lim x1 ⋅ x2 ⋅ x3
e2
n →∞
⎛1 1 1
i ⎜ + 2+ 3+
⎝2 2 2
n →∞
⋅⋅⋅ xn = lim
n →∞
⎡ ⎛ 1 ⎞n ⎤
i ⎢1−⎜ ⎟ ⎥
e ⎢⎣ ⎝ 2 ⎠ ⎥⎦
r
i
3
⋅ xn = lim e 2 ⋅ e 2 ⋅ e 2
= lim e
n →∞
= e2
+
1 ⎞
⎟
2n ⎠
n
Complex Numbers
⎡
⎢
1
⎢ 1 1
⎢∵ 2 + 2 + 3 +
2
2
⎢
⎣
1⎡ ⎛1⎞
⎢1 −
2 ⎢⎣ ⎜⎝ 2 ⎟⎠
1
⋅⋅+ n =
1
2
1−
2
1.19
⎤
⎥
n⎥
⎥⎦
⎛1⎞ ⎥
= 1− ⎜ ⎟ ⎥
⎝2⎠ ⎥
⎦
n⎤
= lim e ip e
⎛1⎞
− ip ⎜ n ⎟
⎝2 ⎠
n→∞
= e ip e
= (cos + i sin )e
−
−
ip
2∞
i
∞
= (−1 + i.0)e0
Hence,
lim x1 ⋅ x2 ⋅ x3 ..... xn = −1
n →∞
Example 24: Prove that e 2 ai cot
Solution:
e 2 ai cot
−1
b
b
⎛ bi − 1 ⎞
⎜⎝
⎟
bi + 1 ⎠
−a
= 1.
−a
=1
bi − 1 bi + i 2 b + i
=
=
bi + 1 bi − i 2 b − i
i = re iq
Let b + i = reiq, then b
r = |b + i| =
⎛ bi − 1 ⎞
⎜
⎟
⎝ bi + 1 ⎠
−1
b 2 + 1 and q = arg (b + i) = tan
1
1
= cot 1b
b
−1
bi − 1 rei
= −i = e 2i = e 2i cot b
bi + 1 re
Substituting in the given equation,
e 2 ai cot
Hence,
e
−1
b
2 ai cot −1 b
⎛ bi − 1 ⎞
⎜
⎟
⎝ bi + 1 ⎠
⎛ bi − 1 ⎞
⎜
⎟
⎝ bi + 1 ⎠
−a
= e 2 ai cot
−1
b
(e 2i cot
−1
b −a
)
= e0 = 1
−a
= 1.
Exercise 1.1
1. Find the modulus and principal
value of the argument
1 + 2i
1 + 2i
(i)
3 i (ii)
(iii)
1 − 3i
1 − (1 − i ) 2
1+ i
(iv)
(v) tan a i.
1 i
−5p
1 3p
⎡
⎢ Ans. : (i) 2, 6 (ii) 2 , 4
⎢
p
⎢
(iii) 1, 0 (iv) 1,
⎢
4
⎢
p
⎢
( v) sec a , a −
⎢⎣
2
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦
1.20
Engineering Mathematics
2. Express in polar form
(i)
(iii)
3 i
1+ i
1− i
(ii)
2 + 6 3i
5 + 3i
.
⎡
⎛
⎞⎤
⎢ Ans. : (i) 2 ⎜ cos − i sin ⎟ ⎥
6
6⎠⎥
⎝
⎢
⎢
⎥
(ii) cos − i sin
⎢
⎥
2
2
⎢
⎥
⎢
⎛
⎞⎥
(iii) 2 ⎜ cos − i sin ⎟ ⎥
⎢
3
3 ⎠ ⎥⎦
⎝
⎢⎣
3. Find the value of
3 4i.
i or 2 + i]
[Ans. : 2
4. Find z if arg ( z + 2i ) =
arg ( z − 2i ) =
4
,
3
.
4
[Ans. : 2]
5. Find the locus of z, if
z 1
is purely
z +i
imaginary.
[Ans. : circle x2 + y2
x
y = 0]
6. Find the locus of z if
(i)
z 1
=1
z +1
(ii) arg
z 1
= .
z +1
4
⎤
⎡ Ans. : (i) x = 0
⎥
⎢
2
2
⎣(ii) circle x + y − 2 y − 1 = 0 ⎦
7. Find two numbers whose sum is 4
and product is 8.
[Ans. : 2 ± 2i]
8. If x + iy = a + ib , prove that
(x2 + y2)2 = a2 + b2.
⎡ Hint : square both the sides and ⎤
⎢
⎥
then take modulus
⎣
⎦
z 1
9. If |z| = 1, z ó 1, prove that
is
z +1
purely imaginary.
⎡ Hint : z = x + iy, | z | = 1 ∴ x 2 + y 2 = 1, ⎤
⎢
⎥
z − 1 ( x − 1) + iy ( x + 1) − iy ⎥
⎢
=
⋅
⎢⎣
z + 1 ( x + 1) + iy ( x + 1) − iy ⎥⎦
z
10. If |z1 + z2| = |z1 z2|, prove that 2 is
z1
purely imaginary.
11. If a = ei2a, b = ei2b, c = ei2g, then prove
ab
c
2 cos(
).
that
c
ab
12. If a = cos a + i sin a, b = cos b +
i sin b, then prove that
1 a b
sin(a
).
2i b a
13. If a = cos a + i sin a, then prove
that
2
1 i tan
(i)
1+ a
2
1+ a
= i cot .
(ii)
1 a
2
14. If a = a + ib, b = c + id,
1
, then
show that if =
+1
a 2 + b2 =
(c 1) 2
d2
(c + 1) 2 + d 2
.
c + id − 1 (c − 1) + id
⎡
⎢ Hint : a + ib = c + id + 1 = (c + 1) + id
⎢
⎢
| (c − 1) + id |
a + ib =
⎢
|
(c + 1) + id |
⎣
⎤
,⎥
⎥
⎥
⎥
⎦
15. If x2 + y2 = 1, then prove that
1 + x + iy
= x + iy .
1 + x − iy
⎡ Hint : x 2 + y 2 = 1, ( x + iy )( x − iy ) = 1,⎤
⎢
⎥
⎢
⎥
x + iy
1
=
⎢
⎥
1
x
−
iy
⎢
⎥
⎢
⎥
x + iy
1
=
⎢ Apply dividendo
⎥
1
+
+
1
+
−
x
iy
x
iy
⎢
⎥
⎢⎣
⎥⎦
Complex Numbers
16. If (1 + ai)(1 + bi)(1 + ci) = p + iq,
prove that
19. Find the value of
(i) x2 6x + 13, when x = 3 + 2i
(i) p tan [tan 1a + tan 1b + tan 1c] = q
(ii) (1 + a2)(1 + b2)(1 + c2) = p2 + q2.
1
17. If (α + i β ) =
, prove that
a + ib
( 2 + 2) (a2 + b2) = 1.
⎡
1 ⎤
⎢ Hint : α + i β =
⎥
a + ib ⎥⎦
⎢⎣
18. If a = cos a + i sin a, b = cos b
+ i sin b, where 0 < ,
find polar form of
<
1 + a2
.
1 − i ab
⎡
⎛p a + b
⎢ Ans. :cos a sec ⎜ −
⎝4
2
⎢⎣
2
1.21
⎡ Hint : ( x − 3) 2 + 4 = ( 2i ) 2 + 4 ⎤
⎥
⎢
= −4 + 4 = 0
⎦
⎣
(ii) x4 4x3 + 4x2 + 8x + 46, when
x = 3 + 2i.
⎡Hint : (x 3)2 = 4, x2 6x + 13 = 0,⎡
⎢
divide given expression by ⎢
⎢
⎢
x2 6x + 13,
⎢
⎢
x4 4x3 + 4x2 + 8x + 46
⎢
⎢
= (x2 6x + 13) (x2 + 2x + 3) ⎢
⎢
⎢⎣
⎢⎣
+7=7
, then
20. If x r = cos
⎡p ⎛ a − b ⎞⎤
⎟
2 ⎠ ⎥⎦
⎞ i ⎢⎣ 4 + ⎜⎝
⎟⎠ e
⎤
⎥
⎥⎦
+ i sin r , prove that
3r
3
(i) x1 . x2 . x3 . . . . . = i
(ii) x0 x1 x2 . . . .
=
i.
1.8 DE MOIVRE’S THEOREM
Statement: For any real number n, one of the values of (cos q + i sin q )n is
cos nq + i sin nq.
Hence, (cos q + i sin q )n = cos nq + i sin nq
Proof: Case I: If n is a positive integer
Let z1 = r1 (cos q1 + i sin q1), z2 = r2 (cos q2 + i sin q2) , …… , zn = rn (cos qn + i sin qn).
z1 z2 = r1 (cos q1 + i sin q1) r2 (cos q2 + i sin q2)
= r1 r2 [(cos q1 cos q2
sin q1 sin q2) + i(sin q1 cos q2 + cos q1 sin q2)]
= r1 r2 [cos (q1 + q2) + i sin (q1 + q2)]
Similarly,
z1 z2……. zn = r1 (cos q1 + i sin q1) r2 (cos q2 + i sin q2)……. rn (cos qn + i sin qn)
= (r1 r2…….. rn) (cos q1 + i sin q1) (cos q2 + i sin q2)….. (cos qn + i sin qn)
= (r1 r2…….. rn)[cos (q1 +q2 ……+ qn) + i sin (q1 + q2……+ qn)] … (1)
If z1 = z2 = ……. = zn = z = r (cos q + i sin q ) , then Eq. (1) reduces to
zn = rn (cos q + i sin q )n = rn (cos nq + i sin nq )
(cos q + i sin q )n = (cos nq + i sin nq ), where n is a positive integer.
1.22
Engineering Mathematics
Case II: If n is a negative integer
Let n = m, where m is a positive integer.
(cos q + i sin q )n = (cos q + i sin q )–m
1
=
(cos q + i sin q ) m
1
=
[ Using Case I]
(cos mq + i sin mq )
(cos mq − i sin mq )
=
(cos mq + i sin mq )(cos mq − i sin mq )
= cos mq − i sin mq
= cos( −m)q + i sin(−m
m)q [ cos ( − q ) = cosq , sin(−q ) = − sin q ]
= cos nq + i sin nq , where n is a negetive integer.
Case III: If n is a rational number
Let n =
p
, where p and q are integers and q
q
0.
q
⎛
⎞
⎛
⎞
Consider ⎜ cos + i sin ⎟ = ⎜ cos q ⋅ + i sin q ⋅ ⎟
q
q⎠
q
q⎠
⎝
⎝
= cos + i sin
Hence,
1
q
(cos + i sin ) = cos
p
q
q
+ i sin
[ Using Casse I and II]
q
⎛
⎞
(cos + i sin ) = ⎜ cos + i sin ⎟
q
q
⎝
⎠
p
⎛
⎞
= ⎜ cos p ⋅ + i sin p ⋅ ⎟
q
q
⎝
⎠
[ Using Case I and II]
(cos q + i sin q )n = (cos nq + i sin nq ), where n is a rational number.
Hence,
(cos q + i sin q )n = cos nq + i sin nq for any real number n.
Example 1: Simplify
π
π
⎛
⎜ 1 + sin 8 + i cos 8
(i) ⎜
⎜⎜ 1 + sin π − i cos π
8
8
⎝
8
⎞
⎟
n
n
⎟ (ii) (1 + cos q + i sin q ) + (1 + cos p - i sin p ) .
⎟⎟
⎠
Complex Numbers
1.23
Solution:
(i)
p
p
⎛
1 + sin + i cos
⎜
8
8
⎜
p
p
⎜ 1 + sin − i cos
⎝
8
8
8
8
⎡1 + cos ⎛ p − p ⎞ + i sin ⎛ p − p ⎞ ⎤
⎞
⎜
⎟
⎜
⎟
⎢
⎝ 2 8 ⎠⎥
⎝2 8⎠
⎟
⎥
⎟ =⎢
⎢
⎥
p
p
p
p
⎛
⎞
⎛
⎞
⎟
⎢⎣1 + cos ⎜⎝ 2 − 8 ⎟⎠ − i sin ⎜⎝ 2 − 8 ⎟⎠ ⎦⎥
⎠
3p
3p
⎛
1 + cos
+ i sin
⎜
8
8
=⎜
3p
3p
⎜ 1 + cos
− i sin
⎝
8
8
⎛
2
⎜ 2 cos
=⎜
2
⎜⎜ 2 cos
⎝
8
3
3
3
cos
+ 2i sin
16
16
16
3
3
3
cos
− 2i sin
16
16
16
3
3
⎛
⎜ cos 16 + i sin 16
=⎜
⎜⎜ cos 3 − i sin 3
16
16
⎝
=
⎞
⎟
⎟
⎟
⎠
8
⎞
⎟
⎟
⎟⎟
⎠
8
⎞
⎟
⎟
⎟⎟
⎠
⎛ 3p ⎞
⎛ 3p ⎞
cos ⎜ 8 ⋅ ⎟ + i sin ⎜ 8 ⋅ ⎟
⎝ 16 ⎠
⎝ 16 ⎠
⎛ 3p ⎞
⎛ 3p ⎞
cos ⎜ 8 ⋅ ⎟ − i sin ⎜ 8 ⋅ ⎟
⎝ 16 ⎠
⎝ 16 ⎠
[Using De Moivre’s theorem]
3p
3p
3p
i
⎛ 6p ⎞
+ i sin
2
i⎜
⎟
e
2
2
=
= 3p = e ⎝ 2 ⎠
3p
3p
−i
− i sin
cos
e 2
2
2
= e i 3p = (cos 3p + i sin 3p ) = −1
cos
(ii) (1 + cos q + i sin q )n + (1 + cos q - i sin q )n
n
⎛
⎞ ⎛
⎞
= ⎜ 2 cos 2 + 2i sin cos ⎟ + ⎜ 2 cos 2 − 2i sin cos ⎟
2
2
2⎠ ⎝
2
2
2⎠
⎝
⎛
⎞
= ⎜ 2 cos ⎟
2
⎝
⎠
n
n
n
⎡⎛
⎞ ⎛
⎞ ⎤
⎢⎜ cos + i sin ⎟ + ⎜ cos − i sin ⎟ ⎥
2
2⎠ ⎝
2
2 ⎠ ⎦⎥
⎢⎣⎝
n
n
n
n
n ⎞
⎛
⎞ ⎛
= ⎜ 2 cos ⎟ ⎜ cos
+ i sin
+ cos
− i sin
⎟
2
2
2
2
2 ⎠
⎝
⎝
⎠
n
n ⎞
⎛
⎞ ⎛
= ⎜ 2 cos ⎟ ⎜ 2 cos
⎟
2⎠ ⎝
2 ⎠
⎝
n
1.24
Engineering Mathematics
n
⎛ 1 + sinα + i cosα ⎞
⎛ nπ
⎞
⎛ nπ
⎞
Example 2: Prove that ⎜
⎟ = cos ⎜ 2 − nα ⎟ + i sin ⎜ 2 − nα ⎟ .
1
+
sin
α
−
i
cos
α
⎝
⎠
⎝
⎠
⎝
⎠
⎞⎤
⎛π
⎞
⎛π
⎡
1 + cos ⎜ − α ⎟ + i sin ⎜ − α ⎟ ⎥
n
⎢
1 + sin α + i cos α ⎞
⎠⎥
⎝2
⎠
⎝2
Solution: ⎛⎜
⎟ =⎢
⎢
⎝ 1 + sin α − i cos α ⎠
⎛π
⎞
⎛π
⎞⎥
⎢ 1 + cos ⎜ − α ⎟ − i sin ⎜ − α ⎟ ⎥
2
2
⎣
⎝
⎠
⎝
⎠⎦
n
α⎞
⎛π α ⎞
⎛π α ⎞⎤
⎡
2⎛π
⎢ 2 cos ⎜⎝ 4 − 2 ⎟⎠ + 2i sin ⎜⎝ 4 − 2 ⎟⎠ cos ⎜⎝ 4 − 2 ⎟⎠ ⎥
⎥
=⎢
⎢
α⎞
⎛π α ⎞
⎛π α ⎞⎥
2⎛π
⎢ 2 cos ⎜ − ⎟ − 2i sin ⎜ − ⎟ cos ⎜ − ⎟ ⎥
⎣
⎝4 2⎠
⎝ 4 2 ⎠⎦
⎝4 2⎠
⎛π α ⎞⎤
⎛π α ⎞
⎡
⎢ cos ⎜⎝ 4 − 2 ⎟⎠ + i sin ⎜⎝ 4 − 2 ⎟⎠ ⎥
⎥
=⎢
⎢
⎛π α ⎞
⎛π α ⎞⎥
⎢ cos ⎜ − ⎟ − i sin ⎜ − ⎟ ⎥
⎣
⎝4 2⎠
⎝ 4 2 ⎠⎦
n
n
n
⎡ i ⎛⎜ π − α ⎞⎟ ⎤
⎡ 2i ⎛⎜ π − α ⎞⎟ ⎤
⎢ e ⎝4 2⎠ ⎥
4
2
⎠⎥
=⎢
⎦
⎥ = ⎢⎣e ⎝
⎢ −i ⎛⎜ π − α ⎞⎟ ⎥
⎢⎣ e ⎝ 4 2 ⎠ ⎥⎦
=e
⎛ nπ
⎞
− nα ⎟
i⎜
⎝ 2
⎠
⎛ nπ
⎞
⎛ nπ
⎞
= cos ⎜
− nα ⎟ + i sin ⎜
− nα ⎟
⎝ 2
⎠
⎝ 2
⎠
(1 + i )8 ( 3 − i )
4
Example 3: Expand in polar form
(1 − i )4 ( 3 + i )
8
.
Solution: Let 1 + i = r1 (cos q1 + i sin q1)
where,
π
⎛1⎞
r1 = 1 + 1 = 2 and θ1 = tan −1 ⎜ ⎟ = tan −1 1 =
4
⎝1⎠
iπ
π
π⎞
⎛
1 + i = 2 ⎜ cos + i sin ⎟ = 2 e 4
4
4⎠
⎝
iπ
−
π
π⎞
⎛
1 − i = 2 ⎜ cos − i sin ⎟ = 2 e 4
4
4⎠
⎝
Let
3 + i = r2 (cos q2 + i sin q2)
where,
r2 = 3 + 1 = 4 = 2
and
⎛ 1 ⎞ π
θ 2 = tan −1 ⎜
⎟=
⎝ 3⎠ 6
n
Complex Numbers
1.25
ip
p
p⎞
⎛
3 + i = 2 ⎜ cos + i sin ⎟ = 2e 6
⎝
6
6⎠
3 − i = 2e
−
ip
6
8
(1 + i )8 ( 3 − i )
4
(1 − i ) 4 ( 3 + 1)
8
=
=
ip
⎛
⎞ ⎛ − i6p ⎞
4
e
2
⎜⎝
⎟⎠ ⎜⎝ 2e ⎟⎠
4
ip
−
⎛
⎞ ⎛ i6p ⎞
4
⎜⎝ 2e ⎟⎠ ⎜⎝ 2e ⎟⎠
8
(
)
) (2 e )
(24 e 2ip ) 24 e −
(22 e − ip
4
8
2 ip
3
4 ip
3
1 3ip − 63ip
e
22
1
= e ip
4
=
Hence,
(1 + i )8 ( 3 − 1)
4
(
8
(1 − i )
4
3 + 1)
=
1
(cos p + i sin p )
4
=
1
(cos p + i sin p )
4
m
m
m
y⎞
⎛m
Example 4: Prove that ( x + iy ) n + ( x - iy ) n = 2( x 2 + y 2 ) 2 n cos ⎜ tan -1 ⎟ .
x⎠
⎝n
Solution: Let x + iy = r (cos q + i sin q )
1
r = x + iy = x 2 + y 2 = ( x 2 + y 2 ) 2
= arg ( x + iy ) = tan −1
m
( x + iy ) n
m
+ ( x − iy ) n
= [r (cos + i sin
=
m
rn
m
)] n
y
x
+ [r (cos − i sin
m
m
m
m ⎞
⎛
+ i sin
+ cos
− i sin
⎜ cos
⎟
n
n
n
n ⎠
⎝
m
m ⎞
⎛
= r n ⎜ 2 cos
⎟
n ⎠
⎝
Substituting the values of r and q,
m
m
m
)] n
m
y⎞
⎛m
( x + iy ) n + ( x − iy ) n = 2( x 2 + y 2 ) 2 n cos ⎜ tan −1 ⎟
x⎠
⎝n
1.26
Engineering Mathematics
Example 5: Prove that if n is any positive integer and (1 + x)n = p0 + p1x + . . . .
. + p x n + . . . . , then
n
n
(i) p0
p2 + p4
. . . . = 2 2 cos n
4
nπ
.
4
Solution: (1 + x)n = p0 + p1x + p2x2 + . . . . . + pnxn + . . . . .
n
−1
(ii) p0 + p4 + p8 + . . . . = 2n-2 + 2 2 cos
... (1)
(i) Putting x = i in Eq. (1),
(1 + i )n = p0 + p1i + p2i2 + p3i3 + p4i4 + p5i5 + . . . . .
= p0 + p1i
= (p0
p2
p2 + p4
p3i + p4 + p5i + . . . . .
. . .) + i (p1
p3 + p5
. . .)
... (2)
Let 1 + i = r (cos q + i sin q )
r=|1+i|=
1 + 1 = 2 , q = arg (1 + i) = tan
1
⎛1⎞ p
⎜⎝ ⎟⎠ =
1
4
p
p⎞
⎛
1 + i = 2 ⎜ cos + i sin ⎟
⎝
4
4⎠
⎡ ⎛
p
p ⎞⎤
(1 + i ) n = ⎢ 2 ⎜ cos + i sin ⎟ ⎥
4
4 ⎠⎦
⎣ ⎝
n
np
np ⎞
⎛
+ i sin
= 2 2 ⎜ cos
⎟
⎝
4
4 ⎠
n
... (3)
From Eqs. (2) and (3), we get
n
np
np
⎛
+ i sin
2 2 ⎜ cos
⎝
4
4
⎞ = (p
0
⎟⎠
p2 + p4 . . . . .) + i (p1
p3 + p5
. . . . .)
Comparing real part on both the sides,
n
2 2 cos
np
= p0
4
p2 + p4 . . . .
n
p0
p2 + p4 . . . . = 2 2 cos
np
4
(ii) Putting x = 1 in Eq. (1),
(1 + 1)n = p0 + p1 + p2 + p3 + p4 + p5 + . . . .
2n = p0 + p1 + p2 + p3 + p4 + p5 + . . . . .
... (4)
... (5)
Putting x = 1 in Eq. (1),
(1
1)n = p0
0 = p0
p1 + p2
p1 + p2
p3 + p4 p5 + . . . .
p3 + p4 p5 + . . . . .
... (6)
Complex Numbers
1.27
Adding Eqs. (5) and (6),
2n = 2 (p0 + p2 + p4 + p6 + p8 + . . . .)
2n 1 = p0 + p2 + p4 + p6 + p8 + . . . . . .
... (7)
Adding Eqs. (4) and (7),
n
np
2 2 cos
+ 2n 1 = 2 (p0 + p4 + p8 + . . . . . . )
4
n
np
−1
p0 + p4 + p8 + . . . . = 2 2 cos 4 + 2n − 2
or
Example 6: If x −
(i)
xyz +
Solution:
1
1
1
= 2i sinθ , y − = 2i sinφ , z − = 2i sin x, prove that
x
y
z
m
1
= 2 cos (p + e + x)
xyz
x−
(ii)
n
x
y
n
+m
⎛θ φ⎞
= 2cos ⎜ − ⎟ .
⎝ m n⎠
x
y
1
1
1
= 2i sin q , y − = 2i sin f , z − = 2i sinY
x
y
z
x 2 − 1 = 2i x sin q
x 2 − 2i x sin q − 1 = 0
x=
2i sin q ± 4i 2 sin 2 q + 4
2
= i sin q ± − sin 2 q + 1
= i sin q ± cos q
Considering the positive sign,
x = i sin q + cos q = cos q + i sin q = eiq
Similarly,
(i)
y = eif, z = eiy
xyz +
1
1
= eiθ eiφ eiψ + iθ iφ iψ
xyz
e e e
= ei (θ +φ +ψ ) + e − i (θ +φ +ψ )
= 2 cos (θ + φ + ψ )
(ii)
m
n
x
+
y
n
y
m
x
=
1
1
(e i ) m
(ei ) n
i
(e )
=e
i
1
n
q f
m n
2cos
+
1
(e i ) m
+e
m
i
q f
m n
n
1.28
Engineering Mathematics
Example 7: If sin ` + sin a + sin f = 0 and cos ` + cos a + cos f = 0, prove that
(i) cos (` + a ) + cos (a + f ) + cos (f + ` ) = 0
(ii) cos 2` + cos 2a + cos 2f = 0
(iii) cos2 ` + cos2 a + cos2 f = 3
2
(iv) sin 3` + sin 3a + sin 3f = 3 sin (` + a + f ).
Solution: sin a + sin b + sin g = 0 = cos a + cos b + cos g
(cos a + cos b + cos g ) + i (sin a + sin b + sin g ) = 0 + i 0
(cos a + i sin a ) + (cos b + i sin b ) + (cos g + i sin g ) = 0
eia + eib + eig = 0
Let x = eia, y = eib, z = eig
x+y+z=0
then
Also
(cos a + cos b + cos g )
(cos a
i sin a ) + (cos b
i (sin a + sin b + sin g ) = 0
i sin b ) + (cos g
e-ia + e
… (1)
i 0
i sin g ) = 0
ib
+ e ig = 0
1
1
1
+ i + i =0
ei
e
e
1 1 1
+ + =0
x y z
yz + zx + xy
=0
xyz
xy + yz + zx = 0
(i) From Eq. (2),
xy + yz + zx = 0
eia eib + eib eig + eig eia = 0
ei(a +b ) + ei(b +g ) + ei(g +a ) = 0
[cos (a + b ) + i sin (a + b )] + [cos ( b + g ) + i sin ( b + g )] + [cos (g + a )
+ i sin (g + a )] = 0 + i 0
Comparing real part on both the sides,
cos (a + b ) + cos ( b + g ) + cos (g + a ) = 0
(ii) From Eq. (1),
x+y+z=0
… (2)
Complex Numbers
1.29
(x + y + z)2 = 0
x2 + y2 + z2 + 2(xy + yz + zx) = 0
x2 + y2 + z2 = 0
[Using Eq. (2)]
(eia )2 + (eib )2 + (eig )2 = 0
e2ia + e2ib + e2ig = 0
(cos 2a + i sin 2a ) + (cos 2b + i sin 2b ) + (cos 2g + i sin 2g ) = 0 + i 0
Comparing real part on both the sides,
cos 2a + cos 2b + cos 2g = 0
… (3)
(iii) From Eq. (3),
cos 2a + cos 2b + cos 2g = 0
1
2 cos2 a + 1
(iv) From Eq. (1),
2 cos2 b + 1
2 cos2 g = 0
3
cos2 a + cos2 b + cos2 g =
2
x+y+z=0
x+y= z
(x + y)3 = ( z)3
x3 + y3 + 3xy (x + y) = z3
x3 + y3 + z3 = 3xy (x + y)
= 3xy ( z) = 3xyz
(eia )3 + (eib )3 + (eig )3 = 3 eia eib eig
e3ia + e3ib + e3ig = 3 ei(a + b + g )
(cos 3a + i sin 3a ) + (cos 3b + i sin 3b ) + (cos 3g + i sin 3g )
= 3 [cos (a + b + g ) + i sin (a + b + g )]
Comparing imaginary part on both the sides,
sin 3a + sin 3b + sin 3g = 3 sin (a + b + g )
1 7i
Example 8: Prove that (4n)th power of
is equal to (-4)n, where n is posi2
(2
i
)
tive integer.
Solution:
1 + 7i
1 + 7i
1 + 7i
=
=
(2 − i ) 2 4 + i 2 − 4i 4 − 1 − 4i
1 + 7i 3 + 4i 3 + 4i + 21i + 28i 2
⋅
=
3 − 4i 3 + 4i
9 + 16
3 + 25i − 28
=
25
= −1 + i
=
1.30
Engineering Mathematics
Let 1 + i = r (cos q + i sin q )
r
| 1 i|
q
tan
1 i
Now,
1
( 1) 2 12
1
1
2 cos
2
3p
4
tan 1 ( 1)
3
4
i sin
⎡ 1 + 7i ⎤
⎢ (2 − i)2 ⎥
⎣
⎦
[ Point ( 1,1) lies in second quadrant]
3
4
4n
⎡ ⎛
3p
3p ⎞ ⎤
= ( −1 + i ) 4 n = ⎢ 2 ⎜ cos
+ i sin
⎟
4
4 ⎠ ⎥⎦
⎣ ⎝
⎡
⎛ 3p
= ( 2 )4 n ⎢cos ⎜ 4 n
⎝
4
⎣
⎞
⎛ 3p
⎟⎠ + i sin ⎜⎝ 4 n
4
4n
⎞⎤
⎟⎠ ⎥
⎦
= ( 2) 2 n (cos 3n p + i sin 3n p )
= ( 4) n ⎡⎣( −1)3n + 0 ⎤⎦ = ⎡⎣ 4 n ( −1)3 ⎤⎦
n
= ( 4) n ⎡⎣( −1) n ⎤⎦
= ( −4) n
Example 9: If ` and a are the roots of the equation x2 - 2x + 4 = 0, then show
that ` n + a n = 2 n+1 cos n and hence find the value of ` 15 + a 15.
3
Solution: Equation x2
2x + 4 = 0 is quadratic in x.
x
2
4 16
2
= 1± i 3
a and b are the roots of the equation.
a=1+i 3,b=1
1
3
i 3 (conjugate of a)
Let 1 + i 3 = r (cos q + i sinq )
= |1 + i 3 | =
1 3
4
2
q = arg (1 i 3)
tan
1
3
3
i
a= 1 i 3
and
2 cos
⎛
⎝
b = 1 − i 3 = 2 ⎜ cos
3
i sin
3
2e 3
i
3
− i sin
−
⎞
3
⎟ = 2e
3⎠
Complex Numbers
( ) + (2e )
ip
n
−
a n + b n = 2e 3
⎛ np
= 2n ⋅ 2 cos ⎜
⎝ 3
ip
3
(
n
= 2n e
1.31
inp
3
+e
inp
3
−
)
⎞
⎛ np ⎞
n +1
⎟⎠ = 2 cos ⎜⎝ ⎟⎠
3
Putting n = 15,
15
15
216 cos
15
3
216 ( 1)
216 cos 5
216
Example 10: If ` and a are roots of z2 sin2 p - z sin 2p + 1 = 0, prove that
` n + a n = 2 cosnp cosecnp , where n is a positive integer.
z sin 2q + 1 = 0 is quadratic in z.
Solution : Equation z2 sin2 q
z=
=
=
sin 2 2
2sin 2
sin 2
2sin cos
4sin 2
4sin 2 cos 2
2sin 2
cos 2
sin
cosec e ± i
cos
1
=
4sin 2
i sin
cos
sin
a and b are the roots of the equation.
a = cosec q · e iq
b = cosec q · e
iq
a n + b n = (cosec q eiq )n + (cosec q e iq )n
= (cosec q )n (einq + e
inq
)
= (cosec q )n · (2 cos nq )
= 2 cos nq cosecn q
Example 11: If z = -1 + i 3 and n is an integer, then prove that z2n + 2nzn + 22n = 0,
if n is not a multiple of 3.
Solution:
Let
z= 1+i 3
1 + i 3 = r (cos q + i sin q )
r
1 i 3
tan
1
3
1
( 3 )2
( 1) 2
tan
1
(
3)
4
2
3
2
[ Point ( −1 + i 3 ) lies in
the second quadrant]
1.32
Engineering Mathematics
z
n
Consider, z
2n
1 i 3
2e
2n
zn
2
3
i sin
2
3
2e
2i
3
n
2i
3
2
2 cos
2n
n
2e
n
2i
3
e
2 in
3
e
2 in
3
2 cos
2n
3
If n is not a multiple of 3, let n = 3k + 1 where k is an integer, then
zn
2n
2n
zn
2 cos
2
(3k 1)
3
2
3
1
2
[ cos (2k
2 cos
2
2
3
2 cos 2k
)
cos ]
1
z n 2n
1
2n z n
z2n + 2n zn + 22n = 0, if n is not a multiple of 3.
n
Example 12: Prove that
n
1 i 3
2
1 i 3
2
has the value -1, if n = 3k
é 1 (not a multiple of 3) and 2 if n = 3k (multiple of 3), where k is an integer.
Solution: Let
1 i 3
= r (cos q + i sin q ), then
2
tan
1 i 3
2
1 i 3
2
and
1 i 3
2
n
1 i 3
2
i 3
2
1
2
r
3
2
1
2
1
2
3
2
cos
3
e
2i
3
tan
2
3
2
i sin
3
i sin
cos
n
1
2
n
e
2i
3
1
1 i 3
= r (cos q
2
2
(
e
e
2
3
2
3
)
1
4
i sin q )
3
4
1
2 cos
2n
3
2
3
2i
3
2i
3
n
e
2 in
3
e
2 in
3
Complex Numbers
1.33
If n = 3k ± 1,
1 i 3
2
n
n
1 i 3
2
2 cos
2
(3k 1)
3
2
3
2 cos
2 cos
2
3
2
3
2 cos 2k
[∵ cos (2k
1
2
2
)
cos ]
1
If n = 3k,
n
1 i 3
2
n
1 i 3
2
2 cos
2
(3k )
3
2 cos 2k
2
Example 13: If z = x + iy = r (cos p + i sin p ), prove that
z
1
[
2
r
x
i r
x ].
z = x + iy = r (cos q + i sin q )
Solution:
cos =
z
(
z)
2
x
y
, sin =
r
r
x iy
x iy
1
( x iy ) 2
z
1
[ r (cos
i sin ) ] 2
1
r 2 cos
1
= ± r2 ±
2
i sin
1 cos
2
[Using De Moivre’s theorem]
2
±i
∵ 1 cos
1
=±
r ±
1
2
(
r
1 cos
2
2 cos 2
x
x
1
r ±i
r
2
2
x i r
x)
2
, 1 cos
2sin 2
2
1.34
Engineering Mathematics
Example 14: Prove that
i cos
i cos
1 sin
1 sin
i cos and hence show that
sin
5
1 sin
5
i cos
i cos
=
i cos
1 sin
Solution:
1 sin
i 1 sin
5
1 cos
1 cos
2 cos 2
=
2 cos 2
cos
=
cos
=e
i sin
2
i sin
2
4
2
4
2
4
2
4
2
⎛p q ⎞
2i ⎜ − ⎟
⎝ 4 2⎠
5
i cos
2
2
2i sin
2i sin
i sin
0.
5
4
2
4
2
4
2
4
2
cos
cos
i
=
e
2
4
2
4 2
e
i
i sin
4
4 2
⎛p
⎞
i ⎜ −q ⎟
⎠
= e ⎝2
⎛p
⎞
⎛p
⎞
= cos ⎜ − q ⎟ + i sin ⎜ − q ⎟
⎝2
⎠
⎝2
⎠
= sin q + i cosq
Putting
=
5
1 sin
1 sin
,
5
5
i cos
i cos
5
sin
5
i cos
5
5
5
1 sin
1 sin
5
5
i cos
i cos
5
5
sin
i cos
5
5
5
5
cos
cos 5
2
3
10
5
i sin
i sin 5
2
3
10
5
[Using De Moivre’s theorem]
Complex Numbers
1.35
3p
3p
+ i sin
2
2
= 0 + i( −1) = −i
= cos
5
1 sin
i cos
5
5
5
1 sin
5
i cos
5
i cos
5
5
i 1 sin
5
5
i 1 sin
i cos
5
0
5
Example 15: Prove that the general value of p which satisfies the equation
4m
, where
(cos p + i sin p )(cos 2p + i sin 2p ) . . . . . (cos np + i sin np ) = 1 is
n( n + 1)
m is an integer.
Solution:
(cos q + i sin q ) (cos 2q + i sin 2q ) . . . . . (cos nq + i sin nq ) = 1
eiq . e2iq . . . . . einq = 1 = cos 0 + i sin 0,
Taking general value of cos q and sin q,
eiq (1+2+3+. . . . . . n) = cos (2 m + 0) + i sin (2 m + 0), where m is an integer
e
iθ
n ( n +1)
2
= cos 2mπ + i sin 2mπ = ei2 m
n (n + 1)
= 2m
2
4m
=
n (n + 1)
General value of q
=
4m
n(n + 1)
Exercise 1.2
1. Simplify
2. Express in polar form
(cos 5 − i sin 5 ) (cos 7 + i sin 7 )
(cos 4 − i sin 4 )9 (cos + i sin )5
2
(i)
⎛
⎜ 1 + cos 9 + i sin 9
(ii) ⎜
⎜⎜ 1 + cos − i sin
9
9
⎝
−3
(i)
18
⎞
⎟
⎟ .
⎟⎟
⎠
[Ans. : (i) 1, (ii) 1]
(ii)
(1 i )6 ( 3 i )
4
(1 i )8 ( 3 i )
5
(1 + i )8 (1 − i 3 )
6
(1 − i )6 (1 + i 3 )
9
.
Ans. : (i)
i
i
, (ii)
4
4
1.36
Engineering Mathematics
3. If z =
1
+
i
, then by using De
2
2
Moivre’s theorem, simplify
(z)10 + ( z )10 , where z is the
complex conjugate of z.
[Ans. : 0]
4. If n is a positive integer, show that
(a + ib)n + (a ib)n = 2 rn cos nq
b
where r2 = a2 + b2 and
tan 1
a
Hence or otherwise deduce that
(1
i 3)
(1
8
i 3)
8
28 .
cos f) + i (sin q
+ [(cos q
cos f)
i (sin q
sin f )]n
sin f)]n.
⎡ Hint : apply cos C − cos D =
⎤
⎢
⎥
C+D
C−D
⎢
−2 sin
sin
and ⎥
⎢
⎥
2
2
⎢
⎥
sin C − sin D =
⎢
⎥
⎢
⎥
C+D
C−D
⎢
⎥
2 cos
sin
⎣
⎦
2
2
Ans. : 2n +1 sin n
2
cos n
2
1
1
= 2 cos , y + = 2 cos ,
x
y
prove that
6. If x +
xr y r +
1
= 2 cos (r + r ) .
x yr
r
[Hint : x = cos q + i sin q ]
7. If 2 cos q = x +
1
, prove that
x
x2n + 1
cos n
=
.
2n 1
+ x cos(n 1)
x
9. If sin a + 2 sin b + 3 sin g = 0 =
cos a + 2 cos b + 3 cos g, prove that
(i) sin 3a + 8 sin 3b + 27 sin 3g
= 18 sin (a + b + g )
(ii) cos 3a + 8 cos 3b + 27 cos 3g
= 18 cos (a + b + g ).
⎡Hint : a = eia, b = 2eib, c = 3eig, ⎡
⎢
a + b + c = 0, (a + b)3 = c3 ⎢⎣
⎣
10. If xn + iyn = (1 + i 3 ) , prove that
n
xn 1 yn
5. Evaluate
[(cos q
8. If sin a + sin b = 0 = cos a + cos b,
prove that
(i) cos 2a + cos 2b = 2 cos ( + a + b )
(ii) sin 2a + sin 2b = 2 sin ( + a + b ).
xn yn 1 = 4n
1
Hint : 1 + i 3 = 2 cos
xn + iyn = 2n cos
3.
p
p
+ i sin
,
3
3
np
np
+ i sin
3
3
np
np
, yn = 2n sin
,
3
3
p
2n 1 cos (n 1)
3
p
2n 1 sin(n 1)
3
xn = 2n cos
xn
1
yn
1
11. Prove that [sin (a + q ) eiasinq ]n
= sinna e-inq.
12. If a and b are the roots of the
equation x 2 2 3 x 4 0 , then
prove that a 3 + b 3 = 0.
13. If a , b are the roots of the equation
x2 2x + 2 = 0, prove that a n + b n
n
= 2 2 2 cos n
4
Hence show that a 8 + b 8 = 32.
14. If a, b are the roots of the equation
3x + 1 = 0, prove that a n + b n =
n
2 cos . Hence show that a 12 + b 12 = 2.
6
x2
Complex Numbers
1.37
1.9 APPLICATIONS OF DE MOIVRE’S THEOREM
1.9.1 Roots of an Algebraic Equations
De Moivre’s theorem can be used to find the roots of an algebraic equation.
General values of cos q = cos (2kp + q ) and sin q = sin (2kp + q ), where k is an integer.
To solve the equation of the type zn = cos q + i sin q, we apply De Moivre’s theorem.
1
z = (cos + i sin ) n
= cos
This shows that cos
n
+ i sin
n
n
+ i sin
n
is one of the n roots of zn = cos q + i sin q. The
other roots are obtained by expressing the number in the general form.
1
z = [ cos( 2kp + q ) + i sin( 2kp + q )] n
⎛ 2kp + q
= cos ⎜
⎝
n
Taking k = 0, 1, 2, . . . . . . n
⎞
⎛ 2kp + q ⎞
⎟⎠
⎟⎠ + i sin ⎜⎝
n
1, we get n roots of the equation.
Note: (i) Complex roots always occur in conjugate pair if coefficients of different
powers of x including constant terms in the equation are real.
(ii) Continued products means product of all the roots of the equation.
Example 1: Find all the values of the following:
1
(i)
2
( 1) 5
(ii) (1 i ) 3
(iii)
3
1 i
+
2
3
1 i
.
2
i . 0 = r (cos q + i sin q )
Solution: (i) 1
r
1,
0
q = tan 1
= tan 1 0 = p
1
1 = cos p + i sin p
= cos (2kp + p) + i sin (2kp + p)
1
( 1) 5
1
[cos (2k
1
1)p
= cos (2k + 1)
i sin(2k 1)p ] 5
p
p
+ i sin (2k + 1)
5
5
1
Taking k = 0, 1, 2, 3, 4, we get all 5 values of ( 1) 5 .
[ Point ( 1,0) lies in
second quadrant]
1.38
Engineering Mathematics
(ii)
1
i = r (cos q + i sin q )
r
q
1 i
tan
1
1 ( 1) 2
1
1
2
p
4
tan 1 ( 1)
[ Point (1, 1) lies in
fourth quadrant]
⎡ ⎛
⎞
⎛
⎞⎤
1 − i = 2 ⎢cos ⎜ − ⎟ + i sin ⎜ − ⎟ ⎥
4
4
⎠
⎝
⎠⎦
⎣ ⎝
⎛
⎞
= 2 ⎜ cos − i sin ⎟
4
4⎠
⎝
2
⎡ ⎛
⎞⎤
(1
1 − i ) 3 = ⎢ 2 ⎜ cos − i sin ⎟ ⎥
4
4 ⎠⎦
⎣ ⎝
2 ⎛
⎡
2p
2p
= ⎢( 2 ) ⎜ cos
− i sin
⎝
4
4
⎣
1
1
p
p ⎞⎤ 3
⎞⎤ 3 ⎡ ⎛
⎟⎠ ⎥ = ⎢ 2 ⎜⎝ cos − i sin ⎟⎠ ⎥
2
2 ⎦
⎣
⎦
1
⎡ ⎧ ⎛
p⎞
p ⎞ ⎫⎤ 3
⎛
= ⎢ 2 ⎨cos ⎜ 2kp + ⎟ − i sin ⎜ 2kp + ⎟ ⎬⎥
⎝
2⎠
2 ⎠ ⎭⎦
⎣ ⎩ ⎝
1
2 3 cos(4 k 1)
p
6
i sin (4 k 1)
p
6
[Using De Moivre’s theorem]
2
Taking k = 0, 1, 2, we get all three values of (1 i ) 3 .
1 i
(iii)
= r (cos q + i sin q )
2
r
1 i
2
1
2
= tan 1 (1) =
1 i
2
1 i
and
3
2
1+ i
2
+3
1− i
cos
cos
4
4
1
2
1
4
i sin
i sin
4
4
1
3
1
p
p⎞ ⎛
p
p ⎞3
⎛
= ⎜ cos + i sin ⎟ + ⎜ cos − i sin ⎟
4
4⎠ ⎝
4
4⎠
2 ⎝
1
⎡
p⎞
p ⎞⎤ 3
⎛
⎛
= ⎢cos ⎜ 2kp + ⎟ + i sin ⎜ 2kp + ⎟ ⎥
⎝
⎝
4⎠
4 ⎠⎦
⎣
1
⎡
p ⎞⎤ 3
p⎞
⎛
⎛
+ ⎢cos ⎜ 2kp + ⎟ − i sin ⎜ 2kp + ⎟ ⎥
⎝
⎠
⎝
4
4 ⎠⎦
⎣
Complex Numbers
p
p
+ i sin (8k + 1)
12
12
p
p
cos (8k 1)
i sin (8k 1)
12
12
p
= 2 cos (8k + 1)
12
1 i
Taking k = 0, 1, 2, we get all three values of 3
2
1.39
= cos (8k + 1)
[Using De Moivre’s theorem]
3
1 i
2
3
⎛1
3 ⎞4
Example 2: Find continued product of all the values of ⎜ + i
.
2 ⎟⎠
⎝2
Solution:
1
3
i
= r (cos q + i sin q )
2
2
r
1
3
i
2
2
= tan
1
3
i
2
2
1
3
i
2
2
cos
1
4
3
4
3
2 = tan
1
2
1
i sin
3
3
(cos
i sin
3
4
1
4
4
Putting k = 0, 1, 2, 3,
4
3
x1 = cos
4
5
x2 = cos
4
7
x3 = cos
4
i sin
4
3
i sin
4
5
i sin
4
7
i sin
4
3
i sin (2k 1)
e4
i
3
4
i
5
4
i
7
4
e
e
e
i sin 3
1
4
3
1
[ cos (2k
i
x0 = cos
3
cos 3
3
i sin )
cos (2k 1)
3=
1
3
3
4
cos
1
) i sin (2k
4
)] 4
1.40
Engineering Mathematics
Continued product is
i
x0 x1 x2 x3
e4 e
e
i
i
3
4
16
4
e
i
5
4
ei 4
e
i
7
4
i
e
4
3
4
5
4
7
4
i sin 4
cos 4
1 i 0
1
3
Hence, continued product of all the values of
i
2
2
3
4
is 1.
Example 3: Show that the nth roots of unity form a geometric progression with
2o
2o
common ratio cos
+ i sin
and show that the continued product of all nth
n
n
roots is ( 1)n+1.
Solution: To find nth roots of unity, consider the equation
xn = 1 = cos 0 + i sin 0 = cos 2kp + i sin 2kp
1
2 kp
2 kp
x = (cos 2kp + i sin 2kp ) n = cos
+ i sin
n
n
Taking k = 0, 1, 2, . . . . . . n 1, we get all n roots as
x0 = cos 0 + i sin 0 = 1
x1 = cos
2 ip
2p
2p
+ isin
= e n = w , say
n
n
4 ip
2 ip
4p
4p
+ isin
= en = e n
n
n
2
x2 = cos
6 ip
2 ip
6p
6p
+ isin
= en = e n
n
n
3
x3 = cos
=w2
= w3
………………………………………………
………………………………………………
xn
1
2( n 1) ip
2 ip
2( n 1)p
2( n 1)p
= cos
+ i sin
=e n = e n
n
n
2
3
Also, the roots are given as 1, w, w , w . . . . . , w
2
2
i sin
with common ratio w = cos
n
n
Continued product of all the nth roots is
x0 x1 x2
xn
1
1 e
=e
=e
2i
n
e
4i
n
n 1
e
n 1
= wn
1
which are in geometric progression
6i
n
e
2( n 1) i
n
2i
{1 2 3 ......( n 1)}
n
2i
n
( n 1)
n
2
[Using sum of A.P.]
Complex Numbers
1.41
= eip (n 1) = einp e ip
= (cos np + i sin np ) (cos p
= ( 1)n ( 1) = ( 1)n+1
i sin p )
Hence, continued product of all nth roots of unity is ( 1)n+1
3m
4m
Example 4: If v is a 7th root of unity, then prove that S = 1 + v + v2m + v + v
5m
6m
+ v + v = 7 if m is a multiple of 7 and is 0 otherwise.
Solution: Taking n = 7 in Example 3, we get 7th root of unity as 1, w, w 2, w 3, w 4,
w 5, w 6 where
w = cos
2p
i
2p
2p
+ i sin
=e 7
7
7
S = 1+ w m + w 2 m + w 3m + ..... + w 6 m =
1[1 − (w m )7 ]
1− w m
1 − w 7m
1− w m
If m is not a multiple of 7, say m = 7k + 1, k is an integer, then
=
e
1
i
2p
7
S=
1
=
1
=
1
e
i
2p
7
7
[Using sum of G.P.]
7 k +1
7 k +1
=
1
(e2ip )7 k +1
1 e
1 (cos 2
2
cos 2k
7
1 (1 i .0)7 k 1
2
2
cos
i sin
7
7
i 2pk +
2p
7
i sin 2 )7 k
1
i sin 2k
=
2
7
1 1
=0
1
Hence, S = 0, if m is not a multiple of 7.
If m is a multiple of 7, say m = 7k, k is an integer, then
S = 1 + w 7k + (w 2)7k + (w 3)7k + . . . . + (w 6)7k
= 1 + (w 7)k + (w 7 )2k + (w 7 )3k + . . . . + (w 7 )6k
= 1 + (1)k + (1)2k + (1)3k + . . . . + (1)6k [∵ w 7 = cos 2p + i sin 2p = 1]
=1+1+1+1+1+1+1=7
Example 5: If v is a complex cube root of unity, prove that (1 - v)6 = -27.
Solution: Taking n = 3 in Example 3, we get cube root of unity i.e. roots of the
2
2
i sin
equation x3 = 1 as 1, w, w 2, where w = cos
3
3
1.42
Engineering Mathematics
w is the root of the equation x 3 = 1. Hence, w 3 =1.
Now
1 + w + w2 =
… (1)
1 w3
1 w
[Using sum of G.P.]
1 1
=0
1
1 + w + w2 = 0
(1 w)6 = [(1 w)2]3 = [1 + w 2
= ( w 2w)3
= ( 3w)3
3
= 27 w
= 27
=
[Using Eq. (1)]
… (2)
2w]3
[Using Eq. (2)]
[Using Eq. (1)]
Example 6: Solve the following equations:
(i) x6 i = 0
(ii) x10 + 11x5 + 10 = 0
(iii) x7 + x4 + i(x3 + 1) = 0
(iv) x4
x3 + x2
x+1=0
(v) (x + 1)8 + x8 = 0.
Solution: (i)
x 6 = i = cos
x
p
p
+ i sin
2
2
cos (4k 1)
2
cos 2k
i sin (4k 1)
2
i sin 2k
2
1
6
2
cos (4k 1)
i sin (4k 1)
[Using De Moivre’s theorem]
12
12
Putting k = 0, 1, 2, 3, 4, 5, we get all the 6 roots of the given equation.
(ii)
x10 + 11x 5 + 10 = 0
x10 + 10x 5 + x 5 + 10 = 0
x 5 (x 5 + 10) + 1(x 5 + 10) = 0
(x 5 + 1) (x 5 + 10) = 0
All the roots of x10 + 11x5 + 10 = 0 are the roots of
x5 + 1 = 0 and x5 + 10 = 0
x5 + 1 = 0
x5 = 1 = cos p + i sin p
= cos (2k1p + p) + i sin (2k1p + p)
1
x = [cos(2k1 + 1)p + i sin(2k1 + 1)p ]5
= cos(2k1 + 1)
=e
i (2 k1 +1)
p
5
p
p
+ i sin (2k1 + 1)
5
5
Complex Numbers
1.43
Putting k1 = 0, 1, 2, 3, 4 we get all the 5 roots of x5 + 1 = 0.
x5 + 10 = 0
x5 = 10 = 10 (cos p + i sin p)
= 10[cos (2k2p + p) + i sin (2k2p + p)]
1
x = [10 cos (2k2 + 1)p + i sin (2k2 + 1)p ]5
1
p⎤
p
⎡
= (10) 5 ⎢cos (2k2 + 1) + i sin (2k2 + 1) ⎥
5
5⎦
⎣
1
= (10) 5 e
i ( 2 k2 +1)
p
5
Putting k2 = 0, 1, 2, 3, 4 we get all 5 roots of x5 + 10 = 0.
All the 10 roots of the equation x10 + 11x5 + 10 = 0 are given by e
where k1 = k2 = 0, 1, 2, 3, 4
i (2 k1 1)
1
5
and (10) 5 e
i (2 k2 1)
5
x 7 + x 4 + i (x 3 + 1) = 0
(iii)
x 4 (x 3 + 1) + i (x 3 + 1) = 0
(x 3 + 1) (x 4 + i ) = 0
All the roots of x7 + x4 + i (x3 + 1) = 0 are the roots of (x3 + 1) = 0 and (x4 + i) = 0.
x3 + 1 = 0
x3 = 1 = cos p + i sin p
= cos (2k1p + p) + i sin (2k1p + p)
x
[ cos (2k1
1
1)
cos (2k1 1)
3
i sin (2k1 1)
i sin (2k1 1)
]3
3
i (2 k1 1)
3
=e
Putting k1 = 0, 1, 2, we get all the 3 roots of x3 + 1 = 0.
x4 + i = 0
x4
x
i
cos
2
i sin
cos (4k2 1)
2
2
cos 2k2
i sin (4k2 1)
i sin 2k2
2
1
4
2
cos (4k2 1)
2
8
i sin (4k2 1)
8
i (4 k2 1)
8
=e
Putting k2 = 0, 1, 2, 3, we get all the 4 roots of x4 + i.
All the 7 roots of the equation x7 + x4 + i (x3 + 1) = 0 are given by e
e
i (4 k2 1)
8
where k1 = 0, 1, 2, and k2 = 0, 1, 2, 3.
i (2 k1 1)
3
and
1.44
Engineering Mathematics
x 4 − x3 + x 2 − x + 1 = 0
(iv)
1 ⎡1 − (− x)5 ⎤⎦
1 − x + x 2 − x3 + x 4 = ⎣
1 − (− x)
[Using sum of G.P.]
1 + x5
1+ x
5
( x + 1) = ( x + 1)(1 − x + x 2 − x3 + x 4 )
=
This shows that all the roots of x5 + 1 = 0 except x =
x + 1 = 0, are the roots of x4 x3 + x2 x + 1 = 0.
1, which corresponds to
x5 1 0
Now, solving
x5
x
i sin
) i sin (2k
1 cos
cos (2k
[cos (2k
)
1
i sin (2k 1)
1)
cos (2k 1)
i sin (2k 1)
5
]5
5
Putting k = 0, 1, 2, 3, 4, we get all the 5 roots of x5 + 1 = 0.
x0
cos
5
3
cos
5
5
cos
5
7
cos
5
9
cos
5
x1
x2
x3
x4
Except x2 =
4
x
(v)
x +x
3
i sin
5
3
i sin
5
5
i sin
5
7
i sin
5
9
i sin
5
cos
i sin
1, remaining roots x0, x1, x3, x4 are the roots of the equation
x +1 = 0
2
1
( x + 1)8 + x 8 = 0
( x + 1)8 = − x 8
8
⎛ x +1⎞
⎜⎝
⎟ = −1
x ⎠
Let x + 1 = z
x
z 8 = −1 = cos p + i sin p
= cos ( 2kp + p ) + i sin (2kp + p )
Complex Numbers
1.45
z = [ cos ( 2k + 1)p + isin (2k + 1 )p
= cos (2k + 1)
=e
i (2 k +1)
p
8
1
8
]
p
p
+ i sin (2k + 1)
8
8
, where k = 0, 1, 2, .... 7
Substituting the value of z,
x + 1 i ( 2 k +1) p8
=e
x
p
x + 1 iq
= e where (2k + 1) = q
8
x
x + 1 = x eiq
x (1 − eiq ) = −1
−1
−1
=
1 − eiq 1 − cos q − i sin q
−1
=
q
q
q
2 sin 2 − i 2 sin cos
2
2
2
x=
q⎞
⎛ q
⎜⎝ sin + i cos ⎟⎠
−1
2
2
=
⋅
q⎞
q⎛ q
q⎞ ⎛ q
2 sin ⎜ sin − i cos ⎟ ⎜ sin + i cos ⎟
2
2⎠
2⎝
2
2⎠ ⎝
q⎞
⎛
− ⎜ 1 + i cot ⎟
⎝
1⎛
q⎞
2⎠
= − ⎜1 + i cot ⎟
=
⎝
q
q
2
2⎠
⎛
⎞
2 ⎜ sin 2 + cos 2 ⎟
⎝
2
2⎠
Substituting the value of q,
x
1
2
i
cot (2k 1)
2
16
Putting k = 0, 1, 2, . . . . . 7 we get all the 8 roots of the equation (x + 1)8 + x8 = 0.
3o
o
⎡
⎤⎡
⎤
Example 7: Show that x 5 − 1 = ( x − 1) ⎢ x 2 + 2 x cos + 1⎥ ⎢ x 2 + 2 x cos
+ 1⎥ .
5
5
⎣
⎦⎣
⎦
Solution:
x5 1 0
x5
x
1 cos 0 i sin 0
(cos 2k
=e
i
2k
5
cos 2k
i sin 2k )
1
5
cos
i sin 2k
2k
5
i sin
2k
5
1.46
Engineering Mathematics
Putting k = 0, 1, 2, . . . 4, we get all the roots of the equation as,
x0 = 1
2p
5
2p
2p
+ i sin
5
5
4p
i
p
p
4
4
x2 = e 5 = cos
+ i sin
5
5
6p
i
6p
6p
x3 = e 5 = cos
+ i sin
5
5
4
p
4p ⎞
⎛
⎞
⎛
= cos ⎜ 2p −
+ i sin ⎜ 2p −
⎟
⎟
⎝
⎝
5 ⎠
5 ⎠
x1 = e
i
= cos
= cos
⎡∵ cos (2p − q ) = cos q ⎤
⎢ sin (2p − q ) = − sin q ⎥
⎦
⎣
4p
−i
4p
4p
− i sin
=e 5
5
5
= x2
x4 = e
i
8p
5
= cos
8p
8p
2p ⎞
2p ⎞
⎛
⎛
+ i sin
= cos ⎜ 2p −
⎟⎠ + i sin ⎜⎝ 2p −
⎟
⎝
5
5
5
5 ⎠
2p
−i
2p
2p
− i sin
= e 5 = x1
5
5
5
x0, x1, x2, x3, x4 are roots of x 1 = 0
= cos
x 5 1 ( x x0 ) ( x x1 ) ( x x2 ) ( x x3 ) ( x x4 )
( x 1) x e
( x 1) x e
2 ip
5
x e
2 ip
5
4 ip
5
x e
2 ip
5
( x 1) x 2
x e
( x 1) x 2
x 2 cos
( x 1) x 2
2 x cos p
( x 1) x 2
2 x cos
e
2p
5
3p
5
x e
4 ip
5
2 ip
5
4 ip
5
x e
1 x2
1 x2
3p
5
1 x2
x e
4 ip
5
x e
x e
x 2 cos
1 x2
2 x cos
2 ip
5
4 ip
5
4p
5
+e
2 ip
5
4 ip
5
+1
1
2 x cos p
p
5
1
∵ cos (p q )
p
1
cos q
5
Example 8: If `, ` 2, ` 3, ` 4 are roots of x5 - 1 = 0, then show that (1 - ` ) (1 - ` 2 )
(1 - ` 3 ) (1 - ` 4 ) = 5.
Solution: One root of x5
a 2, a 3, a 4.
1 = 0 is obviously 1, the remaining roots are given as a,
Complex Numbers
( x 1) ( x 4
x3
1.47
x 5 1 ( x 1) ( x
) (x
2
) (x
3
) (x
4
)
x 1)
) (x
2
) (x
3
) (x
4
)
x2
[∵ xn
( x 1) ( x
1 = (x
1) (xn 1 + xn 2 + xn 3 + . . . . . + 1)]
x + x + x + x + 1 = (x a) (x a 2) (x a 3) (x a 4)
Putting x = 1 on both the sides,
1 + 1 + 1+ 1+ 1 = (1 a) (1 a 2) (1 a 3) (1
(1 a) (1 a 2) (1 a 3) (1 a 4) = 5
4
3
2
2
a 4)
Example 9: If `, a, f, c are the roots of x4 + x3 + x2 + x + 1 = 0, find their values
and show that (1 - ` ) (1 - a ) (1 - f ) (1 - c ) = 5.
Solution:
x 4 + x3 + x 2 + x + 1 = 0
(x 1) (x 4 + x 3 + x 2 + x + 1) = 0
[ ∵ x n 1 = (x 1) (x n 1 ) + (x n 2 ) + . . . . 1 ]
x5 1 = 0
This shows that all the roots of x5 1 = 0, except x = 1, are the roots of
x4 + x3 + x2 + x + 1 = 0.
As solved in Example 7, all roots of x5 1 = 0 are
x0 = 1
x1 = e
2i
5
x2 = e
4i
5
x3 = e
6i
5
x4 = e
8i
5
Except x0 = 1 remaining roots x1, x2, x3, x4 are the roots of x4 + x3 + x2 + x + 1 = 0.
=e
2i
5
,
=e
4i
5
=e
,
6i
5
,
=e
8i
5
Since a, b, g, d are the roots of x4 + x3 + x2 + x + 1 = 0,
x4 + x3 + x2 + x + 1 = (x
a ) (x
b ) (x
g ) (x
Putting x = 1 on both the sides,
1 1 1 1 1 = (1 a ) (1 b ) (1 g ) (1 d )
5 = (1 a ) (1 b ) (1 g ) (1 d )
(1 a ) (1 b ) (1 g ) (1 d ) = 5
Example 10: Find the cube roots of 1 - cos p - i sin p.
Solution:
x3
1 cos q
i sin q
2sin
q
p
cos
2
2
2sin
q
cos 2np
2
2sin 2
q
2
q
q
q
i 2sin cos
2
2
2
i sin
p
2
q
2
p
2
q
2
i sin 2np
q
2
p
2
d)
1.48
Engineering Mathematics
1
⎡
q⎧ ⎛
p q⎞
p q ⎞ ⎫⎤ 3
⎛
x = ⎢ 2 sin ⎨cos ⎜ 2np + − ⎟ − i sin ⎜ 2np + − ⎟ ⎬⎥
⎝
2⎩ ⎝
2 2⎠
2 2 ⎠ ⎭⎦
⎣
1
q ⎞3 ⎡
1⎛
p −q ⎞
1⎛
p − q ⎞⎤
⎛
= ⎜ 2 sin ⎟ ⎢cos ⎜ 2np +
− i sin ⎜ 2np +
⎟
⎟
⎝
2⎠ ⎣
3⎝
2 ⎠
3⎝
2 ⎠ ⎥⎦
1
q ⎞ 3 ⎡ ⎧ ( 4 n + 1)p − q ⎫
⎛
⎧ ( 4 n + 1)p − q ⎫⎤
x = ⎜ 2 sin ⎟ ⎢cos ⎨
⎬ − i sin ⎨
⎬⎥
⎝
2⎠ ⎣ ⎩
6
6
⎭
⎩
⎭⎦
Putting n = 0, 1, 2, we get cube roots of 1
cos q
i sin q .
Example 11: If 1 + i is one root of the equation x4 - 6x3 + 15x2 - 18x + 10 = 0,
find all the other roots.
Solution: Complex roots occur in conjugate pairs. If 1 + i is one root of the given
equation, then 1 i must be another root. Let the remaining roots be a and b.
x4
6 x 3 15 x 2 18 x 10
[x
[( x
(1 i ) ][ x (1 i ) ] ( x
1) i ][ ( x 1) i ) ] ( x
( x 1) 2 i 2 ( x
(x
) (x
)
(x
2
2 x 1 1) ( x
(x
2
2 x 2) ( x
x
4
6x
Hence, remaining roots are 1
) (x
) (x
)
) (x
) (x
)
)
)
2
15 x 18 x 10
x 2x 2
4 x 5)
4x + 5 = 0.
a, b are roots of the equation x2
4
3
)
2
( x2
x=
)( x
16 20 4 i 4
=
= 2±i
2
2
i, 2 + i and 2
i.
Example 12: Show that all the roots of (x + 1)7 = (x - 1)7 are given by
ko
é icot
, k = 1, 2, 3.
7
Solution: (x + 1)7 = (x
x 1
x 1
1)7
7
x 1
x 1
1 cos 0 i sin 0
(cos 2k
cos 2k
i sin 2k )
where, k = 0, 1, 2, 3, 4, 5, 6
1
7
i sin 2k
cos
2k
7
i sin
2k
7
e
i
2k
7
Complex Numbers
1.49
But for k = 0
x 1
cos 0 i sin 0 1
x 1
x 1 x 1
1
Thus,
1, absurd
k
0
2k
i
x +1
= e 7 , where k = 1, 2, 3, 4, 5, 6
x 1
Hence,
x +1 ei
2k
, where =
=
x 1 1
7
Applying componendo—dividendo,
2 x e iq + 1
=
2 e iq − 1
cos q + i sin q + 1
x=
cos q + i sin q − 1
1 + cos q + i sin q
=
− (1 − co
os q ) + i sin q
q
q
q
+ 2i sin cos
2
2
2
=
q
q
2q
−2 sin
+ 2i sin cos
2
2
2
2 cos 2
q ⎛
q
q⎞
⎜ cos + i sin ⎟⎠
2⎝
2
2
=
q ⎛
q
q⎞
2 sin ⎜ i 2 sin + i cos ⎟
2⎝
2
2⎠
2 cos
q⎞
q ⎛
q
⎜ cos + i sin ⎟⎠
2
2⎝
2
q
q
⎛
⎞
i ⎜ i sin + cos ⎟
⎝
2
2⎠
cot
=
= − i cot
Substituting the value of
=
q
2
⎡ 1 1
⎤
⎢∵ i = i 2 = −i ⎥
⎣
⎦
2k
,
7
x = i cot
k
7
Putting k = 1, 2, 3, 4, 5, 6, we get roots of (x + 1)7 = (x
1)7 as,
1.50
Engineering Mathematics
x1 = − i cot
p
2p
x2 = −i cot
7
7
x4
4
7
i cot
x5
i cot
6
7
i cot
i cot
x6
i cot
x3 = −i cot
3
7
5
7
i cot
i cot
i cot
7
3
7
x3 [∵ cot(p
2
7
= i cot
q ) = cot q ]
2
= x2 ,
7
x1
7
1)7 are given by ± i cot
All the roots of (x + 1)7 = (x
3p
7
k
, where k = 1, 2, 3.
7
Example 13: Show that the points representing the roots of the equation
zn = i (z – 1)n on Argand’s diagram are collinear.
zn = i (z
Solution:
1)n
n
z
= i = cos
z 1
p
p
+ i sin
2
2
= cos 2kp +
p
p
+ i sin 2kp +
2
2
1
p
p n
= cos (4k + 1) + i sin (4k + 1)
z 1
2
2
p
p
= cos (4k + 1) + i sin (4k + 1)
2n
2n
z
=e
i (4 k +1)
z
= ei
z 1
z zei
(
z 1 ei
p
2n
= eiq , whereq = i (4k + 1)
p
2n
ei
)
ei
z
ei
1 ei
e
e
i
i
2
i
e2
e
i
2
i
e2
cos + i sin
2
2
e2
2i sin
i
2
2
2i sin
2
1
q
cot
2i
2
1
2
i
q
cot
2
2
1
2
Complex Numbers
z=
1.51
1 i
q
p
− cot , where q = i (4k +1)
2 2
2
2n
Putting k = 0, 1, 2, . . . ., n 1, we get n roots of zn = i (z 1)n. Real part of all the
1
roots remain same as
which means x-coordinate of all the point represented by the
2
1
roots on Argand’s diagram is
i.e. constant. Therefore, all the points lie on the line
2
1
x = and hence are collinear.
2
Example 14: Prove that the roots of the equation x2 - 2ax cos p + a2 = 0 are also
the roots of the equation x2n - 2anxn cos np + a2n = 0.
2ax cos q + a2 = 0,
Solution: x2
4a 2 cos 2
2
Roots of the equation are ae±iq .
x=
2a cos
Putting x = ae±iq in x2n
x2n
4a 2
= a (cos q ± i sin q ) = ae±iq
2anxn cos nq + a2n, we get
2anxn cos nq + a2n = (ae±iq )2n
2an(ae±iq )n cos nq + a2n
= a2n e±inq (e± inq
2 cos nq + e ∓ in )
= a2n e± inq (e± inq + e ∓ in
=a e
2n
Hence, roots of the equation x2
x2n 2anxn cos nq + a2n = 0.
± inq
(2 cos nq
2 cos nq )
2 cos nq )
=0
2ax cos q + a2 = 0 are also the roots of the equation
Example 15: Find all the roots of x12 - 1 = 0 and identify the roots which are
also the roots of x4 - x2 + 1 = 0.
Solution:
x12
1=0
6 2
1=0
(x )
(x + 1) (x
6
6
1) = 0
All the roots of x + 1 = 0 and x6
6
Solving
1 = 0 are the roots of x12
1=0
x +1 =0
6
x6 = 1 = cos p + i sin p
= cos (2k1p + p ) + i sin (2k1p + p )
1
x = [cos (2k1 + 1) p + i sin (2k1 + 1) p ] 6
= cos (2k1 + 1)
6
+ i sin (2k1 + 1)
Putting k1 = 0, 1, 2, 3, 4, 5, we get all roots of x + 1 = 0.
6
6
= ei(2k1+1)
6
1.52
Engineering Mathematics
1=0
x6 = 1 = cos 0 + i sin 0
= cos 2k2p + i sin 2k2p 1
x = (cos 2k2p + i sin 2k2p ) 6
x6
Solving
2k 2
2k
+ i sin 2
6
6
k
i 2
k2
k2
= cos
+ i sin
=e 3
3
3
= cos
Putting k2 = 0, 1, 2, 3, 4, 5, we get all roots of x6
Thus, all the roots of x
2, 3, 4, 5
Now,
1 = 0.
1 = 0 are given by e
12
i (2 k1 +1)
6
and e
i
k2
3
for k1 = k2 = 0, 1,
x6 + 1 = 0
(x2)3 + 1 = 0
(x2 + 1) [(x2)2 x2 + 1] = 0 [∵ a3 + b3 = (a + b) (a2 ab + b2)]
(x2 + 1) (x4 x2 + 1) = 0
This shows that all the roots of x6 + 1 = 0, except x = ± i which corresponds to
2
x + 1 = 0, are the roots of x4 x2 + 1 = 0.
Roots of x6 + 1 = 0 are
i
x0 = e 6
x1 = e
i
x2 = e
7
6
i
9
6
i
11
6
e
x5 = e
i
= e 2 = cos
5
i
6
i
x3 = e
x4
3
6
e
i
3
2
2
cos
+ i sin
3
2
2
i sin
= 0+i = i
3
2
0 i
i
Except x1 = i and x4 = i, the remaining roots x0, x2, x3 and x5 are the roots of the
equation x4 x2 + 1 = 0.
Example 16: Find the roots common to x4 + 1 = 0 and x6 - i = 0.
Solution:
x4 + 1 = 0
x4 = 1 = cos p + i sin p
= cos (2k1p + p) + i sin (2k1p + p)
1
x = [cos (2k1 + 1) + i sin (2k1 + 1) ] 4
= cos (2k1 + 1)
4
+ i sin (2k1 + 1)
4
Complex Numbers
1.53
Putting k1 = 0, 1, 2, 3, we get all the roots of x4 + 1 = 0.
x0 = cos
4
3
cos
4
5
cos
4
7
cos
4
x1
x3
x4
Now,
x6
+ i sin
=
1
4
3
i sin
4
5
i sin
4
7
i sin
4
2
i
+
2
1
i
2
1
2
i
2
2
1
i
2
2
i=0
x 6 = i = cos
2
+ i sin
= cos 2k2 +
+ i sin 2k2 +
2
x = cos (4k2 + 1)
= cos (4k2 + 1)
2
2
12
2
+ i sin (4k2 + 1)
+ i sin (4k2 + 1)
Putting k2 = 0, 1, 2, 3, 4, 5, we get all roots of x6
1
6
2
12
i = 0 as,
x0 = cos
+ i sin
12
12
5
5
+ i sin
x1 = cos
12
12
9
9
3
3
+ i sin
= cos
+ i sin
x2 = cos
12
12
4
4
1
i
2
2
13
13
+ i sin
x3 = cos
12
12
17
17
+ i sin
x4 = cos
12
12
21
21
7
7
+ i sin
= cos
+ i sin
x5 = cos
12
12
4
4
1
i
2
Hence, common roots are
1
2
2
i
2
and
1
2
i
2
.
1.54
Engineering Mathematics
Exercise 1.3
1. Find all the values of the following:
(i) (1 i ) (ii)
1
4
2 + 3i
1+ i
(iii)
5
2z 1
2z 1
= 1,
2z
2z
2k
2k
k =1, 2, 3, 4
= cos
+ i sin
5
5
2z 1
∵ for k 0,
1, 2 z 1 2 z ,
2z
1 0 absurd
Hint :
2
3
1
1+ i 3 + 3 1− i 3
6
, k = 0, 1, 2
.
i (4 k 1)
Ans. : (i) 2 3 e
3
4
3
i.e. equation is of degree 4 and not of 5
,
(ii) 2 cos (6k + 1) 6 k = 0, 1, 2
1
8
13
2
(iii)
1
i 2 k + tan
15
k = 0, 1, 2, 3
2. Find continued product of all the
values of the following :
(i) (1 + i )
(iii)
cos
(ii) (i )
3
+ i sin
2
3
3
4
.
3
(ii) x5 +
3=i
(iii) x + x + x + 1 = 0
7
4
3
8x3 + 4x2
2x + 1 = 0
Hint : (2 x)5 + 1
(2 x 1) (16 x 4
8 x3
4x2
2 x 1)
Solve (2 x) + 1 = 0,
5
2 x = cos (2k + 1)
5
+ i sin (2k + 1) ,
5
k = 0, 1, 2, 3, 4
(vi) (2z
1)5 = 32z5
, k = 0, 1, 2, 3, 4
1
3
,
i
2
2
1
i
2
2
1
(iv)
i cot (2k 1) , k 0, 1,
2
12
2, 3, 4, 5
1 i (2 k +1) 5
, k = 0, 1, 2, 3, 4
e
2
1
k
(vi) 1 + i cos
, k = 1, 2, 3, 4
4
5
( vii) e
(iv) (1 + x)6 + x6 = 0
(v) 16x4
12 k + 5
30
i
3 i
±
2 2
(v)
3. Solve the equations:
1=0
1
(ii) 2 5 e
(iii) 1,
[Ans. : (i) (1 + i) (ii) 1 (iii) 1]
(i) x12
1
3
,±
±i
2
2
(i) ± 1, ± i, ±
1
8
128 = 0.
Ans. :
1
e
(vii) x14 + 127x7
i
2 k1
7
,e
i (2 k2 +1)
7
, k1 = k2 = 0, 1, 2,
3, 4, 5, 6
4. Solve the equations:
(i) x4 x2 + 1 = 0
(ii) x4 + x2 + 1 = 0.
⎡ Hint : (i) Multiply by x 2 + 1,
⎢
(x 2 + 1) (x 4 − x 2 + 1) = 0
⎢
∴ x6 + 1 = 0
⎢
⎢
(ii) Multiply by x 2 − 1,
⎢
(x 2 − 1) (x 4 + x 2 + 1) = 0
⎢
⎢⎣
∴ x6 − 1 = 0
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦
Complex Numbers
p
i ( 2 k +1)
⎤
⎡
6
, k = 0, 2, 3, 5 ⎥
⎢ Ans. : (i) e
kp
⎥
⎢
i
⎥⎦
⎢⎣
(ii) e 3 , k = 1, 2, 4, 5
2
2
5. Solve the equation : x + x = i.
[Hint : x4 i x2 + 1 = 0, x2 = i (1 ± 5) ,
2
⎡ Ans. :
⎤
⎢
⎥
⎢ x = (1 ± 5 ) ⎧cos ⎛ 2k + ⎞ 1 ⎥
⎨ ⎜
⎟
⎢
2
2 ⎠2⎥
⎩ ⎝
⎢
⎥
⎢
⎛
⎞ 1 ⎫⎥
+ i sin ⎜ 2k + ⎟ ⎬⎥
⎢
2 ⎠ 2 ⎭⎦
⎝
⎣
x3) (x4 x) = 5
where x = cos 2 + i sin 2 .
5
5
7. Prove that ( 1 + i)7 = 8 (1 + i).
Hint :
6. Prove that (x2
3
2 cos
4
( 1 i )7
3
i sin
4
7
8. Prove that
(i) 1 + + 2 = 0
1
1
1
(ii)
0,
1+ 2
2+
1+
where is a complex cube root of
unity.
9. Prove that
(i) a 3n + b 3n = 2, n is an integer
(ii) aea x + beb x
x
2
3
3
x cos
x ,
2
2
where a, b are complex cube roots of
unity.
e
3 sin
1.55
10. Prove that
(i) n a ib
n
a ib has n real roots.
m
m
n
(ii) (a ib)
(a ib) n
=2
(
a +b
2
2
)
m
n
cos
m
n
b
.
a
[Hint : Let a + ib = r (cos q + i sin q )]
where q = tan
1
11. Prove that all the roots of (x + 1)6
+ (x
1)6 = 0 are given by
i cot (2k + 1) , k = 0, 1, 2, 3, 4, 5.
12
12. Prove that the points representing the
roots of the equation z3 = i (z 1)3 on
Argand’s diagram are collinear.
13. If a, a 2, a 3, a 4, a 5, a 6 are the
roots of x7 1 = 0, prove that
(1 a ) (1 a 2) (1 a 3) (1 a 4)
(1 a 5) (1 a 6) = 7.
14. If 1 + 2i is one root of the equation
x4 3x3 + 8x2 7x + 5 = 0, find all
the other roots.
Ans.: 1 2i,
1
3
i
2
2
15. Prove that
⎛
⎞
x7 + 1 = (x + 1) ⎜ x 2 − 2 x cos + 1⎟
7 ⎠
⎝
3
5
⎛ 2
⎞⎛
⎞
+ 1⎟ ⎜ x 2 − 2 x cos
+ 1⎟
⎜ x − 2 x cos
7
7
⎝
⎠⎝
⎠.
1.9.2 Expansion of Trigonometric Functions
Type I : Expansion of sinnp, cosnp in terms of sin np, cos np, where n is a positive
integer:
Let x = cos q + i sin q = eiq,
1
= cos q i sin q = e-iq
x
1.56
Engineering Mathematics
1
= 2i sin q
x
1
= 2 cos q and x
x
x+
Hence,
xn = (cos q + i sin q )n = cos nq + i sin nq = einq
Again,
1
= (cos q
xn
xn +
i sin q )n = cos nq
1
= 2 cos nq and xn
xn
i sin nq = e-inq
1
= 2i sin nq
xn
To expand cosnq and sinnq , write cosnq =
and expand R.H.S. using binomial expansion
1
1
x+
n
x
2
n
and sinnq =
n
1
1
x
n
x
(2i )
(x + a)n = xn + nC1xn 1a + nC2xn 2a2 + . . . + an
Example 1: Prove that sin5p =
1
(sin 5p - 5 sin 3p + 10 sin p ).
16
1
= cos q
x
Solution: Let x = cos q + i sin q, then
sin 5
1
1
x
2i
x
i sin q
5
1
x5 5 x 4
(2i )5
1
x
1
x
10 x 3
2
10 x 2
1
x
3
5x
1
x
4
1
x
5
[Using Binomial expansion]
sin 5 q
1
x5
32i 5
5 x 3 10 x
1
32i
1
x5
x5
5 x3
10
x
5
x3
1
x3
1
x5
10 x
1
x
1
(2i sin 5q 5 2i sin 3q 10 2i sin q )
32i
1
(sin 5q 5 sin 3q 10 sin q )
16
Example 2: Prove that cos6p + sin6p =
Solution: Let x = cos q + i sin q, then
[∵ i 5
[∵ x n
1
(3 cos 4p + 5).
8
1
= cos q
x
i sin q
i4 i
1
xn
i]
2i sin nq ]
Complex Numbers
cos
6
1
1
=
x+
2
x
1 6
x
26
=
1
26
6
=
x6 +
6
1
1
x+
6
x
2
1
1
15 x 4 2
x
x
6 x5
20 x3
1
x3
1
x
2i
1
x
1
x6
(2i )6
x6
6x
1
x5
1
x6
1
1
1
+ 6 x 4 + 4 + 15 x 2 + 2 + 20
x6
x
x
1
xn
[∵ x n
20)
2 cos n ]
6
6 x5
15 x 2
1
26
1
x4
15 x 2
1
(2 cos 6 6 2 cos 4 15 2 cos 2
26
1
= 5 (cos 6 + 6 cos 4 + 15 cos 2 + 10)
2
sin 6
1.57
1
x6
1
x
1
x
2
15 x 4
1
x
6x
1
x
1
x4
15 x 2
4
6 x4
1
x
20 x 3
5
1
x
6
1
x2
20
1
(2 cos 6 6 2 cos 4 15 2 cos 2
26
1
(cos 6 6 cos 4 15 cos 2 10)
25
cos6q + sin6q =
3
[∵ i 6
(i 2 )3
20)
1
1
(12 cos 4 + 20) = (3cos 4 + 5)
8
25
Example 3: Expand sin5p cos3p in a series of sines of multiples of p.
Solution: Let x = cos q + i sin q,
1
= cos q
x
i sin q
5
⎡1 ⎛
1⎞ ⎤ ⎡1 ⎛
1⎞ ⎤
sin 5 q cos3 q = ⎢ ⎜ x − ⎟ ⎥ ⎢ ⎜ x + ⎟ ⎥
⎝
⎠
⎝
2
2
i
x
x⎠ ⎦
⎣
⎦ ⎣
2
3
3
1⎞ ⎛
1⎞
1
⎛
⎜⎝ x − ⎟⎠ ⎜⎝ x − ⎟⎠ ⋅ 3
x
x
(2)
=
1
(2i )5
=
1 ⎛
1⎞ ⎛ 2 1 ⎞
⎜ x − ⎟⎠ ⎜⎝ x − 2 ⎟⎠
x
(28 i ) ⎝
x
2
1⎞
⎛
⎜⎝ x + ⎟⎠
x
3
3
[∵ i 5 = i 4 ⋅ i = i ]
1]
1.58
Engineering Mathematics
=
1 ⎛ 2
1⎞ ⎛ 6
3
1⎞
2
⎜⎝ x − 2 + 2 ⎟⎠ ⎜⎝ x − 3 x + 2 − 6 ⎟⎠
8
(2 i )
x
x
x
=
i ⎛ 8
6
2
3
1⎞
1
6
2
4
4
⎜⎝ x − 3 x + 3 − 4 − 2 x + 6 x − 2 + 6 + x − 3 + 4 − 8 ⎟⎠
8
x
x
x
x
x
(2 i )
=
1 ⎡⎛ 8 1 ⎞
⎛ 6 1⎞
⎛ 4 1⎞
⎛ 2 1 ⎞⎤
⎜ x − 8 ⎟⎠ − 2 ⎜⎝ x − 6 ⎟⎠ − 2 ⎜⎝ x − 4 ⎟⎠ + 6 ⎜⎝ x − 2 ⎟⎠ ⎥
x
x
x ⎦
(28 i ) ⎢⎣⎝
x
=
1
(2i sin 8q − 2 ⋅ 2i sin 6q − 2 ⋅ 2i sin 4q + 6 ⋅ 2i sin 2q )
(28 i )
1
(sin 8
27
1
(sin 8
128
2sin 6
2sin 4
2sin 6
6sin 2 )
2sin 4
6sin 2 )
Example 4: If sin4p cos 3p = a1cos p + a3cos 3p + a5cos 5p + a7cos 7p , prove
that a1 + 9a3 + 25a5 + 49a7 = 0.
Solution: Let x = cos q + i sin q
1 = cos q
x
i sin q
4
sin 4 cos3
1 ⎞⎤ ⎡ 1 ⎛
1 ⎞⎤
⎡1 ⎛
= ⎢ ⎜ x − ⎟⎥ ⎢ ⎜ x + ⎟⎥
2
2
i
x
x ⎠⎦
⎠⎦ ⎣ ⎝
⎣ ⎝
1
x
(2i ) 4
1
x
1 1
x
4 4
2 i 23
1
x
27
1
x
1 7
x
27
3x3
1 7
x
27
1
x7
=
1
(2 cos 7q
27
=
1
(cos 7q
64
1
x
x
3
3
1
x
23
x2
1
x2
x6
3x 2
3
x2
1
x6
1
x5
x5
3x
x5
1
x5
2 cos 5q
cos 5q
3
3
1
x
3
x
1
x
3 x3
3
x3
1
x7
1
x3
3 x
1
x
3 · 2 cos 3q + 3 · 2 cosq )
3 cos 3q + 3 cosq )
But, sin4q cos3q = a7 cos 7q + a5 cos 5q + a3 cos 3q + a1 cosq
Complex Numbers
1.59
Comparing Eqs. (1) and (2),
a1 =
3
3
1
1
, a3 = − , a5 = − , a7 =
64
64
64
64
a1 + 9a3 + 25a5 + 49a7 =
3
+9
64
3
64
25
1
64
49
1
64
=0
a1 + 9a3 + 25a5 + 49a7 = 0
Type II : Expansion of sin np, cos np in powers of sin p, cos p :
By De Moivre’s theorem,
cos nq + i sin nq = (cosq + i sinq)n
= cosnq + nC1 cosn 1q · i sinq + nC2 cosn 2q (i sinq )2
+ nC3 cosn 3q (i sinq )3 + . . . . .
= (cosnq - nC2 cos n 2 q sin2q + . . . . )
+ i (nC1 cos n 1 q sin q nC3 cos n 3q sin 3q + . . . . )
Comparing real and imaginary part on both the sides,
cos nq = cosn q + nC2 cosn 2q sin2 q + . . . . .
sin nq = nC1 cosn 1q sin q
Example 1: Expand
C3 cosn 3 sin3 q + . . . . .
n
sin 5
in powers of cosp only.
sin
Solution: (cos 5q + i sin 5q ) = (cos q + i sin q )5 (Using De Moivre’s theorem)
= cos5 q + 5 cos4 q · i sin q + 10 cos3 q (i sin q )2
+ 10 cos2 q (i sin q )3 + 5 cos q (i sin q )4
+ (i sin q )5
= (cos5 q cos3 q sin2 q + 5 cos q sin4 q )
+ i (5 cos4 q sin q 10 cos2 q sin3 q + sin5 q )
Comparing imaginary part on both the sides,
sin 5q = 5 cos4 q sin q 10 cos2 q sin3 q + sin5 q
= sin q (5 cos4 q 10 cos2 q sin2 q + sin4 q )
sin 5
= 5 cos4 q 10 cos2 q (1 cos2 q ) + (1 cos2 q )2
sin
= 5 cos4 q 10 cos2 q + 10 cos4 q
+ 1 2 cos2 q + cos4 q
= 16 cos4 q 12 cos2 q + 1
Example 2: Prove that tan 5 =
that 5 tan 4
10
10 tan 2
10
1 0.
5 tan - 10 tan 3 + tan 5
1 - 10 tan 2 + 5 tan 4
and hence deduce
1.60
Engineering Mathematics
Solution: Comparing real and imaginary part on both the sides in Example 1,
cos 5q = cos5 q 10 cos3 q sin2 q + 5 cos q sin4 q
sin 5q = 5 cos4 q sin q 10 cos2 q sin3 q + sin5 q
tan 5q =
sin 5
5cos 4 sin
10 cos 2 sin 3
sin 5
=
cos 5
cos5
10 cos3 sin 2
5cos sin 4
Dividing numerator and denominator by cos5q,
5 tan
10 tan 3
tan 5
2
4
1 tan
5 tan
tan 5q =
Putting
=
10
,
5 tan
tan 5
tan
10
=
10
10 tan 3
1 tan
5
10
tan
2
10
tan 5
10
5 tan
10
… (1)
4
10
2
Thus, denominator of the R.H.S. of Eq. (1) must be zero.
Hence, 1 tan 2
10
5 tan 4
10
0
Example 3: If sin 6p = a cos5p sin p + b cos3p sin3p + c cos p sin5p, find values
of a, b, c.
Solution: (cos 6q + i sin 6q ) = (cos q + i sin q )6
= cos6q + 6 cos5q (i sin q ) + 15 cos4q (i sin q )2
+ 20 cos3q (i sin q )3+ 15 cos2q (i sin q )4
+ 6 cos q (i sin q )5 + (i sin q )6
= (cos6q 15 cos4q sin2q + 15 cos2q sin4q sin6q )
+ i (6 cos5q sin q 20 cos3q sin3q
+ 6 cos q sin5q )
Comparing imaginary part on both the sides,
sin 6q = 6 cos5q sin q
But
20 cos3q sin3q + 6 cos q sin5q
sin 6q = a cos5q sin q + b cos3q sin3q + c cos q sin5q
Comparing both the expressions ,
a = 6, b = 20, c = 6.
Example 4: Prove that
1 + cos 6
1 + cos 2
= 16 cos4 p - 24 cos2p + 9.
Complex Numbers
1.61
Solution: Comparing real part on both the sides, in Example 3,
cos 6q = cos6 q 15 cos4 q sin2 q + 15 cos2 sin4 q sin6 q
= cos6 q 15 cos4 q (1 cos2 q ) + 15 cos2 q (1 cos2 q )2 (1 cos2 q )3
= cos6 q 15 cos4 q + 15 cos6 q + 15 cos2 q (1 2 cos2 q + cos4 q )
(1 3 cos2 q + 3 cos4 q cos6 q ) = 32 cos6 q 48 cos4 q + 18 cos2 q
6
1 + cos 6
48cos 4
18cos 2
1
= 1 32 cos
2
1 + cos 2
2 cos
= 16 cos4 q 24 cos2 q + 9
Example 5: Using De Moivre’s theorem, prove that 2 (1 + cos 8p ) = (x4 - 4x2 + 2)2,
where x = 2 cos p.
2 (1 + cos 8q ) = 2 . 2 cos2 4q = (2 cos 4q )2
… (1)
cos 4q + i sin 4q = (cos q + i sin q )4
= cos4 q + 4 cos3 q (i sin q ) + 6 cos2 q (i sin q )2
+ 4 cos q (i sin q )3 + (i sin q )4
4
= (cos q 6 cos2 q sin2 q + sin4 q )
+ i (4 cos3 q sin q 4 cos q sin3 q )
Comparing real part on both the sides,
cos 4q = cos4 q 6 cos2 q sin2 q + sin4 q
= cos4 q 6 cos2 q (1 cos2 q ) + (1 cos2 q )2
= cos4 q 6 cos2 q + 6 cos4 q + 1 2 cos2 q + cos4 q
= 8 cos4 q 8 cos2 q + 1
Substituting in Eq. (1),
2 (1 + cos 8q ) = ( 2 cos 4q )2 = (16 cos4 q 16 cos2 q + 2)2
Solution:
= (x4
4x2 + 2)2
x = 2cos q
Example 6: Using De Moivre’s theorem, prove that
1 + cos 9
4
3
2
2
1 + cos = (x - x - 3x + 2x + 1) , where x = 2 cos q.
1 + cos 9
Solution:
1 + cos
2 cos 2
2 cos
9
2
2sin 2
2
2sin
2
2 cos
9
sin
2
2
2sin
cos
=
2
2
2
2
2
2
( sin 5
sin
2
sin 4
2
)
[∵ 2 cos A sin B
sin ( A B)
sin ( A B)]
1.62
Engineering Mathematics
From Example 2, we have
sin 5q = 5 cos4 q sin q 10 cos2 q sin3 q + sin5 q
= sin q [5 cos4 q 10 cos2 q (1 cos2 q ) + (1 cos2 q )2]
= sin q [5 cos4 q 10 cos2 q + 10 cos4 q + 1 2 cos2 q + cos4 q ]
= sin q [16 cos4 q 12 cos2 q + 1]
Comparing imaginary part on both the sides in Example 5,
sin 4q = 4 cos3 q sin q 4 cos q sin3 q
= 4 sin q [cos3 q cos q (1 cos2 q )]
= 4 sin q (cos3 q cos q + cos3 q )
= sin q (8 cos3 q 4 cos q )
sin 5q sin 4q = sin q (16 cos4 q 12 cos2 q + 1 8 cos3 q + 4 cos q )
= sin q (16 cos4 q 8 cos3 q 12 cos2 q + 4 cos q + 1)
1 cos 9
(sin 5 sin 4 ) 2
=
1 + cos
sin 2
2
sin (16 cos 4
8cos3
12 cos 2
=
sin 2
= (x4
x3
4 cos
1) 2
3x2 + 2x + 1)2, where x = 2 cos q
Exercise 1.4
1. Expand cos6q sin6q in terms of
cosine multiples of q.
Ans. :
1
(cos 6q + 15 cos 2q )
16
2. Expand cos8q in terms of cosine
multiples of q.
1
Ans. : 7 (cos 8q + 8 cos 6q +
2
28 cos 4q + 56 cos 2q + 35)
1
3. Prove that cos8q + sin8q =
(cos 8q
64
+ 28 cos 4q + 35).
1
4. Prove that cos4q sin3q =
(sin 7q +
64
sin 5q 3 sin 3q sin q ).
5. Prove that 212cos6q sin7q =
(sin13q sin11q 6 sin 9q + 6 sin 7q +
15 sin 5q 15 sin 3q 20 sin q ).
6. Prove that 256 sin7 q cos2 q =
cos 9q 5 sin 7q + 8 sin 5q 14 sin q.
sin 7
7. Expand
sin
only.
in powers of sin q
⎡ Ans. : 7 − 56 sin 2 q + 112 sin 4 q ⎤
⎢
⎥
− 64 sin 6 q
⎢⎣
⎥⎦
8. Prove that
sin 6
= 32sin5q 32sin3q + 6 sin q.
cos
9. Prove that tan 7q =
7 tan
35 tan 3
21tan 5
tan 7
2
4
1 21tan
35 tan
7 tan 6
and hence deduce 1 21tan 2
14
+
7 tan 6
= 0.
14
14
10. Prove that 1 cos 10q =
2 (16 sin5 q 20 sin3 q + 5 sin q )2.
1 + cos 7
11. Prove that
= (x3 x2
1 + cos
2x + 1)2, where x = 2 cos q.
35 tan 4
Complex Numbers
1.63
Type III : Summation of sine and cosine series
De Moivre’s theorem can also be used to find the sum of sine and cosine series of the
form
… (1)
a0 sin a + a1 sin (a + b ) + a2 sin (a + 2b ) +…………….
… (2)
a0 cos a + a1 cos (a + b ) + a2 cos (a + 2b ) +……………
where, a0, a1, a2, … are either constants or some standard functions.
Working rule:
(i) Denote the given series by S if it is a sine series and by C if it is a cosine series.
(ii) Write cosine series C if sine series S is known, by replacing sine terms
with cosine terms and write sine series S if cosine series C is known, by replacing cosine terms with sine terms in the given series.
(iii) Multiply sine series by i and add to the cosine series to get C + i S. Find the sum
of the series C + i S by using any one of the following series and then separate
its real and imaginary parts to get C and S.
a (1 r n )
, |r|<1
1. Geometric series: (i) a + ar + ar2+ …….. + ar n–1 =
1 r
a
(ii) a + ar + ar2 + …….. =
, | r | < 1.
1 r
x 2 x3
2. Exponential series: 1 + x +
+ + …….. = ex.
2! 3!
3. Logarithmic series: (i) x
2
(ii) x
x
2
3
x
3
……..
x 2 x3
+
2
3
= log (1
……..
= log (1 + x)
x).
4. Trigonometric series:
x3 x5
+
…….. = sin x
3! 5!
x3 x5
+
(iii) x +
+ …….. = sinh x
3! 5!
(i) x
x2 x4
+
……..
2! 4!
x2 x4
+
(iv) 1 +
+ ……..
2! 4!
(ii) 1
= cos x
= cosh x
n(n 1) 2
x + ……….. = (1 + x)n.
2!
x3 x5
+
…………. = tan 1 x
3 5
5. Binomial series: 1 + nx +
6. Gregory series: (i) x
1
1+ x
log
.
2
1 x
1
1
Example 1: Find the sum of the series cos ` + cos 3 + 2 cos 5` + ………
2
2
1
1
Solution: Let C = cos a + cos 3a + 2 cos 5a + ……..
2
2
(ii) x +
x3 x5
+
+
3 5
= tanh 1 x =
1.64
Engineering Mathematics
1
1
sin 3a + 2 sin 5a + ……..
2
2
S = sin a +
C + iS = (cos a + i sin a) +
= eia +
1
1
(cos 3a + i sin 3a) + 2 (cos 5a + i sin 5a) + ……..
2
2
1 i3a 1 i5a
e + 2 e + ……..
2
2
ei
=
1 i2
e
1
2
ei
=
1
[Sum of G.P.]
1
e
2
1
1 i2
e
2
1
i2
ei
1
e
2
=
i2
1
e
2
1
i2
1 i
e
2
1 i2
e
2
1
4
1
1
3
(cos a + i sin a ) − (cos a − i sin a )
cos a + i sin a
2
2
=
= 2
5
5
− cos 2a
− cos 2a
a
4
4
Comparing real parts on both the sides,
1
cos
C= 2
5 4 cos 2
4
n 1
Example 2: Show that
cos
r =1
=
2 cos
.
5 4 cos 2
2r
n
1.
n −1
⎛ 2r ⎞
Solution: Let C = ∑ cos ⎜
⎟
⎝ n ⎠
r =1
2
4
6
⎡ 2(n − 1) ⎤
+ cos
+ cos
+ ............cos ⎢
⎥
n
n
n
n
⎣
⎦
4
6
2
⎡ 2(n − 1) ⎤
+ sin
+ sin
+ ............sin ⎢
S = sin
⎥
n
n
n
n
⎣
⎦
2
2 ⎞ ⎛
4
4 ⎞ ⎛
6
6 ⎞
⎛
+ i sin
+ i sin
C + iS = ⎜ cos
+ i sin
⎟ + ⎜ cos
⎟ + ⎜ cos
⎟ + ...
n
n ⎠ ⎝
n
n ⎠ ⎝
n
n ⎠
⎝
= cos
+ cos
=e
i 2p
n
+e
i 4p
n
+e
i 6p
n
+ ... + e
2(n 1)
2(n 1)
+ i sin
n
n
i 2 ( n −1)p
n
Complex Numbers
e
i 2p
n
1
e
1 e
e
i 2p
n
e
=
1 e
e
i 2p
n
i 2p
n
n 1
[Sum of (n 1) terms of G.P.]
i 2p
n
n
i 2p
n
=
i 2p
n
e
i 2p
n
1 e
cos 2p
1 e
1.65
i sin 2p
i 2p
n
e i 2p
i 2p
n
e
i 2p
n
1 e
1
i 2p
n
1
Equating real part on both the sides,
C= 1
Example 3: Find the sum of the series 1 + x cos ` + x2 cos 2` + x3 cos 3` + …….
n terms, where x < 1. Also find the sum to infinity.
Solution: Let C = 1 + x cos a + x2 cos 2a + x3 cos 3a + …….. n terms
S = 0 + x sin a + x2 sin 2a + x3 sin 3a + …….. n terms
C + iS = (1 + i0) + x (cos a + i sin a) + x2 (cos 2a + i sin 2a) + x3 (cos 3a + i sin 3a)
+ …….. n terms
= 1 + xeia + x2 ei2a + x3 ei3a + …….. n terms
1 ( xei ) n
[Sum of n terms of G.P.]
1 xei
1 x n ei n
=
1 xei
1 x n ei n 1 xe i
1 xei 1 xe i
1 xe i x n ein x n 1ei ( n 1)
=
1 xe i xei x 2
1 x (cos
i sin ) x n (cos n
i sin n ) x n +1 [ cos(n 1)
i sin (n 1) ]
=
2
1 2 x cos
x
=
Equating real part on both the sides,
1 x cos
C=
To find sum to infinity, taking limit n
lim C = lim
n
n
x n cos n
1 2 x cos
x n +1 cos (n 1)
x2
in the above expression,
1 x cos
x n cos n
1 2 x cos
x n +1 cos (n 1)
x2
1.66
Engineering Mathematics
Since x < 1, lim xn = 0 and lim xn+1 = 0
n
n
lim C =
n
1 x cos
1 2 x cos
Hence, 1 + x cos a + x2 cos 2a + x3 cos 3a + ………
x2
=
1 x cos
1 2 x cos
x2
x2
x3
Example 4: Find the sum of the series x sin p +
sin 2p +
sin 3p + …..
2!
3!
x2
x3
sin 2q +
sin 3q + ……..
Solution: Let S = x sin q +
2!
3!
C = 1 + x cos q +
x2
x3
cos 2q +
cos 3q + … [∵ first term = x0 cos(0.q) = cos0 = 1]
2!
3!
x2
x3
(cos 2q + i sin 2q ) +
(cos 3q + i sin 3q ) + ……..
2!
3!
z 2 z3
x 2 i2q x 3 i3q
e + e + …….. = 1 + z + +
+ …….. where z = xeiq
= 1 + xeiq +
2! 3!
2!
3!
C + iS = 1 + x (cos q + i sin q ) +
i
= ez = e xe = ex (cosq +i sin q)
= ex cos q eix sin q = ex cos q [cos (x sin q ) + i sin (x sin q )]
Comparing imaginary part on both the sides,
S = x sin q +
x2
x3
sin 2q +
sin 3q + …….. = ex cos q
2!
3!
sin (x sin q).
Example 5: Find the sum of the series
1
1
sin ` cos a - sin2 ` cos 2a + sin3` cos 3a - ……… .
3
2
1 2
1
Solution: Let C = sin a cos b
sin a cos 2b + sin3 a cos 3b ……..
2
3
1 2
1
sin a sin 2b + sin3 a sin 3b ……..
S = sin a sin b
2
3
C + iS = sin a (cos b + i sin b )
+
= sin a eib
1
sin2 a (cos 2b + i sin 2b )
2
1
sin3 a (cos 3b + i sin 3b )
3
1
1
sin2 a ei2b + sin3 a ei3b
2
3
……..
……..
Complex Numbers
z 2 z3
+
2 3
=z
1.67
…….. where z = sin a eib
= log (1 + z) = log (1 + sin a .eib )
= log [1 + sin a (cos b + i sin b )]
= log [(1 + sin a cos b ) + i (sin a sin b )]
=
sin sin
1
log [(1 + sin a cos b )2 + (sin a sin b )2] + i tan 1 1 + sin sin
2
1
y
∵ log ( x + iy ) = log ( x 2 + y 2 ) + i tan 1
2
x
refer section 1.13
Comparing real part on both the sides,
1
C = log (1 + 2 sin a cos b + sin2 a cos2 b + sin2 a sin2 b )
2
=
1
log (1 + 2 sin a cos b + sin2 a ).
2
Example 6: Find the sum of the series
1
1
a cos2 ` - a3 cos2 3` + a5 cos2 5` - …….. .
3
5
Solution: Let C = a cos2 a
We know that, cos2 a =
C=a
1 + cos 2
2
1 3
1
a cos2 3a + a5 cos2 5a
3
5
, cos2 3a =
(1 + cos 2 )
2
1
a
2
a3
3
a5
5
1 + cos 6
2
……..
etc.
a 3 (1 + cos 6 ) a 5 (1 + cos10 )
+
+
3
5
2
2
.........
a3
cos 6
3
1
a cos 2
2
a5
cos10
5
1
tan 1 a
2
1
a cos 2
2
a3
cos 6
3
∵ tan 1 a
=
1
tan 1 a + C1
2
.........
a5
cos10
5
a
a3
3
a5
5
...
...
… (1)
1.68
Engineering Mathematics
a3
a5
cos 6a + cos 10a + …
3
5
C1 = a cos 2a
where,
Let S1 = a sin 2a
a3
a5
sin 6a + sin 10a + …
3
5
C1 + iS1 = a (cos 2a + i sin 2a)
a3
a5
(cos 6a + i sin 6a) + (cos 10a + i sin 10a)
3
5
= aei2a –
a 3 i6a a 5 i10a
e + e + ……..
3
5
= aei2a
(aei 2 )3 (aei 2 )5
+
3
5
……..
= tan 1 (aei2a)
= tan 1 [a (cos 2a + i sin 2a )]
Let tan 1 (a cos 2a + i sin 2a) = x + iy
tan 1 (a cos 2a ia sin 2a) = x - iy
Adding both the equations,
2x = tan 1 (a cos 2a + ia sin 2a) + tan 1 (a cos 2a
= tan
1
= tan
1
a cos 2
ia sin 2
a cos 2
1 (a cos 2
ia sin 2 )(a cos 2
2a cos 2
= tan
1 a 2 (cos 2 2
sin 2 2 )
ia sin 2a)
ia sin 2
ia sin 2 )
1
2a cos 2
1 a2
1
⎛ 2a cos 2a ⎞
tan −1 ⎜
⎝ 1 − a 2 ⎟⎠
2
C1 + iS1 = tan −1 (a cos 2a + ia sin 2a ) = x + iy
Comparing real part on both the sides,
Hence,
x=
From Eq. (1), we get
1
C1 = x = tan
2
1
2a cos 2
1 a2
1
1
C = tan 1 a + tan
2
2
Example 7: Find the sum of the series a sin ` +
Solution: Let S = a sin a +
and
C = a cos a +
1
2a cos 2
1 a2
.
a3
a5
sin 3` + sin 5` + ……. .
3
5
a3
a5
sin 3a + sin 5a + ……..
3
5
a3
a5
cos 3a + cos 5a + ……..
3
5
Complex Numbers
C + iS = a (cos a + i sin a) +
= aeia +
a3
a5
(cos 3a + i sin 3a) + (cos 5a + i sin 5a) + ……..
3
5
a 3 i3a a 5 i5a
e + e + ……..
3
5
1
1 + aei
log
2
1 aei
1
= [log {(1 + a cos a) + ia sin a}
2
=
=
1.69
1 1
log {(1 + a cos
2 2
)
2
∵
x3 x5
1
1+ x
+ + ... = log
3 5
2
1 x
a cos a)
log {(1
+ a 2 sin 2 } + i tan
1
log{(1 a cos ) 2
2
x+
1
ia sin a}]
a sin
1 + a cos
a 2 sin 2 } i tan
∵ log ( x + iy ) =
1
a sin
1 a cos
1
log ( x 2 + y 2 ) + i tan
2
1
y
x
Comparing imaginary part on both the sides,
1
tan
2
S
=
1
tan
2
1
=
tan
2
=
1
tan
2
1
1
1
a sin
1 a cos
tan
1
a sin
1 a cos
+ tan
1
a sin
1 a cos
a sin
1 a cos
a sin
a sin
+
1
1 a cos
1 a cos
a sin
a sin
1
1 a cos
1 a cos
2a sin
1 a2
Example 8: Find the sum of the series
n sin ` +
n( n + 1)
n( n + 1)( n + 2)
sin 2 +
sin 3 + …….. n terms.
1 2
1 2 3
Solution: Let S = n sin a +
C = 1 + n cos a +
n(n +1)
n(n +1) (n + 2)
sin 2 +
sin 3 + …….. n terms
1 2
1 2 3
n(n +1)
n(n +1)(n + 2)
cos 2 +
cos 3 + …….. n terms
1 2
1 2 3
1.70
Engineering Mathematics
n(n +1)
(cos 2a + i sin 2a)
1 2
n(n +1)(n + 2)
+
(cos 3a + i sin 3a) + …….. n terms
1 2 3
n(n +1) i2a + n(n +1)(n + 2) ei 3
e
+ …….. n terms
= 1 + neia+
1 2 3
1 2
1
= (1 eia) n =
[Using Binomial expansion]
(1 e i ) n
C + iS = 1 + n (cos a + i sin a) +
n
n
⎡ 1
(1 − e − iα ) ⎤
⎡1 − cos α + i sin α ⎤
=⎢
⋅
=⎢
− iα ⎥
− iα
iα
iα
⎥
⎣ 1− e − e +1 ⎦
⎣ (1 − e ) (1 − e ) ⎦
n
n
⎡ α ⎧ ⎛π α ⎞
⎛ π α ⎞⎫ ⎤
α
α⎤
⎡
2 α
⎢ sin ⎨cos ⎜ − ⎟ + i sin ⎜ − ⎟ ⎬ ⎥
⎢ 2 sin 2 + 2i sin 2 cos 2 ⎥
2⎩ ⎝2 2⎠
⎝ 2 2 ⎠⎭ ⎥
⎢
=⎢
=
⎥
α
⎥
⎢
2 − 2 cos α
⎣
⎦
2 sin 2
⎥⎦
⎢⎣
2
n
⎛ nπ nα ⎞ ⎤
⎛ 1 ⎞ ⎡ ⎛ nπ nα ⎞
cos
−
−
=⎜
⎟ + i sin ⎜
⎟ [Using Dee Moivre’s theorem]
α ⎟ ⎢⎣ ⎜⎝ 2
2 ⎠
2 ⎠ ⎥⎦
⎝ 2
⎜ 2 sin ⎟
⎝
2⎠
Comparing imaginary part on both the sides,
1
S = n cosec
2
2
n
sin
n
2
2
Exercise 1.5
Find the sum of the series:
1.
2
4
6
+ ...
+ sin
+ sin
n
n
n
2(n 1)
.
+ sin
n
4. sin
1
1
sin + 2 sin 2 +
2
2
1
1
sin 3 + 4 sin 4 + … .
3
2
2
Ans. :
Hint : Let C = 1 + cos
2sin
5 cos
2. 1 + cos a cos a + cos2 a cos 2a +
cos3 a cos 3a + ……..
[Ans. : sin2 a]
3. 1
cos a cos b +
cos 2
cos 2
2!
cos3 a
cos 3b + ……..
3!
[Ans. : e cos a cos b cos (cos a cos b)]
+ cos
2
n
4
+...
n
[Ans. : 0]
5. cos a + sin a cos 2a +
sin 2
cos 3a + ……..
2!
[Ans. : esin a cos a cos (a + sin2 a)]
6. sin a sin a
1
sin 2a sin2 a +
2
1
sin 3a sin3 a
3
…
Complex Numbers
Ans. : tan
7. x sin a
1
[Ans. : log cot q ]
sin 2
1 + sin cos
13
1
cos a +
cos 2a
2 4
2
13 5
cos 3a + …….. .
2 4 6
10. 1
1
1 2
x sin 2a + x3 sin 3a
3
2
… .
Ans. : tan
8. e cos b
a
+… .
1
1.71
x sin
1 + x cos
Ans. : (2 cos )
1 3a
1
e cos 3b + e5a cos5 b
3
5
11.
1
⎤
⎡
−1 cosb
⎢⎣ Ans. : 2 tan sinh a ⎥⎦
cos
4
1
13
13 5
sin a +
sin 2a +
sin 3a
2
2 4
2 4 6
+ …….. (a np).
1
2
Ans. : 2sin
1
1
9. cos2a + cos 6a + cos 10a + … .
3
5
1
2
2
sin
4
4
1.10 CIRCULAR AND HYPERBOLIC FUNCTIONS
1.10.1 Circular Function
From Euler’s formula, we have
eiq = cos q + i sin q
e iq = cos q i sin q
e
2!
i
eiz e
eiz + e iz
, sin z =
2i
2
These are called circular functions of complex numbers.
iz
cos q =
ei + e
2
i
, sin q =
ei
If z = x + iy is a complex number, then
cos z =
1.10.2 Hyperbolic Function
If z is a complex number, then sine hyperbolic of z is denoted by sinh z and is given as,
e z
2
and cosine hyperbolic of z is denoted by cosh z and is given as,
sinh z =
ez
ez + e z
2
From these expressions, other hyperbolic functions can also be obtained as
sinh z
1
1
1
, cosech z =
, sech z =
.
tanh z =
and coth z =
cosh z
sinh z
tanh z
cosh z
cosh z =
1.72
Engineering Mathematics
z, cosh z, tanh z, we can obtain the following values
of hyperbolic functions.
z
sinh z
cosh z
tanh z
0
0
1
0
1
1
Note: sinh ( z) = sinh z, cosh ( z) = cosh z
1.10.3 Relation between Circular and
Hyperbolic Functions
(i) sin iz = i sinh z and sinh z = i sin iz
Proof : By Euler’s formula,
sin z =
Replacing z by iz,
sin iz =
eiz
ei
2
e
2i
z
(e
=i
e
2i
z
iz
(e
i
i2 z
e
z
ez )
z
2
) = i sinh z
2
1
sinh z = sin iz = i sin iz
i
(ii) cos iz = cosh z
Proof : By Euler’s formula,
cos z =
eiz + e
2
iz
Replacing z by iz,
2
ei z + e
2
(iii) tan iz = i tanh z and tanh z = i tan iz
cos iz =
Proof :
tan iz =
i2 z
=
e z + ez
= cosh z
2
sin iz i sinh z
= i tanh z
=
cos iz cosh z
1
tanh z = tan iz = i tan iz
i
1.10.4 Formulae on Hyperbolic Functions
A.
(i) cosh2 z – sinh2 z = 1
(ii) coth2 z – cosech2 z = 1
(iii) sech2 z + tanh2 z = 1
⎤
⎡ 1
⎢∵ i = −i ⎥
⎦
⎣
Complex Numbers
B. (iv) sinh 2z = 2 sinh z cosh z
(v) cosh 2z = cosh2 z + sinh2 z = 2 cosh2 z – 1 = 1 + 2 sinh2 z
(vi) tanh 2z =
2 tanh z
1 + tanh 2 z
C. (vii) sinh 3z = 3 sinh z + 4 sinh3 z
(viii) cosh 3z = 4 cosh3 z – 3 cosh z
3 tanh z + tanh 3 z
(ix) tanh 3z =
1 + 3 tanh 2 z
Proof : A. (i) For the circular functions, we have
sin2 q + cos2 q = 1
Putting q = iz,
(sin iz)2 + (cos iz)2 = 1
(i sinh z)2 + (cosh z)2 = 1
–sinh2 z + cosh2 z = 1
cosh2 z – sinh2 z = 1
Similarly, (ii) and (iii) can also be proved.
B. (iv) We have
sin 2q = 2 sin q cos q
Putting q = iz,
sin 2iz = 2 sin iz cos iz
i sinh 2z = 2i sinh z cosh z
sinh 2z = 2 sinh z cosh z
Similarly, (v) and (vi) can also be proved.
3 tan
tan 3
C. (ix) We have
tan 3q =
1 tan 2
Putting q = iz,
tan 3iz =
i tanh 3z =
3 tan iz (tan iz )3
1 3(tan iz ) 2
3i tanh z i 3 tanh 3 z
1 3i 2 tanh 2 z
3 tanh z + tanh 3 z
1 + 3 tanh 2 z
Similarly, (vii) and (viii) can also be proved.
tanh 3z =
Similarly, we can prove the following formulae:
D.
(x) sinh (z1 ± z2) = sinh z1 cosh z2 ± cosh z1 sinh z2
(xi) cosh (z1 ± z2) = cosh z1 cosh z2 ± sinh z1 sinh z2
1.73
1.74
Engineering Mathematics
tanh z1 ± tanh z2
(xii) tanh (z1 ± z2) = 1 ± tanh z tanh z
1
E.
z1 + z2
z z
cosh 1 2
2
2
z1 + z2
z1 z2
sinh z1– sinh z2 = 2 cosh
sinh
2
2
z1 + z2
z1 z2
cosh z1+ cosh z2 = 2 cosh
cosh
2
2
z1 + z2
z1 z2
cosh z1– cosh z2 = 2 sinh
sinh
2
2
2 sinh z1cosh z2 = sinh (z1 + z2) + sinh (z1 – z2)
(xiii) sinh z1 + sinh z2 = 2 sinh
(xiv)
(xv)
(xvi)
F.
2
(xvii)
(xviii) 2 cosh z1sinh z2 = sinh (z1+ z2) – sinh (z1 – z2)
(xix) 2 cosh z1 cosh z2= cosh (z1+ z2) + cosh (z1 – z2)
(xviii) 2 sinh z1 sinh z2 = cosh (z1 + z2) – cosh (z1– z2)
1.11 INVERSE HYPERBOLIC FUNCTIONS
If x = sinh u, then u = sinh 1 x is called sine hyperbolic inverse of x, where x is real.
1
x, tanh 1 x, coth 1 x, sech 1 x and cosech 1 x.
The inverse hyperbolic functions are many valued functions but we will consider their
principal values only.
If x is real,
(
(i) sinh 1 x = log x + x 2 + 1
)
(
(ii) cosh 1 x = log x
1+ x
1
(iii) tanh 1 x = log
1 x
2
1
Proof : (i) Let sinh x = y
x = sinh y =
2x = e y
e2y 2x e y
This equation is quadratic in e y.
ey e
2
y
1 e2 y 1
=
ey
ey
1=0
ey =
2x ± 4x2 + 4
2
ey = x ± x 2 + 1
y = log ( x ± x 2 + 1 )
x2 1
)
Complex Numbers
But x – x 2 + 1
(
x = log ( x +
1.75
)
+ 1)
y = log x + x 2 + 1
sinh
1
x2
(ii) Let cosh 1 x = y
ey + e y
2
2y
1 e +1
2x = e y + y =
ey
e
y
2y
2xe = e + 1
e 2y – 2xe y + 1 = 0
2x
4x2 4
ey =
2
x = cosh y =
ey= x
(
x2 1
y = log x
(
)
x2 1 , e y = x
Consider, y = log x
e
1
y
x
y
(
x
2
1 x
log x
y
x
(
log x
(
log x
2
1
)
x2 1
)
... (1)
x2 1
x2 1
x
x2 1
x
2
1
... (2)
x
x2
x2 1
x2 1
)
)
(
x 2 1 = log x
x2 1
(
cosh 1 x = log x + x 2 + 1
)
)
( x 1)
cosh x = ± log ( x + x + 1 )
+ 1 ) = cosh log ( x + x + 1 ) [
2
2
(iii) Let tanh 1 x = y
x = tanh y
x ey e
=
1 ey + e
... (4)
... (5)
2
1
(
x2
... (3)
y = ± log x
x = cosh ± log x + x 2
x
y
y
cosh ( z) = cosh z]
1.76
Engineering Mathematics
Using componendo-dividendo,
1 x ey e y ey
=
1 x ey e y ey
1+ x
e2 y =
1 x
1+ x
2 y = log
1 x
y=
1
1+ x
log
2
1 x
tanh 1 x =
1
1+ x
log
2
1 x
1 + tanh x
Example 1: Prove that
1 tanh x
Solution:
1+ tanh x
1 tanh x
3
y
y
=
2e y
= e2 y
2e y
3
= cosh 6x + sinh 6x.
3
sinh x
cosh
x
=
sinh x
1
cosh x
1+
=
e
e
=
cosh x sinh x
cosh x sinh x
cos ix i sin ix
cos ix i sin ix
3
cos ix i sin ix
cos ix i sin ix
⎡ (cos ix − i sin ix) 2 ⎤
=⎢
⎥
⎣ cos 2 ix + sin 2 ix ⎦
= (cos ix − i sin ix)6
=
cos ix i sin ix
cos ix i sin ix
3
3
= cos 6ix − i sin 6ix
[Using De Moivre’s theorem]
= cosh 6 x − i ⋅ i sinh 6 x
= cosh 6 x + sinh 6 x.
1
Example 2: Prove that
111
Solution:
1
1−
1−
1
1 − cosh 2 x
= cosh 2 x .
1
1
1 - cosh 2 x
1
=
1
1−
1
−sinh 2 x
1
=
1
1−
1 + cosech 2 x
1−
3
[∵ cos h 2 x − sinh 2 x = 1]
Complex Numbers
=
1
[∵ 1 − cot h 2 x = −cosech 2 x]
1
1−
coth 2 x
=
1
1 − tanh 2 x
=
1
sech 2 x
1.77
[∵ 1 − tanh 2 x = sech 2 x]
= cosh 2 x.
5
Example 3: Prove that cosh x =
1
[cosh5 x + 5 cosh 3 x + 10 cosh x ].
16
5
⎛ e x + e− x ⎞
Solution: cosh x = ⎜
⎟
⎝ 2 ⎠
1
= 5 (e5 x + 5e 4 x ⋅ e − x + 10e3 x ⋅ e −2 x + 10e 2 x ⋅ e −33 x + 5e x ⋅ e −4 x + e −5 x )
2
1
= 5 ⎡⎣(e5 x + e −5 x ) + 5 (e3 x + e −3 x ) + 10 (e x + e − x ) ⎤⎦
2
5
1
(2 cosh 5 x + 10 cosh 3 x + 20 cosh x)
25
1
= (cosh 5 x + 5 cosh 3 x + 10 cosh x)
16
=
Example 4: Prove that tanh (log 3 ) = 0.5.
Solution: tanh (log 3 ) =
elog
3
− e − log
3
elog
3
+ e − log
3
1
3 3 −1
=
= 0.5
1
3 +1
3+
3
3−
=
Example 5: Solve the equation 17 cosh x + 18 sinh x = 1 for real values of x.
Solution:
17 cosh x + 18 sinh x = 1
⎛ e x + e− x
17 ⎜
⎝ 2
⎞
⎛ e x − e− x ⎞
⎟ + 18 ⎜
⎟ =1
⎠
⎝ 2 ⎠
35e x − e − x = 2
35e 2 x − 1 = 2e x
35e 2 x − 2e x − 1 = 0
1.78
Engineering Mathematics
This equation is quadratic in ex.
ex =
2 ± 4 + 140 2 ± 12 14 −10
=
= ,
70
70
70 70
For real value, ex should be positive.
ex =
1
1
, x = log = − log 5.
5
5
Example 6: Find tanh x, if sinh x - cosh x = 5.
sinh x − cosh x = 5
Solution:
e − e− x e x + e− x
−
=5
2
2
x
−e − x = 5, e − x = −5, e x =
−1
5
1
− − ( −5)
−1 + 25
5
=
1
−1 − 25
− + ( −5)
5
24
12
=
=−
13
−26
e x − e− x
=
tanh x = x
e + e− x
Example 7: If cos ` cosh
(i) sec (α − i β ) + sec (α + i β ) =
=
x
, sin ` sinh
2
4x
x2 + y2
=
y
, prove that
2
(ii) sec (` - i a ) - sec (` + i a ) = -
4iy
x2 + y2
.
1
1
=
cos(a − i b ) cos a cos i b + sin a sin i b
1
=
cos a cosh b + i sin a sinh b
Solution: sec (a − i b ) =
1
2
=
x
y x + iy
+i
2
2
2
( x − iy ) 2( x − iy )
=
⋅
= 2
( x + iy ) ( x − iy )
x + y2
=
... (1)
Complex Numbers
1.79
1
1
=
cos (a + i b ) cos a cosh b − i sin a sinh b
1
2
2 ( x + iy )
⋅
=
=
=
x
y x − iy x − iy ( x + iy )
−i
2
2
2 ( x + iy )
= 2
x + y2
sec (a + i b ) =
Similarly,
... (2)
Adding Eqs. (1) and (2),
ib ) + sec (a + ib ) =
sec (a
4x
x + y2
2
Subtracting Eq. (2) from Eq. (1),
sec (a
ib )
sec (a + ib ) = −
4iy
x + y2
2
Example 8: If log (tan x) = y, prove that
1
1
(tan n x + cot n x ) (ii) sinh ny = (tan n x − cot n x )
2
2
(iii) sinh (n + 1) y + sinh (n – 1) y = 2 sinh ny cosec 2x
(iv) cosh (n + 1) y + cosh (n – 1) y = 2 cosh ny cosec 2x.
(i) cosh ny =
Solution: (i)
(ii)
e y = tan x ⎫⎪
⎬
e − y = cot x ⎭⎪
cosh ny =
e ny + e − ny 1
= (tan n x + cot n x)
2
2
sinh ny =
e ny − e − ny 1
= (tan n x − cot n x)
2
2
... (1)
(n + 1 + n − 1) y
(n + 1 − n + 1) y
cosh
2
2
y
−y
⎛e +e ⎞
= 2 sinh ny cosh y = 2 sinh ny ⎜
⎟
⎝
2 ⎠
⎛ sin 2 x + cos 2 x ⎞
⎛ tan x + cot x ⎞
=
2
sinh
ny
= 2 sinh ny ⎜
⎟⎠
⎜⎝ 2 sin x cos x ⎟⎠
⎝
2
(iii) sinh (n + 1) y + sinh (n − 1) y = 2 sinh
⎛ 1 ⎞
= 2 sinh ny ⎜
= 2 sinh ny cosec 2 x
⎝ sin 2 x ⎟⎠
1.80
Engineering Mathematics
(n + 1 + n − 1) y
(n + 1 − n + 1) y
cosh
2
2
y
−y
⎛e +e ⎞
= 2 cosh ny cosh y = 2 cosh ny ⎜
⎟
⎝
2 ⎠
⎛ sin 2 x + cos 2 x ⎞
⎛ tan x + cot x ⎞
= 2 cosh ny ⎜
= 2 cosh ny ⎜
⎟
⎝
⎠
⎝ 2 sin x cos x ⎟⎠
2
= 2 cosh ny cosec 2 x
(iv) cosh (n + 1) y + cosh (n − 1) y = 2 cosh
⎡
⎛o p ⎞⎤
Example 9: If u = log ⎢tan ⎜ + ⎟ ⎥ , prove that
⎝ 4 2⎠⎦
⎣
(i) tanh
u
p
= tan
2
2
(ii) cosh u = sec p .
⎡ ⎛p q ⎞⎤
u = log ⎢ tan ⎜ + ⎟ ⎥
⎣ ⎝ 4 2⎠ ⎦
p
q
tan + tan
4
2
eu =
p
q
1 − tan tan
4
2
q
1 + tan
eu
2
=
q
1
1 − tan
2
q⎞
⎛
1 + tan
⎜
2 ⎟ = 2 tanh −1 ⎛ tan q ⎞
u = log ⎜
⎟
⎜⎝
q⎟
2⎠
⎜⎝ 1 − tan ⎟⎠
2
q
u
⎛
⎞
= tanh −1 ⎜ tan ⎟
⎝
2
2⎠
q
u
tanh = tan
2
2
(ii) From part (i),
u
q
tanh = tan
2
2
2 u
2 q
tanh
= tan
2
2
Using componendo-dividendo,
u
1 + tanh 2
1 + tan 2
2 =
2
2 u
2
1 − tanh
1 − tan
2
2
Solution: (i)
Complex Numbers
cosh u = sec
⎡
1 − tan 2
⎢
2
⎢∵ cos =
2
⎢
1 + tan
⎢⎣
2
1.81
u⎤
2⎥
and cosh u =
u⎥
1 − tanh 2 ⎥
2 ⎥⎦
1 + tanh 2
⎛ o x⎞
Example 10: Prove that sinh-1 (tan p ) = log tan ⎜ + ⎟ .
⎝ 4 2⎠
Solution: sinh −1 (tan q ) = log ( tan q + tan 2 q + 1)
⎛ sin q + 1⎞
= log(tan q + secq ) = log ⎜
⎝ cos q ⎟⎠
⎤
⎡
⎡
⎛p q ⎞
⎛p
⎞ ⎤
2 cos 2 ⎜ − ⎟
⎥
⎢
⎢ cos ⎜⎝ 2 − q ⎟⎠ + 1⎥
⎝ 4 2⎠
⎥
⎥ = log ⎢
= log ⎢
⎢ 2 sin ⎛ p − q ⎞ cos ⎛ p − q ⎞ ⎥
⎢ sin ⎛ p − q ⎞ ⎥
⎜⎝
⎟
⎜⎝
⎟
⎜⎝
⎟⎠ ⎥
⎢⎣
⎢⎣
4 2⎠
4 2 ⎠ ⎥⎦
2
⎦
⎡ ⎛p q ⎞⎤
⎡ ⎛p p q ⎞⎤
= log ⎢cot ⎜ − ⎟ ⎥ = log ⎢ tan ⎜ − + ⎟ ⎥
⎝
⎠
4 2 ⎦
⎣
⎣ ⎝ 2 4 2⎠ ⎦
⎡ ⎛p q ⎞⎤
= log ⎢ tan ⎜ + ⎟ ⎥ ⋅
⎣ ⎝ 4 2⎠ ⎦
Example 11: Prove that
(i) tanh -1 x = sinh -1
x
1- x
(ii) sinh -1 x = cosh -1
2
(
)
1 + x2 .
Solution:
(i) sinh −1
(ii) cosh −1
⎛ x
⎞
x2
2
−1
⎡
⎤
= log ⎜
+
+
1
⎟ ⎣⎢∵ sinh x = log x + x + 1 ⎦⎥
1 − x2
⎝ 1 − x2
⎠
1 − x2
1 ⎞
x +1
⎛ x
= log ⎜
+
= log
2
2 ⎟
⎝ 1− x
1− x ⎠
1 − x2
1+ x 1+ x
1+ x 1
1+ x
= log
= log
= log
= tanh −1 x.
1− x
1− x 1+ x
1− x 2
(
x
( 1 + x ) = log ( 1 + x + 1 + x − 1)
= log ( 1 + x + x ) = sinh x.
2
2
2
2
)
(
)
⎡∵ cosh −1 x = log x + x 2 − 1 ⎤
⎢⎣
⎥⎦
−1
⎛p ⎞
Example 12: Prove that sech-1 (sin p ) = log cot ⎜ ⎟ .
⎝ 2⎠
1.82
Solution:
Engineering Mathematics
Let y = sech 1 (sin q )
sech y = sin q
cosh y = cosecq
(
y = cosh −1 (cosecq ) = log cosec q + cosec 2q − 1
⎛ 1 + cos q ⎞
= log(cosecq + cot q ) = log ⎜
⎝ sin q ⎟⎠
q ⎞
⎛
2 cos 2
⎜
2 ⎟
= log ⎜
q⎟
q
⎜⎝ 2 sin cos ⎟⎠
2
2
⎛ q⎞
y = log ⎜ cot ⎟
⎝
2⎠
)
⎛ ⎞
Hence, sech 1 (sin q ) = log cot ⎜ ⎟ .
⎝2⎠
Example 13: Prove that tanh-1 (sin p ) = cosh-1 (sec p ).
−1
Solution: tanh (sin q ) =
1
⎡ (1 + sin q ) (1 + sin q ) ⎤
⎛ 1 + sin q ⎞ 1
log ⎜
= log ⎢
⎥
⎝ 1 − sin q ⎟⎠ 2
2
⎣ (1 − sin q ) (1 + sin q ) ⎦
2
=
1
1
⎛ 1 + sin q ⎞
= ⋅ 2 log (secq + tan q )
log ⎜
⎟
⎝ cos q ⎠
2
2
(
= log secq + sec 2 q − 1
= cosh −1 (secq )
)
⎡ 1 + 1 + z2
Example 14: Prove that cosech-1 z = log ⎢
z
⎢
⎣
0 values of z?
Solution: Let y = cosech 1z
cosech y = z
2
=z
e y − e− y
1 2
ey − y =
z
e
2 y
2y
e − e −1 = 0
z
(
)
⎡∵ cosh −1 x = log x + x 2 − 1 ⎤
⎣⎢
⎦⎥
⎤
⎥ . Is it defined for all
⎥
⎦
Complex Numbers
4
+4
z2
2
1+ 1+ z2
=
z
⎡1 + 1 + z 2 ⎤
y = log ⎢
⎥
z
⎢⎣
⎥⎦
1.83
2
±
ey = z
⎡∵ 1 − 1 + z 2 < 0 and e y cannot be negative ⎤
⎣
⎦
It is not defined for z < 0.
Example 15: If cosh x = sec p, prove that
(i) x = log (sec p + tan p )
(iv) tanh x = sin p
Solution: (i)
(ii) p =
(v) tanh
2
- 2 tan-1 (e-x)
x
p
= é tan .
2
2
(iii)
(
x = cosh −1 (sec q ) = log secq + sec 2 q − 1
= log(secq + tan q )
(ii) From (i),
(iii) sinh x = tan p
)
1 + sin q
cos q
⎤
⎡
⎛p q ⎞
⎛p
⎞⎤ ⎡
2 cos 2 ⎜ − ⎟
⎥
⎢1 + cos ⎜⎝ 2 − q ⎟⎠ ⎥ ⎢
⎝ 4 2⎠
⎥
⎥=⎢
=⎢
⎢ sin ⎛ p − q ⎞ ⎥ ⎢ 2 sin ⎛ p − q ⎞ cos ⎛ p − q ⎞ ⎥
⎜⎝
⎟
⎜⎝
⎟
⎜⎝
⎟⎠ ⎥ ⎢
⎢⎣
2
4 2⎠
4 2 ⎠ ⎥⎦
⎦ ⎣
⎛p q ⎞
= cot ⎜ − ⎟
⎝ 4 2⎠
e x = secq + tan q =
⎛p q ⎞
e − x = tan ⎜ − ⎟
⎝ 4 2⎠
p
q
tan −1 (e − x ) = −
4 2
p
q = − 2 tan −1 (e − x )
2
cosh x = secq
1 + sinh 2 x = 1 + tan 2 q
sinh 2 x = tan 2 q
sinh x = tan q
cosh x = secq
1.84
(iv) from (iii),
(v)
Engineering Mathematics
sinh x = tan q
tan q
tanh x =
= sin q
secq
tanh x = sin q
cosh x = secq
x
q
1 + tanh 2
1 + tan 2
2 =
2
2 x
2q
1 − tanh
1 − tan
2
2
Using componendo—dividendo,
x
= tan 2
2
2
x
tanh = ± tan .
2
2
tanh 2
⎡ ⎛ x - a⎞⎤
i
⎛ a⎞
Example 16: Prove that tan -1 ⎢ i ⎜
⎟⎠ ⎥ = - log ⎜⎝ ⎟⎠ .
⎝
x
+
a
2
x
⎣
⎦
−1 ⎡ ⎛ x − a ⎞ ⎤
Solution: Let tan ⎢i ⎜
⎟⎥ =
⎣ ⎝ x + a ⎠⎦
⎛ x − a⎞
i⎜
= tan q
⎝ x + a ⎟⎠
sin q
x−a
= −i
cos q
x+a
⎤
⎡ 1
⎢⎣∵ i = −i ⎥⎦
Using componendo — dividendo,
x + a + x − a cos q − i sin q
=
x + a − x + a cos q + i sin q
x e − iq
=
= e −2iq
a eiq
a
= e 2iq
x
2iq = log
a
x
a
i
a
1
log = − log
x
x
2i
2
i
⎛ a⎞
−1 ⎡ ⎛ x − a ⎞ ⎤
tan ⎢i ⎜
⎟ ⎥ = − log ⎜⎝ ⎟⎠
2
x
⎣ ⎝ x + a⎠ ⎦
q =
Complex Numbers
1.85
Exercise 1.6
1. Prove that
n
⎛ 1 + tanh x ⎞
⎜
⎟ = cosh 2nx + sinh 2nx.
⎝ 1 − tanh x ⎠
2. Prove that cosec x + coth x = coth
x
.
2
3. Prove that
= − sinh 2 x.
1
1−
1−
1
1 + sinh 2 x
4. If cosh6 x = a cosh 6x + b cosh 4x +
c cosh 2x + d, prove that 5a – 5b +
3c – 4d = 0.
5. If cosh 1 a + cosh 1 b = cosh 1 x,
then prove that
a b −1 + b a −1 =
2
x − 1.
2
⎡ Hint : Let cosh −1 a = p, a = cosh p; ⎤
⎢ cosh −1 b = q, b = cosh q;
⎥
⎢
⎥
−1
⎢ cosh x = y, x = cosh y,
⎥
⎢⎣ ∴ p + q = y,sinh ( p + q ) = sinh y. ⎥⎦
6. If cosh 1 a + cosh 1 b = cosh 1 c,
prove that a2 + b2 + c2 = 2abc + 1.
⎡ Hint : Let cosh −1 a = p, cosh −1 b = q, ⎤
⎢
⎥
cosh −1 c = r , p + q = r ,
⎢
⎥
cosh ( p + q ) = cosh r
⎣
⎦
7. If 6 sinh x + 2 cosh x + 7 = 0, find
tanh x.
3 15 ⎤
⎡
⎢⎣ Ans. : 5 , − 17 ⎥⎦
8. Find the value of tanh log
9. If tanh x =
x
,
2
y
, show that
2
(i) cosec (a – ib ) + cosec (a + ib )
cos a sinh b =
4x
x2 + y 2
(ii) cosec (a ib ) cosec (a + ib )
4iy
= 2
.
x + y2
11. If tan a = tan x tanh y and tan b =
cot x tanh y, prove that tan (a + b ) =
sinh 2y cosec 2x.
12. Prove that
1
sin 1 x = log ix + 1 − x 2 .
i
=
1
2
10. If sin a cosh b =
5.
2⎤
⎡
⎢⎣ Ans. : 3 ⎥⎦
1
, find cosh 2x.
2
4 5⎤
⎡
⎢⎣ Ans. : 3 , 3 ⎥⎦
)
(
⎡
eiu − e − iu ⎤
−1
Hint
:
Let
sin
=
,
=
sin
=
x
u
x
u
⎢
⎥
2i ⎦
⎣
13. Prove that
(
)
cos 1x = −i log x ± 1 − x 2 .
14. Prove that
(
)
sin 1ix = 2np + i log x + 1 + x 2 .
15. Prove that
x
sinh 1(tan x) = log tan ⎛⎜ + ⎞⎟ .
4
2
⎝
⎠
16. Prove that
⎛ tan 2q + tan 2f ⎞
tan −1 ⎜
⎝ tan 2q − tan 2f ⎟⎠
⎛ tan q − tan f ⎞
+ tan −1 ⎜
⎝ tan q + tan f ⎟⎠
−1
= tan (cot q coth f ).
17. Prove that
(i) tanh 1(cos q ) = cosh 1(cosec q ).
(ii) sinh 1(tan q ) = log (sec q + tan q ).
i
⎛ 3i ⎞
18. Prove that cosh −1 ⎜ ⎟ = log 2 + .
2
⎝4⎠
19. If cosh 1(x + iy) + cosh 1(x – iy) =
cosh 1a, prove that 2 (a – 1) x2 +
2 (a + 1) y2 = a2 – 1.
1.86
Engineering Mathematics
⎡ Hint : Let cosh −1 (x + iy ) = a + i b , ⎤
⎢
⎥
−1
⎢cosh (x − iy ) = a − i b , a + i b
⎥
⎢ + a − i b = cosh −1 a, cosh 2a = a,
⎥
⎢
⎥
⎢ 2x = (x + iy ) + (x − iy ) = 2 cosh a cos b , ⎥
⎢ x = cosh a cos b ,
⎥
⎢ 2iy = (x + iy ) − (x − iy )
⎥
⎢
⎥
⎢ = 2 i sinh a sin b , y = sinh a sin b ⎥
⎢ x2
⎥
y2
⎢
+
=1, convert in terms ⎥
2
2
⎢ cosh a sinh a
⎥
⎢⎣of cosh 2a and then in terms of a.
⎥⎦
20. If sinh 1(x + iy) + sinh 1(x – iy) =
sinh 1a, prove that 2 (x2 + y2) a 2 + 1
= a2 – 2x2 + 2y2.
21. Prove that
(i) sinh 1x = cosech
1
x
2x 1 x2
1
1
.
(ii) tanh 1x = cosh
1 x2
1
⎛ x +1⎞
(iii) coth 1x = log ⎜
⎟.
2
⎝ x −1 ⎠
.
1.12 SEPARATION INTO REAL
AND IMAGINARY PARTS
To separate real and imaginary parts of a complex number, following results are used:
(i) sin (x ± iy) = sin x cos iy ± cos x sin iy = sin x cosh y ± icos x sinh y
(ii) cos (x ± iy) = cos x cos iy ∓ sin x sin iy = cos x cosh y ∓ i sin x sinh y
2 sin ( x ± iy ) cos ( x ∓ iy )
(iii) tan ( x ± iy ) =
⋅
2 cos ( x ± iy ) cos ( x ∓ iy )
=
sin 2 x ± sin 2iy sin 2 x ± i sinh 2 y
=
cos 2 x + cos 2iy cos 2 x + cosh 2 y
(iv) sinh (x ± iy) = sinh x cosh iy ± cosh x sinh iy
= sinh x cos iiy ± cosh x ( i sin iiy)
= sinh x cos ( y) ± cosh x [ i sin ( y)]
= sinh x cos y ± i cosh x sin y
(v) cosh (x ± iy) = cosh x cosh iy ± sinh x sinh iy = cosh x cos y ± i sinh x sin y
2 sinh ( x ± iy ) cosh ( x ∓ iy )
(vi) tanh ( x ± iy ) =
⋅
2 cosh ( x ± iy ) cosh ( x ∓ iy )
sinh 2 x ± sinh 2iy sinh 2 x ± i sin 2 y
=
.
=
cosh 2 x + cosh 2iy cosh 2 x + cos 2 y
Example 1: Separate real and imaginary parts of
(i) cos-1 (ix)
(ii) sin-1 (eip)
(iii) sin-1 (cosec p ).
Solution: (i) Let
cos −1 (ix) = α + i β
cos (α + i β ) = ix
cos α cos i β − sin α sin i β = ix
cos α cosh β − i sin α sinh β = 0 + ix
1.87
Complex Numbers
Comparing real and imaginary parts on both the sides,
cos a cosh b = 0,
sin a sinh b = x
But
cosh b
[ 1
… (1)
… (2)
cosh b < ]
From Eq. (1),
cos a = 0 , a =
Putting a =
2
2
in Eq. (2),
sin
2
sinh b
x
sinh b = x
b = sinh 1 ( x) = sinh 1(x)
(
= − log x + x 2 + 1
cos 1 (ix) =
Hence,
2
(
)
)
i log x + x 2 + 1 .
(ii) Let sin 1 (eiq) = x + iy
eiq = sin (x + iy)
cos q + i sin q = sin x cos iy + cos x sin iy
sin x cosh y + i cos x sinh y
Comparing real and imaginary parts on both the sides,
cos q = sin x cosh y
sin q = cos x sinh y
Eliminating y from Eq. (1) and Eq. (2),
cos 2
sin 2
−
sin 2 x cos 2 x
cos 2 cos 2 x − sin 2 sin 2 x
1=
sin 2 x cos 2 x
2
2
2
sin x cos x = cos cos 2 x − (sin 2 )(1 − cos 2 x)
cosh 2 y − sinh 2 y =
(1 − cos 2 x) cos 2 x = (1 − sin 2 ) cos 2 x − sin 2 + sin 2 cos 2 x
cos 4 x = sin 2
cos 2 x = sin
cos x = ± sin
x = cos −1 ( ± sin
From Eq. (2),
sin 2
= cos 2 x sinh 2 y
)
… (1)
… (2)
1.88
Engineering Mathematics
cos 2 x = sin q ,
Putting
sin 2 q = sin q sinh 2 y
sinh 2 y = sin q
sinh y = ± sin q
(
y = sinh −1 ( ± sin q ) = log ± sin q + sin q + 1
(
sin −1 (ei ) = cos −1 ± sin
Hence,
) + i log ( ±
)
sin + sin + 1
)
(iii) Let sin 1 (cosecq ) = x + iy
cosec = sin ( x + iy )
cosec = sin x cos iy + cos x sin iy
= sin x cosh y + i cos x sinh y
Comparing real and imaginary parts on both the sides,
cosec q = sin x cosh y
0 = cos x sinh y
From Eq. (2),
cos x = 0, x =
2
Putting x =
in Eq. (1),
2
cosec q = sin
cosh y = cosh y
2
y = cosh 1 (cosec q )
(
)
= log cosecq + cosec 2q − 1 = log (cosecq + cot q )
q ⎞
⎛
2 cos 2
⎜
⎛ 1 + cos q ⎞
2 ⎟ = log ⎛ cot q ⎞
= log ⎜
= log ⎜
⎜⎝
⎟
q
q⎟
⎝ sin q ⎟⎠
2⎠
⎜⎝ 2 sin cos ⎟⎠
2
2
Hence,
sin −1 (cosec θ ) =
π
θ
+ i log cot
2
2
Example 2: If cos (` é ia) = x + iy, prove that
(i)
x2
cosh 2 β
Solution:
+
y2
sinh 2 β
=1
(ii)
x2
cos 2 α
−
y2
sin 2 α
= 1.
cos (a ± ib ) = x + iy
cos a cos ib ∓ sin a sin ib = x + iy
cos a cosh b ∓ i sin a sinh b = x + iy
… (1)
… (2)
Complex Numbers
1.89
Comparing real and imaginary parts on both the sides,
cos a cosh b = x
sin a sinh b = y
Eliminating a from Eqs. (1) and (2),
cos2 a + sin2 a =
x2
cosh 2
+
y2
sinh 2
x2
cosh 2
+
… (1)
… (2)
y2
sinh 2
= 1.
Eliminating b from Eqs. (1) and (2),
cosh2 b
x2
cos 2
x2
cos 2
sinh2 b
−
y2
sin 2
y2
sin 2
= 1.
Example 3: If cos (u + iv) = x + iy, show that
(i) (1 + x)2 + y2 = (cosh v + cos u)2
(ii) (1 - x)2 + y2 = (cosh v - cos u)2.
Solution:
cos (u + iv) = x + iy
cos u cos iv – sin u sin iv = x + iy
cos u cosh v – i sin u sinh v = x + iy
(i) Consider, 1 + x + iy = 1 + cos u cosh v – i sin u sinh v
Taking modulus on both the sides and squaring,
(1 + x)2 + y2 = (1 + cos u cosh v)2 + sin2 u sinh2 v
1 + 2 cos u cosh v + cos2 u cosh2 v + (1 – cos2 u) (cosh2 v – 1)
1 + 2 cos u cosh v + cos2 u cosh2 v + cosh2 v – 1 – cos2 u cosh2 v + cos2 u
(cosh v + cos u)2.
(ii) Consider, 1 – x – iy = 1 – cos u cosh v + i sin u sinh v
Taking modulus on both the sides and squaring,
(1 x)2 + y2 = 1 – 2 cos u cosh v + cos2 u cosh2 v + (1 – cos2 u) (cosh2 v – 1)
(cosh v – cos u)2 .
Example 4: If cos (x + iy) = cos ` + i sin `, prove that sin ` = ± sin2 x = ± sinh2 y.
Solution:
cos (x + iy) = cos a + i sin a
cos x cos iy sin x sin iy = cos a + i sin a
cos x cosh y i sin x sinh y = cos a + i sin a
Comparing real and imaginary parts on both the sides,
cos x cosh y = cos a
sin x sinh y = sin a
… (1)
… (2)
1.90
Engineering Mathematics
Eliminating y from Eqs. (1) and (2),
cosh2 y
sinh2 y =
1=
cos 2
cos 2 x
cos 2
sin 2
sin 2 x
sin 2 x sin 2 cos 2 x
sin 2 x cos 2 x
sin2 x cos2 x = (1 – sin2 a) sin2 x sin2 a (1 – sin2 x)
sin x (1 – sin2 x) = sin2 x sin2 a sin2 x sin2 a + sin2 a sin2 x
sin2 x – sin4 x = sin2 x sin2 a
sin4 x = sin2 a
… (3)
± sin2 x = sin a
2
sin a = ± sin2 x in Eq. (2),
Putting
sin x sinh y = ± sin2 x
sin2 x sinh2 y = sin4 x
sinh2 y = sin2 x
± sinh2 y = ± sin2 x
… (4)
From Eqs. (3) and (4),
sin a = ± sin2 x = ± sinh2 y .
Example 5: If sinh (p + ie ) = ei`, prove that
(i) sinh4 p = cos2 `
(ii) cos2 e = cos2 ` .
sinh (q + if) = cos a + i sin a
sinh q cosh if + cosh q sinh if = cos a + i sin a
sinh q cos i(if) + cosh q [–i sin i (if)] = cos a + i sin a
sinh q cos ( f) + cosh q [–i sin ( f)] = cos a + i sin a
sinh q cos f + i cosh q sin f = cos a + i sin a
Solution:
Comparing real and imaginary parts on both sides,
sinh q cos f = cos a
cosh q sin f = sin a
(i) Eliminating f from Eqs. (1) and (2),
cos 2 a
sin 2 a
cos 2 f + sin 2 f =
+
sinh 2 q cosh 2 q
cos 2 a
sin 2 a
1=
+
2
sinh q cosh 2 q
sinh2 q cosh2 q = cos2 a cosh2 q + sin2 a sinh2 q
sinh2 q (1 + sinh2 q ) = cos2 a (1 + sinh2 q ) + (1 – cos2 a) sinh2 q
sinh2 q + sinh4 q = cos2 a + cos2 a sinh2 q + sinh2 q – cos2 a sinh2 q
sinh4 q = cos2 a .
… (1)
… (2)
1.91
Complex Numbers
(ii) From Eq. (1),
sinh2 q cos2 f = cos2 a
cos 2 α cos 2 α
cos2 f =
=
cos α
sinh 2 θ
cos2 f = cos a .
[Using (i)]
Example 6: If sin (p + ie) = tan ` + i sec `, show that cos2p cosh 2e = 3.
sin (q + if) = tan a + i sec a
sin q cos if + cos q sin if = tan a + i sec a
sin q cosh f + i cos q sinh f = tan a + i sec a
Comparing real and imaginary parts on both the sides,
sin q cosh f = tan a
cos q sinh f = sec a
Eliminating a from Eqs. (1) and (2),
sec2 a tan2 a = cos2 q sinh2 f sin2 q cosh 2 f
Solution:
… (1)
… (2)
(1 + cos 2θ ) (cosh 2φ − 1) (1 − cos 2θ ) (1 + cosh 2φ )
−
2
2
2
2
4 = cosh 2f 1 + cos 2q cosh 2f cos 2q 1 cosh 2f + cos 2q + cos 2q cosh 2f
2 + 2 cos 2q cosh 2f
3 = cos 2q cosh 2f .
1=
Example 7: If sin (p + ie) = R (cos ` + i sin `), prove that
1
(i) R2 = (cosh 2e - cos 2p )
(ii) tan ` = tanh e cot p.
2
Solution:
sin (q + if) = R (cos a + i sin a)
sin q cosh f + i cos q sinh f R (cos a + i sin a)
Comparing real and imaginary parts on both the sides,
sin q cosh f = R cos a
cos q sinh f = R sin a
Squaring and adding Eq. (1) and Eq. (2),
R 2 = sin 2 θ cosh 2 φ + cos 2 θ sinh 2 φ
(1 − cos 2θ ) (1 + cosh 2φ ) (1 + cos 2θ ) (cosh 2φ − 1)
=
+
2
2
2
2
2 cosh 2φ − 2 cos 2θ 1
=
= (cosh 2φ − cos 2θ ).
4
2
Dividing Eq. (2) by Eq. (1),
tan α =
cos θ sinh φ
= cot θ tanh φ
sin θ cosh φ
… (1)
… (2)
1.92
Engineering Mathematics
u−1
, then show that the argument of u is p + e,
u+1
Example 8: If sin (x + iy) =
where tan p =
Solution:
cos x sinh y
1 + sin x cosh y
and tan e =
cos x sinh y
.
1 sin x cosh y
u − 1 sin ( x + iy )
=
u +1
1
Using componendo – dividendo,
u + 1 + u − 1 1 + sin ( x + iy )
=
u + 1 − u + 1 1 − sin ( x + iy )
nh y
1 + sin x cosh y + i cos x sin
u=
1 − sin x cosh y − i cos x sinh y
arg (u ) = arg (1 + sin x cosh y + i cos x sinh y ) − arg (1 − sin x cosh y − i cos x sinh y )
⎛ cos x sinh y ⎞
⎛ − cos x sinh y ⎞
= tan −1 ⎜
− tan −1 ⎜
⎟
⎝ 1 + sin x cosh y ⎠
⎝ 1 − sin x cosh y ⎟⎠
⎛ cos x sinh y ⎞
⎛ cos x sinh y ⎞
+ tan −1 ⎜
= q +f
= tan −1 ⎜
⎟
⎝ 1 + sin x cosh y ⎠
⎝ 1 − sin x cosh y ⎟⎠
where,
tan q =
cos x sinh y
cos x sinh y
and tan f =
1 + sin x cosh y
1 − sin x cosh y
(
)
p
ip
Example 9: If cos ⎛⎜ + ia ⎞⎟ cosh ⎛⎜ b + ⎞⎟ = 1 , then show that 2b = log 2 + 3 .
⎝4
⎠
⎝
4⎠
i ⎞
⎛
⎞
⎛
Solution:
cos ⎜ + ia ⎟ cosh ⎜ b + ⎟ = 1
4⎠
⎝4
⎠
⎝
i ⎞
⎛
⎞
⎛
cos ⎜ + ia ⎟ cos i ⎜ b + ⎟ = 1
4⎠
⎝4
⎠
⎝
⎛
⎞
⎛
⎞
cos ⎜ + ia ⎟ cos ⎜ ib − ⎟ = 1
4⎠
⎝4
⎠
⎝
⎛
⎞
⎛
⎞
2 cos ⎜ + ia ⎟ cos ⎜ ib − ⎟ = 2
4
4
⎝
⎠
⎝
⎠
⎛
⎞
⎛
⎞
cos ⎜ + ia + ib − ⎟ + cos ⎜ + ia − ib + ⎟ = 2
4⎠
4⎠
⎝4
⎝4
⎡
⎤
cos i (a + b) + cos ⎢ + i (a − b) ⎥ = 2
⎣2
⎦
cosh(a + b) − sin i (a − b) = 2
cosh(a + b) − i sinh(a − b) = 2 + i 0
Complex Numbers
1.93
Comparing real and imaginary parts on both the sides,
cosh(a + b) = 2, sinh( a − b) = 0
a + b = cosh −1 (2) = log ( 2 + 4 − 1)
(
a + b = log 2 + 3
and
)
... (1)
a − b = sinh −1 (0) = log ( 0 + 0 + 1 ) = log 1 = 0
a −b = 0
Subtracting Eq. (2) from Eq. (1),
(
2b = log 2 + 3
)
π⎞
⎛
Example 10: If u + iv = cosh ⎜ α + i ⎟ , find (u2 - v2).
4⎠
⎝
iπ
iπ
Solution: u + iv = cosh α cosh + sinh α sinh
4
4
iπ ⎞
iπ
⎛
u + iv = cosh α cos i ⋅ + sinh α ⎜ −i sin i ⋅ ⎟
4⎠
4
⎝
p
p
+ i sinh a sin
4
4
1
1
=
cosh a + i
sinh a
2
2
= cosh a cos
Comparing real and imaginary parts on both the sides,
cosh
sinh
u=
,v =
2
2
1
1
u 2 − v 2 = (cosh 2 − sinh 2 ) =
2
2
⎛π
⎞
Example 11: If cosec ⎜ + ix ⎟ = u + iv, prove that (u2 + v2)2 = 2 (u2 - v2).
⎝4
⎠
Solution:
⎛
⎞
cosec ⎜ + ix ⎟ = u + iv
⎝4
⎠
1
u − iv
⎛
⎞
sin ⎜ + ix ⎟ =
= 2
2
⎝4
⎠ u + iv u + v
u − iv
sin cos ix + cos sin ix = 2
4
4
u + v2
1
u
v
(cosh x + i sinh x) = 2
−i 2
2
u +v
u + v2
2
... (2)
1.94
Engineering Mathematics
Comparing real and imaginary parts on both the sides,
cosh x
u
= 2
u + v2
2
sinh x
v
=− 2
u + v2
2
cosh 2 x − sinh 2 x =
1=
2u 2
2v 2
− 2
2 2
(u + v ) (u + v 2 ) 2
2
2(u 2 − v 2 )
(u 2 + v 2 ) 2
(u 2 + v 2 ) 2 = 2(u 2 − v 2 )
Example 12: Prove that 2e2x = cosh 2v – cos 2u, where ez = sin (u + iv) and z = x + iy.
Solution:
ez = sin u cos iv + cos u sin iv
e = sin u cosh v + cos u ( i sinh v)
ex eiy = sin u cosh v + i cos u sinh v
ex (cos y + i sin y) = sin u cosh v + i cos u sinh v
x+iy
Comparing real and imaginary parts on both the sides,
ex cos y = sin u cosh v
ex sin y = cos u sinh v
Squaring and adding Eqs. (1) and (2),
e2x (cos2 y + sin2 y) = sin2 u cosh2 v + cos2 u sinh2 v
e2x = (1 – cos2 u) cosh2 v + cos2 u (cosh2 v – 1)
cosh2 v – cos2 u
1 + cosh 2v 1 + cos 2u
−
2
2
2e2x = cosh 2v – cos 2u.
=
Example 13: If log cos (x – iy) = ` + ia , then prove that
1
⎛ cosh 2 y + cos 2 x ⎞
α = log ⎜
⎟ and find a.
2
⎝
2
⎠
Solution:
cos (x – iy) = ea +ib
cos (x – iy) = ea eib
cos x cos iy + sin x sin iy = ea eib
cos x cosh y + i sin x sinh y = ea (cos b + i sin b )
Comparing real and imaginary parts on both the sides,
cos x cosh y = ea cos b
sin x sinh y = ea sin b
… (1)
… (2)
Complex Numbers
1.95
Squaring and adding Eqs. (1) and (2),
cos2x cosh2y + sin2x sinh2y = e2a (cos2b + sin2b )
(1 + cos 2 x) (1 + cosh 2 y ) (1 − cos 2 x) (cosh 2 y − 1)
+
= e2
2
2
2
2
2(cos 2 x + cosh 2 y )
= e2
4
⎛ cos 2 x + cosh 2 y ⎞
log ⎜
⎟=2
2
⎝
⎠
1
⎛ cosh 2 y + cos 2 x ⎞
= log ⎜
⎟.
⎝
⎠
2
2
Dividing Eq. (2) by Eq. (1),
tan b = tanh y tan x
b = tan 1 (tanh y tan x).
Example 14: If sin-1(` + ia ) = x + iy, prove that sin2 x and cosh2 y are the roots
of the equation k2 – (` 2 + a 2 + 1) k + ` 2 = 0.
Solution: sin 1 (a + ib ) = x + iy
a + ib = sin (x + iy)
sin x cos iy cos x sin iy
sin x cosh y i cos x sinh y
Comparing real and imaginary parts on both the sides,
a = sin x cosh y
b = cos x sinh y
Consider,
a 2 + b 2 + 1 = sin2 x cosh2 y + cos2 x sinh2 y + 1
sin2 x cosh2 y + (1 sin2 x) (cosh2 y 1) + 1
sin2 x cosh2 y + cosh2 y 1 sin2 x cosh2 y + sin2 x + 1
cosh2 y + sin2 x
Also,
a 2 = cosh2 y sin2 x
2
2
Then, l – (1 + a + b 2 ) l a 2 l2 (cosh2 y + sin2 x) l + cosh2 y sin2 x = 0
Comparing with l2 – (sum of roots) l + product of roots = 0,
we conclude that cosh2 y and sin2 x are the roots of the given equation.
Example 15: If cos (x + iy) cos (u + iv) = 1, where x, y, u, v are real, then show
that tanh2 y cosh2 v = sin2 u.
Solution: cos (x + iy) cos (u + iv) = 1
cos (x + iy) = sec (u + iv)
sin ( x + iy ) = 1 − cos 2 ( x + iy ) = 1 − sec 2 (u + iv) = − tan 2 (u + iv)
sin ( x + iy ) = i tan (u + iv)
Now,
tan ( x + iy ) =
sin ( x + iy ) i tan (u + iv)
=
= i sin (u + iv)
cos ( x + iy ) sec (u + iv)
1.96
Engineering Mathematics
tan ( x − iy ) = −i sin (u − iv)
tan 2iy = tan [ ( x + iy ) − ( x − iy ) ] =
Then,
=
tan ( x + iy ) − tan ( x − iy )
1 + tan ( x + iy ) tan ( x − iy )
i sin (u + iv) + i sin (u − iv)
=
1 − i 2 sin (u + iv) sin (u − iv)
i ( 2 sin u cos iv )
1
1 + (cos 2iv − cos 2u )
2
2i sin u cosh v
1
1 + (cosh 2v − cos 2u )
2
2 sin u cosh v
2 sin u cosh v
=
tanh 2 y =
2
2
2 cosh v − 1 − 1 + 2 sin u cosh 2 v + sin 2 u
1+
2
Dividing numerator and denominator by cosh2 v,
2 sin u
tanh 2 y = cosh 2v
sin u
1+
cosh 2 v
2 sin u
2 tanh y
2 tanh x ⎤
⎡
= cosh 2v
⎢⎣∵ tanh 2 x = 1 + tanh 2 x ⎥⎦
si n u
1 + tanh 2 y
1+
cosh 2 v
Comparing both the sides,
sin u
tanh y =
.
cosh v
Hence,
tanh 2 y cosh 2 v = sin 2 u.
i tanh 2 y =
Example 16: Separate into real and imaginary parts:
(i) tan (x + iy)
Solution: (i)
(ii) tan-1 (eip ).
sin ( x + iy ) 2 sin ( x + iy ) cos ( x − iy )
=
cos ( x + iy ) 2 cos ( x + iy ) cos ( x − iy )
sin 2 x + sin 2iy sin 2 x + i sinh 2 y
=
=
cos 2 x + cos 2iy cos 2 x + cosh 2 y
tan ( x + iy ) =
sin 2 x
cos 2 x + cosh 2 y
sinh 2 y
Imaginary part =
cos 2 x + cosh 2 y
Real part =
Complex Numbers
(ii) Let x + iy = tan 1 (eiq)
x iy = tan 1 (e iq)
Adding Eqs. (1) and (2),
1.97
… (1)
… (2)
2 x = tan −1 (ei q ) + tan −1 (e − i q ) = tan −1
ei q + e − i q
p
= tan −1 ∞ = np +
2
1 − ei q ⋅ e − i q
np p
+
2
4
Subtracting Eq. (2) from Eq. (1),
x=
ei − e − i
1 + ei ⋅ e − i
2i sin
tan 2iy =
2
i tanh 2 y = i sin
2iy = tan −1
2 y = taanh −1 (sin )
1
1 + sin q 1
= log
= log
2
1 − sin q 2
=
⎛p q ⎞
⎛p
⎞
2 cos 2 ⎜ − ⎟
1 + cos ⎜ − q ⎟
⎝ 4 2⎠
⎝2
⎠ 1
= log
⎛p q ⎞
⎛p
⎞ 2
2 sin 2 ⎜ − ⎟
1 − cos ⎜ − q ⎟
⎝ 4 2⎠
⎝2
⎠
1
⎛p q ⎞
⎛p q ⎞
⎛p q ⎞
log cot 2 ⎜ − ⎟ = log cot ⎜ − ⎟ = − log tan ⎜ − ⎟
⎝ 4 2⎠
⎝ 4 2⎠
⎝ 4 2⎠
2
1
⎛p q ⎞
y = − log tan ⎜ − ⎟
⎝ 4 2⎠
2
Hence,
1 ⎞π i
⎛
⎛π θ ⎞
tan −1 (eiθ ) = ⎜ n + ⎟ − log tan ⎜ − ⎟ .
2
2
2
⎝
⎠
⎝ 4 2⎠
Example 17: If tan (` + ia ) = x + iy, prove that
(i) x2 + y2 + 2x cot 2` = 1
(ii) x2 + y2 - 2y coth 2a = -1.
Solution:
tan (a + ib ) = x + iy
a + ib = tan 1 (x + iy)
a ib = tan 1 (x iy)
(i) Adding Eqs. (1) and (2),
2a = tan −1 ( x + iy ) + tan −1 ( x − iy )
2x
x + iy + x − iy
= tan −1
= tan −1
1 − ( x + iy )( x − iy )
1 − x2 − y 2
2x
tan 2a =
1 − x2 − y 2
… (1)
… (2)
1.98
Engineering Mathematics
1 − x 2 − y 2 = 2 x cot 2
x 2 + y 2 + 2 x cot 2 = 1
(ii) Subtracting Eq. (2) from Eq. (1),
2i b = tan −1 ( x + iy ) − tan −1 ( x − iy ) = tan −1
x + iy − x + iy
1 + ( x + iy )( x − iy )
2iy
1 + x2 + y 2
2iy
i tanh 2 b =
1 + x2 + y 2
n 2i b =
tan
1 + x 2 + y 2 = 2 y coth 2 b
x 2 + y 2 − 2 y coth 2 b = −1.
Example 18: If tan (x + iy) = ` + ia, show that
1−α2 − β 2
1+α + β
2
2
=
cos 2 x
.
cosh 2 y
Solution:
sin ( x + iy ) cos ( x − iy )
⋅
cos ( x + iy ) cos ( x − iy )
sin 2 x + sin 2iy sin 2 x + i sinh 2 y
=
=
cos 2 x + cos 2iy cos 2 x + cosh 2 y
a + i b = tan ( x + iy ) =
a2 +b2 =
sin 2 2 x + sinh 2 2 y
(cos 2 x + cosh 2 y ) 2
Using componendo – dividendo,
1 − a 2 − b 2 (cos 2 x + cosh 2 y )2 − sin 2 2 x − sinh 2 2 y
=
1 + a 2 + b 2 (cos 2 x + cosh 2 y)) 2 + sin 2 2 x + sinh 2 2 y
cos 2 2 x − sin 2 2 x + 1 + 2 cos 2 x cosh 2 y
1 + cosh 2 2 y + sinh 2 2 y + 2 cos 2 x cosh 2 y
2 cos 2 x (cos 2 x + cosh 2 y )
cos 2 x
=
=
2 cosh 2 y (cosh 2 y + cos 2 x) cosh 2 y
=
2x
π
Example 19: If tan ⎛⎜ + iα ⎞⎟ = x + iy, prove that x 2 + y 2 +
=1.
3
⎝6
⎠
Solution:
⎛π
⎞
tan ⎜ + iα ⎟ = x + iy
⎝6
⎠
⎛π
⎞
−1
⎜ + iα ⎟ = tan ( x + iy )
⎝6
⎠
... (1)
1.99
Complex Numbers
⎛π
⎞
−1
⎜ − iα ⎟ = tan ( x − iy )
⎝6
⎠
... (2)
Adding Eqs. (1) and (2),
2p
x + iy + x − iy
= tan −1 ( x + iy ) + tan −1 ( x − iy ) = tan −1
6
1 − ( x + iy )( x − iy )
tan
2x
3 1 − x2 − y 2
2x
3=
1 − x2 − y 2
=
1 − x2 − y 2 =
x +y +
2
2
2x
3
2x
3
= 1⋅
iπ ⎞
⎛
Example 20: If tanh ⎜ α +
= x + iy, prove that x2 + y2 + 2y = 1.
8 ⎟⎠
⎝
Solution:
iπ
tanh ⎛⎜ α + ⎞⎟ = x + iy, tanh
8 ⎠
⎝
iπ ⎞
⎛
⎜α − ⎟ = x
8 ⎠
⎝
iy
ip ⎞
ip ⎞
⎛
⎛
tanh ⎜ a + ⎟ − tanh ⎜ a − ⎟
⎝
⎝
⎡⎛
ip ⎞ ⎛
ip ⎞ ⎤
8⎠
8⎠
tanh ⎢⎜ a + ⎟ − ⎜ a − ⎟ ⎥ =
⎝
⎠
⎝
⎠
i
p
i
8
8
⎣
⎦ 1 − tanh ⎛ a + ⎞ tanh ⎛ a − p ⎞
⎜⎝
⎟
⎜⎝
⎟
8⎠
8⎠
tanh
ip
( x + iy ) − ( x − iy )
=
4 1 − ( x + iy )( x − iy )
−i tan i ⋅
2iy
i
=
4 1 − x2 − y 2
2iy
⎛− ⎞
−i tan ⎜
⎟=
2
2
⎝ 4 ⎠ 1− x − y
2y
tan =
4 1 − x2 − y 2
2y
1=
1 − x2 − y 2
x2 + y 2 + 2 y = 1
1.100
Engineering Mathematics
⎛π
⎞
Example 21: If tan ⎜ + iy ⎟ = reip, show that r = 1, tan p = sinh 2y and
4
⎝
⎠
⎛θ ⎞
tanh y = tan ⎜ ⎟ .
⎝2⎠
⎛π
⎞
tan ⎜ + iy ⎟ = reiθ
⎝4
⎠
Solution:
p
+ tan iy
4
= reiq
p
1 − tan ⋅ tan iy
4
1 + i tanh y
= reiq
1 − i tanh y
tan
r=
where,
1 + tanh 2 y
1 + i tanh y
=
=1
1 − i tanh y
1 + tanh 2 y
⎛ 1 + i tanh y ⎞
q = arg ⎜
⎟ = arg (1 + i tanh y ) − arg (1 − i tanh y )
⎝ 1 − i tanh y ⎠
= tan
n −1 (tanh y ) − tan −1 (− tanh y )
and
= tan −1 (tanh y) + tan −1 (tanh y )
⎛ tanh y + tanh y ⎞
= tan −1 ⎜
⎟
⎝ 1 − tanh y tanh y ⎠
2 tanh y
= sinh 2 y.
tan q =
1 − tanh 2 y
2 tan
2 = 2 tanh y
1 − tanh 2 y
1 − tan 2
2
Comparing both the sides,
tanh y = tan
2
Example 22: If tan (x + iy) = i, where x and y are real, then show that x is indeterminate and y is infinite.
Solution:
tan (x + iy) = i
x + iy = tan–1 (i)
x iy = tan–1 ( i)
… (1)
… (2)
Complex Numbers
1.101
Adding Eqs. (1) and (2),
2x = tan–1 i + tan–1 ( i) = tan −1
= tan–1
i + ( −i )
i −i
= tan −1
1 − i ( −i )
1 + i2
= indeterminate
Subtracting Eq. (2) from Eq. (1),
2iy = tan–1 i tan–1 ( i) = tan–1 i + tan–1 i
2tan–1 i
iy = tan–1 i
tan iy = i
i tanh y = i
tanh y = 1
1
1+1
y = tanh–1 (1) = log
2
1−1
=
1
log
2
=
iπ
⎛
Example 23: If ` + ia = tanh ⎜ x +
4
⎝
Solution:
⎞
2
2
⎟ , prove that ` + a = 1.
⎠
i
tanh x + tanh
i ⎞
⎛
4
a + ib = tanh ⎜ x + ⎟ =
4 ⎠ 1 + tanh x tanh i
⎝
4
ip ⎞
⎛
p
tanh x + ⎜ −i tan i ⎟
tanh x + i tan
⎝
⎠
4
4 = tanh x + i
=
=
p 1 + i tanh x
ip ⎞
⎛
+ i tanh x tan
1 + tanh x ⎜ −i tan i ⎟ 1+
⎝
⎠
4
4
|a + i b | =
tanh 2 x + 1
tanh x + i
tanh x + i
=
=1
=
1 + i tanh x 1 + i tanh x
1 + tanh 2 x
|a + ib |2 = 1
a2+b2=1
Example 24: If x + iy = c cot (u + iv), prove that
x
c
−y
=
=
.
sin 2u sinh 2v cosh 2v − cos 2u
Solution:
x + iy = c cot (u + iv)
x + iy cos (u + iv) 2 sin (u − iv) sin 2u − sin 2iv sin 2u − i sinh 2v
=
⋅
=
=
c
sin (u + iv) 2 sin (u − iv) cos 2iv − cos 2u cosh 2v − cos 2u
1.102
Engineering Mathematics
Comparing real and imaginary parts on both the sides,
x
y
− sinh 2v
sin 2u
=
, =
c cosh 2v − cos 2u c cosh 2v − cos 2u
c
x
−y
=
.
=
sin 2u sinh 2v cosh 2v − cos 2u
Example 25: If
x + iy − c
c sinh u
c sin v
= e u+iv, show that x =
, y=
.
x + iy + c
cosh u cos v
cosh u cos v
x + iy − c
= eu+iv
x + iy + c
Solution:
Applying componendo–dividendo,
2( x + iy ) (1 + eu + iv ) (1 − eu −iv )
=
2c
(1 − eu + iv ) (1 − eu −iv )
[Multiplying and dividing by conjugate of denominator]
x + iy (1 − eu − iv + eu + iv − e 2u ) . e − u
=
c
(1 − eu − iv − eu + iv + e 2u ) e − u
e − u − e − iv + eiv − eu (e − u − eu ) + (eiv − e − iv )
=
e − u − e − iv − eiv + eu (e − u + eu ) − (eiv + e − iv )
−2 sinh u + 2i sin v
=
2 cosh u − 2 cos v
x
y
− sinh u
sin v
+i =
+i
c
c cosh u − cos v cosh u − cos v
Comparing real and imaginary parts on both the sides,
c sinh u
c sin v
x=−
and y =
cosh u − cos v
cosh u − cos v
=
Example 26: If tan (u + iv) = x + iy, find u and v in terms of x and y and show that
u = constant and v = constant are family of circles which are mutually orthogonal.
u + iv = tan 1 (x + iy)
u – iv = tan 1 (x – iy)
Solution:
Adding Eqs. (1) and (2),
2u = tan 1 (x + iy) + tan 1 (x – iy)
2x
2x
= tan −1
1 − ( x + iy ) ( x − iy )
1 − x2 − y 2
2x
tan 2u =
1 − x2 − y 2
= tan −1
… (1)
… (2)
1.103
Complex Numbers
1 x 2 y 2 = 2x cot 2u
x + y + 2x cot 2u – 1 = 0
2
2
If u = constant,
then cot 2u = constant = k1, say
x 2 + y 2 + 2k1 x – 1 = 0
… (3)
which represents family of circles.
Subtracting Eq. (2) from Eq. (1),
2iv = tan 1 (x + iy) – tan 1 (x – iy) = tan −1
x + iy − x + iy
1 + ( x + iy ) ( x − iy )
2iy
1 + x2 + y 2
2iy
i tanh 2v =
1 + x2 + y 2
tan 2iv =
x 2 + y 2 – 2y coth 2v + 1 = 0
If v = constant,
then
tanh 2v = constant
coth 2v = constant = k2, say
x 2 + y 2 2k2 y + 1 = 0
which represents family of circles.
… (4)
Differentiating Eq. (3) w.r.t. x,
2x + 2y
dy
+ 2k1 = 0
dx
(k + x)
dy
=− 1
dx
y
This is the slope of Eq. (3).
Let
dy
= m1 =
dx
(k1 + x)
y
Differentiating Eq. (4) w.r.t. x,
2x + 2y
dy
dx
2k2
(y – k2)
dy
=0
dx
dy
= x
dx
dy
dx
x
k2
y
1.104
Engineering Mathematics
This is the slope of Eq. (4).
Let
dy
x
= m2 =
dx
k2 y
m1 m2 =
Now,
(k1 + x)
x
·
(k2 y )
y
… (5)
Adding Eqs. (3) and (4),
Substituting
2x2 + 2y2 + 2 (k1 x k2 y) = 0
x2 + k1 x = k2 y – y2
x (x + k1) = y (–y + k2)
x (x + k1) = y (k2 – y) in Eq. (5),
y (k2 y )
m 1 m2 =
= –1
y (k2 y )
This shows that family of circles represented by Eqs. (3) and (4) are orthogonal.
Example 27: If tan (x + iy) = sin (u + iv), show that
Solution: tan ( x + iy ) =
=
sin 2 x
tan u
=
.
sinh 2 y tanh v
sin ( x + iy ) 2 sin ( x + iy ) cos ( x − iy )
=
cos ( x + iy ) 2 cos ( x + iy ) cos ( x − iy )
sin 2 x + sin 2iy sin 2 x + i sinh 2 y
=
cos 2 x + cos 2iy cos 2 x + cosh 2 y
sin 2 x + i sinh 2 y
cos 2 x + cosh 2 y
tan ( x + iy ) = sin (u + iv)
tan ( x + iy ) =
Now,
sin 2 x + i sinh 2 y
= sin u cosh v + i cos u sinh v
cos 2 x + cosh 2 y
Comparing real and imaginary parts on both the sides,
sin 2 x
= sin u cosh v
cos 2 x + cosh 2 y
and
sinh 2 y
= cos u sinh v
cos 2 x + cosh 2 y
Dividing Eq. (1) by Eq. (2),
sin 2 x
tan u
=
sinh 2 y tanh v
... (1)
... (2)
1.105
Complex Numbers
Example 28: If
where A =
Solution:
Let r = Aeiq,
1
1
1
=
+ cpi + , where L, p, R are real, prove that q = Ae ip,
r Lpi
R
⎛ 1
⎞
and tanq = R ⎜⎜⎜
- cp⎟⎟⎟ .
⎟⎠
⎝ Lp
⎞⎟
1 ⎛⎜ 1
⎟
+
cp
⎜
⎟⎟⎠
R 2 ⎜⎝ Lp
1
2
1
1
1
i
1
=
+ cpi + = −
+ cpi +
r Lpi
R
Lp
R
1
=
1
e
A
iq
Here,
⎛1⎞
1
1
, q = arg ⎜ ⎟
=
A
⎝ ⎠
Now,
1 1 ⎛
1⎞
= + i ⎜ cp − ⎟
r R ⎝
Lp ⎠
Hence,
1
1
=
A r
2
=
A=
and
1 ⎛
1⎞
+ cp − ⎟ =
Lp ⎠
R 2 ⎜⎝
⎞
1 ⎛ 1
+
− cp⎟
⎠
R 2 ⎜⎝ Lp
1
⎞
1 ⎛ 1
+⎜
− cp⎟
2
⎠
R ⎝ Lp
2
⎛
1⎞
⎜⎝ cp − Lp ⎟⎠
⎛ 1⎞
−q = arg ⎜ ⎟ = tan −1
⎝ r⎠
⎛ 1⎞
⎜⎝ ⎟⎠
R
⎛
1 ⎞
tan (− ) = R ⎜ cp −
⎟
Lp ⎠
⎝
⎛
1 ⎞
− tan = R ⎜ cp −
⎟
Lp ⎠
⎝
⎛ 1
⎞
tan = R ⎜
− cp ⎟
⎝ Lp
⎠
2
1.106
Engineering Mathematics
Exercise 1.7
i
1. Separate into real and imaginary
parts:
(i) cot (x + iy)
(ii) sec (x + iy)
(iii) cosec (x + iy)
(iv) (sin q + i cos q )i.
sin 2 x − i sinh 2 y
⎡
⎤
⎢ Ans. : (i) cosh 2 y − cos 2 x
⎥
⎢
⎥
2 (cos x cosh y − i sin x sinh y ) ⎥
⎢
(ii)
⎢
⎥
cos 2 x + cosh 2 y
⎢
⎥
⎢
2 (sin x cosh y − i cos x sinh y ) ⎥
(iii)
⎢
⎥
cosh 2 y − cos 2 x
⎢
⎥
p
⎢
⎥
q−
2
(iv)) e
⎢⎣
⎥⎦
2. If sin (a + ib ) = x + iy, prove that
x2
y2
(i)
+
=1
cosh 2
sinh 2
(ii)
x2
sin 2
−
y2
cos 2
= 1.
3. If sin (a + ib ) = x + iy, prove that
(i) x2 sech2 b + y2 cosech2 b = 1
(ii) x2 cosec2 a y2 sec2 a = 1.
4. If sin (q + if) = p (cos a + i sin a),
prove that
1
p2 = (cosh 2f cos 2q ),
2
tan a = tanh f cot q .
5. If sin (q + if) = cos a + i sin a,
prove that
(i) cos4 q = sin2 a = sinh4f
1
cos( x − )
(ii) f = log
.
2
cos( x + )
6. If cosh (q + if) = eia, prove that
sin2 a = sin4f = sinh4 q.
7. If sinh (q + if) = x + iy, prove that
x2 cosech2 q + y2 sech2 q = 1 and
y2 cosec2 f x2 sec2 f = 1.
8. If sinh (q + if) = e 3 , prove that
(i) 3 cos2 y – sin2 y = 4 sin2 y cos2 y
(ii) 3 sinh2 x + cosh2 x
= 4 sinh2 x cosh2 x.
9. If cos (x + iy) cos (u + iv) = 1, where
x, y, u, v are real, then prove that
tanh2 v cosh2 y = sin2 x.
10. If x + iy = c sin (u + iv), prove that
u = constant represents a family of
confocal hyperbolas and v = constant represents a family of confocal
ellipses.
Hint
: Separate real and imaginary ⎤
⎡
⎢ parts and then consider
⎥
⎢
⎥
2
2
⎢cosh v − sinh v = 1,
⎥
⎢ x2
⎥
y2
=1, family of ⎥
− 2
⎢ 2
2
2
⎢ c sin u c cos u
⎥
⎢ confocal hyperbolas, since u is
⎥
⎢
⎥
2
2
⎢constant and sin u + cos u = 1, ⎥
⎢
⎥
x2
y2
+ 2
= 1,
⎢ 2
⎥
2
2
⎢ c cosh u c sinh u
⎥
⎢family of confocal ellipses, since ⎥
⎢u is constant.
⎥
⎣
⎦
11. If tan y = tan a tanh b and
tan z = cot a tanh b, prove that
tan (y + z) = sinh 2b cosec 2a.
12. Prove that
u + iv ⎞ sin u + i sinh v
tan ⎛⎜
.
⎟=
⎝ 2 ⎠ cos u + cosh v
13. If A + iB = C tan (x + iy) , prove that
2CA
tan 2 x = 2
.
C − A2 − B 2
14. If tan (x + iy) = eiq, prove that
nπ π
+
(i) θ =
2 4
1
⎛π θ ⎞
(ii) y = log tan ⎜ + ⎟ .
2
⎝ 4 2⎠
1.107
Complex Numbers
15. If tan (q + if) = tan a + i sec a, then
prove that
(i) e2f = cot
2
(ii) 2q = np +
2
+a.
⎡ Hint : tan (q − if ) = tana − i seca , ⎤
⎢
⎥
⎢ tan 2q = tan [(q + if ) + (q − if ) ]
⎥
⎢and tan 2if = tan (q + if ) − (q − if ) ⎥
[
]⎦
⎣
16. Find z satisfying the equation
1
tan z = (1 – i).
2
Hint
:
putting z = x + iy,
⎡
⎤
⎢
⎥
1
⎢ tan ( x + iy ) = (1 − i ), tan ( x − iy ) ⎥
2
⎢
⎥
⎢ 1
⎥
⎢ = (1 + i ) tan 2 x = tan[( x + iy ) ⎥
⎢ 2
⎥
⎢ + ( x − iy )], tan 2iy = tan[( x + iy ) ⎥
⎢
⎥
− ( x − iy )] ⎥
⎢
⎢⎣
⎥⎦
np 1
i
⎤
⎡
−1
⎢⎣ Ans.: z = 2 + 2 tan 2 − 4 log 5⎥⎦
17. Prove that all solutions of the equation
sin z = 2i cos z are given by
n
i
z=
+ log 3.
2 2
[Hint : tan z = 2i]
18. Prove that
⎛ u + iv ⎞ sin u + i sinh v
tan ⎜
.
⎟=
⎝ 2 ⎠ cos u + cosh v
19. Prove that one value of
⎛ x + iy ⎞
tan 1 ⎜
⎟ is
⎝ x − iy ⎠
⎛ x+ y⎞
+ i log ⎜
⎟ , where x > y > 0.
4
⎝ x− y⎠
π
20. If cot ⎛⎜ + iα ⎞⎟ = x + iy, prove that
⎝6
⎠
x2 + y2
2x
3
= 1.
π
21. If cot ⎛⎜ + iα ⎞⎟ = x + iy, prove that
⎝8
⎠
2
2
x + y – 2x = 1.
iπ ⎞
⎛
22. If tanh ⎜ α + ⎟ = x + iy,
6⎠
⎝
prove that x 2 + y 2 +
2y
3
= 1.
23. If tanh (a + ib ) = x + iy, prove that
(i) x2 + y2 – 2x coth 2a = 1
(ii) x2 + y2 + 2y cot 2b = 1.
⎡ Hint : tanh (a − i b ) = x − iy, ⎤
⎢ tanh 2a = tanh [(a + i b )
⎥
⎢
⎥
⎢
⎥
+ (a − i b )]
⎢
⎥
⎢ tanh 2i b = tanh [(a + i b )
⎥
⎢
⎥
− (a − i b )],
⎢
⎥
⎢ tanh 2i b = − i tan i (2i b )
⎥
⎢
⎥
= − i tan ( − 2b )
⎢
⎥
= i tan 2b
⎢⎣
⎥⎦
24. If cot (a + ib ) = x + iy, prove that
(i) x2 + y2 – 2x cot 2a = 1
(ii) x2 + y2 + 2y coth 2b + 1 = 0.
25. Separate real and imaginary parts of
cos 1 (eiq ).
⎡ Ans. : sin −1 sin q +
⎢
⎢
i log 1 + sin q − sin q
⎣
(
26. Prove that
(
)
⎤
⎥
⎥
⎦
)
2
sin 1 (ix) = i log x + x + 1 + 2n .
27. Separate into real and imaginary
parts
⎛ 5i ⎞
(ii) cos −1 ⎜ ⎟
(i) cos 1 (i)
⎝ 12 ⎠
3i
(iii) sin −1 ⎛⎜ ⎞⎟ (iv) sinh 1 (ix)
⎝4⎠
1.108
Engineering Mathematics
(v) tanh 1 (i).
p
⎡
(
)⎤
⎢ Ans. : (i) 2 + i log 2 − 1 ⎥
⎥
⎢
2
p
⎥
⎢
(ii) + i log
⎥
⎢
3
2
⎥
⎢
(iii) i log 2
⎥
⎢
⎥
⎢
p
ip
−1
(iv) cosh x +
⎥
⎢
2
⎥
⎢
ip
⎥
⎢
(v)
⎥⎥⎦
⎢⎢⎣
4
28. Prove that sin (cosec q) =
29. If log cos (x – iy) = a + ib, then prove
that
1
⎛ cosh 2 y + cos 2 x ⎞
= log ⎜
⎟,
2
2
⎝
⎠
tan b = – tan x tanh y.
30. If log sin (x + iy) = a + ib, then prove
that
=
1
⎛ cosh 2 y − cos 2 x ⎞
log ⎜
⎟,
2
2
⎝
⎠
tan b = cot x tanh y.
1
2
+ i log cot
.
2
[Hint : Let sin (cosec q ) = a + ib ]
1.13 LOGARITHM OF A COMPLEX NUMBER
If z and w are two complex numbers and z = ew, then w = log z is called logarithm of
the complex number z.
Let z = x + iy
= reiq
y
where r = z = x 2 + y 2 and q = arg (z) = tan 1
x
log z = log (r eiq) = log r + log eiq
= log r + iq log e = log r + iq
= log x 2 + y 2 + i tan −1 y
x
1
y
log ( x 2 + y 2 ) + i tan −1 x
2
This is called principal value of log (x + iy).
The general value of log (x + iy) is given as
log (x + iy) = log r + i (2np + q )
1
y⎞
⎛
= log ( x 2 + y 2 ) + i ⎜ 2n + tan −1 ⎟ .
2
x⎠
⎝
Hence,
log ( x + iy ) =
Example 1: Find the value of
(i) log i
(ii) log-3 (-2 )
(iii) log (-5)
(iv) log (1 + i).
Complex Numbers
1.109
Solution: (i) log i = 1 log 1 + i tan −1 1 = 0 + i = i
2
0
2
2
1
⎛ 0 ⎞
log 4 + i tan −1 ⎜ ⎟
⎝ −2 ⎠ = log 2 + i
(ii) log −3 (−2) = log e (−2) = 2
0 ⎞ log 3 + i
log e (−3) 1
⎛
log 9 + i tan −1 ⎜ ⎟
2
⎝ −3 ⎠
(iii) log ( −5 ) =
1
⎛ 0 ⎞
log 25 + i tan −1 ⎜ ⎟ = log 5 + i
2
⎝ −5 ⎠
(iv) log (1 + i ) =
1
i
⎛1⎞ 1
log 2 + i tan −1 ⎜ ⎟ = log 2 +
2
4
⎝1⎠ 2
Example 2: Prove that log i i =
4n + 1
, where n, m are integers.
4m + 1
p⎞
1
⎛
log 1 + i ⎜ 2np + ⎟
⎝
log i
2
2⎠
Solution: log i i =
=
p⎞
log i 1
⎛
log 1 + i ⎜ 2mp + ⎟
⎝
2
2⎠
=
i ( 4n + 1) p
i ( 4m + 1) p
=
4n + 1
4m + 1
Example 3: Prove that log (1 + cos 2p + i sin 2p) = log (2 cos p) + ip.
Solution:
log (1 + cos 2q + i sin 2q ) =
1
⎛ sin 2q ⎞
2
log ⎡⎣(1 + cos 2q ) + sin 2 2q ⎤⎦ + i tan −1 ⎜
⎝ 1 + cos 2q ⎟⎠
2
1
⎛ 2 sin q cos q ⎞
log(4 cos 4 q + 4 sin 2 q cos 2 q ) + i tan −1 ⎜
⎝ 2 cos 2 q ⎟⎠
2
1
= log ⎡⎣ 4 cos 2 q (cos 2 q + sin 2 q ) ⎤⎦ + i tan −1 (tan q )
2
1
= log (4 cos 2 q ) + iq
2
= log (2 cos q ) + iq ⋅
=
Example 4: Simplify log (ei` + eia).
Solution: log (eia + eib ) = log [cos a i sin a ) + (cos b + i sin b )]
log [(cos a + cos b ) + i (sin a + sin b )]
1.110
Engineering Mathematics
⎡
⎛ a + b ⎟⎞ ⎛ a − b ⎞⎟⎤
⎛ a + b ⎞⎟ ⎛ a − b ⎞⎟
⎜⎜
⎜⎜
⎥
cos ⎜
i
cos
+
2
sin
= log ⎢ 2 cos ⎜⎜
⎟
⎟
⎢
⎜⎝ 2 ⎟⎟⎠ ⎜⎜⎝ 2 ⎟⎟⎠⎥
⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠
⎣
⎦
⎡
⎤
⎛ a − b ⎞⎟ ⎛ a + b ⎞⎟
⎛ a + b ⎞⎟
⎥
cos ⎜
= log ⎢ 2 cos ⎜⎜
+ i sin ⎜⎜
⎜⎝ 2 ⎟⎟⎠ ⎜⎜⎝ 2 ⎟⎟⎠
⎢
⎜⎝ 2 ⎟⎟⎠⎥
⎣
⎦
⎡ ⎛a + b ⎞
⎤
⎛ a − b ⎞⎟
⎟⎟ + i sin ⎛⎜⎜ a + b ⎞⎟⎟⎥
⎢ cos ⎜⎜
log
= log 2 cos ⎜⎜
+
⎟
⎢ ⎜⎝ 2 ⎟⎠
⎜⎝ 2 ⎟⎠⎥
⎝⎜ 2 ⎟⎠
⎣
⎦
⎛ a + b ⎞⎟
⎟
2 ⎟⎠
i ⎜⎜
⎛ a − b ⎞⎟
⎜⎝
+
log
e
= log 2 cos ⎜⎜
⎟
⎝⎜ 2 ⎟⎠
⎛ a − b ⎞⎟ ⎛ a + b ⎞⎟
+ i⎜
= log 2 cos ⎜⎜
⎜⎝ 2 ⎟⎟⎠ ⎜⎜⎝ 2 ⎟⎟⎠
[ log e = 1]
⎛ x−i ⎞
-1
Example 5: Prove that i log ⎜
⎟ = o - 2 tan x.
x
+
i
⎝
⎠
Solution:
⎛ x −i⎞
i log ⎜
= i [ log( x − i ) − log( x + i ) ]
⎝ x + i ⎟⎠
⎡1
1⎤
⎛ −1⎞ 1
= i ⎢ log( x 2 + 1) + i tan −1 ⎜ ⎟ − log ( x 2 + 1) − tan −1 ⎥
⎝ x⎠ 2
x⎦
⎣2
−
1
−
1
−
1
= i ⎡⎣i ( − cot x − cot x) ⎤⎦ = 2 cot x
⎛p
⎞
= 2 ⎜ − tan −1 x⎟ = p − 2 tan −1 x
⎝2
⎠
Example 6: Simplify tanh-1 (x + iy).
Solution:
tanh −1 ( x + iy ) =
1
⎛ 1 + x + iy ⎞ 1
log ⎜
= [log (1 + x + iy ) − log (1 − x − iy ) ]
2
⎝ 1 − x − iy ⎟⎠ 2
{
}
{
1 ⎡1
y
1
log (1 + x) 2 + y 2 + i tan −1
− log (1 − x) 2 + y 2
2 ⎢⎣ 2
1+ x 2
−y ⎤
−i tan −1
1 − x ⎥⎦
⎤
(1 + x) 2 + y 2
1 ⎡1
⎛ −1 y
−1 y ⎞
= ⎢ log
+
+
i
tan
tan
⎥
⎜
⎟
⎝
2 ⎣⎢ 2
1+ x
1 − x ⎠ ⎦⎥
(1 − x) 2 + y 2
⎡∵ tan −1 ( − x) = − tan −1 x ⎤
⎣
⎦
=
}
Complex Numbers
1.111
⎡
y
y ⎞⎤
⎛
⎢
⎥
+
2
2
1 1
(1 + x) + y
⎜ −1 1 + x 1 − x ⎟ ⎥
= ⎢ log
+
i
tan
⎜
⎟
2 ⎢2
y2 ⎟ ⎥
(1 − x) 2 + y 2
⎜⎜
1−
⎢
⎟⎥
⎝
⎢⎣
1 − x 2 ⎠ ⎥⎦
⎤
1 ⎡1
2y
(1 + x) 2 + y 2
= ⎢ log
+ i tan −1
2
2
2
2⎥
2 ⎣2
1− x − y ⎦
(1 − x) + y
p ⎞ ⎛o p ⎞
⎛ 1 ⎞
⎛1
Example 7: Prove that log ⎜
= log ⎜ cosec ⎟ + i ⎜ − ⎟ .
ip ⎟
⎝1− e ⎠
⎝2
2⎠ ⎝ 2 2⎠
Solution:
⎛ 1 ⎞
log ⎜
= log (1 − eiq ) −1 = − log(1 − eiq )
⎝ 1 − eiq ⎟⎠
q
q
q⎞
⎛
= − log (1 − cos q − i sin q ) = − log ⎜ 2 sin 2 − 2i sin cos ⎟
⎝
2
2
2⎠
⎡ ⎛
q⎞⎛ q
q ⎞⎤
= − ⎢log ⎜ 2 sin ⎟ ⎜ sin − i cos ⎟ ⎥
⎝
⎠
⎝
2
2
2⎠⎦
⎣
⎡ ⎛p q ⎞
q⎞
⎛p q ⎞⎤
⎛
= − log ⎜ 2 sin ⎟ − log ⎢cos ⎜ − ⎟ − i sin ⎜ − ⎟ ⎥
⎝
⎠
⎝ 2 2⎠⎦
⎝
⎠
2 2
2
⎣
q⎞
⎛
= log ⎜ 2 sin ⎟
⎝
2⎠
−1
− log e
⎛p q ⎞
−i⎜ − ⎟
⎝ 2 2⎠
q ⎞ ⎛p q ⎞
⎛1
= log ⎜ cosec ⎟ + i ⎜ − ⎟
⎝2
2⎠ ⎝ 2 2⎠
⎛ π ix ⎞
Example 8: Prove that log tan ⎜ + ⎟ = i tan-1 (sinh x).
⎝4 2⎠
ix
⎛
⎜ 1 + tan 2
ix ⎞
⎛
Solution: log tan ⎜ + ⎟ = log ⎜
⎝4 2⎠
⎜⎜ 1 − tan ix
2
⎝
x
⎛
⎜ 1 + i tanh 2
= log ⎜
⎜⎜ 1 − i tanh x
2
⎝
1
⎛
= log ⎜ 1 + tanh 2
2
⎝
⎞
⎟
⎟
⎟⎟
⎠
⎞
⎟
x⎞
x⎞
⎛
⎛
⎟ = log ⎜1 + i tanh ⎟ − log ⎜1 − i tanh ⎟
2⎠
2⎠
⎝
⎝
⎟⎟
⎠
x⎞
x⎞
x⎞
x⎞ 1
⎛
⎛
⎛
+ i tan −1 ⎜ tanh ⎟ − log ⎜1 + tanh 2 ⎟ − i tan −1 ⎜ − tanh ⎟
2 ⎟⎠
2
2
2
2
⎝
⎠
⎝
⎠
⎝
⎠
1.112
Engineering Mathematics
x
x
⎛
⎜ tanh 2 + tanh 2
⎡ −1 ⎛
x⎞
x ⎞⎤
−1 ⎛
−1
= i ⎢ tan ⎜ tanh ⎟ + tan ⎜ tanh ⎟ ⎥ = i tan ⎜
2⎠
2 ⎠⎦
⎝
⎝
⎣
⎜⎜ 1 − tanh 2 x
2
⎝
x ⎞
⎛
⎜ 2 tanh 2 ⎟
−1
−1
= i tan ⎜
⎟ = i tan (sinh x)
x
⎜⎜ 1 − tanh 2 ⎟⎟
2⎠
⎝
⎞
⎟
⎟
⎟⎟
⎠
⎡ sin( x + iy ) ⎤
-1
Example 9: Prove that log ⎢
⎥ = 2i tan (cot x tanh y).
⎣ sin( x − iy ) ⎦
Solution:
⎡ sin ( x + iy ) ⎤
log ⎢
⎥ = log [sin (x + iy)] – log [sin (x – iy)]
⎣ sin ( x − iy ) ⎦
log (sin x cosh y i cos x sinh y) – log (sin x cosh y – i cos x sinh y)
=
⎛ cos x sinh y ⎞
1
log ( sin 2 x cosh 2 y + cos 2 x sinh 2 y ) + i tan −1 ⎜
⎟
2
⎝ sin x cosh y ⎠
⎛ − cos x sinh y ⎞
1
− log ( sin 2 x cosh 2 y + cos 2 x sinh 2 y ) − i tan −1 ⎜
⎟
2
⎝ sin x cosh y ⎠
i tan 1 (cot x tanh y) + i tan 1 (cot x tanh y)
2i tan 1 (cot x tanh y).
⎡
a + ib ⎞ ⎤ a 2 − b 2
Example 10: Prove that cos ⎢ i log ⎛⎜
=
.
⎝ a − ib ⎟⎠ ⎥⎦ a 2 + b 2
⎣
Solution:
⎡
⎛ a + ib ⎞ ⎤
cos ⎢i log ⎜
= cos[i log (a + ib) − i log (a − ib)]
⎝ a − ib ⎟⎠ ⎥⎦
⎣
⎡1
b 1
⎛ −b ⎞ ⎤
= cos i ⎢ log (a 2 + b 2 ) + i tan −1 − log (a 2 + b 2 ) − i tan −1 ⎜ ⎟ ⎥
⎝ a ⎠⎦
2
a
2
⎣
b⎞
b
b⎞
b⎞
⎛
⎛
⎛
= cos i ⎜ i tan −1 + i tan −1 ⎟ = cos ⎜ −2 tan −1 ⎟ = cos ⎜ 2 tan −1 ⎟
⎝
⎠
⎝
⎝
a⎠
a
a
a⎠
= cos 2q , where tan q =
b
a
b2
2
2
1 − tan q
a2 = a − b
=
=
b2
a 2 + b2
1 + tan 2 q
1+ 2
a
2
1−
Complex Numbers
1.113
2ab
a − ib ⎞
⎛
Example 11: Prove that tan ⎜ i log
.
⎟=
⎝
a + ib ⎠ a 2 − b2
Solution:
⎛ a − ib ⎞ ⎤
⎡ log ⎛⎜ a − ib ⎞⎟
− log ⎜
⎟
⎝ a + ib ⎠
⎢
− e ⎝ a + ib ⎠ ⎥
a − ib ⎞
a − ib ⎞
e
⎛
⎛
=
=
tan ⎜ i log
i
tanh
log
i
⎥
⎢
⎜⎝
⎟
⎟
a − ib
⎛ a − ib ⎞
⎝
a + ib ⎠
a + ib ⎠
− log ⎜
⎥
⎢ log ⎛⎜⎝ a + ib ⎞⎟⎠
⎝ a + ib ⎟⎠
+e
⎥⎦
⎢⎣ e
⎡ ⎛ a − ib ⎞ ⎛ a + ib ⎞ ⎤
⎛ a + ib ⎞ ⎤
⎡ log ⎛⎜ a − ib ⎞⎟
log ⎜
⎟
⎢ ⎜⎝ a + ib ⎟⎠ − ⎜⎝ a − ib ⎟⎠ ⎥
⎢ e ⎝ a + ib ⎠ − e ⎝ a −ib ⎠ ⎥
⎥
= i⎢
⎥ = i⎢
⎛ a − ib ⎞
⎛ a + ib ⎞
⎢ ⎛ a − ib ⎞ ⎛ a + ib ⎞ ⎥
log ⎜
⎢ log ⎜⎝ a + ib ⎟⎠
⎥
⎟
⎢ ⎜⎝ a + ib ⎟⎠ + ⎜⎝ a − ib ⎟⎠ ⎥
+ e ⎝ a − ib ⎠ ⎥⎦
⎢⎣ e
⎦
⎣
⎡ (a − ib) 2 − (a + ib) 2 ⎤
2ab
⎡ −2aib ⎤
= i⎢
= 2
= i⎢ 2
2
2⎥
2⎥
⎣ a − b ⎦ a − b2
⎢⎣ (a − ib) + (a + ib) ⎥⎦
Example 12: If tan [log (x + iy)] = a + ib, where a2 + b2 ñ 1, prove that
2a
tan ⎡⎣ log( x 2 + y 2 ) ⎤⎦ =
.
1 − ( a 2 + b2 )
tan [log (x + iy)] = a + ib
log (x + iy) = tan–1 (a + ib)
log (x – iy) = tan–1 (a – ib)
Adding Eqs. (1) and (2),
Solution:
... (1)
... (2)
log (x + iy) + log (x – iy) = tan–1 (a + ib) + tan–1 (a – ib)
a + ib + a − ib
1 − (a + ib)(a − ib)
2a
−1
log [ ( x + iy )( x − iy ) ] = tan −1
log ( x 2 + y 2 ) = tan
tan ⎡⎣log ( x 2 + y 2 ) ⎤⎦ =
1 − (a 2 + b 2 )
2a
1 − (a 2 + b 2 )
Example 13: If log [log (x + iy)] = a + ib, prove that y = x tan ⎡⎣ tan b log x 2 + y 2 ⎤⎦ .
Solution:
log (x + iy) = ea+ib
1
y
log ( x 2 + y 2 ) + i tan −1 = e a (cos b + i sin b)
2
x
Comparing real and imaginary parts on both the sides,
1
log ( x 2 + y 2 ) = e a cos b
2
... (1)
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Engineering Mathematics
tan −1
and
y
= e a sin b
x
... (2)
Dividing Eq. (2) by Eq. (1),
tan −1
tan b =
tan −1
y
x
1
log ( x 2 + y 2 )
2
y
= tan b log x 2 + y 2
x
y
= tan tan b log x 2 + y 2
x
(
)
(
)
y = x tan tan b log x 2 + y 2 .
Example 14: Separate real and imaginary parts of
(i)log1–i (1 + i)
(ii) (1 + i)i.
Solution: (i) Let x + iy = log1–i (1 + i) =
log (1 + i )
log (1 − i )
ip
1
1
log 2 + i tan −1 1
log 2 +
2
4
2
=
=
ip
1
1
−1
log 2 + i tan ( −1)
log 2 −
2
2
4
2
ip ⎞ ⎛
ip ⎞
⎛
2
+ i log 2
⎜⎝ log 2 + ⎟⎠ ⎜⎝ log 2 + ⎟⎠ (log 2) −
2
2
4
=
=
2
ip ⎞ ⎛
ip ⎞
⎛
(log 2) 2 +
⎜⎝ log 2 − ⎟⎠ ⎜⎝ log 2 + ⎟⎠
2
2
4
Comparing real and imaginary part on both the sides,
x=
(log 2) 2 −
(log 2) 2 +
2
4 ,
2
log 2
y=
(log 2) 2 +
4
2
4
(ii) Let x + iy = (1 + i)i
Taking logarithm on both the sides,
⎛1
⎞
log (x + iy) = i log (1 + i) = i ⎜ log 2 + i tan −1 1⎟
⎝2
⎠
i
π
= log 2 −
2
4
i
x + iy = e 2
log 2 −
e
π
4
= e i log 2 e
−
π
4
=e
−
π
4
⎡⎣cos (log 2 ) + i sin (log 2 ) ⎤⎦
1.115
Complex Numbers
Comparing real and imaginary part on both the sides,
x=e
−
y=e
−
4
cos (log 2 )
4
sin (log 2 )
Example 15: If i`+ia = ` + ia, prove that ` 2 + a 2 = e-(4k+1)oa.
Solution: ia+ib = a + ib
Taking logarithm on both the sides,
(a + ib ) log i = log (a + ib )
p ⎞
pa
pb
⎛
log (a + i b ) = (a + i b ) i ⎜ 2kp + ⎟ = i (4k + 1)
− (4k + 1)
2⎠
2
2
⎝
a + ib = e
i ( 4 k +1)
pa
pb
− ( 4 k +1)
2 e
2
r = a + ib = e
Then,
a2 + b2 = e
− ( 4 k +1)
− ( 4 k +1)
=e
− ( 4 k +1)
pb
pa
i ( 4 k +1)
2 e
2
= reiq , say
pb
2
pb
2
a 2 + b 2 = e − ( 4 k +1)pb .
Example 16: If i
log (1+i)
= A + iB, prove that one value of A is
−π 2
e 8
Solution: A + iB = ilog (1+i)
⎛π
⎞
cos ⎜ log 2 ⎟ .
⎝4
⎠
Taking logarithm on both the sides,
log (A + iB) = log (1 + i) log i
p⎤
⎛1
⎞ ip ip ⎡ 1
= ⎜ log 2 + i tan −1 1⎟ ⋅
log (2) + i ⎥
=
⎝2
⎠ 2
2 ⎢⎣ 2
4⎦
=
p p ip
p 2 −p 2 ip
ip 1
log 2 −
=
+ log 2
⋅ log 2 + i 2 ⋅ =
2 4
4
8
8
4
2 2
p2
8
ip
p2
8
⎡ ⎛p
⎛p
⎞
⎞⎤
⎢cos ⎜⎝ log 2⎟⎠ + i sin ⎜⎝ log 2⎟⎠ ⎥
4
4
⎦
⎣
Comparing real and imaginary part on both the sides,
A + iB = e
−
⋅e 4
log 2
=e
−
A=e
−
2
8
⎛
⎞
cos ⎜ log 2 ⎟
⎝4
⎠
Example 17: By considering only principle value, express (1 + i 3 )1+ i
form of (a + ib).
Solution: Let a + ib = (1 + i 3 )1+ i
3
3
in the
1.116
Engineering Mathematics
Taking logarithm on both the sides,
log (a + ib) = (1 + i 3 ) log (1 + i 3 )
⎡1
⎤
log (a + ib) = (1 + i 3 ) ⎢ log (1 + 3) + i tan −1 3 ⎥
⎣2
⎦
ip ⎞
ip ⎞
⎛
⎛1
= (1 + i 3 ) ⎜ log 4 + ⎟ = (1 + i 3 ) ⎜ log 2 + ⎟
⎝2
⎠
⎝
3
3⎠
= log 2 −
a + ib = e
log 2 −
p 3 ⎛
p⎞
+ i ⎜ 3 log 2 + ⎟
⎝
3
3⎠
3
3 i ⎛ 3 log 2 + ⎞
⎜
3 ⎟⎠
e⎝
3
⎡ ⎛
⎞
⎛
⎞⎤
⎢cos ⎜ 3 log 2 + ⎟ + i sin ⎜ 3 log 2 + ⎟ ⎥
3⎠
3 ⎠⎦
⎝
⎣ ⎝
Comparing real and imaginary parts on both the sides,
=e
3
log 2 −
p 3
3
p⎞
⎛
cos ⎜ 3 log 2 + ⎟
⎝
3⎠
log 2 −
p 3
3
p⎞
⎛
sin ⎜ 3 log 2 + ⎟
⎝
3⎠
a=e
b=e
log 2 −
Example 18: Prove that (1 + i tan `)-i = e2mo +` [cos (log cos `) + i sin (log cos `)].
Solution: Let x + iy = (1 + i tan a )
i
Taking logarithm on both the sides,
log (x + iy) = i log (1 + i tan a )
⎡1
⎤
= −i ⎢ log (1 + tan 2 a ) + i ( 2mp + tan −1 tan a ) ⎥
⎣2
⎦
⎡1
⎤
= −i ⎢ log sec 2 a + i ( 2mp + a ) ⎥
⎣2
⎦
= i log (sec2 a )
−
1
2
+ (2mp + a ) = i log (cos a ) + (2mp + a )
x + iy = ei log cosα e( 2 mπ +α )
= e 2 mπ +α [cos (log cos α ) + i sin (log cos α )]
Hence, (1 + i tan a ) i = e2mp +a [cos (log cos a ) + i sin (log cos a )]
Example 19: Prove that if (1 + i tan `)1+i tan a can have only real values, one of
them is (sec α )sec
2
β
considering only principle value.
1.117
Complex Numbers
Solution: Let x = (1 + i tan a)1+i tan b, where x is real.
Taking logarithm on both the sides,
log x = (1 + i tan b ) log (1 + i tan a)
⎡1
⎤
= (1 + i tan b ) ⎢ log (1 + tan 2 a ) + i tan −1 (tan a ) ⎥
2
⎣
⎦
= (1 + i tan b ) ( log sec a + ia ) = (log sec a − a tan b ) + i (a + tan b log sec a )
Comparing real and imaginary parts on both the sides,
log x = log sec a
x=e
a tan b
and a + tan b log sec a = 0
and a = tan b log sec a
(log sec a a tan b)
Substituting a in x,
x = elog sec a + tan
= e(logseca )sec
2
2
b logseca
b
2
= elogseca (1+ tan
2
b)
b logseca
2
= (sec a )sec b
b
2 tan −1
y
a .
Example 20: If (a + ib) p = mx+iy, prove that =
x log(a 2 + b 2 )
Solution: (a + ib) p = mx+iy
= esec
Taking logarithm on both the sides,
p log (a + ib) = (x + iy) log m
p ⎡1
b⎤
log (a 2 + b 2 ) + i tan −1 ⎥ = x + iy
log m ⎢⎣ 2
a⎦
Comparing real and imaginary parts on both the sides,
x=
p log (a 2 + b 2 )
2 log m
p
b
tan −1
log m
a
b
b
2 tan −1
tan −1
y
a
a .
=
=
2
2
x 1
2
2
log(a + b ) log(a + b )
2
y=
Example 21: If
(1 + i ) x + iy
= a + i a , then by considering only principle values,
(1 − i ) x − iy
⎛β ⎞ πx
prove that tan −1 ⎜ ⎟ =
+ y log 2.
2
⎝α ⎠
1.118
Engineering Mathematics
(1 + i ) x + iy
(1 − i ) x − iy
Taking logarithm on both the sides,
Solution: a + i b =
⎡ (1 + i ) x + iy ⎤
log (a + ib ) = log ⎢
= ( x + iy ) log (1 + i ) − ( x − iy ) log (1 − i )
x − iy ⎥
⎣ (1 − i ) ⎦
⎛1
⎞
⎡1
⎤
= ( x + iy ) ⎜ log 2 + i tan −1 1⎟ − ( x − iy ) ⎢ log 2 + i tan −1 ( −1) ⎥
⎝2
⎠
⎣2
⎦
1
β
iπ
⎛
log (α 2 + β 2 ) + i tan −1 = ( x + iy ) ⎜ log 2 +
⎝
2
4
α
iπ ⎞
⎛
⎞
⎟⎠ − ( x − iy ) ⎜⎝ log 2 − ⎟⎠
4
p y⎞ ⎛
p x⎞
⎛
= ⎜ x log 2 − ⎟ + i ⎜ y log 2 + ⎟
⎝
4⎠ ⎝
4⎠
p y⎞ ⎛
p x⎞
⎛
− ⎜ x log 2 − ⎟ + i ⎜ y log 2 + ⎟
⎝
4⎠ ⎝
4⎠
p x⎞
⎛
= 2i ⎜ y log 2 + ⎟
⎝
4⎠
Comparing imaginary part on both the sides,
px
px
b
tan −1 = 2 y log 2 +
= y log 2 +
.
a
2
2
Example 22: Separate real and imaginary parts of ( i ) i .
Solution: Let a + ib = i
Taking logarithm on both the sides,
log (a + ib) = log i =
1 ip
1
log i = ⋅
2
2 2
i
a + ib = e 4
i=
Hence,
Let ( i )
i
i
e4
= x + iy
Taking logarithm on both the sides,
i log i = log ( x + iy )
ip
e4
ip
log e 4 = log ( x + iy )
p
p ⎞ ip ⎛ 1
i ⎞ ip
ip
p
⎛
log ( x + iy ) = ⎜ cos + i sin ⎟ ⋅
=
+
=
−
⎟
⎝
4
4 ⎠ 4 ⎜⎝ 2
2⎠ 4 4 2 4 2
1.119
Complex Numbers
−p
ip
x + iy = e 4 2 e 4
2
=e
−
p
4 2
p
p ⎞
⎛
+ i sin
⎜⎝ cos
⎟
4 2
4 2⎠
Comparing real and imaginary parts on both the sides,
x=e
y=e
−
−
π
4 2
cos
π
4 2
sin
π
4 2
π
4 2
⎛
i
Example 23: Prove that i i = cos p + i sin p, where p = ( 4n + 1)
Solution: Let ii = x + iy
Taking logarithm on both the sides,
i log i = log (x + iy)
p⎞
⎛
i ⋅ i ⎜ 2mp + ⎟ = log ( x + iy )
⎝
2⎠
( x + iy ) = e
ii = e
1⎞
⎛
− ⎜ 2m+ ⎟ p
⎝
2⎠
1⎞
⎛
− ⎜ 2m+ ⎟ p
⎝
2⎠
i
cos q + i sin q = i i = ie
Taking logarithm on both the sides,
log (cos q + i sin q ) = e
=
1⎞
⎛
−⎜ 2 m + ⎟p
2⎠
⎝
−
1⎞
⎛
⎜⎝ 2 m + ⎟⎠ p
2
log i = e
1⎞
⎛
−⎜ 2 m + ⎟p
2⎠ ⎛
⎝
i 2np
⎜
⎝
1⎞
⎛
−⎜ 2 m + ⎟p
p
2⎠
⎝
e
i (4n + 1)
2
+
p ⎞
2 ⎟⎠
= if , say
cos q + i sin q = eif eiq = eif
Comparing both the sides,
θ =φ = e
Example 24: If i i
(i) z
2
1⎞
⎛
−⎜ 2 m + ⎟π
2⎠
⎝
(4n + 1)
π
.
2
= z, where z = x + iy, prove that
= e − ( 4 n + 1)π y , n ∈ I
(ii) tan
1⎞
o − ⎜⎝ 2 m + 2 ⎟⎠ o
e
.
2
πx y
=
and x 2 + y 2 = e −π y .
2
x
1.120
Engineering Mathematics
= x + iy
ii
ix+iy = x + iy
Taking logarithm on both the sides,
(x + iy) log i = log (x + iy)
Solution:
p⎞
⎛
log ( x + iy ) = ( x + iy ) i ⎜ 2np + ⎟ , n ∈ I
⎝
2⎠
p⎞
⎛
⎛ 4 n + 1⎞
py
= i ⎜ 2np + ⎟ x − ⎜
⎝
⎝ 2 ⎟⎠
2⎠
x + iy = e
i ( 4 n +1)
p x − ⎛ 4 n +1⎞ p y
⎜
⎟
2e ⎝ 2 ⎠
= reiq , say
⎡ − ⎛⎜ 4 n +1⎞⎟ p y ⎤
r = | x + iy | = | x – iy | = ⎢⎣e ⎝ 2 ⎠ ⎥⎦
Then
⎡ − ⎛⎜ 4 n +1 ⎞⎟p y ⎤
| x − iy | = ⎣⎢e ⎝ 2 ⎠ ⎦⎥
2
2
| z |2 = e − ( 4 n +1) p y.
θ = tan −1
and
For n = 0,
tan −1
y
x
=
x
2
y
x
= tan
.
x
2
| z |2 = e −
and
y ⎛ 4n + 1 ⎞
=
πx
x ⎜⎝ 2 ⎟⎠
y
, x 2 + y 2 = e−
y
⋅
Exercise 1.8
1. Find the general value of
2. Considering principal values only,
(i) log ( i)
prove that
(ii) log 3 − i
log 3 + i
i
(iii) log2 5
(iv) sin (log i )
log 2 (−3) =
.
log 2
(v) cos (log i i).
⎡
⎤ 3. Find the general value of log (1 + i) +
p⎞
⎛
⎢ Ans. : (i) i ⎜⎝ 2p n − 2 ⎟⎠
⎥
log (1 – i).
⎢
⎥
[Ans. : log 2]
p⎞
⎛
⎢
⎥
⎢(ii) log 2 + i ⎜⎝ 2p n − 6 ⎟⎠
⎥ 4. Find the general value of
⎢
⎥
2
log 1 + i 3 + log 1 − i 3 ⋅
⎢(iii) [(log5 log 2 + 4p mn)
⎥
⎢ + i ( n log 2 − m log 5)2p ] / [(log 2) 2 + 4p 2 m 2 ]⎥
[Ans. : 2log 2]
⎢
⎥
( v) 0
⎣ (iv) − 1
⎦
(
)
(
)
(
)
1.121
Complex Numbers
5. Prove that sin loge (i i) = 1.
6. Prove that
α −β ⎞
log (eiα − eiβ ) = log ⎛⎜ 2 sin
2 ⎟⎠
⎝
⎛ π +α + β ⎞
+i ⎜
⎟.
2
⎝
⎠
7. Show that log (–log i) = log
2
15. Find the principal value of (x + iy)i
and show that it is entirely real if
1
log ( x 2 + y 2 ) is a multiple of p.
2
i
.
2
8. Prove that log (1 + i tan a) =
log sec a + ia.
9. Prove that log (1 + eiq) =
⎡
⎛ ⎞⎤ i
log ⎢ 2 cos ⎜ ⎟ ⎥ + .
⎝ 2 ⎠⎦ 2
⎣
10. Prove that
⎛ 1 ⎞
⎛1
⎞ i
= log ⎜ sec ⎟ − .
log ⎜
i ⎟
2
2
⎝ 1+ e ⎠
⎝
⎠ 2
11. If sin–1 (x + iy) = log (A + iB), prove
that
x2
y2
= 1 where A2 + B 2 = e 2u .
sin 2 u cos 2 u
12. If (a + ib) p = mx+iy, prove that one of
⎛b⎞
2 tan −1 ⎜ ⎟
y
⎝a⎠ .
the values of is
x
log(a 2 + b 2 )
13. Separate i(1–i) into real and imaginary
parts.
−
p⎞⎤
⎛
⎡
⎜ 2 np + ⎟⎠
2 ⎥
⎢ Ans.: ie⎝
⎥
⎢
⎦
⎣
14. Considering only the principal values,
separate real and imaginary parts of
( x + iy )α + iβ
.
( x − iy )α −iβ
⎡ Ans.: cos 2q + i sin 2q , where ⎤
⎢
⎥
⎢q = a tan −1 y + b log x 2 + y 2 ⎥
⎢⎣
⎥⎦
x
1
⎡
⎤
2
2
⎢ Hint : put 2 log ( x + y ) = n ⎥
⎣
⎦
⎤
⎡
⎛ y⎞
−1
⎢ Ans.: e tan ⎜⎝ x ⎟⎠ [cos log ( x 2 + y 2 ) ⎥
⎥
⎢
⎢⎣
+ i sin log ( x 2 + y 2 ) ⎥⎦
16. If ia+ib = a + ib, prove that
a 2 + b 2 = e–(4n+1)pb .
i
17. If
i
= a + ib, prove that
a +b = e
2
2
πβ
2
.
⎡
Hint :
⎣⎢
( i)
α +iβ
⎤
= α + iβ ⎥
⎦
x
18. If x
= a (cos a + i sin a), prove
that the general value of x is given by
r (cos q + i sin q ) where
(2nπ + α ) sin α + (cos α ) log a
log r =
a
(2nπ + α ) cos α − (sin α ) log a
and θ =
.
a
[Hint : xa (cos a + i sin a) = a (cos a +
i sin a) = aeia ]
⎡ ( a − b) + i ( a + b) ⎤
19. Prove that log ⎢
⎥
⎣ ( a + b) + i ( a − b) ⎦
2ab
⎛
= i ⎜ 2n + tan −1 2
a
− b2
⎝
⎞
⎟.
⎠
[Hint : put a – b = x, a + b = y]
1.122
Engineering Mathematics
FORMULAE
Algebra of Complex Numbers
(i) Addition: z1 + z2 = (x1 + x2) +
i (y1 + y2)
(ii) Subtraction: z1 z2 = (x1 x2) +
i (y1 y2)
(iii) Multiplication: z1 z2 = (x1 x2 y1y2)
+ i (x2y1 + y2x1)
(iv) Division:
(y x − x y )
z1 x1 x2 + y1 y2
= 2
+ i 1 2 2 12 2
2
z2
x2 + y2
( x2 + y2 )
Different Forms of Complex Numbers
(i) Cartesian or Rectangular Form:
z = x + iy
(ii) Polar Form: z = r (cos q + i sinq )
=r q
(iii) Exponential Form: z = reiq
Modulus and Argument (or Amplitude)
of Complex Numbers
Modulus: | z | = r = x 2 + y 2
Argument (or Amplitude):
y
arg( z ) = q = tan −1
x
Properties of Complex Numbers
1
(i) Re (z) = x = (z + z ), Im (z) =
2
1
y = (z z )
2i
(ii) ( z1 + z2 ) = z1 + z2
(iii)
( z1 z2 ) = z . z
1
2
⎛z ⎞ z
(iv) ⎜ 1 ⎟ = 1
⎝ z2 ⎠ z2
(v) z z = |z|2 = | z |2
(vi)
z1 z2 = | z1 | | z2 |
(vii) arg (z1 z2) = arg (z1) + arg (z2)
(viii)
z1
z1
=
z2
z2
⎛z ⎞
(ix) arg ⎜ 1 ⎟ = arg (z1)
⎝ z2 ⎠
arg (z2)
De Moivre’s Theorem
(cosq + i sin q )n = cos nq + i sin nq
where n is any real number
Circular and Hyperbolic Functions
iz
(i) sin z = e
cos z =
e
2i
iz
,
e iz + e − iz
2
e z − e− z
,
2
e z + e− z
cosh z =
2
(ii) sinh z =
Relation between Circular and Hyperbolic Functions
(i) sin iz = i sinh z, sinh z = i sin iz
(ii) cos iz = cosh z
(iii) tan iz = i tanh z, tanh z = i tan iz
Formulae on Hyperbolic Functions
(i) cosh2 z – sinh2 z = 1
(ii) coth2 z – cosech2 z = 1
(iii) sech2 z + tanh2 z = 1
(iv) sinh 2z = 2 sinh z cosh z
(v) cosh 2z = cosh2 z + sinh2 z = 2
cosh2 z – 1 = 1 + 2 sinh2 z
2tanh z
(vi) tanh 2z =
1 + tanh 2 z
(vii) sinh 3z = 3 sinh z + 4 sinh3 z
(viii) cosh 3z = 4 cosh3 z – 3 cosh z
3tanh z + tanh3 z
(ix) tanh 3z =
1 + 3tanh 2 z
1.123
Complex Numbers
(x) sinh (z1 ± z2) = sinh z1 cosh z2
± cosh z1 sinh z2
(xi) cosh (z1 ± z2) = cosh z1 cosh z2
± sinh z1 sinh z2
tanh z1 ± tanh z2
(xii) tanh (z1 ± z2) =
1 ± tanh z1tanh z2
(xiii) sinh z1 + sinh z2 =
z +z
z −z
2 sinh 1 2 cosh 1 2
2
2
(xiv) sinh z1 – sinh z2 =
z +z
z −z
2 cosh 1 2 sinh 1 2
2
2
(xv) cosh z1 + cosh z2 =
z +z
z −z
2 cosh 1 2 cosh 1 2
2
2
(xvi) cosh z1 – cosh z2 =
z1 + z2
z −z
sinh 1 2
2
2
(xvii) 2 sinh z1 cosh z2 = sinh (z1 + z2)
+ sinh (z1 – z2)
(xviii) 2 cosh z1 sinh z2 = sinh (z1 + z2)
– sinh (z1 – z2)
(xix) 2 cosh z1 cosh z2 = cosh (z1 + z2)
+ cosh (z1 – z2)
(xviii) 2 sinh z1 sinh z2 = cosh (z1 + z2)
– cosh (z1 – z2)
Inverse Hyperbolic Functions
(
(iii) tanh
1
x=
)
1
⎛1 + x ⎞
log ⎜
⎝ 1 − x ⎟⎠
2
Separation into Real and Imaginary Parts
(i) sin (x ± iy) = sin x cosh y ± icos x
sinh y
(ii) cos (x ± iy) = cos x cosh y ± i sin x
sinh y
sin 2 x ± isinh 2 y
(iii) tan (x ± iy) =
cos 2 x + cosh 2 y
(iv) sinh (x ± iy) = sinh x cos y ± i cosh
x sin y
2 sinh
2
(i) sinh 1 x = log x + x + 1
(
2
(ii) cosh 1 x = log x + x − 1
)
(v) cosh (x ± iy) = cosh x cos y ± i sinh
x sin y
sinh 2 x ± i sin 2 y
(vi) tanh (x ± iy) =
cosh 2 x + cos 2 y
Logarithm of a Complex Numbers
Principle Value: log (x + iy )
1
y
= log (x 2 + y 2 ) + i tan 1
2
x
1
log r + iq
2
General Value: log (x + iy)
=
=
y⎞
1
⎛
log (x 2 + y 2 ) + i ⎜ 2np + tan −1 ⎟
⎝
x⎠
2
=
1
log r + i (2np + q )
2
MULTIPLE CHOICE QUESTIONS
Choose the correct alternative in each of the following questions:
1. The value of
10
sin
n =1
(a)
(c)
1
i
2n
11
i cos
2n
11
(b) 0
(d) i
is
2. If the area of the triangle on the
complex plane formed by complex
numbers z, iz and z + iz is 50 square
units, then |z| is
(a) 5
(b) 10
(c) 15
(d) none of these
1.124
Engineering Mathematics
2
lies on
z
(a) a circle
(b) an ellipse
(c) a parabola (d) a straight line
4. The region in the Argand’s diagram
defined by z 2i z 2i 5 is the
interior of the ellipse with major axis
along
(a) the real axis
(b) the imaginary axis
(c) y = x
(d) y = x
5. If w is an imaginary cube root of
unity, then (1 + w – w 2)7 is equal to
(a) 128w
(b) 128w
(c) 128w 2
(d) 128w 2
3. If z lies on |z| = 1, then
6. If
z 8i
=
z+6
then z lies on the
curve
(a) x2 + y2 + 6x – 8y = 0
(b) 4x – 3y + 24 = 0
(c) x2 + y2 – 8 = 0
(d) none of these
7. If 2 + i 3 is a root of the quadratic
equation x2 + ax + b = 0, where
a b R then the values of a and b
are respectively,
(a) 4, 7
(b) 4, 7
(c) 4, 7
(d) 4, 7
8. If z1, z2, z3 are vertices of an equilateral triangle inscribed in the circle |z|
then
= 2 and if z1 = + i
(a) z2
(b) z2
z3
z3
(c) z2
(d) z2
i
i
z3
i
i
z3
i
9. The triangle formed by the points,
1+ i
1,
and i as vertices in the Argand
2
diagram is
(a) scalene
(c) isosceles
(b) equilateral
(d) right-angled
1 + c + is
10. If c2 + s2 = 1, then
is equal
1
c
is
to
(a) c + is
(c) s + ic
(b) c – is
(d) s – ic
11. The value of i i is
(a) w
(b) –w2
(c)
(d) none of these
2
12. If x + iy = c sin(u + iv) , then u =
constant represents family of
(a) confocal ellipses
(b) confocal circles
(c) confocal hyperbolas
(d) none of these
1
13. If
then cosh 2x is
x=
2
2
4
(a)
(b)
3
3
1
(c) 1
(d)
3
14. The value of sin(log i i ) is
(a) –1
(b) 1
(c) 0
(d) none of these
15. If log[log(x + iy)] = p + iq, then the
y
value of tan 1 is
x
(a) e pcos q
(b) e qsin p
q
(c) e cos p
(d) e psin q
16. If (a + ib) p = m x+iy then the value of y
is
p
b
tan 1
(a)
log m
a
p
1 a
(b)
tan
log m
b
p
(c)
log(a 2 + b 2 )
2 log m
(d) none of these
17. The general value of which satisfies the equation (cos + i sin 3 )
1.125
Complex Numbers
(cos 3 + isin3 )…[cos(2n – 1) +
isin(2n – 1) ] = 1 is
r
(r 1)
(a) 2
(b)
n
n2
(2r + 1)
2r
(c)
(d)
3
n
n2
18. The smallest positive integer n for
which (1 + i)2n = (1 – i)2n is
(a) 4
(b) 8
(c) 2
(d) 12
(d) an ellipse
23. If the complex number z and its
conjugate z satisfy z z + 2(z – z )
= 12 + 8i, then the values of z are
3 3w 3
6w 4
(a) 1
(c) 0
21. The complex numbers sinx + icos2x
and cosx – sin2x are conjugate to each
other for
(a) x = n
(b) x = n +
2
(c) x =
(d) no value of x
3
b
c
a
b
3.
10.
17.
24.
3 +i
)
= 299 (a + ib), then (a2
2
+ i sin
2
(b) 4
5
3
(c) 4 cos
(c) a parabola y = 2x
2.
9.
16.
23.
(
(a) 4 cos
5
, 0
3
2
Answers
1. d
8. a
15. d
22. a
(d) 2 ± 3i
100
27. If P is a point in the Argand diagram
representing the complex number,
4
4
4 cos
+ i sin
and OP is ro3
3
2
tated through an angle
in the
3
anticlockwise direction, then P in the
new position represents
z +1
22. The locus determined by
=2
z 1
where z 1, z = x + iy is
(b) a circle with centre 0,
(c) 2 ± 2i
26. A root of x3 – 8x2 + px + q = 0 where
p and q are real numbers, is 3 i 3.
The real root is
(a) 2
(b) 6
(c) 9
(d) 12
(b) –1
(d) none of these
(a) a circle with centre
(b) 2 2 ± 2i
+ b2) is equal to
(a) 1
(b) 2
(c) 3
(d) 4
is a complex number such that
25. If
2
+ + 1 = 0, then 31 is equal to
(a)
(b) 2
(c) 1
(d) 0
20. If w is the cube root of unity, then the
value of the
2w 2
4w 3
(a) 2 ± 2 2i
24. If
19. The cube roots of unity lie on a circle
(a) | z | = 1
(b) | z – 1| = 1
(c) | z + 1| = 1 (d) | z – 1| = 2
1 w
2 2w 2
x2 y 2
+
=1
2
3
3
+ i sin
3
(d) 3 + 2i
a
a
d
d
4.
11.
18.
25.
b
d
c
a
5.
12.
19.
26.
d
c
a
a
6.
13.
20.
27.
a
b
c
b
7. c
14. a
21. d
Differential
Calculus I
Chapter
2
2.1 INTRODUCTION
Differential calculus is the study of derivative, i.e., the study of change of functions
w.r.t. the change in inputs. It is the mathematical study of change, motion, growth or
decay, etc. In this chapter, we will study successive differentiation, mean value theorems, such as Rolle’s theorem, Lagrange’s mean value theorem, Cauchy’ mean value
theorem, expansion of functions and indeterminate forms.
2.2 SUCCESSIVE DIFFERENTIATION
If y = f (x) be a differentiable function of x, then its derivative
order derivative of y and is in general a function of x. If
dy
is called the first
dx
dy
is differentiable, then its
dx
derivative is called the second order derivative of y and is denoted by
the derivative of
d2 y
dx 2
d2 y
dx 2
. Similarly,
is called the third order derivative of y and is denoted by
and so on.
The successive differential coefficients of the function y = f (x) are denoted by
dy d 2 y
dn y
, 2 ,… , n , …
dx dx
dx
Alternative methods of writing the differential coefficients are
d
Dy, D2y, D3y, ……, Dny, …
where D =
dx
f (x), f (x), f (x), …, f n(x), …
y (x), y (x), y (x), … , y n(x), …
y1 (x), y2 (x), y3 (x), … , yn (x), …
d3 y
dx 3
2.2
Engineering Mathematics
The value of nth differential coefficient at x = a is denoted by
⎛ dn y ⎞
or
⎜ n⎟
⎝ dx ⎠ x = a
(yn)a
or f n(a)
or
y n(a).
2.2.1 nth Order Derivative of Some Standard Functions
1. y = (ax + b)m, where m is any real number.
Proof: y = (ax + b)m
Differentiating w.r.t. x successively,
y1 = ma (ax + b)m 1
y2 = m (m 1) a2 (ax + b)m
2
y3 = m (m 1) (m 2) a (ax + b)m 3,
...........................................................
............................................................
yn = m (m 1) (m 2) … (m n + 1) an (ax + b)m
3
dn
Hence,
dx n
(ax + b) m = m (m
1) (m
2) … (m
n + 1) an (ax + b)m
n
n
=
m (m − 1)… (m − n + 1)[(m − n)(m − n − 1)… 3 ⋅ 2 ⋅1] a n (ax + b) m − n
(m − n) (m − n − 1)… 3 ⋅ 2 ⋅1
=
a n m !(ax + b) m − n
,
(m − n)!
= n! an ,
= 0,
if n < m
if n = m
if n > m
2. y = (ax + b)−m, where m is any positive integer.
Proof: y = (ax + b) m
Differentiating w.r.t. x successively,
y1 = ( 1) ma (ax + b)
m 1
y2 = ( 1)2 m (m + 1) a2 (ax + b) m 2
y3 = ( 1)3 m (m + 1) (m + 2) a3 (ax + b) m 3
...........................................................
............................................................
yn = ( 1)n m (m + 1) (m + 2) … (m + n 1) an (ax + b)
= ( −1) n
Hence,
dn
dx n
(ax + b) − m = ( −1) n
m n
an
(m + n − 1)… m(m − 1)(m − 2)… 2 ⋅1
(m − 1)(m − 2)… 2 ⋅1
(ax + b) m + n
(m + n − 1)!
an
.
(m − 1)! (ax + b) m + n
Differential Calculus I
dn
Corollary 1: Putting m = 1, we get
(ax + b) −1 = ( −1) n n !
dx n
3. y = log (ax + b)
Proof: y = log (ax + b)
Differentiating w.r.t. x,
a
ax + b
y1 =
Differentiating (n
1) times w.r.t. x,
d n −1
dx
n −1
y1 =
d n −1 ⎛ a ⎞
⎜
⎟
dx n −1 ⎝ ax + b ⎠
n −1
n −1
n −1
⎛ dy ⎞ a ( −1) (n − 1)!a
=
⎜ ⎟
dx n −1 ⎝ dx ⎠
(ax + b) n
d
Hence,
dn
dx n
log (ax + b) =
( −1) n −1 (n − 1)!a n
(ax + b) n
4. y = eax
Proof: y = eax
Differentiating w.r.t. x successively,
y1 = aeax
y2 = a2eax
y3 = a3eax
.................
.................
yn = aneax
Hence,
dn
dx n
(e ax ) = a n e ax
5. y = amx
Proof: y = amx
Differentiating w.r.t. x successively,
y1 = mamx log a
y2 = m2amx (log a)2
y3 = m3amx (log a)3
.............................
.............................
yn = mnamx (log a)n
Hence,
dn
dx n
(a mx ) = m n a mx (log a ) n
2.3
an
(ax + b)1+ n
.
2.4
Engineering Mathematics
6. y = sin (ax + b)
Proof: y = sin (ax + b)
Differentiating w.r.t. x successively,
⎛p
⎞
y1 = a cos ( ax + b) = a sin ⎜ + ax + b ⎟
⎝2
⎠
⎛p
⎞
⎛ 2p
⎞
+ ax + b ⎟
y2 = a 2 cos ⎜ + ax + b ⎟ = a 2 sin ⎜
⎝2
⎠
⎝ 2
⎠
⎛ 2p
⎞
⎛ 3p
⎞
+ ax + b ⎟ = a3 sin ⎜
+ ax + b ⎟
y3 = a3 cos ⎜
⎝ 2
⎠
⎝ 2
⎠
..................................................................................
..................................................................................
⎛ np
⎞
yn = a n sin ⎜
+ ax + b ⎟
⎝ 2
⎠
Hence,
dn
⎛ np
⎞
[sin ( ax + b)] = a n sin ⎜
+ ax + b ⎟
⎝ 2
⎠
dx n
dn
⎛ np
⎞
n
+ ax + b ⎟
Corollary 2: n [cos ( ax + b)] = a cos ⎜⎝
⎠
2
dx
7. y = eax cos (bx + c)
Proof: y = eax cos (bx + c)
Differentiating w.r.t. x,
y1 = aeax cos (bx + c) + (-1) b eax sin (bx + c)
y1 = eax [a cos (bx + c) b sin (bx + c)]
Let a = r cos q, b = r sin q
Then
b
a
y1 = eax [r cos q cos (bx + c)
= reax cos (bx + c + q )
r = a 2 + b 2 , q = tan
1
r sin q sin (bx + c)]
Differentiating w.r.t. x,
y2 = aeax r cos (bx + c + q ) - b eax r sin (bx + c + q )
= reax [r cos q cos (bx + c + q ) r sin q sin (bx + c + q )]
= r2eax cos (bx + c + 2q )
Differentiating n times w.r.t. x,
yn = r neax cos (bx + c + nq ),
b
where r = a 2 + b 2 , q = tan −1 .
a
Differential Calculus I
Hence,
dn
2.5
[eax cos (bx + c)] = rneax cos (bx + c + nq ),
dx n
b
where r = a 2 + b 2 ,q = tan −1 .
a
Corollary 3:
d n ax
[e sin (bx + c)] = rneax sin (bx + c + nq ),
dx n
2
2
−1 b
.
where r = a + b , q = tan
a
Example 1: Find yn if y =
Solution: y =
xn − 1 .
x− 1
x n − 1 ( x − 1)( x n −1 + x n − 2 + x n − 3 + ……… + 1)
=
x−1
( x − 1)
= x n −1 + x n − 2 + x n − 3 + ……… +1
Differentiating n times w.r.t. x,
yn =
dn
dx n
( x n −1 + x n − 2 + x n − 3 + ……… + 1)
⎡ dn
⎤
m
⎢∵ n (ax + b) = 0, if n > m ⎥
⎢⎣ dx
⎥⎦
=0
Example 2: Prove that
d2n
( x2
dx 2 n
1) n = ( 2n)! .
Solution: (x2 - 1)n = (x2)n - nC1 (x2)n 1 + nC2 (x2)n 2 - …........ (-1)n
= x2n - nC1 x2n 2 + nC2 x2n
4
1
- …........
Differentiating 2n times w.r.t. x,
d 2n
dx 2 n
( x 2 − 1) n =
d 2n
dx 2 n
(x2n - nC1 x2n 2 + nC2 x2n
Solution:
y=
=
x
.
( x + 1)4
x
( x + 1)
1
………)
n
⎡ dn
m = m ! a , if n = m
⎢∵ n (ax + b)
if n > m
= 0,
⎢⎣ dx
= (2n) !
Example 3: Find yn, if y =
4
4
( x + 1)3
=
−
( x + 1) − 1
( x + 1) 4
1
( x + 1) 4
⎤
⎥
⎥⎦
2.6
Engineering Mathematics
Differentiating n times w.r.t. x,
( −1) n (n + 2)!
yn =
2 ! ( x + 1) n + 3
−
( −1) n (n + 3)!
3! ( x + 1) n + 4
n+3 ⎤
( −1) n (n + 2)! ⎡
1−
⎥
n+3 ⎢
2 ! ( x + 1)
⎣ 3 ( x + 1) ⎦
( −1) n (n + 2)! ⎡ 3 x − n ⎤
=
⎢
⎥.
2 ! ( x + 1) n + 3 ⎣ 3 ( x + 1) ⎦
=
Example 4: Find yn, if y =
x2 + 4x + 1
.
x3 + 2x2 - x - 2
Solution:
y=
=
x2 + 4 x + 1
x3 + 2 x 2 − x − 2
x2 + 4x + 1
=
=
( x 2 + 4 x + 1)
x 2 ( x + 2) − ( x + 2)
x2 + 4 x + 1
( x + 2) ( x + 1) ( x − 1)
( x + 2) ( x 2 − 1)
A
B
C
+
+
=
x + 2 x +1 x −1
[By partial fraction expansion]
(x2 + 4x + 1) = A(x + 1) (x – 1) + B(x + 2) (x – 1) + C(x + 2) (x + 1)
Putting x = 2,
A= 1
Putting x = 1,
Putting x = 1,
y=
B=1
C=1
1
1
−1
+
+
x + 2 x +1 x −1
Differentiating n times w.r.t. x,
yn = −
( −1) n n !
( x + 2) n +1
Example 5: Find yn, where y =
Solution: y =
x2 + 4
(2 x + 3) ( x − 1) 2
1
1
=
+
2 x + 3 ( x − 1) 2
+
( −1) n n !
( x + 1) n +1
+
x2 + 4
.
( 2 x + 3)( x - 1)2
=
( x − 1) 2 + (2 x + 3)
(2 x + 3) ( x − 1) 2
( −1) n n !
( x − 1) n +1
[Using Cor.1]
Differential Calculus I
2.7
Differentiating n times w.r.t. x,
⎡ (n !)2n
(n + 1)! ⎤
yn = ( −1) n ⎢
+
⎥
n +1
( x − 1) n + 2 ⎥⎦
⎢⎣ (2 x + 3)
Example 6: Find yn, where y =
x4
.
( x − 1) ( x − 2)
x4
x4
= 2
( x − 1) ( x − 2) x − 3 x + 2
15 x − 14
= x 2 + 3x + 7 + 2
x − 3x + 2
Solution: y =
[By dividing]
15 x − 14
( x − 1) ( x − 2)
A
B
= x2 + 3x + 7 +
+
x−1 x− 2
= x 2 + 3x + 7 +
[By partial fraction
expansion]
16
⎛ −1 ⎞
= x 2 + 3x + 7 + ⎜
+
⎝ x − 1⎟⎠ x − 2
Differentiating n times w.r.t. x,
yn = −
( −1) n n !
( x − 1) n +1
+
16 ( −1) n n !
( x − 2) n +1
⎡
⎤
16
1
= ( −1) n n ! ⎢
−
n +1 ⎥
n +1
( x − 1) ⎦
⎣ ( x − 2)
Example 7: If y =
Solution: y =
x3
⎧−( n !) if n is odd ⎫
, then prove that (yn)0 = ⎨
⎬.
x2 - 1
⎩ 0 if n is even ⎭
x3
2
x −1
= x+
= x+
x
2
x −1
1⎛ 1
1 ⎞
1 ⎡ x −1+ x +1 ⎤
x
+
= x+ ⎢
⎥ = x+ ⎜
2 ⎝ x + 1 x − 1 ⎟⎠
( x − 1) ( x + 1)
2 ⎣ ( x − 1)( x + 1) ⎦
Differentiating n times w.r.t. x,
yn =
⎡
⎤
1
1
1
(−1) n n ! ⎢
+
⎥ [Using Cor.1]
+
1
+
1
n
n
2
( x − 1) ⎥⎦
⎢⎣ ( x + 1)
⎡ 1
1 ⎤
! ⎢ n +1 +
⎥
(−1) n +1 ⎥⎦
⎢⎣ (1)
= −(n !), if n is odd.
= 0,
if n is even.
( yn ) 0 =
1
(−1) n n
2
2.8
Engineering Mathematics
Example 8: Find yn, where y =
y=
Solution:
=
=
=
1
.
1 + x + x2
1
1 + x + x2
1
2
1⎞
3
⎛
⎜⎝ x + ⎟⎠ +
2
4
⎛
1
3⎞ ⎛
1
3⎞
⎜x+ 2 +i 2 ⎟ −⎜x+ 2 −i 2 ⎟
⎠ ⎝
⎝
⎠
1
=
⎞
⎛
⎛
1
3 ⎛
1
3⎞
1
3⎞ ⎛
1
3⎞
⎜x+ 2 +i 2 ⎟ ⎜x+ 2 −i 2 ⎟
⎜x+ 2 +i 2 ⎟ ⎜x+ 2 −i 2 ⎟
⎠
⎠
⎠⎝
⎠⎝
⎝
⎝
1 ⎛
1
1
⎞
−
1
3
1
3⎟
i 3 ⎜
⎟
⎜ x+ −i
x+ +i
⎝
2
2
2
2 ⎠
Differentiating n times w.r.t. x,
yn =
Let
1
1
⎤
⎡
−
( −1) n n ! ⎢
n +1
n +1 ⎥
i 3
⎛
1
3⎞
1
3⎞ ⎥
⎢⎛
⎜x+ 2 +i 2 ⎟ ⎥
⎢ ⎜⎝ x + 2 − i 2 ⎟⎠
⎠ ⎦
⎝
⎣
1
1⎞
3
1⎞
3
⎛
⎛
= re iq , ⎜ x + ⎟ − i
= re − iq
⎜⎝ x + ⎟⎠ + i
⎝
⎠
2
2
2
2
3
2
1
3
⎛
⎞
−
1
2
where r = ⎜ x + ⎟ + , q = tan
⎝
1⎞
2⎠ 4
⎛
⎜⎝ x + ⎟⎠
2
x+
1
3
=
cotq
2
2
Substituting in r,
Hence,
r=
3 2
3
3
cot q + =
cosec q
4
4
2
yn =
( −1) n n ! ⎡
1
1 ⎤
⎢ ( re − iq ) n +1 − ( re iq ) n +1 ⎥
i 3 ⎣
⎦
=
( −1) n n! ⎡ e i ( n +1)q − e − i ( n +1)q ⎤
⎥
⎢
⎦
r n +1
i 3 ⎣
[Using Cor.1]
Differential Calculus I
=
2.9
( −1) n n ! 2i sin ( n + 1)q
⎞
⎛ 3
i 3⎜
cosecq ⎟
⎠
⎝ 2
=
n +1
( −1) n n ! 2n + 2 sin n +1 q sin ( n + 1)q
3
n+ 2
2
where q = tan −1
,
3
( 2 x + 1)
1
, find yn.
Example 9: If y = 4
x - a4
Solution:
y=
1
x4 − a4
=
1
( x 2 + a 2 )( x 2 − a 2 )
=
1
2a 2
⋅
( x2 + a2 − x2 + a2 )
( x 2 + a 2 )( x 2 − a 2 )
⎤
1 ⎡
1
1
− 2
2 ⎢ 2
2
2 ⎥
2a ⎢⎣ ( x − a ) ( x + a ) ⎥⎦
⎤
1 ⎡
1
1
= 2⎢
−
⎥
2a ⎣ ( x + a ) ( x − a ) ( x + ia ) ( x − ia ) ⎦
=
1 ⎡ 1 ⎧ ( x + a ) − ( x − a ) ⎫ 1 ⎧ ( x + ia ) − ( x − ia ) ⎫⎤
⎢ ⎨
⎬−
⎨
⎬⎥
2a 2 ⎢⎣ 2a ⎩ ( x + a )( x − a ) ⎭ 2ia ⎩ ( x + ia )( x − ia ) ⎭⎥⎦
1 ⎞
1 ⎛ 1
1 ⎞ 1 ⎛ 1
−
=
−
⎟
⎟− 3 ⎜
3 ⎜
4ia ⎝ x + ia x − ia ⎠ 4a ⎝ x + a x − a ⎠
Differentiating n times w.r.t. x,
(−1) n n ! ⎡
1
1
⎤
yn =
−
Hence,
⎥
3 ⎢
n +1
4ia ⎣ ( x + ia )
( x − ia ) n +1 ⎦
=
−
Let
where
x + ia = reiq, x
ia = re
iq
1
(−1) n n ! ⎡
1
⎤
−
⎢
3
n +1
n +1 ⎥
( x − a) ⎦
4a ⎣ ( x + a )
⎛a⎞
r = x 2 + a 2 , q = tan −1 ⎜ ⎟ .
⎝x⎠
yn =
n
( −1) n n ! ⎡ 1
1
1
1
⎤ ( −1) n ! ⎡
⎤
−
−
−
3
− iq n +1 ⎥
iq n +1
n +1
n +1 ⎥
3
⎢
⎢
( x − a) ⎦
4ia ⎣ ( re )
( re ) ⎦
4 a ⎣ ( x + a)
=
( −1) n n! 1
( −1) n n ! ⎡
1
1
⎤
⋅ n +1 [ e − i ( n +1)q − e i ( n +1)q ] −
−
n +1
n +1 ⎥
3
3
⎢
4ia
4 a ⎣ ( x + a)
( x − a) ⎦
r
=
( −1) n n !
⋅
4ia3
1
(x + a )
2
2
n +1
2
[ −2i sin ( n + 1)q ]
−
=
( −1) n +1 n !
n +1
2 2
2a 3 ( x 2 + a )
1
( −1) n n ! ⎡
1
⎤
−
n +1
n +1 ⎥
3
⎢
4 a ⎣ ( x + a)
( x − a) ⎦
sin( n + 1)q −
1
1
( −1) n n ! ⎡
⎤
−
n +1
n +1 ⎥
3
⎢
4 a ⎣ ( x + a)
( x − a) ⎦
2.10
Engineering Mathematics
Example 10: Find nth order derivatives of
(i) y = sin 2x sin 3x cos 4x
(ii) y = cos4 x
5
3
(iv) y = ex (sin x + cos x)
(iii) y = sin x cos x
x cosa
cos (x sin ` ).
(v) y = e
Solution:
y = sin 2x sin 3x cos 4x
1
= (cos x cos 5x) cos 4x
2
(i)
=
1
(cos x cos 4x
2
cos 5x cos 4x)
1
(cos 5x + cos 3x cos 9x cos x)
4
Differentiating n times w.r.t. x,
1⎡
np ⎞
np ⎞
⎛
⎛
n
yn = ⎢5n cos ⎜ 5 x +
⎟ + 3 cos ⎜⎝ 3 x +
⎟
⎝
4⎣
2 ⎠
2 ⎠
=
np
np ⎞
⎛
⎛
− 9n cos ⎜ 9 x +
⎟ − cos ⎜⎝ x +
⎝
2 ⎠
2
⎞⎤
⎟⎠ ⎥ [Using Cor. 2]
⎦
y = cos4 x
(ii)
⎛ 2 cos 2 x ⎞
=⎜
2 ⎟⎠
⎝
2
1
(1 + cos 2 x) 2
4
1
= (1 + cos 2 2 x + 2 cos 2 x)
4
1 ⎛ 1 + cos 4 x
⎞
+ 2 cos 2 x⎟
= ⎜1 +
⎝
⎠
2
4
1
= (3 + cos 4 x + 4 cos 2 x)
8
=
Differentiating n times w.r.t. x,
yn =
(iii)
1⎡ n
np ⎞
np
⎛
⎛
n
4 cos ⎜ 4 x +
⎟⎠ + 4.2 cos ⎜⎝ 2 x +
⎢
⎝
8⎣
2
2
y = sin5 x cos3 x
= sin2 x (sin x cos x)3
=
sin 2 x
23
sin 3 2 x
⎞⎤
⎟⎠ ⎥
⎦
[Using Cor. 2]
Differential Calculus I
2.11
1 ⎛ 1 − cos 2 x ⎞ ⎛ 3 sin 2 x − sin 6 x ⎞
⎜
⎟⎠ ⎜⎝
⎟⎠
8⎝
2
4
1
= 6 (3 sin 2 x − sin 6 x − 3 cos 2 x sin 2 x + cos 2 x sin 6 x)
2
1
1 ⎡
3
⎤
= 6 ⎢3 sin 2 x − sin 6 x − sin 4 x + (sin 8 x + sin 4 x) ⎥
2
2
2 ⎣
⎦
1 ⎡
1
⎤
= 6 ⎢3 sin 2 x − sin 4 x − sin 6 x + sin 8 x ⎥
2
2 ⎣
⎦
=
Differentiating n times w.r.t. x,
yn =
1
26
⎡ n
np
⎛
⎢ 2 ⋅ 3 sin ⎜⎝ 2 x + 2
⎣
np
⎛
− 6 n sin ⎜ 6 x +
⎝
2
np ⎞
⎞ n
⎛
⎟⎠ − 4 sin ⎜⎝ 4 x +
⎟
2 ⎠
np
⎞ 1 n
⎛
⎟⎠ + 8 sin ⎜⎝ 8 x +
2
2
⎞⎤
⎟⎠ ⎥
⎦
[Using result (6)]
(iv) y = ex (sin x + cos x)
Differentiating n times w.r.t. x,
n
yn = (1 + 1) 2 ⋅ e x [sin ( x + n tan −1 1) + cos ( x + n tan −1 1)] [Using result (7) and Cor. 3]
n
⎡ ⎛
np
= 2 2 e x ⎢sin ⎜ x +
4
⎣ ⎝
np ⎞ ⎤
⎞
⎛
⎟⎠ + cos ⎜⎝ x +
⎟
4 ⎠ ⎥⎦
n
⎡ ⎛
np ⎞
np
⎛p
= 2 2 e x ⎢sin ⎜ x +
+ sin ⎜ + x +
⎟
⎝2
4 ⎠
4
⎣ ⎝
p np
⎛
2x + +
⎜
2 2
= 2 e ⋅ 2 sin ⎜
2
⎜
⎝
n
2
x
n
= 22 ex ⋅
=2
n +1
2
⎞⎤
⎟⎠ ⎥
⎦
⎞
p
⎟
⎟ cos 4
⎟
⎠
⎤
⎡ p
sin ⎢ x + ( n + 1) ⎥
⎣
⎦
4
2
2
p⎤
⎡
e x sin ⎢ x + ( n + 1) ⎥
⎣
4⎦
(v) y = e xcosa cos (x sin a)
Differentiating n times w.r.t. x,
n
sin a ⎞
⎛
yn = (cos 2 a + sin 2 a ) 2 e x cos a ⋅ cos ⎜ x sin a + n tan −1
⎟ [Using result (7)]
⎝
cos a ⎠
= e x cos a cos ( x sin a + na )
2.12
Engineering Mathematics
1
n
n
Example 11: If y(x) = sin px + cos px, prove that yn ( x ) = p [1 + ( -1) sin 2 px ]2 .
1
Hence, find y8(p ), when p = .
4
Solution: y(x) = sin px + cos px
Differentiating n times w.r.t. x,
⎡ ⎛
np ⎞
np
⎛
yn ( x ) = p n ⎢sin ⎜ px +
⎟⎠ + cos ⎜⎝ px +
⎝
2
2
⎣
⎞⎤
⎟⎠ ⎥
⎦
⎡⎧ ⎛
np ⎞
np
⎛
= p n ⎢⎨sin ⎜ px +
⎟⎠ + cos ⎜⎝ px +
⎝
2
2
⎢⎣⎩
[ Using result (66) and Cor.2]
1
2 2
⎞⎫ ⎤
⎟⎠ ⎬ ⎥
⎭ ⎥⎦
⎡
np
np ⎞
np ⎞
np ⎞
⎛
⎛
⎛
2 ⎛
= p ⎢sin 2 ⎜ px +
⎟⎠ + cos ⎜⎝ px +
⎟⎠ + 2 sin ⎜⎝ px +
⎟ cos ⎜⎝ px +
⎝
⎠
2
2
2
2
⎣
n
1
⎞⎤ 2
⎟⎠ ⎥
⎦
1
= p n [1 + sin ( 2 px + np )] 2
1
= p n [1 + sin 2 px cos np + cos 2 px sin np ] 2
1
= p n [1 + ( −1) n sin 2 px ] 2
Putting n = 8, p =
1
and x = p,
4
⎛1⎞
y8 (p ) = ⎜ ⎟
⎝4⎠
8
1
⎡
⎛ p ⎞⎤ 2
8
⎢1 + ( −1) sin 2 ⎜⎝ 4 ⎟⎠ ⎥
⎦
⎣
1
8
31
⎛1⎞ ⎡
⎛ p ⎞⎤ 2 ⎛ 1 ⎞ 2
= ⎜ ⎟ ⎢1 + sin ⎜ ⎟ ⎥ = ⎜ ⎟
⎝4⎠ ⎣
⎝2⎠
⎝ 2 ⎠⎦
( -1) n -1 ( n - 1)!
-1 ⎛ x ⎞
Example 12: If y = tan ⎜ ⎟ , prove that yn =
sin np sin n p ,
⎝a⎠
an
⎛a⎞
where p = tan -1 ⎜ ⎟ .
⎝ x⎠
⎛ x⎞
Solution: y = tan −1 ⎜ ⎟
⎝ a⎠
Differentiating w.r.t. x,
y1 =
1
2
1+
x
a2
⋅
a
a
1
=
=
a a 2 + x 2 ( x + ia )( x − ia )
Differential Calculus I
=
1 ⎡ ( x + ia ) − ( x − ia ) ⎤
2i ⎢⎣ ( x + ia ) ( x − ia ) ⎥⎦
=
1⎛ 1
1 ⎞
−
⎜
2i ⎝ x − ia x + ia ⎟⎠
2.13
Differentiating (n - 1) times w.r.t. x,
yn =
Let
x + ia = reiq, x
( −1) n −1 (n − 1)! ⎡
1
1
⎤
−
⎢
n
n⎥
2i
( x + ia ) ⎦
⎣ ( x − ia )
ia = re
iq
a
⎛a⎞
r = x 2 + a 2 , q = tan −1 ⎜ ⎟ , tan q = , x = a cot q
⎝x⎠
x
where,
r = a 2 cot 2 q + a 2 = a cosecq
Hence,
yn =
=
( −1) n −1 ( n − 1)! ⎡ 1
1 ⎤
− iq n ⎥
− iq n
⎢
2i
⎣ ( re ) ( re ) ⎦
( −1) n −1 ( n − 1)! 1 ( inq
e − e − inq )
2i
rn
( −1) n −1 ( n − 1)! 2i sin nq
⋅
2i
rn
sin nq
= ( −1) n −1 ( n − 1)!
rn
sin nq
= ( −1) n −1 ( n − 1)! n
a cosec n q
( −1) n −1 ( n − 1)!
=
sin nq sin n q .
n
a
=
where,
⎛a⎞
q = tan −1 ⎜ ⎟ .
⎝x⎠
⎛ 1 + x2 - 1 ⎞
-1
Example 13: If y = tan ⎜⎝
⎟⎠ , prove that
x
yn =
[Using Cor. 1]
1
(–1)n−1 (n – 1) ! sinn q sin nq, where q = cot−1 x.
2
⎛ 1 + x 2 − 1⎞
Solution: y = tan −1 ⎜
⎟
⎝
⎠
x
Putting x = tan f,
2.14
Engineering Mathematics
⎛ sec f − 1 ⎞
⎛ 1 − cos f ⎞
y = tan −1 ⎜
= tan −1 ⎜
⎟
⎝ tan f ⎠
⎝ sin f ⎟⎠
f ⎞
⎛
2 sin 2
⎜
2 ⎟ = tan −1 tan f = f
= tan ⎜
f
f⎟
2 2
⎜ 2 sin cos ⎟
⎝
2
2⎠
−1
y=
1 −1
tan x
2
y1 =
1
1
⋅
2 (1 + x 2 )
Differentiating w.r.t. x,
=
1
1
1 ( x + i) − ( x − i)
⋅
= ⋅
2 ( x + i )( x − i ) 4i ( x + i )( x − i )
=
1⎛ 1
1 ⎞
−
⎜⎝
⎟
4i x − i x + i ⎠
Differentiating (n - 1) times w.r.t. x,
yn =
Let
where,
x + i = r eiq, x
i = re
( −1) n −1 (n − 1)! ⎡ 1
1 ⎤
−
⎥
⎢
n
4i
( x + i)n ⎦
⎣ ( x − i)
iq
1
⎛1⎞
r = x 2 + 1, q = tan −1 ⎜ ⎟ , tan q = , x = cot q
⎝x⎠
x
r = cot 2 q + 1 = cosecq
Hence,
yn =
( −1) n −1 ( n − 1)! ⎡ 1
1 ⎤
⎢ − iq n − iq n ⎥
4i
(
)
(
) ⎦
re
re
⎣
( −1) n −1 ( n − 1)! 1 inq
(e − e −inq )
4i
rn
( −1) n −1 ( n − 1)! 2i sin nq
=
⋅
4i
rn
=
( −1) n −1 ( n − 1)!sin nq
2r n
n −1
( −1) ( n − 1)!sin nq
=
2 cosec n q
1
⎛1⎞
= ( −1) n −1 ( n − 1)! sin n q sin nq , where q = tan −1 ⎜ ⎟ .
⎝x⎠
2
=
Differential Calculus I
2.15
-1 ⎛ 1 + x ⎞
, prove that yn = (-1)n−1 (n - 1) ! sinnp sin np
Example 14: If y = tan ⎜
⎝ 1 - x ⎟⎠
⎛1⎞
where θ = tan −1 ⎜ ⎟ .
⎝ x⎠
−1 ⎛ 1 + x ⎞
Solution: y = tan ⎜
⎝ 1 − x ⎟⎠
Putting x = tan f,
⎛ 1 + tan f ⎞
y = tan −1 ⎜
⎝ 1 − tan f ⎟⎠
p
⎛
⎞
tan + tan f
⎜
⎟
4
= tan ⎜
⎟
p
⎜ 1 − tan ⋅ tan f ⎟
⎝
⎠
4
−1
⎡ ⎛p
p
⎞⎤ p
= tan −1 ⎢ tan ⎜ + f ⎟ ⎥ = + f = + tan −1 x
⎠⎦ 4
4
⎣ ⎝4
Proceeding as in Example 13, we get
yn = ( 1)n 1 (n
1)! sinn q sin nq.
⎛ x - x -1 ⎞
.
⎝ x + x -1 ⎟⎠
-1
Example 15: Find the nth order derivative of y = cos ⎜
⎛ x − x −1 ⎞
⎛ 2 ⎞
−1 x − 1
Solution: y = cos −1 ⎜
cos
=
⎟
⎜ 2 ⎟
⎝ x + x −1 ⎠
⎝ x + 1⎠
Putting x = tan f,
⎛ tan 2 f − 1 ⎞
= cos −1 ( − cos 2f )
y = cos −1 ⎜ 2
⎝ tan f + 1 ⎟⎠
= cos −1 [cos(p − 2f )] = p − 2f = p − 2 tan −1 x
Proceeding as in Example 13, we get
⎛ x − x −1 ⎞
.
y = cos −1 ⎜
⎝ x + x −1 ⎟⎠
Example 16: If y = x log (1 + x), prove that yn =
( -1) n 2 ( n - 2)!( x + n)
.
( x + 1) n
Solution: y = x log (1 + x)
Differentiating w.r.t. x,
y1 = log (1 + x) +
x
1
= log (1 + x) + 1 −
1+ x
1+ x
2.16
Engineering Mathematics
Differentiating (n - 1) times w.r.t. x,
yn =
=
=
=
( −1) n − 2 (n − 2)!
( x + 1) n −1
+0−
( −1) n −1 (n − 1)!
( x + 1) n
[Using result (3) and Cor. 1]
( −1) n − 2 (n − 2)! ⎡
1
( −1)1 (n − 1) ⎤
−
⎢
⎥
−1
1
( x + 1) n
⎣ ( x + 1)
⎦
( −1) n − 2 (n − 2) !
( x + 1) n
( x + 1 + n − 1)
( −1) n − 2 (n − 2) ! ( x + n)
( x + 1) n
.
x+n ⎤
⎛ x - 1⎞
⎡ x-n
, prove that yn = (-1)n (n - 2) ! ⎢
,
Example 17: If y = x log ⎜
⎟
n
⎝ x + 1⎠
( x + 1) n ⎥⎦
⎣ ( x - 1)
n ê 2.
⎛ x − 1⎞
Solution: y = x log ⎜
⎝ x + 1⎟⎠
= x log( x − 1) − x log( x + 1)
Differentiating y w.r.t. x,
x
x
−
− log( x + 1)
x −1 x +1
1
1
= log( x − 1) + 1 +
−1+
− log(xx +1)
x −1
x +1
Differentiating (n - 1) times w.r.t. x,
y1 = log( x − 1) +
yn =
( −1) n − 2 (n − 2)!
( x − 1) n −1
+
( −1) n −1 (n − 1)!
( x − 1) n
+
( −1) n −1 (n − 1)!
( x + 1) n
−
( −1) n − 2 (n − 2)!
( x + 1) n −1
[if n – 2 0]
[Using result (3) and Cor. 1]
=
=
(−1) n (n − 2)! ⎡ (−1) −2
(−1) −1 (n − 1) ⎤
+
⎢
⎥
n
−1
1
( x − 1)
⎢⎣ ( x − 1)
⎥⎦
−
1
n
(−1) (n − 2)! ⎡ (−1) (n − 1) (−1) −2 ⎤
−
+
⎢
⎥
1
( x + 1) n ⎢⎣
( x + 1) −1 ⎥⎦
(−1) n (n − 2) !
( x − 1) n
( x − 1 − n + 1) +
(−1) n (n − 2)!
( x + 1) n
x+n ⎤
⎡ x−n
= (−1) n (n − 2)! ⎢
−
⎥ , n ≥ 2.
n
( x + 1) n ⎦
⎣ ( x − 1)
(−n + 1 − x − 1)
Differential Calculus I
2.17
Example 18: If y = x coth−1 x, prove that
yn =
( -1) n ( n - 2)! ⎡ x + n
x-n ⎤
, n ≥ 2.
n
⎢
2
( x - 1) n ⎥⎦
⎣ ( x + 1)
y = x coth−1 x
Solution:
⎛ 1⎞
= x tanh −1 ⎜ ⎟
⎝ x⎠
1
x
1
1−
x
x
⎛ x + 1⎞
= log ⎜
⎝ x − 1⎟⎠
2
1
= x ⋅ log
2
1+
1
[ x log ( x + 1) − x log ( x − 1)]
2
Proceeding as in Example 17, we get
y=
yn =
( −1) n (n − 2)! ⎡ x + n
x−n ⎤
.
⎢ ( 1) n − ( 1) n ⎥ , n ≥ 2
2
x− ⎦
⎣ x+
Example 19: If y = (x – 1)n, prove that y +
Solution: y = (x
y1 y2 y3
y
+
+
+ ... + n = x n .
n!
1! 2 ! 3 !
1)n
Differentiating w.r.t. x successively,
y1 = n(x 1)n 1
y2 = n (n 1) (x 1)n 2
y3 = n (n 1) (n 2) (x 1)n 3
………………………………
...…………………………….
yn = n !
Hence,
y+
= ( x − 1) n +
y1 y2 y3
+
+
+
1! 2 ! 3!
+
yn
n!
n
n(n − 1)
n(n − 1)(n − 2)
( x − 1) n −1 +
( x − 1) n − 2 +
( x − 1) n −3 +
1!
2!
3!
= ( x − 1) n + nC1 ( x − 1) n −1 + nC2 ( x − 1) n − 2 + nC3 ( x − 1) n −3 +
+
n!
n!
+ nCn
= [1 + ( x − 1) ]
n
= xn .
[ Using Binomial Expansion ]
2.18
Example 20: If I n =
Engineering Mathematics
dn
( x n log x ), prove that In = n In−1 + (n - 1) !
dx n
1 1
1⎞
⎛
Hence, prove that I n = n ! ⎜ log x + 1 + + + ... + ⎟ .
⎝
2 3
n⎠
In =
Solution:
For
n=1
I1 =
In =
=
dn
dx n
( x n log x)
d
( x log x) = log x + 1
dx
d n −1 ⎡ d n
⎤
( x log x) ⎥
n −1 ⎢
d
x
⎦
⎣
dx
d n −1 ⎛ n −1
n
⎜ nx log x + x
dx n −1 ⎝
=n
1⎞
⎟
x⎠
d n −1
d n −1
dx
d x n −1
( x n −1 log x) +
n −1
( x n −1 )
I n = nI n −1 + (n − 1)!
In
I
1
= n −1 +
n ! (n − 1)! n
Putting n = 2, 3, 4, ……. in Eq. (1),
I 2 I1 1
= +
2 ! 1! 2
I3 I 2 1
=
+
3! 2 ! 3
I 4 I3 1
= +
4 ! 3! 4
………………
.……………...
I n −1
I
1
= n−2 +
(n − 1)! (n − 2)! n − 1
From Eq. (1),
In − 1
In
1
=
+
n ! (n − 1)! n
... (1)
Differential Calculus I
2.19
Adding all the above equations,
In
1 1 1
1
= I1 + + + + … +
n!
n
2 3 4
1 1 1
1⎞
⎛
I n = n !⎜ log x + 1 + + + + … + ⎟
⎝
n⎠
2 3 4
(
1
)
1
dn
( -1) n e x
n-1 x
x
e
Example 21: Prove that
=
.
x n+1
dx n
Solution:
dn
dx n
(x
1
n −1 x
e
)
⎡ n −1 ⎛ 1
⎤
1
1
1
+
+… +
⎢ x ⎜⎜1 + +
⎥
2
3
−
1
n
3! x
(n − 1)! x
⎥
d ⎢
⎝ x 2! x
= n⎢
⎥
dx ⎢
⎞⎥
1
1
1
+
+
…
+
+
⎟⎟ ⎥
⎢
n ! x n (n + 1)! x n +1 (n + 2)! x n + 2
⎠⎦
⎣
n
=
d n ⎡ n −1
1
1
x n −3
+… +
+
x + xn−2 +
n ⎢
−
2
!
(
1
)!
(
n
x
n !)
dx ⎣
1
1
⎤
+ 2
+
+ …⎥
x (n + 1)! x3 (n + 2)!
⎦
= 0+
=
=
1(−1) n n !
n ! x n +1
+
(−1) n (n + 1)!
(n + 1)! (1!) x n + 2
+
(−1) n (n + 2)!
(n + 2)! (2 !) x n + 3
+ ....
[Using result (1) and (2)]
1
(−1) n ⎛ 1
⎞
1+ +
+ …⎟
n +1 ⎜
2
x
2! x
x
⎠
⎝
(−1) n
x n +1
1
ex.
Exercise 2.1
1. Find the nth order derivative of
x
x +1
(i) y = 2
(ii) y =
.
1
−
4 x2
x −4
⎡
3 ( −1) n n ! 1 ( −1) n n ! ⎤
+
⎢ Ans. : (i)
⎥
4 ( x − 2) n +1 4 ( x + 2) n +1 ⎥
⎢
⎢
⎥
1 ⎡ ( −1) n n ! ( −2) n
⎢
⎥
(ii) ⎢
n +1
4 ⎣ (1 − 2 x )
⎢
⎥
⎢
⎥
n
n
( −1) n ! 2 ⎤ ⎥
⎢
−
⎥
⎢
(1 + 2 x ) n +1 ⎦ ⎥⎦
⎣
2.20
Engineering Mathematics
2. Find the nth order derivative of
x
y=
.
( x − 1)( x − 2)( x − 3)
⎡
⎤
1
⎡
n
⎢ Ans. : ( −1) n ! ⎢ 2( x − 1) n +1
⎥
⎣
⎢
⎥
⎢
2
3
⎤⎥
−
+
⎢
⎥
( x − 2) n +1 2( x − 3) n +1 ⎥⎦ ⎥⎦
⎢⎣
3. Find the nth order derivative of
x
y=
.
1 + 3x + 2 x 2
⎡
⎤
1
⎡
n
⎢ Ans. : (−1) n ! ⎢
n +1 ⎥
⎣ ( x + 1) ⎥
⎢
⎢
⎤ ⎥⎥
2n
⎢
−
⎥
⎢
(2 x + 1) n +1 ⎥⎦ ⎥⎦
⎣
4. Find the nth order derivative of
x2
.
y=
( x + 2)(2 x + 3)
⎡
−4( −1) n n ! 9( −1) n n ! (2) n −1 ⎤
+
⎢ Ans. :
⎥
( x + 2) n +1
(2 x + 3) n +1 ⎥⎦
⎢⎣
5. Find the nth order derivative of
2x + 3
y=
.
( x − 1) 2
⎡
2 ( −1) n n ! 5 ( −1) n (n + 1)!⎤
+
⎢ Ans. :
⎥
( x − 1) n +1
( x − 1) n + 2 ⎥⎦
⎢⎣
6. Find the nth order derivative of
x
y=
.
( x − 1) 2
⎡
( −1) n n ! ( −1) n (n + 1)!⎤
+
⎢ Ans. :
⎥
( x − 1) n +1
( x − 1) n + 2 ⎥⎦
⎢⎣
7. Find the nth order derivative of
x +1
y=
.
( x − 1) n
( x − 1) + 2
⎤
⎡
⎥
⎢ Hint : y = ( 1) n
x
−
⎥
⎢
⎢
1
2 ⎥
=
+
⎥
⎢
( x − 1) n −1 ( x − 1) n ⎥⎦
⎢⎣
⎡
⎡ (2n − 2)! (2n − 1)! ⎤ ⎤
−1) n ⎢
+
⎢ Ans. : (−
⎥⎥
2 n −1
( x − 1) 2 n ⎦ ⎥⎦
⎢⎣
⎣ ( x − 1)
8. Find the nth order derivative of
4x
.
y=
( x − 1) 2 ( x + 1)
⎡
⎤
1
⎡
n
⎢ Ans. : ( −1) n ! ⎢
⎥
n +1
⎣ ( x − 1)
⎢
⎥
⎢
2 (n + 1)
n! ⎤ ⎥
⎢
+
−
⎥⎥
⎢⎣
( x − 1) n + 2 ( x + 1) n +1 ⎦ ⎥⎦
9. Find the nth order derivative of
1
.
y=
(3 x − 2)( x − 3) 2
⎡
⎤
⎡
3n + 2
n
Ans.
:
(
−
1
)
n
!
⎢
⎢
n +1 ⎥
⎣ 49 (3 x − 2) ⎥
⎢
⎢
3
( n + 1) ⎤ ⎥
⎢−
⎥
+
n +1
7 ( x − 3) n + 2 ⎥⎦ ⎥⎦
⎢⎣ 49 ( x − 3)
10. If y =
x2
2 x2 + 7 x + 6
, find y .
n
⎡ Hint : Divide x 2 by 2 x 2 + 7 x + 6,
⎢
1
7x + 6
⎢
y= −
⎢⎣
2 2 ( x + 2) ( 2 x + 3)
⎡
⎤
8
⎡
n
⎢ Ans.. : ( −1) n ! ⎢ − ( x + 2) n +1 ⎥
⎣
⎢
⎥
n
⎢
⎤⎥
9( 2)
⎢
⎥
+
n +1 ⎥
( 2 x + 3) ⎦ ⎥⎦
⎢⎣
4 ⎛
x3 ⎞
11. Prove that d
= 0.
⎜
⎟
dx 4 ⎝ x 2 − 1⎠ x = 0
x
, prove that
12. If y = 2
x + a2
yn = ( 1)n n ! a n 1 (sin q )n+1
cos (n + 1) q.
x
, prove that
13. If y = 2
x +1
yn = (-1) n n! sinn+1 q cos (n + 1) q
⎛1⎞
where q = tan −1 ⎜ ⎟ .
⎝x⎠
⎤
⎥
⎥
⎥⎦
Differential Calculus I
14. Find the nth order derivative of
1
.
y=
1 + x + x 2 + x3
1
⎡
⎤
⎢ Hint : y = (1 + x ) (1 + x 2 )
⎥
⎢
⎥
1
⎢
⎥
=
⎢
(1 + x ) ( x + i ) ( x − i ) ⎥⎦
⎣
⎡
⎤
1
( −1) n n ! ⎡
Ans.
:
⎢
⎥
n +1
⎢
2
⎣ (1 + x )
⎢
⎥
⎢
1
⎤⎥
⎢ + n +1 {sin ( n + 1)q − cos ( n + 1)q }⎥ ⎥
⎦⎦
⎣ 2r
15. Find the nth order derivative of
x
.
y=
1 + x + x2
⎡
⎤ ⎤
⎡cos( n + 1)q
n
⎢ Ans. : ( −1) n ! ⎢
⎥ ,⎥
1
⎢
sin( n + 1)q ⎥ ⎥
r n +1 ⎢ −
⎢
⎥⎦ ⎥
⎢⎣ 3
⎢
⎥
⎢ where r = x 2 + x + 1,
⎥
⎢
⎥
⎛ 3 ⎞
⎢
⎥
−1
q = tan ⎜
⎟
⎢
⎥
⎝ 2x + 1⎠
⎣
⎦
16. Prove that
dn
dx n
tan
1
x
1⎞
⎛
sin ⎜ n tan −1 ⎟
x⎠
= (−1) n −1 (n − 1)! ⎝
.
n
( x 2 + 1) 2
17. Find the nth order derivatives of
⎛ 2x ⎞
y = tan −1 ⎜
.
⎝ 1 − x 2 ⎟⎠
⎡ Ans. : 2( −1) n −1 ( n − 1) ! (sin q ) n sin nq ,⎤
⎢
⎥
⎢ where q = tan −1 ⎛ 1 ⎞ .
⎥
⎜⎝ ⎟⎠
⎢⎣
⎥⎦
x
−1 ⎛ 2 x ⎞
18. If y = sin ⎜⎝
⎟ , prove that
1 + x2 ⎠
2.21
yn = 2 (-1)n−1 (n - 1) ! sinn q sin nq,
1
where q = tan −1 ⎛⎜ ⎞⎟ .
⎝x⎠
2⎞
⎛
−1 1 + x
y
=
sec
19. If
⎜
⎟ , prove that
⎝ 1 − x2 ⎠
yn = 2 (-1)n−1 (n - 1) ! sinn q sin nq.
20. Find the nth order derivative w.r.t. x of
(ii) sin7 x
(i) sin4 x
7
⎡
⎡ 1 ix
7
− ix ⎤
⎢ Hint : sin x = ⎢ (e − e ) ⎥ ,
⎦
⎣ 2i
⎢
⎢⎣expand using binomial expaansion
(iii) sin3 x cos2 x
⎤
⎥
⎥
⎥⎦
(iv) sin3 3x.
⎡
np ⎞ ⎤
⎛
n −1
⎢ Ans. : (i) − 2 cos ⎜⎝ 2 x + 2 ⎟⎠ ⎥
⎢
⎥
⎢
⎥
n
p
⎛
⎞
+ 22 n − 3 cos ⎜ 4 x +
⎢
⎟⎠ ⎥
⎝
2 ⎥
⎢
⎢
⎥
np ⎞
1 ⎡
⎛
⎢(ii) − ⎢7n sin ⎜ 7 x +
⎥
⎟
⎝
2 ⎠
64 ⎣
⎢
⎥
⎢
⎥
⎢ −7.5n sin ⎛⎜ 5 x + np ⎞⎟ + 21.3n sin ⎤ ⎥
⎥⎥
⎝
⎢
2 ⎠
⎥⎥
⎢
⎥⎥
p
n
p
n
⎛
⎞
⎛
⎞
⎢ 3x +
⎟⎠ − 35 sin ⎜⎝ x +
⎟⎠ ⎥ ⎥
⎢ ⎜⎝
2
2
⎦
⎢
⎥
1 ⎡
np ⎞ n
⎛
⎢
⎥
⎢(iii) 16 ⎢ 2 sin ⎜⎝ x + 2 ⎟⎠ + 3 sin ⎥
⎣
⎢
⎥
⎢⎛
np ⎞ n
np ⎞ ⎤ ⎥
⎛
⎢ ⎜ 3x +
⎟ − 5 sin ⎜⎝ 5 x +
⎟ ⎥
2 ⎠
2 ⎠ ⎥⎦ ⎥
⎢⎝
⎢
⎥
3n +1
np ⎞
⎛
⎢(iv)
⎥
sin ⎜ 3 x +
⎟
⎝
4
2 ⎠
⎢
⎥
⎢
⎥
np ⎞
1
⎛
⎢
⎥
− ⋅ 32n sin ⎜ 9 x +
⎟
⎝
4
2 ⎠
⎣⎢
⎦⎥
21. Find the nth order derivative w.r.t. x of
(i) sin 2x cos 6x (ii) sin x cos 3x
(iii) cos x cos 2x cos 3x.
2.22
Engineering Mathematics
⎡
1⎡ n
np ⎞
⎛
⎢ Ans. : (i) ⎢8 sin ⎜⎝ 8 x +
⎟
2
2 ⎠
⎣
⎢
⎢
np ⎞ ⎤
⎛
⎢
− 4 n sin ⎜ 4 x +
⎟
⎝
2 ⎠ ⎥⎦
⎢
⎢
1⎡
np ⎞
⎛
⎢
(ii) ⎢ 4 n sin ⎜ 4 x +
⎟
⎢
⎝
2⎣
2 ⎠
⎢
np ⎞ ⎤
⎛
⎢
− 2n sin ⎜ 2 x +
⎟
⎢
⎝
2 ⎠ ⎥⎦
⎢
⎢
1⎡
np ⎞
⎛
(iii) ⎢6 n cos ⎜ 6 x +
⎢
⎟
⎝
2 ⎠
4⎣
⎢
⎢
np ⎞ n
np
⎛
⎛
⎢ +4 n cos ⎜ 4 x +
⎟⎠ + 2 cos ⎜⎝ 2 x +
⎝
2
2
⎢⎣
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎞⎤⎥
⎟⎠ ⎥ ⎥
⎦ ⎥⎦
22. Find the nth order derivative w.r.t. x of
(i) e5x cos x cos 3x
(ii) ex cos x cos 2x
(iii) eax cos2 x sin x
(iv) 2x sin2 x cos x
(v) 2x sin (3x + 1).
1 5x
⎡
⎤
⎢ Ans. : (i) 2 e
⎥
⎢
⎥
n
⎡
4
⎢
⎥
⎛
⎞
1
−
2
⎢( 41) cos ⎜⎝ 4 x + n tan
⎟⎠ ⎥
⎢
5
⎣
⎢
⎥
n
⎢
⎥
⎤
2
⎛
⎞
−
1
⎢
⎥
+ ( 29) 2 cos ⎜ 2 x + n tan
⎥
⎟
⎝
5 ⎠ ⎦ ⎥⎦
⎢⎣
⎢
⎥
n
⎢
⎥
1 x⎡
(ii) e ⎢(10) 2 cos (3 x +
⎢
⎥
2 ⎣
⎢
⎥
n
⎢
⎥
⎤
n
p
⎛
⎞
⎢
⎥
n tan −1 3) + ( 2) 2 cos ⎜ x +
⎥
⎟
⎝
4 ⎠⎦⎥
⎢
⎢
⎥
n
e ax ⎡ 2
⎢
⎥
2
(iii)
⎢ ( a + 9)
⎢
⎥
4 ⎣
⎢
⎥
⎢
⎥
⎛
−1 3 ⎞
sin ⎜ 3 x + n tan
⎟⎠
⎢
⎥
⎝
a
⎢
⎥
n
⎢
⎥
⎤
1
⎛
⎞
1
−
2
⎢ +( a + 1) 2 sin ⎜ x + n tan
⎥
⎥
⎟
⎝
a ⎠⎦
⎢
⎥
⎢
⎥
1
⎢
⎥
(iv) − r1n 2 x cos (3 x + nf1)
4
⎢
⎥
⎢
⎥
1 n x
+ r2 2 cos ( x + nf 2)
⎢
⎥
4
⎢
⎥
⎢
where r1 = (log 2) 2 + 32 , ⎥
⎢
⎥
⎢
⎥
⎛
⎞
3
f1 = tan −1 ⎜
,
⎢
⎥
⎟
⎝ log 2 ⎠
⎢
⎥
⎢
⎥
2
r2 = (log 2) + 1,
⎢
⎥
⎢
⎥
⎛
⎞
1
⎢
⎥
f 2 = tan −1 ⎜
⎝ log 2 ⎟⎠ ⎥
⎢
⎢
⎥
n
2
x
⎢
⎥
2
( v) 2 [(log 2) + 9]
⎢
⎥
⎢
⎛
3 ⎞ ⎥
−1
sin ⎜ 3 x + 1 + n tan
⎢
⎥
⎝
log 2 ⎟⎠ ⎦
⎣
2.3 LEIBNITZ’S THEOREM
Statement: If y = uv, where u and v are two functions of x whose nth derivatives are
known, then
yn = (uv) n = un v + n C1 un −1v1 + nC2 un − 2 v2 + nC3un − 3 v3 + … + nCn uvn .
Proof:
y = uv
Differential Calculus I
2.23
Differentiating w.r.t. x successively,
y1 = u1v + uv1
= 1C 0 u1v + 1C1u v1
y2 = u2v + 2 u1v1 + uv2
= 2C 0 u2 v + 2C1u1v1 + 2C 2 u v2
This shows that theorem is true for n = 1 and n = 2.
Let the theorem is true for n = m.
ym = (uv) m = mC 0 um v + mC1um −1v1 + mC2 um − 2 v2 + ........... + mCm uvm .
Differentiating ym w.r.t. x,
d
d
ym = (uv) m = mC 0 (um +1v + um v1 ) + mC1 (um v1 + um −1v2 ) + m C2 (um −1 v2 + um − 2 v3 )
dx
dx
+ ........... + m C m (u1vm + uvm +1 )
= mC 0 u m + 1 v +
ym +1 = ( uv )m +1 =
+… +
(
)
C 0 + mC1 um v1 +
m
m +1
C 0 u m + 1v +
m +1
(
)
C1 + mC 2 um −1v2 + ........... + mC m uvm + 1
m
C1 um v1 +
m +1
C 2 um −1v2
⎡∵ mC + mC =
r −1
r
⎣
m +1
C m +1 uvm +1 .
Cr ⎤
⎦
m +1
This shows that theorem is true for n = m + 1 also.
By mathematical induction, theorem is true for all integer values of n.
Hence, yn = (uv) n = un v + n C1 un −1v1 + n C2 un − 2 v2 + n C3un − 3 v3 + … + n Cn uvn .
Example 1: If y = x2 e2x, prove that (yn)0 = 2n-2 n (n - 1).
Solution:
y = x2 e2x
Differentiating n times using Leibnitz’s Theorem,
n (n − 1)
yn = x 2 2n e 2 x + n ⋅ 2 x 2n −1 e 2 x +
⋅ 2 ⋅ 2n − 2 e 2 x
2!
Putting x = 0,
(yn)0 = 2n 2 n (n
1)
Example 2: If u is a function of x, and y = eax u, prove that Dny = eax (D + a)n u,
d
where D =
.
dx
Solution:
y = eax u
Dn (eax u) = (Dn eax) u + nC1 (Dn−1 eax) (Du) + nC2 (Dn−2 eax) (D2u)
+ nC3 (Dn−3 eax) (D3u) + ………. + eax (Dnu)
2.24
Engineering Mathematics
= (an eax) u + nC1 (an−1 eax) (Du) + nC2 (an−2 eax) (D2u)
+ nC3 (an−3 eax) (D3u) + ………. + eax (Dnu)
= eax [an + nC1 an−1 (D) + nC2 an−2 (D2) + nC3 an−3 (D3) + … + Dn] u
= eax (D + a)n u, where D = d
dx
[Using Binomial Expansion]
Example 3: Find the nth order derivative of
(ii) x3 cos x
(iii) x2 ex cos x
(i) y = x2 eax
(iv) x2 tan−1 x.
Solution: (i) y = x2 eax
Let u = eax, v = x2
Differentiating n times using Leibnitz’s Theorem,
yn = un v + nC1 un −1v1 + nC2 un − 2 v2 + … + nCn u vn
= a n e ax ⋅ x 2 + n a n −1e ax ⋅ 2 x +
n(n − 1) n − 2 ax
a e ⋅2
2!
= e ax [ x 2 a n + 2nxa n −1 + n(n − 1)a n − 2 ]
(ii) y = x3 cos x
Let u = cos x, v = x3
Differentiating n times using Leibnitz’s Theorem,
yn = un v + nC1 un −1v1 + nC2 un − 2 v2 + … + nCn u vn
np
⎛
= cos ⎜ x +
⎝
2
( n − 1)p
⎞ 3
⎡
⎟⎠ x + n cos ⎢ x +
2
⎣
⎤ 2
⎥ 3x
⎦
n( n − 1)( n − 2)
( n − 3)p
n( n − 1)
( n − 2)p ⎤
⎡
⎡
cos ⎢ x +
6x +
+
cos ⎢ x +
⎥
2 ⎦
3!
2
2!
⎣
⎣
np
⎛
= x 3 cos ⎜ x +
⎝
2
⎤
⎥6
⎦
( n − 1)p ⎤
⎞
⎡
2
⎟⎠ + 3nx cos ⎢ x +
2 ⎥⎦
⎣
( n − 3)p ⎤
( n − 2)p ⎤
⎡
⎡
+ 3 x n( n − 1) cos ⎢ x +
+ n( n − 1)( n − 2) cos ⎢ x +
⎥
2 ⎥⎦
2 ⎦
⎣
⎣
(iii) y = x2 ex cos x
Let u = ex cos x, v = x2.
n
np ⎞
⎛
un = e x ( 2) 2 cos ⎜ x +
⎟
⎝
4 ⎠
Differentiating n times using Leibnitz’s Theorem,
[Using result (7)]
yn = un v + nC1un −1v1 + nC2 un − 2 v2 + … + nCn u vn
n
np ⎞ 2
⎛
x
= e x ( 2) 2 cos ⎜ x +
⎟⎠ x + ne ( 2)
⎝
4
n −1
2
( n − 1)p ⎤
⎡
cos ⎢ x +
2x
⎣
4 ⎦⎥
Differential Calculus I
+
n( n − 1) n −2 2
( n − 2)p
⎡
( 2) cos ⎢ x +
2!
4
⎣
n
np ⎞ 2
⎛
x
= e x ( 2) 2 cos ⎜ x +
⎟ x + nxe ( 2)
⎝
4 ⎠
+
2.25
n +1
2
⎤
⎥2
⎦
( n − 1)p ⎤
⎡
cos ⎢ x +
4 ⎥⎦
⎣
n( n − 1) 2n
( n − 2)p
⎡
( 2) cos ⎢ x +
2!
4
⎣
⎤
⎥.
⎦
(iv) y = x2 tan−1 x
Let u = tan−1 x, v = x2.
sin nq
1
, whereq = tan −1 , r = 1 + x 2
n
x
r
Differentiating n times using Leibnitz’s Theorem,
un = ( −1) n −1 ( n − 1)!
[ Refer Ex.12, page 12]
yn = un v + nC1un −1v1 + nC2 un − 2 v2 + … + nCn u vn
sin nq 2
sin( n − 1)q
⋅ x + n( −1) n − 2 ( n − 2)!
⋅ 2x
n
r
r n −1
n( n − 1)
sin( n − 2)q
⋅2
+
( −1) n − 3 ( n − 3)!
2!
r n− 2
sin nq
sin( n − 1)q
= ( −1) n −1 ( n − 1)! n ⋅ x 2 + 2nx( −1) n − 2 ( n − 2)!
r n −1
r
sin( n − 2)q
+ n( n − 1) ( −1) n − 3 ( n − 3)!
r n− 2
= ( −1) n −1 ( n − 1)!
Example 4: Find nth order derivative of y = x2 ex and hence, prove that
yn =
1
1
n ( n - 1) y2 - n ( n - 2) y1 + ( n - 1) ( n - 2) y .
2
2
Solution: y = x2 ex
Let u = ex, v = x2
Differentiating n times using Leibnitz’s Theorem,
yn = e x ⋅ x 2 + n e x ⋅ 2 x +
n(n − 1) x
e ⋅2
2!
= e x [ x 2 + 2nx + n(n − 1)]
Putting
n = 1 and 2 successively in Eq. (1),
y1 = e x ( x 2 + 2 x),
Consider ,
y 2 = e x ( x 2 + 4 x + 2)
1
1
n(n − 1) y2 − n(n − 2) y1 + (n − 1) (n − 2) y
2
2
... (1)
2.26
Engineering Mathematics
1
1
n(n − 1) e x ( x 2 + 4 x + 2) − n (n − 2) e x ( x 2 + 2 x) + (n − 1) (n − 2) x 2 e x
2
2
2
x
= e [ x + 2nx + n (n − 1)] = yn
=
Example 5: If y = xn log x, prove that yn+ 1 =
n!
.
x
Solution: y = xn log x
Differentiating w.r.t. x,
1
+ nx n −1 ⋅ log x
x
xy1 = x n + nx n log x
y1 = x n
= x n + ny
Differentiating the above equation n times using Leibnitz’s Theorem,
xyn+1 + nyn = n ! + nyn
yn+1 =
n!
.
x
Example 6: If y = (x2 – 1)n, prove that (x2 – 1) yn+2 + 2x yn+1 - n (n + 1) yn = 0.
Solution: y = (x2
1)n
Differentiating w.r.t. x,
y1 = n( x 2 − 1) n −1 ⋅ 2 x
( x 2 − 1) y1 = n ( x 2 − 1) n 2 x = 2nyx
Differentiating again w.r.t. x,
(x2 1) y2 + 2xy1 = 2 (ny1x + ny)
(x2 1) y2 + (2x 2nx) y1 = 2ny
Differentiating the above equation n times using Leibnitz’s Theorem,
n(n − 1)
( x 2 − 1) yn + 2 + n ⋅ 2 xyn +1 +
⋅ 2 yn + (2 x − 2nx) yn +1 + n.2(1 − n) yn = 2nyn
2!
( x 2 − 1) yn + 2 + 2 xyn +1 − n(n + 1) yn = 0.
Example 7: If y = sin [log (x2 + 2x + 1)], prove that
(x + 1)2 yn+2 + (2n + 1) (x + 1) yn+1 + (n2 + 4) yn = 0.
Solution: y = sin [log (x2 + 2x + 1)] = sin [log (x + 1)2] = sin [2 log (x + 1)]
Differentiating w.r.t. x,
2
x +1
( x + 1) y1 = 2 cos[2 log( x + 1)]
y1 = cos[2 log( x + 1)] ⋅
Differential Calculus I
2.27
Differentiating again w.r.t. x,
( x + 1) y2 + y1 = −2 sin[ 2 log( x + 1)] ⋅
2
( x + 1)
(x + 1)2 y2 + (x + 1) y1 = 4y
Differentiating the above equation n times using Leibnitz’s Theorem,
n(n − 1)
( x + 1) 2 yn + 2 + n ⋅ 2 ( x + 1) yn +1 +
⋅ 2 yn + ( x + 1) yn +1 + n ⋅ yn = −4 yn
2!
(x + 1)2 yn+2 + (2n + 1) (x + 1) yn+1 + (n2 + 4) yn = 0.
Example 8: If y = a cos (log x) + b sin (log x), prove that
x2 yn+2 + (2n + 1) xyn+1 + (n2 + 1) yn = 0.
y = a cos (log x) + b sin (log x)
Solution:
Differentiating w.r.t. x,
1
1
y1 = − a sin (log x) ⋅ + b cos (log x) ⋅
x
x
xy1 = -a sin (log x) + b cos (log x)
Differentiating again w.r.t. x,
1
1
xy2 + y1 = − a cos (log x) ⋅ − b sin (log x) ⋅
x
x
x2y2 + xy1 = -y
Differentiating the above equation n times using Leibnitz’s Theorem,
n (n − 1)
x 2 y n + 2 + n ⋅ 2 x y n +1 +
⋅ 2 yn + xyn +1 + nyn = − yn
2!
x2yn+2 + (2n + 1) x yn+1 + (n2 + 1) yn = 0.
n
⎛ y⎞
⎛ x⎞
Example 9: If cos ⎜ ⎟ = log ⎜ ⎟ , prove that x2 yn+2 + (2n + 1)xyn+1 + 2n2yn = 0.
⎝b⎠
⎝ n⎠
-1
n
Solution:
⎛ y⎞
⎛ x⎞
⎛ x⎞
cos −1 ⎜ ⎟ = log ⎜ ⎟ = n log ⎜ ⎟
⎝b⎠
⎝n⎠
⎝n⎠
⎡
y
⎛ x ⎞⎤
= cos ⎢ n log ⎜ ⎟ ⎥
b
⎝ n ⎠⎦
⎣
Differentiating w.r.t. x,
⎡
⎛ x ⎞⎤
y = b cos ⎢ n log ⎜ ⎟ ⎥
⎝ n ⎠⎦
⎣
⎡
1 1 −bn
⎡
⎛ x ⎞⎤
⎛ x ⎞⎤
y1 = −b sin ⎢ n log ⎜ ⎟ ⎥ ⋅ n ⋅ ⋅ =
sin ⎢ n log ⎜ ⎟ ⎥
x n
x
⎝ n ⎠⎦
⎝ n ⎠⎦
⎣
⎣
n
⎡
⎛ x ⎞⎤
xy1 = −bn sin ⎢ n log ⎜ ⎟ ⎥
⎝ n ⎠⎦
⎣
2.28
Engineering Mathematics
Differentiating again w.r.t. x,
⎡
⎛ x ⎞⎤ 1 1
xy2 + y1 = −bn cos ⎢ n log ⎜ ⎟ ⎥ n ⋅ ⋅
⎝ n ⎠⎦ x n
⎣
n
⎡
⎤
−bn 2
x
⎛ ⎞
=
cos ⎢ n log ⎜ ⎟ ⎥
x
⎝ n ⎠⎦
⎣
⎡
⎛ x ⎞⎤
x 2 y2 + xy1 = −bn 2 cos ⎢ n log ⎜ ⎟ ⎥ = −n 2 y
⎝ n ⎠⎦
⎣
Differentiating the above equation n times using Leibnitz’s Theorem,
n(n − 1)
x 2 yn + 2 + n ⋅ 2 xyn +1 +
⋅ 2 yn + xyn +1 + nyn = − n 2 yn
2!
x2 yn+2 + (2n + 1) xyn+1 + 2n2 yn = 0
1
m
−
1
m
Example 10: If y + y = 2x, prove that
(x2 - 1) yn+2 + (2n + 1) x yn+1 + (n2 - m2) yn = 0.
1
1
m
ym + y
Solution:
1
= 2x
1
ym +
y
= 2x
1
m
2
m
1
y + 1 = 2x y m
2
1
1
Hence,
ym =
1
2x y m + 1 = 0, equation is quadratic in y m .
ym
2x ± 4x2 − 4
= x ± x2 − 1
2
(
y = x ± x2 − 1
)
m
Differentiating w.r.t. x,
(
y1 = m x ± x 2 − 1
(
= m x ± x −1
=
2
(x ±
x −1
m
2
y1 x 2 − 1 = my
( x 2 − 1) y12 = m 2 y 2
)
m −1
)
m −1
⎛
2x ⎞
⎜1 ±
⎟
2
⎝ 2 x −1 ⎠
(
x2 − 1
x2 − 1 ± x
x −1
2
)
m
)
Differential Calculus I
2.29
Differentiating again w.r.t. x,
(x2 - 1) 2y1 y2 + 2x y12 = m2 2y y1
(x2 - 1) y2 + xy1 = m2 y
Differentiating the above equation n times using Leibnitz’s Theorem,
n(n − 1)
⋅ 2 yn + x yn +1 + nyn = m 2 yn
2!
(x2 - 1) yn+2 + (2n + 1) x yn+1 + (n2 - m2) yn = 0
( x 2 − 1) yn + 2 + n ⋅ 2 x yn +1 +
⎛1
⎞
Example 11: If x = cosh ⎜ log y ⎟ , prove that
⎝m
⎠
(x2 - 1) yn+2 + (2n + 1) xyn+1 + (n2 - m2) yn = 0.
Solution:
⎛1
⎞
x = cosh ⎜ log y ⎟
⎝m
⎠
1
cosh −1 x = log y
m
)
(
(
log y = m log x + x 2 − 1 = log x + x 2 − 1
(
y = x + x2 − 1
)
)
m
m
Differentiating w.r.t. x,
(
y1 = m x + x 2 − 1
(
m( x +
=
)
m −1
)
x −1)
= m x + x −1
2
2
m −1
⎛
2x ⎞
⎜1 +
⎟
2
⎝ 2 x −1 ⎠
(
x2 − 1 + x
)
x −1
2
m
x −1
2
=
my
x2 − 1
y1 x 2 − 1 = my
( x 2 − 1) y12 = m 2 y 2
Differentiating again w.r.t. x,
(x2 - 1) 2y1 y2 + 2x y12 = m2 2y y1
(x2 - 1) y2 + xy1 = m2 y
2.30
Engineering Mathematics
Differentiating the above equation n times using Leibnitz’s Theorem,
n(n − 1)
⋅ 2 yn + x yn +1 + nyn = m 2 yn
2!
(x2 - 1) yn+2 + 2nx yn+1 + n2 yn - nyn + x yn+1 + nyn = m2 yn
( x 2 − 1) yn + 2 + n ⋅ 2 x yn +1 +
(x2 - 1) yn+2 + (2n + 1) x yn+1 + (n2 - m2) yn = 0.
Example 12: If y =
sin 1 x
1 x2
y=
Solution:
, prove that (1 - x2) yn+1 – (2n + 1) xyn – n2 yn–1 = 0.
sin −1 x
1 − x2
y 1 − x 2 = sin −1 x
y 2 (1 − x 2 ) = (sin −1 x) 2
Differentiating the above equation w.r.t. x,
2 yy1 (1 − x 2 ) + y 2 ( −2 x) = 2 sin −1 x ⋅
1
1 − x2
= 2y
(1 - x2) y1 - xy = 1
Differentiating the above equation n times using Leibnitz’s Theorem,
n(n − 1)
(1 − x 2 ) yn +1 + n (− 2 x) yn +
(−2) yn −1 − xyn − nyn −1 = 0
2!
(1 - x2) yn+1 - (2n + 1) xyn n2 yn-1 = 0.
Example 13: If y = sec−1 x, prove that
x (x2 - 1) yn+2 + [(2 + 3n) x2 - (n + 1)] yn+1 + n (3n + 1) xyn + n2 (n – 1) yn–1 = 0.
Solution: y = sec−1 x
Differentiating w.r.t. x,
y1 =
x2 (x2
−1
x x2 − 1
1) y12 = 1
Differentiating again w.r.t. x,
2x (x2 - 1) y12 + x2 2x y12 + x2 (x2 - 1) 2y1 y2 = 0
(x2 - 1) y1 + x2 y1 + x (x2 - 1) y2 = 0
Differential Calculus I
2.31
Differentiating the above equation n times using Leibnitz’s Theorem,
n(n − 1)
n(n − 1)
⋅ 2 yn −1 + x 2 yn +1 + n ⋅ 2 x yn +
⋅ 2 yn −1
2!
2!
n(n − 1)
n(n − 1) (n − 2)
+ x ( x 2 − 1) yn + 2 + n (3 x 2 − 1) yn +1 +
⋅ 6 x yn +
⋅ 6 yn − 1 = 0
2!
3!
( x 2 − 1) yn +1 + n ⋅ 2 x yn +
x (x2 - 1) yn+2 + [(2 + 3n) x2
(n + 1)] yn+1 + n (3n + 1) xyn + n2 (n
1) yn−1 = 0
Example 14: If y = sinh−1 x, prove that (1 + x2) yn+2 + (2n + 1) xyn+1 + n2yn = 0.
(
Solution: y = sinh −1 x = log x + x 2 + 1
)
Differentiating w.r.t. x,
⎛
2x ⎞
1
y1 =
⋅
⎜⎜1 +
⎟⎟ =
2
2
x + x + 1 ⎝ 2 x + 1 ⎠ x + x2 + 1
(x2 + 1) y12 = 1
1
(
x2 + 1 + x
x +1
2
Differentiating again w.r.t. x,
(x2 + 1) 2y1 y2 + 2xy12 = 0
(x2 + 1) y2 + xy1 = 0
Differentiating the above equation n times using Leibnitz’s Theorem,
n (n − 1)
⋅ 2 yn + xyn +1 + nyn = 0
2!
(x2 + 1) yn+2 + (2n + 1) xyn+1 + n2yn = 0
( x 2 + 1) yn + 2 + n ⋅ 2 yn +1 +
a sin
Example 15: If y = e
1
x
, prove that
(1 – x2) yn+2 – (2n + 1) xyn+1 – (n2 + a2) yn = 0.
Solution: y = e a sin
−1
x
Differentiating w.r.t. x,
y1 = e a sin
−1
y1 1 − x 2 = ay
(1 - x2) y12 = a2 y2
x
⋅
a
1 − x2
)
2.32
Engineering Mathematics
Differentiating again w.r.t. x,
(1 - x2) 2y1 y2 + (-2x) y12 = a2 2y y1
(1 - x2) y2 - xy1 = a2 y
Differentiating the above equation n times using Leibnitz’s Theorem,
(1 − x 2 ) yn + 2 + n(− 2 x) yn +1 +
n (n − 1)
(− 2) yn − xyn +1 − nyn = a 2 yn
2!
(1 - x2) yn+2 - (2n + 1) xyn+1 - (n2 + a2) yn = 0.
⎛a+ x⎞
, prove that
Example 16: If y = tan -1 ⎜
⎝ a - x ⎟⎠
(a2 + x2) yn+2 + 2 (n + 1) x yn+1 + n (n + 1) yn = 0.
Solution:
⎛ a + x⎞
y = tan −1 ⎜
⎝ a − x ⎟⎠
Putting x = a tan q,
⎛ 1 + tan q ⎞
⎛p
⎞
y = tan −1 ⎜
= tan −1 tan ⎜ + q ⎟
⎝ 1 − tan q ⎟⎠
⎝4
⎠
=
p
p
⎛x⎞
+ q = + tan −1 ⎜ ⎟
⎝a⎠
4
4
Differentiating w.r.t. x,
y1 =
1
1
a
⋅ = 2
2
x a x + a2
1+ 2
a
(x2 + a2) y1 = a
(x2 + a2) y2 + 2x y1 = 0
Differentiating the above equation n times using Leibnitz’s Theorem,
n(n − 1)
( x 2 + a 2 ) y n + 2 + n ⋅ 2 x y n +1 +
⋅ 2 y n + 2 x y n +1 + n ⋅ 2 y n = 0
2!
(x2 + a2) yn+2 + 2(n + 1) x yn+1 + n (n + 1) yn = 0.
1+ x
, prove that y = (1 – x2) y1 and hence, deduce that
1- x
(1 – x2) yn – [2 (n – 1) x + 1] yn–1 – (n – 1) (n – 2) yn – 2 = 0.
Example 17: If y =
Solution:
y=
1+ x
1− x
log y = log
=
1+ x
1− x
1
[log (1 + x) − log (1 − x)]
2
Differential Calculus I
2.33
Differentiating w.r.t. x,
1
1⎛ 1
1 ⎞
1
⋅ y1 = ⎜
+
⎟=
y
2 ⎝ 1 + x 1 − x ⎠ 1 − x2
(1
x2) y1 = y
Differentiating the above equation n times using Leibnitz’s Theorem,
(1 − x 2 ) yn +1 + n ( −2 x) yn +
Replacing n by n
n(n − 1)
( −2) yn −1 = yn
2!
1,
(1 − x 2 ) yn − [2 (n − 1) x + 1] yn −1 − (n − 1) (n − 2) yn − 2 = 0.
Example 18: If f (x) = tan x, prove that
f n (0) − nC2 f n − 2 (0) + nC4 f n − 4 (0) + …… = sin
no
.
2
f ( x) = tan x
sin x
=
cos x
cos x ⋅ f ( x) = sin x
Solution:
Differentiating the above equation n times using Leibnitz’s Theorem,
cos x f n (x) + nC1 ( sin x) f n 1 (x) + nC2 ( cos x) f n 2 (x) + nC3 (sin x) f n 3 (x)
np ⎞
np ⎞
⎛
+ nC4 (cos x) f n 4 (x) + ……… + f ( x ) ⋅ cos ⎛⎜ x +
⎟ = sin ⎜⎝ x +
⎟
⎝
2 ⎠
2 ⎠
Putting x = 0,
np
⎛ np ⎞
f n (0) − nC2 f n − 2 (0) + nC4 f n − 4 (0) + ………… + f (0) cos ⎜ ⎟ = sin
.
⎝ 2 ⎠
2
)
(
2
2
Example 19: If y = ⎡⎢log x + 1 + x ⎤⎥ , prove that yn+2 (0) = – n2yn (0).
⎣
⎦
)
(
Solution: y = ⎡log x + 1 + x 2 ⎤
⎢⎣
⎥⎦
2
Differentiating w.r.t. x,
)
(
1
y1 = 2 ⎡log x + 1 + x 2 ⎤
⎣⎢
⎦⎥ x + 1 + x 2
1
= 2 log x + 1 + x 2 ⋅
1 + x2
(
)
⎛
⎞
1
⋅ 2 x ⎟⎟
⎜⎜1 +
2
⎝ 2 1+ x
⎠
2.34
Engineering Mathematics
(
y1 1 + x 2 = 2 log x + 1 + x 2
)
)
(
2
(1 + x 2 ) y12 = 4 ⎡log x + 1 + x 2 ⎤ = 4 y
⎥⎦
⎣⎢
Differentiating again w.r.t. x,
(1 + x2) 2y1 y2 + 2xy12 = 4y1
(1 + x2) y2 + xy1 = 2
Differentiating the above equation n times using Leibnitz’s Theorem,
n (n − 1)
⋅ 2 yn + xyn +1 + nyn = 0
2!
(1 + x2) yn+2 + (2n + 1) xyn+1 + n2 yn = 0
(1 + x 2 ) yn + 2 + n ⋅ 2 xyn +1 +
Putting x = 0,
(yn+2)0 = -n2 (yn)0.
(
2
Example 20: If y = x + 1 + x
)
m
, prove that
(i) (y2n)0 = [m2 - (2n - 2)2] [m2 - (2n - 4)2] … [m2 - 22] m2.
(ii) (y2n+1)0 = [m2 - (2n - 1)2] [m2 - (2n - 3)2] … [m2 - 12] m.
(
Solution: y = x + 1 + x 2
)
m
Differentiating w.r.t. x,
(
)
(
)
y1 = m x + 1 + x 2
1 + x 2 ⋅ y1 = m x + 1 + x 2
m −1
m
⎛
2x
⎜⎜1 +
2
⎝ 2 1+ x
= my
⎞
⎟⎟
⎠
... (1)
(1 + x2) y12 = m2 y2
Differentiating again w.r.t. x,
(1 + x2) 2y1 y2 + 2xy12 = m2 2yy1
(1 + x2) y2 + xy1 = m2 y
… (2)
Differentiating the above equation n times using Leibnitz’s Theorem,
n(n − 1)
⋅ 2 yn + xyn +1 + nyn = m 2 yn
2!
(1 + x2) yn+2 + (2n + 1) xyn+1 + (n2 m2) yn = 0
(1 + x 2 ) yn + 2 + n ⋅ 2 xyn +1 +
… (3)
Putting x = 0 in Eqs (1), (2) and (3),
and
(y1)0 = m, (y2)0 = m2
(yn+2)0 = (m2 n2) yn(0)
[∵ y(0) = 1]
… (4)
Differential Calculus I
2.35
Putting n = 1, 2, 3, 4, … in Eq. (4),
y3 (0) = (m2 12) y1 (0) = (m2 12) m
y4 (0) = (m2 22) y2 (0) = (m2 22) m2
y5 (0) = (m2 32) y3 (0) = (m2 32) (m2 12) m
y6 (0) = (m2 42) y4 (0) = (m2 42) (m2 22) m2 and so on.
In general,
Even derivative, y2n (0) = [m2 - (2n - 2)2] [m2 - (2n - 4)2] … (m2 22) m2
Odd derivative, y2n+1 (0) = [m2 - (2n - 1)2] [m2 - (2n - 3)2] … (m2 12) m.
(
)
2
Example 21: If y = log x + x + 1 , prove that
2
y2n (0) = 0 and y2n+1 (0) = (-1)n [1 ⋅ 3 ⋅ 5 … ( 2n − 1) ] .
2
(
2
Solution: y = log x + x + 1
2
2
)
Differentiating w.r.t. x,
y1 =
⎛
2x ⎞
⎜⎜1 +
⎟⎟
x + 1 + x ⎝ 2 1 + x2 ⎠
1
2
1 + x 2 ⋅ y1 = 1
(x2 + 1) y12 = 1
Differentiating again w.r.t. x,
(x2 + 1) 2y1 y2 + 2xy12 = 0
(x2 + 1) y2 + xy1 = 0
... (1)
… (2)
Differentiating the above equation n times using Leibnitz’s Theorem,
n(n − 1)
( x 2 + 1) yn + 2 + n ⋅ 2 xyn +1 +
⋅ 2 yn + xyn +1 + nyn = 0
2!
(x2 + 1) yn+2 + (2n + 1) xyn+1 + n2yn = 0
… (3)
Putting x = 0 in Eqs (1), (2) and (3),
y1(0) = 1 and y2(0) = 0
and yn+2(0) = -n2 yn(0)
Putting n = 1, 2, 3, 4, … in Eq. (4),
… (4)
y3(0) = -12 y1(0) =
y4(0) = -22 y2(0) = 0
2
2
( 1) 2 12 32
y5(0) = -32 y3(0)
y6(0) = 0
y7(0) = -52 y5(0) = 52 ( 1) 2 12 32 ( 1)3 12 32 52 etc.
In general,
Even derivative, y2n(0) = 0
Odd derivative, y2n+1(0) = ( 1) n [12 32 52 … (2n 1) 2 ]
Example 22: If y = (sin–1 x)2, prove that (i) (1 - x2) yn+2 - (2n + 1) xyn+1 - n2yn = 0.
(ii) y2n+1(0) = 0 and y2n(0) = 22n–1 [(n – 1)!]2.
2.36
Engineering Mathematics
Solution: (i) y = (sin−1 x)2
Differentiating w.r.t. x,
y1 = 2 sin −1 x ⋅
(1
1
... (1)
1 − x2
x 2)y12 = 4 (sin−1 x)2 = 4y
Differentiating again w.r.t. x,
(1
x 2)2y1 y2 2xy12 = 4y1
(1 x 2) y2 xy1 = 2
… (2)
Differentiating the above equation n times using Leibnitz’s Theorem,
(1 − x 2 ) yn + 2 + n ⋅ (−2 x) yn +1 +
(1
n(n − 1)
(−2) yn − xyn +1 − nyn = 0
2!
x 2)yn+2
(2n + 1) xyn+1
n2yn = 0
… (3)
(ii) Putting x = 0 in Eq. (1), (2) and (3),
y1(0) = 0, y2(0) = 2 = 22–1 [(1
1)!]2
yn+2(0) = n2 yn(0)
Putting n = 1, 2, 3, 4, …, in Eq. (4)
y3(0) = 12 y1(0) = 0
y4(0) = 22 y2(0) = 22 2 = 23 = 24–1 [(2
y5(0) = 0
y6(0) = 42 y4(0) = 42 23 = 26–1 [(3
In general,
Even derivative,
Odd derivative,
y2n(0) = 22n 1 [(n
y2n+1(0) = 0.
… (4)
1)!]2
1)!]2 etc.
1)!]2
Example 23: If y = sin (m sin–1 x), prove that
(i) (1 - x2) yn+2 - (2n + 1) xyn+1 - (n2 - m2) yn = 0.
(ii) (yn )0 = [(n - 2)2 - m2] [(n - 4)2 - m2] … (1 - m2) m, if n is odd
= 0, if n is even.
Solution: (i) y = sin (m sin–1 x)
Differentiating w.r.t. x,
y1 = cos (m sin −1 x) ⋅
1 − x 2 ⋅ y1 = m cos (m sin −1 x)
m
1 − x2
... (1)
Differential Calculus I
2.37
(1 - x2) y12 = m2 cos2 (m sin–1 x)
= m2 [1 – sin2 (m sin–1 x)]
(1 - x2) y12 = m2 (1 – y2)
Differentiating again w.r.t. x,
(1 - x2) 2y1 y2 + y12 (–2x) = m2 ( 2yy1)
(1 - x2) y2 - xy1 = -m2y
… (2)
Differentiating the above equation n times using Leibnitz’s Theorem,
(1 − x 2 ) yn + 2 + n(−2 x) yn +1 +
n(n − 1)
(−2) yn − xyn +1 − nyn = −m 2 yn
2!
(1 - x2) yn+2 - (2n + 1) xyn+1 - (n2
(ii) Putting
m2) yn = 0
… (3)
x = 0 in Eqs (1), (2) and (3),
y1 (0) = cos ( m sin −1 0 ) ⋅
m
1− 0
=m
y1(0) = m
y2(0) = – m2 y(0) = 0
also,
yn+2(0) = (n2 m2) yn(0)
Putting n = 1, 2, 3, … in Eq. (4),
y3(0) = (12 m2) y1(0) = (12
y4(0) = (22 m2) y2(0) = 0
y5(0) = (32
y6(0) = (42
… (4)
m2) m
m2) y3(0) = (32 m2) (12
m2) y4(0) = 0 etc.
m2) m
In general,
yn(0) = [(n 2)2 m2] [(n
= 0, if n is even.
4)2
m2] … (12
m2) m, if n is odd
Example 24: If y = tan−1 x, prove that (i) (x2 + 1) yn+1 + 2nxyn + n (n – 1) yn–1 = 0
(ii) yn (0) = 0, if n is even
= ( -1)
n -1
2 ( n - 1)!,
if n is odd.
Solution: (i) y = tan−1 x
Differentiating w.r.t. x,
1
1 + x2
(x2 + 1) y1 = 1
y1 =
... (1)
Differentiating again w.r.t. x,
(x2 + 1) y2 + 2 x y1 = 0
… (2)
2.38
Engineering Mathematics
Differentiating Eq. (1) n times using Leibnitz’s Theorem,
( x 2 + 1) yn +1 + n ⋅ 2 xyn +
n(n − 1)
⋅ 2 yn −1 = 0
2!
(x2 + 1) yn+1 + 2nxyn + n (n
1) yn 1 = 0
... (3)
(ii) Putting x = 0 in Eqs (1), (2) and (3),
y1(0) = 1, y2(0) = 0
yn+1(0) = n(n
Putting n = 2, 3, 4, … in Eq. (4),
1) yn 1(0)
y3 (0) = −2 (2 − 1) y1 (0) = −2 = −(2 !) = (−1)
y4(0) = 3(3
... (4)
3 −1
2
(2 !)
1) y2(0) = 0
y5 (0) = −4 (4 − 1) y3 (0) = −4 (3) (−2) = (−1) 2 (4 !) = (−1)
5 −1
2
y6(0) = 5(4) y4(0) = 0
y7 (0) = −6 (5) y5 (0) = −6 (5) (−1) 2 (4 !) = (−1)3 (6 !) = (−1)
(4!)
7 −1
2
(6 !)
In general,
yn = 0, if n is even
= ( −1)
n −1
2
(n − 1)!, if n is odd.
-1
Example 25: If y = e m cos x, find ( yn )(0).
−1
Solution: y = e m cos x
Differentiating w.r.t. x,
y1 = e m cos
−1
x
⎛ −m ⎞
⎜
⎟
⎝ 1 − x2 ⎠
... (1)
(1 - x2) y12 = m2 y2
Differentiating again w.r.t. x
(1 x2) 2y1 y2 2xy12 = 2m2 yy1
(1 x2) y2 xy1 = m2 y
Differentiating the above equation n times using Leibnitz’s Theorem,
n (n − 1)
(1 − x 2 ) yn + 2 + n(−2 x) yn +1 +
(−2) yn − xyn +1 − nyn = m 2 yn
2!
(1 x2) yn+2 (2n + 1) xyn+1 (n2 + m2) yn = 0
Putting x = 0 in Eqs (1), (2) and (3),
y1 (0) = − me m cos
−1
0
= − me
yn+2(0) = (n2 + m2) yn(0)
mp
2
, y2 ( 0 ) = m 2 y ( 0 ) = m 2 e
… (2)
... (3)
mp
2
... (4)
Differential Calculus I
2.39
Putting n = 1, 2, 3, 4, … in Eq. (4),
y3 (0) = (12 + m 2 ) y1 (0) = − me
mp
2
y4 ( 0 ) = ( 2 2 + m 2 ) y2 ( 0 ) = m 2 e
mp
2
y5 (0) = (32 + m 2 ) y3 (0) = − me
y6 (0) = ( 4 2 + m 2 ) y4 (0) = m 2 e
(12 + m 2 )
mp
2
mp
2
( 22 + m 2 )
(12 + m 2 ) (32 + m 2 )
( 22 + m 2 ) ( 4 2 + m 2 )
In general,
mp
2
Even derivative,
y2 n ( 0 ) = m 2 e
Odd derivative,
y2 n +1 (0) = − me
( 22 + m 2 ) ( 4 2 + m 2 )…[( 2n − 2) 2 + m 2 ]
mp
2
(12 + m 2 ) (32 + m 2 )…[( 2n − 1) 2 + m 2 ]
Example 26: If x + y = 1, prove that
2
2
dn
( x n y n ) = n ! ⎡⎢ y n - ( nC1 ) y n -1 x + ( nC2 ) y n - 2 x 2
n
⎣
dx
- ( nC 3 ) y n - 3 x 3 +…
……… + (-1) n x n ⎤⎥ .
⎦
2
Solution: x + y = 1, y = 1
x, y1 = 1
Differentiating n times using Leibnitz’s Theorem,
dn n n dn n
x y = n [ x (1 − x) n ]
dx n
dx
n!
n (n − 1) n ! 2
= n !(1 − x) n + n ⋅ x ⋅ n (1 − x) n −1 (−1) +
x n(n − 1) (1 − x) n − 2 (−1) 2
1!
2! 2!
n (n − 1) (n − 2) n ! 3
+
x n (n −1) (n − 2) (1 − x) n −3 (−1)3 + ……… + (−1) n x n
3!
3!
⎡ d n (ax + b) m a n m !(ax + b) m − n ⎤
=
⎢∵
⎥
(m − n)!
dx n
⎣
⎦
2
2
2
n
= n ! ⎡⎢ y n − ( n C1 ) y n −1 x + ( nC2 ) y n − 2 x 2 − ( nC3 ) y n −3 x 3 + ……… + ( −1) x n ⎤⎥
⎣
⎦
[∵ (1 − x) = y ]
Example 27: By finding two different ways the nth derivative of x2n, prove that
1+
n2 n2 ( n - 1)2 n2 ( n - 1)2 ( n - 2)2
( 2n)!
.
+
+
+ …………… =
2
2
2
2
2
2
( n !)2
1
1 2
1 2 3
2.40
Engineering Mathematics
Let y = x2n
Solution:
yn =
=
(2n)! 2 n − n
x
(2n − n)!
⎡ d n (ax + b) m a n m !(ax + b) m − n ⎤
=
⎢∵
⎥
(m − n)!
dx n
⎦
⎣
(2n)! n
x
n!
... (1)
y = xn ⋅ xn
Now,
Differentiating the above equation n times using Leibnitz’s Theorem,
n!
n (n − 1)
n!
x n − ( n − 2)
x n − ( n −1) +
n (n − 1) x n − 2
2!
(n − n + 1)!
(n − n + 2)!
n (n − 1) (n − 2)
n!
+
n (n − 1) (n − 2) x n − 3
x n − (nn − 3) + …………
3!
(n − n + 3)!
yn = x n ⋅ n !+ n ⋅ nx n −1 ⋅
= xn ⋅ n! + n2 ⋅
n ! n n 2 (n − 1) 2
n 2 (n − 1) 2 (n − 2) 2
n
x +
n
!
x
+
n ! x n +…………
1!
(2 !) 2
(3!) 2
⎡ n 2 n 2 (n − 1) 2 n 2 (n − 1) 2 (n − 2) 2
⎤
= x n ⋅ n ! ⎢1 + 2 + 2 2 +
+ …………⎥
2
2
2
1 ⋅2
1 ⋅ 2 ⋅3
⎣ 1
⎦
... (2)
n
Equating coefficients of x in Eqs (1) and (2),
⎡ n 2 n 2 (n − 1) 2 n 2 (n − 1) 2 (n − 2) 2
⎤
n ! ⎢1 + 2 + 2 2 +
+ …………⎥ =
2
2
2
1 ⋅2
1 ⋅ 2 ⋅3
⎣ 1
⎦
2
2
2
2
2
n
n (n − 1)
n (n − 1) (n − 2) 2
Hence, 1 + 2 + 2 2 +
+………………
…y =
1
1 ⋅2
12 ⋅ 22 ⋅ 32
Example 28: If y =
Solution: y =
(2n)!
(n !)
(2n)!
(n !) 2
log x
( -1) n n ! ⎡
1
1 ⎞⎤
⎛
log x - ⎜ 1 + + ... + ⎟ ⎥ .
, prove that yn =
⎝
x
n ⎠⎦
2
x n + 1 ⎢⎣
log x
x
Differentiating n times using Leibnitz’s Theorem,
yn =
(−1) n n !
(−1) n −1 (n − 1)! 1 n (n − 1) (−1) n − 2 (n − 2)! ⎛ 1 ⎞
⋅ log x + n
⋅ +
⎜− 2 ⎟
n +1
2!
x
x n −1
x
xn
⎝ x ⎠
+
n (n − 1) (n − 2) (−1) n −3 (n − 3)! ⎛ 2 ⎞
1 (−1) n −1 (n − 1)!
⋅
…………
+
+
⋅
⎜ 3⎟
3!
x
xn−2
xn
⎝x ⎠
[Using result (2) and (3)]
Differential Calculus I
2.41
(−1) n n !
(−1) n n ! (−1) n n ! (−1) n n !
(−1) n n !
………
x
log
⋅
−
−
−
−
…
…
−
x n +1
x n +1
2 x n +1
3x n +1
nx n +1
n
(−1) n ! ⎡
1 ⎞⎤
⎛ 1 1
=
log x − ⎜1 + + + … + ⎟ ⎥ .
n ⎠⎦
2
3
x n +1 ⎢⎣
⎝
=
Exercise 2.2
3. If y = e ax [a2 x2 - 2nax + n (n + 1)],
prove that yn = an+2 x2 eax.
1. Find the nth order derivative w.r.t. x
(i) xex
(ii) x2e2x
(iii) x log (x + 1)
(iv) x3 sin 2x
2
(v) y = x sin x.
⎡ Ans. : (i) e x ( x + n)
⎢
n 2
n
n −1
2x
⎢ (ii) e [2 x + 2 nx + n ( n − 1) 2 ]
⎢
( −1) n − 2 ( n − 2)!( x + n)
⎢(iii)
( x + 1) n
⎢
⎢
⎢(iv) 2n x 3 sin ⎛⎜ 2 x + np ⎞⎟ +
⎝
⎢
2 ⎠
⎢
p⎤
⎡
⎢
3n x 2 2n −1 sin ⎢ 2 x + ( n − 1) ⎥
⎢
2⎦
⎣
⎢
p⎤
⎡
⎢
+ 3n ( n − 1) x 2n − 2 sin ⎢ 2 x + ( n − 2) ⎥
⎢
2⎦
⎣
⎢
n−3
+ n ( n − 1) ( n − 2) 2
⎢
⎢
np ⎤
⎡
⎢
sin ⎢ 2 x + ( n − 3) ⎥
2 ⎦
⎢
⎣
⎢
⎢( v) x 2 sin ⎛ x + np ⎞
⎜⎝
⎟
⎢
2 ⎠
⎢
p⎤
⎡
⎢
+ 2nx sin ⎢ x + ( n − 1) ⎥
⎢
2⎦
⎣
⎢
p⎤
⎡
⎢
+ ( n2 − n) sin ⎢ x + ( n − 2) ⎥
⎢⎣
2⎦
⎣
2. If y = 7x x7, find y5.
⎡ Ans. : (log 7)5 7 x x 7
⎤
⎢
⎥
4 x 6
+ 35 (log 7) 7 x
⎢
⎥
3 x 5 ⎥
⎢
+ 420 (log 7) 7 x
⎢
⎥
⎢
+ 2100 (log 7) 2 7 x x 4 ⎥
⎢
⎥
+ 4200 (log 7) 7 x x 3 ⎥
⎢
⎢
⎥
+ 2520 7 x x 2
⎣
⎦
4. If y = x2 sin x, prove that
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦
np ⎞
⎛
yn = ( x 2 − n2 + n) sin ⎜ x +
⎟
⎝
2 ⎠
np ⎞ .
⎛
− 2nx cos ⎜ x +
⎟
⎝
2 ⎠
5. If x = tan log y, prove that
(1 + x 2 ) yn +1 + (2nx − 1) yn
+ n (n − 1) yn −1 = 0.
⎡ Hint : log y = tan −1 x, y = e tan
⎣
−1
x
⎤
⎦
6. If y = cos (m sin−1 x), prove that
(1 - x2) yn+2 - (2n + 1) xyn+1
+ (m2 - n2) yn = 0
Hence, obtain yn (0).
⎡ Ans. : yn (0) = ( n2 − m 2 )........... ⎤
⎥
⎢
( 4 2 − m 2 )( 22 − m 2 )( − m 2 ) ⎥⎦
⎢⎣
7. If x = sin q, y = sin 2q, prove that
(1 - x2) yn+2 - (2n + 1) xyn+1 (n2 - 4) yn = 0.
⎡ Hint : y = 2 sin q cos q = 2 x 1 − x 2 ⎤
⎣
⎦
8. If y = x2ex, prove that
yn =
1
n (n − 1) y2 − n (n − 2) y1
2
1
+ (n − 1) (n − 2) y.
2
2.42
Engineering Mathematics
2.4 MEAN VALUE THEOREM
The roots of the given function as well as equality or inequality of any two or more
than two functions can be determined using Mean Value Theorems. These theorems
are Rolle’s Theorem, Lagrange’s Mean Value Theorem, and Cauchy’s Mean Value
Theorem. For better understanding of these theorems, we shall first learn two type of
functions.
2.4.1 Continuous and Differentiable Functions
A function f (x) is said to be continuous at x = a if
(i) f (a) is finite, i.e., f (a) exists.
(ii) lim f ( x) = lim f ( x) = lim f ( x) = f (a)
x→a
x→a +
x→ a −
A function f (x) is said to be differentiable at x = a, if Right Hand Derivative (RHD)
and Left Hand Derivative (LHD) exists and RHD = LHD.
i.e.
or
lim
x→a +
lim
h →0
f ( x) − f (a)
f ( x) − f (a)
= lim
x
→
a
−
x−a
x−a
f ( a + h) − f ( a )
f ( a − h) − f ( a )
= lim
h→ 0
h
h
Note:
(i) If function f (x) is finitely differentiable (derivative is finite) in the interval
(a, b), then it is continuous in the interval [a, b], i.e., every differentiable function is continuous in the given interval. But converse is not necessarily true, i.e.,
a function may be continuous for a value of x without being differentiable for
that value. The interval (a, b) is also called as domain of the function.
(ii) Algebraic, trigonometric, inverse trigonometric, logarithmic and exponential
function are ordinarily continuous and differentiable (with some exceptions).
(iii) Addition, subtraction, product and quotient of two or more continuous and differentiable functions are also continuous and differentiable.
(iv) f (x) and f (x) are differentiable and continuous if f (x) exists.
(v) A function is said to be differentiable if its derivative is neither indeterminate
nor infinite.
2.5 ROLLE’S THEOREM
Statement: If a function f (x) is
(i) continuous in the closed interval [a, b]
(ii) differentiable in the open interval (a, b)
(iii) f (a) = f (b)
then there exists at least one point c in the open interval (a, b) such that f ( c) = 0.
Differential Calculus I
2.43
Proof: Since function f (x) is continuous in the closed interval [a, b], it attains its
maxima and minima at some points in the interval. Let M and m be the maximum and
minimum values of f (x) respectively at some points c and d respectively in the interval
[a, b].
f (c) = M
and
f (d) = m
Now two cases arise:
Case I: If M = m
f (x) = M = m for all x in [a, b]
f (x) = constant for all x in [a, b]
f (x) = 0 for all x in [a, b]
Hence, the theorem is true.
Case II: If M ó m
Since f (a) = f (b), either M or m must be different from f (a) and f (b).
Let M is different from f (a) and f (b).
f (c) is different from f (a) and f (b).
Also,
Hence,
f (c ) ≠ f ( a )
, f (c) ≠ f (b)
∴c ≠ a
∴c ≠ b
a<c<b
Now, since f (x) is differentiable in the open interval (a, b), f (c) exists.
By definition,
f ′(c) = lim
h →0
Since
and
f (c + h) − f (c )
h
f (c) = M, f (c + h)
f (c)
f (c + h ) − f (c )
≤ 0 for h > 0
h
f (c + h ) − f (c )
≥ 0 for h < 0
h
... (1)
... (2)
As h 0, Eq. (1) gives f ( c) 0 and Eq. (2) gives f ( c) 0
Since f (x) is differentiable, f ( c) must be unique.
Hence,
f ( c) = 0 for a < c < b
Similarly, it can be proved that f ( c) = 0 for a < c < b if m is different from f (a) and
f (b).
Note:
(i) There may be more than one point c, such that, f (c) = 0.
(ii) The converse of the theorem is not true, i.e., for some function f (x), f (c) = 0
but f (x) may not satisfy the conditions of Rolle’s theorem.
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2
f (x) = 1
e.g.,
f ′( x) = 1 −
3 ( x 1) 3 in 0
x
10
2
1
( x − 1) 3
f (c) = 0 at c = 9. But f (x) does not exist at x = 1, i.e., not differentiable at x = 1.
Hence, f (x) does not satisfy the conditions of Rolle’s theorem.
2.5.1 Another Form of Rolle’s Theorem
If a function f (x) is
(i) continuous in the closed interval [a, a + h]
(ii) differentiable in the open interval (a, a + h)
(iii) f (a) = f (a + h), then there exists at least one real number q between 0 and 1
such that f ( a + q h) = 0, for 0 < q < 1.
2.5.2 Geometrical Interpretation of Rolle’s Theorem
Let y = f (x) represents a curve with A [a, f (a)] and B [b, f (b)] as end points and
C [c, f (c)] be any point between A and B.
f (c) = slope of the tangent at point C
Thus, geometrically the theorem states that if
(i) curve is continuous at the points A, B and at every point between A and B, i.e.,
in the interval [a, b].
(ii) possesses unique tangent at every point between A and B.
(iii) ordinates of the points A and B are same, i.e., f (a) = f (b), then there exists at
least one point C [c, f (c)] on the curve between A and B, tangent at which is
parallel to x-axis.
y
y
C
A
C3
C1
B
A
B
C2
x=a
x=b
C5
x=a
x
C4
x=b x
Fig. 2.1
2.5.3 Algebraic Interpretation of Rolle’s Theorem
Let f (x) be a polynomial in x. If f (x) = 0 satisfies all the conditions of Rolle’s theorem
and x = a, x = b be the roots of the equation f (x) = 0, then at least one root of the equation f (x) = 0 lies between a and b.
Differential Calculus I
2.45
Example 1: Verify Rolle’s theorem for the following functions:
(i) f (x) = (x - a)m (x - b)n in [a, b], where m, n are positive integers.
(ii) f ( x ) = x ( x + 3) e
-
x
2
in - 3 Ä x Ä 0
(iii) f (x) = | x | in [-1, 1]
(iv) f ( x ) =
sin x
in [p , o ]
ex
⎡ o 5o ⎤
x
(v) f ( x ) = e (sin x - cos x ) in ⎢ ,
⎥
⎣4 4 ⎦
⎡ x 2 + ab ⎤
(vi) f ( x ) = log ⎢
⎥ in [a , b], a > 0, b > 0
⎣ ( a + b) x ⎦
(vii) f (x) = x2 + 1
=3-x
(viii) f (x) = x2 - 2
=x-
0
x
x
-1
1
x
0
x
Solution: (i) f (x) = (x
a)m (x
b)n in [a, b], where m, n are positive integers.
(a) Since m and n are positive integers,
f (x) = (x a)m (x b)n, being a polynomial, is continuous in [a, b].
(b) f (x) = m (x a)m 1 (x b)n + n (x a)m (x b)n 1
= (x a)m 1 (x b)n 1 [m (x b) + n (x a)]
= (x a)m 1 (x b)n 1 [(m + n) x (mb + na)]
exists for every value of x in (a, b). Therefore, f (x) is differentiable in
(a, b).
(c) f (a) = f (b) = 0
Thus, f (x) satisfies all the conditions of Rolle’s theorem. Therefore, there exists
at least one point c in (a, b) such that f ( c) = 0.
(c a)m 1 (c b)n 1 [(m + n) c (mb + na)] = 0
mb + na
m+n
which represents a point that divides the interval [a, b] internally in the ratio of
m : n. Thus, c lies in (a, b).
Hence, theorem is verified.
c=
(ii) f ( x) = x ( x + 3) e
−
x
2
in − 3 ≤ x ≤ 0
−
x
(a) f ( x) = x ( x + 3) e 2 , being composite function of continuous function,
is continuous in [ 3, 0].
(b) f ′ ( x) = ( x + 3) e
−
x
2
+ xe
−
x
2
−
x ( x + 3) − 2x
e
2
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Engineering Mathematics
exists for every value of x in (-3, 0). Therefore, f (x) is differentiable in
(-3, 0).
(c) f (-3) = f (0) = 0
Thus, f (x) satisfies all the conditions of Rolle’s theorem. Therefore, there
exists at least one point c in (-3, 0) such that f ( c) = 0
(c + 3) e
−
c
2
+ ce
−
c
2
−
c (c + 3) − 2c
e =0
2
2 (c + 3) + 2c − c (c + 3) = 0
⎡ − 2c
⎤
⎢∵ e ≠ 0 for anny finite ⎥
⎢
value of c ⎥⎦
⎣
- c2 + c + 6 = 0,
c = –2, 3
c = –2 lies in (–3, 0)
Hence, theorem is verified.
(iii) f (x) = | x | in [-1, 1].
| x | = -x, -1 x 0
= x,
0 x 1
(a) f (x) is continuous in [-1, 1].
(b) Left hand derivative at x = 0
f ′(0 − ) = lim−
x→0
f ( x) − f (0)
−x − 0
= lim−
= −1
x→0
x−0
x
Right hand derivative at x = 0
f ′(0+ ) = lim+
x →0
f ( 0−)
f ( x) − f (0)
x−0
= lim+
=1
x
→
0
x−0
x
f ( 0+)
Thus, function is not differentiable at x = 0 and hence, Rolle’s theorem is not
applicable.
sin x
= e − x sin x
ex
(a) f (x) = e−x sin x, being product of continuous functions, is continuous in
[0, p ]
(b) f (x) = -e−x sin x + e−x cos x
= e−x (cos x - sin x)
exists for every value of x in (0, p ). Therefore, f (x) is differentiable in
(0, p ).
(c) f (0) = f (p ) = 0
Thus, f (x) satisfies all the conditions of Rolle’s theorem. Therefore,
there exists at least one c in (0, p ) such that f ( c) = 0.
(iv) f ( x) =
Differential Calculus I
2.47
f ′(c) = e − c (cos c − sin c) = 0
cos c − sin c = 0
cos c = sin c
tan c = 1, c = np +
[∵ e − c ≠ 0 for any finite value off c]
p
, where n is an integer.
4
n = 0, 1, 2, …
p 5p
c= ,
, …
4 4
p
c = lies in the interval (0, p ).
4
Hence, theorem is verified.
⎡ p 5p ⎤
x
(v) f ( x ) = e (sin x − cos x ) in ⎢ ,
⎥
⎣4 4 ⎦
Putting
(a) f ( x) = e x (sin x − cos x) , being composite function of continuous func⎡ p 5p ⎤
tions, is continuous in ⎢ ,
⎥.
⎣4 4 ⎦
(b) f (x) = ex (sin x cos x) + ex (cos x + sin x)
= 2ex sin x
⎛ p 5p ⎞
exists for every value of x in ⎜ ,
. Therefore, f ( x ) is differentiable
⎝ 4 4 ⎟⎠
⎛ p 5p ⎞
in ⎜ ,
⎝ 4 4 ⎟⎠
(c)
⎛p ⎞
⎛ 5p ⎞
f ⎜ ⎟ = f ⎜ ⎟=0
⎝4⎠
⎝ 4 ⎠
Thus, f (x) satisfies all the conditions of Rolle’s theorem. Therefore, there
⎛ p 5p ⎞
exists at least one point c in ⎜ ,
such that f ( c) = 0
⎝ 4 4 ⎟⎠
2ec sin c = 0
sin c = 0
[∵ ec 0 for any finite value of x]
c = 0, p, 2p, ……………
⎛ p 5p ⎞
.
c = p lies in ⎜ ,
⎝ 4 4 ⎟⎠
Hence, theorem is verified.
⎡ x 2 + ab ⎤
(vi) f ( x) = log ⎢
⎥ in [a, b], a > 0, b > 0
⎣ ( a + b) x ⎦
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Engineering Mathematics
(a) f (x) = log (x2 + ab) log x log (a + b), being composite function of
continuous functions, is continuous in [a, b].
2x
1
−
x + ab x
exists for every value of x in (a, b) [∵ a > 0, b > 0]. Therefore, f (x) is
differentiable in (a, b).
(c) f (a) = log (a2 + ab) log a log (a + b)
= log a + log (a + b) log a log (a + b)
=0
f (b) = log (b2 + ab) log b log (a + b)
= log b + log (b + a) log b log (a + b)
=0
f (a) = f (b)
Thus, f (x) satisfies all the conditions of Rolle’s theorem. Therefore,
there exists at least one point c in (a, b) such that f ( c) = 0
(b) f ′( x) =
2
2c
1
− =0
c + ab c
2c 2 − c 2 − ab = 0
2
c 2 − ab = 0, c = ± ab
Since c = ab lies between a and b [being geometric mean of a and b].
Hence, theorem is verified.
(vii) f (x) = x2 + 1
0 x 1
=3-x
1 x 2
0 x 1
(a) f (x) = x2 + 1
=3-x
1 x 2
is defined every where in [0, 2] and hence, continuous in [0, 2].
(b) Left hand derivative at x = 1
f ( x) − f (1)
x2 + 1 − 2
= lim−
x →1
x →1
x −1
x −1
2
x −1
= lim−
= lim(
x + 1) = 1 + 1 = 2
x →1 x − 1
x →1−
f ′(1− ) = lim−
Right hand derivative at x = 1
f ( x) − f (1)
x −1
f ( x) − f (1)
3− x − 2
= −1
= lim+
= lim+
x →1
x →1
x −1
x −1
f ( 1-) f ( 1+)
f ′(1+ ) = lim+
x →1
Differential Calculus I
2.49
Thus, function is not differentiable at x = 1 and hence, Rolle’s theorem is not
applicable.
(viii) f (x) = x2 - 2
-1 x 0
=x-2
0 x 1
-1 x 0
(a) f (x) = x2 - 2
=x-2
0 x 1
is defined everywhere in [-1, 1], and hence, is continuous in [-1, 1].
(b) Left hand derivative at x = 0
f ( x ) − f ( 0)
x 2 − 2 − ( −2)
= lim−
x→0
x→0
x−0
x
2
2
x −2+2
x
= lim−
= lim−
=0
x→0
x→0 x
x
f ′(0 − ) = lim−
Right hand derivative at x = 0
f ( x) − f (0)
( x − 2) − ( −2)
= lim+
x→0
x−0
x
x
= lim+ = 1
x→0 x
f ′ (0 − ) ≠ f ′ (0 + )
f ′(0+ ) = lim+
x→0
Thus, function is not differentiable at x = 0 and hence, Rolle’s theorem is
not applicable.
Example 2: Prove that between any two roots of ex sin x = 1 there exists at least
one root of ex cos x + 1 = 0.
Solution: Let f (x) = 1
ex sin x
(a) f (x), being composite function of continuous functions, is continuous in a
finite interval.
(b) f (x) = (ex sin x + ex cos x ) = ( 1 + ex cos x )
[∵ex sin x =1]
exists for every finite value of x. Therefore, f (x) is differentiable in a finite
interval.
(c) Let a and b are two roots of the equation, f (x) = 1 ex sin x = 0
Then f (a ) = f (b ) = 0
Thus, f (x) satisfies all the conditions of Rolle’s theorem in [a, b ]. Therefore,
there exists at least one point c in (a, b ) such that f (c) = 0
1 + ec cos c = 0
This shows that c is the root of the equation ex cos x + 1= 0 which lies between
the root a and b of the equation 1 ex sin x = 0.
Example 3: Prove that the equation 2x3 – 3x2 – x + 1 = 0 has at least one root
between 1 and 2.
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Engineering Mathematics
x4
x2
− x 3 − + x [obtained by integratSolution: Let us consider a function f ( x) =
2
2
ing the given equation]
(a) f (x), being an algebraic function, is continuous in [1, 2]
(b) f (x) = 2x3 3x2 x + 1 exists for every value of x in (1, 2). Therefore, f (x) is
differentiable in (1, 2).
(c) f (1) = f (2) = 0
Thus, f (x) satisfies all the conditions of Rolle’s theorem. Therefore, there exists
at least one point c in (1, 2) such that f (c) = 0
2c3 3c2 c + 1 = 0
This shows that c is the root of the equation 2x3 3x2 x + 1 = 0 which lies
between 1 and 2.
Example 4: Prove that if a0 , a1 , a2 , ....... an are real numbers such that
a0
a
a
a
+ 1+ 2 +
+ n -1 + an = 0, then there exists at least one real numn+1 n n 1
2
n
n -1
+ a2 x n - 2 + ....... + an = 0.
ber x between 0 and 1 such that a0 x + a1 x
Solution: Let us consider a function f ( x) =
a0 x n +1 a1 x n a2 x n −1
+
+
+ ....... + an x
n +1
n
n −1
defined in [0, 1].
(a) f (x), being an algebraic function, is continuous in [0, 1].
(b) f (x) = a0 xn + a1 xn−1 + a2 xn−2 + ……. + an
exists for every value of x in (0, 1). Therefore, f (x) is differentiable in (0, 1).
(c) f (0) = 0
f (1) =
a0
a
a
a
+ 1 + 2 + ........ + n −1 + an = 0
n +1 n n −1
2
[given]
f (0) = f (1)
Thus, f (x) satisfies all the conditions of Rolle’s theorem. Therefore, there exists
at least one point c in (0, 1) such that f (c) = 0
a0 cn + a1 cn−1 + a2 cn−2 + ……….. + an = 0
Replacing c by x,
a0 xn + a1 xn−1 + a2 xn−2 + ………. + an = 0
Example 5: If f (x), e (x), Y (x) are differentiable in (a, b), prove that there exists
f (a )
e (a )
Y (a )
at least one point c in (a, b) such that f (b )
e (b)
Y (b) = 0 .
f (c )
e (c )
Y (c )
Differential Calculus I
2.51
f ( a)
f ( a)
y ( a)
Solution: Let us consider a function F ( x ) = f (b)
f ( x)
f ( b)
y ( b)
f ( x)
y ( x)
(a) Since f (x), f (x), Y (x) are differentiable in (a, b), therefore, will be continuous
in [a, b]. F (x), being composite function of continuous functions, is continuous in [a, b].
f ( a)
f ( a)
y ( a)
(b) F ′( x ) = f (b)
f ( b)
y (b) exists for every value of x in (a, b). There-
f ′( x )
y ′( x )
f ′( x )
fore, f (x) is differentiable in (a, b).
(c) f (a) = f (b) = 0
Thus, f (x) satisfies all the conditions of Rolle’s theorem. Therefore, there exists
at least one point c in (a, b) such that f (c) = 0
f ( a)
f ( a)
y ( a)
f ( b)
f ( b)
y ( b ) = 0.
f ′( c )
f ′( c )
y ′(c )
Example 6: If f (x) = x (x + 1) (x + 2) (x + 3), prove that f (x) = 0 has three real
roots.
Solution: f (x) = x (x + 1) (x + 2) (x + 3)
(a) f (x), being polynomial is continuous, in the intervals [-3, -2], [-2, -1],
[-1, 0].
(b) f (x) = (x + 1) (x + 2) (x + 3) + x (x + 2) (x + 3) + x (x + 1) (x + 3)
+ x (x + 1) (x + 2) exists for every value of x in [-3, -2], [-2, -1] and [-1, 0].
Therefore, f (x) is differentiable in [-3, -2], [-2, -1] and [-1, 0].
(c) f (-3) = f (-2) = f (-1) = f (0) = 0
Thus, f (x) satisfies all the conditions of Rolle’s theorem. Therefore, there exists
at least one c1 in (-3, -2), c2 in (-2, -1) and c3 in (-1, 0) such that f (c1) = f (c2)
= f (c3) = 0
Thus c1, c2 and c3 are the roots of f (x) = 0
Hence f (x) = 0 has at least 3 real roots.
Example 7: If k is a real constant, prove that the equation x3 – 6x2 + c = 0 cannot
have distinct roots in [0, 4].
Solution: Let f (x) = x3
Then
6x2 + c = 0 has distinct roots a and b between 0 and 4 i.e.
0 a<b 4
f (a) = 0 = f (b)
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Engineering Mathematics
Also, f (x) being polynomial is, continuous in [a, b] and f (x) = 3x2 12x exists for every
value of x in [a, b]. Therefore, f (x) is differentiable in (a, b)
Thus, f (x) satisfies all the conditions of Rolle’s theorem. Therefore, there exists at least
one point c in (a, b) such that
f (c) = 0
3c2 12c = 0
3c (c 4) = 0
c = 0, 4
But these values of c lies outside the interval (a, b). This is a contradiction to Rolle’s
theorem. Thus, our assumption is wrong.
Hence, f (x) = 0 cannot have distinct roots in [0, 4].
Exercise 2.3
1. Verify Rolle’s Theorem for the following functions:
(i) x 3 − 12 x in ⎡⎣0, 2 3 ⎤⎦
[ Ans. : c = 2]
(ii) x 3
4 x in [ 2, 2]
⎡
2⎤
⎢ Ans. : c = ±
⎥
3
⎣
⎦
3
2
(iii) 2 x + x − 4 x − 2 in ⎡⎣ − 2 , 2 ⎤⎦
2
⎤
⎡
⎢⎣ Ans. : c = 3 , − 1⎥⎦
(iv) x 2 in [1, 2]
⎤
⎡ Ans. : f (1) ≠ f (2),
⎢ theorem is not applicable ⎥
⎦
⎣
2
(v) 2 + ( x − 1) 3 in [0, 2]
⎡ Ans. : not differentiable ⎤
⎢
at x = 1, theorem ⎥⎥
⎢
⎢⎣
is not applicable ⎥⎦
2
(vi) 1 − ( x − 3) 3 in [2, 4]
⎡ Ans. : not differentiable ⎤
⎢
at x = 3, theorem ⎥⎥
⎢
⎢⎣
is not applicable ⎥⎦
(vii)
x2 − 4x
in [0, 4]
x+2
⎡ Ans. : c = 2
⎣
(
)
3 −1 ⎤
⎦
(viii) ( x + 2)3 ( x − 3) 4 in [ −2, 3]
1⎤
⎡
⎢⎣ Ans. : c = 7 ⎥⎦
⎛ x2 + 6 ⎞
(ix) log ⎜
⎟ in [ 2, 3]
⎝ 5x ⎠
[ Ans. : c = 6 ]
⎡ p p⎤
2
(x) cos x in ⎢ − , ⎥
⎣ 4 4⎦
[ Ans. : c = 0]
(xi) sin x in [0, 2p ]
p 3p ⎤
⎡
⎢ Ans. : c = 2 , 2 ⎥
⎣
⎦
(xii) cos x in[0, p ]
⎡ Ans. : not differentiable
⎢
p
⎢
at x = , theorem
2
⎢
⎢
is
not
applicable
⎣
⎡ p p⎤
(xiii) sin x in ⎢ − , ⎥
⎣ 4 4⎦
⎤
⎥
⎥
⎥
⎥
⎦
Differential Calculus I
⎡ Ans. : not differentiable ⎤
⎢
at x = 0, theorem ⎥⎥
⎢
⎢⎣
is not applicable ⎥⎦
(xiv) f (x) = 1
x=0
=x
0<x 1
⎡ Ans. : discontinuous ⎤
⎢
at x = 0, theorem ⎥⎥
⎢
⎢⎣
is not applicable ⎥⎦
(xv) 1
3
( x 1) 2 in [0, 2]
⎡ Ans. : not differentiable ⎤
⎢
at x = 1, theorem ⎥⎥
⎢
⎢⎣
is not applicable ⎥⎦
(xvi) f (x) = x2 + 2 1 x 0
=x+2 0 x 1
⎡ Ans. : not differentiable ⎤
⎢
at x = 0, theorem ⎥⎥
⎢
⎢⎣
is not applicable ⎥⎦
2.53
2. Prove that one root of the equation x
log x 2 + x = 0 lies in (1, 2).
[Hint : Consider f (x) = (x 2) log x]
3. Prove that the equation tan x = 1 x
has a real root in the interval (0, 1).
4. If c is a real constant, prove that the
equation x3 12x + c = 0 cannot have
two distinct roots in the interval [0, 4].
5. If c is a real constant, prove that the
equation x3 + 3x + c = 0 cannot have
more than one real root.
6. If f (x) = a + b 3bx2 4ax3, a 0,
b 0, prove that there exists at least
one value c in (0, 1) such that f (c) = 0.
7. Prove that one root of the equation
sin q
= cos xq lies between 0 and 1.
q
2.6 LAGRANGE’S MEAN VALUE THEOREM (L.M.V.T.)
Statement: If a function f (x) is
(i) continuous in the closed interval [a, b],
(ii) differentiable in the open interval (a, b),
then there exists at least one point c in the open interval (a, b) such that f (b) − f (a) = f ′ (c).
b−a
Proof: Consider a function f (x) = f (x) + A x where A is a constant to be determined,
such that f ( a) = f (b).
f (a ) + A a = f (b) + Ab
f (b) − f (a )
A=−
b−a
Now
(i) f (x) is continuous in the closed interval [a, b], since f (x), x and A are continuous.
(ii) f (x) is differentiable in the open interval (a, b), since f (x), x and A are differentiable.
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(iii) f (a) = f (b)
[by assumption]
f (x) satisfies all the conditions of Rolle’s mean value theorem. Therefore,
there exists at least one point c in the open interval (a, b) such that f (c) = 0
f ′ (c ) + A = 0
f ′ (c ) = − A
f (b) − f (a )
f ′ (c ) =
b−a
2.6.1 Another Form of Lagrange’s
Mean Value Theorem
If a function f (x) is
(i) continuous in the closed interval [a, a + h],
(ii) differentiable in the open interval (a, a + h),
then there exists at least one number q between 0 and 1( 0 < q < 1) such that
f ( a + h) − f ( a)
h
f (a + h) = f (a) + h f (a + q h).
f ′( a + q h) =
2.6.2 Geometrical Interpretation of
Lagrange’s Mean Value Theorem
Let y = f (x) represents a curve with A [a, f (a)] and B [b, f (b)] as end points and
C [c, f (c)] be any point between A and B. Then
f (b) − f (a )
= slope of the chord AB and f (c) = slope of the tangent at point C.
b−a
Thus, geometrically theorem states that if
(i) curve is continuous at the points A, B and at every point between A and B.
(ii) possesses unique tangent at every point between A and B, then there exists at
least one point c on the curve between A and B, tangent at which is parallel to
the chord AB.
y
y
B
C2
B
C3
A
A
C
O
C1
x
O
Fig. 2.2
x
Differential Calculus I
2.55
2.6.3 Algebraic Interpretation of Lagrange’s
Mean Value Theorem
If a function f (x) is defined in the interval [a, b], then f (b) f (a) is the change in the
f (b) − f (a )
function f (x) from x = a to x = b and therefore,
is the average rate of
b−a
change of the function f (x) in the interval [a, b]. Also, f (c) is the actual rate of change
of the function at x = c. Thus, according to the Lagrange’s Mean Value Theorem, average rate of change of a function over an interval is equal to the actual rate of change of
the function at some point in the interval.
2.6.4 Deductions from Lagrange’s Mean
Value Theorem
Increasing Function
Statement: If a function f (x) is
(i) continuous in the closed interval [a, b],
(ii) differentiable in the open interval (a, b),
(iii) f (x) > 0 throughout the interval (a, b), then f (b) > f (a), i.e., f (x) is strictly
(monotonically) increasing function in the closed interval [a, b].
Proof: By Lagrange’s Mean Value theorem
f (b) − f (a )
= f ′ (c )
b−a
Let f (x) > 0 for all x in (a, b).
Then,
f ′(c) > 0,
a<c<b
f (b) − f (a )
>0
b−a
... (1)
[Using (1)]
f (b) f (a) > 0
[∵ b a > 0 being length of the interval]
f (b) > f (a),
b>a
f (x) is strictly (monotonically) increasing function in the closed interval [a, b].
In general,
f (x2) > f (x1) for x2 > x1, for every value of x1, x2 in [a, b].
Decreasing Function
Statement: If a function f (x) is
(i) continuous in the closed interval [a, b],
(ii) differentiable in the open interval (a, b),
(iii) f (x) < 0 throughout the interval (a, b), then f (b) < f (a), i.e., f (x) is strictly
(monotonically) decreasing function in the closed interval [a, b].
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Proof: By Lagrange’s Mean Value theorem
f (b) − f (a )
= f ′ (c )
b−a
Let f (x) < 0 for all x in (a, b).
f ′ (c ) < 0
a<c<b
f (b) − f (a )
<0
b−a
f (b) f (a) < 0
[∵ b a > 0 being length of the interval]
f (b) < f (a),
b>a
f (x) is strictly (monotonically) decreasing function in the closed interval [a, b].
In general,
f (x2) < f (x1) for x2 > x1, for every value of x1, x2 in [a, b].
Then,
Example 1: Verify Lagrange’s Mean Value Theorem for the following functions:
(i) f (x) = x3 in [-2, 2]
(ii) f (x) = lx2 + mx + n in [a, b]
2
(iii) f (x) = x 3 in [-8, 8]
(iv) f (x) = ex in [0, 1]
(v) f (x) = log x in [1, e].
Solution: (i) f (x) = x3 in [-2, 2]
(a) f (x) = x3, being an algebraic function, is continuous in [-2, 2].
(b) f (x) = 3x2 exists for every value of x in (-2, 2). Therefore, f (x) is differentiable
in (-2, 2).
Thus, f (x) satisfies all the conditions of Lagrange’s Mean Value theorem. Therefore,
there exists at least one point c in (-2, 2) such that
f (2) − f ( −2)
= f ′ (c )
2 − ( −2)
(2)3 − ( −2)3
= 3c 2
2 − ( −2)
4 = 3c 2
2
c=±
3
2
c=±
liies in ( −2, 2).
3
Hence, theorem is verified.
Differential Calculus I
2.57
(ii) f (x) = lx2 + mx + n in [a, b].
(a) f (x) = lx2 + mx + n, being an algebraic function, is continuous in [a, b].
(b) f (x) = 2lx + m, exists for every value of x in (a, b). Therefore, f (x) is differentiable in (a, b).
Thus, f (x) satisfies all the conditions of Lagrange’s Mean Value theorem.
Therefore, there exists at least one point c in (a, b) such that
f (b) − f (a)
= f ′ (c )
b−a
(lb 2 + mb + n) − (la 2 + ma + n)
= 2lc + m
b−a
l (b + a ) + m = 2llc + m
b+a
lies in (a, b) being arithmetic mean of a and b.
2
Hence, theorem is verified.
c=
2
(iii) f (x) = x 3 in [-8, 8].
2
(a) f (x) = x 3 , being an algebraic function, is continuous in [-8, 8].
2 −1
2
(b) f ′( x) = x 3 = 1 which does not exists at x = 0.
3
3 x3
Hence, theorem is not applicable.
(iv) f (x) = ex in [0, 1].
(a) f (x) = ex, being an exponential function, is continuous in [0, 1].
(b) f (x) = ex, exists for every value of x in (0, 1). Therefore, f (x) is differentiable in (0, 1).
Thus, f (x) satisfies all the conditions of Lagrange’s Mean Value theorem.
Therefore, there exists at least one point c in (0, 1) such that
f (1) − f (0)
= f ′ (c )
1− 0
e1 − e0
= ec
1− 0
ec = e - 1
c = log (e - 1) = 0.5413 < 1
c = 0.5413 lies in (0, 1).
Hence, theorem is verified.
(v) f (x) = log x in [1, e].
(a) f (x) = log x, being a logarithmic function, is continuous in [1, e].
1
(b) f (x) = , exists for every value of x in [1, e]. Therefore, f (x) is differenx
tiable in [1, e].
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Engineering Mathematics
Thus, f (x) satisfies all the conditions of Lagrange’s Mean Value theorem. Therefore, there exists at least one point c in [1, e] such that
∵
f (e) − f (1)
= f ′ (c )
e −1
log e − log 1 1
=
e −1
c
c=e 1
2<e<3
1<e-1<2
Thus, c = e 1 lies in (1, e).
Hence, theorem is verified.
Example 2: If a, b are real numbers, prove that there exists at least one real
number c such that b3 + ab2 + a2b + a3 = 4c3, a < c < b.
Solution: Let f (x) = x4 is defined in [a, b].
(a) f (x) = x4, being algebraic function, is continuous in [a, b].
(b) f (x) = 4x3 exists for every value of x in (a, b). Therefore, f (x) is differentiable
in (a, b).
Thus, f (x) satisfies all the conditions of Lagrange’s Mean Value theorem. Therefore,
there exists at least one point c in (a, b) such that
f (b) − f (a )
= f ′ (c )
b−a
b4 − a 4
= 4c 3
b−a
(b 2 − a 2 ) (b 2 + a 2 )
= 4c 3
b−a
(b − a ) (b + a ) (b 2 + a 2 )
= 4c 3
b−a
(b − a ) (b3 + ab 2 + a 2 b + a 3 )
= 4c 3
b−a
b3 + ab 2 + a 2 b + a 3 = 4c 3 , a < c < b.
Hence,
Example 3: Using Lagrange’s Mean Value theorem, prove that
cos ap - cos bp
Ä ( b - a ) if p ñ 0.
p
Solution: Let f (x) = cos xq is defined in the interval [a, b].
(a) f (x), being trigonometric function, is continuous in [a, b].
(b) f (x) = -q sin xq exists for all values of x in (a, b). Therefore, f (x) is differentiable in (a, b).
Differential Calculus I
2.59
Thus, f (x) satisfies all the conditions of Lagrange’s Mean Value theorem. Therefore,
there exists at least one point c in (a, b) such that
f ( b) − f ( a)
= f ′( c)
b−a
cos bq − cos aq
= −q sin cq
b−a
cos aq − cos bq
[∵sin x 1]
= sin cq ≤ 1, if q ≠ 0
( b − a) q
cos aq − cos bq
≤ (b − a), if q ≠ 0.
q
Example 4: Find the point on the curve y = log x, tangent at which is parallel to
the chord joining the points (1, 0) and (e, 1).
Solution: Let c be the point on curve y = log x, tangent at which is parallel to the
chord joining the points (1, 0) and (e, 1).
By Lagrange’s Mean Value theorem,
f (b) − f (a )
= f ′ (c )
b−a
Here,
a = x-coordinate of (1, 0) = 1
b = x-coordinate of (e, 1) = e
f (a) = log 1 = 0, f (b) = log e =1
1
f ′( x) =
x
1− 0 1
=
e −1 c
c = e − 1.
Example 5: At what point is the tangent to the curve y = xn parallel to the chord
joining (0, 0) and (k, kn)?
Solution: Let c be the point on curve y = xn tangent at which is parallel to the chord
joining the points (0, 0) and (k, kn).
By Lagrange’s Mean Value theorem,
Here,
f (b) − f (a )
= f ′ (c )
b−a
a = 0, b = k , f (a ) = 0, f (b) = k n
f ′ ( x) = nx n −1
kn − 0
= nc n −1
k −0
k
c= 1 .
n n −1
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Engineering Mathematics
Example 6: Prove that for any quadratic functions f (x) = px2 + qx + r in
1
whatever p, q, r, a, h may be.
2
Solution: (a) f (x) = px2 + qx + r is continuous in [a, a + h] being an algebraic
function.
[a, a + h], the value of q is always
(b) f (x) = 2px + q, exists for every value of x in (a, a + h). Therefore, f (x) is
differentiable in (0, 1).
Thus, f (x) satisfies all the conditions of Lagrange’s Mean Value theorem. Therefore, there exists at least one number q between 0 and 1 such that
f ( a + h) − f ( a )
= f ′ ( a + q h)
h
f (a + h) f (a) = h f (a + q h)
p (a + h)2 + q (a + h) + r pa2 qa r = h [2p (a + q h) + q]
ph2 + 2pah + qh = h [2pa + 2pq h + q]
1
= which is a constant and does not depend on p, q, r, a, h.
2
1
Hence, value of q is always whatever p, q, r, a, h may be.
2
Example 7: Apply Lagrange’s Mean Value theorem to the function
f (x) = log x in [a, a + h] and determine p in terms of a and h. Hence, deduce that
0<
1
1
− < 1.
log (1 + x ) x
Solution: (a) f (x) = log x, being logarithmic function, is continuous in [a, a + h].
1
, exists for every value of x in (a, a + h). Therefore, f (x) is differenx
tiable in (a, a + h).
(b) f ′( x) =
Thus, f (x) satisfies all the conditions of Lagrange’s Mean Value theorem. Therefore,
there exists at least one number q between 0 and 1 such that
f ( a + h) − f ( a)
= f ′( a + q h)
h
f ( a + h) − f ( a) = h f ′( a + q h)
h
a +qh
h
⎛ h⎞
log ⎜1 + ⎟ =
⎝ a ⎠ a +qh
log ( a + h) − log a =
Differential Calculus I
a +qh =
q =
2.61
h
⎛ h⎞
log ⎜1 + ⎟
⎝ a⎠
1
a
−
⎛ h⎞ h
log ⎜1 + ⎟
⎝ a⎠
Putting h = x and a = 1,
1
1
−
log (1 + x ) x
0 <q <1
1
1
0<
− < 1.
log (1 + x ) x
q =
∵
Example 8: Prove that log10 ( x + 1) =
Solution: Let f ( x) = log10 ( x + 1) =
x log10 e
, where x > 0 and 0 < p < 1.
1+p x
log e ( x + 1)
is defined in [a, a + h].
log e 10
(a) f ( x) =
log e ( x +1)
, being logarithmic function, is continuous in [a, a + h].
log e 10
(b) f ′( x) =
1
, exists for every value of x in (a, a + h). Therefore, f (x)
( x + 1) log e 10
is differentiable in (a, a + h).
Thus, f (x) satisfies all the conditions of Lagrange’s Mean Value theorem. Therefore,
there exists at least one number q between 0 and 1 such that
f ( a + h) − f ( a)
= f ′( a + q h)
h
f ( a + h) − f ( a) = h f ′( a + q h)
log e ( a + h + 1) log e ( a + 1)
1
−
=h
log e 10
( a + q h + 1) log e 10
log e 10
0
log e ( a + h + 1) − log e ( a + 1) = h
Putting a = 0, h = x,
1
( a + q h + 1)
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Engineering Mathematics
1
(0 + q x + 1)
x
log e ( x + 1) =
q x +1
log10 ( x + 1)
x
=
log10 e
q x +1
log e (0 + x + 1) − log e (0 + 1) = x
x log10 e
.
1 +q x
1
Example 9: Separate the interval in which f ( x ) = x + is increasing or
x
decreasing.
log10 ( x + 1) =
Hence,
1
x
1 ( x − 1) ( x + 1)
f ′( x) = 1 − 2 =
x
x2
f ( x) = x +
Solution:
(i) f (x) is an increasing function if
f ′( x) > 0,
( x − 1) ( x + 1)
> 0, i.e., ( x − 1) ( x + 1) > 0
x2
Now, (x 1) (x + 1) > 0 if
Case I: (x 1) > 0 and (x + 1) > 0 i.e., x > 1
Case II: (x 1) < 0 and (x + 1) < 0 i.e., x < 1
Hence, f (x) is increasing in (- , 1) and (1, ).
–¥
x
–1
0
1
(ii) f (x) is a decreasing function if f ′( x) < 0,
¥
( x − 1) ( x + 1)
< 0, i.e. ( x − 1) ( x + 1) < 0
x2
Now, (x 1) (x + 1) < 0 if
Case I: (x 1) < 0 and (x + 1) > 0 i.e., 1 < x < 1
Case II: (x 1) > 0 and (x + 1) < 0 i.e., x > 1 and x <
possible.
Hence, f (x) is decreasing in ( 1, 1 ).
1 but this is not
a⎞
b ⎛b ⎞
⎛
Example 10: If 0 < a < b, prove that ⎜ 1 − ⎟ < log < ⎜ − 1⎟ . Hence, prove
⎝
b⎠
a ⎝a ⎠
that
1
1
1
< log (1.2) < and < log 2 < 1.
6
5
2
Differential Calculus I
2.63
Solution: Let f (x) = log x is defined in [a, b] where 0 < a < b.
(a) f (x) = log x, being logarithmic function, is continuous in [a, b].
1
exists for every value of x in (a, b). Therefore, f (x) is differentiable
x
in (a, b).
(b) f ′( x) =
Thus, f (x) satisfies all the conditions of Lagrange’s Mean Value theorem. Therefore,
there exists at least one point c in (a, b), such that
We have,
f (b) − f (a )
= f ′ (c )
b−a
log b − log a 1
=
b−a
c
b b−a
log =
a
c
a<c<b
1 1 1
> >
a c b
b−a b−a b−a
>
>
a
c
b
b
b
a
− 1 > log > 1 −
a
a
b
a
b b
1 − < log < − 1
b
a a
... (1)
[∵ b > a]
[ Using Eq. (1)]
… ( 2)
(i) Putting b = 6, a = 5 in Eq. (2),
1−
5
6 6
< log < − 1
6
5 5
1
1
< log 1.2 <
6
5
(ii) Putting b = 2, a = 1 in Eq. (2),
1−
1
2
< log 2 < − 1
2
1
1
< log 2 < 1.
2
Example 11: Prove that if 0 < a < 1, 0 < b < 1 and a < b, then
b−a
1− a
(i)
2
< sin −1 b - sin −1 a <
b-a
1 - b2
1
1 π
1
π
< sin −1 < 6 2 3
4 6
15
and hence, deduce that
(ii)
o
3
3 o 1
+
< sin -1 < + .
6 15
5 6 8
Engineering Mathematics
2.64
Solution: Let f (x) = sin–1 x defined in [a, b].
(a) f (x) = sin–1 x, being a trigonometric function, is continuous in [a, b].
1
(b) f ′( x) =
, exists for every value of x in (a, b). Therefore, f (x) is
1 − x2
differentiable in (a, b).
Thus, f (x) satisfies all the conditions of Lagrange’s Mean Value theorem. Therefore,
there exists at least one point c in (a, b) such that
f ( b) − f ( a)
= f ′( c )
b−a
sin −1 b − sin −1 a
1
=
b−a
1 − c2
sin −1 b − sin −1 a =
We have,
b−a
1 − c2
a<c<b
a2 < c2 < b2
a2
c2
b2
a2
c2
[∵ a > 0, b > 0]
b2
1 − a2 > 1 − c2 > 1 − b2
1
1− a
b−a
2
1 − a2
b−a
1− a
(i) Putting b =
... (1)
2
<
<
1
1− c
b−a
2
1 − c2
<
<
[∵ 0 < a < 1 and 0 < b < 1]
1
1 − b2
b−a
[∵ b – a > 0]
1 − b2
< sin −1 b − sin −1 a <
b−a
1 − b2
1
1
and a = in Eq. (2),
4
2
1
1
−
−
4 < sin −1 1 − sin −1 1 < 4
4
2
15
3
4
2
−
1
2 3
< sin −1
1 p
1
− <−
4 6
15
1
1 p
1
p
−
< sin −1 < −
.
6 2 3
4 6
15
... (2) [Using Eq. (1)]
Differential Calculus I
(ii) Putting b =
2.65
3
1
and a = in Eq. (2),
5
2
1
1
3
1
10 < sin −1 − sin −1 < 10
5
2 4
3
5
2
1
5 3
< sin −1
3 p 1
− <
5 6 8
p
1
3 p 1
+
< sin −1 < +
6 5 3
5 6 8
p
3
3 p 1
+
< sin −1 < + .
6 15
5 6 8
Example 12: Using Lagrange’s Mean Value theorem, prove that
b- a
< tan −1 b - tan −1 a <
b- a
1+ b
1 + a2
o
3
4 o 1
+
< tan −1 < + .
4 25
3 4 6
2
and hence, deduce that
Solution: Let f (x) = tan−1 x is defined in [a, b] where a > 0, b > 0.
(a) f (x) = tan–1 x, being a trigonometric function is continuous in [a, b].
1
, exists for every value of x in (a, b). Therefore, f (x) is differen(b) f ′( x) =
1 + x2
tiable in (a, b).
Thus, f (x) satisfies all the conditions of Lagrange’s Mean Value theorem. Therefore, there exists at least one point c in (a, b) such that
f (b) − f (a )
= f ′ (c )
b−a
tan −1 b − tan −1 a
1
=
b−a
1 + c2
We have,
a<c<b
a2 < c2 < b2
1 + a2 < 1 + c2 < 1 + b2
1 + b2 > 1 + c2 > 1 + a2
1
1
1
<
<
1 + b2 1 + c2 1 + a 2
b−a b−a b−a
<
<
1 + b2 1 + c2 1 + a 2
b−a
b−a
< tan −1 b − tan −1 a <
2
1+ b
1 + a2
... (1)
[∵ a > 0, b > 0]
[∵ b
a > 0]
... (2) [Using Eq. (1)]
2.66
Putting b =
Engineering Mathematics
4
, a = 1 in Eq. (2),
3
4
4
−1
−1
4
3
< tan −1 − tan −1 1 < 3
16
3
1+1
1+
9
3
4 p 1
< tan −1 − <
25
3 4 6
p 3
4 p 1
+
< tan −1 < + .
4 25
3 4 6
−1
Example 13: Prove that tan x >
Solution: Let f ( x) = tan −1 x −
o
x
0 < tan −1 x < .
2 if
2
x
1+
3
x
x2
1+
3
0 < tan −1 x <
If
p
2
tan 0 < tan (tan −1 x) < tan
p
2
0< x<∞
Now,
⎡ x2
2x ⎤
⎢1 + 3 − x ⋅ 3 ⎥
1
−⎢
f ′( x) =
⎥
1 + x2 ⎢ ⎛ x2 ⎞ 2 ⎥
⎢ ⎜1 + 3 ⎟ ⎥
⎠ ⎦
⎣ ⎝
⎤
⎡ ⎛ x2 ⎞ 2 ⎛ x2 ⎞
⎢ ⎜1 + ⎟ − ⎜1 − ⎟ (1 + x 2 ) ⎥
3⎠ ⎝
3⎠
⎥
⎢⎝
=⎢
⎥
2 2
⎛ x ⎞
⎥
⎢
(1 + x 2 ) ⎜ 1 + ⎟
⎢⎣
3⎠
⎝
⎦⎥
x4 2x2
x2 x4
1+ +
− 1 − x2 + +
9
3
3
3
=
2 2
⎛ x ⎞
(1 + x 2 ) ⎜1 + ⎟
3⎠
⎝
=
4x4
⎛ x2 ⎞
9 (1 + x ) ⎜1 + ⎟
3⎠
⎝
2
i.e., f (x) > 0 for every value of x in (0, )
2
> 0, for every value of x in (0,
Differential Calculus I
2.67
Hence, f (x) is strictly increasing function in (0, ).
f (x) > f (0)
for
x>0
[∵ f (0) = 0]
f (x) > 0
for x > 0
x
tan −1 x −
>0
x2
1+
3
x
tan −1 x >
.
x2
1+
3
Example 14: Prove that x2 - 1 > 2x log x > 4 (x - 1) - 2 log x, for all x > 1.
Solution: (i) Let f (x) = x2 1 2x log x
f (x) = 2x 2 log x 2
It is difficult to decide about the sign of f (x) therefore differentiating again w.r.t. x,
2 2( x − 1)
=
>0
x
x
Hence, f (x) is strictly increasing function for x > 1.
f (x) > f (1) for
x>1
f (x) > 0
for
x>1
[∵ f (1) = 2
f ′′( x) = 2 −
[∵ x > 1]
2 log 1
2 = 0]
Hence, f (x) is strictly increasing function for x > 1.
f (x) > f (1)
for
x>1
f (x) > 0
for
x>1
[ ∵ f (1) = 1 1 2 log 1 = 0]
x2 1 2x log x > 0 for x > 1
x2 1 > 2x log x
for x > 1
… (1)
(ii) Let f (x) = 2x log x
4 (x
1) + 2 log x
f ′( x) = 2 log x + 2 − 4 +
2
x
f (x). Therefore, differentiating again w.r.t. x,
2 2
−
x x2
2( x − 1)
=
>0
x2
Hence, f (x) is strictly increasing function for x > 1.
f ′′( x) =
f (x) > f (1)
f (x) > 0
for
for
x>1
x>1
[∵ x > 1]
[∵ f ′ (1) = 2 log 1 + 2 − 4 +
2
= 0]
1
Hence, f (x) is an increasing function for x > 1.
f (x) > f (1)
f (x) > 0
for
for
x>1
x>1
[∵ f (1) = 2 log 1
4 (1
1) + 2 log 1 = 0 ]
Engineering Mathematics
2.68
2x log x 4 (x
2x log x > 4 (x
1) + 2 log x > 0
for
x>1
1)
for
x>1
2 log x
… (2)
From Eqs (1) and (2), we get
x2
1 > 2x log x > 4 (x
Example 15: Prove that 2 x < log
Solution: (i) Let f (x) = 2 x − log
= 2x
1)
2 log x
for x > 1.
⎡ 1 ⎛ x2 ⎞ ⎤
1+ x
< 2 x ⎢1 + ⎜
in (0, 1).
2 ⎟⎥
1- x
⎣ 3 ⎝ 1 - x ⎠⎦
1+ x
1− x
log (1 + x) + log (1
x)
1
1
2 − 2x − 1 + x − 1 − x
−
=
1+ x 1− x
(1 − x 2 )
2
f ′ ( x) = 2 −
=
−2 x 2
<0
1 − x2
[∵ 0 < x < 1,
1 – x2 > 0]
Hence, f (x) is a decreasing function in (0, 1).
f (x) < f (0) for
x>0
f (x) < 0
x>0
2 x − log
1+ x
<0
1− x
2 x < log
(ii) Let f ( x) = log
[∵ f (0) = 0]
for
1+ x
1− x
... (1)
⎡ 1 ⎛ x2 ⎞ ⎤
1+ x
− 2 x ⎢1 + ⎜
2⎟⎥
1− x
⎣ 3 ⎝1− x ⎠ ⎦
⎡ 1 ⎛ x2 ⎞ ⎤
= log (1 + x) − log (1 − x) − 2 x ⎢1 + ⎜
2⎟⎥
⎣ 3 ⎝1− x ⎠ ⎦
⎡ 1 ⎛ x2 ⎞ ⎤
⎡
2x
2 x3 ⎤
1
1
x
− 2 ⎢1 + ⎜
2
0
f ′ ( x) =
+
−
+
+
⎢
2⎟⎥
2
2 2⎥
1+ x 1− x
⎣ 3 ⎝1− x ⎠ ⎦
⎣ 3 (1 − x ) 3 (1 − x ) ⎦
=
⎡ 3 − 3x 2 + x 2 ⎤
1− x +1+ x
−
2
⎥ − 2x
⎢
2
1 − x2
⎣ 3 (1 − x ) ⎦
=
4x2
4x2
−
2
3 (1 − x ) 3 (1 − x 2 ) 2
=
4 x 2 (1 − x 2 ) − 4 x 2
3 (1 − x 2 ) 2
=
−4 x 4
<0
3 (1 − x 2 ) 2
⎡ 2 x − 2 x3 + 2 x3 ⎤
⎥
⎢
2 2
⎣ 3 (1 − x ) ⎦
Differential Calculus I
2.69
Hence, f (x) is a decreasing function in (0, 1).
f (x) < f (0)
f (x) < 0
for
for
x>0
x>0
[∵ f (0) = 0]
log
⎡ 1 ⎛ x2 ⎞⎤
1+ x
− 2 x ⎢1 + ⎜
<0
2 ⎟⎥
1− x
⎣ 3 ⎝ 1 − x ⎠⎦
log
⎡ 1 ⎛ x2
1+ x
< 2 x ⎢1 + ⎜
2
1− x
⎣ 3 ⎝1− x
⎞⎤
⎟⎥
⎠⎦
... (2)
From Eqs (1) and (2), we get
2 x < log
⎡ 1 ⎛ x2 ⎞⎤
1+ x
< 2 x ⎢1 + ⎜
.
2 ⎟⎥
1− x
⎣ 3 ⎝ 1 − x ⎠⎦
Exercise 2.4
1. Verify Lagrange’s Mean Value theorem for the following functions:
(i)
x
2
⎡ Ans.: c = 5 ⎤
⎣
⎦
1
in[ 1, 1]
x
⎡ Ans.: Discontinuous at x = 0, ⎤
⎢
theorem not applicable ⎥⎦
⎣
1
⎡1 ⎤
(iii) x + in ⎢ , 3⎥
x
⎣2 ⎦
(ii)
⎡
⎢ Ans.: c =
⎣
3⎤
⎥
2⎦
⎤
2⎥
⎦
[ Ans.: c = 1.08 ]
(v) ( x 1)( x 2) in [0, 4]
[ Ans.: c = 2 ]
(vi) ( x 1) ( x 2) ( x 3) in [0, 4]
[ Ans.:
(vii) tan x in [0, 1]
1
⎤
⎥
⎥⎦
1
(viii) x 3 in [ 1, 1]
4 in[2, 3]
1
(iv) log e x in ⎡⎢ ,
⎣2
⎡
16 − p 2
⎢ Ans.: c =
p
⎢⎣
c = 2±2 3 ]
⎡ Ans.: not differentiable
⎢
at x = 0, theorem
⎢
⎢⎣
is not applicable
⎤
⎥
⎥
⎥⎦
(ix) x x 3 in [ 2, 1]
[ Ans.:
c = −1 ]
1
(x) sin x in [0, 1]
⎡
p2 −4
⎢ Ans.: c =
p
⎢⎣
⎤
⎥
⎥⎦
⎡ p⎤
(xi) cos x in ⎢0, ⎥
⎣ 2⎦
⎡
−1 2 ⎤
⎢ Ans.: c = sin p ⎥
⎣
⎦
2. Test whether the Lagrange’s Mean
Value theorem holds for f (x) = 2x2
7x 10 in the interval [2, 5] and if
so, find the value of c.
7⎤
⎡
⎢ Ans.: yes, c = 2 ⎥
⎦
⎣
2.70
Engineering Mathematics
3. Prove that x3 3x2 + 3x + 2 is strictly
increasing in every interval.
⎡ Hint : f ′( x) = 3( x − 1) 2 > 0 for ⎤
⎢
⎥
all values of x except x = 1⎦
⎣
4. Prove that x sin x is strictly increasing in every interval.
5. Separate the intervals in which the
polynomial x3 6x2 36x + 7 is increasing or decreasing:
⎡ Ans.: Increasing in (6, ∞), ⎤
⎢
⎥
(− ∞, − 2) and
⎢
⎥
decreasing in (−2, 6) ⎦⎥
⎣⎢
6. Separate the intervals in which the
following polynomials are increasing
or decreasing:
(i) x3 3x2 + 24x 31
(ii) 2x3 15x2 36x + 40
(iii) 2x3 9x2 + 12x + 5
⎡ Ans.: (i) Increasing in ( − ∞, 4), ⎤
⎢
(2, ∞) and decreasing ⎥⎥
⎢
⎢
⎥
in ( − 2, 4).
⎢
⎥
(ii) Increasing in ( − ∞, − 1), ⎥
⎢
⎢
(6, ∞) and decreasing ⎥
⎢
⎥
in ( − 1, 6).
⎢
⎥
⎢
(iii) Increasing in ( − ∞, 1), ⎥
⎢
⎥
(2, ∞) and decreasing ⎥
⎢
⎢
⎥
in (1, 2).
⎣
⎦
7. Find the value of q in Lagrange’s
Mean Value theorem for the following:
(i) ax2 + bx + c at x = 0
(ii) f (x) = x3, 1 < x < 2
⎡ Ans. (i) interval is (0, h), ⎤
⎢
⎥
1
7⎥
⎢
q =
(ii) − 1 +
⎢⎣
2
3 ⎥⎦
8. Prove that the following functions
are increasing in the given interval:
(i) x3 3x2 + 3x + 1, ( , )
(ii) log x, (a, ), where a > 0
(iii) e x, ( − ∞, ∞)
⎛ p p⎞
(iv) sin x, ⎜ − , ⎟
⎝ 2 2⎠
(v) cos x, ( , 2 )
⎛ p p⎞
(vi) tan x, ⎜ − , ⎟
⎝ 2 2⎠
[Hint : Prove that f
given interval]
(x) > 0 in the
9. Prove that the following functions
are decreasing in the given interval:
2
(i) e − x , (0, ∞)
(ii) cos x, (0, p )
⎛ p⎞
(iii) cosec x, ⎜ 0, ⎟
⎝ 2⎠
⎛
(iv) sin x, ⎜ ,
⎝2
⎞
⎟
⎠
⎛ p
⎞
(v) sec x, ⎜ − , 0 ⎟
⎝ 2 ⎠
(vi) cot x, (0, p )
[Hint : Prove that f (x) < 0 in the
given interval]
10. Prove the following:
(i) log (1 + x) < x for all x > 1
1
< log x < x − 1
x
for all x > 1
(ii) 1 −
(iii) log x < x < tan x
for all x > 1
x
(iv) e > 1 + x for all x > 0
(v) 0 < − log (1 − x) <
x
1− x
for 0 < x < 1
(vi)
1
tan −1 x
<
<1
x
1 + x2
for x > 0
Differential Calculus I
tangent at P is parallel to the x-axis.
sin −1 x
1
<
x
1 − x2
for 0 ≤ x < 1
(vii) 1 <
⎛ e − 1⎞
1
log ⎜
<1
x
⎝ x ⎟⎠
x
(viii) 0 <
2.71
11. Find the point on the curve y = x2,
tangent at which is parallel to the
chord joining the points (1, 1) and
(3, 9).
[Ans. : c = 2]
12. Prove that for the curve y = x2 + 2k1x
+ k2, the chord joining the points
x = a and x = b is parallel to the
a+b
tangent at x =
.
2
13. Prove that the chord joining the
points x = 2, x = 3 on the curve y = x3
is parallel to the tangent to the curve
19
at x =
.
3
14. Prove that on the curve y = 2 sin x
+ cos 2x, there is a point P between
Find the abscissa of P.
p
⎡
⎢ Ans.: c = 6
⎣
y
15. Prove that log ( x + y ) < log x +
x
x > 0, y > 0.
⎤
⎥
⎦
if
⎡ Hint : f ( z ) = log z in [ x, x + y ], ⎤
⎢
⎥
⎢ f ′ ( z ) = 1 > 0,
⎥
z
⎢
⎥
⎢ f ( x + y ) − f ( x)
⎥
⎢
⎥
= f ′ (c),
⎢ ( x + y) − x
⎥
⎢ log ( x + y ) − log x 1 1
⎥
⎢
⎥
= <
y
c x
⎢
⎥
⎢
⎥
(∵ c > x)
⎢
⎥
⎢
⎥
y
⎢log ( x + y ) − log x < ,
⎥
x
⎢
⎥
y
⎢
⎥
log
(
)
l
og
x
x
+
y
<
+
⎢⎣
⎥⎦
x
16. If a, b are real numbers, prove that
there exists at least one real number
⎛p ⎞
(0, 1) and ⎜ , 1⎟ such that the
⎝2 ⎠
c such that b2 + ab + a2 = 3c2, a <
c<b
[Hint : Let f (x) = x3]
2.7 CAUCHY’S MEAN VALUE THEOREM (C.M.V.T.)
Statement: If two functions f (x) and g (x) are
(i) continuous in the closed interval [a, b],
(ii) differentiable in the open interval (a, b),
(iii) g (x) 0 for any x in the open interval (a, b), then there exists at least one
point c in the open interval (a, b) such that
f (b) − f (a ) f ′ (c)
=
g (b) − g (a ) g ′ (c)
Proof: Consider a function f ( x) = f (x) + Ag (x), where A is a constant to be determined such that f (a ) = f (b).
f (a) + Ag (a) = f (b) + Ag (b)
A= −
f (b) − f (a )
g (b) − g (a )
2.72
Engineering Mathematics
Now since f (a) = f (b) and f ( x) being the combination of two continuous and differentiable functions is also continuous in the closed interval [a, b] and differentiable
in the open interval (a, b).
Thus, f (x) satisfies all the conditions of Rolle’s mean value theorem. Therefore,
there exists at least one point c in the open interval (a, b) such that f (c) = 0
f (c) +Ag (c) = 0
f (c) = Ag (c)
where a < c < b and g ′(c) ≠ 0
f ′ (c) f (b) − f (a )
=
,
g ′ (c) g (b) − g (a )
[Substituting value of A]
2.7.1 Another Form of Cauchy’s Mean Value Theorem
If two functions f (x) and g (x) are
(i) continuous in the closed interval [a, a + h],
(ii) differentiable in the open interval (a, a + h),
(iii) g (x) 0 for any x in the open interval (a, a + h), then there exists at least one
number q lying between 0 and 1 such that
f ( a + h) − f ( a ) f ′ ( a + q h)
=
, where 0 < q < 1.
g ( a + h) − g ( a ) g ′ ( a + q h)
Example 1: Verify Cauchy’s Mean Value Theorem for the following functions:
(i) x2 and x4 in [a, b], where a > 0, b > 0
⎡ o⎤
(ii) sin x and cos x in ⎢ 0, ⎥ .
⎣ 2⎦
Solution: (i) Let f (x) = x2, g (x) = x4
(a) f (x) and g (x), both being algebraic functions, are continuous in the closed
interval [a, b].
(b) f (x) = 2x and g (x) = 4x3 exists for all values of x in the open interval (a, b).
Therefore, f (x) and g (x) are differentiable in (a, b), and g (x) = 4x3 0 for any
x in (a, b) since a > 0, b > 0.
Thus, f (x) and g (x) satisfies all the conditions of Cauchy’s Mean Value theorem.
Therefore, there exists at least one point c in (a, b) such that
f (b) − f (a ) f ′ (c)
=
g (b) − g (a ) g ′ (c)
b2 − a 2
2c
1
= 3 = 2
4
4
b −a
4c
2c
(b 2 − a 2 )
1
=
(b 2 + a 2 ) (b 2 − a 2 ) 2c 2
2c2 = b2 + a2
Differential Calculus I
b2 + a 2
2
c=±
c=
2.73
b2 + a 2
2
which lies between a and b.
(ii) Let f (x) = sin x, g (x) = cos x
(a) f (x) and g (x), both being trigonometric functions, are continuous in
⎡ p⎤.
⎢0, 2 ⎥
⎦
⎣
⎛ p⎞
(b) f (x) = cos x, g (x) = sin x exists for all values of x in ⎜ 0, ⎟ and
⎝ 2⎠
g (x) =
sin x
⎛ p⎞
0 for any x in ⎜ 0, ⎟ .
⎝ 2⎠
Thus, f (x) and g (x) satisfies all the conditions of Cauchy’s Mean Value theorem.
⎛ p⎞
Therefore, there exists at least one point c in ⎜ 0, ⎟ such that
⎝ 2⎠
⎛p ⎞
f ⎜ ⎟ − f ( 0)
⎝ 2⎠
=
f ′ (c )
g ′ (c )
⎛p ⎞
g ⎜ ⎟ − g ( 0)
⎝ 2⎠
p
sin − sin 0
cos c
2
=
p
cos − cos 0 − sin c
2
1− 0
= − cot c
0 −1
−1 = − cot c
cot c = 1, c =
p
4
p
.
2
Hence, theorem is verified.
which lies between 0 and
1
, prove that c of Cauchy’s Mean Value
x
x
theorem is the harmonic mean between a and b, a > 0, b > 0.
Example 2: If f ( x ) =
1
2
, and g ( x ) =
Solution: (a) f (x) and g(x) are continuous in the closed interval [a, b] for a > 0, b > 0.
2
1
(b) f ′ ( x) = − 3 and g ′ ( x) = − 2 exists for all x in (a, b) and g (x) 0 for any x
x
in (a, b). x
Engineering Mathematics
2.74
Thus, f (x) and g (x) satisfies all the conditions of Cauchy’s Mean Value theorem.
Therefore, there exists at least one point c in (a, b) such that
f (b) − f (a ) f ′(c)
=
g (b) − g (a ) g ′(c)
1 1
2
−
−
b 2 a 2 = c3
1 1
1
−
− 2
b a
c
(a 2 − b 2 )(ab) 2
=
(a 2 b 2 )(a − b) c
a+b 2
=
ab
c
2 1 1
= +
c b a
1 1⎛1 1⎞
= ⎜ + ⎟
c 2⎝a b⎠
Hence, c is the harmonic mean between a and b.
Example 3: If f ( x ) =
1
, prove that c of Cauchy’s Mean
x
Value theorem is geometric mean between a and b, a > 0, b > 0.
Solution: (a) f ( x) =
(b) f ′ ( x) =
1
2 x
x and g ( x ) =
1
x and g ( x) =
1
, g ′ ( x) = −
2( x)
3
2
are continuous in [a, b] for a > 0, b > 0.
x
exists for all x in (a, b) and g (x)
0 for any x
in (a, b).
Thus, f (x) and g (x) satisfies all the conditions of Cauchy’s Mean Value theorem.
Therefore, there exists at least one point c in (a, b) such that
f (b) − f (a ) f '(c)
=
g (b) − g (a ) g '(c)
1
b− a
= 2 c
1
1
1
−
−
3
b
a
2(c) 2
(
a− b
(
)
a− b
ab
)
=c
c = ab
Hence, c is the geometric mean between a and b.
Differential Calculus I
2.75
Example 4: If f (x) = ex and g (x) = e−x, prove that c of Cauchy’s Mean Value
theorem is arithmetic mean between a and b, a > 0, b > 0.
Solution: (a) f (x) and g (x), being exponential functions, are continuous in [a, b].
(b) f (x) = ex, g (x) = − e−x exists for all x in (a, b) and g (x)
0 for any x in (a, b).
Thus, f (x) and g (x) satisfies all the conditions of Cauchy’s Mean Value theorem.
Therefore, there exists at least one point c in (a, b) such that
f (b) − f (a ) f ′ (c)
=
g (b) − g (a ) g ′ (c)
eb − e a
ec
= −c
−b
−a
e −e
−e
b
a
e −e
= −e2c
1 1
−
eb e a
− (e a − e b ) e b e a
= −e2c
(e a − e b )
ea + b = e2c
a + b = 2c
a+b
c=
2
Hence, c is the arithmetic mean between a and b.
[By comparing ]
Example 5: If 1 < a < b, prove that there exists c satisfying a < c < b such that
b b2 - a 2
log =
.
a
2c 2
Solution: Let f (x) = log x, g(x) = x2 are defined in (a, b).
(a) f (x), being logarithmic function and g(x), being algebraic function, are continuous in [a, b] for a > 1, b > 1.
1
(b) f ′ ( x) = , g ′ ( x) = 2 x exists for all x in (a, b) and g (x) 0 for any x in (a, b)
x
since a > 1, b > 1.
Thus, f (x) and g (x) satisfies all the conditions of Cauchy’s Mean Value theorem.
Therefore, there exists at least one point c in (a, b) such that
f (b) − f (a ) f ′ (c)
=
g (b) − g (a ) g ′ (c)
1
log b − log a c
=
2c
b2 − a 2
Hence,
log
b b2 − a 2
=
.
a
2c 2
Engineering Mathematics
2.76
Example 6: Using appropriate mean value theorem, prove that
sin b − sin a
e −e
b
a
=
cos c
ec
for a < c < b. Hence, deduce that ec sin x = (ex - 1) cos c.
Solution: Let f (x) = sin x, g (x) = ex are defined in (a, b).
(a) f (x), being trigonometric function and g (x), being exponential function, are
continuous in [a, b].
(b) f (x) = cos x, g (x) = ex exists for all x in (a, b) and g (x) 0 for any x in (a, b).
Thus, f (x) and g(x) satisfies all the conditions of Cauchy’s Mean Value theorem.
Therefore, there exists at least one point c in (a, b) such that
f (b) − f (a ) f ′ (c)
=
g (b) − g (a ) g ′ (c)
sin b − sin a cos c
= c
eb − e a
e
... (1)
Putting b = x, a = 0 in Eq. (1),
sin x − sin 0 cos c
= c
e x − e0
e
ec sin x = (ex 1) cos c.
,
Example 7: Using Cauchy s Mean Value theorem, prove that there exists a num1
⎛b⎞
ber c such that 0 < a < c < b and f (b) - f (a) = c f (c) log ⎜ ⎟ . By putting f ( x ) = x n ,
⎝a⎠
1
(
)
deduce that lim n b n − 1 = log b.
n→ ∞
Solution: Let g (x) = log x is defined in [a, b].
(a) Let f (x) is continuous in [a, b] and differentiable in (a, b). Also g (x), being a
1
logarithmic function, is continuous in [a, b] for 0 < a < b. g ′( x) = exists for
x
all x in (a, b) since 0 < a < b and g (x) 0 for any x in (a, b).
Thus, f (x) and g (x) satisfies all the conditions of Cauchy’s Mean Value theorem.
Therefore, there exists at least one point c in (a, b) such that
f (b) − f (a ) f ′ (c)
=
g (b) − g (a ) g ′ (c)
f (b) − f (a ) f ′ (c)
=
1
log b − log a
c
Hence,
⎛b⎞
f (b) − f (a ) = cf ′ (c) log ⎜ ⎟
⎝a⎠
Differential Calculus I
1
Putting f ( x) = x n , f ′ ( x) =
2.77
1 1n −1
x in Eq. (1),
n
1
1
1 1 −1
⎛b⎞
(b) n − (a ) n = c ⋅ (c) n log ⎜ ⎟
n
⎝a⎠
1
1
⎛ 1
⎞
n
⎛b⎞
n ⎜ b n − a n ⎟ = c log ⎜ ⎟
⎝a⎠
⎝
⎠
1
1
⎛ 1
⎞
n
⎛b⎞
lim n ⎜ b n − a n ⎟ = lim c log ⎜ ⎟
n →∞
n →∞
⎝a⎠
⎝
⎠
= c 0 log
b
b
= log
a
a
Putting a = 1,
⎛ 1 ⎞
lim n ⎜ b n − 1⎟ = log b.
n →∞ ⎝
⎠
Exercise 2.5
1. Verify Cauchy’s Mean Value theorem
for the following functions :
(i) f (x) = 3x + 2, g (x) = x2 + 1 in
[1, 4]
(ii) f (x) = x2 + 2, g (x) = x3 − 1 in
[1, 2]
(iii) f (x) = 2x3, g (x) = x6 in [a, b]
(iv) f (x) = log x, g (x)
1
in [1, e]
x
5
⎡
⎢ Ans. : (i) c = 2
⎢
1
⎢
3
3 3
⎢(iii) c = ⎛⎜ a + b ⎞⎟
⎝ 2 ⎠
⎣⎢
14 ⎤
9 ⎥
⎥
⎥
e ⎥
(iv) c =
e − 1 ⎥⎦
(ii) c =
2. Using Cauchy’s Mean Value theo-
rem, find lim
x →1
⎛ p x⎞
cos ⎜ ⎟
⎝ 2⎠
log x
px ⎤
⎡
⎢ Hint : Consider f ( x ) = cos 2 , ⎥
⎢
⎥
⎣ g ( x ) = log x in the interval ( x, 1) ⎦
p c⎤
⎡
⎢⎣ Ans. : − 2 ⎥⎦
3. If f (x) is continuous in [a, b], f (x)
exists in (a, b), prove that there exists
a point c in (a, b) such that
f (b) − f (a ) f ′(c)
(i)
=
2c
b2 − a 2
f (b) − f (a ) f ′(c)
(ii)
=
b3 − a 3
3c 2
[Hint : (i) g (x) = x2 (ii) g (x) = x3]
4. Using appropriate
theorem, prove that
mean
value
sin b − sin a
= cot c, a < c < b.
cos a − cos b
5. If f (x) = sin x and g (x) = cos x in
[a, b], prove that c of Cauchy’s Mean
Engineering Mathematics
2.78
Value theorem is the arithmetic mean
of a and b.
6. If f (x) and g (x) are continuous in
[a, b] and differentiable in (a, b),
g (a) g (b) and g (x) 0 in (a, b),
then there exists at least one c between a and b such that
f ′ (c ) f (c ) − f ( a )
=
, a<c<b
g ′ (c) g (b) − g (c)
⎡ Hint : Let P ( x) = f ( x) g ( x) ⎤
⎢
⎥
⎢
and Q( x) = f ( x) g (b) ⎥
⎢
⎥
⎢
+ g ( x) g (a ), ⎥
⎢
⎥
⎢⎣
⎥⎦
apply CMVT
7. If 0
x
1, prove that
1 − x log (1 + x)
<
<1
1+ x
sin −1 x
⎡ Hint : f ( x) = log (1 + x), g ( x) ⎤
⎢
⎥
= sin −1 x,
⎢
⎥
⎢
apply CMVT in [0, x] ⎥
⎢
⎥
1 1 1
⎢
0 < c < x < 1, > > , ⎥
⎢
c x 1⎥
⎢
⎥
−
c
x
−
1
1
1−1
⎢
⎥
>
>
⎢⎣
⎥⎦
1+ c 1+ x 1+1
8. If f (x), g (x), h (x) are three functions
differentiable in the interval (a, b),
prove that there exists a point c in
(a, b) such that
f ′(c) g ′(c) h′(c)
f ( a ) g ( a ) h( a ) = 0
f (b) g (b) h(b)
Hence, deduce Lagrange’s and
Cauchy’s Mean Value theorem.
⎡ Hint : Consider F ( x )
⎤
⎢
⎥
f ( x) g ( x) h ( x) ⎥
⎢
⎢
= f ( a) g ( a) h ( a) ⎥
⎢
⎥
f ( b) g ( b) h ( b) ⎥
⎢
⎢ Apply Rolle’s theorem. For
⎥
⎢
⎥
⎢ deduction of Lagrange′s
⎥
⎢ MVT g( x ) = x, h ( x ) = 1, Keep ⎥
⎢
⎥
⎢ f ( x) as it is in result. For
⎥
⎢
⎥
⎢ deduction of Cauchy′s MVT ⎥
⎢ take h ( x ) = 1, keep f ( x) and ⎥
⎢
⎥
⎣ g ( x) as it is in the result.
⎦
2.8 TAYLOR’S SERIES
Statement: If f (x + h) be a given function of h which can be expanded into a convergent series of positive ascending integral powers of h, then
f ( x + h) = f ( x) + h f ′( x) +
h2
h3
hn n
f ′′( x) +
f ′′′( x) + ........ +
f ( x) + .......
2!
3!
n!
Proof: Let f (x + h) be a function of h which can be expanded into positive ascending integral powers of h, then
... (1)
f (x + h) = a0 + a1 h + a2 h2 + a3 h3 + a4 h4 +…….……
Differentiating w.r.t. h successively,
... (2)
f (x + h) = a1 + a2 · 2h + a3 · 3h2 + a4 · 4h3 + ……. ………
2
f (x + h) = a2 · 2 + a3 · 6h + a4 · 12h +……. ………
... (3)
f (x + h) = a3 · 6 + a4 · 24h +……. ………
... (4)
and so on
Differential Calculus I
2.79
Putting h = 0 in Eq. (1), (2), (3) and (4),
a0 = f (x)
a1 = f (x)
1
f ′′ ( x)
2!
1
a3 = f ′′′ ( x) and so on
3!
a2 =
Substituting a0, a1, a2 and a3 in Eq. (1),
f ( x + h) = f ( x) + h f ′( x) +
h2
h3
hn n
f ′′( x) +
f ′′′( x) + ....... +
f x + .......
2!
3!
n!
This is known as Taylor’s Series.
Putting x = a and h = x a in above series, we get Taylor’s Series in powers of (x – a) as
f ( x) = f (a ) + ( x − a ) f ′(a ) +
( x − a)2
( x − a )3
f ′′(a ) +
f ′′′(a ) + ... ......
2!
3!
( x − a)n n
+
f (a ) + ......
n!
Example 1: Prove that f (mx) = f ( x ) + ( m − 1) x f ′( x ) +
Solution: f (mx) = f (mx
x + x) = f [x + (m
( m − 1)2 2
x f ′′( x ) + ... .
2!
1) x]
By Taylor’s series,
f ( x + h) = f ( x) + h f ′( x) +
Putting
h2
h3
f ′′( x) +
f ′′′ ( x) + ................
2!
3!
h = (m 1) x,
f [ x + (m − 1) x] = f (mx) = f ( x) + (m − 1) x f ′( x) +
(m − 1) 2 2
x f ′′ ( x) + .........
2!
Example 2: Prove that
⎛ x2 ⎞
x
x2
x3
= f ( x) –
f⎜
f ′ ( x) +
f ′ ′( x ) –
f ′′′( x ) + ... .
⎟
2
⎜ 1+ x ⎟
1+ x
2 !(1 + x )
3 !(1 + x )3
⎝
⎠
x2
x
= x−
,
1+ x
1+ x
Solution:
By Taylor’s series,
f ( x + h) = f ( x) + h f ′( x) +
h2
h3
f ′′ ( x) −
f ′′′ ( x) + ...
2!
3!
Engineering Mathematics
2.80
x
,
1+ x
⎛ x2 ⎞
⎛
x ⎞
f ⎜x −
f
=
⎜
⎟
⎟
1+ x⎠
⎝
⎝1+ x ⎠
x3
x
x2
f
′′
(
x
)
−
f ′′′( x) + ...
= f ( x) −
f ′ ( x) +
1+ x
3! (1 + x)3
2 ! (1 + x) 2
h=−
Putting
Example 3: Expand f (x) = x5 - x4 + x3 - x2 + x - 1 in powers of (x - 1) and find
f (0.99).
Solution: f (x) = x5
x4 + x3
x2 + x
1
By Taylor’s series,
f ( x) = f (a) + ( x − a) f ′ (a) +
( x − a)2
( x − a )3
f ′′ (a ) +
f ′′′ (a ) + ......
2!
3!
Putting a = 1,
f (x) = x5
x4 + x3
x2 + x
1
( x − 1) 2
( x − 1)3
f ′′(1) +
f ′′′ (1)
2!
3!
( x − 1)5 v
( x − 1) 4 iv
f (1) +
f (1) + .......
+
5!
4!
f (1) = 1 − 1 + 1 − 1 + 1 − 1 = 0
= f (1) + ( x − 1) f ′ (1) +
... (1)
Differentiating f (x) w.r.t. x successively,
f (x) = 5x4 4x3 + 3x2 2x + 1,
f (x) = 20x3 12x2 + 6x 2,
f (x) = 60x2 24x + 6,
f i v (x ) = 120x 24,
f v(x) = 120,
f (1) = 5 4 + 3 2 + 1 = 3
f (1) = 20 12 + 6 2 = 12
f (1) = 60 24 + 6 = 42
f iv(1) = 120 24 = 96
f v(1) = 120
Substituting in Eq. (1),
( x − 1) 2
( x − 1)3
( x − 1) 4
( x − 1)5
(12) +
(120)
(42) +
(96) +
5!
2!
3!
4!
= 3( x − 1) + 6( x − 1) 2 + 7( x − 1)3 + 4( x − 1) 4 + ( x − 1)5
f ( x) = 0 + ( x − 1) 3 +
Putting x = 0.99,
f (0.99) = 3 (0.99 1) + 6 (0.99 1)2 + 7 (0.99 1)3 + 4 (0.99 1)4 + (0.99
= 3 ( 0.01) + 6 ( 0.01)2 + 7 ( 0.01)3 + 4 ( 0.01)4 + ( 0.01)5
= 0.02939
1)5
Differential Calculus I
Example 4: Prove that
2.81
1
1 ( x + 2) ( x + 2) 2 ( x + 2) 3
= +
+
+
+ ... .
1– x 3
32
33
34
Solution: Let f ( x) = 1
1− x
By Taylor’s series,
f ( x) = f (a ) + ( x − a ) f ′ (a ) +
Putting a =
( x − a)2
( x − a )3
f ′′ (a ) +
f ′′′ (a ) + ... ..........
2!
3!
2,
1
( x + 2) 2
( x + 2) 3
= f ( −2) + ( x + 2) f ′ ( −2) +
f ′′ ( −2) +
f ′′′ ( −2) + ..........
1− x
2!
3!
... (1)
f ( x) =
f ( −2) =
1
3
Differentiating f (x) w.r.t. x successively,
1
1
, f ′ ( −2) = 2
(1 − x) 2
3
2
2!
f ′′ ( x) =
, f ′′ ( −2) = 3
3
(1 − x)
3
2.3
3!
, f ′′′ ( −2) = 4 and so on
f ′′′ ( x) =
4
(1 − x)
3
f ′ ( x) =
Substituting in Eq. (1),
f ( x) =
1
1 ( x + 2) ( x + 2 ) 2 ( x + 2 ) 3
= +
+
+
+ ............
1− x 3
32
33
34
Example 5: Expand log (cos x) about
o
.
3
Solution: Let f (x) = log (cos x)
By Taylor’s series,
f ( x) = f (a) + ( x − a) f ′ (a) +
Putting
( x − a)2
( x − a )3
f ′′ (a ) +
f ′′′ (a ) + … …
2!
3!
p
,
3
f ( x) = log (cos x)
a=
2
3
p ⎞ ⎛p ⎞ 1 ⎛
p⎞
p⎞
⎛p ⎞ ⎛
⎛p ⎞ 1 ⎛
⎛p ⎞
= f ⎜ ⎟ + ⎜ x − ⎟ f ′ ⎜ ⎟ + ⎜ x − ⎟ f ′′ ⎜ ⎟ + ⎜ x − ⎟ f ′′′ ⎜ ⎟ + … ... (1)
3⎠
3 ⎠ ⎝ 3 ⎠ 2! ⎝
3⎠
⎝3⎠ ⎝
⎝ 3 ⎠ 3! ⎝
⎝3⎠
p⎞
⎛p ⎞
⎛
⎛1⎞
f ⎜ ⎟ = log ⎜ cos ⎟ = log ⎜ ⎟ = − log 2
3
3
⎝ ⎠
⎝
⎠
⎝2⎠
Engineering Mathematics
2.82
Differentiating f (x) w.r.t. x successively,
f ′( x) =
p
⎛p ⎞
f ′ ⎜ ⎟ = − tan = − 3
3
3
⎝ ⎠
p
⎛p ⎞
f ′′ ⎜ ⎟ = − sec 2 = −4
3
⎝3⎠
1
(− sin x) = − tan x,
cos x
f ′′( x) = − sec 2 x,
f ′′′ ( x) = −2 sec 2 x tan x,
p
p
⎛p ⎞
f ′′′ ⎜ ⎟ = −2 sec 2 tan = −2(4) 3 = − 8 3 and so on
3
3
⎝3⎠
Substituting in Eq. (1),
2
p⎞
p⎞
1⎛
⎛
f ( x) = log (cos x) = − log 2 + ⎜ x − ⎟ − 3 + ⎜ x − ⎟ (−4)
3⎠
2! ⎝
3⎠
⎝
(
)
3
+
p⎞
1⎛
⎜ x − ⎟ −8 3 + …
3⎠
3! ⎝
(
)
3
2
4 3⎛
⎛
⎞
⎛
⎞
⎞
= − log 2 − 3 ⎜ x − ⎟ − 2 ⎜ x − ⎟ −
⎜ x − ⎟ −…
3⎠
3⎠
3 ⎝
3⎠
⎝
⎝
Example 6: Obtain tan-1 x in powers of (x - 1).
Solution: Let f (x) = tan 1 x
By Taylor’s series,
f ( x) = f (a ) + ( x − a ) f ′(a ) +
( x − a)2
( x − a )3
f ′′(a ) +
f ′′′(a ) + ……
2!
3!
Putting a = 1,
f ( x) = tan −1 x = f (1) + ( x − 1) f ′(1) +
( x − 1) 2
( x − 1)3
f ′′(1) +
f ′′′(1) + …
2!
3!
p
4
Differentiating f (x) w.r.t. x successively,
f (1) = tan −1 1 =
f ′( x) =
1
,
1 + x2
f ′(1) =
1
2
f ′′ ( x) = −
2x
,
(1 + x 2 ) 2
f ′′′( x) = −
2
8x2
1
+
, f ′′′ (1) =
2 2
2
(1 + x )
(1 + x 2 )3
f ′′ (1) = −
2
1
= − and so on
4
2
... (1)
Differential Calculus I
2.83
Substituting in Eq. (1),
2
3
p
⎛ 1 ⎞ ( x − 1) ⎛ 1 ⎞ ( x − 1) ⎛ 1 ⎞
+ ( x − 1) ⎜ ⎟ +
⎜ ⎟+
⎜− ⎟+
4
2! ⎝ 2 ⎠
3! ⎝ 2 ⎠
⎝2⎠
p 1
1
1
= + ( x − 1) − ( x − 1) 2 +
( x − 1)3
4 2
4
12
f ( x) = tan −1 x =
Example 7: Prove that
log [sin ( x + h)] = log sin x + h cot x –
h2
h3 cos x
cosec2 x +
+
2
3 sin 3 x
Solution: Let f (x) = log (sin x), f (x + h) = log [sin (x + h)]
By Taylor’s series,
f ( x + h) = f ( x) + h f ′( x) +
h2
h3
f ′′( x) +
f ′′′( x) + …………
2!
3!
... (1)
Differentiating f (x) w.r.t. x successively,
1
cos x = cot x
sin x
f ′′( x) = − cosec 2 x
f ′( x) =
f ′′′( x) = 2 cosec2 x cot x =
2 cos x
and so on
sin 3 x
Substituting in Eq. (1),
f (x + h) = log [sin (x + h)]
h2
h3 2 cos x
cosec 2 x +
+ ………
2!
3! sin 3 x
h2
h3 cos x
= log sin x + h cot x − cosec 2 x +
+ ………
2
3 sin 3 x
= log sin x + h cot x −
Example 8: Expand tan-1 (x + h) in powers of h and hence, find the value of
tan-1 (1.003) up to 5 places of decimal.
Solution:
Let f (x) = tan 1 x , f (x + h) = tan 1 (x + h)
By Taylor’s series,
f ( x + h) = f ( x) + h f ′( x) +
h2
h3
f ′′( x) +
f ′′′( x) + …………
2!
3!
Differentiating f (x) w.r.t. x successively,
1
2x
f ′( x) =
,
f ′′( x) = −
1 + x2
(1 + x 2 ) 2
... (1)
Engineering Mathematics
2.84
f ′′′( x) = −
2
2x ⋅ 4x
2(3 x 2 − 1)
+
=
and so on
(1 + x 2 ) 2 (1 + x 2 )3
(1 + x 2 )3
Substituting in Eq. (1),
f ( x + h) = tan −1 ( x + h) = tan −1 ( x + h) ⋅
1
2x ⎤
h2 ⎡
+ ⎢−
⎥
2
2 ! ⎣ (1 + x 2 ) 2 ⎦
1+ x
h3 ⎡ 2(3 x 2 − 1) ⎤
+ ⎢
⎥ +………
3! ⎣ (1 + x 2 )3 ⎦
Putting x = 1, h = 0.0003,
tan −1 (1 + 0.003) = tan −1 (1.0003)
= tan −1 1 +
0.0003 (0.0003) 2
+
2
2!
3
⎛ 2 ⎞ (0.0003) ⎛ 1 ⎞
+
−
⎜
⎟
⎜ ⎟ + ………
3!
⎝ 4⎠
⎝2⎠
p
+ 0.00015 − 2.25 × 10−8 + 2.25 × 10−12 [Considering first 4 terms]
4
= 0.78540
=
Example 9: Prove that
1 + x + 2 x2 = 1 +
x 7 x2
+
2
8
7 x3
+
16
.
Solution: Let f ( x) = x, f ( x + h) = x + h
By Taylor’s series,
f ( x + h) = f ( x) + h f ′( x) +
h2
h3
f ′′( x) +
f ′′′( x) + ………
2!
3!
Putting x = 1, h = x + 2x 2 ,
f ( x + h) = x + h = 1 + x + 2 x 2
= f (1) + ( x + 2 x 2 ) f ′(1) +
f ( x) =
x,
( x + 2 x 2 )2
( x + 2 x 2 )3
f ′′(1) +
f ′′′(1) + ……… ... (1)
2!
3!
f (1) = 1
Differentiating f (x) w.r.t. x successively,
1
f ′( x) =
,
2 x
1⎛ 1⎞ 1
f ′′( x) = ⎜ − ⎟ 3 ,
2⎝ 2⎠ 2
x
f ′′′( x) =
1 ⎛ 1 ⎞⎛ 3 ⎞ 1
⎜ − ⎟⎜ − ⎟ ,
2 ⎝ 2 ⎠ ⎝ 2 ⎠ 52
x
f ′(1) =
1
2
f ′′(1) = −
1
4
f ′′′(1) =
3
and so on
8
Differential Calculus I
2.85
Substituting in Eq. (1),
1
1 ( x 2 + 4 x 3 + 4 x 4 ) 3 ( x3 + …)
1 + x + 2x2 = 1 + ( x + 2x2 ) −
+
+ .........
2
4
2
8
6
x 7 x 2 7 x3
−
+…
= 1+ +
2
8
16
1 + x + 2 x 2 in powers of (x - 1).
Example 10: Expand
1 + x + 2 x 2 = 4 + 2 ( x − 1) 2 + 5 ( x − 1) [Expressing in terms of (x – 1)]
Solution:
Let f ( x) = x, f ( x + h) = x + h
By Taylor’s series,
f ( x + h) = f ( x) + h f ′( x) +
Putting x = 4, h = 2 (x
f ( x + h) =
1)2 + 5 (x
h2
h3
f ′′( x) +
f ′′′( x) + ………
2!
3!
1),
x + h = 4 + 2 ( x − 1) 2 + 5 ( x − 1)
= f (4) + [2 ( x − 1) 2 + 5 ( x − 1)] f ′(4) +
f ( x) = x ,
[2 ( x − 1) 2 + 5 ( x − 1)]2
f ′′(4) + … ... (1)
2!
f (4) = 2
Differentiating f (x) w.r.t. x successively,
f ′( x) =
1
2 x
f ′′( x) =
Substituting in Eq. (1),
,
f ′(4) =
1⎛ 1⎞ 1
⎜− ⎟ ,
2 ⎝ 2 ⎠ 32
x
1
4
f ′′(4) = −
1
and so on
32
1
4
[2 ( x − 1) 2 + 5( x − 1)]2 ⎛ 1 ⎞
+
⎜ − ⎟ + ………
2!
⎝ 32 ⎠
7
5
1 + x + 2 x 2 = 2 + ( x − 1) + ( x − 1) 2 + …………
64
4
4 + 2 ( x − 1) 2 + 5( x − 1) = 2 + [2 ( x − 1) 2 + 5 ( x − 1)]
Example 11: Using Taylor’s theorem, evaluate up to 4 places of decimals:
(i)
1.02
(ii)
25.15
(iii)
9.12
(iv)
10
Engineering Mathematics
2.86
Solution: Let f ( x) = x , f ( x + h) = x + h
By Taylor’s series,
f ( x + h) = f ( x) + h f ′( x) +
(i) Putting
h2
h3
f ′′( x) +
f ′′′( x) + …………
2!
3!
... (1)
x = 1, h = 0.02,
f ( x + h) = x + h = 1 + 0.02
= f (1) + (0.02) f ′(1) +
(0.02) 2
f ′′(1) + ………
2!
f ( x) = x ,
... (2)
f (1) = 1
Differentiating f (x) w.r.t. x successively,
1
1
,
f ′(1) =
2
2 x
1
1
f ′′( x) = − 3 , f ′′(1) = − and so on
4
4x 2
Substituting in Eq. (2) and considering only first 3 terms,
f ′( x) =
1 (0.02) 2
1.02 = 1 + (0.02) +
2
2!
= 1.0099 approx.
⎛ 1⎞
⎜⎝ − ⎟⎠
4
(ii) Putting x = 25, h = 0.15 in Eq. (1),
f ( x + h) =
x + h = 25 + 0.15
= f (25) + (0.15) f ′(25) +
f ( x) = x ,
(0.15) 2
f ′′(25) + …
2!
f (25) = 5
Differentiating f (x) w.r.t. x successively,
f ′( x) =
1
,
f ′(25) =
1
= 0.1
10
2 x
1
1
f ′′( x) = − 3 , f ′′(25) = −
= − 0.002 and so on
500
2
4x
Substituting in Eq. (2) and considering only first 3 terms,
25.15 = 5 + (0.15) (0.1) +
= 5.0150 approx.
(0.15) 2
( −0.002)
2
... (3)
Differential Calculus I
2.87
(iii) Putting x = 9, h = 0.12 in Eq. (1),
f ( x + h) = x + h = 9 + 0.12
= f (9) + (0.12) f ′(9) +
f ( x) = x ,
(0.12) 2
f ′′(9) + …
2!
... (3)
f (9) = 3
Differentiating f (x) w.r.t. x successively,
f ′( x) =
1
f ′(9) =
,
1
6
2 x
1
1
and so on
f ′′ ( x) = − 3 ,
f ′′(9) = −
108
2
4x
Substituting in Eq. (2) and considering only first 3 terms,
2
⎛ 1 ⎞ (0.12) ⎛ 1 ⎞
9.12 = 3 + (0.12) ⎜ ⎟ +
⎜−
⎟
⎝ 6⎠
2 ⎝ 108 ⎠
= 3 + 0.02 − (0.12) (0.06) (0.0093)
= 3.0199 approx.
(iv) Putting
x = 9, h = 1 in Eq. (1),
f ( x + h) = x + h = 9 + 1 = f (9) + f ′(9) +
10 = 3 +
1
f ′′(9) + …
2!
1
1
−
6 216
... (4)
[refer (iii)]
= 3.1620 approx.
Example 12: Find the value of tan (43ç).
Solution: Let f (x) = tan x,
f (x + h) = tan (x + h)
By Taylor’s series,
f ( x + h) = f ( x) + h f ′( x) +
h2
h3
f ′′ ( x) +
f ′′′ ( x) + …………
2!
3!
2p
p
=−
= −0.0349 ,
180
90
tan ( x + h) = tan (45° − 2° ) = tan 43°
Putting x = 45° , h = −2° = −
= f (45° ) + (−0.0349) f ′(45° ) +
f ( x) = tan x ,
(−0.0349) 2
f ′′(45° ) + ……… ... (1)
2!
f (45° ) = tan (45° ) = 1
Engineering Mathematics
2.88
Differentiating f (x) w.r.t. x successively,
f ′ ( x) = sec 2 x,
f ′ (45° ) = sec 2 45° = 2
f ′′(45° ) = 2 sec 2 45° tan 45° = 4
f ′′ ( x) = 2 sec 2 x tan x,
and so on
Substituting in Eq. (1) and considering only first 3 terms,
tan 43° = 1 + (−0.0349)(2) +
(−0.0349) 2
(4)
2!
= 0.9326 approx.
Example 13: Find cosh (1.505) given sinh (1.5) = 2.1293 and cosh (1.5) = 2.3524.
Solution: Let f (x) = cosh x
By Taylor’s series,
f ( x + h) = f ( x ) + h f ′ ( x ) +
h2
h3
f ′′ ( x) +
f ′′′ ( x) + …………
2!
3!
Putting x = 1.5, h = 0.005,
f (x + h) = cosh (x + h) = cosh (1.5 + 0.005)
(0.005)3
(0.005) 2
f ′′(1.5) +
f ′′′ (1.5) +
3!
2!
f ( x) = cosh x, f (1.5) = cosh (1.5) = 2.3524
= f (1.5) + (0.005) f ′(1.5) +
Differentiating f (x) w.r.t. x successively,
f (x) = sinh x, f (1.5) = sinh (1.5) = 2.1293
f (x) = cosh x, f (1.5) = cosh (1.5) = 2.3524
... (1)
and so on
Substituting in Eq. (1) and considering only first 3 terms,
(0.005) 2
coosh (1.5) + …
2!
= 2.3524 + (0.005)(2.1293) + (12.5) (10−6 )(2.3524)
= 2.3631 approx.
cosh (1.505) = cosh (1.5) + (0.005) sinh (1.5) +
Exercise 2.6
1. Expand ex in powers of (x − 1).
⎡
⎛
( x − 1) 2 ⎤
⎢ Ans.: e ⎜1 + ( x − 1) +
⎥
2! ⎥
⎝
⎢
⎢
⎞ ⎥
( x − 1)3
⎢
+
+ …⎟ ⎥
⎢⎣
3!
⎠ ⎥⎦
2. Expand 2x3 + 7x2 + x − 1 in powers of
x − 2.
[Ans.: 45 + 53 (x 2) + 19 (x 2)2
+ 2 (x 2)3]
3. Expand x5 − 5x4 + 6x3 − 7x2 + 8x
in powers of (x 1).
9
⎡ Ans. : − 6 − 3 ( x − 1) − 9 ( x − 1) 2 ⎤
⎢
⎥
− 4 ( x − 1)3 + ( x − 1)5 ⎥⎦
⎢⎣
4. Expand x4 − 3x3 + 2x2 − x + 1 in powers
of (x 3).
⎡ Ans. :16 + 38 ( x − 3) + 29 ( x − 3) 2 ⎤
⎢
⎥
+ 9 ( x − 3)3 + ( x − 3) 4 ⎥⎦
⎢⎣
Differential Calculus I
5. Expand x3 − 2x2 + 3x − 5 in power of
(x 2).
[Ans.: 11 + 7 (x
2) + 4 (x 2)2
+ (x − 2)3]
6. Expand 2x3 + 3x2 − 8x + 7 in terms of
(x 2).
[Ans.: 19 + 28 (x
7. Expand
⎡
⎢ Ans.:
⎣
8. Expand
(x 1).
2) + 15 (x 2)2
+ 2 (x − 2)3]
⎤
a+
−
− …⎥
2 a
8a a
⎦
( x − a )3
1 + x + 2x 2 in powers of
5
7
⎡
⎤
2
⎢ Ans.: 2 + 4 ( x − 1) + 32 ( x − 1) + …⎥
⎣
⎦
9. Expand sin x in powers of (x − a).
⎡ Ans.: sin a + ( x − a )
⎤
⎢
⎥
2
( x − a)
⎥
⎢
cos a −
sin a
⎥
⎢
2!
⎥
⎢
3
( x − a)
⎢
cos a + …⎥
−
⎢⎣
3!
⎦⎥
p
10. Expand cos x in powers of ⎛⎜ x − ⎞⎟ .
2⎠
⎝
⎡
p ⎞ 1⎛
p⎞
⎛
⎢ Ans.: − ⎜ x − ⎟ + ⎜ x − ⎟
2 ⎠ 3! ⎝
2⎠
⎝
⎢
5
⎢
1⎛
p⎞
⎢
− ⎜ x − ⎟ +…
⎢⎣
5! ⎝
2⎠
⎤
3
1 x2
x− ⋅
⎥
2
2 2!
⎥
⎥
3 x3 1 x 4
⋅ + ⋅ +… ⎥
2 3! 2 4 !
⎦
⎛p
⎞
13. Expand tan ⎜ + x ⎟ in powers of x
4
⎝
⎠
upto x4 and hence find the value of
tan (46 36').
x in powers of (x − a).
( x − a)
⎡
1
⎢ Ans.: +
2
⎢
⎢
−
⎢
⎣
2.89
3
⎤
⎥
⎥
⎥
⎥
⎥⎦
p
11. Expand tan x in powers of ⎛⎜ x − ⎞⎟ .
4⎠
⎝
2
⎡
⎤
p⎞
p⎞
⎛
⎛
Ans.:
1
+
2
−
+
2
−
x
x
⎢
⎜
⎟
⎜
⎟ +…⎥
4⎠
4⎠
⎝
⎝
⎢⎣
⎥⎦
⎛p
⎞
12. Expand sin ⎜ + x ⎟ in powers of x
6
4
⎝
⎠
upto x .
⎡
8 3 ⎤
⎛
2
⎢ Ans.: ⎜1 + 2 x + 2 x + 3 x ⎥
⎝
⎢
⎥
⎢
⎥
10 4
⎞
+ x + … ⎟ , 1.0574⎥
⎢
3
⎠
⎣
⎦
14. Using Taylor’s theorem find approximate value of cos 64 .
[Ans.: 0.4384]
15. Using Taylor’s theorem find approximate value of sin (30 30 ).
[Ans.: 0.5073]
16. Expand log x in powers of (x
2).
⎡
1
1 ( x − 2) 2
2
Ans.:
log
+
(
x
−
2
)
−
⋅
⎢
2
2!
4
⎢
⎢
1 ( x − 2) 3
+ ⋅
+…
⎢
3!
4
⎣
⎤
⎥
⎥
⎥
⎥
⎦
17. Expand log sin x in powers of (x 2).
⎡ Ans.: log sin 2 + ( x − 2) cot 2 ⎤
⎢
⎥
1
⎢
− ( x − 2) 2 cosec 2 x + …⎥
⎢⎣
⎥⎦
2
⎛p
⎞
18. Expand log tan ⎜ + x ⎟ in powers
⎝4
⎠
of x.
4 3 4 5
⎤
⎡
⎢ Ans.: 2 x + 3 x + 3 x + … ⎥
⎦
⎣
19. Arrange in powers of x, by Taylor’s
theorem, 7 + (x + 2) + 3 (x + 2)3 +
(x + 2)4.
[Ans.: 49 + 69x + 42x2 + 11x3 + x4]
2.90
Engineering Mathematics
20. Arrange in powers of x, by Taylor’s
theorem, 17 + 6 (x + 2) + 3 (x + 2)3 +
(x + 2)4 − (x + 2)5.
[Ans.: 37 − 6x − 38x2 − 29x3
− 9x4 − x5]
21. Arrange in powers of (x + 1), by
Taylor’s theorem, (x + 2)4 + 5 (x + 2)3
+ 6 (x + 2)2 + 7 (x + 2) + 8.
⎡ Hint : f ( x) = x 4 + 5 x 3 + 6 x 2 + 7 x + 8, ⎤
⎥
⎢
f [( x + 1) + 1] = f (1)
⎥
⎢
⎥
⎢
( x + 1) 2
⎢
f ′′ (1) + …⎥
+ ( x + 1) f ′(1) +
2!
⎦
⎣
⎡ Ans. : 27 + 38 (x + 1) + 27 (x + 1) 2 ⎤
⎥
⎢
+ 9 (x + 1)3 + (x + 1) 4
⎥⎦
⎢⎣
22. Prove that sinh (x + a) = sinh a +
x2
sinh a + …
x cosh a +
2!
Given sinh (1.5) = 2.1293, cosh (1.5)
= 2.3524, find the value of
sinh (1.505).
[Ans. 2.1411]
2.9 MACLAURIN’S SERIES
Statement: If f (x) be a given function of x which can be expanded in positive ascending integral powers of x, then
f ( x) = f (0) + x f ′(0) +
x2
x3
xn n
f ′′ (0) +
f ′′′ (0) + ……… +
f (0) + ………
2!
3!
n!
Proof: Let f (x) be a function of x which can be expanded into positive ascending
integral powers of x, then
f (x) = a0 + a1x + a2 x2 + a3 x3 + a4 x4 +……. …
... (1)
Differentiating w.r.t. x successively,
f (x) = a1 + a2· 2x + a3 · 3x2 + a4 · 4x3 + ……. ………
f (x) = a2 · 2 + a3 · 6x + a4 · 12x2 + ……. ………
f (x) = a3 · 6 + a4 · 24x + ……. ………
... (2)
... (3)
... (4)
and so on
Putting x = 0 in Eq. (1), (2), (3) and (4),
a0 = f (0)
a1 = f (0)
1
a2 = f ′′(0)
2!
1
a3 = f ′′′(0)
3!
and so on.
Substituting a0, a1, a2 and a3 in Eq. (1),
f ( x) = f (0) + x f ′(0) +
x2
x3
xn n
f ′′ (0) +
f ′′′ (0) + ……… +
f (0) + ………
2!
3!
n!
Differential Calculus I
2.91
This is known as Maclaurin’s Series.
This series can also be written as,
y = y (0) + xy1 (0) +
x2
x3
xn
y2 (0) +
y3 (0) + …… +
yn (0) + ………
2!
3!
n!
2.9.1 Standard Expansions
Using Maclaurin’s series, expansion of some standard functions can be obtained.
These expansions can be directly used while solving the examples.
(1) Expansion of ex (Exponential series)
Proof: Let y = ex, y (0) = e0 = 1
dn x
(e ) = e x , yn (0) = e0 = 1 for all values of n .
dx n
Substituting in Maclaurin’s series,
Now yn =
x 2 x3
+ + ………
2 ! 3!
ex = 1 + x +
This series is known as the exponential series.
Note: In the above series
(i) Replacing x by −x,
e− x = 1 − x +
x 2 x3
− + ………
2 ! 3!
(ii) Replacing x by ax,
a 2 x 2 a3 x3
+
+………
2!
3!
eax = 1 + ax +
(2) Expansion of sin x (Sine series)
Proof: Let y = sin x, y (0) = sin 0 = 0
yn =
Now
dn
np ⎞
⎛
(sin x) = sin ⎜ x +
⎟
n
2 ⎠
dx
⎝
⎛ np ⎞
yn (0) = sin ⎜
⎟
⎝ 2 ⎠
Putting
n = 1, 2, 3, 4, 5, ..…..
y1 (0) = 1, y2 (0) = 0, y3 (0) =
1, y4 (0) = 0, y5 (0) = 1, and so on.
Substituting in Maclaurin’s series,
sin x = x −
This series is known as the sine series.
x3 x5
+ − ………
3! 5 !
Engineering Mathematics
2.92
(3) Expansion of cos x (Cosine series)
Proof: Let y = cos x, y (0) = cos 0 = 1
dn
np ⎞
⎛
Now
yn = n (cos x) = cos ⎜ x +
⎟
2 ⎠
dx
⎝
⎛ np ⎞
yn (0) = cos ⎜
⎟
⎝ 2 ⎠
Putting n = 1, 2, 3, 4, …..
y1 (0) = 0, y2 (0) = 1, y3 (0) = 0, y4 (0) = 1,
Substituting in Maclaurin’s series,
x 2 x 4 ………
cos x = 1 − +
−
2! 4!
This series is known as the cosine series.
and so on.
(4) Expansion of tan x (Tangent series)
Proof: Let y = tan x,
y1 = sec 2 x = 1 + tan 2 x = 1 + y 2 ,
y(0) = 0
y1 (0) = 1
y2 = 2 yy1 ,
y2 (0) = 2 y (0) y1 (0) = 2(0)((1) = 0
y3 = 2 y + 2 yy2 ,
y3 (0) = 2(1) 2 + 2(0)(0) = 2
y4 = 4 y1 y2 + 2 y1 y2 + 2 yy3
y4 (0) = 6(1)(0) + 2(0)(2)
2
1
= 6 y1 y2 + 2 yy3 ,
=0
y5 = 6 y2 + 6 y1 y3 + 2 y1 y3 + 2 yy4
2
= 6 y2 + 8 y1 y3 + 2 yy4 ,
2
y5 (0) = 0 + 8(1)(2) + 0
= 16
Substituting in Maclaurin’s series,
x3
x5
tan x = x + (2) + (16) + ………
3!
5!
3
5
x 2x
= x+ +
+ ………
3 15
This series is known as the tangent series.
Note: This series can also be obtained by dividing the sine and cosine series
sin x
since tan x =
.
cos x
(5) Expansion of sinh x
Proof: We have sinh x =
e x − e− x
2
Substituting ex and e x from above exponential series,
⎛
⎞ ⎛
⎞
x 2 x3
x 2 x3
⎜1 + x + + + … ⎟ − ⎜1 − x + − + … ⎟
2! 3!
2! 3!
⎠ ⎝
⎠
sinh x = ⎝
2
x3 x5
= x + + +………
3! 5!
Differential Calculus I
2.93
(6) Expansion of cosh x
e x + e− x
2
Substituting exponential series ex and e x,
Proof: We have sinh x =
⎛
⎞ ⎛
⎞
x 2 x3
x 2 x3
⎜ 1 + x + + + …⎟ + ⎜ 1 − x + − + …⎟
2 ! 3!
2 ! 3!
⎠ ⎝
⎠
cosh x = ⎝
2
x2 x4
= 1 + + +………
2 ! 4!
(7) Expansion of tanh x
Proof: Expansion of tanh x can be obtained by dividing the series of sinh x and
cosh x.
x3 x5 x 7
+ + +…
sinh x
3! 5 ! 7 !
tanh x =
=
cosh x
x2 x4 x6
1+ + + +…
2! 4! 6!
x2 2 5
= x − + x − ………
3 15
x+
Note: This series can also be obtained by using Maclaurin's series (refer tangent
series)
(8) Expansion of log (1 + x) (Logarithmic series)
Proof: Let y = log (1 + x), y (0) = log 1 = 0
yn =
Now
dn
(n − 1)!
[ log (1 + x)] = (−1) n −1 ⋅
dx n
( x + 1) n
yn (0) = ( −1) n −1 ⋅ (n − 1)!
Putting
n = 1, 2, 3, 4, …..
y1(0) = 1, y2 (0) =
1, y3 (0) = 2! and so on
Substituting in Maclaurin’s series,
log (1 + x) = x −
x 2 x3
+ − ………
2
3
This series is known as the Logarithmic series and is valid for 1 < x < 1.
Note: In above series replacing x by −x, we get expansion of log (1 − x)
log (1 − x) = − x −
(9) Expansion of (1 + x)m (Binomial series)
Proof: Let y = (1 + x) m , y (0) = (1 + 0) m = 1
x 2 x3 x 4 x5
− − − −…
2
3
4
5
Engineering Mathematics
2.94
yn = m (m − 1) (m − 2)……… (m − n + 1)(1 + x) m − n
Now
yn (0) = m(m − 1)(m − 2)……… (m − n + 1)
Putting
n = 1, 2, 3, 4, …..
y1(0) = m, y2(0) = m (m − 1), y3(0) = m (m − 1) (m − 2) and so on
Substituting in Maclaurin’s series,
m(m − 1) 2 m(m − 1)(m − 2) 3
(1 + x) m = 1 + mx +
x +
x + ………
2!
3!
This series is known as the Binomial series and is valid for 1 < x < 1.
x
Example 1: Expand 5 up to the first three non-zero terms of the series.
x
f (x) = 5 , f (0) = 50 = 1
x
f (x) = 5 log 5, f (0) = 50 log 5 = log 5
x
f (x) = 5 (log 5)2, f (0) = 50 (log 5)2 = (log 5)2
Substituting in Maclaurin’s series,
Solution: Let
f ( x) = f (0) + x f ′ (0) +
5 x = 1 + x log 5 +
Aliter:
x2
f ′′(0) + ………
2!
x2
(log 5) 2 + ………
2!
x
f (x) = 5x = e log 5 = e x log 5
( x log 5) 2
= 1 + x log 5 +
+ ………
2!
[Using Exponential series]
⎛ 1+ x ⎞
Example 2: Obtain the series log (1 + x) and find the series log ⎜
⎟ and
⎝ 1- x ⎠
⎛ 11 ⎞
hence, find the value of loge ⎜ ⎟ .
⎝ 9 ⎠
Solution: Let y = log (1 + x)
y1 =
1
1
(2 !)
(3!)
etc.
, y2 = −
, y3 =
, y4 = −
1+ x
(1 + x) 2
(1 + x)3
(1 + x) 4
At x = 0, y = 0, y1 = 1, y2 = −1, y3 = 2 !, y4 = −(3!) etc.
Substituting in Maclaurin’s series,
x2
x3
x4
y2 (0) + y3 (0) +
y4 (0) + ………
2!
3!
4!
x4
x 2 x3
= 0 + x − + (2 !) − (3!) + ………
2 ! 3!
4!
x 2 x3 x 4
log (1 + x) = x − + − + ………
2
3 4
y = y (0) + xy1 (0) +
Differential Calculus I
2.95
Replacing x by −x,
log (1 − x) = − x −
x 2 x3 x 4
− − − ………
2 3
4
⎛1+ x⎞
log ⎜
= log (1 + x) − log (1 − x)
⎝ 1 − x ⎟⎠
⎛
⎞
x3 x5
= 2 ⎜ x + + + …⎟
3 5
⎝
⎠
Now,
Putting x =
1
, and considering first three terms,
10
⎡1 1 1
1 1 ⎤
⎛ 11 ⎞
+ ⋅
log e ⎜ ⎟ = 2 ⎢ + ⋅
⎥ = 0.20067
3
9
10
3
5 (10)5 ⎦
(
)
10
⎝ ⎠
⎣
Example 3: If x 3
y3
xy 1
0, prove that y
1
x
3
26 3
x …
81
Solution: x3 + y3 + xy 1 = 0,
Putting x = 0,
y (0) = 1
Differentiating w.r.t. x,
3 x 2 + 3 y 2 y1 + xy1 + y = 0
−1
3
Differentiating Eq. (1) w.r.t. x,
6x + 6yy12 + 3y2y2 + 2y1 + xy2 = 0
...(1)
Putting x = 0, y1 (0) =
Putting x
0, 6
1
3
2
3 y2 (0) 2
1
3
0
y2 (0) = 0
Differentiating Eq. (2) w.r.t. x,
6 + 6y13 + 12yy1y2 + 3y2y3 + 6yy1y2 + 3y2 + xy3 = 0
Putting x = 0,
⎛ −1 ⎞
6 + 6 ⎜ ⎟ + 0 + 3 y3 (0) = 0
⎝ 27 ⎠
− 52
y3 (0) =
and so on.
27
Substituting in Maclaurin’s series,
x2
x3
y2 (0) +
y3 (0) +
2!
3!
x x2
x 3 ⎛ − 52 ⎞
y = 1 − + ( 0) + ⎜
⎟+
3 2!
3! ⎝ 27 ⎠
x 26 3
=1− −
x −
3 81
y = y (0) + xy1 (0) +
… (2)
Engineering Mathematics
2.96
Example 4: If x3 + 2xy2 - y3 + x - 1 = 0, expand y in ascending powers of x.
x3 + 2xy2
Solution:
Putting x = 0,
y3 + x
1=0
y(0) = – 1
Differentiating w.r.t. x,
3x2 + 2y2 + 4xyy1
Putting x = 0,
2
3y2y1 + 1 = 0
… (1)
3y1 (0) + 1 = 0
y1 (0) = 1
Differentiating Eq. (1) w.r.t. x,
6x + 4yy1 + 4yy1 + 4xy12 + 4xyy2
Putting x = 0,
8+6
6yy12
3y2y2 = 0
3y2 (0) = 0
y2 (0)
2
and so on.
3
Substituting in Maclaurin’s series,
x2
y 2 ( 0) +
2!
x2 ⎛ 2 ⎞
y = −1+ x + ⎜− ⎟ +
2! ⎝ 3 ⎠
x2
+
= −1 + x −
3
y = y (0) + xy1 (0) +
Example 5: If x = y (1 + y2), prove that y = x - x3 + 3x5 + … .
Solution:
x = y (1 + y2)
Putting x = 0, y (0) = 0
Differentiating w.r.t. x,
Putting
x = 0,
1 = y1 + 3y2y1
1 = y1 (0)
y1 (0) = 1
Differentiating Eq. (1) w.r.t. x,
0 = y2 + 6yy12 + 3y2y2
Putting x = 0, y2 (0) = 0,
Differentiating Eq. (2) w.r.t. x,
0 = y3 + 12yy1y2 + 6y13 + 6yy1y2 + 3y2y3
0 = y3 (1 + 3y2) + 18yy1y2 + 6y13
Putting x = 0,
0 = y3 (0) + 6
y3 (0) = −6
… (1)
… (2)
… (3)
Differential Calculus I
2.97
Differentiating Eq. (3) w.r.t. x,
0 = (1 + 3y2) y4 + 6yy1y3 + 18y12 y2 + 18yy22 + 18yy1y3 + 18y12 y2
= (1 + 3y2) y4 + 24yy1y3 + 36y12 y2 + 18yy22
Putting x = 0, y4 (0) = 0,
Differentiating Eq. (4) w.r.t. x,
0 = (1 + 3y2) y5 + 6yy1y4 + 24y12 y3 + 24yy2y3 + 24yy1y4 + 72y1y22
… (4)
+ 36y12y3 + 36yy2y3 + 18y1y22
Putting x = 0,
0 = y5 (0) + 24 (−6) + 36 (−6)
y5 (0) = 360 and so on.
Substituting in Maclaurin’s series,
x2
x3
x4
x5
y 2 ( 0) +
y3 (0) +
y 4 ( 0) +
y5 (0) +
2!
3!
4!
5!
x2
x3
x4
x5
= 0 + x.1 + ⋅ 0 + (− 6) +
⋅0+
⋅ 360 +
2!
3!
4!
5!
= x − x3 + 3x5 +
y = y (0) + xy1 (0) +
2
Example 1: Obtain the expansion of 1 + x
1 + x4
1 + x2
Solution:
= (1 + x 2 )(1 + x 4 ) 1
1 + x4
= (1 + x2) (1 − x4 + x8 − x12 + x16 − …)
= 1 + x2 − x4 − x6 + x8 + x10 − …
Example 2: If x
y = x+
y
y2
2
y3
3
y4
4
..... , prove that
x2 x3 x4
+
+
+ ......... and conversely.
2! 3! 4!
Solution:
x = log (1 + y )
1 + y = ex
y
ex 1
= x+
x 2 x3
+ + .......
2! 3!
Conversely,
y
ex 1
ex = 1 + y
x = log (1 + y )
y
y2
2
y3
3
y4
4
.....
2.98
Engineering Mathematics
Example 3: Expand 1 + sin x .
Solution:
1 + sin x = sin
x
x
+ cos
2
2
2
4
⎡ x 1 ⎛ x ⎞3
⎤ ⎡
1 ⎛x⎞
1 ⎛x⎞
= ⎢ − ⎜ ⎟ + ⎥ + ⎢1 − ⎜ ⎟ + ⎜ ⎟ −
4 ⎝2⎠
⎢⎣ 2 3! ⎝ 2 ⎠
⎥⎦ ⎣⎢ 2 ! ⎝ 2 ⎠
2
3
4
x x
x
x
= 1+ −
−
+
−
2 8
48 384
1 4 2 6
x
x ..... .
Example 4: Prove that cos 2 x 1 x 2
3
45
1
Solution:
cos 2 x = (1 + cos 2 x)
2
=
(2 x) 2 (2 x) 4 (2 x)6
1⎡
+
−
+
−
+
1
1
⎢
2⎣
2!
4!
6!
⎤
⎥
⎥⎦
⎤
⎥
⎦
1 4 2 6
x −
x +
3
45
∞
1
32 n + 3 2 n
Example 5: Prove that cosh3 x =
x .
4 n = 0 ( 2n)!
= 1 − x2 +
∑
Solution:
1
(cosh 3 x + 3 cos h x)
4
1 ⎡⎛
(3 x) 2 (3x) 4
= ⎢⎜ 1 +
+
+
4 ⎣⎝
2!
4!
cosh3 x =
⎞
⎛
x2 x4
+
+
⎟ + 3 ⎜1 +
2! 4!
⎠
⎝
=
32 + 3 2 34 + 3 4
1⎡
x +
x +
⎢(1 + 3) +
4⎣
2!
4!
=
1
4
⎞⎤
⎟⎥
⎠⎦
⎤
⎥
⎦
32 n + 3 2 n
x
n = 0 ( 2n)!
∞
∑
Example 6: Prove that sin x sinh x = x 2 −
8 6 32 10
x +
x − ... .
6!
10 !
⎛
⎞ ⎛
⎞
x3 x5 x7 x9
x3 x5 x 7 x9
Solution: sin x sin h x = ⎜ x − + − + − ... ⎟ ⋅ ⎜ x + + + + + ... ⎟
!
!
!
!
!
!
!
3
5
7
9
3
5
7
9!
⎝
⎠ ⎝
⎠
⎡2
1 ⎤ 10 ⎡ 2
2
1 ⎤
= x 2 + x6 ⎢ −
+x ⎢ −
+
+ ...
2⎥
2⎥
⎣ 5! (3!) ⎦
⎣ 9 ! 7 !3! (5!) ⎦
8
32 10
x − ...
= x 2 − x6 +
6!
10 !
Differential Calculus I
2.99
Example 7: Expand log (1 + x + x2 + x3) up to a term in x8.
Solution: log (1 + x + x2 + x3) = log [(1 + x) (1 + x2)]
= log (1 + x) + log (1 + x2)
⎤
⎡
⎤ ⎡
x 2 x 3 x 4 x 5 x 6 x 7 x8
( x 2 ) 2 ( x 2 )3 ( x 2 ) 4
+ ...⎥
= ⎢ x − + − + − + − + ...⎥ + ⎢ x 2 −
+
−
2
3 4
5 6
7 8
2
3
4
⎦
⎣
⎦ ⎣
2
3
5
6
7
x
x 3 4 x
x
x 3 8
= x + + − x + + + − x + ...
2
3 4
5 6
7 8
x2 x3 x4
Example 8: Prove that log (1 + x + x2 + x3 + x4) = x +
+
+
– .
2
3
4
Solution:
⎛ 1 − x5 ⎞
log (1 + x + x2 + x3 + x4) = log ⎜
⎟
⎝ 1− x ⎠
= log (1 − x 5 ) − log (1 − x)
[Using sum of G.P.]
⎛
⎞ ⎛
x10 x15 x 20
x 2 x3 x 4 ⎞
= ⎜ − x5 −
−
−
− ⎟ − ⎜− x −
−
−
⎟
2
3
4
2
3
4 ⎠
⎝
⎠ ⎝
x 2 x3 x 4
=x+
+
+
−
2
3
4
2
3
x
1 x
1 x
Example 9: Prove that log x log 2
1
1
1
2
2 2
3 2
x
Solution:
log x log 2
2
x
= log 2 + log
2
⎡ ⎛x
⎞⎤
= log 2 + log ⎢1 + ⎜ − 1⎟ ⎥
2
⎝
⎠⎦
⎣
2
3
1⎛ x
⎞
⎞
⎛x
⎞ 1⎛ x
= log 2 + ⎜ − 1⎟ − ⎜ − 1⎟ + ⎜ − 1⎟ − …
2
2
2
3
2
⎝
⎠
⎝
⎝
⎠
⎠
⎛ sinh x ⎞ x 2 x 4
Example 10: Prove that log ⎜
−
+ ... .
⎟=
⎝ x ⎠ 6 180
Solution:
⎡1 ⎛
⎛ x2 x4
⎞
⎞⎤
x3 x5
⎛ sinh x ⎞
log ⎜
⎟ = log ⎢ ⎜ x + + + ... ⎟ ⎥ = log ⎜1 + + + ... ⎟
3! 5!
3! 5!
⎝ x ⎠
⎝
⎠
⎠⎦
⎣x⎝
x2
3!
x4
...
5!
x2
6
x4
x2
6
x4
.....
180
1
120
1 x2
2 3!
1
72
x4
...
5!
......
2
.....
... .
Engineering Mathematics
2.100
x2
3
Example 11: Prove that log ( x cot x )
Solution: log ( x cot x)
log
7 4
x
90
.... .
1
x cot x
⎛ tan x ⎞
= − log ⎜
⎟
⎝ x ⎠
⎛ x2 2
⎞
= − log ⎜1 + + x 4 + ... ⎟
3
15
⎝
⎠
2
2
⎡⎛ x
⎤
⎞ 1 ⎛ x2 2
⎞
2
= − ⎢⎜ + x 4 + .... ⎟ − ⎜ + x 4 + ... ⎟ + ...⎥
⎢⎣⎝ 3 15
⎥⎦
⎠ 2 ⎝ 3 15
⎠
⎡ x2
⎤
⎛ 2 1⎞
= − ⎢ + x 4 ⎜ − ⎟ + ...⎥
⎝ 15 18 ⎠
⎣3
⎦
2
7
x
= − − x 4 + ...
3 90
⎛ 1 + e2x
Example 12: Prove that log ⎜
x
⎝ e
⎞
x2 x4 x6
–
–
+
⎟ = log 2 +
2 12 45
⎠
.
⎛ 1 + e2 x ⎞
Solution: log ⎜ x ⎟ = log (e − x + e x ) = log (2 cosh x)
⎝ e ⎠
= log 2 + log cosh x
⎛ x2 x4 x6
⎞
= log 2 + log ⎜1 + + + + ... ⎟
2! 4! 6!
⎝
⎠
2
3
⎞
⎛ x2 x4 x6
⎞ 1 ⎛ x2 x4
⎞ 1 ⎛ x2
= log 2 + ⎜ + + + ... ⎟ − ⎜ + + ... ⎟ + ⎜ + ... ⎟ + ...
⎠
⎝ 2! 4! 6!
⎠ 2 ⎝ 2! 4!
⎠ 3 ⎝ 2!
⎛ x2 x4 x6
⎞ 1 ⎛ x4
⎞ 1 ⎛ x6
⎞
x6
= log 2 + ⎜ + + + ... ⎟ − ⎜ + 2 ⋅ + ... ⎟ + ⎜ + ... ⎟ + ...
48
⎝ 2! 4! 6!
⎠ 2⎝ 4
⎠ 3⎝ 8
⎠
2
x
1
1 ⎞
⎛ 1 1⎞
⎛ 1
= log 2 + + x 4 ⎜ − ⎟ + x 6 ⎜
− + ⎟ + ...
2
24
8
720
48
24
⎝
⎠
⎝
⎠
= log 2 +
x2 x4 x6
− + − ...
2 12 45
Example 13: Prove that log (1 e x )
Solution:
log 2
x
2
x2
8
⎛
x 2 x3 x 4
log (1 + e x ) = log ⎜1 + 1 + x +
+
+
+
2 ! 3! 4 !
⎝
x4
192
⎞
⎟
⎠
...... .
Differential Calculus I
2.101
⎡ ⎛ x x 2 x3 x 4
⎞⎤
= log ⎢ 2 ⎜1 + + + + + ... ⎟ ⎥
⎠⎦
⎣ ⎝ 2 4 12 48
⎞
⎛ x x 2 x3 x 4
= log 2 + log ⎜1 + + + + + ... ⎟
2
4
12
48
⎝
⎠
⎛ x x 2 x3 x 4
⎞ 1 ⎛ x x 2 x3
⎞
= log 2 + ⎜ + + + + ... ⎟ − ⎜ + + + ... ⎟
⎝ 2 4 12 48
⎠ 2 ⎝ 2 4 12
⎠
3
2
4
⎞ 1⎛ x
1 ⎛ x x2
⎞
+ ⎜ + + ... ⎟ − ⎜ + ... ⎟ + ...
3⎝ 2 4
4
2
⎝
⎠
⎠
⎛x⎞
⎛1 1⎞
⎛1 1 1 ⎞
= log 2 + ⎜ ⎟ + x 2 ⎜ − ⎟ + x 3 ⎜ − + ⎟
2
4
8
⎝ ⎠
⎝
⎠
⎝ 12 8 24 ⎠
1
1
1 1 ⎞
⎛ 1
+ x 4 ⎜ − − + − ⎟ + ...
⎝ 48 32 24 16 64 ⎠
= log 2 +
x x2
⎛ 1 ⎞ 4
+ +0+⎜−
⎟ x + ...
2 8
⎝ 192 ⎠
= log 2 +
x x2 x4
+ −
+ ...
2 8 192
1 ⎤
⎡
x 5 x2 x3
251 4
Example 14: Prove that log ⎢ log (1 + x ) x ⎥ = − +
−
+
x + ... .
2
24
8 2880
⎢⎣
⎥⎦
1
1
Solution:
log (1 + x) x = log (1 + x)
x
⎞
1⎛
x 2 x3 x 4 x5
= ⎜x −
+
−
+
− ⎟
x⎝
2
3
4
5
⎠
x x 2 x3 x 4
+
−
+
−
2 3
4
5
⎛ x x 2 x3 x 4
−
+
=1− ⎜ −
+
4
5
⎝ 2 3
= 1− y
=1−
Now,
⎞
⎟
⎠
1
⎡
⎤
y 2 y3 y 4
log ⎢log (1 + x) x ⎥ = log (1 − y ) = − y −
− −
− ...
2
3
4
⎣
⎦
2
⎛ x x 2 x3 x 4
⎞ 1 ⎛ x x 2 x3
⎞
=−⎜ − +
−
+ ... ⎟ − ⋅ ⎜ −
+ − ... ⎟
4
5
4
⎝2 3
⎠ 2 ⎝2 3
⎠
3
4
⎞ 1 ⎛ x x2
⎞
1 ⎛ x x2
− ... ⎟ − ⎜ − + ... ⎟ − ...
− ⎜ −
3⎝2 3
⎠ 4⎝ 2 3
⎠
Engineering Mathematics
2.102
x
⎛1 1⎞
⎛1 1 1 ⎞
⎛1 1 1 1 1 ⎞
= − + x 2 ⎜ − ⎟ − x 3 ⎜ − + ⎟ + x 4 ⎜ − − + − ⎟ + ...
2
⎝3 8⎠
⎝ 4 6 24 ⎠
⎝ 5 18 8 12 64 ⎠
x 5 x 2 x 3 251 4
=− +
− +
x + ...
2 24 8 2880
1+ ex
Example 15: Expand
1
2
2e x
up to the term containing x2.
1
Solution:
1
⎛ 1 + ex ⎞ 2 ⎛ 1 − x 1 ⎞ 2
=⎜ e + ⎟
⎜
x ⎟
2⎠
⎝2
⎝ 2e ⎠
⎞
⎟+
⎠
⎛
x2
1
= ⎜1 − x +
−
2
4
⎝
⎞2
⎟
⎠
⎡ ⎛ x x2
+
= ⎢1 − ⎜ −
⎣ ⎝2 4
⎞⎤ 2
⎟⎥
⎠⎦
1 x
1
2 2
x
4
x
1
4
1
x2
4
Solution: e
=e
x 1
1⎤ 2
⎥
2⎦
1
1
...
x2 1 x2
8 8 4
3 2
x ...
32
Example 16: Prove that e x cos x = 1 + x +
x cos x
1
⎡1 ⎛
x2
= ⎢ ⎜1 − x +
−
2!
⎣2 ⎝
1 1
1
2 2
2!
x
2
x2
4
2
...
...
...
x 2 x 3 11 4 x 5
.
−
−
x −
2
3 24
5
x2 x4
...
2! 4!
2
⎞
⎞ 1⎛
⎞
⎛
1⎛
x3 x5
x3
x3
= 1 + ⎜ x − + − ... ⎟ + ⎜ x − + ... ⎟ + ⎜ x − + .... ⎟
2! 4!
2!
2!
⎠ 3! ⎝
⎠ 2! ⎝
⎠
⎝
4
3
5
⎞
⎞
1⎛
1⎛
x3
x3
+ ⎜ x − − ... ⎟ + ⎜ x − − ... ⎟
4! ⎝
2!
2!
⎠ 5! ⎝
⎠
2
1 ⎞
x
⎛ 1 1⎞
⎛ 1 1 ⎞
⎛ 1 1
= 1 + x + + x3 ⎜ − + ⎟ + x 4 ⎜ − + ⎟ + x5 ⎜ − +
⎟ + ...
2
⎝ 2 6⎠
⎝ 2 24 ⎠
⎝ 24 4 120 ⎠
= 1+ x +
x 2 x 3 11 4 x5
− −
x − + ...
2 3 24
5
Differential Calculus I
2.103
⎛
5x3
ex
2
+
Example 17: Prove that e = e ⎜ 1 + x + x +
6
⎝
ex
Solution: e = e
⎛
x 2 x3
+
+
⎜⎜1 + x +
2! 3!
⎝
= ee
x+
⎞
⎟⎟
⎠
x 2 x3
+
+
2 ! 3!
⎡ ⎛
x 2 x3
= e ⎢1 + ⎜ x +
+
+
2 ! 3!
⎢⎣ ⎝
⎡
⎛1 1⎞
= e ⎢1 + x + x 2 ⎜ + ⎟ +
⎝2 2⎠
⎣
5
⎛
= e ⎜1 + x + x 2 + x3 +
6
⎝
Example 18: Prove that (1
1
Solution:
1
(1 + x) x = e x
=e
=e
1
x) x
e
2
⎞ 1 ⎛
x2
+
⎟ + ⎜x +
2!
⎠ 2! ⎝
⎞
1
3
⎟ + (x + … ) +
3!
⎠
⎛1 1 1⎞
x3 ⎜ + + ⎟ +
⎝6 2 6⎠
⎤
⎥
⎦
⎤
⎥
⎥⎦
⎞
⎟
⎠
e
x
2
11e 2
x
24
...... .
log (1+ x )
⎞
1 ⎛ x 2 x3
⎜ x − + −... ⎟⎟
x ⎜⎝
2 3
⎠
⎛ x x2
⎞
⎜⎜1− + −... ⎟⎟
⎝ 2 3
⎠
= ee
⎛ x x 2 x3 x 4
⎞
⎜⎜ − + − + −... ⎟⎟
⎝ 2 3 4 5
⎠
⎡ ⎛ x x 2 x3 x 4
= e ⎢1 + ⎜ − +
−
+
−
4
5
⎢⎣ ⎝ 2 3
⎡
x
⎛1 1⎞
= e ⎢1 − + x 2 ⎜ + ⎟ +
⎝3 8⎠
⎣ 2
11e 2
e
=e− x+
x +
2
24
Example 19: Prove that sin (e x
Solution:
⎞
⎟.
⎠
1)
⎞ 1 ⎛ x x 2 x3
−
+
⎟ + ⎜− +
4
⎠ 2! ⎝ 2 3
⎤
⎥
⎦
x
x2
2
5 4
x
24
⎛
x 2 x3 x 4
sin (e x − 1) = sin ⎜ x +
+
+
+
2 ! 3! 4 !
⎝
⎞
⎟
⎠
....... .
2
⎞
⎟ +
⎠
⎤
⎥
⎥⎦
Engineering Mathematics
2.104
⎛
x 2 x3 x 4
=⎜x +
+
+
+
2 ! 3! 4 !
⎝
x2
⎛1 1⎞
=x+
+ x3 ⎜ − ⎟ +
2
⎝6 6⎠
=x+
Example 20: Expand
x ex + 1
x2 x4
=1+
+
–
x
12 720
2 e –1
Solution:
x
ex
⎞ 1 ⎛
x 2 x3
+
+
⎟ − ⎜x +
2 ! 3!
⎠ 3! ⎝
⎛ 1 1⎞
− ⎟+
x4 ⎜
⎝ 24 4 ⎠
x2
5 4
−
x +
2 24
up to x4 and hence, prove that
1
.
x
x
=
2
3
e − 1 ⎡⎛
x
x
x 4 x5
+
⎢ ⎜1 + x + + +
2 ! 3! 4 ! 5!
⎣⎝
x
=
2
3
⎛
x
x
x 4 x5
+
+
⎜x+ + +
2 ! 3! 4 ! 5 !
⎝
x
⎡ ⎛ x x 2 x3 x 4
= ⎢1 + ⎜ + +
+
+
⎣ ⎝ 2 ! 3! 4 ! 5 !
⎛ x x 2 x3
x4
=1− ⎜ +
+
+
+
⎝ 2 6 24 120
⎞ ⎤
+ ... ⎟ − 1⎥
⎠ ⎦
⎞
⎟
⎠
−1
⎞⎤
⎟⎥
⎠⎦
⎞ ⎛ x x 2 x3
+
+
⎟+⎜ +
⎠ ⎝ 2 6 24
⎛ x x2
−⎜ +
+
⎝2 6
x
⎛ 1 1⎞
⎛ 1 1 1⎞
= 1 − + x 2 ⎜ − + ⎟ + x3 ⎜ −
+ − ⎟
2
⎝ 6 4⎠
⎝ 24 6 8 ⎠
1
1
⎛ 1
+ x4 ⎜ −
+
+
⎝ 120 36 24
2
4
x x
x
=1− +
+ x 3 ( 0) −
+
2 12
720
x ex + 1 x ⎛
2 ⎞
= ⎜1 + x
⎟
2 ex − 1 2 ⎝
e − 1⎠
x
x
= + x
2 e −1
x
x x2
x4
+1− +
−
+
2
2 12 720
x2
x4
=1+
−
+
12 720
=
3
⎞
⎟ +
⎠
⎞
⎟
⎠
2
3
⎞ ⎛x
⎟ +⎜ +
⎠ ⎝2
−
⎞
⎟
⎠
4
1 1⎞
+ ⎟+
8 16 ⎠
...(1)
[Using Eq. (1)]
Differential Calculus I
2.105
Example 21: Prove that
tan
1
x sin
1 x cos
= x sin +
x2
x3
sin 2 +
sin 3 + .... .
2
3
y = tan
Solution: Let
1
x sin
1 x cos
x sin
1 − x cos
eiy − e − iy
x sin
=
i (eiy + e − iy ) 1 − x cos
tan y =
eiy − e − iy
ix sin
=
iy
− iy
1 − x cos
e +e
Applying componendo dividendo,
eiy 1 x (cos
=
e iy 1 x (cos
e 2iy =
2iy
1 xe i
1 xei
log (1 xe
i sin )
i sin )
i
) log (1 xei )
⎞
⎛
⎞ ⎛
x 2 e −2i
x 3 e −3i
x 2 e 2i
x 3 e 3i
= ⎜ − xe − i −
−
− ... ⎟ − ⎜ − xei −
−
− ... ⎟
2
3
2
3
⎠
⎝
⎠ ⎝
= x (e i − e − i ) +
= x ⋅ 2i sin +
y = x sin +
x 2 2i
x3
(e − e −2i ) + (e3i − e −3i ) + ...
2
3
x2
x3
⋅ 2i sin 2 + ⋅ 2i sin 3 + ...
2
3
x2
x3
sin 2 + sin 3 + ...
2
3
Example 22: Prove that e ax cos bx = 1 + ax +
x cos
cos( x sin ) =
and hence, deduce e
n= 0
(a 2
b 2 ) 2 a ( a 2 3b 2 ) 3
x +
x + ...
2!
3!
xn
cos n .
n!
Solution: e cos bx = e . Real Part of (eibx)
ax
ax
= R.P. of e(a+ib)x
⎡
(a 2 + ib) 2 2 (a + ib)3 3
= R.P. of ⎢1 + (a + ib) x +
x +
x +
2!
3!
⎣
⎤
⎥
⎦
Engineering Mathematics
2.106
⎡
⎤
(a 2 − b 2 + 2aib) 2 (a 3 − ib3 + 3ia 2 b − 3ab 2 ) 3
= R.P ⎢1 + (a + ib) x +
x +
x + ...⎥
2
!
3
!
⎣
⎦
2
2
2
2
(a − b ) 2 a (a − 3b ) 3
x +
x + ...
= 1 + ax +
2!
3!
Putting a = cos a and b = sin a,
e x cos cos ( x sin ) = 1 + x cos +
− sin 2 ) 2 cos3
x +
2!
(cos 2
− 3 cos ⋅ sin 2
3!
x 3 + ...
− 3 cos (1 − cos 2 ) 3
x + ...
3!
= 1 + x cos +
cos 2 2 cos3
x +
2!
= 1 + x cos +
x2
x3
cos 2 + cos 3 + ....
2!
3!
∞
xn
cos n
n=0 n!
=∑
Example 23: Prove that e x = 1 + tan x +
1
1
7
tan 2 x - tan 3 x - tan 4 x + ....... .
2!
3!
4!
Solution: Let e x = a0 + a1 tan x + a2 tan 2 x + a3 tan 3 x + a4 tan 4 x + .........
2
3
...(1)
4
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
x3
x3
x3
x3
= a0 + a1 ⎜ x + + ... ⎟ + a2 ⎜ x + + ... ⎟ + a3 ⎜ x + + ... ⎟ + a4 ⎜ x + + ... ⎟ + ........
3
3
3
3
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎛
⎛
⎞
⎞
2x4
x3
+ ... ⎟ + a3 ( x 3 + ...) + a4 ( x 4 + ...) + ..........
= a0 + a1 ⎜ x + + ... ⎟ + a2 ⎜ x 2 +
3
3
⎝
⎝
⎠
⎠
⎛a
⎞
⎛2
⎞
= a0 + a1 x + a2 x 2 + ⎜ 1 + a3 ⎟ x 3 + ⎜ a2 + a4 ⎟ x 4 + ..........
3
3
⎝
⎠
⎝
⎠
ex = 1 + x +
But
x 2 x3 x 4
+ + + ......
2 ! 3! 4 !
...(2)
...(3)
Thus from Eqs (2) and (3)
1+ x +
x 2 x3 x 4
⎛a
⎞
⎛2
⎞
+ + + ..... = a0 + a1 x + a2 x 2 + ⎜ 1 + a3 ⎟ x 3 + ⎜ a2 + a4 ⎟ x 4 + .......
2 ! 3! 4 !
3
3
⎝
⎠
⎝
⎠
x, x2, x3 and x4 on both the sides,
a0 = 1, a1 = 1, a2 =
1 1 a1
1 1
= , + a3 = =
2! 2 3
3! 6
Differential Calculus I
2.107
1 a1 1 1
1
1
− = − =− =−
6 3 6 3
6
3!
2
1
1
1 2 1
7
7
, a4 =
− ⋅ =−
=−
a2 + a4 = =
3
4 ! 24
24 3 2
24
4!
a3 =
Substituting in Eq. (1),
e x = 1 + tan x +
1
7
1
tan 2 x − tan 3 x − tan 4 x + ...
2!
3!
4!
Example 24: Find the values of a and b such that the expansion of
x (1 + ax )
in ascending powers of x begins with the term x4 and
1 + bx
x4
prove that this term is
36
log (1 x )
Solution: Let f ( x)
log (1 x)
x(1 + ax)
1 + bx
x
x2
2
x3
3
x4
4
...
( x ax 2 )(1 bx)
x
x2
2
x3
3
x4
4
...
( x ax 2 )(1 bx b 2 x 2 b3 x3 ...)
x
x2
2
x3
3
x4
4
...
( x bx 2
1
b a x2
2
1 2
b
3
1
b 2 x 3 b3 x 4
1 3
b
4
ab x 3
ax 2
abx 3
ab 2 x 4
ab3 x 5 .......)
ab 2 x 4 ........
If the expansion begins with the term x4, the coefficients of x2 and x3 must be zero.
1
− + b − a = 0,
2
b =a+
1
2
and
1
− b 2 + ab = 0
3
Substituting b in Eq. (1),
2
1 ⎛
1⎞
1⎞
⎛
−⎜a + ⎟ + a⎜a + ⎟ = 0
3 ⎝
2⎠
2⎠
⎝
1
1
1
− a2 − − a + a2 + a = 0
3
4
2
1
1
1
a= , a=
6
2
12
1 1 4 2
b= + = =
6 2 6 3
... (1)
Engineering Mathematics
2.108
3
2
1
1 ⎛2⎞ 1⎛2⎞
1
x 4 = − + b3 − ab 2 = − + ⎜ ⎟ − ⎜ ⎟ = −
4
4 ⎝3⎠ 6⎝3⎠
36
Hence, the expansion begins with the term
x4
.
36
Exercise 2.7
1. Expand ex sec x in powers of x using
Maclaurin’s series.
[Ans.: 1 + x + x + ...]
2
2. Using Maclaurin’s series, prove that
x2
esin x = 1 + x + + …
2
3. Using Maclaurin’s series, prove that
x2
a x = 1 + x log a + (log a ) 2 + …
2!
4. Prove that
10. Prove that
sin (e x 1)
x
x2
2
5 4
x …
24
11. Prove that
cos n x 1 n
x2
x4
n (3n 2)
…
2!
4!
Hence, deduce that
cos3 x 1
3x 2
2
15 x 4
…
48
12. Prove that
sin 2 x
x4
3
x2
2 6
x …
45
5. Prove that sec x = 1 +
x2 5x4
+
+…
2
24
1
⎤
⎡
−1
⎢ Hint : sec x = cos x = (cos x) = ⎥
⎥
⎢
−1
⎢
⎡ ⎛ x2 x4
⎞⎤ ⎥
⎢
⎢1 − ⎜ − + … ⎟ ⎥ ⎥
⎠ ⎦ ⎥⎦
⎣ ⎝ 2! 4!
⎣⎢
6. Prove that
x cosec x = 1 +
x2 7 x4
+
+…
6 360
7. Prove that
x3
…
3
x3
8. Prove that e x cos x 1 x
…
3
9. Prove that
e x sin 2 x
2x 2x2
cos x cos h x
22 x 4
1
4!
24 x8
…
8!
sinh 3 x = ∑
(3n − 3) − [1 − (− 1) n ] x n
8 ⋅ n!
13. Prove that
e x sin x = 1 + x 2 +
x4 x6
+
+ ………
3 120
14. Prove that
(1 x) x
1 x2
x3
2
5x4
…
6
n (n − 1) 2 ⎤
⎡
n
⎢ Hint : (1 + x) = 1 + nx + 2 ! x ⎥
⎢
⎥
n (n − 1) (n − 2) 3
⎢
⎥
+
x + …⎥
⎢
3!
⎣
⎦
15. Prove that
(1 + x)1+ x = 1 + x + x 2 +
x3
+…
3
(1.01)1.01
[Ans.: 1.0101]
Differential Calculus I
16. Prove that
log x
2.109
23. Prove that
1
( x 1) 2
2
( x 1)
1
( x 1)3 …
3
17. Prove that
e x log (1 + x) = x +
x 2 x3
+ +…
2
3
24. Prove that
2
log (1 x x 2 )
x
2
x
2x
3
3
…
log
xe x
ex 1
x
2
x2
24
x4
…
2880
18. Prove that
[log (1 x)]2
x2
x3
⎞
⎛p
25. Expand log tan ⎜ + x⎟ upto x 5 .
⎠
⎝4
11 4
x …
12
19. Prove that
log cosh x =
1 2 1 4 1 6
x − x +
x −…
2
12
45
20. Prove that
log (1 tan x)
x
x2
2
⎡
⎛p
⎞
⎛ 1 + tan x ⎞
⎢ Hint : log tan ⎜⎝ 4 + x⎟⎠ = log ⎜⎝ 1 − tan x ⎟⎠
⎣
⎤
= log (1 + tan x) − log (1 − tan x) ⎥
⎦
2 x3
………
3
4 3 4 5
⎤
⎡
⎢⎣ Ans. : 2 x + 3 x + 3 x + …⎥⎦
21. Prove that
sin x
log
x
x2
6
x4
180
x6
…
2835
22. Prove that
x3 7
tan x
= + x 4 + ………
log
x
3 90
Example 1: Prove that log (sec x ) =
26. Prove that x = y +
if y
x
x2
2
x3
3
y 2 y3 y 4
+ +
+…
2! 3! 4!
x4
4
…
x2 x4 x6
+
+
+ ........ .
2 12 45
Solution: Let y = log (sec x)
dy
1
=
⋅ sec x tan x = tan x
dx sec x
x3 2
= x + + x 5 + ........
3 15
Integrating Eq. (1),
x2 x4 2 x6
+ + ⋅ + ........
2 12 15 6
x2 x4 x6
log (sec x) = c + + +
+ .........
2 12 45
y = c+
... (1)
Engineering Mathematics
2.110
Putting x = 0,
log (sec 0) = c + 0
c = log 1, c = 0
log (sec x) =
Hence,
Example 2: Prove that sin
x2 x4 x6
+ + + ........
2 12 45
1
x = x+
1 x 3 1.3 x 5 1.3.5 x 7
+
+
+… .
2 3 2.4 5 2.4.6 7
Solution: Let y = sin–1 x
1
−
dy
1
=
= (1 − x 2 ) 2
dx
1 − x2
⎛ 1 ⎞⎛ 3 ⎞⎛ 5 ⎞
⎛ 1 ⎞⎛ 3 ⎞
−
−
⎜ − ⎟⎜ − ⎟⎜ − ⎟
1 2 ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠
2 ⎠⎝ 2 ⎠⎝ 2 ⎠
2 2
= 1+ x +
( − x 2 )3 + …
(− x ) + ⎝
3!
2!
2
= 1+
x 2 1.3 4 1.3.5 6
+
x +…
x +
2 2.4
2.4.6
... (1)
Integrating Eq. (1),
1
2
1
sin −1 x = c + x +
2
y = c+ x+
Putting x = 0,
x 3 1.3 x 5 1.3.5
+
+
3 2.4 5 2.4.6
x 3 1.3 x 5 1.3.5
+
+
3 2.4 5 2.4.6
x7
+ ........
7
x7
+ ........
7
sin 1 0 = c
c=0
sin 1 x = x +
Hence,
–1
Example 3: Prove that cos x =
1 x 3 1.3 x 5 1.3.5 x 7
+
+
+ ........
2 3 2.4 5 2.4.6 7
⎛
1 x 3 1.3 x 5
–⎜x +
+
+
2 ⎝
2 3
2.4 5
Solution: Let y = cos–1 x
dy
1
=−
dx
1 − x2
Proceeding as in Ex. 2, we get
⎛
⎞
1 x 3 1.3 x 5
cos −1 x = c − ⎜ x +
+
+…⎟
2
3
2
4
5
.
⎠
⎝
⎞
⎟.
⎠
Differential Calculus I
2.111
Putting x = 0,
cos −1 0 = c
p
c=
2
⎛
1 x 3 1.3 x 5
−⎜x +
+
+
2 ⎝
2 3 2.4 5
cos −1 x =
Hence,
Example 4: Prove that tan
1
x
x3
3
x
x5
5
x7
7
⎞
⎟
⎠
........ .
Solution: Let y = tan–1 x
dy
1
=
= (1 + x 2 ) −1 = 1 − x 2 + x 4 − x 6 + …
dx 1 + x 2
Integrating Eq. (1),
x3 x5 x 7
y = c + x − + − +…
3 5
7
3
5
x
x
x7
tan −1 x = c + x − + − + ........
3 5
7
... (1)
Putting x = 0,
tan −1 0 = c
c=0
Hence,
tan −1 x = x −
Example 5: Prove that sinh
Solution: Let
1
x
x3 x5 x 7
+ − +…
3 5
7
x3
6
x
(
3 x5
40
y = sin h −1 x = log x + x 2 + 1
........ .
)
⎛
dy
1
2x ⎞
=
⎜⎜1 +
⎟=
2
dx x + x + 1 ⎝ 2 x 2 + 1 ⎟⎠
= (1 + x 2 )
−
1
x2 + 1
1
2
⎛ 1⎞⎛ 3⎞
⎜− ⎟ ⎜− ⎟
1
2⎠⎝ 2⎠ 2 2
= 1 − x2 + ⎝
(x ) − …
2!
2
1
3
= 1 − x2 + x4 − …
2
8
Integrating Eq. (1),
x3 3 x5
+ ⋅ −…
6 8 5
x3 3x5
sinh −1 x = c + x − +
−…
6 40
y = c+ x−
... (1)
Engineering Mathematics
2.112
sinh −1 0 = c, c = 0
Putting x = 0,
sinh −1 x = x −
Example 6: If x = 1 –
Solution:
y2 y4
+
2! 4!
x =1−
x3 3x5
+
−…
6 40
y6
+ ........, find y in a series of x.
6!
y2 y4 y6
+
−
+ ……
2! 4! 6!
= cos y
y = cos –1 x
Proceeding as in Ex. 3, we get
y=
⎛
⎞
x3 3x5
−⎜x +
+
+ …⎟
2 ⎝
6
40
⎠
⎛
x3
3 5
Example 1: Prove that sinh–1 (3x + 4x3) = 3 ⎜ x −
x +
+
6
40
⎝
Solution:
Let y = sinh 1 (3 x + 4 x 3 )
Putting
x = sinh q ,
⎞
⎟.
⎠
y = sinh −1 (3 sinh q + 4 sinh 3 q )
= sinh −1 (sinh 3q )
= 3q
= 3 sinh −1 x
⎞
⎛
x3 3x5
= 3⎜ x − +
−…⎟
6
40
⎝
⎠
⎛
⎛ 2x ⎞
x3 x5 x7
−
−
Example 2: Prove that sin − 1 ⎜
=
2
x
+
+
⎜
⎟
2
3
5
7
⎝1 + x ⎠
⎝
Solution: Let y = sin
Putting
1
2x
1 + x2
x = tan ,
⎛ 2 tan q ⎞
y = sin −1 ⎜
⎟
2
⎝ 1 + tan q ⎠
= sin −1 (sin 2q )
= 2q
= 2 tan −1 x
⎞
⎛
x3 x5 x 7
= 2 ⎜ x − + − +…⎟
3
5
7
⎝
⎠
⎞
⎟.
⎠
Differential
Calculus II
Chapter
3
3.1 INTRODUCTION
In Chapter 2, we have studied a few topics of differential calculus such as successive
differentiation, mean value theorems, expansion of functions and indeterminate forms.
In this chapter, we will study tangents, normals, curvature and envelope of curves.
Curve tracing is covered as the last topic of this chapter. Knowledge of curve tracing
helps in application of integration in finding length, area, volume and surface area.
3.2 TANGENT AND NORMAL
Let P(x, y) and Q(x + h, y + k) be the points on the curve y = f (x). As Q tends to P, the
chord PQ tends to the straight line PQ which touches the curve at point P. This straight
line is called the tangent to the curve at P. The perpendicular drawn to the tangent at P
is called normal to the curve at that point.
y
( y + k ) − y f ( x + h) − f ( x )
Slope of line PQ =
=
( x + h) − x
h
when Q
P, line PQ tends to the tangent at P.
Q(x + h, y + k)
f ( x + h) − f ( x )
Slope of tangent at P = lim
Q→P
h
f ( x + h) − f ( x )
= lim
P(x, y)
h→ 0
h
= f ′( x )
dy
x
=
dx
Fig. 3.1
Equation of the tangent to the curve at any point
(x, y) is given by,
Y − y = f ′( x )( X − x )
1
Slope of normal at P = −
f ′( x )
and equation of the normal to the curve at any point (x, y) is given by,
1
Y−y=−
( X − x)
f ′( x )
where (X, Y ) is any arbitrary point on the tangent (or normal) to the curve.
3.2
Engineering Mathematics
3.2.1 Angle of Intersection of Curves
The angle of intersection of two curves at a point of intersection is defined to be the
angle between the tangents to the curves at that point.
Let m1 and m2 be the slopes of the tangent to the curves y = f1(x) and y = f2(x) respectively at the point of intersection. Angle of intersection of two curves at the point of
intersection is given by,
m − m1
q = tan −1 2
.
1 + m2 m1
3.2.2 Length of Tangent, Sub-tangent,
Normal and Sub-normal
Let P(x, y) be any point on the curve y = f (x). The tangent and normal at the point P meet
the x-axis at T and N respectively. Let PM be the ordinate. PT and PN are the lengths
of the tangent and normal to the curve. TM and MN are the lengths of sub-tangent and
sub-normal to the curve at the point P. Let be the angle which tangent makes with
the x-axis.
dy
tany =
dx
y
Length of tangent = PT = PM cosec y
= y 1 + cot 2 y
⎛ dx ⎞
= y 1+ ⎜ ⎟
⎝ dy ⎠
P(x, y)
2
Length of sub-tangent = TM = PM coty
dx
dy
Length of normal = PN = PM secy
T
=y
M
N
Fig. 3.2
= y 1 + tan 2 y
⎛ dy ⎞
= y 1+ ⎜ ⎟
⎝ dx ⎠
2
Length of sub-normal = MN = PM tany
dy
=y
dx
3.2.3 Length of Perpendicular from the
Origin to the Tangent
Let p be the length of the perpendicular drawn from origin to the tangent.
p=
y − x f ′( x )
1 + [ f ′ ( x)]
2
x
Differential Calculus II
The relation between distance of any point P(x, y) on
the curve from the origin and the length of perpendicular from the origin to the tangent at that point is
called pedal equation of the curve. Pedal equations
can be obtained by eliminating x and y from equations y = f (x),
y − x f ′( x )
p=
and r2 = x2 + y2
1 + ⎡⎣ f ′( x ) 2 ⎤⎦
3.3
y
P(x, y)
r
x
O p
m
Fig. 3.3
Example 1: Find the equations of the tangent and normal to the curve xy = c 2 at
⎛ c⎞
the point ⎜ ct , ⎟ .
⎝ t⎠
xy = c 2
Solution:
Differentiating w.r.t. x,
x
⎛ c⎞
At the point ⎜ ct , ⎟ ,
⎝ t⎠
dy
+y=0
dx
dy
y
=−
dx
x
c
dy
1
=− t =− 2
dx
ct
t
⎛ c⎞
Slope of the tangent to the curve at ⎜ ct , ⎟ is
⎝ t⎠
1
and slope of the normal is t2.
t2
⎛ c⎞
Equation of the tangent to the curve at ⎜ ct , ⎟ is given by,
⎝ t⎠
c
1
Y − = − 2 ( X − ct )
t
t
X + t 2Y = 2ct
⎛ c⎞
Equation of the normal to the curve at ⎜ ct , ⎟ is given by,
⎝ t⎠
c
= t 2 ( X − ct )
t
t 3 X − tY = c(t 4 − 1) .
Y−
Example 2: Find the equations of the tangent and normal to the curve
x2 y2
−
= 1 at the point ( a sec q , b tan q ).
a 2 b2
Solution:
x2 y2
−
=1
a2 b2
3.4
Differentiating w.r.t. x,
Engineering Mathematics
2 x 2 y dy
−
=0
a 2 b 2 dx
dy b 2 x
=
dx a 2 y
dy b 2 ( a secq )
b
At the point ( a secq , b tan q ),
= 2
=
dx a (b tan q ) a sin q
b
Slope of the tangent to the curve at ( a secq , b tan q ) is
and slope of the normal
a
sinq
a
is
sin q .
b
Equation of the tangent to the curve at ( a secq , b tan q ) is given by,
b
Y − b tan q =
( X − a secq )
a sin q
Y
sin 2 q X
sin q −
= − secq
b
cos q
a
⎛X⎞
⎛Y ⎞
2
2
⎜⎝ ⎟⎠ secq − ⎜⎝ ⎟⎠ tan q = sec q − tan q = 1
a
b
Equation of the normal to the curve at ( a secq , b tan q ) is given by,
a
Y − b tan q = − sin q ( X − a secq )
b
Y b
X
a
− tan q = − sin q + tan q
a a
b
b
a b a2 + b2
⎛X⎞
⎛Y ⎞
⎜⎝ ⎟⎠ cos q + ⎜⎝ ⎟⎠ cot q = + =
b
a
b a
ab .
Example 3: Find the equations of the tangent and normal to the curve
y = 2 x 2 - 4 x + 5 at the point (3, 11).
Solution:
y = 2x2 − 4x + 5
Differentiating w.r.t. x,
dy
= 4x − 4
dx
dy
At the point (3, 11),
= 8.
dx
Slope of the tangent to the curve at (3, 11) is 8 and slope of the normal is
Equation of the tangent to the curve at (3, 11) is given by,
Y − 11 = 8 ( X − 3)
8 X − Y = 13
Equation of the normal to the curve at (3, 11) is given by,
1
Y − 11 = − ( X − 3)
8
X + 8Y = 91.
1
.
8
Differential Calculus II
3.5
Example 4: Find the equations of the tangent and normal to the curve x = sin t ,
p
y = cos 2t at t = .
6
Solution:
x = sin t
dx
= cos t
dt
y = cos 2t
dy
= −2 sin 2t
dt
dy dy / dt
2 sin 2t
=
=−
= −4 sin t
dx dx / dt
cos t
dy
⎛p ⎞
= −4 sin ⎜ ⎟ = −2
⎝6⎠
dx
1
p
Slope of the tangent to the curve at t = is 2 and slope of the normal is .
2
6
p
1
1
At t = , x = and y = .
6
2
2
p
Equation of the tangent to the curve at t = is given by,
6
1
1⎞
⎛
Y − = −2 ⎜ X − ⎟
⎝
2
2⎠
4 X + 2Y = 3
p
Equation of the normal to the curve at t = is given by,
6
1 1⎛
1⎞
Y − = ⎜X − ⎟
2 2⎝
2⎠
2 X − 4Y = −1 .
At the point t =
p
,
6
n
n
⎛x⎞ ⎛ y⎞
Example 5: Prove that the curve ⎜ ⎟ + ⎜ ⎟ = 2 touches the straight line
⎝a⎠ ⎝b⎠
x y
+ = 2 at the point (a, b), whatever be the value of n.
a b
n
Solution:
n
⎛x⎞ ⎛ y⎞
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ = 2
a
b
Differentiating w.r.t. x,
n n −1 n n −1 dy
=0
x + n y
dx
an
b
dy
b n x n −1
= − n n −1
dx
a y
At the point (a, b),
dy
b n a n −1
b
= − n n −1 = −
dx
a
a b
3.6
Engineering Mathematics
Equation of the tangent to the curve at (a, b) is given by,
b
Y − b = − ( X − a)
a
bX + aY = 2ab
X Y
+ =2
a b
n
n
⎛x⎞ ⎛ y⎞
x y
Hence, the curve ⎜ ⎟ + ⎜ ⎟ = 2 touches the straight line + = 2 at the point (a, b),
⎝a⎠ ⎝b⎠
a b
whatever be the value of n.
Example 6: Tangents are drawn from the origin to the curve y = sin x . Prove
that their point of contact lies on x 2 y 2 = x 2 - y 2 .
Solution:
Differentiating w.r.t. x,
y = sin x
dy
= cos x
dx
Equation of the tangent to the curve at the origin is given by,
Y − 0 = cos x ( X − 0)
Y
= cos x
X
Let (x1, y1) be the point of contact of the curve and the tangent.
y1 = sin x1
y1
and
= cos x1
x1
Squaring and adding the equations,
y2
y12 + 12 = 1
x1
x12 y12 = x12 − y12
Hence, the point of contact lies on x 2 y 2 = x 2 − y 2 .
Example 7: Show that the line x cos q + y sin q = p will touch the curve
m
m
m
⎛x⎞
⎛ y⎞
m
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ = 1, provided ( a cos q )
a
b
m
1
+ ( b sin q ) m
m
Solution:
m
1
= pm 1 .
m
⎛x⎞
⎛ y⎞
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ = 1
a
b
Differentiating w.r.t. x,
mx m −1 my m −1 dy
+ m
=0
dx
am
b
dy
b m x m −1
= − m m −1
dx
a y
Differential Calculus II
3.7
Equation of the tangent to the curve is given by,
b m x m −1
( X − x)
a m y m −1
x m −1
y m −1 x m y m
X m +Y m = m + m = 1
a
b
a
b
Let (x1, y1) be the point of contact of the curve and the tangent.
sinq
cosq
x1
+ y1
=1
p
p
Y−y=−
x1
Comparing two equations,
x m −1
y m −1
+ y1 m = 1
m
a
b
x m −1 cosq
=
p
am
1
and
⎛ a m cosq ⎞ m −1
x=⎜
p ⎟⎠
⎝
y m −1 sinq
=
p
bm
1
⎛ b m sinq ⎞ m −1
y=⎜
p ⎟⎠
⎝
m
m
⎛x⎞
⎛ y⎞
The point (x, y) lies on the curve ⎜ ⎟ + ⎜ ⎟ = 1
⎝a⎠
⎝b⎠
m
m
1 ⎛ a m cos q ⎞ m −1 1 ⎛ b m sin q ⎞ m −1
+ m⎜
=1
p ⎟⎠
p ⎟⎠
a m ⎜⎝
b ⎝
m
m
m
( a cos q ) m −1 + (b sin q ) m −1 = p m −1
Example 8: Prove that the sum of intercepts of the tangent to the curve
x + y = a on the coordinate axes is constant.
x+ y= a
Solution:
Differentiating w.r.t. x,
1 1 1 1 dy
+
=0
2 x 2 y dx
dy
y
=−
dx
x
Equation of the tangent to the curve at (x, y) is given by,
Y−y=−
y
( X − x)
x
… (1)
3.8
Engineering Mathematics
X intercept is obtained by putting Y = 0 in Eq. (1),
−y = −
y
( X − x)
x
X − x = xy
X = x + xy = x
(
)
x+ y = x a
Y intercept is obtained by putting X = 0 in Eq. (1),
Y − y = xy
Y = y + xy =
(
y
)
x+ y =
y a
Sum of intercepts = X + Y = x a + y a
= a
(
)
x+ y = a
( a)
=a
Hence, sum of intercepts of the tangent to the curve on the coordinate axes is constant.
Example 9: Show that the length of the portion of the normal to the curve
x = a (4cos 3q - 3cosq ), y = a (4sin 3q - 9sinq ) intercepted between the coordinate axes is constant.
Solution:
x = a (4cos3q − 3cosq )
dx
= a ( −12 cos 2 q sin q + 3 sin q )
dq
y = a ( 4 sin 3 q − 9 sin q )
dy
= a (12 sin 2 q cos q − 9 cos q )
dq
dy dy / d q
a (12 sin 2 q cos q − 9 cos q )
=
=
dx dx / d q a ( −12 cos 2 q sin q + 3 sin q )
3 cos q ( 4 sin 2 q − 3)
−3 sin q ( 4 cos 2 q − 1)
cos q [2 (1 − cos 2q ) − 3]
=
− sin q [2 (1 + cos 2q ) − 1]
cos q ( 2 cos 2q + 1)
=
sin q ( 2 cos 2q + 1)
= cot q
sin q
Slope of the normal to the curve = − tan q = −
cos q
Equation of the normal to the curve is given by,
sin q
Y − a ( 4 sin 3 q − 9 sin q ) = −
[ X − a ( 4 cos3 q − 3 cos q )]
cos q
=
… (1)
Differential Calculus II
3.9
Y intercept is obtained by putting X = 0 in Eq. (1),
sin q
Y − a ( 4 sin 3 q − 9 sin q ) =
a ( 4 cos3 q − 3 cos q )
cos q
Y = a sin q ( 4 cos 2 q − 3 + 4 sin 2 q − 9)
= −8a sinq
X intercept is obtained by putting Y = 0 in Eq. (1),
sin q
− a ( 4 sin 3 q − 9 sin q ) = −
[ X − a ( 4 cos3 q − 3 cos q )]
cos q
X = a cos q ( 4 sin 2 q − 9 + 4 cos 2 q − 3)
= −8a cos q
Length of the portion of the normal to the curve intercepted between co-ordinate axes
= X 2 + Y 2 = ( −8a sin q ) 2 + ( −8a cos q ) 2
= 8a
= constant
Example 10: Find the angle of intersection of the curves y 2 = 4ax and x 2 = 4by
at their point of intersection other than origin.
Solution: The points of intersection of the curves are obtained as,
y 2 = 4 ax
⎡⎣∵ x 2 = 4by ⎤⎦
= 4 a ⋅ 2 by
⎛ 3
⎞
y ⎜ y 2 − 8a b ⎟ = 0
⎝
⎠
3
y = 0 and y 2 − 8a b = 0
2
2
1
1 3
⎛
⎞
3 3
y = 0 and y = ⎜ 8ab 2 ⎟ = 4 a b
⎝
⎠
When y = 0, x = 0.
1
1 2
⎛ 2 1 ⎞2
When y = 4 a b , x = 2 by = 2b ⎜ 4 a 3 b 3 ⎟ = 4 a 3 b 3
⎠
⎝
2
3
1
3
1
2
2 1
⎞
⎛ 1 2
Hence, (0, 0) and ⎜ 4 a 3 b 3 , 4 a 3 b 3 ⎟ are the two points of intersection.
⎠
⎝
2
For the curve y = 4 ax,
Differentiating w.r.t. x,
2y
dy
= 4a
dx
dy 2a
=
y
dx
3.10
Engineering Mathematics
1
dy a 3
= 1
dx
2b 3
⎞
⎛
At the point ⎜ 4 a b , 4 a b ⎟ ,
⎠
⎝
1
3
2
3
2
3
1
3
For the curve x 2 = 4by,
Differentiating w.r.t. x,
dy
dx
2 x = 4b
x
dy
=
dx 2b
1
2 1
⎞
⎛ 1 2
At the point ⎜ 4 a 3 b 3 , 4 a 3 b 3 ⎟ ,
⎠
⎝
dy 2a 3
= 1
dx
b3
If m1 and m2 be the slopes of the tangents to the curves, then
m1 =
1
1
a3
2a 3
2b
Angle of intersection = tan
1
1
3
and m2 =
1
b3
m2 m1
1 + m2 m1
= tan
−1
1
1
2a 3
a3
b
1
3
−
1
2b 3
1
1
2a 3 a 3
1+ 1 . 1
b 3 2b 3
1
= tan
−1
1
3a 3 b 3
2
⎞
⎛ 2
2 ⎜a3 + b3 ⎟
⎠
⎝
.
Example 11: Show that the condition that the curves ax 2 + by 2 = 1 and
1 1 1 1
a x 2 + b y 2 = 1 should intersect orthogonally is - = - .
a b a b
Solution: Let P (x1, y1) be the point of intersection of the curves. Hence, ax12 + by12 = 1
and a x12 + b y12 = 1
Solving these two equations,
b′ − b
… (1)
x12 =
ab ′ − a ′b
a − a′
y12 =
… (2)
ab ′ − a ′ b
Differential Calculus II
3.11
For the curve ax 2 + by 2 = 1,
Differentiating w.r.t. x,
2ax + 2by
At the point of intersection (x1, y1),
dy
=0
dx
dy
ax
dx
by
dy
dx
ax1
by1
dy
dx
ax
by
dy
dx
a x1
b y1
For the curve a x 2 + b y 2 = 1,
At the point of intersection (x1, y1),
Since the two curves intersect orthogonally,
⎛ ax1 ⎞ ⎛ a ′x1 ⎞
⎜⎝ − by ⎟⎠ ⎜⎝ − b ′y ⎟⎠ = −1
1
1
aa x12 + bb y22 = 0
aa ′(b ′ − b) bb ′( a − a ′ )
+
=0
ab ′ − a ′b
ab ′ − a ′b
b b a a
+
=0
bb
aa
1
a
1
b
1
a
[Using Eqs (1) and (2)]
1
.
b
Example 12: Show that the curves x 2 = ay and y 2 = 2ax intersect upon the curve
x 3 + y 3 = 3axy and find the angle between each pair at the point of intersection.
Solution: The points of intersection of the curves x 2 = ay and y 2 = 2ax are obtained
as,
1
3
1
x 2 = ay = a 2ax = 2 2 a 2 x 2
1
1 3
⎛ 3
⎞
x 2 ⎜ x 2 − 22 a2 ⎟ = 0
⎝
⎠
1
3
1
3
x 2 = 0 and x 2 − 2 2 a 2 = 0
3.12
Engineering Mathematics
2
1
⎛ 1 3 ⎞3
x = 0 and x = ⎜ 2 2 a 2 ⎟ = 2 3 a
⎝
⎠
When x = 0, y = 0
1
When x = 2 3 a,
2
2ax = 2 3 a
2
⎞
⎛ 1
Hence, (0, 0) and ⎜ 2 3 a, 2 3 a ⎟ are the two points of intersection. On substituting, these
⎠
⎝
points satisfy the equation of the curve x 3 + y 3 = 3axy , and hence they lie on this curve.
For the curve x 2 = ay ,
dy 2 x
=
dx
a
dy
= 0 which indicates that tangent at the origin is x-axis.
dx
2
4
⎞
⎛ 1
dy
At the point ⎜ 2 3 a, 2 3 a ⎟ ,
= 23
dx
⎠
⎝
At the point (0, 0),
For the curve y 2 = 2ax,
2y
At the point (0, 0),
dy
dx
dy
= 2a
dx
dy a
=
dx y
which indicates that tangent at the origin is y-axis.
2
⎞
⎛ 1
dy
At the point ⎜ 2 3 a, 2 3 a ⎟ ,
=2
dx
⎠
⎝
Further, for the curve x 3 + y 3 = 3axy,
3x 2 + 3 y 2
2
3
dy
dy
= 3ay + 3ax
dx
dx
2
dy ay x
=
dx y 2 ax
2
2
⎞ dy
⎞
⎛ 1
⎛ 1
= 0 which indicates that tangent at the point ⎜ 2 3 a, 2 3 a ⎟
At the point ⎜ 2 3 a, 2 3 a ⎟ ,
d
x
⎠
⎠
⎝
⎝
2
⎞
⎛ 13
to the curve is parallel to x-axis. Hence, at the point ⎜ 2 a, 2 3 a ⎟ , angle between the
⎠
⎝
4
2
3
3
−1 ⎛ 3 ⎞
curves x = ay and x + y = 3axy is tan ⎜ 2 ⎟ and angle between the curves y 2 = 2ax
⎝ ⎠
⎛ −2 ⎞
and x 3 + y 3 = 3axy is tan −1 ⎜ 2 3 ⎟ .
⎠
⎝
Differential Calculus II
3.13
Example 13: Find the lengths of the tangent, sub-tangent, normal and suba3
⎛ a⎞
normal to the curve y = 2
at a ,
.
a + x 2 ⎜⎝ 2 ⎟⎠
Solution:
y=
a3
a2 + x2
Differentiating w.r.t. x,
dy
2a 3 x
=− 2
dx
(a + x 2 )2
a
At the point ⎛⎜ a, ⎞⎟ ,
⎝ 2⎠
dy
2a3 ⋅ a
1
=− 2
=−
dx
2
(a + a2 )2
2
⎛ dx ⎞
a
5
⎛ a⎞
Length of the tangent at ⎜ a, ⎟ = y 1 + ⎜ ⎟ =
1 + ( −2) 2 =
a
⎝ 2⎠
⎝ dy ⎠
2
2
dx a
⎛ a⎞
Length of the sub-tangent at ⎜ a, ⎟ = y
= ( −2) = − a
⎝ 2⎠
dy 2
2
2
a
5
⎛ dy ⎞
⎛ 1⎞
⎛ a⎞
Length of the normal at ⎜ a, ⎟ = y 1 + ⎜ ⎟ =
1+ ⎜− ⎟ =
a
⎝ dx ⎠
⎝
⎠
⎝ 2⎠
2
2
4
⎛ a⎞
dy a ⎛ 1 ⎞
a
Length of the sub-normal at ⎜ a, ⎟ = y
= ⎜− ⎟ = − .
⎝ 2⎠
dx 2 ⎝ 2 ⎠
4
Note: Length cannot be negative, therefore depending upon the value of a, consider
always a positive value of length.
Example 14: Find the lengths of the tangent, sub-tangent, normal and subp
normal to the curve x = a (q - sinq ), y = a (1 - cosq ) atq = .
2
Solution:
x = a (q − sin q )
dx
= a (1 − cos q )
dq
y = a (1 − cos q )
dy
= a sin q
dq
a sin q
dy dy / dq
=
=
=
dx dx / dq a (1 − cos q )
Atq =
dy
= 1 and y = a.
2 dx
,
q
q
cos
2
2 = cot q
q
2
2 sin 2
2
2 sin
3.14
Engineering Mathematics
2
⎛ dx ⎞
Hence, length of the tangent = y 1 + ⎜ ⎟ = a 1 + 1 = 2 a
⎝ dy ⎠
dx
=a
dy
Length of the sub-tangent = y
dy
dx
Length of the normal = y 1 +
Length of the sub-normal = y
2
= a 1+1 = 2 a
dy
=a.
dx
Example 15: Prove that the sum of the length of the tangent and sub-tangent at
any point of the curve y = a log ( x 2 - a 2 ) varies as the product of the coordinates
of the point.
Solution:
y = a log ( x 2 − a 2 )
Differentiating w.r.t. x,
dy
dx
a
1
x2
a2
2x
2ax
x2 a2
2
2
⎛ x2 − a2 ⎞
⎛ dx ⎞
( x2 + a2 )
Length of tangent = y 1 + ⎜ ⎟ = y 1 + ⎜
⎟ =y
2ax
⎝ dy ⎠
⎝ 2ax ⎠
Length of sub-tangent = y
dx
( x2 − a2 )
=y
2ax
dy
2x2 1
( x 2 + a2 )
( x 2 − a2 )
+y
=y
= xy
2ax
2ax
2ax 2
Hence, sum of the length of the tangent and sub-tangent varies as the product of the
coordinates of the point.
Sum of the length of tangent and sub-tangent = y
Example 16: Prove that the length of the sub-normal at any point of the curve
x 2 y 2 = a 2 ( x 2 - a 2 ) varies inversely as the cube of its abscissa.
Solution:
x 2 y 2 = a2 ( x 2 − a2 )
Differentiating w.r.t. x,
2 xy 2 + 2 x 2 y
dy
= 2a 2 x
dx
dy 2a 2 x 2 xy 2 a 2 y 2
=
=
xy
dx
2x2 y
… (1)
Differential Calculus II
Length of sub-normal = y
3.15
dy a 2 − y 2 a 4 1
=
= 2⋅
dx
x
x x
[Using Eq. (1)]
a4
x3
Hence, the length of the sub-normal varies inversely as the cube of its abscissa.
=
Exercise 3.1
1. Find the equations of the tangent and
normal to the following curves:
(i) y 2 = 4ax at (a, −2a )
x2 y 2
+
= 1 at (a cos q , b sin q )
a 2 b2
a
(iii) ( x 2 + y 2 ) x − ay 2 = 0 at x =
2
2
2
(iv) x ( x − y ) + a ( x + y ) = 0 at (0, 0)
(ii)
x
(v) y = a cosh
a
(vi) x = 2a cos q − a cos 2q ,
y = 2a sin q − a sin 2q at q =
p
2
2at 3
1
2at 2
,y=
at t =
2
2
2
1+ t
1+ t
(viii) x = a (q + sin q ), y = a (1 + cos q )
(vii) x =
π
.
2
⎡ Ans.: (i) x + y + a = 0, x − y = 3a ⎤
⎢
⎥
x
y
⎢
⎥
(ii) cos q + sin q = 1,
⎢
⎥
a
b
⎢
⎥
ax
by
⎢
⎥
−
= a 2 − b2
⎢
⎥
cos q sin q
⎢
⎥
(iii) 4 x ± 2 y − a = 0,
⎢
⎥
⎢
⎥
2 x ± 4 y = 3a
⎢
⎥
(iv) x + y = 0, x − y = 0
⎢
⎥
⎢
⎥
x
(v) y − y1 = sinh ( x − x1 ) , ⎥
⎢
a
⎢
⎥
⎢
⎥
x
x − x1 + ( y − y1 ) siinh = 0 ⎥
⎢
a
⎢
⎥
at θ =
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎣⎢⎢
(vi) x + y = 3a, x − y + a = 0
(vii) 13 x − 16 y = 2a,
16 x + 13 y = 9a
a
(viii) x + y − p − 2a = 0
2
a
x− y− p =0
2
2. Prove that
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥⎥⎦
x y
+ = 1 touches the curve
a b
x
y = be a at the point where the curve
crosses the y-axis.
3. Prove that all the points of the curve
x
y 2 = 4a x + a sin
at which the
a
tangent is parallel to the x-axis lie on
a parabola.
4. Show that the tangents to the curve
x 3 + y 3 = 3axy at the points where
it meets the parabola y 2 = ax are
parallel to the y-axis.
5. In the curve x m y n = a m + n , prove that
the portion of the tangent intercepted
between the axes, is divided at its
point of contact into segments which
are in a constant ratio.
x
6. In the curve y = a cosh
, prove
a
that the length of the portion of the
normal intercepted between the curve
y2
.
and the x-axis is
a
7. Show that the tangent and normal at any
point of the curve x = aeq (sin cos ),
3.16
Engineering Mathematics
y = aeq (sin + cos ) are equidistant
from the origin.
8. Show that the distance from the origin
of the normal at any point of the curve
q
q⎞
⎛
y = aeq ⎜ cos − 2 sin ⎟ is twice the
⎝
2
2⎠
distance of the tangent at the point
from the origin.
9. Find the angle of intersection of the
following curves:
(ii) x 2 = 4ay and 2 y 2 = ax
(iii) xy = a 2 and x 2 + y 2 = 2a 2
(iv) x − y = a and x + y = a
2
2
2
2
2
(v) y = sin x and y = cos x
(vi) y 2 = ax and x 3 + y 3 = 3ax.
⎡
p
−1 ⎛ 1 ⎞ ⎤
⎢ Ans. : (i) 2 , tan ⎜ 2 ⎟ ⎥
⎝ ⎠⎥
⎢
⎢
p
⎛ 3 ⎞⎥
⎢
(ii) , tan −1 ⎜ ⎟ ⎥
2
⎢
⎝ 5 ⎠⎥
⎢
⎥
(iii) 0
⎢
⎥
⎢
⎥
p
⎢
⎥
(iv)
⎢
⎥
4
⎢
⎥
−1
( v) tan 2 2
⎢
⎥
⎢
⎥
−1 3
( vi) tan 16 ⎥⎦
⎢⎣
10. Show that the curves
x2 y 2
+
=1
a 2 b2
x2
y2
+
= 1 intersect
a2 +
b2 +
orthogonally for all values of .
11. Prove that the curves x2 + 2xy y2 +
2ax = 0 and 3y3 2a2x 4a2 y + a3 = 0
and
9
at the
8
point ( a, a).
[ Ans. : n = −2]
13. Prove that for the curve x m + n = cy 2 n ,
the m th power of the sub-tangent
varies as the n th power of the subnormal.
14. Show that for the curve b y2 = (x + a)3,
the square of the sub-tangent varies
as the sub-normal.
(i) 2 y 2 = x 3 and y 2 = 32 x
2
1
12. Find the value of n so that the subnormal at any point of the curve
xy n = c n +1 is of constant length.
q⎞
⎛ q
x = aeq ⎜ sin + 2 cos ⎟ ,
⎝
2
2⎠
2
intersect at an angle tan
15. Find the lengths of the normal
and sub-normal of the curve
x
x
−
1 ⎛ a
a ⎞
y = a ⎜e + e ⎟ .
2 ⎝
⎠
⎡
⎛ 2x ⎞⎤
2 ⎛x⎞ 1
⎢ Ans. : a cosh ⎜⎝ a ⎟⎠ , 2 a sinh ⎜⎝ a ⎟⎠ ⎥
⎦
⎣
16. Find the sub-tangent, sub-normal,
normal and tangent at the point t on
the curve x = a (t + sin t), y = a (1 –
cos t).
1
⎡
⎤
3 1
⎢ Ans. : a sin t , 2a sin 2 t sec 2 t , ⎥
⎢
⎥
1
1
1
⎢
2a sin t tan t , 2a sin t .⎥
⎢⎣
2
2
2 ⎥⎦
17. Show that the sub-tangent at any
point of the curve x m y n = a m + n varies
as the abscissa.
18. Prove that the length of the subtangent at any point on the hyperbola
xy = c 2 is numerically half the
intercept made by the tangent at that
point on the x-axis.
Differential Calculus II
3.17
3.2.4 Angle between Radius Vector and Tangent
Let P ( r ,q ) be any point on the curve r = f (q ).
Polar co-ordinates can be transformed to rectangular co-ordinates by the relation
y
x = r cos q = f (q ) cos q
y = r sin q = f (q ) sin q
P(r, q )
r f
Let PT be the tangent to the curve at the point
P ( r ,q ). Lety be the angle between tangent PT
and positive x-axis. Let be the angle between
the radius vector OP and tangent PT.
dy dy / dq
Slope of the tangent = tany =
=
dx dx / dq
f ′(q ) sin q + f (q ) cos q
=
f ′(q ) cos q − f (q ) sin q
y
q
O
x
T
Fig. 3.4
Dividing the numerator and denominator by f ′ (q ) cos q ,
tany =
From Fig. 3.4,
tan q + f (q )/ f ′(q )
1 − [ f (q )/ f ′(q ) ] tan q
… (1)
y = q +f
tany = tan (q + f ) =
tan q + tan f
1 − tan f tan q
… (2)
From Eqs (1) and (2),
f (q )
dq
=r
f ′(q )
dr
Corollary: Angle of intersection of curves:
If 1 and 2 are the angles between the common radius vector and the tangents to the
two curves at the point of intersection, then the angle of intersection of two curves is
given by 1 2 .
tan f =
3.2.5 Length of Polar Tangent, Polar Sub-tangent,
Polar Normal and Polar Sub-normal
Let P ( r ,q ) be any point on the curve r = f (q ).
Let PT and PN be the tangent and normal to the curve at the point P. Let NT be a
straight line through the pole O and perpendicular to the radius vector OP. Then OT
and ON are known as the polar sub-tangent and polar sub-normal respectively.
Length of the polar tangent = PT = OP sec f
= r 1 + tan 2 f
⎛ dq ⎞
= r 1+ r2 ⎜
⎝ dr ⎟⎠
2
3.18
Engineering Mathematics
Length of the polar sub-tangent
y
= OT = OP tanf
dq
dr
d
q
= r2
dr
N
= r⋅r
f
P(r, q )
Length of the polar normal
f
r
= PN = OP cosecf
= r 1 + cot 2 f
= r 1+
O
1 ⎛ dr ⎞
⎜
⎟
r 2 ⎝ dq ⎠
⎛ dr ⎞
= r2 + ⎜
⎝ dq ⎟⎠
2
p
f
M
x
T
2
Length of the polar sub-normal
Fig. 3.5
= ON = OP cotf
1 dr
r dq
dr
=
dq
= r⋅
3.2.6 Length of Perpendicular from
Pole to the Tangent
Let p be the length of the perpendicular OM drawn from pole to the tangent PT. From
triangle OPM,
p = r sin f
1
1
1
1
cosec 2f = 2 (1 + cot 2 f )
=
=
p 2 r 2 sin 2 f r 2
r
⎡
1
⎢1 + 2
⎢⎣ r
2
⎛ dr ⎞ ⎤
⎜⎝
⎟⎠ ⎥
dq ⎥⎦
=
1
r2
=
1 1 ⎛ dr ⎞
+ ⎜
⎟
r 2 r 4 ⎝ dq ⎠
2
Pedal Equations
An equation connecting p and r is called pedal equation or p-r equation. The pedal
equation can be obtained by eliminating between polar equation r = f (q ) and equation
1
1 1 dr
= 2+ 4
2
p
r
r d
2
.
Differential Calculus II
3.19
Pedal equation can also be obtained by eliminating and from equations
dq
and p = r sin .
dr
2
2a
q
Example 1: For the curve r =
, prove that (i) f = p - (ii) p = ar .
1 cosq
2
r = f (q ), tan f = r
Solution: (i)
r=
2a
1 cos
=
2a
2sin 2
= a cosec 2
2
2
Differentiating w.r.t. ,
dr
q⎛
q
q⎞ 1
= a ⋅ 2 cosec ⎜ −cosec cot ⎟ ⋅
dq
2⎝
2
2⎠ 2
q
q
q
= −a cosec 2 cot = − r cot
2
2
2
q
q⎞
r
dq
⎛
tan f = r
=
= − tan = tan ⎜ p − ⎟
q
2
2⎠
dr
⎝
−r cot
2
Hence,
(ii)
2
q⎞
q
⎛
p = r sin f = r sin ⎜ p − ⎟ = r sin
2⎠
2
⎝
a
q
p 2 = r 2 sin 2 = r 2 ⋅ = ar
r
2
Example 2: For the curve r = a sin np , prove that
⎛1
⎞
(i) f = tan -1 ⎜ tan nq ⎟
⎝n
⎠
Solution: (i)
(ii) p 2 =
r4
(
r = a sin n
Differentiating w.r.t. ,
dr
= na cos n
d
dθ
a sin nθ
1
=
= tan nθ
dr na cos nθ n
⎛1
⎞
φ = tan −1 ⎜ tan nθ ⎟
⎝n
⎠
tan φ = r
)
n2 a 2 - n2 - 1 r 2
.
3.20
Engineering Mathematics
(ii) We know that,
2
1
1 1 ⎛ dr ⎞
= 2+ 4⎜
⎟
2
p
r
r ⎝ dq ⎠
1
1 1
= + ( na cos nq ) 2
p2 r 2 r 4
=
1 n2 a 2
+ 4 (1 − siin 2 nq )
r2
r
1 n2 a 2 ⎛ r 2 ⎞ 1 n2 2 2
+ 4 ⎜1 − 2 ⎟ = 2 + 4 a − r
r2
r ⎝ a ⎠ r
r
2
2 2
2 2
r +n a −n r
=
r4
r4
.
p2 = 2 2
n a − ( n2 − 1)r 2
(
=
)
Example 3: For the curve r 2 = b 2 sec2q , prove that
(i) y =
p
-q
2
Solution: (i)
(ii) pr = b 2 .
r 2 = b 2 sec 2
Differentiating w.r.t. ,
dr
2r
= 2b 2 sec 2q tan 2q = 2r 2 tan 2q
dq
dr
= r tan 2q
dq
dq
r
⎛p
⎞
= cot 2q = tan ⎜ − 2q ⎟
tan f = r
=
dr r tan 2q
2
⎝
⎠
p
f = − 2q
2
⎛p
⎞ p
y =q + f = q + ⎜ − 2q ⎟ = − q
Now
⎝2
⎠ 2
b2 b2
⎛p
⎞
p = r sin f = r sin ⎜ − 2q ⎟ = r cos 2q = r 2 =
r
r
⎝2
⎠
2
pr = b .
(ii)
Hence,
Example 4: For the parabola
(i) f =
p q
2 2
(ii) p =
1
= 1 + cosq , show that
r
1
q
sec .
2
2
Differential Calculus II
3.21
1
= 1 + cos
r
Solution: (i)
Differentiating w.r.t. ,
−
1 dr
= − sin
r2 d
dr
= r 2 sin
d
d
r
tan = r
= 2
dr r sin
cot
2
2
tan
1
=
r sin
2
1 + cos
=
sin
2 cos 2
=
2sin
2
2
cos
2
2
2
q
cos
1
p
q
⎛
⎞
2 = 1 sec q .
sin ⎜ − ⎟ =
(ii) p = r sin f =
1 + cos q
2
⎝ 2 2 ⎠ 2 cos 2 q 2
2
Example 5: Show that the curves r m = a m cos mq , r m = a m sin mq cut each
other orthogonally.
Solution: For the curve r m = a m cos m ,
Taking logarithm on both the sides,
m log r = m log a + log cos m
Differentiating w.r.t. ,
1 dr
1
( − m sin mq )
=
r dq cos mq
dr
= − r tan mq
dq
dq
r
⎛p
⎞
tan f 1 = r
=
= − cot mq = tan ⎜ + mq ⎟
⎝
⎠
2
dr − r tan mq
p
f 1 = + mq
2
m⋅
For the curve r m = a m sin m ,
Taking logarithm on both the sides,
m log r = m log a + log sin m
3.22
Engineering Mathematics
Differentiating w.r.t. ,
1 dr
1
( m cos mq )
=
r dq sin mq
dr
= r cot mq
dq
dq
r
tan f 2 = r
=
= tan mq
dr r cot mq
f 2 = mq
m⋅
Angle between the curves
m m
2
Hence, the curves cut each other orthogonally.
1
2
2
Example 6: Find the angle of intersection of the curves r = sin q + cos q , and
r = 2sinq .
Solution: The point of intersection is obtained as,
sin + cos = 2sin
cos = sin
tan = 1
=
4
For the curve r = sin q + cos q ,
dr
= cos − sin
d
tan
p⎞
⎛
At the point of intersection ⎜q = ⎟ ,
4⎠
⎝
1
d
sin + cos
=
dr cos
sin
=r
tanf 1= ∞
1
=
2
For the curve r = 2 sinq ,
dr
= 2 cos q
dq
dq 2 sin q
tan f 2 = r
=
= tan q
dr 2 cos q
f 2 =q
p⎞
⎛
At the point of intersection ⎜q = ⎟ ,
4⎠
⎝
f 2=
p
4
Angle of intersection of curves = f 1− f 2 =
p p p
− =
2 4 4
Differential Calculus II
3.23
Example 7: Show that the two curves r 2 = a 2 cos 2q and r = a (1 + cosq ) intersect
1
⎛ 3⎞ 4
at an angle 3sin ⎜ ⎟ .
⎝ 4⎠
Solution: The point of intersection is obtained as,
-1
a 2 cos 2q = a 2 (1 + cos q ) 2
2 cos 2 q − 1 = 1 + 2 cos q + cos 2 q
cos 2 q − 2 cos q − 2 = 0
cos q = 1 ± 3
But, –1 < cos q < 1
Hence,
cos
1 2sin 2
sin 2
sin
2
2
2
2
=
=
1
3
1
3
3
2
3
2
= sin
1
1
2
3
4
3
=
4
1
4
1
4
For the curve r 2 = a 2 cos 2q ,
dr
= −2a 2 sin 2q
dq
dr
a 2 sin 2q
=−
dq
r
r2
a 2 cos 2q
dq
⎛p
⎞
= − cot 2q = tan ⎜ + 2q ⎟
tan f 1 = r
= 2
= 2
dr −a sin 2q −a sin 2q
⎝2
⎠
p
f 1 = + 2q
2
For the curve r = a (1 + cos q ),
2r
dr
= −a sin q
dq
dq a (1 + cos q )
q
⎛p q ⎞
=
tan f 2 = r
= − cot = tan ⎜ + ⎟
dr
−a sin q
2
⎝ 2 2⎠
p q
f2= +
2 2
3.24
Engineering Mathematics
1
4
Hence, angle of intersection of curves = f 1− f 2 = p + 2q − p − q = 3q = 3 sin −1 ⎛⎜ 3 ⎞⎟ .
⎝4⎠
2
2 2
2
Example 8: Show that the curves r 2 cos (2θ - α ) = a 2 sin 2α and r2 = 2a2 sin (2p + `)
cut at right angles at their points of intersection.
Solution: The points of intersection are obtained as,
a 2 sin 2a
= 2a 2 sin ( 2q + a )
cos ( 2q − a )
sin 2a = 2 sin( 2q + a ) cos( 2q − a )
= sin 4q + sin 2a
sin 4q = 0
4q = 0, p , 2p , ……
…
For the curve r 2 cos ( 2q − a ) = a 2 sin 2a ,
Differentiating w.r.t. ,
2r
dr
cos ( 2q − a ) − 2r 2 sin ( 2q − a ) = 0
dq
dr
= r tan ( 2q − a )
dq
dq
r
tan f 1 = r
=
= cot ( 2q − a )
dr r tan( 2q − a )
1
2
(2
tan
2
(2
)
)
For the curve r 2 = 2a 2 sin( 2q + a ),
Differentiating w.r.t. ,
2r
tan f 2 = r
dr
= 4 a 2 cos ( 2q + a )
dq
dr 2a 2
=
cos ( 2q + a )
dq
r
sin ( 2q + a )
dq
r2
= 2
=
= tan ( 2q + a )
dr 2a cos ( 2q + a ) cos ( 2q + a )
f 2 = 2q + a
Angle of intersection
2
At the points of intersection,
1
(2
)
2
(2
)
4
2
p p 3p
, ,
, ……
2 2 2
Hence, the curves cut at right angles at their points of intersection.
f 2 − f 1= −
Differential Calculus II
3.25
Example 9: Find the length of polar sub-tangent and polar sub-normal for the
2a
curve
= 1 - cosq .
r
2a
r
Solution:
1 cos
Differentiating w.r.t. ,
−
Length of the polar sub-tangent = r 2
dq
=
dr
2a dr
= sin q
r 2 dq
dr
r 2 sin q
=−
dq
2a
2a
r2
= −2a cosec q
=−
sin q
r sin q
−
2a
2
Length of the polar sub-normal
dr
r 2 sin q
4a 2
sin q
=
=−
=−
=−
2
dq
2a
(1 − cos q ) 2a
q
q
cos
2
2 = −a cot q cosec 2 q .
q
2
2
4 sin 4
2
2a ⋅ 2 sin
Example 10: Find the length of polar tangent, polar sub-tangent, polar normal
and polar sub-normal to the curve r 2 = a 2 sin 2q .
r 2 = a 2 sin 2
Solution:
Differentiating w.r.t ,
2r
dr
= 2a 2 cos 2
d
dr a 2
= cos 2
d
r
Length of the polar tangent
2
a 4 sin 2 2q
⎛ dq ⎞
= r 1+ r2 ⎜
=
r
1
+
= r 1 + tan 2 2q = r sec 2q = a sin 2 sec 2
⎟
a 4 cos 2 2q
⎝ dr ⎠
Length of the polar sub-tangent
3
3
r
r3
a3 (sin 2q ) 2
dq
=r
= r2 ⋅ 2
= 2
= 2
= a (sin 2q ) 2 sec 2q
dr
a cos 2q a cos 2q
a cos 2q
2
3.26
Engineering Mathematics
dr
d
Length of the polar normal = r 2 +
=
=
2
= r2 +
a 4 cos 2 2
r2
r4
a 4 (1 sin 2 2 )
=
r2
r4
a4
r2
Length of the polar sub-normal =
r4
r4
a4
a 4 sin 2 2
r2
a2
a
=
r
sin 2
=
dr a 2
a 2 cos 2q a cos 2q
=
.
= cos 2q =
dq
r
sin 2q
a sin 2q
Example 11: Find the length of the polar tangent, polar sub-tangent, polar nor-
r 2 - a2
a
- cos -1 .
a
r
mal and polar sub-normal for the curve q =
r2
Solution:
a2
a
cos
1
a
r
Differentiating w.r.t. r,
⎡
⎢
1
1
dq 1 1
⎢
= ⋅
⋅ 2r − ⎢ −
2
2
2
dr a 2 r − a
⎛a⎞
⎢
⎢ 1 − ⎜⎝ r ⎟⎠
⎣
r
a r2
r2
a
a2
r r2
Length of the polar tangent = r 1 + r 2
Length of the polar sub-tangent = r 2
a2
ar r 2
2
d
dr
= r 1+
dq
= r2
dr
⎤
⎥
⎥⎛ a ⎞
⎥ ⎜ − r2 ⎟
⎠
⎥⎝
⎥
⎦
a2
a2
r 2 (r 2 a 2 ) r 2
=
a
a2r 2
r 2 − a2 r 2
=
r − a2
ar
a
2
a2 r 2
⎛ dr ⎞
Length of the polar normal = r 2 + ⎜
= r2 + 2
=
⎟
r − a2
⎝ dq ⎠
Length of the polar sub-normal =
dr
=
dq
ar
r − a2
2
r 2 a2
ar
.
r2
r 2 − a2
Differential Calculus II
3.27
Example 12: Find the length of the polar sub-tangent and polar sub-normal for
the curve r = aeq cot a .
Solution:
r = ae
cot
Differentiating w.r.t. ,
dr
= a cot a eq cot a
dq
Length of the polar sub-tangent
= r2
dq
r2
a 2 e 2q cot a
=
=
= a tan a eq cot a
dr a cot a eq cot a a cot a eq cot a
Length of the polar sub-normal =
dr
= a cot a eq cot a
dq
.
Example 13: Show that in the spiral of Archimides r = aq , the length of the
polar sub-normal is constant and the polar sub-tangent is aq 2 .
r = aq
dr
=a
dq
Solution:
Length of the polar sub-normal =
dr
= a = constant
d
r 2 a 2q 2
d
=
=
= aq 2 .
a
a
dr
Length of the polar sub-tangent = r 2
Exercise 3.2
1. For the curve r = a , prove that
a
(i) cos =
2
a + r2
3. For the curve r 4 = a 4 cos 4 , prove
that
(i)
4
r
.
a + r2
2. For the cardioid r a (1 cos ),
prove that
(ii) p 2 =
(i)
=
2
2
(iii) 2ap 2 = r 3 .
(ii) p = 2a sin 3
2
=
2
+4
(ii) a 4 p = r 5 .
4. For the curve r 3 = a 3 sin 3 , prove
that
(i)
=4
3
4
(ii) pa = r .
3.28
Engineering Mathematics
5. Find the angle between the following
curves:
(i) r = a (1 + cos q ), r = b (1 − cos q )
8. Find the polar sub-tangent to the
curve r 3 = a 3 cos 3 .
[ Ans. : − r cot 3q ]
(ii) r 2 = a 2 cosec 2q , r 2 = b 2 sec 2q
(iii) r (1 + cos q ) = 2a, r (1 − cos q ) = 2b
a
(iv) r = a cos q , r =
2
a
aq
,r=
( v) r =
1+q
(1 + q 2 )
a
( vi) r = a log q , r =
log q
( vii) r = aeq , req = b.
⎡
p
p
p ⎤
(ii)
(iii)
⎥
⎢ Ans. : (i)
2
2
2 ⎥
⎢
p
⎥
⎢
(iv)
(v) tan −1 3
⎥
⎢
3
⎥
⎢
p⎥
⎢
−1 ⎛ 2e ⎞
(vii)
tan
(vi)
⎜⎝ 2 −1 ⎠⎟
⎢
2 ⎥⎦
e
⎣
6. Prove that the length of the polar subtangent for the curve r (1 − cos q ) = 2b
is 2b cosec .
7. Show that for the curve r = a, the
polar sub-tangent is constant and the
r2
polar sub-normal is
.
a
9. Find the length of perpendicular from
the pole on the tangent to the curve
r ( 1) a 2 .
⎡
1
2
2
4
4⎤
a2
= 2 − 3 + 4 − 5 + 6⎥
Ans.
:
⎢
2
p
q
q
q
q
q ⎦
⎣
10. Find the polar sub-tangent for the
l
ellipse = 1 + e cos .
r
length of perpendicular from the pole
to the tangent.
⎡ Ans. :
⎢
⎢(i) l
⎢⎣ e sin
⎤
⎥
1
1 ⎛ 2l
⎞
(ii) 2 = 2 ⎜ − 1 + e 2 ⎟ ⎥
p
l ⎝ r
⎠ ⎥⎦
2
11. Prove that for the curve r = ae m ,
the ratio of polar sub-normal to polar
sub-tangent is proportional to 2 .
12. For the curve r 3 = a 3 cos 3 , show
that the normal at any point to the
curve makes an angle 4 with the
initial line.
3.3 LENGTH OF AN ARC AND ITS DERIVATIVE
3.3.1 Derivative of Arc Length in
Cartesian Form
Let P(x, y) and Q( x + Δx, y + Δy ) be the two neighbouring points on the curve y = f ( x). Let arc AP = s
and arc AQ s
s, where A is a fixed point on the
curve. Then arc PQ
s.
From the right angled triangle PQR,
Q
A
Δy
Δs
P
R
Δx
PQ 2 = PR 2 + RQ 2
= ( Δx ) 2 + ( Δy ) 2
… (1)
M
Fig. 3.6
N
Differential Calculus II
3.29
Dividing by ( x) 2 ,
2
2
2
⎛ PQ ⎞ ⎛ PQ ⎞ ⎛ Δs ⎞
⎛ Δy ⎞
⎜ Δx ⎟ = ⎜ Δs ⎟ ⎜ Δx ⎟ = 1 + ⎜ Δx ⎟
⎝
⎠ ⎝
⎠ ⎝ ⎠
⎝ ⎠
when Q
P,
Also, lim
x
0
Thus as Q
2
… (2)
PQ ⎛ chord PQ ⎞
=⎜
⎟ → 1 and x
Δs ⎝ arc PQ ⎠
s ds
=
and
x dx
y dy
=
x dx
lim
x
0
0
P, Eq. (2) reduces to
2
2
⎛ ds ⎞
⎛ dy ⎞
⎜⎝ ⎟⎠ = 1 + ⎜⎝ ⎟⎠
dx
dx
ds
⎛ dy ⎞
= 1+ ⎜ ⎟
⎝ dx ⎠
dx
2
Similarly, dividing Eq. (1) by ( y ) 2 and taking limits Q
ds
dx
= 1+
dy
dy
P, i.e.,
y
0
2
Corollary: If equation of the curve is given in parametric form
x = x(t ), y = y (t )
ds
dt
ds dx
dx dt
dy
dx
1
=
dx
dt
2
+
2
dx
dt
dy
dt
dx
dt
2
dy
dx
2
dx
dt
2
2
.
3.3.2 Derivative of Arc Length in Polar Form
Let P(r , ) and Q( r + Δr , q + Δq ) be the two neighbouring points on the curve
r = f ( ). Let arc AP = s and arc AQ = s + Δ s, where A is a fixed point on the curve.
Then arc PQ = Δ s.
Draw a perpendicular PN on OQ. From the right-angled triangle PQN,
PQ 2 = PN 2 + NQ 2
= ( r sin Δq ) 2 + (OQ − ON ) 2
= ( r sin Δq ) 2 + ( r + Δr − r cos Δq ) 2
⎡
Δq ⎞ ⎤
⎛
= ( r sin Δq ) 2 + ⎢ Δr + r ⎜ 2 sin 2
⎟
⎝
2 ⎠ ⎥⎦
⎣
2
… (1)
3.30
Engineering Mathematics
y
Q(r +Δr, q +Δq )
N
P(r, q )
r
Δq
A
q
O
x
Fig. 3.7
Rewriting and dividing by (
)2 ,
2
⎛ PQ ⎞
⎛ PQ ⎞
⎜⎝
⎟⎠ = ⎜⎝
⎟
Δ
Δs ⎠
2
⎛ Δs ⎞
⋅⎜
⎝ Δ ⎟⎠
2
Δ
⎛
sin
Δ
Δ
r
sin
⎜
⎛
⎞
2 ⋅ sin Δ
= r2 ⎜
+⎜
+r⋅
⎟
Δ
⎝ Δ ⎠
2
Δ
⎜⎝
2
2
When Q
Also,
Thus as Q
P,
PQ ⎛ chord PQ ⎞
=⎜
⎟ → 1 and
Δs ⎝ arc PQ ⎠
lim
0
sin
s
= 1, lim
0
⎞
⎟
⎟
⎟⎠
2
0
=
ds
r dr
=
and lim
0
d
d
P, Eq. (2) reduces to
ds
d
2
dr
d
= r2 +
ds
dr
= r2 +
d
d
Similarly, dividing Eq. (1) by ( r ) 2 and taking limits Q
ds
d
= 1+ r2
dr
dr
2
.
2
2
P, i.e., Δ r → 0
… (2)
Differential Calculus II
3.31
3.4 CURVATURE
Let P and Q be two neighbouring points on the
curve. Let arc AP = s, where A is a fixed point on
the curve. Lety be the angle made by the tangent at
dy
is called the curvature
P with the x-axis. Then
ds
of the curve at point P . Thus, the curvature is
defined as the rate of turning of the tangent w.r.t.
the arc length.
y
Q
P
A
s
y
x
3.4.1 Radius of Curvature
Fig. 3.8
Cartesian Form
Radius of curvature of the curve at any point is defined as the reciprocal of the curvature at that point and is denoted by .
ds
r=
dy
We know that
dy
= tany
dx
Differentiating w.r.t. x,
d2 y
dy
= sec 2 y
2
dx
dx
dy ds
dy ds
2
= sec y
⋅ = (1 + tan 2 y )
⋅
ds dx
ds dx
⎡ ⎛ dy ⎞ 2 ⎤ 1
= ⎢1 + ⎜ ⎟ ⎥ ⋅
⎢⎣ ⎝ dx ⎠ ⎥⎦ r
1
⎡ ⎛ dy ⎞ 2 ⎤ 2
⎢1 + ⎜ ⎟ ⎥
⎢⎣ ⎝ dx ⎠ ⎥⎦
3
⎡ ⎛ dy ⎞ 2 ⎤ 2
⎢1 + ⎜ ⎟ ⎥
⎢ ⎝ dx ⎠ ⎥⎦
Hence,
r=⎣
d2 y
dx 2
d2 y
Note: The radius of curvature r is positive or negative according as
is positive or
dx 2
negative. This indicates that the curve is either concave or convex.
Since the value of is independent of the choice of axes, interchanging x and y, we get
3
⎡ ⎛ dx ⎞ 2 ⎤ 2
⎢1 + ⎜ ⎟ ⎥
⎢ ⎝ dy ⎠ ⎥⎦
r=⎣
d2 x
dy 2
This formula is useful when
dx
= 0, i.e., the tangent is perpendicular to x-axis.
dy
3.32
Engineering Mathematics
Polar Form
Let r = f ( ) be the curve.
We know that
x = r cos
dx
dr
cos
r sin
d
d
y = r sin
Also,
dy
dr
= sin q
+ r cos q
dq
dq
dr
+ r cos q
sin q
dy dy / dq
d
q
=
=
dr
dx dx / dq
− r sin q
cos q
dq
dr
⎛
⎞
+ r cos q
sin q
d2 y
d ⎜
⎟ dq
d
q
=
⎟ dx
dr
dx 2 dq ⎜
⎜ cos q
− r sin q ⎟
⎝
⎠
dq
2
=
d2r
⎛ dr ⎞
r2 + 2 ⎜ ⎟ − r 2
⎝ dq ⎠
dq
dr
⎛
⎞
− r sin q ⎟
⎜⎝ cos q
⎠
dq
3
3
Now,
⎡ ⎛ dy ⎞ 2 ⎤ 2
⎢1 + ⎜ ⎟ ⎥
⎢⎣ ⎝ dx ⎠ ⎥⎦
r=
d2 y
dx 2
3
2 2
⎡ ⎛
dr
⎞ ⎤
+ r cos q ⎥
⎢ ⎜ sin q
⎟
dq
⎢1 + ⎜
⎟ ⎥
r
d
⎢ ⎜ cos q
− r sin q ⎟ ⎥
⎠ ⎥⎦
⎢⎣ ⎝
dq
=
2
d2r
⎛ dr ⎞
r2 + 2 ⎜
−r 2
⎟
⎝ dq ⎠
dq
3
=
dr
⎛
⎞
− r sin q ⎟
⎜⎝ cos q
⎠
dq
3
2
⎤2
⎡ 2
⎛ dr ⎞
2
2
2
2
(cos q + sin q )⎥
⎢ r (cos q + sin q ) + ⎜
⎝ dq ⎟⎠
⎥⎦
⎢⎣
2
d2r
⎛ dr ⎞
r2 + 2 ⎜
−r 2
⎟
⎝ dq
q⎠
dq
3
⎡ 2 ⎛ dr ⎞ 2 ⎤ 2
⎢r + ⎜
⎟ ⎥
⎝ d ⎠ ⎥⎦
⎢⎣
=
2
d2r .
⎛ dr ⎞
r2 + 2⎜
⎟ −r 2
d
⎝d ⎠
Differential Calculus II
3.33
3.4.2 Newtonian Method to Find Radius of
Curvature at the Origin
If a curve passes through the origin and x-axis is the tangent to the curve at origin, then
the radius of curvature at the origin is given by
x2
x→0 2 y
r = lim
Proof: The curve passes through the origin and x-axis is the tangent to the curve at
the origin,
dy
x = 0, y = 0,
=0
At
dx
0
⎡
⎤
x2
2x
x
lim
= lim
= lim
⎢Indeterminate form 0
⎥
x→0 2 y
x→0
d y x → 0 dy
⎢
⎥
,
,
2
⎣ Applying L Hospital s rule ⎦
dx
dx
1
d2 y
dx 2
1
= lim
[Applying L’Hospital’s rule]
x→0
=
d2 y
dx 2
… (1)
x=0
3
⎡ ⎛ dy ⎞ 2 ⎤ 2
⎢1 + ⎜ ⎟ ⎥
⎢⎣ ⎝ dx ⎠ ⎥⎦
r=
d2 y
dx 2
Now
3
At the origin,
r=
(1 + 0) 2
2
d y
dx 2
=
x =0
1
… (2)
2
d y
dx 2
x =0
From Eqs (1) and (2),
r = lim
x→0
x2
2y
Similarly, if the y-axis is the tangent to the curve at the origin,
y2
x→0 2x
Radius of curvature at the origin can also be found by expanding y in powers of x by
algebraic or trigonometric method. Since the curve passes through the origin, f (0) = 0.
r = lim
Engineering Mathematics
3.34
Let p and q denote the values of
dy
d2 y
and 2 at the origin respectively.
dx
dx
3
2 2
(1 + p )
r (at the origin ) =
q
By Maclaurin’s theorem,
y = f ( x ) = f (0) + xf ′(0) +
= 0+ x⋅ p+
= px +
x2
f ′′(0) + …
2!
x2
⋅ q +…
2!
qx 2
+…
2!
Equating coefficients of suitable powers of x (generally the lowest two), we obtain
equations to determine p and q and hence, is determined at the origin.
Example 1: Find the radius of curvature for the following curves s = f (x )
where x is the angle which the tangent to the curve makes with the x axis.
Solution:
(i) Catenary
Radius of curvature
(ii) Cycloid
Radius of curvature
(iii) Tractrix
Radius of curvature
(iv) Parabola
Radius of curvature
=a
s = c tany
ds
r=
= c sec 2 y
dy
s = 4 a siny
r=
ds
= 4 a cosy
dy
s = c log secy
r=
ds
1
=c
secy tany = c tany
dy
secy
s = a log (tany + secy ) + a tany secy
r=
ds
dy
1
(sec 2 y + secy tany ) + a tany (secy tany ) + a sec3 y
tany + secy
= a secy + a secy (tan 2 y + sec 2 y )
= a secy + a secy (sec 2 y − 1 + sec 2 y )
= 2a sec3 y
Differential Calculus II
s = 8a sin 2
(v) Cardioid
3.35
y
6
Radius of curvature
r=
ds
d ⎛
d ⎡ ⎛
y
y ⎞⎤ 4
2y ⎞
4 a ⎜1 − cos ⎟ ⎥ = a sin
=
⎜ 8a sin
⎟⎠ =
⎢
⎠
⎝
dy d y ⎝
d
3
3
y
6
3 ⎦
⎣
Example 2: Find the radius of curvature of the parabola y 2 = 4ax at any point
(x, y).
y 2 = 4 ax
Solution:
Differentiating w.r.t. x,
2y
dy
= 4a
dx
dy 2 a
=
dx
y
… (1)
Differentiating Eq. (1) w.r.t. x,
d2 y
2a dy
4a 2
=− 2
=− 3
2
dx
y dx
y
3
3
⎡ ⎛ dy ⎞ 2 ⎤ 2 ⎛ 4a 2 ⎞ 2
3
⎢1 + ⎜ ⎟ ⎥
⎜1 + 2 ⎟
y ⎠
⎢⎣ ⎝ dx ⎠ ⎥⎦
( y 2 + 4a 2 ) 2
⎝
=
=
=−
d2 y
4a 2
4a 2
− 3
2
dx
y
3
3
3
3
(4ax + 4a 2 ) 2
( 4a ) 2 ( x + a ) 2
2( x + a ) 2
=−
=
−
=
−
.
4a 2
4a 2
a
Example 3: Find the radius of curvature at any point of catenary y = c cosh
y = c cosh
Solution:
x
c
Differentiating w.r.t. x,
dy
x
= sinh
dx
c
Differentiating again w.r.t. x,
d2 y 1
x
= cosh
2
c
c
dx
3
3
⎡ ⎛ dy ⎞ 2 ⎤ 2
2
⎛
2 x⎞
⎢1 + ⎜ ⎟ ⎥
⎜1 + sinh ⎟
⎢⎣ ⎝ dx ⎠ ⎥⎦
c⎠
=
=⎝
x
1
d2 y
cosh
c
c
dx 2
x
.
c
Engineering Mathematics
3.36
x
x
c
=
= c cosh 2
1
x
c
cosh
c
c
y2
= .
c
cosh 3
Example 4: Find the radius of curvature of the Folium x 3 + y 3 = 3axy at the
3a 3a ⎞
point ⎛⎜ ,
.
2
2 ⎟
⎝
Solution:
⎠
x 3 + y 3 = 3axy
… (1)
Differentiating Eq. (1) w.r.t. x,
dy
dy
= 3ay + 3ax
dx
dx
d
y
d
y
x2 + y 2
= ay + ax
dx
dx
dy ay − x 2
=
dx y 2 − ax
3x 2 + 3 y 2
⎛ 3a 3a ⎞
dy
At the point ⎜ ,
= −1
⎟,
2 ⎠
dx
⎝ 2
Differentiating Eq. (2) w.r.t. x,
2
d2 y
dy
dy
d2 y
⎛ dy ⎞
2x + 2 y ⎜ ⎟ + y2 2 = a
+a
+ ax 2
dx
dx
dx
dx
⎝ dx ⎠
3a 3a ⎞
At the point ⎛⎜ ,
⎟,
2 ⎠
⎝ 2
2
2
2
⎛ 3a ⎞ d y
⎛ 3a ⎞
⎛ 3a ⎞
⎛ 3a ⎞ d y
=
(
−
)
+
(
−
)
+
a
a
a
2 ⎜ ⎟ + 2 ⎜ ⎟ (−1) 2 + ⎜ ⎟
1
1
⎜ ⎟ 2
2
⎝ 2 ⎠ dx
⎝ 2 ⎠
⎝ 2 ⎠
⎝ 2 ⎠ dx
d2 y
32
=−
2
3a
dx
3a 3a ⎞
At the point ⎛⎜ ,
⎟,
2 ⎠
⎝ 2
3
⎡ ⎛ dy ⎞ 2 ⎤ 2
3
⎢1 + ⎜ ⎟ ⎥
⎢⎣ ⎝ dx ⎠ ⎥⎦
[1 + (−1) 2 ] 2
3a
=
=
=−
32
d2 y
8
2
−
2
a
3
dx
… (2)
Differential Calculus II
3.37
Example 5: Find the radius of curvature of the curve x 2 y = a ( x 2 + y 2 ) at
( -2a , 2a ).
Solution:
x 2 y = a( x 2 + y 2 )
… (1)
Differentiating Eq. (1) w.r.t. x,
2 xy + x 2
dy
dy
= 2ax + 2ay
dx
dx
dy 2ax − 2 xy
= 2
dx
x − 2ay
dy
=∞
dx
Hence, differentiating Eq. (1) w.r.t. y,
At the point ( 2a, 2a ),
2 xy
dx
dx
+ x 2 = 2ax + 2ay
dy
dy
… (2)
dx 2ay − x 2
=
dy 2 xy − 2ax
At the point ( 2a, 2a ),
dx
=0
dy
Differentiating Eq. (2) w.r.t. y,
2
2
⎛ dx ⎞
⎛ dx ⎞
d2 x
dx
dx
d2 x
2 xy 2 + 2 x + 2 y ⎜ ⎟ + 2 x
= 2a ⎜ ⎟ + 2ax 2 + 2a
dy
dy
dy
dy
⎝ dy ⎠
⎝ dy ⎠
At the point ( 2a, 2a ),
2(−2a )(2a )
d2 x
d2 x
= 2a (−2a ) 2 + 2a
2
dy
dy
d2 x
2a
1
=
=−
2a
dy 2 4 a 2 − 8 a 2
At the point ( 2a, 2a ) ,
3
⎡ ⎛ dx ⎞ 2 ⎤ 2
⎢1 + ⎜ ⎟ ⎥
⎢ ⎝ dy ⎠ ⎥⎦
=⎣
d2 x
dy 2
= −2a.
Example 6: Find the radius of curvature of the curve
(x2 + y2)2 – 2ax (x2 + y2) – a3 y = 0 at the point (2a, 0).
3.38
Solution:
Engineering Mathematics
( x 2 + y 2 ) 2 − 2ax ( x 2 + y 2 ) − a 3 y = 0
… (1)
Differentiating Eq. (1) w.r.t. x,
dy
dy ⎞
dy ⎞
⎛
⎛
2( x 2 + y 2 ) ⎜ 2 x + 2 y ⎟ − 2ax ⎜ 2 x + 2 y ⎟ − 2a ( x 2 + y 2 ) − a 3
=0
dxx
dx ⎠
dx ⎠
⎝
⎝
At the point (2a, 0),
dy
2(4a 2 )(4a ) − 4a 2 (4a ) − 2a (4a 2 ) − a 3
=0
dx
dy
=8
dx
… (2)
Differentiating Eq. (2) w.r.t. x,
2
2
2
⎡
⎡
d2 y
d2 y
dy ⎞
⎛ dy ⎞ ⎤
⎛ dy ⎞ ⎤
⎛
2( x 2 + y 2 ) ⎢ 2 + 2 y 2 + 2 ⎜ ⎟ ⎥ + 2 ⎜ 2 x + 2 y ⎟ − 2ax ⎢ 2 + 2 y 2 + 2 ⎜ ⎟ ⎥
⎝
⎝ dx ⎠ ⎦
⎝ dx ⎠ ⎦
dx
dx
dx ⎠
⎣
⎣
2
dy ⎞
d y
dy ⎞
⎛
⎛
−2a ⎜ 2 x + 2 y ⎟ − 2a ⎜ 2 x + 2 y ⎟ − a3 2 = 0
⎝
⎠
⎝
⎠
dx
dx
dx
At the point (2a, 0),
2 ( 4 a 2 ) ( 2 + 128) + 2 (16 a 2 ) − 4 a 2 ( 2 + 128) − 8a 2 − 8a 2 − a3
d2 y
=0
dx 2
d 2 y 536
=
a
dx 2
At the point (2a, 0),
3
⎡ ⎛ dy ⎞ 2 ⎤ 2
3
3
⎢1 + ⎜ ⎟ ⎥
⎢⎣ ⎝ dx ⎠ ⎥⎦
(1 + 64) 2 (65) 2 a
=
.
=
=
536
536
d2 y
a
dx 2
Example 7: Find the radius of curvature of the curves x = a log (sec q + tan q ),
y = a sec q .
Solution:
x = a log(secq + tan q )
dx
1
=a
(sec q tan q + sec 2 q ) = a secq
dq
(sec q + tan q )
y = a secq
dy
= a secq taanq
dq
dy dy / dq
a secq tan q
=
=
= tan q
dx dx / dq
a secq
d2 y
dq sec 2 q 1
2
= secq
=
sec
q
=
dx a secq a
dx 2
Differential Calculus II
3.39
3
⎡ ⎛ dy ⎞ 2 ⎤ 2
3
3
⎢1 + ⎜ ⎟ ⎥
(1 + tan 2 θ ) 2 (sec 2 θ ) 2
⎝ dx ⎠ ⎦
⎣
=
=
ρ=
1
1
d2 y
sec
sec θ
θ
2
a
a
dx
= a sec 2 θ .
⎛
Example 8: Find the radius of curvature of the curve x = a ⎜ t ⎝
Solution:
⎛ t3 ⎞
x = a ⎜t − ⎟
3⎠
⎝
t3 ⎞
2
⎟ , y = at .
3⎠
y = at 2
dx
= a (1 − t 2 )
dt
dy
= 2at
dt
2at
2t
dy dy / dt
=
=
=
dx dx / dt a (1 − t 2 ) 1 − t 2
d 2 y 2 (1 − t 2 ) − 2t ( −2t ) dt
=
dx
(1 − t 2 ) 2
dx 2
=
2 (1 + t 2 )
1
2 (1 + t 2 )
⋅
=
2 2
2
(1 − t ) a (1 − t ) a (1 − t 2 )3
3
⎡ ⎛ dy ⎞ 2 ⎤ 2
⎢1 + ⎜ ⎟ ⎥
⎢⎣ ⎝ dx ⎠ ⎥⎦
r=
=
d2 y
dx 2
3
⎡ ⎛ 2t ⎞ 2 ⎤ 2
⎢1 + ⎜
2 ⎟ ⎥
⎢⎣ ⎝ 1 − t ⎠ ⎥⎦
2 (1 + t 2 )
a (1 − t 2 )3
3
3
[(1 − t 2 ) 2 + 4t 2 ] 2 ⋅ a (1 − t 2 )3 a [(1 + t 2 ) 2 ] 2
=
=
2 (1 + t 2 )
(1 − t 2 )3 ⋅ 2 (1 + t 2 )
a
= (1 + t 2 ) 2 .
2
Example 9: Find the radius of curvature of the curve x = e t + e t, y = e t - e t at
t = 0.
Solution:
x = et + e − t
y = e t – e–t
dx
= et − e − t
dt
dy
= et + e − t
dt
dy dy / dt e t + e − t x
=
=
=
dx dx / dt e t − e − t y
d 2 y 1 x dy 1 x x y 2 − x 2
= −
= − ⋅ =
dx 2 y y 2 d x y y 2 y
y3
Engineering Mathematics
3.40
3
3
⎡ ⎛ dy ⎞ 2 ⎤ 2 ⎛ x 2 ⎞ 2
3
⎢1 + ⎜ ⎟ ⎥
⎜⎝1 + y 2 ⎟⎠
⎢⎣ ⎝ dx ⎠ ⎥⎦
( y2 + x2 )2
r=
= 2
=
y − x2
d2 y
( y2 − x2 )
3
2
y
dx
At t = 0, y = 0, x = 2
3
42
r=
= −2 .
−4
Example 10: Find the radius of curvature at any point of the curve x = a cos 3p ,
y = a sin 3p .
Solution:
x = a cos3 q
dx
= −3a cos 2 q sin q
dq
y = a sin 3 q
dy
= 3a sin 2 q cos q
dq
dy dy / d q
3a sin 2 q cos q
=
=
= − tan q
dx dx / dq −3a cos 2 q sin q
d2 y
dq
= − sec 2 q
2
dx
dx
− sec 2 q
=
−3a cos 2 q sin q
1
=
4
3a cos q sin q
3
⎡ ⎛ dy ⎞ 2 ⎤ 2
3
⎢1 + ⎜ ⎟ ⎥
⎢⎣ ⎝ dx ⎠ ⎥⎦
(1 + tan 2 q ) 2
r=
=
= 3a cos q sin q .
1
d2 y
3a cos 4 q sin q
dx 2
Example 11: For the cycloid x = a (q + sin q ), y = a (1 - cos q ), prove that
q
r = 4 a cos .
2
y = a(1 − cos q )
Solution:
x = a(q + sin q )
dy
dx
= a sin q
= a(1 + cos q )
dq
dq
q
q
2 sin cos
dy dy / dq
a sin q
2
2 = tan q
=
=
=
2
dx dx / dq a(1 + cos q )
2q
2 cos
2
Differential Calculus II
d 2 y 1 2 q dq 1 2 q
= sec
= sec ⋅
2 dx 2
2
dx 2 2
1
2a cos 2
q
2
3.41
1
=
4 a cos 4
q
2
3
3
⎡ ⎛ dy ⎞ 2 ⎤ 2
2
⎛
2q ⎞
⎢1 + ⎜ ⎟ ⎥
⎜⎝1 + tan ⎟⎠
⎢⎣ ⎝ dx ⎠ ⎥⎦
2 = 4 a cos q .
r=
=
2
1
2
d y
4 q
dx 2
4 a cos
2
Example 12: Find the radius of curvature of the curve
3a
x=
(sinh u cosh u + u), y = a cosh3 u.
2
3a
Solution:
y = a cosh3 u
x=
(sinh u cosh u + u )
2
dx 3a
dy
= (cosh 2 u + sinh 2 u + 1)
= 3a cosh 2 u sinh u
du 2
du
3a
=
⋅ 2 cosh 2 u
2
= 3a cosh 2 u
dy dy /du 3a cosh 2 u sinh u
=
=
= sinh u
dx dx /du
3a cosh 2 u
cosh u
1
d2 y
du
=
=
= cosh u
2
2
dx 3a cosh u 3a cosh u
dx
3
⎡ ⎛ dy ⎞ 2 ⎤ 2
3
⎢1 + ⎜ ⎟ ⎥
⎢⎣ ⎝ dx ⎠ ⎥⎦
(1 + sinh 2 u ) 2
r=
=
= 3a coshh 4 u .
1
d2 y
3a cosh u
dx 2
Example 13: Find the radius of curvature of the cardioid r = a (1 + cos q ).
Solution:
r = a (1 + cos q )
Differentiating w.r.t. ,
dr
= − a sin q
dq
Differentiating again w.r.t. ,
d2r
= − a cos q
dq 2
3.42
Engineering Mathematics
3
⎡ 2 ⎛ dr ⎞ 2 ⎤ 2
⎢r + ⎜
⎥
⎝ dq ⎟⎠ ⎥⎦
⎢⎣
r=
2
d2r
⎛ dr ⎞
−
r2 + 2 ⎜
r
⎝ dq ⎟⎠
dq 2
3
[a 2 (1 + coss q ) 2 + a 2 sin 2 q ] 2
= 2
a (1 + cos q ) 2 + 2a 2 sin 2 q + a 2 cos q (1 + cos q )
3
2
=
[2a 2 (1 + cos q )]
=
a 2 + 2a 2 + 3a 2 cos q
=
4a
q
cos .
3
2
⎡ 2
⎢⎣ 2a
3
2
⎛
2 q ⎞⎤
⎜⎝ 2 cos ⎟⎠ ⎥
2 ⎦
q⎞
⎛
3a 2 ⎜ 2 cos 2 ⎟
⎝
2⎠
Example 14: Find the radius of curvature of the curve r = aep cot` .
r = ae
Solution:
cot a
Differentiating w.r.t. q,
dr
= a cot α eθ cot α
dθ
Differentiating again w.r.t. q,
d2r
= a cot 2 α eθ cot α
dθ 2
3
⎡ 2 ⎛ dr ⎞ 2 ⎤ 2
⎢r + ⎜
⎟ ⎥
⎝ dθ ⎠ ⎥⎦
⎢⎣
ρ=
2
d2r
⎛ dr ⎞
r2 + 2⎜
⎟ −r 2
dθ
⎝ dθ ⎠
3
[a 2 e 2θ cott α + a 2 cot 2 α e 2θ cot α ] 2
= 2 2θ cot α
ae
+ 2a 2 cot 2 α e 2θ cot α − a 2 cot 2 α e 2θ cot α
3
[a 2 e 2θ cot α (1 + cot 2 α )] 2
= 2 2θ cot α
ae
(1 + cot 2 α )
a 3 e3θ cot α cosec 3α
a 2 e 2θ cot α cosec 2α
= aeθ cot α cosecα
= r cosec α .
=
Differential Calculus II
Example 15: Find the radius of curvature of the curve r =
r=
Solution:
3.43
a
.
p
a
q
Differentiating w.r.t. q,
dr
a
=− 2
dq
q
Differentiating again w.r.t. q,
d 2 r 2a
=
dq 2 q 3
3
3
⎡ 2 ⎛ dr ⎞ 2 ⎤ 2
⎛ a2 a2 ⎞ 2
⎢r + ⎜
⎥
⎟
⎜⎝ q 2 + q 4 ⎟⎠
⎝ dq ⎠ ⎥⎦
⎢⎣
= 2
r=
2
a
a 2 2a 2
d2r
⎛ dr ⎞
2
+
2
−
−
r + 2⎜
r
2
q4 q4
⎝ dq ⎟⎠
dq 2 q
3
⎛ a2 r 2 ⎞ 2
3
3
⎜⎝ q 2 + q 2 ⎟⎠
(a2 + r 2 ) 2 r(a2 + r 2 ) 2
=
=
=
.
a 2q
a3
a2
q2
Example 16: Find the radius of curvature of the curve r m = a m cos mp .
Solution:
r m = a m cos mq
Taking logarithm on both the sides,
m log r = m log a + log cos mq
Differentiating w.r.t. q,
1 dr
1
m⋅ ⋅
=
( − m sin mq )
r dq cos mq
dr
= − r tan mq
dq
Differentiating again w.r.t. q,
d2r
dr
= − r ⋅ m ⋅ sec 2 mq −
tan mq
dq
dq 2
= − mr sec 2 mq + r tan 2 mq
Engineering Mathematics
3.44
3
⎡ 2 ⎛ dr ⎞ 2 ⎤ 2
⎢r + ⎜
⎥
⎝ dq ⎟⎠ ⎥⎦
⎢⎣
r=
2
d2r
⎛ dr ⎞
−r 2
r2 + 2 ⎜
⎟
⎝ dq ⎠
dq
3
( r 2 + r 2 tann 2 mq ) 2
= 2
r + 2r 2 tan 2 mq + mr 2 sec 2 mq − r 2 tan 2 mq
r 3 sec3 mq
r
r
am
=
=
=
=
.
( m + 1) r 2 sec 2 mq ( m + 1) cos mq
r m ( m + 1)r m −1
( m + 1) m
a
Example 17: Prove that the radius of curvature at any point of the curve
r 2 cos2p = a 2 is –
r3
.
a2
Solution:
r 2 cos 2q = a 2
Taking logarithm on both the sides,
2 log r + log cos 2q = log a 2
Differentiating w.r.t. ,
1 dr
1
+
( −2 sin 2q ) = 0
r dq cos 2q
dr
= r tan 2q
dq
2⋅
Differentiating again w.r.t. ,
d2 r
dr
= 2r sec 2 2q +
tan 2q
dq
dq 2
= 2r sec 2 2q + r tan 2 2q
3
⎡ 2 ⎛ dr ⎞ 2 ⎤ 2
⎢ r + ⎜⎝
⎟ ⎥
dq ⎠ ⎦
⎣
r=
2
d2r
⎛ dr ⎞
r2 + 2 ⎜
−r 2
⎟
⎝ dq ⎠
dq
3
( r 2 + r 2 tan 2 2q ) 2
= 2
r + 2r 2 tan 2 2q − r ( r tan 2 2q + 2r sec 2 2q )
3
( r 2 sec 2 2q ) 2
r2
=
= − r sec 2q = − r ⋅ 2
2
2
− r sec 2q
a
r3
=− 2.
a
Differential Calculus II
3.45
Example 18: Prove that at the points in which the Archimedean spiral r = ap
intersects the hyperbolical spiral rp = a, their curvatures are in the ratio 3:1.
Solution: For the curve r = aq,
dr
=a
dq
d2r
=0
dq 2
2
Curvature
k1 =
d2r
⎛ dr ⎞
−r 2
r2 + 2 ⎜
⎟
⎝ dq ⎠
dq
3
=
⎡ 2 ⎛ dr ⎞ 2 ⎤ 2
⎢r + ⎜
⎥
⎝ dq ⎟⎠ ⎥⎦
⎢⎣
a 2q 2 + 2a 2 − 0
3
( a 2q 2 + a 2 ) 2
=
q2 +2
3
a (q 2 + 1) 2
For the curve rq = a ,
r=
a
q
dr
a
=− 2
dq
q
d 2 r 2a
=
dq 2 q 3
2
d2r
⎛ dr ⎞
−r 2
r + 2⎜
⎟
⎝ dq ⎠
dq
2
Curvature
k2 =
3
a2
a 2 2a 2
+
2
−
2
q4
q4 q4 =
=q
3
3
⎛ a2 a2 ⎞ 2
a (q 2 + 1) 2
⎜⎝ q 2 + q 4 ⎟⎠
⎡ 2 ⎛ dr ⎞ 2 ⎤ 2
⎢r + ⎜
⎥
⎝ dq ⎟⎠ ⎥⎦
⎢⎣
The points of intersection of two curves are obtained as,
a
q
2
q =1
q = ±1
aq =
At q = ±1,
3
k1 =
3
a ⋅ 22
1
k2 =
3
a 22
k1 3
=
k2 1
Hence, curvatures are in the ratio 3:1.
Engineering Mathematics
3.46
Example 19: Find the radius of curvature at the origin for the curve
x 3 - 2 x 2 y - 4 y 3 + 5 x 2 - 6 xy + 7 y 2 - 8 y = 0.
Solution: Equating the lowest degree term in the equation to zero, we get y = 0,
i.e., x-axis which is tangent to the curve at the origin.
x2
= lim
x →0 2 y
2 = lim
Hence,
x
0
x2
y
Dividing the equation by y,
x⋅
When x
0, y
x2
x2
− 2x2 − 4 y2 + 5 − 6x + 7 y − 8 = 0
y
y
0.
0 ( 2 ) − 2 ( 0) − 4 ( 0) + 5 ( 2 ) − 6 ( 0 ) + 7 ( 0 ) − 8 = 0
4
= .
5
Example 20: Find the radius of curvature at the origin for the curve
x 3 y − xy 3 + 2 x 2 y − 2 xy 2 + 2 y 2 − 3 x 2 + 3 xy − 4 x = 0.
Solution: Equating the lowest degree term in the equation to zero, we get x = 0,
i.e., y-axis which is tangent to the curve at the origin.
y2
= lim
x →0 2 x
2 = lim
x →0
y2
x
Dividing the equation by x,
x 2 y − y 3 + 2 xy − 2 y 2 +
When y
0, x
2 y2
− 3x + 3 y − 4 = 0
x
0,
0 − 0 + 0 − 0 + 2 (2 ) − 0 + 0 − 4 = 0
=1
Example 21: Find the radius of curvature at the origin for the curve
y = 2 x + 3 x 2 - 2 xy + y 2 .
Solution: Equating the lowest degree terms in the equation to zero, we get
y 2 x 0 which is tangent to the curve at the origin.
Differential Calculus II
Substituting y = px + q
3.47
x2
+ … in the equation of the curve,
2
⎛
⎞ ⎛
x2
x2
x2
px + q + … = 2 x + 3 x 2 − 2 x ⎜ px + q + … ⎟ + ⎜ px + q +
2
2
2
⎝
⎠ ⎝
⎞
⎟
⎠
2
x and x2,
p=2
q
= 3 − 2 p + p2
2
= 3− 4+ 4 = 3
q=6
and
3
3
(1 + p 2 ) 2 (1 + 4) 2 5 5
r=
=
=
.
q
6
6
Example 22: Find the radius of curvature at the origin for the curve
y 2 - 3 xy - 4 x 2 + x 3 + x 4 y + y 5 = 0.
Solution: Equating the lowest degree terms in the equation to zero, we get
y 2 − 3 xy − 4 x 2 = 0 which, indicates that there are two tangents at the origin. Hence,
there will be two values of at the origin.
Substituting y = px + q
x2
+ … in the equation of the curve,
2
2
⎛
⎞
⎛
⎞
⎞
x2
x2
x2
2
3
4 ⎛
…
3
…
4
px
+
q
+
−
x
px
+
q
+
−
x
+
x
+
x
px
+
q
+ …⎟
⎜⎝
⎟
⎜
⎟
⎜
2
2
2
⎠
⎝
⎠
⎝
⎠
5
⎛
⎞
x2
+ ⎜ px + q + …⎟ = 0
2
⎝
⎠
x 2 and x 3 ,
and
From Eq. (1), p = 4,
From Eq. (2),
When p = 4, q = −
When p = −1, q =
2
5
2
5
1
p2 − 3 p − 4 = 0
… (1)
3
pq − q + 1 = 0
2
… (2)
Engineering Mathematics
3.48
2
For p = 4 and q = − ,
5
3
3
(1 + p 2 ) 2 (1 + 16) 2
85
r=
=
=−
17
2
q
2
−
5
For p =
1 and q =
2
,
5
3
3
(1 + p 2 ) 2 (1 + 1) 2
r=
=
= 5 2.
2
q
5
Example 23: Find the radius of curvature at the origin for the curve
x = a (q + sin q ), y = a (1 - cos q ).
Solution: For the cycloid, x = a(q + sin q ), y = a(1 − cos q ), x-axis is the tangent
at the origin.
x2
x→0 2 y
r = lim
a 2 (q + sin q ) 2
q → 0 2a (1 − cos q )
= lim
[∵q = 0 at the originn ]
a
(q + sin q ) 2
⎡0⎤
lim
⎢⎣ 0 ⎥⎦
→
0
q
2
1 − cos q
a
2(q + sin q )(1 + cos q )
= lim
→
q
0
2
sin q
=
[Applying L’Hospital’s rule]
a
⎛ q
⎞
lim 2 ⎜
+ 1⎟ (1 + cos q )
⎠
2 q → 0 ⎝ sin q
a
= ( 2) (1 + 1)(1 + 1)
2
= 4 a.
=
Exercise 3.3
1. Find the radius of curvature of the
following curves:
(i) xy = c 2
a a
(ii) x + y = a at ⎛⎜ , ⎞⎟
⎝4 4⎠
x2 y 2
(iii) 2 + 2 = 1 at (a, 0) and (0, b)
a
b
(iv) x3 + y3 = 2a3 at (a, a)
(v) xy 2 = a 3 − x 3 at (a, 0)
(vi) y =
log x
at x = 1
x
(vii) y = e x at (0, 1)
Differential Calculus II
(viii) x 3
xy 2
6 y2
0 at (3, 3).
3
⎤
⎡
2
2 2
⎢ Ans.: (i) ( x + y ) (ii) a ⎥
⎢
2c 2
2 ⎥
⎥
⎢
2
2
b
a
a ⎥
⎢
(iii)
,
(iv)
⎢
a
b
2 ⎥⎥
⎢
⎢
3a
2 2⎥
( v) −
( vi)
⎥
⎢
3 ⎥
2
⎢
3
⎥
⎢
( vii) 2 2
( viii) 5 2 ⎥⎦
⎢⎣
2. Find the radius of curvature of the
following curves:
(i) x = 1 − t 2 , y = t − t 3 at t = ±1
(ii) x = log t , y =
1 ⎛ 1⎞
⎜t + ⎟
2⎝ t⎠
(iii) x = a (q − sin q ),
y = a (1 − cos q )
(iv) x = a sin 2t (1 + cos 2t ),
y = a cos 2t (1 − cos 2t )
(v) x = 3a cos q − a cos 3q ,
y = 3a sin q − a sin 3q
(vi) x = 3 + 4 cos t ,
y = 4 + 3 sin t at (3, 7)
(vii) x = a( 2 cos q + cos 2q ),
y = a( 2 sin q − sin 2q )
(viii) x = a cos q , y = a sin q .
q
q
1
⎡
⎤
(ii) (1 + t 2 ) 2 ⎥
⎢ Ans.: (i) 2 2
4
⎢
⎥
⎢
(iii) − 4a sin
(iv) 4a cos 3t ⎥
⎢
⎥
2
⎢
⎥
16
⎢
⎥
( v) 3a sin
( vi) −
⎢
⎥
3
⎢
3 ⎥
⎢
3
a (1 + 2 ) 2 ⎥
( vii) 8a sin
( viii)
⎢
⎥
4
2
⎣
⎦
3.49
3. Show that the radius of curvature at any
point of curve x = aeq (sin q − cos q ),
y = aeq (sin q + cos q ) is twice the
distance of the tangent at the point
from the origin.
4. Show that the radius of curvature at
each point of the curve
t⎞
⎛
x = a ⎜ cos t + log tan ⎟ , y = a sin t
⎝
2⎠
is inversely proportional to the length
of the normal intercepted between the
point on the curve and the x-axis.
5. Find the radius of curvature of the
following curves:
(i) r = e 2q at q = log 2
n
n
(ii) r = a sin nq
(iii) r = tanq at q =
3p
4
(iv) r (1 + cos q ) = 2a
r 2 − a2
a
− cos −1
a
r
(vi) r = a(1 − cos q )
(v) q =
q
=b
2
q
r cos = a .
2
(vii) r cos 2
(viii)
⎡
⎤
an
(iii) 5 ⎥
⎢ Ans.: (i) 16 (ii)
n −1
(1 + n)r
⎢
⎥
⎢
⎥
r
2
⎢(iv) 2r
( v) r 2 − a 2 ( vi)
2ar ⎥
a
3
⎢
⎥
⎢
⎥
⎢( vii) 2r ( viii) 2r r
⎥
⎢⎣
⎥⎦
a
b
6. Find the radius of curvature at the
origin for the following curves:
(i) x 4 − y 4 + x 3 − y 3 + x 2 − y 2 + y = 0
Engineering Mathematics
3.50
(viii) 9a 2 x 2 = 4 y 2 ( y − 3a ) 2
(ii) ( x 2 − y 2 )( x 2 + y 2 )
+( x − y )( x 2 + xy + y 2 )
+ ( x − y )( x + y ) + y = 0
(ix) x 3 + y 3 − 3axy = 0
(x) x 2 − 4 xy − 2 y 2 + 10 x + 4 y = 0.
(iii) y = x 4 − 4 x 3 − 18 x 2
(iv) y = x 3 + 5 x 2 + 6 x
4
1
1
17
⎡
⎤
⎢ Ans.: (i) 5 (ii) − 2 (iii) 36 (iv) 10 37 ⎥
⎢
⎥
⎢( v) 1 ( vi) 3 ( vii) ± 2a 2 ( viii) 15 a 5 ⎥
⎢
⎥
4
⎢
⎥
3
29
⎢(ix) a ( x)
⎥
29
⎢⎣
⎥⎦
2
6
(v) x y − xy + 2 x y
3
3
2
+ xy − y 2 + 2 x = 0
(vi) x 3 + 3 x 2 y − 4 y 3 + y 2 − 6 x = 0
2
2
3
(vii) a ( y − x ) = x
3.5 CENTRE AND CIRCLE OF CURVATURE
Let P(x, y) be any point on the curve
y = f ( x). Let the tangent at point P make an
with the x-axis. Let C be the point
angle
on the positive direction of the normal to the
curve at point P such that CP = . Then C
is called the centre of curvature to the curve
at P. The circle with centre C and radius r is
called the circle of curvature at P.
Let C(X, Y) be the centre of curvature.
y
From Fig. 3.9,
O
C(X, Y)
Q
T
y
M
y
r
P(x, y)
N
Fig. 3.9
X = OM = ON − MN
= ON − QP
= x − r siny
3
⎡ ⎛ dy ⎞ 2 ⎤ 2
dy
⎢1 + ⎜ ⎟ ⎥
⎢⎣ ⎝ dx ⎠ ⎥⎦
dx
= x−
⋅
2
d2 y
⎛ dy ⎞
2
1
+
⎟
⎜
dx
⎝ dx ⎠
= x−
dy ⎡ ⎛ dy ⎞
⎢1 + ⎜ ⎟
dx ⎢⎣ ⎝ dx ⎠
d2 y
dx 2
2
⎤
⎥
⎥⎦
dy ⎤
⎡
⎢⎣∵ tany = dx ⎥⎦
x
Differential Calculus II
3.51
Y = MC = MQ + CQ
= y + r cosy
3
⎡ ⎛ dy ⎞ 2 ⎤ 2
⎢1 + ⎜ ⎟ ⎥
⎢⎣ ⎝ dx ⎠ ⎥⎦
= y+
d2 y
dx 2
1
⎛ dy ⎞
1+ ⎜ ⎟
⎝ dx ⎠
2
dy ⎤
⎡
⎢⎣∵ tany = dx ⎥⎦
2
⎛ dy ⎞
1+ ⎜ ⎟
⎝ dx ⎠
= y+
.
d2 y
dx 2
The equation of the circle of curvature of the curve at the point P (x, y) with radius and
centre C (X, Y ) is given by,
( x − X ) 2 + ( y − Y ) 2 = r 2.
3.6 EVOLUTE
The locus of the centres of curvature of the curve is called the evolute of the curve.
The evolute of any curve y = f ( x) is obtained by eliminating x and y from the equation
2
2
dy ⎡ ⎛ dy ⎞ ⎤
⎛ dy ⎞
1
+
⎢ ⎜ ⎟ ⎥
1
+
⎜
⎟
dx ⎢⎣ ⎝ dx ⎠ ⎥⎦
⎝ dx ⎠
, Y = y+
y = f ( x), X = x −
2
2
y
d
d y
2
dx 2
dx
Example 1: Show that the circle of curvature at the origin of the curve
x2
y = mx +
is x 2 + y 2 = a (1 + m2 )( y - mx ).
a
y = mx +
Solution:
x2
a
Differentiating w.r.t. x,
dy
2x
= m+
dx
a
Differentiating again w.r.t. x,
d2 y 2
=
dx 2 a
At the origin,
dy
d2 y 2
= m,
=
dx
dx 2 a
Engineering Mathematics
3.52
Let (X, Y) be the centre of curvature at the origin.
X = x−
dy ⎡ ⎛ dy ⎞
⎢1 + ⎜ ⎟
dx ⎢⎣ ⎝ dx ⎠
2
⎤
⎥
⎥⎦
d2 y
dx 2
= 0−
m (1 + m 2 )
ma (1 + m 2 )
=−
2
2
a
2
⎛ dy ⎞
1+ ⎜ ⎟
⎝ dx ⎠
1 + m 2 a (1 + m 2 )
=
0
+
=
Y = y+
2
2
d2 y
2
a
dx
At the origin,
3
⎡ ⎛ dy ⎞ 2 ⎤ 2
3
⎢1 + ⎜ ⎟ ⎥
⎢⎣ ⎝ dx ⎠ ⎥⎦
(1 + m 2 ) 2
r=
=
2
d2 y
2
a
dx
Hence, the equation of the circle of curvature at the origin is given by,
( x − X )2 + ( y − Y )2 = r 2
2
2
⎡
ma (1 + m 2 ) ⎤ ⎡
a (1 + m 2 ) ⎤
a 2 (1 + m 2 )3
⎢x +
⎥ + ⎢y −
⎥ =
2
2
4
⎣
⎦ ⎣
⎦
x 2 + y 2 = − ma (1 + m 2 ) x + a (1 + m 2 ) y
= a (1 + m 2 ) ( y − mx ).
Example 2: Find the centre and circle of curvature of the curve
⎛a a⎞
, ⎟.
⎝4 4⎠
at the point ⎜
x+ y = a
Solution:
Differentiating w.r.t. x,
1 − 12 1 − 12 dy
x + y
=0
2
2
dx
1
dy
y2
=− 1
dx
x2
Differentiating again w.r.t. x,
1
2
d y
=−
dx 2
x2
1
1 − 12 dy
1 −1
y
− y2 x 2
2
2
dx
x
x+
y= a
Differential Calculus II
⎛a a⎞
At the point ⎜ , ⎟ ,
⎝4 4⎠
3.53
dy
= −1
dx
1
d2 y
=−
dx 2
⎛ a ⎞2 1 ⎛ a ⎞
⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠
4 2 4
−
1
2
1
⎛ a ⎞2 1 ⎛ a ⎞
( −1) − ⎜ ⎟ ⎜ ⎟
⎝4⎠ 2 ⎝4⎠
a
4
−
1
2
=
4
a
a a
Let (X, Y ) be the centre of curvature at the point ⎛⎜ , ⎞⎟ .
⎝4 4⎠
2
dy ⎡ ⎛ dy ⎞ ⎤
⎢1 + ⎜ ⎟ ⎥
dx ⎢⎣ ⎝ dx ⎠ ⎥⎦ a ( −1)(1 + 1) 3a
= −
=
X = x−
4
4
4
d2 y
a
dx 2
2
⎛ dy ⎞
1+ ⎜ ⎟
⎝ dx ⎠
a 1 + 1 3a
= +
=
Y = y+
2
4
4
4
d y
a
dx 2
⎛a a⎞
At the point ⎜ , ⎟ ,
⎝4 4⎠
3
⎡ ⎛ dy ⎞ 2 ⎤ 2
3
⎢1 + ⎜ ⎟ ⎥
⎢⎣ ⎝ dx ⎠ ⎥⎦
(1 + 1) 2
a
r=
=
=
2
4
d y
2
a
dx 2
a a
Hence, the equation of the circle of curvature at the point ⎛⎜ , ⎞⎟ is given by,
⎝4 4⎠
( x − X )2 + ( y − Y )2 = r 2
2
2
3a ⎞ ⎛
3a ⎞
a2
⎛
⎜⎝ x − ⎟⎠ + ⎜⎝ y − ⎟⎠ = .
4
4
2
Example 3: Find the centre and circle of curvature of the curve
y = x3 – 6x2 + 3x + 1 at the point (1, –1).
Solution:
y = x 3 − 6 x 2 + 3x + 1
Differentiating w.r.t. x,
dy
= 3 x 2 − 12 x + 3
dx
3.54
Engineering Mathematics
Differentiating again w.r.t. x,
d2 y
= 6 x − 12
dx 2
At the point (1, 1),
dy
d2y
= −6 ,
= −6
dx
dx 2
Let (X, Y ) be the centre of curvature at the point (1, 1).
X = x−
dy ⎡ ⎛ dy ⎞
⎢1 + ⎜ ⎟
dx ⎢⎣ ⎝ dx ⎠
2
2
d y
dx 2
⎤
⎥
⎥⎦
= 1−
( −6)(1 + 36)
= −36
−6
2
⎛ dy ⎞
1+ ⎜ ⎟
⎝ dx ⎠
1 + 36
43
= −1 +
=−
Y = y+
2
−6
6
d y
dx 2
At the point (1, –1),
3
⎡ ⎛ dy ⎞ 2 ⎤ 2
3
3
⎢1 + ⎜ ⎟ ⎥
⎢⎣ ⎝ dx ⎠ ⎥⎦
(1 + 36) 2
(37) 2
r=
=
=−
−6
6
d 2y
2
dx
Hence, the equation of the circle of curvature at the point (1, 1) is given by
( x − X )2 + ( y − Y )2 =
2
2
43 ⎞
(37)3
⎛
( x + 36) 2 + ⎜ y + ⎟ =
.
⎝
6 ⎠
36
Example 4: Show that the parabolas y = - x 2 + x + 1 and x = - y 2 + y + 1 have
the same circle of curvature at the point (1, 1).
Solution: For the parabola y = − x 2 + x + 1,
Differentiating w.r.t. x,
dy
= −2 x + 1
dx
Differentiating again w.r.t. x,
d2 y
= −2
dx 2
At the point (1, 1),
dy
d2 y
= −1,
= −2
dx
dx 2
Differential Calculus II
3.55
Let (X, Y ) be the centre of curvature at the point (1, 1).
X = x−
2
dy ⎡ ⎛ dy ⎞ ⎤
⎢1 + ⎜ ⎟ ⎥
dx ⎢⎣ ⎝ dx ⎠ ⎥⎦
2
d y
dx 2
= 1−
( −1)(1 + 1)
=0
−2
2
⎛ dy ⎞
1+ ⎜ ⎟
⎝ dx ⎠
(1 + 1)
= 1+
=0
Y = y+
−2
d2 y
dx 2
At the point (1, 1),
3
⎡ ⎛ dy ⎞ 2 ⎤ 2
3
⎢1 + ⎜ ⎟ ⎥
⎡⎣1 + ( −1) 2 ⎤⎦ 2
⎢⎣ ⎝ dx ⎠ ⎥⎦
r=
=
=− 2
−2
d2 y
dx 2
Hence, the equation of the circle of curvature at the point (1, 1) is given by,
( x − X )2 + ( y − Y )2 =
2
( x − 0) 2 + ( y − 0) 2 = ( 2 ) 2
x2 + y 2 = 2
For the parabola x = − y 2 + y + 1,
Differentiating w.r.t. y,
dx
= −2 y + 1
dy
Differentiating again w.r.t. y,
d2 x
= −2
dy 2
At the point (1, 1),
d2 x
dx
= −2
= −1,
dy 2
dy
Let (X, Y ) be the centre of curvature at the point (1, 1).
X = x−
dx ⎡ ⎛ d x ⎞
⎢1 +
dy ⎢⎣ ⎜⎝ dy ⎟⎠
d2 x
dy 2
2
2
⎤
⎥
⎥⎦
= 1−
( −1)(1 + 1)
=0
−2
⎛ dx ⎞
1+ ⎜ ⎟
⎝ dy ⎠
(1 + 1)
Y = y+
= 1+
=0
2
−2
d x
dy 2
Engineering Mathematics
3.56
At the point (1, 1),
3
⎡ ⎛ dx ⎞ 2 ⎤ 2
3
⎢1 + ⎜ ⎟ ⎥
⎢⎣ ⎝ dy ⎠ ⎥⎦
[1 + ( −1) 2 ]2
r=
=
=− 2
−2
d 2x
dy 2
Hence, the equation of the circle of curvature at the point (1, –1) is given by,
( x − X 2 ) + ( y − Y )2 = r 2
( x − 0) 2 + ( y − 0) 2 = ( − 2 ) 2
x2 + y2 = 2
2
2
Thus, the parabolas y = − x + x + 1 and x = − y + y + 1 have the same circle of
curvature at the point (1, 1).
Example 5: Find the evolute of the parabola y 2 = 4ax .
y 2 = 4 ax
Solution:
Differentiating w.r.t. x,
2y
dy
= 4a
dx
dy 2a
=
dx
y
Differentiating again w.r.t. x,
d2 y
2 a dy
4a 2
=− 2
=− 3
2
dx
y dx
y
Let (X, Y ) be the centre of curvature.
X = x−
2
dy ⎡ ⎛ dy ⎞ ⎤
+
1
⎢
⎥
dx ⎢⎣ ⎜⎝ dx ⎟⎠ ⎥⎦
d2 y
dx 2
2a ⎛ 4 a 2 ⎞
⎜1 + 2 ⎟
y ⎝
y ⎠
= x−
2
4a
− 3
y
y 2 + 4 a 2 2ax + 4 ax + 4 a 2
=
2a
2a
= 3 x + 2a
= x+
2
2
4a
⎛ dy ⎞
1+ 2
1+ ⎜ ⎟
y
d
x
Y = y + ⎝2 ⎠ = y +
dy
4a2
− 3
2
dx
y
= y−
y( y 2 + 4 a 2 )
y3
=
−
4a2
4a2
… (1)
Differential Calculus II
3
3.57
3
( 4 ax ) 2
2x 2
=−
=
−
1
4a 2
a2
… (2)
From Eq. (1),
X − 2a
3
x=
From Eq. (2),
Y2 =
4 x 3 4 ⎛ X − 2a ⎞
= ⎜
a
a ⎝ 3 ⎟⎠
3
27aY 2 = 4( X − 2a)3
This is the required evolute of the parabola y 2 = 4 ax.
x2 y2
+
= 1.
a 2 b2
Example 6: Find the evolute of the ellipse
x2 y2
+
=1
a2 b2
Solution:
Differentiating w.r.t. x,
2 x 2 y dy
+
=0
a 2 b 2 dx
dy
b2 x
=− 2
dx
a y
Differentiating again w.r.t. x,
dy ⎞
⎛
y−x ⎟
d2 y
b2 ⎜
dx
=− 2⎜
⎟
dx 2
a ⎜ y2 ⎟
⎜
⎟
⎝
⎠
=−
=−
⎛ b2 x ⎞⎤
b2 ⎡
b2 ⎛ b2 x 2 + a2 y 2 ⎞
y
−
x
−
=
−
⎢
⎥
⎜
⎟
⎜
⎟
2
a2 y 2 ⎣
a2 y 2 ⎝
a2 y
⎝ a y ⎠⎦
⎠
b2 ⎛ a2b2
⎜
a2 y 2 ⎝ a2 y
⎞
b4
⎟= − 2 3
a y
⎠
Let (X, Y ) be the centre of curvature.
2
dy ⎡ ⎛ dy ⎞ ⎤
b2 x ⎛ b4 x 2 ⎞
⎢1 + ⎜ ⎟ ⎥
− 2 ⎜1 + 4 2 ⎟
dx ⎢⎣ ⎝ dx ⎠ ⎥⎦
a y⎝ a y ⎠
X = x−
= x−
2
d y
b4
− 2 3
2
dx
a y
x
x
( a 4 y 2 + b 4 x 2 ) = x − 4 2 [a 2 b 2 ( a 2 − x 2 ) + b 4 x 2 ]
2
ab
ab
a2 − b2 3
x
=
a4
= x−
4
… (1)
Engineering Mathematics
3.58
2
b4 x 2
⎛ dy ⎞
1+ 4 2
1+ ⎜ ⎟
a y
dx
Y = y + ⎝2 ⎠ = y +
b4
d y
− 2 3
2
a y
dx
y
y
( a 4 y 2 + b 4 x 2 ) = y − 2 4 [a 4 y 2 + b 2 a 2 (b 2 − y 2 )]
4
ab
ab
b2 − a2 3
… (2)
=
y
b4
= y−
2
From Eq. (1),
1
⎛ a4 X ⎞ 3
x=⎜ 2
2 ⎟
⎝ a −b ⎠
From Eq. (2),
1
⎛ b 4Y ⎞ 3
y=⎜ 2
2 ⎟
⎝b −a ⎠
Substituting in equation of the ellipse,
2
2
1 ⎛ a 4 X ⎞ 3 1 ⎛ b 4Y ⎞ 3
⎜
⎟ + 2⎜ 2
⎟ =1
a2 ⎝ a2 − b2 ⎠
b ⎝ b − a2 ⎠
2
2
2
( aX ) 3 + (bY ) 3 = ( a 2 − b 2 ) 3
This is the required evolute of the ellipse
x2 y2
+
= 1.
a2 b2
Example 7: Find the evolute of the astroid x = a cos 3 q , y = a sin 3 q .
Solution:
x = a cos3 q
dx
= −3a cos 2 q sin q
dq
y = a sin 3 q
dy
= 3a sin 2 q cos q
dq
dy dy /d q
3a sin 2 q cos q
=
=
= − tan q
dx d x /d q
−3a cos 2 q sin q
sec 2 q
sec 4 q cosec q
d2 y
dq
2
=
−
sec
q
=
−
=
3a
dx
−3a cos 2 q sin q
dx 2
Differential Calculus II
3.59
Let (X, Y ) be the centre of curvature.
X = x−
dy ⎡ ⎛ dy ⎞
⎢1 +
dx ⎢⎣ ⎜⎝ dx ⎟⎠
2
⎤
⎥
⎥⎦
d 2y
dx 2
3a tan q (1 + tan 2 q )
= a cos3 q +
sec 4 q cosec q
= a cos3 q + 3a sin 2 q cos q
… (1)
2
⎛ dy ⎞
1+ ⎜ ⎟
dx
Y = y + ⎝2 ⎠
d y
dx 2
3a(1 + tan 2 q )
= a sin 3 q +
sec 4 q cosec q
= a sin 3 q + 3a cos 2 q sin q
… (2)
Adding Eqs (1) and (2),
X + Y = a (cosq + sin q )3
1
1
( X + Y ) 3 = a 3 (cosq + sin q )
… (3)
Subtracting Eqs (2) from (1),
X − Y = a(cosq − sin q )3
1
1
( X − Y ) 3 = a 3 (cosq − sin q )
… (4)
Squaring and adding Eqs (3) and (4),
2
2
2
( X + Y ) 3 − ( X − Y ) 3 = 2a 3
This is the required evolute of the astroid x = a cos 2 q , y = a sin 3 q .
Example 8: Find the evolute of the curve x = a (cos p + p sinp ),
y = a (sin p – p cosp ).
Solution:
x = a(cosq + q sin q )
dx
= a [− sin q + (q cos q + sin q )] = aq cos q
dq
y = a (sin q − q cos q )
dy
= a [cosq
q − ( −q sin q + cos q )] = aq sin q
dq
dy d y / dq
aq sin q
=
=
= tan q
dx d x /dq
aq cos q
d 2y
dq
1
sec 2 q
2
=
=
=
sec
q
2
dx aq cos q aq cos3 q
dx
Engineering Mathematics
3.60
Let (X, Y ) be the centre of the curvature.
X = x−
dy ⎡ ⎛ dy ⎞
⎢1 +
dx ⎢⎣ ⎜⎝ dx ⎟⎠
d2 y
dx 2
2
⎤
⎥
⎥⎦
= a(cosq + q sin q ) − tan q (1 + tan 2 q )( aq cos3 q )
= a cos q
… (1)
2
⎛ dy ⎞
1+ ⎜ ⎟
dx
Y = y + ⎝2 ⎠
d y
dx 2
= a(sin q − q cos q ) + (1 + tan 2 q ) ( aq cos3 q )
= a sinq
… (2)
From Eqs (1) and (2),
X 2 + Y 2 = a2
This is the equation of circle which is required evolute of the curve.
t⎞
⎛
Example 9: Show that the evolute of the tractrix x = a ⎜ cos t + log tan ⎟ ,
⎝
2⎠
x
y = a sin t is the catenary y = a cosh .
a
t⎞
⎛
Solution:
x = a ⎜ cos t + log tan ⎟
y = a sin t
2⎠
⎝
dy
⎛
⎞
= a cos t
⎜
dx
1
1 2t⎟
dt
= a ⎜ − sin t +
⋅ sec ⎟
t 2
dt
2⎟
⎜⎜
tan
⎟
⎝
2
⎠
⎛
⎞
⎜
⎟
1
= a ⎜ − sin t +
⎟
t
t
⎜⎜
2 sin cos ⎟⎟
⎝
2
2⎠
1 ⎞
⎛
= a ⎜ − sin t +
sin t ⎟⎠
⎝
=a
cos 2 t
sin t
dy dy /dt
a cos t
=
=
= tan t
dx dx /dt
cos 2 t
a
sin t
d 2y
sec 2 t sin t
sin t
2 dt
= sec t
=
=
2
2
dx
dx
a cos t
a cos 4 t
Differential Calculus II
3.61
Let (X, Y ) be the centre of curvature.
X = x−
2
dy ⎡ ⎛ dy ⎞ ⎤
+
1
⎢
⎥
dx ⎢⎣ ⎜⎝ dx ⎟⎠ ⎥⎦
2
dy
dx 2
= x−
tan t (1 + tan 2 t )
sin t
a coss 4 t
t⎞
⎛
= a ⎜ cos t + log tan ⎟ − a cos t
2⎠
⎝
t
= a log tan
2
… (1)
2
⎛ dy ⎞
1+ ⎜ ⎟
1 + tan 2 t
a cos 2 t
dx
Y = y + ⎝2 ⎠ = y +
= a sin t +
sin t
sinn t
d y
4
2
a cos t
dx
a
=
sin t
From Eqs (1) and (2),
tan
X
t
=ea
2
and
2 tan
But
sin t =
sin t =
a
Y
t
2
1 + tan 2
t
2
X
a
2e a
=
2X
Y
1+ e a
2X
⎛
a ⎜1+ e a
Y = ⎜
X
2⎜
⎝ ea
X
= a cosh
h
a
⎞
X
X
⎟ = a ⎛ e− a + e a ⎞
⎜
⎟
⎟⎟ 2
⎝
⎠
⎠
x
Hence, the required evolute of the tactrix is y = a cosh .
a
Example 10: Show that the evolute of the cycloid x = a(p – sin p ),
y = a (1 – cosp ) is another cycloid x = a(q + sin q ), y = - a(1 - cos q ) .
Solution:
x = a(q − sin q )
dx
= a(1 − cos q )
dq
y = a (1 – cos q )
dy
= a sin q
dq
… (2)
Engineering Mathematics
3.62
dy dy/ dq
a sin q
=
=
=
dx dx / dq a(1 − cos q )
q
q
cos
2
2 = cot q
q
2
2 sin 2
2
2 sin
d 2y
1
q dq
= − cosec 2
2
2
2 dx
dx
q ⎡
1
1
⎤
= − cosec2 ⎢
=−
2
2 ⎣ a(1 − cos q ) ⎥⎦
cosec 4
q
2
4a
Let (X, Y ) be the centre of curvature.
X = x−
2
dy ⎡ ⎛ dy ⎞ ⎤
+
1
⎢
⎥
dx ⎢⎣ ⎜⎝ dx ⎟⎠ ⎥⎦
d2 y
dx 2
cot
= a(q − sin q ) +
q⎛
2q ⎞
⎜1 + cot ⎟ 4 a
2⎠
2⎝
q
cosec 4
2
q
q
sin 2
2
2
q
q
= a(q − sin q ) + 4 a sin cos
2
2
= a(q − sin q ) + 2a sin q
= a(q + sin q )
= a(q − sin q ) + 4 a cot
2
⎛ dy ⎞
⎛
2q ⎞
1+ ⎜ ⎟
⎜1 + cot ⎟ 4 a
dx ⎠
⎝
⎝
2⎠
Y = y+
= a(1 − cos q ) −
d2 y
4 q
cos ec
2
dx 2
q
= a(1 − cos q ) − 4a sin 2
2
= a(1 − cos q ) − 2a(1 − cos q )
= −a + a cos q
= −a(1 − cos q )
Hence, the required evolute of the cycloid is x = a (q + sin q ), y = – a (1 – cos q ).
Exercise 3.4
1. Find the centre of curvature of the
following curves:
⎛1 1⎞
(i) y = x 3 at ⎜ 2 , 8 ⎟
⎝
⎠
2
(ii) y = x + 9 at (3, 6)
x
3
3
(iii) x + y = 2a3 at (a, a)
Differential Calculus II
x2 y2
+
= 2 at (3, 2)
9
4
3
3
⎛ 3a 3a ⎞
(v) x + y = 3axy at ⎜ , ⎟
⎝ 2 2 ⎠
x
(vi) y = c cosh at (x, y)
c
2
(vii) xy = c at (c, c)
(viii) y 3 = a 2 x at (x, y)
(ix) x = 3t , y t 2 6 at (a, b)
(x) x (1 at ) cos t a sin t ,
y (1 at ) sin t a cos t.
(iv)
⎤
⎡ Ans. :
⎥
⎢
⎛ 7 31 ⎞
⎛ 15 ⎞
(ii) ⎜ 3, ⎟
⎥
⎢ ( i) ⎜ ,
⎟
⎝ 64 48 ⎠
⎝ 2⎠
⎥
⎢
⎥
⎢
a a
⎛ 5 −5 ⎞
⎥
⎢ (iii) ⎛⎜ , ⎞⎟
(iv) ⎜ ,
⎟
⎥
⎢
⎝6 4 ⎠
⎝2 2⎠
⎥
⎢
⎥
⎢ ( v) ⎛ 21a , 21a ⎞
⎜ 16 16 ⎟
⎥
⎢
⎝
⎠
⎥
⎢
2
2
⎛
⎞
⎥
⎢ ( vi) ⎜ x − y y − c , 2 y ⎟
⎥
⎢
⎜
⎟
c
⎝
⎠
⎥
⎢
⎥
⎢ ( vii) ( 2c, 2c)
⎥
⎢
4
4
4
5
⎛
y a y + 9y ⎞
⎥
⎢ ( viii) ⎜ a + 15
,
⎟
2
4
2a
⎥
⎢
⎝ 6a y
⎠
⎥
⎢
2
⎡
⎤
(
81
+
4
a
)
⎢ (ix) ⎢ −4 a( 20 + a 2 ), b +
⎥
⎥
⎢
18
⎣
⎦⎥
⎢ (x) ( a sin t , − a cos t )
⎥
⎥⎦
⎣⎢
2. Find n so that the centre of curvature
of the curve y = x n at the point (1, 1)
lies on the line y = 2.
[Ans. : n = −1 ]
3. Show that the centre of curvature
at the point P ( a, a) of the curve
x 4 + y 4 = 2a 2 xy divides the line OP
in the ratio 6:1, O being the origin of
co-ordinates.
4. Find the co-ordinates of the centre of
curvature at the origin for the curve
5
x 4 − ax 2 y − axy 2 + a 2 y 2 = 0.
2
⎡
⎛ a ⎞⎤
⎢ Ans. : (0, a) and ⎜ 0, 4 ⎟ ⎥
⎝
⎠⎦
⎣
3.63
5. Find the circle of curvature of the
following curves:
(i) 2 xy + x + y = 4 at (1, 1)
(ii) y 2 = 4 ax at (at2, 2at)
(iii) x 3 + y 3 = 2 xy at (1, 1)
(iv) y = x 3 + 2 x 2 + x + 1 at (0, 1)
(v) xy(x + y) = 2 at (1, 1).
⎡ Ans. :
2
2
⎢
⎢ (i) ⎛⎜ x − 5 ⎞⎟ + ⎛⎜ y − 5 ⎞⎟ = 9
2⎠ ⎝
2⎠
2
⎢
⎝
⎢ (ii) x 2 + y 2 − 6 at 2 x − 4 ax + 4 at 3 y
⎢
= 3a 2 t 4
⎢
2
2
2
⎢
⎛
⎞
⎢ (iii) ⎛ x − 7 ⎞ + ⎛ y − 7 ⎞ = ⎜ 2 ⎟
⎜
⎢
8 ⎟⎠ ⎜⎝
8 ⎟⎠ ⎜⎝ 8 ⎟⎠
⎝
⎢
2
2
⎢ (iv) x + y + x − 3 y + 2 = 0
2
2
⎢⎣ (v) x + y + 5 x − 5 y + 8 = 0
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦
6. Find the evolute of the following
curves:
2
2
(i) x − y = 1
a2 b2
(ii) 2 xy = a 2
3
(iii) x = 4 ay
(iv) x 2 − y 2 = a 2
(v) x = (1 − aq ) cos q + a sin q ,
y = (1 − aq ) sin q − a cos q
(vi) x = a cosh u, y = b sinh u
(vii) x = a cot 2 q , y = 2a cotq .
⎡ Ans. :
⎤
⎢
2
2
2 ⎥
⎢ (i) ( aX ) 3 − (bY ) 3 = ( a 2 + b 2 ) 3 ⎥
⎢
⎥
2
2
2
⎢
⎥
3
3
3
⎢ (ii) ( X + Y ) + ( X − Y ) = 2a ⎥
⎢ (iii) 4(Y − 2a)3 = 27aX 2
⎥
⎢
⎥
2
2
2
⎢
⎥
⎢ (iv ) X 3 − Y 3 = ( 2a) 3
⎥
⎢
⎥
2
2
2
⎢ ( v) X + Y = a
⎥
2
2
2⎥
⎢
2
2
⎢ ( vi) ( aX ) 3 − (bY ) 3 = ( a + b ) 3 ⎥
⎢
⎥
2
3
⎣ ( vii) 27aY = 4( X − 2a)
⎦
Engineering Mathematics
3.64
7. Show that the evolute of the deltoid
x = 2 cost + cos 2t, y = 2 sint – sin 2t
is another deltoid three times the
size of the given deltoid and has the
equations.
X = 3( 2 cos t − cos 2t ),
Y = 3( 2 sin t + sin 2t ) .
8. Show that the evolute of an equian-
gular spiral is an equal equiangular
spiral.
9. Show that the radii of curvatures of
the curve x = axq (sin q − cos q ),
y = aeq (sin q + cos q ) and its evolutes at corresponding points are
equal.
10. Prove that normals to a curve are the
tangents to its evolute.
3.7 ENVELOPES
Consider an equation f ( x, y, a ) = 0. For different values of a we get different curves.
This equation represents a one parameter family of curves with a as the parameter.
Example:
1
(i) The equation y = mx +
represents a family of straight lines where m is the
m
parameter.
(ii) The equation x 2 + y 2 − 2ax = 0 represents a family of circles with their centres
on x-axis and which pass through the origin. Here, a is the parameter.
In a similar way, a two parameter family of curves is represented by the equation
f ( x, y, a , b ) = 0 where and b are the parameters.
Example:
x2 y2
The equation 2 + 2 = 1 represents a family of ellipse, where a and b are the
a
b
parameters.
Envelope of a given family of curves is a curve which touches each member of a
family of curves and at each point is touched by some member of the family of curves.
A family of curve may have no envelope or unique envelope or several envelopes.
Envelope may also be defined as the locus of the limiting positions of the points of
intersection of one member of the family with a neighbouring member when one of them
tends to coincide with the other which is kept fixed.
Determination of envelope
(i) The equation of the envelope of the family of curves f ( x, y, a ) = 0, where a
is the parameter, is obtained by eliminating
between the equations
f ( x, y, a ) = 0
where
and
∂
f ( x, y, a ) = 0
∂a
f
is the partial derivative of f w.r.t. a .
a
Differential Calculus II
3.65
(ii) For two parameter family of curves f ( x, y, a , b ) = 0 with relation g(a , b ) = 0
between the parameters a and b , the equation of the envelope of the family
of curves is obtained by
(a) Writing one of the parameter, say b , in terms of a .
(b) Using this to reduce the equation of two parameter family of curves into
one parameter family of curves.
(c) Proceeding as in step (i).
(iii) Envelope of the family of normals to a given curve is the evolute of the curve.
Example 1: Find the envelope of the family of lines y = mx + am3 , where m is
the parameter.
y = mx + am3
Solution:
Differentiating partially w.r.t. m,
… (1)
0 = x + 3 am 2
1
⎛ x ⎞2
m = ⎜− ⎟
⎝ 3a ⎠
… (2)
Substituting in the Eq. (1),
1
3
⎛ x ⎞2
⎛ x ⎞2
y = ⎜− ⎟ x + a⎜− ⎟
⎝ 3a ⎠
⎝ 3a ⎠
3
⎛ x ⎞
⎛ x ⎞
⎛ x ⎞
y 2 = ⎜ − ⎟ x 2 + a 2 ⎜ − ⎟ + 2ax ⎜ − ⎟
⎝ 3a ⎠
⎝ 3a ⎠
⎝ 3a ⎠
2
27ay 2 = −9 x 3 + ( − x )3 + 6 x 3
27ay 2 = −4 x 3
This is the equation of the required envelope.
Example 2: Find the envelope of the family of lines y = mx + a 2 m2 + b 2 ,
where m is the parameter.
Solution:
y = mx + a 2 m 2 + b 2
… (1)
( y − mx ) 2 = a 2 m 2 + b 2
y 2 + m 2 x 2 − 2mxy = a 2 m 2 + b 2
( x 2 − a 2 )m 2 − 2mxy + ( y 2 − b 2 ) = 0
Differentiating partially w.r.t. m,
2m( x 2 − a 2 ) − 2 xy = 0
m=
xy
x − a2
2
… (2)
… (3)
Engineering Mathematics
3.66
Substituting in Eq. (2),
( x 2 − a2 )
x2 y2
x2 y2
−
2
+ ( y 2 − b2 ) = 0
( x 2 − a2 )2
x 2 − a2
x2 y2
2x2 y2
− 2
+ ( y 2 − b2 ) = 0
2
2
x −a
x − a2
x2 y2
= y 2 − b2
x 2 − a2
x 2 y 2 = ( x 2 − a 2 )( y 2 − b 2 ) = x 2 y 2 − b 2 x 2 − a 2 y 2 + a 2 b 2
x2 y2
+
=1
a2 b2
This is the equation of the required envelope.
x
y
Example 3: Find the envelope of the family of curves cos t + sin t = 1, where
a
b
t is the parameter.
Solution:
x
y
cos t + sin t = 1
a
b
… (1)
Differentiating partially w.r.t. t,
x
y
− sin t + cos t = 0
a
b
Squaring Eqs (1) and (2) and adding,
… (2)
2
2
y
y
⎞ ⎛ x
⎞
⎛x
⎜ a cos t + b sin t ⎟ + ⎜ − a sin t + b cos t ⎟ = 1
⎝
⎠ ⎝
⎠
x2
y2
(cos 2 t + sin 2 t ) + 2 (sin 2 t + cos 2 t ) = 1
2
a
b
x2 y2
+
=1
a2 b2
This is the equation of the required envelope.
Example 4: Find the envelope of the family of circles
x2 + y2 – 2ax cos ` – 2ay sin ` = c2, where a is the parameter.
Solution:
x 2 + y 2 − 2ax cos a − 2ay sin a = c 2
2ax cos a + 2ay sin a = x 2 + y 2 − c 2
Differentiating partially w.r.t. a ,
−2ax sin a + 2ay cos a = 0
Squaring Eqs (1) and (2) and adding,
4a 2 ( x 2 + y 2 ) = ( x 2 + y 2 − c 2 )2
This is the equation of the required envelope.
… (1)
… (2)
Differential Calculus II
3.67
Example 5: Find the envelope of the system of straight lines 2 y - 3tx + at 3 = 0 ,
where t is the parameter.
2 y − 3tx + at 3 = 0
Solution:
… (1)
Differentiating partially w.r.t. t,
−3 x + 3at 2 = 0
x
t2 =
a
Substituting in Eq. (1),
2y – 3t x + at
Substiuting in Eq. (1),
x
=0
a
y
t=
x
y3
⎛ y⎞
2y − 3⎜ ⎟ x + a 3 = 0
⎝x⎠
x
ay 2 = x 3
This is the equation of the required envelope.
Example 6: Find the envelope of family of parabolas
x
+
a
y
= 1, where the
b
parameters a and b are connected by the relation a + b = c.
Solution:
x
y
+
=1
a
b
a+b = c
But
x
y
+
=1
a
c−a
Differentiating w.r.t. a,
… (1)
⎛ 1⎞ 1
⎛ 1⎞ 1
x ⎜− ⎟ 3 + y ⎜− ⎟
(−1) = 0
3
⎝ 2⎠ 2
⎝ 2⎠
2
(a)
(c − a )
3
1
⎛ c − a ⎞2 ⎛ y ⎞2
⎜
⎟ =⎜ ⎟
⎝ a ⎠
⎝x⎠
1
c − a ⎛ y ⎞3
=⎜ ⎟
a
⎝x⎠
1
1
c x3 + y3
=
1
a
x3
1
a=
cx 3
1
1
x3 + y3
… (2)
Engineering Mathematics
3.68
Substituting Eq. (2) in Eq. (1),
1
⎡ ⎛ 13
⎢x ⎜ x + y3
1
⎢ ⎜
⎢⎣ ⎝ cx 3
(
1
⎞⎤ 2
1
⎟⎥ + ⎡ y ⎧
⎢
⎪
1
⎥
⎟
3
⎠ ⎥ ⎢ ⎨⎪ ⎛
⎦ ⎢ ⎜ c − cx
1
1
⎪
⎢ ⎪ ⎜⎝
3
3
x
y
+
⎣ ⎩
)
1
(
1
⎫⎤ 2
⎥ =1
⎞ ⎪⎪⎥
⎟ ⎬⎥
⎟ ⎪⎥
⎠ ⎪⎭⎦
)
1
1
1
1
⎡ 32 13
⎤2 ⎡ 2 1
⎤2
⎢x x + y3 ⎥ + ⎢ y3 x3 + y3 ⎥ = c2
⎣
⎦ ⎣
⎦
1
1 2
⎛ 13
⎞
⎜ x + y3 ⎟
⎝
⎠
1
⎡ 2 12
2 2⎤
1
⎛
⎞
⎛
⎞
⎢ x3 + y3 ⎥ = c2
⎜
⎟
⎜
⎟
⎢
⎥
⎢⎣⎝ ⎠ ⎝ ⎠ ⎥⎦
1
1
1
x3 + y3 = c3
This is the equation of the required envelope.
Example 7: Find the envelope of the family of ellipses
x2 y2
+
= 1 , where the
a 2 b2
parameters a and b are connected by the relation ab = c2.
Solution:
x2 y2
+
=1
a2 b2
ab = c 2
But
x 2 a2 y 2
+ 4 =1
a2
c
… (1)
Differentiating w.r.t. a,
−
y2
2x2
a
+
2
=0
a3
c4
x2 y2
=
a4 c4
x2
a4 = 2 c4
y
a2 =
x 2
c
y
… (2)
Differential Calculus II
3.69
Substituting Eq. (2) in Eq. (1),
x 2 2
c y
x2
y
+
=1
x 2
c4
c
y
xy xy
+
=1
c2 c2
2 xy
=1
c2
2 xy = c 2
This is the equation of the required envelope.
x y
+ = 1 , where
a b
the parameters a and b are connected by the relation a 2 + b 2 = c 2 .
Example 8: Find the envelope of the family of straight lines
Solution:
x y
+ =1
a b
… (1)
Differentiating w.r.t. a,
−
x
y db
− 2
=0
2
a b da
… (2)
a 2 + b2 = c2
… (3)
Also,
Differentiating w.r.t. a,
2 a + 2b
Substituting Eq. (4) in Eq. (2),
x
y
− 2− 2
a
b
db
=0
da
db
a
=−
da
b
… (4)
⎛ a⎞
⎜⎝ − ⎟⎠ = 0
b
x
y
=
a3 b3
x
y
a = b =
a2 b2
x=
a3
c2
and
y=
1
a = (c 2 x ) 3
⎛x⎞ ⎛ y⎞
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ 1
a
b
= 2
a2 + b2
c
b3
c2
1
and
b = (c 2 y ) 3
… (5)
Engineering Mathematics
3.70
Substituting Eq. (5) in Eq. (3),
2
2
(c 2 x ) 3 + (c 2 y ) 3 = c 2
2
2
2
x3 + y3 = c3
This is the equation of the required envelope.
x2 y2
+
= 1, where the
a 2 b2
parameters a and b are connected by the relation a 2 + b 2 = c.
Example 9: Find the envelope of the family of ellipses
Solution:
x2 y2
+
=1
a2 b2
… (1)
Differentiating w.r.t. a,
−
Also
2 x 2 2 y 2 db
− 3
=0
a3
b da
a 2 + b2 = c
… (2)
… (3)
Differentiating w.r.t. a,
2 a + 2b
db
=0
da
db
a
=−
da
b
… (4)
Substituting Eq. (4) in Eq. (2),
−
2x2 2 y2
− 3
a3
b
⎛ a⎞
⎜⎝ − ⎟⎠ = 0
b
x2 y2
=
a4 b4
2
2
x2
y2 ⎛ x ⎞ ⎛ y ⎞
+
⎜
⎟
⎜
⎟
2
2
a2 = b2 = ⎝ a ⎠ ⎝ b ⎠ = 1
c
a2
b2
a2 + b2
2
2
x
1
y
1
=
and
=
a4 c
b4 c
a 2 = cx
Substituting Eq. (5) in Eq. (3),
and
b2 = c y
… (5)
cx + c y = c
x+ y = c
This is the equation of the required envelope.
Example 10: Considering the evolute of a curve as the envelope of its normals,
find the evolute of the parabola y 2 = 4ax.
Solution:
y 2 = 4ax
Differential Calculus II
3.71
Equation of normal to the parabola is
y = mx − 2am − am3 , where m is the parameter.
Differentiating partially w.r.t. m,
0 = x − 2a − 3am 2
1
⎛ x − 2a ⎞ 2
m=⎜
⎝ 3a ⎟⎠
Substituting in equation of the normal,
y = m( x − 2a − am 2 )
1
⎛ x − 2a ⎞ 2
y=⎜
⎝ 3a ⎟⎠
x − 2a ⎞
⎛
⎜⎝ x − 2a − a
⎟
3a ⎠
1
⎛ x − 2a ⎞ 2
⎛2⎞
y=⎜
( x − 2a) ⎜ ⎟
⎝ 3a ⎟⎠
⎝3⎠
27 y 2 = 4( x − 2a)3
This is the equation of the required evolute.
Example 11: Considering the evolute of a curve as the envelope of its normals,
x2 y2
find the evolute of the ellipse 2 + 2 = 1.
a
b
x2 y2
Solution:
… (1)
+
=1
a2 b2
Equation of normal at any point (a cos , b sin ) on the ellipse is
ax
by
a 2 b 2 , where is the parameter.
… (2)
cos
sin
Differentiating partially w.r.t. ,
ax sin q by cos q
+
=0
cos 2 q
sin 2 q
tan 3 q = −
by
ax
1
⎛ by ⎞ 3
tanq = − ⎜ ⎟
⎝ ax ⎠
1
sin q =
−(by ) 3
1
2
2 2
⎡
⎤
3
3
(
)
(
)
+
ax
by
⎢
⎥
⎣
⎦
1
and
cos q =
( ax ) 3
1
2
2 2
⎡
⎤
3
3
⎢( ax ) + (by ) ⎥
⎣
⎦
Engineering Mathematics
3.72
Substituting in Eq. (2),
1
2
2 2
⎡
⎤
ax ⎢( ax ) 3 + (by ) 3 ⎥
⎣
⎦
ax
1
3
1
+
2
2 2
⎡
⎤
by ⎢( ax ) 3 + (by ) 3 ⎥
⎣
⎦
(by )
1
3
= a2 − b2
1
2
2
2
2
2
⎡
⎤ ⎡
⎤
2
2
3
3
3
3
⎢( ax ) + (by ) ⎥ ⎢( ax ) + (by ) ⎥ = a − b
⎣
⎦ ⎣
⎦
3
2
2 2
⎤
⎡
2
2
3
3
⎣⎢( ax ) + (by ) ⎦⎥ = ( a − b )
2
2
2
( ax ) 3 + (by ) 3 = ( a 2 − b 2 ) 3
This is the equation of the required evolute.
Example 12: Find the envelope of the straight lines drawn at right angles to the
radii vectors of the spiral r = aeq cot a through their extremities.
Solution: Let P ( R, f ) be any point on the spiral
y
r = aeq cot a
Q(r, q )
... (1)
R = aef cot a
Let Q (r, ) be any point on the line PQ, which is
drawn through the extremity P of the radius vector
OP and is at right angles to the radius vector. From
OPQ,
R = r cos(q − f )
... (2)
From Eq. (1) and (2),
aef cot a = r cos(q − f ),
... (3)
Taking logarithm on both the sides,
log a + φ cot α = log r + log cos(θ − φ )
Differentiating partially w.r.t. ,
1
sin(q − f )
cos(q − f )
⎛p
⎞
tan ⎜ − a ⎟ = tan(q − f )
⎝2
⎠
π
−α = θ −φ
2
π
φ = θ +α −
2
P (R, f )
q−f
R
f q
x
O
where is the parameter.
cot a =
r
Fig. 3.10
Differential Calculus II
3.73
Substituting in Eq. (3),
π⎞
⎛
⎜ θ +α − ⎟ cot α
2⎠
ae⎝
⎛π
⎞
= r cos ⎜ − α ⎟
⎝2
⎠
⎛ π⎞
⎜α − ⎟
ae⎝ 2 ⎠ eθ cot α = r sin α
This is the equation of the required envelope.
Example 13: Find the envelope of the straight lines drawn at right angles to the
radii vectors of the cardioid r = a(1 + cos p ) through their extremities.
Solution: Let P( R, ) be any point on the cardioid r = a (1 + cos )
R = a(1 + cos f )
… (1)
Let Q(r , ) be any point on the line PQ, which is
drawn through the extremity P of the radius vector
OP and is at right angles to the radius vector. From
OPQ
R = r cos(q − f )
... (2)
y
Q(r, q )
r
a(1+ cos f ) = r cos(q − f ) ,
tan f =
R
... (3)
where is the parameter.
Differentiating partially w.r.t. ,
− a sin f = r sin(q − f )
r sin q cos f − ( r cos q − a) sin f = 0
f q
x
O
Fig. 3.11
r sin q
r cos q − a
r sin q
sin f =
cos f =
and
P(R, f )
q−f
From Eq. (1) and (2),
r + a 2 − 2ar cos q
2
r cos q − a
r + a 2 − 2ar cos q
2
Substituting in Eq. (3),
a (1+ cos φ ) = r (cos θ cos φ + sin θ sin φ )
( r cos q − a) cos f + r sin q sin f = a
( r cos q − a) 2 + r 2 sin 2 q
r 2 + a 2 − 2ar cos q
=a
r 2 + a 2 − 2ar cosq = a
r 2 + a 2 − 2ar cosq = a 2
r = 2a cosq
This is the equation of the required envelope.
[Substituting cosf and sinf ]
Engineering Mathematics
3.74
Example 14: Find the envelope of the circles drawn upon the radii vectors of
the ellipse
x2 y2
+
= 1 as diameter.
a 2 b2
Solution: Any point on the ellipse in the parameteric form is P ( a cos q , b sin q ) with
as the parameter. Hence, equation of the circle on the radius vector to this point as
diameter is given by
x 2 + y 2 − ax cos q − by sin q = 0
... (1)
Differentiating partially w.r.t. ,
ax sin q − by cos q = 0
tanq =
sinq =
cosq =
and
by
ax
by
a x + b2 y 2
2
2
ax
a x + b2 y 2
2
2
Substituting in Eq. (1),
x2 + y2 −
a2 x 2
a2 x 2 + b2 y 2
−
b2 y 2
a2 x 2 + b2 y 2
=0
x 2 + y 2 − a2 x 2 + b2 y 2 = 0
( x 2 + y 2 )2 = a2 x 2 + b2 y 2
This is the equation of the required envelope.
Exercise 3.5
1. Find the envelope of the following
family of curves:
1
(i) y = mx + , the parameter being m. m
(ii) y = mx + 1 + m 2 , the parameter being m.
(iii) y = mx − 2am − am3 , the parameter being m.
(iv) x cos q + y sin q = c sin q cos q ,
the parameter beingq .
(v) x tan q + y secq = c, the parameter being .
(vi) x sin q − y cos q = aq , the
parameter being .
(vii) x cosecq − y cot q = c, the
parameter being .
⎡ Ans. :
⎤
⎢
⎥
2
(
)
i y = 4x
⎢
⎥
⎢
⎥
(ii ) x 2 + y 2 = 1
⎢
⎥
⎢ (iii ) 27ay 2 = 4( x − 2a)3 ⎥
⎢
⎥
2
2
2
⎢
⎥
(iv ) x 3 + y 3 = a 3
⎢
⎥
2
2
2
(v) y = a + x
⎢
⎥
⎢
⎥
(
)
x
a
a
=
cos
q
+
q
sin
q
,
vi
⎢
⎥
⎢
y = a sin q − aq cos q ⎥
⎢
⎥
⎢⎣ ( viii ) x 2 − y 2 = c 2
⎥⎦
Differential Calculus II
2. Find the envelope of the family of
x y
straight lines
+ = 1, where the
a b
parameters a and b are connected by
the relation.
(i) a + b = c
(ii) ab = c 2 .
⎡ Ans. : (i ) x + y = c ⎤
⎢
⎥
(ii ) 4 xy = c 2
⎢⎣
⎥⎦
3. Find the envelope of the family
x2 y2
of curves
+
= 1, where the
y 2 b2
parameters a and b are connected by
the relation
(i) a + b = c
(ii) a + b = c .
2
2
2
2
2
2
⎡
⎤
3
3
3
(
)
Ans.
:
i
+
=
x
y
c
⎢
⎥
⎢
(ii ) x ± y ± c = 0 ⎥⎦
⎣
⎡
⎢ Ans. :
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣
3.75
⎤
⎥
2 ⎥
= (a2 + b2 ) 3 ⎥
⎥
2
2
3
3
(ii ) ( x + y ) + ( x − y ) ⎥⎥
3 ⎥
= ( 2a) 3 ⎥
⎥
⎥
(iii ) y = a cosh x
⎥
a
⎥
2
2
2
(iv ) x 3 + y 3 = ( 4 a) 3 ⎥
⎥
( v) x 2 + y 2 = a2
⎥⎦
2
2
(i ) ( ax ) 3 + (by ) 3
6. Find the envelope of the straight lines
drawn at right angles to the radii vectors of the following curves through
their extremities.
(i) r = a + b cosq
(ii) r n = a n cos nq .
4. Find the envelope of the family of
x
y
+
= 1, where the
curves
a
b
parameters a and b are connected by
⎡ Ans. : (i ) r 2 − 2br cos q + (b 2 − a 2 ) = 0⎤
⎢
⎥
⎛ n ⎞
⎛ n ⎞
⎜⎝
⎟⎠
⎜⎝
⎟⎠
n ⎞ ⎥
⎛
⎢
−
−
1
n
1
n
(ii ) r
cos ⎜
=a
q
⎢
⎝ 1− n ⎟⎠ ⎥⎦
⎣
the relation ab = c 2 .
⎡⎣ Ans. : 16 xy = c 2 ⎤⎦
7. Find the envelopes of the circles described on the radii vectors of the following curves as diameters
2
(i) y = 4ax
l
(ii)
= l + e cos
r
(iii) r 3 = a 3 cos 3 .
5. Considering the evolute of a curve as
the envelope of its normals, find the
evolute of the following curves:
x2 y2
(ii) xy = c 2
−
=1
a2 b2
t⎞
⎛
(iii) x = ⎜ a cos t + log tan ⎟ ,
⎝
2⎠
(i)
y = a sin t
(iv) x = a(3cos t – 2 cos3 t),
y = a(3sin t – 2 sin3 t)
(v) x = a(cos t + t cos t ),
y = a(sin t − t cos t ) .
⎤
⎡ Ans. :
⎥
⎢
2
2
2
( i ) ay + x( x + y ) = 0
⎥
⎢
⎢
( ii ) r 2 (e 2 − 1) − 2ler cos + 2l 2 = 0⎥
⎥
⎢
3
3
⎥
⎢
3
⎛
⎞
4
4
( iii ) r = a cos ⎜ ⎟
⎥
⎢
⎥⎦
⎝ 4 ⎠
⎣⎢
8. Show that family of circles
( x − a ) 2 + y 2 = a 2 has no envelope.
3.76
Engineering Mathematics
3.8 CURVE TRACING
Curve tracing is a procedure to obtain an approximate shape of the curve without
plotting a large number of points on it. In this chapter, we will study the tracing of
cartesian, parametric and polar curves.
3.8.1 Tracing of Cartesian Curves
The points to be taken into consideration while tracing a cartesian curve f (x, y) = 0
are as follows:
(i) Symmetry:
(a) The curve is symmetric about x-axis if the powers of y occurring in the equation are all even i.e., f (x, –y) = f (x, y).
(b) The curve is symmetric about y-axis if the powers of x occuring in the equation are all even i.e., f (–x, y) = f (x, y).
(c) The curve is symmetric about the line y = x, if on interchanging x and y, the
equation remains unchanged i.e., f (y, x) = f (x, y).
(d) The curve is symmetric about the line y = –x if on replacing x by –y and y by
–x, the equation remains unchanged.i.e., f (–y, –x) = f (x, y)
(e) The curve is symmetric in opposite quadrants or about origin if on replacing
x by –x by y and –y, the equation remains unchanged i.e., f (–x, –y) = f (x, y).
(ii) Origin: The curve passes through the origin if there is no constant term in the
equation.
(a) If the curve passes through the origin, the tangents at the origin are obtained
by equating the lowest degree term in x and y to zero.
(b) If there are two or more tangents at the origin, it is called a multiple point. The
multiple point is called a node, a cusp or an isolated point if the tangents at
this point are real and distinct, real and coincident or imaginary respectively.
(iii) Points of Intersection:
(a) The points of intersection of the curve with x and y axis are obtained by putting y = 0 and x = 0 respectively in the equation of the curve.
(b) Tangent at the point of intersection is obtained by shifting the origin to this
point and then equating the lowest degree term to zero.
(iv) Region of Existence: This region is obtained by expressing one variable in terms
of other, i.e., y = f (x) [or x = f (y)] and then finding the values of x (or y) at which
y(or x) becomes imaginary. The curve does not exist in the region which lies between these values of x (or y).
(v) Asymptotes:
(a) Asymptotes parallel to x-axis are obtained by equating the coefficient of
highest degree term of x in the equation to zero.
(b) Asymptotes parallel to y-axis are obtained by equating the coefficient of
highest degree term of y in the equation to zero.
(c) Oblique asymptotes are obtained by the following method:
Let y = mx + c is the asymptote to the curve and f2(x, y), f3(x, y) are the second
and third degree terms in the equation.
Differential Calculus II
3.77
x = 1 and y = m in f2(x, y) and f3(x, y)
f2(x, y) = f2(1, m) or f2(m)
f3 (x, y) = f3(1, m) or f3(m)
f ( m)
Find
c =– 2
f3 ( m)
Solve
f3(m) = 0
m = m1, m2, …
Calculate c at m1, m2, …
Substituting the values of m and c in y = mx + c, we get oblique asymptotes to the
curve.
(vi) Interval of Increasing-decreasing Function:
dy
(a) The curve increases strictly in the interval in which
> 0.
dx
dy
(b) The curve decreases strictly in the interval in which
< 0.
dx
Putting
(c) The curve attains its maximum and minimum values at the points where
dy
= 0.
dx
Example 1: Trace the cissoid y2 (2a – x) = x3.
Solution:
(i) Symmetry: The curve is symmetric about x-axis.
(ii) Origin: The curve passes through the origin.
Equating the lowest degree term i.e. 2ay2 to zero, we get y = 0. Thus, x-axis is the
tangent at the origin.
(iii) Points of Intersection: Putting y = 0, we get x = 0. Thus, the curve meets the coordinate axes only at the origin.
(iv) Region of Existence: From the equation of the curve, y =
x
x
2a
x
which
becomes imaginary when x < 0 or x > 2a. Therefore, the curve does not exist in
the region – < x < 0 and 2a < x < . Thus,
the curve lies in the region 0 < x < 2a.
(v) Asymptotes:
(a) Since coefficient of highest degree term
of x is constant, there is no asymptote
parallel to x-axis.
(b) Equating the coefficient of highest degree term of y to zero, we get
2a – x = 0. Thus, x = 2a is the asymptote
parallel to y-axis.
(vi) Interval of Increasing-decreasing Function:
dy
x 2 (3a x )
=
dx
y ( 2a x ) 2
Fig. 3.12
3.78
Engineering Mathematics
Since the curve is symmetric about x-axis, considering the part of the curve above
x-axis (y > 0),
dy
> 0, when x < 3a i.e., 0 < x < 2a [In the region of existence]
dx
Thus, curve is strictly increasing in this interval.
Example 2: Trace the witch of agnesi xy2 = 4a2 (a – x).
Solution:
(i) Symmetry: The curve is symmetric about x-axis.
(ii) Origin: The curve does not pass through the origin.
(iii) Points of Intersection: Putting y = 0, we get x = a. Thus, the curve meets x-axis at
A(a, 0).
Shifting the origin to A(a, 0) by putting x = X + a and y = Y + 0 in the equation
of the curve, (X + a)Y 2 = 4a2 (–X), (X + a)Y 2 + 4a2 X = 0.
Equating the lowest degree term i.e. 4a2 X to zero, we get X = 0, x – a = 0. Thus,
x = a is the tangent at A(a, 0).
a x
(iv) Region of Existence: From the equation of the curve, y = ± 2a
which
x
becomes imaginary when x < 0 or x > a. Therefore, the curve does not exist in the
region – < x < 0 and a < x < . Thus, the curve lies in the region 0 < x < a.
(v) Asymptotes:
(a) Equating the coefficient of highest
y
degree term of x to zero, we get
y2 + 4a2 = 0 which gives imaginary
values. Thus, there is no asymptote
parallel to x-axis.
(b) Equating the coefficient of highest dex=a
gree term of y to zero, we get x = 0. Thus,
y-axis is the asymptote.
O
x
A(a, 0)
(vi) Interval of Increasing-decreasing Functions:
dy
2a 3
=– 2
dx
x y
Since the curve is symmetric about xaxis, considering the part of the curve
dy
above x-axis (y > 0),
< 0 for all values
dx
of x.
Thus, the curve is strictly decreasing in the
Fig. 3.13
0 < x < a.
Example 3: Trace the strophoid y2 (a + x) = x2 (b – x).
Solution:
(i) Symmetry: The curve is symmetric about x-axis.
(ii) Origin: The curve passes through the origin.
Differential Calculus II
3.79
Equating the lowest degree term i.e. ay2 – bx2 to zero, we get y = ±
b
x. Thus,
a
b
are the two tangents at origin.
a
(iii) Points of Intersection: Putting y = 0, we get x = 0, b. Thus, the curve meets
x-axis at 0(0, 0) and A(b, 0). Shifting the origin to A(b, 0) by putting x = X + b and
y = Y + 0 in the equation of the curve,
Y 2(a + X + b) = (X + b)2 (–X)
2
Y (a + b + X) + X (X 2 + 2bX + b2) = 0
Equating the lowest degree term i.e. b2X to zero, we get X = 0, x – b = 0. Thus,
x = b is the tangent at A(b, 0).
y= ±
(iv) Region of Existence: From the equation of the curve, y = ± x
b−x
which
a+x
becomes imaginary when x < – a or x > b. Therefore, the curve does not exist in
the region – < x < – a and b < x < . Thus, the curve lies in the region – a < x < b.
(v) Asymptotes:
(a) Since the coefficient of highest degree term of x is constant, there is no
asymptote parallel to x-axis.
(b) Equating the coefficient of highest degree term of y to zero, we get x + a = 0.
Thus, x = – a is the asymptote parallel to y-axis.
Since the curve meets x-axis at two points O(0, 0) and A(b, 0), a loop exists in
b
the region 0 < x < b. Also, y = ±
x are the tangents at the origin, therefore after
a
passing through the origin, the curve extends towards the asymptote x = – a in the
region – a < x < 0.
y
y
bx
a
−a
y
b
A(b, 0) x
O
x
x
− ba x
Fig. 3.14
Engineering Mathematics
3.80
Example 4: Trace Folium of Descartes x3 + y3 = 3axy.
Solution:
(i) Symmetry: The curve is not symmetric about the coordinate axes but is symmetric about the line y = x, since after interchanging y and x, equation of the curve
remains unchanged.
(ii) Origin: The curve passes through the origin.
Equating the lowest degree term i.e. xy to zero, we get x = 0 and y = 0. Thus,
x = 0 and y = 0 are the tangents at the origin.
(iii) Points of Intersection:
(a) Putting y = 0, we get x = 0. Thus, the curve meets the coordinate axes only at
the origin.
(b) Putting y = x, we get 2x3 = 3ax2, x = 0, 3a and y = 0, 3a
2
2
3a 3a
,
Thus, the curve meets the line y = x at O(0, 0) and A
.
2 2
( 32a , 32a ) is given by,
( y − 32a ) = ⎛⎜⎝ ddyx ⎞⎟⎠( ) ( x − 32a )
Tangent at A
(
)
y
y=
3a , 3a
2 2
(
⎡ ay − x ⎤
3a
= ⎢ 2
x−
⎥
2
⎣ y − ax ⎦ ( 3a , 3a )
2
(
)
2
)
(
)
F 3a , 3a I
H 2 2K
A
x + y = 3a
O
2
3a
2
Thus, x + y = 3a is the tangent at
3a 3a
,
.
A
2 2
= –1 x −
x
x+y+a=0
Fig. 3.15
(iv) Region of Existence: In the equation of the curve, x and y cannot be negative
simultaneously, otherwise equation of the curve will not be satisfied. Thus, the
curve does not exist in the region where x < 0 and y < 0, i.e., third quadrant.
(v) Asymptotes:
(a) Since coefficients of highest degree term of x and y are constant, the curve
does not have any asymptotes parallel to coordinate axes.
(b) Oblique Asymptotes: Let y = mx + c is the asymptote of the curve.
Putting x = 1 and y = m in the third and second degree terms of the equation
separately
f3(m) = 1 + m3, f2(m) = –3am
Solving
f3(m) = 0, 1 + m3 = 0, m = –1
f ( m)
( 3am)
c=– 2
=–
= a = –a
m
f3 ( m)
3m 2
Thus, y = –x – a i.e. x + y + a = 0 is the asymptote of the curve.
Differential Calculus II
3.81
Since no part of the curve lies in the third quadrant and coordinate axes are the
tangents at the origin, after passing through the origin, the curve extends towards
the asymptote x + y + a = 0 in the second and fourth quadrants.
Example 5: Trace the catenary y = c cosh x .
c
Solution: Rewriting the equation
−x
c cx
(e + e c )
2
(i) Symmetry: The curve is symmetric about y-axis, since on replacing x by –x, equation remains unchanged.
(ii) Origin: The curve does not pass through the origin.
(iii) Points of Intersection: Putting x = 0, we get y = c.
From the equation of the curve,
y=
dy
−x ⎞
⎛ x
= c ⎜ 1 ec − 1 e c ⎟
⎠
dx
2 ⎝c
c
dy
dx
=0
( 0, c )
Thus, the tangent is parallel to x-axis at A(0, c). The curve does not meet x-axis.
(iv) Region of Existence: Since
x
1 cosh
< for – < x < , the curve
c
lies in the region c y < ,
– <x< .
(v) There is no asymptote to the curve.
(vi) Interval of Increasing-decreasing Function:
()
x
dy
1 x
= (e c e c )
dx
2
= sinh x
c
dy
> 0, when x > 0
dx
Thus, the curve is strictly
increasing in 0 < x <
dy
(b)
< 0, when x < 0
dx
Thus, the curve is strictly
decreasing in – < x < 0.
y
(a)
Example 6: Trace the curve y (x2 + a2) = a3.
Solution:
(i) Symmetry: The curve is symmetric about y-axis.
(ii) Origin: The curve does not pass through the origin.
A (0, c)
O
Fig. 3.16
x
3.82
Engineering Mathematics
(iii) Points of Intersection:
(a) Putting x = 0, we get y = a. Thus, the curve meets the y-axis at A(0, a). Shifting the origin to A(0, a) by putting x = X + 0 and y = Y + a in the equation of
the curve,
(Y + a)(X 2 + a2) = a3
(Y + a) X 2 + a2Y = 0
Equating the lowest degree term i.e. a2Y to zero, we get Y = 0, y – a = 0.
Thus, y = a is the tangent at A(0, a).
(b) The curve does not meet x-axis.
y
which
y
becomes imaginary when a – y < 0, i.e., y > 0 and y < 0. Thus, the curve does not
exist in the region a < y < and – < y < 0. Therefore, the curve lies in the region
0 < y < a.
(v) Asymptotes:
(a) Equating the coefficient of highest degree term of x to zero, we get y = 0.
Thus, x-axis is the asymptote parallel to x-axis.
(b) Since coefficient of highest degree term of x is constant, there is no asymptote
parallel to y-axis.
(vi) Interval of Increasing-decreasing Function:
(iv) Region of Existence: From the equation of the curve, x =
a
dy
2 xa3
=– 2
dx
( x + a2 )2
dy
< 0 when x > 0
dx
Thus, y is strictly decreasing in the interval 0 < x < .
dy
> 0 when x < 0
dx
Thus, y is strictly increasing in the interval – < x < 0.
y
A (0, a)
O
Fig. 3.17
y=a
x
a
Differential Calculus II
3.83
Example 7: Trace the curve (x2 – a2)(y2 – b2) = a2 b2.
Solution:
(i) Symmetry: The curve is symmetric about both x and y axes.
(ii) Origin: The curve passes through the origin.
Equating the lowest degree terms i.e. – b2 x2 – a2 y2 to zero, we get imaginary values. Thus, tangents at the origin are imaginary. Therefore, the origin is an isolated
point.
(iii) Points of Intersection: The curve does not meet x and y axes.
ay
bx
(iv) Region of Existence: From equation of the curve, y = ±
,x=±
2
2
2
y b2
x a
y is imaginary when x2 – a2 < 0, i.e., – a < x < a and x is imaginary when y2 – b2 < 0,
i.e., – b < y < b.
Therefore, the curve does not exist in the region where – a < x < a and – < y < – b,
b < y < b. Thus, the curve lies in the region – < x < – a, a < x < and – < y
< – b, b < y < .
(v) Asymptotes:
(a) Equating the coefficient of highest degree term of x to zero, we get y2 – b2 = 0.
Thus, y = b are the asympy
totes parallel to x-axis.
(b) Equating the coefficient of
highest degree term of y to
zero, we get x2 – a2 = 0. Thus,
x = a are the asymptotes
parallel to y-axis.
x −a
x a
y b
(vi) Interval of Increasing-decreasing
O
y −b x
Function:
dy
=
dx
ba 2
3
( x 2 a2) 2
dy
< 0 for all values of x
dx
in the region of existence.
Thus, y is strictly decreasing.
Fig. 3.18
Example 8: Trace the curve y2 = (x – 1) (x – 2) (x – 3).
Solution:
(i) Symmetry: The curve is symmetric about x-axis.
(ii) Origin: The curve does not pass through the origin.
(iii) Points of Intersection: Putting y = 0, we get x = 1, 2, 3. Thus, the curve meets the
x-axis at A(1, 0), B(2, 0) and C(3, 0). Shifting the origin to A(1, 0), B(2, 0) and
C (3, 0) by putting
(a) x = X + 1, y = Y + 0 in the equation of the curve, Y 2 = X (X – 1)(X – 2)
Equating the lowest degree term i.e. 2X to zero, we get X = 0, x – 1 = 0.
Thus, x = 1 is the tangent at A(1, 0).
Engineering Mathematics
3.84
(b) x = X + 2, y = Y + 0 in the equation of the curve, Y 2 = (X + 1) X (X – 1)
Equating the lowest degree term i.e. –X to zero, we get x – 2 = 0. Thus,
x = 2 is the tangent at B(2, 0).
(c) x = X + 3, y = Y + 0 in the equation of the curve, Y 2 = (X + 2) (X + 1) X
Equating the lowest degree term i.e. 2X to zero, we get X = 0, x – 3 = 0.
Thus, x = 3 is the tangent at C(3, 0).
(iv) Region of Existence: From the equation of the curve, y = ± ( x − 1) ( x − 2) ( x − 3)
which becomes imaginary when x < 1, 2 < x < 3. Therefore, the curve does not
exists in the region – < x < 1 and 2 < x < 3. Thus, the curve lies in the region
1 < x < 2 and x > 3.
(v) Asymptotes: Since the coefficients of highest degree of x and y are constants,
there are no asymptotes to the curve.
(vi) Interval of Increasing-decreasing Function:
dy
3 x 2 − 12 x + 11
= ±
dx
2 ( x − 1) ( x − 2) ( x − 3)
= ±
3 ( x − 1.42) ( x − 2.5)
2 ( x − 1) ( x − 2) ( x − 3)
dy
(a)
> 0 when x > 2.5, i.e., 3 < x < [in region of existence] and when
dx
x < 1.42, i.e., 1 < x < 1.42
Thus, y is strictly increasing in both the intervals.
dy
< 0 when 1.42 < x < 2.5, i.e., 1.42 < x < 2 [in region of existence]
(b)
dx
Thus, y is strictly decreasing in this interval.
y
x=1
O
A
(1, 0)
x=2
x=3
B
C
(2,0) (3,0)
Fig. 3.19
x
Differential Calculus II
3.85
Example 9: Trace the curve y2 (x + a) = x2 (3a – x).
Solution:
(i) Symmetry: The curve is symmetric about x-axis.
(ii) Origin: The curve passes through the origin.
Equating the lowest degree term i.e. ay2 – 3ax2 to zero, we get y = ± x 3 . Thus,
y = ± x 3 are two tangents at the origin.
(iii) Points of Intersection: Putting y = 0, we get x = 0, 3a. Thus, the curve meets the
x-axis at A(3a, 0) and O(0, 0). Shifting the origin to A(3a, 0) by putting x = X + 3a
and y = Y + 0 in the equation of the curve, Y 2 (X + 3a + a) = (X + 3a)2 (–X).
Equating the lowest degree term i.e. –9a2X to zero, we get X = 0, x – 3a = 0.
Thus, x = 3a is the tangent at A(3a, 0).
(iv) Region of Existence: From the equation of the curve, y = ± x
3a − x
which
x+a
becomes imaginary when x > 3a or x < – a. Therefore, the curve does not exist in
the region where 3a < x < and – < x < – a. Thus, the curve lies in the region
– a < x < 3a.
(v) Asymptotes:
(a) Since coefficient of highest degree term of x is constant, there is no asymptote
parallel to x-axis.
(b) Equating the coefficient of highest degree term of y to zero, we get x + a = 0.
Thus, x = – a is the asymptote parallel to x-axis.
Since the curve meets the x-axis at two points, due to symmetry a loop exists
between O(0, 0) and A(3a, 0). Also, y = ± x 3 are the tangents at the origin, after
passing through the origin the curve extends towards the asymptote.
y
A(3a, 0)
x
O
x = −a
Fig. 3.20
3.86
Engineering Mathematics
x2 + 2x
.
x2 + 4
Example 10: Trace the curve y =
Solution:
(i) Symmetry: The curve is not symmetric.
(ii) Origin: The curve passes through the origin.
Equating the lowest degree term i.e. 4y – 2x to zero, we get x = 2y. Thus, x = 2y is
the tangent at the origin.
(iii) Points of Intersection: Putting y = 0, we get x = 0, – 2. Thus, the curve meets the
x-axis at A(–2, 0) and O(0, 0). Shifting the origin to P(–2, 0) by putting x = X – 2,
y = Y + 0 in the equation of the curve,
Y (X – 2)2 + 4 = (X – 2)2 + 2(X – 2)
Y (X2 – 4X + 8) = X2 + 6X
Equating the lowest degree term i.e. 8Y – 6X to zero, we get 4y – 3(x + 2) = 0.
Thus, 4y – 3x – 6 = 0 is the tangent at (–2, 0).
(iv) Region of existence: y is defined for all values of x. Thus, the curve lies in the
region – < x < .
(v) Asymptotes:
(a) Equating the coefficient of highest degree of x to zero, we get y – 1 = 0. Thus,
y = 1 is the asymptote parallel to x-axis.
(b) Equating the coefficient of highest degree of y to zero, we get x2 + 4 = 0 which
gives imaginary values. Thus, there is no asymptote parallel to y-axis.
When y = 1, x = 2. This shows that the curve meets y = 1 at B(2, 1).
Thus, the curve approaches the asymptote y = 1 from above when x + and
from below when x – .
(vi) Interval of Increasing-decreasing Function:
dy
− 2( x 2 − 4 x − 4 )
− 2( x + 0.83) ( x − 4.83)
=
=
dx
( x 2 + 4) 2
( x 2 + 4) 2
dy
> 0 when – 0.83 < x < 4.83
dx
Thus, the curve is strictly increasing in this interval.
dy
(b)
< 0, when – < x < – 0.83 and 4.83 < x <
dx
Thus, the curve is strictly decreasing in both the intervals.
(a)
y
x
4.83
B(2, 1)
O
A
(0, −2)
x −0.83
Fig. 3.21
x
Differential Calculus II
3.87
Example 11: Trace the curve x3 + y3 = 3ax2 (a > 0).
Solution:
(i) Symmetry: The curve is neither symmetric about the coordinate axes nor about the
line y = x.
(ii) Origin: The curve passes through the origin.
Equating the lowest degree term i.e. 3ax2 to zero, we get x = 0. Thus x = 0 i.e.,
y-axis is the tangent at origin.
(iii) Points of Intersection: Putting y = 0, we get x = 0, 3a. Thus, the curve meets x-axis
at O(0, 0) and A(3a, 0). Shifting the origin to A(3a, 0) by putting x = X + 3a, y = Y + 0
in the equation of the curve,
(X + 3a)3 + Y 3 = 3a (X + 3a)2
3
2
X + 9a X + 6aX 2 + Y 3 = 0.
Equating the lowest degree term i.e. 9a2 X to zero, we get X = 0, x – 3a = 0.
Thus, x = 3a is the tangent at A(3a, 0).
(iv) Region of Existence: x and y cannot be negative simultaneously, but can take opposite signs. Thus, the curve does not exist in the region where x < 0 and y < 0 i.e.,
third quadrant.
(v) Asymptotes:
(a) Since coefficients of highest degree terms in x and y are constant, the curve
does not have any asymptotes parallel to x and y-axis.
(b) Oblique Asymptote: Let y = mx + c is the asymptote of the curve.
Putting x = 1, y = m in the third and second degree terms of the equation
separately.
f3 (m) = 1 + m3, f2 (m) = –3a
Solving f3 (m) = 0, 1 + m3 = 0, m = –1,
f ( m)
c =– 2
f3 ( m)
3a
=a
3m 2
Thus, y = –x + a or y + x = a is the asymptote of the curve and curve meets
a 2a
,
the asymptote at
3 3
=–
(
)
(vi) Interval of Increasing-decreasing Function:
dy
x ( 2a x )
=
dx
y2
(a)
dy
< 0, when x < 0 and x > 2a
dx
Thus, the curve is strictly decreasing in –
< x < 0 and 2a < x < .
3.88
(b)
Engineering Mathematics
dy
> 0, when 0 < x < 2a
dx
Thus, the curve is strictly increasing in 0 < x < 2a.
y
O
A(3a, 0)
x
x+y=a
Fig. 3.22
Example 12: Trace the curve y3 = a2x – x3.
Solution:
(i) Symmetry: The curve is symmetric in opposite quadrants, since on replacing x by
–x and y by –y, equation remains unchanged.
(ii) Origin: The curve passes through the origin.
Equating the lowest degree term i.e. a2 x to zero, we get x = 0. Thus, x = 0 i.e.,
y-axis is the tangent at origin.
(iii) Points of Intersection: Putting y = 0, we get x = 0, a. Thus, the curve meets the
x-axis at O(0, 0), A(a, 0) and B(– a, 0). Shifting the origin to A(a, 0) and B(– a, 0)
by putting
(a) x = X + a, y = Y + 0 in the equation of the curve,
Y 3 = a2 (X + a) – (X + a)3
Y 3 + X 3 + 3aX 2 + 2a2 X = 0
Equating the lowest degree term i.e. 2a2 X to zero, we get X = 0, x – a = 0.
Thus, x = a is the tangent at A(a, 0).
(b) x = X – a, y = Y + 0 in the equation of the curve,
y3 = a2 (X – a) – (X – a)3
Y 3 + X 3 – 3aX 2 + 2a2 X = 0
Equating the lowest degree term i.e. 2a2 X to zero, we get X = 0, x + a = 0.
Thus, x = – a is the tangent at B(– a, 0).
(iv) Region of Existence: The curve exists everywhere in the region – < x < .
Differential Calculus II
3.89
(v) Asymptotes:
(a) Since coefficients of highest degree term of x and y are constant, the curve
does not have any asymptotes parallel to coordinate axes.
(b) Oblique Asymptotes: Let y = mx + c is the asymptote of the curve.
Putting x = 1 and y = m in the third and second degree terms of the equation
separately
f3 (m) = m3 + 1, f2 (m) = 0
Solving f3 (m) = 0, m3 + 1 = 0, m = –1,
c=–
f 2( m )
=0
f3 ( m)
Thus, y = –x is the asymptote of the curve.
(vi) Interval of Increasing-decreasing Function:
dy
a2 3x 2
=
dx
3y2
(a)
dy
< 0, when x <
dx
a
3
a
3
and x >
Thus, the curve is strictly decreasing in the region –
a
<x<
3
dy
(b)
> 0, when
dx
a
3
<x<
< x <
a
3
Thus, the curve is strictly increasing in the region
a
3
<x<
y
B(−a, 0)
A(a, 0)
O
x
y=
Fig. 3.23
−x
a
.
3
a
and
3
Engineering Mathematics
3.90
Exercise 3.6
Trace the following curves:
1. a 2 x 2
y 3 (2a
4. ay 2 = x(a 2 + x 2 ) .
y) .
Ans.:
Ans.:
y
y
(0, 2a)
x
O
x
O
2. 4ay 2
5. a 2 y 2
x ( x 2a ) 2 .
Ans.:
x 2 ( x a )(2a x) .
Ans.:
y
y
(2a, 0)
x
O
(2a, 0) x
O (a, 0)
3. ay 2 = x 3 .
6. y
Ans.:
y
( x2
x 6)( x 7) .
Ans.:
y
(0, 42)
O
x
(−2, 0)
(3, 0)
O
(7, 0)
x
Differential Calculus II
7. y
x 3 12 x 16.
3.91
10. y 2 ( x a )
Ans.:
Ans.:
x 2 ( x a ), a > 0.
y
y
x
(4, 0)
(−2, 0)
O
−a
y
x−
−a
x
a
x
x
O
(0, −16)
x+
y
−a
11. y 2 (a 2
(2, −32)
8. y =
x2 )
a3 x .
Ans.:
x 2 3x
.
x 1
y
Ans.:
y
x
x
x
−a
a
1
2
x−
O
y
x
O
x
(3, 0)
x2 + 1
.
x2 1
Ans.:
12. y =
9. y 2 x 2
Ans.:
x2
a2 .
y
y
y
x
x
−a
1
a
x
x
−1
1
y
x
O
y
−1
O
1
x
(−1, 0)
3.92
Engineering Mathematics
3.8.2 Tracing of Parametric Curves
The points to be taken into consideration while tracing a parametric curve x = f1 (t),
y = f2 (t), where t is a parameter are as follows:
(i) Symmetry:
(a) The curve is symmetric about x-axis if x is an even function and y is an odd
function of t.
(b) The curve is symmetric about y-axis if x is an odd function and y is an even
function of t.
(c) The curve is symmetric about y-axis if after replacing t by p – t, x becomes
negative and y remains positive.
(ii) Origin: The curve passes through the origin if there exists at least one real value
of t at which x = 0 and y = 0.
(iii) Points of Intersection:
(a) Points of intersection with x-axis: Find the value of t at which y = 0 and then
find x for this value of t.
(b) Points of intersection with y-axis: Find the value of t at which x = 0 and then
find y for this value of t.
(iv) Tangents:
dy
= 0.
(a) Tangent is parallel to x-axis at the point where
dx
dy
(b) Tangent is parallel to y-axis at the point where
.
dx
(v) Maximum and Minimum Values: Determine the maximum and minimum values
of x and y if exists.
(vi) Region: Determine the region where x and y are real. The curve does not exist in
the region, where x or y is imaginary.
(vii) Variation of x and y: Determine the values of x and y for some suitable values
of t.
Note: If x and y are periodic functions of t having the same period, then the curve is
traced for one period only.
Example 1: Trace the hyplocycloid x = a cos3 t, y = b sin3 t.
Solution: x and y are periodic functions of t with period 2p. Therefore, the curve is
traced between 0 to 2p.
(i) Symmetry: The curve is symmetric about x-axis since x is an even function of t
and y is an odd function of t. Also the curve is symmetric about y-axis since after
replacing t by p – t, x becomes negative but y remains positive.
(ii) Origin: The curve does not pass through the origin.
(iii) Points of Intersection:
(a) At t = 0, y = 0 and x = a.
(b) At t = p , x = 0 and y = b.
2
Thus, the curve meets the x-axis at A(a, 0) and y-axis at B(0, b).
Differential Calculus II
3.93
dy
(iv) Tangents: dx = –3a cos2 t sin t,
= 3b sin2 t cos t
dt
dt
dy
dy /dt
3b sin 2 t cos t
b
=
=
=
tan t
2
dx
dx /dt
a
3a cos t sin t
dy
= 0, when t = 0
dx
Thus, the tangent is x-axis at t = 0 i.e., at A(a, 0)
dy
dx
when t = p .
2
Thus, the tangent is y-axis at t = p , i.e., at B(0, b).
2
(v) Maximum and Minimum Values: Maximum values of x and y are a and b respectively since maximum value of cos t and sin t is 1. Minimum values of x and y are
– a and – b respectively since minimum value of cos t and sin t is –1.
(vi) Region: The curve lies in the region – a < x < a and – b < y < b.
(vii) Asymptotes: There is no asymptote of the curve since x and y are finite for all
values of t.
(viii) Variation of x and y:
y
B(0, b)
t
0
6
4
3
2
x a
3 3a
8
a
a
8
0
y 0
b
8
3 3b
8
b
2 2
b
2 2
C
(−a, 0)
A(a, 0) x
O
D(0, − b)
Fig. 3.24
()
⎡
t ⎤
Example 2: Trace the tractrix x = a ⎢cos t + log tan
, y = a sin t.
2 ⎥⎦
⎣
Solution:
(i) Symmetry: The curve is symmetric about x-axis since x is an even function of t and
y is an odd function of t.
Replacing t by p – t,
(
)
⎡
p
t ⎤
x = a ⎢cos (p − t ) + log tan
−
⎥
2
2
⎦
⎣
⎡
t ⎤
= a ⎢ − cos t + log cot ⎥
2
⎣
⎦
Engineering Mathematics
3.94
= a ⎡ − cos t − log tan t ⎤
⎢⎣
2 ⎥⎦
= –x
y = a sin (p – t)
= a sin t
=y
Thus, the curve is symmetric about y-axis.
(ii) Origin: The curve does not pass through the origin.
(iii) Points of Intersection:
(a) At t = 0, y = 0 and x – [∵ log 0 – ]
(b) At t = p , x = 0 and y = a
2
Thus, the curve meets the y-axis at A(0, a) and does not meet y-axis.
t
1
(iv) Tangents: x = a ⎡cos t + log tan 2 ⎤
2
2 ⎦⎥
⎣⎢
dx = a ⎡ − sin t + 1 ⋅ 1 ⋅ 2 tan t sec 2 t ⋅ 1 ⎤
⎢
2 tan 2 t
2
2 2⎥
dt
⎢⎣
⎥⎦
2
a cos 2 t
1 ⎤
= a ⎡ − sin t +
=
sin t ⎦⎥
⎣⎢
sin t
dy
= a cos t
dt
dy
dy / dt
=
dx
dx / dt
a cos t sin t
= tan t
a cos 2 t
=
dy
= tan p
dx
2
Thus, the tangent is y-axis.
(v) Maximum and Minimum Values: Maximum and minimum values of y are a and
– a respectively since maximum and minimum values of sin x are 1 and –1 respectively.
At point A(0, a):
x lies between –
to .
(vi) Region: The curve lies in the region – a < y < a and –
(vii) Asymptotes: lim x = –
t
0
and lim x =
t
p
At t = 0 and t = p, y = 0
Thus, y = 0 i.e., x-axis is the asymptote of the curve.
(viii) Variation of x and y:
<x< .
Differential Calculus II
3.95
y
A(0, a)
p
p
p
p
6
4
3
2
x − ∞ − 0.45 a − 0.17 a − 0.04 a 0
y 0
0.5 a
0.71 a
0.87 a a
t
0
t=0
t=p
O
x
Fig. 3.25
Example 3: Trace the cycloid x = a (t + sin t), y = a (1 – cos t).
Solution:
(i) Symmetry: The curve is symmetric about y-axis since x is an odd function of t and
y is and even function of t.
(ii) Origin: At t = 0, x = 0 and y = 0. Thus, the curve passes through the origin.
(iii) Points of Intersection:
(a) x = 0 only at t = 0
Thus, the curve meets the y-axis only at origin.
(b) If y = 0, cos t = 1, t = 0, 2p, 4p,
Then x = 0, 2ap, 4ap, …
Thus, the curve meets the x-axis at (0, 0), ( 2ap, 0), ( 4ap, 0) …
dy
(iv) Tangents: dx = a (1 + cos t),
= a sin t
dt
dt
dy
dy
= dt
dx
dx
dt
2 sin t cos t
a sin t
2
2 = tan t
=
=
a(1+ cos t )
2 t
2
2 cos
2
dy
(a)
= 0, at t = 0, 2p, 4p, …
dx
Thus, tangent is x-axis at (0, 0), ( 2ap, 0), ( 4ap, 0), …
(b)
dy
dx
, at t = p, 3p, 5p, …
Thus, tangent is parallel to y-axis at ( ap, 2a), ( 3ap, 2a), ( 5ap, 2a), …
Engineering Mathematics
3.96
(v) Maximum and Minimum Values: Maximum and minimum values of y are 2a and
0 since minimum and maximum values of cos t are –1 and 1.
(vi) Region: Curve lies in the region 0 < y < 2a and – < x < .
(vii) Asymptotes: There is no asymptote of the curve since x and y are finite for all
finite values of t.
(viii) Since sin t is a periodic function of period 2p, y is the periodic function of period
2p. Thus curve repeat itself in the intervals [0, ± 2ap], [ ± 2ap, ± 4ap], …
(ix) Variation of x and y:
t
0
x
0
y
0
p
2
p
a
+1
2
p
( )
ap
a
2a
3p
2
3p
a
+1
2
(
)
a
2p
2ap
0
y
(–3ap, 2a)
(–ap, 2a)
(–2ap, 0)
(ap, 2a)
O (0, 0)
(3ap, 2a)
(2ap, 0)
x
Fig. 3.26
Example 4: Trace the curve x = a sin 2t (1 + cos 2t), y = a cos 2t (1 – cos 2t).
Solution: x and y are periodic functions of t with period p, therefore we will discuss
the curve only in the interval 0 t < p
(i) Symmetry: Curve is symmetric about y-axis since x is an odd function of t and y is
an even function of t.
(ii) Origin: At t = 0, x = 0 and y = 0. Thus, origin lies on the curve.
(iii) Points of Intersection: (a) x = 0 at t = 0, ± p , then y = 0, – 2a.
2
Thus, the curve meets the y-axis at (0, 0), (0, – 2a).
(b) y = 0 at t = 0, ± p
4
then
x = 0, ± a
Thus, the curve meets the x-axis at (0, 0), (0, a), (0, – a)
(iv) Tangents: dx = 2a cos 2t (1 + cos 2t) + a sin 2t (– 2 sin 2t)
dt
= 2a cos 2t + 2a cos 4t
= 2a 2 cos 3t cos t
dy
= – 2a sin 2t (1 – cos 2t) + a cos 2t (2 sin 2t)
dt
= 4a sin 2t cos 2t – 2a sin 2t
Differential Calculus II
3.97
= 2a (sin 4t – sin 2t)
= 2a 2 cos 3t sin t
dy / dt
dy
=
dx
dx / dt
=
4a cos 3t sin t
4a cos 3t cos t
= tan t
(a)
dy
= 0, at t = 0
dx
Thus, the tangent is x-axis at (0, 0).
(b)
dy
dx
at t = p
2
Thus, the tangent is y-axis at (0, – 2a).
(v) Asymptotes: There is no asymptote of the curve since x and y are finite for all values of t.
(vi) Variation of x and y:
t
0
p
6
p
3
x
0
3a 3
4
y
0
a
4
a 3
4
3a
4
p
2
0
–2a
2p
3
5p
6
p
a 3
4
3a
4
3a 3
4
0
a
4
0
y
B
–3a 3 a
,
4
4
A
3a 3 a
,
4
4
x
O
C, (0, –2a)
Fig. 3.27
Engineering Mathematics
3.98
Exercise 3.7
Trace the following curves:
1. Astroid x = a cos3 t, y = b sin3 t
4. Cycloid x = a(t – sin t), y = a(1 + cos t)
Ans.:
Ans.:
y
y
(–2ap, 0)
(0, 2a)
(2ap, 2a)
(0, a)
(–ap, 0) O
O
(–a, 0)
(a, 0)
x
5. Parabola x = 1 + sin t, y = 2 cos 2t
Ans.:
(0, –a)
2–1
2. Cycloid x = a(t + sin t), y = a(1 + cos t)
Ans.:
y
(–2ap, 2a)
(–3ap, 0)
(0, 2a)
(–ap, 0)
O
(ap, 0)
x
(ap, 0)
2
y
2+1
,0
2
,0
x
O
(–2, 0)
(2ap, 2a)
x
(3ap, 0)
3. x = a cos t, y = b sin t.
Ans.:
y
(0, b)
O
(a, 0)
x
3.8.3 Tracing of Polar Curves
The points to be taken into consideration while tracing a polar curve r = f (q ) are as
follows:
(i) Symmetry:
(a) A curve is symmetric about the initial line q = 0 (x-axis), if the equation remains unchanged after replacing q by –q.
(b) A curve is symmetric about the line q = p (line through pole perpendicular to
2
the initial line), if the equation remains unchanged after replacing q by p – q.
Differential Calculus II
3.99
(c) A curve is symmetric about the pole (opposite quadrant), if the equation remains unchanged when q is replaced by p + q (or r is replaced by – r)
(d) A curve is symmetric about the line q = p , if the equation remains un4
changed after replacing q by p – q.
2
(ii) Pole: The pole lines on the curve, if for r = 0, there exists atleast one real value of q.
(iii) Points of Intersection: Determine the points where the curve meets the initial line
q = 0, q = p and q = p.
2
(iv) Direction of Tangent: Determine f, i.e., angle between the radius vector and the
dq
tangent at the points of intersection using tan f = r
.
dr
The angle f gives the direction of the tangent at the point of intersection.
(v) Region:
(a) Determine the maximum and minimum value of r if exists. If minimum value
of r is a, then no part of the curve lies inside the circle with radius a and centre at pole. If maximum value of r is b, then the whole curve lies within the
circle of radius b and centre at the pole.
(b) Determine the range of q in which r2 < 0, i.e., r is imaginary, then curve does
not exists in this range.
(vi) Asymptotes: If r
for some q = q1 then the asymptote of the curve may exist
and is given by
r sin (q – q1) = f (q1)
1
where q1 is the solution of
= 0.
f (q )
(vii) Variation of r: Trace the variation of r for some suitable values of q.
dr
If
> 0, then r increases as q increases.
dq
dr
< 0, then r decreases as q increases.
and if
dq
If the curve meets the line of symmetry at two points, then a loop exists between these two points.
Note: Curve of the type r = a sin nq or r = a cos nq consists of (i) n similar loops, if n
is odd (ii) 2n similar loops, if n is even.
If n = 1, then the curve becomes a circle.
Example 1: Trace the cardioid r = a (1 – cos q).
Solution:
(i) Symmetry: The curve is symmetric about the initial line q = 0, since when q is
replaced by – q, equation of the curve remains unchanged.
(ii) Pole: Pole lies on the curve since when q = 0, r = 0. Tangent at the pole is the initial
line q = 0.
Engineering Mathematics
3.100
(iii) Points of Intersection: Putting q = p , p we get r = a, 2a respectively. Thus, the
2
p
curve meets the line q = p and q = p at A a,
and B(2a, p) respectively.
2
2
(iv) Direction of Tangent: r = a (1 – cos q)
dr
= a sin q
dq
dq
tan f = r
dr
2 sin 2 q
a (1 cos q )
q
2
=
=
= tan
q
q
2
a sin q
2 sin cos
2
2
f= q
2
p
At point A a,
: f = p , thus the tangent makes an angle p with the line
2
4
4
q= p .
2
At point B(2a, p): f = p , thus the tangent is perpendicular to the line q = p.
2
(v) Region: Since minimum vale of cos q is –1, the maximum value of r is 2a. Thus,
the whole curve lies within a circle with centre at the pole and radius 2a.
(vi) Asymptote: There is no asymptote of the curve since r is finite for all values of q.
(vii) Variation of r:
( )
( )
q
0
r
0
p
3
a
2
p
2
2p
3
3
2
a
p
2a
q=p
2
p
A a,
2
B
(2a, p)
O
q=0
Fig. 3.28
Example 2: Trace limacon of pascal r = a + b cos q, where a > 0, b > 0.
Solution:
(i) Symmetry: The curve is symmetric about the line q = 0, since when q is replaced
by –q, equation of the curve remains unchanged. Three different cases arise:
Differential Calculus II
3.101
Case I: a > b
(ii) Pole: It does not lie on the curve. If
r = 0, cos q = – a < –1 which is not possible. Thus r 0 for any value of q.
b
(iii) Points of Intersection: Putting q = 0, p , p , we get r = (a + b), a, (a – b) respec2
p
tively. Thus, curve meets the line q = 0, q = p and q = p at A (a + b, 0), B a,
2
2
and C (a – b, p) respectively.
( )
(iv) Direction of Tangent: r = a + b cos q
dr
= –b sin q
dq
d
q
tan f = r
dr
(a + b cos q )
=
− b sin q
At point A(a + b, 0): tan f
,f= p
2
Thus, the tangent is perpendicular to the initial line q = 0
−a ⎞
p
At point B a,
: tan f = – a , f = tan–1 ⎛⎜
= p – tan–1 a
⎝ b ⎟⎠
2
b
b
–1 a
Thus, the tangent makes an angle p – tan
with the line q = p .
b
2
At point C (a – b, p ): tan f
,f= p
2
Thus, the tangent is perpendicular to the line q = p.
(v) Region: Minimum value of r = a – b, since minimum value of cos q = –1. Thus r
is always positive.
(vi) Asymptote: There is no asymptote of the curve since r is finite for all values of q.
(vii) Variation of r:
p
p
2p
q
0
p
3
2
3
b
r
a+b a+
a
a– b
a–b
2
2
( )
q=p
2
p
B a,
2
C
(a – b, p)
A (a + b, 0)
q=0
O
Fig. 3.29
Engineering Mathematics
3.102
Case II: a < b
(ii) Pole: It lies on the curve.
a
If r = 0, cos q =
> –1
b
⎛ −a ⎞
⎛ −a ⎞
Thus at q = cos–1 ⎜⎝
⎟ , r = 0. Therefore, q = cos–1 ⎜⎝
⎟ is the tangent at origin.
b ⎠
b ⎠
(iii) Points of Intersection: The curve meets the line q = 0, q = p and q = p at
2
p
A (a + b, 0), B a,
and C (a – b, p) respectively.
2
(iv) Direction of Tangent: Same as case I
(v) Region: Minimum value of r = a – b < 0, thus r is negative for some values of q.
(vi) Asymptote: Same as case I
(vii) Variation of r: Same as case I. But, here a – b < 0. Therefore, for some values of
q, r is negative. Thus, a smaller loop exists between the points 0 and C.
( )
q=p
2
a, p B
2
O
C {–(b – a), 0}
A (a + b, 0)
q=0
Fig. 3.30
Case III: a = b the r = a (1 + cos q) which is a cardioid.
(ii) Pole: It lies on the curve, since at q = p, r = 0. Tangent at the pole is the line q = p.
p
(iii) Points of Intersection: Curve meets the line q = 0, q = p at A(2a, 0) and B a,
2
2
respectively.
(iv) Direction of Tangent: From Case I
( )
tan f =
a + b cos q
− b sin q
1 + cos q
=
=
− sin q
2 cos 2 q
p q
2
+
= – cot q = tan
q
q
2 2
2
2 sin cos
2
2
(
)
f= p + q
2
2
At point A(2a, 0), f = p . Thus, the tangent is perpendicular to the initial line
2
q = 0.
Differential Calculus II
( p2 ) , f = p2
At point B a,
3.103
+ p = 3p . Thus, the tangent makes an angle 3p
4
4
4
with the line q = p .
2
(v) Region: The maximum value of r is 2a. Thus, the whole curve lies within a circle
with centre at the pole and radius 2a.
(vi) Asymptotes: Same as case I
(vii) Variation of r:
q
0
r
2a
p
3
3a
2
p
2
a
2p
3
a
2
p
0
q=p
2
a, p B
2
O (0, p)
A (2a, 0)
q=0
Fig. 3.31
Example 3: Trace the lemniscate of Bernoulli r2 = a2 cos 2q.
Solution:
(i) Symmetry: The curve is symmetric about the initial line q = 0 and the line
q = p , since when q is replaced by –q and by p – q respectively, equation of the
2
curve remains unchanged.
The curve is also symmetric about the pole since power of r is even.
(ii) Pole: It lies on the curve since r = 0, at q = p . Tangents at the pole are the lines
4
q=± p .
4
(iii) Points of Intersection: The curve meets the initial line q = 0 at A(a, 0) and B(– a, 0).
(iv) Direction of Tangent: r2 = a2 cos 2q
2r
dr
= – 2a2 sin 2q
dq
dr
a 2 sin 2q
=–
dq
r
Engineering Mathematics
3.104
tan f = r dq
dr
(
)
r2
a 2 cos 2q
p
+ 2q
=
–
= – cot 2q = tan
2
2
sin
2
q
a sin 2q
f = p + 2q
2
p
At point A(a, 0), f = . Thus, the tangent is perpendicular to the initial line q = 0.
2
Due to symmetry, the curve is discussed only between q = 0 to q = p .
2
(v) Region:
(a) Since maximum value of cos 2q is 1, the maximum value of r is a. Thus, the
whole curve lies within a circle with centre at the pole and radius a.
=
⎡ Due to symmetry ⎤
⎥
p
p
p ⎢
(b) cos 2q < 0, if
< 2q < p, i.e.,
<q<
considering q
⎢
⎥.
2
4
2 ⎢
p⎥
between 0 and
2 ⎥⎦
⎣⎢
p
p
2
Thus r < 0, when
<q< .
4
2
p
p
Therefore, the curve does not exists in the region
<q< .
4
2
(vi) Asymptote: There is no asymptote of the curve since r is finite for all values of q.
p
and
(vii) Variation of r: Since the curve meets the initial line at two points O 0,
4
A(a, 0) and is symmetric about the initial line, a loop exists between the points O
and A.
( )
q
0
r
a
p
8
a
p
6
a
2
1
( 2) 4
p
4
0
q=p
2
q=p
4
p
O 0,
4
A(0, a)
q=0
q = –p
4
Fig. 3.32
Differential Calculus II
3.105
Example 4: Trace the lemniscate r2 = a2 sin 2q.
Solution:
(i) Symmetry: The curve is symmetric at about
(a) the pole since power of r is even
(b) the line q = p , since on replacing q by p – q, equation remains unchanged.
4
2
(ii) Pole: It lies on the curve since r = 0 at q = 0 and q = p . Tangents at the pole are
2
the lines q = 0 and q = p .
2
p
(iii) Points of Intersection: The curve meets the line q = p at A a,
.
4
4
(iv) Direction of Tangent: r2 = a2 sin 2q
dr
2r
= 2a2 cos 2q
dq
dr
a 2 cos 2q
=
dq
r
d
q
tan f = r
dr
a 2 sin 2q
r2
= 2
= 2
= tan 2q
a cos 2q
a cos 2q
( )
f = 2q
( p4 ) , f = p2 . Thus, the tangent is perpendicular to the line q = p4 .
At point A a,
(v) Region:
(a) Since maximum value of sin 2q is 1, the maximum value of r is a. Thus, the
whole curve lies within a circle of radius a and centre at the pole.
p
(b) sin 2q < 0, p < 2q < 2p, i.e.,
<q<p
2
p
2
Thus, r < 0, when
< q < p. Therefore, the curve does not exists in the
2
p
region
< q < p, i.e., second quadrant and due to symmetry in the fourth
2
quadrant too.
(vi) Asymptote: There is no asymptote of the curve since r is finite for all values of q.
(vii) Variation of r: Since, the curve meets the line q = p at two points O(0, 0) and
4
p
A a,
and is symmetric about this line, a loop exists between the points O and A.
4
( )
q
0
r
0
p
8
a
(2)1 / 4
p
4
a
3p
8
a
(2)1 / 4
p
2
0
Engineering Mathematics
3.106
q=p
2
q=p
4
p
A a,
4
O
q=0
Fig. 3.33
Example 5: Trace the three leaved rose r = a sin 3q.
Solution: Here n = 3 (odd), therefore, the curve consists of three similar loops.
(i) Symmetry: The curve is symmetric about the line q = p , since on replacing q by
2
p – q, equation of the curve remains unchanged.
(ii) Pole: It lies on the curve since r = 0 at q = 0, p , 2p , p. Tangents at the pole are
3 3
the lines q = 0, q = p , q = 2p , q = p.
3
3
p
(iii) Points of Intersection: The curve meets the line q = p at A − a,
.
2
2
(iv) Direction of Tangent: r = a sin 3q
dr
= 3a cos 3q
dq
tan f = r dq
dr
a sin 3q
=
= 1 tan 3q
3a cos 3q
3
p
1
3
p
At point A − a,
, tan f = tan
, f = p . Thus, the tangent is perpen2
3
2
2
p
dicular to the line q =
2
(v) Region: Since, the maximum value of sin 3q is 1, the maximum value of r is a.
Thus, the whole curve lies within a circle of radius a and centre at the pole.
(vi) Asymptote: There is no asymptote of the curve since r is finite for all values of q.
(vii) Variation of r: The curve is symmetric above the line q = p and meets this line at
2
p
A − a,
and also passes through the pole O. Therefore, a loop exists between
2
the points O and A. This curve consists of three similar loops. Therefore, two more
similar loops, exists in the first and second quadrant due to symmetry.
(
(
(
)
)
)
Differential Calculus II
q
0
r
0
p
6
a
q=
3.107
p
3
0
p
2
–a
p
2
B a,
O
p
6
q=0
p
A –a,
2
Fig. 3.34
Example 6: Trace the four leaved rose r = a cos 2q, a > 0.
Solution: Here n = 2 (even), therefore, the curve consists of 2n, i.e., 4 similar loops.
(i) Symmetry: The curve is symmetric about the
(a) initial line q = 0, since on replacing q by – q, equation of the curve remains
unchanged.
(b) line q = p , since on replacing q by (p – q ), equation of the curve remains
2
unchanged.
(ii) Pole: It lies on the curve, since r = 0 at q = p . The tangent at the pole is the line
4
q= p .
4
(iii) Points of Intersection: The curve meets the initial line q = 0 at A(a, 0).
(iv) Direction of Tangent: r = a cos 2q
dr
= –2a sin 2q
dq
tan f = r dq
dr
a cos 2q
=
= – 1 cot 2q
2a sin 2q
2
1
At point A(a, 0), tan f = – cot 0
,f= p .
2
2
Thus, the tangent is perpendicular to the initial line.
(v) Region: Since maximum value of cos 2q is 1, the maximum value of r is a. Thus,
the whole curve lies within a circle of radius a and centre at the pole.
Engineering Mathematics
3.108
(vi) Asymptote: There is no asymptote of the curve since r is finite for all values of q.
(vii) Variation of r: The curve is symmetrical about the initial line q = 0 and meet this
line at A(a, 0) and also passes through origin. Therefore, a loop exists between the
points O and A. This curve consists of 4 similar loops. Hence, three more similar
loops can be drawn using the symmetry about the line q = 0 and q = p .
2
p
p
3p
p
q
0
8
4
8
2
a
a
r
a
0
–
–a
2
2
q=
p
2
q=
p
4
A(a, 0)
O
q=0
Fig. 3.35
Exercise 3.8
Trace the following curves:
1. r = a (1 + sin q )
Ans.:
2. r2 cos 2q = a2
Ans.:
q=p
2
q=p
2
B(0, 2a)
O
O
A(a, 0)
q=0
q=p
4
q=0
q=–p
4
Differential Calculus II
3. r = 2a cos q
3.109
5. r = a sin 2q
Ans.:
Ans.:
q=p
2
O
q=p
2
q=p
4
q=0
(2a, 0)
q=0
O
q = –p
4
4. r = a cos 3q
Ans.:
q=p
2
6. r = 2 (1 – 2 sin q )
Ans.:
q=p
6
O
A(a, 0)
q=0
q = 5p
6
q=p
2
q=p
6
O
A(2, 0)
q=0
B
FORMULAE
Tangent and Normal
Equation of the tangent at any point
(x, y): Y – y = fÄ (x) (X – x)
Length of sub-normal = y
Angle of Intersection of Curves
m2 − m1
1 + m2 m1
Length of Tangent, Sub-tangent, Normal
and Sub-normal
⎛ dx ⎞
Length of tangent = y 1 + ⎜ ⎟
⎝ dy ⎠
dx
dy
⎛ dy ⎞
Length of normal = y 1 + ⎜⎝ ⎟⎠
dx
Equation of the normal at any point
1
(x, y): Y – y = −
(X – x)
f ′( x )
= tan −1
Length of sub-tangent = y
2
2
dy
dx
Length of Perpendicular from Origin to
the Tangent
y − x f ′( x )
p=
1 + [ f ′( x )]2
Angle between Radius Vector and Tangent
f (θ )
dθ
tan φ =
=r
f ′(θ )
dr
Engineering Mathematics
3.110
Length of Polar Tangent, Polar Sub-tangent,
Polar Normal and Polar Sub-normal
Length of polar tangent
2 ⎛d ⎞
= r 1+ r ⎜ ⎟
⎝ dr ⎠
2
Length of polar sub-tangent = r 2
d
dr
Length of polar normal
=
⎛ dr ⎞
r2 + ⎜ ⎟
⎝d ⎠
2
Length of polar sub-normal =
dr
d
Derivative of Length of an arc
(i) Cartesian form
ds
⎛ dy ⎞
= 1+ ⎜ ⎟
⎝ dx ⎠
dx
⎛ dx ⎞
ds
= 1+ ⎜ ⎟
dy
⎝ dy ⎠
2
3
⎡ ⎛ dy ⎞ 2 ⎤ 2
⎢1 + ⎜ ⎟ ⎥
⎢⎣ ⎝ dx ⎠ ⎥⎦
=
d2 y
dx 2
⎡ ⎛ dx ⎞ 2 ⎤
⎢1 + ⎜ ⎟ ⎥
⎢ ⎝ dy ⎠ ⎥⎦
=⎣
d2 x
dy 2
(ii) Polar form
3
⎡ 2 ⎛ dr ⎞ 2 ⎤ 2
⎢r + ⎜ ⎟ ⎥
⎝ dθ ⎠ ⎥⎦
⎢⎣
ρ=
2
d 2r
⎛ dr ⎞
r2 + 2 ⎜ ⎟ − r 2
⎝ dθ ⎠
dθ
X= x−
(ii) Parametric form
2
ds
⎛ dx ⎞ ⎛ dy ⎞
= ⎜ ⎟ +⎜ ⎟
⎝ dt ⎠ ⎝ dt ⎠
dt
2
(iii) Polar form
2
ds
⎛d ⎞
= 1 + r2 ⎜ ⎟
⎝ dr ⎠
dr
2
dy ⎡ ⎛ dy ⎞
⎢1 + ⎜ ⎟
dx ⎢⎣ ⎝ dx ⎠
d2 y
dx 2
2
⎛ dy ⎞
1+ ⎜ ⎟
⎝ dx ⎠
Y= y+
d2 y
dx 2
Choose the correct alternative in each of the following:
1. The equation of the tangent to the
curve y = 2 sin x + sin 2x at x = p is
3
equal to
2
⎤
⎥
⎥⎦
,
Circle of Curvature
Equation of the circle of curvature at
any point (x, y) with radius r and centre
C(X, Y):
(x – X)2 + (y – Y)2 = r2
MULTIPLE CHOICE QUESTIONS
(a) 2y = 3 3
3/ 2
Centre of Curvature
2
ds
⎛ dr ⎞
= r2 + ⎜ ⎟
⎝d ⎠
d
Radius of Curvature
(i) Cartesian form
(b) y = 3 3
(c) 2y + 3 3 = 0
(d) y + 3 3 = 0
Differential Calculus II
2. The sum of the squares of the
intercept made on the co-ordinate
axis by the tangents to the curve
2
2
2
x 3 + y 3 = a 3 is
(b) 2a2
(a) a2
2
(c) 3a
(d) 4a2
3. The equation of the normal to the
curve y = x (2 – x) at the point (2, 0) is
(a) x – 2y = 2
(b) 2x + y = 4
(c) x – 2y + 2 = 0
(d) none of these
4. The length of the normal at t on the
curve x = a (t + sin t), y = a(1 – cos t)
is
(a) a sin t
(b) 2a sin3 t sec t
2
2
(c) 2a sin t tan t
2
2
(d) 2a sin t
2
5. The length of the sub-tangent to the
curve x2 + xy + y2 = 7 at (1, –3) is
(a) 3
(b) 5
(c) 15
(d) 3
5
6. The angle of intersection of the
curves y = 4 – x2 and y = x2 is
4
(a) p
(b) tan–1
3
2
⎛
⎞
4
2
(c) tan–1 ⎜
(d) none of these
⎝ 7 ⎟⎠
()
7. The length of the sub-normal to the
parabola y2 = 4ax at any point is
equal to
2a
a
(c)
2
8. If x =
(b) 2 2a
(a)
(d) 2a
sin q) and
dy
y = a (1 – cos q), then
will be
dx
equal to
a (q
+
3.111
(b) cos q
(a) sin q
2
2
q
q
(c) tan
(d) cot
2
2
9. The radius of curvature at the point
(s, y) on the curve s = c log sec y is
(a) c sec y
(b) c cot y
(c) c sec y tan y (d) c tan y
10. The angle of intersection of the curve
xy = a2 and x2 + y2 = 2a2 is
p
p
(a)
(b)
4
2
(c) 0
(d) p
11. The radius of curvature for the curve
y = ex at (0, 1) is
(a)
2
(b) 2 2
1
1
(c)
(d)
2
2 2
12. If the tangent to a curve at the point
P(x, y) meets the x-axis in T, then
PT is
2
(a) y dx
dy
⎛ dy ⎞
(b) y 1 + ⎝ ⎠
dx
dy
dx
(d) y 1 + ⎛⎜ dx ⎞⎟
⎝ dy ⎠
(c) y
2
x
13. For the curve y = be a , which one of
the following is true?
(a) The sub-tangent is of constant
length and the sub-normal varies as the square of the ordinate.
(b) The sub-tangent varies as the
square of the ordinate and the
sub-normal is of constant length.
(c) The sub-tangent is of constant
length and the sub-normal varies as the ordinate.
(d) The sub-tangent varies as the ordinate and the sub-normal is of
constant length.
14. The envelope of a one-parameter
family of straight lines x cos
Engineering Mathematics
3.112
a + y sin a = a, where a is a parameter is
(b) y2 = 4a2
(a) xy = a2
(c) x2 – y2 = a2 (d) x2 + y2 = a2
15. Match list I and II and select the
correct answer using the codes given
below the list.
List I
(Curves)
List II
(Equations)
x
1. y = c cosh ⎛ ⎞
⎝c⎠
(A) Cubical
parabola
(B) Catenary
(C) Astroid
2. y2 = ax3
3. y = ax3
2
2
2
(D) Semicubical 4. x 3 + y 3 = a 3
parabola
Codes
A
B
C
D
a
3
4
1
2
b
3
1
4
2
c
2
3
4
1
d
4
1
2
3
y
y
y = –x
y=x
O
A (a, 0)
represents the curve given by
(a) y2 (a2 + x2) = x2 (a2 – x2)
(b) x3 + y3 = x
Answers
1. (a)
8. (c)
15. (b)
2. (a)
9. (d)
16. (a)
3. (a)
10. (c)
17. (a)
x
O
16. The following figure
B
(–a, 0)
(c) y2 (a + x) = x2 (3a – x)
(d) y2 = (x – a) (x – b) (x – c)
17. If the equation of the tangent to
y = 3x2 – 4x at (1, –1) is
ax = y + b, then the values of a and b
respectively are
(a) 2 and 3
(b) 3 and 2
(c) 1 and 2
(d) 2 and 1
18. Which one of the following lines
is a line of symmetry of the curve
x3 + y3 = 3 (xy2 + yx2)
(a) x = 0
(b) y = 0
(c) y = x
(d) y = – x
19. The given figure represents the curve
whose parametric equations are
x
(a) x = a (1 – sin t)
y = a (1 – cos t)
(b) x = a (1 + sin t)
y = a (1 – cos t)
(c) x = a (t – sin t)
y = a (1 + cos t)
(d) x = a (t + sin t)
y = a (1 + cos t)
20. The ratio of the sub-tangent to
the sub-normal for any point
on the curve x = a (q + sin q),
y = a (1 – cos q) is
(a) tan2 q
(b) cot2 q
2
2
(c) sin2 q
(d) cos2 q
2
2
4. (c)
11. (b)
18. (c)
5. (c)
12. (d)
19. (d)
6. (c)
13. (a)
20. (b)
7. (d)
14. (d)
Partial
Differentiation
Chapter
4
4.1 INTRODUCTION
We often come across functions which depend on two or more variables. For example,
area of a triangle depends on its base and height, hence we can say that area is the function of two variables, i.e., its base and height. u is called a function of two variables x
and y, if u has a definite value for every pair of x and y. It is written as u = f (x, y). The
variables x and y are independent variables while u is dependent variable. The set of all
the pairs (x, y) for which u is defined is called the domain of the function. Similarly, we
can define function of more than two variables.
4.2 PARTIAL DERIVATIVE
A partial derivative of a function of several variables is the ordinary derivative w.r.t.
one of the variables, when all the remaining variables are kept constant. Consider a
function u = f (x, y), here, u is the dependent variable and x and y are independent
variables. The partial derivative of u = f (x, y) w.r.t. x is the ordinary derivative of u w.r.t.
f
u
x, keeping y constant. It is denoted by
or
or ux or fx and is known as first order
x
x
partial derivative of u w.r.t. x.
∂u
⎡ f ( x + Δx, y ) − f ( x, y ) ⎤
= lim ⎢
⎥⎦
Δ
x
→
0
∂x
Δx
⎣
Similarly, the partial derivative of u = f (x, y) w.r.t. y is the ordinary derivative of u
u
f
w.r.t. y treating x as constant. It is denoted by
or
or uy or fy and is known as first
y
y
order partial derivative of u w.r.t. y.
⎡ f ( x, y + Δy ) − f ( x, y ) ⎤
∂u
= lim ⎢
⎥
Δ
y
→
0
∂y
Δy
⎣
⎦
4.2
Engineering Mathematics
4.2.1 Geometrical Interpretation
The function u = f (x, y) represents a surface. The point
P [x1, y1, f (x1, y1)] on the surface corresponds to the values
x1, y1 of the independent variables x, y. The intersection
of the plane y = y1 (parallel to the zox–plane) and the surface u = f (x, y) is the curve shown by the dotted line in the
Figure. On this curve, x and u vary according to the relation
u = f (x, y1). The ordinary derivative of f (x, y1) w.r.t. x at x1
Fig. 4.1
⎛ ∂u ⎞
⎛ ∂u ⎞
is ⎜ ⎟
. Hence, ⎜ ⎟
is the slope of the tangent to
⎝ ∂x ⎠ ( x, y1 )
⎝ ∂x ⎠ ( x1 , y1 )
the curve of the intersection of the surface u = f (x, y) with the plane y = y1 at the point
P[x1, y1, f (x1, y1)].
⎛ ∂u ⎞
Similarly, ⎜ ⎟
is the slope of the tangent to the curve of the intersection of the
⎝ ∂y ⎠ ( x1, y1 )
surface u = f (x, y) with the plane x = x1 at the point P[x1, y1, f (x1, y1)].
4.3 HIGHER ORDER PARTIAL DERIVATIVES
Partial derivatives of higher order, of a function u = f (x, y), are obtained by partial differentiation of first order partial derivative. Thus, if u = f (x, y), then
∂ 2 u ∂ ⎛ ∂u ⎞
=
⎜ ⎟
∂x 2 ∂x ⎝ ∂x ⎠
∂2 u
∂ ⎛ ∂u ⎞
=
⎜ ⎟
∂y ∂x ∂y ⎝ ∂x ⎠
∂ 2 u ∂ ⎛ ∂u ⎞
=
∂y 2 ∂y ⎜⎝ ∂y ⎟⎠
∂2 u
∂ ⎛ ∂u ⎞
=
∂x ∂y ∂x ⎜⎝ ∂y ⎟⎠
are called second order partial derivatives. Similarly, other higher order derivatives can
also be obtained.
Note:
2
1. If u = f (x, y) possesses continuous second order partial derivatives
∂2 u
∂2 u
u
=
. This is called commutative property.
then
,
∂x ∂y ∂y ∂x
y x
2
u
and
x y
Partial Differentiation
4.3
2. Standard rules for differentiation of sum, difference, product and quotient are
also applicable for partial differentiation.
−
1
Example 1: If u = (1 − 2 xy + y 2 ) 2 , then show that x
u = (1 − 2 xy + y 2 )
Solution:
ìu
ìu
-y
= u3 y 2 .
ìx
ìy
−
1
2
Differentiating u partially w.r.t. x and y,
3
−
∂u −1
=
(1 − 2 xy + y 2 ) 2 (−2 y )
2
∂x
3
−
∂u −1
=
(1 − 2 xy + y 2 ) 2 (−2 x + 2 y )
2
∂y
Hence,
x
3
−
∂u
∂u
−y
= (1 − 2 xy + y 2 ) 2 ( xy − xy + y 2 )
∂x
∂y
3
1
− ⎤
⎡
= ⎢(1 − 2 xy + y 2 ) 2 ⎥ y 2
⎣
⎦
= u3 y 2 .
Example 2: If u = log (tan x + tan y + tan z), then show that
sin 2x
ìu
ìu
ìu
+ sin 2y
+ sin 2z
= 2.
ìx
ìy
ìz
Solution:
u = log (tan x + tan y + tan z)
Differentiating u partially w.r.t. x, y and z,
1
∂u
=
⋅ sec 2 x
∂x tan x + tan y + tan z
1
∂u
=
⋅ sec 2 y
∂y tan x + tan y + tan z
1
∂u
=
⋅ sec 2 z
∂z tan x + tan y + tan z
Hence,
∂u
∂u
∂u
+ sin 2 y
+ sin 2 z
∂x
∂y
∂z
2 sin x cos x sec 2 x + 2 sin y cos y sec 2 y + 2 sin z cos z sec 2 z
=
tan x + tan y + tan z
2 (tan x + tan y + tan z )
=
= 2.
tan x + tan y + tan z
sin 2 x
4.4
Engineering Mathematics
Example 3: If u =
ìu ìu ìu
e x+ y+ z
+
+
= 2u.
, show that
x
y
z
ìx ìy ìz
e +e +e
ex+ y+ z
ex + e y + ez
Differentiating u partially w.r.t. x,
u=
Solution:
∂u
ex+ y+z
ex+ y+z
= x
− x
⋅ ex
y
z
∂x e + e + e ( e + e y + e z ) 2
ex+ y+z
ex + e y + ez
⎛
⎞
ex
−
1
⎜⎝ e x + e y + e z ⎟⎠
... (1)
∂u
ex+ y+z
= x
∂y e + e y + e z
⎛
⎞
ey
−
1
⎜⎝ e x + e y + e z ⎟⎠
... (2)
∂u
ex+ y+z
= x
∂z e + e y + e z
⎛
⎞
ez
−
1
⎜⎝ e x + e y + e z ⎟⎠
... (3)
=
Similarly,
Adding Eqs (1), (2) and (3),
∂u ∂u ∂u
ex+ y+z
+
+
= x
∂x ∂y ∂z e + e y + e z
⎛ ex + e y + ez ⎞
⎜⎝ 3 − e x + e y + e z ⎟⎠
ex+ y+z
(3 − 1)
e + e y + ez
= 2u
=
x
2
⎛ ìu ì u ⎞
⎛
ìu ìu ⎞
x2 + y2
- ⎟ = 4 ⎜1 Example 4: If u (x, y) =
, then show that ⎜
.
⎝
ì
x
ì
y
⎠
⎝
ìx ìy ⎟⎠
x+ y
Solution:
u (x, y) =
x2 + y2
x+y
u (x + y) = x2 + y2
Differentiating Eq. (1) partially w.r.t. x,
∂u
u + ( x + y)
= 2x
∂x
∂u 2 x − u
=
∂x
x+ y
Differentiating Eq. (1) partially w.r.t. y,
u + ( x + y)
∂u
= 2y
∂y
∂u 2 y − u
=
∂y
x+ y
... (1)
Partial Differentiation
2
⎛ ∂u ∂u ⎞
⎛ 2x − u 2 y − u ⎞
⎜⎝ ∂x − ∂y ⎟⎠ = ⎜⎝ x + y − x + y ⎟⎠
⎡ 2( x − y ) ⎤
=⎢
⎥
⎣ ( x + y) ⎦
4.5
2
2
... (2)
⎛ ∂u ∂u ⎞
⎛ 2x − u 2 y − u ⎞
4 ⎜1 −
−
= 4 ⎜1 −
−
⎝ ∂x ∂y ⎟⎠
⎝
x+ y
x + y ⎟⎠
Again,
⎛ 2x − u + 2 y − u ⎞
⎡ 2( x + y )
2u ⎤
= 4 ⎜1 −
= 4 ⎢1 −
+
⎟
⎝
⎠
x+ y
( x + y ) ( x + y ) ⎥⎦
⎣
⎡
⎧ x 2 + y 2 ⎫⎤
⎡ −( x + y ) 2 + 2 x 2 + 2 y 2 ⎤
=
4
= 4 ⎢1 − 2 + 2 ⎨
⎢
⎥
2 ⎬⎥
( x + y)2
⎩ ( x + y ) ⎭⎦
⎣
⎦
⎣
=
4( x 2 + y 2 − 2 xy ) ⎡ 2( x − y ) ⎤
=⎢
⎥
( x + y)2
⎣ ( x + y) ⎦
2
... (3)
From Eqs (1) and (2), we get
2
⎛ ∂u ∂u ⎞
⎛ ∂u ∂u ⎞
⎜⎝ ∂x − ∂y ⎟⎠ = 4 ⎜⎝1 − ∂x − ∂y ⎟⎠ .
−
1
2
− x2
Example 5: If z = ct e
Solution:
4 a2 t
2
ìz
2 ì z
.
=
a
, prove that
ìt
ìx 2
−
− x2
1
2
z = ct e
4 a2 t
Differentiating z partially w.r.t. t,
− x2
− x2
1
−
∂z
1 −3 2
x2 ⎞
2 ⎛
= − ct 2 e 4 a t + ct 2 e 4 a t ⎜ 2 2 ⎟
∂t
2
⎝ 4a t ⎠
− x2
=
4 a2 t
ce
2
t
−
5
2
⎛
x2 ⎞
−
t
+
⎜⎝
2a 2 ⎟⎠
Differentiating z partially w.r.t. x,
− x2
1
−
∂z
2 ⎛ −2 x ⎞
= ct 2 e 4 a t ⎜ 2 ⎟
⎝ 4a t ⎠
∂x
Differentiating
z
partially w.r.t. x,
x
−
1
∂ 2 z −2ct 2
=
∂x 2
4a 2t
−x
⎡ − x2
⎤
2 ⎛ −2 x ⎞
⎢e 4 a t + xe 4 a t ⎜ 2 ⎟ ⎥
⎝ 4a t ⎠ ⎥
⎢⎣
⎦
2
2
... (1)
4.6
Engineering Mathematics
−
1
− x2
ct 2
x2 ⎞
2 ⎛
= 2 2 e 4 a t ⎜ −t + 2 ⎟
2a t
2a ⎠
⎝
− x2
∂ z ce
=
2
∂x 2
4 a2 t
2
a2
⋅t
−
5
2
⎛
x2 ⎞
⎜⎝ −t + 2a 2 ⎟⎠
... (2)
From Eqs (1) and (2), we get
∂z
∂2 z
= a2 2 .
∂t
∂x
Example 6: If u (x, t) = ae–gx sin(nt – gx), where a, g, n are constants, satisfying
ìu
ì2u
1 n
= a 2 2 , prove that g =
the equation
.
ìt
ìx
a 2
Solution:
u (x, t) = ae–gx sin (nt – gx)
Differentiating u partially w.r.t. t,
∂u
= ae − gx [cos ( nt − gx )] n
∂t
Differentiating u partially w.r.t. x,
∂u
= − age − gx sin( nt − gx ) + [ ae − gx cos( nt − gx )] ( − g )
∂x
= − age − gx [sin( nt − gx ) + cos( nt − gx )]
Differentiating
u
partially w.r.t. x,
x
∂2 u
= ag 2 e − gx [sin( nt − gx ) + cos( nt − gx )] − age − gx [− g cos( nt − gx ) + g sin( nt − gx )]
∂x 2
= ag 2 e − gx [2 cos( nt − gx )]
Substituting in
∂u
∂2 u
= a2 2 ,
∂t
∂x
− gx
ae [cos( nt − gx )] n = a3 g 2 e − gx [2 cos( nt − gx )]
g2 =
g=
y
Example 7: If u = e x , find
Solution:
u = ex
y
ì2u
.
ìy ìx
n
2a 2
1 n
.
a 2
Partial Differentiation
Differentiating u partially w.r.t. x,
y
y
∂u
∂ y
= ex
( x ) = e x ⋅ yx y −1
∂x
∂x
Differentiating
u
partially w.r.t. y,
x
y
y
∂ ⎛ ∂u ⎞
∂
xy ∂
( x y ) ⋅ yx y −1 + e x x y −1 + e x y ( x y −1)
⎜⎝ ⎟⎠ = e
∂y ∂x
∂y
∂y
y
y
y
∂2 u
= e x x y log x ⋅ yx y −1 + e x x y −1 + e x yx y −1 log x
∂y ∂x
y
= e x x y −1 ( yx y logg x + 1 + y log x ).
Example 8: If u = e xyz, show that
Solution:
ì 3u
= (1 + 3 xyz + x 2 y 2 z 2 )e xyz .
ìx ìy ìz
u = e xyz
Differentiating u partially w.r.t. z,
∂u
= e xyz ⋅ xy
∂z
Differentiating
u
partially w.r.t. y,
z
∂ ⎛ ∂u ⎞
∂2 u
= xe xyz + x 2 yze xyz
⎜⎝ ⎟⎠ =
∂y ∂z
∂y ∂z
2
Differentiating
u
partially w.r.t. x,
y z
∂ ⎛ ∂2 u ⎞
∂3 u
=
= e xyz + xyze xyz + 2 xyze xyz + x 2 y 2 z 2 e xyz
⎜
⎟
∂x ⎝ ∂y ∂z ⎠ ∂x ∂y ∂z
= (1 + 3xyz + x 2 y 2 z 2 )e xyz .
2
Example 9: If u = x 3 y + e xy , prove that
Solution:
u = x 3 y + e xy
2
Differentiating u partially w.r.t. x,
2
∂u
= 3 x 2 y + e xy ⋅ y 2
∂x
ì2u
ì2u
=
.
ìx ìy ìy ìx
4.7
4.8
Differentiating
Engineering Mathematics
u
partially w.r.t. y,
x
∂ ⎛ ∂u ⎞
2
xy 2
2 xy 2
⎜⎝ ⎟⎠ = 3 x + 2 ye + y e ⋅ 2 xy
∂y ∂x
2
∂2 u
= 3 x 2 + 2 ye xy (1 + xy 2 )
∂y ∂x
... (1)
Differentiating u partially w.r.t. y,
2
∂u
= x 3 + e xy ⋅ 2 xy
∂y
Differentiating
u
partially w.r.t. x,
y
2
2
∂ ⎛ ∂u ⎞
= 3 x 2 + 2 ye xy + 2 xye xy . y 2
∂x ⎜⎝ ∂y ⎟⎠
2
∂2 u
= 3 x 2 + 2 ye xy (1 + xy 2 )
∂x ∂y
... (2)
From Eqs (1) and (2), we get
∂2 u
∂2 u
=
.
∂y ∂x ∂x ∂y
Example 10: If z = x y+ y x, prove that
Solution:
ì2 z
ì2 z
=
.
ìx ìy ìy ìx
z = x y+ y x
y
x
z = e log x + e log y = e y log x + e x log y
Differentiating z partially w.r.t. x,
∂z
y
= e y log x ⋅ + e x log y ⋅ log y
∂x
x
Differentiating
z
partially w.r.t. y,
x
∂ ⎛ ∂z ⎞ 1 y log x
1
x
+ e y log x y log x ) + e x log y ⋅ log y + e x log y ⋅
⎜⎝ ⎟⎠ = (e
∂y ∂x
y
x
y
∂2 z
e y log x
e x log y
=
(1 + y log x ) +
( x log y + 1)
∂y ∂x
x
y
Differentiating z partially w.r.t. y,
∂z
x
= e y log x ⋅ log x + e x log y ⋅
∂y
y
... (1)
Partial Differentiation
Differentiating
4.9
z
partially w.r.t. x,
y
∂ ⎛ ∂z ⎞
y
1
1
x
= e y log x ⋅ log x + e y log x ⋅ + e x log y ⋅ + e x log y log y ⋅
⎜
⎟
∂x ⎝ ∂y ⎠
x
x
y
y
∂2 z
e y log x
e x log y
=
( y log x + 1) +
(1 + x log y )
∂x ∂y
x
y
... (2)
From Eqs (1) and (2), we get
∂2 z
∂2 z
.
=
∂y ∂x ∂x ∂y
3
Example 11: If u = ( 3 xy − y 3 ) − ( y 2 − 2 x ) 2 , show that
ì2u
ì2u
=
.
ìx ìy ìy ìx
3
u = (3 xy − y 3 ) − ( y 2 − 2 x ) 2
Solution:
Differentiating u partially w.r.t. x,
1
1
∂u
3
= 3 y − ( y 2 − 2 x ) 2 ( −2) = 3 y + 3( y 2 − 2 x ) 2
∂x
2
Differentiating
u
partially w.r.t. y,
x
1
−
∂ ⎛ ∂u ⎞
3 2
2
⎜⎝ ⎟⎠ = 3 + ( y − 2 x ) ( 2 y )
∂y ∂x
2
∂2 u
= 3+
∂y ∂x
3y
y − 2x
... (1)
2
Differentiating u partially w.r.t. y,
1
∂u
3
= 3x − 3 y 2 − ( y 2 − 2 x) 2 ( 2 y)
∂y
2
Differentiating
u
partially w.r.t. x,
y
∂ ⎛ ∂u ⎞
1
= 3 − 3y
( −2)
⎜
⎟
2
∂x ⎝ ∂y ⎠
2 y − 2x
∂2 u
= 3+
∂x ∂y
3y
y − 2x
From Eqs (1) and (2), we get
∂2 u
∂2 u
=
∂x ∂y ∂y ∂x
2
... (2)
4.10
Engineering Mathematics
⎛ x⎞
⎛ y⎞
Example 12: If z = x 2 tan -1 ⎜ ⎟ - y 2 tan -1 ⎜ ⎟ ,
⎝ x⎠
⎝ y⎠
prove that
x2 - y2
ì2 z
ì2 z
= 2
.
=
ìx ìy ìy ìx
x + y2
Solution:
⎛x⎞
⎛ y⎞
z = x 2 tan −1 ⎜ ⎟ − y 2 tan −1 ⎜ ⎟
⎝x⎠
⎝ y⎠
Differentiating z partially w.r.t. x,
∂z
⎛ y⎞
= 2 x tan −1 ⎜ ⎟ + x 2 ⋅
⎝x⎠
∂x
= 2 x tan −1
= 2 x tan −1
Differentiating
1
y2
1+ 2
x
⎛ y⎞
⎜⎝ − 2 ⎟⎠ −
x
y2 ⎛ 1 ⎞
x 2 ⎜⎝ y ⎟⎠
1+ 2
y
y
x2 y
y3
− 2
−
x x + y2 x2 + y2
y
−y
x
z
partially w.r.t. y,
x
∂ ⎛ ∂z ⎞
∂2 z
=
= 2x ⋅
⎜ ⎟
∂y ⎝ ∂x ⎠ ∂y ∂x
1 ⎛1⎞
⎜ ⎟ −1
y2 ⎝ x ⎠
1+ 2
x
2
2x
2x2 − x2 − y2
= 2
−1 =
2
x2 + y2
x +y
=
x2 − y2
x2 + y2
... (1)
Differentiating z partially w.r.t. y,
∂z
= x2
∂y
=
1 ⎛1⎞
1
− y2
2 ⎜ ⎟
⎝
⎠
x2
y x
1+ 2
1+ 2
y
x
x3
xy 2
x
+
− 2 y tan −1
2
2
2
2
y
x +y
y +x
= x − 2 y tan −1
Differentiating
⎛ x ⎞
−1 x
⎜⎝ − y 2 ⎟⎠ − 2 y tan y
z
partially w.r.t. x,
y
x
y
Partial Differentiation
∂ ⎛ ∂z ⎞
∂2 z
=
= 1− 2y
∂x ⎜⎝ ∂y ⎟⎠ ∂x ∂y
= 1−
=
4.11
1 ⎛1⎞
x 2 ⎜⎝ y ⎟⎠
1+ 2
y
y2 + x2 − 2 y2
2 y2
=
2
y +x
y2 + x2
2
x2 − y2
x2 + y2
... (2)
From Eqs (1) and (2), we get
∂2 z
∂2 z
x2 − y2
=
= 2
.
∂x ∂y ∂y ∂x x + y 2
Example 13: If u = log (x2 + y2 + z2), prove that x
Solution:
ì2u
ì2u
ì2u
=y
=z
.
ìy ìz
ìz ìx
ìx ìy
u = log (x2 + y2 + z2)
Differentiating u partially w.r.t. x,
∂u
1
= 2
⋅ 2x
∂x x + y 2 + z 2
Differentiating
u
partially w.r.t. y,
x
∂2 u
2x
=− 2
⋅2y
∂x ∂y
( x + y 2 + z 2 )2
z
∂2 u
4 xyz
=− 2
∂x ∂y
( x + y 2 + z 2 )2
... (1)
Differentiating u partially w.r.t. y,
∂u
1
= 2
⋅2y
∂y x + y 2 + z 2
Differentiating
u
partially w.r.t. z,
y
⎞
2y
∂ ⎛ ∂u ⎞ ∂ ⎛
=
∂z ⎜⎝ ∂y ⎟⎠ ∂z ⎜⎝ x 2 + y 2 + z 2 ⎟⎠
=−
x
2y
⋅ 2z
( x + y 2 + z 2 )2
2
4 xyz
∂2 u
=− 2
∂y ∂z
( x + y 2 + z 2 )2
... (2)
4.12
Engineering Mathematics
Differentiating
u
partially w.r.t. z,
x
2x
∂ ⎛ ∂u ⎞
⋅ 2z
⎜ ⎟=− 2
∂z ⎝ ∂x ⎠
( x + y 2 + z 2 )2
y
4 xyz
∂2 u
=− 2
∂z ∂x
( x + y 2 + z 2 )2
From Eqs (1), (2) and (3), we get
x
∂2u
∂2u
∂2u
.
=y
=z
∂y ∂z
∂z ∂x
∂x ∂y
Example 14: If a2 x2 + b2 y2 = c2 z2, evaluate
Solution:
1 ì2 z 1 ì2 z
.
+
a 2 ìx 2 b 2 ìy 2
a2 x2 + b2 y2 = c2 z2
Differentiating partially w.r.t. x,
2a 2 x = 2c 2 z ⋅
∂z
∂x
∂z a 2 x
=
∂x c 2 z
Differentiating
z
partially w.r.t. x,
x
∂2 z a2
=
∂x 2 c 2
2
2
⎛ 1 x ∂z ⎞ a ⎛ x a x ⎞
⎜⎝ − 2 ⋅ ⎟⎠ = 2 ⎜1 − ⋅ 2 ⎟
z z ∂x
c z⎝ z c z⎠
1 ∂2 z
1 ⎛ a2 x 2 ⎞
=
1−
a 2 ∂x 2 c 2 z ⎜⎝ c 2 z 2 ⎟⎠
Similarly,
Hence,
1 ∂2 z
1 ⎛ b2 y 2 ⎞
1−
=
b 2 ∂y 2 c 2 z ⎜⎝ c 2 z 2 ⎟⎠
1 ∂2 z 1 ∂2 z
1 ⎛
a2 x 2 + b2 y 2 ⎞
+ 2 2 = 2 ⎜2 −
2
2
⎟⎠
a ∂x
b ∂y
c z⎝
c2 z 2
=
1 ⎛
1
c2 z 2 ⎞
2 − 2 2 ⎟ = 2 ( 2 − 1)
2 ⎜
c z⎝
c z ⎠ c z
=
1
.
c2 z
Example 15: If u = log (x3 + y3 – x2y – xy2),
prove that
ì2u ì2u
4
ì2u
+2
+
.
=2
ìx ìy ìy 2
ìx
( x + y )2
... (3)
Partial Differentiation
Solution:
u = log (x3 + y3 – x2y – xy2)
= log[( x + y )( x 2 − xy + y 2 ) − xy( x + y )]
= log ( x + y )( x 2 − xy + y 2 − xy )
= log ( x + y )( x − y ) 2
= log ( x + y ) + 2 log( x − y )
Differentiating u partially w.r.t. x,
∂u
1
2
=
+
∂x x + y x − y
Differentiating
u
partially w.r.t. x,
x
∂2 u
1
2
=−
−
2
2
∂x
( x + y) ( x − y)2
Differentiating u partially w.r.t. y,
∂u
1
2
=
−
∂y x + y x − y
Differentiating
Differentiating
u
partially w.r.t. y,
y
∂2 u
1
2
=−
−
∂y 2
( x + y)2 ( x − y)2
u
partially w.r.t. x,
y
∂2 u
1
2
=−
+
∂x ∂y
( x + y )2 ( x − y )2
∂2 u
∂2 u ∂2 u
4
+
2
+ 2 =−
.
2
∂x ∂y ∂y
∂x
( x + y)2
Example 16: If u = log (x3 + y3 + z3 – 3xyz),
2
9
⎛ ì
ì
ì⎞
prove that ⎜
.
u=−
+
+
⎝ ìx ìy ìz ⎟⎠
( x + y + z )2
2
Solution:
⎛∂
⎛∂
∂
∂⎞
∂
∂ ⎞ ⎛ ∂u ∂u ∂u ⎞
⎜⎝ ∂x + ∂y + ∂z ⎟⎠ u = ⎜⎝ ∂x + ∂y + ∂z ⎟⎠ ⎜⎝ ∂x + ∂y + ∂z ⎟⎠
⎛∂
∂
∂⎞
= ⎜ + + ⎟v
⎝ ∂x ∂y ∂z ⎠
where,
v=
∂u ∂u ∂u
+
+
∂x ∂y ∂z
4.13
4.14
Engineering Mathematics
u = log (x3 + y3 + z3 – 3xyz)
Differentiating u partially w.r.t. x, y, and z simultaneously,
∂u
3 x 2 − 3 yz
= 3
∂x x + y 3 + z 3 − 3 xyz
∂u
3 y 2 − 3 xz
= 3
∂y x + y 3 + z 3 − 3 xyz
∂u
3 z 2 − 3 xy
= 3
∂z x + y 3 + z 3 − 3 xyz
v=
∂u ∂u ∂u
+
+
∂x ∂y ∂z
=
3( x 2 + y 2 + z 2 ) − 3( xy + yz + zx )
x 3 + y 3 + z 3 − 3xyz
=
3( x 2 + y 2 + z 2 − xy − yz − zx ) ( x + y + z )
⋅
( x + y + z)
x 3 + y 3 + z 3 − 3 xyz
=
3( x 3 + y 3 + z 3 − 3 xyz )
( x 3 + y 3 + z 3 − 3 xyz )( x + y + z )
=
3
x+ y+z
2
⎛∂
⎛∂
⎞
∂
∂⎞
∂
∂ ⎞⎛
3
⎜⎝ ∂x + ∂y + ∂z ⎟⎠ u = ⎜⎝ ∂x + ∂y + ∂z ⎟⎠ ⎜⎝ x + y + z ⎟⎠
=−
3
3
3
−
−
2
2
( x + y + z ) ( x + y + z ) ( x + y + z )2
=−
9
.
( x + y + z )2
Example 17: If u = 3(ax + by + cz) 2 – (x2 + y2 + z2) and a2 + b2 + c2 = 1, then
show that
ì2u ì2u ì2u
= 0.
+
+
ìx 2 ìy 2 ìz 2
Solution:
u = 3(ax + by + cz) 2 – (x2 + y2 + z2)
Differentiating u partially w.r.t. x,
∂u
= 6 ( ax + by + cz ) a − 2 x
∂x
Differentiating
u
partially w.r.t. x,
x
Partial Differentiation
4.15
∂2 u
= 6a ⋅ a − 2 = 6a 2 − 2
∂x 2
Differentiating u partially w.r.t. y,
∂u
= 6( ax + by + cz )b − 2 y
∂y
u
Differentiating
partially w.r.t. y,
y
∂2 u
= 6b ⋅ b − 2 = 6b 2 − 2
∂y 2
Differentiating u partially w.r.t. z,
∂u
= 6( ax + by + cz )c − 2 z
∂z
Differentiating
u
partially w.r.t. z,
z
∂2 u
= 6c ⋅ c − 2 = 6c 2 − 2
∂z 2
Hence,
∂2 u ∂2 u ∂2 u
+
+
= 6( a 2 + b 2 + c 2 ) − 6
∂x 2 ∂y 2 ∂z 2
[∵ a 2 + b 2 + c 2 = 1]
= 6(1) − 6
=0
Example 18: If u =
Solution:
1
x2 + y2 + z2
u=
, find the value of
ì2u ì2u ì2u
+
+
.
ìx 2 ìy 2 ìz 2
1
x + y2 + z2
2
Differentiating u partially w.r.t. x,
∂u
=−
∂x
Differentiating
1
3
2 2
2( x 2 + y 2 + z )
⋅ 2x = −
x
3
( x2 + y2 + z2 )2
u
partially w.r.t. x,
x
⎡
⎤
∂2 u
1
3x ⋅ 2 x
⎢
⎥
=
−
−
3
5 ⎥
⎢ 2
∂x 2
⎢⎣ ( x + y 2 + z 2 ) 2 2( x 2 + y 2 + z 2 ) 2 ⎥⎦
1
( x 2 + y 2 + z 2 − 3x 2 )
=−
5
( x2 + y2 + z2 )2
4.16
Engineering Mathematics
=
−( −2 x 2 + y 2 + z 2 )
5
( x2 + y2 + z2 )2
Similarly,
∂ 2 u −( x 2 − 2 y 2 + z 2 )
=
5
∂y 2
( x2 + y2 + z2 )2
and
∂ 2 u −( x 2 + y 2 − 2 z 2 )
=
5
∂z 2
( x2 + y2 + z2 )2
Hence,
∂ 2 u ∂ 2 u ∂ 2 u −( −2 x 2 + 2 y 2 + 2 z 2 + 2 x 2 − 2 y 2 − 2 z 2 )
+
+
=
= 0.
5
∂x 2 ∂y 2 ∂z 2
( x2 + y2 + z2 )2
⎛ x⎞
Example 19: If u = z tan -1 ⎜ ⎟ , find the value of
⎝ y⎠
2
2
2
u
u
u
+
+
.
2
2
x
y
z2
⎛x⎞
u = z tan −1 ⎜ ⎟
⎝ y⎠
Differentiating u partially w.r.t. x,
1
1
zy
∂u
=z
⋅ = 2
2
∂x
x y y + x2
1+ 2
y
u
Differentiating
partially w.r.t. x,
x
∂2 u
yz ⋅ 2 x
2 xyz
=− 2
=− 2
∂x 2
( x + y 2 )2
( x + y 2 )2
Solution:
Differentiating u partially w.r.t. y,
∂u
=
∂y
Differentiating
z
x2
1+ 2
y
⎛ x ⎞
− xz
⎜⎝ − y 2 ⎟⎠ = y 2 + x 2
u
partially w.r.t. y,
y
∂2 u
xz ⋅ 2 y
2 xyz
= 2
= 2
2
2 2
∂y
(x + y )
( x + y 2 )2
Differentiating u partially w.r.t. z,
⎛x⎞
∂u
= tan −1 ⎜ ⎟
∂z
⎝ y⎠
Partial Differentiation
4.17
Differentiating u partially w.r.t. z,
z
∂2 u
=0
∂z 2
∂2 u ∂2 u ∂2 u
2 xyz
2 xyz
+
+
=− 2
+
= 0.
∂x 2 ∂y 2 ∂z 2
( x + y 2 )2 ( x 2 + y 2 )2
Hence,
-
1
Example 20: If v = (1 - 2 xy + y 2 ) 2 , find the value of
ì ⎡
ìv ⎤ ì ⎛ 2 ìv ⎞
y
(1 - x 2 ) ⎥ +
.
ìx ⎢⎣
ìx ⎦ ìy ⎜⎝ ìy ⎟⎠
v = (1 − 2 xy + y 2 )
Solution:
−
1
2
Differentiating v partially w.r.t. x,
3
−
∂v
1
= − (1 − 2 xy + y 2 ) 2 ( −2 y )
∂x
2
(1 − x 2 )
3
−
∂v
= y(1 − x 2 )(1 − 2 xy + y 2 ) 2
∂x
3
− ⎤
∂ ⎡
∂v ⎤
∂ ⎡
(1 − x 2 ) ⎥ = y ⎢(1 − x 2 )(1 − 2 xy + y 2 ) 2 ⎥
⎢
∂x ⎣
∂x ⎦
∂x ⎣
⎦
3
5
−
−
⎡
⎤
3
= y ⎢( −2 x )(1 − 2 xy + y 2 ) 2 − (1 − x 2 )(1 − 2 xy + y 2 ) 2 ( −2 y ) ⎥
2
⎣
⎦
−
5
−
5
−
5
= y(1 − 2 xy + y 2 ) 2 [−2 x(1 − 2 xy + y 2 ) + 3 y(1 − x 2 )]
= y(1 − 2 xy + y 2 ) 2 ( −2 x + 4 x 2 y − 2 xy 2 + 3 y − 3 x 2 y )
= y(1 − 2 xy + y 2 ) 2 ( −2 x + x 2 y − 2 xy 2 + 3 y )
Differentiating v partially w.r.t. y,
3
−
∂v
1
= − (1 − 2 xy + y 2 ) 2 ( −2 x + 2 y )
∂y
2
y2
3
−
∂v
= − y 2 ( − x + y )(1 − 2 xy + y 2 ) 2
∂y
3
3
−
−
∂ ⎛ 2 ∂v ⎞
2
2
2
2
2
y
=
−
y
−
x
+
y
−
xy
+
y
−
y
−
xy
+
y
2
(
)(
1
2
)
(
1
2
)
∂y ⎜⎝ ∂y ⎟⎠
+
5
−
3y2
( − x + y )(1 − 2 xy + y 2 ) 2 ( −2 x + 2 y )
2
... (1)
4.18
Engineering Mathematics
−
5
−
5
= y(1 − 2 xy + y 2 ) 2 [2( x − y )(1 − 2 xy + y 2 ) − y(1 − 2 xy + y 2 ) + 3 y( − x + y ) 2 ]
= y(1 − 2 xy + y 2 ) 2 ( 2 x − 4 x 2 y + 2 xy 2 − 2 y + 4 xy 2 − 2 y 3 − y
+ 2 xy 2 − y 3 + 3 yx 2 + 3 y 3 − 6 xy 2 )
−
5
= y(1 − 2 xy + y 2 ) 2 ( 2 x − x 2 y + 2 xy 2 − 3 y )
... (2)
Adding Eqs (1) and (2),
∂ ⎡
∂v ⎤ ∂ ⎛ ∂v ⎞
(1 − x 2 ) ⎥ + ⎜ y 2 ⎟ = 0.
∂x ⎢⎣
∂x ⎦ ∂y ⎝ ∂y ⎠
Example 21: If u = (ar n + br –n)(cos np + sin np ),
ì 2 u 1 ìu 1 ì 2 u
show that
= 0.
+
+
ìr 2 r ìr r 2 ìp 2
Solution:
u = (ar n + br –n)(cos nq + sin nq )
Differentiating u partially w.r.t. r,
∂u
= ( nar n −1 − bnr − n −1 )(cos nq + sin nq )
∂r
Differentiating
u
partially w.r.t. r,
r
∂2 u
= n[a( n − 1)r n − 2 + b( n + 1)r − n − 2 ](cos nq + sin nq )
∂r 2
Differentiating u partially w.r.t. q,
∂u
= ( ar n + br − n )( − n sin nq + n cos nq )
∂q
Differentiating
u
partially w.r.t. q,
q
∂2 u
= ( ar n + br − n )( − n2 cos nq − n2 sin nq )
∂q 2
= − n2 ( ar n + br − n )(cos nq + sin
n nq )
∂ 2 u 1 ∂u 1 ∂ 2 u
+
+
= n[a( n − 1)r n − 2 + b( n + 1)r − n − 2 ](cos nq + sin nq )
∂r 2 r ∂r r 2 ∂q 2
n2
( ar n + br − n ) (cos nq + sin nq )
r2
= (cos nq + sin nq )r n − 2 ( an2 − an + bn2 + bn + an − bn − an2 − bn2 )
+ n( ar n − 2 − br − n − 2 )(cos nq + sin nq ) −
=0
Partial Differentiation
Example 22: Show that
4.19
∂y
1 ∂x
ìx 1 ìy
=–
=
and
and hence, show that
∂r
r∂
ìr r ìp
ì 2 x 1 ìx 1 ì 2 x
r cos p
cos( r sin p ) and y = e r cos p sin( r sin p ) .
= 0 if x = e
+
+
ìr 2 r ìr r 2 ìp 2
Solution:
x = er cos q cos(r sin q )
Differentiating x partially w.r.t. r,
∂x
= e r cosq ⋅ cos q cos ( r sin q ) + e r cosq [− sin( r sin q )]sin q
∂r
= e r cosq [cos q cos( r sin q ) − sin q sin( r sin q )]
= e r cosq cos(q + r sin q )
y=e
r cos q
... (1)
sin( r sin q )
Differentiating y partially w.r.t. r,
∂y
= e r cos q cos q sin( r sin q ) + e r cos q cos( r sin q ) sin q
∂r
= e r cos q sin( r sin q + q )
... (2)
Differentiating x partially w.r.t. q,
∂x
= e r cos q ( − r sin q ) cos( r sin q ) + e r cos q [− sin( r sin q ) ⋅ r cos q )]
∂q
= − re r cos q sin(q + r sin q )
... (3)
Differentiating y partially w.r.t. q,
∂y
= e r cos q ( − r sin q ) sin( r sin q ) + e r cos q cos( r sin q ) ⋅ r cos q
∂q
= re r cos q cos(q + r sin q )
From Eqs (1) and (4), we get
∂x 1 ∂y
=
∂r r ∂
From Eqs (2) and (3), we get
∂y
1 ∂x
=−
,
r ∂q
∂r
Differentiating
x
partially w.r.t. r,
r
∂2 x ∂
=
∂r 2 ∂r
∂y
∂x
= −r
∂q
∂r
2
⎛ 1 ∂y ⎞ −1 ∂y 1 ∂ y
⎜⎝ r ∂q ⎟⎠ = 2 ∂q + r ∂r ∂q
r
... (4)
4.20
Differentiating
Engineering Mathematics
x
partially w.r.t. q,
q
∂2 y
∂2 y
∂2 x
∂ ⎛ ∂y ⎞
r
r
r
=
−
=
−
=
−
⎜
⎟
∂r ∂q
∂q ∂r
∂q 2 ∂q ⎝ ∂r ⎠
∂ 2 x 1 ∂x 1 ∂ 2 x −1 ∂y 1 ∂ 2 y
1 ∂y 1 ∂ 2 y
+
+
=
+
+
−
= 0.
∂r 2 r ∂r r 2 ∂q 2 r 2 ∂ q r ∂ r ∂ q r 2 ∂q r ∂r ∂q
Hence,
n
Example 23: If q = t e
- r2
4t
1 ì ⎛ 2 ìq ⎞ ìq
.
⎜r
⎟=
r 2 ìr ⎝ ìr ⎠ ∂ t
, then find n so that
q =t e
n
Solution:
− r2
4t
Differentiating q partially w.r.t. t,
2
2
2
−r
r
−r
−
⎛ r2 ⎞
∂q
1
⎛
⎞
= nt n −1 e 4 t + t n e 4 t ⎜ 2 ⎟ = e 4 t ⎜ nt n−1 + r 2t n− 2 ⎟
⎝
⎠
∂t
⎝ 4t ⎠
4
Differentiating q partially w.r.t. r,
− r2
∂
⎛ −2r ⎞
= t n e 4t ⎜
⎝ 4t ⎟⎠
∂r
−r
∂ ⎛ 2 ∂ ⎞ ∂ ⎛ t n −1 3 4 t ⎞
r e ⎟
⎜−
⎜⎝ r
⎟⎠ =
∂r
∂r
∂r ⎝ 2
⎠
2
−r
−r
t n −1 ⎡ 2 4 t
⎛ −2r ⎞ ⎤
3
⎢3r e + r e 4 t ⎜
⎥
⎝ 4t ⎟⎠ ⎥
2 ⎢⎣
⎦
2
=−
2
2
−r
⎛ 3 n −1 r 2 n − 2 ⎞
1 ∂ ⎛ 2 ∂ ⎞
4t
=
r
e
⎜⎝ − 2 t + 4 t ⎟⎠
r ⎟⎠
r 2 ∂r ⎜⎝
Substituting in
1 ∂ ⎛ 2 ∂q
⎜r
r 2 ∂r ⎝ ∂r
⎞ ∂q
,
⎟⎠ =
∂t
− r2
− r2
⎛ 3
⎞
1
r2
⎛
⎞
e 4 t ⎜ − t n −1 + t n − 2 ⎟ = e 4 t ⎜ nt n −1 + r 2 t n − 2 ⎟
⎝
⎠
4
4
⎝ 2
⎠
3
− t n −1 = nt n −1
2
3
n=− .
2
Example 24: Find the value of n so that v = r n (3 cos2 p – 1) satisfies the equation
ì
ìr
1
⎛ 2 ìv ⎞
⎜⎝ r
⎟⎠ +
sinq
ìr
ì
ìp
ìv ⎞
⎛
⎜⎝ sin p
⎟ = 0.
ìp ⎠
Partial Differentiation
Solution:
4.21
v = r n (3 cos2 q – 1)
Differentiating v partially w.r.t. r,
∂v
= nr n −1 (3 cos 2 q − 1)
∂r
∂ ⎛ 2 ∂v ⎞ ∂
n +1
2
⎜r
⎟ = [nr (3 cos q − 1)]
∂r ⎝ ∂r ⎠ ∂r
= n( n + 1) r n (3 cos 2 q − 1)
... (1)
Differentiating v partially w.r.t. q,
∂v
= r n ⋅ 6 cos q ( − sin q )
∂q
= −3r n sin 2q
∂
∂q
∂v ⎞
∂
⎛
( −3r n sin
n q ⋅ sin 2q )
⎜⎝ sin q
⎟=
∂q ⎠ ∂ q
= −3r n (cos q sin 2q + 2 sin q cos 2q )
= −3r n [cos q ⋅ 2 sin q cos q + 2 sin q ( 2 cos 2 q − 1)]
1 ∂ ⎛
∂v ⎞
2
2
n
n
2
⎜ sin q
⎟ = −3r ( 2 cos q + 4 cos q − 2) = −6 r (3 cos q − 1)
sin q ∂q ⎝
∂q ⎠
Substituting in
∂
∂r
1 ∂ ⎛
∂v ⎞
⎛ 2 ∂v ⎞
⎜⎝ r
⎟+
⎜ sin q
⎟ = 0,
∂r ⎠ sin q ∂q ⎝
∂q ⎠
n( n + 1)r n (3 cos 2 q − 1) − 6 r n (3 cos 2 q − 1) = 0
n ( n + 1) − 6 = 0
n2 + n − 6 = 0
( n + 3)( n − 2) = 0
n = −3, 2.
Example 25: If x x y y z z = c, show that at x = y = z,
(a)
∂2 z
= − ( x log ex ) − 1
∂x ∂y
(b)
2 ( x 2 - 2)
ì2 z
ì2 z
ì2 z
2
xy
.
+
=
ìx ìy ìy 2 x (1 + log x )
ìx 2
Solution: (a) x x y y z z = c
Taking logarithm on both the sides,
log x x + log y y + log z z = log c
x log x + y log y + z log z = log c
... (1)
Differentiating Eq. (1) partially w.r.t. x,
1
1 ∂z
∂z
x ⋅ + log x + z ⋅
+ log z
=0
∂x
x
z ∂x
[∵ z = f ( x, y )]
4.22
Engineering Mathematics
Differentiating
1 + log x
∂z
=−
∂x
1 + log z
z
partially w.r.t. y,
x
⎡
∂ ⎛ ∂z ⎞
1
1 ∂z ⎤
⋅
⎥
⎜ ⎟ = −(1 + log x ) ⎢ −
2
∂y ⎝ ∂x ⎠
z ∂x ⎦
(
1
+
log
z
)
⎣
(1 + log x )
∂2 z
=
∂y ∂x z (1 + log z ) 2
⎛ 1 + log x ⎞
⎜⎝ − 1 + log z ⎟⎠
(1 + log x ) 2
∂2 z
=−
∂x ∂y
z (1 + log z )3
At x = y = z,
(1 + log x ) 2
∂2 z
1
=−
=−
3
∂x ∂y
x(1 + log x )
x(1 + log x )
= −[ x(log e + log x )]−1
[∵ log e = 1]
= −( x log ex ) −1 .
(b) Differentiating
z
partially w.r.t. x,
x
∂2 z
∂ ⎛ 1 + log x ⎞
=
−
∂x 2 ∂x ⎜⎝ 1 + log z ⎟⎠
=
(1 + log x ) 1 ∂z
1
⋅
−
(1 + log z ) 2 z ∂x x(1 + log z )
=−
(1 + log x ) (1 + log x )
1
⋅
−
2
z (1 + log z ) (1 + log z ) x(1 + log z )
At x = y = z,
∂2 z
2
=−
x(1 + log x )
∂x 2
Similarly,
∂ 2 z − (1 + log y ) 2
1
=
−
2
3
y (1 + log z )
∂y
z (1 + log z )
At x = y = z,
∂2 z
2
=−
2
x(1 + log x )
∂y
Hence,
⎡
⎤ ⎡
⎤
∂2 z
∂2 z ∂2 z
−2
−1
−2
−
+ 2 =
− 2 xy ⎢
+⎢
2
xy
⎥
⎥
2
∂x ∂y ∂y
x(1 + log x )
∂x
⎣ x(1 + log x ) ⎦ ⎣ x(1 + log x ) ⎦
=
2( xy − 2)
2( x 2 − 2)
=
x(1 + log x ) x(1 + log x )
[∵ x = y = z ]
Partial Differentiation
Example 26: If
4.23
y2
x2
z2
+
+
= 1, prove that
a 2 + u b2 + u c 2 + u
2
2
⎛ ìu ⎞
⎛ ìu ⎞
⎛ ìu ⎞ = 2 ⎛ x ìu + y ìu + z ìu ⎞ .
+
+
⎜⎝ ìx
⎜⎝ ⎟⎠
⎜⎝ ⎟⎠
⎜⎝ ìy ⎟⎠
ìy
ìz ⎟⎠
ìx
ìz
2
Solution:
y2
x2
z2
+
+
=1
a2 + u b2 + u c2 + u
Differentiating given equation partially w.r.t. x,
y2
2x
x2
∂u
∂u
z2
∂u
−
−
− 2
=0
2
2
2
2
2
a + u ( a + u ) ∂x ( b + u ) ∂x ( c + u ) 2 ∂ x
⎤
y2
∂u ⎡ x 2
2x
z2
+ 2
+ 2
= 2
⎢ 2
2
2
2 ⎥
∂x ⎣ ( a + u )
(b + u)
(c + u) ⎦ a + u
∂u
2x
⋅p= 2
∂x
(a + u)
y2
x2
z2
p= 2
+
+
( a + u) 2 (b 2 + u) 2 (c 2 + u) 2
∂u
2x
= 2
∂x ( a + u ) p
where,
2y
∂u
= 2
∂y ( b + u ) p
Similarly,
2z
∂u
= 2
∂z ( c + u ) p
2
2
2
⎤
y2
4 ⎡ x2
z2
⎛ ∂u ⎞ ⎛ ∂u ⎞ ⎛ ∂u ⎞
+
+
+
=
+
⎜⎝ ⎟⎠ ⎜ ⎟ ⎜⎝ ⎟⎠
2
2 ⎥
2 ⎢
2
2
2
2
∂x
⎝ ∂y ⎠
∂z
(b + u)
(c + u) ⎦
p ⎣ ( a + u)
=
x
4
4
( p) =
p
p2
∂u
∂u
∂u 2 ⎛ x 2
y2
z2 ⎞
+y
+z
= ⎜ 2
+ 2
+ 2
∂x
∂y
∂z p ⎝ a + u b + u c + u ⎟⎠
=
2
2
(1) =
p
p
From Eqs (1) and (2), we get
2
... (1)
2
2
⎛ ∂u
∂u
∂u ⎞
⎛ ∂ u ⎞ ⎛ ∂u ⎞ ⎛ ∂u ⎞
⎜⎝ ⎟⎠ + ⎜ ⎟ + ⎜⎝ ⎟⎠ = 2 ⎜ x + y + z ⎟ .
∂x
⎝ ∂y ⎠
∂z
⎝ ∂x
∂y
∂z ⎠
... (2)
4.24
Engineering Mathematics
Example 27: If u = f (x + ky) + y (x – ky), show that
Solution:
2
ì2 u
2 ì u
=
k
.
ìy 2
ìx 2
u = f (x + ky) + y (x – ky)
Differentiating u partially w.r.t. x,
∂u
= f ′ ( x + ky ) ⋅1 +y ′ ( x − ky ) ⋅1
∂x
u
Differentiating
partially w.r.t. x,
x
∂2 u
= f ′′( x + ky ) +y ′′( x − ky )
∂x 2
Differentiating u partially w.r.t. y,
... (1)
∂u
= f ′( x + ky ) ⋅ k +y ′( x − ky ) ⋅ ( − k )
∂y
Differentiating
u
partially w.r.t. y,
y
∂2 u
= f ′′( x + ky ) ⋅ k 2 +y ′′( x − ky )( − k ) 2
∂y 2
= k 2 [f ′′( x + ky ) +y ′′( x − ky )]
... (2)
From Eqs (1) and (2), we get
∂2 u
∂2 u
= k2 2 .
2
∂y
∂x
Example 28: If u = xf (x + y) + ye (x + y), then show that
Solution:
ì2 u
ì2 u ì2 u
2
+
= 0.
ìx ìy ìy 2
ìx 2
u = xf (x + y) + yf (x + y),
Differentiating u partially w.r.t. x,
∂u
= f ( x + y ) + xf ′ ( x + y ) + y ′ ( x + y )
∂x
Differentiating
Differentiating
u
partially w.r.t. x,
x
∂2u
= f ′ ( x + y ) + f ′ ( x + y ) + xf ′′ ( x + y ) + y ′′ ( x + y )
∂x 2
= 2 f ′ ( x + y ) + xf ′′ ( x + y ) + y ′′ ( x + y )
u
partially w.r.t. y,
x
∂2u
= f ′ ( x + y ) + xf ′′ ( x + y ) + y ′′ ( x + y ) + ′ ( x + y )
∂x ∂y
Partial Differentiation
4.25
Differentiating u partially w.r.t. y,
∂u
= xf ′ ( x + y ) + f ( x + y ) + yf ′ ( x + y )
∂y
u
Differentiating
partially w.r.t. y,
y
∂2 u
= xf ′′ ( x + y ) + f ′ ( x + y ) + f ′ ( x + y ) + yf ′′ ( x + y )
∂y 2
= xf ′′ ( x + y ) + 2f ′ ( x + y ) + yf ′′ ( x + y )
∂2 u
∂2 u ∂2 u
−2
+
2
∂x ∂y ∂y 2
∂x
Hence,
= 2 f ′ ( x + y ) + xf ′′ ( x + y ) + yf ′′ ( x + y ) − 2 f ′ ( x + y ) − 2 xf ′′ ( x + y )
−2 y ′′ ( x + y ) − 2 ′ ( x + y ) + xf ′′ ( x + y ) + 2 ′ ( x + y ) + y ′′ ( x + y ) = 0
2
2
ì2 u
⎛ x2 ⎞
2 ì u
2 ì u
+
3
xy
2
+
y
= 0.
Example 29: If u = f ⎜ ⎟ , show that x
ìx ìy
ìx 2
ìy 2
⎝ y ⎠
Solution:
⎛ x2 ⎞
u= f ⎜ ⎟
⎝ y ⎠
Differentiating u partially w.r.t. x,
⎛ x2 ⎞ ∂ ⎛ x2
∂u
= f ′⎜ ⎟
∂x
⎝ y ⎠ ∂x ⎜⎝ y
Differentiating
⎞
⎛ x2
′
=
f
⎟⎠
⎜⎝ y
⎞ ⎛ 2x ⎞
⎟⎠ ⎜⎝ y ⎟⎠
u
partially w.r.t. x,
x
⎛ x2
∂2 u 2
= f ′⎜
2
y
⎝ y
∂x
⎞
⎛ x2
⎟⎠ + f ′′ ⎜⎝ y
⎞ ∂ ⎛ x2
⎟⎠ ⋅ ∂x ⎜⎝ y
⎛ x2
2
= f ′⎜
y
⎝ y
⎞
⎛ x2
f
+
′′
⎟⎠
⎜⎝ y
⎞ ⎛ 2x ⎞
⎟⎠ ⋅ ⎜⎝ y ⎟⎠
⎞ ⎛ 2x ⎞
⎟⎠ ⋅ ⎜⎝ y ⎟⎠
2
Differentiating u partially w.r.t. y,
⎛ x2
∂u
= f ′⎜
∂y
⎝ y
Differentiating
⎞ ∂ ⎛ x2
⎟⎠ ⋅ ∂y ⎜⎝ y
⎛ x2
⎞
f
=
′
⎜⎝ y
⎟⎠
⎞ ⎛ − x2 ⎞
⎟⎠ ⋅ ⎜⎝ y 2 ⎟⎠
u
partially w.r.t. y,
y
⎛ x2
∂2 u 2 x 2
= 3 f ′⎜
2
⎝ y
∂y
y
⎛ x2
2x2
= 3 f ′⎜
⎝ y
y
⎞ ⎛ − x2 ⎞
⎛ x2
⎟⎠ + ⎜⎝ y 2 ⎟⎠ f ′′ ⎜⎝ y
2
⎞ ∂ ⎛ x2 ⎞
⎟⎠ ⋅ ∂y ⎜⎝ y ⎟⎠
⎞ ⎛ x2 ⎞
⎛ x2 ⎞
f
+
′′
⎟⎠ ⎜⎝ y 2 ⎟⎠
⎜⎝ y ⎟⎠
4.26
Engineering Mathematics
Differentiating u partially w.r.t. y,
x
⎛ x2 ⎞ 2x
⎛ x2 ⎞ ⎛ − x2 ⎞
∂2 u
2x
= − 2 f ′⎜ ⎟+
f ′′ ⎜ ⎟ ⋅ ⎜ 2 ⎟
∂x ∂y
⎝ y ⎠ y
⎝ y⎠ ⎝ y ⎠
y
x2
2
⎛ x2
∂2 u
∂2 u
2x2
2 ∂ u
+
3
xy
+
2
y
=
f
′
⎜⎝ y
∂x ∂y
y
∂x 2
∂y 2
−
⎞ 4x4
⎛ x2
+
f
′′
⎟⎠ y 2
⎜⎝ y
⎞ 6x2
⎛ x2 ⎞
f
−
′
⎟⎠
⎜⎝ y ⎟⎠
y
⎛ x2 ⎞ 4x2 ⎛ x2 ⎞ 2x4
⎛ x2 ⎞
6x4
f
f
f
+
+
′′
′
′′
⎜⎝ y ⎟⎠
⎜⎝ y ⎟⎠ y 2
⎜⎝ y ⎟⎠ = 0.
y
y2
⎛ xy ⎞
ìu
ìu
xyz
Example 30: If u = e f ⎜ ⎟ , prove that x
= 2 xyzu
+z
⎝ z ⎠
ìx
ìz
ì2 u
ì2 u
ìu
ìu
and y
= 2 xyzu and hence, show that x
.
=y
+z
ìz ìx
ìz ìy
ìy
ìz
⎛ xy ⎞
u = e xyz f ⎜ ⎟
⎝ z ⎠
Solution:
Differentiating u partially w.r.t. x, y and z,
∂u
⎛ xy ⎞
= e xyz yz ⋅ f ⎜ ⎟ + e xyz
⎝ z ⎠
∂x
⎡
⎢f
⎣
⎛ xy ⎞ ⎤ ⎛ y ⎞
′ ⎜ ⎟⎥ ⎜ ⎟
⎝ z ⎠⎦ ⎝ z ⎠
∂u
⎛ xy ⎞
= e xyz xz ⋅ f ⎜ ⎟ + e xyz
⎝ z ⎠
∂y
⎡
⎢f
⎣
⎛ xy ⎞ ⎤ ⎛ x ⎞
′ ⎜ ⎟⎥ ⎜ ⎟
⎝ z ⎠⎦ ⎝ z ⎠
⎡ ⎛ xy ⎞ ⎤ ⎛ − xy ⎞
∂u
⎛ xy ⎞
= e xyz xy ⋅ f ⎜ ⎟ + e xyz ⎢ f ′ ⎜ ⎟ ⎥ ⎜ 2 ⎟
⎝ z ⎠
∂z
⎣ ⎝ z ⎠⎦ ⎝ z ⎠
∂u
∂u
+z
∂x
∂z
⎛ xy ⎞ xy
⎛ xy ⎞
⎛ xy ⎞
⎛ xy ⎞ xy
= e xyz xyzf ⎜ ⎟ + e xyz f ′ ⎜ ⎟ + e xyz xyz ⋅ f ⎜ ⎟ − e xyz f ′ ⎜ ⎟ = 2 xyzu.
⎝ z ⎠
⎝ z ⎠ z
⎝ z ⎠
⎝ z ⎠ z
∂u
∂u
(ii) y
+z
∂y
∂z
⎛ xy ⎞ xy
⎛ xy ⎞
⎛ xy ⎞ xy
⎛ xy ⎞
= e xyz xyz ⋅ f ⎜ ⎟ + e xyz f ′ ⎜ ⎟ + e xyz xyz ⋅ f ⎜ ⎟ − e xyz f ′ ⎜ ⎟ = 2 xyzu.
⎝ z ⎠ z
⎝ z ⎠
⎝ z ⎠ z
⎝ z ⎠
(i) x
u
w.r.t. x,
z
⎡ ⎛ xy ⎞ ⎤ ⎛ y ⎞
∂2 u
⎛ xy ⎞
⎛ xy ⎞
= e xyz yz ⋅ xy f ⎜ ⎟ + e xyz y ⋅ f ⎜ ⎟ + e xyz xy ⎢ f ′ ⎜ ⎟ ⎥ ⎜ ⎟
⎝ z ⎠
⎝ z ⎠
∂z ∂x
⎣ ⎝ z ⎠⎦ ⎝ z ⎠
(iii) Differentiating
⎡ ⎛ xy ⎞ ⎤ ⎛ − xy ⎞
+ e xyz yz ⎢ f ′ ⎜ ⎟ ⎥ ⎜ 2 ⎟ + e xyz
⎣ ⎝ z ⎠⎦ ⎝ z ⎠
⎡ ⎛ xy ⎞ ⎤ ⎛ y ⎞
+ e xyz ⎢ f ′ ⎜ ⎟ ⎥ ⎜ − 2 ⎟
⎣ ⎝ z ⎠⎦ ⎝ z ⎠
⎡ ⎛ xy ⎞ ⎤ ⎛ y ⎞ ⎛ − xy ⎞
⎢ f ′′ ⎜⎝ z ⎟⎠ ⎥ ⎜⎝ z ⎟⎠ ⎜⎝ z 2 ⎟⎠
⎦
⎣
Partial Differentiation
x
∂2 u
= e xyz
∂z ∂x
Differentiating
4.27
2 2
⎡ 2 2
⎛ xy ⎞
⎛ xy ⎞ x y
⎛ xy ⎞ xy ⎛ xy ⎞ ⎤
⋅
+
⋅
−
x
y
z
f
xy
f
f ′′ ⎜ ⎟ − 2 f ′ ⎜ ⎟ ⎥ ... (1)
⎢
⎜⎝ ⎟⎠
⎜⎝ ⎟⎠
3
⎝ z ⎠ z
⎝ z ⎠⎦
z
z
z
⎣
u
w.r.t. y,
z
⎡ ⎛ xy ⎞ ⎤ ⎛ x ⎞
∂2 u
⎛ xy ⎞
⎛ xy ⎞
= e xyz xz ⋅ xy ⋅ f ⎜ ⎟ + e xyz x ⋅ f ⎜ ⎟ + e xyz xy ⎢ f ′ ⎜ ⎟ ⎥ ⎜ ⎟
⎝ z ⎠
⎝ z ⎠
∂z ∂y
⎣ ⎝ z ⎠⎦ ⎝ z ⎠
⎡ ⎛ xy ⎞ ⎤ ⎛ − xy ⎞
⎡ ⎛ xy ⎞ ⎤ ⎛ x ⎞ ⎛ − xy ⎞
+ e xyz xz ⎢ f ′ ⎜ ⎟ ⎥ ⎜ 2 ⎟ + e xyz ⎢ f ′′ ⎜ ⎟ ⎥ ⎜ ⎟ ⎜ 2 ⎟
⎝
⎠
⎝
⎠
z ⎦ z
⎣
⎣ ⎝ z ⎠⎦ ⎝ z ⎠ ⎝ z ⎠
⎡ ⎛ xy ⎞ ⎤ ⎛ x ⎞ ⎛ − xy ⎞
⎡ ⎛ xy ⎞ ⎤ ⎛ x ⎞
⎡ ⎛ xy ⎞ ⎤ ⎛ x ⎞
+ e xyz ⎢ f ′ ⎜ ⎟ ⎥ ⎜ − 2 ⎟ + e xyz ⎢ f ′′ ⎜ ⎟ ⎥ ⎜ ⎟ ⎜ 2 ⎟ + e xyz ⎢ f ′ ⎜ ⎟ ⎥ ⎜ − 2 ⎟
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
⎝
⎠
z ⎦ z
z ⎦ z
z
⎣
⎣
⎣ ⎝ z ⎠⎦ ⎝ z ⎠
y
∂2 u
= e xyz
∂z ∂y
2 2
⎡ 2 2
⎛ xy ⎞
⎛ xy ⎞ x y
⎛ xy ⎞ xy ⎛ xy ⎞ ⎤
⋅
+
⋅
−
x
y
z
f
xy
f
f ′′ ⎜ ⎟ − 2 f ′ ⎜ ⎟ ⎥ ... (2)
⎢
⎜⎝ ⎟⎠
⎜⎝ ⎟⎠
3
⎝ z ⎠ z
⎝ z ⎠⎦
z
z
z
⎣
From Eqs (1) and (2), we get
x
∂2 u
∂2 u
=y
.
∂z ∂x
∂z ∂y
Example 31: If u = r m, r = x 2 + y 2 + z 2 ,
show that
ì2 u ì2 u ì2 u
+
+
= m(m + 1)rm–2.
ìx 2 ìy 2 ìz 2
Solution:
u = rm
Differentiating u partially w.r.t. x,
∂u
∂r
= mr m −1
∂x
∂x
r = x2 + y2 + z2
But
r2 = x2 + y2 + z2
Differentiating r 2 partially w.r.t. x,
2r
∂r
= 2x
∂x
∂r x
=
∂x r
∂u
x
= mr m −1 = mr m − 2 x
∂x
r
Differentiating
u
partially w.r.t. x,
x
4.28
Engineering Mathematics
∂2 u
∂r ⎤
⎡
= m ⎢ r m − 2 + ( m − 2) r m − 3
x
2
∂x ⎥⎦
∂x
⎣
x ⎤
⎡
= m ⎢ r m − 2 + ( m − 2) r m − 3 x ⎥
r ⎦
⎣
= m[r m − 2 + ( m − 2)r m − 4 x 2 ]
... (1)
∂ u
= m[r m − 2 + ( m − 2)r m − 4 y 2 ]
∂y 2
... (2)
∂2 u
= m[r m − 2 + ( m − 2)r m − 4 z 2 ]
∂z 2
... (3)
2
Similarly,
Adding Eqs (1), (2) and (3),
∂2 u ∂2 u ∂2 u
+
+
= 3mr m − 2 + m( m − 2)r m − 4 ( x 2 + y 2 + z 2 )
∂x 2 ∂y 2 ∂z 2
= 3mr m − 2 + m(m
m − 2) r m − 4 ⋅ r 2
= r m − 2 (3m + m 2 − 2m)
= r m− 2 (m + m2 )
= m( m + 1)r m − 2 .
Example 32: If u = f (r) and r2 = x2 + y2 + z2,
prove that
ì2 u ì2 u ì2 u
2
+ 2 + 2 = f ( r ) + f ( r ).
2
r
ìx
ìy
ìz
Solution:
u = f (r)
Differentiating u partially w.r.t. x,
∂u ∂
∂
∂r
∂r
=
f (r) =
f ( r ) ⋅ = f ′( r ) ⋅
∂x ∂x
∂r
∂x
∂x
But
r 2 = x2 + y2 + z2
Differentiating r 2 partially w.r.t. x,
2r
Differentiating
∂r
= 2x
∂x
∂r x
=
∂x r
∂u
x
= f ′( r ) ⋅
∂x
r
u
partially w.r.t. x,
x
Partial Differentiation
4.29
∂2 u ∂ ⎡
x⎤
=
f ′( r ) ⎥
2
⎢
∂x ⎣
r⎦
∂x
∂r x f ′ ( r )
⎛ −1 ⎞ ∂r
⋅ +
+ xf ′( r ) ⎜ 2 ⎟ ⋅
⎝ r ⎠ ∂x
∂x r
r
x
x x f ′( r ) x
− 2 f ′( r ) ⋅
= f ′′( r )
+
r
r
r r
r
= f ′′( r )
= f ′′( r )
Similarly,
and
x 2 f ′( r ) x 2
+
− 3 f ′( r )
r
r2
r
y 2 f ′(r )
∂2 u
=
f
(
r
)
+
−
′′
r
∂y 2
r2
∂2 u
z 2 f ′( r )
= f ′′( r ) 2 +
−
2
r
∂z
r
y2
f ′ (r )
r3
z2
f ′( r )
r3
... (1)
... (2)
... (3)
Adding Eqs (1), (2) and (3),
3 f ′( r ) ( x 2 + y 2 + z 2 )
∂ 2 u ∂ 2 u ∂ 2 u f ′′( r ) 2
2
2
f ′( r )
+
+
=
+
+
+
−
(
x
y
z
)
r
∂x 2 ∂y 2 ∂z 2
r3
r2
f ′′( r ) 2 3 f ′( r ) r 2
⋅r +
− 3 f ′( r )
r
r2
r
2 f ′( r )
= f ′′( r ) +
.
r
=
Example 33: If u = f (r 2) where r 2 = x 2 + y 2 + z 2,
prove that
ì2 u ì2 u ì2 u
+
+
= 4r2 f è(r2) + 6 f (r2).
ìx 2 ìy 2 ìz 2
Solution:
u = f (r 2)
Differentiating u partially w.r.t. x,
∂u ∂
∂
=
f (r 2 ) =
f (l ), where r 2 = l
∂x ∂x
∂x
∂
∂l
∂l
∂r 2
=
= f ′( r 2 )
f ( l ) ⋅ = f ′( l )
∂l
∂x
∂x
∂x
r
∂
= f ′( r 2 ) 2r
∂x
But
r2 = x2 + y2 + z2
Differentiating r 2 partially w.r.t. x,
2r
∂r
= 2x
∂x
∂r x
=
∂x r
4.30
Engineering Mathematics
∂u
x
= f ′( r 2 ) ⋅ 2r
∂x
r
= 2 xf ′( r 2 )
Differentiating
u
partially w.r.t. x,
x
∂f ′ ( r 2 )
∂2 u
2
=
+
2
f
(
r
)
2
x
′
∂x
∂x 2
∂r
∂x
x
= 2 f ′( r 2 ) + 2 xf ′′( r 2 ) ⋅ 2r
r
= 2 f ′( r 2 ) + 4 x 2 f ′′( r 2 )
= 2 f ′( r 2 ) + 2 xf ′′( r 2 ) ⋅ 2r
... (1)
∂ u
= 2 f ′( r 2 ) + 4 y 2 f ′′( r 2 )
2
∂y
... (2)
∂2 u
= 2 f ′( r 2 ) + 4 z 2 f ′′( r 2 )
∂z 2
... (3)
2
Similarly,
and
Adding Eqs (1), (2) and (3),
∂2 u ∂2 u ∂2 u
2
2
2
+
+
= 6 f ′( r 2 ) + 4( x 2 + y 2 + z 2 ) f ′′( r 2 ) = 6 f ′( r ) + 4 r f ′′(r )
∂x 2 ∂y 2 ∂z 2
-
1
Example 34: If f ( r ) = r 2 ( a + log r ), r 2 = x 2 + y 2 + z 2,
ì2 f ì2 f
f (r )
ì2 f
+
+
=- 2 .
ìx 2 ìy 2
4r
ìz 2
prove that
−
1
f ( r ) = r 2 ( a + log r )
Solution:
Differentiating f partially w.r.t. x,
3
∂f
1 −
=− r 2
∂x
2
3
1 −
=− r 2
2
−
−
∂r
( a + log r ) + r 2
∂x
3
−
x
⋅ ( a + log r ) + r 2
r
1
1 ∂r
r ∂x
x
⋅
r
⋅
∂r x ⎤
⎡
⎢ As proved earlier ∂x = r ⎥
⎣
⎦
5
xr 2
( a + log r − 2)
=−
2
f
Differentiating
partially w.r.t. x,
x
−
5
−
5
7
∂2 f
r 2
x 5 − 2 ∂r
xr 2 1 ∂r
(
a
log
r
2
)
r
(
a
log
r
2
)
=
−
+
−
+
⋅
+
−
−
⋅
2
2 2
2 r ∂x
∂x
∂x 2
Partial Differentiation
r
=−
2
−
5
2
4.31
7
( a + log r − 2) +
5
5x − 2 x
x − x
r ⋅ ( a + log r − 2) − r 2
4
2r
r
r
∂r x ⎤
⎡
⎢ As proved earlier ∂x = r ⎥
⎣
⎦
=−
r
−
5
2
−
5
−
5
⎛ 5x 2
( a + log r − 2) ⎜1 − 2
2
⎝ 2r
Similarly,
⎛ 5 y2
∂2 f
r 2
( a + log r − 2) ⎜1 − 2
=−
2
2
∂y
⎝ 2r
and
⎛ 5z 2
∂2 f
r 2
( a + log r − 2) ⎜1 − 2
=−
2
2
⎝ 2r
∂z
Hence,
⎞ x 2 − 52
⎟⎠ − 2r 2 r
⎞ y 2 − 52
⎟⎠ − 2r 2 r
⎞ z 2 − 52
⎟⎠ − 2r 2 r
∂2 f ∂2 f ∂2 f
+
+
∂x 2 ∂y 2 ∂ z 2
=−
r
=−
r
−
5
2
−
5
2
−
5
2
5
2
2
2
5
⎡
⎤ (x + y + z ) −2
( a + log r − 2) ⎢3 − 2 ( x 2 + y 2 + z 2 ) ⎥ −
r
2
2r 2
⎣ 2r
⎦
⎤ ⎛ 5 ⋅ r2
⎡ 12
⎢⎣ r f ( r ) − 2⎥⎦ ⎜ 3 − 2
2
⎝
2r
⎞ r 2 − 52
⎟⎠ − 2r 2 r
−
5
⎡ 12
⎤⎛
5⎞ r 2
=−
⎢⎣ r f ( r ) − 2⎥⎦ ⎜ 3 − ⎟ −
⎝
2
2⎠ 2
r
⎡ 12
⎤
⎢ r f ( r ) − 1 + 1⎥
=−
⎥
2 ⎢ 2
⎢⎣
⎥⎦
r
=−
−
5
2
f (r)
.
4r 2
Example 35: If v = x log (x + r) – r where r 2 = x 2 + y 2, prove that
Solution:
v = x log(x + r) – r
Differentiating v partially w.r.t. x,
∂v
x ⎛ ∂r ⎞ ∂r
= log ( x + r ) +
⎜1 + ⎟ −
∂x
x + r ⎝ ∂x ⎠ ∂x
But
r2 = x2 + y2
2
Differentiating r partially w.r.t. x,
∂r x
=
∂x r
∂2 v ∂2 v
1
+ 2 =
.
2
x+r
∂x
∂y
4.32
Engineering Mathematics
Differentiating r 2 partially w.r.t. y,
∂r y
=
∂y r
∂v
x ⎛
= log ( x + r ) +
⎜1 +
∂x
x+r ⎝
x⎞ x
⎟−
r⎠ r
= log ( x + r ) +
(r + x) x
x
⋅
−
( x + r)
r
r
= log ( x + r ) +
x x
−
r r
= log ( x + r )
v
Differentiating
partially w.r.t. x,
x
∂2 v
1 ⎛ ∂r ⎞
1 ⎛ x⎞ 1
=
⎜1 + ⎟ =
⎜1 + ⎟ =
∂x 2 x + r ⎝ ∂x ⎠ x + r ⎝ r ⎠ r
Differentiating v partially w.r.t. y,
∂v
x ∂r ∂r
x y y
=
⋅ −
=
⋅ −
∂y x + r ∂y ∂y x + r r r
y ⎛x−x−r⎞
⎜
⎟
r ⎝ x+r ⎠
y
=−
x+r
=
Differentiating
v
partially w.r.t. y,
y
y
y y⎞
∂2 v
1
∂r
1 ⎛
=−
+
⋅ =−
1−
⋅
x + r ( x + r ) 2 ∂y
x + r ⎜⎝ x + r r ⎟⎠
∂y 2
1 ⎡ rx + x 2 ⎤
1 ⎡ rx + r 2 − y 2 ⎤
=
−
⎢
⎥
⎢
⎥
x + r ⎣ r( x + r ) ⎦
x + r ⎣ r( x + r ) ⎦
x
=−
r( x + r )
=−
Hence,
∂2 v ∂2 v 1 ⎛
x ⎞
+ 2 = ⎜1 −
⎟
2
⎝
r
x+r⎠
∂x
∂y
=
1⎛x+r−x⎞
1
.
⎜⎝
⎟⎠ =
r
x+r
x+r
Partial Differentiation
4.33
4.4 VARIABLES TO BE TREATED AS CONSTANTS
⎛ ∂r ⎞
⎜⎝ ⎟⎠
∂x y
r
x
x
⎛ ∂x ⎞
⎜⎝ ⎟⎠
∂r q
y
q
r
x is
r
Example 1: If x 2 = au + bv , y 2 = au - bv , prove that
1
⎛ ìu ⎞ ⎛ ìx ⎞
⎛ ìu ⎞ ⎛ ìy ⎞
= ⎜ ⎟ ⎜ ⎟ , where a, b are constants.
⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ =
ìx y ìu v 2
⎝ ìy ⎠ ⎝ ìv ⎠ u
x
x 2 = au + bv
a
⎛ ∂x ⎞
⎛ ∂x ⎞
2 x ⎜ ⎟ = a,
⎜⎝ ⎟⎠ =
⎝ ∂u ⎠ v
∂u v 2 x
Solution:
y 2 = au − bv
⎛ ∂y ⎞
2 y ⎜ ⎟ = − b,
⎝ ∂v ⎠ u
x2 = au + bv,
bv
x 2 + y 2 = 2au, u =
x +y
x
⎛ ∂u ⎞
, ⎜ ⎟ =
⎝ ∂x ⎠ y a
2a
x 2 − y 2 = 2bv, v =
x 2 − y 2 ⎛ ∂v ⎞
y
,⎜ ⎟ = −
2b
⎝ ∂y ⎠ x
b
2
and
b
⎛ ∂y ⎞
⎜⎝ ∂v ⎟⎠ = − 2 y
u
y2 = au
2
Hence,
⎛ ∂u ⎞
⎜⎝ ⎟⎠
∂x y
and
⎛ ∂v ⎞ ⎛ ∂y ⎞
⎛ y⎞⎛ b ⎞ 1
⎜⎝ ∂y ⎟⎠ ⎜⎝ ∂v ⎟⎠ = ⎜⎝ − b ⎟⎠ ⎜⎝ − 2 y ⎟⎠ = 2 .
u
x
x a 1
⎛ ∂x ⎞
=
⎜⎝ ⎟⎠ = ⋅
∂u v a 2 x 2
Example 2: If x = r cos p , y = r sin p , prove that
⎛ ìr ⎞
⎛ ìx ⎞
(a) ⎜ ⎟ = ⎜ ⎟
⎝ ìx ⎠ y ⎝ ìr ⎠ q
(c)
2
2
ì 2 r ì 2 r 1 ⎡ ⎛ ìr ⎞ ⎛ ìr ⎞ ⎤
+
=
+
⎢
⎥
⎜ ⎟
ìx 2 ìy 2 r ⎣⎢ ⎝ ìx ⎠ ⎜⎝ ìy ⎟⎠ ⎥⎦
1 ⎛ ìx ⎞
⎛ ìq ⎞
=
(b) r ⎜
⎝ ìx ⎟⎠ y r ⎜⎝ ìq ⎟⎠ r
(d)
y
ì 2q ì 2q
+
= 0.
ìx 2 ìy 2
q
Engineering Mathematics
4.34
Solution: (a) x = r cos q, y = r sin q,
x2 + y2 = r2
Differentiating r2 partially w.r.t. x keeping y constant,
⎛ ∂r ⎞
2 x = 2r ⎜ ⎟
⎝ ∂x ⎠ y
x
⎛ ∂r ⎞
⎜⎝ ⎟⎠ =
∂x y r
... (1)
x=r
q
r
q
x
⎛ ∂x ⎞
⎜⎝ ⎟⎠ = cos q =
∂r q
r
x
... (2)
From Eqs (1) and (2), we get
⎛ ∂r ⎞
⎛ ∂x ⎞
⎜⎝ ⎟⎠ = ⎜⎝ ⎟⎠
∂x y
∂r q
(b) x = r
q, y = r sin q
x
q
r
∂
x
⎛
⎞
⎜⎝
⎟ = − r sin q
∂q ⎠ r
1 ⎛ ∂x ⎞
⎜
⎟ = − sin q
r ⎝ ∂q ⎠ r
... (3)
y
x
Differentiating tanq partially w.r.t. x keeping y constant,
tanq =
Now,
y
⎛ ∂q ⎞
sec 2 q ⎜
=− 2
⎝ ∂x ⎟⎠ y
x
r 2 ⎛ ∂q ⎞
r sin q
⎜
⎟ =−
x2
x 2 ⎝ ∂x ⎠ y
1 ⎛ ∂x ⎞
⎛ ∂q ⎞
r⎜
= − sin q = ⎜
⎟
⎟
⎝ ∂x ⎠ y
r ⎝ ∂q ⎠ r
(c)
Differentiating
r
partially w.r.t. x,
x
[From Eq. (3)]
∂r x
=
∂x r
∂ ⎛ ∂r ⎞ 1 x ∂r
⎜ ⎟= −
∂x ⎝ ∂x ⎠ r r 2 ∂x
∂2 r 1 x 2 r 2 − x 2 y 2
= −
=
= 3
r3
r
∂x 2 r r 3
⎡ ∂r x ⎤
⎢∵ ∂x = r ⎥
⎣
⎦
Partial Differentiation
4.35
∂2 r x 2
=
∂y 2 r 3
Similarly,
⎡ ∂r y ⎤
⎢∵ ∂y = r ⎥
⎣
⎦
∂2 r ∂2 r y 2 x 2 1 ⎛ x 2 y 2 ⎞
+
=
+
= ⎜ +
⎟
∂x 2 ∂y 2 r 3 r 3 r ⎝ r 2 r 2 ⎠
1 ⎡ ⎛ ∂r ⎞ ⎛ ∂r ⎞
= ⎢⎜ ⎟ + ⎜ ⎟
r ⎢⎣ ⎝ ∂x ⎠ ⎝ ∂y ⎠
2
tanq =
(d)
y
x
x,
∂q
=
∂x
1
y2
1+ 2
x
y
⎛ y⎞
⎜⎝ − 2 ⎟⎠ = − 2
x
x + y2
q
partially w.r.t. x,
x
∂ ⎛ ∂q ⎞
y
2 xy
⋅ 2x = 2
⎜
⎟=
∂x ⎝ ∂x ⎠ ( x 2 + y 2 ) 2
( x + y 2 )2
Differentiating
2 xy
∂ 2q
= 2
2
( x + y 2 )2
∂x
q
y,
∂q
=
∂y
1
1
x
⋅ =
y2 x x2 + y2
1+ 2
x
x ⋅2y
∂ ⎛ ∂q ⎞
=− 2
∂y ⎜⎝ ∂y ⎟⎠
( x + y 2 )2
2 xy
∂ 2q
=− 2
2
∂y
( x + y 2 )2
Hence,
⎤
⎥.
⎥⎦
y
x
q = tan −1
q
2
∂ 2q ∂ 2q
+
= 0.
∂x 2 ∂y 2
Example 3: If ux + vy = 0 and
⎛ ∂v ⎞
x2 + y2
⎛ ∂u ⎞
.
⎜⎝ ⎟⎠ − ⎜ ⎟ = 2
∂x y ⎝ ∂y ⎠ x
y − x2
u v
+ = 1, then prove that
x y
4.36
Solution:
From Eq. (1),
Engineering Mathematics
ux + vy = 0
u v
+ =1
x y
u=−
−vy
x
Substituting in Eq. (2),
− vy v
+ =1
y
x2
v (− y 2 + x 2 ) = x 2 y
v=
x2 y
x2 − y2
Differentiating v partially w.r.t. y keeping x constant,
⎤
⎛ ∂v ⎞
1
y
2 ⎡
⎜⎝ ∂y ⎟⎠ = x ⎢ x 2 − y 2 − ( x 2 − y 2 ) 2 ( − 2 y ) ⎥
⎣
⎦
From Eq. (1),
⎡ x2 − y2 + 2 y2 ⎤ x2 ( x2 + y2 )
= x2 ⎢
=
2
2 2 ⎥
2
2 2
⎣ (x − y ) ⎦ (x − y )
x2 ( x2 + y2 )
=
( x 2 − y 2 )2
ux
v=−
y
Substituting in Eq. (2),
u ux
−
=1
x y2
u( y 2 − x 2 ) = xy 2
u=
xy 2
y − x2
2
Differentiating u w.r.t. x keeping y constant,
⎤
1
x
⎛ ∂u ⎞
2 ⎡
( −2 x ) ⎥
− 2
⎜⎝ ⎟⎠ = y ⎢ 2
2
2 2
∂x y
(y − x )
⎣y −x
⎦
⎡ y2 ( x2 + y2 ) ⎤
=⎢ 2
2 2 ⎥
⎣ (y − x ) ⎦
Hence,
( x 2 + y 2 )( y 2 − x 2 )
⎛ ∂u ⎞ ⎛ ∂v ⎞
−
=
⎜⎝ ⎟⎠ ⎜ ⎟
∂x y ⎝ ∂y ⎠ x
( y 2 − x 2 )2
=
x2 + y2 .
y2 − x2
... (1)
... (2)
Partial Differentiation
Exercise 4.1
1. If u = cos
(
)
(
) (
10. If x = er cos q cos(r sinq ) and
y = er cos q sin (r sinq ), prove that
)
x + y = 0.
∂2 z
3z 2 + x
=−
∂x ∂y
(3 z 2 − x ) 3
3
3. If z = tan ( y + ax ) + ( y − ax ) 2 , show
4. If u = 2(ax + by)2
k(x2 + y2
∂2 u ∂2 u
+
.
∂x 2 ∂y 2
[Ans. : 0]
a2 + b2 = k,
5. If eu =
x+
y
∂u
∂u
sin 2 x
+ sin 2 y
= 2.
∂x
∂y
6. If z3 3yz 3x =
∂z ∂z
=
∂x ∂y
11. If v = ( x 2 − y 2 ) f ( x, y ), prove that
12. If u = f (ax2 + 2hxy + by2
v = f (ax2 + 2hxy + by2), show that
∂ ⎛ ∂v ⎞ ∂ ⎛ ∂v ⎞
u
.
⎜u ⎟ =
∂y ⎝ ∂x ⎠ ∂x ⎜⎝ ∂y ⎟⎠
r
r
13. If x = (eq + e–q ), y = (eq – e–q ),
2
2
∂
∂
x
r
⎛ ⎞
⎛ ⎞
prove that ⎜⎝ ⎟⎠ = ⎜⎝ ⎟⎠ .
∂r q
∂x y
⎡ Hint : x = r cosh q , y = r sinh q ,⎤
⎥⎦
⎢⎣
x2 − y2 = r2
14. If loge q = r – x, r 2 = x 2 + y 2 , show
⎡ ∂ 2 z ⎛ ∂z ⎞ 2 ⎤ ∂ 2 z
(ii) z ⎢
+⎜ ⎟ ⎥ = 2 .
⎢⎣ ∂x ∂y ⎝ ∂x ⎠ ⎥⎦ ∂y
7. If z(z2 + 3x) + 3y =
∂ 2 z ∂ 2 z 2 z ( x − 1)
+
=
.
∂x 2 ∂y 2 ( z 2 + x ) 3
y
8. If u = log (x2 + y2) + tan–1 ⎛⎜ ⎞⎟ , show
⎝x⎠
∂2 u ∂2 u
+
= 0.
that
∂x 2 ∂y 2
1
, find the
9. If u( x, y, z ) = 2
x + y2 + z2
∂u ∂u ∂u
+
+
.
value of
∂x 2 ∂y 2 ∂z 2
2
∂x 1 ∂y ∂y
1 ∂x
=
,
=−
∂r r ∂q ∂r
r ∂q
Hence, deduce that
2
∂ 2 x 1 ∂x 1 ∂ x = 0 .
+
+
2
∂r 2 r ∂r r 2 ∂q
∂2 v ∂2 v
+
= ( x 4 − y 4 ) f ′′( x, y ).
∂x 2 ∂y 2
∂2 z
∂2 z
= a2 2 .
that
2
∂x
∂y
(i) z
⎡
⎤
2
⎢ Ans. :
2
2
2 2⎥
(x + y + z ) ⎦
⎣
x + y , prove that
∂u
∂u
x
+y
∂x
∂y
1
+
x + y sin
2
2. If z3 xz y =
4.37
2
2
that
∂ 2q q ( x 2 + ry 2 )
.
=
∂y 2
r3
⎡
y⎤
r − x ∂r
⎢ Hint : q = e , ∂y = r ⎥
⎣
⎦
15. If u = e ax sin by, prove that
2
u
∂2 u
.
=
x y ∂y ∂x
xy
⎛
16. If u = tan–1 ⎜
2
2
⎝ 1+ x + y
∂2 u
=
∂x ∂y
1
3
2 2
(1 + x + y )
2
.
⎞
,
⎟ prove that
⎠
Engineering Mathematics
4.38
17. If u =
1
y
−
e
( x − a )2
4y
, prove that
22. If x4
z
.
y
u
∂2 u
.
=
x y ∂y ∂x
2
⎡
y 2 − 4 x 3 2 xy − z 2 ⎤
Ans.
:
,
⎢
⎥
2 yz − 4 z 3 2 yz − 4 z 3 ⎦
⎣
3
2
18. If u = tan ( y + ax ) − ( y − ax ) , prove
∂2 u
∂2 u
=
.
∂x ∂y ∂y ∂x
xy
, prove that
19. If u =
2x + z
that
z
x
y2z =
23. If z3 + xy
z
y
1 4⎤
⎡
⎢ Ans.: − 11 , 11⎥
⎣
⎦
3
3
u
u
=
.
2
2
y z
z y
24. Find the value of n for which
20. If u = xm yn
∂u
∂u
=
.
∂ x ∂ y ∂z ∂y ∂x 2
u
u
21. Find
and
for the following
x
y
3
3
−
1
u = kt 2 e
−
x2
na2 t
satisfies the partial
differential equation
∂u
∂2 u
= a2 2 .
∂t
∂x
[Ans. : n = 4]
25. Find the value of n for which
functions:
(i)
z
x
z4 =
xy2 + yz2
x + y −1
(ii) 1 x 2 y 2
(iii) yx
(ax + by)
10
u=t e
n
−
r2
4 kt
satisfies the partial differ-
ential equation
3
( y ax ) 2
⎡
⎤
⎢ Ans. :
⎥
⎢
⎥
1
1
,
⎢( i )
⎥
x + y −1 x + y −1
⎢
⎥
⎢
⎥
−x
−y
⎢(ii)
⎥
,
⎢
1− x2 − y2 1− x2 − y2 ⎥
⎢
⎥
⎢(iii) y x log y, xy x −1
⎥
⎢
⎥
a
⎢(iv )
⎥
,
⎢
⎥
(log e 10)( ax + by )
⎢
⎥
b
⎢
⎥
⎢
⎥
(log e 10)( ax + by )
⎢
⎥
1
1
3a
3
⎢
2⎥
2
⎢⎣( v ) − 2 ( y − ax ) , 2 ( y − ax ) ⎥⎦
26. If x = r sin q
z=r
q,
⎛ ∂ 2 u 2 ∂u ⎞
∂u
=k⎜ 2 +
.
r ∂r ⎟⎠
∂t
⎝ ∂r
⎡ Ans. : n = − 3 ⎤
2 ⎦⎥
⎣⎢
f, y = r sin q sin f,
r q
,
in terms of
x x
r, q, f
⎡ Hint: r 2 = x 2 + y 2 + z 2 ,
⎢
y
⎢
f = tan −1 ,
⎢
x
⎢
x2 + y2
⎢
−1
q
=
tan
⎢⎣
z
⎤
⎥
⎥
⎥
⎥
⎥
⎥⎦
cos q cos f
⎡
,
⎢ Ans. : sin q cos f ,
r
⎢
− sin f
⎢
⎢⎣
r sin q
⎤
⎥
⎥
⎥
⎥⎦
Partial Differentiation
27. If u = x2(y
z) + y2(z
x) + z2(x
y),
y sin y), prove that
29. Prove that f ( x, y, z ) = z tan −1
y
is a
x
harmonic function.
⎡
∂2 f ∂2 f ∂2 f
Hint
:
Prove
that
+
+
⎢
∂x 2 ∂y 2 ∂z 2
⎢
⎢⎣
=0
⎤
⎥
⎥
⎥⎦
30. If z(x + y) = x2 + y2
x
31. If
∂2 z
∂2 z
∂z
+y
=2 .
2
∂x ∂y
∂x
∂x
x2
y2
z2
+
+
= 1, prove that
2+u 4+u 6+u
2
2
r = xy
2
u
= (x2 – y2)[3f (r) + rf (r
x y
∂z
∂z
33. If z = f (x2, y
x
= 2y .
∂x
∂y
34. If z = e ax + by f ( ax − by ), where a, b are
constants, prove that
∂z
∂z
b +a
= 2abz.
∂x
∂y
1
[ f (ct + r) + f (ct – r)]
r
satisfies the partial differential equa-
35. Prove that z =
36. If u + iv = f (x + iy
∂2 u ∂2 u
∂2 v ∂2 v
+
=
+
= 0.
0
,
∂x 2 ∂y 2
∂x 2 ∂y 2
⎡ Hint : u + iv = f ( x + iy ),
⎤
⎢
⎥
u − iv = f ( x − iy )
⎢
⎥
⎢
⎥
1
u = [ f ( x + iy ) + f ( x − iy ) ] , ⎥
⎢
2
⎢
⎥
1
⎢
v
f
(
x
+
iy
)
−
f
(
x
−
iy
)
=
[
]⎥⎥⎦
⎢⎣
2i
37. If u, v, w are function of x, y, z
given as x = u + v + w,
y = u2 + v2 + w2,
z = u3 + v3 + w3,
prove that
u
vw ( w − v )
=
.
x (u − v )( v − w )( w − u )
2
⎛ ∂u ⎞ ⎛ ∂u ⎞ ⎛ ∂u ⎞
⎜⎝ ⎟⎠ + ⎜ ⎟ + ⎜⎝ ⎟⎠
∂x
⎝ ∂y ⎠
∂z
⎛ ∂u
∂u
∂u ⎞
= 2⎜x
+y
+ z ⎟.
⎝ ∂x
∂y
∂z ⎠
32. If u = (x2 y2) f (r
∂ 2 u c 2 ∂ ⎛ 2 ∂u ⎞
=
⎜r
⎟ where c is
∂t 2 r 2 ∂r ⎝ ∂r ⎠
constant.
tion
∂u ∂u ∂u
+
+
= 0.
∂x ∂y ∂z
28. If u = ex (x cos y
∂2 u ∂2 u
+
= 0.
∂x 2 ∂y 2
4.39
[Hint : Differentiate x, y, z w.r.t.
x and solve the equations using
Cramer’s rule]
n
38. If u = ( x 2 + y 2 + z 2 ) 2 , find the value
of n which satisfies the equation
∂2 u ∂2 u ∂2 u
+
+
= 0.
∂x 2 ∂y 2 ∂z 2
[Ans.: 0, –1]
39. If u =
ex + ey
2
⎛ ∂2 u ⎞ ⎛ ∂2 u ⎞ ⎛ ∂2 u ⎞
⎜⎝ ∂x 2 ⎟⎠ ⎜⎝ ∂y 2 ⎟⎠ − ⎜⎝ ∂x ∂y ⎟⎠ = 0.
40. If z = y f (x2
y2
⎛ ∂z ⎞ xz
⎛ ∂z ⎞
y⎜ ⎟+ x⎜ ⎟ = .
⎝ ∂x ⎠
⎝ ∂y ⎠ y
Engineering Mathematics
4.40
4.5 COMPOSITE FUNCTION
4.5.1 Chain Rule
If z = f (u
u
x
y
u = f (x, y)
∂z dz ∂u
df ∂u
∂u
⋅
⋅
=
or
or f ′(u )
∂x du ∂x
du ∂x
∂x
∂z dz ∂u
df ∂u
∂u
=
⋅
⋅
or
or f ′(u ) .
∂y du ∂y
du ∂y
∂y
4.5.2 Composite Function of One Variable or
u
If u = f (x, y
x = f (t), y =
(t
z
t
t
x
du ∂u dx ∂u dy
=
⋅ +
⋅
dt ∂x dt ∂y dt
y
is called total differential of u.
If u = f (x, y, z
of u
t
x = f (t), y =
Fig. 4.2
(t), z = x (t),
du ∂u dx ∂u dy ∂u dz
=
⋅ +
⋅ +
⋅ .
dt ∂x dt ∂y dt ∂z dt
x
y
4.5.3 Composite Function of
Two Variables
If z = f (x, y), where x = f (u, v), y = (u, v), then z is
a function of u, v and is called composite function of
two variables u and v.
∂z ∂z ∂x ∂z ∂y
=
⋅
+
⋅
∂u ∂x ∂u ∂y ∂u
t
Fig. 4.3
z
z
x
∂z ∂z ∂x ∂z ∂y
=
⋅ +
⋅
∂v ∂x ∂v ∂y ∂v
y
x
u
v
dz
.
dt
2
2
2
z = xy + x y, x = at , y = 2at
We know that
y
Fig. 4.4
Example 1: If z = xy2 + x2y, x = at2, y = 2at, find
Solution:
z
z
x
dz ∂z dx ∂z dy
=
⋅ + ⋅
dt ∂x dt ∂y dt
= ( y 2 + 2 xy ) 2at + ( 2 xy + x 2 )2a
y
t
Fig. 4.5
Partial Differentiation
4.41
Substituting x, y and z,
dz
= ( 4 a 2 t 2 + 2at 2 ⋅ 2at ) 2at + ( 2at 2 ⋅ 2at + a 2 t 4 )2a
dt
= 4 a 2 t 2 (1 + t ) 2at + a 2 t 3 ( 4 + t )2a
= 8a3t 3 (1 + t ) + 2a3t 3 ( 4 + t )
= 2a3t 3 (8 + 5t ).
Example 2: If z = sin 1(x - y), x = 3t, y = 4t 3, prove that
Solution:
z = sin 1(x
y), x = 3t, y = 4t3
dz
3
=
.
dt
1 t2
We know that
dz ∂z dx ∂z dy
=
⋅ + ⋅
dt ∂x dt ∂y dt
1
1( −1)
=
⋅3 +
⋅12t 2
2
2
1 − ( x − y)
1 − ( x − y)
=
3 − 12t
z
x
y
2
1 − x 2 − y 2 + 2 xy
t
Fig. 4.6
Substituting x and y,
dz
3(1 − 4t 2 )
3(1 − 4t 2 )
=
=
dt
1 − 9t 2 − 16t 6 + 24t 4
1 − 8t 2 + 16t 4 − t 2 − 16t 6 + 8t 4
3(1 − 4t 2 )
3(1 − 4t 2 )
=
=
(1 − 4t 2 ) 2 − t 2 (1 + 16t 4 − 8t 2 )
(1 − 4t 2 ) 2 − t 2 (1 − 4t 2 ) 2
=
3(1 − 4t 2 )
(1 − 4t ) 1 − t
2
2
=
3
1− t2
.
du
⎛ y⎞
Example 3: If u = tan -1 ⎜ ⎟ , x = e t - e - t , y = e t + e - t , find
.
⎝ x⎠
dt
Solution:
We know that
u
⎛ y⎞
u = tan −1 ⎜ ⎟ , x = e t − e − t , y = e t + e − t
⎝x⎠
du ∂u dx ∂u dy
=
⋅ + ⋅
dt ∂x dt ∂y dt
x
1 ⎛ y⎞ t
1 ⎛1⎞ t
−t
−t
⎜ − ⎟ (e + e ) +
⎜ ⎟ (e − e )
t
y2 ⎝ x ⎠
y2 ⎝ x2 ⎠
Fig. 4.7
1+ 2
1+ 2
x
x
x
x 2 − y 2 (et − e − t ) 2 − (et + e − t ) 2
−y
= 2
⋅y+ 2
⋅x = 2
=
x + y2
x + y2
x + y 2 (et − e − t ) 2 + (et + e − t ) 2
=
=
−4
2
= − 2t
.
2 (e 2t + e −2t )
e + e −2t
y
Engineering Mathematics
4.42
dx
Example 4: If u = x2 + y2 + z2 - 2xyz = 1, show that
1 x
Solution:
We know that
u = x2 + y2 + z2
du =
+
2
dy
1 y
dz
+
2
= 0.
1 z2
2xyz = 1
∂u
∂u
∂u
dx + dy + dz = 0
∂x
∂y
∂z
( 2 x − 2 yz )dx + ( 2 y − 2 xz )dy + ( 2 z − 2 xy )dz = 0
( x − yz )dx + ( y − xz )dy + ( z − xy )dz = 0
... (1)
x 2 + y 2 + z 2 − 2 xyz = 1
We have,
z
x 2 − 2 xyz = 1 − y 2 − z 2
x 2 − 2 xyz + y 2 z 2 = 1 − y 2 − z 2 + y 2 z 2
x
( x − yz ) 2 = (1 − y 2 )(1 − z 2 )
y
x − yz = 1 − y 2 ⋅ 1 − z 2
t
Similarly,
y − xz = 1 − x 2 ⋅ 1 − z 2
and
z − xy = 1 − x 2 ⋅ 1 − y 2
Fig. 4.8
Substituting in Eq. (1),
1 − y 2 ⋅ 1 − z 2 dx + 1 − x 2 ⋅ 1 − z 2 dy + 1 − x 2 ⋅ 1 − y 2 dz = 0
⎛ dx
dy
dz ⎞
1− x2 1− y2 1− z2 ⎜
+
+
⎟=0
2
1− y2
1− z2 ⎠
⎝ 1− x
dx
Hence,
1− x
2
+
dy
1− y
2
+
dz
1− z2
= 0.
Example 5: If u = x2 + y2 + z2, where x = et, y = et sin t, z = et cos t, find
Solution:
du
.
dt
u = x2 + y2 + z2, x = et, y = et sin t, z = et cos t
u
We know that
du ∂u dx ∂u dy ∂u dz
=
⋅ + ⋅ + ⋅
dt ∂x dt ∂y dt ∂z dt
= 2 xe t + 2 y(e t sin t + e t cos t ) + 2 z (e t cos t − e t sin t )
x
z
= 2e t ⋅ e t + 2e t sin t ⋅ e t (sin t + cos t ) + 2e t cos t ⋅ e t (cos t − sin t )
= 2e 2t (1 + sin 2 t + sin t cos t + cos 2 t − cos t sin t )
= 4e 2 t
t
Fig. 4.9
y
Partial Differentiation
4.43
z
o
dz
at t = .
2
dt
z = exy, x = t cost t, y = t sin t
Example 6: If z = exy, x = t cost t, y = t sin t, find
Solution:
We know that
x
dz ∂z dx ∂z dy
=
⋅ + ⋅
dt ∂x dt ∂y dt
y
= e xy [ y(cos t − t sin t ) + e xy x(sin t + t cos t )]
p
p
At
t = , x = 0, y =
2
2
⎡p ⎛
dz
p⎞ ⎤
Hence,
= e 0 ⎢ ⎜ 0 − ⎟ + 0⎥
⎝
dt t = p
2
2⎠ ⎦
⎣
t
Fig. 4.10
2
=−
p2 .
4
Example 7: If z = f (u, v), u = log (x2 + y2), v =
ìz
ìz
ìz
y
-y
= (1 + v 2 ) .
, show that x
ìy
ìx
ìv
x
z = f (u, v), u = log (x2 + y2), v =
Solution:
y
,
x
∂z ∂z ∂u ∂z ∂v ∂z
∂z ⎛ − y ⎞
1
=
⋅
+
⋅
=
⋅
⋅ 2x + ⎜ 2 ⎟
∂x ∂u ∂x ∂v ∂x ∂u x 2 + y 2
∂v ⎝ x ⎠
y
and
x
∂z
2 xy
∂z y 2 ∂z
= 2
⋅
−
⋅
∂x x + y 2 ∂u x 2 ∂v
∂z ∂z ∂u ∂z ∂v ∂z
2y
∂z 1
=
⋅
+
⋅
=
⋅ 2
+
⋅
2
∂y ∂u ∂y ∂v ∂y ∂u x + y
∂v x
z
u
v
∂z
2 xy ∂z ∂z
= 2
⋅ +
∂y x + y 2 ∂u ∂v
x, y
Fig. 4.11
2
Hence, x ∂z − y ∂z = ∂z + y ∂z = (1 + v 2 ) ∂z .
2
∂v
∂y
∂x ∂v x ∂v
Example 8: If w = e (u, v), u = x2 - y2 - 2xy, v = y, prove that
ìw
ìw
equivalent to ( x + y )
+ ( x - y)
= 0.
ìx
ìy
Solution:
w = f (u, v ), u = x 2 − y 2 − 2 xy, v = y
We know that
∂w ∂w ∂u ∂w ∂v ∂w
∂w
=
⋅ +
⋅ =
(2 x − 2 y) +
⋅0
∂x ∂u ∂x ∂v ∂x ∂u
∂v
∂w ∂w
=
(2 x − 2 y)
∂x ∂u
and
∂w ∂w ∂u ∂w ∂v ∂w
∂w
=
⋅ +
⋅
=
( −2 y − 2 x ) +
⋅1
∂y ∂u ∂y ∂v ∂y ∂u
∂v
w
= 0 is
v
w
u
v
x, y
Fig. 4.12
Engineering Mathematics
4.44
∂w
∂w ∂w
= −2( x + y )
+
∂y
∂u ∂v
then
∂w
= 0,
∂v
∂w
∂w
∂w
∂w
= −2( x + y )
and
= 2( x − y )
∂y
∂u
∂x
∂u
∂w
∂w
∂w
∂w
( x + y)
+ ( x − y)
= ( x + y) 2 ( x − y)
− ( x − y) 2 ( x + y)
=0
∂x
∂y
∂u
∂u
Hence,
∂w
∂w
∂w
= 0 is equivalent to ( x + y )
+ ( x − y)
= 0.
∂v
∂x
∂y
If
Example 9: If z = f (x, y) and x = eu + e-v and y = e-u - ev, show that
ìz
ìu
ìz
ìz
ìz
=x
−y
.
ìx
ìy
ìv
Solution:
x = eu + e–v,
z = f (x, y),
z = f ( x, y )
y = e–u – ev
z
We know that
∂z ∂z ∂x ∂z ∂y ∂z u ∂z
=
⋅ + ⋅
=
e + ( −e − u )
∂u ∂x ∂u ∂y ∂u ∂x
∂y
x
y
∂z ∂z ∂x ∂z ∂y ∂z
∂z
=
⋅ + ⋅
= ( −e − v ) + ( −e v )
∂v ∂x ∂v ∂y ∂v ∂x
∂y
and
u, v
∂z ∂z ∂z u
∂z
∂z
∂z
−
= (e + e − v ) − (e − u − e v ) = x − y .
∂x
∂y
∂u ∂v ∂x
∂y
Hence,
Fig. 4.13
Example 10: If z = f (x, y), x = u cosh v, y = u sinh v,
2
2
2
2
1 ⎛ ∂z ⎞
⎛ ∂z ⎞ ⎛ ∂z ⎞
⎛ ∂z ⎞
prove that ⎜ ⎟ − 2 ⎜ ⎟ = ⎜ ⎟ − ⎜ ⎟ .
⎝ ∂x ⎠ ⎝ ∂y ⎠
⎝ ∂u ⎠ u ⎝ ∂v ⎠
Solution:
We know that
z = f (x, y),
x=u
v,
z
y=u
v
∂z ∂z ∂x ∂z ∂y ∂z
∂z
=
⋅ + ⋅
=
cosh v + sinh v
∂u ∂x ∂u ∂y ∂u ∂x
∂y
∂z ∂z ∂x ∂z ∂y ∂z
∂z
=
⋅ + ⋅
=
u sinh v + u cosh v
∂v ∂x ∂v ∂y ∂v ∂x
∂y
and
2
2
2
2
x
y
u, v
Fig. 4.14
⎛ ∂z ⎞
1 ⎛ ∂z ⎞
∂z ∂z
⎛ ∂z ⎞
⎛ ∂z ⎞
2
nh 2 v + 2
cosh v sinh v
⎜⎝ ⎟⎠ − 2 ⎜⎝ ⎟⎠ = ⎜⎝ ⎟⎠ cosh v + ⎜ ⎟ sin
∂u
∂x ∂y
∂x
⎝ ∂y ⎠
u ∂v
2
2
⎛ ∂z ⎞
∂z ∂z
⎛ ∂z ⎞
cosh v sinh v
− ⎜ ⎟ sinh 2 v − ⎜ ⎟ cosh 2 v − 2
⎝ ∂x ⎠
∂x ∂y
⎝ ∂y ⎠
Partial Differentiation
4.45
2
2
⎛ ∂z ⎞
⎛ ∂z ⎞
= ⎜ ⎟ (cosh 2 v − sinh 2 v ) − ⎜ ⎟ (cosh 2 v − sinh 2 v )
⎝ ∂x ⎠
⎝ ∂y ⎠
2
2
⎛ ∂z ⎞ ⎛ ∂z ⎞
= ⎜ ⎟ −⎜ ⎟ .
⎝ ∂x ⎠ ⎝ ∂y ⎠
Example 11: If x = r cos p , y = r sin p , prove that
2
⎛ ∂z ⎞ ⎛ ∂z ⎞
⎜⎝ ⎟⎠ + ⎜ ⎟
∂x
⎝ ∂y ⎠
2
2
⎛ ∂z ⎞
1
=⎜ ⎟ + 2
⎝ ∂r ⎠
r
2
⎛ ∂z ⎞
⎜⎝
⎟ .
∂p ⎠
Solution: Let z = f ( r ,q )
z
x = r cos q , y = r sinq
y
x
x
∂r y
= = cosq and
= = sin q ,
r
∂y r
1 ⎛−y ⎞
−y
− sin q
=
=
⎜ ⎟=
r
y2 ⎝ x2 ⎠ x2 + y2
1+ 2
x
r 2 = x 2 + y 2 , q = tan −1
∂r
∂x
∂q
∂x
∂q
=
∂y
q
x, y
Fig. 4.15
1 ⎛1⎞
x
cos q
=
⎜ ⎟=
r
y2 ⎝ x ⎠ x2 + y2
1+ 2
x
∂z ∂z ∂r ∂z ∂q ∂ z
∂z
=
⋅ +
⋅
= cos q +
∂x ∂r ∂x ∂q ∂x ∂ r
∂q
We know that
r
⎛ − sin q ⎞
⎜⎝
⎟
r ⎠
∂z ∂z ∂r ∂z ∂q
∂z
∂z cos q
=
⋅ +
⋅
= sin q +
⋅
∂y ∂r ∂y ∂q ∂y
∂r
∂q
r
2
2
2
∂z cos q ⎞
∂z sin q ⎞ ⎛ ∂z
⎛ ∂z ⎞ ⎛ ∂z ⎞
⎛ ∂z
⋅
⋅
⎟
⎜⎝ ⎟⎠ + ⎜ ⎟ = ⎜⎝ cos q −
⎟ + ⎜ sin q +
∂x
∂q
r ⎠
⎝ ∂y ⎠
∂r
∂q
r ⎠ ⎝ ∂r
Hence,
2
2
2
1 ⎛ ∂z ⎞
⎛ ∂z ⎞
2
2
= ⎜ ⎟ (cos 2 q + sin 2 q ) + 2 ⎜
⎟ (sin q + cos q )
⎝ ∂r ⎠
r ⎝ ∂q ⎠
2 ∂z ∂z
2 ∂z ∂z
⋅
sin q cos q +
⋅
sin q cos q
−
r ∂r ∂q
r ∂r ∂q
2
2
1 ⎛ ∂z ⎞
⎛ ∂z ⎞
.
=⎜ ⎟ + 2⎜
⎝ ∂r ⎠ r ⎝ ∂q ⎟⎠
Example 12: If z = f (u, v), and u = x2 - y2, v = 2xy, show that
2
2
⎛ ∂z ⎞ ⎛ ∂z ⎞
2
2 2
⎜⎝ ⎟⎠ + ⎜ ⎟ = 4( u + v )
∂x
⎝ ∂y ⎠
1
⎡ ⎛ ∂z ⎞ 2 ⎛ ∂z ⎞ 2 ⎤
⎢⎜ ⎟ + ⎜ ⎟ ⎥ .
⎢⎣ ⎝ ∂u ⎠ ⎝ ∂v ⎠ ⎥⎦
Engineering Mathematics
4.46
Solution: z = f (u, v), and u = x2
We know that
y2, v = 2xy
z
∂z ∂z ∂u ∂z ∂v ∂z
∂z
∂z ⎞
⎛ ∂z
=
⋅ + ⋅ =
⋅ 2x + ⋅ 2 y = 2 ⎜ x + y ⎟
⎝ ∂u
∂x ∂u ∂x ∂v ∂x ∂u
∂v
∂v ⎠
u
∂z ∂z ∂u ∂z ∂v
=
⋅ + .
∂y ∂u ∂y ∂v ∂y
and
x, y
∂z
∂z
∂z ⎞
⎛ ∂z
( −2 y ) + ( 2 x ) = 2 ⎜ − y
=
+x ⎟
⎝ ∂u
∂u
∂v
∂v ⎠
2
Hence,
v
2
2
∂z ⎞
∂z ⎞
⎛ ∂z ⎞ ⎛ ∂z ⎞
⎛ ∂z
⎛ ∂z
+ y ⎟ + 4 ⎜− y
+x ⎟
⎜⎝ ⎟⎠ + ⎜ ⎟ = 4 ⎜⎝ x
⎝ ∂u
∂x
⎝ ∂y ⎠
∂u
∂v ⎠
∂v ⎠
Fig. 4.16
2
2
2
⎡ 2 ⎛ ∂z ⎞ 2
∂z ∂z
2 ⎛ ∂z ⎞
2 ⎛ ∂z ⎞
= 4 ⎢ x ⎜ ⎟ + y ⎜ ⎟ + 2 xy
⋅ +y ⎜ ⎟
⎝ ∂v ⎠
⎝ ∂u ⎠
∂u ∂v
⎢⎣ ⎝ ∂u ⎠
2
∂z ∂z ⎤
⎛ ∂z ⎞
+ x 2 ⎜ ⎟ − 2 xy
⋅ ⎥
⎝ ∂v ⎠
∂u ∂v ⎥⎦
⎡ ⎛ ∂z ⎞ 2 ⎛ ∂z ⎞ 2 ⎤
= 4( x 2 + y 2 ) ⎢ ⎜ ⎟ + ⎜ ⎟ ⎥
⎢⎣ ⎝ ∂u ⎠ ⎝ ∂v ⎠ ⎥⎦
2
2
1 ⎡
⎛ ∂z ⎞ ⎛ ∂z ⎞ ⎤
= 4 ⎡⎣( x 2 + y 2 ) 2 ⎤⎦ 2 ⎢ ⎜ ⎟ + ⎜ ⎟ ⎥
⎢⎣ ⎝ ∂u ⎠ ⎝ ∂v ⎠ ⎥⎦
2
2
1 ⎡
⎛ ∂z ⎞ ⎛ ∂z ⎞ ⎤
= 4 ⎡⎣( x 2 − y 2 ) 2 + 4 x 2 y 2 ⎤⎦ 2 ⎢ ⎜ ⎟ + ⎜ ⎟ ⎥
⎢⎣ ⎝ ∂u ⎠ ⎝ ∂v ⎠ ⎥⎦
1 ⎡
⎛ ∂z ⎞ ⎛ ∂z ⎞
= 4 (u 2 + v 2 ) 2 ⎢ ⎜ ⎟ + ⎜ ⎟
⎢⎣ ⎝ ∂u ⎠ ⎝ ∂v ⎠
2
2
⎤
⎥.
⎥⎦
⎛ y− x z− x⎞
∂u
∂u 2 ∂u
+ y2
+z
= 0.
,
, show that x 2
Example 13: If u = f ⎜
⎟
∂x
∂y
∂z
xz ⎠
⎝ xy
Solution: Let
and
l=
z−x 1 1
y−x 1 1
= −
= − , m=
xz
x z
xy
x y
∂l −1 ∂l = 1 , ∂l
= ,
=0
2
∂x x 2 ∂y y
∂z
∂m −1 ∂m
∂m 1
= 2,
= 0,
=
∂x x
∂y
∂z z 2
⎛ y−x z−x⎞
u= f ⎜
,
= f ( l , m)
⎝ xy
xz ⎟⎠
u
l
m
x, y, z
Fig. 4.17
Partial Differentiation
4.47
We know that
∂u ∂u ∂l ∂u ∂m ∂u ⎛ −1 ⎞ ∂u ⎛ −1 ⎞
=
⋅ +
⋅
=
⎜ ⎟+
⎜ ⎟
∂x ∂l ∂x ∂m ∂x ∂l ⎝ x 2 ⎠ ∂m ⎝ x 2 ⎠
x2
∂u
⎛ ∂u ∂u ⎞
= −⎜ +
⎝ ∂l ∂m ⎟⎠
∂x
... (1)
∂u ∂u ∂l ∂u ∂m ∂u ⎛ 1 ⎞ ∂u
=
⋅ +
⋅
=
+
⋅0
∂y ∂l ∂y ∂m ∂y ∂l ⎜⎝ y 2 ⎟⎠ ∂m
Also,
y2
∂u ∂u
=
∂y ∂l
... (2)
∂u ⎛ 1 ⎞
∂u ∂u . ∂l ∂u . ∂m ∂u
=
⋅0 +
=
+
⎜ ⎟
∂l
∂m ⎝ z 2 ⎠
∂z ∂l ∂z ∂m ∂z
∂u ∂u
=
z2
∂z ∂m
Adding Eqs (1), (2) and (3),
x2
... (3)
∂u
∂u
∂u
⎛ ∂u ∂u ⎞ ∂u ∂u
+ y2
+ z2
= −⎜ +
+
+
= 0.
⎝ ∂l ∂m ⎟⎠ ∂l ∂m
∂x
∂y
∂z
⎛x y z⎞
∂u
∂u
∂u
+y
+z
= 0.
Example 14: If u = f ⎜ , , ⎟ , prove that x
∂x
∂y
∂z
⎝ y z x⎠
Solution: Let
x
= l,
y
∂l 1
= ,
∂x y
y
= m,
z
∂l − x
=
,
∂y y 2
z
=n
x
∂l
=0
∂z
∂m
= 0,
∂x
∂m 1
= ,
∂y z
∂m − y
= 2
∂z
z
∂n
z
=− 2,
∂x
x
∂n
= 0,
∂y
∂n 1
=
∂z x
u
We know that
∂u ∂u ∂l ∂u ∂m ∂u ∂n
=
⋅ +
⋅
+
⋅
∂x ∂l ∂x ∂m ∂x ∂n ∂x
∂u 1 ∂u
∂u ⎛ − z ⎞
=
⋅ +
⋅0 +
⎜ ⎟
∂l y ∂m
∂n ⎝ x 2 ⎠
∂u x ∂u z ∂u
x
= ⋅
− ⋅
∂x y ∂l x ∂n
Also,
∂u ∂u ∂l ∂u ∂m ∂u ∂n
=
⋅ +
⋅
+
⋅
∂y ∂l ∂y ∂m ∂y ∂n ∂y
l
m
x, y, z
Fig. 4.18
n
Engineering Mathematics
4.48
=
y
∂u ⎛ − x ⎞ ∂u 1 ∂ u
+
⋅ + ⋅0
∂l ⎜⎝ y 2 ⎟⎠ ∂m z ∂n
∂u
x ∂u y ∂u
=− ⋅
+ ⋅
∂y
y ∂l z ∂m
∂u ∂u ∂l ∂u
=
⋅ +
∂z ∂l ∂z ∂m
and
=
∂u
∂u ⎛ − y ⎞ ∂u 1
⋅0 +
⎜ ⎟+ ⋅
∂l
∂m ⎝ z 2 ⎠ ∂n x
z
Hence,
x
∂m ∂u ∂n
+
⋅
∂z ∂n ∂z
∂u − y ∂u z ∂u
=
+ ⋅
∂z
z ∂m x ∂n
∂u
∂u
∂u
+y
+z
= 0.
∂x
∂y
∂z
Example 15: If u = f (x2 - y2, y2 - z2, z2 - x2), prove that
1 ∂u 1 ∂u 1 ∂u
+
+
= 0.
x ∂x y ∂y z ∂z
Solution: Let x 2 − y 2 = l , y 2 − z 2 = m, z 2 − x 2 = n
∂l
= 2 x,
∂x
∂m
= 0,
∂x
∂l
= −2 y,
∂y
∂m
= 2 y,
∂y
∂l
=0
∂z
∂m
= −2 z
∂z
∂n
= −2 x,
∂x
∂n
= 0,
∂y
∂n
= 2z
∂z
We know that
∂u ∂u ∂l ∂u ∂m ∂u ∂n
=
⋅ +
⋅
+ ⋅
∂x ∂l ∂x ∂m ∂x ∂n ∂x
∂u
∂u
∂u
=
⋅ 2x +
⋅ 0 + ( −2 x )
∂l
∂m
∂n
Also,
1 ∂u
∂u
∂u
= 2 −2
x ∂x
∂l
∂n
∂u ∂u ∂l ∂u ∂m ∂u ∂n
=
⋅ +
⋅
+
⋅
∂y ∂l ∂y ∂m ∂y ∂n ∂y
∂u
∂u
∂u
=
(−
−2 y ) +
( 2 y ) + ( 0)
∂l
∂m
∂n
1 ∂u
∂u
∂u
= −2 + 2
y ∂y
∂l
∂m
u
l
m
x, y, z
Fig. 4.19
n
Partial Differentiation
4.49
∂u ∂u ∂l ∂u ∂m ∂u ∂n
=
⋅ +
⋅
+ ⋅
∂z ∂l ∂z ∂m ∂z ∂n ∂z
and
∂u
∂u
∂u
( −2 z ) + ( 2 z )
⋅0 +
∂l
∂m
∂n
1 ∂u
∂u
∂u
= −2
+2
z ∂z
∂m
∂n
1 ∂u 1 ∂u 1 ∂u
+
+
= 0.
x ∂x y ∂y z ∂z
=
Example 16: If u = f ( e y − z , e z − x , e x − y ), show that
∂u ∂u ∂u
+
+
= 0.
∂x ∂y ∂z
Solution: Let e y − z = l , e z − x = m, e x − y = n
∂l
= 0,
∂x
∂l
= e y− z = l,
∂y
∂m
= −e z − x = − m,
∂x
∂m
= 0,
∂y
∂m
= ez−x = m
∂z
∂n
= e x − y = n,
∂x
∂n
= −e x − y = − n,
∂y
∂n
=0
∂z
∂l
= −e y − z = −l
∂z
u = f (e y − z , e z − x , e x − y ) = f (l , m, n).
We know that
∂u ∂u ∂l ∂u ∂m ∂u ∂n
=
⋅ +
⋅
+
⋅
∂x ∂l ∂x ∂m ∂x ∂n ∂x
∂u
∂u
∂u
=
⋅0 +
( − m) + ⋅ n
∂l
∂m
∂n
∂u
∂u
+n
= −m
∂m
∂n
∂u ∂u ∂l ∂u ∂m ∂u ∂n
=
⋅ +
⋅
+
⋅
∂y ∂l ∂y ∂m ∂y ∂n ∂y
∂u
∂u
∂u
⋅l +
⋅ 0 + ⋅ ( − n)
∂l
∂m
∂n
∂u
∂u
=l
−n
∂l
∂n
∂u ∂u ∂l ∂u ∂m ∂u ∂n
=
⋅ +
⋅
+
⋅
∂z ∂l ∂z ∂m ∂z ∂n ∂z
∂u
∂u
∂u
∂u
∂u
( −l ) +
=
⋅ m + ⋅ 0 = −l
+m
∂l
∂m
∂n
∂l
∂m
=
and
Hence,
∂u ∂u ∂u
+
+
= 0.
∂x ∂y ∂z
u
l
m
x, y, z
Fig. 4.20
n
Engineering Mathematics
4.50
Example 17: If x = vw , y = wu , z = uv and e is a function of x, y and z,
then prove that x
∂e
∂e
∂e
∂e
∂e
∂e
+y
+z
=u
+v
+w
.
∂x
∂y
∂z
∂u
∂v
∂w
x = vw
∂x
= 0,
∂u
Solution:
∂x 1 w
=
,
∂v 2 v
∂x 1 v
=
∂w 2 w
∂y
= 0,
∂v
∂y 1 u
=
∂w 2 w
f
y = wu
∂y 1 w
=
,
∂u 2 u
x
y
z = uv
∂z 1 v ∂z 1 u ∂z
=
,
=
,
=0
∂u 2 u ∂v 2 v ∂w
We know that
∂f ∂f ∂x ∂f ∂y ∂f ∂z
=
⋅ +
⋅ +
⋅
∂u ∂x ∂u ∂y ∂u ∂z ∂u
=
u
Also,
∂f
∂f 1 w ∂f 1 v
⋅0 +
⋅
+
⋅
∂x
∂y 2 u ∂z 2 u
⎤
∂f 1 ⎡ ∂f
∂f
1 ⎛ ∂f
∂f ⎞
= ⎢
uw +
uv ⎥ = ⎜ y
+z
∂u 2 ⎣ ∂y
∂z
2 ⎝ ∂y
∂z ⎟⎠
⎦
∂f ∂f ∂x ∂f ∂y ∂f ∂z
=
⋅ +
⋅ +
⋅
∂v ∂x ∂v ∂y ∂v ∂z ∂v
=
v
∂f 1 w ∂f
∂f 1 u
⋅
+
⋅0 +
∂x 2 v ∂y
∂z 2 v
∂f 1 ⎛ ∂f
∂f
⎞
= ⎜
vw +
uv ⎟
⎠
∂v 2 ⎝ ∂x
∂z
=
∂f ⎞
1 ⎛ ∂f
+z
⎜x
⎟
∂z ⎠
2 ⎝ ∂x
∂f ∂f ∂x ∂ f ∂ y ∂ f ∂ z
=
⋅
+
⋅
+
⋅
∂w ∂ x ∂ w ∂ y ∂ w ∂ z ∂ w
and
=
w
∂f 1 v ∂f 1 u ∂f
⋅
+
+
⋅0
∂x 2 w ∂y 2 w ∂z
⎞
∂f 1 ⎛ ∂f
∂f
= ⎜
vw +
uw ⎟
∂w 2 ⎝ ∂x
∂y
⎠
=
∂f ⎞
1 ⎛ ∂f
+y
x
⎜
∂y ⎟⎠
2 ⎝ ∂x
u, v, w
Fig. 4.21
z
Partial Differentiation
Hence,
u
4.51
∂f
∂f
∂f
∂f
∂f
∂f
+v
+w
=x
+y
+z
.
∂u
∂v
∂w
∂x
∂y
∂z
Example 18: If f (xy2, z - 2x) = 0, show that 2 x
∂z
∂z
−y
= 4 x.
∂x
∂y
Solution: Let l = xy 2 , m = z − 2 x, f (l , m) = 0
We know that
∂f ∂f ∂l ∂f ∂m
=
⋅ +
⋅
=0
∂x ∂l ∂x ∂m ∂x
∂f 2
∂f ⎛ ∂z
⎞
(y )+
⎜ − 2 ⎟⎠ = 0
∂l
∂m ⎝ ∂x
∂f
∂z
2−
∂l =
∂x
∂f
y2
∂m
∂f ∂f ∂l ∂f ∂m
=
⋅ +
⋅
=0
and
∂y ∂l ∂y ∂m ∂y
[∵ f ( xy 2 , z − 2 x ) = 0]
f
l
x
∂f
∂f ⎛ ∂z ⎞
( 2 xy ) +
=0
∂l
∂m ⎜⎝ ∂y ⎟⎠
∂z
∂f
∂
∂l = − y
∂f
2 xy
∂m
From Eqs (1) and (2), we get
∂z
∂z
∂x = − ∂y
2 xy
y2
2−
4x − 2x
Hence,
2x
m
... (1)
∂z
∂z
= −y
∂x
∂y
∂z
∂z
−y
= 4 x.
∂x
∂y
∂z
∂z
y⎞
⎛ z
+y
= 3z.
Example 19: If f ⎜ 3 , ⎟ = 0, prove that x
∂
x
∂
y
⎝x
x⎠
y
⎛ z ⎞
Solution: Let l = ⎜ 3 ⎟ , m = , then f (l, m) = 0
⎝x ⎠
x
y
x
x
Fig. 4.22
z
y
... (2)
Engineering Mathematics
4.52
We know that
⎤
⎡
⎛ z y⎞
⎢∵ f ⎜⎝ x 3 , x ⎟⎠ = 0 ⎥
⎦
⎣
∂f ∂ f ∂l ∂f ∂m
=
⋅ +
⋅
=0
∂x ∂l ∂x ∂m ∂x
∂f ⎛ −3 z 1 ∂z ⎞ ∂f ⎛ y ⎞
+ 3
⎜
⎟+
⎜− ⎟ = 0
∂l ⎝ x 4
x ∂x ⎠ ∂m ⎝ x 2 ⎠
y
∂f
x2
∂l =
∂f
1 ∂z 3 z
−
∂m x 3 ∂x x 4
=
and
f
l
x2 y
∂z
x − 3z
∂x
m
... (1)
z
∂f ∂f ∂l ∂f ∂m
=
⋅ +
⋅
=0
∂y ∂l ∂y ∂m ∂y
x
x x
y
y
Fig. 4.23
∂f ⎛ 1 ∂z ⎞ ∂f ⎛ 1 ⎞
+
⎜ ⎟=0
∂l ⎜⎝ x 3 ∂y ⎟⎠ ∂m ⎝ x ⎠
⎛1⎞
∂f
−⎜ ⎟
∂l = ⎝ x ⎠
1 ∂z
∂f
∂m x 3 ∂y
=−
x2
∂z
∂y
From Eqs (1) and (2), we get
x2 y
x2
=−
∂z
∂z
x − 3z
∂x
∂y
∂z
∂z
= − x + 3z
y
∂y
∂x
Hence,
x
∂z
∂z
+y
= 3z
∂x
∂y
Example 20: If f (lx + my + nz, x2 + y2 + z2) = 0,
prove that (ly - mx) + (ny - mz)
∂z
∂z
+ ( lz − nx ) = 0.
∂x
∂y
... (2)
Partial Differentiation
4.53
Solution: Let u = lx + my + nz, v = x2 + y2 + z2 then f (u, v) = 0
We know that
∂f ∂f ∂u ∂f ∂v
=
⋅ + ⋅ =0
∂x ∂u ∂x ∂v ∂x
f
∂f ⎛
∂z ⎞ ∂f ⎛
∂z ⎞
⎜ l + n ⎟⎠ + ⎜⎝ 2 x + 2 z ⎟⎠ = 0
∂u ⎝
∂x
∂v
∂x
∂z ⎞
⎛
∂f
2⎜x + z ⎟
∂x ⎠
∂u = − ⎝
... (1)
∂f
∂z ⎞
⎛
⎜⎝ l + n ⎟⎠
∂v
∂x
v
u
x, y, z
∂f ∂f ∂u ∂f ∂v
=
⋅ + ⋅ =0
∂y ∂u ∂y ∂v ∂y
and
x
∂f ⎛
∂z ⎞ ∂f ⎛
∂z ⎞
m + n ⎟ + ⎜ 2 y + 2z ⎟ = 0
⎜
∂u ⎝
∂y ⎠ ∂v ⎝
∂y ⎠
y
Fig. 4.24
⎛
∂z ⎞
∂f
2⎜y + z ⎟
∂y ⎠
∂u = − ⎝
∂f
⎛
∂z ⎞
⎜⎝ m + n ∂y ⎟⎠
∂v
... (2)
From Eqs (1) and (2), we get
⎛
∂z ⎞
∂z ⎞
⎛
2⎜x + z ⎟ 2⎜ y + z ⎟
⎝
∂y ⎠
⎝
∂x ⎠
=
∂z ⎞
⎛
⎛
∂z ⎞
m+n ⎟
⎜⎝ l + n ⎟⎠
⎜
∂x
⎝
∂y ⎠
mx + nx
∂z
∂z
∂z ∂z
∂z
∂z
∂z ∂z
+ mz + nz ⋅ = ly + lz + ny + nz ⋅
∂y
∂x
∂x ∂y
∂y
∂x
∂x ∂y
Hence, (ly − mx ) + ( ny − mz )
∂z
∂z
+ (lz − nx ) = 0.
∂x
∂y
Example 21: If z = f (x, y) where x = log u, y = log v, show that
Solution: z = f (x, y), x = log u, y = log v
∂2 z
∂2 z
= uv
.
∂x ∂y
∂u ∂v
z
We know that
∂z ∂z ∂x ∂z ∂y
=
+
∂u ∂x ∂u ∂y ∂u
∂z 1 ∂z
1 ∂z
=
⋅ + ⋅0 =
∂x u ∂y
u ∂x
x
y
u,v
Engineering Mathematics
4.54
Differentiating
∂z
∂x
z
w.r.t. v,
u
∂ ⎛ ∂z ⎞ ∂ ⎛ 1 ∂z ⎞
⎜ ⎟=
⎜
⎟
∂v ⎝ ∂u ⎠ ∂v ⎝ u ∂x ⎠
=
x
y
1 ⎡ ∂ ⎛ ∂z ⎞ ⎛ ∂x ⎞ ∂ ⎛ ∂ z ⎞ ⎛ ∂ y ⎞ ⎤
⎜ ⎟⋅⎜ ⎟ + ⎜ ⎟⋅⎜ ⎟
u ⎢⎣ ∂x ⎝ ∂x ⎠ ⎝ ∂v ⎠ ∂y ⎝ ∂x ⎠ ⎝ ∂v ⎠ ⎥⎦
∂2 z
1 ⎛ ∂2 z
∂2 z 1 ⎞ 1 ∂2 z
= ⎜ 2 ⋅0 +
⋅ =
⋅
∂u ∂v u ⎝ ∂x
∂x ∂y v ⎟⎠ uv ∂x ∂y
u,v
Fig. 4.25
∂2 z
∂2 z
= uv
.
∂x ∂y
∂u ∂v
Example 22: If x = r cos p , y = r sin p , show that
(i) equation
∂2 u ∂2 u
∂ 2 u 1 ∂u 1 ∂ 2 u
+
=
0
+
+
= 0.
transforms
into
∂x 2 ∂y 2
∂r 2 r ∂r r 2 ∂p 2
2
2
∂ 2 r ∂ 2 r 1 ⎡ ⎛ ∂r ⎞ ⎛ ∂r ⎞ ⎤
+
= ⎢⎜ ⎟ +
(ii)
⎥.
∂x 2 ∂y 2 r ⎢⎣ ⎝ ∂x ⎠ ⎜⎝ ∂y ⎟⎠ ⎥⎦
Solution: (i) x = r cos q , y = r sin q
q = tan −1
x2 + y2 = r2
2 x = 2r
Similarly,
∂r
,
∂x
y
x
∂r x r cos q
= =
= cos q
∂x r
r
∂r y r sin q
= =
= sin q
∂y r
r
∂q
=
∂x
1 ⎛−y ⎞
−y
⎜ ⎟=
y2 ⎝ x2 ⎠ x2 + y2
1+ 2
x
∂q − r sin q − sin q
=
=
r
∂x
r2
Similarly,
∂q
=
∂y
1 ⎛1⎞
x
⎜ ⎟=
y2 ⎝ x ⎠ x2 + y2
1+ 2
x
∂q r cos q cos q
=
=
r
∂y
r2
Let u = f ( r , q ),
r2 = x2 + y2
q = tan −1
y
x
Partial Differentiation
4.55
We know that
u
∂u ∂u ∂r ∂u ∂q
=
⋅ +
⋅
∂x ∂r ∂x ∂q ∂x
∂ 2 u ∂ ⎛ ∂u ∂r ⎞ ∂ ⎛ ∂u ∂q ⎞
=
⋅
⎟
⎜ ⋅ ⎟+ ⎜
∂x 2 ∂x ⎝ ∂r ∂x ⎠ ∂x ⎝ ∂q ∂x ⎠
=
r
∂ ⎛ ∂ u ⎞ ∂ r ∂ u ∂ ⎛ ∂ r ⎞ ∂ ⎛ ∂ u ⎞ ∂q ∂u ∂ ⎛ ∂q ⎞
+
⋅ ⎜
⎟⋅
⎜ ⎟⋅ + ⋅ ⎜ ⎟ + ⎜
⎟
∂ x ⎝ ∂ r ⎠ ∂ x ∂ r ∂ x ⎝ ∂ x ⎠ ∂ x ⎝ ∂ q ⎠ ∂x ∂q ∂x ⎝ ∂x ⎠
q
x, y
⎡ ∂ ⎛ ∂u ⎞ ∂r ∂ ⎛ ∂u ⎞ ∂q ⎤ ∂r ∂u ∂ 2 r
= ⎢ ⎜ ⎟⋅ +
+ ⋅ 2
⎜ ⎟⋅ ⎥
⎣ ∂ r ⎝ ∂ r ⎠ ∂ x ∂q ⎝ ∂ r ⎠ ∂ x ⎦ ∂x ∂r ∂x
∂u
∂r
⎡ ∂ ⎛ ∂u ⎞ ∂r ∂ ⎛ ∂u ⎞ ∂q ⎤ ∂q ∂u ∂ q
+⎢ ⎜
+
⋅ 2
+
⎟
⎜
⎟⋅ ⎥
⎣ ∂r ⎝ ∂q ⎠ ∂x ∂q ⎝ ∂q ⎠ ∂x ⎦ ∂x ∂q ∂x
2
2
∂ 2 u ∂ 2 u ⎛ ∂r ⎞
∂ 2 u ∂q ∂r
= 2⎜ ⎟ +
⋅ ⋅
2
∂x
∂r ⎝ ∂x ⎠ ∂q ∂r ∂x ∂x
r
q
2
+
∂u ∂ 2 r
∂ 2 u ∂r ∂q ∂ 2 u ⎛ ∂q ⎞ ∂u ∂2q
⋅ ⋅ +
⋅ 2+
⎜ ⎟ + ⋅
∂r ∂q ∂x ∂x ∂q 2 ⎝ ∂x ⎠ ∂q ∂x 2
∂r ∂x
We have,
x, y
∂r
= cosq
∂x
∂ ⎛ ∂r ⎞
∂q
⎜⎝ ⎟⎠ = − sin q ⋅
∂x ∂x
∂x
∂2 r
⎛ − sin q
= − sin q ⎜
⎝ r
∂x 2
∂u
∂q
2
⎞ sin q
=
⎟⎠
r
∂q
sin q
=−
∂x
r
Also,
∂ 2q
⎛ cos q ∂q 1 ∂r
⎞
= −⎜
−
sin q ⎟
⎝ r ∂x r 2 ∂x
⎠
∂x 2
r
q
x, y
Fig. 4.26
⎤
⎡ cos q ⎛ − sin q ⎞ 1
= −⎢
⎜⎝
⎟⎠ − 2 cos q ⋅ sin q ⎥
r
r
⎦
⎣ r
2 sin q cos q sin 2q
=
=
r2
r2
2
Substituting in
u
,
x2
∂2 u ∂2 u
∂ 2 u ⎛ − sin q ⎞
∂u sin 2 q ∂ 2 u sin 2 q ∂u sin 2q
2
=
+
+
⋅
+ 2⋅ 2 +
⋅
cos
q
2
cos
q
⎜
⎟
∂r ∂q ⎝ r ⎠
∂r
∂q
r
∂x 2 ∂r 2
∂q
r
r2
2
To get
2
u
p
u
, replace q by + q in
.
2
2
y
x2
Engineering Mathematics
4.56
2
u
y2
2
u
( sin q ) 2
r2
2 2u
( cos q ) ( sin q )
r r q
1 u
1 2u
1 u
cos 2 q
cos 2 q
sin(p
2
2
r r
r q
r2 q
2
2 2u
1 u
u
cos q sin q +
cos 2 q
= 2 sin 2 q +
r r q
r r
r
1 2u
cos 2 q
r2 q 2
2
2
2
u
u
u 1 u 1 2u
+ 2 = 2+
+
2
r r r2 q 2
x
y
r
2q )
1 u
sin 2q
r2 q
Hence, equation
∂ 2 u 1 ∂u 1 ∂ 2 u
∂2 u ∂2 u
= 0.
+
+
transforms
into
+
=
0
∂r 2 r ∂r r 2 ∂q 2
∂x 2 ∂y 2
(ii) We have
∂2 r
∂q cos 2 q
∂ 2 r sin 2 q ∂r
cos
q
=
⋅
=
,
sin
q
,
=
=
r
∂y
r
∂y
∂y 2
∂x 2
Hence,
∂ 2 r ∂ 2 r sin 2 q cos 2 q
+
=
+
r
r
∂x 2 ∂y 2
=
2
2
1 ⎡ ⎛ ∂r ⎞ ⎛ ∂r ⎞ ⎤
⎢⎜ ⎟ + ⎜ ⎟ ⎥ .
r ⎢⎣ ⎝ ∂y ⎠ ⎝ ∂x ⎠ ⎥⎦
Example 23: If z = f (x, y), (x + y) = (u + v)3 and x - y = (u - v)3, show that
⎛ ∂2 z ∂2 z ⎞
⎛ ∂2 z ∂2 z ⎞
( u2 − v 2 ) ⎜ 2 − 2 ⎟ = 9( x 2 − y 2 ) ⎜ 2 − 2 ⎟ .
∂v ⎠
∂y ⎠
⎝ ∂u
⎝ ∂x
Solution:
We have,
x + y = (u + v )3 and x
y
( u v )3
1
⎡(u + v )3 + (u − v )3 ⎤⎦
2⎣
∂x 1
∂2 x
= ⎡⎣3(u + v ) 2 + 3(u − v ) 2 ⎤⎦ = 3(u 2 + v 2 ),
= 6u
∂u 2
∂u 2
2 x = ( u + v )3 + ( u − v )3 , x =
and
∂x 1
∂2 x
= 6u
= ⎡⎣3(u + v ) 2 + 3(u − v ) 2 ( −1) ⎤⎦ = 6uv,
∂v 2
∂v 2
1
2 y = (u + v )3 − (u − v )3 , y = ⎡⎣(u + v )3 − (u − v )3 ⎤⎦
2
∂2 y
∂y 1
= 6v
= ⎡⎣3(u + v ) 2 − 3(u − v ) 2 ⎤⎦ = 6uv,
∂u 2
∂u 2
∂2 y
∂y 1
= 6v
= ⎡⎣3(u + v ) 2 − 3(u − v ) 2 ( −1) ⎤⎦ = 3(u 2 + v 2 ),
∂v 2
∂v 2
Partial Differentiation
4.57
We know that
z
∂z ∂z ∂x ∂z ∂y
⋅ + ⋅
=
∂u ∂x ∂u ∂y ∂u
x
∂ ⎛ ∂z ∂x ⎞ ∂ ⎛ ∂z ∂y ⎞
∂ z
=
⋅
⎜ ⋅ ⎟+
∂u 2 ∂u ⎝ ∂x ∂u ⎠ ∂u ⎜⎝ ∂y ∂u ⎟⎠
y
2
u, v
∂z
∂x
∂ ⎛ ∂ z ⎞ ∂x ∂z ∂ x ∂ ⎛ ∂z ⎞ ∂y ∂z ∂ 2 y
=
⋅ + ⋅
+
⋅
⎜ ⎟⋅ +
∂u ⎝ ∂x ⎠ ∂u ∂x ∂u 2 ∂u ⎜⎝ ∂y ⎟⎠ ∂u ∂y ∂u 2
2
⎡ ∂ ⎛ ∂ z ⎞ ∂x ∂ ⎛ ∂z ⎞ ∂y ⎤ ∂ x ∂ z ∂ 2 x
= ⎢ ⎜ ⎟⋅ + ⎜ ⎟⋅ ⎥ +
⋅ 2
⎣ ∂ x ⎝ ∂ x ⎠ ∂u ∂y ⎝ ∂x ⎠ ∂u ⎦ ∂ u ∂ x ∂ u
x
⎡ ∂ ⎛ ∂ z ⎞ ∂x ∂ ⎛ ∂z ⎞ ∂y ⎤ ∂y ∂z ∂ 2 y
⋅ 2
+⎢ ⎜ ⎟⋅ + ⎜ ⎟⋅ ⎥ +
⎣ ∂ x ⎝ ∂ y ⎠ ∂u ∂y ⎝ ∂y ⎠ ∂u ⎦ ∂u ∂y ∂u
y
u, v
∂z
∂y
2
=
∂ 2 z ⎛ ∂x ⎞
∂ 2 z ∂y ∂x ∂z ∂ 2 x
+
⋅
⋅
+
⋅
⎜ ⎟
∂ x 2 ⎝ ∂ u ⎠ ∂x ∂y ∂u ∂u ∂x ∂u 2
+
2
x
∂ z ∂x ∂ y ∂ z ⎛ ∂ y ⎞ ∂ z ∂ y
⋅
+
⋅
⋅
⎜ ⎟ +
∂y ∂x ∂ u ∂ u ∂ y 2 ⎝ ∂ u ⎠ ∂ y ∂ u 2
2
2
2
u, v
2
∂ 2 z ∂ 2 z ⎛ ∂x ⎞
∂ 2 z ∂x ∂y ∂ 2 z ⎛ ∂y ⎞ ∂ z ∂ 2 x ∂ z ∂ 2 y
= 2 ⎜ ⎟ +2
+
⋅
⋅
⋅
⋅
+
⎜ ⎟ +
2
∂x ∂y ∂u ∂u ∂y 2 ⎝ ∂u ⎠ ∂ x ∂ u 2 ∂ y ∂ u 2
∂x ⎝ ∂u ⎠
∂u
2
Similarly,
y
2
∂ 2 z ∂ 2 z ⎛ ∂x ⎞
∂ 2 z ∂x ∂y ∂ 2 z
=
2
+
⋅
⋅ +
⎜
⎟
∂x ∂y ∂v ∂v ∂y 2
∂v 2 ∂x 2 ⎝ ∂v ⎠
Fig. 4.27
... (1)
2
2
2
⎛ ∂y ⎞ ∂ z ∂ x ∂ z ∂ y
⋅ 2 ... (2)
⋅ 2+
⎜⎝ ⎟⎠ +
∂y ∂v
∂x ∂v
∂v
∂2 z
∂2 z
∂2 z ∂2 z
∂z
∂z
2
2 2
2
2
⋅ 36u 2 v 2 + ⋅ 6u + ⋅ 6v
+
+
⋅
+
+
(
)
(
)
=
u
v
u
v
uv
9
2
3
6
2
2
2
∂x ∂y
∂x
∂y
∂y
∂x
∂u
∂2 z
∂2 z ∂2 z
∂z
∂z
∂2 z
6uv ⋅ 3(u 2 + v 2 ) + 2 ⋅ 9(u 2 + v 2 ) 2 + ⋅ 6u + ⋅ 6 v
= 2 ⋅ 36u 2 v 2 + 2
2
∂x ∂y
∂x
∂y
∂x
∂v
∂y
∂2 z
∂2 z ∂2 z
∂2 z
2
2 2
2 2
⎡(u 2 + v 2 ) 2 − 4u 2 v 2 ⎤⎦
⎡
⎤
)
v
u
v
=
−
u
9
(
+
−
4
−
9
⎦
∂x 2 ⎣
∂u 2 ∂v 2
∂y 2 ⎣
⎛ ∂2 z ∂2 z ⎞
= 9(u 2 − v 2 ) 2 ⎜ 2 − 2 ⎟
∂y ⎠
⎝ ∂x
2
2
∂2 z ⎞
∂2 z ⎞
2
2 3 ⎛∂ z
2
2 ⎛∂ z
Hence, (u − v ) ⎜ 2 − 2 ⎟ = 9(u − v ) ⎜ 2 − 2 ⎟
∂y ⎠
∂v ⎠
⎝ ∂x
⎝ ∂u
⎛ ∂2 z ∂2 z ⎞
= 9(u + v )3 (u − v )3 ⎜ 2 − 2 ⎟
∂y ⎠
⎝ ∂x
⎛ ∂2 z ∂2 z ⎞
= 9( x + y )( x − y ) ⎜ 2 − 2 ⎟
∂y ⎠
⎝ ∂x
⎛ ∂2 z ∂2 z ⎞
= 9( x 2 − y 2 ) ⎜ 2 − 2 ⎟ .
∂y ⎠
⎝ ∂x
Engineering Mathematics
4.58
p
p
Example 24: If x + y = 2e cos e and x − y = 2ie sin e , prove that
(i)
∂v
∂v
∂v
+i
= 2y
∂p
∂e
∂y
(ii)
∂2 v ∂2 v
∂2 v
+ 2 = 4 xy
2
∂x ∂y
∂p
∂e
q
q
Solution: x + y = 2e cos f , x − y = 2ie sin f
2 x = 2eq (cos f + i sin f ), x = eq + i f
∂x
= eq + i f = x,
∂q
∂x
= ieq + i f = ix
∂f
2 y = 2eq (cos f − i sin f ), y = eq − i f
∂y
= eq − i f = y,
∂q
∂y
= −ieq − i f = −iy
∂f
Let v = f (x, y)
We know that
v
∂v ∂v ∂x ∂v ∂y
=
⋅
+
⋅
∂q ∂x ∂q ∂y ∂q
∂v ∂v
∂v
=
x+
y
∂q ∂x
∂y
∂ ⎛ ∂v ⎞ ∂ ⎛ ∂v ⎞
∂2 v
=
y
⎜x ⎟ +
2
∂q ⎝ ∂x ⎠ ∂q ⎜⎝ ∂y ⎟⎠
∂q
∂x ∂v
∂
=
⋅ +x
∂q ∂x
∂q
=
y
q, f
∂ ⎛ ∂v ⎞
⎛ ∂v ⎞ ∂y ∂v
⋅ + y⋅
⎜⎝ ⎟⎠ +
∂q ∂y
∂q ⎜⎝ ∂y ⎟⎠
∂x
Fig. 4.28
⎡ ∂ ⎛ ∂v ⎞ ∂x ∂ ⎛ ∂ v ⎞ ∂ y ⎤
∂x ∂v
+ ⎜ ⎟⋅ ⎥
⋅ + x⎢ ⎜ ⎟⋅
∂q ∂x
⎣ ∂x ⎝ ∂x ⎠ ∂ q ∂ y ⎝ ∂ x ⎠ ∂ q ⎦
+
=x
⎡ ∂ ⎛ ∂v ⎞ ∂x ∂ ⎛ ∂v ⎞ ∂y ⎤
∂y ∂v
⋅ + y⎢ ⎜ ⎟⋅
+ ⎜ ⎟⋅ ⎥
∂q ∂y
⎣ ∂x ⎝ ∂y ⎠ ∂q ∂y ⎝ ∂y ⎠ ∂q ⎦
⎛ ∂2 v
⎞
⎛ ∂2 v
∂2 v
∂v
∂v
∂2 v ⎞
⋅ y⎟ + y + y ⎜
+ x⎜ 2 ⋅x+
⋅ x + 2 ⋅ y⎟
∂x ∂y ⎠
∂x
∂y
∂y
⎝ ∂x
⎝ ∂y ∂x
⎠
2
∂2 v
∂2 v
∂v
∂v
2 ∂ v
xy
y
+
+
+x +y
2
∂y ∂x
∂x
∂y
∂y 2
∂x 2
∂v ∂v ∂x ∂v ∂y
=
⋅
+
⋅
∂f ∂x ∂f ∂y ∂f
⎛ ∂v
∂v
∂v
∂v ⎞
x
= (ix ) + ( −iy ) = i ⎜ x − y ⎟
∂x
∂y
⎝ ∂x
∂y ⎠
= x2
and
x
∂2 v
∂
=
2
∂f
∂f
⎛ ∂v ⎞ ∂
⎜⎝ ix ⎟⎠ −
∂x
∂f
⎛ ∂v ⎞
⎜⎝ iy ∂y ⎟⎠
∂ ⎛ ∂v ⎞
∂x ∂v
∂ ⎛ ∂ v ⎞ ∂y ∂v
⋅ − iy
=i
⋅ + ix
⎜ ⎟−i
∂f ⎜⎝ ∂y ⎟⎠
∂f ∂x
∂ f ⎝ ∂ x ⎠ ∂f ∂y
∂v
∂y
∂v
∂x
y
x
y
q, f
q, f
Fig. 4.29
Partial Differentiation
4.59
⎡ ∂x ∂v
⎧ ∂ ⎛ ∂ v ⎞ ∂ x ∂ ⎛ ∂ v ⎞ ∂ y ⎫ ∂y ∂v
⋅
= i⎢
⋅ + x⎨ ⎜ ⎟⋅
+ ⎜ ⎟⋅ ⎬−
∂
∂
f
x
⎩ ∂ x ⎝ ∂ x ⎠ ∂ f ∂ y ⎝ ∂ x ⎠ ∂ f ⎭ ∂ f ∂y
⎣
⎧ ∂ ⎛ ∂v ⎞ ∂x ∂ ⎛ ∂v ⎞ ∂y ⎫ ⎤
+ ⎜ ⎟ ⋅ ⎬⎥
− y⎨ ⎜ ⎟⋅
⎩ ∂x ⎝ ∂y ⎠ ∂f ∂y ⎝ ∂y ⎠ ∂f ⎭⎥⎦
⎡ ∂v
⎛ ∂2 v
⎛ ∂2 v
∂2 v ⎞
∂v
∂2 v ⎞ ⎤
= i ⎢ix + x ⎜ 2 ix −
iy ⎟ + iy − y ⎜
ix − 2 iy ⎟ ⎥
∂x ∂y ⎠
∂y
∂y ⎠ ⎦
⎝ ∂x
⎝ ∂y ∂x
⎣ ∂x
2
2
2
∂v
∂v
∂v
∂v
∂v
= i 2 x + i 2 x 2 2 − 2i 2 xy
+ i2 y + i2 y2 2
∂x
∂x ∂y
∂y
∂x
∂y
= −x
∂v
∂2 v
∂2 v
∂v
∂2 v
− x 2 2 + 2 xy
− y − y2 2
∂x
∂x ∂y
∂y
∂x
∂y
⎛ ∂v
∂v
∂v
∂v
∂v ⎛ ∂v
∂v ⎞
∂v ⎞
∂v
∂v
+i
= ⎜ x + y ⎟ + i2 ⎜ x − y ⎟ = x + y − x + y
∂y
∂x
∂y
∂q
∂f ⎝ ∂ x
∂y ⎠
⎝ ∂x
∂y ⎠
∂x
= 2y
∂v
.
∂y
∂2 v ∂2 v
∂2 v
+
=
4
xy
.
∂x ∂y
∂q 2 ∂f 2
and
Example 25: If z = f (x, y), where x
` is a constant, show that
2
2
2
⎛ ∂z ⎞ ⎛ ∂z ⎞
⎛ ∂z ⎞ ⎛ ∂z ⎞
(i) ⎜ ⎟ + ⎜ ⎟ = ⎜ ⎟ + ⎜ ⎟
⎝ ∂x ⎠ ⎝ ∂y ⎠
⎝ ∂u ⎠ ⎝ ∂v ⎠
u cos`
v sin` , y = u sin` + v cos` , where
2
Solution: (i) z = f (x, y) and x = u cos a
(ii)
∂2 z ∂2 z ∂2 z ∂2 z
+
=
+
.
∂ x 2 ∂y 2 ∂u 2 ∂v 2
v sin a, y = u sin a + v cos a
z
We know that
∂z ∂z ∂x ∂z ∂y ∂z
∂z
=
⋅
+
⋅
=
cos a + sin a
∂u ∂x ∂u ∂y ∂u ∂x
∂y
x
y
∂z ∂z ∂x ∂z ∂y ∂z
∂z
=
⋅ +
⋅
= ( − sin a ) + cos a
∂v ∂x ∂v ∂y ∂v ∂x
∂y
and
u, v
Hence,
2
2
Fig. 4.30
2
2
⎛ ∂z ⎞
∂z ∂z
⎛ ∂z ⎞ ⎛ ∂z ⎞
⎛ ∂z ⎞
2
2
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ = ⎜⎝ ⎟⎠ cos a + ⎜ ⎟ sin a + 2 ⋅ cos a sin a
∂u
∂x ∂y
∂v
∂x
⎝ ∂y ⎠
2
2
⎛ ∂z ⎞
∂z ∂z
⎛ ∂z ⎞
+ ⎜ ⎟ sin 2 a + ⎜ ⎟ cos 2 a − 2 ⋅ cos a sin a
⎝ ∂x ⎠
⎝ ∂y ⎠
∂x ∂ y
2
⎛ ∂z ⎞ ⎛ ∂z ⎞
= ⎜ ⎟ +⎜ ⎟
⎝ ∂x ⎠ ⎝ ∂y ⎠
2
Engineering Mathematics
4.60
(ii)
∂z ∂z
∂z
=
cos a + sin a
∂u ∂x
∂y
Differentiating
∂z
∂x
z
w.r.t. u,
u
x
⎞
∂2 z
∂ ⎛ ∂z
⎞ ∂ ⎛ ∂z
sin a ⎟
=
⎜⎝ cos a ⎟⎠ +
2
⎜
∂u ∂x
∂u ⎝ ∂y
⎠
∂u
⎡ ∂ ⎛ ∂z ⎞ ∂x ∂ ⎛ ∂ z ⎞ ∂y ⎤
+ ⎜ ⎟ ⋅ ⎥ cosa
=⎢ ⎜ ⎟⋅
⎣ ∂x ⎝ ∂x ⎠ ∂u ∂y ⎝ ∂x ⎠ ∂u ⎦
y
u, v
⎡ ∂ ⎛ ∂z ⎞ ∂x ∂ ⎛ ∂z ⎞ ∂ y ⎤
+⎢ ⎜ ⎟⋅
+ ⎜ ⎟ ⋅ ⎥ sin a
⎣ ∂x ⎝ ∂y ⎠ ∂u ∂y ⎝ ∂ y ⎠ ∂ u ⎦
∂2 z ∂2 z
∂2 z
∂2 z 2
2
=
cos
a
+
sin
a
co
2
s
a
+
sin a
∂x ∂y
∂u 2 ∂x 2
∂y 2
z
Differentiating
w.r.t. v,
v2
⎞
∂2 z
∂ ⎡ ∂z
⎤ ∂ ⎛ ∂z
( − sin a ) ⎥ + ⎜ ⋅ cos a ⎟
=
2
⎢
∂v ⎣ ∂x
⎠
∂v
⎦ ∂v ⎝ ∂y
∂z
∂y
x
y
u, v
Fig. 4.31
⎡ ∂ ⎛ ∂ z ⎞ ∂x ∂ ⎛ ∂z ⎞ ∂y ⎤
= − sin a ⎢ ⎜ ⎟ ⋅ + ⎜ ⎟ ⋅ ⎥
⎣ ∂ x ⎝ ∂x ⎠ ∂v ∂y ⎝ ∂x ⎠ ∂v ⎦
⎡ ∂ ⎛ ∂z ⎞ ∂x ∂ ⎛ ∂z ⎞ ∂y ⎤
+ cosa ⎢ ⎜ ⎟ ⋅ + ⎜ ⎟ ⋅ ⎥
⎣ ∂x ⎝ ∂y ⎠ ∂v ∂y ⎝ ∂y ⎠ ∂v ⎦
⎡ ∂2 z
⎤
⎡ ∂2 z
⎤
∂2 z
∂2 z
= − sin a ⎢ 2 ( − sin a ) +
cos a ⎥ + cos a ⎢
( − sin a ) + 2 cos a ⎥
x
y
y
x
∂
∂
∂
∂
x
∂
∂
y
⎣
⎦
⎣
⎦
2
2
2
2
∂ z ∂ z 2
∂ z
∂ z
=
sin a − 2
cos a sin a + 2 cos 2 a
∂x ∂y
∂v 2 ∂x 2
∂y
∂2 z ∂2 z ∂2 z
∂2 z
+ 2 = 2 (sin 2 a + cos 2 a ) + 2 (sin 2 a + cos 2 a )
2
∂u
∂v
∂x
∂y
Hence,
=
∂2 z ∂2 z
.
+
∂x 2 ∂y 2
Exercise 4.2
⎛x⎞
1. If z = tan −1 ⎜ ⎟ ,
⎝ y⎠
2
y 1 t ,
x = 2t,
dz
2
.
=
dt 1 + t 2
2. If u = x3 + y3
y = b sin t
[Ans.: –3a3
x = a
du
.
dt
2
tsin t + 3b2 sin2 t
t,
t]
Partial Differentiation
3. If u = xeyz
a2
y
2
⎛ ∂z ⎞
∂z
(ii) ⎛⎜ ⎞⎟ + ⎜ ⎟
⎝ ∂x ⎠ ⎝ ∂y ⎠
x2 ,
du
.
dx
z = sin3 x
⎡
⎞⎤
x2
y ⎛
⎢ Ans. : e z ⎜1 − + 3 x cot x ⎟ ⎥
y
⎝
⎠⎦
⎣
4. If u = e
r−x
l
,
r2 = x2 + y2
l is
∂ 2 u ∂ 2 u 2 ∂u u
+
+ ⋅
= .
∂x 2 ∂y 2 l ∂x lr
5. If u =
r
r = ( x − a) 2 + ( y − b) 2 ,
∂2 u ∂2 u
+
= 0 if a, b
∂x 2 ∂y 2
6. If u2 (x2 + y2 + z2) =
∂2 u ∂2 u ∂2 u
+
+
= 0.
∂x 2 ∂y 2 ∂z 2
1⎤
⎡
2
2
2
2
⎢ Hint : Let x + y + z = r , u = r ⎥
⎣
⎦
7. If u = rm
r = x2 + y2 + z2
∂u ∂u ∂u
+
+
.
∂x 2 ∂y 2 ∂z 2
2
2
2
⎡⎣ Ans. : m(m + 1)r m − 2 ⎤⎦
8. If u = f (r
r
x=r
q, y = rsin q
∂ 2 u ∂ 2 u d 2 f 1 df
+
=
+
∂x 2 ∂y 2 dr 2 r dr
9. If z = f (u, v
v = 2xy
x
u
x2
y2 ,
∂z
∂z
⎛ ∂z ⎞
−y
= 2 u2 + v2 ⎜ ⎟ .
⎝ ∂u ⎠
∂x
∂y
10. If z = f (u, v
u = x2 + y2 ,
2
2
v=x −y ,
∂z
∂z
∂z
(i) y + x = 4 xy .
∂x
∂y
∂u
4.61
2
⎡ ⎛ ∂z ⎞ 2 ⎛ ∂z ⎞ 2 ⎤
∂z ∂z
= 4u ⎢ ⎜ ⎟ + ⎜ ⎟ ⎥ + 8v ⋅ .
⎝
⎠
⎝
⎠
∂v ⎥⎦
∂u ∂v
⎢⎣ ∂u
y
⎛
⎞
11. If w = z sin −1 ⎜ ⎟
x = 3u2 + 2v,
⎝x⎠
y = 4u − 2v 3 , z = 2u2 3v2
w
w
.
u
v
12. If w = (x2 + y 2)4 + (x y + 2)3
x = u 2v +
y = 2u + v
w
u = 0, v =
v
[Ans. : 882]
u
13. If w = x + 2y + z2, x = ,
v
y = u2 ev, z = 2u
∂w
∂w
u
+v
= 12u2 + 2vev
∂u
∂v
14. If F
x, y, z
u
∂F
∂F
∂F
+v
+w
∂u
∂v
∂w
∂F
∂F
∂F
=x
+ 2y
+ 3z
,
∂x
∂y
∂z
x = u + v + w, y = uv + vw
z uvw
u
15. If z = f (x, y), x = uv, y = ,
v
∂z
1 ∂z 1 ∂z
=
+
∂ x 2 v ∂ u 2u ∂ u
∂z v ∂z v 2 ∂z
=
−
.
∂y 2 ∂u 2u ∂v
16. If x = u + v, y = uv
F
of x, y
∂2 F
∂2 F ∂2 F
−2
+
2
∂u ∂v ∂v 2
∂u
∂2 v
∂v
= ( x 2 − 4 y) 2 − 2
∂y
∂y
wu,
Engineering Mathematics
4.62
⎡
∂ ⎞ ⎛ ∂F ∂F ⎞ ⎤
⎛ ∂
⎢ Hint : L.H.S. = ⎜⎝ ∂u − ∂v ⎟⎠ ⎜⎝ ∂u − ∂v ⎟⎠ ⎥
⎦
⎣
24. If u = f (x2 + 2yz, y2 + 2zx
∂u
∂u
+ ( x 2 − yz )
∂x
∂y
( y 2 − zx )
17. If u = f (xn – yn, yn – zn, zn – xn
1 ∂u
1 ∂u
1 ∂u
+
+
= 0.
x n −1 ∂x y n −1 ∂y z n −1 ∂z
18. If z = f (x, y
x = u – av,
y = u + av
a2
∂ ⎛ ∂v ⎞
∂ ⎛ ∂v ⎞
⎜⎝ u ⎟⎠ = ∂x ⎜⎝ u ∂y ⎟⎠ .
∂y ∂x
u = lx + my,
2
20. If x = u + av
a
b
v = x + by
2
∂ z
= 0,
∂u ∂v
2
into
of x
21. If z = f (x, y), y = ex, v = ey
∂2 z
∂2 z
= uv
.
∂x ∂y
∂u ∂v
sin u
cos u
,
22. If z = f (x, y), x =
, y=
v
v
∂z ∂z
∂z
∂z
−
= ( y − x)
− ( y + x) .
∂v ∂u
∂x
∂y
23. If u = f (2x – 3y, 3 – 4z, 4z – 2x),
1 ∂u 1 ∂u 1 ∂u
+
+
= 0.
2 ∂x 3 ∂y 4 ∂z
u = x + ay
∂2 f
∂2 f
∂2 f
+2 2 =0
−9
2
∂x ∂y
∂y
∂x
f
f
u v
y
3
⎡
⎤
⎢ Ans. : a = 2 , b = 3⎥
⎣
⎦
b
2⎤
⎡
⎢ Ans. : a = 1, b = 3⎥
⎣
⎦
v
9
∂ z
∂ z
∂ z
−5
+3 2 =0
∂x ∂y
∂x 2
∂y
2
2
a
∂v
∂v
∂v
+i
= 2y .
∂q
∂f
∂y
y = u + bv
2
x – y = 2ieq sin f,
f
27.
⎛ ∂2 z ∂2 z ⎞
= (l 2 + m 2 ) ⎜ 2 + 2 ⎟ .
∂v ⎠
⎝ ∂u
2
26. If x + y = 2eq
∂ z ∂ z
+
∂x 2 ∂y 2
2
ly mx,
∂u
= 0.
∂z
25. If u = f (ax2 + 2hxy + by2),
v = f (ax2 + 2hxy + by2)
∂2 z ∂2 z
∂2 z
− 2 = 4a2
.
2
∂u
∂v
∂x ∂y
19. If z = f (u, v
v
+ (z2 – xy)
28. If x = r
z = f (x, y
(i) ( x
y)
q, y = r
z
x
z
y
r
q
z
r
z
q
⎛ ∂2 z ∂2 z ⎞
(ii) ( x 2 − y 2 ) ⎜ 2 − 2 ⎟
∂y ⎠
⎝ ∂x
= r2
29. If x = ev
z = f (x, y
∂2 z
∂z ∂ 2 z
+r − 2 .
2
∂r ∂q
∂r
u, y = ev
u
⎛ ∂2 z
∂z ⎞
cos u ⎜
− ⎟
⎝ ∂u ∂v ∂u ⎠
⎛ ∂2 z ∂2 z ⎞
∂2 z
.
= xy ⎜ 2 + 2 ⎟ + ( x 2 + y 2 )
∂x ∂y
∂y ⎠
⎝ ∂x
Partial Differentiation
36. If u = ax + by, v
f ( x 2 y 3 , z − 3 x ) = 0,
30. If
∂z
∂z
− 2y
= 9 x.
∂x
∂y
f (y + z, x2 + y2 + z2) =
( y − z)
a2
⎛ ∂u ⎞ ⎛ ∂x ⎞
(ii) ⎜ ⎟ ⎜ ⎟ = 2
.
⎝ ∂x ⎠ y ⎝ ∂u ⎠ v a + b 2
∂z
∂z
−x
= x.
∂x
∂y
37. If u = ax + by, v
32. If f (cx − az , cy − bz ) = 0,
33. If x = a u + b v
a, b
[ Ans. :1]
2
y = a u – b v,
1 ⎛ ∂v ⎞ ⎛ ∂y ⎞
⎛ ∂u ⎞ ⎛ ∂x ⎞
⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ = = ⎜ ⎟ ⎜⎝ ⎟⎠ .
∂x y ∂u v 2 ⎝ ∂y ⎠ x ∂v u
34. If x =
cos q
sin q
, y=
,
u
u
38. If x
x =
y=
[Hint :
⎛ ∂x ⎞ ⎛ ∂u ⎞
2
⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ = cos q .
∂u q ∂x y
35. If x2 = au + bv, y 2
bx ay,
⎛ ∂u ⎞ ⎛ ∂x ⎞ ⎛ ∂y ⎞ ⎛ ∂v ⎞
⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ ⎜ ⎟ .
∂x y ∂u v ∂v x ⎝ ∂y ⎠ u
∂z
∂z
a −b
= c.
∂x
∂y
2
bx ay,
a2 + b2
⎛ ∂y ⎞ ⎛ ∂v ⎞
(i) ⎜ ⎟ ⎜ ⎟ =
⎝ ∂v ⎠ x ⎝ ∂y ⎠ u
a2
3x
31
4.63
y
2 xy 3x 2 y
dx
=2
dt
u = 2 xy − 3 x 2 y,
au bv,
1 ⎛ ∂v ⎞ ⎛ ∂y ⎞
⎛ ∂u ⎞ ⎛ ∂x ⎞
⎜⎝ ⎟⎠ ⎜⎝ ⎟⎠ = = ⎜ ⎟ ⎜⎝ ⎟⎠ .
∂x y ∂u v 2 ⎝ ∂y ⎠ x ∂v u
x = 3, y = 1]
x = 3, y = 1,
dy
du
=0
if
dt
dt
32
⎡
⎤
⎢ Ans.: − 21 cm / sec ( y if decreasing) ⎥
⎣
⎦
4.6 IMPLICIT FUNCTIONS
f (x, y) = c
of x
y
c
dy
∂f / ∂x
=−
dx
∂f / ∂y
Proof: If f (x, y) is a function of x and y, where y is a function of x, then total differential
coefficient of f w.r.t. x is given by
f (x, y) = c
If
df ∂f dx ∂f dy
=
⋅ +
⋅
dx ∂x dx ∂y dx
f (x, y) = c
df
=0
dx
4.64
Hence,
Engineering Mathematics
∂f dx ∂f dy
⋅ +
⋅
=0
∂x dx ∂y dx
dy
∂f / ∂x
=−
.
dx
∂f / ∂y
Example 1: If f (x, y) = 0, e ( x , z ) = 0, show that
Solution:
∂e ∂f dy ∂f ∂e
⋅ ⋅
=
⋅
∂x ∂y dz ∂x ∂z .
f ( x, z ) = 0
f (x, y) =
dz
∂f / ∂x
dy
∂f / ∂x
=−
and
=−
dx
∂f / ∂z
dx
∂f / ∂y
dy / dx
=
dz / dx
∂f / ∂x
∂f / ∂y
∂f / ∂x
−
∂f / ∂z
−
dy ∂f / ∂x ∂f / ∂z
=
⋅
dz ∂f / ∂y ∂f / ∂x
Hence,
∂f . ∂f . dy ∂f . ∂f
=
.
∂x ∂y dz ∂x ∂z
Example 2: If y log (cos x) = x log (sin y), find
dy
.
dx
Solution: Let f ( x, y ) = y log(cos x ) − x log(sin y )
dy
∂f / ∂x
=−
dx
∂f / ∂y
1
( − sin x ) − log(sin y )
cos
x
=−
x
log cos x −
⋅ cos y
sin y
y tan x + log sin y
=
.
lo
og cos x − x cot y
y
du
.
dx
and a2x2 + b2y2 = c2
Example 3: If u = sin (x2 + y2) and a2x2 + b2y2 = c2, find
Solution:
u = sin (x2 + y2)
∂u
= cos( x 2 + y 2 ) ⋅ 2 x
∂x
∂u
= cos( x 2 + y 2 ) ⋅ 2 y
∂y
Partial Differentiation
4.65
f ( x, y ) = a 2 x 2 + b 2 y 2 − c 2
Let
dy
∂f / ∂x
2a 2 x
a2 x
=−
=− 2 =− 2
dx
∂f / ∂y
2b y
b y
We know that
du ∂u dx ∂u dy
=
⋅
+
⋅
dx ∂x dx ∂y dx
⎛ a2 x ⎞
= 2 x cos( x 2 + y 2 ) + 2 y cos( x 2 + y 2 ) ⎜ − 2 ⎟
⎝ b y⎠
⎛ a2 ⎞
= 2 x cos( x 2 + y 2 ) ⋅ ⎜1 − 2 ⎟
⎝ b ⎠
dy
p
=−
dx
q
Example 4: If f (x, y) = constant is an implicit function, show that
and
s=
d2 y
1
= − 3 ( q 2 r − 2 pqs + p 2 t ), if q
dx 2
q
2
0 where p = ∂f , q = ∂f , r = ∂ f ,
∂y
∂x
∂x 2
∂2 f
∂2 f
, t= 2.
∂x∂y
∂y
Solution:
f (x, y) = constant.
∂f
=0
∂x
f
We know that
∂f ∂f dx ∂f dy
=
+
=0
∂x ∂x dx ∂y dx
x
dy
∂f / ∂x
=−
dx
∂f / ∂y
dy
p
=−
dx
q
dy
w.r.t. x,
dx
dp
dq
q⋅ − p
d2 y
= − dx 2 dx
dx 2
q
x
q
p
x
y
y x
y
Differentiating
⎡ ⎧ ⎛ ∂ p dx ∂ p d y ⎞
⎛ ∂q
q dx ∂q dy ⎞ ⎫ ⎤
− p ⎜ ⋅ + ⋅ ⎟⎬ ⎥
⎢ ⎨q ⎜ ⋅ +
⎟
⎝ ∂x dx ∂y dx ⎠ ⎭ ⎥
d y
⎩ ⎝ ∂ x dx ∂ y d x ⎠
=−⎢
2
2
⎢
⎥⎦
dx
q
⎣
2
x
x
Fig. 4.32
Engineering Mathematics
4.66
⎡ ⎧ ∂ ⎛ ∂f ⎞ ∂ ⎛ ∂f ⎞ ⎛ p ⎞ ⎫
⎧ ∂ ⎛ ∂f ⎞ ∂ ⎛ ∂f ⎞ ⎛ p ⎞ ⎫ ⎤
⎢ q ⎨ ⎜ ⎟ + ⎜ ⎟ ⎜ − ⎟⎬ − p ⎨ ⎜ ⎟ + ⎜ ⎟ ⎜ − ⎟⎬ ⎥
⎩ ∂x ⎝ ∂x ⎠ ∂y ⎝ ∂x ⎠ ⎝ q ⎠ ⎭
⎩ ∂x ⎝ ∂y ⎠ ∂y ⎝ ∂y ⎠ ⎝ q ⎠ ⎭ ⎥
= −⎢
2
⎢⎣
⎥⎦
q
⎡ ⎧ ∂2 f p ∂2 f ⎫
⎧ ∂2 f
p ∂2 f ⎫ ⎤
−
⎬ − p⎨
⎢q⎨ 2 −
2 ⎬⎥
q ∂x ∂y⎭
⎩ ∂x
⎩ ∂y ∂x q ∂y ⎭ ⎥
= −⎢
⎥⎦
⎢⎣
q2
1
⎡ q( qr − ps) − p( qs − pt ) ⎤
= − 3 ( q 2 r − 2 pqs + p 2 t ).
= −⎢
3
⎥
q
q
⎣
⎦
Example 5: If x 4 + y 4 + 4a 2 xy = 0, show that (y3 + a2 x)3
d2 y
= 2a2 xy(x2y2 + 3a4).
dx 2
4
4
2
Solution: Let f ( x, y ) = x + y + 4 a xy
4 x 3 + 4a 2 y
∂f / ∂x
dy
=−
=− 3
∂f / ∂y
dx
4 y + 4a 2 x
dy
= −( 4 x 3 + 4 a 2 y )
dx
Differentiating above equation w.r.t. x,
d2 y ⎛
dy
dy ⎞
⎞ dy
⎛
( 4 y 3 + 4 a 2 x ) 2 + ⎜12 y 2
+ 4a2 ⎟
= − ⎜12 x 2 + 4 a 2 ⎟
⎝
⎠ dx
⎝
dx
dx ⎠
dx
( 4 y 3 + 4a 2 x )
2
( 4 y 3 + 4a 2 x)
( 4 y 3 + 4a 2 x)
3
2
3
2
d2 y
2 ⎛ 4 x + 4a y ⎞
2 ⎛ 4 x + 4a y ⎞
2
−
+
12
y
8
a
⎜⎝ 4 y 3 + 4 a 2 x ⎟⎠
⎜⎝ 4 y 3 + 4 a 2 x ⎟⎠ + 12 x = 0
dx 2
d2 y
dx 2
=
−12 y 2 ( x 6 + a 4 y 2 + 2 x 3 a 2 y ) + 8a 2 ( x 3 + a 2 y ) ⋅ ( y 3 + a 2 x ) − 12 x 2 ( y 6 + a 4 x 2 + 2 xy 3 a 2 )
( y 3 + a2 x)2
=
− 12 y 2 x 6 − 12 y 4 a 4 − 24 y 3 x 3 a 2 + 8a 2 ( x 3 + a 2 y )( y 3 + a 2 x ) − 12 x 2 y 6 − 12a 4 x 4 − 24 x 3 y 3 a 2
( y 3 + a2 x)2
( y 3 + a 2 x )3
d2 y
= −3 y 2 x 6 − 3 y 4 a 4 − 6 y 3 a 2 x 3 + 2a 2 x 3 y 3 + 2a 4 x 4 + 2a 4 y 4
dx 2
+ 2a6 xy − 3 x 2 y 6 − 3a 4 x 4 − 6 x 3 y 3 a 2
= −3 x 2 y 2 ( x 4 + y 4 ) − a 4 ( x 4 + y 4 ) − 10 x 3 y 3 a 2 + 2a6 xy
= −3x 2 y 2 ( −4a 2 xy ) − a 4 ( −4a 2 xy ) − 10 x 3 y 3 a 2 + 2a6 xy
= 12a 2 x 3 y 3 + 4 a6 xy − 10 x 3 y 3 a 2 + 2a6 xy
= 2a 2 x 3 y 3 + 6 a6 xy = 2a 2 xy( x 2 y 2 + 3a 4 )
Note: It can also be proved by putting values of p, q, r, s, t in the result of Ex. 4.
Partial Differentiation
4.67
Exercise 4 .3
1. If x 3
y 3 3axy
0,
dy
.
dx
dy
⎡
⎤
⎢ Hint : Find dx at (1, 2) ⎥
⎣
⎦
2⎤
⎡
⎢ Ans. : − 11⎥
⎣
⎦
⎡
ay − x 2 ⎤
Ans.
:
⎢
⎥
y 2 − ax⎦
⎣
dy
.
2. If x3 + 3x2 + 6xy2 + y3 =
dx
⎡
( x 2 + 2 x + 2 y 2 )⎤
⎢ Ans. : −
⎥
( 4 xy + y 2 ) ⎦
⎣
3. If xy = yx
dy y( y − x log y )
=
.
dx x( x − y log x )
4. If f (x, y) = x sin(x – y) – (x + y) = 0,
dy
.
dx
⎡
[sin( x − y )](1 + x ) − 1⎤
⎢ Ans. :
x cos( x − y ) + 1 ⎥⎦
⎣
y
dy
.
5. If y x = sin x,
dx
⎡ Hint : f = x y log y − log sin x, ⎤
⎢
⎥
let x y = z ,
⎢
⎥
⎢
⎥
∂z ∂z
⎢log z = y log x find , and ⎥
∂x ∂y
⎢
⎥
⎢
⎥
∂f
∂f
hen and
⎢ th
⎥
∂x
∂y
⎢⎣
⎥⎦
⎡
−( yx y −1 log y − cot x ) ⎤
Ans.
:
⎢
⎥
x y log x log y + x y y −1 ⎦
⎣
d2 y
.
dx 2
⎡
6 a3 x 2 ( a3 + x 3 )⎤
Ans.
:
⎢
⎥
y9
⎣
⎦
6. If x5 + y5 = 5a3x2
7. If xy 3 − yx 3 = 6
8. Find
d2 y
, if x 1 − y 2 + y 1 − x 2 = a.
dx 2
⎡
⎤
a ⎥
⎢ Ans. :
3⎥
⎢
(1 − x 2 ) 2 ⎥⎦
⎢⎣
9. If u = x
xy
du
.
dx
x3 + y3 + 3xy – 1 = 0,
du ∂ u dx ∂ u dy ⎤
⎡
⎢ Hint : dx = ∂x ⋅ dx + ∂y ⋅ dx ⎥
⎢
⎥
∂f
⎢
⎥
⎢
⎥
dy
∂
x
=−
⎢
⎥
∂f
dx
⎢
⎥
∂y
⎢⎣
⎥⎦
2
⎡
x ⎛ x + ay ⎞ ⎤
⎢ Ans. : 1 + log xy − ⎜ 2
⎥
y ⎝ y + ax ⎟⎠ ⎦
⎣
10. If xm + ym = bm
xm 2
d2 y
m
=
–
(m
–
1)b
.
y 2m 1
dx 2
11. If u = x2y
du
.
dx
x2 + xy + y2 =
12. If x3 + y3 = 3ax2
d2 y
.
dx 2
⎡
2a 2 x 2 ⎤
Ans.
:
−
⎢
⎥
y5 ⎦
⎣
Engineering Mathematics
4.68
4.7 HOMOGENEOUS FUNCTIONS AND
EULER’S THEOREM
f (x, y, z
n
t
f (xt, yt, zt) = t nf (x, y, z)
where, n is a real number.
4.7.1 Euler’s Theorem for Function of Two Variables
Statement: If u is a homogeneous function of two variables x and y of degree n, then
x
∂u
∂u
+y
= nu.
∂x
∂y
Proof: Let u = f (x, y
n
u = f ( X , Y ) = t f ( x, y )
n
X = xt
Y = yt
u = f (X, Y
t
∂u ∂u ∂X ∂u ∂Y
∂u
∂u
=
⋅
+
⋅
=x
+y
∂t ∂X ∂t ∂Y ∂t
∂X
∂Y
At t = 1, X = x
Y=y
∂u
∂u
∂u
=x
+y
∂t
∂x
∂y
u = t n f (x, y
... (1)
u
t,
∂u
= nt n −1 f ( x, y )
∂t
X
Y
At t = 1,
∂u
= nf ( x, y ) = nu
∂t
... (2)
t
Fig. 4.33
From Eqs (1) and (2),
x
∂u
∂u
= nu.
+y
∂y
∂x
4.7.2 Euler’s Theorem for Function of
Three Variables
Statement: If u is a homogeneous function of three variables x, y,
z of degree n, then
∂u
∂u
∂u
x
+y
+z
= nu.
∂x
∂y
∂z
Fig. 4.34
Partial Differentiation
4.69
Proof: Let u = f (x, y, z) is a homogeneous function of degree n.
u = f ( X , Y , Z ) = t n f ( x, y, z )
where, X = xt, Y = yt, Z = zt.
Differentiating u = f (X, Y, Z) w.r.t t using composite function,
∂ u ∂ u ∂ X ∂u ∂Y ∂u ∂Z
∂u
∂u
∂u
=
⋅
+
⋅
+
⋅
=x
+y
+z
∂t ∂X ∂t ∂Y ∂t ∂Z ∂t
∂X
∂Y
∂Z
At t = 1, X = x, Y = y and Z = z
∂u
∂u
∂u
∂u
=x
+y +z
... (1)
∂t
∂x
∂y
∂z
Differentiating u = t n f (x, y, z) w.r.t. t,
∂u
= nt n −1 f ( x, y, z )
∂t
At t = 1,
∂u
= nf ( x, y, z ) = nu
∂t
From Eqs (1) and (2),
∂u
∂u
∂u
x +y +z
= nu.
∂x
∂y
∂z
... (2)
4.7.3 Deductions from Euler’s Theorem
Corollary 1: If u is a homogeneous function of two variables x, y of degree n, then
∂2 u
∂2 u
∂2 u
x 2 2 + 2 xy
+ y 2 2 = n( n − 1)u.
∂x ∂y
∂x
∂y
Proof: Let u is a homogeneous function of two variables x and y of degree n.
By Euler’s theorem,
∂u
∂u
x
+y
= nu
∂x
∂y
Differentiating Eq. (1) partially w.r.t. x,
∂ 2 u ∂u
∂2 u
∂u
+
+
y
=n
∂x ∂y
∂x
∂x 2 ∂x
∂2 u
∂2 u
∂u
x 2+y
= ( n − 1)
∂x ∂y
∂x
∂x
Differentiating (1) partially w.r.t. y,
... (1)
x
x
∂2 u
∂ 2 u ∂u
∂u
+y 2+
=n
∂x ∂y
∂y
∂y
∂y
2
2
∂u
∂u
∂u
x
+ y 2 = ( n − 1)
∂x ∂y
∂y
∂y
... (2)
... (3)
Multiplying Eq. (2) by x and Eq. (3) by y and adding,
x2
⎛ ∂u
∂2 u
∂2 u
∂2 u
∂u ⎞
+ 2 xy
+ y 2 2 = ( n − 1) ⎜ x
+ y ⎟ = ( n − 1)nnu
2
∂x ∂y
⎝ ∂x
∂y ⎠
∂x
∂y
[Using Eq. (1)]
Engineering Mathematics
4.70
x2
2
∂2 u
∂2 u
2 ∂ u
+
2
xy
+
y
= n( n − 1)u.
∂x ∂y
∂x 2
∂y 2
Example 1: Verify Euler’s theorem for
(i)
(iii)
u = x 2 yz - 4 y 2 z 2 + 2 xz 3
u=
x2 + y2
x+ y
y
x
x+ y+z
(ii)
u = x 4 y 2 sin -1
(iv)
u=
x+
.
y+ z
u = x 2 yz − 4 y 2 z 2 + 2 xz 3
Solution: (i)
Replacing x by xt, y by yt and z by zt,
u = t 3 ( x 2 yz − 4 y 2 z 2 + 2 xz 3 )
Hence, u is a homogeneous function of degree 3.
By Euler’s theorem,
∂u
∂u
∂u
… (1)
x
+y
+z
= nu = 3
∂x
∂y
∂z
Differentiating u partially w.r.t. x, y and z,
∂u
∂u
∂u
= 2 xyz + 2 z 3 ,
= x 2 z − 8 yz 2 ,
= x 2 y − 8 y 2 z + 6 xz 2
∂x
∂y
∂z
∂u
∂u
∂u
x
+y
+z
= 2 x 2 yz + 2 xz 3 + x 2 yz − 8 y 2 z 2 + x 2 yz − 8 y 2 z 2 + 6 xz 3
∂x
∂y
∂z
= 4 x 2 yz − 16 y 2 z 2 + 8 xz 3
= 4( x 2 yz − 4 y 2 z 2 + 2 xz 3 )
= 4u
Hence, from Eqs (1) and (2), theorem is verified.
y
(ii)
u = x 4 y 2 sin −1
x
Replacing x by xt and y by yt,
y⎞
⎛
u = t 6 ⎜ x 4 y 2 sin −1 ⎟
⎝
x⎠
Hence, u is a homogeneous function of degree 6.
By Euler’s theorem,
∂u
∂u
x
+y
= nu = 6u
∂x
∂y
Differentiating u partially w.r.t. x,
⎡
∂u
y
x4 ⎛ y ⎞⎤
= y 2 ⎢ 4 x 3 sin −1 +
⎜ − 2 ⎟⎥
∂x
x
y2 ⎝ x ⎠⎥
⎢
1− 2
⎢⎣
⎥⎦
x
⎛
y
yx 3 ⎞
= y 2 ⎜ 4 x 3 sin −1 −
⎟
x
x2 − y2 ⎠
⎝
… (2)
… (1)
Partial Differentiation
4.71
Differentiating u partially w.r.t. y,
⎡
⎤
⎢
⎥
∂u
y
1
1
= x 4 ⎢ 2 y sin −1 + y 2
⋅ ⎥
⎢
∂y
x
y2 x ⎥
1− 2
⎢
⎥
x
⎣
⎦
4 2
y
x
y
= 2 x 4 y sin −1 +
x
x2 − y2
x
∂u
∂u
y
x 4 y3
y
x 4 y3
+y
= 4 x 4 y 2 sin −1 −
+ 2 x 4 y 2 sin −1 +
∂x
∂y
x
x
x2 − y2
x2 − y2
y
= 6u
x
Hence, from Eqs (1) and (2), theorem is verified.
= 6 x 4 y 2 sin −1
u=
(iii)
… (2)
x2 + y2
x+ y
Replacing x by xt and y by yt,
⎛ x2 + y2 ⎞
u=t⎜
⎝ x + y ⎟⎠
Hence, u is a homogeneous function of degree 1.
By Euler’s theorem,
∂u
∂u
x
+y
= nu = u
∂x
∂y
Differentiating u partially w.r.t. x,
∂u
2x
( x 2 + y 2 ) 2 x 2 + 2 xy − x 2 − y 2 x 2 − y 2 + 2 xy
=
−
=
=
∂x x + y ( x + y ) 2
( x + y)2
( x + y)2
… (1)
Differentiating u partially w.r.t. y,
∂u
2y
x2 + y2
y 2 − x 2 + 2 xy
=
−
=
∂y x + y ( x + y ) 2
( x + y)2
x
x
∂u
∂u x 3 − xy 2 + 2 x 2 y + y 3 − x 2 y + 2 xy 2
+y
=
∂x
∂y
( x + y)2
=
x 3 + y 3 + xy 2 + x 2 y
( x + y)2
=
( x + y )( x 2 − xy + y 2 ) + xy( y + x )
( x + y)2
=
( x + y )( x 2 − xy + y 2 + xy )
( x + y)2
∂u
∂u x 2 + y 2
+y
=
=u
∂x
∂y
x+ y
… (2)
Engineering Mathematics
4.72
Hence, from Eqs (1) and (2), theorem is verified.
x+ y+z
(iv)
u=
.
x+ y+ z
Replacing x by xt and y by yt,
1 ⎛
x+ y+z ⎞
u = t2 ⎜
⎟
⎝ x+ y+ z⎠
Hence, u is a homogeneous function of degree
1
.
2
By Euler’s theorem,
∂u
∂u
∂u
1
x
+y
+z
= nu = u
∂x
∂y
∂z
2
Differentiating u partially w.r.t. x,
∂u
=
∂x
∂u
=
∂y
Similarly,
x
x+ y+ z
x+ y+ z
1
x+ y+ z
∂u
∂u
∂u
+y
+z
=
∂x
∂y
∂z
=
1
1
∂u
=
∂z
and
… (1)
x+ y+z
x+ y+ z
−
−
−
x+ y+z
1
2
x+ y+ z
−
x+ y+z
(
x+ y+ z
x+ y+z
(
(
x+ y+ z
x+ y+z
x+ y+ z
( x + y + z)
2
=
(
(
)
)
2
⋅
2
⋅
)
2
⋅
1
2 x
1
2 y
1
2 z
x+ y+ z
x+ y+ z
)
)
2
1
u
2
… (2)
Hence, from Eqs (1) and (2), theorem is verified.
x
⎛ x⎞
y
⎛ x⎞
Example 2: If u = e y sin ⎜ ⎟ + e x cos ⎜ ⎟ , prove that x
⎝ y⎠
⎝ y⎠
x
Solution:
y
⎛x⎞
⎛x⎞
u = e y sin ⎜ ⎟ + e x cos ⎜ ⎟ ,
⎝ y⎠
⎝ y⎠
Replacing x by xt and y by yt,
y
⎡ x
⎛x⎞
⎛ x ⎞⎤
u = t 0 ⎢e y sin ⎜ ⎟ + e x cos ⎜ ⎟ ⎥
⎝ y⎠
⎝ y ⎠ ⎦⎥
⎢⎣
Hence, u is a homogeneous function of degree 0.
By Euler’s theorem,
∂u
∂u
x
+y
= 0.
∂x
∂y
∂u
∂u
+y
= 0.
∂x
∂y
Partial Differentiation
Example 3: Find x
∂u
∂u
+y
where u = (8x2 + y2) (log x – log y).
∂x
∂y
Solution:
u = (8x2 + y2) (log x log y)
⎛x⎞
= (8 x 2 + y 2 ) log ⎜ ⎟
⎝ y⎠
Replacing x by xt and y by yt,
⎛x⎞
u = t 2 (8 x 2 + y 2 ) log ⎜ ⎟
⎝ y⎠
Hence, u is a homogeneous function of degree 2.
By Euler’s theorem,
∂u
∂u
x
+y
= 2u = 2(8 x 2 + y 2 ) (log x − log y ).
∂x
∂y
Example 4: If u =
x2 ⎛ y ⎞ y2 ⎛ x ⎞
f ⎜ ⎟+
g
, prove that
y ⎝ x ⎠ x ⎜⎝ y ⎟⎠
⎡ ∂u
⎛ x ⎞⎤
⎡ ∂u
⎛ y ⎞⎤
− yg ⎜ ⎟ ⎥ = 0.
x2 ⎢ y
− xf ⎜ ⎟ ⎥ + y 2 ⎢ x
⎝ x ⎠⎦
⎝ y ⎠⎦
⎣ ∂x
⎣ ∂y
u=
Solution:
x2 ⎛ y ⎞ y2 ⎛ x ⎞
f ⎜ ⎟+
g
y ⎝ x ⎠ x ⎜⎝ y ⎟⎠
Replacing x by xt and y by yt,
⎡ x2 ⎛ y ⎞ y2 ⎛ x ⎞⎤
u =t⎢ f ⎜ ⎟+
g ⎜ ⎟⎥
⎣ y ⎝ x ⎠ x ⎝ y ⎠⎦
Hence, u is a homogeneous function of degree 1.
By Euler’s theorem,
∂u
∂u
x2 ⎛ y ⎞ y2 ⎛ x ⎞
x +y
= 1⋅ u =
f ⎜ ⎟+
g
∂x
∂y
y ⎝ x ⎠ x ⎜⎝ y ⎟⎠
x2 y
⎛x⎞
∂u
∂u
⎛ y⎞
+ xy 2
= x3 f ⎜ ⎟ + y3 g ⎜ ⎟
⎝x⎠
∂x
∂y
⎝ y⎠
⎡ ∂u
⎛ x ⎞⎤
⎡ ∂u
⎛ y ⎞⎤
x 2 ⎢ y − xf ⎜ ⎟ ⎥ + y 2 ⎢ x − yg ⎜ ⎟ ⎥ = 0.
⎝
⎠
∂
⎝ y ⎠⎦
∂
x
x
y
⎦
⎣
⎣
1
1 log x − log y
Example 5: If u( x , y ) = 2 +
+
, prove that
xy
x2
x
∂u
∂u
+ 2u(x, y) = 0.
x
+y
∂x
∂y
1
1 log x − log y
u( x , y ) = 2 + +
Solution:
xy
x
x2
=
⎛x⎞
1
1
1
+ + 2 log ⎜ ⎟
2
xy x
⎝ y⎠
x
4.73
Engineering Mathematics
4.74
Replacing x by xt and y by yt,
⎡1
⎛ x ⎞⎤
1
1
u( x, y ) = t −2 ⎢ 2 + + 2 log ⎜ ⎟ ⎥
xy x
⎝ y ⎠⎦
⎣x
Hence, u is a homogeneous function of degree 2.
By Euler’s theorem,
x
Hence,
x
∂u
∂u
+y
= −2u( x, y )
∂x
∂y
∂u
∂u
+y
+ 2u( x, y ) = 0.
∂x
∂y
Example 6: If z = log( x 2 + y 2 ) +
x
z
∂z
+y .
x
∂y
x2 + y2
− 2 log( x + y ), find the value of
x+ y
x2 + y2
− 2 log( x + y )
x+ y
x2 + y2
= log( x 2 + y 2 ) +
− log( x + y ) 2
x+ y
z = log( x 2 + y 2 ) +
Solution:
= log
( x2 + y2 ) x2 + y2
+
x+ y
( x + y)2
= u+v
x +y
x2 + y2
where, u = log
, v=
2
x+ y
( x + y)
2
2
Replacing x by xt and y by yt in u and v,
u = t 0 log
⎛ x2 + y2 ⎞
x2 + y2
, v=t⎜
2
( x + y)
⎝ x + y ⎟⎠
Hence, u is a homogeneous function of degree 0 and v is homogeneous function of
degree 1.
By Euler’s theorem,
∂u
∂u
x
+y
= 0⋅u = 0
... (1)
∂x
∂y
∂v
∂v
... (2)
x +y
= 1⋅ v
and
∂x
∂y
Adding Eqs (1) and (2),
⎛ ∂u ∂v ⎞
⎛ ∂u ∂v ⎞
x⎜ + ⎟+ y⎜ + ⎟ = 0+v
⎝ ∂x ∂x ⎠
⎝ ∂y ∂y ⎠
x
∂z
∂z x 2 + y 2
.
+y
=
∂x
∂y
x+ y
Partial Differentiation
∂u
∂u
⎛ y⎞
Example 7: If u = f ⎜ ⎟ + x 2 + y 2 , prove that x
+y
=
⎝ x⎠
∂x
∂y
4.75
x2 + y2 .
Solution: Let u = v + w
⎛ y⎞
v = f ⎜ ⎟ , w = x2 + y2
⎝x⎠
where,
Replacing x by xt and y by yt,
⎛ y⎞
v = t 0 f ⎜ ⎟ and w = t x 2 + y 2
⎝x⎠
Hence, v is a homogeneous function of degree 0 and w is homogeneous function of
degree 1.
By Euler’s theorem,
∂v
∂v
… (1)
x +y
= 0⋅v = 0
∂x
∂y
∂w
∂w
and
x
+y
= 1⋅ w = w
… (2)
∂x
∂y
Adding Eqs (1) and (2),
⎛ ∂v ∂w ⎞
⎛ ∂v ∂w ⎞
x⎜ +
+ y⎜ +
=w
⎝ ∂x ∂x ⎟⎠
⎝ ∂y ∂y ⎟⎠
Hence,
Example 8: If u =
x
x
∂u
∂u
+y
= x2 + y2 .
∂x
∂y
⎛ xy + yz + xz ⎞
x3 y3 z3
+ cos ⎜ 2
, then show that
2
2
2
⎝ x + y 2 + z 2 ⎟⎠
x + y +z
∂u
∂u
∂u
7 x3 y3 z3
+y
+z
= 2
.
∂x
∂y
∂z x + y 2 + z 2
Solution: Let u = v + w
⎛ xy + yz + xz ⎞
x3 y3 z3
, w = cos ⎜ 2
2
2
2
⎝ x + y 2 + z 2 ⎟⎠
x +y +z
Replacing x by xt, y by yt and z by zt,
⎛ x3 y3 z3 ⎞
⎛ xy + yz + xz ⎞
v = t7 ⎜ 2
, w = t 0 cos ⎜ 2
2
2 ⎟
⎝ x + y 2 + z 2 ⎟⎠
⎝x + y +z ⎠
where,
v=
Hence, v is a homogeneous function of degree 7 and w is homogeneous function of
degree 0.
By Euler’s theorem,
∂v
∂v
∂v
x +y +z
= 7v
... (1)
∂x
∂y
∂z
∂w
∂w
∂w
and
x
+y
+z
= 0⋅w = 0
... (2)
∂x
∂y
∂z
Engineering Mathematics
4.76
Adding Eqs (1) and (2),
⎛ ∂v ∂w ⎞
⎛ ∂v ∂w ⎞
⎛ ∂v ∂w ⎞
x⎜ +
+ y⎜ +
+z⎜ +
= 7v
⎝ ∂x ∂x ⎟⎠
⎝ ∂z ∂z ⎟⎠
⎝ ∂y ∂y ⎟⎠
Hence,
Example 9: If v =
x
x
∂u
∂u
∂u
7 x3 y3 z3
+y
+z
= 2
.
∂x
∂y
∂z x + y 2 + z 2
1
f (p ) where x = r cos p, y = r sin p , show that
r
∂v
∂v
+y
+ v = 0.
∂x
∂y
Solution: x = r cos q, y = r sin q
−1 ⎛ y ⎞
r = x 2 + y 2 and q = tan ⎜⎝ ⎟⎠
x
⎡ −1 ⎛ y ⎞ ⎤
1
1
v= f( )=
f ⎢ tan ⎜ ⎟ ⎥
⎝ x ⎠⎦
r
x2 + y2 ⎣
Replacing x by xt and y by yt,
t −1
⎡
⎛ y ⎞⎤
f ⎢ tan −1 ⎜ ⎟ ⎥
⎝ x ⎠⎦
⎣
x +y
Hence, v is a homogeneous function of degree 1.
By Euler’s theorem
∂v
∂y
x +y
= −1 ⋅ v
∂x
∂y
∂v
∂v
x + y + v = 0.
∂x
∂y
v=
2
2
Example 10: If x = eu tan v, y = eu sec v and z = e-2u f (v),
∂z
∂z
prove that x
+y
+ 2z = 0.
∂x
∂y
Solution:
x = eu tan v, y = eu sec v
y 2 − x 2 = e 2 u (sec 2 v − tan 2 v ) = e 2 u
1
e −2 u = 2
y − x2
x tan v
=
= sin v
y sec v
⎛x⎞
v = sin −1 ⎜ ⎟
⎝ y⎠
z = e −2 u f ( v ) =
1
x⎞
⎛
f ⎜ sin −1 ⎟
2
⎝
y⎠
y −x
2
Partial Differentiation
4.77
Replacing x by xt and y by yt,
z=
⎛
⎛
1
1
x⎞
x⎞
f ⎜ sin −1 ⎟ = t −2 2
f ⎜ sin −1 ⎟
2
2
y⎠
y⎠
t (y − x ) ⎝
(y − x ) ⎝
2
2
Hence, z is a homogeneous function of degree 2.
By Euler’s theorem
∂z
∂z
x +y
= −2 z
∂x
∂y
∂z
∂z
x + y + 2 z = 0.
∂y
∂x
Example 11: If u = f (v) where v is a homogeneous function of x, y of degree n,
∂u
∂u
∂u
∂u
prove that x
+y
= nvf ′( v ). Hence, deduce that if u = log v, x
+y
= n.
∂x
∂y
∂x
∂y
Solution:
u = f (v)
∂u
∂v ∂u
∂v
= f ′( v ) ,
= f ′( v )
∂x
∂x ∂y
∂y
⎛ ∂v
∂u
∂u
∂v
∂v
∂v ⎞
x +y
= xf ′( v ) + yf ′( v ) = f ′( v ) ⎜ x + y ⎟
∂x
∂y
∂x
∂y
⎝ ∂x
∂y ⎠
= f ′( v ) ⋅ nv
... (1)
⎡
⎤
∂v
∂v
⎢∵ v is a homogeneous function of degree n, By Euler’s theorem x ∂x + y ∂y = nv ⎥
⎣
⎦
1
If u = log v, f (v) = log v, f ′( v ) =
v
Substituting in Eq. (1),
∂u
∂u 1
x +y
= ⋅ nv = n.
∂x
∂y v
y
⎛ x⎞x
∂2 u
∂2 u
∂2 u
Example 12: If u = ⎜ ⎟ , prove that x 2 2 + 2 xy
+ y 2 2 = 0.
⎝ y⎠
∂x ∂y
∂x
∂y
y
Solution:
⎛ x ⎞x
u=⎜ ⎟
⎝ y⎠
Replacing x by xt and y by yt,
y
⎛ x ⎞x
u=t ⎜ ⎟
⎝ y⎠
Hence, u is a homogeneous function of degree 0.
By Cor. 1
0
x2
∂2 u
∂2 u
∂2 u
+ 2 xy
+ y 2 2 = 0(0 − 1)u = 0.
2
∂x ∂y
∂x
∂y
Engineering Mathematics
4.78
2
2
⎛ x2 + y2 ⎞
∂2 u
2 ∂ u
2 ∂ u
+
xy
+
y
.
Example 13: If u = log ⎜
2
⎟ , find the value of x
∂x ∂y
∂x 2
∂y 2
⎝ x+ y ⎠
⎛ x2 + y2 ⎞
u = log ⎜
⎟
⎝ x+ y ⎠
Solution:
Replacing x by xt and y by yt in u,
⎛ x2 + y2 ⎞
u = t 0 log ⎜
⎟
⎝ x+ y ⎠
Hence, u is a homogeneous function of degree 0.
By Cor. 1
∂2 u
∂2 u
∂2 u
+ y 2 2 = 0(0 − 1)u
x 2 2 + 2 xy
∂x ∂y
∂x
∂y
= 0.
⎛ y⎞
⎛ y⎞
Example 14: If u = xf ⎜ ⎟ + g ⎜ ⎟ , then show that
⎝ x⎠
⎝ x⎠
2
2
2
∂u
∂u
∂u
+ y 2 2 = 0.
x 2 2 + 2 xy
∂x ∂y
∂x
∂y
Solution: Let u = v + w,
⎛ y⎞
⎛ y⎞
v = xf ⎜ ⎟ and w = g ⎜ ⎟
⎝x⎠
⎝x⎠
Replacing x by xt and y by yt,
where,
⎛ y⎞
0 ⎛ y⎞
v = tx f ⎜ ⎟ and w = t g ⎜ ⎟
⎝x⎠
⎝x⎠
Hence, v is a homogeneous function of degree 1 and w is a homogeneous function of
degree 0.
∂2 v
∂2 v
∂2 v
By Cor. 1
... (1)
x 2 2 + 2 xy
+ y 2 2 = 1(1 − 1)v = 0
∂x ∂y
∂x
∂y
x2
∂2 w
∂2 w
∂2 w
+ 2 xy
+ y 2 2 = 0(0 − 1)w = 0
2
∂x ∂y
∂x
∂y
Adding Eqs (1) and (2),
2
⎛ ∂2 v
⎛ ∂2 v ∂2 w ⎞
∂2 w ⎞
∂2 w ⎞
2 ⎛∂ v
+
+
+
x 2 ⎜ 2 + 2 ⎟ + 2 xy ⎜
y
⎜⎝ ∂y 2 ∂y 2 ⎟⎠ = 0
∂x ⎠
⎝ ∂x ∂y ∂x ∂y ⎟⎠
⎝ ∂x
Hence,
Example 15: If z =
x2
2
∂2 u
∂2 u
2 ∂ u
2
+
xy
+
y
= 0.
∂x ∂y
∂x 2
∂y 2
⎛ x⎞
( x 2 + y 2 )n
⎛ y⎞
+ xf ⎜ ⎟ + e ⎜ ⎟ , show that
⎝ x⎠
⎝ y⎠
2n( 2n − 1)
x2
2
∂2 z
∂2 z
2 ∂ z
+
2
xy
+
y
= ( x 2 + y 2 )n .
∂x ∂y
∂x 2
∂y 2
... (2)
Partial Differentiation
4.79
Solution: Let z = u + v + w
⎛x⎞
( x 2 + y 2 )n
⎛ y⎞
u=
, v = xf ⎜ ⎟ , w = f ⎜ ⎟
where,
⎝
⎠
2n( 2n − 1)
x
⎝ y⎠
Replacing x by xt and y by yt in u, v and w
⎛x⎞
t 2n ( x 2 + y 2 )n
, v = t x f ⎛⎜ y ⎞⎟ , w = t 0f ⎜ ⎟
⎝x⎠
2n( 2n − 1)
⎝ y⎠
Hence, u, v and w are homogeneous function of degree 2n, 1 and 0 respectively.
By Cor. 1,
∂2 u
∂2 u
∂2 u
... (1)
x 2 2 + 2 xy
+ y 2 2 = 2n( 2n − 1) u
∂x ∂y
∂y
∂x
u=
2
∂2 v
∂2 v
2 ∂ v
xy
+
2
+
y
= 1(1 − 1) v = 0
∂x ∂y
∂x 2
∂y 2
∂2 w
∂2 w
∂2 w
x 2 2 + 2 xy
+ y 2 2 = 0(0 − 1)w = 0
∂x ∂y
∂x
∂y
x2
and
... (2)
... (3)
Adding Eqs (1), (2) and (3),
⎛ ∂2 u ∂2 v ∂2 w ⎞
⎛ ∂2 u
∂2 v
∂2 w ⎞
+
+
x 2 ⎜ 2 + 2 + 2 ⎟ + 2 xy ⎜
∂x
∂x ⎠
⎝ ∂x
⎝ ∂x ∂y ∂x ∂y ∂x ∂y ⎟⎠
⎛ ∂2 u ∂2 v ∂2 w ⎞
+ y 2 ⎜ 2 + 2 + 2 ⎟ = 2n( 2n − 1)u
∂y
∂y ⎠
⎝ ∂y
2
2
∂ z
∂ z
∂2 z
x 2 2 + 2 xy
+ y 2 2 = ( x 2 + y 2 )n .
∂x ∂y
∂x
∂y
Hence,
⎛ x⎞
⎛ y⎞
Example 16: If z = x n f ⎜ ⎟ + y − n f ⎜ ⎟ , prove that
⎝ x⎠
⎝ y⎠
x2
2
∂2 z
∂2 z
∂z
∂z
2 ∂ z
+
xy
+
y
2
+x
+y
= n2 z .
2
2
∂x ∂y
∂x
∂y
∂x
∂y
Solution: Let z = u + v
⎛x⎞
⎛ y⎞
u = xn f ⎜ ⎟ , v = y−n f ⎜ ⎟
⎝x⎠
⎝ y⎠
Replacing x by xt and y by yt,
where,
⎛x⎞
⎛ y⎞
u = t n xn f ⎜ ⎟ , v = t −n y−n f ⎜ ⎟
⎝x⎠
⎝ y⎠
Hence, u is a homogeneous function of degree n and v is a homogeneous function of
degree n.
By Euler’s theorem and Cor. 1
x2
∂2 u
∂2 u
∂2 u
∂u
∂u
+ 2 xy
+ y2 2 + x + y
= n( n − 1)u + nu
2
∂x ∂y
∂x
∂y
∂x
∂y
= n2 u
... (1)
Engineering Mathematics
4.80
2
∂2 v
∂2 v
∂v
∂v
2 ∂ v
+
2
xy
+
y
+x +y
= − n( − n − 1)v − nv
2
2
∂x ∂y
∂x
∂y
∂x
∂y
= n2 v
Adding Eqs (1) and (2),
and
x2
... (2)
⎛ ∂2 u ∂2 v ⎞
⎛ ∂2 u ∂2 v ⎞
⎛ ∂2 u
∂2 v ⎞
⎛ ∂u ∂v ⎞
+
+ y2 ⎜ 2 + 2 ⎟ + x ⎜ + ⎟
x 2 ⎜ 2 + 2 ⎟ + 2 xy ⎜
⎟
⎝ ∂x ∂x ⎠
∂y ⎠
∂x ⎠
⎝ ∂y
⎝ ∂x
⎝ ∂x ∂y ∂x ∂y ⎠
⎛ ∂u ∂v ⎞
+ y ⎜ + ⎟ = n2 (u + v )
⎝ ∂y ∂y ⎠
x2
Hence,
Example 17: If u =
x3 + y3
y x
+
2
∂2 z
∂2 z
∂z
∂z
2 ∂ z
2
+
xy
+
y
+x +y
= n 2 z.
2
2
∂x ∂y
∂x
∂y
∂x
∂y
2
2
⎞
1
−1 ⎛ x + y
, find value of
sin
2
7
⎜
x
⎝ x + 2 xy ⎟⎠
∂2 u
∂2 u
∂2 u
∂u
∂u
at x = 1, y = 2.
+ 2 xy
+ y2 2 + x
+y
2
∂x ∂y
∂x
∂y
∂x
∂y
Solution: Let u = v + w
x2
2
2
⎡1
⎤
⎛ x3 + y3 ⎞
−1 ⎛ x + y ⎞
v=⎜
⎟ and w = ⎢ x 7 sin ⎜⎝ x 2 + 2 xy ⎟⎠ ⎥
⎝ y x ⎠
⎣
⎦
Replacing x by xt and y by yt,
2
2
3
⎤
⎡1
⎛ x3 + y3 ⎞
−7
−1 ⎛ x + y ⎞
v = t2 ⎜
⎟ and w = t ⎢ x 7 sin ⎜⎝ x 2 + 2 xy ⎟⎠ ⎥
⎝ y x ⎠
⎦
⎣
where,
3
Hence, v is a homogeneous function of degree
and w is a homogeneous function
2
of degree 7.
By Euler’s theorem and Cor. 1,
x2
x2
9
∂2 v
3
∂2 v
∂2 v
∂v
∂v 3 ⎛ 3 ⎞
+ 2 xy
+ y2 2 + x + y
= ⎜ − 1⎟ v + v = v
2
⎝
⎠
2
2
4
x
y
2
∂
∂
x
∂
y
∂
∂x
∂y
... (1)
2
∂2 w
∂2 w
∂w
∂w
2 ∂ w
+
2
xy
+
y
+x
+y
= −7( −7 − 1)w − 7w = 49w ... (2)
∂x ∂y
∂x
∂y
∂x 2
∂y 2
Adding Eqs (1) and (2),
⎛ ∂2 v ∂2 w ⎞
⎛ ∂2 v ∂2 w ⎞
⎛ ∂2 v
∂2 w ⎞
⎛ ∂v ∂w ⎞
+
+ y2 ⎜ 2 + 2 ⎟ + x ⎜ +
x 2 ⎜ 2 + 2 ⎟ + 2 xy ⎜
⎟
⎝ ∂x ∂x ⎟⎠
∂y ⎠
∂x ⎠
⎝ ∂y
⎝ ∂x
⎝ ∂x ∂y ∂x ∂y ⎠
⎛ ∂v ∂w ⎞ 9 v
+y⎜ +
=
+ 49w
⎝ ∂y ∂y ⎟⎠ 4
x2
2
∂2 u
∂2 u
∂u
∂u 9 v
2 ∂ u
2
xy
y
+
+
+x +y
=
+ 49w
4
∂x ∂y
∂x
∂y
∂x 2
∂y 2
Partial Differentiation
At x = 1, y = 2,
v=
1+ 8
2 1
=
9
.
2
w=
and
2
Hence, x
4.81
1
⎛1 + 4 ⎞ p
sin −1 ⎜
=
⎝ 1 + 4 ⎟⎠ 2
(1)7
∂u
∂u 81 49p
∂2 u
∂2 u
∂2 u
= +
.
+ 2 xy
+ y2 2 + x + y
2
∂x
∂y 8
∂x ∂y
2
∂y
∂x
⎛ y⎞
⎛ y⎞
Example 18: If u = x 3 sin −1 ⎜ ⎟ + x 4 tan −1 ⎜ ⎟ , find the value of
⎝ x⎠
⎝ x⎠
2
2
2
∂u
∂u
∂u
∂u
∂u
+ y2 2 + x
+y
x 2 2 + 2 xy
at x = 1, y = 1.
x
∂
∂y
y
∂
∂x ∂y
∂x
Solution: Let u = v + w
3
−1 ⎛ y ⎞
4
−1 ⎛ y ⎞
where, v = x sin ⎜⎝ ⎟⎠ , w = x tan ⎜⎝ ⎟⎠
x
x
Replacing x by xt and y by yt,
⎡
⎡
⎛ y ⎞⎤
⎛ y ⎞⎤
v = t 3 ⎢ x 3 sin −1 ⎜ ⎟ ⎥ and w = t 4 ⎢ x 4 tan −1 ⎜ ⎟ ⎥
⎝
⎠
⎝ x ⎠⎦
x ⎦
⎣
⎣
Hence, v is a homogeneous function of degree 3 and w is a homogeneous function of
degree 4.
By Euler’s theorem,
x2
2
∂2 v
∂2 v
∂v
∂v
2 ∂ v
+
+
+x +y
= 3(3 − 1)v + 3v
2
xy
y
2
2
∂x ∂y
∂x
∂y
∂x
∂y
= 9v
∂w
∂w
∂w
∂w
∂w
+ 2 xy
+ y2 2 + x
+y
= 4( 4 − 1)w + 4 w
∂x ∂y
∂x
∂y
∂x 2
∂y
= 16w
Adding Eqs (1) and (2),
2
and
2
... (1)
2
x2
... (2)
2
⎛ ∂2 v ∂2 w ⎞
⎛ ∂2 v
⎛ ∂v ∂w ⎞
∂2 w ⎞
∂2 w ⎞
⎛ ∂v ∂w ⎞
2 ⎛∂ v
+
+
+ x⎜ +
x 2 ⎜ 2 + 2 ⎟ + 2 xy ⎜
y
+
+ y⎜ +
⎟
2
2 ⎟
⎟
⎜
⎝ ∂x ∂x ⎠
⎝ ∂y ∂y ⎟⎠
∂x ⎠
∂y ⎠
⎝ ∂x
⎝ ∂x ∂y ∂x ∂y ⎠
⎝ ∂y
= 9v + 16 w
x2
At
2
∂2 u
∂2 u
∂u
∂u
2 ∂ u
2
+
xy
+
y
+x +y
= 9v + 16 w
∂x ∂y
∂x
∂y
∂x 2
∂y 2
x = 1, y = 1,
p
p
and w = tan −1 1 =
2
4
2
2
2
∂
u
9
16
17
∂
u
∂
u
∂
u
∂
u
p
p
p
Hence, x 2 2 + 2 xy
+ y2 2 + x
+y
=
+
=
.
2
4
2
∂x ∂y
∂x
∂y
∂x
∂y
v = sin −1 1 =
Engineering Mathematics
4.82
Example 19: If u =
x2
⎛ x2 + y2 ⎞
x4 + y4
+ x 6 tan −1 ⎜ 2
, find the value of
2 2
x y
⎝ x + 2 xy ⎟⎠
2
∂2 u
∂2 u
∂u
∂u
2 ∂ u
xy
y
+
2
+
+x
+y
at x = 1, y = 2.
2
2
∂x ∂y
∂x
∂y
∂x
∂y
Solution: Let u = v + w
x4 + y4
6
−1 ⎛ x + y ⎞
v = 2 2 and w = x tan ⎜ 2
⎝ x + 2 xy ⎟⎠
x y
2
where,
2
Replacing x by xt and y by yt,
⎡
⎛ x2 + y2 ⎞⎤
⎛ x4 + y4 ⎞
v = t 0 ⎜ 2 2 ⎟ and w = t 6 ⎢ x 6 tan −1 ⎜ 2
⎥
⎝ x + 2 xy ⎟⎠ ⎦
⎝ x y ⎠
⎣
Hence, v is a homogeneous function of degree 0 and w is a homogeneous function of
degree 6.
By Euler’s theorem, and Cor. 1
∂2 v
∂2 v
∂2 v
∂v
∂v
+ y2 2 + x + y
= 0(0 − 1)v + 0 ⋅ v
x 2 2 + 2 xy
∂x ∂y
∂x
∂y
∂x
∂y
... (1)
= 0.
∂w
∂2 w
∂w
∂2 w
∂2 w
= 6(6 − 1)w + 6 w
+y
+ y2 2 + x
x 2 2 + 2 xy
∂y
∂x
∂x ∂y
∂y
∂x
... (2)
= 36 w
Adding Eqs (1) and (2),
2
⎛ ∂2 v ∂2 w ⎞
⎛ ∂2 v
∂2 w ⎞
∂2 w ⎞
⎛ ∂v ∂w ⎞
2 ⎛∂ v
+ x⎜ +
+
+
+
x 2 ⎜ 2 + 2 ⎟ + 2 xy ⎜
y
2
2 ⎟
⎜
⎟
⎝ ∂x ∂x ⎟⎠
∂y ⎠
∂x ⎠
⎝ ∂y
⎝ ∂x
⎝ ∂x ∂y ∂x ∂y ⎠
⎛ ∂v ∂w ⎞
+y⎜ +
= 36 w
⎝ ∂y ∂y ⎟⎠
x2
2
∂2 u
∂2 u
∂u
∂u
2 ∂ u
+
2
xy
+
y
+x +y
= 36 w
2
2
∂x ∂y
∂x
∂y
∂x
∂y
At x = 1, y = 2,
p
⎛1 + 4 ⎞
w = tan −1 ⎜
= tan −1 1 =
⎝ 1 + 4 ⎟⎠
4
Hence,
x2
2
∂u 36p
∂u
∂2 u
∂2 u
2 ∂ u
=
= 9p .
+
+x +y
+
y
2
xy
2
2
∂y
∂x
∂x ∂y
4
∂y
∂x
Example 20: If f (x, y, z) = 0 where f (x, y, z) is a homogeneous function of
degree n, then show that
x2
2
∂2 z
∂2 z
2 ∂ z
xy
y
=
−
=
.
∂x ∂y
∂x 2
∂y 2
Partial Differentiation
4.83
Solution: Here z is an implicit function of x and y,
f (x, y, z) = 0,
f
∂f
=0
∂x
Using composite function,
∂ f ∂ f ∂x ∂f ∂y ∂f ∂z
=
⋅ + ⋅ + ⋅ =0
∂ x ∂ x ∂x ∂y ∂x ∂z ∂x
∂f ∂f ∂z
+
⋅
=0
∂x ∂z ∂x
∂f
∂f ∂z
=− ⋅
∂x
∂z ∂x
∂f
∂f ∂z
=− ⋅
∂y
∂z ∂y
Similarly,
f is a homogeneous function of degree n.
By Euler’s theorem.
∂f
∂f
∂f
x +y +z
= nf = 0
∂x
∂y
∂z
f
f
Substituting and from Eqs (1) and (2),
x
y
x
x
y
y
z
x
y
Fig. 4.35
⎤
⎡ ∂y
⎢∵ ∂x = 0 ⎥
⎦
⎣
... (1)
... (2)
[∵ f ( x, y, z ) = 0]
⎛ ∂f ∂z ⎞
∂f
⎛ ∂f ∂z ⎞
x ⎜− ⋅ ⎟ + y ⎜− ⋅ ⎟ + z
=0
⎝ ∂z ∂x ⎠
⎝ ∂z ∂y ⎠
∂z
−x
∂z
∂z
−y +z=0
∂x
∂y
∂z
∂z
x +y
=z
∂x
∂y
... (3)
Differentiating Eq. (3) w.r.t. x,
∂ 2 z ∂z
∂2 z
∂z
x 2 + +y
=
∂x
∂x ∂y ∂x
∂x
x
∂2 z
∂2 z
= −y
2
∂x ∂y
∂x
x2
∂2 z
∂2 z
xy
=
−
∂x ∂y
∂x 2
... (4)
Again differentiating (3) w.r.t. y,
x
∂2 z
∂ 2 z ∂z ∂z
+y 2 +
=
∂y ∂x
∂y ∂y
∂y
∂2 z
∂2 z
x
= −y 2
∂y ∂x
∂y
y2
∂2 z
∂2 z
= − xy
2
∂x ∂y
∂y
... (5)
Engineering Mathematics
4.84
From Eqs (4) and (5), we get
x2
2
∂2 z
∂2 z
2 ∂ z
=
−
=
xy
y
.
∂x ∂y
∂x 2
∂y 2
Exercise 4.4
1. Verify Euler’s theorem for
(i)
⎛ x2 + y2 ⎞
u = tan −1 ⎜
⎟
y
⎝
⎠
⎛ xy + yz + zx ⎞
(ii) u = log ⎜ 2
⎝ x + y 2 + z 2 ⎟⎠
⎛ x2 + y2 ⎞
(iii) u = log ⎜ 2
⎝ x − y 2 ⎟⎠
2
∂2 z
∂2 z
2 ∂ z .
2
+
xy
+
y
∂x ∂y
∂x 2
∂y 2
[Ans. : 6z]
6. If u = x yz 4y z + 2xz , prove that
u
u
u
x
y
z
4 u.
x
y
z
2
2 2
x2 + y2 + z2
x+ y+z
x y z.
(vi) u = + +
y z x
xy + yz + zx
2. If u = cos 2
, find
x + y2 + z2
∂u
∂u
∂u
x +y +z .
∂x
∂y
∂z
u=
[Ans. : 0]
⎛ xy + yz ⎞
3. If u = cos ⎜ 2
⎝ x + y 2 + z 2 ⎟⎠
⎡
⎤
x+ y+ z⎥
+ sin ⎢
,
1
⎢
⎥
4
( xy )
⎢⎣
⎥⎦
∂u
∂u
∂u
+y
+z .
evaluate x
∂x
∂y
∂z
x2
y2
z2
+
+
= 1,
a2 + u b2 + u c2 + u
where u is a homogeneous function
in x, y, z of degree n, prove that
2
2
⎛ x− y⎞
4. If u = sin −1 ⎜
⎟ , show that
⎝ x+ y⎠
∂u
y ∂u
=−
.
∂x
x ∂y
x
y
5. If z = x 3 e , find the value of
2
⎛ ∂u ⎞ ⎛ ∂u ⎞ ⎛ ∂u ⎞
⎜⎝ ⎟⎠ + ⎜ ⎟ + ⎜⎝ ⎟⎠ = 2nu.
∂x
⎝ ∂y ⎠
∂z
x3 y3
, prove that
x3 + y3
∂u
∂u
x +y
= 3u.
∂x
∂y
8. If u =
9. If u =
x
x2 + y2
x+ y
, prove that
∂u
∂u 3
+y
= u.
∂x
∂y 2
10. If u =
x2
xy
, find the value of
x+ y
2
∂2 z
∂2 z
2 ∂ z
2
+
xy
+
y
.
∂x ∂y
∂x 2
∂y 2
[Ans. : 0]
−
3
7. If
(iv) u = 3x 2 yz + 5 xy 2 z + 4 z 4
(v)
x2
[Ans. : 0]
xy + yz
x y z
+ cos 2
,
2
2
x + y2 + z2
x +y +z
2
11. If u =
2 2
2
prove that
x
4x2 y2 z2
∂u
∂u
∂u
+y
+z
= 2
.
∂x
∂y
∂z x + y 2 + z 2
Partial Differentiation
y
y
+y
, prove that
x
x
12. If u = xf
x2
2
∂2 u
∂2 u
2 ∂ u
2
+
xy
+
y
= 0.
∂x ∂y
∂x 2
∂y 2
⎛ y⎞
⎛ y⎞
13. If u = 3x 4 cot −1 ⎜ ⎟ + 16 y 4 cos −1 ⎜ ⎟ ,
⎝x⎠
⎝x⎠
prove that
∂u
∂u
∂u
+ 2 xy
+ y 2 2 = 12u.
∂x ∂y
∂x 2
∂y
2
x2
2
2
y
⎛ y⎞
14. If u = y 2 e x + x 2 tan −1 ⎜ ⎟ , prove that
⎝x⎠
x2
2
∂2 u
∂2 u
2 ∂ u
+
2
xy
+
y
∂x ∂y
∂x 2
∂y 2
∂u
∂u
+x
+y
= 4u.
∂x
∂y
3 2
−1 ⎛ y ⎞
15. If u = x y sin ⎜ ⎟ , prove that
⎝x⎠
∂u
∂u
∂u
+ 2 xy
+ y2 2
∂x ∂y
∂x 2
∂y
∂u
∂u
+x +y
= 25u.
∂x
∂y
2
x2
2
2
⎛3 y−3 x⎞
16. If u = x 2 log ⎜
⎟ , prove that
⎝3 y+3 x⎠
⎛3 y−3 x⎞
∂u
∂u
x
+y
= 2 x 2 log ⎜
⎟.
∂x
∂y
⎝3 y+3 x⎠
⎛ x2 − y2 y2 − z2 z2 − x2 ⎞
,
,
,
17. If u = f ⎜
x2
y 2 ⎟⎠
⎝ z2
prove that
∂2 u
∂2 u
∂2 u
x 2 2 + 2 xy
+ y 2 2 = 0.
∂x ∂y
∂x
∂y
x
∂u
∂u
∂u
+y
+z
= 0.
∂x
∂y
∂z
19. If u = x 2 sin −1
that x
2
y
x
− y 2 cos −1 , prove
x
y
∂2 u
∂2 u
∂2 u
+ 2 xy
+ y 2 2 = 2u.
2
∂x ∂y
∂x
∂y
20. If u = x sin −1
value of x
2
y
y
+ tan −1 , find the
x
x
∂2 u
∂2 u
∂2 u
+ 2 xy
+ y2 2 .
2
∂x ∂y
∂x
∂y
[Ans. : 0]
21. If y = x cos u, prove that
x2
∂2 u
∂2 u
∂2 u
+ 2 xy
+ y 2 2 = 0.
2
∂x ∂y
∂x
∂y
y
⎡
−1 y ⎤
⎢ Hint : cos u = x , u = cos x ⎥
⎣
⎦
y
⎡
⎛ y⎞ y − ⎤
22. If u = x 3 ⎢ tan −1 ⎜ ⎟ + e x ⎥
⎝x⎠ x
⎣
⎦
⎡
⎛x⎞ x
x⎤
+ y −3 ⎢sin −1 ⎜ ⎟ + log ⎥ ,
y⎦
⎝ y⎠ y
⎣
prove that
∂2 u
∂2 u
∂2 u
x 2 2 + 2 xy
+ y2 2
∂x ∂y
∂x
∂y
+x
∂u
∂u
+y
= 9u.
∂x
∂y
23. If z = f (x, y) and u, v are homogeneous
functions of degree n in x, y, then
show that
x
∂z
∂z
∂z ⎞
⎛ ∂z
+y
= n ⎜u + v ⎟ .
⎝
∂x
∂y
∂u
∂v ⎠
2
2
2
24. If u = ( x + y ) 3 , prove that
n
⎛x y z⎞
18. If u = ⎜ + + ⎟ , show that
⎝ y z x⎠
4.85
x2
2
∂2 u
4
∂2 u
2 ∂ u
2
+
xy
+
y
= u.
2
2
9
∂x ∂y
∂x
∂y
Engineering Mathematics
4.86
Corollary 2: If z = f (u) is a homogeneous function of degree n in variables x and y,
∂u
∂u
∂u
f (u)
then x + y + z
=n
.
∂x
∂y
∂z
f ′( u )
Proof: By Euler’s theorem,
∂z
∂z
x +y
= nz = nf (u)
∂x
∂y
∂
∂
x
f (u) + y
f (u ) = nf (u )
∂x
∂y
∂u
∂u
= nf (u )
x f ′(u ) + y f ′(uu )
∂y
∂x
∂u
∂u
f (u)
=n
x +y
∂y
∂x
f ′( u )
Note: If v = f (u) is a homogeneous function of degree n in variables x, y and z, then
x
∂u
∂u
∂u
f (u)
+y
+z
=n
.
∂x
∂y
∂z
f ′( u )
Corollary 3: If z = f (u) is a homogeneous function of degree n in variables x and y,
then
x2
where, g (u ) = n
∂2 u
∂2 u
∂2 u
+ 2 xy
+ y 2 2 = g (u )[ g ′(u) − 1]
2
∂x ∂y
∂x
∂y
f (u )
.
f ′( u )
Proof: By Cor. 2,
x
∂u
∂u
f (u )
+y
=n
= g (u)
∂x
∂y
f ′( u )
... (1)
Differentiating Eq. (1) partially w.r.t. x,
x
∂ 2 u ∂u
∂2 u
∂u
+
+
y
= g ′( u )
∂x ∂y
∂x
∂x 2 ∂x
x
∂2 u
∂2 u
∂u
+
y
= [ g ′(u ) − 1]
2
∂x ∂y
∂x
∂x
... (2)
Differentiating Eq. (1) partially w.r.t. y,
x
∂2 u
∂ 2 u ∂u
∂u
+y 2+
= g ′( u )
∂x ∂y
∂y
∂y
∂y
x
∂2 u
∂2 u
∂u
+ y 2 = [ g ′(u ) − 1]
∂x ∂y
∂y
∂y
... (3)
Partial Differentiation
4.87
Multiplying Eq. (2) by x and Eq. (3) by y and adding,
x2
2
⎛ ∂u
∂2 u
∂2 u
∂u ⎞
2 ∂ u
2
+
xy
+
y
= [ g ′(u) − 1] ⎜ x + y ⎟
2
2
∂x ∂y
⎝ ∂x
∂y ⎠
∂x
∂y
x2
∂2 u
∂2 u
∂2 u
+ 2 xy
+ y 2 2 = [ g ′(u) − 1]g (u)
2
∂x ∂y
∂x
∂y
where, g (u ) = n
[Using Eq. (1)]
f (u )
.
f ′( u )
⎛ x3 + y3 ⎞
∂u
∂u
Example 1: If u = sec −1 ⎜
, prove that x
+y
= 2cot u.
⎝ x + y ⎟⎠
∂x
∂y
⎛ x3 + y3 ⎞
u = sec −1 ⎜
⎝ x + y ⎟⎠
Solution:
Replacing x by xt and y by yt,
⎡ ⎛ x3 + y3 ⎞ ⎤
u = sec −1 ⎢t 2 ⎜
⎟⎥
⎣ ⎝ x + y ⎠⎦
u is a non-homogeneous function. But sec u =
degree 2.
x3 + y3
is a homogeneous function of
x+ y
Let f (u) = sec u
By Cor. 2,
x
∂u
∂u
f (u )
sec u
+y
=n
=2
= 2 cot u
∂x
∂y
f ′( u )
sec u tan u
Example 2: If u = sin–1 (xyz), prove that x
∂u
∂u
∂u
+y
+z
= 3 tan u.
∂x
∂y
∂z
u = sin–1 (xyz)
Solution:
Replacing x by xt, y by yt and z by zt,
u = sin
1
t 3 ( xyz )
u is a non-homogeneous function. But sin u = xyz is a homogeneous function of
degree 3.
Let f (u) = sin u
By Cor. 2,
x
u
u
u
f (u)
sin u
+y +z
=n
=3
= 3tan u.
x
y
z
f (u )
cos u
Engineering Mathematics
4.88
Example 3: If u = log x + log y, prove that x
u
u
+y
= 2.
x
y
u = log x + log y = log xy
Solution:
Replacing x by xt and y by yt,
u = log t 2 ( xy )
u is a non-homogeneous function. But eu = xy is a homogeneous function of
degree 2.
Let f (u) = eu
By Cor. 2,
x
u
u
f (u)
eu
+y
=n
= 2 u = 2.
x
y
f (u)
e
Example 4: If u = log (x2 + y2 + z2), prove that x
u
u
u
+y
+z
= 2.
x
y
z
u = log (x2 + y2 + z2)
Solution:
Replacing x by xt, y by yt, and z by zt,
u = log t 2 ( x 2 + y 2 + z 2 )
u is a non-homogeneous function. But eu = x2 + y2 + z2 is a homogeneous function of
degree 2.
Let f (u) = eu
By Cor. 2
x
∂u
∂u
∂u
f (u)
eu
+y +z
=n
=2 u =2
∂x
∂y
∂z
f ′( u )
e
Example 5: If u = e
⎛ y⎞
x2 f ⎜ ⎟
⎝ x⎠
Solution:
, prove that x
u=e
u
u
+y
= 2u log u .
x
y
⎛ y⎞
x2 f ⎜ ⎟
⎝x⎠
Replacing x by xt and y by yt,
u=e
t 2 x2 f
y
x
2
u is a non-homogeneous function. But log u = x f
degree 2.
y
x
is homogeneous function of
Partial Differentiation
4.89
Let f (u) = log u
By Cor. 2,
u
u
f (u )
log u
+y
=n
=2
= 2u log u.
x
y
f (u)
1/ u
x
xy + yz
+ sin
x y2 z2
Example 6: If u = tan
x
2
u
u
u 1
+y
+z
=
x
y
z 2
(
(
)
x + y + z , show that
) (
x + y + z cos
)
x+ y+ z .
Solution: Let u = v + w
where, v = tan
xy + yz
x y2 z2
and w = sin
2
(
x+ y+ z
)
Replacing x by xt, y by yt, and z by zt,
xy + yz
,
x y2 z2
v = t 0 tan
2
1
w = sin t 2
(
x+ y+ z
)
v is a homogeneous function of degree 0.
By Euler’s theorem,
x
v
x
y
v
y
z
v
z
0 v
0
... (1)
1
w is a non-homogeneous function. But sin w =
function of x, y, z of degree 1 .
2
Let f ( w ) = sin 1 w
(
)
x + y + z is a homogeneous
By Cor. 2,
x
∂w
∂w
∂w
f ( w ) 1 sin −1 w
+y
+z
=n
=
1
∂x
∂y
∂z
f ′( w ) 2
1− w2
=
=
1
2
(
(
x+ y+ z
)
1 − sin 2
) (
x + y + z cos
2
(
x+ y+ z
x+ y+ z
)
)
... (2)
Engineering Mathematics
4.90
Adding Eqs (1) and (2),
⎛ ∂v ∂w ⎞
x⎜ +
+
⎝ ∂x ∂x ⎟⎠
⎛ ∂v ∂w ⎞
⎛ ∂v ∂w ⎞
y⎜ +
+z⎜ +
=
⎟
⎝ ∂z ∂z ⎟⎠
⎝ ∂y ∂y ⎠
Hence,
x
u
u
u
+y
+z
=
x
y
z
(
) (
x + y + z cos
(
) (
x + y + z cos
x2
u
u
+y
x
y
e 2 u (sec 2 v tan 2 v )
e 2u
x
= sin v
y
and
⎛x⎞
v = sin −1 ⎜ ⎟
⎝ y⎠
v is homogeneous function of degree 0.
By Euler’s theorem,
x
x
Hence,
u
u
+y
x
y
Example 8: If u = sin
1
x
(i) x
u
x
y
u
y
1
3
1
3
1
2
1
2
x +y
y
v
v
+y
=0
x
y
v
v
+y
= 0.
x
y
x
1
2
, prove that
1
tan u
12
2
(ii) x 2
2
2
u
u
u tan u
2
+
+
=
2
xy
y
(tan 2 u + 13).
x y
144
x2
y2
Solution:
1
⎛ 13
3
−1 ⎜ x + y
u = sin
1
1
⎜ 2
⎝ x − y2
x+ y+ z
).
2
x = eu tan v, y = eu sec v
y2
)
2
Example 7: If x = eu tan v, y = eu sec v, prove that x
Solution:
x+ y+ z
1
⎞2
⎟
⎟
⎠
x
v
v
+y
= 0.
x
y
Partial Differentiation
u = sin
Replacing x by xt and y by yt,
1
t
1
12
u is a non-homogeneous function. But sin u =
1
3
x +y
1
3
1
1
x2
y2
1
3
1
3
x +y
1
1
x2
y2
4.91
1
2
1
2
is a homogeneous function
1
.
12
with degree
Let f (u) = sin u
By Cor. 2,
(i) x
u
x
u
y
y
n
f (u )
f (u)
1 sin u
12 cos u
1
tan u.
12
By Cor. 3,
(ii) x 2
2
∂2 u
∂2 u
2 ∂ u
= g (u)[ g ′(u) − 1]
+
xy
+
y
2
∂x ∂y
∂y 2
∂x 2
g (u) = n
where,
g (u) =
2
2
Hence, x
u
x2
2
2 xy
u
x y
2
y2
u
y2
(i) x
1
sec 2 u 1
12
1
1 + tan 2 u + 12
tan u
tan u
=
(tan 2 u + 13).
12
12
144
1
x3 + y3
log 2
, find the value of
3
x + y2
u
u
+y
x
y
2
(ii) x 2
1 2
sec u
12
1
tan u
12
=
Example 9: If u =
f (u )
1
=
tan u
f (u ) 12
2
2
u
u
u
2
+
2
xy
+
y
.
2
x y
x
y2
1
x3 + y3
Solution: u = log 2
3
x + y2
Engineering Mathematics
4.92
1
x3 + y3
u = log t 2
3
x + y2
Replacing x by xt and y by yt,
x3 + y3
u is a non-homogeneous function. But e 3u = 2
is a homogeneous function of
x + y2
degree 1.
Let f (u) = e3u
By Cor. 2,
u
u
y
x
y
By Cor. 3,
(i) x
2
2
(ii) x
f (u)
f (u)
n
1
2
u
x2
2 xy
e 3u
3e 3u
1
.
3
2
u
x y
u
y2
y2
g (u )[ g (u ) 1]
g (u) = n
where,
f (u) 1
=
f (u) 3
g (u ) = 0.
2
Hence, x 2
u
x2
2
2 xy
2
u
x y
Example 10: If u = tan
y2
u
y2
1
(0 1)
3
1
.
3
x2 + y2
, prove that
x+ y
1
2
x2
2
2
u
u
u
2
+
2
xy
+
y
= –2 sin3 u cos u.
2
x y
x
y2
Solution:
u = tan
1
u = tan
1
x2 + y2
x+ y
Replacing x by xt and y by yt,
t
x2 + y2
x+ y
x2 + y2
u is a non-homogeneous function. But tan u =
is a homogeneous function of
x+ y
degree 1.
Let f (u) = tan u
By Cor. 3,
2
x2
u
x2
2
2 xy
u
x y
2
y2
u
y2
g (u ) [ g (u ) 1]
where,
g (u)
n
f (u)
f (u)
1
tan u
sec 2 u
sin u cos u
sin 2u
2
Partial Differentiation
4.93
g (u ) = cos 2u
Hence,
2
u
x2
x2
2
2 xy
u
x y
2
y2
u
y2
sin 2u
(cos 2u 1)
2
= sin u cos u ( −2 sin 2 u ) = −2 sin 3 u cos u.
Example 11: If u = sinh
x3 + y3
, prove that
x2 + y2
1
2
x2
2
2
u
u
u
2
+
2
xy
+
y
= –tanh3 u.
x y
x2
y2
Solution:
u = sinh
1
u = sinh
1
x3 + y3
x2 + y2
Replacing x by xt and y by yt,
t
x3 + y3
x2 + y2
x3 + y3
u is a non-homogeneous function. But sinh u = 2
is a homogeneous function
x + y2
of degree 1.
Let f (u) = sinh u
By Cor. 3,
∂2 u
∂2 u
∂2 u
x 2 2 + 2 xy
+ y 2 2 = g (u ) [ g ′(u ) − 1]
∂x ∂y
∂x
∂y
where,
f (u)
sinh u
g (u) n
1
f (u)
cosh u
tanh u
g (u ) = sech 2u.
2
u
x2
2
Hence, x
2
2 xy
u
x y
2
y2
u
y2
tanh u (sech 2 u 1)
= tanh u( tanh2 u) =
Example 12: If u = log
2
u
u
2
xy
2
x y
x
⎛ x+ y ⎞
+ sin −1 ⎜
⎟ , prove that
⎝ x+ y⎠
x2 + y2
x+ y
2
x2
tanh3 u.
2
y2
sin w cos 2w
, where w = sin
4cos 3 w
u
y2
Solution: Let u = v + w
where,
v = log
x+ y
x +y
2
2
,
w = sin
1
x+ y
x+ y
1
x+ y
x+ y
.
Engineering Mathematics
4.94
Replacing x by xt and y by yt,
x+ y
v = t 0 log
1
x +y
2
w = sin
,
2
1
t2
x+ y
x+ y
v is a homogeneous function of degree 0.
By Cor. 1,
2
2
v
x2
x2
2 xy
2
v
v
y2
y2
x y
0 v
... (1)
0
x+ y
and w is a non-homogeneous function. But sin w =
is a homogeneous funcx+ y
1
tion of degree .
2
Let f (w) = sin w
By Cor. 3,
∂2 w
∂2 w
∂2 w
x 2 2 + 2 xy
+ y 2 2 = g ( w )[ g ′( w ) − 1]
∂x ∂y
∂x
∂y
g(w) = n
where,
1 2
sec w.
2
g (w) =
2
2
w
x2
x2
Hence,
2
w
x y
2 xy
w
y2
y2
f ( w ) 1 sin w 1
=
= tan w
f ( w ) 2 cos w 2
1
1
tan w sec 2 w 1
2
2
1
(1 2 cos 2 w )
sin w
2
2 cos3 w
sin w cos 2w
4 cos3 w
... (2)
Adding Eqs (1) and (2),
2
x2
v
x2
2
w
x2
2
Hence, x 2
u
x2
2
2 xy
x y
2
2 xy
u
x y
2
v
2
y2
u
y2
w
x y
2
y2
v
y2
2
w
y2
sin w cos 2w
, where w = sin
4 cos3 w
Example 13: If u = log (x3 + y3 + z3 - 3xyz), then show that
(i)
(ii) x
u
u
u
3
+
+
=
x
y
z x y z
u
u
u
+y
+z
=3
x
y
z
sin w cos 2w
4 cos3 w
1
x+ y
x+ y
.
Partial Differentiation
4.95
2
(iii)
2
2
2
2
2
u
u
u
u
u
u
+
+
+
+
+
=
2
2
2
2
2
2
x y
y z
z x (x
x
y
z
2
u
x2
(iv) x 2
2
u
y2
y2
2
2
u
u
2 xy
2
x y
z
z2
9
y z )2
2
2 yz
u
y z
2
2 zx
u
z x
3.
Solution: u = log (x3 + y3 + z3 3xyz)
(i) Differentiating u w.r.t. x, y and z,
u
1
(3 x 2
x x 3 y 3 z 3 3 xyz
3 yz )
u
y
x3
y3
1
(3 y 2
z 3 3 xyz
3 xz )
u
z
x3
y3
1
(3 z 2
z 3 3 xyz
3 xy )
u
u
u 3( x 2 y 2 z 2 xy yz zx )
+
+
=
x
y
z
x 3 y 3 z 3 3 xyz
3( x 2 y 2
( x y z )( x 2
3
=
.
x+ y+z
=
z2
y2
xy yz zx )
z 2 xy yz zx )
(ii) Replacing x by xt, y by yt and z by zt,
u log t 3 ( x 3 y 3 z 3 3 xyz )
u is a non-homogeneous function. But eu = x3 + y3 + z3 3xyz is a homogeneous function of degree 3.
Let f (u) = eu
By Cor. 2,
x
u
u
u
f (u )
eu
+y +z
=n
= 3 u = 3.
x
y
z
f (u )
e
2
(iii)
2
2
2
2
2
u
u
u
u
u
u
+ 2 + 2 +2
+2
+2
=
2
x y
y z
z x
x
y
z
x
+
y
+
z
⎛∂
3
∂
∂ ⎞⎛
−3
⎞
(1 + 1 + 1)
= ⎜ + + ⎟⎜
=
⎟
⎝ ∂x ∂y ∂z ⎠ ⎝ x + y + z ⎠ ( x + y + z )
=
−9
.
( x + y + z )2
u
u
u
+
+
x
y
z
Engineering Mathematics
4.96
(iv) By Cor. 3,
2
x2
u
x2
2
2 xy
u
x y
2
y2
u
y2
g (u )[ g (u ) 1]
where,
g (u) = n
f (u)
=3
f (u)
g (u) = 0
Hence,
x2
2
∂2 u
∂2 u
2 ∂ u
+
2
xy
+
y
= 3(0 − 1) = −3.
∂x ∂y
∂x 2
∂y 2
Example 14: If u = log r and r2 = x2 + y2, prove that x 2
2
∂2u
∂2u
2 ∂ u
+ 1 = 0.
+
xy
+
y
2
∂x ∂y
∂y 2
∂x 2
u = log r = log x 2 + y 2
Solution:
Replacing x by xt and y by yt,
(
u = log t x 2 + y 2
)
u is a non-homogeneous function of x and y. But e u = x 2 + y 2 is a homogeneous
function of degree 1.
Let f (u) = eu
By Cor. 3,
2
2
2
u
u
u
x 2 2 2 xy
y 2 2 g (u ) [ g (u ) 1]
x y
x
y
g (u) = n
where,
f (u) e u
=
=1
f ′( u ) e u
g ′ (u ) = 0.
2
x2
u
x2
2
2 xy
u
x y
2
y2
u
y2
1(0 1)
1
2
Hence,
x2
2
2
u
u
u
2
2
xy
y
+
+
+ 1 = 0.
2
x y
x
y2
2
Example 15: If u = log r, r = x3 + y3 - x2y - xy2, show that
and x
∂u
∂u
+y
= 3.
∂x
∂y
x
y
2
⎛∂
⎛∂
∂ ⎞ ⎛ ∂u ∂u ⎞
∂⎞
Solution: ⎜ + ⎟ u = ⎜ + ⎟ ⎜ + ⎟
⎝ ∂x ∂y ⎠ ⎝ ∂x ∂y ⎠
⎝ ∂x ∂y ⎠
u = log( x 3 + y 3 − x 2 y − xy 2 )
= log ⎡⎣( x + y )( x 2 + y 2 − xy ) − xy( x + y ) ⎤⎦
u
4
( x + y )2
Partial Differentiation
∂u
∂x
∂u
∂y
4.97
= log( x + y )( x 2 + y 2 − 2 xy ) = log( x + y ) + 2 log( x − y )
1
2
=
+
x+ y x− y
1
2
−
=
x+ y x− y
∂u ∂u
2
+
=
∂x ∂y x + y
2
⎛∂
⎛∂
∂⎞
∂ ⎞⎛ 2 ⎞
⎜⎝ ∂x + ∂y ⎟⎠ u = ⎜⎝ ∂x + ∂y ⎟⎠ ⎜⎝ x + y ⎟⎠
⋅
=−
2
4
2
−
=−
2
2
(xx + y ) ( x + y )
( x + y)2
Replacing x by xt and y by yt in u,
u = log t3 (x3 + y3 x2y xy2)
u is a non-homogeneous function. But eu = x3 + y3 x2y xy2 is a homogeneous function
of degree 3.
Let f (u) = eu
By Cor. 2,
x
u
u
f (u)
eu
+y
=n
= 3 u = 3.
x
y
f (u)
e
Exercise 4.5
1. If u = cos −1
x
u
x
⎛ x+ y ⎞
, prove that
⎜⎝ x + y ⎟⎠
u
1
y
cot u.
y
2
⎛ x2 y2 ⎞
2. If u = sin −1 ⎜
, prove that
⎝ x + y ⎟⎠
u
u
x +y
= 3tan u.
x
y
3. If u = log (x2 + xy + y2), prove that
u
u
x +y
= 2.
x
y
4. If (x − y) tan u = x3 + y3, prove that
u
u
x +y
= sin 2u.
x
y
5. If u = log (x3 + y3 − x2y − xy2), prove
u
u
= 3.
that x + y
x
y
6. If u = sin–1
x3 + y3 + z3
, prove
ax + by + cz
u
u
u
+y
+z
= 2 tan u.
x
y
z
⎛ x3 + y3 ⎞
7. If u = tan −1 ⎜
, prove that
⎝ x − y ⎟⎠
u
u
x +y
= sin 2u.
x
y
that x
1
⎛ 14
4
x
y
+
8. If u = sin ⎜ 1
1
⎜ 6
⎝ x + y6
−1
⎞
⎟ , prove that
⎟
⎠
Engineering Mathematics
4.98
2
∂2 u
∂2 u
2 ∂ u
+
+
2
xy
y
∂x ∂y
∂x 2
∂y 2
1
=
tan u(tan 2 u − 1).
144
x2
9. If
(
)
1
3
that x
11. If
1
3
x + y sin u = x + y , prove
2
(
x
that 4 x
2
∂u
∂2 u
2 ∂ u
+
2
xy
+
y
∂x ∂y
∂x 2
∂y 2
2
that x 2
=
tan u ⎛ 13 tan 2 u ⎞
.
+
12 ⎜⎝ 12
12 ⎟⎠
⎛ x5 − 2 y5 + 6 z5
10. If u = cos −1 ⎜
3
3
3
⎝ ax + by + cz
u
x
y
)
u
y
z
u
z
y cot u x
7
cot u.
2
y
0, prove
u
u
+ 4 y + sin 2u = 0.
x
y
x+ y
Hint : cot u =
x+ y
⎛ ax + by + cz ⎞
12. If u = sin −1 ⎜
⎟ , prove
n
n
n
⎝ x +y +z ⎠
⎞
⎟ , show
⎠
that x
u
u
u
+y
+z
= 2 tan u.
x
y
z
4.8 APPLICATIONS OF PARTIAL
DIFFERENTIATION
4.8.1 Jacobians
If u and v are continuous and differentiable functions of two independent variables x
and y, i.e., u = f1(x, y) and v = f2(x, y), then the determinant
u
x
v
x
u
y
is called the
v
y
Jacobian of u, v with respect to x, y and is denoted as J = (u, v) .
( x, y )
Similarly, if u, v and w are continuous and differentiable functions of three independent variables x, y, z, then the Jacobian of u, v, w with respect to x, y, z is
(u, v, w)
=
( x, y , z )
u u
x y
v v
x y
w w
x y
u
z
v
z
w
z
Jacobian is useful in transformation of variables from cartesian to polar, cylindrical
and spherical coordinates in multiple integrals.
Partial Differentiation
4.99
Properties of Jacobians
1. If u and v are functions of x and y, then
(u , v)
and J * =
( x, y )
J .J* = 1 where J =
x, y )
u, v)
Proof: Let u and v are two functions of x and y.
u = f1(x, y) and v = f2(x, y)
... (1)
Writing x and y in terms of u and v,
x = f1(u, v) and y = f2 (u, v)
... (2)
Differentiating Eq. (1) partially w.r.t. u and v,
∂u
∂u ∂x ∂u ∂y
=1=
⋅
+
⋅
∂u
∂x ∂u ∂y ∂u
∂u
∂u ∂x ∂u ∂y
=0=
⋅ +
⋅
∂v
∂x ∂v ∂y ∂v
... (4)
∂v
∂v ∂x ∂v ∂y
=0=
⋅
+
⋅
∂u
∂ x ∂u ∂y ∂u
... (5)
∂v
∂v ∂x ∂v ∂y
=1=
⋅ +
⋅
∂v
∂x ∂v ∂y ∂v
... (6)
∂u
∂x
∂(u, v ) ∂( x, y )
=
.
J. J* =
∂( x, y ) ∂(u, v ) ∂v
∂x
∂u ∂u
∂x ∂y
=
∂v ∂v
∂x ∂y
∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
∂u ∂x ∂u ∂y
+
∂x ∂u ∂y ∂u
=
∂v ∂x ∂v ∂y
+
∂x ∂u ∂y ∂u
=
1 0
01
∂u
∂y
∂v
∂y
∂x
∂u
∂y
∂u
∂x
∂v
∂y
∂v
⎡Interchanging rows and columns ⎤
⎢of second determinant
⎥
⎣
⎦
∂u ∂x ∂u ∂y
+
∂x ∂v ∂y ∂v
∂v ∂x ∂ v ∂ y
+
∂x ∂ v ∂ y ∂ v
[Substituting Eqs (3), (4), (5), (6)]
=1
Similarly
... (3)
u , v, w)
.
x, y , z )
x, y , z )
=1
u , v, w)
Engineering Mathematics
4.100
2. If u, v are functions of r, s and r, s are functions of x, y, then
∂(u , v) ∂(u , v) ∂( r , s )
=
.
.
∂( x, y ) ∂( r , s ) ∂( x, y )
Proof:
u , v)
.
r , s)
u
r , s)
r
=
v
x, y )
r
u
s
v
s
r
x
s
x
r
y
s
y
u
r
v
r
u
s
v
s
r
x
r
y
s
x
s
y
=
∂u ∂r ∂u ∂s
+
∂r ∂x ∂s ∂x
=
∂v ∂r ∂v ∂s
+
∂r ∂x ∂s ∂x
=
u
x
v
x
⎡Interchanging rows ⎤
⎢
⎥
⎢and columns of
⎥
⎢⎣second determinant ⎥⎦
∂u ∂r ∂u ∂s
+
∂r ∂y ∂s ∂y
∂v ∂r ∂v ∂s
+
∂r ∂y ∂s ∂y
u
y
(u , v)
=
v
( x, y )
y
Similarly, ∂ (u , v, w) = ∂ (u , v, w) . ∂ (r , s, t ) .
∂ ( x, y, z ) ∂ (r , s, t ) ∂ ( x, y, z)
3. If functions u, v of two independent variables x, y are dependent, then
Proof: If u, v are dependent, then there must be a relation f (u, v) = 0
Differentiating Eq. (1) partially w.r.t. x and y,
∂f ∂u ∂f ∂v
⋅
+
⋅
=0
∂u ∂x ∂v ∂x
∂f ∂u ∂f ∂v
⋅
+
⋅
=0
∂u ∂y ∂v ∂y
Eliminating
f
f
and
from Eqs (2) and (3),
u
v
u v
x x
=0
u v
y y
∂ (u , v)
= 0.
∂ ( x, y )
... (1)
... (2)
... (3)
Partial Differentiation
u
x
v
x
u
y
=0
v
y
4.101
⎡Interchanging rows and columns ⎤
⎢of the second determinant
⎥
⎣
⎦
u, v)
= 0.
x, y )
u , v)
for each of the following functions:
x, y )
(ii) u = x sin y, v = y sin x
Example 1: Find the Jacobian
(i) u = x2 – y2 , v = 2xy
y2
y2
, v=
x
x
Solution: (i) u = x2 – y2
u
= 2x
x
(iv) u =
(iii) u = x +
v = 2xy
v
= 2y
x
v
= 2x
y
u
= –2 y
y
u
x
u, v)
=
v
( x, y )
x
(ii)
x+ y
, v = tan–1x + tan–1y .
1 – xy
u = x sin y
u
y
2x – 2 y
=
= 4 (x2 + y2)
v
2 y 2x
y
u
= sin y
x
v = y sin x
∂v
= y cos x
∂x
∂u
= x cos y
∂y
v
= sin x
y
u
x
u, v)
=
v
x, y )
x
u
sin y
x cos y
y
=
y cos x sin x
v
y
= sin x sin y – xy cos x cos y
(iii)
u = x+
y2
x
v=
u
x
y2
x2
v
y2
= 2
x
x
1
y2
x
Engineering Mathematics
4.102
u 2y
=
y
x
v 2y
=
y
x
u
x
u, v )
=
v
x, y )
x
u
y2 2 y
1− 2
y
x
x
=
2
v
−y
2y
y
x
x2
x+ y
1 xy
∂u (1 − xy ) − ( x + y )( − y )
=
∂x
(1 − xy ) 2
1+ y2
=
(1 xy ) 2
(iv) u =
2 y3
x3
2 y3 2 y
=
x
x3
v = tan–1x + tan–1y
1
v
=
x 1+ x2
1
v
=
y 1+ y2
u (1 xy ) ( x y )( x )
=
y
(1 xy ) 2
=
2y
x
1+ x2
(1 xy ) 2
u
1+ y2
1+ x2
2
y
(1 xy ) (1 xy ) 2
=
v
1
1
2
y
1+ x
1+ y2
u
x
( u, v )
=
v
( x, y )
x
1
(1 xy ) 2
1
(1 xy ) 2 = 0.
Example 2: Find the Jacobian for each of the following functions:
(i) x = r cosq ,
y = r sin q
(ii) x = a coshq cosf ,
y = a sinhq sinf .
Solution:
(i)
x = r cosq
x
= cosq
r
y = r sinq
y
= sin q
r
x
q
y
= r cosq
q
r sinq
( x, y )
J=
=
( r ,q )
x
r
y
r
x
cosq − r sinq
q
=
= r cos 2 + r sin 2 = r
y
sinq r cosq
q
Partial Differentiation
(ii)
y = a sinh q sin f
∂y
= a coshq sinf
∂q
y
= a sinhq cosf
f
x = a cosh q cosf
∂x
= a sinh q cos f
∂q
x
a coshq sinf
f
J=
( x, y )
=
(q , f )
4.103
x
q
x
f
y
q
y
f
=
a sinhq cosf − a coshq sinf
a coshq sin f
a sinhq cosf
= a2 (sinh2q cos2f + cosh2q sin2f)
= a2 [sinh2q (1– sin2f) + (1+ sinh2q ) sin2f]
= a2 (sinh2q + sin2f )
a2
(cosh 2q 1 1 cos2f )
2
a2
(cosh 2q
2
cos2f )
Example 3: Find the Jacobian for each of the following functions:
(i) u = xyz,
v = x2 + y2 + z2 ,
w=x+y+z
(ii) x = r sinp cose,
y = r sinp sine,
z = r cosp
wu
vw
uv
,
y
=
,
.
z=
(iii) x =
v
u
w
Solution:
(i)
u = xyz
v = x2 + y2 + z2
w=x+y+z
u
= yz
x
v
= 2x
x
w
=1
x
u
= xz
y
v
= 2y
y
w
=1
y
u
= xy
z
v
= 2z
z
w
=1
z
J=
( u, v , w )
=
( x, y, z )
u u
x y
v v
x y
w w
x y
u
z
yz xz xy
v
z = 2x 2 y 2z
w 1 1 1
z
Engineering Mathematics
4.104
= yz(2y – 2z) – xz (2x – 2z) + xy(2x – 2y)
= 2y2z – 2yz2 – 2x2z + 2xz2 + 2x2y – 2xy2
= 2[x2(y – z) – x(y2 – z2) + yz(y – z)]
= 2(y – z) [x2 – x (y + z) + yz]
= 2(y – z) [y(z – x) – x(z – x)] = 2(y – z) (z – x) (y – x)
(ii)
x = r sinq cosf
x
= sinq cosf
r
y = r sinq sinf
y
= sinq sinf
r
z = r cosq
z
= cosq
r
x
= r cosq cosf
q
y
= r cosq sinf
q
z
q
x
f
y
= r sinq cosf
f
z
=0
f
r sinq sinf
r sinq
∂x ∂x ∂x
∂r ∂ ∂f
J=
∂( x, y, z ) ∂y ∂y ∂y
=
∂( r , ,f ) ∂r ∂ ∂f
∂z ∂z ∂z
∂r ∂ ∂f
sinq cosf r cosq cosf − r sinq sinf
= sinq sinf r cosq sinf
− r sinq
cosq
r sinq cosf
0
sinq cosf
cosq cosf
− sinq sinf
= r sinq sinf
cosq sinf
sinq cosf
2
− sinq
cosq
0
= r2 [cosq (cosq sinq cos2f + sinq cosq sin2f]
+ sinq (sin2q cos2f + sin2q sin2f)]
= r2 (sinq cos2q + sin3q )
vw
u
∂x − vw
= 2
u
∂u
∂x w
=
∂v u
(iii) x =
wu
v
∂y w
=
∂u v
∂y −wu
= 2
v
∂v
y=
uv
w
∂z v
=
∂u w
∂z u
=
∂v w
z=
Partial Differentiation
∂y u
=
∂w v
x v
=
w u
( x, y, z )
=
( u, v , z )
J=
4.105
z
uv
= 2
w w
x
u
y
u
z
u
x
v
y
v
z
v
x
vw
w
u2
y
w
=
w
v
z
v
w
w
vw wu uv
1
wu uv
= 2 2 2 vw
u v w
vw wu uv
w
v
u
u
wu u
v
v2
u
uv
w w2
1 1 1
u2v2w 2
1
1 1
u2v2w 2
1 1 1
= –1(1 – 1) – 1 (– 1 – 1) + 1(1 + 1) = 4.
Example 4: Verify J. J* = 1 for the following functions:
(i) x = eu cos v,
y = eu sin v
(ii) x = u,
y = u tan v,
Solution: (i) x = e cos v
z = w.
u
y = eu sin v
x
= e u cos v
u
y
= e u sin v
u
x
v
y
= e u cos v
v
e u sin v
∂x
∂( x, y ) ∂u
=
J=
∂( u ,v ) ∂y
∂u
= e 2u
∂x
e u cos v − e u sin v
∂v
= u
∂y
e sin v
e u cos v
∂v
cos v − sin v
sin v cos v
= e2u (cos2 v + sin2 v) = e2u
Writing u, v in terms of x and y,
y
x2 + y2 = e2u
= tan v
x
⎛ y⎞
v = tan −1 ⎜ ⎟
⎝x⎠
∂v
−y
=
∂x x 2 + y 2
u=
1
log( x 2 + y 2 )
2
∂u
x
=
∂x x 2 + y 2
Engineering Mathematics
4.106
∂v
x
= 2
∂y x + y 2
∂u
y
= 2
∂y x + y 2
( u, v )
=
( x, y )
J* =
=
J. J* =
Hence,
(ii)
u
x
v
x
u
y
v
y
1
∂ ( x , y ) ∂ ( u, v )
= e 2 u . 2 u =1
.
e
∂ ( u, v ) ∂ ( x , y )
y = u tan v
z=w
x
=1
u
y
= tan v
u
z
=0
u
x
=0
v
∂y
= u sec 2 v
∂v
z
=0
v
J=
y
=0
w
( x, y, z )
=
( u, v , w )
y
x + y2
x
2
x + y2
2
1
1
x2 + y2
=
=
( x 2 + y 2 )2 x 2 + y 2 e 2u
x=u
x
=0
w
x
x + y2
=
y
2
x + y2
2
z
=1
w
x
u
y
u
z
u
x
v
y
v
z
v
x
w
1
0
0
y
2
= tan v u sec v 0 = u sec 2 v
w
0
0
1
z
w
Writing u, v, w in terms of x, y and z,
u=x
tanv =
y y
=
u x
w=z
⎛ y⎞
v = tan −1 ⎜ ⎟
⎝x⎠
∂u
=1
∂x
∂v
−y
= 2
∂x x + y 2
w
=0
x
∂u
=0
∂y
∂v
x
=
∂y x 2 + y 2
w
=0
y
Partial Differentiation
v
=0
z
u
=0
z
J* =
=
Hence, J . J * =
4.107
(u, v, w)
=
( x, y , z )
w
=1
z
u u u
1
0
0
x y z
y
x
−
v v v =
0
x2 + y2 x2 + y2
x y z
0
0
1
w w w
x y z
x
1
1
=
=
x2 + y2
⎡ ⎛ y ⎞ 2 ⎤ u sec 2 v
x ⎢1 + ⎜ ⎟ ⎥
⎢⎣ ⎝ x ⎠ ⎥⎦
∂ ( x , y , z ) ∂ ( u, v , w )
1
.
= 1.
= u sec 2 v.
∂ ( u, v , w ) ∂ ( x , y , z )
u sec 2 v
Example 5: If x = uv and y =
u+ v
( u, v )
, find
.
u v
( x, y)
x
=v
u
u+v
u v
y
2v
=
u (u v ) 2
x
=u
v
y
2u
=
v (u v ) 2
Solution:
y=
x = uv
J=
( x, y )
=
( u, v )
=
x
u
y
u
x
v
u
v
=
2v
2u
y
(u v ) 2 (u v ) 2
v
4uv
(u v ) 2
We know that
( u, v ) ( x , y )
.
=1
( x , y ) ( u, v )
Hence,
Example 6: If u =
∂ ( u, v ) ( u − v ) 2
.
=
4uv
∂( x, y )
3zx
2 yz
4xy
, v=
, w=
, find
y
x
z
( x, y, z )
.
( u, v , w )
Engineering Mathematics
4.108
3zx
y
v 3z
=
x
y
v
3 zx
= 2
y
y
2 yz
x
u
2 yz
= 2
x
x
u 2z
=
y
x
u 2y
=
z
x
v=
Solution: u =
∂v 3 x
=
∂z
y
u u
x y
( u, v , w )
v v
=
( x, y, z )
x y
w w
x y
=
4xy
z
w 4y
=
x
z
∂w 4 x
=
∂y
z
w
4 xy
= 2
z
z
w=
u
−2 yz 2 z 2 y
z
x
x
x2
v
3 z −3 zx 3 x
=
z
y
y
y2
w
4 y 4 x −4 xy
z
z
z
z2
−2 yz ⎛ 12 x 2 yz 12 x 2
−
⎜
yz
x2 ⎝ y2 z2
⎞ 2 z ⎛ −12 xyz 12 xy ⎞ 2 y ⎛ 12 xz 12 xyz ⎞
−
+
⎟− ⎜
⎟+
⎜
⎟
2
yz ⎠ x ⎝ yz
zy 2 ⎠
⎠ x ⎝ yz
= 96
We know that
∂ ( u, v , w ) ∂ ( x , y , z )
⋅
=1
∂ ( x , y , z ) ∂ ( u, v , w )
∂( x, y, z ) 1
= .
∂(u, v, w ) 96
Hence,
Example 7: If u = x2 – y2, v = 2xy, where x = r cosp and y = r sinp, find
Solution:
u = x2 – y2
v = 2xy
u
= 2x
x
u
2y
y
v
= 2y
x
v
= 2x
y
∂u
∂(u, v ) ∂x
=
∂( x, y ) ∂v
∂x
∂u
2x
∂y
=
2y
∂v
∂y
x = r cosq
x
= cos q
r
2y
= 4 (x2 + y2) = 4r2
2x
y = r sinq
y
= sin q
r
( u, v )
.
( r ,q )
Partial Differentiation
x
q
Hence,
y
= r cos q
q
r sin q
x
( x, y )
r
=
y
(r, q )
r
x
q
y
q
4.109
=
cos q
sin q
− r sin q
= r cos2q + r sin2q = r
r cos q
( u, v )
( u, v ) ( x , y )
=
.
= 4r2. r = 4r3.
( r ,q )
( x , y ) ( r ,q )
Example 8: If u = ex cos y, v = ex sin y, where, x = lr + sm and y = mr - ls, verify
chain rule of Jacobians, l, m being constants.
Solution:
u = ex cos y
u
= e x cos y
x
u
e x sin y
y
u
x
( u, v )
=
v
( x, y )
x
u
y
v
y
v = ex sin y
v
= e x sin y
x
v
= e x cos y
y
=
e x cos y
− e x sin y
e x sin y
e x cos y
= e2x (cos2 y + sin2 y) = e2x
x = lr + sm
x
=l
r
x
=m
s
( x, y )
=
( r , s)
( u, v ) ( x , y )
.
( x , y ) ( r , s)
Now, u = elr + ms cos (mr – s l)
∂u
= le lr + ms cos ( mr − sl )
∂r
− me lr + ms sin ( mr − sl )
y = mr – ls
y
=m
r
y
l
s
x
r
y
r
x
l m
s
= – (l2 + m2)
=
y
m l
s
e 2 x (l 2
m2 )
v = elr + ms sin (mr – s l)
∂v
= le lr + ms sin ( mr − sl )
∂r
+ me lr + ms cos ( mr − sl )
… (1)
Engineering Mathematics
4.110
= lex cos y – mex sin y
= lex sin y + mex cos y
∂u
= me lr + ms cos ( mr – sl )
∂s
+ le lr + ms sin ( mr – sl )
∂v
= melr + ms sin (mr − sl )
∂s
− lelr + ms cos (mr − l s )
= mex cos y + lex sin y
(u , v)
=
(r , s)
u
r
v
r
= e2 x
= mex sin y – lex cos y
u
le x cos y me x sin y
s
= x
v
le sin y me x cos y
s
l cos y m sin y
l sin y m cos y
me x cos y le x sin y
me x sin y le x cos y
m cos y l sin y
m sin y l cos y
= e2x [(l cos y – m sin y) (m sin y – l cos y)
– (l sin y + m cos y) (m cos y + l sin y)]
= e2x [lm cos y sin y – l 2 cos2 y – m2 sin2 y + lm sin y cos y
– lm sin y cos y – l2 sin2 y –m2 cos2 y – lm sin y cos y]
= e2x [–l2(cos2 y + sin2 y) – m2(cos2 y + sin2 y)]
Hence,
= – e2x(l2 + m2)
∂ (u , v) ∂ ( x, y ) ∂ (u , v)
.
⋅
=
∂ ( x, y ) ∂ ( r , s ) ∂ ( r , s )
Example 9: If x = vw , y = uw , z = uv and u = r sinp cosf, v = r sinp sinf,
( x, y, z )
w = r cosp, find
.
( r ,q , f )
Solution:
x = vw
y = uw
z = uv
x
=0
u
y 1
=
u 2
z 1 v
=
u 2 u
x 1
=
v 2
w
v
y
=0
v
x 1
=
w 2
v
w
y 1
=
w 2
∂x
∂u
∂ ( x, y, z ) ∂y
=
∂ (u , v, w) ∂u
∂z
∂u
∂x
∂v
∂y
∂v
∂z
∂v
∂x
∂w
∂y
1
=
∂w
2
∂z
1
∂w
2
1
2
0
w
u
z 1 u
=
v 2 v
u
w
w
v
w
u
0
v
u
1 u
2 v
z
=0
w
1
2
v
w
1
2
u
w
0
Partial Differentiation
=−
and
w⎛ 1
⎜−
v ⎝ 4
1
2
∂u
∂r
∂u
∂q
∂ ( u, v , w )
∂v
=
∂ ( r ,q , f )
∂r
∂v
∂q
∂w
∂r
∂w
∂q
v ⎞ 1
⎟+
w⎠ 2
v
w
⎛1
⎜⎝
4
∂u
∂f
sin q cos f
∂v
= sin q sin f
∂f
cos q
∂w
∂f
4.111
w⎞
1
⎟ =
v ⎠
4
r cos q cos f
r cos q sin f
− r sin q
− r sin q sin f
r sin q cos f
0
= sinq cos f (r2 sin2 q cos f) – r cos q cos f (–r sin q cos q cos f)
– r sin q sin f (–r sin2 q sin f – r cos2 q sin f) = r2 sin q
∂ ( x , y , z ) ∂ ( x , y , z ) ∂ ( u, v , w ) 1 2
.
=
= r sin q.
∂ ( r ,q , f ) ∂ ( u, v , w ) ∂ ( r ,q , f ) 4
Hence,
Example 10: Determine whether the following functions are functionally
dependent or not. If functionally dependent, find the relation between them.
(i) u = ex sin y,v = ex cos y
(ii) u = sin–1x + sin–1y, v
(iii) u =
x y
x+z
y=
(iv) u = x + y – z,
x 1 y2
y 1 x2
x+z
y+z
v = x – y + z,
w = x2 + y2 + z2 – 2yz
(v) u = xy + yz + zx,
v = x2 + y2 + z2,
w=x+y+z
(vi) u = x2e–y cosh z,
v = x2e–y sinh z,
w = 3x4e–2y.
Solution: (i)
u = ex sin y
u
= e x sin y
x
u
= e x cos y
y
u
x
( u, v )
=
v
( x, y )
x
u
y
v
y
v = ex cos y
v
= e x cos y
x
v
e x sin y
y
=
e x sin y
e x cos y
e x cos y − e x sin y
= ex(–sin2 y – cos2 y) = –ex
Hence, u and v are functionally independent.
0
Engineering Mathematics
4.112
(ii) u = sin–1x + sin–1y
y 1 x2
⎛ −2 x ⎞
∂v
xy
2
= 1 − y 2 + y ⎜⎜
⎟ = 1− y −
2 ⎟
∂x
1− x2
⎝ 2 1− x ⎠
∂v
− xy
⎛ −2 y ⎞
= x⎜
+ 1 − x2 =
+ 1 − x2
⎟
2
2
⎜ 2 1− y ⎟
∂y
1
−
y
⎝
⎠
1
1
u
1
=
x
1 x2
u
1
=
y
1 y2
( u, v )
=
( x, y )
x 1 y2
v
u
x
v
x
u
1 x2
y =
xy
v
1 y2
1 x2
y
xy
1 x
2
1 y
2
1
1 y2
xy
1 y
2
xy
1
1 x
2
1 y2
1 x2
=0
Hence, u and v are functionally dependent.
Relation between u and v
sin–1 x = a,
sin–1 y = b,
Let
v
x = sin a
y = sin b
x 1 y2
y 1 x 2 = sina cosb + sin b cosa
= sin(a + b )
= sin(sin–1 x + sin–1 y) = sin u
x y
x+z
v=
(iii) u =
,
x+z
y+z
Since number of functions are less than the number of variables, for functional dependence, we must have,
( u, v )
( u, v )
( u, v )
=0,
=0
=0
( x, y )
( y, z )
( z, x)
u ( x z) ( x y)
y z
=
=
2
x
( x + z)
( x + z )2
v
1
=
x y z
u
1
=
y x z
v
y
u
z
v ( y z) ( x z)
y x
=
=
2
z
( y + z)
( y + z )2
( x y)
( x + z )2
u
x
( u, v )
=
v
( x, y )
x
y+z
( x z )2
=
1
v
y
y z
u
y
1
x+z
( x z)
( y + z )2
=0
( x z)
( y + z )2
Partial Differentiation
u
y
( u, v )
=
v
( y, z )
y
( u, v )
=
( z, x)
u
z
v
z
u
z
=
v
z
1
x+z
y x
( x z )2
( x z)
( y + z )2
y x
( y + z )2
y x
u
( x + z)2
x
=
v
y x
x
( y + z)2
=0
y z
( x + z)2
=0
1
y z
Hence, u and v are functionally dependent.
Relation between u and v
x y
x z
u=
,
v=
,
x+z
y+z
u+
(iv)
1 (x
=
v
4.113
y) ( y
x+z
z)
1 y z
=
v x+z
=1
u=x+y–z
u
=1
x
u
=1
y
v=x–y+z
v
=1
x
v
1
y
w = x2 + y2 + z2 – 2yz
w
= 2x
x
w
2 y 2z
y
u
z
v
=1
z
w
z
∂u
∂x
∂(u, v, w ) ∂v
=
∂( x, y, z ) ∂x
∂w
∂x
1
2z 2 y
∂u ∂u
∂y ∂z
1
1
−1
1
1
1
∂v ∂v
= 1
−1
1
= 1
−1
−1
[By ( − 1) c3 ]
∂y ∂z
2x 2 y − 2z 2z − 2 y
2x 2 y − 2z 2 y − 2z
∂w ∂w
∂y ∂z
=0
Hence, u and v are functionally dependent.
Relation among u, v and w
u + v = 2x
u+v
x=
2
u v
2
2
u v
2
2
x2
( y z )2
u – v = 2y –2z
u v
y z
2
Engineering Mathematics
4.114
(v)
1
( 2u 2 + 2v 2 ) = x 2 + y 2 + z 2 − 2 yz
4
u2 + v2 = 2w
u = xy + yz + zx
v = x2 + y2 + z2
u
v
= y+z
= 2x
x
x
u
v
= z+x
= 2y
y
y
w=x+y+z
w
=1
x
w
=1
y
u
= x+ y
z
w
=1
z
v
= 2z
z
( u, v , w )
=
( x, y, z )
u
x
v
x
u
y
v
y
w
x
w
y
u
z
v
z
w
z
y+z z+x x+ y
= 2x
2 y 2z
1
1
1
= 2 (y + z) (y – z) – 2 (z + x) (x – z) + 2 (x + y) (x – y)
=0
Hence, u, v and w are functionally dependent.
Relation among u, v and w
w2 = (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx) = v + 2u
(vi)
u = x2e–ycosh z
u
= 2 xe y cosh z
x
u
x 2 e y cosh z
y
v = x2e–ysinh z
v
= 2 xe y sinh z
x
v
x 2 e y sinh z
y
w = 3x4e–2y
w
= 12 x 3 e 2 y
x
w
6x4e 2 y
y
u
= x 2 e y sinh z
z
v
= x 2 e y cosh z
z
w
=0
z
(u , v, w)
( x, y , z )
u u
x y
v v
x y
w w
x y
u
z
v
z
w
z
2 xe y cosh z
2 xe y sinh z
12 x 3 e
2y
x 2 e y cosh z x 2 e y sinh z
x 2 e y sinh z
6x4e
2y
x 2 e y cosh z
0
= 12 x7e– 4y(cosh2 z – sinh2 z) – 12x7e–4y(cosh2 z – sinh2 z) = 0
Hence, u, v and w are functionally dependent.
Partial Differentiation
4.115
Relation among u, v and w
3u2 – 3v2 = 3(x4e–2ycosh2 z – x4e–2ysinh2 z) = 3x4e–2y = w.
Exercise 4.6
(u, v)
for each
( x, y )
of the following functions:
(i) u = x + y, v = x – y
(ii) u = x2, v = y2
(iii) u = 3x + 5y, v = 4x –3y
y x
(iv) u =
, v = tan–1y – tan–1x
1 + xy
(v) u = x sin y, v = y sin x.
1. Find the Jacobian
⎡ Ans.:
⎢
−1
⎢
( i)
(ii) 4 xy
2
⎢
⎢
(iii) − 29
(iv ) 0
⎢
( v ) sin x sin y − xy cos x cos
⎢⎣
⎤
⎥
⎥
⎥
⎥
⎥
y⎥⎦
2. Find the Jacobian for each of the
following functions:
(i) x = eu cos v,
y = eu sin v
(ii) x = u(1 – v),
y = uv
u+v
y=
(iii) x = uv,
.
u v
Ans. :
(i) e 2u
(ii) u (iii)
4uv
(u v ) 2
3. Find the Jacobian for each of the
following functions:
yz
zx
xy
, v=
, w=
x
y
z
(ii) u = xyz, v = xy + yz + zx,
w=x+y+z
(iii) u = x2, v = sin y, w = e–3z
1 2 2
(iv) x = (u – v ), y = uv, z = w.
2
(i) u =
⎡ Ans. :
⎤
⎢
⎥
(i)
4
⎢
⎥
⎢
(ii) (x − y ) (y − z ) (z − x )⎥
⎢
⎥
(iii) −6e− 3z x cos y
⎢
⎥
⎢
⎥
1
(iv) 2
⎢
⎥
2
u +v
⎣
⎦
.
4. Verify that J J* = 1 for the following
functions:
y2
y2
(i) u = x +
, v=
x
x
(ii) x = u(1 – v), y = uv
(iii) x = sinq cosf, y = sinq sinf.
5. If u = x + y + z, uv = y + z, uvw = z,
( x, y , z )
evaluate
.
(u, v, w)
[Ans. : u2v]
3
3
2
6. If u + v = x + y, u + v2 = x3 + y3,
show that
(u , v) 1 ( y 2 x 2 )
=
.
( x, y ) 2 uv(u v)
(u , v, w)
x
if u =
,
( x, y , z )
1 r2
y
z
v=
, w=
where
2
1 r
1 r2
r2 = x2 + y2 + z2.
5
- ⎤
⎡
2
2
⎢ Ans. : (1- r ) ⎥
⎣
⎦
8. If u = x + y + z, u2v = y + z,
∂ (u , v, w)
u3w = z, show that
= u −5.
∂ ( x, y , z )
7. Calculate
9. Show that
∂ (u , v)
= 6r 3 sin 2q , if
∂ ( r ,q )
Engineering Mathematics
4.116
u = x2 – 2y2, v = 2x2 – y2 and x = r cosq,
y = r sinq .
10. Determine whether the following
function are functionally dependent or not. If functionally depenthem.
x y
(i) u =
,
x+ y
x+ y
v=
y
x2 y 2
2xy
, v= 2
2
2
x +y
x + y2
(iii) u = sin x + sin y, v = sin (x + y)
x y
xy
(iv) u =
,
v=
x+ y
( x + y)2
(ii) u =
(v) u = x + y + z, v = x2 + y2 + z2,
w = x3 + y3 + z3 – 3xyz
(vi) u = xeysin z, v = xeycos z,
w = x2e2y
3x 2
2( y + z )
(vii) u =
, v=
,
2( y + z )
3( x y ) 2
x y
.
w=
x
⎡ Ans.:
⎢
2−v
⎢
(i) Dependent, u =
v
⎢
2
2
⎢
(ii) Dependent, u + v = 1
⎢
(iii) Indepentent
⎢
⎢
(iv ) Dependent, 4v = 1 − u 2
⎢
⎢
( v ) Dependent, 2w = u(3v − u 2 )
⎢
⎢ (vi) Dependent, u 2 + v 2 = w
⎢
2
⎣ ( vii) Dependent, uvw = 1
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎦
4.8.2 Errors and Approximation
Let u = f (x, y) be a continuous function of x and y. If d x and d y are small increments
in x and y respectively and d u is corresponding increment in u, then
u + d u = f (x + d x, y + d y)
d u = f (x + d x, y + d y) f (x, y)
f
f
= dx+ d y
x
y
[expanding by Taylor’s theorem and ignoring higher powers and products of d x and
d y.]
u
u
dx+ d y
x
y
For a function u = f (x, y, z ) of three variables, we have
∂u
∂u
∂u
du = d x+ d y+
dz
∂x
∂y
∂z
du =
or
If d x is the error in x, then
(i) d x is known as Absolute error in x.
(ii) d x is known as Relative error in x.
x
100
d x is known as Percentage error in x.
(iii)
x
Example 1: Find the percentage error in calculating the area of a rectangle
when an error of 3% is made in measuring each of its sides.
Partial Differentiation
4.117
Solution : Let a and b be the side of the rectangle and A is its area.
A = ab
log A = log a + log b
1
1
1
d A= da+ db
A
a
b
100
100
100
dA=
da+
db
A
a
b
Percentage error in measuring each of its sides is 3.
100
d A = 3+3 = 6
A
Hence, percentage error in calculating the area = 6%.
Example 2: Find the percentage error in the area of an ellipse when an error of
1.5% is made in measuring its major and minor axes.
Solution : Let 2a and 2b are the major and minor axes of the ellipse and A is its area.
A = p ab
log A = log p + log a + log b
1
1
1
d A= 0+ da+ db
A
a
b
100
100
100
dA=
da+
db
A
a
b
Percentage error in measuring its major and minor axes is 1.5%.
100d A
= 1.5 + 1.5 = 3
A
Hence, percentage error in area of ellipse = 3%.
Example 3: The focal length of mirror is found from the formula
2
f
Find the percentage error in f if u and v are both in error by 2% each.
Solution:
2
f
1
v
1
u
2
1
1
df
dv
du
f2
v2
u2
2 100
1 100
1 100
1
1
.
df = .
dv+ .
d u = (2) + (2)
f
f
v v
u u
v
u
2
100
d f =2
f
Hence, percentage error in f = 2%.
1
v
1
u
2
2
f
1
v
1
.
u
Engineering Mathematics
4.118
Example 4: If D =
a2 c2
+ , find the percentage error in D if error in measurb 2
1
% and in measuring b and c are 1% each.
2
a2 c2
Solution:
D=
+
b
2
2a
a2
2c
dD
da
db
dc
2
b
2
b
ing a is
a 2 100
100
1 ⎛ 2a 2 . 100
100 ⎞
.
dD = ⎜
da −
d b + c2
d c⎟
⎠
b b
c
D
D⎝ b
a
But
100
da =
a
100
dD =
D
=
Hence, percentage error in D =
1 100
100
,
db =
dc =1
2 b
c
⎞
1 ⎛ 2a 2 1 a 2
. −
+ c2 ⎟
D ⎜⎝ b 2 b
⎠
c2
c2
2bc 2
= 2
= 2
2
D a
c
2a + bc 2
+
b
2
2bc 2
.
2a 2 + bc 2
Example 5: Find the possible percentage error in computing the parallel
resistance R of three resistance R1, R2, R3, if R1, R2, R3, are each in error by 1.2%.
1 1
1
1
= +
+
R R1 R2 R3
Solution:
−
1
1
1
1
d R = − 2 d R1 − 2 d R2 − 2 d R3
2
R
R1
R2
R3
1 . 100
1 100
1 100
1 100
dR= .
d R1 + .
d R2 + ⋅
d R3
R R
R1 R1
R2 R2
R3 R3
100
100
100
d R1 =
d R2 =
d R3 = 1.2
R1
R2
R3
1 100
1
1
1
.
d R = (1.2) + (1.2) + (1.2)
R R
R1
R2
R3
⎛1
1
1 ⎞
= 1.2 ⎜ +
+
⎝ R1 R2 R3 ⎟⎠
⎛1⎞
= 1.2 ⎜ ⎟
⎝R⎠
100
d R = 1.2
R
Hence, percentage error in R = 1.2%
But
Partial Differentiation
4.119
Example 6: The resonant frequency in a series electrical circuit is given by
f =
1
. If the measurement of L and C are in error by 2%
2p LC
and −1% respectively, find the percentage error in f.
Solution:
1
f =
2p LC
1 1
1
− log L − log C
log f = log
2p 2
2
1
1 1
1 1
d f = 0 − . d L − . dC
f
2 L
2 C
100
1 100
1 100
d f =− .
dL− .
dC
f
2 L
2 C
100
dL
L
But
2,
100
df
f
Hence, percentage error in f =
100
dC
C
1
1
1
(2)
( 1)
2
2
0.5
0.5%.
Example 7: If z = 2xy2 − 3x2y and x increases at the rate of 2 cm/s as it passes
through x = 3 cm. Show that if y is passing through y = 1 cm, y must decrease at
32
the rate of
cm / s in order that z remains constant.
15
Solution:
z = 2xy2 − 3x2y
But
z = (2y2 − 6xy) x + (4xy − 3x2) y
x = 3, y = 1, dx = 2, z = 0
0 = (2 − 18) 2 + (12 − 27) y
32
15
32
cm/s.
Hence, y must decrease at the rate of
15
dy
Example 8: If ez = sec x cos y and errors of magnitude h and -h are made in
estimating x and y, where x and y are found to be
corresponding error in z.
3
and
6
respectively, find the
Engineering Mathematics
4.120
e z = sec x cos y
z log e = log sec x + log cos y
Solution:
dz
1
sec x tan x d x
sec x
tan x d x tan y d y
p
p
,y
, d x h, d y
3
6
p
p
d z = tan ( h) − tan ( − h)
3
6
But
x
= 3h −
Hence, error in z =
1
( sin y )d y
cos y
1
3
( − h) = 3h +
h
h
3
=
4h
3
=
4 3
h
3
4 3
h.
3
Example 9: In calculating the volume of a right circular cone, errors of 2%
and 1% are made in height and radius of base respectively. Find the % error in
volume.
Solution: Let r be the radius of base, h height and V volume of the right circular
cone.
1
V = p r2h
3
p
+ 2 log r + log h
3
1
2
1
dV = 0 + d r + d h
V
r
h
100
100
100
dV = 2
dr+
d h = 2 (1) + 2 = 4
V
r
h
log V = log
Hence, percentage error in volume = 4%.
Example 10: The diameter and the altitude of a can in the shape of a 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 possible error in the values
computed for volume and lateral surface.
Solution: Let d and h are diameter and height of the cylinder respectively and V
be its volume.
Partial Differentiation
4.121
2
p
⎛d ⎞
V = p ⎜ ⎟ h = d 2h
⎝2⎠
4
p
log V = log + 2 log d + log h
4
1
2
1
dV = 0 + d d + d h
V
d
h
2
1
1
dV = d d + d h
d
h
V
But d = 4 cm, h = 6 cm, d = 0.1 cm, h = 0.1 cm.
V=
1
dV
V
p 2
p
d h = × ( 4) 2 × 6 = 75.36 cm3
4
4
2
1
0.1
0.1
4
6
V = 75.36 × 0.067
= 5.05 cm3
Hence, error in volume = 5.05 cm
Lateral surface area,
S = 2p rh
3
= p dh
log S = log p + log d + log h
1
1
1
dS = 0+ dd + dh
S
d
h
1
1
1
dS = dd + dh
S
d
h
But d = 4 cm, h = 6 cm, d = 0.1 cm, h = 0.1 cm.
S = p 4 6 = 75.36 cm2
1
1
S=
S
4
0.1) +
1
6
0.1)
S = 75.36 0.0416
= 3.14 cm2
Hence, error in lateral surface area = 3.14 cm2.
Example 11: A balloon is in the form of a right circular cylinder of radius
1.5 m and height 4 m and is surmounted by hemispherical ends. If the radius is
increased by 0.01 m and the height by 0.05 m, find the percentage change in the
volume of the balloon.
Engineering Mathematics
4.122
Solution: Radius of the cylinder, r = 1.5 m
Volume of the cylinder = p r 2 h
Height of the cylinder, h = 4 m
2
Volume of the hemisphere = p r 3
3
Volume of the balloon,
2
2
4
V = p r 2h + p r3 + p r3 = p r 2h + p r3
3
3
3
4
d V = p ( 2rh d r + r 2d h) + p (3r 2d r )
3
r = 1.5 m, h = 4 m, d r = 0.01 m, d h = 0.05 m
But
d V = p [2 × 1.5 × 4 × 0.01 + (1.5) 2 (0.05)] + 4p (1.5) 2 (0.01)
= p (0.12 + 0.1225 + 0.09) = 3.225p
(1.5) 2 4
V
4
(1.5)3
3
= p (9 + 4.5) = 13.5p
Percentage change in the volume,
dV
3.225
× 100 =
× 100
V
13.5
= 2.389%.
Example 12: At a distance 120 feet from the foot of a tower, the elevation of its
top is 60 . If the possible error in measuring the distance and elevation are 1 inch
and 1 minute respectively, find the approximate error in the calculated height of
the tower.
Solution: Let h, x and q are height, horizontal distance and angle of elevation of the
tower respectively.
tan q =
h
x
h = x tan q
log h = log x + log tan q
1
1
1
dh = dx+
sec2 q q
h
x
tan q
60°
Fig. 4.36
1
But x = 120 ft. and q = 60o, h = 120 tan 60° = 120 3, dx = 1 inch = ft.,
12
1 p
radians
dq = 1 minute = .
60 180
1
1 .1
1 .1 . p
.4
dh =
+
120 12
120 3
3 60 180
h = 0.284 ft.
Hence, approximate error in height = 0.284 ft
Partial Differentiation
4.123
Example 13: In estimating the cost of pile of bricks measured 2 m 15 m
1.2 m, the top of the pile is stretched 1% beyond the standard length. If the
count is 450 bricks in 1 cubic m and bricks cost Rs. 450 per thousand, find the
approximate error in the cost.
Solution: Let l, b and h be the length, breadth and height of the pile and V be its
volume.
V
log V
1
dV
V
100
dV
V
= lbh
= log l + log b + log h
1
1
1
= dl + db+ dh
l
b
h
100
100
100
=
dl +
db+
dh
l
b
h
Top is stretched 1% beyond the standard length.
Percentage error in height i.e.
100
d h = 1 and l = 0 , b = 0
h
100
dV = 0 + 0 +1
V
V
l b h 2 15 1.2
dV =
=
=
100
100
100
= 0.36 cubic metre
Hence, error in number of bricks = 0.36
Cost of 162 bricks
162
450
1000
450 = 162
72.9
Hence, error in cost = Rs. 72.90
2
3
Example 14: Evaluate (3.82) + 2 (2.1)
1
5
using theory of approximation.
1
Solution: Let z = (x 2 + 2y 3 ) 5
4
4
−
−
1
1
d z = ( x 2 + 2 y 3 ) 5 ( 2 x )d x + ( x 2 + 2 y 3 ) 5 (6 y 2 )d y
5
5
4
−
1
= ( x 2 + 2 y 3 ) 5 (22 xd x + 6 y 2d y )
5
Consider, x = 4, x = 3.82
4 = 0.18, y = 2, y = 2.1
2 = 0.1
Engineering Mathematics
4.124
Hence,
dz=
4
−
1
(32) 5 [2( 4)( −0.18) + 6( 2) 2 (0.1) = 0.012
5
1
Approximate value = z + d z = (32) 5 + 0.012 = 2.012.
1
2
3
Example 15: Evaluate (1.99) (3.01) (0.98) 10 using approximation.
1
Solution: Let u = x 2 y 3 z 10
1
log z
10
1
2
3
1 1
du = d x+ d y+
dz
u
x
y
10 z
log u = 2 log x + 3 log y +
Consider, x = 2,
x = 1.99
Hence,
y = 3,
y = 3.01
2 = 0.01,
3 = 0.01,
z = 1,
z = 0.98
1 = 0.02
1
10
u = 22 ·33 ·1 = 108
1
1
d u = ( −0.01) + 0.01 + ( −0.02)
108
10
u = 0.216
Approximate value = u + u = 108
= 107.784 .
0.216
1
Example 16: Find the approximate value of [(0.98)2 + (2.01) 2 + (1.94) 2 ]2 .
Solution: Let u = x 2 + y 2 + z 2
u = x2 + y2 + z2
u2 = x2 + y2 + z 2
2u d u = 2 x d x + 2 y d y + 2 z d z
u d u = x d x + yd y + zd z
Consider, x = 1,
x = 0.98
1 = 0.02,
y =2
y = 2.01
and
2 = 0.01,
u = (1) 2 + (2) 2 + (2) 2 = 3
Hence, u u = 1 ( 0.02) + 2 (0.01) + 2 ( 0.06) =
u = 0.04
Approximate value of u = u + u = 3
0.04 = 2.96
0.12
z =2
z = 1.94
2 = 0.06
Partial Differentiation
4.125
Example 17: If the sides and angles of a plane triangle vary in such a way that
a
b
c
its circum radius remains constant, prove that
+
+
= 0, where
cos A cos B cos C
da, db, dc are smaller increments in the sides a, b, c respectively.
Solution: From the sine rule,
a
b
c
=
=
sin A sin B sin C
We know that, circum radius R =
Considering,
a
b
c
=
=
2 sin A 2 sin B 2 sin C
a
R=
2 sin A
1
a cos A
da
dA
2 sin A
2 sin 2 A
R=0
a cos A
da
0
dA
2sin A 2sin 2 A
dR
But R is constant,
da
a
=
d A = 2 Rd A
cos A sin A
db
b
=
d B = 2 Rd B
cos B sin B
dc
c
=
d C = 2 Rd C
cos C sin C
Similarly,
and
da
db
dc
+
+
= 2 R (d A + d B + d C )
cos A cos B cos C
= 2Rd ( A + B + C )
= 2 R d (p ) = 0.
Example 18: If
be the area of the triangle, prove that the error in resultD 1
1
1
1 ⎞
ing from a small error in side c is given by d D = ⎛⎜ +
+
−
⎟ d c,
4 ⎝s s−a s−b s−c⎠
where D
s ( s a ) ( s b) ( s c ) .
Solution:
s( s a)( s b)( s c)
1
log Δ = [log s + log( s − a) + log( s − b) + log( s − c)]
2
1
1 ⎡1
1
1
1
⎤
dΔ = ⎢ ds+
d ( s − a) +
d ( s − b) +
d ( s − c) ⎥
Δ
2 ⎣s
s−a
s−b
s−c
⎦
=
1 ⎡d s d s − d a d s − d b d s − d c ⎤
+
+
+
s − c ⎥⎦
s−a
s−b
2 ⎢⎣ s
Engineering Mathematics
4.126
But
s=
1
( a + b + c), where a and b are constant.
2
Thus, d a = 0, d b = 0, d s =
Hence,
dc
2
dc
⎡
⎤
−d c⎥
Δ ⎢d c
dc
dc
dΔ = ⎢ +
+
+ 2
⎥
2 ⎢ 2 s 2( s − a) 2( s − b)
s−c ⎥
⎣
⎦
1
1
1 ⎤
Δ ⎡1
= ⎢ +
+
−
d c.
4 ⎣ s s − a s − b s − c ⎥⎦
Exercise 4.7
1. In calculating the volume of right
circular cone, errors of 2.75%
and 1.25% are made in height and
radius of the base. Find the % error
in volume.
[Ans. : 5.25%]
2. The height of a cone is H = 30 cm,
the radius of base R = 10 cm. How
will the volume of the cone change,
if H is increasing by 3 mm while R
is decreasing by 1 mm?
⎡ Hint : d h = 3 mm = 0.3 cm, ⎤
⎢
d r = − 1 mm = − 0.1 cm ⎥⎦
⎣
[Ans. : decreased by 10p cm3]
3. How is the relative change in
V = p r2h related to relative change in
r and h? How are percentage changes
related?
dV 2
1 ⎤
⎡
⎢ Ans. : relative change V = r d r + h d h⎥
⎢
⎥
and percentage change in volume ⎥
⎢
⎢
⎥
= (2% change in radius)
⎢
⎥
+ (1% change in height)
⎢⎣
⎥⎦
4. In calculating the total surface area
of a cylinder, error of 1% each are
made in measuring the height and
the base radius. Find % error in
calculating the total surface area.
[Ans. : 2%]
5. In calculating the volume of a
right circular cylinder, errors of 2%
and 1% are made in measuring the
height and base radius respectively.
Find the percentage error in calculating volume of the cylinder.
[Ans. : 4%]
6. Find the percentage error in calculating the area of a rectangle when
an error of 2% is made in measuring each of its sides.
[Ans. : 4%]
7. Find the percentage error in calculating the area of a rectangle when
an error of 1% is found in measuring its sides.
[Ans. : 2%]
8. If R1 and R2 are two resistances in
parallel, their resistance R is giv1 1
1
=
+
en by
. If there is an
R R 1 R2
error of 2% in both R1 and R2, find
percentage error in R.
[Ans. : 2%]
Partial Differentiation
9. One side of a rectangle is a = 10 cm
and the other side is b = 24 cm.
How will the diagonal l of the rectangle change if a is increased by
4 mm and b is decreased by 1 mm?
Ans. :
4
cm
65
10. The resistance R of a circuit was
E
found by using the formula I = .
R
If there is an error of 0.1 ampere in
reading I and 0.5 volts in reading E,
find the corresponding percentage
error in R when I = 15 ampere and
E = 100 volts.
[Ans. : − 0.167%]
11. The radius and height of a cone
are 4 cm and 6 cm respectively.
What is the error in its volume
if the scale used in taking the
measurement is short by 0.01 cm
per cm?
Hint : r = 4 0.01 = 0.04 cm,
h = 6 0.01 = 0.06 cm
3
[Ans. : 0.96p cm ]
12. In estimating the cost of a pile of
bricks measured as 6
50
4,
the top is stretched 1% beyond
its standard length. If the count is
12 bricks per ft3 and bricks cost
Rs. 100 per 1000, find the approximate error in the cost.
[Ans. : Rs. 43.20]
13. Show that the error in calculating
the time period of a pendulum at
any place is zero, if an error of μ %
is made in measuring its length and
gravity at that place.
Hint : T = 2p
l
g
4.127
14. At distance 20 meters from the
foot of a tower, the elevation of its
top is 60°. If the possible error in
measuring distance and elevation
are 1 cm and 1 minute, find the
approximate error in calculating
height.
[Ans. : 0.040]
15. The diameter and the altitude of a
right circular cylinder are measured
as 24 cm and 30 cm respectively.
There is an error of 0.1 cm in each
measurement. Find the possible
error in the volume of the cylinder.
[Ans. : 50.4p cm.]
16. If the measurements of radius, base
and height of a right circular cone
are changed by −1% and 2%, show
that there will be no error in the
volume.
1
17. If f = x 2 y 3 z 10 , find the approximate value of f, when x = 1.99,
y = 3.01 and z = 0.98.
[Ans. : 107.784]
18. If f = x3 y2 z4, find the approximate
value of f, when x = 1.99, y = 3.01,
z = 0.99 .
[Ans. : 68.5202]
1
19. If f (160 x 3 y 3 ) 3 , find the
approximate value of f (2.1, 2.9)
− f (2, 3).
[Ans. : 0.016]
20. If f = e , find the approximate
value of f, when x = 0.01, y = 1.01,
z = 2.01.
[Ans. : 1.02]
xyz
1
21. Find [(2.92)3 + (5.87)3 ]5 approximately by using the theory of
approximation.
[Ans. : 2.96]
Engineering Mathematics
4.128
4.8.3 Maxima and Minima
Let u = f (x, y) be a continuous function of x and y. Then u will be maximum at x = a,
y = b, if f (a, b) > f (a + h, b + k) and will be minimum at x = a, y = b, if f (a, b)
< f (a + h, b + k) for small positive or negative values of h and k.
The point at which function f (x, y) is either maximum or minimum is known as
stationary point. The value of the function at stationary point is known as extreme
(maximum or minimum) value of the function f (x, y).
Working rule: To determine the maxima and minima (extreme values) of a function f (x, y).
∂f
∂f
= 0 and
= 0 simultaneously for x and y.
Step I: Solve
∂x
∂y
Step II: Obtain the values of r =
∂2 f
∂2 f
∂2 f
, s=
,t= 2.
2
∂x ∂y
∂x
∂y
Step III: (i) If rt – s2 > 0 and r < 0 (or t < 0 ) at (a, b), then f (x, y) is maximum at (a, b)
and the maximum value of the function is f (a, b).
(ii) If rt – s2 > 0 and r > 0 (or t > 0) at (a, b), then f (x, y) is minimum at (a, b) and
the minimum value of the function is f (a, b).
(iii) If rt – s2 < 0 at (a, b), then f (x, y) is neither maximum nor minimum at (a, b).
Such point is known as saddle point.
(iv) If rt – s2 = 0 at (a, b), then no conclusion can be made about the extreme values
of f (x, y) and further investigation is required.
Example 1: Show that the minimum value of f ( x , y ) = xy +
Solution:
f ( x, y ) = xy +
a3 a3
+
x
y
Step I: For extreme values,
∂f
a3
= y− 2 =0
∂x
x
∂f
a3
= x− 2 = 0
∂y
y
Solving Eqs (1) and (2),
x2 y = a3
and
xy2 = a3
Solving Eqs (3) and (4),
x=y
Substituting in Eq. (3),
x3 = a3
x=a
y=a
Stationary point is (a, a).
a3 a3
+
is 3a 2 .
x
y
... (1)
... (2)
... (3)
... (4)
Partial Differentiation
4.129
Step II:
∂ 2 f 2a 3
= 3
∂x 2
x
2
∂ f
s=
=1
∂x ∂y
r=
t=
∂ 2 f 2a 3
= 3
∂y 2
y
At (a, a), r = 2, s = 1, t = 2
Step III: At (a, a),
rt – s2 = (2) (2) – (1)2 = 3 > 0
Hence, f (x, y) is minimum at (a, a).
⎛1 1⎞
f min = a 2 + a 3 ⎜ + ⎟ = 3a 2 .
⎝a a⎠
Example 2: Find the stationary value of x3 + y3 – 3axy, a > 0.
Solution:
f (x, y) = x3 + y3 – 3axy
Step I: For extreme values,
∂f
= 3 x 2 − 3ay = 0
∂x
... (1)
∂f
= 3 y 2 − 3ax = 0
∂y
... (2)
From Eq. (1),
y=
x2
a
Substituting in Eq. (2),
x4 – a3x = 0
x (x – a) (x + ax + a2) = 0
x = 0, x = a
Then y = 0, y = a.
2
Hence, stationary points are (0, 0) and (a, a).
Step II:
∂2 f
= 6x
∂x 2
∂2 f
= −3a
s=
∂x ∂y
r=
t =
∂2 f
= 6y
∂y 2
Step III: At (0, 0)
rt – s2 = (0) (0) – (–3a)2 = –9a2 < 0
Engineering Mathematics
4.130
Hence, function f (x, y) is neither maximum nor minimum at (0, 0).
At (a, a)
rt – s2 = (6a) (6a) – (–3a)2 = 27a2 > 0
and
r = 6a > 0
Hence, function f (x, y) is minimum at (a, a).
fmin = a3 + a3 – 3a3
= –a3.
Example 3: Find the extreme values of u = x3 + 3xy2 – 3x2 – 3y2 + 7, if any.
Solution:
u = x3 + 3xy2 – 3x2 – 3y2 + 7
Step I: For extreme values,
∂u
= 3x 2 + 3 y 2 − 6 x = 0
∂x
x2 + y2 – 2x = 0
... (1)
∂u
= 6 xy − 6 y = 0
∂y
and
6y (x – 1) = 0
y = 0, x = 1
Substituting y = 0 in Eq. (1),
x2 – 2x = 0, x = 0, 2
Stationary points are (0, 0), (2, 0)
Substituting x = 1 in Eq. (1),
1 + y2 – 2 = 0, y2 = 1, y = ± 1
Stationary points are (1, 1), (1, –1)
∂2u
= 6 x − 6 = 6( x − 1)
∂x 2
∂2u
= 6y
s=
∂x ∂y
r=
Step II:
t=
∂2u
= 6 x − 6 = 6( x − 1)
∂y 2
Step III:
(x, y)
r
s
t
rt – s2
(0, 0)
−6
0
−6
36 > 0 and r < 0
maximum
(2, 0)
6
0
6
36 > 0 and r > 0
minimum
(1, 1)
0
6
0
−36 < 0
neither maximum nor
minimum
(1, −1)
0
−6
0
−36 < 0
neither maximum nor
minimum
Conclusion
Partial Differentiation
4.131
Hence, u is maximum at (0, 0) and minimum at (2, 0).
umax = 0 + 7 = 7
and
umin = 23 + 3(2) (0)2 – 3(2)2 – 3(0)2 + 7 = 3.
Example 4: Find the extreme values of u = x3 + y3 – 63 (x + y) + 12xy.
Solution:
u (x, y) = x3 + y3 – 63x – 63y + 12xy
∂u
= 3 x 2 − 63 + 12 y
∂x
∂u
= 3 y 2 − 63 + 12 x
∂y
Step I:
For extreme values
∂u
=0
∂x
3x2 – 63 + 12y = 0, 3x2 + 12y = 63
x2 + 4y = 21
... (1)
∂u
=0
∂y
and
3y2 – 63 + 12x = 0, 12x + 3y2 = 63
4x + y2 = 21
Solving Eqs (1) and (2),
x 2 + 4 y = 4 x + y 2 , x 2 − y 2 = 4( x − y )
( x + y )( x − y ) − 4( x − y ) = 0
( x − y )( x + y − 4) = 0
x + y − 4 = 0, x − y = 0
y = 4 − x, y = x
Substituting y = 4 – x in Eq. (1),
x 2 + 4(4 − x) = 21
x 2 − 4 x − 5 = 0, ( x + 1)( x − 5) = 0
x = −1, 5
y = 5, − 1
Stationary points are (–1, 5), (5, –1).
Putting y = x in Eq. (1),
x 2 + 4 x − 21 = 0, ( x + 7)( x − 3) = 0
x = −7, 3
y = −7, 3
Stationary points are (–7, –7), (3, 3).
Hence, all stationary points are: (–1, 5), (5, –1) (–7, –7), (3, 3).
... (2)
Engineering Mathematics
4.132
Step II:
r=
∂2u
= 6x
∂x 2
s=
∂2u
= 12
∂x ∂y
t=
∂2u
= 6y
∂y 2
Step III:
(x, y)
r
s
t
rt – s2
(−1, 5)
−6
12
30
−324 < 0
neither maximum
nor minimum
(5, −1)
30
12
−6
−324 < 0
neither maximum
nor minimum
(−7, −7)
−42
12
−42
1620 > 0 and r < 0
maximum
(3, 3)
18
12
18
180 > 0 and r > 0
minimum
Conclusion
Hence, at (−7, −7), u is maximum.
umax = (−7)3 + (−7)3 – 63 (−7 −7) + 12(−7) (−7) = 2156.
and at (3, 3), u is minimum.
umin = 33 + 33 – 63 (3 + 3) + 12(3) (3) = −216.
Example 5: Find the stationary value of xy (a – x – y).
Solution:
f (x, y) = xy (a – x – y)
= axy – x2y – xy2
Step I: For extreme values,
∂f
= ay − 2 xy − y 2 = 0
∂x
∂f
= ax − x 2 − 2 xy = 0
∂y
From Eqs (1) and (2), we get
y (a – 2x – y) = 0
y = 0, a – 2x – y = 0
and
x (a – x – 2y) = 0
x = 0, a – x – 2y = 0
Considering four pairs of equations
y=0
x=0
y=0
a – x – 2y = 0
a – 2x – y = 0
x=0
a – 2x – y = 0
a – x – 2y = 0
... (1)
... (2)
Partial Differentiation
4.133
Solving these equations, following pairs of values of x and y are obtained.
a a
(0, 0), (0, a), (a, 0), ⎛⎜ , ⎞⎟
⎝3 3⎠
∂2 f
Step II:
r = 2 = −2 y
∂x
∂2 f
= a − 2x − 2 y
s=
∂ x ∂y
t=
∂2 f
= −2 x
∂y 2
Step III:
(x, y)
r
s
t
rt – s2
2
Conclusion
(0, 0)
0
a
0
–a <0
neither maximum nor
minimum
(0, a)
–2a
–a
0
– a2 < 0
neither maximum nor
minimum
(a, 0)
0
–a
–2a
– a2 < 0
neither maximum nor
minimum
2a
3
a
3
2a
3
a2
>0
3
⎛a
⎜⎝ ,
3
a⎞
⎟
3⎠
maximum or minimum
⎛a a⎞
Hence, f (x, y) is maximum or minimum at ⎜ , ⎟ depending on whether a > 0 or
⎝3 3⎠
a < 0.
a a⎛
a a ⎞ a3
f extreme = ⋅ ⎜ a − − ⎟ =
.
3 3⎝
3 3 ⎠ 27
Example 6: Examine the function u = x3 y2 (12 – 3x – 4y) for extreme values.
Solution:
u (x, y) = 12x3 y2
x4 y2 – 4x3 y3
∂u
= 36 x 2 y 2 − 12 x 3 y 2 − 12 x 2 y 3
∂x
= 12 x 2 y 2 (3 − x − y )
∂u
= 24 x 3 y − 6 x 4 y − 12 x 3 y 2 = 6 x 3 y (4 − x − 2 y )
∂y
Step I:
For extreme values,
∂u
=0
∂x
2 2
12x y (3 – x – y) = 0
x = 0, y = 0, x + y = 3
... (1)
Engineering Mathematics
4.134
∂u
=0
∂y
6x3 y (4 – x – 2y) = 0
x = 0, y = 0, x + 2y = 4
Considering six pairs of equations,
x=0
y=0
x=0
x + 2y = 4
y=0
x + 2y = 4
x=0
x+y=3
y=0
x+y=3
x+y=3
x + 2y = 4
and
... (2)
Solving these equations, following pairs of stationary points are obtained
(0, 0), (0, 2), (4, 0), (0, 3), (3, 0), (2, 1)
∂2u
= 72 xy 2 − 36 x 2 y 2 − 24 xy 3 = 12 xy2(6 – 3x – 2y)
∂x 2
∂2u
s=
= 72 x 2 y − 24 x 3 y − 36 x 2 y 2 = 12 x 2 y (6 − 2 x − 3 y )
∂x ∂y
r=
Step II:
t=
∂2u
= 24 x 3 − 6 x 4 − 24 x 3 y = 6 x 3 (4 − x − 4 y )
∂y 2
Step III:
(x, y)
r
s
t
rt – s2
(0, 0)
0
0
0
0
no conclusion
no conclusion
Conclusion
(0, 2)
0
0
0
0
(4, 0)
0
0
0
0
no conclusion
(0, 3)
0
0
0
0
no conclusion
(3, 0)
0
0
0
0
no conclusion
(2, 1)
−48
−48
−96
2304 > 0 and r < 0
maximum
Hence, function is maximum at (2, 1)
umax = (23) (12) (12 – 6 – 4) = 16.
Example 7: Find the extreme values of sin x + sin y + sin(x + y).
Solution:
Step I:
f (x, y) = sin x + sin y + sin(x + y)
∂f
= cos x + cos( x + y )
∂x
∂f
= cos y + cos( x + y )
∂y
Partial Differentiation
4.135
For extreme values,
∂f
= 0, cos x + cos( x + y ) = 0
∂x
∂f
= 0, cos y + cos ( x + y ) = 0
∂y
From Eqs (1) and (2), we get
cos x + cos( x + y ) = cos y + cos( x + y )
cos x = cos y, x = y
... (1)
... (2)
Substituting x = y in Eq. (1),
cos x + cos 2 x = 0,
cos x = − cos 2 x = cos(p − 2 x) or cos(p + 2 x)
x = p − 2 x or p + 2 x
p
x = , −p
3
p
y = , −p
3
⎛p p ⎞
Thus, ⎜ , ⎟ , (−p , − p ) are stationary points.
⎝3 3⎠
∂2 f
r = 2 = − sin x − sin( x + y )
Step II:
∂x
∂2 f
= − sin( x + y )
s=
∂x ∂y
t=
∂2 f
= − sin y − sin( x + y )
∂y 2
Step III:
(x, y)
(
,
3 3
,
r
s
3
)
0
3
2
0
rt – s2
t
3
0
9
> 0 and r < 0
4
0
Conclusion
maximum
no conclusion
Hence, function is maximum at ⎛⎜ p , p ⎞⎟ .
⎝3 3⎠
f max = sin
3
3
3 3 3
2p
p
p
+ sin + sin
=
+
+
=
.
3
3
3
2
2
2
2
Example 8: Find the extreme values of sin x sin y sin (x + y).
Solution:
f (x, y) = sin x sin y sin (x + y)
Engineering Mathematics
4.136
Step I:
∂f
= sin y [cos x sin( x + y ) + sin x cos( x + y )]
∂x
1
= sin y sin (2 x + y ) = [cos 2 x − cos( 2 x + 2 y )]
2
∂f
= sin x [ cos y sin ( x + y ) + sin y cos( x + y )]
∂y
1
= sin x sin ( x + 2 y ) = [ cos 2 y − cos(2 x + 2 y ) ]
2
For extreme values,
∂f
= 0, cos 2 x − cos 2( x + y ) = 0
∂x
and
... (1)
∂f
= 0, cos 2 y − cos 2( x + y ) = 0
∂y
From Eqs (1) and (2), we get
... (2)
and
cos 2x = cos 2y, x = y
Putting x = y in Eq. (1),
cos 2x − cos 2 (x + x) = 0, cos 2x = cos 4x, cos 2x = 2 cos2 2x – 1
2 cos2 2x – cos 2x – 1 = 0
1± 1+ 8
4
1
= 1, –
2
cos 2x = cos 0,
cos 2x =
cos 2x = cos
x = 0,
x=
y = 0,
y=
2
3
3
3
⎛p p ⎞
Thus, (0, 0), ⎜ , ⎟ are stationary points.
⎝3 3⎠
Step II:
∂2 f
= − sin 2 x + sin 2( x + y ) = 2 sin y cos (2 x + y )
∂x 2
∂2 f
s=
= sin 2( x + y )
∂x ∂y
r=
t=
∂2 f
= – sin 2 y + sin 2( x + y ) = 2 sin x cos ( x + 2 y )
∂y 2
Partial Differentiation
4.137
Step III:
(x, y)
r
s
t
rt – s2
(0, 0)
0
0
0
0
⎛
⎞
⎜⎝ , ⎟⎠
3 3
3
3
2
3
9
> 0 and r < 0
4
Conclusion
no conclusion
maximum
⎛p p ⎞
Hence, function is maximum at ⎜ , ⎟ .
⎝3 3⎠
p
p
2p
sin sin
3
3
3
3 3 3 3 3
=
⋅
⋅
=
.
2 2 2
8
f max = sin
Example 9: Find the points on the surface z 2 = xy + 1 nearest to the origin.
Also find that distance.
Solution: Let P (x, y, z) be any point on the surface z2 = xy + 1.
Its distance from the origin is given by
D = ( x2 + y 2 + z 2 )
D2 = x2 + y 2 + z 2
Since P lies on the surface z 2 = xy + 1
D 2 = x 2 + y 2 + xy + 1
Let
f ( x, y ) = x 2 + y 2 + xy + 1
Step I: For extreme values,
∂f
= 2x + y = 0
∂x
∂f
= 2y + x = 0
∂y
Solving Eqs (1) and (2),
x = 0 and y = 0
Stationary point is (0, 0).
Step II:
∂2 f
=2
∂x 2
∂2 f
s=
=1
∂x ∂y
r=
t=
∂2 f
=2
∂y 2
... (1)
... (2)
Engineering Mathematics
4.138
Step III: At (0, 0)
rt – s2 = (2) (2) – (1)2 = 3 > 0
r=2>0
and
Thus, f (x, y) is minimum at (0, 0) and hence D is minimum at (0, 0).
At
x = 0, y = 0
z 2 = xy + 1 = 1
z = ±1
Hence, D is minimum at (0, 0, 1) and (0, 0,
1) .
Thus, the points (0, 0, 1) and (0, 0, –1) on the surface z 2 = xy + 1 are nearest to the
origin.
Minimum distance = 0 + 0 + 1 = 1.
Example 10: A rectangular box open at the top is to have a volume 108 cubic
meters. Find the dimensions of the box if its total surface area is minimum.
Solution: Let x, y and z be the dimensions of the box. Let V and S be its volume and
surface area respectively.
V = xyz
S = xy + 2 xz + 2 yz
V
Substituting z = ,
xy
V
V
2V 2V
S = xy + 2 x ⋅ + 2 y ⋅ = xy +
+
xy
xy
y
x
∂S
2V
= y− 2
∂x
x
Step I:
∂S
2V
= x− 2
∂y
y
For extreme values,
S
=0
x
2V
... (1)
y
0
x2
∂S
=0
∂y
and
x
2V
y2
Solving Eqs (1) and (2),
y=
2V
x2
0
... (2)
Partial Differentiation
4.139
⎛ x4 ⎞
x = 2V ⎜ 2 ⎟ = 0
⎝ 4V ⎠
⎛
x3
x ⎜1 −
⎝ 2V
⎞
⎟⎠ = 0
1
x = (2V ) 3
y=
1
2V
2V
3 [since x
0 being the side of the box]
(
2
V
)
=
=
2
x2
(2V ) 3
1
1
⎡
⎤
Hence, stationary point is ⎣⎢(2V ) 3 , (2V ) 3 ⎦⎥
Step II:
⎡
1
1
⎤
r=
∂ 2 S 4V
= 3
∂x 2
x
s=
∂2 S
=1
∂x ∂y
t=
∂ 2 S 4V
= 3
y
∂y 2
At ⎣⎢(2V ) 3 , (2V ) 3 ⎦⎥ , r =
4V
4V
= 2 > 0, s = 1, t =
=2
2V
2V
1
1
⎡
⎤
Step III: At ⎢⎣(2V ) 3 , (2V ) 3 ⎥⎦ ,
rt s 2
(2)(2) (1) 2
0 and r
3
2
0
1
Hence, S is minimum at
But
x = y = (2V ) 3
V = 108 m3
1
x
and
z=
y
(2 108) 3
6
V
108
=
=3
xy 6 × 6
Hence, dimensions of the box which make its total surface area S minimum are
x = 6, y = 6, z = 3.
Example 11: Show that the rectangular solid of maximum volume that can be
inscribed in a given sphere is a cube.
Engineering Mathematics
4.140
Solution: Let x, y, z be the length, breadth and height of the rectangular solid and
V be its volume.
V = xyz
... (1)
Let given sphere is
x2 + y 2 + z 2 = a2
z2
a2
x2
y2
Substituting in Eq. (1),
V = xy a 2 − x 2 − y 2
V 2 = x 2 y 2 (a 2 − x 2 − y 2 )
Let
Step I:
and
f ( x, y ) V 2
f
x
x 2 y 2 (a 2
x2
y 2 [2 x(a 2
x2
2 xy 2 (a 2
2x2
y2 )
... (2)
y 2 ) x 2 ( 2 x)]
y2 )
∂f
= x 2 [2 y (a 2 − x 2 − y 2 ) + y 2 (−2 y )]
∂y
= 2 x 2 y (a 2 − x 2 − 2 y 2 )
For extreme values,
∂f
= 0, 2 xy 2 (a 2 − 2 x 2 − y 2 ) = 0
∂y
x = 0, y = 0, 2 x 2 + y 2 = a 2
and
f
y
0, 2 x 2 y (a 2
x2
2 y2 )
x = 0, y = 0, x 2 + 2 y 2 = a 2
But x and y are the sides of the rectangular solid, therefore cannot be zero.
Solving 2 x 2 + y 2 = a 2 and x 2 + 2 y 2 = a 2
x=
a
3
, y=
z = a2 −
a
3
a2 a2
a
−
=
3
3
3
Thus, stationary points are ⎛ a , a , a ⎞ .
⎜⎝
⎟
3 3 3⎠
... (3)
... (4)
Partial Differentiation
4.141
∂2 f
= 2a 2 y 2 − 12 x 2 y 2 − 2 y 4
∂x 2
∂2 f
= 4a 2 xy − 8 x 3 y − 8 xy 3
s=
∂x ∂y
r=
Step II:
t=
a ⎞
⎛ a
,
Step III: At ⎜
⎟,
⎝ 3
3⎠
∂2 f
= 2a 2 x 2 − 2 x 4 − 12 x 2 y 2
2
∂y
2a 4 4a 4 2a 4
8a 4
−
−
=−
3
3
9
9
4
4
4
4 a 8a 8a
4a 4
s=
−
−
=−
3
9
9
9
4
4
4
2a
2a 12a
8a 4
t=
−
−
=−
3
9
9
9
r=
rt − s 2 =
64a 4 16a 4 48a 4
−
=
>0
81
81
81
rt – s2 > 0 and r < 0
Therefore, f (x, y) i.e. v2 is maximum at x = y = z and hence, v is maximum when
x = y = z, i.e. rectangular solid is a cube.
Exercise 4.8
1. Examine maxima and minima of the
following functions and find their
extreme values:
(i) 2 + 2x + 2y − x2 − y2
2
(ii) x2 y2
xy
y2
2
2
(iii) x + y + xy + x
y
(iv) x2 + y2 + 6x = 12
x y)
(v) x3 y2
(vi) xy (3a x y)
(vii) x3 + 3xy2 x2 y2 + 4
x y)2
(viii) x4 + y4
4
2
2
(ix) x + x y + y
(x) x4 + y4 − 4a2 xy
(xi) y4 − x4 + 2(x2 − y2)
(xii) x3 + 3x2 + y2 + 4xy
(xiii) x2y − 3x2 − 2y2 − 4y + 3
(xiv) x4 − y4 − x2 − y2 + 1.
⎡ Ans.: (i) max. at (1, 1); 4
⎢
(ii) max. at (0, 0); 0
⎢
⎢
(iii) min. at ( − 2, 3); − 2
⎢
⎤
⎥
⎥
⎥
⎥
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣
⎥
⎥
⎥
⎛ 1 1⎞ 1
(v) max. at ⎜ , ⎟ ;
⎥
⎝ 2 3 ⎠ 432
⎥
3
⎥
(vi) max. at ( a, a)); a
⎥
(vii) max. at (0, 0); 4
⎥
⎥
(viii) min. at 2, − 2
⎥
⎥
and − 2 , 2 ; − 8
⎥
⎥
(ix) min. at (0, 0); 0
⎥
4⎥
(x) min. at ( a, a) and ( − a, a); a
⎥
⎥
(xi) No extreme values
⎥
(xii) No extreme values
⎥
⎥
(xiii) max. at (0, − 1); 5
⎥
(xiv) max. at (0, 0); 1, min at
⎥
⎥
⎛ 1
1 ⎞ 1
⎥
,
±
±
;
⎟
⎜
2
2⎠ 2
⎝
⎦⎥
(iv) min at ( − 3, 0); 3
(
(
)
)
4.142
Engineering Mathematics
2. A rectangular box, open at the top, is
to have a volume of 32 cc. Find the
dimensions of the box requiring least
materials for its construction.
[Ans. : 4, 4, 2]
3. Divide 120 into three parts so that the
sum of their products taken two at a
time shall be maximum.
[Hint : f = xy + yz + zx where x + y +
z = 120]
[Ans. : 40, 40, 40]
4. The sum of three positive numbers
is ‘a’. Determine the maximum value
of their product.
a3
a a a
Ans. : at , ,
27
3 3 3
5. Find the volume of the largest rectangular parallelopiped that can be
inscribed in an ellipsoid
x2 y 2 z 2
+
+
= 1.
a 2 b2 c2
⎡ Hint : Let 2x, 2y , 2z be the sides of ⎤
⎢ the parallelopiped, then its volume ⎥
⎢
⎥
⎢
⎥
x2 y 2
⎢
⎥
v = 8 xyz = 8 xy 1 − 2 − 2
a
b
⎣
⎦
Ans. :
8abc
6. Prove that area of a triangle with constant perimeter is maximum when
the triangle is equilateral.
[Hint :
Area
s ( s a ) ( s b) ( s c )
where 2s = a + b + c, c = 2s a b,
s is constant]
7. Find the shortest distance from origin to the surface xyz2 = 2.
[Ans. : 2]
8. Find the shortest distance from the
origin to the plane x − 2y − 2z = 3.
[Ans. : 1]
9. Find the shortest distance between
the lines
x−3
=
1
x +1
=
7
y−5 z −7
=
and
1
−2
y +1 z +1
=
.
−6
1
⎡ Ans. : 2 29 ⎤
⎣
⎦
10. Find the maximum value of
cos A cos B cos C, where A, B, C are
angles of a triangle.
Ans. :
p
3
p
3
p
3
1
8
3 3
4.8.4 Lagrange’s Method of Undetermined Multipliers
Let f (x, y, z) be a function of three variables x, y, z, and the variables be connected by
the relation
f ( x, y , z ) = 0
... (1)
Suppose we wish to find the values of x, y, z, for which f (x, y, z) is stationary (maximum and minimum)
For this purpose, we construct an auxiliary equation
(x, y, z)
... (2)
F (x, y, z) = f (x, y, z) +
Differentiating partially w.r.t. x, y, z and equating to zero,
∂F ∂f
∂f
... (3)
=
+l
=0
∂x ∂x
∂x
Partial Differentiation
4.143
∂F ∂f
∂f
=
+l
=0
∂y ∂y
∂y
... (4)
∂F ∂f
∂f
... (5)
=
+l
=0
∂z ∂z
∂z
Solving Eqs (1), (3), (4) and (5), we can find the values of x, y, z and l for which
f (x, y, z) has stationary value. This method of obtaining stationary values of f (x, y, z) is
called the Lagrange’s method of undetermined multipliers and Eqs (3), (4) and (5) are
called Lagrange’s equations. The term l is called undetermined multiplier.
Example 1: Find the point on the plane ax + by + cz = p at which the function
f = x2 + y2 + z2 has a minimum value and find this minimum f.
Solution:
f = x2 + y2 + z2
... (1)
ax + by + cz = p
f (x, y, z) = ax + by + cz − p = 0
Lagrange’s equations
f
f
+l
=0
x
x
... (2)
2x + la = 0
x=
la
2
f
f
+l
=0
y
y
2y + lb = 0
y=
lb
2
f
f
+l
=0
z
z
2z + lc = 0
z=
lc
2
Substituting x, y, z in Eq. (2),
⎛ −l a ⎞
⎛ −l b ⎞
⎛ −l c ⎞
a⎜
+b⎜
+c⎜
=p
⎝ 2 ⎟⎠
⎝ 2 ⎟⎠
⎝ 2 ⎟⎠
la2 + lb2 + lc2 = −2p
l=
−2 p
a + b2 + c2
2
4.144
Thus,
The minimum value of
Engineering Mathematics
x=
f =
=
ap
bp
cp
, y= 2
, z= 2
2
2
2
2
a +b +c
a +b +c
a + b2 + c2
2
a2 p2
b2 p 2
c2 p2
+ 2
+ 2
2
2 2
2
2 2
(a + b + c ) (a + b + c ) (a + b 2 + c 2 ) 2
2
p 2 (a 2 + b 2 + c 2 ) 2
p2
=
.
(a 2 + b 2 + c 2 ) 2
a 2 + b2 + c2
Example 2: Find the maximum value of f = x2 y3 z4 subject to the condition
x + y + z = 5.
Solution:
f = x2 y3 z4
x+y+z=5
f (x, y, z) = x + y + z − 5 = 0
... (1)
... (2)
Lagrange’s equations
f
f
+l
=0
x
x
2xy3z4 + = 0
2xy3 z4 = −
... (3)
f
f
+l
=0
y
y
3x2 y2 z4 + = 0
3x2 y2 z4 = −
... (4)
f
f
+l
=0
z
z
4x2 y3 z3 + = 0
4x2 y3z3 = −
From Eqs (3) and (4),
2xy3 z4 = 3x2 y2 z4
2y = 3x
3
y= x
2
From Eqs
2xy3z4 = 4x2 y3 z3
z = 2x
... (5)
Partial Differentiation
4.145
Substituting y and z in Eq. (2),
x+
3
x + 2x = 5
2
9 x = 10
10
x=
9
3
3 ⎛ 10 ⎞ 5
y= x= ⎜ ⎟=
2
2⎝9⎠ 3
⎛ 10 ⎞ 20
z = 2x = 2 ⎜ ⎟ =
⎝9⎠ 9
2
Maximum value of
3
4
(210 ) (59 )
⎛ 10 ⎞ ⎛ 5 ⎞ ⎛ 20 ⎞
.
f =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ =
⎝ 9 ⎠ ⎝3⎠ ⎝ 9 ⎠
315
Example 3: Show that the rectangular solid of maximum value that can be
inscribed in a sphere is a cube.
Solution: Let 2x, 2y, 2z be the length, breadth and height of the rectangular solid.
Let r be the radius of the sphere.
Volume of solid,
Equation of the sphere,
V = 8xyz
2
2
2
… (1)
2
x +y +z =r
… (2)
f (x, y, z) = x + y + z − r = 0
2
2
2
2
Lagrange’s equation
V
f
+l
=0
x
x
8yz + .2x = 0
2 x = −8yz
2 x2 = −8xyz
… (3)
V
f
+l
=0
y
y
8xz + .2y = 0
2 y = −8xz
2 y2 = −8xyz
V
f
+l
=0
z
z
8xy + .2z = 0
… (4)
Engineering Mathematics
4.146
2 z = −8xy
2 z2 = −8xyz
… (5)
From Eqs (3), (4) and (5),
2 x2 = 2 y2 = 2 z2
x2 = y2 = z2
x=y=z
Hence, rectangular solid is a cube.
Example 4: A rectangular box open at the top is to have volume of 32 cubic
units. Find the dimensions of the box requiring least material for its construction.
Solution: Let x, y, z be the dimensions of the box.
Volume
V = xyz = 32
The box is open at the top. Therefore, its surface area
S = xy + 2xz + 2yz
f (x, y, z) = xyz – 32
… (1)
… (2)
… (3)
Lagrange’s equation
S
f
+l
=0
x
x
y + 2z + yz = 0
… (4)
S
f
+l
=0
y
y
x + 2z + xz = 0
… (5)
f
S
+l
=0
z
z
2x + 2y + xy = 0
… (6)
Multiplying Eq. (4) by x,
xy + 2 xz + lxyz = 0
xy + 2 xz + 32l = 0
32l
… (7)
xy + 2 yz + lxyz = 0
xy + 2 yz + 32l = 0
xy 2 yz
32l
… (8)
xy 2 xz
Multiplying Eq. (5) by y,
Partial Differentiation
4.147
Multiplying Eq. (6) by z,
2 xz + 2 y z + lxyz = 0
2 xz + 2 yz + 32l = 0
2 xz + 2 yz = −32l
From Eqs (7) and (8),
… (9)
xy + 2 xz = xy + 2 yz
2 xz = 2 yz
x= y
From Eqs (8) and (9),
xy + 2 yz = 2 xz + 2 yz
xy = 2 xz
y
y = 2 z, z =
2
Substituting x, y, z in Eq. (1),
y y
y
32
2
y 3 = 64
y=4
x= y=4
y
=2
2
Hence, dimensions of the box requiring least material for its construction are 4, 4, 2.
z=
Example 5: Find the maximum and minimum distances from the origin to the
curve 3x2 + 4xy + 6y2 = 140.
Solution: The distance d from the origin (0, 0) to any point (x, y) is given by
d=
x2 + y 2 , d 2 = x2 + y 2
Let f (x, y) = x2 + y2
and
Lagrange’s equations
2x +
f (x, y) = 3x2 + 4xy + 6y2 − 140
f
f
+l
=0
x
x
(6x + 4y) = 0
… (1)
f
f
+l
=0
y
y
2y +
(4x + 12y) = 0
… (2)
Engineering Mathematics
4.148
Solving Eqs (1) and (2),
Substituting
l
x
3x + 2 y
y
2x + 6 y
l
x2
3 x + 2 xy
y2
2 xy + 6 y 2
2
x2 + y 2
3x + 6 y 2 + 4 xy
2
f ( x, y )
140
in Eqs (1) and (2),
f
f
(6 x 4 y ) 0, 2 y
(4 x 12 y )
140
140
(140 3 f ) x 2 fy 0
2x
and
Substituting x =
0
… (3)
… (4)
−2fx + (140 − 6f ) y = 0
2 fy
from Eq. (3) in Eq. (4),
140 3 f
− 4 f 2 + (140 − 3 f ) (140 − 6 f ) = 0
14 f 2 − 1260 f + (1402 ) = 0
f 2 − 90 f − 1400 = 0
f = 70, 20
Thus, maximum and minimum distances are
70, 20.
Example 6: 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.
Solution: Let x and y be two parts of the wire.
x+y=b
… (1)
Let the piece of length x is bent in the form of a square so that each side is
Thus, the area of the square, A 1 =
x x x2
⋅ = .
4 4 16
x
.
4
Suppose piece of length y is bent in the form of a circle of radius r so perimeter of the
circle is y.
y
2p r = y, r =
2p
2
Thus, the area of the circle,
y2
⎛ y ⎞
=
A2 = p ⎜
.
⎝ 2p ⎟⎠
4p
Let sum of the areas is given as
f ( x, y ) =
x2 y 2
+
16 4p
Partial Differentiation
4.149
f (x, y) = x + y – b
and
Lagrange’s equations:
f
f
+l
=0
x
x
2x
l 0, x
16
8l
f
f
+l
=0
y
y
and
2y
l
4p
Substituting x and y in Eq. (1),
0, y
2pl
(–8 ) + (–2p ) = b
l=
Thus,
b
8 + 2p
8b
4b
=
8 + 2p 4 + p
2p b
pb
y=
=
8 + 2p 4 + p
x=
Substituting in f (x, y),
1
4b
f ( x, y ) =
16 4 + p
=
=
b2
(4 + p )
2
2
1+
1
+
4p
pb
4+p
2
b 2p (4 + p )
p2
=
2
4p
4p (4 + p )
b2
4(p + 4)
b2
.
4 (p + 4)
Example 7: A closed rectangular box has length twice its breadth and has constant volume V. Determine the dimensions of the box requiring least surface area.
Hence, the least value of the sum of the areas is
Solution: Let x be the breadth and y be the height of the rectangular box so length
of the box will be 2x.
Volume of the box
V = x . 2x . y = 2x2y
Volume of the box is constant
2x2y = V = constant
Surface area of the box is given by
S = 2 (2x · x + x · y + y · 2x) = 4x2 + 6xy
… (1)
… (2)
Engineering Mathematics
4.150
f (x, y) = 2x2y − V
Let
Lagrange’s equations:
… (3)
∂S
∂f
+l
=0
∂x
∂x
8x + 6y +
(4xy) = 0
2x + 3y +
(2xy) = 0
… (4)
∂S
∂f
+l
=0
∂y
∂y
6x + (2x2) = 0
and
3 x lx 2
Substituting x
3
in Eq. (4),
l
3
2
3y l 2 y
l
3
l
3
l
0, x
0
6
l
2
l
3 y, y
Substituting x and y in Eq. (1),
3
l
2
2
2
l
V
1
−36
⎛ 36 ⎞ 3
, l = −⎜ ⎟
l =
⎝V ⎠
V
3
x
3
l
V
3
36
y
2
l
V
2
36
1
3
1
3
1
3
27V
36
8V
36
1
3
1
3
3V
4
2V
9
1
3
1
1
⎛ 3V ⎞ 3 ⎛ 3V ⎞ 3
Hence, the dimensions of the box requiring least surface area are 2 ⎜ ⎟ , ⎜ ⎟ ,
⎝ 4 ⎠ ⎝ 4 ⎠
2V
9
1
3
.
Example 8: Using the Lagrange’s method find the minimum and maximum
distance from the point (1, 2, 2) to the sphere x2 + y2 + z2 = 36.
Partial Differentiation
4.151
Solution: Given sphere is x2 + y2 + z2 = 36
… (1)
Let the coordinates of any point on the sphere be (x, y, z), then its distance D from the
point (1, 2, 2) is
D
( x 1) 2 ( y 2) 2 ( z 2) 2
Let D2 = f (x, y, z) = (x − 1)2 + (y − 2)2 + (z − 2)2
f (x, y, z) = x2 + y2 + z2 − 36
and
Lagrange’s equations:
f
f
+l
=0
x
x
2 (x − 1) + (2x) = 0
(x − 1) + x = 0
… (2)
f
f
+l
=0
y
y
2 (y − 2) + (2y) = 0
and
(y − 2) + y = 0
… (3)
f
f
+l
=0
z
z
2 (z − 2) + (2z) = 0
(z − 2) + z = 0
… (4)
Multiplying Eq. (2) by x, Eq. (3) by y and Eq. (4) by z and adding,
(x2 + y2 + z2) − (x + 2y + 2z) + (x2 + y2 + z2) = 0
36 (1 + ) − (x + 2y + 2z) = 0 [Using Eq. (1)]
From Eq. (2),
1
x=
1+ l
From Eq. (3),
2
y=
1+ l
From Eq. (4),
2
z=
1+ l
Substituting x, y, z in Eq. (5),
36 (1 l )
1+ 4 + 4
1+ l
0
36 (1 + l ) = 9, (1 + l ) =
2
1
1+ l = ± ,
2
2
1
,
4
… (5)
… (6)
… (7)
… (8)
Engineering Mathematics
4.152
Substituting in Eqs (6), (7) and (8),
x = ± 2, y = ± 4, z = ± 4
Minimum distance = (2 − 1) 2 + (4 − 2) 2 + (4 − 2) 2 = 1 + 4 + 4 = 3
Maximum distance =
(−2 − 1) 2 + (−4 − 2) 2 + (−4 − 2) 2 =
9 + 36 + 36 = 9.
Example 9: Use the method of the Lagrange’s multipliers to find volume of
the largest rectangular parallelopiped that can be inscribed in the ellipsoid
x2 y2 z2
+
+ = 1.
a 2 b2 c 2
x2 y 2 z 2
+
+
=1
a 2 b2 c2
Solution:
… (1)
Let 2x, 2y, 2z be the length, breadth and height of the rectangular parallelopiped
inscribed in the ellipsoid.
Volume of the parallelopiped, V = (2x) (2y) (2z) = 8xyz.
Let
f ( x, y , z )
Lagrange’s equations:
∂V
∂f
+l
∂x
∂x
2x
8 yz + l 2
a
∂V
∂f
+l
∂y
∂y
2y
8 xz + l 2
b
∂V
∂f
+l
∂z
∂z
2z
8 xy + l 2
c
x2
a2
y2
b2
z2
c2
1
=0
= 0, 4 yz + l
x
=0
a2
… (2)
y
=0
b2
… (3)
z
=0
c2
… (4)
=0
= 0, 4 xz + l
=0
= 0, 4 xy + l
Multiplying Eq. (2) by x, Eq. (3) by y and Eq. (4) by z and adding,
⎛ x2 y 2 z 2 ⎞
12 xyz + l ⎜ 2 + 2 + 2 ⎟ = 0
b
c ⎠
⎝a
12 xyz + l = 0
l = −12 xyz
[Using Eq. (1)]
Partial Differentiation
4.153
Substituting in Eq. (2),
x
a2
4 yz 12 xyz
3x 2
a2
1
0, x
0
a
3
Similarly substituting in Eqs (3) and (4),
y=
b
3
, z=
c
3
Volume of the largest rectangular parallelopiped that can be inscribed in the ellipsoid
⎛ a ⎞ ⎛ b ⎞ ⎛ c ⎞ 8abc
V = 8 xyz = 8 ⎜
=
.
⎝ 3 ⎟⎠ ⎜⎝ 3 ⎟⎠ ⎜⎝ 3 ⎟⎠ 3 3
Exercise 4.9
1. Find stationary values of the function
f (x, y, z) = x2 + y2 + z2, given that
z2 = xy + 1.
[Ans. : (0, 0, −1), (0, 0, 1)]
2. Find the stationary value of a3 x2 +
b3 y2 + c3 z2 subject to the fulfillment of
1 1 1
the condition + + = 1, given a,
x y z
b, c are not zero.
1
⎡
⎤
⎢ Ans. : x = a (a + b + c), ⎥
⎢
⎥
1
⎢
y = (a + b + c), ⎥
⎢
⎥
b
⎢
⎥
1
⎢
z = (a + b + c) ⎥
c
⎣⎢
⎦⎥
3. Find the largest product of the numbers x, y and z when x + y + z2 = 16.
4096
Ans. :
4. Find the largest product of the numbers
x, y and z when x2 + y2 + z2 = 9.
Ans. : 3 3
5. Find a point in the plane x + 2y + 3z =
13 nearest to the point (1, 1, 1).
Ans. :
2
, 2,
2
6. Find the shortest distance from
the point (1, 2, 2) to the sphere
x2 + y2 + z2 = 36.
[Ans. : 3]
7.
the origin (0, 0) to the curve
3x2 + 3y2 + 4xy – 2 = 0.
Ans. : 2
8. Decompose a positive number a into
three parts so that their product is
Ans. :
a a a
, ,
3 3 3
xm yn zp
9.
when x + y + z = a.
Ans. :
a m+ n+ p mm nn p p
( m + n + p )m + n + p
10. Find the dimensions of a rectangular
surface area is given when
Engineering Mathematics
4.154
inscribed in the ellipse 4x2 + y2 = 36.
⎡
⎢ Ans. : (i)
⎢
⎢
(ii )
⎢
⎣
s
,
3
s
,
6
s 1 s ⎤
,
⎥
3 2 3 ⎥
s
s ⎥
,
⎥
6
6 ⎦
11. Determine the perpendicular distance
of the point (a, b, c) from the plane
lx + my + nz = 0.
⎡ Ans. : minimum distance ⎤
⎢
⎥
la + mb + nc
⎢
⎥
l 2 + m2 + n2
⎣⎢
⎦⎥
Ans. :
3 2
, 2, Area = 12
2
13. Find the volume of the largest rectangular parallelopiped that can be inscribed in the ellipsoid of revolution
4x2 + 4y2 + 9z2 = 36.
Ans. : 16 3
14. Find the extreme volume of x2 + y2 +
z2 + xy + xz + yz subject to the conditions x + y + z = 1 and x + 2y + 3z = 3.
1 1 5⎤
⎡
⎢⎣ Ans. : 6 , 3 , 6 ⎥⎦
12. Find the length and breadth of a rectangle of maximum area that can be
FORMULAE
Chain Rule
z
z
dz ∂u
dz ∂u
⋅
=
⋅
,
=
y
x
du ∂x
du ∂y
where z = f (u) and u = f (x, y)
Total Differential Coefficient
du
∂u dx
∂u dy
⋅
⋅
=
+
dt
∂x dt
∂y dt
where u = f (x, y) and x = f (t), y = y (t)
du
u dy
(ii)
= u dx +
+ u dz
dt
y dt
x dt
z dt
where u = f (x, y, z) and x = f (t),
y = y (t), z = x (t),
(i)
Composite Function of Two Variables
z
z y
= z x +
u
y u
x u
z
z x
z y
=
+
v
x v
y v
where z = f (x, y) and x = f (u, v),
y = y (u, v)
Implicit Functions
f/ x
dy
=
dx
f/ y
where f (x, y) = c and y is a function of x.
Euler’s Theorem and deductions
u
u
(i) x
+y
= nu
y
y
u
(ii) x u + y
+ z u = nu
y
x
z
2
2
2
u
u
u
(iii) x2 2 + 2xy
+ y2 2
x y
y
x
= n(n – 1)u
f (u )
u
=n
(iv) x u + y
y
f (u )
x
2
2
2
u
u
u
(v) x2 2 + 2xy
+ y2 2
x
y
y
x
= g(u) [g (u) – 1]
f (u )
where g(u) = n
f (u )
Partial Differentiation
4.155
MULTIPLE CHOICE QUESTIONS
Choose the correct alternative in each of the following:
1. If z = f (x + ay) + f (x – ay), then
(b) zxx = a2 zyy
(a) zxx = zyy
2
(c) zyy = a zxx
(d) none of these
2. If x = log (x tan–1 y), then fxy is equal to
1
(a)
(b) 0
x2
(c) 12
(d) none of these
x
x2 y 2 z 2
3. If u = x y z , then
4.
5.
6.
7.
8.
1 1 1
ux + uy + uz is equal to
(a) 0
(b) x y z
(c) x + y + z
(d) none of these
x
x
If z = cos ⎛⎜ ⎞⎟ + sin ⎛⎜ ⎞⎟ , then
⎝ y⎠
⎝ y⎠
x z + y z is equal to
x
y
(a) z
(b) 2z
(c) 0
(d) none of these
3
3
If u = log (x + y + z3 – 3xyz), then
xux + yuy + zuz is equal to
(a) 3u
(b) 2u
(c) 3
(d) none of these
If u = x2 + y2 + z2 be such that
xux + yuy + zuz = lu then, l is equal to
(a) 1
(b) 2
(c) 3
(d) none of these
If f (x, y, z) = 0, then the value of
∂x ∂y ∂z is
⋅
⋅
∂y ∂z ∂x
(a) 1
(b) –1
(c) 0
(d) none of these
y
If u(x, y) = x2 tan–1 ⎛ ⎞
⎝x⎠
x
– y2 tan–1 ⎛⎜ ⎞⎟ , x > 0, y > 0, then
⎝ y⎠
2
x2
2
2
u
u
u
+ 2xy
+ y2 2 is equal to
2
x y
y
x
(a) 0
(b) 2u
(c) u
(d) 3u
2
9. If f (x, y) = e xy , the total differential
of the function at the point (1, 2) is
(a) e(dx + dy) (b) e4 (dx + dy)
(c) e4 (4dx + dy) (d) 4e4 (dx + dy)
dy
is equal to
10. If f (x, y) = 0, then
dx
f
f
y
x
(a)
(b)
f
f
x
y
f
f
y
(c) –
(d) – x
f
f
x
y
11. The function f (x, y) = 2x2 + 2xy – y3 has
(a) only one stationary point at
(0, 0)
(b) two stationary points at (0, 0)
1 1
,
and
6 3
(c) two stationary points at (0, 0)
and (1, –1)
(d) no stationary points
12. If z = f (x, y), dz is equal to
( )
⎛ ∂f ⎞
⎛ ∂f ⎞
(a) ⎝ ⎠ dx + ⎜ ⎟ dy
∂x
⎝ ∂y ⎠
⎛ ∂f ⎞
⎛ ∂f ⎞
(b) ⎜ ⎟ dx + ⎝ ⎠ dy
∂x
⎝ ∂y ⎠
∂f
∂f
(c) ⎛ ⎞ dx – ⎛⎜ ⎞⎟ dy
⎝ ∂x ⎠
⎝ ∂y ⎠
∂f
∂f
(d) ⎛⎜ ⎞⎟ dx – ⎛ ⎞ dy
⎝ ∂x ⎠
⎝ ∂y ⎠
13. The function z = 5xy – 4x2 + y2 – 2x –
y + 5 has at x = 1 , y = 18
41
41
(a) maxima
(b) saddle point
Engineering Mathematics
4.156
14.
15.
16.
17.
18.
(c) minima
(d) none of these
If f (x, y) is such that fx = ex cos y and
fy = ex sin y, then which of the following is true
(a) f (x, y) = ex+y sin (x + y)
(b) f (x, y) = ex sin(x + y)
(c) f (x, y) does not exist
(d) none of these
The percentage error in the area of
a rectangle when an error of 1% is
made in measuring its length and
breadth is equal to
(a) 1%
(b) 2%
(c) 0
(d) 3%
The function f (x) = 10 + x6
(a) is a decreasing function of x
(b) has a minimum at x = 0
(c) has neither a maximum nor a
minimum at x = 0
(d) none of these
If u = f (y + ax) + f (y – ax),
2
2
u
u
2
then
–
a
is
y2
x2
(a) 0
(b) a2
2
(c) a (f – f ) (d) a2 (f + f )
With usual notations, the properties
of maxima and minima under various conditions are,
I
II
(P) Maxima
(i) rt – s2 = 0
Answers
1. (c)
8. (b)
15. (b)
2. (b)
9. (d)
16. (c)
3. (a)
10. (d)
17. (a)
(Q) Minima
(ii) rt – s2 < 0
(R) Saddle
(iii) rt – s2 > 0,
point
r>0
(S) Failure
(iv) rt – s2 > 0
case
r<0
(a) P-i, Q-iii, R-iv, S-ii
(b) P-ii, Q-i, R-iii, S-iv
(c) P-iii, Q-iv, R-ii, S-i
(d) P-iv, Q-ii, R-i, S-iii
(u , v)
for the func( x, y )
tion u = ex sin y, v = (x + log sin y) is
(a) 1
(b) sin x sin y – xy cos x cos y
(c) 0
ex
(d)
x
20. If the function u, v, w of three independent variables x, y, z are not independent, then the Jacobian of u, v,
w w.r.t to x, y, z is always equal to
(a) 1
(b) 0
(c)
(d) Jacobian of x, y, z w.r.t u, v, w
21. The approximate value of f (0.999)
where f (x) = 2x4 + 7x3 – 8x2 + 3x + 1 is
(a) 4.984
(b) 3.984
(c) 2.984
(d) 1.984
19. The Jacobian
4. (c)
11. (b)
18. (c)
5. (c)
12. (a)
19. (c)
6. (b)
13. (b)
20. (b)
7. (b)
14. (c)
21. (a)
Chapter
5
In this chapter, we will learn about the convergence and divergence of an infinite series. There are various methods to test the convergence and divergence of an infinite
series. In this chapter, we will study Comparision Test, D’Alembert’s ratio test, Raabe’s
test, Logarithmic test, Cauchy’s root test and Cauchy’s integral test. We will also study
alternating series, absolute and uniform convergence of the series.
An ordered set of real numbers as u1, u2, u3, ……..un, …… is called a sequence and is
denoted by {un}. If the number of terms in a sequence is infinite, it is said to be infinite
sequence, otherwise it is a finite sequence and un is called the nth term of the sequence.
A sequence is said to be monotonically increasing if un +1 un for each value of n
and is monotonically decreasing if un +1 un for each value of n, whereas the sequence
is called alternating sequence if the terms are alternate positive and negative.
e.g. (i) 1, 2, 3, 4, … is a monotonically increasing sequence.
1 1 1
(ii) 1, , , , … is a monotonically decreasing sequence.
2 3 4
(iii) 1, –2, 3, – 4, … is an alternating sequence.
A sequence {un} is said to be bounded sequence if there exists numbers m and M such
that m < un < M for all n.
A sequence {un} tends to a limt l as n
m such that, | un l |
if for every
0 there exists an integer
for all n > m, i.e., lim un = l.
n
(i) If the sequence {un} has a finite limit, i.e., lim un is finite, the sequence is said
n
to be convergent.
5.2
e.g.
Engineering Mathematics
⎧
⎫
⎪ 1 ⎪
{un } = ⎨
⎬
⎪1 + 1 ⎪
⎩ n⎭
lim un = 1
n
Since limit is finite, the sequence is convergent.
(ii) If the sequence {un} has infinite limit, i.e., lim un is infinite, the sequence is said
n
to be divergent.
e.g.
{un} = {2n + 1}
lim un
n
Since limit is infinite, the sequence is divergent.
(iii) If the limit of the sequence {un} does not exist, i.e., lim un is not unique, the
n
sequence is said to be oscillatory.
e.g.
{un} = ( −1) n +
1
2n
lim un = 1, if n is even
n →∞
= –1, if n is odd
Since limit is not unique, the sequence is oscillatory.
Note 1: Every convergent sequence is bounded but the converse is not true.
Note 2: A monotonic increasing sequence converges if it is bounded above and
diverges to + if it is not bounded above.
Note 3: A monotonic decreasing sequence converges if it is bounded below and
diverges to – if it is not bounded below.
Note 4: If sequence {un} and {vn} converges to l1 and l2 respectively, then
(i) Sequence {un + vn} converges to l1 + l2
(ii) Sequence {un . vn} converges to l1 l2
l
⎧u ⎫
(iii) Sequence ⎨ n ⎬ converges to 1 provided l2
l2
⎩ vn ⎭
0.
If u1, u2, u3, . . . un, . . . is an infinite sequence of real numbers, then the sum of the terms
of the sequence, u1 + u2 + u3 + … + un + … ∞ is called an infinite series.
The infinite series u1 + u2 + u3 + . . . un + . . .
is usually denoted by
un or
n =1
th
un .
The sum of its first n terms is denoted by Sn and is also known as n partial sum
of un .
5.3
Consider the infinite series un u1 u2 u3 …un … and let the sum of the first
, three possibilities arise for Sn:
n terms is Sn = u1 + u2 + u3 + . . . + un. As n
(i) If Sn tends to a finite limit as n
(ii) If Sn tends to
as n
, the series un is said to be convergent.
, the series un is said to be divergent.
(iii) If Sn does not tend to a unique limit as n
series un is said to be oscillatory.
, i.e., limit does not exist, the
1. The convergence or divergence of an infinite series remains unaffected:
(i) by addition or removal of a finite number of its terms.
(ii) by multiplication of each term with a finite number.
2. If two series un and vn are convergent, then (un vn ) is also convergent.
If a positive term series
un is convergent, then lim un = 0.
n
Note: The converse of this result is not true, i.e., if lim un = 0, it is not necessary that
n
series will be convergent.
un = 1 +
e.g.
lim un = lim
n
2
1
n
Sn = 1 +
Now,
1
Sn >
n
1
2
+
1
3
+… +
1
n
+…
=0
+
1
3
+… +
1
n
> 1+
1
n
+
1
n
+… +
1
n
n
n
lim S n = lim n = ∞
n →∞
n →∞
Thus, the series is divergent.
Hence, lim un = 0 is a necessary but not sufficient condition for convergence
n
of un .
5.4
Engineering Mathematics
Consider the geometric series a + ar + ar 2 +
+ ar n 1 +
2
n 1
S n = a + ar + ar +
+ ar
rn )
,
r
1)
,
1
a (1
1
a(r n
r
if r 1
if r 1
lim r n = 0
(i) When | r | 1 ,
n
lim S n =
n
a
is finite.
1 r
Hence, the series is convergent.
lim r n
(ii) When r > 1,
n
lim S n
lim
n
n
a (r n 1)
r 1
Hence, the series is divergent.
(iii) When r = 1
Sn = a + a + a +
= na
lim S n
n
Hence, the series is divergent.
(iv) When r = –1, series becomes a a a
Sn
a a a
( 1) n 1 a
= 0, is n is even
= a, if n is odd
Hence, the series is oscillatory.
(v) When r < –1, let r = –k
lim S n =
n
where k > 0
a[1 − ( −1) n k n ]
a[1 ( k ) n ]
= lim
n →∞
1+ k
1+ k
= – , if n is even
= + , if n is odd
Hence, the series is oscillatory.
From all the above cases, we conclude that the geometric series (1) is
(i) Convergent if | r | 1
(ii) Divergent if r 1
(iii) Oscillatory if r
1
... (1)
5.5
1
1 1
1
= p+ p+ p+
p
1
2
3
n =1 n
(i) Convergent if p > 1
(ii) Divergent if p 1
Note: The p series
(i) lim
n
log n
=0
n
(ii) lim 1 +
n
1
n
is
(vii) lim x n
n
(viii) lim nx n = 0 if x < 1
=e
n
1
n
(iii) lim(n) = 1
n
(ix) nlim
xn
= 0 for all x
n!
(x) lim
an 1
= log a
n
1
(iv) lim(n !) n
n
n!
(v) lim
n
n
n
1
n
if x > 1
n
0
1
an 1
= log a
(xi) lim
n
1
n
1
=
e
(vi) lim x n = 0 if x < 1
n
If un and vn are series of positive terms such that lim
n
un
= l (finite and non-zero),
vn
then both series converge or diverge together.
Proof:
un
=l
vn
lim
n
By definition of limit, for a positive number , however small, there exists an integer
m such that
un
vn
− ∈<
;
for all n > m
un
− l <∈
vn
for all n > m
l
l − ∈<
un
< l+ ∈
vn
for all n > m
5.6
Engineering Mathematics
m terms of
un and
l − ∈<
vn ,
un
< l+ ∈
vn
for all n ... (1)
vn is convergent, then lim(v1 + v2 + v3 +
Case I: If
n
+ vn ) = finite = k , say
From Eq. (1),
un
vn
lim(u1 u2
n
u3
un
(l
)vn
un )
(l
) lim(v1 v2
lim(u1 + u2 + u3 +
n →∞
Hence,
l
for all n
n
v3
vn )
+ un ) < (l + ∈)k (finite)
un is also convergent.
vn is divergent, then
Case II: If
lim(v1 v2
n
v3 ... vn )
... (2)
From Eq. (1),
un
vn
l
lim(u1 u2
n
(l
)vn
un )
(l
) lim(v1 v2
lim(u1 u2
u3
u3
n
Hence,
for all n
un
n
un )
vn )
[From Eq. (2)]
un is also divergent.
Test the convergence of the series
un =
Let vn =
v3
n
=
n2 + 1
n
.
2
n
1
+
n =1
1
3
2
1⎞
⎛
n ⎜1 + 2 ⎟
⎝ n ⎠
1
3
n2
lim
n →∞
un
= lim
vn n→∞
1
1+
1
n2
=1
(finite and non-zero)
5.7
and
1
vn
n
3
2
is convergent since p
3
2
1.
un is also convergent.
Hence, by comparison test,
Test the convergence of the series
1
1
1
3
n 3!n 5!n5
1
1
1
1
2
n
3!n 5!n 4
sin
un
Let vn =
1
sin .
n
n =1
1
n
1
n
un
vn
lim
n
lim 1
n
1
3!n 2
1
5!n 4
= 1 (finite and non-zero)
and
vn =
1
is divergent since p = 1.
n
Hence, by comparison test,
un is also divergent.
1
Test the convergence of the series ( n 1) 3
un
(n 1)
n
1
3
Let vn =
1
2
n3
n
1
1
3n
1
3n
1
3
1
2
3
9n
1
3
n
1
3
1 1
1
3 3
2!
5
5
3
1
1
n
81n
1
3
n
1
1 1
1
1
2
3 3
3
3!
1
n2
1
8
3
1
n3 .
2
3
1 1
3 9n
5
81n 2
1
n3
1
5.8
Engineering Mathematics
lim
n
un
vn
lim
n
=
and
1
vn
n
5
81n 2
1
(finite and non-zero)
3
is divergent since p
2
3
1 1
3 9n
2
1.
3
un is also divergent.
Hence, by comparison test,
n2 1 n
.
np
Test the convergence of the series
1
⎡
⎤
1 ⎞2 ⎥
⎛
⎢
n ⎜1 + 2 ⎟ − 1
⎢⎝ n ⎠
⎥
n2 + 1 − n
⎣
⎦
un =
=
np
np
⎡⎧
1⎛1 ⎞⎛1
1⎛1 ⎞
⎞
⎜ − 1⎟ ⎜ − 2⎟
⎜ − 1⎟
⎢
1
n ⎢ ⎪⎪
2⎝2 ⎠ 1 2⎝2 ⎠⎝2 ⎠ 1
⋅ 4+
⋅ 6+
= p ⎨1 + 2 +
3!
2!
n ⎢ ⎪ 2n
n
n
⎢⎪
⎣⎩
=
Let vn =
n
np
1
1
⎛ 1
−
⎜⎝ 2 − 4 +
2n 8n 16n 6
1
n p +1
lim
n →∞
un
1
1
⎛1
⎞
= lim ⎜ −
+
− …⎟
⎠
vn n→∞ ⎝ 2 8n 2 16n 4
=
and
i.e., p
1 ⎛1
1
1
⎞
−
⎟⎠ = p +1 ⎜⎝ − 2 +
2 8n 16n 4
n
⎫ ⎤
⎪⎪ ⎥
⎬ − 1⎥
⎪ ⎥
⎪⎭ ⎥⎦
⎞
⎟⎠
1
vn
n p +1
1
2
(finite and non-zeero)
is convergent if p + 1 > 1, i.e., p > 0 and divergent if p 1 1,
0.
Hence, by comparison test,
un is also convergent if p > 0 and divergent if p
Test the convergence of the series
0.
14 24 34
+
+
+….
13 23 33
nth term of the numerator = a + (n – 1)d = 14 + (n – 1)10 = 10n + 4
n term of the denominator = n3
th
5.9
nth term of the given series,
Let vn =
10n + 4
1
4
= 2 10 +
3
n
n
n
un =
1
n2
lim
n
un
4
= lim 10 +
n
vn
n
= 10 (finite and non-zero)
1
is convergent since p = 2 > 1.
n2
Hence, by comparison test, un is also convergent.
and S vn =
Test the convergence of the series
nth term of the series, un =
Let vn =
lim
un
= lim
vn n
=
vn
n
=
a n2 b
1
n a+
b
n2
1
n
n
and
1
2
3
+
+
+….
a 12 + b a 22 + b a 32 + b
1
a+
b
n2
1
(finite and non-zero)
a
1
is divergent since p = 1.
n
Hence, by comparison test,
un is also divergent.
Test the convergence of the series
nth term of the series
un =
1
1
1
+ p + p +….
p
3
5
7
1
=
(2n + 1) p
1
np 2 +
1
n
p
5.10
Engineering Mathematics
Let vn =
1
np
lim
n
un
= lim
vn n
=
1
2+
1
n
p
1
(finite and non-zero)
2p
1
is convergent if p > 1 and divergent if p 1.
np
Hence, by comparison test, un is also convergent if p > 1 and divergent if p 1 .
and
vn
Test the convergence of the series
1
1
1
1
1
+
+
+
+
+ … , where x is a positive fraction.
x x 1 x 1 x 2 x 2
Since it is an infinite series, by ignoring the first term, the series can be
rewritten as
un =
1
+
1
+
1
+
1
x 1 x 1
x 2 x 2
2x
2x
= 2 2+ 2
+ ……
x 1 x 22
2x
=
x2 n2
2x
2x
un = 2
=
x n2
x2
n2 2 1
n
Let vn =
+ ……
1
n2
lim
n
un
2x
= lim 2
n
vn
x
1
n2
= –2x (finite and non-zero)
and
vn
1
is convergent since p = 2 > 1.
n2
Hence, by comparison test,
un is also convergent.
5.11
Exercise 5.1
1. Test the convergence of the following series:
(i)
1
n2 + 1
(ii)
(
n 1
(iii)
(
n4 1
n
)
n4 1
)
np
(iv)
n +1 + n
(v)
np
(n + 1) q
(vi)
1
n +1
log
n
n
(vii)
(viii)
1
n
tan
tan
a+
n
b
n
1
n
1
n
1+ 2
.
(ii)
un is divergent if l > 1.
2
3
+
+ …… .
1+ 2 3 1+ 3 4
(1 + a )(1 + b) 1
1
1
1
+ p + p + p + p +…… .
1 2 3
3
5
7
9
[Ans. : Convergent if p > 1,
Divergent if p
n
un is convergent if l < 1.
+
4. Test the convergence of the series
un is a positive term series and lim
(i)
1
2
[Ans. : Divergent]
⎡ Ans. :
⎤
⎢ (i) Convergent ⎥
⎢
⎥
⎢ (ii) Divergent ⎥
⎢
⎥
⎣ (iii) Convergent ⎦
If
⎤
⎥
⎥
1
⎥
Divergent if p ≥ −
⎥
2
⎥
Convergent if p − q + 1 < 0, ⎥
Divergent if p − q + 1 ≥ 0 ⎥
⎥
Convergent
⎥
⎥
Convergent
⎥
Divergent
⎥
⎥
Convergent if a > 1,
⎥
⎥⎦
Divergent if a ≤ 1
Convergent if p < −
2. Test the convergence of the series
(1 + a )(1 + b) (2 + a )(2 + b)
+
+
1⋅ 2 ⋅ 3
2 ⋅3⋅ 4
(3 + a )(3 + b)
+… .
3⋅ 4 ⋅5
[Ans. : Divergent]
3. Test the convergence of the series
1
1
1
(ix)
⎡
⎢ (iv)
⎢
⎢
⎢
⎢
⎢ (v)
⎢
⎢
⎢ (vi)
⎢ (vii)
⎢
⎢(viii)
⎢ (ix)
⎢
⎢⎣
un +1
= l , then
un
1.]
5.12
Engineering Mathematics
Proof:
Case I: If lim
n
un +1
un
l 1.
Consider a number l < r < 1 such that
m terms,
∞
∑u
n
n = m +1
un +1
un
r for all n > m
… (1)
⎞
⎛ u
u
u
= um +1 + um + 2 + um + 3 + … ∞ = um +1 ⎜1 + m + 2 + m + 3 + m + 4 + … ⎟
⎠
⎝ um +1 um +1 um +1
⎛ u
⎞
u
u
u
u
u
= um +1 ⎜1 + m + 2 + m + 3 ⋅ m + 2 + m + 4 ⋅ m + 3 ⋅ m + 2 + … ⎟
⎝ um +1 um + 2 um +1 um + 3 um + 2 um +1
⎠
[ Using Eq. (1)]
< um +1 (1 + r + r ⋅ r + r ⋅ r ⋅ r + …)
1
= um +1 (1 + r + r 2 + r 3 + …) = um +1 ⋅
−
1 r
∞
um + 1
∑ un < 1 − r (finite)
n = m +1
(r < 1)
∞
The series
∑
un is convergent.
n = m +1
The nature of a series remains unchanged if we neglect a finite number of terms in the
∞
beginning. Hence, the series
∑u
n
is convergent.
n =1
lim
Case II: If
n
un +1
un
l 1,
un +1
un
1 for all n
m
... (2)
Neglecting the first m terms,
∞
∑u
n = m +1
n
= u m +1 + u m + 2 + u m + 3 + u m + 4 + … ∞
⎛ u
⎞
u
u
= um +1 ⎜1 + m + 2 + m + 3 + m + 4 + …⎟
⎝ u m +1 u m +1 u m +1
⎠
⎛ u
⎞
u
u
u
u
u
= um +1 ⎜1 + m + 2 + m + 3 ⋅ m + 2 + m + 4 ⋅ m + 3 ⋅ m + 2 + …⎟ > um +1 (1 + 1 + 1 + 1 + …)
⎝ u m + 1 u m + 2 u m +1 u m + 3 u m + 2 u m + 1
⎠
Consider,
(um +1 um + 2 … to n terms)
Sn
lim S n
n
um +1 (1 1 1… to n terms)
um +1 (n)
lim num +1
n
[∵ um +1 is positive]
5.13
∞
∑
The series
un is divergent. The nature of a series remains unchanged if we
n = m +1
∞
neglect a finite number of terms in the beginning. Hence, the series
∑u
n
is divergent.
n =1
un +1
= 1, the ratio test fails, i.e., no conclusion can be drawn about the
un
convergence or divergence of the series.
Note 1: If lim
n
Note 2: It is convenient to use D’Alembert’s ratio test in the following form:
u
If un is a positive term series and lim n = l , then
n
un +1
(i) un is convergent if l > 1.
(ii) un is divergent if l < 1.
(iii) The ratio test fails if l = 1.
Test the convergence of the following series:
2! 3! 4!
(i)
+ + +…
3 32 3 3
(iv)
(ii)
2
2 2 ⋅ 5 2 ⋅ 5 ⋅ 8 2 ⋅ 5 ⋅ 8 ⋅ 11
+
+
+
+…
1 1 ⋅ 5 1 ⋅ 5 ⋅ 9 1 ⋅ 5 ⋅ 9 ⋅ 13
(i)
2
2
⎛ 1 ⎞ ⎛ 1⋅ 2 ⎞ ⎛ 1⋅ 2 ⋅ 3 ⎞
+
+ …∞
(iii) ⎜ ⎟ + ⎜
⎝ 3 ⎠ ⎝ 3 ⋅ 5 ⎟⎠ ⎜⎝ 3 ⋅ 5 ⋅ 7 ⎟⎠
n!
nn
(v)
n!(2) n
.
nn
(n + 1)!
3n
(n + 2)!
= n +1
3
(n + 1)! 3n +1
lim
n
(n + 2)!
3n
un =
un +1
lim
n
un
un +1
lim
n
3
n+2
0 1
Hence, by ratio test, the series is divergent.
un =
(ii)
un +1 =
lim
n
un
un +1
n!
nn
(n + 1)!
(n + 1) n +1
n!
nn
lim
n
(n + 1)!
(n + 1) n +1
lim
n
(n + 1)(n + 1) n
(n + 1)n n
Hence, by ratio test, the series is convergent.
lim 1
n
1
n
n
e 1
5.14
Engineering Mathematics
2
2
2
1
1⋅ 2 ⎞ ⎛ 1⋅ 2 ⋅ 3 ⎞
(iii) The series is given by ⎛⎜ ⎞⎟ + ⎛⎜
+
+…∞
⎝ 3 ⎠ ⎝ 3 ⋅ 5 ⎟⎠ ⎜⎝ 3 ⋅ 5 ⋅ 7 ⎟⎠
⎡ 1⋅ 2 ⋅ 3… n ⎤
un = ⎢
⎥
⎣ 3 ⋅ 5 ⋅ 7 … (2n + 1) ⎦
2
⎡
⎤
1⋅ 2 ⋅ 3 … n(n + 1)
u n +1 = ⎢
⎥
(
2
n
+
3
)
3
5
7
…
(
2
1
)(
⋅
⋅
n
+
⎣
⎦
lim
n →∞
un
u n +1
⎡ 1 ⋅ 2 ⋅ 3… n ⎤
⎢ 3 ⋅ 5 ⋅ 7 … (2n + 1) ⎥
⎣
⎦
2
2
2
3⎞ ⎤
⎡⎛
2+ ⎟ ⎥
⎜
⎢
⎝
⎡ (2n + 3) ⎤
n⎠
⎥ = 4 >1
= lim
= lim ⎢
= lim ⎢
⎥
2
n →∞
n →∞
n →∞ ⎛
⎢ 1+ 1⎞ ⎥
⎣ (n + 1) ⎦
⎡
⎤
1 ⋅ 2 ⋅ 3… n(n + 1)
⎢⎣ ⎜⎝ n ⎟⎠ ⎥⎦
⎢ 3 ⋅ 5 ⋅ 7 …(2n + 1)(2n + 3) ⎥
⎣
⎦
2
Hence, by ratio test, the series is convergent.
2 2.5 2 ⋅ 5 ⋅ 8 2 ⋅ 5 ⋅ 8 ⋅ 11
+
+
+
+…
1 1.5 1 ⋅ 5 ⋅ 9 1 ⋅ 5 ⋅ 9 ⋅ 13
2 ⋅ 5 ⋅ 8 ⋅ 11… (3n − 1)
un =
1 ⋅ 5 ⋅ 9 ⋅ 13… (4n − 3)
(iv) The series is given by
un+1 =
lim
n →∞
un
u n +1
2 ⋅ 5 ⋅ 8 ⋅ 11… (3n − 1)(3n + 2)
1 ⋅ 5 ⋅ 9 ⋅ 13… (4n − 3)(4n + 1)
2 ⋅ 5 ⋅ 8 ⋅ 11… (3n − 1)
1
4+
4n + 1
1 ⋅ 5 ⋅ 9 ⋅ 13… (4n − 3)
n = 4 >1
= lim
= lim
= lim
n →∞ 2 ⋅ 5 ⋅ 8 ⋅ 11…(
(3n − 1)(3n + 2) n→∞ 3n + 2 n→∞
2 3
3+
1 ⋅ 5 ⋅ 9 ⋅ 13… (4n − 3)(4n + 1)
n
Hence, by ratio test, the series is convergent.
n !(2) n
(v)
un =
nn
( n + 1)!( 2) n +1
un +1 =
( n + 1) n +1
u
lim n
n
un +1
n !(2) n
nn
lim
n
(n + 1)!(2) n +1
(n + 1) n +1
lim
n
1 n +1
2 n
n
lim
n
(n + 1)(n + 1) n n !2n
(n + 1)n !2n 2 n n
1
1
lim 1
n
2
n
Hence, by ratio test, the series is convergent.
n
e
2
1
5.15
Test the convergence of 1 +
2p 3p 4p
+
+
+ … ,( p > 0).
2! 3! 4!
np
n!
(n + 1) p
=
(n + 1)!
un =
un +1
lim
n
un
un +1
np
= lim n ! p
n
(n + 1)
(n + 1)!
lim
n
np
(n + 1)!
p
n!
(n + 1)
lim
(n + 1)
n
1+
1
n
p
1
Hence, by ratio test, the series is convergent.
Test the convergence of the series
un =
u n +1 =
un
n →∞ u
n +1
lim
xn 1
.
n
n=1 n 3
xn 1
n 3n
xn
(n + 1) 3n +1
x n −1
n
= lim n ⋅ 3n
n →∞
x
(n + 1) 3n +1
(n + 1) 3 3
⎛ 1⎞
= lim ⎜1 + ⎟
n →∞
x⋅n
x n→∞ ⎝ n ⎠
3
=
x
= lim
Hence, by ratio test, the series is convergent, if 3
x
3
if
1, i.e., x > 3.
x
For x = 1, lim
n
un
un +1
3 1, the series is convergent.
1, i.e., x < 3 and divergent
5.16
Engineering Mathematics
n
Test the convergence of the series
n +1
2
n n
x
n2 + 1
un =
(n + 1)
x n +1
(n + 1) 2 + 1
un +1
lim
n
un
n
(n + 1) 2 + 1 1
lim 2
xn
un +1 n
n +1
n +1
x n +1
lim
n
n
(n 2 + 2n + 2) 1
(n + 1)
x
(n 2 + 1)
1+
lim
n
=
1
1+
n
2 2
+
n n2
1
x
Test the convergence of the series
u n +1 =
lim
n →∞
un
u n +1
x2n
1
x
1
1+ 2
n
Hence, by ratio test, the series is convergent if 1
x
1
1, i.e., x > 1. Ratio test fails for x = 1.
x
un =
xn .
1, i.e., x < 1 and is divergent if
1
2 1
+
x2
3 2
+
x4
4 3
+
x6
5 4
+… .
2
(n + 1) n
x2n
( n + 2) n + 1
1
⎛ 2⎞
1+ ⎟ 1+
⎜
⎝ n⎠
( n + 2) n + 1
x
n 1
= lim
= lim
⋅
⋅ 2
2n
n →∞
n →∞
1
x
x
⎛
⎞
(n + 1) n
⎜⎝1 + ⎟⎠
n
2n− 2
=
1
x2
Hence, by ratio test, the series is convergent if 12
x
1
2
1, i.e., x > 1. Ratio test fails for x = 1.
x2
1, i.e., x2 < 1 and is divergent if
5.17
Exercise 5.2
Test the convergence of the following series:
1. 1
2.
22
2!
32
3!
42
… .
4!
[Ans. : Convergent]
1
2
1 + 2 1 + 22
3
… .
1 + 23
xn
.
(2n )!
10.
[Ans. : Convergent]
11.
n +1 n
⋅ x , x > 0.
n3 + 1
∑
[Ans. : Convergent]
2!
3. 1 2
2
4.
3!
33
4!
… .
44
[Ans. : Convergent]
12. x + 2 x 2 + 3 x 3 + 4 x 4 + … .
Ans. : Convergent for x < 1,
divergent for x > 1
n2
..
3n
[Ans. : Convergent]
5.
Ans. : Convergent for x < 1,
divergent for x > 1
13. 1 +
2n 1
.
3n + 1
x x 2 x3
xn
+ + +… + 2
+… .
2 5 10
n +1
Ans. : Convergent for x < 1,
divergent for x > 1
[Ans. : Convergent]
6.
7.
8.
1
.
n!
14.
[Ans. : Convergent]
Ans. : Convergent for x < 1,
divergent for x > 1
[Ans. : Convergent]
3 2 8 3 15 4
15. x + x + x + x + … .
5
10
17
n2 − 1 n
+ 2
x + …∞
n +1
n 2 (n + 1) 2
.
n!
xn
,x
3 n2
n
x
x2
x3
+
+
+ … ∞.
1⋅ 3 3 ⋅ 5 5 ⋅ 7
0.
Ans. : Convergent for x < 1,
divergent for x > 1
Ans. : Convergent for x < 3,
divergent for x > 3
3n − 2 n −1
⋅ x , x > 0.
9. ∑ n
3 +1
Ans. : Convergent for x < 1,
divergent for x > 3
16.
x
2 3
+
x2
3 4
+
x3
4 5
+…
.
Ans. : Convergent for x < 1,
divergent for x > 1
5.18
If
Engineering Mathematics
un
1
un +1
un is a positive term series and nlim n
(i)
un is convergent if l > 1
(ii)
un is divergent if l < 1
l , then
(iii) Test fails if l = 1
un
Proof: (i) Consider a number p such that p > 1. The series
if p > 1. By comparison test,
un will be convergent if from and after some term
un
un +1
vn
vn +1
un
un +1
1
n
un
1
un +1
lim n
un
1
un +1
n
n
(n + 1) p
np
1
n
p
n
p
1
1
n
1
p
n
p
p ( p 1)
…
2!n 2
p ( p 1)
…
2n 2
lim p
n
l
Hence,
1
is convergent
np
p ( p 1)
…
2n
p 1
un is convergent if l > 1.
(ii) Consider a number p such that p < 1. The series
By comparison test,
vn
1
is divergent if p < 1.
np
un will be divergent if from and after some term
un
un +1
vn
vn +1
Proceeding as above in the case (i), we get
lim n
n
un
1
un +1
lim p
n
l
Hence,
p ( p 1)
…
2n
p 1
un is divergent if l < 1.
(iii) Raabe’s test fails if l = 1 and other tests are required to check the nature of the
series.
Note: When Raabe’s test fails, logarithmic test can be applied.
5.19
Test the convergence of the series
un =
un + 1 =
2 2⋅5
2⋅5⋅8
+
+
+… .
7 7 ⋅ 10 7 ⋅ 10 ⋅ 13
2 ⋅ 5 ⋅ 8…… (3n − 1)
7 ⋅10 ⋅13…… (3n + 4)
2 ⋅ 5 ⋅ 8… (3n − 1) (3n + 2)
7 ⋅ 10 ⋅ 13… (3n + 4) (3n + 7)
un
2 ⋅ 5 ⋅ 8… (3n − 1) 7 ⋅10 ⋅13… (3n + 4)(3n + 7) 3n + 7
=
=
⋅
un+1 7 ⋅10 ⋅13… (3n + 4) 2 ⋅ 5 ⋅ 8…(3n − 1)(3n + 2)
3n + 2
lim n
n
un
1
un +1
lim n
n
3n + 7
1
3n + 2
lim
n
5n
3n + 2
lim
n
5
3+
2
n
5
1
3
Hence, by Raabe’s test, the series is convergent.
Test the convergence of the series
4 7……(3n 1) x n
.
n!
4 ⋅ 7… (3n + 1) x n
n!
4 ⋅ 7… (3n + 1)(3n + 4) x n +1
=
(n + 1)!
un =
un +1
un 4 ⋅ 7… (3n + 1) x n
(n + 1)!
n +1
⋅
=
=
n!
un +1
4 ⋅ 7… (3n + 1)(3n + 4) x n +1 (3n + 4) x
1
1+
un
n = 1
lim
= lim
n →∞ u
n →∞ ⎛
4
3x
⎞
n +1
⎜⎝ 3 + ⎟⎠ x
n
By ratio test, the series is
1
1
1 or x
3
3x
1
1
(ii) Divergent if
1 or x
3x
3
1
(iii) Test fails if x =
3
un
n +1
3n + 3
Then
=
=
1 3n + 4
un+1
(3n + 4)
3
(i) Convergent if
5.20
Engineering Mathematics
Applying Raabe’s test,
⎛
⎞
⎜ −1 ⎟
⎛ un
⎞
1
⎛ −n ⎞
⎛ 3n + 3 ⎞
lim n ⎜
− 1⎟ = lim n ⎜
− 1⎟ = lim ⎜
⎜
⎟ = − <1
⎟ = nlim
n →∞
n
→∞
n
→∞
→∞
3
⎝ 3n + 4 ⎠
⎝ 3n + 4 ⎠
⎝ un +1 ⎠
⎜⎜ 3 + 4 ⎟⎟
n⎠
⎝
1
By Raabe’s test, the series is divergent if x = .
3
1
and is divergent if x
3
Hence, the series is convergent if x
Test the convergence of the series
un =
∑
1
.
3
1 ⋅ 3 ⋅ 5 …… ( 2n − 1) x 2 n+1
2 ⋅ 4 ⋅ 6 …… 2n( 2n + 1) .
1 3 5…… (2n 1) x 2 n +1
2 4 6…… (2n 1)
un + 1 =
1 ⋅ 3 ⋅ 5… (2n − 1) (2n + 1) x 2 n + 3
2 ⋅ 4 ⋅ 6… 2n (2n + 2)(2n + 3)
un
1 ⋅ 3 ⋅ 5…… (2n − 1) x 2 n +1 2 ⋅ 4 ⋅ 6…… 2n(2n + 2)(2n + 3)
=
⋅
un +1
2 ⋅ 4 ⋅ 6…… 2n(2n + 1) 1 ⋅ 3 ⋅ 5…… (2n − 1)(2n + 1) x 2 n + 3
=
(2n + 2)(2n + 3)
(2n + 1) 2 x 2
2⎞ ⎛
3⎞
⎛
⎜ 2 + ⎟⋅⎜ 2 + ⎟ 1
un
n⎠ ⎝
n⎠
lim
= lim ⎝
= 2
2
n →∞ u
n →∞
x
1
⎛
⎞ 2
n +1
⎜2+ ⎟ x
n⎠
⎝
By ratio test, the series is
(i) Convergent if
1
x2
(ii) Divergent if 1
x2
2
1 or x
2
1 or x
1
1
(iii) Test fails if x2 = 1
Then
un
(2n + 2)(2n + 3)
=
un +1
(2n + 1) 2
5.21
Applying Raabe’s test,
lim n
n
un
1
un +1
lim n
n
(2n + 2)(2n + 3)
1
(2n + 1) 2
n(6n + 5)
lim
n
(2n + 1) 2
6+
lim
n
2+
5
n
1
n
2
3
2
1
By Raabe’s test, the series is convergent if x 2 = 1
Hence, the series is convergent if x 2
1 and is divergent if x 2
1.
Test the convergence of the series
∑
a( a + 1)( a + 2) …… ( a + n − 1) ⋅ b( b + 1)( b + 2) …… ( b + n − 1) x n
.
1 ⋅ 2 ⋅ 3 …… n ⋅ c( c + 1)( c + 2) …… ( c + n − 1)
un =
un
un +1
a (a 1)(a 2)…… (a n 1) b(b 1)(b 2)…… (b n 1) x n
1 2 3…… n c(c 1)(c 2)…… (c n 1)
a (a 1)…… (a n 1) b(b 1)…… (b n 1) x n
1 2 3…… n c(c 1)…… (c n 1)
1 2…… n(n 1) c(c 1)...(c n 1)(c n)
a (a 1)…… (a n 1)(a n) b(b 1)…… (b n 1)(b n) x n +1
(n + 1)(c + n)
=
(a + n)(b + n) x
1 c
+1
1+
un
1
n n
= lim
lim
=
n
n
a
b
un +1
x
+1
+1 x
n
n
By ratio test, the series is
(i) Convergent if
1
x
1 or x < 1
1
1 or x >1
x
(iii) Test fails if x = 1
(ii) Divergent if
Then
un
(n + 1)(c + n)
=
un +1 (a + n)(b + n)
5.22
Engineering Mathematics
Applying Raabe’s test,
lim n
n
un
1
un +1
lim n
n
(n + 1)(c + n)
1
(a + n)(b + n)
(c ab) n(1 c a b)
= lim n
= lim
n
n
(a + n)(b + n)
(c ab)
(1 c a b)
n
a
b
+1
+1
n
n
1 c a b
By Raabe’s test, the series is (i) Convergent if 1 c a b 1 or c
(ii) Divergent if 1 c a b 1 or c a b.
a b
Hence, the series is convergent if x < 1 and divergent if x > 1.
For x = 1, the series is convergent if c > a + b and divergent if c < a + b.
Exercise 5.3
Test the convergence of the following series:
1. 1 +
2.
Ans. : Convergent for x 1 and
divergent for x > 1
1
1
1
+
+
+ ... .
2 2 4 2 4 6
[Ans. : Divergent]
12 52 92 …… (4n 3) 2
.
42 82 122 …… (4n) 2
4. 1 +
22 22 4 2 22 4 2 6 2
+
+
+… .
32 32 52 32 52 72
[Ans. : Divergent]
[Ans. : Convergent]
3. (i) 1 +
22
22 4 2
22 4 2 6 2
+
+
+… .
3 4 3 4 5 6 3 4 5 6 7 8
[Ans. : Convergent]
2
(1!)
(2!) 2 2 (3!) 2 3
x+
x +
x +… .
2!
4!
6!
Ans. : Convergent for x < 4 and
divergent for x 4
(ii) 1 +
(iii) 1 + 3 x + 3 6 x 2 + 3 6 9 x 3 + … .
7
7 10
7 10 13
5. a (a + 1) + (a + 1)(a + 2)
2!
3!
(a + 2)(a + 3)
+
+ ….
4!
[Ans. : Convergent for a 0 ]
6.
(n !) 2 2 n
x .
(2n)!
Ans. : Convergent for x < 4 and
2
divergent for x 4
5.23
5.9 LOGARITHMIC TEST
un is a positive term series and if nlim n log
If
(i)
(ii)
un
= l , then
un +1
un is convergent if l > 1.
un is divergent if l < 1.
(iii) Test fails if l = 1.
Proof: Comparing the series
Let
vn
(i) Let
1
,
np
un with the p-series
1
which converges if p > 1 and diverges if p 1.
np
vn is convergent, then un will also be convergent if
un
un +1
vn
vn +1
(n + 1) p
np
un
⎛ 1⎞
> ⎜1 + ⎟
un +1 ⎝ n ⎠
p
⎛ u ⎞
⎛ 1⎞
log ⎜ n ⎟ > log ⎜ 1 + ⎟
u
⎝ n⎠
⎝ n +1 ⎠
⎛ u ⎞
⎛ 1⎞
log ⎜ n ⎟ > p log ⎜1 + ⎟
⎝ n⎠
⎝ un +1 ⎠
p
⎛ u ⎞
1
1
⎛1
log ⎜ n ⎟ > p ⎜ − 2 + 3 −
⎝ n 2n
3n
⎝ u n +1 ⎠
⎛ u ⎞
1
1
⎛
n log ⎜ n ⎟ > p ⎜1 −
+ 2−
⎝
2n 3n
⎝ u n +1 ⎠
⎞
⎟⎠
⎞
⎟⎠
⎛ u ⎞
lim n log ⎜ n ⎟ > p
⎝ u n +1 ⎠
n →∞
l>p>1
Hence,
∵
un is convergent if l > 1.
(ii) Let
vn is divergent, then
un will also be divergent if
Proceeding as above, we get
lim n log
n
un is divergent if l < 1.
un
v
< n
un +1 vn +1
un
<p
un +1
l<p
Hence,
vn is convergent if p > 1
1
∵
vn is divergent if p 1
5.24
Example 1: Test the convergence of the series
12
42
52
82
92
122
132
16 2
… .
2
Solution: un = ( 4 n − 3)
2
( 4 n)
( 4 n + 1) 2
un +1 =
( 4 n + 4) 2
un
un +1
log
un
un +1
⎡⎛
3 ⎞⎛ 1⎞⎤
1 − ⎟ ⎜1 + ⎟ ⎥
⎜
2
2
⎢
⎝
( 4 n − 3) ( 4 n + 4)
4n ⎠ ⎝ n ⎠
⎥
=
⋅
=⎢
2
2
1 ⎞
( 4 n)
( 4 n + 1)
⎛
⎢
⎥
⎜⎝1 + ⎟⎠
⎢⎣
⎥⎦
4n
2 log 1
3
4n
1
n
log 1
⎡⎛ 3 1 32
= 2 ⎢⎜ − − ⋅
−
2
⎣⎝ 4n 2 16n
log 1
2
1
4n
⎞ ⎛1 1 1
⎟+⎜ − ⋅ 2 +
⎠ ⎝n 2 n
∵ log(1 x)
n log
un
u n +1
lim n log
un
u n +1
n
2
3
4
9
32n
3
1
2
2
2
1
2n
1
1
4
⎞⎤
⎟⎥
⎠⎦
⎞ ⎛ 1 1 1
+
⎟−⎜ − ⋅
2
⎠ ⎝ 4n 2 16n
x
x2
2
x3
3
1
32n
0 1
Hence, by logarithmic test, the series is divergent.
Example 2: Test the convergence of the series 1 +
Solution:
un
un +1
un
un +1
x 2! 2 3! 3 4! 4
+ x + 3 x + 4 x +… .
2 32
4
5
n!
xn
(n + 1) n
(n + 1)! n +1
x
(n + 2) n +1
n! xn
(n + 2) n +1
(n + 1) n (n + 1)! x n +1
n !n n +1 1 +
nn 1
1
n
2
n
n +1
n
(n 1)n !
1
x
2
1+
n
1
2
1+
n
1
n
n
1
n
2
1
n
2
1
x
5.25
n
⎤
⎡
a
a
⎛
⎞
⎢∵ lim ⎜1 + ⎟ = e ⎥
⎥
⎢ n→∞ ⎝ n ⎠
⎦
⎣
e2 1
e x
u
lim n
n
u n +1
e
x
By ratio test, the series is
(i) Convergent if
e
> 1 or x < e
x
e
< 1 or x > e
x
e
(iii) Test fails if
1 or x = e
x
(ii) Divergent if
1+
un
un +1
Then
1+
2
n
1
n
n +1
n +1
1
e
Applying logarithmic test,
log
un
u n +1
2
n
(n 1)
1
n
(n 1)
1
lim n log
n
un
u n +1
2
n
(n 1) log 1
lim
n
3
2n
1
2
(n 1) log 1
1 22
2 n2
1 23
3 n3
3
7
2n 2 3n3
7
1
3
3n 2 n 2n 2
5
6n
1
n
7
3n 2
log e
1
n
1
2n 2
7
3n3
1
1
2
Example 3: Test the convergence of the series
1
2n
5
6n 2
7
3n3
1
e.
a + x ( a + 2 x )2 ( a + 3 x )3
+
+
+ ….
1!
2!
3!
Solution:
u n +1
1
1
By logarithmic test, the series is divergent if x = e
Hence, the series is convergent if x < e and is divergent if x
un
1
3n3
(a + nx) n
n!
[a + (n + 1) x]n +1
(n + 1)!
5.26
un
(a + nx)
(n + !)
=
⋅
un + 1
n!
[a + (n + 1) x] n + 1
nx
⎡
⎤
a
a
⎞
⎛
⎢⎜1 + ⎟ ⎥ ⎛⎜1 + 1 ⎞⎟
⎢⎝
nx ⎠ ⎥ ⎝
n⎠
⎣
⎦
n
a ⎞
⎛
(nx) n ⎜1 + ⎟ (n + 1)
1
nx
⎠
⎝
=
⋅
=
n +1
a+x
x
a + x⎞
nx
⎛
⎡
⎤ x
(nnx) n + 1 ⎜1 +
⎟
a + x ⎞ ⎢⎛
a ⎞a + x ⎥
⎛
nx ⎠
⎝
⎟ ⎜1 + ⎟
⎜1 +
⎥
nx ⎠ ⎢⎝
nx ⎠
⎝
⎣
⎦
a
u
lim n
n
u n +1
ex
a
ex
1
ex
+1
n
⎤
⎡
b
⎢∵ lim ⎛⎜1 + b ⎞⎟ = e ⎥
⎥
⎢ n→∞ ⎝ n ⎠
⎦
⎣
1
x
By ratio test, the series is
(i) Convergent if
(iii) Test fails if
1
ex
1
1
> 1 or x <
ex
e
1 or x
1
1
<1 or x >
ex
e
(ii) Divergent if
1
e
un
u n +1
Then
1+
ae
n
1+
n
1
e
n
1+
ae + 1
n
n +1
Applying logarithmic test,
log
un
⎛ ae + 1 ⎞
⎛ ae ⎞
⎛ 1⎞
= n log ⎜1 + ⎟ + log ⎜1 + ⎟ − (n + 1) log ⎜1 +
⎟ + log e
un+1
n ⎠
n ⎠
⎝
⎝
⎝ n⎠
⎛ ae 1 a 2 e 2 1 a 3 e3
= n⎜ − ⋅ 2 + ⋅ 3 −
3 n
⎝ n 2 n
⎞ ⎛1
1
⎟+⎜ − 2 +
⎠ ⎝ n 2n
⎞
⎟
⎠
⎡ ae + 1 1 ⎛ ae + 1 ⎞ 2 1 ⎛ ae + 1 ⎞3
− (n + 1) ⎢
− ⎜
⎟ −
⎟ + ⎜
2⎝ n ⎠ 3⎝ n ⎠
⎢⎣ n
⎛
1 a 2 e 2 1 a 3 e3
+ ⋅ 2 −
= ⎜ ae − ⋅
2 n
3 n
⎝
⎞ ⎛1
1
⎟+⎜ − 2 +
⎠ ⎝ n 2n
⎤
⎥ +1
⎥⎦
⎞
⎟
⎠
2
⎡
1 (ae + 1) 2 1 (ae + 1)3 (ae + 1) 1 ⎛ ae + 1 ⎞
+
+
−
− ⎢(ae + 1) −
⎜
⎟ +
n
n
2
3 n2
2⎝ n ⎠
⎢⎣
⎤
⎥ +1
⎥⎦
5.27
lim n log
n→∞
⎡⎛ a 2 e 2 1 a 3 e3
un
= lim ⎢⎜ −
+
−
un + 1 n → ∞ ⎣⎝
2
3 n
⎞ ⎛
1
+
⎟ + ⎜1 −
n
2
⎠ ⎝
⎞
⎟
⎠
⎧ 1
1 (ae + 1)3
1 (ae + 1) 2
− ⎨− (ae + 1) 2
+ (ae + 1) −
+
n
3
2
n
⎩ 2
⎫⎤
⎬⎥
⎭⎦
1
1
a 2 e2
+ 1 + (a 2 e 2 + 1 + 2ae) − (ae + 1) = < 1
2
2
2
By logarithmic test, the series is divergent if x 1 .
=−
e
1
Hence, the series is convergent if x < and is divergent if x
e
1
.
e
Exercise 5.4
Test the convergence of the following series:
22 x 2 33 x 3 44 x 4 55 x 5
+
+
+
+
2!
3!
4!
5!
1
Ans.: Convergent if x < and
e
1
divergent if x
e
1. x +
2. 1 +
.
1 22 33 4 4 55
+ + + + + .
22 33 4 4 55 66
[Ans. : Convergent]
3.
4. (a + 1)
2
32
43
54
x + x 2 + x3 + x 4 +
2!
3!
4!
5!
x
x2
+ ( a + 2) 2
1!
2!
x3
+ .
3!
Ans. : Convergent if xe < 1 and
divergent if xe 1.
.
+ (a + 3) 2
Ans. : Convergent if xe 1 and
divergent if xe > 1
5.10 CAUCHY’S ROOT TEST
1
If
un is a positive term series and if lim(un ) n
l , then
n
(i)
un is convergent if l < 1.
(ii)
un is divergent if l > 1.
Proof:
1
Case I: If lim(un ) n
n
l 1.
1
Consider a number l < r < 1 such that (un ) n < r for all n > m
un < r n for all n > m
The geometric series
r
n
r r
2
r
3
…
... (1)
5.28
r r2
Sn
r3 … rn
r (1 r n )
1 r
r (1 r n )
lim
n
1 r
r
, which is finite
1 r
lim S n
n
⎡∵ r < 1
⎤
⎢
⎥
n
r = 0⎥
⎢⎣∴ nlim
→∞
⎦
r n is convergent.
Hence, the series
un < r n for all n > m
From Eq. (1),
un < rn
Since
r n is convergent,
un is also convergent.
1
Case II: If lim(un ) n
l 1.
n
1
(un ) n > 1 for all n > m
… (1)
Neglecting the first m terms,
1
un
(um +1 ) m +1
1
Sn
1
1
(um + 2 ) m + 2
(um + 3 ) m + 3 …
1
(um +1 ) m +1
(um + 2 ) m + 2
1
1
(um + 3 ) m + 3 … (um + n ) m + n
lim S n
n
[Using Eq. (1)]
1 1 1…
1 1 1… n terms
n
lim n
n
∞
The series
∑u
n
is divergent. The nature of a series remains unchanged if we
n = m +1
∞
neglect a finite number of terms in the beginning. Hence, the series
∑u
n
is divergent.
n= 1
1
Note: If lim(un ) n
n
1, the root test fails, i.e., no conclusion can be drawn about the
convergence or divergence of the series.
Example 1: Test the convergence of the series
1
2
+
3
5
Solution:
2
+
un
3
7
3
n
+…+
n
2n + 1
n
n
2n + 1
+… .
5.29
1
lim(un ) n
n
2n + 1
lim
n
n
1
lim
n
2+
1
2
1
n
1
Hence, by root test, the series is convergent.
Example 2: Test the convergence of the series
22
12
Solution:
lim(un )
33
23
1
n
n +1
n
1
n
lim
n
n
3
2
44
34
3
4
3
….
n +1
n
1
n +1
1
1
n
2
n
n +1
n +1
n
un
(un )
1
2
1
n +1
n
1
1
n
1
n
1
1
n
1
1
n
1
n
1
1
n
1
1
n
(e 1)
1
Hence, by root test, the series is convergent.
1
Example 3: Test the convergence of the series
1
1+
n
un =
Solution:
⎛ 1⎞
⎜1 + ⎟
⎝ n⎠
1
n2
1
1+
1
n
.
1
(un ) n
lim(un ) n
n2
lim
1
n
n
1
n
1+
1
n
n
1
1
e
Hence, by root test, the series is convergent.
Example 4: Test the convergence of the series
Solution:
un
( n log n) n
.
2 n nn
(n log n) n
2n n n
1
e 1
1
5.30
1
(n log n)
1
lim
n
n
n
2n
2
1
1
1
lim n
1
2 n 1 2
Hence, by root test, the series is convergent.
lim(un ) n
log n
n
lim
[Using L’Hospital’s rule]
( n + 1) n x n
.
nn + 1
Example 5: Test the convergence of the series
Solution:
(n + 1) n x n
n n +1
1
( n + 1) x ( n + 1) x
( un ) n =
=
n +1
1
nn
n.n n
1
1 x
lim(un ) n = lim 1 +
n
n
n 1n
n
=x
un
1
⎡
⎤
n n = 1 using indeterminate form (∞°) method ⎥
⎢∵ lim
x →∞
⎣
⎦
Hence, by root test, the series is convergent, if x < 1 and divergent if x > 1. Root test
fails for x = 1.
1
2
Example 6: Test the convergence of the series
⎛ n ⎞
un = ⎜
⎝ n + 1 ⎟⎠
Solution:
(un )
1
n
n
n +1
2
x
3
1
n
lim
n
2
4
5
x2
3
x3 … .
n −1
x n −1
n 1
n
x
n 1
n
1
lim(un ) n
3
4
1
1+
n
n +1
1
n
( x)
1
n
1
n
1
1
n
1
( x)
1
n
x
Hence, by root test, the series is convergent, if x < 1 and is divergent if x > 1. Root test
fails for x = 1.
5.31
Exercise 5.5
Test the convergence of the following series:
1
1. 1 2
2
1
33
1
44
n
5.
… .
[Ans. : Convergent]
2.
1
.
(log
n) n
n= 2
3.
n +1
.
3n
nx
.
n +1
Ans. : Convergent for x < 1,
divergent for x > 1
2
3
6. 1 + x + x + x + … ( x > 0) .
2 32 43
[Ans. : Convergent]
[Ans. : Convergent]
n
7.
1
1+
n
n2
.
[Ans. : Convergent]
[Ans. : Divergent]
3
1
1+
4.
n2
(1 + nx ) .
nn
Ans. : Convergent if x < 1 and
divergent if x > 1
n
8.
.
n
[Ans. : Convergent]
5.11 CAUCHY’S INTEGRAL TEST
If
1
f ( n) is a positive term series where f (n) decreases as n increases and let
un
f ( x )dx = I , then
(i)
un is convergent if I is finite
(ii)
un is divergent if I is infinite
Proof: Consider the area under the curve y = f (x) from x = 1 to x = n + 1 represented
as
∫
n +1
1
f ( x )dx. Plot the terms f (1), f (2), f (3)…….f (n), f (n + 1).
The area
∫
n +1
1
f ( x)dx lies between the sum of the areas of smaller rectangles and sum
of the areas of larger rectangles
f ( 2) + f (3) +
f ( n + 1) ≤ ∫
n +1
1
f ( x )dx ≤ f (1) + f ( 2) + f (3) +
Sn +1 − f (1) ≤ ∫
n +1
1
As n
f ( x )dx ≤ Sn
first inequality reduces to
∞
lim Sn +1 ≤ ∫ f ( x ) dx + f (1)
n →∞
1
+ f ( n)
5.32
y
∞
This shows that if ∫ f ( x) dx
1
Σ f ( n) = Σun is convergent.
As n
second inequality reduces to
∫
∞
1
f ( x ) dx ≤ lim Sn
n →∞
y¢ = f (x)
∞
lim Sn ≥ ∫ f ( x )dx
n →∞
or
1
f (1) f (2) f (3)
∞
This shows that if ∫ f ( x)dx is infinite,
1
Σ f ( n) = Σun is divergent.
f (n)
1 2
f (n+1)
n n+1
3
x
Fig. 5.1
Example 1: Test the convergence of the series
1
.
n = 2 n log n
Solution:
un =
1
= f ( n)
n log n
1
x log x
∞
m
1
1
dx = lim ∫
f ( x ) dx = ∫
dx
2 x log x
m → ∞ 2 x log x
f ( x) =
∫
∞
2
= lim log log x
m →∞
m
⎡
⎤
f ′( x )
⎢∵ ∫ f ( x ) dx = log f ( x ) ⎥
⎣
⎦
2
= lim (log log m − log log 2) → ∞
m →∞
Hence, by Cauchy’s integral test, the series is divergent.
n2 e
Example 2: Test the convergence of the series
n3
.
n=1
Solution:
∫
∞
1
un = n 2 e − n
3
f ( x) = x 2 e − x
3
∞
f ( x)dx = ∫ x 2 e − x dx
3
1
⎡ 1 m 3
⎤
= lim ⎢ − ∫ e − x (−3 x 2 )dx ⎥
1
m →∞
⎣ 3
⎦
m
3
⎡ 1
⎤
= lim ⎢ − e − x ⎥
m →∞
1 ⎦
⎣ 3
(
)
⎡⎣∵ e f ( x ) f ′( x)dx = e f ( x ) ⎤⎦
3
1
⎡ 1
⎤
= lim ⎢ − e − m − e −1 ⎥ = − (e −∞ − e −1 )
m →∞
3
⎣ 3
⎦
1⎛
1⎞ 1
= − ⎜0 − ⎟ =
(finite)
e ⎠ 3e
3⎝
Hence, by Cauchy’s integral test, the series is convergent.
5.33
Example 3: Show that the harmonic series of order p,
1
1
= p
p
1
n =1 n
1
2p
1
3p
…
Solution:
un
f ( x)
is convergent if p > 1 and is divergent if p
1
np
1
xp
f ( x)dx
1
1.
1
1
dx
xp
m
⎛ m1− p
x − p +1
1 ⎞
= lim
= lim ⎜
−
m →∞ − p + 1
m →∞ ⎝ 1 − p
1 − p ⎟⎠
1
1
,
1− p
→ ∞,
=−
If
p >1
p <1
p = 1,
∫
∞
1
f ( x)dx = ∫
∞
1
1
dx
x
m
= lim log x 1 = lim (log m − log 1)
m →∞
m →∞
= log ∞ → ∞
The integral
∫
∞
1
f ( x)dx
p
p 1.
Hence, by Cauchy’s integral test, the series is convergent if p > 1 and is divergent if
p 1.
Exercise 5.6
Test the convergence of the following series.
1
1.
n =1
n
2
ne n .
3.
.
n =1
[Ans. : Convergent]
[Ans. : Divergent]
2.
1
.
n
+1
n =1
4.
2
[Ans. : Convergent]
1
.
2
n =1 n(log n)
[Ans. : Convergent]
5.34
5.12 ALTERNATING SERIES
An infinite series with alternate positive and negative terms is called an alternating
series.
n 1
Leibnitz’s test for alternating series: An alternating series ( 1) un is convern =1
gent if
(i) each term is numerically less than its preceding term, i.e, un +1 < un or
un > un +1
(ii) lim un
0
n
Example 1: Test the convergence of the series 1
1
2
1
3
4
….
1
un =
Solution:
1
n
(i) The given series is an alternating series.
(ii)
un
1
un +1
1
n 1
n +1
n
n
n n +1
0
un > un +1
(iii)
lim un = lim
n
n
1
n
=0
Hence, by Leibnitz’s test, the series is convergent.
Example 2: Test the convergence of the series x
Solution:
un =
x2
2
x3
3
x4
4
……(if x
1) .
xn
n
(i) The given series is an alternating series.
(ii) un
un +1
xn
n
x n +1
x n [( n + 1) − nx ]
=
n +1
n( n + 1)
=
x n [1 (1 x )n]
> 0,
n( n + 1)
[∵ 0 < x < 1]
un > un +1
xn
=0
n
n
n
Hence, by Leibnitz’s test, the series is convergent.
(iii)
lim un = lim
∵ lim x n = 0 if x < 1
n
5.35
Example 3: Test the convergence of the series
un =
Solution:
1
1p
1
2p
1
3p
1
4p
….
1
np
Case I: If p > 0,
(i) The given series is an alternating series.
1
1
(ii) un un +1
p
( n + 1) p
n
=
( n 1) p n p
> 0,
n p ( n + 1) p
if p > 0
un > un +1
(iii) lim un = lim 1 = 0, if p > 0
n
n
np
Hence, by Leibnitz’s test, the series is convergent if p > 0.
Case II: If
p<0
In this case the conditions (ii) and (iii) of the Leibnitz’s test are not satisfied.
Hence, the given series is oscillatory if p < 0.
Exercise 5.7
Test the convergence of the following series:
4. 1 2 x 3 x 2 4 x 3 … ( x 1) .
1
………
5
[Ans. : Convergent]
[Ans. : Convergent]
x2
x3
x4
5. x
n
n 1
2
3
2.
( 1)
.
1+ x 1+ x 1+ x 1+ x4
2n 1
n =1
+ … (0 < x < 1) .
[Ans. : Oscillatory]
[Ans. : Convergent]
1 1
1
(1
2)
(1
2
3)
3. 2
2 33
43
1
(1 2 3 4) … .
53
[Ans. : Convergent]
1. 1
1
2
1
3
1
4
5.13 ABSOLUTE CONVERGENCE OF A SERIES
The series
n =1
un with both positive and negative terms (not necessarily alternative)
is called absolutely convergent if the corresponding series
terms is convergent.
| un | with all positive
n =1
5.36
Conditional convergence of a series: If the series
divergent, then the series
n =1
un is convergent and
| un | is
n =1
un is called conditionally convergent.
n =1
Note 1: Every absolutely convergent series is a convergent series but converse is not
true.
Note 2: Any convergent series of positive terms is also absolutely convergent.
Example 1: Test the series for absolute or conditional convergence
1
2
3
3
32
4
33
…… .
Solution:
un
( 1) n
n
1
n 1
3
2 3 4
| un| = 1 + + 2 + 3 + …
3 3 3
n =1
n
| un | = n 1
3
n +1
| un +1| = n
3
|u |
n
3n
3
lim n = lim n −1 ⋅
= lim
= 3 >1
n →∞ | u
n →∞ 3
n→ ∞
1
|
n
+
1
n +1
1+
n
| ln | is convergent and hence, the given series is absolutely
By ratio test,
n =1
convergent.
Example 2: Test the series for absolute or conditional convergence
2
3
3 1
4 2
4 1
5 3
5 1
….
6 4
Solution:
un
( 1) n
| un |
n =1
| un | =
Let vn =
1
n
n +1 1
⋅
n+2 n
2
3
1
n +1 1
n+2 n
3 1
4 2
4 1 5 1
….
5 3 6 4
5.37
1
1+
| un |
n +1
n = 1 (finite and non-zzero)
lim
= lim
= lim
n →∞ v
n →∞ n + 2
n →∞
2
n
1+
n
1
and vn = is divergent since p = 1.
n
By comparison test, | un | is divergent.
Hence, the given series is not absolutely convergent.
To check for conditional convergence we need to check the convergence of the given
series.
(i) The given series
un is an alternating series.
(ii)
| un | | un +1|
n +1
n( n + 2)
=
n+2
( n + 1)( n + 3)
n2 + 3n + 3
>0
n( n + 1)( n + 2)( n + 3)
| un | > | un +1 |
lim | un | = lim
(iii)
n
n
n +1
n( n + 2)
1
n =0
= lim
n
2
n 1+
n
1+
By Leibnitz’s test, given series un is convergent. Thus, un is convergent and
is divergent. Hence, the given series is conditionally convergent.
| un |
Exercise 5.8
Test the following series for absolute or conditional convergence:
1 1 1 1 1
4. 1 2 3 4 … .
….
2 5 10 17
2 3 4 5 6
[Ans. : Conditionally convergent]
[Ans. : Conditionally convergent]
1 1 1 1 1
sin x sin 2 x sin 3 x
2. 1 2
….
5.
….
2 32 4 2 52 6 2
13
23
33
[Ans. : Absolutely convergent]
[Ans. : Absolutely convergent]
1. 1
3. 1
1
1
1
….
2
3
4
[Ans. : Conditionally convergent]
5.38
5.14 UNIFORM CONVERGENCE OF A SERIES
un ( x ) of real valued functions defined in the interval (a, b) is said to
The series
n =1
0, there exists a number m
converge uniformly to a function S (x) if for a given
independent of x such that for every x ( a, b),
for all n > m
| Sn ( x ) S ( x ) |
Sn ( x ) = u1 ( x ) + u2 ( x ) + … + un ( x )
where,
Weierstrass’s M-Test: The series
un ( x ) is said to converge uniformly in an
n =1
M n of positive constants such that
interval (a, b), if there exists a convergent series
n =1
| un ( x ) |
Proof: Let
M n for all x
( a, b)
0, there exists a number m such
M n is convergent, then for a given
n =1
that | S − S n | < ∈ for all n > m,
where S
M1
then | M n +1
M n +1
M2
M n+ 2 … |
M n+ 2 …
Now | un ( x ) |
M3 …
and Sn = M1 + M 2 + … + M n
for all n > m
for all n > m
M n for all x
[∵ Mn is positive constant]
( a, b)
| un +1 ( x ) un + 2 ( x ) … | | un +1 ( x ) | | un + 2 ( x ) | …
M n +1
M n+ 2 …
for all n > m
| S ( x ) Sn ( x ) |
where,
for all n > m
Sn (x) = u1(x) + u2 (x) + … + un (x)
Since m does not depend on x, the series
terval (a, b).
Example 1: Test the series
Solution:
un ( x ) converges uniformly in the inn =1
1
for uniform convergence.
3 2
n=1 n + n x
4
un ( x ) =
1
n 4 + n3 x 2
5.39
| un ( x ) | =
<
1 for all x
n4
Mn =
Mn =
n =1
1
n + n3 x 2
4
1
is convergent since p
4
n
n =1
[∵ x 2 > 0]
R
1
n4
4 > 1.
Hence, by M-test, the series is uniformly convergent for all real values of x.
Example 2: Test the series
Solution:
un ( x )
cos( x 2 + n 2 x)
n(n 2 + 2)
cos( x 2 + n2 x ) | sin( x 2 + n2 x ) |
=
n( n2 + 2)
n( n2 + 2)
| un ( x ) | =
Mn =
n =1
cos( x 2 + n2 x )
for uniform convergence.
n( n2 + 2)
n=1
≤
1
n + 2n
<
1
n3
Mn =
1
n3
⎡∵ − 1 ≤ sin ≤ 1⎤
⎢
| sin | ≤ 1⎥⎦
⎣
for all x ∈ R
3
1
is convergent since p = 3 > 1.
3
n
n =1
Hence, by M-test, the series is uniformly convergent for all real values of x.
Example 3: Test the series sin x
sin 2 x
sin 3 x
sin 4 x
2 2
3 3
4 4
convergence.
Solution:
un ( x )
( 1) n
| un ( x ) | =
1
sin nx
n n
sin nx
n n
… . for uniform
5.40
1
for all x ∈ R
3
2
Mn =
n
1
⎡∵ −1 ≤ sin q ≤ 1⎤
⎢
| sin q | ≤ 1⎥⎦
⎣
3
n2
1
Mn =
n =1
n =1
3
2
is convergent since p =
3
>1.
2
n
Hence, by M-test, the series is uniformly convergent for all real values of x.
r n cos n2 x is uniformly
Example 4: Show that if 0 < r < 1, the series
n=1
convergent.
Solution:
un ( x ) = r n cos n2 x
un ( x ) = r n cos n2 x
≤| r n | for all x ∈ r
= rn , 0 < r < 1
⎡∵− 1 ≤ cos ≤ 1⎤
⎢ cos ≤ 1
⎥
⎣
⎦
Mn = rn
Mn =
n =1
r n = r + r 2 + r3 + …
n =1
which is convergent being a geometric series with 0 < r < 1.
Hence, by M-test, the series is uniformly convergent for all real values of x.
Exercise 5.9
Test the following series for uniform convergence:
4. Show that if 0 < r < 1, then the series
sin( x 2 + nx )
1.
; for all real x.
n( n + 2)
n =1
r n sin a n x is uniformly convergent
n =1
[Ans. : Uniformly convergent]
for all real values of x.
1
; for all real x and p > 1.
2.
5. Show that
p
n
+
nq x 2
n =1
1
1
1
1
…
[Ans. : Uniformly convergent]
1+ x2 2 + x2 3 + x2 4 + x2
sin x sin 2 x sin 3 x sin 4 x
converges uniformly in the interval
3.
+ 2 + 2 +
+….
12
2
3
42
x 0 but not absolutely.
[Ans. : Uniformly convergent]
5.41
FORMULAE
Sequence
A sequence {un} is said to be convergent, divergent or oscillatory according
un is finite, infinite or not unique
as nlim
→∞
respectively.
Series
The infinite series un is said to be convergent, divergent or oscillatory accordor not unique
ing as lim S n is finite,
n →∞
respectively.
If a positive term series un is convergent, then lim un = 0 but converse is not
n →∞
true i.e., if lim un = 0, the series may con-
When Raabe’s test fails, Logarithmic test
can be applied.
Logarithmic test: If
un is a positive
u ⎞
⎛
n log n ⎟ = l,
term series and if nlim
→∞ ⎜
u
n +1 ⎠
⎝
then
(i) un is convergent
(ii) un is divergent
(iii) Test fails
Cauchy’s root test: If un is a positive term
1
series and if lim (un ) n = l, then
n→∞
n →∞
verge or diverge. If lim un
n →∞
Comparison test: If un and vn are
series of positive terms such that
u
lim n = l (finite and non-zero),
n→∞ v
n
then both series converge or diverge together.
D’Alembert’s ratio test: If un is a posiu
tive term series and lim n = l, then
n →∞ u
n +1
(i) un is convergent if l > 1.
(ii) un is divergent
if l < 1.
(iii) The ratio test fails if l = 1.
When Ratio test fails, Raabe’s or Logarithmic Test can be applied.
Raabe’s test (higher ratio test): If un is
a positive term series and if
⎛ un
⎞
lim n
− 1⎟ = l, then
n→∞ ⎜ u
⎝ n +1
⎠
(i) un is convergent
(ii) un is divergent
(iii) Test fails
(i)
(ii)
0, the series
is divergent.
if l > 1
if l < 1
if l = 1
if l > 1.
if l < 1.
if l = 1.
un is convergent
un is divergent
if l < 1.
if l > 1.
This test is preferred when un contains nth
powers of itself.
Cauchy’s integral test: If un = f (n) is a
positive term series where f (n) decreases
as n increases and let
(i)
(ii)
∫
∞
1
f (x) dx = I, then
un is convergent if I is finite.
un is divergent if I is infinite.
This test is preferred when evaluation of
the integral of f (x) is easy.
Alternating series: An infinite series
with alternate positive and negative terms
is called an alternating series.
Leibnitz’s test: An alternating series
∞
∑ (−1)
n −1
un is convergent if
n =1
(i) each term is numerically less
than its preceding term, i.e, un +1 <
un or un > un +1
(ii) lim un = 0
n→∞
5.42
Absolute Convergence
∞
∞
The series
∑u
n
with both positive and
n =1
negative terms (not necessarily alternative) is called absolutely convergent if
∞
the corresponding series
∑ |u |
n
with all
∞
∑ | u | is divergent, then the series ∑ u
n
n =1
n
n =1
is called conditionally convergent.
∞
Weierstrass’s M-test: The series
∑ u ( x)
n
n =1
is said to converge uniformally in an
n =1
positive terms is convergent.
interval (a, b), if there exists a convergent
Conditional Convergence
series
∞
∞
If the series
∑u
∑M
n
of positive constants such
n =1
n
is convergent and
that |un(x)|
Mn for all x
(a, b).
n =1
MULTIPLE CHOICE QUESTIONS
Choose the correct alternative in each of the following:
∞
1. The series
∑ cos ⎛⎝ 1n ⎞⎠
is
n =1
(a) convergent
(c) oscillatory
(b) divergent
(d) none of these
∞
2. The series
n
∑ n x+ 1
3
at x = 1 is
n =1
(a) convergent
(c) oscillatory
(b) divergent
(d) none of these
3. The series 1 – 12 + 12 – 12 + …
2
3
4
is
(a) convergent (b) divergent
(c) oscillatory (d) none of these
4. The series 2 – 3 + 4 – 5 + … is
2
3
4
(a) convergent but not absolutely
convergent
(b) divergent
(c) absolutely convergent
(d) oscillates finitely
5. The series 1 + 1 + 1 + 1 +
2
3
4
… is
(a) convergent but not absolutely
convergent
(b) oscillatory
(c) divergent
(d) absolutely convergent
6. In a series of positive terms un if
lim un 0, then series un is
n→∞
(a)
(b)
(c)
(d)
convergent
divergent
not convergent
oscillatory
7. The series 1 – 1 + 1 – 3 + 1 – 7
4
2
8
… is
(a)
(b)
(c)
(d)
convergent
conditionally convergent
absolutely convergent
oscillatory
8. The series
1 – 1 + 1
a +1
a+2
a+3
1 + … convergent if
a+4
(a) a > 0
(b) a < 0
(c) a < –1
(d) none of these
–
9. The series 1 – 2x + 3x2 – 4x3 + …
where 0 < x < 1 is
(a) convergent (b) divergent
(c) oscillatory (d) none of these
10. The series
1
2
+
1 + 2−1
1 + 2− 2
5.43
3
… is
1 + 2− 3
(a) convergent
(c) oscillatory
+
(b) divergent
(d) none of these
11. The series whose nth term is
n3 + 1
– n3 is
(a) convergent
(c) oscillatory
(b) divergent
(d) none of these
1
12. The series whose nth term is sin is
n
(a) convergent (b) divergent
(c) oscillatory (d) none of these
13. The series
2 + 3 + 4 + 5 + n +1 + … is
1
4
9
16
n2
(a) convergent (b) divergent
(c) oscillatory (d) none of these
14. Which of the following is true?
1
1
1
(a) 1 + 1 + 1 + 1 + … is
2 3 33 4 3
convergent
1 1
+ +
2 3
convergent
1
1
(c) 1 + 2 + 2
2
3
convergent
1
1
(d) 1 − k + k
2
3
divergent
(b) 1 +
1
+ … is
4
does not give absolute convergence
of a series?
(a) Root Test
(b) Comparison Test
(c) Ratio Test
(d) Leibnitz Test
17. Which one of the following infinite
series is convergent!
∞
∞
1
1
(b) ∑ 1
(a) ∑ 2
n =1 n − n
n =1 n
n +n
∞
1
(c) ∑
n
n =1 n −
∞
(d)
1
+ … is
4k
15. If un is a series of positive terms,
then
(a) convergence of (–1)nun implies
the convergence of un.
(b) convergence of un implies the
convergence of (–1)nun.
(c) convergence of (–1)nun implies
the divergence of un.
(d) divergence of un implies the divergence of (–1)nun.
3
n =1
n2
− n 2 + 1)
18. The series
2 2
33 x 3
44 x 4
x+ 2 x +
+
+…
3!
4!
2!
is convergent if
(a) a < x <
1
e
(b) x >
2
3
<x<
e
e
19. The series
(c)
+ 12 + … is
4
+
∑ (n
(a)
(b)
(c)
(d)
(d)
3
4
<x<
e
e
3 x 4 + 8 x6 + 15 x8 + …
5
10
17
2
n − 1 2n
+ 2
x + … is
n +1
convergent if x2 1 and divergent if x2 < 1
convergent if x2 1 and divergent if x2 > 1
convergent if x2 < 1 and divergent if x2 1
convergent if x2 > 1 and divergent if x2 1
20. Which one of the following statement hold?
∞
(a) The series
∑x
n=0
16. Which one of the following test
1
e
each x
[–1, 1]
n
converges for
5.44
∞
(b) The series
∑x
n
converges
n=0
uniformly in (–1, 1)
∞
(c) The series
n
∑ xn
converges for
(a) p < q – 1
(c) p q – 1
n =1
each x
[–1, 1 [
∞
(d) The series
21. If p and q are positive real nump
3p
bers, then the series 2q + q +
2
1
4 p + … is convergent for
3q
∑ nx
(b) p < q + 1
(d) p q + 1
n
2
converges
n =1
uniformly in (–1, 1)
Answers
1. (b)
8. (a)
15. (b)
2. (a)
9. (a)
16. (d)
3. (a)
10. (b)
17. (a)
4. (d)
11. (a)
18. (a)
5. (c)
12. (b)
19. (c)
6. (c)
13. (b)
20. (c)
7. (d)
14. (c)
21. (a)
Integral
Calculus
Chapter
6
6.1 INTRODUCTION
Integral calculus helps in developing techniques for determination of the integral of
a given function. In this chapter, we will study applications of integral calculus, such
as finding lengths of arcs of curves, areas of planes, and volumes and surface areas
of solids of revolution. The concept of reduction formula also helps in solving the
integral of a given function. Integral calculus deals with the derivation of formulas for
finding anti-derivatives. It is also useful in solving differential equations.
6.2 REDUCTION FORMULAE
Reduction formulae reduce an integral to a simple integral by repeatedly using integration by parts.
6.2.1 Reduction Formula for ∫ sin n x dx ; n > 0
∫ sin
n
x dx = ∫ sin n −1 x sin x dx
Integrating by parts,
∫ sin
n
x dx = sin n −1x( − cos x ) − ∫ ( n − 1) sin n − 2 x cos x ( − cos x ) dx
= − cos x sin n −1 x + ( n − 1) ∫ sin n − 2 x (1 − sin 2 x ) d x
= − cos x sin n −1 x + ( n − 1) ∫ sin n − 2 x dxx − ( n − 1) ∫ sin n x d x
n ∫ sin n x dx = − cos x sin n −1 x + ( n − 1) ∫ sin n − 2 x dx
1
n −1
x dx = − cos x sin n −1 x +
sin n − 2 x dx
n
n ∫
Note: If n is odd positive integer, the function can be easily integrated using substitution cos x = t.
∫ sin
n
6.2
Engineering Mathematics
6.2.2 Reduction Formula for
∫ cos
n
∫ cos x dx ; n > 0
n
x dx = ∫ cos n −1 x cos x dx
Integrating by parts,
∫ cos x dx = cos
n
n −1
x sin x − ∫ ( n − 1) cos n − 2 x ( − sin x ) sin x dx
= sin x cos n −1 x + ( n − 1) ∫ cos n − 2 x (1 − cos 2 x ) dx
= sin x cos n −1 x + ( n − 1) ∫ cos n − 2 x dx − ( n − 1) ∫ cos n x dx
n ∫ cos n x dx = sin x cos n −1 x + (n − 1)∫ cos n − 2 x dx
n −1
cos n − 2 x dx
n ∫
Note: If n is an odd positive integer, the function can be easily integrated using substitution sin x = t.
1
∫ cos x dx = n sin x cos
n
n −1
x+
6.2.3 Reduction Formula for
∫ tan
n
tan n x dx
x dx = ∫ tan n − 2 x tan 2 x dx
= ∫ tan n − 2 x (sec 2 x − 1) dx
= ∫ tan n − 2 x sec 2 x dx − ∫ tan n − 2 x dx
=
tan n −1 x
− tan n − 2 x dx
n −1 ∫
6.2.4 Reduction Formula for
∫ cot
n
⎡
⎢∵
⎣
f ′( x ) d x =
[ f ( x )]n +1 ⎤
⎥
n +1 ⎦
n
∫ [ f ( x)] f ′( x) dx =
[ f ( x )]n +1 ⎤
⎥
n +1 ⎦
∫ [ f ( x)]
n
cot n x dx
x dx = ∫ cot n − 2 x cot 2 x dx
= ∫ cot n − 2 x (cosec 2 x − 1) dx
= ∫ cot n − 2 x cos ec 2 x dx − ∫ cot n − 2 x dx
=−
cot n −1 x
− cot n − 2 x dx
n −1 ∫
⎡
⎢∵
⎣
6.2.5 Reduction Formula for
∫ sec x dx = ∫ sec
n
n− 2
x sec 2 x dx
sec n x dx
Integral Calculus
6.3
Integrating by parts,
∫ sec x dx = sec
n
n− 2
x tan x − ∫ ( n − 2) sec n − 2 x tan x tan x dx
= tan x sec n − 2 x − ( n − 2) ∫ sec n − 2 x (sec 2 x − 1) dx
(
)
= tan x sec n − 2 x − ( n − 2) sec n x − ∫ sec n − 2 x dxx
[1 + ( n − 2)]∫ sec x dx = tan x sec
n
n− 2
x + ( n − 2) ∫ sec
n− 2
x dx
n− 2
tan x sec x n − 2
+
sec n − 2 x dx
n −1
n −1 ∫
Note: If n is an even positive integer, the function can be easily integrated using substitution of tan x = t.
∫ sec x dx =
n
6.2.6 Reduction Formula for
∫ cosec x dx = ∫ cosec
n
n− 2
cosec n x dx
x cosec 2 x dx
Integrating by parts,
∫ cos ec x dx = cosec
n
n− 2
x ( − cot x ) − ∫ ( n − 2) cosec n − 3 x ( − cosec x cot x )( − cot x ) dx
= − cot x cosec n − 2 x − ( n − 2) ∫ cosec n − 2 x (cosec 2 x − 1) dx
= − cot x cosec n − 2 x − ( n − 2)
(∫ cosec x − ∫ cosec
n
n− 2
x dx
)
[1 + ( n − 2)]∫ cosec n x dx = − cot x cosec n − 2 x + ( n − 2) ∫ cosec n − 2 x dx
− cot x cosec n − 2 x n − 2
+
cosec n − 2 x dx
n −1
n −1 ∫
Note: If n is an even positive integer, the function can be easily integrated using substitution of cot x = t.
n
∫ cosec x dx =
6.2.7 Reduction Formula for ∫ sin m x cos n x dx ; m, n > 0
∫ sin
m
x cos n x dx = ∫ sin m −1 x (sin x cos n x ) dx
Integrating by parts,
m
n
∫ sin x cos x dx = −
cos n +1 x m −1
cos n +1 x
sin x + ∫
( m − 1) sin m − 2 x cos x dx
n +1
n +1
⎡
[ f ( x )]n +1 ⎤
n
⎢∵ ∫ [ f ( x )] f ′( x ) d x =
⎥
n +1 ⎦
⎣
=−
cos n +1 x sin m −1 x m − 1
+
sin m − 2 x cos n x (1 − sin 2 x ) dx
n +1
n +1 ∫
=−
cos n +1 x sin m −1 x m − 1
m −1
sin m − 2 x cos n x dx −
sin m x cos n x dx
+
∫
n +1
n +1
n +1 ∫
6.4
Engineering Mathematics
cos n +1 x sin m −1 x m − 1
⎛ m −1⎞
m
n
x
x
x
1
+
sin
cos
d
=
−
+
sin m − 2 x cos n x dx
⎟
⎜⎝
n +1 ⎠ ∫
n +1
n +1 ∫
m
n
∫ sin x cos x dx = −
cos n +1 x sin m −1 x m − 1
+
sin m − 2 x cos n x dx
m+n
m+n∫
Similarly, it can be proved that
m
n
∫ sin x cos x dx =
(i)
sin m +1 x cos n −1 x n − 1
+
sin m x cos n − 2 x dx
m+n
m+n∫
cos n +1 x sin m −1 x m − 1
+
sin m − 2 x cos n + 2 x dx
n +1
n +1 ∫
This formula is useful when m is positive and n is a negative integer.
∫ sin
(ii)
m
x cos n x dx = −
sin m +1 x cos n −1 x n − 1
+
sin m + 2 x cos n − 2 x dx
m +1
m +1 ∫
This formula is useful when m is negative and n is a positive integer.
m
n
∫ sin x cos x dx =
(iii)
cos n +1 x sin m +1 x m + n + 2
+
sin m + 2 x cos n x dx
m +1
m +1 ∫
This formula is useful when n is a negative integer.
m
n
∫ sin x cos x dx = −
(iv)
sin m +1 x cos n +1 x m + n + 2
+
sin m x cos n + 2 x dx
∫
m +1
m +1
This formula is useful when n is a negative integer.
(v)
m
n
∫ sin x cos x dx =
Example 1: Evaluate
sin 5 x dx.
Solution: Using reduction formula,
1
4
5
4
3
∫ sin x dx = − 5 cos x sin x + 5 ∫ sin x dx
1
4⎛ 1
2
⎞
= − cos x sin 4 x + ⎜ − cos x sin 2 x + ∫ sin x dx⎟
⎝
⎠
5
5 3
3
1
4
8
= − cos x sin 4 x − cos x sin 2 x − cos x
5
15
15
Example 2: Evaluate
sin6 x dx.
Solution: Using reduction formula,
∫ sin
6
1
5
x dx = − cos x sin 5 x + ∫ sin 4 x dx
6
6
3
1
5⎛ 1
⎞
5
= − cos x sin x + ⎜ − cos x sin 3 x + ∫ sin 2 x dx ⎟
⎝
⎠
4
6
6
4
Integral Calculus
6.5
1
5
5⎛ 1
1
⎞
= − cos x sin 5 x −
cos x sin 3 x + ⎜ − cos x sin x + ∫ sin 0 x dx ⎟
⎝
⎠
6
24
8
2
2
1
5
5
5
= − cos x sin 5 x −
cos x sin 3 x − cos x sin x + x
6
24
16
16
cos 3 x dx.
Example 3: Evaluate
Solution: Using reduction formula,
1
2
∫ cos x dx = 3 sin x cos x + 3 ∫ cos x dx
3
2
1
2
= sin x cos 2 x + sin x
3
3
tan6 x dx.
Example 4: Evaluate
Solution: Using reduction formula,
tan 5 x
− ∫ tan 4 x dx
5
⎞
tan 5 x ⎛ tan 3 x
=
−⎜
− ∫ tan 2 x dx ⎟
⎝ 3
⎠
5
5
3
tan x tan x ⎛ tan x
⎞
−
+⎜
− ∫ tan 0 x dx ⎟
=
⎝
⎠
5
3
1
tan 5 x tan 3 x
=
−
+ tan x − x
5
3
6
∫ tan x dx =
Example 5: Evaluate
p
2
p
4
cot 4 x dx.
Solution: Using reduction formula,
cot 3 x
− ∫ cot 2 x dx
3
cot 3 x
cot 3 x ⎛ cot x
⎞
+ cot x + x
=−
− ⎜−
− ∫ cot 0 x dx ⎟ = −
⎠
⎝
3
3
1
4
∫ cot x dx = −
∫
2
4
cot 4 x dx = −
cot 3 x
+ cot x + x
3
2
4
1
3 −8
= + −1− =
2 3
4
12
6.6
Engineering Mathematics
Example 6: Evaluate
sec 4 x dx .
Solution: Using reduction formula,
tan x sec 2 x 2
+ ∫ sec 2 x dx
3
3
tan x sec 2 x 2
=
+ ( tan x + 0 )
3
3
2
1
= tan x sec 2 x + tan x
3
3
4
∫ sec x dx =
Example 7: Evaluate
o
2
o
6
cosec 5 x dx .
Solution: Using reduction formula,
cot x cosec3 x 3
+ ∫ cosec3 x dx
4
4
cot x cosec3 x 3 ⎛ cot x cosec x 1
⎞
+ ∫ cosec x dx ⎟
=−
+ ⎜−
⎠
2
2
4
4⎝
1
3
3
= − cot x cosec3 x − cot x cosec x + log (cosec x − cot x )
4
8
8
5
∫ cosec x dx = −
∫
2
6
2
1
3
3
cosec5x dx = − cot x cosec3 x − cot x cosec x + log(cosec x − cot x )
4
8
8
6
3
1
3
3
= log 1 + ( 3 ) (8) + ( 3 ) ( 2) − log ( 2 − 3 )
8
4
8
8
=
11 3 3
− log (2 − 3 )
4
8
Example 8: Evaluate
sin 2 x cos 4 x dx .
Solution: Using reduction formula,
sin 3 x cos3 x 3
+ ∫ sin 2 x cos 2 x dx
6
6
⎞
1
1 ⎛ sin 3 x cos x 1
+ ∫ sin 2 x cos 0 x dx ⎟
= sin 3 x cos3 x + ⎜
⎝
⎠
6
2
4
4
1
1
1 ⎛ cos x sin x 1
⎞
= sin 3 x cos3 x + sin 3 x cos x + ⎜ −
+ ∫ sin 0 x dx ⎟
⎝
⎠
6
8
8
2
2
1
1
1
= sin 3 x cos3 x + sin 3 x cos x + ( x − sin x cos x )
6
8
16
2
4
∫ sin x cos x dx =
Integral Calculus
6.7
sin 4 x cos 4 x dx .
Example 9:
Solution: Using reduction formula,
∫ sin
4
x cos 4 x dx
cos5 x sin 3 x 3
+ ∫ sin 2 x cos 4 x dx
8
8
⎞
1
3 ⎛ − cos5 x sin x 1
+ ∫ sin 0 x cos 4 x dx ⎟
= − cos5 x sin 3 x + ⎜
⎠
8
8⎝
6
6
=−
⎞
1 ⎛ cos3 x sin x 3
1
1
+ ∫ cos 2 x dx ⎟
= − cos5 x sin 3 x − cos5 x sin x + ⎜
⎝
⎠
16
4
4
8
16
1
1
1
3 ⎛ sin x cos x 1
⎞
= − cos5 x sin 3 x − cos5 x sin x + cos3 x sin x +
+ ∫ cos 0 x dx ⎟
⎜
⎠
8
16
64
64 ⎝
2
2
1
1
1
3
= − cos5 x sin 3 x − cos5 x sin x + cos3 x sin x +
( x + sin x cos x )
8
16
64
128
Exercise 6.1
1. Evaluate
sin 4 x cos 2 x d x
(i)
sin 4 x dx
(ii)
tan 3 x dx
(ii)
(iii)
cot 5 x dx
(iv)
sec5 x dx
1
⎡
⎤
3
⎢ Ans. : (i) − 4 sin x cos x
⎥
⎢
⎥
1
⎢
+ ( x + sin x cos x ) ⎥
⎢
⎥
8
⎢
⎥
1
1
⎢
(ii) sin 5 x − sin 7 x ⎥
⎢⎣
⎥⎦
5
7
(v)
4
cosec x dx
⎡
3 ⎤
cos x sin 3 x 3
− cos x sin x + x ⎥
⎢ Ans. : (i) −
4
8
8
⎢
⎥
1 2
⎢
⎥
(ii) tan x − log sec x
⎢
⎥
2
⎥
⎢
1
1
⎢
(iii) − cot 4 x + cot 2 x + log sin x ⎥
⎥
⎢
4
2
⎥
⎢
1
3
⎥
⎢
(iv ) tan x sec3 x + tan x sec x
4
8
⎥
⎢
⎥
⎢
3
+ log (sec x + tan x ) ⎥
⎢
8
⎥
⎢
⎥
⎢
1
2
2
(
)
−
cot
x
cos
x
−
cot
x
v
ec
⎥
⎢
⎣
⎦
2
3
2. Evaluate
(i)
sin 2 x cos 2 x dx
3. Show that
(i)
(ii)
sin 5 x
1
∫ cos4 x dx = 3 cos3 x
2
−
− cos x
cos x
cos5 x
sin 4 x
=
d
x
∫ sin x
4
− sin 2 x + logsin x
6.8
Engineering Mathematics
4. Evaluate
(i)
sec x tan 5 x dx
(ii)
sin 3 x sec7 x dx
(iii)
cos3 x cosec 4 x dx
2
0
1 5
2 3
⎡
⎤
⎢ Ans.: (i) 5 sec x − 3 sec x + sec x ⎥
⎢
⎥
1
1 6
4
⎢
⎥
(ii) tan x + tan x
⎢
⎥
4
6
⎢
⎥
1
⎢
(iii) − cosec3 x + cosec x ⎥
3
⎣⎢
⎦⎥
sin n x dx and
2
0
cos n x dx , n > 0
Using reduction formula,
∫ sin
n
1
n −1
x dx = − cos x sin n −1 x +
sin n − 2 x dx
n
n ∫
2
n − 1 2 n− 2
1
I n = ∫ 2 sin n x dx = − cos x sin n−1 x +
sin x dx
0
n
n ∫0
0
n − 1 2 n− 2
sin x dx
n ∫0
n −1
I
=
n n− 2
Using this recurrence relation,
= 0+
In
2
In
4
n
n
n
=
n
=
3
In
2
5
In
4
4
6
…
………
…
………
2
I1 , if m is odd
3
1
I 2 = I 0 , if m is even
2
I3 =
Substituting these values,
∫
2
0
n −1 n − 3 n − 5
2
⋅
⋅
… I1 ,
n n−2 n−4
3
n −1 n − 3 n − 5
1
=
⋅
⋅
… I0 ,
n n−2 n−4
2
sin n x dx =
if n is odd
if n is even
Integral Calculus
I1 = ∫ 2 sin x dx = − cos x 02 = 1
Now,
0
p
2
I0
Hence,
6.9
∫
2
0
0
sin n x dx =
=
0
p
2
1 dx
2
n −1 n − 3 n − 5
⋅
⋅
… ,
3
n n−2 n−4
if n is odd
1
n −1 n − 3 n − 5
⋅
… ⋅ ,
⋅
2 2
n n−2 n−4
if n is even
p
2
x
Putting
p
2
sin 0 x dx
dx
y,
dy
⎛
⎞
I n = ∫ 2 sin n x dx = ∫ 2 sin n ⎜ − y ⎟ ( −dy ) = ∫ 2 cos n y dy
0
0
0
⎝2
⎠
= ∫ 2 cos n x dx
0
2
n −1 n − 3 n − 5
n
∫ 02 cos x dx = n ⋅ n − 2 ⋅ n − 4 … 3 ,
=
1
n −1 n − 3 n − 5
⋅
… ⋅ ,
⋅
2 2
n n−2 n−4
if n is odd
if n is even
Corollary: Certain definite integrals can be evaluated using
2
0
cos n x dx, where n is a positive integer.
x = a sin q ,
(i) Putting
a
0
xn
1 x2
dx =
p
2
0
= an
a n sin n q a cos q
dq
cos q
p
2
0
sin n q dq
x = a tan q ,
(ii) Putting
0
1
dx =
2
(a + x 2 )n
=
p
2
0
a sec 2 q
dq
a 2 n (sec 2 q ) n
p
2
1
a2n
1
0
cos 2 n 2 q dq
2
0
sin n x dx and
6.10
Engineering Mathematics
x = a tan q ,
(iii) Putting
∫
0
π
1
∞
(a2 + x 2 )
n+
1
2
dx = ∫ 2
0
sec 2 θ
dθ
sec 2 n +1 θ
π
= ∫ 2 cos 2 n −1 θ dθ
0
∫
π
2
0
sin m x cos n x dx ; m, n > 0
Using reduction formula,
∫ sin
m
x cos n x dx = −
cos n +1 x sin m +1 x m − 1
+
sin m − 2 x cos n x dx
∫
m+n
m+n
I m , n = ∫ 2 sin m x cos n x dx = −
0
cos n +1 x sin m +1 x
m+n
2
0
m − 1 2 m−2
+
sin x cos n x dx
m + n ∫0
m − 1 2 m−2
sin x cos n x dx
m + n ∫0
m −1
I m − 2, n
=
m+n
Using this recurrence relation,
= 0+
m 3
Im
m n 2
m 5
I m 4, n =
Im
m n 4
…
………
Im
2, n
=
…
4, n
6, n
………
I 3, n =
2
I ,
3 + n 1, n
if m is odd
I 2, n =
1
I ,
2 + n 0, n
if m is even
Substituting these values,
∫
2
0
2
m −1 m − 3
⋅
…
⋅ I1, n ,
3+ n
m+n m+n−2
1
m −1 m − 3
⋅
…
⋅ I0, n ,
=
m+n m+n−2
2+n
sin m x cos n x dx =
if m is odd
if m is even
Integral Calculus
I1, n = ∫ 2 sin x ⋅ cos n x dx = −
Now,
0
p
2
cos n +1 x
n +1
p
2
6.11
2
=
0
1
n +1
I 0 , n = ∫ sin x cos x dx = ∫ cos x dx
0
n
0
n
0
2
1
m −1 m − 3
m
n
∫02 sin x cos x dx = m + n ⋅ m + n − 2 … 3 + n ⋅ n + 1 ,
if m is odd and n may be odd or even
m −1 m − 3
m−5
1
n −1 n − 3
2
…
… ,
=
⋅
⋅
×
⋅
m+n n+n−2 m+n−4
2+n
n n−2
3
if m is even and n is odd
m −1 m − 3
m−5
1
n −1 n − 3 1 p
=
⋅
⋅
×
⋅
…
… ⋅ ,
m+n m+n−2 m+n+4 2+n
n n−2 2 2
if m is even and n is even
Hence,
2
Corollary: Certain definite integrals can be evaluated using
0
sin m x cos n x dx,
where m, n are positive integers.
x = a tan q ,
(i) Putting
n
x
a n p2 tan n q
d
x
a sec 2 q dq
0 (a2
x 2 )m
a 2 m 0 sec 2 m q
= a2n
p
2
2m 1
0
sin n q cos 2 m
n 2
q dq
x = a tan q ,
(ii) Putting
∫
x
∞
0
π
n
1
m+
2
2
(a + x )
2
dx = ∫ 2
0
a n sin n θ
cos n θ
1
a
2 m +1
2m + 1
(sec θ )
2
⋅ a sec 2 θ dθ
2
π
= a n − 2 m ∫ 2 sin n θ cos 2 m − n −1 θ dθ
0
x = 2a sin 2 q ,
(iii) Putting
∫
2a
0
2a
x m 2ax − x 2 dx = ∫ x
m+
1
2
0
π
2
0
= ∫ ( 2a)
2a − x dx
1
m+
2
sin 2 m +1θ ⋅ 2a cos θ 4 a sin θ ⋅ cos θ dθ
π
= ( 2a) m + 2 ⋅ 2 ∫ 2 sin 2 m + 2 θ ⋅ cos 2 θ dθ
0
Example 1: Evaluate:
(i)
(iv)
2
0
2
0
sin 5 x dx
(ii)
cos 7 x dx
(v)
2
0
2
0
sin6 x dx
cos 4 x dx
(iii)
2
0
sin 8 x dx
6.12
Engineering Mathematics
Solution:
(i)
∫
0
(iii)
∫
0
(v)
∫
0
2
2
2
sin 5 x dx =
4 2 8
⋅ =
5 3 15
(ii)
∫
0
sin8 x dx =
7 5 3 1
35
⋅ ⋅ ⋅ ⋅ =
8 6 4 2 2 256
(iv)
∫
0
cos 4 x dx =
3 1
3
⋅ ⋅ =
4 2 2 16
2
2
sin 6 x dx =
5 3 1
5
⋅ ⋅ ⋅ =
6 4 2 2 32
cos 7 x dx =
6 4 2 16
⋅ ⋅ =
7 5 3 35
Example 2: Evaluate:
o
4
(i)
(iv)
0
o
∫
0
sin 4 2x dx
sin 2 p
(ii)
1 − cos p
dp
1 + cosp
∫
(v)
⎛ x⎞
sin 5 ⎜ ⎟ dx
⎝2⎠
0
o
4
(iii)
∫
o
2
0
sin 4 p
dp
(1 + cosp )2
cos6 2t dt
0
Solution:
(i)
p
4
0
sin 4 2 x dx
2x = t ,
2dx = dt
x = 0, t = 0
p
p
x= , t=
4
2
Putting
When
p
4
0
(ii)
∫
0
1
2
sin 4 2 x dx
p
2
0
sin 4 t dt
1 3 1 p
2 4 2 2
⎛x⎞
sin 5 ⎜ ⎟ dx
⎝2⎠
1
x = t,
2
1
dx = dt
2
x = 0,
Putting
When
x =p,
∫
0
p
(iii)
0
3p
32
sin 4 q
dq =
(1 + cos q ) 2
p
sin 5
2sin
x
4 2 16
dx = 2 ∫ 2 sin 5 t dt = 2 ⋅ ⋅ =
0
2
5 3 15
q
q
cos
2
2
0
2 cos 2
t=0
p
t=
2
q
2
2
4
dq =
p
0
4 sin 4
q
dq
2
Integral Calculus
q
= t,
2
Putting
1
dq = dt
2
q = 0,
When
t=0
p
t=
2
q =p,
p
0
p
(iv)
0
6.13
sin 4 q
dq
(1 + cos q ) 2
1 cos q
sin 2 q
dq
1 + cos q
p
0
8
q
q
4 sin 2 cos 2
2
2
p
2
sin 4 t dt
0
q
2 dq
2q
2 cos
2
2sin 2
1
dq = dt
2
q = 0,
When
p
0
0
p
2 2
0
3p
2
sin 3
q
dq
2
t=0
p
t=
2
q =p,
p
4
3 1 p
4 2 2
q
= t,
2
Putting
(v)
8
sin 2 q
1 cos q
dq
1 + cos q
4 2
p
2
0
sin 3 t dt
4 2
2
3
8 2
3
cos6 2t dt
2t = x,
2dt = dx
Putting
t = 0,
x=0
p
p
t= , x=
4
2
1
1 5 3 1
5
6
6
∫04 cos 2t dt = 2 ∫02 cos x dx = 2 ⋅ 6 ⋅ 4 ⋅ 2 ⋅ 2 = 64
When
Example 3: Evaluate:
2
(i)
0
sin 5 x cos6 x dx
(ii)
2
0
sin 4 x cos 8 x dx
Solution:
(i)
∫
2
0
sin 5 x cos6 x dx =
4 ⋅ 2 ⋅ 5 ⋅ 3 ⋅1
8
=
11 ⋅ 9 ⋅ 7 ⋅ 5 ⋅ 3 ⋅1 693
(iii)
o
4
0
cos 3 2 x sin 4 4 x dx
6.14
(ii)
Engineering Mathematics
∫
2
0
p
4
(iii)
0
sin 4 x cos8 x dx =
3 ⋅1 ⋅ 7 ⋅ 5 ⋅ 3 ⋅1
7
=
12 ⋅10 ⋅ 8 ⋅ 6 ⋅ 4 ⋅ 2 2 2048
cos3 2 x sin 4 4 x dx
p
4
0
cos3 2 x (2sin 2 x cos 2 x ) 4 dx
p
4
= 16
0
cos 7 2 x sin 4 2 x dx
2x = t
2dx = dt
x = 0, t = 0
p
p
x= , t=
4
2
Putting
When
p
4
0
p
2
cos3 2 x sin 4 4 x dx = 8
0
cos 7 t sin 4 t dt
6 4 2 3
11 9 7 5 3
8
128
1155
Example 4: Evaluate:
(i)
1
0
3
x 2 (1 x ) 2 dx
4
(iv)
0
x3 4x
(ii)
2
0
x4
4 x2
dx
(iii)
0
1
dx
(1 + x 2 )5
x 2 dx
Solution:
(i)
1
0
3
x 2 (1 x ) 2 dx
x = sin 2 q ,
Putting
dx = 2sin q cos q dq
x = 0,
When
q =0
p
q =
2
x = 1,
1
0
p
2
3
x 2 (1 x ) 2 dx
0
=2
2
(ii)
2
0
Putting
x4
4 x2
sin 4 q cos3 q 2sin q cos q dq
p
2
0
sin 5 q cos 4 q dq
4 2 3
9 7 5 3
dx
x = 2sin q ,
dx = 2 cos q dq
16
315
Integral Calculus
When
∫
x4
2
0
4 − x2
x = 0,
q =0
x = 2,
q =
π
p
2
( 2 sin θ ) 4
dx = ∫ 2
4 − 4 sin 2 θ
0
6.15
⋅ 2 cos θ dθ
π
= 16 ∫ 2 sin 4 θ dθ
0
3 1 π
= 16 ⋅ ⋅ ⋅ = 3π
4 2 2
(iii)
0
1
dx
(1 + x 2 )5
x = tan q ,
Putting
dx = sec 2 q dq
When
0
x = 0,
q =0
x
q
1
dx =
(1 + x 2 )5
,
p
2
0
p
2
sec 2 q
dq =
(1 + tan 2 q )5
7 5 3 1 p
8 6 4 2 2
(iv)
4
0
p
2
0
sec 2 q
dq =
sec10 q
p
2
0
cos8 q dq
35p
256
x 3 4 x x 2 dx
x = 4 sin 2 q ,
Putting
dx = 8sin q cos q dq
x = 0,
When
q =0
π
θ=
2
x = 4,
4
0
x 3 4 x x 2 dx
p
2
0
7
1
(4) 2 sin 7 q (4 4 sin 2 q ) 2 8sin q cos q dq
27 16
128
p
2
0
sin8 q cos 2 q dq
16 7 5 3 1 p
10 8 6 4 2 2
128 16
28p
1
10
p
2
0
sin8 q dq
6.16
Engineering Mathematics
Exercise 6.2
1. Evaluate
(i)
(iv)
sin 4 x dx
2
0
2
(ii)
0
3p
Ans. : (i)
16
(ii)
(iii)
(iv)
2a
(v)
8
(ii)
15
1
0
cos5 x sin 4 x dx
2
0
cos 4 x sin 3 x dx
2
0
p
6
cos6 3 x sin 2 6 x dx
0
4. Evaluate
8
8
(ii)
693
315
13
7p
(iii)
(iv)
35
384
(i)
0
1
(ii)
0
3. Evaluate
(ii)
(iii)
1
x 2 (2 x 2 )
0
1 x2
a
0
x4
a2
1
0
x2
3
Ans.: (i)
Ans. : (i)
(i)
3
x 2 (1 x ) 2 dx
5p
3p 4
(ii)
a
16
16
11p
p
(iii)
(iv)
192
32
5p 4
3p
(v)
a (vi)
8
128
sin 6 x cos5 x dx
2
0
x 2 2ax x 2 dx
0
(vi)
2. Evaluate
(i)
0
cos5 x dx
x2
dx
(1 + x 2 ) 4
(iii)
dx
0
1
dx
(1 + x 2 )5
x5
1+ x2
dx
1 x2
x3
dx
(4 + x 2 ) 2
35p
256
1
(iii)
3
Ans.: (i)
dx
x 5 sin 1 x dx
(ii)
3p + 8
24
6.3 RECTIFICATION OF CURVES
The process of determining the length of an arc of a plane curve is known as rectification of curves.
Length of Arc in Cartesian Form We
y = f (x)
y
know from differential calculus that for the curve
y = f (x),
ds
⎛ dy ⎞
= 1+ ⎜ ⎟
⎝ dx ⎠
dx
2
The length of the arc of the curve y = f (x) between
x = a and x = b is given by,
O x=a
x=b
Fig. 6.1
x
Integral Calculus
s=∫
6.17
2
b
ds
⎛ dy ⎞
dx = ∫ 1 + ⎜ ⎟ dx
a
⎝ dx ⎠
dx
b
a
Similarly, the length of the arc of the curve x = f (y) between y = c and
y = d is given by,
s=∫
d
c
2
d
ds
⎛ dx ⎞
dy = ∫ 1 + ⎜ ⎟ d y
c
dy
⎝ dy ⎠
y
y=d
x = f (y)
Length of Arc in Parametric Form
When the equation of the curve is given in parametric form x = f1(t), y = f2(t), we have, from differential
calculus,
2
ds
⎛ dx ⎞ ⎛ dy ⎞
= ⎜ ⎟ +⎜ ⎟
⎝ dt ⎠ ⎝ dt ⎠
dt
y=c
x
O
2
Fig. 6.2
The length of the arc of the curve between the points t = t1 and t = t2 is given by,
s=∫
2
2
⎛ dx ⎞ ⎛ dy ⎞
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ dt
dt
dt
t2
t1
Length of Arc in Polar Form
For the curve r = f (q ), we have, from dif-
ferential calculus,
ds
⎛ dr ⎞
= r2 + ⎜ ⎟
⎝d ⎠
d
q=p
2
2
P (r, q )
r
The length of the arc of the curve r = f (q ) between
the points q = q 1 and q = q 2 is given by,
2
s=∫
2
1
⎛ dr ⎞
r2 + ⎜ ⎟ d
⎝d ⎠
q
q1 q2
q =0
Fig. 6.3
Similarly, the length of the arc of the curve q = f (r) between the points r = r1 and
r = r2 is given by,
s=∫
r2
r1
2
⎛d ⎞
1 + r ⎜ ⎟ dr
⎝ dr ⎠
2
Cartesian Form
Example 1: Show that the length of the arc of the curve 4ax = y 2
y2 a
from (0, a) to any point (x, y) is given by
x.
2a 2
2a 2 log
y
a
a2
Engineering Mathematics
6.18
Solution:
y
− a2
a
a 1
dx
4a
= 2 y − 2a 2 ⋅ ⋅
y a
dy
4ax = y 2 − 2a 2 log
… (1)
y
a
y 2 − a2
dx
=
−
=
2ay
dy 2 a 2 y
For the required arc, y varies from a to y.
s=∫
Length of the arc,
2
⎛ dx ⎞
1 + ⎜ ⎟ dy
⎝ dy ⎠
y
a
2
=∫
=
2
y
⎛ y 2 − a2 ⎞
( y 2 + a2 )
1+ ⎜
dy = ∫
dy
⎟
a
⎝ 2ay ⎠
( 2ay ) 2
y
a
1 ⎛ y
1 y y 2 + a2
y dy + a 2
dy =
∫
y
2a ⎜⎝ ∫a
2a a
∫
y
a
1 ⎞
dy
y ⎟⎠
y
y a2 ⎞
1 y2
1 ⎛ y2
2
=
+ a 2 log y =
⎜⎝ + a log − ⎟⎠
a 2
2a 2
2a 2
a
=
a2 ⎞ a2 ⎤
1 ⎡ y2 ⎛ y2
⎢ + ⎜⎝ − 2ax − ⎟⎠ − ⎥
2a ⎣ 2
2
2
2⎦
… [From Eq. (1)]
y2
a
1 2
( y − 2ax − a 2 ) =
−x−
2
2a
2a
y2 a
=
− −x
2a 2
=
Example 2: Find the length of the arc of the curve y = ex from the point (0, 1)
to (1, e).
y = ex
dy
= ex
dx
For the required arc, x varies from 0 to 1.
Solution:
Length of the arc AB,
s=∫
1
=∫
1
0
0
Putting
2
⎛ dy ⎞
1 + ⎜ ⎟ dx
⎝ dx ⎠
y
y = ex
B(1, e)
A
(0, 1)
x
O
1 + e dx
1 + e2x = t2,
2e2xdx = 2t dt
t
dx = 2
dt
t 1
2x
Fig. 6.4
Integral Calculus
When
Length of the arc,
x = 0,
t= 2
x = 1,
t = 1 + e2
s=∫
1+ e 2
2
t⋅
1+ e 2
=∫
2
t
dt = ∫
t −1
2
1+ e 2
2
6.19
t 2 −1+1
dt
t 2 −1
1 ⎞
1
t −1
⎛
⎜⎝1 + 2 ⎟⎠ dt = t + log
2
t +1
t −1
1+ e 2
2
1⎛
1 + e2 − 1
2 −1⎞
= 1 + e 2 − 2 + ⎜ log
− log
⎟
2⎝
2 +1⎠
1 + e2 + 1
2
⎡ ⎧
⎫
⎧⎪ ( 2 − 1)2 ⎫⎪⎤
1 ⎢ ⎪ ( 1 + e 2 − 1) ⎪
−
log
= 1 + e − 2 + log ⎨
⎬
⎬⎥
⎨
2 − 1 ⎪⎥
2 ⎢ ⎪ 1 + e2 − 1 ⎪
⎪
⎭⎦
⎩
⎭
⎣ ⎩
1
= 1 + e 2 − 2 + log ( 1 + e 2 − 1) − log e 2 − log ( 2 − 1)
2
2
= 1 + e 2 − 2 + log ( 1 + e 2 − 1) − 1 − log ( 2 − 1)
[∵ log e2 = 2 log e = 2]
Example 3: Find the length of the arc of the curve y = log sec x from x = 0 to
o
x= .
3
Solution:
y = log sec x
dy
1
=
⋅ sec x tan x = tan x
dx sec x
For the required arc, x varies from 0 to
p
.
3
2
Length of the arc,
s=∫
3
0
⎛ dy ⎞
1 + ⎜ ⎟ dx
⎝ dx ⎠
p
p
= ∫ 3 1 + tan 2 x dx = ∫ 3 sec x dx
0
= log(sec x + tan x )
= log ( 2 + 3 )
0
p
3
0
⎛ ex − 1⎞
from x = 1 to
⎝ e x + 1 ⎟⎠
Example 4: Find the length of the arc of the curve y = log ⎜
x = 2.
Engineering Mathematics
6.20
Solution:
y = log
ex 1
ex + 1
log (e x
y
dy
dx
e
e
1) log (e x
x
x
e
1 e
x
1)
2e
x
1
e
x
2x
1
For the required arc, x varies from 1 to 2.
Length of the arc,
s=∫
2
=∫
2
2
⎛ dy ⎞
1 + ⎜ ⎟ dx
⎝ dx ⎠
1
1+
1
2x
2 (e
4e 2 x
− 1) 2 + 4e 2 x
dx = ∫
dx
2
1
(e − 1)
(e 2 x − 1) 2
2x
x
−x
2x
2⎛e
2⎛e +e
⎞
+1⎞
= ∫ ⎜ 2x
dx = ∫ ⎜ x
dx
⎟
1 ⎝e
1 ⎝ e − e− x ⎟
−1⎠
⎠
⎡
2
= log (e x − e − x )
⎢∵
1
⎣
= log (e 2 − e −2 ) − log(e − e −1 )
∫
⎤
f ′( x )
dx = log f ( x ) ⎥
f ( x)
⎦
e 2 − e −2
= log (e + e −1 )
e − e −1
⎛ 1⎞
= log ⎜ e + ⎟
⎝
e⎠
= log
Example 5: Show that the length of the arc of the curve ay2 = x3 from the origin
8a
7
to the point whose abscissa is b is
Solution:
9b
4a
1
3
2
1 .
ay 2 = x 3
2ay
y
dy
= 3x 2
dx
P
dy 3 x 2
=
=
dx 2ay
=
3x 2
2a
x3
a
O
x=b
3 x
2 a
For the arc OP, x varies from 0 to b.
Length of the arc OP,
s=
b
0
1+
dy
dx
2
dx
Fig. 6.5
x
Integral Calculus
b
=
9x
dx
4a
1+
0
6.21
2
9x
=
1+
3
4a
3
2
b
4a
9
0
8a
27
9b
4a
1
3
2
1
x
, prove that the length of the arc s,
c
measured from its vertex to any point (x, y), is
Example 6: For the catenary y = c cosh
(i) s = c sinh
x
c
(ii) s 2
y2
c2
y
x
c
y = c cosh
Solution: (i)
(iii) s = c tanx
dy
x
= sinh
dx
c
For the arc AP, x varies from 0 to x.
Length of the arc AP,
x
s=∫
0
=∫
x
0
A
(0, c)
2
⎛ dy ⎞
1 + ⎜ ⎟ dx
⎝ dx ⎠
1 + sinh 2
x
= ∫ cosh
0
= c sinh
(ii)
P(x, y)
y
x
x
dx
c
x
x
dx = c sinh
c
c
x
0
Fig. 6.6
x
c
s 2 = c 2 sinh 2
x
x ⎞
x
⎛
= c 2 ⎜ cosh 2 − 1⎟ = c 2 cosh 2 − c 2
⎝
c
c ⎠
c
= y 2 − c2
(iii) The tangent at point P(x, y) makes an angle y with the x-axis.
dy
x
tan =
= sinh
dx
c
s
=
c
s = c tan
[From (i)]
Engineering Mathematics
6.22
2
1
Example 7: Prove that the length of the arc of the curve y
x 1
x from
3
4
the origin to the point P(x, y) is given by s 2 = y 2 + x 2 . Hence, rectify the loop.
3
2
Solution:
y2
1
x
3
x 1
y
2
P(x, y)
1
1 3
⎛ x⎞
y = x ⎜1 − ⎟ = x 2 − x 2
⎝ 3⎠
3
dy 1
= x
dx 2
−
1
2
A (3, 0) x
O
1 3
(1 − x )
− ⋅ x =
3 2
2 x
1
2
For the arc OP, x varies from 0 to x.
Length of the arc OP,
2
⎛ dy ⎞
1 + ⎜ ⎟ dx
⎝ dx ⎠
s=∫
x
=∫
x
=∫
x
1+ x
0
2 x
0
0
1+
Fig. 6.7
x
(1 − x ) 2
(1 + x ) 2
dx = ∫
dx
0
4x
4x
dx =
1
1 x − 12
2
(
x
x
) dxx
+
2 ∫0
x
1
2
3
2
1 x
x
+
3
2 1
2
2 0
⎛ x⎞
= x ⎜1 + ⎟
⎝ 3⎠
=
2
2
⎛ x⎞
⎛ x⎞ 4
s 2 = x ⎜1 + ⎟ = x ⎜1 − ⎟ + x 2
⎝ 3⎠
⎝ 3⎠ 3
4
= y2 + x2
3
2
1
The points of intersection of the curve y 2 x 1
x and x-axis are obtained as,
3
0
x 1
1
x
3
2
x = 0, 3, 3 and y = 0, 0, 0
Hence, A: (3, 0)
Length of the upper half of the loop = 3 1 +
Length of the complete loop = 4 3
3
=2 3
3
Integral Calculus
6.23
Example 8: Show that the length of the loop of the curve 9ay 2
is 4 3a.
Solution: The points of intersection of the curve 9ay 2
x-axis are obtained as,
( x 2a )( x 5a )2
( x 2a)( x 5a) 2 and
y
0 = ( x − 2a)( x − 5a) 2
x = 2a, 5a and y = 0, 0
Hence, A: (2a, 0) and B: (5a, 0)
9ay 2
( x 2a)( x 5a) 2
B
(5a, 0)
A
(2a,
0)
O
dy
18ay
= ( x − 2a) 2( x − 5a) + ( x − 5a) 2
dx
= ( x − 5a)(3x − 9a)
dy ( x − 5a)( x − 3a)
=
6 ay
dx
x
Fig. 6.8
For the upper half of the loop, x varies from 2a to 5a.
Length of the loop of the curve, s = 2 (Length of upper half of the loop)
= 2∫
= 2∫
= 2∫
=
2
5a
( x − 5a) 2 ( x − 3a) 2
⎛ dy ⎞
dx
1 + ⎜ ⎟ dx = 2 ∫ 1 +
2a
⎝ dx ⎠
36 a 2 y 2
5a
2a
5a
1+
2a
5a
( x − a) 2
x−a
dx = 2 ∫
dx
a
2
4 a( x − 2 a )
2 a ⋅ x − 2a
5a
2a
1
a
∫
5a
(xx − 5a) 2 ( x − 3a) 2
( x − 3a) 2
d
x
=
2
1
+
∫2a 4a( x − 2a) dx
4 a( x − 2a)( x − 5a) 2
5a
2a
( x − 2a) + a
x − 2a
dx =
1
− ⎤
⎡
2
2
(
2
)
x
a
a
x
a
−
+
−
⎥ dx
∫ ⎢
a 2a ⎣
⎦
1
3
1
1 2
( x − 2a) 2 + 2a( x − 2a) 2
=
a 3
5a
5a
=
2a
3
1
⎤
1 ⎡2
2
2
⎢ (3a) + 2a(3a) ⎥
a ⎣3
⎦
= 2 3 a + 2a 3
= 4 3a
Example 9: In the evolute 27 ay 2
4( x 2a ) 3 of the parabola y 2 = 4ax , show
that the length of the arc from one cusp to the point where it meets the parabola
(
)
is 2a 3 3 1 .
Solution: (i) The points of intersection of the parabola y 2 = 4 ax with its evolute
27ay 2 4( x 2a)3 are obtained as,
Engineering Mathematics
6.24
27a ⋅ 4 ax = 4( x − 2a)3
y
(8a, ≥32a)
x − 6 ax − 15a x − 8a = 0
3
2
2
3
B
( x + a ) 2 ( x − 8a ) = 0
x = − a, 8 a
But x
A
a does not lie on the parabola.
O
x = 8a and y = ± 32a
(
Hence, B: 8a, 32a
)
C
(ii) The points of intersection of the evolute
27ay 2 4( x 2a)3 with the x-axis are obtained
as,
x = 2a and y = 0
Hence A: (2a, 0)
Now,
27ay 2 4( x 2a)3
3
2
y
( x 2a) 2
3 3a
1
dy
2 3
=
⋅ ( x − 2a) 2 =
dx 3 3a 2
For the arc AB, x varies from 2a to 8a.
Length of the arc AB,
x
(2a, 0)
s=∫
8a
2a
=∫
8a
2a
Fig. 6.9
x − 2a
3a
2
⎛ dy ⎞
1 + ⎜ ⎟ dx
⎝ dx ⎠
1+
x − 2a
1
dx =
3a
3a
3
2
=
⋅ ( x + a) 2
3a 3
1
=
(8a, −≥32a)
2
3 3a
3
2
8a
=
2a
∫
8a
2a
x + a dx
3
3
⎤
2⎡
⋅ ⎢(9a) 2 − (3a) 2 ⎥
3a 3 ⎣
⎦
1
3
2
(3a) (3 − 1) = 2a (3 3 − 1)
Example 10: Find the length of the parabola x 2 = 4 y which lies inside the circle
y
x 2 + y 2 = 6 y.
Solution: The equation of the circle is
x2 + y2 = 6y
x2 + y2 – 6y = 0
The centre of the circle is (0, 3) and radius
is 3.
The points of intersection of parabola x2 = 4y
and circle x2 + y2 = 6y are obtained as,
A
(−2≥2, 2)
B (2≥2, 2)
O
Fig. 6.10
x
Integral Calculus
6.25
4 y + y2 = 6 y
y2 − 2 y = 0
y( y − 2) = 0
y = 0, 2
When
(
)
y = 0,
y = 2,
(
Hence A : −2 2 , 2 and B : 2 2 , 2
Now,
)
x=0
x = ±2 2
x = 4y
dy x
=
dx 2
2
For the arc, OB, x varies from 0 to 2 2.
s = 2( Length of the arc OB)
Length of the arc OB,
= 2∫
2 2
0
=∫
2 2
0
2
2
⎛ dy ⎞
1 + ⎜ ⎟ dx = 2 ∫
0
⎝ dx ⎠
x 2 + 4 dx =
(
2
1+
x2
dx
4
(
x 2
x + 4 + 2 logg x + x 2 + 4
2
)
2 2
0
)
= 2 ⋅ 12 + 2 log 2 2 + 12 − 2 log 2
= 2 ⎡ 6 + log
⎣
(
)
2+ 3 ⎤
⎦
Example 11: Show that the length of the parabola y2 = 4ax from the vertex to
the end of the latus rectum is a ⎡ 2 + log (1 + 2 ) ⎤ . Find the length of the arc cut
⎣
⎦
off by the line 3y = 8x.
Solution: (i) The points of intersection of the
parabola y 2 = 4 ax and its latus rectum x = a are
obtained as,
y2
4a a
3y = 8x
y
P (a, 2a)
A
4a 2
x=a
y = ± 2a and x = a
Hence, P: (a, 2a) and
Q : ( a,
2a)
O
(a, 0)
x
2
Now,
y
4a
dx
y
=
dy 2a
x=
Q(a, −2a)
y 2 = 4ax
For the arc OP, y varies from 0 to 2a.
Fig. 6.11
Engineering Mathematics
6.26
2
Length of the arc OP,
s=∫
2a
=∫
2a
0
0
⎛ dx ⎞
1 + ⎜ ⎟ dy
⎝ dy ⎠
1+
1 y
=
2a 2
y2
1 2a 2
y + 4 a 2 dy
dy =
2
∫
0
a
2
4a
(
4a 2
y + 4a +
log y + y 2 + 4 a 2
2
2
2
)
2a
0
1 ⎡
a ⋅ 2a 2 + 2a 2 log ( 2a + 2a 2 ) − 2a 2 log 2a ⎤
⎦
2a ⎣
⎛
2a + 2a 2 ⎞
= a ⎜ 2 + log
⎟⎠
⎝
2a
=
= a ⎡ 2 + log (1 + 2 )⎤
⎣
⎦
(ii) The points of intersection of the parabola y 2 = 4 ax and the line 3 y = 8 x are
obtained as,
⎛ 3y ⎞
y 2 = 4a ⎜ ⎟
⎝ 8 ⎠
3a ⎞
⎛
y⎜y− ⎟ = 0
⎝
2⎠
y = 0,
3a
9a
and x = 0,
2
16
9a 3a
,
16 2
3a
For the arc OA, y varies from 0 to
.
2
1 32a 2
y + 4 a 2 dy
Length of the arc OA, s =
2a ∫0
Hence, A :
1 y
=
2a 2
(
4a 2
y 2 + 4a2 +
log y + y 2 + 4 a 2
2
)
3a
2
0
⎧⎪ ⎛ 3a
⎫⎪⎤
⎞
9a 2
1 ⎡ 3a 9a 2
⎢
=
+ 4 a 2 + 2a 2 ⎨log ⎜ +
+ 4 a 2 ⎟ − log 2a⎬⎥
2
4
2a ⎢ 4
4
⎠
⎩⎪ ⎝
⎭⎪⎥⎦
⎣
=
⎫⎤
1 ⎡ 3a 5a
⎛ 3a 5a ⎞
2⎧
g ⎜ + ⎟ − log 2a⎬⎥
⎢ ⋅ + 2a ⎨log
2⎠
2a ⎣ 4 2
⎩ ⎝ 2
⎭⎦
⎞
1 ⎛ 15a 2
+ 2a 2 log 2 ⎟
⎜⎝
⎠
2a 8
15 ⎞
⎛
= a ⎜ log 2 + ⎟
⎝
16 ⎠
=
Integral Calculus
6.27
Exercise 6.3
1. Find the length of the arc of following curves:
(i) y = log tanh
x
2
from x = 1
4. Find the length of the arc of the parabola y 2 = 8 x cut off by its latus rectum.
Find the length of the arc cut off by
the line 3 y = 8 x.
Ans.: 4
to x = 2
(ii) 24 xy = x + 48 from x = 2
(
2 log 2 +
to x = 4
3 y 2 1 from y = 0
to y = 4
(iv) y
x (2 x ) from x = 0
⎡ Ans.: log ⎡ x + (1 + x 2 ) ⎤ ⎤
⎣
⎦ ⎦⎥
⎣⎢
to x = 2
Ans.: (i) log e +
1
17
, (ii)
6
e
8
(82 82 1)
243
1
(iv) log (2 + 5 ) + 5
2
(iii)
2. Find the length of the curve
y 2 (2 x 1)3 cut off by the line
x = 4.
⎡ Ans.: 1022 ⎤
⎢
27 ⎥⎦
⎣
3. Find the arc of the parabola
y 2 4 a( a x ) cut off by the y-axis.
⎡ Ans.: a ⎡ 2 2 − log 3 − 2 2 ⎤ ⎤
⎣
⎦⎦
⎣
(
6. Show that if s is the arc of the curve
9 y 2 x(3 x ) 2
measured
from
the origin to the point P(x, y), then
3s 2 = 3 y 2 + 4 x 2 .
7. Find the length of the loop of the
curve
(i) 3ay 2 x( x a) 2
(ii) 9 y 2 = ( x + 7) ( x + 4) 2
(iii) 9ay 2 x( x 3a) 2
(iv) ay 2 x 2 ( a x )
Ans.: (i)
)
4
3
Example 1: Find the length of the curve x = a(cosp + p sinp ),
y = a(sinp – p cos p ), from p = 0 to p = 2o .
x = a(cosq + q sin q )
dx
= a( − sin + sin + cos ) = a cos
d
y = a (sin − cos )
dy
= a (cos − cos + sin ) = a sin
d
a, (ii) 4 3
(iii) 4 3a, (iv)
Parametric Form
Solution:
15
16
5. Find the length of the arc of the
2
parabola x = 4 ay measured from
the vertex to one extremity of the latus
rectum.
3
(iii) x
)
2 + log 1 + 2 ,
4
4
3
a
Engineering Mathematics
6.28
For the required arc,
varies from 0 to 2 .
Length of the curve,
s=∫
2π
=∫
2p
2
0
( a q cos q ) 2 + ( a q sin q ) 2 dq
0
= a∫
2p
0
2
⎛ dx ⎞ ⎛ dy ⎞
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ dθ
dθ
dθ
q2
q dq = a
2
2p
= 2ap 2
0
p
p
Example 2: Find the length of the curve x = ep sin + 2cos ,
2
2
p
p⎞
⎛
y = ep ⎜ cos − 2 sin ⎟ measured from p = 0 to p = .
⎝
2
2⎠
Solution:
q⎞
⎛ q
x = eq ⎜ sin + 2 cos ⎟
⎝
2
2⎠
dx
= eq
dq
q ⎞ q ⎛1
q
q⎞ 5 q
q
⎛ q
⎜⎝ sin + 2 cos ⎟⎠ + e ⎜⎝ cos − sin ⎟⎠ = e cos
2
2
2
2
2
2
2
q
q⎞
⎛
y = eq ⎜ cos − 2 sin ⎟
⎝
2
2⎠
dy
= eq
dq
q
q⎞ q
⎛
⎜⎝ cos − 2 sin ⎟⎠ + e
2
2
5 q
q⎞
q
⎛ 1 q
⎜⎝ − sin − cos ⎟⎠ = − e sin
2
2
2
2
2
For the required arc, q varies from 0 to p .
Length of the curve
s=∫
π
0
2
2
⎛ dx ⎞ ⎛ d y ⎞
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ dθ
dθ
dθ
π
25 2θ
θ
25 2θ
θ 25
e cos 2 + e 2θ sin 2 dθ = ∫
e dθ
0
4
2 4
2
4
5 π
5 π
= ∫ eθ dθ = eθ
0
0
2
2
5
= (eπ − 1)
2
=∫
π
0
Example 3: Find the length of the cycloid from one cusp to the next cusp
y
x = a(q + sin p ), y = a(1 − cos p ).
Solution:
x = a (q + sin q )
dx
= a (1 + cos q )
dq
y = a (1 − cos q )
dy
= a sin q
dq
B
A
−ap
O
Fig. 6.12
ap
x
Integral Calculus
6.29
For the arc OB, x varies from 0 to ap , hence q varies from 0 to p.
Length of the arc AB,
s = 2 (Length of arc OB)
= 2∫
2
2
π
⎛ dx ⎞ ⎛ dy ⎞
2
2
2
2
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ dθ = 2 ∫0 a (1 + cos θ ) + a sin θ dθ
dθ
dθ
π
0
= 2a ∫
π
0
2(1 + cos θ ) dθ = 4 a
∫
π
0
cos
θ
dθ
2
π
θ
= 4 a 2 sin
20
= 8a
y
Example 4: Find the length of one arc of
the cycloid x = a(p – sinp ), y = a(1 + cosp ).
Solution:
x = a (q − sin q )
dx
= a (1 − cos q )
dq
y = a (1 + cos q )
A
B
O
2ap
x
dy
Fig. 6.13
= − a sin q
dq
For the arc AB, x varies from 0 to 2ap, hence q varies from 0 to 2p.
Length of the arc AB,
s=∫
2π
=∫
2p
0
0
= a∫
2p
= a∫
2p
0
0
2
2
⎛ dx ⎞ ⎛ dy ⎞
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ dθ
dθ
dθ
a 2 (1 − cos q ) 2 + a 2 sin 2 q dq
2 − 2 cos q dq
2 ⋅ 2 sin 2
= 2a −2 cos
q
2
2p
q
q
dq = 2a ∫ sin dq
0
2
2
2p
= −4 a (cos p − cos 0)
0
= 8a
⎡
⎛ t ⎞⎤
Example 5: Find the length of the tractrix x = a ⎢ cos t + log tan ⎜ ⎟ ⎥ , y = a sin t
⎝ 2 ⎠⎦
⎣
o
from t =
to any point t.
2
⎡
⎛ t ⎞⎤
x = a ⎢cos t + log tan ⎜ ⎟ ⎥
Solution:
⎝ 2 ⎠⎦
⎣
Engineering Mathematics
6.30
⎡
⎤
⎢
⎥
dx
1
2 ⎛t ⎞ 1
= a ⎢ − sin t +
sec ⎜ ⎟ ⋅ ⎥
⎝ 2 ⎠ 2⎥
dt
⎛t ⎞
⎢
tan ⎜ ⎟
⎢⎣
⎥⎦
⎝2⎠
⎛
⎞
1
1 ⎞
⎜
⎟
⎛
= a ⎜ − sin t +
= a ⎜ − sin t +
⎟
t
t⎟
⎝
sin t ⎠
2 sin cos ⎟
⎜
⎝
2
2⎠
(1 − sin 2 t )
cos 2 t
=a
sin t
sin t
y = a sin t
=a
dy
= a cos t
dt
For the required arc, t varies from
Length of the curve,
s=∫
=∫
p
to t.
2
2
2
⎛ d x ⎞ ⎛ dy ⎞
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ dt
dt
dt
t
2
t
a2
2
t
cos 4 t
+ a 2 cos 2 t dt = a ∫ cos t cot 2 t + 1 dt
2
sin t
2
t
t
2
2
= a ∫ cot t dt = a log sin t
= a log sin t
Example 6: For the curve x = a ( 2 cos t − cos 2t ), y = a ( 2 sin t − sin 2t ), show
that the length of the arc of the curve measured from t = 0 to the point where the
x
tangent makes an angle x with the tangent, at t = 0 is given by s = 16 a sin 2 .
6
Solution:
x = a ( 2 cos t − cos 2t )
dx
= a ( −2 sin t + 2 sin 2t ) = 2a (sin 2t − sin t )
dt
y = a ( 2 sin t − sin 2t )
dy
= a ( 2 cos t − 2 cos 2t ) = 2a (cos t − cos 2t )
dt
For the required arc, t varies from 0 to t.
Length of the curve,
s=∫
t
0
2
2
⎛ dx ⎞ ⎛ dy ⎞
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ dt
dt
dt
Integral Calculus
=∫
t
6.31
4 a 2 [(sin 2t − sin t ) 2 + (cos t − cos 2t ) 2 dt
0
= 2a ∫
t
0
t
= 2a ∫
0
= 2a ∫
t
0
2 − 2(sin 2t sin t + cos t cos 2t ) dt
2 [1 − cos( 2t − t )] dt = 2a ∫
t
0
2 ⋅ 2 sin 2
= 8a − cos
= 16 a sin 2
t
2
t
0
2 (1 − cos t ) dt
t
t
t
dt = 4 a ∫ sin dt
0
2
2
t⎞
⎛
= 8a ⎜1 − cos ⎟
⎝
2⎠
t
4
… (1)
d y d y / dt
2a (cos t − cos 2t )
=
=
d x d x / dt
2a (sin 2t − sin t )
t
3t
sin
2
2 = tan 3t
=
t
3t
2
2 cos sin
2
2
2 sin
At
y
t = 0, y = 0, d = 0
dx
Hence, the tangent is x-axis at t = 0.
At the point where tangent makes an angle y with the tangent at t = 0, i.e., x-axis,
we get
dy
= tany
dx
tan
3t
= tany
2
3t
2
2y
t=
3
y =
Putting t in Eq. (1),
s = 16 a sin 2
= 16 a sin 2
2
12
6
Engineering Mathematics
6.32
2
2
⎛ x⎞3 ⎛ y⎞3
Example 7: Find the total length of the curve ⎜ ⎟ + ⎜ ⎟ = 1. Hence, deduce
⎝a⎠
⎝b⎠
2
2
2
o
the total length of the curve x 3 + y 3 = a 3 . Also show that the line p =
divides
3
2
2
2
the length of the curve x 3 + y 3 = a 3 in the first quadrant in the ratio 1:3.
Solution: (i) The parametric equations of
the curve
2
y
(0, b) B
2
⎛ x ⎞3 ⎛ y ⎞3
⎜⎝ ⎟⎠ + ⎜⎝ ⎟⎠ = 1 are given by,
a
b
3
x = a cos3 q , y = b sin q
q= p
3
C
p
3
dx
= −3a cos 2 q sin q
dq
dy
= 3b sin 2 q cos q
dq
(−a, 0)
O
A
(a, 0)
(0, −b)
For the arc AB, x varies from a to 0, hence q
p
varies from 0 to .
2
Total length of the curve, s = 4 (Length of the arc AB)
2
p
Fig. 6.14
2
⎛ dx ⎞ ⎛ dy ⎞
dq
= 4∫ 2 ⎜
+
0
⎝ dq ⎟⎠ ⎜⎝ dq ⎟⎠
p
= 4 ∫ 2 9a 2 sin 2 q cos 4 q + 9b 2 sin 4 q cos 2 q dq
0
p
= 12 ∫ 2 sin q cos q a 2 + (b 2 − a 2 ) sin 2 q dq
0
Putting a 2 + (b 2 − a 2 ) sin 2 q = t 2 ,
2(b 2 − a 2 ) sin q cos q dq = 2t dt
sin cos
When
d =
t
dt
b2 − a2
q = 0, t = a
p
q = , t=b
2
b
s = 12 ∫ t ⋅
a
=
t
dt
b2 − a2
12 t 3
b2 − a2 3
b
=
a
4( b 3 − a 3 )
b2 − a2
4( a 2 + ab + b 2 )
=
a+b
x
Integral Calculus
6.33
(ii) Putting b = a,
4( a 2 + a 2 + a 2 )
= 6a
2a
2
2
2
6a 3
= a
(iii) Length of the curve x 3 + y 3 = a 3 in the first quadrant =
4 2
2
2
2
Total length of the curve ( x 3 + y 3 = a 3 ) =
p
Length of the arc AC = ∫ 3 3a sin q cos q dq
0
3a p3
3a − cos 2q
sin 2q dq =
=
∫
0
2
2
2
p
3
0
9a
=
8
Length of the arc BC = length of the arc AB − length of the arc AC
3a 9a 3a
−
=
2
8
8
Length of the arc BC 1
=
Length of the arc AC 3
=
Example 8: Show that the length of the arc of the curve
x sin p + y cos p = f (p ), x cos p – y sin p = f (p ) is given by s = f (p ) + f (p ) + C.
Solution:
x sin q + y cos q = f ′(q )
… (1)
x cos q − y sin q = f ′′(q )
… (2)
Multiplying Eq. (1) by sinq and (2) by cosq and adding,
x = sin q f ′(q ) + cos q f ′′(q )
… (3)
Multiplying Eq. (1) by cosq and (2) by sinq and subtracting,
y = cos q f ′(q ) − sin q f ′′(q )
dx
= cos q f ′(q ) + sin q f ′′(q ) − sin q f ′′(q ) + cos q f ′′′(q )
dq
= cos q [ f ′(q ) + f ′′′(q )]
dy
= cos q f ′′(q ) − sin q f ′(q ) − cos q f ′′(q ) − sin q f ′′′(q )
dq
= − sin q [ f ′(q ) + f ′′′(q )]
2
Length of the arc,
s=∫
2
⎛ dx ⎞
⎛ dy ⎞
⎜⎝
⎟⎠ + ⎜⎝
⎟ dq
dq
dq ⎠
= ∫ (cos 2 q + sin 2 q ) [ f ′(q ) + f ′′′(q )]2 dq
Engineering Mathematics
6.34
= ∫ [ f ′(q ) + f ′′′(q )] dq
= f (q ) + f ′′(q ) + C
Example 9: Show that for the curve 8 a 2 y 2 = x 2 ( a 2 − x 2 ),
s=
arc length
a
(2p + sin p cos p) where x = a sinp and that the perimeter of one of the
2 2
y
oa
loop is
.
2
Solution: When x = a sinq
8a 2 y 2 = a 2 sin 2 q ( a 2 − a 2 sin 2 q )
= a 4 sin 2 q cos 2 q
a
a
y=
sinq cosq =
sin 2q
2 2
4 2
dx
= a cos q
dq
dy
a
=
cos 2q
dq 2 2
B
(−a, 0)
O
A
(a, 0) x
Fig. 6.15
For the upper half of the loop OA, x varies from 0 to a, hence q varies from 0 to
Length of one loop,
s = 2( Length of upper half of the loop OA)
2
2
p
⎛ dx ⎞ ⎛ dy ⎞
dq
= 2∫ 2 ⎜
+
0
⎝ dq ⎟⎠ ⎜⎝ dq ⎟⎠
p
= 2 ∫ 2 a 2 cos 2 q +
0
=
a
=
a
=
a
=
a
p
2
∫
2
0
p
2
∫
2
0
∫
2
p
2
0
2
a
∫
p
2
0
a2
cos 2 2q dq
8
8 cos 2 q + ( 2 cos 2 q − 1) 2 dq
4 cos 4 q + 4 cos 2 q + 1 dq
( 2 cos 2 q + 1) dq
( 2 + cos 2q ) dq
sin 2q
=
2q +
2
2
pa
=
2
p
2
0
=
a
2
(p )
p
.
2
Integral Calculus
6.35
Example 10: Show that the length of one complete wave of the curve y = b cos
a 2 + b 2 and a.
is equal to the perimeter of the ellipse whose semi-axes are
y = b cos
Solution: (i)
x
a
x
a
dy
b
x
= − sin
dx
a
a
x
For one complete wave
varies from 0 to 2p , i.e., x varies from 0 to 2pa.
a
Length of one complete wave, s1 = ∫
2p a
=∫
2p a
0
0
2
⎛ dy ⎞
1 + ⎜ ⎟ dx
⎝ dx ⎠
1+
b2
x
x
1 2p a 2
a + b 2 sin 2 dx
sin 2 dx = ∫
2
0
a
a
a
a
x
= t,
a
Putting
dx = a dt
x = 0, t = 0
x = 2p a, t = 2p
1 2p 2
a + b 2 sin 2 t ⋅ a dt
s1 = ∫
a 0
When
=∫
2p
a 2 + b 2 sin 2 t dt
0
⎡∵
a + b sin t dt ⎢
⎢⎣
p
= 2∫
2
0
= 4∫
p
2
2
2
∫
2a
0
a 2 + b 2 sin 2 t dt
0
(ii) Now, parametric equations of the given
f (t ) dt = 2 ∫ f (t ) dt ⎤
0
⎥
if f ( 2a − t ) = f (t ) ⎥⎦
(1)
y
ellipse are x = a 2 + b 2 cost and y = a sin t
B (0, a)
dx
dy
= − a 2 + b 2 sin t ,
= a cos t
dt
dt
A
For the arc AB, x varies from a + b to 0,
p
hence t varies from 0 to .
2
Perimeter of the ellipse,
s2 = 4 (Length of the arc AB)
2
2
a
x
(√a 2 + b 2, 0(
2
Fig. 6.16
2
p
p
⎛ dx ⎞ ⎛ dy ⎞
= 4 ∫ 2 ⎜ ⎟ + ⎜ ⎟ dt = 4 ∫ 2 ( a 2 + b 2 ) sin 2 t + a 2 cos 2 t dt
0
0
⎝ dt ⎠ ⎝ dt ⎠
p
= 4 ∫ 2 a 2 + b 2 sin 2 t dt
0
… (2)
Engineering Mathematics
6.36
From Eqs. (1) and (2),
Length of one complete wave = perimeter of the ellipse.
Example 11: Show that the perimeter of the ellipse
x2 y2
+
= 1 is
a 2 b2
⎡ e 2 12 ⋅ 3
⎤
12 ⋅ 32 ⋅ 5
2oa ⎢1 − 2 − 2 2 e 4 − 2 2 2 e 6 − …⎥ , where e is the eccentricity of the ellipse.
2 ⋅4 ⋅6
⎣ 2 2 ⋅4
⎦
Solution: The parametric equations of the
given ellipse are x = a cosq and y = b sinq .
dx
= − a sin q
dq
dy
= b cos q
dq
For the arc AB, x varies from a to 0, hence q
p
varies from 0 to .
2
Perimeter of the ellipse = 4(Length of the arc AB)
2
p
2
0
y
B (0, b)
A
(a, 0) x
O
Fig. 6.17
2
p
⎛ dx ⎞ ⎛ dy ⎞
dq = 4 2 a 2 sin 2 q + b 2 cos 2 q dq
⎜⎝
⎟⎠ + ⎜⎝
⎟
∫0
dq
dq ⎠
= 4∫
p
= 4 ∫ 2 a 2 sin 2 q + a 2 (1 − e 2 ) cos 2 q dq
0
p
1
= 4 a ∫ 2 (1 − e 2 cos 2 q ) 2 dq
0
p
2
0
= 4a ∫
⎡
1 ⎛1 ⎞
⎢
⎜⎝ 2 − 1⎟⎠
1
⎢1 + ( −e 2 cos 2 q ) + 2
( −e 2 cos 2 q ) 2
⎣ 2
2!
⎤
1 ⎛1 ⎞ ⎛1
⎞
⎥
⎜ − 1⎟ ⎜ − 2 ⎟⎠
2 ⎝2 ⎠⎝2
( −e 2 cos 2 q )3 + …⎥ dq
+
3!
⎦
1 4
1⋅ 3 6
⎛ 1
⎞
= 4 a ∫ 2 ⎜1 − e 2 cos 2 q −
e cos 4 q −
e cos6 q …⎟ dq
0 ⎝
⎠
2
2⋅4
2⋅ 4 ⋅6
p
1 p
1 4 3 1 p
1 ⋅ 3 6 5 ⋅ 3 ⋅1 p
⎡p 1
⎤
e ⋅ ⋅ −
e ⋅
⋅ – …⎥
= 4a ⎢ − e 2 ⋅ ⋅ −
2 2 2⋅ 4 4 2 2 2⋅ 4 ⋅6
6⋅4⋅2 2
⎣2 2
⎦
2p a 1
e2
22
12 3 4
e
22 4 2
12 32 5 6
e …
22 4 2 6 2
⎡
b2 ⎤
⎢∵ e = 1 − 2 ⎥
a ⎥⎦
⎢⎣
Integral Calculus
6.37
Exercise 6.4
1. Find the length of the following curves:
(i) x = a (2 cos q + cos 2q ),
y = a (2sin q + sin 2q ), from
q = 0 to any point q .
(ii) x a (q sin q ),
y a (1 cos q ) from
q = 0 to q = 2p
Ans. : (i) 8a sin
(ii) 8a
p
(iii)
2(e 2
1)a
(iv) log sec q
(v) log cosh t
(iii) x = aeq sin q ,
y = aeq cos q from
p
q = 0 to q =
2
(iv) x log (secq tan q ) sin q ,
y = cos q from
4b
aq
( a + b) cos
2b
a
4
(vii) a sin q
3
(vi)
2. Prove that the loop of the curve
1
x = t2, y = t – t3 is of length 4 3.
3
3. Show that the length of the arc of
q = 0 to any point q
(v) x a (t tanh t ),
y = a sech t from
t = 0 to any point t.
a+b
(vi) x = (a + b) cosq – bcos
q ,
b
y = (a + b) sinq – b sin ⎛⎜ a + b q ⎞⎟
⎝ b
q
2
the curve x = a(3sinq – sin3 q ),
y = a cos3q measured from (0, a) to
3
a(q + sin q cos q ).
2
4. If ‘s’ be the length of the arc of
the curve x = a(q + sinq cosq ),
y = a(1 + sinq )2, measured from the
p
point q
to a point q , show
2
that s4 varies as y 3 .
any point (x, y) is
⎠
pb
from q =
to any point q.
a
(vii) x = a sin 2q (1 + cos 2q ),
y a cos 2q (1 cos 2q ), from
q = 0 to any point q
Polar Form
Example 1: Find the length of the spiral r = e 2q from q = 0 to p = 2o .
r = e 2q
dr
= 2e 2q
dq
For the required length of the spiral, q varies from 0 to 2p .
Solution:
Length of the spiral, s = ∫
2p
0
2
2p
⎛ dr ⎞
r2 + ⎜
dq = ∫
e 4q + 4e 4q dq
⎟
0
⎝ dq ⎠
6.38
Engineering Mathematics
= 5∫
e 2q
e dq = 5
2
2p
2p
2q
0
0
5 4p
=
(e − 1)
2
Example 2: Find the length of the arc of the equiangular spiral r = aeq cota
from the point corresponding to p = 0 to the point corresponding to q = tana .
Solution:
r = aeq cot a
dr
= a cot a eq cot a
dq
For the required arc, q varies from 0 to tan a .
Length of the arc,
s=∫
tan a
=
tan a
0
0
2
⎛ dr ⎞
dq
r2 + ⎜
⎝ dq ⎟⎠
a 2 e 2q cot a + a 2 cot 2 a e 2q cot a dq
= a 1 + cot 2 a
a cosec a
tan a
0
eq cot a dq
tan a
eq cot a
cot a
0
a cosec a tan a cot a
(e
cot a
1)
a sec a (e 1)
a (e 1) sec a
Example 3: Find the length of the cissoid r = 2a tanq sinq from q = 0 to
p
q = .
4
Solution:
r = 2a tan q sin q
dr
= 2a (sec 2 q sin q + tan q cos q )
dq
= 2a sin q (sec 2 q + 1)
For the required arc length of the cissoid, q varies from 0 to
Length of the curve,
s=∫
p
4
0
=
p
4
0
p
4
0
p
.
4
2
⎛ dr ⎞
dq
r +⎜
⎝ dq ⎟⎠
2
4 a 2 tan 2 q sin 2 q + 4 a 2 sin 2 q (sec 2 q + 1) 2 dq
4 a 2 sin 2 q (sec 2 q 1 sec 4 q
2sec 2 q 1) dq
Integral Calculus
6.39
p
= ∫ 4 4 a 2 sin 2 q sec 2 q (sec 2 q + 3) dq
0
p
= ∫ 4 4 a 2 tan 2 q (tan 2 q + 4) dq
0
p
= ∫ 4 2a tan q tan 2 q + 4 dq
0
Putting
tan q + 4 = t ,
2
2
2 tan q sec 2 q dq = 2t dt
t dt
t dt
t dt
=
=
sec 2 q 1 + tan 2 q t 2 − 3
q = 0, t = 2
p
q = , t= 5
4
tan q dq =
When
5
s = ∫ 2a ⋅
2
= 2a t +
5
t2
3 ⎞
⎛
dt = ∫ 2a ⎜1 + 2
dt
2
⎝ t − 3 ⎟⎠
t −3
2
3
2 3
log
5
t− 3
t+ 3
⎛
3
= 2a ⎜ 5 +
log
2
⎝
2
5− 3
5+ 3
⎡
⎪⎧
3
⎢
= 2a ⎢ 5 − 2 +
log ⎨
2
⎪⎩
⎢⎣
(
−2−
3
2− 3⎞
log
⎟
2
2+ 3⎠
)
(
2
⎧⎪ 2 − 3
5 − 3 ⎫⎪
⎬ − log ⎨
5 − 3 ⎪⎭
⎪⎩ 4 − 3
) ⎫⎪⎬⎤⎥
2
⎪⎭⎥
⎥⎦
⎡
⎤
3
= 2a ⎢ 5 − 2 +
log 4 − 15 − log 7 − 4 3 ⎥
2
⎣
⎦
(
)
(
)
Note: Only positive values of t are considered since q lies in the first quadrant.
Example 4: Find the length of the
whole arc of the cardioid r = a (1 + cosq )
and show that the upper half is bisected
p
by the line q = .
3
Solution:
r = a (1 + cos q )
q= p
2
C
q= p
3
A
D
q=p
dr
a sin q
dq
(i) For the arc BACDO, q varies from 0 to p.
Length of the whole arc of the curve,
s = 2( Length of arc BACDO )
O
Fig. 6.18
B
q=0
Engineering Mathematics
6.40
2
dr
dq
2 r2
p
2
0
a 2 (1 cos q ) 2 ( a sin q ) 2 dq
p
2
0
= 4a
p
a 2 2 cos q dq
p
0
cos
2
0
q
q
dq = 4 a 2sin
2
2
a 2 2 cos 2
q
dq
2
p
= 8a
0
Length of the upper half of the cardioid = 4a
p
intersects the cardioid at point A.
3
p
q
q
Length of the arc, BA = 3 2a cos dq = 2a 2sin
0
2
2
(ii) Let q =
p
3
= 2a
0
Hence, the upper half of the cardioid is bisected by the line q =
p
3
Example 5: Find the length of the cardioid r = a (1 - cosq ) lying outside the
circle r = a cosq .
Solution: The points of intersection of cardioid r
r = a cos q is obtained as,
a(1 cos q ) a cos q
1 = 2 cos q
cos q =
q= p
3
q= p
2
A
1
2
q =±
Hence at A,
a (1 cos q ) and the circle
p
3
q=p B
O
p
q =
3
r a(1 cos q )
Fig. 6.19
dr
= a sin q
dq
p
For the arc of the cardioid lying outside the circle, q varies from
to p .
3
Length of the cardioid lying outside the circle, s = 2 (Length of arc AB)
p
= 2 ∫p
3
p
= 2 ∫p
3
2
⎛ dr ⎞
r2 + ⎜
dq
⎝ dq ⎟⎠
a 2 (1 − cos q ) 2 + ( a sin q ) 2 dq
q=0
Integral Calculus
p
6.41
p
= 2 ∫p a 2 − 2 cos q dq = 2 ∫p a 2 ⋅ 2 sin 2
3
3
q
q
= 4 a ∫p sin dq = 4 a −2 cos
2
2
3
q
dq
2
p
p
p
3
⎛
3⎞
= −8a ⎜ −
⎟
⎝ 2 ⎠
= 4a 3
Example 6: Show that the length of the arc of that part of cardioid
r = a (1 + cos q ) which lies on the side of the line 4r = 3a sec q away from the pole
is 4a.
Solution: The points of intersection of cardioid r = a (1 + cos q ) and the line
4 r = 3a secq are obtained as,
q= p
2
3a
a (1 + cos q ) =
secq
4
4 (1 + cos q ) cos q = 3
4 cos q
(2 cos q
4 cos 2 q
3
0
3) (2 cos q 1)
0
q= p
3
A
B
(2a, 0) q = 0
O
4r = 3a sec q
1
3
and cos q =
(does not exist)
2
2
p
q =±
3
p
Hence at A, q =
3
r = a (1 + cos q )
C
cos q =
Fig. 6.20
dr
a sin q
dq
p
For the arc BA, q varies from 0 to .
3
Length of the arc CBA,
s = 2 (length of arc BA)
= 2∫
2
p
3
⎛ dr ⎞
r +⎜
dq
⎝ dq ⎟⎠
2
0
p
= 2 ∫ 3 a 2 (1 + cos q ) + ( − a sin q ) dq
2
2
0
p
3
p
0
0
= 2 ∫ a 2 + 2 cos q dq = 2 ∫ 3 a 2 ⋅ 2 cos 2
= 4a ∫
p
3
0
= 4a
q
q
cos dq = 4 a 2 sin
2
2
p
3
0
q
dq
2
Engineering Mathematics
6.42
3
Example 7: Find the total length of the curve r = a sin
q= p
2
q
.
3
B
q
r = a sin
3
dr
q
q 1
a 3sin 2 cos
dq
3
3 3
q
q
= a sin 2 cos
3
3
3
Solution:
A
C
D
For the arc OABCD, q varies from 0 to 3p .
2
Length of the curve = 2 ( Length of the arc OABCD )
= 2∫
3p
2
0
= 2∫
3p
2
0
= 2∫
3p
2
0
q=0
O
Fig. 6.21
2
⎛ dr ⎞
r2 + ⎜
dq
⎝ dq ⎟⎠
a 2 sin 6
a sin 2
q
q
q
+ a 2 sin 4 cos 2 dq
3
3
3
3p
q
dq = a ∫ 2
0
3
3
2q
= a q − sin
2
3
2q
⎛
⎜⎝1 − cos
3
⎞
⎟⎠ dq
3p
2
0
3p 3
= a⋅
= pa
2
2
Example 8: Find the perimeter of the
2
2
lemniscate r = a cos 2q .
q= p
2
B
r 2 = a 2 cos 2q
Solution:
2r
dr
dq
dr
dq
A
a 2 ( sin 2q ) 2
O
2
a
sin 2q
r
For the arc OBA, q varies from 0 to
p
.
4
Fig. 6.22
Perimeter of the curve = 4 (Length of the arc OBA)
p
2
⎛ dr ⎞
= 4 ∫ 4 r2 + ⎜
dq
0
⎝ dq ⎟⎠
p
= 4 ∫ 4 a 2 cos 2q +
0
q= p
4
a4
sin 2 2q dq
r2
q=0
Integral Calculus
p
= 4 ∫ 4 a 2 cos 2q +
0
p
= 4 ∫ 4 a 2 cos 2q +
0
p
= 4∫ 4
0
a4
sin 2 2q dq
r2
a 4 sin 2 2q
dq
a 2 cos 2q
p
a4
1
dq = 4a ∫ 4
dq
0
a cos 2q
cos 2q
2
2q = t ,
2dq = dt
Putting
When
6.43
q = 0, t = 0
p
p
q = , t=
2
4
Perimeter of the curve =
4 a p2 dt
2 ∫0 cos t
p
−1
⎛1 1⎞
= 2a ∫ 2 sin 0 t ⋅ (cos t ) 2 dt = aB ⎜ , ⎟
0
⎝2 4⎠
2
⎛ ⎞
1 1 a1 1
a
2 ⎜⎝ 4 ⎟⎠
= 2 4 =
1
1
3
1−
4
4
4
2
2
⎛
⎞
⎛ 1⎞
1 1
a ⎜ ⎟
a p ⎜ ⎟
2 ⎝ 4⎠
⎝ 4⎠
=
=
p
p 2
p
sin
4
2
a ⎛ 1⎞
=
⎜ ⎟
2p ⎝ 4 ⎠
Example 9: Show that for the parabola
2a
= 1 + cos q , the arc intercepted between the
r
vertex and the extremity of the latus rectum is
a
(
)
⎡
p ⎤
⎢∵ n 1− n = sin np ⎥
⎦
⎣
q= p
2
B
O
A
2 + log 1 + 2 .
p
Solution: The latus rectum is the line q = .
2
2a
= 1 + cos q
r
2a
r=
=
1 + cos q
q
= a sec 2
q
2
2 cos 2
2
2a
Fig. 6.23
q=0
Engineering Mathematics
6.44
dr
q
q
= a sec 2 ⋅ tan
dq
2
2
p
For the arc AB, q varies from 0 to .
2
2
p
⎛ dr ⎞
s = ∫ 2 r2 + ⎜
dq
0
⎝ dq ⎟⎠
Length of the arc AB,
p
= ∫ 2 a 2 sec 4
0
p
= ∫ 2 a sec 2
0
Putting
When
tan
q
= t,
2
q
q
q
+ a 2 sec 4 tan 2 dq
2
2
2
q
q
1 + tan 2 dq
2
2
1 2q
q
sec dq = dt , sec 2 dq = 2dt
2
2
2
q = 0, t = 0
p
q = , t =1
2
1
s = ∫ 2 a 1 + t 2 dt = 2 a
0
(
t
1
1 + t 2 + log t + 1 + t 2
2
2
(
)
(
)
1
0
)
1
⎡1
⎤
= 2a ⎢
2 + log 1 + 2 ⎥ = a ⎡ 2 + log 1 + 2 ⎤
⎣
⎦
2
⎣2
⎦
Example 10: Show that the whole length of the limacon r = a cosp + b (a < b) is
equal to that of an ellipse whose semi-axes
q= p
2
are equal in length to the maximum and
B
minimum radii vectors of the limacon.
Solution:
For the arc ABC,
r = a cos + b
q=p
C
(b−a, 0)
dr
a sin
d
varies from 0 to
O
q=0
A
(b + a, 0)
Whole length of the limacon
Fig. 6.24
= 2(length of the arc ABC )
= 2∫
p
0
p
= 2∫
= 2∫
0
p
0
2
⎛ dr ⎞
r2 + ⎜
dq
⎝ dq ⎟⎠
( a cos q + b) 2 + ( − a sin q ) 2 dq
a 2 + b 2 + 2ab cosq dq
… (1)
Integral Calculus
6.45
Maximum radius vector of the limacon = a (1) + b = b + a
[∵ Maximum value of cosq
Minimum radius vector of the limacon
a ( 1) b
b a
[∵ Minimum value of cosq
The parametric equations of the ellipse with
above radii vectors as semi-axes are given as,
x = (b + a) cos q and y (b a) sin q
For the arc AB,
dx
= −(b + a) sin q
dq
dy
(b a) cos q
dq
p
varies from 0 to .
2
= 1]
= −1]
q= p
2
B (0, b- a)
O
A (b+a, 0)
q =0
Fig. 6.25
Whole length of the ellipse = 4(length of the arc AB)
2
2
p
⎛ dx ⎞ ⎛ dy ⎞
dq
= 4∫ 2 ⎜
+
0
⎝ dq ⎟⎠ ⎜⎝ dq ⎟⎠
p
= 4 ∫ 2 [−(b + a) sin q ]2 + [(b − a) cos q ]2 dq
0
p
= 4 ∫ 2 ( a 2 + b 2 + 2ab sin 2 q − 2ab cos 2 q ) dq
0
p
= 4 ∫ 2 ( a 2 + b 2 − 2ab cos 2q ) dq
0
Putting
2q = t ,
dt
2
= 0,
dq =
When
t=0
p
q = , t =p
2
Whole length of the ellipse = 2 ∫
p
= 2∫
p
= 2∫
p
0
0
0
a 2 + b 2 − 2ab cos t dt
a 2 + b 2 − 2ab cos (p − t ) dt
a 2 + b 2 + 2ab cos t dt
⎡∵ a f ( x ) dx = a f ( a − x ) dx ⎤
∫0
⎥⎦
⎢⎣ ∫0
From Eqs. (1) and (2),
Whole length of the limacon = Whole length of the ellipse
… (2)
Engineering Mathematics
6.46
Example 11: Find the length of the arc of the hyperbolic spiral r = a from the
point r = a to r = 2a.
rq = a
Solution:
r
Length of the arc,
s=∫
=∫
=∫
Putting
2a
a
2a
a
2a
a
dq
+q = 0
dr
2
⎛ dq ⎞
dr
1+ r2 ⎜
⎝ dr ⎟⎠
1 + q 2 dr = ∫
a
1+
a2
dr
r2
r +a
dr
r2
2
2
r 2 + a2 = t 2 ,
t dt
2r dr = 2t dt , dr =
When
2a
r = a,
s=∫
=∫
t
t=a 5
t dt
a 2
t 2 − a2
a 5
dt + a 2 ∫
a 2
a2
t=a 2
r = 2a,
a 5
t2
t 2 − a2
a 5
a 2
=∫
a 5
a 2
dt
= t
t − a2
2
t 2 − a2 + a2
dt
t 2 − a2
a 5
a 5
a 2
+
a2
t−a
log
2a
t+a a
)
a⎡
a 5−a
a 2 − a⎤
− log
⎢log
⎥
2⎣
a 2 + a⎦
a 5+a
)
a⎡
5 −1
2 − 1⎤
− log
⎢log
⎥
2⎣
5 +1
2 + 1⎦
)
⎡⎛ 5 − 1⎞ ⎛ 2 + 1⎞⎤
a
log ⎢ ⎜
⎟⎜
⎟⎥
2
⎢⎣ ⎝ 5 + 1 ⎠ ⎝ 2 − 1 ⎠ ⎥⎦
=a
(
5− 2 +
=a
(
5− 2 +
=a
(
5− 2 +
=a
(
⎡
a
5 − 2 + log ⎢
⎢
2
⎢⎣
=a
(
5− 2
)
)
(
2(
+ a log
5 −1
)
2 + 1)
5 +1
5 +1
2
(
)
2
2 +1 ⎤
⎥
2 −1 ⎥
⎥⎦
2
Integral Calculus
6.47
Exercise 6.5
1. Find the perimeter of the following
curves:
(i) r = a cos
(ii) r
a(q 2 1)
(iii) r = a cos3
q
3
(iv) r = ae m
(v) r = aq
q
2
(vii) r = 4 sin 2 q
(vi) r = a sec 2
⎡ Ans. :
⎤
⎢
⎥
(i) p a
⎢
⎥
8a
⎢
⎥
(ii)
3
⎢
⎥
3p a
⎢
⎥
(iii)
⎢
⎥
2
⎢
⎥
1 + m2
⎢
⎥
r
r
−
(iv)
( 2 1) m
⎢
⎥
⎢
⎥
a
⎢
(v) ⎡q 1 + q 2 + sinh −1 q ⎤ ⎥
⎦⎥
2⎣
⎢
⎢
(vi) 2a ⎡⎣ 2 + log ( 2 + 1) ⎤⎦ ⎥
⎢
⎥
4
⎢
⎥
(vii) 8 +
log( 3 + 2)
⎥⎦⎥
3
⎣⎢⎢
2p
bisects the upper half of
3
the cardioid.
[Ans. : 8a]
line q =
3. Find the length of the cardioid
r = a(1 + cos q ) which lies outside
the circle r + a cos q = 0.
⎡ Ans. : 4 3a ⎤
⎣
⎦
4. Prove that the length of the spiral
r = ae cot as r increases from r1 to
r2 is given by (r2 r1 ) sec .
5. Find the length of the cardioid
r a (1 cos ) lying inside the circle r = a cos .
Ans. : 8a 1
m
6. Find the length of the spiral r = ae
lying inside the circle r = a.
Ans. :
a
1 + m2
m
7. Find the length of the arc of parabola
l
= 1 + cos cut off by its latus recr
tum.
2. Find the perimeter of the cardioid
r
3
2
a(1 cos q ) and prove that the
(
6.4 AREAS OF PLANE CURVES (QUADRATURE)
The process of determining the area of a plane region
is known as quadrature.
Area Bounded by the Curve in Cartesian Form
Let y = f (x) be a curve defined in the
interval [a, b]. The area bounded by the curve y = f (x),
the x-axis and the two lines x = a and x = b is given by,
)
⎡ Ans. : l ⎡ 2 + log 1 + 2 ⎤ ⎤
⎣
⎦ ⎦⎥
⎣⎢
Fig. 6.26
Engineering Mathematics
6.48
Area,
A =
b
f ( x ) dx =
a
b
a
y dx
Similarly, the area bounded by the curve x = f ( y ), the y-axis and the two lines,
y = c and y = d is given by,
A=
d
f ( y )dy =
c
d
c
x dy
When the portion of the curve under consideration is
above x-axis, y is positive and hence, area will be positive. When the portion of the curve under consideration is
below x-axis, y is negative and hence, area will be negative. In such a case,
A=
b
a
Fig. 6.27
f ( x ) dx
However, when the curve f (x) crosses the x-axis several times, the total area bounded
by the curve is the sum of the areas above and below the x-axis with the absolute value
taken for the areas when the curve is below x-axis.
Further, the area bounded by the curves y = f1(x)
and y = f2(x) and the lines x = a and x = b is given by,
A
b
a
b
a
b
a
b
f 2 ( x ) dx
[f
2
( y2
( x)
a
f1 ( x ) dx
f1 ( x ) ] dx
y1 ) dx
Fig. 6.28
Area Bounded by the Curve in Parametric Form When the equation of the curve is given in parametric form x = f1 (t ), y = f 2 (t ) with t1
x(t1 ) = a, x(t2 ) = b, the area is given by,
b
b
t2
dx
A = f ( x ) dx = y d x =
y dt
a
a
t1
dt
Area Bounded by the Curve in Polar Form
t
t2 and
Let r = f ( ) be the
equation of the curve and OA, OB be the radii vectors at q = q1 , q = q 2 . The whole
area is divided into small sectors, such as
OPQ subtending an angle
at O. Let
P(r, ) and Q(r + r, + ) be the two
points on the curve. If dA is the area of the
elementary triangular strip OPQ, then
1
δ A = OP ⋅ OQ sin δθ
2
1
= r ( r + δ r ) sin δθ
2
δA 1
sin δθ
= r (r + δ r )
δθ 2
δθ
Fig. 6.29
Integral Calculus
Taking limits as
6.49
0,
A
1
sin
lim r ( r + r )
2 0
dA 1 2
⎡
⎤
sin d q
= r
∵ lim
= 1 and d q → 0, d r → 0 ⎥
⎢
d
q
→
0
d
2
dq
⎣
⎦
1 2 2
2 1
2
∫dA = ∫ 1 2 r d = 2 ∫ 1 r d
1 2
A = ∫ r 2d
2 1
lim
0
=
Cartesian Form
Example 1: Find the area bounded by the ellipse
x2 y2
+
= 1.
a2 b2
Solution: The region is symmetric in all
the four quadrants. For the region in the first
quadrant, x varies from 0 to a.
Area,
A = 4 (Area in the first quadrant)
a
= 4 ∫ y dx =
0
4b a 2
a − x 2 dx
a ∫0
a
⎤
x
4b ⎡ a 2
4b x a 2 − x 2 a 2
+ sin −1
=
=
⎢ ⋅ ⎥
2
2
a
a ⎣ 2 2⎦
a
0
= ab
Fig. 6.30
Example 2: Find the area bounded by the curve a 2 x 2
y 3 (a
y ).
Solution: The region is symmetric about y -axis.
For the region in the first quadrant, y varies from 0 to a.
Area,
A = 2 (Area bounded by the curve in first quadrant)
a
= 2 ∫ x dy =
0
Putting
When
2 a 32
y a − y dy
a ∫0
y = a sin 2 q ,
dy = 2a sin q cos q dq
y = 0, q = 0
y = a, q = p
2
2 p2 32 3
A
a sin q a cos 2 q 2a sin q cos q dq
a 0
p
311 p
4a 2 2 sin 4 q cos 2 q dq 4 a 2
0
6 4 2 2
p a2
=
8
Fig. 6.31
Engineering Mathematics
6.50
Example 3: Find the area enclosed by the curve a 4 y 2
x 4 (a 2
x2 ) .
Solution: The region is symmetric in all the quadrants. For the region in the first
quadrant, x varies from 0 to a.
Area, A = 4(Area in the first quadrant)
a
= 4 ∫ y dx
0
1
a2
= 4⋅
∫
a
0
x 2 a 2 − x 2 dx
Putting x = a sin q ,
dx = a cos q dq
When x = 0,
x = a,
4
a2
A
4a2
=
q =0
p
q =
2
p
2
0
a 2 sin 2 q a cos q a cos q dq
p
2
0
Fig. 6.32
sin 2 q cos 2 q dq
4a2
11 p
4 2 2
1 2
pa
4
Example 4: Find the area enclosed by the curve a 2 y 2
x 2 (2a
x )( x a ).
Solution: The region is symmetric about the x-axis. For the region
above the x-axis, x varies from a to 2a.
Area, A = 2(Area above x-axis)
= 2∫
2a
a
y dx
2 2a
x ( 2a − x )( x − a) dx
a ∫a
2 2a
= ∫ x − x 2 + 3ax − 2a 2 dx
a a
Fig. 6.33
1 2a
= − ∫ ( −2 x + 3a − 3a) − x 2 + 3ax − 2a 2 dx
a
a
2a
1 2a (
− x 2 + 3ax − 2a 2 dx
=− ∫
− x 2 + 3ax − 2a 2 ) ( −2 x + 3a ) dx + 3 ∫
a
a a
=
3
1 2
=−
( − x 2 + 3ax − 2a 2 ) 2
a 3
2a
a
+ 3∫
2a
a
2
2
3a ⎞
⎛a⎞ ⎛
⎜⎝ ⎟⎠ − ⎜⎝ x − ⎟⎠ dx
2
2
n +1
⎡
⎡⎣ f ( x )⎤⎦ ⎤
n
⎢∵ ∫ [ f ( x ) ] f ′( x ) d x =
⎥
n +1 ⎥
⎢
⎣
⎦
Integral Calculus
6.51
2a
3a ⎞
⎛
x−
1 ⎛a⎞ ⎛
3a ⎞
a2
−1 ⎜
2 ⎟
= 3 x ⎜ ⎟ − ⎜ x − ⎟ + sin ⎜
a ⎟
2 ⎝2⎠ ⎝
2⎠
8
⎜
⎟
⎝ 2 ⎠a
2
=
2
3
3a 2
⎡sin −1 1 − sin −1 ( −1) ⎤⎦ = p a 2
8 ⎣
8
Example 5: Find the area included between the curve y 2 ( a - x ) = x 3 and its
asymptote.
Solution: The equation of the curve can be rewritten as,
3
x2
y=
a x
The asymptote is the line, x = a.
The region is symmetric about the x-axis. For
the region, x varies from 0 to a.
Area,
A = 2(Area above x-axis)
=2
a
0
y dx
3
=2
Putting
x2
a
0
a x
dx
Fig. 6.34
x = a sin q ,
dx = 2a sin q cos q dq
2
x = 0,
When
x = a,
q =0
p
q =
2
3
π
2
0
A = 2∫
a 2 sin 3 θ
a cos 2 θ
⋅ 2a sin θ cos θ dθ
π
= 4 a 2 ∫ 2 sin 4 θ dθ = 4 a 2 ⋅
0
=
3 ⋅1 π
⋅
4⋅2 2
3
π a2
4
Example 6: Find the area enclosed by the curve x ( x 2 + y 2 ) = a ( x 2
its asymptote.
Solution: The equation of the curve can be rewritten as,
y=x
a x
a+x
y 2 ) and
Engineering Mathematics
6.52
x
The asymptote is the line,
a
The region is symmetric about the x-axis.
For the region, x varies from a to 0.
Area, A = 2(Area above x-axis)
Putting
When
=2
0
=2
0
a
a
y dx = 2
x( a x )
a2
x2
0
a
a x
dx
a+ x
x
dx
x = a sin ,
dx = a cos q dq
x = 0,
x
Fig. 6.35
q =0
p
2
a sin θ ( a − a sin θ )
⋅ a cos θ dθ
a cos θ
a, q
A = 2∫
0
−
π
2
0 ⎡
0
⎛ 1 − cos 2θ ⎞ ⎤
= 2a 2 ∫ π (sin θ − sin 2 θ ) dθ = 2a 2 ∫ π ⎢sin θ − ⎜
⎟⎠ ⎥ dθ
−
−
⎝
2
⎦
2 ⎣
2
= 2a 2 − cos θ −
θ sin 2θ
+
2
4
0
π
−
2
π⎤
⎡
= 2a 2 ⎢ −1 − ⎥
4⎦
⎣
π⎞
⎛
= −a2 ⎜ 2 + ⎟
⎝
2⎠
A = a2 2 +
Neglecting the negative sign,
p
2
Example 7: Find the area enclosed between the curve y( x 2 + 4a 2 ) = 8 a 3 and
its asymptote.
Solution: The equation of the curve can be
rewritten as,
x2 =
The asymptote is the line,
4 a 2 ( 2a − y )
y
y = 0, i.e., x-axis
The region is symmetric about the y-axis. For the
region in first quadrant, x varies from 0 to .
Area,
A = 2(Area above x -axis in first quadrant)
Fig. 6.36
Integral Calculus
6.53
∞
= 2 ∫ y dx
0
= 2∫
∞
0
∞
x
8a3
1
dx = 16 a3 ⋅
tan −1
2
2
2a
2a 0
x + 4a
= 8a 2 ⋅
2
= 4 a2
Example 8: Find the area included between the curve x 2 y 2
its asymptotes.
a2 ( y2
x 2 ) and
Solution: The equation of the curve can be written as,
ax
y=
2
a x2
The asymptotes are lines, x = a and x
a.
Fig. 6.37
The region is symmetric about both the axes. For the region above the x-axis in the first
quadrant, x varies from 0 to a.
Area, A = 4(Area above x -axis in first quadrant)
a
= 4 ∫ y dx
0
= 4a ∫
x
a
a − x2
0
2
dx
−1
a
= − 2a ∫ ( a 2 − x 2 ) 2 ( −2 x )dx
0
= −4 a a − x
2
2
a
0
= 4a
⎡
⎢∵ ∫
⎢⎣
[ f ( x ) ] f ′ ( x ) dx =
n
[ f ( x)] n+1 ⎤⎥
n +1
⎥⎦
2
Example 9: Find the area enclosed between the curve x 2 = 4 y and the line
x = 4y – 2.
Engineering Mathematics
6.54
Solution: The points of intersection of the curve x 2 = 4 y and line x
4 y 2 are
obtained as,
x2 = x + 2
x2
x 2
( x 1)( x 2)
x
0
0
1, 2 and y =
1
,1
4
Hence, P :
1,
1
4
Area,
A
Area PTQRSP Area POQRSP
and Q : (2, 1)
Fig. 6.38
For the regions PTQRSP and POQRSP, x varies from 1 to 2.
x+2
dx
4
2
A
1
1 x2
4 2
2
2x
1
x2
dx
1 4
2
1 x3
4 3
1
1
2 4
2
4
2
9
=
8
2
1
1 8 1
4 3 3
Example 10: Find the area bounded
by the parabola y = x 2 + 2 and the lines
x = 0, x = 1 and x + y = 0.
Solution: For the regions TPQST and
TOST, x varies from 0 to 1.
Area,
A
Area TPQST
1
0
( x2
1
2) dx
1
=
Area TOST
0
x3
x
+ 2x +
3
2
0
17
=
6
( x) dx
2 1
0
Fig. 6.39
Example 11: Find the area bounded by the parabolas y 2 = 5 x + 6 and x 2 = y.
Solution: The points of intersection of the parabolas y 2 = 5 x + 6 and x 2 = y are
obtained as,
x 4 = 5x + 6
Integral Calculus
x
6.55
y
1, 2 and y = 1, 4
Hence, P : ( 1, 1) and Q : (2, 4)
Area, A
Area PMQRSP Area POQRSP
For the regions PMQRSP and POQRSP, x
varies from 1 to 2.
A= ∫
2
−1
(
)
Q
x2 = y
(−1, 1)
M
(2, 4)
P
2
5 x + 6 dx − ∫ x 2 dx
S
−1
O
R
x
2
(5 x + 6 )
=
3
5⋅
2
3
2
x3
−
3
2
−1
y 2 = 5x + 6
−1
3
⎤ ⎛8 1⎞
2 ⎡
2
⎢(16) − 1⎥ − ⎜⎝ + ⎟⎠
15 ⎣
⎦ 3 3
27
=
5
Fig. 6.40
=
Example 12: Find the area common to the parabola y 2 = x and the circle
x 2 + y 2 = 2.
Solution: The points of intersection of the parabola y 2 = x and circle x 2 + y 2 = 2
are obtained as,
x2 x 2 0
( x 1)( x 2) 0
x 1,
2
When x = 1, y = ± 1 and when x = –2, y2
is negative,
Hence, P : (1, 1) and Q : (1, 1)
The region is symmetric about the x-axis.
Area, A = 2(Area above x-axis)
2(Area ORPSO Area ORPO)
For the regions ORPSO and ORPO, y varies
from 0 to 1.
A = 2 ⎡∫
⎢⎣ 0
1
(
)
Fig. 6.41
2 − y 2 dy − ∫ y 2 dy ⎤
⎥⎦
0
1
1
1
⎡y
y
y3 ⎤
1⎞
⎛1
⎥ = 2⎜ + − ⎟
= 2⎢
−
2 − y 2 + sin −1
⎝ 2 4 3⎠
3 0⎥
20
⎢⎣ 2
⎦
1
= (3 + 2)
6
Engineering Mathematics
6.56
Exercise 6.6
1. Find the area enclosed by the curve
a4 y 2 + b2 x 4 = a2 b2 x 2 .
4
ab
3
2. Prove that the area of a loop of the
16
.
curve y 2 x 2 (4 x 2 ) is
3
3. Find the area of the loop of the curve
y 2 (4 x ) x( x 2) 2 .
10. Find the area between the curve
y2(a + x) = (a – x)3 and its asymptote.
Ans. : 3p a 2
Ans. :
[ Ans. : 2(4
)]
4. Find the area in the first quadrant
bounded by the curve b4y2 = (a2 – x2)3
and the co-ordinate axes.
3p a 4
Ans. :
16b 2
5. Find the area of the loop of the curve
a+ x
y2 = x2
. Also find the area
a x
between the curve and its asymptote.
Ans. : a 2 2
p
p
, a2 2
a
2
6. Prove that area of the loop of the
8a 2
.
curve 3ay 2 x( x a ) 2 is
15 3
7. Find the area enclosed by the curve
y 2 ( x a )(b x), 0 < a, b.
p
( a b) 2
4
8. Find the whole area of the curve
a2 x2
y 2 = x2 2
.
a + x2
Ans. :
Ans. : a 2 (p
2)
9. Show that the area of infinite region enclosed between the curve
x3(1 – y) y = 1 and its asymptote is 2p.
11. Find the area of the loop of the curve
y 2 x + ( x + a ) 2 ( x + 2a ) = 0.
Ans. :
1 2
a (4 p )
2
12. Find the area included between the
y +8
x 2 and the x-axis.
curve
x
[ Ans. : 36]
13. Find the area between the parabola
y 2 = 4 x and line 2 x 3 y 4 0.
1⎤
⎡
⎢⎣ Ans. : 3 ⎥⎦
14. Find the area bounded by the curves
x2 = 4ay and x2 = 4ay.
16
Ans. : a 2
3
15. Find the area enclosed by the curves
8a3
.
x2 = 4ay and x 2 + 4 a 2 =
y
2
Ans. : (3p 2)a 2
3
16. Show that the area enclosed by the
curves xy2 = a2(a – x) and
(a – x)y2 = a2x is (p – 2)a2.
17. Find the area between the ellipses
x2 y2
x2 y2
+ 2 = 1 and 2 + 2 = 1.
2
a
b
b
a
b
Ans. : 4 ab tan 1
a
18. Find the area between the ellipses
x 2 + 2 y 2 = a 2 and 2 x 2 + y 2 = a 2 .
Ans. : 4 2a 2 cot 1 2
Integral Calculus
6.57
19. Find the area above the x-axis included
between the curves y2 = x(2a – x)
and y 2 = ax.
20. Find the area between the curve
xy = 2 and the circle x 2 + y 2 = 5 in
the first quadrant.
⎡
2 ⎞⎤
2 ⎛p
⎢ Ans. : a ⎜⎝ 4 − 3 ⎟⎠ ⎥
⎦
⎣
⎡ Ans. :
⎤
⎢
⎥
⎢ 5 ⎛ sin −1 2 − sin −1 1 ⎞ − 2 log 2⎥
⎟
⎢⎣ 2 ⎜⎝
⎥⎦
5
5⎠
Parametric Form
Example 1: Find the area enclosed between one arch of the cycloid
x = a(p - sin p ), y = a(1 - cos p ) and its base.
Solution:
a(q
x
dx
dq
sin q )
a(1 cos q )
For the region shown, x varies from 0 to 2 a.
x = 0, q = 0
When
x = 2pa, q = 2p
A=
Area,
2p
y
0
=∫
2p
dx
dq
dq
Fig. 6.42
a(1 − cos q ) ⋅ a(1 − cos q )dq
0
2p
= a 2 ∫ (1 − 2 cos q + cos 2 q )dq
0
=a
2
∫
2p
0
1 + cos 2q ⎞
⎛
⎜⎝1 − 2 cos q +
⎟⎠ dq
2
3q
sin 2q
=a
− 2 sin q +
2
4
2p
2
= 3p a
0
2
Example 2: Find the area of the hypocycloid, x = a cos 3 q , y = b sin 3 q .
Solution: x = a cos3 q
dx
dq
3a cos 2 q ( sin q )
For the region in the first quadrant, x varies from
0 to a.
p
When
x = 0, q =
2
x = a, q = 0
Fig. 6.43
Engineering Mathematics
6.58
The region is symmetric in all the quadrants.
Area,
A = 4(Area in the first quadrant)
0
= 4 ∫p y
2
dx
dq
dq
0
= 4 ∫p b sin 3 q ⋅ 3a cos 2 q ( − sin q ) dq
2
p
= 12ab ∫ 2 sin 4 q cos 2 q dq = 12ab
0
3 ⋅1 ⋅1 p
⋅
6⋅4⋅2 2
3p ab
8
=
1
t
Example 3: Find the area bounded by the curve x = a cos t + a log tan 2 ,
2
2
y = a sin t .
1
t
Solution: x = a cos t + a log tan 2
2
2
dx
a 2 1 2t
= − a sin t +
⋅ sec
t 2
dt
2
2
tan
2
a
a sin t
t
t
2sin cos
2
2
a
a
(1 sin 2 t )
a sin t
sin t sin t
a
cos 2 t
=
sin t
For the region in the second quadrant, x varies from − ∞ to 0.
When
x → − ∞, t = 0
p
x = 0, t =
2
The region is symmetric in all the quadrants.
Area,
A = 4(Area in the second quadrant)
p
p
dx
⎛ a
⎞
dt = 4 ∫ 2 a sin t ⎜
cos 2 t ⎟ dt
0
0
⎝
⎠
dt
sin t
p
p
⎛ 1 + cos 2 t ⎞
= 4 a 2 ∫ 2 cos 2 t dt = 4 a 2 ∫ 2 ⎜
⎟⎠ dt
0
0 ⎝
2
= 4∫ 2 y
sin 2t
= 2a t +
2
2
=pa
2
p
2
0
= 2a 2
p
2
Fig. 6.44
Integral Calculus
6.59
Example 4: Find the area bounded by the curve x = 3 + cos q , y = 4sin q .
Solution: x = 3 + cos q
dx
dq
sin q
For the region in the first quadrant, x varies from 3
to 4.
p
x = 3, q =
When
2
x = 4, q = 0
A = 4(Area BCD)
Area,
0
= 4 ∫p y
2
Fig. 6.45
0
dx
dq = 4 ∫p 4 sin q ( − sin q ) dq
dq
2
p
p
= 16 ∫ 2 sin 2 q dq = 8∫ 2 (1 − cos 2q ) dq
0
=8q −
0
sin 2q
2
p
2
0
⎛p ⎞
= 8⎜ ⎟
⎝2⎠
= 4p
Exercise 6.7
1. Find the area enclosed between one
arch of the cycloid x = a(q + sin q ),
y a(1 cos q ) and its base.
Ans. : 3 a 2
2. Find the area of the astroid
2
3
2
3
2
3
x +y =a .
3
Ans. : p a 2
8
3. Find the area bounded by the ellipse
x = a cos t , y = b sin t.
[ Ans. : p ab]
4. Find the area bounded by the curve
x = 2cosq – cos 2q – 1,
y = 2sinq – sin 2q.
[Ans. : 6p ]
5. Show that the area bounded by the
sin 3 t
cissoid x = a sin 2 t , y = a
and
cos t
its asymptote is 3p a 2 .
4
Polar Form
Example 1: Find the area bounded by the cardioid r = a(1 + cos q ) .
Solution: The region is symmetric about the initial line q = 0. For the region above
the initial line, q varies from 0 to p .
Engineering Mathematics
6.60
Area, A = 2(Area above the initial line)
p
1 p 2
2
r dq
a 2 (1 cos q ) 2 dq
0
2 0
p
a2
0
p
a2
a2
(1 2 cos q
0
3
q
2
1 2 cos q
cos 2 q ) dq
1 cos 2q
dq
2
sin 2q
4
2sin q
p
3 2
pa
2
0
Fig. 6.46
Example 2: Find the area bounded by the lemniscate r 2 = a 2 cos 2q .
Solution: The region is symmetric in all the
quadrants. For the region in the first quadrant, q
varies from 0 to p .
4
Area, A = 4(Area in the first quadrant )
p
1 p4 2
4
r dq 2 4 a 2 cos 2q dq
0
2 0
sin 2q
= 2a
2
2
=a
p
4
Fig. 6.47
0
2
Example 3: Find the area bounded by the curve r = a cos 3p .
Solution: For the region in the first quadrant,
varies from 0 to
p
.
6
Area, A = 6(Area in the first quadrant)
1 p6 2
6
r dq
2 0
p
p
⎛ 1 + cos 6q ⎞
= 3∫ 6 a 2 cos 2 3q dq = 3a 2 ∫ 6 ⎜
⎟⎠ dq
0 ⎝
0
2
3a 2
q
2
sin 6q
6
p
6
0
1
= p a2
4
Fig. 6.48
Example 4: Find the area of the curve r = a sin 2p .
Solution: The region is symmetric in all the quadrants. For the region in the first
p
quadrant, varies from 0 to .
2
Integral Calculus
Area,
6.61
A = 4(Area in the first quadrant)
p
2
1
2
4
0
p
2
= 2a 2
0
p
2
= 2a 2
0
r 2 dq
sin 2 2q dq
1 cos 4q
dq
2
sin 4q
4
2
a q
p
2
Fig. 6.49
0
1
= p a2
2
Example 5: Find the area of the smaller loop of the curve r = a ( 2 cos q - 1) .
Solution: The region is symmetric about the initial
line q = 0.
For the region above the initial line,
Area,
varies from 0 to
p
.
4
A = 2(Area above the initial line)
p
2
1 p4 2
4 2(
2
r dq
a
2 cos q 1) dq
0
0
2
a2
a2
p
4
0
p
4
0
(2 cos 2 q
(1 cos 2q
2 2 cos q 1) dq
Fig. 6.50
2 2 cos q 1) dq
1
= a 2q + sin 2q − 2 2 sin q
2
2
p
4
0
1 ⎞
⎛p 1
= a2 ⎜ + − 2 2
⎟
⎝2 2
2⎠
a2
= (p − 3)
2
Example 6: Find the area inside
the cardioid r = 1 + cos q and outside the circle r = 1.
Solution: The points of intersection
of the cardioid r = 1 + cos q and circle
r = 1 are obtained as,
1 + cos q = 1
cos q = 0
q =±
p
2
Fig. 6.51
Engineering Mathematics
6.62
p
2
The region is symmetric about the initial line
p
OACDO, q varies from 0 to .
2
Area, A = 2 (Area above the initial line)
q =
Hence, at C,
= 0. In the regions OBCDO and
⎤
⎡1 p
1 p
= 2( Area OBCDO − Area OACDO ) = 2 ⎢ ∫ 2 (1 + cos q ) 2 dq − ∫ 2 (1) 2 dq ⎥
0
0
⎦
⎣2
2
p
p
1 + cos 2q
⎛
= ∫ 2 (1 + 2 cos q + cos 2 q − 1) dq = ∫ 2 ⎜ 2 cos q +
0
0 ⎝
2
1
sin 2q
= q + 2 sin q +
2
4
⎞
⎟⎠ dq
p
2
0
p
= +2
4
Example 7: Find the area common to the circle r = 3cos q and the cardioid
r = 1 + cos q .
Solution: The points of intersection of the circle r = 3cos q
and the cardioid r = 1 + cos q are
obtained as,
3cos q = 1 + cos q
cos q =
1
2
q =±
p
3
p
3
The region is symmetric about the initial line
q =
Hence, at C,
from 0 to
and in the region, OECFO,
Fig. 6.52
= 0. In the region OACEO,
varies from
3
Area, A = 2(Area above the initial line)
p
p
to .
3
2
= 2( Area OACEO + Area OECFO )
⎡1 p
⎤
1 p
= 2 ⎢ ∫ 3 (1 + cos q ) 2 dq + ∫p2 9 cos 2 q dq ⎥
0
2 3
⎣2
⎦
p
p
⎛ 1 + cos 2q
= ∫ 3 (1 + 2 cos q + cos 2 q ) dq + 9 ∫p2 ⎜
0
⎝
2
3
⎞
⎟⎠ dq
varies
Integral Calculus
p
1 + cos 2q
⎛
= ∫ 3 ⎜1 + 2 cos q +
0 ⎝
2
p
3
sin 2q
3
= q + 2 sin q +
4
2
0
p
1 + cos 2q
⎞
2 ⎛
⎟⎠ dq + 9 ∫p ⎜⎝
2
3
9
sin 2q
+ q+
2
2
p
2
p
3
6.63
⎞
⎟⎠ dq
⎛p
3 ⎞ 9 ⎛p p
3⎞
+ ⎜ − −
= ⎜ + 3+
⎟
8 ⎠ 2 ⎝ 2 3 4 ⎟⎠
⎝2
p 9 3 9p 9 3 15p
+
+
−
=
2
8
12
8
12
5p
=
4
=
Example 8: Find the area common to the circles r = a 2 and r = 2a cos p .
Solution: The points of intersection
of circles r = a 2 and r = 2a cos q
are obtained as,
a 2 = 2a cos q
cos q =
1
q =±
2
p
4
p
4
The region is symmetric about the iniq =
Hence, at C,
tial line
DO,
Area,
= 0. In the region OACEO,
Fig. 6.53
varies from 0 to
varies from
to .
2
4
A = 2(Area above the initial line)
= 2( Area OACEO + Area OECDO )
⎡1 p
⎤
1 p
= 2 ⎢ ∫ 4 ( a 2 ) 2 dq + ∫p2 ( 2a cos q ) 2 dq ⎥
0
2
2
⎣
⎦
4
p
p
⎛ 1 + cos 2q ⎞
= ∫ 4 2a 2 dq + 4 a 2 ∫p2 ⎜
⎟⎠ dq
0
⎝
2
4
= 2a q
2
=
p
4
0
siin 2q
+ 2a q +
2
2
p 2
⎛p p 1 ⎞
a + 2a 2 ⎜ − − ⎟
⎝ 2 4 2⎠
2
= (p − 1)a 2
p
2
p
4
p
and in the region OEC4
Engineering Mathematics
6.64
Example 9: Find the area common to the cardioid r = a(1 – cosp ) and
r = a(1 + cosp ).
Solution: The points of intersection of the cardioid r
r = a(1 + cos q ) are obtained as,
a(1 cos q ) and
a(1 cos q ) a(1 cos q )
2 cos q = 0
cos q = 0
q =±
p
2
Fig. 6.54
The region is symmetric in all the quadrants. In the region OABO,
varies from 0 to
p
.
2
Area, A = 4(Area in the first quadrant)
4
2a 2
= 2a
p
2
1
2
2
0
p
2
0
∫
p
2
0
a 2 (1 cos q ) 2 dq
(1 2 cos q
cos 2 q ) dq
1 + cos 2q ⎞
sin 2q
⎛
2 3
⎜⎝1 − 2 cos q +
⎟⎠ dq = 2a q − 2 sin q +
2
2
4
p
2
0
⎛ 3p
⎞
= 2a 2 ⎜
− 2⎟
⎝ 4
⎠
Example 10: Find the area inside the cardioid r = 2a(1 + cos q ) and outside
2a
the parabola r =
.
1 + cos q
Solution: The points of intersection of the cardioid r = 2a(1 + cos q ) and parabola
2a
r=
are obtained as,
1 + cos q
Integral Calculus
6.65
2a
1 + cos q
1 + cos q = 1
cos q = 0
p
q =±
2
2a(1 + cos q ) =
q =
Hence, at B,
p
2
The region is symmetric about the
initial line q = 0. In the regions OADp
BCO and OABCO, varies from 0 to .
2
Area,
A = 2(Area above the initial line)
= 2( Area OADBCO − Area OABCO )
Fig. 6.55
p
⎡1 p
⎤
1
4a2
= 2 ⎢ ∫ 2 4 a 2 (1 + cos q ) 2 dq − ∫ 2
dq ⎥
0
0 (1 + cos q ) 2
2
2
⎣
⎦
p
p
= 4 a 2 ∫ 2 (1 + 2 cos q + cos 2 q ) dq − 4 a 2 ∫ 2
0
0
p
(
1
2 cos 2 q
2
)
2
dq
p
1 + cos 2q ⎞
⎛
2 2
4 q
= 4 a 2 ∫ 2 ⎜1 + 2 cos q +
dq
⎟⎠ dq − a ∫0 sec
0 ⎝
2
2
3
sin 2q
= 4a
q + 2 sin q +
2
4
2
p
2
0
p
q
q
q⎞
⎛
− a 2 ∫ 2 ⎜ tan 2 sec 2 + sec 2 ⎟ dq
0 ⎝
2
2
2⎠
2
q
q
⎡ 3p
⎤
= 4a ⎢
+ 2⎥ − a 2 tan 3 + 2 tan
3
2
2
⎣ 4
⎦
2
p
2
0
⎡
[ f (q )]n +1 ⎤
n
⎢∵ ∫ [ f (q )] f ′ (q )dq =
⎥
n +1 ⎦
⎣
⎛2
⎞
= a 2 (3p + 8) − a 2 ⎜ + 2 ⎟
⎝3
⎠
16 ⎞
⎛
= a 2 ⎜ 3p + ⎟
⎝
3⎠
Example 11: Find the area of the loop of the curve x 3 + y 3 = 3axy.
Solution: Putting x = r cos q , y = r sin q , equation of the curve becomes,
r 3 (cos3 q + sin 3 q ) = 3ar 2 sin q cos q
3a sin q cos q
r=
cos3 q + sin 3 q
Engineering Mathematics
6.66
p
.
2
For the loop of the curve,
varies from
p
0 to .
2
1 p2 2
Area,
A=
r dq
2 0
1 p2 9a 2 sin 2 q cos 2 q
dq
=
2 0 (cos3 q + sin 3 q ) 2
r = 0 at q = 0 and q =
9a 2
2
=
p
2
0
tan 2 q sec 2 q
dq
(1 + tan 3 q ) 2
1 + tan 3 q = t
Putting
3tan 2 q sec 2 q dq = dt
When
q = 0,
t =1
p
q = ,
2
t→
A=
3a 2
2
3a 2
2
1
Fig. 6.56
1
dt
t2
1
t1
3a 2
2
Example 12: Find the area of the loop of the curve x 4 + 3 x 2 y 2 + 2 y 4 = a 2 xy.
Solution: Putting x = r cos , y = r sin , the equation of the curve becomes,
r 4 cos 4 q + 3r 4 cos 2 q sin 2 q + 2r 4 sin 4 q
= a 2 r 2 cos q sin q
r2 =
a 2 cos q sin q
cos q + 3cos 2 q sin 2 q + 2sin 4 q
4
r = 0 at q = 0 and q =
p
.
2
For the loop of the curve,
varies from 0 to
p
.
2
Area,
1 p2 2
A=
r dq
2 0
cos q sin q
a 2 p2
=
dq
4
0
2
cos q + 3cos 2 q sin 2 q + 2sin 4 q
=
a2
2
∫
p
2
0
tan q sec 2 q
dq
1 + 3 tan 2 q + 2 tan 4 q
Fig. 6.57
[Dividing numerator and denomenator
by cos4 q ]
Integral Calculus
6.67
Putting
tan2 q = t
2 tanq sec q dq = dt
2
When
t=0
q = 0,
p
q = , t
2
1
a2 ∞
a2
dt =
A=
2
∫
4
4 0 1 + 3t + 2t
=
a2
4
∫
∞
0
∫
1
∞
0
( 2t + 1) (t + 1)
dt
∞
2 ⎞
a2
⎛ 1
d
t
−
+
=
− log (t + 1) + log ( 2t + 1) 0
⎜⎝
⎟⎠
t +1 2 t +1
4
⎡
1
∞
2+
2t + 1
a2 ⎢
a2
t
=
=
log
⎢ log
1
t +1 0
4 ⎢
4
1+
⎢⎣
t
2
a
= log 2
4
⎤
2t + 1 ⎥
− log
⎥
t + 1 t =0 ⎥
⎥⎦
t →∞
Example 13: Find the area of the loop of the curve (x2 + y2)(3ay – x2 – y2) = 4ay3.
Solution: Putting x = r cos , y = r sin , the equation of the curve becomes,
r 2 (3ar sin q
r2 )
4ar 3 sin 3 q
a(3sin q 4 sin 3 q )
= a sin 3q
p
r = 0 at q = 0 and q = .
3
p
For the loop of the curve, q varies from 0 to .
3
r
Area,
A=
1 p3 2
1 p3 2 2
r
d
q
=
a sin 3q dq
2 ∫0
2 ∫0
a2
=
4
∫
pa
=
12
p
3
0
2
(1 − cos 6q ) dq = a q − sin 6q
4
6
p
3
0
2
Exercise 6.8
1. Find the area of the loop of the curve
r = a sin 3 .
Ans. :
p a2
12
2. Find the area of the limacon
r = a + b cosq, a > b,
p
Ans. : (2a 2 + b 2 )
2
Engineering Mathematics
6.68
3. Find the area of the limacon
r = a cosq + b, a < b.
Ans. :
p 2
( a + 2b 2 )
2
4. Show that the area of the loop of the
curve r = aq cos q lying in the first
a 2p 2
quadrant is
(p
6).
96
5. Show that the area of a loop of the
a2
curve r cosq = a cos 2q is
(4 p ).
2
6. Show that area of a loop of the curve
1
r = 3 cos q + sin 3q is p .
3
7. Show that the area of the region
enclosed between the two loops
of the curve r = a(1 + 2 cos q ) is
(
)
a2 p + 3 3 .
8. Find the area of the ellipse
l
= 1 + e cos q.
r
Ans. :
Ans. :
a2
e 2 b cot a (e 2g cot a 1)
4 cot a
12. Show that the area contained between
the circle r = a and the curve
r = a cos5q is equal to three fourths
of the area of the circle.
13. Find the area common to two circles
r = a cos and r = a(cos q + sin q ).
Ans. :
a2
(p 1)
4
14. Find the area of the loop of the curve
x 4 + y 4 = 2a 2 xy.
1
Ans. : p a 2
4
p l2
3
(1 e 2 ) 2
9. Show that the area of the loop of the
curve r 2 cos = a 2 sin 3 lying in
the first quadrant is
11. Find the area of the portion of the
curve r = ae cot bounded by the
radii vectors q = b and q = b + g
where g > 2p .
1 2
e3
a log
.
4
4
10. Show that the area bounded by the
spiral r = ae m and two radii is proportional to the difference of the
squares of these radii.
15. Show that the area of a loop of the
5
curve x 5 + y 5 = 5ax 2 y 2 is a 2 .
2
16. Prove that the area of the loop of the
curve x 6 + y 6 = a 2 x 2 y 2 is
p a2
.
12
17. Find the area of the loop of the curve
( x + y )( x 2 + y 2 ) = 2axy.
Ans. : a 2 1
p
4
.
6.5 VOLUME OF SOLID OF REVOLUTION
A solid generated by revolving a plane area about a line in the plane is called a solid
of revolution.
Integral Calculus
6.69
Volume of Solid of Revolution in Cartesian Form
Let y = f ( x) be a curve and the area bounded by the
curve, the x-axis and the two lines x = a and x = b
be revolved about the x-axis. An elementary strip
of width dx at point P ( x, y ) of the curve, generates
elementary solid of volume y 2 dx, when revolved
about the x-axis.
Summing up the volumes of revolution of all such
strips from x = a to x = b, the volume of solid of
revolution is given by,
Volume,
V=
b
a
Fig. 6.58
p y 2 dx
Similarly, if the area bounded by the curve x = f (y),
the y-axis and the two lines, y = c and y = d is
revolved about the y-axis, then the volume of solid of
revolution is given by,
V=
d
c
p x 2 dy
The volume of solid of revolution about any axis
Fig. 6.59
can be obtained by calculating the length of the
perpendicular from point P(x, y) on the axis of revolution. If the area bounded by
the curve y = f ( x) is revolved about the line AB, then the volume of the solid of
revolution is given by,
V = ∫ p ( PM ) 2 d(AM )
with proper limits of integration.
Volume of Solid of Revolution in
Parametric Form
When the equation of the curve is given in parametric
form x = f1 (t), y = f2 (t) with t1 t t2, the volume of the
solid of revolution about the x-axis is given by,
V=
t2
t1
y2
dx
dt
dt
Fig. 6.60
Similarly, the volume of the solid of revolution
about the y-axis is given by,
t2
dy
V = ∫ x 2 dt
t1
dt
Volume of Solid of Revolution in
Polar Form For the curve r = f (q ), bounded
Fig. 6.61
between the radii vectors q = q1 and q = q2 the
volume of the solid of revolution about the initial line is given by,
Engineering Mathematics
6.70
2
V
1
2
=
1
2 2
r r sin d
3
2 3
r sin d
3
Similarly, the volume of the solid of revolution about the line through the pole and
perpendicular to the initial line is given by,
2 2
V
r 2 r cos d
1 3
2 2
r 3 cos d
=
1
3
x2
y2
Example 1: Find the volume generated by revolving the ellipse 2 + 2 = 1
a
b
about the x-axis.
Solution: The volume is generated by
revolving the upper-half of the ellipse about the
x-axis. For the upper half of the ellipse, x varies
from a to a. Due to symmetry about y-axis,
considering the region in the first quadrant where
x varies from 0 to a,
a
Volume, V = 2 ∫ p y 2 dx
0
a⎛
x2 ⎞
= 2p b 2 ∫ ⎜1 − 2 ⎟ dx
0⎝
a ⎠
= 2p b 2 x −
x3
3a 2
a
Fig. 6.62
0
⎛
a ⎞
= 2p b 2 ⎜ a − 2 ⎟
3a ⎠
⎝
3
4
p ab 2
3
Example 2: Find the volume generated by
revolving the area bounded by the parabola
y2 = 8x and its latus rectum about (i) x-axis,
(ii) latus rectum, and (iii) y-axis.
=
Solution: (i) The volume is generated by
revolving the region about the x-axis. For the
region above the x-axis, x varies from 0 to 2.
Volume,
V
2
0
p y 2 dx
p
2
0
8 x dx
8p
x2
2
2
16p
0
Fig. 6.63
Integral Calculus
6.71
(ii) The volume is generated by revolving the region about latus rectum. If P(x, y) is any
point on the curve, its distance from latus rectum is 2 x. For the region shown, y varies
from 4 to 4. Due to symmetry about x-axis, considering the region in the first quandrant
where y varies from 0 to 4,
4
Volume, V = 2 p ( 2 − x ) 2 dy
∫
0
2
4⎛
y2 ⎞
= 2p ∫ ⎜ 2 − ⎟ dy
0 ⎝
8 ⎠
4
4⎛
y2 y4 ⎞
y3
y5
= 2p ∫ ⎜ 4 −
+
dy = 2p 4 y − +
0 ⎝
2 64 ⎟⎠
6 320 0
=
256
p
15
(iii) The volume is generated by revolving the region about the y-axis. For the region
shown in Fig 6.64, y varies from 4 to 4. Due to
symmetry about x-axis, considering the region in
the first quadrant where y varies from 0 to 4,
Volume,
4
V = 2 ∫ p x 2 dy
0
= 2p ∫
4
0
=
y4
dy
64
p y5
32 5
4
⎡
y2 ⎤
⎢∵ x = ⎥
8⎦
⎣
0
32
= p
5
Fig. 6.64
Example 3: Find the volume of the solid generated by revolving the region
bounded by the curve y = log x and x = 2 about the x-axis.
Solution: The volume of the solid is generated by revolving the region about x-axis.
For the region shown, x varies from 1 to 2.
Volume,
2
V = ∫ p y 2 dx
1
2
= p ∫ (log x ) 2 dx
1
2
2
1
⎡
⎤
= p ⎢ (log x ) 2 ⋅ x − ∫ 2 log x ⋅ ⋅ x dx ⎥
1
1
x
⎣
⎦
2
= p ⎡ 2(log 2) 2 − 2 ∫ log x dx ⎤
1
⎦⎥
⎣⎢
2 1
⎡
2
⎛
⎞⎤
= 2p ⎢(log 2) 2 − ⎜ log x ⋅ x 1 − ∫
⋅ x dx ⎟ ⎥
1 x
⎝
⎠⎦
⎣
Fig. 6.65
2
= 2p ⎡(log 2) 2 − 2 log 2 + x 1 ⎤ = 2p [(log 2) 2 − 2 log 2 + 1]
⎣
⎦
2
= 2p (1 − log 2)
Engineering Mathematics
6.72
Example 4: Find the volume of the solid formed by the revolution of the curve
xy2 = a2 (a - x) through four right angles about the y-axis.
Solution: The volume of the solid is formed
by revolving the region about the y-axis. For the
region shown, y varies from
to . Due to
symmetry about the x-axis, considering the region
in the first quadrant, where y varies from 0 to ,
Volume,
∞
V = 2 ∫ p x 2 dy
0
∞
a6
= 2p ∫
dy
0 ( y 2 + a2 )2
⎡
a3 ⎤
⎢∵ x = 2
⎥
y + a2 ⎦
⎣
Putting
When
y = a tanq ,
dy = a sec2q dq
y = 0,
q
q
y
=
2
p
2
V = 2p a 7
Fig. 6.66
0
p
2
2p a 7
a4
0
sec 2 q
dq
( a 2 tan 2 q + a 2 ) 2
cos 2q dq = p a3
sin 2q
=p a q +
2
3
p
2
= p a3
0
p
2
0
(1 + cos 2q ) dq
p
2
p2 3
=
a
2
Example 5: Find the volume of the
solid of revolution of the loop of the curve
y2 =
x 2 (a + x )
about x-axis.
a-x
Solution: The volume of the solid of
revolution is generated by revolving the
upper half of the loop about the x-axis. For
the loop, x varies from a to 0.
Volume,
V=
=p
0
a
p y 2 dx
x 2 (a + x)
dx
a
a x
0
Fig. 6.67
Integral Calculus
Putting
When
a
6.73
x = t,
dx = dt
x = a,
x = 0,
t = 2a
t=a
a (a
t ) 2 (2a t )
p
dt
2a
t
2a 1
p
[2a3 t ( a 2 4 a 2 ) t 2 (2a 2a) t 3 ]dt
a t
V
=p∫
⎛ 2a 3
2
2⎞
⎜⎝ t − 5a + 4 at − t ⎟⎠ dt
2a
a
= p 2a3 log t − 5a 2 t + 2at 2 −
t3
3
2a
a
8 1⎞
⎛
= p a3 ⎜ 2 log 2 − 10 + 5 + 8 − 2 − + ⎟
⎝
3 3⎠
2⎞
⎛
= 2p a3 ⎜ log 2 − ⎟
⎝
3⎠
Example 6: Find the volume of the solid generated by revolving the curve
x3
about its asymptote.
2a - x
Solution: The volume of the solid is generated by revolving the region about its
asymptote.
The asymptote is x = 2a. If P(x, y) is any point on the curve, its distance from the
asymptote is 2a – x. For the region shown, y varies from – to . Due to symmetry
about the x-axis, considering the region in the first quadrant where y varies from 0 to ,
y2 =
Volume,
∞
V = 2 ∫ p ( 2 a − x ) 2 dy
0
3
y=
But,
x2
2a − x
(3a − x) x 2a − x
dx
( 2a − x ) 2
y = 0,
x=0
y
x = 2a
dy =
When
V
Putting
2p
2a
0
(3a x ) x 2a x dx
x = 2a sin2q ,
dx = 4a sinq cosq dq
Fig. 6.68
Engineering Mathematics
6.74
x = 0, q = 0
When
x = 2a, q =
2
p
2
0
V = 2p ∫ (3a − 2a sin 2 q ) 2a cos q sin q ⋅ 4 a sin q cos q dq
p
= 16p a3 ∫ 2 (3 − 2 sin 2 q ) sin 2 q cos 2 q dq
0
p
⎛ p
⎞
= 16p a3 ⎜⎝ 3∫ 2 sin 2 q cos 2 q dq − 2 ∫ 2 sin 4 q cos 2 q dq ⎟⎠
0
0
3 ⋅1 ⋅1 p ⎞
⎛ 1 ⋅1 p
= 16p a3 ⎜ 3 ⋅
⋅ − 2⋅
⋅ ⎟
⎝ 4⋅2 2
6⋅4⋅2 2 ⎠
= 2p 2 a3
Example 7: Find the volume generated
by revolving the area cut off from the
parabola 9y = 4 (9 - x2) by the line
4x + 3y = 12 about x-axis.
Solution: The points of intersection of the
parabola 9y = 4(9 x2) and the line 4x + 3y = 12
are obtained as,
3(12
4x) = 36
4x2
4x2 12x = 0
4x (x 3) = 0
x = 0, 3 and y = 4, 0
Hence, A: (3, 0) and B: (0, 4)
Fig. 6.69
The volume is generated by revolving the
region about the x-axis. For the region shown, x varies from 0 to 3.
3
Volume, V = ∫ p ( y12 − y22 ) dx
where y1 =
0
4
12 − 4 x
(9 − x 2 ) and y2 =
9
3
2
2
3 ⎡⎧ 4
⎫ ⎛ 12 − 4 x ⎞ ⎤
= p ∫ ⎢⎨ ( 9 − x 2 )⎬ − ⎜
⎟ ⎥ dx
0
⎭ ⎝ 3 ⎠ ⎥⎦
⎢⎣⎩ 9
3 ⎡16
16
⎤
= p ∫ ⎢ (81 − 18 x 2 + x 4 ) − (9 − 6 x + x 2 ) ⎥ dx
0 81
9
⎣
⎦
16p
=
81
48
= p
5
16p x 5
27
(
x
−
x
+
54
x
)
d
x
=
− 9 x 3 + 27 x 2
∫0
81 5
3
4
3
2
0
Integral Calculus
6.75
Example 8: Find the volume of the solid formed by revolving the area enclosed
between the curve 27ay2 = 4(x - 2a)3 and the parabola y2 = 4ax about x – axis.
Solution: The points of intersection of parabola y2 = 4ax and the curve
27ay2 = 4(x 2a)3 are obtained as,
3
x
27a (4ax) = 4(x 2a)3
6ax 15a2x 8a3 = 0
(x + a)2 (x 8a) = 0
x = a, 8a
2
But x = a does not lie on the curve. Hence,
x = 8a.
The volume is generated by revolving the region
about the x-axis. For the region shown, x varies
from 0 to 8a for y1 and 2a to 8a for y2.
2
2
where y1 = 4 ax and y2 =
8a
4( x − 2a)
27a
Fig. 6.70
3
8a
2
2
Volume, V =
∫ p y1 dx − ∫ p y2 dx
2a
0
8a
= p ∫ 4 ax dx − p ∫
0
2 8a
x
= 4 ap
2
0
8a
2a
4( x − 2a)3
dx
27a
4p ( x − 2a) 4
−
27a
4
8a
2a
= 128p a3 − 48p a3
= 80p a3
Example 9: A quadrant of a circle of radius a, revolves about its chord. Find the
volume of the spindle thus generated.
Solution: Let the equation of the circle be
x2 + y2 = a2
The volume of the spindle is generated by
revolving the region about the chord AB. Equation
of the chord AB is x + y = a. If P(x, y) is any
point on the circle and M is the foot of the
perpendicular from P on AB, then
x+ y−a
x+ y−a
PM =
=
2
2
2
(1) + (1)
( AM ) 2 = ( AP ) 2 − ( PM ) 2
⎡1
⎤
= [( x − a) 2 + ( y − 0) 2 ] − ⎢ ( x + y − a) 2 ⎥
⎣2
⎦
Fig. 6.71
Engineering Mathematics
6.76
1
= [2( x − a) 2 + 2 y 2 − ( x − a) 2 − y 2 − 2 y( x − a)]
2
1
2
= (x − a − y)
2
x−a− y
AM =
2
1
1 ⎛ dy ⎞
d(AM ) =
(dx − dy ) =
⎜1 − ⎟ dx
2
2 ⎝ dx ⎠
Since P lies on circle x2 + y2 = a2,
dy
2x 2 y
dx
dy
dx
0,
x
y
1 ⎛
x⎞
⎜⎝1 + y ⎟⎠ dx
2
For the region shown, x varies from 0 to a.
d (AM) =
Volume,
V =∫
When
( PM ) 2 d(AM ) =
0
=
Putting
a
2
∫
2
∫
a
0
⎛ x+ y−a⎞
⎜⎝
⎟
2 ⎠
2
1 ⎛ x⎞
⎜1 + ⎟ dx
2 ⎝ y⎠
⎛ x⎞
( x + y − a) 2 ⎜1 + ⎟ dx
⎝ y⎠
a
0
x = a cosq and y = a sinq ,
dx = a sinq dq
p
x= ,q=
2
x = a, q = 0
V=
=
=
=
2
⎛ a cos q
( a cos q + a sin q − a) 2 ⎜1 +
⎝ a sin q
2
0
∫
2
a3
2
∫
2
2
0
a3
2
∫
2
0
⎞
⎟⎠ ( − a sin q ) dq
(cosq + sin q − 1) 2 (cosq + sin q ) dq
( cos q + sin q + cos q sin 2 q + cos 2 q sin q − 1 − 2 cos q sin q ) dq
p a3
1
1
sin q − cos q + sin 3 q − cos3 q − q − sin 2 q
3
3
2
1⎞
p a3 ⎛ 1 p
=
⎜⎝1 + − − 1 + 1 + ⎟⎠
3 2
3
2
p
2
0
Integral Calculus
=
=
6.77
p a3 ⎛ 5 p ⎞
⎜ − ⎟
2 ⎝3 2 ⎠
p a3
6 2
(10 − 3p )
Example 10: Find the volume of the
solid formed by the revolution of a parabolic arc about the line joining the vertex
to one extremity of the latus rectum.
Solution: Let the equation of the parabola
be y2 = 4ax. The volume is generated by
revolving the region about the line OA.
Equation of the line OA is y = 2x.
If P(x, y) is any point on the parabola and
M is the foot of the perpendicular from P
on OA, then
y − 2x
y − 2x
PM =
=
2
2
5
(1) + ( −2)
Fig. 6.72
1
(OM 2 ) = (OP 2 ) − ( PM 2 ) = ( x 2 + y 2 ) − ( y − 2 x ) 2
5
=
OM =
d(OM ) =
5 x 2 + 5 y 2 − y 2 + 4 xy − 4 x 2 ( x + 2 y ) 2
=
5
5
x + 2y
5
1
5
(dx + 2dy ) =
dy ⎞
1 ⎛
⎜1 + 2 ⎟⎠ dx
dx
5⎝
Since P lies on the parabola, y2 = 4ax
dy
2y
= 4a
dx
dy 2 a
=
dx
y
1 ⎛
4a ⎞
⎜⎝1 + y ⎟⎠ dx
5
d(OM) =
For the region, x varies from 0 to a.
Volume,
V =∫
a
=
∫
( PM ) 2 d (OM )
0
a
0
1
1 ⎛ 4a ⎞
( y − 2 x)2
⎜1 + y ⎟⎠ dx
5
5⎝
Differential Equations
⎛ 2x
I.F. = e
=e
10.59
⎞
∫ ⎝⎜ x2 −1 + cot x⎠⎟ dx
log ⎡⎣( x 2 −1) sinn x ⎤⎦
= elog( x
2
−1) + log sin x
= ( x 2 − 1) sin x
Hence, solution is
( x 2 1) sin x y
( x 2 1) sin x cot x dx c
( x 2 1) cos x dx c
( x 2 1) sin x 2 x( cos x) 2( sin x) c
y ( x 2 1) sin x
( x2
3) sin x 2 x cos x c
di
Example 14: Solve L dt + iR = sin t ,
constants.
t
0,
i (0) = 0 where R, v and L are
Solution: Rewriting the equation,
di
dt
R
i
L
sin t
L
The equation is linear in i.
R
, Q
L
P
e
I.F.
R
dt
L
sin t
L
R
eL
t
Solution is
R
R
t
t
eL ⋅i = ∫ eL ⋅
sin t
dt + c
L
⎡
R
t
1 ⎢ eL
e ⋅i = ⎢ 2
L⎢R
+
⎢⎣ L2
R
t
L
R
t
eL L
= 2
R + 2 L2
i=
2
⎤
⎛R
⎞⎥
⎜ sin t − cos t ⎟ ⎥ + c
⎝L
⎠⎥
⎥⎦
⎛R
⎞
⎜ sin t − cos t ⎟ + c
⎝L
⎠
R
− t
1
R sin t − L cos t ) + ce L
2 2 (
R + L
2
Given i(0) = 0
Putting i = 0, t = 0 in Eq. (1),
1
(0 − L) + ce0
R + 2 L2
L
c= 2
R + 2 L2
0=
2
... (1)
Engineering Mathematics
10.60
Hence, solution is
i=
R
− t
1
L
L
(
R
sin
t
−
L
cos
t
)
+
e
R 2 + 2 L2
R 2 + 2 L2
dy
+ y tan x = sin 2 x , y(0) = 0, show that maximum value of y
Example 15: If
d
x
1
is .
2
Solution: The equation is linear in y.
P = tan x, Q = sin2x
e
I.F.
tan x dx
elog sec x
sec x
Hence, solution is
(sec x) y = ∫ sec x ⋅ sin 2 x dx + c = ∫ sec x ⋅ 2 sin x cos x dx + c = 2∫ sin x dx + c
y sec x = −2 cos x + c
y = −2 cos 2 x + c cos x
... (1)
Given y(0) = 0
Putting x = 0, y = 0 in Eq. (1),
0 = − 2 cos 0 + c cos 0 = − 2 + c
c=2
Hence, solution is
2 cos 2 x 2 cos x
y
For maximum or minimum value
dy
0
dx
4 cos x( sin x) 2sin x 0
2sin x(2 cos x 1) 0
sin x = 0, x = 0 and 2 cos x 1 0, cos x
x = 0 and x
3
0,
3
are the points of extreme values.
Now,
When x
1
,x
2
d2 y
dx 2
2
dy
dx
2sin 2 x 2sin x
d2 y
dx 2
4 cos 2 x 2 cos x
0, y is minimum at x = 0.
... (2)
Differential Equations
When x
x
3
3
,
d2 y
dx 2
4 cos
2
3
2 cos
1
2
4
3
10.61
1
2
2
3 0 , y is maximum at
.
Putting x
3
in Eq. (2), we get maximum value of y.
2 cos 2
ymax
3
2 cos
1
1
2
3
1
2
Exercise 10.6
Solve the following differential equations:
1. x 2
dy
dx
3x 2
⎡⎣ Ans. : xe y − tan y + c ⎤⎦
2 xy 1
1⎤
c
⎡
⎢⎣ Ans. : y = x 2 + x + x ⎥⎦
2. (2 y 3 x) dx x dy
dy
dx
y cot x
2y
9.
e
2 x
y
dx
x dy
10. x cos x
dy
dx
[ Ans. :
x2
11. cos 2 x
dy
dx
y
⎡⎣ Ans. : ( y + 1)( x − e ) = c ⎤⎦
e y sec 2 y dy
xy = sin x + c cos x ]
tan x
y4 )
dy
dx
y
⎡
⎤
2x
2
⎢ Ans. : y 2 = y + c ⎥
⎣
⎦
0
y
= 2 x + c ⎤⎦
⎡⎣ Ans. : y = tan x − 1 + ce − tan x ⎤⎦
12. (2 x
y
6. ( y 1) dx [ x ( y 2)e ]dy
x
y ( x sin x cos x) 1
⎤
⎡
x2
x2
=
+ c⎥
Ans.
:
ye
⎢
2
⎦
⎣
7. dx x dy
1
⎡ Ans. : ye 2
⎣
cos x
e
e x ( x 1) 2
⎡⎣ Ans. : y = (1 + x)(e x + c) ⎤⎦
( x 1) 4
2y
⎡
sin 2 x ⎤
+ c⎥
⎢ Ans. : y sin x =
2
⎣
⎦
5. 1 dy
x dx
y
x
⎡
⎤
⎛ x2
⎞
2
⎢ Ans. : y = ⎜ + x + c⎟ ( x + 1) ⎥
⎝ 2
⎠
⎣
⎦
4. dy
dx
dy
dx
0
⎡⎣ Ans. : x 2 y = x 3 + c ⎤⎦
3. ( x 1)
8. (1 x)
13.
a2
x2
⎡ Ans. :
⎣⎢
dy
dx
(x +
y
a2
x2
)
x
x 2 + a 2 y = a 2 x + c ⎤⎥
⎦
Engineering Mathematics
10.62
14.
dy
dx
1
(e x
sin x) x 3
x ey
⎡⎣ Ans. : xe − y = c + y ⎤⎦
15.
2x2
Ans. : y
dy
dx
3
y
x
x 3 , y (1)
Ans. : y
18. If
4
Ans. : y
2 xy
sin x, y
3
0,
show that maximum value of y is
x 3 ( x 3)
19.
2 dy
16. (1 x )
dx
y (0) 1
dy
2 y tan x
dx
dy
dx
y
x
2 x(1 x 2 ),
1
.
8
log x, y (1) 1
x log x x 5 ⎤
⎡
⎢⎣ Ans. : y = 2 − 4 + 4 x ⎥⎦
(1 x 2 )[1 log(1 x 2 )
20.
dy
2 xy
dx
dy
3 y x 4 (e x cos x) 2 x 2 ,
dx
3
y( )
e 2 2
xe
x2
⎡
⎤
x2
x2
+ c⎥
⎢ Ans. : ye =
2
⎣
⎦
17. x
10.3.7 Non-linear Differential Equations
Reducible to Linear Form
Type 1: Bernoulli’s Equation
The equation of the form
dy
dx
Py
Qy n
... (1)
where P and Q are functions of x or constants is a non-linear equation known as
Bernoulli’s equation. This equation can be made linear using the following method:
Dividing Eq. (1) by yn,
1 dy
P
Q
... (2)
n
y dx y n 1
Let
1
yn
1
v
(1 n) dy
y n dx
1 dy
y n dx
dv
dx
dv
(1 n) dx
1
Substituting in Eq. (2),
dv
Pv
1 n dx
dv
(1 n)Pv
dx
1
Q
Q
Differential Equations
10.63
The equation is linear in v and can be solved using the method of linear differential
1
, we get the solution of Eq. (1).
equations. Finally, substituting v
yn 1
Example 1: Solve
dy 2 y
+
= y2 x2 .
dx x
Solution: The equation is in Bernoulli’s form.
Dividing the equation by y2,
Let
1
y
v,
1 dy
y 2 dx
1 dy
y 2 dx
1 2
y x
x2
dv
dx
2
v
x
x2
dv
dx
2
v
x
... (1)
dv
dx
Substituting in Eq. (1),
x2
... (2)
The equation is linear in v.
2
,Q
x
P
I.F.
e
2
dx
x
x2
e
2 log x
elog x
2
x
2
1
x2
Solution of Eq. (2) is
1
1
v = ∫ 2 ( − x 2 )dx + c
2
x
x
= ∫ −dx + c = − x + c
v = − x 3 + cx 2
1
= − x 3 + cx 2
y
Hence,
Example 2: Solve
dy
+ y = y 2 (cos x - sin x ).
dx
Solution: The equation is in Bernoulli’s form.
Dividing the equation by y2,
1 dy
y 2 dx
1
y
cos x sin x
... (1)
Engineering Mathematics
10.64
Let
1
y
v,
1 dy
y 2 dx
dv
dx
Substituting in Eq. (1),
dv
dx
dv
dx
v
cos x sin x
v
cos x sin x
... (2)
The equation is linear in v.
P = −1, Q = − cos x + sin x
I.F. e
dx
e
x
Solution of Eq. (2) is
e − x ⋅ v = ∫ e − x ( − cos x + sin x) dx + c
= − ∫ e − x cos x dx + ∫ e − x sin x dx + c
⎤
⎤ ⎡ e− x
⎡ e− x
( − sin x − cos x) ⎥ + c
= −⎢
( − cos x + sin x) ⎥ + ⎢
2
2
⎦
⎦ ⎣
⎣
−x
−x
e v = − e sin x + c
v = − sin x + ce x
1
= − sin x + ce x
y
Hence,
Example 3: Solve xy(1 + xy 2 )
dy
= 1.
dx
Solution: Rewriting the equation,
dx
dy
xy
dx
dy
xy x 2 y 3
x2 y3
The equation is in the Bernoulli’s form where x is a dependent variable.
Dividing the equation by x2,
1 dx
1
y y3
2
x
x dy
Let
1
x
v,
1 dx
x 2 dy
dv
dy
... (1)
Differential Equations
10.65
Substituting in Eq. (1),
dv
+ vy = y 3
dy
... (2)
The equation is linear in v.
P
y,
I.F.
e
y3
Q
y2
y dy
e2
Solution of Eq. (2) is
y2
y2
e 2 ⋅ v = ∫ e 2 y 3 dy + c
Putting
y2
2
t , y dy
dt
y2
e 2 ⋅ v = ∫ et ⋅ 2t dt + c = 2(et t − et ) + c
y2
2
⎛ y2 ⎞
= 2e (t − 1) + c = 2e ⎜ − 1⎟ + c
⎝ 2
⎠
t
v = y − 2 + ce
2
−
Hence,
Example 4: Solve y 4 dx = x
−
y2
2
−y
1
= y 2 − 2 + ce 2
x
3
4
2
- y 3 x dy .
Solution: Rewriting the equation,
3
dx
dy
x 4
y4
x
y
x 4
y4
x
y
3
dx
dy
The equation is in Bernoulli’s form where x is a dependent variable.
Dividing the equation by x
3
4
x
7
Let x 4
v,
7 34 dx
x
4 dy
dv
dy
3
4
,
dx
dy
7
x4
1
y
1
y4
... (1)
Engineering Mathematics
10.66
Substituting in Eq. (1),
4 dv
7 dy
1
v
y
1
y4
dv
dy
7
v
4y
7
4 y4
... (2)
The equation is linear in v.
P
I.F.
7
,Q
4y
e
7
dy
4y
7
4 y4
7
e4
log y
elog y
7
4
7
y4
Solution of Eq. (2) is
7
7
y4v = ∫ y4 ⋅
7
dy + c
4 y4
⎛ −5 ⎞
7 −49
7 ⎜ 4y 4 ⎟
= ∫ y dy + c =
+c
4
4 ⎜ −5 ⎟
⎝
⎠
7 − 54
y +c
5
5
7
y 3 v = − + cy 4
5
7
y4v = −
Hence,
Example 5: Solve
7
5
7
y 3 x 4 = − + cy 4
5
dr
r2
= r tan p .
dp
cos p
Solution: Rewriting the equation,
dr
d
r tan
r2
cos
The equation is in Bernoulli’s form where r is a dependent variable.
Dividing the equation by r2,
1 dr tan
1
r
cos
r2 d
1
= v,
r
1 dr dv
=
d
r2 d
Let −
... (1)
Differential Equations
Substituting in Eq. (1),
dv
v tan
d
10.67
1
cos
... (2)
The equation is linear in v.
P
1
cos
tan , Q
e
I.F.
tan d
elog sec
sec
Solution of Eq. (2) is
1 ⎞
⎛
sec ⋅ v = ∫ sec ⎜ −
⎟d + c
⎝ cos ⎠
= ∫ − sec 2 d + c = − tan + c
⎛ 1⎞
sec ⎜ − ⎟ = − tan + c
⎝ r⎠
sec
= tan − c
r
Hence,
Type 2: The equation of the form f ( y )
dy
dx
Pf ( y )
Q
... (1)
where P and Q are functions of x or constants can be reduced to the linear form by
dy dv
putting f ( y ) = v, f ′ ( y ) =
in Eq. (1)
dx dx
dv
... (2)
Pv Q
dx
Equation (2) is linear in v and can be solved using the method of linear differential
equation. Finally, substituting v f ( y ), we get the solution of Eq. (1).
Example 1: Solve
dy
+ x sin 2 y = x 3 cos 2 y .
dx
Solution: Dividing the equation by cos2y,
1 dy
cos 2 y dx
2sin y cos y
x
cos 2 y
dy
sec 2 y
2 tan y x
dx
Let tan y
v,
sec 2 y
dy
dx
dv
dx
x3
x3
... (1)
Engineering Mathematics
10.68
Substituting in Eq. (1),
dv
+ 2vx = x 3
dx
... (2)
The equation is linear in v
P = 2 x, Q = x 3
I.F. = e
2 x dx
= ex
2
Solution of Eq. (2) is
e x v = ∫ e x ⋅ x 3 dx + c
2
Putting x 2 = t ,
2 x dx = dt , x dx =
2
dt
2
e x v = ∫ et t
2
2
(
)
1
dt
+ c = tet − et + c
2
2
1 x2 2
e ( x − 1) + c
2
2
1
v = ( x 2 − 1) + ce − x
2
ex v =
tan y =
Hence,
Example 2: Solve x
2
1 2
( x − 1) + ce − x
2
dy
+ y log y = xye x .
dx
Solution: Dividing the equation by xy,
Let log y = v,
1 dy log y
+
= ex
y dx
x
... (1)
dv v
+ = ex
dx x
... (2)
1 dy dv
=
y dx dx
Substituting in Eq. (1),
The equation is linear in v.
P=
1
, Q = ex
x
I.F. = e
1
dx
x
= elog x = x
Differential Equations
10.69
Solution of Eq. (2) is
xv = ∫ xe x dx + c = xe x − e x + c
xv = e x ( x − 1) + c
x log y = e x ( x − 1) + c.
Hence,
dy
+ tan x tan y = cos x sec y .
dx
Example 3: Solve
Solution: Dividing the equation by sec y,
1 dy
+ tan x sin y = cos x
sec y dx
cos y
Let sin y
v, cos y
dy
dx
dy
+ tan x sin y = cos x
dx
... (1)
dv
dx
Substituting in Eq. (1),
dv
dx
tan x v
cos x
... (2)
The equation is linear in v.
P
I.F.
tan x, Q
e
tan x dx
cos x
elog sec x
sec x
Solution of Eq. (2) is
sec x ⋅ v = ∫ sec x ⋅ cos x dx + c
= ∫ dx + c
sec x ⋅ v = x + c
sec x ⋅ sin y = x + c
Hence,
Example 4: Solve
dy
= e x - y (e x
dx
e y ).
Solution: Dividing the equation by e–y,
dy
dx
e2 x
dy
exe y
dx
e2 x
ey
ey
Let e y
v, e y
dy
dx
dv
dx
exe y
... (1)
Engineering Mathematics
10.70
Substituting in Eq. (1),
dv
dx
exv
e2 x
... (2)
The equation is linear in v.
P
I.F.
ex , Q
e
e x dx
e2 x
ee
x
Solution of Eq. (2) is
e e ⋅ v = ∫ e e ⋅ e 2 x dx + c
x
x
Let ex = t, ex dx = dt
e e ⋅ v = ∫ e t t dt + c = e t ⋅ t − e t + c
x
x
= et (t − 1) + c = ee (e x − 1) + c
v = e x − 1 + ce − e
e y = e x − 1 + ce − e
Hence,
Example 5: Solve
e
e −2 x
x
dy
y3
.
= 2x
dx e + y 2
Solution: Rewriting the equation,
Let e −2 x = v, − 2e −2 x
x
2x
dx
dy
e2 x
y3
dx
dy
e
2x
y
1
y
1
y3
... (1)
dx dv
=
,
dy dy
dx
1 dv
=−
dy
2 dy
Substituting in Eq. (1),
1 dv v
2 dy y
dv 2
v
dy y
1
y3
2
y3
... (2)
The equation is linear in v.
P
I.F.
2
,Q
y
e
2
dy
y
2
y3
e 2 log y
elog y
2
y2
Differential Equations
10.71
Solution of Eq. (2) is
⎛ 2⎞
y 2 ⋅ v = ∫ y 2 ⎜ − 3 ⎟ dy + c
⎝ y ⎠
1
dy + c
y
= −2 log y + c
y 2 ⋅ v = −2∫
y 2 e −2 x = −2 log y + c
Hence,
Example 6: Solve 2 xy
Solution: Let y 2
6
3
dy
= ( y 2 + 6) + x 2 ( y 2 + 6)4 .
dx
z, 2 y
dy
dx
dz
dx
Substituting in given equation,
dz
dx
1 dz
z 4 dx
1 dz
1
z 4 dx xz 3
x
1
3 dz dv 1 dz
v, 4
,
3
z
z dx dx z 4 dx
Substituting in Eq. (1),
1 dv
3 dx
dv
dx
The equation is linear in v.
Let
3
x2 z4
z
1
xz 3
1
x2
1
x2
... (1)
1 dv
3 dx
v
x
3v
x
P
1
x2
1
3x 2
3
,
x
I.F. e
3
dx
x
... (2)
1
Q
3x 2
e3log x
elog x
3
x3
Solution Eq. (2) is
1
7
2 9
x 3 v = ∫ x 3 ⋅ 3 x 2 dx + c = 3∫ x 2 dx + c = 3 ⋅ x 2 + c
9
2 92
x + c,
3
9
⎛ 1⎞ 2
x3 ⎜ − 3 ⎟ = x 2 + c
⎝ z ⎠ 3
x3v =
Engineering Mathematics
10.72
3
Hence,
⎛ x ⎞
2 9
−⎜ 2
= x 2 + c.
⎟
3
⎝ y + 6⎠
Example 7: Solve
dz z
z
+ log z = 2 (log z )2 .
dx x
x
Solution: Rewriting the equation,
1
dz
z (log z ) 2 dx
Let
−1
= v,
log z
1 1
log z x
1
x2
... (1)
1
1 dz dv
⋅
=
(log z ) 2 z dx dx
Substituting in Eq. (1),
dv
dx
v
x
1
x2
... (2)
The equation is linear in v.
1
,Q
x
P
I.F.
e
1
dx
x
1
x2
e
log x
elog x
1
x
1
x
1
Solution of Eq. (2) is
1
1 1
x −2
⋅ v = ∫ ⋅ 2 dx + c = ∫ x −3 dx + c =
+c
x
x x
−2
Hence,
1⎛
1 ⎞
1
−
= − 2 +c
⎜
⎟
x ⎝ log z ⎠
2x
1
1
=
−c
x log z 2 x 2
Example 8: Solve x sin p dp + ( x 3 - 2 x 2 cos p + cos p )dx = 0.
Solution: Rewriting the equation,
x sin
d
dx
sin
x 3 (2 x 2 1) cos
d
dx
2x
1
cos
x
0
x2
... (1)
Differential Equations
Let
cos
dv
dx
d
dx
v, sin
10.73
Substituting in Eq. (1),
dv
dx
2x
1
v
x
x2
... (2)
The equation is linear in v.
1
P = 2x − , Q = − x2
x
2x
e
I.F.
1
dx
x
ex
2
log x
2
ex e
log x
2
2
e x elog x
1
2
ex x
1
ex
x
Solution of Eq. (2) is
2
2
ex
ex
⋅v = ∫
⋅ ( − x 2 ) dx + c
x
x
= − ∫ e x ⋅ x dx + c
2
Let x 2
t , 2 x dx
dt
2
dt , x dx
2
ex
dt
et
⋅ v = − ∫ et ⋅ + c = − + c
x
2
2
2
2
ex
ex
⋅v = −
+c
x
2
2
x
v = − + c x e− x
2
Hence,
2
x
− cos = − + cx e − x
2
Example 9: Solve (sec x tan x tan y – e x ) dx + sec x sec 2 y dy = 0.
Solution: Rewriting the equation,
sec x sec 2 y
dy
+ sec x tan x tan y − e x = 0
dx
sec 2 y
dy
dx
tan x tan y
ex
sec x
... (1)
Engineering Mathematics
10.74
Let tan y = v, sec 2 y
dy dv
=
dx dx
Substituting in Eq. (1),
dv
+ (tan x)v = e x cos x
dx
... (2)
The equation is linear in v.
P = tan x, Q = e x cos x
tan x dx
I.F. = e
= elog sec x = sec x
Solution of Eq. (2) is
(sec x)v = ∫ sec x e x cos x dx + c
= ∫ e x dx + c = e x + c
sec x tan y = e x + c
Hence,
Example 10: Solve
dy
+ x ( x + y ) = x 3 ( x + y )3 - 1.
dx
Solution:
dy
dx
x( x
y)
Let x
z, 1
dy
dx
dz dy
,
dx dx
y
x3 ( x
y )3 1
... (1)
dz
1
dx
Substituting in Eq. (1),
dz
1 xz x 3 z 3 1
dx
dz
+ xz = x 3 z 3
dx
Dividing the Eq. (2) by z3,
1 dz x
+
= x3
z 3 dx z 2
Let
... (2)
... (3)
1
2 dz dv 1 dz
1 dv
= v, − 3
= ,
=−
2 dx
z2
z dx dx z 3 dx
Substituting in Eq. (3),
1 dv
xv
2 dx
dv
2 xv
dx
x3
2 x3
... (4)
Differential Equations
10.75
The equation is linear in v.
2 x3
2 x, Q
P
e
I.F.
2 x dx
e
x2
Solution of Eq. (4) is
e − x ⋅ v = ∫ e − x ( −2 x 3 )dx + c
2
Let x 2
t , 2 x dx
2
dt
e − x ⋅ v = − ∫ te − t dt + c = te − t + e − t + c
2
2
= ( x 2 + 1)e − x + c
v = ( x 2 + 1) + ce x
2
Substituting value of v,
2
1
= ( x 2 + 1) + ce x
2
z
2
1
= ( x 2 + 1) + ce x
2
( x + y)
Hence,
Exercise 10.7
Solve the following differential equations:
1.
dy
dx
x3 y 3
⎡
2 3 2⎛2
⎞
2
2⎤
⎢ Ans. : x = − 3 x y ⎜⎝ 3 + log x⎟⎠ + cy ⎥
⎣
⎦
xy
⎡
1
2
x2 ⎤
⎢ Ans. : y 2 = x + 1 + ce ⎥
⎣
⎦
2
3 dy
4
y cos x
2. x y x
dx
⎡⎣ Ans. : x 3 = y 3 (3 sin x − c) ⎤⎦
2
2
3. x(3 x 2 y ) dx 2 y (1 x ) dy 0
dy
6. dx
y2ex
y
⎤
⎡
e− x
Ans.
:
−
= x + c⎥
⎢
y
⎣
⎦
7. x dy
y dx
⎡
2
3
5⎤
⎢ Ans. : y 5 = 5 x + cx ⎥
⎣
⎦
⎡⎣ Ans. : y 2 (1 + x 2 ) = − x 3 + c ⎤⎦
2 2
4. y dx x(1 3 x y ) dy
0
1
− 2 2 ⎤
⎡
6
x y
⎣⎢ Ans. : y = ce
⎦⎥
5. x dy [ y
xy 3 (1 log x)]dx
0
x 3 y 6 dx
8. x
dy
dx
y
y3 xn
1
⎡
n −1
2
n +1 ⎤
⎢ Ans. : y 2 = cx − 2 x ⎥
⎣
⎦
Engineering Mathematics
10.76
2 2
9. xy (1 x y )
dy
dx
1
⎤
⎡
2
− y2
⎢⎣ Ans. : x 2 = ce − y + 1⎥⎦
2 3
3
10. x y dx ( x y 2) dy
19.
0
dy
dx
2
yx cos y
dy
3y
dx
ey
x2
2
21. x
1
x
dy
dx
y2
4
x
3
3x
dy
dx
0
dy
dx
(2 x tan 1 y
x 3 )(1 y 2 )
22.
dr
d
⎡ Ans. : 2 tan −1 y = ( x 2 − 1) + ce − x ⎤
⎣
⎦
2
15. tan y
dy
dx
tan x
[ Ans. :
16. ( y e y
cos y cos 2 x
sec y sec x = sin x + c ]
e x )dx (1 e y )dy
0
⎡⎣ Ans. : y + e y = ( x + c)e − x ⎤⎦
2
17. x cos y
dy
dx
2 x sin y 1
⎡⎣ Ans. : 3 x sin y = cx 3 + 1⎤⎦
cosec y = 1 + cx ]
1
2
x4e x y3
r sin
r2
cos
1
⎤
⎡
⎢ Ans. : r = c cos + sin ⎥
⎦
⎣
23. cos x
0
1
tan y sin y
x2
1
⎤
⎡
⎢⎣ Ans. : cot y = 2 x + cx ⎥⎦
2
4
−
⎤
⎡
2
3
3
⎢ Ans. : y x + 2 x = c ⎥
⎦
⎣
14.
2)3
sin 2 y (sin y cos y ) x
⎡⎣ Ans. : 2 xe − y = 1 + 2cx 2 ⎤⎦
13. y
2) 2(3 y 2
1
⎡
⎞ 6⎤
1 ⎛ x2
⎢ Ans. : 2 = ⎜ e + c⎟ x ⎥
y
⎝
⎠ ⎦⎥
⎢⎣
y
⎤
⎡
⎢⎣ Ans. : x = y sin y + cos y + c ⎥⎦
dy
12.
dx
1
tan y
x
[ Ans. :
20. x
x
3 x(3 y 2
⎡⎣ Ans. : 4 x 9 = (3 y 2 + 2) 2 ( −3 x8 + c) ⎤⎦
3
⎡
⎤
2 2
3
y
⎢ Ans. : x = + + ce ⎥
y 3
⎢⎣
⎥⎦
dx
11. y
dy
dy
dx
18. 4 x 2 y
1
dy
dx
4 y sin x
4 y sec x
⎡ Ans. : y sec 2 x
⎤
⎢
⎥
3
⎛
tan x ⎞
⎢
⎥
= 2 ⎜ tan x +
+ c⎥
⎢
⎟
3 ⎠
⎝
⎣
⎦
24. sin y
dy
dx
cos x(2 cos y sin 2 x)
⎡ Ans. : 4 cos y = 2 sin 2 x − 2 sin x ⎤
⎥
⎢
+ 1 − 4ce −2 sin x ⎦
⎣
y dy
25. e dx 1
ex
⎤
⎡
e2 x
x+ y
+ c⎥
⎢ Ans. : e =
2
⎦
⎣
Engineering Mathematics
6.78
=
=
5
∫
5
a
∫
a
5 5
0
0
⎡ 2
4a 2
2
2 ⎤
⎢ y − 4 xy + 4 x + y ( y − 4 xy + 4 x ) ⎥ dx
⎣
⎦
⎛
ax 2 ⎞
2
8
a
ax
−
16
ax
+
8
ax
x
ax
x
4
8
4
−
+
+
⎜⎝
⎟ dx
ax ⎠
a
=
p
5 5
2
4a ⋅
=
2
x
x
x
4x
+
+ 8a a
− 16 a
3
2
3
2
2
0
p ⎛ 3 4 a 16 a
3⎞
⎜⎝ 2a + 3 + 3 − 8a ⎟⎠
5 5
3
=
3
2
3
2pa3
15 5
3
.
Exercise 6.9
1. Find the volume of the solid of revolution generated by revolving the
plane area bounded by the curves
y = x3, y = 0, and x = 2 about x-axis.
128p ⎤
⎡
⎢ Ans. : 7 ⎥
⎣
⎦
2. Find the volume of the solid of revolution generated by revolving the
region bounded by the curve ex sin x
and x-axis about the x-axis.
p 2p
⎡
⎤
⎢ Ans. : 8 (e − 1) ⎥
⎣
⎦
3. Show that the volume generated
by revolving the loop of the curve
y2(a + x) = x2(3a x) about the x-axis
is p a3(8 log 2 3).
4. Find the volume generated by revolving the curve x (y2 + a2) = a3 about its
asymptote.
1 2 3⎤
⎡
⎢ Ans. : 2 p a ⎥
⎣
⎦
5. The area bounded by the curve
y = x(x 1) (2 x) and the x-axis
between x = 1 and x = 2 is revolved
about the x-axis. Prove that the volume
8p
.
generated is
105
6. The area enclosed by the parabolas
x2 = 4ay and x2 = 4a(2a y) revolves
about the line y = 2a. Find the volume of the solid so generated.
32 3 ⎤
⎡
⎢ Ans. : 3 p a ⎥
⎣
⎦
7. The curve included between the
curves y2 = 4ax and x2 = 4ay revolves
about the x-axis. Find the volume of
the solid of revolution.
96 3 ⎤
⎡
⎢ Ans. : 5 p a ⎥
⎣
⎦
8. Find the volume generated by
revolving the area between the curve
y +8
= x − 2 and the x-axis about the
x
line x + 5 = 0.
[ Ans. : 432p ]
9. Show that the volume of the spindle
formed by the revolution of a parabolic arc about the line joining the
vertex to one extremity of the latus
2p a3
, 4a being the latus
rectum is
15 5
rectum of the parabola.
Integral Calculus
x2 y2
+
= 1 is divided
a2 b2
a
into two parts by the line x = and
2
the smaller part is rotated through
four right angles about this line. Find
the volume generated.
⎡
⎛ 3 3 p ⎞⎤
2
− ⎟⎥
⎢ Ans. : p a b ⎜
3 ⎠ ⎥⎦
⎝ 4
⎢⎣
11. The first quadrant of the ellipse
x2 y2
+
= 1 revolves about the line
a2 b2
joining its extremities. Show that
the volume of the solid generated is
p a2b2 ⎛ 5 p ⎞
⎜ − ⎟.
a2 + b2 ⎝ 3 2 ⎠
10. The ellipse
12. Find the volume of the solid
obtained by revolving the area
between the curves y2 = x3 and x2 = y3
about the x-axis.
5 ⎤
⎡
⎢ Ans. : 28 p ⎥
⎣
⎦
6.79
13. The parabola y2 = 8ax divides the
circle x2 + y2 = 9a2 into two arcs the
smaller of which is rotated about the
x-axis. Show that the volume of solid
28
p a2 .
generated is
3
14. Show that the volume obtained by
revolving the area enclosed between the curves xy2 = a2(a x) and
(a x) y2 = a2x about
a
p a3
is
( 4 − p ).
2
4
15. The loop of the curve 2ay2 = x(x a)2
revolves about the straight line
y = a. Find the volume of the solid
generated.
⎡
⎤
8 2
p a3 ⎥
⎢ Ans. :
15
⎣
⎦
16. The area bounded by the hyperbola
xy = 4 and the line x + y = 5 is
revolved about x-axis. Find the
volume of the solid thus formed.
[Ans. : 9p ]
x=
Parametric Form
Example 1: For the cycloid, x = a (p + sinp ), y = a (1 - cosp ), find the volume
of the solid generated by the revolution of one arch about (i) the tangent at the
vertex (i.e., x-axis), (ii) y-axis, and (iii) the base.
Solution: (i)
x = a (q + sinq )
dx
= a(1 + cos q )
dq
Fig. 6.73
Engineering Mathematics
6.80
The volume of the solid is generated by revolving one arch about the x-axis. For the
region shown, x varies from ap to ap, hence q varies from p to p. Due to symmetry
about the y-axis, considering the region in the first quadrant where q varies from 0 to p,
π
dx
dθ
V = 2∫ π y 2
Volume,
0
dθ
π
= 2π ∫ a 2 (1 − cos θ ) 2 a(1 + cos θ ) dθ
0
θ
θ
⋅ 2 cos 2 dθ
2
2
π
θ
θ
= 16π a3 ∫ sin 4 cos 2 dθ
0
2
2
π
= 2π a3 ∫ 4 sin 4
0
q
=t
2
d q = 2 dt
q = 0, t = 0
Putting
When
q = p, t =
p
2
V = 32 a3 ∫ 2 sin 4 t cos 2 t dt
0
= 32 a3 ⋅
3 ⋅1 ⋅1
⋅
6⋅4⋅2 2
= 2 a3
(ii)
y = a (1 cosq )
dy
= a sin q
dq
The volume of the solid is generated by revolving one arch about the y-axis. For the
region shown, y varies from 0 to 2a, hence q varies from 0 to p.
Volume,
π
dy
dθ
V = ∫ π x2
0
dθ
π
= π ∫ a 2 (θ + sin θ ) 2 a sin θ dθ
0
π
= π a3 ∫ (θ 2 sin θ + 2θ sin 2 θ + siin 3 θ ) dθ
0
π ⎡
1
⎤
= π a3 ∫ ⎢θ 2 sin θ + θ (1 − cos 2θ ) + (3 sin θ − sin 3θ ) ⎥ dθ
0
4
⎣
⎦
2
⎡ ⎛
⎞⎤
1
1
⎞ ⎛θ
= π a3 ( −θ 2 cos θ + 2θ sin θ + 2 cos θ ) + ⎢θ ⎜θ − sin 2θ ⎟ − 1 ⎜ + cos 2θ ⎟ ⎥
⎝
⎠
2
2
4
⎝
⎠⎦
⎣
1⎛
1
⎞
+ ⎜ −3 cos θ + cos 3θ ⎟
⎠
4⎝
3
π
0
Integral Calculus
⎡⎛
= a3 ⎢ ⎜
⎣⎝
⎛3
= a3 ⎜
⎝2
1 3 1⎞ ⎛
1 3 1 ⎞⎤
+ −
− ⎜ 2 − − + ⎟⎥
4 4 12 ⎟⎠ ⎝
4 4 12 ⎠ ⎦
2
2
−2+
2
2
−
2
6.81
−
8⎞
− ⎟
3⎠
(iii) The volume of the solid is generated by revolving one arch about the base which
is the line joining the cusps. If P(x, y) is any point on the curve, its distance from
the base = 2a y = 2a a (1 cosq ) = a(1 + cosq).
For the region shown, x varies from ap to ap, hence q varies from p to p. Due to
symmetry about the y-axis, considering the region in the first quadrant where q varies
from 0 to p,
π
dx
dθ
V = 2 ∫ π ( 2a − y ) 2
Volume,
0
dθ
π
= 2π ∫ a 2 (1 + cos θ ) 2 a(1 + cos θ ) dθ
0
π
= 2π a3 ∫ (1 + cos θ )3 dθ
0
3
π ⎛
θ⎞
= 2π a3 ∫ ⎜ 2 cos 2 ⎟ dθ
0 ⎝
2⎠
π
= 16π a3 ∫ cos6
0
θ
dθ
2
q
= t,
2
d q = 2 dt
Putting
q = 0, t = 0
p
q = p, t =
2
When
p
V = 16p a3 ∫ 2 2 cos6 t dt
0
p
= 32p a3 ∫ 2 cos6 t dt
0
5 ⋅ 3 ⋅1 p
= 32p a3
⋅
6⋅4⋅2 2
= 5p 2 a3
Example 2: Find the volume of the solid generated by the revolution of the loop
1 3
t about the x-axis.
of the curve x = t2, y = t
3
Solution:
x = t2
dx
= 2t
dt
Engineering Mathematics
6.82
The volume of the solid is generated by revolving
the upper half of the loop of the curve about the
x-axis. For the region shown, x varies from 0 to
3, hence t varies from 0 to 3 .
Volume,
V =∫
3
y2
0
=2
∫
=2
0
3
dx
dt =
dt
∫
3
0
2
⎛ 1 3⎞
⎜⎝ t − t ⎟⎠ 2t dt
3
⎛ 2 2 4 1 6⎞
⎜⎝ t − t + t ⎟⎠ t dt = 2
3
9
t 4 2 t6 1 t8
− ⋅ + ⋅
4 3 6 9 8
3
0
∫
0
3
⎛ 3 2 5 1 7⎞
⎜⎝ t − t + t ⎟⎠ dt
3
9
Fig. 6.74
9⎞
⎛9
= 2 ⎜ −3+ ⎟
⎝4
8⎠
3
=
4
Example 3: Find the volume of the solid of revolution generated by revolving
3
3
the curve x = 2t + 3, y = 4t2 - 9 about the x-axis for t = - to t = .
2
2
Solution:
x = 2t + 3
dx
=2
dt
The volume of solid is generated by revolving the curve about the x-axis. For the required region, t varies from
3
Volume,
V = ∫ 23
−
=
=4
y2
2
∫
3
2
3
−
2
∫
3
3
to .
2
2
dx
dt
dt
( 4t 2 − 9) 2 ( 2) dt
3
2
0
(16t 4 − 72t 2 + 81) dt
[∵ (4t 2 – 9)2 is an even function]
3
2
t5
t3
= 4p 16 − 72 + 81t
5
3
0
= 1296p
Example 4: Prove that the volume of solid generated by revolving the cissoid
x = 2a sin2t, y = 2a
sin 3 t
about its asymptote is 2o 2a3.
cos t
Integral Calculus
Solution:
6.83
sin 3 t
cos t
⎛ 3 sin 2 t cos 2 t + sin 3 t sin t ⎞
dy
= 2a ⎜
⎟⎠
dt
cos 2 t
⎝
y = 2a
2
2
⎛ 3 cos 2 t + sin 2 t ⎞
2 ⎛ 3 cos t + 1 − cos t ⎞
= 2a sin 2 t ⎜
=
2
a
sin
t
⎜⎝
⎟⎠
⎟⎠
cos 2 t
cos 2 t
⎝
⎛ 2 cos 2 t + 1 ⎞
= 2a sin 2 t ⎜
⎝ cos 2 t ⎟⎠
The volume of solid is generated by revolving the region
about its asymptote, i.e., the line x = 2a.
If P(x, y) is any point on the curve, its distance from the
asymptote is 2a x = 2a 2a sin2 t = 2a cos2 t. For the
region shown, y varies from
to , hence t varies from
to . Due to symmetry about the x-axis, consider2
2
ing the region in the first quadrant where t varies from
0 to
2
Fig. 6.75
,
Volume,
V = 2 ∫ 2 ( 2a − x ) 2
0
=2
∫
2
0
dy
dt
dt
⎛ 2 cos 2 t + 1 ⎞
4a 2 cos 4 t ⋅ 2a sin 2 t ⎜
⎝ cos 2 t ⎟⎠
= 16 a3 ∫ 2 ( 2 cos 4 t sin 2 t + sin 2 t cos 2 t ) dt
0
1 1 ⎤
⎡ 3 ⋅1 ⋅1
⋅ + ⋅ ⋅ ⎥ = 16
= 16 a3 ⎢ 2 ⋅
⎣ 6⋅4⋅2 2 4 2 2 ⎦
2 3
=2 a
2
⎛6 ⎞
a3 ⎜ ⎟
⎝ 48 ⎠
Exercise 6.10
1. For the cycloid x = a (q
sinq ),
y = a (1 cosq ), find the volume of
the solid generated by the revolution
of an arch about (i) x-axis, (ii) y-axis,
and (iii) the tangent at the vertex.
[Ans. : 5p 2a 3, 6p 3a 3, p 2a 3]
2. Find the volume formed by revolving
one arch of the cycloid x = a(q + sinq ),
y = a(1 + cosq ) about x-axis.
[Ans. : 5p 2a3]
3. Find the volume of the solid formed by
revolving the tractrix
t
x = a cos t + a log tan , y = a sin t
2
about its asymptote.
⎡
2p a3 ⎤
⎢ Ans. :
⎥
3 ⎦
⎣
Engineering Mathematics
6.84
4. If the ellipse x = a cos q, y = b sin q is
revolved about the line x = 2a, show
that the volume of the solid generated
is 4p 2a2b.
5. The area of the curve x = a cos3 q,
y = a sin3 q lying between q = −
p
and
2
p
rotates about the x-axis. Find
2
the volume of solid so generated.
q =
16
⎡
3⎤
⎢ Ans. : 105 p a ⎥
⎣
⎦
Polar Form
Example 1: Find the volume of the solid generated by the revolution about the
initial line of the cardioid r = a (1 - cosp ).
Solution: The volume of the solid is
generated by revolving the upper half of
the cardioid about the initial line q = 0.
For the region above the initial line, q
varies from 0 to p.
p 2
Volume, V = ∫
p r 3 sin q dq
0
3
2p p 3
=
a (1 − cos q )3 sin q dq
3 ∫0
Putting
1 cosq = t,
sinq dq = dt
When
q = 0, t = 0
q = p, t = 2
V=
2p 3 2 3
2p 3 t 4
a ∫ t dt =
a
0
3
3
4
Fig. 6.76
2
0
8
= p a3 .
3
π
Example 2: Find the volume of revolution of a loop about the line θ = of the
2
curve r2 = a2 cos2p.
Solution: The volume of solid is generated by revolving a loop of the curve about
the line q =
p
. For the loop of the curve, q varies from
2
p
p
to . Due to symmetry
4
4
about the line q = 0, considering the loop above the initial line where q varies from 0
to
π
p
2
, Volume, V = 2 ∫ 4 π r 3 cos θ dθ
0 3
4
Integral Calculus
6.85
3
4π π4 3
2
2
θ
a
(cos
)
cos θ dθ
3 ∫0
π
3
4
= π a3 ∫ 4 (1 − 2 sin 2 θ ) 2 cos θ dθ
0
3
=
2 sinq = sin t,
Putting
2 cosq dq = cos t dt
When
θ =0
t=0
π
π
θ= , t=
4
2
V=
=
=
=
4 3 2
1
a ∫ cos3 t ⋅
cos t dt
0
3
2
4
Fig. 6.77
a3 ∫ 2 cos 4 t dt
0
3 2
4
3 2
2 3
a
4 2
a3 ⋅
3 ⋅1
⋅
4⋅2 2
.
Example 3: Find the volume of the solid
generated by revolving the curve r = a + b cosp ,
(a > b) about the initial line.
Solution: The volume of solid is generated by
revolving the upper half of the curve about the
initial line. For the region above the initial line, q
varies from 0 to p.
Volume,
V =∫
p
0
2p
3
2p
=
3
=
2 3
p r sin q dq
3
∫
p
0
Fig. 6.78
( a + b cos q )3 sin q dq
⎛ 1⎞ p
3
⎜⎝ − ⎟⎠ ∫0 ( a + b cos q ) ( −b sin q ) dq
b
p
⎡
−2p ( a + b cos q ) 4
[ f (q )]n +1 ⎤
n
∵
[
f
(
q
)]
f
(
q
)
d
q
=
′
⎢ ∫
⎥
3b
4
n +1 ⎦
⎣
0
p
= − [( a − b) 4 − ( a + b) 4 ]
6b
p
= − [{( a − b) 2 + ( a + b) 2 }{( a − b) 2 − ( a + b) 2 }]
6b
4
= p a( a 2 + b 2 )
3
=
Engineering Mathematics
6.86
Example 4: The arc of the cardioid r = a (1 + cosp ) included between θ = −
and θ =
π
2
π
π
is rotated about the line θ = . Find the volume of the solid of
2
2
revolution.
Solution: The volume of solid is generated
p
by revolving the curve about the line q = .
2
p
For the region shown, q varies from
to
2
p
. Due to symmetry about the initial line,
2
considering the region above the initial line
p
where q varies from 0 to ,
2
π
Volume,
V = 2∫ 2
0
Fig. 6.79
2 3
π a (1 + cos θ )3 cos θ dθ
3
4 3 π2
π a ∫ (cos 4 θ + 3 cos3 θ + 3 cos 2 θ + cos θ ) dθ
0
3
2
4
1 π
⎡ 3 ⋅1 π
⎤
⋅ + 3 ⋅ + 3 ⋅ ⋅ + 1⎥
= π a3 ⎢
3
3
2 2
⎣4⋅2 2
⎦
=
=
π a3
(5 π + 16).
4
Example 5: For the curve r2 = a2 cos 2p, prove that the volume of revolution of
2 3
a
a loop about the tangent at the pole is
.
8
Solution: The volume is generated by
revolving the loop about the tangent at the
p
pole, i.e., the line q = . If P (r, q ) is
4
any point on the curve, its distance from
the line q =
p
4
⎛p
⎞
is r sin ⎜ − q ⎟ , i.e.,
⎝4
⎠
1
r(cosq sin q ).
2
For the region shown, q varies from
to
4
.
4
Fig. 6.80
Integral Calculus
6.87
p
Volume, V = 4 2 p r 2 ⋅ 1 r (cosq − sin q ) dq
∫− p4 3
2
=
When
3 2
∫
p
4
p
−
4
3
(cos 2q ) 2 (cos q − sin q ) dq
=
p
3
3
⎡ p
⎤
2
p a3 ⎢ ∫ 4p (cos 2q ) 2 cos q dq − ∫ 4p (cos 2q ) 2 sin q dq ⎥
−
3
⎣ −4
⎦
4
=
p
3
3
⎡ p
⎤ 2 2
2
p a3 ⎢ 2 ∫ 4 (cos 2q ) 2 cos q dq − 0 ⎥ =
p a3 ∫ 4 (cos 2q ) 2 cos q dq
0
0
3
3
⎣
⎦
=
Putting
2p a3
p
3
2 2
p a3 ∫ 4 (1 − 2 sin 2 q ) 2 cos q dq
0
3
2 sinq = sin t,
2 cosq dq = cos t dt
q = 0,
t=0
p
p
q = , t=
4
2
V=
p
2 2
1
p a3 ∫ 2 cos3t
cos t dt
0
3
2
p
2
2
3 ⋅1 p
= p a3 ∫ 2 cos 4 t dt = p a3 ⋅
⋅
0
3
3
4⋅2 2
p2 3
a.
=
8
Example 6: A solid is formed by rotating the area between two loops of the
curve r = a (1 + 2cosp ) through four right angles. Find the volume generated.
Solution: The volume of solid is generated
by rotating the area between two loops of
the curve through four right angles. For the
2p
curve ACOBA, q varies from 0 to
. For
3
4p
.
the curve BEOB, q varies from p to
3
At the pole,
r=0
1 + 2 cos q = 0
1
cos q =
2
2
q =
3
Fig. 6.81
Engineering Mathematics
6.88
Volume,
⎛ Volume obtained by revolving ⎞
V =⎜
⎟⎠
⎝
the area ACOBA
⎛ Volume obtained by revolving ⎞
−⎜
⎟⎠
⎝
the area BEOB
2p
3
2
p a3 (1 + 2 cos q )3 sin q dq
3
4p
2
− ∫ 3 p a3 (1 + 2 cosq )3 sin q dq
p
3
3
p a 23p
=−
(1 + 2 cos q )3 ( −2 sin q )dq
3 ∫0
p a3 43p
(1 + 2 cos q )3 ( −2 sin q ) dq
+
3 ∫p
=∫
0
p a3 (1 + 2 cos q ) 4
=−
3
4
2p
3
0
p a3 (1 + 2 cos q ) 4
+
3
4
4p
3
p
p a ⎛ 81 ⎞ p a ⎛ 1 ⎞
⎜− ⎟ +
⎜− ⎟
3 ⎝ 4⎠
3 ⎝ 4⎠
20
=
p a3 .
3
3
3
=−
Example 7: Show that if the area lying
within the cardioid r = 2a (1 + cosp )
and outside the parabola r (1 + cosp )
= 2a revolves about the initial line, the
volume generated is 18o a3.
Solution: The points of intersection of
the cardioid r = 2a (1 + cosq ) and parabola
r (1 + cosq ) = 2a are obtained as,
2a(1 + cos q ) =
2a
1 + cos q
1 + 2 cosq + cos2q = 1
cosq = 0, cosq
p
q =±
2
p
q =
Hence, at P,
2
Fig. 6.82
2 (does not exist)
The volume is generated by revolving the region about the initial line. For the regions
p
OBAPO and OBPO, q varies from 0 to .
2
Integral Calculus
6.89
Volume, V = ⎛ Volume obtained by revolving ⎞
⎜⎝
⎟⎠
the area OBAPO
⎛ Volume obtained by revolving ⎞
−⎜
⎟⎠
⎝
the area OBPO
p
=∫2
0
2
p [2a(1 + cos q )]3 sin q dq
3
3
p
2 ⎛ 2a ⎞
−∫2 p ⎜
sin q dq
0 3
⎝ 1 + cos q ⎟⎠
p
16
p a3 ∫ 2 (1 + cos q )3 ( − sin q ) dq
0
3
p
16
+ p a3 ∫ 2 (1 + cos q ) −3 ( − sin q ) dq
0
3
=−
(1 + cos q ) 4
16
= − p a3
3
4
p
2
0
16
(1 + cos q ) −2
+ p a3
3
−2
p
2
0
⎡
[ f (q )]n +1 ⎤
n
∵
d
=
[
(
q
)]
(
q
)
q
f
f
′
⎢ ∫
⎥
n +1 ⎦
⎣
4
8
⎛3⎞
= − p a3 ( −15) − p a3 ⎜ ⎟
⎝4⎠
3
3
= 18p a3 .
Exercise 6.11
1. Find the volume of solid formed by
revolving the curve r = a(1 + cosq )
about the initial line.
8 3⎤
⎡
⎢ Ans. : 3 a ⎥
⎣
⎦
2. Find the volume of solid formed by
revolving the curve r2 = a2 cos 2q
about (i) the initial line, and (ii) the
tangent at the pole.
⎡
⎤
a3 ⎡
3 log 2 + 1 − 2 ⎤ ⎥
⎢ Ans. : (i)
⎣
⎦
6 2
⎢
⎥
⎢
a3 ⎡ 3
⎤⎥
(ii)
log 2 + 1 − 1⎥ ⎥
⎢
12 ⎢⎣ 2
⎦⎦
⎣
3. Prove that the volume generated by
revolving the loop of the curve
(
)
(
)
r = a cos 3q lying between θ = −
to
=
6
π
6
about the initial line.
⎡
19 a 3 ⎤
⎢ Ans. :
⎥
960 ⎦
⎣
4. Find the volume generated by
revolving the curve r = 2a cosq
about the initial line.
⎡
4 a3 ⎤
Ans.
:
⎢
⎥
3 ⎦
⎣
5. Show that the volume of the solid generated by the revolution of
the curve r = a + b secq about its
1 ⎞
⎛2
asymptote is 2 a 2 ⎜ a + b ⎟ .
3
2 ⎠
⎝
Engineering Mathematics
6.90
6.6 SURFACE OF SOLID OF REVOLUTION
Let y = f (x) be a curve included between two lines x = a and x = b. Let P(x, y) be
any point on the curve. When the chord PQ is revolved about the x-axis, a solid
of revolution is generated. The elementary
surface area d S is approximately equal to the
circumference of the circle multiplied by the
PQ.
d S = 2p yPQ = 2p y d s
The total surface area of the solid of revolution
about x-axis is given by,
Fig. 6.83
S = ∫ 2 y ds
Area of Surface of Solid of Revolution in Cartesian Form
Area of surface generated by revolving the arc of the curve y = f (x) about the x-axis
is given by,
b
b
ds
S = ∫ 2 y ds = ∫ 2 y dx
a
a
dx
2
b
⎛ dy ⎞
= ∫ 2 y 1 + ⎜ ⎟ dx
a
⎝ dx ⎠
Similarly, the area of the surface generated by revolving the arc of the curve x = f (y)
about y-axis is given by,
d
d
c
c
S = ∫ 2 x ds = ∫ 2 x
ds
dy
dy
2
=∫
⎛ dx ⎞
2 x 1 + ⎜ ⎟ dy
⎝ dy ⎠
d
c
Area of Surface of Solid of Revolution in Parametric Form
When the equation of the curve is given in parametric form x = f1(t), y = f2(t) with t1 t
t2, the area of surface of solid of revolution about the x-axis is given by,
2
2
t2
ds
⎛ dx ⎞ ⎛ dy ⎞
dt = ∫ 2 y ⎜ ⎟ + ⎜ ⎟ dt
t1
t
1
dt
⎝ dt ⎠ ⎝ dt ⎠
Similarly, the area of surface of solid of revolution about the y-axis is given by,
t2
S=∫ 2 y
t2
S=∫ 2 x
t1
2
2
t2
ds
⎛ dx ⎞ ⎛ dy ⎞
dt = ∫ 2 x ⎜ ⎟ + ⎜ ⎟ dt
t
1
dt
⎝ dt ⎠ ⎝ dt ⎠
Area of Surface of Solid of Revolution in Polar Form
For the curve r = f (q ), bounded between the radii vectors at q = q 1 and q = q 2, the
area of surface of the solid of revolution about the initial line q = 0 is given by,
Integral Calculus
6.91
2
θ2
ds
⎛ dr ⎞
dθ = ∫ 2π r sin θ r 2 + ⎜
⎟ dθ
θ
1
dθ
⎝ dθ ⎠
θ2
S = ∫ 2π y
θ1
=
Similarly, the area of surface of solid of revolution about the line
2
is given by,
2
θ2
ds
⎛ dr ⎞
dθ = ∫ 2π r cos θ r 2 + ⎜
⎟ dθ
θ
1
dθ
⎝ dθ ⎠
θ2
S = ∫ 2π x
θ1
Example 1: Find the area of the surface of revolution generated by revolving
the curve x = y3 from y = 0 to y = 2.
Solution:
x = y3
dx
= 3y2
dy
2
⎛ dx ⎞
ds
= 1 + ⎜ ⎟ = 1 + 9 y4
dy
⎝ dy ⎠
Fig. 6.84
The area of the surface is generated by revolving the region about the y-axis. For the
region shown, y varies from 0 to 2.
Surface area,
2
S=∫ 2 x
0
=
2
36
∫
2
0
2
ds
dy = ∫ 2 y 3 1 + 9 y 4 dy
0
dy
1
(1 + 9 y 4 ) 2 (36 y 3 )dy
2
=
=
3
4 2
(1 + 9 y )
3
18
2
27
(145
⎡
[ f ( y )]n +1 ⎤
n
⎢∵ ∫ [ f ( y )] f ′( y ) dy =
⎥
n +1 ⎦
⎣
0
)
145 − 1
Example 2: Find the area of the surface of revolution of the solid generated by
revolving the ellipse
Solution:
x2 y2
+
= 1 about the x-axis.
16 4
x2
+
16
2x 2 y
+
16 4
y2
=1
4
dy
=0
dx
x
dy
=−
dx
4y
Engineering Mathematics
6.92
ds
⎛ dy ⎞
= 1+ ⎜ ⎟
dx
⎝ dx ⎠
= 1+
2
x2
16 y 2
x 2 + 16 y 2
4y
The area of the surface of solid is generated by
revolving the upper half of the ellipse about the
x-axis. For the region above the x-axis, x varies
Fig. 6.85
from 4 to 4. Due to symmetry about the y-axis,
considering the region in the first quadrant where x varies from 0 to 4,
=
4
S = 2∫ 2 y
Surface area,
0
4
ds
dx
dx
x 2 + 16 y 2
dx =
4y
∫
=
3∫
=
1 64
x 64
x 3
3
− x 2 + ⋅ sin −1
2 3
2 3
8
0
y
∫
4
=4
0
x 2 + 64 − 4 x 2 dx
2
4
0
⎛ 8 ⎞
2
⎜
⎟ − x dx
3
⎝
⎠
4
0
⎡ 64
32
3⎤
− 16 + sin −1
3 ⎢2
⎥
3
3
2 ⎦
⎣
4 ⎞
⎛
= 8 ⎜1 +
⎟
⎝ 3 3⎠
=
Example 3: The part of the parabola y2 = 4ax cut off by the latus rectum
revolves about the tangent at the vertex. Find the surface area of the revolution.
Solution: The points of intersection of the
parabola y2 = 4ax and its latus rectum x = a are
obtained as,
y2 = 4a . a = 4a2
y = ± 2a and x = a
Hence, A: (a, 2a) and B: (a, 2a)
Now,
y2
4a
dx 2 y
y
=
=
dy 4 a 2 a
x=
Fig. 6.86
Integral Calculus
⎛ dx ⎞
ds
= 1+ ⎜ ⎟
dy
⎝ dy ⎠
= 1+
6.93
2
y2
4a 2
The surface area is generated by revolving the region about the tangent at the vertex
i.e., y-axis. For the region shown, y varies from 2a to 2a. Due to symmetry about
x-axis, considering the region in the first quadrant where y varies from 0 to 2a,
Surface area,
S = 2∫
= 2∫
Putting
When
2a
0
2a
0
2 x
ds
dy
dy
2 ⋅
y2
y2
1 + 2 dy
4a
4a
y = 2a tan q,
dy = 2a sec2q dq
y = 0, q = 0
y = 2a, q =
S = 4π ∫
4
4a tan 2 θ
1 + tan 2 θ 2a sec 2 θ dθ
4a
π
4
0
2
= 8π a 2 ∫
= 8π a 2
π
4
0
π
tan 2 θ sec3 θ dθ = 8π a 2 ∫ 4 (sec5 θ − sec3 θ ) dθ
0
1
3
3
tan θ sec3 θ + tan θ sec θ + log(sec θ + tan θ )
4
8
8
π
4
1
1
− tan θ sec θ − log(sec θ + tan θ )
2
2
0
[Using reduction formula]
3
3
1
1
⎡1
= 8 a2 ⎢ 2 2 +
2 + log( 2 + 1) −
2 − log
8
8
2
2
⎣4
= a 2 ⎡3 2 − log
⎣
(
)
(
⎤
2 +1 ⎥
⎦
)
2 +1 ⎤
⎦
Example 4: Find the surface area generated by revolving the loop of the curve
9ay2 = x(3a - x)2 about the x-axis.
Solution: The points of intersection of the curve 9ay2 = x(3a
obtained as,
Hence,
0 = x(3a x)2
x = 0, 3a, 3a and y = 0, 0, 0
A : (3a, 0)
x)2 and x-axis are
Engineering Mathematics
6.94
Now,
9ay 2 = x(3a − x) 2
dy
18ay
= (3a − x) 2 − 2 x(3a − x)
dx
dy (3a − x) 2 − 2 x(3a − x)
=
18ay
dx
(3a − x)(a − x)
=
6ay
2
ds
(3a − x) 2 (a − x) 2
⎛ dy ⎞
= 1+ ⎜ ⎟ = 1+
dx
36a 2 y 2
⎝ dx ⎠
=
36a 2 y 2 + (3a − x) 2 (a − x) 2
36a 2 y 2
Fig. 6.87
1
4ax(3a − x) 2 + (3a − x) 2 (a − x) 2
6ay
1
(3a − x) 2 (a + x) 2
=
6ay
(3a − x)(a + x)
=
6ay
=
The surface area is generated by revolving the loop about the x-axis. For the loop,
x varies from 0 to 3a.
3a
3a
ds
(3a − x)(a + x)
dx
Surface area, S = ∫ 2 y dx = 2 ∫ y ⋅
0
0
dx
6ay
=
3a
3a ∫0
=3 a
(3a 2 + 2ax − x 2 ) dx =
3a
3a 2 x + ax 2 −
x3
3
3a
0
2
Example 5: Find the area of the surface of revolution of a quadrant of a circular arc as obtained by revolving it about a tangent at one of its ends.
Solution: Let x2 + y2 = a2 be the equation of the circle and let AC be the tangent at A.
x2 + y 2 = a2
dy
2x + 2 y
=0
dx
dy
x
=−
dx
y
2
ds
x2
⎛ dy ⎞
= 1+ ⎜ ⎟ = 1+ 2
dx
y
⎝ dx ⎠
=
x2 + y 2 a
=
y
y2
Fig. 6.88
Integral Calculus
6.95
The surface area is generated by revolving the quadrant of circular arc APB about the
line AC. If P(x, y) is any point on the circle, the distance of P from the tangent at A =
a x. For the region shown, x varies from 0 to a.
a
ds
dx = 2
dx
a−x
dx
a2 − x2
Surface area, S = ∫ 2 (a − x)
0
= 2 a∫
a
0
∫
a
0
a
( a − x ) dx
y
a⎛
x
a
= 2 a ∫ ⎜⎜
−
0
2
2
2
a − x2
⎝ a −x
⎞
⎟⎟ dx
⎠
−1
⎤
a⎡
a
1
= 2 a∫ ⎢
+ (a 2 − x) 2 (−2 x) ⎥ dx
0
2
2
2
⎣ a −x
⎦
= 2 a a sin −1
x
+ a2 − x2
a
⎡
[ f ( x)]n +1 ⎤
n
⎢∵ ∫ [ f ( x)] f ′( x)dx =
⎥
n +1 ⎦
⎣
a
0
⎛
⎞
= 2 a⎜a − a⎟
⎝ 2
⎠
2
= a ( − 2)
Exercise 6.12
1. Find the surface area of the solid
generated by revolving the arc of
the parabola y2 = 4ax bounded by its
latus rectum about the x-axis.
⎡
⎤
8a 2
Ans.
:
2 2 −1 ⎥
⎢
3
⎣
⎦
2. Find the area of the curved surface
generated when one loop of the
curve x2(a2 x2) = 8a2y2 is revolved
about the x-axis.
(
)
⎡
a2 ⎤
Ans.
:
⎢
⎥
4 ⎦
⎣
3. Prove that the surface area of the solid
obtained by revolving the ellipse
b2x2 + a2y2 = a2b2 about the x-axis
⎡
⎤
⎛1⎞
is 2 ab ⎢ 1 − e 2 + ⎜ ⎟ sin −1 e ⎥ , e
e
⎝
⎠
⎣
⎦
being the eccentricity of the ellipse.
4. Show that the surface area of the
solid obtained by revolving the
arc of the curve y = sin x from
x = 0 to x = p about the x-axis is
(
)
⎡ 2 + log 2 + 1 ⎤ .
⎣
⎦
5. Show that the area of the surface
formed by rotating the curve y2 = x3
from x = 0 to x = 4 about the y-axis is
128
1 + 125 10 .
1215
6. Find the area of the curved surface
of the cup formed by the revolution
of the smaller part of the parabola
y2 = 4ax cut off by the line x = 3a
about its axis.
2
(
)
56 2 ⎤
⎡
⎢ Ans. : 3 a ⎥
⎦
⎣
Engineering Mathematics
6.96
7. The arc of the parabola y2 = 4ax
between its vertex and an extremity
of its latus rectum revolves about its
axis. Find the surface area traced out.
8
⎡
2⎤
⎢ Ans. : 3 (2 2 − 1) a ⎥
⎣
⎦
8. The arc of the curve a2y = x3 between
x = 0 and x = a is revolved about the
x-axis. Find the area of the surface so
generated.
⎡
⎤
a2
Ans.
:
10 10 − 1 ⎥
⎢
27
⎣
⎦
(
)
9. Find the surface area of the solid
formed by the revolution of the loop
of the curve 3ay2 = x (x a)2 about
the x-axis.
⎡
a2 ⎤
⎢ Ans. :
⎥
3 ⎦
⎣
10. Find the surface area of the solid generated by revolving the area bounded
by the circle x2 + y2 = a2 about the
line y = a.
[Ans. : 4p 2a2]
Parametric Form
Example 1: Prove that the surface generated by the revolution of the tractrix
1
t
x = a cos t + a log tan 2 , y = a sin t about its asymptote is equal to the surface
2
2
of the radius a.
1
t
x = a cos t + a log tan 2
2
2
dx
1
2 t 1
= − a sin t + a ⋅
⋅ sec ⋅
t
dt
2 2
tan
2
a
= − a sin t +
sin t
2
a cos t
=
sin t
y = a sin t
dy
= a cos t
dt
Solution:
2
ds
⎛ dx ⎞ ⎛ dy ⎞
= ⎜ ⎟ +⎜ ⎟
⎝ dt ⎠ ⎝ dt ⎠
dt
Fig. 6.89
2
a 2 cos 4 t
+ a 2 cos 2 t
sin 2 t
a cos t
=
sin t
=
The surface area is generated by revolving the tractrix about its asymptote, i.e., x-axis.
For the region shown, x varies from −∞ to , hence t varies from 0 to p . Due to
Integral Calculus
6.97
symmetry about the y-axis, considering the region in the second quadrant where t varp
ies from 0 to ,
2
p
ds
S = 2 ∫ 2 2p y dt
Surface area,
0
dt
p
a cos t
= 4p ∫ 2 a sin t ⋅
dt
0
sin t
p
= 4p a 2 ∫ 2 cos t dt
0
p
= 4p a 2 sin t 02
= 4p a 2
Example 2: Find the surface area of the solid generated by revolving the astroid
2
2
2
x 3 + y 3 = a 3 about the x-axis.
Solution: The parametric equations of
the astroid are
x = a cos3 q ,
y = a sin 3 q
dx
= −3a cos 2q sinq ,
dq
dy
= 3a sin 2q cosq
dq
2
ds
⎛ dx ⎞ ⎛ dy ⎞
= ⎜
+
⎝ dq ⎟⎠ ⎜⎝ dq ⎟⎠
dq
2
= 9a 2 cos 4 q sin 2 q + 9a 2 sin 4 q cos 2 q
Fig. 6.90
= 3a siin q cos q
The surface area is generated by revolving the upper half of the astroid about the
x axis. For the region shown, x varies from a to a, hence q varies from p to 0.
Due to symmetry about the y-axis, considering the region in the first quadrant, where
p
q varies from 0 to ,
2
p
ds
dq
Surface area,
S = 2 ∫ 2 2p y
0
dq
p
= 4p ∫ 2 a sin 3 q ⋅ 3a sin q cos q dq
0
p
= 12p a 2 ∫ 2 sin 4 q cos q dq
0
sin 5 q
= 12p a 2
5
12
= p a2
5
p
2
0
⎡
[ f (q ) n +1 ⎤
n
⎥
⎢∵ ∫ [ f (q ) f ′(q ) dq =
n +1 ⎦
⎣
Engineering Mathematics
6.98
Example 3: Find the surface area of the solid formed by revolving one arch of
the cycloid x = a(p − sin p ), y = a(1 − cos p )
about the y-axis.
Solution:
x = a(q − sin q )
dx
= a(1 − cos q )
dq
y = a(1 − cos q )
dy
= a sin q
dq
2
Fig. 6.91
2
ds
⎛ dx ⎞ ⎛ dy ⎞
= ⎜
+
= a 2 (1 − cos q ) 2 + a 2 sin 2 q
⎝ dq ⎟⎠ ⎜⎝ dq ⎟⎠
dq
= 2a 2 (1 − cosq
q)
q
2
The surface area is generated by revolving one arch of the curve about the y-axis. For
the region shown, q varies from 0 to 2p .
= 2a sin
Surface area, S = ∫
2p
0
2p x
ds
dq
dq
q
dq
2
2p ⎛
q
q
q⎞
= 4p a 2 ∫ ⎜q sin − 2 sin 2 cos ⎟ dq
0 ⎝
2
2
2⎠
= 2p ∫
2p
0
a (q − sin q )2a sin
⎡
= 4p a ⎢q
⎣
2
q⎞
q ⎞⎤ 4 3 q
⎛
⎛
⎜⎝ −2 cos ⎟⎠ − 1 ⎜⎝ −4 sin ⎟⎠ ⎥ − sin
2
2 ⎦ 3
2
2p
0
⎡
[ f (q )]n +1 ⎤
n
⎢∵ ∫ [ f (q )] f ′(q )dq =
⎥
n +1 ⎦
⎣
= 4p a2(4p )
= 16p 2 a2
Example 4: A circular arc of radius a revolves round its chord. Show that the
surface of the spindle generated is 4π a 2(sin α α cos α ), where 2` is the angle
subtended by the arc at the centre. Find the surface area if the circular arc is a
quadrant of circle.
Solution: Taking the centre of the circle as origin and radius as a, the equation of the
circle is x2 + y2 + a2. The parametric equations of the circle are,
x = a cos q ,
y = a sin q
dx
dy
= − a sin q ,
= a cos q
dq
dq
Integral Calculus
2
6.99
2
ds
⎛ dx ⎞ ⎛ dy ⎞
= ⎜
+
= a 2 sin 2 q + a 2 cos 2 q = a
⎝ dq ⎟⎠ ⎜⎝ dq ⎟⎠
dq
The arc ACB is revolved about the chord AB.
If P(x, y) is any point on the circle and M is
the foot of perpendicular from P on AB, then
PM = ON − OL
= x − a cosa
For the region shown, q varies from a to
a . Due to symmetry about the x-axis, considering the region in the first quadrant where q
varies from 0 to a,
a
Surface area, S = 2 ∫ 2p ( PM )
0
ds
dq
dq
Fig. 6.92
a
= 4p ∫ ( x − a cos a ) adq
0
a
= 4p a∫ ( a cos q − a cos a ) dq
0
= 4p a 2 sin q − q cos a
a
0
= 4p a (sin a − a cos a )
2
p
4
1 p 1 ⎞
⎛
S = 4p a 2 ⎜
−
⎝ 2 4 2 ⎟⎠
When circular arc is quadrant of a circle, a =
=
p a2
2
(4 − p )
Example 5: Show that the total surface area of the solid generated by the revo2
⎛
⎞
lution of an ellipse about its minor axis is 2 a 2 ⎜ 1 + 1 - e log 1 + e ⎟ , where a is
⎝
2e
1- e⎠
the semi-major axis and e is the eccentricity.
Solution: The parametric equations of the ellipse are,
x = a cos q ,
y = b sin q
dx
= − a sin q ,
dq
2
dy
= b cos q
dq
ds
⎛ dx ⎞ ⎛ dy ⎞
= ⎜
+
⎝ dq ⎟⎠ ⎜⎝ dq ⎟⎠
dq
2
= a 2 sin 2 q + b 2 cos 2 q
Fig. 6.93
Engineering Mathematics
6.100
The surface area of the solid is generated by the revolution of the ellipse about its minor
p
p
axis. For the region shown, y varies from b to b, hence q varies from
to .
2
2
Due to symmetry about x-axis, considering the region in the first quadrant where q
varies from 0 to
p
,
2
p
S = 2 ∫ 2 2p x
Surface area,
0
p
ds
dq
dq
= 4p ∫ 2 a cos q a 2 sin 2 q + b 2 cos 2 q dq
0
p
= 4 ap ∫ 2 cos q a 2 sin 2 q + b 2 (1 − sin 2 q ) dq
0
p
= 4 ap ∫ 2 cos q b 2 + a 2 e 2 sin 2 q dq
where e =
0
Putting
When
sinq = t
cosq dq = d t
q = 0,
q =
S = 4 ap ∫
1
0
1−
b2
a2
t=0
p
, t =1
2
b 2 + a 2 e 2 t 2 dt = 4 ap ⋅ ae ∫
1
0
2
⎛b⎞
t 2 + ⎜ ⎟ dt
⎝ ae ⎠
1
⎛
t 2
b2
b2
b2 ⎞
= 4p a e
t + 2 2 + 2 2 log ⎜ t + t 2 + 2 2 ⎟
2
ae
2a e
ae ⎠
⎝
0
2
⎡ 1
b2
1
b2
b⎤
⎛
⎞
= 4p a 2e ⎢
a 2 e 2 + b 2 + 2 2 log ⎜1 +
a 2 e 2 + b 2 ⎟ − 2 2 log ⎥
⎝ ae
⎠ 2a e
ae ⎦
2a e
⎣ 2ae
⎡ 1
b2
a⎞
b2
b⎤
⎛
= 4p a 2e ⎢
⋅ a + 2 2 log ⎜1 + ⎟ − 2 2 log ⎥
⎝
⎠
ae
ae
ae
2
a
e
a
e
2
2
⎣
⎦
⎡∵ b = a 1 − e 2 ⎤
⎣
⎦
⎡
b2
a(1 + e) ⎤
= 2p ⎢ a 2 + log
⎥
e
b ⎦
⎣
⎛
1+ e ⎞
b2
= 2p ⎜ a 2 + log
⎟
⎝
2e
1− e ⎠
⎛ 1 − e2
1+ e ⎞
= 2p a 2 ⎜1 +
log
⎟
⎝
2e
1− e ⎠
⎡∵ b = a 1 − e 2 ⎤
⎣
⎦
Integral Calculus
6.101
Exercise 6.13
1. Find the surface area of the reel
formed by the revolution of the
cycloid x = a(q + sinq ), y = a(1 – cosq )
about (i) the tangent at the vertex,
(ii) y-axis, and (iii) base.
32 2
⎡ Ans. :
⎤
pa
(i)
⎢
⎥
3
⎢
⎥
4 ⎞⎥
⎢
2⎛
p
a
p
−
(ii)
4
2
⎜⎝
⎟
⎢
3 ⎠⎥
⎢
⎥
64 2
⎢
⎥
pa
(iii)
⎥⎦
⎣⎢
3
2. Find the surface area of the solid
generated by the revolution of the
t3
loop of the curve x = t 2 , y t
3
about x-axis.
[Ans. : 3p ]
3. Show that the area of the surface of
the solid generated by revolving the
curve x = a(u − tanh u ), y = a sechu,
about the x-axis is equal to the area
of the surface of a sphere of radius a.
4. Find the area of the surface of
revolution generated by revolving
the cardioid x = 2cosq – cos 2q,
y = 2sinq – sin 2q, about the x-axis.
⎡ Ans. : 128p ⎤
⎢
5 ⎥⎦
⎣
5. Find the area of the surface generated
by revolving the curve x = 3t(t – 2),
3
y = 8t 2 with 0
y-axis.
t
1 about the
[Ans. : 39p ]
6. Find the area of the surface generated
by revolving the curve x = a cos 2 t ,
y = a sin 2 t about x-axis.
⎡ Ans. : 12p 2 ⎤
a ⎥
⎢
5
⎣
⎦
7. Show that the ratio of the areas of
the surface formed by revolving the
arch of the cycloid x = a(q + sin q ),
y = a(1 + cos q ) between two consecutive cusps about the x-axis to
the area enclosed by the cycloid and
64
.
x-axis is
9
Polar Form
q
Example 1: The curve r = e 2 is revolved about the initial line. Prove that the
area of surface of revolution traced out by the part between the points
θ = π is equal to
2
5 ( e + 1).
q
Solution:
r = e2
d r 1 q2
= e
dq 2
2
ds
1
⎛ dr ⎞
= r2 + ⎜
= eq + eq
⎝ dq ⎟⎠
dq
4
=
5 q2
e .
2
= 0 and
Engineering Mathematics
6.102
The surface area is generated by revolving the curve about the initial line. For the
region shown,q varies from 0 to p .
p
ds
dq
Surface area,
S = ∫ 2p y
0
dq
q
p
p
ds
5 q2
dq = 2p ∫ e 2 sin q
= ∫ 2p r sin q
e dq
0
0
dq
2
p
p
= p 5 ∫ eq sin q dq = p 5
0
=
eq
(sin q − cos q )
2
0
p
5 (ep + 1).
2
Example 2: Find the area of the surface of
the solid generated by revolving upper half of
the cardioid r = a(1 − cos ) about the initial
line.
Solution: r = a(1 − cos q )
dr
= a sin q
dq
2
ds
⎛ dr ⎞
= r2 + ⎜
⎝ dq ⎟⎠
dq
Fig. 6.94
= a 2 (1 − cos q ) 2 + a 2 sin 2 q
2
q⎞
q
q⎞
⎛
⎛
= a 2 ⎜ 2 sin 2 ⎟ + a 2 ⎜ 2 sin cos ⎟
⎝
⎠
⎝
2
2
2⎠
= 4 a 2 sin 2
= 2a sin
2
q
2
q
2
The area of the surface of the solid is generated by revolving the upper half of the
cardioid about the initial line q = 0. For the region shown, q varies from 0 to p .
Surface area,
ds
dq
dq
p
ds
= ∫ 2p r sin q
dq
0
dq
p
S = ∫ 2p y
0
q
dq
2
p ⎛
q ⎞⎛
q
q⎞ q
= 4p a 2 ∫ ⎜ 2 sin 2 ⎟ ⎜ 2 sin cos ⎟ sin dq
0 ⎝
2
2⎠
2
2⎠⎝
p
= 4p a 2 ∫ (1 − cos q ) sin q sin
0
Integral Calculus
6.103
q
q
cos dq
2
2
p
q
2
4 q 1
= 32p a ∫ sin ⋅ cos dq
0
2 2
2
p
= 16p a 2 ∫ sin 4
0
= 32p a 2
sin 5
q
2
p
⎡
[ f (q ) n +1 ] ⎤
n
]⎥
⎢∵ ∫ [ f (q ) f ′(q ) dq =
n +1 ⎦
⎣
5
0
32
= p a2
5
Example 3: The arc of cardioid r = a(1 + cos )
included between θ = −
π
to
is rotated about
2
2
π
. Show that the area of the surface
2
48 2
generated is
a 2.
5
Solution:
r = a(1 + cos q )
the line θ =
Fig. 6.95
dr
= − a sin q
dq
2
ds
⎛ dr ⎞
= r2 + ⎜
= a 2 (1 + cos q ) 2 + a 2 sin 2 q
⎝ dq ⎟⎠
dq
2
q⎞
q
q⎞
⎛
⎛
= a 2 ⎜ 2 cos 2 ⎟ + a 2 ⎜ 2 sin cos ⎟
⎝
⎝
2⎠
2
2⎠
= 4 a 2 cos 2
2
q
2
q
2
p
The area of the surface is generated by revolving the cardioid about the line q = .
2
p
p
.
For the region shown, q varies from
to
Due to symmetry about the initial line
2
2
p
considering the region in the first quadrant where q varies from 0 to ,
2
p
d
s
Surface area,
dq
S = 2 ∫ 2 2p y
0
dq
p
ds
= 2 ∫ 2 2p r cos q
dq
0
dq
p
q
= 8p a 2 ∫ 2 (1 + cos q ) cos q cos dq
0
2
= 2a cos
Engineering Mathematics
6.104
p
q ⎞⎛
q⎞
q
⎛
= 8p a 2 ∫ 2 ⎜ 2 cos 2 ⎟ ⎜1 − 2 sin 2 ⎟ cos dq
0 ⎝
2⎠⎝
2⎠
2
p
q ⎞⎛
q⎞
q
⎛
= 8p a 2 ∫ 2 2 ⎜1 − sin 2 ⎟ ⎜1 − 2 sin 2 ⎟ cos dq
0
⎝
2⎠⎝
2⎠
2
Putting
sin
q
= t,
2
1
q
cos dq = dt
2
2
q = 0,
When
t=0
p
1
q = , t=
2
2
1
1
S = 32 p a 2 ∫ 2 (1 − t 2 )(1 − 2t 2 )dt = 32 p a 2 ∫ 2 (1 − 3t 2 + 2t 4 ) dt
0
2
= 32 p a 2 t − t 3 + t 5
5
=
0
1
2
0
1
2 1 ⎤
⎡ 1
= 32 p a 2 ⎢
−
+ ⋅
⎥
5
2
2
2
4 2⎦
⎣
48 2 2
pa
5
Example 4: Find the surface area of the solid formed by the revolution of the
loop about the tangent at the pole of the curve r 2 = a 2 cos 2 .
r 2 = a 2 cos 2q
dr
2r
= −2a 2 sin 2q
dq
a 2 sin 2q
dr
sin 2q
=−
= −a
dq
a cos 2q
cos 2q
Solution:
ds
⎛ dr ⎞
= r2 + ⎜
⎝ dq ⎟⎠
dq
2
= a 2 cos 2q + a 2
=
=
sin 2 2q
cos 2q
a 2 cos 2 2q + a 2 sin 2 2q
cos 2q
a
cos 2q
The surface area is formed by the revolution of the loop about the tangent at the pole
i.e., q =
p
p
. If P ( r , q ) is any point on the curve, its distance from the line q =
is
4
4
Integral Calculus
6.105
Fig. 6.96
1
p
⎛p
⎞
to
r(cosq sin q ). For the region shown, q varies from
r sin ⎜ − q ⎟ , i.e.,
4
⎝4
⎠
2
p
. Due to symmetry about the initial line, considering the region in the first quadrant
4
p
where q varies from 0 to ,
4
p
Surface area,
S = 2 ∫ 4 2p
1
0
=
4p
∫
2
p
4
0
2
r (cosq − sin q )
ds
dq
dq
a cos 2q (cos q − sin q )
a
cos 2q
dq
p
= 2 2p a 2 ∫ 4 (cosq − sin q ) dq = 2 2p a 2 sin q + cos q
0
p
4
0
= 2 2 p a ( 2 − 1)
2
Exercise 6.14
1. Find the area of the surface of the
solid generated by revolving the
curve r 2 = a 2 cos 2q about the initial line.
⎡
1 ⎞⎤
2⎛
⎢ Ans. : 4pa ⎜1 −
⎟⎥
⎝
2 ⎠⎦
⎣
2. Find the area of the surface of the
solid generated by revolving the
curve r = 2a cosq about the initial
line.
⎡⎣ Ans. : 4p a 2 ⎤⎦
3. Find the area of the surface of the
solid generated by revolving the
curve r = 4 cos q about the initial line.
[Ans. : 16p ]
Engineering Mathematics
6.106
FORMULAE
m−2
n+2
+ m − 1 ∫ sin x cos x dx
n +1
Reduction Formulae
sin x dx = – 1 cos x sinn –1 x
n
+ n 1 ∫ sin n − 2 x dx
n
n
1
cos x dx = sin x cosn –1 x
n
n
(i)
(ii)
(iii)
+ n 1 ∫ cos
n
tan n 1 x
n
tan x dx =
n 1
n−2
∫ tan
n−2
–
+ n − 1 ∫ sin m + 2 x cos n − 2 x dx
m +1
x dx
–
sec n x dx =
(v)
= −
x dx
+ m+n+2
m +1
∫ cot
n−2
(b)
+ n −1
m+n
(c)
(viii)
∫ sin
m
p
2
0
x cos n x dx
sin n x dx
= n 1 n
n
n
if n is odd,
= n 1 n
n
n
if n is even
3
2
n 5 … 2,
n 4
3
3 n 5 … 1 p ,
2 n 4
2 2
p
2
sin m +1 x cos n −1 x
m+n
cos n x dx = n 1 n 3
n
n 2
n 5 … 2 , if n is odd
n 4
3
n
1
n
3 n 5 … 1 p,
=
n 2 n 4
2 2
n
x cos n − 2 x dx
if n is even
(ix)
sin m x cos n x dx
=
m+2
=
m −1
x sin x
m+n
m −1
+
sin m − 2 x cos n x dx
m+n ∫
cos
∫ sin
sin m +1 x cos n +1 x
m +1
m
+
n
+
2
+
sin m x cos n + 2 x dx
m +1 ∫
x dx
tan x sec n 2 x
n 1
n +1
cos n +1 x sin m +1 x
m +1
sin m x cos n x dx
(f)
+ n 2 ∫ sec n − 2 x dx
n 1
cot x cosec n 2 x
n
(vi) cosec x dx =
n 1
+ n 2 ∫ cosec n − 2 x dx
n 1
(vii) (a) sin m x cos n x dx
= −
sin m x cos n x dx
(e)
cot x
n 1
cot n x dx =
sin m +1 x cos n −1 x
m +1
=
n 1
(iv)
sin m x cos n x dx
(d)
sin m x cos n x dx
cos n +1 x sin m −1 x
= −
n +1
(x)
0
p
2
0
sin m x cos n x dx = m − 1
m+n
m−3
2
1 ,
…
m+n−2
3 + n n +1
Integral Calculus
if m is odd and n may be odd or even
m−3
m−5 …
= m −1
m+n m+n−2 m+n−4
1
n 1 n 3 … 2,
2+n
n
n 2
3
if m is even and n is odd
m−5 …
= m −1 m − 3
m+n m+n−2 m+n−4
1
n 1 n 3 … 1 p ,
2+n
n
n 2
2 2
if m is even and n is even
Length of Arc
(i) Cartesian form
(a) s = ∫
(b) s = ∫
2
⎛ dy ⎞
1 + ⎜ ⎟ dx
⎝ dx ⎠
b
a
2
c
(ii) Parametric form
s=∫
2
2
⎛ dx ⎞ + ⎛ dy ⎞ dt
⎝ dt ⎠ ⎜⎝ dt ⎟⎠
t2
t1
(iii) Polar form
(a) s = ∫
q2
q1
(b) s = ∫
r2
r1
2
dr ⎞
dq
r2 + ⎛
⎝ dq ⎠
2
dq ⎞
dr
1+ r2 ⎛
⎝ dr ⎠
Areas of Plane Curves
(i) Cartesian form
b
(a) A =
(b) A =
a
d
c
y dx
x dy
(ii) Parametric form
t2
dx
(a) A =
y dt
t1
dt
t2
dy
(b) A =
x dt
t1
dt
(iii) Polar form
1 q2 2
A=
r dq
2 q1
Volume of Solid of Revolution
(i) Cartesian form
b
p y 2 dx (revolution
about x-axis)
(a) V =
a
d
(b) V =
c
p x 2 dy
(revolution about y-axis)
(ii) Parametric form
t2
dx
dt (revolution
p y2
(a) V =
t1
dt
about x-axis)
t2
dy
(b) V =
p x2
dt (revolution
t1
dt
about y-axis)
(iii) Polar form
⎛ dx ⎞
1 + ⎜ ⎟ dy
⎝ dy ⎠
d
6.107
q2
2 3
p r sin q dq
3
(revolution about q = 0)
q2 2
(b) V =
p r 3 cos q dq
q1 3
p
revolution about q =
2
(a) V =
q1
(
)
Area of Surface of Solid of Revolution
(i) Cartesian form
(a) S =
2
⎛ dy ⎞
∫a 2p y 1 + ⎜⎝ dx ⎟⎠ dx
b
(revolution about x-axis)
(b) S =
∫
d
c
2
⎛ dx ⎞
2p x 1 + ⎜ ⎟ dy
⎝ dy ⎠
(revolution about y-axis)
(ii) Parametric form
(a) S =
∫
t2
t1
2
2
⎛ dy ⎞
dx
2p y ⎛ ⎞ + ⎜ ⎟ dt
⎝ dt ⎠ ⎝ dt ⎠
(revolution about x-axis)
(b) S =
∫
t2
t1
2
2
⎛ dy ⎞
dx
2p x ⎛ ⎞ + ⎜ ⎟ dt
⎝ dt ⎠ ⎝ dt ⎠
(revolution about y-axis)
Engineering Mathematics
6.108
(iii) Polar form
2
(b) S =
∫
q2
q1
q2
(a) S = ∫ 2p r sin q r 2 + ⎛ dr ⎞ dq
⎝ dq ⎠
q1
2
dr ⎞
dq
2p r cos q r 2 + ⎛
⎝ dq ⎠
(revolution about q = p2 )
(revolution about q = 0)
MULTIPLE CHOICE QUESTIONS
Choose the correct alternative in each of the following:
p
2
5. The area under the curve y =
7
sin q dq is given by
1. The integral
0
12
16
(a)
(b)
35
35
16p
8
(c)
(d)
35
35
2. The area enclosed between the parabola y = x2 and the straight line y = x is
1
1
(a)
(b)
8
6
1
1
(c)
(d)
3
2
3. The sum of areas of all the loops of
the curve r = 2 sin 3q is
p
3
2
(a) 3 sin 3q dq
0
p
3
(b) 6 sin 2 3q dq
0
p
3
(c) 9 sin 2 3q dq
0
p
3
(d) 12 sin 2 3q dq
0
4. The area bounded by the x-axis,
8
y = 1 + 2 , ordinate at x = 2 and x = 4 is
x
(a) 2
(b) 4
(c) 8
(d) 1
for 0
1
(a)
4
(c) 1
x
x
1 x2
1 is
1
(b)
2
(d) 16
6. The area between the curves y = 1
x
and y = 1 to the right of the line
x +1
x = 1 is
(a) log 3
(b) log 2
(c) 2 log 3
(d) 2 log 2
7. The area in the first quadrant under
is
the curve y = 2 1
( x + 6 x + 10)
(a) p
(b) p – 2 tan–1 3
2
4
p
p
–1
(c)
– tan 3 (d)
+ tan–1 3
2
2
8. The area under y = 2 1 2 for
(x a )
x a + 1 is
1
(a) 1 log (a + 1) (b)
log (a + 1)
2a
a
1
(c)
log (a + 1) (d) none of these
2
9. The area of the region bounded by
y+8
the curve
= x – 2 and the
x
x- axis is
(a) 54
(b) 36
(c) 18
(d) 12
Integral Calculus
10. The area common to the curves
y2 = x and x2 = y is equal to
2
(a) 1
(b)
3
1
(c) 0
(d)
3
11. The length of the arc of the curve
y = log sec x from x = 0 to x = p is
3
(a) log (2 + 3 ) (b) log ( 2 + 3)
(c) log ( 2 + 1) (d) log ( 3 + 1)
( x + 2)
3
2
12. The length of the curve y =
2
3
from x = 0 to x = 3 is
(a) 10
(b) 12
(c) 3p
(d) 6p
13. The whole length of the curve
r = 2a sin q is equal to
(a) p a
(b) 2p a
(c) 3p a
(d) 4p a
14. Let An =
p
4
0
tan n x dx . Then the value
of A10 + A8 is
1
1
(a)
(b)
8
9
1
1
(c)
(d)
10
9
15. The figure bounded by graphs of
y2 = 4x, y = 0 and x = 1 is rotated
round the line x = 1. The volume of
the resulting solid is
(a) 16p
(b) 15p
15
16
16
p
5
(c)
(d) p
5
16
16. The area of the region in the first
quadrant bounded by the y-axis and
the curves y = sin x and y = cos x is
(a)
2
(b)
2 +1
(c)
2 –1
(d) 2 2 – 1
6.109
17. The length of the arc of the curve
6xy = x4 + 3 from x = 1 to x = 2 is
13
17
(a)
(b)
12
12
19
(c)
(d) none of these
12
18. The arc of the sine curve y = sin x
from x = 0 to x = p revolved about the
x-axis. The area of the surface of the
solid generated is
(a) 2p [ 2 + log ( 2 + 1)]
2p 2
[ 2 + log ( 2 + 1)]
3
(c) p [ 2 + log ( 2 + 1)]
3
p2
(d)
[ 2 + log ( 2 + 1)]
3
19. The volume of the solid generated
by revolving the curve x = a cos t,
y = b sin t about the x-axis is
4p ab 2
(a) 4p ab
(b)
3
4p ab 2
(c) 4p ab
(d)
3
20. The volume of solid obtained by revolving the area under y = e–2x about
the x-axis is
p
p
(b)
(a)
2
4
(c) 2p
(d) p
(b)
21. The volume of the solid obtained by
revolution of the loop of the curve
a+x
y2 = x2
about the x-axis is
a−x
(a) 2p a3
2
(b) 2p a3 log 2 −
3
(
(
(c) p a3 log 2 +
2
3
)
)
(d) p a3
22. The volume of solid generated,
when the area of the ellipse
Engineering Mathematics
6.110
(a) 40p
5
36
p
(c)
5
y2
x2
+
= 1 (in the first quadrant)
9
4
is revolved about the y-axis is
(a) 16p
(b) 12p
(c) 8p
(d) 6p
23. The value of the integral
p
2
1
3
0
8
(b) 4
45
45
8
p
(c)
(d) 4p
45
45
24. The area of the surface of the solid
generated by revolving the curve
r = 2a cos q about the initial line is
(a) 2p a2
(b) 4p a2
2
(c) p a
(d) 8p a2
25. The area of the surface of the solid
generated by revolving the curve
x = t3 – 3t, y = 3t2, 0 t 1 about the
x-axis is
(a)
2.
9.
16.
23.
(b)
(b)
(c)
(a)
3.
10.
17.
24.
(b)
(d)
(b)
(b)
x 5 dx is given by
1− x 2
0
4
p
(a)
(b) 4
15
15
8
p
(c)
(d) 8
15
15
27. The value of the integral
26. The integral
sin 2 x cos3 x dx is
Answers
1. (b)
8. (d)
15. (a)
22. (b)
(b) 24p
5
48
p
(d)
5
4.
11.
18.
25.
∞
x2
∫ (1 + x )
2 4
∫
dx is
0
(a) p
(b) p
32
16
2p
(c)
(d) 1
15
32
28. The area of the surface of the solid
generated by the revolution of the
line segment y = 2x from x = 0 to
x = 2 about the x-axis is equal to
(b)
(a)
(a)
(d)
(a) p 5
(b) 2p 5
(c) 4p 5
(d) 8p 5
5.
12.
19.
26.
(c)
(b)
(d)
(d)
6.
13.
20.
27.
(b)
(b)
(b)
(a)
7.
14.
21.
28.
(c)
(b)
(b)
(d)
Gamma and Beta
Functions
Chapter
7
7.1 INTRODUCTION
There are some special functions which have importance in mathematical analysis,
functional analysis, physics or other applications. In this chapter, we will study two
special functions, gamma and beta functions. The beta function is also called the Euler
integral of the first kind. The gamma function is an extension of the factorial function
to real and complex numbers and is also known as Euler integral of the second kind.
Gamma function is a component in various probability distribution functions. It also
appears in various areas such as asymptotic series, definite integration, number theory,
etc.
7.2 GAMMA FUNCTION
Gamma function is defined by the improper integral
by n .
∫
∞
n = ∫ e − x x n −1dx, n > 0
Hence,
0
Alternate form of gamma function
∞
n = 2 ∫ e − x x 2 n −1dx
2
0
Proof: By definition,
∞
n = ∫ e − x x n −1dx
0
Let x = t2,
∞
0
dx = 2t dt
∞
n = ∫ e − t . t 2 n − 2 . 2t dt
2
0
∞
= 2 ∫ e − t ⋅ t 2 n −1 dt
2
0
Changing the variable t to x,
∞
n = 2 ∫ e − x . x 2 n −1dx
0
2
e − x x n −1dx, n > 0 and is denoted
7.2
Engineering Mathematics
7.3 PROPERTIES OF GAMMA FUNCTION
(1) n + 1 = n n
∞
n + 1 = ∫ e − x x n dx
Proof:
0
Integrating by parts,
n + 1 = −e − x x n
∞
0
∞
− ∫ (−e − x ) nx n −1dx
0
∞
= n ∫ e − x x n−1dx
0
=n n
n +1 = n n
Hence,
This is known as recurrence or reduction formula for Gamma function.
Note:
(i)
n + 1 = n!
if n is a positive integer
(ii)
n +1 = n n
if n is a positive real number
(iii)
n=
(iv)
n 1− n =
n +1
n
if n is a negative fraction
sin n
(2) 1 =
2
Proof: By alternate form of Gamma function,
⎛ 1⎞
2.⎜ ⎟ − 1
∞
2
1
= 2∫ e − x x ⎝ 2 ⎠ dx = 2
0
2
∫
∞
0
2
e − x dx
∞
∞
2
2
1. 1
= 2 ∫ e − x dx . 2 ∫ e − y dy
0
0
2 2
= 4∫
∞
0
∫
∞
0
e−( x
2
+ y2 )
Changing to polar coordinates, x = r cosq,
Limits of x
Limits of y
x=0
y=0
dx dy
y = r sinq
dx dy = r dr dq
to
to
x
y
This shows that the region of integration is the first quadrant.
Gamma and Beta Functions
7.3
Draw an elementary radius vector in the region
which starts from the pole and extends up to .
Limits of r
r=0
to
r
Limits of q
q=0
to
θ=
π
2
π
2
1 1
2 ∞
⋅ = 4 ∫ ∫ e − r . r d r dθ
0
2 2
0
π
2
∞
⎛ 1⎞ 2
= 4 ∫ dθ . ∫ ⎜ − ⎟ e− r (− 2r ) d r
⎝ 2⎠
0
0
=
4
θ
−2
= −2 .
π
2
0
2
e− r
2
∞
⎡∵ e f ( r ) . f ′ (r )dr = e f ( r ) ⎤
⎣ ∫
⎦
0
( 0 − 1)
=
1
=
2
Example 1: Find the value of - 5 .
2
n +1
Solution:
n=
n
5
− =
2
−
5
+1
2 3
2
=− −
5
5 2
−
2
2
=− .
5
4 .
=
15
=−
Fig. 7.1
−
−
8
15
3
+1
4 1
2
=
−
3
15
2
−
2
1
+1
8 1
2
=−
1
15 2
−
2
7.4
Engineering Mathematics
Example 2: Given
n=
Solution:
12
−
=
5
8
12
= 0.8935 , find the value of - .
5
5
n +1
n
2
12
7
− +1
+1
− +1
25
5
5
5
.
. 5
=−
=
7
2
12
12
84
−
−
−
5
5
5
−
3
+1
625 8
125 . 5
625
=−
=−
=−
(0.8935) = 1.108
3
504
5
504
168
5
Ç
Example 3: Evaluate
3
e - x dx .
0
1
1 −2
Solution: Let x3 = t, x = t 3 , dx = t 3 dt
3
When
x = 0,
t=0
x
∫
∞
0
,
t
1
∞
−1
3
1 −2
1 1
1 ∞
e − x dx = ∫ e − t . t 3 dt = ∫ e − t t 3 dt =
0
3
3 3
3 0
1
Ç
x
e-
Example 4: Evaluate
x 4 dx .
0
Solution: Let
x = t , x = t2, dx = 2t dt
When
x = 0,
t=0
x
∫
∞
0
e
− x
,
1
4
t
∞
1
x dx = ∫ e ( t ) 4 2t dt
−t
2
0
3
∞
5
3 1 1 3
=2. .
=
p
2
2 2 2 2
= 2 ∫ e − t t 2 dt = 2
0
Example 5: Evaluate
∫
∞
2
( x 2 + 4)e -2 x dx .
0
1
−
1
t 2
t 2
1 1 − 12
Solution: Let 2x = t, x = ⎛⎜ ⎞⎟ , dx =
⋅ t dt =
dt
⎝2⎠
2 2
2 2
2
Gamma and Beta Functions
x = 0,
When
t=0
x
∫
∞
0
( x + 4) e
2
,
=
=
∫
Example 6: Evaluate
Solution: Let
Ç
−
0
=
xne−
ax
1
4
∫
2
∞
0
1
e − t t 2 dt +
∞
xn e−
ax
0
1
2
+
8 2
,
e−t t
−
1
2
dt
0
=
2
17
8 2
∫
Example 7: Evaluate
∞
t
n
⎛t ⎞
2t
dx = ∫ ⎜ ⎟ e − t ⋅ dt
0 ⎝ a⎠
a
2
∞
=
2
a
∫
n +1
2
a n +1
2
x e − x dx .
0
∫
∞
e − t t 2 n +1dt
0
∞
2n + 2
e− x
2
x
0
dx .
1
Solution: Let
x2 = t, x = t 2 , dx =
When
x = 0,
x e − x dx . ∫
∞
0
e
− x2
1 − 12
t dt
2
t=0
,
x
0
∞
dx .
=
∫
∫
2
1 3
2 1
1 .1 1
2 1
+
=
+
4 2 2
2 2 4 2 2 2
2 2
x
2
2
t2
2t
, dx = dt
a
a
x = 0,
t=0
When
∞
1
0
ax = t , x =
∫
t
⎛t
⎞ −t t 2
dt
⎜⎝ + 4⎟⎠ e .
2
2 2
∞
dx = ∫
−2 x 2
7.5
t
∞ e−t
1 −1
1 −1
dx = ∫ t e . t 2 dt . ∫ 1 . t 2 dt
0
0
2
2
x
t4
1
∞
4 −t
=
1 ∞ − t − 14 . ∞ − t − 34
e t dt ∫ e t dt
0
4 ∫0
7.6
Engineering Mathematics
Ç
Example 8: Evaluate
=
1 3. 1 1
1 1
= 1−
4 4 4 4
4 4
=
1
⋅
4 sin
e- x
e− x
x
0
Ç
6
x 4 e - x dx .
0
x
∞
2 2
4
dx .
x = 0,
When
∫
2 =
1
1 −2
x3 = t, x = t 3 , dx = t 3 dt
3
Solution: Let
3
1.
4
3
x
0
=
∞
t=0
,
dx . ∫ x 4 e − x dx = ∫
6
0
∞
0
t
e − t . 1 − 32 . ∞ 34 − t 2 . 1 − 32
t dt ∫ t e
t dt
1
0
3
3
6
t
=
1 ∞ − t − 56 . ∞ − t 2 32
e t dt ∫ e t d t
0
9 ∫0
=
∞
2 2 ⎜ ⎟ −1
1 1 1
⋅ ⋅ 2 ∫ e − t t ⎝ 6 ⎠ dt
0
9 6 2
=
1 1 1 5
⋅
9 6 2 6
⎛ 5⎞
⎡∵ 2 ∞ e − x2 x 2 n −1dx = n ⎤
∫0
⎢⎣
⎥⎦
1 1
1
1
p
1
=
= ⋅
p
18 6
6 18
9
sin
6
1
Example 9: Evaluate
(log x )5 dx .
0
Solution: Let log x =
t, x = e t, dx =
x = 0,
x = 1,
When
∫
1
0
t
t=0
0
(log x)5 dx = ∫ ( −t )5 ( −e − t ) dt
∞
∞
= − ∫ e − t t 5 dt
0
= − 6 = −120
e t dt
Gamma and Beta Functions
Example 10: Evaluate
∫
0
4
⎛ 1⎞
x 3 log ⎜ ⎟ dx .
⎝ x⎠
4
1
⎛1⎞
⎛1⎞
3
∫0 x log ⎜⎝ x ⎟⎠ dx = ∫0 x . 4 log ⎜⎝ x ⎟⎠ dx
1
⎛1⎞
= 4 ∫ x 3 log ⎜ ⎟ dx
0
⎝x⎠
1
Solution:
1
7.7
3
1
⎛1⎞
log ⎜ ⎟ = t , = et , x = e − t , dx = −e − t dt
⎝x⎠
x
x = 0,
t
Let
When
x = 1,
t=0
4
0
⎛ 1⎞
x 3 log ⎜ ⎟ dx = 4∫ e −3t t ( −e − t ) dt
⎠
⎝
0
∞
x
∫
1
∞
= 4 ∫ e −4t t 2 −1dt
0
= 4⋅
=
Example 11: Evaluate
∫
1
0
∫
Solution:
1
0
1
4
dx
.
⎛ 1⎞
x log ⎜ ⎟
⎝ x⎠
1
dx
⎛1⎞
x log ⎜ ⎟
⎝ x⎠
⎡ ∞ − kx n −1
n⎤
⎢∵ ∫0 e x dx = n ⎥
k ⎦
⎣
2
(4) 2
=∫ x
−
1
2
0
⎡ ⎛ 1 ⎞⎤
⎢log ⎜⎝ x ⎟⎠ ⎥
⎣
⎦
−
1
2
dx
1
⎛1⎞
Let log ⎜ ⎟ = t , = et , x = e–t, dx = e–t dt
⎝x⎠
x
When
x = 0,
t
x = 1,
∫
1
0
1
⎛1⎞
x log ⎜ ⎟
⎝x⎠
t=0
0
−
1
1
−1
dx = ∫ ( e − t ) 2 . t
−
∞
∞
=∫ e
0
−
t
2
t 2 dt
1
2
( − e − t ) dt
… (1)
7.8
Engineering Mathematics
1
2
=
⎡ ∞ − kx n −1
n⎤
⎢∵ ∫0 e x dx = n ⎥
k ⎦
⎣
1
2
⎛ 1⎞
⎜⎝ ⎟⎠
2
= 2
Ç
Example 12: Evaluate
0
a
x
dx .
ax
Solution: Let ax = et, x log a = t, dx =
1
dt
log a
x = 0,
x
,
t=0
t
When
∫
∞
0
∫
Example 13: Evaluate
a
∞⎛
xa
t ⎞ . 1 . 1
dx = ∫ ⎜
x
⎟ et log a dt
0 ⎝ log a ⎠
a
∞
=
1
(log a ) a +1
=
1
a +1
(log a ) a +1
=
a +1
(log a ) a +1
∫
e − t t a dt
0
2
3−4 x dx .
0
2
3−4 x = e − t , − 4 x 2 log 3 = −t log e, 4 x 2 log 3 = t
Solution: Let
x=
t
2 log 3
x = 0,
When
x
∫
∞
0
,
t
3−4 x dx = ∫ e − t .
0
=
=
. 1 dt
2 log 3 2 t
t=0
∞
2
1
, dx =
1
4 log 3
1
4 log 3
1
∫
∞
e−t t
−
. 1 dt
t
1
2
dt
0
1
=
4 log 3 2 4 log 3
Gamma and Beta Functions
Example 14:
Prove that
Ç
xe - ax sin bx dx =
0
∫
Solution:
∞
0
7.9
2ab
.
(a + b 2 )2
2
∞
xe − ax sin bx dx = ∫ xe − ax [Imaginary part of eibx]dx
0
Im. part
∫
∞
xe − ax . eibx dx
0
Im. part
∫
∞
e − ( a −ib ) x . x dx
0
= Im. part
Im. part
⎡
⎢∵
⎣
2
(a ib) 2
∞
∫
0
e − kx x n −1 dx =
n⎤
⎥
kn ⎦
1
(a 2 b 2 ) 2iab
2
2
2ab
Im. part ⎡ (a − b ) + 2iab ⎤ = 2
2 2
⎢ 2
2 2
2 2 ⎥
⎣ ( a − b ) + 4a b ⎦ ( a + b )
Exercise 7.1
1. Evaluate the following integrals:
(i)
∫
∞
x e − x dx
3
0
(ii)
∫
∞
e
−
x2
4
dx
0
(iii)
∫
∞
e−
0
1
(iv)
x
x
dx
7
4
( x log x) 4 dx
0
1
dx
x
dx
1
log
(v)
0
(vi)
∫
1
∫
(ix)
3
0
∫
∞
0
1
n
x m e − ax dx =
na
3
4
1
(viii)
∞
0
1⎞
⎛
(vii) ∫ x ⎜ log ⎟ dx
0
x⎠
⎝
1
1
x log dx
x
( m +1)
n
m +1
.
n
3. Prove that
∫
∞
0
∞
x 2 e − x dx . ∫ e − x dx =
4
4
0
8 2
.
4. Prove that
2
5−4 x dx.
⎤
⎥
⎥
4!
⎥
(iv) 5
⎥
5
⎥
⎥
( vi) p
⎥
⎥
4
⎛ 3 ⎞3 4 ⎥
⎥
( viii) ⎜ ⎟
⎝ 4 ⎠ 3⎥
⎥
⎥
⎥⎦
(ii) p
2. Prove that
− log x
0
315
⎡
⎢ Ans. : (i) 16 p
⎢
8
⎢
(iii)
p
⎢
3
⎢
p
⎢
( v)
⎢
2
⎢
⎢
6
⎢
( vii)
625
⎢
⎢
p
⎢
(ix )
⎢⎣
4 log 5
∫
∞
0
xe
− x2
dx .
∫
∞
0
e− x
2
x
dx =
2 2
.
7.10
Engineering Mathematics
5. Prove that
∫
∞
0
8. Prove that
∞
xe − x dx . ∫ x 2 e − x dx =
8
4
0
16 2
.
∫
0
6. Prove that
1
0
∞
x m −1 cos ax dx =
9. Prove that
x m (log x) n dx = (−1) n + 1 .
(m + 1) n +1
∫
∞
m
⎛m ⎞
cos ⎜
⎟.
m
⎝ 2 ⎠
a
x n −1e − ax sin bx dx
0
n
=
7. Prove that
b⎞
⎛
sin ⎜ n tan −1 ⎟ .
a⎠
⎝
( a 2 + b2 )
n
n
2
n
n +1
1⎞
m ⎛
∫0 x ⎜⎝ log x ⎟⎠ dx = (m + 1)n +1 .
1
7.4 BETA FUNCTION
Beta function B(m, n) is defined by
1
B (m, n) = ∫ x m −1 (1 − x) n −1 dx, m > 0, n > 0.
0
B(m, n) is also known as Euler’s integral of first kind.
7.4.1 Trigonometric form of Beta Function
B(m, n) = 2 ∫ 2 sin 2 m −1 x cos 2 n −1 x dx
0
B(m, n) =
Proof:
∫
1
0
Let x
When
sin2q,
dx
x m −1 (1 − x) n −1 dx
2sinq cosq dq
x = 0,
q =0
x = 1,
q =
2
π
2
B(m, n) = ∫ (sin 2 θ ) m −1 (1 − sin 2 θ ) n −1 ⋅ 2 sin θ cos θ dθ
0
π
2
= 2 ∫ sin 2 m −1θ cos 2 n −1θ dθ
0
Changing the variable q to x,
2
2 m −1
x cos 2 n −1 x dx
B(m, n) = 2 ∫ sin
0
Gamma and Beta Functions
Corollary: Putting 2m 1 = p,
2n 1 = q
p +1
,
2
m=
7.11
q +1
2
n=
⎛ p +1 q +1⎞
2
p
q
B⎜
,
⎟ = 2 ∫0 sin x cos x dx
⎝ 2
2 ⎠
7.5 PROPERTIES OF BETA FUNCTION
1. Symmetry
B(m, n) = B(n, m)
Proof:
B(m, n) =
Let 1
x = t,
∫
1
0
dx = dt
When
x m −1 (1 − x) n −1 dx
x = 0,
t=1
x = 1,
t=0
0
B (m, n) = ∫ (1 − t ) m −1 t n −1 (−dt ) =
1
1
∫t
n −1
0
(1 − t ) m −1 dt
B (n, m).
2. Relation between Beta and Gamma Function
m n
B(m, n) =
m+n
Proof: By alternate form of Gamma function,
∞
∞
m n = 2 ∫ e − x x 2 m −1dx . 2 ∫ e − y y 2 n −1dy
2
0
= 4∫
∞
0
2
0
∫
∞
e−( x
2
+ y2 )
x 2 m −1 y 2 n −1dx dy
0
Changing to polar coordinates x = r cosq, y = r sinq
dx dy = r dr dq
Limits of x
x=0
to
x
Limits of y
y=0
to
y
Fig. 7.2
This shows that the region of integration is the first quadrant.
Draw an elementary radius vector in the region which starts
from pole and extends up to .
Limits of r
r=0
to
r
Limits of q
q=0
to
q=
m n=4∫
π
2
0
∫
∞
0
2
2
e − r (r cos θ ) 2 m −1 (r sin θ ) 2 n −1 r dr dθ
7.12
Engineering Mathematics
π
π
2
= 4 ∫ (cos θ )
2 m −1
(sin θ )
2 n −1
0
=4.
2
dθ . ∫ e − r r 2 ( m + n ) −1dr
2
0
1
1
B(m, n) . m + n
2
2
m n
B(m, n) =
m+n
3. Duplication Formula
1
o 2m
= 2 m -1
2
2
m m+
π
2
B(m, n) = 2 ∫ sin 2 m −1θ cos 2 n −1θ dθ
Proof:
0
Putting n = m,
π
2
B(m, m) = 2 ∫ (sin θ cos θ ) 2 m −1 dθ
0
m m
2m
Let 2q
When
t,
1
d = dt
2
q = 0,
q=
2
=
2
22 m −1
∫
π
2
(sin 2θ ) 2 m −1 dθ
0
t=0
t=p
,
m⋅ m
2
1
= 2 m−1 ∫ (sin t ) 2 m−1 ⋅ dt
0
2
2
2m
=
1
22 m −1
2
⋅ 2 ∫ (sin t )
2
=
1
22 m −1
0
(cos t ) dt
0
22 m−1 ∫0
1
⎛ 1⎞
= 2 m −1 B ⎜ m, ⎟
⎝ 2⎠
2
=
2 m −1
2
(sin t ) 2 m−1 (cos t )
1
2
1
m+
2
m
⎛1⎞
2 ⎜ ⎟ −1
⎝2⎠
dt
⎡∵ 2 a f ( x)dx = 2 a f ( x)dx ⎤
∫0
⎢ ∫0
⎥
⎢ if f (2a − x) = f ( x)
⎥
⎣
⎦
Gamma and Beta Functions
m m
2m
=
7.13
1 m
22 m −1
1
m+
2
m m+
1
2m
=
.
2 m −1
2
2
Example 1: Find the value of
(i)
⎛ 4 5⎞
(ii) B ⎜ , ⎟ .
⎝ 3 3⎠
⎛ 3 1⎞
B⎜ , ⎟
⎝ 2 2⎠
Solution: (i)
3 1
⎛3 1⎞ 2 2
B⎜ , ⎟ =
⎝2 2⎠
2
1 1 1
1
= 2 2 2 =
1
2
=
(ii)
.
2
4 5
⎛ 4 5⎞ 3 3
B⎜ , ⎟ =
3
⎝ 3 3⎠
=
1 1
2
1 1 1.2 2
+1 . +1 = .
2 3
3
2 3 3 3 3
=
1 1
1 1
1− = .
9 3
3 9
sin
⎡
⎤
⎢⎣∵ n 1− n = sin n ⎥⎦
3
1 2
2
= .
=
9
3 9 3
1
and n is a positive integer, find the value of n.
60
1
B(n, 3)
60
Example 2: If B(n, 3) =
Solution:
n 3
n+3
n.2
(n + 2) (n + 1) n n
=
1
60
=
1
60
7.14
Engineering Mathematics
n3 + 3n2 + 2n = 120
120 = 0
n3 + 3n2 + 2n
n = 4, 3.5
But n is a positive integer.
Hence, n = 4.
Example 3: Prove that B(n, n) =
Solution:
B(n, n) =
=
=
π .
22 n −1
n .
=
2n
2n
n n
n .
2n
2
2 n −1
1
.
.
1
n+
2
n n+
1
2
1
2
2n
n+
[By Duplication formula]
22 n −1
1
n+
2
.
n
n
n+
1
2
1
1
o
Example 4: Prove that B(n, n) . B ⎛⎜ n + , n + ⎞⎟ = 21- 4 n .
⎝
2
2⎠ n
1
1⎞ n n .
⎛
Solution: B(n, n) B ⎜ n + , n + ⎟ =
2
2⎠
⎝
2n
n+
1
1
n+
2
2
2n + 1
⎛
1⎞
⎜⎝ n n + ⎟⎠
2
2
⎛
1⎞
n n+ ⎟
⎜
1
2⎟
⎜
=
=
⎜ 2n ⎟
.
n
2
2n 2n 2n
⎝
⎠
2
=
1- n
=
Example 5: Prove that n
2
⎞
1 ⎛
⎜ 2 n −1 ⎟ = 21− 4 n
2n ⎝ 2
⎠
n
n
2 .
no
1- p
2 cos
2
o
Solution: We know that
n 1− n =
sin n
2
Gamma and Beta Functions
Replacing n by
7.15
n +1
,
2
n +1
n +1
1−
=
2
2
⎛ n +1⎞
sin ⎜
⎟
⎝ 2 ⎠
n +1 1− n
=
2
2
n ⎞
⎛
sin ⎜ +
⎟
⎝2 2 ⎠
n
n n 1 1− n
2
+
=
n
2 2 2 2
cos
2
n
n 1− n
2
=
n
2
2n −1
cos
2
n
1− n
2
=
n
n
2
1− n
2 cos
2
Exercise 7.2
1. Find the value of
⎛5 3⎞
(i) B ⎜ , ⎟
⎝2 2⎠
⎛1 2⎞
(ii) B ⎜ , ⎟
⎝2 3⎠
p
2p ⎤
⎡
⎢ Ans. : (i) 16 (ii) 3 ⎥
⎣
⎦
1
2. If B(n, 2) =
and n is a positive
42
integer, find the value of n.
[Ans. : n = 6]
3. Prove that
Example 1: Evaluate
∫
1
1
n+
1
1⎞ 1 .
⎛
2
B ⎜ n + , n + ⎟ = 2n
2
2⎠ 2
⎝
n +1
4. Prove that
B(m, n) = B(m, n + 1) + B(m + 1, n).
5. Prove that
3
3
−n +n
2
2
⎛1
⎞
= ⎜ − n 2 ⎟ p sec np, (–1 < 2n < 1).
⎝4
⎠
x 3 (1 − x ) dx.
5
0
Solution: Let
.
x = t , x = t2, dx = 2t dt
7.16
Engineering Mathematics
When
∫ x (1 −
1
x = 0,
t=0
x = 1,
t=1
x ) dx = ∫ t 6 (1 − t ) 2t dt
1
5
3
0
5
0
1
= 2 ∫ t 7 (1 − t ) dt
5
2B(8, 6)
0
=2
Example 2: Evaluate
1
x2
0
1− x
∫
4
8 6
=2
14
∫
dx .
1
0
1
7 ! 5!
=
13!
5148
dx
1 − x4
1
Solution: Let
x4 = t, x = t 4 , dx =
When
x = 0,
t=0
x = 1,
t=1
.
1 − 34
t dt
4
1
∫
1
0
x2
1 − x4
dx .
∫
1
0
dx
1 − x4
=∫
1
0
3
1
−
1 . 1 − 34
. 1 t 4 dt .
∫0 1 − t 4 t dt
1− t 4
t2
=
1
1
1 −3
−
−
1 1 − 14
2
. t 4 (1 − t ) 2 dt
t
(
−
t
)
d
t
1
∫0
16 ∫0
=
1 ⎛3 1⎞ . ⎛1 1⎞
B⎜ , ⎟ B⎜ , ⎟
16 ⎝ 4 2 ⎠
⎝4 2⎠
3 1 1 1
1 4 2 . 4 2
1
. 1
=
=
16 5
16
3
1 1 4
4
4
4 4
Example 3: Evaluate
∫
1
0
1 − y 4 dy .
1
Solution: Let
y4 = t, y = t 4 , dy =
When
y = 0,
t=0
y = 1,
t=1
∫
1
0
1 − 34
t dt
4
1
1
1 −3
1 − y 4 dy = ∫ (1 − t ) 2 . t 4 dt
0
4
3 1
1 ⎛3 1⎞ 1 2 4
= B⎜ , ⎟ =
4 ⎝2 4⎠ 4 7
4
=
p
4
Gamma and Beta Functions
1 1 1
1 2 2 4 1
=
=
6
4 3 3
4 4
p
⎛ 1⎞
⎜⎝ ⎟⎠
4
7.17
2
3 1
4 4
2
2
⎛ 1⎞
⎜ ⎟
p ⎝ 4⎠
p
=
=
p
6
6
1 1
1−
p
4 4
sin
4
⎛ 1⎞
⎜⎝ ⎟⎠
4
2
⎛ 1⎞
⎜ ⎟
1
p ⎝ 4⎠
=
=
6 p 2
6 2
Example 4: Evaluate
∫
2
0
⎛ 1⎞
⎜ ⎟
⎝ 4⎠
2
1
y 4 ( 8 − y 3 ) 3 dy .
−
1
1 −2
Solution: Let y3 = 8t, y = 2t 3 , dy = 2 . t 3 dt
3
When
y = 0,
t=0
y = 2,
∫
2
0
1
t=1
( ) (8 − 8t )
1 4
y 4 ( 8 − y 3 ) 3 dy = ∫ 2t 3
−
1
0
=
−
2a
0
Solution:
∫
2a
0
x 2 2ax - x 2 dx .
2a
5
x 2 2ax − x 2 dx = ∫ x 2 2a − x dx
0
Let x = 2at, dx = 2a dt
When
2 − 32
t dt
3
1
16 1 32
16 ⎛ 5 2 ⎞
−
3 dt =
(
)
t
−
t
1
B⎜ , ⎟
∫
0
3
3 ⎝3 3⎠
5 2
16 3 3
16
=
=
3
3
7
3
Example 5: Evaluate
1
3
x = 0,
t=0
x = 2a,
t=1
2 2
3 3
4. 1
3 3
⎛ 2⎞
2
⎜⎝ ⎟⎠
3
3
=8
1
1
3
3
2
7.18
Engineering Mathematics
∫
2a
0
5
1
x 2 2ax − x 2 dx = ∫ (2at ) 2 2a − 2at . 2a dt
0
1
1 5
⎛7 3⎞
4
= 16a 4 ∫ t 2 (1 − t ) 2 dt = 16a B ⎜ , ⎟
⎝2 2⎠
0
7 3
16a 4 . 5 . 3 . 1 1 . 1 1
= 16a 2 2 =
5
24 2 2 2 2 2 2
4
=
15 a 4
24
3
Example 6: Evaluate
∫
3
0
x2
dx . ∫
3− x
dx
1
0
1− x
1
4
.
3
Solution:
Let x = 3t, dx = 3 dt
When
I1 = ∫
x2
3
3− x
0
dx
x = 0,
t=0
x = 3,
t=1
3
I1 = ∫
(3t ) 2
1
3 − 3t
0
⋅ 3 dt
1 3
= 9 ∫ t 2 (1 − t )
−
1
2
0
⎛5 1⎞
dt = 9 B ⎜ , ⎟
⎝2 2⎠
5 1
9 3 1 1 1
27
=9 2 2 = . .
=
3
2 2 2 2 2
8
1
dx
I2 = ∫
1
0
1− x4
1
Let x 4 = t , x = t4, dx = 4t3 dt
When
x = 0,
t=0
x = 1,
t=1
3
1
4t
0
1− t
I2 = ∫
1
dt
= 4 ∫ t 3 (1 − t )
0
−
1
2
⎛ 1⎞
dt = 4 B ⎜ 4, ⎟
⎝ 2⎠
Gamma and Beta Functions
4
=4
9
2
1
2 =4
7.19
1
128
2
=
7 . 5 . 3 . 1 1 35
2 2 2 2 2
3!
3
∫
Hence,
x2
3
dx . ∫
3− x
0
Example 7: Prove that
∫
0
1− x
dx
a
0
dx
1
1
n n
(a n − x )
1
Solution: Let xn = an t, x = at n , dx =
x = 0,
x = a,
When
∫
a
0
dx
1
n n
(a n − x )
=∫
1
4
=
=
o
⎛o ⎞
cosec ⎜ ⎟ .
⎝ n⎠
n
a 1n −1
t dt
n
t=0
t=1
1
1
0
(a n − a n t )
1
n
1
−1
. a t n dt
n
1 1
−
−1
1 1
1 ⎛ 1
1⎞
n n
1
(
−
)
t
t dt = B ⎜ − + 1, ⎟
∫
0
n
n ⎝ n
n⎠
=
1
= .
n
1−
1 1
n n =1.
n sin
1
⎛ ⎞
= cosec ⎜ ⎟
n
⎝n⎠
Example 8: Prove that
∫
b
∫
9
4
5
Solution: Let (x
When
∫
b
a
⎡
⎤
⎢⎣∵ 1− n n = sin n ⎥⎦
n
( x − a )m (b − x )n dx = (b − a )m + n +1 B( m + 1, n + 1) and
a
hence, deduce that
27 . 128 432 p
=
8
35
35
( x − 5)(9 − x ) dx =
⎛ 1⎞
2⎜ ⎟
⎝ 4⎠
2
.
3
a) = (b
dx = (b
a) t,
x = a,
x = b,
a) dt
t=0
t=1
1
( x − a ) m (b − x) n dx = ∫ [ (b − a )t ] [b − {a + (b − a )t}] (b − a )dt
m
n
0
1
= (b − a ) m + n +1 ∫ t m (1 − t ) n dt
0
= (b − a )
m + n +1
B(m + 1, n + 1)
7.20
Engineering Mathematics
1
1
Putting a = 5, b = 9, m = , n = in the above integral,
4
4
1
1
1 1
9
+ +1
1 ⎞
⎛1
∫5 ( x − 5) 4 (9 − x) 4 dx = (9 − 5) 4 4 B ⎜⎝ 4 + 1, 4 + 1⎟⎠
2
⎛ 1⎞
⎛1 1⎞
5 5
2⎜ ⎟
⎜
⎟
⎝ 4⎠
= 23 4 4 = 8 ⎝ 4 4 ⎠ =
3
5
3.1 1
2
2 2 2
∫
Example 9: Prove that
1
0
∫
1
0
2
x m −1 (1 − x )n −1
B( m , n)
dx =
and hence, evaluate
m+n
(a + bx )
( a + b )m a n
x − 2x + x
dx .
(1 + x )6
2
3
4
Solution: Let x =
When
Also,
a (a + b − bt ) − at (−b)
a ( a + b)
at
dx =
dt =
,
dt
2
(a + b − bt )
(a + b − bt ) 2
a + b − bt
x = 0,
t=0
x = 1,
t=1
1− x = 1−
at
(a + b)(1 − t )
=
a + b − bt
a + b − bt
a + bx = a +
a ( a + b)
bat
=
a + b − bt
a + b − bt
m −1
∫
1
0
m −1
n −1
x (1 − x)
(a + bx) m + n
at
⎛
⎞ ⎡ (a + b)(1 − t ) ⎤
⎟ ⎢
1⎜
⎥
dx = ∫ ⎝ a + b − bt ⎠ ⎣ a +mb+ n− bt ⎦
0
⎡ a ( a + b) ⎤
⎢⎣ a + b − bt ⎥⎦
=
1
( a + b) m a n
1
∫t
m −1
0
(1 − t ) n −1 dt =
Putting a = 1, b = 1, m = 3, n = 3 in the above integral
∫
1
0
∫
1
0
1
x 2 (1 − x) 2
B(3, 3)
dx =
6
(1 + 1)3 .13
(1 + x)
x 2 − 2 x3 + x 4
1 3 3 1. 4
1
dx =
=
=
6
(1 + x)
8 6
8 120 240
3
(1 − x 4 ) 4
1 1 ⎛ 1 7⎞
Example 10: Prove that ∫
dx = . 1 B ⎜ , ⎟ .
4 2
0 (1 + x )
4 4 ⎝ 4 4⎠
2
1
n −1
. a ( a + b ) dt
(a + b − bt ) 2
1
B(m, n)
( a + b) m a n
Gamma and Beta Functions
1
Solution: Let x4 = t, x = t 4 , dx =
1 − 34
t dt
4
x = 0,
x = 1,
When
7.21
t=0
t=1
3
3
−
3
3
1 (1 − t ) 4
(1 − x 4 ) 4
1 −3
1 1 t 4 (1 − t ) 4
∫0 (1 + x 4 )2 dx =∫0 (1 + t )2 . 4 t 4 dt = 4 ∫0 (1 + t )2 dt
1
u
,
2−u
t=
Let
dt =
2
(2 − u ) − u ( −1)
du
du =
2
(
u )2
2
−
(2 − u )
t = 0,
t = 1,
When
u=0
u=1
−
3
3
u ⎞4
⎛ u ⎞ 4⎛
⎜⎝
⎟⎠ ⎜⎝1 −
⎟
1
1 (1 − x )
2−u
2 − u⎠
1
2
.
x
=
du
d
2
∫0 (1 + x 4 )2
4 ∫0
(
2
u )2
−
u ⎞
⎛
⎜⎝1 +
⎟
2 − u⎠
3
4 4
2 1 u
= ∫
4 0
−
3
4
3
( 2 − 2u ) 4
2
2
1 1
du = . 1
4 4
2
∫
1
0
u
−
3
4
3
(1 − u ) 4 du
1 1 ⎛1 7⎞
= . 1 B⎜ , ⎟
4 4 ⎝4 4⎠
2
Exercise 7.3
1. Evaluate the following integrals:
(i)
(ii)
1
∫
0
1
∫
0
1 − x dx
m
dx
1 − x6
2
1 3
⎛
⎞
(iii) ∫ ⎜1 − x 4 ⎟ dx
0
⎝
⎠
1
∫
2
(v)
∫
a
(vi)
∫
(iv)
0
0
1
2
0
−
1
x 2 ( 2 − x ) 2 dx
x 4 a 2 − x 2 dx
x 3 1 − 4 x 2 dx.
⎡ Ans. :
⎤
⎢
⎥
⎢ (i) 1 B ⎛⎜ 1 , 3 ⎞⎟ (ii) 1 B ⎛⎜ 1 , 1 ⎞⎟ ⎥
⎢
m ⎝ m 2⎠
8 ⎝ 8 2⎠ ⎥
⎢
⎥
64 2
⎢(iii) 128
⎥
(iv)
⎢
⎥
1155
15
⎢
⎥
6
1
⎢ ( v) p a
⎥
( vi)
⎢⎣
⎥⎦
32
120
2. Prove that
⎛ 1⎞
2⎜ ⎟
7
(i) ∫ 4 ( x − 3)(7 − x) dx = ⎝ 4 ⎠
3
3
2
7.22
Engineering Mathematics
∫
(ii)
6
5
( x − 5)5 (6 − x)6 dx =
∫
3. Prove that
1
0
−
1
x 3 (1 − x)
(1 + 2 x)
−
5! 6 !
.
12 !
2
3
and hence, evaluate
0
∫
0
dx =
3
.
7
6
π
2
∫
Solution:
π
2
0
x n −1
∫0 (1 + cx)(1 − x)n dx
1
5. Prove that
=
x m −1 (1 − x) n −1
B(m, n)
dx =
(1 + x) m + n
2m
Example 1: Evaluate
1 .
, 0 < n < 1.
(1 + c) n sin n
cot θ dθ .
0
π
1
−
1
cot θ dθ = ∫ 2 (cos θ ) 2 (sin θ ) 2 dθ
0
1 ⎞
⎛1
+1 − +1⎟
1 ⎜2
, 2 ⎟
= B⎜
2 ⎠
2 ⎝ 2
3 1
1 ⎛3 1⎞ 1 4 4
= B⎜ , ⎟ =
2 ⎝4 4⎠ 2 1
=
Example 2: Evaluate
π
4
1
1 1 1
1−
= ⋅
2
4 4 2 sin
=
p
2
4
cos 3 2θ sin 4 4θ dθ .
0
Solution:
∫
π
4
0
π
cos3 2θ sin 4 4θ dθ = ∫ 4 cos3 2θ (2 sin 2θ cos 2θ ) 4 dθ
0
π
= 16 ∫ 4 cos 7 2θ sin 4 2θ dθ
0
Let 2q = t,
When
x3 − 2 x 2 + x
dx.
(1 + x)5
1⎤
⎡
⎢⎣ Ans. : 48 ⎥⎦
4. Prove that
1
∫
1
1
d = dt
2
q = 0,
q=
t=0
,
4
t=
2
Gamma and Beta Functions
∫
π
4
0
7.23
π
1
cos3 2θ sin 4 4θ dθ = 16 ∫ 2 sin 4 t . cos 7 t ⋅ dt
0
2
5
4
1
5
⎛
⎞
= 8. B ⎜ , 4 ⎟ = 4 2
⎠
2 ⎝2
13
2
3 .1 1 .
3!
128
2 2 2
=
= 4⋅
11 9 7 5 3 1 1 1155
⋅ ⋅ ⋅ ⋅ ⋅
2 2 2 2 2 2 2
Example 3: Evaluate
∫
2π
0
sin 2 θ (1 + cos θ )4 dθ .
2π
I = ∫ sin 2 θ (1 + cos θ ) 4 dθ
Solution:
0
2
4
2π
θ
θ⎞ ⎛
θ⎞
⎛
= ∫ ⎜ 2 sin cos ⎟ ⎜ 2 cos 2 ⎟ dθ
0 ⎝
2
2⎠ ⎝
2⎠
2π
= 26 ∫ sin 2
0
Let
2
When
θ
θ
cos10 dθ
2
2
= t , d = 2dt
q = 0,
q = 2p,
t=0
t=p
I = 26 ∫ sin 2 t cos10 t . 2dt
0
= 27 . 2 ∫ 2 sin 2 t cos10 t dt
0
⎡∵
f ( x ) dx = 2 ∫ f ( x ) d x ⎤
⎢ ∫0
⎥
0
⎢ if f (2a − x) = f ( x)
⎥
⎣
⎦
2a
a
3 11
1
3
11
⎛
⎞
7 2
2
= 28 . B ⎜ ,
⎟ =2
2 ⎝2 2 ⎠
7
=
Example 4: Evaluate
∫
o
4
o
4
27 . 1 1 9 . 7 . 5 . 3 . 1 1
21p
.
=
6! 2 2 2 2 2 2 2 2
8
1
(cos p + sin p ) 3 dp .
1
Solution:
∫
π
4
π
−
4
1
3
(cos θ + sin θ ) dθ =
∫
π
4
π
−
4
⎡ ⎛ 1
1
⎞⎤ 3
cos θ +
sin θ ⎟ ⎥ dθ
⎢ 2 ⎜⎝
⎠⎦
2
2
⎣
7.24
Engineering Mathematics
∫
π
4
π
−
4
π
⎛ π
⎞3
2 ⎜ sin cos θ + cos sin θ ⎟ dθ =
⎝
⎠
4
4
Let
+
4
= t,
q=
π
4
π
−
4
∫
⎡ ⎛π
⎞⎤ 3
2 ⎢sin ⎜ + θ ⎟ ⎥ dθ
⎠⎦
⎣ ⎝4
π
4
π
−
4
1
6
dq = dt
q = −
When
∫
1
1
1
6
(cos θ + sin θ ) dθ = 2
1
6
t=0
t=
,
4
1
3
,
4
∫
2
0
2
1
3
(sin t ) dt
1
1
6
1
3
26 ⎛ 4 1 ⎞
2 ∫ 2 (sin t ) (cos t ) dt =
B⎜ , ⎟
0
2 ⎝6 2⎠
=
2 1
2
1
6
3 2 =
3
=
5
5
7
1 1
26
26
6
6 6
1
5
26
Example 5: Prove that
2
Solution:
0
π
2
0
∫
tan n x dx =
tan n x dx =
∫
2
0
0
2
3
1
6
π
⎛ nπ ⎞
sec ⎜
⎟.
2
⎝ 2 ⎠
(sin x) n (cos x) − n dx
1 ⎛ n + 1 −n + 1 ⎞ 1
,
= B⎜
⎟=
2 ⎝ 2
2 ⎠ 2
n + 1 −n + 1
2
2
1
1 n +1
n +1
1−
2 2
2
1
⋅
2
2
Example 6: Evaluate
Solution:
0
0
⋅
⎤
⎡
⎥
⎢∵ n 1− n =
sin
n
⎦
⎣
⎛ n +1⎞
sin ⎜
⎝ 2 ⎟⎠
1
n ⎞
⎛
sin ⎜ +
⎝ 2 2 ⎟⎠
=
1
⎛n ⎞
.
= sec ⎜
⎝ 2 ⎟⎠
2 cos n
2
2
x sin 7 x cos 4 x dx .
x sin 7 x cos 4 x dx =
∫
0
( − x) sin 7 ( − x) cos 4 ( − x)dx
⎡∵ a f ( x)dx = a f (a − x)dx ⎤
∫0
⎣⎢ ∫0
⎦⎥
Gamma and Beta Functions
∫
sin 7 x cos 4 x dx − ∫ x sin 7 x cos 4 x dx
0
2
x sin 7 x cos 4 x dx =
0
7.25
0
sin 7 x cos 4 x dx
0
⎡ 2 7
⎤
4
7
4
2
⎢ ∫0 sin x cos x dx + ∫0 sin ( − x) cos ( − x)dx ⎥
⎣
⎦
⎡∵ 2 a f ( x)dx = a f ( x)dx + a f (2a − x)dx ⎤
∫0
∫0
⎥⎦
⎢⎣ ∫0
2
2
0
sin 7 x cos 4 x dx = 2 ⋅
5
2=
13
2
4
0
x sin 7 x cos 4 x dx =
1 ⎛
B ⎜ 4,
2 ⎝
5⎞
⎟
2⎠
5
2
11 9 7 5 5
2 2 2 2 2
3!
16
1155
Exercise 7.4
1. Evaluate the following integrals:
(i)
π
2
0
(ii)
π
6
0
(iii)
π
3
π
−
6
(iv)
(v)
(vi)
tan θ dθ
2
6
cos 3θ sin 6θ dθ
∫ (
π
2
π
−
2
2π
3 sin θ + cos θ
)
1
4
dθ
∫
cos3 θ (1 + sin θ )2 dθ
∫
sin 2 θ (1 + cos θ )4 dθ
0
0
⎡
⎢ Ans. : (i) 2
⎢
⎢
3
⎢
−
(iii) 2 4
⎢
⎢
⎢
⎢
21
⎢
( v)
⎢⎣
8
7 ⎤
384 ⎥
⎥
⎥
8 ⎥
(iv )
⎥
5 ⎥
⎥
⎥
8 ⎥
(iv )
693 ⎥⎦
(ii)
5
8
9
8
2. Prove that
∫
2
0
x sin 5 x cos6 x dx
(sin x) 2 n dx =
1 ⋅ 3 ⋅ 5 (2n − 1)
⋅ .
2n (n !)
2
7.6 BETA FUNCTION AS IMPROPER INTEGRAL
B(m, n) =
∫
∞
0
x m −1
dx
(1 + x) m + n
Proof: Let x = tan q, dx = 2 tanq sec2q dq
2
7.26
Engineering Mathematics
x = 0,
When
x
∫
∞
0
q=0
π
θ=
2
Ç,
x m −1
dx =
(1 + x) m + n
(tan 2 θ )m −1
⋅ 2 tan θ sec 2 θ dθ
(1 + tan 2 θ )m + n
π
2
0
∫
(tan θ )2 m −1 sec 2 θ
dθ
(sec θ )2 m + 2 n
π
2∫2
0
π
2 ∫ 2 (sin θ )2 m −1 (cos θ )2 n −1 dθ
0
= B(m, n)
Example 1: Prove that
value of
∫
∞
0
∫
x m −1
1
dx = n m B( m , n) and hence, find the
m+n
(a + bx )
a b
∞
0
5
x
.
(2 + 3 x )16
a
dt
b
Solution: Let bx = at, dx =
x = 0,
When
t=0
x
,
t
m −1
∫
x m −1
dx =
(a + bx) m + n
∞
0
∞
∫
0
⎛a ⎞
⎜⎝ t ⎟⎠
a
b
⋅ dt
(a + at ) m + n b
1
a bm
∫
n
∞
0
1
t m −1
dt = n m B(m, n)
a b
(1 + t ) m + n
Putting a = 2, b = 3, m = 6, n = 10 in the above integral,
∫
x5
1
dx = 10 6 B(6, 10)
16
2 3
(2 + 3x)
∞
0
=
Example 2: Prove that
∫
∞
0
Solution:
∫
∞
0
1
6 10
10
6
2
3
16
=
1
10
2
5! 10 !
3 15!
6
x 8 (1 − x 6 )
dx = 0 .
(1 + x )24
x8 (1 − x 6 )
dx =
(1 + x) 24
∫
∞
∫
∞
0
0
∞
x8
x14
x
d
−
∫0 (1 + x)24 dx
(1 + x) 24
∞
x 9 −1
x15 −1
dx − ∫
dx
9 +15
0 (1 + x )15 + 9
(1 + x)
Gamma and Beta Functions
B(9, 15)
7.27
B(15, 9)
0
Example 3: Prove that
x2
5 2
.
dx =
(1 + x 4 )3
128
∞
∫
0
1
Solution: Let x4 = t, x = t 4 , dx =
1 43
t dt
4
x = 0,
When
x
t=0
,
t
1
∫
∞
0
x2
dx =
(1 + x 4 )3
∫
∞
0
t2
1 −3
⋅ t 4 dt
3
(1 + t ) 4
−
1
3
1 ∞ t 4
1 ∞
dt = ∫
3
∫
4 0 (1 + t )
4 0
1 ⎛3
B⎜ ,
4 ⎝4
t4
−1
(1 + t )
3 9
+
4 4
dt
3 9
9⎞ 1 4 4
⎟=
4⎠ 4 3
3 5 1 1
4 4 4 4 = 5 1 1 1
2!
128
4 4
⎤
⎡
5
⎥
⎢∵ 1− n n =
n
sin
128 sin
⎦
⎣
4
5 2
=
128
1
4
Example 4: Prove that
∫
∞
0
Solution:
∫
∞
0
sech 6 x dx =
sech 6 x dx =
∫
∞
0
8
.
15
6
⎛ 2 ⎞
⎜⎝ x
⎟ dx
e + e− x ⎠
26 ⋅
1 ∞
1
dx
x
∫
−
∞
2
(e + e − x ) 6
25 ∫
e6 x
dx
− ∞ (e
+ 1)6
∞
2x
⎡
e x + e− x ⎤
⎢∵ cosh x =
⎥
2 ⎦
⎣
⎡∵ a f ( x)dx = 2 a f ( x)dx ⎤
∫0
⎢ ∫− a
⎥
⎢ if f (− x) = f ( x)
⎥
⎣
⎦
7.28
Engineering Mathematics
e2x = t,
2e2x dx = dt,
x
,
t=0
x
,
t
Let
When
∫
∞
0
sec h 6 x dx = 25
∫
0
∞
0
16
Example 5: Prove that
Solution:
∫
∞
0
When
∫
∞
0
0
t 3−1
dt
(1 + t )3+ 3
33
16
6
24 B(3, 3)
8
2! 2!
= .
15
5!
e 2 mx + e − 2 mx
1
dx = B( n + m , n − m ), n > m .
x
− x 2n
2
(e + e )
e 2 mx + e − 2 mx
dx
(e x + e − x ) 2 n
e2x = t,
Let
∫
∞
1
dt
2t
t3
1
⋅ dt
6
2t
(t + 1)
∞
24 ∫
dx
1 ∞ (e 2 mx + e −2 mx ) e 2 nx
dx
2 ∫− ∞
(e 2 x + 1) 2 n
⎡∵ a f ( x)dx = 2 a f ( x)dx ⎤
∫0
⎢ ∫− a
⎥
⎢ if f (− x) = f ( x)
⎥
⎣
⎦
1 ∞ e2( m + n) x + e2( n − m) x
dx
2 ∫− ∞
(1 + e 2 x ) 2 n
1
2e2x dx = dt,
dx = dt
2t
x
,
t=0
x
,
t
,
e 2 mx + e −2 mx
1 ∞ t m+n + t n−m 1
dx = ∫
⋅ dt
x
− x 2n
(e + e )
2 0 (1 + t ) 2 n
2t
∞
⎤
1⎡ ∞
t ( m + n ) −1
t ( n − m ) −1
t
d
dt ⎥
+
⎢ ∫0
(m+ n)+(n− m)
(n− m)+(n+ m)
∫
0
4 ⎣ (1 + t )
(1 + t )
⎦
1
[ B(m + n, n − m) + B(n − m, n + m)]
4
1
B (n + m, n − m)
[∵ B(m, n) = B(n, m)]
2
Example 6: Prove that
deduce that
∫
o
∫
0
sin x
0
(5 + 3 cos x )
3
2
sin n −1 x
dx
(a + b cos x )n
dx =
⎛
⎜
⎝
3⎞
4 ⎟⎠
2
2 2o
.
=
2n −1
n
2 2
(a 2 − b )
⎛ n n⎞
B ⎜ , ⎟ and hence,
⎝2 2⎠
Gamma and Beta Functions
Solution: Let tan
7.29
x
x
2
= t , = tan −1 t , dx =
dt
2
2
1+ t2
1− t2
2t
,
sin x =
2
1+ t
1+ t2
x = 0,
t=0
cos x =
When
x = p,
∫
0
n −1
sin x
dx =
(a + b cos x) n
Let
(a
∫
t
⎛ 2t ⎞
⎜⎝
⎟
1+ t2 ⎠
∞
0
n −1
⎡
⎛1 − t 2 ⎞⎤
⎢a + b ⎜
⎥
⎝ 1 + t 2 ⎟⎠ ⎦
⎣
a+b ⋅ u
b) t2 = (a + b)u, t =
t = 0,
t
,
When
n
2n ∫
2
dt
1+ t2
⋅
a−b
0
, dt =
t n −1
∞
⎡⎣(a + b) + (a − b) t 2 ⎤⎦
a+b
2 a−b
⋅
1
u
1
2
n
dt
du
u=0
u
n −1
∫
0
⎡ (a + b)u ⎤ 2
n −1
⎢ ( a − b) ⎥
sin x
∞
a+b 1
⎣
⎦
dx = 2n
n
∫0 [(a + b) + (a + b) u ]n ⋅ 2 a − b 1 du
(a + b cos x)
u2
n
2n −1
n
2
( a + b) ( a − b)
Putting a = 5, b = 3, n =
0
sin x
(5 + 3 cos x)
∫
0
−1
(1 + u )
n n
+
2 2
3
in the above integral,
2
3
∫
n
2
u2
∞
3
2
dx =
22
−1
3
2 4
(52 − 3 )
⎛3 3⎞
B⎜ , ⎟
⎝4 4⎠
2
⎛ 3⎞
3 3
2⎜ ⎟
⎝ 4⎠
2 4 4
=
3
2
3
1 1
23 ⋅
2 2
2
=
⎛ 3⎞
⎜ 4⎟
⎝ ⎠
2 2
2
du =
2n −1
n
2 2
(a 2 − b )
⎛n
B⎜ ,
⎝2
n⎞
⎟
2⎠
7.30
Engineering Mathematics
Example 7: Prove that
∫
π
2
cos 2 m −1 sin 2 n −1
Β (m , n )
d = 2m 2n .
2a b
(a 2 cos 2 + b 2 sin 2 )m + n
0
π
2
0
cos 2 m −1 θ sin 2 n −1 θ
dθ
(a 2 cos 2 θ + b 2 sin 2 θ )m + n
π
2
0
cos 2 m−1 θ sin 2 n −1 θ (cosθ )−2 m− 2 n
dθ
(a 2 + b 2 tan 2 θ )m+ n
∫
Solution:
=
=
b2
tan 2
a2
Let
∫
= t,
q=
π
2
0
∫
∫
a 2( m+ n )
(tan θ )2 n −1 sec 2 θ
⎛ b2
2 ⎞
⎜⎝1 + a 2 tan θ ⎟⎠
a
t,
b
tan q =
q = 0,
When
π
2
0
1
m+ n
dθ
sec2 q dq =
a
2b
1
t
dt
t=0
,
2
t
cos 2 m −1 θ sin 2 n −1 θ
1
dθ = 2 ( m + n )
(a cos 2 θ + b 2 sin 2 θ ) m + n
a
2
∫
2a b
2n
2 n −1
(1 + t ) m + n
0
1
2m
⎛ a 12 ⎞
⎜ t ⎟⎠
∞ ⎝b
∫
∞
0
⋅
a 1
dt
⋅
2b t
t n −1
dt
(1 + t ) m + n
1
B(m, n)
2a 2 m b 2 n
Example 8: Prove that B(m, n) =
Solution: We have B(m, n) =
∞
∫
0
∫
1
0
x m −1 + x n −1
dx .
(1 + x )m + n
x m −1
dx
(1 + x) m + n
∞
x m −1
x m −1
x
d
+
∫0 (1 + x)m + n
∫1 (1 + x)m + n dx
1
Consider,
Let
When
I=
∫
∞
1
1
,
y
x =1,
x=
x
,
x m −1
dx
(1 + x) m + n
dx =
y=1
y=0
1
dy
y2
… (1)
Gamma and Beta Functions
I=
⎛1⎞
⎜⎝ y ⎟⎠
0
∫
m −1
⎛ 1⎞
⎜⎝1 + y ⎟⎠
1
7.31
⎛ 1 ⎞
⎜⎝ − y 2 ⎟⎠ dy =
m+n
y n −1
∫0 ( y + 1)m + n dy
1
Substituting in Eq. (1),
B(m, n) =
1
x m −1
y n −1
x
d
+
∫0 (1 + x)m + n
∫0 (1 + y)m + n dy
B(m, n) =
∫
1
Replacing y by x,
1
x m −1
x n −1
dx + ∫
dx
m
+
n
0 (1 + x )
0 (1 + x ) m + n
1
∫
1
0
Example 9: Prove that
Solution: Let tan
2
= t,
∫
π
2
0
2
x m −1 + x n −1
dx
(1 + x) m + n
dθ
1
1 − sin 2 θ
2
=
π
2
0
∫
2
dt
1+ t2
0,
π
θ= ,
2
t=0
=
1
dθ
1
1 − sin 2 θ
2
∫
t=1
1
0
1−
2∫
1
0
1 −3
Let t4 = u, t = u , dt = u 4 du
4
When
.
2t
1+ t2
q
When
2
4
= tan 1 t , d =
sin q =
⎛ 1⎞
⎜ 4⎟
⎝ ⎠
1 ⎛ 2t ⎞
⎜
⎟
2 ⎝1 + t 2 ⎠
1
1
4 2
(1 + t )
1
4
t = 0,
u=0
t = 1,
u=1
dt
2
⋅
2
dt
1+ t2
7.32
Engineering Mathematics
π
2
0
∫
dθ
1−
1
sin 2 θ
2
= 2∫
1
0
(1 + u ) 2
1
1
=
2
1 −3
⋅ u 4 du
4
1
1
∫
1
0
u4
1
−1
1
du =
4
1
(1 + u ) 2
1 ⎛1
B⎜ ,
4 ⎝4
∫
1
0
1
−1
u4 + u4
1
−1
+
(1 + u ) 4
1⎞
⎟
4⎠
du
[From Ex. 8]
⎛ 1⎞
⎜⎜ ⎟⎟
⎝ 4⎠
1 1
1 4 4
=
4 1
2
1
4
2
4
Exercise 7.5
1. Evaluate
∫
∞
0
4. Prove that
dy
.
1+ y4
⎤
⎡
⎢ Ans. : 2 2 ⎥
⎦
⎣
2. Prove that
m −1
∞ x
+ x n −1
∫0 (1 + x)m + n dx = 2B (m, n).
∫
0
∞
0
dx
1
⎛n
B⎜ ,
=
−x n
⎝2
4
(e + e )
x
and hence, evaluate
5. Prove that
3. Prove that
∞
∫
x
dx =
.
(4 + 4 x + x 2 )
4 2
∫
∞
1
x
p +1
∫
∞
0
n⎞
⎟
2⎠
sec h 8 x dx.
16 ⎤
⎡
⎢⎣ Ans. : 35 ⎥⎦
dx
( x − 1) q
= B(p + q, 1 – q), if
p < q < 1.
FORMULAE
Gamma Function
∞
(i) n = ∫ e − x x n − 1dx, n > 0
0
(ii)
∞
n = 2 ∫ e − x x 2 n − 1dx
n +1
(iii) n =
, if n is a negative
fractionn
2
0
Properties of Gamma Function
(i) n + 1 = n! , if n is a positive integer
(ii) n + 1 = n n , if n is a positive real
number
(iv) n 1 − n
(v)
1
= π
2
π
sin nπ
Gamma and Beta Functions
Beta Function
∫
(i) B (m, n) =
m > 0, n > 0
1
0
x m − 1 (1 − x ) n − 1 dx,
Properties of Beta Function
(i) B (m, n) = B (n, m)
2 m −1
x cos 2 n −1 x dx
(ii) B (m, n) = 2 ∫0 sin
∫
∞
0
mn
m+n
(ii) B (m, n) =
π
2
(iii) B (m, n) =
7.33
xm − 1
dx
(1 + x ) m + n
(iii)
m m+
1
π 2m
= 2 m −1
2
2
(Duplication formula)
MULTIPLE CHOICE QUESTIONS
Choose the correct alternative in each of the following:
1. The value of the integral
I=
1
2p
∞
∫e
⎛ − x2 ⎞
⎜⎝
⎟
8 ⎠
dx is
0
(a) 1
(b) p
(c) 2
(d) 2p
2. Match the items in columns I and II
for the following special functions
I
II
(P) b (p, q)
1
2
(i)
∞
(Q)
(R)
(S)
(a)
(b)
(c)
(d)
y p −1
p q
dy
(ii) ∫
(1 + y ) ( p + q )
p+q
0
(iii) b (p, q)
p
p
(iv) p 1 p
sin pp
P-(iv), Q-(iii), R-(i), S-(ii)
P-(ii), Q-(iii), R-(i), S-(iv)
P-(iii), Q-(ii), R-(i), S-(iv)
P-(ii), Q-(iii), R-(iv), S-(i)
∞
3. The value of
∫
3
y e − y dy is
0
p
3
p
(c) p
(d)
6
4. The value of B (m + 1, n) is
n
(a)
B (m, n)
m+n
(a)
p
2
(b)
n
B (m, n)
m +1
m B (m, n)
(c)
m+n
m B (m, n)
(d)
m +1
(b)
5. The value of
p
2
cot q dq is
0
(a)
p 2
2
(b)
p
2
(c)
p 2
4
(d)
p
4
()
3
⎡
1 ⎤
6. The value of ∫ x ⎢log
dx is
0
x ⎥⎦
⎣
(a) 3
(b) 6
325
625
3
(c)
(d) 6
625
325
1
4
7. If B (n, 2) = 1 and n is a positive
6
integer, then the value of n is
(a) 3
(b) – 2
(c) 2
(d) – 3
8. The value of
(a)
p
2
(c) p
2
∫
∞
0
t 2 dt
is
1 + t4
(b)
p
2
(d) p
4
7.34
Engineering Mathematics
9. The value of B (m, m) is
1
(a) 21 – 2m B m,
2
(b) 21 – 2m
(c) 21 – 2m
(d) 21 – 2m
Answers
1. (a)
8. (a)
( )
1
B ( m + 1, )
2
1
B ( m + , 1)
2
3
B ( m, )
2
2. (b)
9. (a)
3. (b)
10. (d)
10. Gamma function is discontinuous for
(a) all p < 0
(b) any p > 0
(c) p = 0 only
(d) p = 0 and negative integers
11. Beta function B (p, q) is convergent for
(a) p > 0, q < 0
(b) p > 0, q > 0
(c) p < 0, q > 0
(d) p < 0, q < 0
4. (c)
11. (b)
5. (a)
6. (b)
7. (c)
Multiple Integral
Chapter
8
8.1 INTRODUCTION
Integration of functions of two or more variables is normally called multiple integration. The particular case of integration of functions of two variables is called double
integration and that of three variables is called triple integration. Sometimes, we have
to change the variables to simplify the integrand while evaluating the multiple integrals.
Variables can be changed by substitution or by changing the co-ordinate system (polar,
spherical or cylindrical coordinates). Integrals can also be solved easily by expressing
them in terms of beta and gamma functions. Multiple integrals are useful in evaluating
plane area, mass of a lamina, mass and volume of solid regions, etc.
8.2 DOUBLE INTEGRAL
Let f (x, y) be a continuous function defined in a closed and bounded region R in the
xy-plane. Divide the region R into small elementary rectangles by drawing lines parallel to co-ordinate axes. Let the total number of complete rectangles which lie inside
the region R is n. Let Ar be the area of rth rectangle and (xr, yr) be any point in this
rectangle.
n
Consider the sum
S=
f ( xr , yr ) Ar
... (1)
r =1
where Ar = xr · yr.
If we increase the number of elementary rectangles, i.e.,
n, then the area of each rectangle decreases. Hence, as
n
, Ar
0. The limit of the sum given by the
Eq. (1), if it exists, is called the double integral of f (x, y)
over the region R and is denoted by
f ( x, y ) dA.
R
Hence,
∫∫
R
where dA = dx dy
f ( x, y ) dA = lim
n
∑ f ( x , y )δA
n→ ∞
δAr → 0 r =1
r
r
r
Fig. 8.1
8.2
Engineering Mathematics
8.2.1 Evaluation of Double Integral
Double integral of a function f (x, y) over region R can be evaluated by two successive
integrations. There are two different methods to evaluate a double integral.
Method-I Let the region R, i.e., PQRS is bounded by the curves y = y1(x), y = y2(x)
and the lines x = a, x = b.
In the region PQRS, draw a vertical strip
AB. Along the strip AB, y varies from y1 to y2
and x is fixed. Therefore, the double integral
is integrated first w.r.t. y between the limits
y1 and y2 treating x as constant.
Now, move the strip AB from PS (i.e.,
x = a) to QR (i.e., x = b) to cover the entire region PQRS. The result of the first integral is
integrated w.r.t. x between the limits a and b.
Hence,
f ( x , y ) dx dy =
R
b
y2
a
y1
f ( x , y ) dy dx
Fig. 8.2
Method-II Let the region R is bounded
by the curves x = x1(y), x = x2(y) and the lines
y = c, y = d.
In the region PQRS, draw a horizontal
strip AB. Along the strip AB, x varies from
x1 to x2 and y is fixed. Therefore, the double
integral is integrated first w.r.t. x between
the limits x1 and x2 treating y as constant.
Now, move the strip AB from PQ (i.e., y
= c) to RS (i.e., y = d ) to cover the entire region PQRS. The result of the first integral is
integrated w.r.t. y between the limits c and d.
f ( x, y ) dx dy =
Hence,
R
d
x2
c
x1
Fig. 8.3
f ( x, y ) dx dy
Note:
(i) If all the four limits are constant, then the function f (x, y) can be integrated w.r.t.
any variable first. But if f (x, y) is implicit and is discontinuous within or on the
boundry of the region of integration, then the change of the order of integration
will affect the result.
(ii) If all the four limits are constant and f (x, y) is explicit, then double integral can be
written as product of two single integrals.
Multiple Integral
Example 1: Evaluate
Solution:
3
1
0
0
3
1
0
0
8.3
( x 2 + 3 y 2 ) dy dx .
( x 2 + 3 y 2 )dy dx =
3
0
1
x 2 y + y 3 0 dx
3
3
=
0
( x 2 + 1) dx =
x3
+x
3
0
= 12
x
1
Example 2: Evaluate
0
0
Solution:
y
x
e dy dx .
1
x
0
0
y
1
e x dy dx =
0
y x
xe x dx
0
1
1
0
x(e 1) dx
x2
(e 1)
2
0
1
(e 1)
2
⎡ 1+ x 2
Solution: ∫0 ⎢ ∫0
⎢
⎣
1+ x2
1
Example 3: Evaluate
0
0
dx dy
.
1 + x2 + y2
1+ x
⎤
1
y
1
−1
⎥ dx =
2
∫0 1 + x 2 tan 1 + x 2 dx
⎥
0
1+ x2 ) + y2 ⎦
1
1
(tan 1 1 tan 1 0) dx
0
1+ x2
1
1
. dx = log x + 1 + x 2
=
0
2 4
4
1+ x
2
dy
1
(
(
=
Solution:
0
0
1 y2
2
1
0
1- y2
2
1
Example 4: Evaluate
0
4
dx dy
1 - x2 - y2
dx dy
1 x
2
y
2
=
=
log (1 + 2)
.
1 y2
2
1
0
1
0
0
sin
dx
(1 y 2 ) x 2
1
1 y2
2
x
1 y2
dy
0
dy
)
1
0
8.4
Engineering Mathematics
1⎛
1
⎞
= ∫ ⎜ sin −1
− sin −1 0 ⎟ dy
0⎝
⎠
2
1
=
Example 5: Evaluate
Solution:
cos 2
4
0
1
2
1
2
0
4
0
4
0
1
2r
dr d
2 (1 + r 2 ) 2
cos 2
4
0
y0 =
r
d dr .
(1 + r 2 )2
cos 2
4
0
4
0
(1 r 2 )
(1 r 2 )
2
cos 2
1
0
2 r dr d
⎡
⎤
n +1 ⎥
⎢
n
f
(
r
)
[
]
⎢∵ [ f ( r ) ] f ′ ( r )dr =
⎥
⎢ ∫
⎥
n +1
⎢
⎥
n ≠ −1
⎣
⎦
d
=−
1 π4 ⎛
1
⎞
− 1⎟ dθ
⎜
2 ∫0 ⎝ 1 + cos 2θ ⎠
=−
1 π4 ⎛ 1
1 π4 ⎛ 1 2
⎞
⎞
1
θ
−
d
=
−
⎜⎝
⎟⎠
⎜⎝ sec θ − 1⎟⎠ dθ
2
∫
∫
0
0
2
2
2
2 cos θ
1 1
=−
tan θ − θ
2 2
π
4
0
1 ⎛1
π π
⎞
= − ⎜ tan − − 0 ⎟
⎝
⎠
2 2
4 4
1
= (π − 2)
8
Example 6: Evaluate
e
0
0
x3
6
y 4 e - y dx dy .
x
Solution: Since both the limits are constant and integrand is explicit in x and y,
integral can be written as, I
Let x 3 = p,
0
e
x3
1
( x ) 2 dx
1 − 32
p dp,
3
When x = 0, p = 0
x → ∞, p → ∞
dx =
e
y6
y 4 dy
y6 = q
1
x = p3 ,
0
1
y = q6
1 − 65
q dq
6
When y = 0, q = 0
dy =
y → ∞, q → ∞
∞
I = ∫ e− p p
0
∞
=
−
1
6
∞
4
1 −2
1 −5
⋅ p 3 d p ⋅ ∫ e − q q 6 ⋅ q 6 dq
3
6
0
∞
5
1
−1
−1
1
e− p p 6 d p ⋅ ∫ e−q q 6 d q
∫
18 0
0
Multiple Integral
1 1
18 6
8.5
5
6
⎡∵ ∞ e − x x n −1dx = n⎤
⎥⎦
⎣⎢ ∫0
1
5 5
1
18
6 6
1
= .
18 sin 5
6
=
Example 7: Show that
⎡
⎤
⎢∵ n 1− n = sin n ⎥
⎣
⎦
9
1
0
dx
1
0
x- y
dy
( x + y )3
1
0
1
dy
0
x- y
dx .
( x + y )3
Solution: Consider,
L.H.S =
1
dx
0
R.H.S =
2x
( x y )3
1
1
0
0
1
x
( x + 1) 2
0
=
x y
dy =
0 ( x + y )3
1
1
0
dx
1
(x
0
2 x ( x y)
dy
( x + y )3
1
dy dx
y)2
1
1
x +1 x
1
1
dx
x
0
1
2( x y ) 2
2x
1
1
dx
x+ y 0
1
1
1
0
( x + 1)2
dx
1
x +1 0
1
2
1
0
dy
1
1
0
0
1
0
1
0
1
0
x y
dx =
( x + y )3
dy
1
0
( x y) 2 y
dx
( x + y )3
2y
dx dy
( x + y )3
1
( x + y)2
1
x+ y
1
0
1
y
dy
( x + y)2 0
1
dy
(1 + y ) 2
1
1+ y
1
0
1
1+ y
y
(1 + y ) 2
1
y
1
dy
y
1
0
1
2
Hence,
1
0
dx
x y
dy
0 ( x + y )3
1
1
0
dy
1
0
x y
dx
( x + y )3
x y
is discontinuous at (0, 0), a point on the boundary of the region (square),
( x + y )3
change of order of integration does not give the same result.
Since
8.6
Engineering Mathematics
Exercise 8.1
Evaluate the following:
1.
2
2 y
1
2 y
2 x 2 y 2 dx dy
5.
Ans. :
2.
1
0
y
0
xy e
dx dy
1
Ans. :
4e
3.
1
x
0
0
10 0
ye xy dx dy
[Ans. : 9(1 e)]
856
945
6.
x 2
1
y
1
log8
log y
1
0
e x + y dx dy
[Ans. : 8(log8 1)]
7.
1
y
0
y2
(1 + xy 2 ) dx dy
e x + y dx dy
1
Ans. : (e 1) 2
2
4.
2
0
a (1+ sin )
0
8.
2
r cos d dr
0
Ans. :
5a3
4
2 ax x 2
2a
0
Ans. :
41
210
Ans. :
2a 4
3
xy dy dx
8.2.2 Working Rule for Evaluation of Double Integral
1. If a region is bounded by more than one curve, then find the points of intersection
of all the curves.
2. Draw all the curves and mark their points of intersection.
3. Identify the region bounded by all the curves.
4. Draw a vertical or horizontal strip in the region whichever makes the integration
easier.
5. Find the variation of y (or x) along the strip and variation of x (or y) in the region.
6. Write the limits of y and x. Lower limit is always obtained from the curve where the
strip starts and upper limit is always obtained from the curve where it terminates.
7. The function is integrated first along the strip (i.e., w.r.t. y first for vertical strip
and w.r.t. x for horizontal strip.)
8. Variation along vertical strip is always taken from lower part to upper part and
along horizontal strip is always taken from left part to right part of the region.
9. If variation along the strip changes within the region, then the region is divided
into parts.
Example 1: Evaluate
ax + by = 1.
e ax + by dx dy , over the triangle bounded by x = 0, y = 0,
Multiple Integral
8.7
Solution:
1. The region of integration is the OPQ.
2. The integration can be done w.r.t. any variable first. Draw a vertical strip AB parallel
to y-axis which starts from x-axis and terminates on the line ax + by = 1.
1 ax
3. Limits of y : y = 0 to y =
b
1
Limits of x : x = 0 to x =
a
1
I = ∫ ∫ e ax + by dx dy = ∫ a e ax ∫
0
1
= ∫ a e ax
0
1− ax
by
b
e
b
=
0
1− ax
b
0
Fig. 8.4
e by dy dx
1 1a ax (1− ax )
e ⎡⎣e
− 1⎤⎦ dx
b ∫0
1 1
1
e ax
= ∫ a (e − e ax )dx = ex −
b 0
b
a
1
a
=
0
1 ⎛e e 1⎞
⎜ − + ⎟
b ⎝a a a⎠
1
=
ab
xy
Example 2: Evaluate
x 2 + y 2 = 1.
1 - y2
dx dy over the first quadrant of the circle
Solution:
1. The region of integration is OPQ.
2. The integration can be done w.r.t. any
variable first. Draw a vertical strip AB
parallel to y-axis which starts from x-axis
and terminates on the circle x 2 + y 2 = 1.
3. Limits of y : y = 0
Limits of x : x = 0
xy
I=
1 y2
1
0
x
1
2
1 x2
0
1
0
to y
1 x2
to x = 1
dx dy
1
1
(1 y 2 ) 2 ( 2 y )dy dx
2
1 x2
1
2 2
x 2(1 y )
dx
0
Fig. 8.5
⎡
[ f ( x )]n +1 ⎤
n
∵
[
f
(
x
)]
f
(
x
)
d
x
=
′
⎢ ∫
⎥
n +1 ⎦
⎣
8.8
Engineering Mathematics
1
2
1
0
2 x ( x 1) dx
1
3
=
x3
3
x2
2
1
0
1
2
1
6
( a - x )2 dx dy , over the right half of the circle x 2 + y 2 = a 2 .
Example 3: Evaluate
Solution:
1. The region of integration is PQR.
2. The integration can be done w.r.t. any
variable first. Draw a vertical strip AB
parallel to y-axis which starts from the
part of the circle x 2 + y 2 = a 2 below
x-axis and terminates on the part of the
circle x 2 + y 2 = a 2 above x-axis.
3. Limits of
a2
y:y
x 2 to y
Limits of x : x = 0
a2
x2
to x = a
I = ∫ ∫ ( a − x ) 2 dx dy
a
a2 − x 2
0
− a2 − x 2
= ∫ (a − x)2 ∫
dy dx
a
a2 − x 2
0
− a2 − x 2
= ∫ (a − x)2 y
dx
a
= ∫ ( a 2 + x 2 − 2ax ) 2 a 2 − x 2 dx
0
a
= 2 ∫ ( a 2 + x 2 − 2ax ) a 2 − x 2 dx
Fig. 8.6
0
Putting x = a sinq, dx = a cosq dq
When x = 0,
q =0
=
x = a,
2
π
I = 2 ∫ 2 ( a 2 + a 2 sin 2 θ − 2a 2 sin θ ) a cosθ ⋅ a cosθ dθ
0
π
= 2a 4 ∫ 2 (cos 2 θ + sin 2 θ cos 2 θ − 2 sin θ cos 2 θ )dθ
0
Multiple Integral
⎡ 3 1 3 3
3⎤
1 ⎥
⎢
= a4 ⎢ 2 2 + 2 2 − 2 2 ⎥
3
5 ⎥
⎢ 2
⎢⎣
2 ⎥⎦
∵2
a4
2
5
8
8
sin p cos q d
p +1 q +1
,
2
2
p +1 q +1
2
= 2
p+q+2
2
4
3
4
3
x2 y2
xy 2 + 2
a
b
Example 4: Evaluate
2
0
=B
2
⎡
⎤
⎛1 1⎞
3 ⎥
⎢1 1 1 ⎜
⎟
⎢
⎥
⎝2 2⎠
= a4 ⎢ 2 2 2 +
−2 2 ⎥
1
2!
3 3⎥
⎢
⎢
2 2⎥
⎣
⎦
a4
8.9
n
2
dx dy, over the first quadrant of the ellipse
x2 y2
+
= 1.
a 2 b2
Solution:
1. The region of integration is OPQ.
2. The integration can be done w.r.t. any variable first. Draw a vertical strip AB parallel to y-axis which starts from x-axis and
2
2
terminates on the ellipse x + y = 1.
2
2
a
b
x2
3. Limits of y : y = 0 to y b 1 2
a
Limits of x : x = 0 to x = a
n
⎛ x2 y2 ⎞ 2
I = ∫ ∫ xy ⎜ 2 + 2 ⎟ dx dy
b ⎠
⎝a
a
b 1−
0
0
= ∫ x∫
x2
a2
Fig. 8.7
n
2
b2 ⎛ x 2 y 2 ⎞ 2 y
+
dy dx
2 ⎜⎝ a 2 b 2 ⎟⎠ b 2
b 1−
n
b2
=
2
∫
a
0
⎛ x2 y2 ⎞ 2
+
x
⎛ n ⎞ ⎜⎝ a 2 b 2 ⎟⎠
⎜⎝ + 1⎟⎠
2
1
x2
a2
+1
dx
0
⎡
[ f ( y)]n+1 ⎤⎥
n
⎢∵ ∫ [ f ( y ) ] f ′( y ) dy =
n + 1 ⎥⎦
⎢⎣
8.10
Engineering Mathematics
b2
( n + 2)
a
0
b2 a2
n+2 2
=
x
a
x 1
1
an+ 2
n+ 2
b2 x 2
( n + 2) 2
dx
an+ 4
n+4
1
an+ 2
a
x n+ 4
n+4 0
a 2 b 2 ( n + 2)
( n + 2) 2( n + 4)
a2 b2
2( n + 4)
Example 5: Evaluate
( x 2 + y 2 ) dx dy over the ellipse 2 x 2 + y 2 = 1.
Solution:
1. The region of integration is PQRS, the ellipse 2 x 2 + y 2 = 1 or
x2
1
2
1
2
2
+
y2
= 1 with
12
and 1 as axes.
2. The integration can be done w.r.t. any variable first. Draw a vertical strip AB parallel to y-axis which starts from the part of the ellipse 2 x 2 + y 2 = 1 below x-axis and
terminates on the part of the ellipse 2 x 2 + y 2 = 1 above x-axis.
Fig. 8.8
3. Limits of y : y
Limits of x : x
1 2 x 2 to y
1
2
to x
1
2
1 2x 2
Multiple Integral
1
I=
( x 2 + y 2 ) dx dy =
2
1
1 2 x2
1 2 x2
8.11
( x 2 + y 2 ) dy dx
2
1
y3
3
x2 y
2
1
2
1 2 x2
1 2 x2
1
4
2
0
x2 1 2x2
Putting 2 x 2 = t , x =
When
1
dx
2
1
2 x2 1 2x2
2
3
1
(1 2 x 2 ) 2 dx
3
3
1
(1 2 x 2 ) 2 dx
3
t
1
, dx =
dt
2
2 2 t
x = 0, t = 0
1
x=
, t =1
2
3
1⎡ t
⎤ 1
1
I = 4 ∫ ⎢ 1 − t + (1 − t ) 2 ⎥
dt
0 2
3
⎣
⎦2 2 t
3
1
1
1⎡1
⎤
⎡ 1 ⎛ 3 3 ⎞ 1 ⎛ 1 5 ⎞⎤
1 −1
= 2 ∫ ⎢ t 2 (1 − t ) 2 + t 2 (1 − t ) 2 ⎥ dt = 2 ⎢ B ⎜ , ⎟ + B ⎜ , ⎟ ⎥
0 2
3
⎣ 2 ⎝ 2 2 ⎠ 3 ⎝ 2 2 ⎠⎦
⎣
⎦
⎡ ⎛ 1 1 ⎞2
⎤
⎡
1 3 1 1⎥
3 3
1 5⎤
⎢ ⎜
⎟
⋅
⋅
⎢1
⎥
⎢1 ⎝ 2 2 ⎠ 1 2 2 2 2 ⎥
1
= 2⎢ ⋅ 2 2 + ⋅ 2 2⎥ = 2⎢ ⋅
+ ⋅
⎥
2
2
3
3 3 ⎥
⎢2 3
⎢2
⎥
⎢⎣
⎥⎦
⎢
⎥
⎣
⎦
⎤ 3 2
⎡1
= 2⎢ ⋅ + ⎥=
4
4
8
16
⎦
⎣
Example 6: Evaluate
(1, 1), (1, 2).
( x 2 - y 2 ) dx dy over the triangle with vertices (0, 1),
Solution:
1. The region of integration is PQR.
2. Equation of the line PQ is y = 1.
Equation of the line PR is
2 1
y 1
( x 0) x
1 0
y = x +1
3. The integration can be done w.r.t.
any variable first. Draw a vertical
strip AB parallel to y-axis which
starts from the line y = 1 and terminates on the line y = x + 1.
Fig. 8.9
8.12
Engineering Mathematics
to y = x + 1
to x = 1
4. Limits of y : y = 1
Limits of x : x = 0
I = ∫ ∫ ( x 2 − y 2 ) dx dy = ∫
1
∫
x +1
0 1
=∫
2
0
1
1⎡
( x + 1)3
1⎤
dx = ∫ ⎢ x 2 ( x + 1) −
− x 2 + ⎥ dx
0
3
3⎦
⎣
1
x
x ( x + 1) 4 x 3 x
1 1 16 1
+ −
− +
= + − +
4
3
12
3 3 0 4 3 12 12
4
=
x +1
y3
x y−
3
1
( x 2 − y 2 ) dy dx
=−
3
2
3
2
e y dx dy over the region bounded by the triangle with
Example 7: Evaluate
vertices (0, 0), (2, 1), (0, 1).
Solution:
1. The region of integration is
OPQ.
2. Equation of the line OQ is
x
y = or x = 2 y.
2
3. Here, it is easier to integrate
w.r.t. x first than with y. Draw
a horizontal strip AB parallel to
x-axis which starts from y-axis
and terminates on the line
x = 2y.
4. Limits of x : x = 0 to x = 2y
Limits of y : y = 0 to y = 1
1
2
e y dx dy =
0
1
0
ey
2
2y
0
dx dy =
Fig. 8.10
1
0
2
2y
e y x 0 dy
2
e y 2 y dy
= ey
2
1
⎡ ∵ e f ( y ) f ′ ( y ) dy = e f ( y ) ⎤
⎣ ∫
⎦
0
e 1
Example 8: Evaluate
(0, 0), (1, 1) and (0, 1).
2 xy 5
1 + x2 y2 - y4
dx dy over the triangle having vertices
Multiple Integral
8.13
Solution:
1. The region of integration is the OPQ.
2. Equation of the line OP is y = x.
3. Here, it is easier to integrate w.r.t. x first than
with y. Draw a horizontal strip AB parallel
to x-axis which starts from y-axis and terminates on the line y = x.
4. Limits of x : x = 0 to x = y
Limits of y : y = 0 to y = 1
2 xy 5
I=
1 x2 y2
1
0
1
0
y3
y
0
y4
(1 x 2 y 2
dx dy
y4 )
1
2
1 y
4 2
y 3 2(1 x 2 y 2
y )
2 xy 2 dx dy
Fig. 8.11
n
∵ [ f ( x )] f ( x )dx =
dy
0
[ f ( x )]n +1
n +1
1
⎡
⎤
= ∫ y 3 ⋅ 2 ⎢1 − (1 − y 4 ) 2 ⎥ dy
0
⎣
⎦
1
1
1
1
= ∫ 2 y 3 dy − 2 ∫ (1 − y 4 ) 2 y 3 dy
0
0
4 1
=2
=
y
4
1
1
− 2 ∫ (1 − y 4 ) 2 y 3 dy
0
0
1
1 1 1
+ ∫ (1 − y 4 ) 2 ( − 4 y 3 ) dy
2 2 0
3
1 1 2
+
(1 − y 4 ) 2
2 2 3
1 1 1
= − =
2 3 6
1
⎡
[ f ( x )]n +1 ⎤
n
⎥
⎢∵ ∫ [ f ( x )] f ′( x )d x =
n +1 ⎦
⎣
=
Example 9: Evaluate
0
y dx dy
( a - x ) ax - y 2
over the region bounded by the
parabola y 2 = x and the line y = x.
Solution:
1. The region of integration is OPQ.
2. The points of intersection of y 2 = x and y = x are obtained as
x2 = x
x( x 1) 0
x = 0, 1 and y = 0, 1
Hence O : (0, 0) and P : (1, 1)
8.14
Engineering Mathematics
3. Here, it is easier to integrate w.r.t. y first than with
x. Draw a vertical strip AB parallel to y-axis, which
starts from the line y = x and terminates on the
parabola y 2 = x.
4. Limits of y : y = x to y = x
Limits of x : x = 0 to x = 1.
I = ∫∫
y dx dy
( a − x ) ax − y 2
1
x ⎛ 1⎞
−
1
− ⎟ ( ax − y 2 ) 2 ( −2 y )dy dx
⎜
∫
0 (a − x) x ⎝
2⎠
=∫
1
=−
1
1 1 1
2 ( ax − y 2 ) 2
∫
2 0 (a − x)
Fig. 8.12
⎡
[ f ( y)]n+1 ⎤⎥
n
⎢∵∫ [ f ( y ) ] f ′( y )dy =
n + 1 ⎥⎦
⎢⎣
x
dx
x
1
1 ⎡
2 2⎤
−
−
−
(
ax
x
)
(
ax
x
)
⎢
⎥ dx
0 (a − x)
⎣
⎦
= −∫
1
= −∫
1
0
1
2
x
a− x
(
)
a − 1 − a − x dx
Putting x = a sin 2 , dx = 2a sin cos d
When
x = 0, q = 0
x = 1,
sin −1
= sin
1
1
a
1
(
a sin
a cos 2
1
sin −1
sin 2
a
= −2 a ∫
0
cos
I = −∫
a
0
= −2 a ∫
sin −1
1
a
0
= −2 a ∫
sin −1
0
1
a
a − 1 − a cos
(
) 2a sin
a − 1 − a cos
cos d
)d
⎡ ⎛ 1 − cos 2 ⎞
⎤
a − 1 − a sin 2 ⎥ d
⎢⎜
⎟
⎠
⎣ ⎝ cos
⎦
⎡
a
(1 − cos 2
⎢ a − 1 (sec − cos ) −
2
⎣
⎤
)⎥ d
⎦
sin −1
= −2 a
= −2 a
a
a sin 2
a − 1 [log (sec + tan ) − sin ] −
+
2
4
⎤
⎡ ⎛ 1 + sin ⎞
a
a sin cos
+
a − 1 ⎢log ⎜
⎟⎠ − sin ⎥ −
⎝
cos
2
2
⎦
⎣
1
a
0
sin −1
0
1
a
Multiple Integral
8.15
⎤
⎡
⎛
⎞
1
1
a sin −1
⎥
⎢
⎜ 1+ a
⎟
1
1
1
a
a
= −2 a ⎢ a − 1 ⎜ log
−
+
⋅
⋅ 1− ⎥
⎟−
⎢
2
2
a⎥
1
a⎟
a
⎜
1−
⎜⎝
⎟⎠
⎢
⎥
a
⎦
⎣
= −2 a( a − 1) log
a +1
a −1
Example 10: Evaluate
+ a − 1 + a sin −1
1
a
y dx dy over the region enclosed by the parabola
x2 = y and the line y = x + 2.
Solution:
1. The region of integration is POQ.
2. The points of intersection of x 2 = y and y
= x + 2 are obtained as
x 2 x 2, x 2 x 2 0
( x 2) ( x 1)
x
0
2, 1and y
4, 1
Hence, P : (–1, 1) and Q : (2, 4)
Fig. 8.13
3. Here, integration can be done w.r.t. any
variable first. Draw a vertical strip AB parallel to y-axis which starts from the
parabola x 2 = y and terminates on the line y = x + 2.
4. Limits of y : y = x 2 to y = x + 2
Limits of x : x = 1 to x = 2
I = ∫ ∫ y dx dy = ∫
=∫
=
2
−1
y2
2
2
∫
x+2
−1 x 2
x+2
dx =
x2
1 ( x + 2)3 x 5
−
2
3
5
y dy d x
1 2
⎡( x + 2) 2 − x 4 ⎤⎦ dx
2 ∫−1 ⎣
2
=
−1
1 ⎛ 64 32 1 1 ⎞
⎜ − − − ⎟⎠
2⎝ 3
5 3 5
36
=
5
Example 11: Evaluate
2
xy ( x + y ) dx dy , over the region enclosed by the par-
2
abolas x = y , y = - x .
Solution:
1. The region of integration is OPQ.
2. The points of intersection of the parabola x 2 = y, and y 2 = − x are obtained as
y 4 y, y 0, 1 and x 0, 1.
Hence, O : (0, 0) and Q : ( 1, 1)
8.16
Engineering Mathematics
Fig. 8.14
3. Here, it is convenient to integrate w.r.t. x first. Draw a horizontal strip AB parallel
to x-axis, which starts from the parabola x 2 = y and terminates on the parabola
y2
x.
4. Limits of x : x = − y to x = − y 2
Limits of y : y = 0 to y = 1
I = ∫ ∫ xy ( x + y ) dx dy
1
− y2
1
x3 x 2 y
= ∫ y ∫ ( x + xy ) d x d y = ∫ y
+
3
2
0 − y
0
− y2
2
1
=∫
0
dy
− y
5
7
⎛ 7
6
3 ⎞
8
7
2
2
−
y
y
y
y
y
y
2
y
y4
⎜
+
+
− ⎟ dy = −
+
+
−
⎝ 3
2
3
2⎠
24 14
21
8
Example 12: Evaluate
1
=0
0
xy dx dy over the region enclosed by the x-axis, the
line x = 2a and the parabola x 2 = 4ay.
Solution:
1. The region of integration is OPQ.
2. The point of intersection of the parabola
x 2 = 4 ay and the line x = 2a is obtained as
4 a 2 = 4 ay, y = a.
Hence, Q : (2a, a)
3. Here, integration can be done w.r.t. any variable first. Draw a vertical strip AB parallel to
y-axis, which starts from x-axis and terminates on the parabola x 2 = 4ay.
Fig. 8.15
Multiple Integral
8.17
x2
4a
to x = 2a
4. Limits of y : y = 0 to y =
Limits of x : x = 0
2a
I = ∫ ∫ xy dx dy = ∫ x ∫
0
x2
4a
0
y d y dx
2
=∫
2a
0
y2
x
2
1 x6
=
32a 2 6
x
4a
2a
dx = ∫ x ⋅
0
0
2a
=
0
x4
dx
32 a 2
1 64 a6
⋅
32a 2 6
a4
=
3
Example 13: Evaluate
2
xy dx dy , over the region enclosed by the circle
2
x + y - 2 x = 0, the parabola y 2 = 2 x and the line y = x.
Solution:
1. The region of integration is OPQRO.
2. The points of intersection of
(i) the circle x 2 y 2 2 x 0
and the line y = x are obtained as
x 2 + x 2 − 2 x = 0, x = 0, 1 and y = 0, 1
Hence, O : (0, 0) and P : (1, 1)
(ii) the circle x 2 y 2 2 x 0 and the
parabola y 2 = 2 x are obtained as
x 2 2 x 2 x 0, x 0 and y = 0
Hence, O : (0, 0)
(iii) the parabola y 2 = 2 x and the line y = x are
obtained as x 2 = 2 x, x = 0, 2 and y = 0, 2
Hence, O : (0, 0) and Q : (2, 2).
Fig. 8.16
3. Here, integration can be done w.r.t. to any
variable first. To integrate w.r.t. y, first we need
to draw vertical strip in the region. But one vertical strip does not cover the entire
region, therefore, we divide the region OPQRO into two subregions OPR and RPQ
and draw one vertical strip in each subregion.
4. In the subregion OPR strip starts from the circle x 2 y 2 2 x 0 and terminates
on the parabola y 2 = 2 x .
2 x x 2 to y
2x
Limits of y : y
Limits of x : x = 0 to x = 1.
5. In the subregion RPQ, strip starts from the line y = x and terminates on the
parabola y 2 = 2 x.
8.18
Engineering Mathematics
Limits of y : y = x to y = 2 x
Limits of x : x = 1 to x = 2
I=
xy dx dy =
xy dx dy +
OPR
=
1
0
=
1
0
2x
x
2 x x2
y2
2
x
1
2
1
0
2
y dy dx +
1
x
2x
dx +
2 x x2
2
1
x
x(2 x 2 x x 2 )dx
1 x4
2 4
1
0
1 2x3
2 3
xy dx dy
RPQ
x4
4
2
1
2x
x
y dy dx
2x
y2
2
dx
x
1
2
2
1
x(2 x x 2 )dx
1 8
1 1
2
8 3
3 8
7
=
12
Example 14: Evaluate
x 2 dx dy , over the region in the first quadrant enclosed
by the rectangular hyperbola xy = 16, the lines y = x, y = 0 and x = 8.
Solution:
1. The region of integration is OPQR.
2. The point of intersection of
(i) the hyperbola xy = 16 and
the line y = x is obtained as
x 2 = 16, x = ±4, and y = ±4
Hence, R : (4, 4) in the first quadrant.
(ii) the hyperbola xy = 16 and line
x = 8 is obtained as 8y = 16, y = 2
Hence, Q : (8, 2)
3. Here, integration can be done w.r.t.
any variable first. To integrate w.r.t. y,
first we need to draw a vertical strip in
Fig. 8.17
the region. But here one vertical strip
can not cover the entire region therefore we divide the region OPQR into two subregions OMR and RMPQ and draw one vertical strip in each subregion.
4. In subregion OMR, strip starts from x axis and terminates on the line y = x.
Limits of y : y = 0 to y = x
Limits of x : x = 0 to x = 4
5. In subregion RMPQ, strip starts from x axis and terminates on the rectangular
hyperbola xy = 16
16
Limits of y : y = 0 to y =
x
Limits of x : x = 4 to x = 8
Multiple Integral
I = ∫ ∫ x 2 dx dy =
8.19
∫∫ x dx dy + ∫∫
2
OMR
x 2 d x dy
RMPQ
4
x
8
16
x
0
0
4
0
= ∫ x 2 ∫ dy dx + ∫ x 2 ∫ dy dx
4
8
16
x
4
0
= ∫ x 2 y 0 dx + ∫ x 2 y
x
0
dx
4
=∫
4
0
x4
x2
16
x dx + ∫ x ⋅ d x =
+ 16
4
x
4 0
2
3
8
8
2
4
= 64 + 8 (64 − 16) = 448
Example 15: Evaluate
dx dy
, over the region bounded by the y
x4 + y2
Solution:
1. The region of integration is bounded by
the line y x 2 i.e., the region above the
parabola x 2 = y and x ≥ 1, i.e., the region
on the right of line x = 1.
2. The point of intersection of x 2 = y and
x = 1 is obtained as 1 = y.
Hence, P : (1, 1)
3. Here, it is easier to integrate w.r.t. y first
than x. Draw a vertical strip AB parallel to
y-axis in the region which starts from the
parabola x 2 = y and extends up to infinity.
4. Limits of y : y = x 2 to y
Limits of x : x = 1 to x
I = ∫∫
=∫
∞
=∫
∞
1
1
=
4
Fig. 8.18
∞ ∞
dx dy
1
=∫ ∫2 4
dy dx
4
2
1
x
x +y
x + y2
∞
∞ 1
1
y
tan −1 2 dx = ∫ 2 (tan
n −1 ∞ − tan −1 1) dx
2
1
x
x x2
x
∞
1 ⎛
1
⎞
⎜ − ⎟ dx = −
x2 ⎝ 2 4 ⎠
4 x1
x2 , x
1.
8.20
Engineering Mathematics
Exercise 8.2
Evaluate the following:
1
dx dy, over the rectangle
1.
xy
1
x
2, 1
y
Ans. :
2.
[ Ans. : (log 2) 2 ]
2.
1
ab
4.
xy 1 x
10.
gle bounded by x = 0, y = 0 and x + y = 1.
5.
xy
16
945
11.
12.
( x + y + a) dx dy, over the region
7.
a3 ]
xy dx dy, over the region bounded
by the x-axis, the line y = 2x and the
x2
parabola y =
.
4a
( x 2 + y 2 ) dx dy, over the area
bounded by the lines y = 4x, x + y = 3,
y = 0, y = 2.
463
48
xy( x + y ) dx dy, over the region
bounded by the parabolas y2 = x,
x2 = y.
3
Ans. :
28
xy ( x + y ) dx dy, over the region
bounded by the curve x2 = y and the
line x = y.
3
Ans. :
56
bounded by the circle x 2 + y 2 = a 2 .
[ Ans. :
1
3
+
3 3 2
Ans. :
y 2 dx dy, over the triangle
having vertices (0, 0), (10, 1), (1, 1).
[Ans. : 6]
6.
y 2 ) 2 dx dy, over the trian-
gle bounded by x-axis, the line y = x
and x = 1.
y dx dy, over the trian-
Ans. :
(4 x 2
Ans. :
4e 3 1)
217
60
1
9.
bounded by the lines x = 0, y = 0,
and x + y = 1.
1
(3e 4
12
y ) dx dy, over the region
Ans. :
e 3 x + 4 y dx dy, over the triangle
Ans. :
(5 2 x
bounded by x-axis, the line x + 2y = 3
and the parabola y 2 = x.
sin ( ax + by ) dx dy, over the triangle bounded by the lines x = 0, y = 0
and ax + by = 1.
Ans. :
3.
8.
2048 4
a
3
13.
xy( x 1) dx dy, over the region
bounded by the rectangular hyperbola xy = 4, the lines y = 0, x = 1, x = 4
and x-axis.
[Ans. : 8 (3 log 4)]
Multiple Integral
8.21
8.3 CHANGE OF ORDER OF INTEGRATION
Sometimes, evaluation of double integral becomes easier by changing the order of
integration. To change the order of integration, first, we draw the region of integration
with the help of the given limits. Then we draw vertical or horizontal strip as per the required order of integration. This change of order also changes the limits of integration.
(I) Change the Order of Integration of the Following
Example 1:
1
0
x
x
f ( x , y )dy dx .
Solution:
1. The function is integrated first w.r.t. y and
then w.r.t. x.
2. Limits of y : y = x to y = x .
Limits of x : x = 0 to x = 1
3. The region is bounded by the line y = x
and the parabola y 2 = x.
4. The points of intersection of y 2 = x and y
= x are obtained as x 2 = x,
x = 0, 1 and y = 0, 1
Hence, O : (0,0) and Q : (1, 1)
5. To change the order of integration, i.e., to
integrate first w.r.t. x, draw a horizontal
Fig. 8.19
strip AB parallel to x-axis which starts from
2
the parabola y = x and terminates on the line y = x.
Limits of x : x = y 2 to x = y
Limits of y : y = 0 to y = 1
Hence, the given integral after change of order can be written as
1
0
x
x
f ( x, y ) dy dx =
1
y
0
y2
f ( x, y ) dx dy
1
Example 2:
1
y3
0
y2
f ( x , y ) dx dy .
Solution:
1. The function is integrated first w.r.t. x and
then w.r.t. y.
1
2. Limits of x : x = y 2 to x = y 3
Limits of y : y = 0 to y = 1
3. The region is bounded by the parabola
y 2 = x and the cubical parabola y = x 3 .
4. The points of intersection of y2 = x and
y = x3 are obtained as x6 = x, x = 0, 1 and
y = 0, 1.
Hence, O (0, 0) and Q : (1, 1)
Fig. 8.20
8.22
Engineering Mathematics
5. To change the order of integration, i.e., to integrate first w.r.t. y, draw a vertical strip
parallel to y-axis which starts from the cubical parabola y = x 3 and terminates on
the parabola y 2 = x.
Limits of y : y = x 3 to y = x
Limits of x : x = 0 to x = 1
Hence, the given integral after change of order can be written as
1
Example 3:
8
0
y
4
y 8
4
1
y3
0
y2
f ( x, y ) dx dy =
1
0
x
x3
f ( x, y ) dy dx
f ( x , y ) dx dy .
Solution:
1. The function is integrated first w.r.t. x and
then w.r.t. y.
y −8
y
2. Limits of x : x =
to x =
4
4
Limits of y : y = 0 to y = 8
3. The region is bounded by the line y = 4x + 8,
y = 4x, y = 8 and x-axis (y = 0).
4. The point of intersection of y = 4x and
y = 8 is obtained as 8 = 4x, x = 2.
Hence, P : (2, 8).
5. To change the order of integration, i.e.,
to integrate first w.r.t. y, divide the region
OPQR into two subregions OQR and OPQ.
Draw a vertical strip parallel to y-axis in
Fig. 8.21
each subregion.
(i) In subregion OQR, strip AB starts from x-axis and terminates on the line y = 4x + 8.
Limits of y : y = 0 to y = 4x + 8
Limits of x : x = 2 to x = 0
(ii) In subregion OPQ, strip CD starts from the line y = 4x and terminates on the
line y = 8.
Limits of y : y = 4x to y = 8
Limits of x : x = 0 to x = 2
Hence, the given integral after change of order can be written as
8
0
Example 4:
y
4
y 8
4
f ( x, y ) dx dy =
y2
a
-a 0
a
0
4 x +8
2 0
f ( x, y ) dy dx +
2
8
0
4x
f ( x , y ) dx dy .
Solution:
1. The function is integrated first w.r.t. x and then w.r.t. y.
f ( x, y ) dy dx
Multiple Integral
8.23
y2
.
a
Limits of y : y = a to y = a.
3. The region is bounded by the parabola y 2 = ax, the y-axis and the line y = a and
y = a.
4. The point of intersection of
2. Limits of x : x = 0 to x =
(i) y 2
ax and y
a is obtained as a 2
ax, x
a
Hence, R : (a, a)
(ii) y 2 = ax and y = a is obtained as a2 = ax, x = a.
Hence, Q : (a, a)
5. To change the order of integration, i.e., to
integrate first w.r.t. y, divide the region into
two subregions ORS and OPQ. Draw vertical strip parallel to y-axis in each subregion.
(i) In subregion ORS, strip AB starts
from the line y = a and terminates
on the parabola y 2 = ax.
Limits of y : y = a to y
ax
Limits of x : x = 0 to x = a
(ii) In subregion OPQ, strip CD starts
from the parabola y 2 = ax and terminates on the line y = a.
Fig. 8.22
Limits of y : y = ax to y = a
Limits of x : x = 0 to x = a
Hence, the given integral after change of order can be written as
a
y2
a
f ( x, y ) dx dy =
a 0
Example 5:
2
2+ 4-2 y
0
y
a
0
ax
a
f ( x, y ) dy dx +
a
0
a
ax
f ( x, y ) dy dx
f ( x , y ) dx dy .
Solution:
1. The function is integrated first w.r.t. x and
then w.r.t. y.
2. Limits of x : x = y to x = 2 + 4 − 2 y
Limits of y : y = 0 to y = 2
3. The region is bounded by the x-axis, the line
y = x and the parabola ( x 2) 2 2(2 y ).
Fig. 8.23
8.24
Engineering Mathematics
4. The points of intersection of y = x and ( x 2) 2 2(2 y ) are obtained as
( x 2) 2 2(2 x ), x = 0, 2 and y = 0, 2
Hence, O : (0, 0) and Q : (2, 2)
5. To change the order of integration, i.e., to integrate first w.r.t. y, divide the region into
two subregions OPQ and PQR. Draw a vertical strip parallel to y-axis in each subregion.
(i) In subregion OPQ, strip AB starts from x-axis and terminates on the line y = x.
Limits of y : y = 0 to y = x
Limits of x : x = 0 to x = 2
(ii) In subregion PQR, strip CD starts from x-axis and terminates on the parabola
( x 2) 2
2(2 y ).
Limits of y : y = 0 to y
2x
x2
2
Limits of x : y = 2 to y = 4
Hence, the given integral after change of order can be written as
2
2
0
y
Example 6:
4 2y
a cos
0
f ( x, y ) dx dy =
a2 - x 2
x tan
2
x
0
0
f ( x, y ) dy dx +
4
2x
2
0
x2
2
f ( x, y ) dy dx
f ( x , y )d y d x.
Solution:
1. The function is integrated first w.r.t. y and then w.r.t. x.
2. Limits of y : y = x tan to y = a 2 − x 2
Limits of x : x = 0 to x = a cos a
3. The region is bounded by the line y = x tan a, the circle x 2 + y 2 = a 2 and y-axis.
Since given limits of x and y are positive, the region lies in the first quadrant.
Fig. 8.24
Multiple Integral
8.25
4. The points of intersection of y = x tan a and x 2 + y 2 = a 2 are obtained as
x 2 + x 2 tan 2 = a 2 , x = ± a cos and y = ± a sin
Hence, P : (a cos a, a sin a ) and P' : (–a cos a, – a sin a)
5. To change the order of integration, i.e., to integrate first w.r.t. x, divide the region
into two subregions OPR and PQR. Draw horizontal strip in each subregion.
(i) In subregion OPR, strip AB starts from y-axis and terminates on the line
y = x tan a.
Limits of x : x = 0 to x = y cot a
Limits of y : y = 0 to y = a sin a
(ii) In subregion PQR, strip CD starts from y-axis and terminates on the circle
x 2 + y 2 = a2 .
Limits of x : x = 0 to x
a2 y2
Limits of y : y = a sin a to y = a
Hence, given integral after change of order can be written as
a2 x 2
a cos
0
x tan
Example 7:
f ( x, y ) dy dx =
4
4x
0
4 x - x2
a sin
0
y cot
0
f ( x, y ) dx dy +
a2 y 2
a
a sin
0
f ( x, y ) dx dy
f ( x , y )dy dx .
Solution:
1. The function is integrated first w.r.t. y and then w.r.t. x.
2. Limits of y : y = 4 x − x 2 to y = 4 x .
Limits of x : x = 0 to x = 4
3. The region is bounded by the circle x 2 y 2 4 x 0, the parabola y 2 = 4 x and
the line x = 4.
4. The point of intersection of
(i) x 2 y 2 4 x 0 and y 2 = 4 x is obtained as x 2 = 0, x = 0 and y = 0
Hence O : (0, 0)
y2
Fig. 8.25
8.26
Engineering Mathematics
(ii) y 2 = 4 x and x = 4 are obtained as y2 = 16, y = ± 4
Hence, Q : (4, 4) and Q : (4, - 4)
5. To change the order of integration, i.e., to integrate first w.r.t. x, divide the region
into three subregions ORT, TPS and RSQ. Draw a horizontal strip parallel to x-axis
in each subregion.
(i) In subregion ORT, strip AB starts from the parabola y 2 = 4 x and terminates
on the circle x 2
y2
4x
0.
2
y
to x 2
4 y 2 (Part of the circle where x < 2)
4
Limits of y : y = 0 to y = 2
(ii) In subregion TPS, strip CD starts from the circle x 2 y 2 4 x 0 and terminates on the line x = 4.
Limits of x : x
Limits of x : x 2
4 y 2 (Part of circle where x > 2) to x = 4
Limits of y : y = 0 to y = 2
(iii) In subregion RSQ, strip EF starts from the parabola y 2 = 4 x and terminates
on the line x = 4.
y2
Limits of x : x =
to x = 4
4
Limits of y : y = 2 to y = 4
Hence, given integral after change of order can be written as
4
0
4x
4 x x2
f ( x, y ) dy dx =
+
Example 8:
2
(4 x )2
0
4 x
2
2
0
y2
4
2
4
0
2
4 y2
4 y2
f ( x, y ) dx dy
f ( x, y ) dx dy +
4
4
2
y2
4
f ( x, y ) dx dy
f ( x , y ) dy dx .
Solution:
1. The function is integrated first w.r.t. y and then w.r.t. x.
2. Limits of y : y = 4 − x to y = ( 4 − x ) 2 .
Limits of x : x = 0 to x = 2
3. The region is enclosed by the parabola y 2 = 4 − x and y = ( 4 − x ) 2 , the lines x = 2
and x = 0.
4. The point of intersection of
(i) x = 2 and y2 = 4 – x are obtained as y 2 (4 2), y
2
Hence, Q : ( 2, 2 ) and Q ′ ( 2, − 2 )
(ii) x = 2 and y (4 x ) 2 is obtained as y (4 2) 2 4
Hence, S : (2, 4)
(iii) x = 0 and y2 = 4 – x are obtained as y 2 = 4, y = ± 2
Hence, P : (0, 2) and P' : (0, 2)
Multiple Integral
8.27
Fig. 8.26
(iv) x = 0 and y (4 x) 2 is obtained as y = 16
Hence, U : (0, 16)
5. To change the order of integration, i.e., to integrate first w. r. t. x, divide the region into
three subregions PQR, PRST and STU. Draw a horizontal strip in each subregion.
(i) In subregion PQR, strip AB starts from parabola y 2 4 x and terminates
on the line x = 2.
Limits of x : x 4 y 2 to x 2
Limits of y : y = 2 to y = 2
(ii) In subregion PRST, strip CD starts from y-axis and terminates on the line x = 2.
Limits of x : x = 0 to x = 2
Limits of y : y = 2 to y = 4
(iii) In the subregion STU, strip EF starts from y-axis and terminates on the
parabola y (4 x) 2 .
Limits of x : x = 0 to x 4
y (Part of the parabola where x < 4)
Limits of y : y = 4 to y = 16
Hence, given integral after change of order can be written as
2
(4 x )2
0
4 x
2
f ( x, y ) dy dx =
+
2
4 y2
2
16
4
4
0
y
f ( x, y ) dx dy +
f ( x, y ) dx dy
4
2
2
0
f ( x, y ) dx dy
8.28
Engineering Mathematics
(II) Change the Order of Integration and
Evaluate the Following
Example 1:
0
x
sin y
dy dx .
y
Solution:
1. The function is integrated first w.r.t. y, but evaluation becomes easier by changing
the order of integration.
2. Limits of y : y = x to y = p .
Limits of x : x = 0 to x = p .
Fig. 8.27
3. The region is bounded by the line y = x, y = p and x = 0.
4. The point of intersection of the line y = x and the line y = p is P : (p, p)
5. To change the order of integration, i.e., to integrate first w.r.t. x, draw a horizontal
strip AB parallel to x-axis which starts from y-axis and terminates on the line y = x.
Limits of x : x = 0 to x = y
Limits of y : y = 0 to y = p
Hence, the given integral after change of order can be written as
sin y
sin y y
sin y y
∫0 ∫x y dy dx = ∫0 y ∫0 dx dy = ∫0 y x 0 dy
=∫
0
Example 2:
x
0
0
xe
x2
y
sin y
⋅ y dy = ∫ sin y dy
0
y
= − cos y 0 = − cos + cos 0
=2
dy dx .
Solution:
1. The function is integrated first w.r.t. y, but evaluation becomes easier by changing
the order of integration.
Multiple Integral
8.29
2. Limits of y : y = 0 to y = x
Limits of x : x = 0 to x
3. The region is the part of the first quadrant
bounded between the lines y = x and y = 0.
4. To change the order of integration, i.e., to
integrate first w.r.t. x, draw a horizontal strip
parallel to x-axis which starts from the line
y = x and extends up to infinity.
Limits of x : x = y to x
Limits of y : y = 0 to y
Hence, the given integral after change of order can be written as
∞
∫ ∫
0
x
0
−
xe
x2
y
dy dx = ∫
∞
0
∫
∞
y
−
xe
x2
y
Fig. 8.28
dx dy
x2
∞⎛
y⎞ ∞ −
= ∫ ⎜− ⎟ ∫ e y
0 ⎝
2⎠ y
x2
⎛ 2x ⎞
⎜⎝ − y ⎟⎠ dx dy
∞
−
1 ∞
=− ∫ ye y
2 0
⎡∵ e f ( x ) f ′( x ) dx = e f ( x ) ⎤
⎣ ∫
⎦
dy
y
=−
=
Example 3:
1- x2
1
0
0
1 ∞
1
y(0 − e − y ) dy = − ye − y − e − y
2 ∫0
2
∞
0
1
2
ey
( e y + 1) 1 - x 2 - y 2
dy dx .
Solution:
1. The function is integrated first w.r.t. y, but
evaluation becomes easier by changing the
order of integration.
2. Limits of y : y = 0 to y
1 x2
Limits of x : x = 0 to x = 1
3. Since given limits of x and y are positive,
the region is the part of circle x 2 + y 2 = 1
in the first quadrant.
4. To change the order of integration, i.e.,
to integrate first w.r.t. x, draw a horizontal strip AB parallel to x-axis which starts
from y-axis and terminates on the circle
x 2 + y 2 = 1.
Fig. 8.29
8.30
Engineering Mathematics
1 y2
Limits of x : x = 0 to x
Limits of y : y = 0 to y = 1
Hence, the given integral after change of order can be written as
0
ey
1 x2
1
0
(e
y
1) 1 x
2
y
2
ey
0 e +1
y
1
1 y2
1
dy dx =
0
(1 y 2 ) x 2
dx dy
1 y2
ey
sin
=
0 ey +1
1
x
1
1 y2
dy
0
ey
(sin 1 1 sin 1 0) dy
0 ey +1
1
ey
dy
0 ey +1 2
1
=
Example 4:
2
(
[log(e + 1) − log 2] = 2 log ⎛⎜⎝
2
a
a − a2 − y2
0
0
)
log e y 1
1
∵
0
f ( y)
dy = log f ( y )
f ( y)
e +1⎞
⎟
2 ⎠
xy log ( x + a )
dx dy .
( x − a )2
Solution:
1. The function is integrated first w.r.t. x, but
evaluation becomes easier by changing the
order of integration.
2. Limits of x : x = 0 to x a
a2 y2
Limits of y : y = 0 to y = a
3. The region is bounded by the circle
(x a)2 + y2 = a2, the lines y = a and x = 0.
4. The point of intersection of
(x a)2 + y2 = a2 and y = a is obtained as
(x a)2 + a2 = a2,
Fig. 8.30
x = a.
Hence, P : (a, a).
5. To change the order of integration, i.e., to integrate first w.r.t. y, draw a vertical strip
AB parallel to y-axis which starts from the circle (x a)2 + y2 = a2 and terminates on
the line y = a.
Limits of y : y
2ax x 2 to y
Limits of x : x = 0 to x = a
a
Multiple Integral
8.31
Hence, given integral after change of order can be written as
a
a − a2 − y 2
0
0
∫ ∫
a x log ( x + a) a
xy log ( x + a)
dx dy = ∫
y dy d x
2
2
0
( x − a)
( x − a) 2 ∫ 2 ax − x
=∫
a
0
x log ( x + a) y 2
2
( x − a) 2
a
2 ax − x 2
dx = ∫
a
0
x log ( x + a) ⎛ a 2 − 2ax + x 2
2
( x − a) 2 ⎜⎝
=
a
2
⎤
a x
1 a
1 ⎡ x2
1
⎢
x
x
+
a
x
=
x
+
a
−∫
log
(
)
d
log
(
)
⋅
dx ⎥
∫
0 2
2 0
2⎢ 2
x+a ⎥
0
⎣
⎦
=
1 ⎡ a2
1 a⎧
a2 ⎫ ⎤
2
a
x
a
log
(
)
−
−
+
⎨
⎬ dx ⎥
⎢
2⎣ 2
2 ∫0 ⎩
x + a⎭ ⎦
⎞
⎟⎠ dx
a
⎤
1 ⎡ a2
1 x2
2
= ⎢ log 2a −
− ax + a log ( x + a) ⎥
2⎢ 2
2 2
0⎥
⎣
⎦
Example 5:
=
⎞
a2
1⎛ 2
a log 2a −
+ a 2 − a 2 log 2a + a 2 log a ⎟
⎜
4⎝
2
⎠
=
⎞ a2
1 ⎛ a2
+ a 2 log a ⎟ =
(1 + 2 log a)
⎜
4⎝ 2
⎠ 8
1
1- 4 y2
2
0 0
1 + x2
1 - x2 1 - x2 - y2
dx dy .
Solution:
1. The function is integrated first w.r.t. x, but evaluation becomes easier by changing
the order of integration.
2. Limits of x : x = 0 to x
1 4 y2
1
2
3. The region is the part of the ellipse in the first quadrant.
4. To change the order of integration, i.e., to integrate first w.r.t. y, draw a vertical strip AB
parallel to y-axis which starts from x-axis and terminates on the ellipse x2 + 4y2 = 1.
Limits of y : y = 0 to y =
1
1 x2
2
Limits of x : x = 0 to x = 1
Limits of y : y = 0 to y
8.32
Engineering Mathematics
Fig. 8.31
Hence, the given integral after changing the order of integration can be written as
1
2
∫ ∫
0
1+ x2
1− 4 y 2
1− x
0
=∫
1
0
=∫
1
2
1+ x2
1− x2
1− x2
2sin 1 x
⋅
a
0
0
1− x
dx =
x
1 x2
2
2
2
1
0
0
⎛
0⎜
⎝
6∫
1
∫
1
1− x 2
2
0
1
(1 − x 2 ) − y 2
dy dx
1 + x 2 ⎛ −1 1
⎞
− sin −1 0 ⎟ dx
⎜ sin
2 ⎝
⎠
2
1− x
⎞
− 1 − x 2 ⎟ dx
⎠
1− x2
2
1
1
sin 1 x
2
0
2
4 2
2
dx = ∫
8
x dy dx
y
0
1+ x2
1
1− x 2
2
1− x2
6
3
sin 1 1
6 2
Example 6:
2
y
sin −1
2 − (1 − x 2 )
0
6
1− x − y
2
1
dx dy = ∫
x 2
⎡
2
2
2⎤
⎢∵ ∫ a − x dx = 2 a − x ⎥
⎢
⎥
a2
⎢
−1 x ⎥
+ sin
⎢⎣
2
a ⎥⎦
.
( a - x )( a - y )( y - x )
Solution:
1. The function is integrated first w.r.t. x, but evaluation becomes easier by changing
the order of integration.
2. Limits of x : x = 0 to x = y
Limits of y : y = 0 to y = a
3. The region is bounded by the line y = x, y = a and x = 0.
4. The point of intersection of y = a and y = x is P : (a, a)
Multiple Integral
8.33
5. To change the order of integration, i.e.,
to integrate first w.r.t. y, draw a vertical
strip AB parallel to y-axis which starts
from the line y = x and terminates on the
line y = a.
Limits of y : y = x to y = a
Limits of x : x = 0 to x = a
Hence, the given integral after changing the order can be written as
a
0
x dy dx
y
0
(a
2
2
x )( a
x
a
=
y )( y x )
0
a
dy
a
2
x
2
x
(a
dx
y )( y x )
Fig. 8.32
2
Putting y x t , dy 2t dt
When
y = x, t = 0
y = a, t
a x
a
∫ ∫
0
x dy dx
y
0
( a − x )( a − y )( y − x )
2
= 2∫
2
0
2
x
a
a2 − x 2
x
a
0
2
a2
x2
2( a 2
∫
a− x
a
x
0
a −x
=∫
2
dt
(a − x) − t 2
0
(sin 1 1 sin 1 0) dx
1 a
2 2
x )
0
1
0
1
x
x
∫
( a − x − t 2 )t 2
x
a
a2 − x 2
0
2
a
2
2t dt
a− x
0
dx = 2 ∫
= a
Example 7:
2
0
1 2
(a
2
sin −1
dx
a− x
t
a− x
dx
0
1
2
x 2 ) ( 2 x ) dx
⎡
[ f ( x)]n+1 ⎤⎥
n
⎢∵ ∫ [ f ( x ) ] f ′( x )dx =
n +1 ⎥
⎢⎣
⎦
y
dx dy .
(1 + xy )2 (1 + y 2 )
Solution:
1. The function is integrated first w.r.t. y, but evaluation becomes easier by changing
the order of integration.
1
2. Limits of y : y = x to y =
x
Limits of x : x = 0 to x = 1
3. The region is bounded by the rectangular hyperbola xy = 1, the line y = x and y-axis
in the first quadrant.
4. The point of intersection of xy = 1 and y = x is obtained as x2 = 1, x = 1 and y = 1
Hence, P : (1, 1)
8.34
Engineering Mathematics
5. To change the order of integration, i.e.,
to integrate first w.r.t. x, divide the region into two subregions OPQ and QPR.
Draw a horizontal strip parallel to x-axis
in each subregion.
(i) In subregion OPQ, strip AB starts from
y-axis and terminates on the line y = x.
Limits of x : x = 0 to x = y
Limits of y : y = 0 to y = 1
(ii) In subregion QPR, strip CD starts
from y-axis and terminates on the
rectangular hyperbola xy = 1.
1
Limits of x : x = 0 to x =
y
Fig. 8.33
Limits of y : y = 1 to y
.
Hence, given integral after changing the order can be written as
1
1
x
0
x
y
dx dy =
(1 + xy ) 2 (1 + y 2 )
1
dy
y(1 + xy ) 0
1
1
1
1
1 dy
2
1 y 1 y2
1
0
1
(1 + y 2 ) 2
y
0
1
dx d y +
(1 + xy ) 2
1
y
1+ y2
1
y
0
1
dx dy
(1 + xy ) 2
1
y
y
0 1 + y2
1
0
y
0 1+ y2
1
1
y
1 + y2
y
1
dy
y(1 + xy ) 0
1
1
1 dy
2
1 y 2
1
1
dy
2
2
1+ y
1
1
dy
1 + y2
Putting y = tanq in the first term of first integral, dy = sec 2 d ,
When y = 0, q = 0
y = 1,
1
∫∫
0
1
x
x
=
4
π
1
∞
y
sec 2 θ dθ
1
dx dy = − ∫ 4
+ tan −1 y 0 + tan −1 y 1
2
4
0
(1 + xy ) (1 + y )
sec θ
2
2
π
= −∫ 4
0
(1 + cos 2θ )
dθ + (tan −1 1 − tan −1 0)
2
1
+ (tan −1 ∞ − tan −1 1)
2
1
sin 2θ
=− θ+
2
2
=
π −1
4
π
4
0
+
3π
π 1 π 3π
= − − sin +
8
8 4
2 8
Multiple Integral
cos -1 x
1- y2
1
Example 8:
0
0
8.35
dx dy .
1 - x2 1 - x2 - y2
Solution:
1. The function is integrated first w.r.t.
x, but evaluation becomes easier by
changing the order of integration.
2. Limits of x : x = 0 to x
1 y2
Limits of y : y = 0 to y = 1
3. Since given limits of x and y are
positive, the region is the part of the
circle x2 + y2 = 1 in the first quadrant.
4. To change the order of integration,
i.e., to integrate first w.r.t. x, draw a
vertical strip AB parallel to y-axis in
the region. AB starts from x-axis and
terminates on the circle x2 + y2 = 1.
Limits of y : y = 0 to y
1 x2
Limits of x : x = 0 to x = 1.
Hence, the given integral after
changing the order can be written as
0
cos -1 x
1- y 2
1
0
1 - x2 1 - x2 - y2
1
cos −1 x
0
1− x2
=∫
(1 − x 2 ) − y 2
0
1
cos −1 x
0
1− x2
=∫
∫
dx dy
1
1− x 2
cos −1 x
−1
⎡(cos 1) − (cos 0) ⎤⎦
4⎣
⎡ ⎛ ⎞2 ⎤
= − ⎢0 − ⎜ ⎟ ⎥
4 ⎢⎣ ⎝ 2 ⎠ ⎥⎦
3
=
16
cos −1 x
0
1− x2
sin
−1
1− x 2
y
1− x2
dx
0
(sin −1 1 − sin −1 0) dx
−1
=−
1
dy dx = ∫
(cos −1 x ) 2
dx = −
=− ∫ −
2
2
2 0
1− x2
1
Fig. 8.34
2
2
1
0
⎡
⎢∵
⎣
n
∫ [ f ( x)]
f ′( x ) dx =
[ f ′( x )] n + 1 ⎤
⎥
n +1 ⎦
8.36
Engineering Mathematics
Exercise 8.3
(I) Change the order of integration of the following:
1.
6
2+ x
0
2 x
f ( x, y ) dy dx
Ans. :
2.
1
2x
0
x
2
6
4
2 y
+
8
6
2
y 2
y
0
y
2
1
y
0
y
1
y
2
1
1
1 x2
f ( x, y ) dx dy
0
2
0
6 x
2
x2 + 4
4
y
a
a
a 0
Ans. :
a
+
5.
a
2a x
0
x2
a
f ( x, y ) dx dy
2
x +6 a
4 x2
a
ay
0
0
+
3
2
y
y
2
6
ax
0
f ( x, y ) dy dx
2
a
0
a
0
f ( x, y )dx dy
Ans. :
x 6
6
+
x 6
x +6
3
2
f ( x, y ) dy dx
x
f ( x, y ) dy dx
f ( x, y )dx dy
f ( x, y )dx dy
6a
2
0
f ( x, y )dx dy
6a+2
2
6a
y 6a
f ( x, y )dx dy
f ( x, y ) dx dy
a2 y 2
2a
2 ax
0
2 ax x 2
Ans. :
f ( x, y ) dx dy
2
y+a
a
Ans. :
11.
6 2y
0
2
a
a2 x 2
0
f ( x, y )dx dy
2a y
3
4 y2
+
2a
f ( x, y )dx dy
2
2
0
+
10.
f ( x, y )dy dx
f ( x, y ) dy dx
Ans. :
f ( x, y ) dy dx
Ans. :
6.
a
2 y 1
+
f ( x, y ) dy dx
a
0
0
0
+
ax
2
1
f ( x, y )dx dy
0
4 x x2
4
4
2
f ( x, y ) dx dy
a
f ( x, y )dy dx
0
f ( x, y ) dy dx
2
4.
x
2
2
+
9.
1
)
Ans. :
f ( x, y ) dx dy
Ans. :
1 y
f ( x, y )dy dx
f ( x, y )dx dy
2
+
21
2y
Ans. :
8.
1
(
1
0
f ( x, y )dy dx
f ( x, y ) dy dx
Ans. :
3.
7.
a
2a
a
+
2a
a
a
x a
f ( x, y ) dy dx
f ( x, y ) dx dy
0
+
f ( x, y ) dy dx
a2 y 2
f ( x, y ) dx dy
a2 y 2
a
a
0
y2
2a
2a
2a
a
y2
2a
f ( x, y ) dx dy
f ( x, y ) dx dy
Multiple Integral
12.
a
a2 − x 2
0
a2 − x 2
4
∫ ∫
Ans. :
f ( x, y ) dy dx
a
2
a2 y 2
a
0
13.
a2
x
0
x
2
4y
a
2
y
0
0
+
b
mx
a
k
x
+
f ( x, y ) dx dy
15.
1
ex
0
a
f ( x, y )dx dy
2
0
ma
b
k
a
a
mb
b
ma
y
m
e
Ans. :
16.
f ( x, y )dx dy
f ( x, y )dx dy
f ( x, y )dx dy
f ( x, y ) dy dx
0 1
f ( x, y )dx dy
a2
y
b
k
y
+
f ( x, y ) dx dy
0
a
k
a
k
b
Ans. :
f ( x, y ) dy dx
Ans. :
14.
2
a2 y 2
a
+
a
8.37
x3
1
1
f ( x, y )dx dy
log y
f ( x, y ) dy dx
0
⎡ Ans. :
⎣⎢
f ( x, y ) dy dx
8
2
0
y3
∫∫
f ( x, y )dx dy ⎤
⎦⎥
1
(II) Change the order of integration and evaluate the following:
1.
1
0
dx
e y y x log y dy
1
Ans. :
2.
1
1
0
y
dy
1
1
e
1
2
0
2y
1
x
0
0
5
2+ x
0
2 x
1
2 x
0
x
0
x
2
0
e 2
x 2 e xy dy dx =
2
cos( x 2 )dy dx =
sin 4
4
a
2 ax
0
0
5
3
2 y
dx dy +
Ans. :
7.
a
0
ay
y
2a
a
0
y2
4a
0
x
dx dy
y
2 2 y x
dx dy
+
1 0
y
= log
4
e
2
y 2
dx dy
a
x
0
x2
a
=
8.
a
2a x
0
x2
a
⎡
⎢ Ans. :
⎢
⎢
⎣
x 2 dx dy =
4 4
a
7
x
dx dy
x + y2
y
0
5
2
Ans. :
1
7
x 2 dy dx
x
dy dx
y
Ans. :
2
= 25
cos( x 2 )dx dy
2
dy dx
Ans. :
6.
Ans. :
4.
0
e y y x log y dx =
x 2 e xy dx dy
Ans. :
3.
1
5.
x
dy dx
x + y2
2
a
log 2
2
xy dy dx
a
∫ ∫
0
ay
0
+∫
2a
a
⎤
⎥
5 ⎥
xy dx dy = a 4 ⎥
6 ⎦
xy dx dy
∫
2a− y
0
8.38
Engineering Mathematics
sin y
x
9.
0
0
(
x )( x
y)
14.
sin y
Ans. :
0
y
(
x )( x
y)
∫∫
0
2
1
a
x
0
0
a
a
0
y
y)
1
x
0
dx dy
( y a) ( a x )( x
y)
15.
1
0
1
y3
a
0
y
2
y
( a x )( x
y )(4 5cos y )
2
dy dx
16.
2
2
10 y 2
3
0
y2
9
y )(4 5cos y ) 2
( a x )( x
2
4 y2
0
2
4 y2
0
0
x + y2
=
4 x x2
4
2
+
1
2
1
3 x
0
0
x
y
0
0
x +y
2
2
y
2
2
2
2
1 x2
2
1
0
18.
2
0
2
dy dx
x2
2
0
1
0
1
+ 5sin
2
1
dy dx
3
10
y2
dx dy
1+ x2
y2
dy dx
2
3
x + y2 +1
2
dy dx
⎤
dx dy ⎥
x + y +1
⎥
⎥
1
= (5 log 5 − 4)
⎥
⎦
4
2
∫ ∫
0
10 x 2
10
1 x2
x
⎡
⎢ Ans. :
⎢
⎢
⎢
⎣
dy dx
2
1 x2
Ans. :
0
x +y
0
0
=
dy dx
2 x2
1
1+ x
2
dx dy
1
2
dy dx = 2
1 4y
e 2
2
dx dy
=
17.
y
2 y2
Ans. :
x
2
e x dy dx =
+
0
y
0
3
dx dy
dx dy
Ans. :
0
x
log 5cos a 4
5
2
1
Ans. :
sin y
dx dy
2
log 2
sin y
1
2
e x dx dy
Ans. :
a
x + y2
0
Ans. :
a
x
2 y2
2
=
11.
13.
2
+
( y a) ( a x )( x
=
0
dx dy
x + y2
0
dy dx
Ans. :
12.
x
y
0
=2
10.
dy dx
x + y2
x
Ans. :
dx dy
x
2− x2
1
dy dx
x
2
2y
2
2
Multiple Integral
8.39
8.4 DOUBLE INTEGRATION IN POLAR
COORDINATES
The double integral can be changed from cartesian coordinates (x, y) to polar
coordinates (r, q) by putting x = r cosq, y = r sinq . Then
f (x, y) dy dx
=
f (r cos q, r sin q) | J | dr dq where J is the Jacobian (functional determinant)
x
( x, y )
r
J=
=
y
(r , )
r
cos
r sin
=
sin r cos
Hence,
f ( x, y )dy dx =
=
x
y
= r (cos 2 + sin 2 ) = r
f ( r cos , r sin ) r dr d
f ( r cos , r sin ) r dr d
8.4.1 Limits of Integration
The limits of integration, if required can be found with the help of the given curves. Let the
region is bounded by the curves r = r1 (q ), r = r2 (q ) and the lines q = q 1, q = q 2.
The region of integration is PQRS. Draw an elementary radius vector AB from origin
which enters in the region from the curve r = r1 (q ) and leaves at the curve r = r2 (q ).
Therefore, limits for r are r1 ( ) to r2 ( ) [i.e., r varies along the AB and q remains
constant]
Fig. 8.35
To cover the entire region PQRS, rotate elementary radius vector AB from PQ to RS.
Therefore, q varies from q1 to q 2.
Hence, double integration in polar form becomes
8.40
Engineering Mathematics
r2 ( )
2
r1 ( )
1
f ( r cos , r sin ) r dr d
Note: The function is integrated first w.r.t. r and then w.r.t. q.
(I) Evaluation of Integral Over a Given Region in Polar
Coordinates:
Example 1: Evaluate
r = a cosp.
r a 2 - r 2 dr d , over the upper half of the circle
Solution:
1. The region of integration is the upper half of the circle r = a cosq .
2. Draw an elementary radius vector OA which starts from the origin and terminates
on the circle r = a cosq .
Limits of r : r = 0 to r = a cosq
Limits of q : q = 0 to
=
2
Fig. 8.36
I=∫
∫r
π
=∫2∫
0
=−
a 2 − r 2 dr dθ
a cosθ
0
1
⎛ 1⎞ 2 2 2
(
)
( −2r ) dr dθ
−
a
r
−
⎜⎝ ⎟⎠
2
3 a cosθ
2 2
1
2( a − r )
3
2∫
π
2
0
2
dθ
∵
[ f (r )] f ( r )dr =
n
0
=−
1 π2 3 3
a3
3
a
sin
θ
−
a
d
θ
=
−
3 ∫0
3
(
)
π
2
0
∫
⎛ 3 sin θ − sin 3θ ⎞
− 1⎟ dθ
⎜⎝
⎠
4
[ f (r )]n+1
n +1
Multiple Integral
a3 1
3 4
cos 3
3
3cos
8.41
a3 3 1
3 4 12
2
0
2
3
a 2
3 3
2
r 4 cos 3 dr d , over the interior of the circle r = 2a cosp .
Example 2: Evaluate
Solution:
1. The region of integration is the interior of the circle r = 2a cos q.
2. Draw an elementary radius vector OA which starts from the origin and terminates
on the circle r = 2a cosq.
Limits of r : r = 0 to r = 2a cosq
Limits of
:
2
to
2
Fig. 8.37
I = ∫ ∫ r 4 cos3 θ dr dθ
π
2
π
−
2
= ∫ cos θ ∫
3
2 a cosθ
0
π
2
π
−
2
r dr d θ = ∫
π
4
r5
cos θ
5
2 a cosθ
3
dθ
0
π
=
1 2
32a5
3
5
cos
(
a
cos
)
θ
θ
d
θ
=
⋅ 2 ∫ 2 cos8 θ dθ
2
π
0
5 ∫− 2
5
=
32a5 ⎛ 9 1 ⎞
⋅B⎜ , ⎟
⎝2 2⎠
5
∵B
p +1 q +1
,
=2
2
2
2
0
sin p cos q d
8.42
Engineering Mathematics
9 1
32a5 2 2
5
5
7 5
a
=
4
7 5 3 1 1 1
32a5 2 2 2 2 2 2
5
24
Example 3: Evaluate
the initial line.
r 2 sin dr d , over the cardioid r = a (1 + cosp ) above
Solution:
1. The region of integration is the part of the cardioid r = a (1 + cosq ) above the
initial line (q = 0).
2. Draw an elementary radius vector OA which starts from the origin and terminates
on the cardioid r = a (1 + cosq ).
3. Limits of r : r = 0 to r = a (1 + cosq )
Limits of : = 0 to =
Fig. 8.38
I=
r 2 sin dr d =
1
3
0
sin
a3 (1 cos )3 d
a3 (1 + cos ) 4
3
4
=−
0
sin
a (1+ cos )
0
a3
3
r 2 dr d =
0
4
a3
(0 − 16) = a 2
12
3
r3
3
a (1+ cos )
d
0
(1 cos )3 ( sin ) d
∵
0
0
sin
[ f ( x)]n
f ( x ) dx =
[ f ( x)]n+1
n +1
Multiple Integral
r dr d
Example 4: Evaluate
r 2 + a2
8.43
, over one loop of the lemniscate r2 = a2 cos2p .
Solution:
1. The region of integration is one loop of the lemniscate r 2 = a 2 cos 2 bounded
to
4
4
2. Draw an elementary radius vector OA which starts from the origin and terminates
on the lemniscate r 2 = a 2 cos 2 .
between the lines
3. Limits of r : r = 0 to r = a cos 2
Limits of :
to
4
4
Fig. 8.39
r dr d
I=
r 2 + a2
1
=
2
=
4
=
a cos 2
4
4
1 a cos 2
2 2
2( r + a )
2
0
4
1
1 2
( r + a 2 ) 2 (2r ) dr d
2
d
π⎞
⎛
= a ⎜2 − ⎟
⎝
2⎠
a
= (4 − π )
2
[ f (r )]
0
π
π
1
⎡
⎤
1 4
4
2
π 2a ⎢(cos 2θ + 1) − 1⎥ dθ = a ∫ π
∫
−
2 −4 ⎣
⎦
4
= a 2 sin θ − θ
∵
n
π
4
π
−
4
(
)
2 cosθ − 1 dθ
⎡ ⎛ π
π⎞ π π⎤
= a ⎢ 2 ⎜ sin + sin ⎟ − − ⎥
⎝
4
4⎠ 4 4⎦
⎣
f
n +1
f (r )]
[
( r ) dr =
n +1
8.44
Engineering Mathematics
r 2 dr d , over the area between the circles r = a sinp
Example 5: Evaluate
and r = 2a sinp .
Solution:
1. The region of integration is the area bounded between the circles r = a sinq and
r = 2a sinq.
2. Draw an elementary radius vector OAB from the origin which enters in the region
from the circle r = a sinq and leaves at the circle r = 2a sinq.
3. Limits of r : r = a sinq to r = 2a sinq
Limits of q : q = 0 to =
Fig. 8.40
r 2 dr d
I
=
7a
3
0
a sin
3
7a 3
12
=
2 a sin
0
sin 3 d =
3cos
7a
3
cos 3
3
r 2 dr d
3
3sin
sin 3
4
0
0
0
r3
3
7a 3
12
3(cos
2 a sin
d
a sin
1
3
0
(8a3 sin 3
a3 sin 3 ) d
d
cos 0)
1
(cos 3
3
cos 0)
7a3 16
28 3
=
a.
12 3
9
(II) Evaluation of Integral by Changing to Polar
Coordinates:
Example 1: Evaluate
x2 + y2 = 1.
1 - x2 - y2
dx dy over the first quadrant of the circle
1 + x2 + y2
Solution:
1. Putting x = r cosq, y = r sinq , polar form of the circle x2 + y2 = 1 is obtained as r = 1.
Multiple Integral
8.45
2. The region of integration is the part of
the circle r = 1 in the first quadrant.
3. Draw an elementary radius vector OA
which starts from the origin and terminates on the circle r = 1.
4. Limits of r : r = 0 to r = 1
Limits of q : q = 0 to
=∫2∫
0
2
1− x2 − y2
dx dy
1+ x2 + y2
I =∫∫
π
=
1
0
1− r2
r dr dθ
1+ r2
Fig. 8.41
Putting r 2 = cos 2t , 2r dr = −2 sin 2t dt
When
π
2
0
∫ ∫
0
π
4
r = 0, t =
4
r = 1, t = 0
π
0
1 − cos 2t
2 sin 2 t
sin 2t dtt dθ
( − sin 2t dt ) dθ = − ∫ 2 ∫π
0
1 + cos 2t
2 cos 2 t
4
π
π
sin t
⋅ 2 sin t cos t dt dθ = ∫ 2 dθ
0
4 cos t
0
= − ∫ 2 ∫π
0
= θ
π
2
0
sin 2t
t−
2
π
4
0
dθ = ∫
π
4
0
∫
(1 − cos 2t ) dt
π ⎛π 1 ⎞ π
⎜ − ⎟ = (π − 2)
2 ⎝ 4 2⎠ 8
4 xy - x2 - y2
e
dx dy , over the region bounded by the
x + y2
circle x2 + y2 - x = 0 in the first quadrant.
Example 2: Evaluate
2
Solution:
1. Putting x = r cosq, y = r sinq , polar form of the circle x2 + y2 – x = 0 is
r2 r cosq = 0, r = cosq.
2. The region of integration is the part of
the circle r = cosq in the first quadrant.
3. Draw an elementary radius vector OA
which starts from the origin and terminates on the circle r = cosq .
4. Limits of r : r = 0 to r = cosq
Limits of q : q = 0 to
=
2
Fig. 8.42
8.46
Engineering Mathematics
4 xy
e
2
x + y2
I=
2
2
0
π
2
x2 y2
dx dy =
cos
cos sin
= −2 ∫ cos θ sin θ e − r
0
π
e
0
(
2
r2
cos
2
0
0
4 r 2 cos sin
e
r2
r2
r dr d
( 2 r ) dr d
cosθ
dθ
0
∵ e f ( r ) f ( r ) dr = e f ( r )
)
= −2 ∫ 2 cos θ sin θ e − cos θ − 1 dθ
0
π
2
0
2
= − ∫ ⎡⎣e − cos θ ( 2 cosθ sin θ ) − sin 2θ ⎤⎦ dθ
=−e
− cos 2 θ
2
cos 2θ
+
2
π
2
0
cos π − cos 0 ⎞
⎛
= − ⎜ e 0 − e −1 +
⎟⎠
⎝
2
⎛ 1 ⎞
= − ⎜1 − − 1⎟
⎝ e ⎠
=
1
e
x2 y2
dx dy , over the region bounded by the circles
( x2 + y2 )
x2 + y2 = a2 and x2 + y2 = b2 (a > b).
Example 3: Evaluate
Solution:
1. Putting x = r cosq, y = r sinq, polar
form of the
(i) circle x 2 + y 2 = a 2 is
r 2 = a 2 , r = a.
(ii) circle x 2 + y 2 = b 2 is
r 2 = b 2 , r = b.
2. The region of integration is the part
bounded between the circles r = a and
r = b.
3. Draw an elementary radius vector OAB
from the origin which enters in the region from the circle r = b and leaves at the
circle r = a.
4. Limits of r : r = b to r = a
Limits of q : q = 0 to = 2
I = ∫∫
4
2
2
2π a r cos θ sin θ
x2 y2
=
⋅ r dr d θ
x
y
d
d
∫0 ∫b
r2
( x2 + y2 )
a
2π
= ∫ cos 2 θ sin 2 θ
0
Fig. 8.43
2
4
4
2π sin 2θ ( a − b )
r4
dθ = ∫
⋅
dθ
0
4 b
4
4
Multiple Integral
=
a4 − b4
16
∫
2π
0
⎛ a4 − b4 ⎞
(1 − cos 4θ )
sin 4θ
dθ = ⎜
θ−
2
4
⎝ 32 ⎟⎠
8.47
2π
0
⎛a −b ⎞
( 2π )
=⎜
⎝ 32 ⎟⎠
4
=
4
π 4
(a − b4 )
16
( x 2 + y 2 )2
dx dy , over the region common to the circles
x2 y2
x2 + y2 = ax and x2 + y2 = by (a, b > 0).
Example 4: Evaluate
Solution:
1. Putting x = r cosq, y = r sinq, polar form of the
(i) circle x 2 + y 2 = ax is r 2 = ar cos , r = a cosq
(ii) circle x 2 + y 2 = by is r 2 = br sin , r = b sinq
Fig. 8.44
2. The region of integration is the common part of the circles r = a cosq and r = b sinq.
3. The point of intersection of the circle r = a cosq and r = b sinq, is obtained as
a
a
b sinq = a cosq, tan = , = tan −1
b
b
a
Hence, at P , = tan −1
b
4. Divide the region into two subregions OAP and OBP. Draw an elementary radius
vector OA and OB in each subregion.
(i) In subregion OAP, elementary radius vector OA starts from the origin and
terminates on the circle r = b sinq.
8.48
Engineering Mathematics
Limits of r : r = 0 to r = b sinq
a
b
(ii) In subregion OBP, elementary radius vector OB starts from the origin and
terminates on the circle r = a cosq.
Limits of r : r = 0 to r = a cosq
a
Limits of q : = tan 1 to =
b
2
Limits of q : q = 0 to q = tan
I=∫∫
=∫
( x 2 + y 2 )2
dx dy
x2 y2
tan −1
a
b
0
=∫
1
∫
b sin θ
0
tan −1
a
b
0
π
a cos θ
r4
r4
2
d
d
⋅
+
r
r
θ
∫tan−1 a ∫0 r 4 sin 2 θ cos2 θ ⋅ r dr dθ
r 4 sin 2 θ cos 2 θ
b
1
r2
sin 2 θ cos 2 θ 2
b sin θ
dθ + ∫
0
π
2
tan −1
1
r2
sin 2 θ cos 2 θ 2
a
b
a cos θ
dθ
0
π
a
=
1 tan−1 b
1
1
1
⋅ b 2 sin 2 θ dθ + ∫ 2 −1 a
⋅ a 2 cos 2 θ dθ
2 ∫0
2 tan b sin 2 θ cos 2 θ
sin 2 θ cos 2 θ
=
b2
2
∫
tan −1
a
b
0
π
sec 2 θ dθ +
a2 2
b2
2
cos
ec
θ
d
θ
=
tan θ
a
2 ∫tan−1 b
2
⎤ a2
b2 ⎡
⎛a⎞
tan tan −1 ⎜ ⎟ − tan 0 ⎥ −
⎢
⎝b⎠
2 ⎣
⎦ 2
ab ab
=
+
= ab
2
b
0
0
e - (x
2
+ y2 )
0
a
b
+
a2
− cot θ
2
dx dy .
Solution:
1. Limits of x : x = 0 to x → ∞
Limits of y : y = 0 to y → ∞
2. The region of integration is the first quadrant.
3. Putting x = r cosq, y = r sinq, the integral
changes to polar form.
4. Draw an elementary radius vector which
starts from the origin and extends up to
infinity.
Limits of r : r = 0 to r → ∞
Limits of q : q = 0 to
I=
0
0
2
0
0
e
e
( x2 y2 )
r2
=
Fig. 8.45
2
dx dy
r dr d
1
2
π
2
tan −1
2
2
⎡ π
⎤ a
⎛ −1 a ⎞ ⎤ b ⎡ a
⎢cot 2 − cot ⎜⎝ tan b ⎟⎠ ⎥ = 2 ⎢ b − 0 ⎥ − 2
⎣
⎦
⎦
⎣
=
Example 5:
tan −1
2
0
0
e
r2
( 2r ) dr d
a
b
b⎤
⎡
⎢0 − a ⎥
⎣
⎦
Multiple Integral
1
2
1
2
=
2
0
2
0
e
r2
8.49
∵ e f ( r ) f ( r ) dr = e f ( r )
d
0
(0
)
1
2
e0 d
2
0
4
dx dy
Example 6:
-
3
2 2
2
.
(1 + x + y )
Solution:
1. Limits of x : x → − ∞ to x → ∞
Limits of y : y → − ∞ to y → ∞
2. The region of integration is the entire coordinate plane.
3. Putting x = r cosq, y = r sinq, integral changes to polar form.
4. Draw an elementary radius vector which starts from origin and extends upto
Limits of r : r = 0 to r → ∞
Limits of q : q = 0 to q = 2p
Fig. 8.46
dx dy
I=
=
3
2 2
(1 + x 2 + y )
=
1
2
1
2
2
0
3
(1 + r 2 ) 2
0
(1 + r 2 ) 2 (2r ) dr d
2
2(1 r )
1
2
d
0
2π
2π
0
0
= − ∫ (0 − 1) dθ = θ
= 2π
0
3
2
0
r dr d
2
0
∵
[ f ( r ) ] f ( r ) dr =
n
[ f (r )]n+1
n +1
8.50
Engineering Mathematics
2 x - x2
1
Example 7:
0
x
( x 2 + y 2 ) dx dy .
Solution:
1. Limits of y : y = x to y
2x x2
Limits of x : x = 0 to x = 1
2. The region of integration is bounded by the line y = x and the circle x2 + y2
3. Putting x = r cosq, y = r sinq, polar form of the
(i) line y = x is r sinq = r cosq, tanq = 1,
(ii) circle x 2
r2
y2
2x
=
2x = 0.
4
0 is
2r cos
0
r = 2 cosq
4. Draw an elementary radius vector OA which starts from the origin and terminates
on the circle r = 2 cosq.
Limits of r : r = 0 to r = 2 cosq
Limits of q :
=
to
4
=
2
Fig. 8.47
2 x x2
1
I
0
=
2
4
x
r4
4
( x2
y 2 )dx dy
2
4
2 cos
0
r 2 r dr d
2 cos
d =4
0
2
4
cos 4 d = 4
2
4
1 + cos 2
2
2
d
Multiple Integral
8.51
1 + cos 4θ ⎞
⎛
= ∫π2 1 + 2 cos 2θ + cos 2 2θ dθ = ∫π2 ⎜1 + 2 cos 2θ +
⎟⎠ dθ
⎝
2
4
4
π
(
)
3
2 sin 2θ sin 4θ
= θ+
+
2
2
8
=
π
π
2
π
4
=
π⎞ 1
3 ⎛π π ⎞ ⎛
⎜ − ⎟ − ⎜ sin π − sin ⎟⎠ + (sinn 2π − sin π )
2⎝2 4⎠ ⎝
2
8
2
2
3π
+1
8
Example 8:
1
1 – x2
0
x – x2
x y e –( x + y )
dx dy .
x2 + y2
Solution:
x x 2 to y
1 x2
1. Limits of y : y
Limits of x : x = 0 to x = 1.
2. The region of integration is the part of the first quadrant bounded by the circles
x2 + y2 x = 0 and x2 + y2 = 1.
3. Putting x = r cosq, y = r sinq, polar form of the
(i) circle x 2 y 2 x 0 is
r 2 r cos
0, r cos
(ii) circle x2 + y2 = 1 is r2 = 1, r = 1
4. Draw an elementary radius vector OAB from the origin which enters in the region
from the circle r = cosq and leaves the region at the circle r = 1.
Limits of r : r = cosq to r = 1
Limits of q : q = 0 to
=
2
Fig. 8.48
8.52
Engineering Mathematics
I=∫
1
0
∫
1− x 2
x − x2
2
2
2
1
r 2 sin θ cos θ e − r
x y e−( x + y )
dx dy = ∫ 2 ∫
⋅ r dr dθ
2
2
0 cosθ
r2
x +y
π
π
1
2
1 2
sin θ cos θ ∫ e − r ( −2r ) dr dθ
∫
0
θ
cos
2
π
2 1
1
⎡∵ e f ( r ) f ′( r )dr = e f ( r ) ⎤
= − ∫ 2 sin θ cos θ e − r
dθ
⎣ ∫
⎦
cosθ
2 0
π
π
2
2
1
1 1 ⎛1
⎞
= − ∫ 2 sin θ cos θ e −1 − e − cos θ dθ = − ∫ 2 ⎜ sin 2θ − e − cos θ ⋅ 2 sin θ cosθ ⎟ dθ
⎠
2 0
2 0 2 ⎝e
=−
(
)
1 1 ⎛ cos 2θ ⎞ − cos2 θ
=−
⎜−
⎟−e
4 e⎝
2 ⎠
π
2
⎡∵ e f (θ ) f ′(θ )dθ = e f (θ ) ⎤
⎣ ∫
⎦
0
π⎞
⎛
− ⎜ cos 2 ⎟
2 ⎤
1⎡ 1
= − ⎢ − (cos π − cos 0 ) − e ⎝ 2 ⎠ + e − cos 0 ⎥
4 ⎢⎣ 2e
⎥⎦
1 ⎡1
1⎤
1⎡ 1
⎤
= − ⎢ − ( −2) − e 0 + e −1 ⎥ = − ⎢ − 1 + ⎥
e
e⎦
4
4 ⎣ 2e
⎣
⎦
=
1 ⎡ 2⎤
1−
4 ⎢⎣ e ⎥⎦
Example 9:
2
1+ 2 x - x2
0
1- 2 x - x2
dx dy
.
( x 2 + y 2 )2
Solution:
1. Limits of
y : y 1 2 x x 2 to y 1
2x x2
Limits of x : x = 0 to x = 2
2. The region of integration is the circle
x2 + y2 2x 2y + 1 = 0 with centre
at (1, 1) and radius 1.
3. Putting x = r cosq, y = r sinq, polar form
of the circle x2 + y2 2x 2y + 1 = 0 is
r 2 2r (cos
sin ) 1 0 .
4. Draw an elementary radius vector OAB
from origin which enters in the region
from the lower part of the circle where
r (cos
sin )
sin 2 and leaves the
region at the upper part of the circle where
Fig. 8.49
r = cos + sin + sin 2 .
Limits of r : r
(cos
Limits of q : q = 0 to
sin )
=
2
sin 2
to r = (cos + sin ) + sin 2
Multiple Integral
I=
=
=
2
1
2 x x2
0
1
2 x x2
(cos
2
r
2
sin )
1
2
1
2
2
0
2
r dr d
r4
where α = cos θ + sin θ
d ,
1
2
0
sin 2
+
2
0
=2
dx dy
( x + y 2 )2
2
(cos + sin ) + sin 2
2
0
8.53
1
(
)
2
(
(
)2
(
0
2
(
)2
)2
2 2
)
and
β = sin 2θ
d
1
2
d
(cos + sin ) sin 2
d =2
(1 sin 2 sin 2 ) 2
4
2
0
(
2
2 2
)
3
2
0
d
1
3
1
2 (cos ) 2 (sin ) 2 + (sin ) 2 (cos ) 2 d
⎡ 1 ⎛ 5 3 ⎞ 1 ⎛ 5 3 ⎞⎤
= 2 2 ⎢ B ⎜ , ⎟ + B ⎜ , ⎟⎥
⎣ 2 ⎝ 4 4 ⎠ 2 ⎝ 4 4 ⎠⎦
⎡ π2 p
1 ⎛ p + 1 q + 1⎞⎤
q
,
⎢∵ ∫0 sin θ cos θ dθ = B ⎜⎝
⎟⎥
2
2 ⎠⎦
2
⎣
5 3
1 1
1
1−
=2 2 4 4 =2 2
4 4
4
2
1 π
=
2 sin π
4
=π
Example 10:
4a
0
y
y2
4a
∵ n1 n
n
sin n
x2 - y2
dx dy .
x2 + y2
Solution:
y2
to x = y
4a
Limits of y : y = 0 to y = 4a.
2. The region of integration is bounded by the line y = x and the parabola y 2 = 4ax.
3. Putting x = r cosq, y = r sinq, polar form of the
1. Limits of x : x =
(i) line y = x is r sinq = r cosq, tanq = 1,
2
2
(ii) parabola y = 4ax is r sin
2
=
4
= 4ar cos , r = 4a cot q cosecq
8.54
Engineering Mathematics
Fig. 8.50
4. Draw an elementary radius vector OA which starts from the origin and terminates
on the parabola r = 4a cotq cosecq.
Limits of r : r = 0 to r = 4a cotq cosecq
: =
Limits of
I=∫
4a
0
2
∫
y
y2
4a
x2 − y2
dx dy = ∫
x2 + y2
π
2
π
4
(1 2sin 2 )
4
8a 2
2
to
4
r2
2
=
2
4 a cot θ cosecθ
∫
0
4 a cot cosec
1
2
d
0
(cot 2 cosec 2
r 2 (cos 2 θ − sin 2 θ )
⋅r dr dθ
r2
2
(1 2sin 2 )(4 a) 2 cot 2 cosec 2 d
4
2 cot 2 ) d
4
8a 2
2
{
}
(cot 2 )( cosec 2 )
2cosec 2
2 d
4
8a
2
cot 3
3
⎡
[ f ( )]n+1 ⎤⎥
n
⎢ ∵ ∫ [ f ( ) ] f ′( ) d =
n + 1 ⎥⎦
⎢⎣
2
2 cot
2
4
8a 2
1
cot 3
3
2
cot 3
4
2 cot
2
cot
4
2
2
4
Multiple Integral
1
( 1) 2( 1) 2
3
4
8a 2
8a 2
8a 2
5
3
8.55
2
5
3
2
a
Example 11:
5 ax - x 2
2 ax
0
x2 + y2
dx dy .
y2
Solution:
5ax x 2
1. Limits of y : y 2 ax to y
Limits of x : x = 0 to x = a
2. Since the limits of x and y are positive, the region of integration is the
part of the first quadrant bounded
by the parabola y 2 = 4ax and the
circle x 2 y 2 5ax 0
3. Putting x = r cosq, y = r sinq, polar
form of the
(i) parabola y 2 = 4ax is
r 2 sin 2 = 4a r cos ,
r = 4a cotq cosecq
(ii) circle x 2 y 2 5a x 0 is
r 2 5a r cos
0,
r = 5a cosq
4. The point of intersection of r = 4a
cotq cosecq and r = 5a cosq is obtained as
4a cotq cosecq = 5a cosq
4
sin 2 = ,
5
2
5
= ± sin −1
Hence, at P,
= sin
Fig. 8.51
1
2
5
5. Draw an elementary radius vector OAB from the origin which enters in the region
from the parabola r = 4a cotq cosecq and terminates on the circle r = 5a cosq.
Limits of r : r = 4a cotq cosecq to r = 5a cosq
2
Limits of q : = sin 1
to =
2
5
I=∫
a
0
=∫
∫
5 ax − x 2
2 ax
π
2
sin
−1
2
5
∫
x2 + y2
dx dy
y2
5 a cos θ
4 a cot θ cosecθ
r
r dr dθ
r sin 2 θ
2
8.56
Engineering Mathematics
π
= ∫ 2 −1
5 a cos θ
2
sin
=∫
5
π
2
2
sin −1
=∫
cosec 2θ r 4 a cot θ cosec θ dθ
cosec 2θ (5a cos θ − 4 a co
ot θ cosecθ ) dθ
5
π
2
2
sin −1
⎡⎣5a cot θ cosecθ + 4 a cosec 2θ ( − cosecθ cot θ ) ⎤⎦ dθ
5
cosec3θ
= −5a cosecθ + 4 a
3
π
2
sin −1
⎡
[ f (θ )]n+1 ⎤⎥
n
⎢∵ ∫ [ f (θ ) ] f ′(θ )dθ =
n + 1 ⎥⎦
⎢⎣
2
5
⎡
π
π 4a
2 ⎞⎤
2 ⎞ 4a
⎛
⎛
= ⎢ −5a cosec + 5a cosec ⎜ sin −1
+ cosec3 − cosec3 ⎜ sin −1
⎟⎥
⎟
⎝
⎝
2 3
2
5⎠ 3
5 ⎠⎦
⎣
3
5
5a 5a
2
(
4a
3
4a
3
⎡
⎛
⎛ −1 2 ⎞
5 ⎞⎤
= cosec ⎜ cosec −1
⎢∵ cosec ⎜ sin
⎥
⎟
⎝
2 ⎟⎠ ⎥
5⎠
⎝
⎢
⎢
⎥
5
⎢
⎥
=
⎢⎣
⎥⎦
2
5
2
)
a
5 5 11
3
Exercise 8.4
(I) Evaluate the following:
− r2
1.
∫ ∫ re a cos sin dr d , over the
2
3.
r sin dA, over
the
cardioid
r = a (1 + cosq ) above the initial line.
4
Ans. : a3
3
4.
∫∫
upper half of the circle r = 2a cosq.
Ans. :
2.
a2
1
3+ 4
16
e
3
r dr d , over the region between
the circles r = 2 sinq and r = 4 sinq.
45
Ans. :
2
r
dr d , over one loop of
r +4
the lemniscate r 2 = 4 cos 2 .
2
[ Ans. : (4
)]
(II) Change to polar coordinates and evaluate the following:
1.
1
2.
dx dy, over the region
xy
bounded by the semi-circle
x 2 y 2 x 0, y 0.
y 2 dx dy, over the area outside
the circle x 2
y2
2
2
the circle x
y
ax
2ax
0 and inside
0.
Ans. :
Ans. :
2
15 a 4
64
Multiple Integral
8.57
sin( x 2 + y 2 ) dx dy, over the circle
3.
3 a4
4
Ans. :
x2 + y 2 = a2 .
(1 cos a 2 )
Ans. :
(
xy x 2 + y
4.
10.
4 xy
e
x + y2
x x2
1
0
( x2 y2 )
2
0
3
2 2
) dx dy, over the first
dx dy
Ans. :
quadrant of the circle x 2 + y 2 = a 2 .
Ans. :
3
5.
a2 y 2
2
y
0
a
x
0
0
3
log 3
2
12.
a
a
0
y
a2 y 2
Ans. :
x2 + y2
Ans. :
(
a
log 1 + 2
4
)
13.
a
a2 x 2
0
ax x 2
7.
a2 x 2
0
0
sin
a
2
(a2
x2
8.
0
0
x2
e
x2 y2
9.
2 ax x 2
2a
0
0
1
2
dx dy
(4 a + x 2 + y 2 ) 2
1
8a 2 4
1
2
tan
1
1
2
dx dy
a
2
x2
y2
[ Ans. : a]
14.
a2 x 2
a
ax x
0
2
xy
e
x + y2
2
( x2 y2 )
dx dy
1 ⎡
⎡
2
− a2 ⎤ ⎤
⎢ Ans. : 4a 2 ⎣1 − (1 + a )e ⎦ ⎥
⎣
⎦
dx dy
Ans. :
a 2 log a
y 2 ) dx dy
a2
Ans. :
2
2
4
2
x 3 dx dy
4
a
log e ( x 2 + y 2 ) dx dy
Ans. :
x2 + y2
0
Ans. :
6.
a
11.
dy dx
3x
0
a7
14
1
e
(1
4
e
2
)
15.
1
x
dx dy
0
x2
x2 + y2
( x 2 + y 2 ) dx dy
Ans. :
2 1
8.5 CHANGE OF VARIABLES OF INTEGRATION
In some cases, evaluation of double integral becomes easier by changing the variables. Let the variables x, y be replaced by new variables u, v by the transformation
x = f1 (u, v ), y = f 2 (u, v ), then
f ( x, y ) dx dy =
f ( f1 , f 2 ) J du dv
... (1)
8.58
Engineering Mathematics
Jacobian, J =
where
( x, y )
( u, v )
=
x
u
y
u
x
v
y
v
Using Eq. (1), the double integral can be transformed to new variables.
Example 1: Using the transformation x - y = u, x + y = v, evaluate
cos
x- y
dx dy over the region bounded by the lines x = 0, y = 0, x + y = 1.
x+ y
Solution: x
y
u, x y v
u v
v u
x=
, y=
2
2
J=
( x, y )
=
( u, v )
x
u
y
u
x
v
y
v
1 1
2 2 1 1 1
=
= + =
1 1 4 4 2
2 2
dx dy = J du dv
1
du dv
2
The region bounded by the lines x = 0, y = 0 and x + y = 1 in xy-plane is a triangle OPQ.
u+v
v−u
Under the transformation x =
,
and y =
2
2
(i) the line x = 0 gets transformed to the line u = –v
=
Fig. 8.52
Multiple Integral
8.59
(ii) the line y = 0 gets transformed to the line u = v
(iii) the line x + y = 1 gets transformed to the line v = 1
Thus, triangle OPQ in xy-plane gets transformed to triangle OP'Q' in uv-plane
bounded by the lines u = v, u = v and v = 1.
In the region, draw a horizontal strip AB parallel to u-axis which starts from the line
u = v and terminates on the line u = v.
Limits of u : u = v to u = v
Limits of v : v = 0 to v = 1
I=
cos
1
2
1
0
x
x
v sin
y
dx dy =
y
1
v2
2sin 1
2
2
=
v
u
v
1
1
2
dv
v
v
0
v
1
0
u 1
du dv
v 2
cos
v [sin 1 sin ( 1) ] dv
1
0
1
sin 1
2
Example 2: Using the transformation x 2 - y 2 = u, 2 xy = v , find
( x 2 + y 2 ) dx dy
over the region in the first quadrant bounded by x2 - y2 = 1, x2 - y2 = 2, xy = 4, xy = 2.
Solution: x2
y2 = u, 2xy = v
It is difficult to express x and y in terms of u and v, therefore we write Jacobian of
u, v in terms of x and y.
∂u
∂ ( u , v ) ∂x
J=
=
∂ ( x , y ) ∂v
∂x
∂u
∂y
=
∂v
∂y
2x
− 2y
2y
2x
= 4 ( x2 + y2 )
du dv = J dx dy
= 4 ( x 2 + y 2 ) dx dy
dx dy =
1
du dv
4 ( x + y2 )
2
The region in xy-plane bounded by the curves x2 y2 = 1, x2 y2 = 2, xy = 4, xy = 2
is transformed to a square in uv-plane bounded by the lines u = 1, u = 2, v = 4, v = 8.
I = ∫∫ ( x 2 + y 2 ) dx dy = ∫
2
1
1 2 8
u v
4 1 4
=1
=
∫
8
4
( x2 + y2 )
1
du dv
4 ( x + y2 )
2
8.60
Engineering Mathematics
Fig. 8.53
Example 3: Using the transformation x + y = u, y = uv, show that
1
1− x
0
0
∫∫
y
e x + y dy dx =
1
( e − 1).
2
Solution: x + y = u ,
x
y = uv
u (1 v),
y
uv
( x, y )
=
J=
( u, v )
1 v
v
u
u
x
u
y
u
x
v
y
v
(1 v ) u u v
u
dx dy = J du dv
= u du dv
Limits of y : y = 0 to y = 1 x
Limits of x : x = 0 to x = 1.
The region in xy-plane is the triangle OPQ bounded by the lines x = 0, y = 0 and
x + y = 1.
Under the transformation x = u (1 v) and y = uv,
(i) the line x = 0 gets transformed to the line u = 0 or v = 1
(ii) the line y = 0 gets transformed to the line u = 0 or v = 0
(iii) the line x + y = 1 gets transformed to the line u = 1
Multiple Integral
8.61
Fig. 8.54
Thus, the triangle OPQ in the xy-plane gets transformed to the square OP'Q'R' in
uv-plane bounded by the lines u = 0, v = 0, u = 1 and v = 1.
In the region, draw a vertical strip AB parallel to the v-axis which starts from the
u-axis and terminates on the line v = 1.
Limits of v : v = 0 to v = 1
Limits of u : u = 0 to u = 1
I=
1
1 x
0
0
ev
1
0
y
e x + y dx dy =
u2
2
1
1
0
0
1
(e1 e 0 )
0
e v u du dv
1
2
1
(e 1)
2
Example 4: Using the transformation x = u (1 + v ), y = v (1 + u), u
evaluate
2
0
y
0
( x - y ) 2 + 2( x + y ) + 1
1
2
dy dx .
Solution: x = u (1 + v), y = v (1 + u )
( x, y )
J=
=
( u, v )
x
u
y
u
x
u
v 1+ v
=
= 1+ u + v
y
v 1 u
v
dx dy = J du dv = (1 + u + v ) du dv
Limits of x : x = 0 to x = y
Limits of y : y = 0 to y = 2.
0, v
0,
8.62
Engineering Mathematics
Fig. 8.55
The region in the xy-plane is the OPQ bounded by the lines x = 0, y = 2 and y = x.
Under the transformation x u (1 v ), y v (1 u ), u 0, v 0
(i) the line x = 0 gets transformed to the line u = 0
(ii) the line y = 2 gets transformed to the curve v (1 + u) = 2
(iii) the line y = x gets transformed to the line u = v
Thus, the triangle OPQ in the xy-plane gets transformed to the region OP'Q' in
uv plane bounded by the lines u = 0, u = v and the curve v (1 + u) = 2.
The point of intersection of u = v and v (1 + u) = 2 is obtained as u2 + u 2 = 0, u = 1,
2 and v = 1, 2.
Hence, P' : (1, 1)
In the region, draw a vertical strip AB parallel to the v-axis which starts from the
line u = v and terminates on the curve v (1 + u) = 2.
2
Limits of v : v = u to v =
1+ u
Limits of u : u = 0 to u = 1
y
2
I
=
0
0
1
2
1+ u
0
u
1
2
1+ u
0
u
1
2
1+ u
0
u
y)2
(x
2( x
(u v ) 2
dy dx
2(u v 2uv ) 1
1
2
(1 u v ) du dv
(1 + u + v ) 1 (1 + u + v ) dv du
dv du
1
0
u2
2 log (1 u )
2
2 log 2
1
2
y) 1
1
2
2
v 1u+ u du
1
0
1
0
2
u du
1+ u
Multiple Integral
Example 5: Evaluate
8.63
xy dx dy by changing the variables over the region in
the first quadrant bounded by the hyperbolas x2 - y2 = a2, x2 - y2 = b2 and the
circles x2 + y2 = c2, x2 + y2 = d2 with 0 < a < b < c < d.
Solution: Let x 2
y2
u, x 2
y2
v
x2 =
u v
,
2
y2 =
v u
2
1
x
x
4
x
u
v
=
1
y
y
4y
u
v
1
dx dy = J du dv =
du dv
8 xy
( x, y )
J=
=
( u, v )
xy dx dy =
1
1
4x
=
1
8 xy
4y
du dv
8
The region bounded by the hyperbolas x2 y2 = a2, x2 y2 = b2 and the circles x2 + y2 = c2,
x2 + y2 = d 2 in xy-plane is the curvilinear rectangle PQRS.
Under the transformation x2 y2 = u and x2 + y2 = v,
(i) the hyperbolas x2 y2 = a2, x2 y2 = b2 get transformed to the lines u = a2,
u = b2 respectively.
(ii) the circles x2 + y2 = c2, x2 + y2 = d 2 get transformed to the lines v = c2, v = d 2
respectively.
Fig. 8.56
8.64
Engineering Mathematics
Thus, the curvilinear rectangle PQRS in the xy-plane gets transformed to the
rectangle P'Q'R'S' in uv-plane bounded by the lines u = a2, u = b2, v = c2 and v = d 2.
In the region, draw a vertical strip AB parallel to v-axis which starts from the line
v = c2 and terminates on the line v = d 2.
Limits of v : v = c2 to v = d 2
Limits of u : u = a2 to u = b2
b2
d2
1
I
du dv
xy dx dy
2
u=a
v = c2 8
1 b2 d 2 1 2
(b a 2 ) ( d 2 c 2 )
u 2 v 2
8 a c
8
( x + y )2 dx dy , by changing the variables over the paral-
Example 6: Evaluate
lelogram with vertices (1, 0), (3, 1), (2, 2), (0, 1).
Solution: The region of integration in xy-plane is the parallelogram PQRS.
Equations of the sides of the parallelogram are obtained as
1 0
( x 1)
3 1
2y x 1
x 2y 1
(i) PQ : y 0
(ii) RS : y 1
2y 2
x 2y
2 1
( x 0)
2 0
x
2
1 0
( x 1)
0 1
x+ y =1
(iii) PS : y 0
(iv) QR : y − 1 =
2 −1
( x − 3)
2−3
Fig. 8.57
y 1
x 3
x+ y = 4
Let x 2 y
x=
u, x
y
v
u 2v
v u
, y=
3
3
∂x ∂x
1
∂ ( x , y ) ∂u ∂v
3
J=
=
=
1
∂ ( u , v ) ∂y ∂y
−
∂u ∂v
3
1
dx dy = J du dv = du dv
3
2
3 1
=
1 3
3
Multiple Integral
Under the transformation x
x+y=v
2y = u, and
(i) the lines x 2y = 1, x 2y = 2 get
transformed to the lines u = 1, u = 2
respectively.
(ii) the lines x + y = 1, x + y = 4 get
transformed to the lines v = 1, v = 4
respectively
Thus, the parallelogram PQRS in the
xy-plane gets transformed to a square
P'Q'R'S' in uv-plane bounded by the lines
u = 1, u = 2, v = 1 and v = 4.
In the region, draw a vertical strip AB
parallel to v-axis which starts from the line
v = 1 and terminates on the line v = 4.
Limits of v : v = 1 to v = 4
Limits of u : u = 2 to u = 1
I=
ellipse
Fig. 8.58
( x + y ) 2 dx dy =
1 1 v3
= u 2
3
3
Example 7: Evaluate
8.65
1
u
4
2 v 1
1
v 2 du dv
3
4
= 21
1
x2 y2
xy 2 + 2
a
b
n
2
dx dy , over the first quadrant of the
x2 y2
+
= 1.
a 2 b2
Solution:
∂x ∂x
∂ ( x , y ) ∂r ∂
J=
=
∂ ( r , ) ∂y ∂y
∂r ∂
a cos
− a r sin
=
= ab r
b sin
b r cos
d x d y = | J | d r d = ab r d r d
Under the transformation x = ar cos q,
x2 y2
y = br sin q, the ellipse 2 + 2 = 1 in
a
b
the xy-plane gets transformed to r2 = 1 or
Fig. 8.59
8.66
Engineering Mathematics
r = 1, circle with centre (0, 0) and radius
1 in the rq -plane.
The region of integration is the part
of the circle r = 1 in first quadrant in the
rq -plane. In the region, draw an elementary radius vector OA from the pole which
terminates on the circle r = 1.
Limits of r : r = 0 to r = 1
Limits of q : q = 0 to
=
2
n
2
2
x
y2
xy 2 + 2
a
b
I=
=
2
0
1
0
dx dy
Fig. 8.60
n
2 2
ab r 2 cos sin ( r ) ab r d r d
= a2b2
2
0
sin 2
2
a 2 b 2 cos 2θ
=
−
2
2
1
0
π
2
0
(r )n+3 d r
1
r n+ 4
n+4 0
2 2
ab
a2b2
1
=
( − cos π + cos 0) ⋅
n + 4 2 ( n + 4)
4
=
Exercise 8.5
1. Using the transformation x + y = u,
x
y = v, evaluate
e
x y
x+ y
dx dy over
the region bounded by x = 0, y = 0
and x + y = 1.
Ans. :
1
1
e
4
e
2. Using the transformation x2 y2 = u,
2xy = v, evaluate ( x 2 y 2 ) dx dy
over the region bounded by the hyperbolas x2 y2 = 1, x2 y2 = 9, xy = 2
and xy = 4.
[Ans. : 4]
3. Using the transformation x + y = u, y = uv
evaluate
0
0
e
( x y)
x p 1 y q 1 dx dy.
Ans. : p q
4. Using the transformation x = u, y = uv,
evaluate
1
x
0
0
x 2 + y 2 dx dy.
⎡
⎤⎤
1⎡ 2 1
+ log 1 + 2 ⎥ ⎥
⎢ Ans. : ⎢
3⎣ 2 2
⎢⎣
⎦ ⎥⎦
(
5. Evaluate
)
( x + y ) 2 dx dy by changing
the variables over the region bounded by
the parallelogram with sides x + y = 0,
x + y = 2, 3x 2y = 0 and 3x 2y = 3.
Ans. :
8
5
Multiple Integral
6. Evaluate
(x
1
y ) 4 e x + y dx dy, by
changing the variables over the region
bounded by the square with vertices at
(1, 0), (2, 1), (1, 2), (0, 1).
Ans. :
8.67
e3
7. Evaluate [ xy (1 x y )] 2 dx dy, by
changing the variables over the region
bounded by the triangle with sides
x = 0, y = 0, x + y = 1.
e
Ans. :
5
2
105
8.6 TRIPLE INTEGRAL
Let f (x, y, z) be a continuous function defined in a closed and bounded region V in
3-dimensional space. Divide the region V into small elementary parallelopipeds by
drawing planes parallel to the coordinate planes. Let the total number of complete
parallelopipeds which lie inside the region V is n. Let dVr be the volume of the r th
parallelopiped and (xr, yr, zr) be any point in this parallelopiped. Consider the sum
n
S = ∑ f ( xr , yr , zr ) Vr
... (1)
r =1
vr = xr ⋅ yr ⋅ z r
where
If we increase the number of elementary parallelopipeds, i.e., n, then the volume of
each parallelopiped decreases. Hence as n → ∞, Vr → 0.
The limit of the sum given by Eq. (1), if it exists is called the triple integral of f (x, y, z)
over the region V and is denoted by
f ( x , y , z ) dV
v
Hence,
∫∫∫
f ( x, y, z ) dV = lim
n
∑ f (x , y , z )
r
n→∞
Vr → 0 r =1
v
r
r
Vr
dV = dx dy dz
where,
8.6.1 Evaluation of Triple Integral
Triple integral of a continuous function f (x, y, z) over a region V can be evaluated by
three successive integrations.
Let the region V is bounded below by a surface z = z1 (x, y) and above by a surface
z = z2 (x, y). Let the projection of region V in xy-plane is R which is bounded by the
curves y = y1 (x), y = y2 (x) and x = a, x = b. Then the triple integral is defined as
I=
b
y2 ( x )
a
y1 ( x )
{
z2 ( x , y )
z1 ( x , y )
}
f ( x , y , z ) dz dy dx
Note: The order of variables in dx dy dz indicates the order of integration. In some
cases this order is not maintained. Therefore it is advisable to identify the order of
integration with the help of the limits.
8.68
Engineering Mathematics
8.6.2 Triple Integral in Cylindrical Coordinates
Cylindrical coordinates r, q, z are used to evaluate the integral in the regions which
are bounded by cylinders along z-axis, planes through z-axis, planes perpendicular to
the z-axis.
Fig. 8.61
Relations between cartesian (rectangular) coordinates (x, y, z) and cylindrical coordinates (r, q, f ) are given as x = r cosq
y = r sinq
z=z
Then
where,
f ( x, y, z ) dx dy dz =
f ( r cos , r sin , z ) J dr d dz
∂x
∂r
∂ ( x , y , z ) ∂y
J=
=
∂ ( r , , z ) ∂r
∂z
∂r
cos
= sin
0
Hence,
f ( x , y , z ) dx dy dz =
r sin
r cos
0
∂x
∂
∂y
∂
∂z
∂
∂x
∂z
∂y
∂z
∂z
∂z
0
0 =r
1
f ( r cos , r sin , z ) r dr d dz
Multiple Integral
8.69
8.6.3 Triple Integral in Spherical Coordinates
Spherical coordinates (r, q, f) are used to evaluate the integral in the regions which are
bounded by sphere with centre at the origin, cone with vertices at the origin and axis
as z-axis.
Relations between cartesian (rectangular) coordinates (x, y, z) and spherical coordinates (r, q, f) are given as
x = r sinq cosf
y = r sinq sinf
z = r cosq
Then
where,
f ( x , y , z ) dx dy dz =
f ( r sin cos , r sin sin , r cos ) J dr d d
∂x
∂r
∂ ( x , y , z ) ∂y
J=
=
∂ ( r , θ , φ ) ∂r
∂z
∂r
sin θ cos φ
= sin θ sin φ
cosθ
= r 2 sin θ
Fig. 8.62
∂x
∂θ
∂y
∂θ
∂z
∂θ
∂x
∂φ
∂y
∂φ
∂z
∂φ
r cos θ cos φ
r cosθ sin φ
− r sin θ sin φ
r sin θ cos φ
− r sin θ
0
8.70
Engineering Mathematics
Hence,
∫∫∫ f ( x, y, z ) dx dy dz = ∫∫∫ f (r sin θ cos φ , r sinθ sin φ , r cosθ ) r
2
sin θ dr dθ dφ ⋅
Note: If the region of integration is a sphere x2 + y2 + z2 = a2 with centre at (0, 0, 0) and
radius a, then limits of r, q, f are
(i) For positive octant of a sphere,
r : r = 0 to r = a
:
= 0 to
=
:
= 0 to
=
2
2
(ii) For hemisphere,
r : r = 0 to r = a
: = 0 to
: = 0 to
=
2
=2
(iii) For complete sphere,
r : r = 0 to r = a
: = 0 to =
: = 0 to = 2
8.6.4 Change of Variable
In some cases, evaluation of triple integral becomes easier by changing the variables.
Let the variables x, y, z be replaced by new variables u, v, w by the transformation
x = f 1(u, v, w), y = f 2(u, v, w), z = f 3(u, v, w).
Then
where,
f ( x, y, z ) dx dy dz =
f ( f1 , f 2 , f 3 ) J du dv dw
∂x
∂u
∂ ( x , y , z ) ∂y
J=
=
∂ ( u, v , w )
∂u
∂z
∂u
∂x
∂v
∂y
∂v
∂z
∂v
∂x
∂w
∂y
∂w
∂z
∂w
8.6.5 Working Rule for Evaluation of Triple Integral
1. Draw all the planes and surfaces and identify the region of integration.
2. Draw an elementary volume parallel to z (y or x) axis.
3. Find the variation of z (y or x) along the elementary volume.
Multiple Integral
8.71
4. Lower and upper limits of z (y or x) are obtained from the equation of the surface
(or plane) where elementary volume starts and terminates respectively.
5. Find the projection of the region on xy (zx or yz) plane.
6. Draw the region of projection in xy (zx or yz) plane.
7. Follow the steps of double integration to find the limits of x and y (z and x or y and z).
Note: 1. If the region is bounded by the cylinders along the z-axis, planes through zaxis, the planes perpendicular to the z-axis, then the variables are changed to cylindrical coordinates.
2. If the region is bounded by the sphere, then the variables are changed to spherical
coordinates.
(I) Evaluation of Integral when Limits are Given
Example 1: Evaluate
2
z
0
1
yz
0
xyz dx dy dz .
Solution: The innermost limits depend on y and z. Hence, integrating first w.r.t. x,
I=
2
z
0
1
1
2
x2
2
yz
yz dy dz =
0
4 z
y
z 3 dz
4 1
2
0
1 z8
8 8
z4
4
2
1
8
2
0
1
2
2
z
0
1
( y 2 z 2 ) yz dy dz
( z 4 1) z 3 dz
1
(32 4)
8
0
7
=
2
Example 2: Evaluate
1
1- x
1- x - y
0
0
0
1
dx dy dz .
( x + y + z + 1) 3
Solution: The innermost limits depend on x and y. Hence, integrating first w.r.t. z,
I=
1
1 x
1 x y
0
0
0
1
2
1
2
(x
1
1 x
0
0
1
0
1 x
2 4
1
y z 1)3
dz dy dx =
y (1 x
2
y ) 1}
1 x
1
dx
x y 10
x2
8
1 x
0
0
1
{x
y
4
1
1
2
1
x
log ( x 1)
2
0
1
0
(x
1 x
4
2( x
1
y z 1) 2
1 x y
dy dx
0
1
dy dx
y 1) 2
1
1
x (1 x ) 1
x 1
1 5
log 2
2 8
dx
8.72
Engineering Mathematics
e
Example 3: Evaluate
1
ex
log y
1
log z dx dy dz .
1
Solution: The inner most limit depends on x and middle limit depends on y.
Hence, integrating first w.r.t. z,
e
I
1
1
x ex
1
ex
log y
1
e
1
e
e
y
1
1
1 y2
y 2
e
1
1
log y
1
z
1
dz dx dy
z
[e x x e x 1]dx dy
e log y (log y 2) log y e (1 2) 1 dy
2) log y e 1] dy
y2
2
ex
ex
z log z 1
dx dy
z1
dy
x1
1
ex
log 1
log y
ex
[ y (log y
log y
e
log z dz dx dy
e x log e x
1
e
e
1
log y
e
1
1
ex
log y
e
1
y dy
[( y
y2
1) log y 2 y e 1] dy
e
1
e
(e 1) y 1
e
2
⎛e
⎞
⎛1 ⎞ y
= log e ⎜ + e ⎟ − log 1 ⎜ + 1⎟ −
+ y − (e 2 − 1) + [(e − 1) (e − 1) ]
⎝2 ⎠ 4
⎝2
⎠
1
2
=
⎛ e2
1 ⎞
e2
+ e − ⎜ + e − − 1⎟ − e 2 + 1 + e 2 − 2 e + 1
2
4 ⎠
⎝4
=
13
e2
− 2e +
4
4
Example 4: Evaluate
0
0
0
dx dy dz
.
(1 + x 2 + y 2 + z 2 )2
Solution:
1. It is difficult to integrate in cartesian form. Putting x = r sinq cosf, y = r sinq sinf,
z = r cosq, the integral changes to spherical form.
2. Limits of x : x = 0 to x
Limits of y : y = 0 to y
Limits of z : z = 0 to z
The region of integration is the positive
octant of the plane.
Limits of r : r = 0 to r
= 0 to
Limits of
:
Limits of
: = 0 to
I=∫
∞
0
∞
∞
0
0
∫ ∫
2
=
2
dx dy dz
(1 + x 2 + y 2 + z 2 ) 2
π
π
∞
2
2
φ =0 θ =0 r = 0
=∫
=
∫ ∫
r 2 sin θ
d r dθ dφ
(1 + r 2 ) 2
Fig. 8.63
Multiple Integral
8.73
Putting r = tan t , dr = sec 2 t dt
When r = 0, t = 0
r
,t
2
π
2
0
π
π
0
0
I = ∫ dφ ∫ 2 sin θ dθ ∫ 2
π
2
0
π
2
0
∫
sin 2 t dt
3 1
1⎞ π 2 2 π 1
= ⋅ π
⎟=
4 2
2⎠ 4 2
π
1 ⎛3
= ⋅1 ⋅ B ⎜ ,
2
2 ⎝2
=
π
tan 2 t
⋅ sec 2 t dt = φ 2 − cosθ
4
0
sec t
π2
8
Example 5: Evaluate
a2 - x 2
a
0
a2 - x 2 - y2
0
0
xy z dx dy dz .
Solution:
1. It is difficult to integrate in cartesian form.
Putting x = r sinq cosf, y = r sinq sinf,
z = r cosq, the integral changes to spherical
form.
2. Limits of z : z
0 to z
a2
x2
2
2
y2
a2 x 2
a
0
0
0
2
2
=0
=0
π
2
0
=∫
a2 x 2 y 2
a
r =0
q
1 21 a
⋅
4 3 6
a6
=
48
=
y
O
φ
x
Fig. 8.64
xyz dx dy dz
r 3 sin 2 cos
cos sin
r 2 sin dr d d
π
a
sin 2φ
1 cos 2φ
dφ ∫ 2 sin 3 θ cos θ dθ ∫ r 5 dr = −
0
0
2
2
2
6
P
r
Limits of y : y 0 to y
a x
Limits of x : x = 0 to x = a
The region of integration is the positive
octant of the sphere x2 + y2 + z2 = a2.
Limits of r : r = 0 to r = a
π
Limits of θ : θ = 0 to θ =
2
π
Limits of φ : φ = 0 to φ =
2
I=
z
π
2
0
1
r6
⋅ B ( 2,1)
6
2
a
0
8.74
Engineering Mathematics
(II) Evaluation of Integral Over the Given Region
Example 1: Evaluate
x 2 yz dx dy dz over the region bounded by the planes
x = 0, y = 0, z = 0 and x + y + z = 1.
Solution:
1. Draw an elementary volume AB parallel to z-axis in the region. AB starts from
xy-plane and terminates on the plane x + y + z = 1.
Limits of z : z = 0 to z = 1 x y.
Fig. 8.65
2. Projection of the plane x + y + z = 1 in xy-plane is OPQ. Putting z = 0 in x + y + z = 1,
we get equation of the line PQ as x + y = 1.
3. Draw a vertical strip A'B' in the region OPQ. A'B' starts from the x-axis and terminates on the line x + y = 1.
Limits of y : y = 0 to y = 1 x
Limits of x : x = 0 to x = 1
I=
1
1 x
1 x y
0
0
0
1
2
1
2
1
2
1
1 x
0
0
1
1 x
0
0
1
1 x
0
0
x 2 yz dz dy dx =
x 2 y (1 x
x 2 y (1 x 2
1
1 x
0
0
x2 y
z2
2
1 x y
dy dx
0
y ) 2 dy dx
y2
( x 2 y x 4 y x 2 y3
2 x 2 y 2 xy ) dy dx
2 x3 y 2 x 2 y 2
2 x 3 y 2 ) dy dx
Multiple Integral
8.75
1− x
1 1
y2 x2 y4
y3
= ∫ ( x 2 + x 4 − 2 x3 ) +
− 2 ( x 2 − x3 )
2 0
2
4
3
=
dx
0
2
x2
(1 − x )3 ⎤
1 1⎡ 2
2 (1 − x )
4
2
+
(
1
−
x
)
−
2
x
(
1
−
x
)
⋅
1
x
(
−
x
)
⋅
⎢
⎥ dx
2
4
3 ⎦
2 ∫0 ⎣
1
1 1 x2
1 (1 − x )5 2 (1 − x )6
(1 − x )7
(1 − x ) 4 dx =
⋅x −
⋅ 2x +
⋅2
= ∫ −
2 0 12
24 −5
30
−210
0
1 ⎛
1 ⎞
⎜0 +
⎟
24 ⎝ 105 ⎠
1
=
2520
=
Example 2: Evaluate
x2 + y2 + z2 = 4.
xyz dx dy dz over the positive octant of the sphere
Solution: Putting x = r sinq cosf, y = r
sinq sinf, z = r cosq
equation of the sphere x2 + y2 + z2 = 4 reduces to
z
r 2 sin 2 θ cos 2 φ + r 2 sin 2 θ sin 2 φ + r 2 cos 2 θ = 4
r 2 = 4, r = 2.
q
The region is the positive octant of the
sphere r = 2.
=
xyz dx dy dz =
2
0
sin
3
cos d
1
sin 4
4
2
=
4
3
sin 0
2
0
2
0
y
O
φ
Limits of r : r = 0 to r = 2
π
Limits of q = 0 to θ =
2
π
Limits of f = 0 to φ =
2
I=
P
r
x
Fig. 8.66
2
0
2
0
( r 3 sin 2 cos cos sin ) r 2 sin dr d d
sin 2
d
2
1
(cos
4
2
0
sin 4
r dr =
4
5
6
cos 0)
2
6
2
0
cos 2
4
2
0
r6
6
2
0
8.76
Engineering Mathematics
dx dy dz
Example 3: Evaluate
2
a - x2 - y2 - z2
over the region bounded by the sphere
x2 + y2 + z2 = a2.
Solution:
1. Putting x = r sinq cosf, y = r sinq sinf, z = r cosq equation of the sphere
x 2 + y 2 + z 2 = a 2 reduces to r = a.
2. For the complete sphere, limits of r : r = 0 to r = a
limits of q : q = 0 to q = p
limits of f : f = 0 to f = 2p
I=∫
2π
0
π
a
0
0
∫ ∫
r 2 sin θ dr dθ dφ
2π
= φ 0 − cos θ
a2 − r 2
π
0
∫
a
0
⎛
⎜⎝
2π
π
a
0
0
0
= ∫ dφ ∫ sin θ dθ ∫
r 2 + a2 − a2
a2 − r 2
dr
⎞
− a 2 − r 2 ⎟ dr
⎠
a −r
a2
2
2
a
= ( 2π )( − cos π + cos 0) a 2 sin −1
r r 2 2 a 2 −1 r
−
a − r − sin
a 2
a0
2
⎞
⎛ a2
a2 π
= 4π ⎜ sin −1 1⎟ = 4π ⋅ ⋅ = π 2 a 2
2 2
⎠
⎝2
Example 4: Evaluate
dx dy dz
1
over the region bounded by the
( x2 + y2 + z2 )2
spheres x 2 + y 2 + z 2 = a 2 and x 2 + y 2 + z 2 = b 2 , a > b > 0.
Solution:
1. Putting x = r sinq cosf, y = r sinq sinf, z = r cosq, equation of the spheres
x2 + y2 + z2 = a2 and x2 + y2 + z2 = b2 reduces to r = a and r = b respectively.
Fig. 8.67
Multiple Integral
8.77
2. Draw an elementary radius vector OAB from the origin in the region. This radius vector enters in the region from the sphere r = b and leaves the region at the sphere r = a.
3. Limits of r : r = b to r = a.
For complete sphere, limits of q : q = 0 to q = p
limits of f : f = 0 to f = 2p
I=∫
2π
0
π
a
0
b
∫ ∫
π
a
2π
r 2 sin θ
dr dθ dφ = ∫ dφ ∫ sin θ dθ ∫ r dr
b
0
0
r
a
2π
= φ 0 − cos θ
r2
(a2 − b2 )
= 2π ( − cos π + cos 0)
= 2π ( a 2 − b 2 )
2 b
2
π
0
z 2 dx dy dz over the region common to the sphere
Example 5: Evaluate
x 2 + y 2 + z 2 = 4 and the cylinder x 2 + y 2 = 2 x .
Solution:
1. Putting x = r cos , y = r sin , z = z , equation of the
(i) sphere x2 + y2 + z2 = 4 reduces to
r2 + z2 = 4
z2 4 r2
(ii) cylinder x 2 + y 2 = 2 x reduces to
r2 = 2r cos q, r = 2 cos q
2. Draw an elementary volume parallel to zaxis in the region. This elementary volume
starts from the part of the sphere z 2 4 r 2 ,
below xy-plane and terminates on the part of
the sphere z 2 4 r 2 , above xy-plane.
Limits of r : z = − 4 − r 2 to z = 4 − r 2
3. Projection of the region in rq plane is the
circle r = 2 cosq.
4. Draw an elementary radius vector OA in
the region (r = 2 cosq ) which starts from
the origin and terminates on the circle
r = 2 cosq
Fig. 8.68
Limits of r : r = 0 to r = 2 cosq
Limits of
:
z 2 dx dy dz
I
to
2
2
2
2
2
2 cos
0
z3
3
2 cos
0
4 r2
r dr d
4 r2
2
4 r2
z 2 r dz dr d
4 r2
1
3
2
2
2 cos
0
3
2(4 r 2 ) 2 r dr d
8.78
Engineering Mathematics
=
π
3
⎤
1 2 2 cosθ ⎡
2 2
−
(
−
r
)
( −2r ) dr ⎥ dθ
4
π ∫
⎢
∫
3 −2 0
⎣
⎦
5 2 cosθ
1
2 (4 − r 2 ) 2
= − ∫ 2π
3 −2
5
f ( r )]
n +1
⎤
⎥
⎥⎦
⎡∵ a f ( ) d = 0, if f ( − ) = − f (
⎢ ∫− a
⎢ sin 5 ( − ) = − sin 5
⎣
)⎤
⎥
⎥
⎦
⎡
⎢∵
⎢⎣
π
dθ
0
2
15
(2
5
sin 5
2
0 25
15
2
2
=
n +1
)
25 d
2
6
[
n
∫ [ f ( r ) ] f ′ ( r ) dr =
2
2
64
=
15
15
Example 6: Evaluate
x y z dx dy dz , over the region bounded by the planes
x = 0, y = 0, z = 0, z = 1 and the cylinder x 2 + y 2 = 1.
Solution:
1. Putting x = r cosq, y = r sinq, z = z, equation of the cylinder x 2 + y 2 = 1 reduces to
r2 = 1, r = 1.
2. Draw an elementary volume AB parallel to z-axis in the region. This elementary
volume AB starts from xy-plane and terminates on the plane z = 1.
Limits of z : z = 0 to z = 1.
3. Projection of the region in rq-plane is the part of the circle r = 1 in the first quadrant.
4. Draw an elementary radius vector OA' in the region in the rq-plane which starts
from the origin and terminates on the circle r = 1.
Fig. 8.69
Multiple Integral
8.79
Limits of r : r = 0 to r = 1
Limits of q : q = 0 to
I
x y z dx dy dz
1
0
z dz
2
0
sin 2
d
2
=
1
2
z =0
1
0
2
=0
3
r dr
1
r =0
z2
2
r 2 cos sin
1
0
cos 2
4
z r dr d dz
2
0
r4
4
1
0
1
=
16
Example 7: Evaluate
x 2 + y 2 dx dy dz , over the region bounded by the
right circular cone x 2 + y 2 = z 2 , z > 0 and the planes z = 0 and z = 1.
Solution:
1. Putting x = r cos , y = r sin , z = z , equation of the cone x 2 + y 2 = z 2 reduces to
r2 = z2, r = z.
2. Draw an elementary volume AB parallel to z-axis in the region, which starts from
the cone r = z and terminates on the plane z = 1.
Limits of z : z = r to z = 1.
3. Projection of the region in rq -plane is the curve of intersection of the cone r = z
and the plane z = 1 which is obtained as r = 1, a circle with centre at the origin and
radius 1.
4. Draw an elementary radius vector OA' in the region in xy (rq ) plane which starts
from the origin and terminates on the circle r = 1.
Limits of r : r = 0 to r = 1
Limits of q : q = 0 to q = 2p
Fig. 8.70
8.80
Engineering Mathematics
x2
I
2
1
0
0
2
0
=
=0
2
1
r 2 z r dr d
r3
3
2
y 2 dx dy dz
r4
4
0
1
2
0
d
1
1
r =0
z=r
1
0
r r dz dr d
r 2 (1 r ) dr
1
12
6
Example 8: Evaluate
( x 2 + y 2 ) dx dy dz , over the region bounded by the
paraboloid x 2 + y 2 = 3 z and the plane z = 3.
Solution:
1. Putting x = r cosq, y = r sinq, z = z, equation of the paraboloid x2 + y2 = 3z reduces to
r2 = 3z.
Fig. 8.71
2. Draw an elementary volume AB parallel to z-axis in the region which starts from
the paraboloid r 2 = 3 z and terminates on the plane z = 3.
Limits of z : z =
r2
to z = 3.
3
3. Projection of the region in rq -plane is the curve of intersection of the paraboloid
r 2 = 3 z and the plane z = 3 which is obtained as r 2 = 9, r = 3, a circle with centre
at the origin and radius 1.
4. Draw an elementary radius vector OA' in the region (circle r = 3) which starts
from origin and terminates on the circle r = 3.
Limits of r : r = 0 to r = 3
Limits of q : q = 0 to q = 2p
Multiple Integral
( x2
I
2
3
0
0
2
y 2 ) dx dy dz
0
2
3
r 3 z r 2 dr d
0
3
3
0
r2
3
3
d
0
3
2
0
=
r6
18 0
35
4
2
r 2 r dz dr d
r3 3
r2
dr
3
36
18
81
2
Example 9: Evaluate
the ellipsoid
3
3r 4
4
8.81
x2 y2 z2
- dx dy dz , where V is the volume of
a 2 b2 c 2
1-
x2 y2 z2
+
+
= 1.
a 2 b2 c 2
Solution: Evaluation of integral becomes easier by changing the variables.
x
y
z
x2 y2 z2
= u,
= v, = w , the ellipsoid 2 + 2 + 2 = 1 gets
a
b
c
a
b
c
transformed to u 2 + v 2 + w2 = 1, which is a sphere of radius 1 and centre at origin.
Under the transformation
∂x
∂u
∂ ( x , y , z ) ∂y
Jacobian, J =
=
∂ ( u , v , w ) ∂u
∂z
∂u
∂x
∂v
∂y
∂v
∂z
∂v
∂x
∂w
a 0 0
∂y
= 0 b 0 = abc
∂w
0 0 c
∂z
∂w
dx dy dz = J du dv dw = abc du dv dw
I = ∫∫∫ 1 −
x2 y2 z2
−
− dx dy dz
a2 b2 c2
= ∫∫∫ 1 − u 2 − v 2 − w 2 abc du dv dw
Using u = r sinq cos f, v = r sinq sinf, w = r cosq and du dv dw = r2 sinq dr dq df, the
equation of the sphere u2 + v2 + w2 = 1 reduces to r2 = 1, r = 1.
For complete sphere, limits of r : r = 0 to r = 1
limits of : = 0 to =
limits of : = 0 to = 2
I
2
0
abc
1
0
0
2
0
d
1 r 2 abc r 2 sin dr d d
0
sin d
1
0
r 2 1 r 2 dr
8.82
Engineering Mathematics
Putting r = sin t, dr = cos t dt
When r = 0, t = 0
r = 1, t =
2
2π
I = abc φ 0 − cos θ
π
0
π
2
0
∫
sin 2 t ⋅ cos t ⋅ cos t dt
1 ⎛3 3⎞
= abc (2π ) (2) ⋅ B ⎜ , ⎟
2 ⎝2 2⎠
1 1
3 3
2 2
= 2 abc 2 2 = 2 abc
2
3
2
abc
=
4
2
x 2 y 2 z 2 dx dy dz , over the region bounded by the surfaces
Example 10: Evaluate
xy = 4, xy = 9, yz = 1, yz = 4, zx = 25, zx = 49.
Solution: Evaluation of integral becomes easier by changing the variables. Under the
transformation xy = u, yz = v, zx = w, surfaces gets transformed to u = 4, u = 9, v = 1, v =
4, w = 25, w = 49.
These equations represent the planes parallel to vw, wu and uv planes in the new
coordinate system.
It is easier to find partial derivatives of u, v, w w.r.t. x, y and z.
( u, v , w )
=
( x, y, z )
Jacobian, J =
y
0
x 0
z y
z
0
u
x
v
x
u
y
v
y
u
z
v
z
w
x
w
y
w
z
y ( zx 0) x (0
x
du dv dw = J dx dy dz = 2 xyz dx dy dz
u v w = x2 y 2 z 2
xyz = u v w
du dv dw = 2 uvw dx dy dz
dx dy dz =
1
2 uvw
d u dv dw
yz ) = 2 xyz
Multiple Integral
8.83
Limits of u : u = 4 to u = 9
Limits of v : v = 1 to v = 4
Limits of w : w = 25 to w = 49
I=
x 2 y 2 z 2 dx dy dz
49
4
9
w = 25 v =1 u = 4
=
=
1
2
49
25
1
w 2 dw
1 2w
2 3
3 49
2
1
uvw
4
1
2v
3
2 uvw
1
v 2 dv
3 4
2
25
9
4
2u
3
1
du dv dw
1
u 2 du
3 9
2
4
4
(343 125) (8 1) (27 8)
27
115976
=
27
Exercise 8.6
(I) Evaluate the following:
1.
1
0
dx
2
0
2
dy
1
x 2 yz dz
5.
4
2 z
0
0
4 z x2
0
dy dx dz
[Ans. : 1]
x
log 2
2.
∫ ∫ ∫
0
0
x+ y
0
Ans. :
3.
a2 r 2
a cos
2
0
0
[Ans. : 8p ]
e x + y + z dz dy dx
0
5
8
6.
a sin
2
0
0
a2 r 2
a
0
r dz dr d
Ans. :
r dz dr d
a3
Ans. :
3 2
2
3
7.
2
y
0
0
x+ y
x y
5a 3
64
( x + y + z ) dz dx dy
[Ans. : 16]
4.
π
a (1+ cos θ )
0
0
∫ ∫
∫
h
0
⎡
⎤
r
2 ⎢1 −
⎥
⎣ a (1 + cos θ ) ⎦
r dz dr dθ
a2 h
Ans. :
2
8.
a
∫ ∫
0
0
a2 − x 2
∫
0
a2 − x 2 − y 2
xyz dz dy dx.
Ans. :
a6
48
8.84
Engineering Mathematics
(II) Evaluate the following over the given region of integration:
bounded by the spheres x2 + y2 + z2 = a2
and x2 + y2 + z2 = b2, a > b > 0.
( x + y + z ) dx dy dz , over the tetra-
1.
hedron bounded by the planes x = 0,
y = 0, z = 0 and x + y + z = 1.
Ans. :
2.
1
8
dx dy dz
, over the tetra(1 + x + y + z )3
hedron bounded by the planes x = 0,
y = 0, z = 0 and x + y + z = 1.
1
5
Ans. :
log 2
2
8
8.
z 2 dx dy dz , over the region common to the spheres x2 + y2 + z2 = a2 and
cylinder x2 + y2 = ax.
Ans. :
9.
octant of the sphere x 2 + y 2 + z 2 = a 2.
( x 2 + y 2 ) dx dy dz , over the
a6
Ans. :
48
Ans. :
10.
x y z ( x 2 + y 2 + z 2 ) dx dy dz ,
4.
over the sphere of radius a and centre
at the origin.
Ans. :
11.
∫∫∫
z2
dx dy dz , over the
x2 + y2 + z2
Ans. :
dx dy dz
7.
3
2 2
(x + y + z )
2
2
12.
8
x2
a2
y2
b2
Ans. :
sphere x 2 + y 2 + z 2 = 2.
2
9
, over the region
a3 b 2 c 2
2520
x y z dx dy dz , over the positive octant of the ellipsoid
4 a7
Ans. :
35
6.
x 2 y z dx dy dz , over the tetrahe-
a8
64
( y 2 z 2 + z 2 x 2 + x 2 y 2 ) dx dy dz ,
5.
16
3
dron bounded by the planes x = 0,
x y z
y = 0, z = 0 and + + = 1.
a b c
over the positive octant of the sphere
x 2 + y 2 + z 2 = a 2.
Ans. :
2 a5
15
region bounded by the paraboloid
x 2 + y 2 = 2 z and the plane z = 2.
x y z dx dy dz , over the positive
3.
a
b
Ans. : 4 log
z2
c2
1.
a2b2c2
48
x2 y2 z2
+
+ dx dy dz over
1
4
9
the region bounded by the ellipsoid
x2 y2 z2
+
+
= 1.
1
4
9
[Ans. : 8p ]
Multiple Integral
8.85
8.7 APPLICATIONS OF MULTIPLE INTEGRALS
8.7.1 Area in Cartesian Form
(i) The area bounded by the curves y = y1(x)
and y = y2(x) intersecting at the points
P (a, b) and Q (c, d) is given as
Area =
c
y2 ( x )
a
y1 ( x )
dy dx
(ii) If equation of the curves are represented as
x = x1 ( y ) and x = x2 ( y ), then
Area =
d
x2 ( y )
b
x1 ( y )
dx dy
Note: Consider the symmetricity of the region
while finding area.
Example 1: Find the area bounded by the ellipse
Fig. 8.72
x2 y2
+
= 1, above x-axis.
a 2 b2
Solution:
1. The region is symmetric about the y-axis.
Total area = 2 (Area bounded by the
ellipse in the first quadrant)
2. Draw a vertical strip AB in the region
which lies in the first quadrant. AB starts
from the x-axis and terminates on the
x2 y 2
+ 2 = 1.
2
b
ellipse a
Limits of y : y = 0 to y = b 1 −
x2
a2
Limits of x : x = 0 to x = a
Area, A = 2
a
a
b 1
0
0
b 1−
= 2∫ y 0
0
x2
a2
x2
a2
Fig. 8.73
dy dx
a
dx = 2 ∫ b 1 −
0
x2
a2
2b a 2
2b x 2
x
a2
a − x 2 dx =
a − x 2 + siin −1
∫
2
a
a 0
a 2
⎞ 2b ⎛ a 2
⎞
2b ⎛ a 2
= ⎜ sin −1 1⎟ = ⎜ ⋅ ⎟
a ⎝ 2
⎠ a ⎝ 2 2⎠
a
=
=
ab
2
0
Engineering Mathematics
8.86
Example 2: Find the area bounded by the parabola y2 = 4x and the line
2x - 3y + 4 = 0.
Solution:
1. The points of intersection of the parabola
y2 = 4x and the line 2x – 3y + 4 = 0 are
obtained as
2x + 4
3
2
= 4x
( x + 2) 2 = 9 x
x2
5x 4 0
x = 1, 4 and y = 2, 4
Hence, P : (1, 2) and Q : (4, 4)
2. Draw a vertical strip AB which starts from
the line 2x – 3y + 4 = 0 and terminates on
the parabola y2 = 4x.
Limits of y : y =
Fig. 8.74
2x + 4
to y = 2 x
3
Limits of x : x = 1 to x = 4
Area, A = ∫
4
1
∫
2 x
2x+4
3
dy d x
4
4⎛
2x + 4 ⎞
2 x
= ∫ y 2 x + 4 dx = ∫ ⎜ 2 x −
⎟ dx
1
1
3 ⎠
3
⎝
4
3
x2
2.2
3
x2
3
4x
3
1
4
1
(8 1)
(16 1)
3
3
4
(4 1)
3
1
3
Example 3: Find the area bounded between the parabolas x2 = 4ay and
x2 = -4a(y - 2a).
Solution:
1. The parabola x 2 = 4ay has vertex (0, 0) and the parabola x 2
(0, 2a). Both the parabolas are symmetric about y-axis.
4a ( y 2a ) has vertex
Multiple Integral
8.87
Fig. 8.75
2
2. The points of intersection of x = 4ay and x 2
4a ( y 2a ) are obtained as
4ay = –4a (y – 2a)
8ay = 8 a2
y = a and x = ± 2a
Hence, P : (2a, a) and R : ( 2a, a)
3. The region is symmetric about y-axis.
Total area = 2 (Area in the first quadrant)
4. Draw a vertical strip AB in the region which lies in the first quadrant. AB starts
from the parabola x2 = 4ay and terminates on the parabola x 2
4a ( y 2a ).
x2
x2
to y = 2a −
4a
4a
Limits of x : x = 0 to x = 2a
Limit of y : y =
Area, A = 2 ∫
2a
= 2∫
2a
0
0
∫
2a −
x2
4a
x2
4a
y
2a −
dy dx
x2
4a
x2
4a
dx
2a ⎛
x2 x2 ⎞
− ⎟ dx
= 2 ∫ ⎜ 2a −
0
4a 4a ⎠
⎝
= 2 2ax −
x3
6a
2a
0
4 ⎞ 16
⎛
= 2 ⎜ 4a 2 − a 2 ⎟ = a 2
3 ⎠ 3
⎝
Example 4: Find the larger area bounded by the circle x2 + y2 = 64a2 and the
parabola y2 = 12ax.
Engineering Mathematics
8.88
Solution:
1. The points of intersection of the parabola y 2 = 12ax and the circle x 2 + y 2 = 64 a 2
are obtained as
x 2 + 12ax − 64a 2 = 0
( x + 16a ) ( x − 4a ) = 0
x = 4a and y = ±4a 3 [∵x = −16a does not lie on the parabola ]
(
)
(
Hence, P : 4a, 4a 3 and T : 4a, − 4a 3
)
Fig. 8.76
2. The region is symmetric about the x-axis.
Total area = 2 (Area above x-axis)
3. Divide the region OPQR above x-axis into two subregions OQR and OPQ. Draw a
vertical strip in each subregion.
(i) In the subregion OQR, the strip AB starts from the x-axis and terminates on the
circle x 2 + y 2 = 64a 2
Limits of y : y = 0 to y = 64a 2 − x 2
Limits of x : x = −8a to x = 0
2
(ii) In the subregion OPQ, the strip CD starts from the parabola y = 12ax and termi2
2
2
nates on the circle x + y = 64a
Limits of y : y = 12ax to 64a 2 − x 2
Limits of x : x = 0 to x = 4a
⎡
⎤
Area, A = 2 ⎢ ∫∫ dy dx + ∫∫ dy dx ⎥
⎢⎣ OQR
⎥⎦
OPQ
Multiple Integral
=2
=2
2
2
64 a 2 x 2
0
8a 0
0
8a
0
8a
y0
64 a 2 x 2
64a 2
dx +
64 a 2 x 2
4a
dy dx +
12 ax
0
4a
y
0
4a
x 2 dx
0
(
64 a 2 x 2
12 ax
x
64a 2
2
x2
dy dx
dx
64a 2
2
64a
x
sin 1
2
8a
8.89
)
x2
0
8a
12ax dx
x
64a 2
2
3 4a
2
2
x2
64a
x
sin 1
2
8a
2 12a
x
3
0
3
⎤
⎡
4a
⎛ 1 ⎞ 2 12a
= 2 ⎢ −32a 2 sin −1 (−1) +
48a 2 + 32a 2 sin −1 ⎜ ⎟ −
( 4a ) 2 ⎥
2
3
⎝2⎠
⎣
⎦
⎡ 64 2 8 3 a 2 ⎤
⎡
32a 2 3 ⎤
⎛
⎞
a −
= 2 ⎢ −32a 2 ⎜ − ⎟ + 8a 2 3 + 32a 2 −
=
2
⎢
⎥
⎥
6
3 ⎦
3 ⎦
⎝ 2⎠
⎣3
⎣
16
= a2 8 − 3
3
(
)
Example 5: Find the area common to the circles x 2 + y 2 - 4 y = 0 and
x 2 + y 2 - 4 x - 4 y + 4 = 0.
Solution:
1. The circles x 2
y2
4y
0 and x 2
y2
4x 4 y 4
0 have equal radii 2.
Fig. 8.77
2. The points of intersection of x 2 y 2 4 y 0 and x 2
obtained as
–4x + 4 = 0
x = 1, y = 2 ± 3
(
Hence, P : 1, 2
)
(
3 and Q : 1, 2
3
)
y2
4x 4 y 4
0 are
Engineering Mathematics
8.90
3. The region is symmetric about the line PQ.
Total area = 2 (Area P C2 Q)
4. Draw a vertical strip AB in the region P C2 Q which starts from the part of the circle
x 2 y 2 4 y 0 below centre line (y < 2) and terminates on the part of the same
circle above centre line (y > 2).
Limits of y : y = 2 − 4 − x 2 to y = 2 + 4 − x 2
Limits of x : x = 1 to x = 2
Area, A = 2 ∫
2
1
2
∫
(
2 + 4 − x2
2− 4− x
2
2
2 + 4 − x2
dy dx = 2 ∫ y 2 −
1
4 − x2
dx
)
= 2 ∫ 2 + 4 − x 2 − 2 + 4 − x 2 dx
1
2
2
1
2 4 x 2 dx
2
x
4
4 x2
2
1
1
3 2 sin 1 1 sin 1
2
2
4 0
3
4
4
x
sin 1
2
21
2
2
2
6
4
2
3
3
2
2
2
2
2
Example 6: Find the area of the loop of the curve x ( x + y ) = a ( x - y ).
Solution: The equation of the curve can be rewritten as y 2 = x 2
a x
a+x
1. The point of intersection of the curve with x-axis (y = 0) is obtained as
x 2 (a x) 0, x 0, x a
The loop of the curve lies between the points O : (0, 0) and P : (a, 0).
2. The region is symmetric about x-axis
Total area = 2 (Area above x-axis)
3. Draw a vertical strip AB in the region
above x-axis. AB starts from x-axis and
a x
terminates on the curve y 2 = x 2
.
a+x
a−x
a+x
Limits of y : y = 0 to y = x
Limits of x : x = 0 to x = a
Area of the loop, A = 2
=2
a
0
a
x
0
0
x
y0
a x
a+ x
a x
a+ x
Fig. 8.78
dy dx
dx = 2
a
0
x
a x
dx
a+ x
Multiple Integral
8.91
Putting x = a cosq, dx = –a sinq dq
π
When x = 0, θ =
2
x = a, θ = 0
0
A= 2
a a cos
( a sin ) d
a + a cos
a cos
2
2sin 2
2a
2
2
0
cos sin
2 cos
2 d
2
2
sin
2a 2
2a 2
2
0
cos
2
2sin cos
2
2
1 + cos 2
2
cos
0
2a 2 1
cos
d
2d
2a 2
2
0
cos (1 cos ) d
2
2a 2 sin
2
sin 2
4
2
0
4
Example 7: Find the area of the loop of the curve 2 y 2 = ( x - 2)( x - 10)2 .
Solution:
1. The points of intersection of 2 y 2
obtained as ( x 2) ( x 10) 2
0, x
( x 2) ( x 10) 2 with the x-axis (y = 0) are
2,10.
The loop of the curve lies between the points
P : (2, 0) and Q : (10, 0).
2. The region is symmetric about x-axis.
Total area = 2 (Area of the loop above x-axis)
3. Draw a vertical strip in the region above
x-axis. AB starts from x-axis and terminates
on the curve 2 y 2 ( x 2) ( x 10) 2 .
Limits of y : y
1
0 to y
2
(10 x) x 2
Limits of x : x = 2 to x = 10
Area of the loop, A = 2
2
10
2
10
(10 x )
2
0
(10 x )
y0
x 2
2
x 2
2
Fig. 8.79
dy dx
dx
2
10
2
(10 x)
x 2
2
dx
Engineering Mathematics
8.92
Putting
When
x − 2 = t 2 , dx = 2t dt
x = 2, t = 0
x = 10, t = 2 2
A =
2
∫
2
2 2
0
2 2
2 2
(−t 2 + 8) t ⋅ 2t dt
2 2
0
( t4
8t 2 ) dt
128 2
5
2 2
128 2
3
t5
5
8t 3
3
2 2
0
1024
15
2
2
3
Example 8: Find the area bounded between the curve x ( y + a ) = a and its
asymptote.
Solution: The equation of the curve can be rewritten as y 2 =
x = 0 is the asymptote of the curve.
a 2 (a − x)
. The line
x
Fig. 8.80
1. The region is symmetric about the x-axis.
Total area = 2 (Area above x-axis)
2. Draw a vertical strip AB in the region above x-axis. AB starts from the x-axis and
a 2 (a x)
,
terminates on the curve y 2 =
x
Limits of y : y = 0 to y = a
Limits of x : x = 0 to x = a
a−x
x
Multiple Integral
Area,
A = 2∫
a
∫
0
a
0
a
= 2∫ a
0
a−x
x
a
a
dy dx = 2 ∫ y 0
a−x
x
0
8.93
dx
a−x
dx
x
Putting x = a sin 2 θ , dx = 2a sin θ cos θ dθ
When x = 0, θ = 0
π
x = a, θ =
2
π
A = 2∫ 2 a
0
= 2a 2
a − a sin 2 θ
⋅ 2a sin θ cos θ dθ
a sin 2 θ
2
0
2a 2
2 cos 2 d = 2a 2
sin 2
2
2
2a 2
0
2
0
2
(1 + cos 2 ) d
1
sin
2
0
= a2
Example 9: Find the area between the rectangular hyperbola 3xy = 2 and the
line 12x + y = 6.
Solution:
1. The points of intersection of the rectangular hyperbola 3xy = 2 and the line
12x + y = 6 are obtained as
3 x (6 − 12 x) = 2
18 x 2 − 9 x + 1 = 0
1 1
x = , and y = 2, 4
3 6
⎛1 ⎞
⎛1
P : ⎜ , 2 ⎟ and Q : ⎜ ,
Hence,
⎝3 ⎠
⎝6
⎞
4⎟
⎠
2. Draw a vertical strip AB in the region
which starts from the rectangular hyperbola
3xy = 2 and terminates on the line
12x + y = 6.
2
Limits of y : y =
to y = 6 − 12 x
3x
1
1
Limits of x : x = to x =
6
3
Fig. 8.81
Engineering Mathematics
8.94
1
6 −12 x
6
3x
Area, A = ∫13 ∫ 2
1
= ∫13 y
6
dy dx
6 −12 x
2
3x
1
2 ⎞
⎛
dx = ∫13 ⎜ 6 − 12 x − ⎟ dx
x⎠
3
⎝
6
1
6x 6x
2
3
2
log x
1
3
(2 1) 6
1
9
1
36
2
1
1
log
log
3
3
6
6
1
2
2
log 2.
3
Example 10: Find the area of the curvilinear triangle bounded by the parabolas
y2 = 12x, x2 = 12y, circle x2 + y2 = 45 and lying outside the circle.
Solution: Required curvilinear triangle is PQR.
Fig. 8.82
1. The point of intersection of
(i) the parabola x 2 = 12 y and the circle x 2 + y 2 = 45 are obtained as
45 y 2
y
2
12 y 45
y
12 y
0
3, 15
But y = –15 does not lie on the parabola x2 = 12y.
Thus, y = 3, x = 6
Hence, P : (6, 3)
(ii) the parabolas x 2 = 12 y and y 2 = 12 x are obtained as
y4
= 12 y, y = 0, 12 and x = 0, 12
144
Hence, Q : (12, 12)
Multiple Integral
8.95
2
2
2
(iii) the parabola y = 12 x and the circle x + y = 45 are obtained as 45 – x2 = 12x,
x2 + 12x – 45 = 0, x = 3, –15 but x = –15 does not lie on the parabola y2 = 12x.
Thus x = 3, y = 6
Hence, R : (3, 6)
2. Divide the region PQR into two subregions PRS and PQS. Draw a vertical strip in
each subregion.
(i) In subregion PRS, strip starts from the circle x 2 + y 2 = 45 and terminates on
the parabola y 2 = 12 x.
Limits of y : y = 45 − x 2 to y = 12 x
Limits of x : x = 3 to x = 6
(ii) In subregion PQS, strip starts from the parabola x2 = 12y and terminates on the
parabola y2 = 12x.
x2
to y = 12 x
12
Limits of x : x = 6 to x = 12
Limits of y : y =
Area PQR, A = Area PRS + Area PQS
=
6
12 x
3
45 x 2
6
3
(
dy dx +
2x
3
x2
12
)
45 x 2 dx
12 x
12
12 x
12
6
3
2
6
dy dx =
12
6
3
y
12 x
45 x 2
dx +
12
6
y
45
sin
2
1
dx
x2
dx
12
12 x
6
x
45 x 2
2
12 x
x2
12
x
12
45
2x
3
12
3
2
3
x
36
3
4 3
6 6 3 3
3
(
)
3 9
3
36
2
4 3
1
12 12 6 6
(123
3
36
45
12
1
42
sin 1
sin 1
2
5
5
(
)
45
sin
2
6
1
2
5
sin
1
1
5
63 )
Example 11: Find the area bounded by the hypocycloid
Solution:
1. The hypocycloid is symmetric in the coordinate plane.
x
a
2
3
y
+
b
2
3
=1 .
8.96
Engineering Mathematics
Fig. 8.83
Total Area = 4 (Area in the first quadrant)
2. Draw a vertical strip AB parallel to y-axis in the region which lies in the
quadrant. AB starts from x-axis and terminates on the curve
3
2 2
⎤
⎡
⎛ x ⎞3 ⎥
⎢
Limits of y : y = 0 to y = b 1 − ⎜ ⎟
⎢ ⎝a⎠ ⎥
⎦
⎣
Limits of x : x = 0 to x = a
3
Area, A = 4 ∫
a
0
∫
a
2 ⎤2
⎡
⎛ x ⎞3
b ⎢⎢1− ⎜ ⎟ ⎥⎥
a
⎢⎣ ⎝ ⎠ ⎥⎦
0
= 4∫ y
0
dy dx
3
2 ⎤2
x ⎞3 ⎥
⎡
⎛
b ⎢⎢1− ⎜ ⎟ ⎥
⎝a⎠ ⎥
⎣⎢
⎦
0
dx
3
2 2
⎡
⎤
3
a
x
⎛
⎞
= 4 ∫ b ⎢1 − ⎜ ⎟ ⎥ dx
0
⎢ ⎝a⎠ ⎥
⎣
⎦
Putting x = a cos3 t , dx = 3a cos 2 t (− sin t ) dt
When x = 0, t =
2
x = a, t = 0
x
a
2
3
y
+
b
2
3
=1
Multiple Integral
0
8.97
3
2
A = 4 ∫ b(1 − cos t ) (−3a cos 2 t sin t ) dt
2
2
= 12ab ∫ 2 sin 4 t cos 2 t dt
0
5 3
3 1 1 1 1
⋅
⋅
1 ⎛5 3⎞
2
2
= 12ab B ⎜ , ⎟ = 6ab
= 6ab 2 2 2 2 2
2 ⎝2 2⎠
3!
4
3
=
ab
8
Exercise 8.7
1. Find the area bounded by y-axis, the
line y = 2x and the line y = 4.
[Ans. : 4]
2. Find the area bounded by the lines
y = 2 + x, y = 2 x and x = 5.
[Ans. : 25]
3. Find the area bounded by the parabola
y2 + x = 0, and the line y = x + 2.
9
Ans. :
2
4. Find the area bounded by the parabola
x = y y2 and the line x + y = 0.
4
Ans. :
3
5. Find the area bounded by the curves
y2 = 4x and 2x 3y + 4 = 0.
1
Ans. :
3
6. Find the area bounded by the parabola
y = x2 3x and the line y = 2x.
125
Ans. :
6
7. Find the area bounded by the
parabolas y2 = x, x2 = 8y.
Ans. :
8. Find the area bounded by the
parabolas y = ax2 and
8
3
y
x2
a
a
Ans. :
4
a
3 a2 + 1
9. Find the area bounded by the curve
y2 (2a x) = x3 and its asymptote.
[Ans. : 3pa2]
10. Find the area of the loop of the
a+ x
curve y 2 = x 2
a x
Ans. : 2a 2
4
1
11. Find the area of one of the loops of
x4 + y4 = 2a2 xy.
a2
Ans. :
4
12. Find the area enclosed by the curve
9xy = 4 and the line 2x + y = 2.
1 4
log 2
Ans. :
3 9
13. Find the area of the smaller region
bounded by the circle x2 + y2 = 9 and
a straight line x = 3 y.
Ans. : 4
2
3
3
2
Engineering Mathematics
8.98
14. Find the area bounded by the x-axis,
circle x2 + y2 = 16 and the line y = x.
[Ans. : 2p]
15. Find the area bounded between the
curves y = 3x2 x 3 and
y = 2x2 + 4x + 7.
Ans. :
45
2
16. Find the area bounded by the asteroid
2
2
2
( x) 3 + ( y ) 3 = (a) 3
3 2⎤
⎡
⎢ Ans. : 8 a ⎥
⎣
⎦
8.7.2 Area in Polar Form
The area bounded by the curves r = r1 (q ), r = r2 (q )
and the lines q = q1 and q = q2 is given as
=
2
1
r2 ( )
r1 ( )
r r
Note: Consider the symmetricity of the region while
finding area.
Example 1: Find the area between the circles
r = 2 sinp and r = 4 sinp.
Fig. 8.84
Solution:
.
2
Total area = 2 (Area in the first quadrant)
1. The region is symmetric about the line
=
Fig. 8.85
2. Draw an elementary radius vector OAB in the region which lies in the first quadrant.
OAB enters in the region from the circle r = 2sinq and leaves at the circle r = 4 sinq.
Limits of r : r = 2 sin θ to r = 4 sin θ
Limits of θ : θ = 0 to θ =
π
2
Multiple Integral
π
2
0
Area, A = 2 ∫
∫
4 sin θ
2 sin θ
π
2
0
r d r dθ = 2 ∫
8.99
r2
2
4 sin θ
dθ
2 sin θ
π
= ∫ 2 (16 sin 2 θ − 4 sin 2 θ ) dθ
0
2
0
6 (1 cos 2 ) d
sin 2
2
6
2
sin
6
2
0
sin 0
3
2
Example 2: Find the area of the crescent bounded by the circles r = 3 and
r = 2cosp.
Solution:
1. The points of intersection of r = 3 and r = 2 cos
are obtained as
3 = 2 cos θ
3
2
π
θ =±
3
cosθ =
Hence, at P,
=
3
.
2. The region is symmetric about the
initial line, q = 0.
Area of the crescent = 2 (Area above
the initial line, q = 0)
3. Draw an elementary radius vecFig. 8.86
tor OAB in the region above the
initial line. OAB enters in the region from the circle r = 3 and leaves at the circle
r = 2 cosq.
Limits of r : r = 3 to r = 2 cos q
p
Limits of q : q = 0 to q =
3
π
Area, A = 2 ∫ 3 ∫
0
π
= 2∫ 3
0
3
0
[ 2 (1
sin
2
3
2 cosθ
3
r2
2
r d r dθ
2 cosθ
π
dθ = ∫ 3 (4 cos 2 θ − 3) dθ
0
3
cos 2 ) 3] d
3
3
2
3
,
2
sin 2
2
3
0
Engineering Mathematics
8.100
But area cannot be negative.
Hence, numerical value of area =
3
3
.
2
Example 3: Find the area which lies inside the circle r = 3a cosp and outside the
cardioid r = a (1 + cosp ).
Solution:
1. The points of intersection of the circle r = 3a cosq and the cardioid r = a (1 + cosq )
are obtained as
3a cos θ = a (1 + cos θ )
1
cosθ =
2
π
θ =±
3
Hence, at R, q =
p
3
Fig. 8.87
2. The region is symmetric about the initial line q = 0.
Total area = 2 (Area above the initial line)
3. Draw an elementary radius vector OAB from the origin in the area above the initial
line. OAB enters in the region from the cardioid r = a (1 + cosq ) and leaves at the
circle r = 3a cosq.
Limits of r : r = a (1 + cos θ ) to r = 3a cos θ
Limits of θ : θ = 0 to θ =
π
3
Multiple Integral
π
3
0
Area, A = 2 ∫
3 a cosθ
∫
a (1+ cosθ )
8.101
π
3
0
r2
2
r dr dθ = 2 ∫
3 a cosθ
dθ
a (1+ cosθ )
π
= ∫ 3 ⎡⎣9a 2 cos 2 θ − a 2 (1 + cos θ ) 2 ⎤⎦ dθ
0
a2
3
0
[4 (1 cos 2 ) 1 2 cos ]d
4sin 2
2
a2 3
a
2
3
3
2
2sin
3
2sin
3
0
2sin
3
= a2
Example 4: Find the area common to the cardioids r = a (1 + cosp ) and
r = a (1 - cosp ).
Solution:
1. The points of intersection of the cardioids r = a(1 + cosq ) and r = a(1
obtained as
a (1 + cosq ) = a(1
cosq = 0
θ =±
Hence, at P, θ =
cosq )
π
2
π
2
Fig. 8.88
cosq ) are
8.102
Engineering Mathematics
2. The region is symmetric in all the quadrants
Total area = 4 (Area in the first quadrant)
3. Draw an elementary radius vector OA from the origin in the region which lies
in the first quadrant. OA starts from the origin and terminates on the cardioid
r = a (1 cosq ).
Limits of r : r = 0 to r = a (1 − cos θ )
Limits of θ : θ = 0 to θ =
π
2
π
a (1− cosθ )
0
0
π
2
0
r2
2
Area, A = 4 ∫ 2 ∫
= 4∫
2a 2
2a 2
2a
2
2
0
r d r dθ
a (1− cosθ )
π
dθ = 2 ∫ 2 a 2 (1 − cos θ ) 2 dθ
0
0
1 2 cos
3
2
2sin
3
4
2
1 + cos 2
2
sin 2
4
d
2
0
Example 5: Find the area inside the cardioid r = 3 (1 + cosp ) and outside the
3
.
parabola r =
1 + cos p
Solution:
1. The points of intersection of the cardioid r = 3(1 + cosq ) and the parabola
3
are obtained as
r=
1 + cos
Fig. 8.89
Multiple Integral
3 (1 + cosθ ) =
8.103
3
(1 + cosθ )
(1 + cosθ ) 2 = 1
cos θ = 0,θ = ±
π
2
π
2
2. The region is symmetric about the initial line q = 0.
Total area = 2 (Area above the initial line)
3. Draw an elementary radius vector OAB from the origin in the region above the
3
initial line q = 0. OAB enters in the region from the parabola r =
and
1 + cos
leaves at the cardioid r = 3 (1 + cosq ).
3
Limits of r : r =
to r = 3 (1 + cosθ )
1 + cosθ
π
Limits of θ : θ = 0 to θ =
2
Hence, at P, θ =
π
Area, A = 2 ∫ 2 ∫
0
3 (1+ cosθ )
3
1+ cosθ
π
r d r dθ = 2 ∫ 2
0
r2
2
3 (1+ cosθ )
dθ
3
1+ cosθ
π
⎡
⎤
1
dθ
= ∫ 2 9 ⎢(1 + cosθ ) 2 −
2⎥
0
(1 + cosθ ) ⎦
⎣
⎤
⎡
⎥
π ⎢
1 + cos 2θ
1
⎥ dθ
−
= 9 ∫ 2 ⎢1 + 2 cosθ +
2
0 ⎢
⎥
2
⎛
2θ ⎞
⎢
⎜ 2 cos ⎟ ⎥
2⎠ ⎦
⎝
⎣
π
⎡3
cos 2θ 1 ⎛
θ⎞
θ⎤
= 9 ∫ 2 ⎢ + 2 cos θ +
− ⎜1 + tan 2 ⎟ sec 2 ⎥ dθ
0
2
4⎝
2⎠
2⎦
⎣2
π
⎡3
cos 2θ 1 2 θ 1
θ ⎛1
θ ⎞⎤
− sec − ⋅ tan 2 ⎜ sec 2 ⎟ ⎥ dθ
= 9 ∫ 2 ⎢ + 2 cos θ +
0
2
4
2 2
2⎝2
2 ⎠⎦
⎣2
θ
tan
sin 2θ 1
θ 1
3θ
2
+ 2 sin θ +
− ⋅ 2 tan −
=9
2
4
4
2 2 3
3
π
2
0
n +1
⎡
f (θ ) ] ⎤
[
n
⎢∵ ∫ [ f (θ ) ] f ′ (θ )dθ =
⎥
n + 1 ⎥⎦
⎢⎣
Engineering Mathematics
8.104
9
=9
3
4
2sin
2
sin
4
1
tan
2
4
1
tan 3
6
4
3
4
+
4 3
Example 6: Find the area common to the circles r = cosp and r = 3 sin .
Solution:
1. The point of intersection of the circles r = cosq and r = 3 sin
3 sin = cos
tan =
=
Hence, at P, q =
is obtained as
1
3
6
p
6
Fig. 8.90
2. Divide the region OQPR into two subregions OQP and ORP. Draw an elementary
radius vector in each subregion.
(i) In subregion OQP, radius vector OA starts from the origin and terminates on the
circle r = 3 sin .
Limits of r : r = 0 to r = 3 sin q
Limits of q : q = 0 to q =
p
6
Multiple Integral
8.105
(ii) In the subregion ORP, the radius vector OB starts from the origin and terminates
on the circle r = cosq.
Limits of r : r = 0 to r = cosq
Limits of
: =
p
p
to q =
6
2
π
Area, A = ∫ 6 ∫
3 sin θ
0
0
π
r2
2
= ∫6
0
π
r dr dθ + ∫π2 ∫
6
3 sin θ
π
dθ + ∫π2
6
0
r2
2
cosθ
0
r d r dθ
cosθ
dθ
0
=
1 π2
1 π6
2
2
3
θ
θ
sin
d
+
π cos θ dθ
2 ∫6
2 ∫0
=
3 π6 ⎛ 1 − cos 2θ
⎜
2 ∫0 ⎝
2
1 π2 ⎛ 1 + cos 2θ
⎞
+
d
θ
π ⎜
⎟
2 ∫6 ⎝
2
⎠
3
4
1
4
3
4 6
5
24
sin 2
2
6
0
1
sin
2
3
sin 2
2
⎞
⎟ dθ
⎠
2
6
1
4 2
1
sin
2
6
1
sin
2
3
4 3
16
Example 7: Find the area common to the circle r = a and the cardioid
r = a (1 + cosp ).
Solution:
1. The points of intersection of the circle r = a and the cardioid r = a (1 + cosq ) are
obtained as
a = a (1 + cosq )
π
cos θ = 0, θ = ±
2
π
Hence, at Q, θ =
2
2. The region is symmetric about the initial line q = 0
Total area = 2 (Area above the initial line)
3. Divide the region OPQR above the initial line into two subregions OPQ and ORQ.
Draw elementary radius vectors in each subregion.
(i) In the subregion OPQ, the radius vector OA starts from the origin and terminates on the circle r = a.
Limits of r : r = 0 to r = a
Limits of θ : θ = 0 to θ =
π
2
Vector Calculus
Chapter
9
9.1 INTRODUCTION
Vector algebra deals with the operations of addition, subtraction and multiplication
of vectors. Vector calculus deals with the differentiation and integration of vector
functions. We will learn about multiplication of vectors in vector algebra and about
derivative of a vector function, gradient, divergence and curl in vector differential
calculus. In vector integral calculus, we will learn about line integral, surface integral,
volume integral and three theorems namely Green’s theorem, divergence theorem and
Stoke’s theorem. It plays an important role in the differential geometry and in the
study of partial differential equations. It is useful in the study of rigid dynamics, fluid
dynamics, heat transfer, electromagnetism, theory of relativity, etc.
9.2 UNIT VECTOR
A unit vector is a vector having unit magnitude. If A is a vector, then unit vector in the
A
direction of A is given as aˆ =
.
| A|
This also shows that A can be represented in terms of unit vector as A = | A | aˆ.
Note:
(i) The unit vectors in the direction of x, y and z- axes are denoted by iˆ, jˆ and k̂
respectively.
(ii) iˆ · iˆ = jˆ · jˆ = k̂ · k̂ = 1, iˆ · jˆ = jˆ · k̂ = k̂ · iˆ = 0, iˆ
jˆ = k̂, jˆ
k̂ = iˆ, k̂
iˆ = jˆ
9.3 COMPONENTS OF A VECTOR
Let OA represent a vector with initial point at the origin O and terminal point at A.
Let (A1, A2, A3) be the rectangular coordinates of the terminal point A. The vectors A1 iˆ,
A2 jˆ, A3 k̂ are called the rectangular component vectors or component vectors of A in
the x, y and z-directions respectively. A1, A2 and A3 are called the rectangular components or components of A in the x, y and z-directions respectively.
9.2
Engineering Mathematics
z
A (A1, A2, A3)
O
y
x
Fig. 9.1
A
A1iˆ + A2 jˆ + A3k̂
A = A=
A12 + A22 + A32
In particular, the position vector of the point (x, y, z) w.r.t. origin is denoted by r and
is written as
r = x iˆ + y jˆ + z k̂
r = r = x2 + y 2 + z 2
9.4 TRIPLE PRODUCT
9.4.1 Scalar Triple Product
Scalar triple product of three vectors a, b and c is a dot product of a vector a and
(
vector b
)
c . It is denoted by a b c and is also known as box product of vectors
a, b and c.
If
a = a1 iˆ + a2 jˆ + a3 k̂,
b = b1 iˆ + b2 jˆ + b3 k̂,
c = c1 iˆ + c2 jˆ + c3 k̂
then
a1
⎡⎣ a b c ⎤⎦ = b1
c1
a2
b2
c2
a3
b3
c3
Vector Calculus
9.3
Note:
(
(i) a b
c
) (a
)
b c
1
(ii) Volume of a parallelogram = ⎡⎣ a b c ⎤⎦ and volume of a parallelopiped = ⎡⎣ a b c ⎤⎦
6
(iii) ⎡⎣ a b c ⎤⎦ = ⎡⎣b c a ⎤⎦ = ⎡⎣c a b ⎤⎦
(iv) If a, b, c are coplanar, then ⎡⎣ a b c ⎤⎦ = 0.
9.4.2 Vector Triple Product
Vector triple product of three vectors a, b and c is a cross product of a vector a and
(
)
vector b
(
)
c or vector a
b and vector c .
( b c) ( a c) b ( a b) c
b) c ( a c) b ( b c) a
a
(a
Note:
(i) a
(ii) a
(b c) ( a b )
(b c) (b c)
Example 1: If a b
to ( b c ) .
Solution:
(a
d)
(b
c
a
c d and a c
c)
(a
(c
b d , show that ( a
d ) is parallel
b)
( a c) ( d b) ( d c)
d ) (b d ) (b d ) (c d )
=0
Hence, ( a d ) is parallel to ( b c ) .
Example 2: If a = ˆi + ˆj – kˆ, b = ˆi – ˆj + kˆ, c = ˆi – ˆj – kˆ, find the vector a
(b c) .
Solution: We know that,
a
( b c) ( a c) b ( a b) c
[(iˆ + jˆ k̂) · (iˆ jˆ k̂)](iˆ jˆ + k̂) [(iˆ + jˆ k̂) · (iˆ jˆ + k̂)] (iˆ jˆ k̂)
= (iˆ · iˆ jˆ · jˆ + k̂ · k̂)(iˆ jˆ + k̂) (iˆ · iˆ jˆ · jˆ k̂ · k̂) (iˆ jˆ k̂)
[∵ iˆ · jˆ = jˆ · k̂ = k̂ · iˆ = 0 ]
9.4
Engineering Mathematics
= (1 1 + 1) (iˆ jˆ + k̂)
= iˆ jˆ + k̂ + iˆ jˆ k̂
= 2iˆ 2jˆ.
(1
1
1) (iˆ
jˆ
k̂)
Example 3: Find the scalars p and q, if ( a b ) c
(b
a
c ) where,
a = 2iˆ + ĵ + pk̂ , b = iˆ – ĵ, c = 4iˆ + q ĵ + 2k̂ .
(a
( a c) b
a (b c) ( a c) b
( a b ) c a (b
( a c) b (b c) a ( a c) b
( b c) a ( a b) c
Solution: We know that,
and
Given,
[(iˆ
(b c) a
(a b ) c
c)
( a b) c
jˆ) · (4iˆ + q jˆ + 2k̂)](2iˆ + jˆ + pk̂) = [(2iˆ + jˆ + p k̂) · (iˆ
(4
8iˆ + 4jˆ + 4p k̂
(8
b) c
2q) iˆ + (4
q)(2iˆ + jˆ + pk̂) = (2
2q iˆ
q jˆ
q) jˆ + (4p
jˆ)](4iˆ + q jˆ + 2k̂)
1) (4iˆ + q jˆ + 2k̂)
pqk̂ = 4iˆ + q jˆ + 2k̂
pq) k̂ = 4iˆ + q jˆ + 2k̂
Equating coefficients of iˆ, jˆ, and k̂ on both the sides,
8 2q = 4
q=2
and
4p pq = 2
4p 2p = 2
p=1
Example 4: Prove that the four points whose position vectors are 3iˆ - 2jˆ + 4kˆ,
6iˆ + 3jˆ + kˆ, 5iˆ + 7jˆ + 3kˆ and 2iˆ + 2jˆ + 6kˆ are coplanar.
Solution: Let A, B, C, D be the four points such that
A = 3iˆ
2 jˆ + 4k̂, B = 6iˆ + 3 jˆ + k̂, C = 5iˆ + 7 jˆ + 3k̂, D = 2iˆ + 2 jˆ + 6k̂
AB
B A (6iˆ + 3 jˆ + k̂)
= 3iˆ + 5 jˆ 3k̂
(3iˆ
2 jˆ + 4k̂)
AC
C A (5iˆ + 7 jˆ + 3k̂)
= 2iˆ + 9 jˆ k̂
(3iˆ
2 jˆ + 4k̂)
AD
D A (2iˆ + 2 jˆ + 6k̂)
= iˆ + 4 jˆ + 2k̂
(3iˆ
2 jˆ + 4 k̂)
Vector Calculus
AB ( AC
3 5
2 9
1 4
AD )
3
1
2
9.5
0
Hence, the four points are coplanar.
Example 5: Prove that ( a b ) , ( b c ) , ( c a ) are non-coplanar if
a , b and c are non-coplanar. Hence obtain the scalars l, m, n such that
a
l ( b c ) m ( c a ) n ( a b) .
Solution: (i) If a, b, c are non-coplanar, then ⎡⎣ a, b, c ⎤⎦ ≠ 0
Consider,
⎡ a × b b × c c × a ⎤ = ( a × b ) ⋅ ⎡⎣(b × c ) × ( c × a )⎤⎦
⎣
⎦
Let
b c
(a
b)
(b c) ( c
a)
p
( p a) c ( p c) a
( a b ) { ( b c ) a } c { ( b c ) c} a
(a
b)
p
(c
{
a)
(a
}
= ( a × b ) ⋅ ⎡⎣b c a ⎤⎦ c − 0
b)
⎡∵ ⎡b c c ⎤ = 0⎤
⎦
⎣ ⎣
⎦
= ⎡⎣b c a ⎤⎦ ⎡⎣( a × b ) ⋅ c ⎤⎦ = ⎡⎣ a b c ⎤⎦ ⎡⎣ a b c ⎤⎦
2
= ⎡⎣ a b c ⎤⎦ ≠ 0
⎡∵ ⎡ a b c ⎤ ≠ 0⎤
⎦
⎣ ⎣
⎦
⎡⎣ a × b b × c c × a ⎤⎦ ≠ 0
Hence, (a b), (b c), (c a) are non-coplanar.
(ii) a
l ( b c ) m ( c a ) n ( a b ) where l, m, n are scalars to be determined.
Taking scalar product with a on both the sides,
a ⋅ a = la ⋅ (b × c ) + ma ⋅ ( c × a ) + na ⋅ ( a × b )
⎡∵ ⎡ a c a ⎤ = 0 = ⎡ a a b ⎤ ⎤
⎦
⎣
⎦⎦
⎣ ⎣
= l ⎡⎣ a b c ⎤⎦ + 0 + 0
l=
a⋅a
⎡⎣ a b c ⎤⎦
Similarly, taking, dot product with b and c,
m=
a⋅b
⎡⎣ a b c ⎤⎦
and n =
a⋅c
⎡⎣ a b c ⎤⎦
9.6
Engineering Mathematics
Example 6: If ⎡⎣a b c ⎤⎦
0, prove that a vector d can be expressed as
⎡ d b c ⎤⎦ a + ⎡⎣ d c a ⎤⎦ b + ⎡⎣ d a b ⎤⎦ c
.
d=⎣
⎡⎣ a b c ⎤⎦
Solution: Since ⎡⎣ a b c ⎤⎦ ≠ 0, a, b, c are non-coplanar vectors, any vector d can be
uniquely expressed as a linear combination of a, b, c.
Let
d = la + m b + nc
where l, m, and n are scalars to be determined.
… (1)
Taking dot product with b c on both the sides,
d ⋅ (b × c ) = la ⋅ (b × c ) + mb ⋅ (b × c ) + nc ⋅ (b × c )
⎡∵ ⎡b b c ⎤ = 0 = ⎡c b c ⎤ ⎤
⎣
⎦⎦
⎦
⎣ ⎣
⎡⎣ d b c ⎤⎦ = l ⎡⎣ a b c ⎤⎦ + 0 + 0
⎡ d b c ⎤⎦
l= ⎣
⎡⎣ a b c ⎤⎦
Similarly, taking dot product with c a and a b on both the sides of Eq. (1),
⎡ d c a ⎤⎦
m= ⎣
, n=
⎡⎣ a b c ⎤⎦
⎡⎣ d a b ⎤⎦
⎡⎣ a b c ⎤⎦
Substituting the values of l, m and n in Eq. (1),
⎡ d b c ⎤⎦ a + ⎡⎣ d c a ⎤⎦ b + ⎡⎣ d a b ⎤⎦ c
d=⎣
⎡⎣ a b c ⎤⎦
Example 7: If a + b + c = O, prove that a b
Solution:
b c
c a.
a+b+c = O
… (1)
Taking cross product with b on both the sides,
(a + b + c)× b = O × b
( a × b ) + (b × b ) + ( c × b ) = O
( a × b ) + O = − ( c × b ) = (b × c )
a b
b c
… (2)
Vector Calculus
9.7
Similarly, taking cross-product with c on both the sides of Eq. (1),
b c
c a
b c
c a.
… (3)
From Eqs. (2) and (3), we get
a b
1 ˆ
Example 8: If â ë (bˆ ë ĉ) =
b , find angles which â makes with bˆ and ĉ.
2
1
Solution: â (b̂ ĉ) = b̂
2
1
b̂
2
Equating the coefficients of b̂ and ĉ on both the sides,
1
and â b̂ = 0
â ĉ =
2
But,
â ĉ = |â| |ĉ| cos q, where q is the angle between â and ĉ.
1.1 cos q
[â and ĉ are unit vectors]
1
cos =
2
(â ĉ) b̂ (â b̂) ĉ =
=
3
â b̂ = 0,
Thus, â is perpendicular to b̂.
Hence, â makes an angle
3
with ĉ and an angle
2
with b̂.
Example 9: A vector x satisfies the equation x b
that x
c
( a c) b
a b
Solution: x b
c b and a x
0, prove
.
c b
Taking cross-product with a on both the sides,
a × ( x × b) = a × (c × b)
(a ⋅ b ) x − (a ⋅ x ) b = (a ⋅ b ) c − (a ⋅ c ) b
(a ⋅ b ) x = (a ⋅ b ) c − (a ⋅ c ) b
(a ⋅ c ) b
x= c−
a⋅b
⎡⎣∵a ⋅ x = 0⎤⎦
9.8
Engineering Mathematics
p
q
r
Example 10: Prove that a p a q a r
p q r
(a
b) .
b p b q b r
Solution: Let p = p1 iˆ + p2 jˆ + p3 k̂, q = q1 iˆ + q2 jˆ + q3 k̂,
r = r1 iˆ + r2 jˆ + r3 k̂, a = a1 iˆ + a2 jˆ + a3 k̂,
b = b iˆ + b jˆ + b k̂
1
2
p1
⎡⎣ p q r ⎤⎦ ( a × b ) = q1
r1
3
p2
q2
r2
p3
q3
r3
iˆ
a1
b1
ˆj
a2
b2
kˆ
a3
b3
Interchanging rows by columns in second determinant,
p p
p iˆ a b
1
⎡⎣ p q r ⎤⎦ ( a × b ) = q1
r1
2
3
q2
r2
1
1
q3 ˆj a2
r3 kˆ a
3
b2
p1iˆ + p2 ĵj + p3 kˆ
= q1iˆ + q2 ˆj + q3 kˆ
r iˆ + r ˆj + r kˆ
1
2
3
p
p⋅a
p ⋅b
= q
q⋅a
q ⋅b
r
r ⋅a
r ⋅b
b3
p1a1 + p2 a2 + p3 q3
p1b1 + p2 b2 + p3b3
q1a1 + q2 a2 + q3 a3
q1b1 + q2 b2 + q3b3
r1a1 + r2 a2 + r3 a3
r1b1 + r2 b2 + r3b3
Interchanging rows by columns,
p
q
r
⎡⎣ p q r ⎤⎦ ( a × b ) = a ⋅ p a ⋅ q a ⋅ r
b⋅ p b⋅q b⋅r
9.5 PRODUCT OF FOUR VECTORS
9.5.1 Scalar Product of Four Vectors
Scalar product of four vectors a, b, c, d is a dot product of vectors ( a b ) and ( c d ) .
(a
b) (c d )
a c
b c
a d
b d
Vector Calculus
9.9
This result is known as “Lagrange’s identity.”
Proof: Let c d
m
( a × b ) ⋅ ( c × d ) = ( a × b ) ⋅ m = a ⋅ (b × m )
= a ⋅ ⎡⎣b × ( c × d ) ⎤⎦
⎡⎣Substituting m ⎤⎦
= a ⋅ ⎡⎣( b ⋅ d ) c − ( b ⋅ c ) d ⎤⎦
( a c )( b d ) ( a d )( b c )
(a
b) (c d )
a c
b c
a d
b d
9.5.2 Vector Product of Four Vectors
Vector product of four vectors a, b, c, d is a cross product of vectors ( a b ) and
( c d ) . The vector product of four vectors a, b, c, d can be expressed in terms of
vectors a and b as well as in terms of vectors c and d .
(i)
( a × b ) × ( c × d ) = ⎡⎣ a c d ⎤⎦ b − ⎡⎣b c d ⎤⎦ a
(ii)
( a × b ) × ( c × d ) = ⎡⎣ a b d ⎤⎦ c − ⎡⎣ a b c ⎤⎦ d
Proof:
(i) Let ( c d ) m
(a
b)
(c
d)
(a
(a
b) m
m)b
a (c
(b m) a
d ) b b (c
d) a
( a × b ) × ( c × d ) = ⎡⎣ a c d ⎤⎦ b − ⎡⎣b c d ⎤⎦ a
(ii) Let a b
n
(a
b)
(c
d) n
(c d )
(n d )c (n c)d
(a b) d c (a
( a × b ) × ( c × d ) = ⎡⎣ a b d ⎤⎦ c − ⎡⎣ a b c ⎤⎦ d
b) c d
9.10
Engineering Mathematics
Example 1: By considering the product ( a b )
(c
d ) in two different ways,
show that ⎡⎣ b c d ⎤⎦ a + ⎡⎣ c a d ⎤⎦ b + ⎡⎣ a b d ⎤⎦ c = ⎡⎣ a b c ⎤⎦ d where a, b, c are non-coplanar vectors.
Solution: We know that,
and
( a × b ) × ( c × d ) = ⎡⎣ a b d ⎤⎦ c − ⎡⎣ a b c ⎤⎦ d
... (1)
( a × b ) × ( c × d ) = ⎡⎣ a c d ⎤⎦ b − ⎡⎣b c d ⎤⎦ a
... (2)
Equating Eq. (1) and (2),
⎡⎣ a b d ⎤⎦ c − ⎡⎣ a b c ⎤⎦ d = ⎡⎣ a c d ⎤⎦ b − ⎡⎣b c d ⎤⎦ a
⎡⎣b c d ⎤⎦ a − ⎡⎣ a c d ⎤⎦ b + ⎡⎣ a b d ⎤⎦ c = ⎡⎣ a b c ⎤⎦ d
⎡∵ ⎡ a c d ⎤ = − ⎡c a d ⎤ ⎤
⎣
⎦⎦
⎦
⎣ ⎣
⎡⎣b c d ⎤⎦ a + ⎡⎣c a d ⎤⎦ b + ⎡⎣ a b d ⎤⎦ c = ⎡⎣ a b c ⎤⎦ d
Example 2: Prove that ( b c )
(a
d)
(c
a)
(b
d)
(a
b)
(c
d)
– 2 ⎡⎣ a b c ⎤⎦ d and hence, show that vector on L.H.S. is parallel to vector d .
Solution:
(b
c)
(a
d)
(c
a)
(b
d)
(a
b)
(c
d)
= ⎡⎣b c d ⎤⎦ a − ⎡⎣b c a ⎤⎦ d + ⎡⎣c a d ⎤⎦ b − ⎡⎣c a b ⎤⎦ d + ⎡⎣ a c d ⎤⎦ b − ⎡⎣b c d ⎤⎦ a
= − ⎡⎣ a b c ⎤⎦ d − ⎡⎣ a c d ⎤⎦ b − ⎡⎣ a b c ⎤⎦ d + ⎡⎣ a c d ⎤⎦ b
= −2 ⎡⎣ a b c ⎤⎦ d
⎡⎣( b × c ) × ( a × d ) + ( c × a ) × ( b × d ) + ( a × b ) × ( c × d )⎤⎦ × d = −2 ⎡ a b c ⎤ d × d = 0.
⎣
⎦
Hence, the given vector is parallel to vector d .
Example 3: Prove that ( b c ) ( a d )
Solution:
(b c ) ( a
=
d)
b a
c a
b d
c d
(c
+
a ) (b d )
c b
a b
c d
a d
+
(c
a) ( b d )
(a
b) (c d )
a c
b c
a d
b d
(a
b) ( c d )
( b a )( c d ) ( c a )( b d ) ( b c )( a d ) ( a b )( c d )
( a c )( b d ) ( b c )( a d )
= 0.
0.
Vector Calculus
9.11
Example 4: Prove that a × ⎡⎣ b × ( c × d ) ⎤⎦ = ( b ⋅ c ) ( d × a ) - ( b ⋅ d ) ( c × a ) .
Solution: a × ⎡⎣b × ( c × d ) ⎤⎦ = a × ⎡⎣( b ⋅ d ) c − ( b ⋅ c ) d ⎤⎦
= (b ⋅ d ) ( a × c ) − (b ⋅ c ) ( a × d )
= ( b ⋅ d ) ⎡⎣ − ( c × a ) ⎤⎦ − ( b ⋅ c ) ⎡⎣ − ( d × a ) ⎤⎦
a × ⎡⎣b × ( c × d ) ⎤⎦ = ( b ⋅ c ) ( d × a ) − ( b ⋅ d ) ( c × a )
{
}
Example 5: Prove that d ⋅ ⎡⎣ a × b × ( c × d ) ⎤⎦ = ( b ⋅ d ) ⎡⎣ a c d ⎤⎦ .
Solution: As proved in Example 4.
a × {b × ( c × d )} = ( b ⋅ c ) ( d × a ) − ( b ⋅ d ) ( c × a )
d ⋅ ⎡⎣ a × {b × ( c × d )}⎤⎦ = d ⋅ ⎡⎣( b ⋅ c ) ( d × a ) − ( b ⋅ d ) ( c × a ) ⎤⎦
= ( b ⋅ c ) {d ⋅ ( d × a )} − ( b ⋅ d ) {d ⋅ ( c × a )}
= ( b ⋅ c ) (0) − ( b ⋅ d ) ⎡⎣ d c a ⎤⎦
{
}
= − ( b ⋅ d ) − ⎡⎣c d a ⎤⎦
⎡⎣ Interchanging c and d ⎤⎦
= ( b ⋅ d ) ⎡⎣ a c d ⎤⎦
Exercise 9.1
1. If a = iˆ 2 jˆ 3 k̂, b = 2iˆ + jˆ k̂,
c = iˆ + 8 jˆ 2 k̂, find a ( b c ) .
[Ans. : 21 iˆ
2. Prove that iˆ ( a
( a jˆ) + k̂ ( a
33 jˆ + 15 k̂]
iˆ) + jˆ
k̂) = 2 a.
3. Prove that ⎡⎣ a + b b + c c + a ⎤⎦
= 2 ⎡⎣ a b c ⎤⎦ and hence, prove that
a, b, c are coplanar if and only if
( a + b) , ( b + c) , ( c + a )
4. Prove that
are coplanar.
a
(b c )
b
(c
a) c
(a
b)
0.
5. Prove that
2
⎡⎣b × c c × a a × b ⎤⎦ = ⎡⎣ a b c ⎤⎦ .
(b c) ( a
prove that ( a c )
6. If a
b ) c, then
b
0.
7. Find the scalars p and q such that
( a b) c a ( b c ) , where
a = p iˆ + jˆ + 2 k̂, b = iˆ
jˆ,
c = 4 iˆ + 2 jˆ + q k̂.
[Ans. : p = 2, q = 4]
9.12
Engineering Mathematics
8. If the vector x and a scalar l satisfy
x = l a + b and
a · x = 2, where a = iˆ + 2 jˆ k̂
and b = 2 iˆ jˆ + q k̂, find x and l.
1
Ans. : x = iˆ + 7 ˆj + 3kˆ, =
6
the equation a
9. If the vector x and a scalar l satisfy the equation a
x =l a + b
and a · x = 1, find the values of l
11. If a = 2 iˆ + 3 jˆ k̂,
b = iˆ + 2 jˆ 4 k̂, c = iˆ + jˆ + k̂,
find ( a b ) · ( a c ) .
12. Prove that
2a2 = | a iˆ|2 + | a
Ans. :
=
2 jˆ and
13. Prove that
⎡⎣( a × b ) × ( a × c ) ⎤⎦ · d
= ( a ⋅ d ) ⎡⎣ a b c ⎤⎦ .
14. Prove that
(i)
(a b) , x = a (a b)
a2
0, x
a2
1 ˆ
(3i 4 ˆj 3kˆ)
5
k̂|2,
where a = | a |.
and x in terms of a and b. Also,
determine them if a = iˆ
b = 2 iˆ + jˆ 2 k̂.
jˆ|2 + | a
(ii)
(a
(c
(a
(c
d ) · (b
c ) + (b d ) ·
a) + (c d ) · (a b) = 0
d)
a) +
(b
(c
c ) + (b d )
d)
(a
b) =
2 ( a b) ( b c) ( c a )
10. If a , b , c are three vectors defined by
q ×r
r×p
p×q
a=
,b=
,c=
, 15. If a, b, c, d are four vectors such that
[p q r]
[p q r]
[p q r]
x a + y b + z c + t d = 0, then prove that
then prove that
x
y
z
−t
=
=
=
.
( p a ) + ( q b ) + ( r c ) = 0.
⎡⎣b c d ⎤⎦ ⎡⎣c a d ⎤⎦ ⎡⎣ a b d ⎤⎦ ⎡⎣ a b c ⎤⎦
9.6 VECTOR FUNCTION OF A SINGLE
SCALAR VARIABLE
If, in some interval (a, b) or [a, b], for every value of a scalar variable t, there
corresponds a value of r , then r is called a vector function of the scalar variable ‘t’
and is denoted by r = f (t ).
9.6.1 Decomposition of a Vector Function
If iˆ, jˆ, k̂ be three unit vectors along the three mutually perpendicular fixed directions
(x, y, and z axes), then r = f (t ) can be decomposed as
r = f (t ) = f1(t) iˆ + f2(t) jˆ + f3(t) k̂
where, f1(t), f2(t) and f3(t) are scalar functions of t. This relation can also be denoted by
f = ( f1, f2, f3 )
| f (t ) | = [ f1 (t )]2 + [ f 2 (t )]2 + [ f3 (t )]2
Vector Calculus
9.13
9.6.2 Differentiation of a Vector Function
Derivative of a vector function f (t ) with respect to a scalar variable t is defined as
df
f (t
= lim
t
0
dt
t)
t
f (t )
where, d t is the change in t.
If f (t)= f1 (t) iˆ + f2 (t) jˆ + f3 (t) k̂ where f1 (t), f2 (t) and f3 (t) are the components
of f (t ) in the direction of x, y, z-axes, then derivative in the component form is
d f df1 ˆ df 2 ˆ df3 ˆ
i+
j+
k.
=
dt
dt
dt
dt
9.7 VELOCITY AND ACCELERATION
Let r (t ) = x (t) iˆ + y (t) jˆ + z (t) k̂ be the position vector of a particle moving along
a curve, at time t. Velocity is the rate of change of displacement with respect to time.
Velocity, v = d r = dx iˆ + dy ˆj + dz kˆ
dt
dt
dt
dt
.
dr
is also denoted by r .
dt
Acceleration is the rate of change of velocity with respect to time.
dv d 2 r
=
dt dt 2
d2 y
d2 z
d2 x
= 2 iˆ + 2 jˆ + 2 kˆ .
dt
dt
dt
Acceleration, a =
9.8 STANDARD RESULTS
Most of the basic rules of differentiation that are true for a scalar function of scalar
variable hold good for vector function of a scalar variable, provided the order of factors
in vector products is maintained.
Let a, b, c are differentiable vector functions of a scalar variable t.
1.
dk
= 0, k is a constant vector
dt
2.
d
( a ± b ) = d a ± db
dt
dt dt
3. d
dt
( a) =
da
d
, f is a scalar function of t.
+a
dt
dt
9.14
Engineering Mathematics
4.
d
( a b)
dt
5.
d
( a b)
dt
6.
⎤ ⎡ db ⎤ ⎡
⎡ da
d ⎡
dc ⎤
b c⎥ + ⎢a
c⎥ + ⎢a b ⎥
⎣ a b c ⎤⎦ = ⎢
dt
dt ⎦
⎣ dt
⎦ ⎣ dt ⎦ ⎣
7.
d
a
dt
da
db
b a
dt
dt
da
dt
(b c)
db
dt
b a
da
dt
(b c)
a
db
dt
c
a
dc
dt
b
9.9 TANGENT VECTOR TO A CURVE AT A POINT
Let P(t) and Q (t + d t) be the two points on the curve r = f (t ) .
The tangent at P is the limiting position of the chord PQ when Q
Let
r + d r = f (t + d t)
PQ
P.
Q
OQ OP
d r = (r + d r ) − r
Since d t is a scalar, vector
parallel to PQ.
As d t 0, Q
of chord PQ is
t)
t
dr
f (t )
P
r
r
is
t
t
0
x
O
P, limiting position
lim
Hence,
r+
r
f (t
=
t
f (t)
dr
f (t + d t)
Fig. 9.2
r dr
=
t dt
dr
is a vector parallel to the tangent at P.
dt
If s is the arc length measured from a fixed point, and d s is the arc length PQ, then
lim
s →0
chord PQ
r
= lim
Q
→
P
arc PQ
s
dr
=1
ds
Hence, d r is a unit vector in the direction of the tangent to the curve at P and is called
ds
unit tangent vector. It is denoted by t̂.
Vector Calculus
d
Example 1: Write down the formula for
dt
A = 5t2 iˆ + t jˆ – t3 k̂, and B = sint iˆ – cost jˆ.
Solution:
Given,
( A ë B)
9.15
and verify the same for
d
( A B) = d A B + A dB
dt
dt
dt
A = 5t2 iˆ + t jˆ t3 k̂,
B = sin t iˆ
iˆ
A B
cos t jˆ
ˆj
kˆ
5t 2
sin t
t3
0
t
cos t
iˆ (0 t3 cos t) jˆ (0 + t3 sin t) + k̂ ( 5t2 cos t t sin t)
( t3 cos t) iˆ (t3 sin t) jˆ (5t2 cos t + t sin t) k̂
(
)
d
A B = ( 3t2 cos t + t3 sin t) iˆ (3t2 sin t + t3 cos t) jˆ
dt
(10t cos t 5t2 sin t + sin t + t cos t) k̂
dA
= 10t iˆ + jˆ
dt
Now,
… (1)
3t2 k̂,
dB
= cos t iˆ + sin t jˆ
dt
ˆj
iˆ
kˆ
dA
dt
B
1
cos t
iˆ(0
3t2 cos t)
iˆ
dB
A
dt
jˆ(0 + 3t2 sin t) + k̂ ( 10 t cos t
5t 2
t
cos t sin t
t3
0
jˆ(0 + t 3 cos t) + k̂ (5t2 sin t
t cos t)
dB
= ( 3t2 cos t + t3 sin t) iˆ (3t2 sin t + t3 cos t) jˆ
dt
(10t cos t + sin t 5t2 sin t + t cos t) k̂
Comparing Eqs. (1) and (2),
B
sin t)
kˆ
ˆj
iˆ(0 + t 3 sin t)
dA
dt
3t 2
0
10 t
sin t
A
(
d
A B
dt
)
dA
dt
B
A
dB
dt
… (2)
9.16
Engineering Mathematics
Example 2: If
d
dt
(u ë v) =
du
dv w v
= w ë u and
=
ë , then prove that
dt
dt
(
)
w ë uëv .
d
( u v ) = du v + u
dt
dt
du
dv
= w u,
= w v
dt
dt
Solution: We know that,
But
d
(u v)
dt
(w
u) v u
dv
dt
( w v)
( v w) u ( v u ) w ( u v ) w ( u w) v
( v w) u ( u w) v ( w v ) u ( w u ) v
w (u v)
1
Example 3: If r = t3 iˆ + 2t 3
5t
dr
jˆ, then show that r ë
= k̂ .
dt
2
1
Solution: r = t3 iˆ + 2t 3
jˆ
5t 2
2 ⎞
⎛ 2
dr
= 3t2 iˆ + ⎜ 6t + 3 ⎟ jˆ
⎝
5
t ⎠
dt
r
dr
dt
t 3iˆ
1
2t 3
5t
3t 5 ( iˆ iˆ )
2t 3
0
6t 5
2
6t 5
1
5t
2
6t 2
2 ˆ
k
5
ˆj
3t 2 iˆ
6t 2
2
5
( iˆ ˆj )
6t 5
2
5t 3
6t 5
2
5t 3
3
5
ˆj
( ˆj iˆ)
( ˆj ˆj )
3
( kˆ) 0
5
[∵ iˆ × iˆ = 0 =
= k̂
Example 4: If a and b are constant vectors and v is constant and
r = a sin v t + b cos v t, prove that r ë
Solution: r = a sin w t + b cos w t
dr
= a w cos w t + b w ( sin w t)
dt
dr
+ v ( a ë b ) = 0.
dt
jˆ × jˆ
]
Vector Calculus
9.17
dr
= ( a sin t + b cos t ) × ( a cos t − b sin t )
dt
r×
= ( a × a ) sin t cos t − ( a × b ) sin 2 t + ( b × a ) cos 2 t
− ( b × b ) cos t sin t
= 0 − ( a × b ) sin 2 t − ( a × b ) cos 2 t − 0 ⎡⎣∵ a × a = 0 = b × b ⎤⎦
= − ( a × b ) (sin 2 t + cos 2 t ) = − ( a × b )
dr
+ ( a b ) w = 0.
dt
Hence, r
Example 5: If r = a sinh t + b cosh t, where a and b are constant, then show
that
d2 r
(i)
dt 2
=r
(ii)
dr
dt
d2 r
dt 2
constant.
Solution: r = a sinh t + b cosh t,
dr
= a cosh t + b sinh t
dt
d 2 r = a sinh t + b cosh t
dt 2
(i)
d2 r
Hence,
(ii)
dt 2
[∵ a and b are constant]
r
=r
dr d2 r (
×
= a cosh t + b sinh t ) × ( a sinh t + b cosh t )
dt dt 2
= ( a × a ) cosh t sinh t + ( a × b ) cosh 2 t + ( b × a ) sinh 2 t + ( b × b ) sinh t cosh t
= 0 + ( a × b ) cosh 2 t − ( a × b ) sinh 2 t + 0
= ( a × b ) ( cosh 2 t − sinh 2 t )
= (a × b)
Hence,
dr
dt
d2 r
dt 2
[∵ cosh 2 t − sinh 2 t = 1]
= constant.
ct
Example 6: If r = a (sin v t) iˆ + b (sin v t) jˆ + 2 (sin v t) k̂, prove that
d2 r
dt
2
+ v2 r =
2c
(cos v t) kˆ.
9.18
Engineering Mathematics
ct
Solution: r = a (sin w t) iˆ + b (sin w t) jˆ + 2 (sin w t) k̂
dr
c
= aw (cos w t) iˆ + bw (cos w t) jˆ + 2 (sin w t + t w cos w t) k̂
dt
d2 r
dt
c
= aw ( w sin w t ) iˆ + bw ( w sin w t) jˆ + 2 [w (cos w t) +
2
w (cos w t) + tw ( w sin w t)] k̂
aw2 (sin w t ) iˆ
bw2 (sin w t) jˆ +
c
2
(2w cos w t
tw2 sin w t) k̂
ct
2c
w2 [a (sin w t) iˆ + b (sin w t) jˆ + 2 (sin w t) k̂] +
(cos w t) k̂
w2 r +
2c
(cos w t) k̂
Example 7: If r = (a cos t) iˆ + (a sin t) jˆ + (at tan a ) k̂, prove that
(i)
d2 r
dr
dt
dt 2
⎡ dr d2 r d3 r ⎤
= a 3 tan α .
(ii) ⎢
2
3 ⎥
⎢⎣ dt dt dt ⎥⎦
a 2 sec
Solution: r = (a cos t) iˆ + (a sin t) jˆ + (at tan a) k̂,
(i)
dr
dt
dr
= ( a sin t ) iˆ + (a cos t) jˆ + (a tan a) k̂
dt
d2 r
= ( a cos t) iˆ + ( a sin t) jˆ + 0 k̂
dt 2
d3 r
= (a sin t) iˆ + ( a cos t) jˆ + 0 · k̂
dt 3
iˆ
jˆ
kˆ
d2 r
a sin t a cos t a tan
dt 2
a cos t
a sin t
0
= iˆ (0 + a2 sin t tan a) jˆ (0 + a2 cos t tan a) + k̂ (a2 sin2 t + a2 cos2 t)
= a2 (sin t tan a) iˆ
dr
dt
(ii)
a2 (cos t tan a) jˆ + a2 k̂
2
d r
a 4 sin 2 t tan 2
dt 2
dr
dt
d2 r
d3 r
2
dt 3
dt
2
2
a 4 = a tan
a 4 cos 2 t tan 2
2
+ 1 = a sec a
=[a2(sin t tan a) iˆ – a2 (cos t tan a) jˆ + a2 k̂]. [(a sin t) iˆ
+ ( a cos t) jˆ + 0 k̂]
= a sin t tan a + a cos t tan a
[∵ iˆ · iˆ = jˆ · jˆ = k̂ · k̂ = 1 and iˆ · jˆ = jˆ · k̂ = k̂ · iˆ = 0]
3
a tan a
3
2
3
2
Vector Calculus
9.19
2
3 ⎤
⎡
Hence, ⎢ d r d r d r ⎥ = a 3 tan .
2
⎣ dt dt dt 3 ⎦
Example 8: If A = (sin t) iˆ + (cos t) jˆ + t k̂, B = (cos t) iˆ – (sin t) jˆ – 3 k̂,
d
C = 2 iˆ + 3 jˆ – k̂, find
A
dt
iˆ
(B
Solution:
(B
A
C)
cos t
2
iˆ
sin t
C)
(B
C ) at t = 0.
ˆj
sin t
3
kˆ
3 = iˆ (sin t + 9) jˆ( cost + 6) +
k̂ (3 cos t + 2 sin t)
1
kˆ
t
ˆj
cos t
sin t 9 cos t 6 3cos t 2sin t
= iˆ (3 cos2 t + 2 sin t cos t t cos t + 6t) jˆ (3 cos t sin t + 2 sin2 t
t sin t 9t) + k̂ (sin t cos t 6 sin t cos t sin t 9 cos t)
⎛3
⎞
2
= (3 cos2 t + sin 2t t cos t + 6t)iˆ ⎜⎝ sin 2 t + 2 sin t − t sin t − 9t ⎟⎠ jˆ
2
+ ( 6 sin t 9 cos t) k̂
d
A
dt
(B
C ) = [6 cos t ( sin t) + 2 cos 2t
(3 cos 2t + 4 sin t cos t
cos t + t sin t + 6] iˆ
sin t t cos t 9) jˆ (6 cos t
9 sin t) k̂
Putting t = 0,
d
dt
A
(B
C ) = 7 iˆ + 6 jˆ
6 k̂.
Example 9: Find the derivative of r
dr
dt
d2 r
dt 2
with respect to ‘t’.
Solution:
⎛ d r d3 r ⎞
⎛ d2 r d2 r ⎞
d ⎡ ⎛ dr d2 r ⎞ ⎤ dr ⎛ dr d2 r ⎞
×⎜ × 2 ⎟ +r×⎜ 2 × 2 ⎟ +r×⎜ × 3 ⎟
⎢r × ⎜ × 2 ⎟ ⎥ =
dt ⎢⎣ ⎝ dt dt ⎠ ⎥⎦ dt ⎝ dt dt ⎠
dt ⎠
⎝ dt dt ⎠
⎝ dt
⎤
⎡ d2 r d2 r
⎛ d r d3 r ⎞
dr ⎛ dr d2 r ⎞
=
×⎜ × 2 ⎟ +r×⎜ × 3 ⎟
⎢∵ 2 × 2 = 0⎥
dt ⎝ dt dt ⎠
dt
⎝ dt dt ⎠
⎥⎦
⎢⎣ dt
Example 10: Find
d ⎛ r×a ⎞
⎜
⎟ , where r is a vector function of scalar variable
dt ⎜⎝ r ⋅ a ⎟⎠
t and r is a constant vector.
d
⎡d(
⎤
r × a )⎥ ( r ⋅ a ) − ( r × a ) ( r ⋅ a )
d ⎛ r × a ⎞ ⎢⎣ dt
d
t
⎦
Solution:
=
⎟
⎜
2
dt ⎝ r ⋅ a ⎠
(r ⋅ a)
9.20
Engineering Mathematics
dr
dt
=
da
dt
a r
( r a) ( r
a)
dr
da
a r
dt
dt
( r a )2
da
= 0, as a is constant.
dt
⎛ dr
⎞
⎛ dr ⎞
× a⎟ ( r ⋅ a ) − ( r × a ) ⎜ ⋅ a⎟
⎜
⎠
⎝ dt ⎠
d ⎛ r × a ⎞ ⎝ dt
=
.
Hence,
⎟
⎜
2
dt ⎝ r ⋅ a ⎠
(r ⋅ a)
But,
df
if f = r2 r + ( a · r ) b where r is a function of t and
dt
a , b are constant vectors.
Example 11: Find
Solution: f = r2 r + ( a r ) b
df
dt
d( 2 )
r r
dt
dr 2
dt
2r
Hence,
dr
dt
d
(a r)b
dt
(r)
(r)
r2
dr
dt
r2
da
dr
r a
b
dt
dt
dr
dt
a
( a r ) db
dt
dr
b
dt
∵
da
dt
db
dt
⎛ dr ⎞
df
dr
dr
= 2rr
+ r2
+ b⎜a⋅ ⎟ .
dt
dt
dt
⎝ dt ⎠
Example 12: If f (t) is a unit vector, prove that f ( t )
Solution: Since f is a unit vector,
f · f =1
Differentiating w.r.t. t,
df
df
⋅f+f⋅
=0
dt
dt
df
2f ⋅
=0
dt
df
f⋅
=0
dt
This shows that f and d f are perpendicular to each other.
dt
d f (t )
dt
d f (t )
.
dt
0
Vector Calculus
Now,
f
df
dt
f
9.21
df
sin n^
dt
df
and nˆ is the unit vector perpendicular to the
where, q is the angle between f and
dt
df
plane of f and
.
dt
df
are perpendicular,
Since f and
dt
f d f = f d f sin nˆ
dt
2
dt
Hence,
f
df
dt
df
nˆ
dt
f
df
dt
df
dt
2
ÈÎ∵ f is a unit vector ˘˚
[∵ | n̂ | = 1]
Example 13: Find the magnitude of the velocity and acceleration of a particle
which moves along the curve x = 2 sin 3t, y = 2 cos 3t, z = 8t at any time t > 0. Find
unit tangent vector to the curve.
Solution: The position vector r of the particle is
r = xiˆ + yjˆ + zk̂ = (2 sin 3t)iˆ + (2 cos 3t)jˆ + (8t)k̂
Velocity,
v=
dr
= (6 cos 3t) iˆ + ( 6 sin 3t) jˆ + 8k̂
dt
v = 36 cos 2 3t + 36sin 2 3t + 64 = 36 + 64 = 10
Acceleration,
a=
d2 r
= ( 18 sin 3t) iˆ + ( 18 cos 3t) jˆ + (0)k̂
dt 2
a = (18) 2 sin 2 3t + (18) 2 cos 2 3t
= 18.
dr
1
Unit tangent vector = dt =
[(6 cos 3t ) iˆ – (6 sin 3t ) jˆ +8kˆ ].
10
dr
dt
Example 14: A particle moves along a plane curve such that its linear velocity
is perpendicular to the radius vector. Show that the path of the particle is a circle.
Solution: Let position vector r of the particle is
r = x iˆ + y jˆ
9.22
Engineering Mathematics
dr
dt
To find path of the particle, we have to develop a relation in x and y. Velocity is perpendicular to the radius vector.
v=
Velocity,
dr
dt
dr
2r
dt
dr dr
r
r
dt dt
d
r r
dt
r r
r
( )
0
0
0
0
c 2 , constant
x2 + y2 = c2
which is a circle with center at the origin and radius c.
Example 15: Find the magnitude of tangential components of acceleration at
any time t of a particle whose position at any time t is given by x = cos t + t sin t,
y = sin t - t cos t.
Solution: Position vector r of the particle is
r = (cos t + t sin t) iˆ + (sin t
Velocity,
t cos t) jˆ
dr
dt
= ( sin t + sin t + t cos t) iˆ + (cos t
= (t cos t) iˆ + (t sin t ) jˆ
v=
d2 r
= (cos t
dt 2
Unit vector in the direction of the tangent is
a=
Acceleration,
cos t + t sin t) jˆ
t sin t) iˆ + (sin t + t cos t) jˆ
dr
(t cos t ) iˆ + (t sin t ) ˆj
tˆ = dt =
= (cos t) iˆ + (sin t) jˆ
t 2 cos 2 t + t 2 sin 2 t
dr
dt
Magnitude of tangential component of acceleration
= a ⋅ tˆ
[(cos t t sin t) iˆ + (sin t + t cos t) jˆ] · [(cos t) iˆ + (sin t) jˆ]
cos2 t t sin t cos t + sin2 t + t cos t sin t
1
Vector Calculus
9.23
Example 16: Show that a particle whose position vector r at any time t is given
by r = (a cos nt) iˆ + (b sin nt) jˆ moves in an ellipse whose center is at the origin
and that its acceleration varies directly as its distance from the center and is directed towards it.
Solution: r = (a cos nt) iˆ + (b sin nt) jˆ
x = a cos nt, y = b sin nt
x2 y 2
+
= cos 2 nt + sin 2 nt = 1
a 2 b2
x2 y 2
+
=1
a 2 b2
which is an ellipse with center at origin.
dr
= ( a n sin nt) iˆ + (b n cos nt) jˆ
dt
d2 r
a = 2 = ( a n2 cos nt) iˆ + ( b n2 sin nt) jˆ
dt
n2 [(a cos nt) iˆ + (b sin nt) jˆ] = n2 r
Now,
Acceleration,
This shows that acceleration of the particle varies directly as its distance r from the
origin (center of the ellipse) and negative sign shows that acceleration is directed
towards the origin.
Exercise 9.2
1. If A = 5t2 iˆ + t jˆ
t3 k̂ and
B = sin t iˆ cos t jˆ, find the value of
(i)
d(
A B)
dt
(ii)
d(
A B)
dt
Ans. :
(i) (5t 2 1) cos t 11t sin t ,
(ii) (t 3 sin t 3t 2 cos t ) iˆ
^
(t 3 cos t 3t 2 sin t ) j
^
(5t 2 sin t sin t 11cos t ) k
2. If A = 4t3 iˆ + t2 jˆ
6t2 k̂ and
B = (sin t) iˆ (cos t) jˆ, verify the
d(
formula of
A B) .
dt
3. If r = A e nt + B e nt, show that
d 2 r n2 r = 0.
dt 2
1
4. If r = t3 iˆ + 2t 3
jˆ, show
5t 2
dr
that r
= k̂.
dt
5. Prove that
d ⎡ dr d 2 r ⎤ ⎡ dr d3 r ⎤
⎥.
⎥ = ⎢r
⎢r
dt ⎣ dt dt 2 ⎦ ⎣ dt dt 3 ⎦
6. Prove that
d 2 ⎡ dr d 2 r ⎤ ⎡ d 2 r d3 r ⎤
⎥
⎥ = ⎢r
⎢r
dt 2 ⎣ dt dt 2 ⎦ ⎣ dt 2 dt 3 ⎦
⎡ dr d4 r ⎤
.
+ ⎢r
4 ⎥
⎣ dt d t ⎦
9.24
Engineering Mathematics
the velocity and acceleration at t = 1
in the direction of iˆ 3 jˆ + 2 k̂
[Hint: unit vector in the direction of
iˆ 3jˆ + 2kˆ is
iˆ 3 ˆj 2kˆ iˆ 3 ˆj 2kˆ
=
nˆ =
,
1+ 9 + 4
14
7. Find the derivatives of the following:
(i) r3 r + a
r
rb
dr
(ii) 2 +
r
a
r
dt
where, r = r , a and b are constant
vectors.
Ans. :
dr
dr
r r3
(i) 3r
dt
dt
2
(ii)
1 dr
r 2 dt
br
( a r )2
2
Find v and a at t = 1, velocity in the
given direction = v n̂ and acceleration
in the given direction = a n̂ ]
d2 r
a
dt 2
r dr
r 3 dt
b
dr
( a r ) dt
dr
a
dt
8. A particle moves along the curve
r = e t (cos t) iˆ + e t (sin t) jˆ + e t k̂.
Find the magnitude of velocity and
acceleration at time t.
Ans. : v = 3e t , a = 5e t
9. A particle moves on the curve
x = 2t2, y = t2 4t, z = 3t 5. Find
Ans. : v
8 2
7
,a
2
7
10. A particle is moving along the curve
r = a t 2 + b t + c, where a, b, c are
constant vectors. Show that acceleration is constant.
11. A particle moves such that its position vector is given by
r = (cos w t) iˆ + (sin w t) jˆ. Show
that velocity v is perpendicular to r.
dr
r 0
Hint: Prove that
dt
9.10 SCALAR AND VECTOR POINT FUNCTION
9.10.1 Field
If a function is defined in any region of space, for every point of the region, then this
region is known as field.
9.10.2 Scalar Point Function
A function f (x, y, z) is called scalar point function defined in the region R, if it
associates a scalar quantity with every point in the region R of space. The temperature
distribution in a heated body, density of a body and potential due to gravity are the
examples of a scalar point function.
9.10.3 Vector Point Function
A function F (x, y, z) is called vector point function defined in the region R, if it
associates a vector quantity with every point in the region R of space. The velocity of a
moving fluid, gravitational force are the examples of vector point function.
Vector Calculus
9.25
9.10.4 Vector Differential Operator Del ( )
The vector differential operator Del (or nabla) is denoted by
iˆ
ˆj
x
kˆ
y
and is defined as
z
9.11 GRADIENT
The gradient of a scalar point function f is written as
iˆ
grad
x
ˆj
or grad f and is defined as
kˆ
y
z
grad f is a vector quantity.
f (x, y, z) is a function of three independent variables and its total differential df is
given as
∂
∂
∂
d =
dx +
dy +
dz
∂x
∂y
∂z
⎛ ∂
∂
∂ ⎞
= ⎜ iˆ
+ ˆj
+ kˆ ⎟ ⋅ ( iˆdx + ˆjdy + kˆdz )
∂
∂
∂z ⎠
x
y
⎝
dr
...(1) ∵ r
xiˆ
yjˆ zkˆ
dr
iˆdx
ˆjdy
ˆjdz
d r cos
where, q is the angle between the vectors
direction, then q = 0,
df =
and d r. If dr and
are in the same
dr
cos q = 1 is the maximum value of cos q. Hence, df is maximum at q = 0.
9.11.1 Normal
Let f (x, y, z) = c represents a family of surfaces for different values of the constant c.
Such a surface for which the value of the function is constant is called level surface.
Now differentiating f, we get
df = 0
But from Eq. (1) of 9.11,
d
dr
dr
0
Hence,
and dr are perpendicular to each other. Since vector dr is in the direction
of the tangent to the given surface, vector
is perpendicular to the tangent to the
surface and hence
is in the direction of normal to the surface.
Thus geometrically
represents a vector normal to the surface f (x, y, z) = c.
9.26
Engineering Mathematics
9.11.2 Directional Derivative
(i) Let f (x, y, z) be a scalar point function. Then
, ,
are the directional
x y z
derivative of f in the direction of the coordinate axes.
f
f
f
are the directional
Similarly, if f (x, y, z) be a vector point function, then
,
,
x y z
derivative of f in the direction of the coordinate axes.
(ii) The directional derivative of a scalar point function f (x, y, z) in the direction
of a line whose direction cosines are l, m, n,
=l
x
+m
y
+n
z
(iii) The directional derivative of scalar point function f (x, y, z) in the direction of
in the direction of a . If â is the unit vector
vector a , is the component of
in the direction of a , then directional derivatives of f in the direction of a
a
aˆ
a
9.11.3 Maximum Directional Derivative
Since the component of a vector is maximum in its own direction, [∵ cos q is maximum
. Since
when q = 0], the directional derivative is maximum in the direction of
is normal to the surface, directional derivative is maximum in the direction of normal.
Maximum directional derivative
cos
cos 0
Standard Results:
(i)
(
)
(ii)
(
(iii)
f (u )
)
(
) (
f (u )
iˆ
x
)
ˆj f (u ) kˆ f (u )
y
z
f (u ) u.
at (1, - 2, 1), if f = 3x2y - y3 z2.
Example 1: Find
ˆj
kˆ
x
y
z
ˆ
ˆ
= i (6xy 0) + j (3x2 3y2z2) + k̂ (0
At x = 1, y = 2, z = 1
= iˆ ( 12) + jˆ (3 12) + k̂ (16)
at (1, 2, 1) = 12 iˆ 9 jˆ + 16 k̂
Solution:
iˆ
2y3z)
Vector Calculus
Example 2: Evaluate
9.27
2
e r , where r2 = x2 + y2 + z2.
Solution: r2 = x2 + y2 + z2
Differentiating partially w.r.t. x, y and z,
r
r x
= 2 x,
=
2r
x
x r
r
r y
= 2 y,
=
2r
y
y r
r
r z
= 2 z,
=
2r
z
z r
2
e
er
iˆ
x
r2
er
iˆ
r
2
2
2
r
ˆj e
y
r
x
er
kˆ
z
r
ˆj e
r
2
x
iˆ(e r 2r )
r
2
ˆj (e r 2
2
er
r
kˆ
r
z
y ˆ r2
z
k (e 2 r )
2r )
r
r
r
y
(
2
2e r xiˆ
yjˆ zkˆ
)
Example 3: If f (x, y) = log x 2 + y 2 and r = xî + yĵ + zk̂, prove that
( kˆ r ) kˆ
( kˆ r ) kˆ r ( kˆ r ) kˆ
r
grad f =
r
.
Solution: f (x, y) = log x 2 + y 2
=
iˆ
f
=
=
1
log( x 2 + y 2 )
2
1
log( x 2
x 2
1
iˆ
2x
2 x2 + y 2
xiˆ + yjˆ
y2 )
ˆj
1
log( x 2
y 2
y2 )
kˆ
1
log( x 2
z 2
y2 )
ˆj
1
2y 0
2 x2 + y 2
x2 + y 2
( xiˆ
xiˆ + yjˆ
yjˆ ) ( xiˆ
yjˆ )
Now, r = xiˆ + yjˆ + zk̂
k̂· r = z
r = xiˆ + yjˆ + (k̂ · r ) k̂
r
(k̂ · r )k̂ = xiˆ + yjˆ
[∵ iˆ · k̂ = jˆ · k̂ = 0, k̂ · k̂ = 1]
9.28
Engineering Mathematics
Substituting x iˆ + y jˆ in
f,
( kˆ r ) kˆ
( kˆ r ) kˆ r ( kˆ r ) kˆ
r
f
r
Example 4: Prove that
rn
nr n 2 r , r = xî + yĵ + zk̂, r = r .
Solution: r = x iˆ + y jˆ + zk̂, r2 = x2 + y2 + z2
∂r x ∂r y ∂r z
= , = , =
∂x r ∂y r ∂z r
∂r n
∂r n
∂r n
∂r n ∂r ˆ ∂r n ∂r ˆ ∂r n ∂r
⋅ +j
⋅ +k
+j
⋅
+k
=i
∂r ∂y
∂x
∂r ∂z
∂y
∂z
∂r ∂x
z
x
y
= iˆ nr n −1 ⋅ + ˆj nr n −1 ⋅ + kˆ nr n−1 ⋅
r
r
r
= nrn 2 (x iˆ + yjˆ + zk̂ )
= nrn 2 r .
∇r n = i
⎛ a ⋅ r ⎞ a n(a ⋅ r )
Example 5: Show that ∇ ⎜ n ⎟ = n − n + 2 ( r ) , where r = xî + yĵ + zk̂,
⎜ r ⎟ r
r
⎝
⎠
r = r , a is constant vector.
Solution: Let a = a1iˆ + a2 ˆj + a3 kˆ, and
a⋅r
=
rn
r = xiˆ + yjˆ + zkˆ
⎛ a ⋅ r ⎞ ⎡ ( a iˆ + a2 ˆj + a3 kˆ ) ⋅ ( xiˆ + yjˆ + zkˆ ) ⎤
= ⎜⎜ n ⎟⎟ = ⎢ 1
⎥
rn
⎦
⎝ r ⎠ ⎣
⎛ a x + a2 y + a3 z ⎞
=⎜ 1
⎟
rn
⎝
⎠
a1 x + a2 y + a3 z
=
x
x
rn
=
=
But,
x
(a1 x a2 y a3 z ) r n
(a1 x a2 y a3 z )
r 2n
a1r n
(a1 x a2 y a3 z )nr n
1
r 2n
r = xiˆ + yjˆ + zk̂, r2 = x2 + y2 + z2
r
x
rn
x
Vector Calculus
9.29
r x r y r z
= ,
= ,
=
x r y r z r
x
Similarly,
y
and
z
=
a1r n
a2 r n
=
(a1 x a2 y a3 z )nr n
y
r
1
r 2n
a3 r n
(a1 x a2 y a3 z )nr n
z
r
1
r 2n
iˆ
=
x
r
1
r 2n
=
=
(a1 x a2 y a3 z )nr n
x
( a1iˆ
ar
n
ˆj
kˆ
y
z
+ a2 jˆ + a3 kˆ ) r n − (a1 x + a2 y + a3 z )nr n − 2 ( xiˆ + yjˆ + zkˆ
r
− ( a ⋅ r ) nr n − 2 r
)
2n
r 2n
⎡⎣∵ a1 x + a2 y + a3 z = ( a1iˆ + a2 jˆ + a3 kˆ ) ⋅ ( xiˆ + yjˆ + zkˆ ) = a ⋅ r ⎤⎦
( )
⎛
⎞
Hence, ∇ ⎜ a ⋅nr ⎟ = an − n an⋅+r2 r .
r
⎝ r ⎠ r
Example 6: If r = xî + yĵ + zk̂ and a , b are constant vectors, prove that
a
3(a r) (b r)
r5
1
r
b
Solution: Let
a b
.
r3
a = a1iˆ + a2 ˆj + a3 kˆ, b = b1iˆ + b2 ˆj + b3 kˆ
∂ ⎛1⎞
∂ ⎛1⎞
∂ ⎛1⎞
⎛1⎞
∇ ⎜ ⎟ = iˆ ⎜ ⎟ + jĵ ⎜ ⎟ + kˆ ⎜ ⎟
∂x ⎝ r ⎠
∂y ⎝ r ⎠
∂z ⎝ r ⎠
⎝r⎠
⎛ 1 ∂r ⎞
⎛ 1 ∂r ⎞ ˆ ⎛ 1 ∂r ⎞
= iˆ ⎜ − 2
⎟ + j ⎜ − 2 ∂y ⎟ + k̂ ⎜ − 2 ∂z ⎟
x
∂
r
r
⎝ r
⎠
⎝
⎠
⎠
⎝
2
2
2
2
ˆ
ˆ
r = xi + yj + zk̂, r = x + y + z
r x r y r z
= ,
= ,
=
x r y r z r
But,
1
r
iˆ
1 x
r2 r
ˆj
1 y
r2 r
kˆ
1 z
r2 r
(
1 ˆ
xi
r3
yjˆ zkˆ
)
r
.
r3
9.30
Engineering Mathematics
1
r
b
xiˆ + yjˆ + zkˆ
r3
(b1iˆ b2 ˆj b3 kˆ)
b1 x + b2 y + b3 z
r3
= , say
1
r
b
iˆ
x
ˆj
x
kˆ
y
z
b1 x + b2 y + b3 z
r3
x
b1r 3
(b1 x b2 y b3 z )
x
r3
b1r 3
r6
r
r
Similarly,
y
and
z
=
=
iˆ
3(b r ) y
r5
b3 r 2
3(b r ) y
r5
ˆj
x
a
Hence,
a
a
b
y
kˆ
b
1
r
1
r
3(b r ) x
r5
(b1iˆ + b2 ˆj + b3 kˆ)
r3
z
3(b r ) r
r5
b
r3
b1r 2
6
b2 r 2
r
x
r6
( b r ) 3r 2 x
b1r 3
(b1 x b2 y b3 z )3r 2
a b
r3
3 ( a r )( b r )
r5
3 ( b r ) ( xiˆ yjˆ zkˆ)
r5
3 ( a r )( b r )
r5
a b
.
r3
Example 7: Find the unit vector normal to the surface x2 + y2 + z2 = a2 at
a
a
a
,
,
.
3
3
3
Solution:
Given surface is
is the vector which is normal to the surface f (x, y, z) = c
x2 + y2 + z2 = a2
f (x, y, z) = x2 + y2 + z2
Vector Calculus
iˆ
x
( x2
y2
ˆj
z2 )
( x2
y
9.31
z 2 ) kˆ
y2
z
( x2
y2
z2 )
= iˆ(2 x) + ˆj (2 y ) + kˆ(2 z )
At the point
a
3
a
,
,
3
a
,
3
∇ =
2a
3
(iˆ + jˆ + kˆ )
a
Unit vector normal to the surface x2 + y2 + z2 = a2 at
=
=
=
a
3
,
a
3
(iˆ + ˆj + kˆ )
2a
4a 2 4a 2 4a 2
+
+
3
3
3
3
(
) = iˆ + ˆj + kˆ .
2a iˆ + ˆj + kˆ
3
3
,
2a 3
3
3
Example 8: Find unit vector normal to the surface x2y + 2xz2 = 8 at the point
(1, 0, 2).
Solution: Given surface is x2y + 2xz2 = 8
f (x, y, z) = x2y + 2xz2
( x 2 y 2 xz 2 ) kˆ ( x 2 y 2 xz 2 )
y
z
2
2
ˆ
= iˆ(2 xy + 2 z ) + ˆj ( x ) + k (4 xz )
At the point (1, 0, 2),
= 8iˆ + jˆ + 8k̂
iˆ
x
( x 2 y 2 xz 2 )
ˆj
Unit vector normal to the surface x2y + 2xz2 = 8 at the point (1, 0, 2)
=
=
8iˆ
ˆj 8kˆ
64 + 1 + 64
=
ˆj 8kˆ
8iˆ
129
.
Example 9: Find the directional derivatives of f = xy2 + yz2 at the point (2, –1, 1)
in the direction of the vector ˆi + 2jˆ + 2kˆ.
Solution:
iˆ
x
( xy 2
yz 2 )
ˆj
y
( xy 2
yz 2 ) kˆ
= iˆy2 + jˆ (2xy + z2) + k̂ (2yz)
z
( xy 2
yz 2 )
9.32
Engineering Mathematics
At the point (2, 1, 1),
= iˆ + jˆ ( 4 + 1) + k̂ ( 2) = iˆ
3 jˆ 2k̂
Directional derivative in the direction of the vector a = iˆ + 2 jˆ + 2k̂
(
(iˆ
a
)
a
(1 6 4)
3
3 ˆj 2kˆ
ˆ
ˆ
ˆ
) ( i + 2 j + 2k )
1+ 4 + 4
3.
1
Example 10: Find the directional derivative of φ =
2
(x + y
P (1, –1, 1) in the direction of a = ˆi + ˆj + kˆ.
2
1
+ z2 )2
at the point
Solution:
∇ = iˆ
∂
∂x
1
1
2 2
( x2 + y 2 + z )
∂
+ ˆj
∂y
2x
⎡
⎤ˆ
= ⎢−
i+
3 ⎥
⎢⎣ 2( x 2 + y 2 + z 2 ) 2 ⎥⎦
=−
1
1
2 2
( x2 + y 2 + z )
∂
+ kˆ
∂z
1
1
( x2 + y 2 + z 2 ) 2
2y
2z
⎤ˆ
⎤ ⎡
ˆj ⎡ −
k
+ ⎢−
3 ⎥
3 ⎥
⎢
⎢⎣ 2( x 2 + y 2 + z 2 ) 2 ⎥⎦ ⎢⎣ 2( x 2 + y 2 + z 2 ) 2 ⎥⎦
( xiˆ + yjˆ + zkˆ )
3
( x2 + y 2 + z 2 ) 2
At the point (1, 1, 1),
−(i − j + k )
∇ =
3
(3) 2
Directional derivative in the direction of a = iˆ + jˆ + k̂
=∇ ⋅
=
a
a
=
−(i − j + k ) ⋅ (i + j + k )
3
(3) 2 1 + 1 + 1
1
−1 + 1 − 1
=− .
2
9
3
Example 11: Find the directional derivative of f = xy2 + yz3 at (2, –1, 1) in the
direction of the normal to the surface x log z – y2 = –4 at (–1, 2, 1).
Solution: Let y = x log z
y2
is normal to the surface x log z
iˆ
x
( x log z
iˆ(log z )
y2 )
y2 = 4
ˆj
y
ˆj ( 2 y ) kˆ x
z
( x log z
y 2 ) kˆ
z
( x log z
y2 )
Vector Calculus
At the point ( 1, 2, 1),
= iˆ (log 1) 4 jˆ k̂
4 jˆ k̂
ˆ
4 j k̂ is a vector normal to the surface x log z
Now,
f = xy2 + yz3
iˆ
x
( xy 2
ˆj
yz 3 )
y
( xy 2
9.33
y2 = 4 at ( 1, 2, 1).
yz 3 ) kˆ
z
( xy 2
yz 3 )
iˆ (y2) + jˆ (2xy + z3) + k̂ (3yz2)
At the point (2, 1, 1),
= iˆ + jˆ ( 4 + 1) + k̂ ( 3) iˆ 3jˆ
3k̂
Directional derivative of f in the direction of the vector 4 jˆ
( 4 ˆj kˆ) 12 3 15
(iˆ 3 ˆj 3kˆ)
16 + 1
17
17
k̂
Example 12: Find directional derivative of the function f = xy2 + yz2 + zx2 along
the tangent to the curve x = t, y = t2, z = t3 at the point (1, 1, 1).
Solution: Tangent to the curve is
dr d ˆ ˆ
= ( xi + yj + zkˆ)
dt dt
d
= (tiˆ + t 2 ˆj + t 3 kˆ) = (iˆ + 2tjˆ + 3t 2 kˆ)
dt
If x = 1, y = 1, z = 1, then t 1
At the point (1, 1, 1), t = 1
T=
T = iˆ + 2jˆ + 3k̂
f = xy2 + yz2 + zx2
iˆ
x
( xy 2
yz 2
zx 2 )
ˆj
y
( xy 2
yz 2
zx 2 ) kˆ
z
( xy 2
yz 2
zx 2 )
iˆ (y2 + 2xz) + jˆ (2xy + z2) + k̂ (2yz + x2)
At the point (1, 1, 1),
= 3iˆ + 3jˆ + 3k̂
Directional derivative of f in the direction of the tangent T = iˆ + 2jˆ + 3k̂ at the point
(1, 1, 1)
T
(iˆ + 2 ˆj + 3kˆ)
18
(3iˆ 3 ˆj 3kˆ)
1
4
9
14
+
+
T
Example 13: Find the directional derivative of e = e2x cos yz at the origin in the
direction of the tangent to the curve x = a sin t, y = a cos t, z = a t at t =
4
.
9.34
Engineering Mathematics
Solution: Tangent to the curve is
dr d
T=
(a sin t )iˆ + (a cos t ) ˆj + (at )kˆ
=
dt dt
(a cos t) iˆ + ( a sin t) jˆ + (a) k̂
a ˆ a ˆ
t
i
j akˆ
At the point
,T
4
2
2
f = e2x cos yz
iˆ
x
(e 2 x cos yz )
ˆj
y
(e 2 x cos yz ) kˆ
z
(e 2 x cos yz )
iˆ (2e2x cos yz) + jˆ ( e2x z sin yz) + k̂ ( e2x y sin yz)
= 2i
At the origin,
Directional derivative in the direction of the tangent to the given curve
⎛ a ˆ a ˆ
⎞
i−
j + akˆ ⎟
⎜
T
2
2
⎠ = 2 a = 1.
=∇
= 2iˆ ⋅ ⎝
2
2
2a
T
a
a
+ + a2
2
2
Example 14: Find the directional derivative of v2, where v = xy2 ˆi + zy2 ˆj + xz2 kˆ
at the point (2, 0, 3) in the direction of the outward normal to the sphere
x2 + y2 + z2 = 14 at the point (3, 2, 1).
Solution: v2 = v · v
( xy 2 iˆ zy 2 ˆj xz 2 kˆ) ( xy 2 iˆ zy 2 ˆj xz 2 kˆ)
x2y4 + z2y4 + x2z4
Let v2 = f
ˆj
kˆ
x
y
z
4
4 ˆ
(2xy + 2xz ) i + (4x2y3 + 4z2y3) jˆ + (2zy4 + 4x2z3) k̂
At the point (2, 0, 3),
= (0 + 324) iˆ + (0 + 0) jˆ + (0 + 432) k̂ = 324 iˆ + 432 k̂
iˆ
Given sphere is x2 + y2 + z2 = 14.
Let y = x2 + y2 + z2
iˆ
Normal to the sphere
At the point (3, 2, 1),
x
ˆj
y
kˆ
z
= 2xiˆ + 2yjˆ + 2zk̂
= 6iˆ + 4jˆ + 2k̂
Directional derivative in the direction of normal to the sphere
( 6iˆ + 4 ˆj + 2kˆ )
∇ψ (
= ∇φ ⋅
= 324iˆ + 432kˆ ) ⋅
∇ψ
36 + 16 + 4
=
1404
14
.
Vector Calculus
9.35
Example 15: Find the directional derivative of e = 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). In what
direction it will be maximum? Find the maximum value of it.
Solution: Position vector of the point P
OP = iˆ + 2jˆ + 3k̂
Position vector of the point Q
OQ = 5iˆ + 0jˆ + 4k̂
PQ = OQ
iˆ
OP = 4iˆ
x
( x2
y2
2jˆ + k̂
2z2 )
ˆj
( x2
y2
2 z 2 ) kˆ
( x2
y2
2z2 )
y
z
ˆ
ˆ
(2x) i + ( 2y) j + (4z) k̂
At the point, (1, 2, 3),
= 2iˆ 4jˆ + 12k̂
Directional derivative at the point (1, 2, 3) in the direction of the line PQ
4iˆ 2 ˆj kˆ
2iˆ 4 ˆj 12kˆ
16 + 4 + 1
8 + 8 + 12
28
=
=
21
7 3
)(
(
=
)
4 7
3
Directional derivative is maximum in the direction of
Maximum value of directional derivative
4 16 144
i.e. 2iˆ
4jˆ + 12k̂
= 164 = 2 41
Example 16: Find the directional derivative of e = 6x2y + 24y2z - 8z2x at
x 1 y 3 z
(1, 1, 1) in the direction parallel to the line
=
= . Hence, find its
2
2
1
maximum value.
Solution:
iˆ
x
(6 x 2 y 24 y 2 z 8 z 2 x)
ˆj
y
(6 x 2 y 24 y 2 z 8 z 2 x)
kˆ
= (12xy
At the point (1, 1, 1),
(6 x 2 y 24 y 2 z 8 z 2 x)
z
8z2) iˆ + (6x2 + 48yz) jˆ + (24y2 16zx) k̂
= 4iˆ + 54 jˆ + 8k̂
9.36
Engineering Mathematics
x 1 y 3 z
=
= .
2
2
1
Direction ratios of the line are 2, 2, 1.
Direction of the line = 2iˆ 2 jˆ + k̂
Directional derivative in the direction of 2iˆ
Given line is
= (4iˆ + 54 ˆj + 8kˆ ) =
2jˆ + k̂ at the point (1, 1, 1)
( 2 iˆ − 2 jˆ + k̂ )
4 + 4 +1
8 − 108 + 8 − 92
=
=
.
3
3
Maximum value of directional derivative
= 4iˆ + 54 ˆj + 8kˆ = 16 + 2916 + 64
= 2996.
Example 17: Find the values of a, b, c if the directional derivative of
e = axy2 + byz + cz2x3 at (1, 2, –1) has maximum magnitude 64 in the direction
parallel to the z-axis.
Solution:
(axy 2 byz cz 2 x 3 ) kˆ (axy 2 byz cz 2 x 3 )
y
z
(ay2 + 3cz2x2) iˆ + (2axy + bz) jˆ + (by + 2czx3) k̂
iˆ
x
(axy 2
ˆj
byz cz 2 x 3 )
At the point (1, 2, 1),
= (4a + 3c) iˆ + (4a
b) jˆ + (2b
2c) k̂
… (1)
The directional derivative is maximum in the direction of
i.e. in the direction of
(4a + 3c) iˆ + (4a b) jˆ + (2b 2c) k̂. But it is given that directional derivative is maxiand
mum in the direction of z-axis i.e., in the direction of 0 iˆ + 0 jˆ + k̂. Therefore,
z-axis are parallel.
4a 3c 4a b 2b 2c
=
=
= l , say
0
0
1
4a + 3c = 0
… (2)
4a b = 0
… (3)
Substituting in Eq. (1),
= (2b
Maximum value of directional derivative is
64
(2b 2c)kˆ
2b
2c = 64,
b
64
c = 32
2c) k̂
. But it is given as 64.
Vector Calculus
9.37
From Eqs. (2) and (3),
4a + 3c = 0,
4a b = 0,
Solving,
b = 3c
Substituting in b c = 32, 4c = 32,
c = 8, b = 24, a = 6
Hence,
a = 6, b = 24, c = 8.
Example 18: For the function e (x, y) =
x
2
x + y2
, find the magnitude of the
directional derivative along a line making an angle 30 with the positive x-axis
at (0, 2).
⎞ ˆ ∂ ⎛ x ⎞ ˆ ∂ ⎛ x ⎞
⎟+ j ⎜ 2
⎟+k ⎜ 2
⎟
∂y ⎝ x + y 2 ⎠
∂z ⎝ x + y 2 ⎠
⎠
x( 2 x ) ⎤ ˆ ⎡
x( 2 y ) ⎤ ˆ
⎡ 1
− 2
i + ⎢− 2
j+0
=⎢ 2
2
2 2 ⎥
+
+
+ y 2 ) 2 ⎥⎦
x
y
x
y
x
(
)
(
⎣
⎦ ⎣
Solution: ∇ = iˆ
=
∂ ⎛ x
⎜
∂x ⎝ x 2 + y 2
2 xy
y 2 − x2 ˆ
ˆj
i− 2
2
2 2
(x + y )
( x + y 2 )2
At the point (0, 2),
y
4 0 ˆ
0
ˆj iˆ
i
=
2
2
(0 + 4)
(0 + 4)
4
Line OA makes an angle 30 with positive
x-axis.
OA = OB + BA
A
Unit vector in the direction of OA
iˆ cos 30 + jˆ sin 30
=
30°
O
3ˆ 1ˆ
i+ j
2
2
3ˆ 1ˆ
i
j
2
2
x
Fig. 9.3
Directional derivative in the direction of
iˆ
4
B
3ˆ 1ˆ
i + j at (0, 2)
2
2
3
8
Example 19: Find the rate of change of e = xyz in the direction normal to the
surface x2y + y2x + yz2 = 3 at the point (1, 1, 1).
Solution: Rate of change of f in the given direction is the directional derivative of
f in that direction.
9.38
Engineering Mathematics
iˆ
( xyz )
x
ˆj
y
( xyz ) kˆ
z
( xyz ) = (yz) iˆ + (xz) jˆ + (xy) k̂
At the point (1, 1, 1),
= iˆ + jˆ + k̂
Given surface is x2y + y2x + yz2 = 3.
Let y = x2y + y2x + yz2
Normal to the surface =
ˆj
kˆ
x
y
z
2 ˆ
2
= (2xy + y ) i + (x + 2xy + z2) jˆ + (2yz) k̂
iˆ
At the point (1, 1, 1),
= 3iˆ + 4 jˆ + 2k̂
Directional derivative in the direction of normal to the given surface
(iˆ
ˆ
ˆ
ˆ
ˆj kˆ) (3i + 4 j + 2k )
9 + 16 + 4
3+ 4+ 2
9
29
29
Example 20: Find the direction in which temperature changes most rapidly
with distance from the point (1, 1, 1) and determine the maximum rate of change
if the temperature at any point is given by e (x, y, z) = xy + yz + zx.
Solution: Temperature is given by f (x, y, z) = xy + yz + zx. Temperature will change
most rapidly i.e., rate of change of temperature, will be maximum in the direction of
.
iˆ
x
( xy
yz zx)
ˆj
y
yz zx) kˆ
( xy
z
( xy
yz zx)
(y + z) iˆ + (x + z) jˆ + (y + x) k̂
At the point (1, 1, 1),
= 2iˆ + 2 jˆ + 2k̂
This shows that temperature will change most rapidly in the direction of 2iˆ + 2jˆ + 2k̂
and maximum rate of change = maximum directional derivative
4 4 4
= 12 = 2 3
Example 21: Find the acute angle between the surfaces x2 + y2 + z2 = 9 and
z = x2 + y2 - z at the point (2, -1, 2).
Solution: The angle between the surfaces at any point is the angle between the
normals to the surfaces at that point.
Let f1 = x2 + y2 + z2, f2 = x2 + y2
z
Normal to f1,
1
iˆ
1
x
ˆj
1
y
kˆ
1
z
= (2x) iˆ + (2y) jˆ + (2z) k̂
Vector Calculus
Normal to f2,
At (2, 1, 2),
iˆ
2
1
= 4iˆ
2jˆ + 4k̂,
2
ˆj
2
x
= 4iˆ
1
1
kˆ
2
y
2jˆ
Let q be the angle between the normals
2
z
= (2x) iˆ + (2y) jˆ
k̂
k̂
and
2
2
1
(4iˆ 2 ˆj 4kˆ) (4iˆ 2 ˆj kˆ)
9.39
2
.
cos
4iˆ 2 ˆj 4kˆ 4iˆ 2 ˆj kˆ cos
(16 4 4)
16 4 16 16 4 1 cos
= 36 21 cos
16 = 6 21 cos
cos =
16
6 21
= cos
Hence, acute angle
1
=
8 21
63
8 21
= 54°251
63
Example 22: Find the angle between the normals to the surface xy = z2 at
P(1, 1, 1) and Q (4, 1, 2).
Solution: Given surface is xy = z2.
Let f = xy
z2
iˆ
Normal to f,
x
( xy z 2 )
y iˆ + x jˆ
ˆj
y
( xy z 2 ) kˆ
2z k̂
Normal at point P (1, 1, 1),
N1 = iˆ + jˆ
2k̂
Normal at point Q (4, 1, 2),
N 2 = iˆ + 4 jˆ
4k̂
Let q be the angle between N1 and N 2 .
N1 · N 2 = N1 N 2 cos
cos q =
=
N1 ⋅ N 2
N1 N 2
1+ 4 + 8
6 33
= cos
1
=
=
(iˆ + jˆ − 2kˆ ) ⋅ (iˆ + 4 jˆ − 4kˆ )
1 + 1 + 4 1 + 16 + 16
13
198
13
198
z
( xy z 2 )
9.40
Engineering Mathematics
Example 23: Find the constants a, b such that the surfaces 5x2 - 2yz - 9x = 0
and ax2y + bz3 = 4 cut orthogonally at (1, -1, 2).
Solution: If surfaces cut orthogonally, then their normals will also cut orthogonally,
i.e., angle between their normals will be 90°.
Given surfaces are 5x2 2yz 9x = 0 and ax2y + bz3 = 4.
Let f1 = 5x2 2yz 9x and f2 = ax2y + bz3
Normal to f1, f1 = iˆ
x
(5x2 2yz 9x) + jˆ
(5x2 2yz 9x) + k̂
z
(5x2 2yz 9x)
9) iˆ + ( 2z) jˆ + ( 2y) k̂
(10x
Normal to f2, f2 = iˆ
y
x
(ax2y + bz3) + jˆ
y
(ax2y + bz3) + k̂
z
(ax2y + bz3)
(2axy) iˆ + (ax2) jˆ + (3bz2) k̂
At the point (1, 1, 2),
f1 = iˆ 4 jˆ + 2k̂
f2 = 2aiˆ + ajˆ + 12bk̂
f1 and f2 are orthogonal.
f1 · f2 = | f1| | f2| cos
2
(iˆ
jˆ + 2 k̂)· ( 2aiˆ + a jˆ + 12bk̂) = 0
2a 4a + 24b = 0
6a + 24 b = 0
a 4b = 0
The point (1, 1, 2) lies on the surface ax2y + bz3 = 4.
a (1)2 ( 1) + b (2)3 = 4
a + 8b = 4
Solving Eqs. (1) and (2), we get
a = 4, b = 1
… (1)
… (2)
Example 24. Find the angle between the surfaces ax2 + y2 + z2 – xy = 1 and
bx2y + y2z + z = 1 at (1, 1, 0).
Solution: Let f1 = ax2 + y2 + z2
xy
f2 = bx2y + y2z + z
The point (1, 1, 0) lies on both the surfaces.
a (1)2 + (1)2 + 0 (1) (1) = 1
a=1
2
and
b (1) + 0 + 0 = 1
b=1
Vector Calculus
9.41
Angle between the given surface is the angle between their normals.
Normal to f1, f1 = iˆ
x
(x2 + y2 + z2
xy) + jˆ
y
(x2 + y2 + z2
xy)
(x2 + y2 + z2 xy)
z
(2x y) iˆ + (2y x) jˆ + (2z) k̂
k̂
Normal to f2, f2 = iˆ
x
(x2y + y2z + z) + jˆ
y
(x2y + y2z + z) + k̂
z
(x2y + y2z + z)
(2xy) iˆ + (x2 + 2yz) jˆ + (y2 + 1) k̂
At the point (1, 1, 0),
f1 = iˆ + jˆ + 0k̂
f2 = 2 iˆ + jˆ + 2k̂
Let the angle between N 1 and N 2 is q.
1
cos
|
=
q =
1|
2 +1
2 9
2
|
=
2
|
=
(iˆ + ˆj ) ⋅ ( 2iˆ + ˆj + 2kˆ )
1+1 4 +1+ 4
1
2
p
4
Hence, angle between the surfaces is
4
.
Example 25: Find the constants a, b if the directional derivative of
e = ay2 + 2bxy + xz at P (1, 2, –1) is maximum in the direction of the tangent to the curve,
r = (t3 – 1) iˆ + (3t – 1) jˆ + (t2 – 1) k̂ at point (0, 2, 0).
Solution: f = ay2 + 2bxy + xz
f1 = iˆ
x
(ay2 + 2bxy + xz) + jˆ
y
(ay2 + 2bxy + xz) + k̂
(2by + z) iˆ + (2ay + 2bx) jˆ + (x) k̂
At the point (1, 2, 1),
1) iˆ + (4a + 2b) jˆ + k̂
Tangent to the curve r = (t3 1) iˆ + (3t 1) jˆ + (t2
f = (4b
d r = (3t2) iˆ + 3 jˆ + (2t) k̂
dt
At the point (0, 2, 0), i.e., at t = 1
dr
= 3 iˆ + 3 jˆ + 2 k̂
dt
1) k̂ is
z
(ay2 + 2bxy + xz)
9.42
Engineering Mathematics
Directional derivative is maximum in the direction of f but it is given that directional derivative is maximum in the direction of the tangent.
Hence, f and
dr
are parallel.
dt
4b 1 4a 2b 1
=
=
3
3
2
4b 1 1
4a 2b 1
= and
= , 8a + 4b = 3
3
2
3
2
5
5 1
and 8a = 3 − 4b = 3 − =
8
2 2
1
a=
16
1
5
a= , b= .
16
8
b=
Hence,
Example 26: The temperature of the points in space is given by e = x2 + y2 – z.
A mosquito located at point (1, 1, 2) desires to fly in such a direction that it will get
warm as soon as possible. In what direction should it move?
Solution: Temperature is given by f = x2 + y2
z
Rate of change (increase) in temperature = f
iˆ
x
(x2 + y2
(2x) iˆ + (2y) jˆ
At the point (1, 1, 2),
f = 2iˆ + 2 jˆ k̂
z) + jˆ
y
(x2 + y2
z) + k̂
z
(x2 + y2
z)
k̂
Mosquito will get warm as soon as possible if it moves in the direction in which rate
of increase in temperature is maximum, i.e., f is maximum. Now, f is maximum in
its own direction, i.e., in the direction of f.
Unit vector in the direction of
=
=
2iˆ 2 ˆj kˆ
4 + 4 +1
2iˆ 2 ˆj kˆ
3
Hence, mosquito should move in the direction of
2iˆ + 2 ˆj − kˆ
.
3
Vector Calculus
9.43
Example 27: Find the direction in which the directional derivative of
=
( x2
y2 )
at (1, 1) is zero.
xy
Solution:
( x, y )
x
y
y
,
x
iˆ
x
x y
y
x
y
iˆ
1
y
x2
x
y y
jˆ
x
y2
y
x
kˆ
x
z y
y
x
1 ˆ
j,
x
At the point (1, 1) f = 2iˆ 2 jˆ.
Let the direction in which directional derivative is zero is r = xiˆ + yjˆ.
(2iˆ
xiˆ
yiˆ
x2
y2
0
2 jˆ) · (xi + y jˆ) = 0
2x 2y = 0, x = y
r = xiˆ + x jˆ
Unit vector in this direction =
x( iˆ + jˆ )
x 1+1
=
iˆ + jˆ
2
Hence, directional derivative is zero in the direction of
iˆ + ˆj
2
.
Exercise 9.3
1. Find f if
(i) f = log (x2 + y2 + z2)
(ii) f = (x2 + y2 + z2) e
x2 y 2 z 2
.
⎡
⎤
2r
−r
⎢ Ans.: (i) 2 (ii) ( 2 − r ) e r ⎥
r
⎢
⎥
⎢
where r = xiˆ + yjˆ + zkˆ , ⎥
⎥
⎢
⎥
⎢
r= r
⎣
⎦
2. Find f and | f| if
(i) f = 2xz4 x2y at (2, 2, 1)
(ii) f = 2xz2 3xy 4x at (1, 1, 2).
⎡Ans. : (i) 10iˆ −4 jˆ − 16kˆ, 2 93 ⎤
⎢
⎥
(ii) 7iˆ − 3 jˆ + 8kˆ, 2 29 ⎥⎦
⎢⎣
3. If A = 2x2iˆ 3yzjˆ + xz2k̂ and
f = 2z x3y find
(i) A · f
(ii) A
f at (1, 1, 1).
[Ans. : (i) 5 (ii) 7iˆ jˆ 11k̂ ]
4. If f = 3x2y, = xz2 2y, find
( f·
).
⎡ Ans. : (6yz 2 − 12x ) iˆ ⎤
⎢
⎥
+ 6xz 2 ˆj + 12xyzkˆ ⎥⎦
⎢⎣
9.44
Engineering Mathematics
5. If r = xiˆ + yjˆ + zk̂, r = | r |, prove that
(i)
(log r) =
r
(ii)
r2
| r |3 = 3r r
(iii)
f (r) = f (r )
6. Prove that
r
.
r
a r
a
n
n
r
r
( )
n a r r
r n+2
,
where a is a constant vector.
7. Find a unit vector normal to the surface
x2y + 2xz = 4 at the point (2, 2, 3).
1 ˆ
⎡
ˆ ⎤
ˆ
⎢⎣ Ans. : 3 (i − 2 j − 2 k ) ⎥⎦
8. Find unit outward drawn normal to
the surface (x 1)2 + y2 + (z + 2)2 = 9
at the point (3, 1, 4).
( 2iˆ + jˆ − 2 kˆ ) ⎤
⎡
Ans.
:
⎥⎦
⎢⎣
3
9. Find a unit vector normal to the surface xy3 z2 = 4 at the point ( 1, 1, 2).
iˆ 3 ˆj kˆ
Ans. :
11
10. Find the directional derivative of
f = x2yz + 4xz2 at (1, 2, 1) in the
direction of 2iˆ jˆ 2k̂.
Ans.:
37
3
11. Find the directional derivative of
f = xy + yz + zx at (1, 2, 0) in the
direction of vector iˆ + 2 jˆ + 2k̂.
10
Ans. :
3
12. Find the maximal directional derivative of x3y2z at (1, 2, 3).
Ans. : 4 91
13. In what direction from the point
(2, 1, 1) is the directional derivative
of f = x2yz3 a maximum? Find its
maximum value of magnitude.
⎡ Ans. : maximum in the direction of ⎤
⎥
⎢
∇ = 4iˆ − 4 ˆj + 12kˆ, 4 11 ⎦
⎣
14. In what direction from the point
(3, 1, 2) is the directional derivative of f = x2y2z4 a maximum? Find its
maximum value of magnitude.
⎡⎣ Ans. : 96 (iˆ + 3jˆ − 3kˆ), 96 19 ⎦⎤
15. In what direction from the point
(1, 3, 2) is the directional derivative
of f = 2xz y2 a maximum? Find its
maximum value of magnitude.
⎡⎣ Ans. : 4iˆ − 6 jˆ + 2kˆ , 2 14 ⎤⎦
16. What is the greatest rate of change of
f = xyz2 at the point (1, 0, 3)?
[Ans. : f = 9]
17. If the directional derivative of
f = ax2 + by + 2z at (1, 1, 1) is maximum in the direction of iˆ + jˆ + k̂,
then find values of a and b.
[Ans. : a = 1, b = 2]
18. If the directional derivative of
f = ax + by + cz at (1, 1, 1) has maximum magnitude 4 in a direction
parallel to x axis, then find values of
a, b, c.
[Ans. : a = 2, b = 2, c = 2]
19. Find the directional derivative of
f = x2y + y2z + z2x2 at (1, 2, 1) in the
direction of the normal to the surface
x2 + y2 z2x = 1 at (1, 1, 1).
4
Ans. :
3
20. Find the directional derivative of
f = x2y + yz2 at (2, 1, 1) in the
direction normal to the surface
x2y + y2x + yz2 = 3 at (1, 1, 1).
13
Ans. :
29
21. Find the directional derivative of
f = x2y + y2z + z2x at (2, 2, 2) in the
direction of the normal to the surface
4x2y + 2z2 = 2 at the point (2, 1, 3).
36
Ans. :
41
Vector Calculus
22. Find the rate of change of f = xy + yz + zx
at (1, 1, 2) in the direction of the normal to the surface x2 + y2 = z + 4.
Ans. :
14
21
23. Find the directional derivative of
f = x2yz2 along the curve x = e t,
y = 2 sin t + 1, z = t cost at t = 0.
1
Ans. :
6
24. Find the directional derivative of
f = x2y2z2 at (1, 1, 1) in the direction of the tangent to the curve x = et,
y = 2 sin t + 1, z = t cos t, at t = 0.
2 3
Ans. :
3
25. Find the directional derivative of the
scalar function f = x2 + xy + z2 at the
point P(1, 1, 1) in the direction of
the line PQ where Q has coordinates
(3, 2, 1).
⎡ Hint : PQ = OQ − OP
⎤
⎢
⎥
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
⎢ = (3i + 2 j + k ) − ( −i − j − k ) ⎥
⎢⎣ = 2iˆ + 3 jˆ + 2kˆ
⎦⎥
Ans. :
1
17
26. Find the directional derivative of
f = 2x3y 3y2z at the point P (1, 2, 1)
in the direction towards Q (3, 1, 5). In
what direction from P is the directional
derivative maximum? Find the
magnitude of maximum directional
derivative.
Ans. :
90
,12iˆ
7
14 jˆ
12kˆ, 22
27. Find the directional derivative of
f = 4xz3 3x2y2z at (2, 1, 2) in the
direction from this point towards the
point (4, 4, 8).
9.45
376
7
Ans. :
28. Find the angle of intersection of
the spheres x2 + y2 + z2 = 29 and
x2 + y2 + z2 + 4x 6y 8z = 47 at
(4, 3, 2).
19
Ans. : cos 1
29
29. Find the angle between the normals
to the surface 2x2 + 3y2 = 5z at the
point (2, 2, 4) and ( 1, 1, 1).
65
Ans. : cos 1
233 77
30. Find the angle between the normals
to the surface xy = z2 at the points
(1, 4, 2) and ( 3, 3, 3).
Ans. :
= cos
1
1
22
31. Find the acute angle between the
surfaces xy2z = 3x + z2 and
3x2 y2 + 2z = 1 at the point (1, 2, 1).
Ans. : cos
1
6
14
32. Find the constant 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).
⎡ Hint : condition for orthogonal - ⎤
⎥.
⎢
ity is ∇f ⋅ ∇y = 0
⎦
⎣
Ans. : a =
5
, b =1
2
33. Find the angle between the two surfaces x2 + y2 + a z2 = 6 and z = 4 y2 + bxy
at P (1, 1, 2).
⎡ Hint : (1, 1, 2) lies on both ⎤
⎢
surfaces, a = 1, b = − 1⎥⎦
⎣
Ans. : cos
1
6
11
9.46
Engineering Mathematics
34. Find the directional derivative of
f = x2 + y2 + z2 in the direction of the
x y z
line = = at (1, 2, 3).
3 4 5
Let the direction in which directional derivative is zero is r = xiˆ + yjˆ
xiˆ + yjˆ
0
x2 + y 2
26
2
5
Ans.
(2iˆ
2 jˆ) · (xiˆ + yjˆ) = 0
2x 2y = 0, x = y
r = xiˆ + x jˆ
unit vector in this direction
x iˆ + ˆj
=
x 1+1
ˆi + ˆj
=
2
35. Find the direction in which the directional derivative of f = (x + y) =
( x2 y 2 )
at (1, 1) is zero.
xy
(
x y
⎡
⎤
⎢ Hint : ( x, y ) = y − x ,
⎥
⎢
⎥
⎢
⎛ 1 y ⎞ˆ ⎛ x 1 ⎞ ˆ ⎥
⎢∇ = ⎜ + 2 ⎟ i + ⎜ − 2 − ⎟ j , ⎥
⎜
x ⎟⎠ ⎥
⎢
⎝y x ⎠ ⎝ y
⎢
⎥
⎣ At (1,, 1),∇ = 2iˆ − 2 jˆ
⎦
)
Hence, directional derivative is zero in
iˆ + ˆj
the direction of
.
2
9.12 DIVERGENCE
The divergence of a vector point function F is denoted by div F or
defined as
iˆ
F
x
ˆj
y
kˆ
z
If
F = F1 iˆ + F2 jˆ + F3 k̂,
then
F
iˆ
=
x
ˆj
y
kˆ
z
F
· (F1 iˆ + F2 jˆ + F3 k̂ )
F
F1
F
+ 2+ 3
x
y
z
which is a scalar quantity.
Note:
(i)
· F
F
(ii)
F · , because
F1
x
F1
x
F
= iˆ
F2
F2
y
F
x
ˆj
· F is a scalar quantity whereas
F3
y
z
is a scalar differential operator.
F3
z
F
y
kˆ
F
z
(if F = F1 iˆ + F2 jˆ + F3 k̂ )
· F and is
Vector Calculus
9.47
9.12.1 Physical Interpretation of Divergence
Consider the case of a homogeneous and incompressible fluid flow. Consider a small rectangular parallelepiped of dimensions d x, d y, d z parallel to x, y and z axes respectively.
Let v = v iˆ + v jˆ + v k̂ be the velocity of the fluid at point A (x, y, z).
1
2
3
The velocity component parallel to x
of the face PQRS
v1 (x + d x, y, z)
v1 +
axis (normal to the face PQRS) at any point
v1
d x [expanding by Taylor’s series and ignoring higher
x
powers of d x]
C
S
D
R
dz
V1 d y
O
Q
B
A
P
x
y
Fig. 9.4
Mass of the fluid flowing in across the face ABCD per unit time
velocity component normal to the face ABCD area of the face ABCD
v1 (d y d z)
Mass of the fluid flowing out across the face PQRS per unit time
velocity component normal to the face PQRS area of the face PQRS
v1
v1
x
x
y z
Gain of fluid in the parallelepiped per unit time in the direction of x-axis
v1
=
v1
x
x
v1
x y z
x
y z v1 y z
9.48
Engineering Mathematics
Similarly, gain of fluid in the parallelepiped per unit time in the direction of y-axis
=
∂v2
x y z
∂y
and gain of fluid in the parallelepiped per unit time in the direction of z-axis
=
∂v3
x y z
∂z
Total gain of fluid in the parallelepiped per unit time
v
v1
v
+ 2+ 3
x
y
z
=
x y z
But, d x d y d z is the volume of the parallelepiped.
Hence, total gain of fluid per unit volume =
v
v1
v
+ 2+ 3
x
y
z
div v =
· v
Note: A point in a vector field F is said to be a source if div F is positive, i.e.,
· F > 0 and is said to be a sink if div F is negative, i.e, · F < 0.
9.12.2 Solenoidal Function
A vector function F is said to be solenoidal if div F = 0 at all points of the function.
For such a vector, there is no loss or gain of fluid.
9.13 CURL
The curl of a vector point function F is denoted by curl F or
iˆ
F
=
x
+ ˆj
y
+ kˆ
iˆ
ˆj
kˆ
x
F1
y
F2
z
F3
iˆ
which is a vector quantity.
F3
y
F2
z
( F1iˆ F2 ˆj F3 kˆ)
z
ˆj
F and is defined as
F3
x
F1
z
kˆ
F2
x
F1
y
Vector Calculus
9.49
9.13.1 Physical Interpretation of Curl
Let
be the angular velocity of a rigid body moving about a fixed point. The linear
velocity v of any particle of the body with position vector r w.r.t. to the fixed point
is given by,
v =w
r
Let w = w1iˆ + w2 jˆ + w3 k̂, r = x iˆ + y jˆ + z k̂
v =w
r
iˆ
=
kˆ
ˆj
1
2
x
3
y
iˆ (w2 z
Curl v =
z
w3 y)
jˆ (w1 z
w3 x) + k̂ (w1 y
w2 x)
v
=
2
z
iˆ
ˆj
kˆ
x
y
z
3
y
3
x
iˆ (w1 + w1) jˆ ( w2
2(w1 iˆ + w2 jˆ + w3 k̂)
z
1
1
y
2
x
w2) + k̂ (w3 + w3)
2w
Curl v = 2 w
Thus, the curl of the linear velocity of any particle of a rigid body is equal to twice
the angular velocity of the body.
This shows that curl of a vector field is connected with rotational properties of the
vector field and justifies the name rotation used for curl.
9.13.2 Irrotational Field
A vector point function F is said to be irrotational, if curl F = 0 at all points of the
function, otherwise it is said to be rotational.
Note: If F = f, then curl F =
Thus, if
F =
f = 0.
F = 0, then we can find a scalar function f so that F = f. A vector
field F which can be derived from a scalar field f so that F = f is called a conservative vector field and f is called the scalar potential.
9.50
Engineering Mathematics
Example 1: If A = x2ziˆ – 2y3z2 jˆ + xy2zk̂, find Æ · A at the point (1, –1, 1).
A
A1
A
+ 2 + 3 , where A = A iˆ + A jˆ + A k̂
1
2
3
x
y
z
· A =
Solution:
· A =
x
(x2z) +
( 2 y3z2) +
z
(xy2z)
6 y2z2 + xy2
2xz
At the point (1, 1, 1),
A
y
2 (1) (1) 6 ( 1) 2 (1) 2 1( 1) 2
6+1
2
3
Example 2: If r = xiˆ + yjˆ + zk̂k, prove that div (grad rn) = n (n + 1) rn–2.
Solution:
grad r n = iˆ
∂r n ˆ ∂r n ˆ ∂r n
+ j
+k
∂x
∂y
∂z
⎛
∂r ⎞
∂r ⎞
∂r ⎞
⎛
⎛
= iˆ ⎜ nr n −1 ⎟ + jˆ ⎜ nr n −1 ⎟ + k̂ ⎜ nr n −1 ⎟
∂y ⎠
∂z ⎠
∂x ⎠
⎝
⎝
⎝
But
r = xiˆ + yjˆ + zk̂,
r2 = r
2
= x2 + y2 + z2
r x r y r z
= ,
= ,
=
x r y r z r
xˆ y ˆ z ˆ
i + j+ k
r
r
r
grad r n = nr n–1
= nr
n 1
( xiˆ + yjˆ + zkˆ)
r
nr n 2 r
div (grad r n) = · (nr n 2 r )
n · rn 2 (xiˆ + yjˆ + zk̂ )
=n
=n x
x
(r n 2 x) +
x
rn
2
+ rn
y
2
(r n 2 y ) +
+y
y
rn
z
2
(r n 2 z )
+ rn
2
+z
z
rn
2
+ rn
2
Vector Calculus
9.51
∂r
∂r
∂r ⎤
⎡
= n ⎢3r n − 2 + x(n − 2)r n − 3
+ y (n − 2)r n − 3
+ z (n − 2)r n − 3 ⎥
∂x
∂y
∂z ⎦
⎣
n 3r n
2
(n 2)r n
3
n 3r n
2
(n 2)r n
3
n (n + 1) rn
( x2
y2
r
r2
r
nr n
z2 )
2
(3
r
x
∵
n
x
,
r
r
y
y
,
r
r
z
z
r
2)
2
Example 3: Prove that for vector function A, Æ ë (Æ ë A ) = Æ (Æ · A ) – Æ2 A .
Solution: Let A = A1iˆ + A2 jˆ + A3 k̂
iˆ
∂
∇× A =
∂x
A1
ˆj
∂
∂y
A2
kˆ
∂
∂z
A3
⎛ ∂A ∂A ⎞
= iˆ ⎜ 3 − 2 ⎟ −
∂z ⎠
⎝ ∂y
(
)
∇× ∇ × A =
ˆj ⎛⎜ ∂A3 − ∂A1 ⎞⎟ + kˆ ⎛⎜ ∂A2 − ∂A1 ⎞⎟
∂y ⎠
∂z ⎠
⎝ ∂x
⎝ ∂x
iˆ
∂
∂x
ˆj
∂
∂y
kˆ
∂
∂z
∂A3 ∂A2
−
∂y
∂z
∂A1 ∂A3
−
∂z
∂x
∂A2 ∂A1
−
∂x
∂y
⎡⎛ ∂ 2 A ∂ 2 A ⎞ ⎛ ∂ 2 A ∂ 2 A
3
2
− 21 ⎟ − ⎜ 21 −
= iˆ⎢⎜
⎜
⎟
⎜
y
x
∂
∂
x
z
∂
∂
∂y ⎠ ⎝ ∂z
⎢⎣⎝
⎞⎤
⎟⎥
⎟⎥
⎠⎦
⎡⎛ ∂ 2 A ∂ 2 A ⎞ ⎛ ∂ 2 A ∂ 2 A ⎞ ⎤
3
2
1
− ĵ ⎢⎜ 22 −
−
⎟⎥
⎟−⎜
2 ⎟
⎜
⎟
⎜
∂
∂
∂
∂
x
y
y
z
∂z ⎠ ⎦⎥
⎢⎣⎝ ∂x
⎠ ⎝
⎡⎛ ∂ 2 A ∂ 2 A ⎞ ⎛ ∂ 2 A ∂ 2 A ⎞ ⎤
3
3
2
1
+ k̂ ⎢⎜
−
−
⎟⎥
⎟−⎜
∂y∂z ⎟⎠ ⎥
⎢⎣⎜⎝ ∂x∂z ∂x 2 ⎟⎠ ⎜⎝ ∂y 2
⎦
Consider
⎡⎛ ∂ 2 A2 ∂ 2 A1 ⎞ ⎛ ∂ 2 A1 ∂ 2 A3 ⎞ ⎤
iˆ ⎢⎜
− 2 ⎟−⎜ 2 −
⎟⎥
∂x ∂z ⎠ ⎥⎦
⎢⎣⎝ ∂y ∂x ∂y ⎠ ⎝ ∂z
⎡ ∂ ⎛ ∂A ⎞ ∂ 2 A ∂ 2 A
∂ ⎛ ∂A ⎞ ⎤
= iˆ ⎢ ⎜ 2 ⎟ − 21 − 21 + ⎜ 3 ⎟ ⎥
∂x ⎝ ∂z ⎠ ⎥⎦
∂z
⎢⎣ ∂x ⎝ ∂y ⎠ ∂y
9.52
Engineering Mathematics
⎡ ∂ ⎛ ∂A ∂A ⎞ ⎛ ∂ 2 A ∂ 2 A ⎞ ⎛ ∂ 2 A ∂ 2 A ⎞ ⎤
= iˆ ⎢ ⎜ 2 + 3 ⎟ − ⎜ 21 + 21 ⎟ + ⎜ 21 − 21 ⎟ ⎥
∂z ⎠ ⎜⎝ ∂y
∂x ⎟⎠ ⎥⎦
∂z ⎟⎠ ⎜⎝ ∂x
⎢⎣ ∂x ⎝ ∂y
⎡
∂ 2 A1 ⎤
⎥
⎢ Adding and subtracting
∂x 2 ⎥⎦
⎢⎣
⎡ ∂ ⎛ ∂A ∂A ∂A ⎞ ⎛ ∂ 2 A ∂ 2 A ∂ 2 A ⎞ ⎤
1
= iˆ ⎢ ⎜ 1 + 2 + 3 ⎟ − ⎜ 21 +
+ 21 ⎟ ⎥
∂y
∂z ⎠ ⎝ ∂x
∂y 2
∂z ⎠ ⎥⎦
⎢⎣ ∂x ⎝ ∂x
∂
= iˆ ( ∇ ⋅ A) − iˆ ∇2 A1
∂x
Similarly,
⎡⎛ ∂ 2 A ∂ 2 A1 ⎞ ⎛ ∂ 2 A3 ∂ 2 A2
−
− ˆj ⎢⎜ 22 −
⎟−⎜
∂x∂y ⎠ ⎝ ∂y∂z ∂z 2
⎢⎣⎝ ∂x
⎞⎤
∂
2
⎟ ⎥ = ˆj ( ∇ ⋅ A ) − ˆj ∇ A2
y
∂
⎠ ⎥⎦
2
2
2
2
⎡⎛ ∂ A1 ∂ A3 ⎞ ⎛ ∂ A3 ∂ A2 ⎞ ⎤
∂
2
and kˆ ⎢⎜
− 2 ⎟−⎜ 2 −
⎟ ⎥ = kˆ ( ∇ ⋅ A ) − kˆ ∇ A3
∂
x
∂
z
∂
y
∂
z
z
∂
∂
x
∂
y
⎠ ⎥⎦
⎠ ⎝
⎣⎢⎝
Hence, ∇ × (∇ × A) = iˆ
x
+ ˆj
y
+ kˆ
z
(∇ ⋅ A)
2
(A1iˆ + A2 jˆ + A3 k̂ )
∇ (∇ ⋅ A) − ∇2 A
Example 4: If A = Æ (xy + yz + zx), find Æ · A and Æ ë A.
Solution:
A =
(xy + yz + zx)
iˆ
x
(xy + yz + zx) + jˆ
y
(xy + yz + zx) + k̂
(y + z) iˆ + (x + z) jˆ + (y + x) k̂
· A =
· [(y + z) iˆ + (z + x) jˆ + (x + y) k̂ ]
x
A
(y + z) +
y
(z + x) +
iˆ
ˆj
kˆ
x
y+z
y
z+x
z
x+ y
z
(x + y) = 0
z
(xy + yz + zx)
Vector Calculus
9.53
∂
∂
⎤
⎡∂
⎤
⎡∂
= iˆ ⎢ ( x + y ) − ( z + x) ⎥ − jˆ ⎢ ( x + y ) − ( y + z ) ⎥
∂z
∂z
⎦
⎣ ∂x
⎦
⎣ ∂y
∂
⎤
⎡∂
+ kˆ ⎢ ( z + x) − ( y + z ) ⎥
∂y
⎦
⎣ ∂x
iˆ (1
jˆ (1
1)
1) + k̂ (1
1)= 0
Example 5: Verify Æ (Æ ë A) = Æ (Æ ⋅ A) – Æ2 A for A = x2yiˆ + x3y2jˆ – 3 x2z2k̂.
Solution:
iˆ
∂
∇× A =
∂x
x2 y
ˆj
∂
∂y
kˆ
∂
∂z
x3 y 2
−3 x 2 z 2
∂
⎤
⎡∂
= iˆ ⎢ ( −3 x 2 z 2 ) − ( x3 y 2 ) ⎥ −
y
z
∂
∂
⎦
⎣
∂
⎤
⎡∂
+ kˆ ⎢ ( x3 y 2 ) − ( x 2 y ) ⎥
∂
x
∂
y
⎦
⎣
0 · iˆ
( 6xz2) jˆ + (3x2y2
ˆj
∇ × (∇ × A) = x
y
6 xz
iˆ (6x2y
(6x2y
· A = iˆ
x
x
z
2
x
+ k̂
2 2
jˆ (6xy2
12xz) iˆ
+ ˆj
(x2y) +
y
(6xy2
+ kˆ
y
z
2x
0) + k̂ (6z2
0)
2x) jˆ + (6z2) k̂
· (x2yiˆ + x3y2jˆ
(x3y2) +
z
3x2z2k̂ )
( 3x2z2)
6x2z
(2xy + 2x3y
z
x2 )
(3 x y
12xz)
2xy + 2 x3y
∇ (∇ ⋅ A) = iˆ
x2)k̂
kˆ
iˆ
0
ˆj ⎡ ∂ ( −3 x 2 z 2 ) − ∂ ( x 2 y ) ⎤
⎢⎣ ∂x
⎥⎦
∂z
(2xy + 2x3y
6x2z) + jˆ
6x2z)
y
(2xy + 2x3y
6x2z)
9.54
Engineering Mathematics
(2y + 6x2y
2
2
A =
x2
12xz) iˆ + (2x + 2x3
(x2yiˆ + x3y2 jˆ
0) jˆ + ( 6x2) k̂
2
3x2z2k̂ ) +
y2
(x2yiˆ + x3y2 jˆ
3x2z2k̂ )
(x2yiˆ + x3y2jˆ
3x2z2k̂ )
2
z2
x
(2xyiˆ + 3x2y2jˆ
(2yiˆ + 6xy2jˆ
∇ (∇ ⋅ A)
2
A = (6x2y
6xz2k̂ ) +
6z2k̂ ) + 2x3jˆ
2
z
( 6x2z k̂ )
6x2k̂ = 2yiˆ + (6xy2 + 2x3) jˆ – 6 (z2 + x2)k̂
12xz) iˆ + (2x
Hence, ∇ × (∇ × A) = ∇ (∇ ⋅ A)
(x2iˆ + 2x3y jˆ) +
y
6xy2) jˆ + (6z2) k̂
A
Example 6: Show that A = 3 y4z2iˆ + 4x3z2 jˆ – 3x2y2k̂ is solenoidal.
Solution: A = 3y4z2 iˆ + 4x3z2 jˆ
· A =
x
(3y4z2) +
y
3x2y2 k̂
(4x3z2) +
z
( 3x2y2) = 0
· A = 0, A is solenoidal.
Since
Example 7: Determine the constant b such that A = (bx + 4y2z) î + (x3sin z – 3y) ĵ
– (ex + 4 cos x2y) k̂ is solenoidal.
Solution: If A is solenoidal, then
· A =0
x
(bx + 4y2z ) +
y
(x3 sin z
3y) +
z
( ex
4 cos x2y) = 0.
b
Example 8: Show that the vector field A =
a ( xiˆ + yjˆ )
x2 + y2
3=0
b=3
is a source field or sink
field according as a > 0 or a < 0.
Solution: Vector field A is a source field if
field if · A < 0.
· A > 0 and vector field A is a sink
Vector Calculus
ax
A
2
x +y
=
2
ay
iˆ
2
x +y
ax
x
2
x +y
+
2
2
ˆj
ay
y
2
x + y2
x ⋅ 2x
1
⎡
= a⎢
−
+
3
2
2
⎢⎣ x + y
2( x 2 + y 2 ) 2
⎡
2
( x2 + y 2 )
= a⎢
−
3
2
2
⎢⎣ x + y
( x2 + y 2 ) 2
=
Since
9.55
1
x +y
2
2
−
y⋅2y
⎤
⎥
2( x 2 + y ) ⎥⎦
3
2 2
⎤
⎥
⎥⎦
a
2
x + y2
x 2 + y 2 is always positive,
· A > 0 if a > 0, and
· A < 0 if a < 0. Hence,
A is a source field if a > 0 and sink field if a < 0.
Example 9: If A = (ax2y + yz) iˆ + (xy2 – xz2) jˆ + (2xyz – 2x2y2) k̂ is solenoidal,
find the constant a.
Solution: If A is solenoidal, then · A = 0,
· [(ax2y + yz) iˆ + (xy2 xz2) jˆ + (2xyz
x
(ax2y + yz) +
y
(xy2
xz2) +
2x2y2) k̂ ] = 0
(2xyz 2x2y2) = 0
z
2axy + 2xy + 2xy = 0
2a = 4
a= 2
Example 10: Find the curl of A = exyz (iˆ + jˆ + k̂ ) at the point (1, 2, 3).
Solution: Curl of A =
iˆ
∂
=
∂x
ˆj
∂
∂y
kˆ
∂
∂z
e xyz
e xyz
e xyz
A
⎛ ∂
⎞
⎛ ∂
⎞
∂
∂
∂
⎛ ∂
⎞
= iˆ ⎜ e xyz − e xyz ⎟ − jˆ ⎜ e xyz − e xyz ⎟ + kˆ ⎜ e xyz − e xyz ⎟
⎝
⎠
∂z
∂x
∂z
∂y
⎝ ∂y
⎠
⎝ ∂x
⎠
(exyz · xz
exyz · xy) iˆ
(exyz · yz
exyz · xy) jˆ + (exyz · yz
exyz · xz) k̂
9.56
Engineering Mathematics
At the point (1, 2, 3),
Curl A = e6 [iˆ (3 2)
e6 (iˆ
jˆ (6
2) + k̂ (6 – 3)]
4 jˆ + 3k̂ )
Example 11: Find curl curl A = x2y iˆ – 2xz jˆ + 2yzk̂ at the point (1, 0, 2).
Solution: Curl A =
A
iˆ
∂
=
∂x
kˆ
∂
∂z
ˆj
∂
∂y
x 2 y −2 xz 2 yz
∂
∂
⎡∂
⎤
⎤
⎡∂
= iˆ ⎢ ( 2 yz ) − ( −2 xz ) ⎥ − ˆj ⎢ ( 2 yz ) − ( x 2 y ) ⎥
x
z
∂
∂
∂
y
z
∂
⎣
⎦
⎦
⎣
∂
⎤
⎡∂
+ kˆ ⎢ ( −2 xz ) − ( x 2 y ) ⎥
x
y
∂
∂
⎦
⎣
(2z + 2x) iˆ
(
)
Curl Curl A =
0) jˆ + ( 2z
(0
(
A)
iˆ
∂
∂x
ˆj
∂
∂y
kˆ
∂
∂z
2( z + x)
0
−( x 2 + 2 z )
=
∂
⎡∂
⎤
= iˆ ⎢ ( − x 2 − 2 z ) − (0) ⎥ −
∂
∂z ⎦
⎣ y
∂
⎡∂
⎤
+ kˆ ⎢ (0) − 2( z + x) ⎥
∂
x
y
∂
⎣
⎦
iˆ (0
0)
jˆ ( 2x
2) + k̂ (0
x2) k̂
ˆj ⎡ ∂ ( − x 2 − 2 z ) − ∂ 2( z + x) ⎤
⎢⎣ ∂x
⎥⎦
∂z
0)
(2x + 2) jˆ
At the point (1, 0, 2),
(
)
Curl Curl A = (2 + 2) jˆ
4 jˆ
Example 12: Prove that F = 2xyz2iˆ + [x2z2 + z cos (yz) ] jˆ + (2x2yz + y cos yz) k̂
is a conservative vector field.
Vector Calculus
F =0
Solution: Vector field F is conservative if
iˆ
F
kˆ
ˆj
x
y
2 xyz
2
iˆ
(2 x 2 yz
y
ˆj
kˆ
9.57
z
2 2
2
x z + z cos yz 2 x yz + y cos yz
y cos yz )
(2 x 2yz
x
( x2 z 2
x
( x2 z 2
z
y cos yz )
z
z cos yz )
z cos yz )
(2 xyz 2 )
(2 xyz 2 )
y
yz sin yz 2x2z cos yz + zy sin yz) iˆ
4xyz) jˆ + (2xz2 2xz2) k̂
(2x2z + cos yz
(4xyz
0
Hence, F is conservative vector field.
Example 13: Determine the constants a and b such that curl of (2xy + 3yz) iˆ +
(x2 + axz – 4 z2) jˆ + (3xy + 2byz) k̂ is zero.
Solution: Let F = (2xy + 3yz) iˆ + (x2 + axz
Curl F =
F =0
iˆ
ˆj
kˆ
x
y
z
2 xy 3 yz
iˆ
y
(3 xy 2byz )
z
kˆ
(3x + 2bz
4z2) jˆ + (3xy + 2byz) k̂
ax + 8z) iˆ
[(3
( x2
x
x2
axz 4 z 2
axz 4 z 2 )
ˆj
axz 4 z 2 )
(3y
3y) jˆ + (2x + az
a)x + 2z (b + 4)] iˆ
3 xy 2byz
x
( x2
y
=0
(3xy 2byz )
(2 xy 3 yz )
2x
0 jˆ + z (a
z
0
3z) k̂ = 0
3) k̂ = 0
(2 xy 3 yz )
9.58
Engineering Mathematics
Comparing coefficients of iˆ and k̂, we get
(3 a) x + 2 (b + 4) z = 0
(a 3) z = 0
Solving both the equations
a = 3, b = 4
Example 14: Show that F = (y2 – z2 + 3yz – 2x) î + (3xz + 2xy) ĵ + (3xy – 2xz + 2z)k̂
is both solenoidal and irrotational .
Solution: If F is solenoidal,
· F =
x
(y2
z2 + 3yz
· F =0
2x) +
y
(3xz + 2xy) +
z
(3xy
2xz + 2z)
2 + 2x 2x + 2 = 0
Hence, F is solenoidal.
F =0
If F is irrotational,
F
y2
iˆ
ˆj
kˆ
(3x
iˆ
ˆj
kˆ
x
y
z
z2
y
x
x
3 yz 2 x 3 xz 2 xy 3 xy 2 xz 2 z
(3 xy 2 xz 2 z )
z
(3 xy 2 xz 2 z )
(3 xz 2 xy )
3x) iˆ
(3y
y
z
(3 xz 2 xy )
( y2
( y2
2z + 2z
z2
z2
3 yz 2 x)
3 yz 2 x)
3y) jˆ + (3z + 2y
2y
3z) k̂ = 0
Hence, F is irrotational.
Example 15: Find the directional derivative of the divergence of F (x, y, z)
= xy iˆ + xy2jˆ + z2k̂ at the point (2, 1, 2) in the direction of the outer normal to the
sphere x2 + y2 + z2 = 9.
Solution: F (x, y, z) = xy iˆ + xy2 jˆ + z2 k̂
Divergence of F (x, y, z) =
=
· F
∂
∂
∂ 2
( xy ) +
( xy 2 ) +
(z )
∂x
∂y
∂z
y + 2xy + 2z
Vector Calculus
Gradient of divergence of F =
9.59
( · F)
= iˆ
x
+ ˆj
y
+ kˆ
z
(y + 2xy + 2z)
= 2y iˆ + (1 + 2x) jˆ + 2k̂
At the point (2, 1, 2),
( · F ) = 2 iˆ + 5 jˆ + 2k̂
+ ˆj + kˆ
(x2 + y2 + z2)
x
y
z
= 2 (xiˆ + yjˆ + zk̂ )
Normal at (2, 1, 2) = 2 (2iˆ + jˆ + 2k̂ )
Normal to sphere = iˆ
Directional derivative in the direction of the outer normal to the sphere x2 + y2 + z2 = 9
4iˆ + 2 ˆj + 4kˆ
= (2iˆ + 5 jˆ + 2k̂ ) ·
=
1
(8 + 10 + 8)
6
=
13
3
16 + 4 + 16
Exercise 9.4
1. Find divergence and curl of
x2 cos z iˆ + y log x jˆ yz k̂.
⎡ Ans. : 2x cos z + log x − y, ⎤
⎢
⎥
ˆ − jˆ x 2 sin z + kˆ y ⎥
⎢
iz
⎣
x⎦
3 2 4
2. If f = 2x y z , prove that div (grad f)
= 12xy2z4 + 4x3z4 + 24x3y2z2.
3. Find curl (curl A ), if
A = x2y iˆ 2xz jˆ + 2yz k̂.
[Ans. : (2x + 2) jˆ]
4. If A = 2yziˆ x2yjˆ + x z2k̂,
B = x2iˆ + yzjˆ
find
xyk̂ and f = 2x2yz3,
(i) ( A · ) f
(ii) A ·
(iii) ( B · ) A
(iv) ( A
(v) A
f
⎡ Ans. : (i) and (ii) 8 xy 2 z 4 − 2 x 4 yz 3 ⎤
⎢
⎥
+ 6 x 3 yz 4 (iii) ( 2 yz 2 − 2 xy 2) iˆ ⎥
⎢
⎢
⎥
− ( 2 x 3 y + x 2 yz ) jˆ
⎢
⎥
⎢
+ ( x 2 z 2 − 2 x 2 yz )kˆ (iv ) and ⎥
⎢
⎥
(v) − (6 x 4 y 2 z 2 + 2 x 3 z 5 ) iˆ ⎥
⎢
⎢
⎥
+ ( 4 x 2 yz 5 − 12 x 2 y 2 z 3 ) jˆ
⎢
⎥
⎢⎣
+ (4 x 2 yz 4 + 4 x 3 y 2 z 3 )kˆ
⎦⎥
5. If A = x2 iˆ + xyex jˆ + sin z k̂, find
·(
[Ans. : 0]
6. If f = tan
f
)f
A ).
7. If f = 2x2
1
y
, find div (grad f).
x
[Ans. : 0]
3y2 + 4z2, find curl (grad f).
[Ans. : 0]
9.60
Engineering Mathematics
8. Prove that for every field A,
div (curl A ) = 0.
9. Prove that gradient field describing a
motion is irrotational.
[Hint: Prove that
f = 0]
10. Prove that A = iˆ yz + jˆ xz + k̂ xy is
irrotational and find a scalar function
f (x, y, z) such that A = grad f.
[Ans. : xyz + c]
11. Prove that A = (6xy + z3) iˆ
+ (3 x2 z) jˆ + (3x z2 y) k̂ is
irrotational. Find the function f
such that A = f.
[Ans. : f = 3 x2y + x z3 yz]
12. Prove that the velocity given by
A = (y + z) iˆ + (z + x) jˆ + (x + y) k̂
is irrotational and find its scalar
potential. Is the motion possible for
an incompressible fluid?
⎡ Ans. : f = yz + zx + xy, motion is ⎤
⎥
⎢
possible because ∇ ⋅ A = 0 ⎦
⎣
13. Prove that A = (z2 + 2xy + 3y) iˆ
+ (3x + 2y + z) jˆ + (y + 2zx) k̂ is
irrotational and find scalar potential f
such that A = f and f (1, 1, 0) = 4.
[Ans. : f = z2x + x2 + 3xy + y2 + yz 1]
14. Prove that A = (z2 + 2x + 3y) iˆ
+ (3x + 2y + z) jˆ + (y + 2zx) k̂ is
conservative and find scalar
potential f such that A = f.
[Ans. : f = x2 + y2 + z2x + 3xy + zy]
15. Prove that A = (y2cosx + z3) iˆ
+ (2y sin x 4) ĵ + (3xz2 + 2) k̂ is irrotational and find its scalar potential.
[Ans. : f = y2 sin x + z3x 4y + 2z]
16. Prove that a = 1 or b = 0, if
(xyz)b (xaî + yaĵ + zak̂ ) is an irrotational
vector.
17. Find the constant a if A = (ax + 3y
+ 4z) iˆ + (x 2y + 3z) jˆ + (3x + 2y z) k̂
is solenoidal.
[Ans. : a = 3]
18. Find the constant a if A = (x + 3y2) iˆ
+ (2y + 2z2) jˆ + (x2 + az) k̂ is solenoidal.
[Ans. : a = 3]
19. Find the constants a, b, c if
A = (axy + bz2) iˆ + (3x2 cz) jˆ +
(3xz2 y) k̂ is irrotational.
[Ans. : a = 6, b = 1, c = 1]
20. 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 = 2x3y2z4.
9.14 PROPERTIES OF GRADIENT,
DIVERGENCE AND CURL
9.14.1 Sum and Difference
The gradient, divergence and curl are distributive with respect to the sum and difference of the functions. If f, g are scalars and A and B are vectors, then
(i)
(ii)
(iii)
(f
g)
(A
(A
B)
B)
f
g
A)
(
(
A)
B)
(
(
B)
Vector Calculus
9.61
∂
∂
∂
Proof: (i) ∇ ( f ± g ) = iˆ ( f ± g ) + ˆj ( f ± g ) + kˆ ( f ± g )
∂x
∂y
∂z
⎛ ∂f
∂f ˆ ∂f ⎞ ⎛ ˆ ∂g ^ˆ ∂g ˆ ∂g ⎞
= ⎜ iˆ
+ jˆ
+k
± i
+j
+k
⎝ ∂x
∂y
∂z ⎟⎠ ⎜⎝ ∂x
∂y
∂z ⎟⎠
= ∇f ± ∇g
(ii) Let A = A1 iˆ + A2 jˆ + A3 k̂, B = B1 iˆ + B2 jˆ + B3 k̂
(A
B)
( A1
=
x
B1 )iˆ ( A2
( A1 ± B1 ) +
y
A
±
B
B1
B
+ 2+ 3
x
y
z
B
(iii) ∇ × ( A ± B ) = ∇ × ( A ± B )
= iˆ ×
∂ (
∂
∂
A ± B ) + jˆ × ( A ± B ) + kˆ × ( A ± B )
∂x
∂y
∂z
iˆ
x
iˆ
(
A)
(A
A
x
B)
B
x
B
x
iˆ
(
A
x
iˆ
B)
9.14.2 Products
If f, g are scalars and A and B are vectors, then
(i)
(ii)
(iii)
(iv)
( f g)
f g
g f or grad ( f g) = f (grad g) + g (grad f )
( A B) ( B ) A ( A ) B B (
or grad ( A B ) ( B ) A ( A ) B
A)
B
( f A) f ( A) ( f ) A
or div ( f A) = f ( div A) + (grad f ) ⋅ A
( A B ) B ( A) A ( B )
or div ( A B ) B curl A A curl B
(
( curl A)
B)
A
A
B3 )kˆ
( A2 ± B2 ) + z ( A3 ± B3 )
A
A1
A
+ 2+ 3
x
y
z
=
B2 ) ˆj ( A3
( curl B )
9.62
(v)
Engineering Mathematics
( f A) f ( A) ( f ) A
or curl ( f A ) f ( curl A ) (grad f )
( A B) ( B
(vi)
or curl ( A B )
)A
B(
(B
A)
)A
A
(A )B
B ( div A )
A(
(A )B
B)
A ( div B ) .
Proof:
(i)
( fg )
iˆ
x
ˆj
( fg )
g
iˆ f
x
g
f iˆ
x
f g
( A B ) iˆ
(ii)
y
g
( fg ) kˆ
z
( fg )
f
x
g
f iˆ
x
g
kˆ
z
f
g iˆ
x
ˆj g
y
f
g iˆ
x
ˆj f
y
f
kˆ
z
g f
x
( A B ) ˆj
iˆ
x
y
( A B ) kˆ
( A B)
iˆ A
B
x
iˆ A
iˆ B
B
x
A
x
z
( A B)
A
B
x
... (1)
Consider,
A
iˆ
B
x
B ˆ
i
x
A
( A iˆ )
⎡∵ a × ( b × c ) = ( a ⋅ c ) b − ( a ⋅ b ) c ⎤
⎣
⎦
B
x
( )
⎛ ^ ∂B ⎞
^⎛
^ ∂B
∂B ⎞
i ⎜⎜ A ⋅
⎟⎟ = A × ⎜⎜ i ×
⎟⎟ + A ⋅ i
∂x
∂x ⎠
⎝
⎝ ∂x ⎠
Similarly, interchanging A and B,
iˆ B
A
x
B
iˆ
A
x
( B iˆ )
A
x
Vector Calculus
9.63
Substituting in Eq. (1),
( A B)
B
x
iˆ
A
B
x
iˆ
A
(
A
B) B
(
A)
( f A)
iˆ
ˆj
x
iˆ
x
(
( A B)
(iv)
= ∑i
^
iˆ
A
x
f
f)
kˆ
y
( A iˆ )
B
x
( A ) B (B
B
x
( B iˆ )
A
x
( B iˆ )
A
x
)A
A1iˆ
B
x
B
A2 ˆj
y
B
A3 kˆ
z
f
(
f
x
A
f
A iˆ
x
(A
ˆj
x
A
x
( A iˆ )
iˆ
A
x
A)
A
x
( f A)
z
( f A)
f iˆ
f
kˆ
y
iˆ
B
iˆ
B
∵
(iii)
B
x
( A iˆ )
iˆ
f
A)
(
A
x
A
f
iˆ
x
f A)
( A B)
z
⎞
∂ (
∂B ∂ A
^ ∂
^ ⎛
⋅ A × B ) = ∑ i ⋅ ( A × B ) = ∑ i ⋅ ⎜⎜ A ×
+
× B ⎟⎟
∂x
∂x
∂x ∂x
⎝
⎠
⎛
⎞
⎛ ∂A
∂B ⎞
= ∑ iˆ ⋅ ⎜⎜ A ×
× B ⎟⎟
⎟⎟ + ∑ iˆ ⋅ ⎜⎜
∂x ⎠
⎝
⎠
⎝ ∂x
∂B
∂A
⋅B
= ∑ iˆ × A ⋅
+ ∑ iˆ ×
∂x
∂x
= −∑ iˆ ×
∂B
∂A
⋅ A + ∑ iˆ ×
⋅B
∂x
∂x
⎡∵ a ⋅ b × c = a × b ⋅ c ⎤
⎣
⎦
⎡
⎤
∂B
in scalar triple product.⎥
⎢ Interchanging A annd
∂x
⎣
⎦
= − ( ∇ × B ) ⋅ A + ( ∇ × A) ⋅ B
= B ⋅ ( ∇ × A) − A ⋅ ( ∇ × B )
(v)
( f A)
iˆ
x
iˆ
ˆj
x
y
kˆ
( f A)
( f A)
z
iˆ
f
A
x
f
A
x
9.64
Engineering Mathematics
f iˆ
A
x
iˆ
A
x
f
f
(vi)
( A B)
(
f
A
x
iˆ
f
iˆ
x
A
A) ( f ) A
iˆ
ˆj
x
iˆ
kˆ
y
( A B)
z
( A B)
x
iˆ
( A B)
x
⎞
⎞
∂B ∂ A
∂B ⎞
^ ⎛
^ ⎛
^ ⎛ ∂A
= ∑ i × ⎜⎜ A ×
+
× B ⎟⎟ = ∑ i × ⎜⎜ A ×
× B ⎟⎟
⎟⎟ + ∑ i × ⎜⎜
∂x ∂x
∂x ⎠
⎠
⎝
⎠
⎝
⎝ ∂x
B
A
x
iˆ
( iˆ A)
B
x
( iˆ B )
A
x
(b
c)
∵a
iˆ
A
B
x
(a c) b (a b) c
⎛ ∂B ⎞
⎛ ∂A⎞
∂B
∂A
= A∑ ⎜⎜ iˆ ⋅
+ ∑ ( B ⋅ iˆ )
⎟⎟ − B ∑ ⎜⎜ iˆ ⋅
⎟⎟ − ∑ ( A ⋅ iˆ )
∂
∂x
∂
x
∂
x
x
⎝
⎠
⎝
⎠
= A ( ∇ ⋅ B ) − B ( ∇ ⋅ A) − ( A ⋅ ∇ ) B + ( B ⋅ ∇ ) A
= ( B ⋅ ∇ ) A − B ( ∇ ⋅ A) − ( A ⋅ ∇ ) B + A ( ∇ ⋅ B )
9.15 SECOND ORDER DIFFERENTIAL OPERATOR
It is a two fold application of the operator
are given below.
(i) Laplacian Operator
iˆ
=
=
ˆj
x
Div (grad f ) =
2
y
kˆ
f
+
x
y
x
2
f
x
2
2
f
+
2
f
y
2
f
+
. Some second order differential operators
f
iˆ
x
z
f
+
y
z
2
f
z
2
ˆj f
y
x
f)
f
kˆ
z
f
z
2
=
.(
2
2
+
y
2
2
+
z2
f
Vector Calculus
9.65
Thus, the scalar differential operator (read as “nabla squared” or “delta”)
2
2
2
2
x
2
y
2
z2
is known as Laplacian operator.
2
2
f
x
2
f
2
f
y
2
2
f
z
2
is known as Laplacian equation.
(ii)
f = curl grad f
⎛ ∂f
= ∇ × ⎜ iˆ +
⎝ ∂x
ˆj
iˆ
∂
∂x
∂f
∂x
=
∂
∂y
∂f
∂y
ˆj ∂f + kˆ ∂f ⎞⎟
∂y
∂z ⎠
kˆ
∂
∂z
∂f
∂z
⎛ ∂2 f
∂2 f ⎞
−
= iˆ ⎜
⎟−
⎝ ∂y ∂z ∂z ∂y ⎠
Hence, curl grad f =
(iii)
(
×
f = 0.
A)
div curl A
A = A1iˆ + A2 jˆ + A3 k̂
Let
iˆ
A
(
A)
y
A2
z
A3
x
A3
y
A2
z
A3
x y
(
kˆ
ˆj
x
A1
2
Hence,
2
2
2
2
ˆj ⎛⎜ ∂ f − ∂ f ⎞⎟ + kˆ ⎛⎜ ∂ f − ∂ f ⎞⎟ = 0
⎝ ∂x ∂y ∂y ∂x ⎠
⎝ ∂x ∂z ∂z ∂x ⎠
2
A2
x z
A3
y
iˆ
y
2
A3
y x
A2
z
A3
x
2
A1
y z
ˆj
A1
z
A3
x
A1
z
A2
x
z
2
A2
z x
kˆ
A1
y
2
A1
z y
0
A ) = div curl A = 0.
Example 1: If r = xiˆ + yjˆ + z k̂, show that div ( r n r ) = ( n + 3 ) r n .
Solution: r n is a scalar and r is a vector.
We know that div
( f A)
f
(
A) ( f ) A
A2
x
A1
y
9.66
Engineering Mathematics
div ( r n r ) = r n ( ∇ ⋅ r ) + (∇r n ) ⋅ r
⎡⎛ ∂
⎤ ⎛ ∂r n
∂
∂r n ˆ ∂r n ⎞
∂ ⎞
= r n ⎢⎜ iˆ
+ jˆ
+ kˆ ⎟ ⋅ ( xiˆ + yjˆ + zkˆ ) ⎥ + ⎜ iˆ
+ ˆj
+k
⎟⋅r
∂y
∂y
∂z ⎠
∂z ⎠
⎣⎝ ∂x
⎦ ⎝ ∂x
∂r
∂r
∂r ⎤
⎡
= r n (1 + 1 + 1) + ⎢iˆ( nr n −1 ) + ˆj ( nr n −1 ) + kˆ( nr n −1 ) ⎥ ⋅ r
∂
x
∂
y
∂z ⎦
⎣
r = xiˆ + yjˆ + zk̂
r2 = x2 + y2 + z2
r x r y r z
= ,
= ,
=
x r y r z r
x
Hence, div ( r n r ) = 3r n + nr n −1 ⎛⎜ iˆ +
⎝ r
ˆj y + kˆ z ⎞⎟ ⋅ ( xiˆ + yjˆ + zkˆ )
r
r⎠
2
⎛ x2 + y 2 + z 2 ⎞
n
n −1 ⎛ r ⎞
n
n
= 3r n + nr n −1 ⎜
⎟ = 3r + nr ⎜ ⎟ = 3r + nr
r
⎝
⎠
⎝ r ⎠
Hence, div ( r n r ) = (n + 3)r n .
Example 2: Find the value of n for which the vector r n r is solenoidal, where
r = xiˆ + yjˆ + z k̂.
Solution: If F = r n r is solenoidal, then
rn r = 0
… (1)
As proved in Ex. 1.,
r n r = (n + 3) r n
Substituting in Eq. (1),
n
(n + 3) r = 0
n= 3
Example 3: Prove that Div (grad r n ) = n (n + 1) rn – 2 , where r = xiˆ + yjˆ + z k̂.
Solution:
Div (grad r n )
( rn )
rn
iˆ
x
nr n
1
nr n
1
n
ˆj r
y
rˆ
i nr n
x
rn
kˆ
z
1
r ˆ
j nr n
y
1
xˆ
y
z
i nr n 1 ˆj nr n 1 kˆ
r
r
r
r ˆ
k
z
Vector Calculus
nr n
n
r
1
2
= n rn
2
( xiˆ + yjˆ + zkˆ )
r
n 2
n rn
9.67
r
(
r ) ( rn 2 ) r
iˆ
ˆj
x
rn 2
iˆ
x
kˆ
y
z
n 2
ˆj r
y
( xiˆ
yjˆ zkˆ )
rn 2
kˆ
z
( xiˆ
yjˆ zkˆ )
⎡
∂r
∂r ˆ
⎧
= n ⎢ r n − 2 (1 + 1 + 1) + ⎨( n − 2) r n − 3 iˆ + ( n − 2) r n − 3
j
∂x
∂y
⎩
⎣
+ ( n − 2) r n − 3
}
⎤
∂r ˆ
k ⋅ ( xiˆ + yjˆ + zkˆ )⎥
∂z
⎦
r = xiˆ + yjˆ + zk̂
r2 = x2 + y2 + z2
r x r y r z
= ,
= ,
=
x r y r z r
Hence, ∇ ⋅ (∇r n ) = n ⎡⎢3r n − 2 + ( n − 2)r n −3 ⎛⎜ x iˆ + y jˆ + z kˆ ⎞⎟ ⋅ ( xiˆ + yjˆ + zkˆ ) ⎤⎥
r
r ⎠
⎝r
⎣
⎦
n 2
n 4
2
2
= n [3r + (n 2) r (x + y + z2)]
= n [3r n 2 + (n 2) r n 4 · r2]
= n [3rn 2 + (n 2) r n 2]
= n (n + 1) rn 2
Example 4: If e and x are two scalar point functions, show that
2
(ex ) = e
2
x+2
x+x
e.
2
Solution:
2
(
)
(
)=
x
2
2
e.
2
(
)
y
2
2
(
)
z2
(
)
2
Consider,
x2
=
x
x
x
(
x
)
+
x
2
=
x
x
+
x2
2
+
x
x
+
x2
... (1)
9.68
Engineering Mathematics
2
Similarly,
y2
2
(
)
(
)
y
y2
y
2
and
z2
2
y
y2
y
2
z
2
z2
z
z
z2
z
Substituting in Eq. (1),
⎛ ∂φ ∂ψ ∂φ ∂ψ ∂φ ∂ψ ⎞
∇ 2 (φψ ) = 2 ⎜
+
+
⎟
⎝ ∂x ∂x ∂y ∂y ∂z ∂z ⎠
⎛ ∂ 2ψ ∂ 2ψ ∂ 2ψ ⎞
⎛ ∂ 2φ ∂ 2φ ∂ 2φ ⎞
+ φ ⎜ 2 + 2 + 2 ⎟ +ψ ⎜ 2 + 2 + 2 ⎟
∂z ⎠
∂y
∂z ⎠
∂y
⎝ ∂x
⎝ ∂x
2
2
= 2∇φ ⋅ ∇ψ + φ∇ ψ + ψ ∇ φ
∇ 2 (φψ ) = φ∇ 2ψ + 2∇φ ⋅∇ψ + ψ∇ 2φ
r
2
Example 5: Prove that
2r 4 , where r = xî + y + z k̂.
r2
⎛ ⎞
Solution: ∇ ⋅ ⎜ r ⎟ = ∇ ⋅ ( r −2 r )
⎜ r2 ⎟
⎝ ⎠
= r −2 ( ∇ ⋅ r ) + (∇r −2 ) ⋅ r
⎡⎛ ∂
∂ ˆ ∂ ⎞ ˆ ˆ
ˆ ⎤
= r −2 ⎢⎜ iˆ
+ jˆ
+k
⎟ ⋅ ( xi + yj + zk ) ⎥
x
y
∂
z
∂
∂
⎠
⎦
⎣⎝
⎛ ∂r −2 ˆ ∂r −2 ˆ ∂r −2
+ ⎜ iˆ
+j
+k
∂y
∂z
⎝ ∂x
⎞
⎟⋅r
⎠
⎡
∂r
∂r ˆ
⎧
= r − 2 ⎢(1 + 1 + 1) + ⎨( − 2 r − 3 ) iˆ + ( − 2 r − 3 )
j
∂
x
∂
y
⎩
⎣
+ (− 2r − 3)
}
⎤
∂r ˆ
k ⋅ ( xiˆ + yjˆ + zkˆ ) ⎥
∂z
⎦
r = xiˆ + yjˆ + zk̂
r2 = x2 + y2 + z2
r x r y r z
= ,
= ,
=
x r y r z r
Hence,
r
r2
r
2
3 2r
3
r
2
3 2r
3
xˆ
i
r
yˆ
j
r
z ˆ
k
r
( x2 + y 2 + z 2 )
r
( xiˆ
y ˆj zkˆ )
Vector Calculus
2
3r
3r
r
r2
2
2r 4 r2
2r 2 = r
2
2
x
2
(r 2 )
2
=
9.69
2
2
(r 2 ) +
y
2
(r 2 ) +
2
z
2
(r 2 )
Now,
∂ ⎡
∂r ⎤
∂ 2 r −2
( −2r −3 ) ⎥
=
2
⎢
∂x ⎣
∂x ⎦
∂x
x⎤
∂ ⎡
∂
=
( −2r −3 ) ⎥ = −2 (r −4 ⋅ x)
⎢
∂x ⎣
∂x
r⎦
∂
r
x
⎛
⎞
⎛
⎞
= −2 ⎜ −4r −5
x + r −4 ⎟ = −2 ⎜ −4r −5 x + r −4 ⎟
⎝
⎠
⎝
⎠
∂x
r
2r
4
2r
4
2r
4
4x2
r2
1
4 y2
r2
1
4z2
r2
1
Similarly,
2
r 2
y2
2
r 2
z2
2
r
r2
2
r 2
x2
2r
4
2
r 2
y2
2
4( x 2
y2
r 2
z2
z2 )
r2
= 2r –4
Hence,
⎡ ⎛ r
∇ 2 ⎢∇⋅ ⎜⎜ 2
⎢⎣ ⎝ r
⎞⎤
−4
⎟⎟ ⎥ = 2r
⎠ ⎥⎦
r
r
Example 6: Prove that
Solution:
r
r
(r r )
1
2
r3
r.
3
2r
4
4r 2
r2
3
9.70
Engineering Mathematics
(
)
= r −1 ∇ ⋅ r + (∇r −1 ) ⋅ r
⎤ ⎛ ∂r −1
⎡⎛ ∂
∂
∂r −1 ˆ ∂r −1 ⎞
∂ ⎞
= r −1 ⎢⎜ iˆ
+ jˆ
+ ˆj
+ kˆ ⎟ ⋅ ( xiˆ + yjˆ + zkˆ ) ⎥ + ⎜ iˆ
+k
⎟⋅r
∂y
∂y
∂z ⎠
∂z ⎠
⎦ ⎝ ∂x
⎣⎝ ∂x
⎛
∂r ˆ −2 ∂r ˆ ⎞
∂r
j−r
k ⎟⋅r
= 3r −1 + ⎜ −r −2 iˆ − r −2
y
x
∂
∂z ⎠
∂
⎝
r = xiˆ + yjˆ + zk̂
r2 = x2 + y2 + z2
r x r y r z
= , = , =
x r y r z r
Hence,
r
r
r
r
3r
1
r
2
3r
1
r
2
3r
1
r
2
iˆ
x
r
r
2
2r
2
= −2r −2
yˆ
j
r
r
2r
ˆj
rˆ
i 2r
x
xˆ
i
r
z ˆ
k
r
3r
r2
r
(2r 1 )
2r
=−
xˆ
i
r
y
1
r
r
2
1
(2r 1 ) kˆ
r ˆ
j 2r
y
2
y ˆ
j
r
(r r )
r
z
2
(2r 1 )
r ˆ
k
z
z ˆ
k
r
r
r
2
r.
r3
Example 7: Show that E =
Solution: Curl E
r
r2
E
r
r2
(r 2 r )
is irrotational.
Vector Calculus
( f A)
We know that,
f
(
A) ( f ) A
(r 2 r ) r 2 (
curl E
r
2
r
2
9.71
iˆ
x
r) ( r 2) r
r 2
iˆ
x
( xiˆ )
( iˆ iˆ )
( 2r 3 )
r
rˆ
i
x
r
r = xiˆ + yjˆ + zk̂
r2 = x2 + y2 + z2
r x r y r z
= , = , =
x r y r z r
Hence,
(r r )
2
0 2r
2r
3
2r
4
xˆ
i
r
3
r
( xiˆ + yjˆ + zkˆ )
(r r )
r
r
=0
Hence, E is irrotational.
(a
Example 8: Prove that
r)
2a , where a is a constant vector.
Solution: Let a = a1 iˆ + a2 jˆ + a3 k̂
r = xiˆ + yjˆ + zk̂
a r
iˆ
a1
ˆj
a2
kˆ
a3
x
y
z
iˆ (a2 z
a3 y)
jˆ (a1 z
iˆ
(a
r)
ˆj
a3 x) + k̂ (a1 y
kˆ
x
y
z
a2 z a3 y a3 x a1 z a1 y a2 x
iˆ (a1 + a1) jˆ ( a2
2(a iˆ + a jˆ + a k̂)
1
2a
2
3
a2) + k̂ (a3 + a3)
a2 x)
9.72
Engineering Mathematics
r
a r
rn
Solution:
(2 n)a
a r
Example 9: Prove that
n
r
n(a r )r
r n+ 2
n
.
( r n A) , where a r A, say
We know that,
( f A)
f
(
( r n A) r n (
r
n
A) ( f ) A
A) ( r n ) A
(a
r
iˆ
r)
n
x
ˆj r
n
y
r
kˆ
n
A
z
As proved in Ex. 8
(a
r)
2a
( r n A) r n ( 2a ) ( nr
n 1
)
rˆ
i
x
r ˆ
j
y
r ˆ
k
z
A
r x r y r z
= , = , =
x r y r z r
As proved earlier,
y
z ⎞
⎛x
Hence, ∇ × ( r − n A ) = 2ar − n − nr − n −1 ⎜ iˆ + ˆj + kˆ ⎟ × A
r
r
r
⎝
⎠
2a
n
−
r n r n +1
n
2a
= n − n+2
r
r
=
=
2a
rn
n
r
2a
rn
na
rn
n+2
r (
× a×r)
r
⎡⎣( r ⋅ r ) a − ( r ⋅ a ) r ⎤⎦ ⎡∵ a × ( b × c ) = ( a ⋅ c ) b − ( a ⋅ b ) c ⎤
⎣
⎦
r2 a
(r a) r
n(a r ) r
r n+2
(2 n)a n ( a r ) r
.
+
rn
r n+2
Example 10: If a is a constant vector, show that a
Solution: Let a = a1 iˆ + a2 jˆ + a3 k̂
r = x1 iˆ + x2 jˆ + x3 k̂
(
r)
(a r ) (a ) r.
Vector Calculus
r
iˆ
ˆj
kˆ
x
r1
y
r2
z
r3
r3
y
iˆ
a
(
r2
z
r3
x
ˆj
iˆ
a1
r)
r1
z
r2
z
r2
x
iˆ a2
r2
x
r1
y
a3
r1
z
r3
x
kˆ a1
r1
z
r3
x
a2
r3
y
r2
z
r
iˆ a2 2
x
a3
r3
x
r
kˆ a1 1
z
a2
r2
z
ˆj a1 r2
x
a3
r2
z
r1
y
kˆ
a3
r3
x
r1
z
r2
x
kˆ
ˆj
a2
r3
y
9.73
r1
x
a1
a3
a1
r3
z
a3
r
kˆ a1 3
x
r1
y
ˆj a1
r1
x
r2
x
ˆj a1 r1
y
r3
z
a2
r
iˆ a2 1
y
r1
y
a3
a3
a3
r3
y
a2
r3
y
r2
y
r2
z
a2
r1
z
r3
y
⎛ ∂
∂
∂ ⎞
= ⎜ iˆ + ˆj + kˆ ⎟ ( a1r1 + a2 r2 + a3 r3 )
∂
∂
∂
x
y
z⎠
⎝
⎛ ∂
∂
∂ ⎞
− ⎜ a1 + a2
+ a3 ⎟ ( r1iˆ + r2 ˆj + r3 kˆ)
∂
∂
x
y
∂
z⎠
⎝
= ∇ ( a ⋅ r ) − ( a ⋅∇ ) r
Example 11: If a is a constant vector such that a = a , prove that
(a r ) a
a2 .
Solution: Let a = a1 iˆ + a2 jˆ + a3 k̂
r = xiˆ + yjˆ + zk̂
r2
y
9.74
Engineering Mathematics
( f A)
We know that,
(a r ) a
Since a is constant,
(a r )
iˆ
= iˆ
x
A) ( f ) A
( a r )(
a
x
(
f
a)
(a r )
a
0
(a r )
ˆj
y
(a r )
kˆ
(a1 x + a2 y + a3 z ) + ˆj
y
z
(a r )
(a1 x + a2 y + a3 z ) + kˆ
z
(a1 x + a2 y + a3 z )
= a1iˆ + a2 ˆj + a3 kˆ
=a
Hence,
(a r ) a
0 a a
= a2 .
Example 12: If F
( a r ) r where a
is a constant vector, find curl F and
prove that it is perpendicular to a .
Solution: Curl F
We know that,
(a r ) r
F
( f A)
(
f
A) ( f ) A
(a r ) r
Curl F
( a r )(
Now,
r
r)
iˆ
ˆj
kˆ
x
x
y
y
z
z
iˆ (0
0
0)
jˆ (0
As proved in Ex. 11
(a r )
(a r ) r
a
0 a r
a r
∇ × ⎡⎣( a ⋅ r ) r ⎤⎦ ⋅ a = ( a × r ) ⋅ a = 0
(a r )
0) + k̂ (0
r
0)
Vector Calculus
(a r ) r
Hence,
is perpendicular to a.
a r
r
Example 13: Prove that
a r
r
Solution:
9.75
r
1
(a
0, where a is a constant vector.
r)
∇ ⋅ ( f A ) = f ( ∇ ⋅ A ) + ( ∇f ) ⋅ A
We know that,
∇ ⋅ ⎡⎣ r −1 ( a × r ) ⎤⎦ = r −1 ⎡⎣∇ ⋅ ( a × r ) ⎤⎦ + (∇r −1 ) ⋅ ( a × r )
⎛ ∂r −1 ˆ ∂r −1 ˆ ∂r −1 ⎞
= r −1 ⎡⎣∇ ⋅ ( a × r ) ⎤⎦ + ⎜ iˆ
+j
+k
⎟ ⋅(a × r )
∂y
∂z ⎠
⎝ ∂x
∇ ⋅ ( A × B ) = B ⋅ ( ∇ × A) − A ⋅ ( ∇ × B )
∇ ⋅ ( a × r ) = r ⋅ (∇ × a ) − a ⋅ (∇ × r )
Since a is constant,
Also,
r
a
0.
0 as proved in Ex. 12.
∇ ⋅(a × r ) = 0
⎛
∂r
∂r ˆ −2 ∂r ˆ ⎞ (
Hence, ∇ ⋅ ⎡⎣ r −1 ( a × r ) ⎤⎦ = 0 + ⎜ −r −2 iˆ − r −2
j−r
k ⎟⋅ a×r)
∂x
∂y
∂z ⎠
⎝
⎛ xiˆ + yjˆ + zkˆ ⎞
−3
= 0 − r −2 ⎜
⎟⎟ ⋅ ( a × r ) = −r ⎡⎣ r ⋅ ( a × r ) ⎤⎦ = 0
⎜
r
⎝
⎠
(r
Example 14: Prove that curl
a) b
Solution: We know that, ( r a ) b
b a , where a and b are constants.
(r b) a (a b) r
Let r ⋅ b = f , say and a ⋅ b = g , say
(r
curl
(r
a) b
f a gr
a) b
(r
( f a)
We know that,
(r
(fa
a) b
gr )
( gr )
( f A)
f
(
A) ( f ) A
a) b
f
(
a) ( f ) a g (
r ) ( g) r
9.76
Engineering Mathematics
Since a is constant,
0. Also
a
r
0
∇ × ⎡⎣( r × a ) × b ⎤⎦ = ⎡⎣∇ ( r ⋅ b )⎤⎦ × a − ⎡⎣∇ ( a ⋅ b )⎤⎦ × r
(r b)
[Substituting f and g]
∵ a and b are constant
a 0
Let b = b1iˆ + b2 jˆ + b3 k̂, r = xiˆ + yjˆ + zk̂
(r b)
iˆ
ˆj
x
kˆ
y
z
(b1 x b2 y b3 z )
= b1iˆ + b2 ˆj + b3 kˆ = b
Substituting in Eq. (1),
(r
Hence, curl
(r
a) b
1
rn
iˆ
=
We know that,
n
(r
n 1
r
ˆj
x
r n+1
z
)
rˆ
i ( nr
x
( nr
n 1
)
xˆ
i
r
yˆ
j
r
.
n
r
n 1
n 1
)
r ˆ
j ( nr
y
n 1
)
r ˆ
k
z
z ˆ
k
r
n
−n r
= − n+ 2 r
n +1
r
r
r
n
r
r
n
(r
( f A)
f
r)
n ( n 2)
n
r
kˆ
y
1
( nr
1
rn
r
b a
b a.
Example 15: Prove that
Solution:
a) b
n
n
n
n
n 1
(
1
r
n+2
n +1
3
r n +1
3
r
n +1
r