Engineering mathematics : a tutorial approach 9780070146150, 0070146152

Table of contents :
Title
Contents
1 Complex Numbers
2 Differential Calculus I
3 Differential Calculus II
4 Partial Differentiation
5 Infinite Series
6 Integral Calculus
7 Gamma and Beta Functions
8 Multiple Integral
9 Vector Calculus
10 Differential Equations
11 Matrices
12 Laplace Transform
13 Fourier Series
14 Fourier Transform
15 Z-transform
Appendix 1 Differential Formulae
Appendix 2 Integral Formulae
Appendix 3 Standard Curve
Index

Citation preview

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 NEW DELHI New Delhi New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto

Tata McGraw-Hill Published by the Tata McGraw Hill Education Private Limited, 7 West Patel Nagar, New Delhi 110 008. Copyright© 2010, by Tata McGraw Hill Education Private Limited. No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, Tata McGraw Hill Education Private Limited ISBN (13): 978-0-07-014615-0 ISBN (10): 0-07-014615-2 Managing Director: Ajay Shukla Head—Higher Education Publishing: Vibha Mahajan Manager: Sponsoring—SEM & Tech Ed: Shalini Jha Editorial Executive: Tina Jajoriya Jr Executive: Editorial Services: Dipika Dey Sr Production Executive: Suneeta S Bohra General Manager: Marketing—Higher Education: Michael J Cruz Sr Product Manager—SEM & Tech Ed: Biju Ganesan Asst Product Manager—SEM & Tech Ed: Amit Paranjpe General Manager—Production: Rajender P Ghansela Asst General Manager—Production: B L Dogra Information contained in this work has been obtained by Tata McGraw Hill, from sources believed to be reliable. However, neither Tata McGraw Hill nor its authors guarantee the accuracy or completeness of any information published herein, and neither Tata McGraw Hill nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that Tata McGraw Hill and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought. Typeset at Tej Composers, WZ-391, Madipur, New Delhi - 110 063 and printed at Gopsons, A-2&3, Sector-64, Noida, U.P. 201 301 Cover Printer: Gopsons 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 [email protected] (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

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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


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 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

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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

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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

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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)]

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(−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

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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.

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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 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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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) =

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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 ⎦

2.48

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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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 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 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 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.

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(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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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 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 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
1 − c2 > 1 − b2 1 1− a b−a

2

1 − a2 b−a 1− a

(i) Putting b =

... (1)

2

<
0, b > 0]

[∵ b

a > 0]

... (2) [Using Eq. (1)]

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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
0, for every value of x in (0,

Differential Calculus I

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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 ]

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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]

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

[∵ f (0) = 0]

log

⎡ 1 ⎛ x2 ⎞⎤ 1+ x − 2 x ⎢1 + ⎜ 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)
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 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

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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

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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 > ⎢⎣ ⎥⎦ 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!

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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 !

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(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!

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(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

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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) +

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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)

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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)

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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 !

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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

.....

... .

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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 ⎞ ⎟ ⎠

...... .

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⎡ ⎛ 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 ⎠

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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

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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) ⎦

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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

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(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)

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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)

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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)

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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)

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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)

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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

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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 ) ⎥ ⎣ ⎦

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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 ]

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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.

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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.

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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

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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.

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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

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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.

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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 < , – 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

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(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

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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).

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(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

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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

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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

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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 < .

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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 < .

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(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 0, when dx

a 3

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:

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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

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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., 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 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

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∂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

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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)

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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)

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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)

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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)

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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

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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

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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 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

− ∈
m

un − l m

l

l − ∈
m

5.6

Engineering Mathematics

m terms of

un and

l − ∈
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 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