Advanced Engineering Mathematics PDF

 Advanced Engineering Mathematics Fourth Edition  R.K. Jain □ S.R.K. Iyengar Advanced Engineering Mathematics Fourth Edition ► Alpha Science International Ltd....

 Advanced Engineering Mathematics Fourth Edition  R.K. Jain □ S.R.K. Iyengar Advanced Engineering Mathematics Fourth Edition ► Alpha Science International Ltd. Oxford, U.K. R.K. Jain (Retd.) Professor of Mathematics Indian Institute of Technology, Delhi New Delhi, India S.R.K. Iyengar (Retd.) Professor & Head Department of Mathematics Indian Institute of Technology, Delhi New Delhi, India Copyright © 2002, 2003, 2007, 2014 First Edition 2002 Second Edition 2003 Third Edition 2007 Fourth Edition 2014 ALPHA SCIENCE INTERNATIONAL LTD. 7200 The Quorum, Oxford Business Park North Garsington Road, Oxford 0X4 2JZ, U.K. www.alphasci.com All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without prior written permission of the publisher. ISBN 978-1-84265-846-8 Printed in India  To Our Parents Bhagat Ram Jain and Sampati Devi Jain & S. T. V. Raghavacharya and Rajya Lakshmi Whose memories had always been an inspiration  Preface to the Fourth Edition We sincerely thank the faculty members and the students of various Institutes and Engineering Colleges for their suggestions to improve the book. Based on these suggestions, we have included the following new material. (i) Condition number of a matrix and Singular Value Decomposition (Chapter 3). (ii) Application of Z-transforms to find the sum of series (Chapter 17). (iii) Cubic splines, B-splines, Romberg integration. Gauss quadrature rules and Two-point boundary value problems (Chapter 18) We hope that the book in the present form includes most of the topics covered in the core courses for Engineering students. We look forward to get more suggestions from the faculty members and the students to improve the book further. R.K. Jain S.R.K. Iyengar  Preface to the First Edition This book is based on the experience and the lecture notes of the authors while teaching mathematics courses to engineering students at the Indian Institute of Technology, Delhi for more than three decades. A number of available textbooks have been a source of inspiration for introduction of concepts and formulation of problems. We are thankful to the authors of these books for their indirect help. This comprehensive textbook covers syllabus for two courses in Mathematics for engineering students in various Institutes, Universities and Engineering Colleges. The emphasis is on the presentation of the fundamentals and theoretical concepts in an intelligible and easy to understand manner. Each chapter in the book has been carefully planned to make it an effective tool to arouse interest in the study and application of mathematics to solve engineering and scientific problems. Simple and illustrative examples are used to explain each theoretical concept. Graded sets of examples and exercises are given in each chapter, which will help the students to understand every important concept. The book contains 682 solved examples and 2984 problems in the exercises. Answers to every problem and hints for difficult problems are given at the end of each chapter which will motivate the students for self- learning. While some problems emphasize the theoretical concepts, others provide enough practice and generate confidence to use these concepts in problem solving. This textbook offers a logical and lucid presentation of both the theory and problem solving techniques so that the student is not lost in unnecessary details. We hope that this textbook will meet the requirement and the expectiations of the engineering students. We will gratefully receive and acknowledge every comment, suggestion for inclusion/exclusion of topics and errors in the book, both from the faculty and the students. We are grateful to our former teachers, colleagues and well wishers for their encouragement and valuable suggestions. We are also thankful to our students for their feed back. We are grateful to the authorities of IIT Delhi for providing us their support. We extend our thanks to the editorial and the production staff of M/s Narosa Publishing House, in particular Mr. Mohinder Singh Sejwal for their care and enthusiasm in the preparation of this book. Last, but not the least, we owe a lot to our family members, in particular, our wives Vinod Jain and Seetha Lakshmi whose encouragement and support had always been inspiring and rejuvenating. We appreciate their patience during our long hours of work day and night. New Delhi R.K. Jain October 2001 S.R.K. Iyengar  Contents Preface to the Fourth Edition vii Preface to the First Edition ix 1. Functions of a Real Variable 1.1 1.1 Introduction 1.1 1.2 Application of Derivatives 1.1 1.2.1 Differentials and Approximations 1.1 1.2.2 Mean Value Theorems 1.2 1.2.3 Indeterminate Forms 1.8 1.2.4 Increasing and Decreasing Functions 1.10 1.2.5 Maximum and Minimum Values of a Function 1.11 1.2.6 Taylor’s Theorem and Taylor’s Series 1.14 1.2.7 Exponential, Logarithmic and Binomial Series 1.22 1.3 Integration and Its Applications 1.30 1.3.1 Areas of Bounded Regions 1.31 1.3.2 Arc Length of a Plane Curve 1.35 1.3.3 Volume of Solids 1.37 1.3.4 Surface Area of a Solid of Revolution 1.44 1.4 Improper Integrals 1.48 1.4.1 Improper Integrals of the First Kind (Range of Integration is Infinite) 1.49 1.4.2 Improper Integral of the Second Kind 1.53 1.4.3 Absolute Convergence of Improper Integrals 1.59 1.4.4 Beta and Gamma Functions 1.60 1.4.5 Improper Integrals Involving a Parameter 1.68 1.4.6 Error Functions 1.71 1.5 Some Properties of Curves and Curve Sketching 1.78 1.5.1 Convexity and Concavity of a Curve 1.78 xn Contents 1.5.2 Curvature, circle of curvature and Radius of curvature 1.81 1.5.3 Evolute and Involute of a Curve 1.91 1.5.4 Envelope of a Family of Curves 1.94 1.5.5 Asymptotes to a Curve 1.96 1.5.6 Curve Sketching 1.108 1.6 Answers and Hints 1.113 2. Functions of Several Real Variables 2.1 2.1 Introduction 2.1 2.2 Functions of Two Variables 2.1 2.2.1 Limits 2.4 2.2.2 Continuity 2.7 2.3 Partial Derivatives 2.12 2.3.1 ' Total Differential and Differentiability 2.17 2.3.2 Approximation by Total Differentials 2.22 2.3.3 Derivatives of Composite and Implicit Functions 2.24 2.4 Higher Order Partial Derivatives 2.32 2.4.1 Homogeneous Functions 2.35 2.4.2 Taylor’s Theorem 2.38 2.5 Maximum and Minimum Values of a Function 2.46 2.5.1 Lagrange Method of Multipliers 2.51 2.6 Multiple Integrals 2.55 2.6.1 Double Integrals 2.55 2.6.2 Triple Integrals 2.65 2.6.3 Change of Variables in Integrals 2.68 2.6.4 Dirichlet Integrals 2.72 2.7 Answers and Hints 2.79 3. Matrices and Eigenvalue Problems 3.1 3.1 Introduction 3.1 3.2 Matrices 3.1 3.2.1 Matrix Algebra 3.3 3.2.2 Some Special Matrices 3.5 3.2.3 Determinants 3.7 3.2.4 Inverse of a Square Matrix 3.9 3.2.5 Solution of n x n Linear System of Equations 3.12 3.3 Vector Spaces 3.18 3.3.1 Subspaces 3.21 3.3.2 Linear Independence of Vectors 3.24  Contents xiii 3.3.3 Dimension and Basis 3.26 3.3.4 Linear Tranformations 3.29 3.4 Solution of General linear System of Equations 3.41 3.4.1 Existence and Uniqueness of the Solution 3.42 3.4.2 Elementary Row and Column Operations 3.42 3.4.3 Echelon Form of a Matrix 3.43 3.4.4 Gauss Elimination Method for Non-homogeneous Systems 3.46 3.4.5 Gauss-Jordan Method 3.50 3.4.6 Homogeneous System of Linear Equations 3.52 3.5 Eigenvalue Problems 3.55 3.5.1 Eigenvalues and Eigenvectors 3.56 3.5.2 Similar and Diagonalizable Matrices 3.64 3.5.3 Special Matrices 3.71 3.6 Quadratic Forms 3.76 3.6.1 Canonical Form of a Quadratic Form 3.80 3.7 Condition Number of a Matrix 3.83 3.8 Singular Value Decomposition 3.85 3.9 Answers and Hints 3.94 4. Ordinary Differential Equations of First Order 4.1 4.1 Introduction 4.1 4.2 Formation of Differential Equations 4.2 4.3 Solution of a Differential Equation 4.3 4.4 Initial and Boundary Value Problems 4.5 4.5 Solution of Equations in Separable Form 4.7 4.5.1 Equations Reducible to Separable Form 4.10 4.6 Exact First Order Differential Equations 4.15 4.6.1 Integrating Factors 4.18 4.7 Linear First Order Equations 4.25 4.8 Some. Special First Order Equations 4.29 4.8.1 Bernoulli Equation 4.29 4.8.2 Riccati Equation 4.31 4.8.3 Clairaut’s Equation 4.32 4.9 Orthogonal Trajectories of a Given Family of Curves 4.34 4.10 Existence and Uniqueness of Solutions 4.40 4.10.1 Picard’s Iteration Method of Solution 4.44 4.11 Answers and Hints 4.47 5. Linear Differential Equations 5.1 5.1 Introduction 5.1 xiv Contents 52 Solutions of Linear Differential Equations 5.2 5.2.1 Linear Independence and Dependence 5.4 5.3 Methods for Solution of Linear Equations 5.9 5.3.1 Differential Operator D 5.9 5.3.2 Solution of Second Order Linear Homogeneous Equations with Constant Coefficients 5.11 5.3.3 Method of Reduction of Order for Variable Coefficient Linear Homogeneous Second Order Equations 5.16 5.3.4 Solution of Higher Order Homogeneous Linear Equations with Constant Coefficients 5.19 5.4 Solution of Non-homogeneous Linear Equations 5.25 5.4.1 Method of Variation of Parameters 5.27 5.4.2 Method of Undetermined Coefficients 5.32 5.4.3 Solution of Euler-Cauchy Equation 5.37 5.5 Operator Methods for Finding Particular Integrals 5.44 5.5.1 Case r (x) = e“' 5.45 5.5.2- Case r (x) = cos (ax) or sin (ox) 5.49 5.5.3 Case r (x) = x“, a > 0 and Integer 5.52 5.6 Simultaneous Linear Equations 5.55 5.6.1 Solution of First Order Systems by Matrix Method 5.58 5.6.2 Method of Undetermined Coefficients to Find the Particular Integral 5.62 5.6.3 Method of Diagonalisation to Find the Particular Integral.5.65 5.7 Answers and Hints 5.68 6. Series Solution of Differential Equation 6.1 6.1 Introduction 6.1 6.2 Ordinary and Singular Points of an Equation 6.1 6.3 Power Series Solution 6.4 6.4 Series solution about a Regular Singular Point: Frobenius Method 6.15 6.5 Answers and Hints 6.28 7. Legendre Polynomials, Chebyshev Polynomials, Bessel Functions and Sturm-Liouville Problem 7.1 7.1 Introduction 7.1 7.2 Legendre Differential Equation and Legendre Polynomials 7.1 7.2.1 Rodrigue’s Formula 7.3 7.2.2 Generating Function for Legendre Polynomials 7.5 7.2.3 Recurrence Relations for Legendre Polynomials 7.7 7.2.4 Orthogonal and Orthonormal Functions 7.9  Contents XV 7.2.5 Orthogonal Property of Legendre Polynomials 7.11 7.2.6 Fourier-Legendre Series 7.13 7.3 Chebyshev Differential Equation and Chebyshev Polynomials 7.16 7.3.1 Chebyshev Polynomials of First Kind 7.18 7.3.2 Chebyshev polynomials of Second Kind 7.26 7.4 Bessel’s Differential Equation and Bessel’s Functions 7.32 7.4.1 Bessel’s Function of the First Kind 7.36 7.4.2 Bessel’ Function of the Second Kind 7.42 7.5 Sturm-Liouville Problem 7.48 7.5.1 Orthogonality of Bessel Functions 7.55 7.5.2 Fourier-Bessel Series 7.57 7.6 Answers and Hints 7.61 8. Laplace Transformation 8.1 8.1 Introduction 8.1 8.2 Basic Theory of Laplace Transforms 8.1 8.3 Laplace Transform solution of Initial Value Problems 8.7 8.3.1 Laplace Transform of Derivatives 8.7 8.3.2 Laplace Transform of Integrals 8.11 8.4 Translation Theorems (Shifting Theorems) 8.14 8.4.1 Heaviside Function or Unit Step Function 8.17 8.5 Laplace Transform of Dirac-delta Function and More Properties of Laplace Transforms 8.25 8.5.1 Laplace Transform of Dirac-delta Function 8.25 8.5.2 Differentiation of Laplace Transform 8.28 8.5.3 Integration of Laplace Transform 8.32 8.5.4 Convolution Theorem 8.34 8.6 Laplace Transform of Periodic Functions 8.39 8.7 Laplace Transform Method for the solution of Some Partial Differential Equations 8.45 8.8 Answers and Hints 8.61 9. Fourier Series, Fourier Integrals and Fourier Transforms 9.1 9.1 Introduction 9.1 9.2 Fourier series 9.1 9.2.1 Fourier Series Expansions of Even and Odd Functions 9.7 9.2.2 Convergence of Fourier Series 9.9 9.3 Fourier Half-range Series 9.16 9.3.1 Complex Form of Fourier Series 9.18 xvi Contents 9.4 Fourier Integrals 9.21 9.5 Application of Fourier Series: Separation of Variables Solution of Linear Partial Differential Equations 9.29 9.5.1 Classification of Linear Second Order Partial Differential Equations 9.29 9.5.2 Separation of Variables Method (Fourier Method) 9.31 9.5.3 Fourier Series Solution of the Heat Equation 9.32 9.5.4 Fourier Series solution of the Wave Equation 9.38 9.5.5 Fourier Series Solution of the Laplace Equation 9.48 9.6 Fourier Transforms 9.54 9.6.1 Fourier Transform Solution of Some Partial Differential Equations 9.67 9.7 Answers and Hints 9.73 10. Functions of a Complex Variable: Analytic Functions 10.1 10.1 Introduction 10.1 10.2 Sets of Points in the Complex Plane 10.1 10.3 Functions of a Complex Variable 10.5 10.4 Elementary Functions 10.13 10.4.1 Exponential Function 10.13 10.4.2 Trigonometric and Hyperbolic Functions 10.15 10.4.3 Logarithm Function 10.20 10.4.4 General Powers of a Complex Number 10.22 10.4.5 Inverse Trigonometric and Hyperbolic Functions 10.24 10.5 Limit and Continuity 10.29 10.5.1 Limit of a Function 10.29 10.5.2 Continuity of a Function 10.35 10.5.3 Uniform Continuity 10.39 10.6 Differentiability and Analyticity 10.42 10.6.1 Cauchy-Riemann Equations 10.46 10.7 Harmonic Functions 10.60 10.8 Answers and Hints 10.66 11. Integration of Complex Functions 11.1 11.1 Introduction 11.1 11.2 Definite Integrals 11.1 11.2.1 Curves in the Complex Plane 11.3 11.2.2 Contour Integrals (Line Integrals in the Complex Plane) 11.6 11.3 Cauchy Integrals Theorem 11.20 11.3.1 Extension of Cauchy Integral Theorem for Multiply Connected Domains ‘ t 11.27 11.3.2 Use of Indefinite Integrals in the Evaluation of Line Integrals 11.32  Contents xvii 11.4 Cauchy Integral Formula 11.36 11.5 Cauchy Integral Formula for Derivatives 11.43 11.6 Answers and Hints 11.50 12. Power Series, Taylor and Laurent Series 12.1 12.1 Introduction 12.1 12.2 Infinite Sequences 12.1 12.2.1 Real Sequences 12.1 12.2.2 Complex Sequences 12.6 12.2.3 Sequences of Functions 12.9 12.2.4 Uniform Convergence 12.11 12.3 Infinite Series 12.13 12.3.1 Tests for Convergence 12.15 12.3.2 Uniform convergence of Series of Functions 12.21 12.4 Power Series 12.24 12.5 Taylor Series 12.29 12.6 Laurent Series 12.38 12.7 Answers and Hints 12.48 13. Zeros, Singularities and Residues 13.1 13.1 Introduction 13.1 13.2 Zeros and Singularities of Complex Functions 13.1 13.3 Residues 13.10 13.4 Evaluation of Contour Integrals Using Residues 13.18 13.5 Evaluation of Real Integrals Using Residues 13.23 13.5.1 Real Definite Integrals Involving Trigonometric Functions 13.23 poo 13.5.2 Improper Integral of the Form f{x)dx 13.27 J—oo 13.5.3 Improper Integral of the Form cos ( 0. 1.2 Engineering Mathematics In the limit, the differential is also written as df{xQ) = dyQ =7(xo) dx. (1.3) Hence, an approximation to f (xq + zi x) can be written as f (xq + zlx) = /(xq) + /'{xq} dx. (1.4) Differentials have application in calculating errors in functions due to small errors in the independent variable. We define \dy\ as the absolute error, dy/y as the relative error and {dyty) X 100 as the percentage error in computations. Example 1.1 Find an approximate value of y= 3(4.02)2 _ 2(4.02)2^2 Solution Let a function be defined as y = f{x) ~. Let Xq = 4 and Zlx = 0.02. Then, we need an approximation to + Zlx) =/(4.02). The approximate value is given by (see Eq. 1.4)). + (0.02)7(4) We have = 48 - 2(8) + 8/2 = 36, 7(x) = 6x - 3x'^2 -4x-2^2 and = 24 - 6 - 4/8 = 3512. Therefore, the required approximation is /(4.02) = 36 4- 0.02 (35/2) = 36.35. Example 1.2 If there is a possible error of 0.02 cm in the measurement of the diameter of a sphere, then find the possible percentage error in its volume, when the radius is 10 cm. Solution Let the radius of the sphere be r cm. Volume of the sphere = V = 4;rz-2/3 ^nd dr = ± 0.01 when r = 10 cm. Differentiating V, we obtain dV = ^Kr^dr. When r = 10, we get from Eq. (1.3), dV = 47t(10)2 (± 0.01) = ± 4;r. Hence, the percentage error in volume is dE ± 12/r 100- 100 = ± 0.3 cubic cm. V 4;r(10)2 1.2.2 Mean Value Theorems We now prove the three basic mean value theorems of the functions of one variable. Theorem 1.1 (Rolle’s theorem) Let a real valued function /(x) be continuous on a closed interval [a, b] and differentiable in the open interval (a, b). iff(a) = /{b}, then there exists ,at least one value c, a c b such that /'{c} - 0.  Functions of a Real Variable 1.3 Proof Since the function /(x) is continuous on the closed interval [a, b], it is bounded and attains its maximum value M and minimum value m at some points in [a, b]. Letf(x) attain respectively its minimum and maximum values at the points c and d e [a, Z?], that is /(c) = m and f{d} = M. If m = M, then the function /(x) is constant over [cz, /?] and therefore, its derivative f'(x} is zero for all X in [a, 6]. ii m M, then both of these cannot be equal to the same quantity /(cz) or /(Zz). We note that f(a') = ffb}. Thus, atleast one of these, say zzz, is different from /(cz) and /{b). Hence, /(c) = m implies c a, f{c') = /{b), implies c b. Therefore, c e {a, b}. We shall now show that at this point c, /'{c} = 0. If f'{c) < 0, then for every x in the interval (c, c + e/, e, 0, < /{c} = m which contradicts the assumption that m is the minimum value of /(x). If f'{c) > 0, then for every x in the interval (c - £2, c), £2 0, fix} < f{c} = m which is again a contradiction. Hence, f\c) = 0. y y b a O I I I I O a c b (b) y I I O a b Fig. LI. Rolle’s theorem. 1.4 Engineering Mathematics Remark 1 (a) Differentiability of /(x) in an open interval (a, Z?) is a necessary condition for the applicability of the Rolle’s theorem. For example, consider the function /(x) = |x|, - 1 < x < 1. Now, /(x) is continuous on [-1, 1] and is differentiable at all points in the interval (-1, 1) except at the point x = 0. Now, 1, X>0 /'(x) = -1, X 0 does not vanish at any point in the interval (-1, 1). This shows that the Rolle’s theorem cannot be applied as the function /(x) is not differentiable in (-1, 1). (b) Rolle’s theorem gives sufficient conditions for the existence of a value c such that f'{c} = 0. For example, the function 0, l’„+, - (zz^ - y„ = 0. For X = 0, we get (0) yn-^2 = (zz^ - zzz^) JZ„(O), zz = O, 1, 2,... Therefore, we find /(O) = To(O) = 0, = Ti(0) = zzz,/"(O) = ^2(0) = 0, /'"(O) = JZ3(O) - zzz(12 - m\ = ^4(0) = 0, r(0) = JZ5(O) = zzz(P - m^} {3^ - m\... Substituting in (1.22), we obtain zzz(l^ - zzz(l^ - zzz^) (3^ —.5 /{x) = mx + x’ + X + 3! 5! Example 1.19 Using Taylor’s series, obtain the value of cos 31° correct to four decimal places. Solution Let /(x) = cos x. We have /'(x) = - sin X, f"{x} = - cos x, = sin x and so on. Using Taylor’s series, we obtain.2 h,3 cos {x + h')= f{x +h}= fix) + hf'(x) + — f'\x) + — ryx} +... 2 h = cos X - h sin X------- cos x + — sin x +... 2 6 1.22 Engineering Mathematics Substituting x = 30° = 7tl6 and A = 1° = ;r/180 = 0.01745, we get cos 31° = cos(7r/6) - (zr/180) sin(7r/6) — (7r/18O)“ cos(.zr/6) + - (;r/180)’ sin(7r/6) +... 2 6 f V3 -^(0.01745)2 J3 + (0.01745)^ 2^ - (0.01745) +... 2 2 7 2 V = 0.86602 - 0.00872 - 0.00013 + 0.00000 +... = 0.85717. Hence, we obtain cos 31° = 0.8572 correct to four decimal places. 1.2.7 Exponential, Logarithmic and Binomial Series Exponential series Consider the Taylor’s polynomial approximation of degree < n about the point x = 0 for the function /(x) = e^. The Taylor’s polynomial approximation is given by 1! 2! «! For /(x) = we obtain f\x} = e^, /*'’(0) = 1, r = 0, !,...,« and 2.1’ , , , X X Hence, fix') = e 1 + X +----- 1----- 1---- 2! n\ Using the Lagrange form of the remainder, we get.n + l y("+l) X.n + l = (C) = ---------- e (« + l)! (n+ 1)! X.n + l or as Rn(x) = e®'-, 0 0 1. (« + !)! 1/1+1 x''^' Therefore, lim 17?„(x) I = lim --------- e e.v = lim =0 n—>oo n—^oo (« + l)! (« + !)! for all X, since e is bounded for a given x. Hence, we obtain the exponential series ,2 n e” = 1 + X + + + — +... (1.25) 2! n\  Functions of a Real Variable 1.23 Example 1,20 For the Taylor's polynomial approximation of degree < n about the point.r = 0 for the function /(x) e\ determine the value of n such that the en’or satisfies |/?„(x)| < 0.005, when - 1 < X < 1. Solution We have the Taylor’s polynomial approximation of e^ as fx} = ~ 1 + x + —H---- 1- —. 2! n\ The maximum error in the interval [-1, 1] is given by Zl+I (n+l) 1^1 e l«„WI = f {c) < max max (« + !)! -i 5. Hence, we will require at least 6 terms in the Taylor's polynomial approximation to achieve the given accuracy. Example 1.21 Obtain the fourth degree Taylor's polynomial approximation to f{x) = about X = 0. Find the maximum error when 0 < x < 0.5. Solution We have 2 3 4 Xx) = /(O) + x/'(0) + /"(O) + y /'"(O) + /'“’(O). For f{x} = e^\ we obtain = 2'e^, /''’(O) = 2’’, r = 0, 1, 2,... and = 1^2 e^\ 4 4a-- 8x’ 16a- 4 2 Therefore, /(x) = e^'‘ = 1 + 2x +------- 1-------- 1- = 1 + 2x -t" 2:^ + —x^ H—x^. 2! 3! 4! 3 3 The error term is given by c X. ^2 lx max x^ max e and l^4WI tt;; 120 0]. (c) Let m and M be the minimum and maximum values of /(x) on [a, Z>]. Then, m{b - a) < [ fix) dx < M{b - a). Ja'a.. (1.33)  Functions of a Real Variable 1.31 (d) fJa flx)dx-lb-a)fl^), ag(x) in [fl, cJ and /(xl^glx) in [r,/?], a c b, then we write the area as (Fig. 1.5) Area = £ [/(x) - g(x)] dx + £ [g(x) - /(x)] dx. (1.41) Area bounded by a curve represented in parametric form Let the curve v /(x) be defined in parametric form as x = 0(t), y=\if{t), a a~ 6. y = ex In x and y = In xl^ex). 1. X = 2z + 1, y = Af' - 1, - 1/2 < z < 1/2 and the x-axis. 8. x = a cos^t, y = b sin^z, Q ’ = ylx , y = 0 from x = 0 to x = 4 about the x-axis. 33. Region bounded byy=Vx,y = 0 from x = 0 to x = 4 about the line y — 2. 34. Region bounded by y = x^ + 1 and y = 3 - x about the x-axis. 35. Region bounded by x = a sin^ t, y = a cos^ t, Q a > 0. In problems 42 to 50, find the surface area of the solid generated by revolving the curve C about the given line. x~ 42. (x - b}^ + y^ = c^, b>a about the y-axis. 43. a>b,y>0 about the x-axis. b^ y^ x"^ 1 44. ^ + Ar- = l,a>6, X > 0 about the y-axis. 45. y= —H---- zr , 1 < X < 2 about the line y = - 1. b^ 4 8x2 46. 1 , 1 < y < 2 about the line x = - 1. X =y--- 1----- 3 4y 41. x = a(k-sint), y = a(l-cos/), Q—00 J p If the limit exists and is finite, say equal to F, then the improper integral converges and has the value a Otherwise, the improper integral diverges. >h (iii) JJ — oo f{x}dx a lim fa}dx + lim f fa}dx (1.63) where c is any finite constant including zero. If both the limits on the right hand side exist separately and are finite, say equal to f and 1^ respectively, then the improper integral converges and has the value F f. If one or both the limits do not exist or are infinite, then the improper integral diverges. Example 1.35 Evaluate the following improper integrals, if they exist. '0 dx r Jo 2 a +x' 2 , a>0. (ii) J dx. 1.50 Engineering Mathematics (iii) f X sin X dx. (iv) J e cos px dx. a 0, p constant. Jo Jo Solution »CX} dx 11 -1 b ;r (i) 2 = lim = lim —tan.2 — ’o + x‘ h—>oo '0 +x' a a 2a Therefore, the improper integral converges to Ttl{2a}. (ii) [ ^dx = e^ dx= lim lim (l-e*)=l. 1 — 0 Therefore, the improper integral converges to 1. (iii) f X sin X dx = lim X sinX dx = lim (sin b~b CO3 b}. Jo Since this limit does not exist, the improper integral diverges. (iv) Using the result —ax —ax e e cos px dx = 2 ' 1 (p sinpx- a CO3px}. a +p 1* b J e e we obtain cos px dx = {p sin px-a cos px} Jo a^ + p 2 Jo 1 0 [e {p sin bp-a cos bp} + a]. Now, sin bp and cos bp have finite values and lim e = 0. Hence, Z) a e-ox COS px dx = lim e cos px dx = h Jo 2 -h p Therefore, the improper integral converges to a/^a^ + p^}. dx Example 1.36 Discuss the convergence of the improper integral Solution We have dx 1 7^ 1 1-/7 K 1-P 1 -1] °o , if /? < 1 Now, lim = 0, if p>l.  Functions of a Real Variable 1.51 Therefore, the improper integral converges if p 1 and diverges if p 1. For 7? = 1, we have rb dx r, ,z> I, —= lim —= lim[lnxl = hm InZ?. Since the limit does not exist, the improper integral diverges. Hence, the given improper integral converges to \l{p - 1) for p > 1 and diverges for p < 1. Example 1.37 Discuss the convergence of the integral J —x.2‘ xe dx. Solution We write 1= xe -X“ —X".2.1 -X' dx = xe dx + xe dx where c is any finite constant. We have /= lim c 'a xe.2 dx+ lim Z?—>oo r xe.2 dx -a ~ -e J! -h' lim + lim e -e a— J. '-e-‘^ ^e’^ = 0. 2 Therefore, the given improper integral converges to 0. It is not always possible to study the convergence/divergence of an improper integral by evaluating it as was done in the previous examples. A simple example is the integral J e"' dx which cannot be evaluated directly. We now present some results which can be used to discuss the convergence or divergence of improper integrals. In this case, we cannot find the value of the improper integral, that is the value to which it converges. However, we may be able to find a bound of the integral. Theorem 1.8 (Comparison Test 1) If 0 < fix) < g(x) for all x, then (i) f ZW dx converges if f g(x) dx converges. Jaa Ja a (ii) g(x) dx diverges if fix} dx diverges. Jaa Theorem 1.9 (Comparison Test 2) Suppose that /(x) and g(x) are positive functions and let fix} (1.64) lim = L, Q L.gW. Then, the improper integrals /(x) t/x and g(x) ^Zx converge or diverge together. Ja a 1.52 Engineering Mathematics Example 1.38 Discuss the convergence of the following improper integrals 2 (ii) j; dx.... dx (iii) I ----- , r e —X’ dx. ' J2 Inx dx X tan 'X 1. Consider the improper integral e dx. We have r e^ dx= lim f e^dx= lim [1 -e ^] = 1. Therefore, the integral e ' dx is convergent. By Comparison Test 1 (i), the given integral is also convergent. Further, its value is less than 1. 1 X Z X 1 (ii) Let /(x) =.2 and g{x) = -r. (e"'^ + l)x X 1 1.Y 1 Now lim = lim - lim = 1. (g-^ + l)x.2 T x->oo e + 1 f~ dx Also, J g(x)oJe x(3 — x) x(3 - x) e: 1.V I - — lim In — lim In 3 0 3-x 3 0 3-x 1 c e I 3-^ c = — lim In -In - lim In e -In 3 3-r 3-f 3 >0 S 3-c Since the limits do not exist, the improper integral diverges. (v) The integrand has infinite discontinuity at.v 0 which lies inside the interval of integration. We write 1.56 Engineering Mathematics I dx dx > dx -E dx ,. I dx x^ f, — X + j;.2 2 X" = lim g^o J-I — -I- lim X" I ,2 X ’1 1 = lim---- 1 + lim — - 1 £->0 e J. ^->4^ Therefore, the improper integral diverges. (vi) The integrand has infinite discontinuities at x = 1 and x = 2, both of which lie inside the interval of integration. We write dx dx 2 dx 3 dx € x^ - 3x + 2 j;'0 (x-l) (x-2) + J,'1 (x-l)(x-2) + I'2 (x-l)(x-2) - + I2 + Iy We find that (a) the integrand in /, has infinite discontinuity at x = 1, (b) the integrand /(x) in /, has infinite discontinuity at both the end points x = 1 and X = 2. We take any point, say x = c inside the limits of integration, at which/(x) is defined. We also find that /(x) < 0 when 1 X 2. We write g(x) = -f{x) so that g(x) > 0 when 1 X < 2. Therefore, we can write dx ‘2 dx f (x-l)(2-x) I (x-l) (2-x)’ (c) the integrand in 1-^ has infinite discontinuity at x = 2. Hence, we can write fi6c 'S dx = lim lim r - 3x -I- 2 f,^o JO (x-l)(x-2) {;,->()JJIi+­ £- (x-i-l)(2-x) 2-£3 , dx , , ,. lim P dx - lim f + ’c Jc (x—l)(2-x) £4-»oJ2-i-e,I4 (x-l)(x-2) lim In £| -l“ 1 -ln2 c- 1 - lim In ------- -In £2 £|—>0 ei 7 e,-»0 2 -c 1-£2 1-£3 C- 1 1 £4 - lim In -In — + lim In — - In £,->0 £3 2 -c £4 —>0 2 £4 + 1 Since the limits do not exist, the improper integral diverges. Note that the improper integral /, diverges. We could have concluded that the improper integral diverges without discussing the convergence/divergence of 4 and E dx dix dx J,' xdx -1 =J,‘ Xylx~ — 1 + I xJx^ — 1 1 s  Functions of a Real Variable 1.57 lim r £/.V + lim I (ft dx = lim [sec ' x]'i+f + lim [sec 1' x]^ £—»0>'l + £ 2 xdx - 1 = lim [sec ' c - sec '"(I + e)] + lim [sec“' b- sec”' c] = sec * c - sec' 1 1 -L -1 n 1 +----- sec c= — 2 2 Therefore, the improper integral converges to TzU. * dx Example 1.40 Discuss the convergence of the improper integral p>Q. {x — a) p ' Ja ’a Solution The integrand has infinite discontinuity at x = a. Wq write Jx dx 1 1 1 r (^-4 'a ^a lim I £->0 Jo+E ,p 1- lim 1^“'.p-i e l/[(l-72) (6-a)'”'] if ;? < 1 if P 1. For/? = 1, we get 'b dx ct’ dx b-a £ X-a lim I £->0 = lim In X — a £-^0 e =_ CX3 Therefore, the improper integral converges for /? < 1 and diverges for /? > 1. ^n/2 Example 1.41 Show that the improper integral J tan X dx is divergent. Solution The integrand has infinite discontinuity at x = ± ;r/2. We write ’ >n/2 tan X dx = lim f tan xdx-^ lim [ tan X dx '-nl2 e—>0 J-(;r/2)+F e->0 Li- r 1 ( lim |-ln(cosx)l + hm [-In(cosx)J {nl2')+e n ln[cos(c)] = lim In cos------ 1- e £->0 2 Jt ln[cos(c)H. - lim < In coa----- 0, 1 x0 ln(l + /z) = lim (1 +h'}^'^ = 1. Therefore, both the integrals /(x) dx and g(x) dv converge or diverge together. *1 *I >2 dx Now, J'g(x)rfr = Jlr' X dxIn X = £->o lim f Ji+e X In X = Jim [ln(ln.T)]J^^ = lim [ln(ln 2) - ln(ln(l + e))] —> o°. E-^O Since Jl g(x)dx is divergent, the given integral J f {x}dx is also divergent by Comparison Test 4. I sinx sinx 1 (ii) We have /(x) -~i=- since sin x/x is bounded and (sin x/x) < XyJX X X \x 1, 0 0 r7l/2 'TT Z Since. g{x)dx is convergent, the given integral f{x)dx is also convergent by ’0 Jo Comparison Test 3 (i). 2 cos"'.1 Example 1.43 Show that the improper integral — dx converges when /? 1. Jo x" cos"' X 1 Solution We have /(x) 00 = lim P„(,b) - P,(c}\ = - e-‘ P„(c) since lim = 0 for fixed A:. >00 The limit exists and the integral I2 converges for a0. Hence the given improper integral (Eq. (1.66)) converges when a> 0.  Functions ofa Real Variable 1.61 This improper integral is called the Gamma function and is denoted by r(a). Therefore, r(a) = j; x^ ' e ’‘dx. a> 0. (1-67) Some identities of Gamma functions 1. e dx = 1. (1.68) 2. r(a+1) = ar(a). (1.69) Integrating Eq. (1.66) by parts, we get.v“ e ' dx = ~ +a£.« 1 e ' dx = ar(a). If a. is negative and not an integer, then we write r(a) = — r(a + 1). a 3. r(/M + 1) = m\, for any position integer m. (1.70) We have r(w + 1) = /wr(zM) = m(m - 1) r(zM - 1) =... = m{m - 1)... ir( 1) = m\. 4. r(l/2)=V;r. (1.71) have.-1/2 jf e “ dx=^ 2 r JQ e-" du. (setx = u^). We write 2 »o« 2 foo fc* e '' Jv = 4 f f e (ir+v-) r 2 2 r e JO du J 0 Jo Jo du dv. Changing to polar coordinates u = r cos G,v = r sin 0, we obtain e/wt/v = r dr dO and 2 f r 2 =4 J’" fJ1=0 /v J0=O i dr d0 = 2jt Jo re^' dr = -K Io K. Hence, r(i/2) = fn. (In Chapter 2, we shall discuss evaluation of double integrals and change of variables.) 5. r(-\/2) = -24n. (1.72) We have r(a) = [r(a + l)]/a. Substituting a = - 1/2, we get I 2J (-^2). Beta function Consider the irnjoroper integral /= £ x"'-' (1-xy.1’ I dx, 0 /«*-!, 0 n 1. (1.73) 1.62 Engineering Mathematics Note that / is a proper integral for m > 1 and n > 1. The improper integral has points of infinite discontinuity at (i).v = 0, when m 1 and (ii) x = 1, when n 1. When 1 and n 1, we take a number, say c between 0 and 1 and write the improper integral as «;-l (1 iW-l 1= X -x)"-' dx+ - xy dx = '+12 ’0" where r x”'-'(i-x)"-' dx and fy ~.m-l X' dx. h = *0 'C /| is an improper integral, since x = 0 is a point of infinite discontinuity, while A is an improper integral, since x = 1 is a point of infinite discontinuity. We consider these two integrals separately. Convergence at x = 0, 0 < wi < 1, of the integral /j In the integral let /(x) = x"’ '(1 -x)',ZJ-1. and g(x) =x,m-1 Zi-I x^~' Now, lim '7-7 = lim.Z)!-l =1.v^O* X *c dx and is convergent only when 1 - m < 1, or m 0. X Therefore, the improper integral converges when m > 0. Convergence at x = 1, 0 < « < 1, of the integral /2 ,m-l In the integral I2, let f (x) = x' (l-x)' and g(x) = (l-x)" Now, lim —= lim' n-l = 1 X->1 (1-4 f g(x)fi?x= f dx and l-n converges when 1 - n < 1, or n > 0. Jc Jc 'c Therefore, the improper integral A converges when n 0. Combining the two results, we deduce that the given improper integral (Eq. (1.73)) converges when m > 0 and n 0. This improper integral is called the Beta function and is denoted by )3(zm, n). Therefore, p{m, n) = £ x"' '(1 -x)" ' dx, ZM > 0, « > 0. (1-74) Some identities of Beta functions 1. n) = P(n, zzz) ' (1.75) (substitute x = 1 - t in Eq. (1.74) and simplify). „ -;r/2 fZi 7r/2 2. P(in,n) — 2\ sin^"' ' (0) cos^” ' ( 0) 6/0 - 2 [ sin^"”' (0} cos'!2”’-'(0) de. Jo Jo (1.76) (substitute x = sin^ d in Eq. (1.74) and simplify).  Functions of a Real Variable 1.63 ,m-l 3. n) = f X m-¥n dx Jo (substitute x — t/(l + t) in Eq. (1.74) and simplify). 4. Pfn, n) = (1-78) r(/7j + n) We can prove this result using double integrals and change of variables. We have r(m) = J“x”-’e-^oo X Then, /= f ------ dx = p-l (1-T) = l-p) = =r(p)r(i-p). Jo y Jo 1 + X '0 r(i) Hence, the result. Example 1.47 Evaluate the following improper integrals ' —X.2‘ xe dx. (ii) e dx in terms of Gamma functions. Solution (i) Substitute x= 4t. We get dx = dtH24t} and.V- , 1 r irf-L 1= rJo Xe Jx = — 2 JO t 2 Jo 1 4 (ii) Substitute x = ? We get ~ dt and 1 1 1 f“’ 1 = Jo e-* Jx=-| t 3 Jo 3 Jo 3 3 Example 1.48 Using Beta and Gamma functions, evaluate the integral /=J'l (1-?)”*, where n is a positive integer. Solution We have (1 +x)"(l -x)" dx. Let l+x = 2tThen, dx^ldt and 1-x = 2(1-/). We obtain I =22"+' £ t\\-ty'dt=2.^''^' P{n+ 1,m + 1) r(n+l)r(w + l)_2^"-^‘(«!)- r(2rt + 2) (2n + l)!  Functions ofa Real Variable 1.65 Example 1.49 Express dx in terms of Beta function and hence evaluate the integral ]' x“(i - 77)'° dx. Solution Let =y. Then dx = dy. We obtain / = y(m-p+\)lp [(m+1)//,-!] zn+ 1 ,n + 1 p P P Now, comparing the integral f r“ 1 X' (1 - vx ) 6/r with the given integral, we find that *0 m = 312, p=\l2 and n = 1/2. Therefore, ( 3A_2r(5)r(3/2) 3A x^'^(l-y[x) 1/2 d!x = 2j8l5,|l = r(i3/2) 13 119753 10395 Now, r(5) = 4! = 24, r — 2 2 2 2 2 2 2 32 r-. 12 J 2(24)(32)r(3/2)_ 1536 512 Hence, 10395 r(3/2) ” 10395 “ 3465 Example 1.50 Using Beta and Gamma functions, show that for any positive integer m >n >Jt/2 2«i-l (2zzi-2)(2zzz-4)...2 J Sin {0}de= {2m - l)(2zzi-3)...3 (2zw-l)(2zzz--3)... 1 n (ii) £ sin2'"( 0)6/0 = {;2n^{^m-2}...2 '2' Solution From Eq. (1.76), we obtain ‘jtH 2»j-l {G) de ( 1 n r.;r/2 sm and P zn + —, — = 2 >0 sin2'”(0) de. >0 < 22; Jo fZr/2 i>l r(z«)r(i/2) sin- dy=-. X T Example 1.55 Using the result J -X- , slit e dx=-----, evaluate the integral e~^ cost^ax) dx. 2 Solution Let 0 on {a, b). Such a function /(x) is also called a convex function. y y I I X X o a b O a b Fig. 1.22. Concave upward curve. Fig. 1.23. Concave downward curve. Concave downward. A curve defined by y = /(x), a <.x < b, is said to be concave downward if and only if the derivative /'U) is a decreasing function on (a, b). In this case, all points on the curve in the interval (a, b) lie below the tangent to the curve at any point in the interval. This means that, as x increases, the tangent at the point (x, /(x)) turns in the clockwise direction (Fig. 1.23). In terms of the second order derivative, we define that a curve is concave downward if/"(x) 0 on (a, b}. Such a function /(x) is also called a concave function. It is possible that a given curve may be concave upward in part of an interval and concave downward in the other part of the interval. In such cases we define the following: Point of inflection. point PCc, f{c)), a c b, on the curve v f{x) at which the curve changes its concavity from concave upward to concave downward or from concave downward to concave upward is called a point of inflection. Since concavity changes at the point of inflection /’(c, /(c)), f'\c - A) and /"(c + A) must be of opposite signs, when /? > 0 is sufficiently small. Hence, if/"(x) is continuous, then we obtain the condition that at the point of inflection, f"{c} = 0. Then tangent line at the point of inflection, having on one side a part of the curve which is concave downward (or upward) and having on the other side a part of the curve which is concave upward (or downward) must cross the curve at the point of inflection (Fig. 1.24).  Functions of a Real Variable 1.79 X O c Fig. 1.24. Point of inflection. A point at which / "(x) does not exist, but / "(x + A) and f "(x - h) are of opposite signs for sufficiently small A > 0, is also called a point of inflection. Therefore, to find the points of inflection, we first find the points at which /'"(x) = 0. Then, we examine whether the sign of /"(x) changes as x crosses these points. It may be pointed out that /"(x) may also vanish at a point which is not a point of inflection. For example, the curve v = /'(x) =.v"* is always concave upwards as y' = f\x) = 4x^ is always an increasing function. Now, /'"(.y) = 12.y“ = 0 gives.y 0 as a possible point of inflection. But /'"(O + h} and /"(O - A) have the same sign. Hence, x = 0 is not a point of inflection. Example 1.57 Find the intervals in which the following curves (i) y = 3x^ + 4x^ - + 12x + 12, (ii)j^= e , (iii)>^ = (x+ 1) 1/3 are concave upward or concave downward. Also find the points of inflection. Solution (i) We have /(x) = 3x'‘ + 4x^ - 6x- + 12x + 12, f'(x) = 12x2 + 12x2 - 12x + 12, /"(x) = 36x2 + 24x - 12 = 12(3x2 2x - 1) = 12(3x - 1) (x + 1). Setting f"{x) = 0, we get x = 1/3 and x = -1. Now, in the interval (1/3 - h, 1/3) we have f"{x) < 0. The curve is concave downward in this interval. In the interval ((1/3), (1/3) + h) we have /"{x} 0. The curve is concave upward in this interval. We find that./'(1/3) = 419/27. Therefore, the point (1/3, 419/27) is a point of inflection. In the interval (-1 - h, -1), we have > 0. The curve is concave upward is this interval. In the interval (-1,-1 + A), we have f"{x) < 0. The curve is concave downward is this interval. We find that /(-I) = - 7. Therefore, the point (-1, -7) is a point of inflection. We find that the curve is concave upward is the intervals ( — oo^ — 1) and (1/3 , ) and concave downward in the interval (-1, 1/3) (Fig. 1.25i) Concave Concave Concave — oo — oo upward -1 downward 1/3 upward Fig. 125 i. Intervals of concavity for Example 1.57 i. 1.80 Engineering Mathematics (ii) We have /(-y) = e f\x) = -Ixe ' , /"(¥) = (4x"“2) e '. Setting = 0, we get x = ± 1 / 72. Now, in the interval (- (l/TJ) - /?, - l/TI), we have /"(x) > 0. The curve is concave upward in this interval. In the interval ( l/v2 , ( 1/a/2 ) + A), we have f'\x} < 0. The curve is concave downward in this interval. We find that /(—1/72) - l/7e. Therefore, the point (-1/72,1/77)is a point of inflection. In the interval ((1/77 ) - h, Myll ), we have f"{x} < 0. The curve is concave downward in this interval. In the interval (1/77, (1/77 ) + /?), we have f"(x} > 0. The curve is concave upward in this interval. We find that /’(1/77 ) = l/7e. Therefore, the point (l/v2 , 1/77) is a point of inflection. We find that the curve is concave upward in the intervals (- 00 - I/V2 ) and (l/v'^, or y —f+(/) a b y a 1 b- b\x- b- c/' v +b X" b- a~b V'.Y b^ V — + 4 a~ ~ a 1 3 V a~ (7 r (7 r a'\-~ a 1’ 1"^, /(x) —> + oo and as X —> 1 ,f{x'} +oo. Therefore, the curve approaches the same end of the asymptote x = 1. (Fig. 1.30) o.v= 1 Fig. 1.30. Vertical asymptote ofy = l/(x- 1)^. Horizontal asymptotes Assume that the given equation g(x, y) = 0 can be solved for x as 1.98 Engineering Mathematics p(t) (1.138) X = h{y} = Let there exist a number b such that lim Zz(y) = ± ©o. (1.139) Then, the line y b is called a horizontal asymptote to the curve. This implies that y Z) is a root of q{y) = 0. Consider the example given in Eq. (1.136). Solving for x, we get ,1/3 1 = J- +1 1/3 or. = 1 ,1'3 1/3 = h(y\ T V V As y 0^, /z(y) —> OO and as y —> 0 , h{y) - (note that y’^^ is defined for j 0). Therefore, the curve approaches the opposite ends of the horizontal asymptote y - 0. Hence, the x-axis or y — 0 is an horizontal asymptote of the curve drawn in Fig. 1.29. Consider the example given in Eq. (1.137). Solving forx, we get 1 1 = ±1 X- 1 = ± or X = i ± —777 = T2 = h(y)- y'^2 ,1/2 T T We note that /z(v) is not defined for y 0. As V —> 0^, A(j’) +oo. Therefore, the curve approaches the opposite ends of horizontal asymptote y = 0. Hence, the x-axis or y = 0 is an horizontal asymptote of the curve drawn in Fig. 1.30. We can give an alternate method for finding an horizontal asymptote. Let the curve be given as y = /(x) = jp(x)/q'(x). Let there exist a number b such that lim f{x} = b. (1.140) Then, the line v b is an horizontal asymptote of the curve. We can verify that v = 0 is an horizontal asymptote of the curves defined in the examples given in equations (1.136) and (1.137). Inclined or oblique asymptotes Assume that the given equation g(x, y) = 0 can be solved for y as y = fix). Let ‘ y = ax + b (1.141) be the equation of the inclined asymptote L. Let P{x, y) be any point on the curve. Draw the lines PS and PR perpendicular to the x-axis and the line L respectively. Then, the coordinates of Q are V|) (Figs. 1.31a, b). Let the asymptote L make an angle 0 k/I) with the positive direction of x-axis. From the triangles TQS and QPR, obtain ^QTS=AQPR=^ e. (1.142)  Functions of a Real Variable 1.99 V Z, 0 /? / ecv v,) /{e o s O T S.r Fig. 1.31(a). Inclined asymptote. Fig. 1.31(b). Inclined asymptote. Using the definition, we have that the line L will be an asymptote if lim PR = ii. (1.143) Now, PR - PQ cos d. Since 0 is a known angle, we obtain lim PR = lim PQ cos 0=0 or lim PQ = 0. (1.144) Now, = |P5 - 051 = -y,| = |/(x) - (ax + d)| since P is a point on the curve y =flx'} and y, is a point on the line L, y = ax + b. Therefore, Eq. (1.144) gives lim PQ = lim |/(x) - (ax + b) | = 0 (1.145) b or lim X -a-----= 0. (1.146) X Now, as X oo, {b/x} = 0. If the limit on the left hand side of (1.146) exists and equals 0, then we must have /(x) /(x) lim -a = 0, or a = lim (1.147) X X Now, using Eq. (1.145), we obtain b = lim [/(x) - ) | = 0. 1.100 Engineering Mathematics In this case, we have again a = lim and b = lim [/(x) - ax]. (1.149) X Alternative Consider the case, when y =/(jf) (1.150) and p{x\ q{x} are polynomials in x. Let the degree of the polynomial p{x} be one unit greater than the degree of the polynomial qix). Then, dividing p{x') by q{x}, we get /(x) = (ax + b) + (1.151) ^(x) where the degree of x in r(x) is less than the degree of x in q{x}. x [r(x)/(7(x)] —> 0 and the distance between the curve y = /(x) and the line y = ax + b, goes to zero. Hence, y = ax + 6 is an inclined asymptote.. For example, consider the curve 3x2 2x Now, as X 0"^, y —> + oo and as x ->0 , y —> - oo, Therefore, x = 0 or j-axis is a vertical asymptote. Now, write y as 1 5 y = — 3x + 1 H—. 1 X Now, (5/x) —> 0 as X —> ± °o. Hence, y = (3x + l)/2 is an inclined asymptote of the curve. Remark 9 1. The graph of a given curve may intersect an horizontal or an inclined asymptote (Fig. 1.33). 2. If in Eq. (1.150), the degree of the polynomial y(x) is two or more units higher than the degree of the polynomial q{x), then there are no horizontal or inclined asymptotes. Example 1.68 Find the asymptotes of the curve (2x + 3)y= (x- 1)2, Solution Write the given equation as (X-1)2 v= 2x + 3 2[x + (3/2)] As X —> (—3/2) , y —> + oo and as X —> (-3/2) , y —> OO Therefore, x = - 3/2 xs a. vertical asymptote.  Functions of a Real Variable 1.93 -12'3 -|2/3 1 a‘^a 1 b^P + —T = 1 a^ b- a^ -b- a2>2, 2/3 + = (,,2 - QT a Hence, the evolute of the given curve is a 2/3 + b^^^ (iii) From x = a 3, y = a sin^ 0, we get dx = - 3a cos^ 6 sin 6, = 3a sin^ 0 cos 3 73 d3 dy _ dy dO _ 3a sin“ 3 cos 3 y' == = - tan 3. dx dxld3 3a cos^ 3 sin 0 d3 1 r = - sec 2 a0 — = dx 3a sin0 cos'* 3 J'' Hence, a = X - ~ [1 + cos ©(cos^ 0-1-3 sin^ 3), r P=y [1 -t- y)^] = a sin 0(sin^ 0-1-3 cos^ 0). y Now, a + P = a (cos^ 3 -l- 3sin^ 0 cos 3 -i- 3cos2 0 sin 0 -i- sin^ 3) = a {cos 3 -F sin 3)^ a - P = a (cos^ 3 - 3cos^ 0 sin 0 4- 3cos 3 sin^ 0 - sin^ 3) = a (cos- 3 sin 3f. Hence, (a -H -i- (a - = a':2^^[(cos 0 -I- sin 3)^ -I- (cos 0 - sin 0)^] 2a- (cos^ 0 -I- sin^ 3} 2a^'^. Therefore, the locus of (a, p) is (^+ Y}^'^ + (X- = 2a^'^ Example 1.64 Show that the normal to a given curve is a tangent to its evolute. Solution The coordinates of the centre of curvature are «='i-4 P=y+ -I [> +()'')’)] T J. These equations are the parametric equations of the evolute, where.v is a parameter. We note that CP (Fig. 1.27) is normal to the given curve at the point P. Differentiating or. f with respect to X, we get 1,94 Engineering Mathematics da _ 1 ,"'12 dx (Z)- 1 [O'Y-y"\y" + 3y)y} + + (vO'ly"'] (/'); y [3/0")’ {1 +(yy (.V ) dp 1 — =y' + [y'\2y'y")-{i+(y')^}y"'] (iv (r")" 1 [y'(y")~ + 2y'(y"f {1 + Cy')‘}y"'] (v ) 1 [3/(y'V - (1 + (yV) y'"] (Z) dp _ dp/dx _ 1 dP Hence, or v' = - 1. da da/dx da Now, dpida is the slope of the tangent to the evolute and v' is the slope of the tangent to the given curve at P. Therefore, the tangent to the evolute and the tangent to the given curve are mutually perpendicular, that is, the normal to a given curve is a tangent to its evolute. 1.5.4 Envelope of a Family of Curves Consider the equation of a family of curves in the form y = g(x, p), or as, f{x, y, p) = (1.128) where is a parameter. We say that Eq. (1.128) represents a one parameter family of curves. For different values of /?, we obtain different members of the family of curves. If each member of an infinite family (one parameter family) of curves is tangent to a certain curve C, and if at each point of the curve C, at least one member of the family is tangent, then the curve C is called the envelope of the given one parameter family of curves. Now, to every value to p in Eq. (1.128), there corresponds a point on the envelope where