Differential Calculus [20 ed.] 9788121904711

Table of contents :
00 Preface
01 Real Numbers
02 Functions And Graphs-Elementary Functions
03 Continuity And Limit
04 Differentiation
05 Successive Differentiation
06 Expansion Of Functions
07 Tangents And Normals
08 Mean Value Theorems
09 Maxima And Minima
10 Indeterminate Forms
11 Partial Differentiation
12 Jacobians
13 Concavity And Points Of Inflexion
14 Curvature And Evolutes
15 Asymptotes
16 Singular Points
17 Curve Tracing
18 Envelopes
19 Change Of Independent Variables

Citation preview

DIFFERENTIAL CALCULUS For B.A. and B.Sc. Students

SHANTI NARAYAN Formerly Dean of Colleges University of Delhi, and Principal Hansraj College, Delhi

Revised by

Dr. P.K. MITTAL M.Sc., Ph.D.

Head Department of Mathematics Government Post Graduate College, Rishikesh Uttarakhand

SHYAMLAL CHARITABLE TRUST RAM NAGAR, NEW DELHI - 110 055

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© S Chand And Company Limited, 1942 All rights reserved. No part of this publication may be reproduced or copied in any material form (including photocopying or storing it in any medium in form of graphics, electronic or mechanical means and whether or not transient or incidental to some other use of this publication) without written permission of the copyright owner. Any breach of this will entail legal action and prosecution without further notice. Jurisdiction: All disputes with respect to this publication shall be subject to the jurisdiction of the Courts, Tribunals and Forums of New Delhi, India only. First Edition 1942 Subsequent Editions and Reprints 1974, 75, 77, 78, 79, 80, 81, 82, 83, 84, 85 (Twice), 86, 87, 88, 90, 91, 92, 93, 94, 96, 97, 98, 99, 2000, 2001, 2002, 2003, 2004, 2005 (Twice), 2007 (Twice), 2008 (Thrice), 2009 (Thrice), 2010, 2011, 2012, 2013, 2014 (Twice), 2016 (Thrice), 2017, 2018, 2019 (Twice) Reprint 2020

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PREFACE TO THE FIFTEENTH EDITION This book originally written about 62 year ago, has during the intervening period, been revised and reprinted several times. Since its last revision, tremendous changes in the trend of studies have taken place which necessiated thorough revision of the book. But, it was really a very difficult task to revise this perfect well written book of late Dr. Shanti Narayan. Since a long time undoubtedly, it was a book loved by teachers and students alike. I took up the task to revise the book with great devotion to meet the rapidly changing demands of students interested in self-study and appearing in different competitive examination. Keeping this in view, a large variety of illustrative solved examples have been included in every chapter. Keeping in mind the syllabi of different Universities, specially that introduced by U.G.C. certain topics have been included which were not present in the original text. This book, in the present form, is a humble effort to make it more useful to students and teachers. I owe my special gratitude to Shri Ravindra Kumar Gupta, Managing Director and Sri Himanshu Gupta, Director of S. Chand & Co. Ltd., for giving me opportunity to revise the books of late Dr. Shanti Narayan, an eminent Indian Mathematician. I would like to express, heart felt gratitude to my friend Mr. Naveen Joshi, General Manager (Sales & Marketing) of S. Chand & Co. Ltd. without whose continuous encouragement and persuation the present book may have not reached to your hands. My sincere thanks are also due to Mr. Shishir Bhatnagar of S. Chand & Co. Ltd. The final tribute and greatest appreciation, however, is reserved for my wife Prabha, without whose persuation, cooperation and timely help, this book would not have seen the light of the day. As the need for improvement is never ending, I will look forward to receive valuable and useful suggestions from our users for further improvement in the book. — Dr. P.K. MITTAL

Disclaimer : While the authors of this book have made every effort to avoid any mistake or omission and have used their skill, expertise and knowledge to the best of their capacity to provide accurate and updated information. The authors and S. Chand does not give any representation or warranty with respect to the accuracy or completeness of the contents of this publication and are selling this publication on the condition and understanding that they shall not be made liable in any manner whatsoever. S.Chand and the authors expressly disclaim all and any liability/responsibility to any person, whether a purchaser or reader of this publication or not, in respect of anything and everything forming part of the contents of this publication. S. Chand shall not be responsible for any errors, omissions or damages arising out of the use of the information contained in this publication. Further, the appearance of the personal name, location, place and incidence, if any; in the illustrations used herein is purely coincidental and work of imagination. Thus the same should in no manner be termed as defamatory to any individual.

(iii)

PREFACE TO THE FIRST EDITION This book is meant for students preparing for the B.A. and B.Sc. examinations of our universities. Some topics of the Honours standard have also been included. They are given in the form of appendices to the relevant chapters. The treatment of the subject is rigorous but no attempt has been made to state and prove the theorems in generalised forms and under less respective conditions as in the case with the modern Theory of Functions. It has also been a constant endeavour of the author to see that the subject is not presented just as a body of formulae. This is to see that the student does not form an unfortunate impression that the study of Calculus consists only in acquiring a skill to manipulate some formulae through constant drilling. The book opens with a brief outline of the development of Real Numbers, their expression as infinite decimals and their representation by points along a line. This is followed by a discussion of the graphs of the elementary functions xn, log x, ex, sin x, sin–1 x etc. Some of the difficulties of inverse functions have also been illustrated by precise formulation of Inverse trigonometrical functions. It is suggested that the teacher in the class needs to refer only a few salient points of this part of the book. The student should, on his part, go through the same in complete details to acquire a sound grasp of the basics of the subject. This part is so present that a student would have no difficulty in an independent study of the same. While the first part of the book is analytical in character, the latter part deals with the geometrical applications of the subject. But this order of the subject is by no means suggested to be rigidly followed in the class. A different order may usefully be adopted at the direction of the teacher. An analysis of the ‘Layman’s’ concept has frequently been made to serve basis for the precise formulation of the corresponding ‘Scientist’s’ concepts. This specially relates to the two concepts of Continuity and Curvature. Geometrical interpretation of results analytically obtained have been given for the benefit of the students. A chapter on ‘Some Important Curves’, has been given before dealing with geometrical applications. This will enable the student to get familiar with the names and shapes of some of the important curves. It is felt that a student will have better understanding of the properties of a curve if he knows how the curve looks like. This chapter will also serve as a useful introduction to the subject of Double points of a curve. Asymptote of a curve has been defined as a line so that the distance of any point on the curve from this line tends to zero, as the point tends to infinity along the curve. It is believed that, of all the definitions of an asymptote, this is the one which is most natural. It embodies the idea to which the concept of asymptotes owes its importance. Moreover, the definition gives rise to a simple method for determining the asymptotes. The various principles and methods have been profusely illustrated by means of a large number of solved examples. (v)

I am indebted to Prof. Sita Ram Gupta, M.A., P.E.S., formerly of the Government College, Lahore who very kindly went through the manuscript and made a number of suggestions. My thanks are also due to my old pupils and friends, Professors Jagan Nath M.A., Vidya Sagar M.A., and Om Parkash M.A., for the help which they rendered in preparing this book. Suggestions for improvement will be thankfully acknowledged. —SHANTI NARAYAN

CONTENTS CHAPTER 1 : Real Numbers 1.1. 1.2. 1.3. 1.4. 1.5.

Introduction Rational numbers and their representation by points along a straight line Irrational numbers Decimal representation of real numbers The modulus or a real number CHAPTER 2 : Functions and Graphs–Elementary Functions

2.1. Functions: domain and range of a function 2.2. Graphs of functions 2.3. Operations on functions 2.4. One-one functions 2.5. Function of a function : Composition of functions 2.6. The function y = x1/n; n being any integer 2.7. Trigonometric and inverse trigonometric functions 2.8. Implicitly defined functions 2.9. Some special functions 2.10. Polar co-ordinates and graphs CHAPTER 3 : Continuity and Limit 3.1. 3.2. 3.3. 3.4. 3.5. 3.6. 3.7. 3.8.

Introduction Limit Theorems on limits Extension of operations on limits Some useful limits Another form of the definition of continuity Some properties of continuous functions Discontinuity CHAPTER 4 : Differentiation

4.1. Introduction 4.2. Derivative of a constant function 4.3. Some general theorems on derivation 4.4. Derivatives of trigonometric functions 4.5. Derivatives of inverse trigonometric functions 4.6. Derivative of y = logax 4.7. Hyperbolic functions 4.8. Derivation of parametrically defined functions 4.9. Logarithmic differentiation 4.10. Derivation of implicity defined functions (vii)

1–10 1 2 4 6 8 11–54 11 14 18 19 24 26 29 36 37 51 55–108 55 61 66 69 76 92 93 95 109–165 109 127 128 134 139 141 142 146 147 150

4.11. Preliminary transformation 4.12. Differentitation of a determinant

154 158

CHAPTER 5 : Successive Differentiation 5.1. 5.2. 5.3. 5.4. 5.5.

Higher order derivatives Calculation of the nth derivative. Some standard results Determination of nth derivative of rational functions The nth derivatives of the products of the powers of sines and cosines Leibnitz’s theorem. The nth derivative of the product of two functions CHAPTER 6 : Expansion of Functions

6.1. Maclaurin’s theorem 6.2. Taylor’s theorem

166 172 173 177 179 191–205 191 200

CHAPTER 7 : Tangents and Normals 7.1 7.2. 7.3. 7.4.

Introduction Equations of the tangent and normal Angle of intersection of two curves Length of the tangent, normal, sub-tangent and sub-normal at any point of a curve 7.5. Pedal equations of p, r equations 7.6. Angle between radius vector and tangent 7.7. Length of the perpendicular from pole to the tangent 7.8. Length of polar sub-tangent and polar sub-normal 7.9. Pedal equations 7.10. Derivative of Arcs 7.11. Differential coefficient of length of Arc 7.12. Derivative of Arc length (polar formula) 7.13. Other formulae CHAPTER 8 : Mean Value Theorems 8.1. 8.2. 8.3. 8.4. 8.5. 8.6. 8.7.

166–190

Rolle’s theorem Lagrange’s mean value theorem Meaning of the sign of derivative Graphs of hyperbolic functions Cauchy’s mean value theorem Higher derivatives Formal expansions of functions

206–246 206 206 219 221 225 227 231 231 232 237 237 238 239 247–280 247 253 257 258 267 270 275

CHAPTER 9 : Maxima and Minima 9.1. Maximum value of a function; minimum value of a function 9.2. A necessary condition for extreme values 9.3. Sufficient condition for extreme value (viii)

281–335 281 282 285

9.4. 9.5. 9.6. 9.7.

Use of second order derivatives Application to problems Maxima and minima of function of two variables Lagrange’s method of undetermined multipliers CHAPTER 10 : Indeterminate Forms

10.1. Indeterminate forms 10.2. The indeterminate form 0/0 10.3. The indeterminate from  /  10.4. The indeterminate from 0.  10.5. The indeterminate form  –  10.6. The indeterminate forms 0°, 1  , 

286 293 310 320 336–355 336 337 341 344 345 346

CHAPTER 11 : Partial Differentiation 11.1. Introduction 11.2. Functions of two variables 11.3. Neighbourhood of a point (a, b) 11.4. Continuity of a Function of two variables, continuity at a point 11.5. Limit of a function of two variables 11.6. Partial derivatives 11.7. Geometrical representation of a function of two variables 11.8. Homogeneous functions 11.9. Theorem on total differentials; composite fucntions; differentiation of composite functions; implicit functions 11.10. Equality of fxy (a, b) and fyx (a, b) 11.11. Taylor’s theorem for a function of two variables CHAPTER 12 : Jacobians 12.1. Definition 12.2. Jacobian of function of function 12.3. Jacobian of implicit functions 12.4. Theorem

356–395 356 356 357 357 358 359 360 369 376 386 388 396–413 396 400 401 409

CHAPTER 13 : Concavity and Points of Inflexion 13.1. Introduction 13.2. Criteria for (i) Concavity upwards, (ii) Concavity downwards (iii) Inflexion at a given point CHAPTER 14 : Curvature and Evolutes 14.1. Introduction. Definition of curvature 14.2. Length of arc as a function. Derivative of arc 14.3. Radius of curvature-cartesian equations 14.4. Newtonian method 14.5. Centre of curvature (ix)

414–420 414 415 421–455 421 423 426 439 441

14.6. Chord of curvature 14.7. Evolutes and involutes 14.8. Properties of the evolute

443 448 452

CHAPTER 15 : Asymptotes 15.1. Definition 15.2. Determination of asymptotes 15.3. Working rules of determining asymptotes 15.4. Asymptotes by inspection 15.5. Intersection of a curve and its asymptotes 15.6. Asymptotes by expansion 15.7. Position of a curve with respect to an asymptote 15.8. Asymptotes in polar co-ordinates CHAPTER 16 : Singular Points 16.1. Introduction. Cusps, nodes and conjugate points 16.2. Definitions 16.3. Tangents at the origin 16.4. Conditions of any point (x, y) to be a multiple point of the curve f(x, y) = 0 16.5. Types of cusps 16.6. Radii of curvature at multiple points CHAPTER 17 : Curve Tracing 17.1. Introduciton 17.2. Procedure for tracing cartesian equations 17.3. Equations of the form y = f (x) 17.4. Equations of the form y2 = f(x) 17.5. Parametric equations 17.6. Tracing of polar curves CHAPTER 18 : Envelopes 18.1. 18.2. 18.3. 18.4. 18.5. 18.6.

One parameter family of curves Consider the family fo straight lines Definition Determination of envelope Theorem To prove that, in general, the envelope of a family of curves touches each member of the family 18.7. If A,B,C are functions of x and y and m is a parameter then the envelope of Am2 + Bm + C = 0 is B2 = 4AC 18.8. Two parameters connected by a relation 18.9. When the equation to a family of curves is not given, but the law is given in accordance with which any member of the family can be determined 18.10. Envelopes of polar curves 18.11. Envelopes of normals (Evolutes)

(x)

456–479 456 456 458 467 468 473 474 476 480–496 480 481 481 484 491 493 497–522 497 497 498 502 510 513 523–542 523 523 524 524 525 526 529 531 534 536 538

CHAPTER 19 : Change of Independent Variables

543–572

19.1. To change the single independent variable into the dependent variable 19.2. To change the independent variable x into another variable t, where x = f(t) 19.3. Case of single independent variable 19.4. Transformation from cartesian to polars and vice-versa 19.5. x and y to be expressed in terms of some third variable 19.6. Two independent variables 19.7. Transformation from cartesian to the polar 19.8. Transformation of

 2V x 2

and

 2V y 2

to polars

543 543 544 549 549 553 554 555

19.9. Transformation of  2V

556

19.10. Orthogonal transformation of  2V

568

(xi)

CHAPTER 1 REAL NUMBERS 1.1. Introduction The subject of Differential Calculus takes its stand upon the set of real numbers and concerns itself with the various properties of the same. It specially introduces and deals with what is called the Limit operation, in addition to being concerned with the Algebraic operations of Addition and Multiplication and their Inverses, Subtraction and Division. The subject is a development of the important notion of Instantaneous rate of change, which is itself a limit idea, as distinguished from the average rate of change and as such, it finds applications to all those branches of knowledge which deal with the same. Thus, it is applied to Geometry, Mechanics and other branches of Theoretical Physics and also to Social Sciences, such as Economics and Psychology. It may be noted here that this application is essentially based on the notion of measurement, whereby we employ real numbers to measure the particular entity which is the object of investigation in any particular department of knowledge. In Mechanics, for instance, we are concerned with the notion of time and, therefore, in the application of Calculus to Mechanics the first step is to correlate the two notions of Time and Real Numbers, i.e. to measure time in terms of real numbers. Similar is the case with other notions such as amount of Heat, Intensity of Light, Force, Demand, Intelligence, etc. The formulation of an entity in terms of real numbers, i.e., measurement must, of course take note of the properties which we intuitively associate with the same. This remark will later on be illustrated with reference to the concepts of Velocity, Acceleration, Curvature, etc. The importance of real numbers for the study of the subject in hand being thus clear, we will, in some of the following articles, see how starting from the set of natural numbers, we arrive at the set of real numbers. It is, however, not intended to give here any logically connected account of the development of the set of real numbers and only a very brief reference to some well-known salient facts will suffice for our purpose. An account of the logical development of number systems is given in the book “Number Systems” by the author. It may also be mentioned here that even though it satisfies a deep philosophical need to base the theory part of Calculus on the notion of number alone to the entire exclusion of every physical basis, but a rigid insistence on the same is not within the scope of the present book and intuitive geometrical notions of Point, Distance etc. will often be applied to for securing simplicity in the context of aims of this book. 1

2

DIFFERENTIAL CALCULUS

1.2. Rational Numbers and their Representation by Points along a Straight Line 1.2.1. Natural Numbers. It was the numbers, 1, 2, 3, 4 etc. that we were first introduced through the process of counting certain sets of objects. The totality of these numbers is known as the set of Natural numbers. While the operations of addition and multiplication are unrestrictedly possible in relation to the set of natural numbers, this is not the case in respect of the inverse operations of subtraction and division. Thus, for example, the symbols 2–3, 2 ÷ 3 are meaningless insofar as the set of natural numbers is concerned. Natural numbers are also referred to as positive integers. The set {1, 2, 3, 4, ......} of natural numbers will be denoted as N. 1 3  1.2.2. Fractional Numbers. At a later stage, another set of numbers like p / q,  e.g ., ,  2 4  where p and q are numbers, was adjoined to the set N of natural numbers. This new set is known as the set of Fractions and it obviously included natural numbers as a sub-set ; q being equal to 1 in this case. The introduction of Fractional numbers is motivated from an abstract point of view, to render Division unrestrictedly possible and from the concrete point of view, to render numbers serviceable for practical needs of measurement also in addition to counting. 1.2.3. Rational Numbers. Still later the set of fractions was enlarged by incorporation in it the set of negative fractions and zero. The entire set of these numbers is known as the set of Rational numbers and is denoted as Q. The introduction of negative numbers is motivated from an abstract point of view, to render subtraction always possible and, from a concrete point of view, to facilitate a unified treatment of oppositely directed pairs of entities, such as gain and loss ; rise and fall, etc. The natural numbers 1, 2, 3, 4, ...... are called positive integers and the numbers –1, –2, ....... are called negative integers. The set {0, –1, 1, –2, 2, –3, 3 ........}, called the set of Integers, is denoted as I. Every non-zero rational number is expressible in the form p/q where p and q are any two integers and q is not zero. 1.2.4. Fundamental operations on rational numbers. An important property of the set of rational numbers is that the operations of addition, multiplication, subtraction and division can be performed upon any two such numbers (with one exception viz., Division by zero which is considered below in (§. 1.2.5) and the number obtained as the result of these operations is again a rational number. This property is expressed by saying that the set of rational numbers is closed with respect to the four fundamental operations (division by zero being excluded). 1.2.5. Meaningless operation of division by zero. It is important to note that the only exception to the above property is ‘Division by zero’ which is a meaningless operation. This may be seen as follows :To divide a by b amounts to determining a number c such that bc = a, and the division will be intelligible, if and only if, the determination of c is uniquely possible. Now, there is no number which when multiplied by zero produces a number other than zero and as such a/0 is no number when a  0. Also any number when multiplied by zero produces zero so that 0/0 may be any number. On account of this Impossibility in one case and Indefiniteness in the other, the operation of division by zero must be always avoided. A disregard of this exception often leads to absurd results as is illustrated below in (i). (i) Let x = 6, we have x = 6  x2 – 36 = x – 6  (x – 6) (x + 6) = x – 6.

REAL NUMBERS

3

Dividing both sides by (x – 6) , we get x + 6 = 1 6 + 6 = 1 12 = 1 which is clearly absurd. Division by x – 6, which is zero, here, is responsible for this absurd conclusion. (ii) We may also remark in this connection that

x 2 – 36 ( x – 6) ( x  6)   x  6, only when x  6 ..(i) x–6 ( x – 6) 2 For x = 6, the left hand expression (x – 36)/(x – 6) is meaningless whereas the right hand expression x + 6, is equal to 12 so that equality ceases to hold for x = 6. The equality (i) above is obtained on dividing the numerator and denominator of the expression (x2 – 36)/(x – 6) by (x – 6) and this operation of division is possible only when the divisor (x – 6)  0, i.e., when x  6. This explains the restricted character of the equality (i). Ex. 1. Show by examples that the set of natural numbers is not closed with respect to the operations of subtraction and division. Also show that the set of positive fractions is not closed with respect to the operation of subtraction. Ex. 2. Show that every rational number is expressible as a terminating or a recurring decimal. To decimalise p/q, we have first to divide p by q and then each remainder, after multiplicaton with 10, is to be divided by q to obtain the successive figures in the decimal expression of p/q. The decimal expression will be terminating if at some stage, the remainder vanishes. Otherwise, the process will be unending. In the latter case, the remainder will always be one of the finite set of numbers, 1, 2, ........., q – 1 and so must repeat itself at some stage. From this stage onward, the quotients will also repeat themselves and the decimal expression will, therefore, be recurring. For a proper understanding of the argument, the student may actually express some fractional 3 3 31 , in decimal notation. numbers, say , , 7 13 123 Ex. 3. For what values of x are the following equalities not valid : (i) (iii)

3.

(i) 0,

x  1, x 1– x

1–

x

1

(ii) a,

x,

(iii) 1,

(ii)

x2 – a2  x  a, x–a

(iv)

1 – cos x x  tan . sin x 2

ANSWERS (iv) n:I

1.2.6. Representation of rational numbers by points along line or by segments of a line. The mode of representing rational numbers by points along a line or by segments of a line, which may be known as number line will now be explained. We start with marking an arbitrary point O on the line and calling it the Origin or zero point. The number zero will be represented by the point O. The point O divides the line into two parts or sides. Any one of these may be called positive and the other, then called negative. Usually, the line is drawn

Fig. 1.1

parallel to the printed lines of the page and the right hand side of O is termed positive and the left hand side of O is negative.

4

DIFFERENTIAL CALCULUS

On the positive side, we take an arbitrary length OA, and call it the unit length. We say that the number 1 is represented by the point A. After having fixed an origin, Positive sense and a unit length on the number line in the manner indicated here, we are in a position to determine a point representing any given rational number as explained below : Positive Integers. Firstly we consider any positive integer m. We take a point on the positive side of the line such that its distance from O is m times the unit length OA. This point will be reached by measuring successively m steps each equal to OA starting from O. This point, then, is said to represent the positive integer m. Negative Integers. To represent a negative integer, – m, we take point on the negative side of O such that its distance from O is m times the unit length OA. This point represents the negative integer, – m. Rational Numbers. Finally, let p/q be any rational number; q being a positive integer. Let OA be divided into q equal parts; OB being one of them. We take a point on the positive or negative side of O according as p is positive or negative such that its distance from O is p times (or, – p times if p is negative) the distance OB. The point so obtained represents the rational number p/q.

1.3. Irrational Numbers We have seen in the last article that every rational number can be represented by a point of a line. Also, it is easy, to see that we can cover the line with such points as closely as we like. The natural question now arises, “Is the converse true ?” “Is it possible to assign a rational number to every point on the number line ?” A simple consideration, as detailed below, will show that it is not so. Construct a square with one of its sides as OA of unit length and take a point P on the line such that OP is equal in length to the diagonal of this square. It will now be shown that the point P cannot correspond to a rational number, i.e., the length of OP cannot have a rational number as its measure.

Fig. 1.2

If possible, let its measure be a rational number p/q so that by the Pythagoras’ theorem, we have (p/q)2 = 12 + 12 = 2  p2 = 2q2. We suppose that the natural numbers p and q have no common factor, for such factors, if any, can be cancelled to begin with. Firstly we notice that (2n)2 = 4n2,

(2n + 1)2 = (4n2 + 4n) + 1

so that the square of an even number is even and that of an odd number is odd. From the equation (i), we see that p2 is an even number. Therefore, p itself must be even. Let then, p be equal to 2n where n is an integer. We have 4n2 = 2q2  q2 = 2n2  q2 is even  q is even. Hence, p and q have a common factor 2 and this conclusion contradicts the hypothesis that they have no common factor. Thus, the measure 2 of OP is not a rational number. There exists, therefore, a point on the number axis not corresponding to any rational number. Again, we take a point L on the line such that the length OL is any rational multiple say, p/q, of OP.

REAL NUMBERS

5

The length OL cannot have rational measure. For if possible, let m/n be the measure of OL so that, p m mq 2   2  q n np which states that 2 is a rational number, being equal to mq/np. Thus we arrive at contradiction. Hence L cannot correspond to a rational number. Thus we see that there exist an unlimited number of points on the number line which do not correspond to any rational number. If we now require that our set of numbers should be such that after the choice of unit length on the line, every point of the line should have a number corresponding to it (or that every length should be capable of measurement), we are obliged to extend our system of numbers further by the introduction of Irrational numbers. We will thus associate an irrational number to every point of the line which does not correspond to a rational number. Real number. A number, rational or irrational, is called a real number. The set of rational and irrational numbers is thus the set of real numbers. This set is denoted as R. Each real number is represented by some point of the number axis and each point of the number axis has some real number, rational or irrational, corresponding to it. The positive real numbers correspond to points on the right of O and the negative real numbers to points on the left of O. We might say that each positive real number is the measure of some length OP and that the set of positive real numbers is adequate to measure every length. A method of representing irrational numbers in the decimal notation is given in Article 1.4. 1.3.1. Number and point. If any real number, say x, is represented by a point P, then we usually say that the point P is x. Thus, the terms, number and point, are generally used in an indistinguishable manner. In the following, unless otherwise stated by a number will be meant a real number. 1.3.2. Closed and open intervals. Let a, b be two given numbers such that a < b. Then the set of numbers x such that axb is called a closed interval denoted by the symbols [a, b]. We generally describe this situation as follows : [a, b] = {x : a  x  b}, so that the closed interval denoted by [a, b] is the set of numbers x satisfying the condition a  x  b. Thus x  [a, b]  a  x  b. The set of numbers x such that a 2. We find that (1.2)3 = 1.728 < 2 and (1.3)3 = 2.197 > 2  1.2  Again consider the numbers

3

2  1.3.

1  2, 1  21, 1  22,........., 1  29, 1  3 which divide the interval [1.2. 1.3] into 10 equal parts. We find that (1.25)3 = 1.953125 < 2 and (1.26)3 = 2.000376 > 2  1.25  3 2  1.26 Again, consider the numbers 1.25, 1.251, 1.252, ........., 1.259, 1.26 which divide the interval, [1.25, 1.26] into 10 equal parts. We find that (1.259)3 = 1.995616979 < 2 and

(1.26)3 = 2.000376 > 2



1.259  3 2  1.26.

Hence

3

2  1.259..........

8

DIFFERENTIAL CALCULUS

Thus, to three decimal places, we have 3

2  1.259.

Ex. 2. Calculate the cube root of 5 to 2 decimal places. Ex. 3. Give any four different irrational numbers in decimal notation and arrange them in order of magnitude. Ex. 4. Consider the following two irrational numbers 0  1020102201022201022220102222201 .........., 0  120121201212120........... which of these two numbers is greater than the other. Insert any three irrational numbers between them. Ex. 5. Consider the following numbers : x =  430340430340430 y =  234223422234222234. Show that .6 < x + y < .8, .66 < x + y < .68, .08 < xy < .15, .09 < xy < .1. Note. The method described above in Ex. 1, which is indeed very cumbersome has only been given to illustrate the basic and elemenrary nature of the problem. In actual practice, however, other methods involving infinite series or other limiting processes are employed. ANSWERS 2.

1.7104,

4. Second > first.

1.5. The Modulus of a Real Number Let x be any given real number. We have the following possibilities : x > 0,

x < 0, x = 0.  x if  0, |x| We write  – x if x  0, and call | x | the Modulus or the Absolute value of the real number x. Thus | 3 | = 3, |– 3 | = – (–3) = 3, | 0 | = 0, | 5 – 7 | = | 7 – 5 | = 2. The modulus of the difference between two numbers is the measure of the distance between the corresponding points on the number line. It would be seen that  x  R, we have | x |  0, | x |  x,  x |  – x. In fact x | = max {x, – x}, where max {x, – x} denotes the greater of the two numbers x and – x. We may also see that | x | denotes the positive square root of x2 so that we have

| x |  ( x 2 ). Some results involving moduli. We now state some simple and useful results involving the moduli of numbers.

REAL NUMBERS

1.5.1.

9

|a+b||a|+|b|

i.e., the modulus of the sum of two numbers is less than or equal to the sum of their moduli. We have a  | a |, b  | b |  a + b  | a | + | b |. Also –a | a |, –b  | b |  – (a + b)  | a | + | b | Now | a + b | = max {(a + b), – (a + b)}. Thus | a + b |  | a | + | b |. In fact, we have | a + b | = | a | + | b |, if a, b have the same sign, and | a + b | < | a | + | b |, if a, b have opposite signs. For example, |7+3|=|7|+|3| |–7–3|=|–7|+|–3| 4=|7–3| = | 7 + (–3) | 2 the expression (1 – x) (x – 2) under the radical sign becomes negative. 5. The domain of the function y 

(1 – x ) ( x – 2) is the closed interval [1, 2].

6. The domains of the functions

y  1/

x,

y  1/

(1 – x)

are ] 0,  [ and] –  1 [respectively. 7. The domain of the function y = x ! is the set of Natural Numbers. 8. The domain of the function y = | x | is R. 9. The domain and range of y = x2 is R and [0,  [ respectively. 10. The domain and range of y = sin x is R and [–1, 1] respectively. 1 p when x is a rational number in its lowest terms q q = 0, when x is irrational. The domain of this function is R – {0}. What is the range ?

11. Let y 

EXERCISES 1. Give the domains of the following functions : (i) y 

2x 2x  7

,

1  (iv) y   – 2  . x  (vii) y = 1/(1 + cos x),

(ii)

y

1 3

x – x,

(v)

 x  y  x – , 1 – x 

(viii)

y  (1  2 sin x).

(iii) y 

( x3 – x ),

 1 – (vi) y    x

 ( x  1)  . 

2. Give the domains of the following functions : (i)

f ( x) 

x 2 – 3x – 1 x–2

(ii) f ( x)  ( x 2 – 1)

(iii)

f ( x) 

1–|x| 2–|x|

14

DIFFERENTIAL CALCULUS

(iv)

(vii)

f ( x)  (– x  6 x – 9) 2

(v) f ( x ) 

x3 ( x 2 – 5 x  4)

f ( x) 

x2  3 x2  | x |

1

(viii) f ( x ) 

f ( x)  (1 – | x |)

(vi)

( – x 2  6 x – 9) 3. Give the differences in the domains of the following pairs of functions :

y

(1 – x) and y  (1 – x 2 ), 1 1 y . (iii) y  1 – x and 1 – x2 4. What is the difference between the domains of the following pairs of functions : x2 (i) y  and y = x. x

(i)

(ii) (iii)

y  [ x 2 ( x  1)] and

yx

y  (1  x 2 )

and

y  ( x 2 – 1)

and

y  ( x – 5) 2

(iv) y = x – 5

( x  1).

ANSWERS 1. (i)

 72

(ii) R  { – 1, 0, 1}

(iii) [ – 1, 0]  [1, +  [

1 2

(v) ] 1, +  [

(vi)

R –

(iv) ] 0,

]

(vii) R  {(2n + 1) : n  I}

(vii)

[ 0,

1 2

( –1

5) ]

 2n –  ,  2n    : n  I 7 6

1 6

2. (i) R  {2},

(ii) ] –  , – 1 ]  [ 1, +  [

(iii) [ – 1, 1]  ] 2, +  [ ]– , – 2[

(iv) {3}

(v) ] –  , 1 [  ] 4, +  [

(vi) R

(vii) [ – 1, 1 ]

(viii) For no value of x. 3. (i) ] – , 1], [ – 1, 1]. The domain of y  (1 – x 2 ) is a proper sub-set of that of y  (1 – x). (ii) R ~ {1}, R ~{ – 1, 1 } While – 1 is a member of the domain of y = 1/(1 – x), – 1 is not a member of that of y = 1/(1 – x2). 4. (i) R ~ {0}, R. (ii) Same domain. (iii) [ – 1, 1 ], R  ] – 1 [. (iv) R, R.

2.2 Graphs of functions Let f be a function with domain [a, b] We have y = f (x),

x  [a, b].

The function f associates to each x  [a, b] a number denoted by f (x). We shall here introduce cartesian graphs of functions.

FUNCTIONS AND GRAPHS–ELEMENTARY FUNCTIONS

15

To represent the function graphically, we take two straight lines XOX and YOY, called coordinate axes, at right angles to each other, as in Plane Analytical Geometry. We take O as origin for both the axes and select unit intervals on OX, OY (usually of the same lengths). Y

To a number x  [a, b] corresponds a point M on Xaxis such that OM = x. Again the function f associates to x  [a, b] a number f (x). We write f (x) = y. We now have a point N on Y- axis such that ON = f (x) = y. Completing the rectangle OMPN, we obtain a point P which is said to correspond to the pair of numbers [x, f (x)] i.e., (x, y).

P(x, y)

N Y X

x

O

The set of points (x, y) obtained by giving different values to x, is said to be the graph of the function f. Also y = f (x) is said to be the equation of the graph.

M

X

Fig. 2.1

Y

While we are considering here Cartesian graphs, we shall later on introduce the notion of Polar Graphs. The graph of a function f is also said to be the graph of the equation y = f (x). EXAMPLES 1. The graph of the function y = x2, x < 0; y = x, x  [0, 1]; y = 1/x, x  [1,  [ is Fig. 2.2. y

y

x

y=

x

P (1, 2) y=

2

o

y = 1x x

x

o

Fig. 2.2

Fig. 2.3

The graph (Fig. 2.3) of the function y = (x2 – 1)/(x – 1) is the straight line y = x + 1, excluding the

2. point P (1, 2). 3. The graph of y = x consists of the set of points (1, 1), (2, 2) (3, 6), (4, 24), etc. 4. The graph of the function f defined as follows :  x, when 0  x  12  f ( x)  1, when x  12 1 – x when 1  x  1,  2 is as given in Fig. 2.4. 5. The graph of the function which associates to each real number x the positive square root of x2 is given in Fig. 2.5 As x 2  x or – x according as x is positive or negative,

Y P ( 12 ,1) A y=x

O

y=1–x

B Fig. 2.4

C

X

16

DIFFERENTIAL CALCULUS

we have  x, when x  0 ; f ( x)    – x, when x  0 . The graph of the function is the same as that of y = | x |. Y

Y

X

O

(Calicut 2004)

Fig. 2.5

X

O

Fig. 2.6

The student should compare the graph (Fig. 2.5) of

y

x

2

Y

with the graph (Fig. 2.6) of y = x.

6. Consider the graph of y = | x | + | x – 1| We have

 – x  1 – x  1 – 2 x when x  0,  y  x  1 – x  1 when 0  x  1, x  x – 1  2 x – 1 when x  1 

(0,1)

(1,1) X

O

The graph (Fig. 2.7) consists of parts of three straight lines, y = 1 – 2x; y = 1, y = 2x – 1

Fig. 2.7

Y

corresponding to the intervals ] –  , 0], ] 0, 1], ] 1,  [. 7. The graph of y = [x] where [x] denotes the greatest integer not greater than x is given in Fig. 2.8.

X

We have

 0, x  [0, 1 [  y   1, x  [1, 2 [  2, x  [ 2, 3 [  and so on. The value of y for negative values of x can also be similarly given. The right hand end-point of each segment is not a point of the graph.

Fig. 2.8

FUNCTIONS AND GRAPHS–ELEMENTARY FUNCTIONS

17

EXERCISES 1. Draw the graphs of the functions with the function values as given : (i)

(iii)

(v)

 1, when x  0 f ( x)    – 1, when x  0

x, when 0  x   f ( x)   1 2 – x, when 2  x  1 1 2

 x 2 , f ( x)    x,

when x  0 when x  0

(ii)

x, when 0  x  12  f ( x)   1 1 – x, when 2  x  1

(iv)

 x, when 0  x  12  1, when f ( x)   x  12 1 – x, when 1  x  1 2 

(vi)

 1/ x, when x  0  f ( x)   0, when x  0  – 1/ x, when x  0 

2. Give the graph of the following functions : (i) (iii)

y

x2 , x

yx

( x – 1)2 , x –1

(ii) y  x 

( x – 1)2  x –1

( x – 2)2 , x–2

( x – 1)2 ( x – 2) 2 ( x – 3) 2   . x –1 x–2 x–3 The positive value of the square root is to be taken in each case. Give also the domain of the function in each case. 3. Give the graphs of the following functions : (i) y = | x |, (ii) y = | x | + | x + 1 | , (iii) y = 2 | x – 1 | + 3 | x + 2 |, (iv) y = [x]2, (v) y = [x] + [x + 1] . 4. Draw the graphs of the following functions : (i) y = [2x], (ii) y = 2 [x] , 2 (iii) y = 2x + [x – 1] , (iv) x = x2 + [x]2. 5. Give the graphs of the following functions for the domain [0, 8]. Give also the range in each case : (iv)

yx

1  y  x   , 2  (iii) y = x – [x], (Calicut 2004) (v) y = [x]2, (i)

(ii) y  [ x] 

1 2

,

(iv) y = x + [x] , (vi) y = [x2],

(vii) y = x. [x],

(viii) y   12 x   1,

(ix) y = x2 + [x], (xi) y = 1/3[x], (xiii) y = [ | x | ],

(x) y = x2 – [x] , (xii) y = 3[3x], (xiv) y = | [x] |.

18

DIFFERENTIAL CALCULUS

6. Sketch the graph of the following function y = sig (x) where sig (x) is defined as follows :

 1 for x  0,  sig ( x)   0 for x  0,  – 1 for x  0.  7. Give the graphs of the following equations : (i) y = sig | x |, (ii) y = sig [x] , (iii) y = sig (x – | x | ).

2. (i) (iii)

ANSWERS (ii) R ~ {1} (iv) R ~ {1, 2, 3}

R ~ {0} R ~ {1, 2}

2.3 Operations on Functions Let f, g be two functions with domains denoted by Df , Dg respectively. We now proceed to give meanings to the symbols. f + g, f – g, fg, f ÷ g. To define f + g, we have to give a meaning to (f + g) (x) where x belongs to the domain of f + g. A natural way of defining (f + g) (x) is as follows : (f + g) (x) = f (x) + g (x). Since we talk of f (x) + g (x), the number x must necessarily belong to the domain of f, as also to that of g and as such x must be a member of the intersection Df  Dg of the domains of the functions f and g. Thus we have (f + g) (x) = f (x) + g (x)  x  Df Dg Similarly, we have the following definitions : (f – g) (x) = f (x) – g (x)  x  Df  Dg. (fg) (x) = f (x) g (x)  x  Df Dg. While we define f ÷ g as follows : (f ÷ g) (x) = f (x) ÷ g (x), it is necessary to exclude those of the members of the domain Dg of the function g for which g (x) is 0. Thus the domain of the function f ÷ g is the set Df  [Dg  {x : g (x) = 0}]. Ex. Given that the functions f and g are defined as follows : (i)

f ( x) 

(ii)

f (x) = x2 – 3x + 1,

(iii)

f ( x) 

(iv)

f ( x)  (sin x),

x,

g (x) = 1/x2, g (x) = 1/(x – 2),

( x 2 – 5 x  4), g (x) = x + 3,

g (x) = 1/(1 – sin x),

what are the domains of the functions fg and f ÷ g in each case.

FUNCTIONS AND GRAPHS–ELEMENTARY FUNCTIONS

(i) (ii) (iii) (vi)

19

ANSWERS ] 0, +  [, ] 0, +  [ R ~ {2}, R ~ {2}. R , R ~ {3} {2n + (2n + 1) ) : n  I} ~ {2n + /2) : n  I} {[2n + (2n + 1) ] : n  I} ~ {(2n + /2) : n  I}.

2.3.1. Polynomial functions and Rational functions. A function of the form f (x) = axn + a1 x n – 1 + ........... + an – 1 x + an where a, a1, a2, ............, an are given real numbers and a  0 is called a Polynomial function of degree n. The domain of every polynomial function is R. A function of the from g ( x) 

a0 x n  a1 x n –1  ...  an –1 x  an b0 x m  b1 x m –1  ...  bm –1 x  bm

where a0 xn + a1 x n–1 + ... + an – 1 x + an, b0 xm + b1 xm – 1 + ... + bm – 1 x + bm, are polynomials is called a Rational function. The domain of a rational function is the set of all those real numbers for which the value of the polynomial in the denominator is not zero. 2.3.2. Constant functions and the Identity function. It is interesting to see that every polynomial function and every rational function can be obtained as a result of the operations of addition, subtraction, multiplication and division performed on the Constant functions and Identity function which we now define. Let a be any given number. Then the function f (x) = a x  R is called a Constant function, also denoted by a permitting an abuse of language. The function f (x) = x,  x  R is called the Identity function which may be denoted by I. Illustration. The polynomial function f (x) = x2 + 2x + 3 can be described as I2 + 2 I + 3 where I denotes the Identity function I2 = I. I and 2, 3 stand for the Constant functions  x  R respectively. Ex. Consider any four rational functions and describe them as obtained through the operations of addition, subtraction, multiplication and division performed on the Identity function and the Constant functions.

2.4 One-one Functions. Consider a function y = f (x) with domain A and range B respectively. To each x  A, there corresponds a y  B. Again if y is a given member of the range B of the function, there will exist at least one member x of the domain A such that f (x) = y. It is possible, however, that in any given case, there may exist more than one member of A, say x1, x2 such that while x1  x2 we have f (x1) = f (x2). Def. A function is said to be one-one if each member of the range of the function arises for one and only one member of the domain of the function. Illustrations. 1. The function y = x3 with domain R is one-one (Fig. 2.9). 2. The function y = x2 with domain R is not one-one. (Fig. 2.10).

20

DIFFERENTIAL CALCULUS

The range of this function is the set of all non-negative numbers and each non-zero member of the range arises from two different members of the domain. As for example, the number 9 of the range arises from two different members – 3 and 3 of the domain (Fig. 2.10).

Fig. 2.9

Fig. 2.10

3. The function y = x2, the domain the set of all non-negative real numbers only, is one-one (Fig. 2.11).

Fig. 2.11

4. The function y = sin x with domain R is not one-one. The range of the function is [–1, 1] and each member of the range arises from an infinite number of members of the domain. For example, the member, 0, of the range arises from each member of the infinite set {n ; n I} (Fig. 2.12).

Fig. 2.12

FUNCTIONS AND GRAPHS–ELEMENTARY FUNCTIONS

21

5. The function y = sin x with domain [–/2, /2] is one-one (Fig. 2.13).

Fig. 2.13

2.4.1. Graph of one-one functions. It may be seen that a function is one-one, if and only if, no line parallel to the x-axis meets the graph of the function in more than one point. For example, the function represented by the graph on the left (Fig. 2.14) is one-one and that represented by the graph on the right (Fig. 2.15) is not one-one.

Fig. 2.14

Fig. 2.15

2.4.2. Invertible functions. Consider a one-one function with domain A and range B. Let y  B. The function f being one-one, the member y  B arises from one and only one member x  A such that f (x) = y. Thus we define a new function, say g, such that g (y) = x  f (x) = y. Also the domain of the function g is the range of the given function f and vice-versa. The function g is said to be an Inverse of the function f. We also say that f is an Invertible function. We thus see that one-one functions are Invertible. Clearly if g is the inverse of f, then f is also the inverse of g. Illustrations. The function : (i) y = x2 with domain R is invertible. (ii) y = x2 with domain R is not invertible. (iii) y = x2 with domain [0, [ is invertible. (iv) y = sin x with domain R is not invertible. (v) y = sin x with domain [–/2, /2] is invertible. 2.4.3. Graph of the inverse of an invertible function. Let y = f (x) be an invertible function and let g denote the inverse of f, we have y = f (x)  x = g (y).

22

DIFFERENTIAL CALCULUS

Let (p, q) be a point on the graph of the function f. Then (q, p) is the corresponding point on the graph of the function g. The segment joining the points (p, q) and (q, p) is bisected at right angles by the line y = x so that the point (q, p) on the graph of the function g is the reflection of the point (p, q) on the graph of the function f in the line y = x (Fig. 2.16).

Fig. 2.16

Fig. 2.17

It follows that the graph of the inverse g of f is obtained by reflecting the graph of f in the line y = x. Fig. 2.17 gives the graph of an invertible function, as also that of the inverse g of f. Instead of writing x = g (y), we may write y = g (x) adopting, as is usual, the symbol x for the independent variable and y for the dependent variable. Ex. the inverse of y = x3 is described as y = x1/3. We have y = f (x) = x3  x = g (y) = y1/3. The domain as well as the range of each of f and g is R. 2.4.4. Criteria for invertibility. Increasing and decreasing functions. We now introduce the notion of increasing and decreasing functions on the basis of which it is possible to give a simple criterion for the invertibility of a function. Def. A function f is called an increasing function in its domain D if x2 > x1  f (x2)  f (x1), x1  D, x2  D. A function f is called strictly increasing if we do not have the equality so that x2 > x1  f (x2) > f (x1). Def. A function is called a decreasing functon in D if x2 > x1  f (x2)  f (x1) ; x1  D, x2  D. A function f is called strictly decreasing if x2 > x1  f (x2) < f (x1). Increasing and Decreasing functions are also called Monotonic functions. Illustrations : The function 1. y = x2 is strictly decreasing in ] – , 0] and strictly increasing in [0,  (Fig. 2.18). 2. y = x3 is strictly increasing in ] –, [ (Fig. 2.19). 3. y = 1/x2 is strictly increasing in ] – , 0[ and strictly decreasing in [0, ] (Fig. 2.20). 4. y = 1/x is strictly decreasing in ] – , 0[ and ] 0, [ (Fig. 2.21).

FUNCTIONS AND GRAPHS–ELEMENTARY FUNCTIONS

Fig. 2.18

Fig. 2.19

23

Fig. 2.20

Fig. 2.21

5. y = [x] is increasing but not strictly increasing in R. 6. y = tan x is strictly increasing in each of ]– /2, /2[, ]/2, 3/2[, ]3/2, 5/2[ etc. (Fig. 2.22).

Fig. 2.22

2.4.5. Invertibility of strictly monotonic functions. Let a function f be strictly monotonic in its domain A and B be its range. Because of the strictly monotonic character of the function f each member of its range B arises from one and only one member of its domain A. Thus the function f is one-one and therefore invertible. It follows that every strictly monotonic function is invertible. Ex. Which of the following functions are invertible? Give also the domains of the inverse of the invertible functions : (i)

y = 1 + x2

;

x  [ – 2, 2],

(ii)

y=1+

x2

;

x  [ –2, 0],

(iii)

y=1+

x2

;

x  [0, 2],

(iv)

y=|x|

;

x  [ –1, 3],

(v)

y=|x|

;

x  [0, 3],

(vi)

y=|x|

(vii) (viii) (ix) (x)

;

x  [ –1, 0],

y=

x4

;

x  ] – , [,

y=

x4

;

x  [0, ],

y=

1/x4

;

x  ] – , 0 [  ] 0, [,

y=

1/x4

;

x [1, 8],

24

DIFFERENTIAL CALCULUS

(xi)

y = 1/x3

;

x  ]–3, 0[  ]0, 5[,

(xii)

y = x + 1/x

;

x  [1, 3], ANSWERS

Ex. (i) (iii) (v) (vii) (ix) (xi) (xii)

Not invertible (ii) Invertible, [1, 5] Invertible, [1, 5] (iv) Not invertible Invertible, [0, 3] (vi) Invertible [0, 1] Not invertible (viii) Invertible, [0, [ Not invertible (x) Invertible, [1/64, 1] Invertible, [–1/27, – [  [1/125, [ Invertible, [2,4/3].

2.5. Function of a Function : Composite of Functions. Consider two functions y = f (x) ; x  Df , and y = g (x) ; x  Dg. We suppose that x  Df  f (x) Dg , i.e., the range of the function f is sub-set of the domain of the function g. We now define a function which we denote by g o f as follows : (g o f) (x) = g [ f (x)] ; x  Df . To each x  Df , the function g o f associates the g-value of the f-value of x. This is possible in that, according to our supposition, f (x) which is a member of the range of f is as well as member of the domain of g. The function g o f is called the Composite of the functions g and f taken in this order. It is also called the function of a function. Illustrations : 1. Consider the two functions f (x) = x2, g (x) = sin x. The domain of the function f is R. Also the range [0, [ of f is a sub-set of the domain R of g. Thus we have (g o f) (x) = g [f (x)] = g (x2) = sin x2, x  R Let us also consider f o g We have (f o g) (x) = f [g (x)] = (sin x2), x  R. The range of the function g being [– 1, 1] we have to restrict the domain of f to [–1, 1] only. It should be noticed that the functions g o f and f o g may not be same. 2. Consider the function y  ( x 2 – 3 x  2).

We shall now express this function as the composite of two functions. Consider f (x) = x2 – 3x + 2.

g ( x) 

x.

FUNCTIONS AND GRAPHS–ELEMENTARY FUNCTIONS

25

Surely we have ( g o f ) ( x)  g [ f ( x) ]  ( x 2 – 3 x  2).

We now consider the domain of the composite g o f of g and f. The domain of f is R. Now only that part of the domain R of f can be considered such that the corresponding range is a sub-set of the domain [0, [ of g. We have x2 – 3x + 2 = (x – 1) (x – 2). Now (x – 1) (x – 2)  0  x ]  – , 1]  [2, [ = D, say. Thus we see that the domain of the composite g o f is the set ] – , 1]  [2, [ = R  ]1, 2 [

x  2 with domain [0, [.

We may also see that ( f o g ) ( x)  ( x – 3)

EXERCISES 1. Express the following functions as the composites of appropriate functions : (i) 2. 3. 4. 5.

y  (sin x),

(ii) y  sin ( x 2  1).

(iii) y = tan (tan x). Show that the function f (x) = (2x + 3)/(x – 3) in invertible. Find the inverse g of f and verify that g o f is the identity function. Repeat the above question with the function f (x) = (4x + 5)/(5x + 4). Given that f is an invertible function with its inverse g, show that gof and fog are identity functions. Consider the three functions : f (x) = x2, g (x) = sin x, h ( x) 

x.

Compute [(g o f) o h] (x), [g o (f o h)] (x), [(f o g) oh] (x), [f o (g o h)] (x). What do you notice? ANSWERS 1. (i) (ii)

f ( x)  sin x, g (u )  u , ( gof ) ( x)  sin x

f ( x)  x 2  1, g (u )  u

h (v)  sin v, ( hog ) of  sin ( x 2  1). (iii) f (x) = tan x, (fof) (x) = tan (tan x). 3  3x g ( x)  ,x2 2. (i) x –2 5 – 4x 4 g ( x)  ,x 3. (ii) 5x – 4 5 4. ((gof) oh) (x) = (go (foh)) (x) = sin x

(( fog ) oh) ( x)  ( fo ( goh)) ( x)  sin 2

x

In the following, we shall study the graphs of the exponential function y = ax and its inverse y = log ax as also the graphs of the trigonometrical functions y = sin x, y = cos x, y = tan x, y = cot x, y = sec x, y = cosec x and their inverses in appropriately chosen domains. The exponential function will be introduced in stages.

26

DIFFERENTIAL CALCULUS

2.6. The function y = x1/n ; n being any integer. The consideration will be given to the four cases which arise when n is a (i) positive even integer,

(ii) positive odd integer,

(iii) negative even integer,

(iv) negative odd integer.

In (i) the function y = xn is strictly decreasing in ] – , 0] and striclty increasing in [0, ] (Fig. 2.23). In (ii) the function y = xn is strictly increasing in ] – 0[ (Fig. 2.24). If follows that the function y = xn is strictly increasing in [0, [ when n is a positive integer, even or odd and as such the function is invertible. We write y = xn  x = y1/n, x [0, [. Thus the inverse of the function y = xn is the function y = x1/n, the domain and range of either function being [0, [.

Fig. 2.23

Fig. 2.25

Fig. 2.24

Fig. 2.26

In (iii) the function y = xn is strictly increasing in ] – [ and strictly decreasing in ] 0, [ (Fig. 2.25). In (iv), the function y = xn is strictly decreasing in ] – [ and also strictly decreasing in ] 0, [ (Fig. 2.26). If follows that the function y = xn is strictly decreasing in ] 0, [ when n is a negative integer, even or odd and as such the function is invertible. We write y = xn  x = y1/n, x  ] 0, [

FUNCTIONS AND GRAPHS–ELEMENTARY FUNCTIONS

27

so that the inverse of the function y = xn is the function y = x1/n, the domain and range of either function being ] 0, [. It is important to note that in each case x1/n, denotes the positive nth root of the positive number x. 2.6.1. Meaning of xa when a is any real number. Case I. Let a be a rational number m/n where m and n are integers and n  0. Then we have xm/n = (xm)1/n. We assume that the reader is already familar with the various properties of the power function y = xa such xa. xb = xa + b where a, b are arbitrary rational numbers. Case II. Let a be an irrational number, say b.a1 a2 a3 a4 ... The number a can be approximated by the sequence of rational numbers b, b . a1, b . a2, b. a1 a2 a3......... This suggests that xa can be thought of as approximated by the sequence of numbers x b , xb . a1 , xb . a1 a2 , xb . a1 a2 a3 ,

the indices being rational numbers. What we have stated here is only an indication of what appears to be natural. The actual treatment requires very sophisticated considerations which is beyond the scope of this book. Power Function. The function y = xa, x  ] 0, [ is called the General power function, a being any given real number. 2.6.2. Exponential Functions. The function y = ax where the base a is a constant and the index, x a variable is called an Exponential function. In the context of what is stated in the preceding, we are not, in this book, concerned with the rigorous justification for the statements which we shall be making. The only basis for these statements will be our natural expectations. To consider the graph of y = ax, we have to consider two cases (i) a > 1, (ii) 0 < a < 1. Let us take a = 2. To have an indication of the nature of variation of the function y = ax, we consider the case when x varies over the sequence ........... –3, –2, –1, 0, 1, 2, 3,........... of integers and the corresponding sequence of the values of ax is ...........1/8, 1/4, 1/2, 1, 2, 4, 8, 16,........... Every member of this sequence is positive. Also the numbers on the left can be made as near 0 as we like and those on the right as large as we like. Now take a = 1/2 < 1. The sequence of values of ax now obtained is ..........., 16, 8, 4, 2, 1, 1/2, 1/4, 1/8,........... In this case the number on the left can be made as large as we like and those on the right as near 0 as we like. The above statements suggest conclusions which we shall now formulate for drawing the graph of the function y = ax. I. y = ax; a > 1. (i) The domain of the function is ] – , [ and the range is ] 0, [. (ii) The domain is strictly increasing.

28

DIFFERENTIAL CALCULUS

We have the graph as given in Fig. 2.27.

Fig. 2.27

Fig. 2.28

ax

It will be seen that if a > 1, the curve y = approaches the x-axis viz. the line y = 0 nearer and nearer if x, while taking negative values becomes greater and greater in absolute value. Also the perpendicular distance of the points of the curve y = ax from the line y = 0 can be made as small as we like if only | x | is sufficiently large so that the curve y = ax approaches the lines y = 0 asymptotically for large negative values of x. We also say that y = 0 is the asymptote of y = ax. II. y = ax; 0 < a < 1. (i) The domain of the function is ] – , [ and the range is ] 0, [. (ii) The function is strictly decreasing. We have the graph as given in Fig. 2.28. Ex. Describe the relationship of the curve y = a x and the line y = 0 if 0 < a < 1. The logarithmic function y = log ax. We have seen that y = ax is strictly increasing when a > 1 and strictly decreasing when 0 < a < 1. Thus this function is invertible. The inverse of this function is denoted by log ax we write. y = ax  x = log ay, where x  ] – , [ and y  ] 0, [. Writing y = log ax in place of x = loga y, we have the graphs of y = log ax as shown in Figs. 2.29, 2.30.

Fig. 2.29

Fig. 2.30

It will be seen that the domain of y = log a x is ]0, [ and range ] – , [. Also a° = 1 loga 1 = 0. Comparing the relationship of the curve y = ax with the line y = 0 with that of the curve y = loga x with the line x = 0, we say that the line x = 0 is the asymptote of y = loga x.

FUNCTIONS AND GRAPHS–ELEMENTARY FUNCTIONS

29

2.7. Trigonometric and Inverse trigonometric functions. 1. The functions y = sin x, y = sin–1 x. The domain of y = sin x is R and range [–1, 1]. It increases strictly from –1 to 1 as x increases from –/2 to /2, decreases strictly from 1 to – 1 as x increases from /2 to 3/2 and so on. We have the graph of y = sin x as given in Fig. 2.31 :

Fig. 2.31

sin x = 0 when x  {0,  ,  2,  3,............}.    It follows that y = sin x is invertible in the domain  – ,  We write  2 2    y  sin x  x  sin –1 y ; x   – ,  , y  [– 1, 1].  2 2 We give below the graphs of the function y = sin x and its inverse y = sin–1 x. (Fig. 2.32 and 2.33 respectively)

Fig. 2.32

Fig. 2.33

   We notice that sin–1 x denotes the number lying between  – ,  whose sin is x.  2 2 –1 Also sin x strictly increases from – /2 to + /2 as x increases from –1 to 1. In particular, we have   , sin –1 (1)  . 2 2 –1 Ex. Give other possible ways of defining y = sin x. sin –1 0  0, sin –1 (–1)  –

30

DIFFERENTIAL CALCULUS

2. The functions y = cos x, y = cos–1 x. The domain of y = cos x is R and range [–1, 1]. Also the function decreasing strictly from 1 to – 1 as x increases from 0 to , increases strictly from – 1 to 1 as x increases from  to 2 and so on. We have the graph of y = cos x as given in Fig. 2.34.

Fig. 2.34



cos x  0 when x  

 , 2



3 , 2





5 ,.... 2

It will thus be seen that in the domain [0, ] the function y = cos x is invertible. We write y = cos x  x = cos–1 y; x  [0, ], y  [–1, 1]. We give below the graphs of the functions y = cos x and its inverse y = cos–1 x. (Fig. 2.35 and 2.36 respectively)

Fig. 2.35

Fig. 2.36

x denotes the number between 0 and  with cosine x. In particular, we have  cos –1 (– 1)  , cos –1 (0)  , cos –1 (1)  0. 2 3. The functions y = tan x, y = tan–1 x. The domain of the function y = tan x is R  {/2,  3/2,  5/2, ............}. So that we exclude from R the numbers for which cos x = 0. The range is ] – [. Thus

cos–1

FUNCTIONS AND GRAPHS–ELEMENTARY FUNCTIONS

31

The graph of y = tan x is given in Fig. 2.37

Fig. 2.37

   The function y = tan x increases strictly from –  to +  as x increases in  – ,  2 2 such invertible. We write    y  tan x  x  tan –1 y; x   – ,  y  ] – , [.  2 2 We give below the graphs of y = tan x, y = tan–1 x.

Fig. 2.38

  and is as 

Fig. 2.39

as shown in Fig. 2.38 and 2.39 respectively, tan–1 x denotes the number between –/2 and /2 whose tangent is x. In particular, we have tan–1 (0) = 0, tan–1 (1) = /4, tan–1 (–1) = –/4. Note. x = –/2, x = /2, x = –3/2, x = 3/2 etc. are the asymptotes of y = tan x. Also y = /2 and y = – /2 are the asymptotes of y = tan–1 x. 4. The functions y = cot x, y = cot–1 x. The domain of y = cot x is R  [0, ,  2,  3,..........}. so that we exclude from R the set of numbers for which sin x = 0. The range is ] – [.

32

DIFFERENTIAL CALCULUS

We give below the graph of y = cot x.

Fig. 2.40

The function y = cot x decreases strictly from +  to –  as x increases in ] 0, [ so that the function is invertible in ] 0, [. y = cot x  x = cot–1 y; x  ] 0, [, y  ] – , [. We give below the graphs of y = cot x and y = cot–1 x. (Fig. 2.41and 2.42 respectively)

Fig. 2.41

Fig. 2.42

cot–1

x denotes the number lying between 0 and  whose cotangent is x. In particular,

cot–1

(0) = /2, cot–1 (1) = /4, cot–1 (–1) = 3/4.

Ex. Give the asymptotes of y = cot x and y = cot–1 x. 5.

[ Ans. {x = n such that n  I}, y = , y = 0.]

The functions y = sec x, y = sec–1 x.

The domain of y = sec x is R  {/2,  3/2,  5/2,..........} so that we have excluded those of the numbers for which cos x = 0. The range is ] – , –1]  [1, [ = R  ] –1, 1[.

FUNCTIONS AND GRAPHS–ELEMENTARY FUNCTIONS

33

Given below is the graph of y = sec x.

Fig. 2.43

The function y = sec x is strictly increasing from 1 to  as x increases in [0, /2 [ and strictly increasing form –  to –1 as x increases in ] /2, ]. Thus y = sec x is invertible in the domain [0, /2 [  ] /2, ] =[0, ]  {/2}. We write y = sec x  x = sec–1 y; x  [0, ]  {/2}, y R ] –1, 1[. We give below the graphs of the functions y = sec x and y = sec–1 x. (Figs. 2.44 and 2.45 respectively)

Fig. 2.44

Fig. 2.45

sec–1 x denotes the number which  [0, /2 [  ] /2, ] whose secant is x. In particular, we have sec–1 (1) = 0, sec–1 (–1) = . 6. The functions y = cosec x, y = cosec–1 x. The domain of y = cosec x is R  {0, ,  2,  3, ............}

34

DIFFERENTIAL CALCULUS

so that we have excluded the set of numbers for which sin x = 0. The range is ] – , –1]  [1, [ = R  ] –1, 1[. We give the graph of y = cosec x on the next page. The function y = cosec x decreases strictly from – 1 to –  as x increases in [–/2, 0[ and decreases strictly from +  to 1 as x increases in ] 0, /2]. The function y = cosec x is thus invertible in the domain [–/2, 0[  ] 0, /2,] = [–/2, /2] ~ {0}.

Fig. 2.46

We give below the graphs of y = cosec x, y = cosec–1 x.

Fig. 2.47

cosec–1

Fig. 2.48

x denotes the number which belongs to [–/2, 0[  ] 0, /2] and whose cosecant is x.

We may note that cosec–1 (–1) = – /2, cosec–1 1 = /2. Ex. Give the asymptotes of y = sec x, y = cosec x, y = sec–1 x, y = cosec–1 x. [ Ans. {x = (2n + 1) /2 : x  I}{x = n : n  I}, y = /2, y = 0}] Asymptotes of a curve. While we have introduced in this chapter the notion of asymptotes of a curve in the context of special curves, we shall take us the general discussion of asymptotes in a later chapter.

FUNCTIONS AND GRAPHS–ELEMENTARY FUNCTIONS

35

Note : The following table gives the domains and ranges of the inverse trigonometric functions: Functions y = sin–1 x y = cos–1 x y = tan–1 x y = cot–1 x y = sec–1 x y = cosec–1 x

Domains

Ranges

[–1, 1] [–1, 1] R R R – ] –1, 1 [ R – ] –1, 1 [

[–/2, /2] [0, ] ]–/2, /2[ ] 0, [ [0, ] ~ {/2} [–/2, /2] ~ {0}

A useful inequality In dealing with Trigonometric functions, we shall find it useful to know that | sin x | < | x | when x  ] – /2, /2 [ except for x = 0 when sin x = x; both being zero.

Fig. 2.49

Consider a circle with centre O and radius 1. Let  AOP = x where 0 < x < /2. (Fig. 2.49) We have Area if  OAP < Area of sector OAP < Area of OAB. 1 1 OA. OP sin x  sin x. 2 2 1 1 Area of sector OAP  .1 . x  x. 2 2 1 1 Area of OAB  .OA . AB  tan x, 2 2 Thus sin x < x < tan x. We have thus seen that sin x < x when x  ]0, /2[ or equivalently | sin x | < | x | in that sin x and x are both positive when x  ] 0, /2[. Let x  ]–/2, 0[ so that sin x and x are both negative. We have sin (– x) < – x for – x  ] 0, /2 [  | sin (– x) | < | – x | | sin x | < | x |. Thus we have seen that | sin x | < | x | when x  ] – /2, 0]  [0, /2[ Note. The inequality is also generally given in books on Elementary Trigonometry.

Area of  OAP 

36

DIFFERENTIAL CALCULUS

EXERCISES 1. Give the graphs of the following functions : (i) y = 2| x | , (ii) y   12  , (iv) y = log 1/2 | x |, (v) y = | log2 x | , (vii) y = | tan x | , (viii) y = | cosec x | , (x) y = | tan–1 x | , (xi) y = | sec–1 x | 2. Give the domains of the following functions : |x|

(iii) y = log2 | x | , (vi) y = | cos x | , (ix) y = | sin–1 x | ,

(i) y = sin (1/x) , (ii) y = log2 (1/x) , (iii) y  log 3 (1/ ( x – 1) ) , (iv) y = log 1/2 sin x , (v) y = sin log2 x , (vi) y = log2 sin–1 x, (vii) y = log3 tan–1 x, (viii) y = log4 (1/tan–1 x) , (ix) y =log1/3 (1 + x) , 3. Give the graphs of the functions with the following function values : 2

2

tan2 x,

2x ,

21/ x ,

2 –1/ x

sin 2x,

cos 3x ,

tan 2x,

cot 3x.

4. Draw the graphs of the following equations : (i) y = x sin x, (ii) y = | log2 | x | |, 5. Give the asymptotes of the following : (i) y = tan 2 x, (iv) y =

(iii) y = tan (tan–1 x). (iii) y  cot

(ii) y = | tan x |,

3x/2,

(v) y =

2x

1 2

x,

+ 1.

ANSWERS 2. (i) R  {0} (ii) ]0, + [ (iv) {]2n, (2n + 1)  [: n  I} (v) ]0, + [ (vii) ]0, /2[ (viii) ]0, /2[

1  {x   n    : n  I} 2  (iii) {x = 2n : n  I}

2

(iii) ]1, + [ (vi) ]0, /2[ (ix) ]– 1, + [

(ii) {x = (2n + 1) /2 : n  I}

5. (i)

(iv) y = 0

(v) y = 1.

2.8 Implicitly Defined Functions The relations y = f (x) is said to be the equation of the graph of the function f. Sometimes, however, we come across equations in x and y which are not of the form y = f (x) so that to some values of x correspond more than one value of y. For example, for the equation x3 + y3 + xy + 2y2 + x + y = 0 x = 0 corresponds to y = 0 and y = – 1. Consider now the equations y 2  4ax,

x2



y2

x2

y2

1 ...(i) a2 b2 a2 b2 which represent a parabola, an ellipse and a hyperbola respectively. These equations are solvable for y and when, so solved, we obtain y   2 (ax ), y  

 1,

b a

–

( a 2 – x 2 ), y  

b a

( x2 – a2 )

FUNCTIONS AND GRAPHS–ELEMENTARY FUNCTIONS

37

so that, in general, to each admissible value of x correspond two values of y. Thus in each case, the given equation defines two different functions. Each point on the graph of the two functions.

y  2 (ax)

y  – 2 (ax)

is a point on the graph of the equation y2 = 4ax. We may similarly describe the situation in the case of the other two equations also. The graphs of the given equations (i) are given in figures 2.50, 2.51, 2.52 respectively. y

y

o

x

Fig. 2.50

y

o

x

o

x

Fig. 2.51

Fig. 2.52

The thick parts of these graphs lying above x-axis are those of the functions b b y  (a 2 – x 2 ) , y  (x2 – a2 ) y  2 (ax), a a and those lying below x-axis are the graphs of the functions b b y– ( a 2 – x 2 ), y– ( x2 – a2 ) y  – 2 (ax), a a We say that the two functions y  2 (ax), y  – 2 (ax) are implicitly defined by the equation y2 = 4ax. Similar statements may be made about the other equations also. The above illustraions do not exhibit the general situation in that while the given equations (i) were explicitly solvable for y in terms of x, it may not be so in every given case. The discussion of the general question as to whether or not to a given equation in x and y correspond one or more functions such that every point on their graphs is a point on the graph of the given equation and viceversa is beyond the scope of this book. In any given case of an equation in x and y, we shall only assume that the equation determines one or more functions. Each of these functions will be said to be Implicitly defined by the given equation.

2.9 Some Special Functions (a) Absolute value function (or modulus function) :  x, x  0 y | x|  – x, x  0 It is the numerical value of x. Properties (i) | x |  a  – a  x  a; (a  0) (ii) | x |  a  x  – a or x  a; (a  0) (iii) | x  y | | x | – | y |  (iv) | x  y |  | x | + | y |

y y=–x

y= x

135° 45° o

Fig. 2.53

x

38

DIFFERENTIAL CALCULUS

(b) Greatest Integer Function : [x] indicates the integral part of x which is nearest and smaller integer for x. It is also known as floor of x. Thus, [3.43201] = 3, [0.325] = 0, [7] = 7, [– 7.392] = – 8, [– 0.8] = – 1. In general n  x < n + 1 (n is an integer) i.e. [x] = n. Properties (i) [x + I] = [x] + I, if I is an integer (ii) [x] = x, if x is an integer

n

n+1

(iii) [– x] = – [x], if x  integer

x Fig. 2.54

(iv) [– x] = – [x] – 1 if x is not an integer (v) [x + y]  [x] + [y] (vi) If (x)  I, then (x)  I If (x)  I, then (x) < I + 1. (c) Fractional part functions y = {x}. It indicates the fractional part of x. if x = I + f, I = [x] and f = {x} Then y = {x} = x – [x]. Properties (i) {x} = 0 if x is an integer (ii) {x} = x if 0  x < 1 (iii) {– 1} = 1 – {x} if x is not an integer (d) Least integer functions

y = (x) indicates the integral part of x which is nearest and greater integer of x. It is also known as ceiling of x. If n < x  n + 1

[x] = n

(x) = n + 1

(n is an integer) i.e. (x) = n + 1 [x] = n, (x) = n + 1

n

x

n+1

Fig. 2.55

Properties (i) (x) = x holds if x is integer. (ii) (x + 1) = (x) + I, I is an integer (iii) Greatest integral function [x] converts x = I + f into (I) while [x] converts to I + 1. (e) Signum function : y = Sgn (x).

FUNCTIONS AND GRAPHS–ELEMENTARY FUNCTIONS

39

It is defined by | x | or  y  Sgn ( x )   x  0 if  1 if x 0     –1 if x  0  0 if x  0 

x ,x0 |x| x0

MISCELLANCEOUS EXAMPLES 1.

2.

Find domain of f (x) = log10 log10 (1 + x3). f (x) = log10 log10 (1 + x3) exists If log10 (1 + x3) > 0 or 1 + x3 > 1 or x3 > 0 or x  (0, ). Thus, domain of given function exists if x > 0.

  x2   Find the domain for y  sin –1  log 2     2      2 x 0 For y to be defined , 2 and Form (i) From (ii)

3.

 1  

x2  21  1  x 2  4 2 –2  x  – 1 or 1  x  2 2 –1 

 Thus we have x  [– 2, – 1]  [1, 2]. Given : y = 2 [x] + 3 and y = 3 [x – 2] + 5, than find value of [x + y]. we have 2 [x] + 3 = 3 [x – 2] + 5  2 [x] + 3 = 3 [x] – 6 + 5 

[x] = 4  4  x < 5

or

x = 4 + f, f is the fractional part



y = 2 [x] + 3 = 11.

Hence 4.

 x2 – 1  log 2   2  x  R – {0}

Solve : we know that 

[x + y] = [4 + f + 11] = [15 + f] = 15. 4 {x} = x + [x] x = [x] + {x} 4 [x] = [x] + {x} + [x]

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

40

DIFFERENTIAL CALCULUS

But

2 [ x] 3 0  {x}  1

So,

0



0  [ x] 



If Thus If So,

{x} 

...(i)

2 [ x] 1 3 3 2

 [ x]  0 or 1.

[x] = 1, then {x} 

2 from (i) 3

5 3 [x] = 0, {x}= 0 x

x=0

Thus, solutions of the given equation are

5.

 5 x  0,   3 If [x] = the greatest integer less than or equal to x, and (x) = the least integer greater than or equal to x, and [x]2 + (x)2 > 25, then find x. Let 

[x]2 + (x)2 > 25



[I + f]2 + (I + f)2 > 25

 

6.

x = I + f where I is an integer and f is fractional part.

I2 + {I + 1}2 > 25 2 I2 + 2 I – 24 > 0



(I + 4) (I – 3) > 0



I < – 4 or I > 3

Here,

x=I+f

So,

x < – 4 + f or x > 3 + f.

Since,

0f 0 Here discriminant is 16 – 4 (3) (5) < 0 and coefficient of x2 = 3 > 0 Hence,

(3x2 – 4x + 5) > 0   x  R



Domain  R.

FUNCTIONS AND GRAPHS–ELEMENTARY FUNCTIONS

41

y = loge (3x2 – 4x + 5)

Now, 

3x2 – 4x + 5 = e y



3x2 - 4x + (5 – e y) = 0

Since x is real thus, (–4)2 – 4(3) (5 – e y)  0 

e y  11/3



7.

y  loge (11/3)

   11  Hence, range is  log   ,   . 3   Find out whether the following function is even, odd or neither even nor odd,

x|x| x –1   f ( x)  [1  x]  [1 – x], –1  x  1  – x| x|, x 1  where | | and [ ] represent modulus and greatest integer function. The given function can be written as (using definitions of modulus and greatest integer function).



  f ( x)  2     2  f ( x)   2  

x  –1 – x2 ,  [ x]  [– x], –1  x  1 x 1

– x2 ,

– x , x –1 – 1  0, –1  x  0 2

2,

x 0

 0 – 1,

0  x 1

2

x 1

– x ,

– x , x –1  –1  x  0  1,  f ( x)   2, x 0  1 0  x 1   – x 2 , x 1 2



8.

which is clearly even as if f (– x) = f (x) Hence, the given function is even. If k be a positive real number such that f (x + k) + f (x) = 0 for all x  R. Prove that f (x) is a periodic function with period 2k. We have

f (x + k) + f (x) = 0,  x  R



f (x + k) = – f (x),  x  R



f (x + 2k) = – f (x + k) = f (x),  x  R

Hence f (x) is periodic with period 2 k.

42

9.

DIFFERENTIAL CALCULUS

Find the period of f (x) = | sin x | + | cos x | | sin x | has period , | cos x | has period , Here, f (x) is even, thus period of 1 | LCM of  and ] 2 = /2

f ( x) 



Period of f (x) = /2.

10. Let f (x) = log x 2 25 and g (x) = logx 5 then f (x) = g (x) holds, then find the interval for x. Domain of f  R – {1, 0} Domain of g  (0, ) – {1} for

f (x) = g (x),

domain of f = domain of g i.e.

f (x) = g (x), if x  (0, ) – {1}

11. Let f : R  R be defined by f (x) =

ex – e– x , Is f (x) invertible? If so, find its inverse. 2

To check the invertibility, we has

e x  e– x 2 2x e 1  which is strictly increasing as e 2x > 0 for all x, Thus one-one 2 ex (ii) on to ; Let y = f (x) (i) one-one : f  ( x) 

ex  e– x where y is strictly monotonic. 2 Hence range of f (x) = ] f (– ), f ()[  range of f (x) = ]–  , [  range of f (x) = co-domain Hence f (x) is one-one and onto, so invertible. To find the inverse, we have y

y

  

e2 x – 1 2 ex

e2x – 2ex y – 1 = 0

ex 

2 y

4 y2  4 2

x = f–1 (y) = log ( y 

Neglecting negative sign, f –1 ( x)  log ( x 

x 2  1)

( y 2  1))

FUNCTIONS AND GRAPHS–ELEMENTARY FUNCTIONS

43

12. Find the domain of single valued function y = f (x) given by the equation 10x + 10y = 10. we have 10 y = 10 – 10x  y = log10 (10 – 10x) Now, y is defined if, 10 – 10x > 0  1 > x. Hence domain of y = f (x) is x < 1  x  ]– , 1[ 13. If x  (0, /2), find the solution of the function f ( x) 

Here x  (0, /2)  we know that

1 – log sin x tan x

0 < sin x < 1

...(i)

loga x < b  x > ab, if 0 < a < 1 x < ab, if a > 1 1

Thus,

f ( x) 

 

tan x > (sin x)0 tan x > 1

– log sin x tan x

...(ii)

exists if, – logsin x (tan x) > 0

  x ,  4 2   x  ,  4 2

 

    ,  is the required solution . 4 2

14. Find all possible values of x satisfying, [ x] [ x – 2] 8 {x}  12 –  [ x – 2] [ x] [ x – 2] [ x] where [ ] donotes the greatest integer function and { } is fractional part. We have 

[ x] [ x – 2] 8 {x}  12 –  [ x – 2] [ x] [ x – 2] [ x]

[ x]2 – [ x – 2]2 8 {x}  12  [ x – 2] [ x] [ x ] [ x – 2]



[x] – [x – 2]) ([x] + [x – 2]) = 8 {x} + 12



([x] – [x] + 2) ([x] + [x] – 2) = 8{x} + 12

 

[ {x + I] = {x} + I}

4([x] – 1) = 8 {x} + 12 [x] – 4 = 2 {x}

...(i)

44

DIFFERENTIAL CALCULUS

Now, we know that 0  {x} < 1  0  2 {x} < 2  0  [x] – 4 < 2  4  [x] < 6  [x] = 4, 5 If, [x] = 4  2 {x} = [x] – 4 become {x} = 0 and if, [x] = 5  2 {x} = [x] – 4 become

{x} 

1 2

...(ii)

...(iii)

From (ii) and (iii) we have x = [x] + {x} i.e., x=4+0=4

x5

and

1 2



11 2

 11  x   4,   2 15. Find the domain and range for 

f ( x)   log (sin –1 For domain,

Now,   and 

( x 2  3 x  2)  where [ ] denotes the greatest integer function.

0 < x2 + 3x + 2  1 x2 + 3x + 2 > 0 (x + 1) (x + 2) > 0 x  (– , –2)  (–1, ) x2 + 3x + 2  1 x2 + 3x + 1  0 – 3 – x 2 



5 – 3  5 ,  2 

 From (i) and (ii) –3– 5   – 3  5 x , – 2    – 2,  2 2     This is the domain for f (x). For range,  0  sin –1 ( x2  3x  2)  2  –1 2 –   log sin ( x  3x  2)  log    2

  

–   log sin –1 log sin –1

( x 2  3x  2)  (a value less than 1)

( x2  3x  2) can take all non-positive integers.

Range of f (x) = set of all non-positive intergers

...(i)

...(ii)

FUNCTIONS AND GRAPHS–ELEMENTARY FUNCTIONS

45

16. The function f (x) is defined in [0, 1]. Find the domain of f (tan x). f (x) is defined in [0, 1] 

x  [0, 1]

For f (tan x) to be defined, we must have 0  tan x  1 n x  n +/4

i.e.,

0  x  /4

or

  domain for f (tan x)   n, n   . 4 



17. If f (x) satisfies the relation, f (x + y) = f (x) + f (y) for all x, y  R and f (1) = 5, find



f (n) 

n 1

f ( n ).

f (r) = f (r – 1) + f (1) f (r) = f (r – 1) + 5 f (r) = [f (r – 2) + 5] + 5 = f (r – 2) + 2.5 = f (r – 3) + 3.5 = f (1) + (r – 1) . 5 = 5 + (r – 1) . 5 =5r

we have 

p



n 1

Also prove that f (x) is odd.



p

p

 (5n) 

n 1

5. p (p  1) 2

Putting x = 0, y = 0 in the given function, we have f (0 + 0) = f (0) + f (0)  f (0) = 0. Putting (–x) for y, we have f (x – x) = f (x) + f (– x)  f (x) + f (– x) = 0  f (– x) = – f (x), Hence f (x) is odd. 18. If the functions f and g be defined as f (x) = ex and g (x) = 3x – 2, where f : R  R and g : R  R. Then find the functions fog and gof. Also find the domain of (fog)–1 and (gof)–1. we have and Now, let 

(fog) (x) = f { g (x)} = f (3x – 2) = e3x – 2 (gof) (x) = g {f (x)}= g (ex) = 3 ex – 2 (fog) (x) = y = e3x – 2 x

log y  2 3

46

DIFFERENTIAL CALCULUS

( fog ) –1 ( x ) 



log x  2 3

and domain of (fog)–1 (x) is x > 0, i.e., x  ]0, [. Again let (gof) (x) = z = 3ex – 2

z2 3  x  2 –1 ( gof ) ( x)  log     3  x2 –1 0 and domain of ( gof ) is 3 i.e., x > –2, i.e., x  ] –2, [, 19. Let f (x + p) = 1 + [2 – 3 f (x) + 3 (f (x))2 – (f (x))3]1/3  x  R, where p > 0, prove that f (x) is periodic and find its period. We have f (x + p) = 1 [1 + {1 – f (x}3]1/3  f (x + p) – 1 = [1 – {f (x) – 1}3]1/3  g (x + p) = [1 – {g (x) }3 }1/3 where g (x) = f(x) – 1. Then g (x + 2p) = [1 – {g (x + p)}3]1/3  g (x + 2p) = [1 – {{1 – {g (x) }3}1/3}3]1/3  g (x + 2p) = [1 – {1 – {g (x)}3 } ]1/3  g (x + 2p) = [{g (x)}3] 1/3 = g (x)  f (x + 2p) = f (x). Hence, f (x) is periodic with period 2p. 

x  log

20. If the terms of the A.P. a – x , x , a  x , ..... are all integers where a, x > 0, then find the least composite odd integral value of a. Since hence 

a – x,

x,

a  x ,.... are in A.P., 2 x 

a– x 

ax

4 x  2a  2 a 2 – x 2

a2 – x2  2 x – a a2 – x2 = (2x – a)2 4a .  x = 0 or 5  The terms of the arithmetic progression are : a a a ,2 ,3 ...... 5 5 5 which are integers.  a = 5n2 ; n  N. For n = 1, a = 5 which is not composite for n = 2, a = 20, which is composite but not odd. for n = 3, a = 45 which is the least composite odd.

or 

FUNCTIONS AND GRAPHS–ELEMENTARY FUNCTIONS

47

EXERCISES 1. Find the domain of the function 1  log10 (1 – x) 2. Find the domain of the function f ( x) 

x2

  3 f ( x)  log log| sin x | ( x 2 – 8 x  23) –  log 2 | sin x |  

3. Find the domain of defintion of f (x)  2 –1  1  x 4. Find domain for f (x) = sin   2x 5. Find the domain of following : (i) 16 – xC2x–1 + 20 – 3xP4x – 5

(ii)

f ( x) 

ecos

–1

f (x) = loge | loge x |

(iv)

f ( x) 

(vi)

.

  . 

log 0.3 ( x – 1) x2 – 2 x – 8

f ( x)  cos –1 f ( x) 

x2  3x  2

(log 4 x 2 )

(iii)

(v)

log 2 ( x  3)

|x| , where [ ] denotes the greatest integer. x

log[ x ]

x2  4 x

C2 x 2  3

  x2  f ( x)  sin –1  log 2    where [.] denotes the greatest integer.  2         6. Find domain for f (x) = [sin x ] cos    [ x – 1]  7. Find the solution set of (x)2 + (x + 1)2 = 25, where (x) is the least integer greater than equal to x. (vii)

8. Find range of y 

x –1

5 – x.

9. Find the range of following : (i)

f (x) = log3 (5 + 4x – x2)

(ii)

f (x) = log [x – 1] sin x, where [.] denotes greatest integer.

(iii)

f ( x)  [sin 2 x] – [cos 2 x] , where [.] denotes greatest integer.

(iv)

f ( x)  log (cos (sin x) )

(v)

f ( x) 

sin x 1  tan x 2

–

cos x 1  cot 2 x

48

DIFFERENTIAL CALCULUS

tan ( [ x 2  x]) (vi) f (x) = 1  sin (cos x ) (vii) f (x) = [| sin x | + | cos x | ], where [.] denotes the greatest integer function.  x 1 . 10. If f is an even function; find the real values of x satisfying the equation f ( x )    x  2 11. Prove that f (x) = x – [x] is periodic function. Also, find its period.

12. Find the period if f (x) = sin x + {x}, where {x} is fractional part of x. 13. Find period of f (x) = cos (cos x) + cos (sin x). 14. Find the period of following functions : (i) (ii)

f ( x) 

1 | sin x | | cos x |     2  cos x sin x  4

f ( x)  ecos

 x – [ x]  cos2  x

x x – cos n! (n  1) ! (iv) f (x) = e ln(sin x) + tan3 x – cosec (3x – 5).

(iii)

f ( x)  sin

15. Consider the real valued function satisfying 2 f (sin x) + f (cos x) = x. Find the domain and range of f (x). 1  x 0  x  2 16. Let f ( x )  3 – x, 2  x  3 , find (fof) (x). 

17. Let f :[ 1 2 , )  [ 3 4 , ), where f (x) = x2 – x + 1. Find the inverse of f (x). 18. Find the set of all solutions of the equation 2 | y | – | 2 y – 1 – 1 | = 2 y –1 + 1.

1  19. Find the domain and range of, f (x)  log cos | x |   ; where [.] denotes the great2  est integer function. 20. Find the domain and range of f (x) = sin–1 (log [x]) = log (sin–1 [x]; where [.] denotes the greatest integer function. 21. If f : R  R is defined by f (x) = x2 + 1, then find value of f–1 (17) and f–1 (–3). 22. If a function is defined as g (x) = | sin x | + sin x,  (x) = sin x + cos x, 0  x  , then find domain for f ( x) 

23.

log  ( x ) g ( x ) .

 x – 1, – 1  x  0 f ( x)   2 and g (x) = sin x. 0  x 1  x , Find h (x) = f ( | g (x) | ) + | f (g (x) ) |.

1 1 1   {x}  , where [.] denotes greatest integral function and {.} denotes [ x] [2 x] 3 fractional part of x. 25. Find the solution of | x2 – 1 + sin x | = | x2 – 1 | + | sin x | belonging to the interval [– 2, 2]. 24. Solve

FUNCTIONS AND GRAPHS–ELEMENTARY FUNCTIONS

49

2 – x2 . 9 27. If the function f : [1, [  [1, [ is defined by f (x) = 2x (x–1) then find f–1 (x). 28. If f (x) be a polynomiol function satisfying 26. Find the range of : f ( x)  3 sin

1 1 f ( x). f    f ( x)  f   and f (4) = 65. Then find f (6).  x x 29. Let f (x) be defined on [– 2, 2] and is given by  – 1, – 2  x  0 and g (x) = f (| x | ) + | f (x) |. Then find g (x). f ( x)    x – 1, 0  x  2 30. Let two functions are defined as :

 x 2 , x 1 –1 x  2  x  1, g ( x)   and f (x)    x  2, 2  x  3 2 x  1, 1  x  2 . find gof. 31. Let f (x) = x2 – 2x, x  R and g (x) = f (f (x) – 1) + f (5 – f (x) ). Show that g (x)  0  x  R. 32. If f is a polynomial function satisfying 2 + f (x). f (y) = f (x) + f (y) + f (x y)  x, y  R and if f (2) = 5, then find f (f (2) ).

33. Check whether the function defined by f ( x  )  1 

2 f ( x) – f 2 ( x)  x  R is periodic or not. If yes, then find its period.

34. Let f (x, y) be a periodic function satisfying the condition f (x, y) = f ((2x + 2y), (2y – 2x))  x, y  R. Define a function g by g (x) = f (2 x, 0). Then prove that, g (x) is periodic function, find its period. 35. Find all the solutions of the equation  3    sin  x –  – cos  x   1 4 4    2 cos 7 x cos 2 x which satisfy the inequality cos 3  sin 3  2 36. Solve the equation : 10( x  1) (3 x  4) – 2.10( x  1) ( x  2)  101 – x – x

2

37. Prove that a2 + b2 + c2 + 2abc < 2, where a, b, c are the sides of triangle ABC such that a + b + c = 2. 38. Show that there exists no polynomial f (x) with integral coefficients which satisfy f (a) = b, f (b) = c, f (c) = a, where a, b, c are distinct integers. 39. Consider a real valued function f (x) satisfying 2 f (x y) = {f (x)}y + {f (y)} x for all x, y  R and f (1) = a, where a  1. n

Prove that (a – 1)



i 1

f (x) = an + 1 – a.

40. Let f : R – {2}  R function satisfies the following functional equation :  2 x  29  2 f ( x)  3 f    100 x  80,  x  R – {2}. Determine f (x).  x–2 

50

DIFFERENTIAL CALCULUS

ANSWERS

 3   3  ]3,  [   ,  ,5  ; 2   2   4. {– 1, 1} ;

1. [– 2, 0[  ]0, 1[ ;

2.

3. ]– 3, [ – {– 1, – 2} ; 5. (i) (iii) (v)

x = {2, 3} x  ] 0, 1[  ]1, [ ;

(ii) (iv)

x  [– 2, – x  ] 2, 4 [

x  ] 2, [ ;

(iv)

] – 8 – 1[  [1, 8] ;

6. R – [1, 2] ;

1

2

]  [ 1 2 , 2]

7. x  ]– 5, – 4]  [2, 3] ;

Range  ]– , log3 9[ {0, 1} ; [– 1, 1] ; {1} ;  – 1  5 – 3  5  , ; 10. x   2 2   12. not periodic ; 14. (i) 2 ; (iii) 2 (n + 1) | ; 9. (i) (iii) (v) (vii)

8. [2, 2 2] ;

(ii) [– , 0] ; (iv) {0} ; (vi) {0} ;

11. period = 1 ; 13. /2 ; (ii) 1 ; (iv) 2 ;

 2  , ; 15. Domain  [– 1, 1], Range   – 3 3   17. x = 1;

 2  x, 0  x  1  16.  2 – x 1  x  2  4 – x, 2  x  3,  18. {y : y  1  y = – 1};

    19. Domain : x    (2n – ), (2n  )  n 1  3 3  Range : {0}

  20. Domain : [1, 2[ ; Range of f ( x)  log  2    22. Domain of f (x) is ;  ,  ; 6 2 2 sin x – sin x  1, – 1  x  0  23. h ( x)   2sin 2 x, 0  x 1   25. x  [– 2, –]  [–1, 0]  [1, ]  {2 } ;  3 3 26. range of f (x)   0,  2  

28. 217 ;

21. x = {+ 4, – 4} ;

24. 29 , 19 , 97 ; 12 6 24 1  1  4 log 2 x

; 2 –2x0  – x,  0  x 1 29. g ( x)   0, 2 ( x – 1) 1  x  2  27.

30. (gof) (x) = {(x + 1)2, – 2  x  1}

32. f (f (2) ) = 26

33. 2  ;

34. period = 12 ;

FUNCTIONS AND GRAPHS–ELEMENTARY FUNCTIONS

35.

x  2n  

3 ; 4

51

36. x  – 1 

(2 x  29) 40. f (x) = 16 – 40 x – 60 ( x – 2) .

1 log10 (1  11) 2

2.10 Polar Co-ordinates and Graphs While we have so far considered the cartesian system of co-ordinates,we now intoroduce to Polar System. In the polar system of co-ordinates, we start with a fixed line called the Initial line and a fixed point on it called the Pole. Let OX be the initial line and O the pole. Let P be a given point. Then the distance OP, called the radius vector, and the  XOP, called the vectorial angle, are called the polar co-ordinates of the point P. Writing OP = r,  XOP, = , we may refer to the point P as (r, ). (Fig. 2.56)

Fig. 2.56

Graph of a function r = f () is the set of points [, f ()] where  belongs to the domain of the given function f. Also we say that r = f () is the polar equation of the graph. Unrestricted variation of polar co-ordinates. If we were concerned with assigning polar coordinates to only individual points in the plane, then it would be enough to consider the radius vector to have positive values and the vectorial angle to lie between 0 and 2 only. While considering graphs of given functions or equivalents it becomes necessary, however, to remove this restriction and to consider both r and  as varying in the interval [– , ]. The necessary conventions for this will be introduced now. The angle, , will be regarded as the measure of rotation of a line which starting from OX revolves round it, the measure being positive or negative according as the rotation is counter-clockwise or clockwise. To find the point (r, ) where r, is negative and,  has any value, we proceed as follows : Let the revolving line starting from OX revolve through, , Then produce this final position of the revolving line backwards through O The point P on this produced line such that OP = | r | is the required point (r, ). Thus if r is negative, the point (r, ) is the same as the point (| r |,  + ). The positions of the points (1, 9/4), (1, – 9/4), (– 1, 9/4), (–1, –9/4) have been marked in the figures 2.57 to 2.60 given below :

Fig. 2.57

Fig. 2.58

Fig. 2.59

Fig. 2.60

52

DIFFERENTIAL CALCULUS

We may notice that of the two points (r, ), (r, – ) each is the reflection of the other in the initial line. Thus the graph of the function f or equivalently that of the equation r = f () will be symmetrical about the initial line, if we have f (– ) = f ()  . In case, therefore, the graph of the function is known to be symmetrical about the initial line, we may, in the first instance, draw only that part of it which corresponds to  [0, ] . The complete graph will then consist of the union of the part already drawn and its reflection in the initial line. Transformation of co-ordinates. It is often found necessary and useful to express a given cartesian equation as a polar equation and vice-versa. For this purpose, we need relations between the cartesian co-ordinates (x, y) and the polar co-ordinates (r, ) of the same point. These relations will naturally assume a simpler from if the reference systems for the two systems of co-ordinates are chosen so as to be related in some suitable manner. We suppose that the initial line of the polar system is the positive direction of x-axis and the pole is the origin. The positive direction of y-axis will then be such that the initial line after revolving through /2 in the counter-clockwise direction coincides with it. Let (x, y) and (r, ) be the Cartesian and Polar co-ordinates respectively of a point P. From the OMP, we get OM/OP = cos  x = r cos 

...(i)

MP/OP = sin  y = r sin 

...(ii)

The equations (i) and (ii) determine the cartesian co-ordinates (x, y) of the point P in terms of its polar-co-ordinates (r, ) and vice-versa. We give below the cartesian forms of a few equations and the corresponding forms obtained by means of the ralations. x = r cos , y = r sin  Cartesian form r

x3 + y3 = 3axy

Polar form 3a sin  cos 

cos3   sin 3  4a sin  cos  r2  sin 4   cos 4  1 r 2  tan 2  2 a 2 cos 2  2 r  cos 4   sin 4 

x4 + y4 = 4a2xy y4 – x4 + xy = 0 x4 + y4 = a2 (x2 – y2)

OBJECTIVE QUESTIONS Note : For each of the following questions four alternatives are given for the answer. Only one of them is correct alternative. Choose the correct alternative. 1.

The domain of the function f ( x) 

1 –| x | is |x|–2

(a) x  ]– , – 1[  ]1, [

(b) x  ]– , – 2[  ]2, [

(c) x  ]– 2, – 1]  [1, 2[

(d) none of these

FUNCTIONS AND GRAPHS–ELEMENTARY FUNCTIONS

2.

The number of roots of the equation

 3   x in  – ,  is 2 2 

cot x  (a) 3 3.

53

(b) 2

(c) 1

(d) infinite

(c) (– , )

(d) none of these

The solution of the equation | 2x – 1 | + | 4 – 2x |  3 is, (a) (– , 0)

4.

(b) (0, 2)

The domain of f ( x)  1 –

x2 is x –1

– 1 – 5 – 1  5 ,  (b)  2 2  

(a) ]– , [ – {1} – 1 – (c)  2 

5.

5 –1 , 2

3 Period of the function sin

(a) 2  6.

7.

5   ]1, [ 

 –1 5  ,1  (d)  2  

x x  cos5 is 2 5

(b) 10 

(c) 8 

 x The domain of f ( x)   log2  ; where [ . ] denotes the greatest integer function is [ x ]  (a) ]– , [ – [0,1[ (b) ]– , 0[ (c) [1, [ (d) none of these

The domain of f ( x)  (tan x – 1)(tan x – 2) is

  –1  (a)  n   , n   tan 2  4     –1  (c) R –  n   , n   tan 2  4   8.

 –1  (b)  , tan 2  4  (d) none of these

The solution set of the equation tan –1

x ( x  1)  sin –1

(a) ]–1, 0[ 9.

(d) 5 

If f ( x ) 

1 x

10. Let f ( x 

 is 2

(b) [–1, 0] 1 1  x2

(c) {–1, 0}

x 2

(d) none of these

, then (f o f o f) (x) is

3x

(a)

x2  x  1 

(b)

1  3x

3x 2

(c)

1 – x2

(d) none of these

1 1 )  x 2  2 , where x  0 ; then f (x) is x x

(a) x2 – 2, x  0

(b) x2 – 2, | x | < 2

(c) x2 – 2, | x |  2 (d) none of these

54

DIFFERENTIAL CALCULUS

11. The period of e cos

4

 x  x – [ x]

 cos 2  x is

 (d) 2 2 12. The range of the function f (x) = sin (sin–1 {x}) where { . } is fractional part of x, is (a) [0, 1[ (b) [0, 1] (c) ]– 1, 1[ (d) none of these x  13. If f ( x)  sin  , x, r  ] 0, [, then range of f (x) is r   (a) ] 0, 1] (b) [0, 1] (c) {0, 1} (d) {0}

(a) 0

(b) 1

(c)

  2 14. Let f :  – ,   [0, 4] be a function defined as f ( x)  3 sin x – cos x  2. Then  3 3  f–1 (x) is given by  –1  x – 2  (a) sin  –  2  6 (c)

2  cos –1 3

 –1  x – 2  (b) sin    2  6

 x – 2    2 

(d) none of these

15. If f is a function such that f (0) = 2, f (1) = 3 and f (x + 2) = 2 f (x) – f (x + 1) for every real x then f (5) is (a) 7 (b) 13 (c) 1 (d) 3  x 2 for x  0 16. If f (x) =  then fof (x) is given by  x, for x  0 (a) x2 for x  0, x for x < 0 (b) x4 for x  0, x2 for x < 0 4 2 (c) x for x  0, – x for x < 0 (d) x4 for x  0, x for x < 0 2 17. If f (x) + 2 f (1 – x) = x + 2,  x  R, then f (x) is given as ( x – 2) 2 (b) x2 – 2 (c) 1 (d) none of these 3 18. If f : I  I be defined by f (x) = [x + 1], where [ . ] denotes the greatest integer function, then f–1 (x) is equal to

(a)

(a) x – 1

(b) [x + 1]

19. Domain of f (x) satisfying 2x + 2f (x) = 2 is (a) ] – 1, 1[ (b) ] – , – 1[ 3  x – 1, x  2 20. Let f (x)  2 , then f–1 (x) is  x  3, x  2

 ( x  1)1/ 3 , x  2 (a)  1/ 2 ( x – 3) , x  2 1. 8. 15.

(c) (c) (a)

2. (a) 9. (b) 16. (d)

1 (c) [ x – 1]

1 (d) x  1

(c) ]– , 1[

(d) none of these

 ( x  1)1/ 3 , x  7  ( x  1)1/ 3 , x  1 (b)  (c)  (d) does not exist. 1/ 2 1/ 2 ( x – 3) , x  7 ( x – 3) , x  1 3. 10. 17.

(b) (c) (a)

ANSWERS 4. (b) 11. (b) 18. (a)

5. (b) 12. (a) 19. (c)

6. (c) 13. (c) 20. (b)

7. (c) 14. (b)

CHAPTER 3 CONTINUITY AND LIMIT 3.1. Introduction The statement that a function is defined in a certain interval means that to each number belonging to the interval, there corresponds a value of the function. The value of the function for any particular number may be quite independent of the value of the function for another number and no relationship in the various values of a function corresponding to the members of the domain is implied in the definition of a function as such. Thus, if x1, x2 are any two members of the domain of a function f so that f (x1), f (x2) are the corresponding values of the function, then | f (x2) – f (x1) | may be large even though | x2 – x1 | is small. It is because the value f (x2) assigned to the function for x = x2. is quite independent of the value f (x1) of the function for x = x1. We now propose to study the change | f (x2) – f (x1) | relative to the change | x2 – x2 | by introducing the notion of Continuity and Discontinuity of a function. In the next article, we first analyse the intuitive notion of Continuity and then state its precise meaning in the form of a definition. 3.1.1. Continuity of a function at an interior point of an interval. Intuitively, continuous variation implies absence of sudden changes so that in order to arrive at a suitable definition of continuity, we have to examine the precise meanings of its implication in relation to a function which is defined in an interval [a, b]. Let c, be a point of ] a, b [ , i.e., an interior point of [a, b ]. The function f will vary continuously at c if, as x changes from c to either side of it, the change in the value of the function is not sudden, i.e., the change in the value f (x) of the function is small if only the change in the value of x is small. We consider a value c + h of x belonging to the interval [a, b]. Here, h, which is the changes in x, may be positive or negative. Then f (c + h) – f (c), is the corresponding change in the value of the function f, which, again, may be positive or negative. For continuity, we require that f (c + h) – f (c) should be numerically small, if h is numerically small. This means that | f (c + h) – f (c) | can be made as small as we like by taking  h  sufficiently small. 55

DIFFERENTIAL CALCULUS

56

The precise definition of continuity should not involve the use of the word small, whose meaning is indefinite, as there exists no absolute standard of smallness. Such a definition would now be given. A function f is continuous at c if, to any positive number, arbitrarily assigned, there corresponds a positive number such that | f (c + h) – f (c) | <  for all values of, h, such that |h| 0 there exists  > 0 such that | f (b – h) – f (b) | <  when 0  h 1 so that 3 (x – 1) is positive and is, therefore, the numerical value of f (x) – f (1). Now | f (x) – f (1) | = 3 (x –1) < .001 if x – 1 < .001/3, i.e., if x < 1 + .001/3 ...(i) (ii) Let x < 1, so that 3 (x – 1) is negative and the numerical value of f (x) – f (1) is 3 (1 – x). Now | f (x) – f (1) | = 3 (1 – x) < .001 if 1 – x < .001/3, i.e., if 1 – .001/3 < x ...(ii) Combining (i) and (ii), we see that | f (x) – f (1) | < .001 for all those values of x for which 1 – .001/3 < x < 1 + .001/3. The test of continuity for 1 is thus satisfied for the particular value .001 of . We may similarly show that the test is true for other particular values of  also. The complete argument is, however, as follows : Let  be a given positive number. We have | f (x) – f (1) = 3 | x – 1 |. Now 3 | x – 1| < x – 1| 0 there corresponds G > 0 such

(iii)

that f  x   l    x  G.

A function f is said to tend to l as x tends to –  , if to any given  > 0, there corresponds G > 0 such that f  x   l    x   G.

(vi) lim f  x   , lim f  x   , x x  A function f is said to tend to  , as x tends to if to given  > 0, there corresponds G > 0 such that (v)

f  x     x  G. A function f is said to tend to +  as x tends to –  if to given  > 0, there corresponds G > 0 such that

f  x     x   G. We may similarly give meanings to

lim f  x    , lim f  x    , x x  Instead of , we may find it more useful to write +  EXAMPLES 1. Show that





lim 1/ x 2  . x0 The function is not defined for 0, i.e., 0 does not belong to the domain of the function f  x   1/ x 2 .

DIFFERENTIAL CALCULUS

68

We write y  1/ x 2 . Case I. Let x > 0. Consider any positive number, say 106.

y  1/ x 2  106 ,

Now

if 0  x  1/103.

Instead of the particular number 106 , we may consider any positive number G. Then

1/ x 2  G if 0  x  1/ G . Thus, there exists an interval  0, 1/ G  , such that for every value of x belonging to it, y is greater than the arbitrarily assigned positive number G, Hence





lim 1/ x 2  . x   0  0 Case II. Let x < 0 be any arbitrarily assigned positive number. Then,  x    1/ G , 0  we have 1/ x 2  G





lim 1/ x 2  . x  0  0 Case III. Combining the conclusions arrived at in the last two cases, we see that corresponding to any arbitrarily assigned positive number G, there exists an interval   1/ G , 1/ G  , around 0, such that for every value of x (other than 0) belonging to this interval, we have 1/ x 2  G. Thus and as such









lim 1/ x 2   or 1/ x 2   as x  0. x0 2. Show that 1 1 lim = , lim =– x x  0 + 0  x  0 – 0  x 1 and lim does not exist. x0 x The function f (x) = 1/x is not defined x = 0. We write y = 1/x.

Case I. Let x > 0 so that y is positive. If G be any positive number taken arbitrarily, then 1/ x  G, if 0 < x < 1/G



lim (1/x) = 

when x  (0 + 0).

Case II. Let x < 0 so that y = 1/x is negative. If G be any positive number taken arbitrarily, then 

1/x < – G

if – 1/G < x < 0

lim (1/x) = – 

when x  (0 – 0).

Case III. Clearly when x  0, lim (1/x) does not exist. 3. Consider f (x) = (2x + 1)/(x – 3) and show that

lim f  x   2, lim f  x   2. x x 

CONTINUITY AND LIMIT

69

Let  > 0 be given. We have

2x  1 7 7 2    for x   3 x3 x3  lim f  x   2. x Again, we have

so that

2x  1 7 –7 2    for x  3 x3 x3  so that

lim f  x   2. x Combining the two results, we see that lim f  x   2. x 

EXERCISES Show that : (i) (ii)

lim 1/  x  2    , x   2  0

lim 1/  x  2    . x   2  0

lim  5 x  2  /  x  3    , x  3  0

lim  5 x  2  /  x  3   . x  3  0









(iii)

lim 5 x 2  3 / x 3   , x   0  0

lim 5 x 2  3 / x3   . x   0  0

(iv)

lim x 2   , x

lim x 2   . x

(v)

lim

x 3   ,

lim ( x3 – x 2 )   .

x c  

x 

3.4 Extension of Operations on Limits Corresponding to the theorems on limits as given in § 3.3. page 66, we also have results pertaining to infinite limits and variables tending to infinity. We give these results without proof in the following tables. These results are valid when x tends to c,

c – 0. If lim f is

c + 0,

+

If lim g is

or

– Then lim f + g is

1

l

+

+

2

l

–

–

3

+

+

+

4

–

–

–

5

+

–

No general conclusion

DIFFERENTIAL CALCULUS

70

If lim f is

If lim gis

Then lim fg is

1

l 0

+

+

2

0

+

No general conclusion

3

+

+

+

If lim gis

Then lim 1/gis

1

0

+

2

+

0

If lim f is

If lim gis

Then lim f | 1g is

1

l 0



0

0

3

l

+

0

4

+

l

+

5

+

+

+

+ No general conclusion

No general conclusion

Indeterminate Forms. From the first table, it is seen that if lim f (x) = +  and lim g (x) = –  we cannot deduce any specific conclusion about lim [f (x) + g (x)]. The fact is that in such a case the behaviour of f (x) + g (x) just does not depend upon the individual behaviours of f (x) and g (x) but on their behaviours relatively to each other. We know consider a few examples to illustrate the different possibilities. EXAMPLES 1.

lim

 x  1  ,

x

 x  1  .

lim x

We examine

lim  

 x  1

 x  1



 x  1  .



x

We have

 x  1





 x  1   x  1  x  1   x  1 2

 x  1   x  1

 0 as x  .

Thus while

lim

 x  1

x

we have

x

lim  

 x  1

x





x

2.

lim x

2

 , lim

 x  1   0.

 1  , lim x

 x  1

 .

 x  1

 .

CONTINUITY AND LIMIT



Now

71



x2  1 

x x

 x  1 

2

2

  1 

 1   x  1

x  x  1



 x  1

lim  x  

x

lim

x

2



 x  1 1

  x  1

so that

 x  1

1   1  2   x  

1  1   2 x x  

 x  1   .



1 



Thus while x

lim  x  

we have lim

3.

x0

1 x

2

2

x

2



 x  1

 1  , lim

 x  1   .



1 

 , lim x0

1 x4

 ,

x



 ,

1  x2  1  1 lim  2  4   lim   . x  x  0 x4 x0x 4.

1  1    lim  2  2   , lim 1  x  2   , x  x  x0 x0  1   1 lim  2  2    1  x  2 x x    x0  In case, we have

lim f

 x   ,

xa

    lim 1  x   1.  x  0

lim g  x   , xa

we say that f (x) – g (x) assumes the Indeterminate form – when x  a.

lim

5.

cot x  ,

x0

lim  cot  1 x 

x

To check lim

x0

 cot x  , put cot x = t, now as x  0; cot x exhibits two values for x  (0 + 0) and

x  (0 – 0), i.e., cot x +  and cot x –  respectively.  we should apply right hand and left hand limit; i.e.,

lim

cot x  , 

x  (0  0)

t   

lim x

 cot x  t  t    as x   0  0  

and

lim x  (0 – 0)

cot x  

lim

t    

x

 cot x  t  t    as x   0  0  

limit does not exist.

DIFFERENTIAL CALCULUS

72

lim cot  1 x  

To find

x

t 

lim t  [0  0]

 cot  1 x  t  t   0  0  as x      

  0  h   0.

0  , . 0  In the context of the tables 2 and 4, we make the following statements: 1. If f (x) = 0 and lim g (x) =  then f (x) g (x) is said to assume the Indeterminate form 0.. Other Indeterminate Forms 0. ,

2. If lim f (x) = 0 and lim g (x) = 0, then f (x)  g (x) is said to assume the indeterminate form

0 . 0

 .  In the following exercises, we shall give several cases of the determination of the limits of expressions which assume one or the other types of the Indeterminate forms.

3. If lim f (x) = , lim g (x) = , then f (x)  g (x) is said to assume the indeterminate form

The general procedure for determining the limits of expressions assuming indeterminate forms will be considered in Chapter IX. EXAMPLES

lim

1. Find

x

We have Now so that



3

8 5 7   7 x3  8 x 2  5 x  7  x3  7   2  3  . x x x   8 5 7 lim  lim  lim 0 2 3 x x x x x x 8 5 7   lim  7   2  3   7. x x x  x  lim x 3  .

Also

x

lim

Thus

x

2. Obtain



2

7x  8x  5x  7 .

7 x

7 x x lim

3

3



 8 x 2  5 x  7  .



 8 x2  5 x  7 .

8 5 7   We have 7 x3  8 x 2  5 x  7  x3  7    3 . 2 x x x   8 5 7 lim  lim  lim  0. Now 2 3 x x x x x x Also

lim

x3   .

x

It follows that

7 x x lim

3



 8 x 2  5 x  7   ,

CONTINUITY AND LIMIT

73

 2 3. Evaluate : lim  x  x  1  x    lim  x 2  x  1  x  

 x2  1  

 2  x  x 1    x 2  1   lim 1 x   

x

x



1 . 2

2

x  x 1 

x2  1

1

 lim

1 11

 x2  1    2 x  1 

x

 lim



  x2  1   x2  x  1    2  x  x 1  

1 1 1   1 2 x x2 x  1  as x  0 as x     

1

 n  2 !   n  1! n    n  2  !   n  1 !

4. Evaluate  lim

G iven limit = lim

x 

( n  1) ! [n  2  1] n3  lim ( n  1) ! [n  2 – 1] x 0 n  1

lim

n

5. Solve : lim

1  3/ n 1  0  1 1  1/ n 1  0

 1   as n  0 as n   

– 3n  (– 1) n

4n – (– 1) n while evaluating the given limit, we come across two cases : when n is even or n is odd. x

Case 1. n is even, say n = 2k lim



k 

– 6 k  (– 1) 2 k 8 k – (– 1) 2 k

1 k  1 8– k –6 –3  .  8 4 Case II : n is odd, say n = 2k + 1 – 6k  1 lim  lim k  k  8k – 1



lim k 

– 3 (2 k  1) – (– 1) 2 k  1 4 (2 k  1) – (– 1)

2k  1

–6

 lim

k 

– 6k – 2 8k  5

DIFFERENTIAL CALCULUS

74

2 k  lim  k  5 8 k –3  4 –6–



– 3 n  (– 1) n

lim n 

4 n – (– 1)

n

–3 . 4



 n 1  Find lim    , where . denotes the greatest integer. r n   r=1 2 

1 1 (1 – n ) n 1 2 2   1 –    2r 1 – 1/ 2 2 r 1 n

1

This tends to one as n   but always remains less than 1.  n 1   r   0 as n   2  1 2 

Thus

n 1 lim   r    . n   r 1 2 



7. Find L = lim

x 

Here, lim

x 

a xp + b xp – 2 + c d xq + e xq – 2 + k

, where a, b, c, d, e and k are constants, and p > 0, q > 0.

a xp  b xp – 2  c d x q  e xq – 2  k

 d  e / x

  k/x 

x p a  b / x2  c / x p  lim

x 

xq

2

q

Now , we have to consider here cases: Case I : p>q b c   x p – q a  2  p  x x    lim x  e k   d  2  q  x x   {Since p – q > 0  x p – q   as x   } =

Case II :

p=q b c   a  2  p  x x   lim  x   e k  d  2  q  x x   a  d

CONTINUITY AND LIMIT

75

p 0, the same also holds when x < 0 in that sin (– x)/(– x) = sin x/x and cos (– x) = cos x. Now since lim cos x  1 it following that x0

lim

x 0

sin x  1. x

EXAMPLES 1. Evaluate: lim

1 – cos x

. x2 1 – cos x 2 sin 2 x / 2 lim  lim x0 x 0 x2 x2 x 0

CONTINUITY AND LIMIT

77 2

lim x0

1  sin x / 2  1 2 1    1  . 2  x/2  2 2 sin 2 x – sin 2 y

2. Evaluate : lim x y

lim

sin 2 x – sin 2 y x2 – y 2

x y

x2 – y 2 sin  x  y  sin  x – y   lim x y  x  y  x – y sin  x  y  sin  x – y   lim . lim x y  x  y  x y  x – y  sin 2 y   1  x  y   x – y   0, but x  y  2 y   2y sin 2 y  . 2y



sin  cos 2 x

3. Evaluate: lim

x

x 0



2

sin  cos x lim

x2

x 0



2

  lim sin  1 – sin x  2

x2

x 0



sin  –  sin 2 x  lim



x2

x 0



sin  sin 2 x  lim

x

x 0



2

2   sin 2  sin  sin x  lim    2 x 0 1 x2   sin x







sin  sin 2 x  lim

2

 sin x =1××1= x 0

4.

lim x  / 4

lim x  / 4

1–

 x  

    lim sin x 0

x

2

x

2

sin 2 x

 – 4x 1–

sin 2 x

 – 4x

lim = x /4  lim x  / 4

1–

sin 2 x

 – 4x

.

1  sin 2 x 1  sin 2 x

1 – sin 2 x 1 . lim x  / 4  – 4x 1  sin 2 x

  sin  – x  4   .1  lim x  / 4   4 – x  4 

which gives R H L at x   Limit does not exists.

 1  1  – and L H L at x   4 4 4 4

DIFFERENTIAL CALCULUS

78

EXERCISES 1. Show that (i)

lim

x 0

sin a x a  , b  0 sin b x b

lim

cos x 1 / 2 – x

(iv)

x  / 2

(vi)

x  0 – 0 

(ix)

3 sin1 x 3  x0 4x 4

lim

(v)

x

–

1 – cos x 

tan x 1 x

(ii) xlim 0

2 2 (vii)

lim

(x)

lim

x  0  0 

x

1 – cos x 



sin x 1 tan x

lim

x 0

lim

(iii) xlim 0 2 2

(viii) xlim 0

sin 1  x  – sin 1 – x  x

x 0

1 – cos x 1  2 x2

tan x – sin x 1  2 x3

 2 cos 1

(In each of the above cases we deal with indeterminate form 0/0) 2. Show that (i)

lim x cot x  1

(ii)

x 0

3. Show that 4. Show that

lim 1 – x  tan x 1

x 2  2 

(iii)

lim x

x 0

 cot x   0

cosec 2 x 1 x 0 cot 2 x lim  cos ec x – cot x   0 lim

x 0

5. Examine whether the following functions are continuous at x = 0. (i)

sin x , when x  0 f  x   1 , when x  0

(ii)

sin 2 x , when x  0 f  x   1 , when x  0

(iii)

 tan 2 x , when x  0   3x f ( x)   2  3 , when x  0

ANSWERS 5. (i)

discontinuous,

(ii) discontinuous, (iii) discontinous.

3.5.2. The number e. To consider n

1  lim  1   n  when n tends to infinity through positive integral values. Step 1. By the Binomial Theorem for a positive integral index, we have n n  n – 1 n – 2  1 1 1 n  n – 1 1  . 2  . 3 1    1  n .  n n 1. 2 1. 2 . 3 n n  n  n – 1 n – 2  ...........  n – n  1 1  ...  . n 1 . 2 . 3 ......... n n

CONTINUITY AND LIMIT

11

79

1  1 1  1  2 1 –    1 –  1 –  1.2  n  1.2.3  n  n  ... 

1  1  2 n – 1  1 – 1 –  ........ 1 – . 1.2... n  n  n n  

The expression on the right is a sum of (n + 1) positive terms. Changing n to n + 1, we get  1  1   n  1 

n 1

11

1  1  1 –   ....... 1.2  n  1

  1 1  2  n  1 –  1 –  ........ 1 – . 1.2...(n  1)  n  1 n  1 n  1  The right hand side consists of the sum of (n + 2) positive terms. Also 

1 1 2 2 3 3 1– ,1 – 1 – ,1 – 1 – ,........ n n 1 n n 1 n n 1 so that each term in the expansion of 1–

n 1

 1  1   n  1  is greater than the corresponding term in the expansion of n

1  1   . n  n 1

n

 1  1   1   Thus we conclude that  1   n  1 n   n for all positive integral values of n, i.e., (1 + 1/n) increases as n increases. Step II. We have n

1 1  1 1  1  2  1    1  1 1 –   1 – 1 –  n 1.2  n  1.2.3  n  n  1 1  2 n –1    ......   1 –  1 –  ........... 1 –  , ...(i) 1.2.3... n  n  n n   1 1 1 11   ...  2 1.2.3 1.2.3...n 1 1 1 11   ...  2 2.2 2.2.2...  n – 1 factors 1 1 1  2  ...  n – 1 2 2 2 n 1 – 1/ 2  1 1  3 – n – 1  3  n. 1 – 1/ 2 2

11

n

...(ii)

Thus we see that 2 < 1  1/ n  increases as n takes up successively positive integral values 1, 2, 3, etc., and remains less than 3 for all values of n. n Hence 1  1/ n  approaches a finite limit as n  .

DIFFERENTIAL CALCULUS

80 n

From (i) and (ii) we see that 1  1/ n  for every value of n so that the limit is a number which lies between 2 and 3. n

It can be proved that lim 1  1/ n  is an irrational number. This number is denoted by e. While we are not giving here any method for computing the value of the number e upto any given number of decimal places, we only state that its value upto 10 significant decimal places is e = 2.7182818284 as can be shown by advanced methods. Note 1. The reader may be able to appreciate the underlying idea of the proof better if he calculates the value of an = (1 + 1/n)n for a new successive value of n. Thus for example a1 = 2, a2, = 2.25, a3 = 2.37, a4 = 2.441,....... Note 2. The proof for the existence of the limit has been based on an intuitively obvious fact that an increasing function which remains less than a fixed number tends to a limit. The formal proof of this fact is beyond the scope of this book. Cor. 1. lim (1 + 1/x)x = e when x tends to infinity taking all the real numbers as its values. If x be any positive real number, then there exists a positive integer n such that nx 1. We write x = 1 + h so that h is positive. By the Binomial theorem for positive integral index, we have n

x n  1  h   1  n h 

n  n – 1 2!

h 2  ..........

n  n – 1 ... 1 n!

hn ,

n

where each term is positive. Thus x  1  n h. Let G be a given positive number. Then we have 1 + nh > G, if n > (G – 1)/h. If  be a positive integer greater than (G – 1) h, then xn > G  n >  n Thus, x > 1  lim x = . (ii) Let x = 1. Here x n  1 for all values of n and therefore, x n  1 in this case. n (iii) Let 0 < 1 < x so that, x n is positive. Let x = 1/y so that y > 1 and lim y  . n 

Let  > 0 be given. As yn  , there exists m such that y n  x n is necessarily positive, it follows that

1  n  m  x n    n  m. As 

DIFFERENTIAL CALCULUS

82

Thus

lim xn = 0 0 < x < 1  lim xn = 0

if 0 < x < 1.

(iv) Let x = 0. Here x n  0 for all n so that xn  0. (v) Let – 1 < x < 0. We write x = – a so that a is a positive number less than 1. We have | xn | = n a . But an  0. Hence xn  0 Thus, – 1 < x < 0  lim xn = 0. (vi) Let x = – 1. Since xn is alternatively – 1 and 1, therefore, it neither tends to any finite limit nor to ± . (vii) Let x < – 1. Here, xn which is alternatively negative and positive, takes values numerically greater than any assigned number, xn does not tend to a limit. Thus, we see that lim x n , when n  , exists finitely if, and only if, – 1 < x  1. Also, it is 0 if x < 1 and is 1 for x = 1. II. Lim xn /n !, when n tends to infinity through positive integral values and x is any given real number. Let x be a positive real number and let m, m + 1 be two consecutive integers between which x lies so that we have m  x < m + 1. We write xn x x x x x x x  . . ...... . . ....... . n! 1 2 3 m m 1 m  2 n x x x Let p  . .......... so that p is a positive number independent of n. 1 2 m x x x x Also, each of m  2 , m  3 , .......... , n is  m  1 . 

0

 x  xn  p  n!  m  1

n–m

–m

 x   p   m  1 n  x   k  ,  m  1

 x     m  1

n

say

where k is a positive number independent of n. n

 x  x 0  1  lim  Now,   0. n  m 1  m  1 Thus, if x be a positive number

xn  0, when n  n! If x be a negative number, say –  , so that  is positive, we have lim

xn  n!

Hence

 –1n  n n!

xn  0  x  R. n  n ! lim



n . n!

CONTINUITY AND LIMIT

83

III. Lim un  0 where un 

m  m – 1 ........  m – n  1

 n – 1!

n

x n , if x  1.

Changing n to n + 1, we get un  1  un  1 un

lim

Thus,

un  1 un

n 



m  m – 1 ........  m – n  n!

x n  1.

m–n m  x  –  1 –  x. n n 

 x.

Let k be a positive number, such that | x | < k < 1. These exists a positive integer p such that un  1 un

 k  n  p. Thus

up 1  k up , up  2  k up  1 ,

......................... ........................ un  k u n – 1 .

Multiplying, we get un  k n –

p

up  kn up k p .

p Now, u p k is a constant independent of n and kn  0 as n 

Thus, lim un  0 if x  1. n 

3.5.4. Some miscellaneous Limits. (i) Logarithmic Limits : These limits are based on expansion of logarithmic series x2 x3   ...... to  2 3 where – 1  x  1and this expansion is true only if the base is e. log 1  x   x –

To evaluate the logarithmic limit we use lim x0

log 1  x  x

 1.

(ii) Exponential limits These limits are based on the expansion

x2  ........  2! To evaluate exponential limit we use the following result, ex  1  x 

ex  1 ax  1  1 and lim  log e a x 0 x 0 x x To evaluate the exponential form 1 we should use the following results: lim

DIFFERENTIAL CALCULUS

84

If

1/ g  x 

lim f  x   lim g  x   0, then lim 1  f  x  

x a

xa

xa

lim

e

x a

f  x gx

lim f  x   1 and lim g  x   ,

or, when

xa

xa

lim  f  x   x a

then

g x

g x

 lim 1  f  x   1 xa

lim  f  x   1 g  x   e x a we should also remember, while evaluating exponential limit, that

x

1/ x

lim 1  x 

x 0

lim 1  x 

x

x 0

1   e ; lim  1    e ; x   x x    e ; lim 1    e  x   x

If we have to calculate the limit O° of form, then lim f  x   1 but f (x) is positive in the xa neighbourhood of x = a. In this case we write

 f  x  

g x



lim  f  x   x a

gx

e

log e  f  x 

g x

lim g  x  log e f  x 

 e xa

Here, it is to be noted that if f (x) is not throughout positive in the neighbourhood of x = a, then lim  f  x   of x = a.

g x

x a

will not exist, because in this case function will not be defined in the neighbourhood EXAMPLES

1. Evaluate: lim

log e 1  2 h  – 2 log e 1  h  h2

h 0

  2 h 2   2 h 3 ......    2  h – h 2  h3 ......   2 h  –   2 3 2 3      lim 2 h 0 h

Given that

 lim

h 2  –1  2 h  .......... h2

h 0

= – 1. 2. Evaluate: lim

x 0

 a bx

– ax – bx  1 x2



.

 



ax bx – 1 – bx – 1

Given limit

 lim

x 0

 lim

x

a

x

–1

2

  lim b

x 0 x = log a × log b.

x 0

x



–1 x

CONTINUITY AND LIMIT

85

e tan x – e x x  0 tan x – x

3. Evaluate: lim





e x e tan x – x – 1  lim

Given limit

[As x  0, tan x – x  0 ]

tan x – x

x0

= e° × 1 = 1 a  4. Evaluate: lim  2 –  xa  x

tan

x 2a

.

  lim 1  xa 

Given limit

e e

a   1 –  x  

tan

x 2a

a x  lim  1 –  tan x a  x 2a x x – a lim   tan x a  x  2a

Let x – a = h      

 a  h  h  lim   tan h0 a  h 2a   e

e e

lim

  h  h tan    ah  2 2a 

lim

  h   h   – cot   a  h   2 a  

h0

h 0

lim

e e e

–h

.

 1 . 2a / 2a

h0

 a  h  tan  h / 2 a 

lim

– 2a h / 2a .   a  h  tan  h / 2a 

h0

– 2a a



e

–2  .

1

5. Evaluate:

   x lim  tan   x   . x0  4   1

The given limit

1  tan x  x  lim   x 0 1 – tan x  1

 2 tan x  x  lim 1   x 0  1 – tan x 



 2 tan x  1 lim   x 0 1 – tan x x e  lim

e

x0

= e 2.

2 tan x 1 . x 1 – tan x

DIFFERENTIAL CALCULUS

86

EXERCISES Evaluate the following limits : lim

1.

log 1   x – a 

2. lim

 x – a

x a

4.

2  lim  1   x   x

7.

 1  5 x2  lim   x 0 1  3 x 2   

 x  6   x 1

x4

x– x

x



1  tan 2 x 6. xlim 0

5. xlim   1/ x 2

lim 1  sin  x 

10.

3. lim

x

x 0

x

ax – bx x0 x

log  5  x  – log  5 – x 

8. xlim  

3  2

9.

lim x  / 4

1/ 2 x



 2 – tan x log tan x

cot  x

x 1

ANSWERS 1.

1;

2.

2/5 ;

3.

log (a/b) ;

4.

e2 ;

6.

e1/ 2 ;

7.

e2 ;

8.

e– 5 ;

9.

1;

5.

e5 ;

10.

e– 1

MISCELLANEOUS EXAMPLES

 sin x  1. Solve lim   , where  .  denotes greatest integer function. x 0   x  We have  sin x  lim   x 0   x  we know:, or i.e.,

sin x  1 as x  0 x

sin x

 1 as x  0 from right or left x at x = 0 + h or x = 0 – h and less than one,

sin x

sin x

 1. x x  from equation (i) whether we find R H L or L H L.  sin x lim  x 0   x 

 0 as

   lim 0  0.  x 0

 sin x lim  x 0   x

  , exists and is 0. 

.....(i)

CONTINUITY AND LIMIT

87

sin  x  2. Solve lim , where [ x ] denotes the greatest integer function. x 0 x sin  x  lim We have to find x 0 x

 



0  0  h  1  0  h  0

Now,

– 1  0 – h  0   0 – h  – 1

and Now,

x=0

R H L at

 lim

sin  0  h 

0  h

h 0

sin 0  limit does not exist. h 0 0

 lim

x=0

L.H.L. at

 lim

sin  0 – h 

0 – h

h 0

 lim

sin  – 1

h 0

–1

 sin1

As RHL  LHL, hence limit does not exist. 3. If and be the roots of ax2 + bx + c = 0; then solve:



lim 1  ax 2  bx  c

x 

2/ x – 



. 2

 lim 1  a  x –    x –    x –  x  2 lim a  x –    x –   x –   ex  

we have, given limit

e

4. If lim

xa sin b x

x 0

sin xc



2a  – 

.

, where a, b, c  R – {0}, exist and has non-zero value. Then show that

a + b = c. We have, lim x0

x a sin b x sin x

c

 sin x   lim x a .    x  x0  lim x x0

a



b

–

c

b

 xc   sin x c 

 b– .x  

b

 xc   sin x c 

 sin x  .   x 

c

   

DIFFERENTIAL CALCULUS

88

It will have non-zero value if and only if it is independent of x, i.e., 

a+b–c=0 a + b = c.





tan x – sin tan –1  tan x 

5. Evaluate:

lim

tan x  cos 2  tan x 

x  /2

we have, right hand limit at x   / 2

lim



x   /2

tan x – sin  x –   tan x  cos 2  tan x  { tan–1 (tan x) = x – , when x >  /2}

1 lim



x   /2

1

sin x tan x

cos 2  tan x  tan x

1 0 1. 1 0



x  /2

L H L at



lim

– x  2

tan x – sin  x  tan x  cos 2  tan x 

 tan – 1  tan x   x, when x   / 2 sin x tan x 2 cos tan x

1– 

lim – 1 x 2





tan x

1 0 1 1 0



 RHL = LHL, hence the given limit = 1. x

x

6. Evaluate lim  cos   –  sin   – cos 2 ,    0,  / 2  x4

Given limit



 lim  cos   x –  sin   x – cos 2  – sin 2  x4

 lim  cos   x –  sin   x – cos 4   sin 4  x4

  cos   sin   2

2

CONTINUITY AND LIMIT

89

 lim  cos   4 x4

cos  

x–4



– 1 –  sin  4 . lim x4

sin  

x–4



–1

= (cos )4. log (cos ) – (sin )4 . log (sin )   ax – 1  lim  log a   x  0 x   1 x  2 x  3 x  .. . .  n x  lim 7. Evaluate: n x  0 

Given limit

   

a/ x

  1x  2 x  ....  n x  lim 1  – 1  n x  0    1x  2 x  ....  n x  n lim  – 1 .  x n x  0   e x lim  x ( n x – 1)  a  (1 – 1) (2 – 1)  ex0    ....   x x  x  n a log1  log 2  log 3  .....  log n n e  log  n !  a / n   n !a / n e n

  1  1  1    n  1   n    n   . . .. .  n    . 8. Evaluate: lim n n 2 2   n  2   2 – 1   n   1 1     n  1  n   .....  n    2    2n – 1    lim   Given limit n   nn   –n

2

1  1 n  n  nn   n  1 2  ....  2n – 1  lim      n n   n   n        2n

1 n  1    lim  1   .  1   n  2n  n

 e1. e1/ 2 . e1/ 4 ...... e1/ 2 1  1/ 2  1/ 4  ......   e

1

e

1–

1 2

 e2 .

n –1

2

     

n

 1  2n – 1 n  ....  1  n –1   2n – 1 2 n     1 an  lim 1    ea  n  n     

DIFFERENTIAL CALCULUS

90

  If the rth term tr of a series is given by t r =

n r , find the value of lim  t r . n  r = 1 1 + r2 + r4

we have    1 2r   tr   2  1  r2  r 1  r2 – r  1  r2  r4      1 1 1   –  2  1  r2  r 1  r2 – r    r







 





Thus,

tr 

 1 1 1 –   2  1  r  r – 1 1  r  r  1 

1 1 1 –  2 3 1 1 1 t2   –  2 3 7 1 1 1 t3   –  2  7 13  -------------------------------------------------

t1 



  1 1 –    1  n  n – 1 1  n  n  1  n   1  lim  tr  lim 1 –  1  n  n  1  n   r 1 n  1 1 1  1 –   . 2  2

tn 



1 2

n

10. If a = min {

x2

+ 2x + 3, x  R } and b = lim

1 – cos θ

θ0

we have 

θ2

x2

a = min { + 2x + 3, x  R } a = min {(x + 1)2 + 2} a is minimum when x = – 1, i.e., a = 2.

2 sin 2  / 2 2 1 .  . 4 2   0   / 2 2

b  lim



n n 1 r n–r   2r .    a b 2 r0 r0



n–r

 1  0 2 4 2n  2  2  2  ....  2  n   2 

, find the value of

 a r .b n  r . r 0

CONTINUITY AND LIMIT

91





Given limit

1 2n

.

4n  1 – 1 2n  3

4n  1 – 1 2n  3

.

EXERCISES Evaluate following limits: 1.

2.

5.

8.

11.

13.

 sin x lim  x  0  x

lim x0 lim x0

sin  x 

 x

  , where [ . ] denotes the greatest integer function. 

.

e – 1  x 1/ x

lim xa

tan x x sin a – a sin x x–a

n –1 lim  cot n  r  1

lim n

3.

lim sin x  cos x  x  /4

4.



lim x cos x x0





lim x0

x cos x ex – e x  sin x

1  3 r – r  r    2    





lim x0

tan

–1

x – sin sin 3 x

–1

an lim , a  R n ! n

y x – yx lim x  y xx – y y

lim x

2  cos x – 1 – x 2





13  23  33  ......  n3 2 n2  1





 sin[ x ] , for  x   0  14. If f  x     x  0, for  x   0 

where [ x ] denotes the greatest integer less than or equal to x, then find

  sin  – x  3   15. Evaluate: lim x   / 3 2 cos x – 1  16. Evaluate: lim  tan x 1 

    log x   4 

1

log x

12 x    2 2 x   32 x   ......   n2 x          17. Evaluate: lim 3 x n

x

lim f  x  . x0

DIFFERENTIAL CALCULUS

92

where [ . ] denotes greatest integral function less than or equal to x.

 1 18. Find lim  2 – x  1  x  , where a   0,  and [ . ] denotes the greatest integer x0  2 function. 2

e x – cos x

19. Evaluate lim

x2 20. If x is a real number is [0, 1]. Then find the value of x0

lim m

lim 1  cos 2 m  n !  x   .   n  

ANSWERS 1. 0 ;

2. limit does not exist ;

6. 1 ;

7. 0 ;

3. – 2

8. sin a – cos a ;

4. –

1 ; 2

9. 0 ;

11.  / 2;

12. 1/4 ;

13. 1/4 ;

14. limit does not exist ;

16. e2 ;

17. x/3 ;

18. 2 ;

19. 3/2 ,

e ; 2 1 – log y 10. 1  log y ;

5.

15. 1/ 3 ;

20. 2 when x  Q , 1 when x  Q ,

3.6. Another form of the definition of continuity. Comparing the definitions of continuity and limit as given in §§ 3.1.1, 3.1.2, we see that f is continuous at an interior point c if and only if.

lim f ( x)  f (c ). xc Thus the limit as x tends to c of a function f which is continuous at c is equal to the value f (c) of the function for c. 3.6.1. Continuity of the sum, difference, product and quotient of two continuous functions. Let f, g be two functions which are continuous for a, so that lim f ( x )  f (a ), lim g  x   g  a  . xa

xa

From the theorems in §3.3, we see that lim ( f  g ) ( x )  lim  f  x   g  x    lim f  x   lim g  x  xa

xa

xa

 f  a   g  a    f  g  a  so that f + g is continuous at a. The continuity of f – g and f g may similarly be proved. for the quotient, we have lim xa

f  x g  x

lim f  x  

xa

, lim g  x   0 lim g  x  when x  a

xa



f a

f  x , = value of g a g  x  for x = a.

xa

CONTINUITY AND LIMIT

93

so that f  g is also continuous at a, provided that lim g  x   g  a   0, when x  a. Thus, the sum, the difference, the product and the quotient of two continuous functions are also continuous (with one obvious exception in the case of the quotient). 3.6.2. Continuity of elementary functions. Consider the polynomial function f  x   a0 x n  a1 x n – 1  .....  an – 1 x  an . n n –1  .....  an – 1 x  an . where a0 ....... an denote As already seen we have f  x   a0 x  a1 x

constant functions and I the Identity function. The result now follows from the preceding theorem and the fact that every constant function and the Identity function are continuous over R. A rational function g  x 

a0 x n  a1 x n – 1  .....  an b0 x m  b1 x m – 1  .....  bm

is continuous for every value of x except for those for which the denominator becomes zero. We have seen that y = sin x is continuous for every value of x. It may be similarly shown that y = cos x is also continuous for every value of x. Hence from § 3.6.1, we see that the functions sin x cos x y  tan x  , y  cot x  , cos x sin x 1 1 y  sec x  , y  cos ec x  cos x sin x are continuous for every member of their domain i.e. for all those values of x for which they are defined. Thus the discontinuities of these four functions arise only when the denominators become zero and for such values of x, these functions themselves cease to be defined. Thus the domain of continuity of each functions with function values sin x, cos x, tan x, cot x, sec x, and cosec x coincides with the corresponding domain of definition. 3.6.3. Continuity of the Inverse of a continuous invertible function. We assume without proof that if f is an invertible continuous function and g denotes its inverse, then g is also continuous. It now follows that the functions

y  sin 1 x, y  cos –1 x, y  tan –1 x, y  cot –1 x, y  sec –1 x, y  cosec –1 x, are continuous in their domains. Finally we assume that the exponential function y = ax is continuous in its domain R. It follows that its inverse y = loga x is also continuous in its domain [ 0,  ]. 3.6.4. Continuity of the composite of two continuous functions. We assume without proof that the composite gof of two continuous functions f and g is continuous.

3.7. Some properties of continuous functions. 3.7.1. If f is a continuous for c and f (c)  0 then there exists an interval [c – , c + ] around c such that f (x) has the sign of f (c) for every value of x in this interval.

DIFFERENTIAL CALCULUS

94

Its truth is obvious if we remember that a continuous function does not undergo sudden changes so that if f (c) is positive for the value c of x and also f is continuous at c, it cannot suddenly become negative or zero and must, therefore, remain positive for values of x in a certain neighbourhood of c. In the formal manner, it may be proved as follows : To given   0, there corresponds   0 such that

f  c  –   f  x   f  c     x   c – , c    . Let f (c) > 0. In this case we take  a positive number less than f (c) so that f (c) –  and f (c) +  are both positive. Thus f (x) which lies between two positive numbers is itself positive when x lies between c –  and c +  Let f (c) < 0. Now f (c) –  is negative and f (c) + will be negative if  1

f  x   lim

log  2  x   x 2 n sin x 1  x2 n

n

log (2 + x) 

for x2 > 1 1

f ( x )  lim

x 2 n log (2  x ) – sin x 1

x

1  x 2n

 – sin x 

lim x 1 0

f ( x)   sin1 and

lim x 1 0

f ( x)  log 3

DIFFERENTIAL CALCULUS

100

f 1 

Also, 

1  log 3  sin1 . 3

f ( x) 

lim x 1 0

lim

f ( x )  f 1 .

x 1 0

Hence f (x) is not continuous at x = 1 11. Discuss the continuity of the function: f (x) = [x] + [– x] at integral values of x. (i) If x is an integer,

 x   x and   x    x  f  x   0 (ii) if x is not an integer, Let x = n + f where n is an integer and f   0, 1 .

 x   n and   x     n  f 



   n  1  1  f      n  1 . f  x   n    n  1   1.



0, if x is an integer f  x     1, if x is not an int eger At x = a, where a is an integer



lim

f  x   1

xa0

and

lim

f  x   1

(As a + 0 and a – 0 are not integers)

xa0

But f (a) = 0 an a is an integer Hence, f (x) has a removable discontinuity at integral values of x. 12. Discuss the continuity of f (x) in [0, 2] where [cos  x], x  1 f  x     2 x  3  x  2 , x  1, where [.] denotes the greatest integral function. First consider x  [0, 1] then f  x    cos  x  is discontinuous where cos  x  I . In  0, 1 , cos  x is an integer at x = 0, 1/2, 1. 

x  0,

1 and 1 may be the points at which f (x) may be discontinuous. 2

Again, consider x  1, 2 If we consider f  x    x  2 2 x  3 then if x  1, 2  ;  x  2   1 and for x = 2 ; [x – 2] = 0

2 x  3  0 at x  3/ 2 Also,  x = 3/2 and 2 may be the points at which f (x) may be discontinuous.

CONTINUITY AND LIMIT

101

Hence, the possible points of discontinuity may be x = 0, 1/2, 1, 3/2, 2. Now, 1; x  0  0; 0  x  1  2  1  1;  x  1 f  x   2  3   3  2 x  ; 1  x  2    2 x  3 ; 3 / 2  x  2  0; x  2

Now, we will consider the cases one by one. f  x   0 and f  0   1

lim x00

 f (x) is discontinuous at x = 0. lim

f  x 

lim

f  x 

x  12  0 x  12  0

lim

00

lim

  1   1

x  12  0 x  12  0

1 . 2   1   1

 f (x) is discontinuous at x  f  x 

lim x 1 0

f  x 

lim

lim x 1 0

x 1 0

 3  2 x    1

lim x 1 0

and f (1) = – 1  f (x) is continuous at x = 1. lim

f  x 

lim

f  x 

x  32  0 x  32  0

lim

 2 x  3  0

lim

3  2 x   0

x  32  0 x  32  0

f (3/2) = 0  f (x) is continuous at x = 3/2. lim x20

f  x 

lim

3  2 x    1

x20

f (2) = 0 f (x) is discontinuous at x = 2.  f (x) is continuous when

x   0, 2  0, 1/ 2, 2 .

DIFFERENTIAL CALCULUS

102

13. Given   cosec x ,    x  0  cos x  sin x  2  f  x    a, x 0  1/ x  e 2 / x  e3 / x e , 0  x  /2  2/x  be3 / x  ae

If f (x) is continuous at x = 0, find a and b. f  x   lim

lim x  (0  0)

e1 / h  e 2 / h  e3 / h

h0

ae 2 / h  be3 / h 1

 lim h0 1  . b

1

e 2 / h  e1 / h  1 1

e1 / h  b

f  x   lim  cos h  sin h  cosec h

lim x  (0 – 0)

h0  1 / sin h

 lim 1   cos h  sin h  1 h0



lim

e



lim eh  0

sin h / 2  cos h / 2 1 e cos h / 2

f (0) = a

and 

 1  2 sin h / 2 cos h / 2  

 2 sin h / 2  2sin h / 2 cos h / 2  

h0

ae

1



1 b



a = 1/e, b = e.

14. Discuss the continuity of the function  2  x    x 1  a ,x0 f  x    2  x    x  log e a, x  0

at x = 0, where [ . ] denotes greatest integral part and { . } denotes fractional part of x. As [x] + {x} = x for any x  R, the given function becomes:  x  x 1  a ,x 0 f  x   x  x   log e a , x  0.

CONTINUITY AND LIMIT

103

 h  h

lim

Now,

a

f  x   lim

x00

1

 h  h

h0

ah  1 h h0

 lim

[0 < h < 1 h] = 0]

 log a

lim

a 1  h  1 h0 1 h

f  x   lim

x00



[ – 1 < 0 – h < 0 [0 – h] = – 1]

a 1  1 1 1 . 1 a

obviously f  0  0   f  0  0  Hence, f (x) is not continuous at x = 0. 15. Discuss the continuity of the function f  x  

x2 . 1 tan  x  2 

f is continuous except possibly at x = – 2.

 x  2 1 x20 x   2  0 tan  1  x  2    x  2 lim f  x  lim  1 1 x20 tan x  2   x 2  0 lim

f  x 

lim

obviously, f (x) is continuous at x  R   2 . EXERCISES 1. Examine the continuity of the following at x = 0: (i)

(iii)

(v)

(1  x)1/ x f ( x)   1

 –1/ x 2 f ( x )  e 1

 e x2  f ( x )   e1/ x2 – 1  1

, when x  0 , when x  0.

, when x  0 , when x  0.

(ii)

(iv)

(1  2 x )1/ x f ( x)   2 e

 e –1/ x  f ( x)  1  e1/ x  1

, when x  0 , when x  0.

, when x  0 , when x  0.

, when x  0 , when x  0.

2. Examine the continuity of the function. f  x 

2 x 3 x   x

at x 

1 and x  1 where [x] denotes the greatest integer not greater than x. 2

DIFFERENTIAL CALCULUS

104

3. Examine the limit of the function f  x  as x  2, where  | x – 2| , x2  f ( x)   x – 2  0 , x2 

4. Discuss the continuity of the function f defined by f  x  x  2  x  3

at x = 2 and x = 3. 5. Prove that the function f defined as  x  f (x)   2– x  2  – 2  3x – x

, x 1 , 1 x  2 ,

x2

is continuous at x = 1 and x = 2 6. Examine the continuity of the function 1 f  x   2 x   x   sin , x  0 x

= 0 when x = 0 where [x] denotes the greatest integer not greater than x; at x = 0 and x = 2. 7. Examine the continuity, the function f defined by 1   2 x  1, 0  x  2  1, x  1  f  x   2  1 1  x,  x  1 2   2 x  2, x  1 1 at the points x  and 1. In case of discontinuity discuss the nature of discontinuity. 2

8. A function f is defined on [0, 1] as follows: f  x 

1 2n

when

1 1 x  n  0, 1, 2, ......  n  1 n 2 2

f (0) = 0 Show that f is discontinuous at the points

1  1  2  1 3 ,   ,   , ...... and examine the nature of 2 2 2

discontinuity. 9. Investigate the points of continuity and discontinuity of the function f defined as follows:  x2 – a for x  a  a f ( x)   a2  a – x for x  a

10. A function f (x) is defined by [ x 2 ] – 1 , for x 2  1  f ( x)   x 2 – 1  , for x 2  1  0

Discuss the continuity of f (x) is continuous at x = 1.

(Avadh 99)

CONTINUITY AND LIMIT

105

11. Let {1  |sin x |}a|1sin x| ; –  / 6  x  0  f ( x)   b ; x0  tan 2 x / tan 3 x ; 0  x  /6  e

Determine a and b such that f (x) is continuous at x = 0. 12. Find a and b so that the function:  x  a 2 sin x ; 0  x  / 4  f ( x)   2 x cot x  b ; / 4  x  / 2 a cos 2 x – b sin x ;  / 2  x   

is continuous for x   0,  ?   1 – cos 4 x  x2  13. Given f ( x)   a  x   16  x – 4 

, x0 , x0 ,

x0

Determine the value of a if possible, so that the function is continuous at x = 0 14. If f  x  

sin 2 x  A sin x  B cos x x3

is continuous at x = 0, find the values of A, B and f (0).

15. Discuss the continuity of the function f (x) = [ [x] ] – [x – 1], where [.] denotes the greatest integral function. n 2 r x , where [.] denotes greatest integral function, discuss the continuity  2 n  r  1 n

16. If f  x   lim of f (x).





n   m  



2m 17. Prove that f  x   lim  lim cos  n! x   is nowhere continuous.

ANSWERS 1. (i) (v)

discontinuous discontinuous

(ii) continuous 1 2

2. Discontinuous at x  , continuous at x   4. Continuous at x = 2, 3; 7.

1 Discontinuous at x  , continuous at x = 1, 2

9. Continuous at x = a. 11.

a  2 / 3, b  e2 / 3;

(iii) discontinuous 1 3. 2

(iv) discontinuous

Limit does not exist

6. Discontinuous 8. Discontinuity of 2nd kind 10. Discontinuous; 12.

a   / 6, b    /12;

13. a = 8;

14. A = – 2, B = 0, f (0) = – 1;

15. Continuous on R;

16. Continuous everywhere.

DIFFERENTIAL CALCULUS

106

OBJECTIVE QUESTIONS For each of the following questions four alternatives are given for the answer, only one of them is correct. Choose the correct alternative: 1. Limit of a function f (x) when x  a exists if: (a) anyone of f (a – 0) and f (a + 0) exists (c)

f  a  0  0  f  a  0

(b)

both f (a – 0) and f (a + 0) exist

f  a  0  f  a  0

(d)

2. A function f (x) is said to be continuous at x = a if: (a) lim f  x  exists (b)

f  a  exists

xa

(c) 3.

lim f  x   f  a  xa

(d)

lim f  x   f  a  xa

lim sin 1/ x  : xa

4.

(a) exists

(b)

is equal to zero

(c) is equal to 

(d)

does not exists

(b)

0

(d)

oscillatory.

(b) (d)

may or may not be continuous in R is continuous in R except at x = 0

lim x sin 1/ x  : is equal to x a

(a) 1 (c)  5. A polynomial function in R : (a) is never continuous in R (c) is always continuous in R

1 6. The function f  x   x sin , x  0 and f (0) = 0 is: x (a) continuous at 0 only (b) discontinuous at 0 only (c) continuous for all values of x (d) discontinuous for all values of x cos x, for x  0 7. The function f  x     cos x, for x  0 is:  (a) continuous for all values of x

(b) continuous for all values of x except at x   / 2 (c) continuous for all values of x except at x = 0 (d) discontinuous for all values of x 8. The function 1  x if x  2 f  x   , is 5  x if x  2 (a) continuous for all values of x (b) continuous for all values of x except x = 2 (c) discontinuous at x = 0 (d) discontinuous at x = 2. 1  9. lim  1   is equal to x x   (a) 1 (b) 0

CONTINUITY AND LIMIT

107

(c) e 10. The function f (x) =  x  is: (a) continuous for all x (c) continuous at x = 0 only

(d)  (b) discontinuous at x = 0 only (d) discontinuous for all x

sin x , Then the value of f (0) so that f (x) becomes continuous at x = 0 is: x (a) 0 (b) 1 (c) e (d)  12. If the function f (x) is defined on R by:

11. Let f  x  

1, when x is rational f  x   , then   1, when x is irrational

(a)

f (x) is continuous at x = 1 only

(c) f (x) is continuous at x   1 only 13.

ax  1 , where a > 0, is equal to x x0 (a) 0

(b) f (x) is continuous at x = – 1 only (d) f (x) is not continuous at any point.

lim

(c)

ea

(b) 1 (d) log a

14. Which two of the functions sin x, e x and log x are continuous on R? (a) sin x and e x (b) sin x and log x (c)

e x and log x

(d) none of these

15. Which of the following is continuous at x = 0? (a) f (x) = 1/x

(b) f (x) =  x/x

(c) f (x) =  x

d x = x/  x

16. Given the function  x , if x  0 f  x   1, if x  0 which of the following is incorrect?

(a)

f (x) is continuous at x = 0

(c) f (x) is continuous at x = 1

(b) f (x) is discontinuous at x = 0 (d)

f (x) is continuous in R – {0}

17. If f (x) = [x] be the greatest integer function then lim f  x  is equal to: x 1

(a) 0

(b) 1

(c) 2

(d) does not exist

18. If f  x  

sin  x 

 x

,  x   0  0,  x   0.

where [x] denotes the greatest integer less then or equal to x, then lim f  x  equals to: x0

(a) 1

(b) 0

(c) – 1

(d) None of these

DIFFERENTIAL CALCULUS

108

19. If function f  x    x  1cot x is continuous at x = 0, f (0) must be defined as: (a (c)

f (0) = 0 f (0) = 1/e

(b) f (0) = e (d) None of these.

 – 1; x   1  20. f  x    x;  1  x  1 1; x  1  is continuous: (a) at x = 1 but not at x = – 1

(b) at x = – 1 but not at x = 1

(c) at both x = 1 and x = – 1

(d) at none of x = 1 and x = – 1. ANSWERS

1. (d)

2. (c)

3. (d)

4. (d)

5. (c)

6. (c)

7.

(c)

8. (a)

9. (a)

10. (c)

11. (b)

12. (d)

13. (d)

14.

(a)

15. (c)

16. (a)

17. (d)

18. (d)

19. (b)

20. (d)

CHAPTER 4 DIFFERENTIATION 4.1 Introduction Rate of change : The subject of Differential Calculus had its origin mainly in the geometrical problem of the determination of the gradient of a curve at a point thereof resulting in the determination of the tangent at the point. This subject has also rendered possible the precise formulations of a large number of physical concepts such as Velocity at an instant, Acceleration at an instant, Curvature at a point, Density at a point, Specific heat at any temperature, etc. each of which appears as a Local or Instantaneous Rate of change as against the Average Rate of Change. The fundamental idea of Local or Instantaneous rate of change pervading all these concepts underlines the introduction of the notion of ‘Derivative’. The ideas underlying Differential Calculus were first conceived by Newton (1643–1727) and Leibnitz (1646–1716). 4.1.1. Derivability and derivative. We consider a function f with some domain D. Let ‘c’ be an interior point of D. We take another member c + h of the domain which lies to the right or left of c (i.e. c + h > c or c + h < c) according as h is positive or negative. The values of the function corresponding to c and c + h are f (c) and f (c + h) respectively . Now, h, is the change in x and f (c + h) – f (c) is corresponding change in f (x). The expression [f (c + h) – f (c)]/h, which is the ratio of these two changes denotes the average rate of change of the dependent variable f (x) as the independent variable x changes from c to c + h. It is possible that this ratio tends to a limit as h tends to 0. The limit, if it exists, is called the Derivative of f at c and denoted as f  (c). Also the function f, then, is said to be Derivable at c. Definition. A function f is said to be derivable at c if, f (c  h) – f (c ) lim h0 h exists and the limit is called the Derivative of the function for c or at c and is denoted by f  (c). The function f is said to be finitely derivable at c if the derivative at c is finite. Right hand and Left hand derivatives. lim

h  (0  0)

f (c  h) – f (c ) h 109

110

DIFFERENTIAL CALCULUS

if it exists, is called the Right hand derivative at c and is denoted by f  (c + 0) or R f  (c) Similarly, f ( c  h ) – f (c ) , h if it exists, is called the Left hand derivative at c and is denoted by f  (c – 0) or L f  (c.) lim

h  (0 – 0)

Clearly f is derivable at an interior point if f  (c + 0) and f ‘ (c – 0) both exist and are equal. EXAMPLES 1. Show that the function f (x) = x2 is derivable for x = 1. f (x) = x2  f (1) = 12 = 1.

Now

To find the derivative for, 1, we change x from 1 to 1 + h so that the function value changes from 1 to (1 + h)2. We have f (1  h) – f (1) 2h  h 2   2  h, [as h  0] h h f (1  h) – f (1)  2  f  (1)  2. h Hence f is derivable at 1 and the derivative f  (1) is 2. lim



h0

2. Show that f (x) = | x | is not derivable at 0. If will be shown that lim [f (0 + h) – f (0)]/h when h tends to 0 does not exist. We have f (h) = h if h > 0, f (h) = – h if h < 0. 

f (0  h) – f (0) f ( h)   1 or – 1. h h

According as h is positive or negative. Thus f  (0 + 0) = 1 and f  (0 – 0) = 1. Now f  (0 + 0) and f  (0 – 0) being different, it follows that f  (0) does not exist and as such the function f is not derivable at 0. 3. If f (x) = x sin (1/x) when x  0 and f (0) = 0, show that f is continuous but not derivable for x = 0. (Osmania 2004) We have 1 1 – 0  x sin x x 1 1 | f ( x) – f (0)|  x sin  | x | sin  | x |.  x x Let  be a given positive number. We have | f (x) – f (0) | <  when | x – 0 | <  f ( x ) – f (0)  x sin



lim f ( x)  0  f (0)

x0

 f is continuous at 0. Again

f ( x) – f (0) x sin (1/ x)   sin (1/ x) x–0 x

DIFFERENTIATION

111

and, as seen in Ex. 3, p. 64 lim sin (1/x) does not exist when x  0, Thus f is not derivable at 0. Ex. Find the derivatives for the given values of x if (i) f (x) = 2x2 + 3x – 4 for x = 5/2. (ii) f (x) = 1/x for x = 5. (iii)

f ( x) 

(iv)

f ( x )  1/ x for x  1.

x for x  1.

ANSWERS (i)

13;

(ii)

– 1/25 ;

(iii)

1/2 ;

(iv)

– 1/2.

4.1.2. Derived Function. While in 4.11, we have defined the derivability of a function at a particular value, we now consider the derivability of a function in its domain. Let f be a function with [a, b] as its domain. We suppose that f is finitely derivable at every point of [a, b] and also that right-hand and left-hand derivatives exist finitely at a function denoted by f  such that f  (x) denotes the derivative of f for x  [a, b] The function f  is called the Derived function or simply the Derivative of the function f. The derived function f  is given as follows : dy  f  ( x ), x  [a, b] dx Also we say that f is derivable in [a, b].

EXAMPLES 1. Find the derivative of f if f (x) = x2. The domain of the function f is the entire set R of real numbers. Let x  R. We have when h  0 f ( x  h ) – f ( x ) ( x  h) 2 – x 2   (2 x  h) . h h

so that lim f ( x  h) – f ( x ) exists and equals 2x. h0 h Thus f  (x) = 2x  x  R. 2. Find the derivative of f if f ( x)  x . The domain of the functon f is the set of all non-negative real numbers i.e., the interval [0, ] : Let x > 0. We have, when h  0 f ( x  h) – f ( h)  h

( x  h) – h

x

 ( x  h) – x   ( x  h)    h    ( x  h)    1    ( x  h)  x    so that lim

h0

f ( x  h) – f ( x ) 1  ;x  0 h 2 x

x  x 

112

DIFFERENTIAL CALCULUS

Thus for x > 0, f ( x) 

x  f  ( x)  1/ 2 x .

We start afresh to examine the existence of the derivative at 0. We have f (0  h) – f (0) h 1    h h h when h  0 through positive values ; h being not defined for negative values of h. We have f  (0 + 0) =  Thus f  ( x)  1/ 2 x ,  x [0, ]. Ex. Find the derived function f  if f (x) is given as follows : (i) 1/(x2 + 3)

(ii) 1/ x (iv) ax2 + bx + c

(iii) x3

ANSWERS (i)

– 2x/(x2 + x3)

(ii)

– 1/2x3/2 ;

(iii)

3x2

(iv)

(2ax + b)

;

4.1.3. Another notation. Consider a derivable function f. We write f (x) = y. Let x be a member of the domain of f. It is usual to denote the change in x by x in place of h and the resulting change f (x +  x) – f (x) in the value of the function by  y so that we have f ( x   x) – f ( x)  y  . x x Thus, we have f ( x   x) – f ( x) y lim  lim  x0  x 0  x x f  ( x)  lim



 x0

y . x

dy y . by dx x Thus f (x) = y  f  (x) = dy/dx. The derivative at c, is, in terms of this notation, is denoted as

It us usual to denote lim

 x0

 dy   dy  so that we have f  (c)       dx  x  c  dx  x  c Ex. (i) Find (dy/dx)x = 0 when y = 2x/(x2 + 1). (ii) If y 

( x 2  1) , find (dy/dx) when x = – 1

ANSWERS (i)

2;

(ii)

– 1/ 2

4.1.4. Derivability implying continuity Theorem. If f is finitely derivable at c, then f is also continuous at c.

DIFFERENTIATION

113

Let f be finitely derivable at c so that the expression

(Banglore 2005, Poorvanchal 2004)

[f (c + h) – f (c)]/h tends to a finite limit as h tends to 0.We write f ( c  h ) – f (c ) h h  f (c  h) – f (c )  lim [ f (c  h) – f (c)]  lim   h h 0 h0  h  f (c  h) – f (c )  lim ( h) = lim h0 h0 h = f  (c) × 0 = 0 f ( c  h ) – f (c ) 



lim f ( x)  f (c)  f is continuous at c.

hc

Thus, we have seen that f is finitely derivable at c  f is continuous at c. Cor. If f is finitely derivable for every point of its domain then it is continuous in the domain. Note 1. If f is derivable at c, then the expression [f (c + h) – f (c)]/h has the Indeterminate form 0/0 when h  0. Note 2. The converse of this theorem is not necessarily true, i.e., a function may be continuous for a value of x without being derivable for that value. For example the function f where f (x) = | x | is continuous but not derivable for 0. EXAMPLES 1. Show that f (x) = x | x | is derivable at the origin. We have

f (x) = x2 if x  0 = – x2 if x < 0

L f  (0)  lim

f ( x) – f (0) (– x 2 )  lim 0 x 0  0 x–0 x

R f  (0)  lim

f ( x) – f (0) x2  lim 0 x 0  0 x x–0

x 0  0

x0  0



L f  (0) = R f  (0)

 f is derivable at the origin. 2. Examine the function f where f ( x) 

x (e –1/ x – e1/ x ) e –1/ x  e1/ x

,x  0

= 0, x = 0. as regards continuity and derivability at the origin.

114

DIFFERENTIAL CALCULUS

 e –1/ x – e1/ x f (0  0)  lim x  –1/ x x0  0  e  e1/ x 

We have

  

 e –2 / x – 1  lim x  –2 / x  x0  0  e  1   0 (0 – 1) = 0 1  0

=

 e –1/ x – e1/ x f (0 – 0)  lim x  –1/ x x0 – 0  e  e1/ x   1 – e2 / x   lim x   x  0 – 0  1  e2 / x    0 (1 – 0)  0 1 0

  

f (0) = 0

Also 

f (0 + 0) = f (0 – 0) = f (0)

 f is continuous at x = 0 f  (0)  lim

Again

x 0

f ( x) – f (0) x–0

= lim

x 0

f ( x) x

 e –1/ x – e1/ x  lim  –1/ x x 0  e  e1/ x  But this limit does not exist as lim

e –1/ x – e1/ x

x  0  0 e –1/ x

 e1/ x



  

e –2 / x – 1

lim

x  0  0 e –2 / x

1

0 –1   –1 0 1

and lim

e –1/ x – e1/ x

x  0 – 0 e –1/ x

 e1/ x



lim

1 – e2 / x

x 0 – 01

 e2 / x

0 –1 1 0 1 Hence f is not derivable at x = 0 3. If a function f is defined by 

f ( x) 

xe1/ x

1  e1/ x = 0, x = 0

,x  0

show that f is continuous but not derivable at x = 0

(Devi Ahilya 2001)

DIFFERENTIATION

We have

115

f (0  0)  lim

x 0  0



xe1/ x 1  e1/ x x

lim

x  0  0 e –1/ x

1

=0 f (0 – 0)  lim

x 0 – 0

xe1/ x 1  e1/ x

=0 f (0) = 0

Also  

f (0 + 0) = f (0 – 0) = f (0) f is continuous at x = 0

Again

f  (0  0) 

lim

x 0  0

f ( x) – f (0) x–0

xe1/ x –0 1  e1/ x  lim x 0  0 x 

=

e1/ x

lim

x 0  0 1

lim

 e1/ x 1

x  0  0 e –1/ x

1

=1 f  (0 – 0) 

lim

x 0 – 0

f ( x) – f (0) x–0

xe1/ x –0 1  e1/ x  lim x 0 – 0 x 

lim

x 0 – 0 1

e1/ x  e1/ x

=0 Since f  (0 + 0)  f  (0 – 0), the derivate of f (x) at x = 0 does not exist. 4. Show that the function f defined by f (x) = | x – 2 | + | x | + | x + 2 | is not derivable at x = – 2, 0 and 2. We have – –  f ( x)   – 

( x – 2) – x – ( x  2) 

– 3x ,

x–2

( x – 2) – x  ( x  2)  – x  4 , – 2  x  0 ( x – 2)  x  ( x  2)  0x2 x4, ( x – 2)  x  ( x  2) 

3x ,

x2

116

DIFFERENTIAL CALCULUS

f  (– 2  0)  

lim

f ( x) – f (– 2) x2

lim

– x4–6  x2

x – 2  0

x – 2  0

=–1 f  (– 2 – 0)  

lim

f ( x) – f (2) x2

lim

– 3x – 6 x2

x – 2 – 0

x – 2 – 0

=–3 

f  (– 2 + 0)  f  (– 2 – 0)

 f is not derivable at x = – 2 Again,

f ( x ) – f (0) x–0 x4–4 = lim x  0 x =1

f' (0 + 0)  lim

x 00

f  (0 – 0)  lim

x 0 – 0



f ( x) – f (0) x–0

lim

x 0 – 0

– x4–4 x

=–1 

f  (0 + 0)  f  (0 – 0)

 f is not derivable at x = 0 Also

f  (2  0) 

=

lim

x 2  0

f ( x ) – f (2) x–2

lim

3x – 6 x–2

lim

f ( x) – f (2) x–2

lim

x4–6 x–2

x 2  0

=3 f  (2 – 0)  

x 2 – 0

x 2 – 0

=1 

f  (2 + 0)  f  (2 – 0)

 f is not derivable at x = 2

lim

x – 2  0

– ( x  2) x2

DIFFERENTIATION

117

5. Discuss the derivability of the functon

at We have

 x x 1 ,  f ( x)   2– x , 1 x  2  2 x2  – 2  3x – x , , x = 1, 2 f ( x) – f (1) x –1 2 – x –1  lim x 1  0 x –1

f  (1  0)  lim

x 1  0

 lim

x 1  0

1– x x –1

=–1 f  (1 – 0)  lim

x 1 – 0

f ( x) – f (1) x –1

x –1 x 1 – 0 x – 1

 lim

=1 

f  (1 + 0)  f  (1 – 0)

 f is not derivable at x = 1 Again

f  (2  0) 

 

lim

x 2  0

f ( x ) – f (2) x–2

– 2  3x – x 2 – 0 x 2  0 x–2 lim

lim

x 2  0

( x – 2) (1 – x) x–2

=–1 f  (2 – 0)  

lim

x 2 – 0

f ( x) – f (2) x–2

2– x–0 x 2 – 0 x–2 lim

=–1 

f  (2 + 0) = f  (2 – 0)

 f is derivable at x = 2. 6. If f (x) = x2 sin (1/x) when x  0 and f (0), show that f is derivable for every value of x but the derivative is not continuous for x = 0. (Calicut 2004) For

1 1  1   x 2 cos    – 2  x x  x  1 1  2 x sin – cos . x x

x  f  ( x)  2 x sin

118

DIFFERENTIAL CALCULUS

x = 0, we have

For

f ( x ) – f (0)  x–0

1 x  x sin 1  f  (0)  0. x x

x 2 sin

Thus the function possesses a derivative for every value of x given by 1 1 – cos when x  0, f  (0) = 0. x x We now show that f  is not continuous for x = 0. f  ( x )  2 x sin

We write

cos

Here

1 1  1 1  2 x sin –  2 x sin – cos  x x  x x

...(i)

1  lim  2 x sin   0. x

x 0 

1 1  In case lim f  ( x ), i.e., lim  2 x sin – cos  had existed, it would follow from (i) that x 0 x 0  x x 1  lim  cos  would also exist. But this is not the case. Hence lim f  ( x) does not exist so that f  is x 0 x 0  x not continuous for x = 0. 7. Examine the continuity and derivability in the interval ]– , [ of the function defined as follows : f (x) = 1 in ] – , 0 [  1  f (x) = 1 + sin x in 0, 2     2 1  1   f ( x )  2   x –   in  ,  . 2    2

The function f is derivable for every value of x except perhaps for x = 0 and x = /2. 1. Firstly we consider x = 0. f (0) = 1 + sin 0 = 1

Now lim

x  (0 – 0)



f ( x)  1 and

lim

x  (0  0)

f ( x) 

lim

x  (0  0)

(1  sin x)  1

lim f ( x)  1  f (0).

x 0

Hence f is continuous for x = 0. Also x < 0  Again

f ( x) – f (0) 1 – 1  0 x–0 x. x0

lim

x  (0 – 0)

f ( x) – f ( x) 0 x–0

f ( x) – f (0) 1  sin x – 1 sin x   x–0 x–0 x

DIFFERENTIATION



lim

x  (0  0)

119

f ( x) – f (0)  x–0 lim

Thus

x  (0  0)

lim

x  (0  0)

sin x  1. x

f ( x) – f (0)  x–0

lim

x  (0 – 0)

f ( x) – f (0) . x–0

Hence the function is not derivable for x = 0. II. Now, we consider x = /2. We have 2

1  1 f ( / 2)  2    –    2 2  2 lim

1  x    – 0 2 

f ( x) 

lim

1  x    – 0 2 

lim f ( x )  lim 1 x   – 2



 0 

(1  sin x)  1  1  2,

2  1     2   x –    2  2    

1 x    – 0 2 

lim f ( x)  2  f ( / 2).

x / 2

Hence, f is continuous for x = /2. for 0  x 

Again,

1 , we have 2

1  f ( x) – f     2   (1  sin x ) – 2  1 – sin x. . 1 1 1 x–  x–  – x 2 2 2 1 Putting  – x  t , we see that 2 1  1 – sin   – t  1 – sin x 2    1 t – x 2 1 1 2sin 2 t sin t 1 – cot t 1 2  sin t 2   t t 2 1t 2 1 – sin x lim  0. 1 1  x   2  – 0   – x   2

For x 

1  2 2

1  1   f ( x) – f    2   x –   – 2 1 2  2    x–  1 1 2 x–  x–  2 2

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DIFFERENTIAL CALCULUS

lim



1  x    0 2 

1  f ( x) – f     2   0. 1 x–  2

1  f     exists and is equal to 0. 2     1 1    , x  0  x exp  –  8. If f ( x)   then test whether (i) f (x) is continuous at x = 0 (ii)   | x | x   0, x0  

f (x) is differentiable at x = 0. We have



We have

 –  1  1  x e  x x , x  0   –  1  1   – x x f ( x)   x e  , x0 0 , x0     x e –2 / x  f ( x)   x  0 

f (0  0) 

lim x e –2/ x

x 0  0

 lim

h 0

f (0 – 0) 

and

, x0 , x0 , x0

h e

2/ h

0

lim x  lim (0 – h)  0

x 0 – 0

h0

f (0) = 0

 f (x) is continuous at x = 0. Again,

f  (0 – 0)  lim

h0

 lim

h0

f  (0  0)  lim

h0

f (0 – h) – f (0) ,h  0 –h

–h–0 1 –h f (0  h) – f (0) h

h e –2 / h – 0  e–  0 h0 h  f (0 – 0)  f  (0 + 0)  lim

So, f (x) is not differentiable at x = 0.

DIFFERENTIATION

121

9. If f (x) = [n + p sin x], x  (0, ), n  z and p is a prime number, where [ . ] denotes the greatest integer function, then find the number of points where f (x) is not differentiable. We have, f (x) = [n + p sin x] is not differentiable at those points where n + p sin x is integer. As p is a prime number  n + p sin x is an integer if sin x = 1, – 1, r/p i. e.

–1 x = /2, – /2, sin

r ,  – sin–1 r/p p

 ,0. 2  Function is not differentiable at

But x  –

r r  sin –1 ,  – sin –1 . , p p 2 0 < r  p – 1. x

where

So, the required number of points are = 1 + 2 (p – 1) = 2p – 1.

 |1 – 4 x 2 |, 0  x  1 10. If f ( x)   Where [ . ] denotes greatest integral function. Discuss 2 [ x – 2 x], 1  x  2, the differentiability of f (x) in [0, 2] As

1  x < 2  – 1  x2 – 2x < 0

 [x2 – 2x] = – 1 when 1  x < 2.







 and

1 – 4 x 2 , 0  x  1/ 2  f ( x)   4 x 2 – 1 , 1/ 2  x  1  –1 , 1  x  2. 

 – 8 x , 0  x  1/ 2   f ( x)   8 x , 1/ 2  x  1  0 , 1  x  2.   – 4 x  1/ 2 and f  ( x)    4 x  1/ 2 8 , x  1 f  ( x)   0 , x  1 1  1  f    0   4, f  – 0  – 4 2  2   f  (1 + 0) = 0, f  (1 – 0) = 8

 f (x) is differentiable,  x  [0, 2] – {1/2, 1}.

122

DIFFERENTIAL CALCULUS

EXERCISES 1. Show that if f (x) = | x | + | x – 1 | , the function f is continuous for every value of x but not derivable for x = 0 and x = 1. 2. Construct a function which is continuous in [1, 5] but not derivable at 2, 3, 4. 3. Show that f (x) = | x |, x  R is continuous at x = – 1 but not derivable at x = – 1. 4. Determine whether f is continuous and has a derivative at the origin where  2  x if x  0 f ( x)    2 – x if x  0

5. Discuss the continuity and derivability of the function 1  x x  f ( x)   2 – x 3 x – x 2 

,

, 1 x  2 ,

at x = 0, 1 and 2. 6. Show that the function defined by  x  f ( x)  1  e1/ x  0 

x0

, 0  x 1 x2

, x0 ,

x0

is continuous at x = 0 but not derivable at that point. 7. Discuss the continuity and the derivable of the function f where 0,  f ( x)   1  q 3, 

when x is irrational or zero, p when x  , a fraction in the lowest terms. q

8. Find the derivative of f where f ( x) 

sin x 2 , when x  0, and f (0) = 0 x

and show that the derivative is continuous at x = 0. 9. Show that the function 1 sin (log x 2 )} 3

; for x  0, f (0) = 0 is everywhere continuous but has no derivative at x = 0. 10. Show that the function defined as f ( x)  x {1 

1 f ( x )  x m sin , x  0, m is a + ve integer x

is continuous at x = 0. Examine its derivability. Determine m when f  is continuous at x = 0. 11. If f (x) = x tan–1 (1/x) when x  0 and f (0) = 0, show that f is continuous but not derivable for x = 0. 12. A function f is defined as 1  (b 2 – a 2 ) 2  x2 a3 1 f ( x)   b 2 – – 6 3x 2  1  b3 – a3     x   3 

for 0  x  a for a  x  b for x  b.

Prove that f and f  are continuous but f  is discontinuous.

DIFFERENTIATION

123

13. Examine for continuity at x = a the function f where  x2 – a, 0  x  a  a  f ( x )  0 , xa  3 a – a , a  x.  x2

Also examine if the function is derivable at a. 14. Is the function f where f ( x)  ( x – a)sin

1 for x  a, f (a) = 0 x–a

continuous and derivable at x = a ? Give your answer with reasons. 15. Discuss the continuity of f in the neighbourhood of the origin when f (x) is defined as follows : (i) f (x) = x log sin x for x  0 and f (0) = 0. (ii) f (x) = e1/x when x  0 and f (0) = 0. 16. f (x) is defined as being equal to – x2, when x  0, to 5x – 4 when 0 < x  1, to 4x2 – 3x when 1 < x < 2 and to 3x + 4 when x  2. Discuss the existence of f  for x = 0, 1 and 2. 17. Show that the function f (x) = 2 | x – 2 | + 5 | x – 3 |  x  R is not derivable at x = 2 and x = 3. ANSWERS 4. f is continuous but not derivable at x = 0. 5. f is continuous at 0, 1 but not derivable. Discontinuous at 2. 7. continuous when x is irrational or zero and discontinuous for other values of x. Differentiable for no value. 8. f  (0) = 0. 13. continuous and derivable at x = a. 14. continuous but not derivable at x = a. 15. (i) continuous (ii) discontinuous. 16. discontinuous at 0; derivable at 1; continuous but not derivable at 2. 4.1.5. Geometrical interpretation of a derivative. To show that f (c), is the tangent of the angle which the tangent line to the curve y = f (x) at the point P [c, f (c)] makes with x-axis. We take two points P [c, f (c)] and Q [c + h, f (c + h)] on the curve y = f (x). Draw the ordinates PL, QM and draw PN  MQ.We have PN = LM = h and NQ = MQ – LP = f (c + h) – f (c)  tan  XRQ = tan  NPQ NQ PN f (c  h) – f (c )  h 

...(i) Fig. 4.1

124

DIFFERENTIAL CALCULUS

Here, XRQ is the angle which the chord PQ of the curve makes with x-axis. As h approcahes 0,the point Q moving along the curve approaches the point P, the chord PQ approaches the tangent line TP and  XRQ approaches  XTP which we denote by . On taking limits, the equation (i) give tan  = f  (c). Thus f  (c) is the slope of the tangent to the curve y = f (x) at the point P [c, f (c)]. The slope of the tangent at a point of a curve is also known as the Gradient of the curve at the point. Cor. The equation of the tangent at the point P [c, f (c)] of the curve y = f (x) is Y – f (c) = f  (c) (X – c). We shall in the following chapter 5 deal with detailed discussion about tangents to curves. Note. The student should note that it is not necessary for every curve to have a tangent line at every point thereof. The existence of the tangent demands the existence of the derivative and we have seen in many examples that every function is not derivable for every value of x. For example, the curve y = | x | does not possess any tangent at (0, 0). This fact may also be seen directly from the graph Fig. 2.5, p. 16. EXERCISES 1. Find the slopes of the tangens to the parabola y = x2 points (2, 4) and (– 1, 1). 2. Show that the tangent to the hyperbola y = 1/x at (1, 1) makes an angle 3/4 with x-axis. ANSWERS 1. 4, – 2. 4.1.6. Differentials. Differential co-efficient. Let a function f be derivable in [a, b]. Let

x  [1, b]

We write

y = f (x).

y  f  ( x). x so that y/x differs from f  (x) by a variable which tends to zero when x  0.

Now when

 x  0, lim

y  f  ( x)     y = f  (x)  x +   x. x The change  y in y consists of two parts, viz f (x) x and  x. So far as contribution of these two parts to y is concerned that of f  (x)  x is more significant than that of  x in that  x is the product of two variables  and x each of which tends to 0. The part f  x is called the Differential of y and is denoted by dy so that we have

We write

dy = f  (x)  x. Considering f (x) = x so that f  (x) = 1 we obtain dx = 1  x =  x. Thus the differential dx of the independent variable x is the same as the increment x. We thus have dy = f  (x) dx. While  y denotes the increment in y, dy stands for the differential of y.

DIFFERENTIATION

125

The derivative f  (x) being the co-efficient of the differential dx is also known as Differential coefficient. The process of finding the derivative of a function is also called Differentiation. EXAMPLES 1. Find the differential dy and the increment  y when y = x3 for (i) arbitrary values of x and  x. (ii) x = 10,  x = 0.1. We have

y +  y = (x +  x)3 = x3 + 3x2  x + 3x ( x)2 + (x)3 y = 3x2  x + 3x (x)2 + (x)3 = 3x2 x + [3x  x + (x)2]  x.

Here,  = 3x  x + (x)2. We have dy = 3x2 dx. For

x = 10 and  x = 0.1, we have 3x2 dx = 300 (0.1) = 30 = dy y = 30 + [30 × 0.1 + (0.1)2] (0.1) = 30 + .301 = 30.301.

Thus considering dy in place of  y, we have an error of .301. 2. Find dy in the following cases : (i) y = x2 + x3 ; for arbitrary x and dx. (ii) y = 2x2 – 2x ; x = 1, dx = 0.01. (iii) y = cos x ; x = /2, dx = 0.1. ANSWERS 2. (i) (2x + 3x2) dx

(ii) .02

(iii) – 0.1

Geometrical interpretation of the differential. From Fig. 4.2 it is clear that x = PN,  y = NQ and dy = NR, so that the differential of a function for given values of x and  x is equal to the increment in the ordinate of the tangent to the curve y = f (x) at [x, f (x)].

Fig. 4.2

4.1.7. Kinematical interpretation of a derivative. Every thinking person is aware of the concepts of Velocity and Acceleration of a moving point. The difficulty, however, arises in assigning precise measures to them. In practice, velocity at an instant is calculated by measuring the distance

126

DIFFERENTIAL CALCULUS

travelled in some short interval of time subsequent to the instant under consideration. This manner of calculating the velocity cannot clearly be considered precise, for different measuring agents may employ different intervals for the purpose. In fact this is only an approximate value of the actual velocity and some approximate value is all that we need in practice. The smaller the interval, the better is, of course, the approximation to the actual velocity. In books not employing the method of Differential Calculus, velocity at an instant, is defined as the distance travelled in an Infinitesimal interval subsequent to the instant. Now there exists no such thing as an infinitesimal interval of time. We can take intervals of time as small as we like and, in fact, an interval with duration smaller than any other is conceivable. The definition, as it stands, is thus meaningless. A meaning can, however, be attached to the above definition by supposing that the words ‘velocity’ and ‘infinitesimal’ in it really stand for approximation to the ‘velocity’ and ‘some short interval of time’, respectively. The precise meaning of the velocity of a moving particle at any instant can only be given by employing the notion of Derivative which we do in the following. Expression for velocity. The motion of a particle along a straight line is analytically represented by an equation of the form s = f (t). Here ‘s’ represents the distance of the particle measured from some fixed point O on the line at time t. Let P be the position of the particle at any time t. Let again Q be its position after some interval  t and let PQ = s. The ratio s/t  is the average velocity over this interval and is an approximation to the actual velocity at P. We know intuitively that better approximations can be obtained by considering smaller values of t. We are thus led to define the measure v of the velocity at time t as lim

t  0

f (t   t ) – f (t )  s ds  lim  . t 0  t t dt

Expression for acceleration. Let v be the velocity at time ‘t’ and let v be the velocity after some short interval of time  t. The ratio v/t is the average acceleration during this interval t and is an approximation to the actual acceleration at time t. The smaller values of t will correspond to better approximations for the acceleration at time t. We are thus led to define the measure of acceleration at time t as  v dv  . t 0  t dt lim

EXERCISES 1. The motion of a particle moving in a straight line being given as follows, find the velocity and acceleration (i) at end of 3 seeconds, (i) initially, in each case :(i) s = t2 + 2t + 3

(ii) s = 1/(t + 1)

(iii)

s  (t  1).

2. A particle moves along a straight line such that ‘s’ is a quadratic function of t; prove that its acceleration remains constant. 3. If s = t3 – 2t2 + 3t – 4, give the position, velocity and acceleration of the particle at the end of 0, 1, 2, seconds.

DIFFERENTIATION

127

ANSWERS 1. (i)

8, 2; 2, 2.

(ii) – 1/16, 1/32; – 1, 2.

(ii) 1/4, – 1/32; 1/2, – 1/4.

3. –4, 3, – 4; –2, 2, 2; 2, 7, 8. Note. The remaining part of this chapter is devoted to developing methods for the determination of the derivatives of the functions. Before, however, taking up some general theorems in § 4.3, we obtain the derivatives of a constant function and of the function f (x) = xn where n is a natural number.

4.2. Derivative of a constant function We consider the constant function y = c x  R where c is a given number. We have y 0  0 x x dy y  lim  0  x  R. dx x Thus, the derivative of a constant function is the zero function which is also a constant function.



Note. Looking at the derivative as the rate of change, this result appears almost intuitive in that the rate of change of anything which does not change is necessarily zero. The result may also be geometrically inferred from the fact that the slope of the tangent at any point of the curve y = c, which is a straight line parallel to x-axis, is 0. 4.2.1. Derivative of y = xn where n is a positive integer. The domain of the function y = xn, n  N is R. Let x denote any given real number. Let y be the change in y corresponding to the change x in x. We have

 y ( x  x) n (– x n )  , x  0 x x = nxn–1 + n (n – 1) xn–2 x + ............. + (x)n–1 

dy  nx n – 1 dx

d ( xn )  nx n –1 ,  x  R, and  n N. dx It follows that when n  N, the function y = xn is derivable at every point of its domain R and its derived function is nxn–1.



Ex. Give the derivatives of the functions (i) y = x3.

(ii) y = x8. ANSWERS

Ex. (i) 3 x2,

(ii) 8 x7

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DIFFERENTIAL CALCULUS

4.3. Some general theorems on Derivation 4.3.1. Derivaive of the sum or difference of two functions : Let u and v be two derivable functions of x. We write y=u+v

...(i)

Let u, v y be the respective changes in u, v, y corresponding to the change x in x so that we have y + y = u + u + v + v

...(ii)

From (i) and (ii) we get  y u v   x x x

so that when x  0, we obtain  u v  y u v  lim    lim .   lim x  x  x  x x   dy du dv   .  dx dx dx We may similarly prove that lim

d (u – v ) du dv  – . dx dx dx The above results may also be seen to be true in the case of any finite number of functions.

Ex. Give the derivatives of the functions with the function values x4 + x7, x5 – x3 + x, x7 + x4 – x3 ANSWERS 4x3 + 7x6, 5x4 – 3x2 + 1, 7x6 + 4x3 – 3x2 4.3.2. Derivative of the product of two functions. Let u, v be two derivable functions of x. We write y = uv. If u, v, y be changes in u, v, y respectively corresponding to the change  x in x, we have y + y = (u + u) (v + v) = uv + uv + vu + u. v 

y = uv + vu + u. v

y v u v u v  u . . x x x x Let x  0. Then u also  0, for u, being a derivable function, is continuous. We have



dy u v   v  lim u v  u .  x x  dx  x  0  x

v  v   u   lim  u   lim  v   lim u .lim x  x   x  dv du dv dv du u v 0 u v dx dx dx dx dx

DIFFERENTIATION

129

d (uv ) dv du u v . dx dx dx Thus we have the theorem.



The product of two derivable fucntions is itself derivable and the derivative of the product of two functions = first function × derivative of the second + second function × derivative of the first. Ex. Show that 

y = u1 u2 u3 dy du du du  u1 u2 3  u1 u3 2  u2 u3 1 . dx dx dx dx

The result can be easily extended to the product of any finite number of derivable functions. 4.3.3. Derivative of y = cu where c is constant and u, is any derivable function of x. Now y = cu 

dy du dc  du  du c u  c  u .0   dx dx dx  dx  dx

d (cu) du c . dx dx Ex. Give the derivatives of functions with function values



(i) (x3 – 2x) (3x2 + 4)

(ii) (x2 – 1)2

(iii) (x3 + 3) (x4 – 4).

ANSWERS Ex.

(i) 15x4 – 6x2 – 8;

(ii) 4x (x2 – 1)

(iii)

x2 (7x4 + 12x – 12).

4.3.4. Derivative of a Quotient. Let u and v be two derivable functions of x. We suppose that v is not zero for the value of x under consideration. We have u v u  u y  y  v  v y

 

y 



y  x

u  u u v  u – u  v –  v  v v v ( v  v )

u v – u. x x v (v  v)

v.

Now, v, being a derivable functions, is continuous, so that x  0  v  0. Thus, u du dv d  v –u dy v    dx 2 dx dx dx v

130

DIFFERENTIAL CALCULUS

so that the derivative of the quotient of two functions. = [(derivative of Numer) (Denominator) – (Numer.) Derivative of Denomi.] ÷ Square of Denominator Note. Being a derivable function v is continuous. Also we have supposed that the value of v is not zero for the value of x under consideration. There is therefore an interval around x such that no value of v is 0 for any point x in the interval. As such we suppose that x + x belongs to the interval around x in question. Cor. Derivative of y = xm where x  0 and m any non zero negative integer. m = – n so that n  N and

Writing

y = xm = 1/xn, x  0, we see that d ( x m ) d (1/ x n ) 0. x n – n . x n –1    nx – n – 1  mx m – 1 . dx dx x 2n In particular, we have d (x– 3 ) d (x– 7 )  – 3x – 4 ,  – 7x– 8 . dx dx Derivative of Polynomial function Consider the polynomial function given by y = a0xn + a1xn + 1 +.............+ an – 1 x + an. Employing the result of § 4.3, we see that  x  R dy  na0 x n – 1  ( n – 1) a1 x n – 2  ............  an – 1 dx so that the derivative of a polynomial function of degree n is a polynomial function of degree (n – 1).

Derivative of a rational function. With the help of the general results obtained so far, we are in a position to find the derivative of every rational function. The domain of derivability of a rational function is the same as its domain of definition. Example. Differentiate y = x/(1 + x). We have dy  dx

(1  x)

dx d (1  x) – x (1  x).1 – x .1 1 dx dx   2 2 (1  x) (1  x) (1  x) 2

EXERCISES 1. Find the derivatives of the following functions : (i) y = 2x2 – 3x3

(ii) y = 5x4 – 3x3 + 1

(iv) y = (1 + 3x2) (2x3 – 1) (vii)

y

(x)

y

x2 1 x

3

x3  1 2

3

( x – 1) ( x – 1)

(iii) `y = – x7 + 2x6 – x3

(v) y = (3x – 4) (x2 – 6x + 7) (vi) (viii)

y

(xi)

y

2x – 4

y = (2x – 1) (3x + 1) y

x –1 ( x – 2) ( x – 3)

x2  1

(ix)

3x  4 4x  5

(xii) y  ( x  2)2

x 1

DIFFERENTIATION

131

2. Given that y

(i)

x2 – 1

(ii)

( x – 1) 2

y

( x  4) 2 ( x – 3)

(iii) y 

x2  1 x 2 – 3x  2

find dy/dx in each case. Describe in each case the domain of the given function as also of the corresponding derived function. 3. Find the derivatives of the functions with the following function values : x2  1

(i)

(ii)

x2 – x – 2

( x – 5) 2

(iii)

x2  1

x4 a 2 – x2

(iv) x (x2 – a2)2. ANSWERS 4x – 9x2 6x (5x3 + x – 1).

1. (i) (iv) (vii)

2x – x4

(viii)

(1  x 3 ) 2 – x2  2 x  1

(ix)

( x – 2) 2 ( x – 3)

(xi)

– (4x – 5)–2. –2

2. (i) (iii) 3. (i) (iii)

( x – 1) 2

(ii) 20x3 – 9x2. (v) 9x2 – 44x + 45.

. 2

– 3x 2  2 x  3 ( x 2 – 3 x  2) 2 2

( x – x – 2)

2

– 2x2  8x  2 ( x 2  1) 2 – 2 x7 – 8x – 6 x 2  6 x2  2 x ( x 2 – 1) 2 ( x 3 – 1) 2

.

( x  4) ( x – 10)

(ii)

( x – 3) 2

, R  {3}.

, R  {1, 2}. 2( x – 5) (5 x  1)

.

2 x3 (2a 2 – x 2 ) (a 2 – x 2 )

7x6 + 12x5 – 3x2. 12x – 1.

(xii) – x (x + 2)–3.

, R  {1}.

– x2 – 6 x  1

(x)

(iii) (vi)

(ii) .

( x 2  1) 2

.

(iv) (x2 – a2) (5x2 – a2).

4.3.5. Derivative of the inverse of an invertible function. Let y = f (x) be a function derivable in its domain [a, b]. We suppose that f admits of an inverse function  so that we have y = f (x)  x =  (y). We have to find a relation between f x and and  (y) for the corresponding values of x and y. Let y be the change in y corresponding to the change x in x, as determined from y = f (x). The change x in x corresponds to the change y in y as determined from x =  (y). We suppose that f  (x)  0. We have y x x  y  .   1   . x y y  x  x  0. dx 1 dy dx   .  1  f  ( x) ' ( y )  1. dy dy / dx dx dy 1

Let  Thus

dx dy and dy are reciprocals of each other. dx

132

DIFFERENTIAL CALCULUS

Ex. Verify the theorem for y = x3 when x = 2. Cor. Derivative of y = x1/n when x > 0 and n  N. The function y = x1/n is the inverse of the function y = xn so that y = x1/n  x = yn. Now



dy dx dy . 1 . ny n – 1  1 dx dy dx

dy 1 1  n –1  dx ny n x1/ n





n 1



1 1/ n – 1 x , x  0. n

d ( x1/ n ) 1 1/ n – 1  x , x  0, n  N. dx n 4.3.6. Derivative of the function of a function. Let f and  be two given derivable functions. We write

Thus

u =  (x),

y = f (u).

Let the range of the function  be a subset of the domain of the function f so that the composite f 0  has a meaning. We have

(f 0 ) (x) = f [ (x)].

Let x be a change in x and u the corresponding change in u as determined from u =  (x). Again correponding to the change u in u, let y be the change in y as determined from y = f (u). We write y y u  . . x u x We suppose that there exists a neighbourhood of x such that for no x + x in this neighbourhood u = 0. Now x  0  u  0.  y   y u  lim    lim  .   x  0  x   x  0  u x  y u  lim . lim  x  0 u  x  0 x dy dy du  . .  dx du dx Thus y = f (u), u =  (x), so that y = f ( (x) )



dy dy du  .  f  (u )  ' ( x ). dx du dx

The result is capable of generalisation to the case of the composites of more that two functions. Note. We have shown that [f (  (x))] = f  ( (x) )   (x) we may also rewrite the result as follows (f 0 )  = (f  0 )  .

Derivative of the function y = xa x > 0 and a  Q. Let

a = p/q ; q  N, p  I.

We write y = xp/q = (x1/q)p

DIFFERENTIATION

133

u = x1/q so that y = up.

and Now

du 1 1/ q – 1 dy  x ,  pu p – 1 dx q du



dy dy du p   u p – 1 . x1/ q – 1 . dx du dx q

Thus

dx a  ax a – 1  a  Q ; x  0 dx

The present result is a generalisation of the corresponding results obtained earlier in §. (4.2 and 4.3.3 (cor.)) Ex. Find the derivatives of the functions defined by the following expression : (1  x 2 ) ,

(i)

(ii)

[(1  x) /(1 – x)]

(i) We write u = 1 + x2, y  u so that y  (1  x 2 ) . du dy 1 –1/ 2 1  2x ,  u  (1  x 2 ) – 1/ 2 dx du 2 2

We have

dy dy du 1   (1  x 2 ) – 1/ 2 2 x  . dx du dx 2 Or, directly without introducing u, we have

Hence

x (1  x 2 )

d (1  x 2 ) d (1  x 2 ) d (1  x 2 )  . dx dx d (1  x 2 ) 

(ii)

We have

1 (1  x 2 ) – 1/ 2 .2 x  x / (1  x 2 ) . 2

y  [(1  x) /(1 – x)] . 1 x , y  u1/ 2 1– x

Let

u

so that

1  x  y  . 1 – x 

We have

du  dx 

(1 – x)

d (1  x) d (1 – x) – (1  x) dx dx (1 – x) 2

(1 – x )1 – (1  x ) (– 1) (1 – x ) 2

dy 1 – 1/ 2 1  1  x   u    du 2 2 1 – x 

Hence

dy dy du 1  1  x   .    dx du dx 2  1 – x 



2 (1 – x ) 2

– 1/ 2

– 1/ 2

2 (1 – x) 2

.

.

134

DIFFERENTIAL CALCULUS



1 1/ 2

(1  x )

(1 – x )3/ 2

or directly 1  x  1  x  1  x  d   d   d  1 – x  1 – x  1 – x  .  dx dx 1  x  d  1 – x   – 1/ 2

1 1  x  2    2 1 – x  (1 – x) 2 1  . 1/ 2 (1  x ) (1 – x )3/ 2 Ex. Find the derivatives of the functions defined by the following expressions : (i) (ax + b)n

(ii) 1/(1 + x2)

( x 2  1) – ( x 2 – 1)

(iv)

(v)

( x 2  1)  ( x 2 – 1) x ( x 2 – 4)

(vii)

(viii)

( x 2 – 1)

(iii)

 a2 – x2   2   a  x2   

(vi)

 1  x2     1 – x2   

(ix)

( ax 2  2bx  c ) 3

1 – a x2 3

1  a x2 2 x2 – 1 x (1  x 2 )

ANSWERS Ex. The derivatives are the function with the following function values : (i) an (a x + b)n – 1. (ii) – 2x/(1 + x2)2. (iii)

(ax  b) / ( ax 2  2bx  c) .

(iv)

– 2 x / ( x 4 – 1) [ x 2  ( x 4 – 1)].

(v)

–

(vi)

2a2 x (a2 – x2 )1/2 (a2  x2 )3/2

– 4a 1/3

3x

(1  ax 2/3 ) 2

x4 – 2 x2  4

(vii) (viii)

( x 2 – 1)3/2 ( x 2 – 4)1/2 2x . 2 1/2 (1  x ) (1 – x 2 )3/2 4 x2  1

(ix)

.

x 2 (1  x 2 )3/2

.

.

4.4. Derivatives of Trigonometric functions : Derivative of y = sin x, x  R. y = sin x,

DIFFERENTIATION



135

y sin ( x  x ) – sin x  x x 1 1 2 cos (2 x   x ) sin  x 2 2  x 1 sin x 1   2  cos  x  x  2  1 x  2

1 sin x dy 1   2  lim cos  x  x  . lim  1  x  0 dx  x  0 2   x 2 As cos x is a continuous function, we have

1   lim cos  x  x   cos x when x  0. 2   1 sin x 2  1 when x  0. lim Also 1 x 2 dy  cos x Thus, we have dx d (sin x) = cos x. dx Derivative of y = cos x, x  R.



y = cos x 

y cos ( x  x ) – cos x  x x 1 1 – 2sin (2 x  x) sin x 2 2  x

1 sin x 1    – sin  x  x  . 2 2  1 x  2 As sin x is a continuous function, we have

1   lim sin  x  x   sin x when x  0. 2  



1 sin x dy  1   2  lim – sin  x  x   lim 1  x  0 dx  x  0  2   x 2 = – sin x. 1 = – sin x.

136

DIFFERENTIAL CALCULUS

d (cos x) = – sin x. dx Ex. Find the derivatives of the functions defined by the following function values :

Thus

(ii) cos3 x

(i) sin 2x

(iii)

(sin x )

y = sin 2x. We write

(i) Let

u = 2x so that y =sin u. dy dy du  .  cos u .2  2 cos 2 x dx du dx

Now or, briefly

d (sin 2 x) d (sin 2 x) d (2 x)  .  cos 2 x .2  2 cos 2 x. dx d (2 x) dx

y = cos3 x = (cos x)3. We write u = cos x so that y = u3.

(ii) Let

dy dy du  .  3u 2 (– sin x )  – 3cos 2 x sin x. dx du dx

Now or, briefly

d (cos3 x) d (cos x)3 d (cos x)3 d (cos x)   , dx dx d (cos x) dx 2 = 3 (cos x) × (– sin x) = – 3 cos2 x sin x (iii) Let

y  (sin x ) . We write

x , v = sin u so that y  v ,

u

dy dy dv du  , . dx dv du dx

Now



1 1/ 2 1 1 cos x 1 v cos u . x – 1/ 2  . . 2 2 4 (sin x x

or, briefly d (sin x ) d (sin x ) d sin x d x  . . dx dx d sin x d x 1 1 (sin x ) – 1/ 2 .cos x . x – 1/ 2 . 2 2 1 cos x  . . 4 x (sin x )



EXERCISES 1. Find the slopes of the tangents to the curves (i) y = sin x at (0, 0), (/2, 1) (ii) y = cos x at (0, 1), (/2, 0). Also indicate the positions of the tangents on the corresponding curves. 2. Find the derivatives of the functions defined by the following function values : (i) sin3 x

(ii)

(cos 2 x )

(iii)

3

sin 2 x

DIFFERENTIATION

137

(v)

sin 2 x 1  cos x

(vi)

cos ( x 2  3x  4)

(vii) sinm x. cosn x. (viii)

sin x 1  cos x

(ix)

( x 2  1) sin 2 x .

x – cos x x  cos x

(xii)

(1  sin 2 x ) x .

(iv)

sin x x

x cos x

(x)

(xi)

(1  x 2 )

ANSWERS 1. (i) 1, 0.

(ii) 0, – 1.

2. (i) 3 sin2 x cos x.

(ii)

– sin 2 x . (cos 2 x )

(iv) (x cos x – sin x)/x2. (vi)

–

2 cos x

(iii) 3 . 3 sin x . (v) sin x.

1 (2 x  3) ( x 2  3 x  4) – 1/ 2 sin ( x 2  3 x  4) . 2

(vii) (m cos2 x – n sin2 x) sin m – 1 x cosn – 1 x. (viii)

cos x

(x)

sin x

1 2x sec . 2 2 2 3/ 2

(1  x )

(ix) –

x sin x (1  x 2 )

( x 2  1)

{2( x 2  1) cos x  sin 2 x}. x sin 2 x  sin 2 x  1

cos x  x sin x

.

(xi) ( x  cos x)2 .

(xii)

2 (1  sin 2 x ) . x

Derivative of y = tan x. y  tan x 



dy  dx 

cos x.

   x  R  (2n  1) ; n  I  2 cos x   sin x

,

d (sin x) d (cos x) – sin x . dx dx cos 2 x

cos x .cos x  sin x .sin x cos 2 x



1 cos 2 x

 sec 2 x .

d (tan x ) = sec 2 x dx for every x which belongs to the domain of tan x.



We may similarly show that d (cot x ) = – cosecc 2 x  x  R  {n  ; n  I}. dx Its proof is left to the reader. Ex. 1. Mark the positions of the tangents on the curves (i) y = tan x at (0, 0) (ii) y = cot x at (/2, 0). 2. Find the derivatives of the functions defined by the following functions values : (i) tan x + cot x (ii) sin x tan 2x (iii) x tan x cot 2x.

.

138

DIFFERENTIAL CALCULUS

(vi)

tan x – cot x tan x  cot x

 1 – tan x     1  tan x 

(v)

(vii) cot2 3x

(vi)

x x   tan  cot  / x . 2 2 

(viii)

 1 – cos x     1  cos x 

(ix) (tan x + cot x)

ANSWERS 2. (i) (ii) (iii) (v)

– 4 cosec 2x cot 2x. cos x tan 2x + 2 sin x sec2 2x. sin 4 x – 8 x sin x

–

1 . [(cos 2 x)] (cos x  sin x )

– 6cos3 x

(vii) (ix)

(iv) 2 sin 2x.

4 cos 2 x sin 2 x

sin 2 3 x

sec2

x–

(vi)

.

– 2(sin x  x cos x )

(viii)

cosec2

1 2x sec . 2 2

( x sin x ) 2

x.

Derivative of y = sec x. We have y

   x  R  (2n  1) ; n  I  2 cos x   1

,

so that cos x  0. dy cos x .0 – 1(– sin x )  tan x sec x.  dx  cos 2 x



d (sec x)  tan x sec x. dx for every x which belongs to the domain of the functions sec x. We may similarly show that

Thus

d (cosec x)  – cot x cosec x. dx Its proof is left to the reader. Ex. Find the derivatives of the functions defined by the following function values : (i) cosec2 3x (iv)

sec ( a  bx )

(ii)

[sec ( ax  b )]

(v) sec (cosec x)

(iii) tan x sec2 x (vi)

ANSWERS Ex. (i) (ii)

– 6 cosec2 3x cot 3x. 1 a [sec ( ax  b)] tan ( ax  b). 2

(iii) sec4 x (1 + 2 tan x). (iv)

1   2 b sec ( a  bx) tan ( a  bx)  / (a  bx) .  

sec (tan x ) .

DIFFERENTIATION

(v)

139

– sec (cosec x) tan (cosec x) cot x cosec x. sec tan x tan tan x

(vi)

2

(sin 2 x) .cos x

4.5. Derivatives of inverse trigonometric functions. The precise definitions of inverse trigonometrical functions as given in § 2.7, will have to be kept in mind to obtain their derivatives. 4.5.1. Derivative of y = sin–1 x. x  [– 1, 1], y  [– /2. /2]. y = sin–1 x  x = sin y.

We have

x  sin y 

Now

dx  cos y. dy

dy 1   y dx cos y if cos y  0 i.e., 2 1 1   (1 – sin 2 y ) (1 – x 2 )

so that

where the sign, plus or minus,is the same as that of cos y, which is positive. We note that y  /2  x  1. Hence,

d (sin – 1 x)  dx

1 (1 – x 2 )

,  x  ] – 1, 1[

4.5.2. Derivative of y = cos–1 x ; x  [– 1, 1], y  [0, ] We have y = cos–1 x  x = cos y Now

x  cos y 

dx  – sin y. dy

dy 1 – if y  0 i.e., y  0 y =  dx sin y –1 –1   2  (1 – cos y )  (1 – x 2 ) where the sign, plus or minus, is the same as that of sin y which is necessarily positive.

Hence,

d (cos –1 x)  dx

–1 (1 – x 2 )

.

x  [– 1, 1] ; cos– 1 x  ]0, [ 4.5.3. Derivative of y = tan–1 x ; x  R, y  ]– /2, /2[ We have y = tan–1 x  x = tan y. Now x = tan y dx  sec 2 y so that dy/dx  0 for any y  dy

140

DIFFERENTIAL CALCULUS

 Hence

dy 1 1 1    . 2 2 dx sec y 1  tan y 1  x 2    d (tan – 1 x) 1  ; x  R ; tan – 1 x   – ,  2 dx 1+x  2 2

d (cot – 1 x) 1 – ; x  R ; cot – 1 x  ]0,  [ 2 dx 1+x Its proof is left to reader. 4.5.5. Derivative of y = sec–1 x ;     x  R  ] – 1, 1[, y   0,    ,   2 2   – 1 We have y = sec x  x = sec y.

4.5.4.

dy  sec y tan y dx dy 1  dx sec y tan y ; if any y  0 i.e., y 0, y  .

Thus 

1



2



sec y sec y – 1)

1 x ( x 2 – 1)

where the sign is the same as that of tan y. Now

 x  ]1, [  y  ]0, [  tan y  0, 2 x  ], – 1[  y  ]

Thus

x  ]1,  [  x  ] , – 1[ 

 ,  [  tan y  0. 2

dy 1  , and dx x ( x 2 – 1) dy 1 – . dx x ( x 2 – 1)

If follows that d (sec – 1 x) 1 for all admissible values of x.  dx | x| (x 2 – 1)

4.5.6. Derivative of y = cosec–1 x ;

   x  R  ] – 1, 1[, y   – , 0   0,  2 2  It may be shown that d(cosec– 1 )x –1  , dx |x| (x2 – 1)

(for all admissible values of x.)

DIFFERENTIATION

141

EXERCISES 1. Find the derivatives of the following functions : (i) (iii) (v) (vii) (ix)

y  sin –1 x ,

(ii)

y  (cot –1 x ) ,

y = tan–1 [(1 + x)/(1 – x)],

(iv)

y  tan –1 (cos x ),

y = sec–1 x2,

(vi)

y = cosec–1 (x–1/2),

 x – x –1  y  cos –1  ,  x  x –1    y  tan –1

4 x , 1 – 4x

x sin –1 x

y

(viii)

(1 – x 2 )

,

 a  b cos x  y  cos –1    b  a cos x 

(x)

2. Indicate the positions of the tangents to the curves : (i)

y = sin–1 x at (0, 0),

y = cos–1 x at (1, 0)

(ii) ANSWERS

1

1. (i)

2

–1

.

(ii)

2 (x – x ) sin x 1 . 2 x 1  cos 2 x

(iv)

–

(vii)

–

(x)

(b 2 – a 2 ) . b  a cos x

2 1 x

. 2

(v) (viii)

x (1 – x 2 )  sin –1 x 2 3/ 2

(1 – x )

4.6. Derivative of y = loga x, x  [0, ], a > 0. We have y log a ( x  x) – log a x  x x

1  x  x  log a   x  x  1 x x    . log a 1   x x x   

1  x  = log a  1   2 x  

x / x

d (log a x ) 1  log a e, x  ]0,  [. dx x Cor. Let a = e. We have y = loge x = log x dy 1 1  log e e   dx x x d (log x ) 1  , x  0.   dx x





.

4 (1  x) x (cot –1 x ) 2 . x ( x 4 – 1) .

1

(iii)

1  x2

.

1

(vi)

2 (x – x2 )

(ix)

2 . x (1  4 x)

.

142

DIFFERENTIAL CALCULUS

Derivative of y = ax, x  ]– , [ , a > 0. We have y = ax  x = loga y. dx 1 x  log a y   log a e . Now dy y dy dx dy 1 . 1 . log a e  1 Also dx dy dx y dy y .  y log e a  a x log e a . dx log a e



Cor, Let a = e so that y = ex. We obtain de x  e x log e e  e x . dx Ex. Find the derivative of the functions defined by the expressions :

(i) log sin x (iii) (v)

(ii) cos (log x)

esin x

(iv) log [sin (log x)]

log ( x 2  x  1)

(vi)

(vii) log (sec x + tan x) (ix) (xi)

(a

x

(x) log10 (sin–1 x2)

)

log[ x  ( x 2  a 2 )]

log

e2 x log x

(viii)

(xiii) a2x sin2 x (xv)

1  1 log tan  x    4  2

(xii) log (emx + e–mx) (xiv)

e

3

ax

a  b tan x a – b tan x

ANSWERS Ex.

(i) cot x, (iv)

(ii)

sin (log x ) , x

2x  1

cot (log x ) , x

(v)

(vii) sec x,

(viii)

2 x.log10 e

(x)

–

(1 – x 4 )sin –1 x 2

2( x 2  x  1)

e 2 x .2 x log x – e 2 x x (log x )

1

,

,

(xi)

( x2  a2 )

+ sin 2x),(xiv)

a e ax , 3 3 (a 2 x 2 )

2

,

(iii)

esin x cos x,

(vi)

sec x, a

,

(ix) (xii)

x

log a

4 4 (a

(xiii)

(sin2

x. log

a2

, )

m (e mx – e – mx ) e mx  e – mx

3

a 2x

x

,

2ab

(xv)

2

2

a cos x – b 2 sin 2 x

4.7 Hyperbolic functions. It is found useful to define sinhx, coshx, tanhx, cothx, sechx and cosechx as follows : ex – e– x ex  e– x sinh x  , cosh x  , 2 2 sinh x e x – e – x cosh x e x  e – x tanh x   x , coth x   , cosh x e  e – x sinh x e x – e – x

.

DIFFERENTIATION

143

1 2 1 2  x , cosech x   x . –x cosh x e – e sinh x e – e – x It may be seen that sinh x, cosh x, tanh x, sech x are defined  x  R and coth x, cosech x for all non-zero values or R. The reader may prove directly that the following results hold good  x  R. sinh (–x) = – sin h x, cosh (– x) = cosh2 x, 2 2 cosh x – sinh x = 1, cosh2 x + sinh2x = cosh 2x sinh 2x = 2 sinh x cosh x, cosh (x + y) = cosh x cosh y + sinh x sinh y, sinh (x + y) = sinh x cosh y + cosh x sinh y. It will be seen that sinh x, cosh x etc. have properties analogous to those of sin x, cos x. The graphs of hyperbolic functions will be given in Chapter 7. 4.7.1. Derivatives of Hyperbolic Functions Derivative of y = sinh x ; x  R sech x 

We have 

e x – e– x 2

y  sinh x 

dy d 1 x  1  (e – e – x )   (e x  e – x )  cosh x  dx dx  2  2

d (sinh x )  cosh x  x  R. dx Derivative of y = cosh x ; x  R.



We have

y  cosh x 

e x  e– x 2

dy e x – e – x   sinh x dx 2 d (cosh x )  sinh x  x  R.  dx Derivative of y = tanh x ; x  R.



y  tanh x 



dy  dx

cosh x.

sinh x ;cosh x  0  x  R cosh x

d (sin x ) d (cosh x) – sinh x. dx dx cosh 2 x



cosh x .cosh x – sinh x.sinh x.



cosh 2 x – sinh 2 x

cosh 2 x 2

cosh x d (tanh x)  sech 2 x. dx



1 2

cosh x

 sec h 2 x ;  x  R

144

DIFFERENTIAL CALCULUS

It may be similarly shown that d (coth x)  – cosech 2 x . dx Its proof is left to reader.

Derivative of y = sech x ; x  R 1 . cosh x cosh x .0 – 1.sinh x

y  sec h x 



dy  dx

–

cosh 2 x sinh x cosh 2 x

 – tanh x sech x ;  x  R.

d (sech x )  – tanh x sech x dx It may be similary shown that





d (cosech x )  – coth x cosech x. dx Its proof is left to the reader. 

4.7.2. Derivatives of inverse hyperbolic functions. Derivative of y = sinh–1 x. Let

y = sinh–1 x so that x = sinh y.



dx  cosh y. dy



dy 1   dx cosh y

1 2



(1  sinh y )

1 (1  x 2 )

,

where the sign of the radical is the same as that of cosh y which we know, is always positive. Hence,

d (sinh –1 x )  dx

1 (1 + x 2 )

.

Derivative of cosh–1 x. Let  

y = cosh–1 x so that x = cosh y. dx  sinh y, dy dy 1   dx sinh y

1 (cosh 2 y – 1)



1 ( x 2 – 1)

where the sigh of the radical is the same as that of sinh y. Now, cosh–1 x i.e., y is always positive so that sinh y is positive. –1 Hence, d (cosh x ) = dx

1 2

( x – 1)

.

.

DIFFERENTIATION

145

Derivative of tanh–1 x. [ | x | < 1 ]. y = tanh–1 x so that x = tanh y.

Let

dx  sech 2 y. dy dy 1 1 1    . 2 2 dx sech y 1 – tanh y 1 – x 2

  Thus

d (tanh –1 x ) 1 = . dx 1 – x2 d (coth –1 x ) 1 =– 2 [| x | > 1]. dx x –1

Its proof is left to the reader. Derivative of sech–1 x. y = sech–1 x so that x = sech y.

Let 

dx  – sech y tanh y. dy



dy 1 – dx sech y tanh y –1





2

sech y . (1 – sech y )

–1 x 1 – x2 )

.

where the sign of the radical is the same as that of tanh y. But we know that sech–1 x, i.e., y is always positive, so that tanh y is always positive. –1 Hence, d (sech x)  – dx x Derivative of cosech–1 x.

1 (1 – x 2 )

.

y = cosech–1 x so that x = cosech y.

Let 

dx  – cosech y . coth y. dy



dy 1 – dx cosech y . coth y –1

 cosech y

–1



2

(cosech y  1)

x

( x 2  1)

,

when the sign of the radical is the same as that of coth y. Now, y, and therefore coth y is positive or negative according as x is positive or negative. 

dy  dx x

–1 2

( x  1)

–1

if x  0 and  – x

x 2  1)

if x  0.

146

DIFFERENTIAL CALCULUS

d (cosech –1 x ) –1 = , dx | x | ( x 2 + 1) for all values of x.

Thus

Ex. Find the derivatives of (i) log (cosh x)

(ii)

esinh

2

(iii) tan x . tanh x.

x

ANSWERS (ii) sinh 2x esinh 2

(i) tanh x

x

(iii) sec2 x tanh x + tan x sech2 x

4.8. Derivation of parametircally defined functions. x = f (t),

Let

y = g (t)

be two derivable functions each defined in some interval. We also suppose that the function f is inveritble and  denotes the inverse of f. Now f being derivable its inverse,  is also derivable. We have x = f (t)  t =  (x). y = g (t),

Also

t =  (x),  y = g [ (x) ].

We say that y is defined as a function of x defined with the help of the parameter t. Now we have dy dy dt dy dx  .   , dx dt dx dt dt where we suppose that dx/ dt  0 for the value of t in question.

Note. It will be seen in Chapter 7 that if f' (t) = dx/ dt is zero for only a finite number of values of t, then the domain of f can be divided into a finite number of intervals in each of which the function f is strictly increasing or decreasing and therefore, invertible. In the following, we shall suppose that this condition is always satisfied so that, we have dy / dx = (dy/ dt)  (dx/ dt) for all those values of t for which dx/ dt  0 EXAMPLES 1. Find dy/ dx when x = 2 cos t – cos 2t,

y = 2 sin t – sin 2t.

We have, dx  – 2 sin t  2 sin 2t , dt

dy  2 cos t – 2 cos 2t dt

dy dy dx cos t – cos 2t 3t     tan dt dt dt sin 2t – sin t 2 for all those values of t for which dx/ dt  0.

Thus

The reader may give the set of values of t for which dx/ dt = 0. dy 2 . 2. If x  e –t and y = tan–1 (2t + 1), find dx

We have and

2 dx  e – t (– 2t ) dt

dy 1  (2) dt 1  (2t  1) 2

DIFFERENTIATION

147

dy dy / dt   dx dx / dt





2 2  1 (4t  4t  1) – 2t / et

– et

2

2

2t (2t 2  2t  1)

EXERCISES 1. Find dy/ dx, when (ii) x = 3 cos t – 2 cos3 t,

(i) x = a (cos t + sin t),

y = 3 sin t – 2 sin3 t,

y = a (sin t – t cos t), (iii) x = a cos3 t, y=a

sin3

(iv) x = a (t – sin t),

t,

y = a (1 + cos t).

2. Find dy/ dx, when (i)

xa

1 – t2 1 t

2at 2

, yb 2

1 t

2at 3

(iii)

x

(iv)

 t2 – 1 xa  2 , y  at  t  1   

1 t

2

, y

2t

1  t2

. 2

x

(ii)

3at 1 t

, y 3

3at 2 1  t2

,

,

 t2 – 1  2   t  1

ANSWERS 1. (i)

tan t,

(ii) cot t,

2

2. (i)

t –1 . 2at

(ii)

(iii) – tan t,

t (2 – t 3 ) 1 – 2t 3

,

(iii)

(iv) – tan t/2.

3 t (1 – t ) 2 2

2at

(iv)

2

(t – 1) (t 2  1)3 / 2

4.9. Logarithmic differentiation : In order to differentiation a function of the form y = uv ; where u, v are both variables, it is found useful to take its logarithm and then differentiate. This process which is known as Logarithmic differentiation is also useful when the function to be differentiated is the puoduct of a number of factors. The following examples illustrate this process. We use the process of logarithmic differentiation for differentiating the function y = xa where a is any real number. Now y = xa  log y = a log x 

1 dy a dy ay     ax a –1 y dx x dx x

EXAMPLES 1. Find the derivative of y = Now 

xsin x.

y = xsin x log y = log (xsin x) = sin x log x.

148

DIFFERENTIAL CALCULUS

Differentiating, we get 1 dy 1 .  cos x log x  sin x . y dx x dy sin x    x sin x  cos x log x  . dx x   2. Differentiate y = [x tan x + (sin x)cos x].



u = xtan x , v = (sin x)cos x

Let

dy du dv   dx dx dx By taking logarithmes, we may show that yuv



du  x tan x dx

tan x   2  sec x log x   x  

 dv cos 2 x   (sin x)cos x  – sin x log sin x   dx sin x   Adding (i) and (ii), we obtain dy/ dx 3. Differentiate

y

x1/ 2 (1 – 2 x) 2 / 3

(2 – 3 x)3/ 4 (3 – 4 x) 4 / 5 Taking lograithms, we obtain 1 2 3 4 log x  log (1 – 2 x ) – log (2 – 3 x ) – log (3 – 4 x ) 2 3 4 5 Differentiating, we obtain log y 

1 dy 1 1 2 –2 3 –3 4 –4 .  .  . – . – . y dx 2 x 3 1 – 2 x 4 2 – 3 x 5 3 – 4 x 



1 4 9 16 –   2 x 3 (1 – 2 x) 4 (2 – 3 x) 5 (3 – 4 x)

dy 4 9 16 1   y –   . dx 2 x 3 (1 – 2 x ) 4 (2 – 3 x ) 5 (3 – 4 x )  

4. If xy . yx = 1, find

dy . dx

Taking logarithm y log x + x log y = 0 Differentiating both sides, we get y.

 

1 dy 1 dy  log x .  1 . log y  x . 0 x dx y dx

x  dy  y  log x  y  dx  –  x  log y    dy ( y  x log y ) y – . . dx ( x  y log x) x

...(i) ..(ii)

DIFFERENTIATION

149

dy . dx u = ycot x and v = (tan–1 x)y, then

5. If ycot x + (tan–1)y = 1, find Let

u+v=1

...(i)

Taking log on both sides for u and v, we obtain log u = cot x log y and log v = y log (tan–1 x) On differentiating, we get 1 du 1 dy .  cot x .  log y . (– cosec2 x) u dx y dx du  cot x dy   y cot x  . – (– cosec2 x) log y  dx  y dx 



Again, 1 dv 1 1 dy .  y . 1 ·  log (tan –1 x ) 2 v dx dx tan x 1  x dv y dy    (tan –1 x ) y   log (tan –1 x ) 2 –1 dx dx   (1  x ) tan x Now, from (i)



du dv  0 dx dx

 cot x dy  y cot x  · – (cosec 2 x ) log y  dx  y 





cot x  y  

y dy    (tan –1 x ) y   log (tan –1 x ) ·   0 2 –1 dx (1  x ) tan x   . cot x  dy  (tan –1 x) y .log (tan –1 x)  y  dx  (tan –1 x) y . y    y cot x . (cosec 2 x) log y –  (1  x 2 ) tan –1 x  

 (tan –1 x) y – 1 . y  ( y cot x . cosec2 x. log y ) –   1  x2 dy    y cot x –1 –1 –1 dx (y .cot x)  (tan x ) . log (tan x)







EXERCISES Find the derivatives of the functions with the following function values : (i) (cos x)log x

(ii) (1 + x–1)x.

(iv) xsin x + (sin x)x (vii)

ex

x

(v) (tan x)cot x + (cot x)tan x (vi) (log x)x + (sin–1 x)sin x.

(1 – x )1/ 2 (2 – x 2 ) 2 / 3 (3 – x 3 )3/ 4 (4 – x 4 )

(iii)

. 4 / 5 (viii)

x3

( x 2  4) ( x 2  3)

(ix) sin x.ex . log x..xx

150

DIFFERENTIAL CALCULUS

ANSWERS Ex. (i)

(cos x)log x [(log cos x) /x – tan x . log x].

(ii) (1 + x–1)x [log (1 + x–1) – (1 + x)–1]. x

(iii)

x x e x log ex.

(iv)

 sin x xsin x   cos x log  x

 x   (sin x) x (log sin x  x cot x) 

(v) (tan x)cot x cosec2 x log (e cot x) – (cot x)tan x sec2 x log (e tan x). (vi)

(vii) (viii)

1   (sin –1 x) sin x  (log x ) x  log log x  log x  

(1 – x)1/ 2 (2 – x 2 )2 / 3 3 3/ 4

(3 – x )

(2x4

+ 15

4 4/5

(4 – x )

x2

 –1  cos x . log sin x   

sin x 2

(1 – x ) sin

–1

  x 

9x2 16 x3  1 4x      – – 3 4  2  2 (1 – x) 3 (2 – x ) 4 (3 – x ) 5 (4 – x ) 

+ 36)/3 (x2 + 3)2/3 (x2 + 4)2/3

(ix) sin x. ex. log x. xx [cot x + (x log x)–1 + log e2 x].

4.10. Derivation of implicitly defined functions Having already introduced the notion of implicitly defined function in chapter 2, we now illustrate the process of derivation of such functions without, even if it were possible, expressing explictly in terms of x. We shall also assume that in any given case the implicitly defined function is also derivable. EXAMPLES 1. Find dy/dx when x3 + y3 = 3axy. Supposing that the given equation determines a function f, we see that x3 + [f (x)3 – 3ax [f(x)] = 0 for every value of x in the domain of the function f. We have here the zero function whose derivative is 0 for every admissible value of x. Using the different rules for derivation, we see that 3x2 + 3[f(x)]2 f  (x) – 3af (x) – 3ax f  (x) = 0  

x2 – af (x) = {ax – [f (x)]2} f  (x) f  ( x) 

x 2 – af ( x)

, ax – [ f ( x)]2 the result holding for all those values of x for which the denominator

ax – [f (x)2]  0. Replace f (x) by y and f  (x) by dy/dx, we obtain While the above discussion is intended to describe the basic situation, in actual practice, however, we proceed as follows, supposing that y denotes an implicitly defined derivable function of x. Making use of the various results on the derivation, we have x3 + y3 = 3axy 

3 x2  3 y2

dy dy    3a  y  x  dx dx  

DIFFERENTIATION

151

dy x 2 – ay .  dx  ax – y 2



the result holding for points (x, y) for which ax – y2  0.

dy sin 2 (a  y )  dx sin a

2. If sin y = x sin (a + y), prove that sin y sin ( a  y )

x

We have

dx cos y sin ( a  y )  cos ( a  y ) sin y  dy sin 2 ( a  y )



sin ( a  y – y )



sin 2 ( a  y )

sin a



2

sin ( a  y )

dy sin 2 (a  y )  dx sin a



dy log x  dx (1 + log x)2

3. If xy = ex–y, prove that

xy = ex–y Taking log of both sides we get y log x = x – y 

y



dy  dx

x 1  log x



1 .x x (1  log x)2

1  log x –

log x (1  log x ) 2

4. If y 1 – x 2  x 1 – y 2  1, find

dy . dx

x = sin , y =sin , then

Let

sin  1 – sin 2   sin  1 – sin 2   1



sin ( + ) = 1



sin–1 x + sin–1 y = sin–1 1

Differentiating both sides, 1 1– x

2



1 1– y

2

.

dy 0 dx

152

DIFFERENTIAL CALCULUS

dy – dx

 5. Find

dy , where y  dx

 1 – y2   2   1 – x 

 x

x x to 

x x to 

 y2  x y y2  xx Taking logarithm, we get 2 log y = y2 log x On differentiation, we obtain 1 dy 1 dy 2.  y 2 .  2 y . .log x y dx x dx We have

2

2  dy y 2  y – 2 y log x  dx  x   dy y3   dx 2 x (1 – y 2 log x) sin x 6. If cos x 1 sin x 1 1  cos x .....  dy (1  y ) cos x  sin x  then prove that dx 1  2 y  cos x – sin x Given function is sin x (1  y ) sin x y  cos x 1  y  cos x 1 1 y or y + y2 + y cos x = (1 + y) sin x. On differentiation, we get dy dy dy dy  2y  y (– sin x )  cos x .  (1  y ) cos x  .sin x dx dx dx dx dy or (1 + 2y + cos x – sin x) = (1 + y) cos x + y sin x dx dy (1  y)cos x  y sin x   dx (1  2 y  cos x – sin x) 

EXERCISES 1. Find

dy if dx

(i) 3x2 – 2y2 = 1 for (1, 1) and (1, – 1) (ii) x4 + y4 – a2 xy = 0 for (x, y) dy

–1

2. If x 1  y  y 1  x  0 prove that dy  (1  x) 2

DIFFERENTIATION

153

dy x x a 2–  log , prove that dx y x– y x– y dy Find if y and x are related by the equation x2 sin y = y2 sin x. dx

3. If 4.

dy if dx (i) xm ym = (x + y)m+n

5. Find

6.

(ii) y = cos (x + y) y  x log

y a  bx

(iii) x2/3 + y2/3 = a2/3

(iv)

(v) x = y log (xy) (vii) (cos x)y = (sin y)x

(vi) xv + yx = c

dy Find when x  e tan dx

 y – x2   –1   x2   

7. If y = x cos y + y cos x, find

dy dx

8. If y  sin x  sin x  sin x  ....... , x

9. If y 

3

x

find

find x

dy . dx

dy . dx

3

x

x

3

x x  ........ dy

10. If 1 – x 6  1 – y 6  a3 ( x3 – y 3 ) , find dx ANSWERS 1. (i) 4.

3 3 ,– . 2 2

y 2 cos x – 2 x sin y x 2 cos y – 2 y sin x

5. (i) (iv) (vii)

(ii) (4x3 – a2y)/(a2x – 4y3) ; .

(ii)

– sin ( x  y ) 1  sin ( x  y )

(iii)

y {(a  bx) y – bx 2 } x ( y – x)( a  bx)

(v)

y ( x – y) x ( x  y)

(vi)

 y –   x ( yx y – 1  x y log y ) x y log x  xy x – 1

log sin y  y tan x log cos x – x cot y

6.

2x + x (2 tan (log x) + sec2 (log x))

7.

cos y – y sin x 1  x sin y – cos x

10.

1/ 3

y x

x2

1 – y6

y2

1 – x6

cos x 8. 2 y –1

5 2/3 x (1 – y ) 9. 3 5/ 3 (x  2 y)

154

DIFFERENTIAL CALCULUS

4.11. Preliminary transformation. In some cases, a preliminary transformation of the function to be differentiated facilitates the process of differentiation a good deal, as is illustrated by the following examples. 1. Differentiate sin –1

2x 1  x2

.

Putting x = tan , we have 2x y  sin –1 1  x2  sin –1



2 tan  1  tan 2 

 sin –1 (sin 2 )  2  2 tan –1 x.

dy 2  . dx 1  x 2

2. Differentiate tan –1

(1  x) –

(1 – x)

(1  x) 

(1 – x)

Putting x = cos , we have

 (1  x)  (1  cos )   2 cos 2 

    2 cos . 2 2

   (1 – x)  (1 – cos )   2sin 2   2 sin . 2 2    cos – sin –1 2 2 y  tan    cos  sin 2 2  1 – tan –1 2  tan  1  tan 2       tan –1  tan  –     4 2     1  –  – cos –1 x. 4 2 4 2 dy 1  .  dx 2 (1 – x 2 ) 3. Find the differential co-officient of tan –1 Let Putting

y = tan –1

2x 2

2x 1– x

, z  sin –1

1– x x = tan , we see that

2

–1 with respect to sin

2x 1  x2

.

2x 1 + x2

.

DIFFERENTIATION

y = tan –1

155

2 tan 

1 – tan 2  –1 2 tan 

 tan –1 (tan 2 )  2   2 tan –1 x.

 sin –1 (sin 2 )  2   2 tan –1 x. 1  tan 2  dy 2 dz 2  ,  .  2 dx 1  x dx 1  x 2 dy dy dx  .  1.  dz dx dz Also otherwise, we have y = z, z = tan

dy  1. dz

so that 4. If

(1 – x 2 ) 

(1 – y 2 )  a ( x – y ), prove that

(1  y 2 ) dy  . dx (1  x 2 )

Putting x = sin  and y = sin , we have (1 – sin 2 ) 

(1 – sin 2 )  a (sin  – sin )



cos  + cos  = a (sin  – sin ) 1 1 2 cos ( – ) cos ( – ) cos   cos  2 2   a 1 1 sin  – sin  2 cos ( – ) sin ( – ) 2 2 1 cot (  –  )  a  2 



1 ( – )  cot –1 a 2  –  = 2 cot–1 a.

sin–1 x – sin–1 y = 2 cot–1 a.

Differentiating, we get 1 – (1 – x 2 ) 

dy  dx

dy  0, (1 – y 2 ) dx 1

(1 – y 2 ) (1 – x 2 )

.

–1  3cos x – 2sin x  Differentiate cos   13    5sin x  4 cos x  sin –1  w.r.t. . 41  

5.

Let

 3cos x – 2sin x  u  cos –1   and 13  

156

DIFFERENTIAL CALCULUS

 5sin x  4 cos x  v  sin –1   41   3 2  cos  and  sin  Now, let 13 13  u = cos–1 (cos  cos x – sin  sin x) = x + 

...(i)

du 1 dx



5

4  cos  and  sin  41 41 then v = sin–1 [cos  sin x + sin  cos x] = x +  Again, let



dv 1 dx

from (i) and (ii)

du  1. dv

...(ii)

EXERCISES Find the derivatives of : x cos –1 x

1.

y

4.

x a y  tan –1  tan –1  a x

7.

y  tan –1

(1 – x

2

.

cos x . 1  sin x

(3x 2 – 1) 1  x 2

2.

y

5.

y  e( x) .

8.

 1 – cos x  y  tan –1    1  cos x 

x2 x

x x2 – 4 a2

3.

y

6.

y  tan –1

9.

y

1/ 2

( x2 – a2 )

.

x1/ 3  a1/ 3 1 – a1/ 3 x1/ 3

(1 – 2 x ) 2 / 3 (1  3 x ) –3/ 4 (1 – 6 x )5 / 6 (1  7 x ) –6 / 7

Obtain the derivatives of the functions given by the following function values : 10. (xx)x.

11.

x( x

x

).

12. (1  x 2 )  (1 – x 2 )

13.

  1  log  tan  x   .   2 

14.

tan –1

15.

 1  tan x   .  1 – tan x 

16.

sin –1 [2ax (1 – a 2 x 2 )]. 17.

18. x log x . log (log x).

19.

tan –1

22.

 e ax – e – ax  sin –1  ax .  e  e – ax   

25.

  x – 2 3/ 4 log e x     x  2 

28.

9x4 sin (3x – 7) log (1 – 5x)

2

21.

e – ax

24.

sin –1

27.

.

sin ( x log x). x2 4

4

.

(x  a )

 x sin a  tan –1  .  1 – x cos a 

(1  x 2 ) – (1 – x 2 )

a  b cos x . b  a cos x

 .  

.

[1 – x2]3/2 .sin – 1 x

.

10log sin x .

20.

(sin x ) cos

23.

tan –1

26.

1  log  1 –

1

x

(1  x 2 ) – 1 . x x x

  . 

.

DIFFERENTIATION

157 x2

29. eax cos (b tan–1 x)

30.

1  1   x 

32. x ax sinh x.

33.

–1 If y  tan

34. Find

cot x coth x.

31. 3a 4 x – x 3 a (a 2 – 3 x 2 )

dy 3a  , prove that dx a 2  x 2

dy when dx

(i) x 

sin 3 t cos3 t ,y at t   / 6 cos 2t cos 2t

(ii) x  sin t cos 2t , y  cos t cos 2t  

1  

(iii) x  a  cos t  log tan t  , y  a sin t 2





 –1  2 35. Differentiate tan  1 x – 1 / x  with respect to tan–1 x.  

36. Differentiate xsin x with respect to (sinx)x –1 37. Differentiate tan

1  x2 – 1 – x 2 1  x2  1 – x2

with respect to cos–1 x2.

38. Differentiate (log x)tan x with respect to sin (m cos–1 x). ANSWERS

1.

3.

3 – x2

cos –1 x – (1 – x 2 )

2.

(1 – x 2 )3/ 2

( x 2 – a 2 )3/ 2 ( x 2 – 4a 2 )1/ 2

.

4.

1 . 2

8.

9. 10.

(1 – 2 x ) 2 / 3 (1  3 x ) –3/ 4 (1 – 6 x )5 / 6 (1  7 x ) –6 / 7

x( x

2

 1)

–

log (ex 2 ).

x (1 – x 4 ) 2a

16.

(1 – a 2 x 2 )

.

.

x

2/3

(1  x 2 / 3 )

.

1 . 2

 5  6 4 9 – –   . 1 – 6 x 1  7 x 3(1 – 2 x ) 4(1  3 x )  

12. 1 – x 2 – 3 x (1  x 2 ) .sin –1 x. 14.

2

1

6.

–

1  x –1 x  2 .  tan 2  a a a  x  2 2 –1 x   a  x  tan  a  a2

x 4 – 2a 2 x 2 – 4a 4

5. exx . log ex. 7.

x 4 (1  x 2 )

x

) x –1

11.

[1  x log x log ex ] x ( x

13.

cosech x.

15.

1 . (cos x – sin x ) (cos 2 x )

17.

10log sin x cot x. log 10.

.

158

DIFFERENTIAL CALCULUS

18. log (e log x) + log x . log log x. –1

x

( a 2 – b 2 )sin x

19.

( a 2  b 2 ) (1  cos 2 x )  4ab cos x

   cos –1 x.cot x – log sin x  .  (1 – x 2 )  

20.

(sin x )cos

21.

2 e –ax [cos (x log x). log ex – 2ax sin (x log x)].

1

22. a sech ax. 25.

.

23.

x2 – 1

24.

a  x4

1 . x (1 – x)

27.

1 – 2 x cos a  x 2

2(1  x )

26.

x2 – 4

2a 2 x

.

2

4

4

.

sin a

5

.



–1 28. 9x4 sin (3x – 7) log (1 – 5x) ×  x – {log (1 – 5 x)} (1 – 5 x)  3cot (3x – 7)  .  

29. eax [a cos (b tan–1 x) – b sin (b tan–1 x) (1 + x2)–1]. 30.

1  1   x 

x2

 1 x    2 x log 1   – . x 1  x   

31. – cosec2 x coth x – cosech2 x cot x. 32. ax sinh x + xax sinh x log a + x ax cosh x 34.

(i) 0.

35.

1 2

37.

–

(iii) tan t.

(ii) – tan 3t.

36. 1 2

38.

x sin x – 1 (sin x  x cos x log x ) (sin x ) x – 1 ( x cos x  sin x log sin x )

– (1 – x 2 ) (log x) tan x [sec 2 x log (log x)  tan x ( x log x) –1 ] m cos (m cos –1 x)

4.12. Differentitation of a Determinant. If a function be given in the form of a determienant, i.e., if p ( x) q ( x) r ( x)

y  u ( x)

then

v ( x)

w ( x)

 ( x)  ( x)  ( x) p  ( x ) q ( x) r ( x) p ( x) q  ( x) r ( x ) p ( x ) q ( x) r  ( x) dy  u  ( x ) v ( x) w ( x)  u ( x) v ( x) w ( x)  u ( x) v ( x) w ( x) , dx  ( x)  ( x)  ( x )  ( x )  ( x)  ( x)  ( x)  ( x)  ( x)

The differentiation can also be done by row-wise. x

Example : If

x2

x3

f ( x )  1 2 x 3 x 2 , find f '(x). 0 2 6x

DIFFERENTIATION

159

On differentiating, we get

d x ( x) x 2 x3 dx d (1) 2 x 3 x 2  1 f  ( x)  dx d (0) 2 6 x 0 dx 1

x2

 0 2x 0

2

x3

x 2x

3x 2  1

2

6x

0

0

d 2 ( x ) x3 x x2 dx d (2 x) 3 x 2  1 2 x dx d (2) 6 x 0 2 dx x3

x2

3x 2

3x 2  1 2 x 6x 0 2

6x 6

x

d 3 x dx d 2 3x dx d 6x dx

= 0 + 0 + 6 (2x2 – x2) = 6x2. MISCELLANEOUS

EXAMPLES

 3 1. If f ( x)  sin  [ x] – x  ,1  x  2, and [x] = the greatest integer  x, then find 2   f  ( 3  / 2).   2  if x  3  / 2 ,[ x ]  1 2   f ( x)  sin  – x3   cos x3 2  f  (x) = – sin x3 . 3x2

We have, 1  

3

2/ 3 2/ 3      f   3   – 3   . sin  / 2  – 3   . 2 2  2 2. Let f (x) be a polynomial function of second degree. If f (1) = f (– 1) and p, q, r are in A.P., then show that f  (p), f  (q), f  (r) are also in A.P.



Let Then Now,

f (x) = a1 x2 + a2x + a3 f  (x) = 2 a1x + a2 f (1) = f (– 1)



a1 + a 2 + a 3 = a 1 – a 2 + a 3



a2 = 0



f  (x) = 2a1x,

and

f  (p) = 2a1 p, f  (q) = 2a1 q, f  (r) = 2a1 r.

as p, q, r are in A.P., so f  (p), f  (q), f  (r) will also be in A.P. 3. If u = f (x3), v = g (x2), f  (x) = cos x and g (x) = sin x, then find We have

du d  f  ( x 3 ). ( x 3 ) dx dx = cos (x3). 3x2 = 3x2 cos x3

du . dv

160

DIFFERENTIAL CALCULUS

and



dv d  g  ( x 2 ). ( x 2 ) dx dx = 2x sin x2

du du / dx 3 x 2 .cos x3   dv dv / dx 2 x .sin x 2

 2x – 1  dy .  and f  (x) = sin x2, find 4. If y  f  2 dx  x  1  2x – 1  d  2x – 1  dy  f   2  .  2  We have dx  x  1  dx  x  1  2

 2 x – 1   ( x 2  1).2 – (2 x – 1).2 x   sin  2  ,   ( x 2  1) 2   x  1   2

 2 x – 1  2 (1  x – x 2 )  sin  2 .  . ( x 2  1) 2  x  1 –1  a1 x –   –1  a2 – a1  5. If y  tan  a   x   tan  1  a a   1   1 2 

 an – an – 1   a – a2  –1 dy  tan –1  3  – tan –1 (an), then find .   .......  tan   1  a a 1  a a dx 3 2  n n –1   

We have

 

   a1 –  x   {(tan –1 a – tan –1 a ) + (tan–1 a – tan–1 a ) + y  tan –1   2 1 3 2  1  a1   x   ........ + (tan–1 an – tan–1 an–1) – tan–1 an}   tan –1 a1 – tan –1 – tan –1 a1 x –1    y  – tan   x dy 1  – – . 2  2  dx 1  ( / x )  x  

 2

x  2

.

d { f n ( x )}. 6. If fn(x) = e f n – 1 ( x ) for all n  N and f0 (x) = x, then find dx We have, ( x) d d { f n ( x )}  e f n1 . { f n – 1 ( x )} dx dx

 f n ( x ).

d { f n – 1 ( x )} dx

...(i)

DIFFERENTIATION

161

From (i), we have d d { f n – 1 ( x )}  f n – 1 ( x ). { f n – 2 ( x )} dx dx

d d { f n ( x )}  f n ( x ). f n – 1 ( x). { f n – 2 ( x)} dx dx

 Similarly

d d  { f n ( x )}  f n ( x ). f n – 1 ( x). f n – 2 ( x)    f 2 ( x) f1 ( x)  f 0 ( x)  dx  dx 

= fn (x) fn–1 (x) fn–2 (x) – – – f2 (x) f1 (x), 7. If for all x, y the function f is defined by : f (x) + f (y) + f (x) . f (y) = 1 and f (x) > 0. Then find f  (x). Putting y = x, we get 2 f (x) + {f (x)}2 = 1 On differentiating w. r. t. x, 2 f  (x) + 2 f (x) f  (x) = 0 2 f  (x) {1+ f (x)} = 0

or 

f  (x) = 0; because f (x) > 0

9. Let f (x) be a real function not identically zero such that f(x + y2n + 1) = f (x) + {f (y)}2n + 1, n  N and x, y are any real numbers and f  (0)  0. Find the value of f (5) and f  (10). x = 0, y = 0, we get

Putting

f (0) = f (0) + {f (0)}2n + 1  f (0) = 0 f  (0)  0 [Given] 

lim

x 0

f ( x) – f (0) 0 x–0



lim

f ( x) 0 x

x0



x > 0 f (x) > 0.

Putting x = 0, y = 1 in the given relation f (1) = f (0) + {f (1)}2n + 1 

f (1) = 0 or 1.

Now, putting y = 1, for all real x, we have f (x + 1) = f (x) + [f (1)}2n + 1. Case I : Then

If f (1) = 0 f (x + 1) = f (x)



f (1) = f (2) = f (3) = ......... = 0

It is not possible. Case II : Then

If f (1) = 1 f (x + 1) = f (x) + 1

162

DIFFERENTIAL CALCULUS



f (2) = f (1) + 1 = 1 + 1 = 2 f (3) = f (2) + 1 = 2 + 1 = 3 f (4) = f (3) + 1 = 3 + 1 = 4 ......................................... ......................................... f (x) = x f  (x) = 1 f  (10) = 1 and f (5) = 5.

and 

EXERCISES 1

dy

1. If 5 f ( x)  3 f    x  2 and y = x f (x) then find at x = 1. dx x 2. If f  (x) = sin x + sin 4x. cos x, find f (2x2 + /2). 3. If f  ( x)  2 x 2 – 1 and y = f (x2) then find

dy at x = 1. dx

4. If f (x) = (x + ) cos x + ( x + ) sin x and f  (x) = x cos x is identity in x, find the values of , ,  and . –1 5. If y  tan

1 2

 tan –1

1 2

 tan –1

x  x 1 x  3x  3 dy 1 1 then show that dx  ( x  n) 2  1 – x 2  1

1 2

x  5x  7

+ ..... to nth term,

f  ( )

6. If f () = cos 1. cos 2. cos 3 ....... cos n, find the value of f () . n

dy

7. If y = (1 + x) (1 + x2) (1 + x4) ....... (1  x 2 ) , find at x = 0. dx 8. If (sin y )

sin  x 2



dy 3 at x = – 1 sec –1 2 x  2 x tan{log e ( x  2)}  0, find dx 2

9. If f r (x), g r (x), hr (x); r = 1, 2, 3 are polynomials in x such that f r (a) = g r (a) = h r (a) ; r = 1, 2, 3 and

f1 ( x)

f 2 ( x)

f3 ( x)

f ( x)  g1 ( x) g 2 ( x) g3 ( x) , h1 ( x)

h2 ( x)

h3 ( x)

then find f  (a). 10. If y  log prove that

x2  x  1 2

x – x 1



1  –1 1  2 x 2 x –1   tan –1  tan  2 3 3 3 

dy 1  4 . dx x  x 2  1

DIFFERENTIATION

163

ANSWERS 1. 7/8 ;

2. (cos 2x2 – sin 8 x2. sin 2 x2). 4x

3. 2 ;

4.  = 0,  = 1,  = 1,  = 0 ;

6. – {tan 1 + tan 2 + ......+ tan n}

7. 1 ;

3

8.

9. 0.

 2 – 3

OBJECTIVE QUESTIONS For each of the following questions four alternatives are given for the answer. Only one of them is correct. Choose the correct alternatives : 1. The value of the derivative of | x – 1 | + | x – 3 | at x = 2 is : (a) – 2 (b) 0 (c) 2 (d) None of these 2. At x = 1, the function 0  x 1  2 ( x – 1) , f ( x)   is ( x – 1) ( x  1) 1 x2 

(a) Continuous and differentiable (b) Continuous but not differentiable (c) Not continuous but differentiable (d) Neither continuous nor differentiable 3. The derivative of f (x) = | x |3 at x = 0 is : (a) – 1

(b) not defined

4. If f  (x0) exists, then lim

h0

(a)

(c) 0

(d) 1/2

f ( x0  h) – ( x0 – h) is equal to 2h

1 f  ( x0 ) 2

(b

f  (x0)

(c) 2 f  (x0)

(d) None of these

5. If for a continuous function f, f (0) = f (1) = 0, f  (1) = 2 and y (x) = f (ex) ef (x), then y  (0) is equal to (a) 1

(b)

2

(c)

0

(d) None of these

(c) 1 + cos2 x

(d) – 1/4

(c)

–1

(d) None of these

(c)

2 sec3 x

(d) 2 cosec3 x.

dy –1  sin x    , then 6. If y  tan   x 1 cos dx   (a) 1/4 (b) 1/2 dy –1  cos x – sin x    , then 7. If y  tan   x x cos sin dx   (a) 0 (b) 1

8. The differential coefficient of log tan x is (a) 2 sec 2 x

(b)

2 cosec 2 x

164

DIFFERENTIAL CALCULUS

2 2   dy –1  1  x  1 – x  y  tan , then 9. If equals dx  1  x 2 – 1 – x2    1 1 (a) (b) – (c) 4 1 – x4 1– x

x 1– x

4

(d)

–

x 1 – x4

10. The derivative of sin–1 x w.r.t. cos –1 (1 – x 2 ) is (a) 1/ 1 – x 2

(b)

cos–1 x

(c)

1

 dy  11. If x = a (1 – cos ), y = a ( – sin), then     dx     / 2 (a) – 2 (b) – 4 (c) 1 12. If x = a{cos t + log tan (t (2)}, y = a sin t, dy/dx = (a) tan t (b) cos t (c) sec t 13. If sin y = x sin (a + y), then

(d) – 1 (d) cosec t

dy  dx

sin 2 (a  y ) (b) sin (a + y) (c) sin2 (a + y) sin a 14. The differential coefficient of f (log x) where f (x) = log x is : (a) x/log x (b) (log x)/x (c) (x log x)–1 (a)

15. If y = log | x |, then

(d) None of these

(d)

sin (a  y ) sin a

(d) None of these

dy  dx

(a) 1/x

(b)

– 1/x

(c)

1 | x|

(d) None of these

(c)

y  (1) = 1

(d) y  (0) = 

16. If f (x) = | log x |, then : 1 1 (b) y  (1 – 0)  x x 17. If logx (log x), then f  (x) at x = e is :

(a) y  (1  0) 

1 e dy 18. If x p y q = (x + y) p + q, then is equal to : dx py y (a) (b) qx x

(a) e

(b)

1  x  dy  , then  19. If y  log  1 – x dx   1 1 (a) (b) x (1  x ) x (1 – x )

(c)

x y

(d) 0

(c)

x y

(d)

(c)

x (1  x)

(d) None of these

(c)

1/e

(d) 1/2 e

qy px

20. If f ( x)  log x2 (log x) then f  (x) at x = e is : (a) 0

(b)

1

DIFFERENTIATION

165

 2x – 1 dy 21. If y  f  2  and f  (x) = sin x2, then at x = 0 equals x  1 dx  

(a)

1 sin1 2

(b)

sin 1

(c)

2 sin 1

(d) None of these

dy at x = y = 1 is : dx (b) 2 (c) 0 dy cos x  cos x  ........  , then  dx sin x – sin x (b) 1 – 2 y (c) 1 – 2y

22. If 2x + 2y = 2x + y, then the value of (a) 1 23. If y 

cos x 

cos x (a) 1 – 2 y

dy  dx –2

(d) – 1

(d)

– cos x 1 – 2y

24. If sin (x + y) = log (x + y), then (a) 2

(b) dy x  e x  ......   25. If y  e , then dx 1 y (a) 1 – y (b) 1 – y

(c)

1

(d) – 1

(c)

2y 1– y

(d) None of these

xx log x

(d) None of these

e

(d) 1/e

y . 2x (loge 2)2

(d) y loge 2.

d x ( x ) is : dx (a) x . xx – 1 (b) xx log e x (c) 27. The derivative of (log x)log x at x = e is equal to : (a) 0 (b) 1 (c)

26. The value of

x

2 28. If y  2 , then

dy  dx

(a) y (log10 2)2

(b)

y (loge 2)2

(c)

29. Let f (x + y) = f (x) f (y) for all x and y. Suppose that f (3) = 3 and f  (0) = 11, then f  (3) is equal to : (a) 22

(b)

dy  30. If xy = ex – y, then dx log x (a) (b) (1  log x ) 2

33

(c)

x –y 1  log x

(c)

28

(d) None of these

x –y (1  log x )

2

(d)

y –x x (1  log x)

ANSWERS 1. 8. 15. 22. 29.

(b) (b) (a) (d) (b)

2. 9. 16. 23. 30.

(a) (d) (a) (b) (a)

3. 10. 17. 24.

(c) (c) (b) (d)

4. 11. 18. 25.

(b) (c) (a) (b)

5. 12. 19. 26.

(b) (a) (b) (b)

6. 13. 20. 27.

(b) (a) (d) (d)

7. 14. 21. 28.

(c) (c) (c) (c)

CHAPTER 5 SUCCESSIVE DIFFERENTIATION 5.1 Higher order derivatives The derivative f '' of a derivable function f is itself a function. We suppose that it also possesses a derivative, which we denote by f '' and call it the second derivative of f. The third derivative of f which is the derivative of f '' is denoted by f''' and so on. Thus, the successive derivatives of f are represented by the symbols, f ', f '' ,......., f n, ....... where each term is the derivates of the preceding one. If we write y = f (x), the symbols y1, y2, y3, ........., yn, .......... dy d y d y dn y , 2 , 3 , ....... n , ........ dx dx dx dx are used to denote the successive derivatives of y. 2

3

or

The symbols dn y f n ( a ),  n  , yn ( a )  dx  x  a

denote the value of the nth derivative at a. EXAMPLES 1. If x = a (cos  +  sin ), y = a (sin  –  cos ), find d2y /dx2. We have

dx/d  = a (– sin  + sin  +  cos ) = a  cos  dy/ d  = a (cos  – cos  +  sin ) = a  sin 

 

dy dy d  dy dx  .    tan  dx d  dx d  d 

d2y dx 2

 sec 2 

d sec 2  sec3    . dx a  cos  a

[Note. The result holds good for all those values of  for which What are these values ?] 166

dx  0. d

SUCCESSIVE DIFFERENTIATION

167

2. If y = sin (sin x), prove that d2y dx

2

 tan x

dy  y cos 2 x  0. dx

dy  cos (sin x ) cos x, dx d2y  – sin (sin x) cos 2 x – cos (sin x) sin x dx 2

We have

d2 y



dx

 y cos 2 x  – cos (sin x) . cos x .

2

d2y

sin x cos x

dy  y cos 2 x  0. dx dx The result holds good for all those values of x for which cos x  0.



2

 tan x

3. If p2 = a2 cos2  + b2 sin2 , prove that p

d2 p d 2



a2 b2

(Avadh 2000)

p3

p2 = a2 cos2  + b2 sin2 

We have

...(i)

Differentiating both sides of (i) w.r.t , we get 2p p



dp  – 2a 2 cos  sin   2b 2 sin  cos  d dp  (b 2 – a 2 ) sin  cos  d

...(ii)

Differentiating both sides of (ii) w.r.t. , we get p

d2 p

2

 dp  2 2 2 2    (b – a ) (cos  – sin ) 2 d  d  

= (b2 cos2  + a2 sin2 ) – (a2 cos2  + b2 sin2 ) = (b2 cos2  + a2 sin2 ) – p2 p



d2 p

 dp   p 2  (b 2 cos 2   a 2 sin 2 ) –   d 2 d

= (b2 cos2  + a2 sin2 ) – 

2

(b 2 – a 2 ) 2 sin 2  cos 2  p2

(from (ii) )

1  2 2 p (b cos 2   a 2 sin 2 ) – (b 2 – a 2 ) 2 sin 2  cos 2   2  p 1

[( a 2 cos 2   b 2 sin 2 ) (b 2 cos 2   a 2 sin 2 ) – (b2 – a 2 ) 2 sin 2 θ cos 2 θ  p2 1  2  a 2 b 2 cos 4   a 2 b 2 sin 4   2a 2 b 2 sin 2  cos 2   p 

168

DIFFERENTIAL CALCULUS

 

a 2b2 p2 a 2b 2

(cos 2   sin 2 ) 2

p2

Dividing both sides by p, we get d2 p a 2b2  p  d 2 p3 4. If y = a cos (log x) + b sin (log x), show that x2

d2 y

dy  y0 dx y = a cos (log x) + b sin (log x)

 x

dx 2

We have

dy 1 1  [– a sin (log x )] .  [b cos (log x )] . dx x x



dy  – a sin (log x )  b cos (log x ) dx Differentiating again w.r.t. x, we get, x



x

d2y

d2y dx

2



dy 1 1  [– a cos (log x ) .  [– b sin (log x)] . dx x x 1  – [ a cos (log x )  b sin (log x )] x y – x

dy  y0 dx dx 5. Change the independent variable to  in the equation d2y 2 x dy y   0 dx 2 1  x 2 dx (1  x 2 )2



x2

2

x

by means of the transformantion x = tan . We have

dy dy dx dy    cos 2  ; dx d d d

dx  sec 2   0 for any value of  d d2y d  dy d 2 y d 2  – 2 cos  sin  . .  cos  . . Again dx d  dx 2 d 2 dx dy d2y  – 2 cos  sin  cos 2  .  cos 2  . 2 .cos 2  d d dy d2 y 3 4  – 2 sin  cos  .  cos  . 2 . d d Substituting the values of x, dy/dx and d2y/ dx2 in the given differential equation, we have

– 2 sin  cos3  .

2 tan  dy y dy d2 y . cos 2   0  cos 4  . 2  2 d  (1  tan 2 ) 2 1  tan  d d

SUCCESSIVE DIFFERENTIATION

– 2 sin  cos3 



d2y



d 2

169

dy dy d2y  2 sin  . cos3   cos 4  . y  0  cos 4  . d  d d

 y  0.

6. If x = f (t) and y = f (t), prove that

d2y dx

w.r.t. t ?



2

f1 2 – f 2 1

, where suffixes denote differentiation

f13

dy dy / dt 1 (t )   dx dx / dt f1 (t ) Again differentiating both sides w.r.t. t, d2y dx

We have

2



f1 (t ) .  2 (t ) – 1 (t ) . f 2 (t ) 2



[ f1 (t )] f1 (t ) 2 (t ) – 1 (t ) . f 2 (t )



f1 2 – 1 f 2

2

[ f1 (t )]

f13

.

.

dt dx

1 f1 (t )

.

xb  a tan –1 (a loge y), a > 0, prove that yy'' – yy' log y = (y' )2. 2  x  b a log y  tan  We have   2a  On differentiating w.r.t. x, we get a  x  b 1 . y   sec2  . y  2a  2a

7. If

2a 2 y  2  1  a  log y  y



Again differentiating both sides w.r.t. x, 2a 2 ( yy  – y  y ) 1  2a 2  log y  . . y  2 y y or yy'' – (y')2 = yy' (log y)  yy'' – yy' (log y) = (y')2.

 log y  8. If 

x  

2

x2 – a2     k log ( x  x 2 – a 2 ) , a  

d2y dy x  2a. 2 dx dx On differentiating w.r.t. x, we get

then prove that ( x 2 – a 2 )

dy  2 log dx

x  

x2 – a2   . a  x

a x2 – a2

.

x2  a2 )

(x 

x2 – a2 k

 (x 

x2 – a2 )

.

(x 

x2  a2 ) x2 – a2

170

DIFFERENTIAL CALCULUS

x . log   x2 – a2

dy  dx

or

2a

dy  2a log dx Again differentiating, we get

or

x2 – a2

x2 – a2

d2y dx 2



dy . dx

x  

x x2 – a2

or

x2 – a2

or

( x2 – a2 )

d2y dx 2 d2y dx

2

x2 – a2    a 

k 2

x – a2

x2 – a2    k a 

2a



x2 – a2

x



x 2

x –a x

2

·

dy  dx

x2 – a2

x

x2 – a2 2a 2

x – a2

dy  2a dx

x  y  f ( x)  f ( y ) 9. If f  , f' (0) = a and f (0) = b, then find f'' (x) where y is independent   2  2 of x.

Since y is independent of x, hence

dy  0. dx

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

dy  1  dy  x  y 1  f  . 1     f  ( x)  f  ( y ) .   2  2 dx  2  dx  

1 x  y 1  f f ( x)  2  2  2

...(i)

Taking x = 0 and y = x in (i), we get 0  x f    f  (0)  2 



x f  a 2

which shows

x x f  a c 2 2



f (x) = ax + c.

let

x = 0,



f (x) = ax + b

On differentiating f ' (x) = a and

f'' (x) = 0

f (0) = b = c

SUCCESSIVE DIFFERENTIATION

171

EXERCISES 1. If y 

log x d 2 y 2 log x – 3 . , show that 2  x dx x3

2 cos x

2. If y = log (sin x), show that y3  sin 3 x . 3. Show that y = x + tan x satisfies the differential equation cos 2 x

d2y – 2 y  2x  0 . dx

4. If y = [(a + bx)/ (c + dx)], then 2y1y3 = 3y22. 1 2

5. If x = 2 cos t – cos 2t and y = 2 sin t – sin 2t, find the value of d2y/dx2 when t  . 6. If x = a sin 2  (1 + cos 2 ), y = a cos 2  (1 – cos 2 ), prove that [1  (dy / dx 2 ]3/ 2  4a cos 3 . d 2 y / dx 2

7. If y = (1/x)x, show that y2 (1) = 0. 2 8. If y = x log [(ax)–1 + a–1], prove that x ( x  1) d 2y  x dy  y – 1.

dx

dx

2

9. If x = sin t, y = sin pt, prove that (1 – x 2 ) 10. If y3 + 3ax2 + x3 = 0, then prove that 11. If ax2 + 2hxy + by2 = 1, show that

d 2 y 2a 2 x 2  0 dx 2 y5

d2y h 2 – ab  2 dx (hx  by )3

12. If y = (tan–1 x)2, prove that ( x 2  1)2 

d y dy –x  p 2 y  0. 2 dx dx

d2y dy –20  2 x ( x 2  1) dx dx 2 dy   

3

13. If ay = sin (x + y), prove that y 1      0 dx 

14. If y = e–x (A cos x + B sin x), prove that 15. If y = Aepx + Beqx, show that

d2y dy 2  2y  0 dx dx 2

d2y dy – ( p  q)  pqy  0 dx dx 2

16. Change the independent variable to in the equation d2y dy  cot x  4 y cosec2 x  0 dx dx 2 x by means of the transformation z  log tan . 2

17. If x = log , y = 2 – 1. Find

d2y dx 2

172

DIFFERENTIAL CALCULUS

18. If y3 – y = 2x, then prove that

d2y 24 y – . dx 2 (3 y 2 – 1)3

19. If 2x = y1/5 + y–1/5 then express y as an explicit function of x and prove that ( x 2 – 1)

20.

d2y dy x  25 y. dx dx 2

x 3 sin x cos x 0 If f ( x)  6 – 1 a2

a

a3

Where a is constant, then prove that

d3 [ f ( x)] at x = 0 is equal to 0. dx3

ANSWERS 5. – 3/2

16.

d2y  4y  0 dz 2

17. 4 2.

5.2. Calculation of the nth Derivative. Some standard results y = (ax + b)m y1 = ma (ax + b)m – 1 y2 = m (m – 1) a2 (ax + b)m – 2 y3 = m (m – 1) (m – 2) a3 (ax + b)m – 3,

5.2.1. Let

so that is general yn = m (m – 1) (m – 2) ....... (m – n + 1) an (ax + b)m – n In case, m is a positive integer, yn can be written as m! a n (ax  b) m – n , (m – n) !

so that, the mth derivative of (ax + b) m is a constant, viz., m ! a m and the (m + 1)th derivative is zero. Cor.1. Putting m = – 1, we get yn = (– 1) (– 2) ....... (– n) an (ax + b) – 1– n]



 1  dn   n n  ax + b  = (– 1) (n !) a . dxn (ax + b)n+1

Cor.2. Let

y = log (ax + b) We have a (– 1) n –1 (n – 1) ! n y1   yn  a ax  b (ax  b) n

5.2.2. Let

y = amx.

y1 =

mamx

log a 

We have y2

= m2 amx (log a)2 ...  yn = mn amx (log a)n. (Bhopal 2000, 2001 ; Jabalpur1994)

Cor. Putting e for a, we get dn (emx ) dx

n

= m n emx .

SUCCESSIVE DIFFERENTIATION

173

y = sin (ax + b). We have

5.2.3. Let

(Banglore 2004)

1   y1 = a cos (ax + b) = a sin  ax  b     2  1  2    y2  a 2 cos  ax  b     a 2 sin  ax  b     2   2  2  3    y3  a 3 cos  ax  b     a 3 sin  ax  b     2   2 

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

. Hence dn sin (ax + b) n

dx 5.2.4. Similarly :

n π  = an sin  ax + b + .  2 

dn cos (ax + b)

n π  = an cos  ax + b + .  2  dx 5.2.5. Let y = eax sin (bx + c). We have n

y1 =

aeax

sin (bx + c) +

eax

(Jiwaji 2000, Gujrat 2005)

b cos (bx + c)

eax

= [a sin (bx + c) + b cos (bx + c)]. In order to put y in a form which will enable us to make the required generalisation, we determine two constants r and  such that a = r cos , b = r sin  r  (a 2  b 2 ),



 = tan–1 (b/a).

Hence, we have y1 = reax sin (bx + c + ). Thus, y1 arises on multiplying y by the constant r and increasing bx + c by the constant . Similary y2 = r2 eax sin (bx + c + 2 ). Hence, in general dn [eax sin (bx + c)] dxn

= r n eax sin (bx + c n )

r = (a 2 + b 2 ),  = tan –1 (b/a).

where 5.26 Similarly :

d n [eax cos (bx + c)] dxn

b  = (a 2 + b 2 )n/2 eax cos  bx + c + n tan –1  . a 

(Kanpur 2003)

5.3. Determination of nth derivative of rational functions Partial Fractions. In order to determine the nth derivative of a rational function, we have to decompose it into partial fractions. Sometimes it also becomes necessary to apply Demoivre’s theoerm, which states that (cos  ± i sin )n = cos n  ± i sin n , where n is any integer, positive or negative and i  (–1).

174

DIFFERENTIAL CALCULUS

EXAMPLES

x2 . ( x  2) (2 x  3) Throwing x2/ (x + 2) (2x + 3) into partial fractions, we obtain

1. Find the nth derivative of y 

(MDU Rohtak 2000)

x2 1  ( x  2) (2 x  3) 2 

8 9   1 – x  2  2 x  3    n  2 n  d x (– 1) ! 8 (– 1) n n !2 n . 9  –    dx n  ( x  2) (2 x  3)  2 ( x  2) n 1 2 (2 x  3) n 1

 9 . 2n  8 – .  n 1 n 1  ( x  2)   (2 x  3) 2. Find the nth derivative of y = x/(x2 + a2). 

x

We have

2

2

x a

(– 1) n n ! 2

x 1  1 1     . ( x  ai ) ( x – ai ) 2  x – ai x  ai 



n dn  x 1 1  (–1) n !     .     n 2 2 n  1 n  1 2 dx  x  a  ( x  ai )   ( x – ai ) To render the result free from ‘i’ and express the same in real form, we determine two numbers r and , such that



x = r cos , a = r sin , r  ( x 2  a 2 ),

so that

 = tan–1 (a/x). 1



( x – ai )

n 1



1 r

n 1



1

and

( x  ai )

n 1

 

Hence. where

. (cos  – i sin ) –( n 1) 1 r n 1 1

r

n 1

r

n 1

1

d n [ x /( x 2  a 2 )] dx n



[cos ( n  1)   i sin ( n  1) ]

(cos   i sin ) –( n 1) {cos ( n  1)  – i sin ( n  1) }.

(– 1) n n ! cos ( n  1)  r n 1

r  ( x 2  a 2 ),

3. Show that the nth derivative of y 

 = tan–1 (a/x). 1 1  x  x 2  x3

is

1 (– 1) n n ! sin n 1   sin ( n  1)  – cos ( n  1)   ( sin   cos ) – n –1  2 where  = cot–1 x

SUCCESSIVE DIFFERENTIATION

(1  x ) (1  x 2 ) 1  (1  x ) ( x  i ) ( x – i )

 dn y n

Put x = r cos , so that 1

(1  x ) 1

n 1

( x  i)

n 1



 

1 ( x – i)

n 1

 

dx n



 i 



x 2  1 and  = cot–1 x

r

and

(i – 1) 1 (i  1) 1  1 1  x  2 . x  i – 2 x– 

1 2

(i  1) 1 1 1 (i – 1) 1   – (– 1) n n !   . n 1  n  1 n  1 2 2 2 ( x – i)  dx ( x  i)  (1  x) 1 = r sin 



dny

1

y

We have



175

1 r

n 1

r

n 1

1 1 r

n 1

1 r

n 1

1 r

n 1

(cos   sin ) – n –1 ,

(cos   i sin ) –( n 1) (cos ( n  1)  – i sin ( n  1) )

(cos  – i sin ) –( n 1) (cos ( n  1)   i sin ( n  1) )

1 (– 1) n n ! 1 . (cos   sin ) – n –1  {(i – 1) (cos ( n  1)  n 1 2 2 r

– i sin (n  1)  – (i  1) (cos (n  1)   i sin (n  1) )} n



Now

1 (–1) n ! (cos   sin ) – n –1  sin ( n  1)  – cos ( n  1)   2 rn  1 rn + 1 = (x2 + 1) (n + 1)/2 = (cot2  + 1) (n + 1)/2 = cosecn + 1  1

 

r

dny dx

n



n 1

 sin n  1 

1 (– 1)n n ! sin n  1  (cos   sin ) – n –1  sin (n  1)  – cos (n  1)  2

4. If y  x log

x –1 , prove that x 1 dn y

xn   x–n  (– 1)n (n – 2) !  – n dx ( x  1) n   ( x  1) x –1 y  x log x 1 n

We have 

dy x –1 2  x  1  log x   2 dx x 1  x – 1  ( x  1)

176

DIFFERENTIAL CALCULUS

d2y



dx 2

 log

x –1 2x  x  1 ( x – 1) ( x  1)

 log

x –1 1 1   x 1 x –1 x 1

 

2 x2 – 1

–

1 ( x – 1) 2

–

1 ( x  1) 2

1 1 1 1 – – – x – 1 x  1 ( x – 1) 2 ( x  1) 2

Differentiating (n – 2) times w.r.t. x, we get dn y dx

n

( n – 2) ! ( n – 1) ! ( n – 1) !   ( n – 2) !  (– 1) n –2  – – – n –1 n –1 ( x  1) ( x – 1) n ( x  1) n   ( x – 1) ( n – 2) !  ( n  2) !  {( x – 1) – ( n – 1)} – {( x  1)  ( n – 1)}  (– 1) n –2  n n ( x  1)  ( x  1)  xn   x–n  (– 1) n – 2 ( n – 2) !  – n ( x – 1) ( x  1) n   xn   x–n  (– 1) n ( n – 2) !  – n ( x  1) n   ( x – 1)

[ (– 1)n – 2 = (– 1)n . (– 1)–2 = (– 1)n . 1 = (– 1)n] EXERCISES 1. Find the nth derivatives, given that (i)

y

x4 ( x – 1) ( x – 2)

(ii)

y

x 1 x2  4

(iii)

y

4x ( x – 1) 2 ( x  1)

(iv)

y

1 (a 2 – x 2 )

(Manipur 2000) (Sagar 97)

2. Find the tenth and the nth derivatives of y

x2  4 x  1 . x  2x2 – x – 2 3

3. Find the nth derivatives of the functions with the following function values : (i) (iii)

x (MDU Rohtak 1999) (ii) 4 1  3x  2 x 2 x 1 (iv) x 2 x2  x  1

1 – a4 x  x 1

4. Prove that the value of the nth derivative of y = x3 / (x2 – 1) for x = 0 is zero, if n is even, and – (n !) if n is odd and greater than 1. 5. Show that the nth derivatives of y = tan–1 x is 1  1  (– 1)n –1 (n – 1) ! sin n   – y  sin n   – y  . 2  2 

(Gorakhpur 1999)

SUCCESSIVE DIFFERENTIATION

177

6. Find the nth derivative of (i)

y  tan –1

1 x 1 – x (Sagar 2001)

–1 (ii) y  tan 1 – x cos  .

x sin 

(iv)

y  sin –1

2x 1  x2

(v) y  tan –1

2x

–1 (iii) y  tan 1 – x 2

1  x2 – 1 x

7. If y = x (x + 1) log (x + 1)3, prove that d n y 3 (– 1)n –1 (n – 3) ! (2 x  n) provided that n  3.  dx n ( x  1)n –1

ANSWERS 

16



1

n 1. (i) (– 1) n !  ( x – 2)n  1 – ( x – 1)n  1  , n  2.  

(– 1)n n !  3 1   ( x – 2)n  1 – ( x  2) n  1  . 4  

(ii)



1

2 (n  1) 

1

n (iii) (– 1) n !  ( x – 1) n  1 – ( x  1)n  1  ( x – 1)n  2  .  



1

1

1



n (iv) 2a (– 1) n !  ( x  a)n  1 – ( x – a)n  1   



1

1

1



–  2. 10 !  11 ( x  1)11 ( x  2)11   ( x – 1) 1 1 1   (– 1)n n !  – – . n 1 n 1 n 1 ( x  1) ( x  2)  ( x – 1)  

3.

 1 2n – . n 1 n 1  x  x  ( 1) (2 1)  

n (i) (– 1) n ! 

(ii) (iii)

(– 1) n n !  1 1 2 sin [( n  1) cot –1 ( x / a )  – –  . 3 n 1 n 1 4a ( x  a) ( a 2  x 2 ) ( n 1) / 2  ( x – a) 

2 (– 1)n n ! sin (n  1) , where r  ( x 2  x  1),   cot –1 [(2 x  1) / 3]. 3 rn  1

6. (i) (– 1)n – 1 (n – 1) ! sinn  sin n  where  = cot–1 x. (ii) (– 1)n – 1 (n – 1) ! cosecn  sinn  sin n , where = cot–1 [(x – cos ) / sin x]. (iii) 2 (– 1)n – 1 (n – 1) ! sinn  sin n  where  = cot–1 x. (iv) Same as (i) (v)

1 (– 1) n –1 ( n – 1) ! sin n  sin n, where  = cot–1 x. 2

5.4. The nth derivatives of the products of the powers of sines and cosines. In order to find out the nth derivative of such a product, we express it as the sum of the sines and cosines of multiples of the independent variable Example 1 will illustrate the process.

178

DIFFERENTIAL CALCULUS

EXAMPLES 1. Find dny/ dxn given that (ii) y = eax cos2 x sin x

(i) y = cos4 x (i) We have

cos 2 x 

1  cos 2 x 2

 1  cos 2 x  cos 4 x    2  



2

1 cos 2 x cos 2 2 x   4 2 4 1 1 1 3 1 1   cos 2 x  (1  cos 4 x )   cos 2 x  cos 4 x 4 2 8 8 2 8



d n (cos 4 x)



dx n



1 n n  1 n n    2 cos  2 x    4 cos  4 x  . 2  2  8  2 

1 (1  cos 2 x ) sin x 2 1 1  sin x  . 2 sin x cos 2 x 2 4 1 1  sin x  (sin 3 x – sin x) 2 4 1 1  sin x  sin 3 x. 4 4

cos2 x sin x 

(ii)

Hence dn dx

n

1 dn 1 dn (e ax sin x)  (e ax sin 3 x) n 4 dx 4 dx n 1 1   ( a 2  1) n / 2 e ax sin  x  n tan –1  4 a 

(eax cos 2 x sin x) 



1 2 3  ( a  9) n / 2 e ax sin  3 x  n tan –1  . 4 a 

2. If y = sin ax + cos ax, prove that yn  a n

{1  (– 1) n sin 2ax}

(Awadh 2001, Kanpur 2001, Reewa 1996, 2000 ; Sagar 99) Differentiating n times, we get n  n    n yn  a n sin  ax    a cos  ax    2   2   an

 n  n       cos  ax   sin  ax  2  2    

 an

{1  sin (2ax  n)}

 an

1  sin 2ax cos n)

 an

(1  (– 1)n sin 2ax)

2

SUCCESSIVE DIFFERENTIATION

179

EXERCISES Find the nth derivatives of the functions with the following function values : 1. sin3 x

2. cos x cos 2x cos 3x

(Calicut 2004)

3. sin 5x sin 3x

4. sin2 x cos3 x

5. eax cos2 bx

6. ex sin4 x

7. ex cos x cos 2x

8. e2x cos x sin2 2x ANSWERS n

1.

3 1 1   3   sin  x  n  –   sin  3 x  n  . 4 2 2   4  

2.

1 4

 n 1 1 1   n n  2 cos (2 x  2 n)  4 cos (4 x  2 n)  6 cos  6 x  2 n   .  

3.

1 2

 n n  n     n  2 cos  2 x  2  – 8 cos  8 x  2    

4.

1 16

5.

e ax  n  2b   a  ( a 2  4b 2 ) n / 2 e ax cos (2bx  n tan –1    .  2   2 

6.

1 x  e 3 – 4n / 2 cos (2 x  n tan –1 2) – 17 n / 2 cos  4 x  n tan –1 4  8

7.

ex  n    (10) n / 2 cos (3 x  n tan –1 3)  (2) n / 2 cos cos  x   2   4  

8.

1 2n  n / 2 e  2.5 cos ( x  n tan –1 / 2) – 13n / 2 cos (3 x  n tan –1 4

 1 1 1       n n  2 cos  x  2 n  – 3 cos  3 x  2 n  – 5 cos  5 x  2 n   .  

3

2)

– 29 n / 2 cos  5 x  n tan –1

5

2



5.5. Leibnitz’s Theorem. The nth derivative of the product of two functions. If u and v be two functions of x possessing derivatives of the nth order, then (uv)n = un + nC1 un – 1 v1 + nC2 un – 2 v2 + ............ + ............ + nCr un–r vr + ......... + nCn uvn. (Reewa 97, 99S; Bhopal 99, 2000; Vikram 99, 2001; Bilaspur 98, 2001; Bhuj 99 ; Jabalpur 99; Indore 2001, Avadh 98, 2002, 2005; Devi Ahilya 2001; Manipur 2000, 2002, Patna 2002, Gujrat 2005, GNOU 2004) This theorem will be proved by Mathematical induction. Step I. By direct differentiation, we have (uv)1 = u1 v + uv1, and

(uv)2 = u2v + u1v1 + u1v1 + uv2 = u2v + 2C1u1v1 + 2C2uv2.

Thus the theorem is true for n = 1, 2. Step II. We assume that the theorem is true for a particular value of n, say m, so that we have (uv)m = um v + mC1 um – 1 v1 + mC2 um – 2 v2 + ......... + mCr – 1 um – r + 1 vr – 1 + mCr um – r vr + ........... + mCm uvm.

180

DIFFERENTIAL CALCULUS

Differentiating both sides, we get (uv)m + 1 = um + 1 v + um v1 + mC1 um v1 + mC1 um – 1 v2 + mC2 um – 1 v2 + mC2 um – 2 v3 + ... + mCr – 1 um – r + 2 vr – 1 + mCr – 1 um – r + 1 vr + mCr um – r + 1 vr +

mC r

um – r vr + 1 + ..... +....... + mCm uvm + 1

= um + 1 v + (1 + mC1) um v1 + (mC1 + mC2) um – 1 v2 + ..... + (mCr – 1 + mCr) um – r + 1 vr + ......... + mCm uvm + 1. We know that mC r–1

+ mCr = m + 1Cr,

1 + mC1 = 1 + m = m + 1C1, mC

m

= 1 = m + 1Cm + 1.

 (uv)m + 1 = um + 1 v + m + 1C1 um v1 + m + 1C2 um – 1 v2 + ........ + m + 1Cr um – r + 1 vr + .... + m + 1Cm + 1 uvm + 1, from which we see that if the theorem is true for any value m of n, then it is also true for the next higher value m + 1 or n. Conclusion. In step I, we have seen that the theorem is true for n = 2. Therefore it must be true for n = 2 + 1, i.e., 3 and so for n = 3 + 1, i.e., 4, and so for every value of n. EXAMPLES 1. Find (x2 ex cos x)n To find the nth derivative, we look upon ex cos x as the first factor and x2 as the second.  (x2 ex cos x)n = (ex cos x)n x2 + nC1 (ex cos x)n – 1. 2x + nC2 (ex cos x)n – 2 . 2 = 2n/2 . ex cos (x + n tan–1 1) . x2 + n 2 (n – 1)/2 ex cos [x + (n – 1). tan–1 1] 2x n (n – 1) ( n –2) / 2 x .2 e cos [ x  ( n – 2) tan –1 1] 2. 2     e x  2 x 2 cos  x  n     4        23/ 2 . nx . cos  x  n – 1   n ( n – 1) cos  x  n – 2   .  4  4  

 2( n –2) / 2

n  y x 2. If cos–1   = log   , prove that b n x2 yn + 2 + (2n + 1) xyn + 1 + 2n2 yn = 0

x  y cos –1    n log n b Differentiating both sides w.r.t. x, we get – y1 n  2 2 x b – y

We have



x2 y12 = n2 (b2 – y2)

Differentiating again w.r.t. x, we get 2x2y1 y2 + 2xy12 = – 2n2 yy1 

x2y2 + xy1 + n2y = 0

(Ravishankar 1999, Jabalpur 2000)

SUCCESSIVE DIFFERENTIATION

181

Applying Leibnitz’s theorem, we get x2yn + 2 + nC1. yn + 1 . 2x + nC2 . yn. 2 + xyn + 1 + nC1 yn + n2 yn = 0 or

x2yn + 2 + (2n + 1) xyn + 1 + 2n2 yn = 0

3. If y1/m + y–1/m = 2x, prove that (x2 – 1) yn + 2 + (2n + 1) xyn + 1 (n2 – m2) yn = 0 (Indore 98 ; Reewa 98 ; Vikram 99 ; Bhopal 2000, 2001; Avadh 99; Rohilkhand 2000; Kanpur 99) y1/m + y–1/m = 2x

We have

y2/m – 2xy1/m + 1 = 0



y1/ m 



4 x2 – 4 2

2x 

x

x2 – 1 x 2 – 1) m



y  (x 



y1  m ( x 

 x 2 – 1) m – 1  1   

  x –1  x

2

Squaring both sides, we get (x2 – 1) y12 = m2 y2 Differentiating again w.r.t. x, we get 2 (x2 – 1) y1 y2 + 2xy12 = 2m2 yy1 (x2 – 1) y2 + xy1 – m2 y = 0



Applying Leibnitz’s Theorem, we get (x2 – 1) yn + 2 + nC1 yn + 1 . 2x + nC2 yn . 2 + xyn + 1 + nC1 yn – m2 yn = 0  (x2 – 1) yn + 2 + (2n + 1) xyn + 1 (n2 – m2) yn = 0 log x , prove that x (–1) n n !  1 1 1 yn  log x – 1 – – – ..... –  (Jabalpur 1995, Jiwaji 98)  n 1 2 3 n  x 1 y  log x We have x Applyling Leibnitz’s Theorem, we get

4. If y 

yn 

(–1) n n ! xn  1

log x  n C1

 n C2

(– 1) n – 1 ( n – 1) ! 1 x xn

(– 1) n – 2 ( n – 2) ! x

n –1

(– 1) .

1 x

2

 n C3

(– 1) n –3 ( n – 3) ! x

n–2

(– 1) (– 2) .

 ......  

(– 1) n n ! x n 1

log x 

n (– 1) n –1 ( n – 1) ! x n 1





1 x3

1 (– 1) n – 1 ( n – 1) ! x xn

n ( n – 1) (– 1) n –1 ( n – 2) ! 1.2 x n 1

n (n – 1) (n – 2) (– 1)n –1 (n – 3) (– 1)n –1 (n – 1) ! ·1 · 2  ......  1. 2 . 3 xn  1 xn  1

182

DIFFERENTIAL CALCULUS



(– 1)n n !  1 1 1 log x – 1 – – – ...... –   n 1 2 3 n  x

n 5. If I n  d ( x n log x ) , prove that In = nIn – 1 + (n – 1) n

(Sagar 1996, Calicut 2004)

dx

In  

dn –1  d  ( x n log x)  n – 1  dx   dx dn –1 dx n – 1

n

 n x n – 1 log x  x n – 1 

dn –1

( x n – 1 log x) 

dx n – 1 = n In – 1 + (n – 1) ! 6. If x + y = 1, then prove that dn dx n

dn –1 dx n – 1

( xn – 1 )

( x n y n )  n !  y n – ( n C1 ) 2 y n – 1 . x ( n C2 ) 2 y n – 2 x 2  ........(– 1) n x n 

(Jabalpur 1995 ; Jiwaji 98) x+y=1  y=1–x xn = u and yn = v, then

Let un = n !, Then

un – 1 = n ! x,

d n (xn yn ) dx n

un – 2 

n! 2 x ...... 2

 D n (uv)

= un v + nC1 un – 1 v1 + nC2 un – 2 v2 + ......... + nCn vn n! 2 = n ! yn + nC1 n ! x ( – nC1 yn – 1) + nC2  2 ! x  nC2 2 ! yn – 2 + ......... + xn (– 1)n n !   = n ! [ yn – (nC1)2 yn – 1 + (nC2)2 yn – 2 x2 + ....... + (– 1)n xn] 7. Prove that d n  sin  dx n  x

x 1 1 1   n 1 [ P sin ( x  n)  Q cos ( x  n)] 2 2  x

where

P = xn – n (n – 1) xn – 2 + n (n – 1) (n – 2) (n – 3) xn – 4

and

Q = n xn – 1 – n (n – 1) (n – 2) xn – 3 + ........ 1   Dn (sin x) = sin  x  n    2  1   Dn–1 (sin x) = sin  x  ( n – 1)    2  1   = – cos  x  n    2  1 1     Dn–2 (sin x) = sin  x  ( n – 2)    sin  x  n   etc.  2   2 

SUCCESSIVE DIFFERENTIATION

183

By Leibnitz Theorem n (n – 1) n – 2 1 1  sin x  1 n n –1 D (sin x) · D 2   Dn  (sin x ). D      D (sin x ) ·  n D 2! x x  x  x 

n (n – 1) (n – 2) n – 3 1 D (sin x ) · D3    ... sin 3! x



n ( n – 1) 2!

 n    2  n (n – 1) ( n – 2)  n    – 6      3      – sin  x   cos  x   2 3 ! 2   x4     x      



1 x



n 1

1 n 1

n  1  n     x   · – n  – cos  x    2  x  2  

n ( n – 1) ( n – 2) ( n – 3)  n     24        .... sin  x  4!  2    x5  

1   [ {xn – n (n – 1) xn – 2 + n (n – 1) (n – 2) (n – 3) xn – 4 + ....} sin  x  n   2  1   + { n xn – 1 n (n – 1) (n – 2) xn – 3 + ...... } cos  x  n    2   n  n        P sin  x  2   Q cos  x  2    

x 8. By finding in two different ways the nth derivative of x2n, prove that 1

n2



n 2 (n – 1) 2



n 2 ( n – 1) 2 ( n – 2) 2

 .... 

(2 n) !

(Rohilkhand 1995) 1 .2 1 .2 .3 ( n !) 2 Dn (x2n) = 2n (2n – 1) (2n – 2) ....... (2n – n + 1) (x)2n – n = 2n (2n – 1) (2n – 2) ..... (n + 1) xn (2n) ! n  x ...(i) n! Also, by Leibnitz’s Theorem, we get Dn (x2n) = Dn (xn . xn) = Dn (xn) . xn + nC1 Dn – 1 (xn) . D(xn) + nC2 Dn – 2 (xn).D2(xn) + nC3 Dn – 3 (xn). D3 (xn) + ....... 2

1

2

2

2

2

2

 n ! x2  n n –2  n (n – 1) x = (n ! ) xn + nC1 (n ! x) nxn – 1  C2   2!  n ! x3  n C3 { n (n – 1) (n – 2) x n –3 }  .... 3!   n2 x 2 ( n – 1) 2 x 2 ( n – 1) 2 ( n – 2) 2  n ! x n 1  2    ... ...(ii) 2 2 2 2 2 1 1 –2 1 .2 .3   Equating the values of Dn (x2n) as given by (i) and (ii) we get   (2n)! x n n 2 n 2 (n – 1) 2 n 2 ( n – 1) 2 ( n – 2) 2 x n n ! 1  2    ...  n! 1 12 . 22 12 . 22 . 32   (2n)! n 2 n 2 (n – 1) 2 n 2 (n – 1) 2 ( n – 2) 2  or 1     ... ( n !) 2 12 12 . 2 2 12 . 22 . 32

184

DIFFERENTIAL CALCULUS

9. Find the value of the nth derivative of y = e m sin 1 x for x = 0. (Gorakhpur 2001, Vikram 97; Jabalpur 97) y = e m sin 1 x

Let

...(i) y1  e m sin

so that

1

x

.

m (1 – x 2 )

...(ii)

(1 – x2) y12 = m2y2



Differentiating, we get (1 – x2) 2y1 y2 – 2xy12 = 2m2 yy1 Dividing by 2y1, we obtain (1 – x2) y2 – xy1 = m2 y

...(iii)

Differentiating n times by Leibnitz’s theorem, we get (1 – x2) yn + 2 – 2nxyn + 1 – n (n – 1) yn – xyn + 1 – nyn = m2yn  (1 – x2) yn + 2 – (2n + 1) xyn + 1 – (n2 + m2) yn = 0 Putting x = 0, we get yn + 2 (0) = (n2 + m2) yn (0)

...(iv)

From (i), (ii) and (iii), we obtain y (0) = 1,

y1 (0) = m,

y2 (0) = m2.

Putting n = 1, 2, 3, 4 etc. in (iv) we get y3 (0) = (12 + m2) y1 (0) = m (12 + m2) ; y4 (0) = (22 + m2) y2 (0) = m2 (22 + m2) ; y5 (0) = (32 + m2) y3 (0) = m (12 + m2) (32 + m2) ; y6 (0) = (42 + m2) y4 (0) = m2 (22 + m2) (42 + m2). In general m 2 (2 2  m 2 ) (4 2  m 2 ).........[( n – 2) 2  m 2 ], when n is even, yn (0)   2 2 2 2 2 2  m (1  m ) (3  m )..........[( n – 2)  m ], when n is odd.

10. Find yn (0) when y = log ( x  1  x 2 ) We have

y  log ( x  1  x 2 )



y1  

...(i)

  x  .1  x  1  x 2  1  x 2  1 1

1  x2

 (1 + x2)y12 = 1 Differentiating again w.r.t. x, we get 2y1 y2 (1 + x2) + 2xy12 = 0  (1 + x2) y2 + xy1 = 0 Applying Leibnitz’s Theorem, we get (1 + x2) yn + 2 + nC1yn + 1 . 2x + nC2 yn . 2 + xyn + 1 + nC1 yn = 0

...(ii)

...(iii)

SUCCESSIVE DIFFERENTIATION

185

 (1 + x2) yn + 2 + (2n + 1) xyn + 1 + n2 yn = 0 Putting x = 0, we get yn + 2 (0) = – n2 yn (0) ...(iv) From (i), (ii) and (iii), we get y = (0) = 1, y1 (0) = 1, y2 (0) = 0 Putting n = 1, 2, 3, 4 etc. in (iv), we get y3 (0) = – 12 y1 (0) = – 12. y4 (0) = – 22 y2 (0) = 0 y5 (0) = – 32 y3 (0) = (– 1)2 12. 32 y6 (0) = – 42 y4 (0) = 0 In general yn (0) = 0 if n is even = (– 1)(n – 1)/2 12. 32. 52 ...(n – 2)2, if n is odd. 11. If y = cos (m sin–1 x) show that (1 – x2) yn + 2 – (2n + 1) xyn + 1 + (m2 – n2) yn = 0 and hence find yn (0). (Kumaon 2002; Gorakhpur 99, 2000 Calicut 2004) –1 We have y = cos (m sin x) ...(i) m –1 y1  – sin ( m sin x )  ...(ii) (1 – x 2 )  (1 – x2) y1 2 = m2 sin2 (m sin–1 x) = m2 (1 – y2) Differentiating we get (1 – x2) 2y1y2 – 2xy1 2 = – 2m2 yy1  (1 – x2) y2 – xy1 + m2 y = 0 ...(iii) Applying Leibnitz’s Theorem, we get (1 – x2) yn + 2 + nC1 . – 2x . yn + 1 + nC2 . – 2 . yn – xyn + 1 – nC1 yn + m2 yn = 0  (1 – x2) yn + 2 – (2n + 1) xy n + 1 + (m2 – n2) yn = 0 Putting x = 0, we get yn  2 (0)  ( n 2  m 2 ) yn (0) From (i), (ii) and (iii) we have y (0) = 1, y1 (0) = 0, y2 (0) = m2 y (0) = m2

Putting n = 1, 2, 3, 4, etc. in (iv) we get y3 (0) = (12 – m2) y1 (0) = 0 y4 (0) = (22 – m2) y2 (0) = m2 (22 – m2) y5 (0) = (32 – m2) y3 (0) = 0 y6 (0) = (42 – m2) y4 (0) = m2 (22 – m2) (42 – m2) In general

yn (0) = 0, if n is odd = m2 (22 – m2) (42 – m2) ........ ((n – 2)2 – m2), if n is even.

...(iv)

186

DIFFERENTIAL CALCULUS

EXERCISES ny/dxn

1. Find d for the following functions : n x (i) x e , (ii) x3 cos x, ax 2 2 (iii) e [a x – 2nax + n (n + 1)],(iv) ex log x. 2. If y = x2 sin x, prove that

(Sagar 93, Gujrat 2005)

dny

n  n     ( x 2 – n 2  n)sin  x   – nx cos  x  . 2 2     dx n

3. If f (x) = x2 tan x, prove that f n (0) – nC2 f n – 2 (0)  nC4 f n – 4 (0) – .............  sin

n . 2

[Write f (x) cos x = sin x and apply Leibnitz theorem.] 4. Differentiate the following equations : (i) (ii)

d2y

dy –x  a 2 y  0. dx dx 2 d2y dy x2 2 – x  (a 2 – m 2 ) y  0. dx dx

(1 – x 2 )

n times with respect to x. 5. If y = (sin–1 x)2, prove that (1 – x 2 )

d2y dx

–x

2

dy – 2  0. dx

Differentiate the above equation n times with respect to x and show that (1 – x2) yn + 2 – (2n + 1) xyn + 1 – n2 yn = 0. (Bilaspur 2000 ; Reewa 2000 ; Ravishankar 2000 ; Gorakhpur 2002, Calcutta 2005, Banglore 2005) 6. If u = tan–1 x, prove that (1 – x 2 )

d 2u dx 2

– 2x

du 0 dx

and hence determine that values of the derivatives of u when x = 0. (Avadh 2000) 7. If y = sin (m sin–1 x), show that (1 – x2) yn + 2 = (2n + 1) xyn + 1 (n2 – m2) yn and find yn (0). (Sagar 2000 ; Bilaspur 2001 ; Jabalpur 2001 ; Indore 99 ; Reewa 2000, 2001; Garhwal 2001; Kanpur 2002 ; Gorakhpur 2003 ; Avadh 2003; MDU Rohtak 99, Poorvanchal 2004) 8. Find yn (0), when y  [ x (1  x 2 ) ]m . (Kanpur 2006, Rohilkhand 2001, Vikram 2000 ; Indore 2001; Gorakhpur 2000, Devi Ahilya 2000 ) –1 9. If em cos x show that (1 – x2) yn + 2 – (2n + 1) xyn + 1 – (n2 + m2) yn = 0 and find yn (0). (Banglore 2004) 10. If y 

sin –1 x 1 – x2

, show that

(1 – x2)yn + 2 – (2n + 3) xyn + 1 – (n + 1)2 yn = 0 11. If y = a cos (log x) + b sin (log x), show that x2 yn + 2 + (2n + 1) xyn + 1 + (n2 + 1) yn = 0 (Ravishankar 1997, 99 ; Bhopal 98 ; Jabalpur 98 ; Indore 96, MDU Rohtak 2000)

SUCCESSIVE DIFFERENTIATION

187

12. If y = (sinh–1 x)2, prove that (1 + x2)yn + 2 + 2 xyn + 1 n2 yn = 0 13. If y = (x2 – 1)n, prove that (x2 – 1) yn + 2 + 2xyn + 1 – n (n + 1) yn = 0 (Kumaon 2003 ; Reewa 97; Avadh 96) (Hint. log y = n log (x2 – 1)  y1 (x2 – 1) = 2nxy) 14. Prove that the nth dirrerential coofficient of xn (1 – x)n is   n2 x n 2 (n – 1) 2 x2 n!(1 – x)n 1 – 2 .   2 2 2 1 1– x 1 .2 (1 – x)  .......  

15. If y  e tan

–1

x

prove that

(1 + x2) yn + 1 + (2nx – 1) yn + n (n – 1) y (n – 1) = 0 (Bilaspur 97; Indore 2000; MDU Rohtak 2001) 16. If y  e a sin

–1

x

= prove that

(1 – x2) yn + 2 – (2n + 1) x yn + 1 – (n2 + a2) yn = 0 (Avadh 2002 ; Ravishankar 2001 ; Jabalpur 97, 99 ; Indore 95 ; Bhopal 95 ; Vikram 97 ; Reewa 99 ; Kumaon 2001 ; Bilaspur 2001 ; Rohilkhand 2003 ; Garhwal 2001; Gujrat 2005; Gauhati 2005) 2

17. If y  log e {x  (1  x 2 )} find (yn)0 (Vikram 1996, Indore 2000, Avadh 99; Rohilkhand 2002, 2004) 2 2 18. If y  log  x  ( x  a )  , prove that (a2 + x2) y2 xy1 = 0. Differentiate this differential  

equation n times and prove that

lim

x 0

yn  2 yn

–

n2

(Rohilkhand 2002)

a2

ANSWERS

1.

 n n 2 n –1 n 2 ( n – 1) 2 n –2 n 2 ( n – 1) 2 ....12  x x  x  ....  (i) e  x  1! 2! n!  

 

1 2

 

 

1 2

 

 

1 2

 

3 2 (ii) x cos  x  n   3nx cos  x  (n – 1)    3n (n – 1) x cos  x  (n – 2)  

1    n (n – 1)(n – 2) cos  x  ( n – 3)   . 2  

(iii) an + 2 . eax . x2. (iv) ex [log x + nc1 x–1 – nc2x–2 + nc3 2 ! x–3 – ........ + (– 1)n – 1 ncn (n – 1) ! . x–n]. 4. (i) (1 – x2) yn + 2 – (2n + 1) xyn + 1 + (a2 – n2) yn = 0. (ii) x2yn + 2 (2n + 1) xyn + 1 + (n2 + a2 – m2) yn = 0. 6. Un(0) = 0, if n is even ; and (–1) (n – 1)/2 (n – 1) ! if n is odd. 7.

0, if n is even yn (0)   2 2 2 2 2 2  m(1 – m ) (3 – m )...( n – 2) – m ), if n is odd

8. y2n (0) = m2 (m2 – 22) (m2 – 42) ... [m2 – (2n – 2)2] ; y2n + 1 (0) = m (m2 – 12) ) (m2 – 32) (m2 – 52) ... [m2 – (2n – 1)2].

188

DIFFERENTIAL CALCULUS

9. y2n + 1 = –em/2 m (12 + m2) (32 + m2) ... ((2n – 1)2 + m2). y2n = em/2 m2 (22 + m2) (42 + m2) ... ((2n – 2)2 + m2). 17. When n is odd 0, when n is even (– 1)

n –2 a

 n – 2   2n –1   !  2  

2

MISCELLANEOUS EXERCISES 1. Show that if x (1 – x) y2 – (4 – 12x) y1 – 36y = 0, then x (1 – x) yn + 2 – [4 – n – (12 – 2n) x] yn + 1 – (4 – n) (9 – n) yn = 0. 2. If yn denotes the nth derivative of eax sin bx and  = tan–1 (b/a), prove that yn = (a cos )n eax sin (bx + n ). Also show that y y + 1 – 2ayn + (a2 +b2)yn – 1 = 0. 3. If y = (a + bx) cos kx + (c + dx) sin kx, prove that d4y dx

4

 2k 2

d2y dx 2

 k 4 y  0. x

–1 4. Find the third derivative of y  tan

d4

(1 – x 2 )

.

12 x 2 – 3

2 . 5. Prove that dx 4 (1  x )  (1  x 2 )7

6. Find the value of the nth derivative of y 

x2 – x ( x 2 – 4) 2

.

for x = 0. 7. If Un denotes the nth derivative of (Lx + M)/(x2 – 2Bx + C), prove that x 2 – 2 Bx  C 2( x – B ) Un  2  U n  1  U n  0. ( n  1) ( n  2) n 1

8. If y = x2 ex, then dny dx

n



1 d2y dy 1 n ( n – 1) – (n – 2)  ( n – 1)(n – 2). y. 2 dx dx 2

(Bhopal 97)

9. If y = (tan–1 x)2, then ( x 2  1)2

d2y dx

2

 2 x ( x 2  1)

dy  2. dx

(Manipur 2002)

Deduce that d n2

( x 2  1) 2

dx n  2

 (4n  2) x ( x 2  1)

d n 1 y dx n 1

 2n 2 (3 x 2  1)

 2n (n – 1)(2n – 1) x

10. If y  e x

2

/2

dny

dx n dn –1 y

dx n – 1

 n (n – 1) 2 (n – 2)

cos x, show that

y2n + 2 (0) – 4ny2n (0) + (2n – 1) 2ny2n – 2 (0) = 0. 11. If y = (1 + x2)m/2 sin (m tan–1 x), show that y2n (0) = 0 and y2n + 1 (0) = (– 1)n m (m – 1) (m – 2).......(m – 2n).

dn – 2 y dx n – 2

 0.

SUCCESSIVE DIFFERENTIATION

189

ANSWERS 4. (1 + 2x2)/(1 – x2)5/2 6. If n is even, yn (0) = 0 ; if n is odd, yn (0)  n !(3n – 5)

1 2

n4

.

OBJECTIVE QUESTIONS Note : For each of the following questions, four alternatives are given for the answer. Only one of them is correct. Choose the correct alternative. d2y 1. If x = t – sin t, y = 1 – cos t, value of at (, 2) will be : dx 2 (a) 0 (b) 1 (c)  (d)  n n 2. Value of D (ax + b) is (Avadh 2002) (a) n an (b) n! an (c) n a bn (d) n ! bn  1  3. Value of D n   is  ax  b  n! an (a) (b) (ax  b ) n

(– 1) n n ! a n ( ax  b) n

(– 1) n n ! a n

(c)

( ax  b) n  1

(b)

b n sin ( ax 

(d)

0

4. Value of Dn{sin (ax + b)} is (a) an sin (ax + b + n)

1 n) 2

1 n ) (d) bn sin (ax + b + n) (Manipur 2000) 2 5. Leibnitz’s theorem is used to find the nth differential coefficient of : (a) trigonometric functions only (b) exponential function only (c) sum and difference of two functions (d) product of two functions (Rohilkhand 2002) n 6. (r + 1) th term in the expression of D (uv) is : n (c) a sin (ax  b 

(a) (c)

n

Cr 1 Dn – r u Dr v

(b)

n

Cr Dn – r u Dr v

(d)

n

(b)

(1 – x2) y2 + xy1 – m2 y = 0

(d)

None of these

n 1

C2 D n – r u D r v

Cr –1 D n – r u D r v (Avadh 2005)

7. If y = sin (m sin–1 x), then (a) (1 – x2) y2 – xy1 + m2y = 0 (c) (1 – x2) y2 – xy1 – m2 y = 0 8. If y =

(tan–1

x)2,

then

(x2

(a) 2

(a)

+ 1)y1, =

(b) 2 –

m2

y–1/m

y

(c)

2y

(d) – 2y

( x 2  1)}]2 , then value of (1 + x2) y2 + 2y1 is equal to

(a) 1 10. If

y2 + 2x

(x2

(b) – 2

9. If y  [log{x  y1/m

+

1)2

= 2x, then

(x2

(c)

–1

(d)

None of these

(c)

 m2 y

(d) None of these

– 1) y2 + xy1 =

(b) –m2 y

11. If y = b cos {n log (x/n)}, then

190

DIFFERENTIAL CALCULUS

(a) x2 y2 – xy1 + x2 y = 0 (c)

x2

y2 – xy1 –

x2

y=0

(b)

x2 y2 + xy1 – x2 y = 0

(d)

x2 y2 + xy1 + x2 y = 0

(b)

(1 + x2) y'' + ky' – xy = 0

(d)

(x2 + 1) y'' + xy' – k2 y = 0

(b)

(– 1) n

(d)

None of these

(b) (d)

nr [1 + sin 2nx]1/2 None of these

k

2 12. Function y   x  x  1  satisfies :   2 2 (a) (x + 1) y' = k y

( x 2  1) y   ky

(c)

13. If y = 1/(a2 – x2), then yn = n (a) (– 1)

n! [( x  a ) – n – ( x – a ) – n ] 2a

n! [( x – a ) – n  ( x  a ) – n ] 2a 14. If u = sin nx + cos nx, then ur = (a) [1 + (–1)r sin 2nx]1/2 (c) nr[1 + (–1)r sin 2nx]1/2 15. Dn(sin x sin 3x) = n (c) (– 1)

(a)

n  n   1   cos  2 x   – cos  4 x    2  2  2   

(b)

n  n   1   cos  4 x   – cos  2 x    2  2  2   

(c)

n  n   1   cos  4 x    cos  2 x   2   2  2   

n! [( x – a ) – n – ( x  a ) – n ] 2a

(d) None of these 16. If y is a polynomial of degree n in x and first coefficient is 2, then D n – 1(y) = (a) 2 (n!) (b) 2 (n!) x (c) 2 (n – 1) ! x (d) None of these 2 x n (n – 1) 17. If y = x e and yn = C2 y2 + ay1 + C2 y, then a = (a) n (n – 1) (b) n (n – 2) (c) – n (n – 1) (d) – n (n – 2) n – 1 18. If y = x log x, then xyn = (a) n ! (b) (n – 1) ! (c) (n – 2) ! (d) None of these 19. If y = e tan

1

x

, then (1 + x2) yn + 2 + [2 (n + 1) x – 1] yn + 1 =

(a) n2 yn

(b) n (n + 1) yn

(c)

– n (n + 1)yn

(d) – n2yn

(c)

– (n2 + 1)yn

(d) None of these

20. If y = sin–1 x, then (1  x 2 ) y n  2  (2 n  1) xyn 1 

(a) n2 yn

(b) – n2 yn ANSWERS

1. (a) 8. (a) 15. (d)

2. (b) 9. (b) 16. (d)

3. (c) 10. (a) 17. (d)

4. (c) 11. (d) 18. (b)

5. (d) 12. (c) 19. (c)

6. (c) 13. (d) 20. (a)

7. (a) 14. (c)

CHAPTER 6 EXPANSION OF FUNCTIONS Introduction. Ordinary algebraic processes, taking limits, differentiation etc, though applicable to the sum of a finite number of terms, may break down for infinite series. In this chapter, we will expand functions in terms of infinite series by using the methods of differential calculus. The infinite series found below are to be regarded as formal expansions, which may not be true in exceptional cases. 6.1. Maclaurin’s Theorem. (Sagar 1997; Bilaspur 97, 2001; Jabalpur 2000 ; Ravishankar 2000 ; Avadh 2003, 2005; Gujrat 2005) Let f (x) be a funtion of x. Let this function is to be expanded in ascending powers of x and let the expansion be differentiable term by term any number of times. Let f (x) = A0 + A1 x + A2 x3 + A3 x3 + A4 x4 + ... Now by successive differentiation, we get f ' (x) = A1 + 2 A2 x + 3 A3 x2 + 4 A4 x3 + ... f '' (x) = 2·1 A2 + 3·2 A3 x + 4·3 A4 x2 + ... f ''' (x) = 3 · 2 · 1 A3 + 4 · 3 · 2 A4 x + ... = ... n f (x) = n (n – 1) (n – 2) ... 2·1 An + (n + 1) n (n – 1) ... 3·2 An + 1 · x + ... Putting x = 0 in each term, we get f (0) = A0, f' (0) = A1, f'' (0) = 2 ! A2 f''' (0) = 3 ! A3, ... fn (0) = n ! An ... with the help of these values of A0, A1, A2, A3, .... we have

f ( x ) = f (0) + x f  (0) +

x2 xn n f  (0) + ... + f (0) 2! n!

This result is konwn as Maclaurin’s Theorem. EXAMPLES 1. Expand cos x by Maclaurin’s series. We have

(Kanpur 2006; Srivenkateshwara 2005)

  y  cos x, y1  – sin x  cos   x  2  2     y2  – sin   x   cos  x   2   2  191

192

DIFFERENTIAL CALCULUS

n  yn  cos  x    2  Now put x = 0; then

then

(y)0 = 1, (y1)0 = 0, (y2)0 = – 1  3  ( y3 ) 0  cos  0  2 

n 2 n = 4; then (y4)0 = cos 2  = 1

In general ( yn )0  cos Now, put

n  0 , if n is odd 2 = (– 1)n/2 if n is even.

Hence

( yn )0  cos

Then

f ( x)  f (0)  x f  (0)  cos x  1 –

x2 f  (0)  ... 2!

x2 x4 x6 xn  –  ...  (– 1) n / 2  ... 2! 4! 6! n!

when n is even. 2. Obtain the expansion of log cosh x in powers of x by Maclaurin’s Theorem. We have

y = log cosh x y1 = tanh x y2 = sech2 x = 1 – tanh2 x = 1 – y21 y3 = – 2 y1 y2 y4 = – 2 (y1 y3 + y22), y5 = – 2 (y1 y4 + y2y3 + 2 y2y3) y6 = – 2 (y1 y5 + 4 y2 y4 + 3 y23)

Now putting

x = 0, we obtain (y)0 = 0, (y1)0 = 0, (y2)0 = 1, (y3)0 = 0, (y4)0 = – 2, (y5)0 = 0 and (y6)0 = 16 etc.

Substitute the values in the Maclaurin’s series

y  ( y )0  x ( y1 )0 

x2 ( y2 )0  ... 2!

x2 x4 x6 –  – ... 2 12 45 3. Expand log (1 + x) by Maclaurin’s Theorem.

we obtain

y  log cosh x 

We have

y = log (1 + x) y1 

1 1 x



(y)0 = 0



(y1)0 = 1

(Sagar 1995, 2000 ; Indore 99)

EXPANSION OF FUNCTIONS

1

y2  –

y3 

193

(1  x ) 2



(y2)0 = – 1



(y3)0 = 2



(y4)0 = – 6

2 (1  x )3

y4  –

6 (1  x ) 4

..................................................................... ..................................................................... yn 



(– 1) n –1 ( n – 1) ! (1  x) n

log (1  x)  x –

(yn)0 = (– 1)n – 1 (n – 1) !



x2 x3 x4 (– 1)n –1 (n – 1) ! xn  –  ...  2 3 4 n!

x2 x3 x4 (– 1) n –1 x n  –  ...  ... 2 3 4 n 4. Use Maclaurin’s Theorem to find the expansion in ascending powers of x of log (1 + ex) to the terms containing x4. (Garhwal 2002; Vikram 97 ; Sagar 98) x–

We have

y = log (1 + ex),

i.e.,

ey = 1 + ex,

...(i)

ey · y1 = ex ey ey

· y2 +

ey

· y3 +

ey

...(ii)

· y1 =

ex

y1 y2 +

ey

2

...(iii) . 2 y1 y 2 =

ex

y1 y3 + y4 + 3 (y2 + y1 y3) + 3 Now put x = 0, in (i), (y)0 = log 2

ey

ey

in (ii), in (iii), or in (iv), or

ey

y1 +

ey

( y1 )0 

1 e

2

log 2

ey

2



y1 y2 +

ey

· y1 +

ey

1 , 4

1 1 1 1 1 1 2 . . ( y3 ) 0  2 · · ·  2 ·  1 · 2 · ·  1 2 2 4 4 2 4 (y3)0 = 0 ( y4 )0  –

log (1  e x )  log 2 

1 1 1 4 x  x2 – x  ... 2 8 192

2

· 3 y1 y2 =

1 1 4

1 6 Hence by Maclaurin’s Theorem,

in (v),

4

1 2

2 ( y2 ) 0  2 · ( y2 ) 0 

...(iv) 2

ex

...(v)

194

DIFFERENTIAL CALCULUS

5. Expand e a sin

We have

–1

in powers of x by Maclaurin’s Theorem and hence obtain the value of e. (Kumaon 2003; Kuvempu, 2005)

x

y  e a sin

–1

x

, y1  e a sin

–1

x

·

a (1 – x 2 )

(1 – x 2 ) y12  a 2 y 2

or

Again differentiating 2 y1 y2 (1 – x2) – 2 x y12 = 2 a2 y y1 y2 (1 – x2) – x y1 = a2 y

or

Differentiating n times more, we have yn + 2 – (2 n + 1) yn + 1 · x – (n2 + a2) yn = 0 Put x = 0, then (yn + 2)0 = (n2 + a2) (yn)0 (y)0 = 1, (y1)0 = a, (y2)0 = a2

Now,

(y3)0 = (12 + a2) (y1)0 = (12 + a2) · a (y4)0 = (22 + a2) (y2)0 = (22 + a2) · a2 Then by Maclaurin’s Theorem

e a sin

–1

x

1 a ·x 

a 2 2 x3 2 x4 2 x  (1  a 2 ) a  (2  a 2 ) a 2  ... 2! 3! 4!  and put a = 1 ; then a

Now put a sin–1 x = , i.e., x  sin

e  1  sin  

sin 2  sin 3  2·  ... 2! 3!

6. Show if y = sin (m sin–1 x), then (1 – x 2 )

d2y d x

2

– x

dy  m2 y  0 dx

Hence or otherwise expand sin m  in powers of sin . y = sin (m sin–1 x),

We have

y1  cos ( m sin –1 x ) .

then or or or

2

(1 – x 2 )

y1 =

m2

(1 –

2 y1 y2 (1 –

x 2)

+ y12 (– 2 x) = – 2 m2 y y1

(1 –

x 2)

m

y 2)

(1 – x2) y2 – x y1 + m2 y = 0

Differentiating n times, we have n (n – 1) yn (– 2) – x yn + 1 – n yn + m2 yn = 0 2! (1 – x2) yn + 2 – (2 n + 1) x yn + 1 + (m2 – n2) = 0. yn  2 (1 – x 2 )  n yn 1 (– 2 x) 

or

EXPANSION OF FUNCTIONS

195

Now, put x = 0, then (yn + 2)0 = (n2 – m2) (yn)0 Now,

(y)0 = 0; (y1)0 = m, (y2)0 = 0 (y3)0 = (12 – m2). m ; (y4) = 0 (y5)0 = (32 – m2) (12 – m2). m etc

hence by Maclaurin’s Theorem, we get

sin (m sin –1 x)  mx  m (12 – m 2 ) Put Hence

x3 x5  m (32 – m 2 ).  ... 3! 5!

sin–1 x = 0 ; then x = sin  1 sin m  = m sin  + 3 ! m (m2 – 1) sin3  + m (m2 – 12) (m2 – 32) sin5  + ....

7. Expand log { 1 – log (1 – x) } in powers of x by Maclaurin’s Theorem as far as the term in x3. x By substituting 1  x for x deduce the expansion of log { 1 + log (1 + x)} as far as the term x3. Let y = log {1 – log (1 – x) }, ...(i) or Then or or

ey = 1 – log (1 – x) 1 (1 – x) (1 – x) y1 = e–y e y · y1 

y2 (1 – x) – y1 =

e–y

...(ii)

y 1,

...(iii)

y3 (1 – x) – 2 y2 = e–y y12 – e–y y2 Put

...(iv)

x=0

in (i)

(y)0 = 0

in (ii)

(y1)0 = 1

in (iii)

(y2)0 = 0

in (iv)

(y3)0 = 1, etc.

Hence

{1 – log (1 – x)}  x 

x3  ... 6

x Now substituting 1  x for x1 we obtain  x  log 1 – log 1 – 1  

or

x 1 x       x   1  x 6  1 

log {1  log (1  x )}  x (1  x ) –1   x (1 – x  x 2  ...)   x – x2 

7 3 x  ... 6

3

  .... x 

1 3 x (1  x ) –3  ... 6

1 3 x (1 – 3 x  ...) 6

196

DIFFERENTIAL CALCULUS

8. Expand (sin–1 x)2 in ascending powers of x. y = f (x) = (sin–1 x)2,

Let

y1  2 sin –1 x.

then

...(i)

1

...(ii)

(1 – x 2 )

(1 – x2) y12 = 4 y.

or

Differentiating again (1 – x2). 2y1 y2 + y12 (– 2 x) = 4 y1 (1 – x2) y2 = x y1 + 2

or

...(iii)

Differentiating n times, we get n ( n – 1) yn (– 2) – x yn + 1 – n yn (1) = 0 2! Putting x = 0 in (i), (ii), (iii) and (iv), we get (1 – x 2 ) yn  2  n yn 1 (– 2 x 

...(iv)

(y)0 = 0; (y1)0 = 0 ; (y2)0 = 2 and (yn + 2)0 = n2 (yn)0 from which we get (y3)0 = 12 . (y1)0 ; (y4)0 = 22 . (y2)0 = 22 . 2; (y5)0 = 32 . (y3)0 = 0; (y6)0 = 42 (y4)0 = 42 . 22. 2 ; etc. Hence by Maclaurin’s Theorem, we get

(sin –1 x) 2  ( y )0  x ( y1 )0 

sin –1 x

9. If y 

2

(1 – x )

x2 x3 ( y2 ) 0  ( y3 )0  ... 2! 3!



x2 x4 2 x6 2 2 (2)  (2 .2)  (4 . 2 )  ... 2! 4! 6!



2 x2 x4 x6  2 . 22 .  2 . 22 . 42 .  ... 2! 4! 6!

, where – 1 < x < 1 and –

(1 – x2)yn + 1 – (2n + 1) x yn

   sin –1 x  , prove that 2 2

– n2 yn – 1 = 0.

Assuming that y can be expanded in ascending powers of x in the form a0 + a1 x + a2 x2 + ... + an xn + .... prove that (n + 1) an + 1 = n an – 1, and hence obtain the general term of the expansion. (Garhwal 97) We have

y

sin –1 x (1 – x 2 )

or (1 – x2) y2 = (sin–1 x)2 Differentiating (i) with respect to x, (1 – x2) . 2 y y1 – 2x y2 = 2 sin– 1 x . or

(1 – x2) y1 – x y – 1 = 0

...(i) 1 1 – x2

 2y

...(ii)

EXPANSION OF FUNCTIONS

197

Differentiating n times by Leibnitz’s Theorem we get (1 – x2) yn + 1 – 2 . nC1 x yn – 2 . nC2 yn – 1 – (x yn + nC1 yn – 1) = 0 or (1 – x2) yn + 1 – (2 n + 1) x yn – n2 yn – 1 = 0

...(iii)

Putting x = 0 in (i), (ii) and (iii), we have (y)0 = 0, (y1)0 = 1 (yn + 1)0 = n2 (yn – 1)0

and

...(iv)

Now by hypothesis, y = a0 + a1 x + a2 x2 + ... + an xn + ...

...(v)

and from Maclaurin’s Theorem

y  ( y )0  x ( y1 )0 

x2 xn ( y2 )0  ...  ( yn )0  ... 2! n!

...(vi)

Equating the two values of y1 we get a0 = (y)0, a1 = (y1)0 ; .... (yn – 1)0 = (n – 1) ! an – 1 (yn)0 = n ! an, (yn + 1)0 = (n + 1) ! an + 1 etc. By putting the values of (yn + 1)0 and (yn – 1)0 in (iv), (n + 1) ! an + 1 = n2 (n – 1) ! an – 1 or

(n + 1) an + 1 = n an – 1

or

an 1 

putting

n an – 1 n 1 n = 1, 3, 5, .... (2 m + 1) in (vii),

...(vii)

a2 = a4 = a6 = .... = a2m = 0, as a0 = (y)0 = 0. putting

n = 2, 4, 6, .... 2m in (vii),

3a3 = 2a1 = 2, as a1 = (y1)0 = 1 5a5 = 4a3, 7a7 = 6a5, ........... ( 2m + 1) a2m + 1 = 2m . a2m – 1 multiplying columnwise, we obtain 3 . 5. 7 .... (2m + 1) a3 . a5 . a7 ... a2m + 1 = (2 . 4 . 6 ... 2m) 1 . a3 . a5 ... a2m – 1 2 . 4 .6 .... 2m a2 m 1   3 . 5 . 7 .... (2m  1) EXERCISES Apply Maclaurin’s Theorem to find the expansion of : 1. (a) sin x ; (Kanpur 2001 ; Ravishankar 98, 99 S, Bilaspur 96; Manipur 99, 2001, 2002; Gujrat, 2005) (b) sec x (MDU Rohtak 2000) (c) sin– 1 x (d) esin x (e) tan x (Mysore 2004) 2. (a) ex cos x

198

DIFFERENTIAL CALCULUS

(b) loge (1 + sin x)

(Bilaspur 1998, 2001 ; Indore 96, 2001; Devi Ahilya 2004; MDU Rohtak 2001)

3. (a) loge (1 + tan x) (b) ex loge (1 + x); (c) e a cos

–1

x

4. (a) ex sec x or

ex cos x

(Kumaon 2000, 2002)

(b) ax 5. Prove that

(Sagar 98)

e ax cos b x  1  ax 

a 2 – b 2 2 a (a 2 – 3b 2 ) 3 x  x 2! 3!  ..... 

(a 2  b2 )n / 2 n x cos (n tan –1 b / a )  ... n! (Garhwal 99)

6. Prove that

e x cos x  1  x –

2 x3 22 x 4 22 x5 23 x 7 – –   ... 3! 4! 5! 7!

(Kumaon 98, Bhopal 98 ; Jabalpur 96, 98 ; Vikram 2001; Garhwal 2004) 7. Prove that 1 2 1 4 1 6 x  x  x  ... (Garhwal 98 ; Kumaon 99) 2 12 45 8. Apply Maclaurin’s Theorem to obtain the term upto x4 in the expansion of log e sec x 

loge (1 + sin2 x)

(Jabalpur 1997 ; Sagar 2002)

9. Prove that 1 3 1 5 x  x – ... 3 5 Hence obtain the expansion of tan –1 x  x –

sin –1

2 x2 1  x4

(Manipur 2000)

, giving the general term. ex

10. By Maclaurin’s Theorem, find three non-vanishing terms in the expansion of

1  ex

11. Prove that for all finite values of x, n/2 2 3 22 5 n  2 x – x – ...  sin  x n  ....  3! 5!  4  n! 12. If y3 – 6xy – 8 = 0, prove that

e x sin x  x  x 2 

1 x3 x 4 .   .... (Kumaon 96) 2 3! 4! 13. Expand loge cos x in powers of x and verify the coefficient of x4 algebraically by the use of well known expansions of cos x and log (1 – x). y2x–

EXPANSION OF FUNCTIONS

14. If y = e m tan

–1

x

199

= a0 + a1 x + a2 x2 + ... + an xn + ...

prove that (n + 1) an + 1 + (n – 1) an – 1 = m an and y  1  mx 

m 2 2 m (m 2 – 2) 3 m 2 (m 2 – 8) 4 x  x  x  ... 2! 3! 4!

15. Prove by Maclaurin’s Theorem, that

[ x  1  x 2 ]m  1  mx 

m2 m (m 2 – 12 ) 3 m 2 (m 2 – 22 ) 4 . x2  x  x  ... 2! 3! 4!

and by using

a m  1  m log e a 

m2 (log e a) 2  .... deduce the expansions of 2!



log e [ x  1  x 2 ] and log e x  



(1  x 2 )  

2

16. If y  (1 – x 2 ) sin –1 .x, prove that (a) y3 (1 – x2) – 3 y2 x + 2 = 0, (b) yn + 3 (1 – x2) – (2 n + 3) x yn + 2 – n (n + 2) yn + 1 = 0; (c)

(1 – x2 ) sin –1 .x  x –

(d)  cot   1 –

x3 2 x5 2.4 x7 – – – ... 3 3 5 3.5 7

sin 2  2 sin 4  2.4 sin 6  – – – .... 3 3 5 3.5 7 2

(e)

3

 1 1 2 1 1 2 4 1 1  1 –   – ·   – · ·   – ... 4 3 2 3 5 2 3 5 7 2 ANSWERS

1. (a) x –

x3 x5  – .... 3! 5!

(c) x  12 · (e) x 

x3 x5 x7  32 . 12  52 . 33 . 12 .  .... 3! 5! 7!

(b) 1 

x 2 5 x 4 61 x 6    .... 2! 4! 6!

(d) 1  x 

x2 x4 –  ... 2 8

23 x5   ... 3 15

2. (a) 1  x 

x2 x3 11 x 4 x5 – – –  ... 2 3 24 5

2 3 4 5 (b) x – x  x – x  x  ... 2 6 12 24

x2 2 x3 9 x5 1 2 2 3    ... (b) x  x  x – ... 2! 3! 5! 2 3  a2 x2 x3 a2 x4  a/2  – a (1  a 2 )  (22  a 2 )  ... 1 – a x  (c) e 2 ! 3 ! 4 !  

3. (a) x –

DIFFERENTIAL CALCULUS

200

2 4. (a) 1  x  x 

2 x3  .... 3!

(b) 1  x log a 

x2 xn (log a)2  ...  (log a)n 2! n!

5 x4  ... 6

8.

x2 –

9.

  x6 x10 x 4 n –2 2  x2 –  – ...  (– 1)n –1  ... 3 5 2n – 1  

10.

1 1 1 x3  x– ·  ... 2 4 8 3!

13.

–

15.

x–

x2 x4 x6 –2 – 16 – ... 2! 4! 6! 2 12 3 22 4  x  x  ..., 2  – x  ... 3!  2 ! 4 ! 

6.2 Taylor’s Theorem

(Ravishankar 2000 ; Bhuj 99; Poorvanchal 2004)

Let f (x + h) be a function of h (x being independent of h) which can be expanded in powers of h and the expansion be differentiable any number of times, then

h2 hn f (x + h) = f (x) + h f ' (x) + f ' (x) + .... + f n (x). + ... 2! n! Let f (x + h) = A0 + A1 h + A2 h2 + A3 h3 + A4 h4 + ... where A0, A1, A2, ... are functions of x alone which are to be determined. Now by successive differentiation with respect to h, we get f ' (x + h) = A1+ 2 A2 h + 3 A3 h2 + 4 A4 h3 + ... f '' (x + h) = 2.1 A2 + 3.2 A3 h + 4.3 A4 h2 + ... f ''' (x + h) = 3.2.1 A3 + 4.3.2 A4 h + ... etc. Putting h = 0 f (x) = A0, f ' (x) = A1 f '' (x) = 2 ! A2, f ''' (x) = 3 ! A3, ... Hence,

f ( x  h)  f ( x )  h f  ( x ) 

h2 h3 hn n f  ( x)  f  ( x)  ... f ( x)  ... 2! 3! n!

Other forms : Putting x = a in Taylor’s Theorem.

h2 hn n f  (a )  ... f (a )  ... 2! n! Now, putting a = 0 and h = x in (i), we get f (a  h)  f (a )  h f  (a ) 

f ( x )  f (0)  x f  ( 0) 

x2 xn n f  (0)  ... f (0)  ... 2! n!

this is nothing but Maclaurin’s Theorem.

...(i)

(Vikram 1996)

EXPANSION OF FUNCTIONS

201

Putting h = x – a in (i), we get the expansion of f (x) in powers of (x – a) which is given below :

( x – a )2 ( x – a )3 f  (a )  2! 3!

f ( x )  f (a )  ( x – a ) f  (a ) 

f  (a )  .... 

( x – a)n n f (a )  ... n!

EXAMPLES 1. Expand log (x + a) in powers of x by Taylor’s Theorem. (Bhopal 1998, 2001 ; Sagar 98 ; Reewa 2000 ; Indore 2002; Avodh 2005) f (x + a) = log (x + a) f (a) = log a

Here 

1 1 2 f  ( a )  ; f  ( a )  – 2 ; f  ( a )  3 a a a Then by Taylor’s Theorem

therefore

log ( x  a )  log a 

... f n ( a ) 

(– 1) n –1 ( n – 1) ! an

.

x x2 x3 –  – ... a 2 a2 3 a3

2. Expand loge cos (x + h) in powers of h by Taylor’s theorem. f (x + h) = loge cos (x + h)

Let

f (x) = loge cos x

so that

f ' (x) = – tan x

then

f '' (x) = – sec2 x f ''' (x) = – 2 sec2 x tan x etc. Hence, loge cos (x + h) = loge cos x + h (– tan x) 

h2 h3 (– sec2 x)  (– 2 sec 2 x tan x)  ... 2! 3!

h2 h3 sec2 x – sec 2 x tan x  ... 2 3 3. Expand log sin x in powers of (x – 2).  log e cos x – h tan x –

(Garhwal, 96, 2001 ; Kumaon 2001 ; Bhopal 96; Poorvanchal 2004) We have

f (x) = log sin x

If can be adjusted as f (x – 2 + 2) = log sin (x – 2 + 2) It is to be expanded in powers of (x – 2) therefore,

f (2) = log sin 2, f ' (2) = cot 2, f '' (2) = – cosec2 2 f ''' (2) = + 2 cosec2 2 cot 2, .... etc.

Hence by Taylor’s Theorem f ( x)  f (2)  ( x – 2) f  (2) 

log sin x  log sin 2  ( x – 2) cot 2 

( x – 2) f  (2)  ... 2!

( x – 2) 2 (– cosec2 2)  ... 2!

202

DIFFERENTIAL CALCULUS

4. Expand 2 x3 + 7 x2 + x – 6 in powers of (x – 2). (Kumaon 97 ; Jiwaji 99, Reewa 2000 ; Jabalpur 2001; Vikram 98, Bilaspur 2000) We have f (x) = 2 x3 + 7 x2 + x – 6. By Taylor’s Theorem , we have

f ( x)  f (2)  ( x – 2) f  (2)  Now, Again

( x – 2)2 ( x – 2)3 f  (2) f  (2)  ... 2! 3!

f (2) = 2·23 + 7·22 + 2 – 6 = 40 f ' (x) = 6 x2 + 14 x + 1, f ' (2) = 53. f '' (x) = 12 x + 14, f '' (2) = 38 f ''' (x) = 12

and Hence

( x – 2)2 ( x – 2)3 · 88  ·12 2! 3! = 40 + 58 (x – 2) + 44 (x – 2)2 + 2 (x – 2)3

f ( x)  40  ( x – 2) · 58 

  5. Expand sin x in powers of  x –  .  2 (Gorakhpur 2001; Bilaspur 95, 99 ; Ravishankar 98 ; Bhopal 97 ; Vikram 97 ; Indore 2000 ; Jabalpur 98, 99 ; Jiwaji 2000, 2001 ; Sagar 2002)

Let

f (x) = sin x

then

 f   1 2

f  (x) = cos x

 f 0 2

f  (x) = – sin x

 f     – 1 2

f  (x) = – cos x

 f     0 2

f iv (x) = sin x

 f iv    1, etc 2

Hence by Taylor’s Theorem,     x–  x –           2   2 sin x  f     x –  f     f     2 2! 2  2 2 3! 2

1–

3

 f     ... 2

4

1   1   x –    x –  – ... 2! 2 4! 2

6. Use Taylor’s theorem to prove that tan –1 ( x  h)  tan –1 x  h sin x ·

where z = cot–1 x.

sin z sin 2 z sin 3 z – (h sin z ) 2 .  ( h sin z )3 ... 1 2 3 (Gorakhpur 2003, Kumaon 2000, 2004;

Reewa 97 ; Jabalpur 97; Indore 99 ; Bilaspur 2001)

EXPANSION OF FUNCTIONS

203

f (x + h) = tan–1 (x + h)

Let 

f (x) = tan–1 x

then

n (x)

f

= (– 1)n – 1 (n – 1) ! sinn z sin n z

1  cot –1 x x f ' (x) = sin z sin z z  tan –1

where 

f '' (x) = – (1 !) sin2 z sin 2 z, f ''' (x) = (2 !) sin3 x sin 3 z, etc. –1 –1  tan ( x  h)  tan x 

h sin z · sin z 1! h2 h3 – (1 !) sin 2 z sin 2 z  (2 !) sin 3 z sin 3 z – ... 2! 3!

 tan –1 x  h sin z ·

sin 3 z sin z sin 2 z  ( h sin z )3 · – ... – (h sin z ) 2 · 3 1 2

7. Prove that  x2  x x2 f  ( x) f  f  ( x)   ...   f ( x) – 2 1 x 2! (1  x) 1  x  (Rohilkhand 2000 ; Garhwal 97)

We can write

 x2   x  f    x – 1  x  1  x   Also, by Taylor’s Theorem f ( x  h)  f ( x )  h f  ( x ) 

h2 f  ( x)  ... 2!

 x  Putting h  –   in this expansion, we get 1  x  x  x x2 1  f x –  f ( x) – f  ( x)  · f  ( x )  ...  2 1 x 1 x (1  x) 2 ! 

or

 x2   x f    f ( x) –  1  x  1 

x2 f  ( x )   f ( x )   ...  2 x 2! (1  x) EXERCISES

1. Prove that log e sin ( x  h)  log e sin x  h cot x –

h2 cosec 2 x 2 

2. Apply Taylor’s theorem to prove ; e x  h  e x  he x 

h2 x e  ... 2!

h3 cosec 2 x cot x  ... 3

204

DIFFERENTIAL CALCULUS

h h2 h3 1 1 (Bhopal 1998)  – 2  3 –  ... xh x x x4 x 4. Expand sin–1 (x + h) in powers of x as far as the terms in x3. (Garhwal 2003)   5. Expand tan–1 x in powers of  x –  . (Sagar 1997 ; Indore 97 ; Bhopal 2002 ;  4 Ravishankar 98, 99 S, 2001 ; Vikram 99) 3. Prove that

  6. Expand sin     in powers of . 4  7. Prove that

f (m x )  f ( x)  ( m – 1) x f  ( x )  

(Rohilkhand 2001 ; Bhopal 96) 1 ( m – 1) 2 x 2 . f  ( x) 2!

1 (m – 1)3 x3 f  ( x)  ... 3!

(Jiwaji 98)

ANSWERS 4.

sin –1 h  x (1 – h 2 ) –1/ 2 

x2 x3 h (1 – h 2 ) –3/ 2  2! 3!

(1 – h )

2 –5 / 2



(1  2h 2 )  ....

2

     x –  x –   4 –  4    ... 2 1 2 4  1   2   1   4 1    16   16  

5.

tan –1

6.

 2 3 4 5  1    1   – 2 ! – 3 !  4 !  5 !  ...   2  OBJECTIVE QUESTIONS

Note : For each of the following questions four alternatives are given for the answer, only one of them is correct. Choose the correct alternative. 1. The (n + 1)th term in Maclaurin’s series is : (a)

x n (n) f (a ) n

(b)

x n (n) f (a ) n!

(c)

x n (n) f (0) n!

(d)

f ( n ) (0)

2. The (n + 1)th term in Maclaurin’s series of tan x is : (a)

xn n!

(b)

x2n  1 2n  1

(c)

(– 1) n x n n

(d)

None of these

EXPANSION OF FUNCTIONS

205

3. f (x) = log x can be expanded in powers of x – 1 by using : (a) Maclaurin’s theorem

(b)

Taylor’s theorem

(c) Leibnitz theorem

(d)

Gregory’s theorem

4. The

nth

term in the expansion of f (a + h) is :

(a)

1 n f (a) n!

(b)

h n f (a) n!

(c)

hn n f (a) n!

(d)

None of these

(b)

a

(d)

(12  a 2 ) a 3!

(a) 0

(b)

–1

(c) 1

(d)

–

5. If ea sin

–1

x

= a0 + a1 x + a2 x2 + ...., then a2 =

(a) 1 (c)

a2 2!

6. If cos x = a0 + a1 x + a2 x2 + ...., then value of a3 = 1 3!

ANSWERS 1. (c)

2. (d)

3. (b)

4. (c)

5. (c)

6. (a).

CHAPTER 7 TANGENTS AND NORMALS 7.1 Introduction. This chapter will be devoted to some elementary applications to Geometry involving only first order derivatives.

7.2 Equations of the tangent and normal. 7.2.1. Cartesian Equations. As shown in Chapter 4 the angel  which the tangent at any point (x, y) on the curve y = f (x) makes with x-axis, is given by dy tan    f  ( x ). dx Thus, the equation of the tangent at a point (x, y) on the curve y = f (x) is Y – y = f  (x) (X – x) where (X, Y) is an arbitray point on the tangent. The normal at a point being the line through the point and perpendicular to the tangent at the point, the equation of the normal at (x, y) to the curve y = f (x) is Y – y = – [1/f  (x)] (X – x)  (X – x) + f  (x) (Y – y) = 0, f  (x)  0. Ex. What happens if f  (x) = 0. 7.2.2. Parametric Cartesian Equations. At any point ‘t’ of the curve x = f (t), y = F (t), where

dx  f  (t )  0, we have dt

/

dy dy dx F  (t )   . dx dt dt f  (t ) Thus the equations of the tangent and the normal at the point ‘t’ of the curve x = f (t), y = F (t) are [X – f (t)] F  (t) – [Y – F  (t)] f  (t) = 0 and [X – f (t)] f (t) + [Y – F  (t)] F (t) = 0 EXAMPLES 1. Find the equations of the tangent and the normal at any point (x, y) of the curve y = c cosh (x/c). Also show that the length of the perpendicular from the foot of the ordinate on any tangent to the curve is constant. 206

TANGENTS AND NORMALS

207

x

The equations of the tangent and the normal to y  c cosh at the point (x, y) are c x x Y – c cosh  ( X – x)sinh , c c x x   Y – c cosh  sinh  X – x  0. c c  respectively.

Fig. 7.1

Also the length of the prependicualr from the foot (x, 0) of the ordinate of (x, y) to the tangent x x x  X sinh – Y   c cosh – x sinh   0 c c c  is

x sinh x / c – 0  (c cosh x / c – x sinh x / c) c cosh x / c   c, cosh x / c (sinh 2 x / c  1)

which is free of x, y and is, therefore, constant. 2. Find the equations of the tangent and the normal at any point (x, y) of the curve xm ym  1 am bm xm y m  –1 0 We have am bm mx m – 1 my m – 1 dy  0 Differentiating am bm dx dy bm xm – 1  – m m –1  dx a y Therefore, the equation of the tangent at (x, y) is Y –y–

bm xm – 1 ( X – x) am y m – 1

X xm – 1 Y ym – 1 xm ym   m  m  1, am bm a b for (x, y) lies on the curve. Also, the equation of the normal at (x, y) is am ym – 1 Y – y  m m – 1 ( X – x), b x X –x Y –y  m m –1. or m m –1 b x a y or

208

DIFFERENTIAL CALCULUS

3. Find the equations of the tangent and normal at  = /2 to the curve x = a ( + sin ), y = a (1 + cos ). dx   a (1  cos )  2a cos 2 , d 2 dy    – a sin   – 2a sin cos d 2 2 dy dy dx  /   – tan  dx d  d  2 dy  – 1 for  = /2.  dx Also, for  = /2, the point on the curve is [a (/2 + 1), a].

We have

Hence the equation of the tangent at  = /2 is  1 1  Y – a  – 1  X – a    1   X  Y – a  – 2 a  0 2 2  

and that of the normal at this point is  1 1  Y – a  1 X – a    1  X – Y – a   0. 2 2   

4. Show that the length of the portion of the tangent to the curve x = a cos 3 , y = a sin 3  intercepted between the co-ordinate axes is constant. dx dy  – 3a cos 2  sin ,  3a sin 2  cos  d d dx dy dx   – tan . /  dx d  d  The tangent at the point ‘’ viz.

We have

y – a sin3  = – tan  (x – a cos3 ) 

x sin  + y cos  – a sin  cos  = 0

meets the co-ordinate axes at the points (a cos , 0), (0, a sin ), Thus the length of the portion of the tangent intercepted between the co-ordinate axes is

(a2 cos2   a2 sin 2 ) = a, which is constant. 5. Prove that the equation of the normal to the Astroid x2/3 + y2/3 = a2/3. may be written in the form x sin  – y cos  + a cos  = 0. Differentiating x2/3 + y2/3 = a2/3, we get 2 –1/ 3 2 –1/ 3 dy x  y 0 3 .3 dx

or

dy y1/ 3  – 1/ 3 . dx x

...(i)

TANGENTS AND NORMALS

209

Therefore the slope of the normal at any point (x, y) = x1/3 / y1/3 But the slope of the given line = tan . x1/3 / y1/3 = tan 

We write

...(ii)

Equations (i) and (ii) have now to be solved to find x and y. We get y2/3 tan2  + y2/3 = a2/3, .y2/3 = a2/3 cos2 , i.e., y = a cos3 .

or

Substituting this value of y in (ii), we get x1/3 = a1/3 sin , i.e., x = a sin3 . Thus tan  the slope of the normal at (a sin3 , a cos3 ,). The equation of the normal at this point is y – a cos3  

sin  ( x – a sin 3 ), cos 

or y cos  – a cos4  = x sin  – a sin4 , i.e., x sin  – y cos  + a (cos2  – sin2 ) (cos2  + sin2 ) = 0, i.e., x sin  – y cos  + a cos 2  = 0, which is the required equation. 6. Find the condition for the line x cos  + y sin  = p,

...(i)

xm / am + ym / bm = 1.

...(ii)

to touch the curve In Ex. 3, (ii) it was shown that the equation of the tangent is

X xm – 1 Y ym – 1   1; am bm

...(iii)

where X, Y are the current co-ordinates. We re-write the equation (i) in the form X cos  + Y sin  – p = 0,

...(iv)

taking X, Y as current co-ordinates instead of x, y. The equations (iii) and (iv) represent the same line. 

xm – 1 m

a cos 



ym – 1 m

b sin 



1 , p 1/( m – 1)



 a m cos   x   p  

1/( m – 1)

 b m sin   ,y     p  

The point (x, y) lies on the given curve. Therefore



1  a m cos     p a m  

m /( m – 1)



1  b m sin     p  b m 

m /( m – 1)

 1,

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DIFFERENTIAL CALCULUS

(a cos )m/(m – 1) + (b sin )m/(m – 1) = pm/(m – 1),



which is the required condition. 7. Show that the length of the portion of the tangent to the astroid x2/3 + y2/3 = a2/3 intercepted between the co-ordinate axes is constant. The equation of the tangent at any point (x, y) (Ex. 5) is Y – y–

y1/ 3

( X – x), x1/ 3 The tangent meets X-axis where X = 0 y

– y1/ 3

( X – x) x1/ 3 X = x1/3 a2/3

 

Again the tangent meets Y-axis where Y = 0 Y – y

– y1/ 3

(– x) x1/ 3 Y = y1/3 a2/3

 

 Length of tangent intercepted between co-ordinate axes 

x2 / 3 a 4 / 3  y2 / 3 a4 / 3



a2 / 3 a4 / 3

 a 2  a. 8. Prove that the portion of the tangent to the curve

x

a2 – y2 a  log a

a2 – y2 y

intercepted between the point of contact and the x-axis is constant.

a2 – y2 a  log a Differentiating w.r.t. x, we get, We have

x

a2 – y2 y – y2

1 1 – a 

y dy   a 2 – y 2 dx  a  –



2

y 2

a – y

2

a



dy y a – y dx 2

  dy 1 y a   –  2 2  dx  a a2 – y2 a y a – y  

2

a – y

2

– (a  y

2

a2 – y2 ) .

dy dx

TANGENTS AND NORMALS

dy  dx



211

– y 2

a – y2

Equation of the tangent at the point (x, y) is – y Y –y ( X – x) a2 – y2 This meets x-axis where Y = 0 so that X x

a2 – y 2

 the portion of the tangent intercepted between the point of contact (x, y) and the point (x 

a 2 – y 2 , 0 ) is

a 2 – y 2  y 2  a = constant.

9. Tangents are drawn from the origin to the curve y = sin x. Prove that their point of contact lies on x2 y2 = x2 – y2 y = sin x ...(i) dy  cos x dx Equation of the tangent at the point (x, y) is Y – y = cos x (X – x) Since it passes through the origin, therefore – y = – x cos x



y  cos x x The point of contact lies on the locus given by these equations (i) and (ii) and that is y2 y2  2  1 x  x2 y2 = x2 – y2 .



...(ii)

10. If ax + bx = 1 is a normal to the parabola y2 = 4x, then prove that a3 + 2a b2 = b2. Let the normal is drawn at the point (x, y) then y12 = 4 x1 from the given curve, we have

2  dy     y1  dx ( x1 , y1 ) Hence, the equation of normal is y y – y1  – 1 ( x – x1 ) 2  

y – y1  –

y1  y2  x – 1  2  4 

4y1 x + 8 y = y13 + 8y1

212

DIFFERENTIAL CALCULUS

Comparing this equation of normal with given equation ax + by = 1 we get

4 y1 8 2 y 3  8 2 y1   1 a b 1 (i) (ii) (iii)

From (i) and (ii), we get y1 

2a  b

From (i) and (iii), we get 4  y12  8  2 a From (ii) and (iii), we get



4  4a 2  2   8 2 2 a b 4 b2 = 4a3 + 8a b2



b2 = a3 + 2a  b2.

11. For the curve xy = c2, prove that the portion of the tangent intercepted between the coordinate axes is bisected at the point of contact. Let the point at which tangent is drawn be (x1, y1) on the curve. Thus x1 y1 = c2 On differentitation given curve, we have dy c2 – 2 dx x



y c2  dy   – – 1   2 x1  dx ( x1 , y1 ) x1

Thus, the equation of tangent is y – y1  –

y1 ( x – x1 ) x1



y x1 – x1 y1 = – x y 1 + x1 y1



x y   –1 2 x1 2 y1

obviously, the tangent live cuts x-and y-axes at (2x1, 0) and (0, 2y1) and the middle point of the line joining these points is (x1, y1). 12. If the tangent at (x0, y0) to the curve x3 + y3 = a3 meets the curve again at (x1, y1) then prove x1 y1 that x  y  1. 0 0 Differentiating the equation of curve 3x 2  3 y 2 .

dy 0 dx



dy x2  –  dx y2

TANGENTS AND NORMALS

213

 Equation of tangent as (x0, y0) is x0 2

y – y0  –

y0 2

( x – x0 )



x x 02 + y y 02 = x 03 + y 03



x x 02 + y y 02 = a 3

This passes through (x1, y1)  Also 

x 1 x 0 2 + y 1 y 02 = a 3 x 13 + y 13 = a 3 = x 03 + y 03 3

3

3

x1 – x0 = y0 – y1

{As tangent passes through (x1, y1)}

3

 (x1 – x0) (x12 + x1 x0 + x02) = (y0 – y1) (y12 + y0 y1 + y02) 

y0 – y1 ( x12  x0 x1  x0 2 – x0 – x1 ( y1 2  y0 y1  y0 2 )

Slope of the line joining (x0, y0) and (x1, y1) y0 – y1 x0 2 = x – x – y0 2 0 1

From these two relations, we have x0 2 y0 2



x0 2  x12  x0 x1 y0 2  y12  y0 y1



x02 y12 – x12 y02 + x0 y0 (x0 y1 – x1 y0) = 0



(x0 y1 – x1 y0) (x0 y1 + x1 y0 – x0 y0) = 0



x0 y1 + x1 y0 – x0 y0 = 0



[ x0 y1  x1 y0]

y1 x  1  1. y0 x0

13. Find the point as the curve 3x2 – 4y2 = 72 which is nearest to the line 3x + 2y + 1 = 0 Slope of the given line is (– 3/2). To find the nearest point we will find a point on the curve at which the tangent is parallel to the given line. Differentiating the given curve, we have dy 3 x  dx 4 y



3x 3  dy   1 –   dx 4 y 2  ( x1 , y1 ) 1

Also the point (x1, y1) lies on the curve, hence 3x12 – 4y12 = 72 Solving these, we get y 12 = 9 

y1 =  3

 Required points are (– 6, 3) and (6, – 3) now, distance of (– 6, 3) from the given line



| – 18  6  1 | 3



11 13

214

DIFFERENTIAL CALCULUS

and distance of (6, – 3) from the given line



| – 18 – 6  1 |



13

13 13  (– 6, 3) is the required point. 14. If the tangent at any point (4m2, 8m3) of x3 = y2 is a normal to the curve at the point where it meets the curve again, then find the value of m. y2 = x3 ...(i) dy 3x 2  dx 2y  Slope at (4m2, 9m3) = 3m  Equation of tangent at (4m2, 8m3) is y – 8m3 = 3m (x – 4m2)  y = 3m x – 4m3 Solving (i) and (ii), we get x = 4m2, m2 2 3  A (4m , 9m ) and B (m2, – m3) are the points of intersection of (i) and (ii). 

Now, Slope of tangent at B  –

...(ii)

3 m 2

2 3m Since tangent at A is normal at B, hence



Slope of normal at B 

2  3m 3m



m2 

2 3

2 . 3 15. Three normals are drawn from the point (c, 0) to the curve y2 = x, show that c must be greater than 1/2. One normal is always the x-axis. Find c for which the other two normals are perpendicular to each other. m



Differentiating equation of the curve, we have dy 1  dx 2 y  m2 m  Let any point on the curve, is  4 , 2    1  dy    m 2 m    dx  ,  m  4 

2 

 m2 m  So, equation of the normal at  4 , 2  is  

y  mx –

m m3 – 2 4

TANGENTS AND NORMALS

215

This passes through (c, 0), then 0  mc –

1 1 m – m3 2 4

m2 1 –c 0 4 2



m  0,



m  0, m   2 (c – 1/ 2)

For m = 0, normal is x-axis. For other two values, it will be real if c > 1/2. For the other two normals to be perpendicular to each other, we must have  1 1  2 c –   – 2 c –   – 1 2 2  c = 3/4.



16. Tangent at a point P1, (other than (0, 0)) on the curve y = x3 meets the curve again at P2. The tangent at P2 meets the curve at P3, and so on, show that the abscissae of P1, P2, P3.......Pn, form a G.P. Let P1 (x1, y1) be a point an the curve y = x3 

y1 = x1

...(i)

3

...(ii)

dy  3x 2 dx  Slope of tangent at P1 = m1 = 3x12

Now,

Equation of the tangent at P1 (x1, y1) is y – x13 = 3x12 (x – x1) 

y = 3x12 x – 2x13

...(iii)

Solving (i) and (ii) we get x3 – 3x12 x + 2x13 = 0  (x – x1)2 (x + 2x1) = 0   

x = x1 or x = – 2x1 x2 = – 2x1, y2 = x23 = – 8x13 P2 (x2, y2) = (–2x1, – 8x13)

Slope of the tangent at P2 = 3x22 = 12x12  Equation of tangent at P2 is y – x23 = 3x22 (x – x2)

...(iv) 3)

3)

Solving (i) and (iv) poin P3 (x3, y3) will be (– 2x2, – 8x2 = (4x1, 64x1 and so on.  Abscissae of P1, P2, P3, ..... are given by x1, – 2x1, 4x1, – 8x1, ..., which is G.P. with comman ratio – 2.

216

DIFFERENTIAL CALCULUS

17. Show that there is no cubic curve for which the tangent lines at two distinct points coincide. y = ax3 + bx2 + cx + d = 0 (a  0)

Let be a cubic curve.

Let us assume that (x1, y1) and (x2, y2), (x1 < x2) are two district points on the curve at which tangents coincide. Then by mean value thereon, there exists x3 (x1 < x3 < x2) such that y2 – y1  y  ( x3 ). x2 – x1

Since, x1, x2, x3 are solutions of the equation 3ax2 + 2bx + c = m. But it is a quadratic can not have more that two roots. Therefore, no such cubic is possible. EXERCISES 1. Find the tangents and normals to the given curves at the given points : (i)

y2 = 4ax at (a, – 2a).

(ii)

x2/a2 + y2/b2 = 1 at (x , y ).

(iii) xy = c2 at (cp, c/p). (iv)

(x2 + y2) x – ay2 = 0 at x = a/2.

(v)

(x2 + y2) x – a (a2 – y2) = 0 for x = – 3a/5.

(vi)

x2 (x – y) + a2 (x + y) = 0 at (0, 0)

(vii) x = 2a cos  – a cos 2 , y = 2a sin  – a sin 2  at  = /2. (viii) c2 (x2 + y2) = x2 y2 at (c/cos , c/sin ). 2at 2

2at 3

1 for t  . 1  t2 1  t2 2 2. Find the equation of the tangent to the curve c2 (x2 + y2) = x2 y2 in the form x cos3  + y sin3  = c.

(ix)

x

, y

3. Prove that the equation of the tangent at any point (4m2, 8m3) of the semi-cubical parabola x3 – y2 = 0 is y = 3mx – 4m3 and show that it meets the curve again at (m2, – m3), where it is normal if 9m2 = 2. 4. Find where the tangent is parallel to x-axis and where it is parallel to y-axis for the following curves : (i) x3 + y3 = a3. (ii) x3 + y3 = 3axy. (iii) 25x2 + 12xy + 4y2 = 1. 5. Tangent at any point of the curve x = a cos3 , y = b sin3  meets the co-ordinates axes in A and B; show that the locus of the point with (OA, OB) as co-ordinates is x2/a2 + y2/b2 = 1. 6. Find the equations of the tangent and the normal to the curve y (x – 2) (x – 3) – x + 7 = 0 at the point where it meets x-axis. 7. Show that the tangent at any point (x, y) on the curve ym = axm ax ay and (m – 1) a  mx m (a  x) on the co-ordinates axes.

–1

+ xm, makes intercepts

8. Prove that the sum of intercepts on the co-ordinates axes of any tangent to constant.

x 

y  a is

TANGENTS AND NORMALS

217

9. Show that the normal at any point of the curve x = a cos  + a  sin , y = a sin  – a  cos  is at a constant distance from the origin. 10. Show that the length of the portion of the normal to the curve x = a (4 cos3  – 3 cos ), y = a (4 sin3  – 9 sin ) intercepted between the co-ordinate axes in constant. 11. Show that the tangent and the normal at any point of the curve x = ae (sin  – cos ), y = ae (sin  + cos ) are equidistant from the origin. 12. Show that the distance from the origin of the normal at any point of the curve

      x  ae  sin  2cos  , y  ae  cos – 2sin  2 2 2 2    is twice the distance of the tangent at the point from the origin. 13. Show that the tangent at the point of the curve 1 (   2t ) with x-axis. 4 14. The tangent at any point on the curve x3 + y3 = 2a3 cuts off lengths p and q on the co-ordinate axes; show that p–3/2 + q–3/2 = 2–1/2 a–3/2.

x = a (t + sin t cos t), y = a (1 + sin t)2 makes angle

15. The tangent at any point P of the curve y = x2 – x3 meets it again at Q. Find the co-ordinates of the mid-point of PQ and show that its locus is y = 1 – 9x + 28x2 – 28x3. 16. Show that the line x cos  + y sin  – p = 0 will touch the curve xm yn – am + n = 0, if

pm + n mm nn = (m + n)m + n am + n cosm  sinn .

17. If p = x cos  + y sin , touches the curve n

prove that

n

 x n – 1  y n – 1   1   a b pn = (a cos )n + (b sin )n n

n

x y  x  y   2 at the point 18. Prove that the curve       2 touches the straight line a b a b (a, b), whatever be the value of n.

19. Find the point on the curve y – exy + x = 0, at which we have vertical tangent. 20. If the line ax + by + c = 0 is normal to the curve xy + 5 = 0, then show that a and b have same sign. 21. Find the equation of normal to the curve x + y = xy, whose it cuts x-axis. 22. Find the equation of normal to the curve y = (1 + x)y + sin–1 (sin2 x) at x = 0. 23. Find the equations of tangents to the ellipse 3x2 + y2 + x + 2y = 0 which are pespendicular to the line 4x – 2y = 1.

218

DIFFERENTIAL CALCULUS

24. Find the equation of tangent and normal to the curve y 2 (x + a) = x 2 (3a – x) at the point where x = a. 25. Find all tangents to the curve y = cos (x + y), – 2  x  2 that are parallel to the line x + 2y = 0 26. Find the point on the curve

xn n



yn n

 2, so that it touches the line

x y  2 a b

a b 27. Prove that the portion of the tangent to the curve xm yn = am + n intercepted between the axes is divided in the ration m : n at the point of contact.   (t )    (t )   and y  a   28. Prove that the equation of the tangent at a point on the curve x  a  f ( t )    f (t )  may be written in the form

x y  (t )  (t )  (t )   (t )

a f (t )  0 f  (t )

29. If p1 and p2 be the perpendiculars from the origin on the tangent and normal respectively at any point (x, y) on curve, then show that dy . If in the above case p 1 = | x sin  – y cos  |, p 2 = | x cos  + y sin  |, where tan   dx 2/3 2/3 2/3 2 2 2 curve be x + y = a , then show that 4p 1 + p 2 = a . 30. If the tangent at (a, b) to the curve x3 + y3 = c3 meets the curve again in (a1, b1) then prove that a1 b1   1  0. a b

31. Show that the angle between the tangents at any point P an the line joining P to the origin O is the same at all points of the curve log (x2 + y2) = c tan–1 (y/x), where c is constant. 32. The tangent to the curve y = x – x3 at a point P meets the curve again at Q. Prove that one point of trisection of PQ lies on the y-axis. Find the locus of the other point of trisection. ANSWERS 1. (i)

x + y + a = 0, x – y = 3a.

(iii) x + p2 y = 2cp ; p3 x – py = c (p4 – 1). (v)

(ii)

b2 x x + a2 y y = 3a.

(iv)

4x ± 2y – a = 0 ; 2x ± 4y = 3a.

31x ± 8y + 9a = 0 ; 8x ± 31y + 42a = 0. (vi)

(vii) x + y = 3a ; x – y + a = 0.

x + y = 0 ; x – y = 0.

(viii) x cos3  + y sin3  = c ;

x sin3  – y cos3  + 2c cot 2  = 0. (ix)

13x – 16y = 2a ; 16x + 13y = 9a.

4. (i) (0, a) ; (a, 0)

(ii) ( 3 2 a , 3 4 a);( 3 4 a , 3 2 a )

5  1 3  3 ,   ,   ,  . (iii)   8  4 8  20 6. x – 20y = 7, 20x + y = 140.

 1 – 2 x1 2 x1 – 7 x12  7 x13   . 15.  2 , 2  

TANGENTS AND NORMALS

219

19. (1, 0) ;

21.

y=x–1;

22. x + y = 1 ;

23.

x + 2y = 0, 3x + 6y + 13 = 0

24. x + 2y + 3a = 0, 2x – y – 3a = 0 ; 25.

x  2y 

26. (a, b)

 3 , x  2y  – 2 2 32. y = x – 5x3.

7.3. Angle of intersection of two curves Def. The angle of intersection of two curves at a point of intersection is the angle between the tangents to the curves at the point. EXAMPLES 1. Find the angle of intersection of the parabolas y2 = 4ax and x2 = 4by, at their point of intersection other that the origin. Now (0, 0) and (4a1/3 b2/3, 4a2/3 b1/3) are the two points of intersection of the parabolas. For

y2 = 4ax,

 dy    at the point (4a1/3 b2/3, 4a2/3 b1/3) is a1/3 /2b1/3  dx  For x2 = 4by,

 dy    at (4a1/3 b2/3, 4a2/3 b1/3) is 2a1/3 /b1/3.  dx  Thus, if m, m be the slopes of the tangents to the two curves, we have m = a 1/3 /2b1/3, m = 2a1/3 /b1/3 so that the required angle a1/ 3 2a1/ 3 – 1/ 3 1/ 3 m – m 3a1/ 3 b1/ 3  tan –1  tan –1 b 1/ 3 2b 1/ 3  tan –1 . 1  mm a 2a 2 (a 2 / 3  b 2 / 3 ) 1  1/ 3 . 1/ 3 b 2b 2. Find the condition for the curves

ax2 + by2 = 1, a1x2 + b1y2 = 1 to intersect orthogonally. If (x, y) be a point of intersection of the two curves, we have For

and for

x2 = (b1 – b)/(ab1 – a1b), y2 = (a – a1)/(ab1 – a1b).

ax   dy  – , ax2 + by2 = 1, we get  dx  by  ( x , y  ) a x  dy  – 1 a1x2 + b1y2 = 1, we have  dx  b 1 y ( x  , y )

For orthogonal intersection, we have – ax – a1 x   –1 by  b1 y 

(Avadh 2001)

220

DIFFERENTIAL CALCULUS

aa1x2 + bb1y2 = 0



aa1 (b1 – b) bb1 ( a – a1 )  0 ab1 – a1b ab1 – a1b

 

b1 – b a – a1  0 bb1 aa1



1 1 1 1 –  – ; b b1 a a1

as the required condition. 3. Find the acute angle between the curve y = | x2 – 1 | and y = | x2 – 3 | at their points of intersection when x > 0. For the intersection of the given curves | x2 – 1 | = | x2 – 3 |  (x2 – 1)2 = (x2 – 3)2   Neglecting Now,

[(x2 – 1) – (x2 – 3)] [(x2 – 1) + (x2 – 3)] = 0 4 [x2 – 2] = 0 

2 as x > 0.

x– y=|

x 2

x2

– 1 | = (x2 – 1) in the neighbourhood of

x 2 and

y = – (x2 – 3) in the neighbourhood of x 



 dy     2x  2 2  dx C1

and

 dy     – 2x  – 2 2  dx C2

Then

tan  

2

2 2 – (– 2 2 1  2 2 (– 2 2) 

4 2 4 2  –7 7

4 2   tan –1  .  7    4. Find the angle of intersection of curves, y = [ | sin x | + | cos x | ] and x 2 + y2 = 5 whose [.] denotes greatest integer function. We know that 1  | sin x | + | cos x |  2 For all real values of y = [ | sin x | + | cos x | ] = 1



Let P and Q be the points or intersection of given curves. Clearly the given curves meet at points where y = 1 so, we get x2 + 1 = 5  x = 2 Now, P (2, 1) and Q (– 2, 1)

TANGENTS AND NORMALS

x2 + y2 = 5

we have

2x  2 y



221

dy 0 dx

dy x – dx y



 dy   dy   – 2 and   2   dx  (2,1)  dx (– 2,1) Clearly, the slope of line y = 1 is zero and the slope of the tangents at P and Q are – 2 and 2 respectively. Thus, the angle of intersection is tan–1 (2). EXERCISES 1. Find the angle between the pairs of curves. (i) x2 – y2 = a2, x2 + y2 = a2

(ii) y2 = ax, x3 + y3 = 3axy.

2.

(iv) 2y2 = x3, y2 = 32x

(iii) y = sin x, y = cos x

2. Prove that the curves x3 + 2xy2 – 10a2y + 3a3 = 0 and y3 + 2xy2 – 5a2 x – a3 = 0, intersection at an angle tan–1 (88/73) at the point (3a – 2a). 3. Prove that the curves x2 + 2xy – y2 + 2ax = 0 and 3y3 – 2a2x – 4a2y + a3 = 0 intersect at an angle 9 tan–1   at the point (a, – a). 8 4. Find the condition for the line y = mx to cut at right angles the conic ax2 + 2hxy + by2 = 1. Hence find the directions of the axes of the conic.

x2 y2 x2 y2   1 will cut orthogonally if a – b = a – b   1 and 1 1 a1 b1 a b 6. Find the angle of intersectin of y = ax and y = bx. 5. Show that the curves

7. Find the condition that the curves ax2 + by2 = 1 and ax2 + by2 = 1 intersect orthogonally and hence show that the curves x2 a2  



y2 b2  

 1 and

x2 a2  



y2 b2  

1

intersect orthoganally. ANSWERS  –1 1 and tan 2 2 2 4. m h + m (a –b) – h = 0. The roots of this quadratic equation in, m, are the slopes of the two axes. log a / b 6. tan   1  log a log b

1. (i) /4.

(ii) tan –1 3 16.

(iii) tan –1 2 2

(iv)

7.4. Length of the tangent, normal, sub-tangent and sub-normal at any point of a curve. Let the tangent and the normal at any point (x, y) of the curve meet the x-axis at T and G respectively. Draw the ordinate PM. Then the lines TM, MG are called the sub-tangent and sub-normal respectively.

222

DIFFERENTIAL CALCULUS

Then tangents PT, PG are sometimes referred to as the lengths of the tangent and the normal respectively. Clearly  MPG = . Also dy . dx From the figure, we have tan  

(i) Tangent

= TP = MP cosec   y (1  cot 2  )

Fig. 7.2

2   dx    = y 1+   .   dy  

dx . dy

(ii) Sub-tangent

= TM  MP cot   y

(iii) Normal

= GP = MP sec   y (1  tan 2 ) 2   dy   = y 1 +    .  dx   

(iv) Sub-normal

dy . dx EXAMPLES

= MG  MP tan   y

1. Show that in the curve y = be x/a, the sub-tangent is of constant length and the subnormal varies as the square of ordinate? Let us consider a point (x1, y1) on the curve. Differentiating the equation of curve w.r.t. x, dy 1  be x / a . dx a



b  dy   e x1 / a    dx ( x1 , y1 ) a

 dx  Length of sub-tangent  y1    dy ( x1 , y1) a  y1 . x / a be =a  dy  Length of sub-normal  y1    dx ( x1 , y1)  b e x1 / a . 

1 2 y . a

b e x1 / a a

TANGENTS AND NORMALS

223

2. Show that in the case of the curve y2 = (x +  )3, the square of the sub-tangent varies as the sub-normal. Differentating the equation of curve 2  y.

dy  3( x  ) 2 dx

dy 3( x  )2  dx 2 y

  Sub-tangent





2 ( x   )3 3( x   )

2



2 ( x   ). 3

dy 3( x  )  y. dx 2 y 2 3( x  )  2

Sub-normal



y 2 y2  (dy / dx ) 3( x   ) 2

 y.

4 ( x   )2 (Sub-tangent) 2 9  (Sub-normal) 3 ( x  )2 2 

=

8β = constant 27

3. If the relation between sub-normal SN and sub-tangent ST at any point S on the curve  by2 = (x + a)3 is (SN) = (ST)2, then find the value of .  Differentiating the equation of the given curve, we get 2 by



dy  3( x  a ) 2 dx

dy 3 ( x  a) 2  . dx 2 by

 Length of sub-tangent ST  

Length of sub-normal SN  y

y dy / dx 2 by 2 3( x  a ) 2

dy 3 ( x  a ) 2  dx 2 b

 ( ST ) 2 (2 by 2 ) 2 .2 b     SN  {3( x  a ) 2 }2 .3( x  a ) 2

 

8b . 27

224

DIFFERENTIAL CALCULUS

EXERCISES 1. Prove that the sub-normal at any point of the parabola y2 = 4ax is constant. 2. Show that the sub-normal at any point of the curve y2x2 = a2 (x2 – a2) varies inversely as the cube of its abscissa. 3. Find the length of the tangent, length of the normal, sub-tangent and sub-normal at the point , on the four cusped hypocyloid x = a cos3 , y = a sin3 . 4. Show that the sub-tangent at any point of the curve xm yn = a m + n varies as the abscissa. 5. Show that for the curve x  a  b log[b  (b 2 – y 2 ) ] –

(b 2 – y 2 )

sum of the sub-normal and sub-tangent is constant. 6. Show that for the curve y  hn

– 1/ 2

e–h

2

n2

,

the product of the abscissa and the sub-tangent is constant. 7. Show that the sub-tangent at any point of the exponential curve, y = aex/b is constant. 8. For the catenary y = c cosh (x/c), prove that the length of the normal is y2/c. 9. Prove that the sum of the tangent and sub-tangent at any point of the curve ey/a = x2 – a2. varies as the product of the corresponding co-ordinates. 10. Show that in the curve y = a log (x2 – a2). the sum of the tangent and the sub-tangent varies as the product of the co-ordinates of the point. [Compare Ex. 9 above]. 11. Prove that in the ellipse x2/a2 + y2/b2 = 1, the length of the normal varies inversely as the perpendicular from the origin on the tangent. 12. In the curve xm + n = am – n y2n, prove the the mth power of the sub-tangent varies as the nth power of the sub-normal. ANSWERS 3. a sin2 , a tan  sin2 , a sin2  cos , a sin3  tan .

TANGENTS AND NORMALS

225

7.5 Pedal equations or p, r equations. Def. A relation between the distance, r, of any point on the curve from the origin (or pole), and the length of the perpendicular, p, from the origin (or pole) to the tangent at the point is called pedal equation of the curve. 7.5.1. To determine the pedal equation of a curve whose Cartesian equation is given. Let the equation of the curve be y = f (x).

...(i)

Equation of the tangent at any point (x, y) is Y –y = f  (x) (X – x) or Xf  (x) – Y + [y – x f  (x)] = 0.

...(ii)

Fig. 7.3

If, p, be the length of the perpendicular from (0, 0) to this tangent, we have p

y – x f  ( x) . [1  f 2 ( x )]

Also r2 = x2 + y2.

...(iii)

Eliminating x, y between (i), (ii) and (iii), we obtain the required pedal equation of the curve (i). EXAMPLES 1. Find the pedal equation of the parabola y2 = 4a (x + a)

(Kanpur 2001; Guwahati 2005; Bangalore 2006)

Tangent at (x, y) is Y – y

or

2a .( X – x) y

2a – yY + (y2 – 2ax) = 0.

...(i)

The length, p, of the perpendicular from the origin to the tangent (i) is given by p





Also

y 2 – 2 ax (4a 2  y 2 ) 4a ( x  1) – 2ax [(4a 2  4a ( x  y 2 )] 2a ( x  2a ) [4a ( x  2a )]

 [ a ( x  2a )].

r2 = x2 + y2 = x2 + 4a (x + a) = (x + 2a)2.

From (ii) and (iii), we obtain p2 = ar, which is the required pedal equations.

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

226

DIFFERENTIAL CALCULUS

2. Find the pedal equation of the curve 2x = 3a cos  + a cos 3  2y = 3a sin  – a sin 3 . We have

dx 3a – (sin   sin 3 ) d 2

dy 3a  (cos   cos 3 ) d 2 3a (cos  – cos 3) dy dy / d     2  – tan  d  dx / d  – 3a / 2 (sin   sin ) 3a a 3a a (– cos   cos 3 ) tan  – sin   sin 3  2 2 2 2  p (1  tan 2 ) – 3a (sin  cos   cos  sin ) a (cos 3  sin  – sin 3  cos ) – 2 cos  2 cos    sec  = – a sin 2   p2 = a2 sin2 2  Now, r2 = x2 + y2 = (9a2 + a2 + 6a2 cos 4)/4 = 10a2 + 6a2 – 12a2 sin2 2 2 2 2 p = 16 a – 12 a . 2 a  r2 = 4a2 – 3p2 is the pedal equation.

EXERCISES 1. Show that the pedal equation of ellipse x2/a2 + y2/b2 = 1, 1

is

p2



1 a2



1 b2

–

r2 a 2 b2

.

(Calcutta 2005)

2. Show that the pedal equation of the astroid x = a cos3 , y = a sin3 , r2

is

=

a2

–

3p2.

3. Show that the pedal equation of the curve x = 2a cos  – a cos 2 , y = 2a sin  – a sin 2 , is 9 (r2 – a2) = 8p2. 4. Show that the pedal equation of the curve x = ae (sin  – cos ), y = ae (sin  + cos ), is

r  2 p. 5. Show that the pedal equation of the curve x = a (3 cos  – cos3 ), y = a (3 sin  – sin3 ), is

3p2 (7a2 – r2) = (10a2 – r2)2.

(Bangalore 2005)

TANGENTS AND NORMALS

227

6. Show that the pedal equation of the curve c2 (x2 + y2) = x2 y2. is 1/p2 + 3/r2 = 1/c2. POLAR CO-ORDINATES

7.6 Angle between radius vector and tangent. Consider the point P (r, ) on the curve r = f (). We transform to cartesian co-ordinates and obtain x = r cos  = f () cos , y = r sin  = f () sin  Considering x = f () cos , y = f () sin  as cartesian paraetric equations of the given curve with () as parameter, we obtain tan  

dy dy dx f  () sin   f () cos     dx d  d  f  () cos  – f () sin 

Dividing the numerator and denminator by f  () cos , we obtain tan   f () / f  () tan   1 – [ f () / f  ()] tan 

...(i)

Denoting the angle between the radius vector and the tangent by , we have  + 

...(ii)

Comparing (i) and (ii), we see that Cor. Angle of intersection of two curves. If 1, 2, be the angles between the common radius vector and the tangents to the two given curves at a point of intersection, then the angle of intersection of the curves is | 1 – 2 |. Note. Precise meaning of . For a point p of the curve, r = f (),

Fig. 7.4

 is precisely defined to be the angle through which the positive direction of the radius vector (i.e., the direction of the radius vector produced) has to rotate to coincide with the direction of the tangent in which  increases. The direction in which  increases is taken as the positive direction of the tangent. EXAMPLES

1. Show that the radius vector is inclined at a constant angle to the tangent at any point on the equiangular spiral r = aeb. ...(i) (Kuvempu 2005) Differentiting (i) w.r.t. , we get dr d 1  b. aeb   br r   d dr b 1 tan   ,  b 1   tan –1 ,  b

228

DIFFERENTIAL CALCULUS

which is a constant. This property of equiangular spiral justifies the adjective ‘Equiangular’. 2. Find the angle of intersection of the cardioides r = a (1 + cos ), r = b (1 – cos ).

(Bangalore 2005)

Let P (r1, 1) be a point of intersection. Let 1, 2 be the angles which OP makes with the two tangents to the two curves at P. r = a (1 + cos ) we have

For the curve

dr  – a sin  d

d a (1  cos ) – 2 cos 2  / 2  –   – cot dr a sin  2sin  / 2.cos  / 2 2



r



  tan   tan    ,  2 2 



   2 2

 1  . 2 2 3. Prove that the normal at any point (r, ) of the curve rn = an cos n  makes an angle (n + 1)  with the initial line. 1 

Hence

The given curve is rn = an cos n  Taking log of both the sides, we obtain n log r = n log a + log cos n  Differentiating, ndr  – n tan n  r d

Hence

tan   r

Fig. 7.5

d    – n cot n   tan   n   dr 2 

  n . 2 Let  be the angle which the normal makes with the initial line.

Therefore



Then

  900   

and



  2

1 1   n     (n  1)  2 2

   (n  1)       = (n + 1)  2 2 4. Find the angle  for the curve a = (r2 – a2)1/2 – a cos–1 (a/r).

Then

Differentiating the given equation w.r.t. , we get

dr a 1 a  (r 2 – a2 )–1/ 2 .2r  d 2 1 – (a / r )2

 a  dr  – 2  r d

TANGENTS AND NORMALS

or

a

d  dr

= 

rd  dr

or

cos  

229

r

–

r 2 – a2

a2 r r 2 – a2

r 2 – a2 r r 2 – a2 a 1 2

2

1  (r – a ) / a = cos–1 (a/r).



tan  



2



r 2 – a2 a

a r

5. Show that the curve rn = an sec (n  + ) and rn = bn sec (n  + ) intersect at an angle which is independent of a and b. Taking logarithm of both sides of first curve, we get

n log r = n log a + log sec (n  + )

Differentiating n dr  n tan (n    ) r d

 

d tan 1  r    cot (n   )  dr  1 

1  – (n   ) 2

Similarly for second curve, we get 2 

1  – (n   ) 2

Angle of intersection of curves = 1 – 2

1  1     – ( n    )  –   – ( n   )  2  2  =  – , which is indenpendent of a and b. 6. If u = 1/r, prove that 2 2 d  u (d u / d  ) –  du / d   d u 2  (du / d )

We have and

tan   r

u

1 r

2

d dr



r

1 u



dr 1 du – 2 d u d

230

DIFFERENTIAL CALCULUS

d  2 d   – u du   – u du  



tan  

1 u

or

tan  

–u du / d 

  du     – tan –1 u /     d  

or

d – d



1   du  1  u /      d   

–



2

(du / d )2 (du / d )2  u 2





d  u    d   du / d  

(du / d ) (du / d ) – u (d 2u / d 2 ) (du / d )2

u ( d 2 u / d 2 ) – (du / d )2 u 2  (du / d )2

EXERCISES 1. Find  for the curves (i) r = a (1 – cos ). (Cardioide) (iii) 2a/r = 1 + cos . (Parabola)

(ii)

rm = am cos m .

(iv)

rm = am (cos m  + sin m ).

2. Show that the two curves r2 = a2 cos 2 and r = a (1 + cos ) intersect at an anlge 3 sin–1 (3/4)1/4. 3. Show that the curves rm = am cos m, rm = am sin m cut each other orthogonally. (Garhwal 2000; Guwahati 2005) 4. Find the angle between the curves (i) (ii)

r = a , r  = a; r = a /(1 + ), r = a/(1 +

(Kanpur 2002) )2;

(iii) r = a cosec2 (/2), r = b sec2 (/2);

(Kuvempu 2005)

(iv) r = a log , r = a/log ; (v)

r2 sin 2  = 4, r2 = 16 sin 2 ;

(vi) r = ae, re = b. 5. Show that in the case of the curve r = a (sec  ± tan ), if a radius vector OPP be drawn cutting the curve in P and P and if the tangents at P, P meet in T, then TP = TP. 6. Show that the tangents to the cardioide r = a (1 + cos ) at the points whose vectorial angles are /3 and 2/3 are respectively parallel and perpenducular to the initial line. 7. The tangents at two points P, Q lying on the same side of the initial line of the cardioide r = a (1 + cos ), are prependicular to each other; show that the line PQ subtends an angle /3 at the pole.

TANGENTS AND NORMALS

231

8. Show that the tangents drawn at the extermities of any chord of the cardioide r = a (1 + cos ) which passes through the pole are perpendicular to each other. 9. If two tangents to the cardioide r = a (1 + cos ) are parallel, show that the line joining their points of contact subtend an angle 2/3 at the pole.

 dy   dy  – y/ y  x . 10. Show that tan    x  dx   dx  ANSWERS 1. (i) /2. (ii) /2 + m . 4. (i) /2. (ii)

tan–1

3.

(iii) /2 – /2.

(iv) /4 + m.

(iii) /2.

(iv) tan–1 [2e/(e2 – 1)]

(v) 2/3. (vi) /2.

7.7. Length of the perpendicular from pole to the tangent. Let OY = p be the perpendicular from the pole to the tangent to the curve r = f () at P (r, ). From the  OPY, we get

(Bangalore 2006)

d dr r dr d

Now

tan   r



sin   



p

r 2

[ r  ( dr / d ) 2 ]

r2 [r 2  (dr / d )2 ]

.

r

or

p  sin  r p = r sin ,

.

O

which we can rewrite as

1

=

r 2 + (dr / d θ )2

=

1

2

+

p2 r4 r2 Yet, another form of p will be obtained, if we write

1  dr    . r4  d θ 

...(i)

(Kanpur 2003)

r = 1/u, so that Substituting these values in (i), we get 2

 du  = u2 +   . 2 p  dθ  1

(Rohilkhand 2002)

7.8. Length of polar sub-tangent and polar sub-normal. Let the line through the pole O, drawn perpendicular to the radius vector OP, meet the tangent and normal at P in points T and G respectively. Then, OT, OG are respectively called the polar sub-tangent and polar sub-normal at P. From the  OPT, we get OT  tan  OP

232

DIFFERENTIAL CALCULUS

OT  OP tan   r 2



d . dr

Hence polar sub-tangent dθ dθ 1 =– where u  . dr du r From the  OPG, we get r2

OG  cot , OP OG = OP cot 



1 dr dr  r. .  . r d d 1 dr 1 du where u   2. r d d θ u Note. The lengths PT and PG are called the lengths of the polar tangents and polar at P. We have

Hence, polar sub-normal 

2   d   PT  OP sec   r [1  tan 2 ]  r 1  r 2     dr   

PG  OP cosec   r [1  cot 2 ] 

 2  dr 2  r    .  d    

7.9. Pedal equations. To obtain the pedal equation of a curve whose polar equation is given. Let r = f () be the given curve.

...(i)

2

1  dr    . 2 2 p r r4  d  Eliminating  between (i) and (ii), we obtain the required pedal equation of the curve. We have

1



1



...(ii)

The pedal equation is somtimes more conveniently obtained by eliminating  and  between (i) d , dr p = r sin .

tan   r

and

EXAMPLES 1. If r = a (1 + cos ), find the polar sub-tangent, polar sub-normal and the length of polar tangent and polar normal when  = tan–1 (3/4). We have

r = a (1 + cos );

Now, at

  tan –1

3 3 4 or sin –1 or cos –1 4 5 5



dr  (– a sin ) d

TANGENTS AND NORMALS

and

233

4 9  r  a 1    a 5   5 dr 3a – d 5

Then, polar sub-tangent  r 2

d dr 2

27 9   5    a – a. – 5  5   3a 

dr 3a – d 5



Polar sub-normal Length of polar tangent

 d   r 1  r   dr 

2

2  9  9  5      a 1   a–  5   5  3 a    

9 a 10 5



Length of polar normal 2   dr    r 2      d     2

2

3 9   – 3a    a     a (10). 5 5   5  2. Find the perpendicular length from the pole on the tangent to the curve r ( – 1) = a 2. The given curve is r  Then

dr 2 a  ( – 1) – a 2  d ( – 1) 2 

Now,

a 2  –1

1 p2







a  ( – 2) ( – 1) 2

1 r2



1  dr    r4  d 

2

1 a 2 4 /( – 1) 2



 a 2 2 ( – 2) 2   . a 4 8 /( – 1)4  ( – 1)4  1

1 1 2 2 4 4 – 3  4 – 5  6 2  2 a a     

234

DIFFERENTIAL CALCULUS

3. Find the pedal equation of the curve rm = am cos m. The given curve is rm = am cos m Then m log r = log a + log cos m and Hence

m dr  – m tan m r d 1 d  – cot m r dr    tan   m  2  

tan  

  m. 2





But

  p = r sin = r sin   m   2 

or

p  r cos m   r

Then

(Bangalore 2004)

rm am

a m p = r m + 1,

which is the required pedal equation. 4. Prove that the locus of the extremity of the polar sub-normal of the curve r = f () is the curve r = f  ( – /2).

Fig. 7.6

Let P be a point on the curve. Let the line drawn through O perpendicular to the radius vector OP meets the normal at P in G. Let be (r1, 1) Then

r1 = OG = polar sub-normal

 Also Hence

dr  f  () d

1 =  GOA = 90° +  or

  1 –

 . 2

r1 = f  (1 – /2)

 Locus of G (r1, 1) is r = f  ( – /2) 5. Prove that the locus of the extremity of the polar sub-tangent of the curve

1    f       0. r 2 

1  f () is r

TANGENTS AND NORMALS

235

Let P (r, ) be any point on the curve OP is the radius vector and PT the tangent to the curve at P. Draw a line OT at right angles to the radius vector OP meeting the tangent in T. Then T is extermity of the polar sub-tangent. Let T be (r1, 1) and we are to find the locus of T (r1, 1). The equation of curve is 1  f () ...(i) r

Fig. 7.7

Differentiating –

1 dr  f  () r2 d 

...(ii)

 Polar sub-tangent OT  r1  r 2 –

d dr

1 f  ()

1  f  ()  0 r1 Also from the figure it is evident that  POA = so we have

or

1   POA – 90   –

or



 2

  1 2

Hence from (iii), we get 1    f    1   0 r1 2   Hence locus of T is

1    f       0. r 2   EXERCISES 1. Show that for the curve  k2 – r2   , 2  r  the length of polar tangent is constant.   cos –1

r – k

236

DIFFERENTIAL CALCULUS

2. Prove that for the curve r–1 = A  + B, the polar sub-tangent is constant. 3. Show that for the spiral r = a , the polar sub-normal is constant.

(Kanpur 2001)

4. Find the polar sub-tangent of (i) r = a (1 + cos ). (Cardioide) (iii) r = a

(ii) 2a/r = 1 + e cos . (Conic)

2/( – 1). 2

5. Show that for the curve : r  aeb  . polar sub-normal varies as θ 2 , polar sub-tangent

6. Find the polar sub-normal of (ii) r = a + b cos .

(i) r = a/; 7. Find p for the curve r = ae/(1 + )2. 8. Find the pedal equations of

(ii) r = ae cot 

(i) r = a/. (iii) l/r = 1 + e cos .

(iv) r = a .

(v) r = a (1 + cos ). (Kovempu 2005, Calcutta 2005) (vi) r = a sin m . 2

1   r 1 – sin    a. (viii) r = a + b cos . 2   (ix) rm = am sin m  + bm cos m .

(vii)

9. Show that the pedal equation of the spiral r = a sech n  is of the form 1 p

2



A

 B.

r2

  2  2  tan . 10. For the curve r  sin , prove that   , p2 = r3, polar sub-tangent  sin 2 2 2 2 ANSWERS

4. (i)

2 a cos3

1 1  cosec . 2 2

6. (i) – a/2. 7.

(ii)

2a e sin 

(iii)

a 3 –2

(ii) – b sin .

p  ae /(1  ) (2  22 ).

8. (i)

1/p2 = 1/r2 + 1/a2. 1

(iii)

p

2



(e 2 – 1) l

2



(ii) p = r sin  2 . lr

(v)

r3 = 2ap2

(vii)

ar3 = 4p4.

(ix)

r m  1  p (a 2m  b 2m )

(iv) 1/p2 = 1/r2 + a2/r4. (vi) r4 = p2 [a2 m2 + (1 – m2) r2]. (viii) r4 = (b2 – a2 + 2ar)p3.

TANGENTS AND NORMALS

237

7.10.Derivative of Arcs We are familar with the determination of the circumference of a circle by finding the limit of sum of the sides of a inscribed polygon when the number of sides is indefinitely increased and length of each side approaches to zero. Thus the length of the circumference of circle is defined as the limit of such a sum of straight line segments. Similarly the lengths of arcs of various curves may be determined. We consider an arc AB, which consists of several arcs. i.e.,

Fig. 7.8

A P1, P1, P2, P2 P3,..., Pn – 1 B

Hence the arc AB = arc AP1 + arc P1 P2 + ... + arc Pn – 1 B and P1, P2, P3, ..., Pn – 1 are points of division. Then we increase indefinitely the number of points of division; thus the consecutive points coincide along the curve and then the sum of chords approaches a limit. This limit is defined as the length of arc AB. Let us again consider arc PQ and chord PQ = c. Chord PQ < arc PQ < PM + MQ where P M and Q M are any two lines enclosing the curve. arc PQ PM MQ 1   Then chord PQ PQ PQ If Q M and P M are at right angles, then arc PQ 1 chord PQ < sin M Q P + cos M Q P If Q  P , the chord P Q tends to the tangent PM and the angle M Q P = 0 arc PQ lim  1. Hence Q  P chord PQ

7.11. Differential Coefficient of Length of Arc Let us consider the curve to be y = f (x). A is some fixed point upon the curve and P anyother point on the curve whose coordinates are (x, y). Let s be the length of arc, AP, i.e., the distance along the curve from A to P and s be the increase in the length of arc A P, i.e., y, when x and y are increased by x and y. Let Q be the point whose coordinates are x + x, y + y and arc P Q be  s.  s  s chord P Q  . Now  x  x arc  x  s chord P Q  . arc P Q x Now chord

P Q  ( P R 2  R Q)2  {( x) 2  ( y ) 2 }

Hence

{( x)2  ( y )2 } s s .   x arc P Q x 

  y s 1  arc PQ   x 

2

Fig. 7.9

238

DIFFERENTIAL CALCULUS

Therefore

lim

 x 0

y    x 

2

s  1 when  x  0 arc P Q

But

and

s x lim 1   lim  x  x  0 arc P Q  x  0

lim

 x0

y d y  x d x

2   dy   1      Then  dx    Similarly we can prove that

ds  dx

ds  dy

7.12.

 1  

 dx     dy 

2

  

Derivative of Arc Length (Polar Formula)

Let the curve be r = f () and P any point (r, ) and Q another point on the curve whose coordinates are (r + r,  + ) which is a nearby point of the curve. We denote the arc P Q by s. PR is perpendicular to OQ. In right triangle PQR, (chord PQ)2 = P R2 + R Q2 and

RP = O P sin P O Q = r sin  

and

R Q = O Q – O R = r +  r – r cos   =  r + r (1 – cos  )

Then (chord P Q)2 = r2 sin2   + [ r + r (1 – cos  )]2  r 2 sin 2    [ r  2 r sin 2 2

2

 s    s   chord P Q  Now            arc P Q    chord P Q lim  1. Since Q  P arc P Q

2

2   2 2 2 r sin     s  s   sin     r 2   r 2             arc P Q                2

Thus

 2 ] 2

2

Taking limits as  0 we get 2 2  ds dr  2   1 r        d  d   

Fig. 7.10

TANGENTS AND NORMALS

239



ds  d

2  2 dr    r     d    

ds  dr

2  2dθ  1  r     d r   

Similarly we can prove that

7.13.

Other Formulae

dy  tan  where  is the angle of the tangent with the x-axis. dx ds  (1  tan 2  )   sec . Then dx dx  cos  Therefore ds dy dy dx  .  tan  .cos  (ii) ds dx ds dy  sin  .  ds (iii) If x and y are given in terms of a parameter t x = f1 (t), y = f2 (t), ds ds dx  . then dt d x dt

(i) Also,



 1  



d x d y d x .     d t   d x dt 



 d x  2  d y 2       dt   d t   

ds = dt

 d x  2  d y 2    +    d t    d t 

d y   dx

2

 dx .  d t

2

 (iv) Also Hence

d dr 1 cos    sec 

2

tan   r

1 2  2d  1  r      d r   

240

DIFFERENTIAL CALCULUS





(d r / d ) 1/ 2

 d r  2  2   r  d   d r/d  d r  d s/d d s

Thus

dr = cos . ds

and

sin  = tan  cos  = r



d dr . dr ds

d = sin  ds We have proved that r

Corollary.

d sin  Gr ds

  Again or

p  r sin   r 2

d ds

d s r2  d p dr cos   . ds (1 – sin 2 ) 

dr ds

 p2  d r 1 – 2   r  ds 

i.e.,

ds = dr

r 2

(r – p2 )

EXAMPLES ds x 1. Find d x for the curve y  a cos h . a The given curve is y  a cosh



dy x  sinh dx a

Thus

ds  dx

x a

x x 1  sinh 2    cosh a a

(Kuvempu 2005)

TANGENTS AND NORMALS

241

ds ds x 2. Calculate d x and d y for the curve y  a log sec and prove that x = a . Also prove a d2 x 1 2x  sin . that 2a a d s2

The given curve is y  a log sec

Then

Hence

Now

x a

dy a x x 1  .sec tan . d x sec x a a a a x  tan a x tan   tan ; i.e. x = a  a

ds  dx

2   d y   1      d x   

x x   1  tan 2   sec a a 

and

ds  dy

 1  

 cosec

dx   d y

2

  2 x    1  cot  a  

x a

Now, again dx x  cos ds a d2 x

then

x 1dx  – sin . a a ds ds 2

x 1 x  – sin . cos a a a

–

1 2x sin 2a a

3. In the curve rm = am cos m ds  a (sec m )( m 1/ m ) prove that d and

a2m

d2 r d s2

 m r2m – 1  0

...(i)

242

DIFFERENTIAL CALCULUS

The given curve is rm = am cos m 

...(i)

Taking logarithms of both sides and then differentiating with respect to , we get m dr  – m tan m  r d

or

dr  – r tan m  d



ds  d

dr r2    d

...(ii) 2

 r 2  r 2 tan 2 m  = r sec m 



a (cos m )1/ m cos m 

= a (cos m )(1 – m)/m = a sec(m – 1)/m m  We have already proved that ds r  r sec m   d cos m 

and

dr  – r tan m  d



dr dr d r tan m   .  – m 1– m ds d ds a r

= – a–m rm tan m 

d2 r d dr d drdr      ds dsds drdsds d  – a – m r m tan m  . – a – m r m tan m  dr





m 2 = – a –2 m r m tan m   m r m – 1 tan m   m r sec m 

d   dr

1  m 2  ma –2 m r m  tan   r m – 1 tan m  – r sec m  . cot m  r 



= m a–2 m r2 m – 1 [tan2 m  – tan m  sec2 m  cot m ] = ma–2 m r2 m – 1 [tan2 m  – sec2 m ] = ma–2 m r2 m – 1 (– 1) 

d 2r ds 2

 m a –2 m r 2 m – 1  0

TANGENTS AND NORMALS

a2m

or

243

d 2r

 m r2m – 1  0

ds 2 4. Prove that for any curve sin 2 

d d 2r r 2 0 d ds

We know that cos  

then – sin  .

dr ds

d  d 2r  2 , differentiating w. r. t. s. ds ds – sin  .

or or

– sin  .

Hence

r

d 2r ds

2

d  d  d 2r .  2 d  ds ds

d  sin  d 2 r .  2 d r ds

(since r

d  sin ) ds

d 0 d EXERCISES

 sin 2 

ds 1. Find d x for the following curves :

(ii) y  log e

(i) 3 a y2 = x ( x – a)2;

ex – 1 ex  1

.

2. For the parabola y2 = 4 a x, prove that

ds  dx

a  x  .  x 

ds for the following curves : dt x = t2, y = t – 1; x = a sec t, y = b tan t; x = a (cos t + t sin t), y = a (sin t – t cos t); x sin t + y cos t = f  (t), x cos – y sin t = f  (t); x = et sin t, y = et cos t.

3. Calculate (i) (ii) (iii) (iv) (v)

4. Find

ds for the curve {4 (x2 + y2) – a2}3 = 27 a4 y2. dy

ds 5. Calculate d  for the following curves :

(i) r = loge sin 3 (iv) r = a ecot ;

(ii) r = a (1 + cos );

(iii) r2 = a2 cos 2;

(v) r = a ( – 1);

(vi) r 

a



2



–1

244

DIFFERENTIAL CALCULUS

6. Prove that for the ellipse

x2



a2

ds a d

y2

 1 , if x = a cos

b2

1 – e

2



cos 2  .

7. Prove that in the curve r  = a ds  dr

r

2

 a2



r

ANSWERS 3x  a

1. (i) 2. (i)

;

2 3ax

(ii)

1  4 r  ; 2

4. 5. (i)

(a2/3

+

e2 x – 1

(ii) (a2 sin2 t + b2)1/2 sec2 t;

(iv) f ´ (t) + f ´´ (t) a2/3

e2 x  1

(iii) a t ;

t

(v)

2e .

2y2/3)/4xy1/3;

 log sin 3 

 ; 2 2 (v) a ( + 1);

(ii) 2 a cos

2



 9 cot 2 3  ; (iii)

a

 sec 2   ;

(iv) a cosec  ecot ;

(vi) a (2 + 1)/(2 – 1)2 OBJECTIVE QUESTIONS

Note : For each of the following questions, four alternatives are given for the answer. Only one of them is correct. Choose the correct allernative. 1. Angle  between the tangent and radius vector is given by: (a) tan  

1 d r dr

(b)

tan  

dr d

(d)

tan   r

(c) tan   r

1 dr r d d dr

2. For the equiangular spiral r = ae cot , the angle  between the tangent and the radius vector at any point on the curve is: 1  2 (c) 2

(a)

(b)



(d)

+

3. The angle of intersection of two curves is defined as the angle between their: (a) normals

(b)

radius vectors

(c) tangents

(d)

none of these

(Avadh 2003)

TANGENTS AND NORMALS

245

4. If 1 and 2 be the angles which the tangets at P (the point of intersection of two curves) to the two curves make with the radius vector OP, then the angle between the curves is given by: –1  tan 1  tan  2  (a) tan  1 – tan  tan    1 2  –1  tan 1 ~ tan  2  (c) tan  1  tan  tan   1 2   5. Polar sub-tangent is equal to :

(a)

d dr

2 (c) r

d dr

(b)

 tan 1 ~ tan 2  tan –1    1 – tan 1 tan 2 

(d)

 tan 1  tan 2  tan –1    1  tan 1 tan 2 

(b)

r

d dr

1 d r dr (Garhwal 2002 ; Awadh 2002 ; Kanpur 2001)

(d)

6. Polar sub-normal is equal to : (a)

dr d

(b)

dr (d) d 7. Which of the following formulae is incorrect?

(c) r

d dr r

d dr

(Awadh 2003)

dx dy  sin   sin  (d) ds ds dx dy  cos   tan  (c) (d) ds ds 8. For the curve r = a (1 – cos ), the angle at which the radius vector cuts the curve is :

(a)

(a) 

(b)

2

(c) /2

(d)

  – 2 2

9. Polar sub-tangent for the curve r = a is : (a) r2/a (b) (c) a/r2 (d) 10. Polar sub-tangent for the curve r = a is : (a) r2/a (b) 2 (c) a/r (d) 11. Polar sub-tangent for the curve l/r = 1 + e cos  is: (a) l/e cos  (b) (c)

e cos  l

(d)

a 1/a a 1/a l/e sin  e sin  l

12. Polar sub-tangent for the curve r = a (1 + cos ) is : 2 (a) 2a cos

  cot 2 2

(b)

2a cos 2

  tan 2 2

2 (c) 2 a sin

  tan 2 2

(d)

2a sin 2

  cot 2 2

246

DIFFERENTIAL CALCULUS

13. Polar sub-tangent for the curve r = a (1 – cos ) is : 2 (a) 2a cos

  cot 2 2

(b)

2a cos 2

  tan 2 2

    tan 2 a sin 2 cot (d) 2 2 2 2 14. Angle of intersection of the curves r = a (1 + cos), r = b (1 – cos) 2 (c) 2 a sin

(a) 

(b)

/2

(c) 0

(d)

– /2

(Garhwal 2001)

15. Angle of intersection of the curves r = a (1 + sin ) and r = a (1 – sin ) is: (a) 0

(b)

/2

(c) 

(d)

–/2

16. Angle of intersection of the curves r = a cos , 2r = a is: (a) /2

(b)

/4

(c) /3

(d)

0

17. For the cardioides r = a (1 – cos ) : (a)  

1  2

(b)



(c)  

1  3

(d)

 = /2

(b)

p = r sin 

(d)

p = r cos n 

(a) p = r sin 

(b)

p = r cos 

(c) p = r

(d)

p = r sin 2

(b)

ds r 2  d p

(d)

ds 0 d

18. Pedal equation of the curve rn = an sin n  is: (a) p = r (c) p = r sin n  19. Pedal equation of the curve r = a

ecot 

is :

20. For any curve, we have ds  (a) d

(c)

r r 2 – p2

ds  const. d

ANSWERS 1. 7. 13. 19.

(d) (a) (c) (a)

2. 8. 14. 20.

(b) (c) (b) (b)

3. (c) 9. (a) 15. (b)

4. (c) 10. (b) 16. (c)

5. (c) 11. (b) 17. (a)

6. (a) 12. (a) 18. (c)

CHAPTER 8 MEAN VALUE THEOREMS Introduction By now, the student must have learnt to distinguish between theorems applicable to a class of functions and those concerning some particular functions like f (x) = sin x, f (x) = log x etc. The theorems applicable to a class of functions are known as General theorems. Of these Rolle’s theorem is the most fundamental.

8.1 Rolle’s Theorem. If a function f is (i) continuous in a closed interval [a, b], (ii) derivable in the open interval ] a, b [and (iii) f (a) = f (b), then there exists at least one value ‘c’ ] a, b [ such that f (c) = 0. (Kumaon 2002, 2004; Calicut 2004; Garhwal 2004, 2005; Mysore 2004; Calcutta 2005) By virtue of the continuity of the function in the closed interval [a, b] it has a greatest value M and a least value m in the interval, (3.7.3 page 117) so that there are two numbers c and d such that f (c) = M, f (d) = m. Now either M=m ...(i) or M m ...(ii) When the greatest value coincides with the least value as in case (i), the function reduces to a constant so that the derivative is equal to 0 for every value of x and therefore the theorem is true in this case. When M and m are unequal, as in case (ii), at least one of them must be different from the equal values f (a), f (b). Let M = f (c) be different from them. The number c being different from a and b, lies within the interval [a, b] and as such belongs to the open interval] a, b [is, in particular, derivable at c, so that f (c  h ) – f ( c ) when h  0 h exists and is the same when h  0 through positive values. Also f (c) is the greatest value of the function, we have f (c + h)  f (c) whatever positive or negative values h has. Thus lim

 f ( c  h ) – f (c )  0 for h  0.  h   f (c  h) – f (c)  0 for h  0.  h 247

....(iii)

....(iv)

248

DIFFERENTIAL CALCULUS

Let h  0 through positive values. From (iii), we get f (c)  0 ...(v) Let h  through negative values. From (iv), we get f (c)  0 ..(vi) The relations (v) and (vi) will both be true if, and only if f  (c) = 0. The same conclusion would be similarly reached if, it is the least value which differs from f (a) and f (b). Hence the theorem. Note 1. When geometrically interpreted, the conclusion of the theorem states that the ordinates of the end points A, B being equal, there is a point on the curve the tangent at which is parallel to the chord AB. (Gorakhpur 2001)

Fig. 8.1

Note.2. The conclusion of Rolle’s theorem may not hold good for a function which does not satisfy any of its conditions. To illustrate this remark, we consider the function given by y = f (x) = | x | in the interval [–1, 1]. The function is continuous in [–1, 1] and f (1) = f (–1). As for the derivative, f being not derivable at 0, the function is not derivable at 0, the function is not derivable in ]–1, 1[. Thus not all the 3 conditions of the Rolle’s theorem are satisfied. Also the conclusion is not valid in that f ' (x) vanishes for no value of x.

Fig. 8.2

EXAMPLES 1. Verify Rolle’s theorem for (i) f (x) = x2 in [–1, 1],

(Avadh 2002; Mysore 2004)

(ii) f (x) = x (x + 3) e–x/2 in [–3, 0]. (i) Let f (x) = x2, x  [–1, 1] so that f (1) = 1 = f (–1). Also the function is derivable in [–1, 1]. The conditions of the theorem being satisfied, the derivative f (x) must vanish for at least one value of x  ]–1, 1[. Directly, we see that the derivative vanishes for x = 0 which belongs to ]–1, 1[. Hence the verification. (ii) Let f (x) = x (x + 3) e –x/2, x  [–3, 0] so that

MEAN VALUE THEOREMS

249

f (–3) = 0 = f (0). Also the function is derivable in the interval [–3, 0]. We have  1 f (x) = (2x + 3) e–x/2 + x (x + 3) e–x/2  –   2

(– x 2  x  6) – x / 2 e 2 f (x) = 0  – x2 + x + 6 = 0. 

Now

The equation –x2 + x + 6 = 0. is satisfied by x = –2, 3. Of these two values of x for which f (x) is zero, –2 belongs to the open interval ]–3, 0[ under consideration. Hence the verification. 2. Prove that if a0, a1, a2, ......, an are real number such that a0 a a  1  ........  n –1  an  0, n 1 n 2

then there exists at least one number x between 0 and 1 such that a0xn + a1 xn – 1 + a2xn –2 + ........ + an = 0 Consider the function f defined as f ( x) 

an – 1 2 a0 n  1 a1 n x  x  ......  x  an x, x  [0,1] n 1 n 2

f being a polynomial satisfies the following conditions : (i) f is continuous in [0, 1] (ii) f is derivable in ]0, 1[ (iii) Since f (0) = 0 and f (1) = 0 by hypothesis,  f (0) = f (1) Hence  some x  ] 0, 1 [such that f (x) = 0 an – 1 a0 a (n  1) x n  1 n x n – 1  .....  .2 x  an  0 i.e., n 1 n 2 a0 x n  a1 x n – 1  ......  an – 1 x  an  0

i.e.,

3. Discuss applicability of Rolle’s Theorem to the function f (x) = | x | in [–1, 1]. Here f (1) = | 1 | = 1 and f (–1) = | –1 | = 1  f (1) = f (–1) = 1. Also we observe that the function f (x) is continuous for all values of x in (–1, 1) except perhaps at x = 0. We find f (0) = | 0 | = 0 Also,

lim f (0  h)  lim | 0  h |  |0|  0

h 0

h 0

lim f (0 – h)  lim | 0 – h |  |0|  0

h 0



h 0

lim f (0  h)  lim (0 – h)  f (0).

h 0

h 0

250

DIFFERENTIAL CALCULUS

Hence the function is continuous at x = 0 and thus the function f (x) is continuous at every point in [–1, 1]. f (0  h) – f (0) h |0  h | – |0|  lim h 0 h h–0  lim 1 h 0 h

Rf  (0)  lim

Also,

h 0

f (0 – h) – f (0) –h |0 – h | – f (0)  lim h0 –h h–0  lim  –1. h0 – h

Lf  (0) lim

and

h 0

 We find that Rf  (0)  Lf (0) and so f  (0) does not exist. Hence the function f (x) is not differentiable in the entire open interval ] –1, 1[ and so Rolle’s Theorem is not applicable to the given function f (x) in [–1, 1] 4. Discuss the applicablility of Rolle’s Theorem to the function f (x) = x2 + 1,

when 0  x  1 when 1 < x  2.

= 3 – x, 02

Solution. Here

f (0) =

+ 1 = 1 and f (2) = 3 – 2 = 1



f (0) = 1 = f (2).

Also we observe that the function f (x) is continuous for all x in the range (0, 2) except perhaps at x = 1. Also Again

f (1) = 12 + 1 = 2

[3 – (1  h)] f (1 + 0)  hlim 0  lim (2 – h)  2; h 0

and

f (1 – 0)  lim (1 – h) 2  1 h 0

 lim (2 – 2h  h 2 )  2 h 0

Hence f (1 – 0) = f (1) = f (1 + 0) and so the function f (x) is continuous at x = 1, and thus continuous in the whole interval (0, 2). Again

f  (x) = 2x,

when 0  x  1

= –1

when 1 < x  2.

 f (x) is differentiable in the interval (0, 2) except perhaps at x = 1. Now

f (1  h) – f (1) h {3 – (1  h)} – 2  lim h0 h

Rf  (1)  lim

h 0

MEAN VALUE THEOREMS

251

2–h–2  lim (–1)  –1 h 0 h

 lim

h 0

And

f (1 – h) – f (1) h

Lf  (1)  lim

h 0

[(1  h)2  1] – 2 –h

 lim

h 0

2h – h 2   lim (2 – h)  2. h0 h0 h  We find that Rf (1)  Lf  (1) and so f (1) does not exist. Hence the function f (x) is not differentiable in the entire range (0, 2) and therefore Rolle’s theorem is not applicable to the given function f (x) in (0, 2).  lim

5. The function f (x) is defined in [0, 1] as follows f (x) = 1

for 0  x 

1 2

1  x  1. 2 Show that f (x) satisfies none of the conditions of Rolle’s Theorem, yet f (x) = 0 for many points in [0, 1].  2 for

Solution. Here we find that 1  1  f  – 0   1 whereas f   0   2 2  2  1  1  1  1 i.e. f  – 0   f   0  , though f   0   2  f   so the function f (x) is discon2  2  2  2 1 tinuous at x  . 2 Also the continuity being a necessary condition for the existence of a finite derivative, the function f (x) does not exist for every point in 0  x  1.

Also

f (0) = 1 and f (1) = 2 (given)

So

f (0)  f (1).

Hence all the three conditions of Rolle’s Theorem are not satisfied by f (x) in [0, 1]. Here f (x) being a function free from x (i.e., constant) in [0, 1], there is possibility of the value of f (x) to be zero at many points in [0, 1]. 6. Verify Rolle’s theorem, for the function: f (x) = (x – a)m (x – b)n; m, n being positive integers; x  [a, b]. Let

f (x) = (x – a)m (x – b)n

f being a polynomial function satisfies the following conditions: (i)

f is continuous in [a, b]

(ii)

f is derivable in ]a, b[

(iii)

f (a) = 0 = f (b)

252

DIFFERENTIAL CALCULUS

Hence there exists some c ] a, b [ such that f (c) = 0 

m (c – a)m – 1 (c – b)n + n (c – a)m (c – b)n – 1 = 0



(c – a)m – 1 (c – b)n –1 (mc – mb + nc – na) = 0

mb  na mn 7. If f (x),  (x),  (x) have derivatives when a  x  b, show that there is a value  of x lying between a and b such that c



f (a)

 (a)

 (a)

f (b)  (b)  (b)  0 f  ()  ()   () Consider the function F (x), where

f (a)  (a)  (a) F ( x)  f (b)  (b)  (b) . f ( x)  ( x)  ( x ) We put x = a in the determinant; we observe that 1st row and third row become identical; thus F (a) = 0. Similarly, F (b) = 0. (i) F (a) = F (b) = 0. (ii) F (x) is continuous in a  x  b, since f (x),  (x),  (x) have derivatives in a  x  b. (iii) It is derivable, a < x < b. All the conditions of Rolle’ theorem are satisfied; then there exists a point x = , where F () = 0.

Thus

f ( a) f (b)

 ( a)  (b)

 ( a)  (b)  0.

f  ( )   ( )   ( )

8. Let f (x) = (x – a) (x – b) (x – c), a < b < c, show that f  (x) = 0 has two roots one belonging to ] a, b [ and other belonging to ] b, c [. Here, f (x) being a polynomial is continuous and differentiable for all real values of x. We also have f (a) = f (b) = f (c) = 0. Then by Rolle’s theorem, f (x) = 0 would have at least one root in ] a, b[ and at least one root in ] b, c [. But f (x) = 0 is a polynomial of degree two, hence f (x) = 0 cannot have more than two roots. Hence exactly one root of f  (x) = 0 would lie in ] a, b [ and exactly one in ] b, c [. EXERCISES 1. Verify Rolle’s Theorem, for the function: (i) f (x) = log [(x2 + ab)/(a + b) x], x  [a, b] (ii) f (x) =2 + (x – 1)2/3 , x  [0, 2] 2. Verify Rolle’s Theorem for the function (i) f (x) = sin x/ex; x  [0, ] (ii) f (x) = ex (sin x – cos x); x  [/ 4, 5/ 4].

(Osmania 2004)

MEAN VALUE THEOREMS

253

3. By considering the function f (x) = (x – 2) log x, show that equation x log x = 2 – x is satisfied by at least one value of x lying between 1 and 2. 4. Show that between any two roots of ex cos x = 1, there exists at least one root of ex sin x – 1 = 0. 5. Show that there is no real number k for which the equation x2 – 3x + k = 0 has two distinct roots in [0, 1]. 6. If f and F are continuous in [a, b] and derivable in ] a, b [ with F (x)  0x  ] a, b [, prove that there exists c  ] a, b [such that f  (c ) f (b ) – f ( a )  F  (c ) F (b ) – F ( a )

[Hint: Consider the function  (x) = f (x) [F (b) – F (a)] – [f (x) – f (a)] [F (x) – F (a)] 7. Show that there is no real number k, for which the equation x3 – 3x + k = 0 has two distinct roots in [0, 1]. 8. If f , g are continuous and differentiable on [a, b], then show that a < c < b. f  (b) – f ( a ) – (b – a ) F  (a ) f  (c)  g (b) – g (a ) – (b – a ) g  (a ) g  (c) [Hint: Apply Rolle’s theorem to the function  (x) = f (x) + (b – x) f (x) + A (g (x) + (b – x) g(x))] 3 ). 4 10. Any of the three conditions of Rolle’s Theorem is not necessary for f  (x) to vanish at some points in (a, b). Illustrate this by considering the function f (x) = sin (1/x) in (–1, 1).

9. Verify Rolle’s Theorem for the function f (x) = sin (1/x) in the interval (0,

8.2. Lagrange’s Mean Value Theorem If a function f is (i) continuous in a closed interval [a, b] and (ii) derivable in the open interval ] a, b [, then there exists at least one value c  ] a, b [ such that f(b) – f (a) = f  (c). b–a (Kanpur 2001; Kumaon 1999, 2001, 2003, 2005; Avadh 2001; Manipur 1999, 2000; Srivenkateshwara 2005, Bangalore 2005, Kuvempu 2005) To prove the theorem, we define a new function  involving f and designed so as to satisfy the conditions of Rolle’s theorem. Let  (x) = f (x) + Ax, x  [a, b] where A is a constant to be determined such that  (a) =  (b). f (b) – f ( a ) f ( a )  Aa  f (b)  Ab  A  – Thus, b–a Now, the function f is derivable in ] a, b [. Also x is derivable and A, is a constant. Therefore,  is derivable in ] a, b [. Thus,  satisfies all the conditions of Rolle’s theorem. There is, therefore, at least one value ‘c’  ] a, b [ such that  (c) = 0, so that we have 0 = (c) = f (c) + A  f (c) = – A 

f (b) – f ( a )  f (c) . b–a

...(i)

254

DIFFERENTIAL CALCULUS

Another form of Statement of Lagrange’s Mean Value Theorem. If a function f is continuous in a closed interval [a, a + h] and derivable in the open interval] a, a + h [, then there exists at least one number ‘’  ] 0, 1 [ such that f (a + h) = f (a) + hf (a +  h). We write b – a = h so that h denotes the length of the interval [a, b] which may now be rewritten as [a, a + h]. The number, c’ which lies between a and a + h is greater than a and less than a + h so that we may write c = a +  h, where  is some number between 0 and 1. Thus the equation (i) becomes



f ( a  h) – f ( a )  f  ( a   h) h f (a + h) = f (a) + hf  (a +  h)

[0 <  < 1]

Geometrically interpreted, the conclusion of the theorem is that there is a point P on the curve, the tangent at which is parallel to the chord AB joining the extermities of the curve. Let P be the point [c, f (c)] on the curve such that f (b) – f ( a )  f  (c ) b–a f (b ) – f ( a ) The slope of the chord AB  and that of the tangent at P [c, f (c)] is f (c). b–a These being equal, by (i), it follows that there exists a point P on the curve, tangent at which is parallel to the chord AB.

Note 1. Now [f (b) – f (a)] is the change in the function f as x changes from a to b so that [f (b) – f (a)]/(b – a) is the average rate of change of the function over the interval [a, b]. Also f (c) is the actual rate of change of the

Fig. 8.3

function for x = c. Thus, the theorem states that the average rate of change of a function over an interval is also the actual rate of change of the function at some point of the interval. In particular, for instance, the average velocity of a particle over an interval of time is equal to the velocity at some instant belonging to the interval. This interpretation of the theorem justifies the name ‘Mean Value’ for the theorem. Note 2. If we draw some curves satisfying the conditions of the theorem, we will realise that the theorem, as stated in the geometrical form, is almost self-evident. EXAMPLE 1. If f (x) = (x – 1) (x – 2) (x – 3); x [0, 4], find c. We have

f (4) = 6,

f (0) = – 6,

(Gorakhpur 2001)

MEAN VALUE THEOREMS

 Also,

255

f (4) – f (0) 12  3 4–0 4 f (x) = (x – 2) (x – 3) + (x – 3) (x – 1) + (x – 1) (x – 2)

= 3x 2 – 12x + 11. f (4) – f (0)  f  (c ) 4–0

We have

3 = 3c 2 – 12c + 11  c =



62 3 6–2 3 , 8 8

Taking 3  1.732........ we may see that both these values of ‘c’ belong to the open interval ] 0, 4 [, so that we have two numbers c in this case. 2. Prove that for any quadratic function px2 + qx+ r, the value of  in Lagrange’s theorem is 1 always whatever p, q, r, a , h may be. 2 Let f (x) = px2 + qx + r, x  [a, a + h] Since f is a polynomial, it satisfies the conditions of Lagrange’s Mean Value Theorem and hence there exists  (0 <  < 1) satisfying f (a + h) – f (a) = h f (a +  h)  p (a + h)2 + q (a + h) + r – pa2 – qa – r = h [2p (a +  h) + q]  2pah + ph2 + qh = 2pah + 2 ph2 + qh 



1 2

3. Let f be defined and continuous in [a – h, a + h] and derivable in ] a – h, a + h [. Prove that there is a real number  between 0 and 1 such that f (a + h) – f (a – h) = h [f  (a + h) + f  (a – h)] Define a function  (x) = f (a + hx) – f (a – hx), x  [0, 1] As x varies in [0, 1], a + hx varies in [a, a + h] and a – hx varies in [a – h, a]. Thus  is continuous in [0, 1] and derivable in ] 0, 1[. Hence by Lagrange’s Mean Value Theorem there exists  (0 <  < 1) such that  (1) –  (0)   () 1– 0

 where

f (a + h) – f (a – h) = h[f  (a + h) + f  (a – h)]  (1) = f (a + h) – f (a – h)

 (0) = 0 4. If a function f is such that its derivative f  is continuous in [a, b] and derivable in ] a, b [ then prove that there exists a number c (a < c < b) such that f (b) = f (a) + (b – a) f (a) +

1 (b – a)2 f  (c) 2

Consider a function  such that  (x) = f (b) – f (x) – (b – x) f  (x) – (b – x)2 A

256

DIFFERENTIAL CALCULUS

where A is a constant to be determined such that  (a) =  (b).  is continuous in [a, b] and derivable in ] a, b [. Thus  satisfies all the conditions of Rolle’s Theorem where A is given by f (b) – f (a) – (b – a) f(a) – (b – a)2 A = 0 ...(i)  there exists c  ] a, b [ such that (c) = 0 We have  (x) = –f (x) + f(x) – (b – x) f(x) + 2 (b – x) A = – (b – x) f (x) + 2 (b – x) A (c) = 0  – (b – c) f (c) + 2 (b – c) A = 0 

(b – c) (2A – f  (c)) = 0 A



1 f (c) 2

{b  c}

From (i), we have 1 (b – a ) 2 f (c) 2 5. Assuming f  to be continuous in [a, b] show that f (b )  f ( a )  (b – a ) f  ( a ) 

f (c ) –

b–c c–a 1 f (a) – f (b)  (c – a ) (c – b) f  () b–a b–a 2

where c and  both lie in [a, b] We define a function F (x)  x  [a, b] as follows: F (x) = (b – a) f (x) – (b – x) f (a) – (x – a) f (b) – (x – a) (x – b) (b – a) A where A is a constant to be determined such that F (c) = 0 i.e. (b – a) f (c) – (b – c) f(a) – (c – a) f (b) – (c – a) (c – b) (b – a) A = 0 ...(i) Also

F (a) = 0 = F (b)

The function F (x) is continuous in [a, c] and [c, b] and derivable in ] a, c [ and ] c, b[ respectively. Thus F satisfies all the conditions of Rolle’s theorem in [a, c] and [c, b]. There exists 1 ] a, c [ and 2  ] c, b [ such that F  (1) = 0 and F  (2) = 0 F  (x) = (b – a) f(x) + f (a) – f (b) – (b – a) A [2x – (a + b)] F is continuous in [a, b] and derivable in ]a, b[, hence it is continuous in [1,2] and derivable in ] 1, 2 [ {a < 1 < c < 2 < b} and F  (1) = 0 = F  (2)  F  satisfies all the conditions of Rolle’s Theorem in [1, 2] There exist some   ] 1, 2 [ such that F () = 0 

(b – a) f  () – 2A (b – a) = 0



(b – a) (f () – 2A) = 0

MEAN VALUE THEOREMS

257

1 f  () 2 Substituting this value in (i), we get: A



1 (c – a) (c – b) (b – a) f  () = 0 2 (b – c ) c–a 1 f (c ) – f (a) – f (b)  (c – a ) (c – b) f () b–a b–a 2

(b – a) f (c) – (b – c) f (a) – (c – a) f (b) – 

EXERCISES 1. Verify the mean value theorem for f (x) = log x in [1, e],

(i) (ii) (iii)

f (x) =

x3

f (x) =

lx2

(Kanpur 2003; Manipur 2002)

in [a, b],

(Kanpur 2001)

+ mx + n in [a, b].

2. Find ‘c’ of the mean value theorem, if  1 f (x) = x (x – 1) (x – 2); x   0,   2

3. Find ‘c’ so that f  (c) = [f (b) – f (a)]/(b – a) in the following cases : (i)

f (x) = x2 – 3x – 1; a = – 11/7, b = 13/7,

(ii)

f (x) = ( x 2 – 4); a = 2, b = 3,

(iii)

f (x) = ex; a = 0, b = 1.

4. Exaplain the failure of the theorem in the interval [–1, 1], when f (x) = 1/x, (x  0); f (0) = 0. 5. Deduce Lagrange’s Mean Value Theorem from Rolle’s Theorem by considering the derivable function  (x) = f (x) – f (a) – A (x – a) where A is a constant to be determined such that  (b) = 0. 6. A twice differentiable function f is such that f (a) = f (b) = 0 and f (c) > 0 for a < c < b. Prove that there is at least one value between a and b for which f () = 0 7. If f , F are continuous and differentiable, then show that for a < c < b f (b) – f ( a ) – (b – a ) f  ( a ) f  (c)   F (b) – F (a ) – (b – a ) F ( a ) F  (c)

ANSWERS 2.

c  (6 –

3. (i)

1 7

21) 6 (ii)

5

(iii) log (e – 1)

8.3. Meaning of the Sign of Derivative. We consider a function f which satisfies the conditions of the Lagrange’s mean value theorem in [a, b]. Let x1, x2 be any two points of [a, b] such that x1 < x2. Applying the mean value theorem to the interval [x1, x2], we see that there exists  ] x1, x2 [ such that

258

DIFFERENTIAL CALCULUS

8.3.1. Let

f (x2) – f (x1) = (x2 – x1) f (). f  (x) = 0  x  ] a, b [.

...(i)

From (i), we get f (x2) = f (x1). As x1, x2 are any two members of [a, b], it follows that every two values of the function are equal and as such f is a constant function is the zero function. This is the converse of the theorem, “Derivative of a constant function is the zero function.” Cor. If two functions f and F have the same derivative  x  ] a, b [, then they differ only by a constant. Now =f–F   (x) = f (x) – F  (x) = 0  x  ] a, b [   is a constant function. 8.3.2. Let f  (x) > 0  x  ] a, b [. From (i) we get f (x2) – f (x1) > 0  f (x2) > f (x1) for, x2 – x1 and f () are both positive. Thus, f  (x) > 0  x  ] a, b [  f is a strictly increasing function in [a, b]. Cor. The function f is strictly increasing in [a, b] if f  (x) > 0  x  ] a, b [ except for a finite number of points where it is zero. We consider the case of the derivative becoming 0 at only one point  ] a, b [. As f  (x) > 0  x  ] a,  [ and  x  ] , b [ the function f is strictly increasing in [a, ] and [, b] and as such also in [a, b]. The result may now be extended to the case when the derivative is 0 at any finite number of points. 8.3.3. Let f  (x) < 0  x  ] a, b [. From (i), we get f (x2) – f (x1) < 0  f (x2) < f (x1) for x2 – x1 is positive and f  () is negative. Thus f  (x) < 0  x  ] a, b [  f is a strictly decreasing function in [a, b]. Cor. The function f is strictly decreasing in [a, b] if f  (x) < 0  x  ] a, b [ except for a finite number of points where the derivative is zero. Note. The results obtained in  8.3.2 and  8.3.3 will be seen to play a very important part in the chapters on Maxima and Minima, Concavity, points of Inflexion and Curve tracing.

8.4. Graphs of hyperbolic functions Graph of y = sin h x. We have  x  R d (sinh x )  cosh x . dx

Now

 x  ] – ,  [, ex > 0 and e–x > 0

so that

cosh x 

1 x (e  e – x )  0 2

MEAN VALUE THEOREMS

259

Thus sinh x is strictly increasing in ] – ,  [. Also sin h 0 = 0 and sin h x < 0  x  ] – , 0 [ and >  0  ] 0,  [. Finally

ex – e– x   x  2

lim sinh x  lim

x 

ex – e– x  – x –  2

lim sinh x  lim

x – 

It follows that the graph of y = sinh x is as given in Fig. 8.4. Graph of y = cosh x. We have  x  R d (cosh x )  sinh x. dx Now sinh x < 0  x  ] – , 0 [ and > 0  x  ] 0,  [. Thus cosh x is strictly decreasing in [– , 0] and strictly increasing in [0, ]. Also

Fig. 8.4

lim cosh x    lim cosh x.

x– 

x 

Finally cosh 0 = 1. The graph of y = cosh x is given in Fig. 8.5. Graph of y = tanh x. We have  x  R d (tanh x )  sec h 2 x dx Thus tanh x is strictly increasing in ] – ,  [.

Also

tanh 0 = 0 and lim tanh x  lim

x 

x 

 lim

ex – e– x ex  e– x 1 – e– 2 x 1 e– 2 x ex – e– x

x 

lim tanh x  lim

x – 

x – 

1

Fig. 8.5

ex  e– x

e2 x – 1

 –1 e2 x  1 The graph of y = tanh x is as given in Fig. 8.6. The line y = x is tangent to y = tanh x at (0, 0). Also y = 1 and y = –1 are the two asymptotes of the curve y = tanh x.

= x lim –

Graphs of y = coth x, y = sech x, y = cosech x. We geve below the graphs without giving the details which are left to the reader. Fig. 8.6

260

DIFFERENTIAL CALCULUS

Fig. 8.7

y = coth x y = sech x y = cosech x Each of these curves possesses asymptotes. Also the asymptotes of y = coth x are x = 0, y = 1 and y = – 1, y = sech x is y = 0, y = cosech x are y = 0 and x = 0. EXAMPLES 1. Show that f (x) = We have

x3

–

3x2

+ 3x + 2 is strictly increasing in every interval.

f  (x) = 3x2 – 6x + 3 = 3 (x – 1)2.

Thus f  (x) > 0 for every value of x except one point viz. 1; where it is zero. Hence the given function is strictly increasing in every interval. 2. Separate the intervals in which f(x) = 2x3 – 15x2 + 36x + 1 is increasing or decreasing. We have Thus Also

f (x) = 6x2 – 30x + 36 = 6 (x – 2) (x – 3). x < 2  f  (x) > 0; 2 < x < 3  f  (x) < 0; x > 3  f (x) > 0 f  (x) = 0 for x = 2 and 3.

The following results now follow. The given function is strictly increasing in ] – , 2 [, and [3,  [; and strictly decreasing in [2, 3]. 3. Prove that if f be defined for all real x such that |f (x) – f (y) | < (x – y)2 then f is constant. We have to show that f  (x) = 0  x  R. Let

c  R.

We have f ( x ) – f (c ) – 0   whenever |x – c| <  x–c  f  (c) = 0

(using given condition) (Taking  = )

 f is a constant function. 4. Find the interval in which the function f (x) = sin (logex) – cos (logex) is strictly increases. Here domain is x > 0 as logex exists when x > 0.

MEAN VALUE THEOREMS

261

f  ( x) 

cos (log e x)  sin (log e x ) x



2 sin 



  cos (log e x)  cos sin (log e x) 4 4 x

  2 sin   log e x  4   x  f (x) strictly increases when f (x)  0, i.e., 

  sin   log x   0 4   2n   log x  (2n  1)  4

2n –

 

e

2 n –  4

   log x  2n   – 4 4

xe

2 n 

3 4

 f (x) is strictly increasing when  2 n –  2 n  3   4, e 4  x   e 

5. Let g(x) = f (x) + f (1 – x) and f (x) > 0,  x  (0, 1). Find the intervals of increase and decrease of g(x). We have Then

g (x) = f (x) + f (1 – x) g(x) = f  (x) – f  (1 – x)

We are given that f  (x) > 0,  x  (0, 1) it clearly means that f  (x) would be increasing on (0, 1) Hence two cases arise: Case I. x > 1 – x and f  (x) is increasing. 1 2



f  (x) > f  (1 – x),  x 

or

g (x) = f  (x) – f  (1 – x) > 0, 

1  x 1 2

1   g (x) is increasing in  , 1 . 2  Case II. x < 1 – x and f  (x) is increasing.

1 2



f  (x) < f  (1 – x),  x 

or

g (x) = f  (x) – f  (1 – x) < 0,  0  x 

 1  g (x) is decreasing in  0,   2

1 2

262

DIFFERENTIAL CALCULUS

6. Show that (a) x/(1 + x) < log (1 + x) < x  x > 0 (b) x –

x2 x2  log (1  x)  x – x0 2 2 (1  x) x ,x0 (1  x) 1 1 x  f  ( x)  – 2 1  x (1  x ) (1  x ) 2

(a) Now f (x) = log (1  x ) –  

f  (x) > 0  x > 0 and is 0 for x = 0

 f is strictly increasing in [0,  [, f (0) = 0.

Also

It follows that f (x) > f (0) = 0  x > 0  Again

x 1 x

log (1  x) 

...(i)

F (x) = x – log (1 + x), x > 0 1 x  1 x 1 x .



F  ( x)  1 –



F  (x) > 0  x > 0 and is 0 for x = 0

 F (x) is strictly increasing in the interval [0,  [ Also

F (0) = 0

F (x) > F (0) = 0  x > log (1 + x)  x > 0 From (i) and (ii) we get the required result. Thus

(b) Consider f ( x)  log (1  x) – x  f  ( x) 

...(ii)

x2 2

1 –1 x 1 x



x2 0x0 1 x

 f (x) is an increasing function for x > 0 

f (x) > f (0) = 0

x2 for x  0 2 Similarly by considering the function 

log (1  x )  x –

F ( x)  x –

x2 – log (1  x) 2 (1  x)

we can show that

log (1  x)  x –

x2 2 (1  x)

...(i)

MEAN VALUE THEOREMS

263

From (i) and (ii) we get the result. 7. Using Lagrange’s mean value theorem show that x  log (1  x)  x x > 0 1 x Hence show that 0 < [log (1 + x)]–1 – x–1 < 1  x > 0

f (x) = log (1 + x) in [0, x]

Consider

f (x) satisfies the condition of Lagrange’s mean value theorem in [0, x], thus there exists  (0 <  < 1) such that f ( x) – f (0)  f  ( x ) x–0



1 1 log (1  x )  x 1  x

Here

0 <  < 1 and x > 0

Now

0 <  < 1,

x > 0 x < x



1 + x < 1 + x



1 1  1  x 1  x



x x  1  x 1  x

Again

0 <  < 1, x > 0



x > 0



1 + x > 1



1 x 1 x 1  x 1  x

...(i)

...(ii)

(i) and (ii) give x  log (1  x)  x 1 x

   8. Show that

1 x 1  [log (1  x )]–1  x x

1 1  1  [log (1  x )]–1  x x –1 –1 1 > [log (1 + x)] – x > 0

v–u 2

 tan –1 v  tan 1 u 

1 v  3  1 –1 4   tan   4 25 3 4 6 Let f (x) = tan–1x, u < x < v



f  ( x) 

v–u 1  v2

,

0 < u < v and deduce that (Gujrat 2005, Nagpur 2005)

1

1  x2 By Lagrange’s Mean Value Theorem, there exists c ] u, v [such that

264

DIFFERENTIAL CALCULUS

f (v ) – f (u )  f  (c ) v–u



tan –1 v – tan –1 u 1  v–u 1  c2

u < c  u 2 < c2  1 + u 2 < 1 + c2 1



1 u

2



1 1  c2



tan –1 v – tan –1 u 1  v–u 1  u2



tan –1 v – tan –1 u 

v–u

...(i)

1  u2

c < v  c2 < v2  1 + c 2 < 1 + v2

Again 1



1  c2



1 1  v2



tan –1 v – tan –1 u 1  v–u 1  v2



tan –1 v – tan –1 u 

v–u

....(ii)

1  v2

From (i) and (ii), we have v–u 1 v

2

 tan –1 v – tan –1 u 

v–u 1  u2

4 ,u=1 3 3 4 1  tan –1 – tan –1 1  25 3 6 v

Let 

3  4 1   –  tan –1   25 4 3 6 4



tan x x  for 0 < x < /2 x sin x We have to show that

9. Show that

tan x sin x – x 2  0 for 0 < x < /2 x sin x  ; it is enough to show that 2 tan x sin x – x2 > 0, 0 < x < /2

Since x sin x > 0 for 0 < x
0 for 0 < x < /2

 f  is strictly increasing in [0, /2]. Also f  (0) = 0 

f  (x) > 0 for 0 < x < /2

 f is strictly increasing in [0, /2]. Also f (0) = 0  

f (x) > 0 for 0 < x < /2 tan x sin x – x2 > 0 for 0 < x < /2

tan x sin x – x 2  0, 0 < x < /2 x sin x tan x x  ,  x sin x 0 < x < /2 2 sin x  1, 0 < x < /2 10. Show that   x 

 sin x ,x0  f ( x)   x Let 1, x  0 f is continuous in [0, /2] and derivable in ] 0, /2 [ x cos x – sin x f ( x )  x2 Let F (x) = x cos x – sin x, x  [0, /2]

F (x) = cos x – x sin x – cos x = – x sin x < 0, x  ] 0, /2[  F is strictly decreasing in [0, /2]  

F (x) < F (0) = 0, x  [0, /2] f (x) < 0, x  ] 0,

 ] 2

 f is strictly decreasing in [0, /2]   

f (0) > f (x) > f (/2) for 0 < x < /2. 1

sin x 1  x /2

2 sin x   1 for 0 < x < /2  x EXERCISES

1. Show that the functions (i) f (x) = sin hx and f (x) = tan hx are strictly increasing in ] –, [. (ii) f(x) = cosh x is strictly decreasing in] –, 0 [and strictly increasing in the [0, [. 2. Examine the functions f (x) = coth x, sech x, cosech x in respect of monotonicity. 3. Separate the intervals in which the polynomials. (i) – x3 + 7x2 – 8x – 18

(ii) x3 + 8x2 + 5x – 2

(iii) (x – 2)2 (x + 1)

(iv) x4 – 13x2 + 36

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DIFFERENTIAL CALCULUS

(v) – 2x4 + 3x2 – 1

(vi) (4 – x2)2

are increasing or decreasing. 4. Find the greatest and least values of the function f (x) = x3 – 9x2 + 24x in [0, 6]. 5. Separate the intervals in which the rational function f (x) = (x2 + x + 1)/(x2 – x + 1) is increasing or decreasing. 6. Determine the intervals in which the function f (x) = (x4 + 6x3 + 17x2 + 32x + 32) e–x is increasing or decreasing. 7. Show that (i) x/sin x increases in the interval ]0, /2[. (ii) x/tan x decreases in the interval] 0, /2[. (iii) the equation tan x – x = 6 has one and only one root in the interval ] –/2, /2[. x if x  ] 0,  [. 1 1  x2 3 8. Show that f (x) = x – sin x is an increasing function throughout every interval. Determine for what values of a, ax – sinx is an increasing function. –1 (iv) tan x 

9. Show that x–1 log (1 + x) decreases as x increases from 0 to . 10. Show that  x > 0

x–

x2 x3 x2 x3   log (1  x)  x –  2 3(1  x) 2 3

11. Show that (i) 1 + x < ex < 1 + xex  x  0 (ii) e–x lies between 1 – x and 1 – x 

x2 ,  x  R. 2

x3 x3 x5 ,  x R. and x – + 6 6 120 12. Let f (x) = x3 + ax2 + bx + 5 sin2x be an increasing function on the set R. Then show that a2 – 3b + 15 = 0.

(iii) sin x lies between x –

13. Find the set of all values of a for which  a4  f ( x)   – 1  x5 – 3 x + log 5 monotonically decreses for all x.  1– a  x 14. Let g (x) = 2f ( ) + f (2 – x) and f  (x) < 0  x  (0, 2). Find the intervals of increase and 2 decrease of g(x).

15. Show that x ] 0, 1 [  x < – log (1 – x) < x (1 – x)–1  1 x 1 x2 x ] 0, 1 [  2 x  log 1 – x  2 x 1  3 . 1 – x2 

  

MEAN VALUE THEOREMS

267

x ] 1,  [  x – 1 > log x > (x – 1) x–1 x] 1, [  x2 – 1 > 2x log x > 4 (x – 1) – 2 log x 16. The derivative of a function f is positive for every value of x in an interval [c – h, c], and negative for every value of x in [c, c + h]; show that f (c) is the greatest value of the function in the interval [c – h, c + h]. ANSWERS 3. (i) Decreasing in ] – , [.

 1  1  (ii) Increasing in ]+, –5] and  – ,    ; Decreasing in  – 5, –  . . 3   3  (iii) Increasing in [– , 0] and [2, Decreasing in (iv) Decreasing in] – , – 3 / 2 ] and [0,

3 / 2 [.

Increasing in [ – 3 / 2 , 0] and [ 3 / 2,  [. (v) Increasing in] – , – 3 / 2 ] and [0,

3 / 2 ].

Increasing in [ – 3 / 2 , 0] and [ 3 / 2,  [. (vi) Decreasing in] – , –2] and [0, 2]. Increasing in [–2, 0] and [2, .[ 4. 36, 0. 5. Increasing in [–1, 1]; decreasing in ] – , –1] and [1, [. 6. Increasing in [–2, –1] and [0, 1]; Decreasing in] – ], –2], [–1, 0] and [1, ]. 13.

 3 – 21  a   – 4,   (1, )  2 

4   4 14. Increasing in  0,  , decreasing in  , 2  3   3

8.5. Cauchy’s mean value theorem. If two functions f (x) and F (x) are derivable in closed interval [a,b] and F’ (x)  0 for any value of x in [a,b],then there exists at least one value ‘c’ of x belonging to the open interval ] a, b[ such that (Kumaon 2000, Gorakhpur 2002; Manipur 2002; Gujrat 2005) f (b ) – f ( a ) f  (c )  F (b ) – F ( a ) F  ( c )

Firstly, we note that [F (b) – F (a) ]  0; for if it were 0, then F (x) would satisfy the conditions of the Rolle’s theorem and its derivative would therefore vanish for at least one value of x and the hypothesis that F’ (x) is never 0 would be contradicted. Now, we define a new function  (x) invloving f (x) and F (x) and designed so as to satisfy the conditions of Rolle’s theorem. Let

 (x) = f (x) + AF (x)

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DIFFERENTIAL CALCULUS

where A is a constant to be determined such that  (a) =  (b). Thus f (a) + AF (a) = f (b) + AF (b)



A–

f (b) – f ( a ) F (b) – F ( a ) , for [F (b) – F (a) ]  0.

Now, f (x) and F (x) are derivable in [a, b]. Also A is a constant. Therefore,  (x) is derivable in [a, b] and its derivative is f  (x) + AF  (x). Thus,  (x) satisfies the conditions of Rolle’s theorem. There is, therefore at least one value’c’ of x belonging to the open interval ] a, b [ such that  (c) = 0  

f  (cF  (c) = 0 f  (c) = – AF  (c),

f (c) f (b ) – f ( a ) , as F’(c)  0. –A F (c) F (b ) – F ( a ) Hence the theorem.



Another form of statement of Cauchy’s mean value theorem. If two functions f (x) and F (x) derivable in a closed interval [a, a + h] and F’ (x)  0 for any x in [a, a + h] then there exists at least one number  belonging to the open interval ] 0, 1 [ such that f ( a  h) – f ( a ) f ( a  h)  F ( a  h) – F ( a ) F ( a  h)

( 0 <  < 1)

The equivalence of the two statements can be easily seen as in the case of Lagrange’s mean value theorem. Note. Taking F (x) = x, we may easily see that Lagrange’s theorem is only a particular case of Cauchy’s. EXAMPLES 1. If in the Cauchy’s Mean Value Theorem, f (x) = ex and F (x) = e–x, show that c is arithmetic mean between a and b. f (b) – f ( a ) eb – e a  –b F (b) – F ( a ) e – e – a

= – ea eb = – ea+b



f ( x) ex  F ( x) – e – x f (c)  – e 2c F (c)



– ea + b = –e2c



c

ab 2

MEAN VALUE THEOREMS

2. Show that Let

269

sin  – sin   cot  , 0 < /2 cos  – cos  f (x) = sin x, F (x) = cos x, x  [].

f and F are continuous in [] and derivable in ] [. Thus there exists ]  [ such that sin β – sin α cos θ =– cos β – cos α sin θ

 3.

sin  – sin β = cot , 0 < /2 cos β – cos α Use Cauchy’s Mean Value theorem to evaluate

1    cos 2  x  lim   x 1  log (1/ x )  From Cauchy’s Mean Value theorem, we have f (b ) – f ( a ) f  (c )   (b ) –  ( a )  (c) where a < c < b

...(i)

1  f (x) = cos   x  ;  (x) = log x 2  a = x , b = 1.

Let

Then from (i)

   cos   – cos  – x  2 2  = log 1 – log x where

1 1  –  sin   c  2 2  1/ c a < c < b, i.e., x < c < 1.

Taking limits as x 1 which inplies that c 1, we have

  1  1  1   – cos  2  x     2  sin  2  c      lim   lim     x 1 c 1 x c log (1/ ) 1/        





 1   cos  2  x   1    , 1  – lim  as sin   c  as c  1  x 1 x log (1/ ) 2 2        1   cos  2  x   1     lim   x 1 x log (1/ )   2   EXERCISES

1. Verify Cauchy’s Mean Value theorem for the functions x2 and x4 in the interval [a, b]; a, b being positive.

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DIFFERENTIAL CALCULUS

2. If in the Cauchy’s mean value theorem, we write for f (x), F (x); (i) x2, x; (ii) sin x, cos x; show that in each case ‘c’ is the arithmetic mean beween a and b. 3. If, in the Cauchy’s mean value theorm, we write for f (x) and F (x),

x and

1 x

reprectively

1 1 and if we write and then, c is the 2 x x harmonic mean between a and b. It is understood that a, and b are both positive.

then, c, is the geometric mean between a and b,

(Kuvempu 2005) 4. Let the function f be continuous in [ a, b] and derivable in ] a, b [ . Show that there exists a number c  ] a, b [ such that 2c [ f (a) – f (b)] = f’ (c) [a2 – b2].

8.6. Higher derivatives. Let f be derivable i.e., f ’ exist in a certain neighbourhood of c. This implies that f is defined and continous in a neighbourhood of c. If the funciton f’ has derivative at c, then this derivative is called the seocnd derivative of f at c, and is denoted by f” (c). In this case f’ is necessarily continuous at c. In general, if f n – 1 (x) exists in a neighbourhood of c, then derivative of f n–1 at c, in case it exists, is called the nth derivative of f at c and is denoted by f n (c). 8.6.1. Generalised Mean Value Theorem : Talylor’s theorem. If a function f is such that (i) the (n – 1)th derivative f n–1 is continuous in [ a, a + h], (ii) the nth derivative f n exists in ] a, a + h [ , and (iii) p is a given positive integer, then there exists at least one number, , between 0 and 1 such that (Manipur 1999; Osmania 2004) n –1

2

f (a  h)  f (a)  hf  (a) 

h h f  (a)  ...  f n – 1 (a) 2! (n – 1) ! 

hn (1 – )n – p n f (a   h). (n – 1) ! p

...(i)

The condition (i) implies the continuity of f, f’, f”,......, f n–2 in [a, a + h]. Let a function be defined by

 ( x)  f ( x)  (a  h – x) f ( x) 

(a  h – x) 2 f  ( x) + ... 2! 

where A is constant to be determined such that (a) =  (a + h).

(a  h – x ) n – 1 n – 1 f ( x )  A (a  h – x ) p (n – 1) !

MEAN VALUE THEOREMS

271

Thus A is given by The function is continuous in [a, a + h], derivable in ] a, a + h [ and  (a) =  (a + h). Hence, by Rolle’s theorm, there exist at least one number, , between 0 and 1 such that ' (a + h) = 0. But

( x) 

(a  h – x )n – 1 n f ( x) – pA (a  h – x ) p – 1 . (n – 1) !

 0 = ' (a + h) =

h n – 1 (1 – ) n – 1 n f (a  h) – pA (1 – ) p – 1 h p – 1 (n – 1) !

hn – p (1 – )n – p . f n (a  h) , for (1 – )  0 and h  0. p . (n – 1) ! Substituting the value of A in (ii), we get the required result (i). 

A

(i) Remainder after n terms. Then term

Rn 

h n (1 – )n – p n f (a  h), p (n – 1) !

is known as the Taylor’s remainder Rn after n terms and is due to Schlomilch and Roche. (ii) Putting p = 1, we obtain

Rn 

h n – 1 (1 – )n – 1 n f (a  h), (n – 1) !

(MDU Rohtak 2000)

which form of remainder is due to Cauchy. (iii) Putting p = n, we obtain

Rn 

hn n f (a  h), n!

(MDU Rohtak 1998, 99)

which is due to Lagrange. Cor. 1. Let x be a point of the interval [a, a + h]. Let f satisfy the conditions of Taylor’s theorem in [a, a + h] so that it satisfies the conditions for [a, x] also. Changing a + h to x i.e., h to x – a, in (i), we obtain

f ( x)  f (a )  ( x – a) f  (a) 

( x – a) 2 n ( x – a )3 f (a )  f (a) +....+ 2! 3!

( x – a)n – 1 n – 1 ( x – a)n (1 – ) n – p n f (a)  f [a +  (x – a)], 0 <  0 x < x  ex < ex Rn 

Now

x < 0  ex < 1. Assuming that for all x

(Refer appendix I)

n

x  0, n  n ! we see that Rn  0 as nx R. lim

Thus, we obtain the expansion ex = 1 + x +

valid x  R.

x2 xn + ......... + + ....... 2! n!

Expansion of sin x. We write

f (x) = sin x so that

1   f n ( x)  sin  x  n    x  R. 2   Denoting by Rn the Lagrange’s form of remainder, we have Rn 

xn n xn n  f ( x )  sin   x   n! n! 2  

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DIFFERENTIAL CALCULUS

Rn 

so that 

xn n!

n   xn  sin   x    n! 2  

Rn  0 as n   x  R.

It follows that xR. x3 x5 + – ...... 3! 5! Expansion of cos x. We may show that sin x = x –

cos x = 1 –

x2 x4 x6 + –  ......  x  R. 2! 4! 6!

Expansion of (1 + x)m. f (x) = (1 + x)m.

We write

Now, m being any real number, (1 + x)m possess continuous derivatives of every order when (1 + x) > 0. i.e., when x > – 1. f n(x) = m (m – 1) (m – 2) ...... (m – n + 1) (1 + x)m –n.

Also,

Considering Cauchy’s form of remainder, we have

Rn  

xn (1 – )n – 1 f n ( x) (n – 1)! xn (1 – )n – 1 m (m – 1) ... (m – n  1) (1  x) m – n (n – 1)!

 xn .

m ( m – 1) ... ( m – n  1)  1 –     ( n – 1)!  1  x 

Let

| x | < 1  – 1 < x < 1.

Now

– 1 < x  –  <  x  1–  < 1+  x



(1 – )/(1 + x) < 1.

 1–   Thus, we have 0    1   x 

n –1

 1.

Let (m – 1) be positive. We have 0 < 1 +  x < 1 + 1 = 2  0 < (1 +  x)m – 1 < 2m. Let (m – 1) be negative. We have x–|x|1+x1–|x| (1 +  x)m – 1  (1 – | x | )m – 1.

 We know that lim

n 

m (m – 1) ... (m – n  1) n x 0 (n – 1)!

we see that Rn  0 as n   if | x | < 1.

n –1

(1  x ) m – 1.

MEAN VALUE THEOREMS

275

Making substitutions, we get m ( m – 1) 2 m ( m – 1) ( m – 2) 3 x  x  ... 2! 3! – 1 < x < 1.

(1  x ) m  1  mx 

when

Note. If m is a positive integer, it may be seen that we get a finite series expansion of (1 + x)m valid  x  R. Expansion of log (1 + x). f (x) = log (1 + x).

Let

We know that log (1 + x) possesses derivatives of every order when (1 + x) > 0, i.e., when x > – 1. f n ( x) 

Also,

(–1)n – 1 (n – 1)! (1  x) n

Taking the Cauchy’s form of remainder, we have

Rn  

xn (1 – )n – 1 f n ( x) (n – 1)! xn (–1)n – 1 ( n – 1)! (1 – ) n – 1 (n – 1)! (1   x) n

 (– 1) n – 1 x n

 1–   1   (1   x )  1   x 

n –1

(Manipur, 2001)

Let | x | < 1. We have –1 < x < 1  – <  x <   1 –  < 1 +  x < 1 + . Thus, we have 0

1–  1 1  x

 1–     1   x 

n –1

 1.

As we have, x  – x  1  x  1 – x 

1 1  . 1  x 1 – x

Thus, n | Rn |  | x |n

1  0 as n  . 1 – | x|

It follows that when | x | < 1,

log (1  x )  x –

x2 x3 x 4 (– 1) n – 1 n – 1  –  ...  x  ... 2 3 4 n –1

By adopting Lagrange’s form of remainder, we may show that the infinite series expansion is valid for x = 1 also.

8.7 Formal Expansions of Functions We have seen that in order to find out if a given functions can be expanded as a infinite Maclaurin series, it is necessary to examine the behaviour of Rn as n tends to infinity. To put down Rn, we

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DIFFERENTIAL CALCULUS

require to obtain the general expression for the nth derivative of the function, so that we fail to apply Maclaurin’s theoram to expand in a power series a function for which a general expression for the nth derivative cannot be determined. Other more powerful methods have, accordingly, been discovered to obtain such expansions whenever they are possible. But to deal with these methods is not within the scope of this book. Formal expressions of a function as a power series may, however, be obtained by assuming that it can be so expanded, i.e., Rn does tend to 0 as n tends to infintiy. Thus we have the result : If f (x) can be expanded as an infinite maclaurin’s series, then x2 f  (0)  ........ ...(i) 2! Such an investigation will not give any information as to the range of values of x for which the expansion is valid. f ( x)  f (0)  xf  (0) 

To obtain the expression of a function, on the assumption that it is possible, we have only to calculate the values of its derivatives for x = 0 and substitute them in (i). EXAMPLES 1. Can the function f (x) defined by f (x) = e1/x for x and f (0) = 0 be expanded in ascending powers of x by Maclaurin’s Theorem ? lim f ( x )  lim e1/ x

x 0

x 0

  1 1  lim 1    ...   , i.e., not finite 2 x 0 x 2x   Hence the given function f (x) is discontinuous at x = 0.

Hence the function f (x) can not be expanded in ascending power of x by Maclaurin’s Theorem. 2. Can the function f ( x)  We have

f ( x) 

x be expanded in ascending powers of x by Maclaurin’s Theorem? x

1 – 1/ 2 1 x , f  ( x)  – x – 3/ 2 , ... 2 4  f  (0), f  (0), f  (0) etc. are all infinite and hence x can not be expanded in ascending powers of x by Maclaurin’s Theorem. f  ( x) 

3. Prove that

sin ax  ax –

a 3 x3 a 5 x5 an – 1 xn – 1 n –1   ...  sin   3! 5! (n – 1)!  2  

Let then

f (x) = sin ax

  f  ( x)  a cos ax  a sin  ax   2 

an xn n   sin  ax  . n! 2  

MEAN VALUE THEOREMS

and so that 

277

    f  ( x)  a 2 cos  ax    a 2 sin  ax  2 .  2 2    ... ... ... ... ... ... ... ... ... ... n – 1   f n – 1 ( x)  a n – 1 sin  ax   2   n    f n ( x)  a n sin  ax   2   n  f n ( x)  a n sin  ax   2   f (0) = sin 0 = 0,

  a, f '' (0) = a2 sin  = 0, 2 3 f  (0)  a 3 sin  – a 3 , f iv (0) = a4 sin 2 = 0, 2  n – 1 .... f n – 1 (0)  a n – 1 sin    2  Putting these values in f  (0)  a sin

f ( x)  f (0)  xf  (0) 

x2 xn – 1 n – 1 xn n f  (0)  ...  f (0)  f ( x), 2! (n – 1)! n!

We get

sin ax  0  xa  0 –

x3 3 x5 5 xn – 1 n – 1 n –1 a 0 a  ...  a sin  3! 5! (n – 1)! 2 

xn n n   a sin  ax   n! 2  

n n a 3 x3 a 5 x5 xn – 1 an – 1  n – 1   a x sin  ax  n  .  ...  sin     n! 2   3! 5! (n – 1)!  2  4. A function f (x) is defined as

or

sin ax  ax –

2

f (x) = e 1/ x , x  0 and f (x) = 0, x = 0. Can it be expanded in ascending powers of x by Maclaurin’s Theorem. Given 

2

f (x) = e 1/ x , x  0  f (0  h) – f (0)  R f  (0)  lim   h 0  h   e –1/ h2 – 0   lim   h0  h    e – 2  lim    1/  

 ,  

where  = 1 / h.

... (i)

278

DIFFERENTIAL CALCULUS

 1      lim  2   lim   2  e      2 e  = 0. [By L Hospital’s Rule]  f (0 – h) – f (0)  L f  (0)  lim   h 0  –h   e1/ h2 – 0   lim  0 as before h 0   – h   f ' (0) = 0 Also from (i) we get

Also,

f  ( x) 

2 x

3

2

e – 1/ x , x  0 2

lim [ f  ( x )]  lim



x 0

x 0

x

3

e – 1/ x

2

2

 lim [2t 3 e – t ], where t  t 

1 x

 2t 3   lim  2   0  f  (0) t  et   Hence f ' (x) is continuous at x = 0. 2

If we find the higher derivatives of f (x) for x  0 we shall get e – 1/ x multiplied by a polynomial 1 in   . Hence all the higher derivatives of f (x) will be zero at x = 0. x Therefore the function f (x) possesses continuous derivative for every value of x.  By Maclaurin’s Theorem, we get

f ( x)  x . f  (0) 

x2 xn – 1 f  (0)  ...   f n – 1 (0)  Rn 2! (n – 1)! x2 xn – 1 . 0  ...  . 0  Rn 2! (n – 1)!

2

–1/ x 0  x.0  i.e., e

Rn  e –1/ x

i.e., 

2

2

lim Rn  0 as e –1/ x does not vanish as n  .

n 

Hence the given function f (x) can not be expanded in ascending powers of x by Maclaurin’s Theorem. 5. If f ( x)  f (0)  xf  (0)  (1 – x) 5/2.

x2 f  (x), find the value of  as x tends to 1, f (x) being 2! (Manipur 2001)

5 15 , f  (x)  (1 – x)1/ 2 2 4 Hence substituting these values in (i), we get

We have

f (0)  1, f  (0)  –

(1 – x)5 / 2  1 –

5 x 2 15 x  (1 – x)1/ 2 2 2! 4

MEAN VALUE THEOREMS

279

Therefore as x  1, we get 0 1– (1 – )1/ 2 

or

4 5



i.e.,

5 1 15  . (1 – )1/ 2 2 2! 4 1–

or

16 25

9 . 25

EXERCISES 1. Can the functions (i) f ( x)  tan –1 ( x / a);

(ii) f (x) = log x

be expanded in ascending powers of x by Maclaurin’s Theorem. 2. If f '' (a), f ''' (a), ...., f n – 1(a) are zero but f n (x) is continuous non-zero at x = a, then show that lim (n – 1) 

h 0

1 , n

where f (a  h)  f (a)  hf  (a)  ...  3. If f (x + h) = f (x) + h f  (x) +

hn – 1 n – 1 f ( x  n – 1 h). (n – 1)!

h2 f ''( x  h), find the value of  as x  a, f (x) being 2!

(x – a)5/2. ANSWERS 1. (i) No, (ii) No;

3.



64 . 225

OBJECTIVE QUESTIONS Note : For each of the following questions four alternatives are given for the answer. Only one of them is correct. Choose the correct alternative. 1. “If f is continuous in [a, b], differentiable in (a, b) and f (a) = f (b), then there exist at least one point c  (a, b) such that f  (c) = 0.” This result is known as (a) Lagrange’s theorem

(b)

Euler’s theorem

(c) Rolle’s theorem

(d)

Cauchy’s theorem

2. To which of the following, Rolle’s theorem can be applied : (a) f (x) = tan x in [0, ]

(b)

f (x) = cos (1/x) in [– 1, 1]

(c) f (x) = x2 in [2, 3]

(d)

f (x) = x (x + 3)e–x/2 in [– 3, 0]

3. The ‘c’ of Rolle’s theorem for the function f (x) = sin x in [0, ] is (a) 0 (c)

1  3

(b) (d)

1  6 1  2

280

DIFFERENTIAL CALCULUS

4. If f is continuous in [a, b] and differentiable in (a, b), then there exists at least one point c in (a, b) such that f  (c) is equal to f (b)  f ( a ) ba

(a)

(b)

f (b) – f ( a ) ba

f (b) – f ( a ) f (b)  f ( a ) (d) b–a b–a 5. Lagrange’s mean value theorem can be proved for a function f (x) by applying Rolle’s theorem to the function

(c)

(a)  (x) = f (x) + kx2

(b)

 (x) = f (x) – kx2

(c)  (x) = f (x) + kx

(d)

 (x) = {f (x)}2 + kx2

6. The function f (x) = x (x + 3)e [– 3, 0]. Then the value of c :

– x/2

satisfy all conditions of Rolle’s theorem in the interval

(a) 0

(b)

1

(c) 2

(d)

–2

7. The value of c of the mean value theorem, if f (x) = x (x – 1) (x – 2); a = 0, b  (a) 1 (c) (1 –

(21) / 6)

(b)

(1 

(d)

none of these.

1 is : 2

(21) / 6)

8. The value of c of the mean value theorem, if f (x) = 2x2 + 3x + 4 in [1, 2] is : (a) 1 (c) 3/2 2x3

x2

(b)

2

(d)

2/3

(Garwhal 2005)

x2

9. The function f (x) = + + – 4x – 2 satisfy all conditons of Rolle’s Theorem at the interval [– 2, 2]. Then the value of c is : (a) – 1

(b)

– 2/3

(c) 1

(d)

3/2

10. Let f (x) = (x – 4) (x – 5) (x – 6) (x – 7); then (a) f  (x) = 0 has four roots (b) three roots of f  (x) = 0 lie in (4, 5)  (5, 6)  (6, 7) (c) the equation f  (x) = 0 has only one real root. (d) three roots of f  (x) = 0 lie in (3, 4)  (4, 5)  (5, 6). 11. Lagrange’s form of remainder after n terms in Tayler’s development of the function ex in a finite form in the interval [a, a + h] is (a)

(c)

hn a h e n! h n (1  ) a h e n!

1. (c) 7. (c)

2. 8.

(d) (c)

3. 9.

(b)

h n h e n!

(d)

h n (1  )n a h e n!

ANSWERS (d) 4. (c) (a) 10. (b)

5. 11.

(c) (b)

(Manipur 2002)

6. (d)

CHAPTER 9 MAXIMA AND MINIMA 9.1 In this chapter we shall be concerned with the application of Differential Calculus to the determination of the values of a function which are greatest or least in their immediate neighbourhoods technically known as relatively greatest and least or maximum and minimum values. It will be assumed that given function possesses continuous derivatives in appropriate intervals of every order that come in question.

Maximum value of a function. We say that f (c) is a maximum value of a function f, if it is the greatest of all its values for values of x in some neighbourhood of c.

Fig. 9.1

Thus f (c) is a maximum value, if there exists some interval ] c – , c + [ around ‘c’ such that f (c) > f (x) for all values of x, other than c, lying in this interval. This is equivalent to saying that f (c) is a maximum value of f if there exists  > 0 such that f (c) > f (c + h)  f (c + h) – f (c) < 0 for 0 < | h | < .

Minimum value of function. f (c) is said to be a minimum value of f, if it is the least of all its values for values of x in some neighbourhood of c. Thus f (c) is a minimum value of f, if there exists some interval ] c – , c +  [ around ‘c’ such that f (c) < f (x) for all values of x, other than c, lying in this interval. This is equivalent to saying that f (c) is a minimum value of f if there exists a positive  such that f (c) < f (c + h)  f (c + h) – f (c) > 0 for 0 < | h | < . The maximum and minimum values of a function are also known as Relatively greatest and least values of the function in that these are the greatest and least values of the function relatively to some neighbourhoods of the points in question. Note 1. The term extreme value is used both for a maximum as well as for a minimum value. Note 2. While ascertaining whether a value f (c) is an extreme value of f or not, we compare f (c) with the values of f for values of x in any neighbourhood of c, so that the values of the function outside the neighbourhood do not come into question. 281

DIFFERENTIAL CALCULUS

282

Fig. 9.2

Thus, a maximum (minimum) values of a function may not be the greatest (least) of all the values of the function in a finite interval. In fact a function can have several maximum and minimum values and a minimum value may even be greater than a maximum value. A glance at the adjoining graph shows that the ordinates of points P1, P3, P5 are the maximum and the ordinates of the points P2, P4 are the minimum values of the corresponding function f and the ordinate of P4 which is a minimum is greater than the ordinate of P1 which is a maximum. Also the reader may perhaps guess that the tangents at the points P1, P2, P3, P4, P5 are parallel to x-axis so that if c1, c2, c3, c4, c5 are the abscissae of these points, then each of f ' (c1), f ' (c2), f ' (c3), f ' (c4), f ' (c5) is zero. In the next section, we shall establish the truth of this result.

9.2. A necessary condition for extreme values. A necessary condition for f (c) to be an extreme value of f is that f ' (c) = 0 so that f (c) is an extreme value  f ' (c) = 0. Here we assume that f is derivable at c. Let f (c) be a maximum value of f. There exists an interval ] c – , c +  [, around c, such that , if c + h is a number other than c, belonging to this interval, we have f (c + h) < f (c). Here, h may be positive or negative. Thus h0

f (c  h ) – f ( c ) 0 h

...(i)

h0

f ( c  h ) – f (c ) 0 h

...(ii)

From (i) and (ii), we have lim

h  (0  0)

f (c  h) – f (c )  0 and h

lim

h  (0 – 0)

f (c  h) – f (c ) 0 h

The relations (iii) will simultaneously be true, if and only if f ' (c) = 0. It can similarly be shown that f ' (c) = 0, if f (c) is a minimum value of f.

...(iii)

MAXIMA AND MINIMA

283

Note 1. The vanishing of f ' (c) is only a necessary but not a sufficient condition for f (c) to be an extreme value as we now show.

Fig. 9.3

Consider f (x) = Now

x3

for x = 0.

x > 0  f (x) > 0 = f (0) ; x < 0  f (x) < 0 = f (0).

Thus, f (0) is not an extreme value even though f ' (0) = 0. Note 2. f (0) = 0 is a minimum value of f (x) = | x | even though f is not derivable at 0. Example. Prove that the function f defined by f (x) = 3 | x | + 4 | x – 1 | x  R has a minimum value 3 at x = 1.

We have

4 – 7 x  4  f (x)   4 – x  3  7x – 4

if x  0 if x  0 if 0  x  1 if x  1 if x  1

From above it is clear that f has minimum value 3 at x = 1. Also Lf ' (1) = – 1

and

Rf ' (1) = 7

Show that function is not derivable at x = 1. Ex. 1. Prove that the function f defined by f (x) = 2 | x – 2 | + 5 | x – 3 | x  R has a minimum value 2 at x = 3. Stationary Values : A function f is said to be stationary for ‘c’ and f (c) a stationary value of f if f ' (c) = 0. The rate of change of a function is zero at a stationary point. While a maximum or minimum value of a stationary value may neither be a maximum nor a minimum value.

DIFFERENTIAL CALCULUS

284

Cor. Greatest and least values of function in an interval. The greatest and least values of a function f in an interval [a, b] are f (a) or f (b) or are given by the values of x for which f ' (x) = 0. The greatest and the least values of a function are also its extreme values in case they are attained at points within the interval so that the derivative must be zero at corresponding point. If c1, c2, ..........., ck be the roots of the equation f ' (x) = 0 which belog to ] a, b [, then the greatest and least values of the function f in [ a, b] are the greatest and least members respectively of the finite set {f (a), f (c1), f (c2), ......., f (ck), f (b) } Hence the result. Ex. 1. Find the greatest and least values of the function f (x) = 3x4 – 2x3 – 6x2 + 6x + 1 in the interval [ 0, 2]. We have

f (x) = 3x4 – 2x3 – 6x2 + 6x + 1



f ' (x) = 12x3 – 6x2 – 12x + 6 = 6 (x – 1) (x + 1) (2x – 1)

1 . 2 The number – 1 does not belong to the interval [0, 2] and is not, therefore, to be considered. Now



f ' (x) = 0 for x = 1, – 1,

 1  39 f (1)  2, f    , f (0)  1, f (2)  21.  2  16 Thus, the least value is 1 and the greatest value is 21.

EXERCISES 1. Find the greatest and least values of the function x4 – 4x3 – 2x2 + 12x + 1 in the interval [– 2, 5]. 2. Find the greatest and least values of the function 2x3 – 15x2 + 36x + 1 in the interval [2, 3]. Also find the greatest and least values in the interval [0, 4]. ANSWERS 1. 136, – 8

2. 29, 28; 33, 1.

Def. Change of Sign. A function is said to change sign from positive to negative as x passes through a number c, if there exists some left-handed neighbourhood ] c – h, c [ of c for every point of which the function is positive and also there exists some right handed neighbourhood ] c, c + h [ of c for every point of which the function is negative. A similar meaning with obvious alterations can be assigned to the statement. A function changes sign from negative to positive as x passes through ‘c’. It is clear that if a continuous function f (x) changes sign as x passes through c’ we must have f (c) = 0. Ex. 1. Show that the function f (x) = (x + 2) (x – 1)2 (2x – 1) (x – 3) changes sign from positive to negative as x passes through 1/2 and from negative to positive as x passes through – 2 or 3; also show that it does not change sign as x passes through 1.

MAXIMA AND MINIMA

285

Ex.2. Show that the function f (x) = (2x + 3) (x + 4) (x – 2) (x – 1)3 changes sign from positive to negative as x passes through – 4 and 1 and from negative to positive as x passes through – 3/2 and 2.

9.3. Sufficient condition for extreme values Theorem. f (c) is an extreme value of f if and only if f ' (x) changes sign as x passes through c. The extreme value f (c) is a maximum value if the sign changes from positive to negative and minimum value in the contrary case. Case I. Let f ' (x) changes sign from positive to negative as x passes through c. In some left-handed neighbourhood ] c – , c [ of c, f ' (x) is positive and so f is strictly increasing implying that f (c) is the greatest of all the values of f in this neighbourhood. In some right-handed neighbourhood ] c, c +  [ of c, f ' (x) is negative and so f is strictly decreasing implying that f (c) is the greatest of all the values of f in this neighbourhood. Thus, f (c) is the greatest of all the values of f (x) when x  ] c – , c +  [, i.e., is the greatest of all the values of f in a certain complete neighbourhood of c and so, by definition, f (c) is a maximum value of f. Case II. Let f ' (x) change sign from negative to positive as x passes through c. It may be shown that f (c) is the least of all the values of f in a certain complete neighbourhood of c and so, by definition, f (c) is a minimum value of f. Case III. If f ' (x) does not change sign, i.e., has the same sign in a certain complete neighbourhood of c, then f (x) is either strictly increasing or decreasing throughout this neighbourhood implying that f (c) is not an extreme value of f. Note. Geometrically interpreted, the theorem states that the tangent to a curve at every point in a certain left-handed neighbourhood of the point P whose ordinate is a maximum (minimum) makes an acute angle (obtuse angle) and the tangent at any point in a certain right-handed neighbourhood of P makes an obtuse angle (or acute angle) with x-axis. In case, the tangent on either side of P makes an acute angle (or obtuse angle), the ordinate of P is neither a maximum nor a minimum. EXAMPLES Ex. 1. Examine the polynomial function given by 10x6 – 24x5 + 15x4 – 40x3 + 108 for maximum and minimum values. We have

f (x) = 10x6 – 24x5 + 15x4 – 40x3 + 108



f ' (x) = 60x5 – 120x4 + 60x3 – 120x2 = 60x2 (x3 – 2x2 + x – 2) = 60x2 (x2 + 1) (x – 2)

 f ' (x) = 0 for x = 0 and x = 2. Thus, we expect extreme values of the function for x = 0 and 2 only. Now, x < 0  f ' (x) < 0, 0 < x < 2  f ' (x) < 0, x > 2  f ' (x) > 0. Here f ' (x) does not change sign as x passes through 0 so that f ' (0) is neither a maximum nor a minimum value even through f ' (0) = 0. Also since f ' (x) changes sign from negative to positive as x passes through 2, therefore, f (2) = – 100 is a minimum values.

DIFFERENTIAL CALCULUS

286

The function has only extreme value, viz., that at 2. Ex. 2. If f ' (x) = (x – a)2n (x – b)2m + 1 where m and n are positive integers is the derivative of a function f. Then show that x = b gives a minimum but x = a is neither a maximum nor a minimum. We have

f ' (x) = (x – a)2n (x – b)2m + 1

Now

f ' (x) = 0  x = a, b

For x < b, (x – a)2n (x – b)2m + 1 < 0 and for x > b, (x – a)2n (x – b)2m + 1 > 0 Thus f ' changes sign from negative to positive as x passes through b as such has minimum at b. Again since 2n is an even integer (x – a)2n (x – b)2m + 1 does not change sign as x passes through a. This function has neither maximum nor minimum for x = a. EXERCISES 1. Find the extreme values of 5x6 + 18x5 + 15x4 – 10. 2. Show that the maximum and minimum values of (x + 1) (x + 4) (x – 1) (x – 4) are – 9 and – 1/9 respectively. 3. Show that 9x5 + 30x4 + 35x3 + 15x2 + 1 is maximum when x = – 2/3 and minimum when x = 0. Also find the greatest and least values in the intervals [– 2/3, 0] and [ – 2, 2]. 4. Show that x5 – 5x4 + 5x2 – 1 has a maximum value when x = 1 a minimum value when x = 3 and neither when x = 0. (Kumaon 2003) ANSWERS 1. max. – 8 ; min. – 10, – 26,

3.

55/27, 1 ; 1109, – 27.

9.4. Use of second order derivatives The derivatives of the first order only have so far been employed for determining and distinguishing between the extreme values of a function. As shown in the present article, the same thing can sometimes be done more conveniently by employing derivatives of second and higher orders. All along the discussion, it will be assumed that f possesses continuous derivatives of second order, that come in question, in some neighbourhood of the point c. Theorem 1. f (c) is a maximum value of the function f if f ' (c) = 0 and f '' (c) > 0. As f '' (c) is positive, there exists an interval [ c – , c + ], around, c for every point x of which the second derivative f '' (x) is positive. This implies that f ' (x) is strictly increasing in [ c – , c + ]. Also since f ' (c) = 0, it follows that f ' (x) < 0  x  [c – , c [ and f (x) > 0  x  ] c, c +  ]. Thus, f is strictly decreasing in [ c – , c ] and strictly increasing in [ c, c + ] so that f (c) is a minimum value of f (x).

MAXIMA AND MINIMA

287

Theorem 2. f (c) is a maximum value of the function f, if f ' (c) = 0 and f '' (c) < 0. As f '' (c) is negative, there exist an interval [c – , c + ] around c for every point x of which the second derivative is negative. Thus, f ' (x) is strictly decreasing in [ c – , c + ]. Also, f ' (c) = 0. It follows that f ' (x) > 0  x  [c – , c [, f ' (x) < 0  x  ] c, c + ]. Thus f is strictly increasing in [c – , c] and strictly decreasing in [c, c + ] so that f (c) is a maximum value of f (x). Note. The theorem, as stated above, ceases to be helpful if for some c, f ' (c) and f '' (c) are both zero. To prove for this deficiency, we need to consider higher order derivatives. It will be shown in Ch. 14 that if each of f ' (c), f '' (c), ..........., f n – 1 (c) is 0 and f n (c)  0, then (i) f (c) is not an extreme value if n is odd, (ii) f (c) is an extreme value if n is even. Also f (c) is a maximum or minimum value according as f n (c) > 0 or f n (c) > 0. EXAMPLES 1. Find the maximum and minimum values of the polynomial function f given by f (x) = 8x5 – 15x4 + 10x2. We have

f ' (x) = 40x4 – 60x3 + 20x = 20x (2x3 – 3x2 + 1) = 20x (x – 1)2 (2x + 1)



f ' (x) = 0 for x = 0, 1, – 1/2.

Again

f '' (x) = 160x3 – 180x2 + 20 = 20 (8x3 – 9x2 + 1)

Now

f ' (0) = 0 and f'' (0) = 20 > 0



f (0) = 0 is a minimum value. f (– 1/2) = 0 and f '' (– 1/2) = – 45 < 0

 While

f (– 1/2) = 21/61 is a maximum value. f ' (1) = 0 and f '' (1) = 0, we see that

as x passes through, 1, f ' (x) does not change sign. Thus f (1) is neither a maximum nor a minimum value. 2. Investigate for maximum and minimum values the function given by y = sin x + cos 2x. (Garhwal 1997, G.N.D.U. Amritsar 2004) dy  cos x – 2 sin 2 x  cos x – 4 sin x cos x dx dy  0 when cos x  0 or sin x  1/ 4.  dx We consider values of x between 0 and 2 only, for the given function is periodic with period 2. Now cos x = 0  x = /2 and 3/2 and sin x = 1/4  x = sin–1 1/4 and  – sin–1 1/4,

We have

DIFFERENTIAL CALCULUS

288

sin–1 1/4 lying between 0 and /2. d2y

Now

dx 2 d2y

 – sin x – 4 cos 2 x

  2 30 2 dx d2y x  3 / 2  2  5  0 dx d2y –1 x  sin 1/ 4  2  – sin x – 4 (1 – 2 sin 2 x )  – 15 / 4  0 dx d2y x   – sin –1 1/ 4  2  – 15 / 4  0 dx Therefore y is maximum for x = sin–1 1/4,  – sin–1 1/4 and is a minimum for x = /2, 3/2. x

Putting these values of x in sin x + cos 2x we see that 9/8, 9/8 are its two maximum values and 0, – 2 are its two minimum values. 3. Show that the maximum value of (1/x)x is (e)1/e. Let

y = (1/x)x



log y = – x log x



1 dy  – (1  log x) y dx



dy  – (1  log x ) (1/ x ) x dx dy  0  log x  1  0  x  e –1 dx

Again

d2y dx 2

x

–

1 1 2 x    (1  log x ) (1/ x ) x x

at x  e –1 ,

d2y 2

 – e (e)1/ e  0

dx  y has maximum for x = e–1 and the maximum value is (e)1/e.

4. Show that the function f defined by f (x) = xp (1 – x)q

xR

where p, q are positive integers has a maximum value for x  We have f (x) = xp (1 – x)q 

f ' (x) = pxp – 1 (1 – x)q – qxp (1 – x)q – 1 = xp – 1 (1 – x)q – 1 (p – x (p + q)) f ' (x) = 0  x = 0, 1, p/p + q

p for all p, q. pq

MAXIMA AND MINIMA

289

Again f '' (x) = (p – 1) xp – 2 (1 – x)q – 1 [p – x (p + q)] – (q – 1) xp – 1 (1 – x)q – 2 [p – x (p + q)] – (p + q) xp – 1 (1 – x)q – 1 p –1

q –1

 p   q  f  ( p / p  q )  – ( p  q )    p  q p  q     < 0 where p and q are integers. Thus the function has a max. value at x = p/p + q for all integers p and q and the max. value is pp qq / (p + q)p + q. 5. Show that the function f defined by f (x) = | x |p | x – 1 |q xR p q p + q has a maximum value p q / (p + q) , p, q being positive. We have

(– 1) p x p (1 – x)q , x0  0 , x 0   f ( x)   x p (1 – x) q , 0  x 1  1 , x 1   x p ( x – 1) q , x 1  p q Consider f (x) = x (1 – x) , 0 < x < 1 By previous example, f has a maximum value at p p x ,0 1 pq pq The max. value is

p p qq

. ( p  q) p  q 6. Find the maxima and minima of the function 1 1 sin 2x + sin 3x  x  [0 ,  ]. 2 3 1 1 f ( x )  sin x  sin 2 x  sin 3 x 2 3 f ' (x) = cos x + cos 2x + cos 3x f (x) = sin x +

We have 

= cos 2x + 2 cos 2x cos x = cos 2x (1 + 2 cos x)  2 2 3 , , , 4 4 3 4 f '' (x) = – sin x – 2 sin 2x – 3 sin 3x f  ( x)  0  x 

 – 4 f     – 2  0  f has a max. at /4. 4 2 3 2  2  f      0  f has a min. at 2 3  3 

DIFFERENTIAL CALCULUS

290

 3  f     2 – 2  4 

2  0  f has a max. at 3 / 4. n

7. If A denotes the arithmetic mean of the real numbers a1, a2, ......., an, show that

 ( x – ai )2 i 1

has a minimum at A. n

 ( x – ai )2

f ( x) 

Let

i 1

f  ( x) 



n

 2 ( x – ai ) i 1

= 2nx – (a1 + a2 ...... + an) = 2nx – 2nA where A 

a1  a2  ...  an n

f ' (x) = 0  x = A. Again

f '' (x) = 2n > 0

Thus the given sum has a minimum for x = A. 8. Investigate the points for maxima and minima, the function x

f ( x) 

3 2 2  2 (t – 1) (t – 2)  3 (t – 1) (t – 2)  dt. 1

f ' (x) = 2 (x – 1) (x – 2)3 + 3 (x – 1)2 (x – 2)2 = (x – 2)2 (x – 1) (5x – 7)  Now,

f ' (x) = 0 for f '' (x) = (x –

Substituting

x = 1, 2, 7/5.

2)2

(10x – 12) + (5x2 – 12x + 7). 2 (x – 2)

x = 1, 2, 7/5, we get 

f '' (1) = – 2 < 0 f '' (2) = 0 f '' (7/5) > 0 9. If a > b > 0 and f () 

f () 

maximum when x = 1

 2

We have

[By Leibnitz-rule]

minimum when x = 7/5. 2

(a – b ) cos  , then find the maximum value of f (). a – b sin 

a 2 – b2 a 2 – b2  , a sec  – b tan   ()

where  () = a sec  – b tan   f () is maximum and minimum as  () is minimum and maximum respectively. Now, ' () = a sec  tan  – b sec2  For maximum and minimum sec  (a tan  – b sec ) = 0 b ...(i) a '' () = sec  tan  (a tan  – b sec ) + (a sec2  – b sec  tan ) sec 



sin  

[as sec   0]

MAXIMA AND MINIMA

291



Putting

sin  

a  a sin 2  – 2b sin  cos3  b a

b2 b – 2b . 2 a a  (b / a )  cos3  a2 – b2 b   0  a  b  0 and sin   is  ve 3 a a cos    aa.

b . a b  f () is maximum when sin   . a 2 ( a – b2 ) (a 2 – b 2 ) · a [ f ()]max   b a – b ·  a

  () is minimum when sin  



(a 2 – b 2 ) 

EXERCISES 1. Find the extreme values of f (x) = 5x6 + 18x5 + 15x4 – 10. 2. Show that the maximum and minimum values of the function defined by (x + 1) (x + 4) / (x – 1) (x – 4) are – 9 and – 1/9 respectively. 3. Find the greatest and least values of the function f (x) = x4 – 8x3 + 22x2 – 24x + 1 in [0, 2], [2, 4] 4. Investigate the maximum and minimum values of the function given by (i) 2x3 – 15x2 + 36x + 10

(ii) 3x4 – 4x3 + 5.

5. Find the values of x for which the function defined by x6 – 6ax5 + 9a2 x4 + a6 has minimum values (a > 0). 6. Determine the values of x for which the function given by 12x5 – 45x4 + 40x3 + 6 attains (i) a maximum value (ii) a minimum value. 7. Find the maximum value of the function f (x) = (x – 1) (x – 2) (x – 3). 8. Find the maxima and minima as well as the greatest and the least values of the function f (x) = x3 – 12x2 + 45x in the interval [0, 7]. 9. Find for what values of x the following expression is a maximum or minimum : 2x3 – 21x2 + 36x – 20. 10. Find the extreme values of the expression x3/ (x4 + 1).

DIFFERENTIAL CALCULUS

292

11. Determine the value for which

x has a maximum value. 1  x tan x

(Kumaon 1997)

12. Find the maxima and minima of the radii vector of the curve c4



a2



b2

. (Garhwal 1999) r 2 sin 2  cos 2  13. Show that the expression (x + 1)2 / (x + 3)3 has a maximum value 2/27 and a minimum value 0.

14. Show that f (x) = x5 – 5x4 + 5x3 – 1 has a maximum value when x = 1, a minimum value when x = 3 and neither when x = 0. 15. Show that xx is a minimum for x = e–1. 16. Show that the maximum value of (x)1/x is e1/e.

(Calicut 2004)

17. Find the maximum value of (log x)/x in ] 0,  [. 18. Find the extreme value of

ax + 1

–

ax

(Bangalore 2005)

– x, (a > 1).

19. Find the maximum and minimum values of x + sin 2x in [0, 2 ]. 20. Find the values of x for which sin x – x cos x is a maximum or a minimum. 21. Find in [ – , ] the maximum and minimum values of x – sin 2 x 

1 sin 3 x. 3

1 . 3 23. Discuss the maxima and minima in the interval ] 0,  [ of

22. Show that sin x (1 + cos x) is maximum when x 

(i)

sin x 

(Kumaon 1998)

1 1 sin 2 x  sin 3 x, 2 3

1 1 cos 2 x  cos 3 x. 2 3 Also, obtain their greatest and the least values in the given interval.

(ii) cos x 

24. Find the minimum and maximum values of (i) sin x cos 2x (iii) sin x

(ii) a sec x + b cosec x, (0 < a < b). (iv) ex cos (x – a).

cos2x

ANSWERS 1. 3. 4. 5.

Max. – 8 ; min. – 10, – 26 Greatest is 0 and the least is – 8 in [0, 2] ; the greatest is 1 and the least is – 8 in [2, 4]. (i) Max. 38; min. 37 (ii) min. 4. Max. for x = 2a and min. for x = 0 and x = 3a.

6. Max. for x = 1, min. for x = 2. 8. Max. 54, min. 50; least 0, greatest 70.

7.

10. Max. for x  4 3 ; min. for x  – 4 3 ;

9. Max. for x = 1 ; min. for x = 6. 11. x = cos x;

12.

rmax 

2 3 / 9.

c

2

( a  b) 2

MAXIMA AND MINIMA

293

17. Min. value : log [(ae – e) log a]/ log a. 18. Max. values : (2  3 3) / 6 ; (8  3 3) / 6. Min. values : (4 – 3 3) / 6 ; (10 – 3 3) / 6. 19. e–1. 20. Max. for x = n where n is an odd positive or even negative integer, and minimum for x = n where n is an even positive or odd negative integer. 1 1 (5  3 3  2), (2 – 3 3  2) 6 6 1 1 Max. value; (–   3 3 – 2), (5  3 3  2). 6 6 4 2 3 3 . ; max. values 23. (i) Min. value 6 4 4 2 3 . Least value 0; greatest value 6 1 5 11 5 (ii) Min. values : – , – ; max . values : , – . 2 6 6 12 5 11 Least value : – ; greatest value : . 6 6

21. Min. values : –

24. (i) (ii)

Min. – 1, – 2 / 3 6; max. 1, 2 / 3 6 . Min (a2/3 + b2/3)3/2 ; max. – (a2/3 + b2/3)3/2.

(iii) Min. 0, – 2 / 3 3 ; max. 0, 2 / 3 3 . (iv) where n is any integer.

9.5. Application to Problems. In the following, we shall apply the theory of maxima and minima to solve problems involving the use of the same. It will be seen that, in general, we shall not need to find the second derivative and complete decision would be made at the stage of the first derivative only when we have obtained the stationary values. In this connection, it will be found useful to determine the limits between which the independent variable lies i.e., the domain of the independent values. Suppose that these limits are a, b. If y is 0 for x = a as well as for x = b and positive otherwise and has only one stationary value, then the stationary value is necessarily the maximum and the greatest. If y   as x  a and x  b and has only one stationary value, then the stationary value is necessarily the minimum and the least. In connection with the problems concerning spheres, cones and cylinders, the following results would be often needed : 1. Sphere of Radius r. 4 3 Surface = 4 r2. r , 3 2. Circular cylinder of height h and radius of base r. Volume 

Volume = r2h,

Curved surface = 2 rh.

The area of each plane face = r2.

DIFFERENTIAL CALCULUS

294

3. Circular Cone of height, h and radius of base, r. Semi-vertical angle = tan–1 (r/h), Slant height  (r 2  h 2 ). 1 rh, , Curved surface  r ( r 2  h 2 ). 3 Here cones and cylinders are always supposed to be right circular.

Volume 

EXAMPLES 1. Show that the height of an open cylinder of given surface and greatest volume is equal to the radus of its base. (G.N.D.U. Amritsar 2004) Let r be the radius of the circular base; h, the height; S the surface and V, the volume of the open cylinder so that S = r2 + 2 rh, ...(i) 2 V = r h. ...(ii) Here, as given, S is a constant and V a variable. Also, h, r are variables. Substituting the values of h, as obtained from (i) in (ii), we get

 S – r 2  Sr – r 2 V  r 2  ,  2  2 r  which gives V in terms of one variable r. As V must be necessarily non-negative, we have

...(iii)

Sr – r 3  0  r 3  Sr  r  ( S / ) Also r is non-negative. Thus r varies in the interval, [0, ( S / )]. Now

dV S – 3 r 2  , dr 2

so that dV/dr = 0 only when r  ( S / 3 ) : negative value of r being inadmissible. Thus V has only one stationary value. Now V = 0 for the end points r = 0 and r. Hence V is greatest for

( S / ) and positive for every other admissible value of

r  ( S / 3 ).

Substituting this value of r in (i), we get h

S – r 2 S –  ( S / 3 )  2 r 2  ( S / 3 )

3   S     .  S  3  Hence h = r for the cylinder of greatest volume and given surface. 2. Show that the radius of right circular cylinder of greatest curved surface which can be inscribed in a given cone is half that of the cone. Let r be the radius OA of the base and h, the height OV of the given cone. We inscribe in it a cylinder, the radius of whose base is OP = x, as shown in the figure. We note that x may take up any value between 0 and r so that x  [ 0, r]. 

2S 1 · 3 2

Fig. 9.4

MAXIMA AND MINIMA

295

To determine the height PL, of this cylinder, we have PL PA  OV OA



PL r – x  h r

h (r – x) . r If S be the curved surface of the cylinder,



PL 

we have

S  2  . OP . PL 

2 xh ( r – x) 2h  (rx – x 2 ), r r

dS 2h  ( r – 2 x )  0 for x  r / 2. dx r Thus S has only one stationary value. Now S is 0 for x = 0 as well as for x = r and is positive for values of x lying between 0 and r. Therefore S is greatest for x = r/2. 3. Find the surface of the right circular cylinder of greatest surface which can be inscribed in a sphere of radius r. We construct a cylinder as shown in the figure 9.5. OA is the radius of the base and CB is the height of this cylinder. Let  AOB = , so that  lies between 0 and . OA  cos  We have OB  OA = OB cos  = r cos . AB  sin  Also OB  AB = OB sin  = r sin . If S be the surface, we have S = 2 . OA2 + 2 . OA.BC = 2 r2 (cos2  + sin 2 ) ...(i)  dS/d  = 2 r2 (– 2 cos  sin  + 2 cos 2 ) = 2 r2 (2 cos 2  – sin 2 )



 dS/d  = 0 for 2 cos 2  – sin 2 = 0  tan 2  = 2

...(ii)

Fig. 9.5

DIFFERENTIAL CALCULUS

296

The reader may established that the equation tan 2  = 2 admits of only one value of   [0, /2] as its solution. Let1  [0, /2] be the root of tan 2  = 2. Now

tan 2 1 = 2  sin 2 1 = 2 / 5 and cos 2 1 = 1/ 5 .

From (i) we see that  = 0  S = 2 r2 

 S 0 2

 1  cos 2 1    1  S  2 r 2   sin 2 1  2  

r 2 (5  5 5) 5 which is greater than 3 r2. 

r 2 (5  5) is the required greatest surface. 5 4. Prove that the least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is 6r 3 .

Hence

We take one vertex A of the triangle at a distance x from the centre O that OA = x. Surely x > r. Let AO meet the circle at P. The two tangents from A and the tangent at P determine an isosceles triangle ABC circumscribing the given circle. We have OL = r.  Also

AL  (OA2 – OL2 )  ( x 2 – r 2 ). BP = AP tan  BAP = AP tan  LAO  AP .

OL  (r  x) AL

r 2

(x – r2 )

.

Fig. 9.6

If, p, denotes the perimeter of the triangle, we have p = AB + AC + BC = 2 AB + 2 BP = 2 (AL + LB) + 2 BP = 2 AL + 4 BP  2 (x2 – r 2 ) 

4 (r  x ) r 2

2

(x – r )

2

(for BL = BP) (x  r)

2

( x2 – r 2 )

...(i)

MAXIMA AND MINIMA



297

2 ( x  r ) ( x 2 – r 2 ) – x ( x  r 2 ) ( x 2 – r 2 ) –1/ 2 dp 2 dx ( x 2 – r 2 )3/ 2 2

( x  r ) 2 ( x – 2r )

( x 2 – r 2 )3/ 2  dp/dx = 0 for x = 2r; negative value, – r, of x being inadmissible. Now, x may take up any value > r only so that it varies in the interval ] r,  [. From (i), we see that x  r  p  . Also x    p  .

Hence p is least for x = 2r. Putting this value of x in (i), we see that the least value of p is 6r 3 . 5. A cone is circumscribed to a sphere of radius r; show that when the volume of the cone is 1 least its altitude is 4r and its semi-vertical angle is sin–1 . 3 We take the vertex A of the cone at a distance x from the centre O of the sphere. (See Fig. 9.6 page 299). By drawing tangent lines from A, as shown in the figure, we construct the cone circumscribing the sphere, Let the semi-vertical angle BAP of the cone be . Now, if v be the volume of the cone, we have 1  . BP 2 . AP. 3 which will now be expressed in terms of x. v

We have Also Again 

AP = AO + OP = r + x. sin  

r ( x2 – r 2 )

BP  tan  AP BP = AP tan   ( r  x ).

v

Thus 

OL r   tan   OA x

(r  x )2 r 2 3 ( x2 – r 2 )

r (x – r2 )

(x  r) 

2

 r 2 ( x  r )2 . 3 (x – r)

dv  r 2 ( x  r ) ( x – 3r )  dx 3 ( x – r )2

 dv /dx is 0 for x = 3r. Here x can take up any value  r. Also x  r 

v   and x    v  .

Thus v is least for x = 3r. Hence, for least volume, the altitude of the cone = AP = r + 3r = 4r –1 and the semi-vertical angle   sin

r r 1  sin –1  sin –1 . x 3r 3

DIFFERENTIAL CALCULUS

298

6. Normal is drawn at a variable point P of an ellipse x2/a2 + y2 /b2 = 1 ; find the maximum distance of the normal from the centre of the ellipse. We take any point P (a cos , b sin ) on the ellipse;  being the eccentric angle of the point. Because of the symmetry of the ellipse about the two coordinate axes, it is enough to consider only those of the values of  which lie between 0 and /2. x cos  y sin    1. a b Hence the slope of the normal at P = a sin /b cos .

The equation of the tangent at P is

Therefore equation of the normal at P is ax sin  – by cos  = (a2 – b2) sin  cos . If, ‘P’ be its perpendicular distance from the centre (0, 0), we obtain p



(a 2 – b 2 ) sin  cos  ( a 2 sin 2   b 2 cos 2 )

.

...(i)

dp b 2 cos 4  – a 2 sin 4   (a 2 – b 2 ) 2 d (a sin 2   b 2 cos 2 )3/ 2

dp b2  0 for tan 4   2 i.e., for tan    b d a a Now, p = 0 when  = 0 or /2 and p is positive when  lies between 0 and /2. Therefore p is maximum when tan  = (b / a ). Substituting this value in (i), we see that the maximum value of p is a – b. 

7. Assuming that the petrol burnt (per hour) in driving a motor boat varies as the cube of its 3 velocity. Show that the most economical speed when going against a current of c miles per hour is 2 c miles per hour. Let v miles per hour be the velocity of the boat so that v – c miles per hour is its velocity relative to water going against the current. Therefore the time required to cover a distance of d miles = d/ (v – c) hours. The petrol burnt per hour = kv3, where k is a constant. Thus the total amount, y, of petrol burnt is given by

y  k.

  Also Thus v 

v2 d v3  kd v–c v–c

dy 3v 2 (v – c) – 1. v3  kd dv (v – c ) 2 dy 3  0 for v  0 and c, of these v = 0 is inadmissible. dv 2 v  c  y  , v    y  .

3 c gives the least value of c. 2

MAXIMA AND MINIMA

299

8. A lane runs at right angles to a road ‘a’ feet wide. Find how many feet wide the lane must be if it is just possible to carry a pole ‘b’ feet long (b > a) from the road into the lane, keeping it horizontal. Let AB be the pole and Q be the angle it makes with the road. Then AC = a cosec . Hence

BC = AB – AC = b – a cosec 

and

BD = BC cos  (length of lane) = (b – a cosec ) cos 

Fig. 9.7

= b cos  – a cot  For maxima or minima, d ( B D)  0, i.e., – b sin  + a cosec2  = 0 d 1/ 3 a a sin 3   ; sin     or b b Hence the required length = b cos  – a cot 

 b 1 – sin 2  – a b

b2 / 3 – a 2 / 3 b2 / 3

= (b2/3 – a2/3)

1 – sin 2  sin 

–a.

b2 / 3 – a2 / 3 a2 / 3

(b 2 / 3 – a 2 / 3 )

= (b2/3 – a2/3)3/2 9. A Person being in a boat ‘a’ miles from the nearest point of the beach wishes to reach as quickly as possible a point ‘b’ miles from that point along the shore. The ratio of his rate of walking to his rate of rowing is sec . Prove that they should land at a distance (b – a cot ) from the place to be reached. Let B is the boat and BC = breadth of the river as ‘a’ miles. Let the person lands, at a distance x from C, i.e., C P = x, P A = b – x and B P 

(a 2  x 2 )

Rate of walking  sec , Rate of rowing W  sec , where W = rate of walking and R = rate of rowing. i.e., R The person walked = b – x b– x t1   R sec 

and the boat rowed

(a 2  x 2 )

DIFFERENTIAL CALCULUS

300

(a 2  x 2 ) R (a 2  x 2 ) b–x   Total time T  R sec  R



t2 

or

T 

1 [(b – x) cos   R

(a 2  x 2 ) ]

Time is least, when dT 0 dx dT 1  1   – cos   ( a 2  x 2 ) –1/ 2 . 2 x  d x R  2  1  x   – cos   2 2  R  ( a  x )   x cos   (a 2  x 2 )

Now,

or

a2 + x2 = x2 sec2  a2 = x2 (sec2  – 1) = x2 tan2 

or 

x = a cot  d2T

Now,

d x2

 (a 2  x 2 ) – x ( a 2  x 2 ) –1/ 2 x      a2  x2 2 2 2 1 a  x – x    2  R  (a  x 2 )3/ 2  a2  R (a 2  x 2 )3/ 2



1 R

d 2T

is positive at x = a cot . Therefore, for x = a cot , time is minimum. dx 2 Hence the required distance = b – x = b – a cot  10. One corner of a long rectangular sheet of paper of width 1 foot is folded over so as to reach the opposite edge of the sheet. Find the minimum length of the crease. Let angle F E C' be . F C' is the edge which has fallen at points F and C'. Then FEC =  Hence

and  C' E D and E C' where l and then ED

Now

= = = = = = AB =

–2 E C = l cos  length of E D. E C' cos C' E D l cos  cos ( – 2 ) – l cos  cos 2  1' given.

Fig. 9.8

MAXIMA AND MINIMA

 Hence, or

301

D C = 1' DE+EC=1 – l cos  cos 2  + l cos  = 1

1  cos  (1 – cos 2 ) l 1 dl – 2  (– sin ) (1 – cos 2 )  cos  (sin 2 ) · 2 l d dl For maxima or minima, d   0  sin  (1 – cos 2 ) – 2 sin 2  cos  = 0



or or or

sin  – sin  cos 2  – 2 sin 2  cos  = 0 sin  – sin  + 2 sin3  – 4 sin  cos2  = 0 2 sin  (sin2  – 2 cos2 ) = 0

Either

sin  = 0

tan  = ±

or

2

2

d l

Then it can be easily shown that

d 2

is negative for tan  

minimum.

2, i.e.,

1 is maximum or l is l

1 1 Hence minimum value of l  cos  (1 – cos 2 )  2 sin 2  cos 

3 3 ft. 4 11. The range R of a shell (in empty space), fired with an initial velocity v0 from a gun inclined to the horizontal at an angle , is determined by the formula 

R

v02 sin 2  g

where g is the acceleration due to gravity. Determine the angle  for which the range will be maximum for a given initial velocity v0. The quantity R is a function of the variable angle . We have to test this function for a maximum  in the interval 0    , 2 2 d R 2 v0 cos 2   d g dR 0 For maxima or minima d

2 i.e. 2 v0 cos 2   0 or g   .  4

d2 R

Also,

d 2

–

cos 2  = 0

4 v02 sin 2  g

DIFFERENTIAL CALCULUS

302 2

2

 d R  – 4 v0  negative , 2 g 4 d  Hence, for the value   the function R has a maximum. 4 v02 Rmax  Then g

At  

x2

y2

 1 and the circle x2 + y2 = a2 at the points where a a2 b2 common ordinate cuts them. Show that if  be the greatest inclination of the tangents, then a–b tan   . 2 ab Let the common ordinate cut the ellipse and the circle in Q and P respectively. If the eccentric angle of Q be  , then from the properties of the ellipse  POM = . Thus the coordinates of Q and P are respectively (a cos , b sin ) and (a cos , a sin ).

12. Tangents are drawn to the ellipse



Tangent to the given ellipse at  is x y cos   sin   1 a b b cot   m1 (say) a Also the equation of the tangent to the given circle at P is

Hence gradient of this tangent  –

x cos  + y sin  = a 

Its gradient = – cot  = m2 (say)

If  be the angle between these tangents, then we have Fig. 9.9 b – cot   cot  a–b ...(i)  tan   a b   b cot  a tan 2 1  cot  a Now  will be maximum when tan  is maximum; i.e. when a tan  + b cot  is minimum. Now, let z = a tan  + b cot  

dz  a sec 2  – b cosec2  d d2 z

 2a sec2  tan   2b cosec2  cot  d 2 For a maximum or a minimum z, dz  0. d

and

This gives tan 2   i.e.

b a

b tan      a

MAXIMA AND MINIMA

303

b d2 z ,  positive, a d 2 because for this , all the trigonometrical ratios are positive.

when

tan  

d2 z b , then  negative   a d 2 Putting this value in (i) we have

When

tan  

b  , a

 is maximum when tan   – a–b

Than tan   –

b a b a

a b



a–b 2 ab

13. A window of fixed perimeter (including the base of the arc) is in the form of a rectangle surmounted by a semi-circle. The semi-circular portion is fitted with coloured glass while the rectangular portion is fitted with clear glass. The clear glass transmits three times as much light per square meter as the coloured glass does. What is the ratio of the sides of the rectangle so that the window transmits the maximum light ? Let 2b be the diameter of the circular portion and a be the lengths of the other side of the rectangle.  Total perimeter = 2a + 4b + b =  (say) ...(1) Let the light transmission rate of the coloured glass be L per square meter and Q be the total amount of transmitted light. Then Q  2ab . 3L 

1  b2 . L 2

L [  b 2  6b ( – 4b –  b)] 2 L  [6b – 24b 2 – 5b 2 ] 2 

 

and

dQ L  [6 – 48b – 10  b]  0 db 2 6 b 48  10  d2 Q db

2



L [– 48 – 10 ]  0 2

Thus, Q is maximum and from (1) and (2) (48 + 10 ) b = 6 = 6 [2a + 4b + b] 

The ratio 

2b 6  . a 6

Fig. 9.10

...(2)

DIFFERENTIAL CALCULUS

304

14. The circle x2 + y2 = 1 cuts the x-axis at A and B. Another circle with centre at B and variable radius intersects the first circle at C above the x-axis and the line segment AB at D. Find the maximum area of the triangle BDC.

Fig. 9.11

The centre of the circle x2 + y2 = 1

...(i)

is (0, 0) and coordinates of B are (1, 0). Let the radius of the variable circle be r. Then its equation is (x – 1)2 + y2 = r2

...(ii)

From (i) and (ii) on subtracting we get 2x – 1 = 1 – r2  x  1 –

r2  OT 2

 r2  CT  OC 2 – OT 2  1 –  1 –   2   Area of  BDC is,

2



1 · BD · CT 2  1 r4  A2  r 2  r 2 –  4  4  1  (4r 4 – r 6 ) 16 A



d ( A2 ) 1 [For maxima, minima]  (16r 3 – 6r 5 )  0 dr 16 2 r2 .  3 2 2 2 d (A ) 1 16 Also, [For r  2 ]  (48 r 2 – 30 r 4 )  – 0 2 3 16 3 dr 4 2 Hence, area is maximum at r  2 and Amax  sq. units. 3 3 3 15. Narendra has x children by his first wife. Pramila has (x + 1) children by her first husband. They marry and have children of their own. The whole family has 24 children. Assuming the two

MAXIMA AND MINIMA

305

children of same parents do not fight, then find the maximum possible number of fights that can take place. Since the whole family has 24 children, those of Narendra and Pramila are 24 – x – (x + 1) = 23 – 2x F = Total number of fights = (number of fights when a Narendra’s child fights a Pramila’s child) + (number of fights when a Narendra’s child fights a Narendra- Pramila child) + (number of fights when a Pramila’s child fights a Narendra- Pramila child) = x (x + 1) + x (23 – 2x) + (x + 1) (23 – 2x) = 23 + 45 x – 3x2 For maximum dF  45 – 6 x  0  x  7.5 dx

But in this case fractional value is not possible. The nearest integral values are x = 7 and x = 8. In either case the total number of fights = 23 + 45 × 7 – 3 × (7)2 = 191. 16. From point A located on a highway a man has to get by bus to his office B located in a lane at a distance l from the highway in the least possible time. At what distance from D should the bus leave the highway when the bus moves in the lane n times slower than on highway? Let AD = s and CD = x, where C is a point where the bus leaves the highway. Let the speed of the bus is v on highway.  Total time taken t

AC BC  v (v / n)

Fig. 9.12 2 2 s– x n l x  v v For extreme, differentiating w.r.t. x, we have



dt 1  n   1  (2 x )   0 2 2 dx v  2 l  x  

DIFFERENTIAL CALCULUS

306



l

n2 x2 = l2 + x2  x 

.

2

( n – 1) l

Thus t is minimum when x 

2

.

( n – 1)

17. Two men are walking on a path x3 + y3 = a3 when the first man arrives at a point (x1, y1), he finds the second man in the direction of his own instantaneous motion. If the coordinates of the second man are (x2, y2) then show that x2 y  2  1  0. x1 y1 Since (x1, y1) and (x2, y2) lie on the curve. x 13 + y 13 = a 3 ...(i) and x 23 + y 23 = a 3 ...(ii) Subtracting (x23 – x13) = – (y23 – y13) ...(iii) Differentiating x3 + y3 = a3 w.r.t. x., we get 3x 2  3 y 2

dy 0 dx

Slope of tangent at ( x1 , y1 )  –

x12

y12  Equation of tangent at (x1, y1) : x12

y – y1  –

y12

( x – x1 )

It passes through (x2, y2)  or

x12

y2 – y1  –

y12

( x2 – x1 )

x12 (x2 – x1) = – y12 (y2 – y1)

...(iv)

Dividing (iii) and (iv), we get x23 – x13 x12 ( x2 – x1 )





– ( y23 – y13 ) – y12 ( y2 – y1 )

x2 2  x12  x1 x2 x12 2





y2 2  y12  y1 y2

2

y12

 x2   y2   y2   x2  x  –y  y –x   1  1  1  1

y 2   x2 y2  x2    x – y   x  y  1  0 1   1 1  1 

MAXIMA AND MINIMA

i.e., But, 

307

x2 y2 x2 y2 either x  y or x  y  1  0 1 1 1 1 x2 y  2 x1 y1 x2 y  2  1  0. x1 y1

EXERCISES 1. Divide a number 15 into two parts such that the square of one multiplied with the cube of the other is a maximum. (G.N.D.U Amritsar 2004) 2. Show that of all rectangles of given area, the square has the smallest perimeter. 3. Find the rectangle of greatest perimeter which can be inscribed in a circle of radius a. (Banglore 2005) 4. If 40 square feet of sheet metal are to be used in the construction of an open tank with a square base, find the dimensions so that the capacity is greatest possible. 5. A figure consists of a semi-circle with a rectangle on its diameter. Given that the perimeter of the figure is 20 feet, find its dimensions in order that its area may be maximum. 6. A, B are fixed points with co-ordinates (0, a) and (0, b) and P is a variable point (x, 0) referred to rectangular axes; prove that x2 = ab when the angle APB is a maximum. 7. A given quantity of metal is to be cast into a half cylinder, i.e., with a rectangular base and semi-circular ends. Show that in order that the total surface area may be minimum the ratio of the length of the cylinder to the diameter of its circular ends is / ( + 2). 8. The sum of the surface of a cube and a sphere is given; show that when the sum of their volumes is least, the diameter of the sphere is equal to the edge of the cube. 9. The strength of a beam varies as the product of its breadth and the square of its depth. Find the dimensions of the strongest beam that can be cut from a circular log of wood of radius a units. 10. The amount of fuel consumed per hour by a certain steamer varies as the cube of its speed. 1 When the speed is 15 miles per hour, the fuel consumed is 4 tons of coal per hour at Rs. 4 2 per ton. The other expenses total Rs. 100 per hour. Find the most economical speed and the cost of a voyage of 1980 miles. 11. Show that the semi-vertical angle of the cone of maximum volume and of given slant height is tan–1 2 . 12. Show that the right circular cylinder of the given surface and maximum volume is such that its height is equal to the diameter of its base. 13. Show that the height of a closed cylinder of given volume and least surface is equal to its diameter. 14. Given the total surface of the right circular cone, show that when the volume of the cone is maximum, then the semi-vertical angle will be sin–1 1/3. 15. Show that the right cone of least curved surface and given volume has an altitude equal to 2 times the radius of its base. (Kumaon 1995) 16. Show that the curved surface of a right circular cylinder of greatest curved surface which can be inscribed in a sphere in one half of that of the sphere.

308

DIFFERENTIAL CALCULUS

17. A cone is inscribed in a sphere of radius r, prove that its volume as well as its curved surface is greatest when its altitude is 4r/3. 18. Find the volume of the greatest cylinder that can be inscribed in a cone of height h and semivertical angle . (Garhwal 2003) 19. Prove that the area of the triangle formed by the tangent at any point of the ellipse x2 /a2 + y2/ b2 = 1 and its axes is a minimum for the point (a / 2, b / 2). 20. Find the area of the greatest isosceles triangle that can be inscribed in a given ellipse, the triangle having its vertex coincident with one extremity of the major axis. 21. A perpendicular is let fall from the centre to a tangent to an ellipse. Find the greatest volume of the intercept between the point of contact and the foot of the perpendicular. 22. A tangent to an ellipse meets the axes in P and Q : show that the least value of PQ is equal to the sum of the semi-axes of the ellipse and also that PQ is divided at the point of contact in the ratio of its semi-axes. 23. N is the foot of the perpendicular drawn from the centre O on to the tangent at a variable point P on the ellipse x2/ a2 + y2/ b2 =1 (a > b). Prove that the maximum area of the triangle OPN is (a2 – b2)/4. 24. One corner of a rectangular sheet of the paper, width one foot, is folded over so as just reach the opposite edge of the sheet; find the minimum length of the crease. 25. A grocer requires cylindrical vessels of thin metal with lids, each to contain exactly a given volume V. Show that if he wishes to be as economical as possible in metal, the radius r of the base is given by 2 r3 = V. If, for other reasons, it is impracticable to use vessels in which the diameter exceeds, threefourths of the height, what should be the radius of the base of each vessel ? 26. A tree trunk, l feet long is in the shape of a frustum of a cone the radii of its ends being a and b feet (a > b). It is required to cut from it a beam of uniform square section. Prove that the beam of the greatest volume that can be cut is al/3 (a – b) feet long. 27. Find the volume of the greatest right circular cone that can be described by the revolution about a side of a right-angled triangle of hypotenuse 1 foot. 28. A rectangular sheet of metal has four equal square portions removed at the corners, and the sides are then turned up so as to form an open rectangular box. Show that when volume contained in the box is a maximum, the depth will be 1  2 2 1/ 2  (a  b) – (a – ab  b )  , 4 where a, b are the sides of the original rectangle. 29. The parcel post regulations restrict parcels to be such that the length plus the girth must not 1 exceed 6 feet and the length must not exceed 3 feet. Determine the parcels of greatest 2 volume that can be sent up by post if the form of the parcel be a right circular cylinder. Will the 3 result be affected if the greatest length permitted were only 1 feet ? 4 30. Show that the maximum rectangle inscribed in a circle is a square. 31. Find the point on the curve 4x2 + a2 y2 = 4a2; 4 < a2 < 8 that is farthest from the point (0, – 2). 32. A despatch rider is in open country at a distance of 6 km from the nearest point P of a straight road. He wishes to proceed as quickly as possible to a point Q on the road 20 km from P. If his maximum speed across country is 40 km/hr, than at what distance from P, he should strike the road?

MAXIMA AND MINIMA

309

33. L L' be latus-rectum of the parabola y2 = 4ax and PP' is a double ordinate between the vertex and the latus-rectum. Show that the area of trapezium PP' L' L is maximum when the distance of PP' from the vertex is a/9. 34. A small disc A slides with zero initial velocity from the top of a smooth hill of height H having a horizontal portion. Determine the ratio h/H so that R is maximum. 35. Two corridors of widths a and b are at right angles. Show that the length of the longest pipe that can be passed round the corner horizontally is (a2/3 + b2/3)3/2. ANSWERS 1. 9, 6. 4.

3. Square with side

(40 / 3),

(40 / 3),

1 2

2a.

(40 / 3).

5. Diameter of the semi-circle = 40/ ( + 4). Height of the rectangle = 20/ ( + 4). 8 4 a. 9. Breadth a, depth 3 3 2/3

2/3

8 3 10. 15   miles per hour, 19800   rupees. 3 5 4  h3 tan 2  . 18. 20. 3 3 ab / 4. 21. a – b. 27

24.

3 3 / 4.

25. (3v/ 8)1/3.

27.

3 3  / 27.

3 1 29. Length 2 ft., girth 4 ft.; yes, length should now be 1 ft. and girth 4 ft. 4 4 31. (0, 2) 32. 8 km 34. 1/2.

Use of third and higher order derivatives for the determination of maximum and minimum values. In, § 9.4, we showed that if f ' (c) = 0 and f '' (c)  0, then f '' (c) > 0  f (c) is a minimum value of f (x), f '' (c) < 0  f (c) is a maximum value of f (x). This result does not refer to what happens if f '' (c) = 0. We now make use of the Higher Mean Value Theorem to obtain a generalisation of this result. As shown in the following article the same thing can sometimes be done more conveniently by employing derivation of the second and higher orders. It will be assumed that f possesses continuous derivatives of every order that appear in the discussion in some neighbourhood of c. We suppose that c + h belongs to this neighbourhood. We suppose that 0 = f '' (c) = f ''' (c) = ...... = f n – 1 (c) and f n (c)  0. We now have

f (c  h)  f (c)  hf  (c) 

h2 f  (c)  ..... 2!  ....... 

h n –1 hn n f n –1 (c)  f (c   h ) (n – 1) ! n!

DIFFERENTIAL CALCULUS

310

so that under the conditions as stated, we have

f (c  h) – f (c ) 

hn n f (c   h), 0    1. n!

Let f n (c) > 0. There exists a neighbourhood ] c – , c +  [ of c for every point x of which f (x) > 0. It follows that if n is even, then for every point c + h of this neighbourhood n

f (c + h) – f (c) > 0 and as such f (c) is a minimum value of the function. Also if n is odd, and

f (c + h) – f (c) < 0

when

c + h  [ c – , c [

f (c + h) – f (c) > 0

when

c + h  ] c, c +  [

and it follows that in this case f (c) is not an extreme value of the function. Let f n (c) < 0. As above it may be shown that f (c) is a maximum value if n is even and not an extreme value if n is odd. EXAMPLE 1. Find the maximum and minimum values of the function f (x) = 8x5 – 15x4 + 10x3. We have

f ' (x) = 40x4 – 60x3 + 30x2 = 20x (x – 1)2 (2x + 1) f ' (x) = 0 for x = 0, 1, – 1/2

We have already discussed the cases when x = 0, – 1/2 for

x = 1, f '' (1) = 0. f ''' (x) = 20 (16x2 – 18x) f ''' (1)  0

 For x = 1 function has neither maximum or minimum.

9.6 Maxima and minima of functions of two variables Let f (x, y) be a function of two independent variables x, y such that it is continuous and finite for all values of x and y in the neighbourhood of their values a and b (say) respectively. Then the value of f (a, b) is called maximum or minimum value of f (x, y) according as f (a + h, b + k) , < or > f (a, b) whatever be the relative values of h and k, positive or negative, provided h and k are finite and sufficienlty small. 9.6.1 Condition for the existence of Maxima or Minima We know by Taylor’s expansion in two variables, that f   f f ( x  h, y  k )  f ( x, y )   h k  y   x 

2 1  2 2 f 2 f 2  f   2 hk  k  h   .... 2 !   x2  x y  y 2 

or f (x + h, y + k) – f (x, y) f   f  h k  terms of second and higher orders  x  y  

...(i)

MAXIMA AND MINIMA

311

Now, from the definition of maxima or minima, it is evident that there shall be a maximum or a minimum for these values of x and y for which the expression f (x + h, y + k) – f (x, y) for all finite by sufficiently small positive or negative values of h and k is of invariable sign. Now if h and k sufficiently small, then the sign of the right hand of (i) will depend on the first degree terms in h, k and therefore the sign of left hand side of (i) will also depend on the first degree terms in h, k on the right hand side of (i). Thus if the sign of both h and k be changed then the sign of the left hand side of (i) will also be changed. Hence the necessary condition for a maximum or minimum values is f f h k 0 ...(ii) x y As (ii) is true for all values of h and k, so we have f f 0, 0 ...(iii) (Sagar 2000; Jabalpur 2002) x y as the necessary condition for a maximum or a minimum value of f (x, y). Note that the above conditions are not sufficient for the existence of a maximum or a minimum value of f (x, y). 9.6.2 Stationary and extreme Points A point (a, b) is called a stationary point if all the first order partial derivatives of the function f (x, y) vanish at that point. A stationary point, if it is a maximum or a minimum is known as an extreme point and the value of the function at an extreme point is known as an extreme value. From above we conclude that a stationary point may be a maximum or a minimum or neither of these two. 9.6.3 Sign of quadratic expressions : Algebraic Lemma We know that 1 2 2 [ a x  2ahxy  aby 2 ] a 1  [( ax  by ) 2  ( ab – h 2 ) y 2 ] ...(i) a If ab – h2 > 0, then from (i) the sign of expression ax2 + 2hxy + by2 will be the same as that of a. ax 2  2hxy  by 2 

If ab – h2 > 0, then nothing can be said about the sign of expression in the square bracket on the right hand of (i) and hence that of the quadratic expression ax2 + 2hxy + by2. 9.6.4 Lagrange’s condition for maximum and minimum values of a function of two variables. (Gorakhpur 2002) If r, s, t denote the values of

2 f  x2

,

2 f 2 f , when x = a, y = b then supposing that the  x  y  y2

necessary conditions for the maximum and minimum are satisfied, i.e.,  f /  x = 0,  f /  y = 0, when x = a, y = b, we can write 1 [ r h 2  2 shk  t k 2 ]  R 2! Where R consists of terms of higher orders of h and k.

f (a + h, b + k) – f (a, b) 

...(i)

Now if h and k be sufficiently small, then sign of the right hand side of (i) depend on the second degree terms on the right hand side of (i) and therefore the sign of the left hand side of (i) depends on

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312

the second degree terms on the right hand side of (i). If for all positive values of h, k this sign is positive then we have a minimum but if it is negative then we have a maximum at (a, b). Now, by Algebraic Lemma of § 9.6.3 the second degree terms rh2 + 2hsk + tk2 shall always be positive if rt – s2 is positive and r is positive. Hence Lagrange’s Condition for minimum is rt – s2 > 0 and r > 0 Also the second degree term rh2 + 2hsk + tk2 in (i) shall always be negative if rt – s2 is positive and r is negative. Hence Lagrange’s Condition for maximum is rt – s2 > 0 and r < 0 But if rt – s2 < 0 ; then there is neither a maximum nor a minimum. If rt – s 2 = 0, then from (i) we have the quadratic expression rh2 + 2hsk + tk 2 1  (rh  ks ) 2 and therefore of the same sign as r. Hence in this case the maximum or minimum r value of the function will depend on the sign of r. h s  –   (say) , then second degree terms of right hand side of (i) vanish k r and we shall have to consider higher degree terms on the right hand side of (i). Now the third degree h terms must collectively vanish when  , otherwise by changing the sign of both h and k, the sign k of the expression f (a + h, b + k) – f (a, b) can be changed and in this case the fourth degree terms must collectively be of the same sign as r and t when h/k = .

If rt – s2 = 0 and

EXAMPLES 1. Discuss the maximum or minimum values of u given by u = x3 y2 (1 – x – y). (Gorakhpur 2000, 2002 ; Kumaon 2000, 2002 ; Sagar 1997 ; Bhopal 96, Jiwaji 2000, 2002 ; Jabalpur 96 ; Bilaspur 98 ; Indore 2001) We have

u  3 x 2 y 2 (1 – x – y )  x 3 y 2 (– 1) x

= 3 x 2 y 2 – 4 x 3 y 2 – 3 x 2 y 3, u  2 x3 y (1 – x – y )  x 3 y 3 (– 1) y

= 2 x3 y – 2 x4 y – 3 x3 y2 r

t

and

s

2 u  x2 2 u y

2

 6 xy 2 – 12 x 2 y 2 – 6 x y 3 ,

 2 x3 – 2 x 4 – 6 x3 y

2 u  6 x 2 y – 8 x3 y – 9 x 2 y 2  x y

Now for maximum or minimum we must have u u  0 and 0 x y

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from  u  0 , we get x 3x2 y2 – 4 x3 y2 – 3 x2 y3 = 0 x2 y2

or

...(i)

(3 – 4x – 3y) = 0,

we get x = 0, y = 0, 4x + 3y = 3

Hence

u  0 , we get y 2 x3 y – 2 x4 y – 3 x3 y2 = 0

Also from

or

x3 y ( 2 – 2 x – 3 y) = 0

Hence we get x = 0, y = 0, 2x + 3 y = 2 4 x + 3 y = 3 and 2 x + 3 y = 2 1 1 x , y we get 2 3 Hence the solutions are Solving

1 1 1 1 , y  , x  , y = 0 and x = 0 y  . 3 2 3 3

x = 0, y = 0, x  when

x

1 1 , y  we have 2 3 3

2

2

3

1 1 1 1 1 1 1 r  6     – 12     – 6      –  2 3  2 3  2 3 4 2

3

2

3

–1 1 1 1 1 1 1 s6   –8   –9      2 3  2 3  2 3 12 3

3

3

1 1 1 1 t  2   – 2   – 6     = – 2/27 2  2 3 3

From these we have rt – s2 > 0 and r < 0, so there is maximum at x 

1 1 , y . 2 3

Similarly we can discuss other solutions. 2. Show that minimum value of u = x y + (a3 / x) + (a3 / y) is 3 a2. (Indore 1996 ; Jabalpur 2002 ; Kanpur 2001; Kumaon 2004) We have

u a3  u a3  y– 2 ; x– 2, x y x y r

t

2 u  x2 2 u  y2





2a 3

;

x3 2a 3 y3

.

s

2 u 1 x y

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Now for maximum or minimum we must have u u  0, 0 x y

So from we get or and from or

u 0 x a3 y– 2 0 x 2 x y = a3

...(i)

a3 u  0 , we have x – 2  0 y y x y2 = a 3

...(ii)

Solving (i) and (ii) we find that x2 y = x y2 or x y (x – y) = 0 or

x = 0, y = 0 and x = y

From (i) and (ii) we see that x = 0 and y = 0 do not hold as it gives a = 0, which is inadmissible. Hence we have x = y and from (i) we get x = a and therefore we have x = y = a. At x = y = a, we have r2

 Also

a3

 2, s  1, t  2 a3 r t – s2 = (2) (2) – 12 = 3 > 0.

r = 2 > 0.

Hence there is a minima at x = y = a. 3 3 Hence the minimum value of u  a  a a  a  a 2  a 2  a 2  3 a 2 . a a 3. Discuss the maximum and minimum values of 1 1 2 sin ( x  y ) cos ( x – y )  cos ( x  y ). 2 2 or

u = sin x + sin y + sin ( x + y) (Bilaspur 2001; Vikram 2001; Jabalpur 98 ; Bhopal 1998, 2000; Indore 2000 ; Kanpur 2002; MDU Rohtak 2001) Let



1 1 ( x  y ) cos ( x – y )  cos ( x  y ) 2 2 = sin x + sin y + cos (x + y)

u  2 sin

u u  cos x – sin ( x  y );  cos y – sin ( x  y ); x y r

2 u  x2

 – sin x – cos ( x  y ); s 

2 u  – cos ( x  y ); x y

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315

t

2 u  y2

 – sin y – cos ( x  y )

For maximum or minimum we must have

u u  0,  0. x y

From

u  0, x

we get cos x – sin (x + y) = 0

...(i)

From

u  0, y

we get cos y – sin (x + y ) = 0

...(ii)

Solving (i) and (ii) we get cos x = cos y This gives x = 2 n  ± y, where n is any interger. In particular x = y. When x = y, from (i) we get

cos x – sin 2 x = 0

This gives

cos x = 0, sin x 

or

cos x (1 – 2 sin x) = 0.

1 . 2

1  , then x  n   (– 1) n ·  y 2 6 And for these values of x and y we have

If sin x 

 1  – cos  2 n   (– 1) n ·   0 2 3  Similarly, t < 0 and s < 0 and r > s, t > s. r–



r t – s2 > 0, r < 0.

Hence there is a maximum when x  n   (– 1) n · If cos x = 0, then x  2 n   From here we get x  y   If

1  y. 6

1  y 2

1 3 5 , ,  etc. 2 2 2

  y then r = – 1 + 1 = 0 2 s = 1, t = 0. x

 r t – s2 < 0. Hence there is neither maximum nor minimum at x  If 

  y. 2

  y , then r = 1 + 1 = 2 = t, s = 1 2 r t – s2 = (2 × 2) – 1 = 3 > 0. Also r > 0. x–

Hence there is a minimum at x  –

  y . In a similar way we may discuss other values too. 2

4. Find the maximum value of u where u

xyz . ( a  x ) ( x  y ) ( y  z ) ( z  b)

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316

We have log u = log x + log y + log z – log (a + x) – log (x + y) – log (y + z) – log (z + b)

1 u 1 1 1 ay – x 2  – –  u x x a  x x  y x (a  x) ( x  y )

Hence, or

u (ay – x 2 ) u  x x (a  x) ( x  y )

Similarly

u ( xz – y 2 ) u  y y ( x  y ) ( y  z )

and

u (by – z 2 ) u  z z ( y  z ) ( z  b)

Hence for maxima or minima of u, we have ay – x2 = 0, xz – y2 = 0 and by – z2 = 0 From the above equation, it follows that a, x, y, z, b are in geometrical progression. Let r be the common ratio, then 1/ 4

b ar 4  b or r    a Also x = ar, y = ar2, z = ar3

Substituting these values, we get u 



ar . ar 2 . ar 3 a (1  r ) . ar (1  r ) . ar 2 (1  r ) . ar 3 (a  r ) 1 a (1  r )

4



1 1/ 4  b  a 1     a  

1 1/ 4

(a

 b1/ 4 ) 4

4

.

To decide whether this value of u is a maximum or minimum, we have

 2u x

2

Therefore or,

–

2 xu   u   (ay – x 2 )  x (a  x) ( x  y ) x  x (a  x) ( x  y )  A–

2u 3

a r (1  r ) 2 A = – ve quantity

, since x = ar, y = ar2, z = ar3

Similarly B, C, ... can be found and we shall finally arrive at a result that this value of u is a maximum. 5. Find a point within a triangle such that the sum of the square of its distances from the three vertices is a minimum. (Ravishankar 1995; Rewa 98, Garhwal 2005) Let (xr, yr), r = 1, 2, 3, be the vertices of the triangle and (x, y) be any point inside the triangle. 3

Let

u

 [( x – xr )2  ( y – yr )2 ] r 1

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317

For maximum or minimum of u, we have

i.e., or Similarly

Now

u   2 ( x – xr )  0 x (x – x1) + (x – x2) + (x – x3) = 0 x1  x2  x3 3

x

u  0 will give y y  y2  y3 y 1 3

r

 2u x

 6, s 

2

 2u

t

 2u  0, y x

 6, so that rt – s2 = 36 = + ve. Also r is + ve.

2

y Hence u is minimum when

x1  x2  x3 y  y2  y3 , y 1 3 3

x

 x1  x2  x3 y1  y2  y3  , Thus the required point is  . 3 3   6. Show that the point such that the sum of the squares of its distances from n given points shall be minimum, is the centre of the mean position of given points. Let the n given points be (a1, b1, c1), (a2, b2, c2), (a3, b3, c3),... etc. and let (x, y, z) be the coordinates of the required point. Then u =  [ (x – a1)2 + (y – b1)2 + (z – c1)2] where the summation extends to each of the n points. For a maximum or minimum of u, we have u  2  ( x – a1 )  2nx – 2  a1  0, x u  2  ( y – b1 )  2ny – 2  b1  0, y u  2  ( z – c1 )  2nz – 2  c1  0, z

hence

 a1  b1  c1 , y ,z . n n n

x

A

2u

 2n. x 2 Since A is positive here, u is a minimum when the point (x, y, z) is the centre of the mean position of the given points. As we can easily show that

Now,

A

H

H

B

A

H

G

, H

B

F

G

F

C

are + ve.

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318

7. Find the maximum value of ( ax  by  cz ) e – 

2 2

x –  2 y – z 2

u  ( ax  by  cz ) e – 

Let

2 2

.

2 2

x –  y – 2 z 2

log u = log (ax + by + cz) – (2 x2 + 2 y2 + 2 z2)

then

Differentiating partially w.r.t. x, we get 1 u a  – 2 2 x  0 u x ax  by  cz

Similarly 1 u b  – 22 y  0 u y ax  by  cz 1 u c  – 2  2  0. u z ax  by  cz

and These give

x ( ax  by  cz ) 

y ( ax  by  cz )  z (ax  by  cz ) 

a 2 2 b

...(1)

2 2 c

...(2)

2 2 Multiplying (1), (2), (3) by a, b, c and adding, we get (ax  by  cz ) 2 

or

1  a2 b2 c2  2  2  2 2  

  

 1  a 2 b 2 c 2     2  2  2        2   = A (say). a b c x , y 2 ,z 2 2 2 A 2 A 2 A

(ax  by  cz ) 

Then

2

1  2u 1  u  a2 –  – – 2 2   Again, u x 2 u 2  x  ( ax  by  cz ) 2    2u a2  – u  2 2   or 2 2 x  ( ax  by  cz )  u 0 since  Hence for these values of x, y, z, u will be maximum.

Maximum value of u is given by



 1   2

1  a 2 b2 c2    2  2   2   

 a2 b 2 c 2   – 4 A2  2  2  2   e     

...(3)

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319



 1   2

 a2 b 2 c 2   –1/ 2  2  2  2   e     



 1   2e

 a2 b 2 c 2    2  2  2       

EXERCISES 1. Discuss the maximum or minimum values of u, when (i) u = x3 + y3 – 3 a x y.

2. 3. 4. 5. 6. 7.

(ii) u = ax2y2 – x4y2 – x3y3.

(Sagar 1998; Ravishankar 2003; Kumaon 99; Indore 96, 98, 2002; MDU Rohtak 2000; Garhwal 2004) Discuss the maximum values of u, where u = 2a2 xy – 3ax2y – ay3 + x3y + xy3. Find the maxima or minima of u, when u = 2 xy – 3x2y – y3 + x3y + xy3. Discuss the maxima or minima of u = x4 + y4 – 2 x2 + 4xy – 2y2. Examine for maximum and minimum values of the function z = x2 – 3 xy + y2 + 2x. Discuss the maximum and minimum values of x4 + 2 x2y – x2 + 3 y2. Find the maximum and minimum values of the function z = sin x sin y sin (x + y).

8. Show that the distance of any point (x, y, z) on the plane 2x + 3y – z = 12, from the origin is given by l

x 2  y 2  (2 x  3 y – 12) 2 .

Hence find a point on the plane that is nearest to the origin. 9. Find a point within a triangle such that the sum of the distances from the angular points may be maximum. 10. Find the maximum and minimum ordinates of the curve u = ax y2z3 – x3y2z3 – xy3z3 – xy3z3 – xy2z4. 11. Show that the maxima and minima of the fraction ax 2  by 2  c  2 hxy  2 gx  2 fy a  x 2  b y 2  c  2h xy  2 g  x  2 f  y

are given by the roots of the equation

a – a u h – h u

h – h u b – b u

g – gu

f – f u

g – gu f – f  u  0. c – c u

12. ABCD is a quadrilateral having no re-entrant angle and P is a point in its plane. Find the position of P for which the sum of its distances from the vertices is a minimum. 13. If x, y, z are angles of a tirangle, then find the maximum value of sin x sin y sin z. (Bhopal 1998; Indore 2001) 14. Find three positive numbers whose sum is 30 and whose product is maximum. (Jiwaji 2003)

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320

ANSWERS 1. (i) No maxima or minima at (0, 0), at (a, a) maximum or minimum according as a < 0 or a > 0. (ii) u is maximum when x = a/2, y = a/3. a a 3 a a a 3 a 2. Maximum at  ,  ,  a, –  ; minima at  , –  ,  a,  , no maxima or minima 2 2 2 2 2 2 2 2 at (a, a), (a, – a). 1 3 1 1 1 3 1 1 3. Maximum at  ,  ,  , –  ; minimum at  , –  ,  ,  ; no mixima or minima at 2 2 2 2 2 2 2 2 (1, 1), (1, – 1).

4. Minima when x  

2, y   2 .

5. Neither maxima nor minima at x 

4 6 , y . 5 5

1 3 – 3 –1 , y , y  – and min. when x  4 2 2 4  7. Max. at x  y  and min., at x = y = 0. 3 6  12 18 8.  , , –  7 7 7

6. Max. when x 

9. Point is such that each side subtends an angle of 120° there. 10. Max. value = 108 a7/77 12. P is the intersection of the diagonals of the quadrilateral. 14. 10, 10, 10

9.7 Lagrange’s Method of Undetermined Multipliers (for several independent variables) Let u =  (x1, x2, x3, .... xn) be a function of n variables x1, x2, x3, ..., xn. Let these variables be connected by following r equation f1 ( x1 , x2 , ...., xn )  0  f 2 ( x1 , x2 , ..., xn )  0   .... .... .... ....  f r ( x1 , x2 , ... xn )  0 

...(i)

Thus there are only n – r independent variables out of these n variables. For maxima or minima of u, we have

Also

du 

u u u dx1  dx2  ...  dxn  0  x1  x2  xn

...(ii)

df1 

 f1  f1  f1 dx1  dx2  ...  dxn  0  x1  x2  xn

...(iii)

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321

 f2  f2  f2 dx1  dx2  ...  dxn  0  x1  x2  xn

df 2 

...(iv)

.................................................. ..................................................  fr  fr  fr dx1  d x2  ...  d xn  0  x1  x2  xn

df r 

...(v)

Multiplying equations (ii), (iii), (iv) ... (v) by 1, 1, 2, ... r respectively and adding, we get A1 dx1 + A2d x2 + ... + Andxn = 0

...(vi)

f f u  fr and i = 1, 2, ... n.  1 1   2 2  ...   r  xi  xi  xi  xi Now, we have at our choice the r quantities 1, 2, ..., r and we choose them in such a way that A1 = 0 = A2 = A3 = ... = Ar ...(vii) Then the equation (vi) is reduced to Ar + 1 d xr + 1 + Ar + 2 dxr + 2 + ... + An d xn = 0 ...(viii) Let us now suppose that out of n variables the following (n – r) variables are independent. xr + 1, xr + 2, ..., xn. Then as dxr + 1, dxr + 2, ..., dxn are independent, so their coefficients are separately zero and we have Ar + 1 = Ar + 2 = ... = An = 0 ...(ix)  From (vii) and (ix), we have

Where

Ai 

A1 = 0 = A2 = A3 = ... = Ar = Ar + 1 = ... An

...(x)

Also from (i) we have f1 = 0 = f2 = ... = fr

...(xi)

From (x) and (xi) we get (n + r) equations which determine the r multipliers 1, 2, ..., r and get the possible values of u. EXAMPLES 1. Find the maxima and minima of x2 + y2 + z2 subject to the conditions ax2 + by2 + cz2 = 1 and lx + my + nz = 0. (Kanpur 96; Vikram 1998; Indore 96, 2000; Sagar 99, 2000 ; Bhopal 96) Let u = x2 + y2 + z 2 ...(i) Given and

ax2 + by2 + cz2 = 1 lx + my + nz = 0

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

From (i) we get du=2xdx+2ydy+2zdz=0

...(iv)

From (ii) and (iii) we have 2 a x dx + 2 b y d y + 2 c z d z = 0 and ldx+mdy+ndz = 0 Multiplying (iv), (v) and (vi) by 1, 1, 2 and adding we get (x d x + y d y + z d z) + 1 (a x d x + b y dy + cz dz) + 2 (ldx + m d y + n d z) = 0

...(v) ...(vi)

DIFFERENTIAL CALCULUS

322

or (x + a 1 x + l 2) dx + (y + b 1 y + m 2) dy + (z + c1z + n 2) dz = 0 Equating to zero the coefficients of dx, dy, and dz we get x  a 1 x  l  2  0   y  b 1 y  m  2  0  z  c 1 z  n  2  0  Multiplying these by x, y, z and adding, we get

...(viii)

(x2 + y2 + z2) + 1 (ax2 + by2 + cz2) + 2 (lx + my + nz) = 0 u + 1 (1) + 2 (0) = 0, from (i), (ii) and (iii)

or

u + 1 = 0

or

1 = – u

or

 From (viii) we have x – a x u + l 2 = 0

y – b y u + m 2 = 0,

z – c z u + n 2 = 0 x (1 – au) = – l2,

or

y (1 – b u) = – m 2,

z (1 – cu) = – n 2. l 2 m 2 n 2 , y , z au – 1 bu – 1 cu – 1 Substituting these values in (iii), we get x

or

l2 m2 n2   0 (a u – 1) (bu – 1) (cu – 1) which gives the maxima and minima of u, i.e., x2 + y2 + z2. Discuss the maxima and minima of the function u = sin x sin y sin z, where x, y, z are the angles of a triangle. 2.

(Kanpur 94 ; Bhopal 1998)

Since x, y, z are the angles of a triangle, hence x+y+z =  Given

...(i)

u = sin x sin y sin z

...(ii)

For max. or min. We have from (ii) d u = cos x sin y sin z dx + sin x cos y sin z dy + sin x sin y cos z dz = 0

...(iii)

Also from (i) dx + dy + dz = 0

...(iv)

Multiplying (iii) by 1 and (iv) by  and adding, we get (cos x sin y sin z + ) dx + (sin x cos y sin z + ) dy + (sin x sin y cos z + ) dz = 0 Equating the coefficients of dx, dy, dz to zero, we get cos x sin y sin z +  = 0, sin x cos y sin z +  = 0 sin x sin y cos z +  = 0 

...(v)

cos x sin y sin z = sin x cos y sin z = sin x sin y cos z = – 

or or

cot x = cot y = cot z, dividing each term by sin x sin y sin z x yz

1 , 3

from (i)

MAXIMA AND MINIMA

323

Let x and y be independent and z be dependent, as the relation (i) exists to find r, s and t. u z  cos x sin y sin z  sin x sin y cos z x x But from (i) we get

From (ii)

z z  0, i.e.,  –1 x x

1

u  cos x sin y sin z – sin x sin y cos z x



 2u

and

 x2

z     – sin x sin y sin z  cos x sin y cos z .  x   z   –  cos x sin y cos z – sin x sin y sin z .   x 

= (– sin x sin y sin z – cos x sin y cos z) – (cos x sin y cos z + sin x sin y sin z) i.e.,

r = –2 (sin x sin y sin z + cos x sin y cos z) 2

 u z  cos x cos y sin z  cos x sin y cos z  yx y

and

z – sin x cos y cos z + sin x sin y sin z  y , z  –1 as before where y s = cos x cos y sin z – cos x sin y cos z – sin x cos y cos z – sin x sin y sin z

or

 , we get 3

Putting x = y = z =

1 1 1 1 1 1 r  – 2 3. 3. 3 . 3.   – 3 2 2 2 2 2 2

1 1 1 1 1 1 1 1 1 1 1 1   3–  3 – 3  – 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2 1 – 3. 2

s

By symmetry 

Also,

t

 2u y

2

 – 3, at x = y = z =

 1  r t – s 2  (– 3) (– 3) –  – 3  2  3 9  3 –    ve 4 4

r  – 3  – ve

Hence there is maximum at x=y=z 

 . 3

2

 3

DIFFERENTIAL CALCULUS

324

3. Show that the maximum and minimum of radii vectors of the section of the surface

x2

( x2  y 2  z 2 )2 

a2



y2 b2



z2 c2

. by the plane x + μy + vz = 0

are given by the equation a2 2

b 2 2



1 – a2r 2

1 – b2 r 2



c2v2 1 – c2r 2

 0.

We have to find the maximum value of radius vector r, where r2 = x2 + y2 + z2. It is given that x2 a2

and

y2

z2

 r4 b2 c2 x + μy + vz = 0. 



Differentiating these equations and putting the condition dr = 0, we have x dx + y dy + z dz = 0 x a

2

dx 

y

z

dz  0 b c2  dx + μ dy +  dz = 0

and

dy 

...(1)

2

...(2) ...(3)

Multiplying (1), (2), (3) by 1, 1, 2 respectively and adding and then equating to zero the coefficients of dx, dy, dz we get x y z

x a2 y b2 z

1    2  0

...(4)

1    2  0

...(5)

1    2  0 c2 Multiplying (4), (5), (6) by x, y, z respectively and adding, we get

r2 + r4 1 = 0 or 1  –

1 r2

Therefore from (4),

1   x 1 – 2 2     2  0 a r   x

or Similarly, y 

b2 r 2  2

a2 r 2  2 1 – a2 r 2

and z 

c2 r 2  2

1 – c2 r 2 1 – b2 r 2 Substituting the values of x, y, z in x + μy + z = 0, we get a2 2 1 – a2 r 2



b2  2 1 – b2 r 2



c2 2 1 – c2 r 2

0

...(6)

MAXIMA AND MINIMA

Find the maximum and minimum values of

4. x

2

a2



325

y

2



b2

Let u 

z

2

c2 x2

x2 a4



y2 b4



z2 c4

, when lx + my + nz = 0 and

 1. Interpret the result geometrically. 

y2

z2



; then a4 b4 c4 2x 2y 2z du  4 dx  4 dy  4 dz  0 a b c l dx + m dy + n dz = 0 2x

2y

x a

y

4

2

dy 

...(2)

2z

dz  0 a b c2 Multiplying (1), (2) and (3) by 1, 1, 2 respectively and adding, and then equating the coefficients of dx, dy and dz to zero, we get 2

dx 

...(1)

 1 l   2

z2 y

 0,

...(4)

0 b2 z  1 n   2 . 2  0 and c4 c Multiplying (4) by x, (5) by y, and (6) by z and adding, we get b4 z

 1 m   2 

x

...(3)

...(5) ...(6)

u + 2 = 0 or 2 = – u, x



a

4

 1 l –

Hence x  –

x a2

 u  0 etc.

1 l a 4

1 – a 2u – 1 n c 4 – x1 m b 4 z Similarly, y  and 1 – c 2u 1 – b2u Substituting these values of x, y, z in lx + my + nz = 0, we get l 2a4 1 – a2 u



m2b4 1 – b2u



n2c4 1 – c 2u

0

...(7)

The equation (7) gives the required maximum or minimum values of u. Geometrical Interpretation. The tangent to the ellipsoid xx  a

2



yy  b

2



zz  c2

x2 a2



y2 b2



z2 c2

 1 at (x, y, z) is

 1. Perpendicular distance to the tangent plane from origin, i.e., p is given by

1 . x 2 y 2 z 2  4  4 a4 b c If the point (x, y, z) lies on the given plane lx + my + nz = 0, the given problem consists of finding out the maximum or minimum value of the perpendicular distance from the origin to the tangent planes to the ellipsoid at the points common to the plane lx + my + nz = 0 and the ellipsoid. p2 

DIFFERENTIAL CALCULUS

326

5. Prove that of all rectangular parallelopiped of the same volume, the cube has the least surface. (Jiwaji 1998, 2000) Then

Let x, y, z be the dimensions of the parallelopiped. Let V denotes its volume and S its surface. V = xyz . ..(1) S = 2xy + 2yz + 2zx

and

...(2)

For maxima and minima of S, we have dS = 2 (y + z) dx + 2 (z + x) dy + 2 (x + y) dz i.e.,

(y + z) dx + (z + x) dy + (x + y) dz = 0

...(3)

Since V is constant, hence yz dx + zx dy + xy dz = 0

...(4)

Multiplying eqn. (3) by 1 and eqn. (4) by  and adding and then equating the coefficients of dx, dy, dz to zero, we get (y + z) + yz = 0

...(5)

(z + x) + zx = 0

...(6)

(x + y) + xy = 0

...(7)

Solving (5), (6) and (7), we get 1 1 1 1 1 1 –       y z z x x y Similarly, y=z

or

1 1 –  0 or x = y y x

Since x = y = z = V1/3, from (1) Let us regard x and y as independent of each other. Then S z z  2y  2y  2z  2x x x x z z z  0 i.e., – . x x x S 2 yz 2 yz  2y –  2z – 2z  2 y – y x x

Also from (1) yz  xy

and

Similarly

and

2 S y

2

2 S y 2



2 yz x

2

–

2 y z 2 yz 2 yz  2  2 = 4 at x = y = z. x x x x

 4 at x = y = z.

2 S 2 y 2 y z 2– – y x x x x 2 z 2 yz  2 x x = 2 – 2 + 2 = 2 at x = y = z. 2–

Hence

rt – s2 = 4 × 4 – 22 = 12 > 0

Since r and rt – s2 are both + ve, S is least when x = y = z, i.e., parallelopiped of least surface is a cube.

MAXIMA AND MINIMA

327

6. Determine the greatest quadrilateral which can be formed with the four given sides , , ,  taken in this order. Let x denotes the angle between  and . y the angle between  and . Then if A denotes the area of the figure, we have 1 (  sin x    sin y ) ...(1) 2 Fig. 9.13 Also, if we draw a diagonal from the intersection of  and  to the intersection of  and , (AC), we have two different values of the length of this diagonal A

2 + 2 – 2  cos x = 2 + 2 – 2 cos y, 2  cos x – 2  cos y + 2 + 2 – 2 – 2 = 0

...(2)

Now for maxima or minima of A, we have 1 dA = ( cos x dx +  cos y dy) = 0 2 i.e.,  cos x dx +  cos y dy = 0

...(3)

i.e.,

Also from (2), – 2  sin x dx + 2  sin y dy = 0

...(4)

Multiplying (3) by 1 and (4) by  and adding and then equating the coefficients of dx, dy to zero, we have  cos x +  sin x = 0  cos y –  sin y = 0

and

–  = cot x = – cot y

These give  Either

cot x = cot (– y) or cot ( – y)

i.e., either x = – y, which is not possible in the present case. x =  –y, i.e.,

or

x+y=

This is the only applicable solution. In this case only one variable, say x is to be regarded as independent. dA 1  dy      cos x    cos y  dx 2  dx 

 Also from (2),

–2 sin x + 2  sin y Hence

dy 0 dx

dA 1 sin x cos y       cos x  dx 2 sin y   1 sin ( x  y )   2 sin y

...(5)

DIFFERENTIAL CALCULUS

328

and  dy  dy   cos ( x  y ) 1  sin y – sin ( x  y ) cos y    d A 1 dx  dx         2 dx 2 sin 2 y   2

    sin x    – 1   sin y  since x + y =    sin y 1 ,       2  2 sin y     (     ) = – , since sin x = sin ( – y) = sin y. 2   sin y = a – ve quantity since sin y is +ve, y being < . Hence the greatest quadrilateral is such that it can be inscribed in a circle. 7. Divide a number n into three parts x, y, z such that ayz + bzx + cxy shall have a maximum or minimum, and determine which it is. Let

u = ayz + bzx + cxy

...(1)

x+y+z=n

...(2)

where

On differentiating we have du = (bz + cy) dx + (cx + az) dy + (ay + bx) dz = 0 and dx + dy + dz = 0. Multiplying (1) by 1 and (2) by , adding and equating the coefficients of dx, dy and dz to zero, we get ...(3) 2 (ayz + bzx + cxy) + (x + y + z)  = 0 or

2u + n = 0 from (1) and (2)

Thus equations (3) may be written as 0  x  cy  bz –

2u 0 n

cx  0  y  az –

2u 0 n

2u 0 n x + y + z – n = 0,

bx  ay  0  z –

and

Eliminating x, y, z from these equations, we get 0 c

b

c 0 a b a 0 1 1 1

2u n 2u n 0 2u n n

...(4)

MAXIMA AND MINIMA

329

This gives maximum or minimum value of u. To see whether it is maximum or minimum, we have u z  bz  cy  ( ay  bx ) , x x where z is regarded as a function of x and y.

= bz + cy – (ay + bx), from (1) 2

and

 u x

2

 –b  b

z  – 2b  r x

 2u z  (c – a )  b  (c – a – b )  s , x y y u z  (az  cx)  ( ay  bx) y y

= ax + cx – ay – bx, 2

 u y

Now,

2

aa

z  – 2a  t y

rt – s2 = 4 ab – (c – a – b)2 = – (a2 + b2 + c2 – 2ab – 2bc – 2ca) = a (b + c – a) + b (c + a – b) + c (a + b – c)

which is +ve if a, b, c are the sides of a triangle. Hence a maximum value is given by (4), where a, b, c are such that a triangle could be constructed with these as sides. 8. Find a maximum value of (x + 1) (y + 1) (z + 1) where ax by cz = A. Interpret the result geometrically. Let u = (x + 1) (y + 1) (z + 1)

..(1)

and x log a + y log b + z log c = log A

...(2)

 du = (y + 1) (z + 1) dx + (z + 1) (x + 1) dy + (x + 1) (y + 1) dz = 0 and dx log a + dy log b + dz log c = 0. Hence

( y  1) ( z  1)   log a  0   ( z  1) ( x  1)   log b  0  ...(3) ( x  1) ( y  1)   log c  0  Multiplying the equations (3) by (x + 1), (y + 1), (z + 1) respectively and adding, we get 3u +  [log a + log b + log c + x log a + y log b + z log c] = 0 or 3u +  [log a + log b + log c + log A] = 0 3u log ( A abc) Also writing (3) as (y + 1) (z + 1) = –  log a etc. and multiplying,

or

–

...(4)

DIFFERENTIAL CALCULUS

330

we get (x + 1)2 (y + 1)2 (z + 1)2 = – 3 log a log b log c or

u2  –

or

u

where Hence

[log ( Aabc)]3

log a log b log c

[log ( Aabc)]3 27 log a log b log c



Now

27 u 3

[log ( Aabc)]3

...(5)

log a3 log b3 log c3

u z  ( y  1) ( z  1)  ( x  1) ( y  1) x x z log a  log c  0, from (2) x u log a  ( y  1) ( z  1) – ( x  1) ( y  1) , x log c

 2u x

2

 ( y  1)

z log a – ( y  1) x log c

log a  – ve if y is + ve. log c Hence if x, y, z are +ve, a maximum value is given by (5).  – 2 ( y  1)

Geometrical Interpretation. a x b y c z = A denotes the surface and (x + 1) (y + 1) (z + 1) will denote the volume of a rectangular parallelopiped, the end point of one of whose diagonals are (–1, –1, –1) and (x, y, z) and whose edges are parallel to the coordinate axes. Hence the problem of finding out the maximum volume of such a solid when the point (x, y, z) moves on the surface a x b y c z = A. 9.

Prove that if x + y + z = 1, ayz + bzx + cxy has an extreme value equal to abc (2bc  2ca  2ab  a 2 – b 2 – c 2 )

.

Prove also that if a, b, c are all positive and c lies between a  b – 2 (ab) and a  b  2 (ab) this value is true maximum and that if a, b, c are all negative and c lies between a  b  2 ( ab), it is true minimum. Let

u = ayz + bzx + cxy

...(1)

and

x+y+z=1

...(2)

on differentiating, we get du = (bz + cy) dx + (cx + az) dy + (ay + bx) dz = 0 dx + dy + dz = 0.

and 

bz + cy +  = 0, cx + az +  = 0, ay + bx +  = 0

Hence

– = bz + cy = cx + az = ay + bx

These give a – c b – c zx y  y   x  b   a 

...(3)

MAXIMA AND MINIMA

or

331

x y z   a (b  c – a ) b ( c  a – b ) ( a  b – c )

=

x y z 2  bc –  a

2



1 2  bc –  a 2 abc

u = ayz + bxz + cxy =

Hence,

(2bc  2ca  2ab – a 2 – b 2 – c 2 )

u z  (bz  cy )  (ay  bx ) , regarding y as constant, x x = bz + cy – ay – bx, from (2)

Now,

 2u



x

2

b

z – b  – 2b  r. x

 2u z b c–ac–a–bs y x x Similarly, 

 2u

 – 2a  t , y 2 rt – s2 = 4ab – (c – a – b)2

= 2 (ab)  c – a – b   2 (ab) – c  a  b  Hence rt – s2

will be positive when c > a + b – 2 (ab) and < a + b + 2 (ab) , whether a, b, are all + ve or all negative. But when a, b, c are all + ve, r and t are negative. Therefore, we have a true maximum in this case and when a, b, c are all – ve, r and t are +ve, so that u is a true minimum in this case. 10. If F (a) = μ  0, F  (a)  0 and x, y, z satisfy the relation F (x) F (y) F (z) = μ3. Prove that the function  (x) +  (y) +  (z) has a maximum when x = y = z = a, provided that  F  ( a ) F  ( a )   ( a )  –    ( a ). F (a)   F  (a ) u =  (x) +  (y) +  (z)

...(i)

F (x) F (y) F (z) = 3  log F (x) + log F (y) + log F (z) = 3 log  For max or min., we have du = ' (x) dx + ' (y) dy + ' (z) dz = 0 And from (iii) F  ( x) F  ( y) F  ( z) dx  dy  dz  0 F ( x) F ( y) F ( z) Multiplying (v) by  and adding to (iv), we get

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

Let

  ( x)  

F  ( x)  0, etc. F ( x)

...(iv)

...(v)

DIFFERENTIAL CALCULUS

332

For the function to be maximum at (a, a, a), we must necessarily have F (a)  (a )   0 F  (a)  (a ) F ( a )  0 , since F' (a)  0. F  (a) and F (a)  0. Now, regard z as a function of x and y so that x, y are independent.

i.e.

–

u z   ( x)   ( z ) . x x F  ( x) F  ( z )  z  · 0 From (iii), F ( x) F (z)  x z i.e.,  – 1 at (a, a, a ) x u F  ( x) F ( z )   ( x) –  ( z )  x F ( x) F  ( z ) 2 u F  ( x) F ( z ) z   ( x) –  ( z ) F ( x) F  ( z ) x  x2



{F  ( x) F ( z )  F ( x) F  ( z )  z } F ( x) F  ( z )  x    –  ( z ) – {F  ( x) F  ( z )  F ( x) F  ( z )  z } F  ( x) F ( z )  x   2    [ F ( x ) F ( z )]    2 F  ( a ) F ( a ) – 2  F  ( a )2     ( a )   ( a ) –  ( a )  F (a) F  (a )     F  (a )  F  (a)  = 2  (a) –  (a)  F (a)  – F (a)       F  ( a ) F  ( a )  – which is – ve if  ( a )     ( a ) F (a )   F  (a ) Hence u is maximum under the given condition.

EXERCISES 1. Determine the maxima and minima of x2 + y2 + z2 when ax2 + by2 + cz2 = 1. (Jiwaji 1999; Indore 96; Bilaspur 97, 2000) 2 2 2 2. Find the minimum value of x + y + z , given that x + by + cz = p. (Sagar 1998; Bilaspur 98; Jiwaji 2000; Ravishankar 96, Rewa 2001; Bhopal 2001; Indore 95, 96, 98, 2000; Jabalpur 2000) 2 2 2 3. Determine the minimum value of x + y + z subject to the condition x + 2y – 4z = 5. 4. Find the minimum value of x2 + y2 + z2 when yz + zx + xy = 3a2. 5. In a plane triangle find the maximum value of u = cos A cos B cos C. 6. If u = x2 + y2 + z2, where ax2 + by2 + cz2 + 2 fyz + 2 gzx + 2 hxy = 1, find the maximum values of u.

MAXIMA AND MINIMA

333

7. If two variables x and y are connected by the relation ax2 + by2 = ab, show that the maximum and minimum values of the functions x2 + y2 + xy will be the values of u given by the equation 4 (u – a) (u – b) = ab. 8. Find the maxima and minima of x2 + y2 + z2 subject to the conditions : ax2 + by2 + cz2 + 2fyz + 2 gzx + 2 hxy = 1 and lx + my + nz = 0. 9.

Find the maximum and minimum values of u = a2x2 + b2y2 + c2z2, where x2 + y2 + z2 = 1 and lx + my + nz = 0. (Sagar 1997; Vikram 2000 )

10. Find the maximum and minimum values of x2 + y2 + z2 subject to the conditions x + y + z = 1 and xyz + 1 = 0. 11. Show that the point within a triangle for which the sum of the squares of its perpendicular distance from the sides is least is the centre of the cosine-circle. 12. Find a point such that sum of squares of its distances from four faces of a given tetrahedron shall be a minimum. Find its value. 13. Find maximum value of f (x, y, z) = x a yb z c subject to the condition x + y + z = 1 where a, b, c are positive numbers. (Kanpur 96) 14. Find the volume of the greatest rectangular parallelopiped inscribed in the ellipsoid whose x2 y2 z2 equation is 2  2  2  1. a b c 15. In a plane triangle, find the maximum value of u = cos A cos B cos C. (Rewa 1999; Ravishankar 2000; Bhopal 97; Bilaspur 2002) 16. Find the maximum and minimum values of ax 2 + by2 + cz2 + 2hxy + 2 gzx + 2 fyz = 0, subject to the conditions lx + my + nz = 0 and x2 + y2 + z2 = 1. 17. Find the triangular pyramid of given base and altitude which has the least surface. 18. Determine the maximum value of OP, O being the origin of coordinates where P describes the curve x2 + y2 + 2z2 = 5, x + 2y + z = 5. 19. Find the shortest distance between the points A (x1, y1, z1) and B (x2, y2, z2) if A lies on the plane x + x2 y2 z2 y + z = 2a and B lies on the ellipsoid 2  2  2  1 . a b c ANSWERS p2

1. u = 1/a, 1/b, 1/c; 5.

ABC

2.

a2  b2  c2

 . 3

a–

1 u

h

h

b–

g

f

6. Root of the eqn.

g 1 u

0

f 1–

1 u

;

3.

25 ; 21

4. 3 a2

DIFFERENTIAL CALCULUS

334

l2

m2



n2

0 u – a2 u – c2 u – c2 gV 2 12. u    i2 16. Max. or Min. value of u are given by a–u h g h b–u f g f c–u

9.



Min. value = 3.

10.

8 abc

14.

3 3

;

15.

1 8

l m 0 n

l m n 0 17. When the side faces are equally inclined to the base.

18. 19.

5 1  2 2 2   ( a  b  c ) – 2a  5

OBJECTIVE QUESTIONS For each of the following questions, four alternatives are given for the answer. Only one of them is correct. Choose the correct alternative. 1. The necessary condition for a function f (x) to have a maxima at x = c is that : (a) f ' (c) > 0 (b) f ' (c) = 0 (c) f ' (c) < 0 (d) None of these 2. A function f (x) has maximum value at x = c if : (a) f  (c) = 0 and f  (c) > 0 (c) f  (c) = 0 and f  (c)  0 3. The maximum value of sin x + cos x is : (a) 2

(b)

f  (c) = 0 and f  (c) < 0

(d)

f  (c)  0 and f  (c) < 0

(b)

2 1

(c) 1 (d) 2 4. If f (x) = | x |, then : (a) f  (0) = 0 (b) f (x) is maximum at x = 0 (c) f (x) is minimum at x = 0 (d) None of these 5. If y = a log x + bx2 + x has its extremum values at x = –1 and x = 2, then : (a) a = 2, b = – 1

(b)

a = 2, b  –

1 2

1 (d) None of these 2 6. The function f (x) = x3 – 6x2 + 24x + 4 has : (a) a maximum value at x = 2 (b) a minimum value at x = 2 (c) a maximum value at x = 4 and a minimum at x = 6 (d) neither maximum nor minimum at any point 7. The necessary conditions for a function f (x, y) to have maximum or minimum value at a point (a, b) are that :

(c) a = –2, b 

MAXIMA AND MINIMA

335

f f at (a, b)  dx y f f (c) at (a, b) 0 x y

(a)

(b) (d)

1 3 x – 2 x 2  3x  1 3 (a) 3/7 (b) (c) 1 (d) 9. sin x (1 + cos x) is a maximum at : (a) x = /3 (b) (c) x = /2 (d) 10. Maximum value of (log x)/x is : (a) e (b) (c) 0 (d)

8. The maximum value of f ( x ) 

11. Function u = sin x sin y sin (x + y) at x = y  (a) Maximum (c) Neither Maxima nor minima 12. The minimum value of xy  (a) a2

 is 3 (b)

(d)

is : 7/3 7

(Garhwal 2001)

x= x=0 1/e 1

Minimum None of these

(Kanpur 2001)

a3 a3  at x = a, y = a is : x y (b) 2a2

(c) 3a2

(d)

13. The minimum value of a where a > b > 0, then :

f f at (a, b)  x y 2 f 2 f  0  at (a, b) x 2  y2

tan2

x+b

cot2

4a2

(Kanpur 2002)

x equals the maximum value of a sin2  + b cos2 ,

(a) a = b

(b)

a = 2b

(c) a = 3b

(d)

a = 4b

(b)

cos 1

(d)

0

14. The maximum value of cos (cos (sin x)) is (a) cos (cos 1) (c) 1 15. Minimum value of

4x2

+ 4x | sin  | –

(a) – 2

cos2

 is : (b)

–1

1 (d) 0 2 16. For any real , the maximum value of cos2 (cos ) + sin2 (sin ) is :

(c) –

(a) 1

(b)

1 + sin2 1

(c) 1 + cos2 1

(d)

does not exist

ANSWERS 1. (b) 7. (c) 13. (d)

2. (b) 8. (b) 14. (a)

3. (b) 9. (a) 15. (b)

4. (d) 10. (b) 16. (b)

5. (c) 11. (c)

6. (d) 12. (c)

CHAPTER 10 INDETERMINATE FORMS 10.1 Indeterminate Forms We know that when x  a lim

f ( x) lim f ( x)  F ( x) lim F ( x ) if lim F(x) 

so that this result fails to give any information regarding the limit of a fraction whose denominator tends to zero as its limit. Now, suppose, that the denominator tends to 0 as x  a. The numerator may or may not tend to zero. Suppose, that the numerator does not tend to zero. Then f(x) / F(x) cannot tend to a finite limit. For, if possible, let it tend to a finite limit, say l. We have in this case, f ( x) f (x) = F ( x) F ( x) 

 f ( x)  F ( x)  lim f (x) = lim   F ( x) 

f ( x) lim F (x) = l.0 = 0. F ( x) Thus, we have a contradiction.

= lim

Three types of different behaviours are possible in this case in that the fraction may tend to +  or – or that the limit may not exist as illustrated below. The limit of the denominator in each of the following three cases is 0 when x 0 : (i) lim (1/x2) = + 

(ii) lim (1/– x2) = – 

(iii) lim (1/x) does not exist. The case when the limit of the numerator is also zero is more important and interesting. A general method of determining the limit of such a fraction will be given in this chapter. As already stated in Chapter 3, we say that a fraction whose numerator and denominator both tend to zero as x tends to a, assumes the indeterminate form 0/0 for x = a. The determination of the derivative dy/dx is itself equivalent to finding the limit of a fraction y/x which assumes the indeterminate form 0/0. Other cases of limits which are reducible to this form will also be considered in this chapter. 336

INDETERMINATE FORMS

337

In what follows it will always be assumed that the functions f and F possess continuous derivatives of every order that come in question in a certain interval enclosing x = a. We shall be making use of the Cauchy’s mean value theorem (Refer  8.5) in this chapter.

10.2 The Indeterminate Form : 0/0

(Gujrat 2004)

If lim F ( x)  0 , lim f ( x)  0 , then x a

xa

f ( x) f ( x)  l.  l  lim x  a F ( x) xa F ( x)

lim

Since the values of the functions f and F at a are not relevant to either the hypothesis or the conclusion of the theorem, we suppose that f (a) = 0, F (a) = 0 and thus render f and F continuous at a. Also, since we are given that lim

x a

f ( x)  l. F ( x)

It follows that there exists a neighbourhood of a for every point x of which f '(x) / F' (x) is defined so that F(x) is not zero. We have by Cauchy’s mean value theorem, f ( a  h) f ( a  h)  f ( a )  F ( a  h) F ( a  h)  F (a ) f (a  h) = F (a  h)

0 0 be a given number. As lim

x 0

] a, a + 1 [,

f '( y )  l , there exists a right handed neighbourhood ] a, a + 1 [ of a such that  y  F '( y ) f '( y ) l   F '( y )

f '( y ) l . F '( y ) If c and x are chosen so as to belong to this neighbourhood, then  also belongs to the same neighbourhood and as such f '() l   l  ...(ii) F '() l  



Keeping c fixed, we see that lim

xa

1  F (c ) / F ( x ) 1 1  f (c ) / f ( x )

so that there exists a neighbourhood ] a, a + 2 [of a such that  x in this neighbourhood 1  F (c ) / F ( x ) 1   1  f (c ) / f ( x )

1  F (c ) / F ( x ) 1  1  f (c ) / f ( x ) Let  be the smaller of the two positive numbers 1 and 2. 1  



...(iii)

From (i), (ii) and (iii), it follows that x  ] a, a +  [, We have (l  ) (1  ) 

f ( x)  (l  ) (1  ) F ( x)

Thus we see that to an arbitrarily assigned  > 0, there corresponds a neighbourhood of a such that x in this neighbourhood (l   ) (1  ) 

f ( x)  (l  ) (1  ) . F ( x)

It follows that lim

xa 0

f ( x) f '( x)  l  lim . xa  0 F ( x) F ( x)

We may similarly consider x tending to a from the left. Thus lim

x a

f '( x) f ( x)  l  lim l. x a F ( x) F '( x)

INDETERMINATE FORMS

343

EXAMPLES log ( x  a) as x  a. log (e x  e a ) Here the numerator and the denominator both tend to as x tends to a.

1. Determine lim

 1   lim   log(e x  ea ) x  a  x  a  log( x  a )

lim



x a

= xlim a = xlim a

e x ( x  a)

0   0

ex e x ( x  a)  e x



ea ea

 1.

tan 3 x as x /2. tan x

2. Determine lim We have

e x  ea

 ex   e x  e a   

tan 3 x 3sec2 3 x  lim x  / 2 tan x x  / 2 sec 2 x lim

3cos 2 x

= lim

2

x  / 2

cos 3 x sin 2 x = lim x  / 2 sin 6 x

= lim

x  / 2

=

   form  0 

2 cos 2 x 6 cos 6 x

1 . 3

3. Discuss the continuity of f at the origin when f (x) = x log sin x for x = 0 and f (0) = 0 lim f ( x)  lim x log sin x x 0

x 0

= lim

x 0

= lim

log sin x 1/ x cot x

1/ x 2 x = lim  x. x 0 tan x x 0

x x  0 tan x

=  lim x . lim x0

=0 = f (0).  f is continuous at x = 0.

   

344

DIFFERENTIAL CALCULUS

EXERCISES 1. Determine the following limits: log sin x , ( x  0) (Mysore 2004) cot x

(i)

log tan x , ( x  0) log x (v) log tan x tan 2 x, ( x  0)

(iii)

tan x   , x  tan 3 x  2 log tan 2 x (iv) log tan x , ( x  0) (ii)

(Gujrat 2004)

 log tan 2 x   Hint : log tan x tan 2 x   log tan x   (vi) log (1 – x) cot (x /2), (x  1).

1. (i) 0

ANSWERS (iii) 1 (iv) 1

(ii) 3

(v) 1

(vi) 0

10.4 The Indeterminate Form 0.  We consider

lim [ f ( x).F ( x)]

x a

lim f ( x)  0 , lim F ( x)   .

when

x a

x a

We write f ( x). F ( x) 

f ( x) F ( x)  1/ F ( x) 1/ f ( x)

so that these new forms are of the type 0/0 and  respectively and the limit can, therefore, be obtained by § 10.2 or by § 10.3. EXAMPLE 1. Determine lim(x log x) as x  0. log x 1/ x log x    lim ( x log x)  lim ,  x 0 x 0 1/ x   

We write 

x log x =

 1 1   x)  0. = xlim   2   xlim( 0  x x  0 The reader may see that by writing x x log x = (1/ log x)

which is of the form (0/0) and employing the corresponding result of § 10.2 would not be of any avail. Note. Since log x is defined for positive values of x only we see that lim ( x log x) really means x 0 lim ( x log x) .

x  (0  0)

INDETERMINATE FORMS

345

EXERCISES 1. Determine the following limits :– (i) x log tan x, ( x  0)

(Gujrat 2005)

(ii) x tan ( /2 – x), ( x  0)

(iii) (a – x) tan

x , ( x  a) 2a

ANSWERS 1. (i) 0

(ii) 1

(iii) 2a/.

10.5 The Indeterminate Form  We consider

lim [ f ( x)  F ( x)]

x a

when

lim f ( x)    lim F ( x) .

x a

x a

We write  1 1  1  f ( x)  F ( x )   ,  f ( x)  f ( x) F ( x)  F ( x) so that the numerator and denominator both tend to 0 as x tends to a. The limit may now be determined with the help of § 10.2.

Note. In order to evaluate the limit of an expression which assumes the form , it is necessary to express the same in such a manner that it assumes the form 0/0 or . EXAMPLES  1  1 1. Determine lim    as x  2  x  2 log( x  1) 

We write 1 1 log ( x  1)  ( x  2)   x  2 log( x  1) ( x  2) log ( x  1)

and see that the new form is of the type 0/0 when x  2. We have  lim 

 1 1 log( x  1)  ( x  2)    log x  2 log ( x  1)  x  2 ( x  2) log ( x  2) The numerator and the denominator are both 0 for x = 2. On using the method of § 10.2 we may 1 show that the required limit is – . 2 x2 

1   1 2. Determine lim  2   as x  0 . sin 2 x  x (Garhwal 2000; G.N.D.U. Amritsar 2004; Gujrat 2005) We have

1   1 lim  2   x sin 2 x 

x0 

 sin 2 x  x 2    = xlim  0  x 2 sin 2 x   

()

0   0

346

DIFFERENTIAL CALCULUS

 sin x  x  = xlim 0  x4   sin 2 x  x 2 lim = x 0  x4  2 cos 2 x  2 = lim x 0 12 x 2 2sin 2 x = lim x 0 6 x2 1 = 3 2

2

 x       sin x   sin 2 x  2 x  = lim 0 x  4 x3  cos 2 x  1 = lim x 0 6x2 2 1  sin x  lim  =  3 x 0  x  2

EXERCISES 1. Determine the following limits :– (i) (ii) (iii) (iv) (v)

1 1    x  , ( x  0)  x e  1 x a   cot  , ( x  0) a x  1  2  2  cot x  , ( x  0)  x    sec x  tan x  ,  x   2  1 1      ( x  0).  x sin x 

x cos ( x / a )    Write cot a  sin ( x / a )   

(Kumaon 97; Garhwal 98; G.N.D.U. Amritsar, 2004)

ANSWERS 1. (i)

1 2

(ii) 0

(iii)

2 3

(iv) 0

(v) 0

10.6. The Indeterminate Forms, 0°, 1°. We consider F x lim  f ( x )    as x  a ,  

when (i) lim f (x) = 0; lim F (x) = 0. (ii) lim f (x) = 1; lim F (x) = . (iii) lim f (x) = ; lim F (x) = 0. We write

y  [ f ( x)]F ( x) 

log y = F(x) .log f(x).

In each of the three cases, we see that the right hand side assumes the indeterminate form 0.  and its limit may, therefore, be determined by the method given in § 10.5.

INDETERMINATE FORMS

lim [ F ( x) . log f ( x)]  l.

Let

xa

lim log y = l,

 



347

log lim y = l  lim y = e1 lim [ f(x)F (x)] = e1.

Since we talk of log f (x), it is necessary that for values of x near a, f (x) is positive. EXAMPLES 1. Determine lim ( x  a ) x  a as x  a . (0o form) y = ( x – a)x – a.

We have 



log y = (x – a) log (x – a) = lim log y  lim

xa

xa

log ( x  a )    1/ ( x  a ) ,   

log ( x  a ) 1/( x  a )

 1 1   = xlim 2 a  x  a ( x  a)

   lim  ( x  a )  0  xa

log lim y = 0  lim y = e0 = 1.



Thus, lim (x – a )x – a = 1 when x  a. Note. Here it is understood that x tends to a through values greater than a , for otherwise the base (x – a) would be negative and (x – a)x – a would have no meaning. 2

2. Determine lim  cos x 1/ x as x  0. y  (cos x )1/ x

We have

log y 

 

 

2

x log cos x

0   0

x2  tan x 1  = lim x0 2x 2 1 log (lim y )   2 x0 2 lim y  e 1/ 2  lim(cos x)1/ x  e1/ 2 . x 0

x 0

x0

x 0

(cos x) 3. Evaluate: xlim 0

We have 

2

log cos x

lim log y  lim

(MDU Rohtak 2001)

cot x

(Kumaon 2001; Avadh 2003)

cot x y = lim (cos x) x 0

log y = lim cot x log cos x x0

= lim

x0

log cos x tan x

0   0

348

DIFFERENTIAL CALCULUS

( sin x cos x)  0 = xlim 0 y = e0 = 1.



1/ x

4. Find lim  tan x  x 0  x 

.

(Garhwal 2003; Patna 2003) 1/ x

We have 

 tan x  y  lim   x0  x  1  tan x  log  loge y = xlim  0 x  x  = lim

x 0

log tan x  log x x

 sec2 x 1  lim = x 0  tan x  x     x  sin x cos x    = xlim 0  x sin x cos x   2 x  sin 2 x  = xlim  0   x sin 2 x    2  2 cos 2 x = xlim    0  sin 2 x  2 x cos 2 x    4sin 2 x = xlim    0  4 cos 2 x  4 x sin 2 x 

( )

0   0 0   0

=0 

y = e0 = 1.

x  5. Determine lim  2   a 

tan

x 2a

as x  a .

(Rohilkhand 2002; Mysore 2004)

x

Let



x  tan  y   2   2a a  x x  log y  tan log  2   2a a   x x  lim log y  lim tan log  2   x a x a 2a a  x   log  2   a   lim = xa x cot 2a 1 1  . a 2 x a = lim  x xa  cosec 2 2a 2a

(1)

0   0

INDETERMINATE FORMS

=  

349

2 

2  x a lim y  e2 / 

log lim y  xa

6. Determine lim (cot x)1/ log x , x  0. Let





y  (cot x)1/ log x . 1 log y  log (cot x) log x log cot x lim log y  lim x 0 x0 log x cosec 2 x cot x  lim  x . 1   1 = lim x0 x 0 1/ x sin x cos x 1 log lim y   1  lim y  e  1/ e. x 0

x0

   x  7. Determine lim  2      y x   2 

Let

tan x

   as x    0 . 2  

tan x

   log y  tan x log  x  2     lim log y  lim tan x log  – x   2  x  (  / 2 – 0) x  (  / 2 – 0)    log  x  2   lim =    cot x x   0   2



– sin 2 x = x   – 0  –   – x  2  2  2 sin x cos x lim = x    0  1   lim

 2

log lim

=0 y0

x (  / 2  0 )



e0  1

lim y  x   / 2  0





   

350

DIFFERENTIAL CALCULUS

EXERCISES 1. Determine the following limits :–

x

1 x

 1

(i) xx, ( x  0)

(ii)

(iii) (cot x)sin 2x, (x  0)

(iv) ( sin x )tan x, ( x  /2) (Calicut 2004, Bangalore 2005, Calcutta 2005)

 2x  1  (v)    x 1   sinh x  (vii)    x 

(viii)

1 xx  1

, ( x  1)

x –1

( x  0)

(vi) ( 1 + sin x )cot x , ( x )

x 2

, (x  0 )

(Patna 2002) 1/ x 2

, ( x  1)

(x) (tan x)tan2x , ( x /4 )  (xi) (sec x)cotx, (x  +0) 2 1  tan x  x 3 lim (xiii)   x 0  x 

 tan x  (ix)  , (x 0)   x  (Garhwal 2001; Kumaon 2002, G.N.D.U. Amritsar 2004) (Gujrat 2004)

(xii) (cot x)sin x, (x ) (Kanpur 2000; Avadh 2001)

ANSWERS e– 1

1. (i) 1 (vii) e1/6 (xiii) .

(ii) (viii) e

(iii) 1 (ix) e1/3

(v) e (xi) 1

(iv) 1 (x) 1/e

MISCELLANEOUS EXERCISES Determine the following limits :– e x  e x  x 1. , (x  0) x 2 sin x log x 2. , (x ) x3 1  x cos x  cosh x  log (1  x ) 3. , ( x ) tan x  x 2 4. log (1  x ) log (1  x )  log (1  x ) , ( x  ) 5.

x4 x  x 1  e    , ( x  0) 2 2 x sinh x   2x

6.

(Manipur 2000)

1  x  log x 1

(vi) e (xii) 1

( 2x  x2 )

, (x )

1/ x 2

7. ( 2x tan x –  sec x ), (x /2 )

9.

( x – 4 x 4 )1/ 2 – ( x / 4)1/ 3 3 1/ 4

, (x ).

 sin x  8. , (x )    x  (Kumaon 2003; MDU Rohtak 1999; Manipur 2002) log

12.

( sin x ) tan

1 – (8 x )

11.

2

(cos ax )b / x , (x )

(1  x sin x ) , ( x ) cos x  1

10.

2

x

, ( x )

INDETERMINATE FORMS

351 1/ x

13.

(1 – x 2 )1/ log (1  x ) , ( x ) sec 2

15. 17.

19. ( 2 – x

21.

)tan x/2 (

ab – ba a

a –b

b

14.

 2 ( cosh x  1)    x2  

16.

asin x  a , ( x ) log sin x

18.

(sec x)cot x, (x )



 2   2 – bx , ( x )  sin  2 – ax   xx – x , ( x ) x – 1 – log x (MDU Rohtak 2001) x )

22.

(MDU Rohtak 1999) log

23.

1 sec( x ) 2

25. Find lim

27. Find lim

, ( x )

(1  x)1/ x – e , ( x ) x

a x – xa

, (x )

24.

x x – aa

, ( x ) (Rohilkhand 2003)

1  ae x  be 2 x  ce 3 x 1  ae x  be2 x  ce3 x

when (i) a = 3, b = –5, c = 4; (iii) a = 3, b = – 3, c = 1. 26. Find lim

1 – x2

(Agra 1998; Kumaon 2004; Avadh 99, 2000)

cos x

1 log sec x cos x 2

, ( x )

1 – 4sin 2 ( x / 6 )

20.

, ( a  b)

2

, (x )

(ii) a = 3, b = – 4, c = 2;

3x log (sin x / x) 2  x3 , ( x ). ( x  sin x ) (1  cos x) (1  x ) e x  (1  x ) e x x ( e x  e x ) – 2 x 2 e – x

, ( x ).

28. Obtain tan x e ax  e  ax 1 lim   (ii) xlim 0 log (1  bx ) x0  x  29. Give the limits of the following when x  log x log x log x log x , 2 , , x x x3 x

(i)

and of the following when x  ( 0 + 0) 2 3 x log x, x log x, x log x, x log x. 2

30. If

f (x) = e 1/ x , x  f (0) = 0,

show that the derivative of every order of f vanishes for x = 0, i.e., f  (0)  0,  n .

x 2 sin1/ x . x 0 tan x

(iii) lim

352

DIFFERENTIAL CALCULUS

31. Find the values of a, b and c such that lim

x ( a  b  cos x )  c sin x x5

x 0

(Agra 2001, Garhwal 1999, Kanpur 2001)

 1.

ANSWERS

1.

1 3

7. – 2.

2. 0

5 3.  . 2

8. e 1/ 6

9.

8 . 9

15. e a

13. e.

14. e1/12.

19. e2/.

20.

3/6.

23. 11e/24.

24.

log ( a / e ) log ae

2 1 . 27. . 5 8 29. 0, 0, 0, 0; 0, 0, 0, 0.

26.

4.

1 12

5.

/ b2

6. – 1.

11. e a

10. – 2 2

1 6

. 16. a log a.

2

b/2

.

12.

17. 2.

18. 1.

21. log (e/b) log be.

25. (i) 1. 28. (i) 1.

e 1/ 2 .

22. – e/2.

(ii) – 1.

(iii)

(ii) 2a/b.

– 1.

(iii) 0.

30. Continuous at 0.

31. a = 120, b = 60, c = 180 OBJECTIVE QUESTIONS For each of the following questions, four alternatives are given for the answer. Only one of them is correct. Choose the correct alternative. 1. Which of the following is not an indeterminate form? (a)  (b) – (c)  (d) 0 

 gra 2001, Kanpur 2001)

2. The formula of L’ Hospital’s rule is :  f ( x)  f ( x)  lim   x  a g ( x)  g ( x)  f ( x) f (a )  (c) xlim a g ( x) g (a)

(a) xlim a

3. Consider (i) lim

x 0

4.

(b)

xa

lim

f ( x) f ( x)  lim g ( x) x  a g ( x)

(d)

lim

f ( x) f '( a )  g ( x) g (a )

xa

sin x tan x  1 , (ii) lim  1 , then : x  0 x x

(a) Both (i) , (ii) are true

(b)

(i) is false, (ii) is true

(c) (i) is true, (ii) is false

(d)

Both (i), (ii) are false.

ax  bx is equal to : x 0 x lim

(a) 0

(b) 

(c) log (a/b)

(d)

log ( a – b)

(Agra 2001, Kanpur 2001)

INDETERMINATE FORMS

5.

353

lim x log x is equal to :

x0

(b) 1

(c) 

log x is : x 1 (b) 

(c) 1

(1  x ) n – 1 is : x (b) n

(c) nn

(d)

(c)

1 6

(d)

(c)

ea  e a log (1  b )

(d)

0

(a) 0 6. The value of lim

x 1

(a) – 1

7. The value of lim

x 0

(a) 1 8. The value of lim

x 0

1  cos x 3x 2

(a) 3

(b)

9. Value of lim

x 0

(a)

(b)

15.

is : 1 3

1 9 (Garhwal 2002)

b 2a

(d)



cos hx  cos x is : x sin x (a) 0 (b) – 1

(c) 0

(d)



(b) – 2

(c) 0

(d)



(a) 1

(b) 0

(c) log (a/b)

(d)

log (b/a)

log x is : cot x (a) 0

(b) 1

(c) – 1

(d)



(b) 1

(c) 1

(d)



lim

x 0

lim

0

sin    cos  is : sin   

(a) 2

14.



(c) 0

x0

13.

(d) 0 (Garhwal 2001, Avadh 2005)

log (1  kx 2 ) is : 1  cos x (b) – 1

(a) 1

12.

–1

eax  e  ax is : log (1  bx)

2a b

10. Value of lim

11.

(d)

 log x  lim   is : x a  x 

lim

x 0

lim

x0

log sin x log x is :

(a) 0

354

DIFFERENTIAL CALCULUS

16. The value of xlim 0

log x 2

(a) 

(b) 1

17. Value of xlim 0

(b) 1

18. Value of

19.

lim

x 0

( 2x2  5 )

–1

(c) – 1

(d)



4

is :

(b)

5 tan 5 x lim is : x  / 2 tan x

3 5

1 (b) 5 5   1   x tan    is : 20. xlim    x  (a) 0 (b) 

4

3 5

(d)

(c) 

(d)

0

(c) 1

(d)

–1

(c)

2

x   lim  (1  x ) tan is : x 1  2  (a) 0

22.

(d)

3x  4

(a)

21.

(c) 0

log tan x log x is :

(a) 0

(a)

is :

cot x 2

(b) 1

(c)

 2

(d)

2 

(b) 

(c) log a

(d)

log (1/a)

(c) 0

(d)

–1

(c) a

(d)

a/2

(c) – 

(d)

1

(c) – 1

(d)



(c) e

(d)

1/e

lim ( a1/ x  1) x is:

x 

(a) 1

23. Value of lim (sin x) tan x is : x  / 2

(b) 

(a) 0

 a  2 x sin  x  is : 24. Value of xlim 0  2  (a)  (b) –  25. Value of lim (sec x  tan x ) is : x  / 2

(a)  26.

(b) 0

lim ( cosec x  cot x ) is :

x 0

(a) 0

(b) 1 1/ x

27. Value of lim (1  x ) x 0

(a) 1

is :

(b) – 1

(Agra 2003; Kanpur 2003)

INDETERMINATE FORMS

355

cot 28. Value of lim ( cos x )

2

x

x 0

(a) e 29.

is:

(b)

1 e

(c)

1

(d)

e

(c) e

(d)

1/e

(c) e

(d)

1/e

6. 12. 18. 24. 30.

(c) (b) (b) (c) (b)

e

lim x x is :

x0

(a) 0

(b) 1 1/ x

1 30. Value of lim   x   x  (a) 0 1. 7. 13. 19. 25.

(a) (b) (b) (a) (b)

2. 8. 14. 20. 26.

(b) (c) (a) (c) (a)

is : (b) 1 3. 9. 15. 21. 27.

(a) (a) (c) (d) (c)

ANSWERS 4. (c) 10. (d) 16. (c) 22. (c) 28. (a)

5. 11. 17. 23. 29.

(a) (a) (b) (a) (b)

CHAPTER 11 PARTIAL DIFFERENTIATION 11.1. Introduction The notion of continuity, limit and differentiation in relation to real valued functions of two real variables, will be briefly explained in this chapter. A few theorems of elementary character will also be given. The subject of functions of two variables is capable of extension to functions of any number n of variables, but the treatment of the subject in this generalised form is not within the scope of this book. Only a few examples dealing with functions of three or more variables may be given.

11.2. Functions of two variables. As in the case of functions of a single variable, we introduce the notion of functions of two variables by considering some examples. (i) The relation z  (1 – x 2 – y 2 ),

...(1)

between x, y, z, determines a value of z corresponding to every pair of numbers x, y, which are such that x2 + y2  1. Denoting a pair of numbers x, y, geometrically by a point on a plane as explained in § 2.2, we see that the points (x, y) for which x2 + y2  1 lie on or within the circle whose centre is at the origin and radius is 1. The region determined by the point (x, y) is called the domain of the point (x, y). We say that the relation (1) associates a real number z to each pair (x, y) such that x2 + y2  1. We say that we have a function whose domain is the set {(x, y) : x2 + y2  1}. We may write z  f ( x, y ) 

(1 – x 2 – y 2 ) : (x, y) Œ {(x, y) : x2 + y2  1}

(i) Consider the relation z  [(a – x ) ( x – b)]  [(c – y ) ( y – d )]

...(2)

where a < b; c < d. Now (a – x) (x – b) is non-negative if a  x  b and (c – y) (y – d) is non-negative if c  y  d. The points (x, y) for which a  x  b, c  y  d determine a rectangular domain bounded by the lines x = a, x = b; y = c, y = d. 356

PARTIAL DIFFERENTIATION

357

We see that the relation (2) associates a number z to every point (x, y) which is a member of the set {(x, y) : a  x  b ; c  y  d}. In general we say that a real valued function f of two variables is defined if to every member (x, y) of a given set D of ordered pairs of real numbers there is associated a real number. The set D is called the domain of the function f. (ii) The relation 2

2

z  e– x – y determines a function of two variables (x, y); the domain of the function being the whole plane i.e., the set of all the ordered pairs of real numbers. We may also write this as follows : z  f ( x, y )  e – x

2

– y2

Ex. Give the domains of the following functions : (i) f (x, y) = 1/[log x + log y] (ii) f (x, y) = log (x + y) (iii) f (x, y) = (xy + yx). ANSWERS (i) {(x, y) : 0 < x, 0 < y} (ii) {(x, y) : 0 < x + y} (iii) {(x, y) : 0 < x, 0 < y}

11.3. Neighbourhood of a point (a, b) Let  be any positive number. The points (x, y) such that a –   x  a + ,b –  y  b +  determine a square bounded by the lines x = a – , x = a +  ; y = b – , y = b + . Its centre is at the point (a, b). This square is called a Neighbourhood of the point (a, b). For every value of , we will get neighbourhood. Thus the set {(x, y) : a –   x  a + , b –   y  b + } is a neighbourhood of the point (a, b).

11.4. Continuity of a function of two variables. Continuity at a point. Let (a, b) be a point of the domain of

(Manipur 2002)

z = f (x, y) As in the case of functions of one variable, we say that f is continuous at (a, b), if for points (x, y) near (a, b), the value f (x, y) of the function is near f (a, b) i.e., f (x, y) can be made as near f (a, b) as we like by taking the points (x, y) sufficiently near (a, b). Formally, we say that a function f of two variables is continuous at a point (a, b), if corresponding to any pre-assigned positive number , there exists a positive number  such that | f (x, y) – f (a, b) | < 

358

DIFFERENTIAL CALCULUS

for all members of the neighbourhood {(x, y) : a –   x  a + , b –   y  b + } of (a, b).

Fig. 11.1

Thus for continuity at (a, b), there exists a square bounded by the lines x = a – , x = a + ; y = b – , y = b +  such that, for any point (x, y) of this square, f (x, y) lies between f (a, b) –  and f (a, b) +  where  is any positive number, however small. 11.4.1. Continuity of a function : A function f is said to be continuous if it is continuous at every point of its domain. 11.4.2. Special case. Let f be continuous at (a, b) and  be any positive number, however small. Then there exists a square bounded by the lines x = a – , x = a +  ; y = b – , y = b + . such that for points (x, y) of the square, the numerical value of the difference between f (x, y) and f (a, b) is less than . In particular, if we consider points of this square lying on the line y = b, we see that for values of x Œ [a – , a + ] the numerical value of the difference between f (x, b) and f (a, b) is less than . Also such a choice of  is possible for every positive number . This is equivalent to saying that f (x) = f (x, b) is a continuous function of a single variable x for x = a. It may be similarly shown that f (y) = f (a, y) is a continuous function of a single variable y for y = b. Thus we have shown that a continuous function of two variables is also a continuous function of each variable separately.

11.5. Limit of a function of two variables. lim f (x, y) = l as (x, y)  (a, b). A function f is said to tend to the limit l, as (x, y) tends to (a, b), if, corresponding to any preassigned positive number , there exists a positive number  such that | f (x, y) – l | <  for all points (x, y), other than possibly (a, b), such that a –   x  a +  ; b –   y  b + . i.e., for all members of the set {(x, y) : x  [a – , a + ], y  [b – , b + ]} ~ {(a, b)}.

PARTIAL DIFFERENTIATION

359

This means that corresponding to every positive number  there exists a neighbourhood of (a, b) such that for every point (x, y) of this neighbourhood, other than possibly (a, b), f (x, y) lies between l –  and l + . 11.5.1. Limit of a continuous function. Comparing the definitions of limit and continuity as given above in § 11.4, and § 11.5, we see that f is continuous at (a, b), if and only if lim f (x, y) = f (a, b) as (x, y)  (a, b). i.e., the limit of the function = value of the function. The same thing may also be expressed by saying that for continuity at (a, b), we have lim f (a + h, b + k) = f (a, b) as (h, k) (0, 0).

11.6. Partial derivatives. Consider

z = f (x, y).

f ( a  h, b ) – f ( a , b ) , h  0, h if it exists, is said to be the partial derivative of f w. r. t. x at (a, b) and is denoted by z    or fx (a, b).   x (a, b )

Then

lim

It will thus be seen that the partial derivative of f w. r. t. x at (a, b) is the derivative of the function f (x) = f (x, b) of a single real variable at x = a. Again f ( a, b  k ) – f ( a, b) as k  0 k if it exists, is called the partial derivative of f (x, y) w. r. to y at (a, b) and is denoted by lim

z    or fy (a, b)   y  (a, b )

so that the partial derivative w. r. t. y at (a, b) is the derivative of the function f (y) = f (a, y) of a single real variable at y = b. If f possesses a partial derivative w. r. to x at every point of its domain, then there arises a new function (a, b)  fx (a, b) and denoted by fx of two variables whose domain is the same as that of the given function f. Thus we have f ( x  h, y ) – f ( x , y ) as h  0. h 11.6.1. Partial derivatives of higher orders. We can form partial derivatives of the first order partial derivatives just as we formed those of f. f x ( x, y )  lim

Thus we have  z  z  ,   or equivalently fxx, fyx x x y x

called the second order partial derivatives of f. Now fxx, may also be denoted by f x 2 .

360

DIFFERENTIAL CALCULUS

We write    z  2 z    z  2 z ,       x   x   x2  y   x   y  x Similarly the second order partial derivatives  z   z   ,   x  y  y  y are respectively denoted by 2z 2 z , or f xy , f y 2 .  x  y  y2 Thus, there are four second order partial derivatives of z at any point (a, b). The partial derivatives 2 z/ y x and 2 z/ x y are distinguished by the order in which z is successively differentiated w. r. to x and y. But, it will be seen that, in general, they are equal. The proof is given in  11.10.1.

11.7. Geometrical representation of a function of two variables We take a pair of perpendicular lines OX and OY. Through the point O, we draw a line Z' OZ perpendicular to the XY plane and call it z-axis. Any one of the two sides of z-axis may be assigned a positive sense. Length, z, will be measured parallel to this axis. The three co-ordinate axes, taken in pairs determine three planes, viz., XY, YZ and ZX which are taken as the co-ordinate planes. Let f be a function defined in some domain in the XY plane. To each point (x, y) of this domain, there corresponds a value of z. Through the point (x, y) we draw the line perpenFig. 11.2 dicular to the XY plane equal in length to z, so that we arrive at another point P denoted as (x, y, z), lying on one or the other side of the plane according as z is positive or negative. Thus to each point of the domain of the function f in the XY plane there corresponds a point P. The set of the points P determines a surface which is said to represent the function geometrically. 11.7.1. Geometrical interpretation of partial derivatives of the first order. Consider

z = f (x, y)

...(i)

We have seen that the functional equation (i) represents a surface geometrically. We now seek the geometrical interpretation of the partial derivatives  z   z  and   .    x   (a, b)   y  (a, b)

The point P [(a, b), f (a, b)] on the surface corresponds to the point (a, b) of the domain of the function.

PARTIAL DIFFERENTIATION

361

If a variable point, starting from P, changes its position on the surface such that y remains constantly equal to b, then it is clear that the locus of the point is the curve of intersection of the surface and the plane y = b. On this curve x and z vary according to the relation z = f (x, b). z Also,  is the ordinary derivative of f (x, b)    x ( a , b ) w.r. to x for x = a.  z  Fig. 11.3 Hence we see that   x  denotes the tangent of the   (a, b) angle which the tangent to the curve, in which the plane y = b parallel to the ZX plane cuts the surface at P [(a, b), f (a, b)], makes with x-axis.  z  Similarly, it may be seen that   y  is the tangent of the angle which the tangent at P [(a, b),   (a, b) f (a, b)] to the curve of intersection of the surface and the plane x = a parallel to the ZY plane, makes with y-axis.

EXAMPLES 2 –1 1. If u  x tan

y x – y 2 tan –1 ; xy  0 x y

prove that 2 u x2 – y2  2 .  x  y x  y2

We have

u  x2 y



(Ravishankar, 1997, 2002; Jabalpur, 1998, 1999; Bhopal, 1994, 1995)

1

.

y2 1 2 x x3

2

x  y

2



1 x – 2 y tan –1  y 2 x y

xy 2 2

x  y

 x – 2 y tan –1

2 u    x y  x

2

– 2 y tan –1

1 x2 1 2 y

.

x y2

x y

x . y

 u     y

 1 – 2y

1 1

x2 y2

.

1 2 y2 x2 – y2 1– 2  . y y  x2 x2  y2

362

DIFFERENTIAL CALCULUS

1

u

2.2. If

2

2

2

; x 2  y 2  z 2  0,

(x  y  z )

show that 2 u

We have



2 u 2



2 u

 0. x y  z2 (Garhwal, 1998; Sagar, 1996, 1997, 2001; Vikram, 1998; Ravishankar, 1996; Jabalpur, 1994; Indore, 2000; Manipur 1999; Banglore 2005) 2

u 1  – ( x 2  y 2  z 2 ) – 3/ 2 2 x  – x ( x 2  y 2  z 2 ) – 3/ 2 , x 2 2 u x

2

 – ( x 2  y 2  z 2 ) – 3/ 2  3 x 2 ( x 2  y 2  z 2 ) – 5 / 2 .

Similarly or by symmetry 2 u  y2 2 u  z2

 – ( x 2  y 2  z 2 ) – 3/ 2  3 y 2 ( x 2  y 2  z 2 ) – 5 / 2 .  – ( x 2  y 2  z 2 ) – 3/ 2  3 z 2 ( x 2  y 2  z 2 ) – 5 / 2 .

Adding, we obtain the result as given. 3. If u = ex y z, show that

3 u  (1  3 x y z  x 2 y 2 z 2 ) e x y z . (Kumaon 2001; Gorakhpur 2003; Avadh 2003)  x y z We have u = e x y z, u  e x y z . yz then (treating y and z constant) x

2 u   u      y x  y   x    y z ex y z y

Again





 z  y e x y z . x z  e x y z 

2 u  x y z2 ex y z  z ex y z  x y

Then and

 z

 2 u       x  y  z    xy

 x y z2 ex y z  z ex y z   

 z2 ex y z . x y  ex y z . 2z   z ex y z . z y  ex y z      = ex y z (1 + 3 x y z + x2 y2 z2)



3 u  1  3 x y z  x2 y2 z 2 ex y z  x y z





PARTIAL DIFFERENTIATION

363

4. If u = log (x3 + y3 + z3 – 3 x y z) show that 2

     9     u–  x  y  z x  y  z)2 (   (Indore, 2002; Jabalpur, 1999; Bilaspur, 1996, 2000; Kanpur, 2000; Kumaon 2004; Avadh 2002)

(a)

and

(b)

2 u x

We have (a)

2



2 u y

2

2 u



3



. z ( x  y  z)2 (Ravishankar, 1998; Indore, 2001; Vikram, 1999; Jabalpur 2000) 2

u = log (x3 + y3 + z3 – 3 x y z) u 3 x2 – 3 x y z  3  x x  y3  z 3 – 3 x y z



...(i) (treating y and z constant)

Similarly, u 3 y2 – 3 x z  3  y x  y3  z3 – 3 x y z

...(ii) (treating x and z constant)

2

and

u 3z – 3x y  3  z x  y 3  z 3 – 3 xyz

...(iii) (treating x and y constant)

2



2

2

 u  u  u ( x  y  z – xy – yz – zx )    x y z x3  y3  z 3 – 3 x y z 

3 x y  z

2

Now,

      u u u           u  x   y  z x   y  z    x  y z

      3      x  y z  x  y  z



     3   3   3       x x  y  z y x  y  z z x  y  z

– –

3 ( x  y  z)

2

9 ( x  y  z )2

–

3 ( x  y  z)

2

–

3 ( x  y  z )2

364

DIFFERENTIAL CALCULUS

(b) Differentiating (i) partially, we get 2 u  x2



6 x ( x3  y 3  z 3 – 3 x y z ) – (3 x 2 – 3 y z ) (3 x 2 – 3 y z ) ( x3  y 3  z 3 – 3 x y z ) 2 3 ( x4 – 2 x y3 – 2 x z 3  3 y 2 z 2 )

–

( x3  y 3  z 3 – 3 x y z ) 2

Similarly, 2 u  y2 2 u

and

 z2

–

–

3 ( y 4 – 2 y z 3 – 2 y x3  3 z 2 x 2 ) ( x3  y 3  z 3 – 3 x y z )2 3 ( z 4 – 2 z x3 – 2 x y 3  3 x 2 y 2 ) ( x3  y 3  z 3 – 3 x y z )2

On adding, we will have 2 u  x2



2 u  y2

– –



2 u  z2

3 ( x 2  y 2  z 2 – xy – yz – zx) 2 ( x3  y 3  z 3 – 3 x y z ) 2 3 ( x  y  z )2

5. If xx yy zz = c show that at x = y = z,

2 z  – ( x log e x) –1 .  x y

(Kumaon, 1998; Rohilkhand, 1999) xx yy zz = c

Given that

Taking log of both the sides, we have x log x + y log y + z log z = log c. Differentiating partially w. r. t. x, we get x.

1 1 z  log x  ( z .  log z ) 0 x z x z (1  log x) – x (1  log z )

i.e.,

Similarly, we have z (1  log y ) – y (1  log z )



2 z (1  log x) 1  z  .  x  y (1  log z )2 z  y –

(1  log x ) 1 (1  log y ) . . (1  log z ) 2 z (1  log z )

PARTIAL DIFFERENTIATION

365

–

(1  log x) 2 1 . (1  log x)3 x

–

1 1 – x (1  log x) x log ex

(since x = y = z)

= – (x log ex)–1. 6. If u = log (tan x + tan y + tan z), prove that (sin 2 x)

u u u  (sin 2 y )  (sin 2 z )  2. x y z

u = log (tan x + tan y + tan z)

Given

u 1  sec 2 x x (tan x  tan y  tan z )

 or

(sin 2 x)

u sec 2 x sin 2 x  x tan x  tan y  tan z

or

(sin 2 x)

u 2 tan x  x tan x  tan y  tan z

(sin 2 z )

and

u 2 tan y  y tan x  tan y  tan z

(sin 2 y )

Similarly

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

u 2 tan z  z tan x  tan y  tan z

...(iii)

Adding (i), (ii) and (iii) we get (sin 2 x)

2 tan x  2 tan y  2 tan z u u u  (sin 2 y )  (sin 2 z )   2. tan x  tan y  tan z x y z

7. If u = 3 (lx + my + nz)2 – (x2 + y2 + z2) and l2 + m2 + n2 = 1, 2 u

show that

x 2



2 u y 2



2 u z 2

 0.

Given u = 3 (lx + my + nz)2 – (x2 + y2 + z2). u  6l (lx  my  nz ) – 2 x x



 2u

Similarly,



 2u x 2

 6m 2 – 2 and

y 2 

 2u y 2



 2u z 2

 2u



 2u x

2

 6l 2 – 2.

 6 n 2 – 2.

z 2

 (6l 2 – 2)  (6m 2 – 2)  (6n 2 – 2)

= 6 (l2 + m2 + n2) – 6 = 6 (1) – 6, = 0.

 l2 + m2 + n2 = 1 (given)

366

DIFFERENTIAL CALCULUS

8. If v = At –1/ 2 e – x

2

/ 4 a 2t

, prove that

v = At –1/ 2 e – x

We have

2

v 2 v  a2 . t x 2

/ 4 a 2t

2 2  – 2x  v  At –1/ 2 . e – x / 4 a t  2  x  4a t  x – v 2a 2t 2 v 1  v  – v x  2 2  x  x 2a t  1   – vx   – v  x  2  2a 2t   2a t   v  4 2 (– 2a 2 t  x 2 ) 4a t





Again,  x2  2 2 1  2 2  – A . t – 3/ 2 e – x / 4 a t 2  4a t  2  x 2 2 1 . e – x / 4a t  2 2 –  2t   4a t

2 2 v  At –1/ 2 . e – x / 4 a t t

 At –1/ 2 

Clearly,

v 2

4a t

( x 2 – 2a 2 t )

v 2 v  a2 . t x 2

EXERCISES 1. Find the first order partial derivatives of (i) tan– 1 (x + y). (ii) eax sin by. 2. Find the second order partial derivatives of y (i) ex – y (ii) ex 3. Verify that (i)

sin –1

log (x2 + y2).

(iii)

tan (tan– 1 x + tan– 1 y).

(iii)

log (y sin x + x sin y).

2 u 2 u  , when u is  x y  y x x . y

xy

(ii) 1 2 z

4. Find the value of when

(iii)

2

a x

2



1 2 z b2  y 2

(1  x 2  y 2 )

,

a2 x2 + b2 y2 – c2 z2 = 0.

5. If z = tan (y + ax) + (y – ax)3/2, find the value of 2 z x

2

– a2

2 z  y2

.

.

PARTIAL DIFFERENTIATION

367

6. If z = tan– 1 (y/x), verify that 2 z  x2



2 z  y2

 0.

7. If z (x + y) = x2 + y2, show that 2

z z   z z  – –    4 1 – . x y x  y  8. If z = 3xy – y3 + (y2 – 2x)3/2, verify that

(Kumaon, 2000)

2

2 z 2 z 2 z 2 z  2 z   and   .  x y  y x  x 2  y 2   x  y  9. If z = f (x + ay) +  (x – ay), prove that 2 z

 a2

2 z

.  y2  x2 10. If z = (x + y) + (x + y)  (y/x), prove that

(Rewa, 2001; Jiwaji, 2002)

 2 z  2 z 2 x  2 z  x 2 –  y 2 –  . x  y  y  x   x  y    11. If u = f (ax2 + 2hxy + by2), v =  (ax2 + 2hxy + by2), prove that  v   u   x  x  

 y

12. If   t n e – r 1 r

2

2

/ 4t

 v  u .  y

, find the value of n which will make

  2   . r  r  r  t

( x 2  y 2 ) , prove that

13. If u = f (r) where r = 2 u x

2



2 u y

2

 f  (r ) 

1 f  (r ). r (Vikram, 1997; Sagar, 1997; Gorakhpur, 2000)

14. If u = log (x2 + y2 + z2), prove that

x

15. If

x2 2

a u



2 u 2 u 2 u  y z .  yz  z x  x y y2 2

b u



z 2

c u

 1 , prove that

u x2  u y2  u z2 = 2 (x ux + y uy + z yz).

16. If tan u 

cos x and tan hv  sin hy u v  and x y

(Kumaon, 2002)

sin x prove that cos hy u v – . y x

(Garhwal 2004)

(Garhwal, 1999)

368

DIFFERENTIAL CALCULUS

17. If u  tan –1

xy

, show that (1  x 2  y 2 ) 2 u 1  .  x  y (1  x 2  y 2 )3/ 2 (Vikram, 2000; Ravishankar, 2003; Bilaspur, 2000; Rewa, 2001)

18. If u =

x2

(y – z) +

y2

(z – x) + z2 (x – y), prove that

u u u    0. x y z

19. If u  log e

( x 2  y 2  z 2 ) , prove that

 2 u 2 u 2 u  ( x2  y2  z 2 )  2     1. x  y 2  z 2   20. If u = ex (x cos y – y sin y), show that 2 u  x2



2 u  y2

(Ravishankar, 1998)

 0.

21. If u = (1 – 2xy + y2)– 1/2, prove that    2 u  2 u (1 – x )  y   0.  x  y  y  22. If a2 x2 + b2 y2 – c2 z2 = 0, show that  x

2 z  x2 x2 23. If u = x 1

 y2 y 1

2 z  y2



a2 b2 c4 z3

(Bilashpur, 2000; Jiwaji, 1995)

( x 2  y 2 ).

z2 z , show that 1

ux + uy + uz = 0. (Bilaspur, 1995; Ravishankar, 2002) 2 2 2 3 3 3 24. If u = x + y + z, v = x + y + z , w = x + y + z – 3 x y z, prove that ux vx wx

uy vy wy

uz vz  0. wz

x y  tan –1 , show that y x u u x  y  0. x y

25. If u  sin –1

(Indore, 1995; Jiwaji, 1995, 2003; Sagar, 1998)

ANSWERS 1 1 , . 1  ( x  y ) 2 1  ( x  y )2 2x 2y , 2 . (iii) 2 2 (x  y ) (x  y2 ) 2. (i) ex – y, – ex – y, – ex – y, ex – y.

1.

(i)

(ii) aeax sin by, beax cos by.

PARTIAL DIFFERENTIATION

(ii)

ye( x

y

)

369

x y – 2 [ yx y  y – 1].

e( x

y

)

x y – 1 [1  (1  x y ) log x ].

y

e( x

y

)

x y – 1 [1  (1  x y ) log x y ].

e( x

y

)

x y (1  x y ) (log x ) 2 .

2 y (1  y 2 )

(iii)

(1 – xy )3

,

4. 1/c2 z.

2 ( x  y) (1 – xy )3 5. 0.

,

2 ( x  y) (1 – xy )3

,

2 x (1  x 2 )

. (1 – xy )3 12. – 3/2.

11.8. Homogeneous Functions

(Manipur 2000)

Consider the function f (x, y) = a0 xn + a1 xn – 1 y + a2 xn – 2 y2 + ..... + an – 1 xyn – 1 + an yn ... (i) We see that the expression, f (x, y) is a polynomial in (x, y) such that the degree of each of the terms is the same viz., n. Because of this f is called a homogeneous function of degree n. This definition of homogeneity applies to polynomial functions only. To enlarge the concept of the homogeneity so as to bring even other functions within its scope, we say that an expression in (x, y) is homogeneous of degree n, if it is expressible as xn f (y/x). The polynomial function (i) which can be rewritten as 2 n  y  y  y  x n  a0  a1  a2    .....  an    x x  x   

is a homogeneous expression of order n according to the new definition also. The functions f (x, y) = xn sin (y/x), f (x, y) = ( y 

x ) /( y  x)

are homogeneous according to the second definition only. The degree of the expression xn sin (y/x) is n. Also

y  x  yx

 y y x 1   1 x  x  –1/ 2 x y y  1 x 1   x x 

1 . 2 11.8.1 Euler’s theorem on Homogeneous Functions.

so that it is of degree –

If z = f (x, y) be a homogeneous function of x, y of degree n then x

z z  y x  y = nz,  x, y Œ the domain of the function.

(Kanpur, 2001, 2003; Kumaon, 1997; Vikram, 1999; Jiwaji, 1997, 1999, 2000; Ravishankar, 1996; Sagar, 1999; G.N.D.U. Amritsar, 2004; Indore, 2000; Avadh 2002; Gorakhpur 2003; MDU Rohtak 2000; Manipur 2002; Kuvempu 2005)

370

DIFFERENTIAL CALCULUS

 y z  xn f   x

We have

z  y  y – y  nx n – 1 f    x n f     2 x x x x  y  y  nx n – 1 f   – yx n – 2 f    x   x z  y 1  y  x n f    .  xn – 1 f    y x x x



and

Thus, we have z z  y  y  nx n f    nz. x y x Hence the result. x

Cor. If z = f (x, y) is a homogeneous function of x, y of degree n, then x2

2 z  x2

 2 xy

2 z 2 z  y2  n ( n – 1) z.  x y  y2

(Kumaon, 1999; Indore, 1995, 2000; Vikram, 2001; Jiwaji, 1995, 2001; MDU Rohtak 1998) Differentiating x

z z  y  nz , x y

...(1)

partially with respect to x and y separately, we obtain z 2 z 2 z z x  y n 2 x  x y x x x

2 z z 2 z z   y n . 2  y x y y y

...(2) ...(3)

Multiplying (2), (3) by x, y respectively and adding, we obtain x2

2 z  x2

 2 xy

2 z 2 z  y2  n ( n – 1) z ,  x y  y2

where we have assumed the equality of 2z/xy and 2z/yx. EXAMPLES 1. If u  cot –1

x y

, show that y u u 1 x  y  sin 2u  0. x y 4

Let z = cot u =

x 

x y

x  y z is a homogeneous function of x and y of degree 1/2. Therefore by Euler’s Theorem, z z 1 x  y  z x y 2

...(1)

...(2)

PARTIAL DIFFERENTIATION

371

From (1), z u  – cosec 2 u x x z u  – cosec 2u y y

and From (2)

u u 1 – y cosec2 u  cot u x y 2 u  u –1 cot u x  y  x y 2 cosec 2 u

– x cosec 2 u



x



2. If u = tan –1

u u 1  y  sin 2u  0 x y 4

x 3  y 3 , x  y show that x– y

u u  y  sin 2u x y 2 2 u 2 u 2  u  2 xy  y2  (1 – 4sin 2 u ) sin 2u. (ii) x 2 2  x  y x y

(i) x

(Gorakhpur, 2002; Kumaon, 2003; Rohilkhand, 2003; Vikram, 2000; Ravishankar, 1997, 1998, 2000; Jabalpur, 1997; Rewa, 2001; Bhopal, 1997; Banglore 2004; 2005) (i) Here u is not a homogeneous function. We, however, write

z  tan u 

x3  y 3 1  ( y / x )3  x2 , x– y 1 – ( y / x)

so that z is a homogeneous function of x, y of degree 2. z z  x  x  y  y  2 z.

...(1)

But z z z u  sec2 u ;  sec 2 u . x x y y

...(2)

Substituting in (1), we obtain  u u   y sec 2 u  x   2 z  2 tan u y  x

u  u 2 sin u  y  cos 2 u  sin 2u. x y cos u (ii) From (2)

 x

2 z  x2

 sec 2 u

...(3)

2

 u   2 sec2 u tan u   , 2 x x 2 u

 u   sec u  2 sec u tan u   2 2 y y y 2 z

2

2 u

2

2

372

DIFFERENTIAL CALCULUS

2 z 2 u u u  sec 2 u  2 sec 2 u tan u  x y  x y x y

and

Also by corollary of Euler’s Theorem, 2 z

x2

 2 xy

 x2

2 z 2 z  y2  2 (2 – 1) z  x y  y2

2  2 2 u 2 u 2 2  u   2 xy  y   sec u  x  x y  x2  y 2   2    u 2 u u 2 2 2  u     2 sec u tan u x   y    2 xy   2 tan u  x x y   y    2



2  u 2 u u  2  u x  2 xy  y  2 tan u  x  y  = 2 sin u cos u 2 2  x  y  x  y x y 



x2

2 u

2

2 u  x2

2 u 2 u  y2 = sin 2 u – 2 tan u sin2 2u  x y  y2 = (1 – 2 tan u sin 2u) sin 2u

 2 xy

(by 3)

= (1 – 4 sin2 u) sin 2u 3. If z = xm f (y/x) + xn g (x/y), prove that x2

2 z x 2

 2 xy

 z 2 z 2 z z   y2  mnz  ( m  n – 1)  x  y . 2  x y y  y  x

u = xm f (y/x) and v = xn g (x/y).

Let

z = u + v.

Then Now u =

xm

...(i)

f (y/x) is a homogeneous function in x and y of degree m.

Therefore with the help of Euler’s Theorem we can x2

Prove that

 2u x 2

 2 xy

 2u 2v  y 2 2  m ( m – 1) u x y y

..(ii) [See cor §. 11.8.1]

Also v =

xn

g (x/y) is a homogeneous function in x and y of degree n, so we have 2v

x2

x

2

 2 xy

2v 2v  y 2 2  n ( n – 1) v. x y y

...(iii)

Adding (ii) and (iii), we have x2

x 2

(u  v)  2 xy

2 2 (u  v)  y 2 2 (u  v) = m (m – 1) u + n (n – 1) v. x y y

2 2 z 2  z  y = m (m – 1) u + n (n – 1) v. ...(iv) x y x 2 y 2 Again from Euler’s Theorem we have for u and v, which are homogeneous functions in x and y of degree m and n respectively, 2 or x

2 z

2

 2 xy

PARTIAL DIFFERENTIATION

373

u u v v  y  mu and x  y  nv. x y x y

x

Adding these two with the help of (i), we get z z  y  mu  nv. x y

x

...(v)

m (m – 1) u + n (n – 1) v

Again

= (m2u + n2v) – (mu + nv) = m (m + n) u + n (m + n) v – mn (u + v) – (mu + nv) = (m + n) (mu + nv) – mn (u + v) – (mu + nv) = (m + n – 1) (mu + nv) – mn (z),

from (i)

Substituting this value in (iv), we get x2

or

x2

2 z



x 2 2 z x

2 z 2 z  y 2 2  ( m  n – 1) ( mu  nv ) – mnz x y y

 2 xy

2

2 z 2 z  y 2 2 + mnz = (m + n – 1) (mu + nv) x y y

 z z   ( m  n – 1)  x   from (v)  x y    (2 x 2  y 2  x z )1/ 2  4. If V = loge sin  2 2 1/ 3  , find the value of  2 ( x  xy  2 yz  z ) 

x

Let

V V  V when x = 0, y = 1, z = 2.  y z x  y z  (2 x 2  y 2  xz )1/ 2 , u  2 ( x 2  xy  2 yz  z 2 )1/ 3

V = loge sin u

then 

V u V u  cot u ,  cot u x x  y  y

and

V u  cot u  z z

 u V V V u  u  y  z  cot u  x  y  z   x  y  z  x  y  z  1 But u is a homogeneous function of degree . 3 u u u 1 x  y z  u  x  y z 3



x

 From (i) x

V V V 1  y z  u cot u . x  y z 3

...(i)

374

DIFFERENTIAL CALCULUS

When x = 0, y = 1, z = 2, we have u  x

1  , and then 4

V V V 1     y z  . cot  . x  y z 3 4 4 12

 ( x 4  y 4 )  5. If u = log  ( x  y )  , show by Euler’s Theorem that   x

u u  y  3. x  y

(Jiwaji, 1997; Vikram, 1999; Rewa, 1999; Jabalpur, 1999, 2002) Here,

eu 

( x4  y 4 ) z ( x  y)

(say)

Now z is a homogeneous function of degree (4 – 1) = 3. Hence by Euler’s Theorem z z x  y  3z x  y   x (e u )  y (e u )  3 e u or x  y u u eu . x  eu . y  3 eu or x  y u u x  y  3.  x  y 6. If u = zeax + by, where z is a homogeneous function in x and y of degree n, prove that u u = (ax + by + n) u.  y x y If z is a homogeneous function of degree n in x and y, then from Euler’s Theorem, z z x  y  nz. we get x y u u   x  y x ( zeax  by )  y ( ze ax  by ) Now x y x y  z ax  by  ax  by   ax  by   z e e  x  e ax  by  z  z    y  y e y x   x     z   ax  by z  ax  by  ax  by   x  y  z x  y e e e  y  y  x  x  ax + by ax + by ax + by = nze + z [xae + bye ] = (n + ax + by) zeax + by = (ax + by + n) u. z z 7. If z = (x + y)  (y/x), where  is any arbitrary function prove that x x  y y  z. z =  (y/x) + (x + y) ' (y/x) (– y/x2) x z  y y  y x  x    – ( x  y )    or x  x x x x

...(i)

...(i)

PARTIAL DIFFERENTIATION

375

z  y  y 1      ( x  y )      y x  x x z  y y  y y  y     ( x  y )    . y x x   x

Also or

...(ii)

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

z z  y  y  ( x  y )     z. y y x

Example 10. If sin v = (x + 2y + 3z) / ( x8  y 8  z 8 ), show that x

v v v  y z  3 tan v  0. x y z

We have x  2 y  3z

sin v 

x8  y 8  z 8

 f (say)

Then f is a homogeneous function in x and of degree 1 – 4 = 3.  By Euler’s Theorem x



x cos v



x

f f f  y z –3f x y z

v v v  y cos v  z cos v  – 3 sin v x y z

v v v  y z  – 3 tan v. x y z

EXERCISES 1. Verify Euler’s theorem for (i) z = ax2 + 2hxy + by2. (Kuvempu 2005) (ii) z = (x2 + xy + y2)– 1. y –1 x y  tan –1 . (iii) z  sin (iv) z  x n log . y x x (v) z 

x1/ 4  y1/ 4

(Ravishankar 1997; G.N.D.U. Amritsar, 2004)

x1/ 5  y1/ 5

2. If u  log

x2  y2 u u  y  1. , prove that x x y x  y

(Kanpur 2006)

u u x2  y2  y  tan u. , show that x x  y x y (Garhwal, 2000; Bhopal, 1996, 1999, 2000; Bilaspur, 1998; Vikram, 2001; Jiwaji, 1997; Jabalpur, 1996, 1998; Rewa, 2000; Gorakhpur, 2003; Patna, 2003; Indore, 1998, 2001)

3. (a) If u  sin –1

(b) If u  sin –1 4. If u  cos –1

u u , show that x  x  y  y  0. y

x –

y

x  x y x 

y

, show that x  u  y  u  1 cot u  0. x  y 2

(Garhwal, 2002)

376

DIFFERENTIAL CALCULUS

–1 5. If z  sec

x3  y 3 z z  y  2 cot z. , show that x x  y x y

(Banglore, 2004; 2005)

 x y  u u 1 6. If u  tan –1  , show that x  y  sin 2 u.  x  y   x  y 4   7. If V = rm where r2 = x2 + y2 + z2, show that 2 V



 x2

2 V



 y2

2 V

 m (m  1) r m – 2 .

 z2

1/ 2

8. If u  sin

–1

1/ 3 1/ 3  x  y   1/ 2 1/ 2   x  y 

,

show that x2

2 u  x2

 2 xy

2 u  2 u tan u  y2  (13  tan 2 u ).  x y 144  y2

(Jiwaji, 1998)

9. If z = xyf (y/x) , show that x  z  y  z  2 z , x y and if z is a constant, then f  ( y / x)  f ( y / x) 10. If u  (a) x (b) x

x2 y 2 x2  y2 2 u  x2

 y

dy   xy  x  dx   . dy   y y – x  dx  

, show that 2 u u  y x x

2 u  2 u u  y  x y y y 2

2 2 u 2  u  y  2 u. (Utkal 2003) x y x 2 y 2 11.9. For the following developments, it will be assumed that f possesses continuous partial derivative w. r. to x and y for every point of domain of the function. 11.9.1. Theorem on Total Differentials. We consider a function z = f (x, y) ...(i) Let (x, y), (x + x, y + y) be any two points so that x, y are the changes in the independent variables x and y. Let  z be the consequent change in z. We have z + z = f (x + x, y + y) ...(ii) 2 (c) x

2 u

 2 xy

From (i) and (ii), we get z = f (x + x, y + y) – f (x, y) = [f (x + x, y + y) – f (x + x, y)] + [f (x + x, y) – f (x, y)] ...(iii) so that we have subtracted and added f (x + x, y). Here the change  z has been expressed as the sum of two differences, to each of which we shall apply Lagrange’s mean value theorem.

PARTIAL DIFFERENTIATION

377

We consider the function f (y) = f (x + x, y) ; where x + x is constant, so that by the mean value theorem, f (x + x, y + y) – f (x + x, y) = y fy (x + x, y + 1 y) We write fy (x + x, y + 1 y) – fy (x, y) = 2 ...(iv) so that 2 depends on x, y and because of the assumed continuity of fy tends to zero as x, y both tend to 0. Again by considering f (x) = f (x, y), keeping y constant, we have by the mean value theorem, f (x + x, y) – f (x, y) = x fx (x + 2 x, y), We write fx (x + 2 x, y) – fx (x, y) = 1 ...(v) so that 1 depends upon x and because of the assumed continuity of fx tends to 0 as x tends to 0. From (iii), (iv) and (v), we get z = x fx (x, y) + y fy (x, y) + 1 x + 2 x 

z z   x  2  y x  y  1 . x y    

Thus the change z in z consists of two parts as marked. Of these the first is called the differential of z and is denoted by dz. Thus dz 

z z x  y x y

...(vi)

Now z = x  dx = dz = 1 . x = x. Similarly by taking z = y we have y = dy. Thus (vi) takes the form dz 

z z dx  dy. x y

It should be carefully noted that the differentials dx and dy of the independent variables x and y are the actual changes  x and  y, but the differential dz of the dependent variable z is not the same as the change  z; it being the principal part of the increment  z. Cor. Approximate Calculations. From above, we see that the approximate change dz in z corresponding to the small changes x and y in x, y is z z dx  dy x y which has been denoted by dz. 11.9.2 Composite functions Let

z = f (x, y)

...(i)

and let

x =  (t)

...(ii)

y =  (t)

...(iii)

so that x, y are themselves functions of a third variable t.

DIFFERENTIAL CALCULUS

378

The equations (i), (ii), (iii) are said to define z as a composite function of t. Again, let

x =  (u, v)

...(iv)

y =  (u, v)

...(v)

so that x, y are functions of the variables u, v. Here the functional equations (i), (iv), (v) define z as a function of u, v, which is called a composite function of u and v. 11.9.3 Differentiation of composite functions. Let z = f (x, y) possess continuous partial derivatives and let x =  (t) y =  (t) possess continuous derivatives. dz  z dx  z dy  .  . Then dt  x dt  y dt Let t, t +  t be any two values. Let  x, y,  z be the changes in x, y, z consequent to the change  t in t. We have x +  x =  (t +  t), y +  y =  (t +  t) z +  z = f (x +  x, y +  y) 

 z = f (x +  x, y +  y) – f (x, y) = [f (x +  x, y +  y) – f (x, y +  y)] + [f (x, y +  y) – f (x, y)].

As in § 7.2, we apply Lagrange’s mean value theorem to the two differences on the right and obtain  z =  x fx (x + 1  x, y +  y) + y fy (x, y + 2  y) 

z x y  f x ( x  1  x, y   y )  f y ( x, y   2  y ) t t t

(0 < 1, 2 < 1). ...(i)

Let  t  0 so that  x and  y  0. Because of the continuity of partial derivatives, we have lim

(  x ,  y )  (0,0)

lim

y 0

f x ( x  1  x, y   y )  f x ( x, y ) 

f y ( x, y   2  y )  f y ( x, y ) 

z , x

z . y

Hence in the limit, (i) becomes dz  z dx  z dy  .  . dt  x dt  y dt

Cor 1. Let

z = f (x, y),

possess continuous first order partial derivatives w. r. to x, y. Let

x =  (u, v) y =  (u, v)

possess continuous first order partial derivatives.

...(ii) (Nagpur 2005)

PARTIAL DIFFERENTIATION

379

To obtain  z/ u, we regard v as a constant so that x and y may be supposed to be functions of u only. Then, by the above theorem, we have z z x z y  .  . . u  x u  y u

It may similarly be shown that z z  x z y  .  . . v x v y v

EXAMPLES 1. Find dz/dt when z = xy2 + x2y, x = at2, y = 2at. Verify by direct substitution. Now

z z  y 2  2 xy,  2 xy  x 2 , x y

dx dy  2at ,  2a. dt dt Substituting these values in (ii), § 11.9.3, we get dz = (y2 + 2xy) 2at + (2xy + x2) 2a dt = (4a2 t2 + 4a2 t3) 2at + (4a2 t3 + a2 t4) 2a

= a3 (16t3 + 10t4) Again

z = x2 y + xy2 = 2a3 t5 + 4a3 t4.

dz = 10a3 t4 + 16a3 t3 = a3 (16t3 + 10t4). dt Hence the verification.



2. z is a function of x and y. Prove that if x = e u + e – v, y = e – u – e v , then

z z z z – x – y . u  v x y

We look upon z as a composite function of u, v. We have z z x z y   . u  x u  y u 

 z u  z –u .e – .e ; x y

z z x z y  .  . v x v  y v –

 z –v  z v .e – .e . x y

380

DIFFERENTIAL CALCULUS

Subtracting, we get z z z u  z –u –  (e  e – v ) – (e – e v ) u  v  x y x

z z – y . x y

3. If H = f (y – z, z – x, x – y), prove that H H H    0. x y z

(Kumaon, 1998, 2000; Garhwal, 2001; MDU Rohtak 1999; Manipur 2000)

Let

u = y – z, v = z – x, w = x – y,



H = f (u, v, w).

We have expressed H as a composite function of x, y, z. We have  H  H u  H v  H  w  .  .  . x u  x v x w x 

H H H .0 (–1)  .1 u v w

–

H H  v w

Similarly H H H –  y w u H H H –  . z u v Adding, we get the result. 4. H is a homogeneous function of x, y, z of degree n, prove that H H H  y z  nz. x y z [Thus is Euler’s theorem for a homogeneous function of three independent variables.] We have x

y z H  x n f  ,  = xn f (u, v) where y/x = u, z/x = v.  x x  But Hence Again

 f u  f  v H  nx n – 1 f (u , v)  x n  .  .  x  u  x  v  x u y v z – ,  – 2. x x x x  f H f  nx n – 1 f (u , v) – x n – 2  y  z . x v  u  f u  f  v  H  xn   . .  y  u  y  v  y

 xn – 1

f u 1  v , for  ,  0. u y x y

PARTIAL DIFFERENTIATION

381

H f  xn – 1 z v

Similarly

H H H  y z  nx n f (u , v)  nz. x y z 5. In a triangle ABC, the angles and sides a and b are made to vary in such a way that the area remains constant and the side c also remains constant. Show that a and b vary by small amounts a, b respectively, then cos A a + cos B b = 0. Let  denote the area of the triangle ABC. Then



x



1 ac sin B 2

where  is constant, Also, we know that b2 = c2 + a2 – 2ac cos B Differentiating (i), we get 1 c [ a cos B B  sin B a ] 2  a cos B B = – sin B a Differentiating (ii), we get 2b b = 2a a – 2c {a (– sin B) B + cos B a}  b b = (a – a cos B) a + ac sin B B

...(i)

...(ii)

0

...(iii)

 – sin B a  = (a – a cos B) a + ac sin B    a cos B   (a – c cos B) cos B – c sin 2 B    a cos B    a cos B – c   a cos B – ( a cos B  b cos A)    a  a   B cos B  cos     b cos A  –  a  cos B   cos A a + cos B b = 0. 6. If the sides and the angles of a plane triangle ABC vary in such a way that its circum radius remains constant, prove that (a/cos A) + (b/cos B) + (c/cos C) = 0, where a, b and c denote small increments in the sides a, b and c respectively. Let R be the circum radius of the  ABC, then a b c R   2 sin A 2 sin B 2 sin C  a = 2R sin A, b = 2R sin B, c = 2R sin C Differentiating, we get a = 2R cos A A, b = 2R cos B B and c = 2R cos C C



a b c    2 R (A  B  C ) cos A cos B cos C = 2R  (A + B + C) = 2R  ()

= 0.

382

DIFFERENTIAL CALCULUS

11.9.4 Implicit functions. Let f be any function of two variables. Ordinarily, we say that, since on solving the equation f (x, y) = 0 ...(i) we can obtain y as a function of x, the equation (i) defines y as an implicit function of x. There arises a theoretical difficulty here. Without investigation we cannot say that corresponding to each value of x, the equation (i) must determine one and only one value of y so that the equation (i) always determines, y, as a function of x and in fact, this is not the case, in general. The investigation of the conditions under which the equation (i) does define y as a function of x is not, however, within the scope of this book. Assuming that the conditions under which the equation (i) defines y as a derivable function of x are satisfied, we shall now obtain the values of dy/dx and d2y/dx2 in terms of the partial derivatives  f/ x,  f/ y, 2 f/ x2, 2 f/ x  y, 2 f/ y2 of ‘f ’ w.r. to x and y. Now, f is a function of two variables x, y and y is again a function of x so that we may regard f as a composite function of x. Its derivative with respect to x is  f dx  f dy  f  f dy .  .   . .  x dx  y dx  x  y dx

As f considered as a function of x alone, is identically equal to 0. As such we have f  f dy  . 0 x  y dx f dy  f / x –  – x if f y  0. dx  f / y fy Differentiating again w.r. to x, regarding  f/ x and  f/ y as composite functions of x, we get

 2 f  2 f dy   f  f  2 f  2 f dy    –     2  y  x dx   y  x   x  y  y 2 dx  d2y x  – 2 dx 2  f   y   2

2 f   f  2 f  f  f 2 f   f  – 2       yx  x  y  x2   y   y2   x  – 3  f   y  

Hence

and

2

f dy – x, dx fy d2y dx 2

–

...(ii)

f x 2 (f y ) 2 – 2f yx f x f y  f y 2 (f x ) 2 (f y ) 3

...(iii) (MDU Rohtak 1999)

Without making use of § 11.9.3, we may obtain dy/dx also as follows : Now

f (x, y) = 0.

Let  x be the increment in x and  y the consequent increment in y so that f (x +  x, y +  y) = 0

PARTIAL DIFFERENTIATION

383

 f (x +  x, y +  y) – f (x, y) = 0  f (x +  x, y +  y) – f (x, y +  y) + f (x, y +  y) – f (x, y) = 0   x fx (x + 1  x, y +  y) +  y fy (x, y + 2  y) = 0 f x ( x  1  x, y   y ) y .  x  – f y ( x, y   2  y ) Let  x  0. We obtain f dy  – x if f y  0. dx fy

EXAMPLES 1. Prove that if y3 – 3ax2 + x3 = 0, then d2y dx 2

Let



2a 2 x 2 y5

 0.

f (x, y) = y3 – 3ax2 + x3 = 0

fx (x, y) = – 6ax + 3x2, fy (x, y) = 3y2 ; fx2 (x, y) = – 6a + 6x, fxy (x, y) = 0, fy2 (x, y) = 6y Substituting these values in (iii), on p. 468, we get We have

d2y dx 2

–

6 ( x – a ) 9 y 4  (3 x 2 – 6ax) 2 6 y

–2 –2 d2y

( x – a ) (3ax 2 – x3 )  ( x 2 – 2ax) 2 y5 a2 x2 y5

.

2a 2 x 2

 0. dx 2 y5 Or, directly, differentiating the given relation w.r. to x,

Thus



27 y 6

3y2



dy  6ax – 3 x 2 dx dy 2ax – x 2  dx y2

d2y



dx 2

(2a – 2 x) y 2 – 2 y (2ax – x 2 ) dy / dx



y4

 

(2a – 2 x) y 2 – 2 y (2ax – x 2 ) (2ax – x 2 ) / y 2 y4 2 ( a – x) y 3 – 2 (2ax – x 2 ) 2

y5 2 2 2. If ax + 2hxy + by + 2gx + 2fy + c = 0, prove that d2 y dx 2



abc  2 fgh – af 2 – bg 2 – ch 2 ( hx  by  f )3

.

–

2a 2 x 2 y5

, as before.

384

DIFFERENTIAL CALCULUS

f (x, y) = ax2 + 2hxy + by2 + 2gx + 2fy + c fx (x, y) = 2 (ax + hy + g) fy (x, y) = 2 (hx + by + f) fxy = 2h = fyx

f x2  2a f y 2  2b d2y



dx 2



{8a ( hx  by  f ) 2 – 16h ( ax  hy  g ) 8 ( hx  by  f )3 ( hx  by  f )  8b ( ax  hy  g ) 2 } 8 ( hx  by  f )3

– ab

(ax2 2



+ 2hxy +

by2

+ 2gx + 2fy) + 2fgh –

2

af2

–

bg2

2

 h (ax  2hxy  by  2 gx  2 fy )



(hx  by  f )3 abc  2 fgh – af 2 – bg 2 – ch 2 (hx  by  f )3

3. If A, B, C are the angles of a triangle such that sin2 A + sin2 B + sin2 C = constant prove that We have

dA tan B – tan C  . dB tan C – tan A sin C = sin ( – (A + B))

= sin (A + B) Let

f (A, B)  sin2 A + sin2 B + sin2 (A + B) – constant fA (A, B) = 2sin A cos A + 2sin (A + B) cos (A + B)



fB (A, B) = 2sin B cos B + 2sin (A + B) cos (A + B) f dA – B dB fA 2sin B cos B  2 sin (A  B) cos (A  B) – 2 sin A cos A  2 sin (A  B) cos (A  B) sin 2B – sin 2C – sin 2A – sin 2C – cos A {sin C cos B – cos C sin B}  – cos B {sin A cos C – cos A sin C} 

cos A sin B cos C – cos A cos B sin C cos A cos B sin C – sin A cos B cos C



tan B – tan C tan C – tan A

EXERCISES 1. If u = x2 – y2, x = 2r – 3s + 4, y = – r + 8s – 5, find  u/ r. 2. If z = (cos y)/x and x = u2 – v, y = ev, find  z/ v.

PARTIAL DIFFERENTIATION

385

sin u cos y cos x ,u  ,v , find  z /  x. cos v sin x sin y 4. If u = (x + y)/(1– xy) ; x = tan (2r – s2), y = cot (r2 s), find  u/ s.

3. If z 

5. Find dy/dx in the following cases:— y

(i) x sin (x – y) – (x + y) = 0.

(ii)

y x  sin x.

(iii) (cos x)y – (sin y)x = 0.

(iv)

x y = yx.

(v) (tan x)y + ycot x = a. 6. If F (x, y, z) = 0, find  z/ x,  z/ y. 7. If z = xyf (y/x) and z is a constant, show that f  ( y / x) x [ y  x ( dy / dx)] .  f ( y / x) y [ y – x ( dy / dx)]

8. If f (x, y) = 0,  (y, z) = 0, show that  f   dz  f   . .  . .  y  z dx  x  y

9. If x (1 – y 2 )  y d2y dx

2



(Kumaon, 1999; Rohilkhand, 2001)

(1 – x 2 )  a , show that a

(1 – x 2 )3/ 2

.

10. If u and v are functions of x and y defined by x = u + e– v sin u, y = v + e– v cos u, prove that

u  v  . y x

11. Find d2 y/dx2 in the following cases :— (i) x3 + y3 = 3axy.

(ii) x4 + y4 = 4a2 xy.

(iii) x5 + y5 = 5a3 xy.

(iv) x5 + y5 = 5a3 x2.

12. If the curves f (x, y) = 0 and  (x, y) = 0 touch each other, show that at the point of contact  f   f  . – .  0. x y y x

ANSWERS 1. 4x + 2y. 2. (cos y – xy sin y)/x2. 3. – (u cot x cos u sin y + z sin v sin x)/(cos v sin y). 4. – (1 + x2) (1 + y2) (2s + r2)/(1 – xy)2. 5. (i)

[y + x2 cos (x – y)]/[x + x2 cos (x – y)].

(ii)

(cot x – yxy – 1 log y) y/xy (y log x log y + 1).

(iii)

sin y ( y sin x  cos x log sin y ) . cos x (sin y log cos x – x cos y )

(iv) y (y – x log y)/x (x – y log x).

386

DIFFERENTIAL CALCULUS

(v)

–

y (tan x) y – 1 sec 2 x – cosec2 x log y ( y )cot x log tan x (tan x ) y  cot x ( y )cot x – 1

.

z F / x z F / y – ; – . x F / z y F / z

6.

11. (i) (iii)

2a 3 xy (ax – y 2 )

, 3

(ii)

6a 3 xy (2a 6  x3 y 3 ) (a 3 x – y 4 )3

,

(iv)

2a 2 xy (3a 4  x 2 y 2 ) ( a 2 x – y 3 )3 6a 3 x 2 ( a 3  x 3 ) y9

,

.

11.10. Equality of fxy (a, b) and fyx (a, b). If has been seen that the two repeated second order partial derivatives are generally equal. They are not, however, always equal as is shown below by considering two examples. It is easy to see a priori also why fyx (a, b) may be different from fxy (a, b). We have Also and 

f xy (a, b)  lim

f y ( a  h, b ) – f y ( a , b ) h

h0

fy (a + h, b) = lim

k 0

fy (a, b) = lim

k 0

f ( a  h, b  k ) – f ( a  h, b ) , k

f ( a , b  k ) – f ( a, b ) . k

fxy (a, b) = lim lim

h 0 k 0

 lim lim

h0 k 0

f ( a  h, b  k ) – f ( a  h , b ) – f ( a , b  k )  f ( a , b ) . hk

 (h, k ) , say. hk

It may similarly be shown that  ( h, k ) . hk Thus we see that fxy (a, b) and fyx (a, b) are repeated limits of the same expression taken in different orders. Also the two repeated limits may not be equal, as, for example

fyx (a, b) = lim lim

k  0 h 0

lim lim

h–k h  lim  1, h  k h 0 h

lim lim

h–k –k  lim  – 1. h  k k 0 k

h 0 k 0

and

k 0 h 0

EXAMPLES 1. Prove that fxy (0, 0)  fyx (0, 0) for the function f given by xy ( x 2 – y 2 ) f (x, y) = ; (x, y)  (0, 0) x2  y2 f (0, 0) = 0. f y (0  h, 0) – f y (0, 0) We have fxy (0, 0) = lim h 0 h Also

fy (0, 0) = lim

k 0

f (0, 0  k ) – f (0, 0) k

...(1) ...(2)

PARTIAL DIFFERENTIATION

387

f (0, k ) – f (0, 0) 0  lim  0 k 0 k k f ( h, 0  k ) – f (h, 0) fy (h, 0)  lim k 0 k  lim

k 0

and

 lim

k 0

hk ( h 2 – k 2 ) k (h 2  k 2 )

 h.

...(3)

Thus from (1), (2) and (3) h–0  1. h 0 h f x (0, 0  k ) – f x (0, 0) fyx (0, 0) = lim k 0 k

fxy (0, 0)  lim Again Also and

f (0  h, 0) – f (0, 0) 0  lim  0 h0 h h f (0  h, k ) – f (0, k ) fx (0, k) = lim h 0 h hk ( h 2 – k 2 )  lim  – k. h0 h ( h 2  k 2 )

fx (0, 0) = lim

h 0

From (4), (5) and (6), –k –0  – 1. k 0 k Thus fxy (0, 0) = 1  – 1 = fyx (0, 0).

fyx (0, 0) = lim

2. Show that fxy (0, 0)  fyx (0, 0) where

f (x, y) = 0 if xy = 0, y x – y 2 tan –1 , if xy  0. x y f y (h, 0) – f y (0, 0)

f (x, y) = x 2 tan –1 We have fxy (0, 0) = lim

h f ( h, k ) – f ( h, 0) fy (h, 0) = lim k 0 k 1 2 k h  lim  h tan –1 – k 2 tan –1  k 0 k  h k h 0

  tan –1 k / h   –1 h  lim  h  – k tan    k 0   k    k /h  = h . 1 – 0 = h, for [tan–1 t/t]  1 as t  0. Also



h  h   , lim tan –1  – k 2 k (0 – 0) k 2 f (0, k ) – f (0, 0)  0. fy (0, 0) = lim k 0 k h–0  1. fxy (0, 0) = lim h 0 h lim

k  (0  0)

tan –1

...(4) ...(5)

...(6)

388

DIFFERENTIAL CALCULUS

We may similarly show that fyx (0, 0) = – 1. Hence the result. 11.10.1. Equality of fxy and fyx. Theorem. If f (x, y) possesses continuous second order partial derivatives fxy and fyx, then fxy = fyx. Consider the expression  (h, k) = f (x + h, y + k) – f (x + h, y) – f (x, y + k) + f (x, y). For the sake of brevity, we write  (x) = f (x, y + k) – f (x, y) 

 (h, k) =  (x + h) –  (x).

By Lagrange’s mean value theorem,  (x + h) –  (x) = h  '(x + 1 h), Also 

0 < 1 < 1.

' (x) = fx (x, y + k) – fx (x, y)  (h, k) = h [fx (x + 1 h, y + k) – fx (x + 1 h, y)].

...(1)

Again applying the mean value theorem to the right side of (1), we obtain  (h, k) = hk fyx (x + 1 h, y + 2 k)

0 < 2 < 1.

 ( h, k ) = fyx (x + 1 h, y + 2 k). hk Again considering F (y) = f (x + h, y) – f (x, y) in place of  (x) and proceeding as before, we may prove that

Thus

 ( h, k ) = fxy (x + 3 h, y + 4 k). hk  fxy (x + 1 h, y + 2 k) = fxy (x + 3 h, y + 4 k).

Let (h, k)  (0, 0). Then, because of the assumed continuity of the partial derivatives, we obtain fyx (x, y) = fxy (x, y). 

fyx = fxy.

11.11. Taylor’s theorem for a function of two variables. As in the case of function of a single variable, we can prove Taylor’s theorem with remainder after n terms. We shall however consider the case of Taylor’s theorem with remainder after 3 terms only. 11.11.1. Theorem. If f possesses continuous partial derivatives of the third order in a neighbourhood of a point (a, b) and if (a + h, b + k) be a point of this neighbourhood, then there exists a positive number  which is less than 1, such that f (a + h, b + k) = f (a, b) + [hfx (a, b) + kfy (a, b)] 

1 2 1 [h f x 2 ( a, b)  2hkf xy ( a, b)  k 2 f y 2 ( a, b)]  [ h 2 f x3 (u , v) 2! 3!

 3h 2 kf y 2 x (u, v)  3hk 2 f y 2 x (u , v )  k 3 f y 2 x (u, v)]

PARTIAL DIFFERENTIATION

389

where u = a +  h, v = b +  k We write z = f (x, y) and x = a + ht, y = b + kt so that z is a function of t which are denote by g (t). We have

(Utkal 2003)

dx dy  f y ( x, y ) dt dt = hfx (x, y) + kfy (x, y) dx dy  dx dy    g (t )  h  f x2 ( x, y )  f yx   k  f xy dt  f y 2 dt  dt dt      h 2 f x2 ( x, y )  2hkf xy ( x, y )  k 2 f y 2 ( x, y )]

g' (t) = fx (x, y)

dx dy  dy dy    g (t )  h 2  f x3 ( x, y )  f yx2  2hk  f x2 y ( x, y )  f y 2 x ( x, y )   dt dt  dx dx    dx dy    k 2  f xy 2 ( x, y )  f y 3 ( x, y )  dt dt   3 2 2 3  h f x3 ( x, y )  3h kf x2 y ( x, y )  3hk f y 2 x ( x, y )  k f y3 ( x, y ) t2 t3 g  (0)  g  (t ), 2! 3! 1 1 g (1)  g (0)  g  (0)  g  (0)  g  (). 2! 3!

Now we have g (t )  g (0)  tg  (0)  

0 0. As (c) = f (c) is positive, there exists an interval [c – , c + ] around c for every point x of which (x) is positive. Thus, (x) is strictly increasing in [c – , c + ]. Also as (c) = 0, we see that (x) < 0  x  [c – , c[, and (x) > 0  x  ] x, c + ].

416

DIFFERENTIAL CALCULUS

Again it follows that  is strictly decreasing in [c – , c] and strictly increasing in [c, c + ]. Also  (c) = 0. Finally, it follows that RQ =  (x) > 0  x  [c – , c [] c, c + ]. Thus, the curvey y = f (x) is concave upwards at [c, f (c)] if f (c) > 0. Case II. Let f (c) < 0. In this case, we may show that there exists an interval [c – , c + ] around c such that RQ =  (x) < 0  x  [c – , c [] c, c + ]. Thus, the curvey = f (x) is concave downwards at [c, f (c)] if f (c) < 0. Case III. Let f (c) = 0. We know that a curve has inflexion at P if it changes from concavity upwards to concavity downwards, or vice-versa, as a point moving along the curve passes through P, i.e., if there is a complete neighbourhood of P such that at every point on one side of P lying in this neighbourhood, the curve is concave upwards and at every point on the other side of P the curve is concave downwards. Thus, the curve has inflexion at [c, f (c)], if f  (x) changes sign as x passes through c. Since f  is assumed to be continuous, it follows that at a point of inflexion f (c) = 0. Note. To determine the criteria for an extreme value for x = c, we have to examine the behaviour of the sign of f (x) – f (c) and for determining the criteria for concavity upwards (downwards), point of inflexion for x = c, we examine the behaviour of the sign of f (x) – f (c) – f (c) (x – c). Thus the curve y = f (x) (a) is rising in [a, b] if is concave upwards in [a, b] if f (x) > 0  x  [a, b] f (x) > 0  x  [a, b]. (b) is falling in [a, b] if is concave downwards in [a, b] if f (x) < 0  x  [a, b] f (x) < 0  x  [a, b]. (c) has an extreme point at has a point of inflexion at [c, f (c)] [c, f (c)] if f (c) = 0 and if f  (c) = 0 and f (x) changes sign f (x) changes sign as x as x passes through c. passes through c. EXAMPLES 1. Find the ranges of values of x for which the curve y = x4 – 6x3 + 12x2 + 5x + 7 is concave upwards or downwards. Also determine the points of inflexion. We have

dy  4 x 3 – 18 x 2  24 x  5, dx d2y  12 x 2 – 36 x  24  12 ( x – 1) ( x – 2) dx 2 d2 y d2y  0  x  ] –  ,1[,  0  x ]1, 2[ dx 2 dx 2

CONCAVITY AND POINTS OF INFLEXION

417

d2y

 0  x ] 2,   [. dx 2 Thus, the curve is concave upwards in the intervals [– , 1] and [2, + ] and concave downwards in the interval [1, 2]. Also the curve has inflexions for x = 1 and x = 2 in that d2y/dx2 changes sign as x passes through 1 and 2. It follows that (1, 19) and (2, 33) are the two points of inflexion on the curve. 2. Show that the curve (a2 + x2) y = a2x has three points of inflexion.

We have

dy a 2 ( a 2  x 2 ) – 2a 2 x 2 a 2 ( a 2 – x 2 )   dx (a 2  x 2 )2 (a 2  x 2 ) 2 d2 y dx 2

 a2 .

– 2 x (a 2  x 2 ) 2 – 4 x (a 2  x 2 ) (a 2 – x 2 )

 a2 .

(a 2  x 2 ) 4 – 2 x (3a 2 – x 2 ) ( a 2  x 2 )3

 2 a2

(x –

3a ) x ( x  3 a ) ( a 2  x 2 )3

d2y

 0 for x  3a, 0, – 3 a. dx 2 It may be seen that d2y/dx2 changes sign as, x passes through each of these three values of x. Hence, the curve has inflexions at the corresponding points. Thus,



( 3 a, 3 a / 4), (0, 0) (– 3 a, – 3 a / 4) are the three points of inflexion on the curve. 3. Find the points of inflexion on the curve y = (log x)3. We have

dy 1  3(log x ) 2 . dx x

d2y dx

2

 6 log x . 

3log x x2

1 x

2

–

3 x2

(log x)2

(2 – log x) 

3log x x2

log

e2 x

d2y

 0 if log x = 0 or log x = 2. dx 2 Now, log x = 0  x = e0 = 1 and log x = 2  x = e2.

Thus,

Thus, we expect points of inflexion on the curve for x = 1 and x = e2. Now, d2y/dx2 changes sign from negative to positive as x passes through 1 and changes sign from positive to negative as x passes through e2. Thus, (1, 0) and (e2, 8) are the two points of inflexion of given curve. EXERCISES 1. Show that y = ex is everywhere concave upwards and the curve y = log x is everywhere concave downwards. 2. Examine the curve y = sin x for concavity upwards, concavity downwards and for points of inflexion in the interval [–2, 2]. Also indicate the positions of the points of inflexion on the curve.

418

DIFFERENTIAL CALCULUS

3. Find the ranges of values of x in which the curves (i) y = 3x5 – 40x3 + 3x – 20 (Banglore 2006) (ii) y = (x2 + 4x + 5) e–x are concave upwards or downwards. Also find their points of inflexion. 4. Find the intervals in which the curve y = (cos x + sin x) ex is concave upwards or downwards; x varying in the interval ] 0, 2 [. 5. Find the points of inflexion on the curves. (i) y = ax3 + bx2 + cx + d (ii) y = (x3 – x)/(3x2 + 1) (iii) x = 3y4 – 4y3 + 5 (iv) x = (y – 1) (y – 2) (y – 3) (v)

y

a 2 (a – x) a2  x2 x

(vi)

(iii) x = 2a cot , y = 2a sin2 

(iv)

y

y

x3 a2  x2 2

(viii) y  be – ( x / a ) x2  2x  2 (ix) xy = a2 log (y/a) (x) y = x2 log (x2 / e3) 2 2 (xi) y = x (x + 1) (xii) a2y2 = x2 (a2 – x2) (xiii) 54y = (x + 5)2 (x3 – 10) Also, obtain the equations of the inflexional tangents to the curves (ii), (iii) and (xi). 6. Show that the curve ay2 = x (x – a) (x – b) has two and, only two points of inflexion. 7. Show that the points of inflexion of the curve y2 = (x – a)2 (x – b) lie on the line 3x + a = 4b. (MDU Rohtak 1998) 8. Find the points of inflexion on the curves : (i) x = a(2 – sin ), y = a(2 – cos ) (ii) x = a tan t, y = a sin t cos t (vii)

(v) y = x2e–x (vii) y = x(x – 3) e2x (ix)

y  e ( x – 1) / x

2

y  xe – x

2

(vi) (viii)

y = ex/2 – 2e–x/2 y = e–2x cos x

(x)

y = x + ex

(xi) y = 2ex + e– x (xii) y = x2 – 6 + ex 9. Find the points of inflexion on the curves : (i) r = a2 / (2 – 1) (ii) r2  = a2  (iii) r = ae / (1 + ) (Transform to the cartesian parametric forms of equations). 10. Show that r = an has points of inflexion if and only if n lies between 0 and –1 and these are given by    [– n (n  1)]. 11. Show that the curve r e = a (1 + ) has no point of inflexion. ANSWERS 2. Concave downwards in [– 2, –] and [0, ]; upwards in [– , 0] and [, 2]. 3. (i) Concave upwards in [–2, 0] and [2,  [; concave downwards in ] – , – 2] and [0, 2]. Inflexions at (–2, 198), (0, –20), (2, –238). (ii) Concave upwards in ] –, –1] and [1, [ ; concave downwards in [–1, 1]. Inflexions at (–1, 2e) and (1, 10/e). 4. Concave upwards in [0, /4] and [5/4, 2] and concave downwards in [/4, 5/4].

CONCAVITY AND POINTS OF INFLEXION

419

5. (i) For x = –b/3a. (ii) (0, 0), (1, 0), (–1, 0).  119 2  , . (iii) (5, 0),  (iv) (0, 2).  27 3  (v) For x = –a and a (2  3) (vi) For x = 0 and  a 3 . (vii) For x = –2 and 1  3 . (viii) For x =  a / 2 .  3 – 3/ 2  , ae3/ 2  . (ix)  ae (x) (1, 3). 2  4  1 . (xi)  3 ,  (xii) (0, 0). 3 3  (xiii) For x = –2, –4  18 . Inflexional tangents to (ii) are x + y = 0, x – 2y ± 1 = 0. Inflexional tangents to (iii) are x = 5 and 3 (9x + 6y) = 151. Inflexional tangents to (xi) are 9 x  3 3 y  1  0.

  2 3  3a    a  4n  ,  3 2  2    2a 3a  , (iii)    3 2   (v) For x  2  2 .

8. (i)

 (ii)   3 a ,  

3a   , (0, 0). 4 

(iv) For x = 0, 

3 . 2

(vi) For x = log 2.

1 1 3  11. (viii) For x : tan x  – . 2 2 4 (ix) For values of x which are the roots of the cubic 2x3 – 5x2 – 4x + 4 = 0. (x) No inflexion. (xi) No inflexion. (xii) No inflexion. 1  9. (i) (3a/2, ± 3 ). (ii)   2a ,  .  2 (iii) (a, 0).

(vii) For x 

OBJECTIVE QUESTIONS For each of the following questions, four alternatives are given for the answer. Only one of them is correct. Choose the correct alternative. 1. A curve is concave upwards if (a)

d2y dx 2 d2y

0 d3 y

0 dx 2 dx3 2. A curve is convex upwards if

(c)

(a) (c)

d2y dx 2 d2y dx 2

 0 and

(b)

0  0 and

(d)

(b) d3y dx3

0

(d)

d2y

0 dx 2 dy becomes infinite dx d2y

0 dx 2 dy becomes infinite dx

CONCAVITY AND POINTS OF INFLEXION

420

3. At the point of inflexion : (a)

d2y dx 2 d2y

0 d3y

d2y dx 2 d3 y

0

0  0. (d) dx 2 dx3 dx3 4. The curve y = x3 – 3x2 – 9x + 9 has a point of inflexion at (a) x = –1 (b) x = 1 (c) x = –3 (d) x = 3. 5. In the curve am – 1 y = xm, origin is a point of inflexion if (a) m is zero (b) m is even and greater than 2 (c) m is odd and greater than 2 (d) Origin is not a point of inflexion.

(c)

 0,

(b)

ANSWERS 1. (a)

2. (b)

3. (c)

4. (b)

5. (c)

CHAPTER 14 CURVATURE AND EVOLUTES 14.1. Introduction. Definition of Curvature In our every day language, we make statements which involve the comparison of bending or curvature of a road at two of its points. For instance, at times, we say, “The bend of the road is, sharper at this place than at that.” Here we depend upon intuitive means of comparing the curvature at the two points and the same may be reliable only when the difference is fairly marked. But we are far from assigning on the basis of intuition only any definite numerical measure to the curvature at any given point of a curve. In order to make ‘Curvature’ a subject of mathematical investigation, we have to assign, by means of some suitable definition, a numerical measure to it in a manner which may be in agreement with out intuitive notion of curvature. Thus we shall now proceed to do. Consider a curve and a point P thereon. We take a point Q on the curve lying near P. Let A be a fixed point on the curve. Let arc AP = s, arc AQ = s + s so that arc PQ = s. Let  be the angles which the tangents at P and Q make with some fixed line. The symbol  denotes the angle through which the tangent turns as a point moves along the curve from P to Q through a distances. According to our intuitive feeling, Fig. 14.1  will be large or small, as compared with s , depending on the degree of sharpness of the bend. This suggests the following definitions :– (i) the total bending or total curvature or the arc PQ is defined to be the angle . (ii) the average curvature of the arc PQ is defined to be the ratios, and (iii) the curvature of the curve at P is defined to be  d   . s ds Thus, by def., d /ds is the curvature of the curve at the point P. lim

QP

421

(Banglore 2005)

422

DIFFERENTIAL CALCULUS

Curvature of a circle. To prove that the curvature of a circle is constant. Intuitively, we feel that the curvature of a circle is the same at every point and that the larger the radius of the circle, the smaller will be its curvature. It will now be shown that these intuitive conclusions are consequences of our formal definition of curvature. Consider a circle with radius, r and centre O. Let P, Q be two points on the circle and let arc PQ = s.

Fig. 14.2

Also let L be the point where the tangents PT, QT at P and Q meet. We have  POQ =  TLT = . From Trigonometry arc PQ  POQ , OP s  1     .  r s r Let Q  P so that in the limit, we have d/ds = 1/r. Thus the curvature at any point of a circle is the reciprocal of the radius of the circle and, is, thus, a constant. Also, it is clear that the curvature, 1/r, decreases as the radius increases. Ex. Show that the curvature of a straight line at every point thereof is zero. Radius of curvature. The reciprocal of the curvature of a curve at any point, in case it is  0, is called its radius of curvature at the point and is generally denoted by, , so that we have  = ds/d. Thus the radius of curvature of a circle at any point, is equal to the radius of the circle. Ex. Find the radius of curvature at any point of the following :– (i) s = c tan  (catenary), (ii) s = 4a sin  ( cycloid), 1 (iii) s = 4a sin  (cardioide), (iv) s = c log sec  (tractrix), 3 (v) s = a log (tan  + sec ) + a tan sec  ( Parabola).

ANSWERS (i)

c sec2 

(ii) 4 a cos 

(iv)

c tan 

(v) 2a sec3 .

(iii)

4 1 a cos  3 3

The expression ds/d for the radius of curvature is suitable only for those curves whose equations are given by means of a relation between s and . We must, therefore, transform the expression so that it may be applicable to the curves whose equations are given in the Cartesian or Polar form.

CURVATURE AND EVOLUTES

423

For this transformation we require the expression for ds/dx in terms of dy/dx which we now proceed to obtain.

14.2. Length of arc as a function. Derivative of arc. Let y = f (x) be the equation of a given curve on which we take a fixed point A. To any given value of x corresponds a value of y viz., f (x) and to this pair of numbers x and f (x) corresponds a point P (x, y) on the curve, and this point P has some arcual length ‘s’ from A. Thus, we have a function s of x for the curve y = f (x). We shall now prove that 2   dy   1      dx    We take a point Q (x + x, y + y ) on the curve near P.

ds  dx

Let arc AQ = s + s so that arc PQ =  s. From the rt. angled  PQN, we have PQ2 = PN2 = NQ2, = (x)2 + (y)2 2



 PQ   y    1    x    x  2

Fig. 14.3

2

2

2

 chord PQ   s   y       1   .  x   arc PQ   x  Assuming, on an intuitive basis that

lim

QP

chord PQ  1, arc PQ

we obtain in the limit 2

2

 ds   dy    1   . dx    dx  We make a convention that for the curve y = f (x), ‘s’ is measured positively in the direction of x increasing, so that , s, increase with x. Hence ds/dx is positive. Thus we have 2   dy   1      dx    taking positive sign before the radical.

ds  dx

Cor. 1. dx/ds = cos , dy/ds = sin . Cor. 2. For parametric cartesian equations with parameter t, 2

2

2

 ds   dx   dy        .  dt   dt   dt 

424

DIFFERENTIAL CALCULUS

EXAMPLES rm

1. In the curve

ds  a sec d

=

m 1 m

am

cos m , prove that

2m m and a

d 2r ds 2

 m r 2m  1  0 .

r ,m  a m cos m

We have

 m log r = m log a + m log cos m 

dr   r tan m d



ds  d =

 dr  r2     d 

r

2

2

 r 2 tan 2 m



= r sec m 1

a ( cos m) m cos m = a sec ( m  1) / m m .

=

Again,

ds r r am  r sec m   m d cos m r m 1 m = a r dr   r tan m d dr dr d  r tan m  .   m 1 m ds d  ds a r =  a  m r m tan m

and 

d 2r ds 2

d  dr  d  dr  dr     ds  ds  dr  ds  ds d m m  a  m r m tan m =  a r tan m . dr d   2 m m r tan m  m r m  1 tan m  m r m sec2 m = a dr   1   2 m m r tan m  r m  1 tan m  r m sec2 m . cot m  = ma r   2 m 2 m  1  2 2  r = ma  tan m  tan m sec m cot m  2 m 2 m  1  2 2  r = ma  tan m  sec m  = m a 2 m r 2 m  1 (  1)







d 2r ds

2

 m a 2 m r 2 m  1  0



CURVATURE AND EVOLUTES

d 2r

a2m

or

2

425

 m r 2 m  1  0.

ds 2. Prove that for any curve

d d 2r  r 2  0. d ds dr We know that cos   ds d  d 2r then  sin  .  2 , differentiating w.r.t.s, ds ds sin 2 

or

 sin  .



 sin 



r

d 2r ds

2

d d d 2r .  2 d  ds ds

d     r ds  sin    

d  sin  d 2 r .  2 d r ds

 sin 2 

d  0. d

EXERCISES 1. Find ds/dx for the curves: (i)

(ii) a log [ a2/(a2 – x2)].

y = cosh x/c,

2. Show that for the parametric equation x = f(t), y = F (t)   dx  2  dy 2        dt     dt  and use this result to find ds/dt for (i) x = a cos3t, y = b sin3 t. (ii) x = a (t – sin t), y = a ( 1 – cos t). (iii) x = a et sin t, y = aet cos t. (iv) x = a (cos t + t sin t), y = a (sin t – t cos t ). 3. Show that for polar equation r = f (), ds  dt

 2  dr  2  r      d     and use this result to find ds/d for ds  d

(i) r = a ( 1 + cos).

(ii) r2 = a2 cos 2. ANSWERS

1. (i)

cosh

x . c

2. (i) 3 sin t cos t

a2  x2

(ii)

a2  x2

( a 2 cos 2 t  b 2 sin 2 t ) .

1 (ii) 2a sin t . 2

(iii)

3. (i) 2a cos /2

(ii)

2 aet .

a ( sec 2).

(iv) at.

(Banglore 2005)

426

DIFFERENTIAL CALCULUS

14.3. Radius of Curvature -Cartesian Equations (Kanpur 2002; Kuvempu 2005) Consider the curve y = f (x). Differentiating dy tan   dx w.r. to s, we obtain ds (1  tan 2  ) d  d 2 y dx ds ds . sec   2 .   2 ds ds  d dx d y dx 2 Also, we have 2

2  ds  dy    1     , dx  dx    where the positive sign is to be taken before the radical.

Hence



ds [1  ( dy / dx) 2 ]3/ 2 (1  y12 )3/ 2   d y2 d 2 y / dx 2

...(A)

Cor. The radius of curvature, , is positive or negative according as d2y/dx2 is positive, or negative, i.e. according as the curve is concave upwards or downwards. Also, (A) shows that the curvature is zero at a point of inflexion. Note. Since the value of  is independent of the choice of x-axis and y-axis, interchanging x and y, we see that, , is also given by 3/ 2

2   dx   d2x  1  .   dy 2   dy   This formula is a specially useful when the tangent is perpendicular to x-axis in which case

dx  0. dy Radius of Curvature—Parametric Equations:

x = f (t), y = F (t). At point where f  (t)  0, we have 2 dy F (t ) d y  f (t ) F  (t )  F (t ) f (t ) . 1 .  , f  (t ) dx f (t ) dx 2  f (t ) 2

Substituting these values in the formula (A) above, we get



[ f 2 (t )  F 2 (t )]3/ 2 f (t ) F  (t )  F (t ) f (t )

...(B)

where the sign is + or – according as f  (t) is positive or negative. Note. The formula (B) shows that the expression for , will retain the same form for the point where f  (t) = 0 but F(t)  0. Radius of Curvature—Polar Equations : Let r = f () be the given curve in polar co-ordinates. Now x = r cos , y = r sin  where r = f (),

(Bhopal 96; Banglore 2005)

CURVATURE AND EVOLUTES

427

may be thought of as the parametric cartesian equations of the curve,  being the parameter. We have

where

dx dy  r1 cos   r sin  ,  r1 sin   r cos  d d r = f (), r1 = f ().

Thus,

y1 

Again,

y2 

=

dy r1 sin   r cos   . dx r1 cos   r sin  d2y dx



2

d  r1 sin   r cos   d   . d   r1 cos   r sin   dx

r 2  2r12  r r2 ( r1 cos   r sin  )3

.

(1  y12 )3/ 2 ( r 2  r12 )3/ 2  2 y2 r  2 r12  r r2

Thus,



where

r1  f  (), r2  f  () .

Radius of Curvature (Pedal Equations) (Rohilkhand 1999; Gorakhpur 2003; Utkal 2003) From the figure,  Differentiating both sides with respect to s, we obtain d d d   ds ds ds

=  

d  d  dr  . ds dr ds

1 1 d  sin   cos  ·  r dr

Fig. 14.4

[

d 1 dr  sin  and  cos  ] ds r ds

1 1 d    sin   r cos  .  dr   r  1 d  ( r sin  ) r dr 1 dp r dr dr r dp 



14.3.1 Radius of curvature when the equation of the curve is given in p and  or to prove that

 p

d2 p d 2

428

DIFFERENTIAL CALCULUS

dp dp dr ds  . . d  dr ds d 

We have

 dr   cos     ds   dr     r  dp  

dp . cos  .  dr dp dr  cos  . r dr dp



= r cos  2

 dp  2 2 2 2 2 2 Now again p     r sin   r cos   r . d    Differentiating with respect to p, we obtain

2p  2. p

or

d2 p d2

dp d 2 p d  dr . . 2r d  d  2 dp dp

 .

This is known as tangential polar formula. 14.3.2 More formulae for Radius of Curvature (i) When x and y are functions of s. 2   dy   1      dx    = sec 

ds  dx

dx . ds Differentiating with respect to s, we obtain cos  



d d 2 x  2 ds ds 1 d2x  sin  .  2  ds

 sin  .

or 



sin  d 2 x / ds 2

while sin  = dy/ds Hence

ρ=–

dy/ds d2x ds 2

(ii) Similarly, we obtain 

by taking

sin  =

dx / ds d2y ds 2

dy . ds

(1  tan 2  )

CURVATURE AND EVOLUTES

429

(iii) Now again, we know 

dy / ds 2

d x / ds

2



sin  2

d x / ds 2

2

sin  d x  2  ds



...(i)

cos  d 2 y  2  ds Squaring and adding (i) and (ii)

and also

we obtain

 d2x  2 2  ds   1

2

...(ii)

  d2y    2   ds

2

  . 

EXAMPLES 1. For the cycloid x = a ( t + sin t), y = a ( 1 – cos t ), prove that

1    4a cos  t  .  2 

(Kumaon 2002; Utkal 2003; Vikram 97; Bhopal 2000; Indore 99, 2000; Jabalpur 97; Sagar 98; Bilaspur 2000, 2001; Devi Ahilya 2001; Banglore 2004)

We have 

dx dy  a (1  cos t ),  a sin t. dt dt dy a sin t 2sin t / 2 cos t / 2    tan t / 2 dx a (1  cos t ) 2 cos 2 t / 2

d2y dx

2

 

1 t dt sec2 . 2 2 dx 1 t dt 1 1 1  sec 2 . . . t 2 2 dx 2a cos 2 4a cos 4 t 2 2

3/ 2   1   1  tan 2 t     2 3/ 2 t   2     [1  (dy / dx) ] a   4 cos .  1 1   2  d 2 y / dx 2 .   4 4a cos (t / 2)   2. For the curve rm = am cos m, prove that am  . ( m  1) r m  1

We have  

r m  a m cos m  m log r  m log a  log cos m m dr sin m . m r d cos m dr r1    r tan m d

(Patna 2003)

430

DIFFERENTIAL CALCULUS



r2 

Hence



d 2r 2

  rm sec 2 m  tan m

d =  rm sec 2 m  r tan 2 m .

dr d

( r 2  r 2 tan 2 m )3/ 2 r 2  2r 2 tan 2 m  r 2 m sec 2 m  r 2 tan 2 m r 3 sec3 m

=

r 2 sec 2 m  mr 2 sec2 m

r 1 am  . m –1 . (m  1) (cos m ) m  1 r  3a 3a  , 3. Show that the curvature of the point   on the Folium x3  y 3  3 axy is 8 2 / 3a .  2 2  (MDU Rohtak 2001) 

Differentiating, we get dy dy 3x 2  3 y 2  3ay  3ax dx dx dy ay  x 2  or dx y 2  ax dy   1 dx   3a , 3a    2

...(i)

2 

Again differentiating (i), we get 2

2 dy dy d2y  dy  2 d y 2x  2 y   a  a  ax   y dx dx dx 2 dx 2  dx  dy 3a 3a , Substituting , – 1 for x , y, respectively, we get dx 2 2 d2y 32  2  3a dx   3a , 3a    2

 2 

32 / 3a d 2 y / dx 2 8 2  3a 3a  ,  Hence the curvature at   = 3/ 2 3/ 2 = 2 2 2 3a (2)     dy    1      dx    x 4. Prove that the radius of curvature at any point of the catenary y  c cosh varies as the c square of the ordinate. (Kumaon 2001; Sagar 99, 2002; Jawaji 2001; Jabalpur 2001) The given curve is x y  c cosh . c Differentiating it with respect to x, we have dy x 1 x  c . sinh .  sinh dx c c c d2y x 1  cosh . and c c dx 2

CURVATURE AND EVOLUTES

431

2   dy   1      dx      d2y dx 2



3/ 2

x  c  1  sinh 2  c   = x cosh c

3/ 2

2 = c cosh x / c  c .

y2 c

2



y2 c

Hence  (ordinate)2. 5. If a curve is defined by the equation x = f (t) and y = (t) prove that ( x 2  y 2 )3/ 2  , (Bilaspur 2001; Garhwal 2001) x y   y  x where dashes denote differentiations with respect to t. We have Now 

dx dy  f  (t )  x  and   (t )  y  dt dt

dy dy / dt y   where y  and x  are functions of t. dx dx / dt x

d2y

d  y   d  y   dt    . dx  x   dt  x   dx dx x  y   y  x  1 . = x x 2 x  y   y  x  = x 3



2



2   dy   1      dx     d 2 y dx 2

3/ 2

3/ 2

 y 2   1  x2    = x y   y x x3

( x2  y 2 )3/ 2 = x  y   y  x x2



y2

 1 , show that the radius of curvature at an end of the major axis is a2 b2 equal to semi-latus rectum of the ellipse. (Ravishankar 2000) The given curve is x2 y2   1. a2 b2

6. In the ellipse

432

DIFFERENTIAL CALCULUS

Differentiating it with respect to x, we get 2x a

or

2



2 y dy 0 . b 2 dx dy b2 x  2 dx a y

  y 2 2 d y b   2  dx 2 a   2 b = 4 3 . ( a2 a y

and

=  = 



Now,

  b2 x x  2  a y y2

      

y 2  b2 x2 )

2 y2 2 2  x . a b    a2 a 4 y3 b2  b4

b2

a 2 y3

2   dy   1      dx   

  

3/ 2

d 2 y / dx 2

 b4 x2   1  4 2  a y  =  4 b / a2 y3

3/ 2

= –

( a 4 y 2  b 4 x 2 )3/ 2 a4 b4

Now,  at one end of major axis, i.e., (a, 0) 

(a 4 · 0  b 4 · a 2 )3/ 2 a 4 b4



b6 a3 a4 b4

b2 = semi-latus rectum. a 7. If CP and CD be a pair of conjugate semi-diameters of an ellipse, prove that the radius of CD 3 curvature at P is , a and b being the lengths of the semi- axes of the ellipse. ab (Ravishankar 1999)

=

We know from coordinate geometry that if P and D are the ends of two conjugate diameters; then their coordinates shall be ( a cos , b sin) and ( – a sin , b cos ) respectively where C x2 y2 is the centre of ellipse 2  2  1 , whose coordinates are (0, 0). a b In example 6, we have just shown

  ( a 4 y 2  b 4 x 2 )3/ 2 / a 4 b 4 

 at P  ( a 4 b 2 sin 2   b 4 a 2 cos 2  )3/ 2 / a 4b 4 

( a 2 sin 2   b 2 cos 2  )3/ 2 ab

CURVATURE AND EVOLUTES

But

CD = =

433

 ( – a sin   0 )

2

 ( b cos   0 ) 2



( a 2 sin 2   b 2 cos 2  )

 (a Therefore  at P 

2

sin 2   b 2 cos 2  )1/ 2

3



ab

CD 3 . = ab 8. Find the value of  for r  a e cot  and show that the radius of curvature subtends a right angle at the pole. (Sagar 98) We are given that r = a e cot  dr  a e  cot  .cot   r cot  d

and

tan   r /

 also

,

dr r   tan  d  r cot 

p = r sin  = r sin .

Now differentiating with respect to r, we have

But

Fig. 14.5

dp  sin  dr r r    r cosec  dp sin  dr

Now from  COP CP r  sin COP sin PCO ’  r  sin 90 sin  180  90 – ( 90 – ) r r   sin  sin  The results (i) and (ii) are the same. Hence CP subtends a right angle at pole. 9. Show that for any curve r d    sin   1  . d    We have d   R. H. S. = sin   1   d   d  d  = r 1  ds  d 

...(i)

...(ii)

434

DIFFERENTIAL CALCULUS

 d d d    .  = r ds d    ds  d  d    = r ds   ds d d r (   )  r  . = r ds ds  10. Find the radius of curvature for the curve r = a ( 1 – cos  ). (Garhwal 96, 2000; Gorakhpur 2003; Ravishanker 98; Indore 97; Vikram 98; Jiwaji 99; Bilaspur 98, 2000) We have

d 2r dr  a cos   a sin  , d d 2 3/ 2

 2  dr  2  r      d      2 d2 r  dr  r2  2   r 2 d  d  3/ 2 a 2 (1  cos  ) 2  a 2 sin 2  = 2 a (1  cos  ) 2  2 a 2 sin 2   a 2 cos  (1  cos )







4  2 a sin  ( 2 a r) 3 2 3 11. Show that for any curve r = f () the curvature is given by

=

 d2 u  u  d 2 

 3  sin  , where u = 1/r.  3/ 2

We have



 2  du 2  u      d       d2 u  u3  u   d 2    d2 u  u 3  u   d   

1  2 3/ 2    du   2  u     d    

u3 2   =  u 2   du      d    

3/ 2

 d2 u  u   d 2  

1/ r 3 2  =  1  1  dr     2 r 4  d     r

3/ 2

 d 2u  u   d 2  

CURVATURE AND EVOLUTES

435

1/ r 3  d 2 u  p3  d2 u  u   u      = 1/ p 3  d 2  r 3  d 2  r 3 sin 3   d2 u  u    = r3 d 2    d2 u  u  sin 3  .  = 2  d    1 d  dy   12. Prove that for any curve  .  dx  ds  We are given that d  dy  d (sin  )   dx  ds  dx d d ( sin  ) . = d dx

(Rohilkhand 2001, 2004)

d  ds . ds dx d 1  . = ds 

= cos  .

13. Show that for the curve s2 = 8 a y,  = 4a

 y  1– . 2a  

The given curve is s2 = 8 a y. Differentiating with respect to s, we get 2s=8a. But

dy ds

dy  sin  ds 2 s = 8 a sin 

 or s = 4 a sin  Now again differentiating with respect to , we have



ds  4a cos  d  = 4 a cos  =4a

(1  sin 2  )

=4a

 s2  1  16 a 2 

y    4 a 1 – . 2 a  

   4 a 

 8a y  1  16 a 2 

  

436

DIFFERENTIAL CALCULUS

EXERCISES 1. Find the radius of curvature at any point on the curves: (i) y = c cosh x/c (catenary),

(Kumaon 2001)

(i) x = a ( cos t + t sin t), y = a ( sin t – t cos t ), (iii) x = ( a cos t )/t, y = ( a sin t )/ t. 2. Show that radius of curvature at any point of the astroid x = a cos3, y = a sin3 is equal to three times the length of the perpendicular from the origin to the tangent. (MDU Rohtak 1999; Banglore 2006) 3. Show that for the curve x = a cos  (1 + sin ), y = a sin  (1 + cos ), the radius of curvature is a, at the point for which the value of the parameter  is – /4. 4. Show that the radius of curvature at any point of the curve x = t – c sinh

t t t cosh , y = 2c cosh . c c c

 t   t  is – 2c cosh 2   sinh   , c   c where t is the parameter. 5. Show that the radius of curvature at a point of the curve x = ae (sin – cos), y = ae (sin  + cos ) is twice the distance of the tangent at the point from the origin. 6. Prove that the radius of curvature at the point ( – 2a , 2a) on the curve x2y = a ( x2 + y2) is , – 2a. 7. Find the radii of curvature of the curve y2 (a – x) = x2 ( a + x) at the origin. 8. Find the radius of curvature for ( x / a ) – ( y / b)  1 at the points where it touches the co-ordinate axes. 9. Show that the ratio of the radii of curvature of the curves xy = a2, x3 = 3a2y, at points which have the same abscissae varies as the square root of the ratio of the ordinates.

1   10. Show that the radius of curvature at each point of the curve x  a  cos t  log tan t  , 2   y = a sin t, is inversely proportional to the length of the normal intercepted between the point on the curve and the x-axis. 11. Show that 3 3 / 2 is the least value of  for y = log x. 12. Show that the radius of curvature of the curve given by

x2 y  a ( x2  a2 / 5 ) is least for the point x = a and its value there is 9a/10. 13. Prove that for the ellipse x2



y2

 1,  

a2 b2

; p3 a2 b2 p, being the perpendicular from the centre on the tangent at the point ( x , y). (Reewa 96, 97, 98; Sagar 99; Kumaon 96, 2005; Garhwal 99, 2002; Indore 98; Jabalpur 99; Jiwaji 97, Avadh 2005)

CURVATURE AND EVOLUTES

437

14. The tangents at two points P, Q on the cycloid x = a (  – sin  ), y = a ( 1 – cos  ), are at right angles; show if 1, 2 be the radii of curvature at these points, then 12   2 2  16 a 2 . 15. Find the points on the parabola y2 = 8x at which the radius of curvature is 7 13 . 16 16. (a) Prove that, if  be the radius of curvature at any point P on the parabola y2 = 4ax and S be its focus, then, 2 varies as (SP)3. (Kumaon 2003) (b) 1,2are the radii of curvature at the extremities of a focal chord of a parabola whose semilatus rectum is l; prove that (1)–2/3 + ( 2)–2/3 = (l)–2/3. 3 a ( sinh u cosh u + u), y = a cosh2 u, if the normal at P (x, y) 2 meets the axis of x in G, the radius of curvature at P is equal to 3 PG. 18. Find the radius of curvature at the point of the curve r = a cos n . Also show that at the point where r = a, its value is a/ ( 1 + n2). 19. Find the radius of curvature at the point ( r, ) on each of the following curves: (i) r = a, (ii) r = a.

17.

Show that for the curve x 

(iii) rn = ansin n,

(iv)

( r 2 – a2 ) a – cos 1 , a r



(v) r (1 + cos) = a. (MDU Rohtak 2000) 20. Find the radius of curvature of the curve r = a sin n at the origin. 21. Prove that for the cardioide r = a (1 + cos ), 2/r is constant. (Kumaon 98; Garhwal 98, 2004; Rohilkhand 2000; Banglore 2004; Bilaspur 96; Avadh 2003) 22. Show that at the points in which the Archimedean spiral r = a intersects the hyperbolical spiral r = a, their curvatures are in the ratio 3 : 1. (Ravishankar 2000, Garhwal 2005) 23. If  1,  2 be the radii of curvature at the extremities of any chord of the cardioide r = a ( 1 + cos ), which passes through the pole, then, Rewa 97; Vikram 2002) 12 + 22 = 16a 24. Find the radius of curvature of the curve r = a (1 + cos) at the point where the tangent is parallel to the initial line. 25. A line is drawn through the origin meeting the cardioide r = a ( 1 – cos ) in the points P, Q and the normals at P, Q meet in C; show that the radii of curvatures at the points P and Q are proportional to PC and QC. 26.

Prove that for the curve r2 = a2 sin 2, the tangent turns three times as fast as the radius vector and that the curvature varies as the radius vector.

27. Show for the curve s = f (x), 2    ds    dx   

2   ds    1   dx 

d2x ds 2

    

.

438

DIFFERENTIAL CALCULUS

28. For any curve, prove that d 2r 2



sin 2 



2

sin  . 

ds r 29. Find the radius of curvature of

x

y =

a , at the point where the line y = x cuts it. (Garhwal 97)

30. If 1, 2 be the radii of curvature at the extremities of two conjugate diameters on an ellipse, prove that (12/3 + 22/3) a2/3 b2/3 = a2 + b2. (Rohilkhand 1999; Gorakhpur 2000; Bilaspur 99) 31. Show that for the curve y  2/3

ax ax 2

2

 x   2   y      .    a   x   y  32. The coordinates of a point on a curve are given by

(Kumaon 2004)

at at , y = a cost – b cos . b b Show that the equation of the tangent at the point whose parameter is t, is x = a sin t – b sin

ab ab ab t – y sin t  ( a  b) sin t=0 2b 2b 2b and the radius of curvature at this point is x cos

4 ab a–b sin t. ab 2b

33. Prove that for the curve s = m (sec3x – 1),  = 3 m tan  sec3, and hence show that 3m . dy d 2 y .  1. dx dx 2 34. Show that for the curve in which s = cex/e, c = s (s2 – c2)1/2. (Kumaon 97; Bhopal 97) ANSWERS 1. (i) y2/c. 4

7.

(ii) at.

19. (i) (iv) 20. an/2.

– a ( t2 + 1)3/2/t4.

8. 2a2/b, 2b2/a.

2a .

9   , 3  . 8 

15.

(iii)

a (1  2 )3 / 2 2

(2   ) r2 – a2 .

18. .

(ii) (v) 24.

( r 2  a 2 n 2  n 2 r 2 )3/ 2 r 2  r 2 n 2  2a 2 n 2 a (1  2 )3/ 2 4

.

.

(iii) an/( n + 1) rn – 1.

(8r 3 / a ) .

2 3 a 3

29.

a/ 2 .

CURVATURE AND EVOLUTES

439

14.4. Newtonian Method. If a curve passes through the origin and the axis of x is tangent at the origin, then x2 , as x  0 2y gives the radius of curvature at the origin. Here we obtain the values of y1 and y2 at the origin.  dy  y1 (0)     dx  (0 / 0) = 0. lim

Now, x2/2y, assumes the indeterminate form 0/0 as x  0.

x2 2x  lim x0 2 y x 0 2 y1

lim

= lim

x0

0   0

1 1 .  y2 y2 ( 0 )

Also, from the formula (,A) of § 14.3, we have at the origin



(1  0)3/ 2 1  . y2 (0) y2 (0)

Thus at the origin where x- axis is a tangent,  x2    lim  . x 0 2y    It can similarly be shown that at the origin where y – axis is a tangent  y2    lim  . x 0 2x    These two formulae are due to Newton. 14. 4.1 Generalised Newtonian Formula. If a curve passes through the origin and x-axis is the tangent at the origin, we have lim

 x2 x2  y2 y  lim    2y  2y 2  x2 2y = , at the origin.  lim

x 2  y 2  OP 2 , is the square of the distance of any point P( x, y) on the curve form the origin O and, y, is the distance of the point P from the tangent x - axis at O. Here,

Fig. 14.6

Interpreted in general terms this conclusion can be stated as follows: ( see Fig. 14.7) If OT be the tangent at any given point O of a curve, and PM, the length of the perpendicular drawn from any point P to the tangent at O, then the radius of curvature at O

OP 2 2PM when the point P tends to O as its limit. = lim

Fig. 14.7

440

DIFFERENTIAL CALCULUS

EXAMPLES 1. Find the radius of curvature at the origin of the curve: x3 – 2x2y + 3xy2 – 4y3 + 5x2 – 6xy + 7y2 – 8y = 0. It is easy to see that x- axis is the tangent at the origin (Refer Chap. 16) Dividing by y, we get x2 x2 x.  2 x 2  3 xy  4 y 2  5  6x  7 y  8  0 y y  x2  Let x  0 so that lim    2 x 0  y  0.2  + 5.2 – 8 = 0    = 4/5 2. Find the radius of curvature at the origin for the curve x3 + y3 – 2 x2 + 6y = 0. The curve passes through ( 0, 0) and x - axis is tangent to the curve at ( 0, 0).  x2  x2 lim 0 lim    2. Hence or x0 2 y x 0 y   y 0 y 0 Now dividing the curve by y, we obtain x2 2 x2 x.  y2  60 y y Taking limit, we get 0.2  + 0 – 4  + 6 = 0 3 . 2 3. Find  at the pole for the curve r = a sin n . (Guwahati 2005) The curve is r = a sin n. Here r and  are 0. Hence initial line is tangent to the curve at origin.  



r 2 a sin n  lim 0 2 sin n n a na  lim .  . 0 n 2 2 4. Apply Newton’s formula to find the radius of curvature at the origin for the cycloid x = a (  + sin ), y = a ( 1 – cos). We know,

  lim

 0

We have

dy dx  a sin   a (1  cos  ) , d d



dy dy / d  a sin    dx dx / d  a (1  cos  ) = tan /2

dy  0 , when  = 0 dx Hence initial line is tangent to origin. 

CURVATURE AND EVOLUTES

Then

  lim

 0

441

x2 a 2 (   sin  ) 2  lim  0 2y 2a (1  cos  )

a  2 (   sin  ) (1  cos  )   2  sin   2 a  2 . (1  cos  )  sin  (   sin  )  lim   =  0 2 cos    lim =  0

= 4a.

[ By Hospital’s rule ] EXERCISES

Find  at origin of the curves: 1. 2x4 + 3y4 + 4x2y + xy – y2 + 2x = 0 2. y = x4 – 4x3 – 18x2 3. 3x3 + y3 + 5y2 + 3yx2 + 2x = 0 4. x4 – y4 + x3 – y3 + x2 – y2 + y = 0. 5. Find the radius of curvature of the cardioid r = a ( 1 – cos ) at the pole (origin). ANSWERS 1. 1;

2.

1 ; 36

3.

1 ; 5

4.

1 ; 2

5. 0.

14.5. Centre of Curvature The centre of curvature at any point P of a curve is the point which lies on the positive direction of the normal at P and is at a distance, , from it. The distance , must be taken with a proper sign. The positive direction of the normal is obtained by rotating the positive direction of the tangent (the positive direction of the tangent is the one in which x increases) through /2 in the anti-clockwise direction.

Fig. 14.8

From an examination of the above figures we see that the centre of curvature at any point of a curve lies on the side towards which the curve is concave. To find the co-ordinates of the centre of curvature for any point P( x, y) of the curve y = f (x). Let the positive direction of the tangent make angle, , with x-axis so that the positive direction of the normal makes angle /2 with x- axis. The equation of the normal at (x, y) is X x Y  y  r, cos (    / 2 ) sin (    / 2 )

442

DIFFERENTIAL CALCULUS

X x Y  y   r,  sin  cos  where X, Y are the co-ordinates of any point on the normal and, r the variable distance of the variable point (X, Y) from (x, y). Thus, the co-ordinates ( X, Y ) of the point on the normal at a distance, r from ( x, y ) are ( x – r sin y + r cos ). For the centre of curvature, we have r = . Hence, if ( x, y) be the centre of curvature, we have X = x –  sin , Y = y +  cos . But we know that y1 1 (1  y12 )3/ 2 sin   cos    . , , y2 [1  y 2 ] [1  y 2 ]



1

1

y1 (1  1  y12 ; Y  y  .  y2 y2 Another method. If C be the centre of curvature for P, we have PC = . X x–

y12 )

Fig. 14.9

Also NCP = XTP = . X = OM = OL – ML = OL – NP = x –  sin .  Y = MC = MN + NC = LP + PC cos  = y +  cos . Substituting the values of sin , cos  and , we can obtain the values of X and Y.

Fig. 14.10

CURVATURE AND EVOLUTES

443

14.6. Chord of curvature If there is any point P at the given curve and a circle having the radius of curvature  is drawn passing through P, then any chord PN is called the chord of curvature. If C be the centre of the circle then PN = R P cos R PN = 2  cos RPN = 2 cos . Different cases of chord of curvature Case I. Chord of curvature through pole (origin). In the figure PN is the required chord of curvature PN = RP cos RPN    = 2  cos      2  Hence chord of curvature through pole = 2  sin  Case II. Chord of curvature through the pole for the curve Fig. 14.11 p = f (r). We know p = r sin ;

 sin  

Hence chord of curvature = 2 . = 2.r

p r

p r

dr p dr .  2p dp r dp

dr = 2 f (r) dp ,

i.e.,

2 f (r ) f  ( r)

Case III. Chord of curvature perpendicular to radius vector. Here PN = chord of curvature perpendicular to radius vector = RP cos RPN = 2  cos  Case IV. Chord of curvature parallel to x - axis and y-axis.

Fig. 14.12

Chord of curvature parallel to x-axis = PN = P R cos RPN   –    2 sin  = 2  cos  2  

(Jabalpur 96, 98)

444

DIFFERENTIAL CALCULUS

and chord of curvature parallel to y-axis = PN= RP cos RPN = 2  cos 

Fig 14.13

Circle of Curvature. The circle with radius equal to radius of curvature  and its centre the centre of curvature (X, Y) is called the circle of curvature. Then the equation of the circle of curvature is ( x – X )2 + ( y – Y)2 = 2. EXAMPLES 1. Find the co-ordinates of the centre of curvature at a point ( x, y) of the parabola y2 = 4 ax. Differentiating, we get dy 2y  4a dx dy 2a d2y 2a dy 4a 2   2  2 .  .  dx y dx y dx y3 If (X, Y) be the centre of curvature, we have 2a  4a 2   1  2  y  y  X x– 2 4a – 3 y y 2  4a 2 2ax  4ax  4a 2 x   3 x  2a 2a 2a 4a 2 1 2 y Y  y 4a 2 – 3 y  y–

y ( y 2  4a 2 )



 y3



( 4ax )3/ 2



2 x3/ 2

4a 2 4a 2 4a 2 a1/ 2 2. Find the centre of curvature of the four cusped hypocycloid x = a cos3, y = a sin3, i.e., x2/3 + y2/3 = a2/3.

...(i)

...(ii) (Guwahati 2005)

CURVATURE AND EVOLUTES

445

dx = – 3a cos2 sin  d dy y = a sin3, then = – 3a sin2 cos  d dy dy / d  y1    – tan  Then dx dx / d  d2y d y2  2   sec 2  and dx dx 1 4  sec  cosec  . 3a Now, if (X, Y) be the coordinates of the centre of curvature, then

We have

x = a cos3, then

y1 (1  y12 ) tan  (1  tan 2  )  a cos3   1 y2 sec 4 .cosec  3a = a cos3 + 3a sin2 cos 

X x–

and

Y  y

1  y12 (1  tan 2  ) 3a  a sin 3   y2 sec 4  . cosec 

= a sin3 + 3a cos2 sin .   Prove that the coordiantes of the centre of curvature at any point (x , y ) can be expressed in the dx dy . form x – and y  d d We know if (X, Y) be the centre of curvature of any point ( x, y) of a curve then 2   dy   dy  1    ·  dx   dx   X x– d2y dx 2 2   dy   y  1      dx    Y  d2y dx 2 dy  tan  and Also we know dx

or 

2   dy   1      dx      d2 y dx 2 2 d y 1 3  sec  d x2 

X x–

3/ 2



sec3  d 2 y / dx 2

(1  tan 2  ) tan   1  3    sec   

446

DIFFERENTIAL CALCULUS

 x –  sin   x –

x–

 ds dy      d  and sin   ds   

d s dy . d  ds

dy d

Similarly, 1  tan 2   y   cos   1 3  sec    ds dx dx  y .  y d  ds d

Y  y

4. Prove that the points on the curve r = f (), the circle of curvature at which passes through the origin are given by the equation f () + f  () = 0. Let P ( r, ) be a point on the given curve, the circle of curvature, with centre C, at which passes through the origin O. Let C be the centre of curvature. Join OP and let PC produced meet the circle of curvature at P in D. Let PT be the tangent to the curve at P, then OPT =  = PDO Now, OP = radius vector r and PD = 2. Hence in POD, we get PO = PD sin or Also  Also

r = 2  sin d r tan   r  , dr r1 sin  



...(i) dr where r1  d

r

...(ii)

2

( r  r12 )1/ 2

( r 2  r12 )3/ 2

...(iii)

r 2  2 r12 – r r2

Hence from (i), (ii) and (iii), we get r



2 ( r 2  r12 )3/ 2 2

r 

2r12

– r r2

Fig. 14.14

.

r (r 2  r12 )

2r ( r 2  r12 ) r 2  2r12 – r r2

or

r 2  2r12 – r r2   2r 2  2 r12

or

r 2 + r r2 = 0

or

r + r2 = 0,

or

f () + f  () = 0

where r = f ()

CURVATURE AND EVOLUTES

5.

447

Show that the chord of curvature through the pole of the equiangular spiral r = a e cot  is 2r. We have r = a e cot  dr  cot   d   a cot  . e

= r cot  Now,  Hence

or

d r   tan  dr r cot  , p = r sin  = r sin .

tan   r

dp  sin  dr

dr r  dp sin   But the chord of curvature through the pole 2r  2  sin   . sin   2r . sin  

r

x 6. Show that the chord of curvature parallel to the axis of y for the curve y  c cosh is double c 2x the ordinate whereas the chord parallel to x-axis is of length c sinh . c (Gorakhpur 2000; Bilaspur 97) We have x dy x  sinh ;  c dx c d2y 1 x  cosh and 2 c c dx 2 3/ 2  3/ 2  dy    2 x   1      1 sinh    dx    c       x 1 d 2 y / dx 2 cosh c c 2 x  c cosh c x Now, tan   sinh c 1 1 x then cos     sech 2 c x 1  tan  ( 1  sinh 2 ) c 2 2 x sin   1 – cos   1 – sech and c x  tanh . c Hence the chord of curvature parallel to y-axis y  c cosh

448

DIFFERENTIAL CALCULUS

 2  cos   2 c cosh 2

x x . sech c c

x  2y , c and the chord of curvature parallel to x-axis is = 2 sin  x x x x  2 c cosh 2 . tanh  2 c cosh sinh c c c c 2x  c sinh . c  2 c cosh

14.7. Evolutes and Involutes If the centre of curvature for each point on a curve be taken, we get a new curve called the evolute of the original one. Also the original curve, when considered with respect to its evolute, is called an involute. The connection between the curve is represented in the adjoining figure. Let P1, P2, P3 .... etc. represent a series of infinitely near points on a curve; C 1 , C 2 , C 3 .... etc. the corresponding centres of curvature; then lines P1 C1, P2 C2 , P3,C3 etc. are normals to the curve and the lines C1 C2, C2 C3, C3 C4 etc, may be regarded in the limit and consecutive elements of the evolute ; also each of the normals P1 C1, P2 C2, P3 C3 etc. passes through two consecutive points on the evolute, they are tangents to that curve on the limit. Again if P1, P2, P3 P4 etc. denote the length of the radii of curvature at the points P1, P2, P3 , etc.; we have

Fig. 14.15

1 = P1C1, 2 = P2 C2, 3 = P3C3 , ...... Therefore 1 – 2 = P1 C1 – P2 C2 = P2 C1 – P2 C2 = C1 C2

( since the points P1, P2 are nearer. )

2 – 3 = C2 C3, 3 – 4 = C3 C4 ..... n – 1 – n = Cn – 1 Cn Hence by addition, we have 1 – n = C1 C2 + C2 C3 + C3 C4 + ....... + Cn – 1 Cn This result still holds when the number n is increased indefinitely and we infer the length of any arc of the evolute is equal, in general to the difference between the radii of curvature at its extremities. In other words, the evolute is the locus of the centre of curvature. Also

14.7.1 To determine the equation of an evolute We know that the coordinates of the centre of curvature are given by (X, Y) whereas

X x–

y1 (1  y12 ) y2

(1  y12 ) . y2 Now we eliminate x and y ( parameters in this case), and then a relation between X and Y is the required equation of the evolute. and

Y  y

CURVATURE AND EVOLUTES

449

EXAMPLES 1. Obtain the evolute of the parabola y2 = 4ax. (Rohilkhand 2002; Kanpur 2000, 2006; Banglore 2004; Garhwal 2004; Kuvempu 2005) Here the point (x, y) may be taken as x = at2, y = 2 at, which satisfy the equation of the parabola. Hence dx dy  2at and  2a dt dt dy 1 d2y 1 dt  and 2  – 2  dx t dx t dx 1 – . 2 at 3 Then if (X, Y) be the coordinates of the centre of curvature, we have 1 1  1 2   y (1  ) t t  X x– 1  at 2 –  1 y2 – 2at 3 1    at 2  2 at 2  1 – 2   3 at 2  2a t   y12

1   1 2  (1  y12 ) t   and Y  y  2 at  1 y2 – 2at 3 = 2 at + ( t2 + 1) ( – 2at) = – 2 at3 Now X = 3 at2 + 2a, Y = – 2 at3 Eliminating t between these relations, we obtain 1/ 2

1/ 3

 X  2a   Y    –   3a   2a  or 4 ( X – 2a )3 = 27 aY2 Hence the locus is 27 ay2 = 4 ( x – 2a)3. 2. Find the evolute of the astroid x = a cos3  , y = a sin3 .

(Rohilkhand, 2000) 2

We have

dy d y 1  – tan ; 2  sec4  cosec  dx 3a dx tan  (1  tan 2  ) X  a cos3   . 3a sec 4  cosec 

= a cos3 + 3a sin2 cos Y  a sin 3  

...(i)

2

1  tan  sec 4  cosec 

= a sin3 + 3a cos2 sin

. 3a

...(ii)

450

DIFFERENTIAL CALCULUS

To eliminate , we separately add and subtract (i) , (ii), so that we have (cos  + sin )3  ( X + Y )1/3 = a1/3 ( cos  + sin )

(X + Y ) = a

(X – Y ) = a (cos  – sin )3  ( X – Y )1/3 = a1/3 ( cos  – sin ) On squaring and adding, we obtain (X + Y )2/3 + (X – Y )2/3 = 2a2/3, as the required evolute. EXERCISES 1. Show that the evolute of the ellipse x = a cos, y = b sin is (ax)2/3 + (by)2/3 = ( a2 – b2 )2/3. 2. (a) Show that for the hyperbola x2/a2 – y2/b2 = 1, a4X = ( a2 + b2) x3, b4Y = – ( a2 + b2) y3 and the equation of the evolute is (ax)2/3 – (by)2/3 = (a2 + b2)2/3. (b) Prove that the evolute of the hyperbola 2xy = a2. is ( x + y )2/3 – ( x – y)2/3 = 2a2/3. 3. Show that the evolute of the tractrix

1   x  a  cos t  log tan t  , y = a sin t 2   is the catenary y = a cosh (x/a). 4. Show that the parabolas y = – x2 + x + 1, x = – y2 + y + 1 have the same circle of curvature at the point (1, 1). 2

2

3  3  1 2   5. Show that  x – a    y – a   a , is the circle of curvature of the curve 4  4  2   x  y  a , at the point ( a/4, a/4). 6. Show that the circle of curvature at the parabola y = mx + x2 is x2 + y2 = ( 1 + m2 ) (y – mx ). 7. (a) Show that the circle of curvature, at the point (am2, 2am) of the parabola y2 = 4ax, has for its equation x2 + y2 – 6am2x – 4ax + 4am3y = 3a2m4. (b) Find the equation of the circle of curvature at the point (0, b) of the ellipse x2/a2 + y2/b2 = 1. 8. Show that the radii of curvatures of the curve x = ae  ( sin – cos) y = ae , ( sin  + cos ) and its evolute at corresponding points are equal. 9. Find the radius of curvature at any point P of the catenary y = c cosh (x/c) and show that PC = PG where C is the centre of curvature at P and G the point of intersection of the normal at P with xaxis. 10. For the lemniscate r2 = a2 cos2, show that the length of the tangent from the origin to the circle of curvature at any point is r 3 / 3 . 11. If P is a point on the curve r2 = a2 cos 2 and Q is the intersection of the normal at P with the line through the pole O perpendicular to OP, prove that the centre of curvature at P is a point of trisection of PQ. 12. If P is a point on the curve r = a ( 1 + cos ) and Q is the intersection of the normal at P with the line through the pole O perpendicular to OP, prove that the centre of curvature at P is a point of trisection of PQ remote from P. 13. The circle of curvature at any point P of the lemniscate r2 = a2 cos 2 meets the radius vector OP at A, show that OP : AP = 1 : 2; O being the pole.

CURVATURE AND EVOLUTES

451

14. 1, 2 are the radii of curvature at the corresponding points of a cycloid and its evolute; prove that 12 +22 is a constant.

 x 15. In the curve y  c cosh   show that coordinates of the centre of curvature are  c  1/ 2

 y2  , Y = 2y. X  x – y  2 –1   c    16. If ( X, Y ) be the coordinates of the centre of curvature of the parabola x  y = a at ( x , y), then prove that X + Y = 3 ( x + y ). 17. Find the equation of circle of curvature at the point (1, –1) of the curve y = x3 – 6x2 + 3x + 1. 18. Find the chord of curvature through the pole for the cardioid r = a (1 + cos).

 x  19. In the curve y = a log sec   prove that the chord of curvature parallel to the axis of y is  a  of constant length. (Bhopal 1998; Jiwaji 96; Vikram 2000) 20. Show that the chord of curvature through the pole of the curve r n = a n cos n is 2r (Gorakhpur 2001; Avadh 2002) n 1 . 21. Find the chord of curvature through the pole of the curve r2 cos2 = a2. 22. If in the cardioid r = a ( 1 + cos), Cr and C be the chords of curvature respectively along and perpendicular to the radius vector, show that 8 a Cr . 3 23. If Cx and Cy be the chords of curvature parallel to the axes at any point of the curve y = aex/a, prove that

Cr + C =

1 C x2



1 C y2



1 . 2aCx

24. Show that the chord of curvature through the focus of a parabola is four times the focal distance of the point, and the chord of curvature parallel to the axis has the same length. 25. Prove that distance between the pole and the centre of curvature corresponding to any point on the curve rn = an cos n is [ a 2 n  ( n 2 – 1) r 2 n ]1/ 2 (n  1) r n – 1

.

ANSWERS 2 2 7. (b) x  y –

2 ( b2 – a2 ) y  ( b 2 – 2a 2 )  0 b 2

43  373   17. ( x + 36 )2 +  y   6  36 

18.

9. y2/c ; 4 r; 3

21. 2r.

452

DIFFERENTIAL CALCULUS

14.8. Properties of the evolute. If P' ( X, Y ) be the centre of curvature for any point P ( x, y) on the given curve, we have X = x –  sin , Y = y +  cos . Differentiating w. r. to x , we obtain dX d d  1 –  cos  – sin  dx dx dx ds dx d  d d 1– – sin   – sin  ...(i) d  ds dx dx dx d d  dY dy  –  sin   cos    dx dx   dx dx

dy dx dy d  d d ...(ii) –  cos   cos  dx d  ds dx dx dx From (i) and (ii), we obtain dY  – cot  ...(iii) dX Now dY/dX is the slope of the tangent to the evolute at P and – cot  is the slope of the normal PP to the original curve at P. By (iii) the slopes of two lines, which have a point P in common, are equal, and as such they coincide. Thus the normals to a curve are the tangents to its evolute. Again, we square (i) and (ii) and add to obtain 

2

2

2

 dY   dX   d  ...(iv)        dx   dx   dy  Let S be the length of the arc of the evolute measured from a fixed point on it upto (X, Y) so that 2

2

 dS   dX   dY         dx   dx   dx  Here, x is a parameter for the evolute. From (iv) and (v) 2

2

...(v)

2

dS d   dS   d    S c,      dx dx  dx   dx  where c is a constant.

Fig. 14.16

Let M1 , M2 be the two points on the evolute corresponding to the points L1, L2 on the original curve. Let  1, 2 be the values of, , for L1 , L2 and S1, S2 be the values of S for M 1, M 2. We have

CURVATURE AND EVOLUTES

453

S1 = 1 + c, S2 = 2 + c  S2 – S1 = 2 – 1, i.e., arc M1 M2 = difference between the radii of curvatures at L1, L2. Thus, we have shown that the difference between the radii of curvatures at two points of a curve is equal to the length of the arc of the evolute between the two corresponding points. Note. We suppose that S is measured positively in the direction of x increasing, so that dS/dx is positive or negative according as, , increases or decreases as x increases. Thus dS d  d  or – ds dx dx according as, , increases or decreases for the values of x, under consideration. It is easy to see that the conclusion arrived at in this section remains the same if we consider dS/dx = – d/dx instead of dS/dx = d/dx. It should, however, be noted that conclusion holds good only for that case of a curve for which , constantly increases or decreases so that d/dx keeps the same sign. EXAMPLE Find the length of the arc of the evolute of the parabola y2 = 4ax which is intercepted between the parabola. The evolute is 27 ay2 = 4 ( x – 2a )3. Let L, M be the points of intersection of the evolute LAM and the parabola. To find the coordinates of the points of intersection L, M, we solve the two equations simultaneously, so that

Fig. 14.17

27 a  4ax  4 ( x – 2a )3 ,  x3 – 6ax2 – 15 a2x – 8a3 = 0. Now, 8a, – a, – a are the roots of this cubic equation of which x = 8a is the only admissible value; – a being negative.  (8a, 4 2a ), (8a, – 4 2a ) are the co-ordinates of L, M. If ( X, Y ) be the centre of curvature for any point ( x, y ) on the parabola, we have X = 3x + 2a, Y = – y3/4a2. Thus A ( 2a, 0 ) is the centre of curvature for O ( 0, 0 ) and L ( 8a, 4 2 a ) is the centre of curvature for P ( 2a, – 2 2a ). The radius of curvature for O = OA = 2a. The radius of curvature at P = PL = 6 3 a .

454

DIFFERENTIAL CALCULUS

arc AL = PL – OA = 2a ( 3



Hence the required length MAL = 4a (3

3 –1) 3 – 1 ).

Ex. Show that the whole length of the evolute of the (i) ellipse x2/a2 + y2/b2 = 1 is 4 ( a2 / b – b2/a ), (ii) astroid x = a cos3, y = x sin3  is 12a. OBJECTIVE QUESTIONS For each of the following questions, four alternatives are given for the answer. Only one of them is correct. Choose the correct alternative. 1. The radius of curvature of any point (x, y) on the curve y = f ( x) is given by (a) (c)

{1  ( dy / dx ) 2 }2 / 3

(b)

d 2 y / dx 2 {1  ( dy / dx ) 2 }3/ 2

(d)

{1  ( dx / dy ) 2 }2 / 3 d 2 y / dx 2 {1  ( dy / dx ) 2 }3

d 2 y / dx 2 2. The intrinsic formula for the radius of curvature is: dy (b) dx 1 ds (c)   (d) s d 3. The pedal formula for the radius of curvature is: dr (a)   r  dp (b) dr (c)   r dp (d) 4. The polar formula for the radius of curvature is:

(a)  

(a)   (c)  

(1  d 2 r / d 2 )3/ 2 2

2

2

2

r  2 ( dr / d  ) – r ( d r / d  )

(b)

d 2 y / dx 2 ds d d  s ds 

(Kumaon 2005)

dp dr 1 dr  r dp

 r



{r 2  (dr / d  ) 2 }3/ 2 2

r  2 ( dr / d  ) 2 – r ( d 2 r / d 2 )

(r 2  d 2 r / d 2 )3/ 2

(d) None of these r 2  2 ( dr / d  ) 2 – r ( d 2 r / d 2 ) 5. The polar tangential formula for the radius of curvature is: 2

d2 p  dp  2 (a)   p   (b)   p    d  d 2 2 dp d p (c)   p  (d)   p  2 d d 6. The radius of curvature of the origin, if y-axis is the tangent at the origin, is given by:

x2 x2 y2 lim (b) xlim (c) (d) 0 y 2y x 0 x 7. The radius of curvature at the point (s, ) on the curve s = c log sec  is : (a) c sin  (b) c cos  (c) c tan  (d) 2 8. he radius of curvature at any point on the hyperbola pr = a is : (a) r3/a2 (b) r2/a2 (c) r/a2 (d) (a) xlim 0

lim

x 0

y2 2x

c cot  r

CURVATURE AND EVOLUTES

455

9. Radius of curvature of any point (s, ) on the curve s = 4a sin  is: (a) 4a cos  (b) 4a sin  (c) 4a sin 2 10. The radius of curvature at ( x, y ) of the curve y = c cos h ( x/c) is: (a) y/c (b) y2/c (c) y3/c 2 11. Radius of curvature of the curve p = ar is: (a) p2/a2 (b) 2p/a2 (c) 2p3/a2 12. For the curve s = aex/a : (a)   s / ( s 2 – a 2 ) (c)  = ( s2 – a2 )1/2

(b)

a = s ( s2 – a2 )1/2

(d)

 = s ( s2 – a2 )1/2

 x 13. The radius of curvature for y  c log sec   is  c  x x (a) c sec (b) c tan (c) c c

(d)

4a cos 

(d)

cy

(d)

p3/ 2a2

 x log sec    c  (Rohilkhand 2002, 2003) 14. The radius of curvature at the point (s,  ) on the curve s = c tan  is : (a) c tan 2 

(b) c sec 2 

(c)

x/c

(d)

c

(d)

None of these

(Avadh 2002, 2005) 15. Radius of curvature of the curve y = (a) 2 2

ex

at the point (0,1) is :

(b) 3 2

(c)

0

(d)

None of these (Avadh 2003)

d 16. For any curve dx

 dx   :  ds  (b) – 1 

(a) 1/

(c)



(d)

2 (Kanpur 2001)

17. Locus of centre of curvature is known as: (a) circle of curvature

(b) chord of curvature

(c) Evolute

(d) Envelope (Rohilkhand 2002, 2004)

18. Normals of a curve are (a) Normals to its evolute

(b) Tangents to its evolute

(c) Chords to its evolute

(d) None of these ANSWERS

1.

(c)

2.

(b)

3. (c)

4. (b)

5.

(c)

6.

(d)

7.

(c)

8.

(a)

9. (a)

10. (a)

11.

(c)

12.

(b)

13.

(a)

14.

(b)

15. (b)

16. (a)

17.

(c)

18.

(b)

CHAPTER 15 ASYMPTOTES 15.1 Definition A straight line is said to be an Asymptote of an infinite branch of a curve, if , as the point P recedes to infinity along the branch, the perpendicular distance of P from the straight line tends to 0. (Kumaon 2004) Illustration. The line x = a is an asymptote of the Cissoid y2 (a – x) = x3 It is easy to see that as P (x, y) moves to infinity, its distance from the line x = a tends to zero. Ex. What are the asymptotes of the curves. y = tan x; y = cot x ; y = sec x and y = cosec x.

15.2 Determination of Asymptotes. We know that the equation of a line which is not parallel to X-axis is of the form y = mx + c ...(i) The abscissa, x, must tend to infinity as the point P (x, y) recedes to infinity along this line. We shall now determine, m, and c, so that the line (i) may be an asymptote of the given curve.

Fig. 15.1

Fig. 15.2

If p = MP be the perpendicular distance of any point P (x, y) on the infinite branch of a given curve from the line (i), we have y  mx  c p (1  m 2 ) Now

p 0 as x 

 lim (y – mx – c) = 0, which means that, when x  lim (y – mx) = c Also,

456

ASYMPTOTES

 or when x 

457

y/x – m = (y – mx). (1/x). lim (y/x – m) = lim (y – mx). lim 1/x = c. 0 = 0 lim (y/x) = m, m = lim ( y / x ) , c = lim ( y – mx ).

Hence,

x 

x 

We have thus the following method to determine asymptotes which are not parallel to y-axis: (i)

Find lim ( y / x) ; let lim ( y / x)  m. x 

x 

(ii) Find lim ( y  mx); let lim ( y  mx)  c x 

x 

Then y = mx + c is an asymptote. The values of y will be different according to the different branches along which P recedes to infinity, and so we expect several values of lim (y/x) corresponding to the several values of y and also several corresponding values of lim (y – mx). Thus a curve may have more than one asymptote. This method will determine all the asymptotes except those which are parallel to Y-axis. To determine such asymptotes, we start with the equation x = my + d which can represent every straight line not parallel to X-axis and show, that when y  m = lim (x/y) and d = lim (x – my). The asymptotes not parallel to any axis can be obtained either way. Ex. 1. Examine the Folium x3 + y3 –3 axy = 0, ...(i) for asymptotes. The given equation is of the third degree. To find lim (y/x), divide the equation (i) by x3, so that 3

y 1  y 1     3a .  0. x x x   Let x . We then get 1 + m3 = 0 or (m + 1) (m2 – m + 1) = 0.  m = –1. 2 The roots of m – m + 1 = 0 are not real. To find lim (y – mx) when m = – 1, i.e., to find lim (y + x), we put y + x = p so that, p is a variable which  c when x  Putting p – x for y in the equation (i), we get x3 + (p – x)3 – 3ax (p – x) = 0, or 3(p + a)x2 – 3(p2 + ap)x + p3 = 0, which is of the second degree in x. Dividing by x2, we get 3( p  a )  3( p 2  ap ).

Let x . We then have 3 (c + a) = 0 or c = – a. Hence, y = – x – a or x + y + a = 0, is the only asymptote of the given curve.

1 1  p 3 . 2  0. x x

458

DIFFERENTIAL CALCULUS

If we start with x = my + d, we get no new asymptotes. Thus x+y+a=0 is the only asymptote of the given curve. Ex. 2. Find the asymptotes of the following curves: (i) x2 (x – y) + ay2 = 0. (ii) x3 + y3 = 3ax2. 3 3 2 (iii) y = x + ax .

(Garhwal 2003)

ANSWERS (i) x – y + a = 0

(ii) x + y = a

(iii)

y = x + a/3

15.3. Working rules for determining asymptotes. Shorter Methods. In practice the rules obtained below for determining asymptotes are found more convenient than the method which involves direct determination of lim (y/x) and lim (y – mx). Firstly we shall consider the case of asymptotes parallel to the co-ordinate axes and then that of oblique asymptotes. 15.3.1 Determination of the aysmptotes parallel to the co-ordinate axes. Asymptotes parallel to Y-axis. Let x=k ...(i) be an asymptote of the curve, so that we have to determine k. Here, y alone tends to infinity as a point P (x, y) recedes to infinity along the curve. The distance PM of any point P (x, y) on the curve from the line (i) is equal to x – k. lim (x – k) = 0 when y , or lim x = k when y , Fig. 15.3 which gives k. Thus to find the asymptotes paralel to Y-axis, we find the definite value or values k1 , k2 , etc., to which x tends as y tends to . Then x = k1, x = k2, etc. are the required asymptotes. We will now obtain a simple rule to obtain the asymptotes of a rational Algebraic curve which are parallel to Y-axis. We arrange the equation of the curve in descending powers of y, so that it takes the form

y m  ( x)  y m 1 1 ( x)  y m  2 2 ( x)  ...  0;

...(ii)

 ( x), 1 ( x), 2 ( x), etc., where are polynomials in x. Dividing the equation (ii) by ym, we get

 ( x)  (1/ y ) . 1 ( x)  (1/ y 2 ). 2 ( x)  ......  0 Let y . We write lim x = k The equation (iii) gives  (k) = 0, so that, k is a root of the equation  (x) = 0.

...(iii)

ASYMPTOTES

459

Let k1, k2 etc., be the roots of  (x) = 0. Then the asymptotes parallel to Y-axis are x = k1, x = k2, etc. From algebra, we know that (x – x1), (x – k2), etc., are the factors of  (x) which is the co-efficient of the highest power ym of y in the given equation. Hence we have the rule : The asymptotes parallel to Y-axis are obtained by equating to zero the real linear factors in the co-efficient of the highest power of y, in the equation of the curve. The curve will have no asymptote parallel to Y-axis, if the co-efficient of the highest power of y, is a constant or if its linear factors are all imaginary. Asymptotes parallel to X-axis. As above it can be shown that the asymptotes, which are parallel to X-axis, are obtained by equating to zero the real linear factors in the co-efficient of the highest power of x, in the equation of the curve. 15. 3. 2 To determine the asymptotes of the general rational algebraic equation U n  U n 1  U n  2  ...  U 2  U1  U 0  0, ...(i) where, Ur , is a homogeneous expression of degree r in x, y. We write Ur in the form U r  x r r ( y / x ) where r (y/x) is a polynomial in y/x of degree, r, at most. So we write (i) in the form  y  y  y  y  y x n n    x n 1 n 1    x n  2 n  2    ...  x1    0    0. x  x x x x ...(ii) n Dividing by x , we get 1 1 1  y 1  y  y  y  y n    n 1    2 n  2    ...  n –1 1    n 0    0. x x x  x x x  x x x ...(iii) On taking limits, as x , we obtain the equation n (m) = 0 ...(iv) which determines the slopes of the asymptote. Let m1 be one of the roots of this equation so that n (m1) = 0. We write p y y  m1 x  p1 , i.e.,  m1  1 . x x Substituting this value of y/x in (ii), we get p  p  p     x n n  m1  1   x n 1 n 1  m1  1   x n  2 n  2  m1  1   x  x  x     p  p    ...  x 1  m1  1   0  m1  i   0 x  x    Expanding each term by Taylor’s theorem, we get   p p2 x n   n (m1 )  1 n ( m1 )  1 2 n ( m1 )  ... x 2x  

p    x n 1 n 1 (m1 )  1 n 1 (m1 )  .... x   p    x n  2 n  2 (m1 )  1 n  2 (m1 )  ....  ...  0 x  

460

DIFFERENTIAL CALCULUS

Arranging terms according to descending powers of x, we get x n  n ( m1 )  x n 1 [ p1 n ( m1 )   n 1 ( m1 )]

p2   x n  2  1 n ( m1 )  p1 n 1 ( m1 )   n  2 ( m1 )   ...  0  2  n – 1 Putting n (m1) = 0 and then dividing by x , we get  p12  1  n (m1 )  p1 n 1 ( m1 )   n  2 ( m1 )   ...  0 [ p1 n (m1 )  n 1 (m1 )]  ...(v)  2  x Let x . We write lim p1 = c1. Therefore, c1 n (m1 )  n 1 (m1 )  0, ...(vi) n 1 ( m1 ) c1   , if n ( m1 )  0. or n (m1 ) n 1 ( m1 ) y  m1 x  Therefore n ( m1 ) is the asymptote corresponding to the slope m1, if n (m1 )  0. n 1 ( m2 ) n 1 ( m3 ) y  m2 x  ; y  m3 x  , etc., Similarly, n ( m2 ) n ( m3 ) are the asymptotes of the curve corresponding to the slopes m2, m3 etc., which are the roots of n (m) = 0, if n (m2 ), n (m3 ), etc., are not 0. Exceptional case. Let n (m1 )  0. If n (m1 )  0 but n 1 (m1 )  0, then the equation (vi) does not determine any value of c1 and, therefore, there is no asymptote corresponding to the slope m1. Now suppose n (m1 )  0  n  1 (m1 ). In this case, (vi) becomes an identity and we have to reexamine the equation (v) which now becomes  p12  1 n ( m1 )  p1 n 1 ( m1 )  n  2 (m1 )   ...  ...  0  2 x   On taking limits, as x  we see that c1 is a root of the equation

(c12 / 2) n ( m1 )  c1 n 1 (m1 )  n  2 (m1 )  0, which determines two values of c1, say c1', c1'', provided that n (m1 )  0. Thus y = m1x + c1', y = m1x + c1'' , are the two asymptotes corresponding to the slope m1, These are clearly parallel. This is known as the case of parallel asymptotes. Important Note. The polynomial n(m) is obtained by putting x = 1 and y = m in the highest degree terms xnn (y/x), and n – 1 (m), n – 2 (m), etc., are obtained from xn – 1 n – 1 (y/x), xn – 2 n – 2 (y/x), etc., in a similar way.

ASYMPTOTES

461

EXAMPLES 1. Find the asymptotes parallel to co-ordinate axes, of the curves: (i)

( x 2  y 2 ) x  ay 2  0

x 2 y 2  a 2 ( x 2  y 2 )  0.

(ii)

(Banglore 2005, 2006) (i)

The co-efficient of the highest power y2 of y is x – a. Hence, the asymptote parallel to y-axis is

x – a = 0. The co-efficient of the highest power x3 of x is 1 which is a constant. Hence there is no asymptote parallel to x-axis. (ii) The co-efficient of the highest power y2 of y is x2 – a2. Also, x 2  a 2  ( x  a) ( x  a ) x  a  0, x  a  0 Hence are the two asymptotes parallel to y-axis. It may similarly be shown that y–a = 0, y + a = 0 are the two asymptotes parallel to x-axis. 2. Find the asymptotes of the cubic curve 2 x3  x 2 y  2 xy 2  y 3  4 x 2  8 xy  4 x  1  0 (G.N.D.U. Amritsar 2004) Putting x = 1, y = m in the third degree and second degree terms separately, we get 3 (m)  2  m  2m2  m3 , 2 (m)   4  8m The slopes of the asymptotes are given by 3 (m)  m3  2m2  m  2  0 or ( m  1) (m – 1) (m  2)  0  m = – 1, 1, 2 Again, c is given by c 3 ( m)  2 ( m)  0 i.e., c (1  4m  3m2 )  ( 4  8m)  0 Putting m = –1, 1, 2, we get c = 2, 2, –4 respectively. Therefore the asymptotes are y = – x + 2, y = x + 2, y = 2x – 4. 3. Find the asymptotes of

x3  x 2 y  xy 2  y 3  2 x 2  4 y 2  2 xy  x  y  1  0 (Rohilkhand 2001) Here

2

3

2

3 (m)  1  m  m  m  (1  m)  m (1  m)  (1  m2 ) (1  m)  (1  m)2 (1  m).

 2 ( m)  2  2 m  4 m 2 . The slopes of the asymptotes, given by 3 (m) = 0, are 1, 1, – 1. To determine c, we have c 3 ( m)  2 ( m)  0; i.e., ...(i) c (1  2m  3m2 )  (2  2m  4m2 )  0. For m = – 1, this gives c = 1 and therefore, y = – x + 1 is the corresponding asymptote. For m = 1, the equation (i) becomes 0. c + 0 = 0 which is identically true. In this exceptional case, c is determined from the equation (c 2 / 2). 3 ( m)  c 2 ( m)  1 ( m)  0.

462

DIFFERENTIAL CALCULUS

i.e.,

(c 2 / 2) ( 2  6m)  c (2  8 m)  1 (1  m)  0. Fore m = 1, this becomes 2c 2  6c  2  0, i.e., c  (3  5) / 2. Hence, y  x  (3  5) / 2 are the two parallel asymptotes corresponding to the slope 1. We have thus obtained all the asymptotes of the curve. 4. Find the asymptotes of x3 + 2x2y – xy2 – 2y3 + xy – y2 – 1 = 0 (Sagar 99; Kumaon 2004; Gorakhpur 2002) The equation of the curve is ...(i) x3  2 x 2 y  xy 2  2 y 3  xy  y 2  1  0. Substituting (mx + c) for y in (i), we get x3  2 x 2 (mx  c)  x (mx  c) 2  2 (mx  c)3  x (mx  c)  (mx  c) 2  1  0 or  and

x3 (1 + 2m – m – 2m3) + x2 (2c – 2mc – 6m2c + m – m2) + ... = 0. m and c are given by 1 + 2m – m2 – 2m3 = 0 2c – 2mc –

6m2

c+m–

m2

=0

From (ii), (1  m2 )  2m (1  m2 )  0 or From (iii),  when when

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

2

(1  m ) (1  2m)  0

or

m  1,  1,  1/ 2. 2c (1  m  3m 2 )  m 2  m or c 

m2  m

2 (1  m  3m3 ) 11 m  1, c   0, 2 (1  1  3) 11 m   1, c    1 and 2 (1  1  3) (1 / 4)  (1 / 2)

(3 / 4) 1  3/ 2 2 2 1  Substituting these pairs of values of m and c in y = mx + c one by one we have required asymptotes as

when

m   1 / 2, c 



1 2

3 4





y  x, y   x  1 and y   12 x  12 or 2 y  x  1. 5. Find all the asymptotes of the curve y3 – 6xy2 + 11x2y – 6x3 + x + y = 0. (Gorakhpur 2000) Let y = mx + c be an asymptote, then substituting mx + c for y in the equation of the curve, we get

(mx  c)3  6 x (mx  c)2  11x 2 (mx  c)  6 x3  x  (mx  c)  0. or

x3 (m3  6m2  11 m  6)  x 2 (3m2 c  12 mc  11c)  x (3mc 2  6c 2  1  m)  (c3  c)  0

Equating the coefficients of x3 and x2 to zero we get

m3  6m 2  11 m  6  0 2

and 3m c  12mc  11c  0 From (i) we get (m  1) (m  2) (m  3)  0 or m  1, 2, 3 Also from (ii) we get (3m2  12m  11) c  0

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

ASYMPTOTES

463

or c = 0, for all the three values of m viz. 1, 2, 3. Hence for these values of m and c we get the required asymptotes as y = x + 0, y = 2x + 0, y = 3x + 0 i.e., y = x, y = 2x and y = 3x. 6. Find the asymptotes of the cubic curve y 3  5 xy 2  8 x 2 y  4 x3  3 y 2  9 xy  6 x 2  2 y  2 x  1. (Kumaon 96; Bhopal 97; Avadh 2000) Putting mx + c for y and rearranging; (m3  5 m2  8 m  4) x3  (3 m2 c  13 m c  8 c  3 m2  9 m  6) x 2 Equating to zero the coefficients of x3 and x2, m3 – 5 m2 + 8 m – 4 = 0 and 3 m2 c – 10 mc + 8 c – 3 m2 + 9 m – 6 = 0 From the first equation, we get (m – 1) (m – 2)2 = 0 Hence m = 1, 2, 2. If m = 1, the second equation gives c = 0 and the corresponding asymptote is y = x. If m = 2, the left hand side of the second equation vanishes identically for all finite values of c. In such cases, to find c, equate the next coefficient to zero. hence 3 m c2 – 5 c2 – 6 m c + 9 c + 2 m – 2 = 0 Putting m = 2, we have c2 – 3 c + 2 = 0, i.e., c = 1 or 2. Hence there are two parallel asymptotes. y = 2 x + 1 and y = 2 x + 2. Hence the required asymptotes are y = x, y = 2 x + 1, y = 2 x + 2 7. Find the asymptotes of the curve x3 – x2y – xy2 + y3 + 2x2 – 4y2 + 2xy + x + y + 1 = 0. Substituting mx + c for y in the given equation, we get

x3  x 2 (mx  c)  x (mx  c) 2  (mx  c)3  2 x 2  4 (mx  c)2  2 x ( mx  c)  x  ( mx  c)  1  0

or

3

2

3

2

2

x (1  m  m  m )  x ( c  2mc  3m c  2  4m 2  2m)  x ( c 2  3mc 2  8mc  2c  1  m)  ...  0.

...(i)

Equating the coefficient of highest power of x in (i) to zero, we have

1  m  m2  m3  0 or (1 – m)  m2 (1  m)  0 or

(1  m) (1  m 2 )  0 or (1  m)2 (1  m)  0 or m  1, 1,  1

Equating the coefficient of x3 in (i) to zero, we get

c (1  2m  3m2 )  2  4m2  2m.

...(ii)

When , m = – 1, from (ii) we have c = 1 and hence the corresponging asymptote is y = – x + 1 or x + y = 1. When m = 1 from (ii) we have 0.c = 0, where from c cannot be determined.  Equating to zero the coefficient of x in (i), we get

464

DIFFERENTIAL CALCULUS

c 2  3mc 2  8mc  2c  1  m  0, which reduces to 2c 2  6c  2  0 or c 2  3c  1  0, when m  1 or c  12 [3  (9  4)]  12 (3  5)  Corresponding asymptotes are y  x  12 (3  5) or 2y  2x  3  5.  The required asymptotes are x  y  1, 2 x  2 y  3  5  0. EXERCISES Find the asymptotes parallel to co-ordinate axes of the following curves: 1. y2x – a2 (x – a) = 0 2. x2y –3x2 – 5xy + 6y + 2 = 0 3. y = x/(x2 – 1) 4. a2/x2 + b2/ y2 = 1 (Banglore 2004) Find the asymptotes of the following curves: 5. x (y – x)2 = x (y – x) + 2 6. x2 (x – y)2 + a2 (x2 – y2) = a2xy (Gorakhpur 2000) 2 2 2 2 2 7. (x – y) (x + y ) – 10 (x – y) x + 12y + 2x + y = 0 8. x2y + xy2 + xy + y2 + 3x = 0 (Banglore 2006) 9. (x – y + 1) (x – y – 2) (x + y) = 8x – 1 10. y3 – x2y + 2xy2 – y + 1 = 0 (Jabalpur 1998, 2000; Ravishankar 1998; Bhopal 1999) 2 11. y (x – y) = x + y 12. x2y2 (x2 – y2)2 = (x2 + y2)2 13. y (y – 1)2 – x2 = 1 14. xy2 = (x + y)2 15. (x + y) (x – y) (2x – y) – 4x (x – 2y) + 4x = 0 16. xy2 – x2y – 3x2 – 2xy + y2 + x – 2y + 1 = 0 17. 2x (y – 3)2 = 3y (x – 1)2 18. (y – a)2 (x2 – a2) = x4 + a4 19. (x + y)2 (x2 + xy + y2) = a2 x2 + a2 (y – x) 20. y2 + 3y2x –x2y –3x2 + y2 –2xy + 3x2 + 4y + 5 = 0 21. (x – y)2 (x – 2y) (x – 3y) – 2a (x3 –y3) –2a2 (x – 2y) (x + y) = 0 22. y2 = x4 /(a2 – x2) Show that the following curves have no asymptotes :– 23. x4 + y4 = a2 (x2 – y2) 24. y2 = x (x + 1)2 4 2 5 25. a y = x (2a – x) 26. x2 (y2 + x2) = a2 (x2 – y2) 27. Find the equation of the tangent to the curve x3 + y3 = 3ax2 which is parallel to its asymptote. 28. An asymptote is sometimes defined as a straight line which cuts the curve in two points at infinity. Criticise this definition and replace it by a correct definition.

1. 3. 5. 7.

x = 0, y = ± a. x = ± 1, y = 0 x = 0, y = x, y = x + 1. y = x – 2, y = x – 3.

ANSWERS 2. x = 2, x = 3, y = 3. 4. x = ± a, y = ± b. 6. y = x ± a, x = ± a. 8. x + 1 = 0, y = 0, x + y = 0

ASYMPTOTES

465

9. y = x – 3, y = x + 2, x + y = 0. 11. y = 0, y = x  12. 13. 15. 16. 17. 19. 20. 21. 22.

10. y = 0, y + x = ± 1.

2

x  y   2, x   1, y   1, y – x   2 y = 0, y ± x = 1. 14. x = 1. x + y = 2, y = x + 2, y = 2x – 4. y + 3 = 0, x + 1 = 0, y = x + 4. x = 0, y = 0, 2y – 3x = 6. 18. x = ± a, y ± x = a. x + y ± a = 0. 4 (y – x) + 1 = 0, 2 (y + x) = 3, 4 (y + 3x) + 9 = 0. x = y + a, x = y + 2a, 2y = x + 14a, 3y = x – 13a. x = ± a. 27. x + y = 4a. 15.3.3 Some deduction from § 15.3.2

(i) The number of asymptotes of an algebraic curve of the nth degree cannot exceed n. The slopes of the asymptotes which are not parallel to y-axis are given as the roots of the equation n (m) = 0 which is of degree n at the most. In case the curve possesses one or more asymptotes parallel to y-axis, then it is easy to see that the degree of n (m) = 0 will be smaller than n by at least the same number. Hence the result. (ii) The asymptotes of an algebraic curve are parallel to the lines obtained by equating to zero the factors of the highest degree terms in its equation. Let, m1 be a root of the equation n (m) = 0, so that the line y – m1 x = 0 is parallel to an asymptote. By elementary algebra, (y/x – m1) is a factor of n (y/x) and hence, y – m1 x is a factor of xn n (y/x), i.e., Un. Also conversely, we see that if, y – m1 x is a factor of Un then m1 is a root of n (m) = 0. In case the highest degree terms contain, x, as a factor, then a little consideration will show that the curve will possess asymptotes parallel to x = 0, i.e., to y-axis. (iii) Case of parallel asymptotes. In this case, m1 satisfies the three equations n (m) = 0, 'n (m) = 0,  n– 1 (m) = 0. Since n (m) and its derivative 'n (m) vanish for m = m1, therefore, by elementary algebra, m1 is a double root of n (m) = 0 and therefore, (y – m1 x)2 is a factor of the highest degree terms Un. Also, since m1 is a root of n – 1 (m) = 0, y – m1 x is a factor of the (n –1)th degree terms Un – 1. Thus, we see that in the exceptional case a twice repeated linear factor of Un is also a nonrepeated factor of Un – 1. There will be no asymptote with slope m 1 , if m 1 is a root of  n (m) = 0, 'n (m) = 0 but not of n – 1 (m) = 0, i.e., if (y – m1 x)2 is a factor of Un and y – m1 x is not a factor of Un – 1 . Note. The results obtained in the paragraphs (ii) and (iii) above enable us to shorten the process of determining the asymptotes as shown in the following examples. The first step will always consist in factorising the expression formed of the highest degree in the given equation.

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EXAMPLES 1. Find the asymptotes of the Folium x3 + y3 – 3axy = 0. (Bhopal 2000; Sagar 1998; Vikram 1998; Gorakhpur 2001) The curve has no asymptotes parallel to co-ordinate axes. Factorizing the highest degree terms, we get (x + y) (x2 – xy + y2) – 3axy = 0. so that, y + x, is the only real linear factor of the highest degree terms. Hence, the curve has only one real asymptote which is parallel to the line y + x = 0 whose slope is – 1. We have, now to find, lim (y + x), when x   and y/x  – 1. 3ayx 3a . ( y / x ) y x 2  . We have x – xy  y 2 1 – ( y / x )  ( y / x ) 2 In the limit, we have – 3a –a ...(i) 111  y = – x – a, i.e., y + x + a = 0, is the only asymptote of the folium. It is easy to see that we could have eliminated the step (i), and simplified the process by saying that the required asymptote is 3a ( y / x ) ( y  x )  lim , 1 – ( y / x)  ( y / x) 2 when x   and y/x  – 1. 2. Find the asymptotes of x3 + 4x2y + 4xy2 + 5x2 + 15xy + 10y2 – 2y + 1 = 0. Equating to zero the co-efficient of the highest power y2 of y, we see that 4x + 10 = 0, i.e., 2x + 5 = 0, is one asymptote. Factorising the highest degree terms, we get x (2y + x)2 + 5x2 + 15xy + 10y2 – 2y + 1 = 0. Here 2y + x is a repeated linear factor of highest degree terms, i.e., 3rd degree. There will, therefore, be no asymptote parallel to 2y + x = 0 if (2y + x) is not a factor of the 2nd degree terms also. But this is not the case. In fact, the equation is x (2y + x)2 + 5 (x + y) (x + 2y) – 2y + 1 = 0. Therefore, the curve has two asymptotes parallel to 2y + x = 0. 1  1    1 We have now to find him  y  x  when x   and y / x  – . Let lim  y  x  = c so 2  2    2 that lim (2y + x) = 2c. Dividing by x, the equation becomes (2y + x)2 + 5 (2y + x) (1 + y/x) – 2y/x + 1/x = 0. In the limit, 1   1 4c 2  5.2c 1 –  – 2  –   0  0, ...(i) 2   2 lim ( y  x) 

ASYMPTOTES

or

467

4c2 + 5c + 1 = 0

1 , – 1. 4 1 1 1 y  – x – and y  – x – 1, Hence 2 4 2 i.e., 4y + 2x + 1 = 0 and 2y + x + 2 = 0, are the two more asymptotes. It is easy to see that we could have eliminated the step (i) and simplified the process by saying that the asymptotes are (2y + x)2 + 5 (2y + x) . lim (1 + y/x) + lim (– 2y/x + 1/x) = 0,



c–

i.e., (2y + x)2 + 5 (2y + x) .

1  1 = 0 or 2 (2y + x)2 + 5 (2y + x) + 2 = 0, 2

which gives 2y + x + 2 = 0 and 4y + 2x + 1 = 0.

3. Find the asymptotes of (x – y)2 (x2 + y2) – 10 (x – y) x2 + 12y2 + 2x + y = 0. The asymptotes parallel to the two imaginary lines x2 + y2 = 0 are imaginary. To obtain the two asymptotes parallel to the lines x – y = 0, we re-write the equation, on dividing it by (x2 + y2), as ( x – y ) 2 – 10 ( x – y )

1

( x – y ) 2 –10 ( x – y ) lim

1



12 ( y / x)2  2 / x  y / x . 1/ x

 0. 1  ( y / x) 1  ( y / x) 2 We take the limits when x   and y/x  1. Therefore the asymptotes are 2

 lim

12 ( y / x) 2  2 / x  y / x .1/ x

1  ( y / x)2 1  ( y / x)2 i.e., (x – y)2 – 5 (x – y) + 6 = 0. i.e., x – y – 2 = 0, x – y – 3 = 0. 4. Find the asymptotes of (x – y + 2) (2x – 3y + 4) (4x – 5y + 6) + 5x – 6y + 7 = 0. The asymptote parallel to x – y + 2 = 0 is 5x – 6 y  7 x – y  2  lim  0, (2 x – 3 y  4) (4 x – 5 y  6) when x   and y/x  1,

 0.

 5 – 6y / x  7/ x 1 .   0. i.e., x – y  2  lim   (2 – 3 y / x  4 / x ) (4 – 5 y / x  6 / x ) x  or x – y + 2 = 0, as the limit is zero because of the factor 1/x. Similarly, we can show that 2x – 3y + 4 = 0, 4x – 5y + 6 = 0 are also the asymptotes of the curve.

15.4. Asymptotes by Inspection. If the equation of a curve of the nth degree can be put in the form Fn + Fn – 2 = 0, where Fn – 2 is of degree (n – 2) at the most, then every linear, factor of Fn, when equated to zero will

468

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give an asymptote, provided that no straight line obtained by equating to zero any other linear factor of Fn is parallel to it or coincident with it. Let ax + by + c = 0 be a non-repeated factor of Fn. We write Fn = (ax + by + c) Fn – 1 where Fn – 1 is of degree (n – 1). The asymptote parallel to ax + by + c = 0 is Fn – 2 ax  by  c  lim  0, Fn – 1 when x   and y/x  – a/b. To determine the limit (F n – 2 /F n – 1), we divide the numerator as well as the denominator by xn – 1 and see that 1/x appears as a factor so that F n – 2 /F n – 1  0 as x  . Thus

ax + by + c = 0

is an asymptote. EXERCISES Find the asymptotes of the following curves :– 1. xy (x + y) = a (x2 – a2) 2. (x – 1) (x – 2) (x + y) + x2 + x + 1 = 0 3. y3 – x3 + y2 + x2 + y – x + 1 = 0 4. x (y2 – 3by + 2b2) = y3 – 3bx2 + b3 5. x3 + 6x2y + 11xy2 + 6y3 + 3x2 + 12xy + 11y2 + 2x + 3y + 5 = 0 6. x2 (3y + x)2 + (3y + x) (x2 + y2) + 9y2 + 6xy + 9y – 6x + 9 = 0 7. (y2 + xy – 2x2)2 + (y2 + xy – 2x2) (2y + x) – 7y2 – 19xy – 28x2 + x + 2y + 3 = 0 8. x (y – 3)3 = 4y (x – 1)3 9. (a + x)2 (b2 + x2) = x2y2 10. y3 – 5xy2 + 8x2y – 4x3 – 3y2 + 9xy – 6x2 + 2y – 2x + 1 = 0 ANSWERS 1. y = a, x = 0, x + y + a = 0. 2. x = 1, x = 2, x + y + 1 = 0. 3. 3 (y – x) + 2 = 0. 4. y – x = 0. 5. x + y + 1 = 0, 3y + x – 1 = 0, 2y + x + 1 = 0. 6. 7. 8. 9. 10.

3y + x = (– 5  106) / 9 . y + 2x + 2 = 0, y + 2x – 1 = 0; y – x + 3 = 0, y – x – 2 = 0. x = 0, y = 0, 2y = 4x + 3, 4x + 2y = 15. x ± y + a = 0, x = 0. y = x, y = 2x + 1, y = 2x + 2.

15.5. Intersection of a curve and its asymptotes. Any asymptote of a curve of the nth degree cuts the curve in (n – 2) points. Let y = mx + c be an asymptote of the curve xn n (y/x) + xn – 1 n – 1 (y/x) + xn – 2 n – 2 (y/x) + ..... = 0. To find the points of intersection, we have to solve the two equations simultaneously. The abscissae of the points of intersection are the roots of the equation xn n (m + c/x) + xn – 1 n – 1 (m + c/x) + xn – 2 n – 2 (m + c/x) + ... = 0.

...(i)

ASYMPTOTES

469

Expanding each term by Taylor’s theorem and arranging according to descending powers of x, we get xn n (m) + [c 'n (m) + n – 1 (m)] xn – 1 + 1 2  n–2  ....  0. ...(ii)  2 c "n (m)  c  n – 1 (m)  n – 2 (m)  x   As y = mx + c is an asymptote, the co-efficients of xn and xn – 1 are both zero. Thus the equation (ii) reduces to that of (n – 2)th degree and, therefore, determines (n – 2) values of x. Hence the result. Cor. 1. The, n, asymptotes of a curve of the nth degree cut it in n (n – 2) points. Cor. 2. If the equation of a curve of the nth degree can be put in the form Fn + Fn – 2 = 0 where Fn – 2 is of degree (n – 2) at the most and Fn consists of, n, non-repeated linear factors, then the n (n – 2) points of intersection of the curve and its asymptotes lies on the curve Fn – 2 = 0. The result follows at once from the fact that Fn = 0 is the joint equation of the, n, asymptotes. At the points of intersection of the curve and its asymptotes, the two equations Fn = 0 and Fn + Fn – 2 = 0 hold simultaneously and therefore at such points we have Fn – 2 = 0. Particular cases (i) For a cubic, n = 3, and therefore the asymptotes cut the curve in 3 (3 – 2) = 3 points which lie on a curve of degree 3 – 2 = 1, i.e., on a straight line. (ii) For a quartic, n = 4, and, therefore the asymptotes cut the curve in 4 (4 – 2) = 8 points which lie on a curve of degree 4 – 2 = 2 i.e., on a conic. EXAMPLES 1. Find the asymptotes of the curve x2y – xy2 + xy + y2 + x – y = 0 and show that they cut the curve again in three points which lie on the line x + y = 0. (Devi Ahilya 2001) The asymptotes of the given curve, as may be easily shown are y = 0, x = 1, x – y + 2 = 0. The joint equation of the asymptotes is y (x – 1) (x – y + 2) = 0, i.e., x2y – xy2 + xy + y2 – 2y = 0. The equation of the curve can be written as x2y – xy2 + xy + y2 – 2y + (x + y) = 0. Here F3 = x2y – xy2 + xy + y2 – 2y, F1 = x + y. Hence, the points of intersection lie on the line F1  x + y = 0. 2. Show that the asymptotes of the quartic (x2 – 4y2) (x2 – 9y2) + 5x2y – 5xy2 – 30y3 + xy + 7y2 – 1 = 0, cut the curve in the eight points which lie on a circle. The asymptotes of the curve are x + 2y = 0, x – 2y + 1 = 0, x – 3y = 0, x + 3y – 1 = 0, so that their joint equation is (x + 2y) (x – 2y + 1) (x – 3y) (x + 3y – 1) = 0, i.e., (x2 – 4y2 ) (x2 – 9y2) + 5x2y – 5xy2 – 30y3 – x2 + xy + 6y2 = 0.

470

DIFFERENTIAL CALCULUS

The equation of the curve can be written as (x2 – 4y2) (x2 – 9y2) + 5x2y – 5xy2 – 30y3 – x2 + xy + 6y2 + (x2 + y2 – 1) = 0. Hence, the points of intersection lie on the circle x2 + y2 – 1 = 0, 3. Find the equation of the cubic which has the same asymptotes as the curve x3 – 6x2 y + 11xy2 – 6y3 + x + y + 1 = 0. (Guwahati 2005) and which passes through the points (0, 0), (1, 0) and (0, 1). We write F3  x3 – 6x2y + 11xy2 – 6y3 = (x – y) (x – 2y) (x – 3y). F1  x + y + 1. The equation of the curve can be written in the form F3 + F1 = 0 where F3 has non-repeated linear factors. Thus F3 = 0 is the joint equation of the asymptotes of the cubic. The general equation of the cubic is of the form F3 + ax + by + c = 0, or x3 – 6x2y + 11xy2 – 6y3 + ax + by + c = 0, where ax + by + c is the general linear expression. In order that it may pass through the points (0, 0), (1, 0) and (0, 1), we must have c=0 1 + a = 0 or a = – 1, – 6 + b = 0 or b = 6. Thus the required cubic is x3 – 6x2y + 11xy2 – 6y3 – x + 6y = 0. 4. Show that asymptotes of the cubic x3 – 2 y3 + x y (2 x – y) + y (x – y) + 1 = 0 (Indore 1998; Jiwaji 1998; Vikram 1999; MDU Rohtak 1998) cut the curve again in three points which lie on the straight line x – y + 1 = 0. The equation of the curve is (x3 – 2 y3 + 2 x2 y – x y2) + (x y – y2) + 1 = 0 ...(i) 3 2 f0 (m) = 1 – 2 m + 2 m – m = 0 i.e., and and Now For

1 2 f0' (m) = – 6 m2 + 2 – 2 m f1 (m) = m – m2. m  1, – 1, –

c–

f1 (m) m – m2 – 2 – 2 m – 6 m2 f 0 (m) m = 1, c = 0; m = – 1, c = – 1; m  –

Hence the asymptotes are (y – x) (y + x + 1) (x + 2 y – 1) = 0 i.e., x3 – 2 y3 + 2 x2 y – x y2 + x y – y2 – x + y = 0 The asymptotes cut the curve in n (n – 2), i.e., 3 (3 – 2) = 3 points.

1 1 ,c 2 2

...(ii)

ASYMPTOTES

471

Subtracting (ii) from (i), we observe that points of intersection of the curve and asymptotes lie on x–y+1=0 5. Find the equation of the straight line on which lie the three points of intersection of the curve (x + a) y2 = (y + b) x2 and its asymptotes. The asymptotes parallel to x-axis and y-axis are y + b = 0; x + a = 0. Now putting y – x = k or x = – k + y, the curve is (– k + y) y2 + a y2 = (y + b) (– k + y)2 or y2 (– k + a – b + 2 k) + .... = 0 If y – x = k is asymptote, the coefficient of y2 is zero, i.e., k = – a + b. Hence third asymptote is y – x = b – a, The combined equation of asymptotes is (x + a) (y + b) (– y + x + b – a) = 0 or x y (y – x) + a y2 – b x2 + a2 y – b2 x + a b (a – b) = 0. Subtracting the combined equation to the asymptotes from the curves, we obtain a2 y – b2 x + a b (a – b) = 0 ...(i) Hence the points of intersection of the curve and asymptotes are 3 (3 – 2) = 3 and they lie on (i). 6. Find the equation of the conic on which lie the eight points of intersection of the quartic curve x y (x2 – y2) + a2 y2 + b2 x2 – a2 b2 = 0, with its asymptote. The given equation of curve is x3 y – x y3 + a2 y2 + b2 x2 – a2 b2 = 0 By putting x = 1 and y = m in the highest degree term, we get f0 (m) = m (1 – m2) and f0' (m) = 1 – 3 m2 Also, f1 (m) = 0 as there is no third degree term in the equation of curve. By putting f0 (m) = 0, we get m = 0, 1, – 1. f ( m) c– 1 0 Also, f 0 ( m)  The asymptotes are y = 0, y = x and y = – x. Also equating the coefficient of highest power of y to zero, we have x = 0 as the asymptote. Thus we have the four asymptotes of curve as x = 0, y = 0, x + y = 0 and x – y = 0. Their combined equation is x y (x + y) (x – y) = 0 or x y (x2 – y2) = 0. Subtracting this equation from the gives equation we find that the point of intersection of the curve and asymptotes lie on a2 y2 + b2 x2 – a2 b2 = 0

or

x2 a

2



y2 b2

1

Which is an ellipse. Also the asymptotes cut the curve in n (n – 2), i.e., 4 (4 – 2) or 8 points. 7. Find the equation of the cubic curve whose asymptotes are x + a = 0, y – a = 0 and x + y + a = 0 and which touches the axis of x at the origin and passes through the point (– 2a, – 2a).

472

DIFFERENTIAL CALCULUS

The combined equation of the asymptotes is F3  (x + a) (y – a) (x + y + a) = 0  The equation of the curve having F3 = 0 as its asymptotes can be written as F3 + Q1 = 0, where Q1 is an expression of degree n – 2, i.e., 3 – 2 or 1. Let Q1  bx + cy + d = 0, then the equation of the curve is (x + a) (y – a) (x + y + a) + (bx + cy + d) = 0. ...(i)  This passes through the origin, so we have d = a3, substituting x = 0 and y = 0 in (i). So the equation of the curve reduces to (x – a) (y – a) (x – y + a) + bx + cy + a3 = 0. or xy (x + y) + a (y2 – x2 + xy) – 2a2x + bx + cy = 0 and the tangent at the origin is (– 2a2 + b) x + cy = 0. obtained by equating the lowest degree terms to zero. But we are given that x-axis, i.e., y = 0 is the tangent at the origin. Hence – 2a2 + b = 0 and the equation of the curve reduces is xy (x + y) + a (y2 – x2 + xy) + cy = 0. ...(ii) Also this passes through (– 2a, – 2a), hence 4a2 (– 4a) + a (4a2 – 4a2 + 4a2) + c (– 2a) = 0 or c = – 6a2. Hence from (ii), the required equation is xy (x + y) + a (y2 – x2 + xy) – 6a2y = 0 8. Find the equation of the quartic curve which has x = 0, y = 0, y = x and y = – x for asymptotes and which passes through (a, b) and which cuts its asymptotes again in eight points that lie on a circle whose centre is origin and radius a. The combined equation of the asymptotes is xy (x – y) (y + x) = 0 or xy (y2 – x2) = 0. ...(i) Also the equation of the circle with origin as centre and radius a is x2 + y2 = a2.

...(ii)

Let the equation of the curve whose asymptotes are given by (i) be xy (y2 – x2) +  (x2 + y2 – a2) = 0.

...(iii)

If it passes through (a, b), then ab (b2 – a2) +  (a2 + b2 – a2) = 0, or  = a (a2 – b2)/b.  The required equation of the curve is bxy (y2 – x2) + a (a2 – b2) (x2 + y2 – a2) = 0, putting the value of  in (iii). EXERCISES 1. Show that the asymptotes of the cubic x3 – xy2 – 2xy + 2x – y = 0 cut the curve again in points which lie on the line 3x – y = 0. 2. If a right line is drawn through the point (a, 0) parallel to the asymptote of the cubic (x – a)3 – x2 y = 0, prove that the portion of the line intercepted by the axes is bisected by the curve. 3. Through any point P on the hyperbola x2 – y2 = 2ax, a straight line is drawn parallel to the only asymptote of the curve x3 + y3 = 3ax2 meeting the curve in A and B; show that P is the mid-point of AB.

ASYMPTOTES

473

4. Show that y = x + a is the only asymptote of the curve x2 (x – y) + ay2 = 0. A straight line parallel to the asymptote meets the curve in P, Q; show that the mid-point of PQ lies on the hyperbola x (x – y) + ay = 0. 5. Find the equation of the straight line on which lie three points of intersection of the cubic x3 + 2x2y – xy2 – 2y3 + 4y2 + 2xy + y – 1 = 0 and its asymptotes. 6. Find the asymptotes of the curve 4x4 – 13x2y2 + 9y4 + 32x2y – 42y3 – 20x2 + 74y2 – 56y + 4x + 16 = 0, and show that they pass through the intersection of the curve with y 2 + 4x = 0. 7. Find all the asymptotes of the curve 3x3 + 2x2y – 7xy2 + 2y3 – 14xy + 7y2 + 4x + 5 = 0. Show that the asymptotes meet the curve again in three points which lie on a straight line, and find the equation of this line. 8. Find the equation of the cubic which has the same asymptotes as the curve x3 – 6x2y + 11xy2 – 6y3 + x + y + 1 = 0, and which touches the axis of y at the origin and passes through the point (3, 2). 9. Find the asymtotes of the curve 4 (x4 + y4) – 17x2 y2 – 4x (4y2 – x2) + 2 (x2 – 2) = 0. and show that they pass through the points of intersection of the curve with the ellipse x2 + 4y2 = 4. 10. Find the asymptotes of the curve (2x – 3y + 1)2 (x + y) – 8x + 2y – 9 = 0 and show that they intersect the curve again in three points, which lie on a straight line. Obtain the equation of the line. ANSWERS 5. 3y + x = 1.

6.

y   x  1, y  

2 4 x . 3 2

7. 6 (y – x) + 7 = 0, 2 (y – 3x) + 3 = 0. 3 (2y + x) + 5 = 0, 106y – 381x + 105 = 0. 3 2 3 2 8. x – 6x y + 11xy – 6y – x = 0. 9. y = 2x3 + 1, y = – 2x – 1, x = 2y, x + 2y = 0. 10. x + y = 0, 2x – 3y – 1 = 0, 2x – 3y + 3 = 0, 4x – 6y + 9 = 0.

15.6. Asymptotes by expansion. To show that y = mx + c is an asymptote of the curve y = mx + c + A/x + B/x2 + C/x3 + ... Dividing by x, we have y/x = m + c/x + A/x2 + B/x3 + C/x4 + ... so that when x  . lim (y/x) = m. Also from (1) we have

...(1)

...(2)

474

DIFFERENTIAL CALCULUS

(y – mx) = c + A/x + B/x2 + C/x3 + ... so that when x  . lim (y – mx) = c. From (2) and (3), we deduce that y = mx + c is an asymptote of the curve (1).

...(3)

15.7. Position of a curve with respect to an asymptote. To find the position of the curve y = mx + c + A/x + B/x2 + C/x3 + ... with respect to its asymptote y = mx + c. Let A  0. Let y1 and y2 denote the ordinates of the curve and the asymptote corresponding to the same abscissa x. We have y1 – y2 = A/x + B/x2 + C/x3 + ... = (1/x) (A + B/x + C/x2 + ...) ...(1) By taking x sufficiently large, we can make B/x + C/x2 + ... as small as we like. We suppose that x is so large numerically that this expression is numerically less than A. Thus, for sufficiently large values of x, the expression A + B/x + C/x2 + D/x3 + ... ...(2) has the sign of A. Thus if A be positive, the expression (1) is positive for sufficiently large values of x so that, from (1) we deduce that when x is positively sufficiently large, then y1 – y2 is positive, i.e., the curve lies above the asymptote and when x is negative but numerically large, y1 – y2 is negative, i.e., the curve lies below the asymptote. Similarly, we may deduce that if A be negative, then the curve lies below the asymptote when, x, is positive but sufficiently large and lies above the asymptote when, x, is negative but sufficiently large numerically. Let A = 0, B  0; we have y1 – y2 = (1/x2) (B + C/x + D/x2 + ...) As above we can show that for numerically sufficiently large values of x, the expression B + C/x + D/x2 + ... has the sign of B. In this case, the curve lies on the same side of the asymptote both for positive and negative values of x; it will be above or below the asymptote according as B is positive or negative. It B = 0 and C  0, we will have a situation similar to that of case (1). Ex. 1. Find the asymptotes of the curve y2 = x (x – a) (x – 2a)/(x + 3a) and determine on which side of the asymptotes the curve lies. We have

y

 x ( x – a ) ( x – 2a )    x  3a   1/ 2

a    x 1 –  x 

1/ 2

2a   1 –  x  

3a   1 –  x  

–1/ 2

ASYMPTOTES

475

  a a2   a a2 3a 27 a 2    x 1 – – 2 ...  1 – – 2 ...   1 –  ...      2 x 8x x 2x 2x 8x2      3a 11a 2    x 1 –  ...   x 2 x 2   Thus we have two values of y, viz., 11 2 y  x – 3a  a x ... 2 11 2 y  – x  3a – a x ... 2 Therefore y = x – 3a, y = – x + 3a are the two asymptotes. The difference between the ordinate of the curve and that of the asymptote y = x – 3a being 11 2 a x – ..., 2 we see that the curve lies above the asymptote when x is positive and below it when x is negative. It may similarly be seen that the curve lies below the second asymptote when x is positive and above it when x is negative. It is easy to see that x = – 3a is also an asymptote of the curve. To find the position of the curve relative to this asymptote, we suppose that x = – 3a + A/y + B/y2 + C/y3 + ... Substituting this value of x in the equation of the curve, we have A   B C    A A A y 2   2  3  ...   – 3a   ...  – 4a   ...  – 5a   ... y y y y y    y   Equating the co-efficients of like powers of y, we have A = 0. B = – 60a3, etc. 60a 3 x  – 3 a –  ...  y2 The difference between the abscissae of the curve and the asymptote x = – 3a, for the same value of y, being – 60a3/y2 + ... , which is negative whether y be positive or negative, we see that the curve lies towards the negative side of x-axis. Ex. 2. Find the asymptotes and their position with regard to the following curves : (i) x3 + y3 = 3ax2. (ii) x3 + y3 = 3axy. (iii) x2 (x – y) + y2 = 0.

ANSWERS (i) x + y = a. The curve lies above or below the asymptote according as x is positive or negative. (ii) x + y + a = 0. The curve lies above the asymptote both for positive as well as negative values of x. (iii) y = x + 1. The curve lies above or below the asymptote according as x is positive or negative.

476

DIFFERENTIAL CALCULUS

15.8. Asymptotes in polar co-ordinates. Lemma. The Polar Equation of a Line. The polar equation of any line is p = r cos ( – ). where, p, is the length of the perpendicular from the pole to the line and , is the angle which this perpendicular makes with the initial line. Let OY be the perpendicular on the given line; Y being its foot. We are given that OY = p; XOY = . If P (r, ) be any point on the line, we have YOP =  – . OY Fig. 15.4  cos YOP. OP  p/r = cos ( – ), i.e., p = r cos ( – ), which is the required equation of the line. To determine the asymptotes of the curve r = f (), we have to obtain the constants, p and, , so that any line p = r cos ( – ), is the asymptote of the given curve. Let P (r, ) be any point on the curve (i). Draw OY  the line (ii). Draw PL  OY and PM  the line (ii). Now, PM = LY = OY – OL Fig. 15.5 = p – OP cos ( – ) = p – r cos ( – ). Now r   as the point recedes to infinity along the curve. Let   1 when r  .

Now,

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

...(iii)

PM p  – cos ( –  ). r r Now when r  , PM  0 so that

We have

PM 1 p  PM .  0 and  0. r r r  lim cos ( – ) = 0 or lim ( – ) = /2, or 1 –  = /2, i.e.,  = 1 – /2. This gives . Again, p = OY is the polar sub-tangent of the point at which the asymptote touches the curve, i.e., the point at infinity on the curve. This may be seen as follows :– Join the pole O to the point at infinity on the curve i.e., draw through O a line parallel to the asymptote. This line is the radius vector of the point at . Draw through O a line perpendicular to the asymptote meeting it at Y. Then, by def. OY is the polar sub-tangent of the point at infinity on the curve.

ASYMPTOTES

Thus

477

1 d  p–  , where u  . du r     1

Note. Without employing the notion of the polar sub-tangent and the point at infinity, the value of p, may also be obtained as follows :– From (iii) we have, when r  , p = lim [r cos ( – )] = lim [r cos ( – 1 + /2)]  lim [ r cos (1 – )]  lim

1

sin (1 – ) 1/ r

which is of the form (0/0). p  lim

1

cos (1 – ) 1 dr r2 d 

 d   lim  r 2  lim 1  dr  Hence, the asymptote is  d lim  –   r cos ( –  )  du     r cos   – 1   2  = r sin (1 – ). where 1 is the limit of  as r   i.e., as u  0.

1  d   – du  where u  r  

Working rule for obtaining asymptotes to polar curves. Change r to 1/u in the given equation and find out the limit of  as u  0. Let 1, be any one of the several possible limits of . Determine (– d /du) and its limit as u  0 and   1. Let this limit be p. Then p = r sin (1 – ). is the corresponding asymptote. To draw the asymptote. 1   Through the pole O draw a line making angle  1 –   with the initial line; on this line take a 2   point Y such that OY = lim (– d /du). The line drawn through Y perpendicular to OY is the required asymptote. EXAMPLES 1. Find the asymptote of the hyperbolic spiral r  = a. Here  = a/r = au so that   0 as u  0. Here 1 = lim  = 0. Since u = /a, we have du/d  = 1/a or d /du = a. Therefore – a = r sin (0 – ) = – r sin . i.e., r sin  = a, is the asymptote.

478

DIFFERENTIAL CALCULUS

2. Find the asymptote of the curve a . r 1 – cos  2 1 1 1  u    – cos   . Here r a 2  1  1 When u  0,  – cos    0 so that cos   . 2 2   1 = ± /3. du 1 d a  sin  or – – . Now, d a du sin  d 2a  – – as   ,  du 3 3

– d 2a   as   – . du 3 3 2a   –  r sin  –   , i.e., 4a  r ( 3 sin  – 3 cos ), 3 3  

and 

    r sin  – –   , i.e., – 4a  r ( 3 sin   3 cos ), 3  3  are the two asymptotes. 2a

and

EXERCISES Find the asymptotes of the following curves : a .  –1 3. r = a sec  + b tan .

1.

r

r

3a sin  cos 

. sin 3   cos3  4. r2 = a2 (sec2  + cosec2 ).

2.

5. r sin 2  = a cos 3 .

6. 2r2 = tan 2 .

7. r  cos  = a cos 2 .

8. r sin n  = a.

9.

rn

sin n  =

an.

10. r = a tan .

11. r = a log .

12. r log  = a.

13. r (1 – e) = a.

14. r (2 – n2) = 2a.

15. r sin  =

ae

16. r ( + ) = ae.

17. Find the equation of the asymptotes of the curve given by the equation rn fn () + rn – 1 fn – 1 () + ... + f0 () = 0 18. Show that all the asymptotes of the curve r tan n  = a, touch the circle r = a/n. 19. Find the asymptotes of the curve r cos 2  = a sin 3  ANSWERS

3. r cos  ± b = a.

1   2r sin       0. 4   4. r cos  ± a = 0, r sin  ± a = 0.

5. 2r sin  = a, 2 = .

6.  = ± /4.

1. – a = r sin (1 – ).

2.

a

(Kumaon 2003)

ASYMPTOTES

479

7. r sin  = a,  (r cos ) + 2a = 0. 9. n = m where m is any integer. 11.

 = 0.

m  a  where m is any integer. r sin   –  n  n cos m  10. r cos  = ± a.

8.

12. a = r sin ( – 1).

13. y + a = 0.

14. y + a = 0. 15. System of parallel lines y = aen where n is any integer or zero. 16. ye + a = 0. 17.

f n – 1 ( ) f n (1 )

 r sin (1 – ) , where 1 is any root or the equation fn () = 0.

19. a = 2r (cos  – sin ), a + 2r (cos  + sin ) = 0. OBJECTIVE QUESTIONS For each of the following questions, four alternatives are given for the answer, only one of them is correct. Choose the correct alternative. 1. The number of asymptotes of a curve of nth degree is : (a) at least one (b) at least n (c) at most n (d) at most 1 (Avadh, 2002, 2005) 2. The parabola y2 = 4ax possesses (a) one real asymptote (b) two real asymptotes (c) many real asymptotes (d) no real asymptote (Garhwal 2004) 3. A closed curve has (a) no asymptote (b) one asymptote (c) infinitely many asymptotes (d) n asymptotes (Kumaon 2005) 4. The asymptotes of the hyperbola xy – 2y = 0 are given by (a) x – 2 = 0, y + 3 = 0 (b) x – 3 = 0, y + 2 = 0 (c) x – 2 = 0, y – 3 = 0 (d) x – 3 = 0, y – 2 = 0 5. Let f (x, y) = 0 be an algebraic equation of degree 8. The asymptotes of this curve will intersect in 18 points only if the number of asymptotes is : (a) 3 (b) 8 (c) 18 (d) 2 6. Which of the following is true ? An algebraic curve of an (a) even degree has at least one real asymptotes (b) even degree may or may not have real asymptote (c) odd degree has at least two imaginary asymptotes (d) odd degree has at least two real asymptotes 7. The asymptotes of the curve x2y2 – a2 (x2 + y2) = 0 form a : (a) circle

(b) triangle

(c) square

(d) pentagon

ANSWERS 1. (c)

2. (d)

3. (a)

4. (c)

5. (a)

6. (b)

7. (c).

CHAPTER 16 SINGULAR POINTS MULTIPLE POINTS, DOUBLE POINTS

16.1 Introduction. Cusps, Nodes and Conjugate points. The cases of curves considered in § 17.4 p. 614 show that curves with implicit equations of the form f (x, y) = 0 exhibit some peculiarities which are not possessed by the curves with explicit equations of the form y = F (x). These peculiarities arise from the fact that the equations f (x, y) = 0 may not define, y, as a single valued function of x. In fact, to each value of x correspond as many values of y as is the degree of the equation in y, and these different values of y give rise to different branches of the curve. We recall to ourselves the following three curves considered in § 13.4.

Fig. 16.1

Fig. 16.2

(i) Origin is a point common to the two branches of the Cissoid y2 (a – x) = x3, and the two branches have a common tangent there. Such a point on a curve is called a cusp. (ii) Origin is a point common to the two branches of the Strophoid (Fig. 16.2) (x2 + y2) x – a (x2 – y2) = 0, and the two branches have different tangents there. Such a point on a curve is called a node. (iii) (– a, 0) is a point common to the two branches of the curve ay2 – x (x + a)2 = 0

(§ 17.4)

(§ 17.5)

and the two branches have imaginary tangents there. There is no point in the immediate neighbourhood of the point (– a, 0) which lies on the curve. Here, a, is positive. Such a point on a curve is a called an isolated or conjugate point. 480

SINGULAR POINTS

481

Fig. 16.3

16.2. Definitions. Double points. Cusp. Node. Conjugate point. A point through which there pass two branches of a curve is called a double point. A curve has two tangents at a double point, one for each branch. The double point will be a node, a cusp or an isolated point according as the two tangents are different and real, coincident or imaginary. Multiple point. A point through which there pass, r, branches of a curve is called a multiple point of the rth order so that a curve has, r, tangents at a multiple point of the rth order. (Banglore 2005) Thus a double point is a multiple point of the second order. A multiple point of the third order is also called a triple point. A multiple point is also, sometimes, called a singular point. (Banglore 2005) 16.3. A simple rule for writing down the tangent or tangents at the origin to rational algebraic curves is obtained in the following article. 16.3.1 Tangents at the origin. The general equation of rational algebraic curve of the nth degree which passes through the origin O, when arranged according to ascending powers of x and y, is of the form (b1 x + b2 y) + (c1 x2 + c2 xy + c3 y2) + (d1 x3 + d2 x2y + ...) + ... = 0 ... (i) where the constant term is absent. Let P (x, y) be any point on the curve. The slope of the chord OP is y/x. Limiting position of the chord OP, when P  O, is the tangent at O so that when x  0 and y  0. lim (y/x) = m, the slope of this tangent. From (i), we have, after dividing by x,

y  y  2  b1  b2    c1 x  c2 y  c3 y .   (d1 x  d 2 xy  ...)  ...  0 x x     On taking limits, when x  0, we get b1 + b2 m = 0 so that m = – b1/b2, if b2  0. Hence y/x = – b1/b2, i.e., b1 x + b2 y = 0 ...(ii) is the tangent at the origin. This may be written down by equating to zero the lowest degree (first degree) terms in the equation (i).

482

DIFFERENTIAL CALCULUS

If b2 = 0 but b1  0, then, considering the slope of OP with reference to Y-axis, it can be shown that the tangent retains the same form. Let b1 = b2 = 0 so that the equation take the form (c1 x2 + c2 xy + c3 y2) + (d1 x3 + d2 x2y + .....) + ... = 0. ...(iii) Dividing by x2 and then taking limits as x  0, we get c1 + c2 m + c3 m2 = 0, ...(iv) which is a quadratic equation in, m, and determines as its two roots the slopes of the two tangents so that the origin is a double point in this case. The equation of either tangent at the origin is y = mx, ...(v) when m is a root of (iv). Eliminating m between (iv) and (v), we obtain c1 x2 + c2 xy + c3 y2 = 0, ...(vi) as the joint equation of the two tangents at the origin. This can be written down by equating to zero the lowest degree terms in (iii). The equation (vi) becomes an identity if c1 = c2 = c3 = 0. In this case the second degree terms, also, do not appear in the equation of the curve. It can now be similarly shown that the equation of the tangents can still be written down by equating to zero the terms of the lowest degree which is third in this case. In general, we see that the equation of the tangent or tangents at the origin is obtained by equating to zero the terms of the lowest degree in the equation of the curve. The origin will be a multiple point on a curve whose equation does not, at least, contain the constant and the first degree terms. Illustrations. (i) The origin is a node on the curve x3 + y3 – 3axy = 0, and x = 0, y = 0 are the two tangents thereat. (ii) The origin is a cusp on the curve (x2 + y2) x – 2ay2 = 0, and y = 0 is the cuspidal tangent. (iii) The origin is an isolated point on the curve a2x2 + b2y2 = (x2 + y2)2, and ax ± iby = 0 are the two imaginary tangents thereat. (iv) The origin is a triple point on the curve 2y5 + 5x5 – 3x (x2 – y2) = 0, and x = 0, x = y, x = – y are the three tangents thereat. EXAMPLE Find the tangent at the origin of the curve x2 (x2 + y2) = a (x – y). The given curve is x2 (x2 + y2) = a (x – y) Equating to zero the lowest degree terms we obtain the equation of the tangents at origin as x – y = 0.

SINGULAR POINTS

483

EXERCISES Find the tangents at the origin to the following curves:– 1. (x2 + y2)2 = 4a2xy.

2. y2 (a2 – x2) = x2 (b – x)2.

3. (x2 + y2) (2a – x) = b2x.

4. a2 (x2 – y2) = x2y2.

5. (x2 + y2)3 = a2 (x2 – y2)2. ANSWERS 1. x = 0, y = 0.

2. bx = ± ay.

3. x = 0.

4 y = ± x.

5. y = ± x.

EXAMPLE Find the equation of the tangent at (– 1, – 2) to the curve x3 + 2x2 + 2xy – y2 + 5x – 2y = 0, and show that this point is a cusp. We will shift the origin to the point (– 1, – 2). To do so we have to write x = X – 1, y = Y – 2, where X, Y are the current co-ordinates of a point on the curve with reference to the new axes. The transformed equation is (X – 1)3 + 2 (X – 1)2 + 2 (X – 1) (Y – 2) – (Y – 2)2 + 5 (X – 1) – 2 (Y – 2) = 0, or X3 – X2 + 2XY – Y2 = 0. Equating to zero the lowest degree terms, we get – X2 + 2XY – Y2 = 0, i.e., (Y – X)2 = 0, which are two coincident lines, and, therefore, the point is a cusp and the cuspidal tangent, i.e., the tangent at the cusp with reference to the new axes is Y – X = 0. To find the equation of the cuspidal tangent with reference to the given system of axes, we write X = x + 1, Y = y + 2. Hence the tangent at (– 1, – 2) is ( y + 2) – (x + 1) = 0, i.e., y = x – 1. EXERCISES Find the equations of the tangents to the following curves : 1. y2 (a2 + x2) = x2 (a2 – x2) at (± a, 0). 2. (x – 2)2 = y (y – 1)2 at (2, 1). 3. x4 – 4ax3 – 2ay3 + 4a2x2 + 3a2y2 – a4 = 0 at (a, 0) and (2a, a). 4. Show that the origin is a node; a cusp or a conjugate point on the curve y2 = ax2 + ax3, according as, a is positive, zero or negative. 1. x = ± a. 3.

 3y 

ANSWERS 2. (y – 1) = ± (x – 2). 2 ( x – a ); 2 ( x – 2a )   3( y – a ).

484

DIFFERENTIAL CALCULUS

16.4. Conditions for any point (x, y) to be a multiple point of the curve f (x, y) = 0. In § 11.9.4, we have seen that at a point (x, y) of the curve f (x, y) = 0, the slope of the tangent, dy/dx is given by the equation dy fx  f y  0. ...(i) dx At a multiple point of a curve, the curve has at least two tangents and accordingly dy/dx must have at least two values at multiple point. The equation (i), being of the first degree in dy/dx, can be satisfied by more than one value of dy/dx, if and only if, fx = 0, fy = 0. Thus, we see that the necessary and sufficient conditions for any point (x, y) on f (x, y) = 0 to be a multiple point are that fx (x, y) = 0, fy (x, y) = 0. To find multiple points (x, y), we have therefore to find the values of (x, y) which simultaneously satisfy the three equations fx (x, y) = 0, fy (x, y) = 0, f (x, y) = 0. 16.4.1 To find the slopes of the tangents at a double point. Differentiating (i), w.r. to x, we have dy  dy  dy d2y   f yx  f y 2   0, f y  dx  dx  dx dx 2 so that at the multiple point, where fy = 0, fx = 0, the values of dy/dx are the roots of the quadratic equation f x2  f xy

2

dy  dy  f y 2    2 f xy  f x 2  0. ...(ii) dx  dx  In case f x 2 , f xy , f y 2 , are not all zero and fx = 0 = fy, the point (x, y) will be a double point and will be a node, cusp or conjugate according as the values of dy/dx are real and distinct, equal or imaginary i.e., according as ( f xy ) 2 – f x 2 f y 2  0,  0,  0. If f x2  f xy  f y 2  0; the point (x, y) will be multiple point of order higher than the second. EXAMPLES 1. Find the multiple points on the curve x4 – 2ay3 – 3a2 y2 – 2a2 x2 + a4 = 0. Also, find the tangents at the multiple point. Let 

f (x, y) =

x4

–

2ay3

–3a2

y2

–

2a2x2

(Gorakhpur 2000, 2001) +

a 4.

fx (x, y) = 4x3 – 4a2x. fy (x, y) = – 6ay2 – 6a2y. fx (x, y) = 0 gives x = 0, a, – a. fy (x, y) = 0 gives y = 0, – a.

Hence, the two partial derivatives vanish for the points (0, 0), (0, – a), (a, 0), (a, – a), (– a, 0), (– a, – a)

SINGULAR POINTS

485

Of these the only points on the curve are (a, 0), (– a, 0), (0, – a). Hence, these are the only three multiple points, on the curve. To find the tangents at the multiple points, we proceed as follows :– We have

f x2 = 12x2 – 4a2, fxy = 0, f y 2 = – 12ay – 6a2. Since at (a, 0), fx2 = 8a2, fxy = 0, fy2 = – 6a2. Therefore, by the equation (ii), the values of dy/dx at (a, 0) are given by – 6a2 (dy/dx)2 + 8a2 = 0. i.e.,

dy/dx =  2 / 3.

The two values being both real, the point (a, 0) is a node. The tangents at (a, 0) are

y   (2 / 3) ( x – a). It may similarly be shown that the tangents at (– a, 0) and (0, – a) are 4 2 ( x  a ), ya x. 3 3 Second method. Differentiating the given equation w. r. to x, we get, y

4x3 – 6ay2y1 – 6a2yy1 – 4a2x = 0, which identically vanishes for the multiple points. Differentiating again, we get 12x2 – 12ayy12 – 6ay2y2 – 6a2y12 – 6a2yy2 – 4a2 = 0. From this we see that (i) for (a, 0), y1 2 =

3 , i.e., y1 = ± 4

3 . 4

3 3 , i.e., y1 = ± . 4 4 2 2 (iii) for (0, – a), y1 2 = , i.e., y1 = ± . 3 3 Knowing the slopes of the tangents, we can now put down their equations.

(ii) for (– a, 0), y1 2 =

Third method. To find the tangents at (a, 0), we shift the origin to this point. The transformed equation is (x = X + a, y = Y + 0) (X + a)4 – 2aY3 – 3a2Y2 – 2a2 (X + a)2 + a4 = 0, or X4 + 4X3a – 2aY3 + 4a2X2 – 3a2Y2 = 0. The tangents at the new origin are 4a2X2 – 3a2Y2 = 0, or

Y 

(4 / 3) X .

486

DIFFERENTIAL CALCULUS

The tangents at the multiple point (a, 0), therefore, are Y 

(4 / 3) ( x – a ).

It may similarly be shown that Y 

(4 / 3) ( x  a ) and y + a = 

(2 / 3) x ,

are the tangents at the multiple points (– a, 0) and (0, – a) respectively. The three multiple points on the curve are nodes. 2. Show that the origin of the curve y2 = b x sin

x , there is a node, or a conjugate point a

according as a and b have like or unlike signs. The given equation of the curve is y2 = b x sin

x a

x  x3 y2  b x  – 3  ...  a a .3!    Then the approximate equation of the curve will be b b y2  x2 – x4 . a 6 a3 Now the tangents at the origin are given by b y2  x2 . a If a and b have like signs, then two tangents at the origin are real and non-concident; so there is a node which can be confirmed by giving small values of positive or negative. If a and b have unlike signs the two tangents at the origin will be imaginary and so we will have a conjugate point there. Hence, origin will be a node or a conjugate point according as a and b have like or unlike signs. 3. Locate the double points of the curve y (y – 6) = x2 (x – 2)3 – 9, and ascertain their nature. (Avadh, 2000; Gorakhpur, 2003) Here f (x, y) = x2 (x – 2)3 – y (y – 6) – 9 = 0 ...(i) f 3 3 2  x = 2 x (x – 2) + x . 3 (x – 2) = x (x – 2)2 (5 x – 4) ...(ii) f  6 – 2y ...(iii) y 2 f = (x – 2)2 (5 x – 4) + 2 x (x – 2) (5 x – 4) + 5 x (x – 2)2 ...(iv)  x2 2 f –2 ...(v)  y2

or

and Now

2 f 0  x y f  0 gives x

...(vi) x  0, 2,

4 5

SINGULAR POINTS

487

f  0 gives y

and

y = 3.

4  Hence possible double points are (0, 3), (2, 3) and  , 3  . 5  4  The point  , 3  does not satisfy f (x, y) = 0, and therefore there are only two double points. 5  2 f 2 f 2 f  – 16,  – 2,  0. Now at (0, 3), 2 2  x y x y 2

 2 f  2 f 2 f . .      x2  y2   x y  Hence (0, 3), is a conjugate point. 2 f

At (2, 3),

 x2

 0,

2 f  y2

 – 2 and

2 f 0  x y

2

 2 f  2 f 2 f .      x2  y2   x y  Therefore the point (2, 3) is a cusp. Further to decide the nature of the cusp, transfer the origin to the point (2, 3). The equation then becomes (y + 3) (y – 3) = x3 (x + 2)2 – 9 i.e., y2 = x3 (x + 2)2. The tangents at the new origin are y2 = 0, i.e., x–axis is the common tangent. Solving for y, we get y = ± (x + 2) x 3 , which shows that when x is negative, y is imaginary and when x is positive, y has two values one positive other negative. Thus near the new origin the curve lies on both sides of x–axis (tangent) and only on one side of y–axis (normal). Hence the new origin (2, 3) is a single cusp of first species. 4. Find the position and nature of the double points on the curve x2y 2 = (a + y)2 (b 2 – y2) distinguishing between the cases a > b or < b. Here f (x, y) = x2y2 – (a + y)2 (b2 – y2) = 0 f f  2 xy 2 , = 2x2y – 2 (a + y) (b2 – y2) + 2y (a + y)2 x y Also at a double point (f / x) = 0, (f / y) = 0 and f = 0.



(f / x) = 0  x = 0 or and

y=0

(f / y) = 0  x2y – (a + y) (b2 – y2) + y (a + y)2 = 0

Putting x = 0 in this result we have (a + y) {y (a + y) – (b2 – y2)} = 0 or (a + y) {2y2 + ay – b2} = 0 1 [– a  ( a 2  8b 2 )]. 4 Putting y = 0 in (i) we do not have any significant result.

or

y  – a or

...(i)

488

DIFFERENTIAL CALCULUS

Hence we get three set of values x = 0, y = – a, x = 0, y = – a satisfies the curve.

1 [– a  4

(a 2  8b 2 )] of which only one set viz.

 (0, – a) is a double point. Shifting the origin to (0, – a) we get equation of the curve as x2 (y – a)2 = y2 {b2 – (y – a)2} or y4 + x2y2 – 2ay2 – 2ax2y + (a2 – b2) y2 + a2x2 = 0

...(ii)

Tangents at the new origin are given by (a2 – b2) y2 + a2x2 = 0, equating the lowest degree terms to zero. If a < b, (iii) gives two imaginary tangents. Hence the new origin is a conjugate point, i.e., (0, – a) is a conjugate point. If a > b, (iii) gives two real and distinct tangents (b 2 – a 2 ) y   ax.

 The new origin may be a node or a conjugate point. Neglecting y4 and y3 from (ii), we get (y2 – 2ay + a2) x2 = (b2 – a2) y2 or (y – a)2 x2 = (b2 – a2) y2 or

x

(b 2 – a 2 ) [ y / ( y – a )].

This shows that the value of x near the origin are real when b > a or a < b. Hence the point (0, – a) is a node. [If a = b, (iii) gives two real and coincident tangents x2 = 0.  The new origin may be a cusp or a conjugage point. Neglecting y4 from (ii), we get (y2 – 2ay + a2) x2 = 2ay3 or

x   [ (2ay 3 )] ( y – a)   y { (2ay )} / ( y – a).

This shows that for small positive values of y, the values of x are real. Hence the curve has real branches through the new origin. Hence (0, – a) is a cusp.] 5. Find the double points of x3 + y3 = 3axy. (Garhwal, 1998, Kuvempu 2005) 3 3 Let f (x, y)  x + y – 3axy = 0. ...(i)  (f / x) = 3x2 – 3ay; (f / y) = 3y2 – 3ax At a double point we know (f / x) = 0, (f / y) = 0, and f = 0 f f  = 0  x2 = ay and y = 0  y2 = ax. x Solving x2 = ay and y2 = ax we get x4 = a2y2 = a3x or x (x3 – a3) = 0 or x = 0, a (only real values being considered)  y = 0, ± a  We have the following sets of values of x and y : x = 0, y = 0; x = 0, y = a; x = 0, y = – a; x = a, y = 0; x = a, y = a and x = a, y = – a. Out of these only x = 0, y = 0 satisfy (i). Hence (0, 0) is a double point. Equating the lowest degree terms in (i) to zero we get xy = 0 or x = 0, y = 0 as the tangents to the given curve at (0, 0). These tangents being real and distinct we conclude that (0, 0) is a node. 6. Show that the curve (a2 + x2) y = a2x has three points of inflexion. Find them.

SINGULAR POINTS

489

The curve is y = a2x / (a2 + x2) 2

2

...(i) 2

2

2

2

2

dy ( a  x ) a – a x (2 x ) a ( a – x )   dx (a 2  x 2 )2 (a 2  x 2 ) 2



2 2 2 2 2 2 2 2  ( a  x ) (– 2 x ) – ( a – x ) . 2 ( a  x ) 2 x   a   dx 2 (a 2  x 2 )4   2 2 2 2  2 2  – 2 x {( a  x )  2 ( a – x )} 2 a x ( x – 3a 2 ) 2  a     ( a 2  x 2 )3 ( a 2  x 2 )3   2 2 2 2 2 2 d3y 2  ( a  x ) ( x – a ) – 2 x ( x  3a )   6 a   Similarly dx3 (a 2  x 2 )4   For the point of inflexion, d2y / dx2 = 0

d2 y

i.e.,

x (x2 – 3a2) = 0 or x  0,  a 3.

For these values of x we find d3y/dx3  0. Also from (i) when x = 0, y = 0 x   a 3, y 

and when

 a3 3 2

2



3 a. 4

a  3a 1   (0, 0),  a (3), (3) a   Points of inflexion are 4   1   and  – a (3), – 4 (3) a  .   7. Find the points of inflexion on the curve x = a (2 – sin ), y = a (2 – cos ). The curve is x = a (2 – sin ), y = a (2 – cos )

 And 



dx d 1  a (2 – cos )   d dx a (2 – cos ) dy  a (sin ) d dy dy / d  a sin  sin     dx dx / d  a (2 – cos ) 2 – cos  d2y d  dy  d  dy  d       2 dx  dx  d   dx  dx dx  d  sin    1    . d  2 – cos  a (2 – cos ) , from (ii)    (2 – cos ) cos  – sin  (sin ) 1  . 2 a (2 – cos ) (2 – cos ) d2y 2



2 cos  – 1

dx a (2 – cos )2 Again differentiating, w.r. to x, we get d3y dx

3



d d

 d 2 y  d d  2 cos  – 1  d   .  2  .  2  dx  dx d   a (2 – cos )  dx

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

490

DIFFERENTIAL CALCULUS



(2 – cos )2 (– 2 sin ) – (2 cos  – 1) . 3 (2 – cos ) 2 sin 

a (2 – cos )6 a (2 – cos ) = – [(1 + 4 cos ) sin ] / {a2 (2 – cos )5}.

,from (ii) ...(iii)

2

d y

to zero, we get 2 cos  – 1 = 0 dx 2  cos  = 1/2   = /3, 5/6 in 0    2 d3 y  0. For these values of  we find from (iii) that dx3 Now obtain the coordinates of the points of inflexion from (i), putting  = /3 and 5/6. Now equating

EXERCISES Find the position and nature of the multiple points on the following curves :– 1. x2 (x – y) + y2 = 0

2. y3 = x3 + ax2

3. x4 + y3 – 2x3 + 3y2 = 0.

4. xy2 – ax2 + 2a2x – a2 = 0.

5. y2 = (x – 1) (x – 2)2.

6. ay2 = (x – a)2 (x – b)2.

7. x4 – 4ax3 + 2ay3 + 4a2x2 – 3a2y2 – a4 = 0.

8. x3 + y3 – 12x – 27y + 70 = 0.

9. x4 + 4ax3 + 4a2x2 – b2y2 – 2b3y – a4 – b4 = 0.

10. x4 + y (y + 4a)3 + 2x2 (y – 5a)2 = 5a2x2.

11. x3 + 2x2 + 2xy – y2 + 5x – 2y = 0. 12. (2y + x + 13. (x + y)3 –

1)2

– 4 (1 –

x)5

(Banglore 2006)

= 0.

2 2 (y – x + 2) = 0.

(Garhwal, 1996)

14. (y2 – a2)3 + x4 (2x + 3a)2 = 0. Find the equations of the tangents at the multiple points of the following curves :– 15. x4 – 4ax3 – 2ay3 + 4a2x2 + 3a2y2 – a4 = 0. 16. x4 – 8x3 + 12x2y + 16x2 + 48xy + 4y2 – 64y = 0. 17. (y – 2)2 = x (x – 1)2. 18. Show that each of the curves (x cos  – y sin  – b)3 = c (x sin  + y cos )2, for all different values of , has a cusp ; show also that all the cusps lie on a circle. ANSWERS 2. Cusp at (0, 0). 4. Node at (a, 0).

Cusp at (0, 0). Node at (0, 0). Node at (2, 0). (a, 0) is a node, cusp or an isolated point according as b/a is less than, equal to or greater than 1. Conjugate point at (a, 0). 8. Conjugate point at (2, 3). Conjugate point at (– a, – b). 10. Cusp at (0, – 4a). Cusp at (– 1, – 2). 12. Cusp at (1, – 1). No multiple point.  3  14. (0, ± a) are triple point ;  – a,  a  are cusps ; node at (– a, 0).  2  1. 3. 5. 6. 7. 9. 11. 13.

SINGULAR POINTS

15.

491

y – a   (2 / 3) x ; y   (2 / 3) ( x – a ); y – a = ± (2/3) (x – 2a).

16. y – 2 = ± (2 / 2) (x – 2).

17. x – y + 1 = 0, x + y = 3.

16.5. Types of cusps. We know that two branches of a curve have a common tangent at a cusp. There are five different ways in which the two branches stand in relation to the common tangent and the common normal as illustrated by the following figures :–

In Fig. 16.4, the two branches lie on the same side of the common normal and on the different sides of the tangent. In Fig. 16.5, the two branches lie on the same side of the normal and on the same side of the tangent. In Fig. 16.6, the two branches lie on the different sides of the normal and on the different sides of the tangent. In Fig. 16.7, the two branches lie on the different sides of the normal and on the same side of the tangent. In Fig. 16.8, the two branches lie on the different sides of the normal but on one side they lie on the same, and on the other on opposite sides of the common tangent. One branch has inflexion at the point. It will thus be seen that the cusp is single or double according as the two branches lie on the same or different sides of the common normal. Also it is of the first or second species according as the branches lie on the different or the same side of the common tangent. EXAMPLES 1. Find the nature of the cusps on the following curves :– (i) y2 = x3.

(ii) y2 – x4 = 0.

(iii) (y – 4x2)2 = x7.

492

DIFFERENTIAL CALCULUS

(i) y = 0 is the cuspidal tangent. Since x cannot be negative, the two branches lie only on the same side of the common normal so that the cusp is single. See Fig. 16.4. Again, y = ± x3/2 so that to each positive value of x correspond two values of y which are of opposite signs and hence the two branches lie on different sides of the common tangent and the cusp is of first species. (ii) Two branches of y2 – x4 = 0, are the two parabolas y – x2 = 0 and y + x2 = 0 which lie on different sides of the common tangent y = 0 and extend to both sides of the common normal x = 0. Thus the origin is a double cusp of first species. See Fig. 16.6. y = 4x2 ± x7/2,

(iii) Here

where the two signs correspond to the two branches; y = 0 is the cuspidal tangent. Since x cannot take up negative values, the two branches lie only on the same side of the common normal and the cusp, therefore, is single. Now, one value of y is always positive and, therefore, the corresponding branch lies above x–axis. Again 4x2 > x7/2, if 4 > x3/2 i.e., if 42/3 > x. Thus, for the values of x lying between 0 and 42/3, the second value of y is also positive and, therefore, the corresponding branch lies above y– axis in the vicinity of the origin. Thus the cusp is of the second species.

Fig. 16.9

2. Show that the curve y2 = 2x2y + x3y + x3, has a single cusp of the first species at the origin. Equating to zero the lowest degree terms, we see that the origin is a cusp and y = 0 is the cuspidal tangent. We re-write the given equation in the form, quadratic in y, y2 – yx2 (2 + x) – x3 = 0, and solve it for y, so that

x 2 (2  x)  [ x 4 (2  x) 2  4 x3 ] 2 For positive values of x, we have x4 (2 + x)2 + 4x3 > x4 (2 + x)2, y

or

[ x 4 (2  x) 2  4 x3 ]  x 2 (2  x),

so that two positive values of x correspond two values of y with opposite signs. Thus the two branches lie on opposite sides of x–axis when x is positive. Again, we have x4 (2 + x)2 + 4x3 = x3 (4 + 4x + 4x2 + x3). For values of x which are sufficiently small in numerical value, 4 + 4x + 4x2 + x3 is positive, for the same  4 when x  0. Thus, for negative values of x which are sufficiently small in numerical value, x3 (4 + 4x + 4x2 + x3), is negative so that the values of y are imaginary. Thus x cannot take up negative values. Hence, the curve has a single cusp of first species at the origin.

SINGULAR POINTS

493

EXERCISES Find the nature of the cusps on the following curves :– 1. x2 (x – y) + y2 = 0.

2. x2 (x + y) – y2 = 0.

3. x3 + y3 – 2ay2 = 0.

4. a4y2 = x5 (2a – x).

5. (y – x)2 + x6 = 0.

6. x6 – ayx4 – a3x2y + a4y2 = 0.

7. x5 – ax3y – a2x2y + a2y2 = 0. 8. Examine the curve x5 + 16x2y – 64y2 = 0 for singularities. 9. Prove that the curve x3 + y3 = ax2 has a cusp of the first species at the origin and a point of inflexion where x = a. 10. Show that the curve y3 = (x – a)2 (2x – a) has a single cusp at (a, 0). ANSWERS Single cusp of first species. Single cusp of first species. Isolated point. Oscu-inflexion.

1. 3. 5. 7.

2. 4. 6. 8.

Single cusp of first species. Single cusp of first species. Double cusp of second species. Oscu-inflexion.

16.6 Radii of curvature at multiple points. The formula for the radius of curvature at any point (x, y) on the curve f (x, y) = 0, as obtained in § 14.4, becomes meaningless at a multiple point where fx = fy = 0. At a multiple point we expect as many values of, , as its order. Of course, these values of  may not be all distinct. The following examples will illustrate the method of determining the values of  at such points. EXAMPLES 1. Find the radii of curvature at the origin of the branches of the curve y4 + 2axy2 = ax3 + x4. Here, 2xy2 = x3, i.e., x = 0, y = ± (1/ 2) x are the three tangents at the origin so that it is a triple point. To find, , for the branch which touches x = 0, we find lim (y2/2x). To do this, we write y2/2x = 1, i.e., x = y2/2 1, and substitute this value of x in the given equation. Lim 1 =  is the radius of curvature of the corresponding branch at the origin. We get y4 

2ay 4 y6 y8 a  2 1 8 13 16 14

a y2 y4 a  . 1 8 13 16 14 Let y  0 so that we have 1 + a/ = 0. Thus, , for this branch = – a. To find, , for the other branches we proceed as follows.

or

1

494

DIFFERENTIAL CALCULUS

Suppose that the equation of either branch is given by

y  f (0)  xf  (0) 

x2 f  (0)  .... 2!

We have, there f (0) = 0. Also we write f ' (0) = p, f '' (0) = q. Thus we have  2 qx  ..... 2 Making substitution in the given equation, we get y  px 

4

1 2    px  qx  ....   2ax 2  

2

1 2   3 4  px  qx  ....   ax  x 2  

Equating co-efficients of x3 and x4, we get 2ap2 = a, p4 + 2apq = 1. 1 3 2 ,q ; These give p  8a 2 p–





1 2

,q–

3 2 . 8a

(1  p 2 )3/ 2   2 3 a, q

for the two branches. 2. Show that the pole is a triple point on the curve r = a (2 cos  + cos 3 ), and that the radii of curvature of the three branches are

3 a / 2, a / 2, 3 a / 2. The radius vector, r, vanishes for the values of, , given by 2 cos  + cos 3  = 0, i.e., or

2 cos  + 4 cos2  – 3 cos  = 0, cos  (4 cos2  – 1) = 0,

1 1 , cos   – . 2 2 Thus, r = 0 when  = /3, /2, 2/3 so that the pole is a triple point. We now proceed to find . We have r1 = a (– 2 sin  – 3 sin 3 ), r2 = a (– 2 cos  – 9 cos 3 ). For  = /3,

or

cos   0, cos  

r1  – 3 a, For  = /2, r1 = a For  = 2/3, r1  – 3 a,

r2 = 8a; r2 = 0; r2 = – 8a.

SINGULAR POINTS

495

Also, r = 0 for each of these branches. Putting these values in the formula 

(r 2  r1 2 )3/ 2 r 2  2r1 2 – rr2

,

we get the required result. EXERCISES 1. Show that the radius of curvature at the origin for both the branches of the curve y2 (a – x) = x2 (a + x)

is

2a.

2. Find the radius of curvature at the point (1, 2) for the curve (y – 2)2 = x (x – 1)2. 3. Find the radii of curvature at the origin of the two branches of the curve given by the equations y = t – t2, x = 1 – t2. (For the origin t = ± 1, so that the two branches correspond to these two values of t). Find the radii of curvature at the origin of the following curves : 4. x3 + y3 = 3axy. (Folium). 5. x2 – 3xy – 4y2 + y3 + y4x + x5 = 0. 6. x5 + ax2y2 – ax3y – 2a2xy2 + a2y2 = 0. 7. Show that (a, 0), in polar co-ordinates, is a triple point on the curve

1   r  a  1  2 sin   , 2   and find the radii of curvature at the point. ANSWERS 2.

2 2.

3. 2 2 for each.

5.

85 17, 5 2. 2

6. a, – a/2, – 53/2 a.

7.

2 3

2a,

2 3

2a,

4. 3a/2 for each.

2 a. 3

OBJECTIVE QUESTIONS For each of the following questions, four alternatives are given for the answer. Only one of them is correct. Choose the correct alternative. 1. If two branches of a curve pass through a point P, and the two tangents at P are real and distinct, then P is a : (a) node (b) cusp (c) conjugate point (d) none 2. An isolated point on the curve is one at which : (a) two branches are real (b) two branches are real and distinct (c) two branches are real and coincident (d) no branch of the curve lies in the neighbourhood of the point.

496

DIFFERENTIAL CALCULUS

3. For the curve y2 (1 + x) = x2 (1 – x), the origin is a : (a) node (c) point of inflexion x3

4. The curve y =

–

3x2

(b)

cusp

(d)

none of these

– 9x + 9 has a point of inflexion at :

(a) x = – 1

(b)

x=1

(c) x = – 3

(d)

x=3

5. A double point on the curve is a cusp if tangents are : (a) real and distinct

(b)

imaginary and distinct

(c) imaginary and coincident

(d)

real and coincident. (Garhwal, 2001; Avadh 2002; 2003)

6. For the curve

x4

–

ax2y

+

axy2

+

a 2y 2

= 0, the origin is :

(a) node

(b)

(c) conjugate point

cusp

(d)

7. Nature of thed origin for the curve

y2

(a2

+

x2 )

(a) Node

=

x2

point of inflexion (a2

– x2) is :

(b)

Cusp

(d)

Point of inflexion

(a) Node

(b)

Cusp

(c) Conjugate Point

(d)

Point of inflexion

(a) Node

(b)

Cusp

(c) Conjugate point

(d)

None of these.

(c) Conjugate point 8. Curve

x3

+

x 2y

=

ay2

has at origin a :

9. Nature of origin on the curve

a2 x

2

–

b2 y2

 1 is :

ANSWERS 1. (a)

2. (d)

8. (b)

9. (a).

3. (a)

4. (b)

5. (d)

6. (c)

7. (a)

CHAPTER 17 CURVE TRACING 17.1. Introduction The general problem of curve tracing in its elementary aspects, will be taken up in this chapter. It will be seen that the equations of curves which we shall trace are generally solvable for y, x and r. Some equations which are not solvable for y or x may be rendered solvable for r, on transformation from Cartesian to Polar system.

17.2. Procedure for tracing Cartesian Equations I. Find out if the curve is symmetrical about any line. In this connection, the following rules, whose truth is evident are helpful:(i) A curve is symmetrical about x-axis if the powers of y which occur in its equation are all even; (ii) A curve is symmetrical about y-axis if the powers of x which occur in its equation are all even; (iii) A curve is symmetrical about the line y = x if , on inter-changing x and y its equation does not change. (iv) A curve is symmetrical about the line y = – x if the equation of the curve remains unchanged when x and y are replaced by – y and – x respectively. (v) The curve is symmetrical in opposite quadrants if the equation of the curve remains unchanged when x and y are replaced by – x and – y respectively. II. Find out if the origin lies on the curve. If it does, write down the tangent or tangents thereat. In case the origin is a multiple point, find out its nature. III. Find out the points common to the curve and the co-ordinate axes if there be any. Also obtain the tangents at such points. IV. Find out the asymyptotes and the points in which each asymptote meets the curve. V. Find out if there is any region of the plane such that no part of the curve lies in it. Such a region is generally obtained on solving the equation for one variable in terms of the other, and find out the set of values of one variable which make the other imaginary. dy VI. Find out and the points where the tangent is parallel to the co-ordinate axes. dx VII. Finally find out the intervals in which y is increasing and decreasing and in particular, the values of x for which y has maximum or minimum values. dy This investigation depends upon the determination of and of the intervals in which it is + ve dx or – ve. Also find the intervals in which the curve is concave upwards or downwards and in particular d2y the values for which the curve has points of inflexion. This investigation depends upon . dx 2 497

498

DIFFERENTIAL CALCULUS

Taking into consideration all the points given above, the curve can be traced out.

17.3. Equations of the form y = f (x). It may be pointed out that a curve y = f (x) where f is a polynomial function has no asymptotes. 1. Trace the curve y = x3 – 12x – 16. We note the following points about the curve. (i) The curve is not symmetric about any line. (ii) Origin does not lie on it. (iii) ( – 2, 0 ), (4, 0) and ( 0, – 16 ) are the only points of intersection with co-ordinate axes.

dy d2y  3x 2 – 12  3( x 2 – 4 ), 2  6 x . dx dx (v) The following is the table of variations of y corresponding to x as deduced from dy/dx. We also note that

(iv)

lim y  –  and lim y    .

x – 

x  

Fig 17.1

Thus, y is strictly increasing in the intervals ] – , – 2] and [ 2,  [. Also y is strictly decreasing in the interval [ – 2, 2 ]. Finally, y is maximum for x = – 2 and minimum for x = 2. (vi) d2y/dx2 = 0 for x = 0. Also, d2y/dx2 < 0  x ] – , 0 [ and > 0  x  ] 0,  [. Thus, the curve is concave downwards in ] – , 0 ] and concave upwards in [ 0,  [. Since d2y/dx2 change sign as x passes through 0, the curve has a point of inflexion for x = 0, i.e., at the point ( 0, – 16 ). We have the curve as given in the following figure 17.2

Fig. 17.2

CURVE TRACING

499

2. Trace the curve 3 4 x  4 x3  3 x 2 – 12 x . 2 We note the following points about the curve. y–

(i) The curve is not symmetrical about any line. (ii) Origin lies on it and it meets co-ordinate axes in origin only. dy  – 6 x 3  12 x 2  6 x – 12  – 6 ( x – 2 ) ( x – 1) ( x  1) , dx d2y  – 18 x 2  24 x  6  6 ( – 3 x 2  4 x  1 ) dx 2 The following is the table of variations: x – –1 1 2 +

(iii)

dy/dx

+

0

–

0

+

0

–

Fig. 17.3

Thus, y is strictly increasing in ] – , – 1 ] and [ 1, 2 ] and decreasing in [ – 1, 1 ] and [ 2,  [. Also, y is maximum for x = – 1 and 2 and minimum for x = 1. (iv) Since

d2y/dx2 = 0 if and only if x =

1 (2  3

7).

7 = 2.6....., we may see that

1 1 ( 2 – 7 ) = – .2..... and ( 2 + 7 ) = 1.5 ... 3 3 Also, d2y/dx2 changes sign as x passes through 1 1 (2– 7 ) and ( 2 + 7 ). 3 3 1 We have d2y/dx2 < 0  x  ] – , ( 2 – 7 ) [ 3 1 2 2 and d y/dx < 0  x  [ ( 2 + 7 ),  [ 3 1 1 2 2 d y/dx > 0  x  ] ( 2 – 7 ), ( 2 + 7 ) [. 3 3 1 Thus, the curve is concave downwards in ] – , ( 2 – 3 1 1  1  3 ( 2 – 7 ) , 3 ( 2  7 )  and downwards in [ 3 ( 2 + 7 ), +  [.   Also the curve has two points of inflexion corresponding to 1 1 x  ( 2 – 7 ) and x  ( 2  7 ) . 3 3 We thus have the curve as drawn in Fig. 17.4.

7 ) ], upwards

in

500

DIFFERENTIAL CALCULUS

Fig. 17.4

3. Trace the curve y

2x  3 2

x – 3x  2 (i) The curve is not symmetric about any line. (ii) It does not pass through the origin. 3  3   (iii) It meets co-ordinate axes in  , 0  and  0, –  only. 2  2   (iv) y is defined for x  R ~ { 1, 2 } = ] –  1 [  ] 1, 2 [  ] 2 +  [ dy – 2 x 2  6 x – 5 – 2 [ ( x – 3 / 2 ) 2  1/ 4 ]  (v) dx  2 ( x – 3x  2 )2 ( x 2 – 3x  2 )2

The following is the table of variations of y: x – 1 dy/dx – –



2 –

Fig 17.5

Thus, y is strictly decreasing in each of the intervals ] – , 1 [, ] 1, 2 [ and ] 2, +  [. Also y has no maximum or minimum (vi) x = 1, x = 2 and y = 0 are the asymptotes to the curve d2y

(vii)



2 ( 2 x – 3) ( x 2 – 3 x  3)

so that ( x 2 – 3 x  2 )3  3  d2y 3 3  0 when x  and changes sign as x passes through . Thus  , 0  is a point of 2 2   2 2 dx inflexion and the slope of the tangent at this point is – 8. dx

2

We have the curve as drawn in Fig. 17.6.

CURVE TRACING

501

Fig. 17.6.

4. Trace the curve y  (i) (ii) (iii) (iv) (v)

8a 3 2

x  4a 2 The curve is symmetrical about y-axis. It does not pass through the origin. It meets y-axis in ( 0, 2a) and y = 2a is the tangent thereat. y = 0 is the only asymptote to the curve y is positive for all values of x. Therefore the curve lies in 1st and 2nd quadrant only.

dy –16 x a 3  (vi) dx . ( x 2  4a 2 ) 2 The following is the table of variations of y x – – 2a –a 0 dy/dx + + 0

a –

2a –



Fig 17.7



dy  0  x 0 dx

dy 0x0 dx Thus, y is strictly increasing in ] – , 0 ] and strictly decreasing in [ 0,  [. Also y is maximum for x = 0 and its value there is 2a. Thus, we have the curve as given in the figure Figure 17.8

and

Fig. 17.8

502

DIFFERENTIAL CALCULUS

EXERCISES 1. Trace the following curves: (i) y = 3x4 – 16x3 + 24x2,

(ii)

y = ( x2 – x – 6 ) ( x – 7 ),

(iii) y = ( x + 1)2 ( x – 3),

(iv)

y = x4 – 24x2 + 80,

(v) y = x4 – 2x + 10,

(vi) y = x4 – 5x3 + 6x2,

(vii) y = 4 – 8x + 5x2 – x3, 1 4 x – 2 x2  2 , 4 1 5 1 3 (xi) y  x – x  2 , 5 3 2. Trace the following curves:

(viii) y = 2x5 – 3x4 + 112,

(ix) y 

(i)

y

(iii) y  (v) y  (vii) y  (ix) y 

x 2 – 3x x –1 1  x2 2 – 3x  x2 x 2 – 12 x  27 x2 – 4 x  5 x –1 x2 3x  6 ( x –1) 2

(x) ( xii)

3 4 1 2 x  x – 4, 2 2 1 6 y  x – x5  x4 . 6

y

(ii)

y

(iv)

y

(vi)

y

(viii)

y

2 x2 – 7 x  5 x2 – 5x  7 2x – 3 3x2 – 4 x x2 – 1

( x – 2 )2 x2  4 x – 5 ( x  2) 2

.

17.4. Equations of the form y2 = f (x) 1. Trace the curve y 2 ( a 2 + x 2) = x 2 ( a 2 – x 2 )

(Garhwal 1999)

We note the following particulars about this curve: (i) It is symmetrical about both the axes. (ii) It passes through the origin and y = ± x are the two tangents thereat. Thus the origin is a node. (iii) It meets x - axis at ( a, 0 ), ( 0, 0) and ( – a, 0 ) and meets y-axis at (0, 0) only. The tangents at ( a, 0) and ( – a, 0) are x = a and x = – a respectively. (iv) The curve has no asymptotes ( v)

y  x

a2 – x2

a2  x2 y is defined for only those of the values of x for which a2 – x2  0  – a  x  a so that the curve lies between the two lines x = a and x = – a.

dy a 4 – 2a 2 x 2 – x 4  2 dx ( a  x 2 )3/ 2 ( a 2 – x 2 )1/ 2 dy   as x  – a and when x  a dx dy  0 when a4 – 2a2 x2 – x4 = 0 Also dx Again a4 – 2a2 x2 – x4 = – [ x4 + 2a2 x2 – a4]

(vi)

CURVE TRACING

503

= – [ x2 – ( – 1 + =–[x – Thus

and

2 2 2 )a ][x –(–1–

( –1  2 ) a ] [ x +

2 ) a2 ]

( –1  2 ) a ] [ x2 + ( 1 +

2 2)a ]

dy  0 for x  ( – 1  2 ) a and x  – ( –1  2 ) a dx dy  0 for x ] – a, – (–1  2) a [] (–1  2) a, a [ dx dy  0 when x  ] – ( –1  2 ) a , [ dx

Thus y is strictly decreasing in [ – a, – increasing in [ –

2) a ,

( –1 

We may note that Thus the curve

( –1 

y2 

( –1 

( –1 

2 ) a ] and [ ( – 1 +

2 ) a , a ] and strictly

2) a ]

2 )  .6

x2 ( a2 – x2 ) a  x2

is as given in the Fig. 17.9

Fig 17.9

2. Trace the curve y2 (x – a) = x2 ( x + a) (Garhwal 2001) (i) The curve is symmetrical about x-axis only. (ii) It passes through the origin and y2 + x2 = 0 i.e. y ± ix = 0 are the two imaginary tangents thereat. Thus the origin is an isolated point. (iii) It meets x-axis ( – a , 0 ) and ( 0, 0 ) and y-axis at the origin only; x = – a is the tangent at (– a, 0). (iv) y = ± ( x + a) and x = a are its three asymptotes. (v)

y2 



x2 ( x  a ) . x–a yx

xa . x–a

y is defined only for those of the values of x for which x2 – a2  0 and x – a  0 or equivalently when x  – a and x > a. Thus, no part of the curve lies between the lines x = – a and x = a except the point (0, 0). (vi)

dy x 2 – ax – a 2  dx ( x – a )3/ 2 ( x  a )1/ 2

504

DIFFERENTIAL CALCULUS

1 dy  0  x2 – ax – a2 = 0 x = ( 1  5 ) a, 2 dx a For x = ( 1 – 5 ) , y is not real. 2  1  dy Now  0  x  ] – , –a ] and  x   2 (1  5 ) a,   so that y is strictly increasing   dx  1  in] – , – a ] and  (1  5 ) a,    2  1 dy    0  x   a, (1  5 ) a  so that y is strictly decreasing in Again 2 dx   1    a, 2 (1  5 ) a  .   Taking the above points into consideration, we can trace the curve as given in the figure.

Fig 17.10

3. Trace the curve y2 x2 = x2 – a 2 (i) The curve is symmetrical about both the axes. (ii) It does not pass through the origin. (iii) It meets x- axis at ( – a, 0 ) but it does not meet y-axis at any point; x = a and x = – a are the tangents at ( a, 0 ) and ( – a, 0) respectively. (iv) y = ± 1 are the two asymptotes of the curve. (v)

x2 – a 2 x x2 y is defined for only those values of x for which x  0 and x2 – a2  0  x  –a or x  a so that no part of the curve lies between the lines x = a and x = – a. y2 

x2 – a2

dy  dx x2 which is never zero.

(vi)

 y

a2 x2 – a 2

dy  0  x  ] – , – a [ and [ a,  [. dx

CURVE TRACING

505

so that y is strictly increasing in ] – , – a ] and [ a,  [.

Fig. 17.11 2 Combining the above facts, we see that the curve y 

x2 – a 2 x2

is as given in the figure.

4. Trace the curve ( x2 + y2 ) x – ay2 = 0 ( a > 0) (i) The curve is symmetrical about x-axis. (ii) Origin lies on it and x-axis is the tangent thereat. (iii) Origin is the only point of intersection with axes. (iv) x = a is the only asymptote to the curve.

x3 , xa a–x

(v)

y2 



y  x

x a– x

x y is defined only for those of the values of x for which a – x is non-negative, i.e., when

0  x < a   x  [ 0, a [ Thus, the curve is situated between the lines x = 0 and x = a (vi)

 3a  – x x  dy 2    dx ( a – x ) a – x

dy dy  0 for x = 0 where y is 0 and 0 < x < a  dx  0 dx so that y is strictly increasing in [ 0, a [.

The curve is as given in Fig 17.11. Note: This curve is known as the cissoid of Diocles. 5. Trace the curve ( x2 + y2) x – a ( x2 – y2) = 0, ( a > 0) (i) The curve is symmetrical about x-axis.

506

DIFFERENTIAL CALCULUS

(ii) Origin lies on it. y = ± x are the two tangents thereat so that origin is a node. (iii) ( 0, 0) and (a, 0) are its only points of intersection with co-ordinate axes. x = a is the tangent at ( a, 0). (iv) x = – a is an asymptote of the curve.  a– x  y 2  x2    a x 

(v)

a– x , x–a ax y is defined only when x  ] – a, a ] so that the curve lies between the lines x = – a and x = a. y  x



dy a 2 – ax – x 2  dx ( a  x) a 2 – x 2 dy = 0  a2 – ax – x2 = 0 dx

(vi)

 Now

–1– 2

x=

5

–1 2

5

a

a is not a member of the interval ] – a, a ].

 –1 5  –1  5 dy a, a  a [ and < 0  x    0  x  ] – a, 2   2 dx   –1  5  –1 5  a  and strictly decreasing in  a, a  Thus y is strictly increasing in  – a, 2 2     –1 5 a gives a maximum value of y. The point corresponding to x = 2 Combining the above facts, the curve is as given in the Fig. 17.12 Also

Fig. 17.12

Note: The curve referred to above is known as strophoid. 6. Trace the curve y2 = ( x – 1) ( x – 2) ( x – 3) (i) The curve is symmetrical about x-axis. (ii) It does not pass through the origin.

CURVE TRACING

507

(iii) It meets x- axis at (1, 0), (2, 0) and ( 3, 0) but it does not meet y-axis; x = 1, x = 2, x = 3 are the tangents at ( 1, 0), (2, 0), ( 3, 0) respectively. (iv) It has no asymptotes. (v) Now (x – 1) ( x – 2) ( x – 3) < 0 when x < 1 and 2 < x < 3 and ( x – 1) ( x – 2 ) ( x – 3) > 0 when 1 < x < 2 and x > 3. Thus no point of the curve lies to the left of the line x = 1 and between the lines x = 2, x = 3. (vi)

y

( x – 1) ( x – 2 ) ( x – 3)

dy  dx 2

3 x 2 – 12 x  11



Thus

( x – 1) ( x – 2 ) ( x – 3) 3( x –  ) ( x –  )

2

( x – 1) ( x – 2 ) ( x – 3)

dy  0 when dx

6– 3  1.42 approx. 3 6 3  2.5 approx. (  ). and x    3 Also  does not belong to the domain of definition of y.

x =  

Now

dy dy  0  x  [1,  ],  0  x  [ , 2 ] dx dx

dy  0 x  3 dx Thus, y is strictly increasing in [ 1, ] and [3,  [ and strictly decreasing in [  , 2]

and

Also y is maximum for x = . dy   as x  so that the curve tends to be || to y-axis as x . Since dx the curve, while departing from (3, 0) where the tangent is parallel to y-axis, must again tend to become parallel to y-axis, we see that it must change its direction of bending for some point whose abscissae is > 3 and where therefore we have a point of inflexion.

We also notice that

We thus have the curve as in Fig. 17.13.

Fig. 17.13

508

DIFFERENTIAL CALCULUS

7. Trace the curve y2 ( x2 – 1) = 2x – 1 (i) The curve is symmetrical about x-axis. (ii) It does not pass through the origin.

1  (iii) It meets x-axis in  , 0  and y-axis in ( 0, 1) and (0, –1) respectively.  2  1  1 The line x = is the tangent to the curve at  , 0  . 2 2   (iv) x = ± 1, and y = 0 are its only asymptotes. (v)

y2 

2x – 1 x2 – 1

For x < – 1 and

1 1  x  1 , y is imaginary, therefore the curve lies in the region – 1 < x < and 2 2

x > 1. (vi)

y

2x – 1 x2 – 1

  dy – x2  x  1    1/ 2 2 3/ 2   dx  (2 x – 1) ( x – 1)  dy  0. Thus y is decreasing for x > 1. For x > 1, dx 1 dy  0, so that y is decreasing in the interval Again for –1 < x < , 2 dx points into consideration, the curve is as shown in the figure.

1   –1, 2  . Taking all these

Fig. 17.14

8. Trace the curve x3 + y3 = 3ax2 (a > 0). (i) It is neither symmetrical about the co-ordinate axes nor about the line y = x. (ii) Origin is a cusp and x = 0 is the cuspidal tangent. (iii) It meets x-axis at (0, 0) and (3a, 0) but meets y-axis at the origin only; x = 3a is the tangent at (3a, 0).

CURVE TRACING

509

 a 2a  (iv) y + x = a is its only asymptote and curve meets the asymptote at  , . 3 3  (v) x and y cannot both be negative. Therefore the curve does not lie in the 3rd quadrant.

(vi)

y2

dy  x (2a – x ) and, therefore dx

dy  0 for x = 2a dx Solving for y, we obtain y = [x2 (3a – x)]1/3 If x = 0 then y = 0 and y-axis is the tangent there. If 0 < x < 3a, then y > 0 and if x increases from dy 0, y also increases; y will go on increasing with x till x = 2a where = 0. When x increases dx beyond 2a, y will constantly be decreasing; y = 0 for x = 3a and is negative for x > 3a. We now consider the negative values of x. If x is negative, y is positive and constantly goes on increasing as x increases numerically, i.e. as x varies from 0 to – . Also x + y = a is the only asymptote of the curve. Taking all the above facts into consideration, we see that the complete curve is as shown in Fig. 17.15

Fig. 17.15

EXERCISES 1. Trace the following curves : (i) 3ay2 = x2 (x – a)

(ii)

y2 = x (x + 1)2

(iii) ay2 = x (a2 – x2)

(iv)

4a4 y2 = x5 (2a – x). (Gorakhpur 2003)

(v) y2 = x2 (4 – x2)

(vi)

x2y2 = (a + y)2 (a2 – y2).

(vii) a2y2 = x2(a2 – x2)

(viii)

(ix) y2x = a2 (a – x) (xi) y2 (a2 + x2) = a2x2 (xiii) y2x2 = x2 – 1 (xv)

y 2x

=a

(x2

–

a 2)

(xvii) a2y2 = x2 (2a – x) (x – a) (xix) y2 (2x – 1) = x (x – 1). (xxi) y2 = x2 (x – 5).

y2 (a2 – x2) = a3x.

(x)

ay2 = x (a2 + x2)

(xii)

y2x = a (x2 + a2).

(xiv)

y2 (a2 – x2) = x4

(xvi)

y 2x 2 = x 2 + 1

(xviii)

y2(x + 2) = x + 1.

(xx)

y2 = x2 (4 – x).

(xxii)

y2 = x (16 – x2).

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DIFFERENTIAL CALCULUS

(xxiii) y2 = (x – 2)2 (x – 3).

(xxiv)

y 2 = x 3 – x 2.

(xxv) y2 = (x – a)2 (x – b) for different values of a and b. (xxvi) y2 (x – 1) = x2.

(xxvii)

y2 = x2 (x2 – 4).

(xxviii) y2 = x2 (25 – 4x2)

(xxix)

y2 = x2 (x2 – 4).

(xxx) y2 = x2 (3 – x).

(xxxi)

y2 (a + x) = x2 (a – x)

(Avadh 96, 98; Garhwal 2001; Gorakhpur 99, 2002; Ravi Shankar 96; Jabalpur 98; Vikram 2000; Sagar 2000) 2. y (a2 + x2) = a2x

3. y3 = a2 x – x3 (Kanpur 2001, Jiwaji 96; Indore 96, 99; Bilaspur 96; Sagar 99)

4. y (x2 – 1) = x2 + 1

5. y (1 – x2) = x2

6. x = (y – 1) (y – 2) (y – 3) 7.

y2

8.

ay2

10.

y2

(a + x) = =

x2

(3a – x).

x3

(a – x) =

(Avadh 97; Ravi Shankar 2001; Bilaspur 2000) (Garhwal 1996, 2000) 9.

x 3,

ay2

= (x – a) (x –

5a)2.

a > 0.

(Banglore 2006)

17.5. Parametric Equations: The following examples will illustrate the process. 1. Trace the curve with parametric equations. x = a cos3,

y = b sin3.

The following table gives the variations of x and y with .

Fig. 17.16

so that we have the following results  varies in [0, /2]  (x, y) starting from (a, 0) moves to the left and upwards to (0, b).  varies in [/2, ]

 (x, y) starting from (0, b) moves to the left and downwards to (– a, 0).

 varies in [, 3/2]  (x, y) starting from (– a, 0) moves to the right and downwards to (0, –b)  varies in [3/2, 2]  (x, y) starting from (0, –b) moves to the right and upwards to (a, 0). Again, we have dx dy  – 3a cos 2  sin ,  3b sin 2  cos . d d Now dx/d = 0 if  {0, /2, , 3/2, 2} It follows that dy dy dx b sin 2  cos  b  –   – tan  dx d  d  a cos 2  sin  a

CURVE TRACING

511

for the values of  for which dx/d  0. Also (i)  0  dy/dx  0, (ii) /2  dy /dx   (iii)   dy/dx  0, (iv)  3/2  dy/dx  Thus the tangent at each of the points (a, 0), (–a, 0) is the x-axis and the tangent at each of the points (0, b), (0, –b) is the y-axis. The curve is as given in the following diagram.

Fig. 17.17

The curve with parametric equations x = a cos3, y = b sin3 is known as Astroid. 2. Trace the curve with parametric equations  x = a ( + sin ), y = a (1 + cos ) The following table gives the variations of x and y with .

Fig. 17.18

so that we have the following results  varies in [–, 0]  (x, y) starting from (–a, 0) moves to the right and upwards to (0, 2a).  varies in [0, ]  the point (x, y) starting from (0, 2a) moves to the right and downwards to (a, 0). Again, we have

Now

dx dy  a (1  cos ),  – a sin  d d dx/d = 0 if  =  or –.

dy dy dx    – tan dx d  d  2 except for the values ±  of  for which dx/d = 0.

Thus

Also it may be seen that at the points given by  =  and  = – , the tangents are parallel to yaxis.

512

DIFFERENTIAL CALCULUS

Thus we have the curve as shown below

Fig. 17.19

We obtain identical portions of curve to the right and to the left if  varies in [, 3], [3, 5] etc. and [–3, –], [–5, –3] etc.......

Fig. 17.20

3. Trace the curve x = a sin 2 (1 + cos 2), y = a cos 2(1 – cos 2). Since, x, y are periodic functions of , with  as their period the values of x, y will repeat themselves as  varies in the interval ]x, 2[, ] 2, 3 [ etc. We thus confine our attention to   [0,  [ only. dx dy  4a cos 3 cos  ;  4a cos 3 We have d d   5 dx  0 for   , , . d 6 2 6 dy   5  tan  if   , ,  dx 6 2 6 Thus, the tangent at any point corresponding to the value  of the parameter makes an angle  with x-axis. Table of Variations of x and y with .





/6

0



/3

/2

2/3



5/6

dx/d

+

–

–

–

–

+

dy/d

+

–

–

+

+

–

Fig. 17.21

CURVE TRACING

513

The curve is as given in Fig. 17.22.

Fig. 17.22

EXERCISES Trace the following curves: (i)

x = a ( – sin ), y = a (1 – cos ).

(ii)

x = a ( + sin ), y = a (1 – cos ).

(iii)

x = a cos3, y = a (sin 3 + sin )

(iv)

x = a (3 cos  – cos3 ), y = a (3 sin  – sin3 )

(v)

x = a (cos  +  sin ), y = a (sin  –  cos )

(vi) (vii)

1 1 sin 3 ), y  a (cos  – cos 3 ) 3 3 x = cos t, y = cot t. x  a (sin  

(viii) x = 1 – e–t, y = t2 + 1, (ix)

x = et + e–t, y = et – e–t.

(x)

x = 1 + sin t, y = 2 cos 2t.

(xi)

x

1 , y  1  cot t. 2cos t 

17.6. Tracing of Polar Curves The following points may be taken into consideration while tracing polar curves of the type r = f (). 1. Symmetry (i) The curve r = f () is symmetrical about the initial line  = 0 if the equation remains unchanged when  is replaced by –. The curve r = a (1 – cos ) is symmetrical about the line  = 0. (ii) The curve is symmetrical about the line  = /2 (i.e. the line through the pole perpendicular to the initial line) if the equation remains unchanged when  is replaced by  – . The curve r = a (1 – sin ) is symmetrical about the line  = /2. (iii) The curve is symmetrical about the pole

514

DIFFERENTIAL CALCULUS

2. Pole The curve passes through the pole if for r = 0, there corresponds a real value of . For example r = a (1 + cos ) passes through the pole. 3. Asymptotes Find out the asymptotes of the curve, if any. 4. Region Find the region in which the curve does not exist. If r is imaginary for some values of  lying between 1 and 2 then there is no portion of the curve between the lines  = 1 and  = 2. If the greatest and the least numerical values of r be a and b respectively then the curve lies entirely within the circle r = a and entirely outside the circle r = b where a > 0 and b > 0. 5. Values of  d . dr If  = 1 when  = 0 then the line  = 1 will be a tangent to the curve at the point  = 1 and if  = 2 when  = /2 then  = 2, the tangent will  to the radius vector  = 2.

Find  with the help of the formula tan   r

6. Special points Trace the variations of r as  varies. Also if

dr  0 then r increases as  increases and if d

dr  0 , then r decreases as  increases. d

Taking the above points into consideration we can trace the requisite curve. EXAMPLES 1. Trace the curve r = a (1 + cos ) (Rohilkhand 1999, 2000, 2002, Bhopal 98 Ravishankar 2000; Rewa 98; Banglore 2005) The curve is symmetrical about the initial line. r = 0   = . Therefore the curve passes through the pole, the tangent at the pole being  = . There are no asymptotes to the curve. The maximum value of r is 2a. Therefore the curve lies within a circle whose centre is at the pole and whose radius is 2a. dr  – a sin  d d (1  cos ) tan   r –  cot  / 2  tan ( / 2   / 2).  dr sin    = /2 + /2. Therefore the tangent to the curve at the point (2a, 0) is perpendicular to the initial line. As  increases from 0 to /2, r decreases from 2a to a. As  increases from /2 to , r decreases from a to 0. The table of variations of r as  varies from 0 to  is as follows: 

:

0

/3

/2

2/3



r

:

2a

3a/2

a

a/2

0

CURVE TRACING

515

Since the curve is symmetrical about the initial line, the complete graph is as shown in the figure:

Fig. 17.23

2. Trace the curve: r = a sin 3. (Kumaon 2003; Gorakhpur 2000) The curve is symmetrical about the line through the pole perpendicular to the initial line.  2 4  5 r  0    0, , , , , . 3 3 3 3 4  5 2 Therefore the curve passes through the pole and  = 0, /3, are the tangents at , , , 3 3 3 the pole. There are no asymptotes to the curve. The maximum value of r is a so that the curve lies wholly within a circle of radius a.    , r increases from 0 to a. As  increases from to , r decreases 6 6 3   form a to 0. As  increases from to , r is negative and numerically increases from 0 to a. This 3 2 portion of the curve lies in the 3rd quadrant. The curve is symmetrical about the line  = /2. Therefore we take the table of variations of r as increases from –/2 to /2

As  increases from 0 to

:

–

r:

a

 2

–

0

 3

–

 6

–a

0

 6

 3

0

a

0

 2 –a

The complete figure is as follows:

Fig. 17.24

3. Trace the curve: r = a cos 2, a > 0 The curve is symmetrical about the initial line as well as about the line through the pole perpendicular to the initial line.

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DIFFERENTIAL CALCULUS

r = 0   = /4,

3 5 7  , , . 4 4 4

 3 5 7  , , , are the tangents at the pole. 4 4 4 4 There are no asymptotes to the curve. The maximum value of r is a so that the curve lies wholly  within a circle of radius a. When  = 0, r = a. As  increases from 0 to , r decreases from a to 0. As 4    increases from to , r is negative and decreases from 0 to – a. This portion of the curve lies in 2 4  3 3 , r increases from – a to 0 and as  increases from the 3rd quadrant. As  increases from to 2 4 4 to , r increases from 0 to a. Since the curve is symmetrical about the initial line the figure is as follows:

Therefore the curve passes through the pole and  

Fig. 17.25

4. Trace the curve: r

a2 1  2

.

Fig. 17.26

We first consider positive values of only. We have dr 2a   , d  (1  2 ) which is always positive so that r, constanly increases as,  , increases. Also, r = 0 when  = 0. Again we have

CURVE TRACING

517

a 2



a

which  a as . 1   1 Thus r, starting from its initial value 0, constantly increases as  increases and approaches, a, as  so that the point (r, ) approaches nearer and nearer the circle whose centre is at the pole and radius equal to a. The circle is shown dotted in the Fig. 17.26. The figure shows the part of the curve corresponding to the positive values of  only, and the part of the curve for negative values is its reflections in the initial line. 2

–2

5. Trace the curve: r = a (sec  + cos )

(Gorakhpur 2001)

1  cos 2   1  r  a  cos    a cos   cos   (i) The curve is symmetrical about the initial line. Here

(ii) r cos  = a, i.e., x = a is its asymptote. dr dr sin 3  a (iii) so that is positive when  lies between 0 and /2. 2 d d cos 

Fig. 17.27

When  = 0, dr/d = 0 so that  = /2 and, therefore, the tangent is perpendicular to the initial line at the point (2a, 0). Also, r = 2a, when  = 0 and r as /2. 1 Hence we see that when  increases from 0 to  , r monotonically increases from 2a to  and 2 the point P (r, ) describes the part of the curve drawn in the first quadrant. 1 When  increases from  to , r, remains negative and decreases in numerical value from  to 2 –2a and so the point P(r, ) describes the part of curve as shown in the fourth quadrant. As the curve is symmetrical about the initial line. no new point will be obtained when  varies from  to 2.

17.7. In the case of some curves, it is found convenient to make use of the polar as well as the Cartesian form of their equations. Some facts are obtained from the Cartesian and the others from the Polar form. Curves whose cartesian equations are not solvable for x and y, but whose polar equations are solvable for r, are generally dealt with in this manner.

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DIFFERENTIAL CALCULUS

6. Trace the curve: x4 + y4 = a2 (x2 – y2). (i) It is symmetrical about both the axes. (ii) Origin is a node on the curve and y = ± x are the nodal tangents. (iii) It meets x-axis at (0, 0), (a, 0) and (–a, 0), but meets y-axis at (0, 0) only; x = a and x = – a are the tangents at (a, 0) and (1, –a, 0). (iv) It has no asymptotes. (v) On changing to polar co-ordinates, the equation becomes r2 

a 2 cos 2  cos 4   sin 4 

.

We see that r

dr 8a 2 sin 3  cos3  – , 2 d  cos4   sin 4  

so that dr/d remains negative as  varies from 0 to /4 and therefore r decreases from a to 0 as  increases from 0 to /4. (vi) As  changes from /4 to /2, r2 remains negative and therefore no point on the curve lies between the lines  = /4. and  = /2.

Fig. 17.28

As the curve is symmetrical about both the axes, we have its shape as shown. (Fig. 17.28). 7. Trace the curve: y4 – x4 + xy = 0.

(P.U.)

(i) It is neither symmetrical about the co-ordinate axes. nor about the line y = x. (ii) It passes through the origin : x = 0, y = 0 are the two tangents thereat so that the origin is a node. (iii) It cuts the co-ordinate axes at the origin only. (iv) y = x, y = –x are its asymptotes. (v) On transforming to polar co-ordinates, we get 1 tan 2 . 2 When  increases from 0 to /4, 2 increases from 0 to /2, and, therefore, r2 monotonically increases from 0 to . r2 

When  increases from /4 to /2, tan 2 and therefore also r2 remains negative and, thus, there is no part of the curve lying between the lines  = /4 and /2.

CURVE TRACING

519

When  increases from /2 to 3/4, r2 increases from 0 to .

Fig. 17.29

When  increases from 3/4 to , between the lines  = 3/4, and  = .

r2

remains negative and so there is no part of the curve lying

We can similarly consider the variations of r2 as  increases from  to 2. Hence we have the curve as drawn. (Fig. 17.29). 8. Trace the Folium of Descartes: x3 + y3 = 3axy. (Gorakhpur 2000, Rohilkhand 2003; Poorvanchal 2004) (i) It is symmetrical about the line y = x and meets it in the point (3a/2, 3a/2). (ii) It passes through the origin and x = 0, y = 0 are the tangents there so that the origin is a node on the curve. (iii) It meets the co-ordinate axes at the origin only. (iv) x + y + a = 0 is its only asymptote. (v) x, y cannot both be negative so that no part of the curve lies in the third quadrant. On transforming to polar co-ordinates, we get 3a sin  cos  r cos3   sin 3  Now, r = 0 for  = 0 and  = /2. dr 3a (cos  – sin ) (1  sin  cos   sin 2  cos 2 )  d (cos 2   sin 2 ) 2 which vanishes only when cos  – sin  = 0, i.e., tan  = 1. i.e.,  = /4 or 5/4.

For

 = /4, r  3a / 2

Thus r monotonically increases from 0 to 3a / 2 , as  increases from 0 to /4, and monotonically decreases from 3a / 2 to 0, as  increases from /4 to /2.

Fig. 17.30

520

DIFFERENTIAL CALCULUS

Again as  increases from /2 to 3/4, r remains negative and numerically increases from 0 to  so that the point (r, ) describes the part of the curve shown in the fourth quadrant. (Fig. 17.30). As  increases from 3/4 to , r remains positive and decreases from  to 0 so that the point describes the part of the curve shown in the second quadrant. It is easy to see that we do not get any new point on the curve when, , increases from  to 2. 9. Find the double point of the curve: x4 + y4 = 4a2xy. and trace it. Let f (x, y) = x4 + y4 – 4a2 xy.  fx (x, y) = 4 (x3 – a2y), fy (x, y) = 4 (y3 – a2x). Putting fx = 0 and fy = 0, we have 3 2 x – a y = 0, y3 – a2x = 0.  y3 – a3y = 0, or y = 0, ±a. Thus we see that fx and fy vanish at the points (0, 0) (a, a) (– a, – a). Of these only (0, 0) is a point of the curve so that the origin is the only double point of the curve. To trace the curve we note the following particulars about it :— (i) It is symmetrical about the line y = x and meets it at

(  2a ,  2a ) .

Fig. 17.31

It passes through the origin and x = 0, y = 0, are the two tangents there. It meeets the co-ordinate axes at the origin only. It has no asymptotes. x and y cannot have values with opposite signs, for such values make 4a2xy negative whereas x4 + y4 is always positive. Thus the curve lies in the first and third quadrants only. (vi) On transforming to polar co-ordinates, we have r2 = 4a2 sin  cos /(sin4 + cos4).

(ii) (iii) (iv) (v)



r

dr (cos 2  – sin 2 ) (cos 4   sin 4   4sin 2  cos 2 )  2a 2 d (sin 4   cos 4 ) 2

so that dr/d = 0 if and only if tan  = 1, i.e.,  = /4 and 5/4.. Thus we have the curve as drawn (Fig. 17.31)

CURVE TRACING

521

10. Trace the curve: x5 + y5 = 5a2 xy2 (i) The curve is symmetrical in opposite quadrants. (ii) The curve passes through the origin and x = 0, y = 0 are the tangents there so that (0, 0) is a node. (iii) It meets the co-ordinate axes at the origin only. (iv) x + y = 0 is an asymptote to the curve. There are no other asymptotes. (v) On transforming to polar co-ordinates we get r2 

5a 2 cos  sin 2 

cos5   sin 5  When  = 0, r = 0. Also when  = /2, r = 0. As  increases from /2 to 3/4, r2 is negative and hence r is imaginary. Therefore no portiion of the curve lies in this region. 3 3 , r = , As  increases from At   to , r decreases from  to 0. 4 4 Since the curve is symmetric in opposite quadrants, the curve is as traced in the figure 17.32.

Fig. 17.32

EXERCISES Trace the following curves: 1. r = a + b cos 

2. r cos3 = a cos 2

3. r cos  = a sin 3

4. r2 cos  = a2 sin 3

5. r = a log 

6. r log  = a

7. r =

ae

sin 

9. r = 2 (1 – 2 sin ) 11. r = a (1 + sin ), a > 0 13. r =

aem

(a > 0, m > 0)

8. r = a ( – sin ) 10. r2 cos  = a2 sin3  12. r = a  sin , a > 0 14. r2 = a2 cos 2 (Kumaon 1997; Gorakhpur 99; Bilaspur 98)

15. r = a (1 – cos ) 17.

x4

+

y4

=4

axy2

19. x5 + y5 = 5a2 x2 y

16. r cos  = a cos 2  18. x5 + y5 = 5 ax2 y2 20. x6 + y6 = 6a2x2y2.

522

DIFFERENTIAL CALCULUS

OBJECTIVE QUESTIONS For each of the following questions, four alternatives are given for the answer. Only one of them is correct. Choose the correct alternative. 1. For the curve y = tan x, which of the following is false? (a) The curve is symmetrical about the x-axis (b) The origin is a cusp (c) x = 2a is an asymptote (d) curve exists for all x  0 2. The number of loops in the curve r = a sin 5 is: (a) 2

(b)

5

(c) 10

(d) 1 (Avadh 2005)

3. The folium of Descartes is given by the equation: (a) x2 + y2 = 3ax2y2

(c)

x3 + y3 = 3 ax2y2

(d) x4 + y4 = 3axy

(c)

y=x

(d) none of these

(c)

4

(d) 6

4

(d) 6

(c)

line  = /2

(d) line  = /4

(a) axis of y is tangent at origin

(b)

curve is symmetrical about x-axis

(c) axis of x is tangent at origin

(d)

Curve is symmetrical about y-axis

(b)

x3 + y3 = 3axy

4. The graph of the curve 1 1 log tan 2 t ), y = a sin t 2 2 is symmetrical about the line : x  a (cos t t 

(a) x = 0

(b)

y=0

5. Number of loops in the curve r = a cos 2 is: (a) 2

(b)

3

6. Number of loops in the curve (a) 2

(b)

r2

=

a2

cos 2 is :

3

(c)

7. The curve r = a + b cos  is symmetrical about : (a) initial line

(b)

y-axis

8. For the curve x = a (t + sin t), y = a (1 – cos t)

ANSWERS 1. (a)

2. (b)

3. (b)

4. (b)

5. (c)

6. (a)

7. (a)

8. (c)

CHAPTER 18 ENVELOPES 18.1 One Parameter family of curves : If (x, y, )  f (x, y, ) be a function of three variables, then the equation f (x, y, ) = 0 determines a curve corresponding to each particular value of . The totality of these curves, obtained by assigning different values to , is said to be a one parameter family of curves. The variable , which is different for different curves is said to be the parameter for the family. Illustration : (i) The equation x2 + y2 – 2 ax = 0, determines a family of circles with their centres on X-axis and which pass through the origin. Here, a is the parameter. (ii) The equation y = mx – 2 am – am3, determines a family of straight lines which are normal to the parabola y2 = 4ax. Here the parameter is m. We now introduce the concept of the envelope of a one - parameter family of curves by means of an example considered in the next article.

18.2 Consider the family of straight lines y = mx + a / m,

...(i)

where, m is the parameter and, a is some given constant. The two members of this family corresponding to the values m1 and m1 +  m of the parameter, m are a y  m1 x  ...(ii) m1 y  ( m1   m) x 

a m1   m

...(iii)

We shall keep, m1, fixed and regard m as a variable which tends towards 0 so that the line (iii) tends to coincide with the line (ii). The two lines (ii), (iii) intersect at the point x, y where 523

524

DIFFERENTIAL CALCULUS

a (2 m1   m) a , y m1 (m1   m) . m1 (m1   m)

x

As m  0, this point of intersection goes on changing its position on the line (ii) and, in the limit, tends to the point

 a 2a   2, m  1  m1 which lies on (ii). This point is the limiting position of the point of intersection of the line (ii) with another line of the family when the latter tends to coincide with the former. There will be a point similarly obtained, on every line of the family. The locus of such points is called the envelope of the given family of lines. To find this locus for the family of lines (ii), we notice that the co-ordinates (x, y) of such a point lying on the line ‘m’ are given by a 2a x 2, y . m m Eliminating m, we obtain y2 = 4ax, as the envelope of the given family of lines.

18.3 Definition. The envelope of a one parameter of family of curves is the locus of the limiting positions of the points of intersection of any two curves of the family when one of them tends to coincide with the other which is kept fixed.

18.4 Determination of Envelope Let

f (x, y, ) = 0,

...(i)

be any given family of curves. Consider the two curves f (x, y, ) = 0 and (x, y,  + ) = 0

...(ii)

corresponding to the values  and  +  of the parameter. The points common to the two curves satisfy the equation f (x, y,  + ) – f (x, y, ) = 0 f ( x, y ,     ) – f ( x, y ,  ) 0 ...(iii)  Let 0. Therefore the limiting positions of the points of intersection of the curves (i) satisfy the equation which is the limit of (iii) viz.,



f (x, y, ) = 0, Thus the co-ordinates of the points on the envelope satisfy the equations f (x, y, ) = 0 and f (x, y, ) = 0. Let the elimination of  between (i) and (iv) lead to an equation  (x, y) = 0. This is, then, the required envelope.

...(iv)

ENVELOPES

525

Rule. To obtain the envelope of the family of curves f (x, y, ) = 0, eliminate, , between f (x, y, ) = 0, and f (x, y, ) = 0, where f (x, y, ) is the partial derivative of f (x, y, ) w.r. to . The equations x =  (), y =  () obtained on solving f (x, y, ) = 0, f (x, y, ) = 0 are the parametric equations of the envelope;  being the parameter. Illustration. To find the envelope of the family of lines y – x – a /  = 0,

...(i)

we eliminate, , between (i) and – x + a/2 = 0, which is obtained by differentiating (i) w.r. to . The eliminant is y2 = 4 x, which is the envelope of the given family of lines. This conclusion agrees with the one already arrived at ‘abinitio’ in § 18.2.

18.5 Theorem. The evolute of a curve is the envelope of its normals. PQ, QR. are the normals and PT, QT, the tangents at two point P, Q of a curve. L is the point of intersection of the tangents.  PRQ =  TLT  =  arc PQ = s Applying the sine formula to the  PRQ, we get PR sin  RQP  PQ sin  PRQ

Fig. 18.1



PQ PR = sin  RQP sin   chord PQ  s   = sin  RQP  arc PQ     sin  

526

DIFFERENTIAL CALCULUS

Let Q  P so that  RQP   RPT =  / 2.  ds lim PR  sin  1   1  .  QP 2 d Thus the limiting position of R which is the intersection of the normals at P and Q is the centre of curvature at P. Hence the theorem.

18.6 To prove that, in general, the envelope of a family of curves touches each member of the family. Let

f (x, y, ) = 0

...(i)

be a given family of curves. Its envelope is obtained by eliminating  between (i) and f (x, y, ) = 0. Let

x =  (), y =  ()

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

be the parametric equations of the envelope obtained by solving (i) and (ii) for x and y in terms of . The equation (iii) satisfy the equation (i) identically, i.e., for every value of . We differentiate (i) w. r. t to , regarding x, y as functions of x so that we obtain  f dx  f dy f     0  x   y d   which, with the help of (ii), becomes f f  ( )    ( )  0 x y We consider points (x, y) such that fx (x, y) and fy (x, y) are not both zero. At such a point the slope of the tangent at the curve ‘’ of the family is  f / x – .  f / y

...(iv)

Also the slope of the tangent at the same point to the envelope (iii) is  () /  (). We see from (iv) that these two slopes are the same. Thus the slopes of the tangents to the curve and the envelope at the common point are equal. This means that the curve at the envelope have the same tangent at the common point so that they touch. This situation may not arise at points of the curve where fx and fy are both zero. This situation does not occur for a line and a conic except when the conic is a pair of straight lines. In the following we shall, in general consider the envelopes of families of lines or conics only. Note. If for any point on a curve,  f /  x and  f /  y are both zero, then the above argument will break down, so that the envelope may not touch a curve at such points. Since at no point on a line and a conic, we have such points, we can say that the envelope of a family of straight lines or of conics touches each member of the family at all their common points without exception. EXAMPLES 1. Find the envelope of the straight lines x cos  + y sin  = l sin  cos ,  being the parameter. (Kumaon 2000; Avadh 99; Rohilkhand 99)

ENVELOPES

527

The given equation of the family of straight lines can be written as x y   l. sin  cos  Differentiating eq. (1) partially w.r.t. , we have –

x 2

sin 

 cos  

y cos 2 

...(1)

 sin   0

x1/ 3 x or tan   1/ 3 y y 1/ 3 x y1/ 3 sin   . so that and cos   ( x2 / 3  y2 / 3 ) ( x2 / 3  y2 / 3 ) Substituting the values of sin  and cos  in (1), we get the required envelope as,

i.e.,

tan 3  

x ( x2 / 3  y2 / 3 )

y ( x2 / 3  y2 / 3 )

l x1/ 3 y1/ 3 or (x2/3 + y2/3)3/2 = l i.e., x2/3 + y2/3 = l2/3 2. Find the envelope of the parabolas y2 = m2 (x – m), m being the parameter. We have, y2 = m2 (x – m). ...(1) Differentiating eq. (1) partially w.r.t. m, we have 0 = 2mx – 3m2, i.e., m = 0 or 2x / 3. If we substitute m = 0 in eq. (1), we obtain y = 0, which is not a part of the envelope since it touches none of the given parabolas. Substituting m = 2x/3 in eq. (1), we obtain 27y2 = 4x3, which is the required envelope of the given parabolas. 3. Find the envelope of x2 sin  + y2 cos  = a2,  is a parameter. The given curve is x2 sin  + y2 cos  = a2 ...(i) Differentiating partially with respect to , we have x2 cos  – y2 sin  = 0,



tan  

x2 y2



,

x4  y 4 x2 y2   sin  cos  1 Putting the values of sin  and cos  in (i) we get or

x4

y4

 a2 ( x 4  y 4 )1/ 2 ( x 4  y 4 )1/ 2  x 4 + y 4 = a 4, is the required envelope. 4. Find the envelope of the family of circles x2 + y2 – 2 a x cos  – 2 a y sin  = c2 where  is the parameter, and interpret the result. 

(Kumaon 2004) (Gorakhpur 1999; Agra 2001)

528

DIFFERENTIAL CALCULUS

We have x2 + y2 – 2 a x cos  – 2 a y sin  – c2 = 0 Differentiating partially with respect to , we get 2 a x sin  – 2 a y cos  = 0 The curve is 2 a x cos  + 2 a y sin  = x2 + y2 – c2 Squaring and adding, we have 4 a2 (x2 + y2) = (x2 + y2 – c2)2 This is the required equation of the envelope. (iii) can also be written as (x2 + y2)2 – 2 (2 a2 + c2) (x2 + y2) + c4 = 0 or

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

(x2 + y2) = [(a2 + c2) + a2] ± 2 a (a 2  c 2 )

=  (a 2  c 2 )  a    represents circle with origin as centre. 5. Find the envelope of the straight line

2

t x cos t + y sin t = a + a cos t log tan , 2

where t is parameter. We are given that t 2 t or x + y tan t = a sec t + a log tan 2 Differentiating partially with respect to t, we obtain 1 1 y sec 2 t  a sec t tan t  a  sec 2 t / 2  tan t / 2 2 a = a sec t tan t  sin t 1   sin t   = a 2  cos t sin t  a y or sin t Putting this value of y in (i), we have

x cos t + y sin t = a + a cos t log tan

x + a sec t = a sec t + a log tan x t  log tan a 2 ex/a = tan t/2.

or or

2 cosh

or

t 2

cosh

Hence y  a cosh

x t t  e x / a  e – x / a  tan  cot a 2 2

x 1 y   a sin t a

x is the equation of the envelope. a

...(i)

ENVELOPES

529

EXERCISES Find the envelope of the following curves, where  is the parameter: 1. y = mx + a/m, m being the parameter. 2. y = mx + am3 3. y = mx + amp 4. a x sec  – b y cosec  = a2 – b2 2

c b cos  – sin   k x y x cosn  + y sinn  = c. x cos3  + y sin3  = c. tx3 + t2 y = a, parameter being t. y = t2 (x – t). Show that the envelope of the straight line

5. 6. 7. 8. 9. 10.

(Kumaon 99)

2

(Kanpur 2003)

  y cos  – x sin  = a – a sin  log tan    , 2 4 where  is parameter, is the curve y = a cosh a/c.

(Gorakhpur 2001)

ANSWERS 1. y2 = 4 a x ; 3. (p – 1)p – 1 xp = – pp a . yp – 1 ; a4 b4  2  k2 ; 5. x2 y 1 1 1  2  2; 7. 2 x y c 9. 4 x3 = 27 y

2. 4 x3 + 27 a y2 = 0 ; 4. a2/3 x2/3 + b2/3 y2/3 = (a2 – b2)2/3 ; 6. x2/(n – 2) + y2/(n – 2) = c2/(n – 2) ; 8. x6 + 4 ay = 0;

18.7 If A, B, C are functions of x and y and m is a parameter then the envelope of Am2 + Bm + C = 0 is B2 = 4 AC. We have A m2 + B m + C = 0 Differentiating with respect to m partially, we get 2mA+B=0 B m– or 2A By putting the value of m in (i), we get A

or

B2 4 A2

–

B2 C0 2A

B2 = 4 A C. EXAMPLES

1. Find the envelope of the family of straight lines a y  mx  . m

(Rohilkhand 98) ...(i)

530

DIFFERENTIAL CALCULUS

The given equation is a m which can be written as quadratic equation in m as m2 x – m y + a = 0 Thus A = x, B = – y, C = a The required envelope is given by B2 = 4 A C or y2 = 4 a x. 2. Find the envelope of the family of curves given by y  mx 

x2 2



y2 2

 k – 2 The given curve is x2

 1, where  is a parameter.

(Garhwal 97; Gorakhpur 2000)

y2

1  2 (k 2 –  2 ) or x2 (k2 – 2) + y2 2 = 2 (k2 – 2) or 4 – (x2 –y2 + k2) 2 + k2 x2 = 0 This is a quadratic equation in 2, A = 1, B = – (x2 – y2 + k2), C = k2 x2 The required envelope is B2 = 4 A C or (x2 – y2 + k2)2 = 4 k2 x2. 3. Find the envelope of the family of trajectories y  x tan  –



g x2

2 u 2 cos 2  The given curve is

, where  is the parameter.

y  x tan  – y  x tan  –

or or

tan 2  

gx

2

2 u2 which is quadratic in tan .

– x tan  

g x2 2u

2

g x2 2u

2

g x2 2u2

sec2  (1  tan 2  )  y0

g x2

g x2 , B = – x, C   y 2 a2 2u2 Hence the required envelope is B2 = 4 A C A

 4 g x2  g x2  2  y  . or 2  2u  2u  On simplification, we obtain u2 g x2   y 2 g 2 u2 x2 

(Avadh 98, 2000)

ENVELOPES

531

EXERCISES x y   1 is the parabola  a– 2 2 2 2. Find the envelope of the family of y  m x  ( a m  b ).

1. Show that the envelope of the lines

x 

y  a.

(Kanpur 2002; Rohilkhand 2002, 01, 04) 3. Find the envelope of the family of y  m x  a (1  m 2 ) where m is the parameter. 4. Find the envelope of the curve (x – )2 + y2 = 4 .

(Banglore 2006)

5. Find the envelope of the family of straight lines y = mx + 1/m.

(Avadh 97; Gorakhpur 2003)

6. Find the envelope of family of straight lines y = m2x + 1/m2. 7. Find the envelope of (x – )2 + (y – )2 = 2 ,  being the parameter. x my  2  1, c is a constant. m c 9. Find the envelope of the ellipses

8. Find the envelope of

x = a sin ( – ), y = b cos  where is the parameter.

(Rohilkhand 97; Agra 99, 2000) ANSWERS

x

2

y

 1; a b2 5. y2 = 4 a x ;

2.

2

8. xy =



2

c2/4

3. x2 + y2 = a2 ;

4. y2 – 4x – 4 = 0 ;

6. y2 = 4x ;

7. (x + y – 1)2 = 8 (x2 + y2) ;

9. x = ± a.

;

18.8 Two parameters connected by a relation When two parameters are involved in the equation of the curve and one parameter is related to the second one, i.e., and

 (x, y, c1, c2) = 0

...(i)

f (c1, c2) = 0

...(ii)

where c1 is independent parameter. Then     c2   0  c1  c2  c1

...(iii)

f f f and  c   c   c  0 1 2 1

...(iv)

Eliminating c1, c2 from (i) and (ii) with the help of (iii) and (iv), we get the required equation of the envelope.

532

DIFFERENTIAL CALCULUS

EXAMPLES 1. Find the envelope of the curve m

m

 x  y       1, a b

when an + bn = cn. The given curve is xm



ym

 1. am bm Differentiating both the sides with respect to t when a and b are taken as functions of some arbitrary parameter t, we have da ym db – m 1  0 dt b dt a a n + b n = cn ,

–

Now,

xm

m 1



...(i)

Differentiating with respect to t, we have da db  n b n –1 0 dt dt Comparing (i) and (ii), we have n an – 1

xm am  1 an – 1



ym bm  1 bn – 1 m

xm ym  x  y     m m a   b a  a  n n n a b a  bn

i..e,

xm

or Therefore



...(ii)

1



m



1 cn

ym

am  n cn bm  n am + n = cn xm and bm + n = cn yn

i.e.,

a = cn/(m + n) xm/(m + n)

and

b = cn/(m + n) ym/(m + n)

Putting these values in a n + b n = cn , we get

(cn/m + n)n [xmn/(m + n) + ymn/(m + n)] = cn xmn/(m + n) + ymn/(m + n) = cmn/(m + n)

or

2. Find the envelope of the straight line Given that x y   1, a b

x y   1, when ab = c2, where c is constant. a b (Kumaon 2001)

ENVELOPES

533

Regarding a, b as functions of t and differentiating with respect to t, we have

–

x a

2



da y db – 2  0 dt b dt

...(i)

ab = c2

Also

da db a 0 dt dt Comparing (i) and (ii), we have b

Differentiating

..(ii)

x / a2 y / b2  b a x/a y/b x/a  y/b 1    ab ab 2 ab 2 c2

or Therefore

x y 1   a b 2

or

2x = a and 2 y = b

Hence the required envelope is 2x  2y = c2

or

xy = c2/4.

or

EXERCISES x y   1, when the parameters a and b are connected by the a b relation a + b = c. (Garhwal 2001, 2003; Avadh 2002) x y 2. Find the envelope of the straight lines   1, where the parameters a and b are connected a b by the relation an + bn = cn. (Rohilkhand 96; Garhwal 98; Gorakhpur 2002)

1. Find the envelope of the lines

x2



y2

 1, where the two parameters a and b a2 b2 are connected by the relation a + b = c, c being constant. (Garhwal 2002; Banglore 2004)

3. Find the envelope of the family of ellipses

4. Find the envelope of the ellipses having the axes of coordinates as principal axes and sum of their semi-axes constant. 5. Find the envelope of the family of parabolas 12

12

12

12

 x  y where an + bn = cn.      1 a b 6. Find the envelope of the family of the parabolas  x  y     a b when ab = c2.

 1,

7. Find the envelope of the straight line parameters and c is constant.

(Garhwal 96, 2000, 2004; Kumaon 2002) x y   1, when am bn = cm + n where a and b are a b (Avadh 96; Gorakhpur 2003)

534

DIFFERENTIAL CALCULUS

8. Find the envelope of a system of concentric and coaxial ellipses of constant area. 9. Find the envelope of the family of curves x y   1, a b when a + b = c.

10. Find the envelope of the straight lines x y   1, if a2 + b2 = c. a b 11. Find the envelope of the circle whose diameter is a line of constant length which slides between two fixed straight lines at right angles.

12. Find the envelope of a straight line of given length which slides with its extremities on two fixed straight lines at right angles. 13. Find the envelope of the ellipse x2



y2

1 a2 b2 (i) When a2 + b2 = c2; (ii) When an + bn = cn;

(Avadh 2003)

(iii) When am bn = cn; where a and b are the parameters and c is a constant. ANSWERS

x 

1. 3. 5. 7. 9. 11.

2. xn/(n + 1) + yn/(n + 1) = cn/(n + 1) ;

y  c ;

x2/3 + y2/3 = c2/3 ; xn/(2n + 1) + yn/(2n + 1) = cn/(2n + 1) ; (m + n)m + n xm yn/mm nn = cm + n ; x1/3 + y1/3 = c1/3 ; x2 + y2 = c2 ;

4. x2/3 + y2/3 = c2/3 ; 6. 16 xy = c2 ; 8. 2 xy = c ; 10. x2/3 + y2/3 = c2/3 ; 12. x2/3 + y2/3 = c2/3 ;

13. (i) x ± y = ± c ; (ii) x2n / (n + 2) + y2n/(n + 2) = c2n/(n + 2) ; (iii)

(m  n)m  n mm nn

x 2m y 2n  c 2m  2n .

18·9 When the equation to a family of curves is not given, but the law is given in accordance with which any member of the family can be determined. EXAMPLES 1. Find the envelope of the circles described on the radii vectors of the parabola y2 = 4 a x as diameter. (Avadh 96) 2 2 The parametric coordinates of a point on parabola y = 4 a x are (at , 2 a t). Equation of the circle joining (0, 0) and (at2, 2 a t) as diameter is x (x – a t2) + y  (y – 2 a t) = 0 or a x t2 + 2 a y t – (x2 + y2) = 0, which is quadratic equation in t.

ENVELOPES

535

Hence The envelope is

A = a x, B = 2 a y, C = – (x2 + y2).

B2 = 4 A C. Hence the required envelope is 4 a2 y2 = – 4 a x (x2 + y2) or x2 + (a + x) y2 = 0. 2. A series of circles have their centres on a given straight line and their radii are proportional to the distances of their corresponding centres from a given point in that line. Find the envelope. Let the x-axis be considered as straight line, on which the centres of the circles lie and (,0) be the centre of the circles where  is a parameter. Then the radii = k . Hence the equation of family of circles is (x – )2 + (y – 0)2 = k2 2 or 2 (1 – k2) – 2 x  + (x2 + y2) = 0, which is a quadratic equation in . Hence A = 1 – k2, B = – 2 x, C = x2 + y2 The envelope is B2 = 4 A C. Therefore the required envelope is 4 x2 = 4 (1 – k2) (x2 + y2) or x2 = (1 – k2) y2, which represents a pair of lines if k2 < 1. 3. Show that the envelope of a circle whose centre lies on the parabola y2 = 4 ax and which passes through its vertex is the cissoid y2 (2a + x) + x3 = 0. 2 Now, (at , 2at) is a point on the parabola other than the vertex (0, 0) so that t  0. Its distance from the vertex (0, 0) is k2

(a 2 t 4  4 a 2 t 2 ) .

Thus, the equation of the given family of circles is (x – at2)2 + (y – 2 at)2 = a2 t4 + 4 a2 t2.  x2 + y2 – 2 at2x – 4 aty = 0. Differentiating (i) w.r. to t, we get – 4 atx – 4 ay = 0 or t = –y/x.

...(i)

Substituting this value of t in (i), we get the required envelope. EXERCISES 1. Find the envelope of the circles which pass through the origin and whose centres lie on x2



y2

1 (Garhwal 2005) a2 b2 (b) x2 – y2 = a2. 2. A circle moves with its centre on the parabola y2 = 4 a x and always passes through the vertex of the parabola. Show that the envelope of the circle is the curve x3 + y2 (x + 2a) = 0. (Garhwal 95)

(a)

536

DIFFERENTIAL CALCULUS

3. Show that the envelope of the family of circles whose diameters are double ordinates of the parabola y2 = 4 a x is the parabola y2 = 4a (x + a). x2

23



y2

 1 perpendiculars are drawn to the axes and the feet a2 b2 of these perpendiculars are joined. Show that the straight lines thus formed always touch the

4. From any point on the ellipse 23

x  y curve       1. a b 5. Show that the envelope of the straight line joining the extremities of a pair of conjugate diameter of an ellipse is a similar ellipse. 6. Show that the envelope of the circles described on the central radii of the rectangular hyperbola x2 – y2 = a2 is the lemniscate r2 = a2 cos 2.

7. Find the envelope of the circles drawn upon the central radii vectors of the ellipse x2 a2



y2 b2

 1, as diameter,

8. Show that the envelope of the polars of the point on the ellipse the ellipse is h2 x2 a4



k 2 y2 b4

(Nagpur 2005) x2 h

2



y2 k2

 1 with respect to

 1.

ANSWERS 1. (a) (b) (x2

7.

(x2 + y2)2 = 4 (a2 x2 + b2 y2), (x2 + y2)2 = 4 a2 (x2 – y2). + y2)2 = a2 x2 + b2 y2.

18.10 Envelopes of Polar Curves EXAMPLES 1. Find the envelope of the straight lines at right angles to the radii vectors to the cardioid r = a (1 + cos ) drawn through their extremities. Let (r, ) be any point on the cardioid. If (R, ) be the polar co-ordinates of any point on the straight line through (r, ) and at right angles to the radius vector, then the equation is R cos ( – ) = r, R cos ( – ) = a (1 + cos )

i.e.,

...(i)

Differentiating logarithmically with respect to the parameter , we get tan ( – ) 

– sin  1  cos 

 – tan

 or

 2

 2  = 2  – 2 .

––

Fig. 18.2

ENVELOPES

537

Substituting this value of  in (i), we get

R cos ( – 2 ) = a (1 + cos 2 )

or

R cos  = a (1 + cos 2 )

or

R cos  = a (1 + cos 2 ) = 2 a cos2  R = 2 a cos .

or

Therefore in current co-ordinates (r, ), the equation of the envelope is r = 2 a cos . 2. Find the envelope of the circles described on the radii vectors of the curve rn = an cos n  as diameter. Let the polar co-ordinates of any point P on the curve rn = an cos n  be (r, ) and the point  which lies on the circle described on OP as diameter (R, ). Now,

OQ = OP cos ( – ),

i.e.

R = r cos ( – )

or

Rn = rn cosn ( – )

But

rn = an cos n ; then (i) becomes

...(i)

Rn = an cos n  cosn ( – ) Taking log of both the sides, we obtain n log R = n log a + log cos n  + n log cos ( – ) Differentiating partially with respect to , we have 0 = – n tan n  + n tan ( – ) tan ( – ) = tan n 

or

–=k+n

or

( – k ) (n  1)  – k k –   R n  a n cos n   cos n    Then  n  1 1  n     Therefore in current co-ordinates (r, ), the equation of the envelope is

Fig. 18.3



or

 – k k –   r n a n cos n  cos n    .   n 1   1 n  EXERCISES 1. Find the envelopes of circles described on the radii vectors of the following curves as diameters : l  1  e cos ; (b) r3 = a3 cos 3 ; (c) r cosn (/n) = a. r 2. Find the envelopes of the straight lines drawn through the extremities of and at right angles to the radii vectors of the following curves :

(a)

(a) r cos ( + ) = p;

(b) r2 cos n  = an;

(c) rn = an cos n .

538

DIFFERENTIAL CALCULUS

ANSWERS 3 (b) r 3 4  a 3 4 cos    ; 4 

1. (a) r2 (e2 – 1) = 2 e l r cos  – l2 ; n –1 (c) r cos 2 2. (a) r cos

  a; n –1

(   )  p; 2

n  n / ( n  1) ; (b) r n /( n  1) cos  a  n  1

 n  n /(1 – n )  a n / (1 – n ) cos  (c) r .  n  1

18.11 Envelope of Normals (Evolutes) The evolute of a curve has been defined as the locus of the centre of curvature, and it has been shown that the centre of curvature is the limiting position of the point of intersection of two consecutive normals. Hence, the evolute of a curve is the envelope of the normals of that curve. Therefore, the normals of a curve touch the evolute. EXAMPLES 1. Assuming that the evolute of a curve is the envelope of its normals, find the evolute of the parabola y2 = 4ax. (Kanpur 2001; Rohilkhand 98, 2002; Banglore 2004) The equation of a normal to the parabola y2 = 4 ax is given by y = mx – 2 am – am3, where m is the parameter Envelope of (1) will be the evolute of the parabola

y2

= 4 ax.

Differentiating (1) partially w.r.t m, we have 0 = x – 2a – 3am2 12

 x – 2a  m   3a  Substituting the value of m in (1), we have



1/ 2

or

 x – 2a  y   3a  27 ay2 = 4 (x – 2a)3,



x – 2a – a 

( x – 2a ) 3a



which is the required equation of the evolute. 2. Prove that the evolute of the tractrix 1 log tan 2 t / 2), y = a sin t, 2 is the catenary y = a cosh (x/a). x  a (cos t 

We have

x = a (cos t + log tan t/2), y = a sin t

Differentiating w.r.t. t, we have

i.e.,

 dx sec 2 t / 2 1  1    a  – sin t     a  – sin t  dt tan t / 2 2  sin t    dx cos 2 t a dt sin t

...(1)

ENVELOPES

539

dy  a cos t dt dy a cos t sin t   tan t.  dx a cos 2 t The equation of normal to the tractrix is tan t (y – a sin t) + x – a cos t – a log tan t/2 = 0 or x + y tan t – a sec t – a log tan t/2 = 0, where t is the parameter. Differentiating (1) partially w.r.t. t, we have

and

y sec 2 t – a sec t tan t – or or

y sec2 t – y 2

–

a sin t 2

cos t a 2

–

a sec2 t / 2 1  0 tan t / 2 2

a 0 sin t

0

or

cos t sin t cos t Substituting for y from (2) in (1), we have x + a sec t – a sec t – a log tan t/2 = 0 x/a = log tan t/2 or ex/a = tan t/2 and e–x/a = cot t/2



...(1)

ex/a + e–x/a = tan t/2 + cot t/2 =

y

a sin t

...(2)

sin 2 t / 2  cos 2 t / 2 1  sin t / 2 cos t / 2 sin t

1 x y x or cosh  cosh     a  sin t a a x or y  a cosh   , a which is the required evolute.

or

3. Prove that the evolute of the cardioid r = a (1 + cos ) is cardioid r 

[from (2)]

1 a (1 – cos ), the pole 3

2  in the latter equation being at the point  a , 0  . 3  Let P (r, ) be any point on the given cardioid r = a (1 + cos ) = 2a cos2 /2

...(1)

2  Let O be the point  a, 0  , C the centre of curvature at P, OC = R, COX = . 3  Draw the perpendiculars CM and OM upon OP. To find out the evolute of (1), we have to find the locus of C.

Now radius of curvature () at 4 1   PC  a cos  3 2 dr  – a sin  Also d d  a (1  cos ) 1 tan   r   – cot   dr – a sin  2

Fig. 18.4

540

DIFFERENTIAL CALCULUS

  tan   tan     2 2     . i.e., 2 2  From the figure,  MPC = /2

or

1 4 1 1 2 CM  PC sin   a cos  sin   a sin  2 3 2 2 3 2 and OM = OO sin   a sin   CM . 3  OC is parallel to OP and consequently  = . Now R = OC = MM = OP – OM – MP 2 4 2 1 = r – a cos  – a cos  3 3 2



or

R  a (1  cos ) –

=

...(2)

2 2 a cos  – a (1  cos ) 3 3

1 a (1 – cos ) on simplification. 3

1 a (1 – cos ), which is the required evolute. 3 4. Show that the whole length of the evolute of the ellipse

Hence the locus of C is r 

x2 a2



 a2 b2  is  1 4 –  . a   b b2 y2

Transferring the origin to (a, 0), the equation of the ellipse becomes ( x  a) 2 2



y2 2

1

i.e.,

x2 2



y2

a b a b Since x = 0 i.e., y-axis is the tangent at origin.

2



2x 0 a

..(1)

 By Newton’s method, the radius of curvature at the origin,   lim

x 0

y2 2x

Eq. (1) can be written as x 1 y2 1  2   0 2a b 2 x a

Taking limit when x  0, we have b2 i.e., (numerically) a b2 Thus the radius of curvature of the ellipse at an end of the major axis is (b2 / a). Similarly, the radius of curvature at an end of the minor axis is (a2/b). 1



1 0 a

 

Hence the length of the arc of the evolute between points corresponding to extremities of the major and minor axes is a2/ b – b2 / a. From the symmetry of the figure the whole length of the evolute is equal to 4 (a2 / b – b2 / a).

ENVELOPES

541

EXERCISES 1. Find the evolute of the ellipse x2



y2

 1. a2 b2 2. Prove that the evolute of the hyperbola 2xy = a2 is

(Agra 2001; Rohilkhand 2004)

(x + y)2/3 – (x – y)2/3 = 2a2/3.

(Rohilkhand 96; Agra 2000)

3. Show that the evolute of an equiangular spiral is an equal equiangular spiral. 4. Prove that the ellipse b2x2 + a2y2 = a2b2 and a2x2 sec4 + b2y2 cosec4  = (a2 – b2)2 are so related that the evolute of the first is the envelope of the second for all values of . 5. Find the evolute of x2/3 + y2/3 = a2/3 or

x = a cos3 t,

y = a sin3 t.

6. Find the evolute of x2 – y2 = a2. 7. Show that in the parabola y2 = 4 ax, the length of the part of the evolute intercepted within the parabola is 4a (3 3 – 1). 8. Prove that the whole length of the evolute of the astroid x = a cos3 , y = a sin2  is 12 a. ANSWERS 1. (ax)2/3 + (by)2/3 = (a2 – b2)2/3

5. (x + y)2/3 + (x – y)2/3 = 2a2/3

6. (ax)2/3 – (ay)2/3 = (2a2)2/3 OBJECTIVE QUESTIONS For each of the following questions, four alternatives are given for the answer. Only one of them is correct. Choose the correct alternative. 1. The equation of the envelope of the family of curves F (x, y, ), where  is a parameter, is obtained by eliminating  between the equations F (x, y, ) = 0 and: (Kumaon 2005)   F ( x, y, 0)  0 F ( x, y ,  )  0 (b) y x  F ( x, y ,  )  0 (c) (d) None of these  2. The envelope of the family of curves A2 + B + C = 0, where A, B and C are functions of x and y is given by:

(a)

– B  ( B 2 – 4 AC ) 0 2A (c) B2 + 4 AC = 0 (a)

(b) (d)

( B 2 – 4 AC ) 0 2A B2 – 4 AC = 0 –B –

(Garhwal 2002; Avadh 2003) 3. Envelope of the family of straight lines y = mx + a / m is : (a) x2 + y2 = a2 (c) y2 = 4 ax

(b) xy = a2 (d) x2 = 4ay (Kanpur 2002; Rohilkhand 2002, 2003; Avadh 2002)

542

DIFFERENTIAL CALCULUS

4. Envelope of x2 sin  + y2 cos  = a2 is: (a) x2 + y2 = a2 (c) x3 + y3 = a3

(b) (d)

x 4 + y 4 = a4 x 2 – y 2 = a2

5. Envelope of the system of circles (x – )2 + y2 = 4  is: (a) y2 – 4x – 4 = a (c) y2 – 4x + 4 = a

(b) (d)

y2 + 4x – 4 = a y2 + 4 x + 4 = a

6. Envelope of the family of curve tx2 + t2y = a is : (a) x2 = 4 ay (c) x2 + 4ay = 0 7.

(b) (d)

Envelope of the family of curves y2 = t2 (x – t) is : (a) x3 = y2 (c) 4x3 = 27y2

1.

x2 + 4 ay = 0 x2 – 4ay = 0

(c)

2. (c)

(b) (d) 3. (d)

ANSWERS 4. (c) 5. (a)

4x3 + 27y2 = 0 x3 = 4y2 6. (c)

7. (c)

CHAPTER 19 CHANGE OF INDEPENDENT VARIABLES 19.1 To change the single independent variable into the dependent variable In y = f (x), let y be dependent variable where x is independent variable; now we want to change the independent variable, into dependent variable, i.e., dy 1  dx     dx dx  dy  dy

d2y dx

2



d  dx  dx  dy 

–1

–1

 dx   –   dy 

–2



d2x dy 2

d 2 x dy  – 3 dy 2 dx  dx   dy   

2 3 2 2  d2x  d 3 x dy  dx   dx   d x  dy     – 3        dy  2 d3 y d  dy 2  dy 3 dx  dy     dy  dx   6 dx   dx 3  dx3  dx      dy    dy     2 2  3 2 d x  dy  dx   d x dx  – 3      dx  dy   dy 3 dy dy 2       5 dx  dx     dy  dy      

 d2x d 3 x dx  – 3  2 dy 3 dy d y  5  dx   dy   

  

2

and similarly we can change these independent variables to dependent variables.

19.2 To change the independent variable x into another variable t, where x = f (t)

We know,

dy dy dy dt    dt dx dt dx dx dt

543

544

DIFFERENTIAL CALCULUS

The operator

Therefore

d 1 d is therefore equivalent to the operator  . dx dx dt dt 2 2 d y dx d x dy  dy   – 2  2 d2y 1 d  dt  dt dt dt dt    2 dx dt  dx  dx 2 dx  dx    dt  dt  dt  dt 

d 2 y dx d 2 x dy  – 2  2 dt dt dt  dt 3  dx     dt 

 d 2 y dx d 2 x dy   – 2   d y 1 d  dt 2 dt dt  dt   3 dx dt   dx3  dx      dt  dt    3

and



1 dx dt

  d 3 y dx d 2 x d 2 y   d 3 x dy d 2 x d 2 y    dx 3  2  2 – 3   2  2     3  dt dt dt dt   dt dt dt    dt    dt 2 2 –6 2 2  dx  d x  d y dx d x dy   dx   –3     –        dt  dt 2  dt 2 dt dt 2 dt   dt  

 dx     dt 

–7

 dx 3  d 3 y dx d 3 x dy  – 3      3  dt dt  dt  dt   dt 2 2 2 2  dx  d x  d y dx d x dy   – 3   2  2  – 2   dt dt    dt  dt  dt dt

 dx  =    dt 

–5

 dx  d 3 y dx d 3 x dy  d 2 x  d 2 y dx d 2 x dy  – 3  – 2     3  –3 2  2  dt dt  dt dt  dt dt  dt dt  dt  dt 2

2  d 2 x  dy dx dy d 3 x d 3 y  dx  d 2 x d 2 y dx      – 3 3 –  2    dt dt dt 3 dt 3  dt  dt 2 dt 2 dt  dt  dt = 5  dx     dt  and similar other higher order differential coefficients can be written. Now x is a function of t; so the differential coefficients of 1st, 2nd and 3rd order are functions of t etc. and can be known.

19.3 Case of Single Independent Variable Let it be required to change x to t, where x = f (t)

...(i)

CHANGE OF INDEPENDENT VARIABLES

We will express

545

dy d 2 y dy d 2 y , 2 etc. in terms of , etc. dx dx dt dt 2

dy dy  dt We know that dx dx dt d 1 d D   Operator dx dx dt dt

...(ii)

...(iii)

 dy  d  dy  1 d  dt       d x 2 dx  dx  dx dt  dx   dt  dt 2 2  d y dx d x dy   – 2    1  dt 2 dt dt  dt  2 = dx  dx    dt  dt  d2y

Now,

d 2 y dx d 2 x dy  –  2 dt dt 2 dt  dt 3  dx     dt 

3

Again,

d y dx

3

d d y 1     dx dx  dx 2  dt 2





 dx 3  dx    7   dx   dt   dt    dt  1

...(iv)

 dx d 2 y dy d 2 x    –  d  dt dt 2 dt dt 2  3 dt    dx       dt    d 3 y dy d 3 x   3 –   dt dt 3  dt 2 2 2 2  dx  d x  dx d y dy d x   –3    2   2 –  2  dt dt    dt  dt  dt dt

d2x d2 y , cancel in the first bracket. dt 2 dt 2 2  2  1  dx  d 3 y dx d 2 x d 2 y dy   d 2 x  dx d 3 x           – 3 3 –    5  3 dt dt 2 dt 2 dt   dt 2  dt dt 3    dx   dt  dt    dt  and so on.

Since

In relations (ii), (iv), (v), we can substitute the value of functions of t. Such transformations can be performed.

...(v)

dx d 2 x d 3 x , , from (i) as they are dt dt 2 dt 3

546

DIFFERENTIAL CALCULUS

EXAMPLES x2

1. Transform the equation

2

d y dx 2

 2x

dy a 2  y  0, dx x 2

into another when z is the independent variable, being given x 

1 . z

dx 1 – 2 dz z dy 1 d dy   ( y)  – z 2 . dx dx dz dz dz 1 dy dy  –z z dx dx

We have  or

x

or From (i),

dy dy  –z dx dz d  dy  d  dy  x x   –z – z  dx  dx  dz  dz 

2 dy dy 2 d y  z z 2 2 dx dz dx dz Adding (i) and (ii), we get

or

x2

d2y

...(i)

...(ii)

2 dy 2 d y  z dx dx 2 dz 2 Hence the transformed equation becomes

x2

z2

or Aliter.

d2y

x

d2y dz 2 d2y

 2x



a2 x2

y0

 a2 y  0 dz 2 dy dy dz dy    – z2 dx dz dx dz

d2 y



d  dy  1 d dy – z2    dx  dx  dx  dz dz dx dz dy d2y 2 – z2 2  = – z – 2 z dz dz   2



4 = z

d2y 2

 2 z3

dy dz



dz Hence, the transformed equation,  d2y dy   dy  x2  z 4 2  2 z3  – 2x  z 2   a2 z 2 y  0 dz dz   dz   2 d y dy dy z2 2  2z – 2z  a2 z2 y  0 or dz dz dz d2y  a2 y  0 . or dz 2

CHANGE OF INDEPENDENT VARIABLES

547

2. Change the independent variable from x to  in the equation (1 – x 2 )

Having given x = cos . We have, x = cos 

d2y dx 2

– x

dy  y  0. dx

dx d2x  – sin ,  – cos   – x. d d 2 dy dy d  1 dy  – , Now, dx dx sin  d  d d2y d  dy  1 d  1 dy      dx  –  2 dx  dx  d   sin  d   dx d  d2 y dy   –sin  2  cos   1  d  d –   2 sin   sin   2 1  d y dy  sin  – cos   = 3  2 d   , sin   d By putting these values the given equation becomes

then

 d2y dy  – cos   sin   2   cos  dy d d (1 – cos 2 )    sin  d   y  0 sin 3   

d2y

or

d

2

–

cos  dy cos  dy   y0 sin  d  sin  d  d2y

or 3.

d 2

Show that the equation d 2s

3

d2x ds 2

 a may be written in the form

 ds   a    0.  dx  dx 2

We have

then,

dx 1  ds ds dx d2x 1 d  1      2 ds dx  ds  ds  dx  dx

d 2s 2  dx 3  ds     dx  –

 y  0.

548

DIFFERENTIAL CALCULUS

Substituting this value in

d2x ds 2

 a,

d 2s dx 2  a 3  ds     dx  –

we get

3

d 2s

 ds   a   0 2  dx  dx

or

EXERCISES Change the independent variable from x to t in the following : 1. 2. 3. 4. 5.

d2y

dy – x 2 y, when x = sin t. dx dx 2 1 dy 2 d y x  x  y  0, when x  . 2 t dx dx 2 d y dy sin 2 2 x 2  sin 4 x  4 y  0, when tan x = et. dx dx d2y dy (a  bx) 2 2  A ( a  bx)  B y  f ( x), when a + bx = et. dx dx d dP (1 – x 2 )  n ( n  1) P  0 when x = cot t. dx dx or (1 – x 2 )

2

x





(1 – x 2 )

d 2P dx

2

– 2x

(ii) when x 

(i) when x = cos t, 6.

x3

d3y dx3

 2 x2

d2 y dx 2

dP  n ( n  1) P  0 dx

 3x

1 1  t  . 2 t

dy  4 y  0, when x = et.0 dx

ANSWERS 1.

d2y 2

dt d2y

 n 2 y  0;

2 2. t

d2 y dt

 y  0; dt 2 2  et – a  2 d y 2 dy  ( Ab – b )  By  f 4. b  ; dt  b  dt 2 d 2P dP  cos t  n ( n  1) P  0; 5. (i) 2 dt dt d 2P 2 2 3 dP – n ( n  1) (t 2 – 1) P  0; (ii) t (t – 1) 2  2 t dt dt d3y d2 y dy – 2 3  4 y  0. 6. 3 dt dt dt

3.

2

t

dy  y  0; dt

CHANGE OF INDEPENDENT VARIABLES

549

19.4 Transformation from Cartesian to Polars and vice-versa It often happens that a result in cartesian is much simplified on reduction to polar or viceversa. x = r cos 

Let

y = r sin . Suppose  to be independent variable, then dy dr sin   r cos  dy d  d    dx dr dx cos  – r sin  d d

 dr  sin   r cos   1 d  d     dx d   dr dx 2 cos  – r sin   d  d 

d2 y

 dr   dr d 2r dr  cos   sin  – r sin   cos     cos  – r sin   =  2 d  d  d    d    d 2r dr  dr   dr – sin   r cos     – sin   cos  – r cos  – sin    2 d  d   d d   dr   cos  – r sin    d 

3

  dr 2 dr d 2r dr 2 sin  cos   sin  cos  = 2   cos  – 3 r 2 d  d  d    d  –r

2

d 2r

dr  dr  2 sin 2   r 2 sin 2   2  sin  cos   sin   3 r 2 d  d    d

3 dr d 2r  dr  2 2 2  – sin  cos  – r 2 cos   r cos     cos  – r sin   d  d 2 d   d

d 2r

2

d 2r  dr  2 2  r –r 2 d =  d  3  dr  cos  – r sin     d  2

d 2r  dr  r2  2  – r 2  d  d = 3 dr   cos  – r sin     d 

19.5 x and y to be expressed in terms of some third variable To show that x

dx dy dr  y r dt dt dt

...(1)

550

DIFFERENTIAL CALCULUS

x

dy dx d – y  r2 dt dt dt 2

2

...(2)

2

 dx   dy   dr  2  d        r    dt   dt   dt   dt  We have x = r cos 

2

...(3)

y = r sin  x2

+ y2 = r2.

Differentitating, We get 2x

dx dy dr  2y  2r dt dt dt

x

or

dx dy dr  y r . dt dt dt 2

 ds  To prove result (3) we will find the value of   in two forms.  dt  2

2

 ds   ds   dx         dt   dx   dt 

2

2 2   dy    dx    1  =      dx    dt   2

 dx   dy  =      dt   dt  2

2

 ds   ds   d          dt   d    dt 

Again,

2

...(i)

2

 2  dr 2   d  2 = r       d     dt   2

 dr  2  d  = r      dt   dt 

2

...(ii)

Equating the two values, 2

2

2

2

 dx   dy   dr  2  d      r      dt   dt   dt   dt  To prove 2nd relation, multiply (3) by r2 and subtracting the square of relation (1) from it. 2 2  dx  2  dy  2   dr   d  ( x 2  y 2 )        r 2    r 4    dt   dt   dt   dt   Squarting 1st relation, 2

2

dx dy  dx   dy   d  x2    x2    2 x y   r2   dt dt  dt   dt   dt  Subtracting (iv) from (iii) we have 2

2

dx dy  dx   dy   d  y 2    x 2   – 2 xy   r4   dt dt  dt   dt   dt 

..(iii)

2

2

...(iv)

CHANGE OF INDEPENDENT VARIABLES

551

2

or or

dy   2 d    dx – x y   r  dt   dt   dt dx dy d y – x  r2 . dt dt dt

2

EXAMPLES dy – y dx

x

1. Transform p 



into polars.

2   dy   1      dx   

Multiplying numerator and denominator by dy dx – y dt dt

x p



  dxdt 

2  dy       dx   d r2 dt

2

  drdt 

2

2  d    r2     dt  

2

 dr  2  d    r   1  dt    dt  2 p2 d   r4    dt 

or

dx , we get dt

2

2

 dr  1 1  dt   2  4 r r  d 2    dt  = 2. Transform tan   r

We have

1 r2

2



1  dr    . r4  d 

d into cartesian form. dr

d r2 d dt  tan   r dr dr r dt d dy dx x – y dt  dt dt = dr dx dy  y x dt dt dt r

552

DIFFERENTIAL CALCULUS

 

dy – y dx = dy x y dx dy x –y  dx dy x y dx

3. Transform x=

v–1,

d2y

dx 2 y = uv.

x

 

dx dt dx dt

to the new variables u and v taking u as the independent variable, given

From the given relations, we have dx 1 dv dy dv ; and vu – 2  du du du du v dy dv vu dy du du   dx 1 dv dx – 2 du v du

Now,

dv vu d 1 du   dx du 1 dv dx 2 – 2 du v du dv   3 v  u v2  v2 d  du    dv du  dv  du du  

d2y

Therefore

2 2  2  dv   dv  2 dv 2 dv 2 d v  3 2 dv  d v       3 v v 2 uv uv – v uv         3 du du  du  du  du 2  d u 2    dv   du     du  2 3 2  u3   dv   dv  2 d v   4 v 2 u – v       3  du  du 2   dv    du     du 



v2

EXERCISES 3/ 2

2   dy   1      dx    1. Transform the formula   d2y dx 2

into polar coordinates.

2. Given x = Given x = a ( + sin ), y = a (1 – cos ), prove that d2y 1   sec 4 . 2 4a 2 dx

CHANGE OF INDEPENDENT VARIABLES

553

3. Show that by putting x2 = s and y2 = t, the equation 2

dy  dy  A x y    ( x 2 – A y 2 – B) – xy  0 dx  dx 

is reduced to

dt B dt ds ts – . ds A dt  1 ds

ANSWERS 32

 2  dr  2  r      d     2 dr d 2r   r2  2  –r 2  d  d

1.

19.6 Two Independent Variables We shall now consider the case in which there are two independent variables x and y. Let

U = f (x, y)

...(i)

x = 1 (u, v)

...(ii)

y = 2 (u, v) be the proposed transformations ; then we have

U U x U    u x u y U U x U    v x v y

y u y  v 

     

U U .  x y U x U y U    – 0 x y y u u

These equations may be solved for

U x U y U    – 0 x y y v v Solving, we get U U y x   U  y U y U x U x – –       v u u v u v v u 1 x y y x –   u v u v U y U y   – U v u  u v x y y x x   – u v u v 



...(iii)

554

DIFFERENTIAL CALCULUS

U x U x   – U u v  v u and x y y x y   – u v u v If, however, we could solve equations (ii) in u, v,

u = F1 (x, y) v = F2 (x, y), u u v v , , , . x y x y When the connecting relations are

then substitute the values of

V =  (u, v), u = F1 (x, y), v = F2 (x, y) Then, and

V V u V v     x u x v x V V u V v     y u y v y

...(iii) ...(iv)

Differentiating (iii) with respect to x, we get  2V x

2



  V u V v     x  u x v x 

v  2u u   V  V  2 v v   V           u x 2 x x  u  v x 2 x x  v  V  2u u   2V u  2V v  V  2 v   2       2  u x x  v x u v x  v x 2 

 To get the bracketted expression, put

v x

  2V u  2V v    2    x   u v x v

V V and for V in (iii) u v

Similarly,  2V y 2



V  2 u u   2V u  2V v   2      2  u y y  u y u v y  

V  2 v v   2V u  2V v      2    v y 2 y  u v y y  v

19.7 Transformation from cartesian to the polar We have,

y = r sin , x = r cos ,

r  ( x 2  y 2 ),   tan –1

r 1 x  cos ,  x  2  r

y , x

2x ( x2  y 2 )



x ( x2  y 2 )

CHANGE OF INDEPENDENT VARIABLES

555

r cos   cos , r y r 1 2y  sin ,    2 r y 2 (x  y2 ) 



y 2

(x  y2 )

r sin   sin  r

 x –  – r sin , x 

1 2



y 1 2 x r sin  sin  – 2 – , r r

 1 y   r cos ,  y y2  1 2 x V V r V      Then x r x  x

y x

2

–

y 2

x  y2

1 x r cos  cos     . x x2  y2 r r2



 cos 

V sin  V – , r r 

V V r V  V cos  V      sin  – , y r y  y r r  V V x V y V V      cos   sin  , r x r y r x y

V V x V y V V      – r sin   r cos  ,  x  y  x y   sin    cos  – x r r    cos    sin  – . y r r 



2  2V 19.8 Transformation of  V2 and 2 to polars

y

x

2

V

We have

x 2

2

 sin     V sin  V         V   cos  – –   cos   r r    r r    x     V sin  V  – sin    cos   V – sin  V   cos  –  cos   r   r r   r  r r  

 cos 2 

 2V r 2 

–

2 cos  sin   2V sin 2   2V   2 r r r2 

sin 2  V 2 cos  sin  V  r r  r2

Similarly,  2V y 2

2

 cos     V cos  V         V   sin      sin   r r    r r     y 

...(i)

556

DIFFERENTIAL CALCULUS

 sin 2 

2v r

2



2 sin  cos   2V 2 sin  cos  – r  r r2 cos 2   r

V  V cos 2   2V  , r r 2 2

...(ii)

Adding (i) and (ii), we get  2V x 2

 or

 2V



 2V



y 2

r 2

(cos 2   sin 2 ) 

 2V  2sin  cos  2sin  cos   –   r   r r 

V  2 cos  sin  2sin  cos   1 V  2V 1 – (cos 2   sin 2 )   (cos 2   sin 2 )   2 2   r r  r r 2 r 2  2V  2V  2V 1 V 1  2V     r r x 2 y 2 r 2 r 2 2

19.9 Transformation of —2 V We know that  2V

 2V

 2V

 2 V x 2 y 2 z 2 If x = r cos , y = r sin , then we know  2V x

2





 2V y

2





 2V r

2

The operator 2 stands for



2

...(i)

1 V 1  2V  2 r r r 2 

2



...(ii)

2

. x 2 y 2 z 2 The transformation formulae are as follows :– x = r sin  cos . If r sin  = u, y = r sin  sin , then x = u cos , z = r cos . y = u sin . By (ii)  2V  2V  2V 1 V 1  2V     u u u 2 2 x 2 y 2 u 2

Where z = r cos , 2



V

u = r sin , 2



V



 2V

u 2 z 2 r 2 Adding (iii) and (iv), we get  2V x 2



...(iii)

 2V y 2



 2V z 2





1 V 1  2V  2 r r r 2  2V u 2



 2V u 2



...(iv)

1 V 1  2V  2 u u u 2 

We know that V V r V      u r u  u z = r cos , u = r sin 

 2V r 2



1 V 1  2V  2 ...(v) r r r 2

CHANGE OF INDEPENDENT VARIABLES



557

r  u

u

 sin . (z  u2 )  z cos    u  2 2 r u  z

r  ( z 2  u 2 ),

2

u   tan –1 , z V V cos  V  sin   u  r r 



1  V  1 V 1 V  2 cot  .   u  u  r r  r Substituting this value in (v), we get

i.e.,

 2V x 2

 2V





y 2

 2V z 2





 2V x 2



 2V y 2



 2V z 2

2 V 1 V 1  2V  2V 1  2V  2 cot   2   r r  u 2 r r 2 r 2 2





 2V



r 2  2V r 2



2 V 1  2V 1 V 1  2V  2  cot   r r  r 2 sin 2  2 r 2 r2

2 V 1  2V 1 V 1  2V  2 2  2 cot   2 2 r r r   r sin  2 r

= —2V. EXAMPLES 1. If V be a function of r alone, when = x2 + y2, show that r2

 2V 2



 2V 2

x y 2 2 r = x + y 2.

We have



 2V r

2



1 V . r r

dr dr x  2 x;   dx dx r V V r x V    , so that x r x r r  x   . x r r  2V   V  x   x V       2 x  x  r r  r r  x 1 V x 2   1 V      r r r r  r r  1 V x 2 V x 2  2V  – 3  2 r r r r r r 2



(Garhwal 2004)

r



...(i)

Similarly,  2V

1 V y 2 V y 2  2V –  r r y 2 r 3 r r 2 r 2 Adding (i) and (ii), we get  2V x 2



 2V y 2





2 V 1 V 2 1  2V 2 – 3 (x  y2 )  2 (x  y2 ) r r r r r r 2

...(ii)

558

DIFFERENTIAL CALCULUS

2 V 1 V  2V –  2 r r r r r  2V 1 V  = r r r 2 2. If u = ex sec y, v = ex tan y and V is a function of u, v, show that

=

2   2V   2V  2V  V  2 2  V cos y  – .   uv  2  2   (u  v ) y  u v v   u  x y We are given that

u = ex sec y, v = ex tan y, 

u2 = e2x sec2 y = e2x (1 + tan2 y) v2 = e2x tan2y



u2 – v2 = e2x. u V u V v     x u x v x

Now,

 e x sec y

or Again, We have 

V V  e x tan y . u x

V V V u v x u v V V u V v     y u y v y v u  e x sec2 y,  e x sec y tan y y y V V V  e x sec y tan y  e x sec2 y y u v



V V u u v V V V cos y v u y u y

= sec y v

or

Now, cos y 



...(ii)

  V       V V   u  u v    v  y  x   u v   u v 

 uv

 2V u

2

 v2

 2V V  2V  2V V v  u2  uv 2  u u v u u v v v

  2V  2V   2V V V  u v  2  2   (u 2  v 2 ) u v u v v u v   u

  2V   2V  2V  V   2V 2 2 cos y  – .   uv  2  2   (u  v ) u v v   u  y x y  3. If x = r cos , y = r sin , prove that 

(i)

...(i)

2 cos 2  x y r2

from (ii)

CHANGE OF INDEPENDENT VARIABLES 2  2u  2 2  u ( x – y ) –  (ii)  x 2 y 2    –1 (i) We have  = tan (y/x)  1  so that y 1  y 2 / x 2

559

 4 xy



 2u  2u u  2 u  r2 2 – r – . x y  2  r

1 x  x x2  y2

2 2 2 2   x y2 – x2  x  y – 2x    2   2 2 2 2 x y x  x  y  (x  y ) ( x  y 2 )2 r 2 (sin 2  – cos 2 ) cos 2  – 2 . 4 r r u u u x  y (ii) We have, r r x y

and

r2

 2u r 2  2u

and 

2 r2

 2u r 2

–r

 x2  y2

 2u x 2

 2 xy

 2u

– 2 xy

x 2

 2u  2u  y2 2 x y y  2u  2u u u  x2 2 – x – y x y x y y

  2u  2u  u  2 u  2u – 2  ( x 2 – y 2 )  2 – 2   4 xy . r  x y y   x

2 2 2 2 4. If V be a function of r alone where r  x 1  x 2  ...  xn , then show that

 2V x 12



 2V x22

 ..... 

 2V xn2



d 2V dr

2



n – 1 dV . r dr

Since V is a function of r alone, differential coefficient of V w.r.t. r will be total.

r 2  x12  x22  ...  xn2

Now  Thus

2r

x1 r r  2 x, i.e., x  r . x1 1 x1 dV V dV r    x1 dr x1 r dr

 2V x12





  2V

Similarly

x22



  V    x1 dV        x1  x1  x1  r dr 

x1 d 2V r x r dV 1 dV    – 12 2 r dr x1 r dr r x1 dr

x12 d 2V r 2 dr 2 x22 d 2V r

2

dr

2



x 2 dV x 1 dV r – 13 , As  1. r dr x1 r r dr



x 2 dV 1 dV – 23 r dr r dr

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

560

DIFFERENTIAL CALCULUS

 2V

and



xn2

xn2 d 2V r 2 dr 2

x 2 dV 1 dV – n3 r dr r dr



Adding these we get x12  x22  ...  xn2  2V  2V  2V x 2  x22  ...  xn2 dV n dV   ....    – 1 2 2 2 2 x1 x2 xn dr r dr dr r3 d 2V

n dV 1 dV – r dr r dr dr 2 d V n – 1 dV   . r dr dr 2



5. If the equation

2

2 x 2





2 y 2

2 2 2 2 as r  x 1  x2  ...  xn

 0 is satisfied when  is a function of x and y, show that it

will also be satisfied when  is the same function of u and v where u  v = tan–1 (y/x).

1 log ( x 2  y 2 ) and 2

1 log ( x 2  y 2 )  i tan –1 ( y / x) 2 = log (x + iy).

We have u  iv   Hence

x + iy = eu + iv = eu (cos v + i sin v). x = eu cos v and y = eu sin v.   x  y      x  y u x u y u x y   x  y      –y x v x v y v x y

Now and

...(1) ...(2)

Now the equation will be satisfied when  is the same function of u and v if But

2 u

2



 2 v

2

2 x 2

 y2

2 y 2

 2 xy

x 2

v 2

 0.

 2   2 2 x  y  y 2 2  x2 2 x y x y x y

2 2  2 2   2 2 ( x  y )  0  = as 2  2  0. 2 2 y   x x y 6. If x = r cos , y = r sin  and r = ez, prove that

 2 xy

2

            u  u  v  v                  x  y  x  y   – y  x  – y x   x  y  x  y  x  y  x y      

– 2 xy

 2u

u 2





 x2

x2

2

 2u  2u         y2 2  r  r – 1 u   – 1 u. x y r  r z  z   y

2   – x  y x y y y

CHANGE OF INDEPENDENT VARIABLES

561

When x = r cos , y = r sin  u u x u y u    cos  r x r y r x u u u so that r r  x y  y y . r        r r – 1 u   x  y  Thus r  r  x  y    x2  x2

 2u x 2  2u x

2

 y2

 2u y 2

 2 xy

 2 xy

 sin 

u y

...(1)      x x  y y – 1 u  

 2u u u u u x  y – x – y x y x y x y

 2u  2u  y2 2 . x y y

This proves one of the relations. Second Part. From the relation r = ez,

dr  ez  r, dz

u u dz 1 u    r z dr r z



u u    or r  . r z r z         – 1 u  – 1  u. There r  r  r  r z  z   This proves the other part of the result. 7. Show that r

Thus

x2

 3u x3

 3 x2 y

where   x

 3u x 2 y

 3 xy 2

 2u x y 2

 y3

 3u y 3

  ( – 1) ( – 2) u

    y  r in polars. x y r

We have  . u   x   y    x u  y u   x y   x y     2 u u   2u  2u  u  2u   x x 2   y   y 2   y x x x y  y   x  x y y  x2  x2

 2u x 2  2u x

2

 2 xy

 2u  2u u u  y2 2  x  y x y x y y

 2 xy

 2u  2u  y 2 2  u x y y

2 2 2 i.e.,  . u – u  x 2  u  2 xy  u  y 2  u x y x 2 y 2 2 2 2 i.e.,  ( – 1) u  x 2  u  2 xy  u  y 2  u x y x 2 y 2

562

DIFFERENTIAL CALCULUS

Now operating on the both sides by operator , we have     2u  2u  2u     2 ( – 1) u   x  y   x 2 2  2 xy  y 2 2  y   x x y y   x 3 2 3 2  u  u  y  u 3   x  3  2 x 2  2 xy 2  2y  y2 2  x y x x y y x   x 3   2u  3u  2u  2u 2  u  y  x2 2  2 xy  2 x  2 y  y  x y x y 2 y 2 x3   x y 3  3u  3u  3u 3  u  x3 3  3x 2 y 2  3 xy 2  y x x y x y 2 y 3

  3u  2u  2u  2  x 2 2  2 xy  y 2 2  x  y y   x  x3

or

x3

 3u x3 x3

or

 3u x3

 3x2 y  3u x

3

 3x 2 y

 3u x 2 y

 3x 2 y

 3u x 2 y

 3 xy 2  3u 2

x y

 3u

 3 xy 2

 3u

 y3

x y 2

 3 xy 2

x y 2  3u y 3

 3u x y

 y3

2

 3u y 3

 2u  ( – 1) u

 2u  ( –1) u   2 ( –1) u

 y3

 3u y 3

= 2 ( – 1) u – 2 ( – 1) u =  ( – 1) ( – 2) u. 8. Transform the equation

2 z x

2

 2 xy 2

z z  2 ( x – y3 )  x2 y2 z  0 x y

by the substitution x = uv, y 

1 and hence show that z is the same function of u and v as of v

x and y. We have from the given relations 1 . y r z u z v z z 1 z    y  0 x u x v x u v v u

u = xy, v 



2 z

1   1 z  1  2 z   x 2 v u  v u  v 2 u 2 z z u z v z 1 z   x  2 y u y v y u y v

 and

y

or



z z 1 z z z  xy – u –v . y u y v u v

Putting these values in the given equation, we get 1 2 z v

2

1  1 z  1   z z   2  2u   –v   2 1 – 2   u u z0 v  v u  v  u v   u  2

...(1)

CHANGE OF INDEPENDENT VARIABLES

or

2 z u

2



563

z z [2u  2u (v 2 – 1)] – 2v (v 2 – 1)  u 2 v 2 z  0, u v 2 z

z z  2 (u – v 2 )  u 2 v 2 z  0, u v u Since this equation is exactly similar to the given equation so z is the same function of u and v as of x and y.  2 9. If u = f (x, y), x2 =  and y  , change the independent variables to ,  in the equation, 

or

 2 uv 2

2

x2

 2u x

2 Solution. We have x2 = , y 

y

2

 y2

 2u y

2

 2y

u  0. y

y . x

u u  u   u y u      y – x   x   x  x 2  

Now

x

i.e., and



 2u

 

 = xy and  

so that

i.e.,

2

– 2 xy

u u y u u u  xy –  – x  x    u u  u   u 1 u    x  y  y   y  x 

u u y u u u  xy    . y  x        x – y  – 2 x y  y

...(1)

...(2) ...(3)

   u u    Now  x x – y y   x x – x y      x2  x2

 2u x 2  2u x 2

– 2 xy

 2u  2u u u  y2 2  x  y x y x y y

– 2 xy

 2u  2u u u u  y2 2  2 y  x – y x y  y  u y y u adding and subtracting y y

so that

x2

 2u x 2

– 2 xy

 2u  2u u  y2 2  2 y x y y y

   u u   u u     x – y  x – y  – x – y  y   x y   x y   x   u   u     – 2   – 2  –  – 2          

564

DIFFERENTIAL CALCULUS

 4 2

 2u 2



 2

u . 

Thus given equation becomes

42

 2u 2

 2

u  0, 

2

i.e.,

 2u 2



u  0. 

10. The position of a point in a plane is defined by the length r of the tangent from it to a fixed circle of radius a and the inclination  of the tangent to a fixed line. Show that the continuity equation

2 x 2



2 y 2

 0 transforms into

1  1  2  a 2   2  1   a   2  1     2 2 – –  – 2 2   0. 2 2 2 r r r  r r  r  r  r r  r r  r Let (x, y) be the coordinates of a point P in the plane referred to Ox, Oy axes through the centre of circle as axes of reference. 2



Draw PR tangent to the circle, then as given r = PR and  is the angle RTO x = ON + LP = a sin  + r cos , y = RN – RL = a cos  – r sin . Now

  x  y    x  y 

= (a cos  – r sin ) and

  + (– a sin  – r sin ) x x

  x  y   r x r y r    cos  – sin  x y

Solving these equations for

...(1)

...(2)

  and simultaneously, we get x 

  a sin    sin     cos   – x  r  r r 

and

  a cos    cos    – sin   – . y  r  r r  1      y may also be obtained by putting 2    for  in x   

CHANGE OF INDEPENDENT VARIABLES

565

2

sin  sin     a sin    – –    cos   r r r      r  x  a sin    sin       cos   –  r r    r 2  a sin    a sin     a sin     cos     cos   2 –  r   r  r r 2 r

Now

2

sin   sin   2   sin   a sin   –  cos    –  r r   r  r r 2  2 2    a cos    sin    cos    – sin   – –   r   r  r r 2 r   

Now putting

1    for , we get 2

2  a cos    a cos     a cos    – sin  – sin  –        r  r 2 y 2  r r 2 r

 2



2 cos   cos   2   cos   a cos     – – – sin      r r   r  r   r r 2  2  a sin    cos    sin    – cos   –  –   r r 2 r     

Adding these, term by term, we get 2 x 2

 a2   2  a 2  a  a 2 a 2 1  1 2   1 –  – –     . y 2  r 2  r 2 r 3 r r 3  r 2 r  r 2 r  r r r 2 2 2



 2

As

x

2



2 y 2

2 r 2



1  1  2  a 2   2  1   a   2  1    2  2 2  –  – 2 2 , 2 r r r  r   r  r  r   r  r

 0,

1  1  2  a 2   2  1   a   2  1     2  2 – –  – 2 2 0 2 2 r r r  r   r  r  r   r  r r 11. Given that f(x, y) has continuous partial derivatives of the first two orders and x + y = (u + v)3, x – y = (u – v)3, show that



2



2  2 f 2 f  2 f  2 2  f 9 ( x 2 – y 2 )  2 –  ( u – v ) –   . y 2  v 2   u 2  x We have x + y = (u + v)3 = u3 + 3 u2v + 3 uv2 + v3

and

x – y = (u – v)3 = u3 – 3u2v + 3 uv2 – v3.

So that on adding and subtracting, we get x = u3 + 3uv2 and y = 3u2v + v3. Now

f f x f y     u x u y u

566

DIFFERENTIAL CALCULUS

 3(u 2  v 2 )

f f  6uv x y

f f f  6uv  3(u 2  v 2 ) . v x y

and

      3(u 2  v 2  2 uv)  3(u 2  v 2  2 uv) u v x y



     3(u  v) 2    x  y         –  3(u – v) 2  – . u v  x y 

and So that

      3  (u  v)      3(u  v)    x v   x y       3( x  y )    x  y  

      3  (u – v)  – –   3(u – v)   x  y   u v        3( x – y )  – .  x  y   Multiplying the above two operators and then operating on f, we have

and

        (u  v )   –  (u – v )  f  u v   u u        3 ( x  y)    3( x – y)  – f  x y   x y  2 2 2  f  2 f  2 2  f 2 2  f ( u – v ) –  9 ( x – y ) –    . or  x 2 v 2  y 2   u 2  12. In cartesian-polar transformation by the formulae x = r sin  cos , y = r sin  sin , z = r cos 

x  y  x r x     1.  , (ii)  r2 , (iii)  x  y r x  x (i) We have from the relation x = r sin  cos  x  sin  cos . r Also from the relation r2 = x2 + y2 + z2

prove that (i)

r r x r sin  cos   x    sin  cos . or x x r r Comparing (1) and (2), we get r

...(1)

...(2)

x r  . r x

(ii) Again from

x = r sin  cos ,

x  r cos  cos  

...(3)

CHANGE OF INDEPENDENT VARIABLES

tan 2  

and from the relation we have or

(iii)

2 tan  sec 2 

567

sin 2  cos 2  2x



x2  y2 z2

,

  x z 2

 2r sin  cos   r2  2 x r cos 2   2 tan  sec 2  x  r cos  cos   from (3).  x  – r sin  sin  We have x = r sin  cos ,   r2

and from the relation y = r sin  sin ,

y  r sin  cos  . 

y –1  y  i.e.,   tan   x x  1 y y r sin  sin      2 2 – 2   – 2 2 2 2 x 1  y / x  x  x  y r sin  sin  – r sin 

Now from the relation tan  

x r sin  cos  cos  1   2   . r sin   x  x  y2 r 2 sin 2  x  y  Substituting these values, we have   dx    y

and

 1  y 1  y 2 / x 2

 sin    cos    (– r sin  sin )  –  (r sin  cos )     r sin    r sin   = sin2  + cos2  = 1.

13. Show that x n

nu

 nx n – 1 y

 nu

x n x n – 1 y     y r where   x in polars. x y r

Let

then

i.e., so that

n  x n

nu x n

 nx –1 y

 ...  y n

nu x n – 1 y

nu y n

 ...  y n

=  ( – 1) ( – 2) ... ( – n + 1) u

nu y n

    nu  nu  nu     n   x  y   x n n  n x n –1 y n n –1  ...  y n n  y   x x y y   x = n + 1 + nn

n + 1 = ( – n) n  n  ( – n – 1)  n – 1  ( – n  1)  n – 1 .

Putting n = 1, 2, 3, ..., n, we have 1 = 0 = u

568

DIFFERENTIAL CALCULUS

2 = ( – 1) 1 =  ( – 1) u 3 =  ( – 1) ( – 2) u ..................................etc. and n =  ( – 1) ( – 2) . ( – n + 1) u. 14. If  = y2 + z–2,  = z2 + x–2,  = x2 + y–2 and u be a function of x, y, z show that x

u u u u u   u  y z  2     x y z       u  u u    4  y2  z2  x2 .     

u u  u  u     x  x  x  x u u u  0 (– 2 x – 3 )  (2 x).    Multiplying both the sides by x, we get We have

x

u u u  – 2 x –2  2 x2 x  

...(1)

u u u  – 2 y –2  2 y2 y   u u u z  – 2 z –2  2z2 . and z   Adding (1), (2) and (3), we get u u u x  y z x y z

Similarly

y

 2 ( y2 – z – 2 )  2 (2 y 2 – ) as

...(2) ...(3)

u u u  2 (z2 – x– 2 )  2 ( x2 – y – 2 )   

u u u  2 (2 z 2 – )  2 ( x2 – )   

 = y2 + z–2 etc.

Suitable transposing, we get x

u u u u u   u  y z  2     x y z       u  u u    4  y2  z2  x2 .     

19.10 Orthogonal Transformation of —2V.  2V

To show that the expression



 2V



 2V

 2V



 2V



 2V

transforms to by changing x 2 y 2 z 2  2 2  2 to any other set of axes O, O, O, mutually at right angles, the origin being the same. Let (l1, m1, n1), (l2, m2, n2), (l3, m3, n3) be the direction cosines of O, O, O, referred to Ox, Oy, Oz as axes. Then l 12  m12  n12  1, l 22  m 22  n 22  1, l 23  m 23  n 23  1

CHANGE OF INDEPENDENT VARIABLES

569

l1l2 + m1m2 + n1n2 = 0 etc. ...(A) Again [l1, l2, l3], [m1, m2, m3], [n1, n2, n3] are the direction cosines of Ox, Oy, Oz referred to O, O, O; 

l 12  l 22  l 23  1, etc. and l1m1 + l2m2 + l2m3 = 0, etc.

...(B)

Now  = projection of line joining (0, 0, 0) and (x, y, z) on O = l1x + m1y + n1z Similarly,  = l2x + m2y + n2z,  = l3x + m3y + n3z. V V  V  V     Now x  x  x  x V V V  l1  l2  l3 ;   

 2V



x 2

   V V V      l1  l2  l3  l1  l2  l3           l12

2

V

Similarly,

y

2

 2V

 2V 

 m12

 n12

 l22

2

 2V 2



 2V

 l32



2

 2 l2 l3

 2V  3V  2V  2 l1 l3  2 l1 l2     

2

V  2

 ... etc.

 2V

 ... etc. z 2  2 Adding these and using results (A) and (B) (namely and

l 12  m12  n12  1 etc. and l2l3 + m2m3 + n2n3 = 0 etc.)

 2V

we get

x 2



 2V y 2



 2V z 2



 2V  2



 2V 2



 2V  2

.

Example. If the co-ordinates x, y be transformed orthogonally to ,  and V be any function of x, y then 2

  2V   2V  2V –   2  2 – 2 2 x y    x y  For orthogonal transformation in two dimensions we have  2V  2V

2

  2V    .    

x =  cos  –  sin , y =  sin  +  cos .  

V V x V y V V     cos   sin  ;  x  y  x y

 2V

   V V     cos   sin   cos   sin  x y  x y    2

 cos 2 

Similarly, and

 2V x 2

 2 sin  cos 

 2V  2V  sin 2  2 . x y y

V V V  – sin   cos   x y 2 2 2 V V  2V 2 2  V  sin   2sin  cos   cos  x y 2 x 2 y 2

... (1) ...(2)

570

DIFFERENTIAL CALCULUS

 2V     V V     cos   sin     – sin   cos     x y   x y    2V  2V   2V  sin  cos   2 – 2   (cos 2  – sin 2 ) x y x   y

Also,

...(3)

2

  2V   –    2  2       2V  2V  2V    cos 2  2  2 sin  cos   sin 2  2  x y x y   2  V  2V  2V   sin 2  2 – 2sin  cos   cos 2  2 x y x y 

 2V

Hence,

 2V

   2V  2V – sin  cos   2 – 2 x  y  

 2V  2



  2V     cos 2 x y  

   2

2

  2V  –  after proper simplification. 2     

 2V

ex

ex

EXERCISES sin y, prove that

1. If u + v = 2 cos y, u – v = 2 i  2V  2V  2V   4 u v , u v where V is a function of u and v. x 2 y 2 2. If u + iv = f (x + iy), when x and y are independent and u, v, x, y are all real, prove that  2V  2V   2V  2V       f  ( x  iy ) f  ( x – iy ). x 2 y 2  u 2 v 2  3. If V be a function of r alone when r2 = x2 + y2 + z2, show that  2V 2



 2V 2



 2V 2



 2V 2



2 V . r r

x y z r 4. In transforming any function u of x, y, z from cartesian to polar by the formulae x = r sin  cos , y = r sin  sin , z = r cos , prove that x  y  x r x    r2 (a) (b) (c)   x    y  1 r x  x 2

2

2

2

2

1  u   1 u   u   u   u   u  (d)                 r     r sin     x   y   z   r  2



2

 0. x 2 y 2 6. If x = r cos , y = r sin , prove that the equation

5. If x = r cos , y = r sin , show that

  2u  2 y   2u xy  2 – 2  – ( x 2 – y 2 ) 0 x y y   x  2u u –  0. becomes r  

2

CHANGE OF INDEPENDENT VARIABLES

571

7. If x = r cos , y = r sin  and r = ez, prove that 2 y  2u  2u  2u u  2 u u x 2 2 – 2 xy  y2 2  2  r   . x y r 2 z y x  8. If x = ae cos , y = ae sin , prove that  2u  2u  2 u  2 u u y 2 2 – 2 xy  x2 2  2  . x y  x y  9. If x = r cos , y = r sin , pove that   2u  2u   2u u  2u r –  xy  2 – 2   ( x 2 – y 2 ) r   x y x   y 10. If x + iy = f ( + i), prove that the equation

 2V

 2V

r 2



 2V y 2

 0 becomes

 0.  2 11. If u is a function of x, y having continuous derivatives upto the second order, and variables x, y are changed to ,  by the transformation x + y = ( + )n, x – y = ( – )n, 2



 2V

2 2  2u  1 2  2u  2 2  u 2  u . prove that ( x – y )  2 – 2   2 ( –  )  2 – y   2     x 12. By putting G = xnV and changing the independent variables x, y to u, v where u = y/x and v = xy, transform the equations V V G G  y  0. x  y  xG , x x y x x Hence show that G = xn  (y/x) and V =  (xy).

OBJECTIVE QUESTIONS For each of the following questions,four alternatives are given for the answer. Only one of them is correct. Choose the correct alternative. d2y  a can be written as : 1. Equation 2 dx 3 d 2 x dx d2x  dx    0  a 0 (a) (b)   dy dy 2 dy 2  dy  d2x 0 (c) (d) None of these dy 2 dy 1 2. When x  , then value of will be : dx z dy dy z2 (a) z (b) dz dz 1 dy 2 dy (c) – z (d) dz z dz dy 3. When x = tan z, then value of will be : dx dy dy sin 2 z (a) cos 2 z (b) dz dz dy dy cot z (c) sec 2 z (d) dz dz

572

DIFFERENTIAL CALCULUS

4. When x = cos ,

dy will be: dx

dy d (c) – sin  dy d

(a) sin 

(b) (d)

5. In polar coordinates x = r cos , value of

dy d dy – cosec  d

cosec 

dy will be: dx

dr d dr – r sin   cos  d dr r cos   sin  d dr r sin  – cos  d r cos   sin 

(a) – tan 

(b)

dr d dr – r sin  cos  d

r cos  – sin 

(c)

(d)

d in cartesian system becomes: dr dy dy x – y  y x dx dx (a) (b) dy dy x y x– y dx dx dy dy x  y x – y dx dx (c) (d) dy dy x y x– y dx dx dx dy  y 7. If x and y be functions of t, then value of x in polar coordinates will be: dt dt  d    dr  r2  (a) r 2 d  (b)    dt   dt  dt (c) r dr (d) None of these dt ANSWERS

6. Polar formula tan   r

1. (b)

2. (c)

3.

(a)

4. (d)

5. (b)

6. (a)

7. (c)