Cambridbe fUMatbematical Series. DIFFERENTIAL CALCULUS FOR BEGINNERS. DIFFERENTIAL CALCULUS FOR BEGIN NERS BY ALFRED LODGE, M.A. ASSISTANT MASTER AT CHARTERHOUSE; FORMERLY FEREDAY FELLOW OF ST. JOHN'S COLLEGE, OXFORD; LATE PROFESSOR OF PURE MATHEMATICS AT THE ROYAL INDIAN ENGINEERING COLLEGE, COOPERS HILL WITH AN INTRODUCTION BY SIR OLIVER LODGE, D.Sc., F.R.S., LL.D. PRINCIPAL OF THE UNIVERSITY OF BIRMINGHAM SECOND EDITION, REVISED LONDON GEORGE BELL & SONS CAMBRIDGE: DEIGHTON, BELL, & CO. 1905 GLASGOW: PRINTED AT THE UNIVERSITY PRESS BY ROBERT MACLEHOSE AND CO. LTD. PREFACE. THE object of this book is to provide an easy introduction to the Calculus for those students who have to use it in their practical work, to make them familiar with its ideas and methods within a limited range. A good working knowledge of elementary Algebra and Trigonometry is assumed, and in the latter part of the book it is supposed that the student will have acquired some knowledge of the methods of coordinate geometry. Stress is laid at the outset on the graphic representation of functions, as by this means a vividness and reality is given to the processes of the Calculus which is not easily obtained without such ocular assistance. In this way the theory of maxima and minima becomes very simple and convincing. It is hoped that the chapter on the graphic solution of equations will be found helpful and interesting. The idea of integration (as the reverse of differentiation) is introduced into the chapter on successive differentiation. Partial fractions have been treated somewhat fully, as the author feels that the acquirement of facility in the various methods of obtaining such fractions is of great educational value, besides being highly interesting. The chapter on curvature is short, but will, it is believed, be found sufficient for ordinary practical purposes. Indeterminate forms have been treated more fully than usual, and have been made to lead up to what appears to be vi PREFACE. the most powerful method of finding asymptotes to plane curves. In the chapter on tracing curves given by cartesian equations this method has been used for finding both linear and curvilinear asymptotes. Polar coordinates are dealt with in the last chapter, but some acquaintance with them is assumed in the latter part of the chapter on partial differentiation. I have adopted the method of differentials at the commencement, and made it the basis of the whole work, having found that this method is most useful and helpful to the student of Physics and Mechanics. It is hoped that matters of primary importance in a first book on the Calculus have been sufficiently dealt with, but it has been difficult in some cases to decide what ought or ought not to be omitted. Numerous illustrative examples will be found throughout the book, some of which are of intrinsic importance, while some are intended to prepare the student for further developments. I have derived much help from Professor Perry's Calculus for Engineers, particularly in dealing with exponential functions. I desire to express my thanks to my brother, Sir Oliver Lodge, for his valuable introduction, which indicates the great range of subjects that are open to the student who has a fair working acquaintance with the Calculus. I cannot express too strongly my deep obligations to Mr. P. J. Heawood, Lecturer in Mathematics in the University of Durham, who has kindly revised all the proof sheets with scrupulous care, and has given many valuable criticisms and suggestions. COOPERS HILL, June, 1902. PREFACE TO THE SECOND EDITION. I have to acknowledge with thanks the receipt of several valuable suggestions and corrections, most of which are incorporated in the present edition. CHARTERHOUSE, April, 1905. CONTENTS. PAGE INTRODUCTION, by SIR OLIVER LODGE,.. xvii CHAPTER I. FIRST PRINCIPLES. GRAPHS OF FUNCTIONS. ART. 1. Explanation of differentials, and the notation for them,. 1 2. Conditions that two quantities shall be equal,.. 2 3, 4. Orders of infinitesimals,....... 3 5. On the relation d(x2) = 2x dx,... 4 6. Graphic representation of the variations of two connected variables,....... 5 7. Gradient of the curve y = 2,... 6 8. Physical meaning of the gradient of the function x2,. 7 9. Function. Explicit and implicit functions,.. 8 Examples,..... 9 10. Notation for 'function,'....... 9 Examples,. 10 11. The graph of a function,....... 10 12. Gradient. Differential coefficient,.... 11 13. Notations for differential coefficients, or first derived functions,......... 11 14. Summary,... 12 15, 16. Algebraic and transcendental functions. Continuous and discontinuous functions,...... 12 Examples,......... 15 17. Discussion on two of the examples, introductory to the next chapter,... 16 18. Differentiation from first principles,.... 17 viii CONTENTS. CHAPTER II. DIFFERENTIATION OF ALGEBRAIC FUNCTIONS, AND GENERAL METHODS. ART. PAGE 19. d(x") =nxn-ldx...... 19 Examples,.........20 20. Three fundamental differential formulae. Examples of differentiation,.......20 21. Remarks on notation,........22 Examples,.........22 22-25. Respecting differentials of products and quotients. d(uv) = v du + u dv,. 24 Examples,.......25 26. Differentiation of a product of many factors,... 26 27. d. 26 ~(M) v du - clv 26 28. D(umvn) = ull- 1v- 1(mvD + nuDv),...27 29. Introductory to next chapter,.. 28 30. Approximations by help of differentials,. 28 31. Recapitulation,......... 29 Examples,...30 CHAPTER III. CIRCULAR FUNCTIONS. 32. Lemma. sin 0 = 0 = tan 0 when 0 is infinitesimal,.. 31 33, 34. d sin x = cos x dx; d cos x = - sin x dx,..32 35. Geometrical proof of the values of d sin x and d cos x,. 34 36. Gradients of the curves y = sin x and y = cos x,.. 35 37. Differentiation of functions involving sin x and cos x,..36 38, 39. d(tan x) = secx dx; d(cot x) = -cosec2x. dx.. 37 40, 41. d(sec x) = sec x tan x dx; d(cosec x) = - cosec x cot x dx,.39 42. Geometrical proof of the formulae for d tan x and d sec x,.39 43. Discussion of the differentials of the circular functions,.. 40 44-50. Inverse circular functions,.. 41 51. Recapitulation,.........43 Examples,........ 44 52. Expansion of a function in powers of x,... 45 CONTENTS. ix ART. 53-55. Expansions of sin x and cos x in powers of x, 56. Respecting the expansions of tan x and sec x, Examples, 57. Uniform circular motion,. 58, 59. Simple harmonic motion,... 60. Combinations of S.H.M.'s of equal periods, Examples,.... PAGE. 47 49 49 50 51 52 52 CHAPTER IV. EXPONENTIAL AND LOGARITHMIC FUNCTIONS. 2 61. d(e) = eXdx, where ex = 1 + x + -2; d(a) = kaxdx, where k= logea, 54 1. 2, Examples, ah -1 62. ha =logea when h - 0, Examples, 63. d(logex) = dx- x, 64. Use of the formula du = ud (logeu),. 65. Some methods of saving labour in differentiation,. 66. d(logax) = dx + kx where k = logea, 67. Modulus of common logarithms. Formula for changing the base of logarithms,. Examples, 68. Compound interest property of exponential functions, 69. Constant proportional growth of exponential functions. Logarithmic increment or decrement, 70, 71. Respecting discontinuous compound interest. It becomes continuous when the intervals of reckoning are infinitely close together. e = limit of 1 + - when p = o,. ( 1 56 57 58 59 60 60 61 62 62 63 65 67 72. Gradient of the curve y = yoekz, Small table of values of ex, 73. Methods of drawing the graph of ysoa~x, 74. Example of battered embankment, 75. The function e-ax sin (bx + c), 76. Recapitulation, Examples,.. 69 70 70 70 73...73 74 x CONTENTS. CHAPTER V. HYPERBOLIC FUNCTIONS. ART. PAGE 77. Definitions of the hyperbolic function,... 76 78. Expansions of cosh x and sinh x,....76 79. Properties of the hyperbolic functions,..... 77 80. Connection between hyperbolic and circular functions,. 78 81. The function X(),........ 79 82. Mode of calculating X(8),.... 80 83. d(X(0))= sec0......... 80 84. The graph of X()....... 80 Table of values of X(0) from 0 =0~ to 90~ at intervals of 1~,. 81 Examples,..... 82 Miscellaneous examples,. o. 82 CHAPTER VI. APPROXIMATE SOLUTION OF EQUATIONS. 85. General methods of approximating to the roots of equations containing one unknown quantity,. 84 86-88. Approximation graphically and by calculation, shown by examples,..... 85 89, 90. Special methods applicable to quadratic equations,.. 91 Examples,.... 94 CHAPTER VII. MAXIMA AND MINIMA OF FUNCTIONS OF ONE VARIABLE. 91, 92. Introduction to the subject of maxima and minima. Definitions,........ 95 93. Possibility of several maxima and minima,... 96 94. Increasing and decreasing functions,.. 97 95, 96. General conditions for maxima and minima,. 97 97. Points of inflexion,.. 98 98-101. Recapitulation of method for finding maxima and minima. Illustrative example. Discussion of the method. Second example,... 98 CONTENTS. xi ART. PAGE 102, 103. Fundamental axiom regarding the position of maximum and minimum values,. 103 104. Maximum and minimum gradient. Point of inflexion,. 105 105. Method of working with a function of two connected variables,.... 106 106, 107. Discrimination between maxima and minima by help of successive differential coefficients,. 107 Examples,.....109 CHAPTER VIII. CURVATURE OF PLANE CURVES. 108, 109. Definition of curvature. Radius and centre of curvature, 110 110. Expression for p at any point of a curve,*... 112 111, 112. Geometrical construction for p,.... 114 113, 114. Special cases. Radius of curvature at the origin when the curve touches one of the axes of coordinates,. 115 115-117. Mode of finding p when the horizontal and vertical scales are different, with examples,... 117 118. Horizontal and vertical chords of curvature,...121 Examples,........ 123 CHAPTER IX. SUCCESSIVE DIFFERENTIATION-INTEGRATION. 119. Successive differential coefficients of a function,... 124 120. Successive integrals of a function,.. 125 121. Formula for Dr(xn) where n is a positive integer,.. 127 122. Formula for Dr(x-)........ 129 123. Formula for Dr(log x) when r is a positive integer,.. 130 124. First integral of log x,........130 125. First integral of xn log x, and the successive integrals of log x, 131 Examples,........ 132 126. Successive differential coefficients and integrals of sin x (other functions given in the examples),... 133 Examples,.........134 * For the case of curves whose equations are in polar coordinates, see Chap, XV., Art. 216. xii CONTENTS. ART. PAGE 127, 128. Methods of finding the successive differential coefficients of tanx and secx,.... 134 129. Formula for Dr(eaxsin bx), and its first integral,... 136 Examples,......... 136 130-132. Leibnitz' theorem. Symbolical representation of Leibnitz' theorem. Example worked out;.. 137 133, 134. Proof that Dn{eaf(x)} =eax(a+D)"f(x). Illustrative examples,......... 139 135. Examples of successive differential coefficients connected by a differential equation (tan x, sec x, and sin-lx),. 141 Examples,......143 CHAPTER X. PARTIAL FRACTIONS. 136. Definition of rational algebraic fractions. Proper and improper fractions,.... 144 Examples,......... 145 137. Simple fractions,...... 145 Examples,..... 146 138. Partial fractions. A given fraction equivalent to the sum of simple fractions,......146 139. Methods of finding the simple fractions in the case of unrepeated linear factors in the denominator of the given fraction,....... 147 140. Repeated factors in the denominator,.... 151 141. Method of procedure in the case of improper fractions,. 153 Examples,.........154 142. Irreducible quadratic factors in the denominator,.. 155 143. Integration of a simple fraction with a quadratic denominator,..... 158 144. Repeated quadratic factor in the denominator,... 159 145. General summary,....... 160 Examples,......161 146. Integration and successive differentiation of 1/(x2 +1),. 162 147. Integration and successive differentiation of x/(x2 + 1),. 162 148. The nth differential coefficients of 1/(x2+ 1)2 and x/(x2 + 1)3, 163 Examples,......... 164 149. List of successive differential coefficients and integrals,. 165 CONTENTS. xiii CHAPTER XI. TAYLOR'S AND MACLAURIN'S THEOREMS. ART. PAGE 150. Taylor's theorem explained,....... 166 Examples,..........167 151. First proof of Taylor's theorem,..... 167 152. Example illustrating the limitations of Taylor's theorem,. 168 153. Second proof of Taylor's theorem,.....171 154. Consideration of the remainder after n terms,... 172 155. Proof of Taylor's theorem with remainder after n terms,. 173 156. Conditions for validity of the infinite series,... 175 157. Convergency of the infinite series,... 176 158. Use of Taylor's theorem for tabulation,.... 176 159. Use of Taylor's theorem for approximating to roots of equations,..... 177 160. Taylor's theorem in connexion with the theory of maxima and minima,....178 161. Special case; f(a+h) -f(a)=hf'(a+ Oh),.. 180 162. Maclaurin's theorem,........ 181 163. Expansion by means of a differential equation,...183 164. Expansion by use of known series,... 185 Examples,....185 CHAPTER XII INDETERMINATE FORMS. 0 165. The indeterminate form......188 166. If f(a)=0, f(x) is divisible by a positive power of x - a,. 189 167. Evaluation of the indeterminate form,.... 190 168. Note on the case when the value is zero or infinite,.. 192 Examples,.....194 169. Other indeterminate forms,......195 170. The form -.....195 171. Treatment of a finite factor,.....197 Examples,....... 197 172. Evaluation of a fractional function of x when x=o,.. 198 xiv CONTENTS. ART. PAGE 173. Relative values of functions of two related variables when the variables are infinite,......199 174. Ratio of two functions of x and y when y=mx + c,..199 Examples,......201 175. Orders of infinite quantities,......201 176. Relative values of terms when x and y are not of the same order of infinity,.. 202 177. Relative values of terms when x and y are infinitesimals of different orders,.... 203 Examples,..... 204 178. Graphic mode of representing the relative importance of terms when xA is comparable with yB, when x and y are both either very great or very small,.. 204 Examples,.........205 CHAPTER XIII. FUNCTIONS OF TWO VARIABLES. 179. Differentiation of a function of two variables. Partial differential coefficients,.. 206 180. Case when the variables are independent,.... 207 Examples,.......... 207 181. Case of related variables. Distinction between total and partial differential coefficient,.... 208 182. Gradient at any point of the curve f(x, y)= constant,.. 209 /du\ ldu\ 183. Meaning of the equations (d)=0, dy-)=0... 211 Examples,........ 212 184. Homogeneous functions. Euler's theorem,... 213 185. Equation of the tangent at any point of u=0,... 214 186. Equation of the normal,...... 215 187. Polar of a point,......... 216 188. Condition that an equation of the second degree shall represent straight lines,....... 216 Examples,......217 189. Successive partial differential coefficients off(x, y),.. 217 190. Homogeneous functions. Extension of Euler's first theorem, 218 CONTENTS XV ART. PAGE 191. Successive total differential coefficients of f(x, y) with regard to a principal variable,.. 219 Examples,.......221 192, 193. Connection between rectangular and polar coordinates. Application to functions,... 222 194. Maximum and minimum values of a function of two variables, 224 195. Extension of Taylor's theorem to a function of two variables, 227 Examples,.... 229 196. Contour lines and lines of flow,..... 230 Examples,.... 232 CHAPTER XIV. CURVE TRACING-RECTANGULAR COORDINATES. 197. Method of tracing when y is an explicit function of x (or vice versa), or when x and y are explicit functions of a third variable,....... 233 198. Algebraic curves. Degree of a curve,..... 234 199. Curve passing through the origin-Mode of determining its shape near the origin-Multiple or conjugate point at the origin-Cusp at the origin,.. 234 Examples,........ 235 200. Double and conjugate points in general,. 236 201. Simple cases of symmetry,...... 237 202. Curves which extend to infinity. Asymptotes,.. 238 203. Parallel asymptotes,. 239 204. Description of the general method of finding linear asymptotes,...... 241 205. Intersections of a straight line and curve. Another method of finding asymptotes,.... 242 Examples,......244 206. Curvilinear asymptotes,......244 Examples,......... 246 207. General hints for drawing a curve whose equation is given in rectangular coordinates,.... 246 Examples,........ 247 207a. Asymptotes parallel to the axes,.. 249 207b. Imaginary points at infinity,.. 249 xvi CONTENTS. CHAPTER XV. POLAR COORDINATES. ART, PAGE 208. Inclination of tangent to radius vector,. 250 209. Polar equations of straight lines and circles,.. 251 Examples,.....251 210. Perpendicular drawn from the origin to the tangent,.. 252 211. Pedal equation of a curve,....... 253 212. Polar sub-tangent,....... 253 213. Tangents at the origin,...... 253 214. Asymptotes,.......... 253 215. Circular asymptotes,........ 254 Examples,.....254 216,217. To find the radius of curvature. Remarks on the method,...... 255 Examples,........ 256 ANSWERS TO EXAMPLES,.... 257 DIAGRAMS OF THE CURVES OF ART. 198,.. 278 INTRODUCTION BY SIR OLIVER LODGE, F.R.S. THE branch of Mathematics called the Differential Calculus was originally invented by Sir Isaac Newton, and later independently by Leibnitz, for the purpose of dealing with variable quantities; that is to say, with any quantity which does not remain constant, of which kind of quantity the whole of Physics and Engineering and daily life is full. For in so far as things remain constant they are stagnant; every kind of activity involves variable quantities, and therefore to every kind of activity the differential calculus is applicable. It is called the differential calculus because it attends not so much to the quantities themselves as to their variations, their increases or differentials: it is the calculus which deals with differentials, especially with the ratio of two differentials to each other. For by a differential is understood an infinitesimal difference or increase or increment (or, for the matter of that, decrease or decrement), and whenever one quantity changes there is always some other quantity which changes too, the two quantities being called connected variables; or either of them is called a function of the other. Although it is customary to attend to very minute increments, and to consider each differential as infinitesimal, yet the ratio of two such microscopic differentials will in general be finite, and may be large. Thus, if a thing advances 1-u part of an inch in the millionth part of a second, it is then going at the rate of 10,000 inches per second, much faster than an express train, and nearly as fast as a bullet. Or if a slope descends the thousandth of a millimetre for each hundredth of a millimetre b xvii xviii DIFFERENTIAL CALCULUS. along, it has a gradient of 1 in 10, and is too steep for a railway without cogs. Or, if a rod expands the millionth part of its length for one-tenth of a degree rise in temperature, it has about the expansibility of iron. In this last example the two connected variables are the temperature and the length of the rod. It may be asked, why deal with infinitesimal quantities at all? Why not attend to appreciable changes of magnitude and take their ratio? If we could depend on quantities varying uniformly, or if they always bore to one another the relation of simple proportion, this would be the natural and sufficient thing to do. But in practice it is only a few quantities which are thus simply connected, and if we were constrained to attend always to finite differences their ratio would in general give us a mere average result, not an actual result at any instant. To know that a bullet has travelled a mile in ten seconds does not tell us with what speed it left the muzzle; and instruments adapted to ascertain this or any other actual velocity must be chronographic instruments able to record extremely small increments of time and the corresponding moderately small distance travelled. In the laboratory it is to be observed that we are bound to deal with finite changes, and thus are limited to a kind of average result: we may make the observed intervals small, but we cannot make them infinitesimal. But in theory we are not so limited, and the theoretical treatment of infinitesimal changes is decidedly simpler and easier than the treatment of finite changes; except when the observed quantities are varying at a steady or a proportional rate. In that case the finite difference becomes as easy to deal with as the differential. It will be observed that these differentials or infinitesimal differences, though simple in theory to deal with, and though they are the quantities primarily considered in the calculus, are not individually important except indeed in connection with approximate calculations. It is the ratio of one differential to another that is for many purposes the important thing: the ratio of the distance travelled to the time taken, or the ratio of ascent to horizontal distance as we climb a hill, or the ratio of the expansion of a body to the rise of temperature INTRODUCTION. xix which causes or accompanies that expansion; in such cases it is the comparison or relation between the changes of the connected variables, to which attention is directed, rather than the changes themselves; and in the limit when the changes are infinitesimal or evanescent, the above ratios denote, respectively, the speed of travel et a given instant, the slope or gradient of the hill at a given/point, and the expansibility of the body at a given temperature. Newton called the rate of variation of a flowing or variable quantity its "fluxion"; now-a-days it is commonly called its "differential coefficient," which is a long and inconvenient term, but is intended to be and io sound more general and inclusive than any of its special cases, such as "speed," or "slope," or "expansibility": diff4rential coefficient is in fact the most general version of thesb terms, and is denoted by some such symbol as F or -dt where dy, dx, and dt denote dx dt, infinitely small changes of any connected variables y and x, or y and t. One of the objects of the present book is to explain the methods by which the ratio between the differentials of connected variables can be obtained, whatever may be the mathematical equation by which they are connected with each other, and to give some illustrations of the use of such ratio when obtained. Integral Calculus. Often it is not the rate of change of a quantity that has to be attended to, but the change itself; not an infinitely small change, but a change of some magnitude. Such finite change is built up of a succession of differential changes. Although each differential is infinitesimal, the sum of a sufficient number of them will become finite. Thus the total expansion of a body for the rise of a good many degrees may be obviously an important quantity; the total height of a hill may be considerable; and so also may the actual distance travelled in a finite time. The connection between these quantities and the rates at which they grow is an important thing to investigate. These whole quantities xx DIFFERENTIAL CALCULUS. are called integrals, and the mode of obtaining them from the given slope or expansibility or speed, by adding the differentials from moment to moment, or from degree to degree, or from step to step, is called integration; and the investigation of the methods by which integration is performed in various cases is the subject of the Integral Calculus. Integration is therefore a kind of addition, and the idea of it is even simpler than the idea of the ratio involved in slopes or expansibilities or speeds. But the actual process of integration is not found in practice to be so simple. It is much easier to differentiate than it is to integrate, just as it is rather easier to multiply than to divide, or to raise to a power than to extract a root. The work of integration is not done by actual addition, but by a kind of backward process, which here can be best explained by an example. Thus: If we are given an expression for the height of a falling body at any instant, we should be able, by means of the differential calculus, to calculate an expression for its speed at any instant. In fact, the expression for the speed is the differential coefficient of the expression for the height. If, on the other hand, we are given the expression for the speed, we should be able, by means of the integral calculus, to obtain the expression for the height at any time. We should, in fact, have to find or recognize the expression whose differential coefficient is given. It is, therefore, a backward process. Some methods of performing this process will be given in this book, but for more general methods the student must wait till he can read a book on the integral calculus. Another way of finding an expression for the finite increase of a variable function, dependent also on the expression for an infinitesimal increase, will be given in this book in the chapter on Taylor's and Maclaurin's series. This method is strictly a method belonging to the differential calculus, though it has interesting applications in connection with approximate integration. Other matters connected with the differential calculus are those connected with the plotting of curves, finding of tangents, normals, asymptotes, curvature, and such like; an important subject, as it is very necessary to be able to plot a curve, and INTRODUCTION. xxi to realize its properties, when its equation is given: more particularly since the relations between any two connected variables can be very conveniently depicted by means of a curve drawn so that the coordinates of any point on the curve represent a pair of values of the variables. Many examples will be found in the following chapters. This graphic method of representing functions is largely used by Physicists and Engineers. Differential Equations. To illustrate further the advantage of being able to treat subjects familiarly by the differential calculus, and to employ operations which are compactly symbolized by its notation, we may proceed to problems involving a little more knowledge of mechanics. Taking the symbol v to represent velocity, it is manifest dv dv that c( must be acceleration, and m A will therefore be force. The force involved in falling stones and such like is constant, viz. their weight. The simplest examples of variable forces are afforded by plucked wires or struck bells or slightly deflected pendulums. In each of these cases there is a restoring force depending on the displacement, and if the displacement is dv called x, the velocity (v) will be,, and the force will be na-, dt' dt or, as it may also be written. dldtj d2x dt or dt2. In the simplest, and all ordinary cases, this force is proportional to the displacement, a law observed in elastic bodies by Hooke. We cannot say that the force is equal to the displacement: that would manifestly be absurd, because they are not quantities of the same kind; but if the displacement is multiplied by an appropriate factor, not a mere numerical factor, but a factor which involves in the elastic case the material and xxii DIFFERENTIAL CALCULUS. dimensions of the spring, or in the stretched string case its length and tension, or in the pendulum case the intensity of gravity and the length of the pendulum; and if this appropriate factor be called c, then it will be legitimate to say that the restoring force is equal to k times the displacement, that is to say, F=kx; or rather, in order to express the fact that it is a restoring force urging the body back to its original position, and not a sort of repulsive force driving it further and further away, we shall write F= - kx. The force, in fact, always acts in a direction opposite to the displacement, and hence must be given the negative sign. So the facts of the case are expressed by the equation mdv d- - kx, or, to avoid the apparent introduction of three variables, of which v is only an intermediate and manifestly connected one v= ct), the equation is usually written dsx m + x =0............................(1) Now this is called a differential equation, and, in fact, a differential equation of the second order, because of the double differentiation involved in acceleration. But, nevertheless, it is a very simple equation: it is the equation of simple Harmonic Motion, the motion of any simply vibrating or sounding body. One of the things we have to learn is how to pass from a differential equation to its integrated form, the form which deals explicitly with the quantities themselves, not with their changes. This we do by the process of integration, not always by straightforward integration but generally by other operations based on the processes of the differential calculus; and the result in the above example is x= acos(pt + ),......................... (2) where p ==/(k/m). This is the integral equation between x and t corresponding to the above given differential equation. INTRODUCTION. xxiii To the mathematician each of these two equations implies the other: they differ only in form; they express precisely the same facts; either may be deduced from the other, one such deduction being made by the process of differentiation, the other or inverse deduction being made by the process of integration. In order further to call attention to the importance and interest of the problems which can thus be treated, I will mention two more examples. The above simple harmonic motion, which might be plotted as a sinuous line with amplitude a, is of the kind known as perpetual. There is no friction or other term to dissipate energy. If such a term be introduced into the differential equation, say a frictional force proportional to the speed, then the amplitude becomes a function of the time, for instance ae-rt, and this would represent a damped vibration, and be plotted as a sinuous curve gradually dying out with logarithmically decaying amplitude, i.e. dying out as a decreasing geometrical progression. Such oscillations occur in electricity, as well as in tuning-forks, and their importance has been emphasized by the discoveries of Hertz. Another illustration of differential equations may be given here in order to emphasize its importance, viz. the case when there are two independent variables, and when you are studying fluctuations of a third variable dependent on these two. For instance, when we are discussing the height of an object above some fixed level: it may vary from place to place, as the ground does in undulating country, or it may vary from time to time like the height of a balloon or cloud, or like the height of the ground if it were subject to earthquake disturbances. And these two modes of variation, viz. time and horizontal distance respectively, are quite disconnected and independent. But if we look at the surface of the sea we shall observe the effect of the two in combination. Imagine the liquid frozen at any instant and you could travel about it as on undulating country. Shut your eyes and let your boat rest on the surface and you are conscious merely of an oscillation up and down. The rate at which you oscillate up and down depends on circumstances beyond your control; the xxiv DIFFERENTIAL CALCULUS. rate at which you row is an independent circumstance which you can determine for yourself. If you attend to the motion of the water itself, it appears to travel horizontally, though in reality each particle merely swings up and down or round and round a vertical circle. But the wave-forms do travel horizontally, and their rate of travel is somehow connected with the rate of swing of the particles up and down. To find out the whole of the circumstances of this complex or wave motion we must analyze its causes and write down a corresponding differential equation and then integrate it. The necessary differential equation is one which expresses the fact that the force urging the water to move may be regarded from two points of view —the active or elastic or gravitational restoring-force mode, which regards the force as dzy proportional to the curvature of the surface, say kd-y (for small waves), and the passive or inertia mode of regarding the d2y reaction to that force, as mass-acceleration, mdt- where y is the height of the element of water surface under consideration. Writing these two expressions as action and reaction equal to one another, we have d2y d2y dX2mdt2' where on the left-hand side y is regarded as changing only with the variable x, as if the surface were frozen, or rather as if an instantaneous photograph were taken of it; and where on the right-hand side we attend only to those changes of y which are due to efflux of time alone, all horizontal motion being ignored, and the particles which come into vertical line from instant to instant alone being contemplated. Such 'equations, where two causes of variation of a single quantity y are independently considered, are called partial differential equations, and their treatment is exceedingly important though rather difficult. The result of integration in this case is to give the solution in the form y =f (x ~ Vt), where V is the velocity of propagation of the waves, and is equal to the square root of the ratio of k to m. INTRODUCTION. xxv It is not to be supposed that these examples will be followed by any one new to the subject; and, moreover, they contain mechanical difficulties as well as difficulties of calculus notation and methods; but they are taken as typical examples to illustrate the kind of power which will be conferred upon the student by acquaintance with the differential or infinitesimal calculus, and to indicate to the student of elementary physics some of the important varieties of physical problems which can by its means be readily and familiarly dealt with. CHAPTER I. FIRST PRINCIPLES-GRAPHIC REPRESENTATION OF FUNCTIONS. THE differential calculus treats primarily of the relative increments of mutually related quantities when such increments are infinitely small. Thus, if a square sheet of metal or a circular metallic plate is slightly expanded by heat, the question as to the mathematical relation between the increase of area and the increase of the side of the square or of the radius of the circle is a question belonging to the differential calculus. These small increments are called 'differentials.' The differential of any quantity x is written for brevity d(x) or dx, the letter d being an abbreviation of the expression differential of, which means infinitely small increment of. Excample. Let x be the length of the side of a square, whose area will therefore be equal to x2, and let the increase of area be required when the side receives an infinitely small increment. If we do not use the notation of the calculus, we should say, let the increased length of side be x', then the increased area is X'2;.'. the increase of area is X'2 - 2, and the increase of side is x' - x. Also, since x'2 - 2 = (' + x) (x' - x), and since, when x' and x differ by an infinitely small amount, x' +x differs infinitely little from 2x, we see that the increment of area is ultimately equal to 2x (x' - x). If we do use the notation of the calculus, we should denote the increased length of side by x + dx, and therefore the L.D.C. A A 2 DIFFERENTIAL CALCULUS. [CHAP. I. increased area would be (x + cl)2 which is 2 + 2x.dx + (dx)2;.. the increase of area is 2x.dx + (dx)2. If we now bring in the condition that dx is to be infinitely small, we see that (dx)2 is infinitely smaller than 2x.dx, since it is the product of two infinitely small factors. Therefore, finally, we have the increase of area= 2x.dx, which is the same as 2x(x' - x) obtained previously. The essential difference between the two methods is that in the first method we give a name (x') to the increased length of side, whereas in the calculus method we give a name (dx) to the increase in the length. This apparently trivial matter is of fundamental importance. 2. Meaning of the sign of equality. Two quantities are equal if their ratio is unity, and their difference is zero. In the case of finite quantities, if one condition is satisfied, the other is also. In the case of infinitely small quantities their difference is ultimately zero, but they are not to be considered equal unless their ratio is unity. Thus when we say that 2xdx+(dx)2 is ultimately equal to 2xdx we mean that their ratio is ultimately = 1, which is evidently the case, since the surplus fraction 2' which is equal to,d ultimately vanishes. The rule (based on the above method of comparison) which will guide us in neglecting relatively small quantities is very simple, and is as follows: To obtain the ultimate value of an expression containing finite and infinitely small quantities, first perform all subtractions that may be possible, and then, if an infinitely small quantity is added to a finite one, the small quantity is relatively unimportant, and is to be omitted. Thus, in the above example, in reducing the increase of area, viz. (x + dx)2 - 2, to its simplest form, the quantity x2 is cancelled by subtraction, leaving the expression 2x.dx + (dx)2 which is equal to (2x + dx)dx, and this in the limit reduces to 2x.dx, since 2x + dx is ultimately equal to 2x. Another way of considering the question is to remember that although dx is to be infinitely small, it is yet considered of sufficient importance to be taken into account, as obviously it is the supposed increase of side on which the whole problem depends, and therefore 2xdx, which is a finite multiple of dx, ARTS. 1-4.] FIRST PRINCIPLES. 3 is also important; but the square of dx, being a product of two infinitely small factors, is unimportant. In fact, 2x.dx and dx are of the same order of smallness, since their ratio is finite, but (dx)2 and 2x.clx are not of the same order of smallness, since the ratio of the first to the second is ultimately zero. All these' small quantities are included in the general name ' infinitesimals,' dx and 2xdx being called infinitesimals of the 1st order, and (dx)2 being called an infinitesimal of the 2nd order. 3. Orders of infinitesimals. The numerical magnitudes of small quantities can be compared only by means of their ratios. If their ratio is finite, they are said to be infinitesimals of the same order, otherwise they are infinitesimals of different orders. Thus, if dx is an infinitesimal of the 1st order, (dx)2,,,, 2nd (dx),,,, 3rd and so on; and in any equation, if infinitesimals of different orders are added together, those which are of high order are unimportant compared with those of lower order; and, lastly, infinitesimals of any order, when added to finite quantities, are unimportant. When, therefore, unimportant terms are cut out, all the terms of an equation will be finite, or all the terms will be infinitesimals of the 1st order, or all will be of the 2nd order, and so on. We shall have no equations in which different orders will be mixed. Thus in the equation area = X2, the terms are finite. In the equation d(area)= 2xdx, both terms are infinitesimals of the 1st order. 4. We may show in a figure the retained, and the omitted, parts of the increment of the area of the square. Thus, let OABC be the original square whose side OA = x. And let O'A'B'C' be the increased square (the increment being infinitely magnified in the figure to enable it to be seen). Then AA' = dx, and the areas of the rectangles BA' and C'B are each =x.dx, so that their sum = 2xdx. 4 DIFFERENTIAL CALCULUS. [CIIAr. T. Lastly, the small square BB' is =(dx)2, and is ultimately infinitely smaller than either rectangle, since it will require an infinite number of such small squares to make up one of the rectangles. C; 8, C B 0 AA Of course, the student must exercise his imagination, and think of AA' as infinitely small, so that the rectangles BA' and C'B are infinitely thin, thinner than the finest ink line that could be drawn, while the square BB' is not only infinitely thin, but also infinitely short. 5. The calculus would record the above result briefly as follows: If y = 2 dy = 2xd:, or, more briefly still, d(rL2) = 2rx c. This result has wider application than merely to the area of a square. It is true, whatever x: may be, that the differential of its square = 2x times the differential of: itself. Thus, find the differential of the area of a circle whose radius, r, receives an infinitely small increment, dr. The increase of area is evidently 7r( + dr)2 - 7rr2, i.e. r {(r+d c)2 -.r2}, and the quantity in the brackets { } is d(r2) which = 2rcdr just as d (2) = 2x. dx;.'. the increase of area = 27r dcr. The steps of the work may be stated as follows: d(r'r2) = r d(, 2) = 7r2rlr = 2r'dr. ARTS. 4-6.] GRAPHS OF FUNCTIONS. 5 If we look at the same problem geometrically we shall see the reasonableness of this result, for 27rr is the circumference of the circle, and dr the width of the added strip, therefore the area of the added strip = 2n7r.dr, if we remember that dr is infinitely small. Of course for any finite increase in r there would be a small error in this result, but the error is an infinitesimal of the second order, and is therefore of less and less relative importance the smaller we take dr. Example. Find the difference between 7rr2 and 7r(r + 1dr)2 when = 4 inches and (1) d-r= 1 inch; (2) dr-= 01 inch; (3) dr- *001 inch. Compare the result in each case with the corresponding value of 2rrrdr. A ns. The ratios of the actual increments to those given by 27rrdr in the several cases are (1) l1v-; (2) ls -o; (3) lsoo. 6. Graphic mode of representing the variations of two mutually connected variables. If two variables are connected, as, for example, the side and the area of a square, their comparative numerical values can be represented artificially by means of a curve, in the following way. Let x and y be the two variables, each measured in terms of convenient units, e.g. in the case of the square, we should have y = x2, when y is measured, say, in square inches, and x in inches. Then if we measure off along a horizontal line, from some fixed point 0, lengths representing the different values of x, and erect perpendiculars to this line, whose lengths shall represent the different values of y, we can draw a curve connecting the ends of these perpendiculars. This curve is called the graphic representation (or shortly, the graph) of y considered as a function of x. We will illustrate by the above example, y = x2. 6 DIFFERENTIAL CALCULUS. [CHAP. I. The following table gives a series of corresponding values of x and y, and these are plotted in the adjoining diagram. The height of the curve at intermediate points gives values of y corresponding to intermediate values of x. X y 80.i I -FT I1~~~~~~~~~~~I 1 3 9 There is no need to take the same scales for x and for y;I4 16.50 5 25 G t of te c e The student should notice that the steepness of the curve 6 36 -401 7 49 I!4,J —+4~AJE!!!vll. 8 64:9 9 81 2 increases as the numbers get larger. This is due to the Scale of x. There is no need to take the same scales for x and for y-em each should be taken on a scale which is most convenient, having regard to the numerical range of the quantity requiring to be shown. The curve representing the function is a parabola. 7. Gradient of th e urve = The student should notice that the steepness of the curve y=x2 increases as the numbers get larger. This is due to the fact that, when z is large, a given increase in x causes a larger proportional increase in its square. Thus, if z increases from 2 to 3, the square increases from 4 to 9, which is an increment of 5, whereas if x increases from 8 to 9, the increment of its square = 81 - 64 = 17, which is much greater, and if x changes from 20 to 21, the increase of the square is much greater still. For any two values of x the average rate of change of x2, compared with the change of x, is obtained by dividing the change ARTS. 6-8.] GRAPHS OF FUNCTIONS. 7 in x2 by the change in x, and will be shown in the diagram by the tangent of the angle of slope of the chord joining the two points. The tangent * of this angle is called the 'gradient' of the chord, a convenient term introduced to avoid the long equivalent phrase, ' the tangent of the angle which the chord makes with the x-axis.' Similarly the gradient of the curve itself at any point is defined as the tangent of the angle which the tangent to the curve at that point makes with the x-axis. It is the limiting case of the gradient of a chord joining two points P and Q when Q moves close up to P. In the above example, if x changes from 6 to 7, the gradient of the chord is 13 vertical units divided by 1 horizontal unit, since 72- 62= 13, and 7 - 6 = 1. We shall call this a gradient of 13, though the tan of the angle depicted in the diagram will not be actually 13 unless the horizontal and vertical scales are the same. All we have to remember is that in any fraction giving the gradient, the numerator is measured in vertical units, and the denominator in horizontal units. Thus a gradient of 13, i.e. 13 - 1 is, as already stated, 13 vertical units to 1 horizontal unit. If this is borne in mind, the difference in the two scales will present no difficulty. The gradient 13 is represented by the ratio QR: PR in the figure. To find the gradient of the curve at any point P, we must find the gradient of the chord PQ when Q is infinitely close to P. Now, we have proved that when x receives an infinitely small increment dx, the change in x2 is 2x.dx, therefore the required gradient is 2x.dx - dx = 2x - 1. For example, when x=6, the gradient of the curve is 12, and when x=7, the gradient is 14. Hence, in this case, while the gradient of the chord joining the two points in the figure is 13, the gradient of the curve itself gradually changes from 12 to 14. 8. It will be well to see what is the physical meaning of the gradient of the curve at any point. The equation y=x2 may be looked upon as a merely numerical relation between a * The numerical value of the tangent of the angle is to be taken as it would be if the vertical and horizontal units of scale were the same length. This numerical value is always meant when the gradient is spoken of. The distorted angle due to unequal scales is a mere accident of the diagram, and is unimportant. 8 DIFFERENTIAL CALCULUS. [CHAP. I. number x and its square. This is how we have just been considering it. But we may also take it as the relation between the side and the area of a square plate. In that case y is so many square inches and x is so many inches. The gradient of the curve is given by the equation: c. y 2z gradient = dy 2x and, if we imagine the area of the square growing by a gradual increase in the length of its variable side x, the value of the gradient for each value of x represents the rate of growth of the area per unit increase in the side, i.e. so many square inches of area per inch growth of side. For example, when = 6 inches, the gradient is 12 square inches per inch, and when x= 7 inches, the gradient is 14 square inches per inch. These are the actual rates at which the square is growing at the two instants when its dimensions are those which correspond to each end of the chord PQ in the figure, and the average rate of growth between these two points is the gradient of the chord, which is 13 square inches per inch. Hence, the gradient of the curve represents the rate of growth of y compared with that of x, and is described as so many vertical units per horizontal unit. We have arrived at this notion of a gradient by considering the special curve y = x2, but the reasoning is perfectly general, and we will therefore state it again below in connection with any two related variables, z and y. But first we will introduce a term of much use in abbreviating the statement that there is a relation between the variables. 9. Function. Explicit and implicit functions. If two variables x and y are so related that a change in one of them necessitates a change in the other, they are said to be futnctions of each other. Thus the area of a square and the length of its side are functions of each other, and so are the volume of a sphere and its radius. If one of the two variables is directly expressed in terms of the other, it is usual to say that the one variable is a function, or an explicit function, of the other: e.g. if y=-2 we say that y is an explicit function of x. We ARTS. 8-10.] GRAPHS OF FUNCTIONS. 9 may of course express x as an explicit function of y, viz. X = /y. If y is expressed as a function of x, y is called the subordinate, or the dependent, variable, and x is called the principal, or the independent, variable. The latter term is not a happy one, as of course it is not independent of y, but so long as the student realizes this, the term does very well as an antithesis to the term 'dependent.' If the variables occur mixed together in an equation, as for example in any of the following cases, 2 + 3y= 6, 2+ y2= 4, y -- C2, each is said to be an implicit function of the other. We may express either of them as an explicit function of the other if we desire it. In some cases, however, such transformation of the equation may be very difficult, and may also be undesirable. EXAMPLES. 1. If y2 - 2ay + x2= 0, make y an explicit function of x. Ans. y=a /(aC2 - x2). 2. If 1 + logy=2 log,,(x +a), make y an explicit function of x. Ans. y = (x +a)2 a. 3. Given tan-ix + tan-ly = a, make y an explicit function of x. tan a - x Ans. = -— tana1 + x tan a 4. Given y=3x _ 2 make x an explicit function of y. A n. 2y - 1 A.s. 3y —2. 10. If we wish to denote that y is some function of x, without specifying what function, it is usual to make use of the first letter of the word function, and to write y =f(). Or we may use other similar letters, and write F(x), or < (x), to denote functions of x. It is merely a convenient notation, with which the student will soon become familiar. The value 10 DIFFERENTIAL CALCULUS. [CHAP. I. of the function f(x) for particular values of x, such as 1, 2, 3, etc., will be denoted by f(l), f(2), f(3), etc. Thus if f(x) = x2 we shall have (l) = 1 f(2)=4 /(3)= 9 and so on. The student should work the following examples, so as to become familiar with the notation. EXAMPLES. 1. If f(x)=x2 - 3x+2, find f(1), f(3), f(O), andlf(a). Anis. 0, 2, 2, - 3a + 2. 2. If f(O)=cos0, write down the values of f(0)), f /(r) /(v). Ans. 1,,0, -1 3. If F(x)=2x, find F(2), F(3), F(0), and F(F(2)). Ans. 4, 8, 1, 16. 4. In the first example write downf(x2), f(/x), f(x + h). Ans. x4-3x2+2, x-3,/x +2, (x+h)2-3(x+h)+2. 5. In the same example, show that h is a factor of f(x + h) -f(x), and show to what the other factor approximates when h is very small. Ans. 2x- 3. 6. If O (x)= ax, prove that {J(x)}2= (2x). 7. If F(x)=logx, show that F(ab) =F(a)+ F(b). 11. The graph of a function. In any case when y is a function of x, say y=f(x), the relation between x and y can be represented pictorially (as stated in Art. 6) by cutting off lengths along a straight line from a given point on it to represent the values of x, and erecting perpendiculars to represent the corresponding values of y. If the ends of these perpendiculars are joined by a curve, the curve is described as the curve y =f(x) and is said to be the 'graph' of the function. The values of x cut off on the given line are called abscissae (meaning cut-off), the point from which they are measured is called the origin, and the perpendicular lengths ARTS. 10-13.] GRAPHS AND GRADIENTS. 11 are called ordinates. The line along which the values of x are measured is called the axis of x, and the line perpendicular to it drawn through the origin is called the axis of y. 12. Gradient. Differential coefficient. If y=f(x), i.e. if y is a function of x, the ratio of the differential of y to the differential of x, or d is called the differential coefficient of y dx with respect to x. If the curve y =f(x) is drawn, the values of d- at the various points are the values of the tangent of the dX angle which the tangent to the curve makes with the axis of x, i.e. are the values of the 'gradient' of the curve at the various points. This is obvious from a figure, for if P, Q are two adjacent points on the curve, PQ is ultimately the tangent at P, and QR - PR is the gradient of the curve. And, if OM=x, MN is dx;.'. PR=dx, and if PM=y, QN-PM is cdy, i.e. QR=dy; QR dy in the limit when both QR PR dx and PR are infinitely small. Q O MN It has been suggested, in consequence of this meaning of dy in connection with the graph of the function, that dy might always be called the gradient of y with respect to x, or, more briefly still, the x-gradient of y, instead of using the long expression, 'the differential coefficient of y with respect to x.' 13. If y is expressed as f(x), its differential coefficient is often expressed as f'(x), i.e. d is denoted by f'(x). When this notation is adopted, f'(x) is often called the first derived function of f(x). 12 DIFFERENTIAL CALCULUS. [CHAP. I. It will often be convenient to draw a graph of f'(x) as well as a graph of f(x). The two curves have a great many interesting relations as we shall see later on. The graph of f'(x) is often called the first derived curve. Another notation sometimes used instead of l is Dy. It is dx an incomplete notation, but has the advantage of brevity. 14. Summary. If I -J() y + dy =f(x + dx) '. dy=f(x + dx) - f(x) or dfc (x) =f(x + dX) - f(x). This is the differential of y, i.e. its infinitely small increment when x receives an infinitely small increment dx. The differential coefficient, or gradient, of y is the ratio of dy to dx in the limit when both are diminished to zero. It is equal to the limiting value of f(x + dx) -f(x) dx or, as it is often put, of f( + h) — /() when h (i.e. dx) = O.. It is expressed in the various notations d-, f(x), Dy, and sometimes even y1 or y'. We may also write, if we please, df(x) dz ', or Df(x), but these expressions are cumbrous. The differential relation may be written df(x) =f'(x)dx, and this explains the reason why f'(x) is called the differential coefficient, since it is the coefficient of the differential of x in the value of df(x). 15. Classes of functions. Functions of any variable x are divided into two great classes, algebraic and transcendental. Such functions as x2 - 3x + 2, 3x - 7 -/(a2- 2), 5X2 -6-2' are called algebraic, the term being ARTS. 13-15.] CLASSES OF FUNCTIONS. 13 used for all functions expressed in powers of x either whole or fractional, and powers and roots of such expressions. All other functions, such as sin x, tan x, ca, log x, sin-lx, are called transcendental. Many of them can be expressed in powers of x, i.e. algebraically, but only by means of an infinite series of such powers. We may also divide functions into single-valued functions, and many-valued functions. For example, x2- 3 + 2 is a single-valued function of x, since there is only one value of the function for each value of x; but ~/(2 - x2) is a two-valued function since every number has two square roots. So also sin x is a single-valued function, but sin-lz is many-valued, there being an infinite number of angles having a given sine. There is also another important method of grouping functions, namely, into continuous and discontinuous functions. For example, the sum of the first x whole numbers is a discontinuous function, since it has no meaning when x is a fraction. Curiously enough, however, the algebraic expression which gives the sum of the first S numbers, namely -x(x+ 1), is a continuous function of x. So that here we have a case where the graph of the algebraic expression representing the function is continuous, though only isolated points on it have reference to the actual function under consideration, viz. the sum of the first x integers. The student should draw this graph by taking different values of x, and plotting. He should make a tabular list of values, as shown, and then plot on a convenient scale. The graph is shown in the adjoining figure. x, j( + 1) 1 1 2 3 3 6 4 10 5 15 6 21 7 28 8 36 9 45 10 55 4-:0 ^3 50 40 so 10 -.n O 2, 4a t6 IU Scale of x 14 DIFFERENTIAL CALCUJLUS. [CHAP. I. The easiest way is to use squared paper, which can be obtained at a cheap rate from Messrs. Partridge & Cooper, Fleet Street, London, E.C., or other wholesale stationers. Another example of a discontinuous function is the product of the first x numbers, known as factorial x, and denoted by j x or x!, whose fundamental properties are I x= x- 1, and I- 1. Here we can draw ordinates for each whole value of x, and we can draw a fair curve through the ends of those ordinates, but we cannot thereby interpret the meaning of the intermediate ordinates of this curve. It is possible, however, to find an expression for factorial x which will be continuous, i.e. will have a value for all intermediate values of x, and which will satisfy the above fundamental laws. The curve, drawn as above, will be the graph of this continuous function, and the isolated points on it corresponding to the integer values of x will represent the product of the first x numbers. This continuous function is of great importance in some kinds of theoretical work. Hence these two examples of apparently discontinuous functions turn out to be, in reality, examples of continuous functions, to isolated values of which we, as a rule, pay exclusive attention. There are many such functions, e.g. the number of spherical shot which can be piled into a pyramid with a square base is such a function. 16. In general, the functions with which we deal will be continuous functions, except for occasional isolated points of discontinuity such as when the function becomes infinite for some finite value of x, as in the case of y = — I- of which the student might usefully draw the graph. He will find the ordinate is positive when x is less than 1, becoming infinitely great when x = 1; then it suddenly becomes negative, and, as x increases, y diminishes in numerical value till finally it becomes zero when x is infinitely great. There is therefore a point of discontinuity when x= 1, but elsewhere the curve is continuous. But points of discontinuity in an otherwise continuous function are an entirely different thing from the discontinuity ARTS. 15, 16.] CLASSES OF FUNCTIONS. 15 of a discontinuous function; the graph of a continuous function is a continuous curve, and its points of discontinuity are merely interruptions to the continuity of the graph, whereas the graph of a truly discontinuous function would consist of isolated points. Such would have been the graph of factorial x if the continuous function whose graph passes through the special points corresponding to the integer values of x had not been discovered. EXAMPLES. 1. The sum of two numbers being 5, find the change in their product, as one of them increases from 0. Draw the graph. 2. Trace in a diagram the values of the reciprocal of a number. 3. Trace the changes in value of the sum formed by adding together a number and its reciprocal. 4. Represent in a diagram the values of 21' for different values of n. 5. Draw the curve y =sin x from 0 to 2rr. [Take horizontal distances representing the values of x, and erect ordinates proportional to the sines of these angles.] 6. Draw the curve y=cosx from 0 to 27r, and show that it is the 7w 5w same as the curve y=sinx from 2 to 5 -7. Draw the curves y =tan x, and y =sec x, from 0 to 27r. 8. Draw the curve y=log1ox. 9. Represent in a diagram the connection between the Fahrenheit and centigrade scales of temperature. 10. Represent by ordinates the values of the coefficients of the different powers of x in the expansion of (1 + x)8. 11. Represent in a diagram the work done by a constant force of 10 lbs. weight, as its point of application moves through varying distances in the direction of the force. 12. Represent in a diagram the force required to stretch an elastic cord whose unstretched length is 6 feet, being given that the force increases uniformly in proportion to the stretch, and that a 10 lb. weight would double the length of the cord. 13. Show that the area between the graph of the last function and the axis of x and the final ordinate represents the work done in stretching the cord. Also draw a graph of the work done. 14. The strength of a beam is proportional to its breadth and the square of its depth. Show the changes in the strength of a rectangular beam of given area of cross-section, as its breadth changes. 16 DIFFERENTIAL CALCULUS. [CHAP. I. 15. Trace the changes in strength of the beam with change of breadth, if the perimeter of the cross-section is constant, and equal to 20 inches. 16. A rectangular beam is cut from a cylindrical log of wood 1 foot in diameter. Trace the change in its strength according to its width. [Taking the breadth (b) and the depth (d) in inches, we have b2 + d = 144, and the strength is proportional to bd2;.'. the strength is proportional to b(144- b2). We have therefore to draw the curve y-=x(144 -x2) from x=0 to x= 12.] 17. Find the dimensions of the above beam when the strength is a maximum. [We have to find for what value of x the ordinate of the above curve is the greatest. This will be the required breadth; and the depth can be obtained from the equation b2 + d2= 144.] 18. Find the dimensions of the beam of Example 15 when the strength is a maximum. 17. The last two problems are of a kind whose solution is most readily effected by the methods of the differential calculus, for evidently the maximum values required will be the ordinates of the respective graphs at points where the gradient of the graph is zero. The points can be approximately discovered by careful drawing of the graph, but they could be more rapidly found if we knew how to find an expression for the gradient at each point, and then solved the equation obtained by putting the gradient=0. There are many problems requiring a knowledge of the gradient of a function, or, as it is generally called, the differential coefficient of the function. The first thing, therefore, to do is to establish methods by which we can with the minimum of trouble find the differential coefficient of any function. To do this we shall first find the differential coefficients or gradients of various simple functions, and then show how to find the gradients of more complicated expressions, which will always be found to be mere combinations of the simpler forms. The process of finding the gradient is called differentiating the function, a loose expression which is only justified by its comparative brevity. We have already discovered that, if y =x2, = 2x, and have seen that we can write the dx result in the form dcy= 2xdx, or d(x2) -2xdx; i.e. the differential of x2 is 2xdx, and its differential coefficient, or gradient, is 2x. ARTS. 17, 18.] DIFFERENTIATION. 17 Our next work will be to find the gradient in the case of any power of x, and after that, to establish general formulae by which the differentiation of more complicated functions of x is very much simplified. 18. Differentiation from first principles. We may, however, find the gradient at any point of either of the above curves without making use of any of the formulae which will be established in the next chapter, and so, by finding when the gradient is zero, solve the problems in the above examples 17 and 18. Indeed, we may in this way find independently the differential coefficient of any function we please. The method is instructive, though much more cumbrous than the systematic methods of the calculus, to be subsequently explained, by which the differential coefficients of more or less complicated functions are deduced from those of a few simple functions, independently obtained once for all. Let (x, y) be the coordinates of any point P on the graph of question No. 16, so that y=144x-x3. Also let (x', y') be the coordinates of an adjacent point Q on the curve, so that y' = 144x' - '3. Now evidently the gradient of PQ is 8'-X, and the limiting value of this gradient when Q is brought close up to P will be the gradient of the curve at P. Now, by subtraction, iy- y = 144 (x'- x) - (x'3 - x3) ' ~Y — 144 - ('2 + ' + x), x, - x which in the limit (when x'=x) becomes equal to 144- 3x2, which is therefore the value of the gradient at the point P, whose abscissa is x. If, now, we find for what value of x this gradient is zero we shall have solved the problem in Question No. 17. The value is x = /48 or nearly 7. Hence the strongest rectangular beam which can be cut from the given log is one whose breadth is./48 inches. The corresponding depth is b/96 inches, so that it is approximately 7 inches broad by 10 inches deep. L.D.C. B 18 DIFFERENTIAL CALCULUS. [ART. 18.] Similarly, in the case of the curve y = (10 - )2, i.e. y= 100x- 20x2 + 3 which is the curve of Question No. 15, we find the gradient = 100 - 40x + 32 =(10 - )(10 - 3x), so that the gradient is zero when x equals either 10 or 3-. The value of the function may be either a maximum or a minimum when its gradient is zero: on consulting the diagram, it becomes evident that the maximum strength is obtained when the breadth is 3: inches. The depth is then 62 inches, i.e. is twice the breadth. This is the answer to Question No. 18. EXAMPLES. Find, from first principles, the gradient at any point of each of the following curves: 1. x2 +y2- -2. 2. x2+y2 + Ax +By+ C 0. 3. y2=4ax. X2 y2 a2 4. 2 + Y1 5. y —. 6. y x3, drawing the curve between x= ~ 2. 7, y2 = 3, drawing the curve between x -0 and x -2 CHAPTER II. DIFFERENTIATION OF ALGEBRAIC FUNCTIONS, AND GENERAL METHODS OF DIFFERENTIATING. 19. 'To find the difterential of xo. Let y = x", then y + dy = (x + dx) d. y = (x + dx)" - x". Expanding (x + dx)Y by the binomial theorem,* the first term will be Xn, which will cancel the term -x'", and the second term will be n.x*-'.dx which is an infinitesimal of the first order, while the other terms will contain higher powers of clx, and will therefore be infinitesimals of higher orders than the first, and therefore negligible. Hence, keeping only the infinitesimal of the first order, dy = nx'2dlx, i.e. d(xn) = nxn-dx, and the gradient, or differential coefficient, is nx'.This is true, whatever is the value of n, so long as it is a constant. Thus d(x3) 3X2dx, d(x'~) = 10xlOcx, 22x2 d(x,/x) = dcx) - 5 - %Ix. dx,,z)= (z2) = 325 = x2v/ - 1), * The Expansion is (x + dx)n= + n - ldx + 1 - d)2 +.... 20 DIFFERENTIAL CALCULUS. [CHAP. II. dI =d(-1)= -I x-)- d:X d? (=d(X-3)= 3. x-az=X -, and so on. EXAMPLES. 1. Find the differential coefficients of the above functions from first principles (see Art. 18). In the case of,/x, having obtained y'-y _ /x' - a yt --; X -X we have to divide numerator and denominator by,/x'-'/x, and then simplify by putting x'=x in the result. Similarly in the case of x/zx. The others will present no difficulty. 2. Find the differential coefficient of rX from first principles. [NOTE. The fraction to be simplified may be written in the form Z/3 _ 13 1 z -, where z=x. Then it is obvious that z'- is the factor to be cancelled.] p 3. Find the differential coefficient of xz from first principles, where p and q are positive integers. 20. Before going on to other functions it will be usefil to see how far-reaching the use of this formula for the differential of x" is, and how many difficult expressions can be differentiated by means of it, with the help of the following general theorems which are applicable to all functions. (i) If u is any function of x, and a is any constant, the differential of au = a.dz, i.e. d(au) = ache. For, if y = au, y + dy = a (u + du);. by subtraction, dy = a.du, i.e. d(au) = adu. For example, d(7rr2) = 7rd(r2), as we have already seen. (ii) If t, v, ZU are any functions of x, the differential of their sum is equal to the sum of their separate differentials, i.e. d (u+v+w)=du+dv+dw. ARTS. 19, 20.] ALGEBRAIC FUNCTIONS. 21 This theorem is self-evident, and both it and the preceding theorem would be equally true if the increments of the functions were not infinitely small. (iii) The differential of a constant is = 0. This also is obvious, as it merely expresses in the language of the calculus that the quantity is constant. Thus, if a is a constant, d(a)=O. NOTE.-Generally, if we use letters to denote constants, and also letters to denote variables, we shall use the early letters of the alphabet, such as a, b, c, or 1, mr, n to denote constants, and the later letters such as x, y, z or t,, v, w to denote variables. This is merely for convenience, in order not to have to state always which quantities are variable, and which are constant. Example. If y = 2x3 - 3x2 + 5x - 7 dy= 2d(xs3) - 3d(x2) + 5dx - O by the theorems just established. Hence, since dx l= nx'-ldx, it follows that dy = 6x2cdx - 6xdx + 5dx = (6 2 - 6x + 5)dx;. =6 _- 6x + 5. dx We will now take a harder example, first premising that the use of x as the principal variable is merely a matter of convenience in establishing our theorems. Any other variable would do equally well, e.g. d (u) == nu"-du where u may be any variable, and may itself be a function of some principal variable x. Example. Find the x-gradient of (3x2 - 2x + 5)3. Let y= (3x2 - 2x + 5)3, then, treating the expression in brackets as a single quantity, we have dy=3(3x2- 2x + 5)2. d(3x2- 2x + 5), on the principle that d(z3) = 3z2d, where z 32 2x + 5; and now, simplifying the last factor, we have finally dy=3(3x2- 2x + 5)2(6xdx - 2dx) and = 3 (3x2 - 2x + 5)2(6x - 2) 6a(3 - )(3x2+ =6(3x- 1)(3x2- 2x+5)2. 22 DIFFERENTIAL CALCULUS. [CHAP. II. We could have differentiated this expression by first cubing out, and then differentiating each term separately, but the process would have been intolerably long, and the form of the result would have been less useful than the factorized form obtained by the simple artifice of initially treating 3x2 - 2x + 5 as a single quantity. Again. Find the differential coefficient of 4/(3x2 - 2x + 5). Let y = /(3x2- 2x + 5). Now d (Jz)=d(z) =; d, (32 - 2x + 5). 2=,/(3X2 - 2x + 5) (6x -2) dx 2^/(3U2 - 2x + 5)' in which 2 is a factor which can be cancelled; and, finally, dy 3x- 1 dx ^/(3x2 - 2x + 5)' 21. In the following examples we shall sometimes denote a function by y, and sometimes by f(x). The corresponding notations for the gradient of the function are, as before stated, dy or Dy, and f'(x), respectively. The student may in every case use which notation he prefers, but he will probably find it best to tuork with the single letter y, and then to please himself as to the form in which he states the final result. EXAMPLES. 1. If y=x(10 - x)2, find Dy, and find for what points the gradient is zero. [See Examples 15, 18, p. 16.] 2. If f(x)=x(144 -2), find f'(x), and find the values of x for which f'(x)=O. [See Examples 16, 17, p. 16.] 2 3. If f(x)=3Jx - 5xJx + 2, find f'(x). Xn dy _ xn-1 4. If y =-, prove that - =, — _ X2 X3 5. If y=ao + alx+ a2 + a — +..., prove that x2 Dy = a + ax + a32 +..., ARTS. 20, 21.] ALGEBRAIC FUNCTIONS. 23 6. Draw the curve y=x3, and also its first derived curve, between the limits x = ~ 2. 7. If y2=x3, find Dy in terms of x. Draw the curve between x=0 and x =4, showing carefully the slope of the curve at the points corresponding to these extreme values of x. 8. Draw the curve y = 2x- 32 + 7x - 2 between the limits x= -2 and x= +3, taking the vertical scale = - of the horizontal scale. 9. Draw the curve y = 6x2 - 6x +7, using the same vertical and horizontal scales as in No. 8. 10. If f(x)=( (2 2x)2, find f (x). Draw the graph between x= - 1 and x =3, finding its gradient at the points where x= - 1, 0, 1, 2, 3 respectively, i.e., find f'(- 1), f'(O), f'(l), f'(2), and f'(3). Draw also the graph off'(x). 11 1 11. Find Dy, when y = 1 - 2+3 12. If 4x2+9y2=36, find dy 13. A ball is thrown up with a velocity of 50 feet per second from the top of a tower 100 feet high. Its height above the ground in t seconds is y=100+50t —16t2. Plot the curve of displacement from the start to when the ground is reached. 14. In the equation of last question find dy, which will be the dt' velocity at each instant, and find at what time the velocity is zero (when the ball will be at its highest point). Plot the diagram of dy velocity, viz., y1=-dy, taking values of t as abscissae and values of yl as ordinates. 15. Find d- in the following cases: dx (1) y=(32 - 5 +6)-2. (2) y= 26 2 (3) y =/(x2- 3x + 2). f4) y — 1 X +J=(z- 1' (5) 3y + 5x= 15. 16. Show that the graph of an equation of the first degree in x and y is a straight line. 24 DIFFERENTIAL CALCULUS. [CHAP. II. dy [Take the equation Ax + By+ C=0, and show that is the same for every point. ] Draw the line 2x - 3y + 6 = 0, and find what its gradient is. 17. Find the gradient of the parabola y = Ax2+Bx + C at the point where x=p, and show that it is equal to the gradient of the chord joining the points on the curve where x=p~q. 22. The only algebraic expressions which at present we have not learned how to differentiate systematically are products, such as (2x2- 3)(5x - 7)3, and quotients, such as 2X2 - 3 22 —7. We could find the differential of the former by 5x- 7' multiplying out and then differentiating each term separately, but it would be a laborious task, and this would moreover be rendered more difficult if any of the factors were raised to fractional powers. A quotient could not without very heavy work be differentiated except by one or other of the methods now to be explained, which are therefore of great importance. 23. Product. If u, v are two functions of a single variable x, find the differential of uv. Let y = uv. Let x receive an infinitesimal increment dx, causing u and v to receive the respective increments du and dv. Then the increased value of the product is given by the equation y + dy = (t + du)(v + dv);. dy = (t +u)(v J+ dv) - uv = vdu + udv + du. dv = vdu + udv in the limit, since du. dv is an infinitesimal of the second order being the product of two infinitely small quantities. Hence d(uv)= v du+u dv. 24. This important rule is the rule for finding the differential of a product. The student should observe that if v is a constant, d(uv)-=du, which is in accordance with Rule (i) Art. 20, and similarly, if xt is a constant, d (uv) = udv. Hence the complete differential of the product when u and v both vary is the sum of the partial differentials obtained on the ARTS. 22-25.] DIFFERENTIAL OF PRODUCT. 25 supposition that first one and then the other of the two factors is constant. If we divide by dx, which we have supposed to cause the increments of the functions, we have, the equation between the differential coefficients, viz.: d (uv) din clv dX dx +V 21-.dx which may be written D(uv)=vDu+uDv. For example, let y=(2X2 - 3)(5x - 7)2. Here din=4xdx; Du-4x, and clvA=3 (5x - 7)2d(5x - 7) 15(5x - 7)2dx Dv =15(5x - 7)2; Dy Ox - 7)3.4x + (2x2 - 3).15(5x - 7)2 (5x - 7)2{20X2 - 28x i- 30X2 - 45} (5x - 7)2(50x2 -'28x - 45). 25. The student should carefully go through this work, and compare it with the result of multiplying out the factors of y and then differentiating. Not only would this latter method be much longer, but it would lose the great advantage of showing that (5x - 7)2 is a factor of the result, which is an important fact worthy of attention. Indeed, always, if any factor in the expression for y occurs raised to the zth dyl power, the same factor will occur also in raised to the 11 -_ 1th power, (See Ex. 4, below.) EXAMPLES. Find cl-in the following cases: dx 1. y=(x - 3)4(2x - 7)5. A ns. (X - 3)3(2x - 7)4(18X - 58). 2. y=(3x -7x+2)8x2~ - 7). Ans. -(32 7x +2) ( 2- 7)'(i~4Sx~- 9 1X2 106 x~+ 147). 26 DIFFERENTIAL CALCULUS. [CHAP. II. 3. y=(2- 2) (7x- 3)-1 Ans. (7x - 6x + 14)(7 -3)-2 4. y -= -uv1. Ans.?u'v1-lv-l(mnv Du + u Dv). 5. If a volume v of a gas contained in a vessel under pressure p is compressed or expanded without loss of heat, the law connecting v and p is that p.vk is constant. Prove that for a slight compression or expansion the connection between the changes of volume and pressure is given by the equation kpdv + vdp = O. 26. If y consists of more than two factors, the preceding method can be used by grouping the factors into two groups: thus Let y = uvw = a. vw then dy = vw. d r + i. dvw = vw. du + u (wdv + vduw) = vwzd + uwdv + uvdw, and Dy = vw De + uw Dv + uv Dw; and similarly in the case of more than three factors. Of course the labour of differentiating in such a case is necessarily heavy, however systematically it may be done. Indeed in the case of simple factors, not repeated, it is probably better to multiply out first and then differentiate. The student might test this in the cases of (x - l)(x-2)(x-3) and (x- 1)(x- 2) (x- 3)(x- 4). 27. Quotient. If y=u+du U +ddy= Vlu - Cd V vdn t udv 2 -- -, in the limit. V2 ARTS. 26-28.] DIFFERENTIAL OF QUOTIENT. 927 (ji) =v du-u dv \v/ V2 d(Y vdf-4 -and *'dx which may be written D(\)=vDu-uDv The student should also obtain this result by writing the fraction in the form uv'1 and differentiating by the rule for differentiating a product. Example. Let, =2X2 -5x -7' dy (5x-7).4x-(2x2-3).5 dx (5x - 7)2 10x2-28x~15 (5x - 7)2 2w2 - 3 Again, let 2J (X 7)3 This is most conveniently differentiated as a product, writing it in the form y=(2X2-3)(5x-7)-3; y= 4x(5x - 7)-3 -15(2X2- 3)_(5x - 7)-4 4x(5x - 7) - 15(2x2-3) (5x - 7)4 45 - 28w- l102 (5x - 7)4 The student should work the same example by the Quotient formula. He will find that a common factor, (5x - 7)2, must be cancelled from numerator and denominator before the result is expressed in its simplest form. 28. Frequently the student might find it adv'antageous, if a factor is repeated, in the case of either a product or a quotient, to use the easily proved formula D (urn VI) = um'lv-l (mv Dzu + nu' Dv). 28 DIFFERENTIAL CALCULUS. [CHAP. II. This was given as an example to be proved in No. 4 of the last set of examples. 29. The student ought now to be able to differentiate any algebraic expression whatever. He cannot yet differentiate transcendental expressions, though in many cases he can do something towards it. E.g. we know that (sin'x)2 + (cos x)2 = 1. 2 sin x.d(sin x) + 2 cos x.d(cos x) = 0. This step is obtained by treating sin and cos x as single quantities, like u2 + v2=1. We do not yet know how to express d(sin x) and d(cosx) in terms of dx. This and other formulae will be the subject of next chapter. Meanwhile we may show one class of problems which may be solved by the help of differentials. 30. Approximations by help of differentials. When a variable quantity x receives an actual increment, the approximate change in any given function of x is given by the value of its differential, provided the change in x is small. Thus, to find (40 1)2 we can use the formula (x+dx)2-=x2+2xdx or (x +h)2 =x2+2xh: which only differs from the true value by h2. Hence (40 1)2=1600+8= 1608, approximately. The true value is 160S-01. Again, the area of a triangle having two sides a, b, and included angle C, is ~ab sin C. If each side is increased by a short length h, the angle being unchanged, the increase in area is approximately ~ (adb + bda) sin C, where da and db are each equal to h, i.e. the increment of area = 1 (a + b) sin C, approximately. Again, evaluate 403 400 1728 1725' From the formula d u)-=- - -d \ v/ v2 we have, approximately, the required difference 3(1725-400) 3975 (1725)2 (l725)2' 3975 The true difference is 12 17 1725 x 1728' ARTS. 28-31.] RECAPITULATION. Again, evaluate 3x2 - 7x +5 when x =302. We have, approximately, f(x + dx) =f(x) +f'(x)dx 3x2 - 7x + 5 + (6x - ) dx...if we take x=3, and dx= 02, we find f(3'02)= 11 +11 x '02-11'22 approximately. Tile true value is 11 2212. Such examples may be multiplied indefinitely. We shall use this method in Chapter VI. to approximate to the roots of equations. NOTE.-If the student prefers to restrict the use of dx to the case when the increment of x is infinitely small, he can use a single letter such as h, above, for a finite increment, or he may use 8x for the finite increment. Similarly k, or 8y, may be used for a finite increment of a variable y, the term dy being kept for the infinitely small or vanishing increment. But in practice the distinction is unnecessary so long as the increments 8x and 8y are small enough for their squares and higher powers to be negligible. 31. Recapitulation. d(constant) = 0, d(au) a[(hin, d(u +-v + r +..) = drt + (dv12 + c..,, d(uv) = ivdi + iud'v, du vdu - udv V~ V2 d (u') = ie 1- du. Additional formula, sometimes useful, -d (ndu")m-i-vl IX dv\ WV/ m = U\ - mv - + nit; when m and n are positive or negative constants. NOTE.-In any of the above, if we wish to use differentials, we employ the letter d; if we wish to express differential coefficients, we write either D, or d divided by dx, as shown in the last formula above. 30 ~DIFFERENTIAL CALCULUS. EXAMPLES. 1. Find dyin the following cases: dx (2) y=(x +a)\/I(x -a). (4) y =(I + X)(l~ 4X2~,X4). (5 =(1+ x2'33 ( I(3)3) 2. If two functions of x are identical, their differential coefficients are also identical. The student may illustrate this by the following two examples (1) Differentiate both sides of the identity (a + 4)= a"~nt nal n - 1)2a X. n Do the same also with a special value of n, say 9?,4. (2) Prove the rule for differentiatin-g a product by making nse of the important identity qV= J(qt j v,)2 _ j(u-t _V)2, and assuming the rule for differentiating a square. 3. Differentiate a+x 4. Differentiate the result of (3), and thenr again the new result, and so on four times. 5. Ev aluate (approxim-ately) x. 4X2 -_llx.+.32, when x= 10-32. 6. Evaluate the same fnnction, when x =2T1 7. Differentiate y 2 x a 8. Differentiate (x -3)2(X~ l)3(2x - 7). 9. Find dywhen X2 y2-a2(x2 _y2). 10, Find 6ywhen y -f (z) and z=px) 11. Find 0-when y(z-2z~,5)-! and 7-=4x2- 8x +7. dyx3 12. Find - when Y/ z2 and z -— " CHAPTER III. DIFFERENTIATION OF THE CIRCULAR FUNCTIONS AND THE INVERSE CIRCULAR FUNCTIONS. 32. Before finding the differentials and differential coefficients of these functions, it is necessary first to establish the important, though simple, theorem that When an angle is very small, its sine and tangent are practically equal to each other and to the circular measure of the angle. By being equal, is meant that their ratio is ultimately one of equality. (See Art. 2.) sin 8 It is easy to prove that t- = 1 when 0 is ultimately made tan 0 infinitely small. sin 8 For, always, ta-n = cos 0, and, when 0 = 0, cos 0= 1, which proves the proposition. With regard to the ultimate equality of 0 with both sin 0 and tan 0, a little consideration will convince the student that the circular measure of a small angle 0 is intermediate in value between sin 0 and tan 0, and therefore when sin 0 = tan 0, i.e., when sin 0: tan 0= 1 by reason of the smallness of 0, it follows that 0 is equal to both of them. If a student turns to a table of values, he will see that these three quantities are very nearly equal when the angle is as much as 1~ or even more. The theorem just proved may be rendered clearer by a figure. Let POA be a small angle whose circular measure = 0. Then if with O as centre and radius r an arc PA is drawn, cutting arc PA the arms of the angle in P and A respectively, 0 =; and 32 DIFFERENTIAL CALCULUS. [CCHAP. III. if PN is drawn perpendicular to OA to meet it in N, and AT is drawn perpendicular to OA to meet OP produced in T, PN TA sin 0=-, and tan =. sin 0 0: tan 0=PN: arcPA: TA. pT O NA Now make 0 infinitely small, which, for our purpose, is best done by making r infinitely great, while T remains fixed. We see that PN, the arc PA, and TA ultimately coincide, so that they are ultimately equal. Therefore sin 0, 0, and tan 0, which are proportional to them, also become ultimately equal, which again establishes the required proposition. Cor. This shows that, to the first order of small quantities, the lines PN, TA, and the chord PA, and the arc PA may all be treated as equal to each other, when the angle is small. This will be a great help in the geometrical treatment of problems dealing with small angles. We also see that ultimately the chord PA is perpendicular to both the radii OP and OA, being indistinguishable from the arc PA. 33. To find the differential of sin x; i.e., if the angle x receives a small increment, dx, find what is the corresponding increment in the sine of the angle. Let y = sin x, then y + dy = sin (x + dx),. dy = sin (x + dx) - sin x, = 2 cos (x + cdxz) sin (1dx), from the formula A+B. A-B sin A- sin B =2 cos -.- sin But sin -dxc is ultimately equal to Idx, by the above theorem. 2 2~ N~~V —VYJ r-. - ARTS. 32-34.1 A 34 IRCULAR FUNCTIONS..33 if the angle is in circular measure. Also cos (x +I dx) is ultimately equal to cos x; finally dy =2 cos x. 1dx = cos x.dx, i.e. d sin x=cos x. dx ~isin,c- and d dinx CosX 34. To find the differential of cos x. We will establish this in several ways, each of which is instructive. (1) Lct y = coSxX, then dy = cos (x + dx) - cos x, 2sin (x + Idx).sin lcdx, -sin x dx, i.e. d cos x=-sin xdx. This first method is differentiation from first principles. The other two methods are based on previous theorems. (2) Let y = cos x, 7rll.~, i Yor:r.:,Sin;-3 _ 1 by the formula of last article, = sin X.d( ) = -sin xzdxr. (3) cos2 X = 1 - sinx, 2 Cos x d Cos x = - 2sin x.d sin x, - 2 sin x.cos x.dx, by last article, d cos x = - sin x dx, as before. Expressing this in the form of a differential coefficient, we have d cosx sinx. dx L.D.C. e __ __I_____ 34 DIFFERENTIAL CALCULUS. [CHAP. III. 35. The formulae for the differentials of the sine and cosine are so important that we will also prove them geometrically. The student should draw his own figure and carefully go through every part of the work, as it is important that he should be able to deal geometrically with infinitely small quantities. In the figure the increments have to be made of a finite magnitude, but they must be imagined to be infinitely small. 0 NMA Let POA =x, and QOA = x + dx, so that QOP = dx. If the angles are in circular measure, dx- Q. r PM. QN Now sin x =, sin ( + dx) QN 9' 9'; d sinxQN-PM QR.'. dsin x = - = - r r QR QP QR - e =__. dx. QP' QP Now - ultimately becomes=cos NQP, and this angle beQP comes equal to the angle at 0, = x;. ultimately equals cos x, QP and, therefore, d sin x = cos x. dx. ON - OM MN RP Similarly, d cos x = - r r' r RP QP r -. _-= sin x.dx. QP r ARTS. 35, 36.] CIRCULAR FUNCTIONS. 35 Interesting examples illustrating the use of the differential coefficients of the sine and cosine of an angle will be found in the articles on uniform circular motion and simple harmonic motion at the end of this chapter. (Articles 57-60.) 36. Gradients of the curves y = sin x and y= cos x. The gradient of the curve y= sinx is given by the equation = cos x. This means that the numerical value of the gradient =cos x, i.e. =cos x measured in vertical units - 1 on the horizontal scale. Moreover, we only proved that dy= cos x dx on the assumption that the angles were measured in circular measure, hence 1 on the horizontal scale must mean 1 radian, which in degrees is 180'+ r, i.e. 574 degrees, approximately. If therefore we turn to the graph of sin x, we shall be able to construct the tangent at each point P by drawing PU horizontally to the right, of length representing 574~ on the horizontal scale, and erecting a perpendicular UT of length equal to cosx on the vertical scale (drawing UT upwards if cos x is positive, downwards if cos x is negative): PT will be the tangent at P. The construction is left to the student, who should verify its truth by drawing the graph of sin x (from 0 to 7r will be sufficient) on a large scale, and actually constructing the tangent by the above method at several points of the graph. By this means he will get a geometrical feeling of the truth of the theorem that d sin x - = cosx. d - He may also obtain a numerical illustration of the truth of the theorem by taking two angles A, B nearly equal, and forming the fraction whose numerator is sin B - sinA, and whose denominator is the circular measure of B-A. On dividing out, so as to express the quotient as a decimal, he will find it is equal to the cosine of some angle intermediate between A and B. Take, for example, A =37~ 20' and B= 37~ 21', so that the sin 37~ 21'- sin 37~ 20' -0002313 fraction circular measure of ' 7950. circular measure of 1' ' 0002909 36 DI FFERENTIAL CALCULUS. [CHAP. III. We can obtain the quotient to only four figures correctly, even if we have used a seven-figure table, since the numerator and denominator do not contain more than four significant figures, but this degree of accuracy is enough to show that the quotient lies between the cosines of A and B, which are ~7951... and -7949.... Of course, our fraction gives the gradient of the chord joining the two given points, and this is obviously parallel to the tangent at some point of the graph intermediate between these two points. We have therefore verified that the gradient at this intermediate point is numerically equal to the cosine of an intermediate angle. Similarly we might verify that the gradient of the chord joining two points on the cosine graph is numerically equal to the sine of an intermediate angle, but is negative (if the angles are acute) because the cosine gets smaller as the angle increases, cos B - cos A. and therefore the fraction - is negative. 37. If we combine the new formulae with the results of the preceding chapter, we shall be able to differentiate any function containing sines and cosines and algebraic expressions. For example, find the differential coefficient of sin2x cos x. Let y = sin2cos x;. y = cos x.d(sin2x) + sin2.d(cos x) = os x. 2 sin x d ( x) +sin2X ( - sin x dx) = 2 cos2x sin x dx - sin3x dx; d. -= s (2co sin(2 os - sin2). An interesting illustration of the formulae will be furnished by taking an identity such as sin 3x =3 sin x - 4 sin3 and differentiating both sides. Since the two sides are identically equal, it follows that their differentials must be equal, and therefore also their differential coefficients must be equal, and the relation between them will form a new identity. Thus: sin 3x = 3 sin x - 4 sin3x;.. cos 3x d (3x) = 3 cos x dx - 12 sin2X cos x dx; ARTS. 36-38.] CIRCULAR FUNCTIONS. 37. 3 cos3x dx= 3 (cos x - 4sin2xcoSx) dx; cos 3x=cos x - 4 sin2x cos x - 4 COS3X - 3 cos x. Examples for practice will be found at the end of this chapter. We will now proceed to find the differentials of the other circular functions and of the inverse circular functions. 38. To find the differential of tan x. Let y = tan x; dy = tan (x + dx) - tan x tan x + tan dx 1 - tan x tand dx (I ~ ta22x) tan dx 1 - tan x tan dx Now, so far, we have made no use of the infinite smallness of dx. When we introduce this simplification, we may simplify the numerator by writing dx for tan dx (Art. 32), and the denominator is ultimately equal to 1. (Art. 2.) finally, d(l =(1 + tan2x)dx -=sec2xdx; i.e. d (tanx)=sec2x dx and dtanx- sec2x. dx ullodification of the above method. We might proceed as follows:dy =tan (x + dx) - tan x sin (x + dx) sin x cos (x + dx) ces x sin (x + dx) cos x - cos(x + dx) sin x cos (x + dx) cos x 38 DIFFERENTIAL CALCULUS. [CHAP. III. sin dx cos (x + dx) cos x dx,in the limit, = sec2x dx, as before. Both the above methods are based directly on first principles. We may also obtain the differential by a purely calculus method, thus, sin x; d tanxz =d -- cos x d sin xx - sndCos by Art. 27, Cos2x y cos2x dx + Sin2x dx cos2X dx - = sec2x dx. COs2x 39. To find the differential of cot x. The most instructive way will be by means of the identity cot x = tan - dcotx=dtanQx) =sec'2 f 7r, which reduces to -cosec2x dx, since dQ x) = - dx. i.e. d cot x= - cosec2x dx and D cot x= - cosec2x. The student may also practise-,finding the differential by cOs x means of the identity cot x = c., and using the quotient formula. ARTS. 38-42.j CIRCULAR FUNCTIONS, 39 40. To find d(sec 4) The easiest way is as follows: d sec x d(cos x)-' (-sin xdx) = SinlX dXI i.e. d seex seex tan x dx. 41. To find d (cosec 4) d (cosec x) d7 sec2- x) ie dcosec x - coseo x cot x dx. 42. Geometrical method of finding d tan x and d sec x in terms of dx. Let POA= XI QOA = x + dx. Draw PR perpendicular to both OP and OQ (Art. 32), cutting OQ in R. P o a A Then dx = P7if dx is in circular measure. 40 DIFFERENTIAL CALCULUS. [CHAPo III. QA PA QP QP RP Also d tan = -- -=- = - a a a RP a QP RP OP RP OP a = sec QPR.dx.sec x. But QPR is ultimately equal to x, being the complement of OPA, since OPR is a right angle;. d tan x = sec2x dx. Similarly, d sec x -, since OP= OR, QR RP OP w ~r or RP OP a = tan x.dx.sec x = sec x tan x dx. 43. The six circular function formulae are therefore d sin x = cos Cdx d cos x = -- sin x dx, d tan tx = sec2x dc d cot t' - - cosec2x dx, d sec = sec x tan x dx d cosec t: = - cosec x cot x dx. The above equations fall naturally into pairs, each of the functions sin x, tan x, sec x, which increase with x (in the first quadrant), being associated with one of the complementary functions cos x, cot x, cosec x which decrease when x increases (in the first quadrant), and whose differentials have accordingly negative signs. This association in pairs will materially assist the student in remembering the formulae. In addition to this complementary pairing of equations, there is also a natural association between the pair of ARTS. 42-45.] CIRCULAR FUNCTIONS. 41 functions which occur in the same equation; it will be found much better to remember d tanx as =sec2xdx than as dx sin x dx co and similarly d sec x = sec x tan x dx rather than cos-x, COS2 coS2X since see x and tan x form a natural pair of connected quantities, just as sin x and cos x form a natural pair. Similarly cosecx and cot x form a natural pair of associated functions. This natural pairing is well shown in the three important formulae: sinx + cos2 = 1, se2e - tan'2= 1, cosec2X - cot2 = 1, which are the three trigonometrical forms of Euc. I. 47. The student may, in fact, derive d secx from dtanx by differentiating the second of these three identities, and he may derive d cosec x from d cot x by means of the third, just as we have already derived d cos x from d sin x by means of the first. 44. Inverse circular functionso The differentials of the inverse functions sin-1x, cos-ix, etc., can be readily deduced from those of the direct functions. For example: 45. To find the differential of sin-lx. Let 0 = sin-lx.sin 0 =x. cos 0 d= dx..d dx d x cos 0 /(I -- x2) i.e. dsin- x-l — The latter form, with the surd, is ambiguous. The sign to be attached to the surd must be + or -, according as cos 0 is positive or negative. In the case of acute angles it is of course positive. 42 DIFFERENTIAL CALCULUS. [CHAP. III. 46. To find the differential of cos-lx. This can be found in the same way by putting 0= cos-1x, dx dx whence finally dO= - sin= -( - ); 7sin (9 /(1 - X2) dx i.e. dcos-1x=- x The ambiguity in the surd must be disposed of in a similar way to that of the last case, only in this instance the sign to be attached to the surd is that of sin 0. The following method is applicable in the case of angles in the first quadrant. The angle whose cosine is x is the complement of the angle whose sine is x, i.e. cos-lx + sin-lx =.. d(cos'x) + d(sinlx) = 0 i.e. the differential of cos-Ix is numerically equal to that of sin-lx, but of opposite sign. 47. The student should draw the graph of sin-z, i.e. the curve y=sin-lx, plotting the values of y in radians, i.e. in circular measure. He will see that y is a multi-valued function of x, and that for each value of x the gradients at the different points of the graph corresponding to the different values of y are alternately + and -, being numerically equal to (1 \/(1 - x2) If yo is the smallest positive value of y corresponding to any value of x, the other values of y for that abscissa are r - yO, 27r + y, 3r- y0, etc., all being included in the expression 7r + ( -1)"y0 where n is any integer, positive or negative. (The use of the factor (- 1)" is a very convenient mode of indicating that a + sign is required when n is even, and a - sign when n is odd.) It would be instructive to also draw the graph of cos-lx, which is merely the sin-ix curve brought down through a distance = on the vertical scale; or, which comes to the same 2 ARTS. 46-51.1 CIRCULAR FUNCTIONS. 43 thing, it is actually the same graph as sin-lx, but with a new origin and x-axis at a height =7r above the old one. 48. To find the differential of tan-lx, i.e. to find dO when tan O = x. We have sec2O dO = dx dO- dx dx since sec20= 1+ tan20. sec20 I1+$x2' ix., dtan-lx=-d I + X2, 49. To find the difl6rential of cot-1x. Proceeding exactly in the same way, we find dx d cot-lx=- d+x2 There is no ambiguity of sign in these expressions. 50. To find the differentials of sec1x and cosec-lx. If O=seex, sec=x; dx dO = see 0 tan O' i.e. ddsec-lxd= xV,(x2 I) Similarly, dcosec-1x=- dx X,,/(X2- 1) In these expressions there is an ambiguity of sign, the sign to be given to the surd being the same as that of tan 0. 51. Recapitulation. dsin0=cosO.dO I d cosO0 = - sin O. dOf.dtan O = sec2O. dO d cotO0= - coscc2O.dO 44 44 ~DIFFEIRENTIAL CALCULUS. [HA.I. rCHAP. III. d see 0 = sec 0 tan 0 dO d cose 0 = - cosec 0cot 0dOf d sinvI } dx d tan-'x) dx d dcot-'f 1~ d see-'x dx = dcosec-x,J~21 EXAMPLES. Find -in the ffive following examples: dx 1.y = cos2X + Sin'2X. 2. y = cos2X - sin'2X. 3. Y =tan 3x -cot 4x. 4.Ysin x -sin 3x 4 cos'x 5. y =sec x +tan x. 6. Obtain another identity by differentiating both sides of the identity sin 2x = 2 sin x cos x. 7. Differentiate both sides of the identity Cos 3x =4 cos3x - 3 Cos X. 8. Differentiate seC4X - 4 sec'x ~ 2 seclx. 9. Differentiate (3scx+ 4 sec2 '+ 8) tan x. 10. Differentiate sin-1(2x'2). 11. Differentiate x sin'lx +, (1 - x2). 12. If f(x) = (x2+ l) tanr'x - x, find f'(x). 13. If f(x) = tan-' X find f'(x).,,/(1- X2) 14. If tan Q =,,/(X + 1)- X, find dO 15. If (1~X2)CosO-1-X2, find dO 16. Draw the curve y = -lx sin 7r, showing that it touches the lines 2-+1 (alternately) when x = 1, 3, 5.., and finding where and at what gradients it crosses the zero line (y =0), and determining also the points where the gradient is zero. ARTS. 51, 52.] CIRCULAR FUNCTIONS. 45 2.rx 17. Draw the curve y =-sin —, showing that it oscillates between 2 x 2 the curves y= - (which should also be drawn). Find where and at what slopes it crosses the zero line, and at what points the gradient is zero. 2 18. Draw similarly the curves y = x sin 7rx, and y = sin 7rx. 2 7rx 19. Draw the curve y = - cos - x 2 20. Draw the curve y= xcos 2 52. Expansion of a function in powers of x. We are now in a position to determine two important series giving the values of sin x and cos x numerically in terms of the circular measure of the angle x. Before proceeding to these series, it may be well to illustrate what is meant by expanding a function of x in a series of powers of x. We know that (1 +x)5= 1 +5x+ 1 02+ 10a3+5 4+ 5. The right-hand expression is a series consisting of positive integral powers of x, the first term being of zero power, and the last of the 5th power. This series, then, is said to be the expansion of (1 + x)5 in positive integral powers of x. We could have determined this series by the help of the calculus in the following way. Any series consisting of positive integral powers of ll must be of the form a, + az +x a2x2 + ax +..., or, if we prefer so to write it, in the form x2 X3 ao + aqx + a%_+ a. +..., where, in either case, a0, a, a2... are numerical coefficients. For the method we are now going to illustrate the second form of the series is usually rather more convenient. What we shall do is to first assume that the given function (1 + x)5 is equal to this series. Then, since differentiating both sides 46 DIFFERENTIAL CALCULUS. [CIIAP III. of an identity leads to another identity, we shall, by differentiating, obtain a second identical equation. Repeating the same operation as many times as we want, we obtain a whole set of equations from which we may discover the values of the coefficients ao, a,.... The whole work is very simple as soon as the student has got into the spirit of it. It is as follows: 2 X3 Let (1 + ~) -= %+c +a + a3-+..,..........(1) Then, differentiating, we have X2 5(1 + )4C(1 +x)=aldx +a. dxdx+a32dx+..., and d(l +x)=dx,. dividing by dx we have X2 X3 5(1 + x)4 - a +a 2t + + Ca............(2) Now differentiate both sides of this identity, and divide by dx, * ~. 5. 4(1 +x)3a2+ax+ax 4+.............. (3) Repeating the operation again and again, we have X2 5.4.3(1 +)2=a3+a4+a5.................. (4) 5. 4. 3. 2(1 +x)=a4 +casx+......................(5) 5.4. 3. 2. l = a5+ +........................(6) 0 = a. + a7x +...................(7) Now since we are assuming that the values of the coefficients a0, al,... are to be such as will make all the above equations identities, it follows that the equations will hold for all values of x. Therefore, put x = 0 in each of them, and we find from (1), ao=1, (2), al = 5, (3), a =5.4, etc.; ARTS. 52, 53.] EXPANSIONS. 47 5. 42 5.4.3x8.'. finally, (1 + )5= 1+5x+ 5 + — +.- +... which reduces to 1 + 5x + 10x2 + 10X3 + 5x4 + X5. The student should also go through the work, assuming that (1 + x)5 = ao + alx + 22 +.... He will obtain the same result, with perhaps rather less labour, but the method above indicated shows the law of formation of the coefficients rather more clearly. He should also expand (l +x)", using either form of series. The expansion found will be a terminating one (ending at x') if n is a positive integer, but for other values of n it will consist of an infinite series. 53. Expansion of sin and cosx in powers of x. Now, let us apply the same method to expand sin x in positive integral powers of x. X2 8X3 Let sinx =aoalx + a2 a + a+ 3-...............(1) Differentiate time after time as we did above, and we get 9:2 X3 COS = a + + a,...,........ *.(2) - sin x= a+ a x+............................. (3) - cosx=a3 +a4x+...,......................,. (4) sinx=a4+...,...............................(5) and so on, the left-hand side repeating itself in the same order as before, viz., sin x, cos x, - sin x, - cos x,.... Now put x= 0 in each of these identities, and we get O=ao=a2 =a4=a6=..., 1=al = -a=a5 = -a=... X3 X5 ~'. finally, sin x- x - + -+ ~.. ad inf. L UL. 48 DIFFERENTIAL CALCULUS. [CHAP. III. We can expand cos x in the same way; but as a matter of fact we have already obtained cos x, for it is given in (2), and is 2 x4 cos =l- + -.... 54. We could have determined these series by two differentiations only. For equation (3) tells us that x2 sin x = - a2 - ax - a42 -.... Hence this series must be identical with the series x2 Cto + alx + a +...; therefore, by equating coefficients, we find that 2 = - (60 3 a= - a1, a4= - a2, etc.; further, by putting x=0 in (1) and (2) we find a%=0, and al = 1;. a2= a = -, a3, 4=0, etc. This determines all the constants, and gives us the series for sin x obtained above. 55. These two series, viz.: sin — =-3+-_ -.., X2 X4 cosX= -_+..., coS - 1 -S+ -,. 2 4 are very important. They are easily remembered: sin x contains all the odd powers of x, each divided by the corresponding factorial, and affected with alternate signs; cos x contains all the even powers (starting with zero power), each divided by the corresponding factorial, and affected with alternate signs. Also if the student differentiates the series for sin x he will obtain the series for cos x, and if he differentiates the series for cos x he will obtain minus the series for sin x, as of course he should. ARTS. 53-56.] EXPANSIONS. 49 cos x is called an even function of x, as its expansion contains only even powers of x, which is in accordance with the fact that cos (- x)= + cos x. sin x is called an odd function of x, since all the powers of x are odd in its expansion. This agrees with the fact that sin( - x)= - sin x, from which it is evident no even powers of x can occur in the expansion, for even powers could not change sign when - x is put for x. 56. We could use the above method to expand tan x and sec x, but we should only succeed in obtaining the successive terms by a good deal of labour because their successive differential coefficients become very troublesome. We can of course help ourselves somewhat by noticing that tan z is an odd function of x, since tan(- z)= - tan x; and by noticing that sec x is an even function of x. We shall return to these expansions in a subsequent chapter. We could obtain a few terms of tan x by dividing sin x by cos x, and a few terms of sec x by dividing 1 by cos x (using, of course, the series just found for these functions). This is left as an exercise for the student. EXAMPLES. 1. Find 3 terms of the series for tan x by dividing I's3 X5 X2 04 x -+I by 1- ~t+4 X2 X4 2. Find 3 terms of the series for sec x by dividing 1 by 1 - +. 3, Find two terms of the series for tanx by assuming it equal to x3 ax + a3 - +..., and differentiating three times. 4, Expand cos 2x in powers of x. [Merely substitute 2x for x in the series for cos x. ] 5. Expand cos2x and sin2x in powers of x. [Make use of the facts that cos2x - sin2 = cos 2x, and cos2x + sin2 = 1. ] x3 6. Draw the graph of x - -, and compare it with the graph of sin x, L-3 a?5 Make a closer approximation by adding the term A. L.D.C. D 50 DIFFERENTIAL CALCULUS. [CIIAP. TIIT. X2 7. Draw the graph of 1 -, and compare it with the graph of cosx. Make a closer approximation by adding the term, and then a closer approximation still by subtracting 6 57. Uniform circular motion. One of the simplest illustrations of the values of the differential coefficients of sin 0 and cos 0 will be found in connection with the motion of a particle travelling with uniform velocity along the circumference of a circle. If the radius is a, and the particle P has moved in time t from an initial position A through an arc aO, 0 being the circular measure of the angle subtended by the arc AP at the centre 0 of the circle, the velocity of the particle is v= -. The position of the particle can be indicated by measuring the base OM and the height MP of the right-angled triangle OMP. ^' A 0 M Denote OM by x, and MP by y; then x=acos0, y=asin0. The component velocities of P in the x and y directions respectively are d and dy dt dt Since x = a cos 0, it follows that dx. dO dt = - a sin 0 -T; dt dt ARTS. 57-59.] SIMPLE HARMONIC MOTION. 51 and since y = c sin 0, it follows that dy a cos dt do t, We can obtain these component velocities in another way, for it is evident from the figure that the component velocities of P are - sin 0 and v cos 0 respectively, and since v is constant, and equal to -t i.e. Vt=act, we have, on differentiating, vdt = adO, whence v = ad. dt' dO Hence, - v sin 0 = - a sin O-d, and v cos 0 = a cos 0O -, agreeing with the component velocities obtained by the first method, viz., by direct differentiation of the values of x and y. 58. If the uniform circular motion of a particle is looked at edgeways from a long way off, that is to say, if it is orthogonally projected on to a line in the plane of the circle, as for instance on to a diameter, the projected motion is a simple oscillation, the simplest kind of oscillation possible, the kind of motion possessed by a point on a musical instrument sounding a pure tone, as for instance a bead on a tuning fork; and it is therefore called a pure or simple harmonic motion, of which the properties can easily be laid down by considering the circular motion above. 59. Simple harmonic motion. If a particle P moves with uniform speed round a circle, and a point M moves along a diameter AA' of the circle with such variable speed that the line PM is always perpendicular to AA', M is said to have a simple harmonic motion. The radius, OA, of the circle is called the amplitude of the simple harmonic motion. Evidently the motion of M is a to-and-fro motion between A and A', the velocity being zero at A and A' and a maximum at the middle point 0, being then equal to the velocity of P. At any intermediate point M, such that OM = x = a cos 0, the velocity 52 DIFFERENTIAL CALCULUS. [CHAP. III. dx do of M = d-= - a sin dt, one complete oscillation being described while 0 changes from 0 to 27r; i.e. in the time during which P describes a whole revolution. This time, T, is called the period of oscillation, and is evidently equal to the length of the circular path described by P, divided by its velocity v; i.e. - v 60. Combinations of simple harmonic motions of equal periods. In the figure, when P moves uniformly round the circle, the points M and N move along their respective diameters with simple harmonic motion, the velocity of M being equal to the horizontal component of the velocity of P, and the velocity of N being equal to the vertical component of the velocity of P. Moreover, when M is at A or A', N is at 0, and when N is at B or B', M is at 0, so that one is a quarter oscillation ahead of the other, while the amplitudes and periods of their oscillations are equal. This gives us the following theorem: Uniform motion round the circumference of a circle can be considered as the resultant of two simple harmonic motions of equal period and amplitude along two intersecting lines at right angles to each other, so arranged that one is a quarter oscillation ahead of the other. If the periods are equal, but either the amplitudes are unequal, or one of the points M, N is not a quarter oscillation ahead of the other, the resultant motion will be not along a circle but along an ellipse. A point moving in such manner is said to have elliptic harmonic motion. EXAMPLES. 1. Draw the curves (1) x=acos 0, y =acos0, (2) x-acos, y =acos(O +18), (3) x=acos, y-=acos( + 36~), (4) x=acos0, y=a cos ( +54~), (5) x=a cos 0, y=a cos ( + 72~), (6) x=acos0, y=acos ( +90~). ARTS. 59, 60.] SIMPLE HARMONIC MOTION. 53 [NOTE.-Take a some convenient length, say 5 inches if the paper is large enough (i.e. 10 inches wide at least), and calculate the values of x and y for values of 0 from 0~ to 360~ at intervals of 18~. Then draw a network of horizontal and vertical lines through the points corresponding to the calculated values of x and y. It will be found that the same network serves for all the above curves.] 2. Show that the tangent to any of the above curves at the point where x=acosO is parallel to the line drawn from the origin to the point on the same curve whose abscissa is a cos( +90~). Test the accuracy of this result by trial on the diagram. 3. Draw the curves x=5cos0, y=3cos(0+a), for the following values of a, viz. 0~, 18~, 36~, 54~, 72~, 90~. [NOTE.-Form a similar network to that in No. 1, except that the distances of the horizontal lines from the origin will be 3, 3cos18~, 3cos36~,..., while the distances of the vertical lines from the origin will be 5, 5 cos 18~, 5 cos 36~,....] 4. Show that each portion of the networks indicated in connection with the above examples is described in the same time, viz., -AT; where T is the period of the complete motion. CHAPTER IV. THE EXPONENTIAL AND LOGARITHMIC FUNCTIONS ax AND log x. 61. The next differentials to be obtained are those of ax and log x. From what we have already done we know that dx will be a factor; the problem in each case will be to find the other factor. Let us first take ax. Let y = a then y + dy = ax+dx dy = a+ddx - cx = c(a~'a- 1). We must try to simplify the factor a d- 1. It is a function of a and dx, and it becomes infinitely small when dx is infinitely small, therefore we may assume that, like all the previous differentials, it contains dx as a factor. It therefore is ultimately of the form A dx (omitting all powers of dx above the first), where A is some function of a, and is therefore only a constant..'. d(a) = ax. A dx, where A is a function of a which has still to be determined. Since the value of A depends on the value of a, let us take a value (at present unknown) of a which will make A = 1. Denote this value of a by the symbol e, and proceed to find its value. It is evidently an important constant number, for the differential of ex will be simpler than that of any other exponential function. [ART. 61.] EXPONENTIAL FUNCTION S. The condition that A = 1 when a = e, gives us the equations dex = exdx and Dex = ex. Thus the differential coefficient of e" is equal to et. The value of e can be determined without much difficulty from this property of its differential coefficient. Expand ex in powers of x by putting Z2 X3 ex = a a+ alx + 2 a t +...,................(1) and differentiating both sides of the assumed identity just as was done in the cases of sin z and cos x, thus obtaining x2 Dex = a1 + ax + aps +.......................(2) But De=e", therefore the series (2) is identical with the series (1); a'. =1=0, a2= C, C 3= Ca, etc. By putting x =0 in (1) we find that a =e~=1, 1=. 1, a2=1, a3= 1,...;.finally e=i+x+X2 +. 3 e@= 1 + x + 1 — + 1. 3 This series is very important. It is known as the exponential series, ex being called the exponential function. If we put x= 1, in this identity, we find e= ++ = + + 1+ + This gives the required value of e. On working it out we find e= 2 718281828.... It can be calculated to as many figures as the patience and accuracy of the computer permit, but the important part of its value, not to be ever forgotten, is 2-7. After that the next eight figures, which are 1828 repeated twice (but then followed by other figures), may also be easily remembered. With this value of e, then, we know that de" = exdx. 56 DIFFERENTIAL CALCULUS. [CHAP. IV. Of course also, if u is any function of x, or any variable whatever, de" = edu.. Hence e is clearly an important number. We shall return to it directly, in connection with logarithms. We can now find out the value of A in the expression for d(ax), viz., a. A dx. For ax= ekx, where ek =a, so that k =logea, Hence d (a) = d (ekx) = ekxd(kx), i.e. d (ax) = cxkdx. Therefore A = k = log a, i.e. ax - 1 =loga dx. We have proved, then, that d (ax) = kaxdx and D (ax) = kax, where k = logea. It will be found that this constant k, is of great importance in connection with the function ax. EXAMPLES. 1. Draw the curves y=2x and y=3x. 2. Draw the curve y-ex, calculating the values of e2, e3,... either by taking e=2'7 approximately, or by help of logarithms. Similarly for e-1, e-2.... [Note that it lies between the two curves of No. 1.] 3. Show that the tangent at any point (x, y) on the curve y=ex cuts the axis of x at the point (x- 1, 0), i.e. at a point situate unit distance to the left of the ordinate of the given point. 4. Draw the curves y= e2 and y=e'x, and find in each case where the tangent at any point (x, y) on the curve cuts the axis of x. 5. Show that the sub-tangent of the curve y = aek is equal to,, whatever the length a may be. [The sub-tangent is the part of the axis of x intercepted between the tangent at any point and the ordinate of that point.] 6. Show that the curves y=aekx, and y=bekx are the same curve, with the origin so placed that in the first curve y = a when x = 0, and, in the second curve, y=b when x=0. 7. Show that the curves y=aekx and y=ae-kx are the same curves with their right and left portions interchanged. ARTS. 61, 62.] EXPONENTIAL FUNCTIONS. 57 8. Draw the curves y = (e+e x) and y =(e - e-). [Measure the required ordinates by help of the curve y =eX.] 9. Draw the curve y= 10TX, and show that it is the same as y=ekx if k =, loge 10= 023.... x2 10. Illustrate graphically the identity ex= 1 + x + 1 +... by drawing the curves (1) y =1 + x, 1.2 (2) y= 1 + x + -f-.2' x2 x3 (3) y=l+-+ x+ +21. 2. 3 between x= + 6. [Each curve can be built on its predecessor by adding to the ordinates the extra portions due to the next term.] 11. Find the series for ax, expanded in powers of x, by assuming x2 a = + ax + a2l- +..., and making use of the relation D (ax) = kax, where k= logea. Also deduce the series from the series for ez, where z=x logea. 12. Differentiate the series for ex, and deduce the fact that de = exdx. 62. Consider again the equation ax - 1 =loge a. dx, i.e., adx I cC - 1 dx =logea, proved at the end of the last article. It is usually written in the form, ah - 1 the limiting value of - is logea, when h = 0. As a particular case, the limiting value of e is 1, when h=0. This is also obvious from the series for eh, viz.: e6- 1 h * h' 1- + -2+ =, when h = 0. These two relations are very important. 58 DIFFERENTIAL CALCULUS. [CHAP. IV. The following examples may help the student to realize their truth. It may be instructive also to view them as approximate formulae when h is small but not quite zero, thus: The student has already learnt that e = 1, and that ac~= 1. He now learns that e7 = 1 + h, where h is a small quantity, so small that its square and higher powers are negligible; and that a7= 1-h logea for similar small values of h, where a is any number whatever. EXAMPLES. ax - 1 1, As a verification of the fact that a - =logea when x=O, plot the 0lox -1 x curve y= -, calculating the ordinates specially for the following values of x, '3, '2, '1, - 1, - 2, - 3. The curve will be found to cross the line x=0 (the axis of y) at the point where y=logel0=2'3 (approx.). ea - 1 2. Calculate the values of e for the same values of x, and draw ex 1 x the curve y=-, showing graphically that y= 1 when x= 0. al -1 3, Putting 1 + x for a in the fraction -h - and expanding (1 + x)7 by the Binomial Theorem, prove that loge(l + x) - X -2 + x3.. [NOTE.-This identity holds only in the range of values of x contained between x= -1 and x =l. This limitation of range will be discussed in Chapter XI.] 4. Draw the curve y = loge (1 + x) between x= - 1 and x =2, on a large scale. (The curve does not exist for negative values of x numerically greater than - 1.) 5, Draw the curves (1) y=x; (2) y=x- x22; (3) y = X - TX2 + 3z3 o (3) y=x-~x2+~x3; (4) y = - x- +Xs3 -4; between the limits x= 2, showing that as more terms are taken the curves lie closer to the curve y=log (1 + x) so long as x is less than 1, but tend to diverge from this curve beyond that limit. ARTS. 62, 63.] LOGARITHMS. 59 63. To find the diffrential of logex. The increase in logx for a given increase in x depends among other things on the base to which the logarithms are taken. The most important case in theoretical work is that in which the base is e, which is often called the natural base of logarithms, as theorems relating to the calculation of logarithms are simpler with that base than any other. It is also called the Napierian base as it, or rather its reciprocal, was used originally by Baron Napier, the discoverer of logarithms (circa A.D. 1624). The common base, 10, was suggested afterwards because of its greater advantage in numerical work. We are, then, taking first the case of logex. Let = logex,.'. e x. edy = dx, dx dx dy=-.. e x x i.e. d (logex) - -x. This is a very important theorem and should be remembered with absolute readiness. We may prove it another way, viz.:-by differentiating x7 - 1 h,and then putting h 0 (see last Article). Thus d h = d(x z -' 1l) = x7l-ldx, which becomes x-1dx when h = 0. The formula is true of course for any variable whatever, e.g. d(loge, )= d where u is any function of x. Thus, for instance, d(logex2)= d(2 ) =2xd =2(logx). Also D log a+bx + 2)= 2b 2cx(lo Also D log(ca + bx cx2) + 2x a + bx + CX2' 60 DIFFERENTIAL CALCULUS. [CHAP. Iv. 64. The above theorem is often useful in the form du = u d (logeu). In fact this formula gives the neatest way of arriving at the differential of a". For, let u = a, so that logu = x logea,.. by the formula just written, we have d ca= cad (x logea), = a logea. dx; and, in particular, the formula d e" = exdx comes out at once,* since x is itself the log of e". Examples, applying the above formula: (1) d(x1) = xd (n loge x) qn dx -- Xn. = nx'- ldx, as proved long before. (2) To take a harder case: d(X sill) =xsilxd(sin x logex) = xsill (logex d sin x + sin x d loge x) =Xsillx(cos. logex+ sin ) dx. 65. Methods of saving labour in differentiation. In many cases if we wish to differentiate a function consisting of a number of factors, it is a practical help to take logarithms first (to base e), and then differentiate. This is called logarithmic differentiation. Thus, let y - vl+x2..log y==logx + log(1-x2)- log(l +x 2); dy dx -d(l-x2) 1d(l+x2)' y = -, C2 1 X2 t12 * The student must guard against thinking that this method gives a proof of the formula for d(ex). It is merely interesting to see how neatly it comes out by this method. (See Ex. 25, p. 63.) ARTS. 64-66.] LOGARITHMS. 61 1 dyl 1 2x 2x y dxx- 2 ' 1- x2 l +X2 1 x x x 1 - X2 1 + t 2; dy (y1 2x dTX \- 1 i- x~ This finishes the differentiation, but, by inserting the value of y, some of the terms will cancel, as shown below. dlyx/- x 2 1- 2x2- X4 dx~ /1 + ' 2 1x(l- x4) 1 - 2x2 - 4 ~, 1 — -. A ns. (1 + x2)2(1 - x2)A-' Another way of simplifying the process in this case would have been to square both sides instead of taking logs. This would have given us 2 2 _ x4 Y 1; 2 y (1 + 2) (2x dx - 4xdx) - (x2 - x4). 2x dx (1 + X2)2 * lyx(1 -2x2 - x4) * dx~ (1+2X2)2 and then dividing by y, and replacing it by its value, the same result as before would have been obtained. These examples illustrate methods of preparing a function for differentiation. Other methods in special cases may, with experience, suggest themselves to the student. If no obvious simplification suggests itself, the function must be carefully differentiated in the ordinary way, as it stands. 66. To find the differential of loggx. Let y =log,; a. CJ=X;.. taking logs to base e, yloga = logex; 62 DIFFERENTIAL CALCULUS, LCHAP. IV. dX.'. logo. dy=; 7 X.*. dy=A-, dy 1 where k logea and = logae. In particular d (log10o) = /where A= log0e =- 43429448.... The student should look out the common log of 2'7182818..., and verify to the first 6 or 7 figures that the above value of /u is correct. 67. The quantity /J is called the modulus of common logarithms, being the multiplier which converts napierian logarithms into common logs, the equation being logl0x = /j logx. The proof of this may be-left as an exercise to the student. He might also prove the following theorem, which is the fundamental theorem connected with the change of base of logarithms, viz., log a where any base whatever may be used on the right-hand side, provided the numerator and denominator are taken to the same base. In words the theorem is The log of a number to any base- the log of the number divided by the log of the base. EXAMPLES. [For Recapitulation of formulae, see Art. 76.] Find d- in the following cases. All the logarithms are to base e. dx 1. y=log sin x. 2. y = tan- lx + log/( 1 x2). ARTS$. 66G, 67.1 EXAMPLES.. 63 3. y=log(x+l)- 2 4. ye 5, yexlogx. 6. y-log(secx+tan x). 7. y=logu ~ 2 8. y=a si I - X +- X2 9, log y = x sinx. 10. log y=-sinx. 11. y x e-x2. 12, y = (sin x) Cos 13. y=tan2x~+ log(cos2x). 14, y =x l log x. 15. y=log(log x)4 16, y = log (e x+ e-"). 17. From the series for ex deduce the series for eOx and e-lx where i denotes,/( - 1), i.e. i2= - 1. 18. Show that the series for I ("ix + e - ix) is equal to the series for cos x. 19. Show similarly that 1 (eix - e - jx sinll X. Tti 20. Deduce that cos x + i sin x = eS"" and that (cos x ~ i sill x)1 - cos nx ~ i sin nx. 21, Differentiate both sidles of the identity log, (I + X) = =j - 1X2 +- R: -. and prove by an elementary miethodl that the resulting ecquation is an identity, provided x lies between the limits ~ 1. 22. Expand loge(l - x) and loge. in power's of x'1. 23. By putting x=- in tbe series for loge, find a series for loge2, and(I evaluate it to a few places of decimals. 1 + i tan x 24. Prove by help of Ex. 20 that loge1 -itan x 2i x and, by nsing the series for the left-hand side, deduce that x z:tani x - ~ tan3x + - tan5x - 25. If y=ex, show that the eqnations clex =exdx, and dy=y d(loge y) are merely different forms of the same eqnation. 26. Differentiate /, and show that the differential coefficient becomes equal to that of x log, x when h 0. 68. Compound interest property of exponential functions. The function ax may be described as a function which grows at a rate proportional to itself, (if a > 1). 64 DIFFERENTIAL CALCULUS. [CHAP. IV. For, if y = ax =ly_ kna"= qly, where the numerical value of the constant k is logg ( io And the finction a-x is a function which diminishes at a rate proportional to itself, since, if y =ady k. where k has the same value as above. The connection of this constant with the rate of growth of a+x will be considered at greater length in the next article. NOTE.-If a is less than 1, the rate of growth of aw is negative, since k is then negative. In fact ac is the same as b-x, where b is equal to -, and is therefore greater than 1 if a is less. We may therefore say that cx is a function which grows or diminishes, always at a rate proportional to itself, according as a is greater or less than 1. In practice it is usually more convenient to take a greater than 1 and to make use of ax or a-" according as we are dealing with an increasing or a decreasing function of x. (It may be noted that if a series of values of ax are calculated for values of x forming an arithmetical progression, such as x= 1, 2, 3,..., the values of ax form a geometrical progression.) There are many such quantities in nature. For example, other things being equal, the population of a country tends to grow at a rate proportional to itself. Again, a hot body cools at a rate proportional to its excess of temperature above that of surrounding bodies. Also a cistern leaks at a rate proportional to its fulness; and a reservoir of compressed air at a rate proportional to the excess of pressure inside it over the external pressure. ARTS. 68, 69.] EXPONENTIAL FUNCTIONS. 65 And a debt subject to compound interest grows at a rate proportional to itself. This is an artificial growth, by jerks as it were, the interest being added to the principal at finite intervals of time, instead of continuously as in natural growths. But the nature of the growth is similar in the case of all such quantities, whence we may call them all compound interest functions. Various problems arise in connection with such functions. We will illustrate some of them by examples. Example. A cup of tea whose temperature 5 minutes ago was 100~ above that of surrounding objects, is now 80~ above them. Find what its temperature will be in another half-hour, assuming it falls by the compound interest law. Let T be the excess temperature, in degrees, t minutes from the time when T was 100. Then T=100a-t, where a is a constant to be determined from the equation 80 = 100a 5, since T = 80 when t = 5. Hence a-5= 8. We then have to find T when t=35. The required equation is T = 100a35, where a-5= '8;.. T=100('8)7. Working this by logarithms, or by actual multiplication, we obtain T=21 nearly; i.e. the temperature will be 21~ above surrounding objects in another half hour. It will be noticed that log T, which equals log 100 -- tlog a, diminishes at a uniform rate, viz. the amount log a per minute, since T = 100a-t, and therefore log T = log 100 - t log a. This fact is connected with an important property of compound interest functions, viz. that their percentage rate of increase or decrease is constant. This we know is the case with money compound interest. We will prove it generally, and also show how to calculate it. 69. Constant proportional growth of exponential functions. Logarithmic increment or decrement. Let us suppose that y =a where a may be considered as known, or at any rate knowable by means of data supplied by experiment as in the case of the 5 minutes observation of the temperature of the tea in last example. Then dy a0loga = ky, say, where k =loga. dy Now d- is the rate of increase of y per unit increase of x; L.D.C. E 66 DIFFERENTIAL CALCULUS. rCHAP. IV. 1 dy and - d is its rate of increase, per unit of y, per unit increase of x, i.e. is its proportional rate of growth; n 100 dy. and 10. - is its percentage rate of growth; i.e., it is the percentage rate of increase of y, per unit increase of x. To illustrate our meaning more clearly, suppose y to be a sum of money growing continuously by compound interest, and that x is so many years from the commencement of the growth. Then is the actual rate of growth per annum of the whole of y, 1 dyx 1 dy is the rate of interest per ~1 per annum, y dx and 10 dy is the rate of interest per ~100 per annum. y dx [NOTE-per annum corresponds to per unit increase of x.] These two latter quantities are constant, but the first of the three grows year by year as y gets larger. It is with the two constant quantities that we have to deal. The expression 1 dy is the rate of increase per unit of y, Y dx and 100 times this is the rate of increase per cent. This last is what we are most used to, but it is more or less artificial, as why should we reckon rate per cent., rather than rate per thousand, or rate per unit? It is in reality merely a numerical convenience, since the rate per unit is generally a small fraction. Of course we do reckon in all three ways as a matter of fact. We talk of 5 %, but the same rate is often spoken of as Is. in the ~, and growth of population is generally reckoned, not per 100, but per 1000. We must consider the rate per unit as the natural mode of reckoning rate, and the rates per 100 or per 1000, or per any other number, as merely convenient practical ways of stating the result. We have now to prove that when y=a", i.e. when y is an exponential or compound interest function, the expression - dy, which is its proportional rate of increase, is constant. dx d This follows at once from the equation d-= ky, which shows A -= ARTS. 69, 70.] EXPONENTIAL FUNCTIONS. 67 that this proportional rate is given by the value of the constant quantity k, which = log ea. Hence logea= the rate per unit of y, per unit increase of x, and 100 logea is the percentage rate of increase of y. It is interesting to note that d=Dlogey. y dx Hence k, or logca, = D logey. For this reason k, or logea, is often called the logarithmic increment of ax. The percentage rate of increase is then 100 times the logarithmic increment. If we choose to put ax in the form of eCx, which we may do, since a=ek, our equation becomes y=e-x, and the logarithmic increment is explicitly shown by this mode of writing the equation. Similarly, if y = a- = e-"k, k is called the logarithmic decrement, and 100k is the percentage rate of diminution of y, per unit increase of x. Example. Determination of the logarithmic decrement in the case of the cup of tea. The equation connecting T with t is T=100a-t=100e-kt, where k = loga = the logarithmic decrement required. To evaluate k, we have the equation a-5='8, so that a5== =1'25, i.e. e5k = 1 25. Hence, taking logarithms, k=log01 l25 - 5, where A/= logloe= '4343, 1 and 1 =2-3026. We shall find k to sufficient accuracy by taking 5 log10ol25=0 02, and 1 -=23, which makes k= '046, and 100k=4'6. Therefore the fall of temperature is 4'6 per cent. per minute. The equation for T may be put in the form T = 100e - 046t, if it is desired to show the value of k explicitly. 70. In the case of money compound interest, we must remember that it is not in practice reckoned continuously, but at intervals, and therefore the above theorems as to rate of interest do not apply without modification. For example at 5 % compound interest, reckoned annually, the amount in n years is P(1 + -),)n if P is the original sum, 68 DIFFERENTIAL CALCULUS. [CHAP. IV. whereas the continuous compound interest method would give the amount as P. e20, which would be too great. If the accretions of interest were made half-yearly, or quarterly, or at still shorter intervals, the amount would approach more n closely to the limit Pe2, and on the contrary, if the accretions were made biennially or at longer intervals, the amount would approach more nearly to the simple interest limit, viz.: (+ 20)It will always lie between those two limits. 71. If k is the rate of interest per ~ per annum, the amount realized per ~ in n years is (1 +k)" if interest is reckoned annually, and is 1 + - if interest is reckoned vn times a year at equal intervals. If in is infinite this becomes el', i.e., the limiting value of (1 +- when m is infinite is ekn, and therefore the limiting value of (1 + - is e; or, to put it I \\P n m/ more simply, (1 + ) becomes equal to e when p = o, where p is put for - which becomes infinite when m does. This is sometimes taken as the definition of e. 15\P If we expand 1 + -) by the binomial theorem, we shall find that when p is infinite, it gives the series already found for e. Thus (1+) 1+()+( - 1)12 + p(p- 1) -2)( + -+ P1 + - +. +. 1++ 1.+2+ 1.2.3 1 1 which, when p = oo, becomes = 1+ l + 1- -+ i -1 +..., which is the series for e. 2. ARTS. 70-72.] EXPONENTIAL FUNCTIONS. 69 72. Gradient of the curve y = yoe. The gradient is the value of d- at each point, which is equal to kyoex, = ky. To draw the tangent at any point P, whose ordinate is PN, measure from N a length NN'= - to the left along the axis of x, 1 then PN' is the tangent at P. For PN + NN' = y =ky, which is the required gradient. The length NN' intercepted on the axis of x between the ordinate and the tangent is called the subtangent. It is constant and equal to the reciprocal of k in these compound interest curves.* The easiest way to obtain a rough drawing of the curve is to measure off lengths ON1, N1N2,... each = l/ along the axis of x, and to draw the ordinates PoO, PiN1, P2N2... =.y, Y, Y2..., where yl, Y2' equal ey0, e2yo... respectively, as is at once seen by putting x = l/k, 2/k... in the given equation. The tangents at P1, P2... will be PO0, PN,... 1 1 The following curve is for yo 5, and k= 20 i.e. = 20. - I - 0 0 20 30 40.50 denotes time, it is called the time-constat. denotes time, it is called the time-constant. 70 DIFFERENTIAL CALCULUS. [CHAP. IV. The subjoined table will facilitate the drawing in all such cases: e= 2-72 e-l 368 e2= 7-39 e-2= 135 e3= 20-08 e-3 = 050 e4= 54-60 e-4 = 018 The student might draw the decreasing curve given by the equation x = 100e-20 in the same way. The successive ordinates will be each - of e the preceding one, and the tangents will be PoN1, PN,.... 73. If many ordinates are wanted, and particularly if the equation is given in the form y=y,a0x, it is probably best to use a table of logarithms, and to work with the equation logo y = logo yo ~ Ax, where A = log0 a. If it is also desired to draw any of the tangents, the value of the subtangent can be readily found, being equal to. If it is remembered k AA that this is the reciprocal of the logarithmic increment or logarithmic decrement as the case may be, and that 100 times this increment or decrement gives the percentage rise or fall of y per unit increase of x, we have the whole theory and practice of these functions. 74. We will finish with another example. Suppose we want to build a wall or embankment with cross-sections so arranged that the pressure per square foot at any part due to the weight of the wall above that part shall not exceed a certain limit. The pressure cannot be everywhere the same, because at the top of the wall the pressure is nothing. There must be an increasing pressure from the top down to some point, and then below this the pressure per square foot can be constant. We must therefore make the condition that the pressure per square foot at a certain point some distance below the top shall not be anywhere exceeded. We will suppose the wall to be vertical on one side and battered on the other. ARTS. 72-74.] EXPONENTIAL FUNCTIONS. 71 Let the pressure at H in the figure =p tons per square foot, and let it be required that the pressure per square foot at any distance x below H shall be the same as at H. Let the material used weigh w tons per cubic foot. Let y feet be the thickness of the wall at this depth (at the point P). Then the weight resting on a horizontal section at P, y feet across x 1 along, will, on our supposition, be py tons. Now take a horizontal section 1 foot along and y+dy feet across AH.*\ x \ at a depth dx below P. The weight resting on this will be greater than the weight on the P section by the weight of a slab y feet across, 1 foot along, and dx feet deep, i.e. containing ydx cubic feet (omitting infinitesimals of the second order). The increase of weight is therefore wydx tons, and the total weight = (py + wy dx) tons. But the area of the section is now y + dy sq. feet, and as the pressure per sq. foot is still to be =p, the weight supported must =p(y + dy) tons. Hence py +wydx=p(y + dy);. wydx=pdy; dy w.'. the thickness y of the wall follows the compound interest law, since its logarithmic increment is constant. 72 DIFFERENTIAL CALCULUS. [CHAP. IV. Let - = k, then =ky, and y = y0ek, where yo is the thickness at H. We can put this result in a simpler form if we suppose the wall above H to be of uniform thickness. Let h feet be the height OH of this part. Then the weight per square foot resting on the horizontal section at H is evidently hw. Hence hw==p;. y= yoe If, then we take points H,, H2,... below H, such that HH1 = HlH2 =... = h, the thickness at these points will be eyo e2y,... and the tangents to the battered slope of the wall at these levels will pass through the points H, H1..., and, similarly, the tangent at the level of H will pass through O. If we wish to obtain intermediate thicknesses by calculation, the easiest plan is to calculate e7, or rather its log1o, and then to calculate the values of y from the equation logo0y = loglo0y + Ax, when A = logloe = Suppose, for example yo= 1, and h = 10 (measured in feet), then - 4343 04343 h- 10 and logly = 04343x. If this wall is 30 feet high, i.e. extends 20 feet below H, the thickness at the base will be 7'4 feet. If it extends 30 feet below H, its base must be 20 feet across. If the wall is battered on both sides, the widths at the various levels will be just the same as in the case considered. For the width depends on the superincumbent weight, and not on the relative amounts of batter. NOTE.-If the upper part OH of the wall is not of uniform thickness, the easiest plan will be to take as h, not its height, but the height of an equivalent weight of wall that is of uniform thickness equal to the thickness at H. The equations will then hold without modification. ARTS. 74, 75.] EXPONENTIAL FUNCTIONS. 73 75. The function e-"Xsin (bz + c). This function is an oscillating function with decreasing amplitude. It is connected with such problems as the extent of swing of a pendulum which is set swinging in a resisting medium and left to come gradually to rest. It is important also in connection with many electrical problems. The student should draw the graph of this function for different values of a, b and c. (See examples Nos. 9-13, p. 75). The function crosses the zero line (y=0) when bx+c=0, wr, 27,.... Its graph oscillates between the curves y = e-ax and y = - e touching them alternately when bx + c = -, T, 2-7' 2.... Its gradient = e-"[b cos(bx + c) - a sin(bx + e)] - /a2 + b2 e-axSin (bx + c - a), b where tan a=-. Its gradient is therefore zero when bx + c = a, a + r, a + 2T,.... 76. Recapitulation. d (ex) = exdx, d(a)= axlogadx, X2 X3 where e" 1 + x + - 2 + ", 1 1 whence + I + 4 + whence e=1 +1+1-2+l 2. +" I. 1 2. 3, =2-718281828.... log0e = = 0'43429448..., and log, 10 = = 2-30258509.... e is also the limit of 1 + -) when = oo P/ which is the same as the limit of (1 + h)h when h= 0. eh- 1 The limit of is 1, when h= 0; ala - 1 and the limit of h is loga, when h= 0. hilgaweh0 74 DIFFERENTIAL CALCULUS. [CHAP. IV. du d logex = d, and, of course, d logeu = -, often used in the form x u du = d logge, where u is any function of x. d logx = loge.-; and, in particular, d logzx =/x. dy 1 dy ldy x 10 x If y=yc+-a~:=ye+-x; ~ - +- ky, and I dy= _, i.e. y grows or diminishes at a rate proportional to itself, i.e. by the law of compound interest; and 100k is the constant rate per cent. at which it grows or diminishes, k being called the logarithmic increment or decrement of the function. The curve has the constant subtangent k. 1 2 Corresponding to the values x = 0,, k,... the values of y are, eyo, e20,... or yo, Y-,,... according to whether y is e e increasing or decreasing, and these values, with the gradients at the points, will often suffice for a rough drawing. If we require intermediate values of y, the working equation is logo0y = logl0y + Ax, where A = yk = logloa. EXAMPLES. 1. Expand (i + 1) by the Binomial theorem, and by putting p = oo in the result, deduce the series for ex. Show that (I + -) has the same limiting value. 2. Differentiate (1 +) x, and deduce the fact that dex exdx. 3. If the population of a county doubles itself in 100 years, find its rate of growth per 1000 per annum, assuming it constant. If the population is a million at the beginning of the century, find what it will be in 20, 50, and 80 years respectively from the beginning. 4, Supposing that if a cable is coiled once round a post a pull on it equal to 50 lbs. weight will balance a resistance of 500 lbs. weight, find what resistance can be balanced by the same pull if the rope is coiled three times round, supposing that the rope is strong enough, and that the advantage of the greater number of turns increases by the compound interest law. ART. 76.1 EXAMPLES. 75 5. In the last question find the equation connecting the pull and resistance in the form R=P. a, where 0 is measured (1) in whole revolutions, (2) in radians, (3) in degrees. 6. Find the ratio of R to P (1) if a half-turn is taken, (2) if only a quarter-turn is taken. 7. Find the equation in the form R= PekO when 0 is in radians. 8. Calculate the thickness at the base of a wall 50 feet high if the top 20 feet have a uniform batter, the thickness of this part increasing from 1 foot at the top to 2 feet at the bottom, and the lower part of the wall being so built that the intensity of pressure at every part of it is the same. Examples of e-ax sin bx: 9. Draw the curve y=e-~x sin rx. 10. Draw the curve y= e-x sin 27rx. 11. Draw the curve y = ex sin (7rx 4 ) 12. Draw the curve y = e- sin (-lrx + ) 13. Draw the curve y=e-x sin (- rx). [In the curves 9-13 find specially where the curves touch the guiding curves y= e-2x, which should be drawn first; also where they cut the zero line, and at what angles; also where their gradients are zero.] Further compound interest problems: 14. A pane of glass destroys or obliterates 5 per cent. of the light falling upon it; how much light gets through 20 such panes one behind the other, assuming that they all act in the same way? 15. In a certain rarefied gas one per cent. of the molecules which start in any given direction are deflected by collision with others after travelling a millimetre; how many are able to travel a centimetre without collision? 16. An electric current left to die out in a certain circuit drops to - of its value in - of a second; how long will it take to drop to a e 10 millionth of its value, assuming that it decreases at a rate proportional to itself? 17. A current left to itself in a very massive conductor diminishes only ten per cent. in five seconds: how soon will it become imperceptible on a galvanometer which can detect a billionth part (10-12) of the initial current? 18. A current is started in the same conductor: find how soon it will rise to within one-thousandth part of its full value, assuming that it rises at a rate proportional to its defect from the maximum? Find the time-constant of this conductor, and draw the curves of rise and fall of the current. CHAPTER V. HYPERBOLIC FUNCTIONS. 77. There have come into use of late years certain functions of e whose properties are very similar in many respects to those of the circular functions. They are connected with a rectangular hyperbola very much in the same way as the circular functions are connected with the circle-hence their name. We shall however here simply give their definitions and more important properties without reference to their hyperbolic connection. The most important of these are (1) -(ex+e-x) which is called the hyperbolic cosine of x, or briefly, cosh x, or ch x, and (2) l(e - e-x), called the hyperbolic sine of x, or sinh x, or shx. Besides these there are tanh x( csh x) and its reciprocal cosh x) coth x; and sech x, which is the reciprocal of cosh x; and lastly, cosech x, which is the reciprocal of sinh x. X2 X3 78. Since e=l 1 + + 1 +1.2.3 1. 2 1. 2.3 X2 X3 and, therefore, e-x= 1 - x+ -1 + ].. 2 1.2.3 _ARTS. ri-i.] HYPERBOLIC FUNCTIONS. 77 x 2 _ _ _ _ _ it follows that cosh x = I + 12+1.2..4+ X3 X5 and sinhx=x + + + 1.2.3 1.2.3.4.5 expansions very similar to those of eos x and sin x, except that all the terms are positive instead of being alternately + and 79. The student can, from the definitions of Art. 77, very readily prove the following properties, and should compare and contrast them with the corresponding properties of the circular functions. (1) d cosh x = sinh x dx. (2) d sinh x = cosh x dx. (3) cosh2x - sinh2x = 1. (4) cosh2x + sinh2x = cosh 2x. (5) 2 sinh x cosh x = sinh 2x. (6) d tanh x = sech2x dx. (7) d coth x = - cosech2x dx. (8) d sech x = - sech x tanh x d.x, (9) d cosech x = - cosech x coth x dx. (10) tanh2x + sech2X = 1. (11) coth2x - cosech2x = 1. And last, but by- no means least, (12) x = log,(cosh x + sinhx). The 3rd and 12th formulae give rise to the inverse formulae: (proved by putting JX = sinh ~u, (13) sinh-lx = loge[x +', I(X2 +1)] whence = Sinh cosh u =,,/(x2 + 1),.. ~=etc. 78 DIFFERENTIAL CALCULUS. [CHAP. V. (proved by putting x = cosh u, (14) cosh-lx = loge[X + ^/(x2 - 1)] whence sinh uM=/(2 - 1), u=. etc. proved by putting (15) tanh-lx= i log — 1+x (15) -X=1 o-and simplifying the righthand side. The most important of these formulae are those relating to coshx and sinhx, viz. (1) to (5), and (12), together with the two inverse formulae (13) and (14). The student should draw the graphs of cosh x.and sinh x, which he can readily do by means of the little table on p. 70. The curve y=coshx is the shape of a uniform chain hanging between two supports, and is therefore of special interest. It is called the common catenary, or, briefly, the catenary. 80. Formula (3), viz. cosh2x - sinh2x = 1, is of great importance in many ways. It is similar in form to the identity sec2 - tan20 = 1, and therefore we are able to make use of a comparative table of secants and tangents if we wish to find sinhx when we know cosh x, or vice versa. Moreover, if we equate cosh x to sec 0, and so make sinh x = tan 0, it follows that tan 0 tanh x = -- = sin 0; sec 0 sn also coth x = cosec 0, sech x = cos 0, and cosech x = cot 0. ARTS. 79-81.] HYPERBOLIC FUNCTIONS. 79 Hence the relations are: sin 0 cosec 0 tan 0 cot 0 sec 0 cos 0 tanh x coth x sinh x cosech x cosh x sech x which are perhaps best put in the following three groups: sin 0 tan 0 sec 0 cos 0 cosec 0 cot 0 O tanh x sinh x cosh x sech x coth x cosech x Of these the quantities in the third group are merely the reciprocals of those in the first, while those in the second group are mutually reciprocal. 81. The function A(0). From formula (12) we see that if cosh x = sec 0, and sinh x = tan 0, the value of x itself is given by the equation x = loge (sec 0 + tan 0). This is a most important function of 0. It comes into very many formulae in mechanics and other applications of mathematics. In particular the spacing of the parallels of latitude in Mercator's projection is connected with this function, their distances from the equator being given by the equation y = a loge(sec 0 + tan 0) for different values of the latitude 0, a being the constant which is used in spacing the lines of longitude from the zero meridian by the equation x=aa5 for different values of the longitude q. The importance of this function has led certain French writers to give it a special name. They call it lambda (), writing it k(0). We may therefore state the connection between x and 0 by saying that if x = X(0), then cosh x = sec 0, sinh x = tan 0, etc. 0 is sometimes called the gudermannian of x, from Gudermann, who first systematically studied these functions, but the name is not very important, as we generally think of x as a function of 0 rather than 0 as a function of x. 80 DIFFERENTIAL CALCULUS. [CHAP. V. 82. Calculation of X(O). It is easy to prove that sec 0 + tan 0 = tan(45~ + ~0). -^For sec0 +tanO~1 + sin 0 sin 90~ + sin 0 For sec 0 + tan 0 = - = cos 0 cos 90 + cos 0 2 sin (45~ + 10) cos (45~ - -0) 2 cos (45~ + 10)cos(45~ - 60) = tan (45~ + 18). Hence X(0) - loge tan(45~ + 0) =l-loglo tan(45~ + 0), where - = 2302585. Therefore, to calculate X(0), look out log0o tan(45~ + -0) and multiply it by -, which can readily be done by logarithms, or by help of a multiplication table provided for the purpose in most books of tables. [For a short table of values of A(0) see p. 82.] We have, then, to remember that X(0) = loge(sec 0 + tan 0) = logtan(45~ + 20). 83. The differential of X(0). d X(O) d(sec 0 + tan 0) (sec 0 tan 0 + sec20)dO sec + tan 0 sec + tan 0 = sec 0. dO. 84. The curve y = AX() will be found instructive. It has real and imaginary regions, being real when cos 0 is positive and imaginary when cos 0 is negative. 1 + sin 8 For sec 8 + tan = sin cos 0 The numerator of this is always positive; therefore the fraction is + or - according as cos 0 is + or -. There is ARTS. 82-84.] HYPERBOLIC FUNCTIONS. 81 no real logarithm of a negative quantity; hence X(0) is real when cos0 is positive, and only then, that is, when 0 is between - 90~ and +90~ or between n360~ - 90~ and n360~+90~. Hence we need only draw the graph between 0= - 90~ and + 90~, as the graph for other values of 0 will be merely a repetition of this part. To draw the graph for negative values of 0, we should notice that X( - )= - X(). For X( - ) = log,(sec 0 - tan 0) and X() = loge (sec 0 + tan 0);.. X( - 0) + X(0)= log,(sec20 - tan20) -log, - 0. If we then first draw the graph from 0~ to 90~, by help of the table, it is easy to draw the part from 0 to - 90~, as it is merely the former part reversed in direction. (See figure.) L.D.C. F DIFFERENTIAL CALCULUS. 82 it = log,(sec 6 + tan 6) =logetan (45~ + 6 ). 6 21 I! 6~ 21 2 0 0 00000:30 0-54931 60 1 31696 1 0-01745 31 0-56956 61 1 35240 2 0-03491 32 0-59003 62 1-38899 3 0 05238 33 0 61073 63 1 42679 4 0-06987 34 0 63166 64 1-46591 5 0 -08738 35 0 65284 65 1 ~50685 6 0-10491 36 0-67428 66 1 54855 7 012248 37 0 69599 67 1'59232 8 0-14008 38 0-71699 68 1 63794 9 0-15773 39 0-74029 69 1,68557 10 11 12 13 14 15 16 17 is 19 20 21 22 23 24 25 26 27 28 29 0 17543 I 40 0-76291 70 1 73542 0-19318 441 0 78586 71 178771 0 21099 42 0-80917 72 1-84273 0-22886 43 0 83284 73 1 9007 9 0-24681 44 0-85690 74 1 96226 0-26484 45 0-88137 75 2-02759 0-28295 46 0-90628 7 6 2-09732 0-30116 47 0-93163 77 2-17212 0-31946 48 0 95747 "d78 2 25280 0-33786 49 0-98381 79 2-34040 0 35638 50 1 01068 so 2 43625 0-37501 0-39377 0-41266 0-43169 0-45088 0-47021 0 48872 0-50939 0-52925 51 50 53 54 55 56 57 58 59 1-03812 1-06616 1-09483 1 12418 1-15423 1-18505 1-21667 1-24916 1-28257 81 82 83 84 85 86 87 88 89 2-54209 2-66031 2-79422 2-94870 3-13130 3-35467 3-64253 4-04813 4-74135 30 054931 60 1-31696 1 90 infinite To find 21 for intermediate values of 6, it will be best to look out logl5tan(45' + g0) and multiply it by 2 302585. HYPERBOLIC FUNCTIONS.. 8. EXAMPLES. Differentiate the following f unctions: 1. cosh x cos x + sinh x sin x. 2, cosh x sin x~+ sinh x cos x. 3. log cosh x. 4. sin1(tvanh x) + tan1(sinh x). 5. If 0 is the gradient angle at any point of the curve y coshlx, show that y=secp and x=X(b). 6. Prove that 2 cosh2x cosh 2x +1 and 2 sinh2x = cosh 2x - 1. 7. Prove that 4 cosh3x = cosh 3x + 3 cosh x. 8. Prove that 4 sinh3x= sinh 3x - 3 sinh x. 9. Show that differentiating the identity of Ex. 7 will give ns the identity of Ex. 8, and vice versa.,MISCELLANEOUS EXAMPLES. Find an expression for y in the following five examples: dx 1. X2 - xy + 2y2 - 2ax -- 6ay + 7a2 =-. 2. y=tan-lx~ log -+. 3. y = sin (ex log x). - (log x)2}. 4. y=e-axcos3x. 5. y=x2sin2x. q2r2 6. If u2 find du (1) if p is variable, and q and r are constants; (2) if q is variable, and p aud 2, are constants; (3) if r is variable, and p and q are constants. [Probably at first it will be found easier to substitute x for the varialle, and then replace the proper variable after differentiating.] 7. If in the last question we denote the respective differential coefficients by and - respectively, find the value of dp' dq' dr du du du P dq Q+r. dr 8. If r~cos2= a2cos 2, find dr dO 9. If x1=cos'l 2(1 - x) du 9. f uC?s1X, find (1~- x~g) x 10. If x =a(O - sin 0), and y=a(I - cos0), find d. dx 3X2 -- 4z2 dl~cr 11. If y = 4 find - when z is constant, and find when x (4x - 7z)4' dx dy d dA is constant, and deduce the value of x + dx- dz CHAPTER VI. APPROXIMATE DETERMINATION OF THE ROOTS OF EQUATIONS. 85. General methods of approximating to the roots of equations containing one unknown quantity. In solving a mathematical problem it is often necessary to find the value or values of an unknown quantity which will satisfy some equation. This is called "finding the root or roots of the equation," or " solving the equation." The student has doubtless often done this work in connection with simple and quadratic equations in one unknown quantity. The object of this chapter is to give methods by which the roots of any equation in one unknown quantity can be readily found, either accurately or approximately. Since all the terms of an equation can, if desired, be put on the left-hand side of the sign of equality, with zero on the right-hand side, the general type of such equations is f(x)=0, where f(x) denotes some function of x. In the case of algebraic equations, it is sometimes possible to split f(x) into factors; in such case the roots are easily found. Thus, if 2 - 3x + 2 = 0, we see that (x- 2)(x- 1)=0,.'. the roots are x= 2 and x=1. This may be called "solution by factorization." This method is only possible in simple cases, and can very seldom be used in solving equations of higher degree than quadratics, and only occasionally even in quadratics. The general method of solving quadratics is, however, well known. The methods which will be suggested in this chapter are applicable to equations of any degree, and to transcendental as well as to algebraic equations. ARTS. 85, 86.] SOLUTION OF EQUATIONS. 85 These methods may be divided into two main classes, namely, graphic and by calculation. These two methods can be made to mutually help each other, and the graphic method will certainly help the student to understand the method by calculation. My object will be, not to attempt in any way an exhaustive treatment of the subject, but merely to give ordinary methods which will be applicable to any equation. I shall assume in the graphic work that the student is able to obtain squared paper to facilitate plotting. 86. Graphic method-First approximation. Every equation in x can be put into the form f(x)= 0. If then we plot the curve y =f(x), and find where this curve cuts the axis of x, we shall have found the roots of the equation. In general we shall not be able to find the roots exactly, but we shall be able to find two near values of x between which a root lies. We shall show how to get nearer and nearer approximations to the root, till any desired accuracy has been obtained. Example. Find the roots of the equation x3 - 4x2 - 11x + 32 = 0. To plot the curve y=f(x) we shall find the values of y for different values of x, and to ensure correct drawing at points near to the axis of x, we will also calculate the gradient at such points. The gradient is given by the equation tan 0 =f'(x) = 3x2 - Sx - 11. Positive values of x. Negative values of x. x y tan P Points. x tan Points. 0 +32 0 +32 1 +18 -1 +38 2 + 2 -15 A -2 +30 3 -10 - 8 B -3 + 2 +40 E 4 -12 + 5 C - 4 -52 +69 F fnot shown in 5 + 2 +24 D diagram, being too far down. etc. This shows that there is a root between 2 and 3, but nearer 2 than 3; because y= + 2 when x= 2, and y= - 10 when x= 3, whence y must = 0 somewhere between. Similarly there is a root between 4 and 5, but nearer to 5 than 4; and another root between - 3 and -4, but much 86 DIFFERENTIAL CALCULUS. [CHAP. VI. nearer to - 3 than to - 4. If we draw the parts of the curve to scale in the neighbourhood of the above points, we shall be able to judge roughly the values of the roots. In the diagram the x-scale is I inch to a unit, and the y-scale is J1 inch to a unit. The curve is drawn through A, B, C, D, and E by help of the gradient values calculated in the table, and the roots are approximately between 2-1 and 2'2, between 4'8 and 5'0, and between - 3-0 and - 3'1. -3 I t I I + tL I _X It will be noticed that in each of the above cases the curve and the tangent are exceedingly close together as they cross the zero line. If we take the point where the tangent crosses as being sufficiently near the true point, we can readily do the calculation. Take, for example, the root between A, B. Let the tangent at A cut the axis in P; the ordinate AL is = 2, and the gradient of AP is tan = - 15. Hence LP= -ALcot = -2-; the length of OP = OL - AL cot k,= 2 + 2, i.e. 213.... Similarly the other positive root = OQ = OM - DM cot2 =5 2- 2 = 492, and the negative root — 3 - = - 305. If these values are not accurate enough for the purpose we need, we can further approximate by using a larger scale ARTS. 86, 87.] SOLUTION OF EQUATIONS. 87 figure (working at each root separately), or by calculation, or both ways; this indeed being the best. (1) Take the root near OP. When x= 2-13, f(x)== +0-085997, f'(z)= -14-43; and when x= 2-14, f(x)= - 0-058056, f'(x)= - 1438; so that we have approximately the points '" _ a, (2-13, 0-086).o05 and /3, (2-14, -. 0058). The curve between these points will be 2,3 - N214 so nearly straight, that we may take the \: crossing point of the chord a/3 as near enough -'5: to the true point. This gives, graphically, 2'1360 for the root. -10 To check this by calculation by means of the tangent at a; the gradient at a is - 14'43: 0'085997. =2'13+ 144 =213596; or we might have taken the gradient at /, which is - 14-38, '0058056 whence =2-14 008056- 2-135963. 14'38 If we had taken the gradient of the chord a/3, which is 14'4053, or say, 14-4, we should have calculated x= 2-13597. Now it is evident that the true value lies between that given by the best tangent value and that given by the chord, so that the true value of x lies between 2-135963 and 2-135970. I have gone into the work in great detail in this case to emphasize the fact that when we have two points very near together, the curve and its chord are nearly coincident, and the chord may be substituted for the curve. This simplification is often of great use. Another point to notice is that we need not calculate the gradient to very many figures. 88 DIFFERENTIAL CALCULUS. [CHAP. VI. (2) Take the root near OQ. When x;=492, f(x)= + 0149888(8), and when x= 4-91, f(x)= - 0-071629 (y), and the gradient of the chord y8 is 22-15. The gradient of the tangent at y is 22-04, or say 22;.x. x is between 4-913238 and 4-913256. The tangent is likely to give a better result than the chord, so we may put x 4 91325. The diagram is left to the student. (3) The negative root is about - 3-05. When x= -3'05, f(x)= -0032625, and f'(x)= 413; 0-032625.the root = -305 + 4-13 --- 41'3 = - 305 + 000079 = - 304921. The three roots should add up to 4, which is the coefficient of - x2 in the equation. We may therefore take them as 2-135961 4'91325 the sum of which is 4. and - 3-04921 These roots have been calculated to excessive minuteness to illustrate the method. In practice much less accuracy is needed, and the roots would therefore be obtained more speedily and with less trouble. For instance, 2-136, 4-913, and - 3'049 would be ample accuracy in most cases. In the work of approximation the diagram would show at once whether the correction was to be positive or negative, but it will be well to work out a formula which will enable us to work correctly, if desired, without a diagram. Suppose that a is the first approximate value of x and that h is the correction, and that we have calculated f(a) and f'(a). ARTS. 87, 88.] SOLUTION OF EQUATIONS. 89 Then f(a + h)= 0 and f(a + h) -f(a) = -f(a); but f(a + h) -f(a) is, when h is small, the differential of f(a), and therefore =hf'(a), if we retain only the first power of h. (h is the same thing as dx, being a small increment of x). hf'(a)= -f(a); *~,= ~ f(a)... f —(a).'. the root is a --- a) f'(a) If we think of the formula from the geometrical point of view in connection with a diagram, if we denote the ordinate f(a) by y, and the gradient f'(a) by tan ~, the root is a-ycot <. If we use the chord joining two points whose abscissae are a and b, the gradient of the chord is f(b) - f(a) ~~-a and the approximate root is b-a af(b) - bf(a) a- - xf(a) f(b)-f() b) - f(a) f(b) -f(a) ' provided f(a) and f(b) are small and of opposite sign. 88. The two modes of approximating may be called the 'tangent' method and the 'chord' method. Both depend for their success on the change of direction not being very rapid near the crossing point. More accurate approximation formulae could be devised, taking the curvature of the curve into account, and their investigation is interesting; but if we make good use of the graphic method of approximation we can always obtain a first approximation to the root good enough to insure l being small, or the change of curvature between two points on either side of the crossing point being small. We may, therefore, be content with the methods already given. 90 DIFFERENTIAL CALCULUS. [CHAP. VI. Let us take another example. Find the angle whose circular measure is equal to its cosine. Let x be the circular measure of the angle; the equation is f(x) = x - cos x = 0. By help of the tables we can find two values of x so near together, on either side of the required value, that the chord of the curve y =f(x) may be taken for the curve itself. 42~ 20' 42~ 21' x= 0-7388561 x= 07391469 cos x= 0-7392394 cos x = 0-7390435 x - cos x = - 0-0003833 (A) x - cos x= + 0-0001034 (B) The diagram shows that the required angle is about 42~ 20' 47". Calculating OP from the figure, we have ~ o3833f 60" t47' seconds. = 47~ seconds. ARTS. 88, 89.] SOLUTION OF EQUATIONS. 91 89. Special graphic representation of the roots of quadratic equations. These equations can be so simply solved in surd form by the ordinary method that a graphic method is not of much use, and I only introduce the following method because of its extreme simplicity. Let the equation be x2 - bx + c = 0. We could of course draw the curve y= x2 - bx + c, and find where it cuts the x-axis. But the method now to be explained is simpler for securing the first approximation to the roots, and the general method can then be used for the final approximation to each root. Draw, as shown in the diagram, OA =, OB= b, and BC=c, the three coefficients being supposed positive in the figure. If b or c are negative they must be measured in opposite directions to those shown. O 2 42 3 4 5 6 Q 7B The angles at 0 and B are right angles. Join AC, and on it as diameter describe a circle cutting OB in P and Q. Then OP and OQ are the roots of the equation. This will be demonstrated if we can show that OP + OQ = b, and OP. OQ = c. Now, if OA cuts the circle again in C', it is obvious that OC' = BC = c; also OP = QB;... (1) OP+OQ=OQ+QB=b, (2) OP. OQ = OA. OC' =c, since OA 1; OP and OQ are the roots. In the figure they are both positive, being measured to the right of 0. If in any diagram either of them is to the left of 0, that root is negative. 92 DIFFERENTIAL CALCULUS. [CHAP. VI. Example 1. Take, as an example, xc2 - 7x + 4 0. The roots are about 0-6 (+) and 6'4 (-). (See figure.) For further approximation, if desired, when x=0-6, f(x)=0 16, f'(x)= - 58; ' x1=06276; and, when x=-64, f(x)=0-16, f'(x)= + 58; x2 =6 3724. This makes x] +2= 7 as it should. The product is 3'9993 which is nearly the correct value 4. If we are not satisfied with these values we can correct by the method suggested in Ex. 2, thus obtaining x1=0-6277 and x2=6 3723. The student may usefully notice that the curve y = x2 - bx + c (not shown in the diagram) goes through the points C, C' in the figure, and has the gradients + b at these points, so that the tangent at C is perpendicular to AB. The curve goes, of course, also through the points P and Q. Example '2. Take, as another example, x2 - 4x - 2=0. The approximate roots are -0'4 and + 44. (See figure.) When x= -0 4, f(x)= -0-24 and f'(x) = - 4-8; h= =- T; and similarly in the case of 4l4, where h= +;.-. the roots are - 0 45 and + 4 45. The sum of these roots is 4, as it should be since the second term in the equation is - 4x. The easiest way to correct further is to note that the product of the roots ought to be - 2, and that any error must be equally shared between the two. Thus (*45+e)(4'45 + e)=20025+ 49e, ARTS. 89, 90.] SOLUTION OF EQUATIONS. 93 if we neglect the square of e;.. e=-0'0005 to 4 decimal places, and the roots are - 04495 and + 4 4495. 90. If the equation is ax2- bx + c =0, we may, instead ot dividing by a, and proceeding as above, do as follows: Draw OA = a, OB= b, [See figure of example below] BC = c, then, drawing the circle on AC as diameter, cutting OB in P and Q, we shall find that OP and OQ are a times the roots. b c For, if x1 and x2 are the roots, x,+X2=-, and xzr.2=-; whereas OP + O = b, and OP. OQ= aac. whereas OP + OQ = b, and OP. OQ= ac. If, then, we draw a line O'P'Q' parallel to OB, at unit distance from A, and take the points 0', P', Q' on it where it is cut by AO, AP, AQ respectively, the roots are O'P' and O'Q'. 94 DIFFERENTIAL CALCULUS. Or, of course, we may, if we please, measure OP and OQ and divide them numerically by a instead of graphically. Or, if we do use the graphic method of reducing OP and OQ we may draw the parallel line O'P'Q' at any convenient distance from A (e.g. 10 units, as in the figure), if we think unit distance too small or too great for convenience, and then reduce finally by easy division or multiplication as the case may be. Example. 7 3x2 - 9 5x+ 24=0. The approximate roots are 0'35 and 0'95. (See figure.) The corresponding values of f(x) are - 003075 and - 003675, and,, f'(x) are - 439 and +4-37..,,,, h are - 00070 and +0-0084... the roots are 0 3430 and 0 9584. We can check this result by adding the roots together. The sum ought to be equal to 9'5+7'3, and will be found correct to 4 decimal places. EXAMPLES. 1. In Ex. 1, p. 92 (viz. x2 - 7x+ 4 -0), having taken the approximate roots in the first instance as being 0'6 (+) and 6'4 (-), so that their sum is 7 as it should be; correct these values by the condition that the errors must be equal (the roots being 06 + e and 6 4-e), and that the product of the roots must be 4; e2 being considered negligible. 2. Find a root of the equation 3= 1 + x. 3. Find the root of the equation x5 = 3 + 2x which lies between 1 and 2. 4. Solve the equation 0 =cot 0. 5. Solve the equation x= l0 loglox. 6. Solve the equation cos 0 = tan 0. 7. If b is the known reciprocal of a given number a, show that the bda correction to be applied to b when a is slightly altered = -. 8. The reciprocal of 3476 is -000287687. Find the reciprocal of 3476 38. 9. Solve the equation x2- 10x+ 7 =0, graphically. 10. Solve the equation x2 - 10x -7=0. 11. Solve the equation 2x2 + x - 7 =0. 12. Show that, in the construction of Art. 90, OP and OQ are the roots of the equation x- bx+ ac=0; and show that Euc. II. 11 and 14 are examples of graphic solution by this method. Show also that PAO and QAO are the angles given by the equation a tan2 0 - b tan 0 + c - 0. CHAPTER VII. MAXIMA AND MINIMA VALUES OF A FUNCTION OF ONE VARIABLE. 91. It is often desirable to know under what conditions some particular function of a given variable shall have a maximum or minimum value, and what that maximum or minimum value may be. For instance, in Examples 15 and 16, p. 16, the question naturally suggested itself; what must be the dimensions of the beams so that their strength should be greatest? We solved these problems graphically by, in each case, expressing the strength (or, rather, a quantity proportional to the strength) as a function of the breadth, and then drawing the graph of the function, and seeing from the figure for what value of the breadth x the function y was greatest. We also suggested a method by which this value of x could have been calculated, as it must be the value of x for which the gradient of the function would be zero. This method is of general application, and applies to minimum values also if the function has any. There are in theory some exceptions, but they seldom or never occur in practice. We will however, in the course of the following investigation, refer to these exceptions. If, then, we want to find the maximum or minimum values of any variable quantity, the first thing to do is to express the value of such quantity (which for definiteness we will call y) in terms of some convenient variable (which we will call x) on which its value depends; i.e. the first thing to do is to form the equation Y =f(), which expresses y in terms of x. 96 DIFFERENTIAL CALCULUS. [CHAP. VII. We have then to find for what value or values of x the function y will have a maximum or minimum value. For this purpose we will carefully define what we mean by such values. 92. Definition of maximum and minimum A maximum value of y is a value which is greater than adjacent values, however near; or, in symbols, if y=f(x), and x = a makes y a maximum, i.e. if f(a) is a maximum value of y, f(a) will be greater than either f(a +h) or f(a-h), however small h may be. (Note-not however great h may be, but however small.) A minimum value of y is a value which is less than adjacent values, however near; or, in symbols, if x= b makes f(x) a minimum, f(b) will be less than either f(b + h) or f(b -h), however small h may be. 93. A function may have several maximum and several minimum values, as, for example, in walking over an undulating road we may go over several hills and through several valleys. On the top of each hill the height above sea-level is a maximum, and at the bottom of each valley the height above sea-level is a minimum. Each maximum is greater than the adjacent minima, but not necessarily greater than all minima. C Thus, in the diagram, the ordinates at A and C are maximum ordinates, and those at B and D are minimum ordinates, and yet the ordinate at A is less than the ordinate at D. In a continuous function maxima and minima values evidently occur alternately. ARTS. 91-96.] MAXIMA AND MINIMA. 97 94. Increasing and decreasing functions. If, as x increases, y also increases, y is called an increasing function of x. It is evident that its 'gradient' dy is positive. If, on the contrary, as x increases, y decreases, y is called a decreasing function. It is evident that in this case dy dx is negative. Most functions are increasing functions for some values of x and decreasing functions for other values. For instance, in Example 15, p. 16, y is an increasing function from x=0 to x= 3, and it then becomes a decreasing function; or, in other words, its gradient is positive between x= 0 and 3., but is negative for greater values of x. When x = 3 the gradient of this curve is zero, and the ordinate is a maximum. 95. General condition that x = a shall make f(x) a maximum. If, as x increases from some smaller value to the critical value a, the function f(x) continually increases, i.e. has its gradient (f'(x)) positive; but when x is greater than a, f(x) begins to decrease, i.e. f'(x) becomes negative: f(a) is a maximum value off(x). General condition that x = a shall make f(x) a minimum. If, as x increases up to a, the function f(x) continually decreases, i.e. f'(x) is negative, and when x is greater than a, f(x) begins to increase, i.e. f'(x) becomes positive, f(a) is a minimum value off(x). In each case the necessity is that f'(x) shall change sign as x passes through the critical value. The distinction between the two cases is in the order of the change. Briefly f(a) is a maximum value of f(x) if, as x increases through the value a, f'(x) changes from + to -; and f(a) is a minimum value of f(x) if, as x increases through the value a, f'(x) changes from - to +. 96. Now it is evident that, if f'(x) changes gradually and does not become infinite, it cannot change sign except by passing through the value 0. Hence, in general, the primary L.D.C. G 98 DIFFERENTIAL CALCULUS. [CHAP. VII. condition for f(a) to be a maximum or a minimum value of f(x) is that f'(c) shall equal 0. The exception occurs when f'(a) is infinite, or when there is an abrupt change of gradient. In practice the latter case never occurs, and the former very seldom. The shape of the curve in such cases is shown in the diagram. 97. Points of inflexion. If f'(a) =0, but the gradient on either side has the same sign, there is a peculiarity in the curve at the point, but f(a) is neither a maximum nor a minimum value off(x). Such cases are shown in the adjoining figures. Such a point on the curve is called a point of inflexion. A B Gradient + Gradient - on both sides of A. on both sides of B. Points of inflexion may occur when the tangent is not horizontal, but they have no connection with the present investigation. The general definition of a point of inflection is that it is a point at which the curve crosses its tangent. 98. From what we have seen it is evident that the first thing to do when we want to find the maximum and minimum values of any function f(x) is to form the auxiliary function f'(x), and to find what values of x make f'(x) = O, i.e. to solve the equation f'(x) = 0. This may be done either graphically or by calculation (see last chapter). The next thing is to see for each value of x so obtained-called the critical values of x, because they have to be critically examined-whether there is ARTS. 96-99.] MAXIMA AND MINIMA. 99 any change in the sign of f'(x) as x increases through the critical value in question, and, if so, whether the change in f'(x) is from + to -, or from - to +. If f'(x) can become infinite for any finite value of x we ought also to see whether there is a change of sign in f'(x) on either side of this value of x. In many practical examples, however, it will be obvious whether the function is a maximum or a minimum without the analytical method of discovering which it is. In such cases we merely put f'(x)=O, and do not investigate the change in its sign on either side. The student can work the examples 1-9 at the end of this chapter in this simple manner. It might be well for him to solve some of these now, before reading the rest of the chapter. [CAUTION.-Beginners, from want of careful thought, often seem to imagine that it is x or f'(x) which becomes a maximum or minimum in the above cases. It is not so: it is f(x) which has a maximum or a minimum value when the above conditions are satisfied. The beginner, too, often finds general statements difficult to follow; he should try and have some definite idea in his mind; probably the best general notion is that furnished by the varying height above sea-level of a man walking along an undulating road, as that most readily lends itself to the conception of several maxima and minima. Other instances, generally of single maximum or minimum values, will gradually accumulate in his mind as he works at the examples at the end of this chapter. He should work at some of them, and then read the above general statements again and again, till they become self-evident.] 99. The following example will illustrate the method: Find the maximum and minimum values of 2x3 - 92 + 12x. Denoting this byf(x), our auxiliary function will be f'(x)=6(2 - 3x+2)=6(x - 1)(x - 2). Hence the critical values of x are 1 and 2, since these values make f'(x)= 0. The neatest way of recording the result of 100 DIFFERENTIAL CALCULUS. [CHAP. VII. examining the sign of f'() for different values of x will be to make a short table as follows: x I f'()I f(x) less than increases 1 (e.g. 0) + with x 1 0 5 (max. value) between decreases as 1 and 2 - x increases 2 0 4 (mim. value) greater increases than 2 with x The statements in italics in the above table are made last. What we first establish (by inspection) is that f'(x) is positive when x is less than 1, that it becomes 0 when x= 1, that then f'(x) becomes negative and remains negative until x= 2 when it becomes 0 again, afterwards becoming positive and remaining so however great x may be. From these fluctuations in the auxiliary function f'(x) we deduce that as x gradually increases from any small value (say, - oo ), f(x) is an increasing function up to x= 1, then f(x) is a decreasing function till x=2, after which it becomes and remains an increasing function. Therefore, finally, f(x) has its maximum value when x= 1 and its minimum value when x= 2. Then by calculation we find that the maximum value is 5, and the minimum value is 4. The student should draw the graph off(x), and also, on the same scale as and immediately under the former diagram, he should plot the graph off'(x), preferably by a dotted curve to show that it is merely an auxiliary function. If he then compares the graphs carefully with the above table he will see that all the statements made in it are visibly true. He will also notice that in this example the maximum value is not greater than all other values, but only greater than adjacent values on either side; and similarly the minimum value is not less than all other values, but only less than adjacent ones. ARTS. 99-101.] MAXIMA AND MINIMA. 101 100. In the above example the roots of f'(x)= 0 are whole numbers. Of course in practice this is not necessarily or even generally the case. The roots can then be obtained by some general method such as those given by last chapter. The graphic method is an excellent one, as very often the values obtained from the diagram will be sufficiently accurate for all practical purposes, since in the neighbourhood of a maximum or a minimum value of f(x) a small error in the determination of the critical value of x makes hardly any appreciable difference in f(x). Indeed, so much is this the case, that in many cases it might seem better not to worry about theory at all, but merely plot f(x) itself and read off its maximum and minimum values. This method in fact was adopted in the cases of strengths of beams in Examples 17, 18, p. 16. The counter advantages of using the auxiliary finction f(x) are (1) that it is often easier to draw the graph of f'(x), being of lower degree, and (2) that we can generally determine the critical values of x more accurately from this function. Graphically, it is easier to determine accurately what values of x make f'(x)=O than it is to determine accurately what values make f(x) a maximum, and, as either method gives the values of x required, it is at any rate well to be able to use the more accurate method even when we intend to work graphically. When we work by calculation, we must use the auxiliary equation f'(x)=0, there is no way of avoiding it. Indeed it is often so quick and simple to use that it hardly needs this defence. 101. We will take as our second problem a geometrical one. Find the maximum cone inscribable in a given sphere. We have now first to find our functionf(x), i.e. we must express the volume of any inscribed cone in terms of some convenient variable x (though of course we need not actually use the symbol x unless we like). Take as variable the height, h, of the cone. The axis AN (=h) of the cone will obviously pass through the centre O of the sphere. Let NP be the radius of the base of the cone, so that the area of the base= r. NP2. The volume is ~ base x height. Let a be the radius of the sphere, so that OA = OP = a. Then NP2=OP2- ON2=a2- (h - a)2=2ah - h;. the volume of the cone= -. 7r(2ah - h2). h = 17r (2ah2- h3). We have, therefore, to make 2ah2 -- 73=a maximum. 102 DIFFERENTIAL CALCULUS. [CHAP. VII. Hence its differential coefficient, viz., 4ah - 3h2, must = 0; 4a h=0O or One of these must make the volume a maximum, and the other must make it a minimum. A Now in a problem of this kind we need no theory to distinguish between maximum and minimum values. It is obvious that h =0 makes the volume of the cone=0, which must therefore be the minimum 4a volume; hence h=43 makes the volume a maximum. Putting this value in the expression 7r. h2(2a - h) we find the maxi16a2 2a 32wa3 mum volume = 7r. -= 81 which is -r of the volume of the sphere (viz. 47ra3). If we want to make a graph of the volume, the neatest way is to take as our function to be plotted the ratio of the conical volume to that of the sphere, as that gets rid of the trouble of the factor ~Tr. 2ah2 - h3 2h2 h3 This ratio is 4a3 4a2 4O,3 If lastly we take as x the ratio of h to the diameter 2a of the sphere, the function reduces to 2x2 - 23 = 2x2(1 - ), which is easy to graph. We have to draw it between x=0 and x=l. We know, moreover, that the function is a maximum when x=@. We may therefore plot the points corresponding to X=0, A, 3, 1, and these, together with the gradients at the same points, will be amply sufficient for a good diagram. The gradient is given by 4x - 6x2. ARTS. 101, 102.] MAXIMA AND MINIMA. 103 The table is f(x) f'(x) Points. 1 -o4 P 1 Sr (max.) 0 Q 1 0 -2 C 0 T A B C If we take each small division on the squared paper=-l-r, for both horizontal and vertical measurements, the figure will be as shown. The tangent PT at P will cut the axis of x in T, such that TA=6 divisions, because then its gradient AP 4 2 TA 6 3 The tangent at C is shown as CU, with a gradient= - 2. Of course we might have reduced the above fractions to decimals and plotted decimally. 102. The following considerations will often enable us to avoid the calculus altogether. Axiom. Between two equal values of a function a maximum or minimum value of the function must occur, and will ultimately coincide with them when they are infinitely near together. B Example. Find the maximum rectangle which can be inscribed in the quadrant AOB of M a circle. Take the rectangles, as shown in the figure, whose angular points P, Q are symmetrically P situated in the arc AB. Then evidently the areas of the two rectangles are equal. When P and Q are infinitely near together they ultimately meet at the middle point M of 0 A the arc. Hence M is the angular point of the maximum rectangle, which is therefore a square. 104 DIFFERENTIAL CALCULUS. [CHAP. VII. This answer is really obvious, for M is in a position of symmetry, and a position of symmetry will always make the function a maximum or a minimum. The following are examples of symmetry: A loop of string lying in a plane surface will enclose the greatest area when it is circular. The solid of greatest volume for a given surface is the sphere. 103. As another example of the axiom take the following: A boatman at A, 3 miles from B the nearest point of the shore, which is straight, wishes to reach C, a point on the shore 6 miles from B, in the shortest time. Where must he land if he can row at 4 miles an hour, and travel on land at 5 miles an hour? A R B P Q C Let APC and AQC be two paths infinitely close together which will take equal times. They will be on either side of the required path and will ultimately coincide with it. Cut off AR= AP. Then PR is ultimately perpendicular to both AP and AR. Now by the APC path, he travels the extra land piece PQ, and by the AQC,,,,,, sea piece RQ. These must take equal times. If these distances are measured in fractions of a mile, the times are - and RQ hours; 5 4 PQ RQ 5 4' i.e. ultimately, cos Q =- [when P and Q ultimately coincide], AB_.. tanQ=, i.e. -B But AB=3 miles;.. BQ =4 miles;.'. he lands 4 miles from B, i.e. 2 miles from C, i.e. he rows 5 miles, and walks 2, and he takes 4+ hours=l hour 39 minutes, not counting the time occupied in landing, which would be the same in any case. If we do the same problem by the calculus method the neatest way ARTS. 102-104.] MAXIMA AND MINIMA. 105 is to express the time in terms of the angle 0 which the path AP makes with AB. AP = AB sec 0 = 3 sec 0, in miles, PC=BC- BP=6- 3 tan 0, in miles; 3 sec 6 - 3 tan. time, in hours, = 4+ 5 =(8 + 5 sec 0 - 4 tanl ). A B P C The differential coefficient of this function has to be zero;.. 5 sec tan0 - 4 sec2 = 0;. sin 0= This leads to the same answer as before. 104. Maximum and minimum gradient. Point of inflexion. A question which often arises in connection with the graph of a function is as to where the gradient is a maximum or minimum. Thus, find the point on the curve y = 22- 23 where the gradient is a maximum, i.e. where the curve is steepest. dy The gradient is given by the equation = 4x - 6x2, or let dx us say, for brevity, y, = 4x - 6x2. We have therefore to make Y1 a maximum. Hence d-y must = 0, i.e. 4 - 12x - 0;.'. x = 1.. dx If we look at the diagram (p. 103) we shall see that this is evidently the case, the point P being the required point. Such a point is always a point of inflexion (Art. 97). There is no difficulty in connection with such a problem except as to the meaning of maximum or minimum when the gradient is negative. In the case of a negative gradient, the point where the curve is steepest is technically a point of minimum gradients 106 DIFFERENTIAL CALCULUS. [CHAP. VII, on the principle that a large negative quantity is algebraically less than a small one. If we like we may call it a negative maximum. The student might notice also the gradients at A and B in the diagram on p. 98. The gradient at each of these points is zero, but at A it is a minimum, because the adjacent gradients are positive, and at B it is a maximum because the adjacent gradients are negative. Of course a zero gradient is in general neither a maximum nor a minimum gradient, as usually the gradient is positive on one side and negative on the other. At such points it is the ordinate which is a maximum or a minimum, not the gradient. The gradient is never a maximum or minimum except at a point of inflexion, and at such a point it always is a maximum or minimum. 105. Occasionally it may be convenient to express the function, whose maximum and minimum values are required, in terms of two variables instead of one only, there being some given relation between the variables. The method of proceeding in such a case is best illustrated by an example. Take the example No. 15, p. 16, where we have a rectangular beam of maximum strength to be cut out of a cylinder of 12 inches diameter. The equations are bh2 = m ax............................... (1) and b2+ h2 = 144,............................... (2) using h to denote the depth or height of the beam, so as not to confuse it with the sign d for differential. Differentiating (1) gives us h2db + 2bhdh =.................................(3) and differentiating (2) gives 2bdb+ 2hdh= 0...............................(4) These two equations between db and dh must be consistent, i.e. they must really be the same equation in different form. Hence the ratio between the coefficients of db must equal the ratio between the coefficients of dh,. h2 2bh i.e. =b; h= 2b2. ARTS 104-106.] MAXIMA AND MINIMA. 107 db Or, we might have found d- from (3), and also from (4), and equated them, leading to the same result. 106. Another mode of discriminating between maximum and minimum values of y. We have seen that y is an increasing function when the gradient (~dy or y,) is positive, and a decreasing function when Yi is negative. Similarly the gradient is itself an increasing function when its differential coefficient dy- is positive, and a decreasing function wh dy1x function when dx is negative. This property can be made use of to facilitate the discrimination between the maximum and minimum values of y. Denote d-J by the symbol for brevity. dx by the symbol y2 for brevity. Now when x increases through a value that makes y a maximum, the value of y, goes through the series of changes +, 0, -. Therefore y, is a decreasing function, and Y2 must be negative. And when x increases through a value that makes y a minimum, the value of y1 goes through the series of changes -, 0, +. Therefore in this case Yl is an increasing function, and Y2 must be positive. Hence, in general, y is a maximum when yl = 0 and Y2 is negative, and y is a minimum when y1 = 0 and Y2 is positive. Therefore, in any given example, after finding the values of x that make yl = 0 we may, if we like, see in each case whether Y2 is negative or positive, instead of going through the fundamental method of seeing whether the change of y, is from + to - or from - to +. Of course if Y2 happens to be zero at any of these points, this new method fails, and we must fall back on the fundamental method. It is probable in such case that we have a maximum or minimum gradient, (i.e. a point of inflexion) and not a maximum or minimum ordinate. Examination of the change of gradient by the fundamental 108 DIFFERENTIAL CALCULUS. [CHAP. VII. method will tell us which of all of them is the case. However, in practical cases, points where y1 and Y2 are both zero seldom occur, and the examination of the sign of Y2 is a neat and convenient mode of discriminating. Take, as example, y= 2x2 - 2x3. Then y=4x- 6x2, =0 when x=O or 3; and Y2=4-12x, which is positive when x = 0, and negative when x= -. Hence x=0 makes y a minimum, and x= - makes y a maximum. 107. Further investigation of the case when y =O and also Y2 =O. We have seen that y is a maximum when y1 =0 and y2 is negative, and y is a minimum when y 0 and Y2 is positive. Similarly if we want to discriminate between maximum and minimum gradients we may form the third auxiliary function dy2 d2 or y3, often called the third derived function, or third differential coefficient. We then see that y1 is a maximum when y, = 0 and y3 is negative. and y1 is a minimum when y2 = 0 and y3 is positive. The method will fail, i e. the result is doubtful, if y3 = 0. From this, if we remember that when y is a maximum or minimum, Y1 is not, and when y, is a maximum or minimum y is not, it follows that y is not a maximum or minimum if y =0, Y2=0, and y3 is not zero, but it may be if y1=0, y2=0, y3=0. Similarly Y1 is not a maximum or minimum if Y2 =, y3 = 0, and Y4 is not zero. Hence in this case y is a maximum or minimum (provided, of course, y1 = 0). However, the reasoning is rather difficult to follow at this stage. It will be much easier when we have learnt a theorem called Taylor's Theorem (ch. xI.), and we shall see then that y is a maximum when Y1 =0, =, y?/3 = 0, and y4 is negative, and a minimum when y, = = y3 =0 and y4 is positive. But the theory is not very useful, as it is far better in such cases to fall back on the fundamental method of examining the sign of y, on each side of the critical point. ARTS. 106, 107.] MAXIMA AND MINIMA. 109 EXAMPLES. 1. Find the greatest cylinder that can be cut out of a conical block, the height of the cone being 3 feet and the diameter of its base being 2 feet. 2. Find the relation between the height (h) and radius (r) of the base of a cylindrical vessel open at the top (such for example as a bushel measure) when the internal surface is a minimum for a given capacity. 3. If the cylinder is closed at both ends (like a tin of preserved meat), find the ratio of h to r when the surface is a minimum for a given capacity (i.e. when it is made most economically). 4. Find the ratio of the height to the radius of the base of a conical vessel fulfilling the same conditions, (1) when it is open; (2) when it is closed by a flat circular plate. 5. Find the maximum cone of given slant height. 6. Find the maximum cylinder inscribable in a given sphere. 7. An open tank is to have a square base and vertical sides, so as to have a given capacity. Find the proportion of its depth to its width so that the expense of lining it with lead may be a minimum. 8. Find the shortest ladder which will reach from the ground to the wall of a house over an obstruction 6 feet high and 5 feet away from the wall. Find the angle of slope of the ladder. 9. Find the maximum and minimum ordinates of the curve y=x3 - 6x2+ 12, and also find the points of maximum gradient. 10. Find the maximum and minimum values of 3x5 - 25x + 60x. 11. Find the minimum value of ax+bx-1, a and b being positive quantities. Show that it occurs when ax is equal to bx-1. 12. Find the points of inflexion of the curve x2y=3(x - y). 13. Find the maximum and minimum values of (x -1)(x- 2)2. Draw a graph of this function between x= - 2 and + 3. Draw also a graph of its first derived function. 14. Find the maximum and minimum values of (x + 1)2(x - 2)3. Draw the graph. 15. Find the maximum and minimum values of (x + 1)3(x - 2)4. Draw the graph. 16. Find the maximum and minimum values of x Draw the graph. x-1 17. Find the maximum and minimum values of 2x - 3x2- 36x +10, and plot a graph of the function. Plot also a graph of its differential coefficient, and show the connections between the two curves. 18. Show that the maximum value of a cos 0 + b sin 0 is,/(a2 + b2). 19, Show that the minimum value of a sec 0 - b tan 0 is,/(a2 - b2). CHAPTER VIII. CURVATURE OF PLANE CURVES. 108. Definition. If two points A, B are taken close together on a curve, the infinitely short arc AB is generally denoted by ds. If AT and BT are the tangents at A and B respectively, the angle between them is dfA, if sb is the gradient-angle of AT, and b + d4 the gradient-angle of BT. Then the curvature of the curve at A is defined to be the limiting value of the ratio d when B is ultimately made to coincide with A, i.e. the curvature equals the ratio of the change of direction between A and B to the length of the arc AB, when AB is infinitely small. If A and B are not infinitely close, this ratio may be called the average curvature of the part of the curve between A and B. From this definition it is evident that a straight line has no curvature, and that a circle has constant curvature, since the ratio is constant, however long the arc may be. This latter fact may be brought out more clearly by the following investigation. 109. Radius and centre of curvature. Let AC and BC be the normals to the curve at two adjacent points A and B respectively, meeting at C. Then if we consider the short arc AB as practically identical with the are of a circle, C will be its centre and CA and CB will be its radius. Denote CA and CB by the letter p. C is called the centre of curvature and p the radius of curvature at A when B coincides ultimately with A. And the circle whose centre is C and radius p is [ARTS. 108, 109.] CURVATURE. 111 called the circle of curvature at A. It is sometimes called the osculating circle, as it more nearly coincides with the curve in the neighbourhood of the point than any other circle does. C B The angle at C=dqb, since the angle between two normals equals the angle between the corresponding tangents. Now, if do is expressed in circular measure, i.e. as a fraction ds of a radian, we have d4 = -, by the definition of circular measure. do 1 Hence -_, ds p i.e., when the gradient angle is expressed in radians, the curvature is numerically equal to the reciprocal of the radius of curvature. If the curve is a circle, C and p are the same for all points, and the curvature of a circle is constant and numerically equal to the reciprocal of its radius. For example, if p= 2 feet, the curvature=, which means that the change of direction is + a radian per foot of arc, or 1 radian per 2 feet of arc. If p = 3 feet, the curvature is 1 radian per 3 feet of arc, or about 19~ per foot, since a radian is a little over 57~. As the equation = is fundamental, we will give a slightly different mode of working it geometrically, which, though not so simple as the foregoing, is in some respects more satisfactory. Let P, Q, R be three equidistant points on the curve, and let the chords PQ, QR make the angles b and f +d4 respectively with the x-axis. 112 DIFFERENTIAL CALCULUS. [CHAP. VIII. Bisect PQ and QR at right angles by AC and BC. Then C will be the centre of the circle described through P, Q, and R. Let p be the radius of the circle. The angle at C = d,), and if AB is called ds, dc = -. P I dq' Hence! = d where dq) is the angle between two conp ds secutive normals, and ds is the element of arc intercepted between them. This is the same as the previous expression for the curvature, since the angle between any two normals is equal to the angle between the corresponding tangents. Moreover, the connection between the above diagrams is very simple, for, without altering the directions of the chords PQ and QR, we can by a slight movement make them the tangents at their middle points A and B respectively. 110. To obtain the radius of curvature as a function of x. If the equation of the curve is y=f(x), its gradient y1 will equal f'(x), and the rate of change of gradient per unit change of x will be dy, which we have denoted by y2, and which is often written f"(x), to signify that it has been obtained from f(x) by two differentiations. We shall be able to express p in terms of Y1 and Y2. For tan b = ye; sec2 = Y2; ARTS. 109, 110.] CURVATURE. 113 Also sec = _ (ds (see figure);. ds= sec <.dx. p -dy / dr 1 y2cos24.dx Therefore co p sec <c dx = y2cos3% and p =- sec3: Y2 i.e. the curvature = y2cos34, and the radius of curvature is its reciprocal. To find the radius of curvature at any point we have therefore to find the value of k at the point from the equation tan = y =f'(x), then we must find the value of Y2 at the point, and multiply the reciprocal of Y2 by sec35. The simple connection between Y2 and the curvature is worth considering. The curvature is the rate of change of the gradient-angle # per unit length of arc; and Y2 is the rate of change of the gradient itself, i.e. tan %, per unit change of abscissa. Tan ~ increases at a greater rate than 4, and the arc increases at a greater rate than the abscissa, hence, for both reasons, Y2 is greater than the curvature. The above investigation shows that we must multiply Y2 by cos35 to obtain the curvature. We may put the details of the argument in words: we want to find the rate of change of k per unit length of arc in terms of the rate of change of gradient (tan 4) per unit change of abscissa. We know that d a - sec2- that is, the rate of change of gradient per unit change of abscissa is sec24 times L.D.C. H 114 DIFFERENTIAL CALCULUS. CCHAP. VIII. the rate of change of < per unit change of abscissa. Also we know that d =sec, i.e. the length of arc between two adjacent points is sec k times the change of abscissa. Therefore, finally, the rate of change of gradient per unit change of abscissa is sec35 times the rate of change of b per unit length of arc. That is, y2 is sec35 times the curvature. The formula for p may be written in the form {1 +y1}, Y2 since y1 =tan B. But it is generally more convenient to use the simpler form - sec3). Y2 111. Geometrical construction for p. To find the radius of curvature at any point P of a given curve. C, M XL — K P /,i _ T N Draw the normal and ordinate at P. On the ordinate take PK, towards the inside of the curve, numerically equal to the reciprocal of y2. Draw KL perpendicular to PK to cut the normal in L, then PL = PKsec 5; draw LM perpendicular to PL to meet the ordinate in M, then PM = PL sec q = PK sec2g; lastly, draw MC perpendicular to the ordinate to cut the normal at C, then CP =PK sec3. Hence, if PK has been drawn to the inside of the curve, C will be the centre of curvature, and CP the radius of curvature, of the given curve at the point P. ARTS. 110-113.] CURVATURE. 115 112. Direction in which PK must be drawn. If the curve has been drawn, it will be obvious in which direction to draw PK, since it must be to the inside of the curve. But we can tell, independently of the figure, in which way to draw PK. If Y2 is positive, K will be above P, i.e., PK will be drawn upward, and if y2 is negative, PK will be drawn downward. For Y= l =sec2, d,.'. when Y2 is positive, d is positive, i.e., 4 will increase with x. This can only be the case if the curve lies above the tangent at P, i.e., if the centre of curvature lies above the tangent. On the other hand, if y2 is negative, ~ decreases as x increases, and the curve must lie below the tangent. The student should convince himself of the general truth of this by taking a part of the curve where the gradient is negative. He must remember that ( is the angle made by the upper part of the tangent with the right-hand part of the x-axis, so that < is obtuse when the gradient is negative. 113. Special cases. (i.) If 4=0, sec3=1;.'. p=-=PK. This is an imY2 portant case in many practical examples, such, for instance, as the curvature of beams, which is in general so slight that f is sufficiently small at each point for us to take sec P = 1 without sensible error at each point. It is also true, of course, in all curves at the points where the gradient is zero, as, for example, when y is a maximum or minimum. (ii.) If Y2 is zero, as at a point of inflexion, PK, and therefore p, is infinite, i.e., at a point of inflection a straight line more nearly coincides with the curve than any circle with finite radius can. Briefly, at such a point the curve is straight, its curvature being zero. (iii.) At a point where y, is infinite, i.e. where the curve is perpendicular to the x-axis, the method fails. To obtain the curvature at such a point, we must invert our treatment and 116 DIFFERENTIAL CALCULUS. [CHAP. VIII. dx dx treat x as a function of y, obtaining xl=, and 2=-. Of 0 1 dy 2 dy course x1 will be zero at such a point, and the curvature will 1 then be x, and p=, the radius being drawn to the right if x2 is positive, and to the left if x2 is negative. We think, in this case, of the gradient angle as being made by the tangent with the y-axis, i.e. by the right-hand part of the tangent with the upper part of the y-axis. 114. Value of p at the origin when the x-axis is the tangent. We may avoid the calculus in this case by the following method. O N The curvature ot a curve at any point is the same as the curvature of its circle of curvature at that point; hence we need only consider the circle of curvature, letting it take the place of the curve. Let the figure show the circle of curvature at 0, and let P be an adjacent point on the circle with coordinates x, y, i.e. ON=x, and NP=y. Then, if NP cuts the circle again in Q, we have NP.NQ = ON2; but NQ is ultimately equal to 2p;.. y.2p=x2, x2 I.e. P in the limit when x and y are infinitely small. Thus, in the parabola y = x2, ARTS. 113-115.] CURVATURE. 117 Similarly, if the curve touches the y-axis at the origin, y2 115. Mode of finding p when the vertical and horizontal scales are different. In the above investigation, and particularly in the construction of Art. 111, we have supposed the figure drawn strictly to one scale, i.e. the same vertical and horizontal scales. But in many cases it has been found convenient to take the vertical scale different from the horizontal. In that case several difficulties arise. First, the circle of curvature, if plotted by points, must also be plotted with its vertical scale different from the horizontal, which would distort it into an ellipse; or if we plot - and try to find the centre of curvature geometrically, the difficulty arises as to what scale we must plot it on. Of course there must be a true circle of curvature at each point-the problem is how to modify our construction so as to find its centre. If we take the numerical equation y =f(x), (to fix the ideas the student might have in his mind the parabola y = x2 plotted on p. 6 with vertical scale 1-tenth of the horizontal), and consider how we plot the various points for calculated numerical values of x and y, we see that we take an arbitrary horizontal unit of scale and plot the abscissae as multiples of that unit, and then we plot the values of y as multiples of an arbitrary vertical unit. Let us take a to denote the length of the horizontal unit, and b to denote the length of the vertical unit, then if (e.g.) we plot the point x= 2, y= 4, we really take x x = 2a horizontally, and y = 4b vertically, so that - = 2, and = 4, when x and y are the plotted lengths. And so in the case of every other point. The equation gives the connection between x - and. Consequently the full expression for the curve plotted is =f ( ) where x and y denote the actual lengths b \ay plotted. [In the case of the parabola the equation from which we plotted was = (-) where a = lOb.] 118 DIFFERENTIAL CALCULUS. [CHAP. VIII. If then we find the radius of curvature at each part of the curve b and then make use of the relation between a and b, we shall have the value of p required. Differentiating, we have Y1_ 1 /x\ af ta) b JV b,/x\, *. y - c-,') =tan d. Again, t2=o2f - *p = -- sec3. To construct for p, we must draw the vertical line, which call k, and then by the zigzag construction find the centre of curvature. If we plot k on the vertical scale, its numerical value is Ick~~~~~~~~ ~a2, which, from the above equation, = - Applying this to the curve y=x2 =f(X), we find f'(x) =2x and f"(x)=2; k a2 Now in the diagram, p. 6, a=10b;.*. a2:b2=100; -. =50, or k=50b. b. ARTS. 115, 116., CURVATURE. 119 Hence, to find p for any point P on the curve, we must draw k vertically upwards to a length of 50 vertical units from P, then draw the normal, and find C by the zigzag construction. In particular, the centre of curvature at the origin is 50 units up on the axis of y. This can be easily tested by the student by actual trial on the diagram with a pair ot compasses. NOTE.-To find the radius of curvature of the parabola at the origin we might have used Art. 114. Since the axis of x X2 y x2 is the tangent there, =; and since =- it follows at once 2 2yb a2' that p= ~. Hence its numerical value on the vertical scale is a 22 b 262 a2 116. The reasoning by which the formula k=bfj was established is a little troublesome at first, as all reasoning in connection with change of units must be; but if we remember that y=f(x) is merely a numerical equation between x and y, and that what we plot are lengths corresponding to those numbers, we see that the whole difficulty consists in ascertaining what is the real connection between the lengths and the numbers represented by them. The fundamental relation is as follows:-If a length x is represented by a number when x is measured in terms of a particular unit, the number expresses the ratio of the length of x to the length of the unit. Thus, if the number is 2 when the length of the unit is a, 2=x:a. E.g., let x=24 inches; then if x is measured in inches, the number representing it is 24, whereas if x is measured in feet, the number representing it is 2, i.e. 24 is the ratio of x to 1 inch, while 2 is the ratio of x to 1 foot. Similarly, if y is measured in terms of a unit whose length is b, the numerical measure of y is the ratio of y to b. Consequently, if y=f(x) is a numerical equation, the curve which we actually plot to represent this equation is Y-f) when a is the length of the horizontal unit of scale and b is the vertical unit. 120 DIFFERENTIAL CALCULUS. [CHAP. VIII. No difficulty of this sort ever arises if b=a, but in many cases such equality is practically inconvenient. 117. We will, as another example, find the radius of curvature of the curve y = x(5 - x), (1) at the highest point, (2) at the points where the curve cuts the x-axis. We will first consider the two scales equal, and then take a = 5b where a, b are the units of the horizontal and vertical scales. y=x(5 - x)=5x - x2, Y1= 5 -2x, Y2= -2;.'. at the highest point, where 0=0, p=~, if the scales are equal. The negative value of Y2 shows that the centre of curvature is below the curve. There is no object in attaching a sign to the value of p. 2 25 If a=5b, we must take the formula P- = -2 2 b - b2y 2.. p= 12- vertical units. At the points where the curve cuts the axis of x, x(5 - x) =0;.x=O, or 5;.'. tan 0 = ~5, and sec 0= +\/26, if the scales are equal; 26,J26.-. p 26 13,/26 units. If the scales are unequal, ~ tan =5 vertical units to 1 horizontal unit 5b = —=1, if a=5b; whence ~ sec 0 = /2 k a2 25 and b byY22 25.. p =2 x 2^/2b = 25,/2b = 5,2a. The student will find it helpful to notice, if he is drawing this curve with the scale relation a=5b, that the centre of curvature of the origin has the abscissa 5a, and the ordinate - 5a, or - 25b, and the centre of curvature of the point (5a, 0) is at (0, - 25b). If he draws arcs of these two circles, and also the arc of the circle of curvature passing through the highest point, (found above), he will, find it very easy to draw in the portion of the curve lying above the x-axis. For a graph of this curve see answer to Ex. 1, p. 15. ARTS. 116-118.] CURVATURE. 121 118. Chord of curvature. If we draw the circle of curvature corresponding to any point P of a curve, the chords of this circle drawn through P are called chords of curvature at P. The chord drawn along the normal is the diameter of curvature; other important chords are the vertical and horizontal chords of curvature. The vertical chord = 2p cos - 2k sec2), where k = -; and Y2 the horizontal chord = 2p sin 9 = 2k sec24 tan 4. The following geometrical investigation of the length of the vertical chord is instructive, and will incidentally throw a good deal of light on the meaning of Y2 Let P, Q, R be three neighbouring points on the given curve such that the ordinate of Q is midway between those of P and R, and let a circle be drawn through P, Q, R. This will ultimately be the circle of curvature at P when Q and R coincide with P. T MN Let the coordinates of P be (x, y), and those of Q be (x + 8x, y + y), i.e. let PM=8x, and MQ =y. The abscissa of R is x + 28x. The ordinate of R exceeds the ordinate of Q by the amount TR, of which TS= y. Now QM=8y where y is the ordinate of P..-. the corresponding notation for RT is &(y + 8y) where y +- y is the ordinate of Q. We may separate this expression into the two parts 8y + 8(8y), of which TS = y and SR = (8y). The notation 8(8y) expresses therefore the excess of the increase of ordinate as we pass from Q to R over the increase 122 DIFFERENTIAL CALCULUS. [CHAP. VIII. of ordinate as we pass from P to Q. It is often written, briefly, 82y (read delta-squared y). Hence 82y denotes the vertical height of R above the chord PQ. Now produce the ordinate of R to meet the circle in U. Then RU will ultimately be the vertical chord of curvature at P, and so also will SU; since both R and S ultimately coincide with P. Let PQ be denoted by 8s, since it is ultimately equal to the small arc PQ. Then SQ = 8s and SP = 28s. Now SR.SU = SP.SQ;. 2y.SU = 2(8s)2 and 2 2v \ and (8-)2. SU =2 \8X Now let Q and R move into coincidence with P, whereupon,- becomes = sec 4 if 4 is the gradient angle at P of the curve and circle, and at the same time (82 is usually written d2y d2y {6X) (d)2 or dY (read d-squared y by dz-squared). d2y Hence. SU = 2 sec2%, dX2 where SU is now the vertical chord of curvature at P. Now, evidently, from the figure, d2y- dtan' for dtand means the excess of the gradient of QR above that of PQ, which is d (Y d(dy)' which is d(, i.e. (d, since dx is the same between Q and R as it was between P and Q, that is, dr is constant. Hence d(dy) means exactly the limiting value of 8(8y) or RS, and dx means the limiting value of 8x or PM. Therefore d tan +. c... (ay) d is the limiting value of ( )2 d2y Hence dSY is what we have before denoted by y2, and we have the formula, 2 sec2( vertical chord of curvature = se- _ = 2k sec2S, as before. Y2 ART. 118.] CURVATURE. 123 EXAMPLES. 1. Find the radius of curvature at the point (2, 1) of the curve y=-, (1) when the vertical and horizontal scales are equal; (2) when the horizontal scale = the vertical. 2. Find the radius of curvature of the above curve at the point (1, 1) under the same scale conditions. 3. Find the radius of curvature at the point (1, 2) of the curve y=x+- when the scales are equal. 4. Find the radius of curvature at any point of the curve y=eX, and also at any point of the curve y = bea. 5. Find the minimum radius of curvature in the curves of No. 4. 6. Find the radius of curvature of the curve y=acosh at any a point, and show that it is equal to the length of the portion of the normal at the point intercepted between the curve and the axis of x. 7. Find the radii of curvature at the points of zero gradient on the curve y = x(10 - x)2. 8. Find the radii of curvature at the points of zero gradient on the curve y = x(144 - x2). 9. Find the radius of curvature of the curve y=x3 - 4x2- 1x +32 at the point (2, 2), (1) when the vertical and horizontal scales are equal; (2) when the horizontal scale is 5 times the vertical scale. 10. Find the radii of curvature of the curve y=x2(6-x), (1) at the origin, (2) at the highest point between x=O and x=6; when the horizontal scale is 5 times the vertical. 11. In any curve y=f(x), prove that y2 + x2= -y13. Exemplify in the case of y=x2. 12. Find the radius of curvature at any point of the ellipse x=acos0, y=bsin0. 13. Find the radius of curvature at any point of the hyperbola x=casecO, y=btan0. 14. Show that the equation of the circle of curvature of the parabola y2= 4ax, at the origin, is x2 + y2 = 4ax. 15. Show that the equation of the circle of curvature of the conic Ax2 + By2 = 2Cx, at the origin, is Bx2 + By2 = 2Cx. 16. Deduce the length of the radius of curvature, at (- a, 0), of the conics b2x2~a2y2=a2b2, by moving the origin to this point. 17. Prove that the line drawn from (a, b) perpendicular to the line bx + ay=ab passes through two of the principal centres of curvature of the ellipse b22 + a2y2 = ab2. [The principal centres of curvature are those which correspond-to the ends of the major and minor axes of the ellipse.] CHAPTER IX. SUCCESSIVE DIFFERENTIATION-INTEGRATION. 119. Successive Differential Coefficients of a Function. If y is a function of x, so also (in general) is its differential coefficient, and consequently this also has a differential coefficient which is also a function of x, and so on. (The only exception occurs when one of the differential coefficients is a constant, and the rest, consequently, become zero, and none of these are functions of x.) These successive differential coefficients are called the first, second, third, etc. Thus, if the function is x5, the first differential coefficient is 5x4, the second is 5.4x3, the third is 5.4.3x2, etc. There are several systems of notation which may be employed to denote these successive functions. The most usual are: dy d2y d3y Y' d-x' d2 dX3'"' y, Dy, D2y, D3y,..., f(x), f'() f"(), f"'(,..., but sometimes, for temporary convenience, a brief, incomplete, notation may be used, such as Y, Y1, Y2, Y3, *"' or y, y" y"', y.... [ARTS. 119, 120.] SUCCESSIVE DERIVATIVES. 125 As an example of successive differential coefficients we may take = X5 Dy = 5x4, D2y = 5.4X3 = 20x3, D3y = 5.4.3x2= 602, D4y= 5.4.3.2x3= 120x, D5y = = 120, and D6y = 0, etc. In this case, therefore, the first four differential coefficients are functions of x, the fifth is a constant, and all the succeeding ones are zero. If, however, the original function had been anything except a positive integral power of x, all its differential coefficients would have been also functions of x. EXAMPLES. 1. If y=15xv4-7x2+3, find d. 2. If f(x) =x2 - 7x2 +Sx, find fiv(x). 3. If y=sinx, show that D"y= sin (x+n ) 4. If y=sin3x, show that D5y=35 sin (3x+5). 5. If y = 55 +3x4 +2x3 + x2 -x + 6, show that D6y = 0. 6. If y=axs-l+bxs 2+..., show that Dsy=O, if s is a positive integer. 120. Successive Integrals of a Function. It is important to notice that such a series as x5 5X4,... can be not only continued forwards, as above, by successive differentiation, but can also be continued backwards, the terms on the left of x5 being such that X5 is one of their differential coefficients. Such an extended series is x58 7 X6 ' 8.7.6' 7.6' 6 ' x6.. 126 DIFFERENTIAL CALCULUS. [CHAP. IX. The terms on the left of x5 are termed its integrals, its first x6 X7 integral being -, its second being 7' and so on. There are several notations for the successive integrals of a function y, but the only one that can usefully be given at this stage is..., D-3y, D-2y, D-y, y. Thus, if y = X5, x6 D-1 =, 7 X7 D-2y 7.6 42 etc. These integral functions are, however, not so definite as the differential coefficients are, for not only is zx6 an integral of x5, but so also is x6 + a, where a is any constant. For, if we take D (x6 + a), the result is x5, since Da = 0;.'. D-1(5)= -x6 + a. Similarly, D-2(x5) = -x7 + ax + b, and D-3(x5) = a, sx8 + ax2 + b + c, (where a, b, c are in each case any constants, not necessarily the same in each case. They are called arbitrary constants.) These results can be verified at once by taking respectively the second and the third differential coefficients of these two functions. It will be seen that the first integral of a function may contain one arbitrary constant, the second may contain two, and so on. These integrals are called the general integrals of x5, and the particular cases in which special values are given to a, b, c,... are called particular integrals, and in especial the integrals in which a, b, c,... are all zero, may be called the principal integrals. Thus -i~xS is the principal third integral of X5. The remaining part, viz., ax2 + bx + c, may be called the complementary part of the third integral. It may be written as D-3(0), for if it be differentiated 3 times, the result = 0; D-3(5) = X8 + D-3(0). ARTS. 120, 121.] SUCCESSIVE DERIVATIVES. 127 121. To find a general formula for D"("). The results indicated in the above example may be easily extended into a general formula for D'r(X), where i is a positive integer, and r is either positive or negative. For, if = x; Dy = Xn-1; D2y=f(n - 1)x-2;..................... Dry= n(n - 1) (n - 2)...(n - r + 1)x"-r, n This is worked on the supposition that r is a positive integer, but it holds equally well if r is a negative integer, as will be seen at once by taking particular cases. Thus, if r= - 1, nD-ly ^ =1=n+1 D- _Ln+l n+ l / n+l \ This only means that D(, = y. Again, D-2y/ = (n+)(2 =-^n+1 & \n+2 (n + 2)(n + 1) n+2 which means that y = D2 +2)(+) the truth of which is (n+2)(n+ 1)' evident on actually differentiating twice. Of course these are only the principal integrals. The general formula is. In D) =-^- D (O) where D"(O)=0 if r is a positive integer, but if r is negative, =- s suppose, Dr(0)= D- s(0) =ax -l + b - 2 +... + a constant, as can be easily verified by seeing that DS (axs- +bx-2...)= 0 (see Ex. 6, p. 125). 128 DIFFERENTIAL CALCULUS. [CHAP. IX. NOTE.-When r is a positive integer, the formula presents no difficulty if r is less than n, but when r is equal to or greater than n, the denominator n —r requires special interpretation, as when r =n, n + 1, n + 2... this factorial becomes 0, -1, [ -2,... Now we know that Dxtn= n, hence on comparing this known result with that given by the formula, we see that we must put 10=1. All the differential coefficients beyond the nth are zero, hence we must put - 1 = o, I- 2 =o, if the formula is to give the correct result. Of course we cannot attach a direct meaning to any of these quantities 0O, [-1,.... All we can do is to treat them as algebraic extensions of the factorial series, and make them obey the formal laws of factorials, which are (1) that [= 1, and (2) that n = n n - 1. This will give us 1' = 1[0,.'. 0=1; and 0=0-1,. l=c00 Also -1=-1 -2,.'. 1-2=c, and so on, which exactly agree with the values required above. Armed with these results we may affirm generally that Dr(X) I- a r + X D(0) in~b - r for all integer values of r, whether positive or negative, provided n is a positive integer. It is possible so to define In that the same formula will hold when n is a negative integer, or even when n is fractional, but it is easy to establish a more convenient formula for the former case, and the latter is too difficult to be attempted here. The student should, however, realize that generalizations of formulae to cover all possible cases are of great mathematical interest, and are continually suggesting themselves for contemplation. Indeed, mathematicians have investigated the case when r is fractional, though we can certainly get no simple idea of what is meant by fractional differentiation. ARTS. 121, 122.] SUCCESSIVE DERIVATIVES. 129 122. To find a general formula for D'(x-"), where n is a positive integer. Let y=x-';. Dy = - nx-v-1, D2yy=n(n + l)x-'-2, Dry =(- 1)'( + 1)... ( + r- 1)x-n-r; i.e. D l I+ '-\ ( -1) This formula also holds good for integrations. For it gives n-i2 I n11 W ~ ln- ( ' ~ ~ -- 1) -" which is evidently a function whose differential coefficient is x-". The complete integral is I (+ - 1- (o1 - D (), (n - )x where D-l(0) denotes a constant. Similarly D_~/1\ [n-s-1 (-1)y D- n,=n-s + DIs(O). This formula holds up to s =n - 1 inclusive, but fails to take us back any further, that is, it holds until we reach a multiple of x-1; but it will not give us the integral of this function, which (as we know from Art. 63) is log x, and cannot be included in the above series. In fact, the series of integrals and differential coefficients of xn and of x-~ when n is a positive integer are two distinct series with an impassable gap between them. Thus, taking n = 2, we have x5 x4 X3 2 2, 2, 0 '"' 6' ' 3' 1 2x, 12, 0, and 1, 1 2 6 24 L.D.C. I 2 X 5 ) - L.D.C. I 130 DIFFERENTIAL CALCULUS. ECHAP. IX. The integrals of logx will never lead us to the first series, neither will the differential coefficients of the first series lead to the second series. If, however, n is fractional, there is one continuous series from one end of the infinite range to the other, and there is no discontinuity. Thus, Dz =,and the integrals of xz give an infinite series of positive powers, while its differential coefficients give an infinite series of negative powers. 123. To find D' log x, when is a positive integer. If y = log x, Dy =x-1; D2y= - 1.-2, D3y= -.- 2.x-3,..................... and Dy = (- 1 )yr- x-'; i.e. Do log x =( 1. The integrals are not obtainable by this formula. One method of finding them will be given in the following articles. For the usual methods, a book on the integral calculus should be consulted. 124. Tofind the integral of logx. We can form the series of integrals of log x by making use of the identity logex= h when h=0. (Art. 62.) Thus, to find D-1logx. If Xh - 1 Y= h ' ARTS. 122-125.] SUCCESSIVE DERIVATIVES. 131 1Xh+ h~ 1i-(Xh ~ - I ) =x(log x - 1) in the limit when h=0. This is the first integral of logx. The second integral will be D - '(x log x) - D - Ix. The most useful way of finding this will be to establish a formula fer DD-l(xezlogx), and then apply it to the case when n - 1, as we shall want the formula for several values of n before we succeed in obtaining the whole series. 125. To find the integral of xn log x, and the successive integrals of log X. Writin Writing h for log x, we have D' h X n+h+l Xnn+l) IZ-1 n (Xh- 12 1)' Xn+1 X7'., putting h = 0, D-l(Xn logX) = (log X n 1n + n + I We can now find the whole series of integrals of log x. We have proved that D -'log x= x log x - x; D 21ogx=D-'(xlogx) —D-lx X2i~ri 1\ X2 = log x - - - - = logx — +I; 2 2\2 132 DIFFERENTIAL CALCULUS. [CHAP. IX. D-31ogx= 2 3(ogx- )- 23( +2) X3 2.3 / 1 1\ H3 1.2.31~gx 1.2.31 +2+3) Proceeding in this way, we find D-rlogx= logx-1 1 ---., EXAMPLES. 1. Write down four of the successive differential coefficients, and four of the successive integrals, of (1) x4 -32 + 2x - 1. (2) 1- +- 1 (x + a)5' (4) xA. 2. Find a formula for the nth differential coefficient of, and also x-a for the nth differential coefficient of a-x 3. Find a formula for the 7nth differential coefficient, and for the nth integral, of (1) /Jx; (2) 1A-. 4. From the identity n=nI n- 1, prove that [~_m. I~zl =(- 1)"-1 Im-l ' when m is a positive integer. (Note that the factorials of negative integers are infinite, but this formula shows that the ratio between two of them is finite.) 5. Deduce the formula for Dr (x-m) from the formula D" (xn) xn-r In- r 6. Can you assign a meaning or a value to n when n is fractional? Assuming it to have a meaning, deduce from the general formula for Dr (xn) the fourth differential coefficient of,,x. ARTS. 125,126.] SUCCESSIVE DERIVATIVES. 133 7. Find the rth differential coefficient, and the rtl integral, of Xn Xr (1); (2). 8. Find by successive differentiation the rth differential coefficient of x"e log x, and show that, if r is less than 9i, it can be put in the form I xn-r{lgx+ )- (i-2r)} if o(-) is used to denote 1 + + +\nj 2 3 it 9. Write down the nth differential coefficient of x'i log x. 10. Form the successive integrals of x' log x, and show that D-x (nlog x)= x +" {log x + () - ( ) } 9I nfl-J-r/j which is of the same form as that of DI' (x log x), with the sign of r changed. 11. Show that the limiting value of hl\ - [ In, when h=0, is - \f - 12. Find the value of Df"(x' log x) by using the algebraic expression x7 - 1 for log x, viz. log x= h- when h =0. 126. General formulae for the successive differential coefficients and integrals of certain other important functions can also be obtained; sinx, cosx, sinmx, cosmx, ex, eax are among the simplest. Thus, if y = sin x, Dy = cos = sin x +, D2y = cos + = sin + 2r = 2sQ+ Dry = sin(x + -) 134 DIFFERENTIAL CALCULUS. [CHAP. IX. This gives for the integrals of sin x, D- sin x = sin(x - ) + a constant........................................ Ds sin x = sin S7 + D-(0) D-s cos os=s x -2)+D-s(0). 2. Show that D' sin mx= ml' sin (mx + -+ ), and verify, by differentiation, that D- sinmx=ml sin(mx- ) +D-1 (0) = - -cos mx -+ a, where a is any constant. m 3. Show that D} cos mx=n cos (mx+ ), and deduce its first integral. 4. Prove that (1) Drex=ex. (2) Dr eax ar eax. (3) D-seax=a-sea +D-(0). 127. To find the successive difTerential coefficients of tan x. The differential coefficients of tan x become very complicated unless the work is done in a very systematic manner. It is not possible to obtain a general formula for the nth differential coefficient. The best way of forming the successive differential coefficients seems to be as follows: Let y =tanx;.. D= sec2x= 1 + y2. (This value of Dy will be continually substituted in the succeeding work.) ARTS. 126-128.] SUCCESSIVE DERIVATIVES. 135 D2y = 2y Dy = 2y + 2y3, D3y = (2 + 6y2) Dy = 2 + 8y2 + 6y4, D4y = (16y + 24y3) Dy = 16y + 40y3 + 24y5, and so on. The work in this way can be done quite rapidly, but each stage must be gone through. We cannot write down Dy?/ without differentiating n times. It will be noticed that all the even orders of differential coefficients have y (i.e. tan x) as a factor, whereas all the odd orders begin with a constant. The integral of tan x is log sec x, as can be seen at once by differentiation. 128. To find the successive differential coefficients of sec x. The differential coefficients of sec x also become complicated unless we work very systematically. We shall be able to express them in terms of see x and powers of tan x. To abbreviate these expressions take = sec x, and y =tanx;.'. Dz = sec x tanx = zy, and Dy = sec2 = 1 + y2. Hence D2z = Z Dy + y Dz =z(1 + 2) + y2 =z(! +2y2), D3=z. 4y Dy + (1 + 2y2)D =z{4y(1+ y) + y( + 2y2)} = z(5y + 6y3), D4z = (5 + 1 y2) Dy + (5y + 6y3) Dz = z {(5 + 23y2 + 18y4) + (5y2 + 6y4)} =z{5 +28y2+24y4}, etc. We cannot obtain a general formula, but we can rapidly obtain any differential coefficient by the above method. It will be seen that secx is a factor of all the differential coefficients, and tan x is a factor of those of odd order. 136 DIFFERENTIAL CALCULUS. [CHAP. IX. 129. To find the r7 differential coefficient of eax sin bx. The function eax sin bx is an important function in connection with electrical theory. We can find a general expression for the rth differential coefficient of this function, which, by putting r= - 1, will give us its integral. Thus, if y = e"x sin bx, Dy = eax (a sin bx + b cos bx). Let a be an auxiliary angle, such that - = tan a, a b Ca so that sin a =, a so that (a2 + b2)' cos a =-(a2 + b2).. Dy = (a2 + b2) ex sin (bx + a) = k eax sin (bx + a), where k = \/(a2 + b2). Similarly, D2y = k2 eax sin (bx + 2a), Dry = kr ex sin (bx + ra) = (a2 + b2)2 ea sin(bx + rtan-1A) This gives, as a particular case, D-ly 1 - 2 ea sin (b - tan-1b Y 102 + b2) a) which is easily verified. EXAMPLES. 1. Write down the rth differential coefficient of e4x sin 3x. 2. Show that D' eax cos bx = k' eax cos (bx + ra) where k =/ (a2 + b2), and tan a=b a. 3. Find the rth differential coefficient of eax sin bx, and deduce its first integral. 4. Find the rth differential coefficient of e -Xcos bx, and deduce its first integral. Verify this last result by differentiation. 5. Deduce the formulae for D' eax and Dr e-ax by putting b=0 in the formulae for Dr e-ax cos bx. 6. Deduce the formula for Dr sin bx by putting a= 0 in either of the formulae for Dr e+ax sin bx. ARTS. 129, 130.] LEIBNITZ' THEOREM. 137 130. Leibnitz' theorem. We have seen how to form the successive differential coefficients of various simple functions, and other examples will be found in the next chapter, which is devoted to methods of dealing with rational algebraic fractions. We will now establish a theorem by means of which the nth differential coefficient of the product of two functions of a variable x can be written down provided we know the successive differential coefficients of the separate factors. This theorem is known as Leibnitz' theoren. If u and v are functions of x, we require to find a formula for D'(zv) in terms of u and v and their successive differential coefficients. By Art. 23, D (zv) = D't.v + u. Dv, and, similarly, D2(uv) = (D2u.v + Dmu.Dv) + (Du.Dv + u.D2v) = D2U.v + 2Du.Dv + i.D2v. Proceeding in the same way, we find successively D3(UV) = D3u.v + 3D2u.Dv + 3DM.D2v + ut.D3v, D4(uv) = D4U.v + 4D3ut,.Dv + 6D2i..D2v + 4Du.D3v + t. D4v, It will be noticed that the numerical coefficients in the above formulae are just the same as those which occur in the expansions of the successive powers of a+b by the binomial theorem, viz.: 1, 1 coefficients of (a + b), 1,2,1,, (a+b)2, 1, 3, 3,1,, (a+b)3, 1, 4, 6, 4, 1, (a+b)4, and it will be readily seen that this is because the terms are formed in a very similar way. Thus (a + b)2 = (a2 + ab) + (ab + b2), = a2 + 2ab + b2 and (a + b)3 = (a3 + 2a2b + ab2) + (cab + 2ab2 + b3) =a3 3a2b + 3ab2 + b. 138 DIFFERENTIAL CALCULUS. [CHIAP. IX. This corresponds exactly to the process of finding D3(uv) from the second differential coefficient, D2(uv), provided we work as follows: we have D2(jvV) = D2u.v + 2DU.Dv + u. D2v; to find D3(uv) we have to differentiate separately both factors of each term; we may, therefore, first differentiate each of the factors D2D, Di, uE, and then each of the remaining factors v, Dv, D2v, the result being D3(uv) = (D3w.v + 2D2%.Dv + Dt.D2v) + (D2u.Dv + 2Du.D2v + u..D3v) = Du. v + 3D2u.Dv + 3Du.D2v + u.D3v. The analogy between this process and that of forming the cube of a + b from a2 + 2ab + b2 by multiplying every term first by a, and then by b, and adding, is thus very apparent. It will be seen that, in each process, viz., the process of finding the successive differential coefficients of uv, and the process of finding the successive powers of a + b, the resulting numerical coefficients in each expansion are obtained by adding together in pairs the numerical coefficients in the preceding expansion, and it is therefore not surprising that the results are the same. Hence the nth differential coefficient of uv is n(n - 1) D'u.v + nD-lu. Dv + D~ -2uL.D2v +... + n(-12u. D.-2v +,D D. Dn-v + u. Dnv. 1.2 This is Leibnitz' theorem. 131. Symbolical representation of Leibnitz' theorem. If D1 be used to denote the process of differentiating uT and its successive differential coefficients, and D2 be used to denote the same operation performed on v and its differential coefficients, the differential coefficient of uv may be represented by (D1 + D2)uv; for D1uv will denote Dit.v by the hypothesis, and D2u,, u.Dv ARTS. 130-133.] ALEIBNITZ' THEOREM. 139 Similarly, the expression found above for D2(uv), viz.: D2U.V ~ 2Du.Dv + U.D2v, may be written in the symbolic form (Di2 + 2D1D2 + D22)uliv, which may be condensed into the form (DID + 2 )2tv; and so on, the general formula being DnUV=(DL+D2)nUV. 132. Let us, by way of illustration, find the 71th differential coefficient of x2ex by means of Leibnitz' theorem. Let y exe.x2; n2 (n2- 1) Dy= D~,lj )(ex). X2 ~nD2rlLl(e")D jx2) +. D?-2 (&2)D_2 (X2) + e ~ + 1).2}, since all the differential coefficients of ex are equal to ex, and DX2=2x, I)2(X2) = 2, and all the remaining differential coefficients of x2 are equal to 0. The result may be, of course, written D??(X2ex) = exxS2 + 2nx + n (n - )}. The above is an example of the following important theorem which results from applying Leibnitz' theorem to a product in which one of the factors is eax. 133. To prove that Dn(u.eax)9 0a"x(D+a)nu. If y= U.e(x, ax 11-1 n2 - 2 n-2nDn22+ ~ D'y = e`x D"Wu + n.ae. D"it + -1) a eaxDn1m + ax, )n,,,1.. ea(DD + )a 140 DIFFERENTIAL CALCULUS. [CHAP. IX Thus, in the above example, D'e(x2e)= ex(l + D))x2 =eX(x2 + n.2x + - ~ 2) = exx2 + 2n. x n(n- 1)}, and, more generally, Dn (x2. ea) = eax(a + D)1?x2 =e'~(a x2++ a-'.2x +2( - (1)a -22) '~ea.+na-1x 1.2 a"'2 = a x, - 2ea x2 + 2nac + n (n - 1)}. 134. The formula of last article will give us the successive integrals of xze" by putting n successively equal to - 1, - 2, etc. Thus D-l(x=2e)=X(1 + D)-lx2 =ex(1 -- D + D2 -...) x2, by the binomial theorem, =e(X2 - 2x + 2), D-2(x2e) = D-l{ex(x2 - 2x + 2)} =e(1 + D)-1(x2- 2x + 2) = e(1 - D + D2 -...) (x2 - 2x + 2) = ex(x2 - 2x + 2) - (2x - 2) + 2} =ex(2 - 4x + 6), or, more rapidly, D - 2(x2e)= ex(1 + D) - 22 =ex( - 2D + 3D2-...)x2 =ex(x2- 2.2x + 3.2) =ex(x2 - 4 + 6). The student can verify these results by successive differentiations, starting with ex(x2 - 4x + 6), so obtaining the series y = ex(x2- 4x + 6), Dy = e(x2 - 4x 6 + 2x - 4)= e(x2 -- 2x + 2), D2y = ex(x2 - 2x + 2 + 2x - 2)= ex. X2, D3y = ex(x2 + 2x) = ex( 2 + 2x) D4y = e(x2 + 2x + 2x + 2) = e(x2 + 4x + 2), D5y = ex(2 + 4x + 2 + 2x + 4)= eX(x2 + 6x + 6), D6y=ex(x2 + 6x+ 6 +2x +6)= e(x2 + 8x + 12), etc., using the formula Dexf(x)=ex(l + D)f(x) each time. ARTS. 133-135.] SUCCESSIVE DERIVATIVES. 141 By using the formula Dnexf(x)=ex(l + D)'f(x), the student may also verify the last result as being the 6th differential coefficient of eX(x2 - 4x + 6), or the 5th,,,, e(2 - 2x + 2), 4th,,,, e2 etc., etc. He will find the time well spent in using the above formula backwards and forwards for different values of n until he is certain of being able to use it without error. The formula is very valuable in connection with the solution of certain classes of differential equations. 135. Successive differential coefficients connected by a differential equation. In some cases where we cannot actually find a formula for the nth differential coefficient of a function, we may be able to establish a relation between its successive differential coefficients which may often be of service. Thus, if y = tan x, Dy =1 + 2, = 1 + y.y. We can write down the result of differentiating this i times, using Leibnitz' theorem for the nth differential coefficient of the product y.y. Thus Dn+y- D Zyy = (DI + D2)yyy =yDny + nDyD1'-y + n(n -l)D2yDn"2 +. n(- l)D_ yD2y + nD_-ly.Dy + D"y.y, the latter half of the series being a repetition of the first half, but in reverse order. It may be interesting to take special cases, writing Y, Y2,..., instead of Dy, D2y,..., for brevity. 112 DIFFERENTIAL CALCULUS. [CHAP.!X. Thus Y2 = YY1 + Y1Y = 2yyl, Y3 = YY2 + 21,2 + Y2Y = 2 (YY2 + y12), y4 = 2 (YY3 + 3YlY2), Y5 = 2 (yy4 + 4yly + 3y22), etc. We shall find these series useful in connection with the expansion of tan x in powers of x (Chapter XI.). The differential coefficients of see x can be connected with each other and with those of tan x in a similar manner. Let z=secer and y =tanx, then z1 =tanxsecx=Yz. Hence, by Leibnitz' theorem,,,f+j = yfl + nyf-l/l +-... +f ylrz,, + yz,. In these two examples all the differential coefficients up to the 7th are required in the expression for the n + Ith differential coefficient, but in the case of many functions a much simpler relation can be found. Thus, if y = sin-lx, 1 Yl/(l - X2)' (I - X2) y12=1. Differentiating this, xe obtain 2(1 - x2)yIy2 - 2xy12 = 0, i.e. I (1 -X') Y2 = XY Hence, applying Leibnitz' theorem to both sides, we obtain n(nz - 1)( ) O1 - e) Yn++ n 2x) 2y,,+, n~n - 1 2) y. =xy.+n+ ny 1.2 nXnif~ (1 -X2)yn+2 = (2n + l)xy,,, + n2y. This series will also be referred to again in Chapter XI. ART. 135.1 SUCCESSIVE DERIVATIVES. 143 EXAMPLES. 1. Prove by actual differentiation of the identity nz(n2 - 1 Dn(uv) = Dn2q. v + 9tDn -D'1i.Dv + Dn - 2u. D2V + 1.2 that D"+1(tv)=l)D"1u.v+ (n~ l)Dnm. Dv~ (+ Ii D2v~. 1.2 [NOTE.-The proof of this will establish the fact that if Leibnitz' theorem holds for it differentiations it will hold for it ~ 1 differentiations, and therefore universally.] 2. Find the qeth differential coefficient of x log x; also of x2log x. 3. Find the 3rd integral of x log x. [Use the formula (D2 + DI)-3(X log x)r=(D2 3 - 3DD2-4) x log x, etc.] 4. If y=e - (x2 -3x4), find D5y and D-3y. Verify by deducing D5y directly from the expression found for D 3y, and vice versa. 5. If y= x2sin 2x, find ]Yy and D-3y. 6. If yrrsin (m sin - Ix), prove that (1 - x2) y2 - xy1 + m2y O. 7. Find the result of differentiating the above differential equation n times. 8. If y = eax(A sin bx + B cos bx), prove that Y2 - 2ay,~+ (e2 ~ b2)y = 0. 9. If y=easin- x, prove that (1 -x2)y,-xy,+a2y, and find the result of differentiating this relation n times. CHAPTER X. SUCCESSIVE DIFFERENTIATION-Continued. RATIONAL FRACTIONS-PARTIAL FRACTIONS. 136. Rational Algebraic Fractions. Proper Fractions. In the preceding chapter it has been shown how to find the successive differential coefficients, and the integrals, of certain simple functions. Another very important class of functions, which are called Rational Algebraic Fractions, will be treated in the present chapter. The simplest form of such fraction A is A where the numerator is a constant and the denominax + b tor is an algebraic function of x of the first degree. In general a rational algebraic fraction is a fraction whose numerator and denominator are rational algebraic functions of x (or some other variable), i.e. are expressions all of whose terms are multiples of integer powers of x. x2 Thus (x 1 +3x and and are all (x- l)(x +x ) 7- 2x x+ 1 examples of rational fractions. The last is a complex fraction, but any complex rational fraction can easily be reduced to a non-complex form; also any rational fraction can be reduced to its lowest terms by cancelling common factors between the numerator and denominator. Moreover, if the degree of the numerator is as high as, or higher than that of the denominator, the fraction can be expressed in the form of a Quotient+ a Remainder whose numerator is of lower degree than the denominator, just as an arithmetical improper [ARTS. 136,137.], PARTIAL FRACTIONS. 145 fraction can be reduced to the sum of a whole number and a proper fraction. In the greater part of what follows we will suppose that this has been done, so that the fractions we shall deal with will be in their lowest terms, and will have their numerators of lower degree than their denominators. Such fractions will be called Proper Fractions. Example. Reduce each of the above three fractions to the form of quotient + proper fraction. 137. Simple Fractions. The simplest form of proper fraction is one in which the denominator is of the first degree. Thus is a simple fraction. ax + b It is easy to write down the nth differential coefficient of such a fraction. For, let y = (ax+ b)-1;... Dy= -.(ax+b)-2D(c+b) = -a(ax+b)-2. Similarly, D2y = ( - a) ( - 2a) (ax + b)-3,.......................... Dny - ( - a)( - 2a)... ( - na)(ax + b)-"-l (-a) nn (ax + b)n+l' The first integral of 1 is log (ax + b), as can easily be ax b a verified by differentiating log (ax + b). 1 Next in order of simplicity come such fractions as ( + b) (a + ).... The differentiation and integration of these (ax+b),. present no new difficulties. The numerators need not be unity; all that is needful is that they shall be constants. Other fractions which present more difficulty, but which must be considered as simple fractions, are Ax +B Ax + B (x+a)2 + b2 { (x +a)2 + 2}2 L.D.C. K 146 DIFFERENTIAL CALCULUS. [CHAP. X. the peculiarity of these fractions being that the quadratic expressions in the denominators are irreducible, i.e. cannot be factorized into real factors, and that the numerators are of the first degree. It will be found that all proper fractions can be expressed as the sum of such simple fractions as are indicated in this article. EXAMPLES. 1. Find the nth differential coefficients, and the first integrals of 1 and 1 (ax + b)' (b - ax)' 2. Write down the first integrals of 1 3 2 4 2x-3' (2x- 3)' 2-5x' (2- 5x)2' 3. State which of the following fractions are simple fractions: 3 3 3 x2+2x' x2+2x+1' x2+2x +2 x+3 1 2 4. Show that 2 2x + can be put into the form (X+l) (x+ )2 X2+ 2x+ ( { x+1) (a+l)2+ x3+ 2X2 3x+4 x+2 x+4 5. Showthat (2 + 4)2 can be put into the form - ( + 4)2 (X2 + 4)2 Jax2+4 (x'2+4)2 [Divide the numerator by x2 + 4; it will be found that the quotient is x + 2 and that the remainder is - x - 4. ] 138. Partial Fractions. To succeed in obtaining the nth differential coefficient or the integral of a proper fraction which is not one of the simple fractions indicated in Art. 137, the fraction must, in general, be broken up into the sum or difference of simple fractions whose denominators will be the several factors of the given denominator. This process is the opposite of the process of adding together a number of simple fractions into one fraction, and like most reverse operations it is not so easy as the direct process. The greater part of this chapter will be devoted to explaining methods of doing this work. For purposes of integration such breaking up into partial fractions is essential, and it is important that the student should acquire some facility in the work. The idea is simple enough, but the details ARTS. 137-139.] PARTIAL FRACTIONS. 147 are in some cases troublesome. The various methods of saving labour in the work will be explained by means of examples. We will start with one simple example and then explain the general theory. Example. Find the integral, and the n1th differential coefficient, of 3 x2-4' The given fraction can be expressed in the form.'. its first integral is 4{log (x- 2) - log (x + 2)}, and its nth differential coefficient 3 ) 1 4 1) {(,x- 2)z+l (x+2)n+1) We proceed to consider the various cases which may arise in the treatment of any proper fraction. If the denominator is of the first degree, the fraction is already in its simplest form; e.g. 1 2 z+3' 3x+4' If the denominator is of degree higher than the first, it can, in general, be broken up into factors. Several cases will arise, according to whether the denominator can be wholly broken up into real factors of the first degree or not, and whether any of its factors are repeated or not. For example, the denominators (x - ) (x - 2)(x + 3), (x - 2)2(x + 3), x2 + x + 1, (x2 + x + 1)3 are all of essentially distinct forms, and will require somewhat different treatment. 139. Unrepeated linear factors in the denominator. The first case which presents itself is that of a fraction whose denominator can be entirely broken up into real factors of the first degree, all different. We have to show how to find the simple fractions whose sum will be equal to the given fraction. Their denominators must, of course, be the several factors of the given denominator: the problem is how to find their numerators. The best way of showing the various methods of doing this will be by means of an example. 148 DIFFERENTIAL CALCULUS. [CIIAP. X. Thus 2X2 - 3 Thus (x - 1) (x - 2) (x +- 3) can be made up of the sum of A B C - + z x-2 + -3' where A, B, C are constants whose values have to be such that when the three simple fractions are added together, the numerator shall be 2x2- 3. Now, if the fractions are added together, the resulting numerator is A(x- 2)(x+3)+B(x- 1)(x+3)+C(x-1)(x- 2) = (A + B + C)x2 + (A + 2B - 3C)x - (6A + 3B - 2C). If this is to become 2x2- 3, we must have A+B+C=2 A+2B-3C=0.. A=- B=1, C=; 6A + 3B-2C= 3J 2X2-3 1/ 1 4 3\ (x- 1)(x- 2)(x +3) 4 x-l Z - 2 + 3 Therefore its nth differential coefficient (- l)'1,r^ f 1 4 3 ] _ '! 4 3 4 {(x - 1)"n+l (x- 2)n+ +(x 3)+ f+ and its integral is {og (- ) + 4 lo g ()+41 - 2)+3 log(x + 3)}, as can be at once verified by differentiation. It will be seen by careful study of the above example that while there will always be as many constants as there are simple fractions, so that the number of constants is equal to the degree of the original denominator, the degree of the resulting numerator when the fractions are added together will be one less than the degree of the denominator. Consequently the number of constants is just sufficient to satisfy all the conditions required to make this numerator ART. 139.] PARTIAL FRACTIONS. 149 identical with the given numerator. Thus in the above example there were three constants and three conditions to be satisfied. The method adopted above for finding the values of the constants is easy to understand, but it is not usually the simplest method to use practically. A better way in practice will be now shown. We had, on comparing numerators, to make A(x- 2)(x+ 3)+ B(x- 1)(x+ 3) +C(x - 1)(- 2)= 22 - 3 identically............................................................ (N ) Now, if the two sides of (N) are identically equal they will be equal whatever value we please to give to x. It is easy to choose special values of x which will enable us to determine A, B, C separately instead of by solving three simultaneous equations. Thus if x= 1, the multipliers of B and C will be zero, and we are left with - 4A = - 1, whence A = -. And if' x-=2, we find that 5B=5;.-. B=l. Finally, if x= -3, we find that 20C = 15;.'. C= 3. We ought to verify the results we have obtained, either by giving another value to x, as for example x=0, or by comparing the coefficients of the different powers of x on each side of (N). If these agree our work is right. Another mode of finding the constants will be found interesting and may be used when desired instead of either of the above methods. Let us consider one factor of the denominator at a time. Suppose the given fraction is f(x) and that it is broken (x-a) (x) up into the simple fraction A- + a number of others whose X-a sum will be of the form k (x) Then f(x) A u (x-a)9 (x) x-a + (x) and, comparing numerators, we have the identity f(x) = A+(x) +u (x - a).....................(N) 150 DIFFERENTIAL CALCULUS. [CHAP. X. In this identity put x= a, and we find f(a) = A (a); f(a).. A-,(a) and therefore the partial fraction A f(a) x-a (x- a)k(a)' i.e. the partial fraction corresponding to x - a is obtainable immediately from the original fraction by the simple plan of putting x=a in every part of the fraction except the factor x - a. We will, however, deduce another formula for A which is sometimes slightly more convenient. Let the original denominator be F(x), so that F (x) = (x - a) (x). Hence F'(x) = (x) + (x - a) +'(x). Now put x = a in this identity, and we obtain '(a) = (a). Hence A f(a) - F'(a)' Applying this to the given example, viz. 2X2 - (x- 1)(x-2)(x+3)' the values of a corresponding to each of the three simple fractions into which it can be split are 1, 2, — 3 respectively. Also f(x)=2x2- 3;.-. f(1)= - 1, f(2)=5, f( - 3)=15, and F(x)= 3 - 7x + 6 (as seen by multiplying out);. F'(x) =3x2- 7; and F'(1)= - 4, F'(2) 5, F'( -3)=20;. A=;, B=l, C = as before. NOTE.-It must be remembered that the above method has been established on the supposition that the simple factors of F(x) are not repeated. If a factor is repeated, the method will fail in the case of that factor, but will still be applicable to such of the factors as are not repeated. The method is valid whether the given fraction is proper or improper, but it must be remembered that in the latter case ARTS. 139, 140.] PARTIAL FRACTIONS. 151 there is a quotient in addition to the sum of the simple fractions. 140. Repeated linear factors in the denominator.-We will now consider the case of a repeated factor in the denominator. Take first 2 3+4 X3 This can evidently be broken up into 2 3 4 X- + — X Similarly, if we have 2X2 - 3x + 4 (x - 1)3 we may break it up into A B C - 1+ ( - 1)2 ( 1)3' where A, B, C are constants to be determined. Adding the fractions together, and equating the numerator so obtained to the given numerator, we have A(x - 1)2+B( - 1)+C= 2a- 3a+4 identically...................................................... (N ) The constants A, B, C are just sufficient in number to enable us to satisfy all the necessary conditions. Several ways of finding the values of A, B, C may be adopted. Thus we may equate the coefficients of the different powers of x, obtaining three simultaneous equations. The solution by this method is left to the student. Or, we may try the effect of putting x= 1 in the above identity. This gives us C= 3. We may now, if we like, transfer C to the other side, and divide both sides by - 1. This gives us A(x - 1)+ B= 2 - 1 identically; whence, by putting az=1, we find B= 1; and, equating the coefficients of x on the two sides, we obtain A= 2. Or, again, having found C=3, we may differentiate both sides of (N), obtaining 2A(x - 1)+ B = 4 - 3; 152! DIFFERENTIAL CALCULUS. [CHAP.. X. whence, by putting x= 1, we find B-i1 and lastly A= 2 as before. Another method may be shown by which all the constants can be determined simply. Thus, taking the same fraction 2X2 - 3x + 4 (X - 1)3 let x-I=y; then 2x2- 3x+4=2(y~ 1)2- 3(y+ 1)~ 4 = 2y2 + y ~3; the fraction= 2+y+3 - - 3 Y3 Y+2 23 which becomes, on replacing y by its equivalent x - 1, 2 1 3 X - 1 (x - 1)2 ( - 1)3' This method is very powerful even in cases where the denominator has other factors besides the repeated factor. Thus, consider the fraction 3X3 - 5X2 + 4 3P+9+ 9+3 -5-10-5 (X - l)3(X2 + I) +4 Putting a - 1 = y, the fraction reduces to 3+4- 1+2 3y3 + 4y2 - y + 2 y5(y2+2y~2) If now we divide the numerator by 2 + 2y+ y2 so as to obtain the quotient in ascending powers of y, till the remainder is divisible by y3, we shall have completely determined the partial fractions required, 2+2+y2y) 2- y ~4y2+3y3( 1 - "y3y2 2+ 2y + y2 viz. + 1 3 3 +3y -3y +3y2 + 3y3 y 3 y2+~ 2 2 _3__ _ 3 Y32y2 y y2+2y+2 -~3y -3 -Tyy 1 3 3 3- 3x 6 2+i7/3 2 6y2 +- 6y3+3y4~ (X - 1)3 2(x- 1)2~x_1~ X2+ '33 -!y -3y4 ARTS. 140, 141.] PARTIAL FRACTIONS. 153 NOTE.-The last of the partial fractions has a quadratic denominator which cannot be split into real factors. We shall consider this kind of fraction more particularly in Article 142. Meanwhile all that is necessary to observe is that it is a proper fraction with a first degree numerator, its denominator being of the second degree and irreducible. It belongs therefore to the class of simple fractions (see Art. 137). 141. Method of procedure in the case of improper fractions. If the fraction is an improper fraction, i.e. has a numerator of degree equal to or greater than that of the denominator, it must be put in the form of Quotient + Remainder, as stated at the beginning of the chapter. This work can often be conveniently performed without actual division in the following manner. Let the fraction be 3 5x2+7x 2 x2 - 3x + 2 The quotient will evidently be of the first degree; call it therefore Ax+ B, where A and B are constants to be determined. The remainder will be a proper fraction, and is C D required in the form + x — We therefore assume the x-1 r-2 C D given fraction to be equal to Ax + B +- +-. x-1 x-2 The identical relation from which A, B, C, and D are to be determined is obtained by adding the assumed equivalent expressions into one fraction and identifying its numerator with the given numerator. The relation is, therefore, (Ax+B)(x- l)(x- 2) +C(x- 2) +D(x- 1) = 3x3 - 5x2 + 7x+ 2..........(N) By comparing the coefficients of x3, we see that A - 3.* x2,, B- 3A — 5,.'. B=4. By putting x=l,,, C= -7.,, X= 2,,, D= 20. * It was obvious in the first instance that the first term of the quotient must be 3x, since it is obtained by dividing 3x3 by x2. We might, therefore, have assumed the quotient to be 3x + B instead of Ax + B. 154 DIFFERENTIAL CALCULUS. rCHAP. X. 7 20. the given fraction = 3x + 4 - + 20 x-1 x- - 2 Alternative method. Make the left-hand side of (N) divisible by x - 2 by subtracting D from both sides, and determine the value of D from the principle that x - 2 must exactly divide the right-hand side also, since both sides are identically equal. This condition will give us D-=20, and the result of the division will leave us with the new identity (Ax + B)(x - 1) + C + 20 = 32+ + 9; (Ax + B) (x — 1) = 3x2 +- (1 +C). Now divide by x- 1, and we find C = - 7, and Ax + B = 3x + 4. The student might practise this division method by adding C to both sides of (N) and dividing through by x- 1. These various methods have been given to show that there are almost innumerable methods of working, all depending on the fact that we are dealing with an identity, and that whatever is true of one side of an identity is necessarily true of the other. EXAMPLES. Reduce the following expressions to a series of partial fractions, and write down their first integrals where possible: 3x- 1 2 2x+5. 2+x- 2. x-x3 x2-2 5x+3 (x+2)(x+ 1)2' x3- 9x 5. (b-c)(c-a)(a-b ) 6 4-2 (x - a) (x - b)(x- c)' (2x - 3)3 x4-3x+7 [ D E] 7. x (x +3) Assume the fraction=x2+Bx+C+ -4+ (X -4)(x + 3) Lsm x-4 x3 3x2 - 5x+2 9 2X2-3x+4 (. x+2)2(x 3)2 (x- 1)(x 2)' 10. 4x - 5 4x3 - 3x + 2 (x - 2)3(2 + x + 3)' (x - 2)3(x2 + 3x 2) 12. 13 (x + 1) (2x + 3) ARTS. 141, 142.] PARTIAL FRACTIONS. 155 142. Irreducible quadratic factors in the denominator. In the previous investigations, methods have been given for facilitating the work of replacing a fraction by a series of equivalent simple fractions when the denominator of the given fraction can be split into factors of the form (x -a)". The simple fractions corresponding to each such factor are of the form A B K (+V - + + (x - a) (X - 6)2 +... (- ) In addition to this we have seen how in some cases to proceed when one of the factors of the denominator was of the second degree, and was not able to be split into two real first degree factors. The general type of such a quadratic factor is (x + a)2 + b2. In such cases the corresponding partial fraction has a numerator of the form Ax + B. (See Art. 137, and the last example of Art. 140, where one of the partial fractions was 2-3 1). In the case referred to, the numerator was found x2+l/ incidentally as it were, in the process of finding the other partial fractions. It is not always possible to proceed in that way, however, and it is desirable to consider other methods of finding the numerators in such cases. These methods we shall show by taking special examples. 3x2 - 5x ~ 1 Take for example the fraction (x2 - 5)(x+ (X2 + 1)(X2 + 4) This can be broken up into two fractions of the form Ax+B Cx+D x2+1 + x24 where the four constants A, B, C and D, are to be found by adding the assumed fractions together, and identifying the numerator so obtained with the numerator of the given fraction. The identity required is thus (Ax + B) (2 + 4) + (C + D) (x2 + l)= *3x2- 5 + 1. (N). *This useful symbol is often used to signify an identity, i.e. an equation in which the two sides are to be equal absolutely, whatever value x may have. We might have used it in the previous articles. 156 DIFFERENTIAL CALCULUS. [CHAP. X. lfethod of equating coefficients. The fundamental method of finding the constants is to equate the coefficients of the different powers of x on each side of (N). Thus Coefficient of x3............ A + C =O. x2............ B+D= 3. x............4A+C= -5. Constant term..........4B + D = 1. These four equations are just sufficient to determine the constants. On solving them, we find 5 2 5 11 A=-, B=-, C=, D=y. -3' 3- t —3' 3 We may, however, just as in the previous articles, adopt other methods of finding the values of these constants, all depending on the above identical equation between the numerator of the given fraction and its assumed equivalent, which we have denoted by (N). Method of forced division. If 3(Cx+D) is added to the left-hand side of (N), it will become divisible by x2 + 4, the quotient being Ax + B + Cx + D; and if the same quantity is added to the right-hand side to preserve the identity, it must also be divisible by x2 + 4. The modified identity is (Ax + B + Cx + D) (x2 + 4) = 32 - 5x + 1 + 3 (Cx + D) 32 + (3C - 5) + (3D + 1). Now, since the right-hand side must be a multiple of x2+4, and its first term is 3x2, it must obviously be 3(x2+ 4), i.e. 3x2 +12, hence 3C-5= 0 5 11 3D+1 =12 ' 3' 3 and, on performing the division, we have the identity Ax+B+Cx+D- =3; A+C=0 5 2 B+D=3 3' 3 This method has taken some time to explain, but it s ery This method has taken some time to explain, but it is very ART. 142.] PARTIAL FRACTIONS. 157 rapid in practice, and it is well worth a little trouble to get used to it. The student might, by way of exercise, make the original identity (N) divisible by x2 + 1, instead of by x2 +4, by subtracting 3 (Ax + B) from both sides.* Method of substitztion. If in (N) we put x2 = -1 (which, it must be noted, makes x imaginary), we shall be able to find A and B without being hampered with C and D, since the coefficient of Cx + D will be equal to zero. Similarly if we put x2= - 4, we can find C and D. (The fact that these substitutions make x imaginary is of great importance, as it enables us to find two coefficients at the same time.) First, to find A and B. Since X2= - 1, (N) becomes 3(Ax+B)= - 5x-2 5 2. Ax+B= — x-. 33 5 Since x is imaginary, we must therefore have A= -5 and B 2 3' Secondly, putting x2= - 4, (N) becomes - 3(Cx +D)= -5- 11; 11 5 11.'. Cu+D=-u+ —,.'. C=,andD=NOTE.-In using this method, it is unnecessary to substitute for x its imaginary value; we keep the symbol x, merely remembering that it is imaginary, but we substitute for x2 wherever it occurs. The fact that x is imaginary makes the resulting equation an identity, so that we might write Ax+B - (5x+2), and Cx+D= (5x+ 11). * It is not necessary to make the two sides exactly divisible: we may divide both sides of (N) as it stands by either x2+4 or x2+1 and note that the quotients on the two sides must be identically equal, and so likewise must the two remainders. Perhaps this is the neatest way of performing the operation. 158 DIFFERENTIAL CALCULUS. [CIHAP X. The original fraction, viz. 3x2 - 5x + 1 (2 + 1)(+2 + 4)' is therefore equal to 1(5x+11 5x+2} 3 X2+ 4 I 1). This is the simplest form to which it can be reduced. 143. The nth differential coefficient of a fraction with a quadratic denominator is not simple, and as it is not practically of great importance we shall only give it at the end of the chapter, to be omitted on first reading if desired. The integral is not very simple, but it is important. It consists of two parts, and depends on the two following formulae which the student already knows, or can easily verify by differentiation. 2x (1) The integral of 2x is log (2 + a2).. 12 + a2 x (12) is -tanlX ",,",, f x2 + a- a 5x 5 Thus the integral of - is log (x2+4). 11 is 11 - 7is 2 tan-~. "9 "a ~X;2 + 2 2 32 - 5x + 1 the integral of 2 - 5 (X2 + I)(2 + 4) 5 a x2+4 11 x 2 is log r2+4I+ tan-2 - tan-Ix In some of the following examples the integration is less simple, but this does not much matter, as it may well be left till the student is working at the integral calculus and has ARTS. 142-144.] PARTIAL FRACTIONS. 159 learnt how to treat all such fractions; the special object of this chapter is to show how to split up any given fractions into a number of simple fractions which shall be integrable by the proper methods. We shall, therefore, not attempt to integrate any of the remaining examples in the chapter. 144. Repeated quadratic factor in the denominator. 3X3 - 5X2 + 7x- 9 Such a fraction as 3(52 + z1 (x 2 ++ 1)3 can be broken up into the sum of fractions of the form A1x + B1 Az + B2 A3x + B3 x2 +-+ 1 (x2+x+ 1)2 (X2 + 1)3; for if we add the three assumed fractions together, the numerator will be of the fifth degree, and the identification of this numerator with the given numerator will lead to just six equations among the six constants, from which they can be determined. The equation of identity between the numerators is (A1x + B1)(x2 + x + 1)2 + (A2x + B2)(x2 + x + 1) + (A3, + B,) =3x - 5X2 + 7x- 9...............oo (N) Using the method of equating coefficients we see at once that A = 0 and B1 = 0, since the coefficients of x5 and x4 are both zero. Equating coefficients of x3 A = 3.,^,,x,, 2 A2 + B2- 5,.5. B - 8.,,,,, x x A2+B2+A3=7,.. A3 =12.,, constant terms B + B3= -9,. B3= - 1. If the given numerator had contained higher powers of x, this method would have been troublesome. Probably the best method then would have been that of forced division, dividing both sides of (N) by x + x + 1. This we will now do. The remainder on the left is evidently A3x + BS, and that on the right will be found to be 12x - 1, the quotient being 3x - 8. Hence A3x +B3 12x-l, 1 /. ~~~3 and (Aix + BI) (x2 + x + 1) + (A2x + B2) 3 -8. 160 DIFFERENTIAL CALCULUS. (CHAP. X. A repetition of the same method shows obviously that A2x+B2 3x- 8, and Aix + B — 0; the fraction may be replaced by 3x - 8 12x-1 (x +X+ 1)2+(x2+ + 1)3 whose numerators are of the first degree only. The sum of these fractions is the simplest equivalent to the given fraction. The student will learn how to integrate such fractions when he comes to work systematically at the integral calculus. NOTE.-The values of A. and B3 above could have been found by the method of substitution, by putting x2= -x-1, and, therefore, 3 = - x2 -x = 1; whence Ax + B = 3 + 5 (x + 1)+ 7x- 9 =12x- 1. This method is really a mere modification of the division method, not quite so good, usually, as actual division, for it gives the relation between the remainders only, instead of the relation between the remainders together with the relation between the quotients, both of which are simultaneously given by the division method. However, the method of substitution is interesting, and occasionally it may save trouble. 145. General Summary. The denominator of every rational fraction can be, in theory at any rate, broken up into first degree factors, or quadratic factors, or a mixture of the two, such factors being repeated or not. This we shall not prove, but take it as a known fact. Any proper fraction can, therefore, always be replaced by a series of partial fractions of the types already considered, and the numerators found by means of one or other of the preceding methods. The number of arbitrary constants which have to be assumed is in every case equal to the degree of the denominator of the given fraction. An improper fraction can be replaced by a quotient + a series of simple fractions. ARTS. 144, 145.] PARTIAL FRACTIONS. 161 The general type of such equivalent quantities may be represented as under. Let the given fraction be f ( ) (x - C)... {(x- b)2 +... Then it may be replaced by Quotient ++ ( - + A.-, )- + + Bix f Cl G Bsx + CS X b (>< _ &)2 + C2 + **'+ {(T - b)2 + 2c"sThe various methods of finding the numerators may be classified as (1) The fundamental method of equating coefficients of the different powers of x in the equation of numerators which we have called (N). (2) The method of substituting special values of x in the same equation. (3) The method of forced division. (4) Differentiation of both sides of the equation (N), followed by use of any of the other methods. (5) The use of a formula in the case of an unrepeated linear factor. (See p. 150.) EXAMPLES. Replace the following fractions by their simplest equivalent partial fractions: 3x3+4x - 2 1 i. 2. x4 +i;2 ' 2 +4 + 1 ' 3x2+2 4 x4 X4 - + ' xx 1' x5 x2 - 7x- 2 5. 6. ~5' + 27' (x- 1)(x2+ 1)2' 2X2 + 3 x2 - 7x - 2 (x3- 5)2 3- 1)(x2+ 1)3 3X5 - 5.x3 + 7x - 2 x4 + 9 9. (-r2 + X + 1)3 10. +<2+x+ 1)03 (X2 +4x + 8)3' L.D.C. L 162 DIFFERENTIAL CALCULUS. [CHAP, X. 146. To find the nth differential coefficient of 1 This can only be obtained in a manageable form by an artifice. Let x= cot 0, and therefore dx = -cosec20 dO, whence DO= - sin20, if D is used for, for brevity. dx Using this substitution, we have 2 = sin2;.. Dy=2sin 0cos0.DO =- sin20sin 20, D2y= - 2(sin 0 cos sin 20+sin20cos 20) D = 2 sin30 sin 30, D3y = 23 sin20(cos 0 sin 30 + sin 0 cos 30)D1 = - 13 sin40 sin 40, D"y = ( - 1)" jlsinn+l 0 sin (2 + 1) 0. This is the formula. The first integral of 1 +2, which we know is tan-x, or, more generally, tan-Ix + any constant, can be deduced from the above formula by putting n = - 1, and remembering that we may replace sin (n + 1) 0 by (n + 1) 0 in this case. (Art. 32.) We shall obtain D-ly= - 0, the general integral being- 0 + a constant, which agrees with the above, since 7r tan-x = - 0. 2 147. To find the nth differential coefficient of x 1 + x2' This can be found by the same substitution, viz. x=cot 0, D= - sin20; ARTS. 146-148.] PARTIAL FRACTIONS. 163 x cot 0 whence, if -1 + 2- co2 — sin 20, 1+ X2 cosec20 2 Dz= - sin20cos 20; D2z= - 2 sin 0(cos 0 cos 20 - sin 0 sin 20)DO = 2 sin3 cos 30;..............ooo.oo.....o o o,............. and, finally, Dnz =( - 1)" Insin"+l cos(n + 1)0. The above result might also have been deduced from the formula for Dn(1 12) by help of Leibnitz' Theorem. 1 x For, putting, as before, y = and = 1 + 2 we have z=x y; '. = xyn + nn-l =(- 1)n lsin"0{cos 0sin(n + 1) sin n0}, which is easily reduced to the previously obtained expression ( - ) [ sin +0 cos (n+ 1) 0, since cos sin(n+ 1) 0 - sin cos(n + 1)0= sin 0. The integral of x is 1 log(l + 2). This cannot be ob1 + X2 tained readily by the artifice of putting n = - 1 in the general expression above. Ax + B 148. Various other expressions of the form -(+ can be (x2 + )r expressed in terms of and and the successive x2 + X1 x2 +1 differential coefficients of these functions, and thus expressions found for their nth differential coefficients and first integrals. But other simpler methods of finding the first integrals of all such expressions will be given in a book on the integral calculus, so that the methods here given are rather interesting than important. Two examples are given below by way of illustration. 164 DIFFERENTIAL CALCULUS. [CHAP. X. (1) To find the nth differential coeficient of 1)2 (W +1)2' By straightforward differentiation we find, after rearrangement, 2 1 Dx2+ 1 (x2+1)2 x2+ 1 1 1/ 1 x \ Hence D(2 )2 (,, D x (X2+-: 1 2s -2 (;T+ + and, in particular, D= 1anx + x ) (2) To find the nt" differential coefficient of ( (W + 1)3' We have at once, by differentiation, 1 4x (I+1)2> W( + 1) 'Y x 1.D'" _ l-}D+i (x2 + 1)3 4 (2 + 1)2 EXAMPLES. Find the nth differential coefficient of 1 1 2. x 3. _ X2 + a2' X 2+a2' (x —a)+b2' 4 x 5 3x+4 (x - a)+b2 x +4x+8' 1 7 x+2 6 (2+4x+8)2' (X2+4x+8)2 8. By differentiating X1 find a formula for the nth differential 1 (X + 1)2 coefficient of (x -1)3, and find its integral. 149. The following list of most of the more important results of this and the preceding chapter may be of use as a summary, and be of help to the student in the process of revision: RECAPITULATION. D-'u D-lu u Du D'u \n xn+n In Ii +r n + 1 lr -r Ax r- 1 + Bx- 2 + + K constant 0 0 0 Izn-r-1 (-1)r 1 1 n {n+r-1 (-l)rr[logx-l-~-... — x(logx- 1) logx - - n-sinx 2- ) sin(x- |. (a 2) sin la - 1I = - cos x =cosx s mill - - - - sin x - m sin x + sin X+ m1' \ 2 7 m \ 22 2 2 / cos mx = -- COS X - COS mx 1 ( rr7\ 1 t 7 7( w ( + r77 - cos(iflX — -S ) cos7T - CO 2! s 7Th iii + yi ) it cos?tmx + 9 ) 'n" 2 m1 2 2 COS CO X a -2e ex e - 2ax O e2 a _ where k= ^/(a2+l2) 1 1 a (-a i)f tan - x 1 - 2x (-1)r r sina lk sin (r~ 1+) or -cot1x x- +1 (x- 1)2 where cot = x _iog(X ) 1) rsin+ I = -cos(rs+r1) (DC + D2) u (DI D2luv uv (DI +D2) (DI+Dx + =Du. v + it. Dv eax(a + D) -7u eax(a + D)-u eax"s kea(a + D) it eai(a + D)ru wahere D-= it(a2B+b) CHAPTER XI. TAYLOR'S AND MACLAURIN'S THEOREMS. 150. Taylor's Theorem. We know that when h is small, the formula f(a + h) -f(a) = hf'(a) is approximately true, being true to the first power of h. [See Chap. VI., also Chap. I., in which dx is used instead of h, and powers of dx beyond the first are neglected.] We desire now to find the complete expression for f(a + h) -f(a), so as to be able to use it for finite values of h. This complete expression will have terms containing h2, h3, etc. Thus (a + h)2 - a2 = h(2a) + h2; and (a + h) - a3 = h (3a2) + h2(3a) + h3. We will establish a formula which will enable us to determine the coefficients of the different powers of h in every case where such a series is available, i.e. in the case of all functions which obey certain conditions. The formula, which is called Taylor's Theorem for the expansion of a function of a + h in powers of h, is f(a+h)=f(a)+hf(a)+hf'"(a)+ f"(a)+ 3f(a)+. The student should verify that this formula holds in the case of the above two examples, and, before proving it, he might make himself familiar with it by working the examples below, [ARTS. 150, 151.] TAYLOR'S THEOREM. 167 EXAMPLES. 1. Expand (a+ h)n in powers of h by Taylor's Theorem. [It will be found that this gives us the Binomial Theorem, which is thus seen to be a particular case of Taylor's Theorem.] 2. If x= a + h, arrange X4 - 3x3 + 72 in powers of h by actual expansion, and show that the terms follow the law given by Taylor's Theorem. 3. Expand ea+^ in powers of h, and compare the result with the expansion of el given in Chap. IV. 4. Expand sin (a + i) in powers of f, and hence deduce the expansions of sin f and cos / in powers of 3. 5. Expand loge(a +h) in powers of h, and deduce the expansion of log, (l + x). 6. Expand log sin (x + h) in powers of h as far as the term involving h3 inclusive. 151. Mode of obtaining Taylor's Theorem. There are several instructive ways of obtaining Taylor's Theorem. The simplest way is to use the method we have before adopted to find the expansions of ex, sin x, and other simple functions, i.e. to assume an expansion in powers of h with unknown coefficients, and then proceed to find what those coefficients must be. This method presupposes that such expansion is possible, which, we shall see later, is not always the case. Assume f(a + h) -f(a) = Alh + A2h2 + A3h3 +..............(1) This, by supposition, is an identity, and therefore if we differentiate both sides the resulting expression will also be an identity. Differentiate both sides with regard to h, keeping a constant.. f'( + h) dh = (A + 2A2h + 3A3h2 +...) dh;. '. f(a + h)= A + 2A2h + 3A3h2 + 4A4h3 +..........(2) This being an identity, we are again entitled to differentiate both sides with regard to h, obtaining another identity; and so on as many times as we please. We thus find f"(a + h)= 1.2A2 + 2.3A3+ 3.4A4h2+...;............(3) f"'(a + h) = 1.2.3A3 + 2.3.4Ah + 3.4.5A5h2 +...;....(4) fiv(a + h) = 1.2.3.4A4 + 2.3.4.5A + +...;.............(5) DIFFERENTIAL CALCULUS. 168 [CHAP. XI. Now, as these are, by supposition, identities, they must hold for all values of h. Hence, putting h=0 in (2), (3), (4),... we find f'(a) = Al; f"(a) = 1.2A2; f"'(a) = 1.2. 3A; fiv(a) = 1.2.3.4A4; These results give us the values of the successive coefficients; and inserting these in (1) we arrive at the required formula, viz.: h2 h3 f(a + h) - f (a) = hf'(a) + 1. /"(a) + 93f('"() +.... NOTE.-In the above work we have assumed that all the differential coefficients of f(a + h) are functions which are finite when h= 0 and are continuous throughout the range of values of h, as if they are not continuous we cannot differentiate them. If they are all finite and continuous throughout the range, and if, in addition, the terms of the series which contain high powers of h become unimportant, the theorem is both true and numerically useful. 152. Example illustrating the limitations of Taylor's Theorem. It is easy to give an illustration of an expansion which is true and intelligible for certain values of h, but not for others. Thus 1 L= 1 + hh2+ h3+..., as can be proved by means of the theorem, or, more simply, by actual division. Now, this apparent identity is true when h is less than unity, but if h is greater than unity we get a negative quantity on the left-hand side and an infinitely great positive quantity on the right-hand side, so that the two sides cannot in this case be equal. It is noteworthy, in connection with the failure of the series when hi is greater than 1, that, when h = 1, both - and its differential coefficients become infinite and bot -— h ARTS. 151, 152.] TAYLOR'S THEOREM. 169 some of them discontinuous, having great positive values when h is just less than 1, and great negative values when h is just greater than 1. This discrepancy between the two sides is very curious in such a simple example. It shows how very careful we must be. We can find out how it arises in the present case, and will be able to apply a correction which will restore the identity of the two sides of the equation. For, if we go through the process of division, we find 1 ( h h =1 +hQ)+:l+hh+/h2+3+. ~h__l+ 1(^h) Here we have an identity whatever h may be. Moreover we have, not an infinite series, but as many terms of the series as we please, followed by a remainder which is not in the form of a series. Now, when h is less than 1, the remainder after n terms of the series can be made as unimportant as we please, by increasing n. But if h is greater than 1, the remainder becomes more and more important the greater n is, and is moreover negative. The terms of the series together with the remainder are equal to the given function, although the sum of the terms by themselves is more and more remote from the value of the function the greater we take n. We are led to the conclusion that the infinite series represents the function properly when the remainder after n terms can be made as small as we please by increasing n, but not otherwise. The conclusion is consequently inevitable that the function and the series 1 + xx2 +... to oo are not 1 -x identical, but are equal to each other through a certain range, and entirely unequal beyond that range. 170 DIFFERENTIAL CALCULUS. [CHAP. XI. In fact the real identity is 1 xn = 1 + z + X2 + + Xn-l 1 - 1 -x being true for all values of n and all values of x whether great or small. If we divide this identity by xn we have the identity 1 1 1 1 _- -+... +. x(l -x) 1 - x ' X From the first of these two identities we see that, when x is numerically less than 1, - = 1 - x+2 + 2... to infinity, and from the second we see that, when x is numerically greater than 1, 1 (1 1 - -(+ + -( + -.+.Q+... to infinity) 1-x \x x3" showing that, when an infinite series of positive powers of x fails to represent a function, it may be possible to expand it in a series of negative powers. In this case the expansion in negative powers could have been obtained by putting z=x in the function, and then expanding in positive powers of z. The student should draw the graph of _-, and also of 1 - x 1 -x, + x+x2, 1 +x+2+ 3, etc. He will see that by taking more and more terms of the series the corresponding graph lies closer and closer to the graph of 1 between the limits x= 1, but not beyond I - x those limits. He should notice also that when x= - 1, the ordinate of the function graph has the value 2, whereas those of the series graphs have alternately the values 1 and 0. In the limit when we take the graph of the infinite series, it will coincide with the graph of between the values 1 -x ARTS. 152, 153.] TAYLOR'S THEOREM. 171 x = 1, but is infinite beyond those limits; and, when x = - 1, its ordinate is not - but either 1 or 0 according to whether we suppose the last term to be an even or an odd power of x. The series therefore is equal to the function throughout the range for which the distant terms of the series are unimportant. The student might also draw the curves 1 1 1 Y =- = --—, etc., from which he will see that the more terms he takes the nearer will the graph approach the graph of the function beyond the limits x= ~ 1. This example has been dwelt on at some length to show clearly that Taylor's infinite series is not alwvays identically equal to the function from which it is derived, but only for a certain range of values of x, and to show that the apparently simple proof given in Art. 151 slurs over a good number of difficulties and limitations. 153. Second proof of Taylor's Theorem. Another instructive mode of arriving at Taylor's series, which will indicate the need that f(x) and all its differential coefficients must be continuous and finite throughout the range of values of x contained between a and a + h inclusive, and that the end of the series must be unimportant, is by gradually building up the finite increment h by a succession of infinitely small equal increments dx. We will first take these increments small but not infinitely so, calling them 8x, and then find what the series becomes in the limit when they are infinitely small and infinite in number, their sum being equal to h. Denoting f(a) by y for brevity, and thinking of the values of the function as ordinates of the graph of f(x), the ordinate corresponding to x = a + 8x will be y + 8y, and the next ordinate, viz. when x= a + 28x, will be y + 8y + 8(y + y)= y + 28y + 8 (8), which is written y + 28y + 82y (see Chap. VIII., p. 121). Similarly, when x = a + 38x, the ordinate = y + 28y + 82y + 8(y + 28y + 82y) =y + 38y + 382y + 83y, 172 DIFFERENTIAL CALCULUS. [CHAP. XI. and, when x= a + m8x, the ordinate m (mn - 1) 82y (m - 1) (m - 2) 3+, 1.2 1.2.3 the coefficients being evidently the same as those which occur in the Binomial Theorem. If, now, we make mx = h, we obtain y 2 l-~ f(a+t)=y + ) = + 2 (8x)2 hA ( 1-)( in) 83 + 1.2.3 (ax)3' Now let 8x be made infinitely small, and consequently m infinitely great: the series becomes f(a + )= y + h c+ 12 +.. = f(a)+ hf'(a) + 12 f"(a) +. We here assume that neither the function nor any of its differential coefficients become infinite or discontinuous between x= a and x = a + h, inclusive, and that the infinitely distant part of the series is unimportant. If these conditions are satisfied, the series is equal to f(a + h). NOTE.-It is not essential that the limiting value of f"(x) shall be finite when n==o, so long as the limiting value of Ah'f/(x) + In is zero. But, of course, f'(x) must not be infinite for any finite value of n. In many cases the conditions will be satisfied for a certain range of values of a and h, but not for all values. So, to repeat the previous caution, we cannot say that the series is the function, but only that through a certain range the two are equal in value. 154. In some cases we may not be able to form all the differential coefficients of a function, so that we may not be ARTS. 153-155.] TAYLOR'S THEOREM. 173 able to tell, except indirectly, whether all of them are finite within the range of values required. But if the function and a certain number (n) of its differential coefficients do fulfil the condition, we can find an approximate expression for the remainder of the series which will enable us to estimate to what extent the sum of the first n terms give an approximate value of f(a + h). We shall find, by help of this approximate remainder, that, in practically all cases, when h is small a few terms of the series will be a real approximation to the value of f(a + h). This discovery is of great importance, as one of the chief uses made of the series is to find, when h is small, but not small enough for us to be content with its first power only, by how much f(a +h) differs from f(a). It is therefore very important to know to some extent the limits of the approximation obtained when we use only a few terms of the infinite series. 155. Taylor's Theorem with the Remainder after n terms. Suppose f(x) and its first n differential coefficients are finite and continuous between x=a and x=a+h. We shall prove that f(a+h)=f(a)+hf'(a)+-h f"(a)+... +hn1 fn- 1(a) +_ fn(a+Oh), where 0 is a positive proper fraction, so that a + Oh is intermediate in value between a and a + h. Assume f(a + h) -f(a) - hf'(a) -... - n-() H. We want to prove that H =f (a + Oh). zn-l Zn Let (z) -f(a z) -(a) - f'(a) -... - "-' (a) — H In- 1 - ' and consider the changes in value of 4(z) and its derivatives as z changes from 0 to h. 174 DIFFERENTIAL CALCULUS. [CHAP. XI. By supposition, (z) and its first n differential coefficients are finite and continuous between the limits z = 0 and z = h. These differential coefficients are en-2 an-_ ' () - f (a + z) -f' (a) - zf" (a) -... (a) el-3 X —2 "() - f( +z) -(a) - ff'(a+) -f -(a) - - H, ~v'-1 (X) -f n-1 (C + z) _-f'n-1 (a) - zH, V (z) _-f (a + z) - H. It is to be observed that k(z) and all its derivatives down to "-~ (z), inclusive, vanish when z = 0. But "n(z) does not as a rule so vanish. Also +(z) vanishes when z=h, by our original assumption. Hence (0) and + (h) are equal (viz. zero) values of (z). But between two equal values of a function it must have a maximum or minimum value, therefore q)(z) must have a maximum or minimum value between z = 0 and z = h. Therefore, since b'(z) is finite and continuous, qb'(z) must be zero for this intermediate value of z. Call this value Olh, where 01 denotes a proper fraction. +'(z) is therefore zero when z= 01h; but it is also zero when z=0; therefore between z=0 and z = Oh, 4'(z) must reach a maximum or minimum value. Hence ("(z) = 0 for this value of z, which we may call 02h. Hence /'"(z)= 0 when z = 0, and also when z= 02h. Therefore +"'(z) must equal 0 for some intermediate value of z, say z= 0,h. Proceeding in this way we find, finally, that n~l(z) = 0 when z=0 and also when z=O _,h, where 0,-_ is a proper fraction. Hence A"(z) vanishes when z= Oh, or say z = Oh, where 0 is a proper fraction..'. H =fn(a + Oh), which was to be proved. It is to be observed that the form of Taylor's series thus proved is not an infinite series at all, but a finite number of ARTS. 155, 156.] TAYLOR'S THEOREM. 175 terms (of the same form as those of the infinite series) followed by a remainder. This remainder is not a perfectly definite expression, like that which was found in connection with the function of Art. 152, as we only know that 0 is some proper fraction, but nevertheless it is an indication of the error to which we are liable if we stop short at n terms of the series. The error cannot be greater than - multiplied by the greatest numerical value of f"(x) between the limits x=a and x=a+h, and it cannot be less than h- multiplied by the least numerical value of fn(x) between those limits. It must not be overlooked that the above proof requires all the differential coefficients of +(z) up to '2(z) inclusive to be finite and continuous, as otherwise we cannot say for certain that each one of them must be zero when the preceding one is a maximum or minimum. This requires that f(x) and all its derivatives as far as f"(x) shall also be finite and continuous throughout the range x = a to x = a + h, since + (z) and its derivatives involve all these quantities. [NOTE.-It would have been sufficient, however, to assert that fn(x) must be finite and continuous throughout the range, for a little consideration will show that if any derivative became discontinuous or infinite, all the succeeding ones must follow suit. Hence, if f"(x) satisfies the conditions, all its predecessors must also satisfy the conditions.] 156. If all the derivatives of f(x) are finite and continuous for all finite values of n, however great, and if, also, by increasing n we can make -nf"(a+ Oh) as small as we please (even though f"(a + Oh) may itself tend to be infinite), we may proceed to the infinite series for f(a + h). [N.B.-If, when x =, the limiting value of f"(a + Oh) is itself finite, the above condition is a fortiori satisfied. The student can take, as examples of the two cases, the series for e2x and for ex.] 176 DIFFERENTIAL CALCULUS. [CHAP. XI. Hence, to recapitulate: If f(x) and all its derivatives to the nth are finite and coitinuous between x = a and x = a + h, f(a + h) =f(a) + hf'(a) +... +- fn(a + Oh); and if, further, by increasing n, we can make In f"(a + Ah) as small as we please, we shall have h2 f(a + h) =f(a) + hf'(a) + 1- f"(a) +... to infinity. The two series are not identical. One is called Taylor's series with a remainder after n terms, and the other, Taylor's series to infinity. 157. Convergency of the series. When, on increasing n, we can make - f'(a + Oh) as small as we please, the infinite series h2 f(a) + hf'(a) + 2 f"(a) +... to infinity is said to be convergent. Its sum is finite and equal to f(a + h). If the condition is satisfied for all values of a and h, the series is said to be universally convergent; if only for a particular range of values, it is said to be conditionally convergent. Thus the series for 1 is conditionally convergent; the series for _-x e", sin x, and cos x are all universally convergent. 158. Use of Taylor's Theorem for Tabulation. Illustrations of the use of Taylor's Theorem will be given in next chapter. We will however give some examples in this and the two following articles. Suppose it required to form a table of values of x(144 - x2) for values of x=1-5, 1'6...to 2-5. We may work as follows: Le f(x)=x(144 - x2);.. f(2)= 280;.. f'() = 144 - 32;.. f(2) = 132; f"(x)= - 6x;.'. f"(2)= - 12 f'"(x)= - 6;.. '"(2)= -6. iRTS. 156-159.] TAYLOR'S THEOREM. 177 Now, by Taylor's Theorem, f(2+h)=f(2)+ f'(2)+ h 2)+..;.. f(2 + h) = 280 + 132h - 6h2 - h3;.. f(2 1)=280+13-2-0 06-0-001=293-139, f(2 2)= 280 + 26 4 - 0 24 - 0008 = 306 152, f(2 3)= 280 +39 6-0 54-0 027=319 033, ind so on. Also f( 1 9) = 280 - 13 2 - 0*06 + 0 001 =266 741, f(1 '8) = 280 - 26 -4 - 024 + 0 008 = 253 368, f(1 7) =280 - 39 6 -0 54+0 027 =239-887................................................. It is obvious that these calculations are much simpler than the disconnected work which would be required by straightforward evaluation without the help of the series. 159. Use of Taylor's Theorem for approximating to roots of equations. As another example, let us find an approximation to the root of an equation f(x)=0 when we know a first approximation x =a. Suppose f(a) = A, where A is small, and that J(a + h)=O, where h is a small correction to be found. The equation is f(a) + hf'(a) + -f"(a) +... = 0. or say, for brevity A + Bh + Ch2 +...= 0, and let us not go beyond the square of h. A The first approximation, neglecting h2, is h= - A This is the approximation given in Chap. VI. An improvement can be found by writing the equation for h in the form Bh= - A -Ch2; A Ch2 *.-B- B L.D.C. M 178 DIFFERENTIAL CALCULUS. [CHAP. XI. Now, on the right-hand side insert for h its first approximate value, - A + B, and we obtain as a closer approximation, A CA2 B B3 i.e. hi- A A2 f"(a) f'(a) 1.2 (f'(a))3' In this way h could be expressed in a series of powers of A. The above is correct as far as A2. The next term would involve A3, but is not simple enough to be of much use. A A2 f(a) The root is a - a-2 f'(a) 1.2 (f'(a))3' where A =f(a). Probably in practice it would be better to express the root as a+h - f,(a) h2, where = =-(a) Example. Find by the above formula the roots of the equation X3 - 4x2 - lx + 32= 0, being given that the approximate roots are 5, 2, and -3. Ans. 4-913, 2'137, and -3'0492. 160. Taylor's Theorem in connection with the theory of maxima and minima. In Arts. 106, 107, a method of determining the conditions that f(a) should be a maximum or a minimum value off(x), depending on the successive derivatives f'(a), f"(a),... was partially indicated. We are now in a position to state the whole theory of the method. If f(a) is a maximum value of f(x), it must be greater than all neighbouring values, i.e. greater than both f(a+h) and f(a -h), for small values of h, and similarly if f(a) is a minimum, it must be less than both f(a + h) and f(a - h). Now f(a + h) =f (a) + hf'(a) + 2- f"(a), d ( a) - and f(a - h)= f(a) - hf(a) + h2 f"(a) -.... ARTS. 159, 160.] TAYLOR'S THEOREM. 179 Hence when h is small we have, approximately, f(a +/h) =f(a) + hf'(a), and f(a - h) =f(a) - hf'(a). One of these will be greater than f(a) and the other less unless f'(a) = 0. Therefore, the first condition that f(a) shall be a maximum or minimum value of f(x) is that f'(a) = O. h2 In that case f(a + h) =f(a) + 1 f"(a), approximately, h2 and f(a - h) =f(a) + f"(), approximately. Hence, if f"(a) is positive, f(a) is a minimum, being obviously less than both f(a + h) and f(a - h). If, on the contrary, f"(a) is negative, f(a) is a maximum. But iff'(a) = 0 and also f"(a)= 0, then f(a + h) =f(a) + 1 f"'(a), approximately, h3 and f(a - h) =f(a) - -- f"'(a), approximately. Hence in this case f(a) cannot be a maximum or minimum value off(x) unless f"'(a) = 0. Proceeding in this way we see that f(a) cannot be a maximum or minimum unless the first derivative which does not vanish is of even order, say f2n'(a). And then we have both h2n f(a + h) and f(a - h) =f(a) + - f 2(a), approximately, and therefore f(a) will be a minimum if f2(a) is positive, and f(a) will be a maximum if f2(a) is negative. This is the theory of the investigation by means of the successive derivatives. 180 DIFFERENTIAL CALCULUS. [CHAP. XI. 161. Special case of Taylor's Theorem. Geometrical illustration. If, in Taylor's Theorem with remainder, we put n= 1, we obtain the equation f(a+h)-f(a)=hf'(a+Oh). This is an interesting equation as it shows that the approximate relation for small values of h, viz. f(a + h) -f(a) = hf'(a), might be improved by taking on the right, not f'(a), but some value of the derivative f'(x) between x = a and x = a + h. This fact can be illustrated geometrically, in such way as to indicate, in the case of a function whose graph is drawn, for what value of x, approximately, the derivative should be chosen. If we draw the curve y=f(x), and take on it two points P and Q whose abscissae are respectively a and a+ h, the ordinates of P and Q will be f(a) and f(a + h), respectively. m <,F^.A:R: — XA fr(a+iz) f(a) (a, o) (a +h,o) Consequently the gradient of the chord PQ will be f(a+h) -f(a) h Now, evidently, if the gradients of the curve between P and Q are finite and vary continuously, i.e. with no break in the back of the curve, the tangent to the curve at some point between P and Q must be parallel to the chord PQ, i.e. the gradient of the curve at this intermediate point will be equal to the gradient of PQ. But f'(a + Oh), where 0 is a proper fraction, just denotes the gradient of the curve at some such intermediate point. Hence ARTS. 161, 162.] MACLAURIN'S THEOREM. 181 the equation with which this article started merely asserts the above geometrical fact, viz. that between P and Q there must be some point on the curve where the tangent is parallel to PQ 162. Maclaurin's theorem. If in Taylor's theorem we put a = 0, and write x for h, we obtain the series f(x) = f(0) + xf'(O) + f"() +... + - fn-1(O)+ n fn(Ox), nwhen f(x) and all its derivatives to the n t", inclusive, are finite and continuous between 0 and x. Also X2 f(x) = f(0) + xf'(O) + 12 f"(O) +... to infinity, if, on increasing n, the remainder -f"(Ox) can be made as small as we please. This theorem, in its two forms, is called Maclaurin's Theorem, which is thus seen to be a special case of Taylor's Theorem. It enables us to express any function of x, which obeys the necessary conditions, in powers of x. The expansions of ex, sinx, cos,... are illustrations of the theorem. A useful, compact, way of stating the theorem is as follows: If y is a function of x, and y, Y2,... are its successive derivatives, and, if a, al, a,... denote the values of y, y, Y2... when x = 0, then x2 y = ao + al, + 1.a22 + * to infinity, provided none of the terms are infinite and the series is convergent. Example. To find the expansion of sin x. Let y=sinx;.'. ao=0; 'Y. Y=os x, Ct= 2= - sin x, a2=0, Y3= - COS, Cx3= - 1. 182 DIFFERENTIAL CALCULUS. [CHAP XI. Continuing, it will be found that Y4, Y5, Y6, Y7 are a repetition of Y, Y1, Y2, and y3, in the same order, and so on. Hence ao=a4=a8 =... =0, c — =9 --... - - 1, ca1=aa=f9 =-..=1, C2 -- = a6 -- l ~ ~ - O0, a3=a7=a11= - -1 a3 = a7( -= all =...= - 1; X3 X5 X7.. finally, y=x- 3 - +.... This series holds for all values of x, since sin x and its derivatives are always finite. Any function whose successive derivatives can be found can be expanded in the same way, provided neither the function nor any of its derivatives become infinite when x=0. The series so found will be a proper representation of the function for such range of values of x, from x=0 onwards, as include no infinite values of the function or its derivatives. The student should form the series for loge(l +x) in the same way, and note that the series will only properly represent this function between x= ~1. As a harder example we may take the following: To expand tan x in powers of x, up to X7 inclusive. One way of forming the successive derivatives in this case has been given in Art. 127, in which, putting y = tan x, we found, in succession, Y1= 1 +y2, Y2 = 2y + 2y, Y3 = 2 + 8y2 + 6y4, Y4 = 16y + 40y3 + 24y5. Continuing in the same way, we find yy5= 16 + 136y2 + 240y4 +..., Y6= 272y +.... Y7= 272 +.... Hence, by putting x=0, and noticing that y=0 when x=0, we find ao=0, a1=l, a2 = 0, a3 =2, a4=0, a5= 16, a6=0, a7=272; 2x3 16x5 272X7. tanx=x+ —+ + +.... ARTS. 162,163.] MACLAURIN'S THEOREM. 183 Sometimes the process of forming the successive differential coefficients becomes very heavy and tedious, so that it would take a long time to find many of them. In some of such cases the difficulty may be avoided or diminished by special artifices. It must be remembered that we only require to find the successive derivatives for the sake of evaluating them when x=0; this will often enable us to save a good deal of labour. Some of the artifices which may be useful will be exemplified in the following articles. 163. Expansion by means of a differential equation. Very often the easiest way of obtaining the successive values of the coefficients in the expansion of a function y is to form an equation connecting y and its derivatives, and then, by Leibnitz' theorem, or otherwise, find the result of differentiating this relation n times. Then find the effect of putting x= 0 in the resulting identity, and we shall obtain an equation by which the terms of the expansion can be determined. To return to the case of tan x. We found that y,=1 +y2. This is a differential equation between y and y,. If we differentiate both sides of this identity n times, using Leibnitz' theorem to differentiate the right-hand side, we find, as in Art. 135, p. 141, n(n- 1) Yn+l = YYn + nYlYn -1 + I 2 Y2Fn - 2 +. 1.2 Hence, when x = 0, n (n -1) an+l = aoan -+ nal -1 + - 1.2 Thus a3 = aa2 + 2a2 + aao = 2aoa2 + 2a12, a5 = aoa + 4a + 6 + 4aa+ 6 4a + a4ao = 2 (aoa4 + 4ala3) + 6a2, a7 = 2 (aOa6 + 6aa5 + 15a2a4^) + 20a32,............................................. Also a1 = 1 (from y, = 1 + y2). If we notice that the expansion of tan x can contain only odd powers of x, since tan( — x)= -tan x (see p. 49), we see that ao, a2 4,,... must be all zero. Hence a = 2a2 = 2, a = 8ala3 = 16, a7 = 12a%1a + 20a32 = 272. 184 DIFFERENTIAL CALCULUS. [CHAP. XI. The student may in a similar way find the expansion of sec x, using the method indicated on p. 142, and making use of the coefficients just found in expanding tanx. To avoid confusion, he had better use bo, bl,... for the sec x coefficients, leaving ao, a,,... for those of tan x. The general relation will be found to be bn+l = anbo + nan- lb, +... + nalbn_ + aob,, from which the successive coefficients can be calculated. As another example, find the expansion of sin-lx, where sin-1x denotes the smallest angle whose sine is x. Let y=sin -x, then, as on p. 142, we find that (1 - X2) Yn+2 - (2n +1) xyn+l = n2y. Now, put x=O, and we obtain an+2=n2aC. We know that %o, a2, a4 are all zero, since sin-lx is an odd function of x. Also al= 1..'. 3=1, a5 = 32a; = 32, a7 = 52. a5 = 52 32,..................... x3 32x5 52.32x7 sin -lx =x+ +- + — - +.... Here the law of formation of the successive coefficients is simple, and we can write down as many of them as we please without difficulty. This is in great contrast to the expansions of tan x and sec x, where each coefficient had to be laboriously formed. Different functions differ very much, and very unexpectedly, in this respect. We might have found the expansion of tan x to a few terms from the equation y = 1 + y2 by assuming y = Alx + A3x3 + A5x5 +... (having no even powers, since tan x is an odd function of x), whence y = A1 + 3A3x2 + 5Ax4 +.... Now, substituting these values in the above differential equation, we must have A1 + 3Asx2 +5A5x4 +... 1 + (A1x + A3x3 + Asx5 +...)2. Hence, equating coefficients, A,=l, 3A3= A12= 1, 5A5 =2A1A3 =, 7A7= A32 + 2A1A5 = + - =, whence tan x = x + ~x3 + 1 x5 + -17 X +. The same method could have been employed also to determine the expansion of sin-ix. ARTS. 163, 164.] MACLAURIN'S THEOREM. 185 In many physical investigations a differential equation is arrived at whose solution is required. In such cases we may find the solution in the form of a series by one of the methods indicated above. Examples will be found at the end of this chapter. In most cases it will be found that the series will involve one or more arbitrary constants. 164. Expansion by use of known series. Very frequently, when the direct use of Maclaurin's theorem would be very tedious, a few terms of an expansion may be readily found by making use of series already found. Thus, to expand esilix in powers of x, to X4 inclusive, we have esin x = 1 + sin x + sinx + sin3x + sinx +... and sin x=x- x3 +...;. esinx=l +(x_-X3)+1(x_ x,3)2+ (x_...)3+ (-...)4+... =l +x - x3 + 1 (x2- X4) +IX3+ 1 X4 + = 1 + 2 - 1 4 + higher powers of x. Other examples in which this method is useful will be found below. EXAMPLES. 1. Find the first four terms in the expansion of sec 0. 2. Expand log sec x to the term involving x6 inclusive. x2 3. If y=sin-lx=a +ax+a21-2+'. ' prove that (1 - x2)y =xyl, and deduce the expansion by the method of equating coefficients. 4. Write down the expansion of 0 in powers of sin 0 [i.e. sin-lx in powers of x]. 5. If y=tanlx=ao+alx+.. prove that an+2= -n(n + 1)a, and write down the expansion of tan-lx; i.e. express 8 in a series of powers of tan 0. 6. Expand easin-' in powers of x. 7. Expand eaXsin bx in powers of x to the term involving x3 inclusive. 8. Expand excos bx to the term involving x3 inclusive. 186 DIFFERENTIAL CALCULUS. [CHAP. XI.] 9. Find the expansion of sin2x in powers of x. 10. Find the expansion of sin x cos 2x in powers of x. 11. Find two terms of the expansion of log(six ). 12. Show that for small values of 0, the equation 0= -(tan 0 + 2 sin 0) is very accurate. Test the equation when 0 is the circular measure of 20~. 13. Show that 0 = (8 sin ~0 - sin 0) and 0=2 (8 sin A0 - sin ~0) are good approximations to 0, for small angles. Test the formulae for an angle of 20~. 14. If c1 is the chord of a circular arc, and c2 is the chord of half the arc, show that, approximately, the arc= -(Sc - cl). 15. F' A, B, C so as to obtain the best approximation to the equation 0= A sin 0 + B sin 20 + C sin 48, when 0 is small. 16. Expand X(0) in powers of 0 up to 05. LNOTE.-X (0) = log (sec 0 + tan 0) = log tan (+) ~ 17. Prove that X(0)=2(tan 0 + tan3 0 + - tan -+...). 18. Prove that X(0) + = 4 (tan 10 + tan5 0+...). 19. Test the approximate formula X(0)=4tan —1 0 when 0 is the circular measure of 20~. 20. Prove that X(0) is approximately (2 tan 0 +sin 0), and that (10 tan 10 - 4 sin 10) is a better approximation. Test the formulae for an angle of 20~. 21. Solve the differential equation (1- x2) 2- =x by assuming y=A + Alx +A2x2 +.... d2y 22. Solve the differential equation + xy=0 by using Leibnitz' and Maclaurin's theorems. [We shall find a2=0, and a,+3= - (n+l)a^. These conditions will enable all the non-zero coefficients to be expressed in terms of a0 and al, which will remain as two arbitrary constants.] 23. Solve the equation d2 + y + y=0. EXAMPLES. 187 d2y dy 24. Solve the equation xc2 + + 0xy=O. 25. Solve the equation dY + n2y=O. 26. Prove that f(a +h -fa h) = 2(h f'(a) + f (a)+ -/f (a) +. and illustrate by expanding sin(a +) - sin (a - 3) in powers of /. 27. Iff(x) is an algebraic function of x of degree not higher than the fourth, prove that h Lf'(a + h) + 4f(a) +f'(a - h)] =f(a + h) -f(a - h). 28. Find the series for loge(l +x) by writing down the expansion of (1 + )-1 and integrating both sides of the equation. 29. Expand (1 - x2)-2 by the Binomial Theorem, and, by integration, deduce the series for sin-Ix. 30. Expand (1 +x2)-1, and deduce the series for tan ix. CHAPTER XII. INDETERMINATE FORMS. 0 165. The indeterminate form ' In evaluating a function of any variable x, it sometimes happens that for a particular value of x the function assumes an indeterminate form. For example. in evaluating the fraction x3- 2x2 - 5x + 6 23- f2-5x+ 6for different values of x, x2 + x- 2 it happens that if we try to insert x= 1, the fraction assumes 0 the form ~, which is indeterminate. It is not the fraction itself which is indeterminate, but merely the particular form in which it is presented. In fact x — 1 is a common factor of both numerator and denominator, and must be cancelled before we can evaluate the fraction for this particular value of x. Another common factor is x + 2, so that the form of the fraction is also indeterminate when x= - 2. If, however, we reduce the fraction to its lowest terms by cancelling the common factors, the form then obtained will not be indeterminate for any non-infinite value of x. The object of the present chapter is to show various methods of evaluating a function when it assumes an indeterminate 0 form, and more particularly when that form is 6. In the present case nothing has to be done but to divide by the G.C.M. of the numerator and denominator, giving us the identity X3 - 2X2 - 5x + 6 X2+ 2 - =x -3. x2+ x-2 [ARTS. 165, 166.] INDETERMINATE FORMS. 189 Hence the value of the fraction is - 2 when x= 1, and -5 when x= - 2. We have already had several important fractions which assume the form - when x= 0, viz., sinx tanx e- 1 ax- 1 x X x x whose values we have found to be respectively 1, 1, 1, logea. In all these cases the evaluation was, or could have been, effected by expressing the functions in powers of x, and then dividing by x. In fact, the fractions are not in their lowest terms, the numerator and denominator having the common factor x. f(x) 0 In general, if a fraction ) becomes of the form 0 when x = a, <p (x) 0 the fraction is not in its lowest terms, but either x- a or some power of x- a is common to both numerator and denominator, and must by some means be divided out before evaluation is possible for this value of x. This theorem follows at once from the following fundamental theorem: 166. If f(x)=O when x=a, either x-a or some positive power of x-a is a factor of f(x). This theorem may be taken as axiomatic, but it may be proved as follows in the case of functions which can be expanded in powers of x - a by Taylor's Theorem. Let x-a=h, so that x=a+h;. f(x) =f(a + h) =f(a) + hf'(a) + l^2f"(a) +. h[f(a) + f"(a) +.. since by supposition, f(a) = 0. Hence h, which is x - a, is a factor of f(x). NOTE.-If f(a) and f'(a) are both zero, (x - a)2 is a factor of f(x), and so on. 190 DIFFERENTIAL CALCULUS. [CHAP. XII. 167. Evaluation of the indeterminate form. Let u and v be two functions of x which both vanish when x = a. It is required to find the value of - when x= a. v (1) Division Method. Evidently, if we can divide both numerator and denominator by x - a, or such power of x - a as is common to them, the indeterminateness disappears. This is therefore one method of evaluation. This method is indeed the foundation of all the methods which can be adopted, though in many cases direct division is not practicable. (2) Differential Method. A method which is often the easiest to adopt is obtained by considering the value of the fraction when x is just greater or less than a, and finding its limiting value as x approaches nearer and nearer, without limit, to the value a. Now, when x=a, u=0, and when x = a dx, u =0 + du; similarly, when x=a+dx, v=0 +dv; u. du.. the limiting value of - is d. v dv x3 - 2X2 - 5x + 6 For example, the value of - 2-2 + when x = 1 X + ' - 2 0 (3x2-4x-5)dx 3x2-4x-5_ O (2x+1 ) dx - 2x+1 Note, that we have here divided both numerator and denominator by dx, which is the value of x - 1 when x = 1 + dx; consequently the essence of this method consists in cancelling the common factor x- 1 and then evaluating the resulting fraction. If the square of x - 1 had been common to both numerator and denominator, the new fraction would have also assumed the form 6, and could be evaluated by using the same method on it. ART. 167.] INDETERMINATE FORMS. 191 (3) Expansion method. In the differential method we assumed a value of x differing infinitesimally from its critical value a, and so obtained for numerator and denominator their infinitesimal increments from zero. These increments were all that were needed for our purpose, but were only partial increments, as we know that differentials are obtained by neglecting powers of dx higher than the first. We may, however, by help of Taylor's theorem, obtain the complete increments, in which case we need not in the first place assume the increase of x to be infinitesimal. Let x =a + h, and for clearness let q =f(x) and v= ((x). Then f(a ()) + hf(a) + 2 "(a) + (a + Ah) S 2 +(a) + h+'(Ca) + ( f"(a) +... * 1)'(a) + -"(a)+... since, by supposition, both f(a) and (S(a) =0. Now we can divide out the common factor h (i.e. x - a), so obtaining the equation f(a + h) f'(a) + -h f"(). A(a + h) () + h"() + Now, put h= 0, and we find f(a) f'(a) dl) (a) a This is exactly the same result as is given by the second method. If both f'(a) and q'(a) = 0, the expansion method enables us to at once divide out by h2, i.e. (x - a)2, obtaining finally in this case f(a).f"(c) In fact, if a high power of xa is common to numerator In fact, if a high power of x -a is common to numerator 192 DIFFERENTIAL CALCULUS. [CHAP. XII. and denominator, the expansion method enables us quickly to cancel the proper power, and evaluate the fraction. Note that the form of any fraction is only indeterminate as long as both numerator and denominator vanish for the particular value of x. If the numerator =0, but not the denominator, the value of the fraction is zero; and if the denominator vanishes, but not the numerator, the value is infinite. 168. Note on the case when the value of the indeterminate form is zero or infinite. In these two special cases, viz. when the true value of the indeterminate form f(a) + ((a) is zero or infinite, there is a f(a) f'(a) peculiarity in the equation = - which, though of no ( (a) (A'(a) importance practically, is yet sufficiently interesting to notice. The test of perfect equality of two expressions is two-fold, viz. that their ratio shall be unity, and their difference zero There is, however, an imperfect equality possible when the quantities themselves are zero or infinite, viz. that they are both zero or both infinite. We shall find that the above equation is of this imperfect nature. Of course if both fractions are zero, their difference is zero; we shall prove that their ratio is not unity but merely finite. When both are infinite we shall prove that their ratio is finite, but not unity, and consequently their difference must be infinite, so that in this case the equation satisfies neither the ratio test nor the difference test, but just merely the condition (which is all we here practically care about) that both are infinite if one is. For suppose that f(a), f'(a),..., fr-(a) are all zero, but not fr(a), then by Taylor's theorem f(a + h)=- f(a) +... and f'(a + h)= fr(a) +... ARTS. 167, 168.] INDETERMINATE FORMS. 193 And suppose that +(a), ('(a),..., k-1(a) are all zero, but not <s(a), so that (+a h) (s(a) +.. IS hs -1 and (a' + h)= i, _s-I _ () +............................... Then, by taking the limiting values of f(a + h) and of '(a + h) (A(a + h) 0'(a + h)' we see that / a) =0 if r > s, and (a) —= c if s > r, and so is 4(a) 4 (a) f'(a) f(a) f'(a) s (a); but the ratio (a) '(a) equals s, which is finite but (^(a)) q(a) 4(a) r not unity. * '(a) f/(a) s - Similarly f'(a) f"(a) s- 1 and so on. 1 '(a) 0'(a) r -1' Hence, the fractions are all of the same order of magnitude, but are not equal in the full sense of the term. Of course, this limited equality is sufficient. All we want to know in such case is that the fraction is zero, or infinite, as the case may be. On the other hand, if s=r, we have the perfect series of equations f(a) f'(a) fr(a) i.e. the equality is complete in the case in which complete equality is essential, viz. when the value of the fraction is finite; and though the equality is incomplete in other cases, it is sufficient for the purpose in view. [N.B.-Of course an equation between finite quantities must be complete: if two finite quantities are equal, their ratio must be unity, and their difference zero. It is only in the case of infinitesimal or infinite quantities that an imperfect equation is possible.] L.D.Co. 194 DIFFERENTIAL CALCULUS. [CHAP. XII. EXAMPLES. sin 8 1. Prove that = 1 when 0=0, if 0 is in radians. 2. Draw the curve y=- between the limits x= 1 radian, tan 6 - sin 0 3. Evaluate tan -. 0 when 0=0. sinlo 4. Evaluate X4 5+ 7- 2 - when x = 2. X4 - 52 + 7 - 10 4 - 4~3 + 7x2 - 12 + 12 5. Evaluate e x when x=2. x8 - 2 - s8x+ 12 ax - bx 6. Evaluate when x =0. X"n - en xn _ ] 7. Evaluate -- and - when n = 0. n h x5 - 32y5 8. Express -. in terms of x when x =2y. x2 - 4y2 a5 ~ 3a4y/ ~ 2ay4 9. Express + y4 in terms of x when x + y 0. x4 - 2x3y - 3y4 10. Evaluate 6 - 3X2 + 2) when ax - 1 /X 2 - 4 11. Evaluate -- when x=2. Jx - \/2+ +a -2 12. Evaluate lo x when x = 1 sin 2x tan2a 13. Evaluate when x=0. (ex- 1)ex - e -x - 2x 14. Evaluate when xa=0. x - sin x 15. Evaluate log sin26 when 0=7 sin 0 + cos 0 - \/2 4 16. Evaluate Cos -x- when n = l. 1 - n= 17. Evaluate sin ax - a sin when a = I. tan ax - a tan x ARTS. 169, 170.] INDETERMINATE FORMS. 195 169. Other indeterminate forms. There are other indeterminate forms besides 0, though this form is by far the most important, and the evaluation of the others depends sooner or later on the evaluation of this form. The other forms are \o sec 0 r log x (2) -, such as --- when 0=- or -o- when x=oo; 1SU /o tan 2 x (3) - oo, such as sec - tan 0, or secn - tan"l, when 0=; (4) 0 x oo, as, for example, x logx when x=O; and the exponential forms (5) 1, {such as (l+) when p= o, or (l+x) when x=O; (6) 0~, such as xx, or Xsinx, when x=O; (7) oo~, such as (tan x).osx, when x=2. The general method of solving cases (2), (3), and (4) consists in reducing them to the standard form 0, and then making use of one of the methods already explained, though in some cases we may deal with these indeterminates in the form - by the method of next article before reducing to the standard form. In the cases (5), (6), and (7) their logarithms are all of the form 0 x cc, and can therefore be evaluated as indicated above. log(I + x) 0 Thus, if tu=(l+x)X, logeu=- l, which is of the form - when x 0 x=0. The limiting value of this fraction is unity when x=0. 1_ Hence u=e. i.e. (1+ x)=e when x=0. 170. The form -; When the numerator and denominator of a fractional ____ ___ _ -- - - - _ ---- --- L 196 DIFFERENTIAL CALCULUS. LCHAP. XII. function of x both become infinite for a particular value of x, the form of the function is indeterminate. In such cases it may be possible to reduce the fraction to the standard form 0, and so evaluate by methods already explained. sec 3x a o r For example becomes - when x=-. sec x o 2 We can write this fraction in the form s which becomes - for cos 3x 0 the required value of x, and on evaluation we find its value to be - 1. Sometimes, however, this transposition may be inconvenient, and the following theorem is useful. If u becomes G when x=a, it may be replaced by du for this value of x. dv 1 1 bU V V2 For - =1 - when x= a, v 1 1 - — du u U2 since the transformed fraction is of the form 0 U u2 dv Hence -=-. v v2 d'e u c du i.e. finally, u v dv log cos x or For example, -, when x=log t- x 7r oo -tanxdx 2 0 dx -o G - x cotx O - cosec2x, dx 2in2= = sin2x = 1. ARTS. 170, 171.] INDETERMINATE FORMS. 197 It is important to notice, however, that before we can succeed in evaluating the fraction, either it or one of its equi0 valent fractions must as a rule be brought into the form - The only exception to this necessity is when the critical value of x is infinite. In such cases the differential method will often be successful without such transposition. -dx log x x 1 o Thus, lo, when x=00, = x -0 x ' dx x 171. If, in the process of evaluating an indeterminate form, one of the factors is finite, the value of this factor may be inserted at once. This will often save considerable labour. sec3O - tan3o IT Thus, to find y= ta when 0=tall 0 2 1 - sin30 Here y= Here sin 0 cos2 1 - sin3.. r = — c20, since sin -, cos2 2 -3 sin2O cos 0 dO - 2 cos 0 sin 0 dO 3 2-' EXAMPLES. secx 1. Evaluate e x and sec x - tan x, when x=-7. tan x 2 2. Evaluate sec3x- tan3x when x=-. 3. Evaluate xlogx when x=0. ex 4. Evaluate - when x==o. Xn secax -- tannx 7r 5, Evaluate ta —x - when x=- at tan" 2x( w 2 6. Evaluate tanx(secx-tanx) when x=-, 198 DIFFERENTIAL CALCULUS. [CHAP. XII. 7. Evaluate, when x=0, and also when x=-, (i) (1 +sinx)cotx; (ii) (1-sinx)cotx; (iii) (cosx)c"tx; (iv) (sinx)taux. 8. Evaluate (1 +cos2x)seC when x= - 1 1 9. Eivaluate (t -) and (s x)x when x=O. 10. Evaluate (logx)x when x=oc. 11. Evaluate (3-x)tan when x=3. (I _ X2) lo (l + X2) 12. Evaluate (1- )log( 2) when x=. cos x. log sec x 172. Evaluation of a rational algebraic fraction (x) when x is infinite. P (x This special case, which is very important, is also very simple, and does not need any of the above methods, the value of such a fraction reducing to the ratio of the highest term in the numerator to the highest term in the denominator, since the other terms are unimportant compared with the highest term when x is infinite. Thus 2 - 3x2 - 5 2 3x3 + 5x2 - x +6 3 when x is infinite. This is too obvious to require formal proof, but such proof can be supplied by putting z=- and multiplying both numerator and denomix nator by z3, bringing the fraction into the form 2 -- 3z + z2 - 5z~ 3 +5z - z2+ 6z which = when z==0, i.e., when x =-.o It is obvious that the value of such a fraction is finite only when the highest terms in the numerator and denominator are of the same degree. If the numerator is of higher degree than the denominator the fraction is infinite, and if the numerator is of lower degree the fraction is zero. NOTE.-It may be noted that the differential method (Art. 170) is legimitate here, and would lead to the right result; ARTS. 172-174.] INDETERMINATE FORMS. 199 but the object of the present article is to show that such method is unnecessary. 173. Relative values of functions of two related variables when the variables are infinite. It is frequently necessary, in the case of a fraction whose numerator and denominator are given as functions of two variables x and y between which there is a given relation, to ascertain the value of the fraction when the variables are infinite. Such fractions present themselves for evaluation in connection with the problem of finding rectilinear and curvilinear asymptotes to a curve, and it is for this reason that they are introduced here. 174. We shall, in this article, consider fractions in which x and y are connected by a linear equation, y = mx + c, and then take other less simple relations. The theory is very simple, but there are some interesting points worthy of consideration. The first example will be the ratio of two homogeneous functions of equal degree. x2 - 3xy + 2y Thus: evaluate X2 - 3xy + 22 when x and y are infinite, 2x2 + xy - y2 being given that y = mx + c. Substituting y = mx + c in the fraction, it becomes (1 - 3m+2m2)x2+(4m - 3)cx+2c2 1 - 3m + 2m2 (2 + m - m2) x2 + (1 - 2m) cx -2 2 + m - m2 in the limit. If m =1 or 1 the value is zero, since the coefficient of x2 in the numerator vanishes; and if m - 1 or 2 the value is infinite, since the coefficient of x2 in the denominator vanishes. In all other cases the ratio is finite. It is very useful to notice that the value of c has no effect on the value of the fraction, and that it would have been just the same if the relation between y and x had been y = mx, i.e., if c had been zero. This is because c, being finite, is unimportant compared with x which is infinite. In fact, although the difference between 200 DIFFERENTIAL CALCULUS. [CHAP. XII. y and mx is not zero, the ratio of y to mx is unity, in the limit when x is infinite. Evidently the ratio of y to mx is all that matters in evaluating the fraction, so that we may take any values of x and y that satisfy the ratio equation - = 1 mx Hence, if we put x= 1, y=, we obtain the value of the fraction, and with the minimum of trouble. This simple method of evaluation is of great use, but we must remember that it applies only to the ratio of two homogeneous functions of equal degree. If the numerator is of higher degree than the denominator, the fraction is in general infinite when x = o, but an exception occurs when the relation between x and y is such as to make the highest power of x vanish in the numerator. Thus: To evaluate 23- 32y + 2Xy2 when y = x + c. x2 +xy-y -2 J=. Here y - x is a factor of the numerator, the other factor being 2xy- x2. 2xy - x2..~. the fraction=(y - x) 2 +xy-_ =2= the value of 2Xy-X2 — 2 I, the value of 2 - (X = ~) being obtained at once by putting x=1, 2x2 + Xy - y2 y in accordance with the method of last article. Similarly, if the numerator is of lower degree than the denominator, the fraction is in general zero when x= co, but an exception occurs when the relation between x and y is such as to make the highest power of x vanish in the denominator. If the numerator is not homogeneous, the fraction can be broken up into a number of separate fractions in each of which the numerator is homogeneous. If the denominator is not homogeneous, the terms of highest degree are the only important ones unless the relation y = mx + c is such as to make them vanish when the value of y is substituted, i.e. unless y-rmx is a factor of the highest degree terms. In such a case the terms of next lower degree become equally important. ARTS. 174, 175.] INDETERMINATE FORMS. 201 The whole theory, in fact, depends on the relative importance of the different terms when x is infinite. The easiest way of discovering whether y - mx is a factor of a homogeneous expression is by finding whether the expression does or does not become zero when we substitute x= 1, y=m. If it does, y - mx is a factor. EXAMPLES. Evaluate for infinite values of x and y, or of one of them: 1. 2ay2 when x=y+c. x(x+y) 2. bx2 -ay2 when x = y + c. xy 3. 2 + 72y 5xy2 - 2y3when 2y = c. 3x2 - 5xy + y2 4. 4xy2 - 6y + 32 -5xy+4 when 2x - 3y = c 2x3 - 3x2y + 3y2 - 5x + 2 5. 2X3 3x2y - y3+4x2 + 2xy - 2y when x +y=c 4x2 + 5xy + y2 and also when 2x - y = c. 6. 4-3y 22 when x=c. 2y2- 3x + y 175. Orders of infinite quantities. We have seen that when y = Mx+c, the evaluation of a fraction consisting of the ratio of two functions of x and y when x and y are infinite is effected by attending only to the terms of highest degree in both numerator and denominator unless y - mx is a factor of such terms, in which case the terms of next lower degree become of equal importance. We may express this systematically by saying that, if x and y are infinities of the first order, the sum of the terms of the first degree in any expression is an infinity of the first order unless y - mx is a factor, in which case the sum is finite; the sum of the terms of the second degree is generally an infinity of the second order, but, if y - mx is a factor, the sum is an infinity of the first order, and if (y - mx)2 is a factor, the sum is finite, 202 DIFFERENTIAL CALCULUS. [CHAP. XII. and so on, the general result being expressed by the statement that the sum of the terms of the nth degree is an infinity of the nth order unless some power of y - mx is a factor, and if (y - mx) is a factor, the sum is an infinity of the (n - r)th order. With this understanding we may say that the ratio of two expressions is obtained by taking the ratio of the sum of the terms of highest order in each, neglecting all those of lower order; and that the ratio is finite, if the 'order of the sum of the highest terms in the numerator and denominator are equal, or, as we may say more concisely, if the numerator and denominator are of the same order of infinity; the ratio is co, if the numerator is of a higher order of infinity than the denominator; and the ratio is zero, if the numerator is of a lower order of infinity than the denominator. NOTE.-The order of any single term in an expression is equal to its degree, but the sum of a number of terms of any degree may be of lower order by reason of some power of y - mx occurring as a factor. 176. In the above articles we have supposed a linear relation to exist between x and y, so that x and y are of the same order of infinity: except in the special case when m=0 or oo, in which case they are not of the same order, and usually one of the variables is finite while the other is infinite. It is important to consider the case in which x and y are not of the same order of infinity, as such cases occur in connection with curvilinear asymptotes. We will suppose that, when x and y are infinite, xA is proportional to yB, i.e., xA and yB are of the same order of infinity. The problem will be to arrange the various terms xmly of any expression in order of magnitude when x and y are infinite. If A and B are unequal, it will no longer be true that the degrees of the various terms give their relative order. It may happen that xy3 is of a higher order of infinity than x3y2, though of course such terms as x2y2 or x2y will be of a lower order of infinity than X3y2 whatever the relation between x and y, as long as x and y become infinite together. ARTS. 175-177.j INDETERMINATE FORMS. 203 Example. If x2=ay, where a is finite, the terms x3y2 and xy3 are of the same order of magnitude, being both comparable with x7, the importance of the terms being estimated by taking the index of x + twice the index of y, and associating together all those terms for which this quantity is the same. Generally, if xA is of the same order of infinity as yB, all terms xmy" for which Bmn+ An is the same are of the same order of infinity. To fix the scale of infinities most conveniently, we will introduce a new variable z such that XA and yB are each comparable with zAB and we will take the different powers of z as measuring the different orders of infinity of the different terms. Then x is of the same order as zB, i.e., is of order B, y is of the same order as zA, i.e., is of order 4, and xmyn is of the same order as zBm+A"n i.e., is of order Bmn + An, i.e., the relative importance of the various terms x"y" in any expression will be measured by the value of Bm + An in each. 177. Relative values of functions of two related variables when the variables are infinitesimal. Another question of equal importance with that of discovering the relative importance of terms when x and y are great, is that of finding their relative importance when x and y are very small. As the two methods of investigation are intimately connected, we will now consider the latter question. If x and y are infinitesimal, and XA is comparable with yB, the different terms x"'y will now be infinitesimals of different orders. Introducing z as before, and taking the powers of z as standards, we find x is an infinitesimal of order B, y A, xmy,,, Bin + An. Now infinitesimals of lower order are more important than those of higher order, so that here the smaller the value of Bi + An the greater will be the importance of x"yn. Hence the order of relative importance when x and y are very small, is the exact reverse of the order when x and y are very great. 204 DIFFERENTIAL CALCULUS. [CHAP. XI. EXAMPLES. 1. If x2 is comparable with y, arrange the terms x6, x3y, y3, xy2 in order of relative importance (1) when x and y are very great, (2) when x and y are very small. 2. If x2 is comparable with y3, arrange the terms x5, x4y, x3y3, xy5 in order of magnitude (1) when x and y are very great, (2) when x and y are very small. 3. If xA is comparable with yB, both being infinitesimal, show that Y x is zero, finite, or infinite according as A is greater than, equal to, or less than B; and that the reverse is true if x and y are very great. 178. Graphic method of determining the relative importance of terms of the form xyn when XA is comparable with yB, when x and y are very great or very small. We have seen that all terms whose indices are connected by the relation Bm + An = constant are of the same order of magnitude. If, then, we represent the terms by points whose abscissae are the values of m, and whose ordinates are the values of n, in the various terms, the terms of equal importance will lie on the straight line Bm + An = constant. 6. --- >. — -..- -------------- 76 "-, 73\ --- —--- 24 4i- —, —. 1 -. - s -.... ----..,^ —... 14 Powers of x. ART. 178.] INDETERMINATE FORMS. 205 The diagram illustrates the relative importance of various terms when x2 is proportional to y3. It shows the terms x2 and y3 connected by a line and the terms x5, X3y3, xy6 connected by a parallel line. It shows also the terms x4y and xy5, which are within the outer line, but x4y is nearer to it than xy5 is; hence they are both of lower order than any of those connected by the outer line, but x4y is of higher order than xy5. Hence when x and y are very great, the terms connected by the outer line are of greatest importance, and those nearer to it are more important than those more remote. If, on the other hand,' x and y are infinitesimal, the terms connected by the lower line are the most important, and those nearer to it are more important than those more remote. We shall refer to this again in connection with curve tracing. EXAMPLES. 1. Write all the terms in an expression of the 4th degree in order of importance when x and y are very large: (1) when y2=ax, (2) when ay2=x3, (3) when ay3=x4, (4) when y4=a-3x, a being finite in each case. 2. Is it possible that x3 and xy3 can be the most important terms in the expression Ax3 + Bx2y + Cx2y2 + Dxy3 when both x and y are either very great or very small? Which pair of terms can be of greatest importance in either case? 3. If 2y2 + 3xy = x3 - x2 + 7, which terms can be of greatest importance? Arrange the other terms in order. 4. If y2: x= 3, when x and y are infinite, find the approximate value of 6xy- when x and y are very great. 3x + 2y 9x2+4Y2 under the same circu4m 5. Find the approximate value of under the same circum stances. [The neatest way of solving (4) and (5) is by ordinary division, continued as far as the term in the quotient which involves y2-x.] CHAPTER XIII. PARTIAL DIFFERENTIAL COEFFICIENTS. 179. Differentiation of a function of two variables. Partial Differential Coefficients. When we have to differentiate a function of two variables, x and y, the most systematic way is to arrange the terms of the differential in two groups, according as they are multiples of dx or of dy. Thus, if t = 3x2 - 5xy + 2y2 - 15x + 6y - 8, instead of writing du = 6xdx - 5 (ydx + xdy) + 4ydy - 15dx + 6dy, and then arranging the terms into the two groups du = (6 - 5y - 15)dx + ( - 5x + 4y + 6)dy, we can, at the first, arrange them in these groups by first differentiating u throughout as if y was a constant, and then differentiating throughout as if x was a constant, the complete differential being the sum of these partial differentials. It is obvious that we may always do this, for every term of the differential must have dx or dy as a factor, and no term can have both, because dx.dy is an infinitesimal of the second order. The coefficient of dx in the above groupings is called the partial differential coefficient of u with regard to x, being evidently what the complete differential coefficient would be it x were the only variable; similarly the coefficient of dy is called the partial differential coefficient of u with respect to y, ARTS. 179, 180.] PARTIAL DIFFERENTIATION. 207 The notation adopted for these differential coefficients is (dt and /d-\ so that the differential equation runs \dxj \, dy) du = d) dx + (d) dy. Similarly, in the case of a function of three or more variables, d, (du+, (d.... 7 d'll\ 7 /d l\ 7 dZe\ 7 \dx/ \dy/ - dz} 180. Case when x and y are independent variables. If x and y are independent of each other, the ratio of du to dx or to dy is an indeterminate one, inasmuch as the increments dx and dy may themselves have any ratio whatever, Thus, if x and y are the coordinates of a point in a plan of a district, and u is the elevation of the actual point above (say) sea-level, then du is the increase in elevation due to a slight movement away from the point, and this change evidently in general depends on the direction of movement, i.e. on the relative values of dx and dy. Thus, if the point is on the side of a hill, one direction (backwards or forwards) would cause no change of level, while any other would involve a more or less rapid alteration of altitude. We shall refer to this later. In Physics and Mechanics also, a quantity may be a function of several variables, some of which may vary independently of the others. Evidently the partial differential coefficients are the rates of change of the function when one variable changes and not the others. The student should practise forming these partial differential coefficients. N.B.-It is unnecessary in this case to put the brackets round the partial differential coefficients, and it is quite usual to omit them, as is done in the following examples. EXAMPLES. x2 - 3xy - y2 d u du 1. If u=, write down and - x+y dx dy 2. If = =ax, write down du and du dx da * Some writers put -x instead of (- etc., as being a less cumbrous distinguishing notation. There is no settled custom: the student can use which he pleases. 208 DIFFERENTIAL CALCULUS. [CHAP. XIII. dx du du a dx 3. If u=xPyq, write down du, du d and d dx dly dp dq' - 2,d dit du 4. If u = x2 - 3xy + 2y2 - 5ax + 6ay - a2, write down du, d dx ~dn da' I x2 y2 du du d cu 5. If L= -+ 2, show that- x+ yy +a- +- b-=O. a b2' dx dy da db 6. In Example 1, show that x d-+y- = t. 181. Case when y is a function of x. Distinction between total and partial differential coefficient. If, however, y is a function of x, as, for example, if = (x2 - 3x)y, where y = sin x, then dxi = y(2x - 3)dx + (X2 - 3x) dy, whence d =x(2 3)(2 - 3)d = (2x - 3) sin x + (x2 - 3x) cos x. Here we have du = (-) dx + () dy, and du /du\ /du\ dy and dX= (d) + (d) d dx \dx} \dy} dz and it becomes important to thoroughly realize the distinction between the complete differential coefficient du and the partial dx one (d The bracket round the partial differential co\dx) efficient is an excellent way of distinguishing it. Another method will be given below, which is often advantageous. As a matter of fact, when y is a function of x, any function zt, of x, might be expressed as a function of x and y in very different ways; and the value of the partial differential coefficients would depend on the way in which y was introduced. Thus, taking another example: ARTS. 181, 182.] PARTIAL DIFFERENTIATION. 209 If y sin x, and u = x2 sin x cos x, we might write = x2y cos x, or u = x2y \/1 - y2, or i = (sin - ly)2 sill X COS X, or in many other ways, in each of which would be different, \dx and so would (dt, though of course ( would be the same in every \dyJ dx case. It is better in such cases to adopt a notation which recognizes that the partial differential coefficients depend on the form presented by the function, and we might say, if i =f(x, y), du = ($f) dx + (df)dy, dd _df df dy and +- dx dx dy dx' where f(x, y) denotes not only the function, but the form in which it is given, and df and df obviously can then denote only dx dy the partial differential coefficients, so that the brackets round them are not absolutely necessary, and can be used or omitted as desired. If, instead of y being given as a function of x, both x and y are given functions of some third variable, t, the equation may be written du, /df\ dx tf\ ddy dt \lx dt Tdyj dt' N.B.-The brackets, though not absolutely necessary, are useful to emphasize the distinction between partial and complete differential coefficients. Similarly, if we are dealing with a function u =f(x, y,,...) of more than two variables, we shall have du _ df\ dx (df\ dy /df \ dz dt ~ \dx) t \dy) dt ' W dt 182. Gradient at any point of the curve f(x, y) = constant. There is another way in which y may be determined as a L.D.C. 0 210 DIFFERENTIAL CALCULUS. [CHAP. XIII. function of x, viz. when the given function of x and y has a constant value; e.g. if u = 0 is the equation of a curve, x, y being the coordinates of a point on the curve. In such case u = 0 is itself the equation giving the relation between x and y. In such case the partial differential coefficients (d) and (-) have no ambiguity, but are perfectly definite in form. This is in fact the kind of case in which we most frequently require to use these derived functions. Since u is constant in value, the complete differential coefficient u is zero, and therefore the symbol for it is not redx quired; hence we may, if we like, omit the brackets from the partial differential coefficients. In what follows we shall sometimes use the brackets and sometimes omit them. If we differentiate u = 0, we obtain (i)dx+ (d) dy =0, and thus obtain the gradient of the curve at any point (x, y), viz. du dy dx =d d-e dy E.g. If the curve is 3x2 - 5xy + 2y2 - 15x + 6y - 8 = 0, we find dy 6x - 5y - 15 dx - -5x + 4y + 6' which gives us the gradient for any assigned point of the curve. It is rather interesting to see what would be given by the gradient equation if we inserted the coordinates of a point which is not on the curve. This may be left to the student, with the remark that u = 0 is only one of a whole family of curves, u= constant, which have the same gradient equation. Example. To what curve of the above family of curves does the point (1, 1) belong, and what is the gradient of that curve at this point? Ans. 3x2- 5xy + 2y2 - 15x+6y+ 9 =0; 24. ARTS. 182, 183.] PARTIAL DIFFERENTIATION. 211 183. Meaning of the equations ( )==, (d =0. du At points on the curve at which cd =0 the gradient of it = 0 is zero. ' du Hence d- = 0 is the equation of a locus which goes through dx all the points on the given curve where the gradient = 0. Moreover, if we consider the whole series of curves given by the equation u= constant, the locus given by the equation du dL = 0 cuts the whole series in points of zero gradient, i.e. it is the locus of points of zero gradient in the given family of *dt curves. Similarly d-=0 is the locus of points of infinite dy gradient. If at any point on the curve both and d are zero, the exdx dy pression for the gradient becomes indeterminate. Such a point is called a singular point on the curve. It may be an isolated point (called a conjugate point), or there may be two branches of the curve passing through it, in which case it is called a double point. (See Arts. 188, 194, 196.) Application to conic sections. If u= 0 is a conic section, i.e. is of second degree, d = 0 is a straight line going through the points of zero gradient on the conic, i.e. through A the points where y is a maximum or a minimum. Thus, in the figure, the / dZ line AB is given by the equation d = 0. hY D Since the tangents at A and B are /B du parallel, it follows that -j =0 is a dia- -I meter. Similarly -=0 is the diameter which goes through the points of infinite gradient, and therefore the centre of 212 DIFFERENTIAL CALCULUS. [CHCAP. XIII. the conic is given by the point of intersection of these two lines. The general equation to a diameter is (dll + = where k is any constant. For example, the centre of the conic 3x2 - 5xy + 2y2 - 15x + 6y - 8 =0 is given by the intersection of the diameters 6x - 5y - 15 = 0, - 5x + 4Y + 6 = 0. Any other diameter can be fonnd by treating these simultaneously, e.g. x - y - 9 = 0 is a diameter, and so are the lines y+39=O and x~30=0. Hence the centre is at the point ( - 30, - 39). Since the tangents at the ends of 6x - 5y - 15 =0 are parallel to the axis of x, it follows that 6x - 5y -15=0 and y+39=0 are a pair of conjngate diameters. Similarly 5x - 4Y -6=0 are conjugate diameters. and x +30=01 The equation of the curve may be written in the form (6x - 5y - 15)2 - (y + 39)2 + 1200 = 0, showing that the curve is a hyperbola whose asymptotes are 6x - 5y - 15 l~(y + 39) = 0, i.e. 6x-4y 24=0 and 6x - 6y - 54= 0, i. e. 3x - 2y~4- 12= 0.) and x-y-9=0.f The curve has a pair of horizontal tangents given by y + 39= %/I1200 ~20,/3= ~3464. From. these data the curve can be drawn. EXAMPLES. 1. Find the centres of the following curves: (i) 4X2~12xy~10y2-4x-4y-2=0; (ii) 3x2-12xy~9y2+6x- 6y-4=0; (iii) xy-3x+2y-5=0. 2. Show that the centre of the curve x2 - 4xy + 4y2 - 2x+ 3y =0 is at an infinite distance, and draw the curve. ARTS. 183,184.] PARTIAL DIFFERENTIATION. 213 3. If u-x2- 4xy+:;y2+2ax - 2ay, draw the lines given by the equations du du du -=0, -y=0, and d-=0. What locus is represented by the equation u=0? du du 4. If u-x + y3 - 3axy, draw the loci represented by 0-= 0 and - =0, and find where they cut the curve u=0. 184. Homogeneous Functions. Euler's Theorem. In the example of last article, the function 32 - 5xy + 2y2 - 15x+ 6y - 8 contains some terms of the 2nd degree, others of the 1st, and the last term is of zero degree. Such a function is really a shortened form of the homogeneous function 3X2 - 5xy + 2y2 - 15ax + 6ay - 8a2, where a denotes the unit of length. For, of course, every expression consisting of a sum of terms must really be homogeneous, i.e. every term must be of the same dimensions, for it is impossible to add square feet, linear feet, and mere numbers together. A function of any degree when rendered visibly homogeneous, as above, possesses certain important properties, of which we will now prove one of the most important, called Euler's theorem. Euler's theorem is this: If u is a homogeneous function of the nlth degree in x, y, and z, then x -+ya-+z is identically equal to nu. For u consists of a number of terms of the form Kxryqz' where p + q + r =n. If we consider the result of operating on the above typical term, we find it becomes x(pKxP-l yr) + y(qKxPyq-1 Zr) + (rKxKysz -1), which reduces to (p + q + r) KxP yq r', i.e. n times Kxzyq2'. 214 2DIFFERENTIAL CALCULUS. [CHAP. XIII. Hence, if we operate on the whole of u, we shall obtain IL times u, which proves the proposition. Take as illustration it = 3x2 - 5xy + 2y2 - 15xZ + 6yz - 8Z2. dit Here x =x(6x - 5y - 15z), dx dit Y dy= =( 5x +f 4y +t 6z), and zdit -=z 15x + 6y - 16z), dz whence, by addition, du dit dit X-+Yd +Zcdx =2u,, x d-Y z which agrees with the predicted result, since here n=2. We shall nake use of this theorem in the following investigation: 185. Equation of the tangent at any point of the curve u=O. du We know that dy dx dx du. dy If, then, we take (x., y) for the point of contact, and (X, Y) as any random point on the tangent, we find for its equation cdu (XY) Y-y dx X-x du dy whence dA dzu du du Xd +Yy-xd-~+yd~ dr dy d dy, This is one form of the equation to the tangent at (X, Y), but it can be improved by help of Euler's theorem. ARTS. 184-186.] PARTIAL DIFFERENTIATION. 215 If u is made homogeneous by taking a as unit of length, we know that du du du x -+ y —+a- = nu, dx dy da and this is equal to 0, since (x, y) is on the curve. Hence, finally, the equation of the tangent at (x, y) is du du du0 Xi-+ Y-+ aa-aO. Example. Find the tangent at any point of the curve 3x2- 5xy + 2y2 - 15x + 6y+20 =0. Making the equation homogeneous by taking a as the unit of length, we have 32 - 5xy + 2y2 - 15ax + 6ay + 20a2=0. The equation of the tangent at any point (x, y) is X[6x - 5y- 15a]+Y[- 5x+4y+6a] + a[ - 15x +6y+40a]=0, i.e. again putting a =1, X(6x -5y- 15)+Y(-5x+4y+6)=15x-6y-40. If we want the tangent at any special point, e.g. (2, 1), we have only to insert the special values of x and y in the above equation. Thus the tangent at (2, 1) is -8X= -16, i.e. X=2. N.B.-It is essential that the point (x, y) should really be on the curve. If not, the equation does not represent a tangent. The student should see that the point (2, 1) does satisfy the required condition. 186. Equation of Normal. Since the normal is perpendicular to the tangent, it follows that if (X, Y) denotes any point on the normal, it satisfies the gradient equation du Y-y dy x-x du' dx or, as it is often written, for symmetry, X-x Y -y (/du (/du\ W~x) \3y) 216 DIFFERENTIAL CALCULUS. [CHAP. XIII. 187. Polar of a Point. The equation of a tangent at any point (z, y) of the curve u=0 is d d X-+ Y-+ -'-=O, dx dyh da if a is the unit of length. If we fix on any special point (X, Y) on the tangent, the above equation is satisfied by the coordinates (x, y) of the point of contact. If more than one tangent can be drawn through (X, Y), the equation is satisfied by the coordinates (x, y) of every point of contact; i.e. the equation is the equation of some curve which passes through the points of contact of all tangents drawn from (x, Y). This curve is called the polar of the point (X, Y) with regard to the curve = 0. It is easy to see that its degree is in general = n - 1 if u= 0 is of the nth degree, so that if u =0 is a conic section, the polar is a straight line; and if u= 0 is of the 3rd degree, the polar is a conic section. 188. Condition that an equation of the second degree shall represent straight lines (real or imaginary). Let u = 0 be the equation of the locus. du du Its centre is given by - = 0, and == 0. dx dy If the centre is on the locus, the locus must be either a single point, or a pair of straight lines intersecting in this point; i.e. in either case, the locus must be a pair of real or imaginary straight lines passing through the point. Now u= 0 is, by Euler's theorem, equivalent to du- du du x- + y ic + = 0. dx dy da The condition that the centre shall lie on the curve is obtained by substituting its coordinates in the equation. du - du These coordinates make -= and = 0 (Art. 183). dx dy Hence, if the centre is on the locus, its coordinates must du also make -= 0. da ARTS. 187-189.] PARTIAL DIFFERENTIATION. 217 The condition, therefore, is that duz du du dx ' dy da shall be simultaneously true. If the equation is ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, the simultaneous equations are, on dividing by 2, ax + hy + g = 0, hx + by +f = 0, gx+fy+ c=0. EXAMPLES. 1. Find the tangent and normal at any point of the following curves (1) X2+y2=a2; (2) y2=4ax; (3) A2 + By2=a2; (4) xy=a2. 2. Find the tangent at any point of the circle 2+ y2+ A.x+ By+ C=0. [Insert a as unit of length.] 3. Show that the curve x3 + yy'=a3 cuts the axes at right angles. Find the polar of (a, a) with regard to this curve. Draw the part of the curve situated in the positive quadrant. 4. Find the polar of (h, k) with regard to the curve xy2 - 2ay2 + Ct2' =0, and, in particular, find the polar of the origin. Find where this polar cuts the curve, and draw the tangents from the origin to the curve. 5. Show that the equation x2 - 4xy+ 3y2+ 2ax - 2ay = 0 represents straight lines, and draw them. 6. Show that the equation x2- xy + y2 -2x - 2y + 4=0 represents a pair of imaginary straight lines, intersecting in the point (2, 2). 189. Successive partial differential coefficients of a function f(x, y) of two variables. Proof that d (df) d Idf) dy dx = dxc-;\~,/ e _ To TtS.'~~~_ _I +fa i 218 DIFFERENTIAL CALCULUS. LCHAP. XIII. The partial differential coefficients of f(x, y) with regard to x and y respectively, namely df and df, are themselves, in -dx dy general, functions of both x and y, and therefore have partial differential coefficients. These will be df) and d (d); dx \dx dy dx/ and d (df\ and /df dxk\dy) dy \dy The first and last of these are called d2f and clf respectively. dX2 dy2 The second and third are equal to each other, as we shall now show, and are called or d - as we please. dxdy dydx Imagine the function to be expanded in powers of x and y, so that it consists of a series of terms of the form Kx}'y'. Then the corresponding terms of df and df respectively will dx dy be pKxP-ly" and qKvyl-1. d d (df)and df ) will Hence the corresponding terms of () and will each be pqKxP'-ly2-l, which proves the proposition. This is easily extended to any number of differentiations, and any number of variables, the result being that if a function of two or more variables be differentiated a number of times with regard to one variable, and a number of times with regard to another variable, the order of performing the work is immaterial. 190. Homogeneous functions of two variables. Extension of Euler's first theorem. If f(x, y) is a homogeneous function, of the nth degree, of two variables x and y, we have proved that df df x - +Y Ck-l2. ARTS. 189-191.] PARTIAL DIFFERENTIATION. 219 Now evidently d{ and df are two homogeneous functions dx ely of x and y, of the n - 1th degree: hence d2f d 2f = 'd and d12f d/2f elf dxey dy2 dy Hence, multiplying the first by x, and the second by y, and adding, we obtain el2f d2f 9(,d2f df df' x2 d-+ 2Xy +e y2 d n2kl)KXd+Yd=n(nll)u. dX2 xdy T2X dyj-10( I V 191. Successive total differential coefficients of a function of two variables x and y, which are themselves functions of a third variable t. Let n =f(x, y), then eli-el d el ely d1 -\,dx/ dt3 \ey,) dl ' Denote, for brevity, dx dy by x,, y,; and similarly dx delby dt2' dht2 by x2, y2; and so on. Then elu elf elf This may be written in the form i ad f w ishtofid the d if indicating that if we wish to find the differential coeflficient of 2.20 DIFFERENTIAL CALCULUS. [CHAP. XIII. a function of x and y with regard to 1, the operation indicated by it must be replaced by the operation indicated by d d X1 + /1d These expressions themselves are called operators, indicating certain definite operations to be performed. There is nothing new in this, except the systematic way of considering and expressing the work which we know has to be performed when we differentiate such a function. dluU / df d dr\ Now let us find i.e. d-( ciF d ct2' g' j it I+ YQ' Evidently this equals [see Art. 24] d d(ItN + df { d (df\+ cdl df\ dt ctd dt ' dx Ft t.f Now dtcj dx +dy)since Ct is a function of x and y. This equals dX d2f d2f x"1x2 + Ydxd' Similarly y) _= (d d+ ) f) d-t \dy I dx- Y d2f d2f' = dxdy + d^y2 Also = 2, and 1=. dt-x2'and c-t Y2' Hence, on substituting, and arranging the terms, we find d2 / 22f d cl9ff d (d df dt2 = (2- +- c 2x1Yl c + y12 ) + (2f 2) (2) Similarly d%, could be formed, but they rapidly increase in complexity in complexity. ART. 191.] PARTIAL DIFFERENTIATION. 221 If we are dealing with the curve u= constant, all these total differential coefficients are zero, and the above equations then give the connections between the partial differential coefficients and the values of x, Y1, yx, y,.... EXAMPLES. 1. If u= x3 - 3x2y + 4xy2 - y, find 7 { dur and d7 du cdxdy ) dy dx) d3u 2. If uI=x3y4, find C —dy in several ways, i.e. by varying the order of the operations indicated, showing that they all lead to the same result. 3. In the function of Example 1, test the formula 2d%2u.. - d%21 2d xd -2 + 2xy -+-2 d + y2dy-2= 6u by going through the operations. 4. Show that x2du + 2xy dta + yc2d2- is not the same as dx2 dxdy dy2 d d 2d ( 4x- +yrY u. Find the values of the two quantities in the case of the function of Example 1. 2d2u d2u d2u d d 2 5. Show that h 2-d +2hk d- +k2 is the same as h- d +k d - u, dx2 dxdy dy2 dx( dy; if h and k are constants. d2u 6. Find what the formula of Art. 191 for d-2 becomes when t-=x, i.e. when y is a given function of x. 7. Find what the same formula reduces to when u —y. 8. Find what the same formula reduces to when d and Y are dt dt constant. Take, for example, x-a=ht, y-b=kt, and compare Ex. 5 above, 9. If u=constant, find the value of d2Y in terms of the successive dx2 partial differential coefficients: (1) by means of the formula obtained above (Ex. 6); (2) by direct differentiation of the equation -(= - ( t-') - (c~ 222 DIFFERENTIAL CALCULUS. [CHAP. xIII. 192. Connection between rectangular coordinates and polar coordinates. The position of a variable point in a plane may be given not only by means of its rectangular coordinates x and y, but also by means of what are called polar coordinates, viz. the distance of the point from the origin, or some other fixed point, and the angle that this line makes with the axis of x. These are called the polar coordinates of the variable point, and are usually denoted by the letters r and 0. If r is measured from the fixed point (a, b), the connection between x, y and r, 0 is Pfx,y) x - a=r cos 0, rA/,^~ y - b = r sin 0, A ^a, and the reverse relations are (a,b) 92 = (x - a)2 + (y - b)2, tanO Y-b y-b x-a _ tan0 = =; sin =0=; cos=-. =x-a r r^ The differential relations between x, y and r, 0, if 0 is expressed in radians, are dx = cos 0 dr - r sin 0 dO, dy = sin 0 dr + r cos 0 d0; and dr = cos 0 dx + sin 0 dy, r4 d0 = - sin 0 dx + cos 0 dy. From these follow the partial differential coefficients: /=cos0, Kdo)- = -rsin0' Cos 0 r sin, and /dr\ /fdr\ (- = cos 0, (- = sin 0, fd\ /d0\ = cos. )= -sin, r - = cos. (dy ARTS. 192,193.] PARTIAL DIFFERENTIATION. 223 It is important to realize exactly what the partial differential coefficients denote. I. (d) and ( denote the rates of change of x and y per unit change in r when x, y and r vary, but not 0. II. (dx and (dy) denote the rates of change of x and y per unit change in 0 when x, y and 0 vary, but not r. III. ( and (d) are formed on the supposition that r, 0 (dx) \dx and x vary, but y does not. IV. (d), (d) assume r, 0, and y to vary, but not x. These changes are shown in the following diagrams, where the point is supposed to move from P to Q, where PQ is infinitely small, but the direction of movement is in each case such that one of the quantities x, y, r, 0 remains unaltered. dr\Qdr Q I -rdx II. p Qd (ai)b) 0 constant A constant dr\ Q dx rd0 dy d. 1rd dP IV. 4 (a,)p y constant fa,b x constant 193. Application to functions. If u is a function of x and y, it can also be expressed in terms of r and 0, 224 DIFFERENTIAL CALCULUS. MCHAP. Xli. Now du = dxLd + ) dy. Hence, if we treat u as a function of r and not of 0, i.e. if we keep 0 constant, we find / = - ) cos 0 + (- sin...................(1) Similarly, if we treat 0 as the variable, r remaining constant, we find /du\ /du\. /du\ ~ (d) = -( d ) sin 0 + r' cos 0.............. (2) Again, since u is expressible as a function of r and 0, du= (dr) dr + (d6-t dO; du\ /duN\ /du\ 1 T*J= tkdr ) cos- - sin.............(3) _,d /clu\ Zulu\ d/du\ 1 and ( ) )sin O+ )- cos 0..............(4) From these relations the second partial differential coefficients can be obtained. Thus, for example, from (1), \d2u,\ [d f / dPu \. A (d 2) = cos 0 { C() cos 0 + (d ) sin 0 \dr2/ \\dx2/ \dxdy + sin 0 adx -y cos 0 + (,2} sin 0a ( d/ \ /' d2. d2 = (d2 cos2 0 + 2 d-) cos 0 sin 0 + (i,- sin20...(5) Idx2/ \dxdy/ y2 This relation is of importance in connection with the maximum and minimum values of a function of two variables, as we will proceed to show. 194. To find the maximum and minimum values of a given function of two variables. Let ie denote the function of two variables x and y, it is required to find for what values of x and y the value of it may be a maximum or minimum. ARTS. 193,194.] PARTIAL DIFFERENTIATION. 225 We can obtain a geometrical picture of the function by supposing x and y to be the rectangular coordinates of points in a horizontal plane, and the values of u to be denoted by the lengths of vertical lines drawn from those points, so that the tops of these vertical lines lie on an undulating surface like a tract of hilly country. At the top of a hill, u will be a maximum, and at the bottom of a valley u will be a minimum. To fix the ideas we will suppose the axis of x is drawn in the east and west direction (positive towards the east), and the axis of y in the north and south direction, positive towards the north. Evidently in general the surface is horizontal at a point where u is a maximum or a minimum. Hence (d) and d) must equal zero at such a point. This is the first condition to be satisfied. We must therefore solve simultaneously the equations ( — )=0, (du) 0 to find the values of x and y for which the surface is horizontal. Let x = a, y = b be a solution of these equations. If these values of x and y make u a maximum, it is evident (from our investigation of the conditions for a maximum value of a function of one variable) that ( d2) and (/ must be negative, as the conditions ensure that an east or west movement from the point, or a north or south movement from the point, shall be downhill. This, however, is not sufficient, for if we are really at the top of a hill, any movement in any direction must begin to take us downhill. To investigate the necessary analytical condition for this to happen, take x- a = r cos 0, and y- b = r sin 0, and consider the movement through a short distance along a direction making any angle 0 with the x-axis. If we are really at the top of a hill we must have (-) =Oand (-dr negative, whatever 0 may be. L.D.C. p 226 DIFFERENTIAL CALCULUS. [CHAP. XIII. Now ()d = d cos 0 + sin 0, so that the first condition, viz. (dr = O, is satisfied whatever 0 may be, since, by supposition, both y- and ) = 0. /___.fd2\ /d2U\, dco\ 02/ \ s n20 Again, \dr2 = cos20 + 2 (a d d -- cos 0 sin 0 + \du2 sin 9f/^A d2f\. /^\ c, 2aI =cos2 {(, + 2 (dxly) tan 0 + (,y2) tan20 Hence, if td, is to be negative for all values of 0, we must have ( + 2 d2u tan 0 + d2 2 tan2 0 \adx2) \dxdy/ \dy2) negative for all values of 0. The needful condition is that du and du shall be negative, dx2 dy2 /d2U\ /d2u\ / dlu \2 and that d') K ) d-, 3) shall be positive. nd, - \,dy2 - \dxdy\ To prove this, denote (d22 d t1, and K/) by A, B, \dx2]' \,d-x'dy C-b y 2,y and C, and denote tan 0 by t. We have to find the condition that A+ 2Bt + Ct2 shall be negative for all values of t. The expression is equal to { (Ct) + 2B (Ct) + AC } { (Ct + B)2 + AC - B2}. This can only be negative for all values of 0 if C is negative and AC - B2 is positive. Q.E.D. ARTS. 194, 195.] PARTIAL DIFFERENTI &TION. 227 Cor. These conditions, of course, involve the condition that A shall be negative as well as C. Similarly, the condition that u shall be a minimum for cdu du values of x and y that make - =0 and d= 0 is that dx dy A and C shall be positive, and AC - B2 positive. These conditions ensure that d2U) shall be positive for all values of 0 We see, therefore, that when (- =0 and (d) =0 and /d2u /d2u\ d2U \2 /d2) ' dy2) d> dy) u will necessarily be either a maximum or a minimum, and, further, that u will be a maximum if (-u and () are,dx2/,dy 2 * —~:-, *. - /d6u,\d,2l\.. negative, and a minimum if (d -2 and (d are positive. \dx2,/ \dy2/ If /ld2% ldd\ / d2u\ i,., ~, ld%~\ If (dx2) (dy- ( dU- ) is negative, we shall have (,) k\dx2Ik\dy22 \dxdy/ bdr' positive for some values of 0 and negative for others, with two intermediate directions for which (d) will be zero, viz. when C-tan = - B + ~/B2- AC. Such points on a hilly surface occur on a neck between two hills, or between two valleys. 195. Extension of Taylor's Theorem to a function of two variables. Let u=f(x, y) denote the function. It is required to find the connection between f(a + h, b + k) and f (a, b) when a, b are any pair of values of x, y, and a+ h, b+k are any other pair. 228 228 ~DIFFERENTIAL CALCULUS. [HP II [CHAP. XIII. Let x-a=rcosO, y-b=rsinO, so that when x=c+h, y=b+k, webhave rcosO==h, rsin6=k. Then u =f (a + r cos 0, b6+ r sin 0), which is a function of r, if we take 6 = constant, i.e. if h, ic are in a given ratio. (With this understanding, we may dispense with the brackets round du d.u in the following work). Hence, by Maclaurin's theorem, din r2 n d-1. 2 dr2 we ndu,, du dren*u*, denote the values of U...,I when r=O, i.e. when x=a, y=b. du d~u*.dm* Now we have seen that -=Cos 0d- + sinm dun dun du drx dy d2u d2u d2u 2 d2in Also -cos 0 - ~2 cos 0sinO~ + sin0 d2 dx2 dxdy dy2' d2U d2u d2u d2U r212-2=h2 —+ 2hle- + k2, dr2 dx2 dx dy dy2 which may be symbolically expressed as r-2din u d d \2 =r dx +kdIU. Similarly rs L= hK d d Hence u-f(a~h,bk),o h +k U h — k O+ *These are obviously partial differential coefficients, so the brackets may be omitted. ART. 195.] PARTIAL DIFFERENTIATION. 229 where on the right-hand side U0 Luo) Ld-, etc., are the values df df ~ d d df df etc., when x =a, =b. of f(X, Y), VdThis may be more conveniently represented in the form f(x+h, y~k)=f(x, Y)+ d and may be briefly designated symbolically as f(x+h, y+k)={l +(h'+kd)~ +k w(h i+< jz+...}(x Y) hd +k d =e x df(XI Y) This formula is the extension of Taylor's theorem to a function of two variables. It may be similarly extended to a function of any number of variables. EXAMPLES. Discuss the maximum and minimum values (if any) of 1. 4X2 +l2xy + 10y2 - 4x - 4y - 2. 2. x2 - 4xy + 3y2 + 2x - 2y + 6. 3. x3 + y3 -6xy. 4. 6 3y2 - x4y2 - x3y3. 5. X4~y4-4X2 + 4xy - 4y22 6. Show that the maximum and minimum values of ( 2U ) i.e. of \dr2j A cos2O + 2B cosO0 sin 0~+ C sin2O, are I{ (A~C) ~,,/A - C)2 +[4B2}1, and 2B that they occur when tan 26 =2 7. Show that these maximum and minimum values have the same sign when AC> B2, and that they are respectively A+ C and 0 when AC = B.2 8. Deduce by means of Ex. 6 that the axes of the ellipse obtained by giving any constant value to the expression in Ex. 1 lie in the directions given by tan 20 = - 2. 9. Show that in Ex. 2 the value of d2) is zero when cos20 - 4 cos 0 sin 0 +3 sin26 = 0. Explain this geometrically in connection with the point where (dki) and (y) are zero. 230 DIFFERENTIAL CALCULUS. [CHAP. XIII. 2x + 3y+6 d -u 2 - 2ux 10. If u= 22+ 1show that d=- and that x2 + y2l d1X2 2 + y2+1' du_ 3-2uy dy-x2 +y2 1' Hence find the maximum and minimum values of u. 11. In the last question, show that when u is a maximum or minimum, (dr)= - 2y 2u, and is therefore the same for all values of 0. 12. By using Taylor's theorem for the expansion of f(x + h, y + k) in powers of h and k, deduce the conditions for a maximum or minimum value of f(x, y). 13. Expand f(x + h, y + k) in powers of h and k by Taylor's theorem whenf(x, y) is any one of the functions considered in Examples 1 to 5. d d d d\ d Operate successively by hd+ kd, (h +h x+kdy) (x +] dy) 14. Expand the functions of Examples 1 and 2 in powers of x and y, and determine h, k so that the coefficients of the first powers of x and y disappear. [Use Taylor's theorem for the expansion, merely interchanging x, y with h, k.] 15. Plot on a diagram values of the function u = x3 +y3- 6xy, for integer values of x and y between +5, using 1 inch as unit of length. Draw (approximately) on the diagram the curves corresponding to iu=O, u=27, and ut= -8, noting that in the last case the locus breaks up into a straight line and a point. 196. Contour lines and lines of flow. If we consider a point anywhere on the side of a hill, there is always one direction in which motion is neither up nor down. If we walk round the hill in such direction that we keep at the same level, we traverse what is called a contour line or line of level. In the case of the surface representing the function u, the level direction at any point is given by () = 0 i.e. cos 0+ du sin 6 =0, du dx i.e. tan 0.-. du dy ART. 196.] PARTIAL DIFFERENTIATION. 231 This, as we have before proved, is the direction of the tangent to the curve u = constant. In fact the curves u=constant are the equations of the contour lines of the given surface. They can be drawn on the map. [For the cases when a contour line reduces to a point, or contains a double, or crossing, point, see Art. 194.] The direction at right angles to the above value of 0 is the direction of steepest slope, the direction in which water would begin to flow from the point down the surface. This direction is given by du tan 0=. du dx The maximum value of (du is thus found to be \dr V dxJ + Vdy) with a + sign if we go uphill, and a - sign if we go downhill. [See Ex. 18, p. 109.] If we travel continuously in the direction of steepest slope, we traverse what is called a line of flow. These lines and their traces on the map are at right angles to the contour lines u = constant. If v = constant is the trace of a line of flow, its direction at any point is given by dv dx dv tan 0= -d. dy Hence the curves satisfy at every point the differential equation v d dv du dx dy -;= - du' dy dx dv du dv du -.e+. =+ 0. d dx dy dy DIFFERENTIAL CALCULUS. [CHAP. X11I. EXAMPLES. 1. Prove that the parabolas y2= a2 - 2ax, and y2= 2 + 2bx intersect at right angles at the points (Q (a - b), ~,.Iab). 2. Draw the parabolas for a=I, 2, 3 and b=l, 2, 3. 3. Prove that the curves 4x2 - 8y2 = a2 and X2y =b3 cut each other at right angles. Draw them for the case a = 5, b = 1. 4. Prove that the curves 2X2 +y2=a2 and y2 bx cut each other at right angles. Draw them for a=l, 2, 3; 6=1, 2, 3. 5. Prove that if it-f(Ax2 - By2), and v=F(XByA), the curves u=coustant, and v=constant, cut each other at right angles. 6. Prove that the confocal conics a" " 2 y2 X2 -1 and x y = 2 a2- C2 Vb - b22 cut each other orthogonally at the points ab 1/ - x=~-, y=~ -v(a2 - c2) ( - b2). C C Draw them for the case a 5, b = 3, c=4. 7. Prove that the cycloid x=a(O-sin6), y=a(l-cos6) cuts the cycloid x=a(O-sinO), y=a(l~-cosO) at right angles. 8. If X2 + y2 + C2 V=X2 +2 - C2 X I show that the curves t = a, v = b cut orthogonally. Draw them for ac =, 2c, 3c; b=c, 2c, 3c. 9. Show that the feet of the normals drawn from any point (h, k) to the curve u= 0 lie on the curve (x-dx du (x - h) -= (/ - ax 10. If u=0 is a conic, and (h, k) its centre, show that the equation given in No. 9 represents the axes of the conic. [This theorem follows at once if we refer the conic to its principal axes.] 11. Show that the axes of the conic 3X2 - 5xy +2y2 - 15x + 6y + 20 = 0 are given by the equation 5X2 + 2xy - 5y2 + 378x - 330y - 765 = 0. 12. Show, by the method of Chap. XII., that the gradients of the A ~~BtanO8 tangent at a douhle point are given by the equation tan= -B ~ C tanG leading at once to the last equation of Art. 194. CHAPTER XIV. CURVE-TRACING. RECTANGULAR COORDINATES. 197. It is important to be able to draw a curve from its equation. For this purpose a certain amount of experience is necessary to enable the important characteristics of the curve, and its general shape, to be readily ascertained. We shall, in this chapter, consider curves given by rectangular coordinates, and, in the next chapter, those given by polar coordinates. In the former case, there is a distinction between those in which y is given as an explicit function of x (or x as an explicit function of y), such as y=x2(6 - x), y = sinx, y=ex... and those in which x and y are implicit functions of each other, such as 3 + y3 - 6xy = 0, x2 + y2 - 2ax = 0,... We already know how to plot a curve when y is an explicit function of x, viz. by making a tabular list of values of x, y and d, and plotting a series of points and gradients, and drawing the curve through those points at the proper gradients, paying particular attention to the points where y is a maximum or minimum, i.e. where the gradient is zero. So also if x is given as an explicit function of y, we must tabulate x, y, - and plot similarly, giving the axis of y the post of honour instead of the axis of x. Or, again, it may be possible to express x and y as explicit functions of some third variable; e.g. in the case of the parabola y2 = 4ax, we may write x = at2, y = 2at, where t is any variable. In such cases, we can easily also express dx 234 DIFFERENTIAL CALCULUS. [CHAP. XIV. in terms of the third variable, and then find as many points and gradients of the curve as we please by simply making a table of values of x, y and -y corresponding to different values of the variable on which they depend. Examples of equations allowing of this method of treatment will be found in the sequel. We shall say no more here about these particular forms of equations, but confine ourselves to the general case (of course including the above simple cases) in which the equation is in the form f(x, y)= O, i.e. in which x and y are, in general, implicit functions of each other. 198. Algebraic curves. Degree of a curve. Equations and curves may be classified into Algebraic and Transcendental; e.g. y = 2x -2 is algebraic, and so are y= /(2x-x2) and y + J(2x - y)= 0, while y = sin x, y = log x, are transcendental. We shall confine our attention solely to algebraic curves whose equations have been cleared of roots and fractions, i.e. in which every term is of the form Axrys, where r and s are positive integers or zero, and A is any numerical coefficient, as for example, (1) x2+y2+3x-2y=0. (2) X -xy +y2 _ 4=0. (3) x3- 6y2 + y=0. (4) x2y2 = a2(x2- /2). (5) x(x- y)+a2(x+y) = 0. The student can have these curves in his mind while reading the following articles, and should endeavour to trace them gradually as he reads. The degree of the highest term or terms in the equation is called the degree of the equation and of the curve. Thus the first is of the 2nd degree; the fourth is of the 4th degree; and the others are of the 3rd degree. 199. Curve passing through the origin. If the equation contains no constant term, the origin is on the curve. In this case, the approximate shape of the curve near the origin can ARTS. 197-199.] CURVE-TRACING. 235 be discovered by neglecting the terms of high degree, and retaining only those of low degree, since, when x and y are small, the lower degree terms are more important than those of high degree (see Arts. 177, 178). And, in particular, if only the lowest degree terms are retained, we obtain the tangent or tangents to the curve at the origin. If the lowest degree terms are of the first degree, there is one branch of the curve passing through the origin; if of the second degree, there are in general two branches passing through the origin, since on equating these terms to zero, we have in general two tangents. In this case the origin is called a double point on the curve. The two tangents thus obtained may be coincident: in such a case there are, in general, two branches passing through the origin and touching each other there; but in special cases the two branches may stop at the origin instead of going through it. The curve is then said to have a cusp at the origin. (Examples will be found below.) If the lowest degree terms cannot be broken up into real factors, the tangents are imaginary, and in this case the origin is an isolated point on the locus, separate from the rest of the curve. Such a point is called a comju gate point. If the equation giving the tangents at the origin is higher than the second degree, there may be several tangents, i.e. several branches of the curve passing through the origin. It is then called a multiple point on the curve. If all the tangents are imaginary, it is, of course, an isolated, or conjugate, point. EXAMPLES. 1. Find the tangents at the origin in the case of the curves (1), (3), (4) and (5) above. 2. Draw the curve x3=6y2 in the neighbourhood of the origin. 3. Draw the curves ay = x2, a2y =x3, y= ax2, y4 = a2x2, between x = f,. 4. Show that the curve y2(3x + 2y) =4y2 + 9x2 has a conjugate point at the origin. 5. Show, by the graphic method of Art. 178, that near the origin the shape of the curve x4+ y4- 2axy2- 3a2x2=0 is approximately given by the equation y4- 2xy2- 3a2x2=0, i.e. by the parabolas y2 - 3ax = 0, and y2+ax =. [Notice that in this case there are two coincident tangents at the origin, given by the equation x2 0, and the curve has two branches, both touching the axis of y at the origin.] _ _____._____. _ _.. _ _._-_ __,. r 236 DIFFERENTIAL CALC JULUS. LCHAP. XIV. 200. Double and conjugate points in general. If we carefully studythe system of contour lines u = constant in connection with Arts. 194, 196, we see that a conjugate point occurs as part of the locus whenever u is a maximum or minimum, i.e. whenever the top of a hill, or the bottom of a valley, is on the level of the particular contour; and that a double point occurs when a neck between two hills or two valleys is on the level of the contour, which then, in general, crosses itself at this point. A triple or higher multiple point is far rarer and is not so easily pictured to the mind. All these points are classified as singular points. We saw in last article how to discover and distinguish between conjugate and double points when they occur at the origin. The investigation in Art. 194 shows that at such du du points, in general, we must have d=0 and d=0; and, further, that the directions of the tangents at such a point are given by the equation /(d2\ d2n d2u\ (d) tan0 + 2 ( cdy tan 0 + (2)= dy2/ \dxdyj \dx2 so that if the roots of this equation are real, it is a double point, which may be a cusp if the roots are equal; and, if they are imaginary, it is a conjugate point. We have only mentioned this matter here because of its geometrical interest in connection with contour lines, but we shall not trouble the student with examples, as it is somewhat beyond the scope of the book. We should note, however, that if, to assist us in drawing du any curve = constant, we draw the auxiliary curves 7=0 du did and -= 0, the gradient is zero where the curve -= 0 cuts /du the given curve, and infinite where -y= 0 cuts the given curve, dy unless it happens that the two auxiliary curves intersect each other on the given curve, in which case the point of intersection is either a double point or a conjugate point on the given curve. ARTS. 200, 201.] CURVE-TRACING. 237 Of course if we are considering the whole series of curves u = constant, the positions of these double or conjugate points are of great importance, as some of the series are bound to contain them. The centre of a hyperbola and the centre of an ellipse are, in this sense, examples of double and conjugate points respectively, the one being the crossing point of the asymptotes of a series of hyperbolas, and the other coinciding with the infinitely small ellipse belonging to a series of ellipses, the series in each case being obtained by giving different values to the constant in some second degree equation u= constant. See Examples 5 and 6, p. 217, and draw several curves of each series, by equating the respective functions to different constants (such as 0, 2, 4, 6). 201. Simple cases of symmetry. If only even powers of y occur in an equation, the line y 0 is a line of symmetry (as for example in the case of the parabola y2 = 4ax), since for every value of x the values of y must be equal in magnitude and opposite in sign, so that for every point (x', y') above the line there must be a point (x', - y') at an equal distance below; hence, if the upper part is folded-over round the axis, it will coincide with the lower part. The knowledge of this symmetry will be of great assistance in drawing the curve. Similarly, if only even powers of x occur, the line x = 0 is a line of symmetry. Thus, in (2) of Art. 198, y=O is a line of symmetry; and, in (4), both axes are lines of symmetry. If all the terms in an equation are of even degree, as in the case of curve No. 4 above, or if all the terms are of odd degree, as in the case of curve No. 5, the origin is what is called the centre of the curve, i.e. all chords passing through the origin are bisected there. To prove this, it is merely needful to see that if a point (x', y') is on the curve, the point ( - x', -y') is also on the curve, which is the case since the equation will be unaltered if ( - x', - y') is substituted for (x', y'). If the equation is unaltered when x and y are interchanged (e.g. x3 - 3axy + y =b3), the curve is symmetrical about the line x = y. For if (x', y') is a point on the curve, (y', x') must also be on it, and these points will be found (by folding or DIFFERENTIAL CALCULUS. 238 [CHAP. XIV. otherwise) to be equidistant from the line x =y, and situate on a line perpendicular to it. 202. Curves which extend to infinity-Asymptotes. An asymptote to a curve is a line which is infinitely close to the curve at points infinitely distant from the origin, i.e. at points whose abscissa and ordinate (or one of them) are infinitely great. Not only must the ratio of y: x be the same for the line as for the curve, but the difference of the ordinates of points on the line and curve must be zero when the abscissa is infinite, and the difference of their abscissae must be zero when the ordinate is infinite. All lines parallel to an asymptote at a finite distance from it will satisfy the condition that the ratio of y: x will be the same as that of the curve, but they will not satisfy the difference condition. The asymptote is the line which satisfies both conditions. The direction of this system of parallel lines is said to be the direction in which the curve extends to infinity. In many cases the curve extends to infinity in more than one direction. The direction or directions of the infinite branches are found by neglecting all the terms in the equation except the highest, since, when x and y are great, their highest powers are more important than their lower powers. By equating the terms of highest degree to zero we obtain a homogeneous equation of the nth degree representing n straight lines (of which an even number may be imaginary) through the origin stretching in the direction of the infinite branches of the curve. If all are imaginary, the curve does not extend to infinity in any direction, and is a closed curve. For example, in the hyperbola b22- a2y2= a2b2, the lines b22 - a2y2 =, i.e. bx ~ ay = 0, stretch towards the infinite parts of the curve, being in fact the asymptotes. In the ellipse b2x2+ ay2 =a2b2 the corresponding lines are b22 + a2y2= 0, and are imaginary. As a rule, the lines through the origin are not the asymptotes themselves, but are parallel to the asymptotes. To find the asymptotes we have to make use of the next highest terms as well as the highest, and approximate by taking in succession the different factors of the highest degree as constant. The method of determining the value of the ARTS. 201-203.] CURVE-TRACING. 239 constant in each case has been already explained in Arts. 173, 174, and will now be illustrated by the following examples (taken from the curves of Art. 198.) The equations so obtained will be the equations of the asymptotes. Examples. (1) Let the equation of the curve be x3 - 6y2 +y3=0. The terms of highest degree are x3 + y3, therefore there is an asymptote in the direction x + y=0. 6y2 Writing the equation in the form x + y-2 - and approximating to the right-hand side by the method of Art. 174, i.e. by putting x=1, y= -1 (equivalent to making the ratio y: x=- 1), we find the value of the fraction is 2;.-. the asymptote is x+y=2. (2) Let the equation of the curve be 2y2 =a2(2 - y2). The infinite directions are y=O and x=0. Considering first the asymptotes in the direction y=0, we find y2 = a2(x2- 2)=a2 when x is infinite (y being finite); y= ~a are a pair of parallel asymptotes. a2(x2- y2) 2 Again, x2= - y2 -a2 when y is infinite. These lines, x2+ a2=0, are imaginary, hence the curve does not go to infinity in the direction x=0. (3) In the curve 2(x-y) + a2(x + y) =0, we have x2= - a2 + y= a2 when y is infinite, x-y and x - y= - a2 X = when x and y are infinite. x x = + a, x - y = 0, are all asymptotes. x3 + a2x NOTE.-By putting the equation in the form y=-2 — we can find xU - a2 as many points and gradients as we please, and so draw the curve. (See Art. 197.) The tangent at the origin is x + y=0. (4) The student can in the same way show that the asymptotes to the curve 3 - y2+y2-4=O are x= 1 and x~y= -a. 203. Parallel asymptotes. In some of the above cases there were a pair of parallel asymptotes, the finding of which was done in the same way as in the case of a single asymptote, with the peculiarity that the parallel asymptotes were sometimes imaginary. 240 DIFFERENTIAL CALCULUS. [CHAP. XIV. The case of parallel asymptotes requires, however, further consideration, as may be shown by the following example. Let the curve be x3 + xZy - xy2 - y + 2.ry + 2y2 - 3x + y = 0, i.e. (x - y)(x + y)2+ 2y(x + y) - 3x + y = 0. There is one asymptote in the direction x-y=0, and a possibility of two asymptotes in the direction x + y = 0. The first presents no difficulty, and is easily seen to be x- y + 1 = 0 by the method explained above. To find the other two, divide by x- y, and approximate by the method of Arts. 174, 175, remembering that, though x and y are infinite where the asymptotes meet the curve, yet x + y is finite. The equation is 2y(x+y) 3x-y (x + y) + = 0.... x-y x-y Now, in the first fraction, the numerator is of higher degree than the denominator, so that, in general, the fraction would be infinite when x and y are infinite. It happens, however, that the numerator contains the finite factor x + y, hence the limiting value of this fraction is - (x +y), since the limiting value of — y under the given conditions is - 1 (Art. 174). x -- Y The limiting value of the second fraction, viz. 3x- y is 2;. the asymptotes are (x + y)2 (x + y) - 2 = 0, i.e. x+y+l=0 and x+y-2=0. N. B.-If x + y had not been a factor of the terms of highest degree but one in the equation, the asymptotes parallel to x + y = 0 would have been at an infinite distance, and therefore useless. It is usual to say in this case that the curve has no linear asymptotes in that direction. The simplest case of this sort is the parabola y2= 4ax, where the highest degree term indicates two asymptotes in the direction y= 0, but on investigation they are found to be at an infinite distance from the origin. We shall recur to this again under the head of curvilinear asymptotes. ARTS. 203, 204.] CURVE-TRACING. 241 The reason why the above difficulty did not arise in the case of any of the specimen curves of Art. 198 was because Nos. 4 and 5, which had repeated factors in the terms of the nth degree, had no terms of the n - 1th degree. 204. Description of the general method of finding linear asymptotes. The general method of obtaining the equations of the linear asymptotes may thus be described. Let the equation of the curve be arranged according to the degrees of the various terms, and let the terms of the rth degree be denoted by u.. Then the equation is of the form un + u,-1 + U-_2 +... + U1 + Uo = 0. Now (1) let ax+ by be a factor (unrepeated) of u O, so that, _= (ax + by) v,_-,. Then, dividing by v,,_, we have Rn-1 __yn-29 ax + by + + - +... = 0. en — IVn-1 The asymptote is ax + by + c = 0, where c is the value of.-' when ax+ by= O (Art. 174). vn-I The remaining fractions all become zero when x and y are infinite. (2) Let (ax + by)2 be a factor of u", so that u, _ (ax + by)2 v,-2. Then (x n-\2) + t-_ ( 2 terms which vanish 0 (vx2,-2 +,\when x and y are infinite/ Now uL — is in general infinite, and there will be no linear n-2 asymptotes in the required direction unless either ax + by is a factor of i__ or the terms represented by ui are nonexistent, but in that case, the method of approximation gives an equation of the form (ax + by)2 + B(ax + by) + C = 0, giving two parallel asymptotes in the required direction. [NOTE.-B will be zero if the u,_ terms are non-existent, and C will be zero if the u,_n terms are non-existent.] L.D.C. Q 242 DIFFERENTIAL CALCULUS. [CHAP. XIV. Similarly, if u, contains (ax + by)' as a factor, there will be no linear asymptotes (i.e. none at a finite distance) unless i,_, contains the factor (ax + by)'-, or is non-existent, and also u_ (ax + by) -2,, and so on. If these conditions are fulfilled, there will be r parallel asymptotes given by an equation of the form (ax + by) + B (ax + by)~-l + C (ax + by)-2 +... o, where any of the coefficients B, C,... will be zero if the corresponding terms are non-existent in the equation of the curve. Of course even in this case some, or even all (if r is even), of the asymptotes may be imaginary: but this is quite different from the failure by reason of any of the terms becoming infinite. If all the asymptotes are imaginary the curve does not go to infinity at all in the specified direction; whereas if a term becomes infinite, the curve does go to infinity, but with no linear asymptote, just as a parabola does in fact. This gives rise to the consideration of what are called curvilinear asymptotes, but before proceeding to the consideration of these, we will approach the theory of linear asymptotes in another way which will be instructive. 205. Intersections of a straight line and curve. Another method of finding asymptotes. A straight line cannot cut a curve of the nth degree in more than n points; for, if we eliminate y by solving simultaneously the equation of the curve and the equation Ax + By + C= 0, we shall obtain (in general) an equation of the nth degree in x whose roots will be the abscissae of the points of intersection of the line and curve. This equation has n roots, of which any even number may be imaginary. Consequently, if a line cuts the curve in less than n real points, the deficiency will be an even number. [Similarly if we eliminate x, we shall obtain an equation in y to which similar remarks apply.] Hence, a curve of odd degree must cut every straight line in at least one point, since it cuts it in an odd number, of which an even number may be imaginary. Consequently a ARTS. 204, 205.] CURVE-TRACING. 243 curve of odd degree must extend to infinity in one direction at least. A curve of even degree may or may not extend to infinity. A curve which does not so extend is called a closed curve. If it does go to infinity it is called an open curve. Infinite roots. If Ax+ By is a factor of the highest degree terms in the equation of the curve, the equation giving the points of intersection of the curve with the line Ax + By + C = 0 will be of the n - 1th degree.* We do not say in this case that there are only n- 1 roots, but that there are n roots, of which one is infinite. (For the only way in which an equation of the nth degree can have an infinite root is by the coefficient of its highest term being zero.) Hence in this case every line parallel to Ax + By = 0 cuts the curve in one point at infinity. If we choose C so that the coefficient of the next highest term is zero, we shall have a line which cuts the curve in two points at infinity (or possibly more). This line will be the asymptote in this direction. [See also Art. 207a on p. 249.] By solving the equation giving the finite points of intersection of the asymptote with the curve, we shall, in general, find n- 2 points, of which an even number may be imaginary. If (Ax + By)2 is a factor of the highest degree terms in the equation of the curve, and if also Ax+ By is a factor of the terms of the n - 1th degree (or if these terms are not present), the equation giving the points of intersection of Ax + By + C = 0 with the curve will be of the n - 2th degree, and every line parallel to Ax+By=0 will meet the curve in two points at infinity. If we then choose C so that a third root is infinite, which in general is possible for two values of C, we shall have two parallel asymptotes, each of which cuts the curve in at least three points at affinity. If (Ax+ By)2 is a factor of the highest degree terms, but Ax+ By is not a factor of the terms of the n- ll' degree we shall have no asymptotes at a finite distance. This case gives rise to curvilinear asymptotes, which we shall investigate in the next article. *In some cases the equation will be of lower degree, say of degree - r. In this case every line parallel to Ax+~ By=0 meets the curve in r points at infinity, and the asymptotes will meet it in r+ 1 points, or possibly more. 244 DIFFERENTIAL CALCULUS. [CHAP. XIV. EXAMPLES. 1. Apply the above method to find the asymptotes to the curves of Art. 198, and to find where the asymptotes cut their respective curves. 2. In the curve (ax+by + c)uc,_, +u,_2+terms of lower degree =0, show that ax +by+c=0 is an asymptote unless u,-_l contains ax +by as a factor. 3. Show that the asymptotes of (x - y + 1)(x + + 1) (x- 2y + 3)=Ax + By + C are x-y+l=0, x+y+l —=0, x-2y+3=0. 4. Show that, in general, if a curve of the 3rd degree has three real asymptotes in different directions, the points in which the asymptotes cut the curve again are collinear. 5. Find the line in which the curve 3 - xy2 + y2 - 4 = cuts its asymptotes. [Find the compound equation of the asymptotes, and solve with the equation of the curve.] 206. Curvilinear asymptotes. In some (not infrequent) cases it will be found that the method of finding the asymptotes in some particular direction gives an infinite value to the constant, instead of a finite one. In such cases there are no straight line asymptotes in that direction. The simplest case is that of the parabola y2 = ax, which as we know goes to infinity, but has no asymptote. Another case is (y -x)2 = a(x +y), which is also a parabola, but with a slanting axis. In such cases we shall usually be able to find some simple curve (e.g. a parabola), which will be asymptotic to the given curve in the required direction, that is, which will approach infinitely close to the given curve when x and y are infinite. Such asymptotic curve is called a curvilinear asymptote. The drawing of the simpler, asymptotic, curve will much facilitate the drawing of the required curve. The discovery of curvilinear asymptotes can be facilitated greatly by making use of the diagram of powers explained in Art. 178. This diagram will also serve to indicate linear asymptotes, and is of great use in many ways. We will illustrate by an example. Let the equation be y2(2y + 3x) = 4y2 + 9x2. On consulting the diagram of powers, we see that, when x and y are infinite, ART. 206.] CURVE-TRACING. 245 (1) y3 may be proportional to xy2 (see line AB), i.e. y proportional to x. This leads to the linear asymptote, viz. A, 4y2+ 9x=2 =. 2y + 2 3x= = =8. D -- (2) xy2 may be proportional to x2 (see line BC), i.e. y2 proportional to x, the first approximation being, therefore, y2= 3x. C Hence the equation, with the terms arranged according to degree, and the terms of each degree arranged in order of importance for this relation between y and x, is y2(3x +2y)=9x2 +4y2, 2 9x2 + 4y2 - 3x +2y' By actual division, and approximating in the third stage of the division by 3x +2y )9 42( 3x -2y+8 making use of the relation y2: 3x=1 9x +6xy (which we have seen holds, for this - 6xy + 4y2 direction, when x and y are infinite), we - 6xy - 4y2 find y2= 3x - 2y + 8, neglecting terms 8 which vanish when x and y are infinite. This equation, therefore, represents a curvilinear asymptote. By arranging it in the form (y+ 1)2=3(x+3) we see that it is a parabola with vertex at (-3, -1), and with the horizontal axis y + 1 = 0, and can easily draw it. NOTE.-In the above example, the repeated factor of the highest degree terms which gave the direction of the curvilinear asymptote was the simple term y2. If the repeated factor were of the form (ax + by)2, or, more generally, (ax + by)'', the work would not be so simple, but by the artifice of putting ax + by = z, and arranging the equation in terms of x and z, or y and z, the same method would be applicable. This is true whether the factor corresponds to a curvilinear asymptote or to parallel asymptotes, though in the latter case the artifice is not needed, as the method given in previous articles is simpler and more straightforward. The student might, however, verify its applicability by putting x + y= z in the equation of Art. 203, and working with x and z by substituting y = z - x in the equation. He will find the division method gives z2 = z + 2 as the equation of the asymptotes. Or he may work with y and z by substituting x = z - y, which will lead to the same result. 246 DIFFERENTIAL CALCULUS. [CHAP. XIV. EXAMPLES. 1. Find the points in which the curve y(2y + 3x) =4y2 + 9x2 cuts its linear and its curvilinear asymptote. 2. Find at what points the gradient of the same curve is (1) zero; (2) infinite; (3) equal to that of the linear asymptote. 3. Show that the origin is a conjugate point of the same curve. 4. Show that no other part of the above curve lies to the left of or below the line 3x + 2y = 4. 5. Show that the line 3x +2y - 4=t2 cuts the curve in points given by the coordinates [3(t+2)' t+2 6. Draw the curve. 207. General hints for drawing a curve whose equation is given in rectangular coordinates. (1) Find its linear and curvilinear asymptotes, if it has any. (2) If it goes through the origin, find its approximate shape there. (3) Find at what points the gradient is zero or infinite. [If the curve is a closed curve, we obtain by this means a rectangle enclosing the curve.] (4) Note whether there are any obvious lines of symmetry. (5) Note whether the origin is a centre of symmetry. (6) Find any special points on the curve: e.g. where it cuts the coordinate axes, and where it cuts its asymptotes. Find the gradients at such points. (7) If y can be expressed as an explicit function of x, or vice versa, or if x and y can both be expressed as explicit functions of a third variable, make a table of values of x, y, and gradients. N.B.-Plot each piece of information as it is obtained, and so gradually arrive at the knowledge of the shape of the curve. Some practice is necessary, as experience makes a little information go a long way. The student will find that when an algebraic curve has linear asymptotes there are always two branches approaching each asymptote. Most frequently these two branches approach the asymptote on opposite sides and at opposite ends, as, for instance, in the case of a hyperbola; but this will not always be the case. ART. 207.] CURVE-TRACING. 247 Sometimes a curve may be derived from a simpler curve by some simple geometrical construction: a few instances will be given in the following examples. Drawings of the curves of Arts. 198 and 206 will be found at the end of the book. The student should compare these carefully with his own attempts at drawing the same curves, and carefully verify his work, before proceeding to the exercises below. His drawings should be on a larger scale than those in the book. EXAMPLES. Draw the following curves: 1. y2(Xz-a)+a2x=0. 2. y2(x-a)-a2x=O. 3. y2(2a-x)=x3. [This curve Is called the cissoid.] 4. y2( - 2a)= x3. 5. y3=x2(x-3a). 6. a2x=y(x - a)2. 7. (x +a)y2=x2(a - x). 8. (x + a)y2=x2(x-a). 9. xy a2(x- a). 10. xy2=a2(a-x). [This curve is called the zitch.] 11. ay2=(x-a)3. [Note that this is the same as ay2=x3, but with every point shifted to the right through the distance a.] 12. ay2=x(x -a)(x - 2a). [Show that it has a curvilinear asymptote ay2=(x- a)3, and that there is no part of the curve between x=a and x=2a.] 13. a2y=x(x-a)2. 14. ay (a - x)=x 3. Show that it has a parabolic asymptote x2+a(x + y) +ca2=0. 15. ay2=x(x - a)2. 16. x3-y3=3axy. 17. a3y2= x3(a2 - x2). 18. Show that in the equation of Ex. 10, we may put x=a cos26, y=a tan 0. Hence show that, if a line be drawn through the origin to cut the circle x2 + y2=ax in P, and the line x=a in Q, the point which has the abscissa of P and the ordinate of Q is on the locus. Plot the locus by this method. 19. Draw the curve (x - a)2y2=x3(2a - x). [This curve is called the conchoid.] 20. Show that, in the conchoid (Ex. 19), if y=x tan0, X-a= a cos. 248 DIFFERENTIAL CALCULUS. [CHAP. xiv.] Hence show that if a line be drawn through the origin in any direction, it cuts the curve in two points P, Q, such that PQ =2a, and the middle point of PQ is on the line x=a. Plot the curve by this method. 21. Plot the curves (1) a2y2 = 2(a2 - x2), (2) a4y2 = 4(a2 - 2). 22. If PN is the ordinate of a point P on the circle x2 +y2=a2, and if PM be drawn parallel to the axis of x to meet the line x= a in M, show that the point P1 where PN is cut by OM (O being the origin) is on the first curve of Ex. 21, and that a similar construction starting from P1 (instead of P) will give a point on the second curve of Ex. 21. Plot the curves by this method. 23. If the point P is on the curve y=f(x), and the points P1 and P2 are obtained by the method of last article, show that the locus of P1 is the curve ay=xf(x), and the locus of P2 is the curve a2y=x2f(x). 24. Show that if a line drawn from any point (x, y) to the point (x+a, 0) cuts the line x=a in a point M, OM will pass through the point x, X2Y) 25. Draw the cycloid x=a( - sin 0), y=a ( — cos 0). Show that the tangent and normal at any point of this curve pass respectively through the points (aO, 2a) and (aO, 0). 26. Draw the curve x (x - y)2 = 2a2(x + y). 27. Draw the curve x (x - y)2 =a (x +- y)2. 28. Draw the curves x2y2=a2(x2 y2). Show that, if (x, Yi) and (x, y2) are points of these curves respectively, 1 1 2. y? 2+2 y Yl/ Y2 Ca provided x2 is not less than a2. 29. Show that the equations y(a2 + x2)=ax2, and y(a2+x2)=a3, both represent the curve of Ex. 10, in different positions. 30. Draw the curve x4 - 5x2y2 + 4y4 a4. 31. Draw the curve -+ /-= 1. 32. Draw the curve ( + (j, =1. 2 2 33. Draw the curve (-X + = 1. 34. Draw the curve (x - 2)y2=(y- 1)x2. ARTS. 207a, 2071.] CURVE TRACING. 249 207a. Linear asymptotes parallel to the axes. By the method of Art. 205 we see that, if there is an asymptote parallel to the axis of y, every line in the direction x = constant will meet the curve in at least one point whose ordinate is infinite. Hence, if we put x= c in the equation of the curve, the resulting equation in y must have fewer than n finite roots, i.e. the highest power of y in the equation must be of less than the nt" degree. If we choose such value or values for x as will make the coefficient of this highest power of y vanish, we shall obtain the asymptotes themselves, since in such case there will be still fewer finite roots to the resulting y equation. Similarly any asymptotes parallel to the axis of x will be obtained by equating to zero the coefficient of the highest power of x in the equation of the curve. Take for example the equation x2y2 4x2 - 9y2 + 3x - 2y = 6. In this equation the coefficient of the highest power of y is X2- 9, and the coefficient of the highest power of x is y2- 4, hence the asymptotes parallel to the axes are x +3=0 and y+ 2 = 0. The student can easily practise the use of this quick method of reading off such asymptotes by studying the curves given in the foregoing set of examples 207b. Imaginary asymptotes. The student may have noticed that in many cases when on first inspection there appear to be two asymptotes in any given direction they turn out to be imaginary [v. Ex. (2) on p. 239]. It is usual merely to say in this case that the curve does not stretch to infinity in this direction. This as a fact is quite true, but is hardly a complete statement, for as the number of finite points in which any line in this direction cuts the curve is less than n, there must be infinitely distant points in sufficient numbers to make up the full total. Hence, the correct algebraical statement is that, in the case supposed, the infinitely distant points are imaginary. CHAPTER XV. POLAR COORDINATES. 208. Inclination of tangent to radius vector. In tracing a curve whose equation is given in polar coordinates, all that is necessary is to make a table of values of r and 0, and to plot the corresponding points of the curve. The labour, however, will often be much diminished, and the accuracy of plotting increased, if we can also find the direction of the tangent at each of the points so found. This is done, not by finding an expression for the gradient of the tangent, but by finding the angle made by the tangent with the radius vector to the point of contact. Let k denote this angle: we shall find an expression for tan ). Let P, Q be two adjacent points on the curve, with the coordinates (r, 0) and (r + dr, 0 + dO), R Q respectively. IDraw, OP, OQ and from OQ cut off OR= OP; join PR. Then, evidently, RQ =dr, and, ultimately, PR= rd0 (Art. 32); therefore rdO 0 tan (OQP) d=-, ultimately. Now evidently the angle OQP is, in the limit, the angle which the tangent at P makes with OP, i.e. OQP =, ultimately. Hence, tan =-r. If we denote dr by r', the formula becomes tan b r + r'. is often convenient to havea symbol r the reciprocal It is often convenient to have a symbol for the reciprocal ARTS. 208, 209.] POLAR COORDINATES. 251 of r. It is usual to denote it by u. It is easy to see that du tan W = -- where u' denotes -. For tan ' d0 F 1, r' ' r' it.,' — U = -- r' r2 r 209. Polar equations of straight lines and circles. It is convenient to recognize the polar equations of straight lines and circles at a glance. Their chief forms are (1) Straight line 0 = 0 (called the initial line), r = a sec 0 (perpendicular to the initial line), 0 = a (any line through the origin), U = A COs 0 + B sin 0 (any line not through the origin). (2) Circle r=a, r = a cos 0, r=a sin 0, r= a cos 0 + b sin 0, r2 - 2r (a cos 0 + b sin 0)+ c = 0. The student should draw these for different values of the constants. He should also find their Cartesian equations by writing x = r cos 0, y = r sin 0, x2 + y2 = r2. EXAMPLES. 1. Show that, in any curve, the tangent is at right angles to the radius vector when r'=0. Find for what value of 0 the radius vector goes through the centre of the circle r= a cos 0 + b sin 0, and find the diameter of the circle. 2. Show that, in the equation r2- 2r(a cos0 + b sin ) + c=0, r is a maximum or minimum when tan 0 = b a, and show that the centre of the circle is at the distance /(a2 + b2) from the origin. 3. Show that the length of the tangent from the origin to the same circle is ^/c, and that the radius of the circle is,/(a2+ b2 - c). 4. In the curve r=a(l+cos ) show that tanb= -cot 0. Draw the curve, which is called the cardioid. 5. Show that if the equation of a curve is of the form f(r, cos 0) =0, the line 0=0 is an axis of symmetry. 6. If the equation is of the form f(r, sin 0) =0, the line 0 - is an axis of symmetry. 252 DIFFERENTIAL CALCULUS. [CHAP. XV. 7. If the equation contains only even powers of r, the origin is a centre of symmetry. 8. Draw the curve 2 = a2cos20 + b2sin20. 9. Draw the curve 2=a2cos20. [This is called a lemniscate.] 10. Show that, in any curve, when the ordinate is a maximum or minimum, tan = - tan 0. Exemplify in the case of the curves r = a ( + cos 0) and r2 a2cos 20. 11. Show that, when the abscissa is a maximum or minimum, tan q = cot 0. Exemplify in the case of r=a(l +cos 0). 12. Draw the curve r=a cos 0 cos 20. 13. Draw the curve r=a sin3~0. 14. Show that, when the focus is the pole, the equation of a parabola can be put in the form r(l - cos 0) =2a. Prove that tan = -tan = - t 0 at each point of the parabola. 15. Obtain the equation of a conic in the form 1 - ecos 0 Prove that, at the highest point of the curve, cos 0 = e. 16. Show that = A + B cos 0 represents a parabola if B2=A2, an ellipse if A2> B2, and a hyperbola if B2> A2. 17. Draw the curve r3=a3sin 30. 18. Draw the curve r=aO [the spiral of Archimedes]. 210. Perpendicular drawn from the origin to the tangent. ~C< Let p denote the length of the per-," =P pendicular OT drawn to the tangent at P.,P Then, evidently, i 'I"<~p = r sin ck. i -/ r /The equation giving p may be put /^ y~ ~ into different forms. 01 1 Thus - cosee2 j \ 7T p r2 (r1 1 1 r'2 U =r2 +r4 ARTs. 210-214.] POLAR COORDINATES 253 This may also be written in the form 1P2 p2-U2+U'2 where t denotes 1, and u' denotes du as before. -1 d as before. 211. Pedal equation of a curve. If 0 is eliminated from the polar equation of the curve by means of the above equation giving the value of p, an equation is obtained which gives the relation between p and r for every point on the curve. This equation is called the pedal equation of the curve. It is useful in connection with questions of curvature, and will be referred to again. 212. Polar sub-tangent. The line drawn from the origin to the tangent at any point, in the direction perpendicular to the radius vector of the point, is called the polar subtangent of the point. If the length of the polar sub-tangent is denoted by q, it evidently satisfies the equation q = r tan ~, which may be put into either of the forms, 1 du q=r2 * ',..o.= -e'* This last is usually the most convenient to use in connection with finding the positions of asymptotes (see below). 213. Tangents at the origin. If the curve goes through the origin, there must be some value or values of 0 which make r= 0. These values of 0 will evidently give the direction or directions in which the curve goes through the origin. 214. Asymptotes. If the curve goes to infinity, there will be some value or values of 0 which will make r infinite, or in other words, will make u=0. These values of 0 give the direction or directions in which the curve goes to infinity, i.e. the directions of the asymptotes. The actual position 254 DIFFERENTIAL CALCULUS. [CIAP. XV. of each asymptote is given by the length of the polar sub-tangent q (which is easier to find than p, and, when r is infinite, coincides with p). It is evident from the formulae of Art. 212 that q is positive when r' is positive, i.e. when r increases with 0. From this it is easy to see that if we stand at the origin looking in the direction, parallel to the asymptote, for which r is positive, q must be drawn to the right if it is positive, and to the left if it is negative. [NOTE.-It is often a good plan to transform the equation to rectangular coordinates, and to find the asymptotes by the method given in the last chapter.] If q is infinite (i.e. if u' = 0) for any value of 0 which makes u = 0, there is no linear asymptote in that direction. There will be a curvilinear asymptote, but polar coordinates are not very convenient for finding curvilinear asymptotes, and it is not usually very important to do so, as the curve itself can in such cases be, as a rule, sufficiently well drawn by means of the table of values of r and 0. 215. Circular asymptotes. In some polar curves, r tends to a finite value, or to zero, when 0 is infinite. In such cases the curve goes round the origin an infinite number of times, gradually approaching a circle whose centre is at the origin and whose radius is the limiting value of r. Such a circle is called an asymptotic circle or a circular asymptote. If the limiting value of r is zero, the asymptotic circle reduces to a point circle at the origin. The curves r = a and (r - c) =a are illustrations of such cases. EXAMPLES. 1. Find the pedal equation of the circle r=a cos0. Prove its truth geometrically. 2. Find the pedal equation of the parabola 2au=1l-cos0. Prove its truth geometrically. 3. Trace the curve r= asec +. 4. Trace the curve r = a sec 0 + b tan 0. 5. Trace the curve 16= 5- 3cos 0. 6. Trace the curve 16=3- 5 cos 0. POLAR COORDINATES. ARTS. 214-216.] 255 a02 7. Trace the curve r = -2, the value of 0 being in radians. 8. Trace the curve r cos 0 = a cos 20. 9. Trace the curve r cos 20= a. 10. Trace the curve r2cos2 =a2cos 20. 11. Trace the curve r2cos20 + a2cos 20 = 0. a02 12. Trace the curve r= 02 1 13. Trace the curve r=aekO. Show that the angle between the tangent and radius vector is constant. [The curve is called the equiangular, or logarithmic, spiral.] Show that different values of a give the same spiral, only altering its position. 216. To find the radius of curvature. Referring to the figure, p. 251, let C be the centre of curvature corresponding to P, so that CP=p. Now, in considering the curvature, we may replace the curve by the circle of curvature for an infinitely small change in the position of P; i.e. dr, dO, and dp will have the same relative values for the circle of curvature as for the curve itself (for dp depends on dk, and the very essence of the circle of curvature is that de shall be the same for it as for the curve). Hence, if we change (r, 0) into (r + dr, 0 + dO) we may consider p and C as unchanged by considering them as belonging to the circle of curvature. Now OC2 = CP2 + OP2 - 2CP.OP cos OPC = p2 - r12 - 2pr sin j =p2 + r2 _ 2pp. Now change r into r + dr, and let p become p + dp we shall have OC2= p2 +(r+dr)2 - 2p(p + d);.'., by subtraction, =2rdr - 2p dp, rdr i.e. p=- d. From this simple formula p can be easily calculated if the pedal equation of the curve is known. 256 DIFFERENTIAL CALCULUS. [CHAP. XV.] 217. Of course OC and p do change as we go from point to point on the cvrve. The above investigation shows that infinitesimal changes in r and p may be considered, regardless of the induced changes in OC and p, by the simple artifice of considering the circle of curvature instead of the actual curve itself. Hence, if we do consider the changes in all the quantities involved in the above equation OC2 =p2+r2- 2pp, we see that there must be two independent differential equations, viz. oc.d (o ) = (p -p) p,.......................(1) and rdr = pdp................................ (2) The first of these is easily proved geometrically if we realize that, to the first order of small quantities, the new centre of curvature must lie on CP; for then d(OC). dp = cos OCP (as can be seen at once from a figure), and so also is (p -p) + OC. The second of them is the equation giving the value of p. EXAMPLES. 2a 1. Draw the parabola r=1 - -si Find its pedal equation, and deduce the length of the radius of curvature at any point. Find the value of p at (1) the vertex; (2) the ends of the latus rectum. 2. Find the radius of curvature at any point of the curve r2=ap. 3. Find the pedal equation of the cardioid r=a(1 + cos 0), and deduce the radius of curvature at any point. 4. Find the pedal equation of the curve r2=a2cos 20, and the radius of curvature at any point. 5. Find the radius of curvature at any point of the curve rn= aucos O. Show that the cardioid and the parabola are particular cases of this curve. ANSWERS TO EXAMPLES. Pages 15, 16. No. L:0 -lo.In I Denoting one of the numbers by x, the product is y=x(5 - x). The table of values is x 0, 1, 2, 3, 4, 5, 6, 7,... y 0, 4, 6, 6, 4, 0, 6, -14,... from which the graph can be drawn, as shown. The greatest value of the product occurs when x=22, i.e. when the two factors are equal. No. 2. C I 2 3 4 5 6 7 8 If y-I, the table of values, from which the graph can be drawn, is x 0, '5, 1, 2, 3, 4, 5,... L.D.C. y, 2, 1, '5, '3, '25, '2, L.D.C. DIFFERENTIAL CALCULUS. No. 3. The curve is y = + - x x i l t c i o a v When x is large, the curve is only a very little above 1 2 the line y=x, since the term I becomes insignificant. 3 32 -x 3 3'33 4 4-25 5 52 The line y=x is called an asymptote. It is shown by the dotted line in the diagram. The curve is a hyperbola. Its other asymptote is the line x=O (i.e. the axis of y). The minimum ordinate occurs when x=1. No. 4. 125 e. 75 50 I" Cj3 S a 4 5 6 Scale of i. ANSWERS (P. 15). 259 If values of n are taken as abscissae, and values of 2n as ordinates, the curve is roughly as shown. The vertical scale has to be very small, as the ordinates grow rapidly. Nos. 5. and 6. o'd 4 N.B. cosx= sin (X+00~) Each horizontal division represents 12~. The student should draw this curve on a larger scale. No. 7. 260 DIFFERENTIAL CALCULUS. No. 8. t 1111 1 11 111 1-111 I H 4.0 COD No. 9. I 1 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 5 ---'1llllll III IIIIIIIIII / i lllllllll I IIIIIllmm- - 41111111 -1111111- IIII II --- —- -11t points; its equation is y -32=9 X. ~~~~~IIf I Ieoe an Illeatr I1 cenIgrI deres IIII II IdenotesII the ~ ~ ~ II111 111 Isame tepeatI IIIn Farnetlllre, llllwhna?, n y=212~~~II IIIIIII II=10 ITe grap III III srIgh lInejIngthe points; '10 Its IIIIIIII IIsIII I II IItI I_ ___L-32 JIII IIII II.t1 The readings on the two scales are equal for a temperature of - 40~. ANSWERS (P. 15). 261 No. 10. 70. T g i a si ln gi to th fl _n g _ _ _. || )|| 60 50 '/ __ 40 so 20 o -:co -,-.. " iX-____ y Im 2 8 4 5 6 7 8 The required graph consists of a series of points corresponding to the integer values of x. No. 11. The graph is a straight line going through the following points: x, measured in feet 0, 1, 2, 3, 4, 5 y, measured in foot-lbs. 0, 10, 20, 30, 40, 50 e If th -se y-sce is of the x-scale, the line will be at 45~ to the axis of x. It passes through O. No. 12. If we measure the stretch in feet along the axis of x, and y in lbs. perpendicular to this axis, then when x= 0, y=0, and when x= 6, y= 10. The graph is a straight line joining these points. Its equation is y= -x. No. 13. Since the force increases uniformly with the amount of stretch, the average force throughout any given stretch x is half the final force, and is therefore -x. Hence the work done= ax2, being the product of the average force into the stretch. This is the area of the triangle contained between the graph of the force, the axis of x, and the final ordinate corresponding to the value of x considered. The area will be so many foot-lbs., the base representing feet, and the height representing lbs. If the work done in causing any stretch x is represented by an ordinate, its graph will be a parabola, and presents no difficulty. Its equation is y =x2. 262 DIFFERENTIAL CALCULUS. No. 14. Since bd is constant, where b is the breadth and d the depth, and the strength is proportional to bd2, it follows, on eliminating d, that the strength is inversely proportional to b. Therefore the graph of - will show the changes in strength. This curve is shown in the answer to No. 2. The theorem does not hold for very thin beams unless they can be prevented from lateral buckling. No. 15. The half-perimeter=b+d=10, and the strength is proportional to bd2, which=b(10 -b)2. The graph is therefore y=x(10 -x\ where x represents the breadth measured in inches. x_^_j^ ~ 150L_ -&/t I x y III 0 0 (20 _ I X I _ I I_ 2 128 4 144 I l Scale of breadth,in inches No. 16. y=x(144-s2). 6 96 YsO 1 11 8 32:: 10 '0 ~ tF"00 2 1280 2 4516 10 No. 17. Fro the diagram we see that the ordinate is a maximum2) 10 440 400 I::::: 12 0 E:::^-::::::::::::;:^::O 100 0 2 4 6 8 10 12 No. 17. From the diagram we see that the ordinate is a maximu= when x is about equal to 7. A more accurate answer can be obtained ANSWERS (pp. 15-24). 263 by equating the gradient to 0, as evidently the curve has zero gradient at the required point. By next chapter, or by Art. 18, we find the gradient is given by the values of 144- 3x2, and this is equal to 0 when x2=48, i.e. when x is just less than 7. The change in length of ordinate is so slight about here, however, that there will be no appreciable error in taking x equal to 7 for the maximum value of y, i.e. the strength is greatest when the breadth is about 7 inches and the depth =/95 or nearly 10 inches. No. 18. From the diagram of No. 15, we see that y is a maximum when x is a little greater than 3'2 inches, which would make the depth a little less than 6'8 inches. So you might guess b=3~, dc=6, i.e. the depth double the breadth. If by help of next chapter, or Art. 18, we equate the gradient to 0, we shall find that this guess is correct. Page 18. x 2x+A 2a b2x 1. 2. - ~ 3. ( =y/2x). 4. -- y 2y +B- yy 2y 3 5. (=-y/). 6. 3x(=. ) 7. 3X 3) Pages 22-24. 1. 32 - 40x + 100; zero gradient when x= 3- and 10. 2. 144 - 3x2; zero gradient when x = ~,/48. 3 3 15^x 1 7 3/x 2,/x 2 x/Jx' 2 10. 4x(x-1)(x-2). x -1 0 1 2 3 f() 9 0 1 0 9 f(x) -24 0 0 0 24 11. - x-2 + -3 - -4. 12. - 9y 13. The graph is a parabola whose highest point is at (1k, 139-). 14. Velocity =-0 when t= 1-. The velocity graph is a straight line whose intercepts on the axes are 1, 50. 3 - 34x 4x 15. (1) -2(6x - 5)(3x2-5x +6)-; (2) - 2 + +2 (x -6)2 (X2+6) (3) 2x-3 2 (4) 1- (2 _ 1) (5) 5 (. 16.. 17. Gradien in each case-2Ap+ x. 16. 2. 17. Gradient in each case =2A +B. 3 264 264 ~DIFFERENTIAL CALCULUS. 1. (x) - (4) l Ox + 28X3- 1 8x5; Page 30. 2)3x-a (3) -a (5) 6x (1-X) (1+ X2)2. (6) a 2 (I1+x2)3 Ca X2~x2) 1 2 -2.3 2.3.4~ Fa -x-)2' (a+x)3' (a_+x)~' (-a+x)5' 5. 588,88. 6. 0. 9. x(a2 - Y 2) 10. f'(z).-O ) y (a + x2) 11. 24 (x- 1) (3z -1)\,/(3Z2 - 2z~5). 12. 9(Z2~+5) (X2 +5) (z - t5)2(X2 - 5)2 — 2.3.4.5 (a~x) x2 -53x~+43). Pages 44, 46. 1. 0. 2. - 2sin 2x. 3. 3 seC23x~4 coSeC24x.. - c~~O5X 5. sec x (sec x + tan x). 6. cos 2x = cos2X -_ Sin2X. 7. sin 3x =4 cos2X sin x - sin x =3 sin x - 4 sin3x. 8. (4 sec4X - 12 sec3X + 4 seC2X) tan x. 9. 15 sec~x. 0. 4x i-X122xalx 13. 114. 1. 2 16. y = 0 when x = 0, 2, 4,.; and at these points = - ~ alternately. dx 4 dy = 0 he tan 7x- WrX The curve is symmetrical on either dx 2 2 side of the axis of y; i.e. the axis of y~is a line of symmetry. 17. y = 7r when =0 (applying Art. 32). ~ 0when tan 7r 7r. dy = F7i y =O when x = 2 4, 6,..;and at these points d The curve cuts the axis of y at right angles (the proof of this is not easy till after we know how to expand sin 7rf and cos -x in powers of x), and this axis is a line of symmetry. 2T 2 ANSWERS (pp. 30-63). 265 18. (1) Similar to No. 16, but on half-scale. (2) The ordinates are double, and tbe abscissae are half, of those which give the corresponding points in No. 17. 19. Y = when x=0, 2, 4, 6, dy0 when cot7rX 7rX x dx 2 2 y=O when x=1, 3, 5,...; and at these points dy IT The left-hand portion is an inverted copy of the right-hand portion, so that the origin is a centre of symmetry. 1 7r ~dX 7rX 20. y== x when x=0, 2, 4, 6,.... =0whencotdx 2 2 y=Owhen =1,3,5,...; ~ dy 7 rIx y = 0 when x = 1, 3, 5,. and at these points dy- 7T* The gradient at the origin is 9. The origin is a centre of symmetry. Pages 49, 60. 1 2 1 5 1. x~x+-X3-x5~2 2-t X2, 4 3 15 2. 2+-4 2 4 2 4. 1-2x2+2x4- - 4x 6 x5-.2... 3 45 315 1 2 1 5. Cos2X-1- X2~IX4 -2x6+6+ X8 5 -3 45 315 6, 7. Calculate the ordinates and gradients for x= 1, \/2, 2, \/6, 3. The corresponding values of sin x are -841, -988, -909, -638, -141; and those of cosx are '540, -156, - -416, - '770, - '990. It will be found that the respective series graphs approicimate to the above more closely as more terms are taken, but that each correction is rather too great. Pages 62, 63. 1. cotx. 2. 1+x x-3 1 +- X2' (X+ -(1)2 -5. 1~logx. 6. secx. 7. 2(1- 2) 1 + X2~+ 4' 4. 2xy. 8. y cos x. log, a. 9. 12. y(x Cos x + sin x). 10. y sin x(cot2x - log sin x). 14. xn-1(1l 9nlog x)- 15. 1 1 22. (1) -x-x2 ---x3-... 2 3 23. 1og02=0'693147.... y(x cos x - sin x) 11. y(1-2x). 13. 2 tan3x. I )X - e-X 1 16. 3C. x logx' ex + eX 1 ( (2) 2 x+ X I+ 266 DIFFERENTIAL CALCULUS. Pages 74, 75. 3. Population in x years = 106. 2Tk Rate of growth per 1000 per annum =6 693( ~ ) 1,148,700; 1,414,200; 1,741,100. 4. 50,000 lbs. weight. 5. ai=10; a2 1 443, or about,A- orW~; a3 =1 0064, or about 15" 6. (1) 3,16...; (2) 1-778,.... 7. k = 366.... [Students of dynamics will know that this is the coefficient of friction between the cable and post.] 8. A little over 40 feet. 14. A little less than 36 per cent. (assuming there is no selective absorption). 15. 90-44 per cent. 16. 12 seconds (nearly). 17. About 22 minutes. 18. 5 minutes 28 seconds (nearly); 47A seconds (nearly). Curve of fal isy=( ~yiif x is measured in seconds; the subtangent is 471 fall is y=(-9)t fx: -n. _9,,~N,+rN,-~~r~\ Nnr seconds. The curve of rise is 1 - (9). It is an image of the curve of fall with respect to the line y=%. Both curves cross this line when. x=33 seconds (nearly). Page 83. 1. 2 sinh x cos x. 2. 2cosh cos x. 3. tanh x. 4. 2 sech x. Miscellaneous Examples. Page 83. 2x- y-2a 2 x -4y+ 6a' 2. 1 -X4 3. {1 - (log X)2}2. e ogx!). cosleologx) -(log Xy2}a. log2i sin(elog x). 4, - eax(a cos3x~+ 3sin 3x). 5. 2x(sin2x + x cos 2x). 2pq2r2(2p2 + q2 + r2). 2p2qrq+ 2p2q2r (p2 +,q2)2(p2 +r)2 (p2+ q2)2(p2 + r2) (p2 q2) (p2 ~ r2)2 7. 0. 8. - a tan 0seceO~/(cos20). 9. (1 - x)I(l+X ~x~410. cot O. - 24x2 - 42xz + 64z2; 84x2 - 32xz - 56z2 (4x - 7z)r (4x - 7z)5 2y. ANSWERS (pp. 74-123). 267 Page 94. 2. 1325. 9. 1P4236. 4. The circular measure of 490 17' 36k". 5. 1-37129. 6. 380 10' 217". 8. -000287656. 9. The roots are 0,75736 and 9-24264. 10. The roots are 10-6569 and - 06569. 11. The roots are 0 7386 and - 4 7386. Page 109. 47r 1. Height=l foot; volume = cubic feet. 2. h=r. 3. h =2r. 4. (1) h2=2r2i; (2) h2=8r2. 5. h2-=l12 r2-12. 2, 4 1 6. r2 = 2a, h2 =a2; volume = of volume of sphere. 3 3 %/3 7. Width= twice depth. 8. 15 feet 6' inches; 46' 44' 24". 9. Max., 12 (when x=0); min., -20 (when x=4). x -2 -1 1 2 10. y - 16 - 38 38 16 max. min. max. mi. 11. 2,.,/(ab). 12. (3, 2), (0, 0), (3, 13. Max., 4 (when x=0); min., 0 (when x=2). 14. Max., 0 (when x= -1); mmin., - 84 (when x= ). Point of inflexion at (2, 0). 15. Max., 18 36 (when x=2); min., 0 (when x=2). Point of inflexion at (- 1, 0). 16. None. The curve is a hyperbola with asymptotes x= y and x= 1. The conjugate hyperbola has a maximum ordinate at (0, - 1), and a minimum ordinate at (2, 3). 17. Max. value=54; min = -71. Page 123. 1. If d, b are the lengths of the horizontal and vertical units of scale; 4a2 b p4sec34, where tan 0 b P=b S 4a Hence, (1) p = 4 37; (2) p = I P4b =2 8a. 2. (1) p=L,/2; (2) p=Il4b=2 8a. 3. 1. 268 DIFFERENTIAL CALCULUS. (a2 + y2)-T 3,/3 y = 1 2 4. p(ay) 5. p= ---; wheny2 a2. _y2 y 1 6. p=a =ysec, since sec2=-. 7. 1 8 113\/226 25ao/10 5a 25b 9. (1) 112; (2) 25 10. p=l2=12, at each point. 2 2 12. (a2sin2 + b2os20) + ab. 13. (a2tan2 +b2sec20)2 -ab. Page 125. 15 3 63 5 15 _ 1. 360. 2. - x 2-x16 1 21 Pages 132, 133. X8 X6 X5 X4 1. (1) D4y=24; D-4y= — +-6 ---0 +D-4(0). 1680 120 60 24 1680 720 24 3 (2) D4y= 9 7 + X; D-4Y =2 + - x logx+D-(0). 1680 1 (3) D4y= -(809; D-4y= 1 15 - 16 9 (4) D4y= — x; D-4y= 2. J-~)n 2. ( and _" -)n+l a (a - X)2+l' 3. (1) Day=(- 1)'.-l 1.3.5 2.. (2n- 3)x- D_ 2nX~- + 21 3.5... (2n + 1)' (2) D"y=( 1) 1.3.5.(2n — 1) Dyn+ __7 - 2n - 2n D-~Y= 1.3.5... (2n - 1) 6. If the graph of In is plotted by drawing a curve through the points corresponding to positive integer values of n, the ordinates corresponding to fractional values of n may be taken as the corresponding values of In. The second part of the question is solved by using the formula i=- In - 1. Thus 4=(- 1)(- 2)(2-3)= -6. xn-r xn+r X2, 1\ 7. (1) +__; (2) 1, 9. Ln log+^ tI-9 Ino gx2r' ~ ANSWERS (PP. 123-146). 269 Page 136. 1. 5'e4,Xsin(3x+rtan-'3). 3. ( 1)i(a2+ b2) e-a sin bx - r tan-'1 D -l a b.(, -b e-" (a sin bx + b cos bx) D-Iy= - ( bSin bx+tan-)"-^ -^a2 )b 2) a a 2 + -b2 4. Deducible from (3), by writing cos bx=sin (bx+2). D-ly=e-ax(b sin bx - a cos bx) (a2 + b2). Page 143. (-l)n-2jn-2 2(-1)"-8-3 3 2. x-; 2-1 Proved either by Leibnitz, or xn-l Xn-2 by direct differentiation. x4(1 1 1 1) 3. xD-3 log x - 3D-4 log x= etc. =,llog x - - - - 4. - 27e-3Z 9- 57x+101}; -27 e3{2-+ 5. (32x2 - 160) cos 2x + 160x sin 2x; { (x2 - 3) cos 2x - 3x sin 2x}. 7. (1 -2) Yn+2-(2n +l)xyn+l=(n2 -m2) Yn 9. ( 1-2)Yn+2-(2n+ 1) xy.+l = (n2 + a2) Y. Page 145. 2x2 - 3x 7x (1) 1- X3 1 (2) 4x+4 -14 (3) {5x+7+5 _ 21 3} Page 146. ( 1 (-a)n r+n-1 -1. r-l(ax+b)+n; a(r-1)(ax+ + b)-l' (2) The same as (1) with the sign of a changed. 1 -3 2 4 2. log (2x - 3); 4(2x- 3)2 log (2- 5x); ( 3. The two last. 270 DIFFERENTIAL CALCULUS. Page 154. 1 2 1. Y=_2++3; y_ i=log(x-2)+21og(x+3). 2. Y-1- x-2 2. y-1=3log(x-2)-log(x+1). 3. y_=-i 1 + 2 log(x + 2) - log(x +l ). 1 2 4. y-1=log(x-3)-3logx- log(x+3). 5. Y-l=(c-b)log(x-a)+(a-c)log(x-b)+(b-a)log(x-c). 1 (x3 3 9 6. y- = log (2 - 3) 2-3 4(2x-3) 1 1 1 7. y- l= x3 + x2+13x +{251 log (x - 4) - 97 log (x + 3)}. 37 x-3 2/ 7 12 \ 8. y- lo $8. Y 125 x+-2 5 x-3+Z+2 2 x-1 1 9. Y- l=og3x+2 2(x- 1)2 1 + 7 44 44x + 69 10 y 3(x-2)3 27(x-2)2 243(x-2) 243(x2+x+3) 7 43 89 11. y-= 6(x-2)2 18(x-2) + 218 log (- 2) 7 log ( + -log (x+2). 12. y=2x + 3 +; yl=21og(2x+3)+3tan-lx-log(x2+ 1). (This integral can be verified by differentiation.) Page 161. 1 1( x-3 5x — I 1 2{x2+X+l+ 2-_+1 22x2 +xa /2+l x-x- 2 +l 1 1 x+2^x3 a —2,/3 2./3 x2 - {^/3+1 x2x +.3 + 1 4 1 x++1} 5 9{ 13+ 2x- 3} 4 3 x2+l x -+l x+30- x2-3x+9.2 2x+2 5x-2 2 x2+1' (X2+)2 X-1' ANSWERS (PP. 154-167). 271 5 2 6x+7 x+3 9(x-1)2 3(x-1) 9(x2+x+1) 3(x2+x+1)2 x +1 2(x+1) 5x-2 1 x2+1 ( 2+1)2 (x2+1)3 x-l 3x - 6 2x - 14 4x - 10 2+x+ l (2+ x+1)2+ (2+x+1)3' 1 8x 55 X2 +4x +8 (x2+4x+ 8)2 (x2+4x+8)3 Page 16 4. 1. () n- sin"+l1 sin (n + 1), where x = a cot 0. 2 1) n+l sinn+l cos (n + 1) 0. 2. an+l 3. (- )nLn sinn+l sin(n + 1)0, where x - a=b cot0. bn+2 4. (- 1), J sin!+10b cos(n + 1) +a sin(n+ 1)0}. bn+2 5. (- 1)n l sinn+1'{3cos(n+ 1)- sin(n+ 1)0}, where x+2=2cot. 2n+1 6. ( - ln sin"+'l{ sin (n + 1) 0 - (n + 1)sin 0 cos(n+ 2) 0}. 2n+5 7.(- 1) I+n - sinn-20 sin(n + 2) 0. 2n+4 ' D-lu=3tan-lx+(+ 1+4 l) 8 -s x + 1) 4 +4Page 17)2 Page 167. h hI2 h3 5. (1) logea+ — 2t+ 3-a8 (2) Put a=l, and h=x, in (1). 6. log sin x + h cot - ih2cosec2x + h3cosec2x cot x -... 272 DIFFERENTIAL CALCULUS. Pages 185-187. 1 5 61 1 1 1 1. i-02+ ~-04 0 e6~2- 02 4 20 6 2 24 720 2 12 45 3. The coefficients of the even powers of x are all zero. Also a, =1. The equations giving the other terms are a3- a, 5=a3(1+ ~5); +~1 etc., whence a5=1, a.=32, a7=52.32, etc. siu30 32sin20 1 4. 0=sin0~-+ +.... 5. 0=tan —Itan30+-1tan5- - L+ 1 5 6. The function satisfies the differential equation 2= c2y ~ xy, + x2 Y,2 whence, finally, a2x:2 (a2+1)ax5 (a2+f4) a2x4 Y + O1 + - + + the general relation being a,+2 = (a2 + n2) a.. 7. x + b abx2 + (3a2b - 3)x+... 1 1 8. 1~ax+2(a2-b2)x2 6(a3-3ab2)x5+.... Both (7) and (8) could have been obtained by equating the real and the imaginary parts in the identity eax(cos bx + i sin bx) e(a+ib)x. LThe general terms of (7) and (8), respectively, may be written (rx)' sin na, and cosna, where r2= a2+b2, and tan a = 9. 1 (I-cs2)=I(2x)2 (2x)4 9.,1~2'e~-[+~t 1 3 -1 31 - 1 10. (sin 3x - sin x)=x - X+ X-5... 11. 1 14. (Express the chords in terms of the radius and the sines of half the angles which they subtend at the centre of the circle, and apply (13).) 1 15. 0= (256 sin 0 - 40 sin 20 + sin 40). 180 16. O+ 0~!+ 0 5 + o 21. See answer to No. 3 (above). 6 24 ANSWERS (PP. 185-204). 273 X3 X6 X9 22. y-ao 1-2.3+2.3.5.6 2.3.5.6.8.9+. a 4 x-3 43 7 - 3.4.6. 79.1+. 23. y=Ao (1-+2-22 -2+ X3) 24. y-=Ao 1-+2 22+42 22' 42 62 +. 25. y=Ao(1-( - I —. )+A (n)n) = A cos nx + B sin nx. Page 194. 3 1.16 5. 7 3.. 4.. 3- 2~ 4- 19' 55 7. (1) logex-log,a; (2) logex. 8. x3. 10. 0. [Evaluate under the root.] 11. 2. 13. 2. 14. 2. 15. 2./2. 16. lxsinx. 6. loga -logeb. 9. -2x. 2 12. -1. 17 x cos x - sin x x sec2x - tan x Pages 197, 198. 1. ratio=l, difference=0. 2. oo. 4. 0. 5. -n. 7. (1) e, 1; (2) e-, 1; (3) 1, 1; (4)1, 1. 9. (1) e; (2) e6-. 10. 1. 11. 6-. 7r 3. 0. 1 6.. 8. 1. 12. 2. C 3.. - 23 c 6. I(1 - 3c). 1. a. 8c-3 9c + 12 Page 201. 2. b-Ca. 5. (l)c+2; (2) c. Page 204. 1. y3 is comparable with x6. x3y and xy2 are comparable with x5. 2. x5 and x3y3 are comparable with z15. x4y is comparable with 14. xy5 is comparable with z13. L.D.C. S 274 DIFFERENTIAL CALCULUS. Page 205. 1. (1) X4, x3y 5x 3+x 2Y2, x2y +Xy3, x2~Xy2+y4, Xy~y3, X ~y2, y. (2) y4, Xy3, X2y2, X3y ~y3, X4 ~Xy2, x2y, X3 +y2, Xy, X2, y, X. (3) y4, Xy3, x2y2, Xly, X4 +y3, Xy2, x2 y, X 3, y2 xy, X2, y, X. (4) XI, Xly, X3, X2y2, Xly, X2, xy3, xy2, xy, x + y4, y 3, y2, y. 2. Assuming that a pair of terms are to be of equal importance, the diagram will show that the only possible pairs are (1) when x and y are great; either x3and X2y2, or x2y2 and xy3; (2) when x and y are sMall; either x3 and x2y, or x2y and Xy3. 3. x 3 and y2 are of greatest importance; then, in order, xy and X2. 4. 2y -8.- 5. 3x-2y~8. Pages 207, 208. X2~+2xy -2y2 4X2 +2xy +y2 1. (X+ y)2 ' (X +y)2 2. ulIog a; ux i-a. 3. pulx; qu/y; uelog,,X; u1og1y. 4. 2x-3y-5a; -3x~4y~6a; -5x~6y —2a. Pages 212, 218. 2. A parabola passing through the origin, whose axis is the line Sx - l0y - 4 =0, and whose vertex is at 28 24y 3. The lines are x -2y +a = 0, 2x -3y +a =O, x y =0. They all meet at (a, a). The locus consists of the lines x - 3y + 2a = 0, and X - Y=O. 4. The loci are the parabolas X2 = ay, and y2 = ax. The first cuts the curve in (0, 0) and (23 a, 4xa), and the second cuts it in (0, 0) and (4 a, 2x3a). Page 21I7. 1. (1) xX-FyY-a2; xY-yX=0. (2) yY=2a(X+x); 2a(Y-y)+y(X-x)=0. (3) AxX~ByY=a2; By(X-x)=Ax(Y-y). (4) yX~xY=2at2; XX-yY-x2-y2. 2. (2x~A)X~(2y~B) Y~(Ax~By~2C)=0. 3. Polar of (a, a) is X2 + y2 == a2. The curve lies outside this circle, touching it at (a, 0) and (0, a). 4. (y2~a2c) h~+(2xy -4ay)k =2ay2 -2a 2X. Polar of (0, 0) is y2 =ax. It cuts the curve at (0, 0) and (a, ~ a). The tangents drawn from the origin are x = 0 and x = ~ y. ANSWERS (PP. 205-230). 275 Page 221. 1. 8y-6x. 2. 24xy3. 4. 6u; 9u. d d2~ d2f 2 d2f dJ d2y 6. dx2=d2 + 2y dxdyY1 d+Y2 dy' 7. dt2 d2 d yf d2f 8. xl2 df + 2xly ddyf + 1 2 In this example, x1=h, and = k. 9 2df df f _(df\2d2f (df \2 d2y} f fdf\3 * dxdydxdy \dx dy2 dy dx2 dyV Pages 229, 230. 1. Minimum at (2, - 1); viz. -4. 2. Critical point at (1, 1), but the function is neither a maximum nor a minimum there. (We shall call such a point a double point.) 3. Minimum at (2, 2); viz. - 8. Double point at (0, 0). 4. Maximum at (3, 2); viz. 108. Double point at (0, 0). 5. Maximum at (0, 0); viz. zero. Minimum at (,/3, - ^/3), and at ( - /3,,/3); viz. - 18. Minimum at (1, 1) and (- 1, - 1); viz. -2. 8. The series of ellipses are concentric, and all have the same axes, and evidently for a very small ellipse of the series the directions in which u increases most rapidly and least rapidly from its minimum value at the centre (2, -1) are the directions of the minor and major axis respectively, and are given by the max. and min. values of d2u du. 2Bd since -- is zero. These occur when tan 20=-C which in this case dr A - C' is equal to -2. 9. u is constant in value (=6) in the assigned directions, starting from the point (1, 1) where d- and L- are both zero. That is, the locus dx dy u=6 consists of two straight lines crossing at (1, 1) and lying in the assigned directions. The lines are x - 3y + 2= 0 and x - y = 0. [For all other values of u, the locus is a hyperbola having these lines as asymptotes. ] 10. Maimum at ( 13' 1); viz.. Minimum at ( - 2, - 3); viz. -. 12. (1) h J+kY must be zero for all values of h and k; i.e. 'dx dy dx and dy must both vanish. (2) h d +kd-) f(x, y) must not be able to change sign, whatever the relation between h and k. 276i DIFFERENTIAL CALCULUS. 13. Let the successive terms in each expansion be denoted by it, U1,?&2,.. Then (1) 12==4h2~ l2hk+ 10k2. (2) u=h2 -4hk~3k2. (3) u2= 3 (h2X + k2y) - 6hk. uC3~=h3+k j3. (4) in5- - 2h4ky ~ /?,3k2 (6 - 4x - 3y) - 3h2k3x. u6 - -h4k2 - h3k3. (5) u2 =6(h2X2 + 1k2y2) - 4h2 + 4hk - 4k2. U= 4 (h3x + k3y). u4:= h4~ k4. 14. (1) 4x2+l2xy~10y2 - 4 when h=2, k= -1. (2) X2-4xy~3y2~6 when h=1, k=1. 15. When u= -8 the equation of the locus can be written in the form (x + y ~ 2) (x2 - xy + y2 - 2x - 2y ~ 4) = 0. [See also Ex. 6, p. 217.] Page 285. 1. (1) 3x -2y=0. (3) y=O. (4) x+y=0; x-y=0. (5) x+y=O. 2. Such a point is called a cusp, and so is 3 (3) O ~~~~below. 3. Page 244. 1. (2) The curve cuts the sloping asymptotes in the points (5, ~'~). It does not cut the vertical asymptote. (3) Cuts the asymptote at (4, 2). (4) In no finite point. (5) In no finite point. 4. If the compound equation of the asymptotes is n = 0, the equation of the curve must be of the form in = Ax + By + C. Therefore the points of intersection lie on the line Ax + By + C=0. (See Examples 3 and 5.) Page 246. 1. Cuts linear asymptote at (4, 2); curvilinear at (0, 2). 2. Zero gradient at (SF: pl62, - 4~44J2), i.e. at (0,46, 1,66), and (15-54, - 966). Infinite gradient at (12-6, - 11-3). The tangent at (0, 2) is parallel to the asymptote. ANSWERS (PP. 229-256). 277 4. The equation can be written y2(3x + 2y - 4)= 9x2, so that 3x + 2y - 4 cannot be negative, unless x and y are both zero. 6. The curve has two branches, on one of which are the points (-1, 5), (0, 2), (4, 2), and (10, 5), and on the other branch are the points (13, -13), (13~, -10), (20, -10). These data, together with the asymptotes, and the points given in Ex. 2 will be sufficient to give a very good idea of the curve, which is shown below. The student should draw it to a larger scale, from his own calculations. Y Page 251. 1. tan-lb/a; /(ar2+b). 4. Points on the curve can easily be found by drawing the circle r =acos 0, and adding the length a to each radius vector. The curve has a cusp at the origin. Page 254. 1. r2=ap. 2. p2=ar. Page 256. 3 1 1. p2=ar. p=2pr-a=2rVa~M. (1) 2a; (2) 4ad/2. 2. la (the curve is a circle). 3. 2ap2=r3. p=4ap+3r=/(2ar). 4. r 3=a2p. p=a2+3r. 5. rn+l=canp. p=an-(n+l)rn-1. For the cardioid, n =; for the parabola, n= -. L.DoC. T 278 DIFFERENTIAL CALCULUS. (I) I (2) II ~ ~ ~ ~ ~, I I I,, (3) (4) ~11 I I GRtAPH~S OF CurtyEs GIVEN IN ARtT. 198. I I I i I1 ~~(5) I 1. I I. I I~ I I I I I I I~~~~~~~~~~~~ GRIAPHS OF CURVES GIVEN IN ART. 198. GLASGOW: PRINTED AT THE UNIVERSITY PRESS BY ROBERT MACLEHOSE AND CO. LTD. A SELECTED CATALOGUE OF EDUCATIONAL WORKS PUBLISHED BY GEORGE BELL & SONS CONTENTS LATIN AND GREEK.. MATHEMATICS ENGLISH... MODERN LANGUAGES. SCIENCE AND TECHNOLOGY MENTAL AND MORAL SCIENCE POLITICAL ECONOMY... HISTORY.. ART... 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