IN MEMORIAM FLORIAN CAJOR1 e cu<^ ESSENTIALS OF CALCULUS BY E. J. TOWNSEND, Ph.D. (Gottingen) Professor of Mathematics University of Illinois and G. A. GOODENOUGH, M.E. (Illinois) Associate Professor of Mechanical Engineering University of Illinois NEW YORK HENRY HOLT AND COMPANY 1910 Copyright, 1910 BY HENRY HOLT AND COMPANY NortoootJ $reBg .7. S. Cashing Co. — Berwick & Smith Co. Norwood, Mass., U.S.A. T(o 6*i PREFACE In the preparation of this volume, the authors have had in mind the needs of those colleges and technical schools in which the time devoted to calculus is limited to a three-hour course for a year, or perhaps to a five-hour course for two terms. The usual division of the subject into differential and integral calculus has been largely disregarded. By the arrangement adopted, the student is early led by easy steps into simple prac- tical applications of the calculus ; and the more difficult topics are postponed until late in the course. The theory of limits has been used exclusively in the develop- ment of fundamental principles. Throughout the book much emphasis is placed upon the applications of the calculus to prac- tical problems. Only such knowledge of physics on the part of the student is assumed as is usually included in an elementary course in that subject. Some problems are introduced that show the use of calculus in discussing well-known applications to physical and engineering phenomena. Such problems are so stated, however, as to require no technical knowledge on the part of the student. The applications to geometry are such as are essential and usually to be found in a first course in calculus. In the selection of material, the authors have departed some- what from the traditional course. Many topics usually included in calculus have been entirely omitted or greatly reduced in extent. Thus, but little attention has been given to special methods of integration ; and reduction formulas for integration, order of contact, envelopes, etc., have been omitted entirely. On the other hand, some parts of the text have been extended beyond the usual limits. Functions of two or more variables, because of their importance in physics, have been discussed more fully than usual. Special stress has been laid on the summation iv PREFACE process, and by numerous examples from physics the student is drilled in the choice of proper elements and in the setting up of definite integrals. Attention is called to the treatment of exact and inexact differentials ; a subject of first importance in the physical applications of calculus, but one that usually receives little or no consideration. The authors take this occasion to express their obligations to their colleagues at the University of Illinois and elsewhere for their helpful suggestions in the preparation of this book, and to the publishers for their cooperation in making its typography of high grade. The authors are under special obligations to Professor H. L. Eietz and to Dr. A. R. Crathorne for their assist- ance in seeing the book through the press. E. J. TOWN SEND. G. A. GOOBENOUGH. Univkrsity of Illinois, July, 1910. CONTENTS CHAPTER I CONSTANTS, VARIABLES, FUNCTIONS ARTICLE PAGE 1. Constants, Variables 1 2. Functions 2 3. Fundamental Problems of Calculus 3 4. Slope of a Tangent 3 5. Speed of a Falling Body 6 6. Given a Derivative to find the Function 7 7. Area under a Curve 7 8. Functional Notation 10 9. Graphs of Functions 11 10. Definition of a Limit . . .12 11. Infinity 14 12. Continuity of Functions 15 13. Laws of Operation with Limits 18 14. Limit of a Monotone Function 20 15. Indeterminate Forms 21 CHAPTER II DERIVATIVES OF ALGEBRAIC FUNCTIONS 16. Increments 27 17. Definition of a Derivative 28 18. Conditions for a Derivative 30 19. Geometrical and Physical Significance of the Derivative . . 31 20 General Theorems on Differentiation .32 21. Derivative of a Constant 32 22. Derivative of a Variable with Respect to Itself .... 32 23. Derivative of a Sum 33 24. Derivative of a Function plus a Constant 33 25. Derivative of a Product 34 26. Derivative of a Constant times a Function 35 27. Derivative of a Quotient 36 28. Derivative of a Constant by a Variable 37 v vi CONTENTS ARTICLK PAGE 29. Derivative of u n 38 30. Algebraic Functions 39 31. Derivative of a Function of a Function .41 32. Derivatives of Inverse Functions 42 33. Derivative of One Function with Respect to Another, when Both are Functions of a Common Variable 44 CHAPTER III ELEMENTARY APPLICATIONS OF DERIVATIVES 34. Slope of a Curve . . .47 35. Derived Curves 49 36. Rolle's Theorem 52 37. Law of the Mean 53 38. Tangents and Normals to Curves 55 39. Length of Tangent, Normal ; Subtangent, Subnormal ... 57 40. Tan ft cot ^ 58 41. Length of Tangent, Normal ; Polar Subtangent, and Polar Sub- normal 60 42. Speed and Acceleration . 61 43. Angular Speed and Acceleration 62 44. Miscellaneous Applications of Derivatives . . . . .64 CHAPTER IV THE DIFFERENTIAL NOTATION 45. The Derivative as a Rate 69 46. Differentials ..." 70 47. Kinematic Interpretation of Differentials 72 48. Differentiation with Differentials 75 49. Differentiation of Implicit Functions 77 50. Applications of Rates and Differentials 79 CHAPTER V DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS 51. Transcendental Functions . .83 52. Differentiation of sin u 83 53. Differentiation of cos u 84 54. Differentiation of tan u 84 55. Differentiation of cot u 85 CONTENTS Vll ARTICLE 56. Differentiation of sec u and esc u 57. Inverse Trigonometric Functions 58. Differentiation of arc sin u and arc cosw 59. Differentiation of arc tan u and arc cot u 60. Differentiation of arc sec u and arc esc u 61. Exponential and Logarithmic Functions 62. Differentiation of a u and e u 63. Differentiation of log a u . 64. Differentiation of log u 65. Logarithmic Differentiation 85 87 88 89 90 92 94 95 95 96 CHAPTER VI INTEGRATION 66. Anti-derivatives and Integrals 67. General Theorems . 68. The Integral (u n du 69. Fundamental Integrals 70. Integration by Inspection. 71. Integration by Substitution 72. Character of the Integration Process 101 102 103 104 105 108 110 CHAPTER VII SIMPLE APPLICATIONS OF INTEGRATION 73. Curves having Given Properties 113 74. Rectilinear Motion 114 75. Rotation about a Fixed Axis 116 76. Motion of a Projectile 116 77. Harmonic Motion . . 118 78. Motion in a Resisting Medium . . ... . . .119 79. Physical Problems involving Exponential Functions . . . 122 CHAPTER VIII SUCCESSIVE DIFFERENTIATION AND INTEGRATION 80. Definition of the nth Derivative 126 81. Successive Differentiation of Implicit Functions .... 127 82. Geometrical and Physical Interpretations of the Second Derivative 128 83. Successive Integration 129 viii CONTENTS 84. Maxima and Minima 132 85. Applications of Maxima and Minima 136 CHAPTER IX CURVES 86. Concavity 142 87. Points of Inflexion 144 88. Asymptotes, Rectangular Coordinates 145 89. Singular Points 149 90. Curve Tracing 152 91. Definition and Measure of Curvature 157 92. Radius of Curvature. Center of Curvature 158 93. Radius of Curvature, Parametric Representation .... 160 94. Roulettes and Involutes 102 95. Curvature of Involutes 163 96. Evolute of a Curve 165 CHAPTER X DEFINITE INTEGRALS 97. Definition of a Definite Integral 169 98. Elementary Properties of Definite Integrals 171 99. Change of Limits 172 100. The Definite Integral as the Limit of a Sum 173 401. Importance of the Summation Process 176 102. Geometrical Representation of a Definite Integral . . . 178 103. Definite Integrals of Discontinuous Functions. Infinite Limits of Integration 179 CHAPTER XI APPLICATIONS OF INTEGRATION TO GEOMETRY AND MECHANICS 104. Plane Areas, Rectangular Coordinates 184 105. Plane Areas, Polar Coordinates . 186 106. Volumes of Solids of Revolution 188 107. Volumes determined by the Summation of Slices . . . . 190 108. Lengths of Curves, Rectangular Coordinates 192 109. Lengths of Curves, Polar Coordinates 195 110. Areas of Surfaces of Revolution 196 111. Mean Value 198 CONTENTS i x ARTICLE p AGE 112. Work of a Variable Force 200 113. Work of Expanding Gases 203 CHAPTER XII SPECIAL METHODS OF INTEGRATION 114. Integration by Parts 207 115. Integration of Rational Fractions 209 116. Integration of Functions containing Radicals .... 213 117. Integration of Special Trigonometric Functions .... 218 118. Use of a Table of Integrals 221 119. Approximate Determination of Integrals 223 120. Simpson's Rules 224 121. Mechanical Integration 227 CHAPTER XIII FUNCTIONS OF TWO OR MORE VARIABLES 122. Definition of a Function of Several Variables . . . . 231 123. Partial Derivatives 232 124. Interchange of Order of Differentiation 236 125. Total Derivatives 237 126. Total Differentials 242 127. Differentiation of Implicit Functions 244 128. Exact and Inexact Differentials 245 CHAPTER XIV MULTIPLE INTEGRALS, APPLICATIONS 129. Multiple Integrals 254 130. Plane Areas by Double Integration ; Rectangular Coordinates . 257 131. Plane Areas by Double Integration ; Polar Coordinates . . 259 132. Volumes by Triple Integration ; Rectangular Coordinates . . 261 133. Volumes by Triple Integration ; Polar Coordinates . . . 265 134. Additional Examples involving Multiple Integrals . . . 267 135. Mass. Mean Density 271 136. First Moments. Centroids 273 137. General Theorems relating to Centroids 276 138. Second Moment. Radius of Gyration 280 139. General Theorems relating to Second Moments .... 282 140. Illustrative Examples 286 CONTENTS CHAPTER XV INFINITE SERIES ARTICLE PAGE 141. Fundamental Definitions 291 142. Tests of Convergence 293 143. Power Series 295 144. Maclaurin's Expansion of a Function in a Power Series . . 296 145. Taylor's Expansion 298 146. Taylor's Theorem. Maclaurin's Theorem 301 147. Integration and Differentiation of Series 303 148. Use of Series in Computation . 306 149. Approximation Formulas 308 150. Maxima and Minima of Functions of a Single Variable . . 311 151. Evaluation of Indeterminate Forms 312 152. Analytic Condition for a Singular Point 317 COURSE IN CALCULUS GREEK ALPHABET Letters Letters Capitals Lower Case Names Capitals Lower Case Names A a Alpha N V Nu B ft Beta E $ Xi r y Gamma o micron A 8 Delta n 7T Pi E e Epsilon p P Rho Z i Zeta 2 4> Phi K K Kappa X X Chi A X Lambda * * Psi M A* Mu o (O Omega Xll CALCULUS CHAPTER I FUNDAMENTAL NOTIONS AND DEFINITIONS 1. Constants, variables. In the applications of mathematics to physical problems, we meet with such magnitudes as velocity, force, mass, length, area, volume, etc. The measure of a magni- tude is expressed by means of a number, represented by a figure or letter, which denotes the ratio that the given magnitude bears to some standard magnitude of the same kind adopted as a unit. For the purposes of calculation, it is the number that is of funda- mental importance. For the sake of brevity, however, we shall often speak of "a velocity v" or "an area A" etc., instead of using the longer but perhaps more precise expressions " a velocity whose measure is v units "or " an area whose ratio to the standard unit of area is A," etc. In any particular discussion, the magnitudes and consequently the corresponding numbers may or may not change. A number that remains unchanged is called a constant. A symbol is then said to represent a constant when it denotes but one value in a given discussion. A symbol that satisfies this condition is itself often called a constant. When a number is permitted to assume different values in the same discussion, it is called a variable. A symbol is said to repre- sent a variable if it has assigned to it different values in the same discussion. Here again it is usual to speak of the symbol itself as the variable. For example, suppose a body falls from rest. The law that gives the relation between the time t and the distance s through which the body falls is expressed by the equation s=lgt*. (1) 1 2 FUNDAMENTAL NOTIONS AND DEFINITIONS [Chap. I. Here the number g is a constant denoting the acceleration at a point on the earth's surface. On the other hand, the time t and the distance s are variables, whose mutual relation is determined by the given equation. Again, in the equation of the circle (x-2y+(y-3y = r>, (2) 2 and 3 are constants that determine the center of the circle, and r is a constant denoting the length of the radius. However, x and y are variables whose corresponding values, as determined by the given equation, fix the various positions of the generating point. To the constant r we may assign any value at pleasure, but when it has been once assigned it must remain unchanged, so long as the circle remains fixed. A constant of this character is called an arbitrary constant or parameter. 2. Functions. In each of the illustrations given in the preced- ing article, it will be observed that the variables involved stand in an intimate relation to each other. For example, in the law of falling bodies, s and t are so related that to any value assigned to t there corresponds a definite value of s. Moreover, to every positive value that we may give s, the given relation between s and t determines corresponding values of t. Again, in the equation of the circle the variables x and y are related in a similar manner. Other illustrations of such relations between variables are familiar to the student from his study of elementary mathematics and physics. Whenever such a relation exists between two variables, we say that the variables are connected by a functional relation, or that the one is a function of the other. We may then define a function as follows : If two variables are so related that for each value that may be assigned to the one there are determined one or more definite values of the other, the second variable is said to be a function of the first. The variable to which we may assign arbitrarily chosen values is called the independent variable. It is also frequently referred to simply as the variable or argument. The variable which is thus determined, that is, the function, is sometimes called the dependent variable. Arts. 2-4] SLOPE OF A TANGENT 3 A function may depend for its value upon two or more inde- pendent variables. For example, the area of an ellipse is a function of the lengths of the major and the minor axes ; the volume of a gas is a function of the temperature and of the pressure to which the gas is subjected. Such functions are said to be functions of two variables. For the present we shall consider only functions of a single variable ; later, some of the properties of functions of two or more variables will be discussed. 3. Fundamental problems of the calculus. In his study of ele- mentary mathematics and in his everyday experiences, the student has frequently had occasion to deal with magnitudes that change. In many instances the changes are abrupt and sometimes periodic. For example, the market price of any commercial product changes abruptly from time to time. When money is placed at compound interest, the amount of interest is usually added to the principal at certain intervals. Among physical phenomena, on the other hand, we encounter numerous illustrations of changes which are evidently continuous. Thus the pressure of the atmosphere varies with the altitude, but the change is gradual, not abrupt ; the speed of a railway train starting from a station changes continuously until the maximum speed is reached; the pressure of a liquid upon a vertical wall increases continuously with the depth. Many other illustrations of both continuous and discontinuous changes will occur to the student. Problems involving discontinuous changes are dealt with for the most part by the ordinary processes of arithmetic and algebra. Problems that involve continuous changes require more powerful mathematical methods and these are the special province of the calculus. To illustrate the methods of the calculus and in a general way give the student some notion of the scope of the sub- ject and of the class of problems to be considered, a few typical problems of fundamental importance are here introduced. 4. Problem i. Slope of a tangent. Required the angle that the tangent to a given curve at a point P 1 (Fig. 1) makes with the X-axis. To make the problem concrete let the equation of the curve be y — 3 V#, and let the abscissa of P Y be x = 4. Through the given point P x let a secant line be drawn cutting 4 FUNDAMENTAL NOTIONS AND DEFINITIONS [Chap. I. the curve in a second point P. Denoting by Ax the increase P X Q, in the abscissa between P l and P, by Ay the corresponding increase Y Fig. 1. QP of the ordinate, and by 6 the angle that the secant makes with the X-axis, we have P,Q Ax As the point P is made to approach the fixed point P ly the secant approaches as a limiting position the tangent at P l and the angle 6 approaches the angle . Hence, by taking Ax smaller and smaller and calculating the ratio — U for the various values of Ax, we can Ax approximate more and more closely the value of tan . The following table gives the calculated values for different assumed values of Ax. Ax ±y Ay AX ■ 2. 1.34847 0.67424 1. 0.708204 0.70820 0.1 0.074537 0.74537 0.01 0.007495 0.74953 Using this arithmetical method, we can arrive at an approximate value of tan . The method, however, is tedious and the result is Art. 4] SLOPE OF A TANGENT 5 at best an approximation. What we need is the limiting value of — as Ax approaches the value zero. This limit can be found Aa; by considering the equation of the curve. Let (xu yi) be the coordinates of the given point P v The co- ordinates of the point P are then (x x + Aa?, y x + Ay). Hence from the equation of the curve, y = 3 VaJ, we have y 1 = S^/x l7 (1) y x + Ay = 3 V^ + Aa;. (2) By subtraction, we have Ay = 3(Vx 1 -\- Ax — Va?i), (3) and therefore it follows that Ay _ g Va?! + Ax — Va^ ,^ Aa; Aa; Rationalizing the numerator, we obtain Ay 3 Aa; 3 ,^ Aa; Ax(Vx x + Aa; + Vxi) V a? x + Aa; + Vol As Aa; approaches zero, the expression — == = approaches ■\/x l -f Aa; -+- Va; x _ , and since at the same time tan 0, which is given by the ratio 2Va; x — y. , approaches tan , we infer that tan = • We have then Aa;' FF ^ r 2Va>! 3 ' the result that for a% = 4, tan = = 0.75. 2V4 The method here developed has the added advantage, that the result is general and the slope of the tangent at any point can be found when once we know the abscissa of the point. All we need to do is to substitute the given value of x in the general formula q — — • The difficulty of applying this method to all problems is 2Vx that in most cases that arise it is not easy to find the limiting value of _^. In the subsequent chapters of the calculus we shall Aa; 6 FUNDAMENTAL NOTIONS AND DEFINITIONS [Chap. I. develop methods by means of which this limit, known as the derivative of y with respect to a?, may be obtained directly from the given functional relation between x and y. 5. Problem 2. Speed of a falling body. To determine the average speed of a moving body during a given time, we simply divide the distance traveled by the time occupied. Thus, an eighteen-hour train from Chicago to New York has an average speed of 905 -f- 18 = 50 T 5 g- miles per hour. The speed at any particular instant is not so easily determined. The obvious way to find this speed approxi- mately is to take small intervals of time, say 5 seconds, 1 second, -j- 1 ^ second, etc., immediately following the instant in question, and divide the distance traversed in the assumed time interval by that time interval. The result is the mean speed for the time interval, and it is evident that the smaller the chosen interval is taken the more nearly the quotient gives the instantaneous speed at the be- ginning of the interval. Such a method is, however, open to the criticism mentioned in the previous problem ; namely, no matter how small the interval taken, the result is merely an approxima- tion. To obviate this objection we make use of the definite law which the motion follows and as in problem 1 find the value of the limit involved. As a concrete case, let us consider the motion of a body falling in a vacuum. From physics, it is known that s = \ gt 2 , where s denotes the distance traversed in t seconds starting from rest, and g is a constant whose value is approximately 32.2. Suppose we wish to find the speed at the end of t x seconds. The distance s 1 traversed in the time t x is given by If M denotes the assumed time interval immediately following t } , and As denotes the distance traversed in this interval of time, we have * 1 + A8 = $gr(« l + AQ 8 . (2) From (1) and (2) we obtain = g tAt + \g(MY. Arts. 5-7] AREA UNDER A CURVE 7 As Therefore, we have — = gt 1 +±gAt. (3) Equation (3) gives the mean speed for the time At, and it is seen that this speed is a variable magnitude depending upon the as- sumed time interval At. As At is taken smaller, this mean speed for the interval ap- proaches the instantaneous speed at the end of t x seconds. How- ever, as At becomes smaller, the expression gt x -\-±g At approaches the fixed value gt v We conclude, therefore, that gt x is the actual speed at the end of t x seconds. At the end of 3 seconds, for ex- ample, the speed is 32.2 x 3 = 96.6 feet per second, and at the end of 10 seconds 322 feet per second. When we have learned how to find the derivative of the function s with respect to t, that is, the A9 limit of the ratio — , directly from the functional relation between At s and t, the process here indicated will be very much simplified. 6. Problem 3. Given a derivative, to find the original function. In problems 1 and 2 we saw the importance of being able to find from the functional relation between two variables the derivative of the one with respect to the other, that is, the limiting values of — and — • In the one case, the result gave the slope of the tan- Ax At & * gent to a curve, and in the other, the speed of a moving body at a particular instant. It is often equally as important to be able to 3 solve the inverse problem ; that is, if we have given as the 2V# slope of the tangent to a curve, or gt as the speed of a moving body, it is of value in certain discussions to be able to say that the curve in question is given by the functional relation y = 3Vx or in the other case that the law of motion for the body is expressed by the equation s = \ gt 2 . A process of the calculus known as in- tegration enables us to solve such problems. 7. Problem 4. Area under a curve. One of the most important problems in calculus, in fact, one of the problems that led to the invention of the calculus, is that of finding the area between a given curve, the X-axis and two given ordinates. To take a con- 8 FUNDAMENTAL NOTIONS AND DEFINITIONS [Chap. I. crete example, let us attempt to find the area between the parabola x = y 2 , the X-axis, the origin, and the ordinate x = 3. See Fig. 2. Let the part of the X-axis between x = and x = 3 be divided into n equal parts, each denoted by Ax. At the subdivi- sions An A. 2 , etc., let ordinates be erected and rectangles be constructed as shown in the figure. The abcissas of the points A x , A 2 , A s , etc., are Ax, 2 Ax, 3 Ax, etc., and the altitudes of the successive rectangles are 0, (Ax) 2 , (2 Ax) 2 , etc. Hence the sum of the areas of the rectangles is 4. = • Aa> + Ax (Ax) 2 + Ax (2 Ax) 2 + [ ... + Aaj (n - 1 Ax) 2 = (Ax) 3 [l 2 + 2 2 + 3 2 + ... +( n _l)*]. From algebra,* we have *■ X Fig. 2. i» +2 « + 3« + - +(»-i)'= n(n - i y n - i) Hence, we have from (1) r V. ; 6 (1) (2) n • Ax 2 j> . Ax) 2 - 3 (n • Ax) Ax + (Ax) 2 \ 6 J (3) But, it will be remembered that n • Ax = segment (L4, Equation (3) then becomes 3. A r = 3 A8-9Ax+(Ax) 2 Y (4) * Rietz & Crathorne's College Algebra, p. 37, Ex. 4. Art. 7] AREA UNDER A CURVE 9 Equation (4) gives the sum of the areas of the rectangles for any assumed value of Ax. As Ax is taken smaller and smaller and the number of rectangles is correspondingly increased, the area A r approaches as a limit the area A under the curve, that is, the area OP^. However, from (4) it follows that as Ax approaches zero, A r has the limiting value — - — = 9, and hence we have as a result ^4 = 9. This problem illustrates a large and important class of problems considered in calculus, namely, those requiring for their solution the summation of an indefinitely large number of indefinitely small elements. All problems in areas and volumes, and many of the applications to mechanics and mathematical physics, are of this character. The preceding problems have been selected as typical of the kind of problems to be discussed later and as indicating some- what the class of problems to which the methods of the calculus particularly apply. The student will have observed that each of these problems depends for its solution upon the finding the limiting value of some function. However, the methods employed in these examples for finding the limit are not of great value, because they become too complicated and tedious when applied to any but the simplest cases. For example, in the last problem, we were able to find the limit easily because the sum of the squares of the first n integers happens to be known. If the curve were given by the equation y = V#, we should have ^ r =(A^) 3 (Vl+V2+V3-h ••• +Vw^T), and the solution would be much more difficult. In the succeeding chapters we shall develop methods for accomplishing this same purpose much more directly and easily. In the meantime it is essential that we call attention to some of the fundamental prop- erties of functions and the laws of limits as applied to the prob- lems that we shall need to consider. EXERCISES 1. By the method of problem 4 find the area between the X-axis, the line y = 2 x, and the ordinate x = 5. 2. Find the slope of the tangent to the curve y = x 2 at the point x = 3. 10 FUNDAMENTAL NOTIONS AND DEFINITIONS [Chap. I. 12 3. Find the slope of the tangent to the curve y — — at the point x = 6. x 4. If a, body is projected vertically upward with an initial speed v , the relation between the distance s and the time t is s = v t — § gt 2 . Find an expression for the speed at the end of t\ seconds. If v = 150, find the speed at the end of 4 seconds. 8. Functional notation. As we have seen (Arts. 2, 4, 5), the relation between a variable and a function of it is expressed by an equation; as y = 3^/x, s = ^gt 2 . However, even when the analytic relation between the independent and dependent varia- bles is known, it is convenient to have a general symbol that shall stand for the function as expressed in terms of its variable. For this purpose the variable is inclosed in a parenthesis and some letter is prefixed. Thus, a function of x may be denoted by such symbols as the following: f{x), F{x), (x), $(x) f etc., which are read "/ function of x " (or simply "/of x "), "^function of x" etc. It is to be understood that the letters / F, <£, \J/, etc., are symbols of functionality and not factors. Whenever two or more different functions are employed in the same discussion, a different symbol must be used for each. If a function depends upon two variables instead of one, it may be expressed symbolically by f(x, y), (x, y), etc. When, in connection with any discussion, we have once defined /(#), then /(a) denotes the same expression with x replaced by a. In a similar way we may form functions f(t), /(4), f(x + h) f etc. Thus, if we have f(x) = 3x 2 + 7» + 9, it follows that /(*) = 3e»+7* + 9, /(4) = 3.4 2 + 7. 4 + 9 = 85, fix + h) = 3(x + hf + lix + h) + 9. Instead of replacing the variable by another variable or by a constant, we may replace it by any function of the same variable or of a different variable. Thus we may write /[*(*)], F[f(y)l (x) are integral rational functions, under what conditions is the function IS^l discontinuous ? 4. Examine for continuity the functions 1 (a) y = arc tan (b)y = 1 +e* 18 FUNDAMENTAL NOTIONS AND DEFINITIONS [Chap. I. 13. Laws of operation with limits. In this section, we shall state without proof * several properties of limits that will be utilized in later discussions. These are given in the following theorems. Theorem I. The limit of a constant times a function is equal to the constant times the limit of the function ; that is, L c>f(x) = c-Lf(x). x=a x=a Theorem II. If each of two functions approaches a limit, the limit of the sum (or difference) of the functions is equal to the sum (or difference) of the limits; that is, L [/(a>) + (x). x=a x=a x = a When Theorem III is extended to the product of a finite num- ber n of equal functions, we have the following : Corollary. If a function has a limit, the limit of the nth power of the function is equal to the nth power of its limit. Theorem IV. If each of two functions has a limit, the limit of the quotient of the functions is equal to the quotient of the limits, pro- vided the limit of the denominator is different from zero; that is, f(x) L AX) x = a4>( x ) L ( x ) x = a x = a Theorem V. If F(y) is a continuous function of y, and y is any function of x, say y = ')]• x±a x=a * For formal proof of these theorems see First Course in Calculus, or Bietz and Crathorne's College Algebra. Art. 13] LAWS OF OPERATION WITH LIMITS 19 The following example will illustrate the use of this theorem : Ex. Find the limit L log (s 2 + 2 x + 1). a- =2 Since the logarithm is a continuous function for all values different from zero, we may apply the above theorem and write L log (z 2 + 2 x + 1) = log L (x 2 + 2 x + 1) = log 9. a?=2 »=2 In fact, we may always write L log0(x)=log L (x) Sl/(x) ^ if,(x), and then L (x) = L ij/(x) = A; x = a x = a Lf(x) = A. x = a Ex. Show that L ?^JL = i, and L ^^- = 1. 0=0 6 0=0 In Fig. 5, an arc AD is described with a radius OA = r, and angle A OD = 0. Then we have, if is expressed in radians, arc AD = r0, AB = r sin 0, AG = rtsm 0. From geometry * AB= log a. 5. i log^' 2_6) =log2. a?=4 X+l 6. L log sin = 0. 2 7. £ [log (x 2 - 1) - -log (35- I)]=l0g3. 21 15. Indeterminate forms. It may happen that for some value of the variable the function takes an illusory or indeterminate form, as -. The evaluation of such an indeterminate form will be of value in finding the derivatives of functions, a process to be considered formally in the next chapter. x 2 — 4 Consider, for example, the function y = . For every x — 2 value of the variable x other than x = 2, the function has a definite value, but for x-—2 it becomes ~~ =-. Strictly ' 2-2 J speaking, the function hence, in order to define the function completely for every value of the variable, we must assign it a value for this particu- lar value of the vari- able. To guide us in our definition, we make use of a simple graphical representa- tion. For x different from x = 2, the equa ti0n *-4 y = — o X 2 -! has no definite value for x = 2; *• X Fig. 6. 22 FUNDAMENTAL NOTIONS AND DEFINITIONS [Chap. I. reduces to y = x -\- 2, (1) whose graph is the line EF, Fig. 6. Hence for values of x other than 2, the values of the function are represented graphically by points on the line EF, as B, C, D. But when x = 2 the function has no definite value, and hence we may represent it by any point whatever on the line x = 2, that is, on MJSf. Of the values which might be assigned to the function for x = 2, there is one value, represented by the point P, which is the limit of the values repre- sented by the points on EF&s x approaches 2 ; and it is convenient to select this value of y as the value of the function when x= 2. We define therefore the value of the function for the critical value of the variable as the limit that the function approaches as the variable approaches the value for which the function becomes x 2 — 4 indeterminate. Thus for x = 2, the value of the function - x — 2 is defined as L — ^— . x ~^x-2 Any definition of a function for those values of the variable for which it becomes indeterminate is of course merely a convention; but the definition adopted in this case is a very useful conven- tion, because by it the function becomes continuous for the values in question. In general, if fix) is indeterminate for any value x = a, the value of the function for x — a is defined by the limit L fix). x = a The complete definition of a function when it assumes an indeter- minate form involves, therefore, the determination of a limit. Among the indeterminate forms that a function may assume are the following : 5 ; ». oo-oo; Oxx; 0°; cc° ; 1*. GO An example of the first form is given by the function — — which x for x — takes the form - . The value of this function for x = is therefore defined by the limit T sii x = 3 L smj =1 (Art. 13) Art. 15] INDETERMINATE FORMS 23 The function x log x for x = has the form Oxx, and the i function (1 + x) x for x = has the form T°. By proper trans- formations all these indeterminate forms may be reduced to the form -. In many cases the limits are easily found by obvious algebraic transformations or by the use of series. In the evaluation of the indeterminate form of a rational function, that is, the quotient of two polynomials, the following principles may be used to advantage. 1. If the function is of the form ^A_l an d takes the form - for if/(x) x = 0, divide both numerator and denominator by the lowest power- of x that occurs in either numerator or denominator. If the fraction becomes — for x = oo , divide both terms of the 00 fraction by the highest power of x in either numerator or denominator. 2. If the function has the form Q£l and takes the form - for $(x) x = a, divide both (x) and i[/(x) by the highest power of (x — a) common to both. The student should note that we do not divide the terms of the fraction by zero or infinity. For example, while the division by the factor (x — a) holds only for x different from a, the values of x may, however, be taken as close as we please to a. Hence, dividing first by this factor and afterwards passing to the limit, we obtain the proper result. Indeterminate forms will be again considered in Chapter XV. The following examples illustrate some methods that may be used in evaluating various indeterminate forms. Ex. 1. Find the value of a;2 ~ 6a; for x = 0. x s — 4 x 2 — 12 x x 2 -6x L . x-6 _ 1 a5:i0iC 3_ 4a .2_ 12x a . = ox 2 -4x- 12 2 9 r 2 -I- S a* 3 Ex. 2. Evaluate the function ^ for x = co. x+5x 3 L 2 x 2 + 3 x s _ L x _ + 3 _3 + 5 -t-52 3 03*00 1 , e -f 5 5 & 24 FUNDAMENTAL NOTIONS AND DEFINITIONS [Chap. I. Ex. 3. Find the value of x 3 - 3 x 2 + 5 x - 3 2 x 3 - 5 x 2 + 8 x - 5 for as x3-3x 2 + 5x-3 = (x 2 - 2 a + 3) (a - 1) _ = a 2 - 2 a + 3 2x 3 -5x 2 + 8x-5 (2 a; 2 - 2x + 5)(x - 1) 2x 2 -3x + 5 Hence, we have i 3 a 2 + 5 x - 3 __ L x? - 2 x + 3 = i 2 x 8 - 5 x 2 + 8 x- 6 1 2 x 2 - 3 x + 5 2 Ex. 4. Evaluate a ~ ^ ~ x ' 2 for x = 0. This function takes the form - . To find its value for x = 0, we multiply- both terms of the fraction by the complementary surd a + Va 2 — x 2 . We then have a - Vx 2 - x 2 _ a 2 - (a 2 - x 2 ) _ x 2 1 . x2 x 2 (a + Va 2 - x 2 ) x 2 (a + Va 2 - x 2 ) a + Va 2 - x 2 ' and L x = Va 2 <«-0a + Va 2 — x' 1 2a Ex. 5. Evaluate 1 ~ cos 6 for = 0. We have Hence, for = 0, 1 2 sin 2 - — cos 2 2 r • 0' sin- 2 2 *=»2 " . 0" sin- 2 2 0=0 2 #«o " . 0" sin- 2 L 2 - 2 J 1 — CO 3 *isd sflned as 0. 0. EXERCISES Find the limiting values of the following functions for the given values of the variables. 1. 4 x 3 - 3 x , x = 0. 2 x 8 — 3 x 2 + 5 x 3. Vl + x — Vx, x = co 5. *MU = 4. 2 x 3 - 3 x 2 - 4 x + 12 x X 2 + 4 ^ _ 21 Vi + x -VT 6. ^- fl5 ,x = a. x = 0. Art. 15] INDETERMINATE FORMS 25 7. Show that L ^^ = n. sin sin nd sin n0 » sin nd sin w0 w0 , n0 n0 1 Suggestion : —. — - = n— — - , L —. — - = ■ — — — t yif sin sin sin T sin 1 8. Show that for = 0, 1 ~ C0S - = 0. sin 10. Show that L isin| = L 6 = - &=QX 4 & 11. Evaluate — + - ^ ~ 5 - for x = and x = oo . 3 a; 4 - 2 x 3 + 6 x 12. Evaluated arctanx . 13. Show that L 1 ~ cos g = - . 85 = X = 2 2 MISCELLANEOUS EXERCISES 1. Given ?/ = x 2 — 3 as + 5. (a) Find the increment Ay corresponding to the increment Ax of the variable. (6) Calculate Ay for Xi = 3 and Ax = 0.1. 2 . Given y = x 8 . (a) Find the expression for the increment Ay for x = Xu (b) Find the limit of the ratio — ^ as Ax = 0. w Ax 3. Given y = sinx. Take X\ = 30°, and assume for Ax the following values: 1°, 30', 5'. Make a table showing the values of sin (xi + Ax), A?/, and— ^. See if the ratio approaches a limit. Ax 4. If (x) = a x , show that O(x)] 2 = 0(2 x). 5. If (y) =e* + e~y, show that 0(3 y) = OQ/)] 3 - 3 (y) and 0(x + y) 0(x - y) = 0(2 x) + 0(2 y). 6. Give a physical illustration of a function that approaches a limit as the variable increases indefinitely. 7. For which of the trigonometric functions is/(0) =/(— 0) ? 8. From geometry or physics give two or more examples of monotone functions. 9. Suppose that air inclosed in a cylinder has an initial volume of 10 cubic feet and an initial pressure of 20 lbs. per square inch. Assuming that air expands according to Boyle's law, viz. pv = constant, calculate the value of -£- for Au = 2, 1, 0.1, 0.01. Determine the same ratio for Av = h, and Av show that as Av = 0, this ratio has the limit — 2. 26 FUNDAMENTAL NOTIONS AND DEFINITIONS [Chap. I. 10. Show that L (sec x — tan x) = 0. 2 I 11. Examine the function e* for continuity. Draw the graph of the func- tion from x=— 2 to x = + 2. 12. Find the limiting values of the following functions for the given value of the variable : , x tan 6 — sin 6 a A /7 N sec 6 — 1 a n («) ^ .* = <>. (6) ^— ,0 = 0. 13. If » is a positive integer show that x = a x- a 14. If the variables in any given functional relation are interchanged (as y — e x , « = €»), show that the graphs of the two functions are symmetri- cal with respect to the line x — y = 0. CHAPTER II DERIVATIVES OF ALGEBRAIC FUNCTIONS 16. Increments. The idea of a variable as employed through- out the calculus is but little used in elementary mathematics. There the symbols x, y, etc., stand usually for unknown but fixed numbers whose values are to be determined. Here, as in the analytic geometry, we associate the same symbols with magnitudes that change or grow according to some law determined by the nature of the problem under discussion. As we have already seen, this idea of growth is of prime importance in the calculus, and it lies at the basis of the discussion in the present chapter. If a variable be assigned some arbitrary value as x x , a function of the variable takes a corresponding value f(x^) ; and if the vari- able changes to any other value x, the function takes a correspond- ing value f(x). The changes x — x 1 and f( x )~f( x i) of the variable and of the function, respectively, are called incre- ments of the variable and of the function and are denoted by Ax and A/(x). See Fig. 7. We have therefore Ax = x — x 1} A/0) =/(*) -/(*i) =f(x 1 + Ax)-f(x l ). If we have y =f(x), then instead of writing A/(#) we can equally well write Ay. Given the increment of the variable, we may calculate the cor- responding increment of the function, as shown in the following examples. /(*)-/(*,) >X Fig. 7. Ex. 1. Given the function Then /(x) = 3x 2 + 4x + 2. = 3(x x + Ax) 2 + 4(aci + Ax) + 2 = 6 cciAx + 3 (Ax)' 2 + 4 Ax. 27 (3 xi 2 + 4 xi + 2) 28 DERIVATIVES OE ALGEBRAIC FUNCTIONS [Chap. II. Ex. 2. Given the law of falling bodies s = \ gt 2 . Calculate As correspond- ing to At = |, h = 4. We have here As = \g {t\ + i) 2 - $ gh 2 Increments are, in general, variables and may be either positive or negative. As was shown in the illustrative problems, Arts. 4 and 5, the ratio of the increment of the function to that of the variable is useful in giving the slope of a secant line, the mean speed of a moving point, etc. In fact, this ratio of the increments always gives a mean value of some kind. It is, however, the limit of this ratio, the derivative, that is of special importance. 1 7. Definition of a derivative. Consider the ratio Ay = Af(x) = /(s 1 + AaQ--/(3 1 > Ax Ax Ax x x being some particular value of x. For Ax = this ratio assumes the indeterminate form -, and to evaluate it for this value of Ax we must find its limit as Ax = 0, according to the principles laid down for the evaluation of indeterminate forms. This limit is called the derivative of f(x) with respect to x for the value x = x v We shall denote derivatives with respect to the variables x, t, etc., by the symbols D x) D t , etc. Thus we have for any value of x, I> x f(a»= L /C* + *f)-fl«0 = L **. (1) Aa;±o Ax A«=oAa? The derivative may therefore be defined as the limit of the ratio of the increment of the function to that of the variable as the latter in- crement approaches the value zero. Instead of the symbol D x , other symbols are often employed. For example, having we may indicate the derivative by any one of the following symbols : fix), yj, y', D y . Art. 17] DEFINITION OF A DERIVATIVE 29 The last two symbols are used only when there is no ambiguity as to the independent variable. The derivative is also referred to as the differential coefficient or as the derived function,* and the process of finding the derivative is called differentiation. This process consists of the following steps : 1. Give to the independent variable an increment. 2. Calculate the corresponding increment of the function. 3. Find the ratio of the increment of the function to that of the variable. 4. Determine the limit of this ratio as the increment of the variable is allowed to approach zero. The following examples illustrate the process of differentiation. Ex. 1. Find the derivative of y =f(x) = 5 x 2 — 3 x + 4. First, giving the variable the increment Asc, we have y + Ay =/(« + Ax) = 5 (x + Ax)' 2 - B(x + Ax) + 4 = 5 x 2 - 3 x + 4 + (10 x - 3) Ax + 5 (Ax) 2 . Subtracting, we obtain as the increment of the function Ay=f(x + Ax) -f(x) = (10 x- 3) Ax + 5 (Ax) 2 . The result of dividing by Ax is the ratio ^ = 10x-3 + 5Ax; Ax and the limit of this ratio as Ax approaches zero is L ^= L (10 x - 3 + 5 Ax) = 10 x - 3. Ax = Ay Aa' = p Hence, /'(x) = D, (5x 2 - 3x + 4) = 10 x - 3. To find the value of the derivative for some particular value of x, we merely substitute that value of x in the derived function /'(x). Thus, for x = xi, /'(xi) = 10 xi - 3 ; for x = 5, /'(5) = 10 x 5 - 3 = 47. ' *The word " derivative " is frequently used to mean the value of the limit (1) for a particular value of x. The term "derived function" refers more properly to the assemblage of all such values. 30 DERIVATIVES OF ALGEBRAIC FUNCTIONS [Chap. II. k Ex. 2. Given the equation expressing Boyle's law, viz. pv = k, or p = - • Find the derivative of p with respect to v. v We have P = k ■ ~ •> V whence p + Ap = k v + Av Subtracting, we get for the increment of p, Ap = k v v{\ Av v -f Av ) + Av) whence Ap _ Av : k v(v 1 + Av) and L & = Av = Av ■■-k L A» = 1 o«(b + Av) v*' Hence D v p =/'(«) =- \ v- For k = 200, and v = 10 (see Ex. 9, p. 25), ft (10) = _200 = _2. EXERCISES Find the derivatives of the following functions, using the general process described in this section. 1. y = sc 4 . 2. y = x' 2 — 4 x + 5. x 2 5. y = x i -x-\-2. 6. y = x-1 7. y = {x- a) s . s. s = \ gt*. 9. s = a« + ifir< 2 . 10. P = ad + bd-h 11. H p (v — b) = fc, find D v p. 18. Conditions for a derivative. Not every function has a derivative for all values of the variable. In order that the ratio /fa + AaQ-Zfa) ^ shall have a limit as Aic = 0, it is necessary that the numerator shall approach zero simultaneously with the denominator. In other words, we must have L /(«, + *») =/(*!), Az = Arts. 18, 19] SIGNIFICANCE OF THE DERIVATIVE 31 which is equivalent to writing L f(x)=f(x x ). X = X\ This is, however, the condition for the continuity of f(x) for the value x = x v It follows that a function cannot have a derivative at a point of discontinuity. Although the numerator of (1) must vanish simultaneously with the denominator, it may happen that as Ax = the ratio becomes infinite for particular values of the variable. In such cases, we say that for the particular value of x in question, say x = x x , the value of the derivative becomes infinite, and write 19. Geometrical and physical significance of the derivative. It A f (x) has been pointed out that the ratio J v ) gives the slope of the Ax secant line passing through the points whose ordinates are y x =f(x 1 ), y =f(x 1 + Ax). If Ax be allowed to approach zero, the secant line approaches as a limit the tangent to the curve y =f(x) at the point (x 1} y r ). Hence, if y =f(x) is represented by a continuous curve, we have f(x^ = tan , where is the angle made with the axis of X by the tangent to the given curve at the point (x l} y x ). For the sake of definiteness we shall take as the acute angle made with the positive direction of the X-axis. It may be either positive or negative. Tan <£ is called the slope or gradient of the tangent to the curve; hence the derivative of the function f(x) for the value x = x 1 represents the slope of the curve y —f(x) at the point (x l} yj. Again, it has been shown that — gives the mean speed of the moving point during the time A£. If A£ approaches zero, the limit, that is the value of the derivative D t s for t = t x , gives the speed of the moving particle at that instant. Similarly, if m denotes the mass and V the volume of a body, the limit of the ratio as A V approaches zero gives the density at a point; and 32 DERIVATIVES OF ALGEBRAIC FUNCTIONS [Chap. II. if AQ denotes the heat entering a body and At the corresponding rise in temperature, the limit of the ratio — — , as At approaches At zero, gives the specific heat at the definite temperature r x . The ratio of the increments gives always a mean value for an interval ; the derivative an instantaneous value or the value at a point. 20. General theorems on differentiation. While the process of finding a derivative described in Art. 16 is sufficient for the dif- ferentiation of all functions that can be differentiated, it becomes inconvenient when the function is complicated, and the labor of differentiation can be abridged by the use of certain general theorems that apply to all classes of functions. These theorems are given in the following Arts. 21-29. Because of their funda- mental importance, the student should have them at his ready command. 21. Derivative of a constant. Let y = c. Then y -f- Ay = c. Subtracting, we have Ay = 0, whence D xV = L *L = 0. or D x c = 0. We have therefore the following theorem : Theorem I. The derivative of a constant is zero. Geometrically the equation y = c is represented by a straight line parallel to the X-axis. The slope of this line is zero in accordance with Theorem I. 22. Derivative of a variable with respect to itself. Given y = x, then y + A?/ = x + A#, and A?/ = Ax. Arts. 20-24] DERIVATIVE OF A SUM 33 Hence D x y = L ^= L — = 1; Ax^oAa,* ^ x = qAx that is, D x a = 1. This result gives the following theorem : Theorem II. Tlie derivative of a variable with respect to itself is unity. Ex. Give a geometrical illustration of Theorem II. 23. Derivative of a sum. Let y =f(x) + 4>(x). Then y + Ay =f(x + Ax) + (x + Aa>), and Ay == /(a; + Aa;)— /(a;) | (x) ^ Ax Ax Ax Taking the limits of both members, we have L ^/ = L f(x + Ax)-f(x) L cf>(x + Ax) - . This process may be extended to the sum of any finite number of functions. Thus, if we have y — u + v -f- 10, where u, v, w are functions of x, we may write D x y = B x (u + v + tu) = D^w + D^v + D^w. We may, therefore, state the following theorem : Theorem III. TJie derivative of the algebraic sum of a finite number of terms is the algebraic sum of their derivatives. 24. Derivative of a function plus a constant. Let y =f(x) + c. By the application of Theorem III, we have D x y = DJ(x) + D x c = D x f(x). (Art. 21) That is, D t [f(x)+c] = DJ(x). 34 DERIVATIVES OF ALGEBRAIC FUNCTIONS [Chap. II. If u be used to denote f(x), this result may be written 2>a(tt + c) = D x u. We have, therefore, as a corollary to Theorem III, the following : Corollary. The derivative of a function is not affected by increasing or decreasing the func- tion by an additive constant. It follows also that two func- tions differing only by a constant term have the same derivative. In the process of differentiating, the constant terms may conse- quently be neglected, f Geometrically, this corollary has an interesting significance. Suppose the function y =f(x) + c to represent some curve. The effect of adding or subtracting a constant term, that is, of chang- ing the value of c, is simply to shift the curve up or down with reference to the X-axis. (See Fig. 8.) As has been shown, the derivative measures the slope of the curve, and it is evident that this slope for any particular value of x, as a, is not changed by shifting the curve as indicated. This corollary has also an important significance in the inverse operation of integration, as we shall see later. 25. Derivative of a product. Given y = f(x) • (x). We have then y + Ay = f(x + Ax) • <£ (x + Ax), Fig. 8. and Ay = f(x + Ax) • (x + Ax)-f(x) • (x) Ax Ax By adding and subtracting f(x) • (x + Ax) in the numerator, this ratio may be written in the form Ax = (x -\- Ax) f(x + Ax) -f(x) 4>(x + Ax)-(x) Ax Ax Arts. 25, 26] DERIVATIVE OF A PRODUCT 35 Since <£(#) is by hypothesis continuous, (x-\-Ax) in the limit becomes (x) ; hence, we have upon passing to the limit, D x y=.D x fi-f.D x . This result can be extended to the product of a finite number of functions ; thus, if y = uvw, where u, v, w are functions of x, we have D x y = X>a. (uviv) = nv • D x w + uw ■ D x v + vw • D^u. We have then the following theorem : xftEOREM IV. TJie derivative of the product of a finite number of factors is the sum of the products obtained by multiplying the derivative of each factor by the product of all the other factors. 26. Derivative of a constant times a function. If we have given y = c- f(%), we have, from Theorem IV, D x y = cD x f(x)+f{x)-D x c = c • D x f{x). That is, D x cf(x) = c . D x f(x). Again, using u to denote f(x), this result may be written t> x (cii) — cD x u. Hence we have the following corollary to Theorem IV : Corollary. The derivative of a constant times a function is the constant times the derivative of the function. We may now combine the results of Theorem I, the corollary to Theorem III, and the above corollary in the one general state- ment that in the process of differentiation constant terms disappear but constant factors remain. Ex. Find the derivative of y = x 3 + 7 x + 8. We have D x y = D x (ar 3 ) + D x (7 x) + D x 8. (Th. Ill) D x (z 3 ) = D x {xxx) = x 2 D x x + x^D^ + x*D x x (Th. IV) = 3 x\ (Th. II) D x 7x = 7D x x (Cor.) = 7. (Th. II) Z> x 8=0. (Th. I) 36 DERIVATIVES OF ALGEBRAIC FUNCTIONS [Chap. II. Combining these results, we have D x y = 3 x 2 -f 7. The student need not write down each step as has been done in this illustrative example ; but he should make himself so familiar with the principles set forth in the preceding theorems that he can write down the results at once. EXERCISES Find the derivatives of the following functions. 1. y = x 2 . 2. */=x 2 -10x + 4. 3. y=x 2 (x-l). 4. y = 5x 3 -x(x + 2). 5. y = x 2 (x 2 - 2) + x (x -f 3). 6. Plot the curve y = 4 x 2 + c, giving c the values 2, 4, 6, 8, and show that each of these curves has the same slope for x = 2. 7. Let the distance traversed by a moving point be given as a function of the time by the equation s — 80 t — 16 t 2 . Derive a general expression for the speed D t s, and find the speed at the end of 3 seconds. 27. Derivative of a quotient. Let y = m, then y + A?/ = ~ — —4 , y - * e£(x + Ax)' and consequently ^ <£ (a; + Aaj) <£ (x) (x + Ax) • <£ (a?) Subtracting and adding (x) • /(a?) in the numerator, we have, after dividing by Ax, . /(x + Ax)-/(x) _ _ (x + Ax) • (x) Kemembering that (x) is continuous and hence, as Ax = 0, <£(x-f Ax) becomes cf>(x), we have for the limiting value (x)-J./(s) -/(«)• -P. »(«) " / " [«#>(*)] 2 Arts. 27, 28] DERIVATIVE OF A QUOTIENT 37 Denoting the two functions by u and v, respectively, this result can be more conveniently written as follows : V JD™U — U T>^V ©- We may state this result in the following theorem : Theorem V. TJie derivative of a fraction is equal to the denomi- nator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. 28. Derivative of the quotient of a constant by a variable. If the numerator of the fraction is a constant, the result just obtained becomes Hence we have the following corollary to Theorem V : Corollary. The derivative of the quotient of a constant by a variable is equal to minus the product of the constant and the deriva- tive of the variable divided by the square of the variable. Ex. Given /(«)= ^^ ; find /(a). JK J x 2 + 3x-f l' J K J From Theorem V, we have fl(x) = (s a + 3 x + 1) D x (x + 2)-(x + 2) D x (x°- + 3 x + 1) J K J (x 2 + 3 x + l) 2 = (a: 2 + 3x + l)l-(x+ 2 ) (2 x + 3) = _ (x 2 + 4 x + 5) ( X 2 + 3 x + 1)2 ( X 2 + 3 x + 1)2 * EXERCISES Find the derivatives of the following functions : 1. */ = 3x 2 -4x-6. 2. ?/ = 4x 3 - 2x 2 + 5. 3. t/ = x 2 (x-5). 4. y = «(a; + 1)(* — 2). 5. y = x\x* — 2)(* + l). 6. ?/ = x 2 (x-4)+x(x 2 + 3). 7. ,-*-*'. a r-I. X 2 X 2 9. y = m . 10. y= ffa; + & . ax + 6 ex + 2_4x + 5 14 as-6 x 2 + 2 * x 2 + 4 a; - 6 a: 2 15. Plot the curve */ = — + c, giving c the values 1, 3, 5, and show that each of the curves has the same slope for x = 3. 16. Show that if the nth power of a variable occurs as a factor in the denominator of a fraction, the n + 1st power of the factor will occur in the denominator of the derivative of the fraction after reduction to lowest terms. 29. Derivative of u n . Let u denote any function of x. To determine the derivative of u n with respect to x, we proceed in the usual manner. Thus, let V = u1 \ then y + Ay=(u + Au) n , and provided Au =£. 0, we have Ay _ (u + A u) n — u n Au -s Ax Au Ax Since u is continuous and therefore Au approaches zero with Ax, we have upon passing to the limit Dj,= L £*. L (» + *")'-«? . L *2. (2) Az = oAa Am = Am Aa=oAa The evaluation of the first limit in the second member requires consideration of the special limit L v n -a n x = a = «• + M*- 2. 13. s = 80-16* 2 . 14 - If =*»(!-»)•. i 15. 2/ = x 8 \/x 2 -5. 16. y = x*(x 2 - a 2 ) », 17. By differentiating — , show that D x x n = nx n ~ l holds for negative inte- gral values of n. x _y 18. Given xe 3 = c ; express ?/ as an explicit function of x. * For a more general definition of an algebraic function, see First Course, p. 9. Arts. 30, 31] FUNCTION OF A FUNCTION 41 19. Given x =f() = a( — sin ) y = F() = a(l — cos 0), express & as an explicit function of y. 31. Derivative of a function of a function. The derivative of u n discussed in Art. 29, where u is a function of x, is a special form of the more general case which we shall now consider. Let us suppose that y=f(u), and u = (x), where these functions have respectively the derivatives f'(u) an( i 4>'( x ) f° r tne corresponding values of the variables u and x. If we wish to find D x y, we can do so by substituting the given value of u mf(u), thus expressing y directly as a function of x ; from this expression we can obtain the desired result by methods already explained. In many cases, however, the derivative can be obtained more easily by the follow- ing method. Let x take an increment ; then u and y will also take increments, and since u and y are continuous functions of x f these increments will approach zero as the increment of x approaches zero. We have then Ay=/(te + A«)— /(«). Ay = f(u + Am) - /(it) t A^ Aw^tO. Ax Au Ax Since Au = as Ax = 0, we have in the limit L *?= L /(" + Am)-/00 x A». Aa;i0Aa? Ait = A?* Ax = oA#' that is, 2>a.2/ = D U V ■ T>^u. (1) This formula expressed in words gives us the following theorem : Theorem. Ify = f(u) and u = (x), the derivative of y with respect to x is the product of the derivative of y with respect to u and the derivative ofu ivith respect to x. This theorem asserts the principle that if u changes m times as rapidly as x, and y changes n times as rapidly as u, then y changes mn times as rapidly as x. For example, if a horse travels twice as fast as a man, and a train four times as fast as the horse, the train travels eight times as fast as the man. 42 DERIVATIVES OF ALGEBRAIC FUNCTIONS [Chap. II Ex. 1. Find D x y when y = Vu and u = 3 x 2 — 4. We have D u y = £ w~ 2 , and Z> x w = 6 x ; hence #**/ =4m"^6x = 3 seiT^ = — V3 x 2 - 4 _3 Ex. 2. Find Z^ when ?/ = (a 2 — x 2 ) *. -3- _5 Put tt = a 2 — x 2 ; then y = u 2 , Z> M ?/ = — § u 2 , and D x u = — 2 x. Therefore A# = 3 xu~% = — (a 2 - x 2 ) 1 - EXERCISES Differentiate the following functions : 1. y = ^x 2 + 5. 2 y =( x 2-2x+5)i *»-<*-*><■ 4 . ,=(«*.*)-* 5. y = (x* + a 3 ) 2 . i , y / x 4 \ 3 7. y = #8a*-4aa. . 6. y = ^^_^J . 9. y = (x*-2x + 6)"*. B y = V (x + m)( , + , } > 13 - P = \Y+J 2 ' ia 2/=3(x 2 -4)^(x + 3)i 13. y = (x+Vx 2 -a 2 )i 12. p=(l + **)"*. 15 . v= * . 14. — * 2 + 2 V a 2 — x 2 Vx 2 + 1 16. By putting f(u) = cu, prove the corollary to Theorem IV, Art. 26. 17. By putting /(w) = w M , develop the law for differentiation of u n given in Art. 29. 18. Write a function which has x 2 as its derivative. 19. Show that the curve for which the slope of the tangent line (that is, tan ) is numerically equal to the abscissa has the general form y = \ x 2 -f C. 20. Find the values of x for which the derivative of /(x) = x 3 — 9 x 2 + 24 x becomes zero. What is the geometrical interpretation ? 21. At what points of the curve y = x 8 — 12 x is the tangent parallel to the X-axis ? 32. Derivatives of inverse functions. If y is given as an explicit function of x, and it is possible to solve for x in terms of y, we may then express x as an explicit function of y. Thus, if Art. 32] INVERSE FUNCTIONS 43 y = ar we have x= ± Vy + 4. In general, let x = (y) be the result of solving the equation y=f(x) for »; then f(x) and £(y) are said to be inverse functions. Examples of inverse functions are sin x and arc sin y, log x and e\ \_ x n and y n , etc. In the process of differentiation we ordinarily express y as an explicit function of x and determine D x y directly. Sometimes, however, it is convenient to express a? as a function of y and find D x y by means of the inverse function. To find the relation between the two derivatives, we may proceed as follows : Given y=f(x); (1) and let the inverse function be » = *(y). (2) By differentiation, we have from (2) by aid of Art. 31, l = D„(y).D x y, or 1 = D y x • D x y. Hence we have, provided D y x ^ 0, This formula states the principle that if y is changing n times as rapidly as x, then x is changing -th as rapidly as y. Ex. Given the equation of the parabola y 2 = 4 px ; find D x y. Solving for x, we have y 2 4p ■^ 2 v V whence D y x = -r 1 - = ~ • From the theorem of inverse functions, we have D x y 1 _2p _ 2p__jp t ~~^Z \ nr. ' D y x y -y/4 px 44 DERIVATIVES OF ALGEBRAIC FUNCTIONS [Chap. II. EXERCISES 1. Find D x y when y 2 = x - 4. 2. Find D x y when y — = x 2 . 2/+ 1 3. If p 2 = a0 + 6, find Z> p. 4. If v 2 = 2 gs, find D.t;. 5. Find functions inverse to the following functions : (a) y = x s ; (6) y = log (x 2 + a 2 ) ; (c) y = Va 2 - z*. 6. Find the function inverse to /(z) = \og(x+Vx 2 + l). 33. Derivative of one function with respect to another, when both are functions of a common variable. Suppose we have given y = f(t), x = (t), and that we wish to find D x y. We have, for At =£ 0, Ay Ay _ At Ax Ax At Because of the continuity of f(t) and (t), Ay = and Aa? = simultaneously as At == 0. Hence, by Art. 17, we have upon passing to the limit. Ex. 1. Given y = z 2 + 1, and x = y/z ; find D x y. We have D z y = 2 *, and D^ = £ 2 _2 \ Hence, X> x y = -^- = 4 z ^4x 3 . Ex. 2. The equations s = ^ #£ 2 and v = gt refer to falling bodies. Find the derivative Z).,v. 2>,* = 22 = £ = I = 2. As gt t v EXERCISES 1. Find 2> x y, when ?/ = 3 £ 2 — t - 10 and z '■ = t + 8. 2. Find D x y, when ?/ = t* — 3 £ 2 + 7 and re = £ 2 — 2 £ + 4. Find the values of this derivative for t = 0, 2, 5. Art. 33] FUNCTIONS OF A COMMON VARIABLE 45 3. Ii P = Vtand 6 = t 2 - 10, find D 6 p. 4. If x = at and y = bt — | c£ 2 , find D x y and D y x. 5. Work examples 3 and 4 by eliminating t before differentiating. 6. Prove the theorem of Art. 33 by making use of the theorems of Arts. 31 and 32. MISCELLANEOUS EXERCISES Differentiate the following functions with respect to the variable indicated. 1. ?/ = 5x 4 + 3.r 3 -8x + 7. 2. y = x 2 (3 x 2 - 4)(x + 1). ill q a v 3. y = x* (x* - a*) 2 . 4. p= ' 2 (x - 2) 2 (L: 7. */ = |(Vx + a) 2 -a(Vx-t-a)^. s. y = ^- 2 >vr+?. 6. y = a + »^^-v 8 v- * M (1 + X)» 10. 12. y = (x + Vx 2 — 1) H . iz + 1 14. .._ (l-z)Vl-x 2 » * x"(a + x)™ 11. */ = Vx + Vl + x 2 . 3 ' 13. 2/ = A/ — * x — 15. y = VaP + l+vte-l + » ' Vx^+l - vx 2 ^ - ! 1 16. y = — ^ 2 ~ a ^ 3 . 1.1. y = x + a x + Vx 2 18. v = 2g±J— .*. is. p= « -. VX 2 + 1 + X 20. Given t> = - m (1 + ap), where m, (7, and a are constants, find 2>»p. 21. Show that the slope of the tangent to the curve y = x 3 + 4 is never negative. Find the slope for x = 0, x = 2. For what values of x does the slope decrease as x increases ? 22. Find the angle which the tangent to the parabola y 2 = 9 x at the point (4, 6) makes with the X-axis. 23. Given y = u + 5 and u = x 2 - 5 ; find D x y. Vw 2 - 15 24. Given s = bt + | a£ 2 and v = at, find D s v in terms of v. 25. The equation pv — C expresses Boyle's law, C being a constant. Find D p and Z?/>v. 46 DERIVATIVES OF ALGEBRAIC FUNCTIONS [Chap. II. 26. From Regnault's experiments, the heat q required to raise the tem- perature of a unit weight of water from 0° C. to a temperature t is given by the equation q = r + 0.00002 r 2 + 0.0000003 r 3 . (a) Find D T q. (b) Calculate the numerical value of D T q for t = 35°. 27. The efficiency of a screw as a mechanical device is given by the relation E = - * — ^ x > , where /x denotes the coefficient of friction and x the tangent of the pitch angle of the screw. Find the derivative D X E, and its values for x = and x = /j.. Find also the value of x for which D X E = 0. 28. Find the slope of the tangent to the circle x 2 + y 2 = 25 for the points x = 0, x = 4, x = — 3. 29. Find D s v, where v 2 = 2 c ( )• \8 a) 30. D x y = x2 and u = - x 2 4- &x. Find D u y. ax + b 2 31. Interpret geometrically the corollary of Art. 26. 32. Find functions which have the following derivatives : (a) 3x 2 ; (5) 4 x* - 1 ; (c) x 2 - x + 2 ; ((t), y = *Kt), (2) where <£(£), \f/(f) are single-valued functions of the variable t. We call these equations the parametric equations of the curve. As examples of parametric equations of curves, we have for the circle the equations y = a sin 6, x = a cos 0, and for the cycloid y = a (1 — cos 6), x = a(0 — sin 6). We may pass from the parametric form of expression to that given by equation (1) by eliminating the common variable. In whichever form the equation of the curve is given, we have 2)^ = tan*, (3) where denotes the acute angle between the tangent to the curve and the positive direction of the X-axis. To determine tan from the parametric equations of the curve, we have by Art. 33 A2/ = g = tan*. (4) Tan cf> has been defined as the slope or gradient of the tangent to the curve. It may also be called the slope of the curve, for the direction of the tangent to the curve is also the direction of the 47 48 ELEMENTARY APPLICATIONS OF DERIVATIVES [Chap. III. curve at the point of tangency. The slope of the curve, therefore, is given by the value of the derivative D x y. The value of the derivative at any point gives still other properties of the curve. Since the direction of the tangent is identical with the direction of the curve at the point of tangency, it follows that so long as is a positive angle the ordinates of the curve are increasing as the abscissa increases. When, how- ever, <£ is positive, tan <£, and hence D x y, is positive ; similarly, when is negative, the ordinates of the curve are decreasing as the abscissa increases. But in this case tan <£, and hence D x y, is negative. If the value of D x y becomes zero for any value of x, then tan is zero and cf> must be zero ; that is, the tangent to the curve at this point is parallel to the X-axis. This occurs when- ever the curve has a turning point, that is, a point where the or- dinates cease to increase and begin to decrease, or vice versa. Turning points are shown at A, C, and E, Fig. 10. If D x y be- comes infinite for a particular value of x, the value of <£ is then \ it or — \ it ; that is, the tangent to the curve at that point is per- pendicular to the X-axis. We may summarize these results as follows : At any point of the curve y =f(x), the ordinate increases or de- creases with increasing x according as D x y is positive or negative. If D x y is zero for any value ofx, then at that point the tangent to the curve is parallel to the axis of abscissas. If D x y becomes infinite for any value of x, the tangent to the curve is perpendicular to the axis of abscissas at that point. Ex. 1. Investigate the curve y = ^ x 3 — x by means of its derivative. Differentiating, we get = 8£ _ j = ( x + 2)(*-2) xJ 12 4 For x > 2 or cc < — 2, D x y is positive, and consequently y increases with x. For values of x between 2 and — 2, D x y is negative, and y therefore decreases as x increases. For x = 2 and for x = — 2, D x y = ; hence at these points the tangent to the curve is parallel to the X-axis. The curve is shown in Fig. 9. Fia. 9. Arts. 34, 35] DERIVED CURVES 49 EXERCISES Investigate the following curves by means of their derivatives. 1. y = x s -Sx 2 +t>ie. 2. y 2 = 8x-10. 3. y= — . X 4. For what values of x does the function x 2 + - increase with x ? For x what value does it decrease as x increases ? 5. The equation t = 53.6 p* -35.7 gives approximately the temperature r of steam as a function of its pressure p. (Here r is in degrees Fahrenheit, and j? in pounds per square foot.) Show how the derivative D p t changes as the pressure increases, and sketch roughly the curve r =f(p). 6. A cylindrical vessel with one end open is to hold 300 cu. in. The superficial area A of this vessel (cylindrical wall plus one base) is a function of the radius of the base. Deduce the equation and examine it by means of the derivative. Interpret the results. 7. The efficiency rj of a hoisting device is a function of the load P raised as expressed by the equation P mP + n where m and n are constants. Show how the derivative varies with the load P and sketch the general form of the curve rj =/(P). 8. By means of the derivative investigate the curves : (a) y = a + b{£- c)» ; (6) y = a + b(x - c) 7 . Show the form of the curves in the vicinity of the point (c, a). 9. Given the continuous curve y =/(:»). Show by means of the graph that, for positive values of Ax, Ay is positive or negative according as the function is increasing or decreasing at a given point. By considering the limit L — ^ deduce the general law given on p. 48. Aa?=oAx 35. Derived curves. The derivative of a function is also, in general, a function of the independent variable, and may be rep- resented by a graph in the same manner and under the same conditions as the original function. This curve, whose equation is y=f'(x), is known as the derived curve. 50 ELEMENTARY APPLICATIONS OF DERIVATIVES [Chap. III. The principles developed in the last article enable us to estab- lish certain general properties of derived curves. Thus the graph of y =f'(x) crosses the axis of x at those points for which the original curve has a turning point. Moreover, we can tell whether the derived curve -crosses the X-axis from above to below or vice versa ; for we know that as x increases, f'(x) is positive or nega- tive according as f(x) increases or decreases. Hence, if the values of f(x) are increasing as the turning point is approached, and de- creasing after it is passed, as at points A and E, Fig. 10, the r p^~\ E A V I \yT B 7 > Si A c JE« \ K Fig. 10. values of fix) pass from positive to negative, and the derived curve passes from the upper to the lower side of the axis, as shown at points A' and E'. At the turning point C, on the other hand, f(x) changes from a decreasing to an increasing function and f'(x) passes from negative to positive values ; that is, the derived curve passes from the lower to the upper side of the X-axis at the point C. Where the derived curve has turning points, as at B' and D\ the curve y = f(x) has points of inflection, as B and D. See Art. 88. For those functions that are to be considered in the present volume the derived curve is, in general, a continuous curve. We shall meet, however, two exceptional cases in which it becomes discontinuous. Art. 35] DERIVED CURVES 51 1. When the tangent to the curve yz=f(x) becomes perpen- dicular to the X-axis, as at M, Fig. 11. Here the value of /'(a?) becomes infinite and the derived curve m has infinite branches. 2. When the curve y = f(x) has an angular point, as at N, Fig. 12. At such a point, the limiting position of the tangent as the point of tangency approaches N from the left is different Fig. 11. Fig. 12. from the limiting position as the point of tangency approaches N from the right. Hence, the value of f'(x) takes a sudden jump, and the derived curve n has a discontinuity. The ordinates of the derived curve are the successive values of D x y of the original curve. Since D x y measures the slope of the tangent to the original curve, it follows that for any value of x the ordinate of the derived curve measures the slope of the original curve. It is evident that the slope of the curve y=f(x) is independent of the position of the axes so long as they remain parallel to their original position. We can shift the curve in the direction of the F-axis, Fig. 8, and the values of f'(x) will remain unchanged. 52 ELEMENTARY APPLICATIONS OF DERIVATIVES [Chap. III. In other words, the curves y =f(x) and y =f(x) + c, where c is any constant, have the same derived curve (see Art. 24). EXERCISES Draw the graphs of the following functions and the derived curve for each of them. Study carefully the combined graphs, and trace out the connection between the slope of the original curve and the ordinate of the derived curve. 1. V = — r- X. 2. y = x* - 2 x 2 ■+ 5. 9 12 J 3. y* = 8x-10. 4 - ocy = 25. 5. y = 3 + v^-4. 6. p = 8 + (as - 4)$. 7. Draw several curves, and by observing the variation in the slope and the turning points sketch in approximately the derived curves. 8. Draw a curve at random, and by observing the variation of the ordi- nate draw roughly the curve of which the first curve is the derived curve. 36. Rolle's theorem. The following proposition, known as Rolle's theorem, is essential in the development of certain other useful theorems. Theorem. If f(x) and f'( x ) are single-valued and continuous for all values of x from x = a, to x = b, and iff(a) =f(b) = 0, then f'(x) vanishes for at least one value of x between a and b* Either f(x) has a constant value zero for all values of x between a and b, or it varies with x. In the first case f'(x) is zero for all values of x. In the second case, since f(a)=f(b) = 0, f{x) must at some point begin to increase and afterwards decrease, or vice versa. It must then have a turning point for some partic- ular value ot x, say x = x u lying between a and b, since, by hy- pothesis, f(x) is continuous. Geometrically, Rolle's theorem means that if a continuous curve cuts the X-axis in two points x = a, x = b } and has a definite direction at every point in this interval (a, b), then at some * Rolle's theorem is stated here in sufficiently general terms for our present purposes. The proof given holds for the theorem as stated. For a more general statement of the theorem and its proof, see Pierpont's Theory of Functions, Vol. I, p. 246. Arts. 36, 37] LAW OF THE MEAN 53 intervening point, say x = x x , the tangent to the curve is parallel to the X-axis (Fig. 13). It is at once evident that, instead of the con- dition f(a)=f(b) = 0, we might have f(a) and f(b) equal to any con- stant so long as they are equal to each other, remains precisely the same in the two cases. +-X The argument EXERCISES 1. By Rolle's theorem show that at least one real root of the equation /' (x) = lies between any two real roots of the equation / (x) = 0. 2. From Ex. 1 show that if two roots of f(x) = are equal, one root of /' (x) = coincides with them. Give a geometric illustration of this statement. 37. Law of the mean. By the use of Rolle's theorem, we may deduce one of the most fundamental theorems of the differential calculus, known as the law of the mean, or the theorem of mean value. Theorem. Let f(x) and f'(x) be single-valued and continuous functions of x in the interval a)-/(«) _ /(a . o) = o, (3) o — a whence /( *% ~{ ( ^ = /W, (4) as the theorem requires. The theorem may be stated in another form which is sometimes convenient. The fact that x lies between a and b can be ex- pressed by the relation x = a + 0(b — a), where is some number between and 1. We may also put b — a = h. Equation (4) then takes the form f(a + h)-f(a) =f(a + Qh ^ (5) ft/ The fraction ^ ' ~~J^ a ) evidently measures the average rate of b — a increase of the function in the interval b — a. Thus, suppose «! and s 2 denote distances traversed by a moving point in the Art. 38] EQUATION OF THE TANGENT 55 times t x and t 2 respectively ; the fraction zSHh gives the average t 2 £1 speed of the point for the time interval t 2 — t x , and the law of the mean asserts that at some instant within this interval the actual instantaneous speed of the point is equal to the average or mean speed for the whole interval. EXERCISES 1. Show geometrically that Kolle's theorem is a special case of the law of the mean. 2. If f(x) = x 2 , find the value of 6 that satisfies relation (5) : (a) when a = 6, h = 1 ; (6) when a = 12, h = 4. 3. If f(x) = x 3 , find the value of 6 in order that (5) shall be satisfied when a = 4, h = 1. 4. A smooth curve is passed through three points (6, 3), (8, 5), and (10, 8.5), which have been determined experimentally, and the slope of the curve at the intermediate point (8, 5) is desired. Find by the law of the mean an approximate value for this slope. 38, Equation of the tangent and of the normal to a curve. Let (x 1} 2/j) be any point on the curve whose equation is y =/(*)• As we have seen, the slope of the tangent to the curve at the point (x l9 ?/i) is given by substituting the value as, for x in the derived function /'(a?); that is, for x = x x , t&n cf>=f'(x 1 ). From analytic geometry, the equation of the tangent at any point ( x d Vi) on * ne curve is V-Vi = rn(x - x x ), where m denotes the slope of the tangent at the point in question. Replacing m by f'( x i)> we have for the equation of the tangent to the curve in terms of the derivative y~y 1 =/ / (aJ 1 )(«-aJ 1 ). (1) The normal to a curve is perpendicular to the tangent to the curve at the point of tangency. The condition that one straight line shall be perpendicular to another is that the slope of the one shall be the negative reciprocal of that of the other. It follows 56 ELEMENTARY APPLICATIONS OF DERIVATIVES [Chap. III. that the equation of the normal to the given curve at the point (»i, 2/i) is *-**-- rbtf*' m * (2) Because of the relation *"ii'm' (Art32) the slope of the normal can frequently be most conveniently obtained by substituting y = y 1 in D v x. Ex. Find the equations of the tangent and normal to the curve y = xi at the point (4, 8). We have /'(») = !«*> and for* = 4, /'(*) = ! x 2 = 3. Substituting this value of /'(as) in (1), the equation of the tangent is y-8 = 3(a:-4), or 3 x — y = 4. At the given point, — = -. Hence, the equation of the normal is f'(xi) 3 or x + 3 y = 28. EXERCISES Find the equations of the tangent and normal to the following curves at the points indicated. 1. y* = 3x + 10, at (-2, 4). 2. xy = 36, at (4, 9). , at (a, a). 2a — x 4. (x - 4)2 + (y + 3)2 = 100, at (10, 5). 5. — + -^=1, at (2, 4\/2l). 25 16 V 3 J 6- y = x* - 3 x + 10, at (2, 12). 7. y = a .i_I, at(2,8|). Find the equations of the tangent and normal to the following curves at the point (xi, yi). 8. y 8 = ax*. 9. 05$ + ?/* = a*. 10. x n + y n = a n . 11. ^--£- = 1. a 2 6 2 Arts. 38, 39] LENGTH OF TANGENT, ETC. 57 39. Lengths of tangent, normal ; subtangent, subnormal. The length of the part of the tangent to a curve that lies between the point of tangency and the axis of abscissas is called the length of the tangent. The length of that part of the normal intercepted between the same point on the curve and the axis of abscissas is called the length of the normal. The projections of these lengths upon the axis of abscissas are called respectively the subtangent and the subnormal. Thus in Fig. 15 let the tangent to the curve at the point P, hav- ing the coordinates x 1} y lf meet the X-axis in the point A, and let the normal at the point P meet it at C. Let BP be a perpendicular let fall from P upon the X-axis. Then for the point P, PA is the length of the tangent, PC is the length of the normal, and AB and BC are the subtangent and subnormal respectively. In the triangles APB and BPC the side BP = y x is given, as is also the angle PAB — CPB = . If the equation of the curve is y = f(x), tan , subtangent = y x cot = Jf 1 * Ex. Find the length of the normal to the curve y 2 = — at the point (4. 4). 4 i From Fig. 15, the required length is y x sec = |, sec = Vl + (f ) 2 , and the length of the normal is 4Vl + (|) 2 = 2 Vl3. 58 ELEMENTARY APPLICATIONS OF DERIVATIVES [Chap. III. EXERCISES 1. Find the length of the tangent, the subtangent, and the subnormal for x H the curve y 2 = — . 4 2. Derive general formulas for the length of the tangent, length of the normal, subtangent, and subnormal in terms of y x and/'(xi). Find the subtangent, subnormal, length of the tangent, and length of the normal for each of the following curves at the points indicated. 3. y 2 =8 *, at (2, 4). 4. xy = a 2 , at (x tl y{). 5. (x - 4)2 + (y + 3)2 = 25, at (7, 1) 6. y = x* - 3 x + 10, at (xi, yi). 7. y% — ax^ a t (« } a ). « r 3 2a — x 9. x ff + y* = o«, at (xi, jfi). 10. Show that the subnormal to the parabola y 2 = 2 mx is. constant. 11. Show that the length of the normal to the curve x 2 + y 2 = a 2 is constant. 12. Using the equations of the tangent and the normal in Art. 38, derive formulas for the length of the tangent, length of the normal, etc., by- finding the intercepts of these lines on the X-axis. 40. Tan ty, cot ^. Let p, be the polar coordinates of any point on a curve, being taken as the independent variable. In order to determine the di- rection of the curve at any point, it is convenient to express the tangent and cotangent of the angle if/ between the radius vector and the tangent to the curve in terms of p, 6, and. their derivatives. In Fig. 16, let P be any point on the given curve, If takes an increment A0, Let P' be the Fig. 16. and let its polar coordinates be p, 0. then p will have a corresponding increment Ap. Art. 40] TAN $, COT + 59 point having the coordinates -f- A0, p 4- Ap, and let MP be drawn perpendicular to OP'. From the figure, we have MP = p sin A0, OM = P cos AO, and MP' = OP' — p cos A0 = p + Ap — p cos A<9. Hence, tan 3fP'P = psinA^ p + Ap — p cos A0 As A0 = 0, the angle MP'P approaches the angle if/. Hence we have L tan MP'P= L n P^M A* = A0 = Ap + p(l - cos A0)' sin Afl or tan ^ = JJ r A0 = Ap / l - cos A(9 A0 P \ AO But we have seen that hence, we have tan* = -^ = pD p 0. (1) From (1) it follows at once that col!^ l_ = 3l>p. (2) T taw \|/ p * Ex. Let the equation of a curve be p 2 = - . Find the angle \p for p = 1. Writing the given equation in the form = ^ = ap-2, P 2 we find D p = - 2 ap" 3 . Hence, tan $ = p D P e = -2 ap~ 2 = - 2 - , P 2 and cot 4> = — ■£— - Y 2a Forp = l, ta,n\(/= — 2a, whence f = arc tan (-2 a). 60 ELEMENTARY APPLICATIONS OF DERIVATIVES [Chap. III. 41. Length of polar tangent and polar normal ; polar subtangent, polar subnormal. Given a curve and the polar coordinates (p l7 2 ) of any point P upon it (Fig. 17). At the point P, draw the tangent and the normal to the curve and through the pole draw a perpendicular to the radius vector OP. Let this perpendicular cut the tan- gent and the normal in the points T and N respectively. PN is called the length of the polar normal, and PT the length of the polar tangent. The projections of these lines upon the perpendicular, namely, NO and OT, are called the polar sub- normal, and polar subtangent, respectively. It will be observed that in the case of rectangular coordinates, the line upon which the tangent and normal were projected to determine the subtangent and subnormal was the X-axis, while in polar coordinates, it is not the initial line but a line through the pole perpendicular to the radius vector. In each of the right triangles OPT and OPN one angle if/ and one side OP=p x are given. Since, from the preceding article, ij/— arc tan^P'^), where P'(0i) is obtained from the equation of the curve p = F(6), it follows that the four lengths in question can be expressed in terms of p x and P'(#i). a 2 Ex. Find the lengths of the polar tangent of the curve p 2 = — ■ Putting the equation in the form p = ad~ 2 , we get J) fl p=-lae *; hence, F'(d l )= 2L-. 2dfi Therefore cotxl/ 1 « 6\t a 1 and tarn// =- 2 0i. From Fig. 17, length of polar tangent = pi sec \f/ = — Vl + 4 0i 2 . Arts. 41, 42] SPEED AND ACCELERATION 61 EXERCISES 1. Derive general formulas for the lengths of the polar tangent and polar normal, the polar subtangent, the polar subnormal, in terms of pi and F'(d{). 2. Find the polar subtangent and polar subnormal of the curve p 2 = — • 8 3. Find the length of the polar normal of the same curve. Find general expressions for the length of the tangent, subtangent, length of the normal, and subnormal of the following curves : 4. P = ad. 5. p = ad 2 . 6. p = 1 + -. 7. p 2 = ad + bd 2 . d 8. Find tan \f/ for the curves of Exs. 4-7. 9. Show that the polar subtangent to the curve pd = a is of constant length. Trace the curve. 42. Speed and acceleration. Let a point move in a given path, straight or curved, and let s denote the variable distance of the point measured along the path from some fixed origin 0. Assum- ing continuous motion from 0, the distance s evidently depends upon the time t ; that is, s = f(t). In a time interval At the point As moves over an element of path of length As, and the ratio — gives the mean speed of the point for this interval. The limiting value of this quotient as the time interval A£ approaches zero gives the instantaneous speed at the beginning of the interval. Denoting this by v, we have therefore v= L ^ = D t s; (1) that is, the speed of a moving point is the time-derivative of the space traversed. The words "speed" and "velocity" are often used as synonyms. An important distinction is, however, frequently made. The term "speed" is used to indicate merely the rate of motion in the path irrespective of direc- tion. Speed, therefore, has only magnitude. The term "velocity" carries with it the additional notion of direction, hence to specify a velocity both magnitude and direction are required. The speed v is, in general, a function of the time t. If Av denotes the change of speed between two points, then the quotient 62 ELEMENTARY APPLICATIONS OF DERIVATIVES [Chap. III. — defines the average tangential acceleration* between those At points ; and the limit of this quotient as At = defines the instan- taneous tangential acceleration. Denoting this by a, we have a= L %-Dp\ (2) that is, the tangential acceleration is the time-derivative of the speed. The ordinary unit of speed is the foot per second, that of accel- eration the foot per second per second. These are abbreviated to ft./sec. and ft./sec. 2 respectively. Ex. In a certain motion the space described is expressed as a function of the time by the following equation : s = at' 2 + bt + c. For the speed, we have v = D t s = 2 at + 6, and for the tangential acceleration, we obtain a — B{0 — 2 a. 43. Angular speed and acceleration. If a body rotates about a fixed axis, any given point of it, not on the axis, moves in a circle whose center lies on this axis, and whose plane is perpendicu- lar to the axis. Let be the center and PAB the circular path of the point, Fig. 18. Dur- ing the motion of the point from P to A, the radius OA sweeps over an angle 6, and in the ad- FlG- 18, ditional interval of time At re- quired for the motion from A to B it sweeps over the angle AOB = A8. The ratio — defines the mean angular speed of the body between the positions OA and OB, and the limit of this * Tangential acceleration is the acceleration in the direction of the motion, that is, tangent to the path. In the case of rectilinear motion, this is the only acceleration ; but in the case of a point moving in a curve, there is another acceleration perpendicular to the tangent. Art. 43] ANGULAR SPEED AND ACCELERATION 63 ratio as B is made to approach A defines the instantaneous angular speed for the position OA of the radius. Denoting this angular speed by a>, we have « o> = L ~ = D t 0; (1) At = A* that is, the angular speed is the time-derivative of the angle swept over. The angular speed may be constant or variable. If variable, the increment between two positions, say A and B, may be de- noted by Aw, and the ratio — — gives the mean angular acceleration between the positions in question. The limit of this mean value as the two chosen positions are made to approach each other is the instantaneous angular acceleration, which is denoted by a. We have then a= L 77-A-3 ( 2 ) At = At that is, the angular acceleration is the time-derivative of the angular speed. With angles measured in radians, the unit of angular speed is the radian per second (rad./sec), and that of angular acceleration is the radian per second per second (rad./sec. 2 ). The relation between the angular speed of a rotating body and the linear speed of any point of the body in its circular path is readily derived. Eeferring to Fig. 18, AB is an element As of the circular path of P. Denoting the radius OA by r, we have therefore As = rAO, whence — = r — , At AT and finally D t s = rD t 6. (3) That is, the speed of any point of a body rotating about a fixed axis is the product of the distance of the point from the axis and the angu- lar speed of the body. 64 ELEMENTARY APPLICATIONS OF DERIVATIVES [Chap. III. A similar law holds for tangential acceleration. Thus from (3) v = rut j whence D t v = rD t io, or a = ra ; (4) that is, the tangential acceleration of the 'point is the angular accel- eration about the fixed axis multiplied by the distance from the point to the axis. EXERCISES 1. If a body is projected vertically upward with a speed of v feet per second, the space traversed in t seconds from the instant of projection is given by the equation 8 = vtf - I gt 2 . (a) Find expressions for the speed and acceleration at the time t\. (b) Take v = 800 and g = 32.2, and find the speed and acceleration at the end of 2 seconds ; (c) at the end of 12 seconds. 2. In Ex. 1, find the whole time occupied by the body rising and falling, also the height to which it rises. Suggestion : Make s = and solve for t. 3. When a body moves in a straight line under the influence of an attractive force that varies as the inverse square of the distance, the motion is given by the equation v 2 = -, in which A: is a constant. Show that the acceleration is — . 2 s 2 4. The angle (in radians) through which a given rotating body turns, start- ing from rest, is given by the equation = t 2 + 5 t. Find the angular speed and angular acceleration at the end of 4 seconds. 5. If the angle is given by the equation d=\\2t- 16*2, find (a) the speed and acceleration at the end of 2.5 seconds ; (6) the time that elapses before the body comes to rest. Derive expressions for w and a from each of the following relations between B and t : 6. 6 = at- bfi. 7. e = aA 8. 6 = a + bt + cfi. 44. Miscellaneous applications. In physics and in chemistry the notion of the derivative is repeatedly encountered. In Art. 44] MISCELLANEOUS APPLICATIONS 65 the case of a moving point, the term " speed " is used for the time-derivative of the distance traversed. In chemistry, likewise, we have the same term used for other time-derivatives ; thus, the speed of reaction and the speed of solution are such derivatives. In physics we meet with a great number of derivatives expressing the rates of change of various physical magnitudes. The heating of substances, variations of pressure, density, and temperature, the variations in velocity and in energy, etc., are such changes. A few of these derivatives are discussed in the following paragraphs. (a) Coefficients of expansion. Let a rod or wire have unit length (1 foot or 1 yard) at some standard temperature, say 32° F. or 0° C. When heated the rod expands by an amount x depending upon the temperature. Thus, denoting the temperature by t , the expansion is x — /(r ) and the new length of the rod is 1 + s = 1+/(t). (1) If now the temperature rises by an amount At, the length of the rod will increase by a corresponding amount Ax. The quotient — is called the At average coefficient of linear expansion for the interval At, and its limit as At is made to approach zero is the coefficient of linear expansion for the tempera- ture t. Denoting this by (7i, we have d= L ^-=D r x. (2) At = 0At In most cases we may assume with sufficient approximation x -/(t) = ar + br 2 , whence (J\ = a + 2 br ; (3) that is, C\ is itself a function of the temperature. In the same way we may arrive at the coefficients of superficial and cubical expansion. Consider a cube with its edge having unit length at 0° C. Assuming equal expansion in all directions, each edge at the temperature r will have a length 1 +/(t). Hence the area of a face will be ^ = [1+/(t)] 2 , (4) and the volume of the cube will be V={l+f{r)y. (5) The derivatives D T A and D T V are the coefficients of superficial and cubical expansion respectively, and may be denoted by C 2 and C 3 . (6) Specific heat. A body, originally at some definite temperature t , is heated and its temperature rises. The heat Q absorbed by the body is in general a function of the final temperature r ; that is, Q=f(r). (6) 66 ELEMENTARY APPLICATIONS OF DERIVATIVES [Chap. III. An increment of heat AQ causes a corresponding rise in temperature At, and the quotient — * is called the average specific heat of the body for the inter- At val At. The limit of this quotient as At approaches zero is defined as the specific heat at the temperature r. Denoting this by c, we have c= L ^ = D T Q. (7) At=0 At EXERCISES 1. Regnault's experiments on the heating of various substances are repre- sented by the following equations : Ether, Q = 0. 5290 t + 0. 000296 t 2 . Chloroform, Q = 0.2:324 r + 0.00005 t 2 . Bisulphide of carbon, Q = 0.2352 t + 0.000082 t 2 . In each case Q denotes the heat required to raise the temperature of the sub- stance from 0° C. to t° C. For each substance find the specific heat at 20° C. 2. It is found by experiment that the volume of water which at 4° C.has unit volume is given by the equation V = 1 + a (t - 4)2, where r denotes the temperature of the water and a = 0.00000838. Find the coefficient of cubical expansion when r = 0° ; also for r = 20°. 3. The electrical resistance B of a wire varies with the temperature r of the wire, the relation being expressed by B =/(f). (a) What is expressed by the derivative D T B ? (6) Find the expression for this derivative from Callendar's formula in which B , a, and /3 are constants. 4. (a) What derivative expresses the rate of the rise of temperature of a gas with respect to the pressure ? (6) With respect to the volume ? (c) What derivative gives the rate of change of the energy IT of the gas with respect to the temperature r ? 5. The pressure of the atmosphere decreases as the distance from the earth's surface increases, and the rate of change of pressure with the height is proportional to the pressure. State this law in the form of an equation. MISCELLANEOUS EXERCISES 1. The equation of a curve is 4 a 2 y = x 3 — 2 ax 2 + a 3 , (a) Find the slope f or x = and x = a. (6) Find the points where the curve is parallel to the X-axis, (c) Find the points at which the slope of the curve is 1. Art. 44] MISCELLANEOUS EXERCISES 67 2. Find the angle at which the circle x 2 -f y 2 = 8 x intersects the curve 9 2-x 3. Find the equation of the tangent to the hyperbola xy = 30 that cuts the X-axis in the point x = 4. Show that the point of tangency of any tangent to this curve lies midway between the intersections of the tan- gent with the X- and F-axes. Investigate the following curves by means of their derivatives : 4. „ = *(«_*)*. s. p = T f^- 6. ? = f^f • 7. Draw curves showing the speed v of a body falling in vacuo (s = \ gt), (a) with values of t as abscissas, (6) with values of s as abscissas. Show that the subnormal of the second curve is numerically equal to the acceler- ation. 8. In the case of the semicubical parabola ay 2 = x s show that 27 subnormal = (subtangent) 2 ■ ■ 8 a 9. (a) Find the equation of a curve whose polar subnormal is constant. (6) Find a curve whose polar subtangent is constant. 10. Show that Rolle's theorem does not hold for f(x) = (x — 1)* — 1 between x = and x = 2. Explain why. 11. Apply Eq. (5), Art. 37, to the function f(x) = X s - 6 x - 8. Find the value of d for a = 4, h =s 1 j also for a = 7, h = 2. 12. Show that for every quadratic function f(x) = ax 2 + bx + c, Eq. (5), Art. 37, is satisfied when = \. From this fact deduce a geometric property of the parabola. 13. If Q denotes quantity of heat, r temperature, s length or distance, and t time, express in words what is meant by the equation D t Q=-k-D s T, which applies to the flow of heat along a bar. 14. Show that the coefficients of superficial and cubical expansion are re- spectively two and three times the coefficient of linear expansion, very nearly. 15. Let m denote the mass of a body moving in a straight line, v the speed, and a the acceleration of the body, F the constant force acting on it, and s the distance traversed. From mechanics we have the following defi- nitions and symbols : mv = momentum of body, (M), ma = force acting, ( F) , Fs = work of force F, ( W). Show that F = D t (mv) = D t M, and also that F = D S W. 68 ELEMENTARY APPLICATIONS OF DERIVATIVES [Chap. III. 16. The kinetic energy of the body (Ex. 15) is given by the expression T=\ mv 2 . (a) Show that the momentum is the v-derivative of the kinetic energy. (&) Show that the time-derivative of the kinetic energy is the product Fv. 17. If for any motion a curve is drawn with the speed v as ordinates and the distances s as abscissas, show that the tangential acceleration for any value of s is given by the subnormal to the curve at that point. 18. The equation of van der Waals, (*+$) («-»)=<* gives the isothermal curves of carbon dioxide. By means of the derivative in- vestigate the general form of the isothermal curves obtained by giving C dif- ferent constant values. 19. The following values of corresponding pressures and temperatures of saturated steam are taken from a standard table : p lb./sq. in. 84 85 86 87 88 t temperature F. 315.19° 316.02° 316.84° 317.65° 318.45° Making use of the law of the mean, find approximately the value of the de- rivative D p t forp = 85 ; also for p = 87. Take 6 = 0.5. 20. Rankine's formula for long columns has the form a y 1 + 6x 2 By means of the derivative investigate the general character of this function, and sketch the curve that represents it. CHAPTER IV THE DIFFERENTIAL NOTATION 45. The derivative as a rate. The fundamental problem of differential calculus is the measurement of the rate of change of the function with respect to the variable. We are not concerned so much with the actual change of the function as with its change per unit increase of the variable, or, in other words, its rate of change. The great importance of the derivative lies in the fact that it gives a precise measure of this rate of change. Thus the derivative D t s measures the rate of change of the distance s with respect to the time, or, more briefly stated, the time-rate of s ; the derivative D t 0, the angular speed, is likewise the time-rate of the angle 0. Again, consider the case of heating a metal bar. The length x of the bar is a function of its temperature t, say x =/(t), and the derivative D r x gives the rate of change of length with respect to the temperature. That the derivative measures the rate of change of the function is easily seen. If Ay denotes the change of the function corre- sponding to a change Ax of the variable, the quotient Ay = f(x + Ax)-f(x) Ax Ax gives the average rate of change for the interval Ax. The rate at which y is changing with respect to x at any particular value of x, as x , is the limiting value of this quotient as Ax approaches zero, that is, It is not necessary that y be expressed directly in terms of x in order that we may discuss their comparative rates of change. In fact, it is often convenient to compare the rates of change of two functions by means of a third variable. For example, we may 69 i 70 THE DIFFERENTIAL NOTATION [Chap. IV. compare the relative rate of change in the market value of wheat in Chicago and steel rails in New York by the use of a third vari- able representing money. The speed of a locomotive may be compared with that of a street car, or the relative speed of two chemical reactions may be found by means of a third variable rep- resenting time. The conception of the derivative as a rate is fundamental in the developments of the present chapter. 46. Differentials. In the preceding chapters we have employed the notation of derivatives explained in Art. 16. We shall now introduce another nota- tion, that of differentials, which for certain pur- poses is more convenient than that of derivatives. Let us consider first the case in which y is ex- pressed directly in terms of x, and let y =f(x) be a continuous function rep- resented by the curve in Fig. 19. At any point P of the curve let a tangent be drawn, and let PS rep- resent an assumed increment Ax of the independent variable x. The corresponding increment Ay of the function is represented by SQ, while the intercept SM between PS and the tangent represents the product Ax • tan = f\x) Ax. Denoting this product by dy, we have therefore dy=f(x)Ax. (1) In order to make the notation symmetrical, it is customary to denote the increment of a; by dx; that is, to write dx for Ax. With this convention (1) becomes dy=f'(x)dx. (2) In equation (2) dx is called the differential of x, dy the differential of y, and f\x) dx the differential of f(x). We have therefore the Fig. 19. Art. 46] DIFFERENTIALS 71 following definitions : The differential of the variable x is merely an assumed increment of x ; and the differential of the function y is the product of the derivative f'(x) of the function and the dif- ferential dx of the variable. Frequently f(x) is called the differential coefficient, that is, the coefficient of the differential of the independent variable. Having assigned a value to dx (= Ax), the values of dy and Ay are determined by the nature of the functional relation y=f(x). Usually these two symbols do not represent the same value. The value of dy is determined by f'(x) dx, that is, by dx • tan cf>, and is represented in the figure by SB. On the other hand, Ay is the change in the function f(x) corresponding to the change Ax of the independent variable, that is, SQ in the figure. The two symbols dy and Ay represent the same value when, and only when, y =f(x) is represented by a straight line. It should be noted that dy is the increment that the ordinate would have if the rate of change of y at the point P were main- tained throughout the interval Ax. It is evident also that if Ax is sufficiently small dy is an approximation to the increment Ay, and that the smaller Ax is chosen, the closer is the approximation. The relation (2) between dy and dx evidently holds when x and y are given as functions of a third variable t. The parametric representation has the advantage that it enables us to show clearly the relation between differentials and rates as measured in terms of a common variable. Suppose we have given «=*(*), 2/ = (t). (4) Hence by Art. 31, we have D t y=f'(x)D t x. (5) As we have seen, the derivatives D t x, D t y express the rates of change of the variables x and y with respect to t, the common variable in terms of which both x and y are expressed. It will be observed that D t x, D t y enter homogeneously into equation (5) 72 THE DIFFERENTIAL NOTATION [Chap. IV. as do dx, dy in equation (2). Moreover, by a comparison of equa- tions (2) and (5), we have J W cto D t x J We may then say that the rates D t x, D t y are either equal to the differentials dx, dy or proportional to them. This relation may be, and often is, made the basis of the definition of differentials ; and it enables us to write the derivatives D t x, D t y in terms of the differentials, or vice versa, whenever it suits our purpose to do so. Since we have ^ it follows that a derivative is equal to the quotient of two differen- tials. It is frequently convenient to write the derivative as such a quotient, and in the future we shall employ either D x y or -2 as dx best suits the problem under discussion. We may read -*- either ax " the derivative of y with respect to »," or " differential y divided by differential xP The distinction between the symbols D x y and -^ should, how- ax ever, be carefully noted. While both lead to the same numerical result and may therefore be used interchangeably, D x y indicates that a certain operation has been performed upon the function ?/ with respect to the independent variable x. It is not possible to separate this symbol into two parts. On the other hand, -* is ax merely a quotient which can be dealt with by the ordinary rules of algebra. 47. Kinematic interpretation of differentials. An instructive illustration of the use of differentials is afforded by the velocity components of a moving point. Suppose a point P to move along a curve m, Fig. 20, and let P x and P y be the projections of P on the X-axis and F-axis respec- tively. As P moves on the curve, the projections P x and P y move on the axes. The velocity of P x along OX is evidently the time- Art. 47] KINEMATIC INTERPRETATION OF DIFFERENTIALS 73 rate of change of the abscissa x, and is therefore given by the derivative D t x. Likewise, the velocity of P y along the F-axis is D t y. The velocity of P along the curve has the direction of the tangent PT, and its magnitude is given by the derivative D c s. Suppose that the velocity of P is such that in a unit of time P would be carried along the tangent from P to T. This displacement PT can be effected in the following manner : Move P along a horizontal line from P to A, and at the same time move the horizontal line PA vertically to BT. Evidently, if the horizontal and vertical motions are made with constant velocity, the point P will move along PT with constant velocity. Since the motion takes place in a unit of time, the displacements PT, PA, and PB respresent respectively the ve- locity in the path, the velocity of the horizontal motion of P x , and the velocity of the vertical motion of P y \ that is, PT=D t s, PA = D t x, and PB = D t y. The two velocities D t x and D t y, which together may replace the velocity D t s in the curve, are called the component velocities in the direction of the axes. From the figure the following relations between the velocity in the curve and the components along the axes are evident : PT 2 = PA 2 + PB 2 , hence (D t sy = (D t xf + (D t y)\ (1) Substituting the differentials ds, dx, and dy for the time-rates, we have Furthermore, since and we have ds = doc + dy PA = PT cos <£, PB = PTsm, dx = ds cos dy = ds sin <£ (2) (3) Ex. 1. A point moves in the parabola y' 2 = 8 x with a constant velocity of 5 units. Find the components along the axes when the point is at (12.5, 10). 74 THE DIFFERENTIAL NOTATION [Chap. IV. From the given equation, we get 2 yDt y = 8 D t x, or 2 y dy = 8 <£r, whence dy=- dx. But ds = Vdx 2 + dy 2 = 5, whence dx\— +1 = 5. * y' 2 We have then denotes the angular speed of OP about the pole 0, PB=OP = 0, whence dy = _Sx 2 y - 4y 2 + 12x. dx x 3 — 8 xy — 3 78 THE DIFFERENTIAL NOTATION [Chap. IV. The values x = 1, y = 1 satisfy the original equation, and for these values the derivative has the value §£} -11- dx. ■1 =11 = 1-1; J i,i io that is, the slope of the curve is 1.1 at the point (1, 1). Ex. 2. The equation representing the adiabatic expansion of air {i.e. ex- pansion without gain or loss of heat) is pv k = C, where k and C are constants. p denotes the pressure and v the volume of the air. Find the rate of change of pressure relative to the volume. Differentiating, we obtain v k dp + kp&~ x dv = 0, or vdp + kp dv = 0, whence -j-=—k-> dv v EXERCISES dv Find the derivative ~ for each of the following functions : 1. x s - xy 2 - c 3 = 0. 2. b 2 x 2 - a 2 y 2 = a 2 b 2 . 3. x 2 -f- y% = c. 4. x 4 — x*y + x 2 y 2 + 2 xy* = 0. 5. Find the general expression for the slope of the tangent to the conic section whose equation is ax 2 + 2 hxy + by 2 + 2 gx + 2 fy -f c = 0. 6. Find the slope of the circle x 2 + y 2 = 100 at the point ( — 8, 6). 7. Find the slope of the ellipse 9 x 2 + 14 y 2 = 50 at the point (2, 1) ; at the point ( — 2, 1) . 8. Find the slope of the curve x*y - 5 x 2 y 2 + 12y* = at the point (2, 1). 9. Derive expressions for the polar sub tangent and polar subnormal of the curves : (a) P 2 -pd^=C. (b) P -p 2 d+ C = 0. 10. Derive general expressions for the derivative =5 when the expansion of a gas follows the following laws, respectively : v (a) p m v n s C, (w, w, and C constants) ; . Art. 50] APPLICATIONS OF DIFFERENTIALS 79 Find 5* in the following : 11. (1 + x) 2 y = (1 - x)y* - «». 12. a^a = 62(6_.a;). 13. ccVy — yy/x = C. 50. Applications of differentials. Suppose we have given y =/(*), and by differentiation we obtain dy=f(x)dx. (1) The differentials cfy and d# may be taken as representing the deriva- tives D t y and D t x, where t denotes any third variable. If we con- sider the variable t as denoting time, then dy and dx represent the time rates of the function y and of the variable x respectively. Hence, if the time rate of x is given, the time rate of y is found by differentiation. Ex. 1. Boyle's law for the expansion of air is expressed by the equation pv — C. At a given instant the pressure is 40 lb./sq. in., the volume of the air is 8 cu. ft., and the volume is increasing at the rate of 0.5 cu. ft./sec. At what rate is the pressure changing? We have pv = C, whence p dv + v dp = 0, or dp = — ® dv. v Since dv = D t v = 0.5, dp=- — x 0.5 =-2.5. 8 Hence, the pressure is decreasing at the rate of 2.5 lb./sq. in. per second. If in equation (1) the differentials are replaced by the incre- ments Ay and Ax, the result is an approximate relation Ay=f'(x)Ax, (2) which is frequently useful in finding the error in the result of a computation due to a small error in the observed data upon which the computation is based. The relative error — ^ is given approxi- mately by the equation * *L = fM± x . (3) y /(*) 80 THE DIFFERENTIAL NOTATION [Chap. IV. Ex. 2. The area A of a circle being determined from a measurement of the diameter D, find the relative error in the calculated area due to an error in the measurement. Since A = \ ttZ)2, we have dA = a irD ■ dD, whence approximately AA =iwD- AD, and AA = l I D AD = 2 Ad A lirD 2 d Hence an error of one per cent in the measurement of the diameter gives approximately an error of two per cent in the calculated area. EXERCISES 1. A point moves in the straight line 5x — 3 y = 30 in such a way that the r-component of its velocity is 6. Find the X-component and the velocity in the path. 2. Find the X- and F-components of the velocity of a point that moves in the line 3 x + 4 y = 12 with a speed of 15 units. 3. Suppose that a straight wire rotates about one end with an angular speed of 3 rad./sec. and that a bead on this wire moves along it with a speed of 8 ft./sec. When the bead is 2 feet from the center of rotation, (a) what is the component of its velocity perpendicular to the wire ? (b) what is its velocity in its path ? 4. Find the polar equation of the curve described by the bead, Ex. 3, (a) when the angular speed w of the wire is a times the linear speed of the bead along the wire ; (6) when « is constant and the speed of the bead on the wire varies inversely as the distance from the center of rotation. 5. If a soap bubble's diameter is increasing at the rate of \ in./sec. when it is 4 inches, at what rate is the inclosed volume increasing ? 6. If a man 6 feet in height is walking at the rate of 3 mi./hr. away from a lamp-post 30 feet high, at what rate is the end of his shadow receding from the lamp-post ? 7. The time of a complete oscillation of a pendulum of length L is given by the formula rr * = 2ttV-. y g Find the rate of change of the time compared with that of the pendulum's length. 8. Find the error in the calculated time if the error in the measurement of the length of the pendulum is 0.5 per cent. 9. Find an expression for the relative error in the volume of a sphere calculated from a measurement of the diameter if there is an error in the measurement. Art. 50] MISCELLANEOUS EXERCISES 81 10. From the formula for kinetic energy T = \ ra» 2 , show that a small change in v involves approximately twice as great a relative change in T. 11. An engine cylinder has a diameter of 12 inches. At what speed is the piston moving when steam is entering the cylinder at the rate of 18 cu. ft./ sec. ? MISCELLANEOUS EXERCISES Differentiate the following functions, using differentials. 1 2 x 2 + 1 ^ ^ 2 Vl+g'+V l 3 x» x 3. 5aji(a^-4)$(x + 7)i. 4. Va — Va — x for the functions defined by the fo] dx Find -^ for the functions defined by the following equations. 5. x s - 5 ax 2 y + 7 y s = 0. 6. x 2 - y 2 = Vl - y 2 7. -**-=<, 8. 1 = *. » + y Vx 2 + y 2 9. Find expressions for the velocity components of a point moving in the parabola y 2 = 20 x with a speed of 12 in. /sec. Find the values for the point (5, 10). 10. A point moves in the straight line 6 x — 8 y — 12 with a velocity of 5 ft. /sec. Find the components of the velocity along the X- and V-axes, respectively. 11. A point moves in the circle x 2 + y 2 = 36 with a velocity of 8 ft. /sec. Find the X- and F-components when the point is at (5, Vll). 12. A point moves along the curve p = — -. When p = 3, the component Ve of the point's velocity along the radius vector is 5 ft. /sec. Find (a) the component perpendicular to the radius vector, and (6) the velocity in the curve. 13. A crank pin moves in a circle 2 feet in diameter with a constant speed of 28 ft. /sec. When the crank makes an angle of 30° with the horizontal, what is (a) the vertical component of the crank pin's velocity ? (6) the horizontal component ? 14. The path of a projectile is the parabola x tan a — 2 vo 2 cos 2 a and the horizontal velocity (D t x) is v cos a. Show that the velocity in the path is Vvq 2 — 2 gy, and that the vertical acceleration is g. 82 THE DIFFERENTIAL NOTATION [Chap. IV. 2 2 2 15. If a point moves in the curve whose equation is X s + y* = c 3 so that the X-component of the velocity is constant, find the acceleration in the direction of the T-axis. 16. Find the equation of the path of a point which moves in such a way that the X-component of the velocity is constant while the y-component is negative and varies directly as the time. Give a physical illustration. 17. Sand or grain, when poured from a height on a level surface, forms a cone with a circular base and a constant angle jS at the vertex, dependent on the material. Let a denote the radius of the base of the cone at a given time, and suppose material is being added at the rate of C cu. ft. /sec. At what rate is the radius increasing ? 18. The velocity of a jet of liquid issuing from an orifice is given by the formula, v = V2 gh, where h is the height of the liquid surface above the orifice. If h = 100 feet and is decreasing at the rate of 0.2 ft. /sec, find the rate at which v is decreasing. Take g = 82.2. 19. Given x = Sy 2 + 7y-\-\) find D x y without first obtaining y as an explicit function of x. 20. Air expands according to the adiabatic law pv 1A = C. When the pressure is 40 lb. /sq. in., the volume is '5 cubic feet and is increasing at the rate of 0.2 cu. ft. /sec. Find the rate at which the pressure is changing. 21. The formula for the electrical resistance of a platinum wire is JB = i?o(l + ar + &r 2 ), where Bo, «, and b are constants, and r denotes the temperature of the wire. Find the rate of increase of resistance at the temperature n, if the tempera- ture is rising at the rate of 0.1° per second. 22. Another formula for the variation of electrical resistance of a metal wire with the temperature is B = Bo (1 — er +/t 2 ) -1 . Find the rate of change of the resistance compared with that of the temperature at tempera- ture T\, 23. Given (x 2 + y 2 ) 2 = a 2 (x 2 - y 2 ). (a) For what values of x has the curve representing this function a turning point!* (b) For what value of x does the tangent make an angle of 45° with the X-axis ? (c) Write the equation of the normal at x = - . 24. A hemispherical bowl 20 inches in diameter has an orifice in the bottom through which the water contained in the bowl is flowing. If, when the surface of the water is 8 inches above the bottom, the rate of flow is 16 cu. in. / sec, at what rate is the water level falling, assuming that the surface remains plane ? CHAPTER V DIFFERENTIATION OF TRANSCENDENTAL FUNCTIONS 51. Transcendental functions. All functions that are not alge- braic (Art. 30) are called transcendental. The transcendental functions include trigonometric, inverse trigonometric, exponential, and logarithmic functions. In a former chapter we considered the differentiation of algebraic functions; in this chapter we shall develop formulas for the differentiation of the more elementary transcendental functions, starting with the trigonometric functions. In the derivation of the following formulas, we shall assume u to be a continuous function of x, and we shall find the derivatives with respect to x. 52. Differentiation of sin u. Let y = sin u. Then y -f A?/ = sin (m + Aw) and Ay = sin (u + Au) — sin u = sin u cos Au + sin Au cos u — sin u. Therefore, we have and A y sin u (1 — cos Au) . sin Au~\ —2- = i '- + COS U — , Au Au Au T Ay . T 1 — cos Au . n T sin Aul L -&■ = — sin u L - — - + cos u L . = Q^u Aw = Am Am = A« J Au = oAu The two limits in the second member have been evaluated ; thus, L 1 " cosAM = 0,and L "^1 (Arts. 13, 14) Aw = A?* Am~0 Au Hence, L — = D u y = cos u. Au = OAu b:J 84 TRANSCENDENTAL FUNCTIONS [Chap. V. Since u is a function of x y we have by Art. 31 D x y = D u y • D x u. Therefore D x y = cos u D x u ; that is, D x (sin u) = cos u D x u. (1) In the differential notation, we have dy — cos u du, or d (sin u) = cos w dw. (2) In the special case where u = x, (1) becomes D x sin 05 = cos a?. (3) The other trigonometric functions may now be differentiated by applying the general laws of differentiation. 53. Differentiation of cos u. Let y = cos u = sin I ^ — Then D x y = cos f | - w J DJ | - wj — — sin w 2> x w ; that is, D x (cos u) — — sin f# D^w, (1) or, in the notation of differentials, d (cos u) =— sin u du, (2) For m = x, D* (cos as) = - sin ac. (3) 54. Differentiation of tan u. T p . sin u If y = tan w = , cos it ,i t, cos ?t ZX. sin u — sin w 7) x cos u then D x ?/ = x - ■ ' ~>S J u D x u cos' u cos 2 w -f- sin 2 u COS' i< — — - D x u = sec 2 w D x u. cos 2 i< Arts. 55, 56] DIFFERENTIATION 85 Hence Dx (tan u) = sec 2 u D x u, (1) or d (tan u) = sec 2 u du. (2) Also D x (tan oc) = sec 2 sc. (3) 55. Differentiation of cot u. -r , . COS U Let 2/ = c °t u then D x 2/ sin w sin u D„ cos u — cos u D„ sin u sin 2 m + cos 2 u t, 2 n = r- 1 - D x ii = — csc^ u D x u. sm 2 w Hence D x (cot t*) = - esc 2 u D x u, (1) or d (cot w) = — esc 2 u du, (2) For u = Xj D x (cot as) = — esc 2 ac. (3) 56. Differentiation of sec u and esc u. 1 If 2/ = secw = cos u we have Z^?/ = D x (cos w) cos 2 u = sec 2 w sin u D x u = sec u tan w Z) x it. Therefore D x (sec w) = sec u tan w 2>a,w, (1) and d (sec w) = sec u tan w rfw. (2) Proceeding as above, we find 2>aj (CSC U) = — CSC w cot w D x u 9 (3) or z tan Va 2 — x 2 = sec 2 Va 2 — x 2 D x Va 2 — x 2 = - sec 2 Va 2 -x 2 Va 2 - x 2 Hence, /'(x) = — — tan Va 2 - x 2 sec 2 Va 2 - x 2 . Va 2 — x 2 EXERCISES Differentiate the following. 1. y = cos ax. 2. y = tan 8 x. 3. y = sin 2 x — cos 2 x. 4. y = sec 2 x. 5. y = sin 2 3 x. 6. y = cos 3 (a 2 — x 2 ). 7. y = x sin x. 8. ?/ = x 2 tan - • — 9. ?/ = cos vx 2 — a 2 . 10. y = x — tan x. 11. 2/ = x 3 csc2x. 12. y = sinx — xcosx. 13. y = 2 x cos x + (x 2 - 2) sin x. 14. y = x — sin x cos x. 15. y = sin x (cos 2 x + 2). 16. p = 2a — • y cos i 17. p = a Vcos 2 0. 18. p = a (sin n0) 19. /o = asec 3 |- 20. /> = cos0 21. s== Acoaut — B sin w£. 22. s=- r cos + 7^1- A Jl-gsin 2 ^. 23. Find the derivative D x y, when x = a(0 — sin0), y = a(l - cos0). 24. Let x = a cos + «0 sin 0, ?/ = a sin — a0 cos 0. Find Ztyx, Z>^^, and D x y. 25. Find the polar subtangent and polar subnormal of the curves : (a) /> = sin0. (6) /> = a(l — cos 0). (c) /o = asec 2 -- 26. From the equation sin 2 = 2 sin cos 0, derive by differentiation formula for cos 2 0. 27. Find expressions for the subtangent and subnormal of the curve y — a sin x. Art. 57] INVERSE TRIGONOMETRIC FUNCTIONS 87 28. Find the angle at which the curves y = cos x and y = tan x intersect. 29. Find the angle which the curves y = sin x and y = cos x make with each other at their points of intersection. 30. Find the value of for which tan is in- creasing twice as fast as 0. 31. When = 22 u , find approximately the changes in sin and cos 0, for a change of 1' in the angle. 32. In a triangle two sides a and b include an angle 0. If the sides remain of constant length and the angle is varied, find the rate of change of the area of the triangle with respect to when = -- 4 33. In a certain type of motion the velocity is given by the expression v = v cos kt. Find an expression for the acceleration. 57. Inverse trigonometric functions. The trigonometric functions are all single- valued functions ; that is, for each value of the variable there corresponds one and only one value of the function. Thus, from x = sin y, x has but one value for each value of y. The inverse trigonometric functions are not single-valued, but mul- tiple-valued functions, for to each value of variable there corresponds any number of values of the function. In Fig. 22 is given the graph of y = arc sin x. For x = a, it will be seen that y may have any one of the values indicated by the points M x , M 2 , M 3 , •••. If, however, the values of y be restricted to the interval [ — ^, ^ ), then within this range the function is single-valued. The graph of y = arc sec x is given in Fig. 23. From the examination of this graph, it will be seen that the function will be single- valued if ^ y <£ it. Ex. Plot roughly the graphs of arc cos x, arc tan cc, and arc cot x, and determine limits within which these functions are single-valued. 88 TRANSCENDENTAL FUNCTIONS [Chap. V. I tit J 7T 7T 2 r~ -/^ > -7T Fig. 23. 58. Differentiation of arc sin u and arc cos u. Let y = arc sin w, then u — sin ?/. Differentiating, we obtain D y u = cos y. 1 Therefore and Since we have D u y = cosy Dj) = D u y>D x u (Art. 31) = J—D x u. cosy cos y = Vl — sin 2 ?/ = Vl — w 2 , 1 Al vr^~ ZU. Arts. 68, 59] DIFFERENTIATION Henc e, D x arc sin u = 1 Vl - u * or in the differential notation, d arc sin u = du Vl-^2 If, as a special case, u = x, D x arc sin oc = 1 89 Z>xU, (1) (2) (3) Proceeding in the same way with y = arc cos u, we obtain Da- arc cos t* = == D^w, (4) dare cos w= ^*__, (5) Da, arc cos a? = . (6) Vl-^2 Since the sign of the radical Vl — x 2 may be either positive or negative, the signs of the derivatives in (3) and (6) are ambiguous. Reference to the graph of arc sin x, Fig. 22, shows how the ambi- guity arises. Thus for x = a, the derivative is positive at M lf M s , M 5 , etc., and negative at points M 2 , M i} etc. When y is restricted to the interval — ^ < y < ^ so as to make the function single- valued, the radical must be taken with the positive sign. The student may show that if fg y ^ 7r, the function y = arc cos x is single-valued and the derivative given by (6) has the proper sign when the radical is given the positive sign. 59. Differentiation of arc tan u and arc cot u. Let y = arc tan u, whence u = tan y, and D y xi = sec 2 y = 1 + u 2 . 1 1 We have then D u y = But by Art. 31, D x y = D u y • D x u = ^—- 2 D x u ; sec 2 y 1 + u 2 1 + " 2 90 TRANSCENDENTAL FUNCTIONS [Chap. V. therefore, !>«. arc tan u = — - — - D x u, „ (1) 1 -\-u* and d arc tan u = du - . (2) 1 + 1* 2 w For u = x D x arc tan x = — ^— . (3) 1 + x 2 To differentiate arc cot u, we may proceed as above, or we may derive the result from (1). Thus, since y = arc cot u = arc tan - , u D ' y = Jlffi °" © = ^T 1 ? ' (~ ^ D ' W = ~ r ^ ? D * U ' Hence, we have D x arc cot u = — - D^U, (4) 1 + U* du 1 + d arc cot W =-_^M_, (5) 2>a, arc cot a? = s . (6) 60. Differentiation of arc sec u and arc esc u. Let . y = arc sec u, then w = sec y, and Z) y w = sec y tan y. Hence, we have D u y — , sec y tan y and D x y = D u y D x u (Art. 31) 1 sec y tan y Dju. Since sec y = u f and tan ?/ = Vsec 2 y — 1 = Vw 2 — 1, we may write D x y = r= Z>,w. w Vw 2 — l Art. 60] DIFFERENTIATION Therefore u V M a _ l End rf nrp hoc 11 — ( ^ u u V w a _ i Also ac Va? 2 - 1 Proceeding in the same way, we obtain 91 n x u 9 (l) (2) (3) D x arc esc u= Ac**, (4) u V w 2 _ i d arc esc u = d u , (5) u V w 2 _ i D x arc esc a? = . (6) a? v^ - 1 From the graph of arc sec jd, Fig. 23, it is readily seen that for 0^y = -, the derivative is positive and therefore the positive a sign of the radical V# 2 — 1 must be taken ; but for --^.y^ir the A derivative is positive while x is negative, and therefore for this interval the negative sign of the radical must be taken. From the graph for arc esc x the student may deduce a similar statement relative to the radical in (6). Ex. y = arc tan VI - 2 x 2 du Let — = u ; then y = arc tan u, dy = VI - 2 x* l + u VI - 2 a? + Vl - 2 a* 1 _ and du = d — == == ^ jr— 5 «£ = - dx. Vl - 2 *2 1 - 2 a* p _ 2 ^1 Substituting these values of u and du, we get 1 1 1 j (fa = 5 r efcc = ax. 1+ * (1-2*2)4 (1-*2)V1-2X2 1 - 2 x* K } 92 TRANSCENDENTAL FUNCTIONS [Chap. V. EXERCISES Differentiate the following : 1. y = arc cos - • 2. y = arc tan -^- • Vk r'2 3. y = arc sec 2L . 4. y - arc s i n Va 2 — x 2 . 5. y = x arc tan jc. 6.0 = arc sec Vp 2 - a 2 1.6 = arc sin ^ ^_ • 8. y = arc tan - + arc tan P a x 9. y = arc cos (cos x). 10. 1/ = arc tan VI — Jc 2 tan 2 x. 1 cc 11. y = x* arc cos x 2 . 12. y = — arc tan _ . Va 2 — x 2 , . x 13. ?/ = h arc sin - • 14. ?/ = Vx 2 — a' 2 — a arc sec - x a a 15. ?/ = - - Va 2 - x 2 + — arc sin - . 2 2 a 16. y = x arc cos x — Vl — x 2 . 61. Exponential and logarithmic functions. The differentiation of the exponential functions a u and e u depends upon the evalua- tion of the limit a x — 1 L > x = % and this limit in turn involves the limit L ( 1 +- j . It will be x) necessary to consider these two limits before the desired formu- las for differentiation can be deduced. Jj ( 1 H — ) . It can readily be shown that L ( 1 + - ) , where n is a positive integer, is some number lying between 2 and 4. The outline of the proof is as follows : Take two numbers a and b such that a > b > and let n be a positive integer. Then we have the identity, qn+l-frn+l _ ^ + ^^ + ^_^ + # + (1) a — 6 from which follows the inequality a — or fl»+i - &1+ 1 < a w (a - b) (n + 1). Art. 61] EXPONENTIAL AND LOGARITHMIC FUNCTIONS 93 The last inequality may be thrown into the form, a n [a - (a - 6) (n + 1) ] < b n +\ (2) Now choose for a and b two sets of numbers, subject to the condition a > b > 0. Let these be : (a) a=l+l, & = 1 + _1_, n n + 1 (6) o = l + ~, 5 = 1. 2 n Substituting these values respectively in (2), we get the two inequalities < I- (4) 2 V 2 w y From (3) it appears that the function (1+-) increases with n ; that is, the function is monotone. For n = 1, the function takes the value 2. From the inequality (4) we obtain, by squaring both members, K-T <4, whence it follows that the function f 1 + - ) cannot exceed 4. The limit required must therefore exist (Art. 14) and it lies between 2 and 4. The exact value of this limit is 2.7182818285 •••, as will be shown here- after in connection with infinite series. This number is denoted by e and is the base of the natural system of logarithms. In the preceding discussion n was given only positive integral values. The limit is the same, however, if instead of n we take a variable x which can take all real values. Hence, in general, we have L * = ae (l + i)" = e. (5) fiX 1 ~x 1 L . To obtain the limit of as x approaches zero, let a? = 35 X = a x — 1. Then x = \og a (1 + y) and as as = 0, y = 0. We have therefore L *L=± = L V = jr 1 X = Q X y=0log«(l+*/) y = 1 log a (l + y)j, 1 1 - 1 l0ga l0g a L (l+|f)» y = 94 TRANSCENDENTAL FUNCTIONS [Chap. V. Since, however, = log a. we have finally log a e OK -t log a. (6) se = SB The substitution of e for a in (6) gives L ^— =-l=-loge = l. (7) 62. Differentiation of «** and e u . Let ?/ = a M ; then y + Ay = a M+A ", and Ay = a M+A " — a M = a M (a AM — 1). Ay u a L Therefore —±=za; Au Au Taking limits, we have Au±0&u Aw = a% But, as shown in the preceding discussion, T a AM -l . L = log a ; Aw = Aw hence, we have D u y = L :_£ = a M log a. Aw = oAw Using the theorem of Art. 31, we have D x y=D u y-D x u =a u log a D x u\ that is, 2>x<*> u = <*> u log a D x u t (1) or da w = a 11 log a d u. (2) If u = x, we have D^a* = a* log a, or da x = a x log a dx. (3) If in (1) we substitute e for a, we obtain (since log e = 1) n x e u = e u n x u; de u = e 1l du. (4) Arts. 63, 04] DIFFERENTIATION 95 For u = x, we have jD^e* = e* ; de x = e x due. (5) It may be noted that e x is a function whose derivative is the function itself. 63. Differentiation of log a u. Let y = log a u, whence u = a y . From the preceding article, we obtain D y u = a y log a, whence D u y = a y log a By the theorem of Art. 31, D x y = D u y • D x u = — ; D x u = — D x u. a y log a u log a We have therefore n a >log a u = -L-2»!i t or dlog a u = -±-*». (1) loga u ' loga u v J If u = x, we have Z>*log a K=r J —^; dlog a ac = -^—^. (2) log am loga ac The fraction is called the modulus of the system of log- log a arithms whose base is a, and may be denoted by m. For the Briggs' or common system, in which a = 10, m = .434294 ••♦. We have, therefore, m D x log a x = -. x 64. Differentiation of log u. By the substitution of e for a in (1) and (2) of the preceding article, the following formulas are obtained (since log e = 1) : n x logu = -D :>l ,u; dlogu = —. (1) u u I> x logx = ±; dlogoc = dv. (2) 90 HC 96 TRANSCENDENTAL FUNCTIONS [Chap. V. Ex. 1. y = log sin x. dy — d sin x = cos x dx = cot x dx. sin x sm x Ex. 2. y = e V i+*\ dy = e^^+J gB u. This completes the demonstration begun in Art. 29, and shows that the formula given there for the derivative of u n holds for all real values of n. EXERCISES Differentiate the following : 1. 2/ = log(*2_3x + 5). 2. y = lQg7* + 5»y \x — m) 4. y = e* 3 + e x . 6. |f = log (log*). 8 . y = « x + e- x . e x — e x 10. y = log tan x. 3. j, = a^ xJ ^\ 5. y = x\ogx. 7. y = e x + e~ x . 9. y — gsin*. 98 TRANSCENDENTAL FUNCTIONS [Chap. V. Vx 2 + 1 - 11. y = log v _ ^ — -• 12. y = log(x + vx 2 -a 2 ). vx 2 + 1 + x i x . 13. y = -log , 14. y = x arc tan x — log V 1 + x 2 . a a+ vx 2 + a 2 15 . y =^(ax-l). 16. ^-A-log^ 6 *" 26 " b In Exs. 17 to 20, differentiate by taking logarithms 17. y = x(l - x) Vl + x 2 . 18. y = x 2 Vx 2 — 4 19., = ^^- 20. , = X S XV 1 — X 2 *1 -fx 21. If z = Ae^' k + Be-*/*, find — . 22. Find an expression for the velocity when the distance traversed by a moving point is expressed as a function of the time by the equation s=(A + Bt)e~ kt . 23. If s = e~ kt [a sin mt + b cos mt~] , find an expression for the velocity. 24. Find the polar subtangent, polar subnormal, lengths of polar tangent and polar normal of the curve p = e aQ . 25. Find the angle between the curve y = log x and the axis of x ; between the same curve and the line y = 2. 26. Find the subtangent, subnormal, and the length of the normal of the catenary y = «(e« + e~«). 2 27. Given log 4.32 = 1.4633, find approximately the value of log 4.33 by means of the theorem Ay =/'(x)Ax. 28. By logarithmic differentiation derive the result da u =a u log a du. MISCELLANEOUS EXERCISES 1. Derive the formulas for the derivatives of cos w, tan u, and sec u directly from the defining equation jr A?/ _ L f(u + Au)-f(u) and the theorem of Art. 31. 2. Obtain the derivative of cot u from the relation cot u = : also the derivative of sec u from sec u = Vl + tan 2 ?«. tan u Art. 65] MISCELLANEOUS EXERCISES 99 3. Write out formulas for the derivatives of the inverse circular functions for the case in which u = - • a 4. Discuss the following curves by means of their derivatives. Find where they cross the X-axis, the angles at which they cross, and the point at which their tangents are parallel to the X-axis. (a) y = sin ( x — - V (&) y = sin2a:. (c) y = sin x + cos x. (d) y = log x. 5. When = 36°, what is the rate of increase of (a) sin ; (6) cos ; (c) tan compared with the rate of increase of ? 6. The cycloid is given by the equations x = a{0 — sin0), y — a{\ — cos0). Derive an expression for the slope of the tangent, and find the value of this slope when a = 3, 6 = f 7r. For what value of is the tangent parallel to the X-axis ? Differentiate the following : 7 . v^r^- a iog « + v <* 2 -* 2 . X o x — 8 a V2 ax + a; 2 + - a 2 log (as + a + V2 ax + x 2 ). - A 9. ^V« 2 -z 2 +^arcsin-. 2 2a 1 1 - 10. 1- arc tan x. 11. log(e* + e~ x ). 12. tana x . x '3 X s 13 . Find the polar subtangents and subnormals of the following curves : (a) p = a(l + cos0). (6) p 2 = e a ». (c) /> = a 2 sin 2 0. (d) p = sec 2 2* 14. Find expressions for tan f (Art. 40) for the curves of Ex. 13. 15. If a point moves in a logarithmic spiral p = e ae , show that the compo- nents of its velocity along and perpendicular to the radius vector have a con- stant ratio. 16. A point moves in the curve p = a(l — cos 0) with a speed of m units. Find the velocity components along and perpendicular to the radius vector. 100 TRANSCENDENTAL FUNCTIONS [Chap. V. 17. The following formulas give approximately the relation between the pressure and temperature of saturated steam : (a) lo gi > = ^ + |+^ 2 - (6) log p = m — n log (T=r +461) T- C (c) logjo = a + ha* - c/3«. (0 = r - 32) From each formula derive an expression for the rate of change of the pres- sure with the temperature. 18. A point moves in accordance with the law expressed by the equation s = ae~ Kt cos 2 ir(bt + c). Derive an expression for the velocity. 19. A radius OP of length a, Fig. 24, rotates with constant angular speed w about . y- the point 0, and the projection M of P on the X-axis therefore moves on this axis. Show that OM = x = a cos (x), to find another function f(x) such that D x f(x) = (x). This inverse operation is called integration, and the resulting function is called the anti-derivative or integral of the given function. The function integrated is called the integrand. Two symbols are in use to denote the inverse operation of integration. The symbol D' 1 may be prefixed to the integrand ; thus, since D x v i = 3x 2 , we have i).- 1 3aj 2 = aj». It is the universal custom, however, to denote integration by plac- ing the symbol I before the differential. Thus, since d(a?)=3a?dx, we write J 3 x 2 dx=x 3 . The direct and inverse operations may therefore be indicated by the symbols D x and D~ l ; or by the symbols d and I . EXERCISES Find anti-derivatives of the following functions : 1. 5x*. 2. 2x. 3. cosz. 4. 1- 5. --• X X 2 Perform the following integrations : 6. fsec20d0. i 7. (2atdt. 8. $ ^Itf' 9 * ^ dx ' 101 102 INTEGRATION [Chap. VI. 67. General theorems. It will be observed that the functions X s and X s -J- C, where C is any constant, have the same differential 3 x 2 dx. Hence the integral I 3 x 2 dx should have the general form x* + C. In general, since the derivative of a function plus a constant is always equal to the derivative of the function, we have the following : Theorem I. The general form of the anti-derivative or integral of a function must involve an additive constant. This constant is arbitrary, and is called the constant of integration. The value of the constant of integration in any particular case may be determined when certain initial conditions in the problem under discussion are known. The determination of this constant from such conditions will be subsequently discussed more fully. An anti-derivative or integral function may be represented by a graph. The geometrical significance of annexing the constant of integration is to increase or decrease each ordinate of this graph by the same length. In other words, as we give various values to the constant of integration, we move the integral curve up or down. See Fig. 8. It has been shown that the derivative of the product of a con- stant and a function is the product of the constant and the de- rivative of the function ; that is, D x c -f(x) = c • D x f(x) or in the differential notation, d(cu) = c du. The inverse operation gives a similar theorem, namely : Theorem II. A constant factor in the integrand may be ivritten either before or after the symbol i ; that is \cdu=c\du. (1) The derivative of the sum of a finite number of functions was obtained by adding the derivatives of the separate functions. The inverse operation gives, therefore, the following theorem : Theorem III. The integral of the algebraic sum of a finite num- ber of functions is equal to the algebraic sum of the integrals of these functions; that is, ((du + dv + dw) = (du + (dv+\dw. (2) Art. 68] THE INTEGRAL 103 . The integral J u tl du. By differentiating the function u n+l , d(u n+1 ) = (n + l)u n du, 68 we obtain whence dl ] = u n du. J 1 + h Therefore, we have \u n du=^ — -+C, J n+1 which holds for all values of n except the value — 1. This for- mula is especially useful in the integration of integral algebraic functions. Ex. 1. (5x B dx = 5(x*dx = %x*+C. Ex. 2. f (4 x 2 - 3) 3 x dx. Put w = 4 x 2 - 3, whence du = 8 x dx. The integral then takes the form £ ( u 9 du. Hence f(4x 2 -3) 3 xdx = i (4 x 2 - 3) 4 + C. EXERCISES Verify the following : 1. f x 4 dx- = \x*>+ C. 2. (x% dx = § x* + 0. 3. f (x 3 -2x + 5)a*x = £x 4 -x 2 + ox + c. 4. f(3»- 2 -5x- 4 )dx=-3x- 1 + fa;- 3 + C. Evaluate the following integrals : 5. f (ax~ 3 + bx-*) dx. 6. f (5 x 6 - 4 x 3 + 2 x + 7) dx. 7. J(3ar* + 4-a£ + 2s*)cfc;. 8. J^l+I + jW 9. j*(»*- 8 «*)<*& 10. f(2x 2 -3x)4«x. 11. §(ax + b)*dx. 12. (*3 x 2 (x 3 + 4) 2 ax. 13. j'xX2x 4 -5) 3 ^. 14. f x(3 x 2 - 7) 4 dx. 15 r (3a; 2 -5)(la; 16. f_*dx_. J Vx 3 - 5 x + 7 ^(x 2 - 5)* 104 INTEGRATION [Chap. VI. 69. Fundamental integrals. From the results obtained in the preceding chapters we have the following list of fundamental integrals. The student should make himself thoroughly familiar with these formulas. C attl+l 1. \u n du = - — -+C. (w=£-l). J n + \ 2. f cos u du - sin u + C. 3. J sin u du = - cos u + C. 4. f sec 2 u du = tan u + C. 5. J esc 2 udu = - cot te + C. 6. f sec u tan w dw = sec u + C 7. J esc ucotudu= — esc w + C. 8. f d ™ =arcsint4 + C 9. f-^- = arctanw+ C = -arc cos w 4- C. =- arc cot m + C. 10. f du = arc sec m + C 11. f « m _i J log« = — arc esc u+C. 12. f e u du = e u + C. 13. f^ = log «* + C. The following additional integrals are important and should also be committed to memory. 14. f_^ = 1 l g^« + O. (**>«*) = J_ log«^+C. (**<«*). 2« a + u is. f dM = iflp(tt4-Vtt» + «*)+(b x y ab > a Ex. 5. f («-*)<*» . J3x 2 -6x + 5 It will be observed that d(S x 2 — 6 x + 5) = fi(x — 1) dx ; hence, except for the factor 6, the numerator of the integrand is the differential of the denomi- nator. By the introduction of this factor the integral reduces to form 13, namely, f — • Thus, we have J u C (x-\)dx = lC 6(x-l)dx = ir d^x 2 -6x+6) = l l ^-2-6 x+5) + C. J3x 2 -6:k + 5 6J3x 2 -6z + 5 (iJ 3z 2 -6a;+5 6 &v In some cases the integral can be reduced to the algebraic sum of several standard integrals. Art. 70] INTEGRATION BY INSPECTION 107 ex.6. f Vl + x dx^f-i±^_ dx = r_^_ + r_^_. J Vl - x J Vl - x 2 J Vl - x 2 J VI - x* Now f — == = arc sin as, (Form 8) •J Vl - x 2 and f g3 2 . 6 ic — 1 Ex. 10. r dx C d(x + 2) •J Vc 2 + 4x + 8 * V(x + 2) 2 + 4 log(x + 2+Vx 2 + 4x + 8)+ C. 108 INTEGRATION [Chap. VI. EXERCISES In the following examples determine by inspection the proper standard form, find the function that replaces u in that form, and perform the integration. dx_ . a- "cos 6 dd 1. ((x + aydx. 2. iJ2x{x 2 -arfdx. 3. (J>^ a To ^ -, «- fcos0d0 r 4. J2xe* dx. 5. j ^^-' 6. J cos 2 0sin0cZ0. _ r are tan x dx g C m dx n ' J 1+* 2 " J Vl-m 2 z 2 ' 9 * J s sec 2 tan **. «./^- 6 . 13. J Va 2 + x 2 16. \a m * x2 xdx. In the following, reduce the integrands to standard forms by writing the denominator as the sum or the difference of squares, and integrate. 17 f dx 18 C dx Jz 2 + 2x-f5 ' J V9 + 8x-x 2 rife 19 r dx 20. f— J >A>2 J.firJ-1ft ^ 11 Vx 2 + 6^ + 10 Jll + 10x-a 2 Verify the following integrations. 21. f ^ = arc sin - + C = - arc cos - + C". J Va 2 - x 2 a a 22. f ^ = 1 arc tan ? + C. J a* + x' 2 a a 23. f — — = larcsec-+ Q = - 1 arccsc-+ C. J xVx 2 - a 2 « « a a The integrals of exercises 21-23 may be regarded as standard forms and should be memorized. 71. Integration by substitution. The reduction of a given inte- grand to a standard form is sometimes most easily effected by the Art. 71] INTEGRATION BY SUBSTITUTION 109 substitution of a new variable. The following examples illustrate this method. Ex. 1. I-?*—. J * _£ e 2 + e 2 x dz Let e 2 — Z' then dx = 2 — j an d tne integral takes the form ' z r 2dz _ 2 C_dz_ _ 2 arc tan ^ _ 2 arc tan g 2 + J 2(2 + 2- 1 ) J Z 2 +l Ex. 2. a J X 2 Vl — X' 2 Let x = cos ; then dx = — sin d0, V 1 — x 2 = sin 0, and the integral takes the form J cos 2 sin J x The substitution x = - may also be used. The student may work out the details. As additional exercises, the derivations of forms 14 to 19 are given. Ex. 3. Since — — - = — f — — ^ , we have, if w 2 > a 2 , u 2 — a 2 2a\a — a u + a) r du _ i r_dw_ _ j_ r_^_ Jw 2 — a 2 2aJ w- a 2fJ«-fa = ±[\og(u - a) - log<> + «)] =~ l0 ^ l± ^' 2a 2 a u + a The student may derive the second integral of form 14 and show that it applies when u 2 < a 2 . Ex. 4. To integrate f du put u 2 ± a 2 = z 2 . Then 2udu=2zdz, -* Vn 2 ± a 2 whence du = dz = du + dz m z u u + z Therefore, C du = f ** = f ^LdL^ = i og ( M + *) ^ Vw 2 ± a 2 J z J u + z = log(t< + Vm 2 ± a 2 ) . Ex. 5. ftaix dx = fBin*cte = _ C djcosx) J J cosx J cosx = — log cos x + G — log sec x + C. 110 INTEGRATION [Chap. VI. Ex. 6. foot x dx = f <^L^ = Cdjsinx) = { ^ % + Q J J sin x J sin x Ex. 7. j esc x dx. Let z = tan - ; then esc x = — — — , and dx = 2 ' 2z 1 + z 2 Substituting these values, we get f csc x dx = f — = log z + C = log tan - + C. Ex. 8. \ sec x dx. Making the same substitution as in Ex. 7, the integral reduces to the form dz — Hence, we have 2 2 1 + tan - (eecxdx = 2(^ — -=log-±-?+ C = log - + j j i-z l-z 1 _ tan ^ = logtan g + j]+ a 2 EXERCISES 1. T_J^_, letz = a + 5x 2. f dx let a; = a sec 0. Ja + bx J x 2 Vx 2 -a* 3. f *5 , let*:=asin0. 4. f '*» • J (a*-xrf J y/x-2 5. f ^ f lets 2 = 2sc+l. • / XV / 2J5+ 1 6. Verify the integrations given in examples 21, 22, and 23 of the preced- ing article by the substitutions x = a sin 0, x — a tan 0, and x — a sec 0, respectively. C dx 7. Evaluate the integral I — by means of the substitution a = a(l-cos0). V2ax-x 2 72. Character of the integration process. There is a funda- mental difference between differentiation and integration. The former is a direct, the latter an inverse process. As we have seen, when once the derivatives of the elementary functions have been found, the derivatives of any functions expressed in terms Art. 72] CHARACTER OF THE INTEGRATION PROCESS 111 of these functions can usually be obtained by one or more direct operations. On the contrary, it is not, in general, possible to determine the integral of a function from the known integrals of the elementary functions in terms of which the given function is expressed. There are no general methods for expressing the in- tegral of a product or a quotient of functions, or of a function of a function in terms of the integrals of the component functions. Hence the process of integration consists, not in a series of direct operations carried out in accordance with a general method, but rather in attempts to reduce the given function to a form, the integral of which is known. Integration, then, is largely a question of judgment in attack- ing the problem. Specific rules cannot be given, but the follow- ing general directions will be found useful. First inspect the integrand carefully and determine whether it is a standard form. Frequently the integrand is merely a disguised standard form and needs only a rearrangement of its terms to be recognized as such. If the integrand is not in a standard form, see if it cannot be made to assume such a form by the introduction of a constant factor. Do not neglect the neutralizing factor outside the in- tegral sign. If this device does not appear to be effective, try the substitution of a new variable. MISCELLANEOUS EXERCISES Integrate the following : 1. froz—3<&. 2. ((3x-x*)~%(S-2x)dx. C x*dx 4 C (x-S)dx (X s - a 3 ) f (*-«)** . 6. (xe*> J Vz 2 - 6 x + 1 J f cos 2 sin 6 dd. 8. ( r (2 + Bx)dx m r dx. cos 6 dd sin 3 dx ±1 C x*dx . \ 12 C 2*3-53+! ^ J 3 x s - 5 J i ' (ax + b)% 2x*-5 x 4 -5x 2 + 2x-7 112 INTEGRATION [Chap. VI. 13 C dx 14 C x 3 dx t J x* + 6 x + 13* ' • J x + 1 " 15. f^f- 1 ^. 16. f- J x s + x J a 17 19 sin x dx x 3 + x J a + b cos as f e cos x sin j. tf x . 18. f ~ x dx. J J 1 + x' 2 f cfo t 20 r (fa Jl+3x + 2z 2 ' ' J VU-Sx-x 2 ' 21. f ; * 22. f_*L ^ Vz 2 + 6 x + 1 J 2 2 - 23. M|-. 24. f_ J 4 - 3 6' 2 J dx 25 xVl — log 2 a; . f ds . 26. f dM — J \/2 as + s 2 ~* v (m + wi) 2 - n 2 27 r_^j^_, 28 f vi-* ^. J Va 8 - x 8 ^ Vl + x 29 f e ~ x ^ 30 f O + 2), we have the inverse relations ( o = D t - 1 a= fa (it, (1) = D t -'«>= Ciodt, (2) and starting with the angular acceleration a as a given function of the time, we can by integration determine the angular speed and the angle swept through by a given radius. For the case in which the angular acceleration is a constant, say a Q , we readily derive the following formulas for = (V + wo- (3) 6 = \a4' i + ^t + { , (4) As in Art. 74, we have also the useful relation a>d. Motion of a projectile. A body is thrown with an initial velocity v at an angle a with the horizontal. Neglecting the resistance of the atmosphere, the path of the body is to be determined. Art. 70] MOTION OF A PROJECTILE 117 Let the plane containing the path be taken as the XF-plane, with the F-axis vertical. Then the X- and F-coinponents of the initial velocity v are respectively v Q cos « and v sin a. The Fig. 25. X-component remains constant throughout the motion, that is, the X-component a x of the acceleration of the body is zero. The F-component a y of the acceleration is — g, as in the case of a body projected vertically upward. We have therefore dv x dt a* 0, _dv 1 a « = a=- g > ■y,= Cj = r oos a. (1) (2) To determine 2 , we have at the beginning of the motion v y = v sin a, when t = ; hence C 2 = v sin a, and v y = v sin a — gt. To find the X- and F-coordinates of the position of the body at the time t, we have ^ v, = — = v cos a, dt x = v t cos a ; y eft 0*j whence and » f = -*■ = i? sin « whence y = vj, sin a — -J- gtf 2 . The constants of integration are readily found to be zero. Eliminating t between (3) and (4), the resulting equation, y = x tan a - — -2 — --, 2 V cos^ a is the equation of the path. (3) (4) (5) 118 SIMPLE APPLICATIONS OF INTEGRATION [Chap. VII. EXERCISES 1. Find an expression for the velocity v and distance s when the accelera- tion is given by the relation : (a) a = m-nt\ (6) a =- nttf cos kt. (c) a =^2 (e«- e~«). 2. If the speed of a point moving in a straight line is given by the relation v = 10 1 — t 2 , find expressions for the acceleration and the distance traversed, starting from rest, at the end of t\ seconds. 3. In Ex. 2 find the time that will elapse before the point comes to rest, the distance the point has moved in that time, and the final acceleration. 4. A hoisting drum has an angular speed w = 24 rad./sec. A brake is applied in such a way as to produce a retardation (negative acceleration) of 3 rad./sec. Find the time required to bring the drum to rest and the number of revolutions made after the brake is applied. 5. Show that the range of a projectile on a horizontal plane is B=^sm2a. 9 6. Show that the velocity of a projectile at any instant is v = vW — 2 gy. 7. A stone is thrown horizontally with an initial velocity Vq from the top of a cliff. Find the path followed. 8. If the top of the cliff is 100 feet above the level of a stream 150 feet wide at its base, what initial velocity is required to carry the stone across the stream ? 77. Harmonic motion. According to the laws of mechanics, the acceleration of a body of given mass is proportional to the force acting on it. Hence, since it is usually the force that is known, the character of the motion is specified by the law ac- cording to which the acceleration changes ; and the velocity and distance traversed are obtained by one and two integrations, respectively. The acceleration a may be given as a function of t, of v, or of s. In this article and those immediately following we shall consider a few important examples of motion following various laws. When a point moves in such a way that the negative accelera- tion is proportional to the distance of the point from a fixed origin 0, the motion is said to be simple harmonic. In this case, we have a = -k 2 s, (1) Arts. 77, 78] MOTION IN A KESISTING MEDIUM 119 whence from the relation vdv = a ds, [Art. 74, Eq. (3)] vdv = — k 2 s ds. Integrating, we obtain v*=C-fcV. To determine the constant C, let v denote the velocity at the origin 0; then v = v when s = 0, and consequently C = V- Hence v 2 = vo is the initial angular speed. 7. Find an expression for /4a;' 2 -7 ' ^ (x-a )V(x-a) 2 -b 2 7. Determine curves whose slopes are respectively (a) 3x + 2; (6) x^ + c ; (c) cos rax; (Zx - £ wx 2 , D^f(x) = 0, when x = L. CHAPTER VIII SUCCESSIVE DIFFERENTIATION AND INTEGRATION 80. Definition of the nth derivative. The derivative of a func- tion is, in general, a function of the same independent variable as the original function, and it may itself have a derivative if it fulfills the necessary conditions. The differentiation of the origi- nal function produces the first derivative ; the differentiation of this first derivative gives the second derivative; the result of differentiating this second derivative is the third derivative ; and so on. If the process is repeated n times, the final result is the nth derivative of the function with respect to its independent variable. As an example, consider the function /(a?)=a^-2aj + 5. The first derivative is 3 a; 2 — 2, which is also a function of x. Differentiating this, the second derivative is 6 x, which is again a function of x. The third deriv- ative is 6, a constant, and the fourth and each successive deriv- ative is zero. If y=f(x) is the original function, the successive derivatives are also denoted by the symbols : D x y, D*y, D x *y, -, D x n y, or by % ^, ^, •-., ^- dx dx 2 ' dx 3 ' ' dx n The symbols —J, — ^, etc., are the ones most frequently used. The form of these symbols may be explained as follows: Since 126 Arts. 80, 81] IMPLICIT FUNCTIONS 127 d f(x) d J v > =f'(x), we may regard — like D x as an operator which, dx dx when applied to f(x), produces the derivative f'(x). If 2/=/0»), we have, therefore, — y = -^ = /"'(#), ' ' dx y dx J v J ' dx\dx) J v h dra_r^\i ,„ () ^ dx\_dx\dxj] J v " For the sake of convenience in writing £-(&) is abbreviated to % ±[£-(&)~\ to % etc. dx\dxj dx~ dx[_dx\dxjj dx 6 A general formula for the nth derivative may be found for cer- tain functions as shown by the following examples. Ex. 1. Let y = x?, where p is a positive integer. Then D x y =pxP~ 1 , D x 2 y=p(p- 1)3*-*, D x s y=p(p-l)(p-2)xps, D x n y=p(p-l)(p - 2) ... (p- n + 1)xp-». For n =p, xP~ n = 1 and D x p(xp)=p !; hence for n>p, the derivative is zero. Ex. 2. Let y = e ax . We have D x y = ae ax , B x 2 y = a 2 e ax , D x *y = a s e ax , Dj*y = a n e ox . 81. Successive differentiation of implicit functions. The follow- ing example illustrates the method of procedure in finding suc- cessive derivatives of an implicit function. 128 DIFFERENTIATION AND INTEGRATION [Chap. VIII. Ex. Given f(x, y) = xhj + 5 y - 3 X = ; find £-2. dx 2 Differentiating with respect to as, we obtain 2^ + ^ + 5^-3 = 0, dx dx (1) whence dy = S-2xy, () dx z* + 5 w A second differentiation with respect to x gives da; 2 (x 2 + 5) 2 - (x 2 + 5)(2?/ + 2a;^)- (3-2a;y)2a; (3) Substituting in (3) the expression for -^ given ! in (2) , we obtain after reduction x (x 2 + 5) 2 >n for ^ da; d 2 y _ 6 x 2 y - 12 a; - 10 y ( 4 ) dx 2 (a; 2 + 5) 2 v J 82. Geometrical and physical interpretations of the second deriva- tive. As has been shown, the first derivative D x y (or -^ J gives \ dx) the rate of change of y with respect to x, and expresses geometri- cally the slope of the curve which is the graph of y =f(x). Evi- dently, the second derivative D x 2 y (or —£ ) gives the rate of change of the slope compared with the rate of change of the ab- scissa x. Take, for example, the function y = mx -J- b, which is represented by a straight line. The slope is constant, and there- fore its rate of change is zero. This is shown by the derivatives ; for D^j = m and D 2 y = 0. In the case of a moving point, the speed v is the derivative D t s, while the tangential acceleration a is D t v. If now D t s be substi- tuted for v, we have a = D t (D t s) = D 2 s, or a = — -, (1) dt 2 ' V J that is, the tangential acceleration is the second time-derivative of the sjmce s. Art. 82] THE SECOND DERIVATIVE 129 In the case of rotation about a fixed axis, the angular speed is dB D t 6 = — = w, and the angular acceleration is (XL a=D t a> = D t (D t 0) = D:-6 = |?; (2) that is, the angular acceleration is the second time-derivative of the angle swept over. EXERCISES Find the second derivatives of the following functions. 1. y = a* + 4a; — 7. 2. y = x x . 3. y = e^cosx. 4. y = arc cos x. 5. y = Va 2 — x' : . 6. y = log(a 2 + x 2 ). Find — -^ for the following. dx :i 7. y = xe x . 8. y = log(x — 3). 9. y = x sin x. 10. If y = sin x + cos x, show that y = £l = ^ • dx 4 dx 8 Find the nth derivatives of 11. y = a*. 12. y = - • 13. y = log x. x 14. Find — ^ from (a) the equation of the parabola y 2 = 'xpx; (6) the dx 2 equation of the ellipse a 2 y 2 + b 2 x 2 = a 2 b 2 . Find — e- for the following implicit functions. dx 2 15. x 8 y + 3 x 2 y 2 + xy s = 0. 16. y 2 - 2axy + x 2 - c = 0. 17. .rev — c = 0. 18. Find — from r dff 2 1 — cos d rfir 19. Find — from r = log sin 0. 20. If jw" = C, show that ^ = n(n + 1) -?• dv 2 v 2 21. If w = e« sin a;, show that ^ - 2 dy + 2 y =0. dx 2 dx 22. If y = log [x + VaM^ 2 ], show that f| + -r*—^ = °- dx 2 a 2 + x 2 dx 130 DIFFERENTIATION AND INTEGRATION [Chap. VIII 83. Successive integration. As the inverse of successive differ- entiation, we have the process of successive integration. Starting with a function y =f(x), considered as an nth derived function, a single integration gives a new function, the integral. The in- tegration of this second function gives a second integral, and so .on. The result of n integrations is the nth integral of the given function. For example, let f(x) = 5 x 2 . Then |6 x 2 dx = f x 8 + Ci is the first integral, I (t £ 2 + Ci)dx = T \ x 4 + C\X + (7 2 is the second integral, ) (A ** + CtoB 4- C 2 )dx = A ^ 5 + i Ci x2 + C^ + G s is the third integral, etc. It will be observed that an integral contains the number of arbitrary constants indicated by its order; thus the third inte- gral has three, the fourth, four, and so on. Let the successive integrals of f(x) be denoted by fi(x), f 2 (x), ••-,/„(#), respectively; we then have / 2 (aO=J/iO»)d«; fs( x ) = JM X ) dx > etc - Ex. 1. Find the successive integrals of y = e ax . /i(x) = f/(x)da; = (e ax dx=-e ax + d, /,(«) = J (Uu + CiW =i«- + Cix + C 2 , /a(x)= f f \e» + dx+ C 2 W = 1*» + J daS"+ C 2 x + C 3 , /„(») = — e ttX + l-ix' 1 - 1 + A: 2 x«- 2 + ... +*„_!* + *„, in which &i, & 2 , ••-, k n involve the constants of integration. Ex. 2. A body falls with a constant acceleration g = 32.2 ft. /sec. 2 . If it starts from rest, through what distance will it fall in 10 seconds ? Art. 83] SUCCESSIVE INTEGRATION 131 We have here —^ = 32.2, (1) ** = (<** dt = f 32.2 dt = 32.2 t + Ci, (2) dt J dt 1 J s = f J| dt = ( (32.2 t + Ci) d* = 16.1 t 2 + G x t + C 2 . (3) We may now determine the constants Ci, C2, from the initial conditions of the problem. Since the body falls from rest, we have for t = 0, s = 0, ^ = 0. dt Hence for t = 0, we have from (2) and (3) & = 0, <7 2 = 0. Therefore, for the time t, the space passed over is given by s = 16.1 t 2 . For t = 10 seconds, s = 1610 feet. EXERCISES Find four successive integrals of the following functions. 1. y=sinax. 2. y=x 2 — 1. 3. y= — • E/L 2 J' 4. Find a curve that passes through the points (0, 0) and (w, n) and for which the second derivative — B at any point is A: times the abscissa at dx 2 that point. 5. In the theory of flexure of beams the following equation occurs : d 2 y dx 2 EI\ E, 7, M, i?,.and w being constants. Derive an expression for y and deter- mine the constants of integration C\ and C% from the conditions y = 0, when x = 0, and y = when x = I. 6. A point has an acceleration expressed by the equation a = — ru> 2 cos ut, where r and o> are constants. Derive expressions for the velocity and the distance traveled. d 2 s We have — = — rw 2 cos /(« ± K), where h is positive and takes all values in the immediate neighbor- hood of zero. Likewise, a function f(x) is said to have a minimum for x = a, if f(a) is less than the values of f(x) for x just preceding and just following x = a ; in other words, if f(a)0, Art. 84] MAXIMA AND MINIMA 135 For this value, the- function has neither a maximum nor a minimum, although /'(x)=0. For the critical value x=— f, we may examine the derivative For x = — f — h, f'(x) is negative, and for x =— $ + h, f'(x) is positive. Since f'(x) changes from negative to positive as x passes through the value — f , the function has a minimum for this value of x. The graph of this function is shown in Fig. 27. Ex. 2. Examine x 3 — 4 x 2 + 5 x - 2 for maxima and minima. f(x)=x s -4x 2 + 5x-2, f'{x)= 3 x 2 -8x+5 = (8x-5)(x-l), Fig. 27. /"(x) = 6x- 8. Putting /'(x) = 0, the critical values are seen to be x = § , x = /"(x) = 6 x § - 8 = 2, and for x = 1, /"(x) = minimum for x = f and a maximum for x = 1. 1. For x = f, 2 ; hence, the function has a Ex. 3. Let Then /(*) = logx logx (logx) 5 Hence for /'(x) = 0, log x 4F r4,5; 1 or x = e gives the critical value. For x < e, logx< 1, and f'(x) is negative, while for x>e, logx>l and f'(x) is positive; hence, /(e) is a minimum value of /(x). Ex.4. Examine(x— 4)3 -f 5 for maxima and minima. Here whence /(*) = /'(X): (x-4)t+6, 2 -^X 3(x - 4) Fig. 28. For x=4, f'(x)=co; hence x = 4 is a critical value. For x > 4, /'(x) is posi- tive, while for x < 4, /'(x) is negative. Hence as x increases, /'(x) changes from negative to positive, and from Theorem I, /(x) has a minimum at x = 4. See Fig. 28. 136 DIFFERENTIATION AND INTEGRATION [Chap. VIII. EXERCISES Examine the following functions for maximum and minimum values. 1. y = x 3 - 10 x 2 + 30. 2. y = x 4 - 6 x 2 + 10. 3. y=x(x 2 -l). 4. J/ = x 2 - 1). Vx - 1 2 * 7. y = x+ 2 Vl + x 2 . 8. y = e x + e~*. x 9. y = xe _x . 10. y = sin * + cos x, 2 C. This shows that p is a minimum f or x = 2 y. Ex. 2. The deflection of a rectangular beam of a fixed length under a given load varies inversely as the product of the breadth and the cube of the depth. From a log a inches in diameter, a beam is to be cut of such dimensions as to make the deflection a minimum. Denoting by z the deflection, and by b and h the breadth and depth, respectively, the equation bh s (6) Fig. 30. expresses the law stated above. There are two variables in this function, but one of them may be elimi- nated by means of a second equation derived from the geometry of the figure, viz. b 2 + h 2 Combining (6) and (7), we have z - ¥ By differentiation of (8) , we obtain dz _ k(i h 2 -S a 2 ) dh h\a 2 -h 2 y CO (8) (9) 138 DIFFERENTIATION AND INTEGRATION [Chap. VIII. This derivative takes the value zero when 4 h 2 — 8 a 2 = 0, that is, when 2 The corresponding value of b is, from (7) , - • The student may show that for these values z is a minimum, not a maximum. EXERCISES 1. Divide a number a into two parts such that the sum of their squares shall be a minimum. 2. The range of a projectile is given by the expression B = v ° Sln — ^ , in 9 which v denotes the initial velocity and the angle which v makes with the horizontal. For what angle

m *u + •■• + vZVw. ^ ! Observe that the coefficients and symbolic exponents follow the law of coefficients and exponents in the expansion of a binomial. This is Leibnitz' theorem. (Compare Gibson's Calculus, § 68.) 8. Using the result of Ex. 7, find (a) D x b y when y = e x cos X ; (6) DJy when y = x 3 log x. 9. If ^L = ± [§£-*& + *£], find */=/(x) knowing that ^ = dx' 2 £7 L 2 2 J dx when x = and when x = I ; also that y = when x = 0. 10. Find curves for which at any point the second derivative ^-2 has the dx 2 constant value 2 a. Which of these curves passes through the origin ? Which has the slope a at the point (—2, 0)? 11. Find three successive integrals of the functions (a) — — + x + 5. (6) e ax — e~ ax . (c) sin (kt + e). x 3 d 2 x 12. Find expressions for the acceleration — for the motions described by the equations (a) x = cie -a * + c 2 e-0< ; (6) *=(Ci + c 2 «)e -a< . d 2 x 13. Find an expression for — when the equation of motion is dt 2 x = e a< (ci cos /3« + c 2 sin /3£) . 14. From (a) of Ex. 12 deduce the relation ^-(<* + /3)^ + a /3x = 0. dt 2 dt Art. 85] MISCELLANEOUS EXERCISES 141 15. The ideal efficiency of a certain type of water turbine is given by the • cu v2 oh *4~ tt^ ii^ equation E — « — — , where u denotes the variable peripheral gh speed of the wheel. Find the value of u for maximum efficiency, also the maximum efficiency. 16. Find maximum and minimum values (if such exist) of the functions (a) ^-2x + 5 (6) sin 0(1 + cos 0). 17. Find — \ for the curve given by the parametric equations dx 2 x = a(0 — sin0), y = a(\ — cos0). 18. Find — \ for the conic section ax 2 + 2 hxy + by' 2 + 2gx + 2fy + c = 0. dx 2 d 2 9 19. Find the acceleration — from the equation s = L + r(l — cos 0) — VL 2 — r 2 sin 2 0, dd knowing that — is a constant w . This is the acceleration of the piston of the steam engine mechanism, L being the length of connecting rod, r that of the crank, and w the angular speed of the crank. 20. Show that the maximum and minimum values of an integral algebraic function occur alternately. 21. The specific heat of superheated steam is given by the equation c = a + pT + p(l + jp\ — — - , where T denotes the absolute temperature, p the pressure, and a, /3, a, O, and n are constants. If p is kept constant, show that c takes a minimum value for some temperature T m and that this value is a + n+ pT m . n + 1 22. A rectilinear motion is such that the acceleration is given by the ex- pression a = e f + e l ; derive expressions for velocity and distance. Show that if the particle starts from rest, the distance is numerically equal to the acceleration. 23. Find expressions for velocity and distance when the acceleration is given by a . = m — nk 2 coskt. Determine the constants of integration Ci and C 2 by taking v = and s = when t = 0. CHAPTER IX CURVES 86. Concavity. In the present chapter we shall discuss some of the applications of differentiation to plane curves, and develop principles that will enable the student to trace a given curve and study its properties. A curve is said to be concave upward at any point if an arc of the curve containing the point lies above the tangent to the curve at the point. It is said to be concave downward if the tangent lies above the arc. Thus the curve shown in Fig. 31 is concave down- ward between the points A and B, and it is concave upward from Bto a > X Fig. 31. The condition for concavity upward or downward can be ex- pressed in terms of the derivatives of the function represented by the curve. When the arc of the curve y =f(x) lies below the tan- dij gent, as between the points A and B, tan cf>, and consequently -~, ctx decreases as x increases. On the other hand, when the arc lies 142 Art. 86] CONCAVITY 143 above the tangent, -^ increases with x. We have therefore the dx general statement: A given curve is concave upward or downward according as the derivative — is an increasing or a decreasing function, dx When the derivative -^ is decreasing, its derivative, namely the dx second derivative, — -, must be either negative or zero; if -f- dx 2 dx is increasing, then the second derivative is either positive or zero. Therefore, if — ^ is different from zero, the preceding statement is equivalent to the following : A given curve is concave upward if the second deHvative is posi- tive, and concave downward if the second derivative is negative. Ex. Given the curve y = x 3 — x 2 + 6, investigate its concavity. The second derivative, — ^ = 6 x — 2, is positive for values of x > h and is dx? negative for values of x < |. Hence, to the left of x = | the curve is concave downward, and to the right of that point it is concave upward. EXERCISES Test the following curves for concavity upward or downward. 1. y = x 3 -2x 2 + 5. 2. y = secx. 3. y = ^ 4. y = ay/x — a. 5. y = \ (e x + e~ x ) . 6. Show that a curve is concave or convex toward the X-axis according &y dx 2 7. Show that the ellipse b 2 x 2 + a 2 y 2 = a?b 2 is everywhere concave toward the X-axis. 8. Test the curves y — e x and y = logic for concavity. 9. Test the expansion curves given by the general equation p m v n = C for concavity. 10. Show that the curve ay 2 = x s has two branches, each of which is con- vex to the X-axis. 144 CURVES [Chap. IX. 87. Points of inflexion. A point at which a curve having a continuous slope ceases to be concave upward and becomes concave downward, or vice versa, is called a point of inflexion of the given curve. The curve shown in Fig. 31 has points of inflexion at A, B, C. It is evident that at such points the curve crosses its tangent. The definition suggests the analytic condition for a point of inflexion. As has been shown, so long as the curve y =/(#) is concave upward the first derivative increases as x increases, and when it is concave downward the derivative decreases as x in- creases. It follows then that, as x passes through a value for which the curve has a point of inflexion, the first derivative passes from an increasing to a decreasing function, or vice versa. See Fig. 10. This is precisely the condition that the derived function shall have a maximum or minimum ; and to determine the points of inflexion of the curve, we need only to examine the derived function for maxima and minima. It is therefore a necessary and sufficient condition for a point of inflexion that the second deriva- tive shall change sign as the independent variable passes through the critical value. This change occurs when the second deriva- tive passes through zero or through infinity; hence the coordi- nates of the points of inflexion of the curve may in general be found by solving the equations f"(x) = 0, /"(*)=<», and determining whether f"(x) changes sign as x passes through the values thus obtained. Ex.- Given the curve whose equation isy = 5x B + 6x — 7. We have /'<*) = £« = 80*, which vanishes for x = 0. Moreover, f"(x) changes sign as x passes from negative to positive values. Hence the curve has a point of inflexion at x = 0. EXERCISES Test the following curves for points of inflexion. 1. y = 2x s -Sx 2 + 4x- 6. 2. y = 3 x 4 + 4x 2 -z + 10. 3. {y— 3) 2 = a;+5. 4. y = sin x. 5. y = cotx. Art. 87] ASYxMPTOTES, RECTANGULAR COORDINATES 145 6. y = x 2 — e x . 7. y — e x — e~ x . 8. y = arc cot x. 9. y = e~*\ 10. * = -^™ 11. y = * 2 + 3 1 + 6a; 2 12. Given a continuous curve. Draw a tangent to this curve, and through some fixed point draw a straight line which shall so change as to remain par- allel to the tangent as the point of tangency changes. Show that as the point of tangency passes through a point of inflexion, the line through the fixed point changes the direction of its rotation. For this reason, the tan- gent at a point of inflexion is sometimes called a stationary tangent. Ex- press in terms of the derivatives of the given function the condition for a stationary tangent. 13. Find the equation of a curve which has a point of inflexion at the point (0, 3) such that the inflexional tangent makes an angle of 45° with the X-axis. How many such curves can be found ? 14. Show that the curve y = has three points of inflexion and that 1 +x 2 these points lie in a straight line. 88. Asymptotes, rectangular coordinates. An asymptote to a plane curve is a straight line, lying partly within the finite region, which is the limiting position of a tangent to the curve as the point of tangency recedes indefinitely along an infinite branch. Two conditions are necessary for an asymptote : (1) The curve must have at least one infinite branch. Thus an ellipse having no infinite branches cannot have an asymptote. (2) The limiting position of the tangent must lie partly within the finite portion of the plane. For example, the tangent to a parabola at infinity is not an asymptote since it lies wholly at infinity. There are two general methods of determining the asymptotes to a curve whose equation is given in rectangular coordinates. Method of limiting intercepts. The equation of the tangent at the point (x lf y{) being y-y^f'ix^-xi), (Art. 38) the intercepts of the tangent on the coordinate axes are re- spectively: . ?/ , Intercept on X-axis = x x u\' W / ( x i) Intercept on F-axis = y x — ^if'i^h)- (2) 146 CURVES [Chap. IX. If one or both of the intercepts has a finite value for a^ = oo or y l = oo , the infinite branch has an asymptote, and the equation of the asymptote may be found from the two intercepts or from one intercept and the limiting value of f'(xi). The following ex- ample will illustrate this method. Ex. 1. Examine the curve whose equation is y z = x 3 — 3 x 2 for asymptotes. Differentiating, we obtain Sy 2 ^ = Sx 2 -6x, dx whence /' (a*) - x ^ ~ 2xi From (1) and (2) we have 2/i X-intercept = x x «3 = — ^ , zi 2 -2zi x{*-2x 1 ' T-intercept = y x - ^ - 2 Xl * ~ -^ For X\ = oo , these intercepts are respectively 1 and — 1. Remembering that the equation of a line in terms of its intercepts a and b is 5 + J-l, a b we have as the equation of the asymptote, x — y = 1. Method of substitution. Let f(x, y) = be the equation of the given curve, and assume the equation of the tangent to be y = mx -f b. (3) Combining these equations, we obtain the equation f(x, mx + b) = 0. (4) Now a tangent to a curve may be regarded as the limiting posi- tion of a secant line as the two points in which this line inter- sects the curve are made to approach coincidence. Hence if the line y = mx -f- b is tangent to the curve f(x, y) = 0, equation (4) must have equal roots, and if the point of tangency recedes in- definitely these roots become infinite. The condition that two of the roots of (4) shall be infinite is that the coefficients of the two highest powers of x in (4) shall vanish.* If, therefore, We * Rietz and Crathorne's College Algebra, Art. 111. Art. 88] ASYMPTOTES 147 equate these coefficients to zero, we can determine the values of m and b, and consequently the equation of the asymptote. Ex. 2. Examine for asymptotes the curve whose equation is x* — x 3 y + xy — y 8 = 0. Substituting y = mx -f &, the resulting equation is a*(l-m) + «*(-»»*-&) + b s = 0. Equating the coefficients of x 4 and X s to 0, we have 1 - m = 0, m* + b = 0, whence m = 1, and 6 = — 1. Substituting these values in (3) , we have as the equation of the asymptote y = x— 1. If in equation (4) the coefficients of x n and ic n_1 contain both m and 5, the terms of degree lower than the (n — l)th do not in general affect the determination of the asymptotes. However, if the coefficient of a" -1 is zero, or if the value of m obtained by equating the coefficient of x* to zero is such as to cause that of x n ~ l to vanish also, we must equate the coefficient of the next lower degree to zero in order to have two equations from which m and b can be uniquely determined. This coefficient, in general, will be of the second degree in b, and hence we have in this case two parallel asymptotes, since for each value of m there are two values of b. Method of inspection. In case the asymptotes are parallel to one of the coordinate axes, we can often determine such asymp- totes by inspection. For this purpose the equation of the curve is written in descending powers of x or of y. It will then have either of the following forms : aar + (by + c)a?- 1 + ••• +£ = 0, (5) *f + (fix + y)tf l ~ 1 + '" +r = 0. (6) If both a and by + c vanish, then two roots of (5) become infinite, and the line by + c = (7) is an asymptote. Similarly, if in (6) both a and f3x + y vanish, theline fix + y = (8) is an asymptote to the given curve. Consequently, if the given equation, when arranged in descending powers of x (or y), does 148 CURVES [Chap. IX. not have the term x n (or y n ), then the coefficient of x n ~ l (or y n ~ l ) equated to zero gives an asymptote to the given curve. If in (5), a, b, and c all vanish, then the coefficient of x n ~ 2 , which is in general a quadratic in y, will, when equated to zero, deter- mine two asymptotes, real or imaginary. A similar statement applies to (6). Ex. 3. The equation 4 x 3 + 2 x 2 y - 6 xy 2 - x 2 + 3 y 2 - 1 = is a cubic in which y s is absent. Hence arranged in descending powers of y, it takes the form o y* - (6 x - 3)y 2 + (2 x 2 )y + (4 x 3 - x 2 - 1) = 0. If we make 6 x — 3 = 0, the coefficients of y* and y' 2 both vanish. Hence the line 6 x — 3 = 0, or x = \ is an asymptote. For x = ^, we have also y = — f , so that the asymptote cuts the curve at the point Q, — f). Ex. 4. The equation x 2 y 2 - 4 xy 2 + 3 x 2 ?/ — 4 x 2 - 5 = is of the fourth degree, but lacks the terms in x 4 , x 3 , y 4 , and y*. It may be arranged in the f0rmS x 4 + x 3 + (y 2 - 4) x 2 - (4 y 2 - 3 y) a; - 5 = 0, y 4 + ?/ 3 -f {x 2 - 4 a;) y 2 + (3 x) y - (4 x 2 + 5) = 0. From the first we have two asymptotes, y + 2 = 0, and y - 2 = 0, determined by placing the coefficient of a; 2 equal to zero. From the second, the asymptotes x = 0, x — 4 = are obtained. The methods given are sufficient to determine all the asymp- totes to an algebraic curve. The second method will be found most convenient in determining asymptotes that make an oblique angle with the axes of coordinates, and the third in finding asymp- totes parallel to the axes. If the equation of the curve involves a transcendental function, the first method may be used. EXERCISES Find the asymptotes of the following curves : 1. y 8 = x 2 (x-b). 2. tf = tf(*±JL\ m \x — a) 3. y 2 = x * • 4. £-* = l. 9 2a -x a 2 b 2 5. y 2 (x-2)= x 2 . 6. y 3 - y 2 - x 2 + X s = 0. 7. x 2 y 2 - c 2 (x 2 + y 2 ) = 0. 8. x 3 + xy 2 -y 2 = 0. Arts. 88, 89] SINGULAR POINTS 149 9. x 8 — xy 2 + ay 2 = 0. 8«3 li. y 3 13. x 3 6 x 2 + x\ y s - x 2 + 2 */ 2 = 0. 15. y + zy 0. 10. V x 2 + 4 a 2 12. a^ = 1. 14. x 2 y + xy 2 = 8. 16. z 2 ?/ 2 - z 3 - ?/ 3 = 0. 89. Singular points. Certain points of a plane curve may have peculiarities not possessed by other points. Thus there may be a multiple point, where two or more branches of the curve intersect, Fig. 32 (a) and (b) ; a tacnode, where two branches of the curve come in contact and have a common tangent at the point, Fig. Fig. 32. 32 (c) ; a cusp, where the two branches terminate at the point of contact and have a common tangent, Fig. 32 (d) ; or a conjugate point, the coordinates of which satisfy the equation of the curve yet through which no branches of the curve pass. All such points are called singular points. At a singular point the derivative -^ has two or more values, real or imaginary, (XX equal or distinct. Geometrically this means that at a singular point the curve has two or more tangents, though some of these may be coincident, and some or all may be imaginary. The character of the curve at a singular point depends therefore upon the values of -* at that point. A general method of evaluating dx 150 CURVES [Chap. IX. -* at singular points cannot be given until later; fortunately, ClOC however, simple special methods are sufficient for. the cases that ordinarily arise. The following examples are illustrative. Ex. 1. Examine the curve y 2 = x 2 — x 4 for singular points. Forming the derivative, we have dy _ 2 x — 4 x 3 _ 2 x — 4 x a dx 2y For 2xVl -x 1 < x < 1, or for — 1 < x < 0, every value of x gives two distinct values of y and two corresponding determinate values of -£. For x = 0, there is one dx value of y, namely y = 0, showing that two branches of the curve pass through the origin. But for x = 0, -^ takes the indeterminate form - and dx this must be evaluated. Using the method of Art. 15, we have 2 x — 4 x 3 r l-2x 2 L - -=±1. tf = 02xVl-X 2 a? = 0Vl-X 2 Hence at the origin -^ = ± 1 and the tangents to the two branches are dx respectively y = x and y = — x. Evidently the curve lies between the limits x = 1 and x =— 1. The student may draw the curve. In relatively few cases can tjie indeterminate expression for -^ dx be evaluated by simple methods as in Ex. 1. Generally the method shown in the following example is effective and is easily applied. Ex. 2. Examine for singular points the curve given by the equation x 4 - 4 x 2 y + if = 0. In this case more than one branch of the curve passes through the origin ; therefore the origin is a point to be examined. Substitute y = mx in the given equation. The result is an equation in x, x 3 (x - 4 wi + m 3 ) =0, (1) which has three roots each equal to zero. Hence there are three branches passing through the origin, that is, the origin is a triple point. From (1) we have also the equation x - 4 m + m* = 0, (2) which gives the relation between the slope m of a secant line y — mx that cuts the curve at the origin and in a second point whose abscissa is x. Since (2) is a cubic in m there are three such lines. As x approaches zero the Art. 89] SINGULAR POINTS 151 points of intersection approach coincidence and each secant line approaches as a limiting position a tangent at the origin. Therefore putting x = in (2) and solving the resulting equation m 3 — 4 m = 0, we get in the three roots m = 0, 2, and — 2 the slopes of the three tangents to the three branches passing through the origin. The curve is shown in Fig. 33. Ex. 3. Examine for singular points the curve ay = (x — by. In general y has two values for any value of x < b, but for x = b, y has the single value ; hence the point (6, 0) is to be examined. For the sake of convenience let the origin be shifted to the point (&, 0). Substituting x—b — x and squaring to re- move the radical, we have a 2 y 2 — x ,b — 0. Letting = mx' i*y 2 ;', we obtain the equation in x', x' 2 {a 2 m 2 - x 3 ) = 0, which has two zero roots, showing that two branches pass through the origin. From the equation a'm' 0, we have f or x = 0, a 2 m 2 = 0, an equa- >X tion in m with equal zero roots. It follows that the branches have a common tangent y — 0. For nega- tive values of x\ that is, for x 2 a or x < 0, y is imaginary ; hence the curve lies wholly between the Y-axis and the line x = 2a, which is an asymptote. For x = 0, ^ = 0, showing that the X-axis dx is a tangent to both branches at the origin. An investigation of the second derivative shows that there is no point of inflexion and that both branches are convex towards the X-axis. The curve has therefore the general form shown in Fig. 38. $a* Ex. 3. The witch y = The curve is symmetrical with respect z : + 4a 2 to the Y-axis, y has a maximum value 2 a for x = 0, and there are points Y Fig. 40. of inflexion at x = ±2aVS. The X-axis is an asymptote. The curve is shown in Fiff. 39. Art. 90] CURVE TRACING 155 Ex. 4. The lemniscate p 2 = a 2 cos 2 0, or (x 2 + y 2 ) 2 = a 2 (x 2 -y 2 ). The curve is symmetrical with respect to both axes, and crosses the X-axis at x = ± a. For values of from \ ir to f ir and from f w to \ w, p is imagi- nary. The origin is a double point, the tangents having the slopes — 1 and 1, respectively. The curve has the form shown in Fig. 40. X X Ex. 5. TJie catenary y = - (e a + c a ) is the curve assumed by a flexible A cord of uniform weight suspended trom two fixed points. The curve is sym- metrical about the T-axis, and cuts the y-axis at a distance a above the origin. The second derivative is positive for all values of x, hence the curve is concave upwards and has no point of inflexion. See Fig. 41. Fig. 42. Ex. 6. The cycloid is the curve described by apoint on the circumference of a circle which rolls on a straight line. Let a denote the radius of the circle, and the angle PCM, Fig. 42, subtended by the arc PM {=■ OM). Then we have for the coordinates of P, x = a(d — sin 0), y = a(l — cos 6). From these equations, we readily obtain dy _ sin d d?y 1_ dx ~ 1 — cos 0' dx 2 ~ a(l - cos 0) 2 The curve has a turning point at x — ira, and since — ^ is always negative, dx 2 the curve is everywhere concave downward. Ex. 7. The astroid, or hypocycloid of four cusps, x' 5 + y* = a? is a curve described by a point on the circumference of a circle of radius \ a rolling within the circumference of a circle of radius a. See Fig. 43. The student may find the derivatives and by their aid study the curve for slope, concav- ity, etc. Ex. 8. The cardioid /> = 2a(l — cos0). The curve is closed, and p is finite for all values of 0. Since cos = cos (—0), the curve is symmetrical with respect to the initial line OX. From the given equation, we get tan \p =z p dd 1 — cos whence sin0 = tan \ 0, (Art. 40) 156 CURVES [Chap. IX. For 6 = 0, \p = 0, that is, OX is a tangent at the origin ; for d r )f = iir, and for d = tt, \p = \ ?r. For the limits of p we have p = when = 0, />= 4 a when = v. See Fig. 44. Fig. 43. Fig. 44. EXERCISES The student should trace the following curves carefully and preserve the graphs for future reference. 1. The group of curves x m y = G{m positive). These are the curves which represent the laws of expanding gases. 2. The curve x- + y* = a?. Show that this curve is the ordinary parabola. 3. The folium of Descartes, X s + y 8 — 3 axy = 0. 4. The exponential curve y = e x . 5. The logarithmic curve y = log x. 6. The logarithmic spiral p = e ae . 7. The spiral of Archimedes, p=a0. 8. The curve p = a sec 2 9. The parabolic spiral p 2 = a 2 0. 10. Rankine's equation for columns, y = 1 + bx* 11. The probability curve y = ke-"* 2 , a < 0. Find turning 'points and points of inflexion. 12. The lituus pH - a 2 . 13. The curves p = a sin nd, p = a cos nd. (Give n values 1, 2, 3, 4.) 14. Show that the cardioid is described by a point on the circumference of a circle of radius a which rolls upon a fixed circle of the same radius. Art. 91] CURVATURE 157 91. Curvature. If a point moves along a plane curve as EF, Fig. 45, the direction of motion at any point is the direction of the tangent to the curve. The direction of the tan- gent continually changes, and a comparison of this change of direction with the distance traversed by the point leads us to the idea of curvature. Thus, as the point moves from P to Q, the tangent turns through the angle A<£, and we say the curvature of the arc PQ is large or small according as the angle Acf> is large or small. Denoting by As the length of the arc PQ, the ratio — * is de- fined as the mean curvature of the arc PQ. Provided the curva- ture is constant, this quotient is the curvature at all points be- tween P and Q ; but if the curvature is not constant, then the curvature at P is defined as the limit A s = As ds (1) The curvature of the curve y =f(x) at the point (a^, ft) may be expressed in terms of the derivatives f'(x) and f"(x). We have the fundamental relation tan =f\x), = arc tan f(x), whence by differentiation Furthermore, we have i + [/'(*)] 2 i + [/'(*)T Dj* = y/l+[f{x)y* (See Art. 47) * A derivation of this relation is given in Art. 108. For the present, the student may make use of Eq. (1), Art. 47. 158 CURVES [Chap. IX. From these equations, we obtain by division At the point (x 1} y^, therefore, the curvature is f"(vi) (2) (3) 92. Radius of curvature. Center of curvature. Suppose that PQ, Fig. 45, is an arc of a circle. Let normals be drawn at P and Q intersecting at M\ then MP= MQ is the radius of the arc, and angle PMQ = A<£. Denoting the radius by r, we have rA<£ = = As ; A<£_ As" 1. - 5 r whence — - = -; (1) As r that is, the curvature of a circle is constant and is the reciprocal of the radius. At any point of a curve y =f(x), conceive a circle drawn tan- gent to the curve, and suppose it to have the same curvature as the curve at that point. This circle is called the circle of curva- ture ; its radius, the radius of curvature ; and its center, the center of curvature of the curve at the given point. Since the circle and curve have the same curvature at the point in question, the radius of curvature R must be the reciprocal of this curvature ; that is, 11 = ^-. d In terms of the derivatives involved, the formula for the radius of curvature at the point (x x , y{) becomes, therefore, R /»(*!) * (2) By hypothesis, the curve and the circle of curvature at the given point have a common tangent; hence the radius of curva- ture has the direction of the normal to the curve. Art. 92] RADIUS AND CENTER OF CURVATURE 159 It is customary in the case of single-valued functions to regard R positive or negative according as the curve is concave upward or concave downward. To establish this convention, it is neces- sary to take the positive value of the radical in the numerator. The sign of R then depends upon that of f"(x), and the required result follows from Art. 87. In many cases it is the numerical value of R alone that is of importance. d 2 v As we have seen,/"(#) = — ^ changes sign at a point of inflexion ; hence R also changes sign at such a point. This property might have been used as the definition of a point of inflexion. Thus a point of inflexion is a point at which the radius of curvature, and conse- quently the curvature itself, changes sign. The coordinates of the center of curvature may be found as follows : If (m, n), Fig. 46, is the center of curvature corresponding to the point ( x d Vi) °f the curve, we have from the figure, x 1 — m = R sin , n — Fig. 46. y x = R cos . But whence cos 1 Vl+L/'Caa)]* tan <£ =f(x 1 ), , sin = /'(*,) Vl +[/'(*,) j Using these expressions for sin <£ and cos , and the expression for R given by (2), we obtain after reduction m = oci — n = yx + /'(^iMi+r/'(^i)] 2 S /"(-l) 2 If a curve is given in polar coordinates, we have x = p cos 0, ?/ = p sin 0. (3) 162 CURVES [Chap. IX. By means of these relations (3) is reduced to the form (5) -(% *+21*b\*-~*£ R = Using the functional symbols for the derivatives, the radius of curvature of the curve p = F{6) at the point (p lf 0^) is R = ^ 1 ^ L v lM (6) P l 2 +2[F>(6 1 )Y-p 1 F>X0 l ) EXERCISES 1. Find the radius of curvature of the ellipse from the equations X = a cos 0, y =b sin 0. 2. Find the radius of curvature of the hypocycloid from the equations x = a cos 3 0, y =a sin 3 0. Derive general expressions for the radii of curvature of the curves given by the following equations in polar coordinates. 3. p = ad. 4. p = a(sin + cos 6). a 5. /> = asin 3 -- 6. p = e a0 . 7. Find the maximum radius of curvature of the cardioid p = 2 a(l — cos 6). 8. Find the minimum radius of curvature of the lemniscate p 2 = a 2 cos 2 0. 9. From the answer to Ex. 1 show that the radius of curvature of the ellipse is greatest at the extremity of the minor axis and least at the extremity of the major axis. 94. Roulettes and involutes. Suppose a plane curve, as AB, Fig. 47, to roll without slipping on a fixed curve MN\ for this purpose we may regard the curves as the boundaries of two disks. Points E, F on the rolling curve describe curves e, f on the fixed plane. Curves generated in this manner are called roulettes. Cycloids and trochoids are examples of roulettes in which the rolling curve is a circle. (See Art. 90.) If the rolling curve is replaced by a straight line, Fig. 48, the roulettes are called involutes of the fixed curve; thus, curves e and /are involutes of the curve MN. Evidently a curve has an infinite number of involutes. Arts. 94, 95] ROULETTES AND INVOLUTES 163 AVe may also consider the involute to be generated by a point on a flexible thread or cord wrapped around the fixed curve. As the cord is unwrapped, any point of it generates an involute. Fig. 47. Fig. 48. Certain properties of involutes are evident from the manner in which they are generated. 1. Any normal to an involute is a tangent of the fixed curve. This property may be seen from purely mechanical principles. Keferring to Fig. 48, the point of contact P, considered as a point of the rolling line AB, has at the instant of contact no motion in the direction of the line, since by hypothesis there is no slipping. Furthermore, since the curves remain in contact, P can have no motion perpendicular to AB. It follows that the point P of AB is at rest, and the curve as a whole is rotating about P as a center. Points E and P are therefore moving in directions perpendicular to the lines PE and PF respectively. Now the point E moves in the direction of the tangent to the curve it describes ; hence the line PE is the normal to the curve e at E, and likewise PF is normal to curve/ at F. In the case of the involute, Fig. 48, the normals PE and PF coincide with the rolling line AB, which is tangent to MN\ hence the normal of the involute is tangent to the curve. 1. Two involutes of the same curve intercept a constant distance on their common normal, This follows at once from the manner in which the involute is generated. Because of this property, involutes are sometimes called parallel curves. 95. Curvature of involutes. Let PQ, Fig. 49, be one position of the rolling straight line, P being the point of contact, and Q sl 164 CURVES [Chap. IX. point on the involute. Let m, n denote the coordinates of the variable point Pon the fixed curve MN f and x, y those of the point lY (*,y) Q on the involute. From the geometry of the fig- ure, we have at once (m— #)tan <£ = n — y. (1) Since QP is tangent to MN, we have tan cf) — dn m dm Fig. 4U. dx but since QP is the * X normal to the involute at the point Q, we have also tan = — — . Substituting this value in (1) we obtain dy (m—x) + (n — y)-^ = 0. (XX (2) Differentiating (2) with respect to n, we have dm dx dn dn But since this equation reduces to dy\dy , . dn dm dn dx dy dy . dn dn dx ^ dn dy dx' i,) dn and multiplying through by — , we obtain finally, (XX whence -'-( dx v 9J da? 1 + n = y + d 2 y dx 2 (3) Art. 96] EVOLUTE OF A CURVE 165 Substituting this value of n in (2), we have dxl \dxj J m = x d 2 y cte 2 (4) Comparing (3) and (4) with (3) of Art. 92, it appears that the point P is the center of curvature and PQ is the radius of curvature at the point Q of the involute. It is not true of roulettes in general that the point of contact of the rolling curves is the center of curvature of the roulette. Thus in Fig. 47, the center of curvature of e at the point E lies somewhere on EP or EP produced, but not at P. Only in the case of the involute does the center of curvature coincide with the center of rotation. 96. Evolute of a curve. From the preceding article it appears that a given curve is the locus of the centers of curvature of each of its involutes. When two curves C x and C 2 are so related that Ci is an involute of C 2 , we call C 2 the evolute of C v Curve C 2 may have other involutes than Ci, but Ci has only the one evolute C 2 . Two properties of the evolute follow from the manner of de- scribing the involute as a roulette, namely : 1. Any normal to the curve Ci is tangent to the evolute C 2 . 2. The difference between two radii of curvature of a given curve is equal to the arc of the evolute between the points of contact of the radii ivith the evolute. Thus in Fig. 50, we have P 2 E 2 -P 1 E 1 = arc P X P 2 Fig. 50. To obtain the equation of the evolute of a given curve, we com- bine the equation of the curve y=m (l) 166 CURVES [Chap. IX. with the equations n = 2/ + _\dx[ dx 2 dx\ m — x — ^fi + (*£ \dx dx 2 (2) (3) which give the coordinates m, n of the center of curvature. If x, y, and the derivatives be eliminated between these equations, the result will be a relation between m and n, the variable coor- dinates of the e volute. Various special expedients may be used in the elimination. Ex. 1. Find the evolute of the rectangular hyperbola xy = — . We have Hence Therefore and dy dx 2x 2 d 2 y = a? dx 2 x s ' 2x 4 a: 4 + a 4 4a 2 x x + 4x* 4 ^ + 3a 4 4 a 2 x 12 x 4 + a i 8x* 8x 3 ra + n 8x* + 12 x 4 a 2 + 6 x 2 a* + « 6 8a¥ \ 2 ax J. 2 ax 2x 2 \ (a 2 -2x 2 \ m — n=a[ ] \ 2 ax J (m + riy — (m — ny which is the desired equation of the evolute. Finally 2d 1 EXERCISES Find the equations of the evolutes of the following curves. 1. The circle x 2 + y 2 = a 2 . 2. The parabola y 2 = 4px. 3. The ellipse ^ + £- * a 2 b 2 1. Art. 96] MISCELLANEOUS EXERCISES 167 4. Find the e volute of the ellipse using the parametric equations x — a cos d, y = b sin 6. 5. Find the e volute of the curve given by the parametric equations x= a (cos 6 + 0sin 0), y = a (sin — 0cos 0). MISCELLANEOUS EXERCISES Find the radius of curvature and coordinates of the center of curvature of the following curves at the points indicated. 1. y 2 = 8x + l, at (6, 7). 2. y = 3 x 3 - 8 x + 4, at (- 2, - 4). 3. y = cos ac, at (0, 1). 4. y = xe x , at (0, 0). 5. Show that the evolute of an arch of the cycloid consists of the halves of an equal cycloid. 6. Show that the radius of curvature at any point of a cycloid is double the length of the normal at the same point. 7. If the equation of a curve can be written in the form y= ± (ax+ b) +0(x), where (#) is a rational fraction the denominator of which is higher degree than the numerator, show that the lines y = ± (ax + b) are asymptotes to the curve. In this way determine the asymptotes to the curve x s — xy 2 + ay' 2 = 0. 8. Show that the ordinates of the curve x s = xy + 1 = and of the parabola y = x 2 approach the same value as x increases. For this reason, the parabola is said to be a curvilinear asymptote to the curve. 9. Trace the curve y + xy — x z = 0. Find its rectilinear and curvilinear asymptotes. 10. Show that at the point (0, 0) the curve y = — * — has two terminat- ing branches with different tangents. , . - Examine for singular points the following curves. 11. x(x - ay + y 2 (x - 2 a) = 0. 12. (x 2 + y 2 )* = 4x 2 + y 2 . 13. x* - 3 xy + 6 xj 2 - y* = 0. 14. By the method of limiting intercepts, find the asymptotes to the curves (a) y» = 4 st* -f «8 ; (b) ** - v - - 1 . a' 2 b 2 15. Find the radius of curvature of the three-cusped hypocycloid from the parametric equations x = o(2 cos B + cos 2 0), y - «(2 sin 6 - sin 2 6). 168 CURVES [Chap. IX. 16. The tractrix is a curve having the property that the length of the tan- gent is a constant a. Find the equation of this curve. 17. Show that a curve whose equation is given in polar coordinates has dfi an asymptote when the subtangent p 2 — is finite for p = • dp 18. Using the result of Ex. 17, find the asymptotes of the following curves : (a) p = a tan 9 \ (b) p = — ; (c) p = a sec 2 d. 19. Prove that the evolute of the logarithmic spiral p = ae k0 is a similar logarithmic spiral. 20. Find the equation of the evolute of the cissoid 2 a — x 21. A circle of radius b rolls on a fixed circle of radius a, a > 6, and a point on the rolling circle generates an epicycloid. Show (a) that the equa- tions of the curve are x = (a +b) cos — 6 cos g "*" 0) dx = D^/ix) =(x)+ a "(l) The difference of the values of the function (x) + C for any two values of the independent variable is called a definite integral. The values of the independent variable substituted are called the limits of integration. We denote a definite integral symbolically by writing the two limits of integration at the extremities of the sign of integration. Thus, if x = a and x = b are the limits of integration in the definite integral formed from (1), we indicate that fact by writing this definite integral as follows : x f{x) dx = lD^f(x)-] a = * (6) - * (a) . (2) This symbol is read : " The definite integral of f(x) between the limits x = a and x= b" or "from a to 6." It is to be noted that in the definite integral the constant of integration disappears. This result follows from the definition ; for, we have [> (x) + 0]\ = [> (b) + 0] - [> (a) + 0] = (b) - *(a). (3) To distinguish definite integrals from those previously discussed, we call the latter indefinite integrals. We may not only pass from the indefinite to the definite integral by the process already indi- cated, but conversely we may pass from the definite to the indefinite form of the integral by assuming the upper limit of integration as variable. Thus, we have s: /(»)(& = <£(»)-<£ (a), where <£ (a) may be taken as the constant of integration. 169 170 DEFINITE INTEGRALS [Chap. X. EXERCISES Verify the following : 1. £ 4 a* =: 1 -cos/3. 4. f — = *. > VY-x 2 « 5. ("»(b)-f(a), (4) £f(x)dx=cf>(c)-cf>(a), (5) £f(x)dx=(b)-(c). • (6) Adding (5) and (6), we obtain f /(a) dx + Cf(x) dx = c£(c) - (a)+ $(b) - <^,(c) =jf >(*)*», (7) 172 DEFINITE INTEGRALS [Chap. X. which establishes the theorem. Evidently the theorem may be extended to include any finite number of values between a and b. EXERCISES 1. Evaluate the integral I (3 x 2 + 5) dx, using successively the limits of integration 1 to 3, 3 to 4, 4 to 6, and 1 to 6, and verify Theorem II by the result. 2. By evaluating the integrals, show that IT ysinddd + ("sinddd = f "sin Odd. 2 3. Show that i 0(w) du = I 0(x) dx. State this result in the form of a Ja Ja theorem. 99. Change of limits. In the process of finding an integral, it is sometimes convenient to change the variable (see Art. 71). In this case, we may by properly changing the limits of integration obtain the definite integral without a second substitution. Sup- pose we have the integral I f(x) dx and make the substitution z = F(x). For x = a, we have z = F(a), and for x = b, z = F(b)\ hence the substitution of F(a) and F(b) as limits of integration in the transformed integral must lead to the same result as the sub- stitution of a and b in the original integral. Ex. Find (* arc Sln x d x , assuming z - arc sin x. Jo VT^ We have dz = dx . whence arc sin x d x = z dz. VI -x 2 Vl - x 2 For x = 0, z= arc sin = 0, the lower limit, and for x = |, z = arc sin \ =^ , the upper limit. Hence /* 2 arc sin x dx . =P »<**=£• J° Vl - x 2 72 EXERCISES r 2 xdx Jo 1 4- x* ' Let z = x 2 . r 2 2 - »r) / 2 (») + •••+(&- ®»_i) /„(«), (1) where Aa^ denotes any one of the subintervals (x t — a^). In each subinterval Ax { , let a { , /3 { denote the smallest and largest numerical values, respectively, of f(x). With Ax f as a base let two rectangles be constructed having a { and ft respectively as their altitudes. Repeating this construction for each subinterval, we have the relation » « » V a^Aa, ^ V /^Aa;, 1 V ftAav (2) i i i The first and the last of these sums have limits as n increases indefinitely since both are monotone functions and limited in magnitude by the conditions placed upon fix). Moreover these limits are equal, say equal to A, and that independently of the manner of subdividing the given interval (a, &).* Since A# = as n becomes infinite, we may now write (see Art. 13) L Vf(x)±x = A. (3) The existence of the limit in (3) being established, we may find the value of the limit by taking the subintervals and the corre- sponding values of the function in any particular manner we please. It is convenient to make the values of Asc equal and to take the value of fix) at the beginning of each interval. We have then for all values of n, nAx=b — a, (4) and the original points of division become a + A#, a -h 2 Ax, • • •, b — Ace. In defining the definite integral, <£ (a?) was taken as a function having f(x) as its derivative. From the law of the mean, we have therefore (x + Aa;) - (x) = Ax • f{x + ■ Aa), < 6 < 1. (5) *See Picard's Traite & Analyse, Vol. 1, p. 4 ; or Veblen and Lennes' In- finitesimal Analysis, p. 150 et seq. Art. 100] THE LIMIT OF A SUM 175 By hypothesis a and /3 are respectively the minimum and maxi- mum values of f(x) in a subinterval ; hence we may write a^f(x + 6- Ax)^/3. By aid of (5) this inequality becomes a • Ax ^ <£> ((b)-cf> (a) = A, (9) and we may therefore write b r* A Jlo]£/0*0 A « = (&)-4>(a)= I f(x)dx. (10) We may henceforth replace at any time the summation pro- cess by the definite integral, and regard the two symbols, A . X an( ^ I as interchangeable whenever the integrand a is a continuous function. From this discussion it also follows that the definite integral 176 DEFINITE INTEGRALS [Chap. X. »6 f fix) dx is represented graphically by the area bounded by the curve y =f(x), the X-axis, and the two ordinates x = a, x = b. lOl. Importance of the summation process. The result just derived and expressed by equation (10) of the preceding article is of the highest importance, for it enables us to replace a difficult and tedious process of direct summation by a process that is iti most cases simple and easily carried out. As was shown in the example of Art. 7, the determination of the area under a curve involves the summation of an indefinitely large number of indefinitely small terms, that is, it requires the limit ^_ Q V f(x) Ax. a But according to (10) the limit is given by the definite integral j f(x) dx. Hence to find the area we have only to find the anti- derivative <£ (x) of the given function fix), substitute the limits a and b, and take the difference (5) — (a). For example, to find the area under the curve y = x 2 from the origin to the ordinate x = 3 (see Art. 7), we have A= Cf(x)dx= Cx*dx = \x*f = §. While the discussion leading to equation (10) was accompanied by a geometrical illustration, and the summation was directed toward the determination of an area, the course of reasoning depends in no way upon geometrical considerations. The method of the definite integral is applied with equal facility to the determination of magnitudes of all kinds — volumes, masses, fluid pressures, heat, work, etc. In the following chapters there will be given examples of the use of the summation process in finding the lengths of curves, the areas of the surfaces, the volumes of solids, etc. In mechanics, determinations of centers of gravity and moments of inertia likewise involve the summation principle. The work done by a variable force is found by the summation of terms of the type FAs, where ^denotes force and s displacement. The impulse of a variable force is the summation of terms of the type F&t. The space over which a moving point travels is found by the summation of terms of the type vAt, where v denotes the Art. 101] IMPORTANCE OF THE SUMMATION PROCESS 177 velocity of the point. If the specific heat c of a substance varies with the temperature r, the heat that must be imparted to the substance to produce a given rise of temperature is determined by the summation 2c At. Other applications will occur to the student. It should be noted that the summation of an infinite number of terms is always necessary when one of the factors entering into the problem varies continuously. As an illustration, take the problem of finding the mass of a body. If V denotes the volume of the body and y its density, the product yV gives the mass, provided the density is constant throughout. The density, how- ever, may differ for different parts of the body, as, for example, when the body is composed of different liquids which arrange themselves in layers or strata. If V lt V 2 , V s , •••, V n denote the volumes of the separate parts, and y lf y 2 , y 3 , •••, y n the correspond- ing densities, the total mass is evidently the sum yiFi + y^ + ^-f •-+7,K. In this case, the number of parts being finite, we need only simple addition. However, the density may vary continuously through- out the body, as in the case of the atmosphere. Here we must have recourse to the summation of an infinite number of indefi- nitely small terms. We divide the total volume V into n parts each equal to AF and multiply each element AV by the density at that part of the body. We thus get n terms of the type 7 A V. If n is finite, the sum of these n terms is not the exact value of the mass because the density varies in the element of volume A V. But as n is taken larger and A V correspondingly smaller, the sum of the n terms approaches more nearly the mass. Hence, to get the exact result, we must increase n indefinitely, thus making AF correspondingly small, and effect the summation of the infinitely large number of infinitesimal terms. That is, we must find L T7AF. As previously shown, the summation is effected most easily by means of the definite integral. The elements to be summed being of the type f(x) Ax, we find the anti-derivative (x) of the function /(#), substitute the limits of integration, say a and b, and take 178 DEFINITE INTEGRALS [Chap. X. Fig. 52. the difference <£(&) — (a). The problem of effecting the sum- mation reduces, therefore, to a problem in integration. Ex. A vertical wall (as a dam) having a height h and breadth 6, Fig. 52, is subjected to water pressure, the intensity of which varies as the depth below the liquid surface. Required the total pressure on the wall. According to the law of liquid pressure, the intensity at the depth x is kx, where k is constant. Let the wall be divided into elements of width Ax and length b ; then if the area of the strip b Ax is multiplied by the intensity of pressure kx at the top of the strip, the product b Ax • kx=kb x Ax gives approximately the pressure on the element of area. The sum of a finite num- ber of terms of the type kbxAx would give a total pressure somewhat smaller h than the actual value; but the limit L 2]kbxAx evidently gives the exact result. This limit is the definite integral ( kbx dx = kb \ xdx = \ kbh 2 . Hence the total pressure on the wall is \ kbh 2 . 102. Geometrical representation of a definite integral. It was pointed out in Art. 100 that a definite integral I f(x) dx is repre- sented by the area bounded by the curve y =f(x) f the X-axis, and the ordinates corresponding to x = a, x = b. Whatever magni- tude the definite integral is used to denote, volume, mass, fluid pressure, work, or moment, this area is the graphical representa- tion of it; that is, the number of units of area is the same as the number of units of the mag- nitude denoted by the integral. In fact, one way of evaluating a definite integral is to measure by some mechanical means the area that represents it. (See Art. 119.) If f(x) becomes negative for certain values of x, the graph of fix) will lie below the X-axis, as from b to c, Fig. 53. In this case the integral J fix) dx is represented by the algebraic sum of the areas I, II, and III, area II being taken as negative. If Fig. 53. Arts. 102, 103] GEOMETRICAL REPRESENTATION 179 the numerical rather than the algebraic sum is desired, we must Xb f*c fix) dx, I f(x) dx, and Jb fix) dx without reference to sign. Ex. Show that the area which represents the total liquid pressure in the example of the preceding article is a triangle whose base is h and whose altitude is kbh. (Note that/(x) = kbx.) EXERCISES 1. Give a geometric interpretation of Theorem II, Art. 98. 2. Give a geometric proof of the following theorem : If M and N are respectively the greatest and least values of the continuous function fix) within the interval (a, b), then, provided b > a, •5 N(b - a)< f f(x) dx < M(b - a). 3. Give a geometric proof of the following theorem : If b > a, and (x), f(x), and \f/(x) are three functions such that for any value of x within the interval (a, 6), (x)dx< Cf(x)dx< C\p(x)dx. 4. Using the theorem of Ex. 3, show that if < n < 2, ( — - lies between 0.5 and 0.7854. J ° 1 + x " 5. Give a geometric proof of the theorem C F(x) dx = (*° F(a - x) dx. 6. Show geometrically that f sin 2 6 dd - \ " cos 2 d dd. 7. Show from geometric considerations that i sin d0 = 0. 8. Show the area that represents the definite integral ( 2 gt dt, which applies to falling bodies. n 103. Definite integrals of discontinuous functions : Infinite limits of integration. So far we have discussed definite integrals of con- tinuous functions with finite limits of integration. We shall now consider cases where one or both of these conditions do not hold. 180 DEFINITE INTEGRALS [Chap. X. Suppose that f(x) has a point of discontinuity at x = c, Fig. 54. The extent of this discontinuity may be finite or infinite. In fix) dx, I f(x) dx, I fix) dx have no meaning according to the defi- nition given for a definite integral. However, the definite integral does have a meaning for the intervals (a, c — e), (c + e, b) ; for within these intervals the function is continuous. We may, therefore, define the value of the definite integral for the intervals (a, c) and (c, b) as the limits L ( f(x) dx, L I f(x) dx, provided these limits exist. The definite integral for the inter- val (a, b) may now be defined as the sum of these two limits. We may not always, however, obtain the proper value of the last integral by evaluating directly the integral I f(x) dx. 1 Ex. 1. Find the area between the curve y = — (x the ordinates for which x = 0, x = 4. v 0, x This function is discontinuous for x nite integral for the given interval is found by taking the sum of the limits 2)2 the X-axis, and 2, as is shown in Fig. 55. The deri- Hence the area in question is infinite. Let us now take the integral directly between the limits and 4. We obtain for the area Rw-K «£&-;=&' i, Art. 103] INFINITE LIMITS OF INTEGRATION 181 an incorrect result. In this case we say therefore that the integral 1 f(x) dx has no meaning. We must also consider the special case in which one of the limits of integration is infinite. This again is a limiting case of the ordinary definite integral, and we define the definite integral fix) dx as the limit L J f(x) dx. Hence, provided this limit exists, the integral I fix) dx has a meaning. The follow- ing example illustrates this case : Ex. 2. Find the area between the X-axis and the witch of Agnesi, whose equation is y= 8a3 x l + 4 a 1 See Fig. 39, Art. 90. The curve is symmetrical with respect to the T-axis and approaches the X-axis as x becomes infinite. Hence the area is given by 2(*yclx= L I6a3p— ^— - = 16 a 8 L P- arc tan -^-1 Jo * 9smao Jo x 2 + 4 a 2 a-=ooL2a 2aJo = 8 a 2 L arc tan — = 4 ira 2 . X=ao 2 d EXERCISES 1. Evaluate f"—^— . 2. Evaluate f e~ ax dx. Jo a 2 + x* J° 3. Evaluate C ^ 4. Evaluate f' 1 — - ^ Jo (1-*)* xVx*=l 5. Of the following definite integrals, which have finite values ? «/■*; <» ff ; (0 JTS; <« ft (?.'• .'•- 6. Show that if ft < 1 , the area under the curve x n y = C from the origin to x = a is finite, while for n > 1 the area under the curve from x = a to x = oo is finite. 7. With the data of Ex. 6 investigate the area when n = 1. 8. Find the area under the curve 2/(z 2 - 1)* = X between x = and x = 4. Draw the curve. 182 DEFINITE INTEGRALS Chap. X. MISCELLANEOUS EXERCISES Evaluate the following definite integrals : l. ["V^db. 2. ^ a (xVa^^ + ^^\dx. \ •2 4. Show that ^r = o«« 3 . f 2 sin 2 6 cos 3 dd. (Put cos 2 6 = 1 - sin 2 6. ) (x(a — x)dx i (a — «) da; ( ^ sin cos Odd \ acos 2 0sm0d0 5. Evaluate (a) Jft _ fl — ; (&) ^- • f ^sinfl cos sin cZ0 6. Find by the summation method of Art. 7 the area between the curve y = e x , the X-axis, and the ordinates x = 1, x = 4. 7. Change the limits of integration of the following integrals when the variable is changed as indicated. /•20 ri (a) \ ( )dx, x = z 2 -j-4; (6) I ( )dx, x = sin0 — cos0; /•log 10 /•« (c) j ( ) dx, e x = 2 2 + 1 ; (d) I ( )dx, x = asin 9. 8. Show from geometrical considerations that («) (* 5 c/(X)dx = c£f(x)dx ; (6) Cf(x)dx =C~ a f(x + a) eta. Evaluate the following : 9. f a* . io. C dx . 11. P— *5 J° V2^ Jo Va 2 - x 2 Jo O - 2 ) 2 12. Prove and interpret geometrically the following theorem : (a) »/(-«) = -/(*), |V(x)dx = 0. (6) If/(-x)=/(x), Cf(x)dx = 2$ o a f(x)dx. 13. Apply the theorems of Ex. 12 to the definite integrals J sin 6 dd ; f cos c?0. 7T ./ 7T _7T 2 2 Art. 103] MISCELLANEOUS EXERCISES 183 14. If n > 2, show that f * dx lies between 0.5 and 0.52. Jo VI -x M fg 15. If /(x) = V8 x + 10, show that I /(x) dx must lie between 3 V50 and 3V74. 16. Find the limits between which the integral f / (x) dx must lie where / (x) = togas. 17. Prove the following theorem geometrically : f 0(x) dx = (6 - a)0[a + 0(6 - o)], where < d < 1. CHAPTER XI APPLICATIONS OF INTEGRATION TO GEOMETRY AND MECHANICS 104. Plane areas, rectangular coordinates. If the equation of a curve in rectangular coordinates is y —f(x), then, as seen in Art. 100, the area A between the curve, the X-axis, and the ordinates x = a and x=b, respectively, is given by the formula /•& rb , ... ^4= J /(as) das = J ydvc. (1) Ex. 1. Find the area included between the parabola y 2 = 8 x, the X-axis, the origin, and the ordinate x = 18. Since y 2 = 8 x, y = V8 x, and .4 = f 18 w dx = V8 f 18 x* dx = V8[| x*] 18 = 144. Jo Jo ° Ex. 2. Find the area under one arch of the curve y = cosx. Fig. 56. From the graph of the curve, Fig. 56, it is seen that y = for x = • for x = - • Hence, we have 2 7T 7T r 2 i 2 area .45(7= I cosxdx = sinx =2. "2 ~2 3ff 3tt Likewise, area QBE = f cos x dx = sin x 1 = - 2. and 184 Art. 104] PLANE AREAS, RECTANGULAR COORDINATES 185 Also, area ABODE — sin x 0, and area BCD = sin x J( The last two results are consistent when the signs of the areas are taken into account. Thus areas ABC and CDE are equal in magnitude but opposite in sign. The numerical magnitude of the area ABCDE is 4. It will be observed that the area ABC is symmetrical with respect to the axis OF; consequently area ABC =2 x area OBC=2 C Likewise, area CDE = 2 J cos x dx = — 2. cos x dx = 2. In general, it is advisable to make use of conditions of symmetry as far as possible, and take the narrowest interval of integration that the problem permits. Frequently it is desirable to find the area between the curve, the Faxis, and the abscissas corresponding to y=a and y = fi, respectively. The formula for this area is = f ocdy. (2) The derivation of form (2) follows precisely that of form (1 ) and need not be given in detail. Ex. 3. Find the area between the curve y s = kx, the F-axis, and the abscissas for y — 2 and y = 3. *-J>*-sJ>*-i-M 65 4k' >X. Ex. 4. Find the area included between the hyperbola x?/=36 and the straight linex+i/ = 15 (Fig. 57). The points of intersection of the two curves are found by solving the equations simultaneously. The solution gives (3, 12) and (12, 3) as the coordinates of A and B, respectively. Area ABCD = area EACBF - area EADBF 'dx x r2 /M2 /»12. j (15-a0 4 k Jo 4k K J EXERCISES 1. Find the area between the line y = 3 x + 4, the X-axis, and the lines x = o, x = a. 2. Find the area between the parabola y — 5 x 2 , the X-axis, and the lines x = 0, x = 4. 3. Find the area under the curve 5 y 2 = x 3 from the origin to the line x=5. 4. Find the area between the parabola y=Sx 2 and the straight line y = 12 x. 5. Take the arc of the equilateral hyperbola xy = C between the points A and B. Show that the area between this arc and the X-axis is the same as the area between the same arc and the F-axis. X X 6. Find the area under the catenary y = - (e a + e B ) between x =— m and x = m. 7. Derive a general expression for the area under the curve xy m = C between x = a and x = b. Discuss the case m = 1 . 8. Find the area swept over by the radius vector of the spiral of Archi- medes p = ad in one revolution. 188 APPLICATIONS OF INTEGRATION [Chap. XL 9. Find the area of the two loops of the lemniscate p 2 = a 2 cos 2 0. Make use of the symmetry of the curve and take narrowest limits of integration. 10. Find the area swept over in two revolutions by the radius vector of the parabolic spiral p 2 = a 2 d. 11. Find the total area bounded by the curve a*y 2 + b 2 x* = a 2 b 2 x 2 . Suggestion : Trace the curve in order to obtain proper limits of integration. 12. Find the area between the curve y 2 (a 2 —x 2 ) = a 2 x 2 , the asymptote x = a, and the X-axis. 13. Show that the area bounded by the spiral pd = C and two radii pi and P2 is proportional to p\ — pi- 14. Find the area of a loop of the curve p 2 — a 2 sin nd. 106. Volumes of solids of revolution. A plane area AEFB, Fig. 59, bounded by the curve EF, whose equation is y ==/(*), the axis of X, and the ordinates AE and BF, rotates about the axis OX and thereby generates a solid of revolution. To de- rive an expression for the volume of this solid we pro- ceed as follows : Let the interval AB be divided into n parts, fT ffp^ +X Fig. 59. x=b and let ordinates be erected at the points of division. In any subinterval Ax t , let ?// and y f " be respectively the smallest and largest numerical values of the ordinate, and construct rectangles having Ax t as a base and y- and y-\ respec- tively, as altitudes. Repeating this construction for each of the subinterval s, we obtain one plane area made up of rectangles lying entirely below the given curve EF and a second plane area made up of rectangles whose upper bases lie above this curve. The solids obtained by revolving these areas about the X-axis have respectively the volumes F' = ]T tt^Az, V" =2} Try" 2 Ax. Art. 106] VOLUMES OF SOLIDS OF REVOLUTION 189 The volume V of the solid generated by the revolution of the area AEFB must lie between V and V". That is, b b ]T Try' 2 Ax ^ V^ ]P Try" 2 Ax. (1) a a It will be seen that V and V" are both monotone functions, the first never decreasing and the second never increasing ; hence since the functions are finite, each has a limit as Ax = (Art. 14). Following the method of Art. 100, it may be shown that the two limits are equal. Consequently we may write b b V=L Viry^Ax^L V rry" 2 Ax. (2) Since y is taken to be a single-valued and continuous function of x, we may replace the common limit by the definite integral and write V=ir£y^dx. (3) By a similar process it can be shown that the volume of the solid generated by a rotation about the F-axis is V=-*§*^dy 9 (4) where c and d are the ordinates of the end points of the curve. Ex. 1. Find the volume generated by the revolution about the X-axis of the area bounded by the line 4 x + y = 12 and the coordinate axes. For y = 0, x = 3 ; hence the limits of integration are x = 0, and x = 3. V=w Cy*dx = TT f 3 (12- 4x) 2 dx = 7r(144x-48x 2 +^V)] 3 = 144 r. Jo jo 3 Jo Ex. 2. Find the volume generated by the rotation about the A"-axis of the area bounded by the segment of the parabola y 2 = 8 x between the origin, the X-axis, and the ordinate for x = 6. V = ir Cifdx = ir f 6 8 x dx = 4 7tx 2 ~| =144 about the F-axis of the area bounded by th< cissa, and the same segment generates the vo V = irC TS * 2 dy=*C rHy - area of the ellipse is — {a 2 - x 2 ). a 2 Hence, we have a 2 Jo 3 rabc. EXERCISES 1. Find the volume of the elliptic paraboloid ^-+ — = 4 x between the planes x = and x = 3. 2. Find the volume of the solid bounded by the surface — + 2- + — = 1. tf b 2 c 2 3. By the method of integration derive the formula for the volume of a pyramid or cone. 4. Derive the formula for the volume of the frustum of a pyramid or cone. 5. A right circular cylinder of base radius a and altitude h has two slices cut from it by planes passing through a diameter of one base and touching the other base. Find the volume of the wedge-shaped solid remaining. 6. A cap for a post is a solid of which every horizontal cross section is a square, and the corners of the squares lie in the surface of a sphere 14 inches in diameter with its center in the upper face of the cap. The depth of the cap is 4 inches. Find the volume. 7. A solid is formed by a variable square whose center moves along the major axis of the ellipse — + *- = 1 in such a way that the side of the square 25 9 is always equal to the double ordinate of the ellipse. Find the volume of the solid. 108. Lengths of curves, rectangular coordinates. The length of a curve may be defined as the limit of the sum of the lengths of the inscribed chords as the number of the chords is increased without limit and the length of each approaches the limit zero. Given the curve AB, Fig. 61, whose equation is y=f(x), where f(x) is a continuous function having a continuous derivative. The length s of this curve between the limits x = a, x = b is required. Divide the given interval b — a, into n equal parts, denoting by Ax one of these equal divisions. Denote by Ac, Ay, respectively, the corresponding values of the chord and Art. 108] LENGTHS OF CURVES 193 of the increment of y. From the definition of the length of a curve, we may now write i±»~*i,*FJE** AX = a Az = W (i) With the restrictions imposed upon y=f(x), the quotient ^, by Ax * X Fig. 61. dy the law of the mean, is equal to the derivative — for some value dx of x within the subinterval Ax. Hence, since by Art. 100 we may take the value of the integrand \1 + ( — ) at any point within this subinterval, we may in (1) replace — by — and write Ax dx ii^RSV Replacing the summation by the definite integral, we have finally ■■£$+( doc/ dx. (2) If y is taken as the independent variable, this formula for the length of a curve becomes L'lMgr* (3) where c, d are the values of y corresponding to x = a, x = 6, respectively. 194 APPLICATIONS OF INTEGRATION [Chap. XI. Ex. 1. Find the length of the curve y 3 = 3 x 2 from the point (0, 0) to the point (3, 3) . From the given equation we have dx _ y 2 dy 2x' •-JCV 1 +(!)"* = 1 £ V¥T ^ dy = I (4 + 8 *>t|! = J r (13^ -4$) =4.32. Ex. 2. Find the length of the cycloid described by a point on the circum- ference of a rolling disk during one revolution of the disk. The equations of the cycloid are x = a(0 — sin0), y = «(1 — cos0), in which denotes the angle through which the disk has turned. Differen- tiating, we have S' = 22„(, ± f)VlT(gjA*. (3) As Ax = 0, the left-hand member of this equation approaches the area of the surface generated by the revolution of the given curve, that is, the required area. In fact, we may define the area of the required surface of revolution as the limit of the sum of the surfaces of these frustums as Aa? = 0. In the risrht-hand member r±f j approaches y, and a /l -f- ( — ) becomes \1 +[ — ] > since Art. 110] AREAS OF SURFACES OF REVOLUTION 197 y —f{x) is assumed to be a continuous function having a continu- ous derivative. We have then or S = 2 7rC b yyjl +fS$doB. (Arts. 100, 108) (5) When OF is taken as the axis of revolution, the formula becomes '-■•r-Vi+K)'*"- <«> From (2) it is evident that we may, if we choose, replace V ^'ti" in (5) and (6) by ^l+ffjdy. ' In some problems the latter form is more convenient. The limits of integration must be changed to the values of y corresponding to x = a, x = b. Ex. 1. Find the surface generated by revolving the hypocycloid %* + y* = a 1 about the .X-axis. From the given equation, we get We have therefore \dy) \y) ■i-J>g)**-^.S*»I !*•?**! =j™*- Ex. 2. Find the area of the surface generated by the. revolution of a cycloid about its base. The equations of the cycloid are x = a(6 - sin0), y = a(l — cos0). Differentiating these equations, we have dx = a(l — cos 0) dd, dy — a sin 6 dd, whence ^i + /<& V dx = VdWTW 2 = aV2(l - cosd)dd. 198 APPLICATIONS OF INTEGRATION [Chap. XI. Substituting in (5), we get S - 2 Tra 2 C w V2(l - cos ey dd -• 'JT-g) - ®-! — TO? 3 EXERCISES 1. Find the area of the surface generated by the revolution of the parabola y 2 = 8 x about the X-axis. Take the limits x = and x = 2 ; also the limits x = 2 and x = 8. 2. A line joins the origin to the point (w, ri). Find by integration the surface of the cone generated by revolving this line about the X-axis. 3. Find the area generated by revolving about the A'- axis the arc of the cubical parabola 2 y = x 8 between x = and x = 2.' 4. Find the area of the surface generated by revolving about the X-axis the X X arc of the catenary y = - (e a + e a ) between x = and x = a. z 111. Mean value. Let y =/(a?) be a continuous function within the interval x — a and x = b, and suppose this interval to be divided into n equal parts each equal to Aa,\ Then b — a = n Ax. Denoting by y lf y 2 , y 3 , •••, y n the values of the function corre- sponding to the values of x at the middle points of these successive subdivisions, let us form the quotient yi + ^ + y 3 + ••• +y» n which evidently is merely the arithmetic mean of the n values of ?/. This quotient will vary with the number of divisions n, and the limit which it approaches as n is indefinitely increased is called the mean value of the function for the interval b — a. Substituting for n its value ~ a ' , (1) may be written Ax y x Aa; + y 2 AaH h y„ Ax b — a The limiting value of this quotient as n becomes infinite, and con- sequently as Ax = 0, is b- a b — a W (1) Art. Ill] MEAN VALUE 199 The numerator of (2) is represented by the area under the curve x=f(x) (as AEFB, Fig. 62) between the ordinates for x = a and x = b. Hence if a rectangle AMNB is constructed having an area equal "to the area under the curve, the altitude of this rectangle represents the quotient i ydx Fig. 62. b-a ' or the mean value of the func- tion. The independent variable may be time, distance, angle, area, volume, or any other geometrical or physical magnitude. Mean values may be taken with reference to different variables. Thus in the case of a moving point, values of the velocity may be taken for equal time intervals or for equal space intervals. In the former case, the mean velocity will be the mean ordinate of the velocity curve on a time base ; in the latter case, it will be the mean ordinate of the velocity curve on a space base. Ex. 1. In simple harmonic motion the velocity of the moving point is v = a sin wt, and the time of a half-oscillation is t = ir/w. Hence, the mean velocity for the time interval to ir/ut is a \ sin cot dt T a ,T COS u>t L o J( --0 Ex. 2. In the case of a falling body we have v = gt and v 2 = 2 gs. For the mean velocity, taking time as the variable, we have for the inter- val to «!, j>« ^l' tdt i = 2 ffh - 200 APPLICATIONS OF INTEGRATION [Chap. XI. Since the velocity at the end of the time h is gh, the mean velocity is one half the final velocity. The mean velocity for the space Si, taking equal space intervals, is ( l vds V'2g I l Vsds si — si 3 d that is, the mean velocity is two thirds the final velocity. EXERCISES 1. Find the mean of the ordinates of the parabola y 2 = 10 x from x = to x = 8. 2. Find the mean ordinate of the curve y = x 2 — 7 x + 5 between x = 1 and x = 5. 3. Find the mean value of the ordinates of the curve y = cos x (a) between x = and x = | it ; (&) between x = and x = 7r. 4. A number n is divided into two parts ; find the mean value of the prod- uct of the parts. 5. During the expansion of steam in an engine cylinder the pressure falls approximately according to the law px = C, where x denotes the distance the piston has moved from the beginning of the stroke. Derive an expression for the mean pressure exerted during the period of expansion. 6. Find the mean ordinate of a semicircle of radius a provided the ordi- nates are drawn at equal intervals on the arc. Suggestion : Use polar coordinates. 112. Work of a variable force. When the point of application of a force is moved in the direction of the line of action of the force, the force is said to do work. For example, work is done by the, drawbar pull of a locomotive when the locomotive moves along the track, thus moving the point of application. If the force is constant in magnitude, the work done is the prod- uct of the force and the distance through which the point of appli- cation moves in the direction of the force. If IT denotes the work, F the force, and s the displacement of the point of application, then W = Fs. (1) For example, the work done in raising a load of 800 pounds a height of 6 feet is 800 X 6 = 4800 foot-pounds. Art. 112] WORK OF A VARIABLE FORCE 201 The magnitude of the force may vary as the point of applica- tion moves, as in compressing a spring. In this case, the work may be expressed as a definite integral as follows : Let s x and s 2 be the initial and final distances of the point of application of the force from some origin, so that s 2 — s x is the displacement. Let s 2 — s x be divided into n subintervals As x , As 2 , etc., and in any subinterval A^ let F- and F" be respectively the smallest and largest values of the force F. Then denoting the work by W, we have ^ F' As ^ W^ 2) F " As - ( 2 ) *2 As in Art. 100, we may show that the functions 2 F'As and 2j F" As have the same limit as As = 0, and we may therefore write at once, since F is a continuous function of s, W= L As = s^=r Fds. (3) From the method of deriving (3), it is clear that the work of a force may be represented by an area. On the displacement S X S 2 let the forces F corresponding to the successive positions of the point of application be laid off as ordi- nates, and let a curve be drawn through the ends of these ordinates (see Fig. 63) ; then the area ^ABSi under this curve will represent the work, of the force be- tween Si and Sf. For the area under the curve is " Fds, F F* Sj s Fig. 63. 8m which according to (3) gives the work W. In order to integrate the expression Fds, the force F must be expressed as a function of the displacement s. The following cases are those occurring most frequently in practice. 202 APPLICATIONS OF INTEGRATION [Chap. XI. (a) When the force is a linear function of the displacement. This is the law in the compression of a spring. We have W = p Qcs + b)ds = l Jcs 2 + bs]* 2 = lfc(8 2 2 - Sl 2 )+K^2-S 1 ). Ex. 1. A spring is 10 inches long, and a force of 48 pounds is required for each inch it is compressed. Find IV -10- the work of compressing the spring — — from 10 inches to a length of 6 inches ; also from a length of 8 inches to a length of 5 inches. Fig. 64. In the first case, 8l = 0, s 2 = 10 - 6 = 4, F = 48 s. Hence, W = f * 2 F ds = 48 f * s ds = 24 iP~Y = 384 in. lb. For the second case, si = 2, s- 2 = 5, and W = 48 C s ds = 24 s 2 T = 504 in. lb. Ex. 2. A bar is stretched from its original length L by a gradually in- creasing load. Denoting by s the amount of stretch for a given force F, Hooke's law gives as the relation between F, s, and L, w _EAs where E denotes the coefficient of elasticity of the material, and A the area of the bar. The work of stretching the bar by an amount s is therefore Jo L Jo 2L We have for wrought iron E — 30,000,000 ; suppose a bar having a cross- section area of 2 square inches be stretched from 60 inches to 60.5 inches. Here ^4 = 2, a = 0.5, and L = 60 ; hence the work is 30000000 x 2 x 0.52 = m in ft 2 x 60 (b) When the force varies inversely as the displacement. In this case, & 8 whence W= fVjbsfc f* 2 - =fc log- 2 - Arts. 112, 113] WORK OF EXPANDING GASES 203 (c) When the force varies inversely as the square of the distance. The law of the inverse square applies to gravitational forces, to forces between electric charges, etc. Ex. Let a positive charge m of electricity be concentrated at a point P, and a unit charge at a point Q at a distance s from P ; then the repulsion of charge m on the unit charge is p — m — ™. PQ The work required to move the unit charge from .«? = a to s = b is Ja Ja S 2 Sj a \a b) 113. Work of expanding gases. A gas is confined by the walls of a cylinder and a movable piston, Fig. 65. By virtue of its pressure, the gas expands or increases in volume, moves the piston, and thus does work against an external re- sistance. Let p denote the pressure exerted by the gas on a unit area (square inch or square foot), and A the area of the piston. Evidently the total force acting on the piston is F =pA ; and for F 1 As, F 2 As, etc., we may write p x A As, p 2 A As, etc. But A As is the volume swept over by the piston in moving through the distance As and may be denoted by Av; hence F As=p Av, and the work done is Fig. 65. v 2 r v °- W= L V p Av = I p Av = ^f J v i dv. In using this formula it must be noted (1) that p denotes pres- sure per unit area, not the total pressure, and (2) that for correct numerical results consistent units must be used, pounds and feet, or pounds and inches throughout; thus, if p is in pounds per square inch, v must be in cubic inches, and the result will be work in incTi-pounds. Ex. 1. Air expands without change of temperature following Boyle's law. pv =p\Vi = const. The work of expansion is W = Pp dv = p lVl C 2 *° = p 1 v l log * Jv x Jv x V Vx w 204 APPLICATIONS OF INTEGRATION [Chap. XI. Ex. 2. Air expanding adiabatically follows the law pv k = i>iVi* = const., where k = 1.41. The work done during the expansion is = \ 2 pdv =pxVi k \ «-* dv = piVi k — J"! Ji 1 — kJv t ~*-iL J~ /fc-i ' since Pi»i* = #*«**• EXERCISES 1. The length of an unstretched spring is 16 inches, and a force of 225 pounds is required to stretch it 1 inch. Find the work required to stretch it from a length of 18 inches to a length of 22.5 inches. 2. In hoisting coal or ore from a mine, the load consists of two parts : (1) the weight M of the car and contents ; (2) the weight of the rope, which is m pounds per foot. Find the work required for hoisting a distance of h feet. Suggestion : Let s denote the distance of the load from the lower level ; then F=M+m(h — 8), and W= ( [M + m(h - *)] ds. 3. In Exs. 1 and 2 draw diagrams showing by areas the work done, and derive the results by elementary geometry. 4. Suppose the force to vary directly as the square of the displacement of its point of application. Derive a formula for the work. 5. Confined air having a volume of 6 cubic feet and a pressure of 80 pounds per square inch expands following the law pv = const, to a final volume of 20 cubic feet. Find the work done. 6. Use the data of Ex. 5 and find the work done if the expansion is adiabatic, i.e. according to the lawpu 1 - 41 = const. 7. Find the work of stretching a round iron bar having a diameter of 1.5 inches from a length of 40 inches to a length of 42.3 inches. Take E = 28,000,000. MISCELLANEOUS EXERCISES 1. Find the areas bounded by the following curves, the X-axis, and the ordinates indicated : (a) y = x 8 + x + 5, from x = to x = 6. (6) y = x s - 3 x 2 + 4, from x = 1 to x = 5. (c) y = e x , from x = to x = 1. Akt. 113] MISCELLANEOUS EXERCISES 205 2. Find the area of one loop of the curve p = a sin 2 0. 3. Find the area between the curve y = tan x and the X-axis from x = to*=-. 3 4. Find the area between the parabola x^ + y^ = cfi and the coordinate axes. 5. Find the volume generated by the revolution of the entire curve x 2 + y*= 1, (a) about the X-axis; (&) about the F-axis. 6. A cylindrical vessel having an altitude of 12 inches and a base diameter of 8 inches is tipped and the contained fluid is poured out until the liquid sur- face coincides with a diameter of the base. Find the quantity of liquid re- maining in the vessel. 7. If e is the eccentricity of an ellipse and

a! m / ^t mm. B HP ^ ft Fig. 57. K)<°- b)=C from an initial volume v\ to a final volume v 2 . 15. Find an expression for the work done by a force that varies as the nth power of the displacement of its point of application. 16. Find the work required to compress 20 cubic feet of air from a pres- sure of 14.5 pounds per square inch to a pressure of 63.5 pounds .per square inch, the equation of the compression curve being jw 1,3 = const. 17. A particle of mass m has simple harmonic motion defined by the equation x = r cos wt. Show that the mean kinetic energy (£ mv 2 ) for a com- plete oscillation is one half the maximum kinetic energy. CHAPTER XII SPECIAL METHODS OF INTEGRATION 114. Integration by parts. Thus far we have made use of only those integrals that could be evaluated by use of the fundamental formulas given in Chapter VI. In some cases we were able to reduce the given function to a fundamental form by a simple transformation. Not all functions, however, can be easily re- duced to those types by the methods already employed, and in this chapter we shall consider some special methods by which this reduction may be effected in cases more complicated than those already discussed. If u and v are two functions of the same independent variable, we have upon differentiating their product, d (uv) = udv -\-v du, or udv — d (uv) — v du. Integrating, we get I udv — uv — \v du. (1) By this formula the integral J u dv is obtained by the evaluation of another integral J v du. This method is called integration by parts, and is one of the most useful of the integral calculus. It is particularly helpful in the integration of the product of two func- tions where the integral of one can be easily found ; also in the integration of logarithmic functions, exponential functions, and inverse trigonometric functions. No general directions can be given as to which of the two functions is to be taken as u and which as dv, except that the selection should be such as will render dv and v du most easily integrable. In case of doubt, first 207 208 SPECIAL METHODS OF INTEGRATION [Chap. XII. try putting dv equal to the most complicated factor that can be easily reduced to a fundamental form. Ex. 1. | xe x dx. Put dv = e x dx, u — x, whence v = e x , du = dx. Substituting these values in formula (1), we have ( xe x dx — xe x — \ e x dx = e x (x - 1) + a Ex. 2. ( x arc tan x dx. Put dv = xdx, u = arc tan x, whence v = - x 2 , du 2 1 + x' 2 Substituting these values in the formula, we have /• i \ C x 2 \ x arc tan x dx = - x 2 arc tan x — \ dx J 2 2 J 1 + x 2 = -x 2 arc tan x - - ( dx + - f — * 2 2J 2J1 + dx_ x 2 — -x 2 arc tan x x + - arc tan x + 2 2 2 = a ^ + 1 arc tan x - - x + C. 2 2 ,,| Ex. 3. \ Va 2 - x 2 dx. Put u = V a 2 — x 2 , dv = dx, whence du = xdx , v = x. Va 2 - x 2 The result of substituting these values in (1) is (Va 2 - x 2 dx = xVa 2 ^ 2 + f- g^g — J J Va 2 - x 2 We may write * = _ a*-x 2 + _^_ = _ V¥ Z^2 + « 2 , y/a 2 -x 2 Va 2 -x 2 y/a 2 -x 2 \fa 2 -x 2 We have therefore f Va 2 -x 2 dx = xy/'a 2 -x 2 - f Va* - x 2 dx + « 2 f — — — . •> J JVa 2 -z 2 whence 2 ( Va 2 - x 2 dx = x Va 2 — x 2 + a 2 arc sin - , J a and f Va 2 '— x 2 dx = - \xVa 2 - x 2 + a 2 arc sin - 1 • Arts. 114, 115] INTEGRATION OF RATIONAL FRACTIONS 209 EXERCISES 1. \x*e x dx. 2. fx' n log x dx. 3. ( arc sin Odd. 4. j arc cot 6 dd. 5. \sin 2 ddd. 6. (e ax sin bxdx. 7. j log x dx. 8. (cos d log sin 6 dd. 9. f x 2 arc tan x dx. 10 . f e a cos - dx. 11. fxcossedx. 12. fsec 8 0d0. 13. CVx* + a*dx. 14. §*x»y/a*-x*dx. 15. (\b 8 log 2 a rife. 16. f* cos cos 2 0<20. 115. Integration of rational fractions. By a rational fraction we mean the quotient of two rational integral functions. We shall consider only those fractions in which the numerator is of lower degree than the denominator; for, if this is not the case, we can always by division reduce the given expression to a ra- tional integral function plus such a fraction. The decomposition of rational fractions is fully treated in algebra, 1 * and a knowledge of the principles and methods involved is assumed here. In the present article, we shall show how the decomposition of rational fractions may be employed in simplifying an integration. We shall consider the following cases. (a) When the denominator is the product of several linear factors, none of which is repeated. The given function can then be written in the form ff x \ = 4M We may now assume /(x)=^- + -B-+ ... +^-, x — «! x — a 2 x — a k * See Rietz & Crathorne's College Algebra, pp. 203-208. 210 SPECIAL METHODS OF INTEGRATION [Chap. XII. and calculate the numerators A, B, •••, K by the principles of undetermined coefficients. The integral of f(x) is then found by taking the sum of the integrals of the separate terms. The following examples illustrate the method. Ex ! f (5s + l) (x — y)'' which can be written in the form »-y 0»-y) 2 0-y) 3 We may calculate the coefficients A, B, C, •••, K, L, M, •••,#, in case (a). The following example illustrates this method. Ex. f ( x-S)dx J x* — 4 x 2 -(- 4 x We have _4 . Jg . X 3 _ 4 X 2 + 4 x x ( X _ 2)2 a; ' a; _ 2 (x - 2) 2 Clearing of fractions, we get the identity x - 8 = A (x - 2) 2 + Bx (x - 2) + Cx. Equating the coefficients of like powers of x, we have -8 = 4 A, 1 = -4A-2B+C, Q = A + B and from these equations we obtain A =-2, B=2 C=-3. 212 SPECIAL METHODS OF INTEGRATION [Chap. XII. Hence f. M^^jfil + I f-*L_ - 3 f ** Jx 3 -4«H4x Ja; J a - 2 J (x - 2)2 = _iog^ + i og(x _ 2 )2 + _§_ + a EXERCISES 1 f yx - 4) c?x 2 f (3 a; - 2) da; J x a - 4 x 2 + 4 x ' J x(x + 3) 2 3 f (2 x - 5) dx 4 f (x- l)c?x J (x - 2) 3 J (x + 3) 2 ' 5 f (4y-3)dy 6 f 6X 3 - 8 x 2 - 4x + 1 ^ '* J v 3 -3i/' 2 'J y 8 -3y 2 J x 4 - 2 x 8 + x s x-1) _ f (3 x — 1 ) dx q C xdx J x a- x 2- X+ \' ' J (x+ 1) 2 ( f x 2 + 2x-3 ^ r J X 3 - x 2 J ( mx 2 dx w + x)* (c) W/ien £/*e denominator contains factors of the second degree. Aside from linear factors, the denominator may contain quadratic factors not decomposable into real linear factors. For the linear factors, we assume a decomposition into partial fractions in ac- cordance with the principles of cases (a) and (b). Corresponding to the quadratic factors, we assume fractions whose numerators are linear in the variable. If any of the quadratic factors appear to a degree higher than the first, then we assume in the decompo- sition as many fractions as the degree of the factor, the numera- tor of each being linear and the denominator having the given factor in increasing powers. The following examples illustrate the method. Ex.1. <" rMx J (x l)(x 2 + l) We write * = -A- + Bx + C . (x-l)(x 2 + l) x-1 x 2 + l Clearing of fractions and equating coefficients of like powers of x, we find Arts. 115, 116] FUNCTIONS CONTAINING RADICALS 213 Hence f *** =1 f_*L +1 f «±1 rf* J (a; - l)(x 2 + 1) 2 J x - 1 2 J x 2 + 1 = I log (x - 1) + \ log (a: 2 + 1) + \ arc tan x + = | log (x - l) 2 (x 2 + 1) + \ arc tan x + C. Ex. 2. f ^±-f (to. J (x- l)(x 2 + l) 2 If we write x + 1 _ ■ A Bx+C Dx+ E (x-l)(x 2 + l) 2 x-1 x 2 + l (z 2 + l) 2 ' we find by clearing of fractions and equating like powers of x the following values for the coefficients : A = \, B=-\, C= -1,D= -1, E = 0. Hence dx J (x-l)(x 2 +l) 2 2 J x-1 2Jx 2 -}-l J (x 2 + l) 2 = | log (x - 1) - | log (x 2 + 1) — £ arc tan x + (- C. 2 (a; 2 + 1) EXERCISES 1. f^+J°*Z^j r dx Jy* + 2y*-3 ' J (a + a) (x* + 6 2 ) 9. f— * 10. f- J x 2 (x 2 + 3) J x 11. f (8^-D^ 12. f_ J (l + x) 2 (l + x + x 2 ) Jx(l + 2x 2 ) x 2 + x + 1 dx is. r **» . J (x 2 + l) 2 116. Integration of functions containing radicals. In the pres- ent article we shall discuss some special methods by which functions containing radicals may frequently be changed into equivalent functions free from radicals. (a) When f{x) contains fractional powers of a + bx, but no other radicals. 214 SPECIAL METHODS OF INTEGRATION [Chap. XII. In this case, f(x) can be transformed into a rational expression by the substitution a + bx = z n , (1) where n is the least common multiple of the denominators of all the fractional exponents of a + bx. This follows from the fact that /(a?) and dx can then be expressed rationally in terms of z and dz, as the following examples will illustrate. Ex.1. Find f y*y . (1 + 80* Put 1 + y = z 3 , whence y = z 3 — 1 , dy=S zHz. Wehavethen f Eft = f g= *) 8 **» = 3 f(t*- ,) dz = s(z 4 dz-s(zdz i)*-i(y + i)*+ c. Ex. 2. Find p ^ + x* + 4) dx x 2 + 1 Put x = 2 4 , whence dx — 4 2 3 dz. We have then f (s* + s* + 4) dx = f (z 2 + g + 4)4g 3 (fe J s 2 + l x- + 1 /• / ft • _ i\ = 4 [}- z 4 + $ * 3 + f z 2 - 2 - | log («« + 1) + arc tan z + O] = z 4 + | z 3 + ^ z 2 - 4 - 6 log (z 2 + 1) + 4 arc tan z + C" = x + f x 4 + 6 x* - 4 x^ - 6 log (x^ + 1) + 4 arc tan x^ + C . (b) Whenf(x) contains the radical Var* + ax + b and no other. In many cases the integral may be made to depend upon one of the fundamental integrals by writing the radical in the form Vw 2 ± a 2 . If this method does not lead to a convenient way of per- Art. 116] FUNCTIONS CONTAINING RADICALS 215 forming the integration, we may rationalize f(x) dx by the substi- tution ■y/x 2 + ax + b = z — x. (2) Squaring (2), we have x 2 + ax + b = z 2 — 2 zx + x 2 , z 2 — b whence a; = — ; a + 2z ds= 2 (f + ^ + *)f 2 VP+IT+T 2Vx 2 + x + l J V(2x + l) 2 + 3 Putting 2 x + 1 = u, we have for the first integral f xdx _ 1 f wriw _ 1 f du J V(2z-f l) 2 + 3 4 ^ Vw^TlJ 4 *^ Vw 2 + 3 = I Vw 2 + 3 - I log [« + Vm 2 + 3] + C. Replacing the value of u and combining, we obtain as the final result \\x+ Vz 2 + x + ij - \ log [2 x + 1;+ 2V« 2 + ic+i] + a (c) TP#en /(#) contains the radical v — x 2 + ax + b and no other. We shall consider only those cases in which the expression under the radical can be broken up into real linear factors. The integration can frequently be most readily performed by putting the radical in the form Va 2 — u 2 and making use of one of the standard forms. If this is not feasible, we may proceed as follows. Write the radical in the form V— x 2 -\-ax + b = V(x — a)(fi — x), and f(x) dx can be rationalized by the substitution V(a? — a)(fi — x) = (ft — x) z [or (x — a) z], 216 SPECIAL METHODS OF INTEGRATION [Chap. XII. as we shall now show. Squaring, we obtain (x -a)(fi- x) = ((3-x) 2 z 2 , whence x = &±*, dx = ^~ a ^ z dz. 1+Z 2 ' (1 + z 2 ) 2 The integrand which results from the substitutions is rational. dx Ex. Find JxV^ xV—x 2 + 4:X — 3 We have here V(se — 1)(3 — x) = (8 — x) z, 3z 2 + l ^_ tedz whence x = "*" , dx l + g 2 ' (1+^)2 Substituting these values in the given integrand, we have f g =2r-i^- = J-arctanV3^(7 JxV-x 2 + 4x-3 J30-+1 V3 = J_ arc tan JIIEEI) + a V3 V 3-aj (d) Integration by trigonometric substitution. Iif(x) contains a radical of the form Va 2 ± u 2 , where u is some function of x, it can often be easily transformed to an equivalent function free from radicals by the substitution of a trigonometric function. All that is necessary is to substitute for u that trigono- metric function which renders the expression under the radical a perfect square. It will be readily seen that this end is accom- plished by the following substitutions : (1) Put u = a sin 6, or a cos in functions involving Va 2 — u 2 . (2) Put u = a tan 6, or a cot 6 in functions involving Va 2 + u 2 . (3) Put u = a sec 0, or a esc in functions involving -\/u 2 — a 2 . Whenever the resulting trigonometric integrand does not fall at once under one of the fundamental formulas, or take one of the special forms discussed in the following article, the student should apply the methods of integration discussed in the preced- ing articles. The following examples illustrate the use of these substitutions. Ex. 1. J Va 2 - x* dx. Put x = a sin z, dx = a cos z dz. Art. 116] TRIGONOMETRIC SUBSTITUTION 217 We have then f Va? — x 2 dx = a 2 \ cos 2 z dz = \ a 2 f (1 + cos 2 z) dz = \ a 2 f dz + \ a 2 f cos 2 z d(2 z) = \a 2 z + \a 2 sin 2z+ C = \a 2 arc sin - + \ xVa? - x 2 + C. a dx Ex. 2. J ax x 2 (x 2 - a 2 )* Put x = a sec z, dx = a sec tan z dz. Then we have dx r a sec 2 tan z dz f- J x 2 s-J (x 2 - a 2 ) 2 ^ a 2 sec 2 2 (a 2 sec 2 z - a 2 )^ = lfcos^^ s -H^+e= Va;2 ~ a2 +a a 2 J a 2 a 2 x 2 Vx dx a J x EXERCISES ! -J; x3 *MVx-3x* dx J l + 3. fx 2 Oix + &)*dx. 4. f dX J J V4 x - 3 - x 2 /» 3dx r dx J V7x-10-x 2 ' ' J Vx 2 + 5 x - 3 r = %L = ± g , and ± dx = ^gEZJE, dx Va 2 - y 2 V From the table of integrals, C ^a 2 -y 2 dy =V ^ Z — 2 _ al a + Va 2 -y 2 ; J y y hence, the required equation is ± x + c = V^^V 2 -a\og a+Va2 ~ y\ y Ex. 3. Evaluate f_^_. J sin 4 6 From the table of integrals, we find C dx _ _ 1 cos a; , m — 2 C dx J sin m x m — 1 sin" 1-1 x m — 1 J Bin" -2 Hence f-*- = - 1 °2i« + ? f_^_. J sin 4 3 sin 8 3 J sin 2 But f _^L - fese 2 dd0 = - cot 0. J sin 2 6 J The required integral is, therefore, _ 1 (9™1 + 2 cot flV-Icot 6>(csc 2 6 + 2). 8\sin»0 / 3 v y EXERCISES Evaluate the following integrals by reference to a table of integrals : 1. (V 2 * sin xdx. 2. f ^ -. J x{x 2 - a 2 )* 3. f "(a 2 — x 2 )^ dx. 4. f sin 4 cos 2 6 dd. Art. 119] APPROXIMATE INTEGRATION 223 5. C' -***— . 6. f_^ Jo V2 ax - x 2 J 3 + 2 cos a: dx C dx 8 C J(l+2x 2 ) 3 ^x 2 (3-2a:) f dx ^ 10 f xcfx J(4-3x + x 2 ) 3 ' ' J(x 2 + 2x~Z>y n 1X ^ rshixdx^ 12 _ C 2 e sin 6 cos 6 dd. 13. f s^njcdx 22 f 2 q J x s ' J° w » f 2 sin 7 dd. 14. f 2 - — Jo Jo l + dx 2 sin a; 15 C x*dx m 16 (V-x 2 +6x-ldx. 17. Vx 2 - 3 x + 7 fviEifc is. | j J x ^ V5 - 4 x + 2 x 2 19. f — -• 20. f sin 2 6 cos 2 ih + « 2 ^ 2 - Art. 120] SIMPSON'S RULES 225 Solving for a and a 2 , we get a o — v* a 2 — 2~p and substituting these values in (3), we have finally area NABCP = | fa + 4 y, + y 8 ). (5) If nowwe assume the part of the curve CDE replaced by a second parabolic arc, we get in the same way, aiea WM = | (y, + 4 y 4 + %)i and thus area NAER = | fa + 4 ft + 2 & + 4 y 4 + y 5 ). This process may be repeated any number of times ; hence, taking an odd number of ordinates n -f 1, dividing the figure into n strips of equal width h, the approximate area is given by the formula ^=|0/i + 4i/ 2 + 22/3+ ••• +2y n _ 1 + 4y n + y n+1 ). (6) This result gives the following rule, known as Simpson's one-third rule: Take the sum of the end ordinates, twice the sum of the intervening odd ordinates, and four times the sum of the even ordinates. Multi- ply the aggregate by one third of the common distance between the ordinates. If we take four ordinates y lf y 2 , y 3 , y A , including three spaces, we may pass through their ends A, B, C, and D, a third degree parabola y = a + a x x + a6 2 ), Ja + &cosx Va 2 -& 2 L*a + 6 J 2' v " VHa + V6 - a tan - los *i (« 2 <& 2 )- V& 2 - a 2 V 6 + a - V&^tan 5 dx 2 arc tan r dx J a + b sin x Va 2 - 6 2 a tan - + b Va 2 , (a 2 >6 2 ), - atan^+6-V6 2 -a 2 = 1 log ? , (« 2 <& 2 ). 25. Find the area of the ellipse — + — = 1. a 2 b 2 26. Find the area bounded by the curve (-) + (-] = 1« 27. Find the area between the cissoid y 2 = x * and its asymptote ■ 2 a — x x = 2a. 28. Find the area under the curve y = \ogx from x = 1 to x = 10. 29. Find the area of the loop of the curve my 2 = (x — a) (x — b) 2 . 30. Find the length (a) of the curve y = e x from x = to x — 2 ; (&) of the curve ?/ = logx from x = 1 to x = 6 ; (c) of the curve 9y 2 = x(x — 3) 2 from x = to x = 3. 31. Find the length of the spiral p = ad from the origin to the point (ira, 7r). 230 SPECIAL METHODS OF INTEGRATION [Chap. XII. sin 2 6 32. Find the length of an arc of the cissoid p = 2 a- from 6 = to COS0 a 33. Find the length of the curve p = asm 3 -- o 34. A circle of radius - rolls on a circle of radius a, and a point on the circumference of the rolling circle traces an epicycloid whose polar equation is 4(p2 _ a 2)3 — 27 a 4 /? 2 sin 2 6. Find the whole length of the curve thus traced. A 2 v' 2 35. Find the volume generated by revolving the ellipse —•+-*- = 1 about the X-axis. . a2 b * 36. Find the volume generated by revolving one arch of the cycloid x = a{6 — sin 0), y = a{\ — cos 0) about the axis OX. 37. Find the volume generated if the cycloid is revolved about OY. 38. Find the volume generated by revolving one arch of the curve y = cos x about the axis OX. 39. Find the volume generated by revolving the cardioid p = 2 a( 1 — cos 6) about the axis OX. 40. Two cylinders, with the same altitude h, have a common upper base of radius a, and the lower bases are tangent to each other. Find the volume common to the two cylinders. 41. Find the surface generated by revolving the ellipse — 4- V- — \ about its major axis ; also about its minor axis. 42. Find the surface generated by revolving the cardioid p = 2 a{\ — cos 6) about the initial line. 43. Find the mean length of the ordinates of a semicircle of radius «, if the ordinates are taken at equidistant intervals on the diameter. 44. Find the mean distance of the points on the circumference of a circle of radius a from a fixed point on the circumference. 45. Zeuner's equation for superheated steam is pv = BT + Cp n . For an isothermal expansion the temperature T is constant. Derive an expression for the work done during an isothermal expansion from v\ to v 2 . Suggestion. Use the equation work = I p dv = pv — \ v dp. 46. The acceleration of a particle that moves under the influence of an attractive force that varies inversely as the square of the distance is a = — - s 2 Derive a relation between v and s, also between t and s, taking s as the initial distance of the particle from the center of attraction. CHAPTER XIII FUNCTIONS OF TWO OR MORE VARIABLES 122. Definition of a function of several variables. Heretofore we have discussed functions of a single independent variable. A function, however, may depend upon two or more variables having no mutual relation, that is, independent of one another. Thus the volume of a gas depends upon the temperature and also upon the pressure to which it is subjected, and the pressure and temperature may vary independently. In general, we may say : z is a function of the independent variables x, y, ••• ichen for each set of values of these variables there is determined a definite value or values of z. A function of two variables »=/(*, y) is represented geometrically by a surface, and to each pair of values of (x, y) there corresponds a point on this surface. The motion of a point on this surface depends upon the manner in which x and y vary. One of these variables may remain constant while the other is allowed to vary, or the two variables may change simultaneously. In the first case, the extremity of the ordinate describes a plane curve lying in a plane parallel to the YZ- or XZ : plane ; and in the second case the extremity of the ordinate may describe a space curve. When one of the variables remains constant, say y = y n , we say that the function f(x, y) is continuous in x at the point (x , y ) if £ /0> 2/o)=/<>o, 2A>). CO x = x Likewise if we have l f(xo,y)=f(%o, 2/0), ( 2 ) y = yo we say that f(x, y) is continuous in y at the point (x , y ). In order 231 232 FUNCTIONS OF TWO OR MORE VARIABLES [Chap. XIII. that the function be continuous in both variables together at the point in question we must have L f(x,y)=f(x ,y ). (3) x = x y = 2/0 As will be seen from these limits, the continuity in (x, y) together involves of necessity continuity in x and in y. In general, when- ever we speak of the continuity of a function of two variables, the continuity with respect to both variables taken together is to be understood unless otherwise stated. 123. Partial derivatives. In the preceding article it was pointed out that a function of two variables may vary either by permitting one of the variables to remain constant while the other changes, or by allowing both to vary simultaneously. The increment of the function /(#, y) due to a change in x alone is /(aj + Aa?, y)-f(x,y). Let us consider the ratio of this increment to the increment of the variable x. The limit of this ratio as Ax = 0, viz. : L /(a? + Aa?, y)-f(x,y) Ax ± o &x is called the partial derivative of f(x, y) with respect to x. Simi- larly, the limit ^ f(x,y + Ly)-f{x,y) Ay ~ ty is called the partial derivative off(x, y) with respect to y. These derivatives are called partial derivatives because they measure only partially the variation of the function as compared with that of the variables. To distinguish these from the derivatives which have been thus far considered, we shall call the latter total deriva- tives. In a subsequent article we shall discuss methods of deter- mining total derivatives of functions of several variables. To distinguish symbolically the two classes of derivatives, we denote the partial derivatives by using the round d instead of d. Thus the partial derivatives of z=f(x, y) with respect to x and y are written dz dz dx' dif Art. 123] PARTIAL DERIVATIVES 233 respectively. They are frequently represented also by the equiva- lent symbols /„'(*, *),'//& 2f> Partial derivatives usually involve both x and y, and "may like- wise have partial derivatives with respect to either variable. Thus we may have JL(§*\ namely, the slope of the tangent to the curve in question. Let z =f(x, y) be the surface shown in Fig. 70. Consider any point on this surface, as P, at the intersection of the curves BPC and EPFj cut from the surface by the planes y = b and x = a, re- spectively. Then the slope of the curve BPC is given by the 234 FUNCTIONS OF TWO OR MORE VARIABLES [Chap. XIII. dz partial derivative —-, and that of the curve EPF by the partial dz derivative — That is, dy tan <£ = --, talli- ca? dy The values of tan <£ and tan if/ for some definite point P on the surface are ob- tained by substituting in dz the expressions for — and dz . dx — , respectively, the corre- sponding values of x and y. Thus if (a, b) is the projec- tion of P on the X F-plane, we substitute a for x and b for y. The process of finding a partial derivative is in all respects the same as that employed in finding an ordinary derivative of a function of a single variable. The following examples illustrate this statement. Ex. 1. Find — and -£, when z = y 2 sin x. dx dy Treating y as a constant and differentiating with respect to x, we have dz Q — = y 1 cos x. dx Likewise, if we consider x constant and differentiate with respect to y, we get We have Fig. 70. dz dy = 2 y sin x. juccessi 1 /e partial derivatives of the functi z = = xe x log y. dx --(l + x) e* logy; f-* = dy xe x _ y frz_ dx 2 (2 + x) e* 1 d^z _xe x y*' d*z _ Jl + x) e* faz (1 + x) e x dydx y dx dy y Art. 123] PARTIAL DERIVATIVES 235 Ex. 3. The surface z — — + 2- is cut by planes x = 4, y = 3. Find the 8 12 slopes of the curves cut from the surface by these planes at the point of intersection of the curves. We have here dz _ x dz _y dx~^ dy~Q At the specified point, therefore, tan = ?~| = 1 ; tan xL = 21 = I. EXERCISES Find the partial derivatives — and — for the following functions. dx dy 1. z = x 2 y 5 . 2. z = sin x cosy. 3. z = ye 2 -f are*. 4. z = xS — az 2 ?/ + &£y 2 -f 2/*. 5. z = y x . 6. z = arc secu- re 7. z = y sin £ + x sin y. 8. z = sin (« + y). fill (17/ (J7t Find — , — , — ■ for the following functions of three variables. dx dy dz 9. u = 3 x?yz~*. 10. u = e* log ?/2. 11. w = sin a cos y + sin y cos 2 + sin z cos a\ 12. «=iog«±*.' 13. M = *L+^±i!. 14. The volume of a cone may be expressed as a function of the altitude and of the radius of the base. Denoting the volume by J 7 , the base radius by r, and the altitude by h, express V in terms of r and h, find the partial de- rivatives — , — , and give interpretations of these derivatives. dr dh 15. Express the area A of a triangle as a function of its base x and alti- tude y. Find the rate at which the area changes : («) when the altitude re- mains unchanged and the base varies ; (6) when the base is constant and the altitude varies. 16. Express the area A of a triangle in terms of two sides m and n and the included angle 6. Find the rate of change of A : (a) when 9 changes, m and n remaining the same ; (6) when m changes, and n remaining the same. 236 FUNCTIONS OF TWO OR MORE VARIABLES [Chap. XIII. 17. Interpret geometrically the equations dx dy Form ^, i^- , i^_, and ^ for the following functions. dx 2 dxdy dy dx dy' 2 18. u = x log y. 19. u = x m y n . 20. u = x 3 + axy — y s . 21. u = e x +y. 22. u = x(l — y s ). 23. The formula H=ks s D* is used to determine the horse power re- quired to drive a steamship, where s denotes the speed and D the displacement. What is the interpretation of the derivative - — ? of - r - ? Derive expres- sions for these derivatives. **8 oD 124. Interchange of order of differentiation. Among the partial derivatives enumerated in Art. 123 were the second derivatives - \ f , \ « In most cases that arise in the applications of the dx dy dy dx calculus to physical problems, it is a matter of indifference in what order the differentiation is performed ; that is, in most cases these two partial derivatives give the same result. Let us consider the limits involved. We have, by definition, §£ = L fjx + Ax,y)-fjx, y) ~v dx Ax = A« V= l f( x >y+*y)-f( x >y) . (2) &J Ay = Ay Consequently, we have f(x + Ax, y -f- Ay) -fjx, y + Ay) f(x + Ax, y) -fjx, y) &f _ T T Ax Ax dydx a?/=OAx = Ay _ L L A x + Ax, y + Ay) -f{x, y + Ay) -/(x -f Ax, y) +/(x, y) A(x,y) = 0. This relation restricts the moving point (x, y) to a particular curve in the XF-plane and consequently the values of z to a particular curve in space, that is, to the intersection of the cylin- der <(>(x, y) = and the surface z=f(x, y). Whenever such a relation as the above exists between x and y, we may choose x itself as the variable t, and formula (a) reduces to the following : dz__dz_.dz_ dy /* x dx doc dy dx 240 FUNCTIONS OF TWO OR MORE VARIABLES [Chap. XIII. dz dz In this formula it is to be observed that — and — have very dx dx n different meanings. In finding the partial derivative — it is ox assumed that x alone varies, that is, that y is constant. On the dz other hand, — expresses the variation of z due to a change dx in both x and y\ for this reason it is called a total derivative. dz Likewise in («), — is the total derivative of z with respect to t. Moreover, the value of — is definitely determined by the value ox of the coordinates of the point in question ; in other words, it is a function of the coordinates only. This follows from the fact that the variation can take place in one direction only. Functions like this, which depend for their values solely upon the coordinates dz of the point, are called point functions. The value of — depends, dx however, not only upon the coordinates of the point, but also upon the direction in which that point is approached. This derivative is therefore not a point function. Formulas (a) and (b) may be extended to functions of any number of variables. Thus, if w=/(a>, y, z), we have du = dudx + dudy + dudz. (c) dt dx dt dy dt dz dt and if further y = (x), z = \f/(x), du _du.dudydu dz^ ,-,^ doc dx by dx dz dx The proof is left as an exercise for the student. Formulas (b) and (d) are useful in the differentiation of somewhat complicated functions of a single variable. The fol- lowing example shows such an application. Ex. 3. Find — , where u = xe v ~^ z ^ sin 8 z. dx Let y = Va 2 — x 2 and z = sin 8 x ; then u = xevz, ox dy dz and ^ = evz, *» = xevz, *» = X0. Art. 125] TOTAL DERIVATIVES 241 Also f* = *- , ^ = 3sin 2 xcosx. dx ^/ a -2 _ £-2 dx Substituting these expressions for the derivatives in (d), we get &Ol = &z x2e * z + 3 x& sin 2 x cos x dx Va 2 - x' 2 = e^^ 1 ^ sins x T i ^ 2 + 3 cot xl • L Va? -a? J EXERCISES Find — in each of the following. 1. u — x 2 + y 2 , and y = e 8inx . 2. u = arc tan - , and y = e x . x 3. u — log {x -J- «/), and y = Va; 2 + a 2 . i 4. w = x 2 e x cosx. 5. u = e ax (y — z), and y = a sin x, « = cos x. 6. A triangle has a base of 10 units and an altitude of 6 units. The base is made to increase at the rate of 2 units, and the altitude to decrease at the rate of £ unit. At what rate does the area change ? 7. A point lying on the ellipsoid — + ^-+ — = lin the position x = 3, 36 25 49 y = — 4, moves so that x increases at the rate of two units per second, while y decreases at the rate of three units per second. Find the rate of change of z. 8. With the same data as in illustrative Ex. 1, suppose the pressure of the gas to be increasing at the rate of 40 pounds per square foot per second, while the temperature is falling at the rate of 1 degree per second. Find the rate of change of the volume. 9. A gas has the equation pv = I2T, and expands following the law pv n — C, where n and C are constants. Derive expressions for the total dT dT derivatives — - and — under these conditions. dv dp 10. Derive the same results by eliminating p and v successively between the characteristic equation of the gas and the equation of the expansion and differentiating the resulting expressions. 242 FUNCTIONS OF TWO OR MORE VARIABLES [Chap. XIII. 126. Total differentials. In formula (a), — , — , and ^ are the v J dt dt dt quotients of two differentials ; hence we may multiply through by the differential dt. The resulting formula dz = ^dx + ^ dy (e) due dy expresses the differential of z in terms of the differentials of x and y. This formula may be extended to any number of variables. U — J \ X D X 2) X 3) ' " 1 X n)> , du , . du 7 . du j . . du , du = — dx x -\ dx 2 -\ dx z -f • • • -\ dx n . dXi dx 2 dx 3 dx n The differential dz in the formula (e) is called the total differen- tial of z. The terms — dx arid — dy are called respectively the dx dy partial ^-differential and partial y-differential of z. These latter are frequently denoted by d x z and d y z, and (e) is then written in the f 0rm J / IN dz — d x z + d y z. (e) Equation (e) therefore expresses symbolically the principle that the total differential of a function of several variables is equal to the sum of the partial differentials. Ex. 1. Let u = e x y 2 sin z. Then ^ = e x y 2 sin z, — = 2 e x y sin «, — = e x y* cos *. dx dy dz Therefore du = e x y 2 sin die + 2 e x y sin z dy + e x ?/' 2 cos z dz. The differential dz of a function of two variables is susceptible of a geometrical representation similar to that for the differential of a function of a single variable (Art. 46). Let PRQS, Fig. 71, be an element of the surface z=f(x, y), and let a tangent plane PEGF be passed through the point P, whose coordinates are ( x n Vh z i)- The arbitrary increments of the independent variables x and y are A# = dx and A?/ = dy, respectively. From the figure we have evidently BD = AE = PA tan EPA = — efo, and 2)6? = CF = — dy. Art. 126] TOTAL DIFFERENTIALS 243 Hence BG = BD + DG = — dx + — ox dy BG = dz. dy, that is, The increment of z corresponding to the increments Ax, Ay of x and y is Az = I?Q ; hence, the total differential dz is usually dif- ferent from the incre- ment Az. The two symbols dz and Az rep- resent the same value when the surface given by z =f(x, y) is a plane. It is worthy of notice that since dz as given by formula (e) is an ap- proximation to Az, we may use that formula ° to calculate approxi- mately the effect on a function z of small errors Ax and Ay in the observed or measured values of the variables x and y. Ex. Given z — x 2 — xy. find approximately the increment of z correspond- ing to the assumed increments Ax = 0.02, Ay = 0.01, when x = 8, y = 5. From formula (e), dz=(2x-y) Ax-x A«/=0.22 - 0.08=0.14. The actual change Az is found to be 0.1402. — *X Fig. 71. EXERCISES Differentiate each of the following functions by means of formula (e) and verify the result by direct differentiation. 2. k=r- x* 1. z = e x y 2 . 3. z = sin x cos y. 4. 2 = « x ey. arc tan : 7. z = x s y* — x 2 1/ 2 . Differentiate each of the following functions. 9. u=x s +y*-2xyz. 10 . u = 0*. 8. z = arc cos ^ + arc tan - 11. u = sin x cos y tan 2. 12. m = arc tan xy 13. u 244 FUNCTIONS OF TWO OR MORE VARIABLES [Chap. XIII. 14. From the perfect gas equation pv = BT, find the change in p produced by simultaneous changes of v and T. 15. The formula for the coefficient of diffusion of a gas is k = Cp T n , where C is a constant, p the pressure, and T the absolute temperature. Find the change in the coefficient due to simultaneous changes in pressure and tem- perature. 16. Taking the" formula H=ks s D* for the horse power of a steamship, derive an approximate expression for the increase in horse power due to an increase As in the speed, and an increase AD in the displacement. 17. If u = x 2 y, find approximately the change in u when x changes from 10 to 10.02 and y from 4 to 4.01. Calculate also the change in u when Ax = 0.002 and Ay = 0.001. In each case compare these approximate changes in u with the actual changes as derived from the original equation. Interpret the results in connection with Fig. 71. 18. Let z = sin x cos y. If x = 22°, y = 37°, find the change of z due to the changes Ax = 10', Ay = 15'. By formula (e), calculate the approximate change and compare the result with the actual change. 19. Let b and h be respectively the breadth and height of a rectangle and A the area ; then A = bh. Interpret geometrically the approximate equation A J. = h Ab + b Ah, and show wherein the approximation lies. 20. Find the relative error in the computed area of an ellipse due to errors in the measurement of its semi-axes a and b. 21. Given a triangle having sides a, &, c, and opposite angles A, B, and C. If a side c is determined by measuring two sides a and b and the included angle C, show that the error Ac is given approximately by the equation Ac = Aa cos B 4- Ab cos A + a A Csin B. 127. Differentiation of implicit functions. Suppose that y is de- fined implicitly as a function of x by means of the relation/(#, y) = 0. Since the function f(x, y) has by definition the constant value zero for all values of x and y, the total differential must also be zero. Hence, we have Af fif dx dy Transposing the first term to the second member of the equation and dividing both members by dx, we have df dp _ 3a? ( f \ by Arts. 127, 128] EXACT AND INEXACT DIFFERENTIALS 245 which may be used as a formula for writing out at once the de- rivative — of an implicit function. This method is easier to dx apply than the earlier one given in Art. 49, and should be used in practice. Ex. Given /(#, y) = x 2 y — xy s = 0. Wehave & = 2xy-y2, ^ = x 2 -Sxy 2 ; ox dv therefore, if dy dx _ 2xy — y s dx~ df~ &-Z%p By EXERCISES For each of the following functions find the first derivative by the method of this article. 1. y 3 — 3 x 2 y + xy 2 = 0. 2. x 3 sin y — y & cosx = 0. 3. xy n =C. 4. (x 2 + y 2 ) 2 + y 2 (x - 2 a) = 0. 5. x 4 — 3 xy 2 + 2y* = 0. 6. e x sin y = C. 7. p(v-b) n =C, find ^- 8. p s -p 2 cosd = C, find ^. dv dd 9. b 2 x 2 + a 2 y 2 = a 2 b 2 . 128. Exact and inexact differentials. In deriving formulas (a) and (e) for the total derivative and total differential, respec- tively, we started with a given function z=f(x, y) of two inde- pendent variables x and y. In the application of calculus to problems in physics and mechanics, we frequently meet with expressions having forms precisely similar to (a) or (e) yet derived by a quite different process. To illustrate this statement let us consider a physical problem, that of heating a gas. The state of a gas is defined by the absolute temperature T and the volume v, and a change in either T or v is accompanied by the absorption of heat. If now the volume v is kept constant, it is known from experiment that a change of temperature AT 7 is accompanied by the absorption of heat A T Q = c m AT, where c m denotes the mean specific heat at constant volume for the interval AT. If we 246 FUNCTIONS OF TWO OR MORE VARIABLES [Chap. XIII. assume that T is a function of the time, say T=(t), the time rate of absorption of heat is T AM T AT T T AT At - At At = At At = At = At that is, d I Q = c dT , dt dt w where c denotes the instantaneous specific heat at the beginning of the interval At. But the specific heat c is the rate of absorption of heat with respect to the temperature when v is constant ; that is, c = — ^- (See Art. 19.) Hence we may write d^Q^dQdT (2) dt dT dt' K } A similar course of reasoning leads to the result d v Q _dQdv^ /nx dt ~ dv dt' ^ } If T and v change simultaneously, the total rate of absorption of heat is therefore dQ^d^dT^ ,dQdo. (4 x dt dT dt dv dt' W Multiplying through by dt, we obtain dQ-QdT+Wdv. (5) Finally, replacing -^ by c and —^ by I, the so-called latent heat of expansion at constant temperature, we have dQ = cdT+ldv. (6) It will be noted that (4) and (5) are in form similar to equations (a) and (e) but that they are derived from observed relations between the increments AT 7 ,' Av, A T Q, A V Q and the coefficients I and c, and not by the differentiation of a previously existing functional relation between Q, T, and v. Art. 128] EXACT AND INEXACT DIFFERENTIALS 247 As a second example, consider the work W of moving a particle in a plane, say the X F-plane. It is shown in mechanics that dW=Xdx + Ydy, (7) where X and Y denote respectively the X- and Y- components of the force acting on the particle. Therefore (see Ex. 15, p. 67), dW dW X= and Y = , and (7) takes the form dx dy w dW^dx+^dy. • (8) dx dy * K } Again (8) is not derived by differentiating a function W=f(x, y) ; it is deduced from the laws of mechanics and the question of a functional relation between W and the coordinates x and y does not enter into consideration in this deduction. When we have given an expression like c dT-\- 1 dv or Xdx + Ydy the question arises : Can this expression be produced by the differ- entiation of some function of the variables involved ; for example, can we find any function as Q = (T, v) that upon differentiation will produce (6) or any function as W=if/(x, y) that will likewise produce (7) ? To state the question more generally, if M and N are any arbitrarily chosen functions of x and y, does a function of the independent variables (x, y) exist that will upon differen- tiation produce Mdx+ Ndy? Slight consideration will show that such a function usually does not exist, and if it does exist, the coefficients M and N must satisfy a certain condition. Let it be assumed that there is such a function, say z = (x, y). Then by differentiation we obtain dz = — dx + — dy. (9) dx dy y w If therefore the differentiation of the given function produces M dx -h Ndy, we must have M te N= dz (10 dx dy that is, M and N must be the partial derivatives of the function z with respect to x and y, respectively. From Art. 124, we have, with proper restrictions regarding continuity, d 2 z = dh Qr ±fdz\ = d_fdz\ dydx dxdy dy\dxj dy\dyj 248 FUNCTIONS OF TWO OR MORE VARIABLES [Chap. XIII. Hence, if M = — - and N= — , we have ox dy dM = dN , 1V) dy dx as the necessary condition that Mdx + Ndy may be produced by the differentiation of a function (x, y). It may also be shown that this condition is sufficient. * If the condition (11) is satisfied, M dx-\- N dy is called an exact differential; if the condition is not satisfied, Mdx + Ndy is called an inexact differential. Ex. 1. Given M dx + iV dy = ^ dx + log x dy. x KereM=y, N= log a, $M = 1 , 2^=1. The condition imposed by x dy x dx x (11) is satisfied and the differential is exact. It is easily seen that the func- tion is y log x. Ex. 2. Given M dx + Ndy = x 2 y dx — 2xy dx. In this case we have ^M = A (^y) = x 2 and — = — ( - 2 xy) = - 2 y. dy dy dx dx The given differential is therefore inexact, and no function of x and y exists, the differentiation of which will produce this differential. The essential difference between exact and inexact differentials appears more clearly when integration is attempted. If the differential Mdx + Ndy is exact, it is then the total differen- tial of some function z = (x, y) of the independent variables x and y. We have therefore f(Mdx + Ndy) =Jdz=(x, y)+ C. (12) If we assign limits of integration, as (x 1} y^ and (x 2 , y 2 ), we have (Mdx+ N dy) = (x 2i y 2 ) - (x ly y x ) . (13) £ The value of the integral therefore depends only upon the initial and final values of the variables. If we represent (a?,, y x ) and (x 2 , y 2 ) by points in the XF-plane, we say that the integral depends on See First Course, Art. 160. Art. 128] EXACT AND INEXACT DIFFERENTIALS 249 the end points only and not at all upon the path by which the variable point moves from one to the other. The integral is thus a, point function of the coordinates x, y. Ex. 3. Given the differential y dx + x dy. This differential satisfies the condition (11) and is therefore exact ; and since by inspection y dx -\-xdy = d(xy), we have Jx v y 1 dx + x dy) I d(xy) mi-2 ~ ^12/1- This integral is represented geometrically by the shaded area, Fig. 72, and is evidently independent y of the path p between the point (cci, y{) and (z 2 , y 2 ). The integration of an exact differential may be effected by the following rule, o which is sufficient for most cases that arise in practice. Integrate Mdx considering y as a constant, then integrate the terms in Ndy that do not contain x, and take the sum of the two integrals. ' Ex. 4. Given dz = (3 x 2 + 2 y 2 )dx + (4 xy - 9 y 2 )dy. Since -^- (3 x 2 + 2 y 2 ) = JL (4 xy - 9 y 2 ) = 4ty, the differential is exact. dy dx Integrating M dx with y constant, we have f Mdx = ( (3 x 2 + 2 y 2 )dx = x* + 2 xy 2 . The part of N that does not contain x is — 9 y 2 , and . Therefore the integral is z = x z + 2 xy 2 — 3 y s + C. j* -9y 2 dy Sy 3 If Mdx + Ndy is an inexact differential, no function (x, y) can be found the differentiation of which will produce this differential. Consequently the integral | 2 (Mdx + Ndy) cannot be expressed as the difference cf>(x 2 , y 2 )— (x 1 , y x ), and to arrive at any definite result we must assume a relation between x and y, as y = F(x). From this relation we can obtain an expression for the derivative -f-; then by means of the identity, dx Mdx + Ndy = (m+ N c ^)dx, \ dx) 250 FUNCTIONS OF TWO OR MORE VARIABLES [Chap. XIII we can express the integrand in terms of x and integrate in the usual manner. The following example illustrates this process. Ex. 5. Investigate the integral, \ (y 2 dx — xdy). Jo, Since — — = 2y,.— — = — 1, the differential is inexact and some relation dy dx between y and x must be assumed. (1) Let the relation be y = 2 x, which is the equation of a line passing through the given end points (0, 0) and (1, 2). From this relation we have ^ = 2, dx whence y 2 dx - x dy = (y 2 - x Sft dx = (4 x 2 - 2 x) dx, and the integral becomes ( (4 x 2 — 2x)dx = \. (2) Let the assumed rela- tion be y 2 = 4 x, which is also satisfied by the end points (0, 0) and (1,2); then dx — \y dy, and the integral becomes \ "(| y* — $ y 2 ) dy = l\, (3) If the relation is y = 2 x 2 , we have dy = 4x dx, and the integral becomes f 4 (x*-x 2 )dx =- T V It appears therefore that the integral of an inexact differential is not determined by the initial and final values of the variables, but requires in addition a relation between those variables, that is, a path between the end points. If dz = M dx + Ndy is an inex- act differential, dz has therefore no definite significance as a total differential so long as x and y remain independent, and assumes definiteness only when a relation between x and y is furnished. Furthermore, z is not a point function of x and y. Returning to the differentials dQ and d W, equations (6) and (7), it' is known from physical considerations that dQ is inexact, therefore Q is not a point function of T and v, and a relation between T and v must be established before the heat absorbed by the gas during a change of state can be determined. If the force acting on a moving particle is the force of gravity alone, or if it is a function of the distance of the particle from a fixed point, then it is found that the force components in (7) satisfy the relation r) "ST c\~Y -r— = -r— • In this case d W= X dx + Ydy is an exact differential dy ox and the work W depends only upon the end points (x lf y r ) and (x 2 , y. 2 ). If frictional forces are taken into account, d Wis not exact Art. 128] EXACT AND INEXACT DIFFERENTIALS 251 and W depends upon the path between the initial and final positions. EXERCISES Determine which of the following differentials are exact, and for such as are exact find the functions that produce them : 1. e x sin y dx + e x cos y dy. 2. v n dp + npv 11 - 1 dv. 3. x 2 y>dx + bx*ydy. 4 . t dx + x log xdy 5. (x 2 — y) dx — x dy. x 6. O 2 - 2 xy — y) dx - (x 2 — 2 xy + x) dy. 7. (x 2 — axy + y 2 ) dx + (y 2 + xy — ax 2 ) dy. 8. (sin y — e x y) dx + (x cos y — e x ) dy. 9. Show a geometrical interpretation of the differential y dx — x dy, and from purely geometrical considerations show that the integral ( 2 ' 2 (ydx — xdy) must depend upon the path. ** Vl 10. For a gas that follows the law pv = BT, we have I = p, whence dQ = cdT + pdv. Show that while dQ is not an exact differential, — & is an exact differential. /»2, 2 11. Find the value of the integral \ (ydx — x dy) for the following paths between (0, 0) and (2, 2) : (a) A path made up of the straight line joining (0, 0) to (2, 0) and the straight line joining (2, 0) to (2, 2). (&) A path made up of the F-axis from (0, 0) to (0, 2) and the line joining (0, 2) to (2, 2). (c) The curve y = x 3 - 3 x. 12. Find the value of the integral \ (2 xy dx+x 2 dy). Show by choos- «/o, o ing two or more paths that the integral is independent of the path. 13. If IF denote the work done by an expanding gas, show that dW = p dv, that dW is an inexact differential, and that consequently IF depends upon the relations of p to v during the expansion. See Art. 113. MISCELLANEOUS EXERCISES Find ^, ^, ^ for each of the following. dx dy dz 1. u = x 2 ?/ 3 5 . 2. u = x s e 2 v cos z. 3. u = (x 3 + -f + zrf. 4. u = arc tan ^J^. z Find D x y for the following functions. 5. x*y 2 - 4 xy* + 3 y* = 0. 6. — - £- - 1 = 0. a 2 b 2 252 FUNCTIONS OF TWO OR MORE VARIABLES [Chap. XIII. 7. ax 2 + by 2 + 2 hxy + 2ex + 2fy + g = Q. 8. Find — in the following cases. dx 9. u = Vx 2 + y 2 , y = arc tan x. 10. u = xH x arc sin x. vx 3 — y 9 = Cy. 11. u = arc tan *■ , ?/ = e _x , 3 = cos x. z 12. w = e ax (y — z) 1 y = a sin x, at = cos x. Form -^-, -^- , and 2J* for each of the following, dx 6y dx 2 dy cty 3 13. w = x 3 - 5 x 2 y + 3 ?/ 4 . 14. u = e x cos y. 15. u = sin x 2 + cos xy. 16. Show that the relation W T^ = ^ 3 cosy. dx ay dx dy 29. Given w = /(x, y) and x = /> cos 0, y = p sin ; show that 5?* , du du x h y — = p — . dx dy dp 30. Integrate the following differentials. (a) 2 xy dx + (x 2 — y 2 ) dy ; (6) 2 x arc tan y dx + ^ dy. 1 + y 2 31. The heat content i and entropy s respectively, of superheated steam are obtained by integrating the following equations, in which T and p are the independent variables. di = (a + pT)dT+Amn)^ Find expressions for i and s. CHAPTER XIV MULTIPLE INTEGRALS. APPLICATIONS 129. Multiple integrals. In previous chapters we considered successive differentiation and successive integration of functions of a single variable and the successive partial differentiation of functions of two or more variables. We now take up the problem of successive integration of functions of several variables. Suppose, for example, that we have given a function f(x, y, z) of three independent variables. We may write J f(p> V, 2) dz = F(x, y, z), (1) §F{x,y,z)dy = F l (x,y,z), (2) JF,{x, y, z) dx = F 2 (x, y, z), (3) where in (1) the integration is taken with respect to z, that is, as if y and x were constants. Likewise in (2) it is taken with respect to y, and in (3) with respect to x. Substituting in (3) the value of F x (x, y, z) from (2), we have F 2 (x, y, z) = JjJ>(z, y, z) dy\ dx, (4) and by the use of (1) this becomes F 2 (x, y, z) = H jT ff(x, y, z) dz~]dy \ dx. (5) The expression (5) may be written more compactly as follows : F 2 (x, y, z) = ffff( x > y> z ) dz d v dx - ( 6 ) 254 Art. 129] MULTIPLE INTEGRALS 255 It is to be understood that the integral is to be taken first with respect to z, then y, and finally x* An expression like the second member of (6), which indicates the result of several successive integrations, is called a multiple integral. If two integrations are involved, the integral is called a double integral; if three, a triple integral; etc. The evaluation of an indefinite multiple integral differs from that of an indefinite single integral in one essential feature; namely, the form of the constant of integration. An example will illustrate this point. Ex. Given i \ e^y 2 dy dx ; find a function u of x and y such that d 2 u = g2 2 dxdy Integrating first with respect to y regarding a; as a constant, we obtain " — = -e 2x y B + constant of integration. dx 3 Since x was considered as constant during the integration, the constant of integration may depend upon x. To make this more clear, we may observe that differentiating either j 2 3 „ or \e^-h(x), with respect to y gives the same result, e^y 2 . Hence, we assume the more general case and write « as =W +#(«). ox o where (x) is an arbitrary function of x. As a special case, (x) may of course be a constant C. Integrating this result, keeping y constant, we obtain u = } e 2x y* + f (x) dx + ^(y). Here again, since y was considered constant during the integration, the inte- gration constant must be taken as an arbitrary function of y. * Books on the calculus differ in the manner of indicating the order in which the integrals are to be taken. Some authors write the differential last which is to be taken in the first integration, etc., that is, the order of the differentials in equation (6) is exactly reversed. The above notation is adopted in this book because it shows best the manner in which the integra- tion has arisen. In other works, the context will usually indicate to the reader the notation employed. 256 MULTIPLE INTEGRALS. APPLICATIONS [Chap. XIV. By assigning limits of integration to each variable, we arrive at the notion of a definite multiple integral. Thus the integral nVy d V dx indicates that e^y 2 is to be integrated first between the limits a and*6 with respect to y, and the result thus found is to be inte- grated with respect to x between the limits and 1. Since x is considered as constant in the integration with respect to y, the y-limits, a and b, may be functions of x. Similarly in a definite triple integral, taken first with respect to z, then y, and finally x, the z-limits may be functions of both y and x, and the ^/-limits functions of x. V Ex. Evaluate \ I I r 2 sin0 dr ddd. 3 Jo Jo Integrating with respect to 0, we obtain 12 > r 6 It should be observed that the upper limit of the r-integral is a function of 0. EXERCISES Evaluate the following integrals. 1. f \x 2 y dy dx. 2. I j e x sin y dy dx. IT c«c°«-«»°\*n nedrde . 6 . re- pdf>de . Jo Jo Jo Ja (1-008 0) t* i C 2 a cos 7. I I p s cos* dp d8. § o a $ Q C *^^- y \z* + y*)dxdydz. Art. 130] PLANE AREAS BY DOUBLE INTEGRATION 257 11. JO Jft t Jo Jo 2 gy dy dx. 10 x 2 y dy dx. • n' Jo Jo P dd dp. x n dy dx. 130. Plane areas by double integration, rectangular coordinates. We have seen (Art. 100) that the area bounded by a curve, the axis of x, and two or- dinates is given by an integral of the form j f(x)dx. When the area is bounded by two curves, as in Fig. 74, the area is given, not by a single integral, but by the difference of the two ^ integrals * X Fig. 74. f b f(x)dx, f h F(x)dx. The same result may be accomplished by successive integration. Consider the element of area Ay Ax. If we sum up these elements with respect to y, we shall have the area of the strip PQ, and then by summing up the strips between the limits a and b we have * « X \X Ax ] Ay = 2 X Ax Ay - a L P J a P Upon passing to the limits first as Ay = and then as Ax = 0, we have the required area given by the double integral A- I f doc dy, J a JFyX) (i) which leads to the same expression for the given area as was obtained in Art. 100. We might equally well have summed up the elements of area 258 MULTIPLE INTEGRALS. APPLICATIONS [Chap. XIV. in the reverse order, namely, first with respect to x and then with respect to y. In this case, we should have obtained (2) where (y) and ij/(y) are the inverse functions of F(x), /(as), respectively. Ex. 1. Find the area between the circle x 2 + y 2 = a 2 and the line x + y = a. See Fig 75. The coordinates of the points of intersection are (0, a) and (a, 0). The lower limit of the y-integral is the value of y found from the equation x + y = a, namely, \y = a — x ; and the upper limit is the value of y deter- mined from the equation x 2 + y 2 = a 2 , namely, y = Va 2 — x 2 . Hence we have »*/Zs .... dx A=\ i a dy< J0 Ja-x * — ( Va 2 — x 2 —(a — x) \dx = [- Va 2 - x 2 + — arc sin - - ax + — 1 L.2 2 a 2 Jo Fig. 75. In some cases the abscissas of the points of intersection of the two curves may not give the proper limits of integration ; in such cases it is necessary to divide the area into two or more parts. Ex. 2. Find the area bounded by the curves x 2 + y 2 = 25, y 2 = *£x, and y = ft x 2 . See Fig. 76. The circle and the first parabola intersect at the point A whose coordinates are (3, 4), and the second parabola intersects at the point C, whose coordinates are (4, 3). From O to A, the equation y 2 = Y ■ * gives the Upper y-limit, but from A to C, this limit is determined from the equation of the circle, x 2 + y 2 = 25. The equation y = ft x 2 of the lower curve OC gives the lower y-limit throughout. Hence we have CC ~ 8 dydx + Jo *)&** dydx = 7.55. Art. 131] PLANE AREAS BY DOUBLE INTEGRATION 259 EXERCISES 1. Find by double integration the area of a parallelogram with one side in the .X-axis. 2. Find by double integration the area of a right triangle with a short side in the X-axis. 3. Find by double integration the area between the parabolas y 2 = 8 x and x 2 = 8 y. 4. Find the area between the circle x 2 + y 2 = a 2 and the line y = b, b < a. 5. Find the area between the circle x 2 + y 2 = 2ax and the parabola y 2 = ax. 6. Find the area between the curve y = x s and the F-axis from the origin to y = 8. 7. Find the area bounded by the parabolas y 2 = - 1 /- x, y 2 = f x and the circle x 2 + y 2 = 25. Consider the first quadrant only. 8. Find the area between the circle x 2 + y 2 = 25 and the line x + y = 7. 9. Find the area bounded by the curves x 2 + y 2 = 169, y 2 = ^ff x s , and 5 x— 12 y = 0, in first quadrant. 10. Find the area of the ellipse — + %- = 1. a 2 & 2 131. Plane areas by double integration, polar coordinates. When it is desired to find the area of a surface bounded by two curves given in polar coordinates, we may em- ploy the method of Art. 105, or we may find the desired area by double integra- tion. The latter method is introduced here as a simple exercise in the use of double integration. Let the polar element of area be ABCD, Fig. "7, bounded by the two radii 0(7, OB, and two circular arcs having their common center at O. Let the polar coordinates of B be (p, 9). From ele- mentary geometry, we have sector AOB = i p 2 A6, (1) sector DOC = Up + Ap) 2 A0. (2) Hence, A.4 = ABCD = \ (p + Ap) * A0 - \ p 2 A0 = (p + iAp)A6kp. (3) 260 MULTIPLE INTEGRALS. APPLICATIONS [Chap. XIV. Using this polar element, we may find the area between two curves p = F(0), p =/(0), Fig. 78, as follows : First keep A0 con- stant and sum the elements of area with respect to p. The result of this summation is an area of the type PQSR, the expression for which is OP A6-L X (p + iAp)Ap Ap = oq Fig. 78. = A p dp. (Art. 100) Joq If now we sum with re- spect to 0, we get the sum of the wedge-like slices, and the limit of this sum is the required area. We have therefore J!l nop /»*, r*op A=L TA0. | pd P =l I pdpdd. A5i0 7 Joq Je x Joq Replacing OQ and OP by F{6) and /(0) respectively, we obtain the formula (4) A — I I pdpdQ The area included between the curve p=/(0) and two radii, as OA and OB, is obtained from (4) by making F(0) = 0. The required area may also be obtained as follows : Summing first with respect to 0, keeping Ap constant, we obtain a segment of a circular ring of the type EFOH. A second summation with respect to p gives the sum of such ring segments, the limit of which sum is the area A. The resulting formula is A Jr*p 8 ru?) p dQ dp, (5) where <£(p) and \f/(p) are the inverse functions of F(0) and /(0), respectively. Ex. Find the area between the circle p = cos 6 and one loop of the lem- niscate p2 — cos 2 6. Fig. 79. If formula (4) is used, the area must be taken in two parts. The upper limit for both integrations is determined from the equation of the circle Art. 131] VOLUME BY TRIPLE INTEGRATION 261 p = cos 0. For values of between and -, the lower limit of the p-integra- 4 tion is obtained from the equation of the lemniscate p 2 = cos 2 0. Since, however, no part of this loop of the lemniscate lies to the left of the line OA, /A Fig. 79. the lower limit for the p-integration for values of greater than - is 0. Hence we have Jo J x V/ C O8 20 pdpdd + 2 n cos0 pdpdd = EXERCISES 1. Find by double integration the area of the circle p = a. 2. Find the area between the cardioid p = 2 a(l — cos 0) and the circle p =— a cos 0. 3. Find the area between two circles tangent internally and having radii ri and r 2 respectively. Work also by single integration. 4. Find the entire area of the cardioid p = 2 a(l — cos 0). 5. Find the area of one loop of the lemniscate p 2 = a 2 cos 2 0. 6. Find the area between the circle p = cos and one loop of the lemniscate p 2 = cos 2 by the use of formula (5). 7. Find the areas between the cardioid p = 2 a(\ — cos 0) and the circle p = 2 a. (The shaded areas OB AD and BDCB, Fig. 80.) Work also by single integration. 8. Work Ex. 3 by formula (5). Fig. 80. 132. Volumes by triple integration, rectangular coordinates. Consider the volume bounded by the coordinate planes and any surface whose equation is z=f(x, y) t where f(x, y) is a 262 MULTIPLE INTEGRALS. APPLICATIONS [Chap. XIV. continuous function. Through two neighboring points P and Q (Fig. 81) within the solid let planes be passed parallel to the three coordinate planes. The six planes will inclose a rectan- gular parallelopiped PQ whose volume is AzAyAx\ for if we Fig. 81. assume the coordinates of P and Q to be respectively (x, y, z) and (x-\-Ax, y-\-Ay, z-j-Az), the edges of the parallelopiped are Ax, Ay, and Az, respectively. We first take the sum of these elementary parallelopipeds with edges lying along the line S'PS. With x, y, Ax, and Ay constant, the limit of this sum as Az is made to approach zero is the volume of the prism whose base is Ax Ay and whose altitude is S'S. Now with x and Ax constant, we take the sum of the prisms lying between the planes AMD and BNC. By an extension of the method of Art. 100, it follows that as Ay is made to approach zero, this sum approaches the volume of the cylindrical slice AB'C'DMN. Finally, we take the sum of the slices parallel to the YZ-plane. As Ax approaches zero, the volume of the cylin- drical slice approaches that of the actual slice ABCDMN. Hence, Art. 132] VOLUMES BY TRIPLE INTEGRATION 263 the limit of the sum of these slices, as Ax approaches zero, gives the volume of the solid. We have, therefore, SS' Volume of prism = Ay Ax L X Az == Ay Ax I dz. MD Volume of cylindrical slice = Ax L X Ay ( j dz) Ay = Oo \y° J = Ax I I dz dy. Jo Jo OE Volume of solid = L 2^Ax( ( dydz) Ax = o \Jo J J j I dzdydx. (1) o Jo Jo Hence the volume is obtained by a triple integration, provided the limits of integration are properly chosen. The first summation, namely, that with respect to z, extends from zero to S/S', which is the value of the ordinate of a point on the surface. The upper limit is, therefore, a variable which is given by z=f(x, y), where x, y are the coordinates of the point S'. The second summation is taken from zero to MD, and hence the limits are zero, and the value of y on the curve FDE, that is, the value of y determined from the equation f(x, y) = 0, it being assumed that y is a single-valued function of x. Let this value be y = cf>(x). Finally, the integration which gives the sum of all the slices between and E has for its limits zero and the constant OE (=«, say). We may therefore write equation (1) in the form F= Joio Jo «*»<*»*«'• (2) It will be observed that the limits for the second and third in- tegration are the same as those that would be used in finding the area OEF, i.e. the projection of the given solid on the XF-plane, by means of the double integral J | dy dx. This fact suggests a 264 MULTIPLE INTEGRALS. APPLICATIONS [Chap. XIV. method of finding the required volume by double integration, namely, by the integral z dy dx, Jo «/o when z is first expressed in terms of x and y. If the given solid is not bounded by the coordinate planes, the lower limits will not be zero, as in (1) and (2) ; they can be readily determined, however, in the same manner as the upper limits. EXERCISES 1. Determine the limits of integration for the triple integral ( I \dzdydx required in finding the volume of the pyramid bounded by the coordinate planes and the plane - + ^ + - = 1. Draw the figure. a b c 2. The cone whose equation is y 2 + z 2 — c 2 x 2 has the X-axis as its axis. Determine the limits of integration when the volume of the cone is found from the triple integral \ \ I dzdy dx. Also when the volume is found from the triple integral \ \ \ dx dy dz. Let h denote the altitude. 3. Find the volumes in Exs. 1 and 2. 4. Find by triple integration the volume of the ellipsoid £ + £ + £= i. a 2 b 2 c 2 5. Find the volume inclosed by the surface x 2 y 2 + c 2 z 2 = a?y 2 and the planes y = and y = c. 6. Show that the volume generated by revolving a plane figure about the X-axis may be found from the formula V = 2 w I i ydydx. JC J f{x) Compare with results in Art. 106. 7. Find the entire volume bounded by the surface whose equation is a* + tr + z = a • 8. Find the volume of the wedge cut from the cylinder x 2 + y 2 — a 2 by the planes z — and z = x tan /S. Art. 133] VOLUMES BY TRIPLE INTEGRATION 265 Fig. 82. 133. Volumes by triple integration, polar coordinates. Let p, 0, cf> denote the coordinates of any point P within a solid, where, as usual, p is the distance OP (see Fig. 82), 6 is the angle ZOP which OP makes with the Z-axis. and is the angle XOP' which the projec- tion of OP on the X F-plane makes with the X-axis. Through P sup- pose three surfaces passed : (1) the surface of a sphere with radius OP and O as a center; (2) a conical surface produced by revolving OP about OZ as an axis ; (3) a plane surface passed through OP and OZ. Now let a second spherical surface y be passed through P lf a second coni- cal surface through S by the rotation of OS/Si about OZ, and a second plane surface through OZ and OQ. The six surfaces inclose a solid element PQRSSxRxQ^ having the two spherical surfaces PQRS and PxQiR^, the two conical surfaces PPiQ x Q and SS^R, and the plane surfaces PPyS^ and QQ^R. By means of such sets of surfaces the entire solid may be divided into elements of this type. Let PP 1= A P , angle POS=A6, angle P'OQ'= angle PHQ= A<£. Then the sides of the given element of volume are PP' = Ap, PQ = PH A = p sin 6 Acf>, and PS = p A<9. Consequently, the volume of this element is p 2 sin 6 A9 Ac£ Ap, plus other terms which vanish when we pass to the limit.* Therefore, denoting the required volume by V, we may write, V= L L L Vy,Vp 2 siTiOApA dd = t f " C sin" d cos* are clearly and 2 w ; hence (1) becomes V= 2 7T f C P 2 sin 6 dp dd. (2) The limits of integration for p and 6 are precisely those that would be employed in finding the area of the plane figure revolved. Ex. 2. Find the volume generated by revolving the cardioid p = 2a(l — cos0) about the initial line. We have in this case Ctt /*2a(l-cos0) F=2 7ri I p*sinedpd6 = l^i 8 f * (l - cos ey sin e do = 6 i «a 3 J» 3 Art. 134] PROBLEMS INVOLVING SUMMATION 267 EXERCISES 1. Find by polar coordinates the volume of a sphere (a) when the origin is taken at the center ; (&) when the origin is taken at the end of a diameter. 2. By passing to polar coordinates tind the entire volume of the solid in- eluded by the surface ^ + y . 2 + ^ = ^ 3. In the same manner find the volume included by the surface (x 2 + y 2 + z 2 ) 3 = 27 a s xyz. 4. Find the volume obtained by revolving the lemniscate p 2 = a 2 cos 2 about the initial line. 5. Find the volume obtained by revolving the loop of the curve cos 2 p = a cos about the initial line. 6. Find the volume obtained by revolving the curve p = 2 a cos + b about the initial line. First plot the curve and determine proper limits for 0. 7. Show the different types of solid elements produced when the summa- tion is effected in the following orders ; (a) with respect to p, then 0, then

*E-S(H)]- 270 MULTIPLE INTEGRALS. APPLICATIONS [Chap. XIV. Ex. 3. Required the volume of a hollow pipe bent as shown in Fig. 85, the axis of the pipe being a circular arc of radius B subtending an angle /3. Take ri and r 2 respectively as the inner and outer radii of the cross section. In the cross section of the pipe we choose an element of area pAdAp, where p is the distance of the element from the center O of the cross section. Evidently the distance of this element from an axis through 0' is B — p sin 6. We may there- fore take as the volume element a rod of cross sec- tion p Ad Ap bent into an arc of radius B — p sin 6 and subtending an angle /3. The length being P(B — p sin 0) , the volume is p(B - psind)pAdAp. The limits for p are ob- viously ri and r 2 ; and one half of the required volume, that on one side of the plane MN, will be obtained by taking — ^ and - as the limits for 6. Hence we have V = 2 f n C 2 (3 (B-p sin 6) pdddp = ttQB (r 2 2 - n s ). 2 A still easier solution is given by the theorem of Pappus (Art. 137). EXERCISES 1. In illustrative example 2, let the quarter cylinder be intersected by a sphere of radius c with its center at the origin. Using the same type of volume element, determine the limits of inte- gration and find the volume of the part of the quarter cylinder intercepted be- tween the sphere and the XF-plane. 2. A conoid is generated by a line kept parallel to a given plane and moved so as to keep in contact with an ellipse and with a straight line AB, Fig. 86, Fig. 80. Art. 135] MASS. MEAN DENSITY 271 parallel to the plane of the ellipse. The semiaxes of the ellipse are a and b, and c is the distance of the line from the plane of the ellipse. Find the volume of the conoid (a) by taking elements parallel to the base j (&) by- taking elements perpendicular to the line AB. 3. Find the volume OABCD, Fig. 87, founded by the three coordi- nate planes and the warped surface generated by the line EF, which re- maining parallel to the XF-plane slides on the lines AD and BC. If OC = a, OD = b, and OA = c, and Az Ay Ax is taken as the element, show that the limits of integration ^ ^ and for z are and c 1 fl flL_Y \ b(a-x)J Also KX" Fig. 87. those for y are and - (a — x). a y* choose an element such that a single integration is sufficient. 4. Show four different elements of area that may be employed in finding the area of an ellipse. Write the integrals and determine the proper limits of integration in each case. 5. Show that the volume of a solid of revolution may be found by means of the double integral 2x f \ ydydx. What type of solid element leads to this integral ? Determine the proper limits of integration. 6. Find the volume of the part cut from a sphere of radius a by a cylinder of radius o, one element of which contains the center of the sphere. 7. A right cylinder whose intersection by the XT-plane gives loop of the lemniscate p 2 = a 2 cos 2 6 is cut by a plane that intersects the XF-plane in the F-axis and makes an angle of 45° with the XF-plane. Find the volume of the solid bounded by the cylindrical surface, the XF-plane, and the oblique plane. 135. Mass. Mean density, If m denote the mass and Fthe volume of a body, the ratio — is called the mean density of the body. If we take a volume element AF, inclosing a point P, and denote the mass of the element by Am, the ratio — — is the mean density of the element; and the limit of this ratio as AF approaches zero (still including the point P) is the density at the 272 MULTIPLE INTEGRALS. APPLICATIONS [Chap. XIV. point P. If the density is the same at all points within a body* the body is said to be homogeneous; otherwise, it is said to be non-homogeneous. Let the mass of a non-homogeneous body be divided into ele- ments Am^ Ara 2 , •••, Am n , whose volumes are respectively AF b AF 2 , •••, AF M . Denoting by y lf y 2 , •••, y n the mean densities of these elements, we have Am 1 = y 1 AF 1 , Ara 2 = y 2 AF 2 , •••, Am n = y n AF M . (1) The mass of the body is therefore m = L VAm k = L Vy k AV k = CydV, (2) n = co i n = go i J and the mean density of the body, which we shall denote by y, is given by the equation CydV y=™=dl (3) 7 V V w If the mass is distributed continuously over a surface, we may replace the element of volume AFby an element of area AA-, and if the mass is distributed along a curve, as, for example, along a thin wire, we replace A V by As, the element of length. The element of volume A V should be so chosen as to lead to the simplest integrations. Usually triple integration will be neces- sary, but in some cases the element may be taken in such a way that a double or even single integration is sufficient. Ex. 1. Find the mean density of a sphere in which the density varies as the square of the distance from the center. Since the distance p of the volume element from the center determines the density, it is evident that a polar element should be chosen. From the given law we may take the density at a distance p from the center as kp 2 , k being a constant ; and for the volume element A V we take p 2 sin d Ap A

V= r . > 7i = ~^ ( 5 ) J ydV J ydV lydV The remarks of Art. 134 relative to the choice of the type of element and the determination of the limits of integration apply with equal force here. Very often the problem can be simplified by care in selecting the element. * The centroid of a mass is often called the center of gravity of the mass. 276 MULTIPLE INTEGRALS. APPLICATIONS [Chap. XIV. 137. General theorems relating to centroids. The following general theorems are useful in the determination of centroids. Theorem I. Tlie first moment of a volume {or mass) with respect to a plane or axis containing the centroid is zero. If the centroidal plane (i.e. plane containing the centroid) is a coordinate plane, the theorem follows at once from (4) or (5). Thus, if the centroid lies in the FZ-plane, x = 0, whence M VM = 0. It can be readily shown that the theorem holds for any centroidal plane ; therefore it holds for the intersection of two such planes, that is, a centroidal axis. Theorem II. A plane of symmetry of a solid or of a homogene- ous mass is a centroidal plane. Likewise an axis of symmetry of a plane figure is a centroidal axis. Take the plane of symmetry as a coordinate plane, say the XFplane. To an element A V above the plane there corresponds an equal element at the same distance below the plane. The moment of the first element with respect to the plane is z x A V, that of the second element is — z x AF, whence the moment of the pair of elements is zero. Since all the elements of the solid can be thus arranged in pairs, the first moment of the solid with re- spect to the plane is zero. Theorem III. If V lf V 2 , •••, V n are n volumes, and ifx x , x 2 , x 3 , • ••, x n are the coordinates of their centroids, then the x-coordinate of the centroid of the system of volumes is __ V l x l +V 2 x 2 + • •• +V n x n m •~ Fi+F.+ - +V H ' (1) A similar theorem holds for n masses, m u m 2 , •••, m n , or for n plane areas in the same plane. The numerator of the second member of (1) % is the moment M ys of the system with respect to the FZ-plane, and the denominator is the total volume; hence the theorem follows from (4), Art. 136. Theorem IV. (TJieor ems of Pappus and Ouldin.) (a) If a plane curve is revolved about an axis in its plane, the area of the surface generated is equal to the product of the length of the curve arid the circumference of the circle described by the centroid of the curve. Art. 137] CENTROIDS 277 (b) If a plane figure is revolved about an axis in its plane, the volume of the solid of revolution generated is equal to the product of the area of the figure and the circumference of the circle (or the length of the circular arc) described by the centroid of the area. From (4), Art. 136, we-have for a plane curve /' ds or sy -s yds, where s denotes the length of the curve. The surface generated by the revolution of the curve about the X-axis is S — 2 ir I y ds. (See Art. 110). Hence equating the two expressions for J y ds, we S = 2*ys. (2) obtain Evidently the theorem holds equally well for an incomplete revo- lution. To prove the second theorem, let the axis of revolution be taken as the axis of x, as before ; then denoting by AA an element of the plane area, we have for the volume generated by a complete revolution of the area, V= L S / 2 7ryAA = 2 7r fydA. AA = 0^ J JydA = Ay; 2 Try A But hence, Ex. 1. Find the coordinates of the centroid of a circular arc of radius a which subtends an angle 2 £ at the center, Fig. 90. Take the line of symmetry as the X-axis ; then y = 0, and the centroid lies in this line. Polar coordinates lead to the simplest integral ; hence taking x = a cos 6, As = a Ad, we get, M v £ 2 cos d dd s Zap For a semicircumference /3 _ 2 « 2 sin ft _ a sin /3 2 a/3 p Fig. 90. - , whence x = — 2 ic 278 MULTIPLE INTEGRALS. APPLICATIONS [Chap. XIV. Ex. 2. Find the coordinates of the centroid of the plate described in illus- trative example 2, Art 135. ( yx dA From the defining equation, x — * kxy for y and dy dx for dA, j y dA b , we have, upon substituting Similarly, V Jo Jo kx *y d v dx b, . fa fa Va '-~ X * Jo Jo kx y d y dx a«. Ex. 3. Find the centroid of a semicircle of radius a. Taking the X-axis as the axis of symmetry, the length of the path de- scribed by the centroid as the semicircle is revolved about the Y-axis is 2 irx. The semicircle by its revolution describes a sphere whose volume is \ rra z ; hence by the second theorem of Pappus, 2 TTX • $ ltd 2 = $ 7m 3 , Ex. 4. Find the centroid of a wedge-shaped solid, Fig. 91, cut from the cylinder x' 2 + y 2 = a 2 by the planes Fig. 91. *=0, J+5 b a 1. The volume of the solid is | wa 2 • 2 & = iva 2 b. The first moment with respect to the XF-plane is given by the in- tegral \ \ (zdzdy dx. The limits of integration, as determined by an inspection of the solid, are as follows : For *, and 6 (l --\. for y, — Va 2 — x 2 and Va 2 — x 2 ; for x, — a and a. Hence, we have fa Sa*-J A l -l) //-' f-a ( l ~ ~Y ^ ~ X% dX = * ****** Art. 137] CENTROIDS 279 For the moment with respect to the FZ-plane, we have ra r v a?-x" /» \ a) ^=J-J-v^J« xdzdydx = 2b( a x (l--\ Va? -x 2 dx=- m 3 b 4 Since the XZ-plane is a plane of symmetry, y = 0. The coordinates of the centroid are therefore ira 2 b va 2 b EXERCISES Find (a) the coordinates of the centroids of the following plane curves ; (&) the coordinates of the centroid of the area between each of the curves and the X-axis. 1. A semicircumference. 2. An arc of the parabola y 2 = 4 ax from x = to x = a. X _x 3. An arc of the catenary y = - (e a + e •) from x = to x = a. 4. One arch of the cycloid x = ad — a sin 0, y = a — a cos 0. a* 3 5. Find the centroid of the area between the cissoid rj 1 = — ■ — and its asymptote x = a. 6. Find the centroid of a segment of the parabola y 2 = 4 ax cut off by the double ordinate corresponding to x = a, assuming that the density at any point is proportional to the distance from the F-axis. 7. Find the centroid of a semicircle of radius a, assuming the density proportional to the distance from the bounding diameter. 8. Find the centroid of a segment of a homogeneous paraboloid of revo- lution between the origin and x = a. 9. Find the centroid of a hemisphere whose density varies as the dis- tance from the base. 10. Find the coordinates of the centroid of one loop of the curve p = a sin 2 6. 11. Find the coordinates of the centroid of the mass of one eighth of a sphere included between the coordinate planes, assuming that the density varies as the distance from the origin. \ ^\ 1 I « 3- — >\ 'ig. 92. 280 MULTIPLE INTEGRALS. APPLICATIONS [Chap. XIV. 12. Erom Theorem III, Art. 137, show that if points P and Q are the centroids of two volumes Vi and V 2 (or areas A\ and A%, or masses mi and m 2 ), the centroid of the whole system V\ + F 2 lies on the line joining P and Q and divides it into segments inversely proportional to V\ and V 2 . 13. A solid is composed of a right circular cylinder and a right cone. See Fig. 92. Find the centroid of the solid. 14. Three masses, mi = 3, m 2 = 4, and m 3 = 5, have their centroids located at the three vertices of an equilateral triangle the side of which has the length a. By means of Theorem III, find the position of the centroid of the system. 15. Using the Theorem of Pappus, find the centroid of the semicircumference of a circle of radius a. 16. Using the Theorem of Pappus, find the volume of the pipe described in the illustrative example 3, Art. 134. 17. Work the illustrative example 4 of this article by taking p Ad Ap Az as the element of volume. 138. Second moment. Radius of gyration. Let each of the elements into which a volume (or area, length, or mass) is divided be multiplied by the square of the distance of some chosen point in the element from a reference line or plane. The limit of the sum of these products as the elements are taken smaller is called the second moment * of the magnitude in question with respect to the line or plane of reference. Formulas for second moments may therefore be derived from those for first moments by squaring the distance factor. Denot- ing the second moment by the general symbol I, we have the fol- lowing formulas corresponding to (1), (2), and (3) of Art. 136. For a plane curve, I x — I y 2 ds. (1) For a plane area, £= I y 2 dA. . (2) For a volume, I xy =fz 2 dV. (3) * The second moment of a mass is sometimes called the moment of inertia of the mass. Art. 138] SECOND MOMENT 281 From its definition, the second moment of a volume must have for its physical dimensions volume x (length) 2 ; hence, the second moment of a volume V can be expressed in the form I=Vk\ (4) in which A; is a length. Similarly the second moment of an area A may be expressed in the form J=^ 2 , (5) and likewise for any magnitude whose second moment is taken. The length k as defined by (4) or (5) is called the radius of gyra- tion of the given magnitude with respect to the reference plane or axis. Ex. 1. Find the second moment and the radius of gyration of the area of a semicircle of radius a with respect to the bounding diameter. Taking the origin at the center, and the bounding diameter as the axis of x, we have, by transforming to polar coordinates, y = p sin 0, A A = p Ad Ap. Therefore, I x = \ " y p 2 sin 2 6 ■ pdddp = \ vat, and *a = I*=l£^:=la», A lira? 4 whence k = \ a. Ex. 2. Find the second moment of a straight rod or wire of length I about one end, assuming that the density varies as the distance from that end. Consider the rod as lying in the X-axis with the end about which the mo- ment is taken at the origin. At a distance x from the origin the density is kx and the mass of an element of length Ax is kx Ax. The second moment of this mass element about the origin is x 2 • kx Ax ; hence the second moment of the rod is 7= L y.x 2 -kxAx= C kx* dx = \ kl*. The mass of the rod is m = f kx dx = \ kl?. Therefore & = - = \ P. m 2 282 MULTIPLE INTEGRALS. APPLICATIONS [Chap. XIV. EXERCISES Find the second moment and the square of the radius of gyration in the following cases. 1. A rectangle of width b and length h about an axis coinciding with the short side b. 2. The same rectangle about an axis through the center parallel to the side b. 3. A circle of radius a about an axis through the center perpendicular to the plane of the circle. 4. The same circle as in Ex. 3, with the density varying as the distance from the center. 5. A triangular area of base b and height h about an axis through the vertex parallel to the base. 6. A semicircumference of a circle of radius a about the diameter. 7. The area between the cycloid x = a(d — sin 0), y = a(l — cos 0) and the X-axis about the X-axis. 8. The cycloidal curve about the X-axis. 9. A solid sphere of radius a with respect to a diametral plane. 10. The same sphere as in Ex. 9, assuming that the density varies as the distance from the diametral plane. 139. General theorems relating to second moments. The follow- ing general theorems are useful in the determination of second moments and radii of gyration. Theorem I. If a solid (or homogeneous mass) is divided by a plane of symmetry, the second moment of the entire solid (or mass) with respect to this plane is double that of the part on one side of the plane of symmetry. This theorem is easily established. For any element AF on one side of the plane at a distance y there is a corresponding equal element on the other side at a distance — y. The second moment of one element is y 2 AV, that of both is y 2 AF+( - y) 2 AV= 2 y 2 AF. Since all the elements of the solid can thus be arranged in pairs, the theorem follows at once. The theorem holds also for the second moment of a plane surface with respect to an axis of symmetry. Art. 139] SECOND MOMENTS 283 Theorem II. The second moment of a solid {or mass) ivith respect to a line is the sum of the second moments with respect to two planes which intersect at right angles in the line. Consider the given line as the X-axis, and take the XZ- and X F-planes as the intersecting planes. A point P in a chosen volume element has the coordinates (x, y, z). Its distance from the X-axis is V# 2 + z 2 , and by the definition the second moment of the solid with respect to this axis is I, --f(y 2 + z 2 )dV. But hence, JV dV= I xz , and J> dV= I xy ; I x = I xz + I xy . (1) A similar theorem applies to plane areas. Referring to Fig. 89, it is clear that the sum of I x and I y , the second moments of the plane area with respect to the axes of x and y, respectively, is the second moment of the area with respect to an axis through the origin perpendicular to the plane of the figure. . Theorem III. The second moment of a solid (or mass) with respect to an axis is equal to its second moment with respect to a parallel axis through the centroid plus the product of the volume (or mass) of the solid and the square of the distance between the axes. Let the given axis pierce the plane of the page at 0, Fig. 93, and suppose O g to be the trace of the parallel axis through the cen- troid. Take the line 00 g as the FlG - 93 - X-axis, and let a denote the distance 00 g . Referred to O g as an origin, the point P in a given volume element has the coordinates (x, y) ; referred to as origin, the coordinates are (x + a, y). The second moment of the solid with respect to the axis through O g is therefore Y Y- a — 4 — *- o X I g =f(x* + f)dV, 284 MULTIPLE INTEGRALS. APPLICATIONS [Chap. XIV. while the second moment with respect to the axis through is I=fi(x + ay+y*]dY = C(x 2 + y 2 )dV+a 2 CdV+2aCxdV = I g + d 2 V+2aCxdV. Now J x d V is the first moment of the solid about the axis through O g \ hence, since this axis contains the centroid, jxdV=0, and we have finally I=I g + a*V. (2) Dividing both members of (2) by V, we get k 2 = k g 2 + a 2 . (3) The student may show that the theorem holds for parallel planes, one of which contains the centroid. Theorem III is specially useful in finding the second moments of volume, as illustrated by Ex. 2 given below Theorem IV. If k x , k 2 , •••, k n are the radii of gyration of n vol- % umes (or areas, or masses) with respect to a line or plane, the radius of gyration k of the entire system with respect to the line or plane is given by k 2 = vjc£ + vjc£± ••• +v n K 2 _ This theorem is useful in finding the second moment of an area or volume which can be conveniently divided into parts. The proof is similar to that of Theorem III, Art. 137. Ex. 1. Find the radius of gyration of a circle of radius a about a tangent. From Theorem I, the radius of gyration about a diameter is the same as that for a semicircle ; hence from Ex. 1, Art. 138, we have k g 2 = \ a 2 . Using Theorem III, we obtain therefore k 2 = \ a 2 + a 2 = | a 2 . Ex. 2. Find the second moment of a right circular cone with respect to a line through the vertex perpendicular to the axis of the solid, Fig. 94. Take the axis of the cone as the Z-axis, and choose the vertex as the Art. 139] RADIUS OF GYRATION 285 origin. Let h be the altitude and b the radius of the base. Consider a slice of the cone parallel to the XF-plane at a distance z from it. The radius of the slice is -z. its thickness is Az, and its h second moment with respect to a diametral axis X', parallel to OX, is approximately z*Az. According to Theorem III, the second mo- ment of this slice with respect to OX is therefore 1 r- z*Az + *V-.sPAz-z* 4 W h* Fig. 94. The second moment of the cone about OX is the limit of the sum of terms ot this type ; that is, 5 A*V4*» J Jo Uh* J The radius of gyration is given by the equation 5 V 4 ^ 2 / 20 v J EXERCISES In the following examples determine the radii of gyration. 1. A square whose side is a : (a) about an axis through the center per- pendicular to its plane ; (6) about a diagonal. Suggestion : Make use of Theorem II. 2. An ellipse with semiaxes a and b : (a) about the major axis ; (6) about the minor axis ; (c) for a tangent at the end of the minor axis ; (d) for a centroidal axis perpendicular to the plane of the ellipse. 3. A ring bounded by circles of radii a\ and a 2 , respectively, about a diameter. Use Theorem IV. 4. A sphere of radius a, about a diameter. 5. A right circular cone : (a) about a plane containing the axis ; (&) about the axis. Take a as base radius and h as altitude. 6. A right circular cylinder : (a) about its axis ; (b) about a generating element ; (c) about a diameter of one base. 286 MULTIPLE INTEGRALS. APPLICATIONS [Chap. XIV. 7. Show that the radius of gyration of a right circular cylinder about its axis is the same as that of the circu- lar cross section about the same axis. Deduce a similar principle for pris- matic solids in general. 8. Find the position of the axis x c passing through the centroid of the cross section shown in Fig. 95. 9. Determine the radius of gyra- tion of the cross section, Fig. 95 : (a) with respect to the axis OX; (&) with respect to the axis x c . 140. Illustrative examples. The following illustrative prob- lems are introduced to give the student further practice in setting up the definite integrals involved in various summations. Ex. 1. Find the total pressure on a circular disk of radius a held in a vertical position below the surface of a liquid. According to the laws of hydrostatics, the intensity of liquid pressure is proportional to the distance below the liquid surface. Tak- ing the polar element of area pAdAp, Fig. 96, the pressure on this element is ky p Ad A/>, where & is a constant ; hence the total pressure is given by the double integral k\ j y p dd dp with appropriate limits of in- tegration. Evidently the limits for p are and a, and the area of the semicircle on one side of the vertical diameter will be included if =. and i 2 2 are Fig. 96. taken as the limits for 6. It should be noted that while the total pressures on the semicircles on the two sides of the vertical diameter are equal, the pressures on the upper and lower semicircles are not equal. Now taking y = h — p sin 0, we have for the total pressure P = 2 k f w (h — p sin 6)p dp dd - ircPkh. Art. 140] ILLUSTRATIVE EXAMPLES 287 The mean intensity of pressure is therefore ircfikh ■*■ ira 2 = kh, which is the intensity at the center of the disk. Ex. 2. Find the volume of liquid that will flow in one second through a rectangular orifice of width b, Fig. 97. It is known from hydromechanics that the velocity v with which a liquid flows through a small orifice at a distance h below the l iquid surface is given by the equation v = V2 gh ; therefore the volume flowing through an orifice having the area A A is AQ = V2~gh A A. When the orifice is large the height h is different at different parts of the orifice and to determine the FlG " 97# total volume flowing we must sum the volumes flowing through the elements into which the area of the orifice is divided. Thus Q = L^P V'2gh AA. For the rectangular orifice in question we naturally take the rectangular ele- ment Ax Ah ; therefore •6 Q Jh. Jo 2 gh clx dh f bV^/thot fti*). Because of contraction of the jet and friction, the discharge determined experimentally is smaller than that calculated. EXERCISES 1. Find the total liquid pressure on a submerged vertical plate of width 6. (See Fig. 97.) Find also the mean intensity of the pressure on the plate. 2. Find the liquid pressure on a vertical triangular plate having a vertex at the liquid surface and the opposite side horizontal. 3. In the preceding examples of liquid pressure, show that the total pressure is equal to the area of the plate multiplied by the intensity of pres- sure at the centroid of the plate. Prove that this statement holds for any submerged plane surface. 4. Write the definite integral with proper limits of integration that gives the total liquid pressure on a submerged vertical disk with an elliptical out- line, the major axis being horizontal. 5. Find an expression for the volume of liquid flowing through a trian- gular notch, that is, a triangular orifice with its base in the liquid surface. Take b and h as the base and altitude respectively. 6. Find the volume of liquid flowing through a semicircular orifice, radius a, with the diameter in the liquid surface. 7. Set up the definite integral that gives the volume flowing through a circular orifice, radius a, the center being at a distance h below the liquid surface. Take (a) rectangular coordinates, (6) polar coordinates. 288 MULTIPLE INTEGRALS. APPLICATIONS [Chap. XIV. 8. Find the definite integral that gives the second moment of the volume n/2 y'2 yi of the ellipsoid — + — H — =1 with respect to an axis parallel to the Z-axis a 2 b 2 c 2 and passing through the point (a, 0, 0) . Suggestion : Take sections parallel to the FZ-plane and use theorem III, Art. 139. 9. A cylindrical vessel is filled with water to a height h. A small orifice of area a is opened in the lower base, the water flows out, and since the water level is falling continuously, the velocity of flow varies continuously. Denoting by A the area of the cylindrical cross section, find an expression for the time required to empty the cylinder. 10. Find an expression for the time required to empty a hemispherical bowl, radius r, through an orifice of area a in the bottom. MISCELLANEOUS EXERCISES Find the following areas : 1. Between the parabola y 2 = 8 x and the circle x 2 + y 2 = 9. 1 2 2. The whole area of the curve l-\ + ( 2 ) = 1. ©' + (8 3. The area of the loop of the curve x 3 + y 3 = 3 axy . 4. One loop of the curve p = a sin 2 6. q a 5. The part of the curve p = a sin 3 ■ — below the X-axis. a 6. The parabola p = a sec 2 - between the vertex and the latus rectum. 7. Fir\d the volume bounded by the cylinder {x — a) 2 + (y — b) 2 = r 2 , the surface xy = cz, and the plane z = 0. 8. Find the volume included between the surface a 2 x 2 + b' 2 z 2 = 2 (ax -f- bz) y and the planes y = ± m. 9. Find the volume generated by revolving the curve p = a(l 4- costf) about the initial line. 10. Find the area of that part of the surface of a sphere x 2 + y 2 + z 2 = 2 az lying within the paraboloid z = mx 2 + ny 2 . 11. By extending the method of Art. 108, show that the length of a curve in space is given by the formula -£++{£>'+{£)'* 12. From the formula of Ex. 11, find the length of the helix z = a arc cos - , z = a arc sin # , b b from z = to z — m. Art. 140] MISCELLANEOUS EXERCISES 289 13. Find the area of the curved surface of a right cone whose base is the astroid x* + y 1 = a J , and whose altitude is c. Suggestion : Taking the origin at the vertex of the cone, the equation of the surface is x J + y* = I - z j • 14. The axis of a right circular cylinder passes through the center of a sphere. The radius of the sphere is a, that of the cylinder is 6, (6 a. 19. A helical screw thread whose cross section is an equilateral triangle is cut on a cylinder. The radius of the cylinder to the root of the thread is r, and the height of the thread is a. Find by the theorem of Pappus the volume of one turn of the thread. 20. Apply the theorem of Pappus to find the surface and volume of a right circular cone. 21. Using the theorem of Pappus, show that an element of the volume of a surface of revolution is 2 wp 2 sin

2 sin0ApAA0. 23. Find the radius of gyration of the hollow rectangle, Fig. 98, about the axis X c , which passes through the center of the figure. 24. In Fig. 99, the thickness of the plates (a, a) and channels (6, b) is f inch ■ 'mam. u t j fjp \ ■ m ■f J«- Fig. Fig 290 MULTIPLE INTEGRALS. APPLICATIONS [Chap. XIV. throughout. Find the length e in order that the radius of gyration with re- spect to the axis Y Y shall be the same as that with respect to the axis XX. 25. A pipe elbow, Fig. 100, is to be provided with a foot a to sup- port it in the position shown. The vertical line MM through the foot should therefore pass through the centroid of the elbow. Find the distance c. 26. The following graphical method is used for finding the radius of gyration of the cross section of a hollow cylindrical column.* Lay off to a scale of 1 inch equal to 4 (or 40) the inner and outer radii as the legs of a right triangle; then the hypotenuse measured to a scale of 1 inch equal to 1 (or 10) is the radius of gyration sought. Give a proof of this construction. 27. By particular examples verify the following rule, due to Routh : For homogeneous masses with axes of symmetry, the square of the radius of gyration is }, J, or I of the sum of the squares of the perpendicular semi- axes, according as the mass is rectangular, elliptic, or ellipsoidal. 28. From the second moment of a sphere about a diameter deduce by differentiation the second moment of a spherical shell about a diameter. 29. Find the centroid of the volume OABCD, Fig. 87. 30. (a) For a surface of revolution show that the radius of gyration about the axis of revolution is given by the equation Fig. 100. * 2 j y* ds \ yds where the X-axis is taken as the axis of revolution, (b) Show that for the corresponding solid of revolution k- !■' A dx y *dx * B. F. LaRue in Engineering News, Feb. 2, 1893. CHAPTER XV INFINITE SERIES 141. Fundamental definitions. From the study of algebra, the student is already familiar with the elementary properties of in- finite series. In the present chapter, we shall recall briefly some of the more important of those properties and develop still others. Let u 1} t%, u Sf ••• be an infinite succession of values following one another according to some fixed law. We denote the sum of the first n of these values by S n , that is, #» = «i.+ *» + ••• +u n - (1) When n becomes infinite, we have the infinite series w 1 4-t« 2 +«3+ •••• (2) If S n has a limit as n becomes infinite, that limit is called the sum of the infinite series. The series (2) is said to be convergent if we have L S n = A, (3) n = oo where A is a definite number ; in all other cases the series is said to be divergent. As n increases indefinitely, S„ may fail to have a limit, and therefore the series may be divergent, either because S n oscillates between two numbers, as in the series 1-1 + 1-1 + 1-1+ •», or because S n ultimately exceeds every finite number. The terms of the series may be functions of some variable x. Then the series is said to converge for any particular value of this variable, say x = x , when, if x is replaced by x in each term, we have L S n (x )=A. n = oo 291 292 INFINITE SERIES [Chap. XV. If the corresponding limits exist for all values of xina certain interval (a, /8), the series is said to converge throughout the inter- val and to define a function in the interval. We may then write, f(x) = L SJx), a£xl the series becomes divergent and does not define a function. Ex. 2. Consider the series 1+ | + | + ... + 2 + .„. This series may be written in the form i+i+fi+iV+fi+i+MU 2 \3 4/ \6 6 7 8/ W w + 1 2(w- 1)/ The sum of the terms in each of these parentheses is greater than J, and as the number m of such groups that can be formed from the given series is indefi- nitely large, we have T a . T , , N ' 6 ' L 8 n > L (m • ^) = go . n = oo m — cc This result leads to an important observation ; namely, that a series is not necessarily convergent because the terms themselves decrease and approach zero as n increases. If a series containing negative terms is still convergent when all of the negative terms are taken positively, that is, when only the absolute values of the terms are considered, the series is said to converge absolutely or unconditionally. If the series is con- vergent when the negative terms are taken with their proper signs, Art. 142] TESTS OF CONVERGENCE 293 but is not convergent when those signs are taken positively, the series is said to converge conditionally. The series 2^3 4^5 is such a series. On the other hand, the series 2 2 2 2 3 is an absolutely convergent series, since it converges when all terms are given the positive sign. 142. Tests of convergence. To test the convergence of a series, certain criteria are necessary. It is to be remembered that whether the terms of the series are constants or functions of a variable, we are concerned only with the limit of the sum of a finite number of constant terms as that number increases indefi- nitely. For, in case the terms are functions of a variable, we either substitute a constant value for the variable and then pass to the limit, or, what amounts to the same thing, we consider for what values of the variable the series has a limit. It is impor- tant, however, to distinguish between convergence at a point, and convergence throughout an interval. The following tests, already considered in algebra, and conse- quently needing no proof here, will be found sufficient for series which arise in ordinary practice.* Theorem I. Comparison test. Given a series of positive terms ^1 + ^2 + ^3+ ••' +**«+ •••• If from some point on the terms of this series are never greater than the corresponding terms of a known convergent series Vi + v 2 -f i's + ••• +v n + ••• of positive terms, then the given series is convergent. If the terms of the given series are from some point on never less than the corre- sponding terms of a known divergent series of positive terms then the given series is divergent. * Compare Rietz and Crathorne's College Algebra, Chapter XVI. 294 INFINITE SERIES [Chap. XV. . Ex. 1. Test the converge ncy of the series From Ex. 1 of the previous article we have the convergent series ( | x | < 1) l+z 2 + z 3 + z 4 + -. + x n + -.. (2) For x = |, this series becomes Since for n > 2, 2"- 1 < n !, and each term of (1) after the second is less than the corresponding term of (3); therefore the series (1) must converge. Ex. 2. Given the series 1+ i + S + f+-. + sfei + "-- (4) Compare the series with the series whose first n terms are 1+1+1+ ... +J-=i(i +!+'!+ ...+iy ( 5 ) 2 4 6 2» 2V 2 3 n) KJ By Ex. 2 of the previous article, the series (5) is divergent. Moreover, since 2 n — 1 < 2 w, each term of the given series (4) is greater than the corre- sponding term of the series defined by (5) ; consequently, the series (4) is divergent. Theorem II. Ratio test. A given series u x -\-u 2 + ... -\-u n - 1 + u n + ••• u is convergent or divergent, according as the limit L — —is less or greater than 1. n = °° Wn ~ x Ex. 3. Given the series 1 + x + ■ — + — + — + a;M ~ 1 + — +— . (6) 2!^3!^ T (*-l)l n! W We have then SB* X -=2-=: x ^r- U Ji- — r Wl T * (11-1)1 for any finite value of x. Hence the series converges for all finite values of x. It is to be noted that this second criterion applies equally well when some of the terms of the series are negative. It gives no test, however, when the limit of the ratio — *- is 1 ; in such cases other methods must be employed. n_1 Abt. 143] POWER SERIES 295 Since each term of a series is finite, the sum of any finite num- ber of terms is also finite, and consequently the convergency or divergency of a series is not affected by omitting a finite number of terms. If the terms of the given series are alternately positive and negative, the following is a convenient test of convergence. Theorem III. An infinite series in which the terms are alternately positive and negative is convergent if its terms decrease numerically and approach zero as a limit ivhen n increases indefinitely.* 143. Power series. By a power series we understand a series of the form ^ + v+ ^ + i ^ + ... +ajf>+ •", (1) where a , a l} •••, a n , etc., are constants. Such series are of importance in mathematics because of the frequency with which they occur and because of the special properties which they possess. For example, it is not always possible to obtain the integral or the derivative of a function by integrating or differentiating term by term the series which de- fines the function. A power series, however, has this important property when the variable is restricted to a proper interval of convergence. Consequently, if f(x) is defined by the power series f ^ = ^ + ^ + ^ + ... + ^ + . . m$ we may then write | f(x)dx= J * l a dx + C Xl ajxdx+ ••• + j i a n x*dx + ••• and df(x) _ d(a ) d(a l x) d(a n x) dx dx dx dx when we place a suitable restriction upon the value of x.f A valuable property involving the convergence of a power series is given by the following theorem. * For proof of this theorem, see Rietz and Crathorne's College Algebra, p. 188. t For a fuller discussion of the conditions for term-by-term differentiation and integration of a series, see First Course, Chap. XV. 296 INFINITE SERIES [Chap. XV. Theorem T. If a power series converges for x = a, it is abso- lutely convergent for all values of x such that | x |< | a |. Let the given series be written in the form «o + aJ-) + a 2 a 2 (*Y + ■ • • + o^WsY + • • • ; (2) then the series of coefficients a , «!«, a 2 a 2 , •••, a n a n , ••• must decrease indefinitely in absolute value since the given series converges for x = a. Let M be equal to or greater than the abso- lute value of any number in this sequence. Then the absolute values of the terms of (2) are less than the corresponding terms of the geometric series \ a er or a n which converges for |aj|<|«|. Hence the given series (1) con- verges absolutely for | x \ < | a | . 144. Maclaurin's expansion of a function in a power series. It is often convenient to express a function in terms of a series. A power series is very serviceable for this purpose, because, as we have seen, such series may be differentiated and integrated term by term, thus obtaining the same result as if the opera- tion had been performed upon the function itself. In the following article we shall discuss two methods of expanding functions in terms of a power series by making use of the prin- ciples of calculus. Suppose we have given a function f(x) which, together with its derivatives, is continuous in the vicinity of the value x — 0. If such a function can be represented by a power series, that series must be of the form f(x) = A» + A l x + A 2 a?+ -. +A n x n + ..., (1) where the coefficients A , A x , A 2 , •••, A n , ••• are to be determined. Since this is a power series, we may find the successive deriva- tives of the function f(x) by term-by-term differentiation of the series. The following identities are thus derived : Art. 144] EXPANSION OF A FUNCTION 297 f\x) = A x + 2 A 2 x + 3 Atf + 4 Atf + ... f"(x) =2 A 2 + 3 • 2 A 3 x + ± • 3 A 4 x> + ... f"'(x) = 3.2A s + . + rm x n + ..., (3) where /"(0) denotes the result obtained by differentiating the function fix) in succession n times and substituting x = in the result. The above series is called Maclaurin's series. For any given function, there still remains to be determined the interval within which the expansion obtained by (3) really repre- sents that function. This question will be discussed in a subsequent article. Assuming that such an expansion is possible, the fol- lowing examples will illustrate the method of computing the successive coefficients in the expansion. When the successive derivatives all become zero from some point on, the expansion has a finite number of terms. Ex. 1. Expand f(x) = (1 + x) m . We have f(x) = (1 + x) m , /'(«)= m(l +*)«-*, f"(x) = m{m - 1) (1 + x)*- 2 , f»(x)=m(m- 1) ...(»— »-M)(l +x) m ~ n , whence /(0) = 1, /'(0) = m, /"(0) = m(m - 1), etc. 298 INFINITE SERIES [Chap. XV. Substituting these values in (1), we have, when m is negative or fractional, the infinite series (1 + *)« = 1 + mx + m(m ~V x* + m ( m ~ *> <> ~ 2 > x* + •• • 2 ! 3 ! t W(W-1)-.(W-71+1) XM _ w ! ' which converges for | x j < 1. If m is a positive integer, the series terminates with first m -f 1 terms, since all of the higher derivatives vanish. Ex. 2. Expand sin as in a power series. /(as) = sin x, /(0) = sin = 0, f'(x)=cosx, /'(0)=1, /"(«) = - sin as, /"(0) = 0, f't'(x)= -C08S, /'"(0) = -l, / IV (z) = sin sb, f™(0) = 0, etc. etc. Hence, sin* = + x + — — + + — + — — + — 3 ! 5! 7 ! -7*3 'Y'h npl = x— — + — — — H — . 3!^5! 7!^ The interval of convergence for this series is (— oo, + qo). EXERCISES Expand in power series the following functions, assuming that such ex- pansions are possible. 1. e x . 2. cos a;. 3. a x . 4. e sinx . 5. log (1 + e x ). 6. log(l+x). 7. arc sin a. 8. arc tan x. 9. (cos 0) n . 10. e 9 sec 6. 11. sec 6 (to four terms). 12. «*»•»*•. 13. log (a; + Vl +ic 2 ). 14. log cos 6. 15. l(e x + e~ x ). 16. log(l — x + x 2 ). 145. Taylor's expansion. In the last article, we studied the expansion of a given function f(x) in the vicinity of the value x = 0. The method may be easily extended to the expansion in the neighborhood of any point x = a, provided the given function and its successive derivatives are continuous. All that is neces- sary is to assume the expansion of the form f(x)=A {) + A l (x-a) + A 2 (x-cif + ... +A n (x-a) n + ••• (1) Art. 145] TAYLOR'S EXPANSION 299 and proceed precisely as in Art. 144. The resulting form of the expansion is +/jj^»-a)» + .« 1 (2) which holds for all values of the variable within the interval of equivalence. The series resulting from this expansion is known as Taylor's series. If in (2) we replace x by (x-\- a), we have f( x + a)=f(a)+f\a)x + £M x ? + ... + £M x « + ... ! (3) Z\ n which is a form in which Taylor's series is frequently written. If x and a are interchanged, the expansion takes the form f(a + x)=f(x)+fXx)a+£^a*+ ... + ££)«• + .... (4) Forms (3) and (4) are useful when it is desired to expand a function of the sum of two numbers in powers of one of them. Ex. 1. Expand e x in powers of x — 1. We have f(x) = e*, /(l) = e, /'(«)=«*, /'(1)= 0, /"(*)=*>, /"(!) = «, etc. etc. Hence, e* = e[~l +(«- 1) + ^=^ + C^zlII 3 + ...]. Ex. 2. Express 3 ar 3 — 5 x 2 + 8 x — 5 in powers of x — 2. In this case /(*) = 3 X s - 5 a- 2 + 8 x - 5, /(2) = 15, /'(as) = 9z 2 -10 x +8, /'(2) = 24, /"(*) = 18* -10, /"(2) = 2G, /'"(«) = 18, /'"(2)=18, • / IV (*)=0, / IV (2)=0. Hence, we have /(*) = /(2)+/'(2)(z - 2) + ^^(x - 2)2 + £^p-(x - 2)8, or 3 x^ - 5 * 2 + 8 x - 5 = 15 + 24(x - 2) + 13(z - 2) 2 + 3 (a; - 2) 3 . 300 INFINITE SERIES [Chap. XV. Ex. 3. Develop log (a; 4- h) in powers of h. We have /(x -f h) = log (x + A), /(*) = logx, /w-i. /"(*)=-£, . Substituting in (3), we obtain log (x + A) = log x + * - ^ + -^ - -^ + ».. x 2x 2 3x 3 4x 4 For x = 1, we have log(l + A) = A-f + f-f + .... Ex. 4. Expand sin (x + y) and derive the formula sin (x + y) = sin x cos ?/ + cos x sin y. We have /(x) = sin x, /' (as) = cos x, /"(*) =-sinx, f'"(x) = — cos x, etc. Substituting in (4), the result is . , . . V 2 . V 8 V 4 . V 5 sin (x -f y) = sin x + y cos x — |-y sin x — ^-j cos x + j7 sin x + ~ cos x — •• • \ 2! ^4! 6!^ ) + co§ «('-fr + gi-n + "") = sin x cos y + cos x sin y. Ex. 5. If /(x) - 5 x 8 - 4 x 2 + 18 x - 7, find/(x - 2) by Taylor's expan- sion. We have here /'(x) = 15 x 2 - 8 x + 18, /"(x) = 30x-8, /'"(*) =80, Hence from (1), /•*(*) = 0. f(x - 2) = /(*) + (- 2) /'(a) + ^|I 2 /"(x) + ^- 8 /'»(«) = 6x 8 -'4x 2 -fl8x-8 - 30 x 2 + 16 x - 36 + 60 x - 16 -40 = 5 x 3 - 34 x 2 + 94 x - 100. Arts. 145, 146] TAYLOR'S THEOREM 301 EXERCISES Develop the following functions. 1. e x + h . 2. (x + y) m . 3. (x + y)«. 4. Arc sin (x + h) to four terms. 5. Log sin (x + h). 6. Find /(a + S), when f(x) = x 3 - 4 x + 7. 7. Find /(x - 1), when /(x) = x 2 + 7 x - 5. 8. Show that log (x + VI + x' 2 ) = x - — + -^ .... oV ' 2-3 2-4.5 9. Expand sin x in powers of x — a. 10. Express 5 x 3 — 6 x 2 + x — 10 in powers of x — 1 ; also in powers of x — 3. Verify the results. 11. Expand logx in powers of x — 1. Find the interval of convergence. 12. Expand - in powers of x — a and determine the interval of convergence. 146. Taylor's theorem. Maclaurin's theorem. In discussing the expansion of a function in terms of a power series, we have as- sumed that the series obtained actually represents the function. We shall now see under what conditions this is true. We may write fix) = S n (x) + R n (x), (1) where /(x) is the given function. If the infinite series given by L S n (x) represents f(x) in a given interval, then for all values n =00 of x in that interval, we must have L B n (x)=0; (2) n = oo for, R n (®) represents the difference between the given function and the sum of the first n terms of the series, that is, the error involved by stopping the expansion with n terms. It is conven- ient to have the value of R n (x) expressed in terms of the deriva- tives of f(x). This value can be obtained for Taylor's expansion as follows. From equation (1), we have f(x)=f(a)+f(a)(x-a)+^(x-ay+ ... 802 INFINITE SERIES [Chap. XV. Since B n (x, a) contains the factor * — *-, we may write it in ft! the form \ x ~ a ) ^^ a \ Replacing R n (x, a) by this expression n ! and transposing, we have f(x)-f(a)-f(a)(x-a)-^(x-af gT^l (*_„)»-> - jj&-2l (x~ay = 0. (4) To find the value of (x, a) in terms of the derivatives of f(x), we shall consider the function F(z)=f(x)-f(z)-f'(z)(x-z)—£^(x-zy-- - f— n (*) , N „_i (x,d), .. ,~. -fc^ijp " J ^Ti {x ~ z) () F(z) satisfies the conditions of Rolle's theorem in that it pos- sesses a derivative for each value of z, where a ^ z (x,a)=r(*i), (8) and BL »-£&&(•'•- o)». ( 9 ) Art. 147] INTEGRATION OF SERIES 303 Replacing R a (x) Dv ^s value as given in (9), we may now write (3) in the following form /(^) = /(a)+/'(a)(a?-tt) + ^^ + ^T (*-«)"• (10) This formula is known as Taylor's theorem. We are now in a position to determine the interval within which the expansion represents the given function by determin- ing the range of values of x for which (2) holds. In the simple cases which will come under consideration this interval will usually coincide with the interval of convergence of the series. We have seen that Maclaurin's expansion is a special case of Taylor's. By putting a = 0, we have as the value of B n (x) in Maclaurin's series M n (x) =£{&&, 0. This integral can be evaluated approximately by expansion in series. Thus we have Vl -e 2 Sin 2 <6= l_l e 2 sm d = (* d/i _ ^2 8. Evaluate VI -x 2 1 *> rvA log(l-M)=*-| + |-|+.... Substituting — x for x, we have 3/ fl» £C By subtraction we get log(l + x)-log(l-x) = logi±| = 2^ + | + J+...\ Let us take x positive and assume x = : then 2y + l l + a? = y + l 1 — x y and we have ^* +i >-^. +s [5^ + i(^ i+ K^y + ^ (7) From this series log (jf + 1) can be calculated when log y is known. Thus, since log 1 = 0, we have i o Jl.ll ,11,11 , \ = 0.693147..-. io g 3=,og2 +2 (H! + !! + ...) = 1.098612-... log 4 = 2 log 2 = 1.386294 • • -. log 6 = log4 + 2 (|+|i + |i + ...) = 1.609438.-. log 6 = log 2 + log 3 = 1.791759 .... 308 INFINITE SERIES [Chap. XV. Evidently it is only necessary to make the computation in the case of prime numbers. Logarithms to the base 10 are obtained from the natural logarithms by means of the following relation : log 10 a = log e a x — !— - = 0.4342945-. • log e a. 10g e 10 V. Computation of trigonometric functions. For all values of x, we have the series x x 3 , x 5 x 7 , sin x = h 1! 3! 5! 7! r*d /«4 /y,6 -I ms . Ms *b . cos*=l-- + --_ + (8) These series converge rapidly, and may be used to compute the natural sine and cosine of any angle. Necessarily x must be ex- pressed in radians. Ex. Find the sine and cosine of 19° 30' correct to five figures. We have x = — — it = .34034. Substituting this value for x in the two 180 -° series, we get sin x = 0.33381, cos £ = 0.94264. 149. Approximation formulas. It is frequently convenient in computation to replace a function by another of approximately the same numerical value but having a more simple form or having a form better adapted to calculation. This substitution may be effected in many cases by expanding the given function and taking a certain number of the first terms of the series. One of the most useful of these approximation formulas is obtained from the binomial formula. Thus let m denote a small fraction, and expand (1 ± m) n . The result is (l±m) w = l±wm + ^P^m 2 ± .... Z ! Since m is small, powers higher than the first may be neglected, and we may write the approximate relation (l±m) n = l±nm. (1) Akt. 149] APPROXIMATION FORMULAS 309 An important special case is that in which n = \ ; for this case we have approximately Vl ± m = 1 ± i ra. (2) From (2) we have the more general formula V¥±l= a (t±£), (3) as may be easily shown. In this formula b is small in compari- son with a. Since e w = l + m + -+- + •••, we have when ra is small the £ . o ! approximate relation e m = 1 + m. (4) Similarly, taking two terms of the series for sin x, cos x, and log(l + x), we obtain the approximate formulas sin ra = ra (1 — | ra 2 ) ; (5) cos ra = 1 — \ ra 2 ; (6) log (1 -f- m) = m -\ ra 2 . (7) When ra is small compared with a the following approximate relations are readily obtained. sin (a ± ra) = sin a ± ra cos a ; (8) log(a + m) = log<.+5»- "t; (9) . _J_ = 1 - T » 2 + ^- (10) a ± m a or ar The degree of error due to the neglected terms may be esti- mated by taking the maximum value of the remainder R. For example consider the approximation (5). If only two terms of the series are used, we have sin ra = ra - %- + ^-f r (m 1 ), where < m x < ra. Since / F (rax) = sin m 1 cannot exceed 1, the difference between sin ra. and the assumed approximation ra — — - 5 5 6 cannot exceed -^- ; hence i? <£ ^-r • If we wish to restrict the 310 INFINITE SERIES [Chap. XV. error to some definite limit r, we have only to write \B\ ) 2 - Q.00083. 2 Ex. 3. In certain problems in surveying the relation between a circular arc and its chord is required. Let s denote the length of the arc, r the radius, and C the chord, Fig. 101. We have s = ra, and C=2rsin — • If a is Arts. 149, 150] MAXIMA AND MINIMA 311 small the approximation (5) may be used. An approximation to G is therefore °= 2 'iH{f)'] = ra — ^ ra s . Hence, s- C = j»j ra 3 , where a is expressed in radians. If a is taken in de- grees, the formula becomes s- G ra s 4514180 The error of the approximations cannot exceed ra 5 2r 120 \2 ) 1920 Fig. 101. EXERCISES 1. Calculate sin 15° and cos 15° to five decimal places. 2. From the series for tana; calculate tan 12°. 3. Find ^1334. 4. From the logarithms given in Art. 148 calculate log 31, also log 73. 5. Using the approximate formula (10) calculate the reciprocal of 102 ; of 97. 6. Find the greatest value of m that will permit the approximation (1 + ra) 4 = 1 + 4 ra with a maximum error of 1 in 1000. 7. Within what limits will three terms of the series for cos x give an error not exceeding 2 units in the 6th decimal place ? 8. Investigate the limits of accuracy of the formula sin (a + ra) = sin a + m cos a. 9. Find the length of the chord of an arc of radius 200 feet subtending an angle of 3° : (a) by trigonometric methods ; (6) by the approximation formula. Compare the results and find the relative error of the approximation. 10. Derive an approximation formula for tan x and show the maximum error involved. 11. Given log 5 = 1.6094, find log 5.01 and log 5.02. ISO. Maxima and minima of functions of a single variable. Taylor's expansion of a function gives a convenient method of developing the condition for maxima and minima. This method is particularly valuable where several of the derived functions 312 INFINITE SERIES [Chap. XV. vanish for the critical value. The condition for a maximum or minimum of a function in such cases may be stated as follows : Theorem. The function f (x) has a maximum (or minimum) for x=a, if the first one of the derived functions f'(x)) f"(x) ••• that does not vanish for x = a, is of even order and negative (or positive). We have from Taylor's theorem, f( X+ h)=f( X ) +f \ X )h+l^h^£^w+ ... + f n ( x + eh ) h % and fix - h) =f(x) -f (x) h + tM n* _ f!!M &»+...+ £fe± eh ) h\ Replacing x by a, we have, after transposing the first term of the second member of the identity, /(a + ft) - /(«) =/' (a) ft + -f^fl h° + £^2) ft» + .. . + /"(« + flft) /t ,, n ! f"(a) h 2 f'"(a) h * 2! 3! I ( ±y f n ( a ~^~ l ^ h n . (2) f(a- ft) - f(a) = - /' (a) ft + LJ£1 ft« _ -£_p ft" + If for x=a the given function has a maximum, then /(a) must exceed the value of the function for all values of the variable in the neighborhood of a ; in other words, the left-hand member of both (1) and (2) must be negative for all values of h sufficiently small. The value of h can be taken so small that hf'(a) will be numerically greater than the sum of the remaining terms of the second member. However f(a -f h) — f(a) and f(a — h) — f(a) cannot both.be negative unless f(a)—0 and f"(a) is negative, assuming that /"(a) does not vanish. It may, however, happen that both/'(«) and f" (a) vanish. In this case, in order to have f(a + h) — f(a) and f (a — h) — f (a) negative, f'"(a) must vanish and/ IV (a) must be negative. In general, in order that/(#) shall have a maximum value for x = a, the first derivative that does not vanish must be of even order and negative. In order that f(x) shall have a minimum for x = a, the two expressions f(a -f h) — f(a) and f(a — h) — f(a) must be positive. Art. 151] INDETERMINATE FORMS 313 This requires that the first derivative that does not vanish shall be even and positive. The argument is similar to that given above and is left to the student. Ex. Examine the function 2 cos x + e x + e~ x for maxima and minima. We have f(x) = 2 cos x + e x + er- ; f'(x) = — 2 sin x + e x — e~ x . For x = 0, f'{x) = ; hence x = is a critical value. We have further f"(x) = - 2 cos x + e x + «-*, whence /"(0) = 0, /'"(«) = 2sina; + e* - e- z , whence /'"(0) = 0, /iv(x) = 2 cosic + e K + e~ x , whence / IV (0) = 4. Since the fourth derivative is positive, and is the first that does not vanish, it follows that f(x) is a minimum for x — 0. EXERCISES Examine for maxima and minima the following functions. 1. tan 2 x — 2 tan x. 2. e x — e~ x — 2 sin x. 3. sinx(l + cosse). 4. pe ax + qe-™. 5. icsinx. 6. 3 cos + tan 2 0. 7. cos z — log cos x. 8. x _m e n,a; . 9. e x — e~ x — a; 2 . 10. cos x(2 — cos x). 151. Evaluation of indeterminate forms. Algebraic methods of evaluating certain indeterminate forms were shown in the examples of Art. 15. For the form -, to which all other forms may be reduced, the differential calculus furnishes a general method of evaluation, which is developed as follows : fix) Let the given function be of the form J \ ' , which reduces to the (x) f (x) form - for x = a. The value of the limit L *-*-£ is required. x ± a (x) Expanding each term of this fraction by Taylor's theorem, we have ,, , f(a)+f\a){x-a)+- f -^-(x-af+ - +€^L(x- a y /W _ «■ JJj (1) 314 INFINITE SERIES [Chap. XV. By hypothesis, /(a) and <£(a) are each equal to zero. The above relation therefore reduces to f(x) f'^ x - a ) + - t ^ li - x - a ^ + - +^(*-«)" ^%'(a)(z-a> + «(*-a)s+---+M (*-<»)• ' (2) Dividing both terms of this fraction by (x — a) and passing to the limit, we have ^ f(») = fjg) x±a4>fr) *'(*)' If /'(a) = and cf>\a)=^0, this limit reduces to zero; if /(a) =£0 and '(a) = 0, it becomes infinite. If /'(a) = and '(a) — 0, the limiting value of the given func- tion can be obtained by dividing the terms of the expanded form of the fraction (x — a) 2 and then passing to the limit. The result is L ^M = /M. a) x = a(x) "(a) V ' Similarly, if /'(a) and <£"(«) are both zero, we divide by (x — a) 3 and again take the limit, and so on. We may therefore state the general law of procedure as follows : To evaluate the indeterminate form -, differentiate the numera- tor and the denominator of the given fraction and substitute the criti- cal value of the variable in the result. The function «^4 ma y ^ so assume the indeterminate form - (x) when x becomes infinite. The limiting value may still be found by considering J } ' \ for, we have upon putting x = -, L ./M = L —gjj ** L -M T t!iA (5) Form 3£ . When the function £iw2. takes the form ^- , it can 00 (x) °° Art. 151] EVALUATION OF INDETERMINATE FORMS 315 1 be reduced to the form -, by writing it in the form Z-j-J. This m form can then be evaluated according to the law just stated. It may be shown as in the case of the form - that if U& has J cf>'(x) a limit as x approaches a definite number or becomes infinite, then fy*f converges to the same limit.* This principle often affords a (x) convenient method of evaluating this indeterminate form ; for we need only to differentiate the numerator and the denominator and then pass to the required limit. Form od — ao . When a function takes the indeterminate form oo — oo it may be reduced to the fundamental form - by writing it as follows : , w -^44, t^i . ( 6) J(X) 0(0) f{x) • *(0) Often, however, a simpler transformation will reduce the function to one of the forms -, — • No general rule can be given, but that transformation should be selected which gives the simplest form. Form x ao. When a function fix) • cf>(x) takes the form X oo for x — a, it may be reduced to the type - by writing it in the form m . m „f!gL (7) (x) Forms 0«, oo°, 1*. The indeterminate forms 0°, oc°, 1" arise from a function of the form [/(a;)]* (x) . This function may be reduced to the type form - as follows : * See Pierpont's Theory of Functions, Vol. I., p. 305. The special student of mathematics would do well to read Arts. 455-459 in the same volume. 316 INFINITE SERIES [Chap. XV. Let y = [mT x \ whence logy = (x) • log [/(#)]• (8) Since, for each of the given forms, (8) takes the form Oxoo, the solution is effected by (7). n •■ *<■ °JL = 12 x 2 - 24 x 4- 10 + y, ^L = x + ll-6y. dx dy The values x = 1, y = 2 satisfy the equations f(x,y) = 0, |£=0, & = 0; ox oty hence the point (1, 2) is a singular point. The character of the singularity dy dx can be determined by evaluating the indeterminate form -^- For con- venience putting -^ = p, we have dx _ _dx_ 12 x 2 - 24 x 4- 10 + y P_ fl£~ x + ll-6y The right-hand member of this equation takes the form - for x = 1, y = 2 ; hence to determine its value at this point, we differentiate both the numera- tor and the denominator with respect to x and have 24 x - 24 4- l-6p £ l Art. 152] SINGULAR POINTS 319 or p = ~ p , * 1-6/ whence 2p — 6p 2 = 0, p = 0, or i. Therefore at the point in question, there are two distinct tangents to the curve and consequently two branches of the curve pass through that point and the singularity is a double point. The slopes of the two tangents are and ^, respectively. Ex. 2. Examine for singular points the curve x 4 + x s y — 4 x 2 y -f y 8 = 0. We have for the partial derivatives & = 4 a* + 3 x 2 y - 8 xy, ^ = x 3 - 4 x 2 + 3 */ 2 . ox dy The values x = 0, x = satisfy the equations /(x, y) = 0, %£ = 0, zL = ; dx dy hence the origin is a singular point. Putting SlU _ p^ we have for the point (0, 0) dx 4 X s + 3 x 2 y - 8 xy l = it-s _ 4 X 2 + 3 y 2 J M = ? whence differentiating numerator and denominator with respect to x, we obtain _ 12 x 2 + 6 xy + 3 x 2 p - 8 y - 8 ay "I 3x 2 -8x + 6*/.p J o,o For x = 0, y = the second member again takes the form 2 ; hence dif- ferentiating numerator and denominator a second time, we have _ 24 x + 6 y 4- 6 xp + 6 xp — 8 p — 8 p ~| = 8j? Jo, o 6x-8 + 6p 2 J 00 3p 2 -4 Therefore jt? (3 p 2 — 4) = 8p, andp = 0, 2, —2. The origin is a triple point, and the three tangents have respectively the slopes 0, 2, and — 2. EXERCISES By the general method of this article examine the following curves for singular points. Additional exercises are furnished by the examples of Art. 89. 1. x 4 - 4 xhf -f y s = 0. 2. x3 - 3 x 2 - 3 y 2 + y 3 = 0. 3. x 5 - 4 x 4 + y s - 4x 2 y = 0. 4. x 3 + 2 x 2 - 4 x«/ + 2 y 2 = 0. 5. x 4 - x 2 y + y 8 - 6 y 2 + 2 x - 12 y - 8 = 0. 6. ay = (x - &)*. 320 INFINITE SERIES [Chap. XV. 7. Trace the curve x 4 + x 8 y — 4 xHj -f- y z = 0, discussed in illustrative Ex.2. Suggestion : Examine for asymptotes, then put y = rax and find values of x for assumed values of the slope m. 8. Trace the curve of Ex. 1. 9. Trace the curve of Ex. 6. MISCELLANEOUS EXERCISES 1. Test for convergence the following series, and determine the interval of convergence : ?a\ i_i_W*\ . 1 l-4/x\ 2 1 1-4-7 /xy, » l+ ! + ii7 + ii5 + -- ; - (c) 1 +xcosa + — cos 2 a H — cos 3 a + •••. 2! 3! (d) cosx + -cos2x + -cos3x + • ••. 2 3 Expand the following functions. x 2. tan0. 3. e*cosx. 4. cos n x. 5. . 6. log sec 2 -. e*-l 2 l 7. Show that (1 + *)« = e (1 - | x + §£ x 2 - T \ x 3 + •.•)• Suggestion: Let u — (1 + x) x , whence log w — _2S_L_±_^2. Make use of the series already found for log (1 + x) to determine the successive de- rivatives. 8. From the expansion for log (n + h) and log (n -J- 1), derive the approxi- mate rule of proportional parts, viz. : log (n +h) — log n _h log (n + 1) — log n 1 ' From this rule find log 7.523, knowing that log 7.52 = 2.0176 and log 7.53 = 2.0189. 9. Develop f(h±K) in powers of h, and determine the values of /(x) =x 2 (16 - x) for the following values of x : 4.7, 4.8, 4.9, 5, 5.1, 5.2, 5.3. 10. Show by development in series that e xS/ ~~^ = cosx + V— lsinx, and e -xV-\ = cos x — V— 1 sin x. 11. Develop the function y = a rcsina? i n series. VI -x 2 Suggestion : Multiply by Vl - x 2 and differentiate. The resulting equa- tion is (1 — x 2 ) =£ — xy = 1. Now assume y = A + Bx + Cx 2 + ••• and deter- dx mine the coefficients. Art. 152] MISCELLANEOUS EXERCISES 321 12. Derive the approximate formula tan (d -f w) = tan 6 + m sec 2 6. 13. The strength of an electric current as shown by a tangent galva- nometer is given by i = C tan , where C is a constant and

ay — j/ 3(x 8 -2x + 5)J 10. 8 Vx 2 — 2 x + 5 2 3. -5x(a 2 -x 2 )i 11. /~ g2 - 2 0*(1 + A 2 )* (5x 2 + 6x-12)J^±4' 'x 2 — 4 12. -0(1 + 4. 5x(a 2 -x 2 ) * 6 2x*(x 2 -2a 2 ) ' 2Vx 2 -a 2 3(x 2 -a 2 )^ u x 3 7. (2x-f a)(3x 2 -4ax)"l (x 2 + 1)^ 8 . m + n + 2x 15. x ( 2 <* ~ gQ . 2 V(x + m)(x + w) (a 2 -x 2 )$ Art. 32. Page 44. 3. -g 4. JJ£. 6. \{ev-fvr 3(*_4)* (l-* 2 ) 2 2V^T& >2s Art. 33. Pages 44, 45. 1. e ,_i. *»£=*-'. s. _L. •.»=». jA, ANSWERS 325 Miscellaneous Exercises. Pages 45, 46. 1. 20^ + 9a;2_8. i* x 2 -6x-5 -8). a*). -3^ 4yu sV 2 . 37. 3. 2(x-3)%Vtf~+~5 2. x(\bx* + \2x 2 -\2x- 3. *af*(4a*-a*)(**- 1d -(2 + aOVl-s 2 (1 + x) 2 15. 2*+ 2 * 3 . 4 2x-5 (a- - 2)3 Vx 4 - 1 1f? 8x4-9a¥ + 3aa* 5. - * . 3(x + a)(a?-a 2 )$ VI + x 2 6. § x*Vx 2 — 1 7 ! 17. 2g~ 2a ^~^.. Vx 2 - i 18. 4x- 2 ( 2 ^ 2 + 1 ). Vx 2 + 1 4 Va+v^ ifl 2a c (1 + x)«+i V 3 (0 - 6)2 q nx + mx + aw c +pma x n+1 (a + x) w +i 22. arc tan (± f). 1n n(x + Vx*-l)n Vx 2 - 1 Q? 10x(2-x 2 ) (x 4 - 10x 2 + 10)^ 11. Vx + Vl + x 2 2Vl+x 2 12. * 24. a • 6 + v 25. _£- - a^vp?. «* ^^ 6 (g + i)V(^ + g) 2 + i ; (#+ ' 1)t . V(0 2 + 0) 2 + i , l u 7 pV4p 4 + (« + 2 60) 2 . 2p« V4 p 4 + (a + 2 60) 2 , aJ-_260 # a + 2 60 ' a + 2 60 ' 2 p 2 p 8. 0; *; 2 + 0; 2 a + K . 2' a + 260 ANSWERS 327 Art. 43. Page 64. 1. (a) v = Vf> — gh \ a = -g. 5. (a) 32 rad./sec.; - 32 rad./sec. 2 . (6) 236.6 ft./sec.; -32.2ft./sec. 2 . (6) 3£ sec. (c) - 86.4 ft./sec. 6. w = a - 3 bt 2 ; a = -Gbt. 4. 13 rad. /sec. ; 2 rad. /sec. 2 . 8. w = 6 + 2 c< ; « = 2c. Art. 44. Page 66. 1. 0.54084; 0.23440; 0.23848. 2. -0.00006704; 0.00026816. Miscellaneous Exercises. Pages 67, 68. 1. ( a ) o ; — £. (6) At x = and as = f a. (c) At x = — f a and x = 2 a. 2. ±450. 3. 15x + 2y = 60. 9. (a) p = kd + c. (b) 6 + - = ft 11. 0.50925 ; 0.51040. Art. 48. Page 77. 1. d|f = 2eZx . 8 dy _ C x ~ 5 ) 2 (2«x-x 2 )* (a2 + x2) Va*-^ _ wrtfc 3. dy= (2x-5a)dx, * V2^Tl 15x*(x-a)* 5. 2x* 3(x 2 +l)* Art. 49. Page 78. 3x 2 -y 2 4 4x«-3x 2 y + 2xy 2 + 2y8 2xy ' ' x»-2x 2 y-6xy 2 tPx 5 ax + hy + g _ 3. -3x?/i 6. |. 328 ANSWERS I" ~* 5 *' 10. (a) -Mj (6) F 8 -«H2«6 »• 1. mv v s (b - v) 9. (a) Subtan. = 4/> 2 v / 0-2 P 0; 11 2y(l + x) + 3x 2 + y 2 Subn.= fi >*!-*>-(*+«)»■ 4,v*-2* i2. - y + 6 . (6) Subtan. = 1 - 2 P ; 2 * y Subn.=^^. 13. *■ ^~ 2 ^ . l-2 P d x Vx-2Vy Art. 50. Pages 80, 81. 1. Vi 1^34. 7 . _JU_. 2. 12; -9. V^Z 3. (a) 6 ft./sec. ; (6) 10 ft. /sec. 8 " " 25 P er cent a PP rox ' 4. (a) 6 = ap; (6) , 2 = **. 9- ^f= 8 ^- 5. 2 7r cu. in. /sec. 70 « 03 • ,u n - — ft./sec. 6. 3| mi./hr. tt Miscellaneous Exercises. Pages 81, 82. 1 gy - dx 5 3 x 2 — 10 axy *Vl-z 2 ' ' 5 ax 2 - 21 j/ 2 " t (fes -^i ? + V1 -^ 2x(l-y 2 )*-3 y (l-y 2 ) x 2 Vl-x 4 (to. ^ _y2 2y(l -y 2 )^ + 3x 3 , 5r 6x8 + 35x 2 -12x-56 1 3 |_x*(x 2 -4)?(x+7)^ J 4. dy= V ^V ^ r ^^ lg < fa. 8. -5. 2Va 2 -x 2 y 9. Vx= _J2iL_; ,„ = 120 .6V2: 6V& VlOO+y 2 V100 + y* 10. 4 ft./sec. ; 8 ft./sec. Ctenjp ft /gec 11. fVIT; 6f. ra 2 '' 12. 17*; IVU05. lg Vl6Tft./sec. 2 . 13. 14 V3; 14. 50 2 20. — 2.24 lb./sq. in. per sec. 15. Oy = c k , where A: = Z)#. 21. 0.1 B (a + 2 &r). 3«V «o flofr-2/rQ 16. y + ex 2 = 0. * (1 - e T1 + r/1 2 ) 2 23. (a) ±?\/0; (6) 0.899 a. 4 (c) y - 1 V2V3 - 3 =-(3 + 2 VS) VWs -Six- ?). 4 24. — in. /sec. 9tt ' ANSWERS 329 Art. 56. Pages 86, 87. 1 . — a sin ax. «/* sin 6 2. 3 tan 2 x sec 2 x. (1 - cos 0) 2 3. 2 [cos 2 x + sin 2 x]. 21. - w(^sinw« + 5 cos (at). 4. 2 sec 2 x tan x. 22 rsin ^r 1 + rcosd 5. 6 sin 3 x cos 3 x. L VZ 2 - r 2 sin 2 0- 6. 6 x cos 2 (a 2 - x 2 ) sin (a 2 - x 2 ) . 23 sin 7. x cos x + sin x. * ~~ cos *P 24. a

Vp 2 - a 2 Va 2 - x 2 330 ANSWERS Art. 65. Pages 97, 98. 1. 2 *- 3 . 12 1 x 2 -3x+5 V x 2 - a 2 2 2m 1 (*-*,») loga' 13 - j^T*' 3 xa^-^loga 14. arc tan x. 15. xe ax . Vx 2 — a 2 16 x 4. 3 x 2 e* 3 + e*. a + & x2 5. 1+logx. ' 17 l-2x + 2x 2 -3x3 6 1 Vl + x 2 xlogx lg x(x 3 +9x 2 -24) 7. e*-e~*. ' Vx^4(x+3)3* 4 (e x — e~ x ) X\t 19. *' + * 9. e 8inx cosx. x (1 -x 2 ) in sec 2 x 20. • 10 - ^nV (l+*)vT=^ n 2 21. Vl-C^e^*-^- ^). Vx^TT* 22. e-M(B-Bkt-Ak). 23. e-** [(ma — &&) cos m£ — (mb + a&) sin mi]. 24. £? : ae«»; 5f!±^L±i ; e .«> V^+T. a a 25. 45° ; arc tan - ■ e 2 e a — e » Miscellaneous Exercises. Pages 98-100. 4. (a) j + wtt; ±J; §tt + wtt. (6) !£; arctan(±2); ? + ?• 4 4 4 2 4 2 (c) ~ + wtt ; arc tan (± V2) ; y + n?r. (d) x = 1 ; - ; nowhere. 4 4 4 5. (a) 0.80902. (6) -0.58779. (c) 1.52786. 6. sin * ; 0.5773; r. 10. 1 l-cos0 x 4 (l+x 2 ). V« 2 - x 2 pX _ P -x 11. cot-; ptan -• z 2 14. (a) 1 + cos ,,, sin0 2 a (e) 2 V' 2 331 16. ?»cos-; msin- 2 2 17. (a) -j^JL + i^V (6) T( ^ n f c y (c)p(ba<> log a- cplogfl). 18. — ae-*{2ir& sin 2 «■(&« + c) + \ cos 2 *•£&* + c)}. Art. 68. Page 103. 6. f x 7 - x 4 + x 2 + 7 x + C. 7 - -i + **-f + X + a 9.^-2*1 + a 10. M^ - 12 x 8 + ^^ - 36 x 6 + §i^f + a 9 7 5 (ax + 6)3 (2^-5)* c 3 a 32 12 (x 3 + 4)« | c 14 . (3^-^ +C , 30 3 15. 2Vx 3 -5x + 7 + O. Art 70. Page 108. 1. $ (x + ay + C. 13. v^T^ 2 + a 2. i{x*-a*y + c. 14 i arcsin ^ + a 3. log (3 x 2 + a 2 ) + 0. c 4 # e x2 _j. (> # 15. arc sec ax + C. 5. --L+c, 16. « w2 ' 3 +a sin 6 m 2 i g a 6. -icos3 0+O. , j 7. Karctanx) 2 +C. 17 - i arc tan -^- + C. 8. - - VI - m 2 x 2 + G. . x -± m 18. arc sin — \- C. 9. i tan 2 + C. 11. A(« x + & ; 7 rt 12. ^log(2x 8 -5)+C. 12 "11-x 9 v 19. log(x + 3+Vx 2 + 6x + 10). 11. ^-(ax + b)*+C. • bK 1(1 20. JL]ogJL±£,tf-l) 2 17. i. 19. 2 cos 2 cot 0. x 2 Art. 83. Pages 131, 132. 1. 2>- 4 (sin ax) = -sin ax + ^x 3 + ^x 2 + c 3 x + c 4 - a 4 6 2 2 * 2> " 4 == ^* a "S** + \ ClxS + l C2x2 + CsX + C4 * 3. I)"* = — | log X + £ CiX 3 + | C2X 3 + c 3 x + c 4 . 4 y = ^ 8 tn^-Bw ^ 6 6 to 5. #/y = \ Mx* + I i?x 3 - & wx 4 + (7ix + C 2 . C 2 = ; C x = -\Ml-\ Bl 2 + & wl\ Art 84. Page 136. 1. Max. for x = ; min. for x = 6$. 11. Min. for x =1 7r, 1 7r, etc. 2. Max. for x=0; min. for x= ± V8. 12. Min. for x = 2. 3. Max. forx = -iV3; min. for u Min . for ,, = _ 2J> . x ss £v3. 3 a 4. 5. 6. 7. 8. 9. Max. for x = ; Min. for x = 2. Min. for x = 1. . Min. for x = v2. Min. for x = 0. Max. for x = 1. min. for x f. 14. Min. for x as — f. 15. Min. for x = — §. 16. Max. for x = e. 17. Min. for x = 3 and for x = — 2. Max. for x = — f . 18. Max. for x = 0. 10. Max. f or x = - • 4 Art. 85 Pages 138, 139. 1. 3. 2 and *. 2 2 - p + Vl + ac 2 . 5. 2. 45° ; max. range = ^Sl . Q 8 in. 7. V15. 4. a 6* 6. Length = diameter. 8. 4 V5l ANSWERS 385 10. Altitude = 2 x diam. of sphere. lg e 12. Base = altitude = rV2. ' 2* 13. a a/2, 6V2. 19. ff. 14. Base = a\% altitude = § a. „ ft , , _ 14 V3 15. Altitude = |x altitude of cone. breadth _ ___ . Radius of base = f x radius of 14 ^g base of cone. depth = — - — 16. Altitude = f a ; radius of base ^ „ , , n _ - _I a ^§ 21. Breadth = 6; depth = 6V3. 17. Radius of base = ^-£-v/— ; 2_V2 7r' altitude = yj— ■ Miscellaneous Exercises. Pages 140, 141. 1. sin *(2 - **) + 4 * cos x. g (a) 4 e , (sin x _ cog %) . (ft) 2. 26*cos(*+j). ^ gffl^ | ' * 24^7 a 3 3. — — — • 10. y = ax* 2 + cix + C2 ; S&M y = ax J + cix ; 4 3a 4 x y = ax 2 + 5 ax -f c 2 . (x 2 - a 2 ) * 11. (a) I)-» = -.|loga; + iaB* + ^ + lc 1 »* + Cjaj + cg. A Z*x O — (6) Z>- 3 = A- (e«* + e-* 8 ) + icix 2 + c 2 x + c 3 . a 3 2 re) 7>" 3 = i cos(kt + e) + ;Ux 2 + c 2 x + c 3 . A/ 2 12. (a) a 2 cie-*' + /3 2 c 2 e-^. (6) ae-«[a(ci + c 2 £) - 2 c 2 ]. 13. e-«'[(a 2 + /S 2 ) (ci cos pt 4- c 2 sin /3<) + 2 «0(ci sin /S< - C2 cos /»)]. 15. J_ ^ -^ i-Vl-c^. * VI - c 2 16. (a) Max. for x = 3 ; max. value = 2. Min. for x = — 1 ; min. value = |. (6) Max. for 6 = £ v ; max. value = £ V3. Min. for = 1 7r ; min. value = — f V3. 17 1 18 a6c + 2fgh - af 2 - bg 2 - ch* a(l-cos0) 2 (hx+by+f)* 19. r»Aco*e + jV±*™y-~\. L (7J 8 - r 2 sin 2 0)£ J 336 ANSWERS Art. 86. Page 143. 1. Concave up if x > § ; concave down if x < §. 2. Concave up if — - + 2 rnr < x < | -f- 2 nr ; 2 2 concave down if - + 2 nr < x <(2 w + l)ir. A 3. Concave up if x < — 1 ; concave down if x > — 1. 4. Concave down at every point. 5. Concave up at every point. 8. (a) Concave up at every point, (b) Concave down at every point. 9. Concave up at every point. Art. 87. Pages 144, 145. 1. 2. 3. 4. x = h No point of inflexion. No point of inflexion. Points of inflexion for x = 7. x = 0. 8. x = 0. 9. % = \y/2. flTT. 10. X = ± 1. 5. Points of inflexion for x = »Z. 11. x = ±J±- 2 V'Sb 6. x = log 2. 13. ?/ = ix 3 + x + 3. Art. 88. Pages 148, 149. 1. 2. 3. 4. 5. 6. 7. 8. y = x-\b. x = a, y = ±(x + a). x = 2a. y = ±^x. a x = 2. 3x + 3y = 2. x = ±c; y as ± c. x = l. 9. x = a; y = x + ^; y = -x- 10. y = 0. 11. y = x + 2. 12. x = ; y = 0. 13. 2/ = x+f 14. x = ; y = ; x + y = 0. 15. x=-l. 16. No asymptotes. ~2 Art. 89. Pages 151, 152. 1. Double point and cusp at (0, 0) ; tangents y 1 = 0, y =x. 2. Cusp at (0, 0) ; tangents y 2 = 0. 3. Tacnode at (0, 0); tangents ?/ 2 = 0. 4. Tacnode at (0, 0); tangents y' 2 = 0. 5. Double point at origin ; tangents y = ± x. 6. Double point at origin ; tangents y = 0, x = 0. 11. Cusp at (a, b) ; y = a the common tangent. 12. Double point at (0, 0) ; tangents y = ± Vax. ANSWERS 337 Art 92. Page 160. 1 (y2 + q 2)l g (4y* + aQ* Vs a2 2 */ 4 — 4 x% 2 (9a 2 y 4 + 4x 2 )* ^ a Va;(8a - 3^ 3 ay (6 ay* - 8 x*) ' 3(2 a -z) + W) 1 f1 (x+j a 4 6 4 "' ~ i fCar+a^ + l] 1 12. -15V3; (20, - 10 V2). 13 Qgff; (21*, 16*). 14. - 1 ; (0, 0). 15. 1 2 ak' 1R 2a 4. x + a 5. sec x. 6. 3(a«y)i T. «?• a a 4 & 4 [l + (2ax+6) 2 ]* 17. z = | log £ =- 0.1534. Art. 93. Page 162. I. J_(a 2 sin 2 ^ + & 2 cos 2 ^)2. 2. 3 a sin cos 0. ab 3. y + rf) 1 . 5. ?asin 2 |. 7 . fa . P 2 + 2a 2 4 3 * a 4. jl. 6 . pvrr^. 8 - i- V2 ° Art. 96. Pages 166, 167. 1. The origin. 2. 4(ro - 2/>) 3 = 27 pn*. 3. faro)* + (&n)$ =(a 2 - 6*)$' 4 . m ^ q2 - fi2 cos«g, n=-^=A 2 sin80. a o 5. m = a cos 0, w = a sin 0, a circle. Miscellaneous Exercises. Pages 167, 168. '•-^(f'-f)- »• ■-»' (0 - 0) - II. Double point at (a, 0) ; tangents y= ±(x-d). 12. Conjugate point at (0, 0). 338 ANSWERS 13. Double point at (0, 0) ; tangents tj = 0,y = *x. 14. (a) |f = * + |; (6) y=±^x. 15. -8asinf0. a 2 16. ±x = a\og a +^-* -vanrj2. y 18. (a) P = a sec ; (6) the initial line 6 = 0. (c) four asymptotes, p = | sec fc ± »\ , /> = | sec (£* ± ^ . 20. 4096 a s m + 1152 a 2 n* + 27 n 4 = 0. Art. 97. Page 170. 9. logV3. 13. 2. 17 - 10. V2-1. 14 ' ^arcsin -• '4* I*, i- 15. t(a*-l)8 18. 10^10-1 12. 1. 16. 2 . 64 Art. 99. Pages 172, 173. 1. | arc tan 4. 4. K* 4 -!)- 7. ilog3 2. 5. arc tan e — * • 8. - « 2 V2 4 3- f 6. 1. Art. 103. Page 182. X 'fa' 2 'i 3 - - *-|- 8.|(1 + ^T5). Miscellaneous Exercises. Pages 182, 183. 1. f\/2~^T3. 9 2x /2 ? l a8 " 3. &, 2 5. (a) i(cos a + cos /8) ; (b) § a. IX. ao. 6. e*-e. Art. 105. Pages 187, 188. L 78 - „ 4 7r3q2 2. 106f. °* ~~ 3 3. 10. 9. «2 4- 32. 10. 4 7r2 a 2. a(e a — e a ). 11. fa&. 12. a 2 . ,2 7. J^( 6 ^_^?) ;01og ». 14. t m-\\ J' ° a n ANSWERS 339 Art. 106. Page 190. 1. (a) 1 7rw 2 m ; (ft) } irm 2 n. 2. 27 *-. 3. 144 r. 5. |-7ra 2 (e c + 2c-e- c ). 7. f *-a 3 (31og2 - 2). 13 g Art. 107. Page 192. 1. 108 7T. 2. fTra&c. 5. a 2 /*(>-A). 6. 698f cu. in. 7. 240. Art. 108. Page 194. 1. 6a. , ■, zh\ 27 LV 4aj J' 4. 2*-a. 5. §*%. Art. 109. Page 196. 1.2. a. 3. Vl + « 2 - 2 . VlS/.y.tX. VS2/^...?\. 4. a. ' ^(.r-i), l^(— .?). Art. 110. Page 198. 1. § 2 _Z ( 2V2-1); ^(5V5-2V2). 3. |j[(87)*- 1]. 27 2. 7rwVw 2 + w 2 . 4. EiL( e 2_ e -2 + 4). 4 1. |V5. Art. 111. Page 200. 4 7rW ' 2 , 6 2. -5f. 5. pi-^- log^ 2 . z 2 — X\ %i 3. (a)^; (6) 0. 6 . 2jr. 7T „ Art. 113. Page 204. 1. 4303$in.-lb. 5. 83,219 ft.-lb. 2 Mh4- mh2 . 2. Mh+—- 6 656g4 ft _ lb< 4. TF=-(s 2 8 -si 3 ). 3 W y 7. 3,273,000 in.-lb. 340 ANSWERS Miscellaneous Exercises. Pages 204-206. 1. (a) 372; (6) 48; (c) e-1. kS. 8 3. log 2. 4. a 2 6' 5. 00 If*; (P) 1*. 6. 128 cu. in. 8. fa 3 . 9. a 2 . 10. 11. 141 . (73)* -27 16 f mn 2 . 12. 2io 13. 2p 14. Clog^- 6 - ^ 2 -^. 15. *( * w+1 - *i w+1 ) w + 1 16. 56,530 ft.-lb. Art. 114. Page 209. 1. e*(x 2 - 2 x + 2) + C. 5. f(* - sin 6 cos 0) + C. 2. a;m+1 i g x _ s OT+1 4. c. ft qea3; sin &a; — beax cos ^ 4. n m + 1 (m + l) 2 a 2 + 6 2 3. arc sin + Vl — 2 + C. 7. x log x - x + C. 4 arc cot + log V 1 + s + C. 8. sin (log sin 6 - 1) + C. 9. g Metm »-g + ?°£ig±31 + g , 3 6 6 10. "VdnS + ooiSUc. 2 V a «/ 11. cos x + x sin x + C. 12. * ["tan sec + log tan (| + ^ 1 + C. 13. f[V2 + log(l+ V2)]- 14. ?~ 15. |[(logx) 2 -|logx + |]+a 16. |. Art. 115. Page 211. 2. log V(x + 5) 5 (x - 3) + C. « logx*(x-4)* ^ , /«TXi ANSWERS 341 i a 2 - x + | log x + H log (x - 5) - § log (a; + 1) + 0. log j-ftii^JQ j + (7. 8. log [* 2 (a - 2)*(* + 2)^ 9 ] + (7. Art. 115. Page 212. \og*^* + -L-+c. 5. iogtri-.i + a * S-2 y J '-vr-^r-^j^.. , log 9 ~ 4x + 0. 7. log 2(X - 2)2 + (a;-!) 8 1 x* x ' x - 1 x-1 + C. x + 1 x-1 C. log(x + 3)+-4_ + a x + 3 logx-^4- C. 8. lie**- 1 4 x + 1 2(x+l) 10. mlog(TO + x)4- 2m * + 0. + x 2 («» + x) 2 + C. Art. 115. Page 213. log (s-l) 3 (a + l) 2 + arc 4 tan j. + (re 2 + 1)» I log * — 1 _ I arc tana; + C. 4 x + 1 2 4 . I log ^±1 + a 4 x 2 + 3 3. 6. 8. 9. 11. 12. 5arctan * ■ + I +0 . 5. 1 log *±I+C. 2 2(l+x 2 ) x^ 6 8 x 2 + 4^ 7. |log^±| + |log(y + l) + -4arctan^+C. 8 & y - 1 8 ' V3 ■ »f« 4-gawtanfUa Vx 2 + 6 2 & & J V3 _lrI + J_ a rctan-*-] + C. 10. log ^S + arc^nx + o. log (l + * + * 2 ) f L__ JL arc tan 2 -^±l + C. (1+X) 1+* V3 log VI +2 a? + 0. V3 13. log v / x 2 TT + 2(x 2 + 1) + G Art. 116. Pages 217, 218. 3[f-^ + |-^ + ^-log(,Ul)J+C. r4x* 4_aj 64xj 128s* 1024^ 4096 1 (3a i_ 4) -] + a L 5 3 27 27 81 243 BV J J 342 ANSWERS 3 2 r (mx + b)% 2 6 (mx + b)$ . b 2 (mx + b)? l G mA. 7 /5 3 _T 4. 2arctan- N f55+6\ 6 ' log [x + | + V^+5x-3] + a 7. log [x - | + Vx 2 - 7 x -f 4] + C. 5. 6 arc tan J^=-^ + 0. 8. J^ arc tan J 3 (^- 2 ) + a '5 — x f 8 ^2(3 — x) 9 , 2 j(^-l) l + §(^) l + (,.l ) l +( ^. 1) l| + . 10. 3 Vx 2 ^ - log [x + Vx 2 - 3] + C. 11. log[x + l-fVx 2 + 2x + 5]+C. 3 12 (i-x 2 )^ _ (1 _ x2) | + 15 log [a . + 1 + Vx^T^] + c . o |8. * 2 + 2 +C. 16. -V^g-K?, Vx 2 + 1 a 2 x 14. _ 2V2x-x 2 _ arc yersx + c 17 x _ arcgin x + c x Vrt 2 — x 2 * (£ - M 2 ) + " 2V2 Vl + x 2 -V2x 25. *■«&. 27. 3 r#. 29. 8(6- a)* . 26. |*j0 Art. 126. Pages 243, 244. 1. e x y 2 dx + 2 e x y dy. 2. y^ log y dx + xy x ~ 1 dy. 3. cos x cos ydx— sin x sin ?/ efy. 5. a x e*(\og adx + dy) . . 2xdx 3 x 2 cZy 6 ydx — xdy y* t x 2 + y 2 7. 3(x 2 2/* + i x" %y*)dx + K*V* ~ 4 x'^y)dy. 8. (ydx-xdy)(— i— + — — V U 2 +2/ 2 xy/x 2 -y 2 > 10. £** log z(y dx-\-x dy) + xy s^ - 1 cte. 11. cos x cos y tan z dx •*■ sin x sin ?/ tan z dy + sin x cos y sec 2 2 dz. 12 zydx + zxdy-xydz M d B (dT_Tdv\ Z 2 + X 2 ?/ 2 ' \ V V 2 / 13 2X^^+X 2 ^ + 3X 2 ^^ . 15 dk=C(T n dp+npT n-l dn a* — y s (a 3 — y 3 ) 2 18. Approx., 0.00117; actual, 0.00116. 20. bAa + aAb . aft ANSWERS 345 Art. 127. Page 245. x 6xy-y 2 4 4 x(x 2 + y 2 ) + y 2 7 _£ra_ 3y 2 -3x 2 + 2xy' ' 4*/(x 2 + y 2 )-\-2y{x- 2 a) ' b - v' „ 3x 2 siny+y 3 sinx g 4 a; 3 — 3 y 2 8 p sin d 3 y 2 cos x— x s cosy ' 6y(x — y) ' 2cos^— 3p 3. -^-- 6. -tan*,. 9. -^. wx « 2 y Art 128. Page 251. 1. Exact. e x sin y. 6. Exact. \ y s + xy 2 — (x 2 + x)y. 2. Exact. pv n . 8. Exact, xsiny — e x y. 5. Exact, ^-xy. 11. («) -4; (&) +4; (c) -8. Miscellaneous Examples. Pages 251-253. 1. 2xy 3 z 5 ; 3 x'ty 2 2 5 ; 5x 2 i/¥. « x 2 y 2 z 2 (a 8 + 2/ :} + z 3 ) ! (x s + y 8 + z 3 ) * (x 3 + y 3 + z 8 )^ 3. 3 x 2 (?y cos z ; 2 xM 2 * cos z ; - x 3 e 2 * sin z. 4. x + y ^ + y)-2 + g? ( X + y y2 + 38 (^ + ^2 + ^2 5 4 y 4 - 3 x 2 y 2 6 ^x ? ax + hy + e 2 x s y - 10 xy z + 15 y 4 ' a 2 y by + hx + f „ Sx 2 y 9 a; + x 3 4- arc tan a; 5y 3 -2x 3 " (l + x 2 )Vx 2 + (arctanx) 2 10. 3 xV arc sin x + x 8 e* arc sin x + - J**— . 11. e ~*( sm x ~ cos x ) -^/l _ x 2 e _2x + cos 2 x 12. e«*sinx(a 2 + l). 13. 0; -10; 72 y. 14. — e z cos y ; — e* sin y ; e x sin y. 15. x 2 ?/ sin xy — 2 x cos xy ; xy 2 sin xy — 2y cos xy ; x 8 sin xy. 19. -*-; ^ *!-. 21. ?-E-; ± 18 on « 2 - & 2 + & 2 - :• 26. V3; 20. cV4& 2 c 2 -(6 2 + c 2 -a 2 ) 2 27. AF = 7rr 2 AA+2 7rr/iAr; M + 2Ar. h r 28. (a) £*y(x 2 +y 2 )+0(x)+vKy). (6) i^ainy + *■(*) +/(y). 30. (a) x 2 y - $ y 3 . (6) (1 + x 2 ) arc tan y — y. si. ( «) gr+ g r ,_ ^yi) ^ 1 + g^ + ,- o . (6) aiogr+zsr-^Biogp-^p^i+lp^ + Jo. 346 ANSWERS Art. 129. Pages 256, 257. 1. \&y*+F(x)+yKy). g yg5c 2 (4 + c 2 ) . 2. — e*cos y + F(x) + 3.549. 9. 60.1+. „ 1t> 8. 0.049. 10. 7ra6. Art. 131. Page 261. 2. ^ 2 . 3. 7r(n 2 -r. 2 2 ). 4. 67m 2 . %, £L 7. 4.8584 a 2 ; 11.1416 a 2 . Art. 132. Page 264. 4. * iraba . 5. ™*- c . 7. i^ 8 - 8. faHan/3. 3 2 86 ¥ Art. 133. Page 267. 3. ?| 2 /Inrr /'a/ 4. TT 1. fir* 2. |L 3. »£. 5. |^(lo g 8-2). -c^^-e-)- fc^^^^jg Art. 134. Pages 270-271. 7ra 3 ttV2 1. Z [ C 3 _ (c 2 - r 2 )2]. 6. I Tra 3 -fa 8 6 2. £ Tra&c. 7. 2 • 16 3. Ja6c. Art. 135. Page 273. 1. f ka*; 7=i*?, if 7 = jty. 4. | Tr&a 4 ; | fca. tt „ „ . 5.1 kab 2 ; 4 kb. 3 6. I of density at base. 3. \kl. ANSWERS 347 Art. 137. Pages 279, 280. £&$;•} (»)(o,|i). 2 (a) / a [6V2 + log(3-2V2)1 ; 4ar.2V2-11 \ /8a §_a\ V 8[V2+log(V2+l)] 3[v'2 + log(V2+l)]y V 5 ' 4^' 3. (a) /Jjt,«(e' ! + 4- = p _ jjt Art. 140. Pages 287, 288. 2 Ah 2. / \ b' 1 /r N ffl 2 . (a) -; (6) j; (c) i& 2 ; (d) ^±-^ 3 4. 5 i(«i 2 + a 2 2 ). 1. \kb(h 2 *-h x *). 5. ^vT^ftfti 7. 2. §*ftft*. 6 * |V2^af 8. 348 ANSWERS Miscellaneous Examples. Pages 288-290. 1. 12.022. 6. fa 2 . Q 8 tu* 2. fTrafc. o 3. fa 2 . 7. 7ra ^ 2 , 1Q gxcg 4. 1 7ra 2 . Vm« , a 6 (27\/3 + 10 7r) ft 4 7rm 5 10 Va 2 6 2 + 1 u. — • o. — - — ■ • ia. 7n 64 ab a 2 13. |a C + 3( a 2 +4c 2 )arctan ^.. _« 2 _Ji«! + 4 ;r 2/ r , a\*\ 14. ^(« 2 -6 2 )*. 23 W-W 12 (Mi - b^s) 18 . f (| + logMV. 2 \2 ac/ 24. 10.814 in. 16. £ of density at vertex. 2 5 c=-( R+ r ^ + rg2 > \ • 17. f of density at base. r a 4 * / 18. 2 7r 2 a 2 6 ; 4 tt 2 «6. 29. « = \ OC, y = f OD, i = £ OA Art. 144. Page 298. 2. l_^ + ^_^+£- + ..., -- D ^CTTW^TT^ UOAN DEPT- W ^H^ 9 " MAY20'69-9PM lA-40m-4,' FEB 17 197 54 6 iViMr 16 (MM fc MHt« 7 1973 4 7 rcmSg'Wf^'HS LD 21A-40m-4,'63 (D6471sl0)476 General Library University of California . cia. kbiih Berkeley U. C. BERKELEY LIBRARIES C0bl35ED35 J5306D69 THE UNIVERSITY OF CALIFORNIA LIBRARY * •..'* J*-'