« "♦ APPLIED CALCULUS PRINCIPLES AND APPLICATIONS ESSENTIALS FOR STUDENTS AND ENGINEERS BY ROBERT GIBBES THOMAS Professor of Mathematics and Engineering at the Citadel, The Military College of South Carolina 45 EXERCISES - 166 FIGURES NEW YORK D. VAN NOSTRAND COMPANY 25 Park Place 1919 Copyright, 1919, BY D. VAN NOSTRAND COMPANY Stanbope [Press H. Gl LSON COM PAN Y BOSTON. USA PREFACE This book as a first course in the Calculus is not designed to be a complete exposition of the Calculus in either its principles or its applications. It is an effort to make clear the basic principles and to show that fundamental ideas are involved in familiar problems. While formulas and alge- braic methods are necessary aids to concise and formal presentation, they are not essential to the expression of the principles and underlying ideas of the Calculus. These can be expressed in plain language without the use of symbols — one writer challenging the citing of a single instance where it cannot be done. The practice is common, at least with " thoughtless think- ers," of blindly using formulas without an} r true conception of the ideas for which they are but the symbolic expres- sion. The formulas of the Calculus are an invaluable aid in economy of thought, but their effective use is dependent upon an adequate knowledge of their derivation. The object of this book is to set forth the methods of the Cal- culus in sflch a way as to lead to a working and fruitful knowledge of its elements, to exhibit something of its power, and to induce its use as an efficient tool. No claim is made for absolute rigor in all the deductions, but confidence is invited in the soundness of the reasoning employed and in the logical conclusions obtained. There are students, and engineers also, who when con- strained to use the Calculus look upon it as a necessary evil. This attitude is without doubt due to their minds having never had a firm grasp upon its principles nor a full realiza- iii 4 iv PREFACE tion of the efficiency of its methods. A student while tak- ing a course in the Calculus usually spends at least one-half his effort in reviewing previous mathematics. In fact, a course in the Calculus is held to involve an excellent review of geometry, algebra, and especially of trigonometry; hence, at the end of a term it is too much to expect of the average student that he have an adequate knowledge of the Calculus. If a choice must be made between the ability to solve equations (including integration processes) and the far more rare ability to set up equations to represent established facts and laws, there can be little question as to which type of ability should be cultivated. The latter is of higher order and is likely to include the former. Engineers, physicists, inventors and men of science generally find it difficult to translate their observations into language which the pure mathematician can understand. In fact, such translation usually involves the writing of the equation: an undertaking beyond the capacity equally of the non-mathematical scientist and the pure mathematician. Integration of the equation, once set up, the mathematician will undertake; conceiv- ably, so might a machine. Fruitful deductions and rules of practice result. The difficulty of realizing these results arises not from difficulties in moving about the symbols, but from inability on the part of nearly all persons to state facts in terms of symbols. It is as if no harmonist knew a melody and no melodist knew a note. This book aims to keep fact and symbol in close association, so that the student will never use the latter without being conscious of the former. It may then be expected that he will ultimately be able to visualize the symbolic expression when the fact is known. Apart from the references in the text and in footnotes, acknowledgment is here made of the clarifying and logical ideas embodied in the books on the Calculus by Gibson, by Taylor, and by Tovmm md and Goodenough; also in Hedrick's paper on the Calculus without Symbols. PREFACE V The introduction to this book ends with a reference to the discoverer of the Calculus. It is deemed not unfitting that the book should close with the Central Forces of the Principia. Robert Gibbes Thomas. The Citadel, Charleston, S. C. February 1st, 1919. CONTENTS INTRODUCTION. PART I. DIFFERENTIAL CALCULUS. CHAPTER I. FUNCTIONS. DIFFERENTIALS. RATES. Article Page 1 . Variables and Constants 5 2. Functions. Dependent and Independent Variables 6 3. Function — Continuous or Discontinuous 7 4. Representation of Functional Relation 7 5. Function — Increasing or Decreasing 8 6. Classes of Functions. Empirical Equations 9 7. Increments 12 Exercise 1 13 8. Uniform and Non-uniform Change 14 9. Differentials 15 10. Illustrations of Differentials 15 11. Rate, Slope, and Velocity 18 12. Rate, Speed, and Acceleration 19 13. Rate and Flexion 21 14. Illustrations 21 Exercise II 23 CHAPTER II. DIFFERENTIATION. DERIVATIVES. LIMITS. 15. Derivative 24 16. Differentiation 25 17. Limits 25 18. Theorems of Limits 26 19. Derivative as a Limit. Function of a Function 26 20. Illustrative Examples 31 21. Replacement Theorem 37 22. Limit of Infinitesimal Arc and Chord 37 vii viii CONTENTS ALGEBRAIC FUNCTIONS. Article Page 23. Formulas and Rules for Differentiation 38 24r-31. Derivation of Formulas 39-43 Exercise III 45 LOGARITHMIC AND EXPONENTIAL FUNCTIONS. 32. Formulas and Rules for Differentiation 47 33. Derivation of Formulas 48 34. Limit ( 1 + -X = e 50 n=oo \ nj 35. Derivation of Formulas 52 36. Limit ( 1 + -Y = e x 52 37. Derivation of Formulas 53 38. Modulus 54 39. Logarithmic Differentiation 56 40. Relative Rate. Percentage Rate 56 Exercise IV 57 41. Relative Error 59 42. Compound Interest Law 60 TRIGONOMETRIC FUNCTIONS. 43. Circular or Radian Measure 64 44. Formulas and Rules for Differentiation 64 45. Derivation of Formulas 65 46. Limit /sin d\ ., a£f °±« \~r) = 1 66 47-50. Derivation of Formulas 68 51. Note on Formula 69 52. Remarks on Formula 70 Exercise V 71 53. The Sine Curve or Wave Curve 72 54. Damped Vibrations 73 r INVERSE TRIGONOMETRIC FUNCTIONS. 55. Formulas and Rules for Differentiation 76 56-59. Derivation of Formulas 77, 78 Exercise VI 78 CONTENTS ix Article Page 60. Hyperbolic Functions 80 61. General Relations 80 62. Numerical Values. Graphs 80 63. Derivatives 81 64. The Catenary 81 65. Inverse Functions 81 66. Derivatives of Inverse Functions 82 . CHAPTER III. SUCCESSIVE DIFFERENTIATION. ACCELERATION. CURVI- * LINEAR MOTION. 67. Successive Differentials 84 68. Successive Derivatives 84 69. Resolution of Acceleration 86 Exercise VII 87 70. Circular Motion 88 71. The Second Law of Motion 90 72. Angular Velocity and Acceleration 92 73. Simple Harmonic Motion 94 74. Self -registering Tide Guage 97 Exercise VIII 97 CHAPTER IV. GEOMETRICAL AND MECHANICAL APPLICATIONS. 75. (a) Tangents and Normals 99 (6) Subtangents and Subnormals 100 76. Illustrative Examples 101 Exercise IX 107 77. Polar Subtangent, Subnormal, Tangent, Normal 108 Exercise X 110 CHAPTER V. MAXIMA AND MINIMA. INFLEXION POINTS. 78. Maxima and Minima Ill 79. The Condition for a Maximum or a Minimum Value. . . Ill 80. Graphical Illustration 113 81. Rule for Applying Fundamental Test 115 CONTENTS Article Page 82. Rule for Determining Maxima and Minima 115 83. Inclusive Rule 116 84. Typical Illustrations 118 85. Inflexion Points 120 86. Polar Curves 123 87. Auxiliary Theorems 124 Exercise XI 125 Problems in Maxima and Minima 127 Determination of Points of Inflexion 132 CHAPTER VI. FRVATURE. EVOLUTES. JUR^AI 88. Curvature V. 133 89. Curvature of a Circle 134 90. Circle, Radius, and Center of Curvature 135 91. Radius of Curvature in Rectangular Coordinates 137 92. Approximate Formula for Radius of Curvature 138 Exercise XII 139 93. Radius of Curvature in Polar Coordinates 140 Exercise XIII 142 94. Coordinates of Center of Curvature 142 95. Evolutes and Involutes 143 96. Properties of the Involute and Evolute 143 97. To find the Equation of the Evolute 144 CHAPTER VII. CHANGE OF THE INDEPENDENT VARIABLE. FUNCTIONS OF TWO OR MORE VARIABLES. 98. Different Forms of Successive Derivatives 149 99. Change of the Independent Variable 149 Exercise XIV 151 100. Function of Several Variables 152 101. Partial Differentials 152 102. Partial Derivatives 153 103. Tangent Plane. Angles with Coordinate Planes 154 Exercise XV 156 104. Total Differentials 156 105. Derivative of an Implicit Function 158 Exercise XVI 158 CONTENTS xi Abticle Page 106. Total Derivatives 159 107. Illustrative Examples 160 Exercise X VII 161 108. Approximate Relative Rates and Errors 162 Exercise XVIII 163 109. Partial Differentials and Derivatives of Higher Orders . 164 110. Interchange of Order of Differentiation 165 Exercise XIX 166 111. Exact Differentials 167 Exercise XX 170 112. Exact Differential Equations 170 PART H. INTEGRAL CALCULUS CHAPTER I. INTEGRATION. STANDARD FORMS. 113. Inverse of Differentiation 171 114. Indefinite Integral 173 115. Illustrative Examples 17-4 116. Elementary Principles 178 117. Standard Forms and Formulas , 180 118. Use of Standard Formulas 182 Exercise XXI 185 Exercise XXII 187 119-121. Derivation of Formulas 187-190 Exercise XXIII 191 122. Reduction Formulas 194 123. Integration by Parts 194 Exercise XXIV 197 124. Reduction Formulas for Binomials 198 Exercise XXV 199 CHAPTER II. DEFINITE INTEGRALS. AREAS. 125. Geometric Meaning of Jf(x)dx 204 126. Derivative of an Area 205 127. The Area under a Curve . 206 Xll CONTENTS Article • Page 128. Definite Integral. . , 206 129. Positive or Negative Areas 208 130. Finite or Infinite Areas — " Limits " Infinite 208 131. Interchange of Limits 210 132. Separation into Parts 210 133. Mean Value of a Function 211 134. Evaluation of Definite Integrals 213 Exercise XXVI.. 213 135. Areas of Curves 214 Exercise XXVII 216 136. To find an Integral from an Area 219 137. Area under Equilateral Hyperbola 221 138. Significance of Area as an Integral 223 139. Areas under Derived Curves 225 CHAPTER III. INTEGRAL CURVES. LENGTHS OF CURVES. CURVE OF A FLEXIBLE CORD. 140. Integral Curves 227 141. Application to Beams 229 142. Lengths of Curves 236 Exercise XXVIII 237 143. Lengths of Polar Curves 239 Exercise XXIX 240 144. Curve of a Cord under Uniform Horizontal Load — Parabola 241 145. The Suspension Bridge 244 146. Curve of a Flexible Cord — Catenary 245 147. Expansion of cosh x/a and sinh x/a 248 148. Approximate Formulas 249 149. Solution of s = a sinh x/a 250 150. The Tractrix 254 151. Evolute of the Tractrix 256 p~\ CHAPTER IV. INTEGRATION AS THE LIMIT OF A SUM. SURFACES AND VOLUMES. 152. Limit of a Sum 259 153. The Summation Process . 261 CONTENTS Xlll Abticle Page 154. Approximate and Exact Summations 262 Exercise XXX 266 155. Volumes 267 156. Representation of a Volume by an Area 268 157. Surface and Volume of any Frustum 270 Exercise XXXI 282 158. Prismoid Formula 283 159. Application of the Prismoid Formula 284 Exercise XXXII 287 160. Surfaces and Solids of Revolution 288 Exercise XXXIII 292 CHAPTER V. SUCCESSIVE INTEGRATION. MULTIPLE INTEGRALS. SURFACES AND VOLUMES. 161. Successive Integration X^^^^^^rT 296 Exercise XXXIV 299 162. Successive Integration with Respect to Two or More Independent Variables 300 163. The Constant of Integration 301 Exercise XXXV 303 164. Plane Areas by Double Integration — Rectangular Co- ordinates 304 Exercise XXXVI 307 165. Plane Areas by Double Integration — Polar Coordi- nates 308 Exercise XXXVII 311 166. Area of any Surface by Double Integration 311 Exercise XXXVIII 316 167. Volumes by Triple Integration — Rectangular Coor- dinates 317 Exercise XXXIX 320 168. Solids by Revolution by Double Integration 321 169. Volumes by Triple Integration — Polar Coordinates .. . 321 170. Volumes by Double Integration — Cylindrical Coor- dinates 324 171. Mass. Mean Density 327 xiv CONTENTS CHAPTER VI. MOMENT OF INERTIA. CENTjER OF GRAVITY. Article ^-^ \_^ Page 172. Moment of a Force about an Axis 331 173. First Moments 331 174. Center of Gravity of a Body 332 175. Center of Gravity of a Plane Surface 333 176. Center of Gravity of any Surface 334 177. Center of Gravity of a Line 334 178. Center of Gravity of a System of Bodies 335 179. The Theorems of Pappus and Guldin 336 Exercise XL 342 180. Second Moments — Moment of Inertia 344 181. Radius of Gyration 345 182. Polar Moment of Inertia 345 183. Moments of Inertia about Parallel Axes 346 184. Product of Inertia of a Plane Area 348 185. Least Moment of Inertia 349 186. Deduction of Formulas for Moment of Inertia 350 187. Moment of Inertia of Compound Areas 352 CHAPTER VII. APPLICATIONS. PRESSURE. STRESS. ATTRACTION. 188. Intensity of a Distributed Force 355 189. Pressure of Liquids 356 Exercise XLI 362 190. Attraction. Law of Gravitation 363 191. Value of the Constant of Gravitation 375 192. Value of the Gravitation Unit of Mass 376 193. Vertical Motion under the Attraction of the Earth .... 376 194. Necessary Limit to the Height of the Atmosphere 378 195. Motion in Resisting Medium 379 196. Motion of a Projectile 380 197. Motion of Projectile in Resisting Medium 383 CHAPTER VIII. INFINITE SERIES. INTEGRATION BY SERIES. 198. Infinite Series 385 199. Power Series 386 CONTENTS XV Article Page 200. Absolutely Convergent Series 388 201. Tests for Convergency 390 Exercise XLII 392 202. Convergency of Power Series 393 203. Integration and Differentiation of Series 394 CHAPTER IX. TAYLOR'S THEOREM. EXPANSION OF FUNCTIONS. INDETERMINATE FORMS. 204. Law of the Mean 401 205. Other Forms of the Law of the Mean 403 206. Extended Law of the Mean 405 207. Taylor's Theorem 405 208. Another Form of Taylor's Theorem 406 209. Maclaurin's Theorem 407 210. Expansion of Functions in Series 408 211. Another Method of Deriving Taylor's and Maclaurin's Series 408 212-214. Expansion by Maclaurin's and Taylor's Theorems 411 215. Examples 412 Exercise XLII I 418 216. The Binomial Theorem 419 217. Approximation Formulas 421 Exercise XLIV 424 218. Application of Taylor's Theorem to Maxima and Minima 424 219. Indeterminate Forms 425 220. Evaluation of Indeterminate Forms 428 221. Method of the Calculus 430 Exercise XLV 434 222. Evaluation of Derivatives of Implicit Functions 435 CHAPTER X. DIFFERENTIAL EQUATIONS. APPLICATIONS. CENTRAL w FORCES. 223. Differential Equations 436 224. Solution of Differential Equations 436 225. Complete Integral 437 xvi CONTENTS Article Page 226. The Need and Fruitfulness of the Solution of Differ- ential Equations 441 227. Equations of the Form Mdx + Ndy = 444 228. Variables Separable 446 229. Equations Homogeneous in x and y 446 230. Linear Equations of the First Order 448 231. Equations of the First Order and nth Degree 449 232. Equations of Orders above the First 449 233. Linear Equations of the Second Order 453 APPLICATIONS. 234. Rectilinear Motion 456 235. Curvilinear Motion 460 236. Simple Circular Pendulum 461 237. Cycloidal Pendulum 464 238. The Centrifugal Railway 466 239. Path of a Liquid Jet 467 240. Discharge from an Orifice 469 CENTRAL FORCES. 241. Definitions 472 242. Force Variable and Not in the Direction of Motion . . . 472 243. Kepler's Laws of Planetary Motion 477 244. Nature of the Force which Acts upon the Planets 478 245. Newton's Verification 480 Index 483-490 What we call objective reality is, in the last analysis, what is common to many thinking beings and could be common to all; this common part . . . can only be the harmony expressed by mathematical laws. — H. POINCARE APPLIED QALCULUS. s INTRODUCTION. The Calculus treats of the rates of change of related variables. The factors of life are ever changing, acting and reacting upon each other. The quantities with which we have to deal in ordinary affairs are for the most part in a state of change. Hence the field in which the principles of the Calculus are directly involved is a wide one. In observing the changes about us we note that they take place at various rates, and the determination of the rapidity of the change may be the controlling factor in many investi- gations. Whenever the rapidity of the change of anything is in question, the methods of the Calculus have appropriate application. In the case of velocity or speed, there is rate of change of distance and time ; in a thermometer we have rate of change of length and temperature, while in the barometer there is rate of change of height and density; in the slope of ground or grade of a road we have rate of change of vertical height and horizontal distance; and in the case of a curve, rate of change of ordinate and abscissa, or slope of the curve. In the case of a body in variable motion, it becomes desirable to determine its velocity at some point of its path or at some instant of time, that is, the instantaneous velocity. This notion of rate of change at an instant is common even to untrained minds. When one says of a train in variable motion, that it is now going at the rate of sixty miles an hour, one means that 1 2 APPLIED CALCULUS at the instant considered the rate is such that, if it were maintained, the train would go sixty miles in an hour, that being the instantaneous velocity. The method of the Calculus in getting the rate of change of a variable at any instant is in accordance with natural procedure: measure the amount of change in a short period of time, then the average rate of change during that period is the ratio of amount of change to length of period; the limit approached by this ratio, as the period of time is diminished towards zero as a limit, is the rate of change at the instant the period began. In determining the greatest and least values of a variable quantity, they are found where the rate of change of the variable is zero. For instance, at the maximum and mini- mum temperature during a day, the rate of change of the temperature is zero. There is a difference, however, in that before the hottest moment the temperature was rising and then afterwards falling, while at the coldest moment the reverse was the case. In both cases the temperature's rate of change was momentarily zero. Here is to be seen the method of the Calculus as to maxima and mimina. A distinguishing feature of the Calculus is that in addition to real sensible quantities it uses ideal hypothetical concepts, which are quantities that exist if certain conditions are maintained. The Calculus connects these two classes of quantities. Passing from the real to the ideal is Differentia- tion, from the ideal to the real is Integration. The advantage of introducing ideal quantities is that in many problems an expression for the ideal is readily formed and from this expression the real quantity is obtained by Integration. In other cases the real quantity being given, the problem is solved by the ideal quantity, obtained from the real by Differentiation. The might of the invisible and intangible forces in Nature, predicated upon a concept (the aether), is generally recognized INTRODUCTION 3 in this day and generation. Therefore, it is not to be wondered at, that in dealing with material things and in seeking the inner law by which they act and react upon each other, we should call to our aid ideal concepts. The exclu- sive realist in his passion for facts is prone to overlook the fact that ideas are the first of facts. It is acknowledged that science is useless unless it teaches us something about reality. Let it be acknowledged that the aim of science is not things themselves, but the relations between things, and the fruitfulness of the ideal quantities of the Calculus is recognized. The differentials employed, when properly defined, are not " ghosts of departed quan- tities," even if in some cases ideal in character. "Airy" perhaps, but never "nothing" they give to the creation of the mind "a local habitation and a name." While the ratio of some real quantities may never equal the ratio of the ideal quantities, nevertheless the former ratio may approach so closely the latter as a limit that the exact value of the ideal ratio can be discerned. So the differentiation of any real variable quantity is possible. On the other hand, the exact integration of every ideal quantity is not possible, for in some cases no corresponding real quantity exists. In this respect there is an analogy with Involution and Evolution. Any number may be raised to a power; but the exact root of every number cannot be found, for no real root exists for some numbers. Differential Calculus deals with the rates of change of continuous variables when the relation of the variables is known. Integral Calculus is concerned with the inverse problem of finding the relation of the variables themselves when their relative rates are known. While some problems to which the Calculus is applied may be solved by other methods, it often furnishes the simplest 4 APPLIED CALCULUS solution; and there are cases in which the Calculus alone gives the solution. The Calculus is a tool for the efficient worker, and in the hands of skillful investigators the Cal- culus has proved to be a powerful instrument in bringing to light the truths of Nature. In reference to the mighty intellect that conceived it, there is pardonable hyperbole in the lines of the Poet : — " Nature and nature's laws lay hid from sight, God said, 'Let Newton be,' and all was light." ERRORS AND OMISSIONS Page 24, on last line, f(x) = m for f(x) = m. 161, in Example 3, 35 for 36. 174, on figure, X misplaced. 196, on second line, integral sign omitted. 212, on fifth line, CP X for CP. 232, on fifth line, integral sign omitted. 251, Example* 1, *Miller and Lilly's Analytic Mechanics. 253, at bottom of page, True should be omitted. 259, on second line, Art. 124 for Art. 128. 263, at end of fourth line, period for comma. 345, in equation, x~ for y 2 . 358, in (2), x for x. 385, in expansion, factor a omitted. 418, in Note, -ir for x in (2) gives e tir = 1, should be, -2 7r for x in (2) or 2tt for x in (1) gives e 2lir = 1, whence e i7r = ±1 hence . . . 456, reference, (Ex. 6, Art. 116) for (Ex. 6, Art. 115). PART I. DIFFERENTIAL CALCULUS. CHAPTER I. FUNCTIONS. DIFFERENTIALS. RATES. 1. Variables and Constants. — A variable is a quantity that changes in value. It is said to vary continuously when, in changing from one value to another, it takes each inter- mediate value successively and only once. If at any value it ceases to vary continuously, it is said to be discontinuous at that value. A constant is a quantity whose value is fixed. If its value is always the same in every discussion, it is an absolute constant. If the fixed value may be different in different discussions, it is an arbitrary constant. In the equation of the circle, x- + y 2 = r 2 , x and y are variables that vary continuously from to =t=r; while r is an arbitrary constant, as its value is fixed only for any one circle. In the ordinary affairs of life we have to deal with con- tinuous variables, such as time; the distance of an object in continuous motion from any point on its path; and with discontinuous variables, such as the amount of a sum of money at interest compounded periodically; the price of cotton; the cost of money orders, etc. In nature we have constants, such as : the jnass of a body, which is an absolute constant; the weight of a body, which is an arbitrary constant, as it is fixed according to latitude 5 6 DIFFERENTIAL CALCULUS and elevation; in mathematics, the ratio of the circum- ference of a circle to its diameter and the base of Naperian logarithms are absolute constants. Variables are represented usually by the last letters of the alphabet; as x, y, z, or p, 6, 4>, etc. The letter A, however, often represents a variable area. Absolute constants are denoted by number symbols, and there are some absolute constants represented by letters, as 7r, e, for the ratio and base just mentioned, each transcen- dental but the most important in mathematics. Arbitrary constants are represented usually by the first letters of the alphabet; as a, b, c, a, 0, y, etc. Particular values of variables are constants and are denoted by X\, y i} z h £2, 2/2, Z2, etc. 2. Functions. Dependent and Independent Vari- ables. — When two variables are so related that the value of one of them depends upon the value of the other, the first is the dependent variable and is said to be a function of the second, and the second is the independent variable, which in connection with the function is usually called simply the variable, or sometimes the argument. The area of a square is a function of the length of a side. The area or the circumference of a circle is a function of its radius. The square, or the square root, or the logarithm of a number, is a function of the number. Any function of x is represented by / (x), F (x), $ (x), etc., and the symbol f (x) denotes any expression involving x, \ whose value depends upon the value of x. In any discussion involving x, f (a) means the value of / (x) when x is replaced \ x by a throughout the expression. In y = fix), x is the \ J independent variable and y is the function or dependent , v I variable. In the equation x 2 + y 2 = r 2 , y =\^/r 2 — x 2 or \f - x =^\/ r 2 _ y2 f so y = f( x ) or x = f(y). If one variable is expressed directly in terms of another, the first is said to be an explicit function of the second. If the relation between REPRESENTATION OF FUNCTIONAL RELATION • 7 the two variables is given by an equation containing them but not solved for either, then either variable is said to be an implicit function of the other. So in x 2 + y 2 = r 2 , y is an implicit function of x and x is an implicit function of y; but in y = Vr 2 — x 2 , y is an explicit function of x, and in x = Vr 2 — y 2 , x is an explicit function of y. A variable may be a function of more than one variable, thus in z 2 — x 2 + y 2 , or in z = xy, z is. a function of x and y. The area of a rectangle is a function of its base and altitude. The volume of a solid is a function of its three dimensions; so in V = xijz, V = f (x, y, z). 3. Function — Continuous or Discontinuous. — A func- tion as/ (x) is said to be continuous between x = a and x = b, if when x varies continuously from a to b, f (x) varies con- tinuously from f (a) to /(&). In other words, f (x) is con- tinuous between x = a and x = b when the locus of y = f(x) between the points (a, f (a)) and (b, f (6)) is an un- broken line, straight or curved. A function is said to be discontinuous at any value when it ceases to vary continuously at that value, even though its variable may be continuous. Some functions are continuous for all values of their variables; others are continuous only between certain limits. For example, sin and cos 6 are continuous for values of 6 from 6 t= to 6 = 2 t; tan 9 is continuous from ,0 = to = *72jind from 6 = tt/2 to 6 = f tt, but when ^passes througIT7r/2 or | r, tan 6 changes from A-oo to !— go , hence tan 6 is discontinuous for 6 = ir/2 or | t. The Calculus treats of continuous variables and functions only, or of variables and functions between their limits of continuity. 4. Representation of Functional Relation. — Often the relation between the function and the argument can be expressed by a simple formula. For example, if s is the distance fallen from rest in time t, then s = f (t) = % gt 2 . 8 DIFFERENTIAL CALCULUS In such cases, the value of the function for any value taken for the variable can be found by simply substituting in the formula; thus, si=/(l) =ig, s 2 =/(2) = ig.2* = 2g, and so for any value. A function is tabulated when values of the argument, as many and as near together as desired, are set down in one column and the corresponding values of the function are set down opposite in another column. For example, in a table of sines, the angle in degrees and minutes is the argument, and the sine of the angle is the function. A function is graphed or exhibited graphically by laying off the values of the argument as abscissas along a horizontal axis, and at the end of each abscissa erecting an ordinate whose length will represent the corresponding value of the function; a curve drawn through the tops (or bottoms) of the ordinates is called the curve, or the graph of the function. If V = f( x ), the curve is the locus of the equation; but it is the length of the ordinate up (or down) to the curve, rather than the curve itself, that represents the function. If p = /(0), the function may be graphed by laying off at a point on a line as axis the various angles, — values of the argument 0, and along the terminal sides of the angles the corresponding values of the function p; a curve through the ends of the vectorial radii will be the graph of the function, and will be a polar curve. Here, too, it is the length of the radius to the curve, rather than the curve itself, that repre- sents the function. The area under a curve may be taken to represent a function while the ordinate or radius repre- sents some other function. (See Art. 139.) 5. Function — Increasing or Decreasing. — An increas- ing function is one that increases when its variable increases, hence, it decreases when its variable decreases. A decreasing function is one that decreases when its variable increases, hence it increases when its variable decreases. Thus a!c and CLASSES OF FUNXTIOXS 9 a x are increasing, and ax and a — x are decreasing functions of x. 6. Classes of Functions. — An algebraic function is one that without the use of infinite series can be expressed by the operations of addition, subtraction, multiplication, division and the operations denoted by constant exponents. The common forms are: (a ± bx), (a =b bx n ), ax, a/x, x n , including x 2 , x 3 , \ // x, 1/Vx, etc. All functions which are not algebraic are called trans- cendental. Of these, the most important are: The exponential functions, y = a x or b x , and y — e*, and their inverse forms, the logarithmic functions, x = log a y or logi y and x = log e y. The trigonometric functions, y = sin 6, x = cos 6, y = tan 0, and the inverse trigonometric functions, d = arc sin y or sin -1 y, 6 = arc cos x or cos -1 x, d = arc tan y or tan -1 y. The hyperbolic functions, sinh x = (e x — e~ x )/2, cosh x = (e x + e- 1 ) /2, tanh x = (e x — e~ x ) e x + er*); and the inverse hyperbolic functions, sinh -1 x = y = log (x + Vx 2 + l), cosh -1 x = y = log (x d= Vx 2 — l), tanh" 1 x = y = \ log (1 + x/1 — x). Note. — The phenomena of change in Nature, in general, have been found to be in accordance with one or the other of three fundamental laws. These have been stated * to be the parabolic law, expressed by the power function y = ax n , where n is constant, positive or negative; the harmonic law, expressed by the periodic function y = a sin (mx) ; and the law of organic growth, or the compound interest law, expressed by the exponential function y = ae bx . It is to be noted that as x increases in arithmetic progression, y of the exponential function increases in geometrical progression; while, as x * In Elementary Mathematical Analysis, by Charles S. Slichter. 10 DIFFERENTIAL CALCULUS increases in geometrical progression, y of the power function increases in geometrical progression also. Examples in Nature of the working of these three laws will be given later. EMPIRICAL EQUATIONS. Very often the form of a function is given only empirically; that is, the values of the function for certain values of the variable are known from experiment or observation, and the intermediate values are not given; for example, the height of the tide read from a gauge every hour. In such cases the Calculus is not of much use unless some known mathematical law can be found which represents the function sufficiently accurately. This problem of finding a mathematical function whose graph shall pass through a series of empirically given points is of great practical importance. The known values of the function and of the variable are plotted on cross-section papeiW logarithmic squared paper" greatly facilitating the solution, and a smooth curve being drawn "to fit" the determined points, the equation of this curve is required. The curve suggested by the plotted points may have for its equation one of the following forms: (straight line) y = a + bx, or y = ?nx; (parabola) y = a + bx + ex 2 , or y = a + ex 2 ; (hyperbola) y = a + c/x + b, y = l/x n , or xy = bx + ay; (sine curve) y = a sin (bx + c) , or y = a sin (mx) ; (power function) y = ax n (n any number) ; (exponential function) y = ae bx . If the curve suggested by the plotted points is a straight line, determine the values of a and b, or of m, from the observed data. The straight line is not likely to pass through all the points plotted, even when the straight-line EMPIRICAL EQUATIONS 11 law is the correct expression of the relation to be determined; for the experimental data are subject to error. If the line fits the points within the limits of accuracy of the experiment, it may be drawn through two of the plotted points, and a and b, or m, may be evaluated from their coordinates. By appropriate treatment of the data many of the laws can be transformed into a linear relation. Thus, when the points plotted suggest a vertical parabola with its vertex on the y-Sixis, the required equation will be of the form, y = a + ex 2 . (1) If t is put for x 2 in (1), and the values of t and y plotted, these values satisfy the relation y = a + ct, that is, a straight- line law. The power function y = ax n may be expressed : log y = log a -\- n log x, (2) that is, the logarithms of the given data satisfy a straight-line law. The straight-line law to fit the logarithms can be determined and compared with (2) to find a and n, which are substituted in y = ax n . The hyperbolic law and the exponential function also can be transformed to the straight-line law, and the constants evaluated. Whether the experimental data can be expressed by a power function or by an exponential function can be determined by a test. When the data show that, as the argument changes by a constant factor, the function also changes by a constant factor, then, the relation can be expressed by a power function. When, however, it is f omuijhat-a-^hange of the argument by a constant increment changes the function by a constant factor, then the relation can be expressed by an equation of the exponential type. (See Note, Art. 6.) A full discussion of this problem of finding the expression of the relation between a function and its argument from limited experimental data involves the theory of least squares, and is out of place in a first course in the Calculus. 12 DIFFERENTIAL CALCULUS This necessarily inadequate treatment of the subject here is warranted by the importance of the problem. * Example. — The amount of water Q, in cu. ft. that flows through 100 feet of pipe of diameter d, in inches, with initial pressure of 50 lbs. per sq. in. is given by the following: d 1 1.5 2^3 4 6 Q 4.88 13.43 27.50 75.13 152.51 409.54 Find a relation between Q and d. Let x = log d, y = log Q ; then the values of x and y are : x = \ogd 0.000 0.176 0.301 0.477 0.602 0.788 y = \ogQ 0.688 1.128 1.439 1.876 2.183 2.612 These values plotted give points in the xy plane very nearly on a straight line; therefore, taking y = a + bx, a and b can be evaluated by measurement on the figure; a = 0.688 = log 4.88, b = 2.473. Hence, log Q = log 4.88 + 2.473 log d = log (4.88 d 2m ) ; whence Q = 4.88 d 2 - m . * (Ziwet and Hopkins.) 7. Increments. — The amount of change in the value of a variable is called an increment. If the variable is increas- ing, its increment is positive; if it is decreasing, its increment is negative and is really a decrement. An increment of a variable is denoted by putting the letter A before it; thus Az, Ay and A (a: 2 ) denote the increments of x, y, and x 2 , respectively. If y = f(x), Ax and ly denote corresponding increments of x and y, and ky = Af(x)=f(x + Ax)-f(x), ... Ay = A/(s) =; /(s + As)-/(s) Ax Ax Ax x denoting any value of x. In the figure, let OPi ... B be the locus of y = f (x) referred to the rectangular axes OX and OY. If when x = INCREMENTS 13 OAfi, Ax = M1M2, then At/ = M 2 P 2 - M^ - DP 2 ; if when x =0M 3 , Ax = M 3 M i} then Aj^MiP^MsPb = -#P 8 . ^ A. In the last case Ay is negative and is what algebraically added to M3P3 gives M4P4. When x = QM X = Xi, f (x) = M 1P1 = / (zi) ; when a; = 0M 2 = Xl + Ax, f (x) = M2P2 =f(xi + Ax) ; hence when x = x h A/ (x) = M2P2 - MiPi = / (a* + Ax) - / (xO. EXERCISE I. 1. One side of a rectangle is 10 feet. Express the variable area A as a function of the other side x. 2. Express the circumference of a circle as a function of its radius r; of its diameter d. 3. Express the area of a circle as a function of its radius r; of its diameter d. 4. Express the diagonal d of a square as a function of a side x. 5. The base of a triangle is 10 feet. Express the variable area A as a function of the altitude y. 6. If y = f(x), y+Ay = f(x+Ax); .-. Ay =Af(x) =f(x+&x)-f(x), and hence, Ay = Af(x) = f(x+Ax) - f (x) Ax Ax Ax If y = mx + 6, find value of Ay and of -—' AX 14 DIFFERENTIAL CALCULUS Aw 7. If y =» x 2 , find value of Ay and of — • 8. If y = x 3 , find value of A?/ and of -£• 9. If y = z n , find value of A?/ and of -^, assuming the binomial theorem. 10. If y = f (x) = mx + 6, write values of /«)), /a), /(-i), /(-^} 8. Uniform and Non-uniform Change. — When the ratio of the corresponding increments of two variables is constant, either variable is said to change uniformly with respect to the other. When y = mx + b, -^- = m (constant). (6, Exercise I.) It follows that any linear function of x changes uni- formly with respect to x; that is, y changes uniformly with respect to x when the point (x, y) moves along any straight line. When the ratio of the corresponding increments of two variables is variable, either variable is said to change non- uniformly with respect to the other. Aw When y = x 2 , -r^ = 2 x + Ax (variable). (7, Exercise I.) L\X Thus the area of a square changes non-uniformly with respect to a side. Any non-linear function of x changes non-uniformly with respect to x, for evidently y changes non- uniformly with respect to x when the point (x, y) moves along any curved line. Since time changes uniformly, any variable will change uniformly when it receives equal increments in equal times; and it will change non-uniformly when it receives unequal increments in equal times. Thus in s = vt, where s is the space passed over in time t ILLUSTRATIONS OF DIFFERENTIALS 15 by an object moving with constant velocity v, s changes uniformly. In s = \ gt 2 , where the object moves with constant accel- eration g, s changes non-uniformly. 9. Differentials. — The differentials of variables that change uniformly with respect to each other are their corre- sponding increments; that is, their actual changes. The differentials of variables that change non-uniformly with respect to each other are what would be their corre- sponding increments if, at the corresponding values con- sidered, the change of each became and continued uniform. As with increments, the differentials will be positive or negative according as the variables are increasing or de- creasing. The differential of a variable is denoted by putting the letter d before it; thus, dx, read " differential x" is the symbol for the differential of x. The differential of a vari- able or function consisting of more than a single letter is indicated by the letter d before a parenthesis enclosing the variable or function; thus, d(x 2 ), d(mx-\-b), d(f(x)), denote the differentials of x 2 , mx + b, and/(V), respectively. 10. Illustrations of Differentials. — (a) Suppose a rec- tangle, with constant altitude, is changing by the base in- creasing. If when the base is A B its increase is BM, then d (base) = BM, and d (rectangle) = BMNC. Here the variables change uni- formly with respect to each other, hence their differentials are their corresponding increments. (b) Conceive a right triangle, with variable base and altitude, is changing by the altitude moving uniformly to the right. If when the base is A B its increment is BM, then the increment of the triangle will be BMDC. But if the N 16 DIFFERENTIAL CALCULUS increase of the triangle became uniform at the value A BC, the increment of the triangle in the same time would evi- dently be BMNC; hence, BMNC and BM may be taken as the differentials of the triangle and of the base, where the base is A B. In this case the triangle changes non-uniformly with respect to its base, so its differential is what would be its increment if, at the value con- sidered, the change became uniform. Since the base changes uniformly, its differential is its actual increment. Here increment of triangle ABC = d (triangle ABC) + triangle CND, while A (base) = d (base). If the change of a variable be uniform, any actual increment may be taken as its differential. If time be considered, the in- terval of time, though arbitrary, must be the same for a function as for its variable. (c) Let the curve OPn be the locus of y = f(x), referred to the axes OX and OY. Conceive the area between OX and the curve as traced by the ordinate of the curve moving uniformly to the right. Let z denote this area, and let MM\ be Ax reckoned from the value OM = x; then MM1P1P = A3. But if the increase of z became uniform at the value OMP, its incre- ment in the same interval would be MM^DP; hence MM X and MM X DP may be taken as the differentials of x and z respectively, when x = OM. ILLUSTRATIONS OF DIFFERENTIALS 17 Hence dz = MM X DP = MPdx = ydx, which shows that area z is changing y times as fast as x. Here Iz = dz + area PDP X . It is seen here that while the actual change in the area does not admit of an exact geometrical expression, the differential of the area, being a rectangle, is exactly and simply expressed. It will be shown further on how by Integration an exact expression for the area itself is obtained from this expression for the differential. Note. — Historically the Calculus originated through the efforts to obtain the exact area of figures boimded by curves, mathematics up to that time having furnished no method applicable to all curves whose equations were known. It is true too that historically the method of Integration was discovered before the method of Differentiation was developed. The Differential Calculus arose through the problem of determining the direction of the tangent at any point of a curve. (See Note, Art. 75.) (d) Let OPn be the locus of y = / 0) and s the length from along the curve. Suppose the point (x, y) to move along the curve to P and thence along the tangent at that point. Then at the value x = OM, the change of x and y would become uniform with respect to each other, as the point (x, y) would be moving along a straight line. The change of s would become uniform also with respect to both x and y. As x is the independent variable it may be taken to vary uniformly, making PD or dx = Az or MMi, the actorchange in x as the point moves along the curve from P to Pi. Then dy is DT, the corresponding 18 DIFFERENTIAL CALCULUS uniform change of y, and ds is PT, the corresponding uniform change of s. It is evident that while dx = Ax, dy is not equal to Ay and ds is not equal to As. When, and only when, the locus is a straight line will dy = Ay and ds = As, after dx has been taken equal to Ax. It should be noted that it is not essential that dx should be made equal to Ax, for dx may be taken as any value other than zero, and then dy will be the perpendicular distance from the end of dx to the tangent and ds will be the distance from the point (x, y) along the tangent to end of dy. From figure, (ds) 2 = (dx) 2 + (dy) 2 . 11. Rate, Slope, and Velocity. — The differential triangle PDT in figure for (d) Art, 10, gives —- = tan = slope of the dy curve y= f(x) at point (x, y), and -j- is the ratio of the change of y to the change of x at the point (x, y) , or for any corresponding values of x and y; and ~ is called the rate of y with respect to x. Aw -r— is the average slope or the average rate of change of y with respect to x, while the point (x, y) moves over As on the curve or while x and y take successive values over any range. If s =f(i), where s denotes distance from some origin, ds and t, the time elapsed, then -j- is the rate of change of s with respect to t, what is called velocity, speed, or rate of ds motion: v = -=-• at In the case of uniform motion in a straight or curved path, s As ds = 7 = tt = TT=a constant. In the case of non-uniform t At dt ds or variable motion, v = ^ • — a variable. RATE, SPEED, AND ACCELERATION 19 In figure for (d) Art. 10, it is seen that (ds) 2 = (dx) 2 + (dy) 2 ; dividing byW , (fH$)V(g)' ; .*. velocity of a point in its path is resultant velocity, ds . ffdxy frhi\- + ($-<•* + *, dx x-component is v x = ~rr = = velocity parallel to x-axis, ^/-component is v y = -~ = : velocity parallel to ?/-axis; i^rk~ dy - d y / dx - • tan -dx--Tt/Tt> " dy dx -rr = ~tt tan 0, or v y = v x tan 0; . dy dy /ds dy ds . , ^ = ^sm 0, or y y = y sin 0; dx dx /ds cos = Ts = dt/dt> •"• -77 = -77 COS 0, Or Ps = H COS 0. It appears that -j- , the rate of y with respect to x, is the'ratio of the time rate of y to the time rate of x. These expressions for velocity and their relations include the case in which the motion is uniform or variable along a straight line. 12. Rate, Speed, and Acceleration. — Acceleration is rate of change of speed or velocity. Hence, if the speed is changing, -=- , the time rate of change of speed, is called the acceleration along the path, or the tangential acceleration, and will be denoted by a t . The total acceleration a is equal to a t , when the path is a straight line; otherwise, they are not equal. It is desirable to distinguish between speed and velocity. A body is in motion relative to some other body when its position is changing with respect to that other. 20 DIFFERENTIAL CALCULUS Change of position involves change of distance or of direction or of both distance and direction. If a point moves con- tinuously in the same direction, the path is a straight line; if the direction is continuously changing, the path is a curved line. The direction of motion at any point of a curvilinear path is the direction of the tangent at that point, and from one point to another the direction of motion changes through the angle between the two tan- gents. Thus from Pi to P 2 the direction changes through angle . When the position of a point changes the displacement takes place along some continuous path, straight or curved, and a certain time elapses. The rate at which the change of posi- tion takes place is the velocity of the point. If the point moves so that equal distances are passed over in equal intervals of time, the motion is uniform and the point has constant speed, whether the path is straight or curved. If the direction also is constant, that is, if the path is a straight line, the point has constant velocity. Thus there is uniform motion with constant speed either in a straight line or in a curved line, but there is uniform or constant velocity in a straight line only. The extremity of either hand of a clock moves in a circular path over equal distances in equal intervals of time, but its direction is continuously changing. The motion is uniform and the speed constant, but the velocity is not constant since the direction is variable. Hence, a body may move in a circle with constant speed and yet its velocity is variable. In this case the acceleration a t along the tangent is zero, while the total acceleration a, the rate of change of the velocity, is normal, directed towards the center, and has a constant value depending upon the speed and radius. (This value will be derived later.) ILLUSTRATIONS 21 The term speed thus denotes the magnitude of a velocity. However, the term velocity itself is ordinarily used in the sense of speed as well as in the strict sense of speed and direction. In the great majority of cases the direction is assumed to be known, and the magnitude of the velocity is what is in question. Note. — A velocity having both magnitude and direction is what is called a vector quantity and can be represented by a straight line having the direction of the velocity and a length denoting its magnitude. Hence the sides of the tri- angle PTD in figure for (d) Art. 10, may be taken to repre- sent the resultant velocity v and its components v x and v y . 13. Rate and Flexion. — Flexion has been adopted by some writers as a term for the rate of change of slope. Hence, when the slope changes, and it always does except for a straight line, -r- , the rate of change of the slope with respect to x will be called the flexion of the curve and will be denoted by b, from the word bend. When the velocity and the slope are uniform, there is no acceleration and no flexion; that is, dv A dm -rr = and -r- = 0. at ax 14. Illustrations. — Consider the established equations of motion: s = vt or v = - > v = gt = 32t(a t = g = 32 ft. per sec. per sec. approximately); s = \ gt 2 = 16 1\ When the motion is uniform the velocity or speed is the whole distance divided by the whole time ; or any increment of the distance divided by the corresponding increment of the time is the velocity at any point, and it is the same as at any other point, since it is constant. s As ds .'. v = - = -r-=^- = Si constant. t At dt 22 DIFFERENTIAL CALCULUS In the case of variable motion the whole distance divided by the whole time gives the average velocity over the whole distance; or any increment of the distance divided by the corresponding increment of the time gives the average velocity over that increment of the distance. The velocity at any point is now given by the distance that would be gone over in any time divided by that time, if at the point the motion became and continued uniform or the velocity became constant. ds Thus v = -t = 32 t gives v = 32 ft. per sec. at the end of the first second ; and means that the distance in the next second would be 32 ft., if at the end of the first second the velocity became constant. As a matter of fact, s = 16 t 2 gives 16 feet for the distance in the first second, and 48 feet for the distance actually passed over in the next second. This variation of distance is of course due to the velocity being constantly accelerated. So when it is said that a train at any point is moving at sixty miles per hour, it is not asserted that it will actually go sixty miles in the next hour; but what is implied is, that the train would go sixty miles in any hour if from that point it continued to move with unchanged velocity. Therefore, in ordinary language, variable velocity is ex- pressed by the differential of the distance divided by the ds differential of the time; that is, by -tt • In the case above, ds _ 60 miles _ 1 mile _ 88 feet . dt ~ 1 hour 1 min. 1 sec. thus dt may be taken as any value other than zero, if the corresponding value of ds is taken. EXERCISE II 23 EXERCISE II. 1. u = 2 x. Show graphically the change in u when x is given an increment, by taking x as the base of a variable rectangle of altitude 2, and u as the area. Is the change uniform for u ? 2. u = x 2 . (a) Show graphically the change in u when x is given an increment, by taking x as the side of a variable square, and u as the area. Show graphically the change in u if the change became uniform. (6) Show same when x is taken as the base of a variable right triangle of altitude 2 x, and u as the area. Show the change in u if the change became uniform. 3. V = x 3 . Show graphically the change in V when x is given an increment, if the change in V became uniform; x being the side of a variable cube, and V the volume of the cube. 4. If a body is moving with uniform velocity and passes over 1000 feet in 10 seconds, what is its velocity at any point? If distance is taken as axis of ordinates and time as axis of abscissas, what would the slope of the graph be ? 5. s = 16 1 2 . Compute the values of s when t = 1, 2, 1.1, 1.01, 1.001. Get the average velocity between t = 1 and t = 2, between t = 1 and t = 1.1, between t = 1 and t = 1.01, between t = 1 and t = 1.001. From v = 32 t, get the velocity at t = 1 and compare average velocities with it. What would be the distance passed over in the second second, if at the end of the first the velocity became uniform? What is the actual distance passed over in the second second ? Which is the increment? Which is the differential? 6. If a ship is sailing northeast at 10 knots, what is its northerly rate of motion? What is its easterly rate? If it is sailing S. 30° W. at 10 knots, what is its southerly rate ? What is its westerly rate? 7. If the grade of a road is such that the rise is 52.8 feet in every mile, what is the slope? 8. If the grade of a road is continuously changing, the average slope is given by what ? The slope at any point would be the slope of what ? CHAPTER II. DIFFERENTIATION. DERIVATIVES. LIMITS. 15. Derivative. — The ratio of the differential of a function of a single variable to that of the variable is called dv the derivative of the function. Thus -/ denotes the deriva- dx tive of y as a function of x. Since the derivative may vary with x, as the slope of a curve varies from point to point, it is, in general, itself a function of x; hence, the derivative of f(x) is appropriately denoted by /'(#), and is often called the derived function. So .*. dy = f'(x)dx. Since dy = /' (#) dx, the derivative is also called the differ- ent/ ential coefficient. The derivative ~r is sometimes denoted u n dx by Dx2/. In the case of a curve the derivative is the slope, in the case of motion it is the velocity, speed, or rate of motion; m every case it is the rate of change of the function with respect to the argument or variable. Examples. — dy erivauve 24 If V — f (x) = mx + b, the derivative -p = / (x) = m. LIMITS 25 ds If s = / (0 = vt> the derivative 37 = /' (0 = v - If y = /(*) = 0*, the derivative ^ = f (t) = g. Here m., v, and g are constants. 16. Differentiation. — The operation of finding the differential of the function in terms of the differential of the argument, or the equivalent operation of finding the deriva- tive, is called differentiation. The sign of differentiation is the letter d; thus d in the expression d (x 2 ) indicates the operation of finding the differential of x 2 , and in -7- (x 2 ), that of finding the derivative. D x y, -j- , and f (x) each denote the derivative of y as a function of x. The general method of getting the derivative of y = / (x) is by finding the limit of the ratio of the increments of y and x as they are diminished towards zero as a limit; for the limit which the ratio approaches, when defined to be the dy derivative, can be shown to be ™ 17. Limits. — The student has been made acquainted in Geometry with the notion and use of limits; for exam- ple, the area or the circumference of the circle, as the limit of the area or the perimeter of the inscribed and cir- cumscribed polygons, when the number of sides increases without limit, or when the length of the side approaches zero as a limit. A precise statement of a limit as used in the Calculus is as follows : When the difference between a variable and a constant becomes and remains less, in absolute value, than any assigned positive quantity, however small, then the constant is the limit of the variable. If x is the variable and a is the limit, the notation is lim x = a, or x = a, or lim (a — x) = 0, or (a — x) — 0; 26 DIFFERENTIAL CALCULUS in which = is the symbol for approaches as a limit. When the limit of a variable is zero, the variable is an infinitesimal. The difference between any variable and its limit is always an infinitesimal. 18. Theorems of Limits. — The elementary theorems of limits are: 1. If two variables are equal, their limits are equal. 2. The limit of the sum or product of a constant and a variable is the sum or product of the constant and the limit of the variable. 3. The limit of the variable sum or product of two or more variables is the sum or product of their limits. 4. The limit of the variable quotient of two variables is the quotient of their limits, except when the limit of the divisor is zero. (See Note, Art. 20, for proof.) Note. — The Differential Calculus solves such limits as the exceptional case just stated. 19. Derivative as a Limit. — The limit of a variable, as z, is often written It (z). Lim t^ , or It-^-, denotes It(-r^) when Ax = 0. Az =o LAzJ Ax' \AxJ dv A?/ In defining -^ as a rate (Art. 11), it is stated that -^ is the average slope, or average rate of change of y with respect to x, over the range Ax. As has been given (Art. 7), Ay = Af(x) = f(x + Ax)-f(x) ^ Ax Ax Ax where y = f (x) and x is any value of x. It remains to be shown that lim r^i = lim /(*+A*)-/w = & Ax -oLAsJ ax=o Ax dx (a) By rates voithout the aid of a locus. Let time rates be used and let t, x, and y denote any corresponding values of REMARKS. FUNCTION OF A FUNCTION 27 t, x, and y, from which At, Ax, and A?/ are reckoned. Since -rj is the average rate of y over interval Ay, -~ is the time rate of y at a value of y between y and y + Ay; Ay _ ( time rate of ?/ at the value of y { m A£ — ( from which Ay is reckoned. $ Ax _ ( time rate of x at the value of x ) , . A£ ~" ) from which Ax is reckoned. S Dividing (1) by (2), there results, Ax =o \_Ax\ the time-rate of x dt j dt dx (Compare Art. 11.) _ the time-rate of y _ dy /dx _ dy Thus in showing the derivative as a limit, it appears that the derivative of a function expresses the ratio of the rate of change of the function to that of its variable. It is evident that a function is an increasing or a decreasing function according as its derivative is positive or negative. In the above derivation in place of time-rates, the rate of any other variable of which x and y are functions could be used. Remarks. Function of a Function. — It should be noted that x and y being taken as functions of a third variable t, to every value of this auxiliary variable there corresponds a pair of values of x and y, so y is indirectly determined as a function of x. The derivative of y as a function of x mediately through t is, as shown; (fy _ dy I dx dx dt J dt a t . dy dy dx Solving gives Tt= Tx' Hi and this gives the formula for the derivative of the func- tion of a function. For if y is directly given as a function of x, and x as a function of t, then y is said to 28 DIFFERENTIAL CALCULUS be a function of a function of t, as it is given as a function of t mediately through x. If y is a given function of z, and z a given function of x, then y is a function of a function of x, and the formula for the derivative of y is dy _ dy dz dx dz dx Functions of functions often occur and there may be several intermediate variables such as z in above case. A function, as/ (x), is defined to be continuous for the value a of x, or, .more simply, continuous at a, if / (a) is a definite finite number, and if lim f(x) = f (a) ; that is, if lim / (x) = x=a f (lim x) . By this definition the elementary functions of a single variable are continuous for all values of the variable except those for which a function becomes infinite. A concrete case in everyday experience of a function of a function is the change in length of a metal bar as the tem- perature changes with time. Here the length is a function of the temperature, and the temperature is a function of the time; hence the length is a function of a function of the time. The length, being directly a function of the temperature, is indirectly a function of the time through the temperature, which is directly a function of the time. The rate of change of length per second is equal to the product of the rate of change of length per degree and the rate of change of temperature per second. If /, T, and t denote the length, temperature, and time, respectively, then, in accordance with the formula, K, (2) pA7 + 7Ap + ApA7 = 0, or (7 + A7)Ap = -pA7, (3) A? = V__ (4) A7 7+A7' w average rate of change from 7 to 7 + A V, v % = hm r^l = lim (- vl—) = - £, (5) d7 af^oL^^J at'=o\ 7 + AT/ 7 rate of change for any corresponding values of p and V> ILLUSTRATIVE EXAMPLES 35 .'. dp = — T?dV, showing that the . of pressure is V increase ^ times the , of volume at any corresponding values of pressure and volume. Example 6. — Let M be the mass of a body, V its volume, and p its density; then, AM M AV V p ' the density at any point, when the body is of uniform density; .. AM dM Iim ~jr-== = -p^r = p, the density at any point, when the density varies from point to point. Here when the body is not homogeneous, the density being variable, -z-y. is the average density of the portion of mass, AM; while the derivative, -^, expresses the density at a point of the body whether the density is variable or uniform. Note. — In regard to lim \-r^ \ = -j- , the derivative of y as a function of x, it is important to note that, since the limit of the divisor is zero, it is wrong to write lim Ay iX = LAzj in ax=o I Ax | lim Ax This case is specially excepted in Theorem 4, Art. 18. To prove the Theorem 4, Art. 18: Since y = - • x, x lim y =r lim - • lim x, by Theorem 3, ,. y lim?v ..,. \ lim - = r-. — - , if lim x is not zero, x lima; 36 DIFFERENTIAL CALCULUS When lim x is zero, division by it is inadmissible by the laws v of Algebra. If lim x were zero and lim y not zero, then - x is infinite and has no limit; hence the exception in Theorem 4. The notation lim - = oo , if so written, means that, as x x=Q X approaches zero as a limit, - increases without limit; that is, x the limit is non-existent. Infinity or an infinite quantity is not a limit, and the symbol oo means a variable increasing without limit. In Example 4, where y = -, - = — . Here where lim x xxx- v is zero and lim y is not zero, - is infinite, having no limit. X "• In Example 5, the limit — £ is finite for finite values of p and V. From V — t? an d tJ = ™, p — °° as F = 0; hence as lim V is zero and lim p is not zero, ^ is infinite; At? p •'• A7 = ~ y + AF is infinite as V ~ °> [Apl xr= is non-existent when V is infinitesimal. A7J Ai/ In Example 3, where y = mx -\-b, Ay = mAa;, and ~- = m. If Ax = 0, Ay = 0, but their ratio is constant and ap- proaches no limit. Since Ay = mAx, the law of change of the variables is known and the ratio of two infinitesimals is a finite constant. In Example 1, where- y = x 2 , Ay = 2xAx + Atf, ^ = 2z + A:r, .% lim T^l = 2x. Here the limit of the ratio of two infinitesimals is a finite constant for any particular finite value of x; but, as x may LIMIT OF INFINITESIMAL ARC AND CHORD 37 have any value, the limit of the ratio may be zero, finite, or non-existent. Thus it is seen that, no matter how small two quantities may be, their ratio may be either small or large; and that, if the two quantities are variables both with zero as their limit, the limit of their ratio may be either finite, zero, or non- existent, but is not 0. (See Art. 219.) To find lim U-^ , as in the illustrative examples, the limit Ax -o LA x\ of an equal variable is found; which limit is, in general, determinate and not identical with the indeterminate expres- sion 0. In certain cases the limit of the ratio of two infinitesimals is found by finding the limit of some other variable which, though not equal to the ratio, has the same limit. Examples of such cases will be given further on. (See Art, 70, Art. 77.) 21. Replacement Theorem. — The limit of the ratio of two variables is the same when either variable is replaced by any other variable the limit of whose ratio to the one replaced is unity. Let 0, 0i, 0, and fa, be any four variables, so that lt^ = \, U4-=1, and lt S - = c. (1) 0i fa 6 6\ 4>\ _ 6\ fa . 01 01 fa 0i ... ul = H^.lt e -.lt*l = lt^, by(l) fa 01 01 in which is replaced by 0i, and by fa, but the limit of the two variables is the same. 22. Limit of Infinitesimal Arc and Chord. — The limit of the ratio of an infinitesimal arc of any plane curve to its chord is unity. Since s (Art, 19 (6), figure) is a function of x, 38 DIFFERENTIAL CALCULUS But It chord PPl = u sec DPP, = sec DPT = ^- (2) Ax dx Dividing (1) by (2), Mm [^_] = l. (3) It follows from Art. 21 that in a limit an infinitesimal arc may be replaced by its chord. ALGEBRAIC FUNCTIONS. 23. Formulas and Rules for Differentiation. — By the general method any function can be differentiated, but it is usually more directly done by formulas or rules established by the general method or by other methods. In the following formulas u, v, y, and z denote variable quantities, functions of x; and a, c, and n, constant quan- " d " tities. If in the formulas ~r or "Dj" be substituted for ax "d" and in the rules " derivative" be substituted for differ- ential, they are still valid. [I] If y = x, dy = dx. The differentials of equals are equal. [II] d (a) = 0. The differential of a constant is zero. [III] d (v + y + • • • - z + c) = dv + dy + • • • - dz. The differential of a polynomial is the sum of the differentials of its terms. [IV] d (ax) = a dx. The differential of the product of a constant and a variable is the product of the constant and the differential of the variable. [Vo] d (uy) = ydu-\- udy. The differential of the product of two variables is the sum of the products of each variable by the differential of the other. [V 6 ] d (uyz . . . ) = (yz . . . ) du + (uz . . . ) dy + (uy . . . ) dz + • • • • DERIVATION OF [I] 39 The differential of the product of any number of variables is the sum of the products of the differential of each by all the rest. Y\ D-dN-N- dD [VI] D D 2 the differential of a fraction is the denominator by the differential of the numerator minus the numerator by the differential of the denominator, divided by the square of the denominator. [VII] d(x n ) = nx^dx. The differential of a variable with a constant exponent is the product of the exponent and the variable with the exponent less one by the differential of the variable. 24. Derivation of [I], — If i/ is continuously equal to x, it is evident that y and x must change at equal rates; that is, dy _ dx dx dx' dx Since -=- = h the rate of x is the unit dx rate, so in general the rate of a variable with respect to itself is unity, or the derivative of / (x), when / (x) is x, is one. Geometrically the locus of y = x is the straight line through origin making angle = j with z-axis. V At/ tan 4> = - = -~ = 1, x AX dy = dx. i • Av . L Aw dy and smce -r- 2 is constant, -r 2 - — —- = 1. Az Az dx dy = dx. For examples of [I], if y 2 = 2 px, d(y 2 ) = d(2px); or if x 2 + y 2 = a 2 , d (x 2 + y 2 ) = d (a 2 ) = 0. 40 DIFFERENTIAL CALCULUS 25. Derivation of [II]. — By definition the value of a constant is fixed, therefore the rate of a constant is zero; that is, ^.= 0, .'. da = 0. • dx Aw . Aw . Aw = 0, .*. i-^ = 0, and since -r^ is constant, - *. Ax Ax If w = a, a change in x makes no change in w, hence Ay Ax ^ = ^ = 0, .-. dy = da = 0. Ax dx ' * Geometrically the slope of w = a (a line parallel to z-axis) is at every point zero. 26. Derivation of [III]. — It is manifest that the rate of the sum of v + y + ■ ■ ■ — 2 + c is equal to the sum of the rates of its parts, v, y, . . . — z and c; that is, d (v -f y + • • • — g + c) _ dv dy _dz dc dx dx dx dx dx Multiplying by dx, since dc = 0, the result is [III]. The rule shows that differentials are summed like any other algebraic quantities. For example, d (b 2 x 2 ± a 2 y 2 - a 2 b 2 ) = d (b 2 x 2 ) ± d (a 2 y 2 ) - d (a 2 b 2 ). 27. Derivation of [IV]. — Since A (ax) = a Ax, the ratio of the increments is constant and ax changes uniformly with respect to x. Hence by definition - — f of differentials d (ax) = adx. If y = ax, -j- = a, slope of line. Geometrically, if dz o «r MdxM, z = ax be area of a rectangle of base x and altitude a, then the rectangle MPPiMi is the change of z made by a change Ax ( = dx) of x, and being a uniform change is the differential of z, .'. dz = d (ax) = a dx. DERIVATION OF [Y t ] 41 For examples: d(2px) = 2pdx, and d(~) = dl-x) = —- [V a ] will be seen to include [IV] as a special case. 28. Derivation of [VJ. — Let 2 = uy; then z, a function of u is a function of y also. Geometrically, let u and 1/ be the base and altitude of a variable rectangle conceived as generated by the side y moving to the right and the upper base u moving upward; then z = uy is the area. If at the value OMPN, du = MJkfi, and dy = dy *—4 1 Z=uy NNi, the differential of the area ° « M M i is MMiDP + NPBN h as that sum would be the change of the area of the rectangle due to the change of u and y, if at the value OMPN the change of its area became uniform. Hence dz = d (uy) = y du + udy. Here Az = A (uy) = d (uy) + P DPiB, since that sum is the actual change of the area due to the change of u and y. It is to be noted that iiy = u, then the rectangle is a square and area z = u 2 , .'. d (u 2 ) = udu + udu = 2udu. If y = a, z = au, dz = adu + uda = adu, since da = 0. Hence [IV] is a special case of [VJ. 29. Derivation of [VJ. — To prove d(uyz)=yzdu + uzdy + uy dz. If in [VJ, yz is put for y; d (uyz) = yzdu + ud (yz) = yz du + u (z dy + y dz) = yzdu + uz dy + uy dz. By repeating this process the rule is proved for any number of variables. If y = z = u, then d (uyz) = d (u 3 ) = u 2 du + u 2 du+'u 2 du = Zu 2 du. To derive d(uyz) geometrically, let V = xyz = uyz. 42 DIFFERENTIAL CALCULUS 1 ife *i—X If x, y, and z be the edges of a variable right parallelopiped conceived as generated by the face yz moving to the right, the face xz moving to the front, and the face xy moving upward, then the volume is the product of the three __^yr edges; that is, V = xyz. If at the value OP, Ax = AA h dy =BB h and dz = CC h the differential of the volume is PAi + PB 1 + PCi; as that sum would be the change of \ the volume of the parallel- opiped due to the change of x, y, and z, if at the value OP the change of its volume became uniform. Hence dV = d (xyz) = yzdx + xz dy + xy dz. ^ Here AV = dV + PNi + PLi + PMi + PPi, since that sum is the actual change of the volume due to the change of x, y, and z. If y = z — x, then the parallelopiped is a cube and V = x z , .*. d (x z ) = x 2 dx + x 2 dx + x 2 dx = 3 x 2 dx. dent), then zx = y. :. d (zx) = xdz + zdx Derivation of [VI]. — Let z = - (x and y indepen- x dy, by[VJ. Solving, Corollary. — dy zdx dz = — — di*) = ^ x x) _ xdy — ydx x x 2 , /a\ a dx DERIVATION OF [VII] 43 - , /a\ xda — a dx adx for d[-}= 5 = — , since da = 0. \x/ x- x- - , /x\ adx — x da dx 7 for a - = ; = — j since aa = ; \a/ a- a hence, for a fraction with constant denominator, use [IV]. For another derivation of [VI], see Corollary of next Art. 31. 31. Derivation of [VII]. — I. When the exponent is a positive integer. (a) If n is a positive integer, x n = x • x • x • ton factors; hence, c? (x n ) = d(x-x - x to n factors) = x n ~ l dx + x n ~ l dx + x n ~ l dx + • • • to n terms, by [V&], (b) By the general method. Let y = x n . y + Ay— (x + A#) n = x n + 7?x n_1 Ax + (terms with common factor Ax"), by A?/ = nx 1l ~ l Ax + (terms with factor Ax~) [ Binomial Ay „ . ,. .., „ ". , Theorem. = nx n ~ l + (terms with factor Ax), -j- = lim | ~^L | = nx*- 1 , .'. dy = nx n ~ l dx. ax az=o m II. When the exponent is a positive fraction. Let y = x n , then y n = x m , :. d(y n ) = d(x m ), ny n ~ l dy = mx m ~ l dx, 1 m x m l , mx m ~ 1 y dy = rdx = -. n y n ~ l n y n 44 DIFFERENTIAL CALCULUS (*)- = m j.m — 1 /v. n iyy\ _ J dx = — x n dx* n x m n III. When the exponent is negative. Let y = x~ n , n being integral or fractional; then y = — ; •'• dy = d fcr») = ~ i£~ ** by [VI] - Cor " Art - 30, dy = d (x~ n ) = — nx _n_1 dx. Corollary. — d ( - J = d (xy~ l ) = y~ x dx — xy~ 2 dy _ dx _xdy _ ydx — xdy , VT , y y 2 y 2 Note. — A general proof of [VII] by logarithms, given further on (Art. 37), includes the case where the exponent is incommensurable. So the Formula or Rule is valid for any constant exponent. It is called the Power Formula and is of most frequent application. Examples. — d (Vx) = d (x 2) = jz x~% dx = 2 2Vx • dl— 7- j = d (x~%) = — = x~*dx = =• Wx) 2 2 Vx 3 d(x v *)= V2x v ' 2 - 1 dx(= 1.414 x AU dx, approximately). d (x T ) = tx*- 1 dx (= 3.1416 x 2Am dx, approximately). d ((ax + b) n ) = n (ax + 6) n-1 d (ax -\-b) = na (ax + b) n_1 dx, .*. 3- ((ax + b) n ) = na (ax + 6) n_1 . ax Note. — The last example may be seen to be an application of the formula for the derivative of a function of a function. For let y = (ax + b) n and put z = ax + b, then y = z n ; now y is a function of z, and z is a function of x; that is, y is EXERCISE III 45 a function of a function of x. The formula given in Remarks, „_ . dy dy dz . Art. 19, is / = /-^-' ' dx dz dx .-. ~r = -r ((ax + 6) n ) = -^-^ • v , J = nz n ~ l • a dx dx dz dx = na(ax + 6) n_1 . In applying Rule [VII], if all within the parenthesis, as (ax + b) , is regarded as the variable, the actual substitution of z may be dispensed with in getting the derivative of such functions. EXERCISE m. By one or more of the formulas I-VII differentiate: l. y 6 ^- 2 + _4 2 3 Vx 3 x & dy =d(6 si) + d (4 x~§) - d (2 x" 1 ) + d (3 x~ 4 ). §y = J_ 6 2 12 dx ; x^- x^ / 1\« lim 6 ** = b dx = lim 1 + -) =e (denoting the limit by that letter) ; ... x f^lo & e, ort = !2S ^ = - ax dx x x (m = logb e = the modulus) ; 50 DIFFERENTIAL CALCULUS Hence, d(\og b x) = -dx. [VIII C X d (loge x) = - dx, since log e e = 1 = the modulus. [VIII 6 ] x The limit of ( 1 + - ) as n is increased without limit is e, the Napierian base. (See next Art. 34.) 34. Lim (l + -Y* = e. (See Ex. 7, Art. 221.) ™=« \ it) When n is a positive integer, by the Binomial Theorem, 1 w(w-l) 1 , n(n-l)(n-2) 1 1+n 'n + 1-2 V + 1.2-3 V + '*' = 1 , 1 , __V n) V n/V n/ "^ i " 1 " L2 "^ 1-2- 3 "^ " ' In the expansion there are (n + 1) terms in all, and every term after the second can be written in the form given to the 3rd and 4th terms. As n = oo , - = 0, n K)' nj ^ ^2^2.3^2-3.4 = e = 2.7182818 .... Note 1. — The limit is denoted by e, which is an irrational number, and was proved by Hermite, in 1874, to be trans- cendental or non-algebraic. The number e was the first number to be proved transcendental. Not until 1882 was the attempt to prove the number -k transcendental successful. This was finally done by Lindemann. The proofs consist in showing that neither of the two numbers is the root of an algebraic equation with integers for coefficients. Algebraic real numbers are defined as those real numbers which are roots of such an equation. The importance of these two numbers, considered the most important in mathematics, DERIVATION OF VALUE OF e 51 warrants some notice. They are connected by the remark- able relation, e* s ~ x = 1. (See Ex. 10, Exercise XLIII.) Note 2. — The above derivation of the limit is not com- plete, for the result is true not only when n is "a positive integer," but also for n positive or negative, integral, frac- tional, or incommensurable. The value of the limit, e, can be easily computed to any desired degree of precision by taking a sufficient number of terms of the series. Twelve terms gives the result correct to seven decimals; that is, e = 2.7182818 . . . By comparing the sum of (n + 1) terms of the series with the sum, I + I + 2 + 92 + ' * " o^i ' wmcn is greater than the othor, and equal to 3 — | n_1 , it is manifest that no matter how great n may be, the sum of the (n + 1) terms is certainly finite and less than 3. The ^( 1+l+ k+h+ ■■+£} may be considered as c, or as usually written, to infinity. (See Ex. 5, Art. 215.) Without expanding, the lim ( 1 + - ) can be computed to n = x \ 11/ any desired number of decimals by giving increasing values to n; thus, (1 + -A) 10 = 2.59374. (1.01) 100 = 2.70481. (1.001) 1000 = 2.71692. (1.000001) 1000000 = 2.71828. The last number agrees with the value of e, the required limit, to five decimals. 1 Corollary. — Lim (1 + n) n = e. (See Ex. 8, Art. 221.) n=0 52 DIFFERENTIAL CALCULUS 35. Derivation of [IX a ] and [IXJ. — (i) Let y = b*, then log e y = x log e b. d (loge y) = d(x log e 6), or — = log e b dx, :. dy = d (b x ) = b x log c b dx (b being positive). [IX a ] Hence, d{e x ) = e x dx (since log e e = 1). [IX&] (ii) \iy = b x ,x = \og h y. dx = d(\og b y) = 1 ^^. ""' * = lo&e ^ = 6 * l0ge 6 ^ ( since log^e = l0ge T ' IXa ' Hence, d(e x ) = e x dx. [IX*,] (iii) By the general method of limits. Let y = e x . y + Ay = e x+Ax . Ay = e x+Ax — e x = e x (e Ax - 1). Ay _ e x (e Ax — 1) Ax ~ Ax Ax=o|_A:r_| dx ~ a*=o.L \ Ax )\~ 6 A S V Ax / = e* (since lim ^ "" * ) = l) {Cor., Art. 36); .*. dy = d(e x ) = e x dx. [IX*] Corollary. — d (&*) = b x log e 6 dx. [lX a ] 1 i I For if — = log e b, 6 X — e m , since 6 = e m : d(b x ) = d[e m J; :. d(b*) = e ™ — = te- rn m :. d (b x ) = b x log e b dx. 36. Lim (l + -V = e*. \ DERIVATION OF [X] 53 In limfl +-) , if x ^ 0, by putting n = Nx, when n = ao \ 71/ n = oo so is N = oo ; hence HH'+*H('+*)T and 5i( 1+ 9' -UK 1 +*)'? = H 1 +^)T = e *' since lim/ (z) = / (lim a;), the case of a function of a function. (See Remarks, Art. 19.) By exactly the same method as in Art. 34, it may be shown that 1+ »+Ii+a + -" + 5i)' 1 + - J for positive integral values of n. It can be shown that the limit of this series is a finite number for all finite values of x no matter how great n may be. (See Art. 213 and Ex. 5, Art. 215.) Corollary. — Limf ) = 1, which may be put in the form, lim ( — ; ] = 1. 37. Derivation of [X]. — Let z = y x , then log e z = x log e y. (y positive and inde- pendent of x.) d(\og e z) = d(x\og € y), dz , 7 . dy — = \og e ydx + x-^; & y .*. dz = d (y x ) = y x log e y dx + xy x ~ l dy (y positive). [X] Note. — Formulas [VII], [IX a ], [IX&] are seen to result from [X] as special cases. 54 DIFFERENTIAL CALCULUS Let y = x n , then \og e y = n\og e x, d(\og e y) = d(n\og e x), dy _ dx < y x ' .'. dy = d (x n ) = nx n ~ l dx. [VII] If x were negative, to avoid logarithms of negative num- bers, both members of y = x n are squared before differ- entiating. This derivation of [VII] includes the case where n is incommensurable. 38. Modulus. — In Art. 33 (ii), it appears that log 6 e is the modulus of the system of logarithms whose base is b. Honce, when the base is 10, as in the common system, and the value of e is known, a table of logarithms will give the value of the modulus of the common system to as many decimals as the table gives. The modulus of the common system, denoted by M, is logio e = 0.43429 ... If this value of M is deduced independently of any knowledge of the value of e, which can be done; then the value of e can be gotten from a table of logarithms; for, since M = logio e, then e = 10 M ; that is, e is the number whose common logarithm is 0.43429. . . . In Art, 35 (i) and (ii), it appears that log e b is equal to log e 10 = .— *— = —5 = 2.3026 log*?' fee log 10 e .434 approximately. (See Ex. 6, Art. 215.) To get these results independently, let x be any number whose logarithm in the system with base 10 is I, and in that with base e is V; then 10' = x and e 1 ' = x; .*. 10' = e r . (1) Let W 1 = e; (2) /. 10'= 1QM; /. 1= Ml' or p = M , (3) MODULUS 55 and since 10 and e are constant, so also is M. From (2), M = logi e, or from (1) I = logi e r = l f \og w e; j, = logi e = M; or in general, log&e = ra. Since I = Ml', or logio x = M log e x, it follows that the com- mon logarithm of any number is equal to M times the Napierian logarithm of that number. Now d (log 10 x) = Md (log e x) or d ^ wX ^ = M, d (log e x) or m for base b, and dividing [VII I a ] by [VII l b ] gives , n v = m. the modulus; d (l0g e X) /. M = logic ^ = modulus of common system (b = 10), and m = log e e = 1 , modulus of natural system (b = e). From (1) above, Z log e 10 = V or j, = ^jk = M, by (3) ; •• logl ° e = io^To' or loge 10 = Fi = oaMTT. = 2 ' 3026 ' approximately. To summarize in two equations: Common log = 0.434 times natural log. Natural log = 2.3026 times common log. Note. — Since the modulus of the natural system is unity the differentials of logarithms are simpler when the logarithms are in that svstem; hence, in the Calculus and in most analytic work, Napierian logarithms are employed for the most part. Any finite number except one could be made the base of a system of logarithms. For computation the common logarithms are the best, as having the base 10 affords rules for the integral part of the logarithms and obviates the necessity of that part appearing in the tables. It is usual in writing log for logarithm to omit the subscript 56 DIFFERENTIAL CALCULUS indicating the base, when no ambiguity results. Hereafter, when no subscript to log appears, e will be understood. 39. Logarithmic Differentiation. — Exponential func- tions and also those involving products and quotients are often more easily differentiated by first taking logarithms. This method which is used in the last two derivations (Art. 35 and Art. 37) is called logarithmic differentiation. To derive [V a ], let z = uy, then log z = log u + log y, d(\ogz) = f = ^ + ? = d(logu) +d(\ogy); z u, y :. dz = d (uy) = ydu + udy. To derive [V&], let V = uyz, then log V = log u + log y + log z, j/i t/\ dV du dy dz d(logV) = T =- + 7 + 7; .'. dV = d (yz) = yzdu + uz dy -f- uy dz. To derive [VI], let z = y/x, then log z = log y — log x f d (log z) = -^ = -;-v = f/ ( lo S2/) - d (log a;); Z // O/ h -*©- |m = ^_ ydx = xdy-ydx x x 2 x 2 40. Relative Rate. Percentage Rate. — The logarithmic derivative of a function may be defined as the relative rate of increase of the function. Thus, when y = f (x), dy — = • x is the relative rate of y. V I {x) Hence, when z = xy and therefore, log z = log x -f logy; dz dx dy dx _ dx dx. z " x y ' EXERCISE 57 that is, the relative rate of increase of a product is the sum oj the relative rates of increase of the factors. If the logarithmic derivative is multiplied 100 times, the product expresses the percentage rate of increase. Thus when dy ^■=Ky, 100- = 100 K ax y is the percentage rate of increase, and is here constant. EXERCISE IV. By one or more of the formulas [I] to [X], differentiate: i , , ,, dy 3 log6 e 3 ra 1. y = log!, x 3 = 3 logft x. , -j- — — - — = — • (tx x x n n \s dy Slogioe,, ,, 1.302... 2. y = (log w x)\ -£ = — ^— (logio xY = f x = -f^ (logio*) 2 = • x "- (logic*)'. |=log, + l. dy _ 1 + log z 3. y = s log x. ~- = log a; + 1. 4. w : zlogz dx (zlogz) 2 6 - » = 1o « srl ■ log {ax - 6) - log (a * + b) - %~ 5CT 6. y = log 1 + _ = log (1 + Vx) - log (1 - Vx). 8. 7/ =6*e* &- (1+ log 6)fc*e*. o i / i i irx dy a x log a + 6* log 6 9. y = log (a« + IF). J! = g a . + 6 x • 10. y = x>5 x . ^ = z*5 4 (5+*log5). 11. y=a* ^ = x*Qogz + l). 12. y = ^. # =x evl±l^. ax x * In some of the examples logarithmic differentiation is employed to advantage; that is, take logarithms first and then differentiate. 58 DIFFERENTIAL CALCULUS Here log y = e x log x :. ^ = e x — + log xe x dx. y x 13. y = x ] °£ x . dy = 2 x log «-» log x • dx. 14. y = log (logs). -£ = — dx x log x 15. z ,oga = a logz . (Differentiate^both members and verify results.) 16. (x + e x Y =2^ + 4 xH x + 6 x 2 e 2X + 4 ze 3X + e 4X . (Do as in 15.) 17. (e x + e~ x y = e 2X + 2 + e~* x . (Do as in 15.) e x — e~ x dv 4 18. 19. y = log e x + e~ x dx (e x + «- x ) 2 e x dy 1 1 + e x dx l+e x 20. y = (log x) x . d £ = (log x) x [^- x + log log x} • 21. Find the slope of the curve y = logio x, or x = lO 2 ', showing that the results are identical. What is the value of the slope at (1, 0)? What is the slope of the curve y = log e x, or x = e y , and its value at (1,0)? 22. Find the slope of the curve x = log e y, or y = e x ; and note that the slope at any point has the value of the ordinate at that point. Value of the slope atz = 0? At x = 1? Atz=-oo? 23. Find the slope of the curve y = - \e " + e <*/ at x = 0. Ans. 0. What is the abscissa of the point where the curve is inclined 45° to the ar-axis ? Ans. x = a log* (l + V2). 24.* Find the value of x when logio x increases at the same rate as x. Ans. x = log 10 e = 0.4343 . . . dx * Since d (logio x) = logio e • — ; x dx=x> d . (1 ° glo3:) = 2.3026 x • d (logio x) ; logio e hence, any number N increases about 2.3 A T times as fast as logio N. When N = 0.4343 . . . ,dN = 0.4343 X 2.3026 d (logio N) = d (logio A). Find how much faster x is increasing than logio x for x = 1. Ans. 2.3026 x = 2.3026 . . . RELATIVE ERROR 59 25. When the space passed over by a moving point is given by s = ae l + be* 1 , find the velocity and the acceleration, showing that the acceleration is equal to the space. 26. Find the slope of the curve y = -\e" — e «/ at x = 0. Ans. 1. X _x ga g a \ 27. Find the slope of the curve y = at x = 0. Ans. -• e a + e a 28. Find the derivative of the implicit function y in (?+ v = xy. Passing to logarithms : i i , dy y (1 - x) 29. x y = y x . Passing to logarithms : y log x = x log y. ^| 30. Find the slope of the probability curve y = e~* 2 . .«~-»>™ t-&g& Ans. -2xe-x\ What is the value of the slope at z =0? Ans. 0. 2 At I = 1? Ans. e 41. Relative Error. — Since when y = f(x), the relative rate of increase of y is dy df(x) . dx _ dx _ f (x) where dy = /' (x) dx ; hence, Ay = f (x) kx, (1) are approximate relations. The relation (1) is 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 relation (2) gives approximately the relative Aty error — -• y 60 DIFFERENTIAL CALCULUS 1. Thus, to find an expression for the relative error in the volume of a sphere calculated from a measurement of the diameter when there is an error in the measurement. Here . A 7 (I \ AD AD Hence, an error of one per cent in the measurement of the diameter gives approximately an error of three per cent in the calculated volume. 2. Again, from the formula for kinetic energy K = \ mv 2 , to show that a small change in v involves approximately twice as great a relative change in K. Here AK v Av n Av K § mv 2 v 3. If a square is laid out 100 ft. on a side and the tape is 0.01 ft. too long, an error of T J^y of one per cent, the relative error in the area is, approximately, M_ ox^ -200-^1- 0002 A ~ 2X x 2 - 2UO (100) 2 "' UUU2 ' or T i o of one per cent. 42. The Compound Interest Law. — The limit (l+-j = e x , in Art. 36, arises in a variety of problems. When a function has the general form y = ae bz , (1) then -^ = bae bx = by; that is, the rate of ch^ge* of the function is proportional to the function ifself. Many of the changes that occur in nature are in accordance with this law, called by Lord Kelvin the compound interest law. It is so called because of the fact that the amount of a sum of money at compound interest has a rate of change at any value proportional to that value, when the change is continuous. THE COMPOUND INTEREST LAW 61 Let A = amount, r = rate per cent, P = principal, and t number of years; then, A = P(l + r) 1 , when interest is compounded yearly; at n equal intervals each year; K" A = lim P (l+^ n J = Pe'<(byArt. 36) (2) is the amount when the interest is compounded continuously. dA Here A = Pe rt and -4r = rPe rt = rA, hence, the rate of dt change of the A is proportional to the value of A , the factor of proportionality being the rate per cent at which the interest is reckoned. As a comparison, it may be noted that $1.00 will amount to $2,594, in ten years with interest at 10 per cent, compounded yearly; while the amount will be $2,718 when compounded continuously. If in A = Pe rt , t increase in any arithmetical progression, whose common difference is h, A will increase in a geometrical progression whose common ratio is e rh ; for if t become t + h, A will become Pe r ( t+h \ that is, Ae rh . Hence A is a quantity which is equally multiplied in equal times. The density of the air towards the sea level from an eleva- tion is a quantity which is equally multiplied in equal distances of descent, for the increase in density per foot of descent is due to the weight of that layer of air which is itself proportional to the density. Many s other instances occur in physics. -~f When bacteria grow freely the increase per second in the number in a cubic inch of culture is proportional to the number present. The relation between the number N and the time t is expressed by the equation, N = Cekt > •'• w = kCekt = kNi (3) 62 DIFFERENTIAL CALCULUS where N is the number of thousand per cubic inch, and k is the rate of increase shown by a colony of one thousand per cubic inch. So many instances of this kind are found in organic growth — where the rate of growth grows as the total grows — that the law is called the law of organic growth, as well as the compound interest law. When a quantity has a rate of change which is proportional to the quantity itself, if the functional relation is expressed by an equation, it must be of the form (1). In the case of the density of the air, the relation (see Art. 226) between the density p and the height h above the sea level is expressed by P = po6 ~ kh > '"' a% = ~ k P» e ~ kh= ~ k P> W where p is the density at the sea level and k is a constant to be determined by experiment. From barometric observa- tions at different altitudes, it has been found that at the height of 3J miles above the earth's surface, the air is about one-half as dense as it is at the surface. Hence, to deter- mine k, p -s.bk — I ; 6 2 5 -3.5 k = log 0.5, or k = log 0.5 -3.5 0.198; •*• P = Poe-°- 198 \ where h is > in miles. (5) Here, as h increases in arithmetical progression, p decreases in geometrical progression, the force of gravity and the temperature being taken constant. The varying density at different heights is found by giving values to h ; thus, making h = 35, gives — =0.001; hence, according to this law at the Po height of 35 miles the density of the air is about one-thou- sandth of the density at the sea level. As the pressure p is THE COMPOUND INTEREST LAW 63 proportional to the density, p = k'p; and the same law holds for the pressure of the air; hence, p = poe- k ' h , :. -| = -kpoe- k ' h = -k% (6) where k f is a constant to be found by experiment. Knowing the pressure at the sea level and observing the pressure at some height, k f is determined; or it can be determined from the value of the pressure at any two differing heights. When the pressure is expressed in inches of mercury in a barometer, the pressure in lbs. per square inch = 0.4908 X barometer reading in inches. Taking p = 30" when h = 0, and p = 24", say, when h = 5830 ft., k' is readily computed. In millimeters the equation is p = 760 e~ h/mo , where h is in meters. The relation between the decomposition of radium and time is expressed by the equation q = qtfr"; .'. J = -kq^~ kt = - kq, (7) where q is the original quantity and q is the quantity remain- ing after a time t. The constant k can be found from the fact that half the original quantity disappears in 1800 years. The relation between the varying difference of tempera- ture of a body and that of the surrounding medium and the time of cooling is expressed, according to Newton's Law, by b = 8oe- kt ; .'. ~ = -k8 e~ kt = -kd, (8) where 8 = r — r , the difference in the temperature of the body and that of the medium, 5 = t\ — r , the difference when t = 0, k a constant; that is, r = r + (ti — r ) e~ kt , where — kt indicates the body is cooling. 64 DIFFERENTIAL CALCULUS TRIGONOMETRIC FUNCTIONS. 43. Circular or Radian Measure. — The formulas for differentiation of trigonometric functions are simpler when the angle is measured in radians than in degrees. Hence, in the formulas that follow, the angle will be in radians. A radian, the unit of circular measure, is an angle which when placed at the center of a circle intercepts an arc equal in length to the radius. 180° Since 2 wr is arc of 360°, a radian equals , or 57.3° ♦ 7T nearly. In circular or radian measure, an angle in radians is equal to the length of the intercepted arc divided by the radius; = -, where 6 is angle in radians, s is number of units in arc, and r is the number of units in radius. Hence, s = rd; that is, in any circle the length of an arc equals the product of the measure of its subtended central angle in radians and the length of the radius. If r = 1, then s = d; that is, the arc and the angle have the same numerical measure. Trigonometric functions are called circular func- tions. 44. Formulas and Rules for Differentiation. — [XI] d* sin 8 = cos 6 d$. The differential of the sine of an angle is the cosine of the angle by the differential of the angle. [XII] dcos6 = - sin0c/e. The differential of the cosine of an angle is minus the sine of the angle by the differential of the angle. [XIII] dtanG = sec 2 6d9. The differential of the tangent of an angle is the secant squared of the angle by the differential of the angle. [XIV] d cot 9 = - cosec 2 9 by sin = 2 sin sin A0 A0 2 cos (•+¥> Trig.; [(Art. 46). = 1, as A0 = 0; 66 DIFFERENTIAL CALCULUS A0^ sin -7T = COS i d0 a^oL^J a^o A? I V 2/ ,\ cfy = d sin = cos dd. Corollary. — d covers = d (1 — sin 0) = — cos dd. Now let z = cos = sin (~ — 0j ; d# = dcos0 = dsinl^ — 0j = cosf^ — djdl- — 6j; '. d cos = — sin d0. Corollary. — d vers = d (1 — cos 0) = sin c?0. LJj (=!) = , Let be the number of radians in the angle NO A, where the angle is taken acute; by Ge- ometry, if AT and BT are tan- gents at A and B, then, chord AB < arc AB < AT + BT, and therefore MA < arc iVA -"->cos0. sin cos Thus the ratio in sin lies between 1 and cos 0. LIMIT OF RATIO OF INFINITESIMALS 67 When approaches as its limit, cos approaches 1 as its limit; therefore, also sin 0/0 approaches 1 as its limit. (See Art. 215, Ex. 1.) Corollary. — Since lim — — = 1, 0=0 .. 2 MA .. chord AB t ,. M , , hm A ^ = hm j^- = 1. (Art. 22, also.) It may be noted that, MA sin „,,..«, since -ryF = t — « = cos 0, when = and cos 0=1, A T tan sin0 and tan approach equality; and, since the arc is intermediate in value between sin and tan 0, the three functions approach equality as the angle nears zero. So lim ( - — - ) = 1 and lim [— — ) = 1, as well as hm 6 =o \tan0/ 0=0 V / 0=0 /sin 11 V~T = 1. These are fundamental examples of the ratio of infini- tesimals approaching a constant value as a limit. Con- sider again the equality of ratios, -t-=- = -^nr' Suppose the points A and B approach N; so long -as A and B are not coincident, that is, so long as AB is really a chord, the equality still exists. The ratio MA : A T may be considered a function of OM, or equally well a function of the angle NO A. As 03/ approaches ON as its hrnit, or as the angle NO A approaches zero as a limit, the ratio MA : AT ap- proaches 1 as its limit. The nearer OM gets to ON, or the nearer A gets to N, the nearer does the ratio MA : A7 7 get to unity. The crucial fact is that the reasoning is vitiated if OM becomes actually equal to ON; for then the triangles will cease to exist, the terms of the one ratio will be zero and those of the other will be identical, and the equation on which the reasoning is based could not be established. 68 DIFFERENTIAL CALCULUS 47. Derivation of [XIII]. — Since tan0 = -, dtsaid = d( -J; Vcos0/' d tan = COS0' cos d sin — sin d cos cos 2 (cos 2 + sin 2 0)d0 cos 2 48. Derivation of [XIV]. - Since cot0 = tan = sec 2 dd. %->)■■ /. dcot0 = dtan(| - 0) = sec 2 (| - fl)d fe - d) = — cosec 2 0d6 49. Derivation of [XV]. — Since sec0 = COS0 sin c?0 d sec = d = — cos 2 = sec tan d0. \cos 0/ = £ 50. Derivation of [XVI]. - Since cosec = sec (^ — 0] ; /. dcosecfl = dscc(| - 0) = sec/| - 0) tan fe - d) d fe - d) = — cosec cot dd. Note. — In the derivations of the formulas for the cosine, cotangent, and cosecant, as given, it may be noted that, as in the last example of Art. 31, the formula for the derivative of the function of a function has appropriate application. Thus for cos0, let x = cos0 = sin(~ — 0J and = ^ — 0, NOTE ON [XI] 69 dx dx d then, or Is .'. -r n cos 6 = — sin 6 or d cos = — sin B dQ* dd In practice the actual substitution of the auxiliary symbol may be dispensed with. The formula applies to such functions as y = sin (ax + b) . Thus put z = ax + b, making y = sin z; then dy _ dy dz dx dz dx or -=- sin (ax + 6) = -=- sin 2 X -=- (ax -f- 6) = cos 2 • a: .*. -=- sin (ax + 6) = a cos (ax + 5) or d sin (ax + b) = a cos (ax + 6) dx. Again, let the function be y = sin 2 (ax + 0) and put z =» sin (ax + b) ; then ^ sin 2 (ax + 6) = — (z 2 ) . — sin (ax + 6) = 2 2 X a cos (ax + 6) = 2 a sin (ax + 6) cos (ax + 6) ; .'. d sin 2 (ax + 6) = 2 a sin (ax + 6) cos (ax + b) dx. 51. Note on [XI]. — If the angle is measured in degrees, then d sin 6 = — cos 6 d0, since degrees is r-^r radians lot) loU and sin0° = sin^^j; d sin C = dsin &) 7T /7T0 r \ -„ 7T „ ,„ = T80 COS (lW cW= 180 COS< ' d( >- 70 DIFFERENTIAL CALCULUS It is thus seen that the formulas for differentiation of the trigonometric functions are simpler when the angle is measured in radians than when measured in degrees. For the same reason that Napierian or natural logarithms are employed in differentiation, radian or circular measure is used for angles of the trigonometric functions, when differ- entiation is to be done. 52. Remarks on [XI]. — The fundamental limit of Art. 46, lim(-T— )= 1 means that when is a small number 0=0 V / sin is approximately equal to 0. For angle of 1°, = j^: = 0.0174533 . . . , sin = 0.0174524 . . . ; so "tney are equal to five decimals. Of course for angle of 1' or 1", they are equal to a great many more decimals, but theyaTe never exactly equal however small the angle may be, since the sine is always less than the arc. Since d sin = cos dd, if the value 0° is taken for and dsind = cos0°~ = .0174533 ... or ^^ = 1, A sin = sin (0° + A0) - sin 0° - .0174524 . . . = sin 1°; A sin B _ .0174524 . . . .. A sinfl = rising _ 1 A0 " .0174533 . . . ' US A0 dd 6=0° From , = cos0= cos0° = 1, it is seen that, at = 0°, do the sine of is changing at the same rate as is changing; so the slope of the curve y = sin is unity at the origin, and the tangent to the curve at that point makes an angle of 45° with 0-axis. The conditions are the same at = 2 -k. As " do IT = cos0 = cos 90° = 0, at = ^, the sine of is not changing, REMARKS OX [XI] 71 the rate being zero, and the tangent to the curve at that point is parallel to 0-axis. At 6 = it, cos 180° = — 1, so the sine of is decreasing, at that value of 0, at the same rate as is increasing, and the tangent to the curve at that point makes angle of 135° with 0-axis. Thus the rate of change of the sine of 0, at any value of 0, can be found ; and the difTeren- sin 6 tial of the sine, the change if the change became uniform, will always differ from the increment, the actual change of the sine, when the angle is given an increment. In taking sines or other functions from tables by interpolation, the changes are assumed as uniform within allowable limits of error. EXERCISE V. 1. y = sin x-. 2. y = sin 2 x. 3. y = cos ax. 4. y = /(0) = tan w 0. 6. f{6) = tan30 +sec30. 6. / (as) = sin (log ax) . 7. f(x) = log (sin ax). Q sin x + cos x 8. y= 9. /(0) = log (tan ad). 10. f(d) = log (cot 00). 11. /(0) = tan (log 0). 12. f(d) = log (sec 0). dy = 2 x cos x 2 dx. dy = 2 sin x cos x dx = sin 2 x dx. dy = — a sin ax dx. -M =f (0) =m tan™" 1 sec 2 0. fix) fix) dy = dx r® f(e) f'(0) = 3 sec 2 3 + 3 sec 3 tan 3 0. = 1/xcos (log ax). = a cot ax. 2sinz 2a 2a ~ 2 sin (off) cos (ad) ~~ sin (2 ad) = - 2 a/sin (2 a0). = l/0sec 2 (log0). /' (e) = sec tan i sec0 = tan0. 72 DIFFERENTIAL CALCULUS 13. / Or) = x sin x . f (x) = x sin x (sin x/x + log x • cos x). 14. / (re) = (sin x) x . f (x) = (sin x) x {x cot x + log sin x). 15. / (0) = (sin 0) tan e . f (0) = (sin 0) tan e (1 +sec 2 log sin 0) . 16. /(0) = | tan 3 0- tan 0+0. /' (0) = tan 4 0. „ , , -(sec Vl - x) 2 3 {X) 2Vl-x r . f{x) =« tan Vl -x. >•/«»= log \/f^- /' (0) = esc 0. By differentiation derive each of the following pairs of identities from the other: 19. sin 2 = 2 sin cos 0, cos 2 s cos 2 - sin 2 0. on . na 2tan0 . 1 - tan 2 •20. sin20 = — — — — , cos20 = r— — — ~ 1 + tan 2 1 + tan 2 21. sin 3 = 3 sin - 4 sin 3 0, cos 3 = 4 cos 3 — 3 cos 0. 22. sin (m + n) s sin to0 cos ri0 + cos md sin ri0, cos (m -\-n)6 = cos md cos n9 — sin m0 sin n0. 23. If vary uniformly, so that 360° is described in w seconds, show that the rates of increase of sin 0, when = 0°, 30°, 45°, 60°, 90°, are respectively, 2, V3, VS, 1, 0, per second. (See figure, Art. 52.) 53. The Sine Curve or Wave Curve. — The locus of the equation y = sinx, (1) where x is an angle in radians, is called the sine curve, from its equation, or the wave curve, from its shape. The maximum value of y is called the amplitude, being unity in (1); and, since the curve is unchanged when x + 2 it is substituted for x, the curve y = sin x is a periodic curve with a period equal to 2 ir. (See figure, Art. 52, and figure, Art. 73.) The more general form of the equation is y = a sin mx, (2) where a is the amplitude and — is the period, m a constant. The curve is called the sinusoid also, and is of great importance, since it is the type form of the fundamental waves of science; such as, sound waves, vibrations of rods, DAMPED VIBRATIONS 73 wires, plates and bridge members, tidal waves in the ocean, and ripples on a water surface. The ordinary progressive waves of the sea are not of this shape, as they have the form of a trochoid. 54. Damped Vibrations. — When a body vibrates in a medium like a gas or liquid, the amplitude of the swings get smaller and smaller, or the motion slowly (or rapidly in some cases) dies out. Thus, when a pendulum vibrates in the air the rate of decay of the amplitude is quite slow; but when in oil the rate is rapid. The ratio between the lengths of the successive amplitudes of vibration is called the damping factor or the modulus of decay. In all such cases the amplitude of the swings differ by a constant amount or the logarithmic decrement is constant. Hence the amplitude must satisfy an equation of the form A = ae~ bt , (1) where A is the amplitude and t the time. The actual motion is given by an equation of the form v y = ae~ bt sin cot, where a> = - is a constant. (2) (See Art, 73.) * In plotting a curve whose equation is of this form, say, 7T y = e-< x sin-x, (3) much is gained by the following considerations: 1. Since the numerical value of the sine never exceeds unity the values of y in (3) will not exceed in numerical value the value of the first factor e~* z . As the extreme values of sin|7rx are +1 and —l,y has the extreme values e~ ix and — e~* x . Hence, if the curves y == g-k* and y = —e~ ix (4) * This illustration is given substantially in Smith and Gales's New Analytic Geometry. 74 DIFFERENTIAL CALCULUS are drawn, the locus of (3) will lie entirely between these curves. They are called boundary curves, and they are plotted by three or more points, the second being symmetri- cal to the first with respect to the z-axis. 2. When sin f irx = 0, then in (3) y = 0, since the first factor is always finite. Hence, the locus of (3) meets the a:-axis in the same points as the sine curve y = sin | irx. (5) 3. The required curve is tangent to the boundary curves when the second factor, sin J irx, is +1 or — 1; that is, when the ordinates of the curve (5) have a maximum or a minimum value. The tangency is proven by finding the derivative of y in (3) and noting that, when sin \ irx is +1 or — 1, it will be the same as the derivatives of y in (4) . Hence, the slopes of the curves and the ordinates being equal for the same values of x, the required curve is tangent to one or the other of the boundary curves for those values of x that make sin \ irx = + 1 or —1. Thus, differentiating (3) and (4) gives dy 1 , . 1 . v . 1 j- = — t e_i sin 7T irx + - e _i x cos s irx dx 4 2 2 2 = — i e -i z , when sin \ irx = 1, = \ e _i x , when sin \ wx = — 1 . For the sine curve (5) the period is 4 and the amplitude is 1. This curve is the broken line of the figure. The locus of (3) crosses the x-axis at x = 0, ±2, ±4, ±6; DAMPED VIBRATIONS 75 etc., and is tangent to the boundary curves (4) at x = ±1, ±3, ±5, etc. The discussion having disclosed these facts, the curve is readily sketched, as in the figure; that is, the wind- ing curve between the boundary curves (4). A more general form of the equation of a damped vibration is y = ae~ bt sin (ut — a), where a = — is constant. (3') This equation may be written either (see Art. 73) y = e~ bt (A sin kt + B cos kt), where A and B are constants, or y = A sin (cot — a), where A = ae~ ht . (3") Here A is a variable decreasing amplitude, whose relative rate of decrease is — dA/dx -s- A = b; that is, the relative rate of decrease of A is constant. The successive derivatives from (3') are (by Art. 68) : dy -JL = ae~ ht [— b sin (ut — a) + co cos (at — a)], d 2 y —jL = ae -bt [52^2 sm ( w j _ a ) _ 2 oo) cos (ut — a)], whence it follows that Equations which contain derivatives or differentials are called differential equations. The equation (4') is the funda- mental differential equation for damped vibrations. The term in -r-, or v, proportional to the velocity, occurs in equa- tions for vibration only when damping is considered. Vibra- tions are cases of simple harmonic motion — damping being caused by resistances, such as friction, etc. Simple har- monic motion is treated in Art. 73. 76 DIFFERENTIAL CALCULUS INVERSE TRIGONOMETRIC FUNCTIONS. 55. Formulas and Rules for Differentiation. — The direct trigonometric functions are single-valued but the angle has to be restricted to a certain range in order that the inverse functions may be single valued. To make the inverse functions single-valued, the angle denoted by- sin -1 x, cosec -1 x, tan -1 x, cot -1 x, covers -1 x, is taken to lie between — - and ~ , and the angle denoted by cos -1 x, sec -1 x y vers -1 x } to lie between and t. Thus ^=5 =oos_1 ("2 ? ) ; sinr " 1 (— I) — I- V 2 / 6 ' sin -1 x + cos -1 x = ■= ; tan -1 x + cot -1 x = = , if x be positive, = -r, if x be negative. These restrictions will be assumed in the following formulas, and all will be expressed in terms of the letter x. While the symbols sin -1 x and arc sin x are both used to denote the angle whose sine is x, in writing the formulas the notation sin -1 x is preferable. [XVII] d sin -1 x = dx ■ Vl- x 2 The differential of an angle in terms of its sine is the differen- tial of the sine divided by the square root of one minus the square of the sine. [XVIII] dcos -1 *- — — V1-.T 2 The differential of an angle in terms of its cosine is minus the differential of the angle in terms of its sine. [XIX] < /tan -1 ^ = T 4 :L -2- 1 + .Y* 5 DERIVATION OF [XVII] AND [XVIII] 77 The differential of an angle in terms of its tangent is the differential of the tangent divided by one plus the square of the tangent. dx [XX] dcotr 1 *- -TT^' 1 + x 2 The differential of an angle in terms of its cotangent is minus the differential of the angle in terms of its tangent. [XXI] d sec" 1 x = — = . xVx 2 -1 The differential of an angle in terms of its secant is the differ- ential of the secant divided by the secant and the square root of the square of the secant minus one. [XXII] d cosec" 1 x = dx . The differential of an angle in terms of its cosecant is minus the differential of the angle in terms of its secant. dx [XXIII] rivers" 1 * = V2x-x 2 The differential of an angle in terms of its versine is the differential of the versine divided by the square root of twice the versine minus the square of the versine. dv [XXIV] d covers" 1 x = ax V2x-x 2 The differential of an angle in terms of its coversine is minus the differential of the angle in terms of its versine. 56. Derivation of [XVII] and [XVIII]. — Let 6 = sin -1 x ; then sin 6 = x, the differential of which by [XI] is cos dd = dx; dx dx dx dd = cos 6 Vl - sin 2 d Vl -x< dx Now d cos -1 x = d ( - — sin -1 x ) = — 78 DIFFERENTIAL CALCULUS 57. Derivation of [XIX] and [XX]. — Let = tan -1 x; then tan = x, the differential of which by [XIII] is sec 2 Odd = dx; fiR - dx dx dx " ~ sec 2 !? ~~ 1 + tan 2 ~ 1 + x 2 ' Now d cot -1 x = d ( - — tan -1 x) = — .. . 2 - 58. Derivation of [XXI] and [XXII]. — Let = sec -1 x; then sec 6 = x, the differential of which by [XV] is sec0tan0d0 =dx; ... dd = dx = dx sec tan x Vsec 2 0—1 dx d cosec -1 x = d f s — sec -1 x) = x Vx 2 - 1 Now \2 I x Vx 2 - 1 59. Derivation of [XXIII] and [XXIV]. — Let = vers -1 x; then vers d = x, the differential of which by Cor., Art. 45, is sin0d0 = dx; in — dx _ dx _ dx " sin0 " Vl - cos 2 ~ Vl - (1 - vers 0) 2 dx dx Vl-(l-x) 2 V2x-x 2 Now d covers -1 x = d [ = — vers -1 x ) = . \2 / V2x-x 2 EXERCISE VI. - , • _, x d (x/a) 1. d sin *- — ■ , = = dx « Vl - (x/a) 2 Va 2 - x 2 , ,x — dx ,, _.x adx . 2. d cos -1 - = , ; a tan J - = ,, , ., a Va 2 — x 2 a a 2 -f x- , ,x —adx , _, x adx d cot -1 - = -=—, — s ; d sec a a 2 + x 2 ' « x Vx 2 - a 2 ' . , x —adx , _, x dx d esc -1 - = , ; a vers x - = a ^ Vx 2 - a 2 ' a V2 ax - x 2 Note. — These may be considered standard formulas. EXERCISE VI 3. y = tan x tan -1 x. * -i 2x 4. y =tan» r ^ ;• 5. ?/ = sin l 15. ?/ = tan 1 16. ?/ = arc sin Vl- 2 dx ' sec 2 x tan -1 x -\- tanx 1+x 2 dy _ dx 2 (1 - x 2 ) 1 + 6 x 2 + x 4 dy _ 1 V2 dx Vl-2x-x ■ ,/-■ — d V Vl + 6. y = arc sin V sin x. ^- = ^ 1. y = x sin_1 x . . cscx. dx dy dx = 3*11-1 x / sin tg logx V V x Vl-x 2 / di/ n 8. y = tan"* (n tan x) . _ - ____. e * _ e -* d y _ -2 9. y = arc cos ^j—r x - ^ - ^+^' 2x 2 dy 2 10. y = arc vers j-— ^ = r+ — 2 - 3a; -x 3 dy _ 3 11. y = arc tan ^ _ »^ » 12. j/ ^arcsinj-q^- x + a 13. w = arc tan ■= 9 1 — ax dy 14. = arc tan -j- • dx 1+x 2 dy _ dx -2 1 + x 2 dy _ dx d 1 1+x 2 d*y dx 2 dx dy "Off 1 dx Vl -x 2 dy _ -2 e z + e~ x dx. e x + e" 17. y = arc tan (sec x + tan x). ~d = 2 = 1. . /sinx — cosx\ dj/ 18. y -arc sin ^ ^= J- ^ ,3*-2 , . ,3x-12 dy rt A e^ 4- e _ax ^ _ 2a 20. y = arc cot ^ x _ e -a« ' ^x " e 20X + e^ 2 "*" 80 DIFFERENTIAL CALCULUS 21. What is the slope of the curve y = sin x ? Its inclination lies between what values? What is its inclination at x =0? What at x=v/2? The slope = cos x; hence, at any point, it must have a value between — 1 and +1, inclusive. Hence, the inclination of the curve at any point is between and 7r/4 or between 3 7r/4 and r, inclusive. (See figure, ,Art. 52.) 60. Hyperbolic Functions. — These are certain functions, recognized as far back as 1757, that have been introduced in recent years, and that are coming more and more into use. As the trigonometric functions are called circular because of their relation to the circle, the hyperbolic are so called because of their relation to the rectangular hyperbola, the relations being in some respects the same. The functions are analogous to the trigonometric functions and their names are the same. They are the hyperbolic sine, cosine, tangent, etc., and they are defined as follows: sinh x = = (e 2 e~ x ), cschz = 1 coshz tanh x = 2 (e* + e-»), e x _ e -x sechz coth x = sinhz 1 cosh x ' e x + e~ x e x _|_ e - x e x _ e - 61. General Relations. — Besides the reciprocal rela- tions given above, the same as those between the circular functions, there are analogous relations : cosh 2 x — sinh 2 a: =1; 1 — tanh 2 x = sech 2 z; coth 2 x — 1 = csch 2 x; sinh 2 x = 2 sinh x cosh x; cosh 2 x = cosh 2 x + sinh 2 x = 2 cosh 2 -1 = 1+2 sinh 2 x. 62. Numerical Values. Graphs. — The sine may have any value from — oo to oo the cosine any value from 1 to / j/= tanh j? INVERSE FUNCTIONS 81 oo ; the tangent any value between —1 and 1, and the lines whose equations are y = ± 1 are asymptotes to the graph of tanh x. The graphs of the sine and cosine are both asymp- totic to the graph of y = J e x . 63. Derivatives. — Since -j- e* = e* and -5- e -1 = — e -1 , ax ax by differentiating the hyperbolic functions as functions of x, the several derivatives are readily found to be as follows: j- sinh x = cosh x; -7- cosh x = sinh x; -j- tanh x = sech 2 x; -7-cothx = — cosech 2 x; -r-cosechx = — cosechxcothx: ax -j- sech x = — sech x tanh x. ax The differentials are given at once by the derivatives, or vice versa; thus, d sinh x = cosh x dx, and so for the others. 64. The Catenary. — The curve y = cosh x = \ (e x +e~ x ) is called the catenary and is important because it is the curve of a perfectly flexible and inextensible cord between two points, and is the curve that a material cable when hung between two supports is assumed to take. Since -j- cosh x = sinh x, -p = 75 (e x — e~ x ) = sinh x is the slope of the catenary. The general equation of the catenary is ?/ = a cosh - = 5 \e a + e a /, where a is the distance from the origin to the lowest point of the curve. (See Art. 146.) 65. Inverse Functions. — The inverse functions are useful when expressed as logarithms. If y = sinh -1 x, the logarithmic form of y is found from x = sinh?/ = h(e y - e - *), 82 DIFFERENTIAL CALCULUS which is reduced to e 2 V - 2xe» - 1 = 0; solving as a quadratic gives e v = x ± vV + 1; but as e y is always positive, e v = x + Vx 2 + 1 ; .*. sinh -1 x = y = log (x + Vz 2 + 1 ) . In the same way is found, cosh -1 x = log (x =b Vx 2 — 1 ) . Since (a - V^=l) - (aj + v? _ I) i /. log (a - Vs 2 - 1 ) = -log (x + Vx 2 - l). For each value of z greater than 1 there are two values of cosh -1 x, equal numerically but of opposite sign. In the same way again is found, 1 1 + x tanh" 1 x = ~ log , if x 2 < 1 ; COth -1 £ = jrlOff 7, if X 2 > 1. 2 x — 1 66. Derivatives of Inverse Functions. — The derivatives of the inverse functions are found by differentiating their logarithmic forms, using formula -r- log x — — The derivatives, taking - instead of z, are: d . . , x 1 t- sinh -1 dx a Vx 2 + a 2 ' d u i x ! -s- cosh -1 - = ± , ; dx a V^ 2 - a 2 3- tanh -1 - = -= r, (x 2 < a 2 ) ; dx a a 2 — x 1 -7- coth -1 - = —„ 5 , (x 2 > a 2 ). dx ax 2 — a 2 DERIVATIVES OF INVERSE FUNCTIONS 83 Since the inverse cosine is not single-valued, for the positive ordinate of cosh -1 '- c is used instead of x, X X ordinate of cosh -1 L , the + sign must be taken. When - a a . , , x , x + Vx- + a 2 sinn -1 - = log a & a = log (x + Vx 2 + a 2 ) — log a, x so the derivative of sinh -1 - is the same as that of log Qj (x + Vx 2 + a 2 ), since d (log a) = 0. The divisor a occurs in the logarithmic form of cosh -1 - also, so its presence should be borne in mind when comparing the same result expressed in logarithms and in inverse hyperbolic sines or cosines. The relation of the inverse hyperbolic sine to the equi- lateral hyperbola is shown in Art. 137, and the inverse func- tions are again considered in Art. 120. CHAPTER III. SUCCESSIVE DIFFERENTIATION. ACCELERATION. CURVILINEAR MOTION. 67. Successive Differentials. — It is often desired to differentiate the differential of a variable or to get the deriv- ative of a derivative. For, while the differential of the in- dependent variable, being arbitrary, is usually supposed to have the same value at all values of the variable and hence to be a constant, the differential of the dependent variable, except when the function is linear, is a variable, subject to differentiation. The differential of dy is called the second differential of y; the differential of the second differential of y is called the third differential of y; and so on. d(dy) is written d 2 y; d(d 2 y) or dd dy, is written d 3 y; and so on. The figure written like an exponent to d denotes how many times in succession the operation of differentiation has been performed, dy, d 2 y, d 3 y, . . . d n y are called the successive differentials of y. Example. — The successive differentials of y when y = ax 3 : dy = 3 ax 2 dx ; d 2 y = Sadx • d (x 2 ) = 6 ax dx 2 ; d 3 y = Qadx 2 • dx = 6 a dx 3 ; d A y = d (6 a dx 3 ) = 0. The independent variable being x, dx is treated as a constant. Note that according to the notation adopted d 2 y = ddy; dy 2 =(dy) 2 ; d(y 2 ) = 2ydy. 68. Successive Derivatives. — The derivative of the first derivative of a function is called the second derivative of the function; the derivative of the second derivative is called the third derivative; and so on. 84 SUCCESSIVE DERIVATIVES 85 When x is independent, d dy = d 2 y d_ d 2 y __ dfy d d n ~ l y _ d n y dx dx ~ dx 2 ' dx dx 2 ~ dx 3 ' ' dx dx n ~ l ~ dx n The successive derivatives of / (x) are denoted by fix), /"«, r(x), r(*)> ■ . . ,f n (x)> Thus if / (x) = x\ f (x) = 4 x 3 , f" (x) = 12 x 2 , }"'{x) = 24 a;, / IV (x) = 24, /* (x) = 0. Hence, if y — J (x) and x is independent; dx J {X) ' dx 2 J Wi ' ' ' ' dx n J {X) ' The nth derivative of some functions can be easily found by inspection of a few of the derivatives. Example 1.— f(x) = e x ,f'{x) = e*,f"(x) = e*,/'"(x) = e* . . . , /. /» (x) = e*. This function e x is remarkable in that its rate of change, or derivative, is equal to the function itself. Example 2.-/(0) = sin 0, /' (0) = cos 0, f" (0) = -sin0, /'" (0) = - cos 0, / IV (0) = sin 0. /' (0) = cos = sin (# + 1) , r (0) = cos (* + !) = sin (* + 2 • |) , /'" (0) = cos (0 + *) = sin(0 + 3.^) . . . ; /./"(0) = sin(0 + n.g- Each of the successive derivatives of f (x) equals the x-rate of the preceding derivative, for f n (x) = -p f n ~ l (x) = the x-rate off~*(x). Corollary. — / n_1 (x) is an increasing or a decreasing function of x according as f n (x) is positive or negative, and conversely. Note. — The tangential acceleration is, _ dv _ d ds _ d 2 s at ~~dt~dtdt~M> 86 DIFFERENTIAL CALCULUS and the flexion is , _ dm _ d dy _ dfy dx dx dx dx 2 ds (See Art. 12 and Art. 13.) Hence, the speed -r- is increasing d 2 s or decreasing according as the acceleration -p is positive or negative, and the slope is increasing or decreasing according as the flexion -=2 is plus or minus. When the second derivatives are equal to zero, the first derivatives are constant, or conversely. (See Art. 13.) 69. Resolution of Acceleration. — An acceleration, like a velocity, being a quantity which has magnitude and direction, may be represented by a straight line, that is, by a vector. In general the acceleration a at any point (x, y) of a curvi- linear path may be resolved into two components in given directions. The directions usually taken are along the tangent and normal at the point, and in directions parallel to rectangular axes OX, OY. With the notation of the figure for (d), Art. 10, the com- ponents parallel to the axes being the rates of change of dx/dt and dy/dt will be denoted by. d 2 x/dt 2 and d 2 y/dt 2 , respectively. The rate of change of the velocity is the resultant acceleration To find the component acceleration a t along the tangent at P; resolve the axial accelerations along the tangent, giving for the sum of tangential components, d 2 x d 2 y . d 2 x dx d-y dy d 2 s a I = ^cos4> + ^sin0=^.^ + (W2 dg ^ EXERCISE VII 87 by differentiating, (dA^/dxY./dyV \dt) \dtj ^{dtj at = di? = Tt = rate °^ cnan & e °^ tne s P ee d. Hence, the tangential component is the same as for recti- linear motion. EXERCISE VII. Find dy, d 2 y, dhj, when: 1. y = 2 x 5 - 5 x 4 + 20 x 3 - 5 x 2 + 2 x. dhj = 120 (x 2 - x + 1) dx 3 . 2. y = a? log (x - 1). d*y = 2 ^ ~ 3 * + 3) dx\ (x - l) 3 3. y = (x 2 - 6 x + 12) e x . d 3 */ = x 2 e* dx 3 . 4. i/ = log sin x. d 3 ?/ = 2 cos x sin -3 x dx 3 . 6. ?/ = tan x. d 3 ?/ = (6 sec 4 x — 4 sec 2 x) dx 3 . Find the successive derivatives: 6. /(x) =x 6 + 4x 4 + 3x + 2. / VI (s)=[6, /▼* fcc) = 0. 7./ (x) = log (1 + x); find nth derivative. f (x) = (1 + x)~\ f" (x) = (-1) (1 + x)" 2 , f'"(x) = (-1)2[2_(1 +X)" 3 , / IV (X) = (-1) 3 [3(1 +*)"«, . . . .'. r (x) = (-l)"" 1 ln-1 (1 + x)-\ 8. / (x) = x 3 log x. / IV (x) = 6 x" 1 . 9. y = log (e* + e~ x ). p{ = -8 , C * ~ e ~* . y & v ' dx 6 (e x + e - *) 3 10. Find formula, known as Leibnitz's theorem, for d n (uv). Let u and v be functions of x; then d (uv) = dwv + u dv, (1) d 2 (uv) = d 2 u • v + dudv -\-dudv -\-u d 2 v = d 2 wv-{-2dudv -\- ud 2 v; (2) .'. d 3 (uv) = d*wv + 3d 2 udv + 3 du d 2 v + u d 3 v. (3) The coefficients and exponents of differentiation are according to the Binomial theorem, however far the differentiation is continued; /. d n (uv) =d n u-v + nd n ~ l udv+ n(w j~ 1} d'^u d 2 v + • • . + udu d n ~ l v + ud n v. 88 DIFFERENTIAL CALCULUS 11. Ifz° + 2,* = a°, g- 13 - *h+& dx 2 1 **- ' dx 2 a* Z/ 3 a 2 ?/ 3 12. If 2/2 = 2p x, 14. If-„-^ = l 6- dx 2 d 2 ^ dx 2 ' y 3 ¥ a 2 y 3 ' 70. Circular Motion. — When a point describes a circle of radius r with constant speed v, it has a constant accelera- tion v 2 /r directed towards the center of the circle. Let PT be the velocity at P, and P\T\ that at P±. A velocity being a directed quantity may be represented by a vector; that is, by a straight line whose length denotes magnitude and whose direction is the given direction. Hence from a common origin o, the vectors op and opi are drawn equal to the vectors PT and PiTi, respectively. Since the speed is constant each vector is increment, denoted by At;. The Ay A* directed along pp h and is laid off as pm. As A£ approaches zero, Pi approaches P, and pi approaches p along the circular arc indicated by the dotted line; pm approaches a vector pt directed along the tangent to the arc v, and ppi is the vector average acceleration for the interval of time A£ is the lim A<=0 m represents the accelera- ppi at p. This vector, dv tion 37 of the point P moving in the circle of radius r; and since the direction is at right angles with the tangent at P, the acceleration is directed towards the center 0, is normal acceleration, therefore, denoted by a n . To find the magni- tude of the normal acceleration a n \ since the sectors popi and POPi are similar, the angles at o and being equal, CIRCULAR MOTION 89 arc ppi arc PP\ arc ppi As . op OP v r arc ppi = ?£, and UmP^Ulimrffl r At' A< -oL At ] r M^lAtj re ppi by its chord, ,. Tchorclppil _ dv _ v a<=o L <^t dt r At replacing the arc ppi by its chord, (Art. 22.) ds dt dv v 2 Otherwise, it may be seen that while the point P describes the circle of radius r, the point p describes the circle of radius v, the velocity of p in its path being the acceleration of P in its path. Since the circles are described in the same time, the velocities are to each other as the paths of the two points, or as the radii of the circles. velocity of p velocity of P . v r velocity of p is -j , rate of change of velocity v of P, a n v v 2 — = - i a n = — • v r r Since the speed is constant, the rate of change a t is zero, v" cit 2 + a n 2 = a n = - ; that is, the total acceleration r is the normal acceleration, the change of velocity being change of direction only. Since 4?rV s = 2irr = vt; a = -^-, (2) where T is time of a revolution. Note. — By Newton's Second Law the measure of the W force on a moving body is — a (Art. 71); hence, the force acting on a body weighing W lbs. revolving in a circle of 90 DIFFERENTIAL CALCULUS Wv 2 radius r is pounds of force, is directed towards the gr center of the circle, and is called centripetal force. The reaction of the body to this force is by the Third Law equal in magnitude and opposite in direction. It acts upon the axis or upon whatever deflects the body from its otherwise rectilinear path, and has been called the centrifugal force, although a misnomer. The centripetal force is the active force, the other is the equal and opposite reaction and should be called the centrifugal reaction, since it is the resistance which the inertia of the body opposes to the force acting upon it. 71. The Second Law of Motion. — According to New- ton's Second Law of Motion the rate of change of momentum of a body is proportional to the resultant of the impressed forces acting on the body. Let a body of standard weight W be moving with velocity v, then Wv = momentum of the body; • tne acceleration of gravity. 92 DIFFERENTIAL CALCULUS The absolute unit of force is thus, that force, which acting on the unit of mass (or weight) for the unit of time, generates the unit of velocity. The absolute unit of force is thus 1/g of a pound avoirdupois, about \ of an ounce, and F is given in this unit when in F = ma, m is expressed in pounds, the unit being a pound. The ordinary unit of force, sometimes called the Engineer's unit, is one pound and is g times the absolute unit used in Physics. Newton's Second Law of Motion gives as a definition of force: force is the time-rate of change of momentum. Using the much abused term "mass," the definition is: the force is the product of the mass times the acceleration. From W F = ^a; (2) 9 Wd?s Wv* 6 dt*> tn gr w WdH W-rate of E is momentum; dE dv „ that is, the time-rate of E is product of force and velocity. 72. Angular Velocity and Acceleration. — When a body is rotating about an axis the amount of rotation depends upon the time; so if 6 is the angle through which any line in the body, intersecting the axis at right angles, turns, then ANGULAR VELOCITY AND ACCELERATION 93 gives the amount of rotation and is a function of the time t. Thus in the case of a wheel the rotation is measured by the angle through which a spoke turns in a time t. The rota- tion is uniform if the bod}- rotates through equal angles in equal intervals of time. The rate of rotation or the rate of change of the angle is the angular velocity or speed and is denoted by co. If the rotation is uniform, the angular velocity is constant a and co = - , 9 being in radians ; hence, if the uniform rate of rotation is co radians per second, the body rotates through cot radians in t seconds of time. If the rotation is not uniform the rate at which the body is rotating at any instant is the angular velocity at that time, . .. A0 dd and co = lrm — = -=r • a<=o & dt This expression for angular velocity is general and is applic- able when the rotation is uniform also; for then, _ 6 _ Id _ dd W ~ t ~ \t ~ dt' although, the ratios being constant, no limit is involved Similarly, the a acceleration ; and Similarly, the angular acceleration is a = - , for constant = dco = d (d&\ = ) dt or d?s = m dt 2 r dF i tangential acceleration a t = ra, the relation between tangential acceleration and angular acceleration when a is the angular acceleration of a particle at a distance r from the axis of rotation. dv 2,.,2 Since = ro)- dt~r normal acceleration a n = rar. 73. Simple Harmonic Motion. — If a point move uni- formly on a circle and the point be projected on any straight line in the plane of the circle, the back-and-forth motion of the projected point on the given straight line is called simple harmonic motion. It is denoted by the letters S. H. M. Let the point P move upon the circumference of a circle of radius a with the uniform velocity of v feet per second, so SIMPLE HARMONIC MOTION 95 that the radius OP rotates with uniform angular velocity at the rate of - = co radians per second. The projection, P', a of P on the vertical diameter, moves up and down. Let 6 be the angle that the radius makes with the z-axis, then if the point P was at A when t = 0, the displacement OP' = y is given by y = a sin 6 = a sin ut. If the point P was at P when t = 0, and at A when t = t , vt then y = a sin (ut — a), where a = ut = — • (1) When the displacement at time t is given by (1) the motion is S. H. M. Hence, the point P' describes S. H. M. The velocity of a point describing S. H. M. is, from (1), dy , ± x -j7 = aw cos (cot — a) , and the acceleration is d 2 y 2 . , , • N -~nf 2 ~ ~ aoi sm (^ — a ) = -uhj, from (1), or d 2 y dP l + ^y = 0. (2) (3) (4) (5) It should be noted that equation (1) may be written in the form y = a sin (cot — a) = a [sin ut cos a — cos wt sin a] or y = A sin kt + B cos kt, where A = a cos a and B = — a sin a are constants. This equation and y = a sin (A;£ — a) are the general formulas for S. H. M. The acceleration of a particle describing S. H. M., as shown by (4), is proportional to the displacement and 96 DIFFERENTIAL CALCULUS oppositely directed. It is oppositely directed since the motion is one of oscillation about a position of equilibrium. When the body is above this position the force is directed downward, and when it is below, the force is upward. In the figure the point P' has a negative acceleration when above and a positive acceleration when below 0. The acceleration is zero at 0, a maximum at B' and a minimum at B; while the corresponding velocity, as given by (2) , has its maximum numerical value as P' passes through in either direction, and is zero at B and B' ', the ends of the vertical diameter. The factor of proportionality or is connected with the period 2 7T T by the relation T = — , where the period of the S. H. M. CO y = a sin ait is the time T required for a complete revolution of the point P; that is, wT = 2t. The time t = to make part of a revolution is called the phase, a being epoch angle. The number of complete periods per unit of time is N = 7„ = ?r~ > where N is the frequency of the S. H. M. Let P' be a tracing point capable of describing a curve on a uniformly translated sheet of paper, SS', then if the sheet be moved with the same speed as the point P moves on the circumference of the circle of radius a, P' describing S. H. M. on the vertical diameter will trace the sinusoid P'BP'B' on the moving paper. The sinusoid will have as its equation v x y — a sin - 1 = a sin - , a a where x is the abscissa of any point of the sinusoid referred EXERCISE VIII 97 to an origin (as (V) moving with the paper. The circle is shown in the figure in several positions corresponding to the different angles through which the radius OP has revolved, or the different positions of the projected point P' on the vertical diameter BOB'. The amplitude of the S. H. M. is the same as that of the sinusoid; that is, the radius a of the circle. The period of the sinusoid is 2 wa, corresponding to the period, T = — , of the S. H. M. of the point P' on the CO vertical diameter. 74. Self -registering Tide Gauge. — The principle by which the up-and-down motion of a point is represented by a curve is utilized in the self -registering tide gauge for record- ing the rise and fall of the tide. Such a gauge consists essentially of a float protected by a surrounding house or tube, and attached by suitable mechanism to a pencil that has a motion proportional to the vertical rise and fall of the float. The pencil bears against a piece of graduated paper fastened to a drum that is revolved by clockwork. There will thus be drawn on the paper a curve where the horizontal units are time, and the vertical units are feet of rise and fall. The stage of the tide is given for any time. EXERCISE Vm. 1. The angle (in radians) through which a rotating body turns, starting from rest, is given by the equation 6 = | at - + u t + is the angle XTQ, measured from the positive direction of the z-axis to the tangent TP h and mi is the slope of the curve at the point Pi (x h yi) . Hence from the equation of a line through a given point ( x h Vi)i V — 2/i = wi(£ — Xi); the equation of the tangent at the point Pi (x h yi) is y — y x = \-r\ (x — Xi), in which the subscript denotes that the quan- tity is taken with the value which it has at the point Pi. Since the sign of the derivative of a function indicates whether the function is increasing or decreasing, when mi is positive the curve is rising at Pi, and when mi is negative the curve is falling there. If mi is zero the tangent is horizontal, parallel to, or coincident with the z-axis; and if mi is infinite, the tangent is vertical, parallel to, or coincident with ?/-axis. Points where the slope has any desired value can be found 99 Tii = = — cot A mi 100 DIFFERENTIAL CALCULUS by setting the derivative equal to the given number and solving the resulting equation for x. The slope of the normal NP h being the negative reciprocal of the slope of the tangent TP h is \ _dx\ L dy\; Hence, the equation of the normal is = [~ d il (x ~ Xi) or x - Xi+iy - yi) Mr°- (b) Subtangents and Subnormals. — The subtangent and the subnormal are the projections on the z-axis of the part of the tangent and normal, respectively, between the point of tangency and the x-axis. From the figure: dx Subtangent TM = yi cot 4> = y\ y-yi Subnormal MN = yi tan = y± dy. 'dy dx Tangent TP, = VmP* + TM* = \/y^ + yf [j| J 2/i V 1 + hr =2/i cosec <£• Normal NP X = VmP* + MN 2 = \J y? + tf \^-~f = 2/iVl+[JJ 1 = 2/isecaxis. It is evident dy that at the origin where y = 0, -~ = oo ; that is, the y-axis is tangent at the vertex. It is seen also that, as y increases without limit, the tangent at its extremity becomes more and more nearly parallel to the x-axis. 3. On the circle x 2 + y 2 = 1, to find the points where the slope is 1, 0, or oo . r^/i = _ *1 = i ... \_dxji 2/1 2/i = -xi; ILLUSTRATIVE EXMAPLES 103 substituting in x 2 + 2/ 2 = 1, Xi = =b£ V2 and 2/1 = =F=J V2. B1--2-* •'• XI = and * =±1 ' by substituting in x 2 + 2/ 2 = 1- [21=-| =go ' •■• »-° and *- ±i ' by substituting in x 2 + ?/ 2 = 1. 4. To find at what angle the circle x 2 + y 2 = 8 and the parabola y 2 = 2x intersect. Making the two equations simultaneous, the points of intersection are found to be (2, 2) and (2, —2). =Fl. For circle, mi = ~d?/~ xi _ 2 _dx_ i " yi " ±2 For parabola, mi = ~dy~ _dx_ i 2/i ±2 For angle of intersection, 2 1 . mi- tan 4> = r— — - m mirr 2 ±2 d=2 2 1-i = =F3, or = tan" 1 (=F3), and from table of tangents, = 108° 26' or 71° 34'. 5. The path of a point is the arc of a parabola y 2 = 2 px, and its velocity is v; find its velocity parallel to each axis. Let s denote the length of the path measured from any point on it; then -r- = v. From y 2 = 2 px, dy _ p dx dt y dt Substituting these values in (Shift' HW <-•»». \dt) + y*Ut)' 104 DIFFERENTIAL CALCULUS /dx\ 2 _ v 2 y 2 v 2 dx _ yv IT eft i/ oft Vw 2 + p 2 6. A comet's orbit is a parabola, and its velocity is v; find its rate of approach to the sun, which is at the focus of its orbit. Let p denote the distance from the focus to any point on y 2 = 2px; then p = x + \ p, from point to directrix; dp _ dx " di~di' p being constant. Hence, the comet approaches or recedes from the sun just as fast as it moves parallel to the axis of its orbit; dp dx y dt dt \/y1 v. (Example 5.) P At the vertex, y = 0; hence, at the vertex -7- is zero. When y = V, <1. and dp It ~ = |V2, dp dx ~dt < v, s y dt Vy 2 + p 2 dy _ dt pv Vy 2 + p 2 > (Example 5) lim —7 pv . -0: dx . ds -di = dt =V > SmCe (ds\ 2 = (dx\\ (dy\* \dt) " \dt) r \dt) hi limit as y increases without limit. Hence -r- = -=7 is always less than v and approaches v as a ILLUSTRATIVE EXAMPLES 105 7. To compare the velocity of a train moving along a horizontal tangent with the velocity of a point on the flange of one of the wheels, and to compare also the horizontal and vertical components of the flange point. Let a wheel whose radius is a roll along a horizontal line with a velocity v\ find the velocity of any point P on its rim, also the velocity of P horizontally and vertically. ) = a vers 6, ) (i) The path of P is a cycloid whose equations are: x = a (6 — sin y = a (1 — cos0) where 6 denotes the variable angle DCP, and a the radius CD. Since the center of the wheel is vertically over D, v = the time-rate of OD d (ad) di dd dt' cW dt (2) Differentiating equations (1) gives, by (2), v vers 6 = the velocity horizontally, dx , dO a dB 37 = a (1 — cos 6) -r ± = a vers -y- at at at and dy . dd f- = a sin jt dt dt v sin 6 = the velocity vertically. (3) (4) 106 DIFFERENTIAL CALCULUS = velocity of P along its path. (5) The velocity of P may be considered as the resultant of two velocities each = v, one along PT tangent to the circle and the other along PH parallel to the path of C. The resultant PB must bisect the angle HPT; :. DPB = 90° and PB is tangent to the cycloid, the path of P, making PD the ds di = V ' normal. At 0, e = o : MP, e " 3 At Pi, ,e 7T ~ 2 d dx dy ds _ ft dt dt dt dx 1 dy 1 /~ Tt=2 V ' J = 2^ dx dy ds /x Tt = Tt = v > di = vV2 ' tk-^-2v dy -0 dt dt ' dt A ds di : v = V2a-y : a. At P 2 , 9 = ir, From (5) is obtained Hence, the velocity of P is to that of C as the chord DP is to the radius DC; that is, P and C are momentarily moving about D with equal angular velocities. (See Art. 72.) When 6 = 60°, their linear velocities also are equal, as shown above. 8. Find the equation of the tangent and the Values of the subnormal and normal of the cycloid. Dividing (4) by (3), Example 7, gives sin0 V(2 a — 3 y)y/a> vers 6 PH _ VHB- 1)11 and dy = sinfl = V(2a - y)y/a = . / (2a- y) dx vers 6 y/a V y sin 6 = -777S = and vers 6 = - from (1) ; CP a a EXERCISE IX 107 is the equation of the tangent at point (x h yi). m i i dy sin 6 sin0 . _ „„ ,,_ The subnormal = y-r =y z = y — j- = a sin = PH = MD. J dx y vers B 9 y/a Thus the normal at P passes through the foot of the per- pendicular to OX from C. Hence, to draw a tangent and normal at P, locate C, draw the perpendicular DCB equal to 2 a, and join P with B and D; then PB and PD will be respectively the tangent and normal at P. Normal = DP = VDB • DH = V2a-y. 9. Eliminating in equations (1) of Example 7 , equation of cycloid is x = a • arc vers y/a =F V2 ay — y~, since = arc vers y/a and a • sin = d= V(2 a — ?/) a. EXERCISE IX. Deduce the following equations of the tangent and the normal: 1. The ellipse, x 2 /a 2 + y 2 IIP- = 1, xix/a 2 + yiy/b 2 = 1, 2. The hyperbola, x 2 /a 2 - y 2 /b 2 = 1, Xix/a 2 - y iy /b 2 = 1, y-y, = -^{x- x:). 3. The hyperbola, 2xy = a 2 , Xi?/ + y\X = a 2 , y x y — X\X = yi 2 — Xi 2 . 4. The circle, x 2 + y 2 = 2 ax, y — y% = {x — Xi) (a — Xi)/y h y — yi = (x- xi) yi/(xi - r). 5. Find the equations of the tangent and normal at (3/2 a, 3/2 a): x 3 -f y 3 = 3 ax?/. A?is. x + ?/ = 3a, x = ?/. 6. x + ?/ = 2c*-* at (1, 1). Ans. 3y = x + 2, 3 x + y = 4. 7. (x/a) n + (y/b) n = 2, at (a, 6). Ans. x/a + ?//& = 2, ax — by = a 2 — b 2 . 108 DIFFERENTIAL CALCULUS 8. Show that the sum of the intercepts of the tangent to the para- bola x* + V* = a*, is equal to a. 9. Show that the area of the triangle intercepted from the co- ordinate axes by the tangent to the hyperbola, 2 xy = a 2 , is equal to a 2 . 10. Show that the part of the tangent to the hypocycloid x 3 +y * = a *, intercepted between the axes, is equal to a. 11. Find the slope of the logarithmic curve x = log6 y. The slope varies as what ? What is the slope of the curve x = log y ? 12. Find the normal, subnormal, tangent, and subtangent of the catenary y = a/2 (e*/ a + e~ x ' a ). Ans. y 2 /a; a/4 (e 2X / a - e - 2X ' a ): ■ ^ \ , ay Vy* - a 2 V2/2 - a 2 13. At what angles does the line Sy — 2x — 8 = cut the parabola = 8 x? Ans. arc tan 0.2; arc tan 0.125. 77. Polar Subtangent, Subnormal, Tangent, Normal. — Let arc mP = s, and arc PQ = As; then £ POQ = Ad, T circular arc PM = p&6, and MQ = A p. The chords PM and PQ, the tangents RPH and TPZ, are drawn; and ZH is drawn perpendicular to PH, Z being any point on the tangent PZ. When As = 0, the limiting positions of the secants PM and PQ are the tangents RPH and TPZ, respectively; hence, It ( Z PMQ) = / RPK = tt/2 = z PHZ, U(Z OQP) = z OPT = + = z #ZP, and It ZMPQ= zHPZ. POLAR SUBTAXGEXT 109 Now in a problem of limits the chord of an infinitesimal arc can be substituted for the arc, since the limit of their ratio is unity (Art. 22 and Cor., Art. 46) ; so Ap MQ sin MPQ . tC As chord PQ sinPJ/Q' , . ,/pAfl u chord .VP ..sin.VQP Aga.n, B-^- = ft chord pQ = B 3"^^^^ i ••• S -««•-£■ (2) From (1) and (2), it follows that, if PZ is taken as ds, ds = PZ, dp = HZ, and p d0 = #P. Drawing OT perpendicular to OP, and PA and (XV perpen- dicular to the tangent TP, the length P7 7 is the polar tan- gent; PA, the polar normal; OA, the pofar subnormal; and 07 7 , the pofa?* subtangent. From the right-angled HPZ ds 2 = dp 2 + p 2 dd- 2 ; (3) . , pde , dp , . pde ,.. sm ^ = ¥' cos ^ = 5? tan * = ^T (4) Polar subt. = OT = OP tan ^ = p 2 dd/dp. (5) Polar subn. = OA = OP cot ^ = dp/dd. (6) Polar tan. = PT = VOP 2 + OT 2 = p \J 1 -+ dfl 2 dp 2 Polar norm. = AP - VOP 2 + OA 2 = y p 2 + ^- (8) p = ON = OP sin ^ = p 2 dd/ds P 2 Vp^ + (d P /^) 2 ' (9) = ^ + d. (10) Corollary. — If PZ represents the velocity at P of a moving 110 DIFFERENTIAL CALCULUS point (p, 0) along its path, PK ( = HZ) and PR will repre- sent its component velocities at P along the radius vector and a line perpendicular to it. If the path is a circle with center at 0, \J/ is 90°; and • , • ™o -. pdd , fAS ds dd sm^ = sin 90 ° = 1 = —-, from (4), or -=- = p-j-> .*. v = rw; that is, the linear velocity = radius times angular velocity. (See Art. 72.) EXERCISE X. 1. Find the subtangent, subnormal, tangent, normal, and p of the spiral of Archimedes p = ad. _____ Ans. subt. = p 2 /a; subn. = a; normal = Vp 2 + a 2 ; tangent = p Vl + p 2 /a 2 ; p = p 2 /(p 2 + a 2 )*- 2. In the spiral of Archimedes show that tan ^ = 0; thence find the values of \p, when 6 = 2-w and 4 tt. Ans. 80° 57' and 85° 27'. 3. Find the subtangent, subnormal, tangent, and normal of the logarithmic spiral p = a e . Ans. subt. = p/loga; subn. = ploga; tan. = p Vl + (log a) -2 ; norm. = p Vl + (log a) 2 . 4. Show why the logarithmic spiral is called the equiangular spiral, by finding that \f/ is constant. If a = e, $ = ir/4, subt. = subn., and tan. = norm. 5. Find the subtangent, subnormal, and p of the Lemniscate of Bernouilli p 2 = a- cos 2 6. Ans. subt. = — p 3 /a 2 sin2 0; su bn. = — a 2 sin2 0/p; p = p 3/Vp4 + a 4 sin 2 2 6 = P z /a 2 . 6. In the circle p = a sin 6, find \p and <£. Ans. ^ = 0, and = 20. The angle between two polar curves is found as for the other curves. 7. Find the angle of intersection between the circle p - 2a cos 0, and the cissoid p = 2 a sin tan 0. Ans. arc tan 2. CHAPTER V. MAXIMA AND MINIMA. INFLEXION POINTS. 78. Maxima and Minima. — One of the principal uses of derivatives is to find out under what conditions the value of the function differentiated becomes a maximum or a minimum. This is often very important in engineering questions, when it is most desirable to know what conditions will make the cost of labor and material a minimum, or will make efficiency and output a maximum. A maximum value of a function or variable is defined to be a value greater than those values immediately before and after it, and a minimum value to be one less than those immediately before and after it. It follows that the function is increasing before, and decreasing after reaching a maxi- mum value; while it is decreasing before, and increasing after reaching a minimum value. The points on the graph of y = f (x) at which the function ceases to increase and begins to decrease, or ceases to de- crease and begins to increase, are maxima or minima points ; and the values of the function at those points are maxima or minima values. It is to be noted that a maximum value is not necessarily the greatest value the function can have nor a minimum the least; f (a) is a maximum if it be greater than any other value of/ (x) near/ (a) and on either side of it; and/ (a) is a mini- mum if it be less than any other value of / (x) near / (a) and on either side of it. 79. The Condition for a Maximum or a Minimum Value. — If / (x) is a function of an increasing variable x; then 111 - 112 DIFFERENTIAL CALCULUS for / (a) to be a maximum, / (x) must be increasing just before / (a) and therefore /' (x) must be positive ; on the other hand / (x) must be decreasing just after / (a) and therefore /' (x) must be negative. Hence, as x increases through the value a, f (x) must change from a positive to a negative value. Conversely, if as x increases through the value a, f (x) changes from a positive to a negative value, / (a) will be a maximum value of / (x) . Hence / (a) will be a maximum value of / (x) if, and only if, /' (x) changes from a positive to a negative value as x increases through the value a. In the same way it may be seen that/ (a) will be a minimum value of / (x) if, and only if, /' (x) changes from a negative to a positive value as x increases through the value a. This condition has been called the fundamental condition or test. For the cases of most frequent occurrence; when f (a) is a maximum or a minimum, f (a) = 0. In most cases it is a necessary condition for a maximum or a minimum value of a function that the first derivative at that value shall be zero. For in most cases the first derivative /' (x) is con- tinuous; and, when continuous, it changes sign by passing through the value zero only. But if /' (x) is not continuous, as is the case for some functions, then it may change sign by becoming infinite for some finite value of x; for if/' (x) is a fraction whose denominator becomes zero for some finite value of x, f (x) changes sign as x increases through that value. For example, when /' (x) = 5 , for x = a = 2, X Zi f (a) = - - = 00 ; here /' (x) is negative before, and positive after x increases through the value 2; hence, / (2) is a minimum according to the fundamental test. Again, there are exceptional non-algebraic functions for which/' (x), as x increases through some finite value a, changes sign GRAPHICAL ILLUSTRATION 113 without becoming either zero or infinite. (See Note, Art. 80.) Excepting such rare functions, a theorem may be stated thus: For all algebraic Junctions any value of x which 7tiakes f (x) a maximum or a minimum is a root off (x) = or f (x) = go . The converse of this theorem is not true ; that is, any root of /' (x) — or /' (x) = oo does not necessarily make / (x) either a maximum or a minimum. These roots are called critical values of x, and each root may be tested by rule. 80. Graphical Illustration. — Let P ... P 3 ... P? be the locus of y = f (x) . Then / (x) will be represented by the ordinate of the point (x, y), and /' (x) by the slope of the locus at the point (x, y). By definition, the ordinates MP, MzP 2 , and M 4 P 4 represent maxima of / (x) ; while 0, MiPi, M 3 P 3 , and M b P b represent minima. (Art. 78.) The slope /' (x) is positive immediately before a maximum ordinate, and negative immediately after; while the slope is negative immediately before a minimum ordinate and posi- tive after. The slope /' (x) is or 00 at any point whose ordinate / (x) is either a maximum or a minimum. The slope f (x) is discontinuous at the points P4 and P 5 , where it changes sign by becoming infinite as x increases through the values OMt and 0M- ) that is, /' (x) = go . 114 DIFFERENTIAL CALCULUS The slope f (x) is at P 6 and oo at P 7 ; but it does not change sign at either point, and neither M 6 Pe nor M 7 P 7 is a maximum or a minimum ordinate; it does, however, change in value at each point, P 6 being a point where the slope f (x) is a minimum and P 7 one where it is a maximum. The points P 6 and P 7 are inflexion points, at which the curve changes from being concave downward to upward, or vice versa. Note. — Points such as P 4 and P 5 occur on railroad "Y's," and such points where branches of a curve end tangent to each other are called cusps. At a point on a non-algebraic curve where branches end and are not tangent to each other, called a shooting point, f (x) may change abruptly from a positive finite value to a nega- tive value, or vice versa; hence, / (a) would be a maximum or a minimum without /' (x) becoming either zero or infinite. The supplementary figure shows a shooting point at which / (a) is a minimum; /' (x) becoming — 1 as x increases to a, and + 1 as x decreases to the same value a, thus changing from a negative to a positive value as x increases through the value a. It is to be noted that, while on an exceptional curve like the one shown the tangents at a maximum or a minimum point may have various directions, on any algebraic curve the tangent is parallel to one or other of the two rectangular axes; that is, the tangent at a maximum or a minimum point is horizontal, the slope being continuous; otherwise it is vertical ; and on only exceptional non-algebraic curves will it have any other direction. It may be noted also, as in the graphical illustration given, that maxima and minima occur alternately; that is, a minimum between any two consecutive maxima and vice versa. It may be seen that a maximum may be less than RULE FOR APPLYING FUNDAMENTAL TEST 115 some minimum not consecutive, since by definition it is necessarily greater than those values only immediately before and after it. It may be seen also that when the slope is continuous at least one inflexion point must occur between a maximum and a minimum point. The only inflexion points marked on the curve are P 6 and P 7 , occurring where /' (x) = and oo , but /' (x) may have any value at an inflexion point, although its rate, j" (x), must change sign there, becoming or oo . Hence, at any inflexion point, a point where the slope f (x) is a maximum or a minimum, 5" (x) — or oo . The converse is not true, for J" (x) may be or oo at other points. 81. Rule for Applying Fundamental Test. — Let a be a critical value given by either f (x) = or f (x) = oo , or, in general, any value of x to be tested, and Ax a small positive number; then: If f ( a ~ Ax) is positive and f (a + Ax) is negative, f (a) is a maximum off (x) . (Art. 79.) Iff ( a ~~ Ax) is negative andf {a + Ax) is positive, f (a) is a minimum of f (x). (Art. 79.) If f ( a ~ Ax) and f (a + Ax) are both positive or both negative, f (a) is neither a mvaximum nor a minimum of fix). This rule is general and is valid for all functions that are continuous one-valued functions, which comprise all those usually encountered in this connection. 82. While the rule just stated applies in every case; when /' (x), as well as / (x), is continuous and therefore the criticalyalues of x are roots of f (x) = 0, a rule usually easier to apply may be deduced from the fundamental test or condition. Let a be a critical value of x given by /' (x) =0. If / (a) is a maximum value of / (x) , /' (x) changes from a positive to a negative value as x increases through a ; therefore, near a, f (x) is a decreasing function, and therefore, its derivative, 116 DIFFERENTIAL CALCULUS /" (x), must be negative near a. But if /" (a) is not zero, then near a the sign of /" (x) is that of J" (a). Hence f" (a), if it is not zero, will be negative when/ (a) is a maxi- mum value of / (x). In the same way it is seen that/" (a), if it is not zero, will be positive when / (a) is a minimum value of / (x) . Conversely, / (a) will be a maximum or a minimum value of / (x) according as /" (a) is negative or positive. Hence this rule for determining the maxima and minima values of/ (x) when/ (x), f (x) are continuous: The roots of the equation f (x) = are, in most cases, the values of x which make f (x) a maximum or a minimum. If a be a root of f (x) = 0; then f (a) will be a maximum value of f (x), if f" (a) is negative, but a minimum, if f" (a) is positive. 83. While the above rule is all that is needed in most cases, it does not provide for the case when the critical value a makes/" (x) become zero. When/" (a) =0,/ (a) may be either a maximum or a minimum, or it may be neither, and the point on the graph of/ (x) may, or may not, be a point of inflexion; so an extension of the rule is needed to provide for cases where /" (x) and the succeeding derivatives may in turn become zero for the value, a. If no derivative is found that does not become zero when a is substituted for x, then recourse may be had to the funda- mental test, that rule applying in every case. But if/' (a), /" (a), . . . , f n ~ l (a) all are found to be zero, and/ n (a) not zero; then the following rule, inclusive of the preceding, applies. Let a be a critical value of x given by f (x) = 0, and let a be substituted for x in the successive derivatives of f (x). If the order n of the first of the derivatives that is not zero is an even integer, f (a) will be a maximum or a minimum off (x) according as this derivative is negative or positive. If the order n of the first of the derivatives that is not zero is RULE FOR DETERMINING MAXIMA AND MINIMA 117 an odd integer,/ (a) will be neither a maximum nor a minimum of f (x) regardless of the sign of this derivative. Note. — This conclusion can be deduced by examining the signs of the derivatives near a; thus, as follows: If /' (a) and /" (a) be zero but /'" (a) not zero; since f" (x), the rate of /" 0), has when a: is a a value not zero; f" (x), the rate of /' (x), is then increasing or decreasing according as /'" (a) is positive or negative, and, since it is zero when x is o, it must change sign as x increases through a; therefore, /' (x), the rate of / (x), must be either decreasing before and increasing after, or increasing before and decreas- ing after x is a, and so, continuing to be positive or negative according as f" (a) is positive or negative, does not change sign as x increases through a; hence/ (a) is neither a maxi- mum nor a minhnmu of/ (x) regardless of the sign of /'" (a). Now if /'" (a) also is zero but / IV (a) not zero; since f™ (x), the rate of /'" (x), has when x is a a value not zero, /"' (x), the rate of f" (x) , is then decreasing or increasing according as / IV (x) is negative or positive and, since it is zero when x is a, it must change sign as x increases through a; therefore, /" 0), the rate of/' (x), must be either increasing before and decreasing after, or decreasing before and increasing after x is a, and so, continuing negative or positive according as / JV (a) is negative or positive, does not change sign as x increases through a; /' (x), the rate of / (x), must then be either decreasing, or increasing before and after x is a, and, as it is zero when x is a, it changes from a positive to a nega- tive value, or from a negative to a positive value, as x in- creases through a, according as / IV (a) is negative or positive; hence / (a) is a maximum or a minimum of / (x) according as / IV (a), the first of the derivatives that is not zero, is neg- ative or positive. In the same way it follows that, if f* (a) is the first of the derivatives that is not zero, / (a) is neither a maximum nor a minimum of / (x) regardless of the sign of f* (a) ; and that, 118 DIFFERENTIAL CALCULUS if / VI (a) is the first, / (a) is a maximum or a minimum of/ (x) according as / VI (a) is negative or positive ; and so on for the succeeding derivatives: hence the inclusive rule given. (For proof by Taylor's Theorem, see Art. 218.) y=ffxh4x* ffa)=ffoJ=o, neither a max.noraminoff(x) Y] y=f(x)=x* ffa) =f(o)=o, a mm. of (x) r f"(x)=-12x 2 f(x)=-Z4x -2S L ffa) ff/oJ=o, neither a rruix. nor a mfn , offfxj fJxJ-24 V=f(x) = -x4 ffa) -SlpJ =o,a max. off(x) 84. Typical Illustrations. — The foregoing deductions may be verified by the graphs of the successive derivatives of x 3 } —X s , x A , and — x 4 , referred to the same axes as those of the graphs of the functions. The usual case when /' (a) is TYPICAL ILLUSTRATIONS 119 zero and /" (a) not zero, is well illustrated by the graphs of the function sin and its derivatives. f(e)~sinf) f(w; » x = c; ax 3(x — c) 3 dy here it can be seen that /' (x) or j~ does not change sign as x passes through c, and, therefore, the function has neither max. nor min. 8. Examine (x - l) 4 (x + 2) 3 for max. and min. f(x) = (3-l)"(x + 2)*(7s + 5)=0; ... x = 1, -2, -f. /' (1 - Ax) is -, /' (1 + Ax) is + ; .*. / (1) = is a min. /' (-f - Ax) is +, /'(-$+ Ax) is - ; .\ /(-f) is a max. /' (-2 - Ax) and f (-2 + Ax) are both +; hence / (-2) is neither max. nor min. (q x^ 9. Examine ^r~ for max. and min. a — 2x /' (x) = (a - x) 2 (4 x - a) /(a - 2 x) 2 ; /' (x) = gives x = a, a/4; /' (x) = oo gives (a — 2 x) 2 = 0, or x = a/2. /' (a/4) changes from — to + as x passes through a/4; .'. /(a/4) is min. When x = a, or a/2, /'(x) does not change sign; .'. /(a) and / (a/2) are neither max. nor min. x 2 _ 7 x _i_ 6 10. Examine rr; — for maxima and minima. x — 10 Ans. / (4) is a max.; / (16) is a min. (2 _i_ 2) 3 11. Examine ; =rs for maxima and minima. (x - 3) 2 Ans. / (3) is a max; / (13) is a min. 12. When/ (x) = (x -_1) (x - 2) (x - 3), / (2 - I/V3) . § V3, is a max., and/ (2 + I/V3) = -§ V3 is a min. 13. Show that the maximum value of sin + cos 6 is V2. 14. Show that the maximum value of a sin + b cos is Va 2 -f- o 2 . 15. Show that e is a minimum of x/log x. 16. Show that 1/ne is a maximum of log x/x n . 17. Show that e 1/c is a maximum of x 1 x . 18. Show that 1 is a maximum of 2 tan — tan 2 0. 19. Find the maximum value of tan -1 x — tan -1 x/4, the angles being taken in the first quadrant. Ans. tan -1 1. 20. Show that 2 is a maximum ordinate and —26 is a minimum ordinate of the curve y = x 6 —-5x i + 5x 3 + l. EXERCISE XI 127 PROBLEMS IN MAXIMA AND MINIMA. 1. Find the maximum rectangle that can be inscribed in a circle of radius a. Let 2 i = base and 2 y = altitude; then area A = 4 xy = 4 x Va 2 — x 2 . Take / (x) = x 2 (a 2 - x 2 ) = a 2 x 2 - x 4 [by Art. 87]; f (x) = 2 a 2 x - 4 x 3 = 2 x (a 2 - 2 x 2 ) = 0; .-. x = 0, a/V2] f"(x) = 2 a 2 - 12 x 2 ; /" (0) = 2 a 2 ; .'. / (0) = is min. ' f" (a/y/2) = 2a 2 - 6a 2 = -4a 2 ; .'. /(0/V2) = a 4 /4. .*. A = 4 VaV4 = 2 a 2 is the area of the maximum rectangle, which is a square. Note. — By Geometry without the Calculus method: A = 2 ay = 2aVa— 'i 5 ]^ = 2 a 2 , since the radical quantity is evidently greatest when x = 0. 2. The strength of a beam of rectangular cross section varies as the breadth b and as d 2 , the square of the depth. Find the dimensions of the section of the strongest beam that can be cut from a cylindrical 128 DIFFERENTIAL CALCULUS log whose diameter is 2 a. Strength oc 6d 2 ; /. strength = kbd 2 A; is a constant; let fff>)~ b (4 a 2 -6 2 ) = 4a 2 6 -¥; r— f (&) = 4a 2 - 362 = = 0; /. b = 2a V3' d = yi( 2 °)- f(b) = 3(< :a 2 - 4a 2 \ 3 J 2a |(4i 3V3 where Hence, the rectangle may be laid off on the end of the log by drawing a diameter and dividing it into three equal parts; from the points of division drawing perpendiculars in opposite directions to the circum- ference and joining the points of intersection wfth the ends of the diameter, as in the figure. The strength of the beam is about 0.65 of that of the log, but it is the strongest beam of rectangular section. 3. The stiffness of a rectangular beam varies as the breadth b and as d 3 , the cube of the depth. Find the dimensions of the stiffest beam that can be cut from the log. \fc- v- Stiffness <* bd 3 ; ,% stiffness = kbd 3 (k constant); let stiffness = b (4 a 2 - 6«)3. Take f(b) = 4a%* - 6 § ; f (b) = f (a 2 6"" - 6 s ) » 0; .-. 6 2 = a 2 or b = a) :. d = (4 a 2 - a 2 )* = a V3. To draw the rectangle, lay off from ends of a diameter chords at angles of 30° with diameter and join^nds_of_chords with ends of diameter. 4. A square piece of pasteboardis!^o~T5e~niaa v e i n t o a bc rc~by cutting out a square at each corner. Find the side of the square cut out, so that the remainder of the sheet will form a vessel of maximum capacity. Let a be side of square sheet and x side of square cut out; then ■ EXERCISE XI 129 fix) = x(a-2x) 2 . f (x) = - 4 x (a - 2 x) + (a - 2 x) 2 = (a - 6 x) (a - 2 z) = 0; .'. a = 1/6 a, 1/2 a. /(l/6a) = 1/6 a. (a - 1/3 a) 2 = 2/27 a 3 , maximum capacity. / (1/2 a) = 1/2 a (a — a) =0, minimum. cl-Zjc tZ X 6. A rectangular sheet of tin 15" X 8" has a square cut out at each corner. Find the side of the square so that the remainder of the sheet may form a box of maximum contents. Arts. If". 6. A channel rectangular in section, carrying a given volume of water, is to be so proportioned as to have a mini- mum wetted perimeter. Find the proportions of the channel. Let x be the width of the bottom, and y the height of the water surface. Since the given volume is proportional to the cross section, xy = V, where V is constant. 2V p *= x + 2y = x + -—, from (1); A — -— y — — — — — . — (1) that dx x 2 ' 2V V2~V = V2 2 xy, or x = 2 y. xy] To show that this makes p a minimum; note that for x 2 < 2 V, •£■ is dx dp negative, and for x 2 > 2 V, j- is positive, therefore for x 2 = 2 V, or x = 2 y, p is a minimu m. 130 DIFFERENTIAL CALCULUS 7. Find the dimensions of a conical tent that for a given volume will require the least cloth. y = i Trr 2 h', .'. h = S V/irr 2 , where r is radius of base and h altitude. (1) S = Trr W 2 + h* = irr (r 2 + 9 V 2 /* 2 ^)' = (tt 2 /- 4 + 9 F 2 /r 2 )*, $ denoting lateral surface; let / ( r ) ^ ^r 4 + 9 F 2 A 2 (by Art. 87), 18 V 2 n V2 "/6J. V 7T /'(r) =4ttV3--^ =0, /" (r) = 127r 2 r 2 + 54 F 2 //- 4 , positive for any r; hence r = -=- y — makes S a minimum. From (1), A = V — = r V^; and slant height = r V3. f 7T 8. From a given circular sheet of tin, find the sector to be cut out so that the remainder may form a conical vessel of maximum capacity. Ans. Angle of sector = (l- Vf) 2tt = 66° 14'. 9. The work of propelling a steamer through the water varies as the cube of her speed; show that her most economical rate per hour against a current running n miles per hour is 3 n/2 miles per hour. Let v = speed of the steamer in miles per hour; then cv 3 = work per hour, c being a constant; and v — n = the actual distance advanced per hour. Hence, cv z /v — n = the work per mile of actual advance. Find the most economical speed against a current of 4 miles per hour. 10. The cost of fuel consumed by a steamer varies as the cube of her speed, and is $25.00 per hour when the speed is 10 miles per hour. The other expenses are $100 per hour. Find the most economical speed. Let C = cost per hour for fuel at speed of v miles per hour; EXERCISE XI 131 then C : $25 = f 3 : (10) 3 ; .*. C = $25 yVQO) 3 ; F , s $25 r 3 d , SlOOd . . . ,. . , d . , f d-) = — — — • - -4 : where a is distance and - is hours. * (10) 3 v v v .,.. 50 , 100 tf n . t-3 = 2000, or v = ^2000 = 12.6 miles per hr. / (12.6) = $12 d (approximately); hence cost for one hour about SI 50; cost for running 10 miles at 12.6 miles per hour about SI 20, while the cost for running the 10 miles at 10 miles per hour is SI 25, and at 15 miles per hour the cost for running 10 miles is about S123. 11. The amount of fuel consumed by a steamer varies as the cube of her speed. When her speed is 15 miles per hour she burns 4| tons of coal per hour at S4.00 per ton. The other expenses are S12.00 per hour. Find her most economical speed and the minimum cost of a voyage of 2080 miles. Ans. 10.4 mi. per hr.; S3600. 12. A vessel is anchored 3 miles off shore. Opposite a point 5 miles farther along the shore, another vessel is anchored 9 miles from the shore. A boat from the first vessel is to land a passenger on the shore and then proceed to the other vessel. Find the shortest course of the boat. Let hi be the distance the boat goes from first vessel to shore and h 2 the distance from the shore to the other vessel; then / (x) =h+h = (32 + x*)* + [9^ + (5 - z)]*; ft f \ % «3 % r\ . J W " (9~+^ " (81 + (5 - x¥)? " ' whence x = ± f ; h + h 2 = -^ + V = 13 miles. 13. Find the number of equal parts into which a given number n must be divided that their continued product may be a maximum. Let /»-©■' ™=(i)H- l H :. log - = 1 ; - = e, and x = - , & x ' x e hence the number of parts is n/e, and each part is e. 132 DIFFERENTIAL CALCULUS DETERMINATION OF POINTS OF INFLEXION. 1. Examine y = x 3 — 3 x 2 — 9 x + 9; for points of inflexion. P = 3x 2 -6z-9, dx pi = 6x - 6 = 6 (x - 1) = 0, .\ x = 1, is abscissa of an inflexion point. The point is (1, —2), to the right of which the curve is concave upward. 2. Examine x 3 — 3 bx 2 + a?y = 0, for points of inflexion. Ans. (b, 2 b 3 /a?) is a point of inflexion, or of maximum slope, to the right of which the curve is concave downward. 3. Examine y = c sin x for points of inflexion. Ans. (0, 0), (±7r, 0), (±2tt, 0) . . . . 8 a 3 4. Examine the Witch of Agnesi, y = . , . — r, for inflexion points. x 2 + 4 a 2 Ans. (±2a/\ / 3, 3 a/2); concave downward between these points, concave upwards outside of them. Find the points of inflexion of the following curves: 5. (:r/a) 2 + (y/b)i = 1. Ans. x = ±a/V2. 6. y = (x 2 + x) e~ x . Ans. x = and x = 3. tj -ax ' bx a 2 (log a — 1°S &) 7. y = e ax — e~ bx . Ans. , • a — b 8. ?/ = z 3 /a 2 + x 2 . Ans. (0, 0), (a V3, 3 a Vf), (-a V3, -3 a VJ). CHAPTER VI. CURVATURE. EVOLUTES. 88. Curvature. The flexion (Art. 13), b dm d?y dx = dx 2} being the rate of change of the tangent of the angle made with the z-axis by the tangent to a curve, is one measure of the bending of the curve at the point of tangency. This measure, however, is dependent upon the position of the axes and would change if the axes were rotated. There is a measure of the bending called the curvature, which does not depend upon the choice of the axes, as it is expressed in terms that are the same after the axes are rotated, or even before any axes are drawn. The curvature is denoted by K = — , the rate of change of the angle of inclination = tan -1 m, with respect to the length of arc s. Thus, let P and Pi be two points on a plane curve, and + A$ the angles which the tangents at P and Pi make with the x-axis, s the arc AP meas- ured from some fixed point A on the curve up to P, and As the arc PPi. The angle is in radians, and A is evidently the angle between the two tangents. The angle A<£ is the total curvature of the arc As, as it is a measure of the deviation from a straight line of that portion of the curve between the points P and Pi. The sharper the 133 134 DIFFERENTIAL CALCULUS bending of the curve between the two points the greater is A0 for equal values of As. The average curvature of the arc As is denned as -r— , and is, therefore, the average change per unit length of arc in the inclination of the tangent line. The limit of -r— , when Pi approaches P as its limiting position, is called the curvature of the curve at P; that is, the curvature at a point on a curve is K = lim -r— = -=- • As =o As as Otherwise, by rates, the curvature of any curve, as APPi at any point, as P, is the s-rate at which the curve bends at P, or the s-rate at which the tangent revolves, where s de- notes the length of the variable arc AP. If 4> denotes (in radians) the variable angle XTP as P moves along the curve APPi, then, evidently, the curvature of APPi at P equals the s-rate of ; that is, K = ~~- as 89. Curvature of a Circle. — For a circle of radius R As = PA0 and therefore, A0_ 1 d$_ l_ As ~ R ] ds ~ R' since the ratio of the increments is con- stant; that is, the average curvature of any arc of a circle is equal to the curva- ture at any point of that circle. In other words, a circle is a curve of constant curvature and its curvature is equal to the reciprocal of its radius; that is, the curvature of a circle equals 1/R radians to a unit of arc. For example, if R = 2, the circle bends uniformly at the CIRCLE, RADIUS, AND CENTER OF CURVATURE 135 If R = J, the curvature of the circle is 2 radians per unit of arc. If R = 1, for the circle of unit radius the curvature is evidently a radian per unit of arc. 90. Circle, Radius, and Center of Curvature. — The curvature of any curve except the circle varies from one point to another. A circle tangent to a curve and having the same curvature as the curve at the point of contact, therefore, having a radius equal to the reciprocal of the curvature at that point, is called the circle of curvature at that point; its radius is called the radius of curvature; and its center, the center of curvature. If R denotes the radius of the circle of curvature at any point of a curve, then, since the curvature of the curve is -=- and equals the curvature of the circle, it follows that, d 1 , ds ck=W and B = d* If at P (Art. 72, figure) the direction of the path of (x, y) became constant, (x, y) would trace the tangent at P, and ds might be represented by a length on the tangent ; while if at P the change of direction of the path became uniform with respect to s, (x, y) would trace the circle of curvature at P, and ds would represent an arc of the circle, since it equals R d = l would be the constant change of angle at center of the circle of curvature. Thus it is that the curvature is uniform when, as the moving point passes over equal arcs, the tangent turns through equal angles; or conversely; and, as this is the case with the circle only, it is the only curve of uniform curvature. For any curve the measure of the curvature at a point is the limit of the ratio between the angle described by the tangent and the arc described by the point of contact, as that arc approaches zero; and this limit -=- equals the 136 DIFFERENTIAL CALCULUS reciprocal of the radius of curvature at the point; hence, ds the radius of curvature is -7- and equals the reciprocal of the curvature. The figure shows the circle of curvature for the point P (x, y) of the ellipse; C is the center of curvature, and CP the radius of curvature. It is to be noticed that the circle of curvature crosses the ellipse at P, and this must be so; for at P the circle and ellipse have the same curvature, but towards A the curvature of the ellipse increases, while that of the circle remains the same, being constant. Hence on the side of P towards the vertex A the circle is outside of the ellipse. From P towards B the curvature of the ellipse decreases, and, therefore, on the side of P towards the vertex B the circle is inside of the ellipse. So, in general, the circle of curvature crosses the curve at the point of contact. The only exceptions to this rule are at points of maximum and minimum curvature, as the vertices of the ellipse. From A along the curve in either direction, the curvature of the ellipse decreases; hence the circle of curvature at A lies entirely within the ellipse. From B the curvature of the ellipse increases in each direction and so the circle of curva- ture at B lies entirely without the ellipse. RADIUS OF CURVATURE IN COORDINATES 137 91. Radius of Curvature in Rectangular Coordinates. — Since = tan -1 m, and since ds 2 = dx 2 + dy 2 ; dd> = =— ; — ; and ds = Vl + m 2 dx, where m = -¥- ; v 1 + m 2 dx ' hence the radius of curvature ds Vl + m 2 dx (1+ra 2 )* R d(ji dm dm 1 + m 2 dx (1 +m 2 )? b ~ d?i dx (i) d 2 w # will be positive or negative according as -^ is positive or negative, if is always taken as the acute angle which the tangent makes with the x-axis; for then, whether $ is posi- tive or negative, 1 + ( ~) = sec 3 will be positive and d 2 y the sign of R will be that of ■-— . Hence the sign of R will be plus or minus according as the curve at the point is concave sec 3 cb upward or downward. R may be in the form — r — • If the reciprocals of the members of (1) are taken, then K - t = 3/t 1 + (!)?' which - ma y be in the f0 ™ b cos 3 4>. d 2 v 1 . If -~ is zero at any point of a curve, then K = -^is zero and R is infinite. Thus at a point of inflexion R is infinite. It may be noted that as a curve approaches being a straight line, its curvature approaches zero and its radius of curvature becomes infinite, that is, it increases in length without limit. So a straight line is the line that the arc of a circle of curva- 138 DIFFERENTIAL CALCULUS ture approaches as the radius of the circle increases without limit. On the contrary as the radius of curvature at a point approaches zero, the curvature at the point becomes infinite and the curve will approach a mere point, since the circle of curvature will diminish with zero as a limit for its radius. 92. Approximate Formula for Radius of Curvature. — Since K = ['+©■] 3' it is seen that the flexion when multiplied by the factor 1/(1 + m 2 )^ gives a measure of the bending of a curve independent of the position of the axes. The flexion is the rate of change of the tangent of the inclina- tion of the curve at a point with respect to the abscissa, while the curvature is the rate of change of the inclination of the curve at a point with respect to the arc, where the inclination of the curve is that of the tangent line at the point. However, when the curve deviates but slightly from a horizontal straight line, the curvature is approximately the same as the flexion, since the slope m — -j- being small, f-p) is very small compared with 1, and therefore the formula becomes approximately This approximation for the curvature is used to advantage in the flexure of beams and columns. The approximate formula for the radius of curvature is consequently dx 2 EXERCISE XII 139 EXERCISE XH. 1. To find R and the curvature of the ellipse — + — =1. a 1 b l dy _ _ fr^c d?y _ ¥ dx a 2 y' dx 2 a 2 y 3 Substituting these values in (1), Art. 91, gives \ *aYJ \ V ) ^ ' v d 1 a 4 6 4 ds R ( a y + 5V)! The maximum curvature is a/b 2 , at A (a, 0), where R = 6 2 /a is a minimum, and the minimum curvature is b/a 2 . at B (0, 6), where # = a 2 /b is a maximum. (See Art. 90, figure; Art. 97, figure.) 2. To find R and the curvature of the parabola y 2 = 4 px. dy = 2_p (Py _ _ 4p 2 ete 2/ ' dx 2 y 3 Substituting these values in (1) of Art. 91 gives ds B 2 (x + p) § ' At the vertex (0, 0), R = 2 p, the minimum radius; and the maxi- mum curvature is (1/2 p) radian to a unit of arc. (See Example 1, Art. 97.) d 2 y 4 » 2 Since -r^ = j- is negative for positive values of y and posi- tive for negative values, the curve is concave downward at points whose ordinates are positive, and concave upward at points whose ordinates are negative. The sign of R may be neglected, since the sign of -j~ will indicate whether the curve is concave upward or down- ward at any point . 3. To find R of the cycloid x = a vers -1 (y/a) T V2 ay — y 2 . dy = V2 ay - y 2 d 2 y = a t dx y ' dx 2 y 2 140 DIFFERENTIAL CALCULUS Substituting these values in (1) of Art. 75 gives -( i + s V Jff ) , (-9-C^ , (-S)-* VI * At the highest point, y = 2 a, and, therefore, R = 4 a, the maximum of R. At the vertex (0, 0), R = 0, and also at other points where y is zero; therefore, R being zero, at those points, which are cusps, the curvature is infinite. (See Example 3, Art. 97, figure.) Find R and the curvature of each of the following curves. 4. The equilateral hyperbola 2 xy = a 2 . R = (x 2 + y 2 ) l /a 2 . 5. The cubical parabola y z = a?x. 6. The logarithmic curve y = b x . 7. The catenary y = | (e x / a + e~ x / a ). 8. The hypocycloid x* + y 3 = a 3 . 9. The curve x* + ?/* = ai 93. Radius of Curvature in Polar (4) and (10) of Art. 67, = A _j_ ^ ^ = tan -1 d 6 a 4 */ JV - ~ ds ( 9 yi + a 4)f d my ds {in 2 + y 2 )* d4> ds a R = ■■ 3 (axy)*. ds a- 2 (.f + y) j Coor ding ites. — From p c/p/d0' d4 = cbp cU = (c/p/^) 2 - p • d 2 p/rffl 2 . ""' dd " + d0 ' c/0 p 2 + (dp/d0) 2 d^_ p 2 + 2 (dp/dfl) 2 - p • d 2 p/dd\ '"" d0 ~ p 2 + (dp/d0) 2 . ds/d0 _ [p 2 + (dp/dd)^ , , * * a d/dd p^ + 2 (rfp/^) 2 - P • dr-p/de 3 ' w Corollary. — Since R = oo at a point of inflexion, p? + 2(dp/d0)*-p.g=O is a necessary condition for such a point. RADIUS OF CURVATURE IN POLAR COORDINATES 141 Example. — To find R for the curve p = sin 0. Here dp a d 2 P • a R (p 2 + COS 2 0)* (sin 2 + cos 2 0) 1 p 2 +2cos 2 0-p(-sin0) sin 2 0+2cos 2 0+sin 2 2 This curve p = sin 0, a circle with unit diameter, in connection with the formula for polar curves, tan \f/ = —jr, furnishes a derivation of the dp/dd d (sin 6) . Since for circle yp = d, tan 6 = dp/dd dp sin0 tan 6 tan 6 that is, d (sin 6) = cos 6 dO. Also from figure: OP P dd = cosddd; tan0 OA cos0' and since tan0 = dp dd = dp/dd' cos = subnormal OA ; so again, d (sin 0) = cos dd. This curve serves as an illustration of maxima and minima in polar coordinates. Thus, p = sin will be a maximum or a minimum when dp = cos = 0, when = - or 3tt and since dd 2 TV — sin 0, is negative when = s <2 2 p p = sin ;r = 1 is a maximum, while r 2 «0 2 — sin is posi- 142 DIFFERENTIAL CALCULUS trve when = 3tt 2 ' p = sin — = — 1 is a minimum. As the denominator of the fractional value of R is 2 for any value of 0, there is no inflexion point, R not being infinite at any point. EXERCISE Xm. Find R in each of the following curves : 1. The Cardioid p = a (1 — cos0). 2. The Lemniscate p 2 = a 2 cos 2 0. Where is an inflexion point ? 3. The Spiral of Archimedes p = a 4. The Logarithmic Spiral p = a e . R = 2V2a P /3. fl = a 2 /3 p. R = a (d 2 + D § 2 + 2 R =pVl + (log a) 2 . 94. Coordinates of Center of Curvature. — Let P (x, y) be any point on the curve ab, and C (a, 0) the corresponding center of curvature. Then PC is R and is perpendicular to the tangent PT. Hence Z BCP = z XTP = 4>, OA = OM - BP, AC = MP + BC; that is, a = x — R sin cf>, (3 = y -f R cos 0; or a = x R fs- f . r>dx (i) (2) PROPERTIES OF THE INVOLUTE AND EVOLUTE 143 Substituting in (2) . the values of R and ds, gives a = x — dx 2 = 2/ + dx 2 (3) 95. Evolutes and Involutes. — Every point of a curve, as in, has a corresponding center of curvature. As the point (x, y) moves along the curve in, by equation (3) above, the point (a, /3) wall trace another curve, as ev. The curve ev, which is the locus of the centers of curvature of in, is called the evolute of in. To express the inverse relation, in is called an involute of ev. The figure shows an arc of an involute of a circle. 96. Properties of the Involute and Evolute. — I. Since dx/ds = cos 4>, dy/ds = sin , and ds = R d(j>, dx = cos 4>ds = R cos d, (1) and dy = sin ds = R sin d — sin 4> dR = — sin dR, (3) d(3 = dy - R sin d + cos dR = cos dR. (4) 144 DIFFERENTIAL CALCULUS Dividing (4) by (3) gives d(3/da — — cot = — dx/dy. That is, the normal to the involute at (x, y),asP (Art. 95, figure), is tangent to the evolute at the corresponding point (a, (3), as C. II. Squaring and adding (3) and (4) gives da 2 + dp 2 = dR 2 . Let s denote the length of an arc of the evolute; then, da 2 + dp 2 = ds 2 . Hence, ds = ±dR; :. As = ± A#. That is, the difference between two radii of curvature, as C3P3 and C1P1 (Art. 95, figure), is equal to the corresponding arc of the evolute, C1C3. These two properties show that the involute in can be described by a point in a string unwound from the evolute ev. From this property the evolute receives its name. It may be noted that a curve has only one evolute, but an unlimited number of involutes, as each point on the string which is unwound would describe an involute. 97. To Find the Equation of the Evolute of a Given Curve. — Differentiating the equation of the given curve and using equations (3) of Art. 94, a and will be expressed in terms of x and y. These two equations and that of the given curve furnish three equations between a, jS, x and y. Elimi- nating x and y from these equations, a relation between a and jS is obtained, and this relation is the equation of the evolute of the given curve, which would itself be an involute of the curve found. Examples. — Find the equation of the evolute of the following curves: 1. The parabola y 2 = 4px. (1) dy _ 2 p d?y _ 4p 2 dx y ' dx 2 y 3 EQUATION OF THE EVOLUTE Substituting these values in (3) of Art. 94 gives a = Sx + 2p; 0= -y*/±p 2 ; /. x - (a - 2 p)/3, 7/ = - ^402? Substituting these values of z and y in (1) gives /3 2 = 4(a-2p) 3 /27p, as the equation of the evolute of y 2 = 4 px. The locus of (2) is the semi- cubical parabola. Thus, if iOn is the locus of (1), F being the focus, then eAv is the locus of (2) , where OA = 2 p = 2 OF is the minimum radius of curvature at 0, the point of maximum curvature on the parabola. (Example 2, Exercise XII.) 2. The ellipse a 2 y 2 + &V = a 2 6 2 . (1) aj/ _ _ b 2 x d 2 y _ 6 4 dx a 2 y } dx 2 a 2 y z Substituting these values in (3) of Art. 94 gives (a 2 - }f) x* 145 (2) a = = (a 2 -6 2 )i/ 3 \a 2 - b 2 ) ' y \a 2 - b 2 ) Hence, the equation of the evolute of the ellipse a 2 y 2 + b 2 x 2 = a 2 b 2 is (aa)3 + (6/3)i = (a 2 - 6 2 )l The evolute is C1C2C1C2. Ci is center of curvature for 146 DIFFERENTIAL CALCULUS A; C for P; C 2 for B; &' for A'; C 2 ' for B'. In the figure shown a = 2b; when a = 6 V2, then the center of curvature for B is at B' and for B' at £. When a < b V2, the centers for B and B' are within the ellipse. The points Ci, C2, CV, and C2 are cusps. The length of the evolute is evidently four times the difference between R at B (a, b), and R at A (a, 0); that is, (1, Exercise XII), 4(a 2 /6-& 2 /a) =4(a 3 -6 3 ) /a&. Corollary. — For circle, since a = b, the evolute is a point, the center of the circle. The involute of the circle is given by the equations, n a x = a (cos0 + 0sin0), ij = a (sin0 — 0cos0). AP is the arc of an involute of the circle. 3. The cycloid x — a vers -1 (y/a) =F ^2 ay — y 2 . dy _ V2 ay dx M //- dx 2 y ax" y Substituting these values in (3) of Art. 94: y = - ft x = a = 2 V - 2 a/3 .*. a = a vers -1 ( — j3/a) d= V /3 2 ; 2ap-0*. (1) The locus of (1) is another cycloid equal to the given one, the EQUATION OF THE EVOLUTE 147 highest point being at the origin; that is, the evolute of a cycloid is an equal cycloid. Thus, the evolute of the arc OP\ is the arc OCi, which equals Pix; and the evolute of P\X is Cix, which equals OP\. Since R = 2 V2ay (3, Exercise XII), CiPi = 4 a. Then OP x X = 2 • OCi = 2 • CiPi = 8a. Hence, the length of one branch of the cycloid is eight times the radius of the generating circle. (See Example 3, Art. 142.) If the figure shown be inverted, the principle of the cy- cloidal pendulum may be perceived. A weight, suspended from G by a flexible cord, may be made to oscillate in the arc OPiX, by means of some surface shaped like the arcs CiO and C\X causing a continuous change in the length of the cord as it comes in contact with the surface. The cycloidal pendulum is isochronal, as the time of an oscillation is independent of the length of the arc. (See Art. 237.) 4. The hyperbola b 2 x 2 — a 2 y 2 = a 2 b 2 . (aa)3 - (&/3)* = (a 2 + b 2 )*. 5. The equilateral hyperbola 2 xy = a 2 . (a + (3p - (a - /3)3 = 2 a*. 6. Find the length of an arc of the evolute of the parabola y 2 — 4 px in terms of the abscissas of its extremities. 148 DIFFERENTIAL CALCULUS Arc AC =CP-AO = 2 (* + P) s _ 2p (Example 1, figure) 7. Show that in the catenary y = a/2 \e a + e a ), a = x — y/a Vy 2 — a 2 , (3 = 2y. 8. Find the equation of the evolute of the hypocycloid X Z -J- y3 = a z, (a + 0)* + (a-/3)* = 2 a*. CHAPTER VII. CHANGE OF THE INDEPENDENT VARIABLE. FUNCTIONS OF TWO OR MORE VARIABLES. 98. Different Forms of Successive Derivatives. — As given in Arts. 67, 68, where x is independent dx may be taken as having always the same value and is accordingly treated as a constant; hence, d dy _ d 2 y d_ d dy _ d 3 y dx dx dx 2 ' dx dx dx dx 3 ' When neither x nor y is independent, -j- is a fraction with both numerator and denominator variable, and d dx = d 2 x } etc., hence, d dy _ dx d 2 y — dy d 2 x , . dx"dx~ dx* ' {) d_d_dy_ dx 2 d z y - dx dy d z x -3dx d 2 x d 2 y + 3dy (d 2 x) 2 dx dx dx dx b ' When y is independent, d 2 y = 0, d 3 y = 0, . . . ; hence, d dy _ dy d z x dx dx dx 3 ' d d dy 3 dy (d 2 x) 2 — dx dy d?x dx dx dx dx 1 do (2') 99. Change of the Independent Variable. — In some ap- plications of the Calculus it is necessary to make a differen- tial equation depend on a new independent variable in place of the one originally selected; that is, there is need to change the independent variable. 149 150 DIFFERENTIAL CALCULUS When x = 4>(z) and it is desired to change the independent d~y d 3 y variable from x to z; for -r\, -r\, . . . , respectively, the second members of (1), (2), . . . , above, are substituted; and in the resulting equation, for x, dx, d 2 x, . . . , their values gotten from the equation x = 4> (z) are substituted. Example 1. — Given y d 2 y + dy 2 + dx 2 = 0, in which x is independent, to find the transformed equation in which neither x nor y is independent; also the one in which y is independent. d 2 y Dividing both members by dx 2 , substituting for -r^ the second member of (1) Art. 98, and multiplying both mem- bers by dx 3 , gives y (d 2 y dx — d 2 x dy) + (dy 2 dx + dx 3 ) = 0, in which neither x nor y is independent. Putting d 2 y = 0, and dividing by — dy 3 , gives d 2 x dx 3 dx __ n y ay~ay~dy~ ' in which the position of dy indicates that y is independent. Example 2. — To change the independent variable from x to 6 in p [l + (dy/dx) 2 }\ d?y dx 2 given x = p cos 6,y = p sin 0, p being a function of 6. From the data, , dy = sin dp + p cos tf#, dx = cos dp — p sin dd, d 2 y = sin 6 d 2 p + 2 cos ddddp- p sin d0 2 , and d 2 x = cos dd 2 p- 2 sin ddddp- p cos (9 d0 2 . EXERCISE XIV 151 Substituting these values in value of R and simplifying, gives = \f + (dp/dey]t the ^^ o{ ^ . n ^ 9 ^ To change the independent variable from x to i/; for d 2 y/dx 2 , (Py/dx*, . . . , respectively, the second members of (1'), (2'), • • • , above, are substituted; or in the general result, as in example 1, make d 2 y = 0, d?y = 0, etc. Example 3. — Change the independent variable from x to y in 3 AW _dy<$y_ d 2 y/dy\ 2 = Q \dx 2 ) dx dx? dx 2 \dx) d 2 v d^v Substituting for -r\ and -r\ , respectively, the second mem- bers of (1') and (2') gives after reduction d 3 x d 2 x _ n ¥ 3 + ¥ 2 " ' in which the position of dy shows that y is independent. EXERCISE XIV. 1. Given x = cos 9, change the independent variable from x to in i& ^_ ^y , y _ A d2y _ dx 2 1 - x* dx + 1 - x> - U * Ari ^ d9» + y ~ °' 2. Given x = -, change the independent variable from z to z in 3.2^4.0x^4-^7; -0 y1r?9 ^4_„2„_o * e^ + ^dx + x 2 *'- - An5, ^+ a ^/-°- 3. Given x 2 = 4 2, change the independent variable from x to 2 in 4. Given x = cos 2, change the independent variable from x to 2 in 152 DIFFERENTIAL CALCULUS 5. Change the independent variable from x to y in g + «,--. )(g)* = o. a* J*.-* 6. Given 2 = , , , , to find the transformed equation when ydy -\-xdx' x = p cos 6, y = p sin 0, and p is independent. Arcs, z = p — - dp 100. Function of Several Variables. — A function may depend upon two or more variables having no mutual rela- tion, that is, independent of each other. Thus the volume of a gas depends upon the temperature and also upon the pressure to which it is subjected, and the temperature and the pressure may vary independently. A variable z is a function of the independent variables x, y, . . . when for each set of values of these variables there is determined a definite value or values of z. A function of two variables z — f (x, y), where x and y are independent, is represented geometrically by a surface, plane or curved according to the form of the functional relation ; and to each pair of values of (x, y) there corresponds a point on this surface. When x and y vary, the point takes another posi- tion, and it will take the new position either by x and y varying simultaneously or by one remaining constant while the other changes. 101. Partial Differentials. — A partial differential of a function of two or more variables is the differential when only one of the variables is supposed to change. Let z = f (x, y) be the surface shown in the figure, and P(x, y, z) the moving point; then if y is constant while x changes, P will move on the plane curve PA and dx may be represented by PM or P'M'-y on the other hand, if x is constant while y changes, P will move on the plane curve PB and dy may be represented by PN or P'N'. The differential of 2 as a PARTIAL DIFFERENTIALS 153 function of x, y being regarded as a constant, is denoted by d x z; and the differential of z when y alone is variable is denoted by d y z. These differentials are the partial differ- entials of z with respect to x and y, respectively. Note. — The partial d x z may in the figure be represented by the distance on the ordinate from the point M to the tangent TP, and so too, the partial d y z may be represented by the distance from N to the tangent T'P, both being negative in this case as z is decreasing. The A^ and the A v z are the distances from the points M and N to the surface curves through P 2 . 102. Partial Derivatives. — The partial derivatives of z with respect to x and y are denoted by ■=- and -7- , respectively, and they may be represented by the equivalent notation, f x '(x, y) and f v '(x, y). In the figure of Art. 101, consider P as the intersection of the curves CPA and C'PB, cut from the surface by the planes y = b and x = a, respectively; then the slope of the 154 DIFFERENTIAL CALCULUS dz curve CPA is given by the partial derivative -7- , and that of the curve C'PB by the partial derivative -p ; that is, the partial derivatives are the tangents of the inclination of the tangent lines at P to the axes of X and Y, respectively. The values of the slopes for some definite point P on the surface are gotten by substituting in the expressions for -r- and -j- , respectively, the corresponding values of x and y. Thus in this case (a, 6), or P' , being the projection of P on the xy- plane, a is substituted for x and b for y. 103. Tangent Plane. Angles with Coordinate Planes. — In the figure of Art. 101, let P be the point (x h y h Zi) ; PT, the tangent to CPA in the plane y = y\\ and PT ', the tangent to C'PB in the plane x = Xi. The equations of PT are Sl = [fi (a si)i y = Vh (i) and of PT', z-z l = L^J (?/ - 2/1), z = zi. (2) The plane tangent to the surface at P has for its equation, -*-[£l«» J *a +[*!<*-»>• (3) since it is determined by the two intersecting tangents, is of the first degree with respect to x, y, z, and is satisfied by (1) and (2). The equations of the normal through P are those of a line through (x h y h z x ) perpendicular to (3). Its equations are -* /[bI-'VKI--*-*- (4) The angles made by the tangent plane with the coordinate planes are equal to the inclinations of the normal to the axes. TANGENT PLANE 155 The direction cosines of the line perpendicular to (3) are proportional to [|], [g],-l. Hence, if a, (3, y, are the inclinations of the normal to OX, OY, OZ, respectively, °" B /[sl = ""/[SI = C0ST/ " l (5) Also cos 5 a + cos 2 /3 + cos 2 7 = 1. (6) From (5) and (6), in general, at any point (x, y, z), — ©r/'+e? + eif- -MS)V'+(£HI)" For the inclination of the tangent plane to XY, from (7), — ©T+6ST- (9) From (9), calling the tangent of the angle made by the tangent plane with the plane XY the slope, **-- vewt)* Example. — Find the equations of the tangent plane and normal, to the sphere x 2 + y 2 + z 2 = a 2 , at (x h y h 2i). 3« _ __ x $i - _y. dx~ z' dy~ z' |"^1 - _ £! r^il = - y -l. 156 DIFFERENTIAL CALCULUS Substituting in (3), z- Zi = - f* (x - xi) - v -r (y - 2/1), xxi + yyi + zzi = x x 2 + y x 2 + z Y 2 = a 2 . Arts. From (4) for the normal : (x - xi) ^ = (y - yi) ^ = z - si, xi yi Z\ xi 2/1 zi EXERCISE XV. 1. Find the equation of the tangent plane and its slope, for the ellipsoid, x 2 + 2j/ 8 + 3z 2 = 20, at (3, 2, +z,). Arcs. 3 x + 4 2/ + 3 z = 20; f. 2. Find the equation of the tangent plane to the elliptic paraboloid, z = 3 x 2 + 2 1/ 2 , at the point (1, 2, 11). Ans. 6 x -\- 8y — z = 11. 3. Find the equations of the tangent plane and normal to the cone, 3x 2 -2/ 2 + 2z 2 =0, at {x h y h z x ). a o 10 » x - xx y - y x z - Zi Ans. 3 xxi -2/2/1 + 2 ggi = 0; = - — -f- = -x- — 0X1 —2/i Z Zi iVote. — The equations of the tangent plane and normal are illusory if formed for the origin. Every tangent plane to the cone goes through the origin and there is no definite normal at the origin. When at special points on a surface the three partial derivatives of the function with respect to each of the three variables are all zero, there is no definite tangent plane or normal at the point. Such points are called conical points, the vertex of a cone being the typical case. 104. Total Differentials. — When z = f (x, y) is differ- entiated, both x and y varying, the total differential dz or df (x, y) is gotten. The derivations of the formulas for differentiation of al- gebraic, logarithmic and exponential functions, given *in TOTAL DIFFERENTIALS 157 Chapter II, hold when u, v, y, and z denote functions of two or more independent variables; hence the total differential of / (x, y) may be gotten by the principles established in those derivations. The total differential of a function of two or more variables is equal to the sum of its partial differentials. If z = / (x, y) , then dz = d x z + d y z = -^dx + ^.dy; and if v = fix, y, z), then, dv = dj) + d v v + d z v = -j- dx + -7- dy -f- -5- dz> y dx dy u dz where the last form of the partial differentials is another convenient notation. In the figure of Art. 101, dz is repre- sented by the distance on the ordinate from D to the tan- gent plane at P and is there negative, Iz being DP 2 , which is negative. The truth of this theorem has been illustrated geometrically in the derivations of d(uy) and d(xyz) in Arts. 28 and 29, and the theorem is readily established analytically. Thus, it has been found that all the terms of d(f(x, y)) are of the first degree in dx and dy; hence, if z = f (x, y), dz = (x, y) dx + 4>i(x, y) dy, (1) where \{x, y) denote, respectively, the sums of the coefficients of dx and dy in the several terms of dz. When x alone varies, (1) becomes d x z = (x, y) dx. (2) When y alone varies, (1) becomes d y z = 0i (x, y) dy. (3) Hence, from (1), (2) and (3), dz = d x z + d y z r= -£dx +^dy. 158 DIFFERENTIAL CALCULUS 105. If . = ,(,,,) = c, * -g* (!) for j|-3a*j,-0. ^rfj^ag. ax ax — y- 10. a* - y* = 0. ^ = ?/ ~^ logy - dx x 2 — x?/ log x TOTAL DERIVATIVES 159 n. *iog»-»iog*-o. *v. = y. ***v-v . ° * v ° dx x y log x — x 106. Total Derivatives. — If u = f (x, y, z), y = (x), and z = 0i (x), u is directly a function of x and indirectly a function of x through y and 2. The total differential, du = -r-dx +-j-dy + -r- dz (by Art. 104) becomes by dividing by dx, du _ du du dy du dz n . dx dx dy dx dz dx' du where -=- is the total derivative of u as a function of x. dx Corollary 1. — If u = f (y, z), y = (x), and z = #1(2), du _ du dy du dz ... dx dz/ dx d? dx Corollary 2. — If u = f (y) and y = 0(x), du _ du dy , . dx dy dx' where -7- is the derivative of a function of a function, and dx (3) is the formula that is the subject of Remarks in Art. 19. Corollary 3. — If u = f (x, y, z) and x, y and z are inde- pendent of each other, they may be regarded as functions of time t; hence, the expression for the total differential du above becomes by dividing by dt, du _ du dx du dy du dz ... du where -7- is the total time-derivative or rate of change of u. Similarly, when z = f (x, y), x and y being functions of t, dz _dz_ dx .dz_ dy ,_. dt~dx ~dt + dy~dt' ^ ' 160 DIFFERENTIAL CALCULUS If y is a function of x, as y = 4>{x), putting x for t in (5), gives dx ete cfa/ eta 107. Illustrative Examples. — Example 1. — The edges of a right parallelopiped are 6, 8 and 10 feet. They are increasing at the rate of 0.02 foot per second, 0.03 foot per second and 0.04 foot per second, respectively. Show at what rate the volume is increasing. Let volume — u = xyz, then by (4), Art. 106: du dx , dy . dz -17 = VZ -77 + XZ -77 + XV -T7 dt u dt dt * dt = 80 X 0.02 + 60 X 0.03 + 48 X 0.04 = 1.60 + 1.80 + 1.92 = 5.32 cubic feet per second. See Art. 29 where du (= dV) is shown geometrically by figure. Example 2. — Given the formula for gas, pV = KT, where p is pressure, V is volume, T is temperature, and K is a constant. Let K = 50, and let the volume and tempera- ture at a given time be V = 5 cu. ft. and T = 250°. The corresponding pressure is po = 5 ° X - 25 ° = 2500 lb. per sq. ft. o If in this state the temperature is rising at the rate of 0.5 degree per minute and the volume is increasing at the rate of 0.2 cu. ft. per minute, required the rate at which the T pressure is changing. Here p = 50 ^> , dp 50 dp _ ft T whenC e _£ = y , -£__»_. Hence, in the given state, 1 50 T=T *'l ILLUSTRATIVE EXAMPLES 161 n- V=V and %\= -™^° = Given ^ = 0.5 and ^ = 0.2. dt at Then by (5) Art. 106; that is, the pressure is decreasing at the rate of 95 lb. per sq. ft. per min. x^ iP" z^ Example 3. — A point on an ellipsoid — + — + t^ = 1, in the position x = 3, y = — 4, moves so that x increases at the rate of t two units per second, while y decreases at the rate of three units per second. Find the rate of change of z. Here dz_ 7_x dz_ 7y y 2 dx dy _ ~dt~ A ~dt~ 6 > dz 14x 21 y dt ~ 36 \/-S-S »^-S-SJ (by (5) Art. 106) 3, -4 -r- = 7= units per sec, the rate of change of z. dt ' 15 vn EXERCISE XVH. 1. w = z 2 + y 3 + zy, 2 = sin a:, y = e x ; find •=— ■ Ans. ■r- = 3e 3I +« I (sin x + cos x) + sin 2 x. 2. u = vx2 + t/ 2 , ?/ = rax + c. dw _ (1 + m 2 ) x + rac dx Vx 2 + (rax + c) 2 162 DIFFERENTIAL CALCULUS 3. u = sin -1 (y — z), y = 3 x, z = 4 x 3 . 4. u = tan" 1 ^ , x 2 + y 2 = r 2 . du dx du dx du dx 5. it = log (z + y), y = Vx 2 + a 2 . Vi -x 2 1 Vr 2 1 Vx 2 + a 2 6. With the same data as in illustrative Ex. 2, when the pressure of the gas is increasing at the rate of 40 lb. per sq. ft. per sec. and the temperature is falling at the rate of 1 degree per sec, find the rate of change of the volume. A dV ai ft Ans. -rr = — 0.1 cu. ft. per sec. 7. A point on a elliptic paraboloid z = 2 x 2 + 5 t/ 2 , in the position x = —3,y = l, moves so that the rate of change of x is 3 units per sec, and that of y is 2 units per sec Find the rate of change of z. Ans. -r = —16 units per sec 108. Approximate Relative Rates and Errors. — The method of Art. 41 for rinding the errors or small differences in a function, due to slight variations or inaccuracies in the independent variable, is applicable to a function of two or more variables. Since when an area A = / (x, y), the rela- tive rate of increase of A is dA dA dl_,dy fx 0, y) f y '(x,y) = f x , y '(A) . A "*" A ~ A '"*" A A ' hence, AA^Az + ^Ay (1) AA dA Ax , dA Ay /n , and "d^X + ^T (2) are approximate relations. When, for example, the area of a rectangle is given by A = xy, and therefore, dA = xdy + y dx, when x and y are the measurements and dx and dy the errors or inaccuracies, then dA gives the approximate error in area due to the errors dx and dy. APPROXIMATE RELATIVE RATES AND ERRORS 163 If a rectangle is laid out 1000 ft. on one side and 100 ft. on the other, and the tape is 0.01 ft. too long; then. by (1) AA =y>kx + x-ky = 100 X 0.1 + 1000 X 0.01 = 10.00 + 10.00 = 20 sq.ft. is the approximate error and the exact error is 20.001 sq. ft., found by more laborious computation. The approximate relative error is by (2) AA = 20 1 A " 100,000" 5000' making the percentage error 100 AA 1 ft no * i -. — = ptt or 0.02 of 1 per cent. A 50 * EXERCISE XVIIL 1. In the illustrative Example 1 of Art. 107, suppose the error in measuring the edges was 0.02 ft., 0.03 ft., and 0.04 ft., respectively, find the approximate error in the volume computed with 6, 8 and 10 ft. as the edges. Ans. 5.32 cu. ft. (Exact error, AV = 5.339624 cu. ft.) 2. The total surface of a cylinder with diameter equal to altitude is to be gilded at a cost of 10 cents per square inch. If the altitude is measured as 24 in., find the maximum error in cost, measurement being accurate to ^ in. 3. The period of a pendulum is T = 2tt y — . Find the greatest error in the period if there is an error of dzxV ft. in measuring a 10 ft. L, and g, taken as 32 ft. /sec 2 , may be in error ^ ft. per sec 2 . Find the percentage error: Ans. 0.0204 sec, || per cent. 4. In estimating the number of bricks in a pile, if the pile is measured to be 8 X 50 X 5 ft., and the count is 12 bricks to the cubic foot, find the cost of the error when the tape is stretched 2 per cent beyond the standard length, bricks being sold at S10 per thousand. 5. If the side c of a triangle ABC is determined by measuring the sides a and 6 and the included angle C, show that the error Ac, due to inaccurate measurements, is given approximately by the equation, Ac = Aa cos B + A6 cos A + aAC sin B. 164 DIFFERENTIAL CALCULUS 6. If the horse power of a steamship is given by the formula H = Kifi D*, show that the increase in horse power, due to an increase Av in the speed and an increase AD in the displacement, is given approxi- mately by the equation, AH = SKifiD 1 • Av + % KifiD-i - AD. 7. Show that the relative error in the area of the ellipse due to inaccurate measurements of the semi-axes a and b is given approxi- , . , AA b' Aa + a' Ab matelyby -j- = ~ b 8. The equation for the length L and the period T of a pendulum being 4 tt 2 L = T 2 g, if L is calculated taking T = \ and g = 32 ft. /sec 2 , while the true values are T = 1.02 and g = 32.01 ft. /sec 2 , show that the approximate error in L is AL = 0.0326 . . ft., and the percentage error about 4 per cent. 9. In determining specific gravity by the formula s = A/ A — W, where A is the weight in air and W the weight, find (a) approximately the maximum error in s if A can be read within 0.01 lb. and W to 0.02 lb., the actual readings being A = 9 lb., W = 5 lb., find (6) the maximum relative error. Ans. (a) As = 0.0144; ,.. As 23 23 (6) T = 3000 = 36 perCent - 109. Partial Differentials and Derivatives of Higher Orders. — If only one of the independent variables is supposed to vary at the same time, by successive differen- tiations there are formed the successive partial differentials d x 2 u, d y 2 u, d x 3 u, d y 3 u, ... or d 2 u 1 , d 2 u 7 „ d 3 u . , d 3 u , , w dx ' df dy > a?^- w d * For example, if u = x 2 + xy 2 + y 2 , (1) d x u = (2 x + y 2 ) dx, d x 2 u = 2 dx 2 , d x s u = 0; d y u = (2 xy + 2 y) dy, d 2 u = (2 x + 2) dy 2 f d y z u = 0. If u is differentiated with respect to x, then the result with respect to y, there is gotten the second partial differential^ d ™ u or £r y dxdy - INTERCHANGE OF ORDER OF DIFFERENTIATION 165 For example, if u = x z + x 2 y 2 , (2) d x u = (3 x 2 + 2 xy 2 ) dx, d xy 2 u = 4 xy dx dy. Similarly, the third partial differential d v Ju or 2 d?/ da; 2 denotes the result gotten by differentiating u once with respect to y, then this result twice successively with respect to x. The symbols for the partial derivatives are: d 2 u d 2 u d 2 u d z u d 3 u » ~; — r~ i "r - ^ ' ~r~?> dx 2 dx dy dy 2 dx z dy dx 2 ' In getting the successive partial differentials and deriva- tives of u or / (x, y) } dx and dy are treated as constants, since x and y are independent variables, varying by uniform increments. The equivalent symbols for the higher partial derivatives by another notation are for/ (x, y), /."(*,*), Uy\x,y), /„"(*, y), h"\x,y), f,*'"(x,y) t 110. Interchange of Order of Differentiation. — r* t( \ &u d 2 u , . If »-/fe»). d^y^dtTx' (1) d 3 it _ d 3 w. _ d 3 w da: 2 d?/ ~ dxdydx~ dydx 2} '' that is, (f w is differentiated successively m times with respect to x and n times with respect to y, the result is independent of the order of these differentiations. It can be shown that the order is always a matter of indifference if f x '(x, y), f xy "{x, y) or f y '(x, y), f yx "(x, y) are continuous functions of the two variables (x, y) taken together. In most cases that call for the application of the methods of the Calculus to physical problems the partial derivatives give the same result in whatever order the differention is done. 166 DIFFERENTIAL CALCULUS For example, to verify the theorem in some cases: Example 1. — Given u = e x cos y; du d 1 Tx = emsy ' dy\ K dxj | d 2 U = -j — r- = — e x sin 1 ' dydx du . d I -=-e* S my, ^( 4yj I - dhi - e'sin ' dxdy mple 2. — Given u = iz. du log z d 2 U 1 . dx y dz dx yz ' du X dz yz 1 d 2 u 1 . dx dz yz ' du x log z d 2 u log z . dy y 2 dx dy y 2 du log z d 2 u log z . dx y dy dx y 2 du x d 2 u x . dz yz 1 dy dz y 2 z ' du x log z d 2 w x dy y 2 dz dy y 2 z EXERCISE XIX. Verify the identities (1) and (2) of Art. 110 in each of the following nine examples: 1. u = cos (x + y). 2. u = e x smy. 3. u = cosxy 2 . 4. u '= x z y 2 + ay 3 . 5. u = log (x 2 + y 2 ). 6. u = ?/ x . 7. u = xy cos (x + w). 8. m = tan -1 -. 9. u = sin 2 x cos y. 10. llu = {x + y)\ *_ + „_=_. 11. ifM = (x 2 + M *g +9 *aS+*3- a 1 d 2 M a 2 M d 2 U _ 12, Uu = (x 2 + 2/2 -f-^i' dx 2 + y) an d the question is whether it could be so 168 DIFFERENTIAL CALCULUS gotten. In general, if M and N are any chosen functions of x and y, does a function of the independent variables (x, y) exist that will upon differentiation give Mdx + Ndy? If there is such a function u = F (x,y), then dw = Tx dx + fy dy - I (4) Now if the differentiation of the given function gives Mdx + Ndy, (1) a comparison with the exact differential given in (4) gives M = d~x' N = Ty' (5) that is, M and N must be the partial derivatives of the function u with respect to x and y, respectively. According to the theorem of Art. 110, differentiating (5) gives dM = d 2 u = 5iV cfy dx dy dx Hence, if M = -j- and JV = -r- , it is manifest that 2M = -^ (6) di/ dx is the necessary condition that M dx + N dy may be gotten by the differentiation of a function F (x, y), and it may be shown that it is a sufficient condition. • When the condition (6) is satisfied, M dx + JV dy is an exact differential; when the condition is not satisfied, M dx + AT di/ is an inexact differential. Example 1. — Given M dx + N dy = ydx + xdy. TT IT AT dM , dN ^ Here M = y, N = x, -j- =1, -j- = 1. * a?/ ax The condition (6) is satisfied and ydx + xdy is an exact differential. The test is hardly needed in this simple case, EXACT DIFFERENTIALS 169 as it may be seen at once that the function sought is xy+C, where C is a constant, positive or negative, or zero. Example 2. — Given M dx -\- N dy = y dx — x dy. Here, since -=— = 1 and -=— = — 1, the condition (6) is not satisfied ; hence, y dx — x dy is an inexact differential, and no function of (x, y) exists, the differentiation of which will give this differential . Note. — If the equation ydx — xdy = is given, it may be changed to an exact differential equation; M dx + N dy = 0, being called an exact differential equation when M dx + N dy is an exact differential. Thus, multiplying by y~ 2 , the equation given becomes ydx - xdy _ V 2 which is exact, and the function F (x, y) is given by x/y = C. Again, multiplied by 1/xy, the equation given becomes *? _ c !e = o x y which is exact, and the function F (x, y) is given by log x/y = log C: Either of these results evidently implies the other. Multiplying the equation given by —x~ 2 gives F (x, y) by y/x = Ci. Example 3. — Given Mdx + N dy = -dx + log x dy. XJ JLT V XT 1 BM \ dN 1 Here M = -, N = log x, -^ = -, —- = -■ x dy x dx x The condition (6) is satisfied and the differential is exact. It is easy to recognize that F (x, y) is in this case y log x. Example 4. — Given M dx + N dy = sin y dx + x cos y dy. Here M = sin y, N = x cos y, -=— = cos y, -j- — cos y. The condition (6) is satisfied and the differential is exact. The function F {x, y) may be seen to be x sin y. 170 DIFFERENTIAL CALCULUS Example 5. — Given x dy — y dx. Change to value in polar coordinates, by x = p cos 0, y = p sin 0; x dy — y dx = p 2 dd. Dividing by x 2 , xdy-ydx = _ldB ^ X 2 p 2 cos 2 where the differentials are exact, and the function F (x, y) is y/x = tan 0. (See Note Example 2.) EXERCISE XX. Determine which of the following differentials are exact, and for such as are exact find the functions that differentiated would give them : 1. y sin 2 x dx + sin 2 x dy. Ans. y sin 2 x. 2. (ye* + e v) dx + (e x + xe v ) dy. Ans. ye x + xe y . 3. (y 3 - 2 xy) dx 4- (3 xy 2 - x 2 ) dy. Ans. y z x - x 2 y. 4. v n dp + npv n - l dv. Ans. pv n . 5. e x sin y dx + e* cos y dy. Ans. e x sin y. v 2 6. — da: + x log z d?/. Ans. Inexact. x 7. (x 2 — y)dx - xdy. Ans. \ x 3 — xy. 8. e x (x 2 + y 2 + 2 x) dx + 2 e x 2/di/. Ar?s. it is a natural inference that an inverse operation to differen- tiation should yield the function. This inverse operation, the opposite of differentiation, is called integration and to integrate any given function (which when continuous is always the derivative of some other function) means to find that other tunction whose derivative is the given function. The function to be found is called an integral of the given function, which is called the integrand; that is, a function is an integral of its differential. The process of finding an integral of a given function is integration, the inverse of differentiation; that is, integration is anti-differentiation and an integral is an anti-differential. 171 172 INTEGRAL CALCULUS When dy = d{f(x)); d~ l (dy) = d~ l (df (x)) , (read "the anti-differential of dy equals the anti-differential of d{J(x))" is the inverse expression, reducing to y = f (x) } as the two symbols neutralize each other. The sign of integration is, however, / , an elongated S; and this symbol indicates that the differential expression before which it appears is to be integrated, the whole expression denoting the integral itself. Thus Cdy^d-^dy) and f d (f (x)) = d~ l (df (x)) ; the sign of integration and the symbol of differentiation indicating inverse operations here neutralize each other, so / d(f(x))=f(x). There is here a close analogy with the algebraic signs of evolution and involution; for example, Vx 2 = x, the two symbols indicating inverse operations neutralizing each other. The analogy extends further to the fact that, while the operation of raising a given number to the second or other power is a direct operation and involves no difficulty in any case, the inverse operation of extracting a root may not be done so directly and in many cases can be done approximately only. While it has been shown that every continuous function has an integral,* this integral may not be expressible in terms of the elementary functions. In such cases, however, an approximate expression for the integral may be obtained by infinite series or by the measure- ment of an area representing the integral. Most of the functions that occur in practice can be integrated in terms of elementary functions, either directly by the knowledge acquired from differentiation, by reversing the rules of differentiation, or by reference to a table of integrals. ' * By Picard, in Traite d' Analyse. INDEFINITE .INTEGRAL 173 Except for simple differential expressions the process of integration is less simple and easy than the process of differ- entiation. Just as any finite number can be raised to a power, so can any finite continuous function be differentiated; and as the roots of some numbers can be expressed approxi- mately only, so the integrals of some functions can be expressed approximately only. There is one function whose integral is not some other function but is the function itself. This is the function e x , whose derivative is e x . As I e x dx = e I e x dx = e x , so Vl = 1; the particular analogy in this exceptional case is manifest. 114. Indefinite Integral. — When g = /' (x) or dy = f Or) dx, y = Jf (x) dx, read "y is equal to an integral of f (x) dx." An integral of dy is evidently y, and / (x) is an integral of its differential f'(x)dx. Thus integrals of many simple differential expressions are known directly, by merely recalling the function which differentiated results in the given expression. However, since the differential of any constant term of a function is zero, the function sought may contain a con- stant or constants no indication of which appears in the given differential or derivative. Hence, the integral of a differential expression is in general indefinite, owing to the lack of knowledge as to the existence or value, if existent, of constant terms of the function sought. If F (x) is a function whose derivative is / (x), then / / (x) dx = F (x) + C, is the indefinite integral, where C is a general constant, called the constant of integration, de- noting a value either positive or negative or zero. 174 INTEGRAL CALCULUS When dy dx 115. Illustrative Examples. — Example 1 f (x) = m, is given as the constant slope of y = / (x) ; then y = I mdx = mx + C. The result is indefinite be- m, is the slope of y = mx + C, y = mx cause dy dx Cor y = mx. The constant C added to right member of the equation includes all constant terms, if any, of the function; for if the result be written y + C" = mx + C", then y = mx + C" -C = mx+C. The letter C, often omitted, should be written as part of the result of the integration. Data may be available in some cases to make the value of C known, or to eliminate it, and thus to make the result determinate. In this example, if it is known that the function has the value b when x is zero, then y = mx + b, since C is equal to b when x is zero. If y is —b, or if y = 0, when x = 0; then y = mx — 6, or y = mx. As shown in the figure the function is a straight line making an angle ( = tan -1 m) with the X-axis, the con- stant of integration being the F-intercept. The indefinite or general integral is y = mx + C, any straight line with slope m. ILLUSTRATIVE EXAMPLES 175 Note. — When the value of C is determined the integral is called a particular integral. Example. 2. — When -~ = 2 x is given as the rate of change of y with respect to x, then y = I 2 x dx = x 2 + C, where x- + C is the general integral of 2 x dx, since the differential of (x 2 -f C) is 2 a; efo. Here z 2 + C is a function whose rate of change is 2 x, and if the rate of change is that of the ordinate to the abscissa, or the slope of a curve, then the integral, y = x 2 + C, is the equa- tion of the curve. The locus of the equation is a parabola with its vertex at a distance C above or below the origin, or at the origin, according as the value of C is positive, negative, or zero. If y is known for some value of x, then the value of C is easily determined. For instance, if it is known that the point (a, 6) lies on the curve, then y = x 2 + C, must be satis- fied by the coordinates (a, b), giving b = a- + C, and, there- fore, C — b — a 2 . Hence, the particular parabola is y = x 2 + b — a 2 . If the curve is known to pass through the origin, then, since C is zero, y = x 2 is the parabola with vertex at the origin. dA Example 3. — If a given derivative -r- = 2 x represents the rate of change of an area A to a length x, then A = j 2 x dx = x 2 + C, is an area where x 2 may represent the area of a variable triangle formed by the straight line y = 2 x, the ordinate at any value of x, and X-axis. In the figure 176 INTEGRAL CALCULUS shown the area A is zero when x is zero, and therefore C is zero. The area of any triangle being one-half the product of base and altitude, the result of the integration, A — x 2 , is seen to be true. The area A = x 2 + C may represent a square of side x and some additional area represented by C, undetermined, as it might be positive, negative, or zero — ■ the derivative of the area in either case being the given rate 2 x. dA Example 4. — If the given derivative is -p- = x 2 , then -/ X A x 2 dx = -= + C, since d (f +c ) = x 2 dx. Here the area ■=■ is that bounded by the curve, a parabola y = x 2 , the ordinate at any value of x, and the X-axis.* In the figure shown the area A is zero when x is zero, and there- fore C is zero. It is seen that the area OPM is exactly one-third of the area of the circumscribed rectangle. Hence, the area OPN between the curve, the abscissa at the end of any ordinate, and the F-axis, is two-thirds the area of the same rectangle. ILLUSTRATIVE EXAMPLES 177 Example 5. — The acceleration of a falling body being nearly constant near the earth's surface, it is required to find the velocity and the distance after any time. If s denotes the distance along a straight line positive upward d 2 s and t the time, then -== is the acceleration. at 2 dv d 2 s di a when t = 0. 8 = Jds= j -32.2 tdt + 2000 dt= -lQ.lt 2 +2000 1 + (C = so = 0), s = so = 0, when t = 0. To find the time of rising, make v = = -32.2 t + 2000; /. t = 62.1 sec. 178 INTEGRAL CALCULUS To find the height i\ will rise, s = -16.1(62.1) 2 + 2000 (62.1) = 62,112 ft. To find the time of flight, s = = -16.1 t 2 + 2000 t; :. t = 124.2 sec. and t = 0. Hence, the time of falling is the same as that of rising, since the time of flight is twice that of rising. The height it will rise may be found, by making v = in 2 2 o„ . • c v * ( 20QQ ) 2 R9 no* v z = v 2 -2gs; .. s = ^- = 64 4 = 62,112 It., the same as above. Remarks. — These examples illustrate the important fact that the knowledge of the rate of change of a quantity to- gether with the knowledge of its original value, makes possible the complete determination of the value of that quantity at any time. This must be so, since two different quantities with the same rate of change always have a con- stant difference, the rate of change of their difference being zero. This is in accordance with the undoubted fact that if the rate of change of a quantity decreases to zero and remains zero, the quantity ceases to change at all, being then constant. The fact is formulated in principle (iv) of Art. 116, and is the converse of the fact that if a quantity is constant, its rate is zero. 116. Elementary Principles. — While there is a general method of differentiation, for the inverse process of integra- tion no general method has been devised. For the integra- tion of the various differential expressions, rules have been formulated and special methods have been found, one or more of which provide for every case in which integration is possible. These rules or formulas are derived or disclosed through knowledge of the rules of differentiation; in fact, the rules most used are merely directions for retracing the steps taken in differentiation. ELEMENTARY PRINCIPLES 179 Elementary principles that apply in integration may be expressed as follows: (i) Jf(x)dx^F(x)+C, if dF(x)=f(x)dx. — = log x + C, since d (log x) = — x x This principle furnishes the most direct proof of formulas for indefinite integration, and provides a decisive test of the correctness of the result of any integration. Thus, lx n dx = — — r -f- C, since d\ — —) = x n dx\ J n + 1 \n-\-lJ i /Ifdx == — r + C, since dU — A=b x dx. log b \\og bj In this manner the test can be applied to prove any formula, or to verify the result of the integration of any expression. (ii) A constant factor can be transposed from one side of the sign of integration to the other, and a constant factor can be introduced on one side, if its reciprocal is introduced on the other, without changing the value of the integral. For, if a is a constant, I ay dx = a I y dx, since the differentiation of the equation gives, ay dx = ay dx. Hence, fydx = \faydx = af\ydx. (iii) The integral of a polynomial is equal to the sum of the integrals of its several terms. For / (<£+ x — x 2 ) dx = / adx + I xdx — I x 2 dx, since the differentiation of the equation gives a dx + x dx — x 2 dx = a dx + x dx — x 2 dx. 180 INTEGRAL CALCULUS (iv) fo = C, since dC = 0. The integral of zero is a constant. ds Thus, if -=7 = v, where v is constant velocity, d^s dv -Tp = -r = 0; that is, acceleration is zero, hence . js- /.-«-. 117. Standard Forms and Formulas. — There follows a list of standard integrable forms, that is, differential functions whose integrals can be expressed in finite forms involving no other than algebraic, trigonometric, inverse trigonometric, exponential, or logarithmic functions. To integrate a func- tion that is not expressed in terms of an immediately inte- grable form, it is reduced if possible to one or more of such forms and the formula applied. The formulas in general are gotten by merely reversing theiormulas for differentiation, and each can be proved by the principle (i) of Art. 116. The list will be found to contain the one or more than one integral to which every integrable form is reducible. These forms may, therefore, be called fundamental although only the first three are really fundamental, since each of the others by substitutions can be reduced to one of the three. While the list is of standard integrable forms, it may be supple- mented by other integrable forms; but no list of forms is exhaustive, even when extended into tables of integrals. After the acquirement of familiarity with the rules of differ- entiation, and the common methods of reduction with the standard integrals, the use of tables of integrals for the com- plicated forms is recommended; much time otherwise given to formal work in integration being thereby saved. STANDARD FORMS AND FORMULAS 181 / x n+l x n dx = — j-rr + C, where n is not —1. tt + 1 /dx — = log x + C = log x + log c' = log (c'x). til. fb*dx = r ^- 1 - + C. J log 6 IV. f(?dx = e x + C V. I sin z dx = — cos x + C, or vers x + C. VI. / cos x dx = sinx + C, or — covers a; + C. VII. / sec 2 xdx = tanx + C. VIII. / esc- x dx = — cot x + C. IX. I sec x tan x dx = sec x + C. X. / esc x cot x dx = — esc z + C. XL / tan z dx = log sec x + C = — log cos x + C. XII. / cot xdx = log sin x + C = —log esc x + C. XIII. / esc xdx = log tan ■= + C = log (esc x — cot x) -f C. XIV. J\ecs(fc = logtan(| + |) + C = log (sec x + tan x) + C XV. f- r ^ = -tan- 1 - + C, or --cof^ + C. J a 2 + x 2 a a a a XVI. f-^ = ^log^+C = -tanh- 1 -+C'. (x 2 a 2 ) 2a °jc— a a a 182 INTEGRAL CALCULUS XVII. r^ = ^log^+C=-icoth-^+C'. (x 2 >a 2 ) J x 2 —a 2 2a x+a a a = 2-a^S +C =4 tanh -1 +C - <*<*> XVIII. f^= 2 = ^l + C, or -«ri5 + (T. XIX f 7 ^ = log(x+V^+^)+C,orsinh- 1 --f-C'. J vf + a 2 a XX f / dX = log(z+ViW)+C,orcosh- 1 -- r -C'. J Vx 2 — a 2 a XXI / — 7= = - sec -1 - + C, or — csc^-H-C. J x Vx 2 - a 2 a a a a XXII. f a/o * 2 = vers" 1 ? + C, or - covers" 1 ^ + C". J V2ax — x 2 a a The two forms of the integral in several of the formulas correspond to different values of the constants denoted by C and C. Thus in formula V, vers x -f- cos x = 1 = C — C", and similarly in XVIII, sin- 1 - + cos- 1 - = ^=C , -C. a a 2 When the differentials of two functions are equal, their rates are equal; therefore, the functions will be equal or differ by a constant; hence the variable parts of the indefi- nite integrals of the same or equal differentials are equal or differ by a constant. 118. Use of Standard Formulas. — When a given func- tion to be integrated is not expressed in a form immediately integrable, by various algebraic and trigonometric trans- formations or substitutions, the effort is made to reduce it to one or more of the standard forms so as to apply the formula. When different methods are used the results may have USE OF STANDARD FORMULAS 183 different forms, but upon reduction they will always be found to differ (if at all) only by a constant, in accordance with the statement above. Formula I is the standard formula for the Power Form. It is of most frequent application, and it may be expressed in words as follows : The integral of the product of a variable base with any con- stant exponent (except — 1) and the differential of the base is the base, with its exponent increased by 1, divided by the in- creased exponent, and a constant. The proof has been given in (i), Art. 116; it may be derived thus: since f2xdx = x 2 + C, I 3x 2 dx = x 3 + C, etc., (ft + \)x n dx = x n+l + C; hence, in general, /< / /v.n+1 x n dx = — r-r + C- ft -+- 1 When a given integrand is a fraction with denominator to a power, it may become this form by bringing up the variable quantity with change of exponent's sign; but, since the variable quantity is represented by x in the formula, it is essential that the differential of the variable quantity, and not merely dx, be present in the integrand before the formula is applicable. If a constant factor is lacking, it may be supplied in ac- cordance with (ii), Art. 116; but it should be noted that the principle is only for constant factors. The value of an integral is changed when a variable factor is transferred from one side of the sign / to the other; thus, Cx 2 dx = i x s + C, but x I x dx = | x 3 + C. 184 INTEGRAL CALCULUS When a change of sign is needed, the constant factor — 1 effects the change. Thus, for an example, r xdx = _1 f( a 2_ x 2yh(- 2x dx)=-V~tf^ + C. J v a 2 — x 2 ^ J To verify : fd (-Vo 2 -a^ + C) = f- \ (a 2 - x 2 )-* (-2 xdx) i xdx ■. , as given. Va 2 — x 1 When a variable factor is lacking, resort may be had to expansion and then application of the formula to each term of the polynomial. Thus, for an example: f2(l+x 2 ) 2 dx = 2 f(l + 2x 2 +x*)dx = 2(x+ix*+%x b )+C. When the numerator is of higher power than denominator, reduce by division and then apply formula or formulas, thus : = ix 2 + x + 2\og(x-l)+C. If n = — 1, formula I gives a result that is not finite; but when n = — 1, the form reduces to form II and that formula applies. Thus, I x~ l dx = / — = log x + C. Formula II may be stated in words as follows: The integral of a fraction whose numerator is the differential of its denominator is the Napierian logarithm of the denomina- tor, and a constant. The result will be real only when x is positive. When X > a> J x~^a = l0g (x ~ a) + C; but, if x < a, / — — = / ■ * = log (a — x)+ C. J x- a J a- x &v J USE OF STANDARD FORMULAS 185 Formula III may be stated in words as follows: The integral of the product of a constant base with a variable exponent and the differential of the exponent is the base, with exponent unchanged, divided by the Napierian logarithm of the base, and a constant. Here the base b must be positive and not unity. Formula 77 is the special case of III, the Napierian logarithm of base e being unity. In applying these two formulas to given integrands, it is essential that the differential of the variable quantity, and not merely dx, be present. Thus, for examples : /*** = wS h "^ h ^ dX = I&6 + C - I e x ' n dx = n I e x ' n — = ne x ' n + C. C^*dx = ^— Cir^\0gTT^dx = ^- + C. J 3 log 7T J 3 log 7T The following are examples of integration by one or more of the first four standard formulas. EXERCISE XXI. In these examples the results may be verified by (i), Art. 116, and the verification should be made where the result is not given. 2. f(ax + b) n dx = - C(ax + b) n adx = {aX . + , 6) ** + C. J a J a (/! t 1) 3. C(2a + 3bxydx. 4. C ^ dx a . J J (a 2 + z 3 )* 6. jYi+^Ydx. 6. fr-v-tzydt. 7. / 2 »*(£ + l) d V- 8 - f^2Y+lds. 9. C \ r 2~pxdx = V2p Cx* dx = \ vTpz* (= f x V2px) + C. 10. f {ax n +b) p x n ^dx = — C (ax n +b) p nax n - l dx = ^^r^+C. 186 INTEGRAL CALCULUS 1. f5xVl-2x*dx=-$j'a-2&)t(-±xdx)=-$(l-2a*)*+C. 2. jVT^J x e x dx= - f(l-e x )H-e x dx) = - § (1 - e x )% + C. „ /"adx f - n j axi ~ n . /^ 3. I — =- = a I x n dx = ^ h C. J x n J 1 — n /£ a/vsa- a: 2 ) dx. 6 - iVS? dx " 3 JVt5? d * - a log (a + te "> + c - /• - (2 qg - x 2 ) dx _ - (3 ax 2 - x 3 ) 1 J (Sax* -x*)* 2 22. /^tto-.-f + f-logCr + D + Cf. 23 ' /ifc -I* <**>+* 24. f !?*" 1 "? dx = x + log (3x - 1) § + C. ./ o x — 1 or f dx r e~ x dx J e x (3+e- x ) ~ J 3+e~ x ' nn C sin 2 x , __ /* cot x , 26. I — — r-^- dx. 27. ( i -. — dx. J 1 + sin 2 x J log sin x 28. f{e x -e- x ¥dx= ^ X ~^ -2x+C. 29. JV + e~«J dx. 30 - J?fif - /(- & + StS) = 2l0g (e * + 2) " * + a -j^r-rdx. 32. fa«fe*(te = 1 ° . , + C. e z + 1 J log a + log 6 33. f£»|* JVco9x sina . d;r 22. JV an(ax) sec 2 (ax)dx. 120. Derivation of Formulas XVI, XVII, XIX, and XX. - By the application of Formula II, the following results: DERIVATION OF FORMULAS XVI, XVII, XIX, AND XX 189 For XVI, put — x 2 2 a \a + x a — Xj dx r dx j_ r dx j^ r z J a 2 — x 2 2aja + x 2a J a = ^ log(a + x) _ _L log(a _ x) + c = s- log h C 2a a — x = - tanh" 1 - + C". (x 2 < a 2 ) (XVI) a & or r dx r -dx = i ^ a 2 ) 2a * x — a a a y ' The first or second of these results is used according as a — x or x — a is positive; that is, the form of the result which is real is to be taken. For XVII, put 1 = 1 / 1 LA x 2 — a 2 2 a \x — a x + a J r dx j_ r dx l r dx J x 2 — a 2 2ajx — a 2a J v z + a = 2^ log (x- a) - — log (x + a) + C 1 , x — a n = - - coth" 1 - + C, (x- > a 2 ) (XVII) a a / v / or r dx r -dx = i r -dx 1 r J x 2 — a 2 J a 2 — x 2 2a J a — x 2a J a dx + x J-log^^ + C= -itanh-^+C. (x 2 < a 2 ) 2a & a + £ a a v J 190 INTEGRAL CALCULUS The first or second of these results is used according as x — a or a — x is positive. For XIX, let Vx 2 + a 2 = z — x; or z = x + Vx 2 + a 2 , (1) /. a 2 = z 2 — 2 xz. d(a 2 ) = = 2zdz - 2zdz- 2zcfc; (2 — x) dz = zdx; dz dx dx t /-.x — = = , — , by (1) 2 z-x Vx 2 + a 2 J Vx 2 + a 2 «/ « or sinh" 1 - + C". (XIX) a For XX, similarly, on letting Vx 2 — a 2 = z — x, i dx _ = logGr+\ / z 2 -a 2 ) + C, or cosh" 1 - + C"; (XX) Vx 2 — a 2 a The logarithmic form of cosh -1 - is log ( x "*" x ~ a ) , but its derivative or differential is the same as that of log (x + Vx 2 — a 2 ), the constant a disappearing in the differ- x entiation; and so too with the sinh -1 -. (See Art. 66.) a 121. Derivation of Formulas XV, XVIII, XXI, and XXII. — These formulas are merely the reverse of the differential forms given in Examples 1 and 2, Exercise VI. They may be derived from the forms for the inverse trigo- nometric functions of x . Thus: r dx + x 2 a I , . x 2 a I , . fx\ 2 a a since DERIVATION OF FORMULAS XV, XVIII, XXI, AND XXII 191 Since tan- 1 - = I - cot" 1 -, d (tan- 1 -) = d(- cot" 1 -V a 2 a' \ a/ \ a) Hence / C t 2 = - tan- 1 - + C, or -- cot" 1 - + C. (XV) a 2 + x 2 In the same way the second forms follow for formulas XVIII, XXI, and XXII. The standard forms are given in terms of -, because they are of more use than those in terms of x; the latter, being special cases where a = 1, are often given as the standard forms. Integrals may be obtained by reduction to either form. EXERCISE XXIII. ~b J 6 2 + c 2 x 2 c~Jb> + (ex) 2 ~ be b + U A Jft*-c»j« cJ6 2 -(cx) 2 26c 10g 6-cx + L tcz <&) ^tanh-^ + C OC cJ(cz) 2 -o 2 2oc g cz-o + C lCX>0j oc 6 3 f ^ f__^_ J_4 -t ^ + 3 r * Jx 2 + 6^ + 12 J(x + 3) 2 + 3 V3 V3 + 4 f d:c f J* _ 1 lntr (* + 3) - 2 ' J x* + 6 x + 5 J (z + 3) 2 - 4 4 g (x + 3) + 2 " l " 1, z + 1 . ~ = -. log — —^ + C. - r x 2 dx 1 , a; 3 — 1 . _ . r dx „ r xdx 1 a: 2 _ a C dx 7 - jF+^ = 2^ tan ^ + C - 8 - Jl6*T 192 INTEGRAL CALCULUS 9 - /s^p - Ac ><* 13 + c - - £ coth - " +c - ^ > 62) 10 ?nr log ^-^ + 6 2 ) v 4 ac - 6 2 v 4 ac - 6 2 1 . 2 ax + 6-V 6 2 - 4 ac . n .. . „. log + C (4 ac < 6 2 ) v 6 2 - 4 ac 2 ax + 6 + V 6 2 - 4 ac 11 f 2x + 7 W2x + 4)dx f __^_ Jx 2 + 4x + 5 J x 2 + 4x + 5 + J (x + 2) 2 + l = log (x 2 + 4x + 5) + 3 tan" 1 (x + 2) + C. J x 2 + 2x + 1 ~ J (x + l) 2 ^ J (x + l) 2 = \og(x + l)+j±- [ + C. * « C dx If & dx 1 . _ bx . n 13. I — = j- I = r sin L — + (7, •J VoV - 6 2 x 2 b J V(ac) 2 - (bx) 2 ° ac ac 14. f . X = J log (&x + Vb 2 x 2 + a 2 c 2 ) + C, or sinh" 1 - + C". J Vb 2 x 2 + a 2 c 2 b ac 15. f . dx = I log (bx + vW _ a2c 2) + c, or cosh" 1 — + C J Vb 2 x 2 -a 2 c 2 ° ac 16. C~ 7 M= = -L f- 7-^— = 4= sin- x \/l + C. ^ v a - 6x 2 v 6 ^ v a/6 - x 2 V6 y a 17 . f—^= = -1= sin-xV/f + C. 18. f~J^=- J Vs - 2 x 2 V2 V 3 ^ V3 - 4 x 2 19. f , ^ = f # <** = sin- 2 -±p- + C. on T dx • _, x - 1 . „ 20. I — — = sin l — ; h C. ^ V2 + 2 x - x 2 2i. c d * _ . i r ^ v ax 2 + 6x + c v a ^ V3 2adx V(2ax + 6) 2 + 4ac - 6 2 = -^ log (2 ax + 6 + 2 Va Vax 2 + 6x + c) + C f . dx = log (x + a + V x 2 + 2ax) + C. ^ vf -}- 2 ax , EXERCISE XXIII 193 23. f ; dx = -^log (x Vo~ + Vax 2 - b) + C. J Vax 2 — b v a dx 1 /* 2adx 24. r d * - = — f — J V - ax 2 + bx + c V a ^ V4 ac + 6 2 (2 ax - 6) 2 1 . , 2ax -b , _ — r- sin x r + C. Yd V4ac + 6 2 25. p^dx = f4±^^=r-i^+ ({i-x^xdx. J Vl-x J vl - X 2 J Vi - x 2 J = sin" 1 x - Vl - x 2 + C. 26. p£±» fc = rfe + S* = VJ5TT+ log (, + V5^D + C. ^ v x - 1 ■* v x 2 - 1 f dx r cdx 1 , ex . ~ 27. I . I . == -rsec- x -7- + C. J x Vc 2 x 2 - a 2 b 2 J ex V(cx) 2 - (a&) 2 a » a ° 28. C 5dx . J x V3x 2 -5 29. f V *±JJl dx = f X . +g dx = sec- 1 -+ log (x+ Vx^O+C. J xVx— a J xVx 2 — o? a so. f2*EEI*. r ";z^L d^ r^^-f-^^ J x J x Vx 2 — a 2 ^ vx 2 — a 2 ^ x vx 2 - a 2 = Vx 2 - a 2 - a sec" 1 - + C. a 81. f— *-,- r-^*L 5 -f ( ^- 1 )-l^. & - _* + c. J (1 _ X 2)§ J ( X "2 _ i)s J Vl-x 2 «« f dx If 6 dx 1 6x ~ 32. j ——== = T ( = T vers * h C. J V2 a&x - 6 2 x 2 h J V2 a (6x) - (6x) 2 ° a oo C —dx 1 _, bx , „ 33. I = =» r covers x -r + C. J V8 6x-6 2 x 2 & 4 oa f dx _.2x . n 34. I _ = vers L \-C. J Vax — x 2 a „_ /* — xdx _ f a — 2x — a , _ r (a — 2 x) dx a /* dx Vax - x 2 /• — xdx _ f a — 2x — a , _ /* (a — 2 x) dx a r J Vax - x 2 J 2 Vax - x 2 J 2 Vax - x 2 2 ^ , a _, 2 x . n Vax - x 2 - 5 vers * h C. 36. f - F ^ = -(•— ** - sin-! ( 2 -^) + C J Vax - x 2 J Va 2 /4 - (x - a/2) 2 V a / z 2 /4 - (x - a/2) 2 194 INTEGRAL CALCULUS Note. — It may be noticed that the result for Example 34 differs from the last result for Example 36. The difference is accounted for by the values of the constants of inte- gration. As may be seen in the figure, (¥-) . 7T 2X + ~ = vers l — , _ CL that is, 2X a a= 2 '■-.radius = 1 arc BP + £ arc OBP. Each result may be verified by dif- ferentiation according to (i) Art. 116. These results illustrate the statement at the end of Art. 117. 122. Reduction Formulas. — A formula by which a differential expression not directly integrable can be re- duced to a standard form or a form easier to integrate than the original function, is called a reduction formula. A general reduction formula that has a wide application and is most useful in the reduction of an integral to a known form is the formula for integration by parts. Many special formulas of reduction are obtained by apply- ing this general formula to particular forms. 123. Integration by Parts. — Differentiation gives d (uv) - udv -{-vdu. Integrating, = I d (uv) = I udv + / vdu; transposing, I udv = uv — I vdu. (1) The formula (1) may be used for integrating udv when the integral of v du can be found. This method of integra- tion by parts may be adopted when / (x) dx is not directly integrable but can be resolved into two factors; one, as dv, directly integrable; and the other, as v du, a standard form or a form less difficult than u dv. uv INTEGRATION BY PARTS 195 No rule can be given for choosing the factors u and dv other than the general direction that the factor of / (x) dx taken as dv is first chosen as that part directly integrable, and then what remains whether one or more factors must be taken as u. When the function given to be integrated contains more than one factor that is directly integrable, there is some choice to be exercised in selecting the factor dv; and in some cases a different choice may be necessary, if the first choice results in v du being non-integrable. It may be that one or more applications of the formula to I v du will be effective. The use of the formula is illustrated in the following examples, the formula being written, dv I f(x)dx = j u = uv — J vdu. (1) sin -1 - dx = x sin -1 - + v d 2 — x 2 + C. a a x Let dv = dx: then w = sin -1 -> a 7 dx v = x, du = . Va 2 - x 2 Substituting in (1) : sm _i - dx = x sin -1 I , a a J Va 2 - x 2 = zsin- 1 - + Va 2 - x 2 + C. (Compare example in Art. 118.) Example 2. — / x • cos x dx = x • sin x — f sin x dx = x sin x + cos x + C. 196 INTEGRAL CALCULUS Example 3. — / x 2 sin x dx = 2 x sin x — (x 2 — 2) cos x + C / x 2 • sin x dx = x 2 (— cos a;) + 2 J cos x • x dx = x 2 ( — cos x) + 2 (x sin x+cos x) + C, by Ex. 2, = 2 # sin x — (x 2 — 2) cos x + C. Example 4. — JV log x -a 2 - j=aHog(x+Vx>-a?)+e (3) = \x Vx^tf- I a 2 cosh" 1 - +C. (3') — Z a Note. — The integrals of the three last examples are of sufficient importance to be considered as standard forms. Exam-pie 9. — f x I — - — jdx = j x • sinh xdx = x cosh x — sinh x + C. / x • sinh xdx = X- cosh x — I cosh x>dx = x cosh z — sinh x + C. EXERCISE XXIV. Verify the following by j udv = uv — j vdu. 1. f cos" 1 - dx = x cos" 1 - - Va 2 - x 2 + C. J a a 2. f tan" 1 - dx = x tan" 1 - - \ a log (a 2 + x 2 ) + C. 3. f cot" 1 -dx = x cot" 1 - + I a log (a 2 + x 2 ) + C. 4. j log x dx = x (log x — 1) + C. 5. f x log x dx = \ x 2 (log x-h) +C. 6. f x 3 log x dx = I x 4 (log x - I) + C. * By same method as in Example 6. 198 INTEGRAL CALCULUS 7. f x ( log(x+V^3tf) + C > by (A), J Vx 2 — a 2 & * . | V^^ + £ cosh- 5+ C. ( Co 7 ar « Ex - 8 >) 2 2 a \ Art. 123. / /* o^r^—, — 5j x 3 Vx 2 ± a 2 , a 2 /• x 2 dx, , m , 5. I x 2 vx 2 ± a 2 dx = ^ ± -r I , , by (B), J 4.4./ Vx 2 ± a 2 = (f ±1^)"^^ -^2 lo 8^+ ^±^)+C, byExs. 3,4. 6. f> Vx^dx = ^ ^fp ± * f-^==, by (B). J m + 2 m 4- 2 »/ Vx 2 ± a 2 7 f rfa; Vx 2 d= a 2 J x 2 Vx 2 ± a 2 T a 2 x "•" JW± *>-**- £1^±^ - (-1-2 + 2 + 1 r ± a2) _j rfx> by (C)< T a 1 J «/ t2 V/i2 _ ^2 GTX i x 2 Va 2 - x 2 x" 1 (a 2 - x 2 ) 1 JV 2 (a 2 - x 2 ) _i dx = — _ - -K-l-2 + 2 + 1) r^ _ -, dx> (C)# —a 2 J REDUCTION FORMULAS FOR BINOMIALS 201 C dx Vx 2 — a 2 1 _, x ~ = (:C2 ~ q ! )3 + ^ sec" 1 - + C, by Standard Form XXI. 2a 2 x 2 2 a 3 a in C dx 1 1 _,*., /, 10. | j = . r sec 1 - •+ C. •J x (x 2 - a 2 ) 5 a 2 Vx 2 — a 2 a a JV 1 (x 2 - a 2 )"^x = - ^fr 2 " 02 ) ' - | JV 1 (x 2 - a 2 r"dx, by(D), _l se c-i-+C. a 2 Vx 2 _ a 2 a 3 a H C dx = x + c J (a 2 - x 2 )* a 2 Va 2 - x 2 C(n2 *\-*L x(a 2 -x 2 )~* J(a 2 -x 2 )idx=- 2fll( _ 4) + " 3 2 + a 2 °(-') +1 / a;0(a2 -^^ by(D) - 12. fV2ax-x 2 dx = ?-^L V2ax-x 2 + % sin" 1 ^^ + C, J A _ a or — — v 2 ax — x 2 + — vers x - + C £ a CL fVfZ^dx . p(2a -*)»«&. C A PP'y (A) and (B) ) »/ ^ \ in succession. / Or fV2 ax - x 2 dx = f^a 2 - (x - a) 2 dx a- . ,x — a V2 ax - x 2 + % sin" 1 — - + C, by Ex. 2, 2 ' 2 a x — a ~~ 2" V 2ax-x 2 + g- 2 vers- 1 g + C f See Ex. 36, \ V 2 ax x + 2 vers o + C . ^ Exerdge X XIIlJ 13. f x™ V2ax-x 2 dx = f z*H V2a-x dx = - xm ~ l ( 2ax ~ x ^ J J m + 2 + (2m + ^ )a fx™" 1 V2ax-x 2 dx, by (A). 202 INTEGRAL CALCULUS x m dx x TO_1 V2 ax - x 2 "•/ i,/ V2, (2m -1) a r x m ~ 1 dx J V2ax - x 2 dx V2 ax — x 6 x m V2ax-x 2 (2m-l)oaJ» m — 1 /* dx + oJ r w-i i,/ dx (2 m - 1) a J z m-i V2ax - x 2 x Va 2 - x 2 . 1 . + — , log , by(C). + C. x*V a 2 - x 2 2 °> 2x2 2 a3 v a 2 - x 2 + a - " 1( " 1 r 2 3 a2 +2 + 1) p- 1 ^-x 2 )-^x ) by(C), + i f /* . See Ex. 18. Va 2 - x 2 , 1 2a 2 x 2 Vx 2 + a 2 1 / dx _Vx 2 _+a 2 1_, x c x*Vx 2 + a 2 2 °< 2x2 2 ° 3 a + V&+ a 2 - ll ~ 1 ~_lt 2+1) S** (x2 +ai) ~* dx ' by (C) - V X 2 + gl 1 /• dX 18. /: dx = -log V a 2 - x 2 « v a 2 - x 2 + a + C. Here m + 1 = 0, therefore Formula (C) fails. Let a 2 — x 2 = 2 2 ; .'. —xdx = zdz, x 2 = a 2 — z 2 . / dx _ r dz 1 , a — z ~ x Va 2 - x 2 ~~J 2 2 -o 2 "2o g o+z + 1 . a - Va 2 - x 2 n = rr- log , + C 2 a a 4. Va 2 - x 2 -log ; a a + v a 2 — x 2 + C. 19 REDUCTION FORMULAS FOR BINOMIALS 203 • f 7 ^_ = V 7 _* +C. J x V x 2 + a 2 a Vx 2 + a 2 + a Again m + 1 = 0, and Formula (C) fails. • Let a 2 + x 2 = z 2 ; :. xdx = z dz, x 2 = z 2 - a 2 . / axis, y = / (x) , some undetermined fixed ordinate as MqPq or M2P2, and the moving ordinate MP. Let Ax = dx be MM\\ then, while A A the actual incre- ment of the area A is MPPiMi, dA is MP DM h the incre- ment that A would get if, at the value M P PM, the change of A became uniform and so continued while x increased uniformly from the value OM to OMi. Hence 204 &**]% DERIVATIVE OF AN AREA 205 dA = MPDMi = ydx=f(x) dx, :. A = J ydx = I f(x) dx, where A 'is indeterminate so long as the fixed ordinate M P or M2P2 is indeterminate. , dA 126. Derivative of an Area. — Since dA that is, the derivative of the area with respect to x is the ordinate of the bounding curve. This important result may be obtained by the method of limits also, taking the increments infinitesimal. Thus, A A = MPP 1 M 1 , AA > y Ax, and AA < (y + Ay) Ax, .*. y Ax < AA < (y + Ay) Ax, AA dA v A A .'. -j- = lim -r — = y, dx A x=o Ax *' since lim (y + Ay) Ax=0 y, Ay = as Ax = 0. In case y decreases as x increases, the curve falls from P to Pi, and the inequality signs are reversed, but the result is the same. Let A be the area between the t/-axis and the curve; then, dA dy lim -r — Ay=o Ay = x, or dA — x dy, = I xdy. Here, the derivative of the area with respect to y is the abscissa of the bounding curve. 206 INTEGRAL CALCULUS 127. The Area under a Curve. — Let the curve y = f (x) of Art. 125 bey = x 2 . Let OM = a and OMi = b; then MoPoPM = A= fx 2 dx = ^ + C. As the area is measured from x = a, ... 4 = o = ! + c, c=-|, £ 3 tt 3 •'• A ^ = 3-3' (1) where A x is the variable area M P J°M. Making x *= 6 in (1) gives MoPoPiM, - A 6 - | -r J- (2) The usual notation is , /" 6 2 , z 3 1 & 6 3 a 3 , oA 128. Definite Integral. — In general, when b > a the increment produced in the indefinite integral F (x) -f- C by the increase of x from a to 6 is F(b)+C- (F(a) + C) =F(6) -F(a). This increment of the indefinite integral of /(#) dx is called "the definite integral of f(x) dx between the limits a and 6," and is denoted by f f (x) dx. Hence, s. b f(x)dx~F(b) -F(a), a The operation is that of finding the increment of the indefinite integral of/ (x) dx from x = a to x = b, where b is called the upper or superior limit, and a the lower or inferior limit, although they are more precisely termed "end values" of DEFINITE INTEGRAL 207 the variable, as they are not "limits" in the usual sense of the word. If the upper end value is variable, then f X f(x) dx = F (z)T = F(x)- F(a). When the lower end value a is arbitrary, — F (a) may be represented by an arbitrary constant C, hence £ f(x)dx = F(x)+C. *J a Since r 'f(xydx = F(x)+C, /■ an indefinite integral is an integral whose upper end value is the variable and whose lower end value is arbitrary. Hence, when the integral is represented by an area and the area is known for some value a of x, A= ff(x)dx = F(x)-F(a), where C is —F (a), and the area A is determinate. If the area under the curve y = x- (Art. 127) be reckoned from z = 0; when x is zero, A is zero, therefore, C is zero and A x = j , the area of OPM. (3) (See Example 4, Art. 115.) A v = fy* dy = lyi = l x\ the area of OPX. (3') Making x = a in (3) gives a 3 A a = -w , the area of OP M 0) (4) o e and making x = b, gives A b = ^, the area of OPJIl (5) o 208 INTEGRAL CALCULUS x 3 ~| a a 3 x 3 ~] h H-3' " si 'I ¥ - a 2 , As definite integrals, A = I x 2 dx If, in Example 3 of Art. 115, the area A is from x = a to x = 6; then A = / 2xdx the area of a trapezoid. It may be noted that, when the integral is " between limits," it is not customary to write the constant, as it will be eliminated. 129. Positive or Negative Areas. — The area under or above a curve y = f (x), from x = a to x = b, will be posi- tive or negative according as y is positive or negative from x = a to x = b; hence, when the curve crosses the x-axis, the areas are gotten separately, otherwise the result will be the algebraic sum of the areas and may be zero, since the areas above and below the axis may for some curves be equal. For example, the areas for the curve of sines or of cosines as shown in Art. 140 illustrate the principle. 130. Finite or Infinite Areas — ' 'Limits " Infinite. — From the geometrical meaning of an integral it follows that / (x) dx has an integral whenever / (x) is continuous, hence the end values a and b are in general taken so that/ (x) will be finite, continuous, and have the same sign, from x = a to x = b. If x = b = oo , then Jrv> r*x=b f(x)dx = lim I }{x)dx, a 6=oo *Jx = a - where the limit may, or may not, exist. When the limit of the in- tegral is finite the total area is found; but if as b becomes infinite the integral becomes infinite, then no limit exists and the area up to x = b becomes infinite as b becomes infinite. FINITE OR INFINITE AREAS — LIMITS INFINITE 209 Example 1. — A = / — dx = lim / -- cfo = lim = 1. J l X 1 b=*> Jl X £ &=« L ZJl Limit exists although as x approaches zero, f(x) = — 2 be- comes infinite. A = I -= dx = hm / — da; = lim > Jo x 2 b=*Jo x 2 b=« L ^Jo which does not exist, since = ao ; that is, the area up to x = b becomes infinite as b becomes infinite. A = I -= dx = lim / -rrfx = lim > Jo X 2 a=oJa X 2 a = L ^Ja which does not exist, since = go , the area becoming infinite as x approaches zero. -2.0 -1.5 -10 -0.5 Example 2. — A = / X e* dx = Hm I ^dx = lim e x = e a J — x o= — x t/a a = — ^L Ja is total area under curve up to ordinate at P (x, y). 210 INTEGRAL CALCULUS Area to right of y-axis = OMPB = A= f e x dx = e*T = e x Jo Jo Area to left of y-axis = / e x dx = lim / e x dx = lim e x %J — co a= — oo *J a a = — oo |_ J = 1. Note. — When y = / (x) = e*, ^/, the function, is the ordi- nate, equals the slope at the end of the ordinate, and may represent the total area under the curve up to the ordinate. (See Art. 138.) 131. Interchange of Limits. — Since the definite integral £ b f(x)dx = F{b)-F(a); it follows that I f(x)dx= - I f 0) dx, Ja Jb since the second member is — [F (a)— F (&)]= F(b) — F(a). It follows also that the definite integral is a function of its limits, not of its variable; thus Jf (y) dy has the same value as / / (x) dx, a J a each being F(b) - F(a). The algebraic sign of a definite integral is changed by an interchange of the limits of integration, and conversely. 132. Separation into Parts. — A definite integral may be separated into parts with other limits or end values. Thus, f b f(x)dx= f C f(x)dx+ f b f(x)dx. (1) Let the curve y = / (x) be drawn and the ordinates AP X , BP h CP 2 , be erected at the points for which x = a, x = b, X = c. MEAN VALUE OF A FUNCTION 211 Since area APP^B «= area APP,C + area CP 2 P 1 B, (1) follows. A C B C If OC = c', and C r P$ is the corresponding ordinate, then, area APP^B = area APP Z C - area £PiP 3 C"; and hence Pf(x)dx = f C f(x)dx- f C f(x)dx = ff (x) dx + f l J(x) dx, by Art. 131. Note. — It may be seen that Jf(x)dx= I f (a — x) dx, t/0 for each is F(a) - F(o). Thus, - Pf(a -x)d{a-x)= -F(a- x)T = F(a)-F (o) = Pf(x)dx. Jo 133. Mean Value of a Function. — The mean value of / (x) between the values / (a) and / (b) is ./>> dx Let area APPiB represent the definite integral / / (x) dx. 212 Then INTEGRAL CALCULUS f b f(x)dx = areaAPPi# = area of a rectangle with base A B and height greater than AP but less than BPi = AB • CP 9 = (b — a)f(c), where OC = c. Hence f(c) = f>< x) dx b — a where/ (c) is the mean value of f (x) for values of x that vary continuously from a to b. The mean value may be denned to be the height of a rectangle which has a base equal to b — a and an area equi- valent to the value of the integral. Example 1. — To find the mean value of the function Vx from x = 1 to x = 4. Let OP Pi be the locus of y = v'z, /(c) OA = 1 and OB = 4. 2 »7 I x 2 dx ix^\ -> H = 14 4-1 3 9 1 ~ ~ J 9 ' 1£ = 1.56| = CP', mean value. x = (V) 2 = W = 2.42 = OC. EVALUATION OF DEFINITE INTEGRALS 213 Example 2. — To find the mean value of sin as varies from to 7r/2, or from to w. T f sind dO "Ho 1 2 tt/2-0 7T/2 7r/2 7T / * sinddd — cos -*° — — fi^fifi 7T-0 — — u.uouu. 7T 7T 0.6366. Example 3. — To find the mean length of the ordinates of a semi-circle of radius a, the ordinates for equidistant in- tervals on the arc. J a sin dd — a cos n o Jo 2 a -0 0.6366 a. Example 4. — To find the mean length of the ordinates of a semi-circle of radius a, the ordinates for equidistant inter- vals along the diameter. a -(-a) 2 2 aJ- 2 2a Va 2 - z 2 dx = 7}X Va 2 — x "2a = ja = 0.7854 a. 4 134. Evaluation of Definite Integrals. EXERCISE XXVI. .n+i -| b frn+i _ a n+i i. f^ci*-— T-— ^- Ja n + ljo n -f- 1 2. J 2 4x 3 dx = x 4 ] 2 = 16- 1 = 15. 3. j^(f - ^) = log x + log (2 - z)]* = log (2 x - a?) J* = log(2x -x 2 ). 214 INTEGRAL CALCULUS X*> 1 "l°° 1 e~ ax dx = - -e~ ax = -• a Jo a TT TT IT 5. C ^r- n dd = C cos- 2 0sin0d0 = sec0~| = V<2 - 1. Jo cos 2 Jo Jo a r a dx . xl a 7T 6. I , = arc sin- = =• Jo Va 2 — x 2 a -Jo 2 „ r a dx 1 x~\ a ir 8. r 2 ^»-Rr. Jo v 2 r — y r<*> xdx 1 , ."I 00 x 10 - J. r+^ = 2 arctan *l -r 11. f , e , ^L = arc tan e x = arc tan e x — \- Jo 1 +e 2X Jo 4 -~ r 1 . •-- , r^l - cos 2 .„ it 12. J sin 2 Odd = J ■ = rf(? = T -« f* „ „ r*l + cos 2 ,„ TT 13. J cos 2 Odd = J — E - ^ rf0 = j- Note. — Considering the areas between the axes and the graphs: TT IT sin n xdx = \ cos n a; cfo, where n is positive, o Jo Jsin n x dx = 2 | sin n x dx, where n is positive, o Jo IT J cos" .t dx = 2 I cos" x dx, if n is an even integer, ^0 but = 0, if n is an odd integer. 135. Areas of Curves. — As has been shown, the formu- las in rectangular coordinates are A = I ydx and A = j xdy. AREAS OF CURVES 215 (a) Let A denote the area between the curves y = f (x) and y = F(x); let x = 031, dx = PE; then, the variable area A = P,P'P and dA = PP'DE = (/ (x) - F (x)) dx; :. area P.PT.P = A= I *\f (x) - F (x)) dx, where the points of intersection are (x , y ) and (xi, y{). If the locus of y = F (x) is the x-axis and x and X\ are a and b, this formula reduces to 4 = J> x) dx. (b) In polar coordinates, A = h s p-dd. For let P be any fixed point (p , O ) and P (p, 0) any variable point. Consider the area P G OP to be generated by the radius vector p as increases from O , and denote it by A. With OP as a radius draw the arc PD and let dd = Z POPi ; then dA = OPD = ip-pdd = i p 2 dd, the increment of A, if uniform, as in a circle. /. A = | /Vd0, or i = | Tp 2 ^, if ; dx = a cob d; then, j Va 2 — x 2 dx = a 2 C cos 2 d<£, where is the complement of above, = f / (1 + COs2)d) + C = | (0 + sin*cos*)+ C f^ + IVT^ + c, the indefinite integral. Compare Ex. 6, Art. 123. Area - 4 f ° V a 2 -x 2 dx = 4 1"^ sin" 1 -+£ Va 2 -.r 2 + C~\ ° = «*, as above. ./o L^ « ^ Jo 6 7ra 2 Corollary. — Area of Ellipse = 4 - f Va 2 - x 2 dx ■ 4- ^f- TO FIND AN INTEGRAL FROM AN AREA 219 136. To find an Integral from an Area. — An integral may be found from an area, when it can be gotten geomet- rically from the figure. Example 1 . — Find / Va 2 — x 2 dx from the figure of the circle y = Va 2 — x 2 . Area = BOMP = BOP + OMP = \ a 2 tf> + i xy M A If the initial ordinate is not OB and is undetermined, then, Area = /v* x 2 dx = - \^a? x 2 + — sin-^ + C, as above, C being independent of x and indefinite when the initial ordinate is undetermined. Example 2. — Find I V2ax — x 2 dx, using circle y = V2ax — x 2 . Area = OBPM = OBPC + PCM a a" V 2 VerS_1 a+ 2 x — a / V2ax 220 INTEGRAL CALCULUS or Area = CBPM = BPC + PCM a 2 . , x — a x — a /= = — v 2 ax — x 2 . = 2 sin If the initial ordinate from which area is reckoned is undetermined, then fV2ax-x 2 dx = ^=-? V2ax-x 2 + \ a 2 sin" 1 ^—^ + C, *7 Z ad ra l where C = —r- , if Area = 0, when x = 0; or x — a V2ax-x 2 + \ a2 vers_1 * + C '> where C" = 0, if Area = 0, when x = 0. As may be seen in the figure, sin- 1 - — - + s = OCP = vers- 1 - \ a 2 a that is, a 2 I . . x — a , 7r\ a 2l Sln ^T~ + 2 J = 2 rc — a . 7ra- a' ^ sin -1 2 a + vers' (See Note at end of Exercise XXIII.) Either result gives lV2ax- x 2 dx = lira 2 = areaofOBA. AREA UNDER EQUILATERAL HYPERBOLA 221 Example 3. — Find / {mx + b) dx, by means of line y = mx + b. Area = OMPB = BDP + OJ/DB = \ x • ??l£ + # • 6 rar 2 . , = —zr- + bx. If the initial ordinate is not OB, and is undetermined, then / mx 3 (mx + b)dx = '^r- 137. Area under Equilateral Hyperbola. — As in the figure of the circle y = Va 2 — x 2 , X x 1 1 Va 2 — x 2 dx — jz x Va 2 — x 2 + ■= a 2 sin -1 - expresses the area BOMP and a 2 sin -1 - is represented by twice the area of the circular sector BOP ; so 1 1 PVtf + tfdx =^x Va^+tf + ^ a 2 sinh- 1 - (Ex. 7, Art. 123) may be shown to express the area AOMP under the equi- lateral hyperbola y = v a 2 + x 2 , and a 2 sinh -1 - to be repre- sented by twice the area of the hyperbolic sector A OP. 222 INTEGRAL CALCULUS To get jVa 2 -\-x 2 dx; let x = asinh<£, dx = acoshd; then, / *Va 2 + x 2 dx = a 2 I cosh 2 d » ^ ( + sinh cosh 0) = ^VaHs 2 +- fl 2 sinh" 1 -: 2d £i o> as also in Ex. 7, Art. 123. If x = a cosh and dx = a sinh d<£ be substituted in l Vx 2 - a 2 dx; len, £ Vx 2 — a 2 dx = ~x Vx 2 — a 2 — 7T coslr 1 x (as in Ex. 8, Art. 123), and ^a 2 cosh -11 will be represented by the area of a sector of the equilateral hyperbola y = Vx 2 - a 2 . SIGNIFICANCE OF AREA AS AN INTEGRAL 223 138. Significance of Area as an Integral. — The units of the number represented by A as the measure of an area will depend upon the units chosen for the abscissa and the ordinate. If the unit for x be one inch and that for y be ten inches, then a unit of A would represent ten square inches; if on the graph the unit of x is one-tenth of an inch and the unit of y is one inch, these representing one and ten inches respectively, an area on the graph of one-tenth of a square inch will represent the area of ten square inches. The integrals represented by areas may be functions of various kinds, such as lengths, surfaces, volumes, velocities, accelerations, weights, forces, work, etc. Hence, the physi- cal interpretation of the area will depend upon the nature of the quantities represented by abscissa and ordinate. (a) If, in the figure of Art. 125, the ordinate represents velocity and the abscissa represents time, then the area represents distance; and, since velocity ds where a is acceleration, dA -—= v = at. dt Hence, A = f atdt = hat 2 + C = M P PM, and since s = \ at 2 -+• s 0) C is s , initial distance or area; and the number of square units of A ( = M P PM) will equal the number of linear units of distance passed over by a moving point in the time t = M M. (b) If the ordinate represents acceleration and the ab- scissa still represents time, then the area represents velocity; and since acceleration _ dv dA _ _ dv a ~dt' dt~ a ~dt' 224 INTEGRAL CALCULUS Hence, A = Cdv = fadt = at + C = MoPoPM, where a is constant acceleration, and since v = at + v , C is Vq, initial velocity or area; and the number of square units of A will equal the number of units of velocity acquired by a moving point in the time t = M M. (c) If the ordinate represents a force acting in a constant direction, and if the abscissa represents the distance through which the force has acted, then the area A = MoPoPM will represent the work done by the force acting through the distance represented by M M. If the force is constant in magnitude as well as in direction, the area will be a rectangle, dA r since -=- = F, constant, gives A = l Fds = Fs + C. Whether the force is constant or variable the area A = I F ds represents the work done, the area being that under the graph of the equation y = f (F), representing the force. If the force is not constant in direction, the area will still represent the work, provided the ordinate represents the component of the force along the tangent to the path of its point of application. By means of certain contrivances the curve y = f (F) may be plotted mechanically by the force itself, as, for example, in the steam engine by means of the indicator. Having the curve, the mean force may be easily found; it is given Fds MqM, the area divided by the distance through which the force acts. The area may be read off at once by the polar planimeter, and the work done found directly. It is manifest that a function may be represented graphi- cally either by the ordinate of a curve or by the area under a curve; if the ordinate is made to represent the function, the by/ AREAS UNDER DERIVED CURVES 225 slope of the curve is the derivative of the function; if the area under a curve is taken to represent the function, then the derivative of the function is the ordinate of the curve, since the ordinate is the derivative of the area. It is usually preferable to represent by the ordinate * that which in the investigation is mainly under examination; therefore, if this is the derivative, the latter method, where the area is the function and the ordinate is the derivative, should be used rather than the former method, which is to be used when the function is mainly under consideration. The function e* is exceptional, in that the ordinate repre- sents the function, the slope of the curve, and the area under the curve. (See Example 2, Art. 130.) 139. Areas under Derived Curves. — It has been shown (Art. 84, figures) by drawing the graphs of a function and its successive derivatives that the variation of the function is exhibited to advantage. It may now be seen that the area under any derived curve is represented by the ordinate of its primitive curve. Thus the area under the graph of y = f (x) is represented by the ordinate of y = / (x), that under the graph of y = J" (x) by * Irving Fisher, A Brief Introduction to the Infinitesimal Calculus. 226 INTEGRAL CALCULUS the ordinate oiy = f 0*0 > aDCl so on for the successive derived curves. Drawing the graphs of y = f (x) = ^ and y = /'" (x) = 2 together with those of y = f (x) = x 2 and y = J" (x) = 2 x (shown in Examples 3 and 4, Art. 115), it is seen that the areas are represented as stated. It may be seen also that j f(x)dx= j x 2 dx = -~ = A, being represented by the ordinate of y = — , is an integral function of x 2 and the graph an integral curve of x 2 . If y = 2 be the fundamental curve, then y = 2 x is the x z first integral curve; y = x 2 , the second; y = -5-, the third; x* y — — , the fourth; and so on. CHAPTER III. INTEGRAL CURVES. LENGTH OF CURVES. CURVE OF A FLEXIBLE CORD. 140. Integral Curves. — If F (x) has / (x) for its deriva- tive, then F(x) is called an Integral Function or simply an Integral of fix). The General Integral is / / (x) dx = F (x) + C, called also the Indefinite Integral. The graph of an integral function is called an integral curve. If the original or fundamental function is y=f(x), (i) then y = F(x) (2) is the first integral curve of the curve (1), where F (x) is that integral of / (x) which is zero when x is zero. In the general figure of Art. 125, F (x) is the area OPM under the curve y = f (x); in the figure of Art. 139, if y = / (x) = x 2 , then F(x) is the area under y = x 2 and is the ordinate of the integral curve y = — o It is manifest that for the same abscissa x, the number that indicates the length of the ordinate of the first integral curve is the same as the number that represents the area between the original curve, the axis (or axes for some func- tions), and the ordinate for this same abscissa. Hence, the ordinates of the first integral curve may represent the areas of the original curve bounded as stated. It may be seen also that for the same abscissa x, the number that expresses the slope of the first integral curve is the same as the number that measures the length of the ordinate of 227 228 INTEGRAL CALCULUS the original curve. Hence the ordinates of the original curve may represent the slopes of the first integral curve. The integral curve of the curve of equation (2) is called the second integral curve of the original curve of equation (1). The integral curve of the second is called the third integral curve of the original curve (1), and so on. Thus for any given curve there is a series of successive integral curves.* The function cos and the first and second integral curves are shown with their graphs. Jt/2 7t 3Jy2 2Jt Let y = cos 8 be the fundamental function, then y = f cos 6 dd = sin + C, where C is zero, as y is zero when is zero; and y = I sinddd = -cos0 + C, * The statements in the three paragraphs above with some difference of words are givon in Murray's Integral Calculus, where a fuller treat- ment will be found in the Appendix. APPLICATION TO BEAMS 229 where C is one, as y is zero when is zero. Hence, y = sin and y = 1 — cos = vers are the first and second integral curves of the curve z/ = cos0. It is seen that the ordinate of the first integral curve at $ = x/2 is +1, that number being the same number that measures the area under the fundamental curve for the same abscissa; the ordinate being zero at & = it indicates that the algebraic sum of the areas of the fundamental curve is zero and hence that the area below the axis from = x/2 to t is exactly equal to that above from = to x/2; the ordinate being zero again at = 2 x indicates that the areas of the fundamental curve above and below the axis are exactly equal up to = 2w. It is manifest that the ordinates of the second integral curve indicate the corresponding areas for the first integral curve, the number being +2 from = to x and 2 — 2 = up to = 2 w. In the case of this function the series of integral curves can be extended indefinitely without any difficulty. It is manifest also that the ordinate at any point on the fundamental curve gives the slope at the corresponding point on the first integral curve, the ordinate for the first gives the slope of the second, and so on. The subject of successive integral curves has useful appli- cation to problems in mechanics and engineering. Illustrative examples follow, showing the application to the expression and graphical representation of the shearing force and bending moment throughout the length of a loaded beam; also to the slope and deflection of the elastic curve, the curve of the mean fiber of the material of the beam. 141. Application to Beams. — As given in the Mechanics of Beams, the Vertical Shear at any section of a loaded beam is the algebraic sum of the vertical external forces on either side of the section, and the Bending Moment is the algebraic 230 INTEGRAL CALCULUS sum of the moments of those forces about a point in the section. The moment of a force about a point is the product of the force and the length of the perpendicular from the point to the line of action of the force. (See Art. 172.) The Elastic Curve is the curve assumed by the mean fiber along the axis of a longitudinal section through the centers of gravity of the cross sections of the beam, the Slope is the slope of the tangent to the curve at any point, and the Deflection is the ordinate at that point. Taking abscissas to denote as usual horizontal lengths, the ordinates will represent the quantities to be depicted by the curves. The fundamental curve is the curve for L, the load; the Shear S is represented by the first integral curve; the Moment M by the second integral curve; EI upon the Slope m, by the third integral curve; and EI upon the De- flection d, by the fourth integral curve, where E and I are constants denoting the Modulus of Elasticity and Moment of Inertia, respectively. Example 1. — Let a beam of length I between supports be simply supported at each end and loaded with a uniform load of w lbs. per linear ft. I, = y = —w, w taken with negative sign as a downward force. S = y = I —wdx = —wx + [So = -w,S being zero when x = -A • M = y = /(t " wx ) dx = y x ~ if + (Mo = °' M when x = 0). „ T/ s C (wl wx 2 \ , wl 2 wx 3 EI(m -If) - J (y * " -2) dx = T* " X -j- ( mo = — — wl 3 , m when x = 0, m being zero at x = ^ J • EI APPLICATION TO BEAMS dx 231 r/wlx 2 ivx* wl*\ Wlx z IVX* wl 3 X . , , nix n\ ~12" ~ 2T ~ "2T + ( ° = ' )# EXAMPLE 1. Load Line 7 5 wl^ m „^ do sSSIrlr mtaR - EXAMPLE 2 *==ZZ1 Inflexion Point d=-j^^,max. Inflexion Point — ■ .,.,,.,,,; n XDe-flection Curue ^Deflection Curve *-0grfJ&-3f ~-~y 3 V3l= 0.58-1 -A^^^-J XSlope Curue M= 2 /24 wl?max. M=- 2 /J2 wl 2 ,max. tieg. Shear and Load Lines the same as Example!. XMoment Curve 232 INTEGRAL CALCULUS Example 2. — Let a beam of length I between supports be fixed at each end and loaded with a uniform load of w lbs. per linear ft. L = y = —w. S = y = — wdx=—wx + [So = ~k f S being zero when x = ■= J- M = y = J {2~ wx ) dx = j x ~^ wl 2 + (M . M at x = 0), Mo = - -yz , see below. EJ (m - ?/) = J [^-x - -g- + Moj dx = -j-z 2 - — + MqX + (m = 0, m at z = 0) on7 1^3^ /Wl 2 \ = -j- x 2 ^ — ( jo x = — M x j , from m = at x = Z. nr /j \ C ( W l i WX * wl2 \ J »tf-iF)-J( T ^-- r "i27 = °> d at x = 0). Example 3. — Cantilever Bridge.* In the cantilever bridge the joints are placed at the inflexion points, where the Bending Moment would be zero if the bridge were con- tinuous over the whole span. Let the beam of Example 2 have joints at the inflexion points, and let the length of each cantilever arm be denoted by a and the length of the suspended span by 6. The Shear line and the Moment curve will be unchanged but the Slope and Deflection curves will not be continuous as they were without joints in the beam. For slope and de- * The essential features of this example are given by H. E. Smith in his " Strength of Material." APPLICATION TO BEAMS 233 flection each part is to be considered as a separate beam, the arms having in addition to their uniform load a concentrated load at their jointed ends, equal to half the uniform load on the middle span b, since that part is supported at the joints by the arms. L = — w. S being zero when x = ~ r = f(^(2a + b)~wx)dx = ^(2a + b)x- + (m = -™(a + b)\ M = 0atx = a. -^ ,, n w fe% . ,v vox 1 wa , , . From M = = ^ (2a + b) x - ^ g- (a + &), M wrr a: = a and rr = a + 6, that is, x=(l-i v / 3)!/2 and a; = (l+ i V® 2/2, and & = 1 V3 J, from M = = -^ 2~~T2 Exam P le 2 * 234 INTEGRAL CALCULUS It is seen that the maximum negative moment T V wl 2 at the fixed ends is twice the maximum positive moment ^ wl 2 at the middle. For equal strength, the bending moments at the ends and middle of the beam should have equal numeri- cal values; this requires that wa , . , N wb 2 T (a + 6) = -g-, that is, 4a 2 -f46a = 6 2 ; or 4a 2 + 46a + b 2 = 2b 2 ; or (2a + 6) 2 = 26 2 , /. b = i V21 and a = (l - J V2) 1/2. It follows that wb 2 wl 2 , wa , , , N wl 2 __ = __ and __ (a+6) = __, giving equal strength at ends and middle of beam. Example 4. — Let a beam of length I between supports be fixed at the right end and simply supported at the left end; and let it be loaded with a load uniformly increasing from zero at the left end to w lbs. per linear ft. at the right end. T w L = y= -jx. /w w — jxdx = — tyj x 2 + (£0 = S, when x = 0). M = y = f(s - ~x 2 ^j dx = Sox - ^x* + (Mo = 0, M when x = 0). EI {m _ y) _ j( SfrT _ • *y x = *£ - £, + (mo=o7 — ^- , m at x = J , from m = at a: = Z. ett /j \ T/rf S l 2 . S0X 2 w A , rf fiW 2 aSox 3 wx b , , , . . APPLICATION TO BEAMS 235 __., . ft wl* S Q l* Sol* wP , , EI (d = y) = - 24 - T + T " S51 1 when * = *' .*. £ = 77y supporting force at 0. w£ it; „~| 4 7 •'• s = » = io-2i a;2 L = -io wZ ' 4 Si at right end; supporting force =- + — wl. °r° mmmm ^^ Load Line ^liiiiiiim i iij ii iiin V-—y s Vsl=0.447l ^^^^miiiiii^ Shear Curoe M=, 0.03 >wl? max. ■ TTTTTrmTT^ at right end, M= 0. 06 z / 5 wl 2 , max. nee/. '. nea. ^\ ^TT-mnTTTTTT HI VsVmsI =0.77-1 ^uiff ppp^— Moment Curoe X Slope Carve XDeflection Curve 0.916 wl^ 384 EI ^ U 10 2^ ' x = i a/5 Z = 0.447 . . . I, when the shear is zero. , , wl wx 3 ~\ v 5 10 A no 79 M = y = io x - eiL^r n* = 003 wP > 236 INTEGRAL CALCULUS maximum positive moment where shear is zero, since ax M -y = ro x -t\,r-k wl2 =-° Miwl2 ' maximum negative moment. ,, _ wl wx 3 M = 0= id x -JT> .-. x = Vfl ^ i V3X 5 = 0.77 , .-. . I , where moment is zero and where there is~an inflexion point on the Elastic Curve, the fourth integral curve, since M is second derivative of d with respect to x. „ T/ N wl 3 . Wlx 2 WX 4 ~] wl 3 r.r . EI ( m = y) = - — +— - —^= -—,E1 upon slope at left end. n _ _ wl 3 wlx 2 _ wx* ~ 120 + "20" Wl' .'. x = £ V 5 I and I, when slope is zero. ,-, T , . N wl 3 wlx 3 'wx b l 0.016 r= 71 *nd=v) = - m x +w-miU«i=-^- VEv * = —0.002385 wl 4 = — ^-r-wl 4 , EI upon max. deflection. The deflection is zero at x = and x = I, as used above. 142. Lengths of Curves. — Rectangular coordinates. — It has been shown in Art. 10, (d) that ds 2 = dx 2 + dy 2 ; now let s denote the length of the arc whose ends are the points (xo, 2/0) and (x, y), then ds = Vdx 2 + dy 2 ; whence •=£[■+ ($*]'-■ or according as ds is expressed in terms of x or of y. LENGTHS OF CURVES 237 In getting the length of any curve, that formula which gives the simpler expression to integrate is the preferable one to use. EXERCISE XXVm. 1. Find s of the circle x 1 + y- = a 2 , and the circumference. Here dy _ _ x . /dy V = z 2 , dx y' \dxj y 2 ' hence s= PTl + ^~P dx = f * & + *)* dXf using (1), J" x dx . .X~] x . . x . ,xo . = a sin x - = a sin -1 a sin -1 — x Va 2 — x 2 °-J x o « a Circumference, s = 4 a sin -1 - = 4 - a = 2 7ra. aJo 2 2. Find s of the semi-cubical parabola ay 2 = x 3 . Here fdyY = 9x m \dxj 4a' """ S = J V 1 + Ta) dx ' "^g (1)> From the origin, s = ^ [(* + 4^)* ~ x ] ■ 3. Find s of the cycloid x = a arc vers - =F V2 ay — y 2 . Here / = H and T 7 sin = wx give by division , . wx dy , , tan 4> = -rr = ~r, as before. H ax TJ Also. T — = VH 2 + (wx) 2 gives the tension at any 1 COS

\ wl ~2H~ COS 0i where approximately. (10) 244 INTEGRAL CALCULUS When S and H are given to find I, a cubic equation results, the solving of which may be avoided by putting *S 3 for P, since they are nearly equal when d is small; then, w 2 S 3 1 = S - 247/2 > approximately. (11) 145. The Suspension Bridge. — The cables of a suspen- sion bridge are loaded approximately uniformly horizontally, since the roadway is horizontal, or nearly so; and the extra weight of the cables and the hangers near the supports is a small part of the total load carried by the cables. As shown in Art. 144, under the conditions stated, the curve of the WX" cables is the parabola whose equation is y = ^tj, where H is the horizontal tension. Example 1. — The Brooklyn Bridge is a suspension bridge which has stays and stiffening trusses to prevent oscillation. The span between main towers is 1595 feet. If the sag OD is 128 feet and the weight per foot supported by each cable is 1200 lbs., without considering the stays or stiffening trusses, to find the terminal and horizontal ten- sions, substitute the numerical values in (3) of Art. 144. For terminal tension, Ti = H sec 0! or T x VWU+- For horizontal tension, H = |^ = 2,981, 279 lbs. Ti = H sec 0i id Sd VP + 1Q

y), or P; then ws, the weight of the arc OP, is in equi- librium with the tangential tensions at and P. Denote 246 INTEGRAL CALCULUS the horizontal tension, which is the same at all points, by H—wa. If c • PT represents the tangential tension T at P, c • PD and c • DT will represent, respectively, the horizontal and vertical tension at P. Hence, v+- — > r fi~ -iC ■AK- X z --- ---> < N p f ' u \Z r & ysj> 2/z\ n — [~- X a T /< ITS ifl F i M z °i i» r «i - dy _ C' DT _ws _s dz c • PD wa a ' s _ v ds 2 - dx 1 a dx dx a ds Va 2 + s 2 ' + Va 2 + s 2 " «_ p * , r.+v*+*i a Jo Va 2 + s 2 L a J t 2 + 5 2 The exponential equation is e a = + Va 2 + ac" + V^+7 2 , which solved for s gives for length of OP, S = |(^-e")=asinh5. (Compare Ex. 5, Art. 142.) (1) (2) (3) CURVE OF A FLEXIBLE CORD — CATENARY 247 dii Substituting in (3), s = a -p. from (1), gives 'hi dx if - --\ x = 2Y~ e I =sinh a' (4) ••• J!* = l£( e * - •"■)* = I sinh ^" ; 2/ + a = | f e~ a + e a J = a cosh- (5) is the equation of the catenary referred to the axes OX, OY. If the origin of coordinates is taken at a distance a below the lowest point of the curve and the curve be referred to the axes O1X1 and OiY, its equation is y = a cosh- = ^(e° + e a j- (5i) The horizontal line through Oi is called the directrix of the catenary, and Oi is called the origin. Corollary I. — Since a = OiO is the length of the cord whose weight is equal to the horizontal tension, and there- fore the tension at the lowest point 0, it follows that if the part AO of the curve were removed and a cord of length a, and of the same weight per unit length as the cord of the curve, were joined to the arc OP and suspended over a smooth peg at 0, the curve OPB would still be in equihbrium. Corollary II. — Since the sides of the triangle PTD are proportional and parallel to the three forces under which the arc OP is in equihbrium, it follows that : tension at P = c-PT __ T__d±_y tension at ~ c • PD ~ wa ~ dx~~ a' from (3) and (5i), by differentiating (3); - T = wy; that is, the tension at any point of the catenary is equal to the 248 INTEGRAL CALCULUS weight of a portion of the cord whose length is equal to the ordi- nate at that point. Therefore, if a cord of uniform density and cross section hangs freely over any two smooth pegs, the vertical portions which hang over the pegs must each terminate on the direc- trix of the catenary. Corollary III. — Subtracting the square of (3) from the square of (5i) gives y 2 = s 2 + a 2 ; (6) dx 2 and from (6), after substituting a 2 = y 2 ^ from Corollary II, s = y Ts - (7) From M, the foot of the ordinate at P, draw the perpendicu- lar MT U then PTi = y cos MPT^ = y-£, which in (7) gives PT t = s = the arc OP; (8) and since y 2 - P7Y + 7W 2 , from (6) and (8), TiM = a. (9) Therefore, the point Ti is on the involute of the catenary which originates from the curve at 0; 7W is a tangent to this involute; and TJ?, the tangent to the catenary, is normal to the involute. As 7W is the tangent to this last curve, and is equal to he constant quantity a, the involute is the equitangential curve, or tractrix. 147. Expansion of cosh x/a and sinh x/a. — Expanding X _x ef 1 and e <* in series and taking the sum of the two series term by term gives for (5i), = ° + fa + 2&+---. (10) APPROXIMATE FORMULAS 249 Taking the difference of the two series gives for (3), s = a sinh - = a - + -j^ + -^i + ■ • • a \_a a 3 3! a°o! .-. for (4), X For these expansions, put - for x in Examples 8 and 7 of Exercise XLIII. 148. Approximate Formulas. — The series in Art. 147 are rapidly convergent when x is small compared with a. x Near the origin - is a small quantity, s is nearly the same as CL sly o ■£ x } and -j- = - = - ; hence, neglecting the terms with the higher powers of x ; y = a ( l+ ^2\) = a+ fa' 0I y - a = fa' (1) (X x^ \ f = tan* = E + ~^7 - - + A' (3) dx a a 3 3! a 6 a 3 Equation (1) is the equation of a parabola with its vertex at (0, a), the lowest point of the catenary. Where the cord is very taut with small sag, s is very nearly the same as x, and as they are nearly equal in any case near the vertex where -^ is small, if x is put for s in (1) of Art. 146, then ax + Wa 2 ' (2) the equation of the parabola with its vertex at (0,0) or 0. dy _ x m . _ C* x dx _ x 2 dx~ a' V ~ Jo ~a~ ~ 2a 250 INTEGRAL CALCULUS Hence, near its lowest point the catenary approximates in shape a parabola. When xi = i I, y\ = a + d, d denoting the sag; hence, for the sag at the lowest point of the curve, <* = £ + »+- -(from (10), Art. 147), or, approximately, d = j- a = |J; /. H = g [(2) and (3), Art. 144]. (4) Also at the supports tan i is, approximately, tan^ = ^ = ^ = ^ [(4), Art. 144]. (5) When Zi = o> Sl "2" t "48a 2i " " " ~2~ t ~48# 2 ~ 1 ~' * ' ~2~ 1 ~6Z' i ~' "' W Hence, when the supports are at the same elevation, 73 7 „273 Total length = 2si = I + xts = 1 + 24a 2 ' 24# 2 = I + ^, approx. [(10), Art. 144]. (7) Note* — When the sag is 1 per cent of the span, the error in H or d, from using these approximate formulas, compared with the value from those for the true catenary, is about ^ ? of 1 per cent. For a sag of 10 per cent of the span, the error is about 2 per cent. 149. Solution of s = a sinh x/a. — While in practice the approximate parabolic formulas of the preceding article are x generally used, the curve y = a cosh - appropriately mag- nified fits any cord hanging under its own weight, the con- stant a depending upon the tautness of the cord. When the horizontal distance x h from the lowest point of * Poorman's Applied Mechanics. SOLUTION FOR THE CATENARY 251 the curve to one of the supports B, and the length of the arc OB = si are given, then approximate values of a, y h T h H, and d = yi — a can be found. For putting xi for x in (3), then Si = 7i\e a — e a ) = asinh—> (1) 2 a where Si and X\ being given, a is found by solving the trans- cendental equation. The solution is by approximations and use of tables of the hyperbolic functions. When the supports are at the same elevation and I denotes their horizontal distance apart or the span, I = 2 x i} and for total length of cord, £ = 2si = a\e ra -e~^) } (2) where S and I being given, a is found by solving the equation. The curve heretofore considered is called the Common Catenary, the cord being of uniform cross section. In the Catenary of Uniform Strength, the area of the cross section of the cord at any point is varied so there is a con- stant tension per unit area of cross section. The equation is y = clog sec (x/c). Example 1. — A chain 62 feet long, weighing 20 lbs. per ft., is suspended at two points in a horizontal line 50 ft. apart. Find a; the horizontal tension; the terminal tension; the sag of the chain. Here 8 = |(e« - e~*) = |(e« - e"°) = 31. (1) S = x + 6^ = 25 + ff~ = 31, by (2) ' Art ' 148; (25) 2 = 6X6 . "a 2 25 ■ ' — = - = 1.2, approximately. a o 252 INTEGRAL CALCULUS Now let — = 2; .-. a = — , which substituted in (1) gives d Z Let /(2) = e z — e~ z — 2.48 2 = 0, where 1.2 is approx- imate value for z. Reference to a table of hyperbolic sines will show that the value of z is between 1.1 and 1.2. The following is a method of getting a closer approxima- tion to the value of the root z without the use of tables. Let z = Z\ + h, where z x is an approximate root differing from the root z by a small quantity h. By Taylor's Theorem : /(«) -/(ft + h) = /(ft) + V W + f /"to + 0. Neglecting higher powers of h,f(zi) + hf (ft) = 0; .*. z = ft — ^ . , ■ , as first approximation for z. Let this value be 2 2 , differing from 2 by k < h; ■ ■ /(«) =/fe + fc) =/(*)+¥'<*) + ^/"(*) + • • • = 0. Neglecting higher powers of k, f (« 2 ) + fc/' fe) = 0; as second approximate value for 2. By repeating this process a value which approximates closer and closer to the true value of the root can be gotten. Applying the method to this example : /to e 1 - 2 ~ e- 1 ' 2 - 2.48 X 1.2 _ 0.04 _ nnor /'to" e 1 - 2 + e-^ -2.48 1.14" UAWD ' /. Z2 = ft - h = 1.2 - 0.035 = 1.165. SOLUTION FOR THE CATENARY 253 Now _ gi-165 _ e-i.165 _ 2.48 X 1.165 _ 0.005 _ _ nm - * " e 1 - 165 + e- 1 - 165 - 2.48 ~ " 1.038" °- UU ° ; .*. z = z 2 -k = 1.165 - 0.005 - 1.16, an approximation close enough, the value found from the table by interpolation being the same. Hence, 25 25 01 „ u a = — = — — = 21.oo ft. z 1.16 H = wa = 20 X 21.55 = 431 lbs. Vl = Vs{ 2 + a 2 = V(31) 2 + (21.55)- = 37.75 ft. (6) Cor. Ill, Art. 146. d = y l - a = 37.75 - 21.55 = 16.20 ft, T = wy t = 20 X 37.75 = 755 lbs. Cor. II, Art. 146. Example 2. — To get the correction for the sag in measur- ing horizontal distances with a steel tape unsupported except at the ends where pull is applied. A 100 ft. steel tape was weighed and its weight found to be If lbs., making w = 0.0175 lb. Using the equation of the parabola, wx~ y = x-p, where P = H is the pull in pounds; d = ^-5, exact if I is span, approx. if I is arc or tape. o 1 Length of tape = s = / + -=-= =-=- + • • • O I O I ?/ ,273 7/ ,475 , _Sd 2 _ iv 2 l s { approx. correction for one I ,. X " S ~ ~3l " 2TP (tape length, (10), Art. 144 J ' () True horizontal distance = l = s — KTpzi approx., s 3 being put for? 3 . (11), Art. 144. (2) 254 INTEGRAL CALCULUS In (1) when s and P are given to find I a cubic equation results, the solving of which may be avoided by putting s 3 for Z 3 , since they are nearly equal, the correction being small. Substituting the pull P in lbs. in (1) or (2), the correction or the horizontal distance, respectively, for each tape length will be found. The pull may be measured with a spring balance. Since sag shortens the distance between tape ends and pull lengthens it by stretching the steel, there is some pull at which the two effects are balanced. The stretch is given by where P is pull in lbs., I is length of tape, A is area of the cross section, and E is modulus of elasticity, usually expressed in lbs. per sq. in.; for steel, E = 30,000,000 lbs. per sq. in. To find the pull that will just balance the effect of sag, put the values of x from (1) and y from (3) equal; w 2 V PI . . 24P~ 2 = AE' a PP roximatel yi whence, P = \/ — sr — ) approximately. In this example the width of the tape was fV in. and the thickness F V in., making area A = 0.00525 sq. in. V 2 !5 (100) 3 30 = 271bs. 24 (1000) This pull to balance effect of sag can be determined experi- mentally by marking on a horizontal surface two points 100 feet apart, and then noting the pull on a spring balance when the ends of the suspended tape are exactly over the points marked. 150. The Tractrix. — The characteristic property of the tractrix is that the length of its tangent at any point is THE TRACTRIX 255 constant. Denote the constant length of the tangent PT by a. If a string of length a has a weight attached at one end while the other end moves along OX in a rough horizontal plane XOY, the point P of the weight, as it is drawn over the plane, will trace the tractrix APPi .... Let AO be the initial position of the string and PT any intermediate position. Since at every instant the force exerted on the weight at P is in the direction of the string PT, the motion of the point P must be in the same direction; that is, the direction of the tractrix at P is the same as that of the line PT, which is, therefore, a tangent to the curve. To find the length and equation of the curve: let PD = ds; then -PN = dy and ND = dx; ds _ PD _ a m •'* dy~ ~PN~ y U Hence, if sis reckoned from A (0, a), then the length s=-a pft-afcgi (2) Ja y y Also, from the figure, dy _ y nr dx _ Va 2 - y 2 . ,. -j- — / -i or j— — , {o) dx Va 2 - y 2 dy y 256 INTEGRAL CALCULUS /. x = f V - Va 2 -y 2 ^ = - Va 2 - y 2 J a y + a log I — — fl2 ~ y j (See Ex. 20, Exercise XXV.) is the equation of the tractrix. Example 1. — To find the area bounded by the tractrix in the first quadrant, the z-axis, and the ?/-axis. A = f°°ydx= - f°Va 2 -y 2 dy (from (3)) [y -. /—* — 5 . °> 2 • i vl a m 2 = [2 Va - y+ 2 Sm airir- Example 2. — To find the area bounded by the catenary, the x-axis, the i/-axis, and any ordinate y. A=% PV + e~ V dx = % \e« - e~») - a 2 sinh - (1) Z Jo £ a — a\-\e a — e a ) \ = as, where s = length of arc. (See (3), Art. 146.) For area up to x = a, ..ft! -.-:)];- f(.-I) ■ Hence, by (1), A= ('a Jo cosh -dx = a 2 sinh - ■ a a 151. Evolute of the Tractrix. — It has been shown in Art. 146, Corollary III, that the involute of the catenary is the tractrix. Conversely, it may be shown that the evolute of the tractrix is the catenary. Let (a, 0) be the coordinates of C, the point of intersection of the normal at any point P (x, y) of the tractrix and the perpendicular to .r-axis at T, end of tangent, PT = a. From the figure (Art. 150), OT = a = x + Va 2 - y 2 , (1) (3) EVOLUTE OF THE TRACTRIX 257 and (TC and PC being drawn) m ^ n a 2 (3 a f from similar triangles 1 /e% . TC = V = -, since - = -[ PTC &nd PTM Equation of tractrix, /-s s i i [~ a + Va 2 — w 2 l x = - va 2 - ?/ 2 + alog — — • Eliminating x and y from these three equations gives a [ (3 + Va 2 - a 2 ] . ^n + VFJ, (4) a Solving (4) for /3 gives for the relation between a and j3, which is the equation of the catenary with origin at 0; and its lowest point at A, one end of the tractrix. The normal PC to the tractrix, the involute, is the tangent to the catenary, the evolute, and is equal in length to the arc AC (not drawn) of the catenary. If the equations for coordinates, a and (3, of the center of curvature, given in Art. 94, be used, the values found in (1) and (2) will result. Note. — It may be shown independently of Art. 146, Corollary III, that the catenary is the curve which has the property that the line drawn from the foot of any ordinate of the curve perpendicular to the corresponding tangent is of constant length a. Thus, let 6 be the angle which the tangent CP makes with the a:-axis and it is evident from the figure (CP being drawn) that o.l 1 = COS0 = Vl+tan 2 vMS 258 INTEGRAL CALCULUS da d/3 l -f"da=r d& CI t/0 J a V(3 2 -a 2 ' , « = lo g ( /3 + V^^)T=logf-±^Z); a ja \ a / ... e i = g+^^, (4) a Hence, /3 = = \e a + e V , as before. CHAPTER IV. INTEGRATION AS THE LIMIT OF A SUM. SUR- FACES AND VOLUMES. 152. Limit of a Sum. — A definite integral has been defined (Art. 12|) as an increment of an indefinite integral. It will now be shown that a definite integral equals the limit of the sum of an infinite number of infinitesimal increments or differentials. Many problems in pure and applied mathematics can be brought under the following form : Given a continuous function, y = f (x), from x = a to x = b. Divide the interval from x = a to x = b into n equal parts, of length Ax = (b — a)/n. Let x h Xz, Xz, . . . x n be values of x, one in each interval;- take the value of the function at each of these points, and multiply by Ax; then form the sum: f(x,)Ax+f(x 2 )Ax + • • • +f(xn)Ax. (1) Required, the limit of this sum, as n increases indefinitely and Ax approaches zero.' This problem may be interpreted geometrically as the problem of finding the area under the curve y = / (x), between the ordinates x = a and x = b; each term of the sum representing the area of a rectangle whose base is Ax and whose altitude is the height of the curve at one of the points selected. Let the area M1P1BX be denoted by A; let OM x = a, OX = b, and PiB be the locus of y = f (x). Let Az be one of the equal parts of MiX, although the parts need not be made equal provided the largest of them approaches zero when n is made to increase indefinitely. 259 260 INTEGRAL CALCULUS It is easily seen that the difference between the sum of the rectangles as formed and the area A is less than a rectangle whose base is Ax and whose altitude is a constant, / (6) — / (a) . Since this difference approaches zero as Ax = 0. the sum of either set of rectangles approaches the area A as a M y M z M 5 M 4 M 5 M n X limit. It is evident that the sum of the rectangles which are partly above the curve is greater than A, while the sum of those which are wholly under the curve is less than A. By the notation of a sum, letting T be the difference, ^ = 2 f(x)&x±T, where T, < /(b) Ax, = 0, asAx = 0; a :. limit £* / (x) Ax - A = f * / (x) dx. (2) The equation (2) is true, for it has been already shown that A = f b f(x)dx= ff(x)dx] - ff(x)dx\ =F(6)-F(o), da *J Jx=b *J Jr=a where f f(x)dx = F (x). It follows that the limit of (1) is limr/(x,)Ax+/(x 2 )Ax+- • • +/(x n )Ax] = F(6)-F(a), (3) where a and b are end values of x and I f(x)dx = F(x). THE SUMMATION PROCESS 261 X >4X<- ira z The theorem of this Article summarized in equation (2) may- be said to be the fundamental theorem of the Integral Calculus. As a simple example of the determi- nation of an area by getting the limit of a sum of an indefinite number of in- finitesimal elements of area, let a circle of radius a be divided into concentric rings of width Ax; then for the area A, A = limit V 2ttx&x = 2ir ( a xdx = 2irx 2 /2\ = Ax=0 ^T Jo JO Here A A = 2 w (x + \ Ax) Ax and dA = 2 irx dx. 153. The Summation Process. — On account of the frequency of the occurrence of the summation process, it may be said that an integral means the limit of a sum, the limit being in most cases most easily found as an anti- differential or anti-derivative; that is, by the inverse process to differentiation, namely, by integration. The symbol / for integration, the elongated S, is derived from the initial letter of summa, the integral being originally conceived as a definite integral, the limit of a sum. Accord- ing to Art. 128, the indefinite integral also may be regarded as the limit of a sum. The fact that the summation of an indefinitely large number of indefinitely small terms is in most cases easily effected by a comparatively simple process is of the highest importance. Thus integration replaces the tedious and often difficult process of direct summation and gives an exact result, while the other often gives but an approximation at the best. While the process of summation has been illustrated geometrically by the determination of an area, the reason 262 INTEGRAL CALCULUS of the process by no means depends upon geometrical con- siderations. The method is applicable to the determination of the limit of the sum of small magnitudes of all kinds — volumes, masses, velocities, pressures, heat, work, etc. For an example of finding the limit of the sum of small volumes, consider the volume V generated by revolving the area' M1P1BX of the figure of Art. 152 about OX as an axis. Each of the rectangles PiM 2 , . . . , PnX will generate a cylinder whose volume will be expressed by w (/ (x)) 2 Ax; hence, V =^\{f(x)) 2 Ax + T, a where T, < w (/ (6)) 2 Ax, =0, as Ax = 0; /. limit V 6 7T (/ Or)) 2 Ax = V = fV (f(x)) 2 dx. (1) Ax=0 a *J a Example. — Find volume of a sphere by revolution of = a 2 — x 2 : V = I 7r (a 2 — x 2 ) dx = f to?. For another example of finding the limit of the sum of small volumes, find the volume of the sphere considered as made up of concentric shells of thickness Ap. V = lim y.^Trp 2 -A P = 4tt f f?dp=Sira*. Ap=0 «/0 154. Approximate and Exact Summations. — When the rate of change (or the derivative) of a variable quantity is given, the total amount (or the integral of the rate) can be obtained approximately by direct summation, and exactly by finding the limit of a sum; that is, by integration. For example, suppose the speed of a train is increasing uniformly from zero to 60 miles per hour, in 88 seconds; that is, from zero to 88 ft. per sec. in 88 seconds, the increase in speed each second (the acceleration) is 1 foot per second. APPROXIMATE AND EXACT SUMMATIONS 263 Hence the speeds at the beginnings of each of the seconds are 0, 1, 2, 3, ... , etc. Taking the speeds as approximately the same during each second as at the beginnings, the total distance. 07.00 s=0+l+2+3+ ■ • • + 86 +87 = ^^=3828 ft., which is evidently less than the true distance. Taking the speed at the end of a second as that during the second, s=l+2 + 3+4+--. + 87 + 88 = E^H = 3916 ft., which is evidently greater than the true distance. These values for the distance differ by 88 ft. and it is certain that the true distance is between 3828 ft. and 3916 ft. When the length of the interval during which the speed is taken as constant is reduced more and more, the result will be more and more accurate, nearer and nearer to the true distance. Manifestly, the exact distance is the limit approached by this summation of small distances as the interval of time At approaches zero: 1=88 l =3 /»f=S8 /2~|*=88 = limYf>A*= / tdt = -\ = 3872ft. =0 M=0 ^ Jt=0 4j t =0 In general, s = limit X v\t= f atdt = § at 2 , At=0 t=0 J 1 = a being constant acceleration. In mechanics, the determinations of centers of gravity, centers of pressure, moments of inertia, varying stress, etc., involve the summation principle; and the greater number of the integrations in practice appear more naturally as limits of sums than as reversed rates, anti-derivatives, or anti-differe ntials. The summation of an infinite number of terms is always 264 INTEGRAL CALCULUS involved when one of the factors entering into the problem varies continuously. For example, in the problem of finding the mass of a body, defined as the product of density and volume; when the density p varies continuously, m limit V P A7= fpdV, A7=0 J where the integral taken between " limits," that is, with end values for the independent variable, is the limit required. Hence the mass is given by a definite integral, which can be evaluated when the density p is a known function of the volume V, that is, of the variables x, y, z or r, 6, , in terms of which the volume may be expressed. When the density p is constant, it is evident that the mass is m = 5) P A7 = p CdV = P V. Thus, when the body is composed of different liquids of varying densities in the layers or strata, the total mass is found by the addition of a finite number of terms. For if Vi, V 2 , 7 3 , . . . V n denote the volumes of the separate parts, and pi, p 2 , p 3 , • . • p n the corresponding densities, then m = pi V l + p 2 V 2 + P3V3 + • • • + PnVn, where the summation is made without integration. The above will give an approximate result even when the density varies throughout the whole mass. When, however, the density varies continuously as in the atmosphere, the total volume is divided into n parts each equal to AV and each part is multiplied by the density at that part of the body. There are then n elements of the form pAF, and when n is finite their summation will be an approximation to the mass of the whole; but to get the exact value, the limit of the sum, as n becomes infinite and A.V = 0, must be found, and hence the exact value of the whole mass is deter- mined by the process of integration. 7 APPROXIMATE AND EXACT SUMMATIONS 265 Example 1. — If y = x 2 , find V x 2 Ax for different values 2 2 of Az, and get lim V x 2 Ax. Get lim V x 2 Ax. A^-^n "i A~-^n ^"^a Ax=0 '1 When Az = 0.2, Ax=0 ]£"x 2 Az = (l 2 + F2 2 + O 2 + ]L6 2 + D?) 0.2 = 2.04. When Az = 0.1, 2) 2 x 2 Az = (l 2 "+ LI 2 + V L2 2 + • • • + L9 2 ) 0.1 = 2.18. When Ax = 0.05, V% 2 Az = (l 2 + L05 2 + Li 2 + • • ■ + L95 2 ) 0.05 = 2.26. **\ Lim T a; 2 Aa;= / x 2 dx = -=■ = — - — = 2.33J square units in MiP 1 P 2 M 2 . x-^ 1 C l x z ~\ l 1 Lim 2j x 2 Ax = I x 2 dx = — = - ^o Jo ojo o = 0.33 J = J of rectangle OilfiP^i. 1 x-n 4 Ax Example 2. — If y = -, find 2 — f° r different values of x ^^ i x ^Ax l Ax Ax. and get lim V — Get V - — as Ax = 266 INTEGRAL CALCULUS When Ax = 1, 2rNi + ^) (1) = 1 - 833 - X 4 At — = 1.593. i x When Ax = 0.1, V — = 1.426. Lim T — = / — = logrc = log4 - log 1 = log 4 = 1.386 = Area AfiPJFW*. LimT — = / — = log# =logl-loga= -loga = oo ; Ax=0 a x *Ja x Ja Ja=0 hence, when a = 0, the limit does not exist, asY — = oo . ' **0 X Jax=0 (Compare Ex. 11, Art. 135.) Note. — For examples of application see Art. 189. EXERCISE XXX. 1. If y = x, find ^ x Ax > when Ax = 1; when Ax = 0.5; when '3 ,7 Ax = 0.2. Get lim V x Ax. Ans. 18; 19; 19.6. Ax=0 *"*3 Arcs. 20. 2. If y = tan 0, find ^ tan A0, when Ad = ^ ; when A0 = ^ ; when Ad = r^. Get lim V 3 tan Afl. Ans.' 0.316; 0.328; 0.340. 180 A0i °^ An*, log. V2 = 0.346. Determine the following quantities (a) approximately by summation of a limited number of terms; (b) exactly by finding the limit of the sum of an infinite number of terms by integration. 3. The area under the curve y = x 3 , from x = to x = 2; from X m — 1 tO X ■■ 1. 4. The distance passed over by a body falling with constant accelera- tion g = 32.2 per sec. 2 , from / = 1 to / = 4, v = gt being the relation of v and t. VOLUMES 267 5. The increase in speed of a body falling with acceleration of g = 32.2 per sec. 2 , from t = tot = 3. 6. The number of revolutions made in 5 minutes by a wheel which revolves with angular speed u = t-/ 1000 radians per second. 7. The time required by the wheel of Ex. 6 to make the first ten revolutions. 155. Volumes. — The volumes of most solids may be found approximately by the summation of a finite number of parts and exactly by finding the limit of the sum of an infinite number of terms by integration. Example. — To find the volume of the right circular cone whose altitude is h and the radiiis of whose base is a. Divid- ing the volume into parts, each AT', by passing planes Ax apart parallel to the base A h , and denoting a section at a distance x from the vertex at the origin by A x , then, since A x 'Ai_ = x 2 7?.' 2 , V is given approximately by and exactly by . x' A kJ r 2 \x V = lim = Ak r* h 1 3 * x o h- " fe 2 Jo x 2 dx (i) (2) (3) While AT is a frustum of the cone, dV may be represented by the cylinder PMMi = A x • Ax = iry 2 dx. It is to be noted that the equations all apply to a pyramid with any plane base Ah as well as to the cone. 268 INTEGRAL CALCULUS For another example : to find the volume of a sphere with radius a, divide by planes perpendicular to OX; then, since V = lim X^o^—^ Az = ^ f a (a 2 - z 2 ) dx Ax=0 -a & a J- a = M a2x ~ f L = $ • h 3 = r a3 > where 4 » ira 1 Otherwise; 2} AF = 2) A x Aa: = X ^ A- T > where A x = iry 2 ; :. V = lim V 7T?/ 2 Ax = 7r / (a 2 — x 2 ) dx = « 7ra 3 . Ax=0 ^ J-a u 156. Representation of a Volume by an Area. — In Art. 138 on the significance of an area as an integral it was stated that the integrals represented by areas might be functions of various kinds. To show an example of a volume as an integral represented by an area under a curve, let the volume of the paraboloid of revolution, between x = and x = 4, be first found as the limit of the sum of the parts between REPRESENTATION OF A VOLUME BY AN AREA 269 the parallel planes Ax apart, as Ax = and the number of the parts increases without limit. The equation of the generating parabola being y 2 = J x, V = limit V wy 2 Ax = ■? I x dx = £ tt = 2tt cubic units. ax=o fie, 4 Jo 4 2 Jo *-X To represent this volume graphically by an area, the line OP' is drawn by the equation y = jx, this being the function which was integrated to get the volume of the solid P 2 OP' 2 . Producing the ordinate M 2 P 2 to P", the area OM 2 P" graphically represents the volume of the solid P 2 OP' 2 . For, ydx=-r I xdx=- i - F r\ = 2 7T square units. 4 Jq 4 Zjo 270 INTEGRAL CALCULUS The last result may be verified by noting that the ordinate M%P" , for x = 4, being w, the area of the triangle is 2 t. In the same way it may be seen that any part of the area, as OMP', represents the corresponding part of the volume of the solid ; that is, there is the same number of square units in the one as there are cubic units in the other. If OP'P'", the first integral curve of OP'P", whose equa- tion is y = / ^rxdx = -=- (see Art. 140) Jo 4 o be drawn, its ordinates will represent both the areas of the parts of OM 2 P" and the volumes of the parts of the parab- oloid measured from 0; that is, the measure of the ordinates in linear units will be the same as that of the areas in square units and that of the volumes in cubic units. Length of M 2 P'" = y = -«- = 2tt linear units. Note. — The volume of the cone of Art. 155 may be graphi- cally represented by the area under the parabola y = -jjx 2 , and the volume of the sphere by the area under the parabola y = 7T (a 2 — x 2 ). If the first integral curves, V = g-p a? and y = v \o~x - |- j , be drawn, their ordinates will represent both the areas and the volumes in the two cases, respectively. 157. Surface and Volume of Any Frustum. — A solid bounded by two parallel planes is, in general, called a frustum. One or both of the truncating planes may in special cases, as in the sphere, touch the frustum \n only one point and be tangent planes. The method of dividing the solid into thin slices and taking the sum of the approximate expressions for the small parts as an approximate expression for the whole, and taking the SURFACE AND VOLUME OF ANY FRUSTUM 271 limit of the sum as an exact expression for the whole, may be applied to any solid even when the solid is not regular and the sections not regular plane curves. Let the solid represented in the figure be divided into slices by planes perpendicular to an axis OX; then, taking A x \x as an approximate expression for the volume of the slice P — N1M1R1, A x being the area of the section PNMR at a distance x from plane ZOY, the approximate expression is x=h 2)A7 = 2)^*^i x=0 where h(= OA) is the distance between the truncating or bounding planes. The exact expression is V = Urn V. A x Ax = / A x dx. (1) Ax=0 £* Q Jx=Q When the area of a section is a function of the distance x from one of the bounding planes and hence A x can be ex- pressed in terms of x, the limit may be found by integration. The frustum formula for volumes is, therefore, -x x=h F (x) dx or V Jx = h x F{x)dx, (!') 272 INTEGRAL CALCULUS where A x is F(x), some function of x; the one form giving the whole volume and the other a segment or any part thereof. To get expressions for the area of the surface S, let P be the curve NPR, then A£ = NPRRiP'N h and the approxi- mate expression is X^S= £PAs, s=0 where As = NNi and s is the length of CN. The exact expression for the surface is £ = limit £P As = / Pds. (2) As=0 s=0 Js=0 When the curve P is a function of s, the bounding curve in XZ plane, and can be expressed in terms of s, or when ds can be expressed in terms of P, with change of end values, the limit can be found by integration. If the surface S is conceived as generated by the curve NPR as it moves with its plane always perpendicular to OX, when its plane is in the position as shown, at a distance x from plane YZ, let iViV'i be drawn equal to ds but parallel to OX; then since the surface is cylindrical, the increase of S, if the increase became uniform, is dS = P ds, the surface NPRR'P'Ni'; 8 Jr* s =s h Pds. (2) s=0 If the curve NPR is a circle, as in solids of revolution, with the center at M on the z-axis, then P = 2 iry and A x = iry 2 , (2) and (1) becoming yds (3) =0 Jn*~h y 2 dx, (4) x=0 SURFACE AND VOLUME OF ANY FRUSTUM 273 where dV = wy 2 dx is the volume of the cylinder generated by the area of the circle wy 2 , as it moves uniformly through Az = dx. Note. — In deriving (1) and (4), in the figure, NNi = Az = dx; while in deriving (2) and (3), ds, the uniform change of s along a tangent to the curve CN at the point A 7 , is drawn parallel to OX and represented in length by NNi, although it is not the same as Az = dx but is really longer. Example 1. — To find the lateral surface of the cone of Art. 155: by (3), S = 2 7r / yds = 27rj -y sds, where s = -y = OP, Jo I/O ' a as 2 ~\ l -= 2Trj 7 r\ = wal, where I = OPh, an element. I Z Jo Again, y ds = 2 7r I y-dy, since c?s = d ( - y ) = - dy, o Jo fl \a / a limit or end value being changed from 2 to a. Example 2. — To find the surface of the paraboloid of Art, 156: = 2 t £ y [1 + 64 jf ]% from y 2 = \x, dy 2 128 |[l + (64^]J =^((65)1-1) o^ (65 V65 — l) square units. 274 INTEGRAL CALCULUS Example 3. — To find the surface of the sphere of Art. 155, or any part of it, as a zone. For a change take origin at A' on the circumference, making y = V2 ax — x 2 and 2 iry the curve P bounding the section A x ; then by (2) or (3), S= \Pds = 2ir j yds = 2tt \ adx JO Jo Jx (where yds = adx, from similar triangles, OMP and PDT) ]x ~|2a = 2 7ra (x — Xo) or 2 wax = 4 ira 2 . x JO Ddx T D' T' Drawing PT = ds from P parallel to rr-axis, 2 wy ds is the lateral surface of the cylinder PT', which is equal in area to that of the cylinder DT' ', Which is 2 ira dx. The volume is again, with origin at A', by (4), 1 V=ir J y 2 dx = ir f (2 ax - x 2 ) dx = tt — ^ ^ = 7r(7 , \ Example 4. — To find the lateral surface of a quadrangular pyramid. Let P h = perimeter of base and I = 0P h = slant height. Let PMN be the position of the generating perim- eter P when s = OP. Since P and Ph are similar, SURFACE AND VOLUME OF ANY FRUSTUM 275 P OP s , „ P h . ,.. p h = 0F h = V hence ' p = T s ' m(2); that is, the convex surface of any pyramid or cone (Ex. 1) is measured by half the product of perimeter of base and slant height. Y For the volume, V Jo A x dx = AuC h " ft 2 Jo " = Ah ff 1 h? 3 Jo 3 since -r- = A, a? ft 2 ' A x = areaPMiV, aAhh; that is, the volume of any pyramid or cone ((3), Art. 155) is measured by one-third the product of its base and altitude. Note. — The foregoing, for the purpose of illustration, have been for the most part examples of elementary solids whose surfaces and volumes are known from solid geometry. The fruitfulness of the method is seen in the determination of the surfaces and volumes of the frusta of unfamiliar and complex solids. The following are some examples: Example 5. A monument is to be built in horizontal rec- tangular sections, one side of a section to vary as the dis- tance below the top and the other as the square of this 276 INTEGRAL CALCULUS distance. The base is to be a square 30 feet on a side, and the height of the monument is to be 20 feet. Find the vol- ume when it is made up of rectangular blocks with vertical sides; and also the volume when the sections vary contin- uously from top to base. Taking origin at top and z positive downward, let A z = 4xy, 2y = az, 2x = bz 2 ; 2y = 20a = 30; .'. a = f: ; 2x = 400 b = 30; .*. b = ^ ; .'. y = %z, for line OB; x = & z 2 > for curve OA. A z = 4xy = abz z = fa z 3 , 2=20 -i V = X ^Az =5.^[0 + 5 3 + 10 3 + 15 3 -|-20 3 ] 3 J A *= 5 = 7031i cu. ft. V = lim T — z 3 Az = — / z 3 dz 9^ [VI 2 80 |_4 J c 4500 cu. ft. Hence, as Az, the thickness of the blocks, is made less and less, the volume will approach 4500 cu. ft., the volume when sections vary continuously. The plan is shown on reduced scale. The projection of OD, the curve of intersection of the JO plane surface OBD and the curved surface OAD, is O'D on plane parallel to IF plane. The equation of O'D is SURFACE AND VOLUME OF ANY FRUSTUM 277 y 1 = 15 x, by eliminating z from y = \z and x = A z 2 . The plan shows the corners of the blocks on this curve. Example 6. Find the lateral surface of the monument of Ex. 5. When built of rectangular blocks, the sum of the rectangular areas gives the area of the surface. When the stone is shaped to make sections vary continuously, or when this is effected by using concrete in shaped forms, find the areas of the surfaces OBD and OAD separately. i(OAD) - ^P|^] - 2 f x 38.8 - 862 „. ft. = 8oJo * L 1+ 16j ^SO'U "-aLsJi 3 8000 10R ft = 64'-3- =12 ° Sq - ft - 4 (OBD) = 4 X 125 = 500 sq. ft. Total surface = 1362 sq. ft. Note. — Since OBD is a plane surface, its area may be found by A= i X dS = Lo 125 S ' ds " 125 ' 4 = 125 Sq - ft - as above. Here z = T f 5 s 2 is the equation of curve OD in the oblique plane of sx, for since OB = 25, y = f a, and 2/ 2 = 15 x becomes ? 9 : - 7 s 2 = 15 z, or x = T fy s 2 . In Ex. 5 above, ?/ 2 = 15 x is given as equation of pro- jection of OD on xy plane, or any plane parallel thereto. 278 INTEGRAL CALCULUS While OD is given -as a line in space by two equations, by rotating axis OY about OX through tan -1 § % = cos -1 f , it is given by one equation, s 2 = 1 % & x, in plane of sx. Example 7. Find the volume common to two right circular cylinders of equal radius a, whose axes intersect at right angles. Let the two cylinders be x 2 -\- z 2 = a 2 and y 2 + z 2 = a 2 ; then A z = LMPN xy = a 2 z 2 , and V = 8 f a A 2 dz = 8 /"(a 2 - z 2 ) dz = s\a 2 z - |T = |a 3 . The total volume common, being 8 times Z — OACB, is -^ a 3 . Example 8. A dome has the shape of the figure of Ex. 7, find the area of the curved surface. The surface ZBC is equal in area to the surface ZAC, and is one-eighth part of the surface of the dome, which surface is the upper half of the surface of the common volume of Ex.7. Hence the surface of the dome of eight equal parts is given by SURFACE AND VOLUME OF ANY FRUSTUM 279 S = 8 ZBC = 8 f Sh P ds = S f Sh NPds Jo Jo -°I«h(t)T* = 8 a I -dz = Sa I dz = 8a 2 . Jo 2/ Jo The result shows that each of the curved surfaces of the solid Z — OACB is equal in area to its base OACB; the surface of the dome being just twice that of its base. Note. — Another determina- tfa tion of the area of ZBC may- be made by developing the curved surface upon a plane and finding the area as a plane area, Thus, developing ZBC ?--, J c as the plane area Z'BC, with B as origin; area Z'BC = area ZBC TTU, f 2 x'dz', where x = x' = a cos 6, and z' = ad IT TT = I a cos Bd {ad) = a 2 sin = a 2 . Jo Jo Example 9. Given a right cylinder of altitude h, and radius of base a. Through a diameter of the upper base two planes are passed, touching the lower base on opposite sides. Find the volume included between the planes. 280 INTEGRAL CALCULUS V = r A x dx = 4 f° (MNPR) dx = 4 Pyz dx t/0 t/0 «/o = 4 P(a 2 -z 2 )^-(a-z))* dx-— P (a 2 - xrf x dx a Jo a Jo = 4 f[}vy=^+}*r*5|;+g^ - *>»]; ira?h— - a 2 /i = (vol. of cylinder) — (vol. outside the planes), o Here NP MR = OZ MA OA ' NP z = - (a — x) ; a MN = RP = y = (a 2 - x 2 )*. It may be noted that, when h is equal to a, the volume outside of the planes being % a 3 , is one-fourth of the volume common to the two cylinders of Ex. 7. Example 10. Two cylinders of equal altitude h have a circle of radius a, for their common upper base. Their lower bases are tangent to each other. Find the volume com- mon to the two cylinders. V= f hi A v dy= P(PMM')dy = Pxzdy J h t J —a J— a = / x-xdy = - I x 2 dy = - I (a 2 — y 2 ) dy J —a 0, d J —a Q> J— a = a[ ay ~^L = 3 ah - SURFACE AND VOLUME OF ANY FRUSTUM 281 Here, PMM f being similar to ZAA', Ar „ OZ-NM h , NP = — tti — > or z = -x, where x 2 = a 2 — y 2 . OA a u P is on curve of intersection of the cylinders. It is seen that the volume found is equal to the volume outside the planes of Ex. 9. Example 11. A torus is generated by a circle of radius b revolving about an axis in its plane, a being the distance of the center of the circle from the axis. Find the volume by means of sections perpendicular to the axis. V = f ki A v dy= f Xh [ T (a + x) 2 -Tr(a-x) 2 ]dy Jh x Jx=-b - 7T f U b [(a + Vb 2 - y 2 ) 2 - (a- Vb 2 - y 2 ) 2 ] dy = tt / 4a Vb 2 - y 2 dy = = 47ra = 47ra "Trfe 2 " 2 2w 2 ab 2 = 2ira-Trb 2 . 282 INTEGRAL CALCULUS Note. — The last form of the result shows that the .volume is the product of the area of the cross section and the length of the circumference described by the center of the revolving circle, radius a being mean of a + b and a — b. EXERCISE XXXI. 1. Find the volume of the right conoid whose base is a circle of radius a, and whose altitude is h. (a) With origin at 0, on the circumference; y 2 = 2 ax - x 2 . a 2 - x 2 . Aiis. — • 2. An isosceles triangle moves perpendicular to the plane of the ellipse x 2 /a 2 + y 2 /b 2 = 1, its base is the double ordinate of the ellipse, and the vertical angle 2 A is constant. Find the volume generated by the triangle. . 4 ab 2 cot A Ans. 3 x 2 y 2 s 2 3. Find the volume of the ellipsoid i + jb + "^ = 1 by considering the volume generated by moving a variable ellipse along the axis of X. Area of ellipse = irab. From result get volume of a sphere. Ans. %irabc. 4. A football is 16 inches long and a plane section containing a seam is an ellipse the minor axis of which is 8 inches in length. Find the volume (a) if the leather is so stiff that every cross section is a square; (b) if the cross section is a circle. Ans. (a) 341 § cu. in. ,.. 512 7T (6) — ^— cu. in. 6. To fell a tree 2 a feet in diameter, a cut is made halfway through from each side. The lower face of each cut is horizontal; the inclined face makes an angle of 45° with the horizontal. Find the volume of the wood cut out. Compare Ex. 9 of illustrative examples. Ans. f a 3 cu. ft. y 2 z 2 6. Find the volume of the elliptic paraboloid 2 x = — H — cut off by the plane x = h. Ans. ir "vpq h 2 . 7. Find the volume of Ex. 9 by moving the trapezoidal section along the F-axis. Note that the triangular section of the volume outside the PRISMOID FORMULA 283 cutting planes will at the same time generate that volume, the same as the volume of 5 above, when h = a. 8. A cap for a post is a solid of which every horizontal section is a square, and the corners of the square lie in the surface of a sphere 12 inches in diameter with its center in the upper face of the cap. The depth of the cap is 4 inches. Find the volume of the cap. Compare Ex. 7 of illustrative examples. Ans. 490f cu. in. 9. Find the surface of the cap of 8, above. Compare Ex. 8 of illustrative examples. Ans. Curved surface = 192 sq. in.; surface of top = 72 sq. in. 10. Show that the volume of the frustum of any pyramid or cone is equal to ^ (A + Ah + VAoAh) where A and Ah are the bases, and h is its height. 158. Prismoid Formula.* — If two solids contained between the same two parallel planes have all their corre- sponding sections parallel to these planes equal, that is, if the area A/ of the one is the same as the area A s " of the other, then their total volumes are equal, since the two volumes are given by the same integral. Let the distance between the bounding planes be, in general, s = x, or y, or z. If the area A 8 is a section of a solid included between two parallel planes and is a quadratic function of s, A s = as 2 + bs + c, (1) where s is the distance of the section A s from one of the two parallel planes, then the volume is given by (as 2 -{- bs + c) ds = \a - + b - + cs\ ah 3 bh? , , . = -y + -7r + ch, (2) Jo Js=0 where h is the distance of the terminal plane from the initial plane of reference; that is, the height, or length, of the solid, as the case may be. * This derivation of the formula is substantially that given in Davis's Calculus. 284 INTEGRAL CALCULUS The area A = A 8 the area A h =A S and the area A m = A e ■i = as 2 + bs + c = = c; s = = as 2 + bs + c =a/i 2 + ^ + c; =^ Js=/i = as 2 + 6s + c] =^-+ — + c, where A m is the area of a section midway between the end sections, A and Ah. The average of A , A A , and 4 times A m , is g (A + Aa + 4 A TO ) = -y + -^ + c; and this average section multiplied by /i is the total volume : ,+ A,+4A m _ o^ 3 ta 2 6 Xn ~ 3 + 2 Js=0 This is the Prismoid Formula, so called because it holds not only for every solid whose volume is given in elementary geometry but for any prismoid, that is, for a solid with any end sections whatever, with sides formed by straight lines joining points of one end section with points of the other end section. 159. Application of the Prismoid Formula. — The for- mula holds even for many solids that are not prismoids, for example, spheres and paraboloids. It holds for all solids defined by equation (1), Art. 158, and even for all cases where A 8 is any cubic function of s: A 8 = as* + bs 2 + cs + d. (1) When / (x) is a quadratic or a cubic function of x; then, in general, JT|V 0») dx = [/(a) + 4/(^±- 6 ) +/(6)]^> (2) in accordance with the prismoid formula. The practical application of the formula is mainly for the close approxi- mation it gives to the volume of objects in nature; for any APPLICATION OF THE PRISMOID FORMULA 285 elevation or irregularity of the crust of the earth can be approximated to quite closely, either by the frustum of a cone, sphere, cylinder, pyramid, paraboloid, wedge, or prism; and as the formula holds for these solids as well as for any combination of them, it can be applied without determining which of the solids actually approximates most nearly to the object whose volume is desired. While it is thus used to approximate to the volumes of irregular solids, it is to be remembered that it gives the exact volume, when the area of a section A s is either a quadratic or a cubic function of s, including of course a linear function as a special case of the quadratic or cubic function. Example 1. — In the case of the cone or pyramid, Art. 155, x 2 it is seen that A x = Ah u is & quadratic function of x, and hence V = ^(o + A h + 4^/i = ±A h h. Example 2. — In the case of the sphere, n% — 1* 2 A X = A { = T (a 2 — x 2 ), hence, V = J (0 + + 4 A ) 2 a = f A a = A m = ira 2 . Example 3. — In the case of the paraboloid of revolu- tion, about the axis OY, of the curve y = x 2 ; 7=i(0 + A, + 4A m )/i Here A 8 = iry is a linear function of the distance y, for by (1), Art. 158, a = 0, b = 7r, c = 0; hence the formula holds. |7ra 3 , where A 286 INTEGRAL CALCULUS Example 4. — The prismoid shown in figure is composed of a prism, a wedge, and two pyramids. Let A be the smaller end section, A h the larger, and A m the mid section. V = A h = g (A + A + 4 A ), for prism, F - A' h \ - J(o + A; + 4^), for wedge, V = A); I = 1(0 + A;' + ii*) , for pyramid. The formula is seen to hold for the three forms of solids composing the prismoid. As a practical case, let the figure represent a section of a railway embankment 100 feet in length. A fill of 10 ft. with side slopes lj to 1, makes A = 250 sq. ft. A fill of 20 ft. with side slopes 1^ to 1, makes A h = 800 sq. ft. A fill of 15 ft. with side slopes li to 1, makes 4 A m = 1950 sq. ft. Hence V= H* (250 + 800 + 1950) = 50,000 cu. ft. APPLICATION OF THE PRISMOID FORMULA 287 Here V = h/2 (A + A h ) = H - (250 + 800) = 52,500 cu. ft., by average end areas. V = luAm = 100 X 487.5 = 48,750 cu. ft., by mean area. It is seen that the error of the approximation by the average end areas is twice that by mean area and of opposite sign. Since the errors vary as the square of the difference in dimen- sions of the two end areas, when the end areas are very different, the true prismoid formula should be used, but when the end areas are alike, or nearly so, the approximate formulas may give results as nearly exact as may be desired. EXERCISE XXXII. 1. Get the volume of a frustum of a solid included between the planes s = and s = h, when the area A 8 of a parallel cross section is a cubic function, as 3 + bs 2 + cs + d, of the distance s from one of the bounding planes; first by direct integration using the frustum formula, then by the prismoid formula. Thus prove the statement at the beginning of Art. 159. 2. Show according to (2) Art. 159, as in the case of volumes, that the area under any curve y = f (x), where / (x) is any quadratic or cubic function of x, between x — a and x = b, is — g— (Va + ?/6 + 4 y m ), (1) where y a , yb, ym represent the values of y at x = a, x = b, and x = §(« + &). 3. Find, first by direct integration, and then by (1) of Ex. 2, the areas under each of the following curves. (a) y = x 2 , between x = and x = 2. (6) y = x 2 + 2 x + 3, between x = 1 and x = 5. 4. Show that, when (1) of Ex. 2 is used to get area under the curve y = x* between x = 1 and x = 3, the error is about 4.2 per cent. 6. Find the volume made by revolving the area between the curve y = x 2 and the z-axis about the z-axis, between x = and x = 2. See Ex. 3, Art. 159. Find first by (1) of Art. 153; then by the prismoid formula show that the result by that formula is in error about 4.2 per cent. Note. — The prismoid formula is not applicable for exact results, when A 8 is given by a higher function than a cubic; in that case, it and the general, formula (2), Art. 159, for / (x), give approximations. 288 INTEGRAL CALCULUS 160. Surfaces and Solids of Revolution. — To get an expression for the area of a surface made by the revolution of a curve y = f (x) //| s An about the axis OX, let A (xo, y ) be a fixed point and P (x, y) a va- riable point on the curve OP Q P. Let P P = s, and PP' = As, and let PD and P'R be drawn each parallel to OX and equal in length to As. Let S denote the surface generated by the revolution of P P about the z-axis; then A$ equals the surface generated by PP'. It is evident that surface PD < A£ < surface P'R) that is, 2 iry As < ^S < 2 w {y + ^y) As; A*S dividing by As, 2 iry < -^ < 2 t (y + Ay) ; = -T- = 2tt?/, since A?/ = 0, as As = 0; /. dS = 2 iry ds or 5 = 2 tt / S ?/ ds. (See (3) Art, 157.) (1) Jo Here dS = 2iry ds may be represented by the lateral surface of a cylinder MPT', the circumference of whose base is 2 7r?/ and whose length is PT' ', drawn parallel to OX and equal to PT, which represents ds along the tangent at P. This is so, for this surface is what the change of S would' he, if at P the change became uniform, ds being the uniform change of s as x increases uniformly from that point. The surface S may be considered as generated by the circumference of a circle of varying radius y and hence the point P moving on the curve according to the law expressed by its equation y=f(x). Since hence, limit , As=0 L^S TA/S SURFACES AND SOLIDS OF REVOLUTION 289 ds = (dx> + A 7 > M'MP; that is, 7T (y + Ay) 2 Ax > A V > iry 2 Ax ; AV dividing by Ax, t (y + Ay) 2 > -r— > wy 2 ; hence, lim -r— = -=- = iry 2 , since Ay = 0, as Ax = 0; .*. dV = iry 2 dx or V = w I y 2 dx. (6) If the revolution is made about a line y = b, then V = rr f X (y-b) 2 dx, (8) and when the revolution is about a line x = a, then (x-a) 2 dy. (9) V =7T f y Note. — It may be noted that the cone and the sphere of Art. 155 and the paraboloids of Art. 156 and Art. 159 are all solids of revolution, and hence the formulas of this Art. 160 are applicable to the determination of their surfaces and volumes. SURFACES AND SOLIDS OF REVOLUTION 291 Example 1. — Find the volume generated by the revolu- tion of the area of the equilateral hyperbola xy = 1 about OX. V = it I y 2 dx = ir I — 9 dx=—7r - =tt ; ko=o = oo; hence, the entire volume has no limit. r 1 ll Xo=1 y = 7r =7r cubic units; L^o xj x=a0 1M X hence the limit of the volume, from the section at x = OM = 1, extending indefinitely to the right, is the same as the volume of the cylinder generated by the revolution of NP , the abscissa of P , about OX. Thus, while the area under the curve y = 1/x, from the ordinate MqP at x = 1, in- definitely to the right, is unlimited (as shown in Ex. 11, Art. 135), the volume made by its revolution about OX has a definite limit. According to Art. 156, if the curve y"=ir/x 292 INTEGRAL CALCULUS is drawn, any one of its ordinates in linear units will represent the volume of the solid extending indefinitely to the right of that ordinate; thus, in the figure the ordinate M Po" = tt represents the volume to the right of P MoP ', and the ordinate MiPi" = J w, the volume to the right of PiMiPi. In general, the ordinate MP" at x = OM represents the volume of the solid to the right of the section at any distance x from the origin, and it represents also the area under the curve y = tt/x 2 to the right of the ordinate to that curve. Example 2. — Find the volume to the left of the ?/-axis of the solid generated by the revolution of the exponential curve y = e x about the z-axis. Jr*y=i fx=0 7T 1° 7T y 2 dx = tv j e 2x dx= -e 2x \ = » cubic units. (See Ex. 2, Art. 130, for figure.) EXERCISE XXXIH. In these examples, a segment of a solid of revolution means the portion included between two planes perpendicular to its axis, the solid or its segment being, in general, & frustum; and a zone means the convex surface of a segment. 1. Find the area of a zone of the paraboloid of revolution about the x-axis. y 2 = 2 px, the plane curve. Ans. 5— [(p 2 +yrf — (p 2 +2A> 2 )*]. 6 p See Ex. 2, Art. 157, where p = f , y = 0. 2. Find the area of a zone of the ellipsoid of revolution about the x-axis; that is, a zone of the prolate spheroid. Get entire surface. 6 2 y 2 = — (a 2 — x 2 ) = (1 — e 2 ) (a 2 — x 2 ), where e is the eccentricity. y ds = - v a 2 — e 2 x 2 dx. a :. S = 2tt- C Va 2 - e 2 x 2 dx a Jx h r /—■> n ■ « 2 • ,^1 x = ir - \ x v a 2 — e 2 x 2 -\ sin -1 — a\_ e a J T The entire surface = 2irb [b + (a/e) sin -1 e]. SURFACES AND SOLIDS OF REVOLUTION 293 The surface of a sphere = limit 2 irb [6+ (a/e) sin -1 e] = 2ra[a +a] = 4 ira 2 , since for circle e = 0, limit — — - = lim - — = 1 ; a = b. e=o L e J e=o LsinflJ 3. Find the area of the surface generated by the revolution of the cycloid about its base. Taking the parametric equations of the cycloid, x = a (0 — sin 0), y = a (1 — cos0); dx = a (1 — cos 0) dd, dy = asmddd; ds = Vdx 2 + dy 2 = a V2 (1 - cos 0) dd. 64 „ 3'° S = 2w (yds = 2wa 2 C*V2(l-cosd) 3 M = lQwa 2 Csm 3 (^d(^\ = 4. Find the surface generated by revolving the catenary about the y-axis, from x = to x = a. Also about the rr-axis. Here y = %\ Jo 8. Find the volume of a segment of the prolate spheroid, and the entire volume. Find the latter to be two-thirds the volume of the cir- cumscribed cylinder of revolution. Am. ~ \a 2 (x - xo) -■£ (x z - xo 3 )l- 9. Find the volume of the oblate spheroid, that is, the ellipsoid of revolution about the minor axis which is on the ?/-axis. Find the volume to be two-thirds of that of the circumscribed cyl- inder of revolution. 10. Find the volume of the paraboloid made by x 2 = 2 py about the ?/-axis. (Compare Ex. 3, Art. 159.) Find the volume to be one-half that of the circumscribed cylinder of revolution. 11. Find that the volume of the solid generated by revolving an arch of the cycloid about its base is five-eighths of the circumscribed cylinder. Here V = 2* f *' , V * dy or V = to* C* (1 - cosd) 3 dd. J o V2ay — y 2 J * 12. Find the volume generated by the catenary revolving about the x-axis, from x = a to x = —a. Also find the volume by the area with the same arc revolving about the ?/-axis. Here V = * j_Jj \e" + e ~ a ) dx = 51 \°-e« + 2x - |e a J = x (c2 + 4 ~ e_2) = 8 - 83 a '"'- And V = tt f ° x 2 dy = £ C x 2 \f - e~ a ) dx. J & J SURFACES AND SOLIDS OF REVOLUTION 295 Integrating by parts gives V =\ \ax 2 \e a +e a ) -2a 2 x\e a - e a ) + 2 a 2 \e* + e~ A = ^ (e + 5 e" 1 - 4) = 0.878 a 3 . 13. Find the volume of the solid generated by the revolution of the tract rix about the x-axis. y 2 dx = -ir I Va 2 - ifydy = —■ Ja 6 14. Find the volume generated by the revolution of the hypocycloid about the x-axis. V =tt ( y-dx =t C (a 1 - x l )*dx = ^tto 3 . J J -a lOo 15. Find the volume generated by the revolution about the ?/-axis of the equilateral hyperbola xy = 1, from x = to x = 1. V = ir | x*dy — x I ~ = — - =7r cubic units. (Compare Ex. 1, Art. 160.) 16. Find the volume of the segment of the solid generated by the revolution of the equilateral hyperbola x°- — y 2 = a 2 about the z-axis, the altitude of the segment being a, measured from the vertex. Ans. jira z . 17. Find the volume generated by revolving about either axis the part of the parabola x 5 + \p = a? intercepted by the axes. Ans. T ; jra 3 . 18. Find the volume of the solid generated by the quadrant of a circle revolved about a tangent at one extremity. V = 7rf\a - x) 2 dy =Trf\a- V^=7)' dy = ^ (| - |) 19. Find the volume generated by the revolution of the cissoid x 3 y 2 = - — — — about the x-axis, from the origin to x = a. Ans. |7ra 3 (31og2 -2). 20. Find the volume generated by the revolution of the cissoid about its asymptote x = 2 a. Ans. 2jr 2 a 3 . CHAPTER V. SUCCESSIVE INTEGRATION. MULTIPLE INTEGRALS. SURFACES AND VOLUMES. 161. Successive Integration. — As the inverse of succes- sive differentiation there is successive integration. If a start is made with a function y = f (x), considered as an nth derived function, a single integration gives another function, the integral; the integration of this function gives a second integral, and so on. The result of n integrations is the nth integral of the given function. In Art. 140 on Integral Curves successive integration was indicated, and in Art. 141 the process was employed in application to beams. For successive integration with respect to a single independent variable, in general; let fi(x)=ff(x)dx, (1) f*(x)=ffi(x)dx, (2) h (x) = jh Or) dx. (3) Since f 2 (x) = / [fi(x)]dx, it follows from (1) that ft(x)=f[ff(x)dx]dx; (4) and since / 3 (x) = I [f 2 (x) ] dx, 296 SUCCESSIVE INTEGRATION 297 it follows from (4) that fs W = / 1 / [f f {X) dX \ dX \ dX ' (5) The integral in (4) is called a double integral and is written ff> Similarly, the integral in (5) is called a triple integral and is written /// f{x) dx\ If an integral is evaluated by two or more successive in- tegrations, it is called a multiple integral. For example, to evaluate the multiple integral / / / e x dx?; e x + Ci is the first integral, e x + Cix + C 2 is the second integral, C x 2 e x + —jr- + Cix + C 3 is the third integral. iff ix* = e x + ^f + C 2 x + C 3 . If limits are given for each successive integration, the integral is definite; if limits are not given, it is indefinite. d 2 s Example 1. — Given the acceleration -r^ = — g to find s. This is Ex. 5 of Art. 115, and may be written thus: s = I ( — gt+vo) dt, where v is the constant of integration, s = — \ gt 2 + vat -\- So, where s Q is the constant of integration. Example 2. — Determine the curve for every point of which the rate of change of the slope is 2. 298 INTEGRAL CALCULUS Here ^y = A.(^y) = ^ = 2 - dx 2 dx \dx) dx ' .'. y=Jj2dx\ y = I (2x + Ci) dx, where 2 x + d is the first integral, y = x 2 + CiX + d, the second integral. This is the equation of any parabola that has its axis parallel to the y-axis and drawn upwards, and its latus rectum equal to 1 . All such parabolas may be gotten by giving all possible values to C\ and Ci, the arbitrary constants of integration. d 2 v Example 3. — Determine the locus of the equation -~ = 0. y=ffodx 2 , y = I mdx, where m is the constant of integration, y = mx + b, where b is the constant of integration. The locus is the system of straight lines, the arbitrary con- stants m and b representing the slope and ?/-intercept, respectively. Example 4. — In the theory of flexure of beams where E, I, M, R, and w are constants. Get an expression for y and determine the constants of integration from the con- ditions, y = when x = 0, and y = when x = I. »~TlJ[ Mx+ -2---6- + Cl ) dx ' 1 [Mx 2 . Rx* WX 4 , „ t „~\ n . n n SUCCESSIVE INTEGRATION 299 n 1 [Ml 2 , RP id* , r 7 1 . . = El [~Y + "6" " 24 + U \ ' when X = Z; ■"' ' £/|_ 2 6 + 24_T J_ \Mtf TW _wtf _m _IW wF\ y £/[ 2 + 6 24 2 6 + 24 J r 2 r 3 n Example o. — Evaluate III x 3 dx 3 . Jo J I J 2 Letting I denote the integral and making the integrations in order from right to left; t P PM 4 ^ 2 «n P Pa 2 ' = 60 r [YP dx = 120 pdx = 240. Example 6. — A point has an acceleration expressed by the equation a t = — roo 2 sin cot, where r and a; are constants. Get expressions for the velocity and the distance or space passed over. Here a t = -7-7 = — roo 2 sin cot and s = I I — rco 2 sin cot dt 2 , ds Cd 2 s u .f. .j. +- ..„ . . v = -j = I -=■=■ dt = — roo 2 J sin oot dt = roo cos cot + Ci, s = / -r dt = roo I cos co£ eft -f / Ci dt, s = r sin oot + &£ + C 2 , which is the law of simple harmonic motion. (See Art. 73.) EXERCISE XXXIV. 1. Evaluate f jf (x 2 - 1) dx 3 . Ans. £ - | + ^ + C 2 x + C* 2. Evaluate J J f j ^ dx *- Ans. - 1 log x + J x 3 + * C>x 2 + C 3 x> C 4 . 300 INTEGRAL CALCULUS 3; Evaluate \ j ( ( sinaxdx 4 . Ans. 1/a 4 sin ax + £ Gx 3 + \ C 2 x 2 + C s x + C t . 4. Evaluate f f f x*dx*. Ans. 16. Ji J\ Jo 6. Evaluate f C ( x 3 dxK Ans. 80. 6. Find the curve at each of whose points the rate of change of the slope is four times the abscissa, and which passes through the origin and the point (2, 4). Ans. 3 y = 2 x (x 2 - 1). 7. Evaluate f 2 f C r sind d&. Ans. t(0- a). Jo J a Jo d?s 8. The differential equation of falling bodies is -r=r = — g] show that s = - ~ + Ci< + C 2 ; and find G and C 2 , if s = and v = 100, when t = 0. 9. A point has an acceleration expressed by the equation at = — rco 2 cos cot, where r and co are constants. Find expressions for the velocity and the distance passed over. Find G and C 2 , if s = r and v = 0, when t = 0. Ans. v = —rco sin ut + G; s = r cos coi + G£ + Cs 162. Successive Integration with Respect to Two or More Independent Variables. — In the preceding Article successive integration was of functions with respect to a single independent variable. Successive integration of functions of several independent variables are now to be considered. Suppose there is given a function / (x, y, z) of three independent variables. Let /, (x, y, z) = J f (x, y, z) dz, (1) h fo V, z) = J /i 0, V, z) dy, (2) /s fa V, z) = J h fa V, z) dx, (3) where in (1) the integration is with respect to z, that is, as if x and y were constants. Likewise in (2) it is with respect THE CONSTANT OF INTEGRATION 301 to y, as if x and z were constants, and in (3) with respect to x, as if y and z were constants. Equation (2), by substitution from (1), becomes h fo y,z)=j \Jf fe y, «) (x), or it may be simply C. This is so, since differentiating either 2 xy 2 + C, or 2 xy 2 + (x), with respect to y gives the same result, 4 xy. Hence, g = 2a^ + *(s), where (x) is an arbitrary function of x and may be a con- stant C. Integrating this result, with y constant, gives u = xhf + f (x) dx + F (y), (2) where, since y was regarded as constant during the integra- tion, the integration constant is an arbitrary function of y and may be C with a constant value, possibly zero. THE CONSTANT OF INTEGRATION 303 By referring to Art. 109, (2), it will be seen that if u = x 3 + x 2 y 2 ; -r—j-dxdy = 4:xydxdy; that is, 4>(x) =3 x 2 and F (y) = 0, for that function u of 0, y). The indefiniteness of the result in (2) is manifest, for u = I I \xydxdy = x 2 y 2 , if both constants of integration are zero, that is, <£ (x) =0 and F {y) = 0. The indefiniteness is removed when limits for the variables are given, the integral being then a definite integral. Example. — I I xyz dxdydz= I I xydxdy\-\ a Jo Jy Ja t/0 |_~ Jv EXERCISE XXXV. Evaluate the following integrals : 1. ff#V -V). ^2bcosd 6 Xb fy 5 3 — O? I p 2 sin Bdpdd = — ^— (cos /3 — cos 7). ^* J J ^~* 2 ^ds = 6& 3 . J«3 /»2 /»5 I J xy 2 dxdydz = 17£. 2 •'1 •'2 /»3 /»2 fh 10. J I I x?/ 2 dz dy dx = 2\\. 11. J* J* f*xy*dzdy dx = 17%. 12. f f f * x>y*z dx dydz = \ a?V{b* - a 3 ). J a Jo Jb J~2 r x r x+y e 8 — 3 3 e* ./o »/o o 4 14. f b f h V2jy dxdy = l ^¥g (h 2 * - hj) b. Jo J hi 4* r 1 * r a JrtJ a 2 (8-7r) 15. I I pdddp = — ^ -■ Jo J a(l-cos0) o 16. f f ° (w + 2v)dvdw = W ^. I I x 2 2/2 dx dy dz = 32 a 7 . Jo Jly 164. Plane Areas by Double Integration — Rectangular coordinates. — It has been shown in Art. 135, that the area between two curves y = f (x) and y = F (x) is given by A= r\f(x)-F(x))dx, (1) t/x where the points of intersection are (x , y ) and (x h y\). The area is thus given not by a single integral but by the differ- ence between two integrals, / f(x)dx — I F (x) dx. The «A «/x result is gotten also by double integration, finding the limit of two sums. Let the element of area be ly lx, (x, y) being any point P of the area. If the elements are summed PLANE AREAS BY DOUBLE INTEGRATION 305 up with respect to y, with the limits MD and MN, or F(x) and / (x), x being constant, the area of the strip DN' is gotten. If the strips are summed up with the limits a and b for x, then V %±x\Ay=2 %AxAy x=a\_F(x) J a F(x) is the expression for the sums. Taking the limits of the sums, first as Ay = and then as Ax = 0, the area ADBN is given by the double integral nf{x) dxdy, (2) (x) which integrated first with respect to y gives A= £u(x)-F(x))dx. (1) If the elements are summed up in reverse order, first with respect to x with the limits H'H and H'S, or /- 1 (y) and F^iy), y being constant, and then with respect to y with limits c and d, there results dydx, (3) . Hv) 306 INTEGRAL CALCULUS where/ -1 (y) and F _1 (y) are the inverse functions of f(x) and F (x), respectively. Integrating (3) the area ADBN is gotten, as given by (2). Hence, in general, A= f fdx dy (4) is the formula for area by double integration, the limits being taken so as to include the required area. The order of integration is indifferent provided the limits be adapted to the order taken. Corollary. — d xy 2 A = dx dy and d yx 2 A = dy dx. Example 1. — Find the area bounded by the parabolas y 2 = 2 px and x 2 = 2 py. The parabolas intersect at the points (0, 0) and (2 p, 2 p). J^2 P rVjpl I dxdy = f p 2 , by formula (2) . J 12 f dy dx = f p 2 , by formula (3). ■2p 2py Example 2. — Find the area bounded by the circle x 2 + y 2 = 12, the parabola y 2 = 4 x, and the parabola x 2 = 4 y. For the part OPP 2 the limits for x are and 2, while for the part PP1P2, they are 2 and Vs, the point P being (2, Vs) and the point P x (Vs, 2). For both parts the lower limit for y is the ordinate of x 2 = 4 y; for OPP2 the upper limit for y is the ordinate of y 2 = 4 x, and for PPiP 2) that of x 2 + y 2 = 12. A = OPPiP* = OPP 2 + PPiP-2 = ( 2 [''dxdy Jo «7/2 4 /»\8 /»v'i2-j2 + / / dxdy = 3.92. «/2 t/j* PLANE AREAS BY DOUBLE INTEGRATION 307 Example 3. — Find the area between the parabola y 2 = ax and the circle y 2 = 2 ax — x 2 . nv2az-z 2 dxdy 'ax Jo Z 6 Note. — It may be seen that in finding some areas there is no advantage in using double integration, as after the first integration with the limits substituted, the remaining in- tegral is what might have been formed at first. There are, however, cases where double integration furnishes the only method of solution; hence the need for some practice in its application. EXERCISE XXXVI. 1. Find the area between the circle x 2 + y 2 = a 2 and the line y A- f f^dzdy-'-^*. •'o Ja—x 4 2. Find by double integration the area between the parabolas y 2 = 8 x and x 2 = 8 y. Ans. 21§. 3. Find the area bounded by the circle x 2 + y 2 = 25, the parabola y* = -L 6 x, and the parabola y = t& x 2 . Ans. 7.55. 4. Find by double integration the area of the x 2 v 2 ellipse - 2 + £- 2 = 1. 5. Find the area of any right triangle, using double integration. a C h fk j j C ha j a x2 l b i ah. -x+a A - /*/ dxdy = £(- %z + a)dx = -£ f + ax] * = \ ab. 308 INTEGRAL CALCULUS 165. Plane Areas by Double Integration — Polar coor- dinates. — As has been shown in Art. 135(b), the area in polar coordinates of P\OP 2 , generated by the radius vector p as increases from 0i to 2 is given by -*£■ de. (i) To find the area between two polar curves by double integration, let the element of area be PDD'P', bound- ed by the two radii OP', OD', and the two circular arcs, concentric at 0. Let the coordinates of P be (p, 0) ; then from geometry, sector POD = \ p 2 A0, sector P'OD' = \ (p + Ap) 2 A0. Hence, A A = PDD'P' = i (p + Ap) 2 AS]- J p 2 A0 = (p + iA P )A0Ap. Keeping A0 constant and summing the elements of area with respect to p gives an area AA'B'B, expressed by OA r* QA A0 • lim y. (p + | Ap) Ap = A0 / pdp. Ap=0 oa> Jo A' Making the summation now with respect to 0, the sum of the radial slices is gotten, and the limit of this sum is _.0 2 roA re 2 roA A = lim V A0- / P dp= I I pdpdd. A0=O ^0 y J AO' Jo, JOA' Replacing OA' and OA by F(d) and / (0) respectively, the formula is £".. rue) I pdddp. - (2) . J F (0) When F (0) = 0, the area PiOP 2 between the curve p = / (0) PLANE AREAS BY DOUBLE INTEGRATION 309 and the radii is (1), A = \ p 2 dd, where (2) has been in- tegrated as to p. If the summing of the elements of area be made first with respect to 0, keeping Ap constant, RSDP, a segment of a circular ring, is gotten. 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 J Pl t/F-i(p) dpdd, (3) where F _1 (p) and/ _1 (p) are the inverse functions of F(0) and / (0), respectively. Corollary. — d dp 2 A = pdddp and d p0 2 A = pdpdd are rec- tangles with sides p dd and dp. Example 1. — A simple case of the application of the formulas is in finding the area of the circle p = a. (2) (3) Jr»2ir Pa P27T ~~la / pdOdp = \ / p'ddl t/0 «/o Jo = ia 2 ^T 7r = 7ra 2 n2ir Pa "]2tt pdpdd = / pdpdl Jo Jo = 27r^T = 7ra* 310 INTEGRAL CALCULUS In (2) the sectors are summed, while in (3) the rings are summed. In this case of the circle it is to be noted that no limit need be invoked, since the integral is the sum in each case, the in- crements being the differ- entials, the variables all * increasing uniformly. Example 2. — Find the area between the two tan- gent circles p = 2 a cos and p = 2 b cos 0, where a > b. TT 7T JP* /*2acos0 /»* / pdddp = 4t(a 2 -b 2 ) I cos 2 6 */2bcos0 *Jo dd = ir{a 2 -b 2 ). Example 3. — Find the areas between the cardioid p 2 a (1 — cos 0) and the circle p = 2 a. pdOdp _a(l-cos0) = 4 a 2 /[I - (1 - cos0) 2 ]c?0 7T = 4a 2 J (2cos0-cos 2 0)d0 = 8a 2 -7ra 2 . AREA OF ANY SURFACE BY DOUBLE INTEGRATION 311 J r V /»2«(l-cos0) / pdddp w nJ2a 2 = 4a 2 P[(l -cos0) 2 - l]dd \ «=4a 2 J(-2cos6 + cos 2 e)dd = 8a 2 +Tra 2 . I EXERCISE XXXVH. 1. Find by double integration the entire area of the cardioid p = 2a(l— cos0). Ans. 6?ra 2 . 2. Find the area (1) between the first and the second spire of the spiral of Archimedes p = ad; (2) between any two consecutive spires; (3) the area described by the radius vector in one revolution from = 0, and the area added by the nth revolution. Ans. (1) *#■ Tr*a 2 ; (2) (n 2 + 2n+ f) tH^ 2 ; (3) f 7r 3 a 2 , f (n 3 - 1) tt 3 ^. 3. Find by double integration the area of one loop of the lemniscate p 2 = a 2 cos 2d. Ans. \ a 2 . 4. Find by double integration the area between the circle p = cos and one loop of the lemniscate p 2 = cos 2 6. Get the area between the circle and the line 6 «= tt/4 and then between that line, the lemniscate, and the circle. . ir — 2 -k — 2.x — 2 166. Area of any Surface by Double Integration. — Let the surface be given by an equation between the rectangular coordinates, x, y, z. Let the equation of the given surface be z =f(*,y)- Passing two series of planes parallel, respectively, to XZ and YZ, will divide the given surface into elements. These planes will at the same time divide the plane XY into ele- mentary rectangles, one of which is P'Pi', the projection upon the plane XY of the corresponding element of the surface PP 2 . Let x, y, z be the coordinates of P and x + Ax, y + Ay, z + Az, those of P 2 , x and y being independent; then P'M' = Ax and P'N' = Ay. The planes which cut the element PP 2 312 INTEGRAL CALCULUS from the surface will cut a parallelogram from the tangent plane at P, the projection of which on the plane XY is P'P 2 ' = Ax Ay, the same as the projection of the element PP 2 . The projection is the product of the area of the parallelogram and the cosine of the angle made by the tangent plane with the plane XY; hence, denoting the angle by 7 and the parallelogram cut from the tangent plane by PT, area PT = area P'P 2 ' • sec 7 = Ax Ay sec 7. As Ax and Ay approach zero, the point P 2 approaches the point P, and the areas PT and PP 2 approach equality; that is, the element of surface approaches coincidence with the parallelogram, a portion of the tangent plane at P; hence, area PP 2 = A xy 2 S = Ax Ay sec 7, approximately; that is, area PP 2 = A xy 2 S = Ax Ay sec 7 ; Um ^^ \ = sec 7 ; ( 21 j .*. d xu 2 S = sec 7 • dx dy. (1) AREA OF ANY SURFACE BY DOUBLE INTEGRATION 313 F^A rt .103(8) >8 ec,4l + (|)V(|)^( o ^^) hence from (1), //[■+©'+(£)' dx dy, (2) the limits being so taken as to include the desired surface. Let £ denote that part of the surface z = f(x, y), z being a one-valued function, which is included by the cylindrical surfaces y = O (x), y = 4> (x), and the planes x = a, y = b; - s -f •CHINS)"]'** <*> In finding the area of the given surface a more convenient form of the equation of the surface may be either x = f (y, z), or y = / (z, x). The formula for the area will be then either //['+(!)*+< dx" \dzj dydz, (3) or //['+@H2)7^ with the proper limits of integration. In applying the formulas, the values of the partial deriva- tives are gotten from the equation of the surface the area of which is sought; hence, when there are two surfaces each of which intercepts a portion of the other, the partial derivatives in each case are taken from the equation of that surface whose partial area is being sought. This will be illustrated in the following examples. Example 1. — To find the surface of the sphere whose equation is x 2 + y 2 + z 2 = a 2 . 314 INTEGRAL CALCULUS Let - ABC of the figure (Art. 166) be one-eighth of the sphere. dx~ z 1 dy~ z* i-iVfeY.i f*Y-i i * 2 i y 2 = a2 = a * 1 "*" \dx) ~*~\dy) l ' r #~ r # z 2 a 2 - x 2 - y 2 Area = 8 = 8 ABC = 8a P f * , ^^ by (2) Jo Jo Va 2 -z 2 -?/ 2 = 8 a / "sin- 1 , y ds Jo Va 2 - x 2 Jo a C a a a 2 / Compare Ex. 2, \ = 4:ira I dx = 4ira 2 . ^ . v ' Jo V Exercise XXXIII. / Here the integration was over the region OAB, the projec- tion of the curved surface ABC on IF plane. The first integration with respect to y summed all the elements in a strip LL'K'K, y varying from zero to NL', that is, between limits and Va 2 — x 2 , the equation of the intersection of the surface with the XY plane being x 2 + y 2 = a 2 . Integrating next with respect to x, the surface ABC is gotten by sum- ming all the strips from x = to x = a. Example 2. — Find the area of the portion of the surface of a sphere which is intercepted by a right cylinder, one of whose edges passes through the center of the sphere, and the radius of whose base is half that of the sphere. Note. — This is the celebrated Florentine enigma, pro- posed by Vincent Viviani as a challenge to the mathemati- cians of his time. (Williamson's Integral Calculus.) Taking the origin at the center of the sphere, an element of the cylinder for the 2-axis and a diameter of a right section of the cylinder for the x-axis, the equation of the sphere will be x 2 + y 2 -f- z 2 = a 2 , and the equation of the cylinder, x 2 _|_ y2 _ ax AREA OF ANY SURFACE BY DOUBLE INTEGRATION' 315 The area of APCD is one-fourth of the area sought, and since this surface is a portion of the surface of the sphere, the par- tial derivatives -7- , -j- must be taken from x 2 + y 2 + z 2 = a 2 , ax dy v ' giving, as in Ex. 1, using formula (2), S= f f adxdy J J Va 2 — x 2 — y 2 to be integrated over the region OP' A. Hence, The limits for y are from x 2 + y 2 = ax, the equation of the curve OP' A, the boundary of the projection of the surface APCD on the XY plane. Example 3. — Find the surface of the cylinder of Ex. 2, intercepted by the sphere. The area of APCOP' is one-fourth of the area sought, and since it is a part of the lateral surface of the cylinder x 2 + y 2 = ax, the partial derivatives in formula (2) must be 316 INTEGRAL CALCULUS dz taken from this equation. But from this equation ^ = oo , — = oo, and formula (2) does not apply, which is, more- dy over, evident since the element of surface is dx dz in the strip P'P, and the area of the surface A PC OP' cannot be found from its projection on the XY plane, for this projection is the arc AP'O. The projection is made on the XZ plane and formula (4) used. The partial derivatives are found to be dy = a-2x dy = Q dx y ' dz Since P is on the sphere, pp 2 = z 2 = a 2 - (x 2 + y 2 ) = a 2 - ax, since P is on the cylinder. Hence, dx dz rt C a ^a 2 — cix ~2« r f /"* =2«r Jo Jo V ax — x 2 Jo = 2a f a \/^dx = 4:a' Vc dx Here the integration is over the region 0AP"C, AP"C being the projection of A PC on XZ plane. The first integration sums up the elements of surface in the strip P'P and the next integration sums up the strips from x = to x = a. By eliminating y from x 2 + y 2 + z 2 = a 2 and x 2 + y 2 = ax, z> = a 2 — ax (as found above), which is the equation of AP"C 1 from which the limits of z are taken. EXERCISE XXXVHI. Find by double integration the areas of the surfaces given in the following examples: 1. The zone of the sphere, x 2 -f- if + z- = r 2 , included between the planes x = a and x = b. A us. 2 wr (6 — o). VOLUMES BY TRIPLE INTEGRATION 317 2. The surface of the right cylinder x 2 -f z 2 = a 2 intercepted by the right cylinder x 2 + y 2 = a 2 . Compare Ex. 8, Art. 157. Ans. 8 a 2 . 3. The part of the plane - + t + - = 1, in the first octant, inter- a o c cepted by the coordinate planes. Ans. §V ' a 2 \f- -f a 2 c 2 -+- b 2 c 2 . 4. The surface of the cylinder x 2 + y 2 = a 2 , included between the plane z = mx and the XY plane. Find by both formula (3) and formula (4), and show why formula (2) does not apply. Ans. 4 ma 2 . 5. The surface of the paraboloid of revolution y 2 + z 2 = 4 ax, intercepted by the parabolic cylinder y 2 = ax and the plane i = 3a. 6. The surface of the cylinder of Ex. 5, intercepted by the parabo- loid of revolution and the given plane. I [ % r rfx d« = 2 V3 ( (4 ax + a 2 )^ dx = (13VT3-1)^. 167. Volumes by Triple Integration — Rectangular Co- ordinates. — Let the volume be that of a solid bounded by the coordinate planes and any surface given by an equation between the coordinates x, y, and z. Let P be any point (x, y, z) within the solid — ABC, the surface being given by z = f (x, y) , where / (x, y) is a con- tinuous function. Let K' be the point (x + Ax, y + \y, z + Az), the diagonally opposite corner of the rectangular parallelopiped formed by passing planes through P and K', the planes being parallel to the coordinate planes. Let more planes be passed. Taking first the sum of the elementary parallelopipeds whose edges lie along the line NT, the limit of this sum, as A z is made to approach zero, is the volume of the prism whose base is Ix Ay and whose altitude is NT, x, y, Ax, and Ay remaining constant during the summation. Next with x and Ax constant, sum the prisms between the planes MHL and SDR. The limit of this sum as Ay is made to approach zero is the volume of the cylindrical slice 318 INTEGRAL CALCULUS LR'D'HMS. Finally, when taking the sum of the slices parallel to the YZ plane, as Ax approaches zero, the volume of the cylindrical slice approaches that of the actual slice LRDHMS; hence, the limit of the sum of the slices, as Ax approaches zero, is the volume of the solid. Jr*OA PMH (*NT I I dx dy dz, o Jo Jo where V is the volume of — ABC. Let V denote the volume bounded by the curved surfaces z = f {x, y), z = f(x, y); the cylindrical surfaces y = o(x), y = 4>(x); and the planes x = a, x = b; then I dxdydz. (1) . j(j) Jfu(x,y) Corollary. — d xyz W = dxdydz, d y JV = dydzdx, . . . If z is expressed in terms of x and y, and f (x, y) = 0; a J,j) (x) z dx dy. VOLUMES BY TRIPLE INTEGRATION 319 Note. — The formula, V — I j I dx dy dz, may be de- rived from the figure by the definition of differentials. Thus, the variables x, y, z, being independent, dx, dy, dz may be taken as finite constants, the parallelopiped PK' being dx dy dz. When x and y are regarded constant, PK' is the differential of the prism NK. Hence, integrating dx dy dz between the limits z = and z = NT gives the prism NT dx dy, which is the differential of the solid MSR'L - T. Integrating NT dx dy between the limits y = and y = MH gives the cjdinder MLH — D' ', or MLH dx, which is the differ- ential of the solid OBC — M. Integrating MLH dx between the limits x = and x = OA gives the volume OBC — A, or V. Hence, T T = / l f dx dy dz, the limits being so chosen as to include the volume sought. Example. — Find the volume of the ellipsoid t + yl + t= ! a z o z c The entire volume is eight times that in the first octant, where the limits are: z = 0, z = cVl - x 2 /a 2 + y 2 /b 2 ; y = 0, y = bV\ - x 2 /a 2 ; x = 0, x = a; / dxdydz = Jo Corollary. — For sphere, a = b = c; .'. 1 3 4 7ra 3 / dy dx dz *Jo naVl_ 2/ c 2 />bVl_ X 2/ a 2_ 2 2/ c 2 / dz dx dy. Jo 320 INTEGRAL CALCULUS EXERCISE XXXIX. Find by triple integration the volumes required: 1. The tetrahedron bounded by the coordinate planes and by the plane a b c o See Ex. 3, Exercise XXXVIII, for the surface of the plane. 2. The volume bounded by the cylinder x 2 + y 2 = a 2 and the planes . 4 ma 3 z = and z = mx. Ans. — — 3. A cylindrical vessel with a height of 12 inches and a base diam- eter of 8 inches is tipped and the contained liquid is poured out until the surface of the remaining liquid coincides with a diameter of the base. Find the volume remaining in the vessel. Ans. 128 cu. in. Note that the volume is one-half that given by Ex. 2, above. 4. The volume included between the paraboloid of revolution yi + z 2 = 4 ax, the parabolic cylinder y 2 = ax and the plane x = 3 a. See Exs. 5 and 6, Exercise XXXVIII, for the surfaces. J-3a /"Vax /»(4ox-y 2 )^ , ,- N ( dxdydz = {Gir + 9V3)a 3 . o J o «A) 5. The volume included between the paraboloid of revolution x t -\- y i = az, the cylinder x 2 + y 2 = 2 ax, and the XY plane. . x*+& Xia pV^ax-x* r a I I dxdydz = f ira 3 . 6. The entire volume bounded by the surface Ans - V =Ll I dxdydz - w 7. The entire volume bounded by the surface ? i i . % i a 4 ira 3 x s + ?/ J +z s = a ! . Ans. -qH - * 8. The volume of the part of the cylinder intercepted by the sphere. The radius of the sphere is a and it has its center on the surface of a right cylinder, the radius of whose base is a/2. See Exs. 2 and 3, Art. 166. J -a r \/ax-x* /•Va»— »«— «l I I dxdydz = i(ir- f)a». SOLIDS OF REVOLUTION BY DOUBLE INTEGRATION 321 168. Solids of Revolution by Double Integration. — In the figure of Art. 164, where P(x, y) is any point in the area ADBN, x and y being independent, Ax Ay is the element of area. Conceive the area ADBN to revolve through radians about OX as an axis; then By ■ Ax ly < A xy W < {y + Ly) ■ Ax Ay; ■■ ey< tS- y hence," hm \ \= - r - j- = dy; A X ,Ay=olAxAyj dxdy .'. d 2 V = By dxdy, / ydxdy. (1) Similarly, about OY, Jrvi rrny) I x dy dx. (2) 2/0 JF-Hy) Putting 6 = 2 7r, the formulas give the volumes generated by a complete revolution of the area. Corollary. — If the axis of revolution cuts the area, (1) or (2) will give the difference between the volumes generated by the two parts. Hence 7 = 0, when these two parts generate equal volumes. Integrating (1) first with respect to y, and (2) first with respect to x, the upper limits being the variables y or x and the lower limits zero, gives V = ir f Xl y 2 dx (l r ) and V = tt f m x 2 dy, (2') the formulas for solids of revolution, single integration. 169. Volumes by Triple Integration — Polar Coordinates. — Let the point P (p, 6, ) be any point within a por- tion of a solid bounded by a surface and the rectangular planes. As usual, p is the distance OP from the pole at the origin, 6 is the angle ZOP which OP makes with the z-axis, 322 INTEGRAL CALCULUS and is the angle XOP' which the projection of OP on the XY plane makes with the x-axis. Let the solid be divided into elementary volumes like PDD1Q1 by the following means. (1) Through the 2-axis pass a series of consecutive planes, dividing the solid into wedge-shaped slices such as COAB. (2) Round the 2-axis describe a series of right cones with their vertices at 0, thus dividing each slice into elementary pyramids like - RSTV. (3) With as a center describe a series of consecutive spheres. Thus the solid is divided into elementary solids like PDDiQi, whose volume is given approximately by the product of three of its edges, PP lf PPi, and PQ. Let edge PQ = Ap, angle POP 2 = A0, angle AOB = angle PO'Pi = A(j>; then edge PP X = p sin A, and edge PP 2 =» P A0. Hence, the volume of the elementary solid is given ap- proximately by p 2 sin A0 A<£ Ap. It can be expressed VOLUMES BY TRIPLE INTEGRATION 323 exactly but the additional terms vanish when the three in- crements are made to approach zero. Therefore, the volume of the solid is given by the limit of 2X2 P 2 sin0A0A0A P ; A0=O A0=O Ap=0 ■SIS p 2 sin dd d dp, (1) each integral to be taken between the limits required to find the volume sought. The summation can be made in any order so long as the volume is continuous. For a solid of revolution with the 2-axis as the axis of revolution, the formula (1) for the volume becomes ■// P 2 smdd6dp, (2) since the limits for $ are then evidently and 2 w. The limits for p and are then the same as are used in getting the area of the plane figure revolved. Corollary. — d p e^V = p 2 sin dd d dp is an elementary rectangular parallelopiped; and d p6 2 V = 27rp 2 sin Odd dp is a circular ring with rectangular section. Example 1. — Find the volume of a sphere of radius a, using polar coordinates, pole at end of a diameter. By formula (1); or by (2), if the volume is considered as generated by revolving the semicircle about the initial line, the line from which is measured, I p 2 sin d dd dp t/0 x X r_ cos 4 o y L 4 Jo 167ra 3 r 3 . . - Aa -r — / cos 3 sin 0(20 O t/0 167ra 3 f war. 324 INTEGRAL CALCULUS Example 2. — Find the volume generated by revolving the cardioid, p = 2 a (1 — cos 6) about the initial line. JrV /*2a(l-cos0) / p 2 sin 6 dd dp o Jo 167ra ; j: (1 - cos 0) 3 sin Odd 64 ira 6 . Example 3. — Find the volume made by revolving the lemniscate p 2 = a 2 cos 2 6 about the initial line. *a v cos 2 V = 2tt I f p 2 sinddddp Jo Jo = ^- f T (cos20)*sin0d0 = ^- r(2cos 2 0-l)tsin<9 o Jo 6 Jo V 2 V2 6/ ctt 170. Volumes by Double Integration — Cylindrical Co- ordinates. — In finding the volume of some solids the integration is performed more readily with the use of cylin- drical coordinates. In this system of coordi- nates the position of a point is given by the cylindrical co- ordinates (r, , z), where (r, 0) are the polar coordinates of the projection (x, y, 0), on the XY plane, of the point (x, y, z). It is evident that the equa- tions of transformation from rectangular to cylindrical co- ordinates are: - p J? A 7 f 2 \5t \T p > p' Q' r cos 4>, y = r sin , z = z; (i) VOLUMES BY DOUBLE INTEGRATION 325 and those from cylindrical to rectangular, r = Vx 2 + y 2 , = cos" 1 - = sin" 1 ^ = tan" 1 ^, z = z. (2) II X To derive a formula for volume the differential element of area in the XY plane may be taken as the rectangular base of an elementary right prism with altitude z, the base of the actual prisms into which the solid may be divided being bounded by lines two only of which are right lines, the other two being circular arcs, and the altitude of possibly only one edge being z, since the surface of the solid may be curved, or not parallel to the XY plane, even when plane. The expression for the volume of the solid is -// zr d dr, (1) where z must be expressed in terms of r or in order to effect the integration, and where the limits are to be such as will give the volume sought. Corollary. — d r

dr is a right prism with rectangu- lar base. The double integral in (1) is the limit of the sum of the elementary solids into which the given solid is conceived to be divided, or it is to be considered simply as the anti- differential of a second partial differential, when the differ- entials are taken as finite constants. Either way of regard- ing the differentials leads to the same result. Example 1 . — To find the volume of a sphere of radius a. Taking the pole at the center of the sphere, by (1), 7 = 2 / f a Va 2 -r 2 rddr Jq Jo <*r ddr Jo Jb = 2 £* d*[- \{a 2 - r 2 )*]" = f7r(a 2 -& 2 )*; /. Vol. of core = y (a 3 - (a 2 - 6 2 )*). Example 3. — Find the volume in first octant cut from a right cylinder, with its base of radius ri on IF plane and axis X TJ Z the 2-axis, by the plane - + j- + - = 1 . Here CL C * = *(l-f-|)-«(l -£«»*- jjsin*). V = c J I ( 1 cos — r sin 4) J r d$ dr rw n 3 f cos , sindr = § (tt - J) a 3 . Example 5. — Find the volume of the segment of the right cylinder which has its base a loop of the lemniscate r 2 = a 2 cos 2 in the XY plane and its upper surface a plane which intersects the XY plane in the ?/-axis at an angle of 45°. MASS 327 Here z = x = rcos; cos 2 Jo Jo 3 Jo r cos 0r d dr cos 5 2 cos d 7T = ^"-f (l-2sin 2 0)t C os0^ 7T V2 a 3 16 171. Mass. Mean Density. — As stated in Art. 154, the mass of a body, being defined as the product of density and volume, when the density * varies continuously, m = limitT7A7= fydV, (1) which becomes m = j J j ydxdydz, (2) or m = j J j t p 2 sin 6 ddd(f> dp, (3) according as rectangular or polar elements of volume are used. In these expressions y denotes the varying density at the different points within the body. The mean density of the body, denoted by y, is given by the equation fydV - m y=y = —y- ' (4) When the mass is considered as distributed continuously over a surface, the element of volume dV is replaced by dA = dx dy or pdd dp; and when the mass is considered as * Density will now be denoted by 7, instead of p used'in Art. 154. 328 INTEGRAL CALCULUS distributed along a line, straight or curved, dV is replaced by ds, the element of length. When the integral is considered as an anti-differential the elements are expressed directly in terms of the finite differen- tials; when, however, the integral is considered as the limit of a sum or sums, the elements are expressed in terms of the infinitesimal increments, the differentials appearing under the integral sign. Example 1. — Find the mean density of a sphere in which the density varies as the square of the distance from the center. Here the distance p of a volume element determining its density, the polar element should be used. Taking the density at a distance p from the center as kp 2 , k being a constant, and the volume element as p 2 sin A0 A<£ Ap, from (4) , I I J ftp 4 sin B dO d^> dp — Jo Jo Jo o 7 9 7 = — ■ — = = ■= ka~. 1 7ra 3 5 Again, since the density is the same for all points at the same distance p from the center, taking for the volume element a spherical shell of thickness Ap, A V = 4 irp 2 Ap, whence 7r I kp 4 dp = |fca 2 . o The mean density is thus shown to be three-fifths the density at the surface of the sphere. [The mean density of the earth according to the best determinations is very nearly 5.527 times that of water, while the average density of rocks at or near the surface is only about two and a half times that of water; hence, the mean density of the earth is about twice the average density at the surface.! See Corollary at end of Art. 190. mass 329 Example 2. — Find the mass and mean density of a semi- circular plate of radius a, whose density varies as the distance from the bounding diameter. Here 7 = ky, m / kydxdy = §/ca 3 ; ■aJo - _ I ka? _4:ka \ TCL 2 3 IT Or m = I I kp 2 sin ddddp = § ka 3 , Jo Jo where ky — kr sin 6. By a single integration, the element of area being x • Ay = Va 2 — y 2 &y, m = 2 JkVa 2 -y 2 ydy = %ka\ Example 3. — Find the mean density of a straight wire of length I, the density of which varies as the distance from one end. -f 7 = ks ds 7 , _ Id 1 ~ 2 ' Example 4. — Find the mass and mean density of a hemi- spherical solid, radius a, the density varying as the distance from the base. m fkzTX 2 dz = I kir (a 2 -z 2 )zdz = \ irka 4 ; Jo Jo lirka 4 3 Itto 3 8 ka. Here the element of volume is a spherical segment, ttx 2 A2 = -k {a 2 — z 2 ) Az, at a distance z from the base. Example 5. — Find the mean density of a right circular 330 INTEGRAL CALCULUS cone of height h, in which the density varies as the distance from a plane through the vertex perpendicular to the axis. /a 2 f h kz-wx 2 dz kir r- I z? dz h 2 J = 3 , , 7 ~ iira 2 h ~ iira 2 h 4 Here origin is taken at the vertex and element of volume is TTX 2 Az = a being radius of base; y = kz. of TTX 2 AZ = 7T t; Z 2 . h 2 CHAPTER VI. MOMENT OF INERTIA. CENTER OF GRAVITY. 172. Moment of a Force about an Axis. — The moment of a force about an axis perpendicular to its line of direction is the product of the magnitude of the force and the length of the perpendicular from the axis to the line of action of the force. The moment is the measure of the tendency of the force to produce rotation about the axis. The moment of a force about a point is identical with the moment about an axis through the point, perpendicular to the plane containing the point and the line of action of the force. 173. First Moments. — Let a line, surface, or solid be divided into elements; let each element (As, AA or AT) be multiplied by the distance of a chosen point within the element from a reference line or plane. The limit of the sum of these products as the elements are taken smaller and smaller is called the first moment of the line, surface, or solid. For the first moment M x of a plane curve about the x-axis, M x = \im Vy\s = I yds; (1) As=0 ** J and for the first moment of a plane area about the same axis, M, = lim T?/AA = fydA. (2) The first moment of a solid with respect to one of the co- ordinate planes, say the XY plane, is given by the equation, M xy = lim 331 %z±Y =fzdV. (3) 332 INTEGRAL CALCULUS For A A and AF appropriate expressions for the area and volume elements are to be used and the values corresponding for dA and dV substituted, in order to effect the integrations. The elements may be so taken that a single integration is sufficient, but double or triple integration will in general be required. 174/ Center of Gravity of a Body. — Let a given mass be referred to a system of rectangular coordinates, and let M xy , M yz , M xz , denote the first moments with respect to the three coordinate planes. The first moments of the mass of a solid are derived from those of the geometrical solid by the introduction of a density factor. There will be a point G (x, y, i), given by the equations, JyxdV fyydV fyzdV m = f ydV; x = — , y = -» -, z = -^ -, (4) I ydV J ydV J ydV in which the letter y denotes the density. The point G thus determined is the centroid of the mass. It is also the center of gravity of the weight W. Since W = mg, the masses of particles of a body are directly proportional to their/Weights; hence, the center of gravity is the same as the center of mass. The force of gravity acting on any mass is an example of a force distributed through a volume. If w denote the weight per unit of volume, at any point in a given mass, W its entire weight, and x, y, z, the coordinates of the center of gravity, the point where the resultant force exerted by gravity would act; then, from (4) or (3), // wxdV j ivydV I wzdV wdV; x=^ , y=4- , z = ^ . (5) JwdV wdV IwdV If in (4) and in (5), y and w are constant, that is, if the mass CENTER OF GRAVITY OF A PLANE SURFACE 333 is homogeneous, they may be taken from under the integral sign and canceled; whence, JxdV fydV CzdV m=yV,W = wV; x= — y — ,y = — — , z = , (6) the coordinates of the centroid of a volume, or of a homo- geneous body. The quantities / xdY = zV, j y dV = yV, j zdV =W, equal the moments of the volume with respect to the YZ, XZ, and XY planes, respectively. 175. Center of Gravity of a Plane Surface. — If in the formulas (6) for the centroid of a volume dV is replaced by t dA and i taken equal to zero, then fxtdA CxdA fytdA fydA V = tA,x = + p = ^ , y=^- = ^- , (1) J tdA I dA t dA / dA where the point (x, y) is in the XY plane, and in winch dA is the area of an element of the surface of a thin plate of uniform thickness and material, making t dA = dV. If A be the area of the middle layer, it is evident that xA = I xdA and yA = f ydA i (2) which are called the moments of the area A with respect to the y-axis and the x-axis. By the center of gravity of a plane surface is meant that point which is the center of gravity of a thin plate of uniform thickness and material whose middle layer is the surface given. The formulas for its coordinates are, therefore, those given in (1), j xdA I ydA * = — a — > y = — a — 334 INTEGRAL CALCULUS It is evident that the moment of an area about an axis through its center of gravity will be equal to zero. 176. Center of Gravity of any Surface. — The formulas (6), Art. 174, become for any surface, I xdA j ydA j zdA 5=-X-, v = ^r~ ' i = ^— (1) By the same method as in Art. 174, it can be shown that the coordinates of the center of gravity of any surface, plane or curved, are given by the equations (1). 177. Center of Gravity of a Line. — If in the formulas (5), Art. 174, for the coordinates of the center of gravity of a body of weight W, w dV is replaced by w ds, and 2 = 0; then, / wxds j xds j wyds lyds i=— i = ' , y = — = — , (1) j wds I ds I wds I ds where the point (x, y) is in the XY plane, and in which w ds is the weight of an elementary length of a slender rod of uniform section and material whose weight per unit of length is equal to w. If s be the length of the center line of the rod, it is evident that = I xds and ys = j xs = j xds and ys = I y ds, (2) which are called the moments of the line with respect to the 2/-axis and the z-axis. The rod may be straight, in which case the center of gravity will be at the middle point of its center line. If the center line of the rod is a plane curve in the plane XY, the coordinates of the center of gravity are given by (1). By the expression, center of gravity of a line, is meant the point which is the center of gravity of a slender rod of uniform section and material, of which the given line CENTER OF GRAVITY OF A SYSTEM OF BODIES 335 is the center line. The coordinates of a line are, therefore, those given in (1). It is evident that the moment of a line about an axis through its center of gravity will be equal to zero. If the center line of the rod, or any given line, is not a plane curve, from (5) as before, the equations are j xds I yds I zds x = -~ ' y = ~p ' 5 = -p I ds j ds j ds The moments of the line with respect to the YZ, XZ, and XY planes, respectively, will be xs = f xds, ys = I yds, zs = I z ds. 178. Center of Gravity of a System of Bodies. — If, instead of a single body, there is a system of bodies whose volumes are Vi, V 2 , V 3 , . . . V n , the coordinates of their centers of gravity being, respectively, (x h y h zi), etc.; and, if (x 0j 2/0, io) denote the coordinates of the center of gravity of the system and Vo its total volume, then V = Vx + V 2 H- Vz + • • • + V n ', M yg = VlX! + V 2 X 2 + • • • V n X n = J) Vx= V&). Similarly, M xz = J) Vy -= V yo, and M xy =^Vz= V z . Hence, -_%Vx _ _%Vy _ _^V~z Xo ~~v^> yo ~~v^> Zo ~~v^' where the numerators are the sum of the moments of the system with the respective coordinate plane, the equalities following from equations (6) of Art. 174. Similar equations hold for weights or masses upon substituting W or m for V, and for any group of surfaces by substituting A, where A 336 INTEGRAL CALCULUS is the sum of the areas of the several surfaces. If the sur- faces are plane, _ ^Ax _ _ *%Ay The last equations are useful in getting the center of gravity of plane figures composed of parts, the centers of gravity of which are known or easily found. Similar equations hold for a system of lines; s being the sum of the lines, X s ^ - X s 2/ _ X s * £o = — — , 2/= — > z = — — i So So So if the lines are not all in one plane; and zo = when they are in one plane, taken as the plane of XY. 179. The Theorems of Pappus and Guldin — First Theorem. — An arc of a plane curve revolving about an axis in the plane of the curve, but not intersecting it, generates a surface of revolution, the area of which equals the product of the length of the revolving arc and the length of the path described by its center of gravity. Second Theorem. — A plane area, bounded by a closed curve, revolving about an axis in its plane but outside the area, generates a solid, the volume of which equals the product of the revolving area and the distance traveled by its center of gravity. To prove the first theorem; let the z-axis be the axis of revolution, then the surface generated by the revolution of the curve about the z-axis is r f V ds. (1) Art. 160. From (2), Art. 177, for a plane curve, ys = f y ds. Hence S = 2irys. (1) THE THEOREMS OF PAPPUS AND GULDIN 337 It is evident that, if only part of a revolution is made, the area of the surface generated will be given by Si = 6ys, (2) where 8 denotes the angle in radians through which the plane containing the curve is turned. It is to be noted that the theorem and proof include the case of a segment of a straight line revolving about any axis. To prove the second theorem; let the z-axis be the axis of revolution, as before; then, denoting by A A an element of the plane area, the volume generated by a complete revolution of the area is V = lim y.2iry\A = 2w fydA. Now JydA = yA; (2) Art, 175; hence, V = 2wyA. (3) It is evident that if only part of a revolution is made, the angle turned through by the plane of the area being radians, the volume generated will be given by Vi = By A- (4) Example 1. — Find the center of gravity of a semicircle of radius a. Find it for the semicircular arc also. Taking the diameter along the ?/-axis, the length of the path described by the center of gravity as the semicircular area is revolved about the y-axis is 2 irx. The semicircle by its revolution describes a sphere whose volume is $ ira 3 ; hence, by the second theorem of Pappus, \ irx • \ wa 2 = $ 7ro 3 , _ 4a & = =—■ 7T 338 INTEGRAL CALCULUS Also the arc describes the surface of a sphere, 47ra 2 ; hence, by the first theorem of Pappus, 2 ttx - ira = 4 ira 2 ; _ 2a .*. x = Example 2. — Find the center of gravity of the semi- ellipse, - 2 + ^ = 1. Taking the z-axis as the axis of symmetry and applying the second theorem of Pappus as in Ex. 1, it is found that x = 5— , also ; hence, as y = 0, the centers of gravity of the two areas are identical. Example 3. — Find the center of gravity of the ^4 x quadrant of the ellipse. Let AOB be the quadrant of the ellipse, and the element of area a narrow strip parallel to !/-axis. dA = ydx = - Va 2 * a x 2 dx. x = /xdA - I x Va 2 — x 2 dx a Jo A j Tab a\_ Jo _ 4 a /The same abscissa\ i 7ra6 ~ 3 7r \as for semi-ellipse. / _46 Similarly, it is found that y = ^— • Corollary. — For the circle, x 2 + y 1 2 - - 4a a 2 ;x = y = ^. EXAMPLES OF CENTER OF GRAVITY 339 Example 4. — Find the center of gravity of a circular arc. Let AB be a circular arc, whose radius is a and whose center is at origin 0. Let 0i and 2 be the angles with x-axis made by the radii OB and OA to the ends of the arc. From . / /«• xds I yds and y = * I ds using polar coordinates, x = a o cos 6, y = a sin 6, ds = a dd; \J a 2 cos dd • T a sin / a dO e\ Je 2 Je 2 a (sin di — sin 2 ) di — &2 and 2/ = re, 10! I a 2 sin Odd — a cos , „ „ N Jg 2 J^ _ a (cos 02 — cos 0i) a dd d 01 — 02 When 2 = 0, the equations reduce to a sin 0i x = a (1 — cos0i) »■ — * — Corollary. — When 0i = 0i, 2 = — 0i; x When X = 90°, 02 = 0; x = a sin 0i 2a y = o. 2_a_ 7T Example 5. — Find the center of gravity of a triangle. Let the triangle ABC have base b and altitude A, and let the x-axis be through the vertex parallel to the base, and the 2/-axis positive downwards. Take dA = L dy, where L is 340 INTEGRAL CALCULUS the length of an elementary strip, parallel to the base and at a distance y from z-axis. dA = ^ydy; hence L :y = b :h; -i- A ib) Similarly, by taking strips parallel to h or the 2/-axis, I xdA _t J x 2 dx x = hbh b. A Y" X 1 /I 1 /l i h i i Y -i»y ////M/M D / i /! 1% / ! m C Y t* -b > Hence the perpendicular distance of the center of gravity from the base will be \ h, and, since the center of gravity of each elementary strip will be its middle point, the center of gravity of the triangle will be on the median line Am, at one-third the distance from m to A. Similarly, it is on the median line from B; hence it is at the intersection of the medians. Example 6. — Find the center of gravity of a semicircular plate of radius a, whose density varies as the distance from the center. Here, the density being determined by the distance from the center, the polar element is used. EXAMPLES OF CENTER OF GRAVITY 341 Let the density y = kp, and the z-axis as in Ex. 1 ; then x = JyxdA fS k r 3cosededp f ydA J jkp 2 ded P 4j . 4f. cos 6 dd 3a and y = 0. de Example 7. — Find the center of gravity of the volume cut from a right cylinder, the radius of whose base is a, by the planes z = and z = mx, the volume above the plane of XY alone con- sidered. Here h x x dx dy dz nvtf-^ p a / a Jo vV __ Tra^h'S _ Ta 3 h/S X ~ V ia 2 h 2 „2 x 2 dx 3_ 16 TQ, 7=f a 2 h from Ex. 2, Exercise XXXIX. I zdxdy dz ./-Vfli-jiJo = -= / x 2 Va 2 — x 2 dx = a 2 Jo iraW 16 _ Tra 2 h 2/ 1Q Tra 2 h 2 /\(S Z = 77 = o o, - 32 irh; and y = 0. 342 INTEGRAL CALCULUS EXERCISE XL. Find the coordinates of the center of gravity: 1. Area of the parabolic half segment, y- = 4 ax, x = to x = a. Ans. (f a, \a). 2. The area under one arch of the cycloid, x = a (0 — sin 0), y = a (1 — cos0). Ans. (wa, $ a). 3. The area under the catenary, y = - ^e« + e a J,x = 0tox = a. AnS ' \e + V 8(e-e-i) ) 4. A semicircular plate of radius a, the density varying as the distance from the bounding diameter. Ans. ( T %7ra, 0). 5. A homogeneous right circular cone, radius of base, a, and altitude h. The axis on the z-axis. Ans. x = %h,y = z = 0. 6. A homogeneous paraboloid of revolution from the origin to x = h. Ans. x = f h, y = 0. 7. A hemisphere whose density varies as the distance from the base whose radius is a. Ans. (0, 0, ^ a.) 8. The eighth part of a sphere in the first octant, the density of the mass varying as the distance from the pole or origin at the center. Ans. x = y = | a. 9. The circular sector subtending the angle 6 h radius a. t \ t? n a ( \ a /2a(sin0i) 2a(l — cos0i)\ (a) Fore, = 0, (a) Am. (3—^—, 3 5- - 2 } (b) For 0, = 90°, the quadrant. (6) Ans. x = y = P^- O TV 4 a (c) For 0i = 180°, the semicircle. (c) Ans. x = 5— , y = 0. 10. Find the center of gravity of a T-section. Using the method of Art. 178, with the dimensions on figure, taking the moments of the three rectangles composing the section, %Ay 4X7^ + 6X4 + 8X} _58_ 29 y = sr = — 4 + 6 + 8 — - - is - 9 - 3 - 22|h19 - x = 0, as the center of gravity will be on the axis of symmetry. Here advantage is taken of the knowledge that the center of gravity of each of the rectangles is at the center of the rectangle. Note. — This is the method of finding the centers of gravity of the various shapes, or built-up sections, used in constructions. EXAMPLES OF CENTER OF GRAVITY 343 v\ <7|- y? < 8" 11. Find the center of gravity of the trapezoid OAao. Let the upper base be b and the lower base B, the altitude h. Divide the trapezoid into two triangles by diagonal Oa, then the distances of the centers of gravity from OA" are | h and f h for the triangles, respectively. Then W A X 3 (B + b) h/ B + 2b \ 3\B+b J The center of gravity of each strip of area parallel to the bases will be its middle point, hence the center of gravity of the whole area is on the median line mhi of the trapezoid. Graphically, the center of gravity is located by again dividing the trapezoid into two other triangles by the diagonal oA, and drawing lines connecting each pair of centers of gravity; the intersection of these connecting lines is the center of gravity G of the trapezoid. 344 INTEGRAL CALCULUS 12. From Art. 178, show that the center of gravity of two volumes, masses, areas, or lines, lies on the line joining their separate centers of gravity and divides that line into segments inversely proportional to the two magnitudes. In Ex. 11, the point G, the center of gravity of the trapezoid divides the line GiG 2 connecting the centers of gravity of two triangles, inversely as the areas of the triangles, and it divides the line G'G" in the same way. In this way the common center of gravity of the Earth and the Sun is found. Taking the distance from the Earth to the Sun as 92,400,000 miles, and the mass of the Sun as 327,000 times that of the Earth, makes the distance of the common center of gravity from the center of the Sun ' ' miles, or only about 280 miles; so that the Sun is considered practically at rest relative to the Earth. 180. Second Moments — Moment of Inertia. — The term moment of inertia is applied to a number of expressions which are second moments of lines, of areas, or of solids. Let each of the elements of length, of area, or of volume (As, AA, or AF), into which a line, a surface, or a solid may be supposed to be 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 and smaller is called the secorid moment of the given line, surface, or solid with respect to the reference line or plane. Formulas for second moments are derived from those for first moments by squaring the distance factor. Denoting the second moment by /, the general symbol for moment of inertia, the following formulas correspond to (1), (2), (3), Art. 173. For a plane curve, I x = I y 2 ds. (1) For a plane area, I x = f y 2 dA. (2) For a volume, I xy =fz 2 dV. (3) POLAR MOMENT OF INERTIA 345 As applied to an area, the moment of inertia is a numerical quantity entering into a large number of engineering com- putations and takes its name from the analogy between the mathematical expression for it and that for the moment of inertia of a mass or solid. It is evident from the form of the expression that the moment of inertia is always a positive quantity, being unlike the first moment in that respect. In distinction from the moment of inertia, the first moment is sometimes termed the statical moment. 181. Radius of Gyration. — The radius of gyration of a solid is the distance from the reference line, called the in- ertia axis, to that point in the solid at which, if its entire mass could be concentrated, its moment of inertia would be unchanged. Thus, if m, the entire mass of a body, be con- sidered as concentrated at a point, and k denote the distance from the inertia axis to that point, the expression for the moment of inertia, / r 2 dm, will be equal to k 2 m. Therefore, I = k 2 m and k = y — , k being the radius of gyration of the solid of mass m. In the case of the plane area, by analogy, I x = Cx 2 dA = k 2 A, and k = y^-, where A is the entire area and k is its radius of gyration with respect to the a>axis. The radius of gyration of both the solid mass and the plane area will evidently be expressed in linear units. 182. Polar Moment of Inertia. — The polar moment of inertia of a plane area is the moment of inertia about an axis perpendicular to its plane. The moments of inertia about any two rectangular axes in the plane of the area are called rectangular moments of inertia when they are mentioned in connection with the polar moment. It is evident that 346 INTEGRAL CALCULUS for an area in the plane of XY, the moment of inertia about the 2-axis is a polar moment of inertia, and that it is equal to the sum of the two rectangular moments. Y x__m p ~o Thus, let the point P (x, y, 0) be in the element of area, then, L= fr 2 dA is the polar moment of inertia of the area about the z-axis or with respect to the point in its plane. But since r 2 = x 1 + y 2 , I z = f (y 2 + & )dA> hence, I z = I x + I v . The symbol for the polar moment of inertia is, in general, the letter J. 183. Moments of Inertia about Parallel Axes. — Theorem. — The moment of inertia of an area about an axis in its plane, not passing through its center of gravity, is equal to its moment of inertia about a parallel axis, passing through its center of gravity, increased by the product of the ana and the square of the distance between the two axes. MOMENTS OF INERTIA ABOUT PARALLEL AXES 347 Let the origin be at O g , the center of gravity of the area, and take the y-axis parallel to the inertia axis through in the plane. Let the point P (x, y) be in the element of area, <^a~> \° o 9 then its coordinates with respect to the axis through are (x + a, y). The moment of inertia I G = j x 2 dA, and that about the a:ds through is 7= f(x + a) 2 dA = J(x 2 + a 2 + 2ax)dA = fx 2 dA+ Ca 2 dA+ ^2axdA = I G + Aa 2 + 2a CxdA. (1) The quantity I x dA must be equal to zero, since the y-zxis passes through the center of gravity (Art. 175). Therefore, I = I G + Aa 2 . (2) It is evident that equation (1) gives the relation between the moments of inertia with respect to any two parallel axes in the plane of the area, O g Y being replaced by an axis through any point in the plane. In the same way it can be proved that the polar moment of inertia of the area, with respect to any point 0, is equal to its polar moment of inertia with respect to its center of 348 INTEGRAL CALCULUS gravity plus the product of the area and the square of the distance between the point and the center of gravity. Corollary 1. — It follows from (2), that, of all parallel axes, the axis through the center of gravity, called the gravity axis, has the least moment of inertia. Corollary 2. — When the inertia axis is a gravity axis, the radius of gyration, then called the principal radius of gyra- tion, is the least radius for parallel axes; from (2), Ak 2 = Ak G 2 + Aa 2 , ,\ k 2 = k G 2 + a 2 , (3) where kg is the principal radius of gyration and a is the distance between it and a parallel axis. 184. Product of Inertia of a Plane Area. — The product of inertia of a plane area is a numerical quantity which is of value only as it is found to enter into the determination of the relations between moments of inertia with respect to different axes. The product of inertia of an area A with respect to the axes of x and y is a second moment, / xydA and may be defined as the limit of the sum of the products of the elementary areas and the product of their distances from the two coordinate axes. Unlike the moment of inertia, the product of inertia may evident^ be either posi- tive or negative, depending upon its distribution in the different quadrants; and the area may be so located that its product of inertia will be zero. The axes may be so chosen as to make the product of inertia of an area zero, and such axes are called principal axes, the corresponding moments of inertia being called principal moments of inertia. It can be shown that for any point of an area (or body) there exists a pair of rectangular axes for which I xy dA = 0, and that the moment of inertia is a maximum when taken with respect to one of the principal LEAST MOMENT OF INERTIA 349 axes, and a minimum when taken with respect to the other. The relation between the polar moment and the two rec- tangular moments (Art. 182) shows that if one of the two is a maximum the other is a minimum, and vice versa. 185. Least Moment of Inertia. — In designing a column an engineer needs to know the least radius of gyration, and consequently the least moment of inertia, of the cross section, since the resistance to bending is least about that axis which has the least moment of inertia. It has been stated that the moment of inertia of an area is least about a principal axis through the center of gravity, and it is necessary, there- fore, to determine those principal axes which pass through the center of gravity. In many cases the position of the principal axes are known at once, for all axes of symmetry are principal axes, / xy dA being equal to zero for such axes, since for every point (x, y) there is another (x, —y). For example, any diameter of a circular area, the axis of a parabola, either axis of the ellipse or of the hyperbola is a principal axis. For a rectangle it is obvious that the lines through the center parallel to the sides are principal axes, but the diagonal of a rectangular plate is not a principal axis at its middle point. The gravity axes parallel and perpendicular to the web of the cross section of an /-beam, channel, or T-beam are principal axes. When the section is unsymmetrical, it is necessary to evaluate the integral, I xy dA, in determining the principal axes. Remark. — A full treatment of the subject of moment of inertia and product of inertia is beyond the scope of tins book. What has been given has been confined for the most part to areas, as that part of the subject has more immediate application in engineering. 350 INTEGRAL CALCULUS 186. Deduction of Formulas for Moment of Inertia. 1. Rectangle of base b and altitude h: h I x = ftfdA = f 2 by*dy = ^bh*. Gar-i la'v = h + Aa 2 = j2 W + hh P 3 - fx 2 dA = F hx 2 dx = &¥h. ~2 Again, iv = / by 2 dy t/0 = \bh\ h = tV& 4 > J = lb A ; for square. Y A ^9 X'—- A' _j 5' 1 (1) (2) (3) J = I x + I v = ~bW + ^Vh = b ^ + W). (4) (2') 2. Triangle about the axes: (a) Through the apex parallel to the base. (6) Through the center of gravity parallel to the base. (c) Through the base. DEDUCTION OF FORMULAS Let the base be 6 and the altitude h. (a) I x = ftf d A=£\y>dy = h -^- (b) I = I x -Aa* = T - Y (^-) = bh? 36' fM t . a 2 bh*bh /W\ bh* 3. Circle: (a) Polar moment of inertia, axis through center. (6) Moment of inertia about a diameter. 351 (1) (2) (3) (a) J = fr* dA = J r 2 to 3 dp = ~ Trrf 4 32 (1) 352 INTEGRAL CALCULUS For a sector with angle e, J -£•*-? (2) J 0r 4 ] ?rr 4 = 4 if 8 ' (3) J 0r 4 ~| Trr 4 (4) (b) Since for the circle the moments of inertia about all diameters are equal, I x + I y =2I x = J = — ■ • 7=7= 2 , .. lx ly 7rr 4 4 Trd 4 " 64* (5) For a circular quadrant, I x = I v = ~ J = 7rr 4 = 16* (6) For a semicircle, I -I ~ X -J- 7rr 4 = 8" (7) 4. Ellipse: Iy= Cx 2 dA = — f a x 2 (a 2 -x 2 )* a J - a dx = ira 3 b ~4~' (1) I x =Jy*dA = ^f b /(b 2 -y 2 )*dy = irab 3 ~4~' (2) J = I x + I y = T -f(a 2 + V). (3) 187. Moment of Inertia of Compound Areas, — Since the moment of inertia is always positive, the moment of inertia of an area about any axis is equal to the sum of the MOMENT OF INERTIA OF COMPOUND AREAS 353 moments of inertia, about that axis, of the parts into which the area may be divided. In some cases the area being considered the difference of two areas, its moment of inertia will be equal to the difference of the moments of inertia of the two areas. Example 1 . — Find the moment _ of inertia of the T-section shown with the dimensions on figure. (a) About the axis of X through top of section. (6) About the axis X through - the center of gravity. (a) 7 I = p/ i 3 = J(4-l)l 3 + Hlx4 3 ) = H-^=¥(ms.) 4 . X A y 3X|+4X2 K -a i"*7 14 * X r (b) 2/o = 9i 19. 7 = l4 ms - A 3 + 4 7o = /*-Ai/o 2 = -y— 7(H) 2 = 22.33-9.32= 13.01 (ins.) 4 . b Example 2. tion: hollow square. Hollow rectangular sec- &! 4 ) for ^(Vh-b^h). Example 3. — Hollow circular section I — t (V2 4 — ri 4 ) , about a diameter ; 7T J = 9 ( r 2 4 — ri 4 ) , about axis through center. Or in terms of the diameters: 64 7T 32 W - d,*). 354 INTEGRAL CALCULUS Example 4. — Find the moment of inertia of the section shown. l * - T2 W ~ 2 (? + f ° 2 - 2a T (fc))' b ^ »■ Art ' 183 = ^ (6) 4 - ^ - t (2)2 (3) 2 + 2.3-4^= 46.33 (ins.) 4 . h = I* j = I x + l y = 92.66 (ins.) 4 . Example 5. — Find the moment of inertia of the trapezoid (a) about its lower base; (b) about the gravity axis. t Itu, lt U 4 . 6 3 . 8 • 6 3 J.-4W + I2W-— + -12- - 216 + 144 = 360 (ins.) 4 . I G = I X - Aa 2 = 360 - 36 (§) 2 = 104 (ins.) 4 . CHAPTER VII. APPLICATIONS. PRESSURE. STRESS. ATTRACTION. 188. Intensity of a Distributed Force. — A distributed force is one that acts on a surface, such as the pressure of water against the surface of contact, the pressure of a weight upon the surface of its support; or, one that acts through a given volume, such as the attraction of the earth on a body. All forces are really distributed forces since no finite force can act at a point of no area; although this is true, in some cases it is convenient to regard a force, whose place of appli- cation is small, as though it were applied at a point. Such a force is called a concentrated force. A distributed force is conceived as " equivalent to" a concentrated force called the resultant force, when the force of gravity acting on every particle of a body is taken as acting at a point within the body, called the center of gravity. A distributed force is regarded as the limiting case of a system of concentrated forces whose number becomes larger as their individual magnitudes become smaller. It is thus that a force is re- garded as having a definite point of application and a definite line of action: when so regarded it is a localized vector quantity. When a force is distributed over an area, the intensity of the force at a point is the number of units of force acting on a unit of area including that point. Briefly the intensity is defined as the force per unit of area. If the force is uniformly distributed, the intensity p will be equal to the force P, acting on the entire area, divided by the area A ; that is, 355 356 INTEGRAL CALCULUS If the force is not uniformly distributed, the intensity at any point of the area will be given by '-&Ih1-S the limit of the ratio of the force, acting on a small element of the area, to that element as it approaches zero as a limit. When the intensity varies from point to point over any area, the force on that area divided by that area gives the average intensity on the area. In any case the entire force is given by (3) = j pdA, where p, if variable, must be expressed in the same terms as dA in order to get P by integration. If p is constant, P ; = p I dA = pA, and P ~ ~\' (1) 189. Pressure of Liquids. — The pressure of a liquid on a surface is normal to the surface, and the intensity of pressure varies as the depth of the point below the free surface of the liquid. The intensity is given by p = wh, (1) where w is the weight of a cubic unit of the liquid and h, called the head, is the depth of the point below the free surface. If w is expressed in pounds per cubic foot, h should be in feet, and p will then result in pounds per square foot. For water w is usually taken as 62^ lbs. per cubic foot, and the intensity of pressure given in pounds per square foot. When the intensity p is constant on any horizontally immersed plane surface the total pressure P is, by (1) Art. 188, p = Awh = 62.5 Ah lbs. (2) PRESSURE OF LIQUIDS 357 When the surface under pressure is not horizontal, by (3), Art. 188, wxdA, (3) where the limits of x are the least and greatest heads on the area. When the area extends to the surface of the liquid, the lower limit becomes zero and the upper may be taken as h. Since in (3), fxdA = xA, by (2), Art. 175, wxdA = Awx = 62.5 Ax lbs. / Hence, the total pressure on an immersed area is the product of that area, the weight of a cubic unit of water, and the head upon its c-enter of gravity. In general, the pressure of any liquid upon an area is equal to the weight of a column of liquid whose base is the area pressed and whose height is the depth of the center of gravity of the area below the surface. Example 1. — The vertical face of a dam subjected to the pressure of water is h ft. in height and b ft. in breadth. The pressure of the water varies as the depth; the intensity at a depth x is wx, w being the constant weight of a cubic unit of water. Required the total pressure on the face of the dam, and the location of the center of pressure. Let the area of pressure be divided into strips of width Ax and length b, then wx • b Ax is approximately the pressure on the element of area — for wx is the intensity of pressure at the top of the strip. The sum of a finite number of terms of the form wbx Ax would give a result for the total pressure less than the actual value; but the exact value is t, ,. ^ A 7 a i C h j wbx 2 ~\ h wbh 2 wh ,. , . P= hm V wbxlx = wb I xdx= -=- =—=- = -=- -Oh. (1) Ax = 0^0 «/0 Z JO Z Z 358 INTEGRAL CALCULUS The intensity of pressure is a uniformly varying force having zero value at the surface of the water and value wh at the bottom. The center of pressure, being the point of applica- tion of the resultant pressure, is given by taking the moment xS/ 3 h of P about the surface line equal to the limit of the sum of the moments of the elementary pressures about that line : wbx 3 ~] h wbh? x-P s: wbx 2 dx = Jo wbx 2 dx j; wbxdx wbh*/S 2 wbh 2 /2 ~ 3 (2) In general, the center of pressure of a rectangle with a side at the surface is two-thirds the height of the rectangle below the surface. When the top of the area is hi below the surface and the bottom is h 2 below, the total pressure is xdx = ~pr (h 2 2 — hx 2 ), and wb x = hi x 2 dx wb J ht dx I M~~ X 6 3_ = x 2 ~ ft, 2 W - ^! 3 * 3 W - h 2 2. hi (3) PRESSURE OF LIQUIDS 359 Note. — That P It may be noted that the second moment in the numerator is the moment of inertia of the area, and the first moment in the denominator is the statical moment. J wbx dx in (1) is the reversal of a rate may be seen by considering the rate of change of the total pressure when the depth x is increased by Ax, for then the pressure P on the area is increased by AP = wx • b Ax, approximately, and AP/Ax = wbx (nearly). r AP dP lim -= — = -;— = wbx, ax=o Ax dx Hence the rate of change of P; and P wbx dx as in (1); so the total pressure is a function of h and its rate of change is wbx = pb, where p = wx is the pressure on a unit of area. Note. — Whenever an external force acts on a body it induces a resisting force within the body. This is in accord- ance with Newton's third law of motion. This in- ternal resistance is due to the molecular forces or stresses within the body. A stress is a distributed force acting on a surface Example 2. — A verti- cal rectangular section ABBiAi of a beam of breadth b and depth d is subjected to a stress of tension and compression uniformly varying in intensity from zero at the middle fiber to S t and *S C at the outside fibers, at the distance, y x = \ d, from the neutral axis or middle fiber of the section. 360 INTEGRAL CALCULUS Find the total tensile and compressive stresses and the centers of stress on each half of the section. The intensity of stress at the distance y from the neutral axis is S/yitjj hence S/yiyb Ay is approximately the stress on a strip Ay in depth — the intensity at the edge of the strip being taken. The sum of a finite number of such terms would give a result less than the total stress on the half section; but the exact value is given by p = hm2, -y b ^y = —l y d y d Sb ?/ 2 > Sb f Sbd For the center of stress, the point of application of the total stress, d - t> C mSh 2.7 Sby^ Sb 2 T 2 Sbd*. -_Sbd 2 /Sbd_ 1 " y ~ 12 / 4 ~3 d ' When S = S t = S c , P — Pi] hence, the total tensile and compressive stresses form a couple with arm § d, the moment being —t- • ~d = — ^— , called the section modulus. By the mechanics of beams, if M denote the moment of the external forces acting on the beam, M = — ^ — Example 3. — A vertical circular section of a beam is subjected to a stress of tension and compression uniformly varying in intensity from zero at the horizontal diameter 2 a of the section to S t and $ c at the top and bottom fibers, respectively. PRESSURE OF LIQUIDS 361 Find the total stresses on the upper and lower semicircles and the centers of stress on the semicircles. Denoting by P the total stress on either semicircle, and taking S t = S c = S; P=\imy\ a -y2xAy = — V ^a 2 -y 2 ydy Ay=0 ^0 CL d Jo = 2S Tra 4 = Swa* Sird 3 . a " 16 8 64 ' - = Srd? /Sd 2 ^Zird " y 64 / 6 == 32 ' Hence the couple formed by the forces P has an arm, 2y = Ts^d, and M = ^Sd 2 '—ird = -7^-, the section modulus. O It) oZ Example 4. — Find the total water pressure upon the end of a circular right cylinder immersed lengthwise, one element of the cylinder just at the surface of the water. Find the center of pressure of the circular area. 362 INTEGRAL CALCULUS The intensity of pressure at a depth y being wy, the approximate pressure on a strip is wy • 2 x A?/. o The total pressure is P = lim ^ 2 wz?/ A?/ Ay=0 (2ay - y 2 Yydy = 2wa I (2 ay - y 2 )*dy o Jo = 2toa~ = W7ra 3 ; (Ex. 13, Exercise XXV.) f*2a r*2a 2w I (2ay-y 2 )y 2 dy $a-2w I (2ay-y 2 )*ydy — Jo Jo y " P ~ WTTO? = ^4- = 7« = |d. (Ex. 13, Exercise XXV.) wira? 4 8 EXERCISE XLI. 1. (6) The pressure upon one side of the gate of a dry dock, the wetted area being a rectangle 80 ft. long and 30 ft. deep, is to be found exactly. Take w = 62£ lb. for the weight of a cubic foot of water. (c) Find the depth of the center of pressure. Ans. (c) 20 ft. (a) Find the pressure approximately by a limited number of terms. (See Art. 154.) Ans. (b) 112^ tons. 2. The pressure on the gate that closes a water main half full of water, the diam- eter of the main being 8 ft. Get the exact (6) pressure only, (c) Find the center of pressure. Ans. (b) P = 1 ^w lbs. (c) y - |» ft, 3. Find the exact pressure on a cir- cular disk 10 ft. in diameter, submerged below water with its plane vertical and its center 10 ft. below the surface. Here ATTRACTION. LAW OF GRAVITATION 363 p = w C =a (10 -y) 2xdy = 2w f° (10 - y) (a 2 - yrf dy J— 5=— a •/— 5 = — a = 20 16' f (a 2 - y 2 ) h dy-2wj (a 2 - y 2 ) * y cfy = 20 w; • - 2 / 7T + | (a 2 - ?/ 2 ) f l = 250ttw J— 5 = — a = 2507r-62ilb. 4. Find the pressure on the face of a temporary bulkhead 4 ft. in diameter closing an unfinished water main, when water is let in from the reservoir. The center of bulkhead is 40 ft. below the surface of the water in the reservoir. Ans. Nearly 16 tons. 6. Find the pressure on the end of a parabolic trough when it is full of water. The parabola has its vertex downward, the latus rectum is in the surface and is 4 ft. long. Here P = lim V 2 w (1 - y) x Ay «= 2 w C 2 (1 - y) y* dy = 4 w ( y*dy - f y* dy = 4 w [f j/* - \ 2, 1 ]* =Hw = 66| lbs. jr Y --4- " "1 x Ij, jy^« sV^s^^//j£^ ^ ijf 6. A horizontal cylindrical tank is half full of oil weighing 50 lb. per cubic foot. The diameter of each end is 4 ft. Find the pressure on each end. Find the pressure when the tank is full also. Ans. 266f lb.; 1256 1b. 190. * Attraction. Law of Gravitation. — Every portion of matter acts on every other portion of matter with forces of attraction or repulsion. According to Newton's Law of Universal Gravitation, every particle of matter attracts every other such particle with a force which acts along the line joining the two particles, and whose magnitude is pro- * This article is based on a discussion in Fuller and Johnston's Applied Mechanics. 364 INTEGRAL CALCULUS portional directly to the product of their masses and in- versely to the square of the distance. between them. If the masses of the particles are m and m,\ and the distance between them is r, the law may be expressed algebraically by F-K**, (1) where F is the attractive force between the particles and K is a constant, determined by experiment, its numerical value depending on the units in which F, m, mi and r are expressed. The value of K having been determined in one case is then known for all cases. While formula (1) expresses the law of gravitation, the general algebraic expression for the law of attraction would be F = K wr (2) where (r) is some function of the distance between the particles, depending on the nature of the attractive force, K is a constant, and m and mi other quantities than the masses of particles. In interpreting formula (1), it is to be noted that it applies strictly only to particles; for the particles having finite masses must have finite dimensions and hence, as the distance between them is diminished, r cannot be less than a certain finite quantity and the maximum value of F, when the particles are in contact, will be a finite quantity. If r were taken to be zero in any case, F for finite values of m and m\ would become oo which would be impossible under the conditions. The formula, while applying strictly only to particles, gives, to a close approximation, the attraction between two bodies of finite size, whose linear dimensions are small compared to the distance between them. In the application of the law the attraction of one particle on another may be ATTRACTION. LAW OF GRAVITATION 365 regarded as acting at a point. It will be shown that any sphere attracts any outside particle as if the whole attraction was towards a point at the center of the sphere, but, in general, the attraction of bodies on exterior particles is not always towards the center of gravity of the attracting body. Attraction of gravitation is a mutual action between two particles or bodies; that is, each exerts an attractive force upon the other, the two forces being equal in magnitude and opposite in direction. This is implied in the Law, and it is also in accordance with the law of "action and reaction," Newton's third law of motion. It is evident that, in formula (1), K is equal to the force with which two particles of unit mass at a unit distance apart attract each other. If the equation is divided by mi, then — = a = K-^j (3) mi r 2 where a is the acceleration which would be produced in the mass mi by the attraction of the mass m at a distance r. The quantity K — 2 would also equal the force of attraction exerted by the mass m on a mass unity at a distance r. Briefly this is called the attraction at the point, at which the unit mass is situated, exerted by the mass m. The attraction at a point exerted by any mass is called the strength of the field of force, or briefly, the strength of field, by which the space through which the attraction of the mass is exerted is expressed. Electrostatic and magnetic attraction and repulsion are other examples of forces, which are governed by laws similar to that of gravitation. The following examples are based on the law F-K±, r 366 INTEGRAL CALCULUS where F is the attraction of a particle of mass m for a par- ticle of unit mass, the body being taken as homogeneous, of uniform density; that is, each cubic unit having the same weight. Example 1. — Attraction of a Rod of Uniform Section. — (a) Let the rod of small section be in the form of a circular arc; to find the attraction at the center of the circle. Let r be the radius, a the angle subtended at the center, and m the mass of a unit length of the rod. Take the axis OX bisecting the angle a, and let be the angle which the radius to any point P makes with OX. The attraction at of a particle at P is AF Km As Km Ad (1) r* r Since all the elementary forces of attraction are directed to the point 0, the resultant R is found from the sum of the components of the elementary forces. Xx = Kpf cos Bad = sin-> « * r 2 2 a 2 the attractions being neutralized. Hence, When a = ir, R = 2 Km 2Km . a sin-- r 2 (2) (3) ATTRACTION. LAW OF GRAVITATION 367 When a = 2t, R = 0; since the arc being a circumfer- ence of a circle, the attractions neutralize each other. (6) Let the rod be straight; to find the attraction at a point. Let r be the shortest distance from a point to the rod. Taking as origin, the equation of the rod is x = r (constant) . When the rod is B'AB the angles may be taken as in (a) for the circular arc. The attraction at of a particle at P' on the rod is . ,-, Km 4 Km cos 2 6 A /1A AF = wy y = — ? — y ' ( } The resultant attraction is found as in (a) ; since y = r tan 6, rdd Xx = ^f comedy = ^f cosede 2 Km . a 2 sin - , as in (a) . £ Y = ^JcosHsmOdy = ^p f* sin add = 0. _ 2 Hence, # = sin-, as in (a). (2') r J It is thus shown that if the straight rod B'AB is of the same mass per unit length as P 2 Pi, the resultant attraction of B'AB at is the -same as the attraction of P%Pi, since the sum of the attractions of the elementary masses m Ay and the sum of the attractions of the elementary masses m As have the same limit. (&') Let the rod be still straight but A X B, the angles with OX of the lines from to the ends being «i and a 2 ; then, V X = — pcos 6de = ~ (sin a 2 - sin a,), S Tr Km f a * . a ln Km . v Y = I sin Odd = (cos ai - cos a 2 ). r J ai r 368 INTEGRAL CALCULUS Hence, R r — COS (aj ~«l)J = and tan0 r %Y_ cos a. (2) From the geometry of the figure, r 2 = a 2 + d 2 - 2adcosd, 370 INTEGRAL CALCULUS which differentiated gives t dv rdr = ad sin Odd: .'. sin 6 and from the figure, cos = ad-dd' d — a cos 6 r Substituting these values in (2) gives exactly, at I d 2 - a 2 + r 2> d 2 \ r 2 dF = Kyirf 2 {--^-)dr; (3) hence, F = K yir j2 J ^ ( ? )dr AKyiraH KM' d 2 d 2 (4) It follows from (4) that the attraction is the same as though the mass of the shell were concentrated at its center. It follows also that a sphere, which is either homogeneous or consists of concentric shells of uniform density, attracts a particle without the sphere as if the mass of the sphere were concentrated at its center. This law holds almost exactly, for bodies slightly flattened at the poles, if the particle is not too close to the attracting body. Since both these con- ditions exist in the case of the Earth and other members of the Solar System, this law has important applications. (6) Let the point P' inside the shell be the position of the particle. The equation (3) in case (a) is true for this case too, but the limits for r are now a — d and a -f- d. Hence, „ Kywat [a 2 — d 2 \ a 2 -d 2 , >+" n ... that is, the resultant of all the attractions of the elementary masses of the spherical shell on a particle within the shell is zero. (c) Let the point where the particle is be on the surface of the shell. ATTRACTION. LAW OF GRAVITATION 371 In (4) making d = a gives F = 4Kywt = ^— (6) Corollary. — If a particle be inside a homogeneous sphere at a distance d from its center, all that portion of the sphere at a greater distance from the center than the particle has no effect on the particle, while the remaining portion attracts the particle in the same way as if the mass of the remain- ing portion were concentrated at the center of the sphere. Thus the attraction of the sphere on the particle is „ %Kiryd* IKiryd d- (7) that is, within a homogeneous sphere the attraction varies as the distance from the center. The attraction of a sphere of mass M on a particle at the surface is from (7), making d = a, F = - s Kirya = -^-- (8) Hence, the attraction for an external particle is '-¥■ where d is the distance from the particle to center of sphere. Note. — The propositions respecting the attraction of a uniform spherical shell on an external or internal particle were given by Newton (Principia, Lib. I, Prop. 70, 71). It was in 1685, nineteen years after he had conceived" the theory of universal gravitation, that he completed the veri- fication of the theory, by proving that a sphere in which the density depends only upon the distance from the center attracts an external particle as if the mass of the sphere were concentrated at its center. Thus was the great induction by this supplementary proposition finally established. 372 INTEGRAL CALCULUS Example 3. — Attraction of the Earth.* — I. Find the relation between the attraction of the Earth on a body at the surface and at a point h feet above the surface. Taking the Earth as a sphere whose density is a function of the distance from the center, R as the radius, and F and F' as the Earth's attraction upon the body at the surface and at h feet above the surface, F/F' = (R + hy/R 2 (by Ex. 2, (8) and (9)), or F' = FR 2 /(R + h) 2 . (1) If h is a small fraction of R, then approximately, F' = F (1 + h/R)~ 2 = F (1 - 2 h/R). (2). Since the "weight" of a body is the force with which the Earth attracts it, the equations (1) and (2) give the relation between the weight of a body at the surface and at a height h feet above the surface. And, if g and g' are the values of the acceleration of gravity at the surface and at the point h feet above the surface, since F/F' = g/g', the equations give the relation between g and g' also. (a) Find approximately at what height above the surface will the weight of a body be T V of one per cent less than at the surface. Taking the mean radius of the Earth as 20,902,000 ft., F'/F = l-2h/R = 1- 1/1000; . R 2 0,902,000 1AilRf ", , " * = 2005 = 2000 = 10 > 451feet - Corollary. — A mass which at the surface weighs one pound at 10,451 ft. will weigh 0.999 lb. (b) Find how much the value of g is changed by a change of elevation of one foot above the surface. * This example is based on examples in Hoskins's Theoretical Me- chanics. ATTRACTION. LAW OF GRAVITATION 373 The value of g for different latitudes and elevations is given by the following formula, in which g is in feet per second, I is the latitude, and h the elevation in feet above sea level : g = 32.0894 (1 + 0.005243 sin 2 1) (1 - 0.000.0000957 h). This gives g = 32.0894 at the equator at sea level, and g = 32.174 at 45° latitude at sea level; this latter value, g = 32.174 ft. per sec. per sec. is the standard value. II. Find the relation between the attraction of the Earth on a body at the surface and at a point h below the surface, (a) Taking the Earth as a sphere of uniform density of radius R, F"/F = (R- h)/R = 1 - h/R (by Cor. Ex. 2), (3) where F and F" denote the attraction at the surface and at h below the surface. Corollary. — Under these conditions, the weight of a body and the value of g would decrease with the depth h below the surface. (b) Taking the Earth as a sphere whose density is a function of the distance from the center, let y denote the mean density of the whole Earth and 70 the mean density of the outer shell of thickness h. Let M be the mass of the whole Earth, M" that of the inner sphere of radius R — h, m the mass of the attracted body, F and F" the attraction at the surface and at h below the surface. Then F is equal to the attraction between two particles of masses M and m whose distance apart is ~R, and F" is equal to the attraction between two particles of masses M" and m whose distance apart is R — h. That is, F = KMm/R 2 , F" = KM"m/(R - h) 2 ; . F" M"( R \« hence > T = ir{R=h)' (4 J 374 INTEGRAL CALCULUS Now M = t7r# 3 7; M -M" = J 7r 7 o [R s - (R - h)*]; •■• ¥ =>-?H^)H-?) + ?(t)' <» which substituted in equation (1), gives £=('-?)G^H-(V> « If /i is a small fraction of #, equation (6) may be reduced to the approximate formula, Corollary. — If the mean density of the outer layer of the Earth is less than two-thirds the mean density of the whole Earth, the weight of a body increases as it is taken below the surface of the Earth. (See Ex. 1, Art. 171.) The mean density of the Earth being taken as 5.52 and that of the layer near the surface as 2.76, about the density of the rocks, makes 70/7 = \ and equation (7), F"/F = W"/W = q"/q = 1 + h/2R. (8) That the weight of a body increases as it is taken below the surface has been shown by actual trial. From (8), the depth to which a body must be taken in order that it gain T <^ of one per cent in weight is approximately, , 2X20,902,000 . 1Qnn h= 10,000 = 4180ft -; that is, a mass weighing a pound at the surface will weigh 1.0001 lb. at a depth of 4180 ft, below the surface. Compared with case I, it may be seen that, under the conditions, for the same value of h, the gain in weight is one- fourth as much as the loss in weight when the body is above the surface, the same ratio of change applying to the value of g also. VALUE OF THE CONSTANT OF GRAVITATION 375 191. Value of the Constant of Gravitation.* — From the foregoing as to the attraction of a sphere, it follows that the formula for the attraction of two particles, F = K rm^_ [( 1)Art<19 o] will apply to two spheres, which are either homogeneous throughout or composed of a series of concentric shells, each one of which is of uniform density, m and m' being the masses of the spheres and r the distance between their centers. By measuring the force of attraction between two spheres of known mass and distance apart, the value of K the constant of gravitation has been found. As stated in Art. 190, its numerical value will depend on the units used for the other quantities in the equation. The relation between the constant K and the mass of the Earth, taking the Earth as a sphere whose density is a function of the distance from the center may be shown as follows. Let the units be the British gravitation units, and let R be the radius of the Earth in feet, M its mass, y its mean density. Consider the attraction of the Earth on a body of mass m at the surface. By the formula (1) of Art. 190, the value of the attraction is KM?n/R 2 ; but (since the unit force is the weight of a pound mass) expressed in pounds force, its measure is m. Hence, m = KMm/R 2 or KM = R\ (1) Since the value of R is known, either K or M can be found when the other is known. Putting for M its value in terms of 7, K-UR*y = R 2 or ^7 = ~- (2) *^Arts. 191 and 192 are based on Articles in Hoskins's Theoretical Mechanics. 376 INTEGRAL CALCULUS Taking y = 345 lbs. per cu. ft. and R = 20,900,000 ft. gives K = -r\- = 3/(4 7T X 20,900,000 X 345)= 3.31 X 10" 11 . 4 7T/L7 Otherwise, if the value of K, found by direct measurement of the attraction of two spherical bodies, is substituted in (2) the value of 7, the mean density of the Earth is found to be 5.527. The density of water being unity, and its weight 62.4 lbs. per cu. ft., the mean density of the Earth is about 345 lbs. per cu. ft., as used above. 192. Value of the Gravitation Unit of Mass.* — As stated in Art. 190, the force with which two particles of unit mass at a unit distance apart attract each other is equal to K, the constant of gravitation; this is evident from the equation, F = R m^_ [(1)Art< m] Let m pounds be the mass of each of two particles which, when one foot apart, attract each other with one pound force. Substituting K = 3.31 X 10 -11 , as given above, putting F =. 1, m = m f , and r = 1; gives m = l/VE = 173,800 lb. If a mass equal to 173,800 pounds be taken as the unit mass, the constant K becomes unity and the formula for attraction is then F = mm' The gravitation unit of mass is thus shown to be a mass equal to about 173,800 pounds, distance being in feet and force in pounds-force. 193. Vertical Motion under the Attraction of the Earth. — Let the Earth be taken as in Example 3, Art. 190, r as the radius and s the distance of the moving particle from * See Footnote on page 375, VERTICAL MOTION 377 the center 0. Taking distance, velocity, and acceleration as positive outward, then, as in (1) Ex. 3, F' F g Since dv di = a an gives v dv = ads. or a' = gr- -r = v, eliminating dt Hence, a => — g' = — ^-, neglecting air re- S" sistance, which gives I vdv= j — gr 2 s~ 2 ds; v 2 l r /l 1\ integrating, ^ = ^ s ~ l I = ^ [j ~ J ) J then, v 2 = 2gr 2 ^--^j (1) gives the velocity of a particle towards the Earth from any distance s. For the velocity acquired by a body in falling to the surface from a height h, put s = r + h in (1), giving, = 2 K^h) = 2gh, approximately, (2) if h is small, which is the formula when g is constant, as it is taken near the surface. When - is small, putting (h/r)' +•••]. (3) r + (2) becomes v 2 = 2 gh [1 - (h/r) + (h/r) 2 378 INTEGRAL CALCULUS By taking any number of terms of this series, an approxi- mate result may be g otte n as nearly correct as desired. If, in (1), s = oo , v = V2 gr; so if a body fell toward the Earth from an infinite distance, its velocity, neglecting air resist- ance, would be V2 gr = 6.95 miles per second, for r = 3960 miles. If falling from a finite distance s, the velocity must be less than this. Hence, a body can never reach the earth with this velocity; and if air resistance is considered, the velocity for s = co is less than V2 gr. If projected outward with velocity V2 gr and air resistance be neglected, the body would go an infinite distance. This velocity is called the critical velocity or velocity of escape, for under the conditions it is supposed that certain particles of the atmosphere may escape from the attraction of the Earth. In this connection, it is to be recalled that due to the Earth's rotation, there is at its surface a 'centrifugal force mg/289, exerted by a particle of mass m, which lessens the value that g would otherwise have. 194.* Necessary Limit to the Height of the Atmosphere. — The centrifugal force of a particle of mass m on the surface 7710 of the Earth is muftr = ^r, and at a distance s from the 7TIQS center it would be mo) 2 s = *„ . The Earth's attraction at 289 r Trior 2 that distance being — ^-, in order that the particle be re- tained in its path these two forces must equal each other; mgs _ mgr 2 '"' 289r ~ - ^~ , or s 3 = 289 r 3 , hence s = ^289 r = 6.6 r = 26,000 miles approximately; that is, a height above the surface of about 22,000 miles. The actual height of the atmosphere is probably much less than * Bowser's Hydromechanics. MOTION IN RESISTING MEDIUM 379 this. The estimates of the height by various scientists have been very divergent — from 40 miles to 216 miles; but the latter appears to be the most likely, for meteors have been observed at an altitude of more than 200 miles and, as they become luminous only when they are heated by contact with the air, this is evidence that some atmosphere exists at that height. It is supposed that at a height much less than 5.6 r, the air may be liquefied by extreme cold. 195.* Motion in Resisting Medium. — Consider the motion of a body near the surface of the Earth under the action of gravity taken as a constant force and the air taken as a resisting medium of uniform density, the resistance varying as the square of the velocity. Let a particle be supposed to descend towards the Earth from rest, and let s be the distance of the particle from the starting point at any time t, gk 2 the resistance of the air on a particle for a unit of velocity — gk 2 being the coefficient of resistance. The resistance of the air at the distance s from the origin will be gk 2 [-n) , acting upwards, while g acts downwards, the mass being a unit. The equation of motion is d 2 s T „ fds\ 2 dt 2 9 or g dt = Integrating, <(f) I , 1, dt gt = 2k l0g —J s ' 1 h dt % = 0, v = 0, giving C = 0. * Bowser's Analytic Mechanics.' 380 INTEGRAL CALCULUS Passing to exponentials, ds _ 1 e kot — e~ kgt dt~ k e kgt +e~ kgti (2) which gives the velocity in terms of the time. To get it in terms of the space, from (1), .-. log[l-fc 2 ^y]= -2gk\s = 0, v = 0;d = 0, (3) f$-h* -**">• (4) which gives the velocity in terms of the distance. Also, integrating (2); gkh = log (e kgt + e~ kgt ) - log 2; /. 2 e QkH = e fc(7< + e-* ff S (5) which gives the relation between the distance and the time of falling through it. As the time increases the term e~ kgt diminishes and from (5) the space increases, becoming infinite when the time is infinite; but from (2) as the time increases the velocity becomes more nearly uniform, and when t = oo , the velocity = 1/k; and although this state is never reached, yet it is that to which the motion approaches. 196. Motion of a Projectile. — If a body be projected with a given velocity in a direction not vertical and be acted on by gravity only, neglecting the resistance of the air, it is called a projectile. The path, called the trajectory, will result from a combination of the motions due to the velocity of projection and to g, the vertical acceleration of gravity. Let the plane in which a particle is projected with a velocity MOTION OF A PROJECTILE 381 v be the plane of XY, and let the line of projection be inclined at an angle a to the z-axis, making v cos a and v sin a the resolved parts of the velocity of projection along the axes of x and y. Y Let (x, y) be the position of the particle P at the time t; then, since the horizontal acceleration is zero and the vertical acceleration negative, d 2 x _ d 2 y _ dt?~ ; ~df 2 ~~ g ' Taking the first and second integrals of these equations, determining the value of the constants of integration corre- sponding to t = and t = t, gives dx = z;cosa; dy = v sin a — gt; dt" w "~> dt~ w ~"~ yv > (1) x = vcos at; y = v sin at — | gt 2 . (2) Equations (1) and (2) give the coordinates of the particle and its velocity parallel to either axis at any time t. Eliminating t between equations (2), gives y = x tan a gr (3) 2 v 2 cos 2 a' which is the equation of the trajectory, and shows that the path of the particle is a parabola. Putting equation (3) in the form 2 v 2 sin a cos a 9 x = — 2 v 2 cos 2 a 9 y> 382 INTEGRAL CALCULUS or / v 2 sin a cos a\ 2 2 v 2 cos 2 at t> 2 sin 2 a\ , AS and comparing this with the equation of a parabola, (x-h) 2 = -2p(y-k), it is seen that : v sin a. cos a. the abscissa of the vertex = ; (5) 9 v 2 sin 2 at the ordinate of the vertex = — = ; (6) *9 ,, , , 2 v 2 cos 2 a fP9S the latus rectum = (7) 9 By transferring the origin to the vertex, (4) becomes * 2 =-^f%, (8) which is the equation of a parabola with its axis vertical and the vertex the highest point of the curve. The distance between the point of projection and the point where the projectile strikes the horizontal plane, called the Range, is ^ D v 2 sin 2 a rtv . OB = x = , (9) 9 when y = 0, from (3), which is evident geometrically, since OB = 2 OC; that is, the rangt is equal tojtwice the abscissa of the vertex. It follows from (9) that the range is greatest for a given velocity of projection, when a = 45°, in which ease the v 2 range = - . It. appears from (9) that the range is the same 1/ for the complement of a as for a. The greatest height CA is given by (6) which, when a = 45°, becomes v 2 /4:g. The height of the directrix, CD = CA + AD = "-^ + i ?i^!^ = |L 20 4 g 2g MOTION OF PROJECTILE IN RESISTING MEDIUM 383 Hence, when a = 45° the focus of the parabola is in the horizontal line through the point of projection, for then CA = \ CD. To find the velocity V at any point of the path, from (1), = v 2 cos 2 a + (v 2 sin 2 a — 2 v sin agt + g 2 t 2 ) = v 2 -2gy, or JJ - f - y = MS - MP = PS. V 2 Since — is the height through which a particle must fall from rest to acquire a velocity V, it follows that the velocity at any point P on the curve is that which the particle falling freely through the vertical height SP would acquire; that is, in falling from the directrix to the curve; and the velocity of projection at is that which the particle would acquire in falling freely through the height CD. For the time of flight, put y = in (3) and solve for 2 v 2 sin a cos a , . , ,. . n , , . ,. . x = , which divided by v cos a gives, time of flight = ; or in (2) put y = and solve for t, giving r\ j 4 2 y sin a , . t = and t = , as before. 9 197. Motion of Projectile in Resisting Medium. — If the resistance of the air is taken to vary as the square of the velocity and the angle of projection is very small, the pro- jectile rising but a very little above the horizontal, the equation of the trajectory above the horizontal line can be found. Thus the equation gx 2 gkx 3 " ~" 2 v 2 cos 2 a 3v 2 cos 2 a may be derived under such conditions; where the first two 384 INTEGRAL CALCULUS terms represent the trajectory neglecting air resistance, as found in (3), Art. 196. For the high velocities of cannon-balls the trajectory is found to be very different from the parabolic path and the range much less than that deduced for it. Experiments show that the angle of projection for greatest range is about 34°, rather than 45°, as deduced for the para- bolic path. The simplest formula for making out a range table is Helie's: gx 2 / 1 . kx\ y = x tan a- * I— » H )» " 2 a \Vo 2 Vo J 2 cos 2 where k = 0.0000000458 -, d being the diameter of the projectile in inches, and w its weight in pounds. In addition to the resistance of the air, allowance has to be made in firing for the drift, that is, the tendency for most projectiles to bear to the right upon leaving the gun, due to the right-handed rotation given to the projectile. CHAPTER VIII. INFINITE SERIES. INTEGRATION BY SERIES. 198. Infinite Series. — When a series consists of a succession of terms whose values are fixed by some law and the number of its terms is unlimited, it is an infinite series. Let Ui, M2, W3, . . . be an infinite succession of such values and let the sum of the first n of these values be denoted by S n , that is, S n = Ml + Uz + ' • • + U n . (1) When n becomes infinite, then the infinite series is Mi + Ui + Us + • • • . (2) If S n has a definite limit as n becomes infinite, that limit is called the value of the infinite series, and the series is said to be convergent. If S n has no definite limit, either oscillating between two finite values or increasing in value beyond any finite value, the series is said to be divergent. Thus the geometrical series, a + ar + ar* + ar*+ • • • (3) is convergent only when r is between — 1 and +1, for S n = a + ar-\-ar 2 -\- • • • +ar n ~ l = — - = - , 1 — r 1 — r 1 — r and lim S n = lim a ( * ~ rn) = -?—, when -1< r < 1. n=oo n=ao L r 1 T ail — r n ) Since *S n=3C = — ^ = x, when — 1 > r = 1, 1 — r and oscillates between and a when r = —1, the series for those values of r is divergent, *S^= X having no definite limit. 385 386 INTEGRAL CALCULUS The terms of a series may be functions of some variable x; then the series is said to converge for any particular value of this variable, say x = a, when, if x is replaced by a in each term, lim S n (a) exists. When the corresponding limits exist n=oo for all values of x in a certain interval, say from x = a — h to x = a + h, the series is said to converge throughout the interval and to define a function in the interval. Example 1. — Let the given series be 1 + x + x 2 + x 3 + • • • + x"- 1 + • • • . (1) \ — %n J Here lim S n (x) = lim -— — = _ , for — 1 < x < 1 ; n=oo n=» -L X 1 X and within the interval x = — 1 to x = +1, not including the end values, the series defines the function fix) = z 1 — x For the end values x = — 1 and x = -\- \, and for all values beyond, the series is divergent and does not define a function. Example 2. — Similarly the given series, 1 - x + x 2 - x 3 + • • • + (- 1) V- 1 + • • • (2) has lim>Sn(V) = ., , , for — 1 < x < 1, n= oo -»■ *~ r" X and within the interval x = —1 to x = +1, the series de- fines the function fix) = t— J 1 -f- x For the end values x = —1 and x = +1, and for all values beyond, the series is divergent and does not define a function. 199. Power Series. — An infinite series of the form a -f aix + a 2 x 2 + a 3 :r 3 + • • • + a n x n + • • ■ , (1) where a , ai . . . , a n , etc., are constants, is called a power senes in x. One of the form a + ai — a) + a 2 (x — a) 2 + a 3 Or — a) 3 + • • • -r-M*- a)»+ • • • (2) is called a power series in (x — a). POWER SERIES 387 The series (1) and (2) of the preceding examples (Art. 198), are power series in x, representing the functions and 1 — x T—r — for certain values of x. Such series are of importance because of the frequency with which they 'occur and the special properties which they possess. For instance, the sum of a few terms of an infinite series representing some function may be a very close approxima- tion to the value of the function. Thus, if in the series (1), (Ex. 1, Art. 198), x = §, the well-known series converging to the value 2 results: 1^=2=1+1 + 1 + 1+...+^+.... (3) If the terms in (1) (Ex. 1, Art. 198) after x k ~ l be neg- lected, the error would be, .&+: x k _|_ x k+* _|_ . . . _|_ x n _j_ 1 This error, ——— , would be very small compared with the j. x value of the function, ; , and would decrease as k was 1 — x increased; that is, a closer and closer approximation would be made to the value of the function the greater the number of the terms retained. For the particular value x = }, it may be noticed that the error made in stopping with any term is exactly the value of that term. For smaller values of £ a very much closer approximation would be made even when only a few of the terms are taken. This method of approximation is practically useful when the exact value of the function is unknown or does not admit of exact numerical expression, for examples, the numbers e and 7r, the logarithms of numbers, and the trigonometric func- tions of angles in general. In Articles to follow power series for such functions will be given. 388 INTEGRAL CALCULUS 200. Absolutely Convergent Series. — A series the abso- lute values of whose terms form a convergent series is said to be absolutely convergent ; other convergent series are said to be conditionally convergent. For example, the series 1 - I + i - i + ' • • (Ex. 2, Art. 198), (1) is an absolutely convergent series, since it converges when all terms are given the positive sign, as (3), Art. 199. On the other hand, the series i-i + i-i + 1... (2) is conditionally convergent, since the series resulting from making all terms positive, is divergent. The series (3) may be seen to be divergent in the form, 2 ' V3 + S+d^+---+2F=-i)+---> for the sum of the terms in each of the parentheses is greater than |, and as the number m of such groups that can be formed in the given series is unlimited, Sn=oc > (m X i) = oo. This result reveals the important fact that while the defini- tion of an infinite convergent series, requiring S n =ao to have a definite limit (Art. 198), makes lim u n = Q a necessary n=oo condition, that condition is not sufficient to insure conver- gency. In other words, a series is not necessarily convergent when the terms themselves decrease and approach zero, as the number of terms increases without limit. For the series (3), the condition is fulfilled; the nth term approaches zero as a limit, as n increases without limit, and yet the sum of the first n terms has no limit and the series is divergent. ABSOLUTELY CONVERGENT SERIES 389 When, however, the terms of the decreasing series are alternately positive and negative the condition is sufficient to prove convergency. Thus the series (2) being such a series is convergent, though not absolutely so. This series may be put in the forms, (i - i) + (i - i) + tt - i) + • • • , 01' 2 "h T 1 2 + 3"V + * * ' , i - (i - i) - (i - « • . . , or 1 - i - 2 ] o - • ' * , where it can be seen that the sum of n terms of the series is greater than \ and less than 1. It will be shown further on that the limit of the sum is log 2 =;0.69. . . (Ex. 1, Art. 203). While it is thus seen that the series (2) is conditionally convergent, the following series is absolutely convergent : 1 - I I I 4- 1 - . (A\ 2 2 3 2 4 2 5 2 ' since 1 + ^ + ^ + ^ + - 2 + • • • (5) is convergent, as may be shown by comparison with the series 2 2 2 2 3 2 4 ' or 1 + * + } + *+••-, (3), Art. 199, known to be convergent. The series (5) is the more general series when p = 2; and the series (3) of this Article, called the harmonic series, is the series (6) when p = 1. This series (6) is, therefore, divergent for p = 1; for p < 1, every term after the first is greater than the corre- sponding term of the series (3), hence (6) is divergent in this case also. 390 INTEGRAL CALCULUS For p > 1, compare I? T ^ 3"/ V4» 5" T 6» 1") + (P+-- - + lfc) + '-- (6) with i+zi+iwi+i+i+ri 1p ' y2p 2 P / \4 P 4 P 4 P 4 P / ' +(£+••.•+£)+•■■ m There is the same number of terms in the corresponding groups of the two series and the sum of the terms in those of (6) is in each group less than the sum in those of (7). Now (7) may be put in the form 1+2,1,8 2 a geometrical series, whose ratio, — , is less than unity. Hence by (3) of Art. 198, (7) is convergent and consequently (6) is convergent. The series (6) and the geometrical series are useful as standard series, with which others may often be compared to test convergency. 201. Tests for Convergency. — It has been seen that a conditional test is that in every convergent series the nth term must approach zero as a limit, as n is increased without limit. This condition is involved in and may be deduced from the definition (Art. 198), that the infinite series Ui + u 2 + m + • • • (1) is convergent, if lim S n exists. n=co For since S n = Sn-i + u n , lim S n = lim £ n _i + lim u n = lim S n -i, # = oo n=oo n=oo n=oo if lim S n exists ; .*. lim u n = 0. The converse is not true. TESTS FOR CONVERGENCY 391 In the same way it may be shown that the Remainder after n terms of a convergent series must approach zero as a limit as n becomes infinite; thus, letting R n denote the series of terms after the nth, Rn = Uu+l + Un+2 + Un+3 + ' ' ' . Now if S denotes the value of the series (1), it is the limit of the sum of the terms in the series; that is, S = lim (S n + R n ) =\miS n ; :. lim R n = 0. For example, in (3), Art. 199, R n = (i)"" 1 ; .*• lim R n ' = 0. It was seen in (3), Art. 200, that although the nth term approaches zero, the series is not convergent, hence the remainder after n terms does not approach zero as a limit in that series, as it is a divergent series. For convergency, lim R n = is a decisive test; Km u n = n=oo 71= x is a decisive test only when the terms alternately have dif- ferent signs; it is a conditional test when all the terms have the same sign. For divergency, lim u n not equal to zero is a n=x decisive test. Hence, when lim u n = 0, unless the terms alternately have different signs, the test is indecisive. The Comparison Test. — It may often be determined whether a given series of positive terms is convergent or divergent, by comparing its terms with those of another series known to be convergent or divergent. The method of applying this test and the standard series useful for com- parison have been given in the preceding Article 200. This test is often available when other tests fail to be decisive. The Ratio Test. — A given series Wl + "2 +«•+•'•+ Un~l + Un + • • • u is convergent or divergent, according as lim ■ — — is less or greater n= Un—l than 1. 392 INTEGRAL CALCULUS This test applies when some of the terms of the series are negative as well as when they are all positive. It is no test, however, when lim — — = 1 ; in that case other tests must n=oo Un—1 be applied. For example, let the given series be Here lim — - = lim ( — . / -, r^r ) — nm - = n = ooU n -i „=oo\w!/ (n—\)\J n = ooTl for any finite value of x. Hence the series (2) converges for all finite values of x. This series (2), as given in Art. 36, is the expansion of e*; and, when x = 1, the limit of the series is the number e, in Art. 34 (see Ex. 5, Art. 215, also). The ratio test is found to be true by comparing the given series with the geometrical series; hence, when lim — — = 1 and the test fails to be de- n=oc 1l> n —i cisive, recourse is to the comparison test, as shown in Art. 200, for the harmonic series 1+ 2+3+ n^ EXERCISE XLII. 1. Find whether the following series are convergent or divergent: (i) l +* + * +!+••• . (2) I+J + I + J+---. (3) F2 + 2^3 + 3^1 + 4^5 + ' ' * ' (4) 1+I + I + I + I+.... (5) ^f! + f! + f!+---- 02 Q3 44 (6) l + 2! + 3] + 4!+'--- CONVERGENCE OF POWER SERIES 393 wl+l+T0 + :+5 n 2 + l (9) 1 +-^+ * 1 + VI 1 + V2 1 + V3 (10) l + ^ + | + ^ + ^ + 2. Examine the following series for convergency: (1) logf-logf +log|-logf + • • • . (2) sec = — sec -. + sec = — sec 3 + • • • . 6 4 5 o (3) sin 2 £ + sin 2 ^ + sin 2 ? + sin 2 ? + • • • . 2, 6 4 o 202. Convergence of Power Series. — A power series in x may converge for all values of x, as in (2), example of the Ratio Test ; but generally it will converge for certain values of x and diverge for others, as in Examples 1 and 2, Art. 198. Applying the ratio test to the power series, a + aix + (hx 2 + asx 3 + • • • + a n x n + • • • . (1) (Art, 199.) u n -\ a n -i ' im n=ao Un = lim ■ 71=00 1 &n— 1 = \x\] 1 im a n The series (1) is convergent or divergent according as | x | lim n=oo a n dn-1 < 1, or | x | lim 7i = SC a n fln-l >i; that is, according as | x \ < lim Gn-1 Cl n , or 1 3 1 > lim -^ The case | x | = iim a n -i a n is undecidec lb y thi 3 tes 3t. When, however, a power series is convergent for any value of x, say a, it is absolutely convergent for all values of x such that | x | < | a |. For example, given the series l+2z + 3x 2 + 4z 3 + • • • nx n ~ l + (n + 1) x" H ; (2) 394 INTEGRAL CALCULUS . a n -i n ,. n ,. 1 ■ here = — r-r , lim — — = lim r = 1 ; a n w + 1 n=00 n + l n=a0 ^ , 1 n hence (2) is convergent or divergent, according as |z|l; that is, (2) is convergent when — 1 < x < 1, and the interval from —1 to +1 is the interval of convergence. Here the interval does not include the end values,. (2) being divergent when | x | == 1. 203. Integration and Differentiation of Series. — A power series has the important property that, when the variable of the function is restricted to the interval of con- vergence, it is possible to get the integral or the derivative of the function by integrating or differentiating term by term the series which defines the function. Hence, if / (x) is defined by the power series, f(x) = a + aw + a 2 x 2 -f • • • + cinX n + • • • ; (1) then Jf(x)dx= f a dx+ I x aixdx-\- • • • + / a n x n dx+ • • • , X *J Xq t/x J Xq and df(x) = d(a ) d (aix) d (a n x n ) dx dx ■*" dx > ' ' "*" ~llx h ' ' ' ' when the restriction necessary to insure convergence is placed upon the value of x. Example 1. — For — 1 < x < 1, ^^ = 1 - x + x 2 - X s + • • • . (Ex. 2, Art. 198.) (2) Hence, Jn — = / dx — I xdx+ \ x 2 dx — I x z dx + • • • , o 1 + x Jo Jo Jo. Jo thatis, log(l+a:) = z-*! 2 + f- J+ • • • . (1) INTEGRATION AND DIFFERENTIATION OF SERIES 395 This is the logarithmic series with base e, and is true for — 1 < x = 1; that is, it is true for values of x within the original interval of convergence, including the end value 1; but for the other end value —1, it decreases without limit. On putting x = 1 in (1), log2=l-J + i-i+---= 0.69 .... (See Art. 200.) On putting x = — 1 in (1), log0= - (l + i + i + i+ • • ■ ) = -». (See Art. 200.) In the same way, for — 1 < x < 1, the integration of 1 — — = 1 + x + X 2 + x* + ■ 1 — 3! • • (Ex. 1, Art. 198) (1) gives log (1 — x) = — x — \ x 2 - . i r 3_i x 4 } (2) which may be gotten by putting —x for x in (1). This logarithmic series is true for — 1 = x < 1; that is, it is true for values of x within the interval of convergence, including the end value —1; but for the other end value 1, it decreases without limit. For x = — 1, (2) gives log 2 as above, and for x = 1, log as above. Hence by neither series can the logarithm of a number greater than 2 be found. Bv a combination of the two series the logarithm of any number can be found. By subtracting (2) from (1), 1 + x log (l + x} - log (i - x) = log yzt^ = 2(z+! + !°+- • •), for | x | < 1. (3) For x = \, logi±| = log2 = 2g + 3^ + ^+^+. • .) = 0.6931 .. .. For x = ^ ^[±1 = ^3 = 2(1+3^+^+.^+...) 1.0986 . 396 INTEGRAL CALCULUS This series (3) converges very much more rapidly for values of x less than 1 than the series (1), which converges so slowly that 100 terms give only the first two decimals correctly for the log 2, while (3) gives four decimals correctly taking only four terms of the series. Any number may be 1 + x put in the form , but it is necessary to calculate directly j. x the logarithms of the prime numbers 2, 3, 5, 7 only, as the others can be expressed in terms of these. Thus, 1 + - 5 5 log Yz^i = lo S 3 > and then lo § 5 = log - + log 3; and again, I _i_ i 7 7 log - f- = log -, and then log 7 = log - + log 5. 1 — 6 o o To get the common logarithms whose base is 10, multiply these natural logarithms by 0.4343 . . . , the modulus of the common system. (See Art. 38 and Ex. 6, Art. 215.) Example 2. — For — 1 < x < 1, — — = 1 + x + x 2 + x s + • • • . (Ex. 1, Art. 198.) (1) 1 — x By differentiation, 1+2z+3x 2 +4.t 3 H . (See (2), Art. 202.) (2) (1 - x) 2 By differentiation again = £(1. 2 + 2-33 + 3. 4s*+ • • • ). (3) (1 - xf 2 Hence, the general series, 1 t\ \ » 1 1 , m (m + 1) 9 = (1 — x)~ m = 1 + mx H ^-^tt 1 — -x 2 (1 - x) m v ' ' ' 2! + w (nt + l)( w + 2) g , + >> (4) INTEGRATION AND DIFFERENTIATION OF SERIES 397 Example 3. — For -1 < x < 1, by (4) of Ex. 2, or by division, rq^=i-z 2 +* 4 • (i) Hence, / — - — r = / dx — / x 2 dx + / x 4 dx— • • • , Jo 1 + x? . Jo Jo Jo that is, arc tan x = x — ^- + — — •••. (2) o o This is Gregory's series, named after its discoverer, James Gregory. Although series (1) oscillates when x = 1, series (2) is convergent and defines arc tan x even when x = 1. On putting x = 1, arctanl = j=l-g + -- 7 + • • • ; , ^(,-1+1-1+...). While the value of w may be found approximately from this series, the series converges so slowly that it is better to use other more rapidly convergent series, such as, and 7T 1 1 - = 4 arc tan - — arc tan ^r— (Machin's Series), 4 O Zoo - — arc tan - + arc tan - • (Euler's Series.) Example 4. — For - 1 < x < 1, by (4) of Ex. 2, = (l-x 2 )-* = l+^x 2 +-- i x*+ 7r - r -;xs + Vl-x 2 2 ' 2-4 ' 2-4-6 hence, C * dx . , 1 x 3 , 1 • 3 x 5 . 1 - 3 • 5 x 1 , J VT^ =arCSmX = a:+ 2-3 + 2T4'5 + ^4T6'7 + "-- This series, due to Newton and used by him to compute the value of T approximately, converges rapidly for x < 1. 398 INTEGRAL CALCULUS When x = J, this series gives . 1 7T 1 , * 1 1-3 1-3-5 arc sin - = - = -+ o » a + r> , r os + 2 6 2'2-3-2 3, 2-4.5.2 5, 2.4.6.7-2 71 To ten places, tt = 3.1415926536 .... By means of series the value of tt has been carried to 700 decimal places. J fa x fa (a 2 — e 2 x 2 ) 2 . This o v a 2 — x 2 can- not be integrated directly, but on expanding (a 2 — e 2 x 2 )* by the binomial theorem the terms of the resulting convergent series can be integrated separately. Thus, i P X P T (a 2 — e 2 x 2 )* = a — t— — ^-z — • • • , where e < 1, (1) 2 a 8 a 3 / 2 o 2\i t*»t/ " e X j Va 2 - z 2 (See Ex. 1, Exercise XXV.) • x J "<>> dx e 2 P* x 2 dx e 4 C a _^dx_ o Va 2 -x 2 2aJ Va?-x 2 8a 3 J Va 2 -f v - = Wi _ 6 _! _ /L3Yi 4 - / i-3-5 y c « \ m " 2 V 2 2 \2-4/ 3 V2-4.6/ 5 '/ w = a.T (1 -e 2 sin 2 0)*d0; T ? /i i • • «« i a • i* n ^ /See Ex. 6, Exer-\ = a\ (l-|e 2 sin 2 0-ie 4 sin 4 0- • • •) d0 • vvn Jo V cise XXII. / " 2( 1 -(2J g2 -(2T4J3-(2-T4T6J5 } (2) When a; = IT a sin 0, - e 2 z 2 )* — e 2 sin dx Vo 2 - 2 6)* dd .r 2 INTEGRATION AND DIFFERENTIATION OF SERIES 399 Example 6. — Given P , 2adX (See Art. 236.) Jo V2g{h-x)(2ax-x 2 ) This does not admit of direct integration, but on expanding it into a power series in x/2a the integral can be evaluated approximately. Thus, f h 2adx = i~a f h dx L _ x\-% Jo V2g(h-x)(2ax-x 2 ) VaJo Vhx - x 2 \ 2a/ by (4) of Ex. 2, dx m^m^Mh-i Vhx — x 2 by integrating, =^MI)' 2 V(H)'fe)' T h dcf>; [Art. 236 T Q Jo then, ,— «■ 2V- P(l -fc 2 sin 2 )"* o> y Q J « _ r [by (4) of Ex. 2.] = 2y^^l+i/c 2 sin 2 + |.|fc 4 sin 4 0+ . . . ) 0>, Ity integrating, 400 INTEGRAL CALCULUS When k is small, the approximate value of the integral is i l a again -k y -• Note. — The integral forms in Examples 5 and 6 are called elliptic integrals. The forms F - 7 =M== and f Vl - e 2 sin 2 d4> Jo Vl - /c 2 sin 2 Jo are known respectively as " elliptic integrals of the first and the second kind." a In the first kind k = sin ^ ; and, in Ex. 6, a is the angle each side of the vertical through which a pendulum of length a vibrates, the approximate time of a vibration being t y - , as found. (See Art. 236.) Tables give values for varying values of a. In Ex. 5, e is the eccentricity of an ellipse and 6 is the complement of the eccentric angle. By taking a few terms of the final series, when e is small, an approximate value of an elliptic arc of a quadrant's length is obtained. When e = the result is the length of a quadrant of the circumference of a circle. CHAPTER IX. TAYLOR'S THEOREM — EXPANSION OF FUNC- TIONS. INDETERMINATE FORMS. 204. Law of the Mean. — The mean value of / (x) between the values / (a) and / (b) is, by Art. 133, / f(x)dx f(c)- b — a where c is some value between a and b. If the function of x is (x) = f (x) and x x is some constant value between a and x, then x — a x — a or f(x)=f(a)+f(x 1 )(x-a), (1) the Law of the Mean, or Theorem of Mean Value. If the curve in the figure be the graph of y = f(x) ; then the ordinate at Pi will be f'(xi), the mean value of f'(x) between/' (a) at P a and/'(z) at any value x, the integral being represented by the area under the curve from x = a to x = x. If the curve is y = f (x) , it may be seen that there must be at least one point Pi between the points (a, / (a)) and (x, f (x)) at which the slope of the tangent is equal to the slope of the secant through those points; that is r(g j- /w-/w -|g,' J v ' x- a Ax and hence (1). 401- 402 INTEGRAL CALCULUS This may be put in the form, which may be used to determine increments approximately, and is the Theorem of Finite Differences. The theorem may be extended so as to express in terms of the second derivative the difference made in using the first derivative at x = a in place of its value at x = X\. Thus, if the function of x is (x) = f" (x), and x 2 is some constant value between a and x, then, J J" (x) dx _a f[ (x) - /' (a) /by mean value, \ J KX2) ~ x-a x-a ' \ Art. 133, ) or f'(x)=f'(a)+f"(x 2 )(x-a). Integrating this equation between the limits x = a and x = x, f (a) and /" (x 2 ) being constants, gives /(*) =f(a) +f (a) (x - a) +/" fe) ii^!, ( 2 ) a second Theorem of Mean Value, or the Law of the Mean. OTHER FORMS OF THE LAW OF THE MEAN 403 If the tangent at P a meet the ordinate MP produced at R, then, MR=f (a) + /' (a) (x - a) ; MP = f (x) , and, therefore, both in sign and in magnitude, RP = MP - MR = /" ta) ( * ~ a)2 - Here the deviation of the curve at P is below the tangent at P a , f" fe) being negative, and, measured along the line of ( x _ a )2 the ordinate MP, is equal to /" (x 2 ) - — s-^-. When the curve is above the tangent, }" (x 2 ) will be positive and RP upward. 205. Other Forms of the Law of the Mean. — The theorems (1) and (2) may be given in the following forms. In the theorems x, the symbol for the argument in general has been used for any value of the argument, a definite value but not constant. Now, if Xi be any number between a and x, then xi — a and x — a are of the same sign whether a is less or greater than x; therefore, (xi — a)/ (x — a) is a posi- tive proper fraction, 6 say, and xi = a + 6 (x — a) will denote any number between a and x. Letting x = a -\- h, x — a = h; then theorem (1) will become f(a + h) =f(a)+hf(a + dh), (l fl ) and theorem (2) will become / (a + h) = / (a) + hf (a) + |/" (a + eji) . (2 a ) The 0i of (2 a ) is not necessarily the same as the 6 of (1 ). If a is replaced by x, the forms become f(x + h) =f(x)+hf(x + dh), (h) f(x + h)=f(x) + hf (x) + |/" (* + M) • (26) 404 INTEGRAL CALCULUS If a is made zero and then x is put for ft, the forms are f(x)=f(0)+xf(dx), (l c ) f(x)=f(0)+xf(0) + p"(B l x). (2c) Example 1. — To find the value of 0, if f(x) = x 2 . Here f(x) = 2x; /'(a + 0ft) = 2(a + 0ft); and (a + ft) 2 = a 2 + 2aft + ft 2 = a 2 + ft • 2 (a + Jft); also, by (1 ), (a + h) 2 =/(o) + ft/' (a + 0ft) = a 2 + ft • 2 (a + 0ft); hence in this case 6 = \. To find at what point on the parabola y = x 2 , the tangent is parallel to the secant through the points where x = 1 and x = 3. By theorem (1), .*. xi = 2; or by theorem (1 ), Xi = a + 0ft = 1 + \ (2) = 2, since = ■§. Example 2. — To find at what point on the curve y = sin x, the tangent is parallel to the secant through the points where x = 30° and x = 31°. Here Mt N sin31°-sin30° 0.51504-0.5 n0£M _„ / (ft) = cos* = 31 o_ 30 o - 0.01745 = °' 86177; /. * = cos- 1 0.86177 = 30° 29'; hence y x = sin 30° 29' = 0.50729, = |J. Example 3. — To show that sin a; is less than a: but is greater than x — \x 2 . f(x) = sin a:; f'(x) = cos a;; /"(a;) = — sin x; /(0) = 0; f (0) = 1; /" (0!*) = - sin (0^). By theorem (l c ), sin a: = + x cos (0a:), < x, since cos (Ox) < 1. TAYLOR'S THEOREM 405 By theorem (2 C ), sin x = + x — Tr-sin (Six) > x — — , since | sin (Six) \ < 1. Example 4. — To show that cos x is greater than 1 — \ x 2 . f(x) = cos a:; f'(x) = — sinz; f"(x) = - cos a;; /(0) = 1; /' (0) - 0; /" (M = - cos (d lX ). By theorem (2 C ), cos a; = 1 — | z 2 cos (^io;) > 1 — \ x 2 , since | cos (fiix) | < 1. 206. Extended Law of the Mean. — The law of the mean or the theorem of mean value in its several forms may be used to obtain approximate expressions for a given function in the neighborhood of a given point x = a. Still closer approximations may be obtained from the law when extended in the form of a series arranged according to ascend- ing powers of x — a with the successive derivatives as con- stant coefficients. For values of x near to a, the higher powers of x — a may then become negligible. The most convenient theorem for this purpose is the one which follows. 207. Taylor's Theorem. — If f (x) is continuous, and has derivatives through the nth, in the neighborhood of a given -point x = a, then, for any value of x in this neighborhood, f(x)=f(a)+ f -^-(x-a)+ f -^f(x-ay+ ■ ■ ■ where X is some unknown quantity between a and x. The last term is the error in stopping the series with the nth term, the term in {x — a) n ~ 1 ; and the formula is of practical use only when 406 INTEGRAL CALCULUS this error becomes smaller and smaller as the number of terms is increased. The form of the remainder R n (x) is seen to differ from the general term of the series only in that the derivative in the coefficient of the power of (x — a) is taken for x = X instead of for x = a. The simplest proof of this theorem is the extension by integration of the law of the mean — a further extension than already used for the theorems of mean value. Thus, if the function of x is <£ (x) = /'" (x), and X is some unknown constant value between a and x, then, x /'" (x) dx = /'" (X) (x - a) (by mean value, Art. 133) = f" (x) - f" (a) = f" (X) (x - a). Integrating this equation between the limits x = a and x — Xj J 1 (x) - f (a) - f" (a) (x - a) = /'" (X) ^^, /" (a) and /'" (X) being constants. Integrating again, /' (a) also being constant, gives f(x) = f(a) +f (a) (x - a) +/" (a) ^-^ +/'" (X) (a 2-3 As this process can be continued to include the nth deriv- ative, by induction Taylor's Theorem results. 208. Another Form of Taylor's Theorem. — If in the form (1) x is put for a and (x + h) for x, it becomes six + h) =/(x) +/' (x) \+r (x>| + • • ■ +/ "" 1(a;) (S)! +/ " ( * + x +«*-+ ••• /-HO)*"- 1 , f n (ex)x n "*" (w- 1)! ** n\ ' w where the last term is the remainder after n terms, and is some positive proper fraction. Cauchy's form of the remainder is x n (l - 0)"- 1 R n (x) = /» (ex) (n-1)! 408 INTEGRAL CALCULUS 210. Expansion of Functions in Series. — It has been shown in examples of Chapter VIII on Infinite Series how useful it is to be able to represent a function by means of a series. Apart from the purpose of computation such rep- resentation may be an aid to an understanding of the proper- ties of functions. Taylor's Theorem and Maclaurin's Theorem furnish a general method of expanding or develop- ing any one of a numerous class of functions into a power series. For when the error term in Taylor's and in Maclaurin's Theorem approaches zero as n increases, each becomes a convergent infinite series, called Taylor's series and Mac- laurin's series for the given function, about the given point x = a. Some functions may be expanded by division, some by the binomial theorem, others by the logarithmic or the exponential series. All of these series are but particular cases of Taylor's Theorem. 211. Another Method of Deriving Taylor's and Mac- laurin's Series. — 1. Maclaurin's Series. — If a function of a single variable is expanded or developed into a series of terms arranged according to the ascending integral powers of that variable, and the constant coefficients found, the development will be the form of Maclaurin's Theorem with- out the remainder. Thus, let / (x) and its successive deriv- atives be continuous in the neighborhood of x = V say from x = —a to x = a, and assume that for values of x within that interval, f(x) = A + Bx + Cx 2 + Dx z + Ex" + • • • (1) If equation (1) is identically true, then the equation resulting from differentiating both its members, viz., f'(x) = B + 2Cx + 3Dx 2 + 4:Ex :i + also is identically true for values of x in some interval that includes zero. For similar values of x, the following equa- ANOTHER METHOD OF DERIVING SERIES 409 tions resulting from successive differentiations are identically true: /"(*) = 2C + 2'SDx + S'±Ex 2 + f'"(x) = 2'3D + 2>S'4;Ex+ • • ♦ . P y (x) = 2-3-4E + Putting x = in each of these equations gives : 4 =/(0), B =/mC=qp,D=qf^=«ete. Hence, on substituting these values in equation (1), /(z)=/(0)+/'(0)*+^^ + ^g!£!+ • •• + / ^r+--^ ' (2) which is Maclaurin's series as in (3), Art. 209, without the remainder. If f{x) is not continuous, no development according to powers of x is possible. Thus if / (x) = log x, f (0) = — oo . A power series represents a continuous function, hence no power series in x can be expected to develop log x. It is evident that, whenever the function or any one of its derivatives is discontinuous for x = 0, the function cannot be developed in a Maclaurin's series. 2. Taylor's Series. — Let / (x) and its successive deriva- tives be continuous in the neighborhood of x — a, say from x = a — h to x = a -\- h, and assume that for values of x near x = a, f(x) =A+B(x-a)+C(x-a) 2 + D(x-ay + E(x-ay+..., (3) is an identically true equation. Then the following equations resulting from successive differentiations are identically true for values of x near x = a. 410 INTEGRAL CALCULUS f (x) = B + 2C (x- a) +3D(x - a) 2 +4E (x - a)* + • ■ ■ . f"(x) = 2C + 2-SD(x-a)+3'4:E{x-a) 2 + • ■ ■ . /"'(a;) =+2-3D + 2.3-4E(a;-a)+ • • • . pv(x) = +2-3-4E + Putting x = a in each of these equations gives Hence, on substituting these values in equation (3), /m -/(«) + / -rf (* - a ) +*^r (* - °) s + • • • + £M(x-a)"+ ■ ■ ■ , (4) which is Taylor's series as in (1), Art. 207, without the remainder. If in (4) x is put for a and (x + h) for x, it becomes f(x + h)=f(x)+f'(x)^+f"(x)^+ ■ ■ ■ which is Taylor's series as in (2), Art. 208, without the remainder. Here the development is not according to powers of x, but of some value (x — a) or h near to the value x = a. Hence, when the values of the function and all its derivatives are known or can be found for some one value of x, say a, then the value of the function for x = a + h can be found from the development. Thus, when / (x) = logo:; / (1) = 0, f {1 ) = 1, /"(I) = -1, /"'(l) = 2!, . . . /(«)(!) . (-l)"+i (n — 1)!; and the series will be h 2 h 3 h A log(l + fc)=fc-J + J-j+ which agrees with (1), Ex. 1, Art. 203. EXPANSION BY THEOREMS 411 212. Expansion by Maclaurin's and Taylor's Theorems. — If, on applying to a given function any one of +hese formulas, the last term becomes 0, or approaches as a limit when n becomes infinite, the formula develops this function; if not, the formula fails for this function. That is, if R n (x) = when n = x , Maclaurin's or Taylor's series is the devel- opment of /(x) or/(x + h), respectively. If/' 1 (x) increases (or decreases) from / n (0) to/ n (x), and the sum of the first n terms in Maclaurin's series is taken as the value of / (x) , the error, being f n (6x) — . , lies between /•(0)Jj and />(*)fj- If f n (x) increases (or decreases) from f n (x) to f n (x + h), and the sum of the first n terms in Taylor's series is taken as the value of /(x + h), the error, being/" (x -\- 6h) —., lies between f(x) h ^ and f(x + h)^- SY*7l /v» /*• /y* jy* /y /y» 213. Since — = -.-.- . . -.-, n\ 12 3 n — 2 n — 1 n x n — i = w/ien x is finite and n = oo , for last factor approaches zero. Hence, #„=* (x) = when f n (dx) is finite, or when /" (x + Bh) is finite, in Maclaurin's or Taylor's Theorems, respectively. 214. If / (x) = / ( — x), the expansion of/ (x) will contain only even powers of x; while if / (x) = —f(—x) the expan- sion of/ (x) will involve only odd powers of x. For examples, see the expansions of sin x and of cos x following. 412 INTEGRAL CALCULUS 215. Examples. — 1. Sin x. — Expansion by Maclaurin's Theorem : f(x) = smx, /(0) =sinO = 0, /'(*) = cos z, /'(0) = 1, /"(*) = -sin*, f(0)=0, f'"(x) = -cosz, / ,,, (0) = -1, f"(x) = sinz, f* (0) =0, fn(x) = sm(x+ 7 fj, /«(0)=sing), f»(6x)=8m(ox + ™y Since sin f -^ J is or ±1 according as n is even or odd, the coefficients of the even powers of x will be zero, and only odd powers of x will occur, the terms being alternately positive and negative. Thus, x 3 , x 5 x 1 , , x n . I n , nir\ smx = x- - + --_+ • • • +_sin(to + -J. Here, R n (x) = — sin (dx + ~ ) , x n not numerically greater than —., which has zero for limit; .'. R (x) = 0. Hence the series n=oo /y»3 /y»5 /y»7 /v«9 S1M = I_ 3! + 5!"7! + 9!" " " ' (1) is absolutely convergent for every finite value of x. The series converges rapidly and may be used for comput- ing the natural sine of any "angle expressed in radians. Thus, for the sin (5°43'46".5 = T V radian), sin (0.1 radian) = 0.1 - ^^- + %r- 5 = 0.09983 o! 5! EXAMPLES 413 For the sin(l° = ^ radian = 0.017453 . . . Y sinl ° = sin (iio) = iio-(iio) 3 3 L ! + (ifo) 5i-"-= a017452 ---; .*. sin 1°= arc 1°, to five places of decimals. (See Art. 40.) ' 2. Cos x. — Expansion may be made by Maclaurin's Theorem as is done for the sin x, or the differentiation of the sine series term by term gives the series x 2 . x* x 6 , x 8 cos * =1 -2!+4!~6! + 8!~ " " ' (2) which is absolutely convergent for every finite value of x. For the cos (5°43'46".5 = T V radian), CO l) 2 (0 1Y cos (0.1 radian) = 1-^^- + ^^-- ■ • - = 0.995007 For the cos(l° = ^radian = 0.017453 . . . cosl° = cos^) = 1 - ( I ^)^ + ( i | g ) 4 ^ = 0.999847 . . . ; /. cos 1° = 1 - 0.00015 . . . 3. Sin (x + h). — Expansion by Taylor's Theorem. Here / (x -f h) = sin (x + h) ; f(x) = sinz, f(x) = cosx, f"(x) = — sin.r, etc. Hence, h 2 h 3 h 4 sin(x-\-h) = sinz+ftcosx — — f sinx— ^cosx+ — f sinx+ • • • . A h 2 h± fc« \ = ™ a5 ( 1 -2! + 4!~6! + • " ') + COBa: ( A "3! + 5!~7! + ' ' ') (1) = sin x cos h + cos x sin h, (h for x in Exs. 1 and 2) the well-known relation true for all values of x and h. 414 INTEGRAL CALCULUS h 3 h 5 h 7 When x = in (1), sin h = h — oy + Fj - 7J + as in Ex. 1 for x. The series (1) is rapidly convergent for small values of h. r . = 34'22".65; then, 7T 1 small values of h. Thus, let x = - and h = -r^ of a radian 6 100 . /f, iv ■ *7i (0.0: Sm U + 100; = Sm 6V 1 2! (0.01) 2 (0.01) 4 -(0.01 -»! + «!-... ); 4! 6 sin 30° 34'22".65 = 0.5 (0.99995 . . . )+ 0.86603 . . . (0.00999 ....)= 0.50863 . . . = 0.5 .+ 0.00863 . . . 4. Cos (x + h). — Expansion may be made in the same way as for sin (x -f- h), or the differentiation of the series (1), h being constant, gives * cos (x + A) = cosa; ^1 - 2j + jj - gj + • • • 1 sin a; A ft 3 ft 5 ft 7 , \ , Q . = cos x cos ft — sin x sin ft, the well-known relation. ft 2 ft 4 ft 6 When x = in (2), cos/i=1_ 2l+4|-g]+ ' " * > as in Ex. 2 for a\ When x = ! and ft = j~of a radian = 34' 22 ,, .65, in (2); then cos60°34'22".65 = 0.5 (0.99995 .'..)- 0.86603 . . . (0.00999 . . . ) =0.49132 . . . =0.5-0.00868 5. a x and e x . — Expansion by Maclaurin's Theorem. Here * /(*)-*■, /(0) = a°=l, f(x) = a* log a, /'(0) = loga, /"(x) = a*(loga) 2 , /"(0) = (log a) 2 , f'"(x)=a>(\ogay, f"'(0) = (log a) 3 , f n (x) = a* (log a) », /» (0) = (log a) », f n (6x) = a fe (loga)\ EXAMPLES 415 a log a , (a; log a) 2 (a: log a) 3 •' a=1+_ T" + 2! + 3! + '" n\ Since f n (x) = a x (log a) n , when a is positive, f(x) and all its successive derivatives are continuous for all values of x. When a; is finite, a te is finite. By Art. 213, v "° } = 0, (xloga) n ±_ ~n\ when n = oc and x log a is finite. Hence R n (x) = when n = ob' and z is finite. Therefore, the exponential series xbga (xlogo)^ (slog a)"" 1 "*■ 1 ^ 2! " r " " r (n-1)! - 1 l ; is the development of a x when a is positive and z is finite, being then absolutely convergent. Value of e x . — Putting a = e in (1), gives (since log e = 1) 5^1)1 Fcrfwe 0/ e. — Putting x = 1 in (2), gives * =1 + 1 + 2l + l! + i + -'-+(^!+--- = 2.718281 .... (See Art. 34.) 6. LogJ(l + #). — Expansion by Maclaurin's Theorem. Here / (x) = log a (1+ x), f (0) = log a 1 = 0, .., . m '-i+f + 21 + il + li- Nlf^lTT- w 1 w_ l+*' ./ \vj - in, /"W = -(TT^' /"(0) - -m, >"'(*> = (1 +V /'"(<)) = 2 m, 416 INTEGRAL CALCULUS (—l)n-l ( n _ l)\ m f n (x) - (i+ g )» ' /" (°) = ^" 1 ) n_1 ^ ~ W™* ; {m (i + dx) n :. loga (1 + x) ( x 2 , z 3 a; 4 . . (x-6x\ n f(-l) n - l \ When x < — 1, log a (1 + x) has no real value. When x = — 1, the odd derivatives are discontinuous. When x > —l,f(x) and all its successive derivatives are continuous. Cauchy's form of remainder, fl,(»)=/-(te) 8 ^_g | > gives B .( X ) = ^__J --^^-m, in which the second factor is finite, and the first factor = 0, when n = oo and x > — 1 and < 1 or x = 1. Hence the logarithmic series l0g a (1 + X) ( X*X* X*. , (-l)*- g 3»-l , \ is the expansion of log (1 + x) when # > — 1 and < 1 or x=l. Putting —x for x in (1), gives log a (l-z) = m(-z- |-|-| 4 + • • • ). (2) Subtracting (2) from (1), gives . l+x / . a: 3 a: 5 . x 7 . \ /Compare (3), \ , Q * log a j-j = 2m(x+ -+-+-+... J. ^ ^ 203 J (3) T i ! +1 1 +Z Z+ 1 fAS Let a: = 27+T ; then r^ = — • (4) EXAMPLES 417 Substituting in (3) the values in (4), gives log/-±i = 2m (^ + 3(2 / +1) , + •••); (5) •• log a (^+l) = loge g +2m(^ I + 3(22 1 +1)3 + ...). (6) When 2>0, 0<£<1; hence the series in (6) is conver- gent for all positive values of z. When 1 is put for z, log a 2 is found ; then 2 for z, and log a 3 is found; thus log a (z + 1) can be readily computed when loga z is known. (See Ex. 1, Art. 203.) Whenra = 1, a = e, the Napierian base; thus (5) becomes log "-T i = 2 (2iTl + sW+W + " ' • )• 2 + 1 Dividing (5) by (7) and denoting — — by N, gives (7) logaAVlogN = m; that is, log a N = mlogiV. (8) Valve of m. — Putting N = a in (8), gives 1 = mloga; that is, m = 1/loga, (9) or N = e, gives, log a e = m\ :. 1/loga = loga c, or 1 = log a • log a e. (10) Valve of M. — Denoting the modulus of the common system whose base is 10 by M; from (9) and (10), M = io^io = logloe 0.434294 .... (Compare Art. 38.) 2.302585 Note. — When a = 10 and m = M, or when a = e and m = 1, all the series in this example become common loga- rithmic series, or Napierian logarithmic series, respectively. 418 INTEGRAL CALCULUS 0-i4 6. e* = l+z + ! | + 3 l + ! ] +.... o»2 /j*3 ^t^4 6. e~ x = 1 - x + — - ^ + -j - • • • , by replacing x by -x in 5. 7. sinh x = ^ — = x + q"| + 51 + * e x + e~ x x 2 x 4 8. cosh x = = = ^ + o7 + 41 + ' ' ' > ^y combining terms of 5 and 6, or by differentiating terms of 7. A , sinhx x 3 . 2X 5 17 x 7 . 9 - tanha; = J35^ = :c -3 + T5--^l5- + ---- 10. From 5 and (1) and (2) of Examples 1 and 2, making x = V— 1 • x = ix, get ei* = cos x + i sin x, (1) and e -ix = cos x — i sin x. (2) Note. — Putting t for x in (1) gives the remarkable relation, e l7r = — 1 ; while putting — -w for x in (2) gives e l7r = 1, hence iir is an imaginary value of log 1, the real value being zero. 11. From (1) and (2) of 10, by subtraction and by addition get e ix _ e -ix e ix _|_ e -ix sin x = — (3), cosx = ^ ( 4 ) 12. Evaluate f e^ dx = f (1 - x 2 + ix 4 - ix 6 + • • • ) dx. Get result when end value is 1. When end value is 00 the value of the integral is \ v^.* This integral is important in the theory of probability. * Williamson's Integral Calculus, Ex. 4, Art. 116, also Gibson's Calculus, Ex. 3, Art. 136. THE BINOMIAL THEOREM 419 216. The Binomial Theorem. — I. The binomial theorem is seen to be a special case of Taylor's Theorem by expanding (x + h) m in a power series in h by Taylor's formula. Thus, f(x + h) = (x + h)™; .'. f(x)=x m , f (x) = mx m ~ 1 , f " (x) — m (m — 1) x m ~ 2 , /'" (x) = m (m - i) ( m - 2) x m ~ 3 , f n ~ 1 (x)= m (m — 1) . . . (m — n + 2) x m ~ n+1 . Substituting these values in Taylor's series (5), of Art. 211, (x + h) m = x m + mx m ~ l h + m( ra~ 1 ) a; m-2^2 + m (m - 1) (m - 2) ^ m _ 3/i3 + . . . o . . , . m(m-l) . .(m-« + 2) x „_„ +1/t „_ 1+ . . . ^ (n — 1)! is the resulting Binomial Theorem. Here f n (x) = m (m — 1) . . . (m — n + 1) z m-n . Hence, / (x) and all its successive derivatives are continu- ous for all values of x. h n (\ — 6) n ~ l Cauchy's form of remainder, R n (x) =f n (x+dh) . _ n> , gives m (m - 1) . . .(m-n + 1) ( h - dh \ n (x + dh) m Kn W- ( n - 1)! \x + 0h) ' 1-0 When | X \ > h and n = oo , the product of the first and second factors = 0, and the last factor is finite; hence, R (x) = 0. 71=00 Hence, the binomial theorem holds true when the first term of the binomial is greater absolutely than the second. When m is a positive integer, the series (1) stops with the (m + l)th term, since/ 71 (x) = when n > m, and is therefore a finite series of m + 1 terms, 420 INTEGRAL CALCULUS If | h | > x, the expansion may be a power series in x ; thus, /7 , n 7 , 71 t m(m— 1) h m ~ 2 . . (h + x) m =h m + nth™' 1 x -\ x? x + * ' ' m(m-l). . . (w - n + 2) fe"-" +1 , , . ...+ ^— ^ x + ... (2) is a true expansion when \h\> x. II. (1 -f x) m may be expanded in a power series in £ by Maclaurin's Theorem, giving /-, , n -, , i m(m—l) , . m(m— l)(m — 2) , . (1 + a:)» = 1 + ?nx + 2| , V + — * ^ -x* + • • ' . . ■ ^(m- 1) . . . (m-n + 2) .-.+ (n _i)! x + -W As in case I, when m is a positive integer, the series (3) stops with the (rn + l)th term and is therefore a finite series of m + 1 terms. If m is negative or fractional, the series is infinite. The ratio test shows that the infinite series con- verges absolutely when | x \ < 1 and diverges when \x\> 1 ; therefore R n (x) needs examination only f or | x | = 1. Here f n (x) = m (m — 1) . . . (m — n + 1) (1 + x) m ~ n , f n {Bx) = m (m - 1) . . . (m - n -f 1) (1 + Bx) m - n . x n n _ 0)n-l Cauchy's form of remainder, R n (x) =f n (dx) — t -y. — , gives For values of x between and ±1, the last factor is finite for every n ; the second factor is always positive and cannot exceed unity; the first factor approaches zero as a limit as n increases without limit, since it is the expression for the nth term of the convergent series l + (m-l), + (w - 1) 2 j m - 2) ^+---. APPROXIMATION FORMULAS 421 Hence, R (x) = 0, and the infinite series converges to n=oo (1 + x) m for every value of m, when | x | < 1. For x = dbl, the following results may be found proved in Chrystal's Algebra. These cases are not so important. When x = +1, the series converges absolutely if m > 0, but conditionally if > m > — 1, oscillates if m = — 1, and diverges if ?n < — 1 . When x = — 1, the series converges absolutely if m > 0, and diverges if m < 0. If a > b, (a + b) m can be written a m (1 + b/a) m and expanded by Maclaurin's formula, since b/a < 1 may take the place of x in (1 + x) m . Hence, in this case, (a + b) m =a m + ma m ~ l b + - ^^ 1} a m ~ 2 b 2 + ■ • • , (4) which agrees with (1) and is the Binomial Theorem, proved true for a > b whether m is positive or negative, whole or fractional. If a < b, interchange them in (4) and the result will agree with (2) and be a true expansion of (b + a) m . 217. Approximation Formulas. — Often a function may be replaced by another having approximately the same numerical value but a form better adapted for computation. In such cases the given function may be expanded in a series and a certain number of terms, beginning with the first, taken as an approximate value of the function; the number of the terms taken being according to the precision desired for the result. The binomial theorem furnishes one of the most useful of the approximation formulas. Thus, if m denotes a small fraction, expanding (1 ± m) n gives (1 dbf»)» = 1 ± nm + n {n ~ 1} m 2 ± • • • , 422 INTEGRAL CALCULUS where, since m is small, neglecting powers higher than the first, the approximate relation, (l±m) n = 1 ±nm (1) results. For the special case n = J, Vl±m = 1 ± | m. (2) For 6 small in comparison with a, the general form is V?±& = a(l±^)- (3) For examples: Vl +x = l + Jx— • • • ; = 1 — x + • • • ; , = 1 — -x + • • • . 1+z r Vl+x 2 For extraction of roots in general i (a n ± b) n = ail ±— J =a(l±x) n , (4) 7 l where x = — . Expanding (1 ± x) n gives (l.^T-ljA (n-D^ x (n-l)(2n-l) ^ (i±xj -i± n « n2 2! ± n3 3! • (s; Example. — v'lOOO = ^1024 - 24 = 4 (1 - T yi Substituting T f ^ for x in the series ( )0 5 5 10 5 10 15 gives to six figures 0.995268; hence, v / T000 = 4 X 0.995268 = 3.981072. Since e x = 1 + x + — . + —.+ • • • , when x is small e* = 1 + x (6) is the approximate relation. APPROXIMATION FORMULAS 423 From the series for sin x, cos x, and log (1 + x), sin x = x (1 — \x 2 ) } cos x = 1 — \ x 2 , log(l +z) = x - \x 2 , are the approximate relations when x is small. When x is small compared with a, sin (a ± z) = sin a d= £ cos a, log(a + x) =1 °g a +^-2^' 1 1 # a; 2 - -F -^ H — ^' a a 2 a 3 (7) (8) (9) (10) (11) (12) a ±x are approximate relations when succeeding terms of the respective series are neglected. -^ In all these cases the error made in taking the approxima- tion for the value of the function may be closely estimated from the value of R n (x), the error term, for the particular series em- ployed. Example. — In considering the length of a circular arc and its corresponding chord in railway surveying, use may be made of the approximate relation (7). Thus, letting s denote length of the arc, r the radius, c the chord, a the angle in radians; s = ra and When a is small, -»'*i-»i|>-gS)l 2 r sin — ra — 3*5 ra 6 or for a in degrees, s — c = s — c ra A 4514180' s 4 re? Since the error of the approximation cannot exceed ra° 1920 424 INTEGRAL CALCULUS EXERCISE XLIV. 1. Expand (x + y) m . 2. Expand (x + yf. 3. Expand e x+h . 4. Expand log sin (x + h). 6. Expand sin" 1 (x + h). 6. Expand e ainx . 7. Given / (x) = x 3 - 4 x + 7, find / (x + 3) and / (x - 2) by Taylor's series. Then find / (x + 3) and / (x — 2) by usual algebraic method and thus verify results. 8. Using the approximation formula (12) compute the reciprocal of 101; and of 99. Compare results with those obtained by division. 9. Find the length of the chord of an arc of radius 5729.65 feet subtending an angle of 1°: (a) by trigonometric methods; (6) by the approximation formula (7). Find results when the radius is 5729.58 feet. Compare results and find error of approximation. 10. Find the length along the slope of a road that rises 5 ft. in a horizontal distance of 100 ft. by the approximation formula (3). Deter- mine to how many places of decimals is the result correct. 218. Application of Taylor's Theorem to Maxima and Minima. — This Article is supplementary to Art. 83, being an additional proof of the rule given in that Article for the determination of whether a critical value x = a, a root of f'(x) = 0, makes f(x) a maximum, a minimum, or neither. Let / (x) be a function of x such that / (a + h) and f (a — h) can be expanded in Taylor's series, and let / (a) be the value to be tested. Developing f (a — h) and / (a + h) by formula (2), Art. 208: f(a-h)=f(a)-hf(a)+^f'>(a)-p'"(a)+ • . . f(a + h) - /(a) + hf (a) + g/(a) + J/'". (a) + • • • + y$lf(a + (a - 6,h), (3) /(a + *)-/(«)-j^(a + «k). (4) Since / (z) and its successive derivatives are assumed to be continuous at and near x = a, the signs of /" (a — 0i/i) and /" (a -+- 2 ft), for very small values of ft, are the same as the sign of f n (a) . It is manifest that if n is an even integer, / (a) will be a maximum or a minimum according as f n (a) is negative or positive; and if n is odd, f (a) will be neither a maximum nor a minimum whether the sign of /" (a) is negative or positive. These conclusions are manifest because when n is even and f n (a) is negative, the left members of (3) and (4) are both negative, and hence f (a) > f (a — h), f (a) > f (a + h); that is, / (a) is a maximum. When n is even and f n (a) is positive, the left members are both positive, and hence/ (a) < / (a — h),f (a) < / (a + h) ; that is, / (a) is a minimum. When n is odd, whether f n (a) is negative or positive, the left members have different signs, and hence/ (a) >/ (a — /*), / ( a ) >/ ( a + ft) ; that is, / (a) is neither a maximum nor a minimum regardless of the sign of f n (a). 219. Indeterminate Forms. — It was noted at the end of Art. 20 that the derivative of f(x), may be finite, zero, or non-existent, but not 0/0. The symbol 0/0 is called an indeterminate form, and when / (x) takes that form for some value of x, say a, then / (x) is really undefined for x = a, although it may be defined for any other value of x. It is possible, however, that / (x) 426 INTEGRAL CALCULUS may have a definite limit A when x converges to a; it is customary then to call / (a) = 0/0 an indeterminate form, and to define A as the value of / (x) when x = a, calling it the true value of / (x) for x = a. The advantage of having this "true value" assigned by definition is that/(V), being in general continuous, thereby becomes continuous up to and including the value a. x 2 — 4 Take, for example, the function y = — . For every X z value of x other than x = 2, the function has a definite value, 4 — 40 but for x = 2 it becomes - - = ^ . Since the function has £ Ll U no definite value when x = 2, the limit which the function approaches as x converges to the value 2 is assigned as the value of the function when x = 2. If h=o I -\- h — A h±Q x 2 — 4 .\ lim = 4. x ±2 x - 2 Thus the true or limiting value of this function which takes the indeterminate form 0/0 is 4. For values of x other than 2, .-. X —± = lim (x + 2) = 4. X — 1 J J= 2 x=2 On the graph of y = x + 2, the ordinates of points for values of x other than 2 represent the values of the function, but for x = 2, the function having no definite value may be represented by any ordinate lying along the line x = 2. Of the values that may be assigned to the function for x = 2, there is one value represented by MP = 4, which is the limit of the values represented by the ordinates of points on INDETERMINATE FORMS 427 y = x + 2 as x approaches 2; and it is desirable to select this value of y as the value of the function when x = 2. By this selection the function is defined for x = 2 and thus becomes continuous through that value of the variable x. — x In general, lim / (x) defines the value of the function when x—a fix) is indeterminate for x = a. The expression f (x)] a denotes the value of f(x) when x = a. The value of a function of x for x = a usually means the result obtained by substituting a for x in the function. When, however, the substitution results in any one of the indeterminate forms, 0, oo /oo, 0.x, x - oo, 0°, x 1 the definition must be enlarged; thus, the value of a function for any particular value of its variable is the limit which the function approaches when the variable approaches this particular value as its limit . This definition need be used only when the ordinary method of getting the value of the function gives rise to an indeterminate form. 428 INTEGRAL CALCULUS 220. Evaluation of Indeterminate Forms. — In many cases the limits desired are easily found by simple algebraic transformations or by the use of series. When the function that assumes the indeterminate form is the quotient of two polynomials, or can be put in that form, the following direc- tions may be of service. fix) 1. If the function is of the form t-t-( and becomes 0/0 for 4>(x) x = 0, divide both numerator and denominator by the lowest power of x that occurs in either. If the fraction becomes oo/oo for x = oo, divide both terms of the fraction by the highest power of x in either. fix) 2. If the function has the form , ; and becomes 0/0 for 4>(x) x = a, divide both terms by the highest power of (x — a) common to both EXAMPLES x 3 + 3 x 2 — 5 x _5 3x 4 - 2x 3 + 6x|o 6' When x = 0, this fraction takes the indeterminate form 0/0. Hence to evaluate it for x = 0, its limit when x = 0. must be found. For values of x other than 0, x 3 + 3 x 2 - 5 x x 2 + 3 x - 5 . 2. 3x 4 - 2x 3 + 6x 3x 3 - 2x 2 + b' x 3 + 3x 2 -5x ,. x 2 + 3x-5 S 3x 4 - 2x 3 + 6x ~ S2J 3x 3 - 2x 2 + 6 5 6 x 3 + 3 x 2 - 5 x 1 3x 4 -2x 3 + 6x_L 13 5 x 3 + 3 x 2 — 5 x x x 2 x 3 . 3x 4 -2x 3 + 6x 2 ,6' X X 3 EVALUATION OF INDETERMINATE FORMS 429 x 3 + 3 x 2 - 5 x £s,3 i + i_5 .. x^x 2 x 3 n ■ lun — 2~T =3 = °' 3 - - + - 3 x x 3 3. vrr vr Jo By rationalizing numerator, Vl+x- Vl-x _ / Vl+x- Vl-aA / Vl+a + Vl-sA 2 vi + x + vi V lim r x=o L Vl +x- VI J x=0 - o Vl + a; + 1-aJ 4. Vl + x - Vx 1 = 0. By changing form, vT+~x - Vx = (Vi+x - Vx) 1 + x — X (Vl+x + Vx) (Vl+x + Vx) Vl +x + Vx' . lim (Vl+x - Vx) = lim /T-r- , — 7= = °- x = oo * = x|_Vl + £ + VxJ a; — sin x~] 1 "Jo = 6 x 3 By expanding sin x in series, x — sin x x 3 x 6 5! + 7! lim' i=0 x 3 sinx ?[ x -{ x -i + i. )] • • • , if * * 0; 430 INTEGRAL CALCULUS 6. smx-x cos x 1 x c JO By expanding sin x and cos x in series, sin x — x cos "~a?L\ 3! + 5! ( x 4 .or 2! + 4! z=0 L X COS 5 ]= )] lim x=0 = lim x=0 -( X 3 \ 30 + 30 + )] B 221. Method of the Calculus. — For the form 0/0, to which all other indeterminate forms may be reduced, Tay- lor's Theorem furnishes a general method of evaluation. I. When/(#) and (x) are continuous functions of x and reduces to the form 0/0 for x = a, the value of lim 2 is desired. (1) That is, if £/ie ratio of two functions of x takes the form 0/0 when x = a, then the ratio of these functions when x = a is equal to the ratio of their derivatives when x = a. If / (x + h) and (x + h) can be expanded by Taylor's formula in the neighborhood of x = a, it is seen that »- (MWI = 9f\ ■< ^t is, \tm = pi . (1) By Taylor's Theorem [Art. 208, formula (2)], putting a for z, /(o + fe) = f W+hf'ja + Oji) = /'(a + fl.ft) (a + /*) (a) + /i0' (a + 6 2 h) 0' (a + 2 /i) ' since /(a) = 0, 4>(a) = 0; [See also (1„), Art, 205] METHOD OF THE CALCULUS 431 + W + 2 h). fM. (1) If /'(a) and 0' (a) are both zero, then [(2 a ), Art. 205] fc -o |> (a + h) J ft=o U (a + Oji)] " (a In this way it is seen that if, for z = a, f (x) and (x) and their successive derivatives, including their nth derivatives, are zero, while f n+1 (a) and (f> n+1 (a) are not both zero, then lim r/j£)] =lim r/!^Mi ; that is , \mi jgm (2) fix) If the function , \ takes the form 0/0 when x is infinite, {x) by putting x = - the problem is reduced to the evaluation z of the limit for z = 0, and hence the method applies to this case also. fix) II. Form go /go . — When the function : ^-4 takes the form oo /qo, it can be reduced to the form 0/0, by writing it in the form —7-^ /tt-t. This form can be evaluated as before. 4>(x)/ fix) Thus, let f(a) = go and (a) = 00, a being finite or in- Now 2-^7 = —p-r / 77-r , which is in the form 0/0. Ap- (a) 0(a)/ /(a)' plying formula (1), 432 INTEGRAL CALCULUS r/Ml [0W1 2 _ r/W T £M = [7MT R^l • U(*)J.~ fw U(A)J '/'(«) Uwl'L/'^J.' [/(«)? r/wi . jl = rrwi . Uc*oJ« kw i U'wJa If ,, { is indeterminate, continue according to formula (2) until two derivatives are obtained whose ratio is deter- minate, which ratio is the limiting value sought for the function. III. Other Forms. — The evaluation of the other indeter- minate forms may be made to depend upon the preceding. (a) Form • oo . — When a function f(x) • (x) takes the form • oo for x = a, it may be reduced to the form 0/0 or oo/oo ; thus, /».#«- *g or *g. (6) Form oo — oo . — By some transformation and simpli- fication, a function taking the form oo — oo may be reduced to a definite value, or to one of the preceding indeterminate forms. (c) Forms 0°, oo°, l 00 . — These forms arise from a func- tion of the form [f(x)]^ x) . This function may be reduced to the form #. Thus let y = [f (x)]* (x) , whence \ogy = 4>(x).\og[f(x)]. (3) Since for each of the given forms, (3) takes the form O.oo, the evaluation is effected as in (a), the value of y being found from log y. METHOD OF THE CALCULUS 433 EXAMPLES. x 2 -4 x-2 1- 4. x 2 -4 1 = 0. x-2j 2 0' x-2 2x1 1 J 2 4. (As in Art. 219.) x-sinx Jo 6 sin xl _ . . x — sin xl 1 — cos xl ? Jo ~ ' " x 1 Jo ~ 3X 2 " J _ sin xl _ cos xl _ 1 6 x Jo 6 Jo 6 - a n l . = na"- 1 . - a ] a n — a n l x n — a n l nx n ~ l l . = n^ •'• = — \ — = na"- 1 . x — a J a x — a Ja 1 J a . a x - ¥1 . a 4. = logr- £ Jo O a* - fr x J ° • a* - 6*1 . x . , . 1 ■ = - , .'. = log a • a 1 - log 6 • b* \ X Jo X Jo Jo x rt — z log a — log b = log gx _ e - x _ x — e — e~ x — 2 x x — sin x 1^£T = 2. sin x Jo 1 _ ? . ^ - e~ z - 2 x ] _ e* + e~ x - 2 1 Jo ' " x — sin x Jo 1 — cos x J ^ gx - e~ x l J&j±jrf\ sinx Jo cosx J Um Llog(l+«)l=?5S(i±^l __•_] = ffl) W here,= i. X = ocL \ X l\ Z Jo 1+02 Jo X 434 INTEGRAL CALCULUS /Compare \ e ' l Art. 34. / log(l+-)*] = zlog(l + j)] = a (by Ex. 6); - H)1>- - K)1 8. (1 + xY 1 = e. (1+^=1-; ••• ^(1+^=^^(1+4=^=1; .*. (1 + z) x - e. (Compare Cor., Art. 34.) 9. (1 - a;)*] = - or e-K Jo « iog(i-^] o =i.io g (i-*)] o =^y o =-i. .-. (l-x)^ = e~ x or EXERCISE XLV. Evaluate the following indeterminate forms: 1 — cos x x 1 Jo 2" 9. x- 1 lJi n 3. tan a: — sin 4. (sinx) tar 2. sec x — tan x L = 0. - sin af| _ 1 *X Jo~2* Jr 1 - 6. z sin:c l = 1. Jo 6. sin x log z]o = 0. 7. s*]o = 1. 8. (l+x^] = l. 12 . £l^!l = 2 . sm.T Jo a; — sin" 1 .c ~| _ 1 sin 3 x Jo 6 14. tang-g-1 = 2 a; — sin x Jo 15. (log *)*]„ = 1. 16. aF^-e" 1 . EVALUATION OF DERIVATIVES 435 222. Evaluation of Derivatives of Implicit Functions. — 1. Find the slope of a; 4 - a 2 xy + b 2 y 2 = at (0,0). Here ?1 "TT^Vll -J" ( Art - 105 )' cfojo.o a 2 a; - 2 6 2 2/Jo,o .2^ Hence -/ ax in.o dec 12 x 5 da; 2 6 2*/ da; dx 1.0 a*- 26*^ dx. 0,0 ... * („, _ 2 &*^U «. *] =0; whence *] = or £ 2. Find the slope of a; 3 - 3 axy + y 3 = at (0,0). Ans. -rM = or oo . aa;Jo,o CHAPTER X. DIFFERENTIAL EQUATIONS. APPLICATIONS. CENTRAL FORCES. 223. Differential Equations. — A differential equation is one that involves one or more differentials or derivatives. An ordinary differential equation is a differential equation which involves one independent variable only. The deriv- atives in such an equation are therefore ordinary derivatives. The order of a differential equation is the order of the highest differential or derivative which it contains. The degree of a differential equation is that of the highest power of the highest differential or derivative which it con- tains, after the equation is freed from fractions and radicals. For examples : dy = mdx, (1) are ordinary differential equations. Equation (1) is of the first order and first degree, (2) is of the second order and first degree, and (3) is of the second order and second degree, after being rationalized. 224. Solution of Differential Equations. — It has been seen in foregoing chapters how when an equation expressing a functional relation between two variables is given, the differentiation of the equation gives a differential equation expressing the rate of the function. On the other hand, it has been seen that the rate of a function being given a differ- 436 COMPLETE INTEGRAL 437 ential equation is thereby formed, the integration of which yields the function, indeterminate though it may in general be. It is this rinding of the function from an equation involving the derivative that constitutes the solving of a differential equation. The general solution of a differential equation is the most general equation free from differentials or derivatives, from which the given equation may be derived by differentiation. The general solution, for example, of the equation g = C, is y - Cx + Ci, and of ■— = 0, is y = Cx + C h also. dv d 2 v Thus, when -~ represents the slope and -t~ the flexion, this function is any non-vertical line in the plane. Here, y = Cx, y = Ci, y = 0, y = x, y = 2 x + 1, . . . are par- ticular solutions, which are included in the general solution. (See Ex. 1, Art. 115.) [Note. — The general solution of a differential equation may not include all possible solutions. A solution not in- cluded in the general solution is called a singular solution. The discussion of such solutions is beyond the scope of this book.] The general solution of a differential equation of the nth. order contains n arbitrary constants of integration (for ex- amples, see Arts. 140, 141, 161); to determine these constants, n conditions connecting the function, the variable, and the successive derivatives must be known. The general solution is called the complete integral or primitive of the differential equation. 225. Complete Integral. — When a differential equation is given, passing by integration to the complete integral is 438 INTEGRAL CALCULUS solving the equation. It has been proved that every differ- ential equation has a complete integral, and that when the equation is of the nth order the integral contains n arbitrary constants. The complete integral then contains one, two, ... or n constants that do not appear in the differential equation, when that equation is of the first, second, ... or nth order. If the complete integral be differentiated these constants are eliminated, whatever may be their particular values, hence they are called arbitrary. Example 1. — Let a given equation be y = f p + k; (i) dy x differentiating, -j- = - ; (2) differentiating again, -^ = - = - ■£ (from (2)) ; (3) ••• *S-§ = °> « 2 ) Art - 223) W is the differential equation. To find a general form for the complete integral of (4), du d?y dm solve by letting m = -~- ; then -=-* 2 = -7- ; substituting in ... . dm _ dm dx (4) , gives x -; m = 0, or — = — ; dx m x integrating, log m = log x + log c = log ex, where c is a constant; dy /rk ,x .'. m = ex, or -j- = cx\ (2 ) integrating again gives 2/ = !* 2 + C!, (I') the complete integral, in which c and C\ are the arbitrary constants; and which has the form of equation (1). Whatever be the value of C\, equation (l 7 ) represents a COMPLETE INTEGRAL 439 2 parabola on the y-axis with latus rectum - ; hence (2') is the c differential equation of all such parabolas. Equation (4) is the differential equation of all parabolas whose axes are on the y-Sixis. Suppose (4) is given, and the problem is to find a function y that shall satisfy that equation, have its first derivative equal to 1/p when x = 1, and be equal to k when x - 0. These conditions give, from (1') and (2'), fc = + ci; l/p = c, and hence the function, y = x 2 /2 p + fc. (Compare Ex. 2, Art. 161.) (1) In this way the constants can be determined when the necessary conditions are known. N.B. — Equations of the second order with one variable absent, of the form j '(yf , ~r, x) = 0, may often be solved by the method used in this example. (Art. 232, III.) Example 2. — Let the equation be g + 2 b g + (V + <*) y = 0. ((4) Art. 54.) (1) It is seen in Art. 54 that this differential equation results from the differentiation of the equation y = ae~ bt sin (ut -a), or y = e~ bt (A sin ut + B cos ut) . (2) To show that (2) is the solution of an equation of the form of (1), let y = e~ ht u and (1) becomes C ^ + ^u = 0, (3) where co 2 = b 2 + co 2 - (i»2 6) 2 ; then by Ex. 3, following, u = A sin mt + B cos wt, and y = e~ bt (A sin ut + B cos ut). (2') (Compare Art. 233, III.) 440 INTEGRAL CALCULUS Example 3. — Let the equation be ^|+co 2 u = 0. ((3) of Ex. 2.) Multiplying by 2 du gives integrating, /^Y = -^ ( M a + Ci) = co 2 (a 2 - w 2 ), taking ft = -a 2 , extracting root, -r- = to Va 2 — w 2 . Integrating, / M = l udt |/ gives sin -1 - = co£ + C 2 , or, solving for w, w = a sin (co£ + Cg) = A sin co£ + 5 cos ut, where A = a cos C 2 and 5 = a sin C 2 are arbitrary constants. Example 4. — Let the equation be Multiplying by 2 du gives /ciiA 2 integrating, i-^j = co 2 (u 2 + Ci); extracting root, -rr = co Vu 2 + Ci. Integrating, / . = / co eft, gives log (w + Vu 2 + Ci) - co£ + C 2 , or, solving for u, u = Ae" 1 + £e^, THE NEED AND FRUITFULNESS 441 where 2 A = e°* and 2B = — de _C2 are arbitrary constants. By means of the hyperbolic functions this result may be written in the form u = a sinh (coZ) + b cosh (co£), where b + a = 2 A and b — a = 2 B. Hence, in this case, a solution of (1) of Ex. 2 is y = Ae-^-^ 1 -^ Be-^+^K Example 5. — Let the equation be ^ = dt 2 u ' resulting from (3) of Ex. 2, when b 2 + w 2 = 6 2 , or co 2 = 0. Integrating gives -77 = &, t* = Cit + C 2 . Hence in this case, the solution of equation (1) of Ex. 2, is 2, = e -u [p xi + c 2 ), where C\ and C 2 are constants. Note. — The foregoing examples are solutions of important differential equations, that of Ex. 2, as shown in Art. 54, being the typical form for damped vibrations. 226. The Need and Fruitfulness of the Solution of Differential Equations. — Attention has heretofore been called to the need of finding the inverse of a rate, in solving many problems that arise in everyday life as well as in science. In fact the inverse problem is more often the real question demanding solution. It has been shown (Ex. 5, Art. 115) how, when the acceleration, the rate of change of the speed of a moving body, is known, the velocity and the distance for any time are found by the solving of a differential equation. (Ex. 1, Art. 161.) It has been shown (Art. 42), that when a function has the general form y = ae bx , the rate of change is proportional to the function itself, and that so many changes in Nature 442 INTEGRAL CALCULUS occur in this way that the law of change, known as the Compound Interest Law, is also called the Law of Organic Growth. Now, if it is known that some function changes at a rate proportional to itself, expressing this by the differen- tial equation, | = %, or kdx = d f, then, kx = f < ^ = \og e y + c, or y = e kx ~ c = Ce kx , J y where C = e~ c is an arbitrary constant. The only function whose rate of change is proportional to itself is thus shown to be of the form Ce kx (or ae bx ), where C and k (or a and b) are arbitrary, and k (or b) is the factor of proportionality. This may be expressed also by the statement, that the only function, whose relative rate of change (logarithmic derivative) is constant, is Ce kx (or ae hx ). It has been shown (Art. 73), that when a point has simple harmonic motion its relative acceleration is a negative con- stant. Thus, when the displacement of a point is given by the equation y = a sin (cot — a), there results the differential d?u equation —; = —o> 2 y, where co is constant, and hence the relative acceleration is -^ / y = — co 2 . Conversely, when the motion, as in a vibration, is due to a force that increases with the distance from the central position, the acceleration, being according to Newton's second law of motion proportional to the force, is d 2 s 72 at = de = ~ ks > where, as the force acts towards the origin, the acceleration is negative when s is positive and positive when s is negative. THE NEED AND FRUITFULNESS 443 From the relation v dv = a t ds, gotten by eliminating dt in dv/dt = a t and ds/dt = v; J vdv= J -k 2 s ds; :. v 2 = d - k 2 s 2 ; putting Ci = k 2 a 2 , whence sin -1 1-\ = kt + C 2 or s = a sin (kt + C 2 ) = A sin kt + B cos kt, where A = a cos C 2 and B = a sin C 2 are arbitrary con- stants. This equation for s is the characteristic equation of simple harmonic motion; the amplitude of the motion is a, the period is 2 w/k, and the phase is — Ci/k. Thus, it is found that, when the acceleration along a straight line is a negative constant times the distance from a fixed point, the only motion resulting is the simple har- monic motion. In general, it has been shown that, whenever the rate of change of a function of a single independent variable is known and also the value of the function for some one value of the variable, it is possible to find by integration the value of the function for any value of the variable. Hence it has followed that the solution of a differential equation gives the area under any curve whose equation is known, thus solving the problem that had baffled the mathematicians of the ages before the discovery of this general method of effecting the quadrature of curves of any degree. When it is recalled that the magnitude of any quantity whatever, whether of volume, mass, weight, force, work, etc., may be represented by an area under a curve, the fruit- fulness of the solution of many differential equations is recognized. 444 INTEGRAL CALCULUS Note. — In this chapter and in the foregoing chapters the differential equations that have been solved have been for the most part ordinary, involving the function and one independent variable. While a general discussion of differ- ential equations, including those other than ordinary, is too large a subject for a first course in the Calculus and is beyond the scope of this book, some of the special equations are so important that their solution has been given, and some more will follow. In Art. Ill, the solution of some differential equations that have the form M dx + N dy = was effected, and in Art. 112, the definition of an exact differential equa- tion of that form was given. 227. Equations of the Form Mdx + N dy = 0. — In this form M and N are functions of x and y. The variables are said to be separated when M, or the coefficient of dx, contains x only, and N contains y only. When M dx + N dy is an exact differential (as defined in Art. Ill), the total differential of some function of x and y, then Mdx + Ndy = 0, (1) is an exact differential equation. After applying the test dM dN „ a . . , 111N -*y=Tx> W) Art. Ill), and finding the condition satisfied, integrate the coefficient of dx regarding y as constant, putting u= fMdx+f(y), (2) and then determining / (y) so that ¥ = N. ■ (3) dy Or regarding x first as constant, put u = fNdy+f(x), (4) EQUATIONS 445 and so determine / (x) that du dx - M - < 5 > These equations involve the conditions, , r du %T du , rt% M = d~x> N = W . < 6 > Example. — Solve (3 x 2 + 4 xy) dz + (2 z 2 + 2 y) cfy = 0. — — = 4-x = -^-, hence, the condition is satisfied. ay ax u = J{Zx 2 + ±xy) dx +f(y) = x* + 2x 2 y +f(y); ^=2a*+f'(y) = 2x* + 2y, :. f(y)=2y and f(y)=y 2 ; hence u = x 3 + 2 x 2 y + ?/ 2 ; /. x 3 + 2 z 2 ?/ + ?/ 2 = C is a solution and is the complete integral. When the equation, M dx + N dy = 0, is not exact, it may be multiplied by some factor that will make it exact in some cases. This factor is called an inte- grating factor. Rules have been given for finding an in- tegrating factor, but in many cases a factor, or several factors, that will make the equation become exact, may be found by inspection. For Example, see Ex. 2, Note, Art. Ill, y dx — x dy = is made an exact differential equation by either the factor £ _2 > V~ 2 , or (%y)~ l i and a solution effected in each case. For another example, (1 + xy) y dx +{1- xy) x dy = 0, ydx-\-xdy + xy 2 dx — x 2 y dy = 0, or d (xy) -+- xy 2 dx — x 2 y dy = 0; 446 INTEGRAL CALCULUS dividing by x 2 y 2 gives d{xy) dx dy _ Q> (xy) 2 x y ' 1 x 1 f- log - = log c, or x = eye?*. xy y 228. Variables Separable. — An equation of the form Mdx+Ndy = 0, in which the variables are separated can be solved by in- tegrating its terms separately. The variables are separable when the equation can be put in the form f(x)dx-+F(y)dy = 0. Example 1. — cos x dx — sin ydy = 0. Here sin x + cos y = C is evidently the general solution. Example 2. — Va 2 - y 2 dx + Va 2 - x 2 dy = 0. Dividing by V a 2 - y 2 Va^-lc 2 ; . dx + , dy = ; Va 2 — x 2 Va 2 — y 2 x n integrating, sin -1 - + sin -1 - = C; Co Q/ taking sine of each member, the first member being a sum, gives x Va 2 — y 2 + y Va 2 — x 2 = a 2 sin C = C\. Example 3. — (1 — x) dy — (1 -f y) dx = 0. Dividing by (1-*)(1+V); =^L - ^ = ; 1+7/ 1 — x integrating, log (1 + y) + log (1 - x) = log c or c h or (1 + ?/) (1 — x) = c or ^ r >. The final equation may be gotten at once by inspection. 229. Equations Homogeneous in x and y. — If an equa- tion is homogeneous in x and y, the substitution of y = vx EQUATIONS HOMOGENEOUS IN X AND Y 447 fVT.ll give a differential equation in which the variables are easily separable. Example 1. — Solve (x 2 + y 2 ) dx — 2 xy dy = 0. Putting y — vx and dividing by x 2 gives (1 + v 2 ) dx - 2 v (x dv + v dx) = 0; separating the variables, _ 2 = 0; ntegrating, log [x (1 — v 2 )] = logc; Dutting y/x for y, the solution becomes x 2 — y 2 = ex. Example 2. — Solve (x 2 + y 2 ) dy — xy dx = 0. Putting y = vx and dividing by z 2 gives (1 + v 2 ) (x dv + v dx) — v dx = 0; separating the variables, 1 (- -=■ = ; v x v 3 ntegrating, log v + log x — § v~ 2 = C = log c ; y y x 2 3utting - for v gives log - = 7 —^ > x c _ y~ )r y — ce 22/ \ Example 3. — Find the system of curves at any point of ;vhich as (x, y), the subtangent is equal to the sum of x and y. ' dx From the conditions, the subtangent being y -z- > dx lence, ydx — xdy = y dy; ,. . ,. , ydx — xdy dy dividing by ?/ 2 , * ^ lf ; x ntegrating, - = log y + log Ci = log dy, X Dr ?/ = ce lJ is the general equation of the system of curves. 448 INTEGRAL CALCULUS 230. Linear Equations of the First Order. — A linear differential equation is one in which the dependent variable and its differentials appear only in the first degree. The form of the linear equation of the first order is dy + Pydx = Qdx, (1) where P and Q are functions of x or are constants. The linear equation occurs very frequently. The solution of dy + Pydx = 0, or dy/y + P dx = 0, , . , fPdx t fPdx is log y + log e J = log c, or ye J = c. Differentiating the latter form gives ef Pdx {dy + Pydx) = 0, /P dx is an integrating factor of (1). Mul- tiplying (1) by this factor gives eJ fJ (dy + Py dx) = eJ x Qdx; and this, on integration, gives ye f Pdx =fef P, <*Qdx. (2) The equality expressed in (2) may be used as a formula for solving any linear equation in the general form (1). Example 1 . — Solve xdy — ydx — x 3 dx = 0. Putting it in the general form (1), it becomes v dy — -dx = x 2 dx. u x Hence, I Pdx = — I — =— log x = log - ; fPdx log] 1 x and / eJ Pdx Q dx = I xdx = ^- + C. EQUATIONS OF ORDERS ABOVE THE FIRST 449 Substituting these values in formula (2) gives y - = \x 2 + C, or y = \x* + Cx. XL J Example 2. — Solve dy + y dx = e^dx. y = (x + C) e - *. Example 3. — Solve cos x • dy + y sin x • dx = dx. y = sin x + C cos x. Example 4. — Solve (1 + x 2 ) dy — yxdx = adx. y = ax + C Vl + x 2 . Ti a Example 5. — Solve dy -\--ydx = —dx. x n y = ax-\-C. 231. Equations of the First Order and nth Degree. — An equation of the first order and nth degree, which is resolvable into n equivalent rational equations of the first degree, may be solved by the solution of the equivalent equations. To illustrate, let p = ~ in the examples. Example 1. — Solve (W + (x + y) ^ + xy = 0. The given equation is p 2 + (x + y) p + xy = 0; factoring, (P + y) (P + #) = 0> which is equivalent to the equations, p -f 7/ = 0, p + £ = 0, of which the solutions are, log y + x + C = 0, 2 y + x 2 + 2 C = 0. The combined solution is (\ogy + x + C)(2y + x 2 + 2C) = 0. Example 2. — p 2 - ar 3 = 0. 25 (y + C) 2 = 4 ax 5 . Example 3. — p 2 — 5 p + 6 = 0. (y-2x-C)(y-Sx-C) = 0. Example 4. — p 3 - ax 4 = 0. 343 (y + C) 3 = 27 a 7 . Example 5. — p 3 + 2 xp 2 - y 2 p 2 - 2 xy 2 p = 0. (2/ - C) (y + x 2 - C) (xy + Cy + 1) = 0. 232. Equations of Orders above the First. — Examples will be given of solving four special forms of such equations. 450 INTEGRAL CALCULUS d n y I. Equations of the form j— n = f(x). The solutions of equations of this type can be gotten by n successive integrations. Examples have already been given in Art. 161 and in other Articles. Example. — d 4 y = x z dx\ y = ^ H — ^- H — ^- + C 3 x + C 4 . d 2 y II. Equations of the form -j-^ = }{y). For these equations 2 dy is an integrating factor. Example. — Solve ^ + a 2 y = 0. (1) Multiplying by 2 dy gives integrating, ^V = _ a y + Ci = a 2 ( Cl 2 - ?/ 2 ), where d = aV- (2) From (2), dy/V Cl 2 - y 2 = adx; (3) integrating (3) sin -1 ?//ci = ax + C 2 , (4) or y = Ci sin (az + C2), (5) which may be written, y = A sin ax + £ cos ax. (6) See Ex. 3, Art. 225, where this solution was obtained, and in Art. 226, the equation for simple harmonic motion results; and there, for the differential equation, d 2 s _ M dt 2 " K S > 2 ds might be used as an integrating factor. (d n y dy \ -r-^, . . . , -j-, xj = 0; that is, equations of the nth order with y absent. EQUATIONS OF ORDERS ABOVE THE FIRST 451 The solution of a differential equation of this form was given in Ex. 1, Art. 225. r> 4. dy . 4-u d 2 y dp d n y d n ~ 1 p Put p = -r-, then ji = t 1 , . . • , -7-t = -3 — t ■ • da; dx 2 dx dx n dx n ~ l Substituting these values in the general form gives /(£?.•■•.&»«)-* (1) which is an equation of the (n — l)th order between p and x. Example 1 . — To show that the circle is the curve for which the expression for radius of curvature is constant. ? \dx, , = & dx 2 Substituting p for dy/dx, inverting the fractions, and sepa- rating the variables, gives dp _ dx < (l+p 2 )*~ #' integrating, v = ±^A V 1 + p K a being arbitrary constant; solving for p, p - dx VR 2 - (x-a) 2 ' whence y — b = ± VR 2 — (x — a) 2 , b being arbitrary constant; hence, (x — a) 2 + (y — b) 2 = R 2 , for all circles of radius R. When n is 2, the equation being of the second order, the substitution dy dp = fy V dx' dx dx 2 ' reduces the equation given to an equation of the first order in dp/dx, p, x. Solving, if possible, gives a relation of the 452 INTEGRAL CALCULUS form / (p, x, C) = 0. This is still of the first order, in x and y, and may be integrated. d 2 s 1 1 Example 2. — Solve -^ + -r + ^ = 0. _,,. ds . dp 1 1 Putting p = ^ gives _ + 7 + - = 0. Separating variables, — dp = —^ — dt; integrating, — p = — - + log t + Ci. Integrating again, s = log*- t\ogt + (l -COt- C 2 , which gives in the case of motion the relation between the space or distance and the time. (d n y\ dy \ dW' * * ' ' dx' V ) = °* Put v-^' then *y- v &, tl-tftE + JW? etc rut V- dx , then dx2 -p dy} dxZ - p dy2 + p ^j , etc. Substituting these values in the general form gives an equation of the (n — l)th order between p and y. Thus when n is 2, the equation being of the second order, the substitution _ dy dp _ d 2 y dx ' d?/ dx 2 reduces the given equation to an equation of the first order in y and p. This is solved, if possible; and then dy/dx put for p, giving an equation in x and y of the first order to be integrated for y. d 2 s Example. — Solve a t = -p = f(s). -n i. rfs ., d 2 s vdv ., s Put P-«- 5 . then ^-.gj. -/(.), giving the known relation vdv = a t ds. LINEAR EQUATIONS OF THE SECOND ORDER 453 Integrating gives - = I f(s)ds + C, the energy integral, called so from the relation Fs = \mv 2 y the equation of kinetic energy, where Fs is work done by a force F through a distance s. When / 0) is given, v is replaced by ds/dt and the integra- tion of the resulting equation gives the solution in terms of s and t and the equation may be solved for s. Special examples under this case were given in II and in Art. 225, Exs. 3 and 4. 233. Linear Equations of the Second Order. — The general form of the equation of the second order is where P, Q, R are functions of x alone or constants. The complete integral of all linear equations is the sum of two functions, called the complementary function and the particular integral. The complementary function is the complete integral of the equation when R, the term independent of y and its derivatives, is zero. This function will contain two arbitrary constants, when the equation is of the second order. The particular integral is any solution of the equation as it is in the general form, and contains no arbitrary constant. To consider the complementary function in which P, Q are constants, let the equation be I. Let y = e kx , k constant; then substituting in (1) gives (k 2 + ak + b) e k * = 0. If 'k is a root of the quadratic equation, fc 2 + ak + b = 0, (2) 454 INTEGRAL CALCULUS called the auxiliary equation, e kx will satisfy (1). The two roots fci, k 2 of (2) are fcj = -ia + V|a 2 - 6, fe 2 = -Ja- Vj a 2 - 6, and e fclX , e* 2 * are two solutions of (1). Hence the complete integral of (1) is y = £&* + Be^ = er hax (Ae nx + 5e" nx ), (3) where u = v J a 2 — 6. For special cases: II. If a 2 = 4 6, equation (2) has two equal roots, k\ = k 2 = — \ a. In this case (3) becomes y = (A+£)e-*« where (A + B) might be replaced by one constant C. When a 2 = 4 b, let y = e~? ax u and (1) becomes, without the factor e~* ax , ^ = dx ' of which the complete integral is u = A + Bx. Examples of this solution have been given in Arts. 224 and 225. The complete integral of (1) when the auxiliary equa- tion has two equal roots, each — \ a, is y = (A + Bx) e-* ax . (4) III. If a 2 < 4 b, the roots of (2) are imaginary. Again, let y = e~* ax u and equation (1) becomes g + OT V=0, (5) where \a 2 — b = — m 2 and m is real. Now (5) is satisfied by it = cos mx, u = sin mx; its complete integral is then u = A cos mx -f B sin mx, and therefore the complete integral of (1), when a 2 < 4 6, is y = e -* ax u = 6 _iax (A cos mx + B sin wu). (6) To show how (3) and (6) are written when the roots of (2) LINEAR EQUATIONS OF THE SECOND ORDER 455 are known: when the roots of (2) are real, let \ a 2 — b = n 2 ; the roots are then, — \a-\-n,—\a — n) and the solution is y = e -hax (A e nx _|_ Be~ nx ); when the roots of (2) are imaginary, let \ a 2 — b = — n 2 , and the roots are then, —\a-\-ni, — \a — ni; and the solution is y = e -\ax ^ cos nx -\- B sin nx), so that instead of e nix , e~ nix , there are cos nx, sin nx. It may be noted that the auxiliary equation is written by putting k 2 for -y-| , k for -j- , and omitting y. The solving of a linear equation of the second order in- cluding these three cases has been given in Art. 225, Ex- amples 2, 3,4,5; for the equation for damped vibrations, Example 1. — Solve § + 8 ^ + 25 2/ = - The auxiliary equation is k 2 + 8 k + 25 = 0, and its roots are h = — 4 + 3 i, Ifa — —4 — 3 *. Hence, the complete integral is y = e~ ix (A cos 3 x + B sin 3 x) . Example 2. — Solve ^-2^-35z/ = 0. k 2 - 2k -35 = 0. ax 2 ax The roots of the auxiliary equation are ki = —5, k 2 = 7. Hence, the complete integral is y = Ae~ bx + Be 7x = e x (Ae~ 6x + Be &x ). 456 INTEGRAL CALCULUS APPLICATIONS. 234. Rectilinear Motion. — I. When the acceleration is constant, da d?s n d 2 s ds s = i at 2 + vot + s . For bodies falling freely towards the earth from moderate heights, the acceleration g being taken constant, v = v - gt, s = tfct — | gt 2 + «o; from rest, v= -gt, s= -\gt 2 . (Ex. 5, Art. 115.) ' Projected outward from rest, h = v t — \ gt 2 (Ex. 6, Art. 116), where h is height from point of projection. II. When the acceleration varies as the distance. Let d 2 s a = -« = — ks, where k, a constant, is the acceleration at a unit's distance from the origin ; and let the body be of unit mass at an initial distance r; then, *-(*)'- C\ — ks 2 = kr 2 — ks 2 , where kr 2 = C\\ t = k~i f , ds = h~+ cos" 1 - ( + C 2 - 0), J Vr 2 - s 2 r s £ = 0, when s = r; whence r cos (kh) = r sin (kh+^\ where the last result is gotten if the positive sign of the radical is taken in integrating. RECTILINEAR MOTION 457 Putting s = 0, gives v - r Vk, the velocity at the origin, and t = 7r/2 fc~ a , f irk'*, f wk~*, . . . , or s = r, gives t = 0, 27I-/&2, 4 7i7&2 , . . . Hence the motion is periodic, the period being 2 *•/&», which is independent of the initial distance. Differentiating s = r cos (h*t), gives v = jT = — A^r sin (&^) ; which expresses the velocity in terms of the time that the body has been moving. It is seen that this is a case of simple harmonic motion. (Arts. 73 and 226.) It has been shown in Art. 190, Cor., that the attraction of a sphere of uniform density for an internal particle varies as the distance from the center. Hence, a particular case of the periodic motion just considered would be that of a body which could pass freely through the earth, taken as a homo- geneous sphere. Such a body would vibrate through the center from surface to surface. To find the time of this half period : t = Trfc-* = 3.1416 V20900000/32.17 sec. = 42 min. about. III. When the acceleration varies inversely as the square of the distance. Let k be the. acceleration at unit distance from origin; .. d?s k then, „ = _=_-, where s is to be taken always positive; multiplying by 2 ds, 2 @)<:)=- 2fo - 2ds; integrating, * = (|J = 2k g -f), (1) 458 INTEGRAL CALCULUS where s = r, when v = 0, which gives the velocity of a particle at any distance s. For the time / 2fc —sds Vrs — s 2 negative, since s decreases as t increases, _ [1 r- 2s _ r__J- I , . |_2 y/rs — s 2 2 Vrs — s 2 J integrating between limits corresponding to t = t and t = 0, gives = V^[ Vrs - s2 -^ vers ~ 1 T + ?} (2) When the particle arrives at the origin, s = 0, therefore, the time to the origin from the point where s = r is / r \i 1 Vk\2 It is seen from (1) that the velocity = when s = r, and = oo when s = 0; hence the particle approaches the origin with increasing velocity. While the attractive force causing the acceleration is very great near the origin, there can be no attraction at the origin itself; therefore, the particle must pass through the origin; and the conditions being the same on either side of the origin the motion must be retarded as rapidly as it was accelerated; hence, the particle will go to a point at a distance r equal to that from which it started and the motion will continue oscillatory. An illustration of this general case has been given in Art. 193, where the attraction of the Earth for an external particle was considered as the cause of motion. IV. When the acceleration varies as the distance and the motion is away from the origin. RECTILINEAR MOTION 459 Let k again be the acceleration per unit mass at unit distance from the origin; then d 2 s . a = de = ks '> using 2 ds as an integrating factor gives 2 jL s . d (^) = 2ksd S ; /ds\ 2 integrating, v 2 = 1-=-) = ks 2 + v 2 , where v is the initial velocity; whence, s = 2 k ( e kh _ e -kh) m ( See Ex 4? Art 225.) Here, as t increases s also increases, and the particle recedes further and further from the origin; and the velocity also increases and becomes oo when s = t = oo. Thus in this case the motion is not oscillatory. V. When the acceleration is constant and the motion is in a medium whose resistance varies as the square of the velocity. In Art. 195 the case where the motion was towards the Earth has been given. Let now the particle be projected outward with a given velocity v . Using again gk 2 , as the coefficient of resistance, the resistance of the air on a particle for a unit of velocity, and taking the particle of unit mass, with g constant, ('9 whence, ' ; v> = —kgdtj integrating, tan -1 (A- --J = tan -1 (kv ) — kgt, 460 INTEGRAL CALCULUS where C = tan -1 (kv ) ; solving, ds 1 kv Q v = tan kgt dt k 1 + kv tan kgV ^ ' which gives the velocity in terms of the time. To get it in terms of the space ; from ( 1 ) , integrating, log 1 + kW where c = log (1 + k 2 v 2 ); whence, = -2gk 2 ds; -2gk 2 s, '-sr- = n = v 'e- id e -2gk*s\ Writing tan %Z in (2) in terms of sine and cosine and integrating, s = r— log (kv sin kgt + cos kgt), k g which gives the space described by the particle in terms of the time. 235. Curvilinear Motion. — Let a body slide without friction down any curve ab. The acceleration caused by gravity at any point P is g sin a, where a = PTD, PT being a tangent to the curve. Let PT = ds; then -PD = dy; hence, d 2 s . dy ... - = gSlna =-g-. (1) Let y be the ordinate of the initial point on the curve; then v = when y = y . SIMPLE CIRCULAR PENDULUM 461 Integrating (1) gives v= j t = ^2g(yo- y)- (2) It follows from (2) that the velocity of a body acquired by moving freely down any frictionless path is the same, and is what it would acquire in falling freely through the vertical height between the initial and terminal points. — , the time will depend upon the path. 236. Simple Circular Pendulum. — Consider the motion of a particle on a smooth circular arc under the action of gravity as the only force. Taking the axis of x as vertical, the equation of the circle is y 2 = 2 ax - x 2 . (1) Let K (h,k) be the point where the particle starts from rest, and P (x, y) where it is at the time t. Then the particle will have fallen through the height h — x, and hence from (2), Art. 235, v = f t = V2g(h-x). (2) It is seen from (2) that the velocity is a maximum when x = and a minimum when x = h; so the particle will pass 462 INTEGRAL CALCULUS through following the curve to the point K' where x = h and will oscillate between K and K'. To find the time in passing from K to K' ) from (2), ii _ — - , negative since s decreases as t increases, dt ~ V2g(h-x) — adx , /-.N j adx :, from (1) ds = — • (3) V2g{h- x) (2 ax - x 2 ) While this expression does not admit of direct integration, approximate values of the integral can be gotten as shown in Ex. 6, Art. 203. When the arc is small, the approximate value of the time can be gotten from (3) thus : T = 2*= 2 P ° adx gJh V2gJ h V(h-x)x(2a-x) h dx Vh x — X 1 by taking (2 a — x) = 2 a, -v&-*¥I-'\/r » If instead of moving on a curve, the particle is assumed to be suspended by a rod or cord of no weight, it becomes a simple pendulum, existing in theory. To reduce (3) to the form known as "an elliptic integral" of the first kind; let h = a vers a, x = a vers 6, a = KCO, being constant and 6 variable ; then h — x = a (vers a — vers 0) = a (cos 6 — cos a) ; dx = a sin dd; and (2 ax — x 2 ) = a 2 sin 2 6. Substituting in (3) : 2 a • a sin 6 dd r=-i-r_ V2gJ y/a (cos 6 — cos a) a 2 sin 2 dd gJo Vsin 2 a/2 - sin 2 0/2 SIMPLE CIRCULAR PENDULUM 463 -v/*P d$ Jo 2 V Vsina/2/ 2 sin a/2 cos d = Ji r v J sin a/2 Vl -sin 2 <£ Vl - sin 2 a/2 sin 2 = 2 V - / , ., by cancellation. (5) V 9 J Vl -sin 2 a/2 sin 2 Here is denned by sin = - — 4r, giving by differenti- sin a/ _ ation, 4jJ cos B/2de/2 COS0C?0 = — r 1 — 7? p— , sm a/2 whence, J,, _ 2 sin a/2 cos d _ 2 sin a /2 cos d<£ cos 0/2 " Vl - sin 2 a/2 sin 2 For change of limits, 6 = a when = 7r/2, and = when = 0. Putting k = sin a/2 in (5), it becomes ,— * T=*2\/- T(l - k 2 sin 2 0)"*d0 = 2 V^X" (* + ^' 2 Si " 2 + l'I fc4 Si " 4 + ' ' ' ) d * "^h@'^(H)V + (liI)V + .} <6 , ?r The form / (1 — k 2 sin 2 <£)~^ d = JaY ^M V gJv Vvny-V 2 V a VoJv V 0L 2/oJ t vy Q y-y 2 * Q V» Try-, when y = 0, and 2t=T 9 the time of one oscillation of a pendulum if it swings in the arc of a cycloid. The time of an oscillation being independent of the length of the arc, the cycloidal pendulum is isochronal. The pendulum is described in Art. 97, Ex. 3. The cycloid is the curve of quickest descent, the Brachysto- chrone; that is, the curve down which without friction gravity will cause a particle to fall in the shortest time. CYCLOIDAL PENDULUM 465 The following is a comparison between the times down the chord of a circular arc, a circular arc, and the arc of a cycloid. For the time down the chord I of a circle y 2 = 2 rx — x •• -Vj d 2 s I From (1), Art. 235, -p = gsina = g — , from the circle, V = Jt = '2r t ( +C = °' since y = °' when ' = °^ S = hr f ~ ( +C = °> since s = 0j when l = 0)t < =\ / ¥ =2 v / ^ whens=L c X ^sT s B ta \ (, J J^^y^ V" For the time down the circular arc AO; 2 V sf L 4 V 2 + 4/ ^ 64 W 2 +4/ ^ J Urt. 236/ = l\- [1+0.07+0.01+ • • • ]=0.54x\A-(1.7...)V^ Z f gr f g V g For the time down the arc of a cycloid from A to 0; 466 INTEGRAL CALCULUS From the figure, 7T 2 +4 r = — - — a or a = 4r hence fc 2 7T 2 +4 ' or I 2 = (2r-2a). V^=a-6...)^<(i.7...)V^<^- Hence, It is at once evident that It may be seen that the approximate value lv/^(i. 5 7...)v/-; from (5); 474 INTEGRAL CALCULUS and as at P , p = R sin 0, 7=— t-; h=VRsmp. Rsm(3 As the force varies inversely as the square of the distance, TC 1 F = -=■ = Ku 2 , where u = - ; P 2 P d 2 u K hence, ^- + u = -p, from (8). (8') Using 2 dw as integrating factor and integrating, when I - 0, t* = - = ^ and ^j +^ 2 = F by * = - 2 and Art. 77 (9); T^ 2 _ 2X = V 2 R-2K " h 2 h 2 R h 2 R ; substituting this gives »W |U ,_ H-2JC 2fa Hence which shows that the velocity is greatest when p is least, and least when p is greatest. Changing the form of (9) to du 2 V 2 R-2K . K 2 K 2 IK V + TF-fc— )• (10) dd 2 h 2 R and simplifying by letting u = b, and jjiy— + jl = gives [c 2 - (u - &)•]« FORCE VARIABLE 475 the negative sign of the radical being taken. Integrating gives _ t u — b , cos 1 = 6 — c, c where c' is an arbitrary constant; /. u = 6 + ccos(0- c'). (11) Replacing in (11) the values of b and c and the value of h, and dividing both terms of the second member by K, gives = 1 = WW/3/if P u l + [l/K 2 (V 2 R-2K)RV 2 sm 2 p+l]2cos(^-c') , which is the polar equation of the path and is the equation of a conic section, the pole being at the focus, and the angle (6 — c') being measured from the shorter length of the major axis. For if e is the eccentricity of a conic section, p the focal radius vector, and the angle between p and that point of a conic section which is nearest the focus, then 1 a(l-e 2 ) a(e 2 -l) /10 , u 1 + e cos <{> 1 + e cos Comparing (12) and (13), it is seen that, e 2 = l/K 2 (V 2 R -2K) KV 2 sin 2 + 1 ; (14) = e - c'. (15) Since the conic section is an ellipse, parabola, or hyper- bola, according as e is less than, equal to, or greater than unity, and from (14), e is thus, according as V 2 R — 2 K is negative, zero, or positive; therefore, it is seen that V 2 < -5-, e < 1, and the orbit is an ellipse, 2 K V 2 = -5-, e=l, and the orbit is a parabola, it 9 js V 2 > -5- , e > 1, and the orbit is a hyperbola. K 476 INTEGRAL CALCULUS Corollary 1. — By (1) of Art. 234, III, it is seen that the square of the velocity of a particle falling through an infinite distance to a point R distant from the center of force is under the law of attraction now considered 2 K/R. Hence the conditions above may be expressed by stating that the orbit, described about this center of force, will be an ellipse, a parabola, or a hyperbola, according as the velocity of pro- jection is less than, equal to, or greater than the velocity through an infinite distance. Corollary 2. — From (15) it is seen that 6 — c' is the angle between the focal radius vector p, and that part of the principal axis which is between the focus and the point of the orbit which is nearest the focus; that is, it is the angle POA in the figure; and hence, if the principal axis is the initial line, c' = 0. 2 K Corollary 3. — If the orbit is an ellipse, V 2 < —^- ; hence, K e 2 =l- 1/K 2 (2K - V 2 R) RV 2 sin 2 p, from (14). (16) The polar equation of the ellipse is a (1 - e 2 ) . P = 1 + e cos comparing it with (12), corresponding terms give n - R 2 V 2 sm 2 (S . a (I - e 2 ) = ^ , substituting for 1 — e 2 its value from (16) and solving for a; KR 2K- V 2 R' (17) which shows that the major axis is independent of the direction of projection. In the figure, P is the point of projection; FP = R; P T is the line along which the particle is projected with velocity V; FP Q T = j8, the angle of projection; FP = p; PFA = d; KEPLER'S LAWS OF PLANETARY MOTION 477 FT = p = R sin &; if /3 = 90°, the particle is projected from an apse, A or A', an end of the major axis. To determine the apsidal distances, FA and FA'; -^ = 0; dd •'• " 2- ¥ + ll _ fc 2 = ' from(9) ' (18) the two roots of which are the reciprocals of the two apsidal distances, a (1 — e) and a (1 + e). Since the coefficient of the second term of (18) is the sum of the roots with their signs changed, 1 1 2K . n _ h 2 nm + „m i ^ = "IT i • • « (1 - e2 ) = ^. (19) a(l-e) ' a(l+e) /i 2 7 v J K which is one-half the latus rectum. From (5) p 2 dd = h dt; and area swept over by the radius vector is 2A h which shows that the areas swept over by the radius vector in different times are proportional to the times, and equal areas will be described in equal times. If t = 1, A = § h\ hence, h = twice the sectorial area described in a unit of time. For the time of describing the ellipse, calling the time, T; A = \J?d0 = \£hdt = \ht; .: t = T _ 2 area of ellipse _ 2 TTOb 2 7ra 2 Vl - e h (from (19)) VKa (1 - e 2 ) which is the periodic time. 243. Kepler's Laws of Planetary Motion. — From a long series of observations of the planets, especially of Mars, Kepler deduced the following three laws which completely describe planetary motion. 478 INTEGRAL CALCULUS I. The orbits of the planets are ellipses, of which the sun occupies a focus. II. The radius vector of each planet describes equal areas in equal times. III. The squares of the periodic times of the planets are as the cubes of the major axes of their orbits. The statement of these laws marked an epoch in the development of mechanics, for the investigations of Newton as to the nature of the attractive force led to his discovery of the law of universal gravitation. The conclusions deduced by Newton from Kepler's three laws will be briefly shown. 244. Nature of the Force which Acts upon the Planets. — (1) From the second of Kepler's Laws, it follows that the planets are retained in their orbits by an attraction tending towards the Sun. Let (x, y) be the position of a planet at the time t, referred to rectangular axes through the Sun in the plane of the motion of the planet; X, Y, the component accelerations due to the attraction acting on it, resolved parallel to the axes; then the equations of motion are, ^1 = y ^1 = v dt 2 ' dt 2 ' ■•■ xC 3- yd i = xY - yX - (1) By Kepler's second law, if A be the area described by the radius vector, dA/dt is constant, ••• f or Vtr\ (from (5), Art. 242) 1 / dy dx\ „ = 7i\ x -ji — 2/~n =a constant, from 2{ di M) (3), Art. 242. FORCE WHICH ACTS UPON THE PLANETS 479 Differentiating gives x C ^- x dt 2 d 2 x y W 2 = = 0; . xY -yX = : (from (i)) 7 * X • Y X ■ -, or y y = X which shows that the axial components of the acceleration, due to the attraction acting on the planet, are proportional to the coordinates of the planet; and therefore by the parallelogram of forces, the resultant of X and Y passes through the origin. Hence, the forces acting on the planets all pass through the Sun's center. (2) From the first of Kepler's laws it follows that the central attraction varies inversely as the square of the distance. The polar equation of the ellipse, referred to its focus, is a (1 - e 2 ) 9 1 + e cos B 1 1 -\- e cos 6 or - = u= —Tz. ~-> p a (1 - e 2 ) which by differentiation gives, d*u dd' 2 " t " 1 U ~ a(l-e 2 )' and, Art. therefore, if F is the attraction to the focus, 242, by (8), h 2 1 a (1 - e 2 ) p 2 Hence, if the orbit be an ellipse, described about a center of at- traction at the focus, the law of intensity is that of the inverse square of the distance. (3) From the third law it follows that the attraction of the Sun (supposed fixed) which acts on a unit of mass of each 480 INTEGRAL CALCULUS of the planets, is the same for each planet at the same distance. By Art. 242 (20), T 2 = ^-a\ Since by the third law, T 2 varies as a 3 , K must be constant ; that is, the strength of the attraction of the Sun must be the same for all the planets. Hence, not only is the law of force the same for all the planets, but the absolute force is the same. The third law shows also that the law of the intensity of the force is that of the inverse square of the distance. Since the planets move in ellipses slightly different from circles, assume for simplicity that their orbits are actually circles. If Ri, R2, Rs are the radii and 7\, T 2 , T s , the respec- tive times of revolution of the planets, Kepler's third law may be written as follows: #i 3 R2 3 #3 3 , , T? = T7 2= T? = ' ' ' = a constant ' The expression for the central acceleration of motion in a circle is a = v 2 /R = 4t 2 R/T 2 , or T 2 = 4^/a (Art. 70 (2)). Substituting this value gives ai^Ri 2 = CI2R2 2 = (I3R3 2 = constant; or a = constant/7? 2 . Note. — Arts. 242, 243, 244, are based on the discussion in Bowser's Analytic Mechanics. For a fuller discussion, see Tait and Steele's Dynamics of a Particle, and Percival Frost's translation of Newton's Principia, Sec. I, II, III. 245. Newton's Verification. — The greatest of Newton's achievements is considered an achievement of the imagina- tion, his conception of the universality of natural law. At an early age he (in 1666) conceived with his far-reaching mind the then daring idea that the sublime, inscrutable, central force was nothing but commonplace gravity, known to exist on and near the earth. He verified his idea first in the case of the moon. He discovered that the same acceleration that controls the motion of an object near the earth also pre- NEWTON'S VERIFICATION 481 vented the moon from moving away in a rectilinear path from the earth, and that its tangential velocity prevented it from falling to the earth. Assuming that the moon's orbit is circular, its acceleration towards the earth is (by Art. 70 (2)), a = v - = ^^ = 0.0089 ft. /sec. 2 , where R = 238,800 miles and T = 27.32 days. From the law of inverse squares : a _ r 2 _ r 2 - 1 ~g~R 2 ~ (60.267 r) 2 ~ 3632 ' a 32.089 3632 3632 0.0088 ft./sec 2 ., where 32.089 is the value of g on the earth at the equator, and R is 60.267 times r, the radius of the earth. As these results differ by only T o£ooth of a foot, the conclusion is that the centripetal force on the moon in its orbit is due to the earth's attraction, acting according to the law of inverse squares. Owing to an inaccurate value of the earth's radius which was in use at the time Newton first made the computation, the result then obtained seemed to show that the law of attraction was not that of inverse squares. Records show that Newton, although unshaken in his belief, laid aside his calculations; and it was not until thirteen years afterwards that, a new determination of the radius having been made, he repeated the investigation and found the verification sought for. Five years later (in 1684) he was induced to consider the whole subject of gravitation; and then he solved the supple- mentary problems in regard to the attraction of a sphere for an external particle, which established his theory — now known as Newton's Law of Universal Gravitation. INDEX (Numbers refer to pages) Acceleration, 19, 21 angular, 92 gravity, 90 normal, 20, 89, 94 resolution of, 86 tangential, 19, 86, 94 total, 19 Agnesi, witch of, 132 Algebraic functions, 9, 38 Anti-derivatives, 263 Anti-differential, 171 Application of Taylor's theorem, 424 Applications, 99, 355, 456 beams, 229 prismoid formula, 284 Approximate formulas, 138, 249 relative rates and errors, 162 Approximation formulas, 421 Arc, differential of, 17 infinitesimal, 37 length of, 64 Archimedes, 110, 142 Area, any surface, 311 as an integral, 223 by double integration, 304, 308 derivative of, 205 finite or infinite, 208 hyperbola, equilateral, 221 positive or negative, 208 under a curve, 206, 214 under derived curves, 225 Argument, 6 Atmosphere, limit of, 378 Attraction, 363 Auxiliary equation, 454 theorems, 87 Base, 6, 48, 56 common, 54, 55 Xaperian or natural, 49, 50, 54 Beams, 229 Binomial theorem, 419 Brachystochrone, 464 Cable, 81 Calculus, 1, 2, 17 Differential, 3, 17, 100 Integral, 3, 171 Cardioid, 142, 217, 310, 311 Catenary, 81, 140, 148, 245 surface generated by, 293 uniform strength, 251 volume generated by, 294 Cauchy's remainder, 416 Center of curvature, 135, 142 of gravity, 332, 333, 334, 335 Central forces, 472 Centrifugal force, 90, 378 railway, 466 Centripetal force, 90 Centroid, 332 Change of independent variable, 149 483 484 INDEX Change, rate of, 1, 2, 5 uniform and non-uniform, 14 ChrystaFs Algebra, 421 Circle, area of, 217, 218, 261, 309 curvature of, 135 equation of, 5 length of circumference, 237 Circular functions, 64 measure, 64, 70 motion, 20, 88 pendulum, 461 Cissoid, length of, 237 volume generated by, 295 Coefficient, differential, 24 of contraction, 469 of velocity, 468 Comparison test, 391 Complementary function, 453 Complete integral, 437 Compound interest law, 9, 60, 442 Concavity, 122, 123 Cone, 267, 270 Conic section, 475 Conoid, 282 Constant, 5 absolute, 5, 6 arbitrary, 5, 6 of integration, 173, 301 Continuity, 7 Continuous functions, 7 variables, 5, 7 Convergence, tests for, 390, 393 Convergent series, 388 Cosine, differential of, 64, 65 graph of, 119, 228 Cubical parabola, 140, 237 Curvature, 133 center of, 135 circle of, 135 radius of, 135, 137, 140 Curve, 8 of cord with load uniform hori- zontally, 241 of flexible cord, 245 polar, 123 Curvilinear motion, 460 Curves, derived, 118, 225 integral, 227 lengths of, 236, 239 Cusps, 114, 140, 146 Cycloid, 105 area of, 217 equation of, 107 evolute of, 147 length of, 147, 237 surface generated by, 293 volume generated by, 294 Cycloidal pendulum, 464 Cylinder, 267, 279, 280, 288, 290, 291 Cylindrical coordinates, 324 slice, 317 surface, 318 Damped vibrations, 73, 441, 455 Decreasing function, 8 Definite integral, 206, 213 Deflection of beams, 141 Density of air, 62 mean, 327, 374 Dependent variable, 6 Derivative, 24, 25 as a limit, 26 as slope of curve, 29 of area, 205 partial, 153, 164 successive, 84, 149 total, 159 Derived curves, 118, 225 function, 24 Differential Calculus, 3, 17 coefficient, 24 INDEX 485 Differential equations, 436, 441 of sine in degrees, 69 Differentials, 3, 15 exact, 167 inexact, 168 partial, 152, 164 successive, 84 total, 156 Differentiation, 2, 25 of series, 394 rules of, 64 Discharge from an orifice, 469 Discontinuous function, 7 variables, 5 Divergency of series, 391 Double integration, 304, 308, 324 E, modulus of elasticity, 230, 254 e, Napierian base, 6, 50, 418 Eccentricity, 475 Elastic curve, 230 Elasticity, modulus of, 230, 254 Elementary principles, 178 Elements, 261, 331 Ellipse, area of, 218 length of quadrant, 238, 398 Ellipsoid, 292, 294 Elliptic integral, 400, 462, 463 Empirical equations, 10 Energy integral, 453 kinetic, 92 Equation of evolute, 144 of involute of circle, 146 of involute of cycloid, 146 of normal, 100 of tangent, 99 Equations, differential, 436 exact differential, 444 homogeneous, 446 linear, 448 of first order, 449 of order above first, 449 Equilateral hyperbola, 140, 147, 221 Error term, 421 Errors, percentage, 163 Euler's series, 397 Evaluation of definite integrals, 213 of derivatives of implicit func- tions, 435 of indeterminate forms, 428 Evolute, 143, 144, 256 Evolution, 3 Exact differential equations, 167, 170, 444 Examples, illustrative, 31, 76, 160, 174, 412 Expansion by Maclaurin's and by Taylor's Theorems, 411 of cosh x/a and sinh x/a, 248 of functions in series, 408 Explicit function, 6 Exponential function," 9, 47 Extended law of the mean, 40." Factor, integrating, 445 Falling bodies, 21, 177, 297 First moment, 331 Flexion, 21, 133 Force, central, 472 centrifugal, 90, 378 centripetal, 90 concentrated, 355 definition of, 92 distributed, 355 variable, 472 Forms, indeterminate, 425, 428 standard, 180, 181, 183, 203, 403 Formula, prismoid, 283 projectile, Helie's, 384 Formulas, 38, 47, 64 approximation, 421 reduction, 194, 198 486 INDEX Frustum, surface and volume of any, 270 Function, 6 algebraic, 9, 38 continuous, 7 discontinuous, 7 exponential, 9, 47 hyperbolic, 9, 80 inverse, 76, 82 logarithmic, 9, 47 of a function, 27 power, 9, 10 transcendental, 9 trigonometric, 9, 64 several variables, 152 Functional relation, 7 Fundamental conception, 30 condition or test, 112 rule for applying test, 115 theorem, 261 g, acceleration of gravity, 90 Gas, formula for, 160 Gauge, self-registering, 97 Geometric meaning of integral, 204 Geometric progression, 9, 10 series, 385 Grade, 23 Graphical illustration, 113 Graphs, 8, 80, 118 cosine, 119, 228 cycloid, 105 sine, 71, 119, 228 versine, 228 Gravitation, law of, 363, 371, 481 unit of mass, 376 Gravity, acceleration of, 90, 91 center of, 332, 333, 334 Gregory's series, 397 Guldin's theorems, 336 Gyration, radius of, 345 Harmonic law, 9 motion, 94, 442, 457 series, 389, 392 Helie's formula, 384 Homogeneous equations, 446 Hyperbola, equilateral, 140, 221 Hyperbolic functions, 9, 80 logarithms, 216 Hypocycloid, 140, 148 Ideal quantity, 2 Ideas, 2 Illustrations, 21 typical, 118 Illustrative examples, 31, 76, 160, 174, 412 Implicit function, 7 Increasing function, 8 Increment, 12 Indefinite constant, 173, 301 integral, 173, 301 Independent variable, 6, 149 Indeterminate forms, 425 Inertia, 91 moment of, 230, 345, 346, 349, 350, 352 product of, 348 Inexact differential, 168 Infinite series, 385 Infinitesimal, 30, 36, 37 arc and chord, 37 Infinity, 36 Inflexion, point of, 114, 120, 132, 140 Integral, complete, 437, 453 Calculus, 3, 171 definite, 206 energy, 453 from an area, 219 general, 227 indefinite, 173, 301 multiple, 297 INDEX 487 Integral, particular, 175, 453 Integrals, elliptic, 400, 462, 463 Integrand, 171 Integrating factor, 445 Integration, 2, 17, 170, 171 double, 304, 308, 324 constant of, 173, 301 parts, 194 series, 394 successive, 296, 303 triple, 317, 321 Intensity, 355 Interchange of limits, 210 of order of differentiation, 165 of order of integration, 299, 301, 306 Interest, compound, law of, 9, 60 Inverse functions, 9 hyperbolic functions, 81 of differentiation, 171 trigonometric functions, 76 Involute, 143 of the catenary, 248 Involution, 3 Jet, liquid, path of, 467 Kepler's laws, 477 Kinetic energy, 92 Lagrange's remainder, 407 Law, compound interest, 9, 60, 442 extended, of the mean, 405 of organic growth, 9, 62, 442 of the mean, 401, 402, 403 of motion, 90, 359, 365 parabolic, 9 planetary motion, 477 universal gravitation, 363, 371, 481 Lemniscate, 110, 142, 217, 326 Lengths of curves, 236, 239 Limit of a sum, 259 of height of atmosphere, 378 of infinitesimal arc, 37 Limits, 25, 26, 30, 206, 208, 210 Linear equations, 448, 453 Liquid jet, 467 pressure, 356 Lituus, 124 Locus, 8 Logarithmic differentiation, 56 curve, 140 functions, 47 series, 395 spiral, 142 Logarithms, 54, 55 common, 55 hyperbolic, 216 natural, 55 Machin's series, 397 Maclaurin's theorem, 407 series, 408 Mass, 91, 327 unit of, 376 Maxima and minima, 111 application of Taylor's theorem, to, 424 problems, 127 Maximum and minimum, 111 Mean density, 327, 374 law of the, 401, 402, 403 value of a function, 211 Method of the Calculus, 430 of limits, 30 Modulus, 54 of elasticity, 230, 254 Moment, 331, 345, 355 of inertia, 230, 350 least moment of inertia, 349 of inertia for parallel axes, 346 principal moment of inertia, 348 488 INDEX Moments of area, 333, 352 first, 331 of line, 334 second, 344 of volume, 333 Momentum, 92 Motion, circular, 88 curvilinear, 460 in resisting medium, 379, 383, 459 planetary, 477 rectilinear, 456 second law of, 90 simple harmonic, 94, 442, 457 third law of, 90, 359, 365 Napierian or natural base, 49, 50, 54 logarithms, 55, 70, 216 Newton, 4, 90, 359, 363, 371, 480 Normal acceleration, 20, 89, 94 Normal, 99 polar, 109 Notation for functions, 6 Number e, 6, 50, 418 7T, 6, 50, 398, 418 Orbit, 472, 475 Order above the first, 449 first, 449 of a differential equation, 436 Ordinary differential equation, 436, 444 Organic growth, 9, 62, 442 Orifice, discharge from, 469 Oscillating series, 385, 397 7T, 6, 50, 398, 418 Pappus, theorem of, 336 Parabolic cable or cord, 241 law, 9 Partial derivations, 153, 164 differentials, 152, 164 Particular integral, 175, 453 values, 6 Parts, integration by, 194 Path of a projectile, 381 of a liquid jet, 467 Pendulum, simple circular, 461 cycloidal, 464 Percentage rate, 56 error, 163 Period, 96 Periodic time, 477 Plane, tangent, 154 Plane areas, 304, 308 Planetary motion, 477, 478 Points of inflexion, 114, 120, 132, 140 Polar curve, 8, 123 moment of inertia, 345 subtangent, subnormal, 108 Power form, 183 formula, 44 function, 9, 10 series, 386, 393 Pressure of air, 62 liquid, 356 Primitive of differential equation, 437 Principles, 371, 480 Prismoid formula, 283, 284 Probability integral, 418 Problems, maxima and minima, 127 Process, summation, 261 Product of inertia, 348 Progression, 9, 10 Projectile, path of, 381 Quadrature of curves, 443 Quotient, differential of, 39, 42 limit of, 26, 35 INDEX 489 Radian, 64, 70 Radium, 63 Radius of curvature, 135, 137, 140 of gyration, 345 Railway, centrifugal, 466 Range formula, Helie's, 384 Rate, 18, 19, 21 of change, 1, 2 percentage, 56 relative, 56, 162 Rectification of curves, 237 Rectilinear motion, 456 Relative error, 59, 162 rate, 56, 162 Remainder, Cauchy's, 416 Lagrange's, 407 Remarks, 27, 70 Replacement theorem, 37 Representation of functional re- lation, 7 volume by area, 269 Resisting medium, 379, 383, 459 Rotation, 92 Secant, 29, 65, 68 Second derivative, 84, 86 moments, 344 Semi-cubical parabola, 145, 237 Separable variable, 446 Separated differential equation, 444 Separation into parts, 210 Series, infinite, 385 absolutely convergent, 388 convergent, 385, 393 divergent, 385 Euler's, 397 for e, 50 Gregory's, 397 geometric, 385 harmonic, 389, 392 Series, integration and differen- tiation of, 394 logarithmic, 395 Machin's, 397 Maclaurin's, 408 Newton's, 397 oscillating, 385, 397 power, 386, 393 Taylor's, 409 Shear, 229 Shooting point, 114 Significance of area, 223 Simple circular pendulum, 461 harmonic motion, 94, 442, 457 Sine curve, 72 differential of, 64, 65 graph of, 71, 119, 228 ratio to arc, 66, 67 Slope, 18, 23, 99 rate of change of, 21 Solids of revolution, 288 by double integration, 321 Solution of a differential equation, 436 of s = a sinh x/a, 250 Speed, 1, 18, 19 Sphere, 268, 274, 285, 313, 323 Spherical shell, 368, 371 Spheroid, 294 Spiral of Archimedes, 110, 142 logarithmic, 110, 142 Standard forms, 180, 182, 203 Statical moment, 345, 359 Stiffness of beams, 128 Strength of beams, 127 Subnormal, 100, 108 Subtangent, 100, 108 Successive derivatives, 84, 149 differentials, 89 differentiation, 84 integration, 296, 300 Summation process, 261 490 INDEX Summation process, approximate and exact, 262 Surface of any frustum, 270 of revolution, 288 Suspension bridge, 244 Tangent, equation of, 99 plane, 154 polar, 109 Tangential acceleration, 19, 86, 94 Taylor's theorem, 405, 406, 407 series, 409 Test, comparison, 391 for convergence, 390 ratio, 391 Theorem, replacement, 37 binomial, 419 finite differences, 402 Maclaurin's, 407 Taylor's, 405, 406, 407 Theorems, auxiliary, 124 of limits, 26 of mean value, 401, 402 Theorems of Pappus and Guldin, '336 Tide gauge, 97 Time, periodic, 477 rate of change, 19 Total derivative, 159 differential, 156 Tractrix, 254 Transcendental functions, 9 numbers, 6, 50 Trapezoid, 343 Trigonometric functions, 9, 64 Triple integration, 317, 32K Typical illustrations, 118 Uniform change, 14 speed or velocity, 20 Unit of force, 91 of mass, 91, 376 Universal gravitation, 363, 371, 481 Use of standard formulas, 182 Value, mean, 211 Variable, 5 continuous, 5 dependent, 6 discontinuous, 5 independent, 6, 149 separable, 446 Vector quantity, 21, 355 Velocity, 18 angular, 92 average, 22, 23 constant, 20 instantaneous, 1 on a curve, 94 tangential, 94 uniform, 20 Verification, Newton's, 480 Vibrations, damped, 73 Volume by an area, 268 of frustum, 270 Volumes, 267 by double integration, 324 by triple integration, 317, 321 Water pressure, 357 Wave curve, 72 Wetted perimeter, 129 Witch of Agnesi, 132 Work, 167 I £= • + h.i y -i) 3, ■■I ■ 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. RENEWALS ONLY— TEL. NO. 642-3405 This book is due on the last date stamped below, or on the date to which renewed. Renewed books are subject to immediate recall. fiEC'D LD D lt'2 9 Q Ay , r-t S mil d ^ \Q?^ 7 ^^ OAJVI47 5 m JUN i ^' 1 ' REC'DLD JUN3 71-5?£fc» ■ ^K ■m» rf^s 82 1939 LD21A-60m-3,'70 (N5382sl0)476-A-32 General Library University of California Berkeley (MOOOSlUfllOti Berkeley U. C. BERKELEY LIBRARIES II llll UN III I HUM Ml f [ II III 47, •i UNIVERSITY OF CALIFORNIA UfcRARY