ELEMENTARY CALCULUS BY FREDERICK S. WOODS AND FREDERICK H. BAILEY PIlOlfMSSOUS 0V MATHEMATICS IN THK MASSACHUSETTS INSTITUTE OF TECHNOLOGY GINN AND COMPANY ItORVOK NI3W YORK OHtCAGO LONDON ATLANTA AM.A8 COLUMBUS SAN PBANOISOO COPYRIGHT, 1922, BY FREDERICK 8 WOODS AND FREDERICK H BAILEY ALL BIGHTS RESERVED PRINTED IN THB UNITED STATES OF AMERICA I 82611 153 GINN AND COMPANY . PRO- PRIETORS BOSTON > V.S.A. PKEFACB This book is adapted to the use of students in the first year in technical school or college, and is based upon the experience of the authors m teaching calculus to students in the Massa- chusetts Institute of Technology immediately upon entrance. It is accordingly assumed that the student has had college- entrance algebra, including graphs, and an elementary course in trigonometry, but that he has not studied analytic geometry. The first three chapters form an introductory course in which the fundamental ideas of the calculus are introduced, including derivative, differential, and the definite integral, but the formal work is restricted to that involving only the poly- nomial. These chapters alone are well fitted for a short course of about a term. The definition of the derivative is obtained through the concept of speed, using familiar illustrations, and the idea of a derivative as measuring the rate of change of related quantities is emphasized. The slope of a curve is introduced later. This is designed to prevent the student from acquiring the notion that the derivative is fundamentally a geometric concept. For the same reason, problems from mechanics are prominent throughout the book. With Chapter IV a more formal development of the subject begins, and certain portions of analytic geometry are introduced as needed. These include, among other things, the straight line, the conic sections, the cycloid, and polar coordinates. The book contains a large number of well-graded exercises for the student. Drill exercises are placed at the end of most sec- tions, and a miscellaneous 'Set of exercises, for review or further work, is found at the end of each chapter except the first. ui iv PREFACE Throughout the book, the authors believe, the matter is pre- sented in a manner which is well within the capacity of a first- year student to understand. They have endeavored to teach the calculus from a common-sense standpoint as a very useful tool. They have used as much mathematical rigor as the student is able to understand, but have refrained from raising the more difficult questions which the student in his first course is able neither to appreciate nor to master. Students who have completed this text and wish to continue their study of mathematics may next take a brief course in differential equations and then a course in advanced calculus, or they may take a course m advanced calculus which includes differential equations. It would also be desirable for such stu- dents to have a brief course in analytic geometry, which may either follow this text directly or come later. This arrangement of work the authors consider preferable to the one for a long time common in American colleges by which courses in higher algebra and analytic geometry precede the calculus. However, the teacher who prefers to follow the older arrangement will find this text adapted to such a program. F. S. WOODS F H BAILEY CONTENTS CHAPTER I. EATES SECTION PAGE 1 Limits . 1 2 Average speed . . . .... 3 3. True speed ... . 5 4. Algebiaic method . .... . 8 5 Acceleiation . ....... 9 G Rate of change ........ .11 CHAPTER II. DIFFERENTIATION 7 The derivative . . . . . .... .15 8. Differentiation of a polynomial ... 18 9 Sign of the derivative ... . . . .... 20 10 Velocity and acceleration (continued) . ... 21 11 Rate of change (continued) . ... ... 24 12 Graphs . .... .... 27 13 Real roots of an equation 30 14 Slope of a straight line . .... .31 15 Slope of a curve ... . 36 16 The second derivative . . , .39 17. Maxima and minima . . ... 41 18 Integration ... ... 44 19 Area . 47 20 Differentials ...*... 50 21. Appioximations ... . ... 53 General exercises .... . . . . - . .55 CHAPTER III. SUMMATION 22. Area by summation . . . ... 00 23. The definite integral .... . 62 24. The general summation problem .66 25. Pressure 68 26 Yolume . 71 General exercises ... 76 T vi CONTENTS CHAPTER IV. ALGEBRAIC FUNCTIONS SECTION PAGE 27 Distance between two points ... . . . . 79 28. Circle ... . .... ... 79 29. Parabola ... 81 30 Parabolic segment . 83 81. Ellipse ... 85 32 Hyperbola . . 87 33. Other curves 91 34 Theorems on limits . . .93 35. Theorems on derivatives .... 94 36. Formulas 101 37 Differentiation of implicit functions ... 102 38 Tangent line ... . 104 39. The differentials dx, dy, ds 106 40. Motion in a curve ... ... 107 41. Related velocities and rates Ill General exercises . 113 CHAPTER V. TRIGONOMETRIC FUNCTIONS 42. Circular measure . .... ... 119 43 Graphs of trigonometric functions . . ... 121 44. Differentiation of trigonometric functions 124 45. Simple harmonic motion . .... 127 46 Graphs of inverse trigonometric functions 130 47 Differentiation of inverse trigonometric functions . . 131 48. Angular velocity . . ... . ... . 185 49. The cycloid . . .... ... . 137 50. Curvature 189 51 Polar coSrdinates , 142 53. The differentials dr, dO, ds, in polar coordinates . . . 146 General exercises . . . . 149 CHAPTER VI. EXPONENTIAL AND LOGARITHMIC FUNCTIONS 53. The exponential function .... 154 54. The logarithm . . 154 55. Certain empirical equations 159 56. Differentiation . . 163 57. The compound-interest law 166 General exercises 168 CONTENTS vii CHAPTER VII. SERIES SECTION PAGE! 58. Power series . . . . . 172 59. Maclaurm's series . 173 60. Taylor's seuos . . . 177 General exercises . . ... 179 CHAPTER VIII. PARTIAL DIFFERENTIATION 61 Partial differentiation . . . 181 62. Higher partial denvatives . 184 63. Total differential of a function of two variables . . . 185 64 Rate of change ... .189 General exercises . . . 191 CHAPTER IX. INTEGRATION 65. Introduction . . . 194 66. Integral of u n . . 195 67-68 Other algebraic integrands 199 69 Integrals of trigonometric functions ... 205 70. Integrals of exponential functions . . ... 207 71-72. Substitutions . . . 208 73-74 Integration by parts . . . 212 75 Integration of rational fractions ... . 216 76. Table of integrals . . 217 General exercises ... .... . 220 CHAPTER X. APPLICATIONS 77. Review problems . ... 225 78. Infinite limits, or integrand 229 79. Area in polar cobrdmates . 230 80. Mean value of a function . .... 233 81. Length of a plane curve 235 82. Work 237 General exercises ... . . ... 239 CHAPTER XI. REPEATED INTEGRATION 83. Double integrals ... . . . . 244 84. Area as a double integral . . . . . . 246 85. Center of gravity . 249 yiii CONTENTS SECTION PAGE 86 Center of gravity of a composite area . . . 255 87 Theoiems . . . 257 88. Moment of inertia 260 89 Moments of inertia about parallel axes . 266 90 Space cooidinates 269 91. Certain sui faces . .... 271 92. Volume . . . .... .277 93 Center of gravity of a solid . 282 94 Moment of inertia of a solid . . 283 General exeicises . . . . . 286 ANSWERS 291 INDEX 315 ELEMENTARY CALCULUS * CHAPTER I RATES 1. Limits. Since the calculus is based upon the idea of a limit it is necessary to have a clear understanding of the word. Two examples already familiar to the student will be sufficient. In finding the aiea of a circle m plane geometry it is usual to begin by inscribing a regular polygon in the circle. The area of the polygon differs from that of the circle by a certain amount. As the number of sides of the polygon is increased, this difference becomes less and less. Moreover, if we take any small number 0, we can find an inscribed polygon whose area differs from that of the circle by less than e\ and if one such polygon has been found, any polygon with a larger number of sides will still differ m area from the circle by less than e. The area of the circle is said to be the limit of the area of the inscribed polygon. As another example of a limit consider the geometric progres- sion with an unlimited number -of terms The sum of the first two terms of this series is 1, the sum of the first three terms is If, the sum of the first four terms is 1-J, and so on. It may be found by trial and is proved in the algebras that the sum of the terms becomes more nearly equal to 2 as the number of terms which are taken becomes greater. Moreover, it may be shown that if any small number e is assumed, it is possible to take a number of terms n so that the sum of these terms differs from 2 by less than e. If a value of n has thus been found, then the sum of a number of terms 1 2 RATES greater than n will still differ from 2 by less than e. The number 2 is said to be the limit of the sum of the first n terms of the series. In each of these two examples there is a certain variable namely, the area of the inscribed polygon of n sides in one case and the sum of the first n terms of the series in the other case and a certain constant, the area of the circle and the number 2 respectively. In each case the difference between the constant and the variable may be made less than any small number e by taking n sufficiently large, and this difference then continues to be less than e for any larger value of n. This is the essential property of a limit, which may be defined as follows: A constant A is said to be the limit of a variable Xif, as the vari- able changes its value according to some law, the difference between the variable and the constant becomes and remains less than any small quantity which may be assigned. The definition does not say that the variable never reaches its limit. In most cases in this book, however, the variable fails to do so, as in the two examples already given. For the polygon is never exactly a circle, nor is the sum. of the terms of the series exactly 2. Examples may be given, however, of a variable's becoming equal to its limit, as in the case of a swinging pendulum finally coming to rest. But the fact that a variable may never reach its limit does not make the limit inexact. There is nothing inexact about the area of a circle or about the number 2. The student should notice the significance of the word " remains" in the definition. If a railroad train approaches a station, the difference between the position of the train and a point on the track opposite the station becomes less than any number which may be named ; but if the train keeps on by the station, that difference does not remain small. Hence there is no limit approached in this case. If X is a variable and A a constant which X approaches as a limit, it follows from the definition that we may write X**A + e, (1) SPEED 3 where e is a quantity (not necessarily positive) which may be made, and then will remain, as small as we please. Conversely, if as the result of any reasoning we arrive at a formula of the form (1) where X is a variable and A a constant, and if we see that we can make e as small as we please and that it will then remain just as small or smaller as X varies, we can say that A is the limit of X. It is in this way that we shall determine limits in the following pages. 2. Average speed. Let us suppose a body (for example, an automobile) moving from a point A to a point B (Fig. 1), a distance of 100 mi If the automobile takes 5 hr. for the trip, we are accustomed to say that it has traveled at the rate of 20 mi. an hour. Everybody knows A p Q B that this does not mean that the ' ' ' automobile went exactly 20 mi. in each hour of the trip, exactly 10 mi. in each half hour, exactly 5 mi in each quarter hour, and so on. Probably no automobile ever ran in such a way as that. The expression " 20 mi. an hour " may be understood as meaning that a fictitious automobile traveling in the steady manner* just described would actually cover the 100 mi. in just 5 hr. ; but for the actual automobile which made the trip, " 20 mi. an hour " gives only a certain average speed. So if a man walks 9 mi. in 3 hr., he has an average speed of 3 mi. an hour. If a stone falls 144 ft. in 3 sec., it has an average speed of 48 ft. per second. In neither of these cases, however, does the average speed give us any information as to the actual speed of the moving object at a given instant of its motion. The point we are making is so important, and it is so often overlooked, that we repeat it in the following statement: If a "body traverses a distance in a certain time, the average speed of the body in that time is given ly the formula, , distance average speed = time but this formula does not in general give the true speed at any given time. 4 KATES EXERCISES 1. A man runs a half mile m 2 mm and 3 sec. What is his average speed m feet per second ? 2. A man walks a mile m 25 min. What is his average speed in yards per second ? 3. A train 600 ft. long takes 10 sec. to pass a given milepost. What is its average speed in miles per hour ? 4. A stone is thrown directly downward from the edge of a vertical cliff Two seconds afterwards it passes a point 84 ft down the side of the cliff, and 4 sec after it is thrown it passes a point 296 ft. down the side of the cliff What is the average speed of the stone in falling between the two mentioned points ? 5. A railroad train runs on the following schedule : Boston 10 00 A M. Worcester (45 mi.) 11.10 Springfield (90 mi ) 12.35 P M. Pittsfield (151 mi.) 225 Albany (201 mi.) 355 Find the average speed between each two consecutive stations and for the entire trip. 6. A body moves four times around a circle of diameter 6 ft in 1 min. What is its average speed in feet per second ? 7. A block slides from the top to the bottom of an inclined plane which makes an angle of 30 with the horizontal. If the top is 50 ft. higher than the bottom and it requires f mm. for the block to slide down, what is its average speed in feet per second ? 8. Two roads intersect at a point C B starts along one road toward C from a point 5 mi. distant from C and walks at an average speed of 3 mi an hour. Twenty minutes later A starts along the other road toward C from a point 2 mi. away from G At what average speed must A walk if he is to reach C at the same instant that B arrives ? 9. A man rows across a river $ mi. wide and lands at a point mi. farther down the river. If the banks of the river are parallel straight lines and he takes ^ hr. to cross, what is his average speed in feet per minute if his course is a straight line ? SPEED 5 10. A trolley car is running along a straight street at an average speed of 12 mi. per hour. A house is 50 yd. back from the car track and 100 yd. up the street from a car station. A man comes out of the house when a car is 200 yd away from the station What must be the average speed of the man m yards per minute if he goes in a straight line to the station and arrives at the same instant as the car ? 3. True speed. How then shall we determine the speed at which a moving body passes any given fixed point P in its motion (Fig. 1) ? In answering this question the mathema- tician begins exactly as does the policeman in setting a trap for speeding. He takes a point Q near to P and determines the distance PQ and the time it takes to pass over that distance. Suppose, for example, that the distance PQ is ^ mi. and the time is 1 min. Then, by 2, the average speed with which the distance is traversed is mi. = 30 mi. per hour. hr - This is merely the average speed, however, and can no more be taken for the true speed at the point P than could the 20 mi. an hour which we obtained by considering the entire distance A3. It is true that the 30 mi. an hour obtained from the interval PQ is likely to be nearer the true speed at B than was the 20 ini. an hour obtained from AS, because the interval PQ is shorter. The last statement suggests a method for obtaining a still better measure of the speed at P ; namely, by taking the interval PQ still smaller. Suppose, for example, that PQ is taken as fa mi. and that the time is 6J sec. A calculation shows that the average speed at which this distance was traversed was 36 mi. an hour. This is a better value for the speed at P. Now, having seen that we get a better value for the speed at P each time that we decrease the size of the interval PQ, we can find no end to the process except by means of the idea of a limit denned in 1. We say, in fact, that the speed of a moving body at any point of its path u the limit approached ly the average 6 RATES speed computed for a small distance beginning at that point, the limit to be determined by taking this distance smaller and smaller. This definition may seem to the student a little intricate, and we shall proceed to explain it further. In the case of the automobile, which we have been using for an illustration, there are practical difficulties in taking a very small distance, because neither the measurement of the distance nor that of the time can be exact. This does not alter ^ the fact, however, that theoretically to determine the speed of the car we ought to find the time it takes to go an extremely minute distance, and the more minute the dis- tance the better the result. For example, if it were possi- ble to discover that an automobile ran ^ in. in -^^-g- sec., we should be pretty safe in saying that it was moving at p a speed of 30 mi. an hour. _p Such fineness of measurement is, of course, impossible ; but if an algebraic formula connecting the distance and the time is known, the calculation can be made as fine as this and finer. We will therefore take a familiar case in which such a formula is known ; namely, that of a falling body. Let us take the formula from physics that if s is the distance through which a body falls from rest, and t is the time it takes to fall the distance s t then s = 16* 3 , (1) and let us ask what is the speed of the body at the instant when t SB 2. In Fig. 2 let be the point from which the body falls, % its position when t 2, and 7 its position a short time later. The average speed with which the body falls through the distance PP Z is, by 2, that distance divided by the time it takes to traverse it. We shall proceed to make several succes- sive calculations of this average speed, assuming 7^ and the corresponding time smaller and smaller. In so doing it will be convenient to introduce a notation as follows : Let ^ represent the time at which the body reaches 2?, and t a the time at which it reaches J%. Also let ^ equal the distance OP^ and a the distance OP a . Then s^-s^P^ and SPEED 7 tf 2 ^ is the time it takes to traverse the distance P^. Then the average speed at which the body traverses Ufa is So Si (2) Now, by the statement of our particular problem, Therefore, from (1), s 1= 16 (2) 2 = 64. We shall assume a value of t a a little larger than 2, compute s a from (1), and the average speed from (2). That having been done, we shall take t z a little nearer to 2 than it was at first, and again compute the average speed This we shall do repeatedly, each time taking t a nearer to 2. Our results can best be exhibited in the form of a table, as follows : t a -L 2.1 70.56 .1 656 656 201 646416 01 .6416 64.16 2 001 64.064016 .001 064016 64 016 2.0001 64 00640016 .0001 .00640016 64 0016 It is fairly evident from the above arithmetical work that as the time i? 3 ^ and the corresponding distance s s s x become smaller, the more nearly is the average speed equal to 64. Therefore we are led to infer, in accordance with 1, that the speed at which the body passes the point J$ is 64 ft. per second. In the same manner the speed of the body may be computed at any point of its path by a purely arithmetical calculation. In the next section we shall go farther with the same problem and employ algebra. EXERCISES 1. Estimate the speed of a falling body at the end of the third second, given that s = 16 t z , exhibiting the work in a table. 2. Estimate the speed of the body in Ex. 1 at the end of the fourth, second, exhibiting the work in a table. 8 RATES 3. The distance of a falling body from a fixed point, at, any th is given by the equation s = 100 + 16 t*. Estimate the speed of t body at the end of the fourth second, exhibiting the work in a tab 4. A body is falling so that the distance traversed in the tiuu is given by the equation s = 16 1 2 + 10 1 Estimate tho speed of 1,1 body when t = 2 sec , exhibiting the work in a table. 6. A body is thrown upward with such a speed that at. any tin its distance from the surface of the earth is given by tho equsitu s = 100 1 16 t a . Estimate its speed at the end o a sexumd, exhibi ing the work in a table. 6. The distance of a falling body from a fixed point a,t any tiin is given by the equation s = 50 -f- 20 1 + 16 1*. Estimate its spee at the end of the first second, exhibiting tho work in a table. 4. Algebraic method. In this section we shall show how it i possible to derive an algebraic formula for the speed, still con fining ourselves to the special example of the falling body whos< equation of motion is ., , 2 ^ s = i.o t . (1 Instead of taking a definite numerical value for f t , wo dial keep the algebraic symbol t r Then Also, instead of adding successive small qimntit,ie,s to t to get 2 , we shall represent the amount added by the al symbol^. That is, , . , . and, from (1), s z== 16 ( Hence * 2 - ^ = 16^+ A) 2 - 10 tf. 82 This is a general expression for the distance / t V* in Ffj. Now t z -t : = Ti, and therefore the average speed with body traverses %P S is represented by tho expression It is obvious that if Ji is taken smaller and smidH Iho nver- age speed approaches 32 ^ as a limit, In i'aufc, tho quantity 83^ ACCELEKATION 9 satisfies exactly the definition of limit given in 1. For if e is any number, no matter how small, we have simply to take 10 Ti < e in order that the average speed should differ from 32 ^ by less than e ; and after that, for still smaller values of h, this d iff 01 once remains less than e. We have, then, the result that if the space traversed by a 1 ailing body is given by the formula the speed of the body at any time is given by the formula It may be well to emphasize that this is not the result which would be obtained by dividing s by t. EXERCISE Find the speed in each of the problems in 3 by the method explained in this section. 5. Acceleration. Let us consider the case of a body which is supposed to move so that if s is the distance in feet and t is the time in seconds, s = t s (V) Then, by the method of 4, we find that if v is the speed in feet per second, = 3 i 2 We see that when t = 1, v = 3 ; when t = 2, v 12 ; when t = 3, v = 27 ; and so on. That is, the body is gaming speed with each second. We wish to find how fast it is gaining speed. To find this out, let us take a specific time (,-4. The speed at this time we call v lt so that, by (2), i= = 3 (4) 2 = 48 ft. per second. Take * 2 =5; then v = 3 (5) 2 ~ 75 ft. per second. 10 RATES Therefore the body has gained 75 48 = 27 units of speed in 1 sec. This number, then, represents the average rate at which the body is gaining speed during the particular second con- sidered. It does not give exactly the rate at which the speed is increasing at the beginning of the second, because the rate is constantly changing. To find how fast the body is gaining speed when ^ = 4, we must proceed exactly as we did in finding the speed itself. That is, we must compute the gain of speed in a very small interval of time and compare that with the time. Let us take t.= 4.1. a Then 2 = 50.43 and ^^=2.43. Then the body has gained 2.43 units of speed hi .1 sec., which 2 43 is at the rate of -~- 24.3 units per second. Again, take 2 =4.01. Then v 2 = 48.2403 and ^^=.2403, A gain of ,2403 units of speed in .01 sec. is at the rate of ' . =24.03 units per second. We exhibit these results, and one other obtained in the same way, in a table: 4,1 60.43 .1 2.48 24.3 4.01 48.2403 .01 .2403 24,08 4.001 48.024003 001 .024003 24.003 The rate at which a body is gaining speed is called its acceleration. Our discussion suggests that in. the example before us the acceleration is 24 units of speed per second. But the unit of speed is expressed in feet per second, and so we say that the acceleration is 24 ft. per second per second. KATE OF OHANG-E 11 By the method used in determining speed, we may get a general formula to determine the acceleration from equation (2). We take , , r Then v a = and v z ~Vi= The average rate at which the speed is gamed is then h, and the limit of this, as h becomes smaller and smaller, is obviously 6t t . This is, of course, a result which is valid only for the special example that we are considering. A general statement of the meaning of acceleration is as follows: . , ,. ,. ., , change in speed Acceleration = limit of . 5 change in time EXERCISES 1. If s = 4 1*, find the speed and the acceleration when t = # r 2. If s = rf 8 + 1*, find the speed and the acceleration when t = 2. 3. If s = 3 t z + 2 1 + 5, how far has the body moved at the end of the fifth second ? With what speed does it reach that point, and how fast is the speed increasing ? 4. If s = 4 i 8 + 2 1 3 + 1 + 4, find the distance traveled and the speed when t = 2. 5. If s = $ t 6 + 1 -t- 10, find the speed and the acceleration when t=z>2 and when t = 3. Compare the average speed and the average acceleration during this second with the speed and the acceleration at the beginning and the end of the second. 6. If s = at + &, show that the speed is constant. 7. If at 2 + to + G, show that' the acceleration is constant. 8. If s = at* + bt* + ct +/, find the formulas for the speed and the acceleration. 6. Rate of change. Let us consider another example which may be solved by processes similar to those used for determining speed and acceleration. L2 KATES m the A stone is thrown into still water, forming ripples which ,ravel from the center of disturbance in the form of circles Tig. 3). Let r be the radius of a circle and A its area. Then J "J /-I N A=irr. (1) We wish to compare changes in the area with changes adius. If we take r t = 3, then A^ 9 TT; and if we take ,hen ^t a = 167T. That is, a change )f 1 unit in ?*, when r 3, causes i change of 7 TT units in A. We are ^empted to say that A is increas- ng TTT times as fast as r. But jefore making such a statement it s well to see whether this law holds 'or all changes made in r, starting 'rom r 1 = 8, and especially for small jhanges in r. We will again exhibit the calcu- ^ IQ 3 ation in the form of a table. Here ^=3, A^Qnr, and r z is a variously assumed value of r not nuch different from 3. 1 01 .001 9.61 TT 9.0601 TT 9,006001 f .1 .01 001 A,,- A! .01 TT .0001 7T 000001 IT r a - r x ITT 6.01 TT 6 001 TT The number in the last column changes with the number a r^ Therefore, if we wish to measure the rate at which A is 3hanging as compared with r at the instant when r = 3, we must bake the limit of the numbers in the last column. That limit is abviously 6 TT. We say that at the instant when r 8, the area of the circle is changing 6 TT times as fast as the radius. Hence, if the radius Is changing at the rate of 2 ft. per second, for example, the area is changing at the rate of 12 TT sq. ft. per second. Another way of expressing the saine idea is to say that when, r ts 3, the rate RATE OF CHANGE 13 of change of A with respect to r is 6 TT. Whichever form of expres- sion is used, we mean that the change in the area divided by the change in the radius approaches a limit 6 TT. The number 6 TT was, of course, dependent upon the value r = 3, with which we started. Another value of ^ assumed at the start would produce another result. For example, we may compute that when r x = 4, the rate of change of A with respect to r is 8-rr; and when r^= 5, the rate is 10 ?r. Better still, we may derive a general formula which will give us the required rate for any value of r^. To do this take Then A 2 = TT (r* + 2 rji + A 2 ) and A z - A 1 = TT (2 r x h + 7i 3 ) ; A A SO that 2 l = 2 TH* + 7i7T. T V 'z 'l The limit of this quantity, as Ti is taken smaller and smaller, is Hence we see that from formula (1) we may derive the fact that the rate of change of A with respect to r is 2 TIT. EXERCISES 1. In the example of the text, if the circumference of the circle winch bounds the disturbed area is 10 ft and the circumference is increasing at the rate of 3 ft. per second, how fast is the area increasing ? 2. In the example of the text find a general expression for the rate of change of the area with respect to the circumference. 3. A soap bubble is expanding, always remaining spherical. If the radius of the bubble is increasing at the rate of 2 in. per second, how fast is the volume increasing ? 4. In Ex 3 find the general expression for the rate of change of the volume with respect to the radius. 5. If a soap bubble is expanding as in Ex. 3, how fast is the area of its surface increasing 9 14 BATES 6. In Ex. 5 find the general expression for the rate of change of the surface with respect to the radius 7. A cube of metal is expanding under the influence of heat. Assuming that the metal retains the form of a cube, find the rate of change at which the volume is increasing with respect to an edge. 8. The altitude of a right circular cylinder is always equal to the diameter of the base. If the cylinder is assumed to expand, always retaining its form and proportions, what is the rate of change of the volume with respect to the radius of the base ? 9. Find the rate of change of the area of a sector of a circle of radms 6 ft with respect to the angle at the center of the circle. 10. Find the rate of change of the area of a sector of a circle with respect to the radius of the circle if the angle at the center 7T of the circle is always j. What is the value of the rate when the radius is 8 in. ? CHAPTER II DIFFERENTIATION 7. The derivative. The examples we have been considering in the foregoing sections of the book are alike in the methods used to solve them. We shall proceed now to examine this method so as to bring out its general character. In the first place, we notice that we have to do with two quantities so related that the value of one depends upon the value of the other. Thus the distance traveled by a moving body depends upon the time, and the area of a circle depends upon the radius. In such a case one quantity is said to be a function of the other. That is, a quantity y is said to be a function of another quantity, x, if the value of y is determined ly the value of x. The fact that y is a function of x is expressed by the equation y=/<v>, and the particular value of the function when x has a definite value a is then expressed as /(a). Thus, if f(x) = x*- 3 aM-42j + l, /(2)=2-3(2) 3 +4(2) + l = 5, /(0)=0-3(0) + 4(0) + 1 = 1. It is in general true that a change in x causes a change in the function y, and that if the change in x is sufficiently small, the change in y is small also. Some exceptions to this may be noticed later, but this is the general rule. A change in a; is called an increment of x and is denoted by the symbol Ao? (read " delta x "). Similarly, a change in y is called an increment of y and is denoted by Ay. For example, consider 15 16 DIFFERENTIATION When = 2, # = 12. When a? =2.1, y = 12.71. The change in x is .1, and the change m y is .71, and we write Aa = .l, Ay = .71. So, in general, if a^ is one value of a;, and x z a second value of #, then Ax = xs, i x l1 or #2=0^4- Are; (1) and if y x and y 2 are the corresponding values of y, then ty=y z -Vv <* y 2 =y a 4Ay. (2) The word increment really means "increase," but as we are dealing with algebraic quantities, the increment may be nega- tive when it means a decrease. For example, if a man invests $1000 and at the end of a year has $1200, the increment of his wealth is $200. If he has $800 at the end of the year, the increment is $200. So, if a thermometer registers 65 in the morning and 57 at night, the increment is ,8. The incre- ment is always the second value of the quantity considered minus the first value. Now, having determined increments of x and of y, the next step is to compare them by dividing the increment of y by the increment of x. This is what we did in each of the three problems we have worked in 3-6. In finding speed we began by dividing an increment of distance by an increment of time, in finding acceleration we began by dividing an increment of speed by an increment of time, and in discussing the ripples in the water we began by dividing an increment of area by an increment of radius. . The quotient thus obtained is - That is, Ax Ay _ increment of y _ change in y AJC increment of x change m x An examination of the tables of numerical values in 3, 5, 6 shows that the quotient ^ depends upon the magnitude of Asc, and that in each problem it was necessary to determine its limit DEEIVATIVE 17 as A# approached zero. This limit is called the derivative of y with respect to x, and is denoted by the symbol -~- We have then fy v -j. Ay v , change in v ~ = limit oi ~ = limit of . . a ax Ax change in x At present the student is to take the symbol -^ not as a dx fraction, but as one undivided symbol to represent the deriva- tive. Later we shall consider what meaning may be given to dx and dy separately. At this stage the form -^ suggests simply A (%>*Xs the fraction ^, which has approached a definite limiting value, The process of finding the derivative is called differentiation, and we are said to differentiate y with respect to x. From the definition and from the examples with which we began the book, the process is seen to involve the following four steps : 1. The assumption at pleasure of Ax. 2. The determination of the corresponding Ay. Aw 8. The division of Ay by Aa: to form -*- Ax 4. The determination of the limit approached by the quotient in step 3 as the increment assumed in step 1 approaches zero. Let us apply this method to finding -J*- when y - - Let x t be a definite value of x, and y,- the corresponding value of y. x i 1. Take Ax = 7i. Then, by (1), x^x^h. 2. Then ^l^-J.; 11 h whence, by (2), Ay- 3. By division, T^ = . . 7 J Ax asf+AiS 18 DIEFEBENTIATION 4. By inspection it is evident that the limit, as h approaches zero, is 5 which is the value of the derivative when x-=x*. xl But x^ may be any value of x ; so we may drop the subscript 1 and write as a general formula dx 3? EXERCISES Pind from the definition the derivatives of the following ex- pressions : 6. y-rf + i. 2 2. y = *+2a*+i. e -y=2+* 3. y = a: 4 ~a 8 . 7. y = "+ \x*+ x - 5. i =i 8 = 3a;2 + 1 . y ~~ x s y ~~ x 8. Differentiation of a polynomial. We shall now obtain for- mulas by means of which the derivative of a polynomial may be written down quickly. In the first place we have the theorem : The derivative of a polynomial is the sum of the derivatives of its separate terms. This follows from the definition of a derivative if we reflect that the change in a polynomial is the sum of the changes in its terms. A more formal proof will be given later. We have then to consider the terms of a polynomial, which have in general the form #af. Since we wish to have general formulas, we shall omit the subscript 1 in denoting the first values of x and y. We have then the theorem: If y =s ax*) where n is a positive integer and a is a constant, then (1) dx ^ ' To prove this, apply the method of 7 : 1. Take Aa?=A; whence jc a = x + h. POLYNOMIAL 19 2. Then y 3 =ax2 = a(z + A)"; whence Ay == a (x + A)" are" .4 3. By division, ^ = a(nx^- l + n ^ n ~ 1 ^ tf-'h + . . . + A"" 1 )- ZA3/ j > 4. By inspection, the limit approached by > as Ji approaches zero, is seen to be anx n ~ l . Therefore ~ = anx n ~\ as was to be proved. dx The polynomial may also have a term of the form ax. This is only a special case of (1) with w = l, but for clearness we say explicitly: If y = ax, where a is a constant, then - (2) dx Finally, a polynomial may have a constant term. c. For this we have the theorem: If y = G, where c is a constant, then f^O. (3) ax The proof of this is that as c is constant, A<? is always zero, no matter what the value of A is. Hence and therefore = 0. ax As an example of the use of the theorems, consider y We write at once ^ ax 20 DIFFERENTIATION EXERCISES Find the derivative of each of the following polynomials : 1. 3.2 + a- 3. 6. x 1 + 7 x s + 21 8 - 14 as 2. o; 8 +2a; + l. 7. x*- x* + 4a - 1 3. o: 4 + 4a5 8 +6a; 2 +4a; + l. 8. 3 4. x 6 + |a: 4 +2a; 2 +3. 9. aa 5. a; 6 -4a: 4 + a: 2 -4a; 10. a + bx z + ca; 4 + ex 6 . i 9. Sign of the derivative. If ~ is positive, an increase in the dx , dy . value of x causes an increase in the value of y. If ~ is negative, dx an increase in the value of x causes a decrease in the value of y. To prove this theorem, let us consider that -jj- is positive. Then, since ~- is the limit of -> it follows that is positive dx A A for sufficiently small values of Aa;; that is, if A# is assumed positive, ky is also positive, and therefore an increase of x causes an increase of y. Similarly, if -j- is negative, it follows Aw that is negative for sufficiently small values of Aa: ; that is, if Arc is positive, A?/ must be negative, so that an increase of x causes a decrease of y. In applying this theorem it is necessary to determine the sign of a derivative. In case the derivative is a polynomial, this may be conveniently done by breaking it up into factors and considering the sign of each factor. It is obvious that a factor of the form x a is positive when x is greater than a, and negative when x is less than a. Suppose, then, we wish to determine the sign of There are three factors to consider, and three numbers are im- portant ; namely, those which make one of the factors equal to zero. These numbers arranged in order of size are 3, 1, and 6. We have the four cases : 1. x< 3. All factors are negative and the product is Dative. VELOCITY AND ACCELERATION 21 2. 3 < x < 1, The first factor is positive and the others are negative. Therefore the product is positive. 3. 1 < x < 6. The first two factors are positive and the last is negative. Therefore the product is negative. 4. x > 6. All factors are positive and the product is positive. As an example of the use of the theorem, suppose we have # = X s - 3; y*-Q x + 27, and ask for what values of x an increase in y will cause an increase in y. We form the derivative and factor it. Thus, ^==3a a -6a3-9==3(a: + r)(a;-3). //* Proceeding as above, we have the three cases: 1. x< 1. -^ is positive, and an increase in x therefore dx increases y. _ 2. 1 < x < 3. -^ is negative, and therefore an increase in x , dx decreases y. _ 3. a; > 3. -^ is positive, and therefore an increase in x in- creases y. These results may be checked by substituting values of x in the derivative. EXERCISES Pind for what values of x each of the following expressions will increase if x is increased, and for what values of x they will decrease if x is increased : 1. a! > -4aj + 6. 6. 2. 3a a +10a; + 7. 7. x a - x*- 5x + 5 3. l + Ssc-a: 2 . 8. 1 + 6x +12a: 2 + 8oj 8 . 4. 7_3a._3a; 2 9. 6 + 60; + 6a a - 2 3 - 3 5. 2 je 8 +3a; a - 12 + 17. 10. 12 - 12a - 6a; 2 + 4a; 8 + 10. Velocity and acceleration (continued). The method by which the speed of a body was determined in 4 was in reality a method of differentiation, and the speed was the derivative of the distance with respect to the time. In that discussion, how- ever, we SQ arranged each problem that the result was positive 22 DIFFERENTIATION and gave a numerical measure (feet per second, miles per hour, etc.) for the rate at which the body was moving. Since we may now expect, on occasion, negative signs, we will replace the word speed by the word velocity, which we denote by the letter v. In accordance with the previous work, we have da -, N * a) The distinction between speed and velocity, as we use the words, is simply one of algebraic sign. The speed is the numer- ical measure of the velocity and is always positive, but the velocity may be either positive or negative. From 9 the velocity is positive when the body so moves that s increases with the time. This happens when the body moves in the direction in which s is measured. On the other hand, the velocity is negative when the body so moves that s decreases with the time. This happens when the body moves in the direc- tion opposite to that in which s is measured. For example, suppose a body moves from A to B (Fig. 1), a distance of 100 mi., and let P be the position of the body at a time t, and let us assume that we know that AP = 4 1. If we measure s from A, we have ds whence v = - = 4. dt On the other hand, if we measure s from J5, we have whence v - = 4. dt "We will now define acceleration by the formula dv *=w in full accord with 5 ; or, since v is found by differentiating s, we may write VELOCITY AND ACCELERATION 23 where the symbol on the right indicates that * is to be differ- entiated twice in succession. The result is called a second derivative, A positive acceleration means that the velocity is increasing, but it must be remembered that the word increase is used in the algebraic sense. Thus, if a number changes from 8 to 5, it algebraically increases, although numerically it decreases. Hence, if a negative velocity is increased, the speed is less. Simi- larly, if the acceleration is negative, the velocity is decreasing, but if the velocity is negative, that means an increasing speed. There are four cases of combinations of signs which may occur : 1. v positive, a positive. The body is moving in the direction in which s is measured and with increasing speed. 2. v positive, a negative. The body is moving in the direction in which s is measured and with decreasing speed. 8. v negative, a positive. The body is moving in the direction opposite to that in which s is measured and with decreasing speed. 4. v negative, a negative. The body is moving in the direc- tion opposite to that in which 8 is measured and with increasing As an example, suppose a body thrown vertically into the air with a velocity of 96 ft. per second. From physics, if s is meas- ured up from the earth, we have From this equation we compute v = 96 - 32 t, When t< 3, v is positive and a is negative. The body is going up with decreasing speed. When t > 3, v is negative and a is negative. The body is coming down with increasing speed. On the other hand, suppose a body is thrown down from a height with a velocity of 96 ft. per second. Then, if 8 is measured 24 DIFFERENTIATION down from the point from which the body is thrown, we have, from physics, 8 = Wt+lQt*, from which we compute v = 96 + 32 1, = 32. Here v is always positive and a is always positive. There- fore the hody is always going down (until it strikes) with an increasing speed. EXERCISES In the following examples find the expression for the velocity and determine when the body is moving in the direction in which s is measured and when in the opposite direction . 1. s = t* 3t + 6. 3. s = t* Qt* + 24* + 3. 2. s = 10* - tf 4. 's = 8 + 12* - 6i 2 + < 8 . 5. s = 25 4 - In the following examples find the expressions for the velocity and the acceleration, and determine the periods of time during which the velocity is increasing and those during which it is decreasing . 6. s = 3z5-4i + 4 8. s = %t* 2 2 7. s = 1 + 5t - #. 9. s = t* - 5t* + St + 1. 10. s = 1 + 4 + 2t* t 8 . 11. Rate of change (continued). In 6 .we have solved a problem in which we are finally led to find the rate of increase of the area of a circle with respect to its radius. This problem is typical of a good many others. Let x be an independent variable and y a function of x. A change Aa; made in x causes a change Ay in y. The fraction compares the change in y with the change in x. For exam- ple, if Aa; = .001, and Ay = .009061, then we may say that the change in y is at the average rate of ' = 9.061 per unit change in x. This does not mean that a unit change in a? would actually make a change of 9.061 units in y, any more than the RATE OF CHANGE 25 statement that an automobile is moving at the rate of 40 mi. an hour means that it actually goes 40 mi. in an hour's time. The fraction then gives a measure for the average rate at which y is changing compared with the change in x. But this measure depends upon the value of A#, as has been shown in the numerical calculations of 6. To obtain a measure of the instantaneous rate of change of y with respect to x which shall not depend upon the magnitude of A#, we must take the limit of ~ t as we did in G. Ax We have, therefore, the following definition : Tlie derivative -j- measures the rate of change ofy with respect to a\ ax Another way of putting the same thing is to say that if -^ has the value m, then y is changing m times as fast as x. Still another way of expressing the same idea is to say that the rate of change of y with respect to x is defined as meaning the limit of the ratio of a small change in y to a small change in x. We will illustrate the above general discussion, and at the same time show how it may be practically applied, by the following example, which we will first solve arithmetically and then by calculus. Suppose we have a vessel in the shape of a cone (Fig 4) of radius 3 in. and altitude 9 in. into which water is being poured at the rate of 100 cu. in. per second. Re- quired the rate at which the depth of the water is increasing when the depth is 6 in. From similar triangles in the figure, if h is the depth of the water and r the radius of its surface, r -. If Fis the volume of water, w , . i , ,^ FlG. 4 We are asked to find the rate at which the depth is increasing when h is C in. Let us call that depth so that 7^= 6. Then 26 DIFFERENTIATION y = 8 TT. Now we will increase \ by successive small amounts and see how great an increase in V^ is necessary to cause that change in \ ; that is, how much water must be poured in to raise the depth by that amount. The calculation may be tabulated as follows . A h AF AT A /i .1 .407 IT 4 07 ir 01 04007 TT 4.007 v .001 0040007 TT 4.0007 TT The limit of the numbers in the last column is evidently 4 TT. Therefore the volume is increasing 4 IT times as fast as the depth. But, by hypothesis, the volume is increasing at the rate of 100 cu. in. per second, so that the depth is increasing at the rate of - = 7.96 in. per second. 47T We have solved the problem by arithmetic to exhibit again the meaning of the derivative. The solution by calculus is much quicker. We begin by finding dV 1 , 3 aTe**' This is the general expression for the rate of change of V with respect to A, or, in other words, it tells us that V is instan- taneously increasing \ rf times as fast as li for any given A. Therefore, when A = 6, V is increasing 4 TT times as fast as A, and as V is increasing at the rate of 100 cu. m. per second, A is increasing at the rate of - = 7.96 in. per second. 4 7T EXERCISES 1. An icicle, which is melting, is always in the form of a right circular cone of which the vertical angle is 60 Find the rate of change of the volume of the icicle with respect to its length. 2. A series of right sections is made in a right circular cone of which the vertical angle is 90 How fast will the areas of the sec- tions be increasing if the cutting plane recedes from the vertex at the rate of 3 ft. per second ? GRAPHS 27 3. A solution is being poured into a conical filter at the rate of 5 cc per second and is running out at the rate of 1 ce. per second The radius of the top of the filter is 10 cm. and the depth of the filter is 30 cm Find the rate at which the level of the solution is rising in the filter when it is one fourth of the way* to the top. 4. A peg in the form of a right circular cone of which the ver- tical angle is 60 is being driven into the sand at the rate of 1 m. per second, the axis of the cone being perpendicular to the surface of the sand, which is a plane. How fast is the lateral surface of the peg disappearing in the sand when the vertex of the peg is 5 in. below the surface of the sand? 5. A trough is in the form of a right prism with its ends equi- lateral triangles placed vertically The length of the trough is 10 ft It contains water which leaks out at the rate of cu ft per minute. Find the rate, in inches per minute, at which the level of the water is sinking in the trough when the depth is 2 ft. 6. A trough is 10 ft. long, and its cross section, which is vertical, is a regular trapezoid with its top side 4ft. in length, its bottom side 2 ft , and its altitude 5 ft. It contains water to the depth of 3 ft , and water is running in so that the depth is increasing at the rate of 2 ft. per second. How fast is the water running in ? 7. A balloon is in the form of a right circular cone with a hemi- spherical top. The radius of the largest cross section is equal to the altitude of the cone. The shape and proportions of the balloon are assumed to be unaltered as the balloon is inflated. Find the rate of increase of the volume with respect to the total height of the balloon. 8. A spherical shell of ice surrounds a spherical iron ball concen- tric with it. The radius of the iron ball is 6 in. As the ice melts, how fast is the mass of the ice decreasing with respect to its thickness ? 12. Graphs. The relation between a variable x and a function y may be pictured to the eye by a graph. It is expected that students will have acquired some knowledge of the graph in the study of algebra, and the following brief discussion is given for a review. Take two lines OX and 07 (Fig. 5), intersecting at right angles at 0, which is called the origin of codrdinates. The line called the axis of as, and the line 0rthe axis of y ; together 28 DIFFERENTIATION they are called the coordinate axes, or axes of reference. On OX we lay off a distance OM equal to any given value of x, measur- ing to the right if x is positive and to the left if x is negative. From M we erect a perpendicular MP, equal in length to the value of y, measured up if y is positive and down if y is negative. The point P thus determmed is said to have the coordinates x and y and is denoted by (x, y). It follows that the numerical value of x measures the distance Y of the point P from OY, and the numerical value of y measures the distance of P from OX. The coor- dinate x is called the abscissa, and the coordinate y the ordinate. It is p evident that any pair of coordinates (x, y) fix a single point P, and that any point P has a single pair of M -^ coordinates. The point P is said to _, g be plotted when its position is fixed 111 this way, and the plotting is conveniently carried out on paper ruled for that purpose into squares. If y is a function of x, values of x may be assumed at pleasure and the corresponding values of y computed. Then each pair of values (x, y) may be plotted and a series of points found. The locus of these points is a curve called the graph of the function. It may happen that the locus consists of distinct portions not connected in the graph. In this case it is still customary to say that these portions together form a single curve. For example, let - ... _ r y~bz ar. (1) We assume values of x and compute values of y. The results are exhibited in the following table : -i o l 2 8 4 6 ' o 2/-60 4 6 6 4 0-0 These points are plotted and connected by -a Smooth curve* giving the result shown in F-ig. 6. This -curve should have the GEAPHS 29 property that the coordinates of any point on it satisfy equa- tion (1) and that any point whose coordinates satisfy (1) lies on the curve. It is called the graph both of the function y and of the equation (1), and equation (1) is called the equation of the curve. Of course we are absolutely sure of only those points whose coordinates we have actually computed. If greater accuracy is desired, more points must be found y by assuming fractional values of x. For instance, there is doubt as to the 7 . shape of the curve between the points 6 / \ (2, 6) and (3, C). We take, therefore, 4 ' > #=2^- and find / = 6|. This gives 3 ( us another point to aid us in draw- 2 ing the graph. Later, by use of the 1 calculus, we can show that this last -1 0^1 2 8 4 5 point is really the highest point of " -2 the curve. -s The curve (Fig. 6) gives us a ~ 4 graphical representation of the way "^ in which y varies with x. We see, for example, that when x varies from 1 to 2, y is increasing; that when x varies from 3 to 6, y is decreasing; and that at some point between (2, 6) and (3, 6), not yet exactly determined, y has its largest value. It is also evident that the steepness of the curve indicates in some way the rate at which y is increasing with respect to x. For example, when # = 1, an increase of 1 unit in x causes an increase of 6 units in y ; while when a =1, an increase of 1 unit in x causes an increase of only 2 units m y. The curve is therefore steeper when x = 1 than it is when x = 1. Now we have seen that the derivative - measures the rate ax of change of y with respect to x. Hence we expect the derivative to be connected in some way with the steepness of the curve. We shaft tihmwfnro /iianr.. *Hr. ^ QTL m gs \%. and" 15. USc Lib B'lore Q 515N22 ! 13838 in IIIIIIIIIIIIIIIIIMIII ~ 30 DIFFERENTIATION 1 EXERCISES Plot the graphs of the following equations : 1. y = 2x -|-3. 4. ?/ = a; 2 -5a; + 6. 7. y = zc 8 . 2. y = 2x + 4. 5. y = a; 2 + 4 a; + 8 8. y = x* 4aj a 3. y 5. 6. y 9 3 a; a; 2 . 9. j/ = a: 8 1. 10. What is the effect on the graph of y mx + 3 if different values are assigned to m ? How are the graphs related ? What does this indicate as to the meaning of m ? 11. What is the effect on the graph of y = 2x + J if different values are assigned to b ? What is the meaning of & ? 12. Show by similar triangles that y = mx is always a straight line passing through 0. 13. By the use of Exs. 11 and 12 show that y = mx + b is always a straight line 13. Real roots of an equation. It is evident that the real roots of the equation f(x) determine points on the axis of x at which the curve y=f(x) crosses or touches that axis. More- over, if x 1 and 2 ( t <* 2 ) are two values of x such that f(x^) and/( 2 ) are of opposite algebraic sign, the graph is on one side of the axis when x = x i , and on the other side when ic = a; 2 . Therefore it must have crossed the axis an odd number of times between the points x = z l and x = x z . Of course it may have touched the axis at any number of intermediate points. Now, if f(x) has a factor of the form (x a) 1 , the curve y =/(#) crosses the axis of x at the point x = a when k is odd, and touches the axis of x when k is even. In each case the equation /(#) = is said to have Jc equal roots, x a. Since, then, a point of crossing corresponds to an odd number of equal roots of an equation, and a point of touching corresponds to an even number of equal roots, it follows that the equation f(x) = has an odd number of real roots between x^ and x z if /(#,) and / (:c 2 ) have opposite signs. The above gives a ready means of locating the real roots of an equation in the form /(z) = 0, for we have only to find two values of x, as x l and a? 2 , for which f(x) has different signs. We then know that the equation has an odd number of real roots STEAIGHT LINE 31 between these values, and the nearer together x l and # 2 , the more nearly do we know the values of the intermediate roots. In locating the roots in this manner it is not necessary to con- struct the corresponding graph, though it may be helpful. Ex. Find a real root of the equation X s + 2 x 17 = 0, accurate to two decimal places. Denoting x 3 + 2 x 17 by f(x) and assigning successive integial values to x, we find/(2) = 5 and/(3) = 16 Hence there is a leal root of the equation between 2 and 3. We now assign values to x between 2 and 3, at intervals of one tenth, as 2 1, 2 2, 2.3, etc., and we begin with the values nearei 2; since /(2) is nearer zeio than is/(3). Proceeding in this way we find jf (2.3) = 233 and/(2 4) = 1.624 , hence the root is between 2.3 and 2.4. Now, assigning values to a? between 2 3 and 2.4 at intervals of one hun- diedth, we find /(2.31) = - .054 and /(2.32) = 127, hence the root as between 2 31 and 2.32. To determine the last decimal place accurately, we let x = 2.315 and find /(2.315) = .037. Hence the root is between 2 31 and 2 315 and is 2.31, accurate to two decimal places If /(2 315) had been negative, we should have known the root to be between 2 315 and 2.32 and to be 2 32, accurate to two decimal places. EXERCISES Find the real roots, accurate to two decimal places, of the follow- ing equations : 1. a 8 +2a;-6 = 0. 4. j*- 4a 8 + 4 = 0. 2. a 8 + o;+ll = 5. a: 8 - 3ic 2 + 60: - 11 = 0. 3. *-lla: + 6 = 0. 6. 14. Slope of a straight line. Let LK (Figs. 7 and 8) be any straight line not parallel to OX GS OY, and let J? (a^, y^ and P 2 (# 2 , y 3 ) be any two points on it. If we imagine a point to move on the line from P 1 to P^ the increment of x is # 2 x and the increment of y is y^y^ We shall define the slope as the ratio of the increment ofy to the increment ofx and denote it by m. We have then, by definition, 82 DIFFERENTIATION A geometric interpretation of the slope is easily given. For if we draw through JFJ a line parallel to OJf, and through J^ a line parallel to OF, and call R the intersection of these lines, then x z x l = P 1 E and y^y^RP^ Also, if $ is the angle which the line makes with OX measured as in the figure, then : tan 0. C ) It is clear from the figures as well as from formula (2) that the value of m is independent of the two points chosen to define it, provided only that these are on the given line. We may there- fore always choose the two points so that y^y^ is positive. EIG. 7 FIG. 8 Then if the line runs up to the right, as in Fig. 7, # 2 a^ is positive and the slope is positive. If the line runs down to the right, as in Fig. 8, x z o^ is negative and m is negative. There- fore the algebraic sign of m determines the general direction in which the line runs, while the magnitude of m determines the steepness of the line. Formula (1) may be used to obtain the equation of the line. Let m be given a fixed value and the point I^(x^ y^) be held fixed, but let Ji^ be allowed to wander over the line, taking on, therefore, variable coordinates (a?, y). Equation (1) may then be written y-y^m(x- ^). (3) This is the equation of a line through a fixed point (a^, y^ with a fixed slope m, since it is satisfied by the coordinates of any point on the line and by those of no other point. STRAIGHT LINE 33 In particular, P^ (x^ y^) may be taken as the point with coor- dinates (0, i) in which the line cuts OY. Then equation (3) becomes y = mx + l. (4) Since any straight line not parallel to OX or to OY intersects OY somewhere and has a definite slope, the equation of any such line may be written in the form (4). It remains to examine lines parallel either to OX or to OF. If the line is parallel to OX, we have no triangle as in Figs. 7 and 8, but the numerator of the fraction m (1) is zero, and we there- fore say such a line has the slope 0. Its equation is of the form y = , (5) since it consists of all points for which this equation is true. If the line is parallel to OY, again we have no triangle as in Figs. 7 and 8, but the denominator of the fraction in (1) is zero, and m accordance with established usage we say that the slope of the line is infinite, or that m=oo. This means that as the position of the line approaches parallelism with OY the value of the fraction (1) increases without limit. The equation of such a line is x = a. (6) Finally we notice that any equa- tion of the form ^ + 5y + (7=0 (7) always represents a straight line. This follows from the fact that the equation may be written either as (4), (5), or (6). Tbe line (7) may be plotted by locating two points and drawing a straight line through them. Its slope may be found by writing the equation in the form (4) when possible. The coefficient of x is then the slope. If two lines are parallel they make equal angles with OX. Therefore, if m l and m z are the slopes of the lines, we have, from (2), m z =m,. (8) If two lines are perpendicular and make angles <^ and ^> s respectively with OX, it is evident from Fig. 9 that < a = 90+ ^ ; 34 DIFFERENTIATION whence tan $ == ~ cot $ = Hence, if m^ and m a are th slopes of the lines, we have It is easy to show, conversely, that if equation (8) is satis fied by two lines, they are parallel, and that if equation (9) i satisfied, they are perpendicular. Therefore equations (8) am (9) are the conditions for parallelism and perpendiculant respectively. Ex. 1. Find the equation of a straight line passing through the pom (1, 2) and parallel to the straight line determined by the two points (4, 2 and (2, - 3) By (1) the slope of the line determined by the two points (4, 2) am _ g _ 2 5 (2, 3) is = - Therefore, by (3), the equation of the requiiei line is n R / i\ y - 2 = i (a; - 1), which i educes to x 2y 1 = 0. Ex. 2. Find the equation of a straight line through the point (2, 3 and perpendicular to the line 2 a: 3 ?/ + 7 = The equation of the given straight line may be written in the forn y = 3 x + , which is form (4). Therefore m = . Accordingly, by (0) the slope of the required line is By (3) the equation of the requim line is y + 3 = -|(a;-2), which reduces to 3 x + 2 y - 0. Ex. 3. Find the equation of the straight line passing through the poinl ( 3, 3) and the point of intersection of the two lines 2 a? y 3 = anr The coordinates of the point of intersection, of the two given lines musl satisfy the equation of each line. Therefore the codrdinates of the poinl of intersection are found by solving the two equations simultaneously, The result is (1, - 1) We now have the problem to pass a straight line through the points (- 3, 3) and (1, - 1). By (1) the slope of the required line is J* + 1 = - 1. Therefore, by (3), the equation of the line is which reduces to x + y = 0. STRAIGHT LINE 35 EXERCISES 1. Find the equation of the straight line which passes through (2, - 3) with the slope 3 2. Find the equation of the straight line which passes through ( 3, 1) with the slope $ 3. Find the equation of the straight line passing through the points (1, 4) and (f , ). 4. Find the equation of the straight line passing through the points (2, 3) and (- 3, - 3). 5. Find the equation of the straight line passing through the point (2, 2) and making an angle of 60 with OX. 6. Find the equation of the straight line passing through the point Q, |) and making an angle of 135 with OX. ^7. Find the equation of the straight hue passing through the point ( 2, 3) and parallel to the line x + 2 y + 1 = 0. 8. Find the equation of the straight line passing through the point ( 2, 3) and perpendicular to the line 3 a; + 4 ?/ 12 =0. 9. Find the equation of the straight line passing through the point (, ) and parallel to the straight line determined by the two points (f, |) and (J, - ) 10. Find the equation of the straight line passing through (i ~ i) an( ^ perpendicular to the straight line determined by the points (2, 1) and (- 3, 5). 11. If /? is the angle between two straight lines which make angles #j and < 2 (<j!> 2 > <j) respectively with OX, prove from a diagram similar to Fig 9 that /3 = <f> 2 < r If tan t = m^ and tan tj> z = w a , prove by trigonometry that tan ft = . f l 1 + m 2 Wj 12. Find the angle between the lines # 2?/-fl=;0 and 13. Find the angle between the lines 2x 4^ + 5 = and y 6= 0. 14. Find the angle between the lines y = Sas + 4 and a?H-3y-f-7=0. 15. The vertex of a right angle is at (2, 4) and one of its sides passes through the point (- 2, 2). Find the equation of the other side. 16. Find the foot of the perpendicular from the origin to the line 36 DIFFERENTIATION 15. Slope of a curve. Let An (Fig. 10) be any curve serving as the graphical representation of a function of a\ Let 1\ be any point on the curve with coordinates y; l = OM^ t //,= yl/j/J. TU\o l^xM^ and draw the perpendicular J/ a /j, fixing the point / a ' on the curve with the coor- M in Q^OQ *T* (iT(/T Q/ " 1\/T T* * s O LiiixajUco ^jj"""" '-'' 2* ,yjj """ -*'*j*j" jr ^r> Draw ^J? parallel to OX. /R Then PR W.M".. = ArK. p V" -K 7jj J5 = Jlf.,Pa M, 1 1 = A?/, ^ \ a , r>t> / and Ax * / Draw the straight line / ^ ^ j, prolonging it to form a JT IU . to secant J? Then, by 14, -^ is the slope of the secant J>8, and may be called the average slope of tho curve between the points JJ and P v To obtain a number which may be used for tho actual slope of the curve at the point / it is necessary to uso tho limit process (with which the student should now bo familiar), by which we allow kx to become smaller and smaller and the point P z to approach P l along the curve. Tho result is tho derivative of y with respect to #, and we have tho following result : The slope of a curve at any point u yiven, ly the value of tfo derivative -^ at that point. dx f As this limit process takes place, tho point / approaching t!u point Py it appears from the ligure thai, the secant 7A' approaches a limiting position 1{T. The line 1\T is called a tanymt to 1,lu> curve, a tangent Iriny then by deflnitwn the Una approached an tt limit by a secant throuc/h two point* of the curve a tht* two point* approach coincidence. It follows that tho slope* of tho tangent is the limit of the slope of the secant, Therefore, slope of a curve at any point in the mme an the dope of tfta *angent at that point, SLOPE OF A CUEVE 37 From this and 9 AVO may at once deduce the theorem : If the derivative is positive, the curve runs up to the right. If the derivative is negative, the curve runs down to the right. Jf the derivative is zero, the tangent to the curve is parallel to OX. If the derivative is infinite, the tangent to the curve is perpendic- ular to OX. dif The values of x which make ~~ zero or infinite are of par- dx ticular interest m the plotting of a curve. If the derivative changes its sign at such a point, the curve will change its cliiec- tion from down to up or from up to down. Such a point will be called a turning-point. If y is an algebraic polynomial, its derivative cannot be infinite; so we shall be concerned in this chapter only with turning-points for which ^ = 0. Ax They are illustrated m the two following examples: Ex. 1. Consider equation (1) of 12, y = 5 x a; 8 . Plere ^ = 5 - 2 a: = 2^- dx \2 Equating ~ to zero and solving, we have x = - as a possible turning- point It is evident that when x<- , is positive, and when #>- ~ is 2 dx 2 dx negative Therefoie x = \ corresponds to a turning-point of the curve at which the lattei changes its direction from up to down It may be called a high point of the cmve Ex. 2. Consider Here ^ = |(* a - 2* - 3) = |(a: - 3) (a + 1). (IX o o Equating f -~- to zero and solving, we have x = 1 and x = 3 as possible dv turning-points From the factored form of -^, and reasoning as in 9, we see that when y < 1, ~ is positive ; when 1< c < 8, ^r is negative ; 38 DIFFERENTIATION when x > 3, is positive. Therefore both x 1 and x = 3 give turmng- dx points, the former giving a high point, and the latter a low point- Substituting these values of x in the equa- tion of the curve, we find the high point to be ( lj 4f ) and the low point to be (3, |). The graph is shown in Fig 11 It is to be noticed that the solu- tions of the equation do not ^ dx always give turning-points as illus- trated in the next example. Ex. 3. Consider Here dx = x z - 6 x + 9 = (x 3) 2 Solving = 0, we have x = 3 ; but since the derivative is a perfect dx square, it is never negative Therefore x = 3 does not give a turning-point, although when x = 3 the tangent to the curve is parallel to OX. The curve is shown in Fig. 12. The equation of the tangent to a curve at a point (a^, y x ) is easily written down We let ( -^ ) represent the value of -^ at VaaJ/! ,, . dx the point (a^, y x ). Then m = f-^j, and, n I, I X from (3), 14, the equation of the tangent is Ex. 4. Find the equation of the tangent at (1, 1) to the curve We have and dx Therefore the equation of the tangent is wljich reduces to 1 = 0. SECOND DERIVATIVE From (2), 14, it also follows that if $ is the angle which the tangent at any point of a curve makes with OX, then EXERCISES Locate the turning-points, and then plot the following curves : 1. 2/ = 3a a -f-4a: + 4. 4. 2/ = a- 8 - 6 # a + 9 a; + 3. 2. ?/=3 + 3a!-2a; a . 5. 2/ = (2o; 8 + 3a: 2 -12a;- 20) 3. 2/ = ce 8 -3a: 2 -j-4. 6. ?/ = 2 + 9ce -f 3a; a x* ' 7. Find the equation of the tangent to the curve y 3 2x -f- a 2 at the point for which x = 2 8. Find the equation of the tangent to the curve 2/=l+3a: as 2 3a; 8 at the point for which x = 1. 9. Find the points on the curve y = a 8 + 3a: 2 3a; + 1 at which the tangents to the curve have the slope 6 10. Find the equations of the tangents to the curve y = x s + 2x* - x + 2 which make an angle 135 with OX 11. Find the equations of the tangents to the curve y =a5 8 -f-a; a 2x which are perpendicular to the line 3: + 2y + 4 = 12. Find the angle of intersection of the tangents to the curve y = oj 8 -}- as 2 2 at the points for which x = 1 and x = 1 respectively. 16. The second derivative. The derivative of the derivative is called the second derivative and is indicated by the symbol -=-(] or -, We have met an illustration of this in the ax \dx/ ax* case of the acceleration. We wish to see now what the second derivative means for the graph. Since ~ is equal to the slope of the graph, -we have dx d*y d , , ^ j& ^ -=- (slope). dor dx 40 DIFFERENTIATION From this and 9 we have the following theorem : If the second derivative is positive, the slope is increasing as x increases ; and if the second derivative is negative, the slope is de- creasing as x increases. We may accordingly use the second derivative to distinguish between the high turning-points and the low turning-points of a curve, as follows: If, when x a, - = and -r is positive, it is evident that /7 du -^ is increasing through zero ; hence, when x < a, -~ is nega- dx d <lr tive, and when x > a, -j- is positive. The point for which x = a Cu3s is therefore a low turning-point, by 9. d'ii d^y Similarly, if, when x = a, -^- and -~ is negative, it is du evident that -& is decreasing through zero ; hence, when x < a, j CbX J -Z- is positive, and when x > a, -?- is negative. The point for dx dx which x = a is therefore a high turning-point of the curve, by 9. These conclusions may be stated as follows: If -j- and -=-* is positive at a point of a curve, that point ^ "^ " rlai fj^itl is a low point of the curve. If -J- = and -~ is negative at a CliK ClX point of a curve, that point is a high point of the curve. In addition to the second derivative, we may also have third, d?u d^ti fourth, and higher derivatives indicated by the symbols -,'{> '^ etc. These have no simple geometric meaning. EXERCISES Plot the following curves after determining tlieir high and low- points by the use of -^ and -r : dx dor = 3a; 8 x 2 3. y = 7 18x 3a- a H- 4,r 8 , MAXIMA AND MINIMA 41 17. Maxima and minima. If f(a) is a value of /() which is greater than the values obtained either by increasing or by decreasing x by a small amount, /(ff) is called a maximum value of f(p). If /() is a value of /(#) which is smaller than the values of f(j) found either by increasing or by decreasing x by a small amount, /(a) is called a minimum value of /(a?). It is evident that if we place and make the graph of this equation, a maximum value of /(of) occurs at a high point of the curve and a minimum value at a low point. From the previous sections we have, accordingly, the following rule for finding maxima and minima: To find the values of x which give maximum or minimum lvalues of y, solve the equation ^ = 0. dx If x = a is a root of this equation, it must bo tested to see whether it gives a maximum or minimum, and which. We have two tests : TEST I. If the sign of -^ changes from + to as x increases Cv*& through a, then & = a (jives a maximum value of y. If the sign of -~- changes from to + as x increases through a, tJien x = a gives (K& a minimum value of y. H'U d^ii TEST II. If x = a makes -^ = and -~~ netHttive. then # = # dx do* , , /y y / ff *aj gives a maximum value of y, Jf x a mal(e -~ = and -~~. ty^fj cfj positive, then x a gives a minimum value of y. Either of these tests may be applied according to convenience, It may be noticed that Test I always works, while Test II fails d\i to give information if ---^ = when x a. It is also frequently CtJs possible by the application of common sense to a problem to determine whether the result is a maximum or minimum, and neither of the formal tests need then be applied. 42 DIFFERENTIATION Ex. 1. A rectangular box is to be formed by cutting a square from each, corner of a rectangular piece of cardboard and bending the resulting figure. The dimensions of the piece of cardboard being 20 in by 30 m., required the largest box which can be made. Let x be the side of the square cut out. Then, if the cardboard is bent along the dotted lines of Fig 13, the dimensions of the box are 30 - 2 x, 20 2 x, x. Let V be the volume of the box. Then p = (20 - 2 a:) (30 - 2 *) 8Q&3D 7V = 600 - 200 x + 12 a; 2 . "j dx fiy I - . ------- 1 - Equating ___ to zero, we have 13 whence x = 25 5 = 3.9 or 12 7. The result 12.7 is impossible, since that amount cannot be cut twice fiom the side of 20 in The result 3.9 corresponds to a possible maximum, and the tests are to be applied. dV To apply Test I we write in the factored form dx dV when it appears that - changes from + to , as x increases through 3.9 ax Hence x = 3.9 gives a maximum value of V. d s V To apply Test II we find - = - 200 + 24 x and substitute x = 3.9. dyr The result is negative. Therefore x = 3.9 gives a maximum value of V. The maximum value of V is 1056 + cu. in., found by substituting x =s 3.9 in the equation for V. Ex. 2. A piece of wood is in the form of a right circular cone, the altitude and the radius of the base of which are each equal to 12 in. What is the volume of the largest right circular cylinder that can be cut from this piece of wood, the axis of the cylinder to coincide with the axis of the cone ? Let x be the radius of the base of the required cylinder, y its altitude, and V its volume. Then V= vx*y. (1) We cannot, however, apply our method directly to this value of V, since it involves two variables x and y. It is necessary to find a connection MAXIMA AND MINIMA 43 between x and y and eliminate one of them. To do so, consider Fig. 14, which is a cross section of cone and cylinder. From smiilai triangles we have FE _AD t EC DC' that18 ' whence y 12 x. Substituting in (1), we have whence dV dx 24 vx 8 irx*. dV Equating to zero and solving, we find dx B x = or 8. The value x = is evidently not a solution of the problem, but x = 8 is a possible solution. Applying Test I, we find that as x increases through the value 8, dx changes its sign from + to . Applying Test II, we find that 24 ir 6 irx is negative when x = 8. Either test shows that x = 8 (IX corresponds to a maximum value of V. To find V substitute x = 8 in the expression for V. We have V = 256 ir cu. in. EXERCISES 1. A piece of wire of length 20 in. is bent into a rectangle one side of which is a:. Find the maximum, area. 2. A gardener has a certain length of wire fencing with which to fence three sides of a rectangular plot of land, the fourth side being made by a wall already constructed. Required the dimensions of the plot which contains the maximum area. 3. A gardener is to lay out a flower bed in the form of a sector of a circle. If he has 20 ft. of wire with which to inclose it, what radius will he take for the circle to have his garden as large as possible ? 4. In a given isosceles triangle of base 20 and altitude 10 a rec- tangle of base x is inscribed. Find the rectangle of maximum area. 6. A right circular cylinder with altitude 2 as is inscribed in a sphere of radius a. Find the cylinder of maximum, volume, 44 DIFFEEENTIATION 6. A rectangular box with a square base and open at the top is to be made out of a given amount of mateiial. If no allowance is made for the thickness of the material or for waste in construction, what are the dimensions of the largest box that can be made ? 7. A piece of wire 12 ft. in length is cut into six portions, two of one length and four of another Each of the two former portions is bent into the form of a square, and the corners of the two squares are fastened together by the remaining portions of wire, so that the. completed figure is a rectangular parallelepiped Find the lengths into which the wire must be divided so as to produce a figure of maximum volume 8. The strength of a rectangular beam, varies as the product oE its breadth and the squaie of its depth Find the dimensions of the strongest rectangular beam that can be cut from a circular cylindri- cal log of radius a inches 9. An isosceles triangle of constant perimeter is revolved about its base to form a solid of revolution. What are the altitude and the base of the triangle when the volume of the solid generated is a maximum ? 10 The combined length and girth of a postal parcel is 60m. Find the maximum volume (1) when the parcel is rectangular with square cross section , (2) when it is cylindrical 11. A piece of galvanized iron 5 feet long and a feet wide is to be bent into a U-shaped water drain I feet long. If we assume that the cross section of the drain is exactly represented by a rectangle on top of a semicircle, what must be the dimensions of the rectangle and the semicircle in order that the drain may have the greatest capacity (1) when the drain is closed on top 9 (2) when it is open on top ? 12. A circular filter paper 10m. in diameter is folded into a right circular cone. Find the height of the cone when it has the greatest volume 18. Integration. It is often desirable to reverse the process of differentiation. For example, if the velocity or the acceleration of a moving body is given, we jftay wish to find the distance traversed ; or if the slope of a curve is given, we may wish to find the curve. INTEGKATION 45 The inverse operation to differentiation is called integration, and the result of the operation is called an integral. In the case of a polynomial it may be performed by simply working the formulas of differentiation backwards. Thus, if n is a positive integer and , then The first term of this formula is justified by the fact that if it is differentiated, the result is exactly aa?. The second term is justified by the fact that the derivative of a constant is zero. The constant may have any value whatever and cannot be determined by the process of integration. It is called the constant of integration and can only be determined in a given problem by special information given in the problem. The examples will show how this is to be done. Again, if ^ = , dx then ;/ = ax + C. (2) This is only a special case of (1) with n = 0. Finally, if = a Q r n + a^" l + + a n ^x + a n , ax C8) Ex. 1. The velocity v with which a body is moving along a straight line AB (Fig. 15) is given by the equation How far will the body move in the J'IG. 15 time from * = 2 to * = 4? If when t = 2 the body is at P v and if when t = 4 it is at P B , we are to find P,P S . (IV By hypothesis, -j = 10 1 + 5. Therefore s - B i* + 5 1 + C. (1) 46 DIFFERENTIATION have first to determine C. If s la measured from P v it follows that when t _ 2> s - o. Therefore, substituting in (1), we have = 8(2) 2 + 5 (2) + C; whence C=-42, and (1) becomes s = 8 i 3 + 5 1 - 42. (2) This is the distance of the body from P l at any time /. Accordingly, il> remains for us to substitute t = 4 m (2) to find the loquiied distance 1\I\, Thei e results _ 2 _ 42 _ 106> If the velocity is in feet per second, the required distance is in foot. Ex. 2. Required the curve the slope of which at any point is twice the abscissa of the point. By hypothesis, -j- = 2 x CIX Therefore y = JB* + C (1) Any curve whose equation can be derived fiom (1) by giving C a definite value satisfies the condition of the problem (Fig. 16) If it is required that the curve should pass through the point (2, 3), we have, from (1), 3 = 4 + <?; whence C = 1, and therefore the equation of the curve is PICK 10 But if it is required that the curve should pass through ( 8, 10), we have, from (1), 10 = 9 + C; whence C = 1, and the equation is y = 2 + 1 EXERCISES In the following problems v is the velocity, in feet per second, of a moving body at any time t l. If v = 32 1 + 30, how far will the body move in the time from AEEA 47 2. If v = 3 1* + 4 1 + 2, how far will the body move in the time from t = 1 to t = 3 f 3. If v = 20 + 25, how far will the body move in the fourth second ? 4. If v = t 2 2 1 + 4, how far will the body move in the fifth and sixth seconds ? 5. If v 192 32 1, how far will the body move before v 9 6. A curve passes through the point (1, 1), and its slope at any point (x,y) is 3 more than twice the abscissa of the point. What is its equation ? 7. The slope of a curve at any point (aj, y) is Go; 2 + 2x 4, and the curve passes through the point (0, 6) What is its equation ? 8. The slope of a curve at any point (x, y) is 4 3# jc 2 , and the curve passes through the point ( 6, 1). What is its equation ? 9. A curve passes through the point (5, 2), and its slope at any point (05, y) is one half the abscissa of the point What is its equation ? 10. A curve passes through the point ( 2, 4), and its slope at any point (x, y) is a; 2 x + 1. What is its equation ? 19. Area. An important application of integration occurs in the problem of finding an area bounded as follows : Let US (Fig. 17) be any curve with the equation y =/(&), and leitkED and BC be any two ordinates. It is required to find the area bounded by the curve RS, y the two ordinates J3D and J3C, and the axis of x. Take -MP, any variable or- dinate between MD and J3<7, and let us denote by A the area EMPD bounded by the curve, the axis of #, the fixed ordinate J8I>, and the variable Q jj jf jy" B -^ ordinate MP. jUlG J7 It is evident that as values are assigned to a; =OJf, different positions of MP and correspond- ing values of A are determined. Hence A is a function of x for dA which we will find -r ax 48 DIFFERENTIATION Take MN=Ax and draw the corresponding ordinate NQ. Then the area MNQP=&A. If L is the length of the longest ordinate of the curve between MP and NQ, and s is the length of the shortest orduiate in the same region, it is evident that s A# < Au4 < L Aa;, for L&x is the area of a rectangle entirely surrounding AJ, and is the area of a rectangle entirely included m A^t. Dividing by A&, we have . A.1 r s < < i. Aa; As Aa; approaches zero, JVQ approaches coincidence with MP, and hence * and L, which are always between NQ and MP, approach coincidence with MP. Hence at the limit we have ~ = MP^ y =/(*). (1) Therefore, by integrating, A=F(x) + C, (2) where F(x) is used simply as a symbol for any function whose derivative is /(#) We must now find C. Let OE = a. When MP coincides with ED, the area is zero. That is, when x = a, A = 0. Substituting in (2), we have whence C = F(a), and therefore (2) becomes ^=^(00-^(0). (8) Finally, let us obtain the required area JSBCD. If OB ft, thin will be obtained by placing x = b in (3). Therefore we have, finally> JFta). (4) AEEA 49 In solving problems the student is advised to begin with formula (1) and follow the method of the text, as shown in the following example: Ex. Find the area bounded by tho axis of x, the curve y = ^x z , and the ordmates x = 1 and x = 3. In Fig 18, BE is the line x = l, CD is the line x = 3, and the required area is the a.re&J3CDE Then, by (1), ~dx = S X *' whence A ~ \ x 8 + C ' When a; = 1, A = 0, and therefore whence C = -J, and A = J a: 3 % Finally, when ^ x s=3, EXERCISES 1. Find the area bounded by the curve y = 4 a 1 a; 2 , the axis of x, and the lines x = 1 and a: = 3. 2. Find the area bounded by the curve y = s* + 8 a; + 18, the axis of x, and the lines x = 6 and x = 2. 3. Find the area bounded by the curve y = 1C + 12 x x s , the axis of x, and the lines x = 1 and a; = 2 4. Find the area bounded by the curve y + a; 2 9 = and the axis of x. . 5. Find the area bounded by the axis of x and the curve y = 2 x a; 2 . 6. Find the smaller of the two areas bounded by the curve y = 5 x* x a , the axis of x, and the line x = 1. 7. Find the area bounded by the axis of x, the axis of y, and the curve Ayx 3 605 + 9. 8. Find the area bounded by the curve y = x*~ 2 3 <taj + 8 and the axis of x. 50 DIFFERENTIATION 9. If A denotes the area bounded by the axis of y, the curve a; =/(y), a fixed line y ~ b, and any variable line parallel to OX, prove that 7 10. If A denotes an area bounded above by the curve //=/(*")> below by the curve y = F(x), at the left by the fixed line x = (t, and at the right by a variable ordinate, prove that 20. Differentials. The derivative has been denned as the limit of and has been denoted by the symbol ~. This symbol Aa? clx is in the fractional form to suggest that it is the limit of a fraction, but thus far we have made no attempt to treat it as a fraction. It is, however, desirable in many cases to treat the derivative as a fraction and to consider dx and dy as separate quantities. To do this it is necessary to define dx and dy in such a manner that their quotient shall be the derivative. We shall begin by defining dx, when x is the independent variable; that is, the variable whose values can be assumed independently of any oilier quantity. We shall call dxth& differential of x and define it as a change in x which may have any magnitude, but which is generally regarded as small and may be made to approach aero as a limit. In other words, the differential of the independent variable x is identical with the increment of x ; that is, dx sss A*. (1) After dx has been defined, it is necessary to define dy so that its quotient by dx is the derivative. Therefore, if y =/(V) and -j- =/'(;), we have dx dy t That is, the differential of the function y is equal to the derivative times the differential of the independent variable x. DIFFERENTIALS 51 In equation (2) the derivative appears as the coefficient of dx. For this reason it is sometimes called the diff&renticd coefficient. It is important to notice the distinction between dy and Ay. The diff erential dy is not the limit of the increment A#, since both dy and ky have the same limit, zero. Neither is Ay equal to a very small increment A#, since it generally differs in value from Ay. It is true, however, that when dy and Ay both become small, they differ by a quantity which is small compared with each of them. These statements may best be under- stood from the following examples : Ex. 1. Let A be the area of a square with the side x so that If a; is increased by Ao; = dx, A is increased by A.4, where A4 = (a? + dxf - x* = 2 x dx + (dx) a . & ^ Now, by (2), dA = 2zdx, FNJ. ig so that A.I and dA differ by (dx)' 2 Referring to Fig 19,. we see that dA is represented by the rectangles (1) and (2), while A.4 is represented by the rectangles (1) and (2) together with the square (3) , and it is obvious from the figuie that the square (8) is very small compared with the rectangles (1) and (2), provided djc is taken small. For example, if x = 5 and dx = .001, the rectangles (1) and (2) have together the area 2 z die = .01 and the square (3) has the area .000001. Ex. 2. Let s - 16 t s , where $ is the distance traversed by a moving body in the time t If t is increased by Al = dt, we have As = 10 (t + dt)* - 1C & - 32 tdi + 16 (dt}\ and, from (2), ds = 32 1 dt ; so that As and ds differ by 16 (dt) z The terra 16 (rf/) a is very small com- pared with the term BZtdt, if dt is small. For example, if f = 4 and dt - .001, then 32 tdt = ,128, while 16 (eft) 2 .000016 In this problem As is the actual distance traversed in the time dt, and da is the distance which would have been traversed if the body had moved throughout the time dt with the same velocity which it had at the begin- ning of the time dt. 52 DIFFERENTIATION In general, if y=f(%) and we make a graphical representa- tion, we may have two cases as shown in Figs. 20 and 21. In each figure, WN = PR &v = dx and RQ=ky, since HQ is the total change in y caused by a change of dx = M.N in x. If PT is the tangent to the curve at P, then, by 15, so that, by (2), dy = (tan HPT) (P7?) = R T. R M N C M N FIG. 20 PIG 21 In Fig. 20, dy < A#, and in Fig. 21 dy >&y; but in each case the difference between dy and Ay is represented m magnitude by the length of QT. This shows that RQ = Ay is the change in y as the point JP is supposed to move along the curve /=/(^') while RT dy is the change in the value of y as the point P is supposed to move along the tangent to that curve. Now, as a very small arc docs not deviate much from its tangent, it is not hard to see graphi- cally that if the point Q is taken close to P, the difference between RQ and RT, namely, QT, is very small compared with RT. A more rigorous examination of the difference between the increment and the differential lies outside the range of this book. EXERCISES 1. If y = x 6 - 3ic 2 + ix + 1, find dy. 2. If y = x 4 + 4 8 - x* + 6.-B, find dy. 3. If V is the volume of a cube of edge x, find both AF and dV and interpret geometrically. APPROXIMATIONS 53 4. If A is the area of a circle of radius r, find both Avl and dA. Show that A.4 is the exact area of a ring 1 of width dr, and that dA is the product of the inner circumference of the ring by its width. 5. If V is the volume of a sphere of radius r, find AF and dV. Show that A V is the exact volume of a spherical shell of thickness dr, and that dV is the product of the area of the inner surface of the shell by its thickness. 6. If A is the area described in 19, show that dA = ydx Show geometrically how this differs from A/i 7. If s is the distance traversed by a moving body, t the time, and v the velocity, show that ds = vdt. How does ds differ from A& 9 8. If y = x z and x = 5, find the numerical difference between dy and A?/, with successive assumptions of dx = 01, dx, = .001, and dx = .0001. 9. If y = x 8 and x = 3, find the numerical difference between dy and Ay for dx = .001 and for dx = 0001 10. For a circle of radius 4 in. compute the numerical difference between dA and AJ. corresponding to an increase of r by .001 in. 11. Eor a sphere of radius 3 ft. find the numerical difference between dV and A V when r is increased by 1 in. 21. Approximations. The previous section brings out the fact that the differential of y differs from the increment of y by a very small amount, which becomes less the smaller the incre- ment of x is taken. The differential may be used, therefore, to make certain approximate calculations, especially when the ques- tion is to determine the effect upon a function caused by small changes in the independent variable. This is illustrated in the following examples: Ex. 1. Find approximately the change in the area of a square of side 2 in. caused by an increase of .002 in. in. the side. Let x be the side of the square, A its area. Then A = x 2 and dA = 2 xdx. Placing x 2 and dx =* .002, we find dA = .008, which is approximately the required change in the area. If we wish to know how nearly correct the approximation is, we may com- pute A4 = (2.002) 8 - (2) 2 = .008004, which is the exact change in A. Our approximate change is therefore in error by .OOOQOAj a very email amount. 54 DIFFERENTIATION Ex. 2. Find approximately the volume of a sphere of radius 1.9 in. The volume of a sphere of radius 2 in. is -^ TT, and the volume of the required sphere may be found by computing the change in the volume of a sphere of radius 2 caused by decreasing its radius by .1. If r is the radius of the sphere and V its volume, we have F= Tjr 8 and d V = 4 vr*dr. Placing r = 2 and dr .1, we find dV= 1.6 TT. Hence the volume of the required sphere is approximately To find how much this is in error we may compute exactly the volume of the required sphere by the formula V= ^7r(1.9) 8 = 9.1453 ir. The approximate volume is therefore in error by .0786 TT, which is less than 1 per cent of the true volume. EXERCISES 1. The side of a square is measured as 3 ft. long. If this length is in error by 1 in , find approximately the resulting error in the area of the square. 2. The diameter of a spherical ball is measured as 2 in., and the volume and the surface are computed. If an error of ^ in. has boon made in measuring the diameter, what is the approximate error in the volume and the surface ? 3. The radius and the altitude of a right circular cone are meas- ured as 3 in. and 5 in. respectively. What is the approximate error in the volume if an error of ^ in. is made in the radius ? What is the error in the volume if an error of ^ in. is made in the altitude ? 4. Find approximately the volume of a cube with 3.0002 in. on each edge. 5. The altitude of a certain right circular cone is the same as the radius of the base. Find approximately the volume of the cono if the altitude is 3.00002 in. 6. The distance * of a moving body from a fixed point of its path, at any time t, is given by the equation $ = 16 t a -f- 100 1 50. Find approximately the distance when t s= 4.0004. 7. Find the approximate value of v? + x 2 when x =s 1,0003. 8. Find approximately the value of as* + * a + 4 when = .99989. GENERAL EXERCISES 55 9. Show that the volume of a thin cylindrical shell is approxi- mately equal to the area of its inner surface times its thickness. 10. If V is the volume and S the curved area of a right circular cone with radius of its base r and its vertical angle 2 a, show that V s= I 7n <s ctn a and S = irr 1 esc a Thence show that the volume of a thin conical shell is approximately equal to the area of its inner surface multiplied by its thickness. GENERAL EXERCISES Find the derivatives of the following functions from the definition 3 + 2x 1 6. VJB> ' ' * 2. ! 4. a x 8. Prove from the definition that the derivative of - is , 1 C C 9. By expanding and differentiating, prove that the derivative of (2 x + 5) 8 is 6(2a; + 5) 2 . 10. By expanding and differentiating, prove that the derivative of (a 2 + 1) 8 is 6 a; (a 2 + 1) 2 . 11. By expanding and differentiating, prove that the derivative of (a; + a) n is n(x + a)"" 1 } where % is a positive integer. 12. By expanding and differentiating, prove that the derivative of (a; 2 + a s ) n is 2 nx(x i + a 2 )"- 1 , where n is a positive integer. 13. Find when x*+ 8 a; 8 4- 24 a 8 + 32 a- + 16 is increasing and when decreasing, as x increases. 14. Find when 9 x* 24 of 8 x* -f- 32 x + 11 is increasing and when decreasing, as x increases. 15. Find a general rule for the values of x for which ace 2 + bx + e is increasing or decreasing, as x increases. 16. Find a general rule for the values of x for which x* a?x + 1 is increasing or decreasing, as x increases. 17. A right circular cone of altitude x is inscribed in a sphere of radius a. Find when an increase in the altitude of the cone will cause an increase in its volume and when it will cause a decrease, A-B * HINT, In these examples make use of the relation vA'vIt = 56 DIFFERENTIATION 18. A particle is moving in a straight line in such a manner that its distance x from a fixed point A of the straight line, at any time t, is given by the equation x t s 9 & + 15 1 + 100. When will the particle be approaching A ? 19. The velocity of a certain moving body is given by the equation v = t z 7 1 -f- 10. During what time will it be moving in a direction opposite to that in which s is measured, and how far will it move ? 20. If a stone is thrown up from the surface of the earth with a velocity of 200 ft. per second, the distance traversed in t seconds is given by the equation s = 200 1 16 z5 a . Find when the stone moves up and when down 21. The velocity of a certain moving body, at any time t, is given by the equation v = t 2 8 1 + 12 Find when thu velocity of the body is increasing and when decreasing. 22. At any time t, the distance s of a certain moving body from a fixed point in its path is given by the equation s = 16 24 1 -f 9 tf 2 f. When is its velocity increasing and when decreasing ? When is its speed increasing and when decreasing ? 23. At any time t, the distance of a certain moving body from a fixed point in its path is given by the equation s t 6 6 1* -j- 9 1 + 1 When is its speed increasing and when decreasing ? 24. A sphere of ice is melting at such a rate that its volume is decreasing at the rate of 10 cu in. per minute. At what rate is the radius of the sphere decreasing when the sphere is 2 ft. in diameter ? 25. Water is running at the rate of 1 cu. ft. per second into a basin in the form of a frustum of a right circular cone, the radii of the upper and the lower base being 10 in and 6 in. respectively, and the depth being 6 in. How fast is the water rising in the basin wluw it is at the depth of 3 in. ? <. 26. A vessel is in the form of an inverted right circular cone the vertical angle of which is 60. The vessel is originally filled with liquid which flows out at the bottom at the rate of 3 cu. in. per minutw. At what rate is the inner surface of the vessel being exposed when the liquid is at a depth of 1 ft. in the vessel ? 27. Find the equation of the straight line which passes through the point (4, 6) with the slope . 28. Find the equation of the straight line through the points (2, - 3) and (- 3, 4). GENERAL EXERCISES 57 29. Find the equation of the straight lino determined by the points (2, -4) and (2, 4). 30. Find the equation of the straight line through the point, (1, 3) and parallel to the line a? 2 y -f- 7 = 31. Find the equation of the straight lino tli rough the point (2, 7) and perpendicular to the lino 2 . -f 4 // + 9 = 32. Find the angle between the straight lines 2,r -|- 3// + H = and y + 3 x -f- 1 = 0. Find the turning-points of the following curves and draw the graphs 33. y=3 as-ai" 34. 2/ = 16c a -40a- + 25 35. y } (x 2 - 4 x - 2). 36. y = a* - 6 a* - 15 x + 5. 37. y = 3- 3 + 3 as* 9 a: 14. 38. Find the point of intersection of the tangents to the curve y = x + 2 a; 2 x s at the points for which y = 1 and a? = 2 respectively 39. Show that the equation of the tangent to the curve y = o" 2 + 2bx + G at the point (x 1} y^) is y = 2 (ca^-f /;), a,r, a + 0. 40. Show that the equation of the tangent to the curve y a* a + ax + fi at the point (a^, y a ) is y = (3 a a + )*" 2 , -|_ />. 41. Find the area of the triangle included between the coordintitc axes and the tangent to the curve y = a; 8 at the point (2, 8). 42. Find the angle between the tangents to the curve //=-2./'" -f- 4 a; 2 x at the points the abscissas of which arc 1 and 1 respectively. 43. Find the equation of the tangent to the curve y =s x n 3 a; 9 -+ 4.x 12 which has the slope 1. 44. Find the points on the curve ;// = 3 .r, B 4 a*, 2 at which the tangents are parallel to the lino x y = 0. V 1 46. A length I of wire is to be cut into two portions which aw to be bent into the forms of a circle and a square respectively. Show that the sum of the areas of those figures will bo least when the wire is cut in the ratio IT : 4. 58 DIFFERENTIATION 46. A log in the form of a -frustum of a cone is 10 ft. long, the diameters of the bases being 4 ft. and 2 ft. A beam with a square cross section is cut from it so that the axis of the beam eoiuuidos with the axis of the log. Find the beam of greatest volume that can be so cut 47. Required the right circular cone of greatest volume which can be inscribed in a given sphere. 48. The total surface of a regular triangular prism is to be Jt. Find its altitude and the side of its base when its volume is a maximum. 49. A piece of wire 9 in. long is cut into five pieces, two of ouo length and three of another Each of the two equal pieces is bent into an equilateral triangle, and the vertices of the two triangles are connected by the remaining three pieces so as to form a regular triangular prism. How is the wire cut when the prism has the largest volume ? 50. If t is time in seconds, v the velocity of a moving body in feet per second, and v = 200 32 1, how far will the body move in the first 5 sec. ? 51. If v = 200 32 1, where v is the velocity of a moving body in feet per second and t is time in seconds, how far will the body move in the fifth second ? 62. A curve passes through the point (2, 8), and its slope at any point is equal to 3 more than twice the abscissa of the point. Find the equation of the curve. 53. A curve passes through the point (0, 0), and its slope at any point is oj a 2 x + 7. Find its equation. 54. Find the area bounded by the curve y + s a 16 = and llio axis of ao, 5 5 . Find the area bounded by the curve y = 2 as 8 15 a? a 4. 3G + 1 , the ordinates through the turning-points of the curve, and OX. 56. Find the area between the curve y x a and the straight lint? y x -f- 6. 57. Find the area between the curves y =* jc a and y = 18 a 8 . 58. The curve y = ax* is known to pass through the point (h, Jc), Prove that the area bounded by the curve, the axis of a-, and the line x = h is hk. 59. Compute the difference between &A and dA for the area 4 of a circle of radius 5, corresponding to an increase of .01 in the radius. GENERAL EXERCISES 59 60. Compute the difference between AF and dV for the volume V of a sphere of radius 5, corresponding to an increase of 01 in the radius 61. If a cubical shell is formed by increasing each edge of a cube by dx, show that the volume of the shell is approximately equal to its inside surface multiplied by its thickness. 62. If the diameter of a sphere is measured and found to be 2 ft , and the volume is calculated, what is the approximate error in the calculated volume if an error of in has been made in obtaining the radius ? 63. A box in the form of a right circular cylinder is 6 in deep and 6 in. across the bottom. Find the approximate capacity of the box when it is lined so as to be 5 9 in deep and 5 9 in across the bottom. 64. A rough wooden model is in the form of a regular quadran- gular pyramid 3 in. tall and 3 in. on each side of the base. After it is smoothed down, its dimensions are all decreased by .01. What is the approximate volume of the material removed ? 65. By use of the differential find approximately the area of a circle of radius 1.99. What is the error made in this approximation ? 66. Find approximately the value of a; 6 + 4 C 8 + a when x = 3.0002 and when x = 2.9998. 67. The edge of a cube is 2.0001 in. Find approximately its surface. 68. The motion of a certain body is defined by the equation s = i 8 + 3 i* + 9 1 27. Find approximately the distance traversed in the interval of time from t = 3 to t = 3.0087. CHAPTER III SUMMATION 22. Area by summation. Let us consider the problem to find the area bounded by the curve y ^ # 2 , the axis of r, and the ordmates x = 2 and x 3 (Fig. 22). This may be solved by the method of 19 ; but we wish to show that it may also be considered as a problem in summation, since the area is approximately equal to the sum of a number of rectangles constructed as follows : We divide the axis of x between x = 2 and x 3 into 10 parts, 3 2 each of which we call A#, so that Ax = r =.1. If x l is the first point of division, x a the second point, and so on, and rectangles are constructed as shown in the figure, then the altitude of the first rectangle is (2) 2 , that of the second rec- tangle is z 2 =|(2. 1) 2 =.882, and so on. The area of tho first rectangle is ^(2) 2 Ao;=.08, that of the second rectangle is J 1 2 Aa; = |(2.1) 2 A=.0882, and so on. Accordingly, we make the following calculation: s = 2, -|(2) 3 Aa;= .08 0^=2.1, %(xy&x= .0882 # 2 =2.2, (z 2 ) 2 Az = .0968 z 8 =2.3, l(xy&x= .1058 z 4 =2.4, %(xy&x= .1152 aj a =2.5, O 6 ) 2 Aa= .1250 a; 6 =2.6, %(x^&x= .1352 z 7 =2.7, |(r 7 ) 2 A^= .1458 z 8 =2.8, (z 8 ) 2 Aa;= .1508 tf=2.9, f Ar= -1G82 . ^ i his is a first approximation to the area. For a better approximation the axis of x between #= 2 and 05= 3 may be divided into 20 parts with Aa;= .05. The result is 1.241 8. 60 ABJEA 61 FIG. 22 If the base of the required figure is divided into 100 parts with Az = .01, the sum of the areas of the 100 rectangles constructed as above is 1.26167. The larger the num- ber of parts into , which the base of the figure is divided, the more nearly is the required area ob- tained. In fact, the re- quired area is the limit approached as the number of parts is indefinitely "~ Q increased and the size of A approaches zero. We shall now proceed to generalize the problem just handled. Let LK (Fig. 23) be a curve with equation y =/(), and let OEa and OB b. It is re- quired to find the area bounded by the curve LK, the axis of #, and the ordmates at E and B. For convenience we assume in the first place that a < b and that f(x) is positive for all values of x between a and b. We will divide the line EB into n equal parts by placing and laying off the E MI M e Jf J/ Mj M t B FIG. 28 n lengths EM^- Let _ 1 J5=Aa: (in Fig. 23,n=9). . ., Jf^A* parallel to OX. /(a)Aa;=:the area of the rectangle /(#].) AOJ = the area of the rectangle /(* 2 ) As? = the area of the rectangle _ 1 ), and Then alsc the area of the rectangle 62 SUMMATION The sum /(a) A* +/(aOA* +/(aA* + . . . +/(^_ 1 )Aa; (1) is then the sum of the areas of these rectangles and equal to the area of the polygon EDR^R^ . . . R^^^li^B. It is evident that the limit of this sum as n is indefinitely increased is the area bounded by ED, EB, BC, and the arc DC. The sum (1) is expressed concisely by the notation where S (sigma), the Greek form of the letter S, stands for the word " sum," and the whole expression indicates that the sum is to be taken of all terms obtained from/(,)Aa; by giving to i in succession the values 0, 1, 2, 3, ., n 1, where x ~a. The limit of this sum is expressed by the symbol C b Ja where /is a modified form of S. Hence / /(#)<fa; = Lim^/(a; f )A:c=: i/a isO area It is evident that the result is not vitiated if ED or J3C is of length zero. 23. The definite integral. We have seen in 19 that if A is the area EBCD of 22, where F(x) is any function whose derivative is/(). Comparing this with the result of 22, we have the important formula r h J a The limit of the sum (1), 22, which is denoted by C /(*) dx, Jn is called a definite integral, and the numbers a and b are called DEFINITE INTEG-KAL 68 the lower limit and the upper limit*, respectively, of the definite integral. On the other hand, the symbol I f(x)dx is called an indefinite integral and indicates the process of integration as already de- fined in 18. Thus, from that section, we have + 0, G, and, in general, \ f(x)dx=F(x)+ C, where F(x) is any function whose derivative is /(#) We may therefore express formula (1) in the following rule : To find the value of \ f(x)dx, evaluate \f(x)dx, substitute x = b and x: = a successively, and subtract the latter result from the former. It is to be noticed that in evaluating I f(x)dx the constant of integration is to be omitted, because if it is added, it dis- appears in the subtraction, since In practice it is convenient to express F^ F^a) by the symbol [^ ()]> so that Ex. 1. The example of 22 may now be completely solved. The required area is 27 8 10 * The student should notice that the word "limit" is here used In a quite different from that in which it is used when a variable is said to approach a limit (1). 64 SUMMATION The expression /(a;) dx which appears in formula (1) is called the element of integration. It is obviously equal to dF(x). In fact, it follows at once from 19 that dA=ydx = f(x) dx. In the discussion of 22 we have assumed that y and dr are positive, so that dA is positive. If y is negative that is, if the curve m Fig. 23 is below the axis of x and if dx is positive, the product ydx is negative and the y area found by formula (1) has a nega- tive sign. Finally, if the area required is partly above the axis of x and partly below, it is necessary to find each part separately, as in the follow- ing example: Ex. 2. Fmd the aiea bounded by the curve y = x 3 a; 2 6 x and the axis of a- Plotting the curve (Fig 24), we see that it crosses the axis of x at the points B (-2, 0), 0(0, 0), and C(3, 0). Hence part of the area is above the axis of x and part below. Accoidingly, we shall find it necessary to solve the pioblem in two parts, first finding the aiea above the axis of x and then finding that below To find the first area we proceed as in 22, dividing the area up into elementary rectangles fox- each of which dA = ydx = (a. 8 a; 2 - 6 *) dx , whence A- C\x*- x*-Qx)dx = [I* 4 - jj 3? o FIG. 24 Similarly, for the area below the axis of x we find, as before, dA = tlx = x s x s But in this case y = x s jc z G x is negative and hence dA is negative, for we are making x vary fiom to 8, and therefore dx is positive, Tlioio- foie we expect to find the result of the summation negative. In faot, we have -s A - I (X s - x* - 6 x^dx = [| a: 4 ~ J as* ~ J o -^(3)-3(3)]-0 = -15i DEFINITE INTEGRAL 65 .8 As we aie asked to compute the total aiea bounded by the ciuve aud the axis of x, we discard the negative sign in the last summation aud add 5J and 15 f, thus obtaining 21 ^ s as the required result. If we had computed the definite integral i 8 x z 6 a.) dx, we should have obtained the icsult 10 fy, which is the algebraic sum of the two portions of area computed separately. Ex. 3. Find the area bounded by the two curves y = .r 2 and y = 8 a; 2 . We diaw the curves (Fig 25) ?/ = r a f\ "\ y x (.*-} and y = 8 - a; 8 , (2) and by solving their equations we find that they mtcisect at the points P x (2, 4)aud> 3 (-2,4) The reqiuiod area OP^JIP^O is evi- dently twice the area OJ\liO, since both ciuves aie symmetiical with le- spect to OY. Accoidmgly, wo shall find the area OP^BO and multiply it by 2. This lattei area may be found by subtracting the aic>a ON^P^O from the area ON^P-^BO, each of these areas being found as in the pievious example ; or we may proceed as follows : Divide ON^ into n parts dx, and through the points of division draw stiaight lines paiallol to OY, intersect- ing both curves. Let one of those lines be M^ Q l Q 2 Through the ] 101 nts Q : and (2 2 draw straight linos parallel to OX until they meet the next vortical line to the right, thcnoby forming the rectangle Q^SQy, The aroa of such a rectangle may be takon as dA and may be computed as follows: its base is ih % and ids altitude IH Q^ = A/^ M^Q,! = (8 j 8 ) j" u =8 2 .c a ; for M^Q^ is tho ordinato of a point on the curve (2) and J/j^ the ordinate of a point ou (1). Thcicfore , , , A Finally, tho required area is 2(10) = 21 } = C \R - 2 r a ) rl,r =[83-- ?, 3*] 66 SUMMATION EXERCISES 1. Find the area bounded by the curve 4 y as 2 2 = 0, the axis of x, and the lines x 2 and x = 2. 2. Find the area bounded by the curve y = a? 7 X* + 8 a + 16 the axis of x, and the lines x = 1 and as = 3 3. Find the area bounded by the curve y = 25 x 10 a; 2 -f- oi 8 and the axis of a. 4. Find the area bounded by the axis of x and the curve y = 25 - a 2 . 5. Find the area bounded by the curve y=4aj 2 4aj -3 and the axis of x. 6. Find the area bounded below by the axis of x and above by the curve y = # 8 4 a 2 4 x + 16. 7. Find the area bounded by the curve y = 4 as 8 Sec 2 9cc+18 and the axis of #. 8. Find the area bounded by the curve x* + 2y 8=0 and tho straight line x-)-2y 6 = 0. 9. Find the area bounded by the curve 3y aj 2 = and the straight line 2x + y 9 = 10. Find the area of the crescent-shaped figure bounded by the two curves y = x* + 7 and y = 2 x z -f- 3 11. Find the area bounded by the curves 4y=sa3 a 4 a; and a 2 4o: + 42/ 24 = 0. 12. Find the area bounded by the curve o; + 3 = 2/ a 2y and the axis of y. 24. The general summation problem. The formula F(a) (1) rf J a has been obtained by the study of an area, but it may be given a much more general application. For if f(x) is any function of x whatever, it may be graphically represented by the curve y =/(). The rectangles of Fig. 23 are then the graphical rep- resentations of the products f(z)dx, and the symbol f f(x)dx J a GENERAL PROBLEM 6T represents the limit of the sum of these products. We may accordingly say: Any problem which requires the determination of the limit of the sum of products of the type f(x) dx may be solved by the use of formula (1). Let us illustrate this by considering again the problem, already solved in 18, of determining the distance traveled in the time from t = t t to t = 3 by a body whose velocity v is known. Since ds "*#' we have ds = vdt, which is approximately the distance traveled in a small interval of time dt. Let the whole time from t = t^ to t = t z be divided into a number of intervals each equal to dt. Then the total dis- tance traveled is equal to the sum of the distances traveled in the several intervals dt, and hence is equal approximately to the sum of the several terms vdt This approximation becomes better as the size of the intervals dt becomes smaller and their number larger, and we conclude that the limit of the sum of the terms vdt is the actual distance traveled by the body. Hence we have, if s is the total distance traveled, a = I vdt. A If, now, we know v in terms of t, we may apply formula (1). Ex. If v = 16 1 + 5, find the distance traveled in the time from t = 2 to * = 4. We have directly s = f *(16 1 + 5) dt = [8 1* + 5 *] = 106. EXERCISES 1. At any time t the velocity of a moving body is 3 & 4- 2 1 f t. per second. How far will it move in the first 6 sec.? 2. How far will the body in Ex. 1 move during the seventh second ? 3. At any time t the velocity of a moving body is 6 + 5 1 & ft. per second, Show that this velocity is positive during the interval from t = 1 to t f= 6, and find how far the body moves during that interval. 68 SUMMATION 4. At any time t the velocity of a moving body is 4)5 3 24 f +11 ft. per second. During what interval of time is the velocity negative, and how far will the body move during that interval ? 6. The number of foot pounds of work done in lifting a weight in the product of the weight in pounds and the distance in feet through which the weight is lifted. A cubic foot of water weighs 021 lb. Compute the work done in emptying a cylindrical tank of depth 8 ft. and radius 2 ft , considering it as the limit of the sum of the jnort'rt of work done in lifting each thin layer of water to the top of the tank. 25. Pressure. It is shown m physics that the pressure on one side of a plane surface of area A, immersed in a liquid tit a, uniform depth of h units below the Kin-face of the liquid, is equal to wJul, where w is the weight of a unit volume of the liquid. This may be remembered by noticing that wliA is the weight of the column " of the liquid which would be supported by the area A. Since the pressure is the same in all directions, we can also determine ^ IQ 2Q the pressure on one side of a plane surface which is perpendicular to the surface of the liquid and hence is not at a uniform depth. Let ABC (Fig. 26) represent such a surface and RS the line of intersection of the plane of ABC with the surface of the liquid. Divide ABC into strips by drawing straight lines parallel to Jttf. Call the area of one of these strips dA, as in 28, and the depth of one edge h. Then, since the strip is narrow and horizontal, tho depth of every point differs only alightly from 7i, and tho pres- sure on the strip is then approximately wMA, Talcing P m tho total pressure, we write ^p_ j 3* The total pressure P is the sum of tho pressures on the several strips and is therefore the limit of the sum of terms of tho form whdA, the limit being approached as the number of tho strips is indefinitely increased and the width of each indefinitely decreased. Therefore / P= I wlidA, PEESSUKE 69 where the limits of integration are to be taken so as to include the whole area the pressure on which is to be determined. To evaluate the integral it is necessary to express both h and dA 111 terms of the same variable. Ex. 1. Find the pleasure on one side of a rectangle BCDE (Fig 27), wheie the sides BC and ED aiu each 4 ft. long, the sides BE and CD are each 8 ft. long, immersed in watei so that the plane of the rectangle is peipendiculai to the surface of the water, and the side BC is paiallel & to the surface of the watei and 2 ft. below at In Fig. 27, LK is the line of inter- section of the surface of the water and the plane of the rectangle. Let M * N be the point of intersection of LK and BE produced Then, if x is measured downward from along E D BE, x has the value 2 at the point B ^ and x has the value 5 at the point E. We now divide BE into parts dx, and through the points of division draw straight lines parallel to BC, thus dividing the given rectangle into elemental y rectangles such as MNP.S Therefore dA = area of MNR S = MN - MS - 4 dx. Since MN is at a distance x below LK, the pressure on the elementary rectangle MNRS is approximately wx(4: dx'). Accoidmgly, we have and P = f G 4 nxdx = [2 vxffe* 2 w(6) - 2 w(2) 2 = 42 20 t/2 ^ For water, w - 02 J Ib = ^ T. Hence we have finally P = 2625 Ib. = 1 T % T Ex. 2. The base CD (Fig. 28) of a triangle BCD is 7 ft., and its altitude from B to CD is 5 ft This triangle is immersed in water with its plane perpendicular to the suiface of the water and with CD parallel to the sur- face, and 1 ft below it, B being below CD Find the total pressure on one side of this triangle. Let LK represent the line of intersection of the plane of the triangle and the surface of the water. Then B is 6 ft below LK Let BX be per* pendicular to LK and intersect CD at T We will measure distances from B in the direction BX and denote them by x. Then, at the point B, x has the value , and at T, x has the value 5. TO SUMMATION Divide the distance BT into parts dx, and through the points of divi- sion draw straight lines parallel to CD, and on ouch of these linen n lower base construct a rectangle such as MNJR8, where J'l and F sire two consecutive points of division % onBX. L K Then BE = x, G \ EF = dx, and, by similar triangles, whence CD BT MN x and Then dA = the area of MNRS = J xdx, Since B is C ft. below LK, and BE = x, it follows that E is (fl - a;) ft. below LK. Hence the pressure on the rectangle is approximately dP = (& a" ?*) (6 or) w = (-V 1 war $ ?/JJK S ) rfj?, /*" /o = (105 to - iJA w) - = JL^ - to - 2010^ Ib. = Ii4 T. EXERCISES s/ 1. A gate in the side of a dam is in the form of a square, 4 ft. on a side, the upper side being parallel to and 1C It. bolow the surfaoo of the water in the reservoir. What is the pressure on the gate '( v* 2. Find the total pressure on one side of a triangle of base 6 ft. and altitude 6 ft., submerged in water so that the altitude is vertical and the vertex is m the surface of the water. </ 3. Find the total pressure on one side of a triangle of base 4 ft. and altitude 6 ft., submerged in water so that the base is horizontal, the altitude vertical, and the vertex above the base and 4 ft, from tlio surface of the water. e base of an isosceles triangle is 8 ft. and the equal sidas 6 ft. The triangle is completely immersed in water, its baae allel to and 6 ft. below the surface of the water, its alti- g perpendicular to the surface of the water, and its vertex uc.ijmg ,uove the base. Find the total pressure on one side of the triangle. VOLUME 71 v 6. Find the pressure on one side of an equilateral triangle, 6 It. on a side, if it is partly submerged in water so that one vortex is one foot above the surface of the water, the corresponding altitude being perpendicular to the surface of the water. \s. The gate in Ex. 1 is strengthened by a brace which runs diagonally from one corner to another. Find the pressure on each of the two portions of the gate one above, the other below, the brace. 7. A dam is in the form of: a trapezoid, with its two horizontal sides 300 and 100 ft. respectively, the longer side being at the top ; and the height is 15 ft. What is the pressure on the dam when the water is level with the top of the dam ? 8. What is the pressure on tho dam of Ex. 7 when the water reaches halfway to the top of the dam ? 9. If it had been necessary to construct the dam of Ex. 7 with the shorter side at tho top instead of the longer side, how much greater pressure would the dam have had to sustain when the reservoir is full of water ? 10. The center board of a yacht is in the form, of a trapezoid in which the two parallel sides are 3 ft and 6 ft., respectively, in length, and the side perpendicular to these two is 4 ft. in length. Assuming that the last-named side is parallel to the surface of the water at a depth of 2 ft., and that the parallel sides are vertical, find the pressure on one side of the board. 11. Where shall a horizontal lino be drawn across tho gate of Ex. 1 so that tho pressure on the portion above tho lino shall equal the pressure on the portion below ? 26. Volume. The volume of a solid may be computed by di- viding it into n elements of volume, dV^ and taking the limit* of the sum of these elements as n is increased indefinitely, tho magnitude of each element at the same time approaching aero. The question in each case is the determination of the form of the element dr. We shall discuss a comparatively simple case of a solid such as is shown in Fig, 29. In this figure let Olf be a straight line, and let tho distance of any point of it from be denoted by h. At one end tho solid is bounded Tby a plane perpendicular to OH at <7, where 00 ~ a> 72 SUMMATION and at the other end it is bounded by a piano perpendicular to OH at B, where OS = 5, so that it has parallel bases. The solid is assumed to be such that the area A of any plane section made by a plane perpendicular to Oil at a point distant h from can be expressed as a func- tion of h. To find the volume of such a solid we divide the distance CB into n parts dh, and through the points of division pass planes perpendicular to OH. We have thus divided the solid into slices of which the thickness is dh Since A is the area of the base of a slice, and since the volume of the slice is approximately equal to the volume of a right cylinder of the same base and thickness, we write dV=Adli. The volume of the solid is then the limit of the sum of terms of the above type, and therefore /"' F= I Adh. Jo. It is clear that the above discussion is valid even when one or both of the bases corresponding toh=a and 7t = fi, respectively, reduces to a point. Ex.1. Let OY (Fig. 30) be an edge of a solid such that all its sections made by planes perpendicular to OFaie rectangles, the sides of a rectangle in a plane distant y fiorn being re- spectively 2 y and y z We shall find the volume in- cluded between the planes j, IO go # = and?/ = 2 J. Dividing the distance from # = 0toy = 2& into n parts <li/, and passing planes perpendicular to OY, we form rectangles such as MffttS, wlioro, if VOLUME 73 OM = y, MN - y* and MS = 2 y Hence the area MNRS = 2 y s , and the volume of the elementary cylinder standing on MNRS as a base is thatls ' Therefore V = Ex. 2. The axes of two equal light circular cylinders of radius o inter- sect at right angles. Required the volume common to the two cylinders. Let OA and OB (Fig. 31) be the axes of the ^ cylmdeis and OY the common perpendicular to OA and 013 at then point of intersection 0. Then OA D and OBD are quadrants of two equal chcles cut from the two cylinders by the planes through OY perpendicular to the axes OB and OA, and OD = a Then the figure represents one eighth of the requned volume. We divide the distance OD into n parts dy, and through the points of division pass planes per- pendicular to OY Any section, such as LMNP, is a square, of which one side NP is equal to V OP Z ON* OP = a, being a radius of one of the cylinders, and hence, as ON= y, _ NP = Va 2 - f Accordingly, the area of LMNP = a z y z , and the volume of the ele- mentary cylinder standing on LMNP as a base is whence V = f "(a a - y^ dy - [a 2 // - J y] = a 8 . Hence the total volume is - 1 / a 8 . t This method of finding volumes is particularly useful when the sections of the solid made by parallel planes are bounded by circles or by concentric circles. Such a solid may be gen- erated by the revolution of a plane area around an axis in its plane, and is called a solid of revolution. We take the following examples of solids of revolution : Ex. 3. Find the volume of the solid generated by revolving about OX the area bounded by the curve ?y 2 = 4 *, the axis of x, and the line x = 8. The generating area is shown in Fig. 32, where AB is the line a- = 8. Hence OA s= 3, 74 SUMMATION Divide OA into n parts dx, and through the points of division pass straight lines parallel to OY, meeting the curve. When the area is revolved about OX, each of these lines, as MP, JVQ, etc., generates a circle, the plane of which is perpendicular to OX The area ^ of the circle generated by MP, for example, is irMP z , which is equal to iry z = 7r(4 x), if Hence the area of any plane section of the solid made by a plane perpendicular to OX can be expressed in terms of its dis- tance from 0, and we may apply the pievious method for finding the volume. Since the base of any elementary cylinder is 4 TTX and its altitude is dx, we have Hence V=f*4 vxdx = [2 ** = 18 . Ex. 4. Find the volume of the ring surface generated by revolving about the axis of x the area bounded by the line y = 5 and the curve y = 9 x~ The line and the curve (Fig 33) are seen to intersect at the points P l (2, 5) ^ and P z (2, 5), and the ring is generated by the area P^BP^P^ Since this area is symmetrical with respect to OY, it is evi- dent that the volume of the ring is twice the volume generated by the area AP Z BA. Accordingly, we shall find the latter volume and multiply it by 2. We divide the line OM Z = 2 (M z being the projection of P 2 on OX) into n parts dx, and through the points of division draw straight lines parallel to OY and intersect- ing the straight line and the curve One of these lines, as M QP, will, when revolved about OX, generate a circular ring, the outer radius of which is MP = y = Q~x* and the inner radius of which is MQ *=.y = 5 Hence the area of the ring is jjf M 'M Accordingly, = TT (56 - 18 x 9 + a?<). FIG- dV = TT (56 - 18 w 2 + a;*) dx S3 Accordingly, the volume of the ring is 2 (70$ w) = 140$ TT. VOLUME 75 EXERCISES 1. The section of a certain solid made by any plane perpendicular to a given line Oil is a circle with one point in OH and its center on a straight line OB intersecting Oil at an angle of 46 If the height of this solid measured from along Oil is 4ft., find its volume by integration. 2. A solid is such that any cross section perpendicular to an axis is an equilateral triangle of which each side is equal to the square of the distance of the plane of the triangle from a fixed point on the axis. The total length of the axis from the fixed point is 5. Find the volume. 3. Find the volume of the solid generated by revolving about OX the area bounded by OX and the curve y = 4 as a 2 . 4. Find the volume of the solid generated by revolving about OX the area included between the axis of x and the curve y 2 3 x* 5. Find the volume of the solid generated by revolving about the line y = 2 the area bounded by the axis of y, the lines x = 3 and y = 2, and the curve y = 3 a: 2 . 6. On a spherical ball of radius 5 in. two great circles are drawn intersecting at right angles at the points A and B* The material of the ball is then cut away so that the sections perpendicular to AB are squares with their vertices on the two great circles. Find the volume left. 7. Find the volume generated by revolving about the line x = 2 the area bounded by the curve if 8 a?, the axis of a?, and the line x = 2. 8. Any plane section of a certain solid made by a plane perpen- dicular to OF is a square of which the center lies on OY and two opposite vertices lie on the curve y = 4 as 2 . Find the volume of the solid if the extreme distance along OYis 3. 9. Find the volume generated by revolving about OY the area bounded by the curve y 2 8 x and the line x = 2. 10. Find the volume of the solid generated by revolving about OX the area bounded by the curves y = 6 x as 2 and y s= flj 8 6 a? + 10. 11. The cross section of a certain solid made by any plane perpen- dicular to OX is a square, the ends of one of whose sides are on the curves 16 y = y? and 4 y = y? 12, Find the volume of this solid between the points of intersection of the curves. T6 SUMMATION GENERAL EXERCISES 1. The velocity in feet per second of a moving body at any time t is t* 4 1 + 4 Show that the body is always moving in the direction in which s is measured, and find how far it will move during the fifth second 2. The velocity in feet per second of a moving body at any time t is t* 4 1. Show that after t = 4 the body will always move in the direction in which s is measured, and find how far it will move in the time from t = 6 to t = 9. 3. At any time t the velocity in feet per second of a moving body is t z 6 1 + 5 How many feet will the body move in the direction opposite to that in which s is measured? 4. At any time t the velocity in miles per hour of a moving body is t* 2 1 3. If the initial moment of time is 12 o'clock noon, how far will the body move in the time from 11.30 A M. to 2 p M. ? y 5 . Find the area bounded by the curves 9 y = 4 x z and 45 9 y = a: 3 . 6. Find the total area bounded by the curves ?/ 2 =4a; and T/ 2 = 4 a 8 4 ax. '' 7. Find the total area bounded by the curve y = x B and the straight line y = 4 x. 8. Find the total area bounded by the curve y = x (x 1) (03 3) and the straight line y = 4 (a; 1). 9. ABCD is a quadrilateral with A 90, J3 = 90, AB 5 ft., BC = 2 ft., A D 4 ft. It is completely immersed in water with AB in the surface and AD and BC perpendicular to the surface. Find the pressure on one side 10. Prove that the pressure on one side of a rectangle completely submerged with its plane vertical is equal to the area of the rectangle multiplied by the depth of its center and by w (consider only the case in which one side of the rectangle is parallel to the surface). 11. Prove that the pressure on one side of a triangle completely submerged with its plane vertical is equal to its area multiplied by the depth of its median point and by w (consider only the case in which one side of the triangle is parallel to the 1 surface). 12. The end of a trough, full of water, is assumed to be in the form of an equilateral triangle, with its vertex down and its plane vertical. What is the effect upon the pressure on the end if the level of the water sinks halfway to the bottom? GENERAL EXERCISES 77 13. A square 2 ft on a side is immersed in water, with one vertex in the surface of the water and with the diagonal through that vertex perpendicular to the surface of the water. How much greater is the pressure on the lower half of the square than that on the upper half? 14. A board is symmetrical with respect to the line AJB, and is of such a shape that the length of any line across the board perpendic- ular to AB is twice the cube of the distance of the line from A. AD is 2 ft. long The board is totally submerged in water, AB being perpendicular to the suiface of the water and A one foot below the surface. Find the pressure on one side of the board. 15. Find the pressure on one side of an area the equations of whose boundary hues are x = 0, y = 4, and v/ 2 = 4 x respectively, where the axis of x is taken in the surface of the water and where the positive direction of the y axis is downward and vertical. 16. Find the volume generated by revolving about OX the area bounded by OX and the curve 4 y = 16 a; 2 . I,- 17. Find the volume generated by revolving about OX the area bounded by the curve y = a 2 + 2 and the line y = 3. 18. Find the volume generated by revolving about OX the area bounded by OX and the curve y = 3 x x 8 . 19. Find the volume generated by revolving about the line y = 1 the area bounded by the curves 9 y = 2 x* and 9 y = 36 2 a; 2 . 20. An axman makes a wedge-shaped cut in the trunk of a tree. Assuming that the trunk is a right circular cylinder of radius 8 in., that the lower surface of the cut is a horizontal plane, and that the upper surface is a plane inclined at an angle of 45 to the horizontal and intersecting the lower surface of the cut in a diameter, find the amount of wood cut out. 21. On a system of parallel chords of a circle of radius 2 there are constructed equilateral triangles with their planes perpendicular to the plane of the circle and on the same side of that plane, thus forming a solid. Find the volume of the solid. 22. Show that the volume of the solid generated by revolving about OY the area bounded by OX and the curve y = a Ix* is equal to the area of the base of the solid multiplied by half its altitude. 23. In a sphere of radius a find the volume of a segment of one base and altitude /*. 78 SUMMATION 24. A solid is sucli that any cross section perpendicular to an axis is a circle, with its radius equal to the square root of the distance of the section from a fixed point of the axis. The total length of the axis from the fixed point is 4. Find the volume of the solid. 25. A variable square moves with its plane perpendicular to the axis of y and with the ends of one of its diagonals respectively in the parts of the curves y* = 16 x and ^ = 4 x, which are above the axis of x Find the volume generated by the square as its plane moves a distance 8 from the origin. 26. The plane of a variable circle moves so as to be perpendicular to OX, and the ends of a diameter are on the curves y = a; 2 and y = 3 cc 2 8 Find the volume of the solid generated as the plane moves from one point of intersection of the curves to the other. 27. All sections of a certain solid made by planes perpendicular to OF are isosceles triangles. The base of each triangle is a line drawn perpendicular to OY, with its ends in the curve y = 4 a: 2 . The altitude of each triangle is equal to its base Find the volume of the solid included between the planes for which y = and y 6. 28. All sections of a certain solid made by planes perpendicular to Y are right isosceles triangles One leg of each triangle coincides with the line perpendicular to OY with its ends in OY and the curve y* 4 x. Find the volume of the solid between the sections for which y = and y = 8. 29. Find the work done in pumping all the water from a full cylindrical tank, of height 15 ft. and radius 3 ft., to a height of 20 ft. above the top of the tank. 30. Find the work done in emptying of water a full conical receiver of altitude 6ft and radius 3ft., the vertex of the cone being down. CHAPTER IV ALGEBRAIC FUNCTIONS 27. Distance between two points. Let P 1 (a^, / x ) and P z (r z , / a ) (Fig. 34) be any two points in the plane XOY, such that the straight line P^P Z is not parallel either to OX or to OY. Through P! draw a straight line parallel to OJT, and through P 2 draw a straight line parallel to OY, and denote their point of intersection by R. iPlTATl T^ 7? Al 1 * "-" 1" - ty JL UOll JL^J.V * A-l*t/ t'g " " *t*j and In the right triangle P^RP Z whence If C 1 ) is parallel to 0JT, and the formula reduces to J?J = JBa- ! (2) In like manner, if ^-x^ P^ is parallel to OF, and the formula reduces to 28. Circle. Since a ciVc?e is the locus of a point which is always at a constant distance from a fixed point, formula (1) 27, enables us to write down immediately the equation of P circle. Let C(h, 7c) (Fig. 35) be the center of a circle of radius r. Then, if P(x, y) is any point of the circle, by (1), 27, a? and y must satisfy the equation -^ a =^ (1) 79 80 ALGEBKAIC FUNCTIONS Moreover, any point the coordinates of which satisfy (1), must be at the distance r from C and hence be a point of the circle. Accordingly, (1) is the equation of a circle. If (1) is expanded, it becomes ! -2Aa;-2^ + A 2 +F r 2 =0, (2) an equation of the second degree with no term in xy and with the coefficients of a; 2 and / a equal. Conversely, any equation of the second degree with no xy term and with the coefficients of a; 2 and y* equal (as O where A, G, F, and are any con- ^ 1Q g5 stants) may be transformed into the form (1) and represents a circle, unless the number cor- responding to r z is negative (see Ex. 3, page 81), in which case the equation is satisfied by no real values of x and y and accordingly has no corresponding locus. The circle is most readily drawn by making such transfor- mation, locating the center, and constructing the circle with compasses. Ex.1. This equation may be written in the form (x*-2x ) + (y2-4y ) = 0, and the terms in the parentheses may be made perfect squares by adding 1 in the first parenthesis and 4 in the second parenthesis As we have added a total of 5 to the left-hand side of the equation, we must add an equal amount to the right-hand side of the equation The result is Cr a - 2a: + 1) + (f - 4?/ + 4) = 5, which may be placed in the form (-l) + <y-a)-5, the equation of a circle of radius V5 with its center at the point (1, 2). CIRCLE 81 Ex. 2. 9 a: 2 + 9 f - 9 x + 6 y - 8 = 0. Placing 8 on the right-hand side of the equation and then dividing by 9, we have ** + ,*-*+ 3, = |, which may be treated by the method used in Ex. 1. The result is the equation of a circle of ladius JV5, with its center at (\, i). Ex.3. 9a; 2 + 9y 2 -6a;-|-12y + ll = Proceeding as in Ex. 2, we have, as the transformed equation, an equation which cannot be satisfied by any real values of js and y, since the sum of two positive quantities cannot be negative Hence this equation corresponds to no real curve. EXERCISES 1. Find the equation of the circle with the center (4, 2) and the radius 3 2. Find the equation of the circle with the center (0, 1) and the radius 5. 3. Find the center and the radius of the circle 9 = 0. 4. Find the center and the radius of the circle - 6v - 15 = 0. 5. Find the equation of the straight line passing through the center of the circle and perpendicular to the line 2aj + 3v/ 4 = 6. Prove that two circles are concentric if their equations differ only in the absolute term 29. Parabola. The locus of a point equally distant from a fixed point and a fixed straight line is catted a parabola. The fixed point is called the focus and the fixed straight line is called the directrix, 82 ALGEBRAIC FUNCTIONS Let F (Fig. 36) be tho focus and A\V the directrix of a parabola. Through F draw a straight line perpendicular to ftS, intersecting it at JP, and let this lino be the axis of jr. Let the middle point of T>F be taken as 0, tho origin of coordi- nates, and draw the axis QY. Thon, if tho distance, PA* is 2 <, Urn coordinates of F are (0, 0) and tho equation of H$ is sr -~ - r. Let /^(.r, y) be any point of tho parabola, and draw Uu straight line FP and tho straight lino NP gy perpendicular to US. Then NP = + , and, by 27, FP- V(' <0 2 + //* /J whence, from tho definition of tho parabola, (a-*)** if- Or +<0" which reduces to ,?/ 2 = 4 6'a;. (1 ) ** p x(li ,w Conversely, if the coordinates of any point /* satisfy (1), it can be shown that the distances FP and NP aru equal, and hence P is a point of the parabola. Solving (1) for y in terms of a-, we havo y = 2VS, (2) We assume that e is positive. Thon it is evident that if a negative value is assigned to #, y is imaginary, and no correspond- ing points of the parabola can bo located. All poKitivo vahu'M may be assigned to JR, however, and hence tho parabola IICH entirely on the positive sidtj of tho axis OF. Accordingly, wo assign positive valuos to #, compute tho nor- responding values of y, and draw a smooth curve through tho points thus located. It is to be noticed that to every value tWHigncd to w lhro aro two corresponding valuos of ?/, equal in magnitude and opposite in algebraic sign, to which there correspond two points of tho parabola on opposite sides of OX and equally dintant from it. Hence the parabola is ttymmMoal with rospect to CUT, and ac- cordingly OX is called the gunk of the parabola. The point at which its axis intersects a parabola is called the w tew of the parabola. Accordingly, is the vertex of the parabola. PARABOLA 83 Returning to Fig. 36, if F is taken at the left of with the coordinates ( c, 0), and RS is taken at the right of with the equation x = c, equation (1) becomes f = -kcx (3) and represents a parabola lying on the negative side of OY. Hence we conclude that any equation in the form f=fa, (4) where k is a positive or a negative constant, is a parabola, with (k Z' and its directrix the straight line x'-^* Similarly, the equation x*= Icy (5) represents a parabola, with its vertex at and with its axis coin- ciding with the positive or the negative part of OY, according as & is positive or negative. The focus is always the point 0, -^ j and the directrix is the line y = - whether k be positive ^"Y "i or negative. 30. Parabolic segment. An important property of the parabola is contained in the following theorem : The square of any two chords of a parabola which are perpen- dicular to its axis are to each other as their distances from the vertex of the parabola. This theorem may be proved as follows : Let %(%!, #j.) and P z (x^ 7/ 3 ) be any two points of any parab- ola /= Tex (Fig. 37). Then y*= lex^ and yl 7c a ; y\ X* whence , ~ yl ** whence 84 ALGEBRAIC FUNCTIONS From the symmetry of the parabola, 2# 1 = But a? = OM^ and x,== OM Z , and hence (1) becomes and 2# 2 = FIG. 37 and the theorem is proved. The figure bounded by the parabola and a chord perpendicular to the axis of the parabola, as Q^OP^ (Fig. 37), is called a parabolic segment. The chord is called the lase of the segment, the vertex of the parabola is called the vertex of the segment, and the distance from, the vertex to the base is called the altitude of the segment. EXERCISES Plot the following parabolas, determining the focus of each : 1. !/* = 8x 3. 2/ 2 =:6a5. 2. cc 2 =42/. 4. aj 2 = -7y. 5. The altitude of a parabolic segment is 10 ft., and the length of its base is 16 ft. A straight line drawn across the segment perpen- dicular to its axis is 10 ft. long. How far is it from the vertex of the segment ? 6. An arch in the form of a parabolic curve, the axis being vertical, is 50 ft. across the bottom, and the highest point is 15 ft. above the horizontal. What is the length of a beam placed horizon- tally across the arch 6 ft. from the top ? 7. The cable of a suspension bridge hangs in the form of a parabola. The roadway, which is horizontal and 400ft. long, is supported by vertical wires attached to the cable, the longest wire being 80 ft. and the shortest being 20 ft. Find the length o a supporting wire attached to the roadway 75 ft. from tho middle. 8. Any section of a given parabolic mirror made by a plane passing through the axis of the mirror is a parabolic segment of which the altitude is 6 in. and the length of the base 10 in. Find the circumference of the section of the mirror made by a piano perpendicular to its axis and 4 in. from, its vertex ELLIPSE 85 9. "Find the equation of the parabola having the line x = 3 as its directrix and having its focus at the origin of cooidmates. 10. Find the equation of the parabola having the line y 2 as its directrix and having its locus at the point (2, 4). 31. Ellipse. The locus of a point the sum of whose distances from two fixed points is constant is called an ellipse. The two fixed points are called the foci. Let F and F' (Fig. 38) be the two loci, and let the distance F'F be 2 c. Let the straight line determined by F' and F be taken as the axis of x, and the middle point of F'F be taken as 0, the origin of coordinates, and draw the axis OY. Then the coordinates of F' and F are respectively ( c, 0) and (*, 0). Let P(x, y) be anj' point of the ellipse, and 2 a repre- n' sent the constant sum of its 00 Ju ICr uO distances from the foci. Then, from the definition of the ellipse, the sum of the distances F'P and FP is 2 a, and from the triangle F'PF it is evident that 2 a > 2 G ; whence a > c. By 27, and FP= whence, from the definition of the ellipse, V(s-M) 2 +# 2 W(a-c) a +/= 2 - (1) Clearing (1) of radicals, we have O 2 - c a > 2 + ay = a 4 - aV. (2) Dividing (2) by a* aV, we have 86 ALGEBRAIC FUNCTIONS But since ><?, a* c* is a positive quantity which may be denoted by J 3 , and (3) becomes Conversely, if the coordinates of any point P satisfy (4), it can be shown that the sum of the distances Jf'P and FP ia 2 a, and hence P is a point of the ellipse. Solving (4) for y in terms of #, we have y=s >/-** (5) It is evident that the only values which can be assigned to x must be numerically less than a; for if any numerically larger values are assigned to r, the corresponding values of y are imaginary, and no corresponding points can be plotted. Hence the curve lies entirely between the lines x = a and x = a. We may, then, assign the possible values to #, compute the corresponding values of y, and, locating the corresponding points, draw a smooth curve through them. As in the case of the pa- rabola, we observe that OX is an axis of symmetry of the ellipse. We may also solve (4) for x in terms of y, with the result as = 5^-^. (0) From this form of the equation we find that the ellipse lies entirely between the lines ,y = --iandy = 5 and is symmetrical with respect to OY. Hence the ellipse has two axes, A' A and B'JB (Fig. 88), which are at right angles to each other. But A 1 A = 2 a and JR'13 = 2 1) ; and since a > 5, it follows that A' A > B'B, Hence A' A is called the major axis of the ellipse, and fl'JJ is called the minor axis of the ellipse. The ends of the major axis, A' and A, are called the wrtwes of the ellipse, and the point midway between the vertices is called the center of the ellipse ; that is, is the center of the ellipse, and it can be readily shown that any chord of the ellipse which passes through is bisected by that point. ELLIPSE 87 From the definition of />, c=* Va a #*, and the coordinates of the foci are ( V<r A 9 , 0). OF The ratio ----- (thai is, the ratio of the distance of the focus from the center to the distance of either vertex from the center) is called the eccentricity of the ellipse and is denoted by t>. lint M'^VtfCjS, (7) Vrt a i* and hence c - - - > (8) \ / whence it followa that the eccentricity of an ellipse is always less than unity. Similarly, any equation in form (4), in which I' 2 > a 2 , represents an ellipse with its center at 0, its major axis on 6>J", and its minor axis on CLY. Then the vertices are the points (0, &), ~ the foci are the points (0, vV *), and e = In either case the nearer the foci approach coincidence, the smaller e becomes and the more nearly b = a. Hence a drde inat/ le considered an an ellipse with dointxdfnt fovi and equal axes. Its eceentrieity is, of course, zero. EXERCISES Plot tho following ellipses, finding the vertices, the Toci, and the eceentrieity of eaoh : 1. 9 ^ a + 1 if 144. 3. 3 tc 8 + 4 ?/ 2. 2. 9 w! 8 + 4 ?/ 30. 4. 2 a; 3 + 3 y a s= 1. 6. Find tho equation of the ellipse winch lias its foci at tlie points ( 2, 0) and ((>, 0) and which has the sum of the dis lances of any point on it from the foci equal to 10. 6. Find the equation of the ellipse having its food at the points (0, 0) and (0, ) and having the length of its major axis equal to 7. 32. Hyperbola. The locus of a point the difference of whose , distances from two faced points is constant is called a hyperbola. The two fixed points are called the foci* 88 ALGEBRAIC FUNCTIONS Let F and F' (Fig. 30) be the two foci, and lot tho distance F'F bo 2 0. Let the straight line determined by F' and F bo taken as the axis of .P, and the middle point of F'F be taken as 0, the origin of coordinates, and draw the axis O Y. Then the coordinates of F 1 and F are respectively ( <?, 0) and (c, 0). Let jP (a;, ?/) be any point of the hyperbola and 2 a represent the constant dif- ference of its distances from the foci. Then, from the definition of the hyperbola, . , the difference of the dis- ' tances F'P and FP is 2 a, and from the triangle F'PFii is evident that 2 < 2 , for the difference of any two sides of a triangle IH loss than the third side ; whence a < a. 1?ia * 80 By 27, and whence either ;*+; 2 (1) or V(* + 0'+^-V(? ::: ^+7-2a, (2) according as JRP or ^"/* in the groatcsr diHtance. Clearing either (1) or (2) of radicals, wo obtain tho aamo result : Dividing (3) by a* V, we have i it _ "1 fA.\ ' "T a * * \*J cr r But since <c, 2 * is a negative quantity whiuh may bo denoted by ft a , and (4) becomes HYPJEEBOLA 89 Conversely, if the coordinates of any point P satisfy (5), it can be shown that the difference of the distances F'P and FP is 2 a, and hence P is a point of the hyperbola. Solving (5) for y in terms of #, we have y = -^i?-d\ (6) In this equation we may assume for x only values that are numerically greater than a, as any other values give imaginary values for y. Hence there are no points of the hyperbola be- tween the lines t = a and x = a. The hyperbola is symmetrical with respect to OX. As the values assigned to x increase numerically, the corre- sponding points of the hyperbola recede from the axis OX. We may, however, write (6) m the form 5? (7) Now if y x and i/ a are the ordinates of points of (7) and of the straight lines y = -x respectively, then w 7 oa whence Lim (# a Hence, by prolonging the straight lines and the curve indefi- nitely, we can make them come as near together as we please. Now, when a straight line has such a position with respect to a curve that as the two are indefinitely prolonged the dis- tance between them approaches zero as a limit, the straight line is called an asymptote of the curve. It follows that the lines y = -x and y => as are asymptotes of the hyperbola (Fig. 39). (t CL If we had solved (5) for a in terms of y, the result would have been (8) 90 ALGEBRAIC FUNCTIONS from which it appears that all values may be assigned to #, and. that OT is also an axis of symmetry of the hyperbola. The points A' and A in which one axis of the hyperbola inter- sects the hyperbola are called the vertices, and the portion of the axis extending from A' to A is called the transverse axis. The point midway between the vertices is called the center ; that is, is the center of the hyperbola, and it can readily be shown that any chord of the hyperbola which passes through O is bisected by that point. The other axis of the hyperbola, which is perpendicular to the transverse axis, is called the conju- gate axis. This axis does not intersect the curve, as is evident from the figure, but it is useful in fixing the asymptotes and thus determining the shape of the curve for large values of x. From the definition of 5, c = Va 2 + 6 2 , and the coordinates of the foci are (V 2 +5 2 , 0). Therefore If we define the eccentricity of the hyperbola as the ratio OF , , _ > we have a = ^+* ( 1 0) a quantity which is evidently always greater than unity. Similarly, the equation 2 is the equation of a hyperbola, with its center at 0, its trans- verse axis on OF, and its conjugate axis on OX. Then the ver- tices are the points (0, 5), the foci are the points (0, V& 2 +a 2 ), the asymptotes are the straight lines y = -#, and e = 7" Cv If 5 = a, in either (5) or (11), the equation of the hyperbola assumes the form a 2 -2/ 2 =a 2 or y-a^=:a a , (12) and the hyperbola is called an equilateral "hyperbola. The equa- tions of the asymptotes become y =* x ; and as these lines are perpendicular to each other, the hyperbola is also called a rectangular hyperbola. CURVES EXERCISES 91 Plot the following hyperbolas, finding the vertices, the foci, the asymptotes, and the eccentricity of each . 2. 9o; a 4?/ a =36. 5. 3. 32/ a -2jc a =6 6. 7. Find the equation of the hyperbola having its foci at the points (0, 0) and (4, 0), and the difference of the distances of any point on it from, the foci equal to 2. 8. The foci of a hyperbola are at the points ( 4, 2) and (4, 2), and the difference of the distances of any point on it from the foci is 4. Eind the equation of the hyperbola, and plot. 33. Other curves. In the discussion of the parabola, the ellipse, and the hyperbola, the axes of symmetry and the asymptotes were of considerable assistance in constructing the curves ; more- over, the knowledge that there could be no points of the curve in certain parts of the plane decreased the labor of drawing the curves. We shall now plot the loci of a few equations, noting in advance whether the curve is bounded in any direc- tion or has any axes of symmetry or asymptotes. In this way we shall be able to anticipate to a con- ^ siderable extent the form of the curve. Ex. 1. (y + 3) a = (x - 2) a (a; + 1). Solving for y, we have In the first place, we see that the only values that may be assigned to x are greater than 1, and hence the curve lies entirely on the positive side of the line x = 1. Further- more, corresponding to every value of a?, there are two values of y which determine two points at equal distances from the line y = 8. Hence we conclude that the line y 8 is an axis of symmetry of the curve. Assigning values to x and locating the points determined, we draw the curve (JFig. 40). < 49 92 ALGEBKAIO FUNCTIONS Ex. 2. xy = 4. 4. Solving foi y, we have y = - x It is evident, then, that we may assign to x any real value except xero, in which case we should be asked to divide 4 by 0, a process that cannot be carried out. Consequently, there can be no point of the curve on the line # = 0; that is, on OY. We may, however, assume values for x as near to zero as we wish, and the nearer they are to zero, the nearer the corresponding points are to OF; but as the points come nearer to OY they recede along the curve. Hence OF is an asymptote x^ of the curve. If we solve for x, we have Fm. 41 and, reasoning as above, we conclude that the line y = (that is, the axis OX) is also an asymptote of the curve. The curve is drawn in Fig. 41. It is a special case of the curve xy = k, where k is a ical constant which may be either positive or negative, and is, in fact, a i octangular hyperbola leferiod to its asymptotes as axes. It is customaiy to say that when the denominator of a fraction is SMTO, the value of the fraction becomes infinite The curve just constructed shows graphically what is meant by such an expiession. Ex. 3. sry + 2 or + y 1 = 0. Solving for y, we have y from which we conclude that the line a? = 1 is an asymptote of the curve. Solving for x, we have Fio. 42 2+y from which we conclude that the line ?/= 2 is also an asymptote of the curve. We accordingly draw these two asymptotes (Fig. 42) and the curve through the points determined by assigning values to either x or y and computing the corresponding values of the other vai table. The curve is, in fact, a rectangular hyperbola, with the lines x 1 and y SB 2 as its asymptotes. CUEVES Ex.4. a- 8 X Solving for y, we have Vx* 5 ' 2a~a; whence it is evident that the curve is sym- metrical with respect to OX The lines x and x = 2 a, corresponding to tlie values of a? which make the numuiatoi and the denomi- nator of the fraction under the ladical sign lespectively zero, divide the plane into three stups; and only values between and 2 a can be substituted for y, since all other values make y imaginaiy. It follows that the curve lies entirely in the strip bounded by the two lines x and x 2 a. By the same reasoning that was used in Exs 2 and 3, it can be shown that the line x = 2 a is an asymptote of the curve. The curve, which is called a cuboid, is drawn in Fig. 43. EXERCISES Plot the following curves : 1. 2/ a =o; 8 2. 2/ a =aj a (te + 4). 3. 2/ a =4(o;-8). 4. y a =a: a 5cc-f 6. 6. y z = a) (a? 2 4). 6. / a =aj 8 FIG. 43 7. y a = 8. 7/ 2 =4 a?. 9. xy 5. 10. 3y oj// = 11. xij 2aj + 34. Theorems on limits. In. obtaining more general formulas for differentiation, the following theorems on limits will be assumed without formal proof : 1. The limit of the sum of a finite number of variables is equal to the. sum of the limits of the variables. 2. The limit of the product of a finite number of variables is egual to the product of the limits of the variables. 94 ALGEBRAIC FUNCTIONS 3. The limit of a constant multiplied by a variable is equal to the constant multiplied by the limit of the variable. 4. The limit of the quotient of two variables is equal to the quotient of the limits of the variables, provided the limit of the divisor is not zero. 35. Theorems on derivatives. In order to extend the process of differentiation to functions other than polynomials, we shall need the following theorems : 1. The derivative of a constant is zero. This theorem was proved m 8. 2. The derivative of a constant times a function is equal to the constant times the derivative of the function. Let u be a function of x which can be differentiated, let c be a constant, and place ,. _ . * U -^ G&6i Give x an increment Ax, and let AM and Ay be the corre- sponding increments of u and y. Then Ay = c (u + AM) cu =s c AM. TT Ay AM Hence - = <, Ax Ax and, by theorem 3, 34, T . Ay T . AM Lim -T- C Lim -- Ax Ax Therefore ~i~ c T~ ax dx by the definition of a derivative. Ex. 1. y = 5 (a; 3 + 3 x z + 1). 3. The derivative of the sum of a finite number of functions is egual to the sum of the derivatives of the functions. Let M, v, and w be three functions of x which can be differen- tiated, and let DERIVATIVES 95 Give x an increment Aa;, and let the corresponding increments of u, t>, w, and y be AM, Av, Aw, and A#. Then Ay = (M H- Aw + v + At) + w + Aw) (w + v + w) = Aw + Aw + Aw ; 1 Ay AM , Aw , Aw whence ^ = -7 + T~~ + 7 A# Aa; Aa; Aa; Now let Aa; approach zero. By theorem 1, 34, , . Ay T . AM , T A0 . 7 Aw Lini ^ = Lim - - + Lim - + Lim - ; Aa; Ax Ax Ax that is, by the definition, of a derivative, dy __ du dv dw duo dx dx dx The proof is evidently applicable to any finite number of functions. Ex. 2. y = x* - 3 x s + 2 x z - 7x. 4. The derivative of the product of a finite number of functions is equal to the sum of the products obtained "by multiplying the derivative of each factor by all the other factors. Let u and v be two functions of x which can be differentiated, andlet y = uv. Give x an increment Aa;, and let the corresponding increments of u, v, and y be Aw, Av, and A#. Then Ay (it + Aw) (v + Av) uv = u Av + v AM + AM Av , Ay Av Au . AM A and = w * + T~ + T-" ^ Aa? Ax Ax Ax 96 ALGEBRAIC FUNCTIONS If, now, Ax approaches zero, we have, by 34, r Aw T . Aw , T Aw , T . AM T . A Lim -~ = u Lim - + v Lim - + Lim - Lim Aw. Ao; Aa; Aa; Are But Lim Aw = 0, -. ,, P dy dv , du and theretore -^- = w 4- w ax ax ax Again, let y = uvw. Regarding uv as one function and applying the result already obtained, we have dy dw , d(uv) -JL = m - + w -+ dx ax at dw \ du . < = uv-=--)-w\u + v dx \_ dx t tu? dw dv du = UV + UW -f VW -r- ax dx dx The proof is clearly applicable to any finite numbers of factors. Ex. 3. y = 3a:- dx = (3 x - 5) (z" + 1) (3 a; 2 ) + (3 x - 5)a; 8 (2 a;) + (a: 2 + I)* 8 (3) = (18 a; 8 - 25 a; 2 + 12 a: - 15)a: 2 5. The derivative of a fraction is equal to t7ie denominator times the derivative of the numerator minus the numerator times the deriva- tive of the denominator, all divided ty the square of the denominator. nt Let y -> where u and v are two functions of x which can be v differentiated. Give x an increment Aa;, and let Aw, Av, and A?/ be the corresponding increments of w, v, and y. Then . _ u + Aw _ u _ v Aw w Aw v-f-A'y w w a +'uAw An Az> w- -- u- and % = A.; A^^ Ao: v a + w Aw DERIVATIVES 97 Now let A* approach zero. By 34, T . A-M T . A?J v Liin - -- u Lim - A/ A Aa; Lim - = ,, T . , > Aa; tr+vLmiAv du dv 01 _________ rilj _ l rty , efa/ <fcc difo whence - = dx v* T 8 1 Ex.4. y = ~ ~ J x* + l (^_(a, 2 + l)(27:)-(a; 2 -l)2r_ 4 .1? dx (x* + I) 3 (a,' J + I) 2 6. TJie derivative of the nth power of a function is obtained ly multiplying n times the (nT)th power of the function ly the derivative of the function. Let y = u n , where u is any function of x which can be differ- entiated and n is a constant. We need to distinguish four cases : CASE I. When n is a positive integer. Give x an increment Aa, and let AM and A?/ be the corre- sponding increments of u and y. Then Ay = (u 4- Aw)" u n ; whence, by the binomial theorem, A?/ n _iAw , n(n 1) B _ 2A AM , ^L = nu n l ~ H ft y w w 2 Au- h . Aa; Aa? 2 Aa; ' Aa; Now let Aa;, AM, A?/ approach zero, and apply theorems 1 and 2, 34. The limit of ^ is ^ the limit of ~ i s ^*, and the Aa; a* Aa; ajc limit of all terms except the first on the right-hand side of the last equation is zero, since each contains the factor AM. Therefore d du ^. = nw n ~ 1 - T - < ax dx 08 ALGEBKAIC FUNCTIONS CASE II. When n is a positive rational fraction. ny Let n = where p and q are positive integers, and place By raising both sides of this equation to the gth power, we have y=u. Here we have two functions of x which are equal for all values o x. Taking the derivative of both sides of the last equation, we have, by Case I, since p and q are positive integers, ^-1^==*-^. yy dx P dx Substituting the value of y and dividing, we have f^itf- 1 **. * dx q dx Hence, in this case also, dy _ ,du -^- nu n ~ l dx ax CASE III. When n is a negative rational number. Let n m, where m is a positive number, and place ys=U~ m s= -- 9 u m Then ^= ** (by 5) dx u zm mu m ~ l = (by Cases I and II) u mu~ m ~^~r dx Hence, ui this case also, d_ _i^w dx dx DERIVATIVES 99 CASE IV. When n is an irrational number. The formula is true in this case also, but the proof will not be given. It appears that the theorem is true for all real values of n. It may be restated as a working-rule in the following words: To differentiate a power of any quantity, bring down the exponent as a coefficient, write the quantity with an exponent one less, and multiply ly the derivative of the quantity. Ex. 5. y = (x s + 4 x* - 5 x + 7) s . ^ = 3 (a: 8 + 4 a 2 - 5 x + 7) 2 -f (a; 3 + 4 x z - 5 x + 7) tlJG U3S = 3 (So; 2 + Bx - 5) (.e 3 + 4a: 2 - 5 x + 7) 3 . Ex. 6. y = Vtf + -. = ^ + x- * dx 3 2 3 *' Ex. 7. y = (a: + l)Va; 8 = (z + 1) [ J (a; 2 + I)" * - 2 ar] + (a 2 + 1 (a 2 + !) 2 a 8 + x 4- 1 Y /Li 100 ALGEBRAIC FUNCTIONS 7. If y is a function of x, then x is a function of y, and the derivative of x with respect to y is the reciprocal of the derivative of y with respect to x. Let A# and A?/ be corresponding increments of x and y. It is immaterial whether b*.x is assumed and A?/ determined, or A# is assumed and Aa; determined. In either case A.r 1 A^~"Ay' Aa; . _. A,r 1 whence Lun = A ; Ay T A?/ ' Lim - Aas dx 1 that is, -- = - 8. If y is a function of u and u v' function of a*, then y is a function of x, and the derivative of y with respect to x is equal to the product of the derivative of y with respect to u and the derivative of u with respect to x. An increment Aa; determines an increment Aw, and this in turn determines an increment Ay, Then, evidently, A?/ __ A?/ Aw Ax Aw A:# , T . Aw , . A?/ T . AM whence Lim ~ = Lim - Lim . ; A:B A?t Arc , , , . dy dy du that is, ^ .X. . dx du dx Ex. 9. w a u 3 + 8 u + 1, whoro w = - 1+ The same result is obtained by fmbHtituting in the expression for y the value of u in terms of a; and then differentiating. DERIVATIVES 101 36. Formulas. We may now collect our formulas of differen- tiation in the following table : ^ = 0, (1) dx rf co = tf *, C2 ) dx dv d(u4-v) _du dv n dx dx dx d Cuv) dv , du , , , \ - U T + V 1T' ( 4 ) dx dx dx , /?/\ du dv d(-) v- -- M \v/ dx dx dy^y. d* t /g-) dx du dx ^ ^ dy ^ = (9) dx dx du Formula (9) is a combination of (7) and (8). The first six formulas may be changed to corresponding for- mulas for differentials by multiplying both sides of each equation by dx. They are 7 n _< - v J J da = 0, (10) (11) Zw, (12) d (uv) udv + v du, (1 3) w\ vdu udv 102 ALGEBRAIC FUNCTIONS EXERCISES Find -~ in each of the following oases : 2. y s (a; 2 - 2 1 + 3) (a: 3 +60: + 9). 14. H-l "** 7 . 8 X?/ = (4 jr fl + 3 x + I) 3 - 4 X* + 1. 16. > re V<.) 17 ' '/ = (' + 18. 37. BijEferentiatlon of implicit functions. Consider any equa- tion containing two variables a? and ,?/, If one of them, EH jr, in chosen as the independent variable and a value is assigned in it, the values of y are determined, Hence tho given equation defines y as a function of rr. If the equation is solved for y in terms of a, y is called an earpKoit function of x, If the equation is not solved for y, y is (jailed an implidt function of #* For example, ^+8aM-4y + 4^ + 2^ + 4^0, which may be written # 9 + (4 aj + 2)y + (8 .-K 9 -|- 4 4- 4) 0, defines y as an implicit function of a?. If the equation is solved for y, the renult # 2j lVa^T8 expresses y as an explicit function of #. IMPLICIT FUNCTIONS 103 If it is required to find the derivative of an implicit function, the equation may be differentiated as given, the result being an equation which may be solved algebraically for the derivative. This method is illustrated in the following examples : Ex. 1. z 2 + 2 = 5. If x is the independent variable, that is, 2 x + 2 y- = 0, dx whence Jl = -. dx y Or the derivative may be found by taking the differential of both sides, as follows: + /) = d(5) = 0; that is, 2 xdx + 2 ydy = 0, whence ^=_. dx y It is also possible first to solve the given equation f or y, thus . y = V5-ff 2 ; whence rfa; V5 a: 8 a result evidently equivalent to the result previously found. The method of finding the second derivative of an implicit function is illustrated in the following example : Ex. 2. Find ^f if a a + f = 5 (IJS We know from Ex. 1 that ^L = _ 2 . dx y Therefore fSUfLft dor ax \y/ .*-*(!) _ ?y a + a- 8 _ 5 f 2/ 8 ' since ^ 2 + a; 2 = 5, from the given equation 104 ALGEBRAIC FUNCTIONS EXERCISES "Find ~ from each of tho following equations : CtJC 1. a* + / - 3 *y = 0. 3. v/ = ~~- j -|- // 2. aj 2 // -f- 4 V = 8 a 9 . 4. V // + ' + V ^ ,r -- . Pind ^ and l -~. from each oJ' tho following (wuiationa : 6. 2x a 6. 4 a- 3 9v/ a =3C. 7 . ai+jf-a. io - 8. a* + * = <A 11. 38. Tangent line. Let J^C^'j, ^O bo a clioHou point, of any curve, and lot (^) bo the value of '/- when aj=sw ( . and ;/-://,. V ^V v / // >* ^ ^ /x..\ VWi <M Then f-^J is the slopo oC tho curve at, iho point J^ and also \dx/\ the slopo of tho tan gout line ( 1H) to tho ourvo at that point, Accordingly, tho equation of the tang-out line at J{ m ( 15) Ex. 1. Find tho equation (if tho lungojit lino to tho parabola / 8 ss JJ .r at tho point (3, ). By differentiation wo have whence //SB.'-., . ti // Hence, at tho point (3, JJ), tho Hlopn of tho tangont lino is J, and itH equation is y-3-1 (*-) or ar 2 ?/ + 3 = 0. The angle of mteraeotion of two curves is tho angle betwoon their respective tangontu at the point of iutenwotiun. The method of finding tho anglo of intersection i illustrated in the example on. tho following pago* TANGENT LINE 105 Ex. 2. Find the angle of intersection of the cncle # 2 + # 2 = 8 and of the parabola ar 2 = 2 y. v The points of intersection are P l (2, 2) and />(- 2, 2) (Fig 44), and fioiu the sym- ^ nietry of the diagram it is evident, that the angles of intersection at P l and P z aie the same. Diffeientiating the equation of the circle, we have 2 JT + 2 y = 0, whence = ; ' (Ijc ux y and differentiating the equation of tlTb pa- labola, we find = i. Fid 44 Hence at P i the slope of the tangent to the circle 1 is 1, and the slope of the tangent to the parabola is 2 Accordingly, if /? denotes the angle of intersection, by Ex 11, p 35, or y = tan~ 1 3. EXERCISES 1. Find the equation of the tangent lino to the ourve x 8 + 16 y - 8 = at the point (2, 2). 2. Find the equation of the tangent lino to the p,urve 5 a; 2 4 a- 2 // = 4 y 8 at the point (2,1). 3. Find the point at which the tangent to the curve 8 y = a; 8 at (1, &) intersects the curve again. 4. Find the angle of intersection of the tangents to the curve y z = x 9 at the points for which x = 1 9T If 5. Show that the equation of the tangent to the ellipse ~j + ^ = 1 at the point (a^, y,) is + -M = L 6. Show that the equation of the tangent to the hyperbola | - |j = 1 at the point fo, ^) is ^ - %& = I. Y. Show that the equation of the tangent to the parabola ?/ = kx 1: at the point (x lt y^ is y$ = ^ (as + a 1 ,). Draw eacli pair of the following curves in one diagram and deter- mine the angles at which, they intersect : 106 ALCKEBRAK! FUNCTIONS 8. a: 9 + y j = S, a- 9 + if - U ^ H- 4 >/ - f> 0. 9. a; a = 3 #, U // a 8 a-. 10. y fl ~ 4 '; <* 'I' //" -- fi 11. y = 2.r, .<//- 18. 12. x 4 // J 0, .r - - -1 j' 4 if 0. 13. aj a +y 9 ~2, l >, .<' a -//- ". 39. The differentials <?#, dy, ds. On any given <uw li-t tin- distance from somo Jixod initiul point moasmvd alnnjjf the cur\i to any point P bo donoiwd by , whcro is posit.ivt* if /' lifs in one direction from the initial point, and negative if /' lies in I In- apposite direction. Tho (jlit)ico of tho ^ positive direction is purely arbitrary. ./ We shall take as the positive direc- tion of the tangent that which shown the positive direction of the ourvo, and shall denote the angle between // the positive direction of (L\ and the ' positive direction of the tangent by 0, Now for a fixed curve and a fixed initial point the position of a point /' p, (Ji is determined if a is given. Hence & and y, the coordinates of J\ are functions of which in are continuous and may bo differentiated. \V Hhall now Mhuw tlmt dx , dy . , -7- COS G>, - aa BUI O, ds </ Let arcP0=Afl (Fig. 45), whenj 7* and Q are so <hnm thai As is positive. Then 1'R AJJ and A*<y A^/, and '/f As arc /'( tiro /%; chord /</ " 00 Aa chord PQ MOTION IN A CURVE 107 We shall assume without proof that the ratio of a small chord to its arc is very nearly equal to unity, and that the limit of "=1 as the point Q approaches the point P along the curve. At the same time the limit of RPQ = <f>. Hence, taking limits, we have dx d ^ = cos & ^ = sm ^ (1) ttrO l*O If the notation of differentials is used, equations (1) become dx = ds cos 0, dyds' sin $ ; whence, by squaring and adding, we obtain the important equation ds = dx + dy . (2) This relation between the differentials of #, y, and s is often rep- resented by the triangle of Fig. 46. This figure is convenient as a device for memorizing formulas (1) and (2), but it should be borne m mind that RQ is not rigorously y equal to dy ( 20), nor is PQ rigor- ously equal to ds. In fact, RQ = Ay, and PQ = As ; but if this triangle is regarded as a plane right triangle, we recall immediately the values of sin^>, cos<, and tan< which have been previously proved. 40. Motion ia a curve. When a o JL , , . .IT FIG. 46 body moves m a curve, the discus- sion of velocity and acceleration becomes somewhat complicated, as the directions as well as the magnitudes of these quantities need to be considered. We shall not discuss acceleration, but shall notice that the definition for the magnitude of the velocity, or the speed, is the same as before (namely, ds V=S 3I' dt where s is distance measured on the curved path) and that the direction of the velocity is that of the tangent to the curve. 108 ALGEBRAIC I UNCTIONS Moreover, as the body moves along a curved path through a distance PQ=&s (Fig. 47), x changes by an amount PII&K, and y changes by an amount RQ=&y. We have then Lim = -^ = <y = velocity of A at the body in its path, Ax dx , Lim = -7- = v x = component At dt of velocity parallel to OX, Lim ^ = ^ = v v = component At dt J r FIG 47 of velocity parallel to Y. Otherwise expressed, v represents the velocity of P, v r the velocity of the projection of P upon 0-3T, and v v the velocity of the projection of P on OY. Now, by (8), 36, and by 39, _ dx _ dx ds x dt ds dt = V COS <, (1) , dy dy ds and v. . ~ = -* y dt ds dt = v sm <p. (2) Squaring and adding, we have v*=v*+ V ;. (3) Formulas (1), (2), and (3) are of especial value when a par- ticle moves in the plane XO F, and the coordinates x and y of its position at any time t are each given as a function of t. The path of the moving particle may then be determined as follows : Assign any value to t and locate the point corresponding to the values of x and y thus determined. This will evidently be the position of the moving particle at that instant of time. In this way, by assigning successive values to t we can locate other points through which the particle is moving at the corre- sponding instants of time. The locus of the points thus deter- mined is a curve which, is evidently the path of the particle. MOTION IN A CURVE 109 The two equations accordingly represent the curve and are called its parametric representation, the variable t being 1 called a parameter.* By (9), 36, the slope of the curve is given by the formula j al dx d v x dt In. case t can be eliminated from the two given equations, the result is the (#, /) equation of the curve, sometimes called the Cartesian equation; but such ehmmation is not essential, and often is not desirable, particularly if the velocity of the particle in its path is to be determined. Ex. 1. A particle moves m the plane XOY so that at any time t, a; = a + bt, y = c + dt, where a, &,' c, and d are any real constants Determine its path and its velocity in its path To determine the path we eliminate t from the given equations, with the result d f .. y c = -(x d), the equation of a straight line passing through the point (a, c) with the i d slope - In this case the path may also be determined as follows Fiom the given equations we find dx l dt, and dy ddt\ whence = '- As the slope of / rA *' the path is always the same (that is, -I, the path must be a straight line which passes through the point (a, c) the point determined when t = 0. To determine the velocity of the particle in its path we find, by differ- entiating the given equations, dx , di/ . v *=jr l > v = dt = d > whence, by (3), v = V& 2 H- d\ Hence the particle moves along the straight line with a constant velocity. * It may be noted m passing that the parameter in the parametiic represen- tation of a curve is not necessarily time, but may be any third variable in terms of which a and y can be expressed. 110 ALGEBRAIC FUNCTIONS Ex. 2. If a projectile starts with an initial velocity v in an initial direc- tion which makes an angle a with the axis of x taken as horizontal, its position at any time t is given by the parametnc equations x v t cos a, y = v t sin a \ gt z . JTind its velocity in its path. dx We have v x = = v cos a, at dy v v = ~f t ~' Hence = Vi> 2 gv t sin a + g*P. EXERCISES 1. The coordinates of the position of a moving particle at any time t are given by the equations x = 2 1, y t s Determine the path. of the particle and its speed in its path 2. The coordinates of the position of a moving particle at any time t are given by the equations x t 2 , y = t + 1 Determine the path of the particle and its speed in its path. 3. The coordinates of the position of a moving particle at any time t are given by the equations x = 2 1, y == f t $ t*. Determine the path of the particle and its speed in its path. 4. At what point of its path will the particle of Ex. 3 be moving most slowly ? 5. The coordinates of the position of a moving particle at any time t are given by the equations x = ? 3, y t 9 + 2. Determine the path of the particle and its speed in its path 6. The coordinates of the position of a moving particle at any time t are given by the equations x = 4 &, y 4 (1 t) 2 . Determine the path of the particle and its speed in its path. 7. Find the highest point in the path of a projectile. 8. Find the point in its path at Trhich the speed of a projectile is a minimum. 9. Find the range (that is, the distance to the point at which the projectile will fall on OX), the velocity at that point, and the angle at which the projectile will meet OX. 10. Show that in general the same range may be produced by two different values of a, and find the value of which produces the greatest range. 11. Find the (a, y) equation of the path of a projectile, and plot. VELOCITIES AND RATES 111 41. Related velocities and rates. Another problem of some- what different type arises when we know the velocity of one point iii its path, which may be straight or curved, and wish to find the velocity of another point which is in some way con- nected vviUi tho first but, in general, describes a different path. The method, in general, is to form an equation connecting the distances traveled by the two points and then to differentiate tho equation thus formed with respect to the time t. The result is an equation connecting the velocities of the two points. Ex. 1 A lump is 00 ft. above tho ground. A stone is let drop from a point on tho same lovol us the lamp and 20 ft, away from. it. Find the speed of the stone's shadow on the ground at the end of 1 HOC., assuming that the distance traversed by a falling body in the time t is 1(5 < a . Lot A C (Fig, 48) be tho surface of tho ground which is assumed to be a homontal plane, L the position of the lamp, the point from which the stone was dropped, and S the position of the _ stone at any time t. Then Q is the posi- ^ I0 4g tion of the shadow of S on the ground, LSQ being a straight line. Let OS = x and BQ, = ?/. Then L = 20, BO = 60, and BS = 60 a?. In the similar triangles LOS and SBQ, whence y SB ~ 20. (2) 7 ^ 3 We know x =16 < a , whence ~ = 82 1 ; and wish to find ?, the velocity of Q. at at Difterontiating (2) with respect to t, we have dt x* dt When / s= 1 sec,, x =s 16, and ~ =s 82 ; whence, by substitution, we find at ~& ss 150 ft. per second. dt The result is negative because y is decreasing as time goes on. In 6 and 11, if the rate of one of two related quantities was known, we were able to find the rate of the other quantity. 112 ALGEBRAIC FUNCTIONS This type of problem may also be solved by the same method by which the problem of related velocities has been solved. We shall illustrate by taking the same problem that was used in 11. Ex. 2. Water is being poured at the rate of 100 cu. in. per second into a vessel in the shape of a right circulai cone of radius 3 in. and altitude 9 iu. Required the rate at which the depth of the water is increasing when the depth is 6 in. As in 11, we have V = ^ irh a ; dV 1 , z d1i whence -^ = * ^ We have given = 100, h = 6 ; from which we compute ^ = 25 =796 . dt v EXERCISES 1. A point is moving on the curve y 2 = sc s The velocity along OX is 2 ft. per second What is the velocity along OY when x = 2 ? 2. A ball is swung in a circle at the end of a cord 3 ft long so as to make 40 revolutions per minute. If the cord breaks, allowing the ball to fly off at a tangent, at what rate will it be receding from the center of its previous path 2 sec after the cord breaks, if no allowance is made for the action of any new force ? 3. The mside of a vessel is in the form of an inverted regular quadrangular pyramid, 4 ft square at the top and 2 ft deep. The vessel is originally filled with water which leaks out at the bottom at the rate of 10 cu. in. per minute. How fast is the level of the water falling when the water is 10 in. deep ? 4. The top of a ladder 20 ft long slides down the side of a ver- tical wall at a speed of 3 ft per second. The foot of the ladder slides on horizontal land. Find the path described by the middle point of the ladder, and its speed in its path. 5. A boat with the anchor fast on the bottom at a depth of 40 ft. is drifting at the rate of 3 mi per hour, the cable attached to the anchor slipping over the end of the boat At what rate is the cable leaving the boat when 50 ft of cable are out, assuming it forms a straight line from the boat to the anchor ? GENERAL EXERCISES 113 6. A solution is being pouied into a conical filter at the rate of 5 cc per second and is running out at the rate of 2 cc. per second. The radius of the top of the filter is 8 cm and the depth of the filter is 20 cm. Find the rate at which the level of the solution is rising in the filter when it is one third of the way to the top. 7 . A trough is in the form of a right prism with its ends isosceles triangles placed vertically. It is 5 ft long, 1 ft across the top, and 8 in. deep It contains water which leaks out at the rate of 1 qt. (57| cu. in ) per minute Find the rate at which, the level of the water is sinking in the trough when the depth is 3 in. 8. The angle between the straight lines AB and BC is 60, and AB is 40 ft. long. A particle at A begins to move along AB toward B at the rate of 5 ft per second, and at the same time a particle at B begins to move along BC toward C at the rate of 4 ft per second. At what rate are the two particles approaching each, other at the end of 1 sec * 9. The foot of a ladder 50ft. long rests on horizontal ground, and the top of the ladder rests against the side of a pyramid which makes an angle of 120 with the ground. If the foot of the ladder is drawn directly away from the base of the pyramid at the uniform rate of 2 ft. per second, how fast will the top of the ladder slide down the side of the pyramid ? GENERAL EXERCISES Plot the curves : 1. 3 a 2 + 7 f = 21 9. ? / 2 (4 + or 2 ) = cc 2 (4 - a 2 ). 2. 4?/ a == 9.r. z a, x * 10. ?/- = a; a - . 3. 9a; 2 y*=lG. a + x 4. 2/ a - 2 y = a; 8 + 2 a? - 1. n _ 8 a 8 12. y -rf+4a"' IS. 6. 7, (7/-a J ) 2 =9-a ! 2 > == ^T4* 8. (* + yf f(y + 2). 15. xY + 36 = 16 f Find the turning-points of the following curves and plot the curves 16. y = (2 + oO(4 - a-,) 2 . = (x - I) 3 17. y = (x + 3) 2 (a: - 2). ' J JB + 1 ' a 03 , - at x 114 ALGKEBBAIC FUNCTIONS 20. Find the equation of the tangent to the curve f = x z ~- / 3 a 6a\ the point ( -g-> ~6/ 21. Find the equation of the tangent to the curve x* + y* = * at the point (a;.,^, 3^). 22. Prove that if a tangent to a parabola ?/ 2 = 7o has the slope ?;?, its point of contact is (j- 5, ^-) and therefore its equation is A 4 VTE 7/t' J& tfvj a 2 JE ?/ 23. Prove that if a tangent to an ellipse + ~ = 1 has the slope m, ft U its point of contact is ( * m ,^ , & 2 ) and therefore ^ \ Vrt 2 ?^ 2 + 6 2 V arm 2 + 67 its equation is y = wcc Va 2 m 2 + i 2 24. Show that a tangent to a parabola makes equal angles with the axis and a line from the focus to the point of contact. 25. Show that a tangent to an ellipse makes equal angles with the two lines drawn to the foci from the point of contact Find the angles of intersection of the following pairs of curves . 26. , = ~ 28. a; 2 = 29. a: 2 -4y-4=0, a? + 12y - 36 = 0. 30. 2/ ! =o; 8 } 2/ ! =(2-a;) 8 . 31. 32. 33. The coordinates of a moving particle are given by the equa- tions x tf 3 , y = (1 tf 2 )^. Find its path and its velocity in its path. 34. A particle moves so that its coordinates at the time t are 2 x = 2t, y = . . Find its path and its velocity in its path. 35. A projectile so moves that x = at, y = bt ^g^. Find its path and its velocity in its path. GENERAL EXERCISES 115 36. A body so moves that x = 2 + tf$, y = 1 -f t. Find its path and its velocity in its path. 37. A particle is moving along the curve if 4. x; and when as = 4, its ordmate is increasing at the rate of 10 ft. per second. At what rate is the abscissa then changing, and how fast is the particle moving in the curve ? Where will the abscissa be changing ten times as fast as the ordmate ? 38. A particle describes the circle o; a +y a =a 2 with a constant speed v . Find the components of its velocity. 39. A particle describes the parabola y 2 = 4 ax in such a way that its a-component of velocity is equal to ct Find its y-component of velocity and its velocity in its path. 40. A particle moves so that x = 2 1, y = 2 "Vt & Show that it moves around a semicircle in the time from t = to t = 1, and find its velocity in its path during that time. 41. At 12 o'clock a vessel is sailing due north at the uniform rate of 20 mi. an hour. Another vessel, 40 mi. north of the first, is sailing at the uniform rate of 15 mi an hour on a course 30 north of east. At what rate is the distance between the two vessels diminishing at the end of one hour ? What is the shortest distance between the two vessels ? 42. The top of a ladder 32 ft. long rests against a vertical wall, and the foot is drawn along a horizontal plane at the rate of 4 ft. per second in a straight line from the wall. Find the path of a point on the ladder one third of the distance "from the foot of the ladder, and its velocity in its path. 43. A man standing on a wharf 20 ft. above the water pulls in a rope, attached to a boat, at the uniform rate of 3 ft. per second Find the velocity with which the boat approaches the wharf 44. The volume and the radius of a cylindrical boiler are expand- ing at the rate of .8 cu. ft. and .002 ft. per minute respectively. How fast is the length of the boiler changing when the boiler contains 40 cu. ft. and has a radius of 2 ft. ? 45. The inside of a cistern is in the form of a frustum of a regular quadrangular pyramid. The bottom is 40 ft. square, the top is 60 ft. square, and the depth is 10 ft, If the water leaks out at the bottom at the rate of 5 cu. ft. per minute, how fast is the level of the water falling when the water is 5 ft, deep in the cistern ? 116 ALGEBRAIC FUNCTIONS 46. The inside of a cistern is in the form of a frustum of a right circular cone of vertical angle 90. The cistern is smallest, at the base, which is 4 ft in diameter. Water is being poured in at the rate of 5 cu ft. per minute. How fast is the water rising in tho cistern when it is 2 ft deep ? 47. The inside of a bowl is in the form of a hemispherical sur- face of radius 10 in If watei is running out of it at the iat,o of 2 eu in. per minute, how fast is the depth of the water decreasing when the water is 3 in deep ? 48. How fast is the surface of the bowl in Ex 47 being exposed ? 49. The inside of a bowl 4 in deep and 8 in. across the top is in the form of a surface of revolution formed by revolving a parabolic segment about its axis Water is running into the bowl at the rate of 1 cu in per second How fast is the water rising in the bowl when it is 2 in deep ? 50. It is required to fence off a rectangular piece of ground to con- tain 200 sq ft , one side to be bounded by a wall already constructed. Find the dimensions which will require the least amount of fencing. 51. The hypotenuse of a right triangle is given. Find the other sides if the area is a maximum 52. The stiffness of a rectangular beam varies as the product of the breadth and the cube of the depth. Find the dimensions of the stiffest beam which can be cut from a circular cylindrical log of diameter 18 in. 53. A rectangular _plot of land to contain 384 sq. ft. is to be in- closed by a fence, and is to be divided into two equal lots by a fence parallel to one of the sides What must be the dimensions of tho rectangle that the least amount of fencing may be required ? 54. An open tank with a square base and vertical sides is to havo a capacity of 500 cu ft Fiud the dimensions so that the cost of lining it may be a minimum 55. A rectangular box with a square base and open at the top is to be made out of a given amount of material. If no allowance is made for thickness of material or for waste in construction, what are the dimensions of the largest box which can be made ? ^ 56. A metal vessel, open at the top, is to be cast in the form of a right circular cylinder. If rt is to hold 27 TT cu in , and the thickness of the side and that of the bottom are each to be 1 in, what will be the inside dimensions when the least amount of material is used ? GENERAL EXERCISES 117 57 . A gallon oil can (231 cu in ) is io be made in the form, of a right chcular cylinder. The material used for the top and the bottom costs twice as much per square inch as the material used for the side What is the radius of the most economical can that can be made if no allowance is made for thickness of material or waste in construction ? 58. A tent is to be constructed in the form of a regular quadran- gular pyiauucl Find the ratio of its height to a side of its babe when the air space inside the tent is as great as possible for a given wall surface 59. It is required to construct from two equal circular plates of radius a a buoy composed of two equal cones having a common base. Find the radius of the base when the volume is the greatest 60. Two towns, A and I>, are situated respectively 12 mi. and 18 mi. back from a straight river from which they are to get thoir water supply by means of the same pumping-station. At what point on the bank of the river should the station be placed so that the least amount of piping may be required, if the nearest points 011 the river from A and B respectively are 20 mi. apart and if the piping goes directly from the pumping-station to each of the towns 9 61. A man 011 one side of a river, the banks of which are assumed to be parallel straight lines mi apart, wishes to reach a point on the opposite side of the river and 5 mi. further along the bank If he can row 3 mi. an hour and travel on land 5 mi. an hour, find the route he should take to make the trip in the least time. 62. A power house stands upon one side of a river of width I miles, and a manufacturing plant stands upon the opposite side, a miles downstream. Find the most economical way to construct the con- necting cable if it costs m, dollars per mile on land and n dollars a mile through water, assuming the banks of the river to be parallel straight lines. 63. A vessel A is sailing due east at the uniform rate of 8 mi. per hour when she sights another vessel B directly ahead and 20 mi. away. B is sailing in a straight course S. 30 W at the uniform rate of 6 mi per hour. When will the two vessels be nearest to each other? 64. The number of tons of coal consumed per hour by a certain ship is 0.2 + 0.001 v* } where v is the speed in miles per hour. Find an expression for the amount of coal consumed on a voyage of 1000 mi. and the most economical speed at which to make the voyage. 118 ALGEBRAIC FUNCTIONS 65. The fuel consumed by a certain steamship in an hour is pro- portional to the cube of the velocity which would be given to the steamship in still water. If it is required to steam a certain distance against a current flowing a miles an hour, find the most economical speed. a 2 66. An isosceles triangle is inscribed in the ellipse ^ 4- ~~ = 1, Ct () (a > 6), with its vertex in the upper end of the minor axis o the ellipse and its base parallel to the major axis Determine the length of the base and the altitude of the triangle of greatest area whiuli can be so inscribed. CHAPTER V TRIGONOMETRIC FUNCTIONS 42. Circular measure. The circular measure of an angle is the quotient of the length of an arc of a circle, with its center at the vertex of the angle and included between its sides, divided by the radius of the arc. Thus, if 6 is the angle, a the length of the arc, and r the radius, we have -~ CD The unit of angle in this measurement is the radian, which is the angle for which a r in (1), and any angle may be said to contain a certain number of radians. But the quotient - in r formula (1) is an abstract number, and it is also customary to speak of the angle 6 as having the magnitude - without using the word radian. Thus, we speak of the angle 1, the angle -|, the angle > etc. In all work involving calculus, and in most theoretical work of any kind, all angles which occur are understood to be ex- pressed in radians. In fact, many of the calculus formulas would be false unless the angles involved were so expressed. The student should carefully note this fact, although the reason for it is not yet apparent. From this point of view such a trigonometric equation as y = sin x (2) may be considered as defining a functional relation between two quantities exactly as does the simpler equation y a; 3 . For we may, in (2), assign any arbitrary value to x and determine the corresponding value of y. This may be done by a direct 119 120 TRIGONOMETRIC FUNCTIONS computation (as will be shown in Chapter VII), or it may be done by means of a table of trigonometric functions, in which case we must interpret the value of x as denoting so many radians. One of the reasons for expressing an angle in circular measure is that it makes true the formula *T bill fv *4 xrtx Lim - = 1, (3) A-+0 fl where the left-hand member of the equation is to be read " the limit of ^-y as h approaches zero as JB a limit." To prove this theorem we proceed as follows: ->. Ti ' Let h be the angle A OB (Fig. 49), r the ^ % \ x j // radius of the arc AB described from as "" x ^j/ a center, a the length of AB, p the length B ' of the perpendicular BC from B to OA, and t the length of the tangent drawn from B to meet OA produced in D. Revolve the figure on OA as an axis until B takes the position B'. Then the chord BGB'=2p, the arc BAB' =2 a, and the tangent B f D=tliG tangent BD. Evidently BD + DB' > BAB' >BCB'; whence t>a>p. Dividing through by r, we have r r r that is, tan h > h > sin h. Dividing by sin A, we have cos h sin A ' or, by inverting, cos h < ^ ^ < 1. h GRAPHS 121 Now as h approaches zero, cos h approaches 1. Hence r ft which lies between cos h and 1, must also approach 1 ; that is, A - o n This result may be used to find the limit of k as It approaches zero as a limit For we have l c\ a fl n /I 'I/ 2 sin 3 - sin 2 - sm- 1 cos h _ 2 2 _ h 2 A ~ h I 2 7t 2 2 , . h sm- Now as h approaches zero as a limit, approaches unity, by (3). Therefore h r 1 cos h A . , . Lira =0. (4) A-O 7i V ' 43. Graphs of trigonometric functions. We may plot a trigo- nometric function by assigning values to x and computing, or taking from a table, the corresponding values of y. In so doing, any angle which may occur should be expressed in circular measure, as explained m the previous section. In this connec- tion it is to be remembered that TT is simply the number 3.1416, and that the angle w means an angle with that number of radians and is therefore the angle whose degree measure is 180. The manner of plotting can be best explained by examples. Ex. 1. y = a sin fa- it is convenient first to fix the values of x which make y equal to aero Now the sine is zero when the angle is 0, IT, 2 w, 3 IT, TT, 2 TT, on , iu general, kv, where k is any positive or negative integer. To make y = 0, therefore, we have to place bx = kv, whence 2w IT n TT 2ir 8w * > r- > U, ~) -- , ~j > '. b b b b b The sine takes its maximum value + 1 when the angle has the values if 5 TT OTT TT 5 TT TT e>'~2"' ~n~' eifcc< a * * s > * n *k is case > "when x = > , ~, etc. For these values of or, w = a 122 TRIGONOMETRIC FUNCTIONS O rr . The sine takes its minimum value 1 when the angle is '-, - . etc. , q 17 _ ii 2 that is, in this case, when x = ' - etc Foi these values of r, y = a. 20 2o These values of x for which the sine is 1 lie halfway between the values of x for which the sine is 7 FIG 50 These points on the graph are enough to determine its general shape Other values of x may be used to fix the shape more exactly The graph is shown in Fig 50, with a = 3 and I = 2 The curve may be said to repre- sent a wave. The distance from peak to peak, -, is the wave length, and the height a above OX is the amplitude Ex 2. y = a cos Ix. As in Ex 1, we fix first the points for which y = Now the cosine of an angle is zero when the angle is -> , , etc ; that is, any odd multiple of \> We have, therefore, y = when At 26' STT 26' Y 2b FIG. 61 Halfway between these points the cosine has its maximum value + 1 or its minimum value - 1 alternately, and y = a. The graph is shown in Fig 51, with a = 3and& = 2 GKRAPHS 128 Ex. 3. y = a sin (bx + c). We have y = when bx + c = 0, TT, 2 TT, 3 rr, c>tc.; that is, when 7 T 2ir ]?i&. 52 Halfway between these values of x tho HHIQ has its maximum value + 1 and its minimum value 1 alternately, and y db a. Tho tmvvo is tho same as in Ex. 1, but is shifted ~ units to the loft (Fig. 52). Ex, 4. y sin a; + 4 sin 2 ,x. The graph is found by adding the ordinatcs of tho two curves y = sin x and y ^ sin 2 x, as shown in Fig. 58. Y sin x+i sin 8*0 3?io. 68 EXERCISES Plot the graphs of the following equations : 1. y us 2 sin 3 a:. 6. y 2. 7/ = 3oos 7 - y 8, y 3. Ssinfaj ~V 9. 2/ ^x 10. y -j. 11. y tan 2 SB, SOOCB. 4. ^2008 vers jr. 6. y *a 3 sin (as 2), 12, gin 4. s i n 124 TKIGONOMETKIC FUNCTIONS 44. Differentiation of trigonometric functions. The formulas for the differentiation of trigonometric functions are as follows, where u represents any function of x which can be differentiated : d . du sinM = cosM - dx dx d du cosM = sinw > dx dx d . 2 du tan u = sec u -> dx dx A ct n W = -csc 2 w^, (4) dx dx d , du XCN sec u = sec u tan u ( o) dx dx d , du X /J N esc u = esc u ctn u (6) dx dx These formulas are proved as follows: 1. Let y sin w, where u is any function of a; which may be differentiated. Give x an increment A and let AM and Ay be the corresponding increments of u and y. Then Ly = sin (u -f- Aw) sin w = sin w cos AM + cos u sin A% sin u = cos u sin AM (1 cos Aw) sin u ; . AV sin AM 1 cos AM . whence - = cos M sm M. Au AM AM Now let Ar and therefore AM approach zero. By (3), 42, T sinAit ^ , , , A ^ , T . 1 cosAw A m . . Lim - = 1, and, by (4), 42, Lim A = 0. Therefore AM Aw <fo/ ~ = cos u. du But by (8), 36, ^ = K dx dudx and therefore -^ = cosw-^ aa: dx 3 2. To find cos w, wo write = sin u rrn d d / Then -7- cos u = -~ sm ?23 tfo /7T \ ?/ ) \2 / '7T = coal 3. To find --- tan 7/, wo write djc , sin ?/ tan?< = cos ?/, ,, . a Then - tan u = ~ dx cos u d , d cos u - sin ? sin u -~ cos ?* eos 4 tt VVH . cos 2 w ^ y f?. ;i 4. To find ~^~ cinu. wo write j? , COS 7^1 ctn u = -i sin u rn , d . d cosw I hen ctn ?/ = .- -: d.r, ax Bin u , d d . sin u - cos w cos u -=- fiin M 126 TRIGONOMETBIC FUNCTIONS 5. To find sec u, we write ax 1 , ^1 secw = - = (COSM) \ cos u Then sec w = - (cos )~* cos u (by (6), 36) sum <2 j 6. To find esc w, we write aa; esc u = - = (sin. u)~\ sin M 3 7 Then esc u = (sin w)~ 2 sin u (by (6), 36) - Ex. 1. y = tan 2 a; tan 3 a; = tan 2 x (tan a:) 8 . ^ = sec 2 2 x-j- (2 a) - 2 (tana) tana; dx dx^ ' v J dx = 2 seo 2 2 a; 2 tan a; sec 2 a:. Ex. 2. y = (2 sec*ar + 3 sec a a;) sin x. = sin a: ^8 sec 8 *-! (see a) + 6 sec x ~- (sec a) J + (2sec*a: + 3 sec 2 a:)-^(sina;) = sin x (8 sec*a; tan x + Q sec 2 a; tan a:) + (2 sec*a; + 3 sec s a:) cos x = (1 cos s a:) (8 sec 6 a; + 6 sec 8 a;) + (2 sec s a: + 3 sec a;) = 8 sec 6 a: 3 sec ar. EXERCISES nd in each of the following cases : y = 2 tan|- 6. y = & sin 8 5a! - sin5 . sw.*2x. 7. ^rs s a 5aj. 8 . . SIMPLE HARMONIC MOTION 127 9. 2/ = cos 8 -2cos. 11. y 2 . , o * x ,o sec a* + ton as 10. y - ctn- + 2 ctn -. 12. y = 13. y = sin (2 a? + 1) cos (2 a; 1). 14. y = tan 8 3 a; 3 tan 3 x + 9 ie. 15. y = see 2 a: tan 2 x. 16. y = ^ (3 cos 6 2 8 5 cos B 2 os). 17. sin 2 a + tan 3 ?/ = 0. 18. asy + ctn xij = 0. 45. Simple harmonic motion. Let a particle of mass m move in a straight line so that its distance s measured from a fixed point in the line is given at any time t by the equation s s=a G sin Si, (1) where c and b are constants. We have for the velocity v and the acceleration a v ^ cbcos ^ (2 ) a sss c6 2 sin Ztf. (3) When 0, 8 a and the particle is at (Fig. 54), When t = -j^r* s =s c and the particle is at A t where OA** o, 2 o When t is between and v is positive and a is negative, >a y so that the particle is moving from to A with decreasing speed. When t is between and T v is i A 25 o , , ,. ,-, , ]?i<*. 64 negative and a is negative, so that the particle moves toward with increasing speed. When 71" ==--, the particle is at 0. 8 As * varies from ? to -^?? the particle moves with decreasing b 26 speed from to J?, where OJ? =* o. Finally, as varies from |^ to -, the particle moves back 2o o from J5 to (9 with increasing speed. 128 TRIGONOMETRIC FUNCTIONS The motion is then repeated, and the particle oscillates between B and A, the time required for a complete oscillation being, as 27T we have seen, =- The motion of the particle is called simple harmonic motion. The quantity c is called the amplitude, and tho 2 7T interval -= after which the motion repeats itself, is called the period. Since force is proportional to the mass times the acceleration, the force F acting on the particle is given by the formula F kma = kmcb s sm bt= Jcmb*s. This shows that the force is proportional to the distance s from the point 0. The negative sign shows that the force pro- duces acceleration with a sign opposite to that of s, and there- fore slows up the particle when it is moving away from arid increases its speed when it moves toward 0. The force is there- fore always directed toward and is an attracting force. If, instead of equation (1), we write the equation s = c8mb(t-t Q ), - (4) the change amounts simply to altering the instant from which the time is measured. For the value of s which corresponds to t t in (1) corresponds to t = t l +t in (4). Hence (4) represents simple harmonic motion of amplitude c and period But (4) may be written * s = G cos fa sin bt o sin bt cos bt, which is the same as s = A sin bt + B cos bt, where A = c cos bt , B = c sin bt . A and B may have any values in (5), for if A and B are given, we have, from the last two equations, c=V^ a +5 2 , tanfa = --, which determines G and t in (4). SIMPLE HARMONIC MOTION 129 Therefore equation (5) also represents simple harmonic motion 2 7T with amplitude VA z +jB a and period -7 In particular, if in (5) A = and 33 = c, we have 8 = G COS It. (6) If in (4) we place t = - ' , it becomes *j & * =<! cos fi (-*), (7) which differs from (6) only in the instant from which the time is measured. EXERCISES 1. A particle moves with constant speed v around a circle. Prove that its projection on any diameter of the circle describes simple harmonic motion 2. A point moves with simple harmonic motion of period 4 sec. and amplitude 3 ft Find the equation of its motion. 3. Given the equation s = 5 sin 2 1 Find the tune of a complete oscillation and the amplitude of the swing. 4. Find at what time and place the speed is the greatest for the motion defined by the equation s = G sin It Do the same for the acceleration. 5. At what point in a simple harmonic motion is the velocity zero, and at what point is the acceleration zero ? 6. The motion of a particle in a straight line is expressed by the equation s = 5 2 cos 3 *. Express the velocity and the acceleration in terms of s and show that the motion is simple harmonic. 7. A particle moving with a simple harmonic motion of amplitude 5 ft has a velocity of 8 ft. per second when at a distance oj! 3 ft. from its mean position. Find its period. 8. A particle moving with simple harmonic motion has a velocity of 6 ft. per second when at a distance of 8 ft. from its mean position, and a velocity of 8 ft. per second when at a distance oC 6 ft. from its mean position. Find its amplitude and its period. 9. A point moves with simple harmonic motion given by the equation *= # sin et. Describe its motion, 130 TRIGONOMETRIC FUNCTIONS 46. Graphs of inverse trigonometric functions. The equation x = sin y (1) defines a relation between the quantities x and y which may be stated by saying either that x 'is the sine of the angle y or that the angle y has the sine x. When we wish to use the latter form of expressing the relation, we write in place of equation (1) the equation y = sm -^, (2) where 1 is not to be understood as a negative exponent but as part of a new symbol sin" 1 . To avoid the possible ambiguity formula (2) is sometimes written y = arc sin x. Equations (1) and (2) have exactly the same meaning, and the student should accustom himself to pass from one to the other without difficulty. In equation (1) y is considered the independent variable, while in (2) x is the independent variable. Equation (2) then defines a function of x which is called the anti-sine of x or the inverse sine of x. It will add to the clearness of the student's thinking, however, if he will read equation (2) as " y is the angle whose sine is x." Similarly, if a = cos#, then y = GOS~' L x; if x ioxiy, then y tan" 1 a; ; and so on for the other trigonometric functions. We get in this way the whole class of inverse trigonometric functions. It is to be noticed that, from equation (2), y is not completely determined when x is given, since there is an infinite number of angles with the same sine. For example, if # = - #=> > STT 18 IT , _.. . t 2 . & -~> ft > etc. 1ms causes a certain amount of ambiguity in using inverse trigonometric functions, but the ambiguity is re- moved if the quadrant is known in which the angle y lies. We have the same sort of ambiguity when we pass from the equa- tion x = y* to the equation y = Va;, for if x is given, there are two values of y. To obtain the graph of the function expressed in (2) wo may change (2) into the equivalent form (1) and proceed as GRAPHS 131 in 43. In this way it is evident that the graphs of the inverse trigonometric functions are the same as those of the direct func- tions but differently placed with reference to the coordinate axes. It is to be noticed particularly that to any value of x corresponds an infinite number of values of y. X FIG. 55 FIG 56 Ex. 1. y = sin- 1 * From this, x = sin y, and we may plot the graph by assuming values of y and computing those of a (Fig. 55). Ex. 2. y = tan- 1 *. Then x == tan y, and the graph is as in Fig. 56. EXERCISES Plot the graphs of the following equations : 1. y tan- 1 2 x. 3. y = sin" 1 (a: 1). 5. y =1+ cos" 1 :*;. 2. y =3 ctn-^cc. 4. y = tan- 1 (* + 1). 6. y = ^tan" 1 ^ 7. ?/ = cos-^jc 2). 8. y sin-^S x 4. 1) - - 47. Differentiation of inverse trigonometric functions. The formulas for the differentiation of the inverse trigonometric functions are as follows: 1. -7- sin"" 1 ^ = . -r- when sin"" 1 ^ is in the first or the ax vl w a <*x ,. , , , , x u lourth quadrant ; i diL i when sin-^f is in the second or A/ "I M_ QI^ QiX ,t i t i x w the third quadrant. 132 TKIGONOMETKIC FUNCTIONS 2. - cos" 1 ^ = , -7- when cos" 1 ** is m the first or the ^ x vlw second quadrant; i (Jtjf = when cos" 1 ^ is m the third or the VI - u 2 dx f ourth quadrant d L. 1 du O ~z dx 1 + w z ax A d . _j 1 du 4. ctn u= 3 dx 1 + 5. sec" 1 ?* = when sec" 1 ^ is in the first or the dx u^/u 2 -l dx = when sec" 1 u is in the second or uvu I ^ ie f our tli quadrant. fj J. el/Mi 6 CSC" I M= . -7- when csc" 1 ^ is in the first or x uvu 1 |.j ie third quadrant; = , -r- when csc"" 1 ^ is in the second or The proofs of these formulas are as follows: 1. If y = sin~X then sin y = u. TT i P A A dy du Hence, by 44, co S2 ,^ = -; cfc 1 du , whence -- -- dx cos y dx But cos y = Vl u 2 when # is m the first or the fourth quad- rant, and cosy = Vl u 2 when y is m the second or the third quadrant. 2. If y = cos" 1 ^, then cos y u- Hence _ sin ^ = ^ ; 9 dx dx whence * eta sin # DIFFERENTIATION 183 But sin y Vl M a when y is in the first or the second quad- rant, and am y = Vl u 2 when y is in the third or the fourth quadrant. 3. It' y tan' 1 it, then tan y = u. it 2 d?t d u 1 1 ence sec u ~^~ ; /yif>- /yVp dv*t Lt*v , dy 1 du whence -~ = dx 1 + u dx 4. If yssctir 1 !/, then ctn ?/ = w. V T 41 (iV (/ ( V UV , (??/ 1 du whence ~ = = ; -7- a 1 + w" are 5. If yssseo'X then sec ?/ = M. 1 1 ence sec y tan ?y -~ = -r- ; ti*y* ft'V . d'i/ 1 <??* whence - = , -=-* djc sec ;y tan y dx But Rcoyss?/, and tan,y = V?^ a --l when ?/ is in the first or the third quadrant, and tan ?/:=-~-vV 1 when y is in the second or the fourth quadrant. M T$ /j/ rtiart '""'J'^f then esc y = ?/,. _ r . dy du Hence esc y otn y j~ =s y ; rvM/ \4iJu whence -f- = , -r- ?a; esc y ofo&y dx 134 TRIGONOMETRIC FUNCTIONS But cscy = w, and ctn^=Vw 2 1 when y is in the first or the third quadrant, and ctn# = Vw 2 1 when y is in the second or the fourth quadrant. If the quadrant in which an angle lies is not material in a problem, it will be assumed to be in the first quadrant. This applies particularly to formal exercises m differentiation. Ex. 1. y = sin" 1 Vl x z , where y is an acute angle dv 1 . .fLn. a>\| ^ L-(l-a 2 ) dx dx Vl - This result may also be obtained by placing sin- 1 Vl x z = cos" 1 a:. Ex. 2. y = sec" 1 V4a: a + 4ar + 2. dy dx dx V4 x* + 4 x + 2 V(4 a,- 2 + 4 x + 2) - 1 ~ ~2a; 2 + 2a,-- EXERCISES Find -7^ in each of the following cases : 2. y = sin" 1 -. _ -ift 3 . y _ sin g i 3a; 4. y = cos -1 11. y = COS~ 2 5. y = tan~ 1 6. y = tan" 1 Vcc 2 2 x. 7. y = ctn- 1 ^- 8. y = sec" 1 5 a. 9. y = csc~ 1 2aj. , . a; + 6 10, y-tan-> 13. y = tan" 1 Va 3 - 1 + 16. y = 2\a a;/ 18. y ss Vl aj 9 +a5003~ 1 Vl w a ANGULAR VELOCITY 135 48. Angular velocity. If a line OP (Fig. 57) is revolving in a plane about one of its ends 0, and in a time t the line OP has moved from an initial position OM to the position OP, the angle MOP = 6 denotes the amount of rotation. The rate of change of with respect to t is called the angular velocity of OP. The angular velocity is commonly denoted by the Greek letter < ; so we have the formula JQ In accordance with 42 the angle 8 is taken in radians ; so that if t is in seconds, the angular velocity is in radians per second. By dividing by 2-Tr, the angular velocity may be reduced to revolutions per second, since one revolution is equivalent to TT radians. A point Q on the line OP at a distance r from describes a circle of radius r which intersects OM at A. If s is the length of the arc of the circle A Q measured from A, then, by 42, s = rO. (2) ds Now -r- is called the linear velocity of the point $, since it dt measures the rate at which s is described ; and from (2) and (1), ds d0 showing that the farther the point Q is from. the greater is its linear velocity. Similarly, the angular acceleration, which is denoted by oc, is denned by the relation , , 2 ,, J aft> a o This is connected with the linear acceleration -=-5 by the formula 136 TRIGONOMETRIC FUNCTIONS Ex 1. If a wheel revolves so that the angular velocity is given by the formula u> = 8 1, how many revolutions will it make in the time from t = 2 to t = 5 ? We take a spoke of the wheel as the line OP Then we have dQ = 8 tdt Hence the angle through which the wheel revolves in the given time is B = C 6 B tdt = [4 fif =100 - 16 = 84. 1/2 The result is in radians. It may be reduced to revolutions by dividing by 2 ir. The answer is 13.4 revolutions Ex. 2. A particle traverses a circle at a uniform rate of n revolutions a second Determine the motion of the projection of the particle on a diameter of the circle Let P (Fig 58) be the particle, OX the diameter of the circle, and M the projection of P on OX Then x = a cos 6, where a is the radius of the circle. By hypothesis the angular velocity of OP is 2 nir radians per second Therefore , o> = = 2 mr ; dt whence FIG. 68 If we consider that when t = 0, the particle is on OX, then (7 = 0. Therefore /, x = a cos 8 = a cos 2 rart = a cos <at. The point M therefore describes a simple harmonic motion In fact, simple harmonic motion is often defined in this way EXERCISES 1. A flywheel 4ft in diameter makes 3 revolutions a second. Find the components of velocity in feet per second of a point on the rim when it is 6 m above the level of the center of the wheel. 2. A point on the rim of a flywheel of radius 5 ft which is 3 ft. above the level of the center of the wheel has a horizontal component of velocity of 100 ft per second Find the number of revolutions per second of the wheel. CYCLOID 1 3. If the horizontal and vertical projections of a point descr simple harmonic motions given by the equations x = 5 cos 3t, y = 5 sin 3 1, show that the point moves in a circle and find its angular velocity 49. The cycloid. If a wheel rolls upon a straight line, ea point of the rim describes a curve called a cycloid. Let a wheel of radius a roll upon the axis of #, and let (Fig. 59) be its center at any time of its motion, JV its point N FIG. 59 contact with 6L3T, and P the point which describes the cycloi Take as the origin of coordinates, 0, the point found by rollu the wheel to the left until P meets OX. Then ON= arc PN. Draw MP and CN, each perpendicular to OX, PR parallel OX, and connect C and P. Let angle NGP <j>. Then x = OM = 0-ZV- JfJV = a<j> a sin <. a a cos $ Hence the parametric representation ( 40) of the cycloid a&=a(<f> sin<), # = (! -cos<). 138 TRIGONOMETRIC FUNCTIONS If the wheel revolves with a constant angular velocity to = -^ we have, by 40, i) = a, C\. cos <f>") -*- = Q>* (1 cos <p), ?; = a sm <j> -If- = am sw. <f> ', at whence v a = a 2 co 2 (2-2 cos <) = 4 a 2 6) 2 sin 2 1 v = 2 aco sm ^> as an expression for the velocity in its path of a point on the run of the wheel. EXERCISES A 1. Prove that the slope of the cycloid at any point is ctn ^- 2. Show that the straight line drawn from any point on the rim of a rolling wheel perpendicular to the cycloid which that point is describing goes through the lowest point of the rolling wheel. 3. Show that any point on the run of the wheel has a horizontal component of velocity which is proportional to the vertical height of the point 4. Show that the highest point of the rolling wheel moves twice as fast as either of the two points whose distance from the ground as half the radius of the wheel 5. Show that the vertical component of velocity is a maximum when the point which describes the cycloid is on the level of the center of the rolling wheel. 6. Show that a point on the spoke of a rolling wheel at a distance b from the center describes a curve given by the equations x = a<f> b sin <, y = a b cos <j>, and find the velocity of the point in its path. The curve is called a trochoid. 7. Find the slope of the trochoid and find the point at which the curve is steepest. 8. Show that when a point on a spoke of a wheel describes a trochoid, the average of the velocities of the point when in its highest and lowest positions is equal to the linear velocity of the wheel CUJftVATUKE 139 50. Curvature. If a point describes a curve, the change of direction of its motion may be measured by the change of the angle < ( 15). For example, in the curve of Fig. 60, if AJ%= s and ./J^ = As, and if fa and fa are the values of <j& for the points P l and P z respectively, then </> 2 ^ is the total change of direction of the curve between J? and P y If Y fafa = A<, expressed in circular measure, the ratio is the average change As of direction per linear unit of the arc PJ\. Regarding as a function of s, and taking the limit of as As As approaches zero as a limit, we have -f-> which is called the curvature of the curve at as the point P. Hence the curvature of a curve is the rate of change of the direction of the curve with respect to the length of the arc. If is constant, the curvature is constant or uniform ; other- ds wise the curvature is variable. Applying this definition to the circle of Fig. 61, of which the Y center is C and the radius is a, FIG CO we have A< As = a Arf>. di -- P^CP^, and hence Therefore - = - -j As a Hence - and the circle is as a a curve of constant curvature equal to the reciprocal of its radius. The reciprocal of the curva- ture is called the radius of cur- vature and will be denoted by p. Through every point of a curve we may pass a circle with its radius equal to p, which shall have the same tangent as the curve at the point and shall lie on the FIG. 61 140 TRIGONOMETRIC FUNCTIONS same side of the tangent. Since the curvature of a circle is uniform and equal to the reciprocal of its radius, the curvatures of the curve and of the circle are the same, and the circle shows the curvature of the curve in a manner similar to that in which the tangent shows the direction of the curve. The circle is called the circle of curvature. From the definition of curvature it follows that _ ds If the equation of the curve is in rectangular coordinates, ds by (9), 36, p = ~ dx To transform this expression further, we note that ,2 , 2 . ,2 whence, dividing by dx* and taking the square root, we have Since <j> = tan- 1 , (by 15) dx 3 < dx ^ . fdy\ \dx, Substituting, we have p = ^ 2 x d y dx* In the above expression "for p there is an apparent ambiguity of sign, on account of the radical sign. If only the numerical value of p is required, a negative sign may be disregarded. CUBVATUEE 141 3,2 y2 Ex. 1. Find the radius of curvature of the ellipse + *j = 1. dy & a a: Here -f = - -=- da; a a y Therefore P Ex. 2. Find the radius of curvature of the cycloid ( 49). We have -^ = a(l cos<) = 2asin 2 2, a<p <s d?/ . n A i ^ = a sin /> =s 2 a sm J cos ^ d<p it & Therefore, by (9), 36, f^ = ctn^. dx 2 and EXERCISES 1. Find the radius of ourvature of the curve y* -fa a*. 222 2. Find the radius of curvature of the curve x* + y * = 3. Find the radius of curvature of the curve y taa-^a; 1) at the point for which x 2 / 7T\ 3 4. Show that the circle ( x -r- ) + y 2 = 1 is tangent to the curve y = sin a; at the point for which x = -jr, and has the same radius of curvature at that point. B. Find the radius of curvature of the curve x = cos t, y = cos 2 1, at the point for which t = 0, 6. Find the radius of curvature of the curve x = a cos < -{- a< sin A, y =s a sin <]!> a<j! cos <. 7. Prove that the radius of curvature of the curve as = o.cos 8 <, lias its greatest value when <j> = TT 142 TRIGONOMETRIC FUNCTIONS 51. Polar coordinates. So far we have determined the posi- tion of a point in the plane by two distances, x and y. We may, however, use a distance and a direction, as follows : Let (Fig. 62), called the origin, or pole, be a fixed point, and let OM, called the initial line, be a fixed line. Take P any point in the plane, and draw OP. Denote OP by r, and the angle M OP by 6 Then r and 6 are called the polar coor- dinates of the point P(r, 0), and when given will completely determine P. For example, the point (2, 15) is plotted by laying off the angle MOP =15 and meas- uring OP= 2. OP, or r, is called the radius vector, and 6 the veotorial angle, of P. These quantities may be either positive or negative. A negative value of 6 is laid off in the direction of the motion of the hands of a clock, a positive angle m the opposite direction. After the angle 6 has been constructed, positive values of r are measured from along the terminal line of 0, and negative values of r from along the backward extension of the terminal line. It follows that the same point may have more than one pair of coor- dinates. Thus (2, 195), (2, -165), (-2, 15), and (- 2, - 345) refer to the same point. In practice it is usu- ally convenient to restrict to positive values. Plotting in polar coordinates is facili- tated by using paper ruled as m Figs. 64 and 65. The angle 6 is determined from " ElG 63 the numbers at the ends of the straight hnes, and the value of r is counted off on the concentric circles, either toward or away from the number which indicates 6, according as r is positive or negative. The relation between (r, 0) and (x, y) is found as follows : Let the pole and the initial line OM of a system of polar coordinates be at the same time the origin and the axis of # of a system of rectangular coordinates. Let P (Fig, 63) be any point POLAR COORDINATES 143 of the plane, (a;, #) its rectangular coordinates, and (r, 0) its polar coordinates. Then, by the definition of the trigonometric functions, XJ X cos 6 = - sm = 2-. Whence follows, on the one hand, x = r cos y = r sin i and, on the other hand, sin0 = cos = (1) (2) By means of (1) a transformation can be made from rectangular to polar coordinates, and by means of (2) from polai to rectangular coordinates When an equation is given in polar coordinates, the corre- sponding curve may be plotted by giving to 6 convenient values, computing the corre- sponding values of r, plotting the resulting points, and drawing a curve through them. Ex. 1. r a cos $ a is a constant which may be given any con- venient value We may then find from a table of natural cosines the value of r which corresponds to any value of 9. By plotting the points cor- J * xesponding to values of 6 FIG 04 from to 90, we obtain the arc ABCO (Fig. 64). Values of 6 from 90 to 180 give the arc ODEA. Values of 6 from 180 to 270 give again the arc ABCO, and those fiom 270 to 860 give again the arc ODEA. Values of 6 greater than 360 can clearly give no points not already found. The curve is a circle. aw MS 144 TRIGONOMETRIC FUNCTIONS Ex. 2. r= a sin 35. As 6 increases from to 30, r increases from to a ; as 9 increases from 30 to 60, r decreases from a to , the point (r, 0) traces out the loop 040 (Fig. 65), which is evidently symmetrical with respect to the radius OA. As 6 increases from 60 to 90, r is negative and decreases from to a ; as increases from 90 to 120, r increases from isoj a to ; the point (r, 0) traces out the loop OBO As & increases from 120 to 180, the point (r, 6} traces out the loop OCO. Larger values of & give points already found, since sin 3 (180 + 5) = - sm 3 6. The three loops are congiu- ent, because sm,3 (60 + 0) = sin30 This curve is called a rose of three leaves. r = i a V2 cos 2 0. Ex. 3. r 2 = Solving for r, we have Hence, corresponding to any values of 9 which make cos 2 9 positive, there will be two values of r numerically equal and opposite in sign, and two corresponding points of the curve symmetrically situated with respect to the pole If values are assigned to 9 which make cos 2 9 negative, the corresponding values of r will be imaginary and there will be no points on the curve. Accordingly, as 9 increases | ^ | $f from to 45, r decreases numer- ically from aV2 to 0, and the portions of the curve in the first and the third quadrant are con- structed (Fig 66) ; as 9 increases from 45 to 135, cos 2 9 is negative, and there is no portion of the curve between the lines 9 = 45 and 9 = 135 , finally, as 9 increases from 135 to 180, r increases numerically from to aV2, and the portions of the curve in the second and the fourth quadiant are constructed The curve is now complete, as we should only repeat the curve already found if we assigned further values to 0, it is called the lemniscate. FIG 06 GRAPHS 145 Ex. 4. The spiral of Archimedes, r = a8. In plotting, 6 is usually considered in cncular measuie When &= 0, r = , and as 6 mci eases, ? inci eases, so that the cuive winds an infinite uiuubei of times around the ongin while leced- ing from it (Fig 67) In the flgme the heavy line represents the poitiou of the spiral coriespondmg to positive values of 0, and the dotted line the portion coriespondmg to negative values of 6 EXERCISES Plot the graphs of the following curves a 1. r = a sm 6 9. r = a sin 8 - 2. r = asm 20 10. ?' 2 = a 2 sin 3. r = a cos 30 11. ?* = 4. r = a sin-- 12. ,, = a (l _ cos 2 0). 13. r = a(l+2cos20) 5 - r=acos 2 ' 14. r =atan0 6. 7- = 3 cos + 5 15. r = a tan 2 7. ?' = 3 cos + 3.* 1 16 r = t 8. r = 3 cos + 2 ' 1 + cos Find the points of intersection of the following pairs of curves : 17. r = 2 sm 0, ? = 2-x/S cos 0. 18. r 2 = a 2 cos 0, ;- 2 = a 2 sin 2 19. r = 1 + sin 0, r = 2 sin 20. r 2 =a 2 sin0, ' 2 =a 2 sm30 Transform the following equations to polar coordinates . 21 . .r?/ = 4. 23. x z + y*-2ay = 0. 22 . a; 2 + f - 4 </a? - 4 ay = 24. (a* + 2/ 2 ) 2 = a a ( a - Z/ 2 ) Transform the following equations to rectangular coordinates : 25. r= ft sec 6 27. r = atan0. 26. r = 2 n cos 28. r = a cos 2 * The curve is called a cardioid t The curve is a parabola with the ongui at the focus. 146 TRIGONOMETRIC FUNCTIONS 52. The differentials <?r, d9, ds, in polar coordinates. We have seen, in 39, that the differential of arc in rectangular coordinates is given by the equation (1) If we wish to change this to polar coordinates, we have to place x = r cos 0, y r sin 6 ; whence dx = cos 6dr r sin 6 d6, dy = sin Qdr + r cos 6d6. Substituting in (1), we have ds* = dr z + r*d6 z . (2) This formula may be remembered by means of an " elemen- tary triangle " (Fig. 68), constructed as follows : Let P be a point on a curve r =/(0), the coordinates of P being (r, 0), where OP = r and MOP = 6. Let Q be increased by an amount d0 t thus determining another point Q on the curve. From as a center and with a radius equal to r, describe an arc of a circle intersecting OQ in R so that O.E = OP = r. Then, by 42, PR, = rdd. Now EQ is equal to Ar, and PQ is equal to As. -^ 6g We shall mark them, however, as dr and da respectively, and the formula (2) is then correctly obtained by treating the triangle PQJR as a right triangle with straight-line sides. The fact is that the smaller the triangle becomes as Q approaches P, the more nearly does it behave as a straight-line triangle ; and in the limit, formula (2) is exactly true. Other formulas may be read out of the triangle PQR. Let us denote by i/r the angle PQJR, which is the angle made by the curve with any radius vector. Then, if we treat the triangle PQR as a straight-line right-angle triangle, we have the formulas : . (3) dr k J DIFFERENTIALS 147 The above is not a proof of the formulas. To supply the proof we need to go through a limit process, as follows: We connect the points P and Q by a straight line (Fig. 69) and draw a straight line from P per- pendicular to OQ meeting OQ at S. Then the triangle PQS is a straight- line right-angle triangle, and therefore , chord PQ SP arcPg arc PQ ' chord PQ' FIG. 69 Now angle POQ = A0, arc PQ = As, and, from the right triangle OSP, SP = OP sin POQ = r sm A0. Therefore rsuiA * arcP r Ag chord P<2 A0 As chord PQ Now let A0 approach zero as a limit, so that Q approaches P along the curve. The angle SQP approaches the angle OPT, where PT is the tangent at P. At the same time ap- A/9 rlfi proaches 1, by 42 ; - approaches by definition ; and T>/~) ^ aic '' approaches 1, by 39. In this figure we denote the CllOlXl JL ty angle OPT by >/r and have, from (4), . . dB ,CN smf = r , (5) which is the first of formulas (3). It is true that in Fig. 69 we have denoted OPT by -^ and that in Fig. 68 ^ denotes OQP. But if we remember that the angle OQP approaches OPT as a limit when Q approaches P, and that in using Fig. 68 to read off the formulas (3) we are really anticipating this limit process, the difference appears unessential. The other formulas (3) may be obtained by a limit process similar to the one just used, or they may be obtained more 148 TEIGONOMETEIC FUNCTIONS quickly by combining (5) and (2). For, from (2) and (5), we have whence cos -\lr = . (6) ds By dividing (5) by (6) we have rdd /7N CO dr In using (7) it may be convenient to write it in the form tan i/r = , (8) dr d0 since the equation of tho curve is usually given in the form t?rt r =/(#), and - is found by direct differentiation. cLu Ex. Find the angle which the cmve 1 = a sin 4 makes with the radius vector 6 = 80 Here ^ = 4 a cos 4 0. Therefore, fiom (8), tan ^ = " sin 4 ^ = i tan 'I 6 do ^ T 4 cos 4 6 4 Substituting 6 = 30, we have tan ^ = * tan 120 = - | V3 = - 4 MO Therefore j/r = 156 35'. EXERCISES 1. Find the angle -which the curve r = a cos 3 makes with the radius vector 6 = 45 2. Find the angle which the curve r = 2 + 3 cos makes with the radius vector 6 = 90 A 3. Find the angle which the curve r = a 2 sin a makes with the initial line. 9 6 4. Show that for the curve r = a, sin 8 5 j i/r = - o 3 5. Show that the angle between the cardioid r = a(l cos 0) and any radius vector is always half the angle between the radius vector and the initial line. GENEHAL EXERCISES 149 6. Show that the angle between the lemniscate r 2 = 2 . 2 cos 2 9 TT and any ladms vector is always plus twice the angle between the radius vector and the initial line 7. Show that the curves r a = a a sin 2 6 and ?- 2 = a 2 cos 2 6 inter- sect at right angles GENERAL EXERCISES Find the graphs of the following equations : a 1 4- 1 * o / L ^ 1. ?/ = 4sm - 5. y = Ssmlas + TT ) 2. y = cos (2 x 3) 6. vy a = tan x. 3 . y = tan- 7- y = 2cos2(o! - 2) 4. ys= Jsin2aj + ^8in8a5. 8. ?/ = 3cos3h + ^ Find -f^ in the following cases : CtttXj 9. y = 2x 10. y = ,J tan (3 K + 2) + J tan 8 (3 x + 2). 12. tan (x + ?/) + tan (x y) = 0. 13. y=3ctn fi | + 5ctn 8 |. 21. y = 14. ?/ = csc a 4 x + 2 ctn 4 x 22. ?/ = 16. ?/ = sm a 4 x cos*2 a; 23> V cos ~ a .a_|_ 4 > - cos 8 - 2 cos ~ 24. v = ctn" 1 Vsc 2 2x. 17. ?/ = 2 18. ?y=atan 8 2a! ^tan2C+03. 25. 2/ = csc" 1 1 t SC "y" X 19. y = sin 20. ?/ = cos" 1 3 28. A particle moves in a straight line so that s = 6 Show that the motion is simple harmonits and find the center about which, the particle oscillates and the- amplitude of the motion. 150 TRIGONOMETRIC FUNCTIONS 20 ?/ 29. A particle moves on the ellipse i + ^ = 1 so that its projeq- tion upon OX describes simple harmonic motion given by x = a cos kt. Show that its projection upon OF also describes simple harmonic motion and find the velocity of the particle in its path. 7T 30. A particle moving with simple harmonic motion of period -r has a velocity of 9 ft per second when at a distance of 2 ft. from its mean position. Find the amplitude of the motion. 31. A particle moves according to the equation s = 4 sin \t + 5 cos t. Show that the motion is simple harmonic and find the amplitude of the swing and the time at which the particle passes through its mean position. 32. Find the radius of curvature of the curve y = x sin- at the 2 x point for which x = 7T 33. Find the radius of curvature of the curve ?/ = at the 7!! point for which x = TT 34. Find the radius of curvature of the curve y a sin' 1 ' V, 2 .r* CL at the point for which x = r 35. Find the radius of curvature of the curve x => a cos <, y = I sin <{>, IT at the point for which <j> = -r- Plot the graphs of the following curves . a 36. r ! =a 2 sm|- 41. r =1-20. 42. r a = 37. r a = a 2 sin40 .. , 43. r*= 38. r =a(l-sin0). . v ' 44. r =l + sm s . 39. r =a(l+cos20). ^ 30 40. r = a(l + 2 sin 0). 45. r =* 1 + sin -7p Find the points of intersection of the following pairs of curves ; 46. r*=3cos20, ?- s = 2cos a 0. 47. r BS a cos 0, r 2 BB a 2 sin 2 0. 48. r 2 sin 0, 7-' B* 4 sin 2 0. 49. r ~a(l + sin 20), r 2 =? 4 a'siii 2 0. GENERAL EXERCISES 151 Transform the following curves to polar coordinates : 50. = " Transform the following curves to ^-coordinates 52 ?' 2 =2a a sm20. 53. r = a(l cos0) 54. Find the angle at which the curve r = 3 + sin 2 6 meets the circle r 3 55. Find the angle of intersection of the two curves r = 2 sin 6 and r 2 = 4 sin 2 56. Find the angle of intersection of the curves r = a cos and r = a sin 2 B. 57. If a ball is fired from a gun with the initial velocity v , it as? describes a path the equation of which is i/ = x tan a ' r i J where a is the angle of elevation of the gun and OX is horizontal What is the value of a when the horizontal range is greatest ? 58. In measuring an electric current by means of a tangent galva- nometer, the percentage of error due to a small error in reading is proportional to tan x + ctn x. For what value of x will this percent- age of error be least ? 59. A tablet 8 ft high is placed on a wall so that the bottom of the tablet is 29 ft. from the ground. How far from the wall should a person stand in order that he may see the tablet to best advantage (that is, that the angle between the lines from his eye to the top and to the bottom of the tablet should be the greatest), assuming that his eye is 5 ft. from the ground ? 60. One side of a triangle is 12 ft. and the opposite angle is 36 Find the other angles of the triangle when its area is a maximum 61. Above the center of a round table of radius 2 ft is a hanging lamp. Plow far should the lamp be above the table in order that the edge of the table may be most brilliantly lighted, given that the illumination varies inversely as the square of the distance and directly as the cosine of the angle of incidence ? 62. A weight P is dragged along the ground by a force F. If the coefficient of friction is k, in what direction should the force be applied to produce the best result ? L52 TRIGONOMETRIC FUNCTIONS 63. An open gutter is to be constructed of boards in such a way ,hat the bottom and sides, measuied on the inside, are to be each 3 in wide and both sides are to have the same slope How wide should the gutter be across the top in oider that its capacity may je as great as possible ? 64. A steel girdei 27 ft long is to be moved on rollers along a Dassageway and into a corridor 8 ft. in width at right angles to the Dassageway. If the horizontal width of the girder is neglected, how vide must the passageway be in order that the girder may go around -he corner 9 65. Two particles are moving in the same straight line so that iheir distances from a fixed point are respectively x = a cos kt and e' = acosud -f qOj & and a being constants Find the greatest hstance between them 66. Show that for any curve in polar coordinates the maximum Liid the minimum values of r occui in general when the radius vector s perpendicular to the curve. 67. Two men aie at one end of the diameter of a circle of 40 yd adius. One goes directly toward the center of the circle at the miform rate of 6 ft. pei second, and the other goes around the .ircumference at the rate of 2 TT ft per second How fast are they eparatmg at the end of 10 sec. 'f 68. Given that two sides and the included angle of a triangle are > ft , 10 ft , and 30 respectively, and are changing at the rates of - ft , 3 ft , and 12 per second respectively, what is the area of the riangle and how fast is it changing ? 69. A revolving light in a lighthouse mi offshore makes one evolution a minute If the line of the shore is a straight line, how ast is the ray of light moving along the shore when it passes a ioint one mile from the point nearest to- the lighthouse ? 70. BC is a rod a feet long, connected with a piston rod at C, and t B with a crank AB, b feet long, revolving about A. Find C's elocity in terms of All's angular velocity. 71. At any time t the coordinates of a point moving in the ajy-plane re x = 2 3 cos t, y = 3 + 2 sin t Find its path and its velocity in ,s path. At what points will it have a maximum speed? 72 . At any time t the coordinates of a moving point are x = 2 sec 3 1, = 4 tan 3 1. Find the equation of its path and its velocity in its path. GENERAL EXERCISES 153 73. The parametric equations of the. path of a moving particle are O3 = 2cos 8 <, 2/=2sin 8 < If the angle < increases at the rate of 2 radians per second, find the velocity of the particle in its path 74. A particle moves along the curve y = smo3 so that the a-component of its velocity has always the constant value a Find the velocity of the particle along the curve and determine the points of the curve at which the particle is moving fastest arid those at which it is moving most slowly 75. Find the angle of intersection of the curves y = smx and y = cos x 76. Find the angle of intersection of the curves y = sma; and / , ^ 2/ = sm(a: + ^ ) 77. Find the angle of intersection of the curves y since and y = cos 2 x between the lines x = and ce = 2 TT. 78. Find the points of intersection of the curves T/ = since and y = sin 3 x between the lines x = and x = TT Determine the angles at the points of intersection. 79. Find all the points of intersection of the curves y = cos x and y = sin 2 x which, lie between the lines x = and x = 2 TT, and determine the angles of intersection at each of the points found. CHAPTER VI EXPONENTIAL AND LOGARITHMIC FUNCTIONS 53. The exponential function. The equation / = *, where a is any constant, defines y as a function of x called the exponential function. If x = n, an integer, y is determined by raising a, to the nth power by multiplication. If #=*- a positive fraction, y is the jth root of the plli power of a. If x is a positive irrational number, the approximate value of y may be obtained by expressing oc approximately as a fraction. If x = 0, y = a=l. If x =- m, y = a~ m = The graph of the function is readily found. Ex. Find the gi aph of y = (1 5y. By giving convenient values to x we obtain the curve shown in Fig 70 To determine the shape of the curve at the extreme left, we place a equal to a large negative number, say x = - 100 Then y = (1.5)- = ^ J , which is very small It is obvious that the larger numencally the negative value of x becomes, the smaller y becomes, so that the curve appi caches asymptotically the negative _. portion of the a:-axis. O On the other hand, if # is a large positive number, y is also large. FIG. 70 54. The logarithm. If a number N may be obtained by placing an exponent L on another number a and computing the result, then L is said to be the logarithm of .2V to the base a. That is, if N= d>, (1) then L = log a JV. (2) 154 LOGARITHMS 155 Formulas (1) and (2) are simply two different ways of ex- pressing the same fact as to the relation of ,JV" and L, and the student should accustom himself to pass from one to the other as convenience may demand. From these formulas follow easily the fundamental properties of logarithms; namely, M = log =log a N, (3) log a l=0, log a - = -log JV. Theoretically any number, except or 1, may be used as the base of a system of logarithms. Practically there are only two numbers so used. The first is the number 10, the use of which as a base gives the common system of logarithms, which are the most convenient for calculations and are used almost exclusively in trigonometry. Another number, however, is more convenient in theoretical discussions, since it gives simpler formulas. This number is denoted by the letter e and is expressed by the infinite series where 21=1x2, 31=1x2x8, 41=1x2x3x4, etc. Computing the above series to seven decimal places, we have e = 2,7182818.... An important property of this number, which is necessary in finding the derivative of a logarithm, is that 156 EXPONENTIAL AND LOGARITHMIC FUNCTIONS To check this arithmetically we may take successive small values of h and make the following computation : When &=.!, (1+ A)*= (1.1) 10 = 2.59374. When h =.01, (1+ A)*=(1.01) 100 = 2.70481. When h = .001, (1 + hy= (1.001) 1000 = 2.71692. When h =.0001, (1+ hy= (1.0001) 10000 = 2.71815. i Working algebraically, we expand (1+A) A by the binomial theorem, obtaining ... . _._. ... - 11 if i~1 V - 2 A* 2! 8! where JS represents the sum of all terms involving A, 7i a , 7i 8 , etc. Now it may be shown by advanced methods that as h approaches zero, B, also approaches zero ; so that When the number e is used as the base of a system of loga- rithms, the logarithms are called natural logarithms, or Napierian logarithms. We shall denote a natural logarithm by the symbol In*; thus, N ~*> (4) then L = In N. Tables of natural logarithms exist, and should be used if possible. In case such a table is not available, the student * This notation is generally used by engineers. The student should fenow that the abbreviation "log" is used by many authors to denote the natural logarithm. In this book " log " is used for the logarithm to the base 10. LOGAEITHMS 157 may find the natural logarithm by use of a table of common logarithms, as follows- Let it be required to find In 213. If x = In 213, then, by (4), 213 = e v ; whence, by (3), log 213 = a? log e, or log 213 2.3284 Iog2.7183 0.4343 Certain graphs involving the number y e are important and are shown in the examples. o Ex. 1. y = In or. Giving x positive values and finding y, we obtain Fig 71. Ex. 2. y = e-*? I FIG. 71 The curve (Fig 72) is symmetrical with respect to OY and is always above OX. When a; = 0, y = 1 As a: increases numerically, y decreases, approaching zero. Hence OX is an asymptote ^ FIG. 72 FIG. 73 Ex.3. y = This is the curve (Fig. 73) made by a cord or a chain held at the ends and allowed to hang freely. It is called the catenary. Ex. 4. y e ~ *" sin Ix. The values of y may be computed by multiplying the ordmates of the curve y = <s~ oa! by the values of sin bx for the corresponding abscissas. Since the value of sinZw oscillates between 1 and -1, the values of e- 158 EXPONENTIAL AND LOGARITHMIC FUNCTIONS cannot exceed those of e-** Hence the graph lies in the portion of the plane between the curves y = e~ ax y and y = e ~ ax . When a: is a mul- tiple of -, y is zero. The graph therefore crosses the axis of x an infinite number of times Fig. 74 shows the graph when a = 1, b = 2 TT Ex. 5. y e x When x approaches zeio, being positive, y increases without limit -1 '' When x approaches zero, being neg- ative, y approaches zeio , for example, when x = Ttnjff' y = e1000 ' and when x ~ ~~ "W 11 = e - 1000 = - The function is therefore J glOOO discontinuous for x The line y - 1 is an asymptote (Fig. 75), for as x increases without limit, being posi- tive or negative, approaches 0, and y appioaches 1. Ex. 6. r = e a . The use of r and 6 indicates that we are using polar coordinates. When 6 = 0, r 1. As increases, r in- creases, and the curve winds around the origin at increasing distances from it (Fig 76). When 6 is negative and increasing numer- ically without limit, r approaches zero. Hence the curve winds an infinite number of times aiound the origin, continually ap- proaching it The dotted line in the figure corresponds to negative values of 6 The curve is called the logarithmic spiral EXERCISES Plot the graphs of the following equations ! y = ()* 5 - y = x&x - ?a* 74 X 75 M 3. y= B 4. y 7. y = log2o:. 1 'a: 8. = FIG. 76 9. y = log sin x. 10. y = log tan ar. 11. y = e~ 2ar sin4ai. 12. y = e~ x cos 3 a;. 13. r=e-* e . EMPIKICAL EQUATIONS 159 55. Certain empirical equations. If x and y are two related quantities which are connected by a given equation, we may plot the corresponding curve on a system of ^-coordinates, and every point of this curve determines corresponding values of x and y. Conversely, let x and y be two related quantities of which some corresponding pairs of values have been determined, and let it be desired to find by means of these data an equation con- necting x and y in general. On this basis alone the problem cannot be solved exactly. The best we can do is to assume that the desired equation is of a certain form and then endeavor to adjust the constants in the equation in such a way that it fits the data as nearly as possible. We may proceed as follows : Plot the points corresponding to the known values of x and y. The simplest case is that in which the plotted points appear to lie on a straight line or nearly so. In that case it is assumed that the required relation may be put in the form y = mx + *, (1) where m and I are constants to be determined to fit the data. The next step is to draw a straight line so that the plotted points either lie on it or are close to it and about evenly dis- tributed on both sides of it. The equation of this line may be found by means of two points on it, which may be either two points determined by the original data or any other two points on the line. The resulting equation is called an empirical equation and expresses approximately the general relation between x and y. In fact, more than one such equation may be derived from the same data, and the choice of the best equation depends on the judgment and experience of the worker. Ex. 1. Corresponding values of two related quantities x and y are given by the following table : ai 1 2 4 6 10 y 1.8 2.2 2.9 3.9 6.1 Find the empiucal equation connecting them. 160 EXPONENTIAL AND LOGARITHMIC FUNCTIONS We plot the points (a;, y) and draw the straight line, as shown, in Fig. 77. The straight line is seen to pass through the points (0, 1) and (2, 2). Its equation is therefore T7" ?/ = 5 x + 1, which is the required equation 111 many cases, however, the plotted points will not appear to lie on or near a straight line. We shall con- sider here only two of these ~o i cases, which are closely con- - FIG ?7 nected with the case just considered. They are the cases in which it may be anticipated from previous experience that the required relation is either of the form y=db*, (2) where a and b are constants, or of the form y=atf, (3) where a and n are constants. Both of these cases may be brought directly under the first case by taking the logarithm of the equation as written. Equa- tion (2) then becomes log y = log a, + x log 6. (4) As log a and log I are constants, if we denote log y by y', (4) assumes the form (1) in x and y\ and we have only to plot the points (#, y'~) on an o^'-system of axes and determine a straight line by means of them. The transformation from. (4) back to (2) is easy, as shown in Ex. 2. Taking the logarithm of (3), we have log y = log a + n log x. (5) If we denote logy by y' and logo; by x\ (5) assumes the form (1) in. a/ and /, since log a and n are constants. Accordingly we plot the points (V, y r ~) on an a/?/-system of axes, determine the correspond ing straight line, and then transform back to (8), as shown in Ex. 3. EMPIKICAL EQUATIONS 161 Ex. 2. Corresponding values of two related quantities x and y are given by the following table . a; 8 10 12 14 16 18 20 y 3.2 40 73 98 152 240 364 Find an empirical equation of the form y = dbf* Taking the logarithm of the equation y = ob*, and denoting log y by y', we have ' if = log a + x log i. Determining the loganthm of each of the given values of y, we form a table of corresponding values of x and y', as follows : x 8 10 12 14 16 18 20 if = logy 5051 6028 8633 9912 1.1818 1.3909 15611 We choose a large-scale plotting-paper, assume on the y'-axis a scale four times as laige as that on the or-axis, plot the points (x, /), and 20 J, point. Its equation is 10 > tour times as laige as tnat on tne * or-axis, plot the points (x, /), and i j draw the straight line (Fig 78) 20 I H-iivM-in-li +.Vo -fifaf. o.nrl tliA HTvKh i e through the first and the sixth 15 ' = .08858 x - .20354. 5 m 2 4 6 8 10 12 14 16 18 20 * Therefore log a = 20354 = FlG ?g 9 7965 - 10, whence a = .626 ; and log & = .08858, whence & = 1 22. Substituting these values in the assumed equation, we have as the required empirical equation The result may be tested by substitxit- ing the given values of x in the equation. The computed values of y will Le found to agree fairly well with the given, values. Ex. 3. Corresponding values of pressure and volume taken from an indicator card of an air -compressor are as follows: p 18 21 205 33.5 44 62 v .035 .656 .476 .897 .321 .243 Find the relation between them in the tormpv" ~ c, 162 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Writing the assumed relation, in. the form p = cv~ n and taking the logarithms of both sides of the equation, we have log p =n log v + log c, or where y = log jtjj x = log v, and 6 = log c. The corresponding values of x and y are - 4012 -.4935 - 6144 1 5250 1 6435 1 7994 x = logu 1972 y = logp 1 2553 - 2549 - .3233 1 3222 1 4232 For convenience we assume on the ar-axis a scale twice as large as that on the y-axis, plot the points (x, ?y), and diaw the straight line as shown in Fig 79 The construction should be made on large-scale plotting- paper. The line is seen to pass through the points (- 05, 1 075) and ( .46, 1.6) Its equation is therefore y=- 128 x +101 Hence n 1 28, log c = 1 01, c = 10 2, and the required rela- tion between p and v is pv i as = 10 2 ) -.55-50- lt5-W-SS-SO~25-0-15 10- OS FIG. 79 EXERCISES 1. Show that the following points lie approximately on a straight line, and find its equation : 4 21 46 13 7 20 12 22 12 9 25 14,5 30 18.2 2. For a galvanometer the deflection D, measured in millimeters on a proper scale, and the current /, measured in microamperes, are determined in a series of readings as follows : D 291 48.2 72.7 92.0 118.0 140.0 165.0 109.0 I 0.0493 0.0821 0123 0.154 0197 0234 0.274 0328 Find an empirical law connecting D and /. EMPIRICAL EQUATIONS ' - , 163 3. Corresponding values of two related quantities x and y are given in the following table x 01 03 0.5 0.7 0.9 1.1 1 3 1.5 y 3316 4050 4046 0.6041 0.7379 9013 1 1008 1.3445 Find an empirical equation connecting x and y in the form y = ab*. 4. In a certain chemical reaction the concentration c of sodmm acetate produced at the end of the stated number of minutes t is as follows : * 1 2 8 4 5 c 00837 00700 00586 00492 00410 Find an empirical equation connecting c and t in the form c = atf 5. The deflection a of a loaded beam with a constant load is found for various lengths I as follows Z 1000 900 800 700 600 a 7.14 5.22 3.64 242 1.50 Find an empirical equation connecting a and I in the form a = nlf 1 6. The relation between the pressure j and the volume v of a gas is found experimentally as follows : p 20 23.5 31 42 59 78 v 0019 0.540 0.442 0358 0277 0219 Find an empirical equation connecting p and v in the form pv n = c. 56. Differentiation. The formulas for the differentiation of the exponential and the logarithmic functions are as follows, where, as usual, u represents any function which can be differen- tiated with respect to #, In means the Napierian logarithm, and a is any constant: d , log a e du ,1 . -r-log a tt = fi a-- r (1) das u da d , 1 du ^ o ^ --- In u = - 3-, (2) aa? u ax 164 EXPONENTIAL AND LOaAEITHMIC FUNCTIONS dx" ~- ~dx (3) d u B du dx dx' ^ -' The proofs of these formulas are as follows: .j To find - log a M place y = log M. Then, if u is given an increment AM, y receives an increment Ay, where Ay = log a (M + AM) - log M Aw, A , AM\ A ) ' u / the transformations being made by (3), 54. Then AM Now, as AM approaches zero the fraction may be taken A of 54. u as A of 54. u Hence Lim fl 4- Y "= e. A-0\ M / and 2. If y = hi u, the base a of the previous formula is e ; and since log e=l, we have dy _ 1 du dx u dx DIFFERENTIATION 165 3. If # = ", we have In y = In a u = u In a. Hence, by formula (2), 1 dy , du -- = lna ; y ax ax . dy ul du whence ~ = a In a - ax ax 4. If y = e w the previous formula becomes ^ = e ^. dx dx Ex. 1. y = In (a; 2 -4 a + 5). dy _ 2 x 4 da; a; 2 4 a; + 5 Ex.2. y = '- Ex. 3. y = e- ^ = cos bx (e~ dx dx EXERCISES Find in each of the following cases : dx -- I~sin2aj 2. y*= 3. JJ^-. . 4. y = a sln ""X 12. y = e" a>e sin 3 cc. 5. / . y^ln 6. y = lnV2a; a +6a; + 9. * , a; - 3 14. y - 6 8a (9 a 2 - 6ai + 2). 7 . w == -J In - 5 x + 3 _ 16. y e* x (2 sin a: - cos ). 8. = ln(flj + Vaj 8 + 4). a: 9 9. 3/ = 17. y = sec x tan 03 + In (sec a? -f- tan *). . VaTTl-1 18. v In 166 EXPONENTIAL AND 'LOGARITHMIC FUNCTIONS 57. The compound-interest law. An important use of the ex- ponential function occurs in. the problem to determine a, function whose rate of change is proportional to the value of the function. If y is such a function of #, it must satisfy the equation where ~k is a constant called the proportionality factor. We may write equation (1) in the form ld V-7 f . r /fc , ydx whence, by a very obvious reversal of formula (2), 56, we have In y = kx 4- (7, where C is the constant of integration (18). From this, by (1) and (2), 54, Finally we place e c =A, where A may be any constant, since C is any constant, and have as a final result (2) The constants A and Jc must be determined by other condi- tions of a particular problem, as was done in 18. The law of change here discussed is often called the compound- interest law, because of its occurrence in the following problem : Ex. Let a sum of money P be put at interest at the rate of r% per annum. A The interest gained in a time Ai is -Pr^ Ai, where A* is expressed in years. But the interest is an increment of the principal P, so that we have In ordinary compound interest the interest is computed for a certain interval (usually one-half year), the principal lemaming constant during that interval The interest at the end of the half year is then added to the principal to make a new principal on which interest is computed for th.9 COMPOUND-INTEREST LAW 167 next half year. The principal P therefore changes abruptly at the end of each half year. Let us now suppose that the principal changes continuously ; that is, that any amount of interest theoretically eained, in no matter how small a time, is immediately added to the puncipal. The average rate of change of the principal in the peuod Ai is, fiorn 11, * = lL m Ai 100 ( } To obtain the rate of change we must let A< approach zero in equation (1), and have dp ? , From this, as in the text, we have r P=Ae lo . (2) To make the problem conci ete, suppose the original principal were $100 and the rate 4%, and we ask what would be the principal at the end of 14 yr. We know that when t = 0, P = 100. Substituting these values in (2), we have A = 100, so that (2) becomes JL, 1 P = 100e 100 =100e 20 Placing now t = 14, we have to compute P = 100 ei e The value of P may be taken from a table if the student has access to tables of powers of e In case a table of common logarithms is alone available, P may be found by first taking the logarithm of both sides of the last equation. Thus logP = loglQO + it loge = 24053; whence P = $254, approximately EXERCISES 1. The rate, of change of y with respect to x is always equal to \ y, and when x = 0, y = 5. Find the law connecting y and x, 2. The rate of change of y with lespect to x is always 0.01 times ?/, and when x = 10, y = 50. Find the law connecting y and x. 3. The rate of change of y with respect to x is proportional to y, When x = 0, y = 7, and when x = 2, y = 14 Find the law connect- ing y and x. 4. The sum of $100 is put at interest at the rate of 5% per annum under the condition that the interest shall be compounded at each instant of time. How much will it amount to in 40 yr.? 168 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 6. At a certain date the population of a town is 10,000. Forty years later it is 25,000. If the population increases at a rate which is always proportional to the population at the time, find a general expression for the population at any time t. 6. In a chemical reaction the rate of change of concentration of a substance is proportional to the concentration at any time. If the concentration is y^ when t = 0, and is T T when t = 6, find the law connecting the concentration and the time. 7. A rotating wheel is slowing down in such a manner that the angular acceleration is proportional to the angular velocity. If the angular velocity at the beginning of the slowing down is 100 revolu- tions per second, and in 1 min. it is cut down to 50 revolutions per second, how long will it take to reduce the velocity to 25 revolutions per second ? GENERAL EXERCISES Plot the graphs of the following equations : i i 1 - y = ($)"* 4 - y el ~ x - ? y e?. 2. y = e 1 -*. 5. y = (e* + -*). 8. y = aser*. 3. y = e S GOSX. 6. y= ~ 9. y=*y?&-*. J e'+e"* J 10. For a copper-nickel thermocouple the relation between the temperature t in degrees and the thermoelectric power p in micro- volts is given by the following table : t 50 100 160 200 P 24 25 26 26.9 27.5 Find an empirical law connecting t and p. 11. The safe loads in thousands of pounds for beams of the same cross section but of various lengths in feet are found as follows : Length 10 11 12 18 14 16 Load 123.6 121.5 111.8 107.2 1018 90.4 Find an empirical equation connecting the data. GENERAL EXERCISES 169 12. In the following table s denotes the distance of a moving body from a fixed point in its path at time t t I 2 4678 s 10 4 6400 0.1024 0410 0164 Find an empirical equation connecting s and t in the form s = ab*. 13. In the following table c denotes the chemical concentration of a substance at the time t . t 2 4 6 8 10 c 00060 00048 0.0033 0.0023 00016 Find an empirical equation connecting c and t in the form c = ah*. 14. The relation between the length I (in millimeters) and the time t (in seconds) of a swinging pendulum is found as follows : I 634 805 90.4 1013 107.3 1406 t 0.806 0892 0.960 1010 1038 1.198 Find an empirical equation connecting I and t in the form t Td n . 15. For a dynamometer the relation "between the deflection 8, 2ir when the unit 6 = -r-rri and the current I, measured in amperes, is as follows : 6 40 86 120 160 201 240 280 320 362 I 0.147 0215 0.252 0293 0.329 0360 0.390 0.417 0442 Find an empirical equation connecting J and 6 in the form I k$" 16. In a chemical experiment the relation between the concen- tration y of undissociated hydrochloric acid and the concentration x of hydrogen ions is shown in the table x 1.68 1.22 784 426 0.092 047 0096 0.0049 y 1 82 676 216 074 0085 0.00815 0.00036 00014 Find an empirical equation connecting the two quantities m the form y = kx n . 170 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 17. Assuming Boyle's law, pv = c, determine o graphically from the following pairs of observed values : 39.92 42.17 45.80 48 62 51 80 (J0.47 CMS 1)7 40.37 38.32 85.32 33.29 31.22 2080 Tind -J- in each of the following cases : CCSs . Bx-2 18. = 19. y = In sin a-. e _ e - 20. ?/ = tan" 1 & 21. y = In (2as + V4 aj a l) + 2a osc"^ a;. _a 22. y = x z e x . 23. = l 24. y = I tan 2 oas + In cos ax. 25 . y = x tan- 1 ^ $ In (1 + a; 2 ). 26. A substance of amount x is being decomposed at a rate which is proportional to a;. If x = 3 12 when t = 0, and cc == 1.30 whun t = 40 min., find the value of x when t = 1 hr. 27. A substance is being transformed into another at a rate whiith is proportional to the amount of the substance still imtranafovnuid, If the amount is 50 when t = 0, arid 15.6 when t = 4 hr., lind how long it will be before y^ of the original substance will rumain. 28. According to Newton's law the rate at which tho temperature of a body cools in air is proportional to the difference botwoou tho temperature of the body and that of the air. If the tempwraturo of the air is kept at 60, and the body cools from 130 to 120" in 300 sue., when will its temperature be 100 ? 29. Assuming that the rate of change of atmospheric prnssuro p at a distance h above the surface of the earth is proportional to tho pressure, and that the pressure at sea level is 14.7 Ib. per square inch and at a distance of 1COO ft. above soa level is 18.8 Ib. per square inch, find the law connecting^ and h. GENERAL EXERCISES 171 30. Prove that the curve y = e~ Zx sm Sec is tangent to the curve y = e- 2 * at any point common to the two curves. 31. At any time t the coordinates of a point moving in a plane are x = a~' 2t cos 2 1, y o~' 2t sin 2 1. Find the velocity of the point at any time t Find the rate at which the distance of the point from the origin is decreasing. Prove that the path of the point is a loga- rithmic spiral. 32. Show that tho logarithmic spiral r = e" 9 cuts all radius vectors at a constant angle 33. Find the radius of curvature of the curve y = e~ Za sin 2x at 7T the point for which x = ft ( 5 34. Show that tho catenary y n^+e n and the parabola 1 ^ y = a -f- - a; 3 have tho same slope and the same curvature at their i Q> common point. 35 . Find the radius of curvature of the curve x =s e t sin t, y = e* cos t. 36. Show that the product of the radii of curvature of the curve X y =s ae " at the two points for which x = a is a?(e + e" 1 ) 8 . 37. Find the radius of curvature of the curve ?/ = In a; and its least value. 38. Find the radius of curvature of the curve y = e r cosaj at the 7T point for which x = -= \ n ) CHAPTER VII SERIES 58. Power series. The expression a fl + a : x + ajK? + a$+ ajf-\ ---- , (1) where a Ql a^ a z , are constants, is called a power series in x. The terms of the series may be unlimited in number, in which case we have an infinite series, or the series may terminate after a finite number of terms, in which case it reduces to a polynomial. If the series (1) is an infinite series, it is said to converge for a definite value of x when the sum of the first n terms approaches a limit as n increases indefinitely. Infinite series may arise through the use of elementary opera- tions. Thus, if we divide 1 by 1 a; in the ordinary manner, we obtain the quotient and we may write LX (2) ^ ' Similarly, if we extract the square root of 1 + x by the rule taught in elementary algebra, arranging the work as follows; l+x 2 + - V 1 4 a+ I a 2 4 2? a; 8 x* 4" "8 + 64' 172 POWER SERIES 173 the operation may be continued indefinitely. We may write The results (2) and (3) are useful only for values of x for which the series in each case converges. When that happens the more terms we take of the series, the more nearly is their sum equal to the function on the left of the equation, and in that sense the function is equal to the series. For example, the series (2) is a geometric progression which is known to con- verge when x is a positive or negative number numerically less than 1. If we place x = % in (2), we have "- which is true in the sense that the limit of the sum of the terms on the right is f . If, however, we place x 3 in (2), we have which is false. A reason for this difference may be seen by considering the remainder in the division which produced (2) but which is neglected in writing the series. This remainder is after n terms of the quotient have been obtained ; and if l-x x is numerically less than 1, the remainder becomes smaller and smaller as n increases, while if x is numerically greater than 1, the remainder becomes larger. Hence in the former case it may be neglected, but not in the latter case. The calculus offers a general method for finding such series as those obtained by the special methods which led to (2) and (3). This method will be given in the following section. 59. Maclatirin's series. We shall assume that a function can usually be expressed by a power series which is valid for ap- propriate values of x, and that the derivative of the function may be found by differentiating the series term by term. The proof of these assumptions lies outside the scope of this book. Let us proceed to find the expansion of sin x into a series. We begin by writing $oLv s *A+Bx + Cx*+Dx*+Ex*+Fx*+ - - -, (1) where -4, J?, (7, etc. are coefficients to be determined. 174 SERIES By differentiating (1) successively, we have - 5 . 4 cosa;=3.2. J> + 4.3 . 2.Je + 5. sin z = 4- 3- 2- J+5. 4- 3 . 2 . Jfc -3 - By substituting x = in equation (1) and each of the fol- lowing equations, we get .4 = 0, 5=1, C=0, 3.2.D = -1, J?=0, 5.4.3.2.^ = 1; whence JL = 0, 5 = 1, tf= 0, -O = -^ T #=<>, ^=^- o I o Suhstitutuig these values in (1), we have 8 , X 6 SQ^ sma; = a; __ + __..., (2) and the law of the following terms is evident. The above method may obviously be used for any function which may be expanded into a series. We may also obtain a general formula by repeating the above operations for a general I auction /(a;). We place f(x) = A + Bz + C3?+J)x*+lSz l + - (3) and, by differentiation, obtain in succession f"(x) = 3 ! Z> + 4 3 2 . Ex+ . ., where f f (x), /"()/'"(); and/ v (a;) represent the first, second, third, and fourth derivatives of /(#). We now place a; = in these equations, indicating the results of that substitution on the left of the equations by the symbols MACLAUKIN'S SERIES 175 > f" ()> etc - We thus determine A, B, C, D, E, etc., and, substituting in (8), have A*) =AOHAO>+^AoX+^/"(OX+ ^<oy+- - - (4) This is called Madaurin's scries. Ex. 1. Find the value of sin 10 to foui decimal places. Wo tuny use sones (8), but have? to remembei ( 42) that x must be in eiivuhii measure. Hence wo i>lace x = -~~ 17458, where wo take five loO significant figmes in ordei to insure accuracy in the fomth significant iigiU'O of the result.* Subntibutmg in (2), we have . TT ,_,_ f.17458) 8 , sin = .17451}- v ' + - ia O = .17458 - 00089 = .17804. Ilonco to four decimal places am 10= .1736. We have u.scd only two terms of the series, since a rough calculation, which may bo made with a = 2, shows that the third term of the series will not affect the fourth decimal place. Ex. 2. Find the value of sin 61 to four decimal places. In radians the angle 01 is ~-.TT= 1.0647. If this nuxnbei were sub- luO Htituted in the sonos (2), a great many terms would have to be taken to include all which affect the first four decimal places. We shall therefore find a series for 8in(~+oA and afterwards place o: = j~(=l). We chow the angle ~^(~ (10) because it is an angle near 61 for which we " know the nine and cosine. The series may b(5 obtained by the method by which (2) was obtained, For variety we shall use the general formula (4). Wo have then . \ i _. /(O) . tnn| = |V; This JH not a general rule. In other cases the student may need to cany two or men three more significant figures in the calculation than are needed in the result. 176 SERIES Therefore, substituting in (4), we have In this we place x = -^- = .01745 and perform the arithmetical calcula- loO . . tion. We have sin 61= sin(^ + ~}= 8746. \o 180/ Ex. 3. Expand In (1 + a) The function In x is an example of a function which cannot be expanded into a Maclaurm's series, since if we place /(a;) = In a;, we find /(0),/'(0), etc to be infinite, and the series (4) cannot be written We can, however, expand ln(l + a;) by series (4) or by using the method employed in obtain- ing (2). The latter method is more instructive because of an interesting abbreviation of the work We place ln(l + a;) = A + Bx + Cx z + Dx 3 + Eat + . Then, by differentiating, = B + 2 CJB + 3 Dx* + 4 Ex s + - . 1 + x But we know, by elementary algebra, that - - =1 x + x z x s + 1 + x Hence, by comparing the last two series, we have JB=1, C = -k Z>=, E=-%, etc. By placing x = in the first series, we find In 1 = A, whence .4 = 0. We have, therefore, ^ 8 ^ In (1 + a;) = x -_ + _-_+.... EXERCISES Expand each, of the following functions into a Maclaurin's series : 1. d*. 2. cossc. 3. tan a;. 4. sin" 1 ^. 5. tan" 1 ^ 6. sinf-j + a?). 7. ln(2 + aj). 8. Prove the binomial theorem t . ... i , n(n 1) . n(n 'IVw 2) , . (a + o;) n =a' l +wa n - 1 aj4- v / a n ~ 2 a5 2 + v O y . v / a n - 8 a5*4--- . i\ O\ 9. Compute sin 5 to four decimal places 10. Compute cos 62* to four decimal places. TAYLOR'S SERIES 177 60. Taylor's series. In the use of Maclaurin's series, as given in the previous section, it is usually necessary to restrict our- selves to small values of x. This is for two reasons. In the first place, the series may not converge for large values of x\ and in the second place, even if it converges, the number of terms of the series which it is necessary to take to obtain a required degree of accuracy may be inconveniently large. This difficulty may be overcome by an ingenious use of Maclaurin's series as illustrated in Ex. 2 of the previous section. We may, however, obtain another form of series which may be used when Maclaurin's series is inconvenient. Let f(x) be a given function, and let a be a fixed value of x for which the values of /(#) and its derivatives are known. Let * be a variable, or general, value of x which does not differ much from a ; that is, let x a be a small number, positive or negative. We shall then assume that/(a;) can be expanded in powers of the binomial x a ; that is, we write f(&=A+(x-a) + C(x-a)*+I)(x-a) 8 -\ , (1) and the problem is to determine the coefficients -4, B, C, . We differentiate equation (1) successively, obtaining In each of these equations place x = a. We have f(a)~A, /< = , /"(a) =2,1 (7, etc.; f" (ch f" (a\ whence 4=/(a), B =/'(), C^^f 2 ' #= 8 , ' et <>- Substituting in equation (1), we have as the final result a)^ - - .. (2) This is known as Taylor's series. Since, as has been said, it is valid for values of x which make x a a small quantity, the SERIES function /(a?) is said to be expanded in the neighborhood of x a. It is to be noticed that Taylor's series reduces to Maclaurin's series when a = 0. Maclaunn's series is therefore an expansion in the neighborhood of x 0. Ex. Expand In a; in the neighborhood of x = 3. Here we have to place a = 3 in the general formula. The calculation of the coefficients is as follows : /():= In a, /(3) = ln3, and therefore 1 f~ QNfl BT ( - 3 ) This enables us to calculate the natural logarithm of a number near 3, provided we know the logarithm of 3. For example, let us have given In 3 = 1 0986 and desire In 3. Then x - 3 = , and the series gives = 1.0986 + 1667 - .0139 + 0015 - .0002 + . . . = 1 2527. The last figure cannot be depended upon, since we have used only four decimal places in the calculation. EXERCISES Expand each of the following functions into a Taylor's series, using the value of a given in each case : 1. In*, a = 5. M TT 4. cos a;, a - 2. j a = 2. . TT 3. sin x, a = 8. Compute sin 46 to four decimal places by Taylor's series. 9. Compute cos 32 to four decimal places by Taylor's series. 10. Compute e 1 - 1 to four decimal places by Taylor's series. GENERAL EXEECISES 179 GENERAL EXERCISES Expand each of the following functions into series in powers of a; 1. ln(l a?). ,1+as . fir ' 4. ln= 6. sm(v + 2. seco;. 1 -a? \b 3. ,~ 6. cos(-7r + ). 7. /z 3' 8. Verify the expansion of tan a? (Ex 3, 59) by dividing the series for sin x by that for cos x. 9. Verify the expansion of sec x (Ex. 2) by dividing 1 by the series for cos x. 1 x 10. Expand 1 by Maclaurin's series and verify by dividing JL ~T~ tC the numerator by the denominator 11. Expand e* cos x into a Maclaurin's series, and verify by multiplying the series for & by that for cosaj. 12. Expand e^sime into a Maclaurin's series, and verify by multiplying the series for tf* by that for sin a;. 13. Expand e T ln(l+aj) into a Maclaurin's series, and verify by multiplying the series for e 00 by that for ln(l-|- K). 14. Compute cos 15 to four decimal places 15. Compute sin 31 to four decimal places 16. Compute e* to four decimal places by the series found in Ex. 1, 59. 17. Using the series for ln(l+ x~), compute Inf to five decimal places. 18. Using the series found in Ex. 4, compute In 2 to five decimal places, and thence, by aid of the result of Ex. 17, find In 3 to four decimal places. 19. Using the series found in Ex. 4, compute In | to five decimal places, and thonce, by aid of the first result of Ex. 18, find In 6 to four decimal places. 20. Using the series found in Ex. 4, compute ln$ to four decimal places, and thence, by aid of the result of Ex. 18, find In 7 to three decimal places. 21. Compute the value of TT to four decimal places, from the ex- 1 IT pansion of sin- 1 a5 (Ex. 4, 59) and the relation sin- 1 ^ = -g 180 SEEIES 22. Compute the value of IT to four decimal places, from the ex- pansion of tan- 1 ;*; (Ex. 5, 59) and the relation tan- 1 - + 2 tan" 1 - = 23. Compute -v/17 to four decimal places by the binomial theorem (Ex. 8, 59), placing a = 16, x = 1. 24. Compute "\/26 to four decimal places by the binomial theorem (Ex. 8, 59), placing a = 27, x =- 1. /"* X o-\r\ /* 25. Obtain the integral / dx in the form of a series pansion. J x 26. Obtain the integral I e'^dx in the form of a series pansion. ^ f*x C&C 27. Obtain the integral / in the form of a series expansion. Uo * ~T" * 28. Obtain the integral / 5 in the form of a series expansion. i/o x expansion CHAPTER VIII PARTIAL DIFFERENTIATION 61. Partial differentiation. A quantity is a function of two variables x and y when the values of x and y determine the quantity. Such a function is represented by the symbol /(#, #). For example, the volume V of a right circular cylinder is a function of its radius r and its altitude h, and in this case Similarly, we may have a function of three or more variables represented by the symbols /(a;, #, z),f(x, y, 2, w), etc. Consider now /(*, jr), where x and y are independent varia- bles so that the value of x depends in no way upon the value of y nor does the value of y depend upon that of x. Then we may change x without changing y, and the change in x causes a change in /. The limit of the ratio of these changes is the derivative of / with respect to x when y is constant, and may / JTJ?\ be represented by the symbol ( ) v^ \ dx J v Similarly, the derivative of / with respect to y when x is constant, is represented by the symbol i-j- Y These derivatives \dy/x are called partial derivatives of / with respect to x and y re- spectively, The symbol used indicates by the letter outside the parenthesis the variable held constant in the differentiation. When no ambiguity can arise as to this variable, the partial de- ftf flf rivatives are represented by the symbols and ^-, thus: dx \dx A*-*!) Aa; + A ^> Ay 181 182 PARTIAL DIFFERENTIATION So, in general, if we have a function of any number of variables f(x, y, ., 2), we may have a partial derivative with respect to each of the variables. These derivatives are expressed by the sym- QJ? rvj* A / ' ' ' ' *' orsometimesb y/*C*> y> - - * To compute these derivatives we have to apply the formulas for the derivative of a function of one variable, regarding as constant all the variables except the one with respect to which we differentiate. Ex. 1. Consider a perfect gas obeying the law v = We may change P the temperature while keeping the pressure unchanged. If A* and AJ> are corresponding increments of t and v, then P P P , 8v c and = - . dt p Or we may change the pressure while keeping the temperature un- changed If Ap and Ai? are corresponding increments of p and v, then A and *L = -. dp p* Ex. 2. /= a; 8 - 3 x*y + y, Ex. 3. /= sin (re 2 + y 2 ), ?L = 3 a; 2 - 6 xy, %- = 2 a; cos (x* + w 2 ), -- v ^ y Ex. 4. In differentiating in this way care must be taken to have the functions expressed in terms of the independent variables. Let x r cos 9 t y = r sin 6 Then , or or fa A fy A - = r sin 6, ~ = r cos 6, c9 dd where r and 6 are the independent variables. PARTIAL DERIVATIVES 183 Moreover, since r 8r Sx where x and y are the independent variables. fix = sin0, (2) It is to be emphasized that in (1) is not the reciprocal of in (2). ftr j /ON r i and,in(2), = ( T . j /-IN In fact, in (1), = ^ ' dr and because the variable held constant is differ- ent in the two cases, there is no reason that one should be the leciprocal of the other. It happens in this case that the two are eqiial, but this is not a general rule. Graphically (Fig. 80), if OP = r is increased by PQ = Ar, while 6 is constant, then PR = Aa: is determined Then 9j, fdx\ PR = ( ) \drja Sr T n Lim. - - = cos 0. PQ, Moreover (Fig. 81), if OM = x is increased by MN-PQ, = Aa;, while y is constant, then R Q= Ar , . -, m, is determined Then dr /dr\ T . = ( ) = Lim 8x \dx/v ST f^y fix cos0. It happens here that = But -5? - or ox co in (1), and , in (2), are neither equal nor reciprocal. EXERCISES n Jx 3-1 , t J35 ZS.M and ~ in each of the following cases 6 . , = C y 2ajv 6. sm - Ji 9, If * = ln<X - 2xy + 2/ a + 3aj - K 2 10. If = Va; 2 + 2/ 2 e- r , prove a; + y prove ^ *. = 184 PAETIAL DIFFERENTIATION 62. Higher partial derivatives. The partial derivatives of /(#, y) are themselves functions of x and y which may have partial derivatives, called the second partial derivatives of /(a-, y~). rrn, d /3A /2A 2 /9A 2 /9A T> 4. -4. vi They are r ( )' ( TT- ) TT-( )' ^ ut ^ mav b shown 17 dx\dxj 8y\dx/ 8x\dy/ dy\dy/ J that the order of differentiation with respect to x and y is imma- terial when the functions and their derivative fulfill the ordinary conditions as to continuity, so that the second partial derivatives are three in number, expressed by the symbols a/8/\ 3/3A a 2 / dx\Zy) dy\dx] dxcy Similarly, the third partial derivatives of /(#, /) are four in number; namely, \ j 3 /gA "" " So, in general, T; signifies the result of difEerentiating y) p times with respect to a?, and g times with respect to , the order of differentiating being immaterial. In like manner, p 9r signifies the result of differentiating /(#, y, g) jp times with respect to #, q times with respect to y, and r times with respect to 2, in any order. TOTAL DIFFERENTIAL 185 EXERCISES 1. If = (x z + 2/ 2 ) tan- 1 ^ find x & 2. If e"sm(x y), find ^-3 3V 3. If = Verify ^-^] = -^-[^- m each of the following cases : 17 &e\<ty/ By\faJ i x 4. * = a:?/ 2 + 2 ye- 1 . 6. s = sm- 1 -. y K g - y + y 7 - * O. w I is> , te y Vflj 2 4- j/ 2 , 82 &* 8. If^ = tan- 9. If In (a; 2 a 2 /), prove a 2 ji g-g = 0. O2 / -r/\ 10. If V = i cos w<, prove n*r ^^ ^ + w (w + ^)ii = ^- 63. Total differential of a function of two variables. In 20 the differential of a function of a single variable, y =/(), is defined by the equation ay=f'(x)dx, (1) where /* (#) is the derivative of y. But /(aO=Lim; (2) and hence, according to the definition of a limit ( 1), |2=/0) + e, (8) where e denotes the difference between the variable ^ and its limit f (x) and approaches zero as a limit as A > 0. Multiplying (3) by As, we have Ay=/(a;)Aa? + Ac. (4) But Az = dx and Ay =/( + Aas) -/(), so that (4) may be written in the form -/(SB) =/ () * + e <n (5) 186 PARTIAL DIFFERENTIATION In the case of a function of two variables, /(, y), if x alone is changed, we have, by (5), f(x + Az, y) -/(a, y) = ^dfc + e^ (6) the theory being the same as in the case of a function of one OJJ variable, since y is held constant. The term dx may be denoted by the symbol d x f. Similarly, if x is held constant and y alone is changed, we have /(a, y + Ay) -/(*, y) = ^ dy 4- e 2 cZy, (7) df and dy may be denoted by the symbol d f. dy Finally, let x and y both change. Then. A/=/CB+AZ, y+Ay)-/(a;, y) ==/(+ Aas, y+ Ay)-/(z+ As, y)4-/(z+ Az,y) -/(,y> (8) Then, by (6), /(a + As, y) -/O, y) = | <fc + ^db; (9) and similarly, by (7), /(as + As, y + Ay) -/<> + As, y) = ^dy + e/rfy, (10) OJ? * where is to be computed for the value (a? + Aa;, y). But if of y is a continuous function, as we shall assume it is, its value oy for (x + A#, y) differs from its value for (x, y) by an amount which approaches zero as dx approaches zero. Hence we may write, from (8), (9), and (10), A/= dx + y + ^rfg + e 3 rfy, (11) cj? ay? where both ^ and ^- are computed for (x, y\ dx dy ^ aj We now write df = j dx + % dy, (1 2) ox dy so that A/= 4f+ e^ + e t dy, (13) and ?f is called the total differential of the function, the expres- sions dfc/ and ^ being called the partial differentials. TOTAL DIFFERENTIAL 187 It is evident, by analogy with the case of a function of a single variable, that a partial differential expresses approximately the change in the function caused by a change in one of the independent variables, and that the total differential expresses approximately the change in the function caused by changes in both the independent variables. It is evident from the defini- tion that , ,. , ,, , ,. ._, A ,. v f. (14) Ex. The period of a simple pendulum with small oscillations is whence g Let I = 100 cm with a possible error of ^ mm in measuring, and T = 2 sec. with a possible error of ^ $ y sec. in measuring. Then <ll = -^ Moreover, dg = - dl =j- dT, and we obtain the largest possible error in g by taking dl and dT of oppo- site signs, say dl = ^, dT = -j^. Then dg = ^ + *r a = 1.05 v* as 10.36. The ratio of error is <!l = ^ _ 2 ~ = .0005 + .01 - .0105 = 1.05%. g I T EXERCISES 1. Calculate the numerical difference between A and dz wlien s=s 4 icy a 8 y a , a = 2, ?/ = 3, Aas = cZaj s= 01, aud Ay = dy =.001 2. An angle < is determined from the formula < = tan" 1 ^ Toy x measuring the sides x and y of a right triangle. If x and, y are found to be 6 ft. and 8 ft. respectively, with a possible error of one tenth of an inch, in measuring each, iind approximately the greatest possible error in <. 3. If C is the strength of an electric current due to an electro- motive force E along a circuit of resistance R, by Ohm's law 188 PARTIAL DIFFERENTIATION If errors of 1 per cent are made in measuring E and R, find approximately the greatest possible percentage of error in com- puting C 4. If F denotes the focal length of a combination of two lenses in contact, their thickness being neglected, and / x and / 2 denote the respective focal lengths of the lenses, then I = l , i * A /.' If / x and/ 2 are said to be 6 in and 10 in respectively, find approx- imately the greatest possible error in the computation of F from the above formula if errors of .01 in. in / t and 0.1 in. in f z are made 5. The eccentricity e of an ellipse of axes 2 a and 2b (a > I) is given by the formula . The axes of an ellipse are said to be 10 ft. and 6 ft. respectively. Find approximately the greatest possible error in the determination of e if there are possible eriors of .1 ft. m a and .01 ft. in 5. 6. The hypotenuse and one side of a right triangle are respectively 13 in and 5 in. If the hypotenuse is increased by .01 in., and the given side is decreased by 01 in , find approximately the change in the other side, the triangle being kept a right triangle 7. The horizontal range R of a bullet having an initial velocity of v , fired at an elevation a, is given by the formula Ji = 9 Find approximately the greatest possible error in the computation of R if v = 10,000 ft per second with a possible error of 10 ft per second, and a 60 with a possible error of 1' (take g = 32). 8. The density D of a body is determined by the formula * , w w' where w is the weight of the body in air and w 1 the weight in water. If w = 244,000 gr. and w'= 220,400 gr., find approximately the largest possible error in D caused by an error of 5 gr. in w and an error of 10 gr. in w', HATE OF CHANGE 189 QJ7 64. Rate of change. The partial derivative - gives the rate ox of change of / with respect to x when x alone varies, and the partial derivative -j- gives the rate of change of / with respect to y when y alone varies. It is sometimes desirable to find the rate of change of/ with respect to some other variable, t. Ob- viously, if this rate is to have any meaning, x and y must be functions of t, thus making/ also a function of t. Now, by 11, the rate of change of / with respect to t is the derivative . at To obtain this derivative we have simply to divide df, as given by (12), 63, by dt, obtaining m this way dt 8x dt By dt The same result may be obtained by dividing A/, as given by (11), 63, by At and taking the limit as At approaches zero as a limit. Ex. 1 If the radius of a right circular cylinder is increasing at the rate of 2 in per second, and the altitude is increasing at the late of 3 in per second, how fast is the volume increasing when the altitude is 15 in. and the radius 5 in ? Let V be the volume, r the radius, and h the altitude. Then _ , y W ' dt ~ dr dt dh dt dr . nd7i + 7ir a -~. dt dt By hypothesis, ~ = 2, ~ = 3, = 5, 7t = 15. Therefore ~ = 375 IT cu. in. T dt dt dt per second. The same result may be obtained without partial differentiation by ex pressing V directly in terms of t. Foi, by hypothesis, r = 5 + 2 1, 7i =15 + 3 1 if we choose t = when r = 5 and 7i = 15. Therefore 7= (8 75 + 375 t + 120 i a +12 * 8 )7r; whence ~ = (375 + 240 1 + 36 2 ) *-. When t 0, *- = 375 TT cu. in. per second, as before. dt 190 PARTIAL DIFFERENTIATION Ex. 2. The temperature of a point in a plane is given by the formula 1 The rate of change of the temperature in a direction parallel to OX is, accordingly, &*__ 2* to ~"~ (a: 2 + jr 2 ) 2 ' which gives the limit of the change in the temperature compared with a change in x when x alone varies Similarly, the rate of change of u in a direction parallel to OF is Jw _ 2 ?/ Suppose now we wish to find the rate of change of the tempei ature in a direction which makes an angle a with OX From Pig. 82, if P^(x v y^) is a fixed point, and P (x, y) a moving point _ on the line through P x making an angle a with OX, and s is the distance P^P, we have p P! R = x x 1 jo whence x = x t + s cos a, " y = y^ + s sin a, ^. , dx dy and - = cos a, ^ = sma. PIGI 82 Replacing t by s in formula (1), and substituting the values of -- and dii d s ~ which we lust found, we have as du du du = cos + sin a as dx dy _ _ 2 x cos ex + 2 y sin ex Formula (1) has been written on the hypothesis that x and y are functions of t only. If x and y are functions of two vari- ables, t and s, and (1) is derived on the assumption that t alone varies, we have simply to use the notation of 61 to write at once a which may also be written as dt dx dt %y dt GENERAL EXERCISES 191 EXERCISES 1. If 2 = e tan x } x = sin tf, y = cos t, find the rate of change of with respect to t 1 -f- x 2. If = tan" 1 -: ~ ) x = sin t, y cos t, find the rate of change of with respect to t when t jL 3. If V= (a * 6""*) cos ay, prove that V and its deiivatives m any direction are all equal to zero at the point ( 0, ) 4. If F= ) find the rate of change of V at the point Va; 2 + y z (1, 1) in a direction making an angle of 45 with OX. 5. If the electric potential F-at a,ny point of a plane is given by the formula V= In Vcc a + y 2 , find the rate of change of potential at any point (1) in a direction toward the origin ; (2) in a direction at right angles to the direction toward the origin B. If the electric potential F at any point of the plane is given -\/(g> _ ff*) 2 + W 2 by the formula F= In , * ? find the rate of change of V(o;H-a) 2 -|-y 2 b potential at the point (0, a) in the direction of the axis of y, and at the point (a, a) in the direction toward the point ( a, 0). GENERAL EXERCISES , TJ! xy 1 80 fa A 1. If = sin , . > prove x -z -- y 5- = 0. xy + 1 i ox ^ oy "i /?/ /i 2 ' If = S 3. If at = f + ye* prove oj 2 -f y r\2 O2 4. If # = e-y cos a (7c cc), prove that jr-j 4- ^ = 0. B. If 3 = e-c-wx s i n / CCj p rov e that f t = <*.- 1 6. If * = e-**sin (TO?/ + x Vft 2 , 2 - 7c 2 ). prove that v ' x * 2 P a W' rtrr fc /IN *T. * ^^ , ! Sfr , 1 A 7. If F= <?**cos (a In r), prove that -5-5- 4- -~ H- 1 -5-75 == 0. N ' or r 0>' ir 0* 192 PAETIAL DIFFEKENTIATION 8. A right circular cylinder has an altitude 8 ft. and a radius 6 ft. Find approximately the change in the volume caused by de- creasing the altitude by .1 ft. and the radius by .01 ft. 9. The velocity v, with which vibrations travel along a flexible string, is given by the formula where t is the tension of the string and m the mass of a unit length. of it Find approximately the greatest possible error in the compu- tation of v if t is found to be 6,000,000 dynes and m to bo .005 gr. per centimeter, the measurement of t being subject to a possible error of 1000 dynes and that of m to a possible error of .0005 gr. 10. The base AB of a triangle is 12 in. long, the side AC is 10 in., and the angle A is 60. Calculate the change in the area caused by increasing A C by 01 in and the angle A by 1. Calculate also the differential of area corresponding to the same increments. 11. The distance between two points A and B on opposite sides of a pond is determined by taking a third point C and measuring AC = 90 ft , BC = 110 ft , and BCA = 60. Find approximately the greatest possible error in the computed length of AB caused by possible errors of 4 in. in the measurement of both AC and BC. 12. The distance of an inaccessible object A from a point B is found by measuring a base line B C 100 ft , the angle CBA =s<x= 45, and the angle BCA = 0= 60. Find the greatest possible error in the computed length of AB caused by errors of 1' in measuring both a and 13. The equal sides of an isosceles triangle are increasing at the uniform rate of .01 in. per second, and the vertical angle is increas- ing at the uniform rate of .01 radians per second. How fast is the area of the triangle increasing when the equal sides are each 2 ft. long and the angle at the vertex is 45 ? 14. Prove that the rate of change of * = In (a; -J-Vjc a + f) in the direction of the line drawn from the origin of coordinates to any point P(x, y) is equal to the reciprocal of the length of OP. 15. The altitude of a right circular cone increases at the uniform rate of .1 in per second, and its radius increases at the uniform rate of .01 in. per second How fast is the lateral surface of the cone increasing when its altitude is 2 ft. and its radius 1 ft.? GENERAL EXERCISES 193 *t - ffi ~\ JL* > 16. Given = tan" 1 ; 1- tan" 1 Find the general expres- sion for the derivative of along the line drawn from the origin of coordinates to any point. Find also the value of this derivative at the point (1, 1). 17. In what direction from the point (3, 4) is the rate of change of the function tt kzy a maximum, and what is the value of that maximum rate ? 18. Find a general expression for the rate of change of the func- tion u s= &-v sin x + - e-*v sin 3 cc at the point ( -77 ) ). Find also the 3 \ 3 / maximum value of the rate of change. CHAPTER IX INTEGRATION 65. Introduction. In 18 and 23 the process of integration was defined as the determination of a function when its deriva- tive or its differential is known. We denoted the process of integration by the symbol /; that is, if then Cf(x) dx = F(x) + C, where C is the constant of integration (18). The expression f(x) dx is said to be under the sign of inte- gration, and/(a;) is called the integrand. The expression J<\J'} + (' is called the indefinite integral to distinguish it from the definite integral defined in 23. Since integration appears as the converse of differentiation, it is evident that some formulas of integration may bo found by direct reversal of the corresponding formulas of differentia- tion, possibly with some modifications, and that the correctness of any formula may be verified by differentiation. In all the formulas which will be derived, the constant C will be omitted, since it is independent of the form of the integrand; but it must be added in all the indefinite integrals found by means of the formulas. However, if the indefinite integral is found in the course of the evaluation of a definite integral, the constant may be omitted, as it will simply cancel out if it has previously been written in ( 23). The two formulas / * I cdu = c I du (1) ft and i >, - - .^ / 194 ](du + dv + dw-\ ) = Cdu + Cdv + Cdw + . (2) INTEGEAL OF w* 195 are of fundamental importance. Stated in words they are as follows : (1) A constant factor may be changed from one side of the sign of integration to the other. (2) The integral of tJie sum of a finite number of functions is the sum of the integrals of the separate functions. To prove (1), we note that since c du d(cu), it follows that I cdu = I d (cu) cu = c I du. In like manner, to prove (2), since du -H dv + dw + . . . = d(u + v + w +...), we have / (du + dv + dw H- ) = I d (u + v + + ) u + v + w+ I du -H I dv + I dw + . . . The application of these formulas is illustrated in the follow- ing articles. 66. Integral of u". Since for all values of m except m = /w m \ or d )=w m ~ \m/ it follows that Cu m ~ l du = . J m Placing m == n +1, we have /or aZZ values of n except n s= 1. In the case %= 1, the expression under the sign of inte- gration in (1) becomes , which is recognized as Therefore T = lnw. (2) 196 INTEGRATION In applying these formulas the problem is to choose for u some function of x which will bring the given integral, if pos- sible, under one of the formulas. The form of the integrand suggests the function of x which should be chosen for u. Ex. 1. Find the value of JY aa; 2 + Ix + - + J dx Applying (2), 65, and then (1), 05, we have fft J \ = afx z (lx + bj'xflx + cf ( ~- + k JV The fiist, the second, and the fourth of these integrals may be evaluak'tl by formula (1) and the thud by foirnula (2), wheie u = x, the icsults beiny 11 & respectively - ax 9 , - bx*, -- , and c In a.. - , o & Therefore C lax* + lx + - + -^ } dx = ~ ax* + J lx z + c In i - - + C J \ x a,*/ 3 2 x Ex. 2. Find the value of J*(z 2 + 2)ar<fo. If the factors of the integrand are multiplied together, we have f (a; 2 + 2) xdx = f(x s + 2 a;) dx, which may be evaluated by the same method as that used in Ex 1, the result being z 4 + x z + C. Or we may let x 2 + 2 = , whence 2xdx = du, so that xdx = % du. Hence 22 Comparing the two values of the integral found by the two methods of integration, we see that they differ only by the constant unity, which may be made a part of the constant of integration. Ex. 3. Find the value of C(ox z + 2 bz) a (ox + &) dx. Let GKC 2 + 2 bx = u Then (2 ax + 2 fydx = du, so that (ox + &) dx = J rfw. Hence J" (oa; 2 + 2 fcc) 8 (aa: + l)dx = C %u*du 1 / 7 1 u 4 , =s / M B du = . + C 2J 24 INTEGRAL OF M" 197 Ex. 4. Find the value of C 4 < + *] dx J ax 2 + 2 fa; As in Ex. 3, let ax z + 2bx = u. Then (2 ax + 2 b) dx =* du, so that Ilence = 2 In w + C s= 2 In (az 2 + 2 to) + Ex. 5. Find the value of C (e + Let e * + & = u. Then e ax adx = tfu. Hence J(e + 6)cdiB J^ jf If the integrand is a trigonometric expression it is often pos- sible to carry out the integration by either formula (1) or (2). This may happen when the integrand can be expressed m terms of one of the elementary trigonometric functions, the whole expression being multiplied by the diffeiential of that function. For instance, the expression to be integrated may consist of a function of smo; multiplied by cosEefe, or a function of cos a; multiplied by ( sma^a;), etc. Ex. 6. Find the value of [ Vsm x coa s xdx. Since d(smx) = oosxrfx, we will separate out the factor cosrcrfar and express the rest of the integrand in terms of sin a:. Thus -Vsinxcos s xdx = Vsirue (1 sm s ;c) (cosxdx). Now place sin x = M, and we have 198 INTEGRATION Ex. 7. Find the value of Jsec 6 2xdz. Since d(tan 2 a:) = 2 sec 2 2 xdx, we separate out the factor sec 8 2 rdc and expiess the rest of the integrand in terra a of tan 2 a: Thus sec 2 x dx = sec* 2 x (sec 3 2 = (1 + tan 8 2 a:) 2 (sec 3 2 a:r/ = (1 + 2 tan 2 2 ar +*tan*2 a;) (se Now place tan 2z=su, and we have fsec 6 2 xdx = = Jtan2a: EXERCISES Find the values of the following integrals *6 M 4- sec 3 ace tan ax + C. . C -j=\dx. C( r 1 \^ J [x-vx j=]dx. J \ x~vx/ r?**- C tffa "J^T i. C(x* + l)*xdx. I Vo 4 + 4x 8 t7a;. /. c J 2 a; -f sin 2 a;' in f 1-cosa; !. I . . .,dx. J (aj sinaj) 4 -/: 12. 13 dx. &r. sin ax rfa;. e 8 *^ e* x + & _ , 1 + COS 2 G , 9. I - - - ote. fif J 1 + cos aa 14. I cos 8 2 as sin 2 OJ$B. 16. I sin 3 3 a; cos 3 xdx. 16. / sin (x + 2) cos (a -|- 2) Ja;. 17. / cos^3 x sin 3 *c?a;. 18. I sec 4 3ic^a5. J 19. rctn 2 (2a;-|-l)cso !1 (2aj "*/' ALG-EBKAIC INTEGRANDS 199 67. Other algebraic integrands. From the formulas for the differentiation of sm~ 1 M, tan" 1 ^, and sec-'w, we derive, by re- versal, the corresponding formulas of integration : du == sin" 1 ?/, du and These formulas are much more serviceable, however, if u is / replaced by - (a > 0). Making this substitution and evident fit reductions, we have as oar required formulas du . ,u .^ N == sin" 1 - (1) <J <J /y NX a u a L _, - ON = -tan- 1 -> (2) 2 s y and + a 2 a a du /du 1 *u s Q ~ ^-sec" 1 -- (3) wV w a - a* a- Referring to 1, 47, we see that sin" 1 - must be taken in the Uf first or the fourth quadrant; if, however, it is necessary to u have sin" 1 - in the second or the third quadrant, the minus sign must be prefixed. In like manner, in (3), sec" 1 - must be taken in the first or the third quadrant or else its sign must be changed. /fix __ Letting 2 Vj) 4 a: 2 r fa _. f \& u J -y/9 _ 358 J V9 ~ M 3 /fix __ Letting 2 x = u, we have du = 2 dx ; whence dx = ^ du, and Vj) 4 a: 2 200 INTEGRATION dx Ex.2 Find the value of f , * Jf we let V:JW~ M, UH-J J jt V;} j, a - 4 t du = V3 rfj; ; whence <ir = -= f/, and Va /^j _ r __ <lu x V3 x 2 -4 ^ w vV~-~4 / rfr Ex. 3. Find the value of I /. 17 Since V4 a; js z = Vd (r 2) 2 , we may lot w =s a 2 ; whttiusH ^/.r - */, and C ^ C _ ~ t J - 2 "^ ^* S1J1, -i + C' 6 x -~ 2 Ex. 4. Find the value of C <lx J 2 .r + 3 u, + 5 We may first write the integrand in the form 1 1 = 1 1 2 aJ+x+ 2'* and let u = x + I Then du = dj., J 2 2 + 3 x + 5 = <>J (.r + ^) B +"Y = ~f 2J i 1 X s ton-'JL 4 4 _ 2 t 4tt -~r~ au v" f 5f +r - Vai / ALGEBBAIC INTEGRANDS 201 5z-2 , Separating the integrand into two fractions 5z _ 2 2o; a 4-3 2a; 2 +3' and using (2), 65, we have 2 / 5xdx r 2dx __ / J If we let rz = 2 a; 2 + 3, then du = 4 artfo 5a;c?a; 5 /vM 5, 5 and if we let u V2 x, then rfu = V2 dx , r 2dx AT / du /= 1 . . it Vo. ,a;Vo and I i4 = V2 I - = V2 - tan- 1 ^ = tan- 1 - J2a; 2 +3 Ju*+$ V3 3 3 Ex. 6. Find the value of There is here a certain ambiguity, since tan- 1 V and tan~ 1 ( 1) have each an infinite number of values. If, however, we remember that the graph of tan~ 1 a; is composed of an infinite number of distinct parts, or Iranches (Fig 56, 46), the ambiguity is removed by taking the values of tan- 1 Vs and tan- 1 ( 1) from the same branch of the graph For if we consider - 1 ^ tan- x o and select any value of tan~ 1 a, then if & = a, / | Ja a X + 1 tan~ l 6 must be taken equal to tan- 1 ^, since the value of the integral is then zero. As & varies from equality with a to its final value, tan- x & will vaiy from tan -1 a to the nearest value of tau- a &. The simplest way to choose the proper values of tan- 1 ?* and tan~ 1 is to take them both between and Then we have /'Va dx _ v __ / TT\ _ 7 ./-i a 2 + 1 ~ 3 \ i/~ 1 TT 12* The same ambiguity occurs in the determination of a definite integi al by (1), but the simplest way to obviate it is to take both values of sin- 1 7T 7T between -^ and The proof is left to the student. B 202 EXERCISES Find the values of the following integrals : r dx r dx l ' J Vl6 - 9 a? ' J V5 x - 3 a/ dx dx /dx C B a - 12. /, - /dx C (?SK 13 I a:V4a; 2 -9 ' J 3-e 3 - 4 a; + 2* ra.r + 11 x 1^ 16 ' 7 ' a ^ J 10 ' r / * 19 r J VGas-a* 19 ' J_ T ^ o J V6*-4a* 68. Closely resembling formulas (1) and (2) of the last section m the form of the integrand are the following formulas ; +o. (1) /du> -. / / ^r^^ (u+ ^ u --^ (2) and These formulas can be easily verified by differentiation and this verification should be made by the student ' ALGEBRAIC INTEGRANDS 203 /f/1" . Letting V2 a; = u, we have du = "v2dx ; whence dx = du, and ,;= du = _JL T rfu = -i- In [M + Vu*-8] + C Va = 4= ln C' ^2 + Va x a - 3] + a rfe /(/ ~ ~ Vy a B As in Ex. 4, 07, we may write tho integrand in the form Va Vx a -f- 3 a? Va V(j; + ^) a ~ $ and lot M = x + S ! whence rfw = rfx, /__.4!L _JL T - ^- ; V3 a: a + 1 x VJJ ^ V(,^ + *j) a - J V8^ Vu a ^ " vl n w "vl ~ -Lin (3 + 2 + VO a; 2 + 12 *) +K> Vi) where C = -- In 8 + A". Va / o i IK z a;" 8 + * JLO Writing the integrand in the form I ^j 1 ,..^ "S we let * x Hh 1 ; whence rf ss rfa?. 204 INTEGRATION r dx _ 1 f* <lf J 2s* + a: -15 ~ 2 J (* + i) a W" 2 2(-V) + V llnljli + C 11 a + 3 It I >1 ' 11 X + i) where C = -j^ In 2 + K. EXERCISES Find the values of the following integrals : 1 r *" 11 r~ /* ~ j.i - . * J.J.. i , - J V^T2 J 3^ + 6 2 c dx 12 r ^ J V9rf-l' Jo*-3(B + 3 r *" 13 r rfa! ' J V3 2 -4 ' J *+- * C dx C <fa ' J Va^a" ' J 4if-2* - 5. f ^ IB f* rf f- ' J V3jc a +2a3 + 3 ' J V5-""4' 6 C dx r* dx J***-M ' J VOSTT' r dx ' I oa -i' 1 ^ 2a ^- 1 '-a- 8' TRIGONOMETRIC FUNCTIONS 205 69. Integrals of trigonometric functions. Of the following for- mulas for the integration of the trigonometric functions, each of the first six is the direct converse of the corresponding for- mula of differentiation ( 44), and the last four can readily be verified by differentiation, which is left to the student. / sin udu cos M, (1) I cos udu = sin w, (2) I seu"ud (, = tan u, (8) /Qscfudu = ctn w, (4) v J sec u tan u du = sec it, (5) esc u ctn udu = esc u, (6) I tan udu = hi sec u, (7) I ctn udu = In sin it, (8} | sec udu = In (sec u -h tan it), (9^ I esc udu = In (esc it ctn u) . Ex. 1. Find the value of Tsui 7a.v7a;. If we let 11=7,1:, then du7dr; whence dx = \ du, and Cs\n7x dx = T sin (| f/u) =3 ^ I sin udu = \ cos + C* = J-cos7a? + C, 206 INTEGRATION Ex. 2. Find the value of f sec (2 x + 1) tan (2 x + 1) dx If we let w = 2 a; -f 1, then du = 2 dx, and Tsec (2 x + 1) tan (2 x + 1) cfo = f sec M tan du = ^secu + C = sec (2 a; + 1) + (7. Often a trigonometric transformation of the integrand facili- tates the carrying out of the integration, as shown in the following examples: Ex. 3. Find the value of f coa z axdx Since cos s ax - (1 + cos 2 ax), Ccos*ax dx = f ( J + J cos 2 aar) dx = Tdk + ^ Tcos 2 aacfo J i / = o x + T~ sin 2 aa: + C 1 , A T CZ the second integral being evaluated by formula (2) with M = 2 ax. Ex. 4. Find the value of f Vl + cos arcfo. Since cos x = 2 cos 2 - 1, ii Vl + cosa;=v / 2cos|, - ib and f Vl + cos xdx ~ C V^ cos - </j; Ex. 5. Find the value of Aan 2 3 xdx, Since tan 2 3a; sec a 3 - 1, Jtan 2 3 arcfa: = J*(sec 2 3 a: - 1) da: = Csec*3xdx Cdx = J-tan3a; a;, bhe first integral being evaluated by formula (3) with v*= TRIGONOMETRIC FUNCTIONS 207 EXERCISES Find the values of the following integrals : ,, 1. /sin (3 a; 2) die. 13. I cos^dte. t 2. | cos (4 2x)dx. /y 3. jpxa 14. I (sin^ + cos-j dx. 3. 1 sec(3a: l)tan(3a; l)da; -/(-- * 2aA s tan T j i 4. / sec a Tdo;. J 4 16. fain 2 ! cos a | dx. * f+ 3a; j ^6. 1 tan -%- dx. 17./^ 1+ eos = 5x 6. I ctno'ajcfa;. 18. 1 Vl COS ' 7 .Jcse ( 2* + s)<fc. c* . 19. I sin oxdx. Jo S-Jcscfctnl^ C\ ai 20. I tan 75 aa;. Jo 2 9. I sec(4a; + 2)<c. Tw / 21. J tan ( cc "|~ v/O ^ .=)*. 22. a;. v'10. Jose 9 (3 2 a:) da rr ,, /*cos2o; , 23. I ( .yll. I : dx. ] ir * J smaj J~i r . * r*i 12. I sin a 7rdaj. 24. J J 2 Jis 70. Integrals of exponential functions. The formulas / It J M XI \ and | a v du*=-. a* (2) J In a are derived immediately from the corresponding formulas of differentiation. 208 INTEGRATION Ex. 1. Find the value of C$ x djc If we let 3 x = u, we have Ce 8x dx /*/B JLrrfa;. x z x i i If we place V 5 = 5"" and let = u, we have . _/*. EXERCISES !Fmd the values of the f ollowing integrals : 1. ie^^dx. 6. f(a + -)<&. 9. f 10* dx. //g2x _ a-2as / e^ajtfo;. 6. / _ a , rf. 10. / 2 00 11. C\**-*dx. . f V ~ *"'rfaj. Jo ^+0"* 8. 4. e^f+txfa Qt +e"dx. 12 71. Substitutions. In all the integrations that have been made m the previous sections -we have substituted a new vari- able M for some function of x, thereby making the given integral identical with one of the formulas. There are other cases in which the choice of the new variable u is not so evident, but in which, nevertheless, it is possible to reduce the given integral to one of the known integrals by an appropriate choice and wh> stitution of a new variable. We shall suggest in this section a few of the more common substitutions which it is desirable to try, I. Integrand involving powers of a + U. The substitution of some power of 2 for a -f- bx is usually desirable. SUBSTITUTIONS 209 Ex. 1. Find the Value of f ^^ , . J (1+2*)* Here we let 1 + 2 x = s? ; then x = ^ ( 8 - 1) and dx = % z*dz. Therefore C x * dx = - f(z< - 2 * + z) tls 17 (1+2 a?)* 8J Eeplacing a by its value (1 + 2 #)* and simplifying, we have = (1 + 2 B) * ( " 12 * + 2 ^ + c ' II. Integrand involving powers of a + ?;#". The substitution of some power of for a + frr 71 is desirable if the expression under the integral sign contains x n ~ l d& as a factor, since = bnx n ~ 1 cfo. /x/j.2 J. x 2 rfa;. a; We may write the integral in the form and place a a + o 2 = a . Then xdx = sdz, and the integral becomes Replacing s by its value in terms of x, we have x 2 Va a + a 2 + a Ex. 3. Find the value of fjs 6 (l + 2 spfidx. We may write the integral in the form and place 1 + 2 a; 8 = 2 . Thou xhlx = ifisdz, and tho now integral in z ia fr J(*< - * a ) ds = n^ s (3 * 2 ~ 5) + C. Replacing z by its value, we have Cx 6 (1 + 2 a)icfa ~ ^ (1 + 2 3 8 )8 (3 a 8 - 1) + C. X 210 INTEGRATION III. Integrand involving Va 2 ar 2 . If a right triangle is constructed with one leg equal to x and with the hypotenuse equal to a (Fig. 83), the substitution x = a sin is suggested. Ex. 4. Find the value of fVa 8 a: 2 da;. _, gg Let a; = a sin <. Then dx = a cos <f> d<f> and, from the triangle, V 2 a; a = a cos <. Therefore C^/a?x*dx = a 2 f cos 2 But and for, from the triangle, sin d> = - and cos <& = a a Finally, by substitution, we have J 2 \ a/ IV. Integrand involving V^+a 3 . If a right triangle is constructed with the two legs equal to x and a respectively (Fig. 84), the substitu- tion x = a tan <f> is suggested. Ex. 5. Find the value of f dx i/ /* o * ^o\ $ Let i = a tan < Then dx = a sec 8 <tf< and, from the triangle, V# a + a a = a sec <f>. Therefore f But, from the triangle, sin <f> = * ; so that, by substitution, Va; 8 + flr a * SUBSTITUTIONS 211 V. Integrand involving V# a a 2 . If a right triangle is constructed with the hypotenuse equal to x and with one leg equal to a (Fig. 85), the substitution x = a sec <#> is suggested. -*f r r a Ex. 6. Find the value of / a 8 Vo: 2 a?dx " ^^ ge Let as = a sec <. Then dx = a sec < tan < d< and, from the triangle, Va; 2 - a 2 = a tan Therefore fa 8 Vj; 2 - a 2 dx = a = a G J"(tan 2 </> + tan 4 <) sec 2 .J. _ j Exit, from the triangle, tan <j> = , so that, by substitution, we have J> V* 2 - a a rf.c = T V (2 a a + 3 a; 2 ) V(a; 2 - a 2 ) 8 + C We might have written this integral in the form fa; 2 Va; fl cPfxdx) and solved by letting z a = a; 2 a 2 . 72. If the value of the indefinite integral is found by substitu- X6 f(x)dx may be performed in two ways, differing in the manner in which the limits are substituted. These two ways are shown in the solutions of the following example: Ex. Find f Va"--.ida;. Jo By Ex. 4, 71, Va x*dx = i/a Va - x* + Therefore C " 212 INTEGRATION Or we may proceed as follows Let x = a sin < When x = 0, = ; and when x = a, < = , so that < varies from to - as x vanes fiom to a. Accordingly, " v 4 The second method is evidently the better method, as it obviates the necessity of replacing z in the indefinite integral by its value in tenna of x before the limits of integration can be substituted. EXERCISES Find the values of the following integrals : 1. \ X . * 6. I ~~ . 11 '/(*-* 7. r **< J (3- /, \J & 9. f * J (V 5 - 5 - I . . , 10. I aV2a; $dx. 15 73. Integration "by parts. Another method of importance in the reduction of a given integral to a known type is that of integration by parts, the formula for which is derived from the formula for the differential of a product, d (uv) = udv + v du. From this formula we derive uv = / udv + f vdUy which is usually written in the form I udv = uv I vdu. INTEGEATION BY PAETS 213 In the use of this formula the aim is evidently to make the original integration depend upon the evaluation of a simpler integral. Ex. 1. Find the value of Cxe*dx, If we let x = u and cPdx = dv, we have du = dx and u = e*. Substituting in our formula, we have j x&dx = are 1 j e x dx It is evident that in selecting the expression for dv it is desirable, if possible, to choose an expression that is easily integrated. Ex. 2. Find the value of Csm Here we may let sin- 1 x = u and dx = do, -whence du = rx and v = x. Substituting in our formula, we have "^ 1 ~~ x * xdx sin-i-xdx = x sin -1 a: { J Vl-o; 2 = x sin- 1 ^ + l a; 2 + C, the last integral being evaluated by (1), 66 Sometimes an integral may be evaluated by successive inte- gration by parts. Ex. 3. Find the value of Cx*e*dx. Here we let x z = u and eFdx = do. Then du ^xdx and = e 31 . Therefore C The integral f xeF dx may be evaluated by integration by parts (see Ex. 1), so that finally Ja;V= dx = x*<*> - 2 (ar - 1) e* + C = eP (a; 2 - 2 x + 2) + (7. Ex. 4. Find the value of je" sin&zda:. " sin bx = u and' c** rfz = dv, we liave = - *<"* sin &K fc 8 * cos Ba; ?. a v 214 INTEGRATION In the integral Ce 01 cos bxdx we let cos bx = u and e ax dx dv, and havi /I b /* e"* cos bxdx - e"* cos fa + - / e** sin Sards. a a w Substituting this value above, we have /e"* sin bxdx = - e 01 sin &e -- ( - eP* cos &r + - fc * sin bx dx\ a a\a aJ ) Now bringing to the left-hand member of the equation all the terms containing the integral, we have (1 + ) fe * sin bxdx = - e"* sin bx -- -e 031 cos bx, \ a?/ J a a? t. r __ , , fi** (a sin bx b cos bx) whence I e"* sin bxdx = v ' J a z +b* Ex. 5. Find the value of fVa; 2 + a?dx. g Placing Va; 3 + c 2 = w and <?a; = fo, whence rf = we have Va; B + o a and w = a, Since a; 2 = (a^ + a 2 ) a 2 , the second integral of (1) may be written as /a; 2 + a 2 which equals z 2 + a s /* Evaluating this last integral and substituting in (1), we hare whence JVa; 2 + a?dx =s $ [a Var 2 + a a + a a In (x + 74. If the value of the indefinite integral f/O) ** is found by integration by parts, the value of the definite integral r* I f(z)dv may be found by substituting the limits a, and I in va the usual manner, in the indefinite integral. INTEGRATION BY PARTS 215 IT Ex. Find the value of C*x*sinxdx. Jo To find the value of the indefinite integral, let a; 2 = and sin a; da; = dv. Then J a; a sm xdx = a? 2 cos x + 2 J x coaxdx. In / a; cos a, dx, lot a; = u and cos xdx = do. Then \x cos x dx x sin x i sin x dx = x sin a; + cos x. Finally, \ve have I x*ainxdx = ar a cosa, + 2 a? sin a: + 2 cos a; + C. IT "" P Ha Hence C*j? sinarda; a; 2 cos x + 2 x sin x + 2 cos a; = -2. The better method, however, is as follows: . l$.f(x)dx is denoted by udv, the definite integral \ f(x)dx stb Jo. may be denoted by I udv, where it is understood that a and b t/a are the values of the independent variable. Then /& . s*b I udv=z\uv~\ n I vdu. Jo, v a To prove this, note that ib follows at once from the equation , /^6 />fc /*! /& ** / / / / i/a yo /o vo Applying this method to the problem just solved, we have IT v r -13 ir r'5'a; a sina?afa;= a; 2 cosa; + 2 C^xciOBxdx /* * j Jo ["|J W 2a;sina: 2 C*&inxdx w STT-f 2 cos a: I L Jo 216 INTEGRATION EXERCISES Find the values of tho following intftfra . CynP'tlr. 5. I .rscc" 1 !}^//^. 9. I ./ *-V.r. . I stP<i* L (?x- 6. I (In Hindoos ;'(/.r. 10, / ,i"'lu.i'f/>. . COOB- I X<?JK. 7. I/' 9 "Von .r /.i'. il. / Mr "J, !/,!. | tair~ 1 3a:</J3 8. / ,r uos 3 '^ <Ar. 18, / VPOH JJuv/.i". 4. 75. Integration of rational fractions, A rtttwixtt fwrfiHit i.s it fraction whose numerator and denominator ur polynomials. Il can often be integrated by expressing it UH tho twin of fractions whoso denominators are faot-oi'H of th of the original fraction. Wo .shall illustrate only tin* inw in which the degree of the numerator is lews thun tho th'gm* of tin* denoiuinator and in which the ftu'torH of tho tlmitmunutur m* all of the first degree and all different, Ex. Find tho value of f-~^ f ^}'\" </'' The factors of tho denominator arc x + tt, x 2, a<l x + 2, Wo ftHwtmo (x + 8)(j- 4) ""a + a-B^ + a whojo J, /?, and ^7aro constants to bo (lotonniucd. Clearing (1) of fmclions by uiuUiplyuig by (ja + n)(x 9 4), w or l, /J, and (7 firo to bo datidrmmnd so thai Iho righWrnml memhr of (3) .shall bo identical with th k'ft-hand nuwnbcr, lhn <'<*iHj5iHtu( wf powers of a: on the two sides of tho equation muni Iw tujual. Therefore, equating the ooeffieimitH of liku powors of / in (Jl), theequatzons - 1 A + <t 7i ~ G ~ li, whence we find ^t = ~ 2, Ji 2, r, Y = 1, RATIONAL FRACTIONS 217 Substituting these values in (1), we have a? 8 + 11 x +14 _ 2 2 1 = - 2 In (x + 3) + 2 In (a; - 2) + In (a, + 2) + C EXERCISES Find the values of the following integrals i r a;+i 7 A r *' J ?-"o7+8 '^ 4 ' J ( 2) (*- Bas+l T __ 2 -5a! + 5 T ' J ( iC -.l)(a J ~2)( a! -3) rfa; - 6> J 76. Table of integrals. The formulas of integration used in this chapter are sufficient for the solution of most of the prob- lems which occur in practice. To these formulas we have added a few others. In some cases they represent an integral which has already been evaluated, and in other cases they are the result of an integration by parts. In all cases they can be verified by differentiating both sides of the equation. These collected formulas form a brief table of integrals which will aid in the solution of the problems in this book. It will be noticed that some of the formulas express the given integral only in terms of a simpler integral. I. FUNDAMENTAL 1. | cdu=sc I du. 2. I (du + Av + dw ) = I du + | dv + I dw . 8. / udv sauv I vdu. 213 INTEGRATION II. ALGEBRAIC 4. / u n du = . (n = 1) J n + l C du 5. I = mu. du 1 p^ = ln r A I ' J _ r du _ 1 , u a ' \ ~o 5 o ^ ! ' J u a za u-\-a 8. / vfls u du = ( u ~\a? J V 2V 9. JWa 2 -w a dM = -i.O a _t ju^a*-u*du=>- w + 2 sm~- 19. w 10. 12. 13. I __ / t>. n n / . _ 14. 15. 16. /du Va 2 - fl 2 /* L s* I u^/u z a z du = % (w 2 TABLE 219 _. du 1 ,M 20. I . =-seo- 1 - /, . I V2 au u*du = - (w a) V2 au w 2 + a 2 sin" 1 /. 22. / t ""* = sm- 1 - V2 aw - w 2 a III. TRIGONOMETRIC 23. I smudu=x cosw. ft . /* . , , u 1 24. I sm waw == - -r sm 2 w. J 24 /I w, 1 /* sm n udu = sin""^ cos w H I sin"" 2 wcZM. (n 3= 0) w w J ^ J 26. | cos udu SB sin w. /w 1 cos 2 wdw == - + T sill 2 w. 2 4 /I 7i _ i / coa*udu BBS - cos n - 1 w sin w H I cos" ~ a w^. (w ^ 0) n n J 29. I tan udu In sec w. /tan'*"''^ C H /\\ tan w w?w s= rf I tan w "" udu. (n l^* 0) n-1 J 81. f ctnw rfw sss In sin w. /r*i"n " *** i/ /** ctn n w^ - r - I ctn 11 - 3 wc?w. (n - 1 + 0) n-1 J 88. I sec wrfti a= In (sec t* + tan w). 34. Jseo'wtfw =s tan M, 85, Jcso WC^M In (cso w - ctn ). 86, I cso a wc?'M! ctaa w. 220 INTM! RATION 37. I sec u tau u du HUC w. 38. I oso ? ctn w tft* = <IBO ?f. /" / /Hin mfl ?M'0,s" 1 U H - I r . Hill m WCt)S n ?/,(Zjts=S" '-'-.-,} / ,sh'"M riLS* "'/0/, m + w / 1 ><</ ( i i it a ) I || I U 1 |'ll"*1/ I"*/ VU*' O/ //')/ " - - j., U4_ i I Wtll // 4*1 i 1 !* J/ */Jjf * v I niix c* ^v'Ci (v vvCv "*-"*^ T i Ji i * F**' I ~ *i** J si , n I , j M111 41 . I Hill"' W C50S" U dlt,*x 42. Tsii m + 1 IV. EXPONENTIAL r<'"?MM". 4. Ca*du**-~a H . J In a 43. 44. 45. 46. A* , I WCOH" H M < H- 1 {/ f 1 / > GENERAL EXERCISES Find the vaJuos of tho following mtgralH t I. J (3 aj 8 4- 4 - ^ 3 ~ ^ flfaj. 4, f(j f . f2aj a J GENERAL EXERCISES 221 7. f(2- 9. f i+ * 2 * w- r J (3x + xrf J r^-. 27. r J as - 1 J ( // n-(8.-l)orf(8.-l)*, " J ( IS. 13. dx dx /C dx GSG^4; C l/C 30 I * / "\/ 1\ 'V* r - - 11 '>* i/ v ti a/ " t*/ 11. fsec 8 (-2Han 8 (:-2)daJ. 31. C , dx J V ' J V34-2aj-o; a 15. / ctn (a; 1) sec 4 (a; 1) dx. 32 . | J J W . 1C. f oso 8 2ajctu s 2a;^sB. 33. J 17. J tan'Saj-yseoSa-rZa-. 34 . T- & /H -U-Jlu-.iniLJ.UI (itn 2 IB V(!HO '2xdx. 35. /-.. -.-.IIU.J. rt CMC* fiaJ"V(!t,u 5 "/?. 3g / _ * I . .<!() J ' /* **" *? *"" $$, o*y I siu*4aj " J 4. s /CHC 4 5 C /* ^ 38t / v tan 8 Sir J >!' M.J- *. v 39 '/9> + 1 19 20. 222 INTEGRATION 58. * 54. 55, / 42. I 69 ' / 9 a -7' /dx x Vcc* 6 60. / f . ,y oar 4 c dx J x Vce 6 4 61 -/4*6^ 45. I J Vl~12a; 2 -4aj 4 "/Wl"* 46. fl^T^dx r dx ., r 4-3oj , 47. 1 . dx J V9-2a; 2 -fl7C /dx &5 'J9x*-(>x- Va: 2 -7 49. f dx J V4a: 2 +3 661 J 4a; 2 -4aj- O.J V3 * . 67. r 2 da; ' J V4a; 4 -5' i dx 52 f f*^ . J Vo; 8 +7 69. | / 5 2 + 2x '. 53. / --'*,&, 70t / ^ / /^fal "71 I V3a: 2 + l ' /J "J /<a; ^ V2a; a - 3aj' 72 ' J seo(a!- |)<fa. 56. /^ dx _ ^8^43;^ / v4.'r a _L /Lo _l_7 ' d< I ?T~ <W5> * yv * a; T*a:-f-< ^y cos 2 x /OnK SI t . . w . i / sin 2 a/ cos /w a / 4 I I -- _^ j_ "V n <T^ _I_ /Lrw'i * ^ * I I , ***"' ' " voa+4a;-l J \ Sinas COSJB 75. Cam*~dx GENERAL EXEECISES 91. Cx5*+*dx. 223 . ri 008405, 76. / da;. J l+cos4aj 92. 1 xVda;. /" cos4a; 7 77. j da;. J cos 2 x sin 2 a; /cc a? tan" 1 ^*- 78. da;. x x 94. 1 aj 3 sm3fl3daj. i cso ~ H~ ctn jr 79. 1(860*2 a? tan*2 a;) <fcc. 80. I Vl + sm 2 a; da;. 81. 1 Ve s daJ. 95. I (In 2a:) a da5. 96. | In (3 ; + V9o; a 4) 97. r a xdx J ' 82. I xtyefidx. M -/"i-". * 88 -/i- ,/ "i s*. r "^ '. 100. 1 J (ox -j~l)* 86. 1 a; 6 *^a5 8 + 2 das. /"a; 8 -6 j a -9 x + 24 i/ "~" """ ow 86. 1 ^ - 102. r a y -?~-- Ja V aj a - 9 87. r 2 ^ 1 - 103 ' Ji V4 - a; 8 * 8s - r r J (l-aj 3 ) 8 ^jTe^.-.- / 3 daj 105. 1 no' H i ' ft' 89. 106. 224 107. INTEGEATION ptan(as- J \ Q na 108. 109. 110. 111 112 Bx , Bx 114. | i/S * . a; V a; 8 4- 9 115 - 116. 118. ofo: JV t/Q I ~~zz" J* e J^J* g tan-la 119. 1 *sQ 120. 1 Jo l+x* dX /a" JP^tC 4 123. I acHarr^cfaj. . 1 Jo CHAPTER X APPLICATIONS 77. Review problems. The methods in Chapter III for de- termining areas, volumes, and pressures are entirely general, and with our new for- y mulas of integration we can now apply these methods to a still wider range of cases. Ex. 1. Find the area of the x* i/" ellipse -^ + ~ = 1. It is evident from the sym- metry of the curve (Fig. 86) that one fourth of the required aiea is bounded by the axis of p IG< gg y, the axis of x, and the curve. Constructing the rectangle MNQ,P as the element of area dA, we have dA**ydx = --VtP-aPdx. a Hence 5ST Ex.2. Find the area bounded by the ~ axis of x, the parabola y 9 =a kx, and the straight line y + Sff-frssO (Fig. 87), c The straight line and the parabola intersect at the point C ( j s ) > and /, \ \4 -i/ the straight line intersects OX at B ( - , j . Draw CD perpendicular to 0Z. If we construct the elements of area as in Ex, 1, they will be of different 226 APPLICATIONS form according as they aie to the left or to the right of the line CD] fc on the left of CD we shall have dAydx k*x*dx, and on the right of CD we shall have dA = ydx = (k 2 a:) dx It will, accordingly, be necessary to compute sepaiately the areas ODi and DBC and take their sum. Area Area DBC = J* 3 (X: - 2 x) dx = \kx - a; 2 ! 2 = & k*. 1 4 Hence the required area is 7 S k s It is to be noted that the area DBC since it is that of a right triangle, could have been found by the formulas ^ of plane geometry ; for the altitude D C = | and the base DB = l~^-~, 2 z 2 4 4 and hence the area = 16 Or we may construct the element of area as shown in Fig 88 Then, if x 1 and z 2 are the abscissas respectively of P a and P t , . 2 4 3k 48 Ex. 3, Let the ellipse of Ex 1 be represented by the equations x = a cos <f>, y = l sin <. Using the same element of area, and expressing y and dx in terms of we have , , ,, . .... dA = (b sin </>) ( a sin As x vanes from to a, </ varies from to ; A hence =4 f t/O =- 4 EEVIEW PROBLEMS 227 It is evident from foimula (1), 23, that the sign of a definite integral is changed by interchanging the limits. Hence A = irab Ex. 4. Find the volume of the ring solid generated by revolving a ciicle of radius a about an axis in its plane b units fiorn its center (b > a) Take the axis of i evolution Y as OY (Fig. 89) and the line through the center as OX. Then the equation of the ciicle is K (r-&) 2 + y 2 =a 2 . ' M"-- A straight line parallel to OX meets the circle in two points . P v where a, = x i b Va 2 y*, and P 2 , where x = x z = b + Va 2 ij\ A section of the requned solid made by a plane through P 1 P 2 perpendicular to OY is bounded by two concentric circles with radii MP l = x v and M P z = x z respectively. Hence, if dV denotes the PIG. 89 element of volume, dy= The summation extends from the point L, wheie y = a, to the point If, where y = a. On account of symmetry, however, we may take twice the integral from y = to y = a. Hence V- 2 C Jo PIG. 90 Ex. 5. Find the pressure on a parabolic segment, with base 2 b and altitude a, submerged so that its base is in the surface of the liquid and its axis is vertical Let RQ.C (Fig. 90) be the parabolic segment, and let CB be drawn through the vertex C of the segment perpendicular to RQ, an the surface of the liquid. According to the data, RQ = 2 ft, CB-a. Draw LN parallel to Tti, and on LN as a base construct an element of area, dA. Let CM=x. 228 APPLICATIONS Then dA *=(LN)ilje. r> , * But, from 30, whence a 2/; I and therefore dA = - <to. * The depth of Zi^V below the surface of the liquid is CH ( W - ,r ; hence, if 10 is the weight of a unit volume of tho liquid, tip sz 1 $* (a x) wdxt a* , .. r a 2biii i, x , and P - { r j? 1 ( - .r) of.c Jo EXERCISES 1. Find the area of an arch of tho curve y = sin *. w / * _ > \ 2. Find the area bounded by tho catenary y = A v'"Hh " % fcliw axis of a?, and the lines r = /*. 3. Find the area included between tho curve ?/ = - .. * . ., and .... * 7 r + -1 rt^ its asymptote. 4. Find the area of one of the closed figurus boundtul by Uit* curves if = 16 a? and ?/ 2 = sc 8 . 6. Find the area bounded by the curve 2/ a sss2(ce 1) and tilio Iine2a; 3?/ = 6. Find the area between the axis of SB and ono arch of tht cycloid a; = a(j!> sin ^), y = a(l COH 0). 7. Find the volume of tho solid generated by revolving about OY the plane surface bounded by OY and the curve* a$ -j- ^ s 3, 8. Any section of a certain solid made by a piano pM'pwulinulnr to OX is an isosceles triangle with tho ends of its busa r8tint? on Ihw ellipse 1 4- '4 = 1 an d its altitude equal to tho distance of tho plane from the center of the ellipse. Find the total volume o tho Bulid, 9, Find the volume of the solid formed by revolving about tho line 2 y + a the area bounded by 0110 arch of tho curve y m sin as and the axis of x. REVIEW PROBLEMS 229 10. Find the volume of the solid formed by revolving about the line y + a = the area bounded by the circle a 8 + t/ 2 = a 2 . 11. Find the volume of the solid formed by revolving about the line x a the area bounded by that line and the curve ay 2 = x 3 . 12. A right circular cone with vertical angle 60 has its vertex at the center of a sphere of radius a Find the volume of the portion of the sphere included in the cone. 13. A trough 2 ft. deep and 2 ft. broad at the top has semielliptical ends If it is full of water, find the pressure on one end. 14. A parabolic segment with base 18 and altitude 6 is submerged so that its base is horizontal, its axis vertical, and its vertex m the surface of the liquid Find the total pressure. 15. A pond of 15 ft. depth is crossed by a roadway with vertical sides. A culvert, whose cross section is in the form of a parabolic segment with horizontal base on a level with the bottom of the pond, runs under the road. Assuming that the base of the parabolic segment is 4 ft. and its altitude is 3 ft., find the total pressure on the bulkhead which temporarily closes the culvert. 1. Find the pressure on a board whose boundary consists of a straight line and one arch of a sine curve, submerged so that the board is vertical and the straight line is in the surface of the water. 78. Infinite limits or integrand. There are cases m which it may seem to be necessary to use infinity for one or both, of the limits of a definite integral, or in which the integrand becomes infinite. We shall restrict the discussion of these cases to the solution of the following illustrative examples : Ex. 1. Find the area bounded by the curve y = -^ (Fig. 91), the axis of x, and the ordinate x = 1. x It is seen that the curve has the axis of x as an asymptote ; and hence, strictly speaking, the required area is not completely ^ bounded, since the curve and its asymptote do not intersect. Accordingly, in Fig. 91, let OM=l and 0JV = ft (& > 1) and draw the ordinates MP and ' If the value of b is increased, the boundary line NQ, moves to the right; and the greater b becomes, the nearer the area approaches unity. 230 APPLICATIONS We may, accordingly, define, the area boundt'd by the eune, Ihe u\i* of r, and the oidinate .r = 1 as tlw hunt of the area Mj\Ql*\w It tm'tcnst's indeti nitely, and denoto it by the symbol -'f = Lim f ft '('' .- 1. Ex. & Find the, area bounded by the eurvt' // - ^ (KU?. tiiJ), tlw axis of X) and the oidinatoH JK and u' -- . v "" " '" Sinco the line u = <t is an jwyinptntc' of Ihc eum*, //~~- * when i fmthtiimoH 1 , ilic area in not, ntneUy speukiuj-, hounded. We IIH ( \, lun find the area bounded on thci ri^hl by the tirdinuto a; = a Ji, where ft is a Htuall quantity, with the lesult If A M), Hin~ l a -A ' (I 7T Hence wo may regard J as tlui vahu of th> area required, and express it by tlw if **(/" "*~~* M ""jj" 1 ii * Ftu. UiJ Ex. 3. Find the valuo of J ' "f, Proceeding as in Ex, 1, w( plaoo / f(JG * . /* * w J* / - -7- -a Lim / , . ^i V / h **<' i Vj But an expression which increases Indeflnifcely m * Voo } hrn^o th* givt*n integral has no finite value, We accoidmgly conclude that in ranli aawt w rntiut dftermiuo n limit, and that the problem has no solution if wt* cannot find a 79. Area in polar coordinates, Let (Fig, 98) he thn po! mid OH the initial line of a system of polar noSniinatai (r, &}< nf> and 0% two fixed radius vectors for which 0*0 and Bmi respectively, and P& any curve for which the aquation Ift Requirod the area JJOE AREA IN POLAR COORDINATES 231 To construct the differential of area, dA, we divide the angle into parts, dd. Let OP and OQ be any two consecutive radius vectors ; then the angle POQ = d0. With as a center and OP as a radius, we draw the arc of a circle, intersecting OQ at R. The area of the sector It is obvious that the re- quired area is the limit of the sum of the sectors as their number is indefinitely in- creased. Therefore we have and FIG. 93 This result is unchanged if P l coincides with 0, but in that case OP l must be tangent to the curve. So also P z may coincide with 0. Ex. 1. Find the area of one loop of the curve r = a sin 3 (Fig 65, 51) As the loop is contained between the two tangents 0=0 and 6 = , the required area is given by the equation 8 = f Jo 12 Ex. 2. Find the area bounded by the lines 0= ~ and 0=~t the curve ~ 44 r = 2 a cos 0, and the loop of the curve r = a cos 2 6 which is bisected by the initial line. Since the loop of the curve r = a cos 2 Q is tangent to the line OL (Fig. 94), for which = - ~, and the line ON, for which 6 f it is evi- dent that the required area can be found by obtaining the area OLMNO, bounded by the lines OL and ON and the curve r 2 a cos 6, and subtract- ing from it the area of the loop. The area may also be found as follows : Let OPjPg be any radius vector cutting the loop r = a cos 2 Q at P t and the curve r = 2 a cos 6 at P a . Let OP l = r t and 0P 2 r v Draw the radius 232 APPLICATIONS vector OQ^jj, making an angle dd with OP^P^ With OP l and OP S as- radii and as a center, constiuct arcs of circles niti'iaoetouft O(^l^ at A\ and J2 2 respectively. Then the area ot the sector J\01fi IH ^ rfilQ and tho area of the sector P Z OR S is ^ r| dd. We ^y may now take the area P 1 / > 3 JB a /' and have Then ^ = / __ 7T ,_ 4 01, since the required area is symmetrical with respect to the line OM, wo may place (>2-*i)0. FIG, 04 From the curve r = 2 a cos0, we have rf = 4 a 8 cos s 0, and from the curve r = a cos 2 0, we have r a = a 2 cos 2 2 0; so that finally - a a cos 2 2 ^) <W e-?-'22 2 8 EXERCISES 1. Find the total area of the lemniscato ?' a = 2 2 cos 2 0, 2. Find the area of one loop of the curve r =s a sin n 0, 3. Find the total area of the cardioid r = a(l-f- COB 0). 4. Find the total area bounded by the curve ? a 6 + 3 <uw 0. 5. Find the area of the loop of the curve r* t a" cos 2 non 3 which is bisected by the initial line. 6. Find the area bounded by the curves r = a cos 8 and r == ., 7. Find the total area bounded by the curve r = 3 + 2 COB 4 0. 8. Find the area bounded by the curve ?cos a ^=sl and the 7T * lines = and = 75 . 9. Find the area bounded by the curves r ^ 6 + 4 cos and r s= 4 cos 10. Findtheareaboundedbythecurvesr=acos0andr a =:a 8 cos2^ MEAN VALUE 233 B 80. Mean value of a function. Let f(x) be any function of x and let y =/() be represented by the curve AB (Fig. 95), where OM=a and ON=b. Take the points M^ M z , , M n _ r so as to divide distance MN into n equal parts, each equal to dx, and at the points M, M^ M z , ' -, M n _ l erect the ordinates 2/0' y t y> ' &.-! Tlien the average, or mean, value of these n ordinates is M This fraction is equal to ndx , dx-\ b a If n is indefinitely increased, this expression approaches as a limit the value This is evidently the mean value of an " infinite number " of values of the function / (x) taken at equal distances between the values x = a and x = 6. It is called the mean vafote of the function for that interval. Graphically this value is the altitude of a rectangle with the base MN which has the same area as MNBA which equals / /a We see from the above discussion that the average of the function y depends upon the variable x of which the equal intervals dx were taken, and we say that the function was averaged with respect to x. If the function can also be averaged with respect to some other variable which is divided into equal parts the result may be different. This is illustrated in the examples which follow. 234 APPLICATIONS Ex. 1. Find the mean velocity of a body falling from rest duimg the tune t i if the velocity is averaged with respect to the tune. Here we imagine the time from to ^ divided into equal intervals dt and the velocities at the beginning of each interval averaged. Proceeding as in the text, we find, since v gt, that the mean velocity equals 1 Since the velocity is gt^ when t = t v it appears that in this case the mean velocity is half the final velocity. Ex. 2 Find the mean velocity of a body falling from rest through a distance s 1 if the velocity is averaged with respect to the distance. Here we imagine the distance from to s 1 divided into equal intervals ds and the velocities at the beginning of each interval aveiaged Pro- ceeding as in the text, we find, since v = V2 y$, that the mean velocity is Since the velocity is v^2 gs v when s = s v we see that in this case the mean velocity is two thirds the final velocity. EXERCISES 1. Find the mean value of the lengths of the perpendiculars from a diameter of a semicircle to the circumference, assuming the perpendiculars to be drawn at equal distances on the diameter. 2. Find the mean length of the perpendiculars drawn from the circumference of a semicircle to its diameter, assuming the perpen- diculars to be drawn at equal distances on the circumference 3. Find the mean value of the ordmates of the curve y = sm x 7T between x = and x = -5- , assuming that the points at which the ordmates are drawn are at equal distances on the axis of x. 4. The range of a projectile fired with an initial velocity V Q and v^ an elevation a is sin 2 a. Find the mean range as a varies from TT ff to 7p averaging with respect to a. 6. Find the mean area of the plane sections of a right circular ' cone of altitude h and radius a made by planes perpendicular to the axis at equal distances apart. LENGTH OF PLANE CURVE 235 6. In a sphere of radius a a series of right circular cones is inscribed, the bases of which are perpendicular to a given diameter at equidistant points. Find the mean volume of the cones 7. The angular velocity of a certain revolving wheel varies with the time until at the end of 5 mm. it becomes constant and equal to 200 revolutions per minute If the wheel starts from rest, what is its mean angular velocity with respect to the time during the interval in which the angular velocity is variable ? 8. The formula connecting the pressure p in pounds per square inch and the volume v in cubic inches of a certain gas is pv = 20. Find the average pressure as the gas expands from 2^- cu in. to 5 cu. in. 9. Show that if y is a linear function of aj, the mean value of y with respect to x is equal to one half the sum of the first and the last value of y in the interval over which the average is taken. 81. Length, of a plane curve. To find the length of any curve AB (Fig. 96), assume n 1 points, J?, J?, , -_ r be- tween A and B and connect each pair of consecutive points by a straight line. The length of AB is then defined as the limit of the sum of the lengths of the n chords AQ, %%, ]%PV -, X^.iB as n is increased without limit and the length of each chord approaches zero as a limit. By means of this definition we have already shown ( 39 and 52) that " FIG. 96 d8**^dy?+df (1) in Cartesian coordinates, and <2 = V,fr a 4-rW (2) in polar coordinates. Hence we have s = I ^dx*+ dy* (3) and s => C-Vdr* + r*d6\ (4) To evaluate either (3) or, (4) we must express one of the variables involved in terms of. the cither, or both in terms of a third* The limits of integration may then be determined. 236 APPLICATIONS Ex. I. Find the length of the parabola y 2 = kx from the vertex to the point (a, 6). From the equation of the parabola we find 2 ydy = kdx Hence formula (3) becomes either Either integral leads to the result Ex. 2. Find the length of one arch of the cycloid x = a(<jf> sm<), y = a (1 cos <). We have dx a (1 cos <) dtp, dy a sin <j> d$ ; whence, from (1), ds = a V2 2 cos <j>d<j> = 2a sin 2 rf<ji>. /* 27r <i Therefore s 2 a | sin d<& = 8 a. /o 2 EXERCISES v 1. Find the length of the curve 3y*(x I) 8 from its point of intersection with OX to the point (4, 3"). af - --^ w 2. Find the length of the catenary y -~\e a + e ") from x = to ai = h. v 3. Find the total length of the curve 4. Find the total length of the curve x = a cos 8 ^>, y = a sin 8 <. 5. Find the length of the curve x = a cos <f> -}- a<l> sin <, y = a sin < a< cos <, from < = to j! = 4 TT. 6. Find the length of the curve x e~*cos t, y= e-*sin#, between 7T the points for which t = and t = -^ . 7. Find the length of the curve r = a cos 4 j from the point on the curve for which 6 ~ to the pole 8. Find the total length of the curve r = a (1-|- cos 5). 9. Show that the length of the logarithmic spiral re a between any two points is proportional to the difference of the radius vectors. of the- points, WORK 237 82. Work. By definition the work done in moving a body against a constant force is equal to the force multiplied by the distance through which the body is moved. If the foot is taken as the unit of distance and the pound is taken as the unit of force, the unit of measure of work is called & foot-pound. Thus the work done in lifting a weight of 25 Ib. through a distance of 50 ft. is 1250 ft-lb. Suppose now that a body is moved along 0JT(Fig. 97) from A (x = a) to B (x = b") against a force which is not constant but is a function of a;, expressed by /(). Let the line AB be divided into intervals each equal A i j/W J> x to dx, and let one of these inter- vals be MN, where OM x. Then the force at the point M is /(a?), and if the force were con- stantly equal to f(x) throughout the interval MN, the work done in moving the body through MN would be/ (x) dx. This expression therefore represents approximately the work actually done, and the approximation becomes more and more nearly exact as MN is taken smaller and smaller. The work done in moving from A to B is the limit of the sum of the terms f(x) dx computed for all the intervals between A and B. Hence we have and /> = / Jit Ex. The force which resists the stretching of a spring is propor- tional to the amount the spring has been already sti etched. Foi a cer- tain spring this force is known to be 10 Ib. when the spring has been stretched & in. Find the work done in stretching the spring 1 in. from its natural (unstretched) length. If F is the force reqmied to stretch the spring through a distance x, we have, from the statement of the problem, and since F ~ 10 when x = %, we have k = 20. Therefore F = 20 x. Reasoning as in the text, we have w~C\ Jo 238 APPLICATIONS EXERCISES 1. A positive charge m of electricity is fixed at 0, The repulsion on a unit charge at a distance x from is -3 Find tho work done 33 m bringing a unit charge from infinity to a distance a from 0. 2. Assuming that the force required to strotdi a wiro from the *t* length a to the length a + x is proportional to -'j and that a foixio if* of 1 Ib stretches a certain wire 36 in in lungth to a length .0,'i in. greater, find the work done in stretching that wiro from M in, to 40 in. 3. A block slides along a straight line from O against a resistance ka? equal to . a ? where 7c and a are constants and or, is the distance (K j" Cb of the block from at any time. Find the work done in moving the block from a distance a to a distance a V from 0. 4. Find the foot-pounds of work done in lifting to a height of 20 ft above the top of a tank all the water contained in a full cylin- drical tank of radius 2 ft. and altitude 10 ft. 5. A bag containing originally 80 Ib. of sand is liftod through a vertical distance of 8 ft. If the sand leaks out, at such a rato that while the bag is being lifted, the number of pounds) of sand lost is wjual to a constant times the square of the number of foot through which the bag has been lifted, and a total of 20 Ib. of sand i& lost during the lifting, find the number of foot-pounds of work done in lifting tho bag, 6. A body moves in a straight line according to the formula c ef, where x is the distance traversed in a time t. If tho resistance of the air is proportional to the square of tho velocity, find tho work done against the resistance of the air as the body moves from != to ft* a. 7. Assuming that above the surface of the earth the form* of the earth's attraction varies inversely as tho square of the diBtanca from the earth's center, find the work done in moving a weight of w pounds from the surface of the earth to a distance a miles above the surfaca. 8. A wire carrying an electric current of magnitude C is bent into a circle of radius a. The force exerted by the current upon a unit magnetic pole at a distance a from the center of the circle in a straight line perpendicular to the plane of the circle is known to be 27rCa s (a?+ ^ ' done in brin g in 8 a ttnit magnetic pole from infinity to the center of the circle along the line just mentioned, GENERAL EXEKClSES 239 9. A piston is free to slide in a cylinder of cross section S The force acting on the piston is pS, where p is the pressure of the gas in the cylinder, and is 7.7 Ib. per square inch when the volume v is 2 5 cu in. Find the work done as the volume changes from 2 cu. in. to 6 cu. in., according as the law connecting p and v is (1) pv k or (2)X* = * GENERAL EXERCISES 1. Find the area of the sector of the ellipse 4oj 2 +9y 2 =36 cut out of the first quadrant by the axis of x and the line 2 y = x. 2. Find the area of each of the two parts into which the area of the circle a; 3 + if = 36 is divided by the curve y 2 = a; 8 . 3. Find the area bounded by the hyperbola xy = 12 and the straight line x + y 8 = 0. 4. Find the area bounded by the parabola a; 2 =4 ay and the 8 a 8 5. Find the area of the loop of the curve ay 2 = (x a) (x 2 a) z . 6. Find the area of the two parts into which the loop of the curve y* = cc 2 (4 a;) is divided by the line x y = 0. 7. Find the area bounded by the curve ary* + a 2 6 2 = a 2 2/ a and its asymptotes. 8. Find the area bounded by the curve 2/ 2 (aj a + & 2 ) = <&& and its asymptotes. 9. Find the area bounded by the curve a? = a cos 6, y = b sin 8 10. Find the area inclosed by the curve x = a cos 8 0, y = a sin 8 0. 11. Two parabolas have a common vertex and a common axis, but lie in perpendicular planes. An ellipse moves with its plane perpen- dicular to the axis and with the ends of its axes on the parabolas. Find the volume generated when the ellipse has moved a distance h from the common vertex of the parabolas. 12. Find the volume of the solid formed by revolving about the line a; = 4 the figure bounded by the parabola y*= 4 x and the line x = 1. 13. A right circular cylinder of radius a, is intersected by two planes, the first of which is perpendicular to the axis of the cylinder and the second of which makes an angle with the first. Find the volume of the portion of the cylinder included between these two planes if their line of intersection is tangent to the circle cut from the cylinder by the first plane. 240 APPLICATIONS fn O ., w. u^ U.UU.UJLU w*.^c* U w ~~ ~ + y* a? as a base, an isosceles triangle is constructed with its altitude equal to the ordinate and its plane perpendicular to the plane of the curve. Find the volume generated as the triangle moves from x = a to x = a. 15. Find the volume of the solid generated by revolving about 8 a 8 the line OF the figure bounded by the curve y = 2 . and the line y o,. 16. Find the volume of the solid formed by revolving about the line x = 2 the plane area bounded by that line, the parabola y z = 3x, and the lines y = 3. 17. Find the volume formed by revolving about the line x = 2 the plane figure bounded by the curve y 3 = 4 (2 a;) and the axis of y. 18. The sections of a solid made by planes perpendicular to OF aie circles with one diameter extending from the curve ^ 4 a; to the curve y 2 = 4 4 x. Find the volume of the solid between the points of intersection of the curves. 19. The area bounded by the circle x* + $ 2 ax = is revolved about OX, forming a solid sphere Find the volume of the two parts into which the sphere is divided by the surface formed by revolving 3.8 the curve y 2 = about OA' y 2a x 20. Find the volume of the two solids formed by revolving about Y the areas bounded by the curves ar 8 -f # 2 = 5 and g/ 2 = 4 ai. 21. Find the volume of the solid formed by revolving about OX the area bounded by OX, the lines x = and x == a, and the curve z y = x + ae a . 22. The three straight lines OA, OB, and OC determine two planes which intersect at right angles in OA. The angle A OB is 45 and the angle AOC is 60. The section of a certain solid made by any plane perpendicular to OA is a quadrant of an ellipse, the center of the ellipse being in OA, an end of an axis of the ellipse being in OB, and an end of the other axis of the ellipse being m OC. Find the volume of this solid between the point and a plane perpendicular to OA at a distance of two units from 0. 23. The section of a solid made by any plane perpendicular to OX is a rectangle of dimensions a? and sm x, x being the distance of the plane from O. Find the volume of this solid included between the planes for which x = and x = IT. GENERAL EXEECISES 241 24. An oil tank is in the form of a horizontal cylinder the ends of which are circles 4 ft. in diameter. The tank is full of oil, which weighs 50 Ib. per cubic foot Calculate the pressure on one end of the tank. 25. The gasoline tank of an automobile is in the form of a hon- zontal cylinder the ends of which are plane ellipses 20 in high and 10 in broad. Assuming w as the weight of a cubic inch of gasoline, find the pressure on one end of the tank when the gasoline is 15 in deep. 26. A horizontal gutter is U-shaped, a semicircle of radius 3 in , surmounted by a rectangle 6 in wide by 4 in. deep If the gutter is full of water and a board is placed across the end, how much pressure is exerted on the board ? 27. The end of a horizontal gutter is in the form of a semicircle of 3 in. radius, the diameter of the semicircle being at the top and horizontal. The gutter receives water from a roof 50 ft above the top of the gutter. If the pipe leading from the roof to the gutter is full, what is the pressure on a board closing the end of the gutter ? 28. A circular water main has a diameter of 5 ft. One end is closed by a bulkhead, and the other is connected with a reservoir in which the surface of the water is 20 ft above the center of the bulkhead Find the total pressure on the bulkhead. 29. Find the area of a loop of the curve ?* a = a? sin n6. 30. Find the area swept over by a radius vector of the curve IT r a tan as $ changes from to 4 4 31. Find the area inclosed by the curve r = H .. and the . J 1 cos $ curve r = H , 1 + cos 6 32. Find the area bounded by the circles r=a cos 6 and r= a, sin 6. 33. Find the area cut off from one loop of the curve r 2 = 2 a 2 sin 2 6 by the circle r a 34. Find the area of the segment of the cardioid r = a (1+ cos 0) cut off by a straight line perpendicular to the initial line at a dis- tance | a from the origin 0. 35. Find the area cut off from a loop of the curve r = a sm 3 6 by . , . , the circle r s= 242 APPLICATIONS 36. Find the area cut off from the lemniscate r 2 = 2 a 2 cos 2 by _/*> the straight line r cos = 75 ' 37. Find each of the three areas bounded by the curves r = a and r = a (1 + sin 0). 8 38. Find the mean height of the curve y 2 . 2 between the lines x = 2a and aj = 2 a. 39. A particle describes a simple harmonic motion defined by the (792i?/ \ ~9~) during a complete vibration is half the maximum kinetic energy iL the average is taken with respect to the time. 40. In the motion defined in Ex 39 what will be the ratio of the mean kinetic energy during a complete vibration to the maxi- mum kinetic energy, if the average is taken with respect to the space traversed? 41. A quantity of steam expands according to the Iawj0v 08 =; 2000, p being the pressure in pounds absolute per square foot Find the average pressure as the volume v increases from 1 cu. ft. to 5 cu. ft. a 2 42. Find the length of the curve y = a In 2 . from the origin a ^ to the point for which x = ~ 43. Find the length of the curve y = Li' x - between the points for which x =sl and x = 2 respectively. ~~ 44. Find the total length of the curve x = a cos 8 <, y = b sin*<. a 45. Find the total length of the curve r = a sin 8 ^ 46. Find the length of the spiral r = aO from the pole to the end of the first revolution. JA 47. If a center of force attracts with a magnitude equal to r> SC# where x is the distance of the body from the center, how much work will be done in moving the body in a straight line away from the center, from a distance a to a distance 8 a from the center ? 48. A body is moved along a straight line toward a center of force which repels with a magnitude equal to 7ccc when the body is at a distance x from the center. How much work will be done in moving the body from a distance 2 a to a distance a from the ter? GENERAL EXERCISES 243 49. A central force attracts a body at a distance x from the center 7c by an amount -j Find the work done in moving the body directly away from the center from a distance a to the distance 2 a. 50. How much work is done against hydrostatic pressure in rais- ing a plate 2 ft. square from a depth of 20 ft. to the surface of the water, if it is kept at all times parallel to the surface of the water ? 51. A spherical bag of radius 5 in. contains gas at a pressure equal to 15 Ib. per square inch. Assuming that the pressure is in- versely proportional to the volume occupied by the gas, find the work required to compress the bag into a sphere of radius 4 in. CHAPTER XI REPEATED INTEGRATION 83. Double integrals. The symbol nv t f(x, y) dxdy, (1) _ 4 in which a and b are constants and y^ and y^ are either con- stants or functions of x, indicates that two integrations are to be carried out in succession. The first integral to be evaluated is where x and dx are to be held constant. The result is a func- tion of x only, multiplied by dx; let us say, for convenience, ;) dx. The second integral to be evaluated is, then, F(x) dx, which is of the familiar type. Similarly, the symbol 6 /*x s f(x, y) dydx, (2) where a and b are constants and ^ and # g are either constants or functions of #, indicates first the integration in which y and dy are handled as constants, and afterwards integration with respect to y between the limits a and b. 244 ft n 2 Ex. 1. Evaluate / I xythdy. The first integral is j scydxily The second integration is DOUBLE INTEGRALS 245 s /.a Ex. 2. Evaluate f f "' O 2 + y*)<lr.dii Jo Ji-x ^ ' The ni*st integration is The second integration is Ex.3. Evaluate f^" f**jpdyt?jc. Jo i/o The first integration is v 3 r n j/ a ( u y*<lytlx = ?/ 2 ^/y * = /-.//. 1/0 L Jo <l ' The second integration is r 2a ^ y= rir a = ! a 4 Jo 4a '' LaOoJo 6 A definite integral in one variable has been shown lo be the limit of a sum, from which we infer that formula (1) involves first the determination of the limit of a sum with respect to ;/, followed by the determination of the limit of a sum with respect to x. The application of the double integral comes from it interpretation as the limit of a double summation, How, such forms arise in practice will be illustrated in the following sections, EXERCISES Find the values of the following integrals ; a"%/ / 8 r* ^dydx, 3. / / Ji J,, n\ xydoady. 4. ' 246 5. REPEATED INTEGRATION dydx 8. / /-as | I c/O ^/o rf r Jo /o r 8 a(l + coi S) ? sin Odddr. r*d$dr. r cos 6dOdr 84. Area as a double integral. Let it be required to find an area (such as is shown in Fig. 98) hounded by two curves, with the equations y^~f^(x) and y 2 =/ 2 () intersecting in points for which x a and z=b respectively. Let the plane be divided into rectangles by straight lines parallel to OX and OT respectively. Then the area of one such rectangle is where dx is the distance between two consecutive lines parallel to O Y 3 and. where dy is the distance between, tw,o consecutive lines parallel to OX. The sum of the rectangles which are either AKEA AS DOUBLE INTEGRAL 247 wholly or partially within the required area will be an approx- imation to the required area, but only an approximation, because the rectangles will extend partially outside the area. We assume as evident, however, that the sum thus found becomes more nearly equal to the required area as the number of rectangles becomes larger and dx and dy smaller. Hence we say that the required area is the limit of the sum of the terms dxdy. The summation must be so carried out as to include every rectangle once and only once. To do this systematically we begin with any rectangle in the interior, such as PQJRS, and add first those rectangles which lie in the vertical column with it. That is, we take the limit of the sum of dxdy, with x and dx constant and y varying from y^f^(x) to 2/ 2 =/ fl (X). This is indicated by the symbol /" vt/. x = [/ 2 (x) -/,()] dx. (2) This is the area of the strip TUVW. We are now to take the limit of the sum of all such strips as dx approaches zero and x varies from a to 5. We have then ^ <= /"<>,- ^ ^ = f V, (x) -/ x <V>] dx. (3) Jo. J a If we put together what we have done, we see that we have n s dxdy. (4) _. This discussion enables us to express the area as a double integral. It does not, however, give us any more convenient way to compute the area than that found in Chapter III, for the result (2) is simply what we may write down at once for the area of a vertical strip (see Ex. 3, 28). If it should be more convenient first to find the area of a horizontal strip, we may write n* dy dx. (5) 248 BEPEATED INTEGRATION Consider a similar problem in polar coordinates. Let an area, as in Fig. 99, be bounded by two curves r^/X^O and r 2 =/ s (0), and let the values of corresponding to the points B and be l and a respectively. The plane may be divided into four-sided figures by circles with centers at and straight lines radiating from 0. Let the angle between two consecu- tive radii be d6 and the distance between two consecutive circles be dr. We want first the area of one of the quadrilaterals such as PQRS. Here OP = r, = dr, and the angle POS=d0. By geom- etry the area of the sector POS=^r 2 d0 and A the area of the sector FIG. 99 therefore PQRS = J (r + dr)*d6 - J r*d0 = rdrd0 + % (dr)*d0. Now as dr and d0 approach zero as a limit the ratio of the second term in this expression to the first term also approaches zero, since this ratio involves the factor dr. It may be shown that the second term does not affect the limit of the sum of the expression, and we are therefore justified in writing as the differential of area dA = rd0dr. (6) The required area is the limit of the sum of these differ- entials. To find it we first take the limit of the sura of the quadrilaterals, such as PQRS, which lie in the same sector UO V. That is, we integrate rd0dr, holding and dd constant and al- lowing r to vary from r^ to r. We have / ^r. (7) which is the area of the strip TUVW. CENTER OF GRAVITY 249 Finally we take the limit of the sum of the areas of all such strips in the required area and have (8) If we put together what we have done, we may write dOdr. 9 /*s /"a = I I r t/0 t J>i It is clear that this formula leads to nothing which has not been obtained in 79, but it is convenient sometimes to have the expression (9). 85. Center of gravity. It is shown in mechanics that the cen- ter of gravity of n particles of masses m^ w 2 , -, m n lying m a plane at points whose coordinates are^, y x ), (x z , / 2 ), , (#, y^) respectively is given by the formulas = ss \-m n This is the point through which the resultant of the weights of the particles always passes, no matter how the particles are placed with respect to the direction of the earth's attraction, We now wish to extend formulas (1) so that they may be applied to physical bodies in which the number of particles may be said to be infinite. For that purpose we divide the body into n elementary portions such that the mass of each may be considered as concentrated at a point (or, y). Then, if m is the total mass of the body, the mass of each element is dm. We have then to replace the m's of formula (1) by dm and to take the limit of the sums involved in (1) as the number n is indefinitely increased and the elements of mass become indefinitely small. There result the general formulas / xdm I ydm - X = J , 9 ,J . (2) I dm I dm 250 REPEATED INTEGRATION To apply these formulas we consider first a slender wire so fine and so placed that it may be represented by a plane curve. More strictly speaking, the curve may be taken as the mathe- matical line which runs through the center of the physical wire. Let the curve be divided into elements of length ds. Then, if c is the area of the cross section of the wire and D is its density, the mass of an element of the wire is Dads. For con- venience we place DC = p and write where p is a constant. If this is substituted in (2), the constant p may be taken out of the integrals and canceled, and the result may be written in the form sx = ixds, sy = I yds, (3) where a on the left of the equations is the total length of the curve. These formulas give the center of gravity of a plane curve. Ex. 1. Find the center of gravity of one fourth of the circumference of a circle of radius a. Here we know that the total length is \ rrax = Cxds, vay = Cyds. To integrate, it is convenient to in- troduce the central angle <j> (Fig 100), whence x = a cos <j>,y = a sin <f>, ds = ad<j>. 7T Then |irox= l*a*cos<f>dd>, Jo , whence we , so that, from (3), we have 2a FIG. 100 Consider next a thin plate, which may be represented by a plane area. Strictly speaking, the area is that of a section through the middle of the plate. If t is the thickness of the plate and _D its density, the mass of an element of the plate with the area A A is DtdA. For convenience we place Dt = p and write dm pdA, CENTER OF GEAVITY 251 where p is a constant. If this is substituted in (2) and the p's are canceled, we have i = I xdA, AyI ydA, (4) where A is the total area. These formulas give the center of gravity of a plane area, Ex. 2. Find the center of gravity of the area bounded by the parabola y* = kx, the axis of x, and the ordmate through the point (a, b) of the parabola (Fig 101) We place dA= dxdy in (4) and have A js = C Cx dx dy, Ay = Cfy dx dy. To evaluate, we choose the element dxdy inside the area in a general position, and first sum with respect to y along a vertical strip. We shall denote by y l the value of y on the _ i . parabola, to distinguish it from the general ^ values of y inside the area The first integra- tion gives us, therefore, respectively C Vi xdxdy = xy 1 dx and C l ydxdy \y* dx, so that we have Ax Cxy^dx, Ay = C^y* dx- J / On examination of these results we see that each contains the factor y^dx (which is the area ( 22) of an elementary vertical strip), multiplied respectively by a? and ^ y v which are the coordinates of the middle point of the ordmate y r These results are the same as if we had taken dA = y t dx in the general foimula (4), and had taken the point (x, y) at which the mass of dA is concentrated as (z, y x ), which is in the limit the middle point of dA. In fact this is often done in computing centers of gravity of plane areas, and the first integration is thus avoided. Kow, from the equation of the parabola y% = kx, and to complete the integration, we have to substitute this value for y t and integrate with respect to x from x = to x = a We have AS = * from the /** yfco" Ay i i kxdx <- of the curve, k = and, by 23, A ~ - < fl O 252 REPEATED INTEGRATION Substituting these values and i educing, we have finally x = | a, y=&b In solving this problem we have carried out the successive mtegratioii.s separately, in oider to show clearly just what has been done. If now we collect all this into a double integral, we have Ax=f a f ''xdxdy. /ly=f"f 'ytlxda. Jo Jo Jo Jo Ex. 3. Find the center of gravity of a sextant of a circle of radum a To solve this problem it is convenient to place the sextant so that the axis of v bisects it (Fig. 102) and to use ._ polar coordinates. From the symmetry of the figure the centei of gravity lies on OX, so that we may write at once y = 0. To find a take r an element of area ? dddr in polar cooi- dinates and place x = r cos We have then, from (4), a /" r ^ / r 2 cos 6 dddr, rr Jo ~6 where A = lira?, one sixth the area of a ciicle In the first integration 6 and dO are constant, and the summation takes place along a line radiating from with r varying from to a The angle then varies from - - to ~ , and thus the entire area is covered The solution is as follows . 6 liraPx =J 8 a 3 cos Odd whence 6 _ 2a Consider now a solid of revolution formed by revolving the plane area (Fig. 103) ABCD about Y as an axis. It is assumed that the equation of the curve CD is given. It is evident from symmetry that the center of gravity of the solid lies on OY, so that we have to find only y. Let dFbe any element of volume. Then dm = pdV, where p is the density and is assumed constant. Substituting in (2), we have (5) CENTER OF GRAVITY 253 Let the solid be divided into thin slices perpendicular to as was done in 26, and let the summation first take place over one of these slices. In this summa- tion y is constant, and the result oL the summation is therefore y times the volume of the slice. It is therefore u(^r'^^y)- We have now to extend the summation over all the slices. This gives the result r b Vy = / ir^ydy, (G) / TIG. 103 where OA = a and 07? = &. It is to be noticed that this result is what we obtain if we interpret dm in (2) as the mass of the slice and consider it concentrated at the middle point of one base of the slice. Ex. 4. Find the centei of gravity of a light cucular cone of altitude 6 and radius (Fig. 104) This is a solid of revolution formed by revolv- ing a light triangle about OF. However, the equation of a straight line noed not be used, as X CL similar triangles are simpler. We have - = - > y b whence x -y. The volume V is known to be b J 7r 8 6. Therefore, from (G), we have f*mP aj = J - fdy = whence FIG. 104 EXERCISES 1. Show that the center of gravity of a semicircunifereuce of o a radius a lies at a distance of from the center of the circle on 7T the radius which bisects the semicircumference. 2, Show that the center of gravity of a circular arc which subtends an. angle <x at the center of a circle of radius a lies at a distance sin ~ from the center of the circle on the radius which, bisects the arc, a 254 EEPEATED INTEGRATION - 3. A wire hangs so as to form the catenary y ~ (/* -1- "). Find the center of gravity of the piece of the curve between Iho points for which x = and x = a. 4. Find the center of gravity of the arc of the cycloid x = a(<[> sm$), y = a(l coB<f>), between the first two sharp points. 5. Find the center of gravity of a parabolic segment of base, 2 b and altitude a. 6. Find the center of gravity of a quadrant of the area of a circle. 7. Find the center of gravity of a triangle. 8. Find the center of gravity of the area bounded by the ourvo y = sin x and the axis of x between x = and x = IT. 9. Find the center of gravity of the plane area bounded by tho two parabolas y* = 20 x and a? = 20 y. 10. Find the center of gravity of a figure which is composed of a rectangle of base 2 a and altitude I surmounted by a semicircle of radius a 11. Find the center of gravity of the area bounded by the Ihvst arch of the cycloid (Ex. 4) and the axis of aj. 12. Show that the center of gravity of a sector of a circle lies at 4 a a a distance ~ sin- from the vertex of the sector on a line bisecting the angle of the sector, where a is the angle and a the radius. 13. Find the center of gravity of the area bounded by the cardioid 14. Find the center of gravity of the area bounded by the curve r = 2 cos & + 3. 15. Find the center of gravity of a solid hemisphere. 16. Find the center of gravity of a solid formed by revolving about its altitude a parabolic segment of base 26 and altitude a. 17. Find the center of gravity of the solid formed by revolving about OF the plane figure bounded by the parabola y*** fee, the axis of y, and the line y = k. 18. Find the center of gravity of the solid bounded by the sur- faces of a right circular cone and a hemisphere of radius & with the base of the cone coinciding with the base of the hemisphere and the vertex of the cone in the surface of the hemisphere CENTER OF GRAVITY 255 86. Center of gravity of a composite area. In finding the cen- ter of gravity of a body which is composed of several parts the following theorem is useful: If a body of mass M is composed of several parts of masses M^ Mf - , M n , and if the centers of gravity of these parts are respectively (x f ^), (z a , / ? ), .,(# w , #), then the center of grav- ity of the composite body is yiven by the formulas + J C ; We shall prove the theorem for the % coordinate. The proof for y is the same. By 85 we have, for the original body, MX - fxdm, ' (2) where the integration is to be taken over all the partial masses M v Jf a , , M n into which the body is divided. But we have also where the subscripts indicate that the integration in each case is restricted to one of the several bodies. But formula (2) can be written r C C MX = / x^m^ I x^dm^-}- + / ^dm n ; and, by substitution from (3), the theorem is proved, x. Find the center of gravity of an area bounded by two circles one of which is completely inside the other. Let the two circles be placed aa in Fig. 105, where the center of the larger circle of radius a is at the origin, and the center of the smaller circle of radius o is on the axis of $ at a distance o from the origin* 256 REPEATED 1NTE( 1 RAT U )K The area which can be considered as composed of two ]>jut.n i that of the larger circle, the two paits being, fust, the smaller eirde and, wroucl, UN- irregular ring whose centei of gravity we wish to find Now the center of gravity of a circle is known to be at its center, Theiefore, in the formula of the theoiem, we know (T, J/), which is on the left of the equation, to bo (0, 0), and (~c v yj to be (c, 0), and wish to find (7 2 , T/ z ). Since we are dealing with aieas, we take the masses to be equal to the ai eas, and have, accordingly, M = ira z (the mass of the larger ciiclc), M = irb z (the mass of tho smallei cncle), and My, = if ( a ft a ) (the mass of tho ring). Substituting in the formula, we havo jor> = IT/A- + TT (<r - A- 1 ) r a ; whence, by solving for ,r a It is unnecessary to find y i} since, by Hyiniuctiy, Mi coutiM of lies on OX EXERCISES 1. Show that if there are only two component masses fl/^ mul. JHf^ in formulas (1) of the theorem, the center of gravity of tho oomjxmto mass lies on the line connecting the centers of gravity of tho com- ponent masses at such a point as to divide that hue into sogmtmlH inversely proportional to the masses. 2. Prove that if a mass M^ -with center of gravity (.r^ //j) haw <m(, out of it a mass M z with center of gravity (i? 9 , y a ), tho c-oiiUu 1 of gravity of the remaining mass is Mi-Mt J/ t - M t and r s aro tangtnit oxtonuJly. Find 3. Two circles of radii their center of gravity 4. Find the center of gravity of a hemispherical hol3 boundod by two concentnc hemispheres of radii r^ and r^ 5. Place r 9 = ^ + A^ m Ex. 4, let A?' approach xero, and thus iind the center of gravity of a hemispherical surface. CENTER OF GRAVITY 25Y 6. Find the center of gravity of a hollow right circular cone bounded by two parallel conical surfaces of altitudes 7^ and A 2 respectively and with their bases in the same plane. 7. Place \= /^-H A/i in Ex 6, let A/t approach zero, and thus find the center of gravity of a conical surface 8. Find the center of gravity of a carpenter's square each aim o which is 15 in on its outer edge and 2 in wide 9. From a square of edge 8 in a quadrant of a circle is cut out, the center of the quadrant being at a corner of the squaie and the radius of the quadrant being 4 in. Find the center of gravity of the figure remaining 10. Two iron balls of radius 4m and 6 in respectively are con- no uted by an iron rod of length 1 in. Assuming that the rod is a cylinder o radius 1 in., find the center of gravity of the system. 11. A cubical pedestal of side 4 ft. is surmounted by a sphere of radius 2 ft Find the center of giavity of the system, assuming that the sphere rests on the middle point of the top of the pedestal. 87. Theorems. The folio wing theorems involving the center oi' gravity may often ho used to ad- Q vantage in finding pressures, volumes x c>I solids of revolution, or areas of surfaces of revolution. I. The total pressure on a plane sur- faae immersed in liquid in a vertical position is equal to the area of the sur- face multiplied ly tlw pressure at its center of gravity. Let the area he placed a in Fig 106, ETO< 106 where the axis of K is in the surface of the liquid and where the axis of y is measured downward. Then, by 25, - /* I yO p a- fl! i)^> '*' which may be written IIH a double integral in the form p = CCwydydx => w nydxdy, (2) 258 BEPEATED INTEGRATION In fact, this may be written down directly, since the pressure on a small rectangle dxdy is its area, dxdy, times its depth, T/, times w. Moreover, from 85, we have Ay --CCydxdy. (3) By comparison of (2) and (3) we have P = wyA. But wyis the pressure at the center of gravity, and the theorem is proved for areas of the above general shape. If the area is not of this shape, it may be divided into such areas, and the theorem may be proved with the aid of the theorem of 86. Ex.1. A circular bulkhead which closes y="b the outlet of a reservoir has a radius 8 ft , and its center is 12 ft below the surface of the water. Find the total pressure on it Here A = 9 it and the depth of the center of gravity is 12. Therefore P = 108 irw = -V-Trtons = 10.6 tons. II. The volume generated by revolving a plane area about an axis in its plane not intersecting the area is equal to the area of the figure multiplied by the circumference of the circle described by its center of gravity. To prove this take an area as in Fig. 107. Then, by 26, if V is the volume generated by the revolution about OY, (4) which can be written as a double integral in the form F=27rf C\dydx. (5) t/ i/, By 85, J&= C C\dydx; Ja u/jTj and, by comparison of (4) and (5), we have V~ which was to be proved. THEOREMS OF PAPPUS 259 Ex. 2. Find the volume of the nag surface formed by revolving about an axis in its plane a circle of radius a whose center is at a distance c from the axis, where c > a. We know that A = -sra 2 and that the center of gravity of the circle is at the center of the circle and therefore describes a circumference of length 2 ire. Therefoie F = 2ir 2 ffl 2c. III. The area generated by revolving a plane curve about an axis in its plane not intersecting the curve is equal to the length of the curve multiplied by the circumference of the circle described by its center of gravity. To prove this we need a formula for the area of a surface of revolution which has not been given. It may be shown that if S is this area, then r S^Zirlxds. (6) A rigorous proof of this will not be given here. However, the - student may make the formula seem plausible by noticing that an element ds of the curve will generate on the surface a belt of width ds and length 2 irx. The product of length by breadth may be taken as the area of the belt. Moreover, by 85, we have sx I xds; (7) and comparing the two equations (6) and (7), we have which was to be proved. Ex. 3. Find the area of the ring surface described in Ex. 2. We know that s - 2 TTO and that the center of gravity of a circumfer- ence is at its center and therefore describes a circumference of length 2 ire Therefore 5 = 4 ^ac. Theorems II and III are known as the theorems of Pappus. EXERCISES 1. Find by the theorems of Pappus the volume and the surface of a sphere. 2, Find by the theorems of Pappus the volume and the laterax surface of a right circular cone. 260 .REPEATED INTEGRATION 3. Find by the theorems of Pappus the volume generated by revolving a parabolic segment about its altitude. 4. Find by the theorems of Pappus Iho volume generated by revolving a parabolic segment about its base. 5. Find by the theorems of Pappus the volume geru>ratod by revolving a parabolic segment about the tangent at its vortex 6. Find the volume and the surface generated by revolving a square of side a about an axis in its piano perpendicular to one. of its diagonals and at a distance b\b> ;= } from it,s center. V V2/ 7. Find the volume and the area generated by revolving' a right triangle with legs a and b about an axis in its plane ]>urullel to the leg of length a on the opposite side from the hypotenuse and at a distance c from, the vertex of the right angle. 8. A circular water main has a diameter of 4 ft. One end IH closed by a bulkhead, and the other is connected with a reservoir in which the surface of the water is 18 ft. above the center of the bulkhead Find the pressure on the bulkhead. 9. Find the pressure on an ellipse of semiaxos a. and I completely submerged, if the center of the ellipse is o units below the surface of the liquid. 10. Find the pressure on a semiellipse of semiaxes a and I (a > />) submerged with the major axis in the surface of the liquid and the minor axis vertical. 11. Find the pressure on a parabolic segment submerged with the base horizontal, the axis vertical, the vertex above the base, and the vertex c units below the surface of the liquid. ^ 12. What is the effect on the pressure of a submerged vertical area in a reservoir if the level of the water in the reservoir is raised by c feet ? 88. Moment of inertia. The moment of inertia of a particle about an axis is the product of its mass and the square of its distance from the axis. The moment of inertia of a number of particles about the same axis is the sum of the moments of inertia of the separate particles abont that axis. From these definitions we may derive the moment of inertia of a thin plate. Let the surface of the plate be divided into elements o area cU Then the mass of each element is pdA, where p is the product of the thickness of the plate and ita density. Lot M bo MOMENT OF INEETIA 261 the distance of any point in the element from the axis about which we wish the moment of inertia. Then the moment of inertia of clement is approximately I?p dA. We say " approximately " because not all points of the element are exactly a distance R from the axis, as R is the distance ot some one point in the element. However, the smaller the ele- ment the more nearly can it be regarded as concentrated at one point and the limit of the sum of all the elements, as their size approaches zero and their number increases without limit, is the moment of inertia of the plate. Hence, if I represents the moment of inertia of the plate, we have = f (1) If in (1) we let /o = l, the resulting equation is 1= C&dA, (2) where I is called the moment of inertia of the plane area. When dA in (1) or (2) is replaced by dxdy or rdrdd, the double sign of integration must be used. Ex. 1. Find the moment of inertia of a rectangle of dimensions a and 6 about the side of length & Let the rectangle be placed as in Fig. 108. Let it be divided up into elements dA ~ dfdg. Then a- is the distance of some point in an element from OY. Hence, in (2), we have y JR SB x and dA = fatly. Therefore y=b We first sum the rectangles in a vertical strip, w y ranges from to &. We have This is the moment of ineitia of j. IG< IQ% the strip MN, and might -have been written down at once, since all points on the left-hand boundary of the strip are at a distance x from OY and since the area of the strip is Idx 262 KEPEATED INTEGRATION The second integration gives now Jo If, instead of asking for the moment of inertia of the area, we had asked for that of a plate of metal of thickness t and density D, the above result would be multiplied by p = Dt. But m that case the total mass M of the plate is aab, so that we have I \ Ma? Ex. 2. Find the moment of inertia of the quadrant of an ellipse r 2 7/ 2 }. >L 1 (a > 6) about its major axis If we take any element of area as dxtly, we find the distance of its lower edge from the axis about which we wish the moment ot ineitia to be y (Fig 109). Hence R = y and ^ I It will now be convenient to sum first with respect to x t since each point of a horizontal strip is at the same distance from OX "We therefore write J-ffffy**. FIG. 109 Now, indicating by a^ the abscissa of a point on the ellipse to distinguish it from the general x which is that of a point inside the ellipse, we have for the first integration For the second integration / To integrate, place y = I sin <j> Then TT = afc 8 f * s /o 16 If, instead of the area, we consider a thin plate of mass M, the above result must be multiplied by p, where M = irabp , whence The polar moment of inertia of a plane area is defined as the moment of inertia of the area about an axis perpendicular to its plane. This may also be called conveniently the moment MOMENT OF INERTIA 263 of inertia with respect to the point in which the axis cuts the plane of the area, for the distance of an element from the axis is simply its distance from that point. Thus we may speak, for example, of the polar moment of inertia with respect to an axis through the origin perpendicular to the plane of au area, or, more concisely, of the polar moment with respect to the origin. If the area is divided into elements dxdy, and one point in the element has the coordinates (#, #), the distance of that point from the origin is Vi?+]A That is, in (2), if we place (lA dxvly and J2 a =o; 2 +y 2 , we shall have the formula for the polar moment of inertia with respect to the origin. Denoting this by J , we have (3) This integral may be split up into two integrals, giving (4) where the change in the order of the differentials in the two integrals indicates the order in which the integration may be most conveniently carried out. The first integral in (4) is the moment of inertia about OY and may be denoted by / ; the second integral is the moment of inertia about OX and may be denoted by I K . Therefore formula (4) may be written as so that the problem of finding the moment of inertia may be reduced to the solving of two problems of the type of the first part of this section. Ex, 3. Find the polar moment of inertia of an ellipse with respect to the origin, In Ex. 2 we found I K for a quadrant of the ellipse. For the entire ellipse it is four times as great, since moments of inertia are added by definition. Hence 264 REPEATED INTEGEATION By a similar calculation I y - Theiefoie ^0=4 ffa * ( 2 + &*) If the area is leplaced by a plate of mass Af, this result gives / = \. M(a* + I") If polar coordinates aie used, the element of area is rd6d and the distance of a point in an element from the origin is Hence, in (2), dA = rdBdr and R = r. Therefore r- (6 In practice it is usually convenient to integrate first wit respect to r, holding 8 constant. This is, in fact, to find tr polar moment of inertia of a sector with vertex at 0. Ex.4. Find the polar moment of inertia of a cncle with lespect to point on its circtimference Let the circle be placed as in Fig 110 Its equation is then (Ex 1, 5] r = 2 a cos 0, where a is the radius If we take any element rdQdr an find I for all elements which lie in the same sector with it, we have to add the elements i*dQdr, with r ranging from to t v where r t is the value of ? on the cncle , and therefore > l = 2 a cos We have f Vrffltft = i 7 * (IB = 4 a 4 /o 1 We have finally to .rm these quantities, with 1 anging from to + - We have ** * I =J*la*cQ&6d6= I ~I TIG. 110 If M is the mass of a ciicular plate, this result, multiplied by p, Ex. 5. Find the polar moment of inertia of a cucle with respect to i centei. Heie it will be convenient first to find the polar moment of inertia of ring (Fig 111). We integrate first with respect to 6, keeping r constan We have MOMENT OF INERTIA 265 which is the approximate area of the ring Swrdr multiplied by the square of the distance of its inner cucumference from the center We then have, by the second integration, ^ = f a Jo If M is the mass of a circular plate, this result, multiplied by p, gives I / / f\ S \ U \\ \ X The moment oE inertia of a solid of revolution about the axis of revo- lution is the sum of the moments of inertia of the circular sections about the same axis ; that is, of the polar ]FIG< m moments of inertia of the circular sections about their centers. If the axis of revolution is OY, the radius of any circular section perpendicular to OF is x and its thickness is dy. Its mass is therefore pirx*dy ; and therefore, by Ex. 5, its moment of inertia about Y is \ pirtfdy. The total moment of inertia of the solid is therefore X\pTr \ x*dy. Ex. 6. Find the moment of inertia of a circular cone about its axis. Take the cone as in Ex. 4, 85. Then we have But, if M is the mass of the cone, we have M = Therefore / = fa Ma*. EXERCISES 1. Find the moment of inertia of a rectangle of base & and alti- tude a about a line through its center and parallel to its base. 2. Find the moment of inertia of a triangle of base b and altitude a about a line through its vertex and parallel to its base. 3. Find the moment of inertia of a triangle of base & and alti- tude a about its base. 4. Find the moment of inertia of an ellipse about its minor axip and also about its major axis, 266 REPEATED INTEGRATION 5. Find the moment of inertia of a trapezoid about its lower base, taking the lower base as b, the upper base as a, and the altitude as 7*. 6. Find the moment of inertia about its base of a parabolic seg- ment of base b and altitude a. 7. Find the polar moment of inertia of a rectangle of base I and altitude a about its center. 8. Find the polar moment of inertia about its center of a circular ring, the outer radius being r z and the inner radius r r 9. Find the polar moment of inertia of a right triangle of sides a and b about the vertex of the right angle. 10. Find the polar moment of inertia about the origin of the area bounded by the hyperbola xy=6 and the straight line aj+y 7=0. 11. Find the polar moment of inertia about the origin o the area bounded by the curves y=x z and y = 2 as 2 . 12. Find the polar moment of inertia about the origin of the area of one loop of the lemniscate r 2 = 2a? cos 2 0. 13. Find the moment of inertia of a right circular cylinder of height Ji, radius r, and mass M, about its axis. 14. Find the moment of inertia about its axis of a hollow right cir- cular cylinder of mass M, its inner radius being r v its outer radius r a , and its height h. 15. Find the moment of inertia of a solid sphere about a diameter. 16. A ring is cut from a spherical shell whose inner and outer radii are respectively 5 ft. and 6 f t , by two parallel planes on the same side of the center and distant 1 ft and 3 ft respectively from the center Find the moment of inertia of this ring about its axis 17. The radius of the upper base and the radius of the lower base of the frustum of a right circular cone are respectively r^ and r# and its mass is M. Find its moment of inertia about its axis. 89. Moments of inertia about parallel axes. The finding of a moment of inertia is often simplified by use of the following theorem ; The moment of inertia of a body about an oasis is equal to its moment of inertia about a parallel axis through its center of gravity plus the product of the mass of the body ly the square of the distance between the axes. MOMENT OF INEKTIA 267 We shall prove this theorem only for a plane area, in the two cases in winch the axes lie in the plane of the figure or are perpendicular to that plane. We shall also consider the mass of the area as equal to the v KT * area, as in 88. Case I. Wlien the axes lie in tJie plane of the Jiyure. Let the area be placed as in Fig. 112, where the center of grav- ity (^ #0 lf} taken as the origin (0, 0) and where the axis of y is taken parallel to the axis LK, L about which we wish to find the moment of inertia. Let x be the distance of an element dxdy from OF, and ^ its distance from LK. Then, if I g is the moment of inertia about OF, and J z the moment of inertia about LK, we have ^^ 1=11 x*dxdy, It i I xldxdy. (1) JJ JJ Moreover, if a is the distance between OF and LK, we have so that, by substituting from (2) in the second equation of (1), we have rf * **. np Ji- x*dxdy + 2 a I I xdxdy + a* I dxdy. (3) Now, by 84, ndxdy^Ai by 85, \( xdxdy = Ax = 0, since by hypothesis x = ; and, by (1), the first integral on the right hand of (8) is I g . Therefore (8) can be written 7,-7,+ oU, (4) which proves the theorem for this case. Case II. When the axes are perpendicular to the plane of the figure. We have to do now with polar moments of inertia. Let the area be placed as in Fig. 118, where the center of gravity is 268 REPEATED INTEGRATION taken as the origin, and P is any point about which we wish the polar moment of inertia. Let I g be the polar moment of inertia about 0, and I p the polar moment of inertia about P. Draw through P axes PX' and , PY 1 parallel to the axes of coor- dinates OX and OY. Let I x and I tl be the moments of inertia about OX and OY respectively, and let 1^ and J tf / be the moments of inertia about PX' and PY'. Then, by (5), 88, / / J J.-- Moreover, if (a, 6) are the coordinates of P, we have, by Case I, 1^ = 1^ a *A, /,=/ + VA. (6) Therefore, from (5), we have /,==/,+ <V+&V> (7) which proves the theorem for this case also. The student may easily prove that the theorem is true also for the moment of inertia of any solid of revolution about an axis parallel to the axis of revolution of the solid. Ex. Find the polar moment of inertia of a circle with respect to a point on the circumfeience. The center of gravity of a circle is at its center, and the distance of any point on its circumference from its center is a. By Ex 5, 88, the polar moment of a circle about its center is |ira*. Therefore, by the above theorem, I, = i* + o (F) t *. This result agrees with Ex 4, 88, where the required moment of inertia was found directly EXERCISES 1. Find the moment of inertia of a circle about a tangent. 2. Find the polar moment of inertia about an outer corner of a picture frame bounded by two rectangles, the outer one being of dimensions 8 ft. by 12 ft, and the inner one of dimensions 5 ft. by 9 ft. SPACE COORDINATES 269 3. Find the moment of inertia about one of its outer edges of a carpenter's square of which the outer edges are 15 in. and the inner edges 13 in 4. Find the polar moment of inertia about the outer corner of the carpenter's square in Ex. 3. 5. From a square of side 20 a circular hole of radius 5 is cut, the center of the circle being at the center of the square. Find the moment of inertia of the resulting figure about a side of the square. 6. Find the polar moment of inertia about a corner of the square of the figure in Ex. 5. 7. Find the moment of inertia of a hollow cylindrical column of outer radius r a and inner radius 1\ about an element of the inner cylinder. 8. Find the moment of inertia of the hollow column of Ex. 7 about an element of the outer cylinder. 9. Find the moment of inertia of a circular ring of inner radius ^ and outer radius ?- a about a tangent to the outer circle. 10. A circle of radius a has cut from it a circle of radius 75 tangent i to the larger circle Find the moment of inertia of the remaining figure about the line through the centers of the two circles 11. Find the moment of inertia of the figure in Ex 10 about a line through the center of the larger circle perpendicular to the line of centers of the two circles and in the plane of the circles. 90. Space coordinates. In the preceding pages we have be- come familiar with two methods of fixing the position of a point in a plane ; namely, by Cartesian coordinates (x, y), and by polar coordinates (r, 0). If, now, any plane has been thus supplied with a coordinate system, and, starting from a point in that plane, we measure another distance, called 0, at right angles to the plane, we can reach any point in space. The quan- tity will be considered positive if measured in one direction, and negative if measured in the other. We have, accordingly, two systems of space coordinates. 1. Cartesian coordinates. We take any plane, as -3T0F, in which are already drawn a pair of coordinate axes, OX and OF, at right angles with each other, Perpendicular to this plane at 2TO BEPEATED INTEOKATION z the origin we draw a third axis OZ (Fig. 114). If P is any point of space, we draw PM parallel to OZ, meeting the plane XO Y at Jf, and from M draw a line par- allel to OF, meeting OX at L. Then for the point P (#, y, z), OL x, LM y, and MP = z. It is to be noticed that the three axes determine three planes, XOY, YOZ, and ZOX, called the coor- dinate planes, and that we may just as readily draw the line from P perpendic- ular to either the plane YOZ or ZOX and then complete the construction as above. These possibilities are shown in Fig. 115, where it is seen that x = OL = NM= SE = TP, with similar sets of values for y and s. 2. Cylindrical coordinates. Let XOY be any plane in which a fixed point is the origin of a system of polar coor- dinates, and OX is the initial line of that system (Fig. 116). Let OZ be an axis perpendicular to the plane XOY at 0. If P is any point in space, we draw from P a straight line parallel to OZ until it meets the plane XOY at M. y Then, if the polar coordinates of M in the plane XOY are r = OM, 6 = XOM, and we denote the dis- tance MP by s, the cylindrical coordinates of P are (r, 0, g). It is evident that the axes OX and OZ determine a fixed plane, and that the angle 6 is the plane angle of the dihedral angle between that fixed plane and the plane through OZ and the point P. If SP is drawn in the latter plane perpendicular to OZ, it is evident that OM= SP = r and OS = MP = z, The coordinate r, therefore, measures the distance of the point P from the axis OZ, and the coordinate 2 measures the distance of P from the plane Z 116 SUEFACES 271 If the line OX of the cylindrical coordinates is the same as the axis OX of the Cartesian coordinates, and the axis OZ is the same in both systems, it is evident, from 51, that a; = r cos 0, y == r sin 0, = z. (1) These are formulas by winch we may pass from one system to the other. It is convenient to notice especially that (2) 91. Certain surfaces. A single equation between the coordi- nates of a point in space represents a surface. We shall give examples of the equations of certain surfaces which are impor- tant in applications. In this connection it should be noticed that when we speak of the equation of a sphere we mean the equation of a spherical surface, and when we speak of the volume of a sphere wo mean the volume of the solid bounded by a spherical surface. The word sphere, then, indicates a sur- face or a solid, according to the context. Similarly, the word cone is used to denote either a conical surface indefinite in extent or a solid bounded by a conical surface and a plane base. It is in the former sense that we speak of the equation of a cone, and in the latter sense that we speak of the volume of a cone. In the same way the word cylinder may denote either a cylindrical surface or a solid bounded by a cylindrical surface and two piano bases. This double use of these words makes no con- fusion in practice, as the context always indicates the proper mean- ing in any particular case. 1. Sphere will center at origin. Con- sider any sphere (Fig. 117) with its center at the origin of coordinates and its radius equal to a. Let P be any point on the surface of the sphere. Pass a plane through P and OZ, draw PS perpendicular to OZ, and connect and P. Then, using cylindrical co6rdinates, in the right triangle OPS, OS=z, SP^r, and OP as a, Therefore . 8s= a 9 . (1) FIG. 117 272 REPEATED INTEGRATION This equation is satisfied by the cyhnducal coordinates of any. point on the surface of the sphere and by those of no other point Jt is therefore the equation of the sphere in cylindrical coordinates. By means of (2), 90, equation (1) *"** *+? + *,* (>) which is the equation of the sphere in Cartesian coordinates. 2. Sphere tangent at origin to a cooi 1 - dinate plane. Consider a sphere tangent to the plane XOY at (Fig 118). Let P be any point on the surface of the sphere Let A be the point in which the axis OZ again meets the sphere Pass a plane through P and OZ , connect A. and P, and P , and diawPS perpendicular to OZ Then, using cylindrical coordi- nates, OS = z, SP = i, and OA = 2 a, where a is the radius of the sphere Now OA P is a right triangle, since it is inscribed in a semicircle, and PS is the perpendicular from the vertex of the right angle to the hypotenuse. Therefore, by elementary plane geometry, ~SP*=OS' SA = OS(OA - 08) Substituting the proper values, we have X FIG. 118 Z (3) which is the equation, of the sphere in cylindrical coordinates. B J ( 2 ) 90 > equation (3) becomes # a + f y a +2 2 2 a? = 0, (4) which is the equation of the sphere in Cartesian coordinates. 3 Right circular cone. Consider any right circular cone with its vertex at the origin and its axis along OZ (Fig 119). Let a be the angle which each element of the cone makes with OZ. Take P any point on the surface of the cone, pass a plane through P and OZ, and draw PS perpendicular to OZ. Then SP SP = r and OS = z. But = tan SOP - tan a Therefore we have C/o r = gtano (5\ \ V J as the equation of the cone in cylindrical coordinates, FIG. 119 SURFACES 273 By 2, 90, equation (.">) becomes 0, (6) as the equation of the cone in Cartesian coordinates, As explained above, we have heie used the word cone in the sense of a conical surface If the cone is a solid with its altitude h and the radius of its base a, then tan a - In this case equation % li (5) or (6) is that of the curved surface of the cone only. 4. Surface of revolution. Consider any sur- face of revolution with OZ the axis of revolution (Fig 120). Take P any point on tho smface and pass a plane through P and OZ In the piano POZ draw OR peipendicular to OZ and, fiom P, a straight line perpendicular to OZ meeting OZ in S If we regard OR and OZ as a pair of i octangular axes foi the plane POZ, tho equation of the cuive CD in which the plane POZ cuts the surface is FIG 120 =/(r) f (7) exactly as y ~f(x) is the equation of a curve in 12 But CD is the same curve in all sections of the surface through OZ Therefore equation (7) is true for all points P and is tho equation of the sur- face in cylindrical coordinates. Whan the plane POZ coincides with the plane X OZ t r is equal to x, and equation (7) becomes, z for that section, z= .ff y .\ > /-g\ Hence we have the following theorem : The equation of a surface of revolution formed by revolving about OZ any curve in the plane XOZ may be found in cylindrical coordinates by writing rfor x in the equation of the curve. Tho equation of the surface in Cartesian coor- dinates may thtn be found by placing r = Vr* + y*. For example, the equation of the surface formed by revolving the parabola 2 =4o: about OZ as an axis is z a as 4 ? in cylindrical co&rdinates, or s*sBlO(.e fl + y 8 ) in Cartesian coordinates. 5. Cylinder. Consider first a right circular cylinder with its axis along OZ (Fig. 121). From any point P of the surface of the cylinder draw 7 J S perpendicular to OZ. Then 8P is always equal to , the radius of the cylinder. Therefore, for all points on the surface, reo, (9) 121 274 REPEATED INTEGRATION which is the equation of the cylinder in cylindrical cobrdinates. Reduced to Cartesian coordinates equation (9) becomes x*+y*=a?, (10) the equation of the cylinder in Cartesian coordinates. More generally, any equation in x and y only, or in r and 6 only, repre- sents a cylinder. In fact, either of these equations, if interpreted in tho plane XOY, represents a curve, but if a line is drawn from any point in this curve perpendicular to the plane XOY, and P is any point on tins lino, the coordinates of P also satisfy the equation, since z is not involved in the equation. As examples, the equation y a =4:x xepresputs a parabolic cylinder, and the equation r = a sin 3 & represents a cylinder whoso base is a rose of three leaves (Fig. 65, p. 144). 6 Ellipsoid Consider the surface defined by the equation c 2 (11) If we place z = 0, we get the points on the surface which Ho in tho XO I' plane These points satisfy the equation . * - + *= = ! (12) _.^ and therefore form an ellipse. Similarly, the points in the ZOX plane lie on the ellipse S + 7 3 = 1 > (13) and those in the YOZ plane lie on the ellipse 6 s c 2 (14) FIG. 122 The construction of these ellipses gives a general idea of the shape of the surface (Fig 122). To make this more precise, let UH place * = j in (11), where z x is a fixed value. We have which can be written c 2 ==1, (16) SURFACES 275 As long as z < c 3 , equation (16) represents an ellipse with semiaxes a v'l " and b A/1 ^ By taking a sufficient number of these sections we may construct the ellipsoid with as much exactness as desired. If z = c 8 in (16), the axes of the ellipse reduce to zero, and we have a point If Sj a > r a , the axes are imaginaiy, and there is no section. 1 Elliptic paraboloid. Consider the surface where we shall assume, foi defimteness, that c is positive. If we place z = 0, we get 2 - 2 + f 8 = > ^ ft A /ift (17) (18) which is satisfied in real quantities only by x and y = 0. Therefore the XOY plane simply touches the surface at the origin If we place z = c, we get the ellipse 1 ' (19) which lies in the plane c units distant from the XOY plane If we place y = 0, we get the parabola and if we place a? = 0, we get the parabola FIG. 123 The sections (10), (20), and (21) determine the general outline of surface. For more detail we place z = s x and find the ellipse c c so that all sections parallel to the XOY plane and above it are ellipses (Fig. 123). 8. Elliptic cone. Consider the surface , I 3 -!! 0. (23) Proceeding as in 7, we find that the section z = is simply the origin and that the section, z = cis the ellipse 276 REPEATED INTEGRATION If we place x 0, we get the two straight lines i =-t*- and if we place y = 0, we get the two stiaight lines a = -z c (25) (20) The sections we have found suggest a cone with an elliptic baso. To prove that the surface really is a cone, we change equation (SJiJ) to cylin- cotirdmates, obtaining ^ + E^\ r a = ~- nm 2 /i2 / /(2 V <* B I/ / C Now if 6 is held constant in (27), the coefficient of r z is constant, and thi equation may be written T 1 J r = kz, (28) which is the equation of two straight hues in the plane through OK Z Z FIG 124 determined by & = const Hence any plane trough OZ cuts the surface two straight lines, and the surface is a cone (Fig. 124). 9. Plane Consider the surface n Ax + ED + Cz -f D = 0. /29\ The section z = is the straight hne ffff (P lg 1 25 ) with the equation Ax + Vy + DssQ, (80) the secfcon y = is the straight line LH with the equation At + a+DszO, (31) and the section , = ia the straight hne L/C with the equation By + Cz + T) = 0. VOLUME 277 The two lines (31) and (32) intersect OZ in the point L (O, 0, J , unless C = 0. Assuming for the present that C is not zero, we change equation (29) to cyhndncal cooidmates, obtaining (.4 cos 6 + B sin d)r + Cz + D = 0. (33) This is the equation of a stiaight line LN in the plane B = const It passes through the point L, which has the cyhndncal coordinates r = 0, 2 _ ~i ; and it meets the line KH, since when z = 0, equation (33) is the same as equation (30). Hence the sui face is covei ed by straight lines which pass through L and meet KH. The locus of such lines is clearly a plane We have assumed that C in (29) is not zero. If C = 0, equation (29) is Q. (34) The point L does not exist, since the lines corresponding to HL and KL are now parallel. But, by 5, equation (34) lepresents a plane parallel to OZ intersecting XOYm the line whose equation is (34) Theiefoie we have the following theorem Any equation of the first degree represents a plane. 92. Volume. Starting from any point (#, y, z) in space, we may draw linos of length dx, dy, and dz in directions parallel to OX, OF, and OZ respectively, and on these lines as edges construct a rectangular parallelepiped. The volume of this fig- ure we call the element of volume dV and have (1) For cylindrical coordinates we construct an element of volume whose base is rd6dr( 84), the element of plane area in polar coordinates, and whose altitude is dz. This figure has for its volume dV the product of its base by its altitude, and we have d7=rd0drdz. ( 2 ) The two elements of volume dV given in (1) and (2) are not equal to each other, since they refer to differently shaped figures. Each is to be used in its appropriate place. To find the volume of any solid we divide it into elements of one of these types. , To do this in Cartesian coordinates, note that the a-coordinate of any point will determine a plane parallel to the plane YOZ 278 REPEATED INTEGRATION and x units from it, and that similar planes correspond to the values of y and & We may, accordingly, divide any ruquirotl volume into elements of volume as follows: Pass planes through the volume parallel to Y0% ami <ta* units apart. The result is to divide the required volume into slices of thickness dx, one of which # is shown in Fig. 126. Secondly, pass planes through the volume parallel to JTO^and dy units apart, with the result that each slice is divided into columns of cross sec- tion dxdy. One such column is shown in Fig. 126. Finally, pass planes through the required volume parallel to XQY and dz units apart, with the result that each column is divided into rectangular parallelepipeds of dimensions das, dy, and fe, One of these is shown in Fig. 126. It is to be noted that the order followed in the above explanation is not fixed and that, in fact, the choice of be- ginning with either a or y or 4 and the subsequent order depend upon the particular volume z considered. A similar construction may be made for cylindrical coordinates. In this case the coordinate & determines a plane through OZ. We accordingly divide the volume by means of planes through OZ making the angle d0 with each other. The result is a set of slices one of which is shown in Fig. 127 Fro. 127 VOLUME 279 Finally, these columns are divided into elements of volume by planes parallel to XOY at a distance dz apart. One such element is shown in Fig. 127. When the volume has been divided in either of these ways, it is evident that some of the elements will extend outside the boundary surfaces of the solid. The sum of all the elements that are either completely or partially in the volume will be approximately the volume of the solid, and this approximation becomes better as the size of each element becomes smaller. In fact, the volume is the limit of the sum of the elements. The determination of this limit involves in principle three in- tegrations, and we write = I | | dxdyds (3) or V= CCCrd6drdz. (4) In carrying out the integrations we may, in some cases, find it convenient first to hold z and dz constant. We shall then be taking the limit of the sum of the elements which lie in a plane parallel to the XOY plane. We may indicate this by writing (3) or (4) in the form V=* Cdz CCdxdy or V= Cdz CCrdOdr. (5) But, by 84, 1 1 dxdy = A and \\rd6dr~A, where A is the area of the plane section at a distance from XO Y. Hence (5) is r F= / Adss, (6) in agreement with 26. ** Hence, whenever it is possible to find A by elementary means without integration, the use of (6) is preferable. This is illus- trated in Ex. 1. In some cases, however, this method of evaluation, is not convenient, and it is necessary to carry out three integrations. This is illustrated in Ex. 2. 280 REPEATED INTEGRATION Ex. 1. Fmd the volume of the ellipsoid + =1. By 6, 91, the section made by a plane parallel to XOY is an ellipse with semiaxes a %/! ^ and Z> %/! - ^ Therefore, by Ex. 1, 77, its area / z a \ c is waft (1 1 Hence we use formula (6) and have \ c 2 / r e l s a \ 4 V = irab I 1 1 -]ds=:- irdbc. J_ c \ c 2 / 3 Ex. 2. Find the volume bounded above by the sphere a: 2 -f ?/ a + s 2 = 5 and below by the paraboloid a; 2 + y z = 4 z (Fig 128). As these are stu faces of i evolution, this example may be solved by the method of Ex. 1, but in so doing we need two integrations one foi the sphere and the other foi the paraboloid We shall solve the example, however, by the other method in order to illustrate that method We fii st reduce our equations to cyhn- ^ dncal coordinates, obtaining lespectively r* + z* = 5 (1) and r 1 = 4 z (2) The surfaces intersect when r has the same value in both equations; that is, when z* + 4 z = 5, (3) which gives z=loiz = 5 The latter value is impossible , but when z = 1, we have r = 2 in both equations Theiefore the surfaces intersect in a circle o radius 2 in the plane z = 1 lies duecHly above the circle r = 2 in the XOY plane. We now imagine the element tdQdrdz inside the surface and, holding r, 0, dd, dt constant, we take the sum of all the elements obtained by varying z inside the volume These elements obviously extend from 2 = s l in the lower boundary to z = z 2 in the upper boundary. From (2), z 1 = and, from (1), 2 2 V5 r 8 . The first integration is therefore This circle rdddr MOdr. We must now allow 6 and r so to vary as to cover the entire circle ? == 2 above which the required volume stands. ' If we hold Q constant, r varies from to 2. The second integration is therefore VOLUME 281 Finally, 9 must vary from to 2 IT, and the third integration is /5V5 If we put together what we have done, we have />2ir /.a ft F=f / / Jo Jo t/, EXERCISES 1. Find the volume bounded by the paraboloid = y? + ^ and the planes x = 0, y = 0, and g = 4. & C^ 9*2 2. Prove that the volume bounded by the surface - = -5 4- fa and the plane * = c is one half the product of the area of the base by the altitude. 3. Find the volume bounded by the plane = and the cylinders SB B -f f = 2 and y* = a* az. 4. Find the volume cut from the sphere r 2 4- 2 = a 2 by the cylinder r = a cos 6. 5. Find the volume bounded below by the paraboloid r* = a and above by the sphere r 2 4- 2 2 a# = 6. Find the volume bounded by the plane XOY, the cylinder a-a {> 2/a _ 2 ax = 0, and the right circular cone having its vertex at 0, its axis coincident with OZ, and its vertical angle equal to 90 7. Find the volume bounded by the surfaces r a = &, = 0, and y sa a COS 0, 8. Find the volume bounded by a sphere of radius a and a right circular cone, the axis of the cone coinciding with a diameter of the sphere, the vertex being at an end of the diameter, and the vertical angle of the cone being 90 . 9. Find the volume of the sphere of radius a and with its center at the origin of coordinates, included in the cylinder having for its base one loop of the curve i* ;= a 2 cos 2 6. 10, Find the volume of the paraboloid a? 4- f = 2 * cut off by the plane #~x 4-1. 11. Find the volume of the solid bounded by the paraboloid 1 and the plane * = *. dm 282 EEPEATED INTEGRATION 93. Center of gravity of a solid. The center of gravity of a solid has three coordinates, x, % z, which are defined by the equations I xdm I ydm I *= J ' ' v = J r ' - I dm I dm I where dm is the mass of one of the elements into which the solid may be divided, and #, y, and a are the coordinates of the point at which the element dm may be regarded as concentrated. The derivation of these formulas is the same as that in 85 and is left to the student. When dm is expressed in terms oi space coordinates, the integrals become triple integrals, and the limits 'of integration are to be substituted so as to include the whole solid. We place dm = pdF, where p is the density. If p is constant, it may be placed outside the integral signs and canceled from numerators and denominators. Formulas (1) then become 7x= CxdV, Vy=CydV, Vz*= CzdV. (2) Ex. Find the center of gravity of a body bounded by one nappe of a right circular cone of vertical angle 2 a and a sphere of radius a, the center of the sphere being at the vertex of the cone. If the center of the sphere is taken as the origin of coordinates and the axis of the cone as the axis of 2, it is evident from the symmetry of the solid that x = y = Q. To find z we shall use cylindrical coordinates, the equations of the sphere and the cone being respectively r 2 + z 2 =s o a and r z tan a. As in Ex. 2, 92, the surfaces intersect in the circle r a sin a in the plane z = a cos a. Therefore />2r a a tin a />Va 8 r 8 V=\ I / rdOdrdz = S Tra 8 (1 - cos a) JO /o t/rotnar o \ ./ n nZv n a sin a /-\/a 2 r* and \zdV-\ \ \ rzdOdrdz = Iwa 4 sin a cr. t/ i/O /0 t/rctnrt w Therefore, from (2), 5 s? f a (1 + cos a). CENTER OF GRAVITY 283 EXERCISES 1. Find the center of gravity of a solid bounded by the paraboloid y 7 / - = ' -f 7; and the plane = c c a 2 1r L 2. A ring is cut from a spherical shell, the inner radius and the outer radius of which are respectively 4 ft. and 5 ft., by two parallel planes on the same side of the center of the shell and distant 1 ft and 3 ft. respectively from the center. Find the center of gravity of this ring. 3. Find the center of gravity of a solid in the form of the frustum of a right circular cone the height of which is h, and the radius of tho upper base and the radius of the lower base of which are respectively r^ and ?' 2 . 4. Find the center of gravity of that portion of the solid of Ex. 2, p. 73, which is above the plane determined by OA and OB (Fig. 31). 5. Find the center of gravity of a body in the form of an octant Qn A a i * ej* of the ellipsoid ~ + 73 + -= = 1. e a? o* <? 6. Find the center of gravity of a solid, bounded below by the paraboloid az r* and above by the right circular cone * + f = 2 a. 7. Find the center of gravity of a solid bounded below by the cone * SB r and above by the sphere r* + # 2 = 1. 8. Find the center of gravity of a solid bounded by the surfaces JB = 0, i* + * a ** i 3 , and r = 94. Moment of inertia of a solid. If a solid body is divided into elements of volume c?F, then, as in 88, the moment of inertia of the solid about any axis is 1= CtfpdV** p C&dT, (1) where JB is the distance of any point of the element from the axis, and p is the density of the solid, which we have assumed to be constant and therefore have been able to take out of the integral sign. If M is the total mass of the solid, p may be determine^ from the formula JtfwpK 284 REPEATED INTEGRATION If the moment of inertia about OZ, which we shall call 1^ is required, then hi cylindrical coordinates JR = r and dV= r$6drdz, so "that (1) becomes I z = p CCCr*d0 drds. (2) If we use Cartesian coordinates to determine /,,, we have and dV= dxdydz, so that 'xdydz. (3j Similarly, if I y and I x are the moments of inertia about OY and OX respectively, we have f\ f\ SI (4) In evaluating (2) it is sometimes convenient to integrate with respect to z last. We indicate this by the formula I a = p Cdz CCr 9 dO dr. (5) But I I i^dddr is, by 88, the polar moment of inertia of a plane section perpendicular to OZ about the point in which OZ intersects the plane section. Consequently, if this polar moment is known, the evaluation of (5) reduces to a single integration. This has already been illustrated in the case of solids of revolution. A similar result is obtained by considering (3). In fact, the ease with which a moment of inertia is found depends upon a proper choice of Cartesian or cylindrical coordinates and, after that choice has been made, upon the order in which the integra- tions are carried out. Equation (3) may be written in the form >WCy*dxdydt, (6) and the order of integration in the two integrals need not be the same. Similar forms are derived from (4). The theorem of . 89 holds for solids. This is easily proved by the same methods used in that section. MOMENT OF INERTIA 285 Ex. Find the moment of inertia about OZ of a cylindrical solid of altitude h whose base is one loop of the curve r a sin 3 6, The base of this cylinder is shown in Fig 65, p. 144. We have, from formula (2), - In8fl wheie the limits are obtained as follows: First, holding r, 6, d6, dr constant, we allow z to vary from the lower base s = to the nppei base z = li, and integrate. The result phr a dOdi is the moment of inertia of a column such as is shown in Fig 127. We next hold and d& constant and allow r to vary from its value at the origin to its value on the curve r = a sin 3 0, and integrate. The result \ /7i 4 ain 4 3 Qd& is the moment of inertia of a slice as shown in Fig. 127. Finally, we sum all those slices while allowing to vary from its smallest value to its largest value ^ The result is zz The volume of the cylinder may be computed from the formula a" 1 *- /idBinSO nh V = I f i rdBdrdz = A Jo Jo Jo Therefore Af= -faphePir and J s =JI/ 2 . EXERCISES 1. Find the moment of inertia of a rectangular parallelepiped about an axis through its center parallel to one of its edges. 2. Find the moment of inertia about OZ of a solid bounded by the surface t= 2 and $ = r. 3. Find the moment of inertia of a right circular cone of radius a and height h about any diameter of its base as an axis. 4. Find the moment of inertia aboxit OZ of a solid bounded by 2 Q the paraboloid * =* ~ + ^ and the plane = e. 5. Find the moment of inertia of a right circular cone of height li and radius a about an axis perpendicular to the axis of the cone at its vertex. 6. Find the moment of inertia of a right circular cylinder of height h and radius a, about a diameter of its base. 7. Find the moment of inertia about OZ of the portion of the sphere ?* 4- *" ~ a a out out by the plane and the cylinder r ** a cos 6. 286 REPEATED INTEGRATION 8. Find the moment of inertia about OX. of a solid 'bounded by the paraboloid r* and the plane = 2. 9. Find the moment of inertia about its axis of a right elliptic cylinder of height h } the major and the minor axis of its base being respectively 2 a and 2 b. 10. Find the moment of inertia about OZ of the ellipsoid t+t + t =i I ' 7,2 ' 2 ^ GENERAL EXERCISES n n n 1. Find the center of gravity of the arc of the curve x* -I- y* = *, which is above the axis of x. 2. A wire is bent into a curve of the form 9y 2 = a? 8 . Find the center of gravity of the portion of the wire between the points for which x = and x = 5 respectively. 3. Find the center of gravity of the area bounded by the curve ay* = a? and any double ordmate. 4. Find the center of gravity of the area bounded by the axis of x, the axis of y, and the curve j/ 2 = 8 2 x> 5. Find the center of gravity of the area bounded by the curves y = x 9 and y = -5 > the axis of x, and the line x = 2 6. Find the center of gravity of the area bounded by the axes of x and y and the curve x = a cos 8 ^, y = a sin s < 7. Find the center of gravity of the area bounded by the ellipse jg2 & ~z + a =1 ( a > *) tlie <Hfle as 2 + y 2 = a , and the axis of y. 8. Find the center of gravity of the area bounded by the parabola v? = 8 y and the circle cc a + f = 128 9. Find the center of gravity of the area bounded by the curves ** ~ a (y - &) = 0, a 2 ay = 0, the axis of y, and the line = o, ^ 10. Find the center of gravity of an area in the form of a semi- circle of radius a surmounted by an equilateral triangle having one of its sides coinciding with the diameter of the semicircle. 11. Find the center of gravity of an area in the form of a rec- tangle of dimensions a and I surmounted by an equilateral triangle one side of which coincides with one side of the rectangle which is b units long. GENERAL EXERCISES 287 12. Find the center of gravity of the segment of a circle of radius a cut off by a straight line b units from the center. 13. From a rectangle b units long and a units broad a semicircle of diameter a units long is cut, the diameter of the semicircle coinciding with a side of the rectangle. Find the center of gravity of the portion of the rectangle left. 14. Find the center of gravity of a plate in the form of one half of a circular ring the inner and the outer radii of which are respectively ' x and > 15. In the result of Ex. 14, place r z = r^ + AT- and find the limit as Ar vO, thus obtaining the center of gravity of a semicircumference. 16. Find the center of gravity of a plate in the form of a T-square 10 in across the top and 12 in. tall, the width of the upright and that of the top being each 2 in. 17. From a plate in the form of a regular hexagon 5 in. on a side, one of the six equilateral triangles into which it may be divided is removed. Find the center of gravity of the portion left. 18. Find the center of gravity of a plate, in the form of the ellipse r o/ ~ + 75 = 1 (a > i), in which there is a circular hole of radius c, the center of the hole being on the major axis of the ellipse at a distance d from its center. 19. Find the center of gravity of the solid formed by revolving !K 2 ?/ 2 about OY the surface bounded by the hyperbola -j j$ =1 and the lines ?/ = and y = &. a 20. Find the center of gravity of the solid generated by revolving about the line % = a the area bounded by that line, the axis of a;, and the parabola y a = hx. 21. Find the center of gravity of the segment cut from a sphere of radius a by two parallel planes distant respectively \ and h z (h z > 7^) from the center of the sphere. 22. Find the moment of inertia of a plane triangle of altitude a and base b about an axis passing through its center of gravity parallel to the base. 23. Find the moment of inertia of a parallelogram of altitude a and base b about its base as an axis, 288 BEPEATED INTEGRATION 24. Find the moment of inertia of a plane circular ring, tho inner radius and the outer radius of which are respectively 3 m and 5 in., about a diameter of the ring as an axis. 25. A square plate 10 in. on a side has a square hole 5 in. on a side cut in it, the center of the hole being at the center of tho jilafco and its sides parallel to the sides of the plate. Find the moimmt of inertia of the plate about a line through its center parallel to ouo side as an axis. 26. Find the moment of inertia of the plate of Ex. 25 nbout one of the outer sides as an axis. 27. Find the moment of inertia of the plate of Ex. 20 about one side of the hole as an axis. 28. Find the moment of inertia of the plate of Ex 2tf about ono of its diagonals as an axis 29. A square plate 8 in. on a side has a circular hole 4 in. in diameter cut in it, the center of the hole coinciding with tho cuntor of the square Find the moment of inertia of the plate about a lino passing through its center parallel to one side as an axis. 30. Find the moment of inertia of the plate of Ex. 29 about a diagonal of the square as an axis 31. Find the moment of inertia of a semicircle about a tangent parallel to its diameter as an axis. 32. Find the polar moment of inertia of the plate of Ex. 25 about its center. 33. Find the polar moment of inertia of the entire area bounded by the curve T- 2 = a? sm 3 6 about the pole. 34. Find the polar moment of inertia of the area bounded by tho cardioid r = a ( 1 + cos ff) about the pole. 35. Find the polar moment of inertia of that area of tho circle r = a which is not included in the curve r a, sin 2 6 about tho pole. 36. Find the moment of inertia about OF of a solid bounded by the surface generated by revolving about OY the area bounded by the curve </ x, the axis of y, and the line y 2. 37. A solid is in the form of a hemispherical shell the inner radius and the outer radius of which are respectively and r , Find its moment of inertia about any diameter of the base of the shell as an axis. GENERAL EXERCISES 289 38. A solid is in the form of a spherical cone cut from a sphere of radius a, the vertical angle of the cone being 90. Find its moment of inertia about its axis. 39. A solid is cut from a hemisphere of radius 5 in by a right circular cylinder of radius 3 in, the axis of the cylinder being perpendicular to the base of the hemisphere at its center Find its moment of inertia about the axis of the cylinder as an axis. 40. An anchor ring of mass M is bounded by the surface generated by revolving a circle of radius a about an axis in its plane distant b(b > <i) from its center. Find the moment of inertia of this anchor ring about its axis. oj2 ?/" 41. Find the moment of inertia of the elliptic cylinder + 77 =1 a* (r (a > ft), its height being h, about the major axis of its base. 42. Find the center of gravity of the solid bounded by the cylinder = 2 a cos 0, the cone ss = r, and the plane x = 43. Find the moment of inertia about OZ of the solid of Ex. 42. 44. Find the volume of the cylinder having for its base one loop of the curve r = 2 a cos 2 0, between the cone = 2 r and the plane = 46. Find the center of gravity of the solid of Ex. 44. 46. Find the moment of inertia about OZ of the solid of Ex. 44. 47. Find the volume of the cylinder having for its base one loop of the curve r ss a cos 2 & and bounded by the planes = and ft =s x + 2 a. 48. Find the moment of inertia about OZ of the solid of Ex. 47 49. Find the volume of the cylinder r = 2 a cos included between the planes * = and * =s 2 x + a. 60. Find the moment of inertia about OZ of the solid of Ex. 49. 61. Through a spherical shell of which, the inner radius and the outer radius are respectively r l and r a , a circular hole of radius a (a < fj) is bored, the axis of the hole coinciding with a diameter of the shell. Find the moment of inertia of the ring thus formed about the axis of the hole. ANSWERS [The answers to some problems are intentionally omitted.] CHAPTER I Page 4 ( 2) 1. 21^. 4. 100ft per second 7. 2J 2. l^jj. 5. 88 07 mi. per hour for entire trip. 8. l^mi per hour. 3. 40J\>. 6. 1,26. 9. 08.4. Page 5 ( 2) 10. 106 8. Page 7 ( 3) 1. 06 ft. per second. 2. 128 ft. per second. Page 8 ( 3) 8. 128 ft. per second. 5. 68 ft per second. 4. 74 ft pci bccoud. 6. 52 ft. per second. Page 11 ( 5) 1, 12 ii ; 24 j. 8. 85 ; 82 ; 6. 5. 5, 4, when t = 2 ; 10, 6, when t = 3 2. 16 j 14. 4, 42 5 57. 8. 8ai a + 2M + c ; 6at + 26. Page 13 ( 6) 1. i- sci. ft. per second. 2, 0. 3. 87rr a cu. in. per second. TT 2?r 4, 4wr a . 5, 167irsq. in. per second. Page 14 ( 6) 6. 8irr. 7. 8 (edge) 3 . 8, birr*. 9. 18. 10. 2ir. CHAPTER II Page 18 ( 7) 3. 4fc B -2. 4. :- 5. 2 ; 8, -,_ S . 7. tf + B+1. 8, 8-~ 201 292 ANSWERS Page 21 ( 9) 1. Increasing if x > 2 ; decreasing if x < 2. 2. Increasing if x > g , decreasing if x < |. 3. Increasing if x < ; decreasing if x > . 4. Increasing if a < , decreasing if sc > . 5. Increasing if aj < 2 or x > 1 , decreasing if 2 < K < 1. 6. Increasing if x < 5 or x > S , decreasing if 5 < x < 8. 7. Increasing if < 1 or a: > ; decreasing if 1 < x < fj. 8. Always increasing. 9. Increasing iftc< lor ^<a<l; decreasing if 1 < x < J or a; > 1. 10. Increasing if x > 1 , decreasing if x < 1. Page 24 ( 10) 1. When t < - 1 or t > I , when !<<!. 2. When 5 ; when t > 5. 3. When t < 2 or t > 4 ; when 2 < i < 4. 4. Always moves in dnection in winch a is measxirod. 5. When t > f , when < $. 6. Always increasing. 7. Always decreasing 8. Increasing when t > 2 , decreasing when t < 2. 9. Increasing when t > , decreasing when i < $ 10. Increasing when < < f , decreasing when < > |. Page 26 ( 11) 1. TrA 8 . 2. 6 irh sq. ft. per second. Page 27 ( 11) 8. 2 cm. per second. 5. 0.20 TT 4. 20 9 sq in. per second. 6. 64 cu. ft per second. Tl 3 8 (t tal hol ff ht >*' 8. 4 TT (2 + 12 ( + 36) , t is thickness. Page 31 ( 13) 1. 1.46. 3 0.46; 2.05. 5. 2.41. 2. - 2 07. 4. 1.12 ; S 93. G. 2.52. Page 35 ( 14) 1. 8a;-y_9 = 0. 6, + y + l = *12. tan-ij. 2. 2x + Sy + 3 = 7. aj + 2y + 8 = 0. 18, tan-*12. 3. 21aj-2y-13 = 0. 8. 4x- Sy-1 = 0. 14.*. 4 - J/ + S = ,_ -M.-4y-.6-0. W. i.-Sy-M-o. 6. V3a;-y-2V3-2= ! 0. 10. 5a: - Qy - 4 = 0. W, (-Itfr, 2ft)* *The symbol tan- 1 jj- represents the angle whose tangent is \ (cl. 46). ANSWEES 293 Page 39 ( 15) 1. (- I >|). 2, (2, 4J). 3. (0, 4), (2, 0). 4, (1, 7), (3, 3). 6. (- 2, 0), (1, - 0). 6. (-1, - 3), (3,29). Page 43 ( 17) 1. aSsq.in. 2. Length is twice breadth. Page 44 (5 17) 6, Depth is one-half side of base. 7, 2 portions 4 ft. long ; 4 portions 1 ft. long -o jjt. 8, Breadth = 9. (- 3, 10), (1, 2) 10. 278 + 27^-86 = 11. 18a5 18. tan-ijj 3. 5ft. 4. 50. - - 28 = 0. 5. , .. ; depth = 9, Altitude = ^-f- 5 base - ? (P - perimeter) 4 4 10, 2000 cu. in. ; 2547 cu. in 1L Height of rectangle = radius of semicircle , semicircle of radius, 5 12. -7= in. VS Page 46 ( 18) 1, 426ft. Page 47 ( 18) 8. 40ft. 5. 676ft. 7. y = 2x 8 + x a -4x+6. 3, 05ft. 4, 88$ ft. Page 49 ( 19) 1. 7$. 2. 1 Page 53 ( 20) 8. 0.0001 ; 0.000001 ; 0.00000001 9. 0.000009001 ; 0.000000090001. 3, 62J. 4. 36. 5. Page 54 1. 72 sq. in. 8, 16 Sir cu. in. 6. 10. 000003 sq. in. 11, 456 58 ou. in. 4. 27.0054 ou. in. 5. 28.2749 cu m. 6. 606 0912. 7. 0012. 8. 5.99934 294 ANSWERS Page 55 (General Exercises) _ ' o 2a A 4 x g _ j t 13. Increasing if x > 2 , ' (a - x) 2 " ' (x 2 + I) 2 ' ' 2 Vajs ' decreasing if < - 2. 14. increasing if f<o;<ora;>2, decreasing if a; < f or <j < x < 2. 15. Increasing if x > - , decreasing if y, < - . 2 o, 2(t 16. Increasing if x < ^= or x > ~ , decreasing if ^= < x < = V3 V3 V3 V3 4/7 4 ft 17. Increase if x < - , decrease if x > - 3 o Page 56 (General Exercises) 18. 1< t < 5 19. 2 < * < 5 , 4. 20. Up when < i < 6 J , down when 6 < < 12. 21. Increasing when t > 4 , decreasing when t < 4. 22. D increasing when < 3, v deci easing when t > 3 , speed increasing when 2<i<3ori>4, speed decreasing when !S<2or3<i<4 23. Increasing when !<<2ori>3, decreasing when *<lor2<i<3 24. 0055 in. per minute 25. 8 6 in. per second 27. x + 2 y + 6 = 0. 26. 1 sq in per minute. 28. 7x+ Page 57 (General Exercises) 29. x- 2 = 31. 2x-y + 3 = 0. 83. (-, 8|). 80. x - 2y - 7 = 0. 32. tan-l 34. (lj, 0) 35. (2, - 2). 41. 10. 36. (- 1, 13), (5, - 95). 42. tan-i T V 37. (- 3, 13), (1, - 19). 43. x - y - 11 = 38. (- 4, 20). 44. (1, - 1), (- J, - Page 58 (General Exercises) 46. 6f ft. long 52. y - x 2 + Bx - IB 47. Altitude of cone is | radius of sphere. 53. y = x 8 x* + 7 x 54.85^. i** a ^ *u 48. Altitude= ; sideof base =^_ M 2g 49. 2 pieces 3 in long, 3 pieces 1 in. long 56. 20 1 50 600ft. 57. 72. 51. 56ft 59. 0.0003. Page 59 (General Exercises) 60. 0.00629. 64. 0.09 cu in 67. 24.0024 sq. in. 62. 288 TT cu in. 65. 0.0003. 69, 0.4698. 63, 161. 16 cu. in, 66. 854.1028j 353.8972, Page 66 ( 23) 1. 8J 3. 52 T S . 2. 23,} 4. 166f r Page 67 ( 24) 1. 160ft Page 68 ( 24) 4. When m Page 70 ( 25) 1. 8$ T. Page 71 ( 25) 6. Approx. 2418 Ib. 6. ANSWEES 29i CHAPTER IH 6. 5 7. 36i 6. 42 1 8. 2J. |. 9. 96. 11. 42 J 10. 10|. 12. 10. 2. 140ft. ; 83ft. 3. 57Jft. 5, 8000irft.-lb. 2. 2JT. 3. 3T. Page 75 ( 26) 1. 21 TT. 625V8 ~~ 7. 585}T. 11. 2 1 ft. from upper side 8. 84 A *. 4. l^w. 5. 6577r. cu in. 7. 8 8. 2| Page 76 (General Exercises) 1. 6Jft. 5. 20. 2. 81ft. 3. 10 ft "' ~3~ 4. 8JJ mi- ' 8. Page 77 (General Exercises) 13. Twice as groat. 17. 14, Jft T. 16. 16 to. 35 16. 68 ^TT 19. 96 IT. Page 78 (General Exercises) 24. STT. 26. 115J. 26. 29. 728,049 ft.-lb. Page 81 ( 28) 1. & a + i/ 8 - 8as -f CHAPTER IV 11 cs 0. 0. 6, 8a~ 9. 234| T 10. 2 T. 9. 25f IT 10. 213 JTT. 11. 38| 8. 9. JfT 12. Reduced to | original pressure 20. 34lcu.m. 21.^-1 28. (aft 2 - Jft")ir 27. 9 28. 204| 80. 5301 ft Jb. 3. (-3, 5); 5. *.(- *.!); 296 ANSWERS Page 84 ( 30) M-2,o). MO, -if) 7. as/, ft a- (o.i). 5. 8H ft. . iom/5 MH,0). 6. lOVlOft. 8. i m Page 85 ( 30) 9. y 2 + 6a;-9 = 0. 10. x*-4x-12y + 16= 0. Page 87 ( 31) 2- (0, 3), (0, V6), 5. 9x 2 + 25^a - 36x - 189 = 6. 49a; 2 + 24 y2 _ 12 0y _ 144 _ . Page 91 ( 32) I- (3,0), ( 4 . o, f 2 - (2,0), (Vl3, 0), 3a;2y=0; 1 3. (0, V2), (0, V5), 4. (2V,0); (4,0), Page 102 ( 36) 1. 18s 2 + 22a;_3 8). 9- _ ' ~(a; 2 -9) 2 ' 11. -. 21 12. 13. 7 2a! _i + ^ * 2 ^' 14 . 8. ' ANSWERS 297 15 -" . -. 18. ' 3i tr - i>) \' (x ~"a) (JR - 6) V(a a + x 2 ) 8 ii . M 1 jr8 11 *"' . 19. , 22. v'W-.tfi (x + l)Vx a -l V(x 2 +9) 8 ' r ~ ] . 20. v*--* vV + 8)* V(l + x)* Page 104 ( 37) <tt/ je a Vi/ + x Vy x __x 2 2a 8 x * ""' U , 1/& ai a/// g !_. 8> " i' ' '* *" --J i ^ ,,'S* ""ftw* 8u 8 ' xa 2X2 e ' -3 ! Page 105 (38) Sa O. 2, x~7y + 5 = 0. 3. (-2.-1). 4. PAgel06(88) -. 10, tan-18. 12. 8.*". . 5 5 ton" 1 A- 11. tan-i|. 13. "i * Page 110 (40) 1. x-r 8 5 V44.9tK ,. fo a Bin ) a, x^d/-!)' 4 , V4tf+i. \ 20 20 3. x*-0x + = 0j SVll7-48t + l6i a . B a sill 2a 4. (8,1). 9 ' '4* 11. yaa \ "aj 8V i Moond. 2. 12.6 ft, per second. 3, ^u, perminute. 4, Circle; ! 5 -fl, per secona (x- distance of point from *dl). x , 2,64 ft. per second. Pftge 113 (41) permi nute. 8. 6.6 ft. per second. 6. 0.18 cm. por second. ?, 0.21 in. per mumi*. , ^ti^t.per 8 ccond, where, is the distance of top of ladder, and IB ^dtonoe of foot of ladder, fromDase of pyramid. 298 ANSWERS Page 114 (General Exercises) 20. 3 21. x 8 27. tan- ' ^ " 28. -; tan- 17. 2i 30. tan-i . 5 35. 81. -, tan- 1 -- 2' 2 ~ 7T On i n 11 32. -,lan 33. x$ + ifi = 2 ato - <W8? ' OJ 2 -1- 4 ' (i 2 + I) 2 Page 115 (General Exercises) 37. 20 ft per second , lOVEft per second, (100,20). 1 a ' v a 39. Velocity in path = VCKC + a 5 40. Page 116 (General Exercises) 46. 08 ft. per minute 47. 01 in. per minute. 48. |f sq in per minute 49. 04 in. per second. 50. Length is twice breadth 51 Other sides equal. Page 117 (General Exercises) 57. 2 64 in 58. 41. 6 8 mi per houi , 28 8 mi 42. 43. t 4 |250-3i a _ ' 3 \ 04- 1 3 ft pei second, whore 3s Vs 2 - 400 s is length of lope from man to boat. 44. 06 ft per minute 45. sJu ft per minute. 52. Breadth, 9 m , depth, 9V3 in. 53. Length = | bieadth 54. Side of base, 10 ft , depth, 5 ft 55. Depth = ^ side of base. 56. Radius, 3 in. , height, 3 in. 1 Vi' 60. 8 mi from point on bank nearest to A. bm , , bn , a mi. on land , mi in water 59. 61. 4 1|} mi tiavel ou land Vn 2 m 2 63. l^f-hr. Page 118 (General Exercises) Vn 2 - mr 64. v/lOO mi per hour. Page 126 ( 44) 1. 15 cos 5 x 2. sec 2 -. 3. 2sm 2 2zcos2x. 4. 5 sin 10 x. 65. Velocity m still water = mi per hour 66. Base = aVI , altitude = | b CHAPTER V 6. 5 sin 2 5 x cos 8 5 aj. _ o60!, &X 7. 5 sec 2 tan . 2 2 8. -8csc 8 3a;ctn3x. ANSWERS 299 Page 127 ( 44) 9. hin- 1 . 10. 11. 12. 2 sec x (sou x + tana-) 2 . Page 129 ( 45) . S_. Mil--- 3. 7T, 5. 4. At moan point of motion ; at extreme points of motion. 2 13. 2 cos 4 x. 14. 9tan 4 3aj. 15 . 2 sec 2 x (sec 2 2 x + tan 2 2 x) 16 sm 8 2 a; cos 2 2 a; 17. - ? } 18. -?. 5. At extieme points of motion, at moan point of motion. 6. 2V(fc - 3) (5 - s) ; 4(4-8) 7. TT. 8. 10, 2?r. Page 134 ( 47) 3 7. JfL. 1 8. ( * 1 0. - V !L 8 " ' Vx-x a x v 4 x a 8 4 ' Vl2x- ( .)x a 10 ' n i 5, - 1 11. - 2 + 2 x + a (x + 1) "V 1 n * 12. - , l 18. Page 136 ( 48) 1. 1)3. = 9.42 ft. per second ; % = =F 86.40 ft. per second. Page 137 ( 48) 3. 8 radians per unit of time. Page 138 ( 49) a b cos <t) ' Page 141 ( 50) 17V17 2. 63 6. 300 ANSWERS Page 145 ( 51) 17. Origin , (V3, j). 18. Origin , ( a^j|, |) . 19. Origin , (2, | 21. ^8^20 = 22. r 23. r 24. r 25. x - a = 26. x 2 + V* - 2 ax = 27. 28. Page Page 9. 10. 11. 18. 19. 148 ( 52) 149 (General Exercises) 2(l + sec 2 2x). sec 4 (3x + 2). cos 2 (2- 3 x). sec 2 (x y) + sec 2 (x + y) sec 2 (x y) sec 2 (x + y) 1 .x 1. IT tan-* 1 3. 0. 21. 20. - tan 4 2 a;. l _ (x + 1) Vx - 1 V2 + x - x 2 27. 2(x + l)Vx 1 xV49x 2 -l 4 ig t 5 5 14 8csc 2 4x(ctn4x + l) 15. a tan ox sec 2 ox. 16. 8 cos 8 2 x sin 4 a; cos 6 x. 1 . x)Vx 3 2x 2 V3-4x-4x 2 2~3x Vo x 2 24. 25. 26. Page 29. 30. 31. (x 2 -l)Vx 2 -2 150 (General Exercises) fc Va 2 sm 2 ^ + 6 2 cos 2 ^. 2^_ V41, 5 = 3; 2 35. j 47. Origin ; / ^ ""*" O * / A. 34. 2aVi. 48. Origin; (^,tan-i 2 ). 49. Origin; Page 151 (General Exercises) 51. - = COS(? 52. 53. (x 8 + y 2 ) 2 + 2 aa;(x a + y 2 ) - a 54, tan-i. s= 0. ANSWERS 301 65. 0) tan-12 57. -. 59 IflVift 56. 0, :,ton-i8V5. Kg* 6 ' 72 ' 2 68 '4 61. V2ft 62. At an angle tan- 1 /*; with ground. Page 152 (General Exercises) 63.12m. 65. a. 68. 15 sq ft , 10.04 sq. ft. per second. 64. sVBft. 67. 0.1 ft per second. 69. 26 7 mi. per minute. nn /i /i , & 2 8in0COS(? \ . . ,.., , 70. / b sin & + , ) times angular velocity of AS. where = angle \ Va a -6 a sin 2 0/ CL4.B. 71, <* ~ 2 ) 2 + fa - 3 ) 2 - 1 ; V9 sin 2 * + 4 cosat , where I = (2 k + 1) | 78. 5^-^ = 1 ; OsecSiVtan 2 3< + 4sec 2 3<. Page 153 (General Exercises) 73. sin 20. 75. tan-^Vi 74. aVl+cos 2 o5, fastest when 75. tan- 1 ^ x = far; most slowly when 77. 0, tan- 1 3 Vs. a5 = (2A! + l)-. 78. tan-i^, tan-i4V2 2 79. tan-i3; tari-ij. CHAPTER VI Page 162 ( 55) (The student is not expected to obtain exactly these answers , they are given merely to indicate approximately the solution.) 1. y =0.62 x- 0.70. 2. 1 = 0.00172) Page 163 ( 55) 3. ^=s0.80(2.7)^. 4. c=0.010(0,84). 5. a=0.0000000048Z3'o. 6 j" = 10. Page 165 ( 56) 1 1 12. e- 2ac (3cos3a; 2sin8x). " ~* < n j. 1 13, Ctll"" 1 !!/. 14. 3. 2 a; a** -Una. "' Vi^"+~4* 16, 4. a n ^ _., 9 16. e" + e- 5| rf j. A -. i ' 10. - 4 sec 2*. W- 2 see's. 20:4.8 n 2(e 8a! e- 2a? ) ig. J; . - - - "' ' eaa+e-is* ' asVx + l 302 Page 167 ( 57) ANSWERS 2. ?/ = 45.22e oola: . 4. Page 168 ( 57) 5. P = 10000 e 022 9* 6. c = 0.01e- n44B <. 7, 2 mm, Page 168 (General Exercises) *10. p = 0.018 1 + 24. *11. Load = 102 - 6 6 length. Page 169 (General Exercises) *12. s = 25 (0 40)' *14 t = 1 Vl. *13. c = 010 (0.83)* *15. 1= 0.023 V0 Page 170 (General Exercises) 18. ?/ = "17. pv = 1620 21. 2 csc- 1 2 a;. 25. tan~^x. 18 l . 2 22. 2 (x + 1) e 26 0.898 ' 9 a: 2 - 4 19. ctn x 23. * . 27. 16 8 hr o g2 i . fijX 28. 1000 sec 20. 24. a tan 3 ox. 29. p = 147i Page 171 (General Exercises) 31. 2"\/2e~ 2i , 2e~ 2 * 33. 87. (l + 1) * 35. Vic*. x ' 2 38. Page 176 ( 59) CHAPTER VII 3.8 - 2. , , a; 8 . 2s 5 , 17x7 3. a; + H --- h 8 15 ^ 315 2i 41 61 4- ^ "4" "" -L. - ____ - ____ __ JL 2 32-4 6 + 2 . 4 6 7 e. V2 - 2 2 2 s 3 3 T 5 7 T ' X ~2l~3! + " j * Statement in regard to answeis to exeicises in 55 is true of tins answer, 9. 0.0872 10, 4695, ANSWERS 303 Page 178 ( 60) 5 2 5 2 8 5 8 ___ + .___ _____ + . 8 / ir\ 8 1 V V , oy 2 21 2 si 6 E + i to + 2 2! 2 31 8. 0.7193 9, 0.8480 10. 8.0042 Page 179 (General Exercises) _ + ' 5 ~ 2 * 5 ' 2 1 2_ ' 8 I a j.^ - 1 ^!_I^ ?!. 1 z %~ x a '21 2*31 10. 1 - 2cc + 2 2 - 2fc 8 + . . 14. 9659. 11 i j. 2aj 8 4a* 15. 0.61CO. u . a H . x -- + ... T 21 8! 5!^ ' 18. 0.69815; 10986. 13 3.4.?! + ! + ^!+..., 19> - 22 314; 1.6094. Page 180 (General Exercises) 88, 2.0805. 2e a;8 a 6 _ a 7 M, 2.9625. ' "*" 3 5(2 I) "" 7(8 !) 304 ANSWERS CHAPTER VIII Page 183 ( 61) 8j/ 2 , . xy x 2 ' xy y 3 5. y X 6. ---, (K + y)* x + y ' s._L_, *_. 7 i * x 2 +2/ 2 ' x 2 +i/ 2 y ' tf y x 1 4. , > , s. , ; VI x a y 2 Vl x 2 ^ 2 v x 2 + y a Page 185 ( 62) x 2 y* 1. -5 5- 2. e" sm (x y) 3, Page 187 ( 63) 1. 000061 2. 0012. 3, 2^ Page 188 ( 63) 4.0.018m 5.00105 6.0015m. 7. 6320ft. 8.0.0064. Page 191 ( 64) 1. -2. 2. -| 4. - 5. - J . o. 6. 0;0 Vx 2 + y 2 Page 192 (General Exercises) 8. - 14.33 cu ft. 10. 0.5655 sq. in.; 11. 3.0 in 13. 2.206 sq. in. per second 9l 1735 05756sq.m 12. 035m, 15. 4.4 uq. in. per second. Page 193 (General Exercises) 16. _iyJ5. 17 /y-tan-ia fit ia J V5 . 5 a tan A , &*. is. -cos a . 'Sina, 1. CHAPTER IX Page 198 ( 66) . -p if* -p uiiiij,. 7. i(x* + 4)a. 14 1 , . 2. g \(3x+7)x4 8. llnW-im. ' ~a ln ( 1 + c ^). 9. ^In(2x + sin2x) 14. ~Jcos*2x. 10. - 1 W- 4. i nz __! L. 3(x-smx) 16 a: 2. 758 .1 1)]. a 17. _ 12. ln(x 8 + Sx 2 + 1). ig, i. 20. ANSWERS 805 Page 202 ( 67) , 1 ,3x 1. -sin- 1 - 3 4 i .sin- 1 Vs 4. sec -1 xV8. i a 5. ' V7 V? V21 4x-3 3 6x-5 11. -^-sm-i Vs 5 V3 *5 -=-. VlO 12. 13. V5 V2 V2 16.? 17. ~. 4 Page 204 ( 68) 3. 5. . Jln(8a+Vo"-l). -p n -r- 15 5sm- 1 --2V4-x 2 . ft 18. v. 19. ~- 2C 36 11. In (3 1 . 2x-6 20 U 2x + 5' 7. 8. 9. - 2 VlC 8X+V15 -- 7 : 2 Vi5 ~ w, 4 to -.. Page 207 ( 69) 1, - ^ COB (S 05-2). 2, ~Jsin(4~2x). 3, Jsec(8x 1). 4, 4tan?. 4 2 8x 6. $ln sin 6 as, 7. 3a; + 5 12. ^Lln-*"- 7- V5 2 x - 8 + V5 ' V33 2 x + 5 -t- V33_ 1 . 4x-l-Vl3 14. =rln _ 2 VlS 4x-l + Vl3 8 + V5 15. 16, 18. -4=1] V2 19. ln; 20. iln 8. 2 esc - 9. ^ln [sec (4 x + 2) + tan (4 a + 2)]. 10. ctn(8-2a). 11. In (esc x etnas) + 2 cosx. 13. 14, a cosx. 306 ANSWERS 15. 3 tan - sec - \ O d 1 X. io> -waaui, aa, -in V2 2 1 + V2 19 s - ^. 16. |(x sin x) \J* ii 7J* 20. In j S3< 4* , w 4V2 3x 17. - Sill 8 4 21. V3-l-^- 24. 3 iii Page 208 ( 70) 1, l e 2a: + 5 5. ^(eaae-a^+aa;. 3 Sai -5 * 6. eF + e- 35 . ' 2 In 2 S^iic i 7. ln(#-l)-x. n. i (e-1) e+l g. 2(e^ fi~^). c a j, i go + &z c + to 99 12- I" --- ' ' ft(H-lnc) ' In 10 Page 212 ( 72) _ 4. Vx 2 1 tan- 1 V.e 2 1 8. - 5. x ,x 6 sin- 1 -- V4-X 2 2 9. - 4Vx 2 -4 10. |. 11.. 12.- 18. ^(9\/8-10V5). 14.. 15. 2 Page 216 ( 74) 1. (8 x 1) e 3 * 7. J (2 cos + sin a) e 2 *. 8. J(x 2 + 2xsincc + 2cosx). 9. 4-2Ve 10. J (81 In 8 -96). 11. i(-2). 12. l(7T-2). 2. 3. zcos-ix Vl x 2 4. xtan- 1 3a-j^ln(l+ 9a 2 5. la^Beo-^x 6. sin z (In sin a; 1) Page 217 ( 75) 4. In X 2 -l 2) 2 2. Page 220 (General Exercises) --. a; 2 as 2 5. In' ANSWERS 307 Page 221 (General Exercises) T. /, (2 + C ST ) S . 9. -\/8 a; + x a 8. ij (1 + 2s + a;*)*. 10. z a + a: + In (a - 1). 11. n \, [B fdu (2 z - 1) - 3 sm (2 1 1) ] . 12. - un - /3 COB* - + 4 cos 2 - + B\. 8 f>\ 5 5 / 13. a 1 ,) ctn 4 x (] 5 + 10 ctn 2 4 x + 3 ctn 4 4 a) . 14. ,V, t 8 HucB (- c - 2) - 5 suc (a- - 2)]. 15. i tau (jt - 1) + In tan (.r - 1) 16. Ju (7 c\sc fi 2 x 5 c'sc 7 2 tr) . 11 i(MoSa--7)v / 8ee8j. 18. Vcac2x 19. ~?% (8 ctn 5 x + 8 ctn 3 5 a:) Vctn 6 x 20. T jt,j(lB3 34sin a 4o; + Own 4 4 a) \/wn4a5. 21. - 7 ' fl (9 ctn 5 x + 4 ctn 6 a 1 ) \/ctn 8 5 x. 23, sin- 3 V5 1 . 2 sr + 8 - sec- 1 -! 5 24. ^sin-i-^-. s aVa ,. JL O 35 25 - - i ppf**" J- i , * ' V6 V6 4 2 21 i B ec-i^ti. 2 2 28. J-sec-i^li. 'sVs Vs 5 2;r-5 ,2J!-1 OB 35. Vl6 30. sin- 1 31. 00 i ,23; + 8 33. = sin- 1 = V2 V8 36. _ Vo 38 39. Vn Vc Vn 2V6 V5 Page 222 (General Exercises) 41. -ft 1 . a -2 42 - Rl Tl **" 1 -i"T rrm-i 2 2 1 1-2 43. -i-soc-i^-. aVo V6 1 9-S . - JL nv ** "^T SGC "*"* * 48. ln(oj-|-Vx a -7). 49. 45, 4 1 46.1 VlO 51. 62. i Inj^-h Va lfi + 7). _ 53, 2 V(c a + 4 + In (a; 4- V a + 4). 54. V8ic a +l -- 4r1 V21 55, -i= In 8 + -Ox). 308 ANSWERS 1 57. =ln(5x + 2 + V25x 2 + 20 x 5) VH 58. ln(3x-4 + V9x 2 - 24x + 14) 4 V6 x V& - 2 i ^ ^ 62. 59. - 6V? 8x+V7 ^in^?- 2 . 4 Vo x V5 + 2 2V21 63. ln-. 6 x .. 1, 2x-5 64. -In 5 x .^ 2V6 70. l __ 1 , x-1 bo. in 12 3x + 4V6 2x- 67. 15 5" x + 2 Page 223 (General Exercises) 75 3 3 .a; x 9 . 2a; -x -sm 8 -cos- sin . 8 4 3 3 16 3 68. ta 71. (tan3x ctnSx). 72. In [sec(x-g) + tan ( aj - )] 73. cos2x. 74. ln(secx + tanx). Gx-3) 76 Jtan2x x as. ^ (x a + 4) taiv-i- a i. 77. |(sm2x cos2x). 94. ^ (2-9 x 2 ) cos 3 a; + x sin 3 x 78. x -f- 2 ( ctn esc 1 \ 9 O / 95. x[(ln2x) 2 -21n2x- f2] \ / 79. tan2x x 96. x In (8 x +V9 x 2 - 4)- J V x a - 4. 80 smx cosx 1 4/ 0\ / * 81. 2Ve= 98 In '* ' 82. f v^ 1 . '2 (x + 8) 8 ' 83. x Jln(l + e 2z ). 99. In fa" 2 ) 2 . V4x 2 -9 85. ^ (2 x 8 + x s - 6) -v^x 8 + 2. 100. ^ i n I 2 x + J ) ( 2 x + s ) : 8 2r 1 86. f Vx + 3 u At ifi ~~" JL 87. ^x- 1 l s ln(2x + l). 101. lin^-^fa + S)^ 88. Z ~ 2 (x - 3) 2 3(l-x 2 )! 102.^ 12 89. a8 27(4x 2 +9)f 103. *. 12 90 (2x 2 + 25)Vx 2 -25 104 -^Infl. 1875 x 8 105. 4 In 4 91. (In 5)8 ' 106. ANSWERS 809 Page 224 (General Exercises) 107. V5-2 112. 6? 118. V5 ' 113. 7. 119 . 108 '0 114.^11X2. 109>c2 ~ e - 1 9 + 4V2 WO- 110. ~^j- . ' ^^/g n 14 ' 121. 2 - In 3. , I / ! \ 116. STT 122. Til ... ........ V A" i t ft*/ ' lu 5 ' 117. UvS. 123. CHAI?TEB X Page 228 ( 77) 1. 2. 4. (h A\ 5. e e y 6. 3, 47ra a 7. Page 229 ( 77) 10. 2w 2 a 8 . 2 ?H^V2 A/8^ 13> T ^ T 16< 11. Jjfira 8 ' ~8" 14. 259jw. 16. JTTW. Page 232 ( 79) 1. 2a 2 . T 2 5 a a V2 7, HTT. ^ 2 7m_ 2 'JJ," 3 8. J. 10. 4-(7T-2). 4n' 4. ~~. 6. |7ra 2 9. 407T. * 2 Page 234 ( 80) 1. ^L. ; ft is radius of semicircle. 2. ; a is xadius of semicircle. 2 2u 3 IT 7T0 Page 235 ( 80) 6. 7ra 8 . 7. 100 revolutions per minute. 8. 5.64 Ib. per square inch, Page 236 ( 81) 8), 4. Ca. 7. I 5. STr'a. 310 ANSWERS Page 238 ( 82) l m vka 5 686 1 ft.-lb 'a '12 6. 2kca 2 , k is the con- 2, 22f ft.-lb 4. 196,350 ft.-lb stant ratio 7. - mi.-lb ; B is radius of earth in miles 8. 2 irC R + a Page 239 ( 82) 9. 1.76ft.-lb., 1 56ft.-lb. Page 239 (General Exercises) 1 Ssm-i| 3. 16 -12 In 3 6. 1^ , 15 T \ ~ , 5_ _. 9V3 247T + - . 5 15 10 ^TTCE 2 _ 11. ^ Trfc 2 Vi^jj , fe t and k z are the values for k in the equation j/ 2 = 12. -.. 13. 15 Page 240 (General Exercises) 14. dfrrf. 15.4FGn4-l) 16 17. 19. 5 ., 15 15 ' 23. _ 4 Page 241 (General Exercises) 24. 400irlb. 28. 123 T 31. 21 J. . 26. 441b. n 8 34. -(8ir+9V3) 35. ' J.D ' Page 242 (General Exercises) 861 -4 37. ^(8+7T), (57T-8), (8-7T) 88 ' T* ^ * 41.950 42. -(In 9-1). ANSWERS 311 48, | 45. -"" 2 8a3 Page 243 (General Exercises) q T. 49. ~. 50. 50,000 ft.-lb. 51, 438,1 ft.-lb. a a* CHAPTER XI Page 245 ( 83) 1.80 -In 8. 2. In 3. 8. 2Jf. 4. IT 1 Page 246 ( 83) 6. 14. ' T" ' 4 "~ ' ' ~6~" 10, ~ (22 - TT). 11, Va (IT V2 - 4). 12. . y 30 Page 254 ( 85) (2 cz (i (f^ 4- 4 c" 1^\ 4 a 'N/S ~~r> . /0 . ) 6. On axis of qxxadrant, from e+l 4e(e 8 --l) / ? . . ' Sv ' . v 7 center of circle. 4. lira, ~ r -\ 7. Intersection ot medians. 8. /-, -\. 5, On axis, ~ from vertex. \ 2 8 / 10. On axis, distant 4 a * + 3 f + 0fca fr0 m L!!' 16. On axis, distant (radius) from base. ir. On axis, distant - from base. a 16. On axis, distant - from base. 18. Middle point of axis. Page 256 ( 86) 3, On line of centers, distant (ri *"*,? from center of circle of radius r,. r x 2 + r| 4, On axis of shell, distant ^ a ~~ *' from common base of spherical surfaces. 5, Middle point of axis. 8 ( r *" ^ ) Page 257 ( 86) 6, On axis, a 1 - distant from base, 7, On axis, $ of distance from vertex to base. 8, (4$, 4|), the outer edges of the square being; taken as OJTand OF. 312 ANSWERS 9. On axis, distant 4 8 from corner of square 10. On axis, distant 3.98 m from center of cylinder in direction of larger ball. 11. On axis, distant 8.4 ft. from base of pedestal. Page 260 ( 87) 3. base x altitude 4. ^ , a = altitude and 26 = base of segment. 15 5- 7rg ; a = altitude and 26 = base of segment. 5 6. 27ra 2 6, 8ira6. 7. ^(b + Sc), 7r[2ca + 26c + 6 2 + (6 + 2c) Va 2 + 6 2 ] ii 4u>a6 /Kr4-q\ 3 a = altitude and 26 = base of segment 12. cw (area) . 11. (0 C -t- O CM , 15 Page 265 ( 88) 1. j\Ma*. 2. %Ma z . 3. Ma 2 4iJfaMJft Page 266 ( 88) ^.A 2 (6+3a) 6(6 + a) 9. 10. JJf(a 2 +6 2 ). 14. ^Jlf(r 2 2 +r 2 ). 145 f- 15. Jfa 2 ,a=radms. 6. ^.Ma 2 11 8 1J' 16. 6757r. 7. 1 J 3 3f(a2 + 6 2 ) 12. 13. T~ 17 S 3f ^ ~ TI iJfr 2 . ' 10""r- ri 8 Page 268 ( 89) 1 McP, a = radius. Page 269 ( 89) 3. 2284$. 4. 4569. 5. 44,990 Page 281 ( 92) 1. 2?r. S . 4 6. 89,980 7 ^Jf(r 3 2 - 8. llf^H B 77ra 8 o. 6 9 10. 11. 4) 6 32a8 9 7 Sffffl4 326 8. Tra 8 9. (37T + 20 16V2). 10. u, ANSWERS 313 Page 283 ( 93) 2t On axis of ring, distant 2ft. from / Q a * center of shell. 6- f^i -^ V 3. On ax 1S , distant 7i ( r ' + 2 Vt + '.") / 6 + 8 4W + V. + T/) ?. (0,0, + from upper base , , 8/00 8 2 ( 2b2 - 2 ) \ 4. On axis, from base. Ot ( u u . ( ' 8 \ 8[b 8 -(b 2 -a a )&]/ Page 285 ( 94) 1. ^5 Jlf (a 2 + 6 2 ), where Jf is mass, and a and & are the lengths of the sides perpendicular to the axis. 2. Atf M. 8, jf M(3 a 2 -1- 2 W) 5. ^ M(a + 4 /i) 2 a 2 (15 TT - 26) 4. J Jfcf (a 8 + 6 2 ). 6. ^ M (8 a 2 + 47i). ' 25 (3 TT - 4) Page 286 ( 94) 14.9T 8. i^.. 3 Page 286 ( General Exercises) . 8. .. 9. }flf(a 2 + & 8 ). 10. 3 . f , oV a; = \ 7 / 6 is the ordhiale. 852 10. On axis, distant _ from base of triangle and away from . semicircle. 11, On axis, distant 4a *+ 2ab -^ + & * from base. Page 287 (General Exercises) 18, On axis of segment, distant ^ T.IL1'" ^ from center of circle, 8 ( 7m8 314 ANSWERS 18. On axis, distant , , ~~ from center of semicircle. 8(86 -Tra) . ! J- 7* f J- f^\ 14. On axis of plate, distant * v ' a 1J from center of circles. gp V 2""" I/ 15. On axis, distant 1 from center of circle 7T 16. On axis of square, 8 in. f lorn bottom. 17. On axis, distant - in. from center of hexagon o c 2 d 18. On axis, distant from center of ellipse ab ~ C 96 19. On axis of solid, distant from smaller base 16 20. On axis, distant from base ^ of distance to top. 21. On axis of segment, distant l ~ * 2 ~" ll '-" fmrn ofmtnr 4-T^n^fh Ji ~\ f] 8 'ii **"* \jouuv3i of sphere 21 2 22. T^lfa 2 23, Page 288 (General Exercises) 24. *j-2l. a* 25. JJL5 u 31> S7 ( 15 ^ - 32 )J 34. 16 26. *fM. a 1S radius. 27. &&M 35. 32. %JiM. 30 29 XPJli_4 4 86 ' 30, uyi_4Tr] 33. ~. 37t Page 289 (General Exercises) 38- 3Q-18-5V2;. 42. (^,0, 39. J/i&M 43. 40. 1 1 ly/OLO . y in, 44 48 ' 32 + 3465 ' 49 ' 3ira8 80. 51. ^ __ ^13838 103 INDEX (The numbeis refer to the pages) Abscissa, 28 Acceleration, 9, 21, 186 Algebraic functions, 70 Amplitude, 128 Angle, between curve and radius vector, 140 between curves, 104 between straight lines, 35 vectonal, 142 Angular velocity and acceleration, 185 Anti-sine, 180 Approximations, 53, 187 Arc, differential of, 106, 146 Archimedes, spual of, 145 Area, as double integral, 246 of ellipse, 225 of plane curve, 47, 225 in polar coordinates, 280 by stimulation, 60 of suiface of revolution, 259 Asymptote, of any curve, 80, 02 of hyperbola, 90 Average value See Mean value Axis, of cooidmates, 28 of ellipse, 86 of hyperbola, 00 of parabola, 82 Cardioicl, 145 Cartesian equation, 100 Cartesian space coordinates, 260 Catenary, 157 Center of _ gravity, of any solid, 282 of circular arc, 258 ' of composite area, 255 of half a parabolic segment, 25f of plane area, 251, Center of gravity, of plane curve, 250 of quarter circumference, 250 of right circular cone, 253 of sextant of circle, 252 of solid of revolution, 252 Circle, 79, 148 of cuivature, 140 Circular measure, 119 Cissoid, 98 Compound-interest law, 166 Cone, circular, 272 elliptic, 275 Constant of integration, 45, 194 Coordinates, 27 cylindrical, 270 polar, 142 space, 269 Curvatuie, 189 Curves, 91 Cycloid, 137 Cylinder, 278 Cylindrical codrdinates, 270 Definite integral, 62, 194 Derivative, 15 higher, 40 partial, 181 second, 39 sign of, 20, 40 Differential, 50 of arc, 106, -146 of area, 64 total, 185 Differential coefficient, 51 Differentiation, 15 of algebraic functions, 94 815 316 INDEX Differentiation, of exponential and logarithmic functions, 163 of implicit functions, 102 of inverse trigonometric functions, 131 partial, 181 of polynomial, 18 of trigonometric functions, 124 Directrix of parabola, 81 Distance between two points, 79 Double integration, 244 e, the number, 155 Eccentricity, of ellipse, 87 of hyperbola, 90 Element of integration, 64 Ellipse, 85 area of, 225 Ellipsoid, 274 volume of, 280 Elliptic cone, 275 Elliptic paraboloid, 275 Equation of a curve, 29 Equations, empirical, 159 parametric, 109 roots of, 30 Equilateral hyperbola, 90, 92 Exponential functions, 154 Palling body, 6, 8 Focus, of ellipse, 85 of hyperbola, 87 of parabola, 81 Force, 128 Formulas of differentiation, 101, 124, 131, 163 of integration, 195, 199, 202, 205, 207, 217 Fractions, partial, 216 Functions, 15 algebraic, 79 exponential, 154 implicit 102 inverse trigonometric, 130 logarithmic, 154 trigonometric, 119 Graphs, 27 of exponential functions, 157 of inverse tiigonometnc functions, 130 of logarithmic functions, 157 in polar coordinates, 142 of trigonometric functions, 121 Hyperbola, 87 Implicit functions, 102 Increment, 16 Indefinite integral, 63, 194 Infinite integrand, 229 Infinite limits, 229 Integral, 45, 194 definite, 62, 194 double, 244 indefinite, 68, 194 Integrals, table oJ, 217 Integrand, 194 Integration, 45, 194 collected formulas, 217 constant of, 45, 194 by partial fractions, 216 by parts, 212 of a polynomial, 45 repeated, 244 by substitution, 208 Inverse sine, 130 Inverse trigonometric functions, 130 Lemniscate, 144 Length of curve, 235 Limit, 1 of of (1 + h)\ 156 theorems on, 93 Limits of definite integral, 63 Line, straight, 81 Linear velocity, 135 Logarithm, 154 Napierian, 156 INDEX 317 Logarithm, natural, 156 Logarithmic spual, 168 Maclaurm's series, 173 Maxima and minima, 41 Mean value, 233 Measure, circular, 119 Moment o inertia, 200 of circle, 204 polar, 202 ol quadrant of ellipse, 262 of rectangle, 201 of solid, 283 of solid of revolution, 265 Moments of inertia about parallel axes, 260 Motion, in a curve, 107 simple harmonic, 127 Napierian logarithm, 166 Ordinate, 28 Origin, 27, 142 Pappus, theorems of, 259 Parabola, 81, 146 Parabolic segment, 83 Paraboloid, 275 Parallel lines, 38 Parameter, 109 Parametric representation, 109 Partial differentiation, 181 Partial fractions, 216 Parts, integration by, 212 Period, 128 Perpendicular linos, 84 Plane, 276 Polar coordinates, 142 Polar moment of inertia, 202 Pole, 142 Polynomial, derivative of, 18 integral of, 46 Power series, 172 Pressure, 68 theorem on, 267 Projectile, 110 Radian, 119 Eadms of curvature, 139 Eadius vector, 142 Bate of change, 11, 189 Revolution, solid of, 73 surface of, 259, 273 Hoots of an equation, 30 Rose of three leaves, 144 Second derivative, 89 sign of, 40 Segment, parabolic, 83 Senes, 172 Maclaurm's, 173 power, 172 Taylor's, 177 Sign of derivative, 20, 40 Simple harmonic motion, 127 Slope, of curve, 86 of straight line, 31 Solid of revolution, 73 Space coordinates, 269 Speed, average, 3 true, 6 Sphere, 271, 272 Spiral, logarithmic, 158 of Archimedes, 145 Straight line, 81 Substitution, integration by, 208 Summation, 66 Surfaces, 271 of revolution, 273 Table of integrals, 217 Tangent line, 88, 104 Taylor's series, 177 Total differential, 185 Trigonometric functions, 119 Trochoid, 138 Turning-point, 37 Value, mean, 238 Vector, radius, 142 Vectonal angle, 142 Velocity, 21, 107 angular, 135 318 Velocities, related, 111 Vertex, of ellipse, 86 o hyperbola, 90 ot paiabola, 82 of parabolic segment, 84 INDEX Volume, of any solid, 277 of solid with parallel bases, 71 of solid of revolution, 73 Work, 237