Digitized by the Internet Archive in 2008 with funding from IVIicrosoft Corporation http://www.archive.org/details/elementarycalculOOwoodrich ELEMENTARY CALCULUS BY FREDERICK S. WOODS AND FREDERICK H. BAILEY PROFESSORS OF MATHEMATICS IN THE MASSACHUSETTS INSTITUTE OF TECHNOLOGY GINN AND COMPANY BOSTON . NEW YORK • CHICAGO • LONDON ATLANTA • DALLAS • COLUMBUS • SAN FRANCISCO COPYRIGHT, 1922, BY FREDERICK S. WOODS AND FREDERICK H. BAILEY ALL RIGHTS RESERVED gfte fltftcngum gre« GINN AND COMPANY • PRO- PRIETORS • BOSTON • U.S.A. "^l^ PREFACE 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 in 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. iii ivi305109 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 liave 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 in 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. RATES SECTION PAGE 1. Limits 1 2. Average speed 3 3. True speed 5 4. Algebraic method 8 5. Acceleration 9 6. 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. Approximations 53 General exercises , ... 55 CHAPTER III. SUMMATION 22. Area by summation 60 23. The definite integral 62 24. The general summation problem 66 25. Pressure 68 26. Volume 71 General exercises 76 vi CONTENTS CHAPTER IV. ALGEBRAIC FUNCTIONS SECTION PAGE 27. Distance between two points 79 28. Circle 79 29. Parabola 81 30. Parabolic segment ' 83 31. 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 135 49. The cycloid *" 137 50. Curvature 139 51. Polar coordinates 142 52. The differentials dr, dd, 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. Maclaurin's series 173 60. Taylor's series 177 General exercises 179 CHAPTER VIII. PARTIAL DIFFERENTIATION 61. Partial differentiation 181 62. Higher partial derivatives 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 m« 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 coordinates 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 viii CONTENTS SECTION PAGE 86. Center of gravity of a composite area 255 87. Theorems 257 88. Moment of inertia 260 89. Moments of inertia about parallel axes 266 90. Space coordinates 269 91. Certain surfaces ... 271 92. Volume 277 93. Center of gravity of a solid 282 94. Moment of inertia of a solid 283 General exercises 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 area of a circle in 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 e, 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 in 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 i + J + i + U--- The sum of the first two terms of this series is 1|, the sum of the first three terms is 1|, the sum of the first four terms is 1|, 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 he 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 ^ 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 . FOB 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 6 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. iii 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 by the formula , distance average speed = —-. > hut this formula does not in general give the true speed at any given time. EATES EXERCISES 1. A man runs a half mile in 2 min. and 3 sec. What is his average speed in feet per second ? 2. A man walks a mile in 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 (99 mi.) 12.35 p.m. Pittsfield (151 mi.) 2.25 Albany (201 mi.) 3.55 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 § min. 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 C. 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 6 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 in 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 P^ is i mi. and the time is 1 min. Then, by § 2, the average speed with which the distance is traversed is i mi. i mi. . ■f — : — = -^ — - — = dO mi. per hour. 1 mm. Jq 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 AB, It is true that the 30 mi. an hour obtained from the interval P^ is likely to be nearer the true speed at B than was the 20 mi. an hour obtained from AB^ because the interval P^ 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 i^ taken as y^g mi. and that the time is 6| 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 defined in § 1. We say, in fact, that the speed of a moving body at any point of its path is the limit approached by the average 6 RATES speed computed for a small distance heginning 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 q 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 yL- in. in -^gV'o ^^^•' we should be pretty safe in saying that it was moving at \p a speed of 30 mi. an hour. . jd 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 ^ „ Fig. 2 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, then s = Ut\ (1) and let us ask what is the speed of the body at the instant when ^ = 2. In Fig. 2 let be the point from which the body falls, ij its position when f = 2, and ij its position a short time later. The average speed with which the body falls through the distance P^P^ 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 ijij and the corresponding time smaller and smaller. In so doing it will be convenient to introduce a notation as follows : Let t^ represent the time at which the body reaches i?, and ^2 the time at which it reaches ij. Also let s^ equal the distance OP^, and s^ the distance OP^, Then s^— 8^ = T[Il, and SPEED 7 t^— t^ is the time it takes to traverse the distance P^P^, Then the average speed at which the body traverses P^P^ is (2) «.-«! Now, by the statement of our particular problem, Therefore, from (1), s^ = 16 (2)^= 64. We shall assume a value of t^ a little larger than 2, compute 8^ from (1), and the average speed from (2). That having been done, we shall take t^ 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^ nearer to 2. Our results can best be exhibited in the form of a table, as follows : h h k - h h-h 2.1 2.01 2.001 2.0001 70.56 64.6416 64.064016 64.00640016 .1 .01 .001 .0001 6.56 .6416 .064016 .00640016 65.6 64.16 64.016 64.0016 It is fairly evident from the above arithmetical work that as the time t^ — t^ and the corresponding distance s^ — s^ 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 ij 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 5 = 16 ^^ 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 time is given by the equation s = 100 -h 16 ^^. Estimate the speed of the body at the end of the fourth second, exhibiting the work in a table. 4. A body is falling so that the distance traversed in the time t is given by the equation s = 16 ^^ + 10 1. Estimate the speed of the body when t = 2 sec, exhibiting the work in a table. 5. A body is thrown upward with such a speed that at any time its distance from the surface of the earth is given by the equation s = 100 1 — 16 1^. Estimate its speed at the end of a second, exhibit- ing the work in a table. 6. The distance of a falling body from a fixed point at any time is given by the equation s = 50 -f- 20 ^ + 16 ^^ Estimate its speed at the end of the first second, exhibiting the work in a table. 4. Algebraic method. In this section we shall show how it is possible to derive an algebraic formula for the speed, still con- fining ourselves to the special example of the falling body whose equation of motion is s = 16t^ Cl^ Instead of taking a definite numerical value for t^^ we shall keep the algebraic symbol t^. Then s, = 16t^. Also, instead of adding successive small quantities to t^ to get t^, we shall represent the amount added by the algebraic symbol h. That is, t =t 4-h and, from (1), 8^ = 16 (t^-\- hy. Hence s.,- s^ = lQ(t^+ hy-16tl=^2tji+161i\ This is a general expression for the distance P^I^ in Fig. 2. Now t^— t^= /i, and therefore the average speed with which the body traverses I^I^ is represented by the expression h It is obvious that if h is taken smaller and smaller, the aver- age speed approaches 32 ^^ as a limit. In fact, the quantity 32 t^ ACCELERATION 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 16 A < g in order that the average speed should differ from 32 ^^ by less than e ; and after that, for still smaller values of A, this difference remains less than e. We have, then, the result that if the space traversed by a falling body is given by the formula the speed of the body at any time is given by the formula speed = 32 ^. It may be well to emphasize that this is not the result which would be obtained by dividing s by ^. EXERCISE Eind 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, ^3 ..,. s = r. (1) Then, by the method of § 4, we find that if v is the speed in feet per second, _ q /2 (<^\ We see that when f = 1, v = 3 ; when ^ = 2, v = 12 ; when t = 3, V = 27 ; and so on. That is, the body is gaining 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^, so that, by (2), t,^= 3(4)'= 48 ft. per second. Take t^—b\ then V = 3(5y=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 t^ = 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. Then and v^= 50.43 V. = 2.43. Then the body has gained 2.43 units of speed in .1 sec, which -^—- = 24.3 units per second. is at the rate of Again, take Then and ^^=4.01. 2;^= 48.2403 i; =.2403. A gain of .2403 units of speed in .01 sec. is at the rate of .2403 .01 = 24.03 units per second. We exhibit these results, and one other obtained in the same way, in a table : k ^2 ^2-i^ «2 - ^1 4.: 4.01 4.001 50.43 48.2403 48.024003 .1 .01 .001 2.43 .2403 .024003 24.3 24.03 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. EATE OF CHANGE 11 By the method used in determining speed, we may get a general formula to determine the acceleration from equation (2). We take . ^ , z. Then v^=^(t^+hy and v^ — v^= 6 t^h+ 3 h^. The average rate at which the speed is gained is then — i— = 6 ^j + 3 A, and the limit of this, as h becomes smaller and smaller, is obviously 6 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: . 1 ^. ^^ '^ £ chancre in speed Acceleration = limit of — - — ^ — -. — ^, change in time EXERCISES 1. If s = 4 1^, find the speed and the acceleration when t =^t^. 2. If s = ^^ -}- ^^, find the speed and the acceleration when t = 2. 3. If s = 3 ^'-^ + 2 ^ + 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 ^^ + 2 ^2 ^ # + 4, find the distance traveled and the speed when t = 2. 5. If s = J f^ + ^ -|- 10, find the speed and the acceleration when t = 2 and when ^ = 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. li s — at -^h, show that the speed is constant. 7. li s = at^ -\- ht -\- c, show that the acceleration is constant. 8. If s = at^ -\- W^ -\- ct -{• f, 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. 12 RATES A stone is thrown into still water, forming ripples which travel from the center of disturbance in the form of circles (Fig. 3). Let r be the radius of a circle and A its area. Then A=TTr\ (1) We wish to compare changes in the area with changes in the radius. If we take 7-^= 3, then ^^= 9 tt; and if we take r^= 4, then ^2 = 16 TT. That is, a change of 1 unit in r, when r = 3, causes a change of 7 tt units in A, We are tempted to say that A is increas- ing Ztt times as fast as r. But before making such a statement it is well to see whether this law holds for all changes made in /-, starting from r^ = 3, and especially for small changes in r. We will again exhibit the calcu- -p^^ o lation in the form of a table. Here 7-^=3, ^^=9 7r, and r^ is a variously assumed value of r not much different from 3. ^2 ^2 ^2-^1 A,-A, A,-A, 3.1 3.01 3.001 9.61 TT 9.0601 TT 9.006001 IT .1 .01 .001 .61 TT .0601 TT .006001 TT 6.l7r 6.01 TT 6.001 TT The number in the last column changes with the number r^— r^. Therefore, if we wish to measure the rate at which A is changing as compared with r at the instant when r = 3, we must take the limit of the numbers in the last column. That limit is obviously 6 tt. We say that at the instant when r = 3, the area of the circle is changing 6 ir 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 7r sq. ft. per second. Another way of expressing the same idea is to say that when r = 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 r^ assumed at the start would produce another result. For example, we may compute that when r^= 4, the rate of change of A with respect to r is Stt; and when r^= 5, the rate is 10 tt. Better still, we may derive a general formula which will give us the required rate for any value of r^. To do this take , 7 Then A^ = tt (r^ -\-2r,h-\- A^) and A^-A^=7r(2r^h-hh^); so that —^ = 2 7rr, + hw. '•.-'■1 * The limit of this quantity, as h is taken smaller and smaller, is 2 7rr^, 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 irr. EXERCISES 1. In the example of the text, if the circumference of the circle which 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? 14 KATES 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 radius 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 77" of the circle is always — . 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 movmg 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 by the value of x. The fact that y is a function of x is expressed by the equation y =f(^}^ and the particular value of the function when x has a definite value a is then expressed as /(«). Thus, if /(2)= 2^- 3(2)^+4(2)4-1 = 5, /(0)=0-3(0) + 4(0) + l = l. 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 x is called an increment of x and is denoted by the symbol Ax (read " delta X "). Similarly, a change in y is called an increment of y and is denoted by Ay. For example, consider y = a^-\-Sx-\-± 15 16 DIFFERENTIATION When a;= 2, y = 12. When a:= 2.1, ?/ = 12.71. The change in X is .1, and the change in y is .71, and we write A:r = .l, Ay = .71. So, in general, if x^ is one value of x^ and x^ a second value of x^ then Lx = x^— x^^ or x^=x^-{-Ax; (1) and if ?/^ and y^ are the corresponding values of i/, then l^y=y^-yv ^r y^=^y^-^Ay, (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 ?/, 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, Ay _ increment oi y _ change in y Ax increment of x change in x An examination of the tables of numerical values in §§ 3, 5, 6 shows that the quotient — ^ depends upon the magnitude of Ax, Ax and that in each problem it was necessary to determine its limit DERIVATIVE 17 as Arr approached zero. This limit is called the derivative of y with respect to x, and is denoted by the symbol -^« We have then ax ^ = limit of ^ = limit of 2h^5g?4>Ly. ax Ax change m x At present the student is to take the symbol -^ not as a ax 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 the fraction — » which has approached a definite limiting value. A^ 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, All 3. The division of Ay by Ax to form — ^' 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 -^ when y = -' Let x^ ^ ax X be a definite value of x, and y^=— the corresponding value of y, 1. Take Ax = h. Then, by (1), x^=x^-i-h. 2. Then y=- ^ " x^+h whence, by (2), A^ = ^ - i = - ^^1^- 3. By division, — ^ = z — ^ — Ax x^ + h^x 18 DIFFERENTIATION 4. By inspection it is evident that the limit, as h approaches zero, is -t which is the value of the derivative when x = x.. xl But x^ may be any value oix] so we may drop the subscript 1 and write as a general formula ^ = _i. dx x^ EXERCISES Find from the definition the derivatives of the following ex- pressions : 1. 2/ = 4(0^2 + 1). 5. y=.x' + -^' 2. y = x'+2x'' + l. 6. 2/ = 2^' 3. y = x^ — x\ 7.y = ^x^-^^x^ + x — 5. 1 •^ 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 aof. 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: Ify = aaf , where n is a positive integer and a is a constant, then -^ = an7^-^. (1) dx To prove this, apply the method of § 7 : 1. Take A:c = A ; whence x^—x-\-h. POLYNOMIAL 19 2. Then y^= ax^ = a(x-\-hy\ whence ^y = a(x + hy — aaf 3. By divison, ^=^a(nx^-^+ ^^'^~^^ af'-''h+... + 1f-^\ Ax 2 . A?/ 4. By inspection, the limit approached by —-^ as A approaches zero, is seen to be anx""'^. Therefore -~ = anx""'^, as was to be proved. ax The polynomial may also have a term of the form ax. This is only a special case of (1) with n = 1, but for clearness we say explicitly: If y = ax, where a is a constant, then ^ = a. (2) dx Finally, a polynomial may have a constant term c. For this we have the theorem: If y = c, where c is a constant, then ^ = 0. (3) dx The proof of this is that as c is constant, Ac is always zero, no matter what the value of Ax is. Hence Ax dc and therefore -^- = 0. dx As an example of the use of the theorems, consider y = 6a:'H-4a;«-2a:4-7. We write at once ^ = 24a:«+12:r^-2. dx 20 DIFFERENTIATION EXERCISES Find the derivative of each of the following polynomials : 1. ic2 + £c-3. 6. x'-^7x^-\-21x^-Ux. 2. ^£c'-f2ic + l. 7. x}-x^+4:x-l. 3. x^-{-^x^+6x^-\-4.x-hl. 8. 3 + 2x2 + 7cc*-lla;^ 4. ^x^-{-lx*-\- 2x^+3. 9. ax^ + bx'' + cx-\-f. 5. x^-4.x*-^x^-4.x. 10. a + te^ ^ cic^ + ea;^ 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 -r- i^ negative, an increase in the value of x causes a decrease in the value of y, du To prove this theorem, let us consider that -^ is positive. Then, since — is the limit of — ^ ? it follows that — is positive dx Ax Ax for sufficiently small values of A^:; that is, if Ax is assumed positive. Ay is also positive, and therefore an increase of x causes an increase of y. Similarly, if -^ is negative, it follows that — is negative for sufficiently small values of Ax ; that is, Ax if Ax is positive. Ay 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 (x + S}{x-l}(x-6). 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<—S. All factors are negative and the product is negative. VELOCITY AND ACCELERATION 21 2. — 3 < a: < 1. The first factor is positive and the others are negative. Therefore the product is positive. 3. l 6. All factors are positive and the product is positive. As an example of the use of the theorem, suppose we have and ask for what values of x an increase in x will cause an increase in «/. We form the derivative and factor it. Thus, ^ = 3 rr' - 6 a: - 9 = 3 (a: + 1 ) (a; - 3) . cix Proceeding as above, we have the three cases 1. x<- increases «/. 1. X < — l. -^ is positive, and an increase in x therefore ax 2. — 1 < a: < 3. -^ is negative, and therefore an increase in x decreases y. 3. a: > 3. -^ is positive, and therefore an increase in x in- dx creases y. These results may be checked by substituting values of x in the derivative. EXERCISES Find 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;2_4a;-f5. 6. \x^-\-x^-\hx-\-\\. 2. 3a;2+10a; + 7. 7. x'^-x^-hx^-h. 3. \^-^x-x-. 8. l + 6x -f 12a;2+8a;». 4. 7_3a^_3a^2 9. 6H-6a; + 6a;2-2x«- 5. 2a:3_|_3^2_;12a^ + 17. 10. 12-12x-6a;2+4a 10. Velocity and acceleration (continued). The method by which the speed of a body was determined in § 4 was in reahty a method of differentiation, and the speed was the derivative of the distance with respect to the time. In that discussion, how- ever, we so 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 ds ,^. " = 71 (^> 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 ^ to ^ (Fig- 1), a distance of 100 mi., and let P be the position of the body at a time ^, and let us assume that we know that AP = 4:t. If we measure s from A^ we have 8=AP = 4:t; whence v = —- = 4:. dt On the other hand, if we measure s from B^ we have whence v = -- = — 4. dt We will now define acceleration by the formula dv a = —"> dt in full accord with § 5 ; or, since v is found by differentiating s, we may write ^2 VELOCITY AND ACCELERATION 23 where the symbol on the right indicates that 8 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. For instance, if — 8 is changed to — 5, it is algebraically increased, although numerically it is smaller. Hence, if a negative velocity is increased, the speed is less. Similarly, 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. 3. 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 s is measured and with increasing speed. 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 8= 96^-16^^ From this equation we compute v = ^Q- 32 t, « = -32. When ^ < 3, V is positive and a is negative. The body is going up with decreasing speed. When ^ > 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 s is measured 24 DIFFERENTIATION down from the point from which the body is thrown, we have, from physics, s = 96t+16t% from which we compute v = 96-\-S'2t, a = 32. Here v is always positive and a is always positive. There- fore the body 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^-St-\-e. 3, s = t^- 9^2 -f- 24:t + 3. 2. s = 10t-t\ 4. s = 8 + 12t - et"" -f- t\ 5. s = t'-f-h2. 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 = 3^2 _ 4^ ^ 4 s. s = ^t^ - 2t^ + 3^ + 2. 10. s = l-\-At-\- 2^ - t\ 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 ^x made in x causes a change A^ in y. The fraction ~- compares the change in y with the change in x. For exam- ple, if A2: = .001, and A?/ =.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 x would actually make a change of 9.061 units in y, any more than the KATE 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 Aa:, 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 Aa:, we must take the limit of — ^, as we did in § 6. Aa; We have, therefore, the following definition : The derivative -j- measures the rate of change of y with respect to x. Another way of putting the same thing is to say that if dy dx 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, ^ = o of water. Fig. 4 7r7 T,\^h'. If V is the volume (1) We are asked to find the rate at which the depth is increasing when h is 6 in. Let us call that depth \, so that h^= 6. Then 26 DIFFERENTIATION F = 8 TT. Now we will increase h^ by successive small amounts and see how great an increase in V^ is necessary to cause that change in h^ ; that is, how much water must be poured in to raise the depth by that amount. The calculation may be tabulated as follows : ^h AF AF Ah .1 .01 .001 .407 TT .04007 TT .0040007 TT 4.07 TT 4.007 TT 4.0007 TT The limit of the numbers in the last column is evidently 4 tt. Therefore the volume is increasing 4 tt 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. 47r ^ 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 ,. This is the general expression for the rate of change of V with respect to 7i, or, in other words, it tells us that V is instan- taneously increasing i irh'^ times as fast as h for any given h. Therefore, when A = 6, F is increasing 4 tt times as fast as A, and as F is increasing at the rate of 100 cu. in. per second, h is increasing at the rate of - — = 7.96 in. per second. 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 2T 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 cc. 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 in. 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 4 ft. 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 OY (Fig. 5), intersecting at right angles at 0, which is called the origin of coordinates. The line OX is called the axis of x, and the line OF the axis of y ; together 28 DIFFERENTIATION M X 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 JfP, equal in length to the value of i/, measured up if ?/ is positive and down if ^ is negative. The point P thus determined is said to have the coordinates X and 1/ and is denoted by (x, y). It follows that the numerical value of X measures the distance y of the point P from OF, 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 ordiiiate. It is evident that any pair of coordinates (x, y) fix a single point P, and that any point P has a single pair of coordinates. The point P is said to be plotted when its position is fixed in this way, and the plotting is conveniently carried out on paper ruled for that purpose into squares. If «/ 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, «/) 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 2 ^1^ ^ ' y=^bx — x^. (1) We assume values of x and compute values of y. The results are exhibited in the following table : Fig. 5 X -1 1 2 3 4 5 6 y -6 4 6 6 4 -6 These points are plotted and connected by a smooth curve, giving the result shown in Fig. 6. This curve should have the GRAPHS 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 by assuming fractional values of x. For instance, there is doubt as to the shape of the curve between the points (2, 6) and (3, 6). We take, therefore, x=2^ and find y = 6^. This gives us another point to aid us in draw- ing the graph. Later, by use of the calculus, we can show that this last point is really the highest point of the curve. The curve (Fig. 6) gives us a graphical representation of the way in which y varies with x. We see, for example, that when x varies from — 1 to 2, 1/ is increasing ; that when x varies from 3 to 6, ?/ 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 i/ is increasing with respect to x. For example, when x=—l^ 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 in y. The curve is therefore steeper when x= —1 than it is when x = l. 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 shall therefore discuss this connection in §§14 and 15. Fig. 6 80 DIFFERENTIATION ' EXERCISES Plot the graphs of the following equations : 1. y = 2x-}-3. 4. 7j = x'^-5x + 6. 7.y = x^ 2. ?/ = - 2a; + 4. 5. y = x^ + 4:X -{- S. 8. y = x^ - 4x\ 3. y = 5. 6. y=9-Sx-x\ 9. y = x^-l. 10. What is the effect on the graph of y = mx -\- S 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 oi y = 2x + b ii different values are assigned to ^ ? What is the meaning of b? 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 + bis 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^ and x^ (x^ < x^) are two values of x such that /(a:^) and f(^x^} are of opposite algebraic sign, the graph is on one side of the axis when x = x^, and on the other side when x = x^. Therefore it must have crossed the axis an odd number of times between the points x = x^ and x = x^. 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 — «)*, the curve y =f(x) crosses the axis of x at the point x = a when h is odd, and touches the axis of x when k is even. In each case the equation f(x) = is said to have k 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^ if f(x^ and / (x^ have opposite signs. The above gives a ready means of locating the real roots of an equation in the form f(pc) = 0, for we have only to find two values of x^ as x^ and x^^ for which /(a:) has different signs. We then know that the equation has an odd number of real roots STRAIGHT LINE 31 )etween these values, and the nearer together x^ and x^^ 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^ •{■ 2a: — 17= 0, accurate to two decimal places. Denoting x^+ 2 a; — 17 hj f(x) and assigning successive integral values to X, we find/(2) = — 5 and/(3) = 16. Hence there is a real 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 nearer 2, since /(2) is nearer zero than is /(3). Proceeding in this way we find /(2.3) = — .233 and/(2.4) = 1.624 ; hence the root is between 2.3 and 2.4. Now, assigning values to x between 2.3 and 2.4 at intervals of one hun- dredth, we find /(2.31) = - .054 and /(2.32) = .127 ; hence the root is 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. li /(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. a^* - 4cc^-f 4 = 0. 2. a!^ + a^-Ml = 0. 5. x^ - Sx^ -{- 6x - 11 = 0. 3. x'-llx-\-5 = 0. 6. x^+3x^-^4.x-\-7 = 0, 14. Slope of a straight line. Let LK (Figs. 7 and 8) be any straight line not parallel to OX or OY, and let I^ (x^j ^/i) ^^^ ij (0^2, y.2) be any two points on it. If we imagine a point to move on the line from ^ to ^, the increment of x is x^—x^ and the increment of 1/ is y^— y^- We shall define the do'pe as the ratio of the increment of y to the increment of x and denote it by m. We have then, by definition, x^—x. Ax 32 DIFFERENTIATION A geometric interpretation of the slope is easily given. For if we draw through ij a line parallel to OX, and through i^ a line parallel to OY, and call R the intersection of these lines, then x^—x^ = P^R and y^—y^ = RB^. Also, if is the angle which the line makes with OX measured as in the figure, then m RB, P^R tan (^. (2) 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. : T. K ^^ y^ 1 ■\r- ^ L Ji. L 1 ^2 it; "\< . \# Fig. 7 Fig. 8 Then if the line runs up to the right, as in Fig. 7, x^ — x^ is positive and the slope is positive. If the line runs down to the right, as in Fig. 8, x^ — x^ is negative and w 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 Pi(x-^^ y^ be held fixed, but let i^ be allowed to wander over the line, taking on, therefore, variable coordinates (x^ ?/). Equation (1) may then be written y-y=m(x-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, i^ (x^^, y^) may be taken as the point with coor- dinates (0, 6) in which the line cuts 0¥, Then equation (3) becomes y = mx-it-h (4) Since any straight line not parallel to OX ov 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 OY, If the line is parallel to OX, we have no triangle as in Figs. 7 and 8, but the numerator of the fraction in (1) is zero, and we there- fore say such a line has the slope 0. Its equation is of the form y = h, (5) since it consists of all points for which this equation is true. If the line is parallel to OF, again we have no triangle as in Figs. 7 and 8, but the denominator of the fraction in (1) is zero, and in 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 Ax + By + C=0 (7) ' ' ^'""'^ always represents a straight line. This follows from the fact that the equation may be written either as (4), (5), or (6). The 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^ and m^ are the slopes of the lines, we have, from (2), m=m^, (8) If two lines are perpendicular and make angles <^j and (f>^ respectively with OX, it is evident from Fig. 9 that ^= 90°-f <^j ; 34 DIFFERENTIATION whence tan (t>^= — cot (j)^= — • Hence, if m^ and m^ are the slopes of the lines, we have ^^ m,= --' (9) It is easy to show, conversely, that if equation (8) is satis- fied by two lines, they are parallel, and that if equation (9) is satisfied, they are perpendicular. Therefore equations (8) and (9) are the conditions for parallelism and perpendicularity respectively. Ex. 1. Find the equation of a straight line passing through the point (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) and — 3 — 2 5 (2, — 3) is — — = -• Therefore, by (3), the equation of the required line is o 5 / i \ which reduces to 5x — 2 7/ — l = Q. Ex. 2. Find the equation of a straight line through the point (2, — 3) and perpendicular to the line 2a; — 3?/4-7 = 0. The equation of the given straight line may be written in the form y = § X + ^, which is form (4). Therefore vi = §. Accordingly, by (9), the slope of the required line is — f . By (3) the equation of the required line is y + S = -%(x-2), which reduces to S x + 2 y = 0. Ex. 3. Find the equation of the straight line passing through the point (— 3, 3) and the point of intersection of the two lines 2x — y — 3 = and 3x-\-2y-l = 0. The coordinates of the point of intersection of the two given lines must satisfy the equation of each line. Therefore the coordinates of the point 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+1 (— 3, 3) and (1, — 1). By (1) the slope of the required line is — = — 1. Therefore, by (3), the equation of the line is "" ~" y + l = -(x-l), 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 (^, — I) and making an angle of 135° with OX. 7. Find the equation of the straight line 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 cc + 4 z/ — 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 (^, - \). 10. Find the equation of the straight line passing through ih ~ \) ^^^ perpendicular to the straight line determined by the points (2, — 1) and (— 3, 5). 11. If )8 is the angle between two straight lines which make angles <^j and <^2(^2 > ^i) respectively with OX^ prove from a diagram similar to Fig. 9 that /3= ^— ^. If tan c^^ = m^ and tan ^ = m^, prove by trigonometry that _ tan/8 = T-f ^• 12. Find the angle between the lines x — 2 ij + 1=0 and 2a;-32/ + 7 = 0. 13. Find the angle between the lines 2a; — 4y + 5 = and 5x + 2y-6 = 0. 14. Find the angle between the lines y = 3ic + 4 and x + Sy+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 2x- 32/ + 10 = 0. 36 DIFFERENTIATION 15. Slope of a curve. Let AB (Fig. 10) be any curve serving as the graphical representation of a function of x. Let i^ be any point on the curve with coordinates x^ = OM^, y^=M^P^, Take ^=M^M^ and draw the perpendicular M^^, fixing the point j? on the curve with the coor- dinates X.-^ = OM^, ^2 = ^2^' Draw F^E parallel to OX. Then ^E = M^M2=Ax, Draw the straight line ^i^, prolonging it to form a secant P^S. Then, by § 14, ^x Fig. 10 is the slope of the secant P^S^ and may be called the average slope of the curve between the points JP and i^. To obtain a number which may be used for the actual slope of the curve at the point i^, it is necessary to use the limit process (with which the student should now be familiar), by which we allow A^: to become smaller and smaller and the point ij to approach ij along the curve. The result is the derivative of y with respect to x^ and we have the following result: The slope of a curve at any point is given hy the value of the derivative -j- at that point, (XX As this limit process takes place, the point ij approaching the point ij, it appears from the figure that the secant I^S approaches a limiting position P^T. The line iJT is called a tangent to the curve, a tangent being then by definition the line approached as a liinit by a secant through two points of the curve as the two points approach coincidence. It follows that the slope of the tangent is the limit of the slope of the secant. Therefore, The slope of a curve at any point is the same as the slope of the tangent at that point. SLOPE OF A CURVE 37 From this and § 9 we 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. If 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. The values of x which make -^ zero or infinite are of par- dx ticular interest in the plotting of a curve. If the derivative changes its sign at such a point, the curve will change its direc- tion from down to up or from up to down. Such a point will be called a turning-point. If g is an algebraic polynomial, its derivative cannot be infinite; so we shall be concerned in this chapter only with turning-points for which ^ = 0. dx They are illustrated in the two following examples: Ex. 1. Consider equation (1) of § 12, Equating — to zero and solving, we have x = - as a possible turning- point. It is evident that when x<'-, ~ is positive; and when x>-, -^ is ^ (Ix *!i (IX negative. Therefore x = f corresponds to a turning-point of the curve at which the latter changes its direction from up to down. It may be called a high point of the curve. Ex. 2. Consider 3, = 1 (x3 - 3 ^2 - 9 a: -f- 32). Here g = ?(x2 - 2x - 3) = ?(:r - 3) (x -f 1). Equating — to zero and solving, we have x = — 1 and x = 3 as possible turning-points. From the factored form of -^ , and reasoning as in § 9, dx J we see that when x< — l, — is positive ; when — 1< x < 3, -^ is negative ; dx dx Here ^ = 5-2x = 2 dx 38 DIFFERENTIATION dy when a; > 3, -^ is positive. Therefore both x = — 1 and a: = 3 give turning- 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 (— 1, 4|) 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 ax always give turning-points as illus- trated in the next example. Ex.3. Consider y = \{x^^ -^ x"^ + 21 x -1%). Here dy dx 6 a: -f 9 = (x - 3)2. dv Solving -^ = 0, we have x = 3 ; but since the derivative is a perfect square, it is never negative. Therefore x = 3 does not give a turning-point, although when x = Z 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 (x^ -, y^) is easily written down. We let \-f] represent the value of -j- at . ^^1 /dy\ the point (x^^ y^). Then m = { — )', and, from (3), § 14, the equation of the tangent is Ex. 4. Find the equation of the tangent at (1, — 1) to the curve y = x^ — ^ X ■\- 2. We have and Fig. 12 dx Therefore the equation of the tangent is ^4-l=-2(x-l), which reduces to 2x -\- y — \ = 0. SECOND DERIVATIVE 39 From (2), § 14, it also follows that if is the angle which the tangent at any point of a curve makes with OX, then J = tan<^. (2) EXERCISES Locate the turning-points, and then plot the following curves : 2. 2j = S-\-Sx-2x\ 5. 2/ = ^(2;:c3 + 3aj2_i2ic-20). S. y = x^-Sx'^-\-4:. 6. y = 2 -\-9x-\-Sx:'-x^ 7. Find the equation of the tangent to the curve y = S — 2x -\- x^ at the point for which x = 2. 8. Find the equation of the tangent to the curve y=l-\-3x—x^—3x^ at the point for which x = — 1. 9. Find the points on the curve y = x^ -{- 3x^ — 3a; -f- 1 at which the tangents to the curve have the slope 6. 10. Find the equations of the tangents to the curve y = x^+2x^-x-\-2 which make an angle 135° with OX. 11. Find the equations of the tangents to the curve y = x^-\-x^—2x which are perpendicular to the line Sx-\-2y-\-4: = 0. 12. Find the angle of intersection of the tangents to the curve y = x^+x^—2 at the points for which x = —1 and x=l respectively. 16. The second derivative. The derivative of the derivative is called the second derivative and is indicated by the symbol ^;-( , ) or -^o' We have met an illustration of this in the dx \dx/ dx case of the acceleration. We wish to see now what the second derivative means for the graph. Since -j- is equal to the slope of the graph, we have 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 —^ is positive, it is evident that d d -^ is increasing through zero ; hence, when x < a, -^ \^ nega- dx ^ dx tive, and when x > a, -^ is positive. The point for which x = a dx is therefore a low turning-point, by § 9. Similarly, if, when x = a, ^ = and -— f is negative, it is dy . ^^ "^^ evident that -^ is decreasing through zero ; hence, when x < a, dy dy ~ is positive, and when x> a, -^ is negative. The point for ax ax which x = ais therefore a high turning-point of the curve, by § 9. These conclusions may be stated as follows: dy d^y If -^ = and — ^ is positive at a point of a curve, that point . dy d\ is a low point of the curve. If -— = and -— | is negative at a ax ax point of a curve, that point is a high point of the curve. In addition to the second derivative, we may also have third, fourth, and higher derivatives indicated by the symbols -r-f' -^^ etc. These have no simple geometric meaning. EXERCISES Plot the following curves after determining their high and low points bv the use of ~ and ^ : ax dx^ 1. 2j = Sx^-x-2. 3. y = 7-lSx-3x^-^4.x\ 2. y^S-\-Sx-x'-x\ 4. 7/ = 6-15x-\-lSx^-4:X^ MAXIMA AND MINIMA 41 17. Maxima and minima. If f(a) is a value of f(x) which is greater than the values obtained either by increasing or by decreasing a; by a small amount, f(a) is called a maximum value of f(x). If /(a) is a value of f(x) which is smaller than the values of f(x) found either by increasing or by decreasing x by a small amount, /(a) is called a minimum value oif(x). It is evident that if we place y =/(^) and make the graph of this equation, a maximum value oif(x) 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 values of y^ solve the equation ax If a: = a is a root of this equation, it must be 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 ax through a, then x = a gives a maximum value of y. If the sign of -J- changes from — to -\- as x increases through a, then x — a gives a minimum value of y. dv d^v Test II. If x = a makes -^ = and —~ negative, then a: = a dx dxr 2 gives a maximum value of y. If x = a makes -^= and -j^ 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 1/ to give information if — f = when x = a. It is also frequently 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 in., 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 a:, 20 — 2 a:, x. Let V be the volume of the box. Then F = a:(20 - 2 x) (30 - 2 x) = 600 a; - 100 x^ + 4 x\ dV dx = 600-200a: + 12a-2. dV Equating to zero, we have dx X X 30 -2 X 1 whence 3 a:2 - 50 ar + 150 = ; 25±5V7 Fig. 13 3.9 or 12.7. The result 12.7 is impossible, since that amount cannot be cut twice from 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 ilV dx = 12 (X- 3.9) (x- 12.7), dV when it appears that — changes from + to — , as x increases through 3.9. Hence t = 3.9 gives a maximum value of V. dW To apply Test II we find -— -200 + 24 a: and substitute x - 3.9. The result is negative. Therefore a: = 3.9 gives a maximum value of F. The maximum value of V is 1056 + cu. in., found by substituting a: = 3.9 in the equation for F. 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 \ — TTX^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 similar triangles we have FE ^AD, EC~ DC' V 12. that is, 12 -X 12 whence y = \2 — x. Substituting in (1), we have F - 12 7rx2 - Tra:^ dV whence — = 24 ttx — 3 ttx^. dx dV Equating — to zero and solving, we find dx B< a: = or 8. The value a: = is evidently not a solution of the problem, but x = 8 is a possible solution. Applying Test T, we find that as x increases through the value 8, dV — — changes its sign from + to — . Applying Test II, we find that dx d^V — - = 24: TT — Q TTx is negative when x = 8. Either test shows that x = 8 dx^ corresponds to a maximum value of V. To find V substitute a: = 8 in the expression for V. We have V = 256 rr cu. in. EXERCISES 1. A piece of wire of length 20 in. is bent into a rectangle one side of which is x. 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. 5. A right circular- cylinder with altitude 2x is inscribed in a sphere of radius a. Find the cylinder of maximum volume. 44 DIFFEKENTIATION • 6. 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 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 of its breadth and the square 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 60 in. 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 Z> feet long and a feet wide is to be bent into a U-shaped water drain h 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 ? (2) when it is open on top ? 12. A circular filter paper 10 in. 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 may wish to find the distance traversed ; or if the slope of a curve is given, we may wish to find the curve. INTEGRATION 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 , -f- = ax"", ax then y-f^l^^'- (1> The first term of this formula is justified by the fact that if it is differentiated, the result is exactly ax"". The second term is justified by the fact that the derivative of a constant is zero. The constant C 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 y = ax^-C. (2) This is only a special case of (1) with n = 0. Finally, if ^ = ^V^+...+^Va„^ + ^. .(3) n -\-l n I Ex. 1. The velocity v with which a body is moving along a straight line AB (Fig. 15) is given by the equation A B How far will the body move in the Fig. 15 time from / = 2 to / = 4 ? If when < = 2 the body is at P^, and if when i = 4 it is at Pj, we are to find P,P,. ^^^ By hypothesis, -^ = 16 ^ + 5. Therefore s = 8 ^2 + 5 ^ + C. (1) 46 DIFFERENTIATION We have first to determine C If s is measured from P^ , it follows that ^^^^ t = 2, s = 0. Therefore, substituting in (1), we have = 8(2)2 + 5(2) + C; whence C = ~ 42, and (1) becomes s = S t^ -{■ 6 t - 4:2. (2) This is the distance of the body from P^ at any time t. Accordingly, it remains for us to substitute / = 4 in (2) to find the required distance PiP^- There results p^p^ ^ g ^^^2 + 5 (^4) _ 42 = 106. If the velocity is in feet per second, the required distance is in feet. Ex. 2. Required the curve the slope of which at any point is twice the abscissa of the point. dy By hypothesis, Therefore dx 2x. j/ = x^ + a (1) Any curve whose equation can be derived from (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 + C; whence C = — 1, .and therefore the equation of the curve is y = x^ — 1. Fig. 16 But if it is required that the curve should pass through (— 3, 10), we have, from (1), 10 = 9 + C- whence C = 1, and the equation is y = x^ + 1. EXERCISES In the following problems v is the velocity, in feet per second, of a moving body at any time t. 1 . If V = 32 ^ + 30, how far will the body move in the time from t = 2 tot = 6? AREA 47 2. If V = 3 ^^ + 4 ^ 4- 2, how far will the body move in the time from t = ltot = 3? 3. li V = 20 1 -\- 25, how far will the body move in the fourth second ? 4. It V = t^— 2t -\- 4:, how far will the body move in the fifth and sixth seconds ? 5. If V = 192 — 32 t, how far will the body move before v = 0? 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 (x, 2/) is 6 £c^ + 2 ic — 4, and the curve passes through the point (0, 6). What is its equation ? 8. The slope of a curve at any point (cc, ?/) is 4 — 3 x — a^^, 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 (cc, 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 (a^, ?/) is ic^ — a? 4- 1. W^hat is its equation ? 19. Area. An important application of integration occurs in the problem of finding an area bounded as follows : Let RS (Fig. 17) be any curve with the equation y —f{x), and let EB and ^C be any two ordinates. It is required to find the area bounded by the curve US^ y the two ordinates ED and BC^ and the axis of x. Take -MP, any variable or- dinate between ED and BC^ and let us denote by A the area EMPD bounded by the curve, the axis of a;, the fixed ordinate ED, and the variable ordinate MP. It is evident that as values are assigned to x=OM, different positions of MP and correspond- ing values of A are determined. Hence ^ is a function of x for which we will find dA dx 48 DIFFERENTIATION Take MN=Ax and draw the corresponding ordinate KQ, Then the area MNQF=AA. If L is the length of the longest ordinate of the curve between MF and NQ, and s is the length of the shortest ordinate in the same region, it is evident that sAx £J. Then, by (1), / dx 3 ' . L D / whence A = lx^ + C. / When x = l, ^ = 0, and therefore / = l + C; / f whence C = — ^, / and A = lx^-^. V eX Finally, when x = Z, \ Vs ^^.^ B C ' ^ = h(^y-h = 2|. Fig. 18 EXERCISES 1. Find the area bounded by the curve y = 4tx — x^, the axis of x, and the lines x = 1 and £c = 3. 2. Find the area bounded by the curve y = x^ -{- % x -\- IS, the axis of X, and the lines cc = — 6 and x = — 2. 3. Find the area bounded by the curve ?/ = 16 -f 12ic — x', the axis of Xy and the lines x = — 1 and x — 2. 4. Find the area bounded by the curve y + cc^ — 9 = and the axis of X. 5. Find the area bounded by the axis of x and the curve y z= 2 X — x^. 6. Find the area bounded by the curve y — ^x^ — x^, 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 4:y = x^ — 6^ + 9. 8. Find the area bounded by the curve y — x^ — 2 x^ — A^x -{• % and the axis of x. 50 DIFFERENTIATION 9. If ^ denotes the area bounded by the axis of y^ the curve X =/(^), a fixed line y = h, and any variable line parallel to OX, prove that 10. If A denotes an area bounded above by the curve y =f(x), below by the curve y = F(x), at the left by the fixed line x = a, and at the right by a variable ordinate, prove that 20. Differentials. The derivative has been defined as the limit of — ^ and has been denoted by the symbol -^. This symbol 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 other quantity. We shall call dx the 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 zero as a limit. In other words, the differential of the independent variable X is identical with the increment of x; that is, dx = Ax, (1) After dx has been defined, it is necessary to define dy so that its quotient by dx is the derivative. Therefore, if y —f(x) and -^ =f(x), we have ^^ dy=f{x)dx. (2) That is, the differential of the function y is equal to the derivative times the differential of the independent variable x. dx (1) (3) X aj« (2) DIFFERENTIALS 51 In equation (2) the derivative appears as the coefficient of dx. For this reason it is sometimes called the differential coefficient. It is important to notice the distinction between dy and A^. The differential dy is not the limit of the increment A?/, since both dy and A«/ have the same limit, zero. Neither is dy equal to a very small increment Aj/, since it generally differs in value from A?/. It is true, however, that when dy and A?/ 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 A =x\ If a: is increased by ^x = dx, A is increased by A^, where A^ = (a: + dxy -x'^ = 2xdx+ (dxf. ' x "dx Now, by (2), dA = 2 xdx, -pio. 19 so that A^ and dA differ by (rfa:)^. 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 figure that the square (3) is very small compared with the rectangles (1) and (2), provided dx is taken small. For example, if a: = 5 and dx = .001, the rectangles (1) and (2) have together the area 2 xdx = .01 and the square (3) has the area .000001. Ex. 2. Let s = 16 f^, where s is the distance traversed by a moving body in the time L If t is increased by At = dt, we have " As = lQ(t + dty -16fi = 32tdt-{- 16 (dty, and, from (2), ds = ^2tdt', so that As and ds differ by 16 (dty. The term 16 (dt)^ is very small com- pared with the term S2tdt, if dt is small. For example, if < = 4 and dt = .001, then S2tdt = .128, while 16 (dty = .000016. In this problem As is the actual distance traversed in the time dt, and ds 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(pc) and we make a graphical representa- tion, we may have two cases as shown in Figs. 20 and 21. In each figure, M]V = PE = Ax = dx and IiQ=Ay, since EQ is the total change in y caused by a change of dx = MN in x. If FT is the tangent to the curve at P, then, by § 15, !=/« tan EPT so that, by (2), dy = (tan EFT) (FE) = ET Fig. 21 In Fig. 20, dy < Ay, and in Fig. 21 dy >Ay; but in each case the difference between dy and Ay is represented in magnitude by the length of QT. This shows that EQ =Ay is the change in y as the point F is supposed to move along the curve y=f(x), while ET=zdy is the change in the value of y as the point F is supposed to move along the tangent to that curve. Now, as a very small arc does not deviate much from its tangent, it is not hard to see graphi- cally that if the point Q is tal^en close to P, the difference between EQ and ET, namely, QT, is very small'compared with ET. A more rigorous examination of the difference between the increment and the differential lies outside the range of this book. EXERCISES 1. If 2/ = a;«-3cc2 4-4£c + l, find c?2/. 2. li y = x^ ^ ^x^ - x'^ -\- Qx, find dy. 3. If V is the volume of a cube of edge x, find both AV and dV and interpret geometrically. APPROXIMATIONS 53 4. If A is the area of a circle of radius r, find both A.4 and dA. Show that A^ is the exact area of a ring of width dr, and that dA is the product of the inner circumference of the ring by its width. 5. If F is the volume of a sphere of radius r, find AF and dV. Show that A F is the exact volume of a spherical shell of thickness d)', 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^. 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 As ? 8. It y = x^ and x = 5, find the numerical difference between 6??/ and Ay, with successive assumptions of dx = .01, dx = .001, and dx = .0001. 9. If. y = x^ 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 AA corresponding to an increase of r by .001 in. 11. For a sphere of radius 3 ft. find the numerical difference between dV and AF 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 ^ 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 a change of .002 in. in the side. Let X be the side of the square, A its area. Then A = x^ and dA = 2x dx. 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 A^ = (2.002)2- (2)2= .008004, which is the exact change in A. Our approximate change is therefore in error by .000004, a very small 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 ^^- ir, 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 Y its volume, we have V=%irr^ and dV=^Trr'^dr. Placing r — 2 and dr = — .1, we find dV = — 1.6 tt. Hence the volume of the required sphere is approximately Aj2 TT- 1.6 7r = 9.0667 TT. To find how much this is in error we may compute exactly the volume of the required sphere by the formula ' V= 4 TT (1.9)3 =9.1453 77. 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 2i in., and the volume and the surface are computed. If an error of ^^ in. has been 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 error in the corre- sponding 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 cone if the altitude is 3.00002 in. 6. The distance s of a moving body from a fixed point of its path, at any time t, is given by the equation s = 16t'^ -\- 100 1 — 50. Find approximately the distance when t = 4.0004. 7. Find the approximate value oi x^ -\- x — 2 when x = 1.0003. 8. Find approximately the value of cc* + ^^ + 4 when x = .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 F 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 = ^ irr^ ctn a and S = in-^ 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 5. vi;* • 1-x ' • x^ + l' « 1 * 6. — ^• a-\-x rr^ _ 1 Va; 2. -—^ — 4. a-x ' x' + l 7. Vx^-M.* 8. Prove from the definition that the derivative of — is - — tt- 9. By expanding and differentiating, prove that the derivative of (2ic + 5)8is 6(2a^ + 5)2. 10. By expanding and differentiating, prove that the derivative of {x^+lfi^ 6cc(cc2 + l)l 11. By expanding and differentiating, prove that the derivative of (x -j- a)" is n{x -\- a)"~^, where ti is a positive integer. 12. By expanding and differentiating, prove that the derivative of (x^ -\- a^y is 2 nx(x^ + a^)"~\ where ti is a positive integer. 13. Find when x'* + 8 03^ + 24 a?^ -+- 32 a? + 16 is increasing and when decreasing, as x increases. 14. Find when 9 ic^ — 24 a:;* — 8 a;^ -|- 32 a: + 11 is increasing and when decreasing, as x increases. 15. Find a general rule for the values of x for which ax^ -{-hx -{- c is increasing or decreasing, as x increases. 16. Find a general rule for the values of x for which x^ — a^x + h 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. * Hint. In these examples make use of the relation VA—Vb= — =: =• ■Va + Vb 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^ — ^t"^ -\-l^t -{- 100. When will the particle be approaching A ? 19. The velocity of a certain moving body is given by the equation V = t"^ — 7 t -\-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 ^ — 16 ^^. 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"^ — ^t -\-12. Find when the 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 ^ -|- 9 if ^ — ^^ When is its velocity increasing and when decreasing ? When is its speed increasing and when decreasing ? 23. At any time #, the distance of a certain moving body from a fixed point in its path is given by the equation s = ^* — 6 ^^ -f 9 ^ + 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 when 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 minute. 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, — 5) 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 line 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 ic — 2?/-|-7 = 0. 31. Find the equation of the straight line through the point (2, 7) and perpendicular to the line 2x -\- 4:y -{- 9 = 0. 32. Find the angle between the straight lines 2x -\- Sy -[- 5 = and y-\-Sx-{-l = 0. Find the turning-points of the following curves and draw the graphs : 33. y = S-x-x\ 34. y = 16x^-4:0x-^25, S5. ij = ^(x^- 4.x-2y 36. y = x^-6x^-15x-h5. 37. y = x^-\-Sx^-9x-U. 38. Find the point of intersection of the tangents to the curve y = x -\- 2x^ — x^ at the points for which x = — 1 and x = 2 respectively. 39. Show that the equation of the tangent to the curve y = ax"^ -\- 2bx -^ c Sit the point (xi, yi) is y = 2 (ax^ -\-h)x — ax^ -\- c. 40. Show that the equation of the tangent to the curve y = x^ -\- ax -\- b at the point (x^ , y^) is y = (Sx^ -\- a)x — 2 xf + b. 41. Find the area of the triangle included between the coordinate axes and the tangent to the curve y = x^ Sit the point (2, 8). 42. Find the angle between the tangents to the curve y = 2x^ ■i- 4:x^ — X at the points the abscissas of which are — 1 and 1 respectively. 43. Find the equation of the tangent to the curve y = x^^Sx'^ 4- 4 ic — 12 which has the slope 1. 44. Find the points on the curve y = Sx'^ — Ax^ at which the tangents are parallel to the line x — y = 0. 45. A length I of wire is to be cut into two portions which are to be bent into the forms of a circle and a square respectively. Show that the sum of the areas of these figures will be least when the wire is cut in the ratio tt : 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 coincides 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 k. 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 one 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 t, how far will the body move in the first 5 sec. ? 51. li V = 200 — 32 t, 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 ? 52. A curve passes through the point (2, — 3), 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 ic^ — 2 a? -h 7. Find its equation. 54. Find the area bounded by the curve ?/ + a?^ — 16 = and the axis of X. 55. Find the area bounded by the curve y = 2 x^ — 1^ x^ -\- ^Q> x +1 and the ordinates through the turning-points of the curve. 56. Find the area between the curve y — x? and the straight line 2/ = X -f 6. 57. Find the area between the curves y = x'^ and y — 1% — x^. 58. The curve y = ax? is known to jiass through the point (A, Iz). Prove that the area bounded by the curve, the axis of x, and the line x = li\^ \ hk. 59. Compute the difference between A^ and dA for the area A of a circle of radius 5, corresponding to an increase of .01 in the radius. GENERAL EXERCISES 59 60. Compute the difference between A F 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 i 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 x^+ 4 ic^-f- x 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 = f -\- 3f -\- 9t — 27. Find approximatel}^ the distance traversed in the interval of time from ^ = 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 = ^x^, the axis of x^ and the ordinates x=2 and a; = 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 2^ = 3 into 10 parts, 3 — 2 each of which we call Aa:, so that A2: = --— — =.1. If x^ is the first point of division, x^ the second point, and so on, and rectangles are constructed as shown in the figure, then the altitude of the first rectangle is ^(2)^ that of the second rec- tangle is ^ xl = \ (2.iy = .^^% and so on. The area of the first rectangle is i(2)^Aa7 =.08, that of the second rectangle is ^j^Aa; = 1(2-1)"^ = -0882, and so on. Accordingly, we make the i following ca Iculatii ^ = 2, \(.^y^-= .08 x=2.\. •(^,)^Ax = .0882 x = 2.2, w.r^= .0968 0^3= 2.3, i(x^yAx= .1058 x= 2.4, K^,yA^ = .1152 x^= 2.5, i(^5)'^ = .1250 ^e=2-6, K^e)'^^ = .1352 ^,= 2.7, l(x^fAx = .1458 ^,= 2.8, U^fAx^ .1568 x,= 2.9, lCx^yAx = .1682 1.2170 This is a first approximation to the area. For a better approximation the axis of x between x= 2 and x= 3 may be divided into 20 parts with A.3:;=.05. The result is 1.2418. 60 AREA 61 If the base of the required figure is divided into 100 parts with Aa:=.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 increased and the size of Ax approaches zero. We shall now proceed to generalize the problem just handled. Let LK (Fig. 23) be a curve with equation i/ =f(x)^ and let OE=a and OB = h. It is re- quired to find the area bounded by the curve LK^ the axis of x^ and the ordinates at E and B. For convenience we assume in the first place that a ^X,-i=^„-i- Draw JS-i) =/(«), ^/„_i^_i=/(^„-i). and^C; also lengths EM^ =M^M^= M^M^ = Let 0M^= x^, 0M^= x.^, • M,P,=f(x,^.M,P,=f(x,), . DR^, P^R^, P^R,, . . ., Pn-^Rn^ parallel to OX. Then /(a)Aa; = the area of the rectangle EDR^M^, f(x;)Ax = the area of the rectangle M^P^R^M.;,, f(x^Ax = t\\Q area of the rectangle M^P^R^M^, /(:r„_i)Aa: = the area of the rectangle M„_,P„_,R,B. 62 SUMMATION The sum /(«)A:r +f(x,)^x +f(x.^^x + . . . +f(x,^^,)^x (1) is then the sum of the areas of these rectangles and equal to the area of the polygon EDE^I^R.^ . • . Ii^^_^]^^_^R^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 1=0 where 2 (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 /(a:.) 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 /(x^dx, Ja where / is a modified form of S. f(x)dx = Lim V/(a:.)A:r = the area EBCD. a «=» (=0 It is evident that the result is not vitiated if ED or BC is of length zero. 23. The definite integral. We have seen in § 19 that if A is the area EBCD of § 22, A = E(b}-E(a), where E(^x^ is any function whose derivative is /(a:). Comparing this with the result of § 22, we have the important formula f''f(xyx=FCb)-FC P= j whdA, PRESSURE 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 in terms of the same variable. Ex. 1. Find the pressure on one side of a rectangle BCDE (Fig. 27), where the sides BC and ED are each 4 ft. long, the sides BE and CD are each 3 ft. long, immersed in water so that the plane of the rectangle is perpendicular to the surface of the water, and the side ^C is parallel to the surface of the water and 2 ft. below it. In Fig. 27, LK is the line of inter- section of the surface of the water ^ ^ Fig. 27 X and the plane of the rectangle. Let ~ M be the point of intersection of ^ LK and BE produced. Then, if a: is measured downward from along E BE, X has the value 2 at the poi^t B axid 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 elementary rectangles such as MNRS. Therefore dA = area of MNRS = MN • M^ = 4 dx. Since MN is at a distance x below LK, the pressure on the elementary rectangle MNRS is approximately wx(i: dx). Accordingly, we have dP = 4 wxdx and P= f^iwxdx = [2icx^f=2w(dy-2w(2y = i2w. For water, w = 62 1 lb. = ^l t. Hence we have finally . P = 2625 1b. = 1/^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 surface 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. 70 SUMMATION Divide the distance BT into parts dx, and through the points of divi- sion draw straight lines parallel to CD, and on each of these lines as lower base construct a rectangle such as MNRS, where E and F are two consecutive points of division -^ on BX. Then BE = x, EF = dx, and, by similar triangles, MN ^ BE ^ CD ~ BT' MN X whence -— — = - 7 5 and MN =lx. Then Since £ is 6 ft. below LK, and BE below LK. Hence the pressure on the rectangle is approximately dP = (Ix dx) (6 — x)w = (*/- wx — I icx^) dx, and ^= f VV wx- I tvx^) dx = [V" w?^^ - 75 w^^]^ Jo = (105iv - ^^- tv) - = i|o_ ,^ :^ 2916§ lb. = l|i T. dA = the area of MNRS = J xdx. it follows that ^ is (6 - x) ft. EXERCISES 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 15 ft. below the surface of the water in the reservoir. What is the pressure on the gate ? 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 in 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 the surface of the water. 4. The base of an isosceles triangle is 8 ft. and the equal sides are each 5 ft. The triangle is completely immersed in water, its base being parallel to and 5 ft. below the surface of the water, its alti- tude being perpendicular to the surface of the water, and its vertex being above the base. Find the total pressure on one side of the triangle. VOLUME 71 5. Find the pressure on one side of an equilateral triangle, 6 ft. on a side, if it is partly submerged in liquid so that one vertex is one foot above the surface of the liquid, the corresponding altitude being perpendicular to the surface of the liquid. 6. 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 the 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 the 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 line be drawn across the gate of Ex. 1 so that the pressure on the portion above the line 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, the magnitude of each element at the same time approaching zero. The question in each case is the determination of the form of the element dV. We shall discuss a comparatively simple case of a solid such as is shown in Fig. 29. In this figure let OH be a straight line, and let the distance of any point of it from be denoted by h. At one end the solid is bounded by a plane perpendicular to OH dX C, where OC=a^ 72 SUMMATION and at the other end it is bounded by a plane perpendicular to OH at J5, where OB = 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 OH 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=Adk The volume of the solid is then the limit of the sum of terms of the above type, and therefore V= C Adh. It is clear that the above discussion is valid even when one or both of the bases corresponding ioh = a and h = h, respectively, reduces to a point. Ex.1. Let OF (Fig. 30) be an edge of a solid such that all its sections made by planes perpendicular to OF are rectangles, the sides of a rectangle in a plane distant y from O being re- spectively 2 y and y\ We shall find the volume in- cluded between the planes y = and 3/ = 2^. Dividing the distance from y = Q io y = 2\ into n parts dy, and passing planes perpendicular to OY, we form rectangles such as MNRS, where, if VOLUME 73 OM = ij, MN = 3/2 and MS = 2 y. Hence the area MNRS = 2y\ and the volume of the elementary cylinder standing on MNRS as a base is 2y^dy\ *^^*'^' dV=2fdy. Therefore V = C^^y^dy = [^ y^f^= 19^|. Ex. 2. The axes of two equal right circular cylinders of radius a inter- sect at right angles. Required the volume common to the two cylinders. Let OA and OB (Fig. 31) be the axes of the cylinders and OY the common perpendicular to i OA and OB at their point of intersection 0. Then ^| OAD and OBD are quadrants of two equal circles 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 -p L"'' of the required volume. /^ We divide the distance OB into n parts dy, / ,^^^^ \ ^./^"^ and through the points of division pass planes per- [yi V^ pendicular to OY. Any section, such as LMNP^ ^ ^ is a square, of which one side NP is equal to Fig. 31 WOP^— ON^. OP = a, being a radius of one of the cylinders, and hence, as 0N= y, NP = Va2 - /. Accordingly, the area of LMNP = a^ — y^, and the volume of the ele- mentary cylinder standing on LMNP as a base is dV= (a'^-y'^)dy; whence V = J \a^ - y^) dy = [a^y - ^ y^]-= i a^ Hence the total volume is ^^ a'. 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^ = 4:Xf the axis of x, and the line x = 3. The generating area is shown in Fig. 32, where ^ i5 is the line x = 3. Hence OA = 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 AIP, NQ, etc., generates a circle, the plane of which is perpendicular to OX. The area of the circle generated by MP, for example, is ttMP ^, which is equal to tt?/^ = tt (4 x), if OM=x. 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 previous method for finding the volume. Since the base of any elementary cylinder is 4 TTX and its altitude is dx, we have dV = 4:Trx dx. Hence V= C^^Trxdx = [2 7rx^f^ = lSTr. ^^^ 32 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 = Q — x^. The line and the curve (Fig. 33) are seen to intersect at the points Pi(— 2, 5) and P^i^, 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^BA. Accordingly, we shall find the latter volume and multiply it by 2. We divide the line OM^ = 2 (M^ being the projection of P^ on OX) into n parts dx, and through the points of division draw straight lines parallel to OF and intersect- ing the straight line and the curve. One of these lines, as MQP, will, when revolved about OX, generate a circular ring, the outer radius of which is MP = y = 9 — x^ and the inner radius of which is MQ=y = 5. Hence the area of the ring is ttMP^- ttMQ^ = 7r(9 - x^ - 7r(5)2 = '7r(56-18a:2-f24). Fig. 33 Accordingly, 'dV = v (oQ — 18 x^ + x^) dx and V= f^Tr(5Q -18 x^ ■{■ x^)dx = 7r[5Q X - Q x^ + la:^]^ = 70|ir. t/O Accordingly, the volume of the ring is 2 (70 § tt) = 140| ir. VOLUME 75 EXERCISES 1. The section of a certain solid made by any plane perpendicular to a given line OH is a circle with one point in OH and its center on a straight line intersecting OH at an angle of 45°. If the height of this solid measured from along OH is 4 ft., 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 = A:X — x^. 4. Find the volume of the solid generated by revolving about OX the area included between the axis of x and the curve y = 2x^ — x^. 5. Find the volume of the solid generated by revolving about the line ?/ = — 2 the area bounded by the axis of y, the lines a: = 3 and ?/ = — 2, and the curve y = 3x^. 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 ^jB 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 y^z= Sx, the axis of x, 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:X^. Find the volume of the solid if the extreme distance along OY is 3. 9. Find the volume generated by revolving about *0Y the area bounded by the curve y^= Sx and the line x = 2, 10. Find the volume of the solid generated by revolving about OX the area bounded by the curves y = 6x — x^ and y = x^ — 6x -\-10. 11. A variable square moves with its plane perpendicular to OX, so that the ends of one of its sides are on the curves 16y = x" and 4:y = x^ — 12. Find the volume of the solid generated as the square takes all possible positions under the given conditions. 76 SUMMATION GENERAL EXERCISES 1. The velocity in feet per second of a moving body at any time t is ^^—4^ + 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 ^ is ^^ — 4 ^. Show that after j^ = 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 ^ = 6 to ^ = 9. 3. At any time t the velocity in feet per second of a moving body is ^^—6^ + 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 ^^ — 2 ^ — 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. ? 5 . Find the area bounded by the curves 9y = ^x^ and 45 — 9 ?/ = cc^. 6. Find the total area bounded by the curves y^= ^ax and 7/^ = 4 a^ — 4 ax. 7. Find the total area bounded by the curve y = x^ and the straight line y = A:X. 8. Find the total area bounded by the curve y — x(x — l)(x — ^) and the straight line ?/ = 4 (x — 1). 9. ABCD is a quadrilateral with A = 90°, B = 90°, ^^ = 5 ft., EC = 2 ft., ^ Z> = 4 f t. 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 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 AB^ 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. ^-B is 2 ft. long. The board is totally submerged in water, AB being perpendicular to the surface 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 lines are ic = 0, i/ = 4, and y^ = 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 — ic^. 17. Find the volume generated by revolving about OX the area bounded by the curve y = x^ + 2 and the line y = S. 18. Find the volume generated by revolving about OX the area bounded by OX and the curve y = S x — x^. 19. Find the volume generated by revolving about the line y = — l the area bounded by the curves 9y = 2x^ and 9 y = S6 — 2 x'^. 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 — bx^ 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 h. 78 SUMMATION 24. A solid is such 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^ = lQx and y^ = 4 ic, 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 OXj and the ends of a diameter are on the curves y = x^ and y = Sx^ — 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 Y, with its ends in the curve y = Ax"^. 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 = Q. 28. All sections of a certain solid made by planes perpendicular to OF are right isosceles triangles. One leg of each triangle coincides with the line perpendicular to OF with its ends in OF and the curve y^ = 4:X. Find the volume of the solid between the sections for which y = and y = S. 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 6 ft. and radius 3 ft., the vertex of the cone being down. CHAPTER IV ALGEBRAIC FUNCTIONS 27. Distance between two points. Let ij (x^, j/^) and I^x^, y^) (Fig. 34) be any two points in the plane XOF, such that the straight line iJTJ is not parallel either to OX or to Y, Through ij draw a straight line parallel to OX, and through i^ draw a straight line parallel to OF, and denote their point of intersection by iJ. Then P^R = A.r = x^— x^ and RP, In the right triangle PfiP^ p^p^ = ^pX+rp^' ^ p. / ^ I I Fig. 34 whence P,P, = y/(x,-x,y-^iy,-y,y •(1) If y^^y^^ P^P^ is parallel to OX, and the formula reduces to P,P, = x^-x^. (2) In like manner, if x^=x^^ P^P^ is parallel to OF, and the p.p. = y.-y.' (3) formula reduces to 28. Circle. Since a circle 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 a circle. Let 0(A, ^) (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 Qc-hy+iy-hf^r'. (1) 79 -X Fig. 35 80 ALGEBRAIC 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 x''-hf-2hx-2Jcf/-\-h'-\-k'-r'=0, (2) an equation of the second degree with no term in xi/ and with the coefficients of x^ and y^ „ equal. Conversely, any equation of the second degree with no xi/ term and with the coefficients of x^ and «/^ equal (as Aa^-{-Af-{-2Gx-{-2Fi/-\-C=0, (3) _ where A, G, F, and C are any con- stants) may be transformed into the form (1) and represents a circle, unless the number cor- responding to r^ is negative (see Ex. 3, page 81), in which ease 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. x^ + y^-2x-iy = 0. This equation may be written in the form (x^-2x )4-(/-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 (3^2 - 2 a: -f- 1) -f (y2 _ 4 y -1- 4) = 5, which may be placed in the form (x - 1)2 + (y- 2)2 = 5, the equation of a circle of radius Vo with its center at the point (1, 2). CIRCLE 81 Ex. 2. 9x^ + 9 f -Ox + Q7j-S = 0. Placing 8 on the right-hand side of the equation and then dividing by 9, we have 212 . ^ s which may be treated by the method used in Ex. 1. The result is the equation of a circle of radius ^VE, with its center at (i, —J). Ex.3. 9^:2 + 9/ -6a; +12?/ + 11 = 0. Proceeding as in Ex. 2, we have, as the transformed equation, an equation which cannot be satisfied by any real values of x 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 ic' + 2/'+6ic-10?/ + 9 = 0. 4. Find the center and the radius of the circle 5x'-\-5y^+Sx-6i/-15 = 0. 5. Find the equation of the straight line passing through the center of the circle x' + f+2x-y + l = and perpendicular to the line 2a; + 3?/-4 = 0. 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 called 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 the focus and ^*S^ the directrix of a parabola. Through F draw a straight Hue perpendicular to RS^ intersecting it at Z>, and let this line be the axis of x. Let the middle point of DF be taken as 0, the origin of coordi- nates, and draw the axis OY. Then, if the distance DF'i^ 2 c, the coordinates of i^ are (c?, 0) and the equation of RS is x=—c. Let P(x^ y) be any point of the parabola, and draw the straight line FP and the straight line NF SY perpendicular to RS. Then NF = x-\- c, N p^ (f D \F and, by § 27, FF= ^ {x - cf^- if-, ~^ ^ whence, from the definition of the parabola, ^^^cf^f={x^c)\ which reduces to i/=4:cx. (1) ^ Fig. 36 Conversely, if the coordinates of any point P satisfy (1), it can be shown that the distances FP and NP are equal, and hence P is a point of the parabola. Solving (1) for y in terms of x, we have ^ = ±2V^. (2) We assume that c is positive. Then it is evident that if a negative value is assigned to x^ y is imaginary, and no correspond- ing points of the parabola can be located. All positive values may be assigned to x^ however, and hence the parabola lies entirely on the positive side of the axis OY. Accordingly, we assign positive values to a:, compute the cor- responding values of y^ and draw a smooth curve through the points thus located. It is to be noticed that to every value assigned to x there are two corresponding values of «/, equal in magnitude and opposite in algebraic sign, to which there correspond two points of the parabola on opposite sides of OX and equally distant from it. Hence the parabola is symmetrical with respect to OX, and ac- cordingly OX is called the axis of the parabola. The point at which its axis intersects a parabola is called the ver- tex 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 a; = c?, equation (1) becomes y''=-4:cx (3) and represents a parabola lying on the negative side of OY. Hence we conclude that any equation in the form y''=lcx, (4) where A: is a positive or a negative constant, is a parabola, with its vertex at O, its axis on OX, its focus at the point (-» 0), and its directrix the straight line x = ~ -• 4 Similarly, the equation x^='ky ' (5) represents a parabola, with its vertex at and with its axis coin- ciding with the positive or the negative part of OF, according as Jc is positive or negative. The focus is always the point / k\ k . (0, -h and the directrix is the line ?/ = —-, whether k be positive 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 F^(x^, ?/i) and P^Cx^^ y.2^ be any two points of any parab- ola y''=kx (Fig. 37). Then y^ = kx^ and y2 = kx^\ whence whence 11^ = :^. (1) yf = ?!, (2yJ' ^2 84 ALGEBRAIC FUNCTIONS From the symmetry of the parabola, 2?/^= Q^P^ and 2^/^= Q^I^. But X, OM^ and x = OM^ and hence (1) becomes and the theorem is proved. The figure bounded by the parabola and a chord perpendicular to the axis of the parabola, as QfiP^ (Fig. 37), is called a 'parabolic segment. The chord is called the ha%e 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. Fig. 37 EXERCISES Plot the following parabolas, determining the focus of each : 1. if = — Sec. 3. 2/2= ^x. 2. x' 42/. 4. x^ 7 2/. 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 400 ft. 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 of a supporting wire attached to the roadway 75 ft. from the 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 plane perpendicular to its axis and 4 in. from its vertex ELLIPSE 85 9. Find the equation of the parabola having the line ic = 3 as its directrix and having its focus at the origin of coordinates. 10. Find the equation of the parabola having the line y = — 2 as its directrix and having its focus 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 i\\Q foci. Let F and F' (Fig. 38) be the two foci, 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 y middle point of F'F be taken as 0, the origin of coordinates, and draw the axis OF. Then the coordinates of F' and F are respectively (— c, 0) and a'\ f^ (c,0). Let P(x^ y) be any point of the ellipse, and 2 a repre- sent the constant sum of its distances from the foci. Then, from the definition of the ellipse, the sum of the distances F'F and FF is 2 a, and from the triangle F'FF it is evident that 2 a > 2 (? ; whence a> c. By § 27, and F'F=^{x^cy^y' FP=^(x-cy+y'', whence, from the definition of the ellipse, ^(x -h cy+ y''+^(x - cy+f= 2 a. Clearing (1) of radicals, we have Dividing (2) by a^— c^c^^ we have 2^.2 a^e (1) (2) ^. = 1. (3) 86 ALGEBEAIC FUNCTIONS But since a>c, a^—c^ is a positive quantity which may be denoted by i^ and (3) becomes Conversely, if the coordinates of any point P satisfy (4), it can be shown that the sum of the distances F'P and FP is 2 a, and hence P is a point of the ellipse. Solving (4) for y in terms of x, we have h y = ±-\(f—x^, (5) a 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 x^ 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 rr, 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 ?/, with the result a: = ±^V6^i:7. (6) From this form of the equation we find that the ellipse lies entirely between the lines y =—h and y = h and is symmetrical with respect to OY. Hence the ellipse has two axes. A' A and B'B (Fig. 38), which are at right angles to each other. But A' A = 2 a and B'B = 1h\ and since a > 6, it follows that A' A > B'B. Hence A' A is called the major axis of the ellipse, and B'B is called the minor axis of the ellipse. The ends of the major axis, A' and A, are called the vertices 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 8T From the defin ition oi h, c = Va^— h'\ and the coordinates of the foci are (± Va^— b% 0). OF The ratio — — (that is, the ratio of the distance of the focus from the center to the distance of either vertex from the center) is called the eccejitricity of the ellipse and is denoted by e. But OF=^a'-b'\ (7) and hence e = ; (S^ a ^ ^ whence it follows that the eccentricity of an ellipse is always less than unity. Similarly, any equation in form (4), in which h^ > a^, represents an ellipse with its center at 0, its major axis on OF, and its minor axis on OX. Then the vertices are the points (0, ± 5), the foci are the points (O, ± V^^— d\ and e = Li either case the nearer the foci approach coincidence, the smaller e becomes and the more nearly h = a. Hence a circle may he considered as an ellipse with coincident foci and equal axes. Its eccentricity is, of course, zero. EXERCISES Plot the following ellipses, finding the vertices, the foci, and the eccentricity of each : 1. 9a;2 + 16/=144. 3. Sx'' -\- 4.y'' = 2. 2. 9x^-\-Atf = 36. 4. 2x^+Si/ = l. 6. Find the equation of the ellipse which has its foci at the points (— 2, 0) and (6, 0) and which has the sum of the distances of any point on it from the foci equal to 10. 6. Find the equation of the ellipse having its foci at the points (0, 0) and (0, 5) and having the length of its major axis equal to 7. 32. Hyperbola. 77ie locus of a point the difference of whose distances from two fixed points is constant is called a hyperbola. The two fixed points are called the foci. 88 ALGEBRAIC FUNCTIONS Let F and F' (Fig. 39) be the two foci, 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 OF. Then the coordinates of F' and F are respectively (— c?, 0) and (c?, 0). Let P (x^ y) 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'F and FF is 2 «, and from the triangle F'PF it is evident that 2 a < 2 where u and v are two functions of x which can be V differentiated. Give x an increment Ax, and let Au, Av, and Ay be the corresponding increments of u, v, and y. Then . w + Au u v Au — uAv Ay = = v-\-Av V v^-\-vAv and Au Av V u — Ay Ax Ax Ax V -\-v Av DERIVATIVES 97 Now let A^ approach zero. By § 34, J . Au ^ , Av V Lim u Lim — - ^ . Ay Ax Ax Lim-^= — p — ; Ax t; + ?^ -L-im Av whence Ex.4. y = ^ du dv dy dx dx dx ~ v^ x^ + 1 dy _ (■r2 + l)(2a:)-(a:^-l)2a: ^ 4 a: dx (2:2 + 1)2 ~(a:« + l)2' 6. The derivative of the nth power of a function is obtained by multiplying n times the (n—l^th power of the function by the derivative of the function. Let y = t^", 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 Ax^ and let Au and Ay be the corre- sponding increments of u and y. Then Ay = (u + Auy — u"" I whence, by the binomial theorem. At/ = n?/"-^ Au + '^^^~^^ u''-\Auy+ ... -^(^Auy. Ay „_iAm , nCn — l^ Au ' , ,. ,„ .Au Ax Aa: 2 Ax Ax Now let Ax, Au, Ay approach zero, and apply theorems 1 and 2, § 34. The limit of —^ is -^, the limit of -^ is -^, and the Ax dx Ax dx limit of all terms except the first on the right-hand side of the last equation is zero, since each contains the factor Au. Therefore ^ ^^ dx dx 98 ALGEBEAIC FUNCTIONS Case II. When n is a positive rational fraction. Let n = —-, where p and q are positive integers, and place y = u\ By raising both sides of this equation to the ^th power, we have Here we have two functions of x which are equal for all values of x. Taking the derivative of both sides of the last equation, we have, by Case I, since p and q are positive integers, n \dy „ ^du Substituting the value of y and dividing, we have dy __p '^-'^du dx q dx Hence, in this case also, dy „ T du — ^ = nu • dx dx Case III. When % is a negative rational number. Let n=—m, where m is a positive number, and place „. 1 y=-- =-.■ div") dy dx len -/ = dx u^-^ (by 5) mu — - dx (by Cases I and II) = — mu —-• dx ence, in this case also. dy n \du 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 by the derivative of the quantity. Ex. 5. y = (x-3 + 4 x2 - 5 a: + 7)^. ^ = 3 (x3 + 4 a:2 - 5 a: + 7)2 — (a:^ + 4 a:2 - 5 a: + 7) dx dx = 3 (3 a:2 + 8 a; - 5) (x^ + 4 a;2 - 5 a: + 7)2. Ex. 6. ?/ = v^ + i = a:! + x-^. X'' -^ = -X 3— 3a:- dx 3 3^^ Ex. 7. ?/ = (a: + l)Va;2 + l. dx dx dx = (a: + 1) [I (x2 + 1)- i . 2 a:] + (x^ + 1)^ = ^i^ + (.^ + l)^ (a:2 + 1)2 2 a:2 + a: + 1 Va;2 + 1 Ex.8. y='l1IZ = (-I-)\ id dx~3\x^-\-l) dxW-hl) ^ 1 / a:^ + 1 \ H - 2 x^ ~3\ X I (x^ + iy 1 - 2 a:^ 3a:^(x3 + l)^ 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 Ax and Ay be corresponding increments of x and y. It is immaterial whether Ax is assumed and Ay determined, or Ay is assumed and Ax determined. In either case whence that is, Ax 1 Ay '"Ay Ax T irv* — 1 Ay Lim Ax dx 1 Ty~~ ~'dy dx ^. If y is a function of u and u is a function of x, 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 Ax determines an increment Au, and this in turn determines an increment Ay. Then, evidently. whence Lim ^^^ = Lim ^^ . Lim — Ax that is, Ex. 9. 7/ Ay_ Ax Ay Au Au Ax L . Ay Ax •- Lira Au dy_ dx dy du' du dx' = u^ ' + 3 w + l, where u _ 1 x^ dy _ dx .{2u + 3)( -a- 2 + 3a:2 x^ 4 + 6a:2 The same result is obtained by substituting in the expression for y the vahie of u in terms of x and then differentiating. DERIVATIVES 101 36. Formulas. We may now collect our formulas of differen- tiation in the following table : %'"■ (1) d(cu) du dx dx (2) d(u-\-v') du dv dx dx dx (3) d{uv) dv du dx dx dx (-t) , /u\ du dv \vl dx dx (5) dx v^ d(iC) „ ,du dx dx (6) dx 1 dy dy (7) dx dy dy du dx du dx (8) dy dy du dx dx (9) 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 hydx. They are ^^^^^^ ^^^^ d(^eu) = cdu, (11) d(u + v) = d^w + dv, (12) d(uv') = udv -\-vdUy (13) ,/w\ vdu — udv ,^ .. d(ii'') = 7iu''-^du, (15) 102 ALGEBRAIC FUNCTIONS EXERCISES Find -^ in each of the following cases : 1. 7/ = (3a: + l)(2x2+3a;-2). ^^^+9 * *^ ^ x-1 ' y x^ + x-j-l' 16. ?/ = X V9 - a 6. 2/ = "v^aj^ 1 17. v/ = (cc + 1) Vx^ - 1. ■- Ex. 1. Find the equation of the tangent Une to the parabola ?/2 = 3 ^ at the point (3, 3). By differentiation we have 24^ = 3; dx whence -r — tt-' dx 2y Hence, at the point (3, 3), the slope of the tangent line is ^, and its equation is y_3=,i(^_3) or x-2y + S =0. The angle of intersection of two curves is the angle between their respective tangents at the point of intersection. The method of finding the angle of intersection is illustrated in the example on the following page. TANGENT LINE 105 Ex. 2. Find the angle of intersection of the circle x^ + 3^2 _ g and of the parabola a;^ = 2 y. The points of intersection are Pi(2, 2) and 7^2 (~ 2» ^) (^^S- 44), and from the sym- metry of the diagram it is evident that the angles of intersection at P^ and P^ are the same. Differentiating the equation of the circle, we have 2a: + 2?/— = 0, whence -^ = ; ax dx y and differentiating the equation of the pa- rabola, we find -^ — X. dx Hence at P^ the slope of the tangent to the circle 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, — 1 — 2 tan ^ Fig. 44 iS 1-2 tan -13. EXERCISES 1. Find the equation of the tangent line to the curve a;'— 83/^ + 16 2/ - 8 = at the point (2, 2). 2. Find the equation of the tangent line to the curve 5 a;* — ^iX^y = 4 ?/' at the point (2, 1). 3. Find the point at which the tangent to the curve 8?/ = ic^ at (1, \) intersects the curve again. 4. Find the angle of intersection of the tangents to the curve y^ = x^ at the points for which x = 1. x^ y^ 5. Show that the equation of the tangent to the ellipse ^ + 7^ = 1 X X 1/ 1/ do at the point (x^, y^'i^ ~T + "ri^ = 1- 6. Show that the equation of the tangent to the hyperbola 2 = 1 at the point (x^, y^)is^-^ = l. x' ?/ a 7. Show that the equation of the tangent to the parabola y^—kx k at the point {x^, y^ is y^y = 2 (^ + ^i)- Draw each pair of -the following curves in one diagram and deter- mine the angles at which they intersect : 106 ALGEBRAIC FUNCTIONS 8. x''-^y''=5, x^-{-y''-2x-\-4:2j-5 = 0. 9. x^ = 3ij, 9y^=Sx. 10. y^=4.x, x^-\- f=5. 11. y = 2x, icy = 18. 12. 07 — 4?/ — 4=0, x^— 4:X — 4:1/ = 0. 13. x^-{-i/==25, x^=Sy-3, 39. The differentials dx, dy, ds. On any given curve let the distance from some fixed initial point measured along the curve to any point P be denoted by s, where s is positive if F lies in one direction from the initial point and negative if P lies in the opposite direction. The choice of the positive direction is purely arbitrary. We shall take as the positive direc- tion of the tangent that which shows the positive direction of the curve, and shall denote the angle between the positive direction of OX and the positive direction of the tangent by (/>. Now for a fixed curve and a fixed initial point the position of a point P is determined if 8 is given. Hence x and «/, the coordinates of P, are functions of s which in general are continuous and may be differentiated. We shall now show that dx -— = cos dy _ ds = sin <^. Let 2LTC FQ = As (Fig. 45), where P and Q are so chosen that As is positive. Then PP = Ax and RQ = Ay, and Ax As PR chord P^ PP arcP^ diVcPQ chord P^ „„„ = ^ • cos EFQ, Ay _ RQ _ chord P() As arcP^ arcP^ chord P^ chord P^ RQ chord Pg arcP^ sinPP^. MOTION IN A CURVE lOT 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 EPQ = <^. Hence, taking limits, we have ^^ di/ _ = cos 6, -^ as sm ; whence, by squaring and adding, we obtain the important equation ds = dx -\- dy' (2) dx E This relation between the differentials of x^ 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 in mind that RQ \s, not rigorously y equal to dy (§ 20), nor is PQ rigor- ously equal to ds. In fact, BQ = 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 )\ 2/^=16(:r-3). 32. 2if=x\ x'-\-y^-4.x = 0. 33. The coordinates of a moving particle are given by the equa- tions x = f, y = (1— fy^. 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 = ht — \gf. Find its path and its velocity in its path. GENERAL EXERCISES 115 36. A body so moves that x = — 2 -^ t\ ij = 1 -^ t. Find its path and its velocity in its path. 37. A particle is moving along the curve 1/^== Ax; and when cc = 4, its ordinate 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 ordinate ? 38. A particle describes the circle x^ -\- y^ = a^ with a constant velocity v^. Find the components of its velocity. 39. A particle describes the parabola if = A^ax in such a way that its ic-component of velocity is equal to ct. Find its ?/-component of velocity and its velocity in its path. 40. A particle moves so that x = 2t, y = 2 V^ — f. Show that it moves around a semicircle in the time from ^ = to ^ = 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 f t. ? 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 the 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 water is running out of it at the rate of 2 cu. 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 the rectangle that the least amount of fencing may be required ? 54. An open tank with a square base and vertical sides is to have a capacity of 500 cu. ft. Find 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 it 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 to be made in the form of a right circular 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 pyramid. Eind the ratio of its height to a side of its base when the air space inside the tent is as great as possible for a given wall surface. 69. 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 B, are situated respectively 12 mi. and 18 mi. back from a straight river from which they are to get their 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 on 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 ? 61. A man on 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 h 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 -f 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. 66. An isosceles triangle is inscribed in the ellipse ~^ + f^ = 1, (a > b), with its vertex in the upper end of the minor axis of 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 which 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 ^=-/ (1) 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 J, 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 = ^mx (2) may be considered as defining a functional relation between two quantities exactly as does the simpler equation y — x^. 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 Lim?i^=l, (8) A-»0 H where the left-hand member of the equation is to be read " the limit of — -— as h approaches zero a's B a limit. ^^ \ To prove this theorem we proceed as ^/^T) '^\^^\\ follows : ^^ I ] ) Let h be the angle AOB (Fig. 49), r the ""^-.^ j // radius of the arc AB described from as ^^-V a center, a the length of AB, p the length ^^^ ^' 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 BCB'=2p, the arc BAB' =2 a, and the tangent B'D= the 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 h, we have cos h sm h or, by inverting, , ^ sin A ^ ^ cos h < —7— < 1. GRAPHS 121 Now as h approaches zero, cos A approaches 1. Hence — — > h which lies between cos A and 1, must also approach 1; that is, T . sin ^ ^ Lim— z— =1. This result may be used to find the limit of as h approaches zero as a limit. For we have r» • 2^ . 2^ I • h 2 suT - sin^ - , I sm - 1-cosA 2 2 A 2 h 2 h 2\ h 2 1 2 / . h sm- Now as h approaches zero as a limit, approaches unity, by (3). Therefore ^ A -^. 1 — cosA ^ ,.. Lim = 0. (4) 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 in the previous section. In this connec- tion it is to be remembered that tt is simply the number 3.1416, and that the angle tt 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 hx. It is convenient first to fix the values of x which make y equal to zero. Now the sine is zero when the angle is 0, tt, 2 tt, 3 tt, — tt, — 2 tt, or, in general, k-K, where k is any positive or negative integer. To make y = 0, therefore, we have to place bx = kw; whence 27r TT - TT 27r Stt x=-", — , - -, 0, -, --, — -, •••. b b b b b The sine takes its maximum value + 1 when the angle has the values 7r.5 7r97r^ ^, ^ . . ^,. , ''■5 7r9ir,„., - > — - > — , etc. ; that is, in this case, when ar = -— , -— , — - , etc. h or these 222 2b 2b 2b values oi X, y = a. 122 TKIGONOMETRIC FUNCTIONS The sine takes its minimum value —1 when the angle is — ^ — -> etc. ; that is, in this case, when x — — » — > etc. For these values oi x, y =— a. These values of x for which the sine is ± 1 lie halfway between the values of x for which the sine is 0. Y 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 b = 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 cosbx. As in Ex. 1, we fix first the points for which y = 0. Now the cosine of an angle is zero when the angle is -> -— -j -— -» etc.; that is, any odd multiple of - • We have, therefore, y = when Fig. 51 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 = 3 and 6 = 2. GRAPHS 123 Ex.3, y = asin(bx + c). We have y = when bx + c = 0, tt, 2 tt, 3 rr, etc.; that is, when Fig. 52 Halfway between these values of x the sine has its maximum value + 1 and its minimum value — 1 alternately, and y =± a. The curve is the same as in Ex. 1, but is shifted - units to the left (Fig. 52). Ex.4. ?/ = sinx + ^ sin 2x. The graph is found by adding the ordinates of the two curves y = sin x and y = \ sin 2 x, as shown in Fig. 53. Y i^-ysinSx ^---~rl o /S^x „.. ■""'X>/ 2 --' \ \ y= sin X + -J sin 2x ~'=sinx Fig. 53 EXERCISES Plot the graphs of the following equations 1. ?/ = 2 sin 3 a?. 2. 2/ = 3cos^. 3. ?/ = 5 sin (^ — t)' 4. ?/ = 2 cos (^ + t) 5. 2/ = 2 sin (x - 2). e. y = tan 2 a;. 7. y = ctn 3 a:;. 8.7/ = sec 07. d. y = CSC 2 a:. 10. ?/ = vers 0!. 11. ?/ = 1 + sin 2 a;. 12. y = sin^x -\- sin 2 x. iU TKIGONOMETHIC t^UNCTlOKS 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 ^-, . -— sinu = co&u—-'> (1) dx dx (2) (3) (4) (5) (6) These formulas are proved as follows: 1. Let 2/ = sin u^ where u is any function of x which may be differentiated. Give x an increment Ax and let Au and A?/ be the corresponding increments of u and ?/. Then At/ = sin (u + Aw) — sin w = sin u cos Aw + cos u sin A?i — sin u = cos u sin Aw — (1 — cos Aw) sin w ; Aw sin Aw 1— cosAw . whence — ^ = cos w — sm w. Aw Aw Aw d dx COS w = du — smw-—» dx d dx tan w = 2 du sec^w -— , dx d dx ctnw = 2 du — CSC W —-5 dx d dx sec w = du sec w tan W-— J dx d dx CSC w = du — CSC w ctn w— - dx Now let Ax and therefore Aw approach zero. By (3), § 42, di/ _ Urn ^^^^ = 1, and, by (4), § 42, Lim -*• ^^^^"^ = 0. Therefore Aw ' J V y 5 » ^^ cos w. du But by (8), § 36, ^ = ^^, •^ ^ ^ ^ dx dudx and therefore -^ = cos w — — dx dx DIFFERENTIATION 125 2. To find — - cos u, we write ax cos u = sin ing-«). Then -;- cos u = —-sm(— — u] ax ax \2 / = — sin t* -—■ 3. To find — - tan w, we write ax sm w tan u = cos w rr^, d ^ c? sin w 1 hen -— tan u = - ax ax cos u d . . d cos ^ — sin ?^ — sin u — cos u 't _ £f (by (5), § 36) (o . o ■- au cos 2^ + sin^w)— - dx (by (1) and (2)) = sec M-— dx 4. To find — - ctnw, we write dx cosw ctn u — — sm w m, d ^ d GQSU 1 hen — - ctn u = - — -. dx dx sin u d d . sinw — cos w — cos w — sm 1^ ff ff (by (5), § 36) -^m'u-GO^'u du ^^y ^^^ ^^^ ^2)) sin^ifc . dx du dx 126 TRIGONOMETRIC FUNCTIONS 5. To find — - sec w, we write ax sec u = = (cos u) \ cosw Then — sect^ = — (cos w)"^ — cosw (by (6), § 36) sin 2^ du GO^^u dx du ^QGUidillU- — dx (by (2)) 6. To find -— CSC u^ we write dx CSC u = -: — = (sm U) \ smw Then — esc u = — (sin w)~^ — sin u (by (6), § 36) (XX (XX — — CSC u ctn i^ — • (By (1)) (XX Ex. 1. y — tan 2 a: — tan^a: = tan 2 x — (tan a:)^. — = 860^2 a; — (2 x) — 2 (tan x) — tan x dx dx dx — 2 860^2 a; — 2 tan x sec^a:. Ex. 2. y = (2 sec^a: + 3 sec^a:) sin x. — = sin a: Ssec^a; — (seca:)+ 6 sec a: — (sec a:) + (2sec*a: + Ssec^a:) — (sin a:) dx L dx dx J dx = sin a: (8 sec* a: tan a: + 6 sec^a: tan a;) + (2 sec* a: + 3 sec^a:) cos x = (1 — cos^a;) (8 sec^a: + 6 sec^a:) + (2 sec^a; + 3 sec x) — 8 sec^x — 3 sec x. EXERCISES dti Eind "T- in each of the followincr cases : dx 1. y = 3 sin 5x. 2. y = 2tan|. 3. y = J sin'2jr. 4. y = cos^Scc. 5. X 1 . ^ 2/ = ---sm2a:. 6. y =^\ sin^5 £c — ;^ sin^5 x. 7. J>x 8. 2/ = J csc^3 2c. SIMPLE HAKMONIC MOTION 127 2 ^x ^ X ■ , . X ^ X 9. 2/ = -cos^- — 2cos-- 11. i/ = 4:sm-— 2xco8^' 2 , „x , ^ , X , ^ sec ic + tan x 10. 2/ = o ctn^o + 2 ctn -. 12. ?/ = '-- ,3 2 2 -^ sec ic — tan a; 13. ?/ = sin (2 X + 1) cos (2x — 1). 14. y = tan^3cc- 3tan3x + 9a;. 15. ?/ = sec 2 X tan 2 cc. t^ 16. y = 5V(3cos52x- 5cos»2a:). 17. sin2a;H-tan3?/ = 0. 18. xy -\- ctn xt/ = 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 = c sin bt, (1) where c and h are constants. We have for the velocity v and the acceleration a i i^ ^^.^ v = eo cos btj (2) a = — (?5^sin 5^. (3) When ^ = 0, 8 = and the particle is at (Fig. 54). When t^-^r-p s = e and the particle is at A, where 0A= c. When t is between and — -> t; is positive and a is negative, M SO that the particle is moving from to ^ with decreasing speed. When t is between — - and -rfvis> ~b J Zt b negative and a is negative, so that the particle moves toward with increasing speed. When TT ^ = — , the particle is at 0. As ^ varies from ^ to -— ^» the particle moves with decreasing speed from to B^ where OB = — c. Finally, as t varies from 9T ^o — » the particle moves back from ^ to with increasing speed. 128 TRIGONOMETRIC FUNCTIONS The motion is then repeated, and the particle oscillates between B and J, the time required for a complete oscillation being, as 2 TT we have seen, — — The motion of the particle is called simple harmonic motion. The quantity c is called the amplitude, and the 2 TT interval —r-^ 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 = — hmcV^mi ht= — Jcmh^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 and 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 = c sin h(t— t^), (4) the change amounts simply to altering the instant from which the time is measured. For the value of s which corresponds to ^ = ^^ in (1) corresponds to t=t^-{-t^ in (4). Hence (4) represents 2 TT simple harmonic motion of amplitude c and period -— — But (4) may be written s = c cos 5^^ sin bt — c sin bt^cos bt, which is the same as s=A sin bt-\-B cos bt, (5) 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=^A^ + B\ tanJ^ =--, ° A which determines e and t^ in (4). SIMPLE HARMONIC MOTION 129 Therefore equation (5) also represents simple harmonic motion with amplitude y/ A^-\-B'^ and period — — In particular, if in (5) A—0 and B = c^ we have « = c cos hU (6) If in (4) we place t^ = —- — tl^ it becomes 8=:CG0^h(t-t[), (7) which differs from (6) only in the instant from which the time is measured. EXERCISES 1. A particle moves with constant velocity 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 5 = 5 sin 2 t. Find the time 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 = c sin ht. 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^^. 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 of 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 of 6 ft. from its mean position. Find its amplitude and its period. 9. A point moves with simple harmonic motion given by the equation s = a — ^ sin ct. Describe its motion. 130 TRIGONOMETRIC FUNCTIONS 46. Graphs of inverse trigonometric functions. The equation ir = 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 = ,iu-^^, (2) where — 1 is not to be understood as a negative exponent but as part of a new symbol sin~^ 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 a:." Similarly, if x = cosy, then y = GOS~'^x; if x = ts^ny, then y = tan~^ x ; 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 1 TT of angles with the same sine. For example, if a: = -» ^ = ^' CIO Z b — 1 —-—■, etc. This 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) we may change (2) into the equivalent form (1) and proceed as I GRAPHS 131 m § 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. Fig. 56 Ex. 1. ?/ = sin-^x. From this, x = sin y, and we may plot the graph by assuming values of y and computing those of x (Fig. 55). Ex. 2. y = tan-^a:. Then x = tan y, and the graph is as in Fig. 56. EXERCISES Plot the graphs of the following equations : l.y = t^n-^2x. 3. 1/ = sm-^(x — l). 5. ?/ =l-f- cos-^rc. 2. 7/ = ctn-i3ic. 4. 2/ = tan-^ (x -f 1). 6. ij = ^tsLU-^x. 7. 7/ = cos- ^(a: — 2). 8. ij = sm-\2x i-1) - -- 47. Differentiation of inverse trigonometric functions. The formulas for the differentiation of the inverse trigonometric functions are as follows: du 1. -— sm ^u = -—= dx Vl- ,2 dx Vl when sin-^tt is in the first or the fourth quadrant; — when sin~^w is in the second or ^ ^ the third quadrant. 132 TRIGONOMETRIC FUNCTIONS 2. — cos"^w = — when cos~^w is in the first or the -^ ~" ^ second quadrant ; — - when cos"^w is in the third or the ^ fourth quadrant. g c? _j 1 du dx \-\-u^ dx ^ c? 1 _ \ du ' dx 1-^u^ dx 5. — -sec"^i^ = — —- when sec~^w is in the first or the dx n^u^-ldx third quadrant; -r- when sec~^«^ is in the second or uy/u^-1 dx the fourth quadrant. d . 1 du ^ 1 . . 1 o — CSC ^w = — when csc~^w is in the first or dx u^u'-ldx the third quadrant; — - when csc~^M is in the second or u^u""-! dx The proofs of these formulas 1. If y = the fourth quadrant, are as follows: : sin~^w, then siny = ■u. Hence, whence by § 44, cos y-^ = dx dx du "dx' 1 du cos y dx But cos ?/ = Vl — \ ^ when y is in the first or the fourth quad- rant, and cos ?/ = — Vl — i^ when y is in the second or the third quadrant. 2. If y = cos~^u, en cos y = u. Hence dy du ^ dx dx lence dy _ 1 du dx sin y dx I DIFFERENTIATION 133 But sin y = Vl — u^ when «/ is in the first or the second quad- rant, and sin 1/ = — Vl — u^ when y is in the third or the fourth quadrant. 3. If 1/ = tan~^2^, then tan y = w. Hence sec^ v -^ = — - ; ax ax whence dy _ 1 du dx 1+u^ dx 4. If y = ctn~^it, then ctn y — u. Hence — csc^ ?/ -^ = -- ; whence dy _ 1 du dx l-{-u^ dx 5. If y = sec ^1^, then sec y = u. TT ^ c?y du Hence sec ^ tan v -^ = -7- ; dx dx whence dy _ 1 du dx sec y tan y dx But seQy = u, and tany = Vtt^— 1 when y is in the first or the third quadrant, and tan«/ = — Vm^— 1 when y is in the second or the fourth quadrant. 6. If y = csG~^u, then CSC y — u. TT ^ <^y ^^^ Hence — csc y ctn y-^ = -—; dx dx , (?V 1 (??/ whence -r^ = ;-• dx csc y ctn ^ aa; 134 TRIGONOMETRIC FUNCTIONS But csGi/ = Uy and ctn ?/ = Vw^ — 1 when y is in the first or the third quadrant, and ctn y — — ^u^ — 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 in differentiation. Ex. 1. 2/ = sin-i Vl — x^, where y is an acute angle. dx Vl-(l-x2) dx Vl-a:2 This result may also be obtained by placing sin-^vl Ex. 2. 2/ = sec -1 V4 x2 + 4 a; + 2. d V 4 a:2 4- 4 a: + 2 dy ilx ^^^ dx V4 :c2 + 4 a; + 2 V(4 x-^ + 4 a: + 2) - 1 4x4-2 1 (4 a;2 + 4 x + 2) (2 a; + 1) 2 a:^ + 2 a: + 1 EXERCISES Find -^ in each of the following cases : 1. y = ^i\\-^^x. ^■\/x'-\-2x . 11. v = cos~^- — • 2. 2/ = sin~^-. y = sm 1—^—. ^ X 3. y = 4. 3/ = cos-^ ^^^~^ - 13- 2/ = tan-W^^^l + cos-^^. 5. 2/ = tan-i(^ + l). 14- 2/ = ^Vrir^^ + sin-ix. 6. y = tan- Vi^32^. 15 _^ + tan-^. 1 -^ x'2 + 4 2 7. 2/ = ctn-i-^- ^ 16. 2/= Vx2-4-2sec-i-. 8. 2/ = sec~^5a!. ^ 9. 2/ = csc-i2a;. 17. 2/ = tan-i-(^---j. • 2/= an ^^_ 2- ^g^ y= Vl— ic^-l-^'cos-i Vl— ic^. I 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 6 with respect to ^ is called the angular velocity of OP. The angular velocity is commonly denoted by the Greek letter (o ; so we have y the formula ^q In accordance with § 42 the angle 6 /- is taken in radians ; so that if t is in ^^^ 57 seconds, the angular velocity is in radians per second. By dividing by 2 tt, the angular velocity may be reduced to revolutions per second, since one revolution is equivalent to 2 7r 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 AQ measured from J, then, by § 42, s = rO. (2) Now — is called the linear velocity of the point Q^ since it measures the rate at which s is described ; and from (2) and (1), d% dO ^ox showing that the farther the point Q is from the greater is its linear velocity. Similarly, the angular acceleration, which is denoted by a, is defined by the relation , ,2/, •^ d(o du ^.^ dh This is connected with the linear acceleration — by the formula ^^ ^,^ 136 TRIGONOMETRIC FUNCTIONS Ex. 1. If a wheel revolves so that the angular velocity is given by the formula w = 8 <, how many revolutions will it make in the time from t = 2 to < = 5 ? We take a spoke of the wheel as the line OP. Then we have de = 8tdt. Hence the angle through which the wheel revolves in the given time is e= f%tdt = [4 t^t = 100 - 16 = 84. «/2 The result is in radians. It may be reduced to revolutions by dividing by 2 TT. 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 whence dd ^ (0 = - — = 2 nir: dt e = 2n7rt + C. Fig. 58 If we consider that when / = 0, the particle is on OX, then C = 0. Therefore ^ „ , X = a cos a = a cos 2 mrt = a cos o>^ The point M therefore describes a simple harmonic motion. In fact, simple harmonic motion is often defined in this way. EXERCISES 1. A flywheel 4 ft. 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 in. 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. I CYCLOID 137 3. If the horizontal and vertical projections of a point describe simple harmonic motions given by the equations x = 5 cos St, 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, each point of the rim describes a curve called a cycloid. Let a wheel of radius a roll upon the axis of x, and let C (Fig. 59) be its center at any time of its motion, .JV its point of Fig. 59 contact with OX, and P the point which describes the cycloid. Take as the origin of coordinates, O, the point found by rolling the wheel to the left until F meets OX. Then ON=^tgFN, Draw MP and ON, each perpendicular to OX, PR parallel to OX, and connect C and P. Let angle NOP = <^. Then x=^OM=ON- MN =^^vcPN-PR = a(t> — a sin <^. i/ = MP=NC-PC = a — a cos (j). Hence the parametric representation (§40) of the cycloid is x = a(^ — sm )-— = a(o(\ — cos <^), v^^ = a sni 9 ~ = aca sni 9 ; whence v^ = aV (2 — 2 cos <^) = 4 aV sin^ ?» and ?; = 2 «ft) sin ^? as an expression for the velocity in its path of a point on the rim of the wheel. EXERCISES 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 rim 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 is 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 h from the center describes a curve given by the equations X = a — h sin , y = a — h cos <^, and find the velocity of the point in its path. The curve is called a frochold. 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. CUIIVATUIIE 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 AI^= s and J^I^ = As, and if <^i and (f)..^ are the values of for the points j^ and li respectively, then .^— c^^ is the total change of direction of the curve between J^ and ^. If y (^2 — <^i = A(/>, expressed in circular measure, the ratio ^ is the average change of direction per linear unit of the arc ij^. Regarding (^ as a function of s, and taking the limit of — as As As approaches zero as a limit, we have -— ? 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. y X yy A — ^^ X^2 ^ Fig. 60 If # . is constant, the curvature is constant or uniform ; other- as wise the curvature is variable. Applying this definition to the circle of Fig. 61, of which the center is C and the radius is a, we have A<^— iJCiJ, and hence As = a A(f). Therefore -~- — - • d^ \ As a Hence -p- = -? 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 /?. Through every point of a curve we may pass a circle with its radius equal to /o, which shall have the same tangent as the curve at the point and shall lie on the Fig. 61 140 TRIGONOMETKIC 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 If the equation of the curve is in rectangular coordinates, ds by (9), §36, p = ^. dx To transform this expression further, we note that ds = dx 4- dy ; whence, dividing by dx and taking the square root, we have dx Since 4> = tan- > {^\ (by § 15) d(^ _ dx^ Substituting, we have p = -r^ • 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. CURVATURE 141 and Ex. 1. Find the radius of curvature of the ellipse H — = 1. Here -j. = - —- ax a^y tPy__ b* Therefore (oV+M^t. '^ a*b* Ex. 2. Find the radius of curvature of the cycloid (§ 49). We have -— = a (1 — cos ) = 2a sin^ — , dcp 2 dy • . o • <^ ^ -— = a sin

2 CSC' ^ 4a 2 EXERCISES 1. Find the radius of curvature of the curve i/ — -^^ x•^ 2. Find the radius of curvature of the curve x^ -\- y^ = a7'. 3. Find the radius of curvature of the curve y = tan~^(ic — 1) at the point for which x = 2. 4. Show that the circle (x — -~\ + ?/ = 1 is tangent to the curve y = sin X at the point for which ic = — , and has the same radius of curvature at that point. 5. Find the radius of curvature of the curve x = cos t^y = cos 2 t^ at the point for which ^ = 0. 6. Find the radius of curvature of the curve a? = a cos <^ + a^ sin cos . 7. Show that the product of the radii of curvature of the curve y = ae ''at the two points for which cc = ± « is a^{e -\- e"^)^ 142 TRIGONOMETRIC FUNCTIONS Pir,e) Fig. 62 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 F any point in the plane, and draw OF. Denote OF by r, and the angle MOF by 6. Then r and 6 are called the polar coor- dinates of the point F(^r, 6), and when given will completely determine F, For example, the point (2, 15°) is plotted by laying off the angle MOF =15° and meas- uring 0F= 2. OF, or r, is called the radius vector, and 6 the vectorial angle, of P. These quantities may be either positive or negative. A negative value of is laid off in the direction of the motion of the hands of a clock, a positive angle in the opposite direction. After the angle 6 has been constructed, positive values of r are measured from along the terminal line of 6, 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 6 to positive values. Plotting in polar coordinates is facili- tated by using paper ruled as in Figs. 64 and 65. The angle is determined from the numbers at the ends of the straight lines, 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, 6) 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 rr of a system of rectangular coordinates. Let F (Fig. 63) be any point Fig. 63 POLAR COORDINATES 143 of the plane, (x, y) its rectangular coordinates, and (r, 0) its polar coordinates. Then, by the definition of the trigonometric functions, cos ^ = -7 r sin r Whence follows, on the one hand, x = r cos ^, «/ = r sin ^ ; and, on the other hand, sin e = ^ ^x'-\- y cos Q = — (1) (2) By means of (1) a transformation can be made from rectangular to polar coordinates, and by means of (2) from polar to rectangular coordinates. When an equation is given in polar coordinates, the corre- sponding curve may be plotted by giving to Q convenient values, computing the corre- sponding values of r, ^^o ^^^° plotting the resulting points, and drawing a curve through them. Ex. 1. r = acos^. 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 Q. By plotting the points cor- responding to values of from 0° to 90°, we obtain the arc ABCO (Fig. 64). Values of B from 90° to 180° give the arc ODEA. Values of Q from 180° to 270° give again the arc ABCO, and those from 270° to 360° give again the arc ODEA . Values of Q greater than 360° can clearly give no points not already found. The curve is a circle. 144 TKIGONOMETRIC FUNCTIONS Ex. 2. r = a sin 3^. As 6 increases from 0° to 30°, r increases from to a ; as ^ increases from 30° to 60°, r decreases from a to ; the point (r, 6) traces out the loop OAO (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 6 increases from 90° to 120°, r increases from — a to ; the point (r, 6) traces out the loop OBO. As 6 increases from 120° to 180°, the point (r, 6) traces out the loop OCO. Larger values of 6 give points already found, since sin 3 (180° + ^) = -sin 3 a The three loops are congru- ent, because sin;3 (60° -\- B) = — sin 3 0. This curve is called a rose of three leaves. Ex. 3. r2 = 2 a2 cos 2 0. Solving for r, we have ± a V2 cos 2 e. Hence, corresponding to any values of 6 which make cos 2 6 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 6 which make cos 2 6 negative, the corresponding values of r will be imaginary and there will be no points on the curve. Accordingly, as 6 increases from 0° 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 $ increases from 45° to 135°, cos 2 ^ is negative, and there is no portion of the curve between the lines 6 = 45° and = 135° ; finally, as 6 increases from 135° to 180°, r increases numerically from to a V2, and the portions of the curve in the second and the fourth quadrant are constructed. The curve is now complete, as we should only repeat the curve already found if we assigned further values to ^; it is called the lemniscate. GRAPHS 145 Fig. 67 Ex. 4. The spiral of Archimedes, r = aO. In plotting, is usually considered in circular measure. When 6=0,r = 0', and as $ increases, r increases, so that the curve winds an infinite number of times around the origin while reced- ing from it (Fig. 67). In the figure the heavy line represents the portion of the spiral corresponding to positive values of 6, and the dotted line the portion corresponding to negative values of 6. EXERCISES Plot the graphs of the following curves : 9. r = a sin^-- o 10. 7'^=a'^smO. 11. 7^= a'^sinSe. 12. r =a(l-cos2^). 13. r =a(l + 2cos2^). 14. r = a tan 0. 15. r = atsin2e. 1 1. r = a sin 6. 2. r = asm2 0. 3. r = a cos 3 6. 4. . e r = a sin-. o 5. e r = a cos-- 6. r = 3cos^ + 5. 7. r = 3 cos ^ + 3.* 8. r = 3 cos ^ + 2. 16. r t 1 H- cos ^ Find the points of intersection of the following pairs of curves : 17. r =2 sin B, r = 2 Vs cos Q. 18. r^ = ^2 (jog Q^ ^.2 ^ ^2 gjj^ 2 Q. 19. r = 1 -f sin ^, r = 2 sin 6. 20. r2=ci2sin^, r'' = a^ ^\\^?> Q. Transform the following equations to polar coordinates : 21. ry = 4. 23. x^-\-if—2 ay = 0. 22. x^-\-7/-4.ax-4.ay= 0. 24. (x^ -f y^ = a\x^ - t/^. Transform the following equations to rectangular coordinates : 25. r=asec^. 27. r = a tan ^. 26. r = 2acos^. 28. r = acos2$. * The curve is called a cardioid. t The curve is a parabola with the origin at the focus. 146 TEIGONOMETRIC FUNCTIONS 52. The differentials dr, d$y ds, in polar coordinates. We have seen, in § 39, that the differential of arc in rectangular coordinates is given by the equation ds'=dx'-^df. (1) If we wish to change this to polar coordinates, we have to place x = r cos 6, y = r sin Q ; whence dx = Q.Q'&Odr — r mvQdQ^ dy = sin 6dr -\-r cos Qdd. Substituting in (1), we have ds'=dr'^r^dd\ (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 =/(^), the coordinates of P being (r, ^), where OP = r and MOP = 6. Let 6 be increased by an amount dO, thus determining another point Q on the curve. From (9 as a center and with a radius equal to r, describe an arc of a circle intersecting OQ m R so that OR = OP = r. Then, by § 42, PE = rdd. Now BQ is equal to Ar, and P^ is equal to As. We shall mark them, however, as dr and ds respectively, and the formula (2) is then correctly obtained by treating the triangle PQE 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 PQE. Let us denote by ^|r the angle PQE, which is the angle made by the curve with any radius vector. Then, if we treat the triangle PQE as a straight-line right-angle triangle, we have the formulas : , rdO , dr , , rdO ,„. sm '»ir = -— -, cos'Ur = -— , tan lir = — — (6) ds ds dr 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 F and ^ by a straight line (Fig. 69) and draw a straight line from P per- pendicular to 0^ meeting OQ at S, Then the triangle PQS is a straight- line right-angle triangle, and therefore sin SQP = , "^f — chord PQ SP arc PQ arc PQ chord PQ Fig. 69 Now angle POQ = Ad^SiTc PQ = As, and, from the right triangle OSP, SP = OP sin POQ = r sin A^. Therefore • ci^^ rsinA^ arc PO sinA^ M sltg PQ sin kjCJP = ■ • — ^ . , ^ . ( 4 ) As chord Pg ^0 As chord PQ ^ ^ Now let AQ 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- kQ tq Au proaches 1, by § 42 ; -— approaches — -> by definition ; and arcPO -r — , approaches 1, by § 39. In this figure we denote the angle OPT by i|r and have, from (4), sin^ = 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 a/t 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 TRIGONOMETRIC FUNCTIONS quickly by combining (5) and (2). For, from (2) and (5), we have -IMM ds I \ds whence cos ^ir = -— . (6) ds By dividing (5) by (6) we have tan lir = — — (7) dr In using (7) it may be convenient to write it in the form tan yjr = — ? (8) dr Td since the equation of the curve is usually given in the form dv r =f(6^^ and — is found by direct differentiation. do Ex. Find the angle which the curve r = a sin 4 6 makes with the radius vector 6 = 30°. Here — ^ = 4 a cos 4 ^. Therefore, from (8), tan il/ = _^L^15 = _ tan 4 0. (W 4 a cos 4 ^ 4 Substituting 6 = 30°, we have tan i/^ = | tan 120° = - | V3 = - .4330. Therefore ij/ = 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 -j- 3 cos makes with the radius vector 6 = 90°. 3. Find the angle which the curve ?' = a^sin^- makes with the initial line. B 4. Show that for the curve r = a sin^ - ? xp = j.- o o 5. Show that the angle between the cardioid r = a(l— cos 6) and any radius vector is always half the angle between the radius vector and the initial line. GENERAL EXERCISES 149 6. Show that the angle between the leinniscate r^ = 2 d^ cos 2 6 TT 2 77" and any radius vector is always — plus twice the angle between the radius vector and the initial line. 7. Show that the curves i^ — ar sin 2 and i^ = a^ cos 2 6 inter- sect at right angles. GENERAL EXERCISES Find the graphs of the following equations : , . .r -h 1 ., . / 7r\ 1. t/ = 4:sm— — 5. i/ = Ssm\^x + ^j- 2. y/ = cos (2 a? — 3). 6. / = tan x. 3. 1/ = tan^. 7. y = 2 cos 2(x- - 2). 4. y = ^sin 2a; -f- ^sin 3a:. 8. // = 3 cos 3(x -1- j ). Find -r^ in the following cases : dx 9. v/ = 2 a; + tan 2 0-. 10. y = itan(3a; + 2)-|-itan«(3£c + 2). 11. ij = ^x — ^^^\\i{^ — *6 x). 12. tan (x -\-i/) + tan (x — */) = 0. 13. y/ = 3ctn^'^ + o ctn^^- 21. // = ctn-^ \i-- o o yx 14. y/ = csc^4a: -|- 2ctn 4 a;. 22. y = sec~^7a-. 15. fy = ^tan'^aa;. ir- - 4 16. y/ = sinHa;cos*2a;. 23. .y = cos-^^qp^- 2 „a; a; 17. y = -cos*7 — 2cos 7- 24. y = ctn"^ Va;^ — 2a;. ^34 4 ^ 2 18. 2/ = ^tan^2x — ^tan2a; + a:. 25. y = csc-*^ r- 19. y = sin-^^^-^. 2g^ .y=:3V9^:^^-f 2sin-- 20. y = cos-^ ^'^~ ■ 27. y = tan'^o; Va;^ - 2. i5 28. A particle moves in a straight line so that s = 5 — 4 sin^— • Show that the motion is simple harmonic and find the center about which the particle oscillates and the amplitude of the motion. 150 TRIGONOMETRIC FUNCTIONS 29. A particle moves on the ellipse — -f- ^- = 1 so that its projec- tion upon OX describes simple harmonic motion given hy x = a cos kt. Show that its projection upon OY also describes simple harmonic motion and find the velocity of the particle in its path. 30. A particle moving with simple harmonic motion of period ~ o 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 Jy ^^ + 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 ^ point for which x — — TT sin X 33. Find the radius of curvature of the curve // = at the X point for which ,t = tt. X I 34. Find the radius of curvature of the curve y = a sin~^ - — ^d^ — x^ a ^ at the point for which x =^ -• 35. Find the radius of curvature of the curve x = a cos = —- Plot the graphs of the following curves : 36. 1^ = a"^ sin -• 41. r =1 — 26. 42. 7-2 =4 sec 2^. 37. r2 = a^ sin 4:0. .^ 2 2^ a 43. r^= a^ta,nO. 38. r = a(l — sin 6). $ ^ ^ 44. /• =l + sin-. 39. r =a(l +cos2^). ^ 40. r = ^t(l + 2 sin 6). 45. r = 1 -f- sin -7-- Find the points of intersection of the following pairs of curves : 46. A-2 = 3 cos 2 6, t^ = 2 cos" 6. 47 . r = a cos 6, r^ = a^ sin 2 B. 48. /• = 2 sin e, ?-2 = 4 sin 2 B. 49. r = «(1 + sin 2 B), r^ = 4 a^ sin 2 B. GENERAL EXERCISES 151 Transform the following curves to polar coordinates : 50. 1/ = . 51. v/ + 7/V-^ - ^V == 0. Transform the following curves to .xy-coordinates : 52. /•- = 2 a' sin 2 0. 53. /• = a (1 — cos 6). 54. Find the angle at which the curve r = 3 -f- sin 2 6 meets the circle r = 3. 55. Find the angle of intersection of the two curves r = 2 sin d and t^ = 4: sin 2 0. 56. Find the angle of intersection of the curves r = a cos and 7' = a sin 2 $. 57. If a ball is fired from a gun with the initial velocity v^, it QX describes a path the equation of which is y = xtsma — - — r --) ^ ^ "^ 2?;(fcosvj: 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 bottbm 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 liis 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. How 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 ? 152 TRIGONOMETRIC FUNCTIONS 63. An open gutter is to be constructed of boards in such a way that the bottom and sides, measured on the inside, are to be each 6 in. wide and both sides are to have the same slope. How wide should the gutter be across the top in order that its capacity may be as great as possible ? 64. A steel girder 27 ft. long is to be moved on rollers along a passageway and into a corridor 8 ft. in width at right angles to the passageway. If the horizontal width of the girder is neglected, how wide must the passageway be in order that the girder may go around the corner ? 65. Two particles are moving in the same straight line so that their distances from a fixed point are respectively x = a cos kt and a;' = acoslH 4- — ), k and a being constants. Find the greatest distance between them. 66. Show that for any curve in polar coordinates the maximum and the minimum values of r occur in general when the radius vector is perpendicular to the curve. 67. Two men are at one end of the diameter of a circle of 40 yd. radius. One goes directly toward the center of the circle at the uniform rate of 6 ft. per second, and the other goes around the circumference at the rate of 2 tt ft. per second. How fast are they separating at the end of 10 sec. ? 68. Given that two sides and the included angle of a triangle are 6 ft., 10 ft., and 30° respectively, and are changing at the rates of 4 ft., — 3 ft., and 12° per second respectively, what is the area of the triangle and how fast is it changing ? 69. A revolving light in a lighthouse \ mi. offshore makes one revolution a minute. If the line of the shore is a straight line, how fast is the ray of light moving along the shore when it passes a point 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 at B with a crank AB, h feet long, revolving about A. Find C's velocity in terms of .45's angular velocity. 71. At any time t the coordinates of a point moving in the ic^z-plane are x = 2 — 3 cos t, y = 3 -\- 2 sin t. Find its path and its velocity in its path. At what points will it have a maximum velocity ? 72 . At any time t the coordinates of a moving point are x = 2 sec 3 /, ?y = 4 tan 3 f. Find the equation of its path and its velocity in its path. I GENERAL EXERCISES 153 73. The parametric equations of the path of a moving particle are x = 2cos^, y = 2sin^. 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 ?/ = sina; so that the £c-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 and those at which it is moving most slowly. 75. Find the angle of intersection of the curves 2/ = sin a; and y = cos X. 76. Find the angle of intersection of the curves y = sina; and y = smlx + ^y 77. Find the angle of intersection of the curves ?/ = sina; and y = cos 2 X between the lines x = and a; = 2 tt. 78. Find the points of intersection of the curves y = sin a; and y = sin 3 x between the lines x = and x = 7r. Determine the angles at the points of intersection. 79. Find all the points of intersection of the curves y = cosx and 2/ = sin 2 ic which lie between the lines x = and cc = 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 y = «^ where a is any constant, defines ^ as a function of x called the exponential function. If x = n, an integer, j/ is determined by raising a to the nth power by multiplication. p If x = -i a positive fraction, y is the ^-th root of the pih power of a. If a: is a positive irrational number, the approximate value of 2/ may be obtained by expressing x approximately as a fraction. li x= 0, 1/ = a^ = l. If a; = — w, y = a ~ "* = — . The graph of the function is readily found. Ex. Find the graph oi y = (1.5)\ By giving convenient values to .r we obtain the curve shown in Fig. 70. To determine the shape of the curve at the extreme left, we place x equal to a large negative number, say x=- 100. Then y = (1.5)-ioo = 1 — , J if K J (1.5)100 which is very small. It is obvious that the larger numerically the negative value of x becomes, the smaller y becomes, so that the curve approaches asymptotically the negative portion of the a;-axis. On the other hand, if a: is a large positive number, y is also large. Fig. 70 54. The logarithm. If a number JS/" 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 N to the base a. That is, if . 2^= «^ (1) then L = \og^]Sr. (2) 154 LOGARITHMS 155 Formulas (1) and (2) are simply two different ways of ex- pressing the same fact as to the relation of N and X, 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, log«^+ log„ Jf = log„ JfiV, log„i\^-log„Jf=log,-, n\og,N=log„N\ (3) logj=0, 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 ,1111 '==^"^1^21 + 3! ■^^^•••' where 2 ! =1 x 2, 3 ! =1 x 2 x 3, 4 ! =1 x 2 x 3 x 4, etc. Computing the above series to seven decimal places, we have ^ = 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 A=.l, (1 + A)*= (1.1)^°= 2.59374. When ^=.01, (1-f- A)^=(1.01)^°°= 2.70481. When h = .001, (1 + hy= (1.001)^°°^= 2.71692. When A =.0001, (1+ A)^= (1.0001)^'^°'^'^= 2.71815. 1 Working algebraically, we expand (1-j-hy by the binomial theorem, obtaining ^ lA-lVl-2\ 1 (1-h) (1-A)(1-2A) 12! 3! where E represents the sum of all terms involving A, h% h^, etc. Now it may be shown by advanced methods that as h approaches zero, H also approaches zero ; so that L™(l+A)^=l+l+i, + l + ... = . 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, if N=e\ . (4) then X = lniV; 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 know 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. LOGARITHMS 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 :r = ln213, then, by (4), 213 = e^; whence, by (3), log 21S = x log e, log 213 2.3284 or log2.7183 0.4343 5.36L Certain graphs involving the number y e are important and are shown in the examples. Ex. 1. !/ = Inx. Giving X positive values and finding ij, we obtain Fig. 71. Fig. 71 y The curve (Fig. 72) is symmetrical vrith respect to F and is always above OX. When x = 0, y = 1. As a: increases numerically, y decreases, approaching zero. Hence OX is an asymptote. Fig. 72 Fig. 73 Ex. t. y = _(e«4.e «). 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. 2/ = e-"^sinbx. The values of y may be computed by multiplying the ordinates of the curve y = e-'^^hj the values of sin hx for the corresponding abscissas. Since the value of smhx oscillates between 1 and —1, the values of g-"^ sin 6a: 158 EXPONENTIAL AND LOGARITHMIC FUNCTIONS cannot exceed those of e~* plane between the curves y = e-^^ and y= — e~^. When x 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, & = 2 tt. 1 Ex. b. y = e^. When X approaches zero, being positive, y increases without limit. When X approaches zero, being neg- ative, y approaches zero ; for example, when ^ = TTJ^iya» y = ^^°°^ ^^^ when x = - roVu* y = e -"^^^^ = • The function is therefore ^ glOOO discontinuous for x = 0. The line r/ = 1 is an asymptote (Fig. 75), for as X increases without limit, being posi- tive or negative, — approaches 0, and y approaches 1. Ex. 6. r = e«^. The use of r and 6 indicates that we are using polar coordinates. When ^ = 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 around 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. Hence the graph lies in the portion of the Y Tig. 75 Fig. 76 EXERCISES Plot the graphs of the following equations : 1. y = (1)-. h, y = xe\ 6. y = \{e^-e-% 7. 7/ = log 1x. , 1 8. ]/ = log-- 2. y = (^y. 3. y = e^ + i 4 y e 9. y = log since. 10. y = log tan ic. 11. y = e~^^sin 4,x. 12. y = e"^cos Sx. 13. r=e-2^ EMPIRICAL 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 a;^-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-\-b, (1) where m and h 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 Ime. 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 : X 1 2 4 6 10 y 1.3 2.2 2.9 3.9 6.1 Find the empirical equation connecting them. 160 EXPONENTIAL AND LOGARITHMIC FUNCTIONS We plot the points (x, rj) 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 y = .5 ar + 1, which is the required equation. In 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 cases, which are closely con- 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 Fig. 77 y = ah'', where a and h are constants, or of the form (2) (3) y = ax-\ 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 h. (4) As log a and log h are constants, if we denote log y by y\ (4) assumes the form (1) in x and ?/', and we have only to plot the points (x, ?/') on an xy'-^y^tQm. 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 log y by y' and log x by x\ (5) assumes the form (1) in x' and y', since log « and ti are constants. Accordingly we plot the points (x', y) on an x' y' -sy^tQVLi of axes, determine the corresponding straight line, and then transform back to (3), as shown in Ex. 3. EMPIRICAL EQUATIONS 161 Ex, 2. Corresponding values of two related quantities x and y are given by the following table : X 8 10 12 14 16 18 20 V 3.2 4.6 7.3 9.8 15.2 24.6 36.4 ,_ ., Find an empirical equation of the form y = alF. Taking the logarithm of the equation y = aZ/*", and denoting log y by y\ we have y' = log a ■\- X log h. Determining the logarithm 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 y' = \ogy .5051 .6628 .8633 .9912 1.1818 1.3909 1.5611 We choose a large-scale plotting-paper, assume on the ?/'-axis a scale four times as large as that on the X-axis, plot the points (x, /), and draw the straight line (Fig. 78) 2.0 through the first and the sixth 1.5 1.0 .5 point. Its equation is v' = . 08858 a: -.20354. 2 4 6 8 10 12 14 16 18 20 Fig. 78 Therefore log a = - .20354 = 9.7965 - 10, whence a = .626 ; and log b = .08858, whence h = 1.22. Substituting these values in the assumed equation, we have ^^^ ^ ^^ ^ ' y = .626 (1.22)^ as the required empirical equation. The result may be tested by substitut- ing the given values of x in the equation. The computed values of y will be 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 26.5 33.5 44 62 V .635 .556 .475 1 .397 .321 .243 Find the relation between them in the form pv** 162 EXPONENTIAL AND LOGARITHMIC FUNCTIONS Writing the assumed relation in the form p = cv-^ and taking the logarithms of both sides of the equation, we have \ogp =— n log V + log c, or y =— nx + b, "where y = log/>, x = log v, and h = log c. The corresponding values of x and y are a; = log u - J972 - .2549 - .3233 - .4012 - .4935 - .6144 y = logp 1.2553 1.3222 1.4232 1.5250 1.6435 1.7924 For convenience we assume on the a:-axis a scale twice as large as that the 3/-axis, plot the points (x, ?/), and draw the straight line as shown 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 1/ =- 1.28 x + 1.01. Hence n = 1.28, logc = 1.01, c =:10.2, and the required rela- tion between p and v is ,1-28 = 10.2. on in pv^ -5o-50-k5-l*O-35-.30-25-20':15-10'.O5 Fig. 79 EXERCISES 1. Show that the following points lie approximately on a straight line, and find its equation : X 4 9 13 20 22 25 30 y 2.1 4.6 7 12 12.9 14.5 18.2 2. For a galvanometer the deflection D, measured in millimeters on a proper scale, and the current 7, measured in microamperes, are determined in a series of readings as follows : D 29.1 48.2 72.7 92.0 118.0 140.0 165.0 199.0 I 0.0493 0.0821 0.123 0.154 0.197 0.234 0.274 0.328 Find an empirical law connecting D and /. I EMPIRICAL EQUATIONS 163 3. Corresponding values of two related quantities x and y are given in the following table : X 0.1 0.3 0.5 0.7 0.9 1.1 1.3 1.6 V 0.3316 0.4050 0.4946 0.6041 0.7379 0.9013 1.1008 1.3446 Find an empirical equation connecting x and y in the form y = ah^. 4. In a certain chemical reaction the concentration c of sodium acetate produced at the end of the stated number of minutes t is as follows : t 1 2 3 4 5 c 0.00837 0.00700 0.00586 0.00492 0.00410 Eind an empirical equation connecting c and t in the form c = ab\ 5. The deflection a of a loaded beam with a constant load is found for various lengths I as follows : I 1000 900 800 700 600 a 7.14 5.22 3.64 2.42 1.50 Eind an empirical equation connecting a and I in the form a = nl^. 6. The relation between the pressure p and the volume v of a gas is found experimentally as follows : p 20 23.5 31 42 59 78 V 0.619 0.540 0.442 0.358 0.277 0.219 Eind an empirical equation connecting p and v in the form/^y" = c. 56. Differentiation. The formulas for the differentiation of the exponential and the logarithmic functions are as follovi^s, where, as usual, u represents any function which can be differen- tiated with respect to x, In means the Napierian logarithm, and a is any constant: rf Inor /> rft/ d , loff„ e du dx ^" u dx d ^ _\ du dx u dx (2) 164 EXPONENTIAL AND LOGAEITHMIC FUNCTIONS (4) dx dx — 6« = e« — ^ dx dx The proofs of these formulas are as follows: l.By(8>§36,£log„. = |-log.».|. To find — log„w place y= log^u. Then, if u is given an increment Am, y receives an increment Ai/, where ^y = log«(^ + Aw) - log^w the transformations being made by (3), § 54. Au Then ^=liogil + 42f Aw u ^ \ u XAm Aw Now, as Aw approaches zero the fraction — may be taken as ^ of § 54. ^ / Aw\^ Hence Lim ( 1 H ) = e. Therefore ^ = lw g du u ^" and ^^log^^^ dx u dx 2. If ?/ = In w, 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^ = In a" = m In a. Hence, by formula (2), 1 dy _^ du ^ y dx dx dy „ , du whence -^ = a" In a -— • dx dx 4. li y = ^" the previous formula becomes dy_ dx '1- Ex. 1. y = ln(a:2- 4a: + 5). dy_ 2x- - 4 dx x2-4x + 5 Ex. 2. y = e-\ dy _ dx - 2a:e- ■< Ex. 3. y = : e-«^cos6ar. -^ = COS hx — {e- «^) + e- "^ — (cos 6a:) = — ae- "^ cos 6a: — he- «^sin 6a: ^^ ^^ ^-^ =-c-«^(acos6a: + 6sin6x). EXERCISES Find -r in each of the following cases : dx . 1 — sin 2x 10. 2/ = In:; r— ^^ — ^ 1 + sm 2 ic 11. ?/ = ln(e2^+e-2^). 12. y — e-2*sin 3 a-. 1. 2/ = e ^. 2. 2/ = :K^"+^"')- 3. 2/ = : a^-\ 4. ?/ = : a"""X 5. 2/ = :ln(a:2+4a!- ■1> 6. :lnV2x2+6 ic + 9. 7. 2/ = = ^^% + 3- 13. 2/ = In Vl4-ic^ + a:ctn-^a!. 14. 2/ = e«^(9a;2_ 6a; -f 2). 15. 2/ = e2^(2 sin a; — cos a;). 8. 2/ = In (x + Va; ^+4). 9. 2/ = In (3a3 + VOa^^ + l). 16. 2/ = sin ^^ _^ ^ 17. 2/ = seca:tancc 4- ln(seca; + tanx). , V^Tl-1 18. y = In - . -• 166 EXPONENTIAL AND LOGAKITHMIC 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 x^ it must satisfy the equation |=%. m where A: is a constant called the proportionality factor. We may write equation (1) in the form y dx whence, by a very obvious reversal of formula (2), § 56, we have In y = kx-\-C, where C is the constant of integration (§18). From this, by (1) and (2), § 54, Finally we place e^=A, where A may be any constant, since C is any constant, and have as a final result • y=AeK (2) The constants A and h 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. The interest gained in a time A^ is P A^ where A^ is expressed in years. But the interest is an increment of the principal P, so that we have AP=P-^A/. 100 In ordinary compound interest the interest is computed for a certain interval (usually one-half year), the principal remaining 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 the 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 earned, in no matter how small a time, is immediately added to the principal. The average rate of change of the principal in the period A< is, from § 11, AP Pr M 100 ^ ^ To obtain the rate of change we must let A/ approach zero in equation (1), and have ,_, ^— p -L-. dr~ lOO' From this, as in the text, we have P=Ae^o\ (2) To make the problem concrete, 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 < = 0, P = 100. Substituting these values in (2), we have A = 100, so that (2) becomes P = 100eioo =100e25. Placing now i = 14, we have to compute P = 100 e^'S. 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 log P = log 100 + it log e = 2.4053 ; 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 respect to x is always 0.01 times y, 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, ?/ = 14. Find the law connect- ing y and X. 4. The sum of 1 100 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 5. 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 ^ = 0, and is ^^3^ when t = h^ 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 : 1- y = (i)- 2. 2/= e>-» X ~2 3. y e "COS a:!. 4. ?/ = e^""*". 5. 2/=i(^"+e"")- 6- 2/ = :t e^+^' 7. y = xe^. 9. y = x — ^2„^a; 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 50 100 150 200 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 13 14 15 Load 123.6 121.5 111.8 107.2 101.3 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 1 2 4 6 7 8 s 10 4 0.6400 0.1024 0.0410 0.0164 Find an empirical equation connecting s and t in the form s = abK 13. In the following table c denotes the chemical concentration of a substance at the time t : t 2 4 6 8 10 c 0.0069 0.0048 0.0033 0.0023 0.0016 Find an empirical equation connecting c and t in the form c = ab\ 14. The relation between the length I (in millimeters) and the time t (in seconds) of a swinging pendulum is found as follows : I 63.4 80.5 90.4 101.3 107.3 140.6 t 0.806 0.892 0.960 1.010 1.038 1.198 Find an empirical equation connecting I and t in the form t = kl^. 15. For a dynamometer the relation between the deflection 6, 27r when the unit = -r— > and the current /, measured in amperes, is as follows : e 40 86 120 160 201 240 280 320 362 I 0.147 0.215 0.252 0.293 0.329 0.360 0.390 0.417 0.442 Find an empirical equation connecting / and in the form / = kG^. 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 0.784 0.426 0.092 0.047 0.0096 0.0049 y 1.32 0.676 0.216 0.074 0.0085 0.00315 0.00036 0.00014 Find an empirical equation connecting the two quantities in the form y = kx". k 170 EXPONENTIAL AND LOGARITHMIC FUNCTIONS 17. Assuming Boyle's law, ^z; = c, determine c graphically from the following pairs of observed values : p 39.92 42.17 45.80 48.52 51.89 60.47 65.97 V 40.37 38.32 35.32 33.29 31.22 26.86 24.53 Find -~ in each of the following cases : 3x-2 19. y = In sinar. 20. y = tan-i — 21. 2/ = In (2a^ + V4 a;2 - 1) + 2 x csc-i2 a;. / _2 22. y = x^e * 23. y = ln^^^^-e-H 24. y = ^ tan^ aa; + In cos aa;. 26. y = a; tan-^a; — ^ In (1 + a;^). 26. A substance of amount x is being decomposed at a rate which is proportional to x. If a; = 3.12 when ^ = 0, and x = 1.36 when ^ = 40 min., find the value of x when ^ = 1 hr. 27. A substance is being transformed into another at a rate which is proportional to the amount of the substance still untransformed. If the amount is 50 when ^ = 0, and 15.6 when ^ = 4 hr., find how long it will be before ^^^ of the original substance will remain. 28. According to Newton's law the rate at which the temperature of a body cools in air is proportional to the difference between the temperature of the body and that of the air. If the temperature of the air is kept at 60°, and the body cools from 130° to 120° in 300 sec, when will its temperature be 100° ? 29. Assuming that the rate of change of atmospheric pressure p at a distance h above the surface of the earth is proportional to the pressure, and that the pressure at sea level is 14.7 lb. per square inch and at a distance of 1600 ft. above sea level is 13.8 lb. per square inch, find the law connecting p and // . GENERAL EXERCISES 171 30. Prove that the curve y = e~^^ sin 3 ic is tangent to the curve y = e-2a: at any point common to the two curves. 31. At any time t the coordinates of a point moving in a plane are x = e~^' cos 2 t, y = e~'^^ sin 2 t. 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 the logarithmic spiral r = e"^ cuts all radius vectors at a constant angle. 33. Find the radius of curvature of the curve y = e~^'' sin 2cc at the point for which x — —- 34. Show that the catenary y = -\e'* -\- e "/ and the parabola 1 "^ y = a -\- ^— x"^ have the same slope and the same curvature at their Z a common point. 35. Find the radius of curvature of the curve x — e^ sin t,y = e^ cos t. 36. Find the radius of curvature of the curve x = a cos^ ^, 77" y — a sin^ <^, at the point for which <^ = j-- 37. Find the radius of curvature of the curve y = \nx and its least value. 38. Find the radius of curvature of the curve y — e^ cos a; at the point for which cc = — • CHAPTER YII SERIES 58. Power series. The expression % + ct^x -\r a^r + a^x^ + a^^ + (1) where a^^ a^ a^, • • • 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 l+x + x'+x^-] , and we may write 1 ......... ^2) 1 + x + x^+x^+x^^- Similarly, if we extract the square root of 1 + a^ by the rule taught in elementary algebra, arranging the work as follows: l-\-x 1 1 + £_^_|. 2 8 ^-i X X 4 2 + x x' 8 x' 4 4 x^ x^ 8 "^64' 172 POWER SERIES 173 the operation may be continued indefinitely. We may write -2 I X X ' ViT^ = l + |-|+--- . (3) 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 ic is a positive or negative number numerically less than 1. If we place x — \\xi (2), we have 3— 14-1-|_1_|_ 1 -L 1 J_... which is true in the sense that the limit of the sum of the terms on the right is |. If, however, we place rr = 3 in (2), we have -i-=l-|-3 + 9 + 27+..., 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 a:" after n terms of the quotient have been obtained ; and if \—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. Maclaurin'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 &mx=A + Bx + Cx''-{-Dx''+Ex' + Fx''^- ' ' ', (1) where A^ B, C, etc. are coefficients to be determined. 174 SERIES By differentiating (1) successively, we have cosx = B-\-2Cx + SDx''+^ Ex^-{- 5 Fx'-\- • . ., - sin X = 2 C + 3 . 2 . D^ + 4 . 3 . ^V+ 5 . 4 . Fx^'^ • • ., - cos a; = 3 . 2 . i) + 4 . 3 . 2 . ^o: 4- 5 . 4 . 3i Vl+ x" V3 / -Wl-x" 8. Verify the expansion of tana; (Ex. 3, §59) by dividing the series for sin x by that for cos x. 9. Verify the expansion of sec a; (Ex. 2) by dividing 1 by the series for cos x. 1— X 10. Expand by Maclaurin's series and verify by dividing the numerator by the denominator. 11. Expand e^cos a? into a Maclaurin's series, and verify by multiplying the series for e^ by that for cos x. 12. Expand e^sina; into a Maclaurin's series, and verify by multiplying the series for e^ by that for since. 13. Expand e^ln(l+a;) into a Maclaurin's series, and verify by multiplying the series for e^ by that for In (1 -|- a^). 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(l4-a:;), compute In | 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 thence, by aid of the first result of Ex. 18, find In 5 to four decimal places. 20. Using the series found in Ex. 4, compute In ^ 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 TT pansion of sin~^a; (Ex. 4, § 59) and the relation sin~^- = - • 180 SERIES 22. Compute the value of tt to four decimal places, from the ex- 1 1 TT pansion of tan~^x (Ex. 5, § 59) and the relation tan~^- + 2 tan~^- = — • 23. Compute •\/ll 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, a; =- 1. ' dx in the form of a series 26. Obtain the integral / e'^^dx in the form of a series expansion. ^^ ' in the form of a series expansion. JC^ dx . I ij ^ in the form of a series expansion. CHAPTER Vni 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 /(a;, y). For example, the volume F of a right circular cylinder is a function of its radius r and its altitude A, and in this case Similarly, we may have a function of three or more variables represented by the symbols /(a:, y, z^,f(x, y, z, u), etc. Consider now /(a;, ^), 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 be represented by the symbol /-r^ J • Similarly, the derivative of / with respect to y when x is constant, is represented by the symbol l^\ These derivatives 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- rivatives are represented by the symbols ^ and --, thus: ^/^m _^._fix + ^x, y^ -f(x,y) \dx), I , ; Lim ex \ax/„ Ax-o Ax dy \dylx Ay-^o Ay 181 182 PARTIAL DIFFERENTIATION So, in general, if we have a function of any number of variables f(x^ ?/,..., 2), we may have a partial derivative with respect to each of the variables. These derivatives are expressed by the sym- To compute thfese 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. ct Ex. 1. Consider a perfect gas obeying the law v = — • We may change P the temperature while keeping the pressure unchanged. If At and At; are corresponding increments of t and v, then . _ c(t + At) ct _ cAt and p P P dv _ c dt~p' Or we may change the pressure while keeping the temperature un- changed. If Ap and Ay are corresponding increments of p and v, then ctAp Av- '^ - ct p + Ap P - dv ct and — = . dp p^ ■ Ex,2. /=a:3-3a:2?/ + ?/3. 1 = 3.-6.,, 2f = -3x2 + 32^2. ^2 +^AP Ex. 3. f=Bui{x'^ + y'^), 2 X cos (.2 + 3/2-j, y - o ^ .^o r^2 , „2> dx ^ = 2 3/ cos (.2 + y^). vy dy 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 6y y = r sin 0- Then — = cos 6, — = sin 6, dr dr dx ... . /, dy ^ where r and 6 are the independent variables. (1) PARTIAL DERIVATIVES 183 Moreover, since r = Var^-j. y'^ and 6 = tan~^ -> = cos 6, _ sin 5 r y dr_ S5 _ a: _ cos ^ sin^, (2) where x and y are the independent variables. dx dr . It is to be emphasized that — in (1) is not the reciprocal of — in (2). InfacMn(l),|=g)^anj:(.),|=@. and because the variable held constant is differ- ent in the two cases, there is no reason that one should be the reciprocal of the other. It happens in this case that the two are equal, but this is not a general rule. Graphically (Fig. 80), if OP = r is increased by PQ = Ar, while 6 is constant, then PR = Ax is determined. Then dx _ /dx\ dr \drj e Moreover (Fig. 81), if OM = a: is increased by MN= PQ = Aa:, while y is constant, then RQ = Ar RQ ^ PQ dx J. PR ^ Lim — — = cos a. PQ determined. Then — = ( — ) = Lim dx \dxly cos 0. It happens here that — = — But dd ^'' ^-^ in (1), and — , in (2), are neither equal nor .^ ^ , aa; ^ ^ reciprocal. EXERCISES Find 77- and 75- in each of the following cases : ex cy 1. z== x^—4:X^y + ^xy'^-\-by^. 5. z = \n xy 2. z _ ^y 3. z = ctn 4. z = sm~'^xy. 6. z 7. z sin x — y 2xy x-\-y S. z = ln(a; -f- Vx^ -f f). 9. If ;^ = ln(^2 _ 2xy ^^-\-Sx -Sy), prove -^ -^ — = 0. / ^ dz dz 10. If ^ = Wx^ + i/^e-^, prove x-^ -\- y-^ — z. 184 PARTIAL DIFFERENTIATION 62. Higher partial derivatives. The partial derivatives of f(x^ y) are themselves functions of x and y which may have partial derivatives, called the second partial derivatives oif(x, ?/). T'-^y '"^ £(!)' 4©' £©' 4(1)' ^"* '' "'^ ^' ''°"" 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 dx \dx) z^ -^^ \dy) dy\dx) =~«- •^''" dx\dy/ dy\dx/ 8xdy dy\dy) df ■'"• Similarly, the third partial derivatives of f(Xj y') are four in number; namely. dxKdx") ex''' dy\dx^) dx\dxdyl d3?\dy/ da^dy d m\_ d / dy\ c' /df\ ay ^ dx\dy^/ dy\dxdy/ dy^\dx/ dxdy^ dy\df)~dy'' ^P + Qf So, in general, - — ~ signifies the result of differentiating if f(xj y) p times with respect to x, and q times with respect to y, the order of differentiating being immaterial. In like manner, ^^ signifies the result of differentiating if f(x, y, z) p times with respect to x, q times with respect to y, and r times with respect to 2, in any order. TOTAL DIFFERENTIAL 186 EXERCISES 1. Uz = (x^-{- if) tan-i^, find ^'^ 2. If z = eP sin (a; — y), find ^• 3. If;. = ln(a;2+2/^), find^. df Verify Yy^]~Y\^] ^"^ ^^^ ^^ *^^ following cases 1 X 4. z = xy'^-\-2ye''. 6. ^ = sin 5. . = i^^+i^. 7..= * 8. If ;^ = tan-i-, prove ^ + ^-^ = 0. 9. If ^ = In {x" - ahf), prove «^' ^ - ^2 = 0- ^2 (7.7) d'^V 10. If F = r"' cos n<^y prove wV — ) ^^ -{■ 7n{m -\- 1)^t^ = 0. 63. Total differential of a function of two variables. In § 20 the differential of a function of a single variable, y =fQjc)^ is defined by the equation ay=f(x)dx, (1) where f {x) is the derivative of y. But /(rr)=Lim(^); (2) and hence, according to the definition of a limit (§ 1), g=/(.) + e. (3) where € denotes the difference between the variable -^ and its limit /'(a;) and approaches zero as a limit as Lx—^ 0. Multiplying (3) by Aa:, we have A^ =/ {x) A2: + e b.x. (4) But A3: = dx and A?/ =f{x + A2;) -f(x), so that (4) may be written in the form f{x + Aa:) -/(^) =/ (2-) ^2: + € dx. (5) 186 PARTIAL DIFFERENTIATION In the case of a function of two variables, /(a;, y)^iix alone is changed, we have, by (5), f(x + A2:, y) -f(x, y) = -£dx^- e^dx, (6) the theory being the same as in the case of a function of one variable, since y is held constant. The term -^ dx may be denoted by the symbol d^f. Similarly, if x is held constant and y alone is changed, we have f(x, y + Ay) -fix, y)JI^dy + e^dy, (7) and -^dy may be denoted by the symbol dj^f. Finally, let x and y both change. Then A/=/(:?^+ A2:, 2/ + A^) -f{x, y) =/(ic+ Aa;, y + Ay)-f(xi-Ax, y)+f(x + Ax,y^-f(x,y^. (8) Then, by (6), f(x + Ax, y) -f(x, y') = £dx + e^dx ; (9) and similarly, by (7), f(x + Ax, y + Ay) -f(x + Ax, y) = ^dy + edy, (10) df ^ where ~ is to be computed for the value (x-{-Ax, y'). But if df . y . r^ is a continuous function, as we shall assume it is, its value dy for (x + Ax, y) differs from its value for (x, «/) by an amount which approaches zero as dx approaches zero. Hence we may write, from (8), (9), and (10), Af=%dx + ^^dy-^e^dx + e^dy, (11) ex cy where both -^ ^^^ ^ ^^^ computed for {x, y). We now write df= — dx+ — dy, (12) -^ dx dy ^' ^ ^ so that Af=df+ e^dx + €^dy, (1 3) and df is called the total differential of the function, the expres- sions d^f and dj^ 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- ti<»^*hat df=.dj+dj. (14) Ex. The period of a simple pendulum with small oscillations is >// 4 ir'H whence g = -—- • Let I = 100 cm. with a possible error of ^ mm. in measuring, and T = 2 sec. with a possible error of ^^^ sec. in measuring. Then dl = ± i^ andrfr=±xk- 4^2 8^2; Moreover, dg=—dl- -yg- dT, and we obtain the largest possible error in g by taking dl and dT oi oppo- site signs say dl = 5'^, dT = — ^^q. Then dg = ^ + Tr^ = 1.05 ir^ = 10.36. The ratio of error is 'IR = €-2 — = .0005 + .01 = .0105 = 1.05%. g I T EXERCISES 1. Calculate the numerical difference between A^ and dz when z ~ ^xy — x^ — y'^, X = 2, y = ^, ^.x = dx = .01, and A^/ = dy =.001. 2. An angle is determined from the formula <^ = tan~^^ by X measuring the sides x and 3/ 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, find 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 i?, by Ohm's law 188 PARTIAL DIFFERENTIATION If errors of 1 per cent are made in measuring E and 72, 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 f^ and f^ denote the respective focal lengths of the lenses, then 1=1+1. F A f. If /j and f^ 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 f^ and 0.1 in. in/^ are made. 5. The eccentricity e of an ellipse of axes 2 a and 2^ (a > Z») is given by the formula . e = a 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 errors of .1 ft. in a and .01 ft. in h, 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 7t of a bullet having an initial velocity of v^, fired at an elevation a, is given by the formula v^ sin 2 cr R = • 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 ^/;' the weight in water. \i w = 244,000 gr. and w^ = 220,400 gr., find approximately the largest possible error in B caused by an error of 5 gr. in w and an error of 10 gr. in w'. RATE OF CHANGE 189 64. Rate of change. The partial derivative ^ gives the rate ex • of change of / with respect to x when x alone varies, and the partial derivative ^ gives the rate of change of / with respect to 1/ when ?/ 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 f, thus making / also a function of t Now, by § 11, the rate of change of / with respect to t is the derivative —. civ To obtain this derivative we have simply to divide df, as given by (12), § 63, by dt, obtaining in this way <¥^^dxdf^dy dt dx dt dy dt ' ^ ^ The same result may be obtained by dividing A/, as given by (11), § 63, by A^ and taking the limit as A^ 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 rate 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 V=7rrVi. ^ ^^' dt ~ dr dt'^ dh dt = 2 7rrh— + irr^—- dt dt By hypothesis, — = 2, — = 3, r = 5, 7i = 15. Therefore — -= 375 tt cu. in. ^ *^j dt dt dt per second. The same result may be obtained without partial differentiation by ex- pressing V directly in terms of t. For, by hypothesis, r = 5 + 2 f, ^ =15 + 3 /, if we choose t = when r= d and h — 15. Therefore F=(375 + 375 < + 120 ^2+12 ^3)7r ; whence — = (375 + 240 i + 36 f") it. dt ^ When / = 0, - — = 375 ir 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, ^ _ ^ 2x dx ~ (a;2 + yy ' 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 m in a direction parallel to F is du _ 2 y J^'' (x^+yY Suppose now we wish to find the rate of change of the temperature in a direction which makes an angle a with OX. From Fig. 82, if Pi{x^, y^) is a fixed point, and P(x, y) a moving point y. on the line through P^ making an angle a with OX, and s is the distance P^P, we have whence and P,R-- - X - x^= s COS a, RP = ^y - y^ = s sin a X - = ^1 + s cos a, y^ = yi + s sin a, dx — = cos or, ds dy ~ = sm c ds i? R Fig. 82 dx dy Replacing < by s in formula (1), and substituting the values of — and ds — which we just found, we have ds du du . du . — - = — cos a -] sm a ds dx dy _ _2 X cos a + 2 y sin a ~ (x'-^yy Formula (1) has been written on the hypothesis that x and ?/ are functions of t only. If x and i/ are functions of two vari- ables, t and 8, and (1) is derived on the assumption that t alone varies, we have simply to use the notation of § 61 to write at once m=^f(^\.^f(il\, (2) \dtj. dx\dt)7 dy\dt), which may also be written as ¥^^fdx_^dldy^ dt dx dt dy dt (3) GENERAL EXERCISES 191 EXERCISES 1. li z = e Xj X = sin t, y = cos t, find the rate of change of z with respect to t. \-\- X 2. li z = tan- ^3 > x = sin ^, ?/ = cos t, find the rate of change ~ y TT of z with respect to t when ^ = «■ * 3. If F= (e**^ — e'"^) cos ay, prove that F and its derivatives in any direction are all equal to zero at the point ( 0, ^ ) • 4. If V= — ? find the rate of change of V at the point (1, 1) in a direction making an angle of 45° with OX. 5. If the electric potential V at any point of a plane is given by the formula F= In Vic^+ y^, 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, 6. If the electric potential V at any point of the plane is given V(cc — af-^-iP- by the formula F= In — , ' — ~ > find the rate of change of ■>J{x^af^f ^ 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 . xy —1 dz dz 1. it z = sm -^ ) prove x-:: ?/ ^ = U. xy -\-l ^ dx ^ cy 1 ' dz dz - dz dz 3. li z = y^-{- ye"", prove x^ -^ + y -^ = ^ y^- d^z d^z 4. If ^ = e-"^ cos a(k — x), prove that — + — ^ = 0. 5. liz = e-(^^+«='^-^)' sin kx, prove that ^ = a^ _J _ ^2^^ 6. If s = e-^^'sin {my + a? -Wa^m^— k^), prove that ^ + 2 A; ^ = ^2 d?f d^V 1 ^F 1 d'^V 7. If F= e'^'^cos (a In r), prove that -^ + - ^ + -^ ^ = 0. 192 PARTIAL DIFFERENTIATION 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 r— 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 ^ is found to be 6,000,000 dynes and m to be .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 J. C is 10 in., and the angle A is 60°. Calculate the change in the area caused by increasing ^ 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 ^C = 90 ft., BC = 110 ft., and BCA = 60°. Find approximately the greatest possible error in the computed length oi 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 5C = 100 ft., the angle CBA = a= 45°, and the angle BCA = p = 60°. Find the greatest possible error in the computed length oi AB caused by errors of 1' in measuring both a and p. 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 ^ = \n{x -\-^x^-\- y^) in the direction of the line drawn from the origin of coordinates to any point P (a;, 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 ■^ ^ 1 4- a; 16. Given z = tan"^— ^ h taii"^ Find the general expres- sion for the derivative of z 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 z = kxi/ 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 = e-v sin x + - e-sv sin 3 a; at the point ( ^ » ^ )• Find also the 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 i; that is, if f(x)dx^dF(x), then Cf(x) dx = F(x) + (7, 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 F(x} -f C 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 be 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 C may be omitted, as it will simply cancel out if it has previously been written in (§23). The two formulas •» ^ / cdu = c I du (1) and j{du -\-dv-{-dw-{ ) = jdu + fdv -f fdw -|- • • • (2) 194 INTEGRAL OF u- 195 are df fundamental importance. Stated in words they are as follows : (!') A constant factor may he changed from one side of the sign of integration to the other. (2) The integral of the sum of a finite number of functions is the sum of the integrals of the separate functions. To prove (1), we note that since cdu = d(^cu), it follows that I cdu = I d (cu) = cu = c I du. In like manner, to prove (2), since du-{- dv -{-dw -\- • * ' = d(u-\-v -\-w -{- ' * '^y we have I (du -{- dv -\- dw -{- - ' ') = id(u-\-v + w-\-»''') = Cdu 4- Idv 4- \dw -\ . The application of these formulas is illustrated in the follow- ing articles. 66. Integral of i/". Since for all values of m except m = d(tr} = mu"'~^du, or d(—j=u"'-^du, u"'~^du = — . m Placing w = n 4-1, we have u-du = ^^ (1) for all values of n except n = — 1. In the case n = — l, the expression under the sign of inte- gration in (1) becomes — ? which is recognized as c?(lnM). u Therefore /^ = ln«. (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 j (ax^ + bx -\ h — ) dx. Applying (2), § 65, and then (1), § 65, we have I I ax^ + bx •] h —\dx = j ax^ dx + j bxdx -\- j -dx ■{■ j -^dx = a I x'^dx ■}■ b I xdx -\- c j [■ k j x~^ dx. The first, the second, and the fourth of these integrals may be evaluated by formula (1) and the third by formula (2), where u = x, the results being 1 1 it respectively - ax% - hx% » and c In x. o Jd x Therefore C lax'' ■\- hx ^- ■{■ !^dx = \ax^ ■\-\hx'^ ^- c\nx -- -V C. •/ \ a: xV 3 2 x Ex. 2. Find the value ofj(x^+ 2)xdx. If the factors of the integrand are multiplied together, we have f(x^+2)xdx=j(x^+2x)dxy which may be evaluated by the same method as that used in Ex. 1, the result being ^ x* -\- x^ •}■ C. Or we may let x^-^2 = u, whence 2xdx = du, so that xdx—\ du. Hence f(x^+ 2)xdx=f^udu = ifudu 2 2 = i(a:2+2)2+C. 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 f(ax^ + 2 hxy(ax + h)dx. Let ax^ + 2hx = u. Then (2 ax + 2 b) dx = du, so that (ax + b) dx = | du. Hence f (ax^ + 2bxy (ax -{■ b)dx = f ^u^du = l(ax^ + 2bxy+C. INTEGRAL OF ir 197 Ex.4. Find the value of rli^f±^. As in Ex. 3, let ax^ -\-2bx = u. Then (2ax + 2b)dx = du, so that (ax + b)dx = Idu. ' Hence r i(ax + b)dx ^r2_du^^r±i J ax^ + 2bx J u J u = 2 In M + C = 2 In (ax2 + 2 6a:) + C = ln(aar2 + 2 6x)2+C. Ex. 5. Find the value of f (e«^ + b^e^dx. Lete«* + 6 = M. Then c«^a 0). Making this substitution and evident reductions, we have as our required formulas du . ,u f- V«2- u^ a (1) du 1 , _^u ,^. - tan ^ - » (2) and / du 1 _^u^ „ u^ + a^ a a du 1 _j w u Referring to 1, § 47, we see that sin~^- must be taken in the first or the fourth quadrant; if, however, it is necessary to u have sin~^ - in the second or the third quadrant, the minus sign u must be prefixed. In like manner, in (3), sec~^ - must be taken in the first or the third quadrant or else its sign must be changed. /dx — ■ . Letting 2 a; = w, we have du = 2dx', V9-4a:2 /dx _ r ^du whence dx = ^ du, and ^ ^ '* ^ V9 - 4 a;2 -^ V9 - m^ _ 1 /» du ~ 2 J V9 - w2 1 . ,2a: = - sin 2 3 ^ + C. 200 INTEGRATION dx Ex. 2. Find the value of f — , If we let Vs x - u, then ^ ^ xVSx^-i — zzdu, and V3 dx r du _ ^ - xvo = a, X X2+1 tan~^6 must be taken equal to tan~^a, since the value of the integral is then zero. As b varies from equality with a to its final value, tan-^6 will vary from tan-^a to the nearest value of tan- ^6. The simplest way to choose the proper values of tan-^Z; and tan-^a is to take them both between — - and - • Then we have rVs dx _ ^ _ / ir\ _ 7j J-i x2 + l~3 V 4/ li 12 ' The same ambiguity occurs in the determination of a definite integral by (1), but the simplest way to obviate it is to take both values of sin-^ — between and - . The proof is left to the student. 2 2 ^ 202 INTEGEATION EXERCISES Find the values of the following integrals : dx r dx 1. / dx r dx r dx /dx r / dx ' r d^ r d^ r 3^ + ii dx r 2x-\-^ V2 + 4a;-3a;2* ^^* J V4 - x' /a dx r^ dx ^ r dx r^ dx . 4 3 6^0; 10, 9 dx r^ dx J V6a;-4x2 r ^a;V4ic2_9 68. Closely resembling formulas (1) and (2) of the last section in the form of the integrand are the following formulas : r-^ = ln(«+Vi?T7^), (1) r-^ = ln(«+Vi?3^), (2) / du _ 1 , u— a g and These formulas can be easily verified by differentiation, and this verification should be made by the student. ALGEBKAIC INTEGRANDS 203 /dx Letting '\/2 x = u, we have du = V2 dx ; whence dx = — =• and, from the triangle, Va- — x'^ — a cos <^. Therefore fVa^— x^dx = a^ Ccos^cf)dcf> = ^a^ C (1 + cos 2 (fi)d = la^(<}> + I sin 2 <^) + C, x But <^ = sin- ^ - , a and sin 2 ^ = 2 sin <^ cos <^ 2 x V a2 _ a;2 a' X V flt^ — 37^ for, from the triangle, sin cf> = - and cos ch = a a Finally, by substitution, we have r Va2 _ xhlx = 2 (-^ ^^^ " ^'^ + a^sin-i-j + C. IV. Integrand involving y/x'^ + a^ If a right triangle is constructed with the two legs equal to a: and a respectively (Fig. 84), the substitu- v/o^>" tion x = a tan <^ is suggested. Ex. 5. Find the value of / dx (x^ + a2)i Let X = a tan <^. Then dx = a sec^ff>d and, from the triangle, Vx^ + a?- — a sec . Therefore / =— f — ^ = — f cos (f>dd> = — sin th + C. (x2 + a^)l « *^ sec — — — = ; so that, by substitution, Vx^ + a2 SUBSTITUTIONS 211 • ^^ V. Integrand involving y/x^— (f. If a right triangle is constructed with the hypotenuse equal to X and with one leg equal to a (Fig. 85), J^ the substitution ic = « sec <^ is suggested. Ex. 6. Find the value oifx^Vx^ - d^dx. Let X = a sec <^. Then dx = a sec <^ tan d and, from the triangle, Vx^ — a^ = a tan . Therefore | x^ Vx'-^ — d^dx = a^ j tam^ffy sec* (f>d = a^ j (tan^*^ + tsin* Va:2 — a^ But, from the triangle, tan varies from to - as a; varies from to a. Accordingly, " ^ = g(*+|sin2.^)] 2 - _ _ Tra^ / The second method is evidently the better method, as it obviates the necessity of replacing z in the indefinite integral by its value in terms of X before the limits of integration can be substituted. EXERCISES Find the values of the following integrals : r x^dx r x'dx r^ dx (4 dx \\ r^ dx J Wx-l J (3-x2)^ J-i 3. fx(2x-S)^dx. 8. f ^=. 13. r J J xWl-\-4.x' Ji i^+iy 5. / — 7- 10. / x-\/2x-3dx. 15. r xVT^^hlx. 73. Integration by parts. Another method of importance in the reduction of a given integral to a known type is that of integration hy parts, the formula for which is derived from the formula for the differential of a product, d(uv) = udv + vdu. From this formula we derive uv= I udv + I vduy which is usually written in the form j udv = uv — I V du. INTEGRATION 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 a: = M and e'dx = dv^ we have du = dx and r = c*. Substituting in our formula, we have ixe'^dx — xeF— \ e^dx = xe^- e^ + C = (X -1)€^+ p. 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 isin.-'^xdx. dx Here we may let sin-^ a: = u and dx = dv, whence du = and v = x. Substituting in our formula, we have V 1 — x I sin~^xdx = X sin-^x — i - xdx VI -x2 = X sin-ix + Vl - 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 fx^e^dx. Here we let x^ = u and e'^dx = dv. Then du = 2 xdx and v = e^. Therefore Cx^e^dx — a;V _ 2 Cxef^dx. The integral I xe^dx may be evaluated by integration by parts (see Ex. 1), so that finally fx^e^'dx = a:V - 2 (a: - 1) e^ +C = ^^(x^ - 2 ar + 2) + C. Ex. 4. Find the value of j e*^ sin hxdx. Letting sin&x = u and e"^(/x = dv, we have /e"^ sin hxdx = - e^ sin bx I e°^ cos bxdx. a a J 214 INTEGEATION In the integral fe"^ cos hxdx we let cos bx = u and e"^dx — dv, and have /e«-» cos hxdx = - e^ cos bx + - ( e"^ sin bx dx. a a J Substituting this value above, we have /e^ sin bxdx — - e'^ sin 6x ( - e'^ cos 6x + - | e«^ sin &a:! 117. r 109. / V^TT?7^^^ Ji JV2V(X2-1) '^ 118. r^_-^ 119. I xsin~^cc(ix. r^ • 1 /'^ __xdx__ ^^ 2 -) = — ab sin^ . As X varies from to a, varies from — to ; we have hence A =4: C ydx =— 4 r ab sin^d. REVIEW PROBLEMS 227 It is evident from formula (1), § 23, that the sign of a definite integral is changed by interchanging the limits. Hence Jo Ex. 4. Find the volume of the ring solid generated by revolving a circle o^ radius a about an axis in its plane b units from its center (b > a). Take the axis of revolution Y as OY (Fig. 89) and the line through the center as OX. Then the equation of the circle is (x - by +f= a2. A straight line parallel to OX meets the circle in two points : P^,wh.ere x=x^ = b—Va'^—y'^, and Pg, where x = X2= b + Vd^ — y'^. A section of the required solid made by a plane through P^Pg perpendicular to OF is bounded by two concentric circles with radii MP^ = x^ and MP^ = x^ respectively. Hence, if d V denotes the element of volume, ,Tr , o ox t dV= (tt^I — TTx}) dy 4 7r6Va2- y^dy. The summation extends from the point L, where y = — a, to the point K, where y =^ a. On account of symmetry, however, we may take twice the integral from y = to y = a. Hence T- V=2 r"4 7r6 Va2. Jo yhly = 2 7r'^a%. 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 RQC (Fig. 90) be the parabolic segment, and let CB be drawn through the vertex C of the segment perpendicular to PQ in the surface of the liquid. According to the data, RQ — 2byCB = a. Draw LN parallel to TS, and on LN as a base construct an element of area, dA. Let CM=x. 228 APPLICATIONS Then dA =(LN)dx. But, from § 30, ln' cm whence a and therefore dA= —x^dx. The depth of LN below the surface of the liquid is CB — CM = a — a; ; hence, if w is the weight of a unit volume of the liquid, dP = — x* (a — a:) wdx, a* and P = \ — r ^ {a — x) dx EXERCISES 1. Find the area of an arch of the curve y = sin x. a( - --\ 2. Find the area bounded by the catenary y = -\e'' -\- e "/, the axis of X, and the lines x = ^h. 8 a' 3. Find the area included between the curve y = -7, — 5 and X -\- 4 a its asymptote. 4. Find the area of one of the closed figures bounded by the curves 1/^ = 16 cc and ?/^ = x^. 5. Find the area bounded by the curve 7f=2{x — V) and the line 2x-3y = 0. 6. Find the area between the axis of x and one arch of the cycloid X = a(<^ — sin <^), y = a(l — cos ). 7. Find the volume of the solid generated by revolving about OY the plane surface bounded by OF and the curve x^ -{■ y^ = a^. 8. Any section of a certain solid made by a plane perpendicular to OX is an isosceles triangle with the ends of its base resting on the x^ -»2 ellipse -^ 4- T^ = 1 and its altitude equal to the distance of the plane from the center of the ellipse. Find the total volume of the solid. 9. Find the volume of the solid formed by revolving about the line 2 ?/ + a = the area bounded by one arch of the curve y = sin x 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 x^ -\- 1/^= c?. 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^ = x\ 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 in 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. 16. 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 in 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. 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 0M=1 and ON = b(b>l) and draw the ordinates MP and NQ. Then r^dx r n^ 1 area MNQP = f 'if=-i=l-i. Ji x^ L a:Ji 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. k 230 APPLICATIONS We may, accordingly, define the area bounded by the curve, the axis of x, and the ordinate a: = 1 as the limit of the area MNQP as b increases indefi- nitely, and denote it by the symbol >^dx Ji x^ t,^^Ji x^ Ex. 2. Find the area bounded by the curve y = (Fig. 92), the / 2 i axis of X, and the ordinates x = and x = a. v a — a;^ Since the line x = a is an asymptote of the curve, y — >- oo when x — >► a ; furthermore, the area is not, strictly speaking, bounded. We may, however, find the area bounded on the right by the ordinate X = a — h, where h is a small quantity, with the result -h r<^-^ dx r • 1 xl"-^ . , a — I — == = sm-^ - = sin~^ Jo Va^ — x^ L aJo u If ^ . , a — li . n TT 0, sm-i -^ sm-i 1 = - a 2 Hence we may regard — as the value of the area required, and express it by the integral dx ^ . r»-f^ dx /*« a. Jo ^/'^. Fig. 92 dx _^ . r^ dx Vx 6 -» ^•^i Vx Va2 - x^ h-.o'^o Va^ x^ 1 Vx Proceeding as in Ex. 1, we place r But f'^ = \2VxT=2Vb-2, •^1 -y/x L Ji an expression which increases indefinitely as b — >- go ; hence the given integral has no finite value. We accordingly conclude that in each case we must determine a limit, and that the problem has no solution if we cannot find a limit. 79. Area in polar coordinates. Let (Fig. 93) be the pole and OM the initial line of a system of polar coordinates (r, 6), OP^ and OiJ two fixed radius vectors for which = 6^ and = 6^ respectively, and I^I^ any curve for which the equation is r=f(6). Required the area I^OI^. AREA IN POLAR COORDINATES 231 To construct the differential of area, dA, we divide the angle J^OI^ into parts, d6. Let OF and 0^ be any two consecutive radius vectors ; then the angle P0Q = d6. With as a center and OP as a radius, we draw the arc of a circle, intersecting OQ at E. The area of the sector FOIt=^(OFyd6 = ^r'd0. 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 dA = l7^de and A = '^r'de. This result is unchanged if ij coincides with 0, but in that case 01^ must be tangent to the curve coincide with 0. Fig. 93 So also ij may 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 ^ = and 6 = —, the required area is given by the equation Jo = ^p(i-cosQe)de 12 Ex. 2. Find the area bounded by the lines B= — - and 6= -t the curve ^ 4 4 r = 2 a cos $, and the loop of the curve r = a cos 2 which is bisected by the initial line. Since the loop of the curve r =^ a cos 2 ^ is tangent to the line OL (Fig. 94), for which $ = — j, and the line ON, for which ^ =-T, it is evi- 4' 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 OP^Pr^ be any radius vector cutting the loop r = a cos 2 ^ at Pj and the curve r = 2 a cos ^ at P^. Let OP^ - r^ and OP^ = r^. Draw the radius 232 APPLICATIONS vector OQ^Q^, making an angle dO with OP^P^. With OP^ and OP^ as radii and as a center, construct arcs of circles intersecting OQ^Q^ at R^ and /?2 respectively. Then the area of the sector P^OR^ is \r^dd and the area of the sector P^OR^ is I r^dd. We may now take the area P^P^^R^R^ as dA^ and have dA = l{rl-rl)d6. Then A=pl(^rl-rl)d6', ~ i or, since the required area is symmetrical with respect to the line OM,we may place A=:2nh(r.!-r^)de Jo. =!:<-'■' \')M. From the curve r = 2 a cos 6, we have 4 a^ cos^ 6, and from the curve r = a cos 2 $, we have r^ = a2cos2 2 6; so that finally ^= r4^4a2cos2^- a^cos^2e)d0 '■•[ 26+ sin 26-^- 6 sin 4^-1* 1 (3 TT + 8). EXERCISES 1. Find the total area of the lemniscate r^=2a'^ cos 2 $. 2. Find the area of one loop of the curve r = a sin ti ^. 3. Find the total area of the cardioid r = a(l -f- cos ^). 4. Find the total area bounded by the curve r = 5 + 3 cos ^. 5. Find the area of the loop of the curve r'^ = a^ cos 2 ^ cos ^ which is bisected by the initial line. 6. Find the area bounded by the curves r = a cos 3 ^ and r — a. 7. Find the total area bounded by the curve r = 3 + 2 cos 4 ^. Q 8. Find the area bounded by the curve rcos^-=l and the 77- ^ lines ^ = and 6= — - 9. Find the area bounded by the curves r = 6 -f 4 cos ^ and r = 4 cos Q. 10. Find the area bounded by the curves r = a cos ^ and r*= a?- cos 2 B. MEAN VALUE 233 80. Mean value of a function. Letf(x) be any function of x and let ?/ =f(x) be represented by the curve AB (Fig. 95), where OM=a and ON=h. Take the points Jf^, Jf^' * * '» ^«-i so as to divide distance MN into 71 equal parts, each equal to dx^ and at the points Jf, ilf^, ilS/^, • • •, iJ/l , erect the ordinates ^0' y^ y^ yn -11 Then the average, or mean, value of these n ordinates is This fraction is equal to +yn-ddx Fig. 95 {yo+yi+yi+" ndx y^dx-\-y^dx-^y^dx-\- h — a ■^yn-idx If n is indefinitely increased, this expression approaches as a limit the value 1 r^ I f(x)dx. This is evidently the mean value of an " infinite number " of values of the function f{x) taken at equal distances between the values x = a and x = h. It is called the mean value 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 f(x)dx. 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 during the time t^ if the velocity is averaged with respect to the time. Here we imagine the time from to i?^ 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 -±-j\,d< = iff,. Since the velocity is gt^ when t = t^, 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^ if the velocity is averaged with respect to the distance. Here we imagine the distance from to s^ divided into equal intervals (Is and the velocities at the beginning of each interval averaged. Pro- ceeding as in the text, we find, since v = V2 gs, that the mean velocity is 1 /"i. ?i - Jo '2 OS (Is = |V2 gsas= %^'^(JSy Since the velocity is V2 gs^, when s = s^, 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 circumferen-ce 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 ordinates of the curve y = sin x TT between x = and x = —, assuming that the points at which the ordinates are drawn are at equal distances on the axis of x. 4. The range of a projectile fired with an initial velocity v^ and an elevation a is — ^ sin 2 a. Find the mean range as a varies from TT . ^. to — > averaging with respect to a. 5. 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 min. 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 jpv = 20. Find the average pressure as the gas expands from 2\ cu. in. to 5 cu. in. 9. Show that if 3/ is a linear function of x, 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, ij, -^, • • •, -?_i, 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 JiJ, ^j^, ^7J, . . ., J^_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 in Cartesian coordinates, and Jl Fig. (1) in polar coordinates. Hence we have and ds = Vdr^-hr^de'' s= CVdx^-^df 8 = Cy/d7^-itr'dd\ (2) (3) (4) To evaluate either (3) or (4) we must express one of the variables involved in terms of the other, or both in terms of a third. The limits of integration may then be determined. 236 APPLICATIONS Ex. 1. Find the length of the parabola -(p- = kx from the vertex to the point {a, b). From the equation of the parabola we find 2ydij = kdx. Hence formula (3) becomes either =XVn^-=X\'H^-' Either integral leads to the result k, 2 ?> + V4 ?>2 + y^2 s = -f-V462_^P + jln 2k 4 k Ex. 2. Find the length of one arch of the cycloid X = a(<}> — sin ), y = a(l— cos ). We have dx = a(l — cos ) d ; whence, from (1), ds — a V 2 — 2 cos = 2 a sin ~ 6/^. J»27r J, sin ^d = 8 a. EXERCISES 1. Find the length of the curve 3y^=(x — iy from its point of intersection with OX to the point (4, 3). 2. Find the length of the catenary y = ^Ve" + e °/ from x = to X = h. 3. Find the total length of the curve x^-{-y^= a*- 4. Find the total length of the curve x = a cos^<^, y = cc sin^<^. 5. Find the length of the curve X = a cos -{- a sin <^, y = a sin — a and that a force of 1 lb. stretches a certain wire 36 in. in length to a length .03 in. greater, find the work done in stretching that wire from 36 in. to 40 in. 3. A block slides along a straight line from O against a resistance equal to —^ :;' where k and a are constants and x is the distance ^ x^-{- of of the block from at any time. Find the work done in moving the block from a distance a to a distance a Vs 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 801b. of sand is lifted through a vertical distance of 8 ft. If the sand leaks out at such a rate that while the bag is being lifted, the number of pounds of sand lost is equal to a constant times the square of the number of feet through which the bag has been lifted, and a total of 20 lb. of sand is lost during the lifting, find the number of foot-pounds of work done in lifting the bag. 6. A body moves in a straight line according to the formula x — cf^ where x is the distance traversed in a time t. If the resistance of the air is proportional to the square of the velocity, find the work done against the resistance of the air as the body moves from ic = to cc = a. 7. Assuming that above the surface of the earth the force of the earth's attraction varies inversely as the square of the distance from the earth's center, find the work done in moving a weight of ic pounds from the surface of the earth to a distance a miles above the surface. 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 x from the center of the circle in a straight line perpendicular to the plane of the circle is known to be ' Find the work done in bringing a unit magnetic pole from infinity to the center of the circle along the line just mentioned. GENERAL EXEKCISES 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 7.71b. 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 j^ and v is (1) pv = k or (2) pv'-^ = k. GENERAL EXERCISES 1. Find the area of the sector of the ellipse 4 cc^ -|- 9 y^ = 36 cut out of the first quadrant by the axis of x and the line 2y = x. 2. Find the area of each of the two parts into which the area of the circle x^ -}- y^ = 36 is divided by the curve y^ = x^. 3. Find the area bounded by the hyperbola xy = 12 and the straight line x + y — S = 0. 4. Find the area bounded by the parabola x?—^ay and the 0^2 + 4 a^ 5. Find the area of the loop of the curve at/^ = (a; — a) (x — 2 of. 6. Find the area of the two parts into which the loop of the curve if- = x^ {^ — X) is divided by the line x — y = ^. 7. Find the area bounded by the curve xh/ -f- a%'^ — ahf and its asymptotes. 8. Find the area bounded by the curve y'^ix^ + a?) = a^x? and its asymptotes. 9. Find the area bounded by the curve x = aco^6,y = h sin^^. 10. Find the area inclosed by the curve x = a cos^^, y = a sin^^. 1 1 . 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 A from the common vertex of the parabolas. 12. Find the volume of the solid formed by revolving about the line x = ^L the figure bounded by the parabola y'^ = ^x and the line x = l. 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 14. On the double ordinate of the curve x'^ 4- 1/3 = 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' the line OY the figure bounded by the curve i/ = .^ ^ and the line y = (i' 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^=zSXf and the lines y = ±S. 17. Find the volume formed by revolving about the line x = 2 the plane figure bounded by the curve y^= 4:(2 — x) and the axis of y. 18. The sections of a solid made by planes perpendicular to OF are circles with one diameter extending from the curve y^ = 4:x to the curve y^= A — 4.x. Find the volume of the solid between the points of intersection of the curves. 19. The area bounded by the circle x^ -{- y^— 2ax = 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 the curve ?/ = 7, about OX. ^ 2a — X 20. Find the volume of the two solids formed by revolving about Y the areas bounded by the curves x^ -\- y^ = 5 and y^=z 4:X. 21. Find the volume of the solid formed by revolving about OX the area bounded by OX, the lines a; = and x = a, and the curve X y = X -\- ae^. 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 in 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 x'^ and sin x, x being the distance of the plane from 0. Find the volume of this solid included between the planes for which cc = and x = ir. GENERAL EXERCISES 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 lb. 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 hori- 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 r^= a^ sin nO. 30. Find the area swept over by a radius vector of the curve TT r — a tan ^ as ^ changes from to — • 4 31. Find the area inclosed by the curve r = ;: and the . ^ 1 — cos ^ 4 curve r = ;: ♦ 1 + cos ^ 32. Find the area bounded by the circles r = a cos ^ and r = a sin B. 33. Find the area cut off from one loop of the curve r^ = 2 a^ sin 2 Q by the circle r — a. 34. Find the area of the segment of the cardioid r = a (1 + cos ^) cut off by a straight line perpendicular to the initial line at a dis- tance I a from the origin 0. 35. Find the area cut off from a loop of the curve r = a sin 3 ^ by the circle r = — - — • 242 APPLICATIONS 36. Find the area cut off from the lemniscate t^ = 2o? cos 2 ^ by the straight line r cos B = ' 37. Find each of the three areas bounded by the curves r = a and r = a (1 + sin &). 38. Find the mean height of the curve y = ^^ ^ ^ — ^ between the lines X = — 2a and x = 2a. 39. A particle describes a simple harmonic motion defined by the during a complete vibration is half the maximum kinetic energy if 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 \2jw pv^-^ = 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. 42. Find the length of the curve y — a In-^ ^ from the origin a ^ — "^ to the point for which x — -- 43. Find the length of the curve y = In^ — r between the points for which x =1 and x = 2 respectively. 44. Find the total length of the curve x — a cos^<^, y = h sin^<^. Q 45. Find the total length of the curve r = a sin'-* o 46. Find the length of the spiral r — ad from the pole to the end of the first revolution. k 47. If a center of force attracts with a magnitude equal to — » 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 lex 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 center ? GENERAL EXERCISES 243 49. A central force attracts a body at a distance x from the center k by an amount — • 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 lb. 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 I j f(x,y)dxdy, (1) in which a and h 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 f(x, y) dxdy, 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, F(x)dx. The second integral to be evaluated is, then, X b F(x)dx, which is of the familiar type. Similarly, the symbol f(x,y)dydx, (2) Xi where a and h are constants and x^ and x^ are either constants or functions of y, indicates first the integration X f(x, y) dydx. in which y and dy are handled as constants, and afterwards integration with respect to y between the limits a and 5. 244 DOUBLE INTEGRALS ' 245 3 /»2 Ex. 1. Evaluate j C xydxdy. The first integral is I xy dx dy = \\ xy'^ dx\ =2xdx. The second integration is f%xdx= [x2]' = 9. Ex.2. Evaluate f C "" {x^ ^^ y'^)dxdy. Jo J\—x The first integration is f~ (a:2 + y'^)dxdy = [(x'^y + i/)^ix]J-f = (:r - 2 x^ + ^x^-^x^)dx. The second integration is f\x-2x^-^ ^x^- \x^)dx= ^V Jo 2a i^ Ex. 3. Evaluate f " [^''y^dydx. The first integration is fj^y^dydx=: \y'xdyY~^=^dy. The second integration is Jo 4a ^ 120 aJo 5 A definite integral in one variable has been shown to 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 y, followed by the determination of the limit of a sum with respect to X. The application of the double integral comes from its 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 : 1. r r^^dydx. 3. j yx'dydx. 2. I {''xydxdy. 4, j j cos -dxdy. 246 EEPEATED INTEGRATION 8. n^' dydx J [ \x^ ■\- f)dxdy. Jq n Jo Jo .a(l + co8 0) 'dOdr. a cos '2 6 rdOdr 9 10 11 12 .^ ^iaBind r^dOdr. Jo Jo ■n. 3 rl r« dOdr 4 ir Jo Jo r sin Odd dr. COB 2d ,!r r,V2 8in2( r cos Odd dr. 84. Area as a double integral. Let it be required to find an area (such as is shown in Fig. 98) bounded by two curves, with the equations ^i=/i(^) and y^-=f^ix) intersecting in points for Fig. 98 which x—a and x—h respectively. Let the plane be divided into rectangles by straight lines parallel to OX and Y respectively. Then the area of one such rectangle is dA= dxdy, (1) where dx is the distance between two consecutive lines parallel to OY^ and where dy is the distance between two consecutive lines parallel to OX. The sum of the rectangles which are either AREA 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 PQRS^ 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^=zf^(x) to y^=f^{^- This is indicated by the symbol rdxdy = iy-y^dx=^U^i^~)-fSx)-\Ax. (2) This is the area of the strip TXJVW. We are now to take the limit of the sum of all such strips as dx approaches zero and X varies from a to h. We have then ^= r(%-y,)''^= r'[/.(^)-/.(^)] dx, (5) 248 REPEATED INTEGRATION Consider a similar problem in polar coordinates. Let an area, as in Fig. 99, be bounded by two curves r^=f^{6) and r^=f^(0^, and let the values of 6 corresponding to the points B and C be 6^ and 6^ 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 dO and the distance between two consecutive circles be d7\ We want first the area of one of the quadrilaterals such as PQRS. Here OF = r, PQ = dr, and the angle FOS=de. By geom- etry the area of the sector FOS= i r^ dd and the area of the sector QOR = l(r-^drydd', therefore FQRS = J (^ + t^r) Vl9 - \ r^dO = rdrdd + i (drydd. Now as dr and dO 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 = rdedr, (6) The required area is the limit of the sum of these differ- entials. To find it we first take the limit of the sum of the quadrilaterals, such as FQBS, which lie in the same sector UOV. That is, we integrate rdddr, holding 6 and dO constant and al- lowing r to vary from r^ to r^. We have £ rdedr = \(rl-rl)de, which is the area of the strip TUVW, (7) 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 A=f\cr!-r^)de. (8) If we put together what we have done, we may write rdOdr, (9) J 9, Jr. 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^, • • •, w„ lying in a plane at points whose coordinates are (x^^ y^^ (x^^ y^^ . . ., {x^^ y^ respectively is given by the formulas - ^ m^x^-\-m^x^^ . . . + m^x^ ^ m^y^Jrm^^^r"'^m^y^ m^-\- m^-\- ' ' ' ■\- m^ 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 (x^ 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 I xdm I ydm -^ = 1- yJ (2) \ 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 Dcds, For con- venience we place Dc = p and write dm = p ds, where /9 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 = j xds, s?/ = I i/ds, (3) where s 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 | Tra ; so that, from (3), we have Y ^ irax = I xds, \ iray — ( yds. To integrate, it is convenient to in- troduce the central angle (Fig. 100) ; whence x = a cos , ds = ad<}>. ■JT Then I irax z= C^a^ cos ^ d, Jo \iray = /o whence 2 2a a^ sin ^ (/ ; _ 2a y = — 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 dA is Dt dA. For convenience we place Dt = p and write , , , dm = pa A, CENTER OP GRAVITY 251 where /o is a constant. If this is substituted in (2) and the p's are canceled, we have Ax = / xdA, Ap= I 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 ^2 = kx, the axis of x, and the ordinate through the point (a, b) of the parabola (Fig. 101). We place dA = dxdy in (4) and have A X = ffx dx dy, A y = ffy 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^ the value of y on the parabola, to distinguish it from the general values of y inside the area. The first integra- tion gives us, therefore, respectively Fig. 101 r ^xdxdy — xy^dx and j ^ ydxdy — \y\ dx, so that we have Ax = Cxy^dx, Ay = fhVi dx. 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 x and \ y^, which are the coordinates of the middle point of the ordinate y^ These results are the same as if we had taken dA = y^ dx in the general formula (4), and had taken the point (x, y) at which the mass of dA is concentrated as (x, ly^), 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. Now, from the equation of the parabola y^ = kx, and to complete the integration, we have to substitute this value for y^ and integrate with respect to x from x = to x = a. We have Ax=pk^x^dx= I ArM, Ay=P \kxdx = ^- &2 2 But, from the equation of the curve, k= — and, by § 23, J = - ah. a o 252 REPEATED INTEGRATION Substituting these values and reducing, we have finally In solving this problem we have carried out the successive integrations Separately, in order to show clearly just what has been done. If now we collect all this into a double integral, we have /» a /. •\/kj: P^ n "^kx Ax = I I xdxmj. Aij = \ | ydxdy. Jo Jo Jo Jo Ex. 3. Find the center of gravity of a sextant of a circle of radius a. To solve this problem it is convenient to place the sextant so that the axis of X bisects it (Fig. 102) and to use polar coordinates. From the symmetry of the figure the center of gravity lies on OX, so that we may write at once y = 0. To find x take an element of area rdOdr in polar coor- dinates and place a: = r cos B. We have then, from (4), TT Ax= f^ fr^ cos SdOdr, J n Jo where A = } ira^, one sixth the area of a circle. 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 6 then varies from ~ ^ to ^ ' ^^^ thns the entire area is covered. The solution is as follows : Fig. 102 ira^x f[ la^cosede=^a^-y whence 2a Consider now a solid of revolution formed by revolving the plane area (Fig. 103) ABCD about OF 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 OF, so that we have to find only y. Let ^Fbe any element of volume. Then dm = pdV, where p is the density and is assumed constant. Substituting in (2), we have Vy =jya (5) CENTER OF GRAVITY 253 Let the solid be divided into thin slices perpendicular to OF, 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 of the summation is therefore y times the volume of the slice. It is therefore y(jrx^dy'). We have now to extend the summation over all the slices. This gives the result Vy -i: irx^ydy, where 0A = a and OB = h. 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 center of gravity of a right circular cone of altitude h and radius a (Fig. 104). This is a solid of revolution formed by revolv- ing a right triangle about OY. However, the equation of a straight line need not be used, as similar triangles are simpler. We have - = - ; a y ^ vv'hence x = - y. The volume V is known to be \Tra%. Therefore, from (6), we have whence y = 1 6. Pig. 104 EXERCISES 1. Show that the center of gravity of a semicircumference of radius a lies at a distance of — from the center of the circle on TT the radius which bisects the semicircumference. 2. Show that the center of gravity of a circular arc which subtends an ansrle a at the center of a circle of radius a lies at a distance — sin - ^ a 2 from the center of the circle on the radius which bisects the arc. 1 254 REPEATED INTEGRATION n — — — 3. A wire hangs so as to form the catenary ?/ = - (e" -h e "). Find the center of gravity of the piece of the curve between the points for which x = and x = a. 4. Find the center of gravity of the arc of the cycloid X == a( — sin <^), y = a(l— cos <^), 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 curve 1/ = sin X and the axis of x between a; = and x = tt. 9. Find the center of gravity of the plane area bounded by the two parabolas y^= 20 ic and x^ = 20 y. 10. Find the center of gravity of a figure which is composed of a rectangle of base 2 a and altitude b surmounted by a semicircle of radius a. 11. Find the center of gravity of the area bounded by the first arch of the cycloid (Ex. 4) and the axis of x. 12. Show that the center of gravity of a sector of a circle lies at 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 r = a(l-{- cos 6). 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 2 b and altitude a. 17. Find the center of gravity of the solid formed by revolving about OY the plane figure bounded by the parabola y^ = kx, 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 a, 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 256 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 J[fj, iHfg, • • 'i M^^ and if the centers of gravity of these parts are respectively (x^, y^, (x^, y^, - - .,(^„, yj, then the center of grav- ity of the composite body is given by the formulas Mi = M^\ + M^^+ . . . + Jf„^„, My = M^y^+M^y^+... + M^^. ^^ We shall prove the theorem for the x coordinate. The proof for y is the same. By § 85 we have, for the original body, Mx = I xdm, (2) where the integration is to be taken over all the partial masses J/j, M^, • • •, Jf„ into which the body is divided. But we have also = jxjm^, (3) ^2^2 /' i£ic„ = / x_ dm^, 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 r r Mx = I x^dm^-\- I x^dm^-^ \- j x^dm^ ; and, by substitution from (3), the theorem is proved. Ex. 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 as 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 b is on the axis of a: at a distance c from the origin. W 256 REPEATED IKTEGRATION The area which can be considered as composed of two parts is that of the larger circle, the two parts being, first, the smaller circle and, second, the irregular ring whose center of gravity we wish to find. Now the center of gravity of a circle is known to be at its center. Therefore, in the formula of the theorem, we know (x, y), which is on the left of the equation, to be (0, 0), and (x^, y^) to be (c, 0), and wish to find (Xg, y^)- Since we are dealing with areas, we take the masses to be equal to the areas, and have, accordingly, M = tto^ (the mass of the larger circle), M^= nh^ (the mass of the smaller circle), and M^ = TV (a^ — 6^) (the mass of the ring). Substituting in the formula, we have Fig. 105 whence, by solving for x^, 7ra2 . = irh'^c + ir {a^ - h'^) x^ It is unnecessary to find lies on OX. since, by symmetry, the center of gravity EXERCISES 1. Show that if there are only two component masses M^ and M^ in formulas (1) of the theorem, the center of gravity of the composite mass lies on the line connecting the centers of gravity of the com- ponent masses at such a point as to divide that line into segments inversely proportional to the masses. 2. Prove that if a mass M^ with center of gravity (x^^, y^ has cut out of it a mass M^ with center of gravity (x^^ y^, the center of gravity of the remaining mass is M^x^ M,-M^ y M^ - M^ r^ and r^ are tangent externally. Find 3. Two circles of radii their center of gravity. 4. Eind. the center of gravity of a hemispherical shell bounded by two concentric hemispheres of radii r^ and r^. 5. Place r^ = r J + Ar in Ex. 4, let Ar approach zero, and thus find the center of gravity of a hemispherical surface. CENTER OF GRAVITY 257 6. Find the center of gravity of a hollow right circular cone bounded by two parallel conical surfaces of altitudes h^ and h^ respectively and with their bases in the same plane. 7. Place 7^2= h^-\- ^h in Ex. 6, let AA 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 arm of 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 square and the radius of the quadrant being 4 in. Find the center of gravity of the figure remaining. 10. Two iron balls of radius 4 in. and 6 in. respectively are con- nected by an iron rod of length 1 in. Assuming that the rod is a cylinder of 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 gravity of the system, assuming that the sphere rests on the middle point of the top of the pedestal. 87. Theorems. The following theorems involving the center of gravity may often be used to ad- vantage in finding pressures, volumes of solids of revolution, or areas of surfaces of revolution. I. The total pressure on a plane sur- face immersed in liquid in a vertical position is equal to the area of the sur- face multiplied hy the pressure at its center of gravity. Let the area be placed as in Fig. 106, p^^ j^g where the axis of x is in the surface of the liquid and where the axis of y is measured downward. Then, by § 25, F=jwy{x^-x;)dy, (1) which may be written as a double integral in the form p = I / wydydx z=w \ i ydxdy. (2) 258 KEPEATED INTEGRATION In fact, this may be written down directly, since the pressure on a small rectangle dxdy is its area, dxdy, times its depth, y, times w. Moreover, from § 85, we have Ay -- j j ydxdy. (3) y=a Fig. 107 By comparison of (2) and (3) we have P = wyA. But wy is 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 3 ft., and its center is 12 ft. below the surface of the water. Find the total pressure on it. Here ^ = 9 tt and the depth of the center of gravity is 12. Therefore P = 108 TTw = -2/- '^ ^ons = 10.6 tons. II. 77ie 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, V=iT [ (xl-x^dy, which can be written as a double integral in the form V— 2 7r I I xdydx. By §85, Ax= j I xdydx; and, by comparison of (4) and (5), we have V^^irxA, which was to be proved. (4) (5) THEOREMS OF PAPPUS 259 Ex. 2. Find the volume of the ring 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 = Tra^ and that the center of gravity of the circle is at the center of the circle and therefore describes a circumference of length 2 TTC. Therefore V=2 tt Vc. III. The area generated hy revolving a plane curve about an axis in its plane not intersecting the curve is equal to the length of the curve multiplied hy 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 ^ S=2ir\xds. (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 S=27rsx, which was to be proved. Ex. 3. Find the area of the ring surface described in Ex. 2. We know that s = 2 7ra and that the center of gravity of a circumfer- ence is at its center and therefore describes a circumference of length 2 ire. Therefore g = 4^ ^z^c. 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 lateral surface of a right circular cone. 260 REPEATED INTEGRATIOK 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 the volume generated by revolving a parabolic segment about its base. 5. Find by the theorems of Pappus the volume generated by revolving a parabolic segment about the tangent at its vertex. 6. Find the volume and the surface generated by revolving a square of side a about an axis in its plane perpendicular to one of its diagonals and at a distance h from its center. 7. Find the volume and the area generated by revolving a right triangle with legs a and h about an axis in its plane parallel to the leg of length a and at a distance c from the vertex of the right angle on the opposite side from the hypotenuse. 8. A circular water main has a diameter of 4 ft. One end is 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 semiaxes a and h completely submerged, if the center of the ellipse is c units below the surface of the liquid. 10. Find the pressure on a semiellipse of semiaxes a and h {a > h) 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 about 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 of area dA. Then the mass of each element is pdA^ where p is the product of the thickness of the plate and its density. Let R be MOMENT OF INERTIA 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 element is approximately R'^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 of 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 / represents the moment of inertia of the plate, we have /= CR^pdA. (1) If in (1) we let /o = 1, the resulting equation is =/^' dA, (2) where / is called the moment of inertia of the plane area. When dA in (1) or (2) is replaced by dxdi/ or rdrdd, the double sign of integration must be used. Ex. 1. Find the moment of inertia of a rectangle of dimensions a and b about the side of length b. Let the rectangle be placed as in Fig. 108. Let it be divided up into elements dA = dxdy. Then x is the distance of some point in an element from OY. Hence, in (2), we have y R = X and dA = dxdy. Therefore / = iix^dxdy. We first sum the rectangles in a vertical strip, as y ranges from to b. We have I x^dxdy = X'bdx. This is the moment of inertia of 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 bdx. 262 REPEATED INTEGRATION The second integration gives now 1= Cxndx = \a%. 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 in that case the total mass M of the plate is pah, so that we have ^ ^ 1=1 Ma\ Ex. 2. Find the moment of inertia of the quadrant of an ellipse — 4- ^ = 1 (a>h) about its major axis. a^ b^ If we take any element of area as dxdy, we find the distance of its lower edge from the' axis about which we wish the moment of inertia to be y (Fig. 109). Hence R = y and . CCy'^dxdy. It will now be convenient to sum first with respect to x, since each point of a horizontal strip is at the same distance from OX. We therefore write ■.ffy^dydi Fig. 109 Now, indicating by x^ 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 I ^y'^dydx = y^x-^dy = -y^Vb^ — y^dy. For the second integration I = j f\''Vb^-y^dy. b^ To integrate, place y = bsm . Then I = ah^ f~^sm^cos'^d>d = Jo If, instead of the area, we consider a thin plate of mass M, the above result must be multiplied by p, where M = I irabp ; whence I=^Mb^. 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 Trah- 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 an 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 (x^ y)^ the distance of that point from the origin is \/x^-^y^. That is, in (2), if we place dA = dxdy and R'^=x^+ y"^^ we shall have the formula for the polar moment of inertia with respect to the origin. Denoting this by /jj, we have I.^fJCx'+fydxdy. (3) This integral may be split up into two integrals, giving I,= ff^dxdi/+fffdydx, (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 I^ ; the second integral is the moment of inertia about OX and may be denoted by I^. 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^ 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 4 = 1 7ra6«. 264 KEPEATED INTEGRATION By a similar calculation Therefore ^y = i 'r«'^^- If the area is replaced by a plate of mass M, this result gives If polar coordinates are used, the element of area is rdOdr, and the distance of a point in an element from the origin is r. Hence, in (2), dA = rdddr and E = r, Therefore =//" dOdr. (6) In practice it is usually convenient to integrate first with respect to r, holding 6 constant. This is, in fact, to find the polar moment of inertia of a sector with vertex at 0. Ex. 4. Find the polar moment of inertia of a circle with respect to a point on its circumference. Let the circle be placed as in Fig. 110. Its equation is then (Ex. 1, § 51) r = 2acos^, where a is the radius. If we take any element rdddr and find Iq for all elements which lie in the same sector with it, we have to add the elements r^dOdr, with r ranging from to Tj, where r^ is the value of r on the circle ; and therefore r^ = 2a cos $. We have 'r^dBdr = \ r^dO = 4 a^ cos^OdO. We have finally to sum these quantities, with $ ranging from — ^ to ^ We have IT lQ=n4: a* co8*dde = f ira^ Fig. 110 If M is the mass of a circular plate, this result, multiplied by p, gives /o = i Ma'. Ex. 5. Find the polar moment of inertia of a circle with respect to its center. Here it will be convenient first to find the polar moment of inertia of a ring (Fig. 111). We integrate first with respect to 6, keeping r constant. We have i: r^drde = 2Trr^dr, MOMENT OF INERTIA 265 which is the approximate area of the ring 2 vrdr multiplied by the square of the distance of its inner circumference from the center. We then have, by the second integration, C^irr^dr l7ra'- Fig. Ill If Af is the mass of a circular plate, this result, multiplied by p, gives /o = h Ma\ The moment of 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 moments of inertia of the circular sections about their centers. If the axis of revolution is OF, the radius of any circular section perpendicular to Or is a; and its thickness is dy. Its mass is therefore pirx^dy ; and therefore, by Ex. 5, its moment of inertia about F is i pirx^dy. The total moment of inertia of the solid is therefore 1=19-^ i ^^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 1 r ^ a* 1 But, if M is the mass of the cone, we have M = \ pira^b. Therefore EXERCISES 1. Find the moment of inertia of a rectangle of base b 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 h 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 h and alti- tude a about its base. 4. Find the moment of inertia of an ellipse about its minor axis and also about its major axis. 266 KEPEATED INTEGRATION 6. Find the moment of inertia of a trapezoid about its lower base, taking the lower base as h, the upper base as a, and the altitude as h. 6. Find the moment of inertia about its base of a parabolic seg- ment of base h and altitude a. 7. Find the polar moment of inertia of a rectangle of base b and altitude a about its center. 8. Find the polar moment of inertia about its center of a circular ringj the outer radius being r^ and the inner radius r^. 9. Find the polar moment of inertia of a right triangle of sides a and h 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 = Q> and the straight line cc + y— 7 = 0. 11. Find the polar moment of inertia about the origin of the area bounded by the curves y = x^ and y = 2 — x^. 12. Find the polar moment of inertia about the origin of the area of one loop of the lemniscate r^ = 2 a^ cos 2 6. 13. Find the moment of inertia of a right circular cylinder of height hy 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^, its outer radius r^, and its height h. 15. Find the moment of inertia of a solid sphere about a diameter. 16. A ring is cut fram a spherical shell whose inner and outer radii are respectively 5 ft. and 6 ft., 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 axis 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 by the square of the distance between the axes. MOMENT OF INERTIA 267 We shall prove this theorem only for a plane area, in the two cases in which 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 area, as in § 88. Case I, When the axes lie in the plane of the figure. Let the area be placed as in Fig. 112, where the center of grav- ity (x, ^) is taken as the origin (0, 0) and where the axis of ^ is taken parallel to the axis LK, about which we wish to find the moment of inertia. Let x be the distance of an element dxdi/ from OF, and x^ its distance from LK. Then, if / is the Fig. 112 moment of inertia about OF, and I^ the moment of inertia about LK, we have ^^ ^^ : / i x'^dxdy. Moreover, if a is the distance between F and LK, we have x^=x-{-a\ (2) so that, by substituting from (2) in the second equation of (1), we have ^^ ^^ Ii= I f x^dxdi/ -\-2a I I xdxdy -{• a^ I I dxdy, (3) Now, by § 84, CCdxdy = ^ ; by § 85, CC xdxdy = Ax=^, smce by hypothesis 2: = ; and, by (1), the first integral on the right hand of (3) is I^. Therefore (3) can be written Z = J + aM, (4) which proves the theorem for this case. Case IL 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. 113, 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 Ig be the polar moment of inertia about 0, and I^ the polar moment of inertia about P. Draw through P axes PX' and P Y' parallel to the axes of coor- dinates OX and Y, Let I^ and /y be the moments of inertia about OX and OY respectively, and let I^, and I^, be the moments of inertia about PX' and PY', Then, by (5), § 88, Fig. 113 Moreover, if (a, 5) are the coordinates of P, we have, by Case I, r_r_L^2j /, = /-f h'^A, i,= r + d^A, (6) Therefore, from (5), we have I, = I,+ (ia'+h')A, (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 circumference. 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 ^tto*. Therefore, by the above theorem, ^ , a , o ^ o^ ^ a 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. SPACB 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^ and inner radius r^ 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 r^ and outer radius r^ about a tangent to the outer circle. 10. A circle of radius a has cut from it a circle of radius - tangent z 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, ^). 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 «, at right angles to the plane, we can reach any point in space. The quan- tity z 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 XOY^ in which are already drawn a pair of coordinate axes, OX and OF, at right angles with each other. Perpendicular to this plane at 270 KEPEATED INTEGRATION 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 M^ and from M draw a line par- allel to OY^ meeting OX at L, Then for the point P (x, y^ 2), 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= 0L = NM= SB = TP, with similar sets of values for ?/ and z. 2. Cylindrical coordinates. Let XOY be any plane in which a fixed point O 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 31, Then, if the polar coordinates of M in the plane XO Y are r = OM, 6 = XOM^ and we denote the dis- tance MP by 2, the cylindrical coordinates of P are (r, 6, 2;). 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 z measures the distance of P from the plane XOY, Fig. 116 SURFACES 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 x — r cos ^, 1/ = r sin ^, 2 = z. (1) These are formulas by which we may pass from one system to the other. It is convenient to notice especially that r^=2^+/. (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 we 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 plane 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 with 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 coordinates, in the right triangle OPS, OS=z, SP=r, Fig. 117 and OP = a. Therefore r^+ z^= a\ (1) 272 EEPEATED INTEGRATION Fig. 118 This equation is satisfied by the cylindrical coordinates of any point on the surface of the sphere and by those of no other point. It is therefore the equation of the sphere in cylindrical coordinates. Z By means of (2), § 90, equation (1) becomes a:2 + 2/2 + 22 = a\ (2) which is the equation of the sphere in Cartesian coordinates. 2. Sphere tangent at origin to a coor- dinate plane. Consider a sphere tangent to the plane XOY at O (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 draw PS perpendicular to OZ. Then, using cylindrical coordi- nates, OS = z, SP = r, and OA =2a, where a is the radius of the sphere. Now OAP 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 -OS). Substituting the proper values, we have r^=2az-z'^, (3) which is the equation of the sphere in cylindrical coordinates. By (2), § 90, equation (3) becomes x^-\- f-\- z^-2az = 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 /-* any point on the surface of the cone, pass a plane through P and OZ, and draw PS perpendicular to OZ. Then SP Fig. 119 SP = r and 05 = z. But —- = tan SOI OS tana. Therefore we have r = z tan a (5) as the equation of the cone in cylindrical coordinates. SURFACES 273 By 2, § 90, equation (5) becomes x^-{- y^-z^tan^a = 0, (6) as the equation of the cone in Cartesian coordinates. As explained above, we have here 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 h (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 the surface and pass a plane through P and OZ. In the plane POZ draw OR perpendicular to OZ and, from P, a straight line perpendicular to OZ meeting OZ in S. If we regard OR and OZ as a pair of rectangular axes for the plane POZ, the equation of the curve CD in which the plane POZ cuts the surface is Fig. 120 ^ =f(r), (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 the equation of the sur- face in cylindrical coordinates. When the plane POZ coincides with the plane X OZ, r is equal to x, and equation (7) becomes, y for that section, ^ _ f(x). (S) Hence we have the following theorem : The equation of a surface of revolution formed hy revolving about OZ any curve in the plane XOZ may be found in cylindrical coordinates by writing r for x in the equation of the curve. The equation of the surface in Cartesian coor- dinates may then be found by placing r = Vx^ -t- y'^. For example, the equation of the surface formed by revolving the parabola z^= ix about OZ as an axis is z^= ir in cylindrical coordinates, or z* = IQ (x^ ■{■ y^) 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 PS perpendicular to OZ. Then SP is always equal to a, the radius of the cylinder. Therefore, for all points on the surface, r = a, (9) 274 KEPEATED INTEGRATION which is the equation of the cylinder in cylindrical coordinates. Reduced to Cartesian coordinates equation (9) becomes X^+y2=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 the 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 this line, the coordinates of P also satisfy the equation, since z is not involved in the equation. As examples, the equation y'^= ^x represents a parabolic cylinder, and the equation r = a sin 3 ^ represents a cylinder whose base is a rose of three leaves (Fig. 65, p. 144). 6. Ellipsoid. Consider the surface defined by the equation + P + ? = 1. ai) If we place 2 = 0, we get the points on the surface which lie in the XOY plane. These points satisfy the equation o ,,2 and therefore form an ellipse. Similarly, the points in the ZOX plane lie on the ellipse ^^ + 1 = 1' (1^) and those in the YOZ plane lie on the ellipse '/>2 ^ £.2 1. (14) .--^ — J- — .^^-^ ,__ \^_^ "^ — ' 1^ ^^^ — -^-^^ \y I '_— -^' — Fig. 122 The construction of these ellipses gives a general idea of the shape of the surface (Fig. 122). To make this more precise, place z — z^ in (11), where z^ is a fixed value. We have which can be written a' IP- l-^> «^(^-S) ^^(i-S) let us (15) (16) which is satisfied by all points which lie in the plane at the distance from the A' OF plane. SURFACES 275 As long as z^ < c^, equation (16) represents an ellipse with semiaxes ^■2 a v'l — ^ and ^^ \'l — -^ • ^J taking a sufficient number of these sections we may construct the ellipsoid with as much exactness as desired. If z^ — c^ in (16), the axes of the ellipse reduce to zero, and we have a point. If z^ > c^ the axes are imaginary, and there is no section. 7. Elliptic paraboloid. Consider the surface ? = ^ + ^, c a^ ¥ where we shall assume, for definiteness, that c is positive. If we place 2 = 0, we get a'' V' ' (17) (18) which is satisfied in real quantities only by a: = and y = 0. Therefore the XOY plane simply touches the surface at the origin. If we place z = c, we get the ellipse a^^ b'~^' (19) which lies in the plane c units distant from the XOF plane. If we place y = 0, we get the parabola ^ 2 z = — x^; a^ (20) and if we place a: = 0, we get the parabola c (21) Fig. 123 The sections (19), (20), and (21) determine the general outline of surface. For more detail we place z = z^ and find the ellipse x^ V- ^'^^"^' (22) c c so that all sections parallel to the XOY plane and above it are ellipses (Fig. 123). 8. Elliptic cone. Consider the surface - + ^ - - = 0. (23) . Proceeding as in 7, we find that the section r = is simply the origin and that the section z = c is the ellipse 62 1. (24) 276 REPEATED INTEGRATION If we place a; = 0, we get the two straight lines h and if we place y = 0, we get the two straight lines X = ± -z. c (25) (26) The sections we have found suggest a cone with an elliptic base. To prove that the surface really is a cone, we change equation (23) to cylin- drical coordinates, obtaining fco^ sin^^' \ 2 ^^ (27) Now if 6 is held constant in (27), the coefficient of r^ is constant, and the equation may be written ± kz, (28) which is the equation of two straight lines in the plane through OZ Z Z Fig. 124 Fig. 125 determined by ^ = const. Hence any plane through OZ cuts the surface in two straight lines, and the surface is a cone (Fig. 124). 9. Plane. Consider the surface Ax + By-{- Cz ■\- D = 0. (29) The section 2 = is the straight line KH (Fig. 125) with tlie equation Ax-\- By\- D = 0, (30) the section ^ = is the straight line LH with the equation Ax + Cz + D = 0, (31) and the section x = is the straight line LK with the equation B,j+Cz + D^Q. (32) VOLUME 277 The two lines (31) and (32) intersect OZ in the point L (o, 0, - -), unless C = 0. Assuming for the present that C is not zero, we change equation (30) to cylindrical coordinates, obtaining (.4 cos ^ + 5 sin B)r-\- Cz^ D = 0. (33) This is the equation of a straight line LN in the plane Q — const. It passes through the point Z, which has the cylindrical coordinates r = 0, 2 = — — ; and it meets the line KHy since when 2 = 0, equation (33) is the same as equation (30). Hence the surface is covered 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 ^x + £?/ + i) = 0. (34) The point L does not exist, since! the lines corresponding to HL and KL are now parallel. But, by 5, equation (34) represents a plane parallel to OZ intersecting XOY va. the line whose equation is (34). Therefore we have the following theorem : Any equation of the Jirst degree represents a plane. 92. Volume. Starting from any point (x^ y^ z) in space, we may draw lines of length dx, dy^ and dz in directions parallel to OXj OY, 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 dV=dxdydz. (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 d V the product of its base by its altitude, and we have dV=rdedrdz. (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 EEPEATED INTEGRATION Fig. 126 and X units from it, and that similar planes correspond to the values of y and z. We may, accordingly, divide any required volume into elements of volume as follows: Pass planes through the volume parallel to YOZ and dx 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 XOZ 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 XOY and dz units apart, with the result that each column is divided into rectangular parallelepipeds of dimensions dx^ dy^ and dz. 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 x oy y oy z^ and the subsequent order, depend upon the particular volume considered. A similar construction may be made for cylindrical coordinates. In this case the coordinate 6 determines a plane through OZ. We accordingly divide the volume by means of planes through OZ making the angle dQ with each other. The result is a set of slices one of which is shown in Fig. 127. The coordinate r determines a cylinder with OZ as its axis. We accordingly divide each slice into columns by means of cylinders with radii differing by dr. One such column is shown in Fig. 127. Fig. 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 F= CCCdxdydz (3) V= CCCrdOdrdz, (4) or 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= fdz CCdxdy or F= C dz ffrdddr. (5) But, by § 84, I / dxdy = A and j I rd6dr = A, where A is the area of the plane section at a distance z from XOY. Hence (5) is ^ V= Adz, (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 -p2 2/2 2" Ex. 1. Find the volume of the ellipsoid -;, + ^ •¥ — — l. a^ b^ c^ By 6, § 91, the section made by a plane parallel to XOY is an ellipse V" ^ \ ^2 1 and i a/1 ^ • Therefore, by Ex. 1, § 77, its area is TTob |1 ^1- Hence we use formula (6) and have V = irab j (l -\ dz = - irabc. Ex. 2. Find the volume bounded above by the sphere x^i- y^+ z^= 5 and below by the paraboloid x^+ y^ = 4iZ (Fig. 128). As these are surfaces of revolution, this example may be solved by the method of Ex. 1, but in so doing we need two integrations — one for the sphere and the other for the paraboloid. We shall solve the example, however, by the other method in order to illustrate that method. We first reduce our equations to cylin- drical coordinates, obtaining respectively r2 + 22 and = iz. (1) (2) The surfaces intersect when r has the same value in both equations; that is, when 1 , A ' /ox 2^ + 4 2 = o, (3) which gives z =1ot z = — 6. The latter value is impossible ; but when z = 1, we have r = 2 in both equations. Therefore the surfaces intersect in a circle of radius 2 in the plane z lies directly above the circle r = 2 in the XOY plane. We now imagine the element rdOdrdz inside the surface and, holding r, B, dO, dr constant, we take the sum of all the elements obtained by varying z inside the volume. These elements obviously extend from 2 = Zj in the lower boundary to z = z^ in the upper boundary. From (2), z^ = — and, from (1), ^2 = V 5 — •dSdrf^ The first integration is therefore Uddr We must now allow 6 and r so to vary as to cover the entire circle r = 2 above which the required volume stands. If we hold 6 constant, r varies from to 2. The second integration is therefore - VOLUME 281 Finally, must vary from to 2 tt, and the third integration is If we put together what we have done, we have /I 2 JT /» 2 /."v 5 — r* 9 TT / y— N F=/ (I rrf^c/rcf2 = =-(5V5-4). Jo ./O ./r2 3 £X£RCIS£S 1 . Find the volume bounded by the paraboloid z = x^ -\- y^ and the planes x = 0, y = 0, and z = 4. Z CC 7/^ 2. Prove that the volume bounded by the surface - = — + f- c a^ ¥ 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 z — and the cylinders x^ -\- y'^= V? and y'^= c? — az. 4. Find the volume cut from the sphere r^-\-z'^ = a? by the cylinder r = a cos B. 5. Find the volume bounded below by the paraboloid r^^az and above by the sphere r^ -{- z^ — 2 az = Q. 6. Find the volume bounded by the plane XOY, the cylinder ic^ -j- 2/^ — 2 ax = 0, and the right circular cone having its vertex at Oy its axis coincident with OZ, and its vertical angle equal to 90°. 7. Find the volume bounded by the surfaces r^ =&«,« = 0, and r = 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 r^ = a^ cos 2 9. 10. Find the volume of the paraboloid x^-\-y'^ = 2z cut off by the plane z =^x -^-1. 11. Find the volume of the solid bounded by the paraboloid z = x^ -\- y"^ and the plane z = x. 282 REPEATED INTEGRATION 93. Center of gravity of a solid. The center of gravity of a solid has three coordinates, ^, ^, 2, which are defined by the equations I xdm j ydm I zdm [dm j dm t dm where dm is the mass of one of the elements into which the solid may be divided, and rr, ?/, and z 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 of 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 — pd F, 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 Vx= 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 z, it is evident from the symmetry of the solid that x = y = 0. To find z we shall use cylindrical coordinates, the equations of the sphere and the cone being respectively r^ •{■ z^ = 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 I f rdOdrdz = % ira^(l - cos a) Jo Jrctna zdV=j I I rzdddrdz = I Tra"^ sin- a. Jo Jo Jrcttxa Therefore, from (2), 2 = § (1 + cos a). CENTER OF GRAVITY 283 EXERCISES 1. Find the center of gravity of a solid bounded by the paraboloid - = — , 4- 7? and the plane z ■= c. 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 A, and the radius of the upper base and the radius of the lower base of which are respectively v^ and r^. 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 of the ellipsoid -^ + 75 + — , = 1. ^ (V- If c^ 6. Find the center of gravity of a solid bounded below by the paraboloid az = 7^ and above by the right circular cone z -\- r = 2a. 7. Find the center of gravity of a solid bounded below by the cone z = r and above by the sphere r^ -{- z^ = 1. 8. Find the center of gravity of a solid bounded by the surfaces z = 0, 7^ -\- z^ = 1/, and r = a{a< b). 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 /= CR^pdV=^p CR^dV, (1) where E 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 determined from the formula M=pV, 284 REPEATED INTEGRATION If the moment of inertia about OZ, which we shall call /^, is required, then in cylindrical coordinates R = r and dV= rdOdrdz, so that (1) becomes I, = p fCCr^dddrdz. (2) If we use Cartesian coordinates to determine /^, we have R^—^-^-'if- and dV= dxdydz^ so that Similarly, if 7^ and I^ are the moments of inertia about OY and OX respectively, we have /, = pfff(^^'-^ ^') ^^^^^^' ^. = ^jjj^^'^ ^'^ dxdydz, (4) In evaluating (2) it is sometimes convenient to integrate with respect to z last. We indicate this by the formula ='HI r'dddr, (5) But j I r^dOdr 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 I^= p III x^dxdydz + P I I I y^dxdydz, (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 d. The base of this cylinder is shown in Fig. 65, p. 144. We have, from formula (2), t „.,„,a , \ y /» _ /» a sin 3 nh where the limits are obtained as follows: First, holding r, B, d9, dr constant, we allow z to vary from the lower base 2 = to the upper base z = h, and integrate. The result pht^dBdr is the moment of inertia of a column such as is shown in Fig. 127. We next hold 6 and d$ constant and allow r to vary from its value at the origin to its value on the curve r = a sin 3 6, and integrate. The result ^pAa*sin^3 Odd is the moment of inertia of a slice as shown in Fig. 127. Finally, we sum all these slices while allowing 6 to vary from its smallest value to its largest value - • The result is — ph a%. 3 32 The volume of the cylinder may be computed from the formula V = j^J J rdedrdz= j\ AaV. Therefore M = ^\^ ph a% and I^ = | Afa^. 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 z = 2 and z"^ = 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 about OZ of a solid bounded by z _x'^ . ?/^ c a^ 5. Find the moment of inertia of a right circular cone of height h 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 r^-\- z^= a} cut out by the plane « = and the cylinder r — a cos 0. the paraboloid ~ = -^ + ^ and the plane ^ = c. 286 REPEATED INTEGRATION 8. Find the moment of inertia about OX of a solid bounded by the paraboloid z — r^ and the plane z = 2. 9. Find the moment of inertia about its axis of a right elliptic cylinder of height ^, the major and the minor axis of its base being respectively 2 a and 2 h. 10. Find the moment of inertia about OZ of the ellipsoid GENERAL EXERCISES 1. Find the center of gravity of the arc of the curve x^ ■\-y'^ = a^, which is above the axis of x. 2. A wire is bent into a curve of the form 9?/^ = x^. 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^ = x^ and any double ordinate. 4. Find the center of gravity of the area bounded by the axis of X, the axis of ^/, and the curve i/^ = 8 — 2 a:-. 5. Find the center of gravity of the area bounded by the curves 2/ = x^ and i/ = —^ 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 ?/ and the curve x = a GOS^, 1/ = a sin^^. 7. Find the center of gravity of the area bounded by the ellipse x^ ip' -^ + y^ = 1 (a> h)^ the circle x'^ -{- tf = a^, and the axis of y. 8. Find the center of gravity of the area bounded by the parabola x''=%y and the circle ic^ + t/^ = 128. 9. Find the center of gravity of the area bounded by the curves x^ — a (y — b) = 0, x^ — ay = 0, the axis of y, and the line x = c. 10. Find the center of gravity of an area in the fqrm 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 b 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 h units from the center. 13. From a rectangle h 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 r^ and r^. 15. In the result of Ex. 14, place r^= r^-\- Ar and find the limit as Ar— vO, thus obtaining the center of gravity of a semicircum fere nee. 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 -^ + 7^ = 1 (a > J), in which there is a circular hole of radius c, (V- If ^ ■'' 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 x^ if about OY the surface bounded by the hyperbola "i — 7^ =1 and the lines y = and y = b. 20. Find the center of gravity of the solid generated by revolving about the line x = a the area bounded by that line, the axis of a-, and the parabola y"^ = kx. 21. Find the center of gravity of the segment cut from a sphere of radius a by two parallel planes distant respectively h^ and h^ (h^ > h^ 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 REPEATED INTEGRATION 24. Find the moment of inertia of a plane circular ring, the inner radius and the outer radius of which are respectively 3 in. 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 the plate and its sides parallel to the sides of the plate. Find the moment of inertia of the plate about a line through its center parallel to one side as an axis. 26. Find the moment of inertia of the plate of Ex. 25 about one of the outer sides as an axis. 27. Find the moment of inertia of the plate of Ex. 25 about one side of the hole as an axis. 28. Find the moment of inertia of the plate of Ex. 25 about one 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 the center of the square. Find the moment of inertia of the plate about a line 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 r^ = c^ sin 3 Q about the pole. 34. Find the polar moment of inertia of the area bounded by the cardioid r = a ( 1 + cos 0') about the pole. 35. Find the polar moment of inertia of that area of the circle r = a which is not included in the curve r = a sin 2 B about the pole. 36. Find the moment of inertia about OF of a solid bounded by the surface generated by revolving about OF the area bounded by the curve y^=Xy 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 r^ 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 > a) from its center. Find the moment of inertia of this anchor ring about its axis. 41. Find the moment of inertia of the elliptic cylinder — + 7- =1 (a > ^), its height being 7i, about the major axis of its base. 42. Find the center of gravity of the solid bounded by the cylinder r =2a cos 6, the cone z = r, and the plane ^ = 0. 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 — 2a cos 2 6, between the cone z = 2r and the plane z = 0. 45. 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 = a cos 2 B and bounded by the planes ^ = and z = 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 = 2a cos included between the planes z = and z = 2x -\- a. 50. Find the moment of inertia about OZ of the solid of Ex. 49. 51. Through a spherical shell of which the inner radius and the outer radius are respectively r^ and r^, a circular hole of radius a{a < r^ 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|f. 4. 106 ft. per second. 2. 1^|. 5. 33.97 mi. per hour for entire trip. 8. 1^ mi. per hour. 3. iO\^. 6. 1.26. - -- . 7. 21. 9. 98.4. Page 5 (§ 2) 10. 196.8. Page 7 (§ 3) 1. 96 ft. per second. Page 8 (§ 3) 3. 128 ft. per second. 4. 74 ft. per second. 2. 128 ft. per second. 5. 68 ft. per second. 6. 52 ft. per second. Page 11 (§ 5) 1. 12«i2; 24«i. 3. 85; 32; 6. 6. 5, 4, when « = 2 ; 10, 6, when i = 3. 2. 16; 14. 4. 42; 57. S. Sat^ + 2bt + c ; Qat + 2b. Page 13 (§ 6) 1. — sq. ft. per second. TT 4. 4 7rr2. Page 14 (§ 6) 6. 8 7rr. 7. 3(edge)2. 27r 3. 8 Trr^ cu. in. per second. 6. 16 Trrsq. in. per second. 8. 67rr2 9. 18. 10. 2 Page 18 (§ 7) 1. 8x. 2. 3x2 + 4x. 6. - ' (2 + x)2 CHAPTER II 3. 4x3-2x. 7. x^ + x+1. 291 --!. 6. 2x x8 8.3--^, 292 ANSWERS Page 21 (§ 9) 1. Increasing if x > 2 ; decreasing if a; < 2. 2. Increasing if x > — ^ ; decreasing if a; < — ^. 3. Increasing if x < | ; decreasing if x > ^. 4. Increasing if x < — ^ ; decreasing if x > — |^. 6. Increasing ifx< — 2orx>l; decreasing if — 2 < x < 1. 6. Increasing ifx<— 5orx>3; decreasing if — 5 < x < 3. 7. Increasing ifx< — lorx>f; decreasing if — 1 < x < ^. 8. Always increasing. 9. Increasing ifx< — lor— ^l. 10. Increasing if x > 1 ; decreasing if x < 1. Page 24 (§ 10) 1. When t<—lort>l; when —l 5. 3. Wlien t<2ort>4; when 2 ^; when t < |. 6. Always increasing. 7. Always decreasing. 8. Increasing when t> 2 ; decreasing when t <2. 9. Increasing when t> ^ ; decreasing when t < ^. 10. Increasing when t < § ; decreasing when i > §. Page 26 (§ 11) 1. ^ Trh'^. 2. 6 7r/i sq. ft. per second. Page 27 (§ 11) 3. 0.2 cm. per second. 5. 0.26. ^-(tt\h' ht\^ 4. 20.9 sq. in. per second. 6. 64 cu. ft. per second. '8 & ^ • 8. 4 TT (i2 + 12 « + 36) ; t is thickness. Page 31 (§ 13) 1. 1.46. 3. 0.46; 2.05. 6. 2.41. 2. -2.07. 4. 1.12; 3.93. 6. -2.52. Page 35 (§ 14) 1. Sx-y-9 = 0. 6.x + 2/ + 1 = 0. *12. tan-i^. 2. 2 X + 3 y + 3 = 0. 7. X + 2 2/ + 8 = 0. 13. tan-il2. 3. 21 X- 22/ -13 = 0. 8. 4x-3?/-l=0. H. |. 4.2/4-3 = 0. _ 9. 12x-42/-5 = 0. U. 2x-3y-16 = 0. 5. V3x-2/-2V3-2 = 0. 10. 5x - Oy - 4 = 0. 16. (-l^^^, 21*3). *The symbol tan-^ I represents the angle whose tangent is ^ (cf. § 46). ANSWERS 29: Page 39 (§ 15) 1. (- §, 2|). 1, 2x-y-l = 0. 2. i'l Hh 8. 4x + y + 4 = 0. 8. (0, 4), (2, 0). 9. (- 3, 10), (1, 2). 4. (1, 7), (3, 3). 10. 27x + 272/-86 = z0;x + y-2 = 0. 6. (- 2, 0), (1, - 9). 11. 18 X- 27 2/ + 80 = = 0; 18x-27y-28 = 6. (- 1, - 3), (3, 29). 12. tan-i§. Page 43 (§ 17) 1. 25sq.in. 3. 6 ft. ^ ina^VS 2. Lerxgth is twice breadth. 4. 50. ^' 9 Page 44 (§ 17) 6. Depth is one-half side of base. 7. 2 portions 4 ft. long ; 4 portions 1 ft. long. 2aV6 8. Breadth = ; depth = ^;base=^ 4 ' 4 ; 2547 cu. in. 9. Altitude = 10. 2000 cu. in . 11. Height of rectangle 5 12. -7=in. V3 (p = perimeter) radius of semicircle ; semicircle of radius, Page 46 (§ 18) 1. 426 ft. Page 47 (§ 18) 2. 46 ft. 3. 95 ft. 4. 38|ft. 5. 576 ft. 6. 2/ = x2 + 3 a; - 6. 7. y = 2x^-\-x^-ix + 6. 621. Page 49 (§ 19) 1. 7^. 2. 13^. Page 53 (§ 20) 8. 0.0001 ; 0.000001 ; 0.00000001. 9. 0.000009001 ; 0.000000090001. Page 54 (§ 21) 1. 72 sq. in. „ 57r . TT . 2. — cu. m. ; - sq. in. 16 ' 2 ^ « 27r 37r 3. — cu. in. ; — cu.in. 5 25 8. y = 7+4a;-|x2- i xs. 9. y=l(x2-33). 10. 2/=^x3- ^x2 + x+ |. 4. 36. 6. IJ. 6. lj%. 7. 2^. 8. 2^. 10. 0.000003 sq. in. 11. 456.58 cu.in. 4. 27.0054 cu. in. 6. 28.2749 cu. in. 6. 606.0912. 7. 0.0012. 8. 6.99934. 294 Page 55 (General Exercises) 5 ^ 2x (1 - X)2 (X2 -f 1)2 2a ^ 4x 4. ANSWERS . 6. 1 . 2Vx 7. --^ Vx2 + 1 6 ' . 13. Increasing if x > — 2 ; 2Va;3 decreasing if x < — 2. (a - x)2 (x2 + 1)2 14. Increasing if — |2; decreasing if x < — | or | < x < 2. 16. Increasing if x > ; decreasing if x < 2a 2a 16. Increasing if x < :^ or x > — r: ; decreasing if — -• 3 3 Page 56 (General Exercises) 18. 1< « < 5. , 19. 2 < i < 5 ; 41 20. Up when < f < 61 ; down when 6\ 4 ; decreasing when i < 4. 22. V increasing when i < 3, u decreasing when ^ > 3 ; speed increasing when 2 < i < 3 or f > 4, speed decreasing when i < 2 or 3 < i < 4. 23. Increasing when l<£<2ori>3; decreasing when i < 1 or 2 < i < 3. 24. 0.0055 in. per minute. 25. 8.6 in. per second. 27. x-\- 2y + 6 = 0. 26. 1 sq.in. per minute. 28. 7x + 5?/ + 1 = 0. Page 57 (General Exercises) 29. X- 2 = 0. 31. 2x-?/ + 3 = 0. 33. (-^, 3|). ZO. x-2y-7 = 0. 32. tan-i |. 34. (11, 0). 35. (2, -2). 41. 102. 36. (- 1, 13), (5, - 95). 42. tan-i j%. 37. (- 3, 13), (1, - 19). 43. X - ?/ - 11 = 0. 38. (-4,20). 44. (1, -1),(-^, -,V\). Page 58 (General Exercises) 46. 62 ft. long. 52. ?/ = x2 + 3x - 13. 47. Altitude of cone is | radius of sphere. 63. y = ^ x^ — x^ + 7x. AQ AU-. ^ *l4^ -A ,u '1'^^ 64.85^. 48. Altitude = \ ——; side of base = A -_ ooi \243 \ 27 56. 28|. 49. 2 pieces 3 in. long ; 3 pieces 1 in. long. 56. 20^. 60. 600 ft. 67. 72. 51. 56 ft. ' 69. 0.0003. Page 59 (General Exercises) 60. 0.00629. 64. 0.09 cu. in. 67. 24.0024 sq. in. 62. 288 TT cu. in. 66. 0.0003. 68. 0.4698. 63. 161.16 cu. in. 66. 354.1028 ; 353.8972. ANSWERS 295 CHAPTER III Page 66 (§ 23) 1. 3,^. 3. 52^15. 6. 5.;^. 7. 36^i. 2. 23^. 4. 166§. 6. 42§. 8. 21. 9. 96. 11. 42§. 10. 10§. 12. 10§. Page 67 (§ 24) 1. 150 ft. 2. 140 ft. 3. 57^ ft. Page 68 (§ 24) 4. When ^ < « < -i^ ; 83^ ft. 6. 8000 TT ft.-lb. Page 70 (§ 25) 1. 8^T. 2. 21 T. 3. 3T. 4. l^T. Page 71 (§ 25) 6. Approx. 24131b. 7. 585 j^T. 6. 4^\T.; 4^\T. 8. 117, \T. 11. 2.1 ft. from upper side. 9. 234 § T. 10. 2/^ T. Page 75 (§ 26) 1. 21 ^TT. 3. 34^2^ TT. 6. 3331 c'u 625 V3 4. l^TT. t.S^^TT. 4 ' 6. 557f7r. 8. 2i. L. in. 9. 25f TT. 10. 213^ 11. 761". Page 76 (General Exercises) 1. 6|ft. 5. 20. 8. 2. 81ft. 4a2V2 9. 3. lO^ft. 3 ' 12. 4. 8l|mi. 7. 8. 21tV. , Reduced to | original pressure. Page 77 (General Exercises) 13. Twice as great. 17. 6if7r. 14. l^T. ^^ 144 7rV3 15. mw. ^^' 35 ■ 16. 68T^7r. 19. 96 TT. 20. 341^ cu. in. 21 32 V3 3 23. (a/i2-i;i8)7r. Page 78 (General Exercises) 24. 8 7r. 26. 1151. 26. 34^^^^ tt. 29. 728,949 ft.-lb. 30. 27. 9. 28. 204 1. 5301 ft.-lb. CHAPTER IV Page 81 (§ 28) 1. x2 + ?/2_8x-l-4i/ + ll = 0. 2. x^ + y^ + 2y-24 = 0. 6. 3x- 22/ -1- 4 = 0. 3. (-3,5); 5. 4. (-1,^-); 2. 296 ANSWERS Page 84 (§ 30) 1. (-2,0). . 4.(0,-13). 2. (0, 1). 6. SUft. ^ IOttVc. 3. (1^,0). 6. lOVlOft. 8 Page 85 (§ 30) Q.y^ + 6x-9 = 0. 10. x2-4x-12i/ + 16 = 0. Page 87 (§ 31) l.(..0);(.V7,0);^. 3.(.f,0>(4«,0);l. V5 . / V2 ^\ . / Ve .\ Vs 6. 9x2 4. 25^2 _ 36a, _ 139 = 0. 6. 49x2 + 24 1/2 - 120?/ - 144 = 0. 2.(0,±3);(0,±V6);4l 4.(.f,o);(±-f,o); Page 91 (§ 32) _ 1. (±3,0); (±Vl3, 0); 2x±3y = 0', -^. 2. (±2, 0); (± Vl3, 0); 3x±22/ = 0; — . _ 3. (0, ± V2); (0, ± V5); V2x±V3i/ = 0; ^. 4. (± 2 V2, 0) ; (± 4, 0) ; X ± 2/ = ; V2. 6. (0, ± IV (^0, ± -^^; X ± 2y = ; Vs. 3x2 + 8a; 7. 3x2-2/2 -12x + 9 = 0. 8. 3x2-1/2 + 4?/ - 16 = 0. Page 102 (§ 36) 1. 18x2 + 22x-3. 2. 4x2(x + 3). 3. 6x(x4-l). 4 36 X (X2 - 9)2 6. x2 + 4 X + 1 (X2 + X + 1)2 6. 3\Vx VxV 7. 2.-1+4-4- 2Vx3 + 4x2 + 1 10. 3xVx2 + 4. 11. - 3xVa2-x2. X 13. 14. V(9 - X2)3 3x(x8 + 4) V(x6 + 10x2 + 3)2 1 8. 2(8x + 3)(4x2 + 3x + l). (x-l)Vx2-l 16. 2(x-6)V(a;- 16.^-^^^ -a)(x-6) V9-X2 ^^ 2.X2 + X-1 Vx2-1 ige 104 (§ 37) J ay-x2 4. ANSWERS 297 d-h ^Q «' 2j (x + l)(3xHx+2) 19. V(a2 + X2)3 Vx2 + 1 1 ^» X8 (X + 1) VX=^ - 1 V (X2 + 9)8 3 x2 + 6 „„ 1 V(x3 + 3)4 ^(1 + xV •\Jy + X — V?/ — X _ x^ _ ^ 2 a^x y^ - ax ' v^Tx + ^v-x ' y'^' y^ g 2xy 5_?^._A. 8.-^;^. x2 + 4a2" ■ Sy' 3y^' x^ 2x^ 3. y . 4x. 16 9. -^■ 2* ' 9?/' 9?/3* xi Sx^y-s x + y{x + y) y + 2, 2y + 4 ^j 2x + ^ . 6«^ * x + 3'(x + 3)2' • x-{-2y' {x-{-2yf' Page 105 (§ 38) 1. 3x-4 2/ + 2 = 0. 2.x- 7?/ + 5 = 0. 3. (- 2, - 1). 4. tan-Uj^. Page 106 (§ 38) TT ^-19 ^0" tan-13. 12. tan-if ; tan-if. 8. -. 9. -; tan ^^. ^^ tan-i|. 13. tan-i^- Page 110 (§ 40) 29 2g 1. x8 = 8 ?/ ; V4 + 9^4. ^ (v^ sin 2 a Uq^ sin2 a^ 2. x = (?/-l)2; V4t2+i. 3. x2-6x + 9?/ = 0; | Vll7- 48 i + 16 ^2. 4. (3, 1). 9. ~ ; uo ; a. 5. {y - 2)2 = (X + 3)3 ; « V4 + 9 ^2. ^9 6. x^ + 2/^ = 2 ; 8 Vl - 2 t + 2 £2. 10- ;^- 7 /V^f^^, ^Ji^in^X 11. y = xi2.na ^ \ 2g 2g }' ^^o 2vn cos2 a Page 112 (§ 41) 1. 3V2ft. per second. 2. 12.5 ft. per second. 3. ^^^ in. per minute. 15 4. Circle; — ft. per second (x = distance of point from wall). X 5. 2.64 ft. per second. Page 113 (§41) 6. 0.13 cm. per second. 7. 0.21 in. per minute. 8. 6.6 ft. per second. 2 X I d 2/ 9. — - — - ft. per second, where x is the distance of top of ladder, and y is 2x+y distance of foot of ladder, from base of pyramid. 298 ANSWERS Page 114 (General Exercises) 20. 31x + 8?/ + 9a = 0. 21. x{~^x + Vi'^y = a^. 28. -: tan- 17. 2 27. Page 36. 37. 38. -; tan-i- 4' 3 tan-if. 39. !r. '' 12 30. tan-i — 5 IT 1 31. -; tan-1- 2' 2 32. -; tan-i- 2 2 34. y 8 ^ 2V(f-^ + l)*+4t2 x'-^+ 4 33. x*+ */^= 1; 3^ 36. 2 a^y = 2abx- gx^ ; 115 (General Exercises) (X + 2)2 = (y - 1)3 ; |V9i + 4. 41. 5.8 mi. per hour; 28.8 mi 20 ft. per second ; 10 VS ft. per second ; (100, 20) v^ = ±^-^; v, = T a a 43. — z===^ ft. per second, where 40. Page 46. 47. 48. 49. 50. 51. Page 67. 60. Page Velocity in path 1 2ct Vax + a'^. 116 (General Exercises) 0.08 ft. per minute. 0.01 in. per minute. 1^ sq. in. per minute. 0.04 in. per second. Length is twice breadth. Other sides equal. 117 (General Exercises) 2.64 in. 68. 8 mi. from point on bank nearest to A bm . , J bn . . ^ a ::=== mi. on land ; — -z=== mi. in water. Vs2 _ 400 s is length of rope from man to boat. 44. 0.06 ft. per minute. 45. -g^ Q ft. per minute. 52. Breadth, 9 in. ; depth, 9V3in. 63. Length = | breadth.' 54. Side of base, 10 ft. ; depth, 5 ft. 55. Depth = ^ side of base. 56. Radius, 3 in. ; height, 3 in. 1 ,^ aVd —-. 59. — — . V2 3 61. 4 II mi. travel on land. Vw2 m" Vw2 63. l^f hr. 118 (General Exercises) 64. V 100 mi. per hour. 3a 65. Velocity in still water = — mi. per hour. 66. Base = aV3 ; altitude = § 6. ge 126 (§ 44) CHAPTER V 1. 15 cos 5 X. 5. sin2x. 2. S. 2sin2 2xcos2x. 6. 7. 5sin2 5xcos3 6x. 5sec2^tan5^. 2 2 4. — 5 sin 10 X. 8. - 3csc33xctn3x. ANSWERS 299 Page 127 (§ 44) 9. sin^ 10. — CSC* 11. X sin - . 2 12. 2 sec X (sec x + tan x)^. Page 129 (§ 45) 13. 2 cos 4 X. 14. 9 tan* 3 X. 15. 2sec2x(sec22x+ tan"'^2x). 16. sin3 2xcos2 2x. 17. — § cos2xcos2 32/. ^. X 18. 2. s = 3 sin Trt 3. TT ; 5. 4. At mean point of motion ; at extreme points of motion. Page 134 (§ 47) Vl-9x2 1 X Vx2 - 1 1 V6 X - x2 3 Vl2x-9x-^ 1 2 + 2 X + x2 1 6. At extreme points of motion ; at mean point of motion. 6. 2 V(s - 8) (5 - s) ; 4(4 -s). 7. -TT. 8. 10; 2 7r. 13. (x-1) Vx'-^ — 2x 7. 8 2x x4 + l 1 9. 10 xv'25x2-l 1 xV4x2-l 2 11 x2+4 1 12. (x+l)Vx2 + 2x 1 X \ X- — 1 14. 2 Vl - xK 16 15. 16. 17. (x2+4)« Vx2-4 X 2a x2 + a2 Vl-X2 18. cos-iVl -x2. Page 136 (§ 48) 1. Va:=± 9.42 ft. per second ; Vy = T 36.46 ft. per second. Page 137 (§ 48) 3. 3 radians per unit of time. Page 138 (§ 49) „ 6 sin ,6 7. — ; = cos- 1 - • a — b cos a 2. 5.3. Page 141 (§ 50) 1. §(x + 3)Vx2 + 3x. 2. 3(axy)i 5V5 5. itVit 6. atp. 300 ANSWERS Page 145 (§ 51) 17. Origin ; ^Vs, 1^ 18. Origin ; ^± a^?, ^j 20.0rigin;(±a,^);(±^,g;(i^_,^). 21. r2sin2^ = 8. 22. r = 4a(cos^+ sin^). 23. r = 2asin^. 24. r2 = a2 cos 2 ^. 19. Origin (^•1) 25. X — a = 0. 26. x^ + 2/2 - 2 ax = 0. 27. x* + x22/2 = a2?/2. 28. (X2 + 2/2)3 = a2(x2- 2/2)2. Page 148 (§ 52) 1. tan-i ;\. Page 149 (General Exercises) 9. 2(l + sec22x). 10. sec*(3x + 2). 11. cos2(2-3x). sec2 (x — 2/) + sec2 (x + y) sec2 (x — y)— sec2 (x + y) 21. - 2. TT- tan-i-§. 3. 0. 12. 1 r 17. -Sin8-. 2 4 18. tan4 2x. 19. ' 13. -3ctn2-csc*^- 5 5 14. - 8csc24x(ctn4x + l). 15. a tanaxsec2ax. 16. 8cos3 2x sin4x cos6x. 1 20. (x + 1) Vx 1 2(x + l)Vx 1 xV49x2-l 4 V2 + X — x2 27. (x- ■ i)V^ -2x 95 2 V3- -4x- 4x2 26. 2- -3x x2 + 4 (x2-l)Vx2-2 Page 150 (General Exercises) 29. fcVa2sin2fci + b^ cos^ kt. 30. 21^ 31. V4I; 7r + 2tan-if. -. 16 V2 V9- s = 3; 2. x2 35. (a2 + 62)i£ 2a6V2 7r3 (7r2 + 1)1 27r 34. 2 a Vs. 33 ^ . . / V6 7r\ / V6 57r\ 46.0ngm;(±-,-);(±-,-j. /^,tan-ilV Vv5 y 47. Origin ; 48. Origin; (^7^, tan-i2V 49. Origin ; (2 a, ^\ ; ^2 a, 5^V Page 151 (General Exercises) _ 2asin2^ 52. (x2 + y^ - 4 a^xy = 0. "■ " cos^ ' 53. (x2 + 2/2)2 + 2 ax(x2 + ?/2) - a22/2 = 0. 61. r=actn^. 54. tan-i§. ANSWERS i. 0; tan-12. 57. '^ 4 1. 0; -; tan-isVs. 58.? 4 301 69. IfiVsft. 60. 72°. 61. V2 ft, 62. At an angle tan-^A; with ground. Page 152 (General Exercises) 63. 12 in. 65. a. 68. 15 sq.ft.; 10.94 sq. ft. per second. 64. sVSft. 67. 6.1 ft. per second. 69. 26.7 mi. per minute. 70. (bsm6 -\ — ) times angular velocity of AB, wliere - angle V Va^-b^sin^e/ CAB. 71. ^^ ~ ^^" + ^-^ ~ ^^^ = 1 ; V9sin^i4-4cos2'f ; where i = (2k-\-l)-. 9 4 ' ^ '2 72. "i _ L = 1 ; 6 sec 3i Vtan-'^ St + 4 sec^ 3 1. 4 16 ' Page 153 (General Exercises) 73. 6 si n 20. 75. tan-i2V2. 74. a Vl + cos2 X ; fastest when 76. tan-i|. x = k7r; most slowly when 77. 0; tan-isVs. x = {2k + l)-. 78. tan-i^; tan-i4V2. 2 79. tan-13; tan-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. ?/ =0.62 a; -0.76. 2. Z=0.0017D. Page 163 (§ 55) 3.^=0.30(2.7)'. 4. c = 0,010(0.84)<. 5. a =0.0000000048 /^.oe. e. pv^-^5 = l0. Page 165 (§ 56) 12. e-2^(3cos3ic-2sin3a;). 1 1 -I ^' x^^ ''• x2-9 13. ctn-ia;. 2. ^(ex_e-a;). g 1 14. 27x2 e8^. 8. 2xa-^-ilna. ' Vi^T4' 15. 5 e^- sin x. 4. „,sip-^x ^"ct g _ 2 6, 2^ + ^ V9x2 + 1 x2 + 4a;_i 10. —4 sec 2 X. 2x + 3 2(e2a:-e-2^) 2 x2 + 6 X + 9 ' ■ e2a: + e-2'f * xVx + 1 e^-f e-^ 17. 2 sec^x. 18. 1 302 ANSWERS Page 167 (§ 57) a: 1. 2/ = 5e2. 2. ?/ = 45.22eO-oi^. 3. ?/-7eO-847^. 4. $739. Page 168 (§ 57) 6. P = 10000 eOo^^9'. 6. c = 0.01 e-o.0446<. 7, 2 min. Page 168 (General Exercises) no. p = 0.018 1 + 24. *11. Load = 192 - 6.6 length. Page 169 (General Exercises) *12. s = 25(0.40)'. *14. t = OA Vz. _ *16. y = OAOx^-^. *13. c = 0.010(0.83)'. *15. 1= 0.023 V^. Page 170 (General Exercises) *17. pu=1620. 21. 2 CSC- 12a;. 26. tan-ix. "•9^4* 22. 2(x+l)e-^-. ^6. 0.898. 19. etna;. no 2 27. 15.8 hr. 2 e2-^- e^^ 28. 1090 sec. e^ + e- ^ 24. a tan^ ax. 29. p = 14. 7 e-o-oooo4 a. Page 171 (General Exercises) 31. 2V2e-2«; 2e-2^ 33. (e^^ + ^)"\ 35. V2e'. 36.—. 8 62^^ 2 3^ (a;2 + l)t 3V3 gg (1 + e^)l X ' 2 ' ' n ' 2e2 CHAPTER VII Page 176 (§ 59) . ^ a;2 a;3 7.2 7.4 ,.6 3 15 318 4.x + L^ + ll?.^,llU.5!+... 2 32-4 6+2.4.6 7 2 2 22^3 23^ 9. 0.0872. 10. 0.4695. * Statement in regard to answers to exercises in § 55 is true of this answer. ANSWERS 308 Page 178 (§ 60) l.ln5 + -^ 2 52 +3^^5^+ • ■ 2 2^6/ 2 ' 2! 2' 3! "■■ 4 2^ ^2 2! 2 3! W2 + -L(x-l)+J__.(^-4..(^+'.... V2 2V2 2! ' 4V2 3! 8. 0.7193. 9. 0.8480. 10. 3.0042. Page 179 (General Exercises) / a;2 , x3 X* \ ^ , 1 „ 1 . 3 ^ 1.3.5. '' -(^+ 2 + 3 +I--7- '' ^-2"' + 2— 4^^-274:6^^+ -•• 2! 4! 6! \ 3 57/ , 1 V3 1 X2 '2*2! ^ 2 3!^ ••• « 1^^^ 1 x2 V3 x8 '•2+ 2^- "2"2! 2 3! * * • 7.1 + 1x2 + ; 1.3 ^ 1.3.5 « 3.4^'+2.4.6^' + ... 10. l-2x + 2x2-2x8+ .... 14. 0.9659. 11. 2x3 4x4 16. 0.5150. 16. 1.6487. 12. 2x2 2x3 4a;5 ^+2! +3! 6.. + -- 17. 0.40546. 18. 0.69315 ; 1.0986. 13. age x^ , x^ , 3x« , ^+2 + 3 + i0 + -- 180 (General Exercises) 19. 0.22314; 20. 0.8473; 1.6094. 1.946. 23. 2.0305. 2.9625. 26,x-| + x6 ^' I 24. 5(2!) 7(3!) 25. ^ x8 x6 a;7 3(3!) 5(5!) 7(7!) 28.X + I 27. a: - ^ + x3 X* 3 4 + •.••. 304 ANSWERS CHAPTER VIII Page 183 (§ 61) 1. 3x2-8 x?/ + 82/2; . y x 0. — — 4 aj2 + 16 x?/ + 15 2/2. xy ~x^' xy — y^ y^-x^y x^-yH « 2 j/2 ^^^ 2x2/ 2x2 2x?/ (^, * a 0« " CO^ , ■ • COS ■ (X2+ 2/2)2' (X2+ 2/2)2 (X + 2/)2 X + ?/ (X + 2/)2 X + y q2/ X i2Lx- jj, • — • 7 py • py a;24.2/2' ^2+2/2 '• ^ ' y1^ ' y X 1 ?/ 4. Vl - x22/2 ' Vl - x22/2 ■ Vx2 + 2/2 ' Vx2+~^ (x + Vx2 + 2/2) Page 185 (§ 62) 1.?!ji1!. 2. -6.sin(x-2/). 3. ^i^!-Il^ . Page 187 (§ 63) 1. 0.000061. 2. 0.0012. 3. 2%. Page 188 (§ 63) 4. 0.018 in. 6. 0.0105. 6. 0.015 in. 7. 6320 ft. 8. 0.0063. Page 191 (§ 64) 1. -z. 2. - f . 4. -\. 5. ^ ; 0. 6. ; 0. Vx2 + 2/2 Page 192 (General Exercises) 8. - 14.33 cu. ft. 10. 0.5655 sq. in. ; 11. 3.9 in. 13. 2.206 sq. in. per second. 9. 1735. 0.5756 sq. in. 12. 0.35 in. 15. 4.4 sq. in. per second. Page 193 (General Exercises) 3V2 ,„ . 1 •» K7 ,C 1 V3 . 16. — . 17. a=tan-i^; 5^. ig, _ cosa —sin a; 1. CHAPTER IX Page 198 (§ 66) 1. 2x3 + 2x2 + 61nx. 7. i(^* + 4)i 13. _lln (1 + cosax). 2. 34^(3x+7)x*. 8. iln{e3^4-6). « 2(x3 + 5) 9. ^ln(2x + sin2x). 14. — |cos4 2x. 5Vx 10. ____J___. 1^- ?Fsin33x. 2_ J__ ' 3(x-sinx)3 16. |sin2(x + 2). ^ 2x2 2i_ -ln(e«^+ tan ax). ,„ , |„ 5. ^[x2 + ln(x2-l)], a ^ ^ 17. -^cos33x. 8. ^(x2 + l)4. 12. ^In(x3 + 3x2+ 1). 18. ^ (3tan3x+tan33x). 19. -^ctn3(2x+ 1). 20. l[cos3(2x-3)- 3cos(2x- 3)]. 3. 4. Inx ANSWERS 305 Page 202 (§ 67) 1. Isin-i?^. 6. JLtan-i^. 10. ^in-il^^ 3 4 V7 V7 2 3 1 . ,xV21 1 . .6x-6 8. - sec- ' 1 r-u^ io-i^ + 2 4. sec- ^x Vs. 3 8. :^tan-i-^i:^. 12. sm - lan ' • — /r 2 2 V5 6 sin-i— — -. 9. sin-i? — ^. 18. -—tan- 1-—^ V3 VlO 3 V2 V2 14. ir31n(x2 + 4)4-lltan-i-l. 15. 5sin-i- - 2 V4 - x2. 2L 2j 2 16.^. 17.-. 18. TT. 19.—. 20.^ 6 4 36 18 Page 204 (§ 68) 1. In (x 4- V.g'-^ +2). 11, 1 In "^ . 2. iln(3x+ V9x2-l). ^ 3x + 5 1 1 2x-3- V5 3. ;^-ln(3x + V9x^-12). ^2. —In-——-^^ 4. ln(x+l + Vx2+2x). 13. L. In ^ ^ "^ ^ ~ ^^ . 1 , /— . ' VSS 2X + 5 + V33 5. -^-ln(3x + l + V9x2 + 6x+9). ^ 4a,_l_Vi3 ^"^ 14. =^ln z=' 6 J_ln?-^^ ^^^^ 4x-l + Vl3 * 20 2x+5' 3 + V5 y 1 ^^ xV2-l 15. In— ^.^ ' 2V2 XV2+1I 16. ^ In (3 + VlO). ^ 1 , 3x-Vl5 ,^ , , 4 + 2V3 In =z' 17. A In 2V15 3x+Vl5 _ 3 + V5 9. 1 In (x2 - 5) + -^ In ^-^! ' 18. -^ In (2 + Vs). ' ^ ^ 2V5 X + V5 V2 ^'-'^%-Tl- 20. fin i. Page 207 (§ 69) 1. _icos(3x-2). 8. -2 CSC?. 2. -|sin(4-2x). 2 8. isec(3x-l). 9. ' ln[sec(4x + 2) + tan(4x + 2)], x 10. ^ctn(3-2x). ^•^^^"4' • 11. ln(cscx-ctnx)+ 2cosx. ^2, 3x 12. i(x-sinx). 5. -In sec — . ^^ . ? 2 18. -(x + 3sin-). 6. ^InsinSx. 2\ 3/ 7. ^ln[csc(2x + 3)-ctn(2x + 3)]. 14. x-cosx. 306 15. 3 1 ( tan — — sec — ) — \ 3 3/ ANSWERS 1 18. V2 cos 2 X. 16. |(x — sinx). ,„ 4V2 . 3x 17. — -— sm-— • 3 4 Page 208 (§ 70) 1. ^e2» + 5. 2. ^e»^. 3. e-4-— -. e+ 1 ga + 6x ga + 6a; * 6(1 + Inc) * Page 212 (§ 72) 19. §. 20. In 4. 21. V3 - 1 5. i(e2x_e-2^) + 2x. 6. e^ + e-*. 7. In(e2^-l)-x. 8. 2(e*-e-^). 99 1, 2 + V3 - In 2 I + V2 23. -. 4 24. 1. 10. 2 In 2 11. lie -I). e2 + l 9. 12. In In 10 2e 1. -x2-4x + 121n(x + 2) + — ^• 2 x+2 2. T«5(3x2 + 4x + 8)Vx-2: 3. ^ f ^ (16x 2 _ 6x - 2 7) v^2x - 3. 4. Vx2 - 1 - tan- 1 Vx2 - 1. V4x2 + 1 X24-8 Vx2 + 4 X • .^ — ===- — sin- 1 - V4 - x2 2 15(3 -x2)t 10. I 4 Vx2 - 4 11.:^. 12 12. 13. Jj(9v'^-10V2). 14. — ?. 15. 22 Page 216 (§ 74) 1. ^(3x-l)e3^. 2. I(2x2-2X + I)e2 a:. 3. X cos- ix — Vl — x2. 4. xtan-i3x- j^ln(l + 9x2) . \ x2 sec- 1 2 X |V4; 6. sin X (In sin X — 1). Page 217 (§ 75) (x-4)7 7. \ (2 cos X + sin x) e2^. 8. ;i (x2 + 2 X sin X + 2 cos x). 9. 4 - 2Ve. 10. ^(81 In 3 -26). 11. i(7r-2). 12. |(7r-2). 1. In (X - 2)6 2. ln(x+3)2V 2x- l. 3.1n(x-2)^|5i. Page 220 (General Exercises) 1. x3 + 2x2 + l + -i^. ■ X 2x2 2. |xl+ |x^ + 3xi 3. ^ xf - I xi 4. In 5. In x2-l (x + 2)2' 2x2 + x x-1 6. ^ln(x-l)(x + 2)2(x-3)3. 4. ia;4+ 43.84.2x2. 5. i(x2-4)2. 6. \ (x8 + 3)1 ANSWERS 307 Page 221 (General Exercises) 7. I (2 + e2^)8. 9. \/3x + x». 8. 1(1 -I- 2x + x*)^. 10. ^ x2 + X + ln(x - 1). 11. ^\- [5 sin8 (2 X - 1) - 3 sinS (2 x - 1) ]. 12. - sin-Zs COS*? + 4 cos^- + 8^ • 3 5\ 5 5 / 13. _gi^ ctn4x(15 + 10ctn24x + 3ctn*4x). 14. yij [3 sec^ (X - 2) - 5 sec^ (x - 2)]. 16. I tan2 (X - 1) + In tan (x - 1). 16. j^ (7 csc5 2 X - 5 csc7 2x ) . 17. } (se c^ 3 X - 7) v^sec 3 x. 18. — Vcsc2x. 19. — ^^ (8 ctn 5 X + 3 ctn^ 5 x) \/ctn5x. 20. ^i3(153-34sin24x + 9 sin* 4 x) \/sin 4 x. 21. - j\ (9 ctn 5 X + 4 ctn^ 5 x) {/ctn^Sx. «« 1 . n2x 22. -sin-i 2 5 «« . t « «^ 1 ,2x + 3 „e 1 . ,xVr6 23. sin-i— ^. 29. -sec-i ^— . 35. -^^ tan-i • V5 5 5 Vl5 3 „. 1 . t 3x 30. sin- 1^^^. ,« 1 , ,x-2 24. -sin-i — -. 5 36. -—tan- 1 — — . 3 2V2 , Ve V6 „, 1 ,3x 31. sin-i «« 1 2x + 2 26. — nsec-i — ^- 2 37. =tan-i = — V5 V5 1 2x-l 2V3 V3 «« 1 ,xV5 32. -sin- 1 „„ 1, 3x + l 26. -sec-i^. 2 2V2 ^^•2^^TTr- o^ 1... 1^ + 1 33. ^sin-i^^±^. 09 2 18X + 5 27. -sec-i-— ;— . ./q ^/q 39' --zz::tan ^ — r— 2 2 V2 V3 ^jl ^jl ««1 ,3x — 1 oiilxi2x 1 3.2 28. — -^sec-i — — -. 34. ^tan-i-^. 40. itan-i-. 3V3 V3 2V5 V5 6 3 Page 222 (General Exercises) 41. _i_tan-i?!^- 47. 2 V2 sin- 1 ^-^ + ? Vq-^^^ . 3V2I 7 3^ 2 1 . x2 - 2 48. In (x + Vx2 - 7). 42. -sin-i 2 2 ' > 49. ^ln(2x + V4x2 + 3). 43. J_sec-i— . 50. -^ln(2x+ V4X2 4-10). 2V6 V6 V2 J j^8 61. |ln(2x2 + V4x*-5). - sec- J — 6 2 44. -sec-i - . 62. ^ In (x^ + Vx« + 7). J 2x2 + 3 63. 2 Vx2 + 4 + ln(x + Vx2 + 4). 46. -sin-i . g^ V3^^ i.ln(3x+V9x2+3). ^^" — V3 46. lln(3x2 + 7) Ltan-i?^^ 66. -l:ln(4x - 3 + 2 V4x2 - «x). 3 ^ ^ ^ V2I 7 V2 308 ANSWERS 56. iln(2x+ 1 + V4x2 + 4x+ 7). ^^ 1 , 3x — V? ^ ^ 69. — —In -. 67. A.ln(5x + 2 + V25x« + 20a: - 5). ^^ ^^+"^7 -^ , 60. -^in-^^-^ 68. ^ln(3x-4 4- v9x2- 24x+ 14). 4V5 xV6 + 2 61. JLln^^^+^ 4 Ve X Ve - 2 _ «« 1 1 /o o -Tx 11 , 3x — V21 62. \ In (3 x2 _ 7) _ in _ . 2V2I 3X+V21 ^^ 1, 2x-3 ^^ 1, x-1 ^„ 1, 5x-2 63. -In 65. —In 67. —In 6 X 12 3x + l _ 15 5x + l ^^ 1, 2x-5 ^^ 1 , 2x-l-\/6 ^„ 1, x-2 64. -In 66. ^In — 68. -In 5 X ^ 4V6 2X-1 + V6 8 3x + 2 1 5x + l-Ve 71. ^(tan3x- ctn3x). * 2V6 6X + 1+V6 72. ln[sec(x-§) + tan(x- §)]. „^ 1, 2x-l 73. -cos2x. 70. -In 5 X + 2 74. In (secx + tanx). Page 223 (General Exercises) „^ 3 3 . ,x X 9 . 2x 92. i e2^ (4x3- 6x2+ 6x - 3). 75. - X Sim - cos sin — . s v / « 4 ^ ^ 1^ ^ 93. ^-(x2 + 4)tan-i^_x. 76. ^tan2x-x. ^^ ' 2 77. ^(sin2x-cos2x). 94. ^^ (2-9a;2)cos3x + | xsin3x. 78. x + 2(ctn^-csc?V 96. x[(ln2x)2_ 21n2x + 2]. \ 2 2/ 96. xln(3x + V9x2-4)-^V9x2-4. 79. tan 2 X - x. 97. ^1^ In (2 x - 1 ) (2x + 3)3. sin X — cos X. 1, x*(x-3) fii Oa/I^ 98. -In 81. 2 V^^ 2 (x + 3)3 82. iver. (x-2)2 83. x-iln(l + e2a:). 99. ln--^==. 84. ^(3x-l)(5x + l)i . (2x + l)(2x + 3)2 86. x\ ( 2x6+x » - 6) ^x8 + 2. 100. - In ^^ ^^^^^ ^ 86. fVx^ + 3. 1 (x2_4){x + .3)2 87. ^x3_^i. In (2x3 + 1). 101. -In J_ . «« 3x2-2 ^ ^ 88. ,^» TT 3(l-x2)l 102. 12 103. 27(4x2+ 9)t ' 12 ^^ (2 x2 + 25) Vx2 - 25 104. yj 1" ff- ^^' 1875^^ 105. ^In4. 91. (X In 5-1) 5-+ 2 ,^^ I + V3 106. (In 5)2 4V3 ANSWERS 309 Page 224 (General Exercises) 107. V3-2 112. 6^?. 118. 5V(l8^2-8V3). V3 * 113. 7f 119. ^ig(3V3-,r). lO®' 0- 114. ^In2. ^°®*!!~f ,,. 1 , 9+4V2 120. i(e^+l). ,,„ e^— 1 115. ^In T^" 4^2 14 121. 2 -In 3. 1 / !I n 116. 3 7r. 122. l tt^. 111. _V53-5V. ,- In 5 117. ^iV3. 123. ^^j(7r+21n2-2). CHAPTER X Page 228 (§ 77) L. 2. { - --\ 2. a2Ve«-e «/. 6. 3 7ra2. o ^ 1. 2. 4. 8^%. 8. pa2. 5. A. 9. -(7r + 4a). 3. 4 7ra2. 7. ^^'^ira^. 2 Page 229 (§ 77) 10. 2 TT^a^. ^2 TToB _ 13. ^1^ T. 15. S^% 1 11. ^|7ra3. '3 ^' 14. 259^ w;. 16. | ttw. Page 232 (§79) 1. 2a2. 0^9 ,; «^^ 7. llTT. „ 3. S7ra2. 6. 59 TT 3 8. 4. 10. -(7r-2) 4n 4. 6. |7ra2. 9. 407r. Page 234 (§ 80) 1. — ; a is radius of semicircle. 2. — ; a is radius of semicircle. 4 TT 3. ^. 4. £^. 6. J.a^. TT TTf/ Page 235 (§ 80) 6. fTra^. 7. 100 revolutions per minute. 8. 5.54 lb. per square inch. Page 236 (§ 81) 1. i(l3VT3-8). 2. -Ve«-e «/. 2 3. 6 a. 4. 6 a. 6. STT^a. 6. V2(l-e"2). 8. 8 a. 310 ANSWERS Page 238 (§ 82) , m - Trka 1. — • St — • a 12 6. 5862 ft.-lb. 6. 2A;ca2; Jc is the con- 2. 22|ft.-lb. 4. 196,350 ft.-lb. stant ratio. 7. .^^mi.-lb.; R is radius of earth in miles. B + a 8. 2 7rC. Page 239 (§ 82) 9. 1.76 ft.-lb. ; 1.56 ft.-lb. Page 239 (General Exercises) 1. 3sin-if. 3. 16- 12 In 3. 2.12, »f; 4.^(3' 2). 6. l§g;15,'V 7. 2Trab. 8. 4a2. 24. + «f. 5.«f. 5 15 - 3 7ra6 9- 4 • 10. f 7ra2. 11. \ irh?' Vk^k^ ; k^ and ATj are the values for k in the equation i/2 = kx. 272 TT 12. 13. ira^tan^. 15 Page 240 (General Exercises) 14. j^t^aK 16. 47ra3(ln4-l). 16. ^^^'^ 5 :. 17. «'"^^^. 5 19. •7ra3(6-81n2); ^(121n2-7). .0.^';i|(2aVB-.2, 21. !^(ll + 3e2). 3 23. 7r2 - 4. Page 241 (General Exercises) 24. 400 IT lb. 28. 12.3 T. 31. 21^. 25. ^^ (8,r + 9 V3) w. gg, ^ . 32. ^ .^ _ 2) 26. 4.41b. n 8 27. 307 lb. 30. - (4 - tt). 33. ^ (3 V3 - tt). 8 6 34. ~(8 7r-h9V3). 35. -(3V3-^). ID i ^ Page 242 (General Exercises) a^-Vs a^ r/2 r/2 37. _(8-F7r); --(57r-8); ^(8-7r). 4 4 ^ • '' 4 ^ ^' 4 38. ^. 40. §. 41. 050. 42. - (In 9 - 1). ANSWERS 311 * 2 8 a3 Page 243 (General Exercises) 3k_ 8a2 49.—. 60. 50,000 ft.-lb. 51. 438.1 ft.-lb. CHAPTER XI Page 245 (§ 83) 1. 36 -In 3. 2. In 3. 3. 2|f. 4. tt-I. Page 246 (§ 83) 6. K'^-SVS). 7^ 7m«; 8^ «(,r-2). 9. I^'. 6. 14. 4 4^ '^ 6 10. ^ (22 - it). 11. Va(7rV2-4). 12.1—?. 9 ' 30 Page 254 (§ 85) 2a a(e*4-4e2-l)\ ^ ^„ _:_* .„„^.„.. 4 a V2 ^ / 2a a(e*4-4e2-l)\ ^ ^ . , , 4aV2. 3. ( , — ^ — — — '). 6. On axis of quadrant, from A / 4a\ 4. (.a, -). center of circle. 7. Intersection of medians. 8. /-, -V 5. On axis, — from vertex. \2 8/ 5 9. (9, 9). 10. On axis, distant from base. 3 7ra + 12 6 k 16. On axis, distant ^ (radius) from base. 17. On axis, distant - from base. a ^ 16. On axis, distant - from base. 18. Middle point of axis. Page 256 (§ 86) (r x !• ) r 2 3. On line of centers, distant ~ ^-— ? from center of circle of radius r,. 4. On axis of shell, distant _L? Li from common base of spherical surfaces. 6. Middle point of axis. ® W " ''1 ) Page 257 (§ 86) 6. On axis, — distant from base. 4ihi-h!) 7. On axis, ^ of distance from vertex to base. 8. mi, 4f I), the outer edges of the square being taken as OJT and OY. 312 ANSWERS 9. On axis, distant 4.8 from corner of square. 10. On axis, distant 3.98 in. from center of cylinder in direction of larger ball. 11. On axis, distant 3.4 ft. from base of pedestal. Page 260 (§ 87) 3. ^ base x altitude. 4. — ; a = altitude and 2 6 = base of segment. 6- ; a = altitude and 2 6 = base of segment. 5 6. 2 7ra26; 87ra6. 7- ^(^ + 3c); 7r[2cd + 26c + 62 + (6 + 2c) Va^ + 62]. 3 8. 7.07 T. 9. Trabcw. 10. ^^^^^ . 4 woh 11. (5 c + 3 a) ; a = altitude and 2 6 = base of segment. 12. cw (area). Page 265 (§ 88) 1. ^\Ma'^. 2. IMa^. 3. IMa^. 4. IMa^- iMb"^, Page 266 (§ 88) ^ ^^ h^{b+3a) 9. ^iW(a2+ 62). 14. liW-(r|+ rf). 6(6 + a) * 10. 145f. 15, § Jlfa2^ a = radius. 6. i^Ma^. 11- H]' 16. 1151 iTT. 7. Ti5JW-(a2+62). 12.^. 3-r5-rf ^^ '^ 4 17. -^M- -. %. IM (r 2 + r 2) . 13. 1 .¥r2. 10 r| - r f Page 268 (§ 89) 1. I JVfa2 ; a = radius. 2.3918^. Page 269 (§ 89) 3. 1608§. 6. 89,980. 4. 45691. 7. lM(ri + Srf). 6. 44,990. 8. ^M(Vi2+3r|). Page 281 (§ 92) 1. 8 7r. 4. ^(37r-4). « 3 7ra3 77ra8 o. — : — • 0. 4 6 8. Tra*. 9. — (37r + 20-16V2). 10 9. lM(r? + 5r2). 10. ^v^'^'- 11. l^Ma'i. e.- 3 7ra4 326 97r "1 4 ANSWERS 313 Page 283 (§ 93) 2. On axis of ring, distant 2 ft. from , g ^. center of shell. 6. /O, 0, — V 3. On axis, distant ^ (^i + ^ ^t^2 + 3 ^2 ) / 6 + 3V2\ 4(ri^ + r,r, + r,0 7. (o, 0, — ^^— j . from upper base. 3a a ((\ (^ _3a2(262-a2) \ 4. On axis, from base. *• ( "' "' "T" I" 8 \ 8[63-(62_a'2)l]/ Page 285 (§ 94) 1. ^^M{a?- + 62), where Jlf is mass, and a and 6 are the lengths of the sides perpendicular to the axis. 2. ^ M. 3. XjW(3a2 + 2/i2). 6. ^3^i|f(a2 + 4/i2). ^^ 2a2(157r - 26) 4. ^Jtf(a2 + 62). 6. ^\JW{3a2+ 4^2). * 25(37r-4) Page 286 (§ 94) 8 11^ 3 9. ^M(a2 + 62). 10. lM{a^-\-b'^). Page 286 ( General Exercises) 1.(0,^). ^ /4(l + 51n2) 73X \ 5 / \ 15 252/ 7_LQ^/^\ « /256 a 256 a\ o 253 2 , ^•\^' 19 / » (^^ ^^« + ^)^ 3. ■"' ^ /4a 4(a4-6) \ /56 \ '^^''' ^'^ ^ ( — , ) ; X = & is the ordinate. ^ /^ 352 \ ^' ^ *• I'' 5(2 + 3 J- /S 3V2\ « /^ & c2\ V5'"i"/' „ ^•i2'2 + 3^j- 4. \5 4 / 2 CL 10. On axis, distant =:- from base of triangle and away from . . , 3(7r + 2V3) semicircle. „ „ . ^- . ,4a24-2a6V3 + 62 11. On axis, distant from base. 2(4a,+ 6V3) Page 287 (General Exercises) 4(a2_52\| 12. On axis of segment, distant i — from center of circle. 3(rf - 2 6 VS^ -"P - 2 «= sin- > J 314 ANSWERS 13. On axis, distant ^]^ ^ ~^ ^ from center of semicircle. 3 (8 6 — Tra) 14. On axis of plate, distant ^V "^/^^^"^^^^ from center of circles 2r 3 7r(r24-rj) 15. On axis, distant — ^ from center of circle. TT 16. On axis of square, 8 in. from bottom. Vs 17. On axis, distant -—in. from center of hexagon. o 18. On axis, distant — from center of ellipse. "^^-^ 96 19. On axis of solid, distant — from smaller base. 16 20. On axis, distant from base j% of distance to top. 21. On axis of segment, distant ^l^ ^' V^j ' h?) - (h,^ - k!)] ofsphere. 4[3aHA, - ..) _ (.3_ .a^j *- center 22. j\Ma^. 23. ^Ma^. Page 288 (General Exercises) 24. -\^M. a^ 25.-\YM. 31. -(15 TT- 32); 34 35 24^ - — 16 5 7ra* 26. ^^^M. a is radius, 27. %^-M. ,„ 5000 625 TT 36. 28. J^^Z-Jkr. ^^--^ 1^ -.--4.. 33^^.^ 47r . iP^_4_4^. ^^'-T' 37. -^(r|-r/). 16 Page 289 (General Exercises) 30 (8-5V2). 42. (^, 0, ?1^V 46. Z??^ ^^'^^^ V 5 128 ; V 35 ' "' ~^ ' 39. -W/^. 43. ||3fa2. 46. ^s^a^. 40. 1 Jlf (4 62 + 3 a2). 32 ^3 ^3 41. ^i^Jtf(362 + 4/i2). **• -^- 47. ^(I05 7r+64V2). dft 3 7ras 256a5V2 ^2" "^ 3465 • ^^- ^''"'- «0. -y-Jlfa2. 51' yj [(2jl + 3 a2) (,| _ «2)f _ (2 r'2 ^ 3 ^2) (,2 _ ^2)f ] . INDEX (The numbers refer to the pages) Abscissa, 28 Acceleration, 9, 21, 135 Algebraic functions, 79 Amplitude, 128 Angle, between curve and radius vector, 146 between curves, 104 between straight lines, 35 vectorial, 142 Angular velocity and acceleration, 135 Anti-sine, 130 Approximations, 53, 187 Arc, differential of, 106, 146 Archimedes, spiral of, 145 Area, as double integral, 246 of ellipse, 225 of plane curve, 47, 225 in polar coordinates, 230 by summation, 60 of surface of revolution, 259 Asymptote, of any curve, 89, 92 of hyperbola, 90 Average value. See Mean value Axis, of coordinates, 28 of ellipse, 86 of hyperbola, 90 of parabola, 82 Cardioid, 145 Cartesian equation, 109 Cartesian space coordinates, 269 Catenary, 157 Center of gravity, of any solid, 282 of circular arc, 253 of composite area, 255 of half a parabolic segment, 251 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, 143 of curvature, 140 Circular measure, 119 Cissoid, 93 Compound-interest law, 166 Cone, circular, 272 elliptic, 275 Constant of integration, 45, 194 Coordinates, 27 cylindrical, 270 polar, 142 space, 269 Curvature, 139 Curves, 91 Cycloid, 137 Cylinder, 273 Cylindrical coordinates, 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, 94 315 316 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 Falling 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 trigonometric 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, 63, 194 Integrals, table of, 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 , sin/i ,„^ of . , 120 h of 1 — cos h 121 h 1 of (1 + A)^ 156 theorems on, 93 Limits of definite integral, 63 Line, straight, 31 Linear velocity, 135 Logarithm, 154 Napierian, 156 INDEX 317 Logarithm, natural, 156 Logarithmic spiral, 168 Maclaurin's series, 173 Maxima and minima, 41 Mean value, 233 Measure, circular, 119 Moment of inertia, 260 of circle, 264 polar, 262 of quadrant of ellipse, 262 of rectangle, 261 of solid, 283 of solid of revolution, 265 Moments of inertia about parallel axes, 266 Motion, in a curve, 107 simple harmonic, 127 Napierian logarithm, 156 Ordinate, 28 Origin, 27, 142 Pappus, theorems of, 259 Parabola, 81, 145 Parabolic segment, 83 Paraboloid, 275 Parallel lines, 33 Parameter, 109 Parametric representation, 109 Partial differentiation, 181 Partial fractions, 216 Parts, integration by, 212 Period, 128 Perpendicular lines, 34 Plane, 276 Polar coordinates, 142 Polar moment of inertia, 262 Pole, 142 Polynomial, derivative of, 18 integral of, 45 Power series, 172 Pressure, 68 theorem on, 257 Projectile, 110 Radian, 119 Radius of curvature, 139 Radius vector, 142 Rate of change, 11, 189 Revolution, solid of, 73 surface of, 259, 273 Roots of an equation, 30 Rose of three leaves, 144 Second derivative, 39 sign of, 40 Segment, parabolic, 83 Series, 172 Maclaurin's, 173 power, 172 Taylor's, 177 Sign of derivative, 20, 40 Simple harmonic motion, 127 Slope, of curve, 36 of straight line, 31 Solid of revolution, 73 Space coordinates, 269 Speed, average, 3 true, 5 Sphere, 271, 272 Spiral, logarithmic, 158 of Archimedes, 145 Straight line, 31 Substitution, integration by, 208 Summation, 66 Surfaces, 271 of revolution, 273 Table of integrals, 217 Tangent line, 38, 104 Taylor's series, 177 Total differential, 185 Trigonometric functions, 119 Trochoid, 138 Turning-point, 37 Value, mean, 233 Vector, radius, 142 Vectorial angle, 142 Velocity, 21, 107 angular, 135 318 INDEX Velocities, related, 111 Vertex, of ellipse, 86 of hyperbola, 90 of parabola, 82 of parabolic segment, 84 Volume, of any solid, 277 of solid with parallel bases, 71 of solid of revolution, 73 Work, 237 /■; 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. 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