THE CALCULUS A SERIES OF MATHATEATICAL TEXTS EDITED BY EARLE RAYMOND HEDRICK THE CALCULUS By ELLERY WILLIAMS DAVIS and WILLIAM CHARLES BRENKE. ANALYTIC GEOMETRY AND ALGEBRA By ALEXANDER ZIWET and LouIS ALLEN HOPKINS. ELEMENTS OF ANALYTIC GEOMETRY By ALEXANDER ZIWET and Louis ALLEN HOPKINS. PLANE AND SPHERICAL TRIGONOMETRY By ALFRED MONROE KENYON and LOUIS INGOLD. ELEMENTS OF PLANE TRIGONOMETRY By ALFRED MONROE KENYON and LOUIs INGOLD. ELEMENTARY MATHEMATICAL ANALYSIS By JOHN WESLEY YOUNG and FRANK MILLETT MORGAN. PLANE TRIGONOMETRY By JOHN WESLEY YOUNG and FRANK MILLETT MORGAN. COLLEGE ALGEBRA By ERNEST BROWN SKINNER. MATHEMATICS FOR STUDENTS OF AGRICULTURE AND GENERAL SCIENCE By ALFRED MONROE KENYON and WILLIAM VERNON LOVITT. MATHEMATICS FOR STUDENTS OF AGRICULTURE By SAMUEL EUGENE RASOR. THE MACMILLAN TABLES Prepared under the direction of EARLE RAYMOND HEDRICK. PLANE AND SOLID GEOMETRY By WALTER BURTON FORD and CHARLES AMMERMAN. CONSTRUCTIVE GEOMETRY Prepared under the direction of EARLE RAYMOND HEDRICK. JUNIOR HIGH SCHOOL MATHEMATICS By WILLIAM LEDLEY VOSBURGH and FREDERICK WILLIAM GENTLEMAN. A BRIEF COURSE IN COLLEGE ALGEBRA By WALTER BURTON FORD. THE CALCULUS BY ELLERY WILLIAMS DAVIS PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF NEBRASKA AND WILLIAM CHARLES BRENKE PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF NEBRASKA REVISED EDITION ettu lork THE MACMILLAN COMPANY 1923 All rights reserved PRINTED IN THE UNITED STATES OF AMERICA COPYRIGHT, 1912 AND 1922 BY THE MACMILLAN COMPANY Set up and electrotyped Revised Edition published September, 1922 PREFACE TO THE FIRST EDITION THE significance of the Calculus, the possibility of applying it in other fields, its usefulness, ought to be kept constantly and vividly before the student during his study of the subject, rather than be deferred to an uncertain future. Not only for students who intend to become engineers, but also for those planning a profound study of other sciences, the usefulness of the Calculus is universally recognized by teachers; it should be consciously realized by the student himself. It is obvious that students interested primarily in mathematics, particularly if they expect to instruct others, should recognize the same fact. To all these, and even to the student who expects only general culture, the use of certain types of applications tends to make the subject more real and tangible, and offers a basis for an interest that is not artificial. Such an interest is necessary to secure proper attention and to insure any real grasp of the essential ideas. For this reason, the attempt is made in this book to present as many and as varied applications of the Calculus as it is possible to do without venturing into technical fields whose subject matter is itself unknown and incomprehensible to the student, and without abandoning an orderly presentation of fundamental principles. The same general tendency has led to the treatment of topics with a view toward bringing out their essential usefulness. Thus the treatment of the logarithmic derivative is V vi PREFACE vitalized by its presentation as the relative rate of change of a quantity; and it is fundamentally connected with the important "compound interest law," which arises in any phenomenon in which the relative rate of increase (logarithmic derivative) is constant. Another instance of the same tendency is the attempt, in the introduction of the precise concept of curvature, to explain the reason for the adoption of this, as opposed to other simpler but cruder measures of bending. These are only instances, of two typical kinds, of the way in which the effort to bring out the usefulness of the subject has influenced the presentation of even the traditional topics. Rigorous forms of demonstration are not insisted upon, especially where the precisely rigorous proofs would be beyond the present grasp of the student. Rather the stress is laid upon the student's certain comprehension of that which is done, and his conviction that the results obtained are both reasonable and useful. At the same time, an effort has been made to avoid those grosser errors and actual misstatements of fact which have often offended the teacher in texts otherwise attractive and teachable. Thus a proof for the formula for differentiating a logarithm is given which lays stress on the very meaning of logarithms; while it is not absolutely rigorous, it is at least just as rigorous as the more traditional proof which makes use of the limit of (1 +1/)n as n becomes infinite, and it is far more convincing and instructive. The proof used for the derivative of the sine of an angle is quite as sound as the more traditional proof (which is also indicated), and makes use of fundamentally useful concrete concepts connected with circular motion. These two proofs again illustrate the tendency to make the subject PREFACE vii. vivid, tangible, and convincing to the student; this tendency will be found to dominate, in so far as it was found possible, every phase of every topic. Many traditional theorems are omitted or reduced in importance. In many cases, such theorems are reproduced in exercises, with a sufficient hint to enable the student to master them. Thus Taylor's Theorem in several variables, for which wide applications are not apparent until further study of mathematics and science, is presented in this manner. On the other hand, many theorems of importance, both from mathematical and scientific grounds, which have been omitted traditionally, are included. Examples of this sort are the brief treatment of simple harmonic motion, the wide application of Cavalieri's theorem and the prismoid formula, other approximation formulas, the theory of least squares (under the head of exercises in maxima and minima), and many other topics. The Exercises throughout are colored by the views expressed above, to bring out the usefulness of the subject and to give tangible concrete meaning to the concepts involved. Yet formal exercises are not at all avoided, nor is this necessary if the student's interest has been secured through conviction of the usefulness of the topics considered. Far more exercises are stated than should be attempted by any one student. This will lend variety, and will make possible the assignment of different problems to different students and to classes in successive years. It is urged that care be taken in selecting from the exercises, since the lists are graded so that certain groups of exercises prepare the student for other groups which follow; but it is unnecessary that all of any group be assigned, and it is urged that in general less than viii * * V111 PREFACE half be used for any one student. Exercises that involve practical applications and others that involve bits of theory to be worked out by the student are of frequent occurrence. These should not be avoided, for they are in tune with the spirit of the whole book; great care has been taken to select these exercises to avoid technical concepts strange to the student or proofs that are too difficult. An effort is made to remove many technical difficulties by the intelligent use of tables. Tables of Integrals and many other useful tables are appended; it is hoped that these will be found usable and helpful. Parts of the book may be omitted without destroying the essential unity of the whole. Thus the rather complete treatment of Differential Equations (of the more elementary types) can be omitted. Even the chapter on Functions of Several Variables can be omitted, at least except for a few paragraphs, without vital harm; and the same may be said of the chapter on Approximations. The omission of entire chapters, of course, would only be contemplated where the pressure of time is unusual; but many paragraphs may be omitted at the discretion of the teacher. Although care has been exercised to secure a consistent order of topics, some teachers may desire to alter it; for example, an earlier introduction of transcendental functions and of portions of the chapter on Approximations may be desired, and is entirely feasible. But it is urged that the comparatively early introduction of Integration as a summation process be retained, since this further impresses the usefulness of the subject, and accustoms the student to the ideas of derivative and integral before his attention is diverted by a variety of formal rules. PREFACE ix Purely destructive criticism and abandonment of coherent arrangement are just as dangerous as ultra-conservatism. This book attempts to preserve the essential features of the Calculus, to give the student a thorough training in mathematical reasoning, to create in him a sure mathematical imagination, and to meet fairly the reasonable demand for enlivening and enriching the subject through applications at the expense of purely formal work that contains no essential principle. E. W. DAVIS, W. C. BRENKE, E. R. HEDRICK, EDITOR. JUNE, 1912. PREFACE TO THE REVISED EDITION THE Davis Calculus was very favorably received by the mathematical world at the time of its original appearance in 1912. The necessity for some revision arose from the usual exhaustion of the old lists of exercises by repeated use of them in class-rooms, and from suggestions of minor changes of forms and of arrangement of the textual matter as a result of actual experience in its use. Professor Davis was intending such a revision at the time of his death, and it has remained for Professor Brenke, in collaboration with Professor E. R. Hedrick, to carry it out. The lists of exercises have been thoroughly revised. Most of the old formal exercises have been replaced by new ones, the lists have been extended or shortened, as experience indicated they should be, and some of the applications formerly contained in the lists of problems have been transferred to the body of the text. The spirit of the original text was to bring out to the student the real significance of the Calculus; and this was accomplished in an unusually effective manner. In the revision, every effort has been made to retain and to amplify this spirit. The technique, and mechanical drill, have not been neglected, but the reasons for learning this technique have been demonstrated to the student unmistakably. Some rearrangement of topics has occurred. Thus integration as a summation has been postponed until after the technique of integration has been mastered. The latter half of the book has been somewhat simplified, and a few more xi xii PREFACE difficult topics that were not reached by many classes have been omitted. It is hoped that the revision will appeal to many and that it will do justice to the great teacher who was the principal author of the original edition. W. C. BRENKE, E. R. HEDRICK. CONTENTS (Page numbers in roman type refer to the body of the book; those in italic type refer to pages of the Tables.) CHAPTER PAGE I. Functions- Slope — Speed....... 1 II. Limits - Derivatives... 14 III. Differentiation of Algebraic Functions.... 24 IV. Implicit Functions- Differentials..... 39 V. Tangents-Extremes..... 49 VI. Successive Derivatives........ 60 VII. Reversal of Rates - Integration..... 79 VIII. Logarithms - Exponential Functions.... 99 IX. Trigonometric Functions....... 119 X. Applications to Curves - Length- Curvature..133 XI. Polar Coordinates......... 147 XII. Technique of Integration....... 156 XIII. Integrals as Limits of Sums....... 192 XIV. Multiple Integrals - Applications..... 208 XV. EmpiricalCurves-Increments-Integrating Devices 234 XVI. Law of the Mean - Taylor's Formula - Series. 247 XVII. Partial Derivatives- Applications.. 274 XVIII. Curved Surfaces -Curves in Space... 289 XIX. Differential Equations........ 311 TABLES TABLE I. Signs and Abbreviations....... 1 II. Standard Formulas......... 3 III. Standard Curves.9....... 19 IV. Standard Integrals.......... 35 V. Numerical Tables.... 51 Index............61 xiii THE CALCULUS CHAPTER I FUNCTIONS - SLOPE - SPEED 1. Variables. Constants. Functions. A quantity which changes is called a variable. The temperature at a given place, the annual rainfall, the speed of a falling body, the distance from the earth to the sun, are variables. A quantity that has a fixed value is called a constant. Ordinary numbers, such as 2/3, \/2, -7,?r, log 5, etc., are constants. If one variable, y, depends on another variable, x, in such a way that y is determined when x is known, y is called a function of x; the variable x is called the independent variable, and y is called the dependent variable. Thus the area A of a square is a function of the side s of the square, since A = s2. The volume of a sphere is a function of the radius. In general, a mathematical expression that involves a variable x is a function of that variable. 2. The Function Notation. A very useful abbreviation for a function of a variable x consists in writing f (x) (read f of x) in place of the given expression. Thus if f(x) = x2 + 3x + 1, we may write f(2) = 22 + 3.2 + 1 = 11, that is, the value of x2 + 3 x +1 when x - 2 is 11. Likewise f(3) = 19, f(- 1) = - 1, f(O) = 1, etc. f(a) = a2+3a+. f(u+v) = (u+v)2+3(u+v)+ 1. Other letters than f are often used-, to avoid confusion, but f is used most often, because it is the initial of the word func1 2 THE CALCULUS [I, ~2 tion. Other letters than x are often used for the variable. In any case, given f (x), to find f (a), simply substitute a for x in the given expression. 3. Graphs. In our study of variables and functions, much use will be made of graphs. To draw the graph of an equation that contains two variables, we may determine by trial several pairs of numbers which satisfy the equation and plot these number pairs as points of the graph, as in elementary analytic geometry. Shorter methods for drawing certain graphs are indicated in some of the following exercises. Thus to draw the graph of the equation y = sin x + cos x, first draw on the same sheet of paper the graphs of the two equations y = sin x and y = cos x, and add the corresponding ordinates. Certain standard graphs are shown in the tables at the back of this book. EXERCISES Calculate the values of each of the following functions for a suitably chosen set of values of x, and draw the graph. Estimate the values of x and f (x) at points where the curve has a highest or a lowest point. Also determine graphically the solutions of the equation f(x) = 0. 1. f(x) -x2 -5x +2. 2. f(x) =x3-2x+4. 3. f(x)= x4-5x3 +3x2-2x +3. 4. f(x) = (x + 1)/(2x- 3). 5. f(x) = logio x. 6. f (x) = (loglo x)2. 7. f(x) = sin x. 8. f(x) = csc x. 9. f(x) = cos x. 1f(x) = sin x + cos x. 11. f(x) - x + sin x. 12. f(x) = sin 2 x. I, ~ 31 FUNCTIONS - SLOPE - SPEED 3 13. Iff (x) = sin x and 'p (x) = cos x, show that [f (X)12 + [0 (X)]2 = 1; f(x)- p(x) = tan x; f(x + y) =f (x) '(y) +f (y) k (x); 4p(x + y) =?; f (x)= (p/2 - x); ' (x) =f(r/2 - x)=-c (r - x); f(-)X= -f( + x); ' X(-X) = W(x). 14. If f(x) = logio x, show that f (x) +f(Y) =f (x.y); f (x) = 2f(x); f(m/n) - f (n/m) = 2 f(m) - 2 f (n); f (m/n) + f (n/m) = 0. 15. If f(x) = tan x, ' (x) = cos x, draw the curves y =f(x), y = ' (x), y = f (x) - ' (x). Mark the points where f (x) = ' (x) and estimate the values of x and y there. 16. Taking f(X) = X2, compare the graph of y = (x) with that of y =f (x) + 1, and with that of y =f (x + 1). 17. Taking any two curves y = f(x), y = ' (x), how -can you most easily draw y =f(X) - 'p(x)? y =f(x) +'p(x)? Draw y = x2 + 1/x. 18. How can you most easily draw y = (x) + 5? y =f (x + 5)? assuming that y = 1(x) is drawn. 19. Draw y = x2 and show how to deduce from it the graph of y = 2 x2; the graph of y = - x2. Assuming that y = 1(x) is drawn, show how to draw the graph of y = 2f(x); that of y = -ff(x). 20. From the graph of y = X2, show how to draw the graph of y = (2 X)2; that of y = X2 ~ 2; that of y = (x + 2)2;that h f oy = (2 x - 3)2. 21. What is the effect upon a curve if, in the equation, x and y are interchanged? Compare the graphs of y = 1(x), x = f(y). Plot each of the following curves: 22. y+1=sin(3x-2). 23. y = 2x + sin x. 24. y=2x+22x. 25. y = 2-x cos x. Plot each of the following curves, using polar coordinates. 26. r =sin 6. 27. r = sin 2O. 28. r = cos 3 O. 29. r = 3 + 2 sinO0. 30. r = 2 + 3 sinO0. 31. r = 1 + cos0. 32. r=O. 33. r = 1/O. 34. r = 20. 35. r = 2-0. 4 THE CALCULUS [I, ~4 4. Rate of Increase. Slope. In the study of any quantity, its rate of increase (or decrease), when some related quantity changes, is a very important consideration. Graphically, the rate of increase of y with respect to x is shown by the rate of increase of the height of a curve. If the curve is very flat, there iT is a small rate of increase; if steep, a large l rate. _ The steepness, or slope, of a curve shows y-xI2 the rate at which the dependent variable is _ r _ 'i increasing with respect to the independent variable. When we speak of the slope of a curve at any point P, we mean the slope __0 1 23 i of its tangent at that point. To find this, we must start, as in analytic geometry, with F _. 1 _. a secant through P. Let the equation of the curve, Fig. 1, be y = x2, and let the point P at which the slope is to be found, be the point (2, 4). Let Q be any other point on the curve, and let Ax represent the difference of the values of x at the two points P and Q.* Then in the figure, OA = 2, AB = Ax, and OB = 2 + Ax. Moreover, since y = x2 at every point, the value of y at Q is BQ = (2 + Ax)2. The slope of the secant PQ is the quotient of the differences Ay and Ax: tan L MPQ = MQ (2+ x)2-4 +Ax. Ax PM Ax The slope m of the tangent at P, that is tan Z MPT, is the * Ax may be regarded as an abbreviation of the phrase, "difference of the x's." The quotient of two such differences is called a difference quotient. Notice particularly that Ax does not mean A X x. Instead of "difference of the x's," the phrases "change in x" and "increment of x" are often used. I, ~ 41 FUNCTIONS - SLOPE - SPEED 5 limit of the slope of the secant as Q approaches P. But it is clear that this limit as Q approaches P is 4, since Ax approaches zero when Q approaches P. Hence the slope m of the curve is 4 at the point P. At any other point the argument would be similar. If the coordinates of P are (a, a2), those of Q would be [(a + Ax), (a + Ax)2]; and the slope of the secant would be the difference quotient Ay + Ax: Ay (a + Ax)2-a2 2 aAxx + a Ax. - - - =- -- -2a+Ax. Ax Ax Ax Hence the slope of the curve at the point (a, a2) is * m = lim Ay,/Ax = lim (2 a + Ax) = 2 a. Ax —)0 Ax —0 On the curve y = x2, the slope at any point is numerically twice the value of x. When the slope can be found, as above, the equation of the tangent at P can be written down at once, by analytic geometry, since the slope m and a point (a, b) on a line determine its equation: y-b = m (x- a). The normal to a curve is defined to be the line through a point on the curve perpendicular to the tangent line at that point. Hence, if the slope of the tangent at a point (a, b) is m, the slope of the normal is - 1/m, and the equation of the normal is 1 - b =- (x - a). Thus, in the preceding example, at the point (2, 4), where we found m=4, the equation of the tangent is (y-4)=4(x-2), or 4x-y =4. The equation of the normal is y-4 = - (x - 2), or x + 4 y=18. * Read "Ax- O" "as Ax approaches zero." A discussion of limits is given in Chapter II. 6 THE CALCULUS [I, ~5 5. General Rules. A part of the preceding work holds true for any curve, and all of the work is at least similar. Thus, for any curve, the slope is m = lim (Ay/Ax); that is, the slope m of the curve is the limit of the difference quotient Ay/Ax. The changes in various examples arise in the calculation of the difference quotient, Ay - Ar, and of its limit, m. This difference quotient is always obtained, as above, by finding the value of y at Q from the value of x at Q, from the equation of the curve, then finding Ay by subtracting from this the value of y at P, and finally forming the difference quotient by dividing Ay by Ax. 6. Slope Negative or Zero. If the slope of the curve is negative, the rate of increase in its height is negative, that is, the height is really decreasing with respect to the independent variable.* If the slope is zero, the tangent to the curve is horizontal. This is what happens ordinarily at a highest point (maximum) or at a lowest point (minimum) on a curve. EXAMPLE 1. Thus the curve y = x2, as we have just seen, has, at any point x = a, a slope m = 2 a. Since m is positive when a is positive, the curve is rising on the right of the origin; since m is negative when a is negative, the curve is falling (that is, its height y decreases as x increases) on the left of the origin. At the origin m = 0; the origin is the lowest point (a minimum) on the curve, because the curve falls as we come toward the origin and rises afterwards. * Increase or decrease in the height is always measured as we go toward the right, i.e. as the independent variable increases. I, ~ 6] FUNCTIONS - SLOPE - SPEED 7 EXAMPLE 2. Find the slope of the curve (1) y = X2 + 3 x-5 at the point where x = - 2; also in general at a point x = a. Use these values to find the equation of the tangent -- at x = 2; the tangent at any point. - _ / When x = - 2, we find y = - 7 (P in i y=X2 + 3r- -5 Fig. 2); taking any second point Q, (- 2 + Ax, - 7 + Ay), its coordinates must satisfy the given equation, therefore -7 + Ay = (-2+ Ax)2 - - +- 3 (- 2 +Ax) - 5, - or __ Ay = —4 Ax + a2 + 3 Ax \ _ _ Ax+ Ax, A I. where Ax2 means the square of Ax. Hence A t-L the slope of the secant PQ is _ Pl A _ Ay/Ax = - 1 + ax. _ _ lT The slope m of the curve is the limit of Ay/Ax _ as Ax approaches zero; i.e. FIG. 2. m= lim Ay= lim (- 1 + Ax) =-1. Ax —0 Ax Ax-o It follows that the equation of the tangent at (- 2, - 7) is (y +7) = - (x +2), or x +y +9 = 0 Likewise, if we take the point P (a, b) in any position on the curve whatsoever, the equation (1) gives b = a2 + 3 a -5. Any second point Q has coordinates (a + Ax, b + Ay) where Ax and Ay are the differences in x and in y, respectively, between P and Q. Since Q also lies on the curve, these coordinates satisfy (1): b + Ay = (a + Ax)2 + 3 (a + Ax) - 5. Subtracting the last equation from the preceding, Ay = 2 a Ax + Ax2 + 3 Ax, whence Ay/Ax = (2 a +3) + Ax, and m = lim A = lim [(2 a + 3) + x] = 2 a + 3. Ax —O Ax Ax-0 Therefore the tangent at (a, b) is y —(a2+ 3 a - 5) = (2 a + 3) (x- a), or (2 a + 3) x-y =a2 + 5. _ ___ S THE - CALCULUS I, ~ 6 The slope m is zero when 2 a + 3 = 0, i.e. when a = - 3/2. Hence the tangent is horizontal at the point where x = - 3/2. EXAMPLE 3. Consider the curve y = X3 - 12 x + 7. If the value of x at any point P is a, the value of y is a3 - 12 a - 7. If the value of x at Q is a + Ax, the value of y at Q is (a+ Ax)3 - 12 (a + Ax) + 7. Hence Ay [(a + Ax)3 - 12 (a + Ax) + 7] -'[a3 - 12 a + 7] Ax Ax = (3 a2 + 3 a Ax +Ax -12, and m = lim = 3 a2 - 12. a-so AX --.I I - - I I - I I 11 I I F I.- - 01.-Z — I i I I I i I I I I I I I I I iI I I -II I ( II I I I I I I I 1 18 T I!k I I I I I - I I I I I I I I tt[t~rfi El M I I I -.1-i -1:1 -i -9 I A PI I t I I{4 1j I I I I I I VI _ _ _ _ _ _ I I I I I I FIG. 3. For example, if x = 1, y = - 4; at this point (1, - 4) the slope is 3.12 - 12 = - 9; and the equation of the tangent is (y.+4) =-9-(x-1), or 9x y-5=0. The points at which the slope is zero would be determined by the equation m = 3 a2 - 12 = 0, which gives a = ~ 2. When x has either of these values, the tangent is horizontal. From the equation of the curve, if x = + 2, y = - 9; when x = - 2, y = 23. Hence the horizontal tangents are at the points (2, - 9) and (- 2, 23). (See Fig. 3.) I, ~ 71 FUNCTIONS - SLOPE - SPEED 9 EXERCISES Find the slope and the equation of the tangent line to each of the following curves at the point indicated. Verify each result by drawing the graph of the curve and the graph of the tangent line from their equations. 1. y =x2-2; (1, —1). 2. y=x2/2; (2, 2). 3. y=2x2-3; (2,5). 4. y=x2-4x+3; (2, -1). 5. y = 3; (1, 1). 6. y =X3 - 9; (2, - 10). 7. y = x3 - 3 x + 4; (0,4). 8. y = 2 x3 - 3 2; (1,- 1). Draw each of the following curves, using for greater accuracy the precise values of x and y for which the tangent is horizontal, and the knowledge of the values of x for which the curve rises or falls. 9. y = 2 + 5 -5. 12. y = 3 - 2. 15. y = x4. 10. y=3x-x2. 13. y =x3-6X. 16. y=x4-2x2. 11. y = 3-3 x + 2. 14. y =3-6 x + 5. 17. y = 4 - x3. 7. Speed. An important case of rate of change is the rate at which a body moves, - its speed. Consider the motion of a body falling from rest under the influence of gravity. During the first second it passes over 16 ft., during the next it passes over 48 ft., during the third over 80 ft. In general, if t is the number of seconds, and s the entire distance it has fallen, s= 16 t2 if the gravitational constant g be taken as 32. The graph of this equation (see Fig. 4) is a parabola with its vertex at the origin. The speed, that is the rate of increase of the space passed over, is the slope of this curve, i.e. As lim - At->O At This may be seen directly in another way. The average speed for an interval of time At is found by dividing the difference between the space passed over at the beginning and at 10 THE CALCULUS [I, ~ 7 the end of that interval of time by the difference in time: i.e. the average speed is the difference quotient As - At. By the speed at a given instant we mean the limit of the average speed over an interval At beginning or ending at that instant as that interval approaches zero, i.e. speed = lim s At-,O At Taking the equation s = 16 t2, if t = 1/2, s = 4 (see point P in Fig. 4). After a lapse of time At, the new [ I_ I I _ I I/ i Ii.values are -— ]-64- - 1-!/ and = 1/2 +At, --- s =16 t.- - / I and s = 16(1/2 + At)2 _ -_ _-_- - - (Q in Fig. 4). Then As = 16 (1/2+ At)2- 4 = 16 At + 16 At2, _ ^ _16 _ __ As/At = 16 + 16 At. ___\^-.-p^ — -- Whence: ~ t AS~s speed = lim A- lim (16 + 16 At) = 16; FIG. 4. At —o that is, the speed at the end of the first half second is 16 ft. per second. Likewise, for any value of t, say t = T, s = 16T2; while for t = T + At, s = 16 (T + At)2; hence As 16 (T + At2 - 16T32 T At average speed = = 32 T + 16 A A t and speed = lim A = 32 T. At-O At Thus, at the end of two seconds, T = 2, and the speed is 32.2 = 64, in feet per second. 8. Component Speeds. Any curve may be regarded as the path of a moving point. If a point P does move along a curve, both x and y are fixed when the time t is fixed. To I, ~ 9] FUNCTIONS - SLOPE - SPEED 11 specify the motion completely, we need equations which give the values of x and y in terms of t. The horizontal speed is the rate of change of x with respect to the time. This may be thought of as the speed of the projection M of P on the x-axis. Likewise, the vertical speed is the rate of change of y with respect to the time. It would be the speed of the projection of P on the y-axis. Hence, by ~ 7, we have horizontal speed = limy A At->o At and P vertical speed = im A (xy) at-+o At Since Ay _Ay. Ax O- M - x Ax At At FIG. 5. it follows that m = (vertical speed) - (horizontal speed); that is, the slope of the curve is the ratio of the rate of increase of y to the rate of increase of x. 9. Continuous Functions. In ~~ 4-8, we have supposed that the curves used were smooth. All of the functions which we have used could be represented by smooth curves. Except perhaps at isolated points, a small change in the value of one coordinate has caused a small change in the value of the other coordinate. Throughout this text, unless the contrary is expressly stated, the functions dealt with will be of the same sort. Such functions are called continuous. (See ~ 10, p. 14.) The curve y = 1/x is continuous except at the point x = 0. The curve y = tan x is continuous except at the 12 THE CALCULUS [I, ~ 9 points x = - 7r/2, + 3 r/2, etc. Such exceptional points occur frequently. We do not discard a curve because of them, but it is understood that any of our results may fail at such points. EXERCISES 1. From the formula s = 16 t2, calculate the values of s when t = 1, 2, 1.1, 1.01, 1.001. From these values calculate the average speed between t = 1 and t = 2; between t = 1 and t = 1.1; between t = 1 and t = 1.01; between t = 1 and t 1.001. Show that these average speeds are successively nearer to the speed at the instant t = 1. 2. Calculate as in Ex. 1 the average speed for smaller and smaller intervals of time after t = 2; and show that these approach the speed at the instant t = 2. 3. A body thrown vertically downwards from any height with an original velocity of 50 ft. per second, passes over in time t (in seconds) a distance s (in feet) given by the equation s = 50 t + 16 t2 (if g = 32, as in ~ 7). Find the speed v at the time t = 1; at the time t = 2; at the time t = 4; at the time t = T. 4. In Ex. 3 calculate the average speeds for smaller and smaller intervals of time after t = 0; and show that they approach the original speed vo = 50. Repeat the calculations for intervals beginning with t = 2. Calculate the speed of a body at the times indicated in the following possible relations between s and t: 5. s = t2; t= 1, 3, 20, T. 7. s = - 16t2 + 80 t; t = 0, 2, 5. 6. s = 16t2 - 50t; t = 0, 2, T. 8. s = t3-6t +4; t =0, 1/2, 1. 9. The relation in Ex. 7 holds (approximately, since g = 32 approximately) for a body thrown upward with an initial speed of 80 ft. per second, where s means the distance from the starting point counted positive upwards. Draw a graph which represents this relation between the values of s and t. In this graph mark the greatest value of s. What is the value of v at that point? Find exact values of s and t for this point. 10. A body thrown horizontally with an original speed of 8 ft. per second falls in a vertical plane curved path so that the values of its hori I, ~ 9] FUNCTIONS - SLOPE - SPEED 13 zontal and its vertical distances from its original position are respectively, x = 8 t, y = 16 t2, where y is measured downwards. Show that the vertical speed is 32 T, and that the horizontal speed is 8, at the instant t = T. Eliminate t to show that the path is the curve 4 y = x2. 11. Find the component speeds and the resultant speed when the path is given by the equations x = t + 1, y t2 - 1. Calculate their numerical values when t = 1; when t = 0; when t = 2. Plot the path. 12. Proceed as in Ex. 11 when the path is given by the equations x = 2 + t, y = 2 -t2. CHAPTER II LIMITS - DERIVATIVES 10. Limits. Infinitesimals. We have been led in what precedes to make use of limits. Thus the tangent to a curve at the point P is defined by saying that its slope is the limit of the slope of a variable secant through P; the speed at a given instant is the limit of the average speed; the difference of the two values of x, ax, was thought of as approaching zero; and so on. To make these concepts clear, the following precise statements are necessary and desirable. When the difference between a variable x and a constant a becomes and remains less, in absolute value,* than any preassigned positive quantity, however small, then a is the limit of the variable x. We also use the expression "x approaches a as a limit," or, more simply, "x approaches a." The symbol for limit is lim; the symbol for approaches is -*; thus we may write lim x = a, or x —a, or lim (a - x) =0, or a- x -0.t When the limit of a variable is zero, the variable is called an infinitesimal. Thus a - x above is an infinitesimal. The difference between any variable and its limit is always an infinitesimal. When a variable x approaches a limit a, any continuous function f (x) approaches the limit f(a): * When dealing with real numbers, absolute value is the value without regard to signs, so that the absolute value of-2 is 2, for example. A convenient symbol for it is two vertical lines; thus 13 —71=4. t The symbol — is often used in place of -I. 14 II, ~ 11] LIMITS- DERIVATIVES thus, if y = f (x) and b =f (a), we may write lim y = b, or lim f (x) = f(a). X ---a [X --->a This condition is the precise definition of continuity at the point x=a. (See ~ 9, p. 11.) 11. Properties of Limits. The following properties of limits will be assumed as self-evident. Some of them have already been used in the articles noted below. THEOREM A. The limit of the sum of two variables is the sum of the limits of the two variables. This is easily extended to the case of more than two variables. (Used in ~~ 4, 6, and 7.) THEOREM B. The limit of the product of two variables is the product of the limits of the variables. (Used in ~~ 4, 6, and 7.) THEOREM C. The limit of the quotient of one variable divided by another is the quotient of the limits of the variables, provided the limit of the divisor is not zero. (Used in ~ 8.) The exceptional case in Theorem C is really the most interesting and important case of all. The exception arises because when zero occurs as a denominator, lhe division cannot be performed. In finding the slope of a curve, we consider lim (Ay/Ax) as Ax approaches zero; notice that this is precisely the case ruled out in Theorem C. Again, the speed is lim (As/At) as At approaches zero. The limit of any such difference quotient is one of these exceptional cases. THEOREM D. The limit of the ratio of two infinitesimals depends upon the law connecting them; otherwise it is quite indeterminate. Of this the student will see many instances; for the Differential Calculus consists of the consideration of just such limits. In fact, the very reason for the existence of the Differ 16 THE CALCULUS [II, ~ 11 ential Calculus is that the exceptional case of Theorem C is important, and cannot be settled in an offhand manner. The thing to be noted here is, that, no matter-how small two quantities may be, their ratio may be either small or large; and that, if the two quantities are variables whose limit is zero, the limit of their ratio may be either finite, zero, or non-existent. In our work with such forms we shall try to substitute an equivalent form whose limit can be found. Theorem D accounts for the case when the numerator as well as the denominator in Theorem C is infinitesimal. There remains the case when the denominator only is infinitesimal. A variable whose reciprocal is infinitesimal is said to become infinite as the reciprocal approaches zero. Thus Y = l/x is a variable whose reciprocal is x. As x approaches zero, y is said to become infinite. Notice, however, that y has no value whatever when x = 0. Likewise y = sec x is a variable whose reciprocal, cos x, is infinitesimal as x approaches 7/2. Hence we say that sec x becomes infinite as x approaches 7r/2. In any case, it is clear that a variable which becomes infinite becomes and remains larger in absolute value than any preassigned positive number, however large. The student should carefully notice that infinity is not a number. When we say that "sec x becomes infinite as x approaches 7r/2," * we do not mean that sec (7r/2) has a value, we merely tell what occurs when x approaches 7r/2. * Or, as is stated in short form in many texts, "sec (7r/2) = oo." II, ~ 11] LIMITS - DERIVATIVES 17 EXERCISES 1. Imagine a point traversing a line-segment in such fashion that it traverses half the segment in the first second, half the remainder in the next second, and so on; always half the remainder in the next following second. Will it ever traverse the entire line? Show that the remainder after t seconds is 1/21, if the total length of the segment is 1. Is this infinitesimal? Why? 2. Show that the distance traversed by the point in Ex. 1 in t seconds is 1/2 + 1/22 +... + 1/2t. Show that this sum is equal to 1 - 1/2t; hence show that its limit is 1. Show that in any case the limit of the distance traversed is the total distance, as t increases indefinitely. 3. Show that the limit of 3 - x2 as x approaches zero is 3. State this result in the symbols used in ~ 10. Draw the graph of y = 3 - x2 and show that y approaches 3 as x approaches zero. Evaluate the following limits: 4. lim (7-5 +3x2). 7. lim 3-2 lO im2 —3+2 -2 4 + 2 2 x-2 X2 + 2 x + 3 2x a + bx 5. lim (7-5x-+3x2). 8. lim 2. 11. lim +x. x->2 x —>o 4 - X ax->l + dx 1 - x a + bx + CX2 6. lim (k2 +kx -2x2). 9. lim l x. 12. lim bxcx2 x-Ok x->2 X x-o m + nx + lx2 If the numerator and denominator of a fraction contain a common factor, that factor may be canceled in finding a limit, since the value of the fraction which we use is not changed. Evaluate before and after canceling a common factor: 13 lim (x -2) (x - 1) 14. lim x- 2) x-,-+(2 x — 3) (x — 1) x-2 (X2 + 1) (x-2),.x2 x2 — 4 x +3 X2 +x - 2 15. lim. 16. lim 7. lim x — 0 X 1 x2 - 17 — 2 x2 + - 3,. _iVaTT + 1 X2 (X - 1)2 Xn 0,n>, 18. lim'X- 19 - 20. lim x ={ 0 ' 1 x->-1 x + 1 x-o0 x3 - 2x2 2x-o0 x 1, n = 1. 21. Show that 3 x2 d-5 lim = 3. X-ooX2 +4 4x+5 [HINT. Divide numerator and denominator by x2; then such terms as 5/x2 approach zero as x becomes infinite.] 18 THE CALCULUS [II, ~ 11 Evaluate: 2 x — 1 52- 4 ax + b 22. lim 2 23. lim 24. lim x —oo3 x - 2 x-oo3 X2 + 2 x —>omx +- n 22 x /1 + x2 VJax 2+ bx + c 25. lim 26. lm im 27. lim + x-*Ao x/1 + x2 oo /2 — 4 x-oo px q 28. Let 0 be the center of a circle of radius r = OB, and let a = Z COB be an angle at the center (Fig. 6). Let BT be perpendicular to OB, and let BF be perpendicZ/ G2 XF cT ular to OC. Show that OF approaches OC as a o 0\<. o S / approaches zero; likewise are CB-O0, arc DB-A0, and FC — 0, as a- 0. 29. In Fig. 6, show that the obvious FIG. 6. geometric inequality FB < arc CB <BT is equivalent to r sin a < r a < r tan a, if a is measured in circular measure. Hence show that a/sin a lies between 1 and 1/cos a, and therefore that lim (a/sin a) = 1 as a-O0. (Cf. ~ 72.) 30. In Fig. 6, show that FB OF BT FC are CB lim = 0; lim- = 1; lim - 0; lim-= 0; lim r- C= 0. a-O r a->o r a-O a —o r a->o r 12. Derivatives. While such illustrations as those in the preceding exercises are interesting and reasonably important, by far the most important cases of the ratio of two infinitesimals are those of the type studied in ~~ 4-8, in which each of the infinitesimals is the difference of two values of a variable, such as Ay/Ax or As/At. Such a difference quotient Ay/Ax, of y with respect to x, evidently represents the average rate, of increase of y with respect to. x in the interval Ax; if x represents time and y distance, then Ay/Ax is the average speed over the interval Ax (~ 7, p. 9); if y =f (x) is thought of as a curve, then Ay/Ax is the slope of a secant or the average rate of rise of the curve in the interval Ax (~ 4, p. 4). II, ~ 13] LIMITS - DERIVATIVES 19 The limit obtained in such cases represents the instantaneous rate of increase of one variable with respect to the other, - this may be the slope of a curve, or the speed of a moving object, or some other rate, depending upon the nature of the problem in which it arises. In general, the limit of the quotient Ay/Ax of two infinitesimal differences is called the derivative of y with respect to x; it is represented by the symbol dy/dx: dy derivative of y with respect to x= lim Ay. dx Ax —,o Ax Henceforth we shall use this new symbol dy/dx or other convenient abbreviations; * but the student must not forget the real meaning: slope, in the case of curve; speed, in the case of motion; some other tangible concept in any new problem which we may undertake. In every case the meaning is the rate of increase of y with respect to x. Any mathematical formulas we obtain will apply in any of these cases. We shall use the letters x and y, the letters s and t, and other suggestive combinations; but any formula written in x and y also holds true, for example, with the letters s and t, or for any other pair of letters. 13. Formula for Derivatives. If we are to find the value of a derivative, as in ~~ 4-7, we must have given one of the variables y as a function of the other x: (1) y=f (x). If we think of (1) as a curve, we may take, as in ~ 4, any * Often read "the x derivative of y." Other names frequently used are differential coefficient and derived function. Other convenient notations are Dxy, yx, f'(x) y', y; the last two are not safe unless it is otherwise clear what the independent variable is. 20 THE CALCULUS [II, ~ 13 point P whose coordinates are x and y, and join it by a secant PQ to any other point Q, whose coordinates are x + Ax, y + Ay. P A FIG. 7. Here x and y represent fixed values I_ of x and y. This will prove more I,, convenient than to use new letters, 9+ each time, as we did in ~~ 4-7. Since P lies on the curve (1), its B coordinates (x, y) satisfy the equation (1), y =f (x). Since Q lies on 0 (1), x + Ax and y + Ay satisfy the same equation; hence we must have (2) y + y = f (x + Ax). Subtracting (1) from (2) we get (3) Ay =f(x + Ax)-f (x). Whence the difference quotient is (4) Ay f(x + Ax)-f(x), average slope over PM, Ax Ax and therefore the derivative is ( dy = lim Ay - lim f (x + Ax)- f () = slope at P.* ) dx Ax-O Ax Ax-o ax 14. Rule for Differentiation. The process of finding a derivative is called differentiation. To apply formula (5) of ~ 13: (A) Find (y + Ay) by substituting (x + Ax) for x in the given function or equation; this gives y + Ay = f(x + Ax). (B) Subtract y from y + Ay; this gives Ay = f (x + x)-f (x). * Instead of slope, read speed in case the problem deals with a motion, as in ~ 7. In general, Ay/Ax is the average rate of increase, and dy/dx is the instantaneous rate. II, ~ 14] LIMITS - DERIVATIVES 21 (C) Divide Ay by Ax to find the difference quotient Ay/Ax; simplify this result. (D) Find the limit of Ay/Ax as Ax approaches zero; this result is the derivative, dy/dx. EXAMPLE 1. Given y = f (x) _ x2, to find dy/dx. (A) f (x + x) = (x + A)2. (B) Ay = f (x + Ax) -f (x) = (x + AX)2 -2 2 x Ax A2. (C) Ay/Ax = (2 x ax + AX). Ax = 2 x + Ax. (D) dy/dx = lim Ay/Ax = lim (2 x + Ax) = 2 X. Ax —0 Ax —0O Compare this work and the answer with the work of ~ 4, p. 5. EXAMPLE 2. Given y =f (x) x3 - 12 x + 7, to find dy/dx. f (x + A) = (x + x)3 - 12 (x + AX) + 7. (B) Ay =(f (x +Ax) - f (x) = 3 X2 Ax + 3 X 2+ 3 -- 12Ax. (C) Ay/Ax = 3 x2 + 3 x AX + A2 - 12. (D) dy/dx = lim Ay/Ax = lim (3 x2 + 3 Ax Ax A2 - 12) = 3 x2 - 12. Ax —0 Ax —0+ Compare this work and the answer with the work of Example 3, ~ 6. EXAMPLE 3. Given y = f (x) 1/x2, to find dy/dx. (A) f (x + x) = (x + AX)2 (B) Ay =f ( +A x) -f () = - - 2 ( + 2Ax (x + aX)2 a2 a2 (a + AP)2 (C7) ~ 2/Ay _ 2 x + Ax (C) (xAx X2 (X + AX)2 Ax x2 (x +Ax)2 (D) dy/dx = lim y = lim [- 2 Ax] A2 = Ax —O AX Ax-OL 2 X( + Ax)2J 4 x3 EXAMPLE 4. Given y =f (x) Vx, to find dy/dx, or df (x)/dx. (A) f (x + Ax) = V + Ax. (B) Ay =f (x +A) -f (x) = x +A axF a V. (C) =Ay V +Ax- Vx Vx+Ax- D-V Vx + aX + V' Ax Ax Ax Vix A+ + VX + Ax + Vx (D) dy lim Q= lm a 1 --- dx Aax-^)oAx Ax-,o VxNl Ax~ V + N x 2\ 22 THE CALCULUS [II, ~ 14 EXAMPLE 5. Given y =f (x) - x7, to find df (x)/dx. (A) f (x + Ax) = (x + Ax)7 = x7 + 7 x6 Ax + (terms with a factor Ax2). (B) Ay = f (x + Ax) - f (x) = 7 X6 Ax+ (terms with a factor Ax2). (C) Ay/Ax = 7 x6 + (terms with a factor Ax). (D) dy/dx = lim Ay/Ax = lim [7 x6 + (terms with a factor Ax)] = 7 x6. SAx-0 Ax —0 EXERCISES Find the derivative, with respect to x, of each of the following functions: 1. x2-4x+3. 5. 8x-x4. 9. x2 2. x3 + 2 x2. 6. x4 +x+2 + 1 0. 0. I - 13x —2 3. 3x-x3. 7 1 11. 3 2 x —1 x + 2 4. x4+2x2. 8. x-3 12. / - 1. 13. Find the equation of the tangent to the curve y = 4/x at the point where x = 3. 14. Determine the values of x for which the curve y = x3 - 12 x + 4 rises, and those for which it falls. Draw the graph accurately. Proceed as in Ex. 14 for each of the following curves: 15. X3 - 15x +3. 17. 2x4 —64x. 16. x3-3 2. 18. X4-32x2. 19. If a body moves so that its horizontal and its vertical distances from a point are, respectively, x = 15 t, y = - 16 t2 + 15 t, find its horizontal speed and its vertical speed. Show that the path is y =-16x2/225+x, and that the slope of this path is the ratio of the vertical speed to the horizontal speed. [These equations represent, approximately, the motion of an object thrown upward at an angle of 45~, with a speed 15/2.] 20. A stone is dropped into still water. The circumference c of the growing circular waves thus made, as a function of the radius r, is c = 2 7rr. Show that dc/dr = 2 r-, i.e. that the circumference changes 2 7r times as fast as the radius. II, ~ 14] LIMITS - DERIVATIVES 23 Let A be the area of the circle. Show that dA/dr = 2 irr; i.e. the rate at which the area is changing compared to the radius is numerically equal to the circumference. 21. Determine the rates of change of the following variables: (a) The surface of a sphere compared with its radius, as the sphere expands. (b) The volume of a cube compared with its edge, as the cube enlarges. (c) The volume of a right circular cone compared with the radius of its base (the height being fixed), as the base spreads out. 22. If a man 6 ft. tall is at a distance x from the base of an arc light 10 ft. high, and if the length of his shadow is s, show that s/6 = x/4, or s = 3 x/2. Find the rate (ds/dx) at which the length s of his shadow increases as compared with his distance x from the lamp base. CHAPTER III DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 15. Classification of Functions. For convenience it is usual to classify functions into certain groups. A function which can be expressed directly in terms of the independent variable x by means of the three elementary operations of multiplication, addition, and subtraction is called a polynomial in x. Thus, 2 x3 + 4 x2 - 7 x + 3, x3 -4 x + 6, etc., are polynomials. The most general polynomial is aox" + a1xn-I +... an,-x + a,, where the coefficients ao, a,,..., a, are constants, and the exponents are positive integers. A function which can be expressed directly in terms of the independent variable x by means of the four elementary operations of multiplication, division, addition, and subtraction, is called a rational function of x. Thus, 1/x, (x3 -3 x)/(2 x+7), etc., are rational. The most general rational function is the quotient of two polynomials, since more than one division can be reduced to a single division by the rules for the combination of fractions. All polynomials are also rational functions. If, besides the four elementary operations, a function requires for its direct expression in the independent variable x at most the extraction of integral roots, it is called a simple algebraic function * of x. Thus, /x, (/2 + - 2)/(3 - x), * Since the expression "algebraic function" is used in the broader sense of ~ 25 in advanced mathematics, we shall call these simple algebraic functions. 24 III, ~ 161 ALGEBRAIC FUNCTIONS 25 etc., are simple algebraic functions. All rational functions are also simple algebraic functions. Simple algebraic functions which are not rational are called irrational functions. A function which is not an algebraic function is called a transcendental function. Thus sinx, log x, ex, tan-l (1 + x) + x2, etc., are transcendental. In this chapter we shall deal only with algebraic functions. 16. Differentiation of Polynomials. We have differentiated a number of polynomials in Chapter I. To simplify the work to a mere matter of routine, we need four rules: The derivative of a constant is zero: [I dc = o. dx The derivative of a constant times a function is equal to the constant times the derivative of the function: ~~[II] ~d(c.u) = c du dx dx The derivative of the sum of two functions is equal to the sum of their derivatives: ~[III] d(u + v) du dv dx dx dx The derivative of a power, xn, with respect to x is nxn-1: [IV] = nxn[We shall prove this at once in the case when n is a positive integer; later we shall prove that it is true also for negative and fractional values of n.] Each of these rules was illustrated in Chapter II, ~ 14. To prove them we use the rule of ~ 14. THE CALCULUS [111, ~ 16 Proof of [I]. If y =c, a change in x produces no change in y; hence Ay = 0. Therefore dy/dx = lim Ay/Ax = lim 0 = 0 as Ax approaches zero. Geometrically, the slope of the curve y = c (a horizontal straight line) is everywhere zero. Proof of [II]. If y = cc eu where u is a function of x, a change Ax in x produces a change Au in u and a change Ay in y; following the rule of ~ 14 we find: (A) Y + Ay = cu (U + Au). (B) Ay = c-Au. Au (C) Ax - dy rAul Au du (D) =lim ic c-lim c ---dx Ax-,O L Ax Ax-*O Ax dX Thus d(7x2)/dx = 7. d(x2)/dx = 72 x = 14 x. (See ~~4, 15.) Proof of [III]. If y = u ~ v, where u and v are functions of X, a change Ax in x produces changes Ay, Au, Av in y, u, v, respectively, hence (A) + Ay = (u + Au) + (v + AV); (B) Ay = Au +AV; (C) Ay Au AV Ax Ax Ax' dy Au Au V du dv (D) = lim -+ lim - - dx aXOAx Ax,oAx dx dx Thus d(X3 -12 x + 7) -d(X3) d(12 x) ~d(7) 3 212~o dx dx dx dx by applying the preceding rules and noticing that dx3/dx = 3 X2. III, ~ 161 ALGEBRAIC FUNCTIONS 27 Proof of [IV]. If y = Xn' we proceed as in Example 5, ~ 14. (A) y + Ay = (X + Ax)n = ~n + nx"n Ax ~ (terms which have a common factor Ax2). (B) Ay = nx"-'Ax+(terms with a common factor Ax). (C) Ay/Ax = nxn-1 + (terms which have a factor Ax). (D) dy/dx = lim (Ay/Ax) = nx"'. Ax-~O This proof holds good only for positive integral values of n. For negative and fractional values of n, see ~ ~ 17, 21. EXAMPLE 1. d (x9)ldx = 9 x8 EXAMPLE 2. dxldx = 1 x0 = 1, since x0 = 1. This is also evident directly: dx/ldx = irn zx/Ax = 1. Notice however that no new rule is necessary. Ax-'0 EXAMPLE 3. d(X4-7X2~3x-5) =4x3-14 x+3. dx dz EXAMPLE 4. d (Axm ~ Bxn + C) = mAxm-lI + nBxn-1. dx EXERCISES Calculate the derivative of each of the following expressions with respect to the independent variable it contains (x or r or s or t or y or it). In this list, the first letters of the alphabet, down to n, inclusive, represent constants. 1. y = 3X4. 5. s= 10t5+ 50 t. 9. q=6t3~3t6. 2. y = X5/5. 6. s=- 4t 2 t3. 10. q =t7 - 8 t5+O 10. 3. y=4 X5+5. 7. s =3(t4 -- 2 t2). 11. q = 6(r3 + r2 + 1). 4. y =5(X6 - 3). 8. s = mt2+ nt3. 12. q = Ar10 + Br5. 13. u =v(v- 1). 1 t4. i = (1 - V2) (2+ V2). 15. u = v6(v5 + 2). 16. u = (v+ 2) (V6 - 3). 17. z = ay5(ys - by4). 18. z = lo(2 jt7 - 5 U9 + 8). 19. z = kxm + hxn. 20. r =CX 3n - dX2m. 21. r = -n(t2n - t2). 22. t = A - By M2. 28 THE CALCULUS [III, ~ 16 Locate the vertex of each of the following parabolas by finding the point at which the slope is zero. 23. y=x2 + 4 x + 4. 25. y=5+2x —x2. 24. y =x2-6x+2. 26. y=ax2+2bx+c. 27. What is the slope of the curve y = 2 x3 - 3 x2 +- 4 at x = 0, ~ 2 + 4? Where is the slope 9/2? - 3/2? Where is the tangent horizontal? Draw the graph. 28. What is the slope of the curve y =x4/4 - 2 x3 + 4 x2 at x =0, 1,- 1, - 2? Where is the tangent horizontal? Where is the slope equal to eight times the value of x? 29. Show that the function 3 - 3 x2 + 3 x - 1 always increases with x. Where is the tangent horizontal? 30. Find the angle between the curve y = x3 and the straight line y = 9 x at each of their points of intersection. 17. Differentiation of Rational Functions. In order to differentiate all rational functions, we need only one more rule, -that for differentiating a fraction. The derivative of a quotient NID of two functions N and D is equal to the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator: [V]d D) dx dx [V] dx D2 To prove this rule, let y = N/D, where N and D are functions of x; then a change Ax in x produces changes Ay, AN, AD in y, N, and D, respectively. Hence, by the rule of ~ 14: N+AN (A) y + Ay=D N+ + + DAD, N+AN N D AN-N.AD (B) AY=D+ADDD - D D(D+ AD); III, ~ 17] ALGEBRAIC FUNCTIONS 29 (C) AN AD D — N Ay _ Ax Ax Ax D(D+AD); dN dD. D -N N dy A.y dx dx d=lim - D2 dx Ax-o Ax D2 (D) (3 x- 7) d2 + 3) -(X2 +3) (3 x - 7) EXAMPLE 1. k — (3-7)2 Wx- \3 x - 7 - (3x- 7)2 (3x-7).2x-(x2+3).3 3x2-14x-9 (3x- 7)2, (3x-7)2 2 dl d(x2) EXA E d /1 W da x 0-2x 2 EXAMPLE2. 2j = = X ' -—, ---'= x Eds X2 (X2)2 X4 X3 (Compare Example 3, ~ 14.) d d l\ O- kxk-1 -k 1 EXAMPLE 3. d (x-k) =d (- ) = 2 -kx Note that formula IV holds also when n is a negative integer, for if n=- k, formula IV gives the result we have just proved. EXERCISES Calculate the derivative of each of the following functions. 1. q1-x. 6. y=2x5(=5). y -- x 2. y =xa-2 3. y = x +2. 3x-4 4. y- - 3 5. y —s x 2 7. y=4x-7+ 2. 3t5+5 t2 9. s't3 1. 10. s = ht5 -kt-5. 30 THE CALCULUS [III, ~ 17 11. V = U2 - U4 + U2 - 12. v U 16 u3 +4 u2- 3 u 13. v = u 3 U2 14. q = ar-rn + br-n. r2 - r3 15. q r3 -r2' 16. W ab4~. bZ2 + C 17. w 2+ 3 18. W 22 + 19. r2 ~ r+ r2-r+1 0 - r + 20. z 2y 3 Y2 - 2 y + 6 2yJ-l- 5 21. x =2 1 y y3 - I Y2 + yJ + 22. q = ~7 4 +4 2r3 - 6 2 r2/7 23. s = (t- 3 + 4)(t-3 - 5). 24. r= a+ b (u+!). 25. u =v V2 - I1 -25.~~~~~v (bm - an) )(at, + b) 2 ax I-f-b 27. y = 2 a2(ax + b)2' 28. y (X:3 + 3 x +~ 1)2' 2.z 1 +- 3 y - 3 Y2 29. a 3(y - 1)3 a;6 30. z X6 31. Compare the slopes of the family of curves y = Xn, where n = 0, + 1, + 2, etc., - 1, - 2, etc., at the common point (1, 1). What is the angle between y = a2 and y = x1'? See Tables, III, A. 18. Derivative of a Product. The following rule is often useful in simplifying differentiations: The derivative of the product of two functions is equal to the first factor times the derivative of the second plus the second factor times the derivative of the first: [VI] d(u ~ v) dv du dx dx dx If y=u - v where u and v are functions of x, a change Ax in x produces changes Ay, Au, Av in y, u, and v, respectively: III, ~ 19] ALGEBRAIC FUNCTIONS 31 (A) y+Ay = (u+ Au) (v +Av); (B) Ay = (u + Au) (v + Av) - u v~ = u Av + v Au + Au A,:; Av (C) Ay/Ax = u(Av/zx) + v(Au/Ax) + Aur-; (D) dy/dx = lim (Ay/Ax) = u(dv/dx) + v(du/dx). AX-*0 EXAMPLE 1. To find the derivative of y = (x2 + 3)(x3 + 4). Method 1. We may perform the indicated multiplication and write: dy d d [(X2 +3)(X3+4)] =d[x5_-3x3+4x2+12] = 5x4+9 2 +8x. dx d dx Method 2. Using the new rule, we write: dy d d d (2+3)(x3+ 4) + (x3 + 4) d (x2 +3) = (2 + 3)3x2 + (3 +4)2x = 5x4+ 9 2 + 8x. In other examples which we shall soon meet, the saving in labor due to the new rule is even greater than in this example. 19. The Derivative of a Function of a Function. Another convenient rule is the following: The derivative of a function of a variable u, which itself is a function of another variable x, is found by multiplying the derivative of the original function with respect to u by the derivative of u with respect to x. dy _ dy du [VII] dx du dx If y is a function of u, and u is a function of x, a change Ax in x produces a change Au in u; that in turn produces a change Ay in y; hence: Ay Ay Au Ax Au Ax Taking limits on both sides, we have formula [VII]. 32 THE CALCULUS [III, ~ 19 EXAMPLE 1. To find the derivative of y = (x2 + 2)3. Method 1. We may expand the cube and write: dy d d dy = d [(x2 + 2)3] =d (x6 + 6 x4 +12 X2 +8) = 6 x5 + 24 x3 + 24 x. Method 2. Using the new rule, we may simplify this work: let u = x2 + 2, then y = (x2+ 2)3 = u3; rule [VI] gives dy dy du d(u3) d(x2 + 2) = 3 u2(2 x) dx du dx du dx = 3(x2 + 2)2. (2 x) = 3(x4 + 4 x2 + 4) (2x) = 6x5 + 24 x3 + 24 x. 20. Parameter Forms. If x and y are given as functions of a third variable t, in the form x =f(t), y = 0(t), the variable t is called a parameter, and the two given equations are called parameter equations. Elimination of t between these equations would give an equation connecting x and y, from which dy/dx might be found. But it is often desirable to find dy/dx without elimination of t. Since y_ Ay Ax Ax At At we have, in the limit as Ax approaches zero, dy dy_ dx [VIIa] dx dt dt This formula is essentially the same as the result of ~ 8, p. 11. If we replace t by y in [VII a], we obtain the following important special case: dy dx [VIIb] dx 1 d dx - dy EXAMPLE. Ify = t2 +2and = 3 t + 4, to finddydx. Method 1. We may solve the equation x = 3 t + 4 for t and substitute this value of t in the first equation; (x-42+, -- x2 8 + 34 III, ~ 20] ALGEBRAIC FUNCTIONS 33 Method 2. we write: dy 2 8 2 8 2t d- = 2-x = (3 t + 4) = - dx9 9 9 9 3 Using the new rule (with letters as used in ~8, p. 11), dy dy.dx d(t2+ 2). d(3t+ 4) 2 dxdtdt2t- +3=- t. dx =d - d-T - d- dt 3 EXERC 1. y = 2x(x2-1). 2. y x2(x3 + 3). 3. y = (2 x - 3) (x + 3). 4. y= ( +x) (1-x+x2). 5. y =(4-x2) (1 + 3). 6. y = (2x-x2) (2-3x -x3). 7. y= (x2- 1)2. 8.' y = (2 2 - 3)2. 9. y = (3 -2)2. 10. y = (x3 + 2)3. 11. y = (3 2 + 5)4. 12. y (5 X3 —7)5. 13. s = (1 X 2 t - 3 t2)2. 14. s = (t2 + 3 t + 7)3. Determine dy/dx in each of the fqo 29. Y= u 3 x - 4. 6 z -4 3 Y= 5 z 31. z =2 -z= 2 - 4 x. ISES 15. 16. 17. 18. 19. s = (t3 - t- 4)3. u = (7-5 v + 2 3)4. u = (v4+ 3 v2 - 2)4. u = (a + by + cv2)5. 1 = (t3 + 1)2 20. y = (t2+t+ 1) X3 + 3 21. y ( + 2)3 -22. y (2 X2 - 5)3 (x3 + 3)4 23. r = (2 s2- 3)-2. 24. r =(1 - s2 +4)-3. 25. u = (2 + 2)3 (3 v - 5)2. 26. u = (3 - 2)2 (2 v- 1)3. 27. s = t(t2 + 3) (t3 + 4). 28. s = (1 -t) (1 - 5 t2) (3-4t3). [lowing pairs of equations: 30. y = 6u2-7u- 1, u = x2 - 1/2. r 1 — } y = — ) 32. 1 1 -- x I-a 34 THE CALCULUS [III, ~ 20 Draw each of the curves represented by the following pairs of parameter equations and determine dy/dx. 33*. { y- t+2 34. { x-2 t + 3 t2, Y 33 t + 2. ~- 2 t [ 4. What is the slope in each case when t = 1? Show this in your graphs. Find the value of the slope in each case at a point where the parameter has the value 2. 35. Draw the graph of the function y = (2 x- 1)2(3 x +4)2. Determine its horizontal tangents. 36. Proceed as in Ex. 35, for the function y = (2 x- 1)2. (3 x + 4)2. 21. Differentiation of Irrational Functions. In order to differentiate irrational expressions, we proceed to prove that the formula for the derivative of a power (Rule [IV]) holds true for all fractional powers: dxn P [IVa] =nxn- n= dx q where n is any positive or negative integer or fraction. The formula has been proved in ~ 17 for the case when n is a negative integer. Suppose next that (1) y = xpl/ where p and q are any positive or negative integers. If we set (2) x = tq, y = tP, which together are equivalent to y = xp/, and apply formula [VIIa], we find: dy = dy. dx =ptl qtq-l = dx dt d pt q but since t = x1q, substitution for t gives dy = p (xl/q)p_- = P Xp/q-1 dx q q This proves [IV] for all fractional values of n. III, ~ 22] ALGEBRAIC FUNCTIONS 35 The rule also holds when n is incommensurable; for example, given y = xv2, it is true that dy/dx = /2 x2-"1; we shall postpone the proof of this until ~ 85, p. 85. 22. Collection of Formulas. Any formula may be combined with [VII], for in any example, any convenient part may be denoted by a new letter, as in ~ 20. For example, Rule [IV] may be written du' dun du dx du. d, by [VII], = nu-,. du by [IV]. dx du dx dx The formulas we have proved are collected below: [I]dc 0. dx d (c ~ u) du [II] =C dx dx d (u ~ v) du dv [III] d(- ) dx + -. Holds for subtraction also. dx dx dX dun du [IV] d nu - du d(N Dd -N dD dC- - dx dx d [V] D = d Special casedx =dx D2 d u2 [VI] d (u v) ud + du dx dx dx [VII] dy _ dy du dx du dx' [y = f(u), u = 4 (x)]. dy dy dx ~[VIIa] dx dt * dt [y - f (t), x = (t)]. [VIIb] dy = dx dx dy These formulas enable us to differentiate any simple algebraic function. 36 THE CALCULUS [III, ~ 23 23. Illustrative Examples of Irrational Functions. In this article the preceding formulas are applied to examples. d' dX112 1 1 EXAMPLE 1. d d 1/2/-1 = -X-1 1 dx dx 2 2 2 Vx (See Ex. 4, p. 21.) EXAMPLE 2. Given y = -V3 x2 +4, to find dyldx. Method 1. Set it = 3 x2 ~4, then y = Vu; by Rule [VII], dy dy du 1 6x 3 xV3X2+4 dx - *= 6x= dr du dx 2V\U 2 2\3X2+4 3X2-j —4 Method 2. Square both sides, and take the derivative of each side of the resulting equation with respect to x. d(y2) d(3X2.+ 4)6 dx dx But by Rule [IV], d(y2) = d(y2). dy2 dy dx dy dx dx' hence, 2 y-L = 6 x ordoy= 3x - 3 x 3 3x\3X2+ 4 dx dx y VX2 3~4 3X+ 4 This method, which is excellent when it can be applied, can be used to give a third proof of the Rule [IV] for fractional powers. The next example is one in which this method cannot be applied directly. EXAMPLE 3. Given y = x3 - 2V3X2+ 4, to find dyldx. y d (X3) d 6 x~13 6xV + 4 _ -x- -2 x V3x~4 = 3 X2 _ dx dx dx 3X2+ 4 EXAMPLE 4. Given y = (X3 -2) V/3x2 — 4, to find dy/dx. dyd d dx - dx 2) W3- (d\/3 d [by Rule VI] 3 x,\13X2 $ 4 = V3x2+4 2 + (X3- 2) 3X2 4 [by Example 2] 3 x2 + 4 =~x~2I 3Lx -i24=A 3X2+ 4Jxa3x2+ 4 III, ~-23] ALGEBRAIC FUNCTIONS 37 x - 1- ~ /x EXAMPLE 5. Given y = - -, to find dy/dx. Vx+1 +-Vx First reduce y to its simplest form: -VT +- - ~V-x V;x _+1 - -V;- 2 x + 1 - 2 Ni2 4 -xa /Vx + + Vx Vx+ -Vx (x + )-x = 2 x + 1- 2 /X2 + X. Then dy d d d 2)- du _= (2 x + 1) - 2 - \/2 + x = 2- 2 -dx dx dx du dx where u = x2 + x; hence dY=2-2 — d=2- 1 (2 -+1). dx 2x/u dx /x2 + x This example may be done also by first applying the rule for the derivative of a fraction [Rule V]; but the work is usually simpler, as in this example, if the given expression is first simplified. EXERCISES Calculate the derivatives of A _ 1 y 43. = 4/3. =2 -x2.. y 2. s = 10t5/2. 4. v = /4. 6. s:.5 lO10 x3 7. yX2 lvX 14. s = 3t-4. 6 2 8. 2- v- 15. v =uV/2-+3 9. s = 3(2 t2/3 + 3 t - 2/3). 16. s = /t2- 1. 10. y = 2 x(xl/3 + x5/3). 17. s = /t2 -3 t. 2 x5/'3 5+3t 11. = -- + 2 18. = — 5 12. s t3(=l2 t2V t7 319. y = 2. 23 319 +X/. = V7X/4 = 7 t - 3/4. U. 13. y = /2 +3ax. 20. y 2 x2+4x. 38 THE CALCULUS [III, ~ 23 21. y = X2V\3 x-4. 27. v- Va2 - u U 22. y = (5 + 3 x)V6 x - 4. 28. y = (9 - 6 x + 5 x2) j(1 +X)2. 23. v =1 X + X2. 29. s = (1 + t2) \1 - t2. 24. s= t+-t 30. y X 5 + 3 t Vi + X ++ V~1-X (First rationalize the denominator.) N/1 + X2 + X 25. y l+/1Vx 31. yN/1 + X2 X 26. s- i - t2 32. ___a2 - X 4/1 + P z\/a2 + X2 Draw the graphs of the equations below, and determine the tangent at the point mentioned in each case. 33. y = 1 -_X2, (X = ). 36. y = V(1I +x) (2 + 3 x), (x = 2). 34. y = V1 + X2, (X =. 37. y = x1 + x, (x=). 35. y =,(x = 2). 38. y = xl/2 - x1/3,(x=1). 39. Find the angle between the curves y = X2/3 and y = X3/2 at (1, 1), 40. Find the angle between the curves y = xP/q and y = xq/P at (1, 1). 41. In compressing air, if no heat escapes, the pressure and volume of the gas are connected by the relation pv'41= const. Find the rate of change of the pressure with respect to the volume, dp/dv. 42. In compressing air, if the temperature of the air is constant, the pressure and the volume are connected by the relation pv = const. Find dp/dv, and compare this result with that of Ex. 41. CHAPTER IV IMPLICIT FUNCTIONS -DIFFERENTIALS 24. Equations in Unsolved Form. An equation in two variables x and y is often given in unsolved form; i.e. neither variable is expressed directly in terms of the other. Thus (1) x2 +y2 = 1 represents a circle of unit radius about the origin. Such an equation often can be solved for one variable in terms of the other; thus (1) gives (2) y = /1 -2, or y = -1-x2. The first solution represents the upper half of the circle, the second the lower half. Now we can find dy/dx as in ~ 23: dy -x dy =-x x (3) or d dx /1 - o' dx N/1-x2' where the first holds true on the upper half, the second on the lower half, of the circle. By Rule [VII] such a derivative may be found directly without solving the equation. From (1) (X2 +y2) = _ 0; dx dx _d dx_ (x2) d(y2) d(y2) dy but d-(X2 + y2) + d 2 x dx by VII; dx dx dx dy dx hence dy (4) 2 x + 2 = 0, 39 39 40 THE CALCULUS [IV, ~ 24 or (5) dy_ x dx y This result agrees with (3), since y = Vi/ - x2. This method is the same as that used in the second solution of Ex. 2, p. 36. It may be used whenever the given equation really has any solution, without actually getting that solution. Such a formula as (4) is much more convenient than (3), since it is more compact, and is stated in one formula instead of in two. But the student must never use (5) for values of x and y without substituting those values in (1) to make sure that the point (x, y) actually lies on the curve; and he must never use (5) when (5) does not give a definite value for dy/dx.* Thus it would be very unwise to use (4) at the point x= 1, y = 2, for that point does not lie on the curve (1); it would be equally unwise to try to substitute x = 1, y = 0, since that would lead to a division by zero, which is impossible. 25. Explicit and Implicit Functions. If one variable y is expressed directly in terms of another variable x, we say that y is an explicit function of x. If, as in ~ 24, the two variables are related to each other by means of an equation which is not solved explicitly for y, then y is called an implicit function of x. Thus, (1) in ~ 24 gives y as an implicit function of x; but either part of (2) gives y as an explicit function of x. Definition. If the original equation is a simple polynomial * These precautions, which are quite easy to remember, are really sufficient to avoid all errors for all curves mentioned in this book, at least provided the equation like (4) [not (5)] is used in its original form, before any cancelation has been performed. IV, ~ 25] IMPLICIT FUNCTIONS —DIFFERENTIALS 41 in x and y equated to zero, any explicit function of x obtained by solving it for y is called an algebraic function. See ~ 15. EXAMPLE. X3 + y3- 3 xy = 0. (Folium of Descartes: Tables, III, 15.) This equation is difficult to solve directly for y. Hence, as in ~ 24, we find dy/dx by Rule [VII]; differentiating both sides with respect to x, we find: 3x2 +3y2y —3y-3xY -=0; dy y -2 whence dx - dx y2 _-x At the point (2/3, 4/3), for example, dy/dx =4/5; hence the equation of the tangent at (2/3, 4/3) is (y - 4/3) = (4/5) (x - 2/3) or 4 x - 5 y + 4 = 0. Verify the fact that the point (2/3, 4/3) really lies on the curve. Note that this formula is useless at the point (0, 0) although that point lies on the curve. EXERCISES In each of these exercises the student should take some point on the curve, and find the equation of the tangent there. 1. From the equation x2y = 1 find dy/dx by the two methods of ~ 24, first solving for y, then without solving for y. Write the result in terms of x and y; and also in terms of x alone, when possible. Find dy/dx in the following examples by the two methods of ~ 24. 2. x3y = 10. 7. X3 +y3 =a3 3. x2 - xy=5. 8. x4-4y2 =4. 4. 2 xy +x + y = 0. 9. X3 +y2- 3 x =0. 5. x2-4y2 =36. 10. (x + y)2-2x=4. 6. X3-y3 =1. 11. y2-x2 + y2 = 0. Find dy/dx in the following examples without solving for y: check the answers when possible by the other method of ~ 24. 12. x2 + 2 xy + y2 = 2. 15. ax2 + 2 hxy + by2 = k. 13. x2y2 —8xy + 7=0. 16. y4-2y2x + x2 =0. 14. ax2 + 2 by +cy 2 dx + 2 ey+f =0. 17. Vx -+ V /y = Va. 18. x3/2 + y3/2 = a3/2. 42 THE CALCULUS [IV, ~ 25 In the following pairs of parameter equations, find dy/dx by ~ 20: when possible eliminate t to find the ordinary equation, and show that the derivative found is correct. Regarding each pair of equations as defining the position of a point (x, y) at the time t, find the horizontal speed and the vertical speed at the time t. (See ~ 8.) Find also the total speed from the relation total speed = V(horizontal speed)2 + (vertical speed)2. 19. x = 48 t' 20 x =- 3 t + 5, 21 = 3 2 + 1, y = [=8t2. y = 4 t —2. y = 2yt3. 3 t t2 - 1 t2 -+ 1 22 l+3 2 = t2 x2 = t2, _ 22. ___+_ 23.. 24. 1 3t2 — 2t 2t Y=l+f'3 Y=t2+l' y=t2_l' 25. On a circle of unit radius about the origin dy/dx = -x/y; this is positive when x and y have different signs, negative when x and y have the same sign. Show that this agrees with the fact that the circle rises in the second and fourth quadrants and falls in the first and third quadrants as x increases. 26. Show that the curve xy = 1 is falling at all its points. 27. Show that the curve x2y = 1 is rising in the second quadrant and falling in the first quadrant. 28. The equation x1/2 + y1/2 = 1 is the equivalent of the equation x2 - 2 xy + y2 2 x - 2 y + 1 = 0, if the radicals x1/2 and y1/2 be taken with both signs. Show that the values of dy/dx calculated from the two equations agree. By methods of analytic geometry, it is easy to see that the curve is a parabola whose axis is the line y = x, with its vertex at (1/4, 1/4). 29. The curve of Ex. 28 is also represented by the parameter equations 4 x = (1 + t)2, 4 y = (1 - t)2. Test this fact by substitution, and show that the value of dy/dx obtained from these equations agrees with the value obtained in Ex. 28. [The curve is most easily drawn from the parameter equations.] If t denotes the time in seconds since a particle moving on this curve passed the point (1/4, 1/4), find the total speed of the particle at any time. IV, ~ 26] IMPLICIT FUNCTIONS-DIFFERENTIALS 43 26. Differentials. Let the curve PQ (Fig. 8) be a part of the graph of the equation y = f(x). Let P be any point (x, y) on the curve, and let Q be a second point (x + Ax, y + Ay) on it. The change Ax = PM in x causes a change Ay = MQ in y. The slope of the tangent PT at the point P is given by the formula Q T (1) m = tana a M lim Ay dy Ax-Ao Ax dx __ 0wr A B If this slope were main- FIG. 8. tained over the interval Ax, the change produced in y would be (2) MK = m Ax. When Ax is small, the change in y, MQ, will usually be nearly equal to MK. In many problems, it is a sufficiently exact approximation to Ay. Moreover, its value may be found readily by (2), whereas the actual calculation of Ay itself might be tedious or impracticable. This quantity MK, which is an approximation to Ay, is called the differential of y, and is denoted by the symbol dy. Hence we may write (3) dy = m Ax = A x. In particular, if the curve is the straight line y = x, we find m = 1; hence the differential of x is (4) dx=1. Ax. If we divide (3) by (4) we find (5) dy - dx= m, where dy +- dx now denotes a real division, since dy and dx 44 THE CALCULUS [IV, ~ 26 are actual quantities defined by the equations (3) and (4), and dx (= Ax) is not zero. Since m stands for the derivative of y with respect to x, it follows that that derivative is equal to the quotient of dy by dx, (6) ldy= dy ' dx; this fact is the reason for our use of the symbol dy/dx to represent a derivative originally. In the figure all quantities here mentioned are shown: dx = Ax = AB, dy = MK, Ay = MQ, = tan, d tan a. Ax dx = tan a. The quantities dx (= Ax), dy (= mAx), Ay, Ay - dy (= KQ), are infinitesimal when Ax approaches zero, i.e. they approach zero as Ax approaches zero. 27. Differential Formulas. For any given function y = f(x), dy can be computed in terms of dx(= Ax), by computing the derivative and multiplying it by dx. Every formula for differentiation can therefore be written as a differential formula; the first six in the list in ~ 22, p. 35, become after multiplication by dx: [I] dc = 0. (The differential of a constant is zero.) [II] d (c. u)= c du. [III] d(u+v) =du+dv. [IV] d (un) = nun-du. VN d () DdN-NdD [VI] d (u - v)= udv + vdu. IV, ~ 271 IMPLICIT FUNCTIONS - DIFFERENTIALS 45 Rules [VII], [VIla], and [VIIb], of ~ 22, p. 35, appear as identities, since the derivatives may actually be used as quotients of the differentials. From the point of view of the differential notation Rule [VII] merely shows that we may use algebraic cancelation in products or quotients which contain differentials. Rules [I]-[VI] are sufficient to express all differentials of simple algebraic functions. A great advantage occurs in the case of equations not in explicit form, since all applications of Rule [VII] reduce to algebraic cancelation of differentials. EXAMPLE 1. Given y = x3 - 12 x + 7, to find dy and m. dy = d (X3 - 12 x + 7) =d (X3) - d (12 x) + d (7) = 3 x2 dx - 12 dx, whence m=dy+ dx= 3 X2 - 12 as in Example 2, p. 21. x2 +3 EXAMPLE 2. Given y 3 to find dy (Example 1, p. 29). 3x-7' (3 d - 7) d (X2 + 3) - (X2 + 3) d (3 x - 7) (3 x -7)2 3 x2 - 14 x - 9d (3 x - 7)2 EXAMPLE 3. Given y = (X2 + 2)3, to find dy (Example 1, p. 32). dy d[(X2 + 2)3] = 3 (X2 + 2)2 d (x2 + 2) 3(X2 + 2)2. 2 x. dx. EXAMPLE 4. Given y = x3 - 2V/3 22 +4, to find dy (Example 3 p. 36). dy =d (x ) - 2 dV- x2+i 4 3x2dx - 2 d (3 X2 + 4) 2V/3 X2 + 4 EXAMPLE 5. Given x2 + y2 = 1, to find dy in terms of dx (~ 24, p. 39). d (X2 + y2) = d (1) = 0; but d (X2 + y2) = d (x2) + d (y2) = 2 xdx + 2 ydy; hence 2 x dx + 2 y dy = 0, or dy = - (xy) dx, orm =dyldx = -x/y. 46 - THE CALCULUS [IVI ~ 27 EXAMPLE 6. To find dy and m when X3 + y3 - 3 xy = 0. (Example, p. 41.) d (X3) +d(y3)- 3 d(xy) O, or 3X2 dx +3 y2dy- 3 xdy- 3 ydxO0, or (x2 -y) dx +(y2 -x) dy 0, whence dy =YXdx, or m= dxy 2- X y2 - Xdx yEXAMPLE 7. To find dy in terms of dx whenx 3 t +4, y= t2 +2. (Example, p. 32.) We find dx= d(3 t +4) = 3dt; dy =d (t2 +2) = 2 tdt; hence m =dy — dx = (2/3) t,or dy =(2/3) tdx; but since t = (x - 4)/3, this may be written: dy =(2/9) (x -4) dx, or m = dy = (2/9) (x -4). dx EXERCISES [These exercises may be used for further drill in differentiation, and for reviews. It is scarcely advisable that all of them should be solved on first reading.] Calculate the differentials of the following express-ions. 1. y = ax2 +bx +c. 15. u=1/V'2v+v.2 2. y = (a2 + X2)2. 16. u-= (I1-2 V2)/(2-V2). 3. y =(ax2+bx+ C)3. 17. u v2 +2 v+1I V3 -1 4. y =(a- bX2)5. 18. - v22 v+3' U V 2 1 5. y = 1/(ax + b). 19. z = 1 /-,/(2 -y34 6. y = 1/(ax + b)2. 20. z = 1/~/(y -. a)3. 7. s= (1+2t) (1-3t). 21. z=y/V1+y. 8. s= (2- 3t) 2(3 +2 t2). 22. z= Va +by/y. 9. = t2 (a -t) 3. 23. r =(a +bs")P'. 10. = t4(3 -2 t3)2 24 =/abV 12. = Vt - 1. 26. r =l1/<Ya +b4. ~~~3/(2t2+1\2 v/ a + bx IV, ~ 271 IMPLICIT FUNCTIONS - DIFFERENTIALS 47 Determine dy in terms of dx from the equations below. 29. xy - y = 4. 33. (I - ax) (X2 + y2) = 4. 30. x2- 2xy - 3y2 = 0. 34. x2 ~ y2= (ax + b)2. 31 + Y2 2 y2 a ~ bx 3 _ =X. 35.- x2 - y2 X2 a - bx 32. y4- 2y2x -l = 0. 36. (X + y)3'2 + (X - y)3/2-a3/2. 37. Obtain the equation of the tangent at (2, - 1) to the curve 4 X2 -2 xy - 5 y2 -6 x -4 y -7 = 0. 38. Obtain the equation of the tangent at (2, 1) to the curve x3'- 7 X2y - 5 Y3 - 4 X2 - 10 xy + 8 x - 5 y + 18 = 0. Obtain the equation of the tangent at (xo, Yo) to each of the following curves: CURVE 39. y2 = 4 ax; YY( 40. x2 + y2 = XX X2 2 L2XX( 41. a2 _2 - 2-a Find the derivative dy/dx for the curves pairs of parameter equations given below. _t 2 X = X, - %/2 0), 42. 1+t' 1 t I 2 lt 2t2 TANGENT ) =2 a(x+xo). ) + YYo =.a2 )+YY_0=1 -1 defined by each of the F = 3a - 2t at 44. 4 (a - t)3 "r = 2t2 1 = 4 7rr2 Y = 4 7rr3 iiY = 0-1 + 0-2. x4 { 4rr22 y -r 48. Calculate the x and y components of the speed (vx and Vy) at any time t, and the resultant speed VvV2 + Vy2, for the motion 2 t 1 - t2 49. If a particle moves so that its coordinates in terms of time are x = 1 - t + t2, y = 1 + t + t2, show that its path is a parabola. Show that from the moment t = 0 its speed steadily increases. 48 THE CALCULUS [IV, ~ 27 50. The electrical resistance of a platinum wire varies with the temperature, according to the equation R = Ro (1 -a + b02)-l; calculate d R in terms of do. What is the meaning of d R/dO? 51. Van der Waal's equation giving the relation between the pressure and volume of a gas at constant temperature is (P+ a)(v-b)=c. Draw the graph when a =.0087, b =.0023, c = 1.1. Express dv in terms of dp. What is the meaning of dv/dp? 52. The crushing strength of a hollow cast iron column of length 1, inner diameter d, and outer diameter D, is T = 46.65 D3.55 - d3.55) r= 46.65 — )11 -Calculate the rate of change of T with respect to D, d, and 1, when each of these alone varies. CHAPTER V TANGENTS - EXTREMES 28. Tangents and Normals. We have seen in ~ 4, p. 4, that if the equation of a curve C is given in explicit form: (1) Y =f (x), the derivative at any point P on C represents the rate of rise, or slope, of C at P: (2) [IdYlt = [slope of Cat p= slope of PT = tan a = mp, where a is the angle XHT, counted from the positive direction of the X-axis to the tangent PT, and where mp denotes the slope of C at P. Hence (~ 4, p. 5) the equation of the tangent is (3) (y - dyp) x p(X-x), where the subscript P indicates that' the quantity affected is taken with the value which it has at P. y T 9 x FIG. 9. 0 If the slope mp is positive, the curve is rising at P; if mp is negative, the curve is falling; if mp is zero, the tangent is horizontal (~ 6, p. 6). Points where the slope has any desired value can be found by setting the derivative equal to the given number, and solving the resulting equation for x. Since, by analytic geometry, the slope n of the normal PN is the negative reciprocal of the slope of the tangent, we have, (4) np = slope of PN =- - - - -[yd]' mp [dy/dx]p 49 50 THE CALCULUS [V, ~ 28 hence the equation of the normal is: (5) ( -[YP) d (x - X) [dy/dx]p 29. Tangents and Normals for Curves not in Explicit Form. The equation of the curve may be given in the implicit form (1) F (x, y) = 0, as in ~~ 24-25, pp. 39-41; or the equations in parameter form may be given: (2) x=f (t),y = (t), as in ~ 20, p. 32. In either of these cases, dy/dx can be found, and this value may be used in the formulas of ~ 28. No new formulas are necessary. 30. Secondary Quantities. In Fig. 9, ~ 28, since tan a (= mp = [dy/dx]p), and AP (= yp), are supposed to be known, the right triangles HAP and PAN can both be solved by trigonometry, and the lengths HA, AN, HP, PN can be found in terms of mp and yp: [Subtangent]p = HA = AP - tan a = yp - mp = [y/m]p. [Subnormal]p = AN = AP tan a = [y m]p, since a = Z APN. [Length of tangent]p = HP = V/AP2 + HA2 = ^/y2 + [y/m]2 = [y /1 + (1/m)2]p. [Length of normal]p = PN = /AP + AN2 = /y2 + (y. m)2 = [y + m2]p. It is usual to give these lengths the names indicated above; and to calculate the numerical magnitudes without regard to signs, unless the contrary is explicitly stated. V, ~ 311 TANGENTS - EXTREMES 51 31. Illustrative Examples. In this article, a few typical examples are solved. EXAMPLE 1. Given the curve y = 3 - 12 x + 7 (Ex. 2, p. 21), we have m = dy/dx = 3 x2- 12. (1) The tangent (T) and the normal (N) at a point where x = a are (T) y- (a3- 12a +7) = (3 a2- 12) (x- a), — 1 (N) y - (a3 -12a +7) = a2-12 ( - a); thus, at x = 3, the tangent and normal are (T) y+2 =15(x- 3), (N) y+2=- - (x-3). (2) The tangent has a given slope k at points where 3x2- 12 =k, i.e.x = ~+ k+12; there are always two points where the slope is the same, if k > - 12; thus if k =0, x = 2; if k = -9, x = + 1; if k =- 12, x = 0; if k < - 12, no real value for x exists (see Fig. 15, p. 68). (3) The secondary quantities of ~ 30 may be calculated without using the formulas of ~ 30. Thus, at the point where x = 3, the tangent (T) cuts the x-axis where x = 47/15; the normal (N) cuts the x-axis where x =- 27. If the student will draw a figure showing these points and lines, he will observe directly that the subtangent is 2/15, the subnormal 30, the length of the tangent /22 + (2/15)2, the length of the normal '/302 + 22. These values agree with those given by ~ 30. EXAMPLE 2. Given the circle x2 + y2 = 1, we have m = dy/dx = - x/y [see ~ 24]. (1) The tangent (T) and normal (N) at a point (xo, yo) are (T) (y - yo) =-z0 (x - xo), (N) (y - yo) - = Y (x - xo); Yo xo or, since xo2 + y02 = 1, (T) xxo + yyo = 1, (N) yxo - yox; thus, at the point (3/5, 4/5), which lies on the circle, we have (T) 3x +4y =5, (N) 3 y = 4x. (2) The tangent has a given slope k at points where -~ = k, i.e. xo + kyo = 0. Yo 52 THE CALCULUS [V, ~ 31 The coordinates (xo, yo) can be. found by solving this equation simultaneously with the equation of the circle, or by actually drawing the line xo + kyo = 0. Thus the points where the slope is + 1 lie on the straight line x +y = 0; hence, solving x +y = 0 and x2 +y2 = 1, the coordinates are found to be x = + 1/V2, y = T 1/V/2; but these points are most readily located in a figure by actually drawing the line x +y = 0. EXERCISES Find the equation of the tangent and that of the normal, and find the four quantities defined in ~ 30, for each of the following curves at the point indicated: 1. y =x3-12 x +7; (1,- 4). 5. x=y -3y2 + 5; (3, 1). 2. y=3x_2, (- 1 3) 6. y= (1+t)2; (1, 1). 3. 9 x-y2+ = 25; (1,4). 7 x = t2+4t-1{ (t 1 4. xy+y2- 2x =5; (- 4, 1). ly=t3-3t+5f' ( 8. The curves of Exs. 1 and 3 pass through the point (1, - 4); at what angle do they cross? Determine the equation of the tangent and that of the normal to each of the following curves at any point (xo, Yo) on it. 9. y = kx2. 13. b2x2 ~ a2y2 = a2b2. 10. y2 = 2 px. 14. ax2 + 2 bxy + cy2 = f. 11. 2 + y2=a2. 15. ax2 + 2 bxy+ cy + 2 dx + 2 ey+f=0. 12. y = kx3. 16. y = (ax + b)/(cx + d). 32. Extremes. In ~ 6, and in numerous examples, we have found points on a curve at which the tangent is horizontal, i.e. at which the slope is zero. If the slope of the curve y = f (x) is zero at the point where x = a, the curve may go through the point in any one of the ways illustrated in Fig. 10. In case (a), f (a) is called a maximum of f (x). In case (b), f (a) is called a minimum of f (x). V, ~ 32] TANGENTS - EXTREMES 53 m i naximum) The value of f (x) at a point where x = a is a minimum value if it is less than any other value of f(x) for values of x sufficiently near to x = a. A maximum or a minimum is called an extreme value, or an extreme of f (x). A value of x for which the slope m is zero is called a critical value. The corresponding point on the curve is Y Y fx 7.Tr I!if/) fia) /) if(a) ifa) a- x a x a x a x (a) (b ) () (d) FIG. 10. called a critical point. At such a point, f(x) may be a maximum or a minimum, but it is not necessarily either. Thus, in cases (c) and (d), Fig. 10, the value off (a) is neither a maximum nor a minimum of f (x). On the other hand, extremes may also occur at points where the derivative has no meaning, or at points where the function becomes meaningless. Thus, the curve y = 2/3 gives Y m = 2/(3 x1/3): hence m is meaning- y=x less when x = 0; in fact, the curve has a vertical tangent at that point. > It is easy to see that this is, however, F x the lowest point on the curve. Again, if a duplicating apparatus costs $150, and if the running expenses are ic. per sheet, the total costs of printing n sheets is t = 150 +.01 n. This equation represents a straight line; geometrically there are no extreme values of t; but practically t is a minimum when n = 0, since negative values of n are meaningless. Such cases are usually easy to observe. 54 THE CALCULUS [V, ~ 33 33. Fundamental Theorem. We proceed to show that a function f (x) cannot have an extreme except at a critical point; that is, assuming that f (x) and its derivative have definite meanings at x = a and everywhere near x = a, no extreme can occur if the derivative is not zero at x = a. We are supposing that all our functions are continuous; if, then, the derivative m is positive at x = a, it cannot suddenly become negative or zero. Hence m is positive on both sides of x = a, and there can be no extreme there. Likewise if m is negative, the curve is falling near x = a on both sides of x = a; there can be no extreme. 34. Final Tests. It is not certain that f (x) has an extreme value at a critical point. To decide the matter, we proceed to determine whether the curve rises or falls to the left and to the right of the critical point: it rises if m > 0; it falls if m < 0. (See Fig. 10 in ~ 32.) Near a maximum, the curve rises on the left and falls on the right. Near a minimum, the curve falls on the left and rises on the right. If the curve rises on both sides, or falls on both sides, of the critical point, there is no extreme at that point. 35. Illustrative Examples. EXAMPLE 1. To find the extreme values of the function y = f (x) = x3 - 12x + 7. (SeeEx. 3, p. 8.) (A) To find the Critical Values. Set the derivative equal to zero and solve for x: m = dy = 3 x2- 12; 3 x2- 12 = 0; x = 2 or x = -2. dx (B) Precautions. Notice that f(x) and its derivative each have a meaning for every value of x; hence x = + 2 and x = - 2 are the only critical values. V, ~ 351 TANGENTS - EXTREMES 55 (C) Final Tests. m = 3 2 - 12 = 3 (x2 - than 2, negative if x is slightly less than 2; hence the curve rises on the right and falls on the left of x = 2, therefore f (2) = - 9 is a minimum of f (x). The student may show that f(- 2) = 23 is a maximum of f(x). (See Fig. 3, p. 8.) 4) is positive if x is greater __ I I I L A I~~~~~~, ___. I I I I~~~~~~~~~~~~~~~~~~~~~, EXAMPLE 2. To find the extremes of the function y =f (x) =3 x4-12x3 + 50. (A) Critical Values. Setting dy/dx = 0, -2 and solving, we find: m = =12 x3-36 2; 12 3 -36x2 = 0; O dx y- - 0x\'teo x = 0, or x =3. _ 1 (B) Precautions. y and dy/dx have a _ 2 meaning everywhere; the only critical - --- values are 0 and 3. - _ _ (C) Final Tests. Near x = 0, m = 12 x2 (x - 3) is negative on both sides; _____ hence there is no extreme there, though ______ the tangent is horizontal. FIG. 12. Near x = 3, m = 12 x2(x - 3) is positive on the right, negative on the left; hence f(3) = - 31 is a minimum. The information given above is of great assistance in accurate drawing. EXAMPLE 3. Two railroad tracks cross at right angles; on one of them an eastbound train going 15 mi. per hour clears the crossing one minute before the engine of a southbound train running ylJl/~ ^^at 20 mi. per hour reaches the crossing. D Find when the trains were closest together. Let x and y be the distance in miles of the rear end of the first train and the en~-6' ---- x '- gine of the second one from the crossing, FIG 13. respectively, at a time t measured in minutes beginning with the instant the first train clears the crossing; then 15 20 1 2 25 x=-t, y=- (1-t), D2=x2 + y2 = - t where D is the distance between the trains in miles. where D is the distance between the trains in miles. 56 THE CALCULUS[ [V, ~ 35 Since D is a positive quantity, it is a minimum whenever D2 is a minimum; hence we write: d(D2) 2 25 2 25 t 16 dt 9 9+72t 9+72t=; 25' when t < 16/25, m < 0; if t> 16/25, m> 0; hence D2 is diminishing before t = 16/25 and increasing afterwards. It follows that D is a minimum when t = 16/25. Substituting this value for t, we find the values x = 4/25, y = 3/25, D2 = 1/25; hence the minimum distance between the trains is 1/5 of a mile, and this occurs 16/25 of a minute after the first train clears the crossing. EXAMPLE 4. To find the most economical shape for a pan with a square bottom and vertical sides, if it is to hold 4 cu. ft. Let x be the length of one side of the base, and let h be the height. Let V be the volume and A the total area. Then V = hx2 = 4, whence h = 4/x2; and A =x2+4hx =x2 + 1 whence we find dA 16 16 m=- =2x --- 2x —~-=o x3=8 x=2. dx x2' x2 8, =2. When x < 2, m = 2(x3 - 8)/x2 is negative; when x> 2, m is positive; hence A is decreasing when x is increasing toward 2, and A is increasing as x is increasing past 2; therefore x = 2 gives the minimum total area A = 12. Notice that the height is h =4/x2 =1. The correct dimensions are x = 2, h = 1 (in feet). EXERCISES Determine the maximum and minimum values of the following functions and draw the graphs, choosing suitable scales. 1. y = x3 - 6 2 + 2. 2. s= 2 t3 - 6 t2 - 18 t + 15. 3. p =q3 - 6 q2-15 q. 4. y X3 + 2 ax2+ a2x. 5. x=y4-4y2 +2. 6. =U4- 4u3+4u2 +3. 7. m =n5-5n4+ 5 n3+ 1. 8. A =r6-6r4+4r3 +9r2-12r+4. 9. s = (2 t-1) (1 - t)2. 10. V=h(h —1)2. 11. r = (s2 - 1) (2-4). 12. x =(y - 2)3 (y + 3)3. (x + 2)2 13. y = (X +1)2 a2 14. v=u —. u V, ~ 35] TANGENTS- EXTREMES 57 x-2 h 15. Y= y 2 16. K= x2 - 2 1 + 4K ah2 + bh + c x3 - X 17. Y x4 -x2+ 18. Q = k + 1-k. 19. D=r/8 -r2. 20. R= 7x-+6-x. 21. What is the largest rectangular area that can be inclosed by a line 80 feet long? 22. What must be the ratio of the sides of a right triangle to make its area a maximum, if the hypothenuse is constant? 23. Determine two possible numbers whose product is a maximum if the sum of their squares is 98. Is there any minimum? 24. Determine two numbers whose product is 100 and such that the sum of their squares is a minimum. Is there any maximum? Did you account for negative possible values of the two numbers? 25. What are the most economical proportions for a cylindrical can? Is there any most extravagant type? Mention other considerations which affect the actual design of a tomato can. Is an ordinary flour barrel this shape? What considerations enter in making a barrel? 26. What are the most economical proportions for a cylindrical pint cup? (1 pint = 287 cu. in.) Mention considerations of design. i 27. Determine the best proportions for a square tank with vertical sides, without a top. Is there any most extravagant shape? 28. The strength of a rectangular beam varies as the product of the breadth by the square of the depth. What is the form of the strongest beam that can be cut from a given circular log? Mention some other practical considerations which affect actual sawing of timber. 29. The stiffness of a rectangular beam varies as the product of the breadth by the cube of the depth. What are the dimensions of the stiffest beam that can be cut from a circular log? 30. Is a beam of the commercial size 3" X 8" stronger (or stiffer) than the size 2" X 12" (1) when on edge, (2) when lying flat? [Commercial sizes of lumber are always a little short.] 31. What line through the point (3, 4) will form the smallest triangle with the coordinate axes? Is there any other minimum? maximum? 58 THE CALCULUS [V, ~ 35 32. Determine the shortest distance from the point (0, 3) to a point on the hyperbola x2 - y2 = 16. Show that it lies on the normal. [HINT. Use the square of the distance.] 33. The distance D from the point (2, 0) to any point of the circle x2 + y2 = 1 is given by the equation D2 = 5 - 4 x. Discover the maximum and minimum values of D2, and show why the rule fails. 34. Show that the maximum and minimum on the cubic y = x3 - ax + b are at equal distances from the y-axis. Compute y at these points. 35. Show that the cubic x3 - ax + b = 0 has three real roots if the extreme values of the left-hand side (Ex. 34) have different signs. Express this condition algebraically by an inequality which states that the product of the two extreme values is negative. [Any cubic can be reduced to this form by the substitution x = x' + k; hence this test may be applied to any cubic.] 36. Show that if the equation x3 - ax + b = 0 has two real roots, the derivative of the left-hand side (i.e. 3 x2 - a) must vanish somewhere between the two roots. Show that the converse is not true. 37. The line y = mx passes through the origin for any value of m. The points (1, 2.4), (3, 7.6), (10, 25) do not lie on any one such line: the values of y found from the equation y = mx at x = 1, 3, 10 are m, 3 m, 10 m; the differences between these and the given values of y are (m - 2.4), (3 m- 7.6), (10 m - 25). It is usual to assume that that line for which the sum of the squares of these differences S = (m - 2.4)2 + (3 m - 7.6)2 + (10 m - 25)2 is least is the best compromise. Show that this would give m = 2.50 (nearly). Draw the figure. 38. In an experiment on an iron rod the amount of stretching s (in thousandths of an inch) and the pull p (in hundreds of pounds) were found to be (p = 5, s = 4), (p =10, s = 8), (p =20, s = 17). Find the best compromise value for m in the equation s = m p, under the assumption of Ex. 37. Ans. About 5/6. 39. A city's bids for laying cement sidewalks of uniform width and specifications are as follows: Job No. 1: length = 250 ft., cost, $110; Job No. 2: length, 600 ft., cost, $250; Job No. 3: 1500 ft., cost, $630. Find the price per foot for such walks, under the assumption of Ex. 37. How much does this differ from the arithmetic average of the price per foot in the three separate jobs? V, ~ 351 TANGENTS - EXTREMES 59 40. The amount of water in a standpipe reaches 2000 gal. in 250 sec., 5000 gal. in 610 sec. From this information (which may be slightly faulty) find the rate at which water was flowing into the tank, under assumption of Ex. 37. 41. The values 1 in. = 2.5 cm., 1 ft. = 30.5 cm. are frequently quoted, but they do not agree precisely. The number of centimeters c, and the number of inches i, in a given length are surely connected by an equation of the form c = ki. Show that the assumptions of Ex. 37 give k = 2.541. Is this the same as the average of the values in the two cases? Which result is more accurate? 42. In experiments on the velocity of sound, it was found that sound travels 1 mi. in 5 sec., 3 mi. in 14.5 sec. These measurements do not agree precisely. Show that the compromise of Ex. 37 gives the velocity of sound 1088 ft. per second. How does this compare with the average of the two velocities found in the separate experiments? 43. A quantity of water which at 0~ C. occupies a volume vo, at 0~ C. occupies a volume v = v0 (1- 10-4 X.57580 + 10-5 X.756 02 - 10-7 X.351 03). Show that the volume is least (density greatest), at 4~ C. (nearly). 44. Determine the rectangle of greatest perimeter that can be inscribed in a given circle. Is there any minimum? 45. What is the largest rectangle that can be inscribed in an isosceles triangle? Is there any minimum? 46. Find the area of the largest rectangle that can be inscribed in a segment of the parabola y2 = 4 ax cut off by the line x = h. 47. Determine the cylinder of greatest volume that can be inscribed in a given sphere. Is there also a minimum? 48. Determine the cylinder of greatest convex surface that can be inscribed in a sphere. Is there a minimum? 49. Determine the cylinder of greatest total surface (including the area of the bases) that can be inscribed in a given sphere. 50. What is the volume of the largest cone that can be inscribed in a given sphere? 51. What is the area of the maximum rectangle that can be inscribed in the ellipse x2/a2 + y2/b2 = 1? CHAPTER VI SUCCESSIVE DERIVATIVES 36. Time-rates. In all the applications, derivatives are rates of increase (or decrease) of some quantity with respect to some other quantity which is taken as the standard of comparison, or independent variable. Among all rates, those which occur most frequently are time-rates, that is, rate of change of a quantity with respect to the time. 37. Speed. Thus the speed of a moving body is the timerate of increase of the distance it has traveled: (1) v = speed* = lim As = ds At —O At dt' as in ~ 7, p. 9, and in numerous examples. 38. Tangential Acceleration. The speed itself may change; the time-rate of change of speed is called the acceleration along the path, or the tangential accelerationt. Av dv (2). jT = tangential acceleration t = lim - - dt At->O At dt * The speed v is distinguished from the velocity V by the fact that the speed does not depend on the direction; when we speak of velocity we shall always denote it by v (in black-faced type) and we shall specify the direction. t The general acceleration j is also a directed quantity; when we speak of the acceleration j (not tangential acceleration j,) we shall denote it by j, and give its direction. As in the case of speed, the letter j, in italic type, denotes the value of j without its direction. 60 VI, ~ 39] SUCCESSIVE DERIVATIVES 61 Thus for a body falling from rest, if g represents the gravitational constant, s = i gt2; hence ds v = =gt, and dv JT = =; it follows that the tangential acceleration of a body falling from rest is constant; that constant is precisely the gravitational constant g.* In obtaining the tangential acceleration, we actually differentiate the distance s twice, once to get v, and again to get dv/dt or jr, hence the tangential acceleration is also said to be the second derivative of the distance s passed over. 39. Second Derivatives, Flexion. It often happens, as in ~ 38, that we wish to differentiate a function twice. In any case, given y = f (x), the slope of the graph is dy. Ay m = lim dx ax-o \x The slope itself may change (and it always does except on a straight line); the rate of change of the slope with respect to x will be called the flexion t of the curve: dm. b = flexion= = lim - 2 dx Ax —o Ax and will be denoted by b, the initial letter of the word bend. Thus for y = x2, we find m = 2 x, b = 2; t * The value of g is approximately 32.2 ft. per second per second=981 cm. per second per second. t The word curvature is used in a somewhat different sense. See ~ 86, p. 139. t The flexion for this parabola is constant; note that this means the rate of change of m per unit increase in x, not per unit increase in length along the curve. 62 THE CALCULUS [VI, ~ 39 for y =x3, m = 3 x2, b = 6 x; for y =x3-12x+7, m=3x2-12, b=6x; for any straight line y = kx + c, m = k, b = 0. The value of b is obtained by differentiating the given function twice; the result is called a second derivative, and is represented by the symbol: d2y d (d\ dm= b dx2 dx \dx dx Likewise, the tangential acceleration in a motion is d2s d (ds dv dt2 dt dtdt) T' If the relation between s and t is represented graphically, the speed is represented by the slope, the tangential acceleration by the flexion, of the graph. Thus if s= gt2/2 be represented graphically, as in Fig. 4, p. 10, the slope of the (t, s) curve is ds m = slope = d = gt = speed = v, and the flexion of the (t, s) curve is dm d2s dv b = flexion = -dt 2 = dt =g = tangential acceleration =jT. 40. Speed and Acceleration. Parameter Forms. Let the equations x=f(t), y=4(t) represent the position of a moving point P in terms of time t as variable. Then, as in ~ 8, dx (1) Vx = = horizontal component of the speed of P; (2) v, =d = vertical component of the speed of P; (2 d~=at= VI, ~ 401 SUCCESSIVE DERIVATIVES 63 (3) v = \/vx2 + Vy2 = total speed of P, in the direction tan-' (vy/Vx). Further, d2X (4) jx = - = horizontal acceleration of P; d2y (5) jy = 2 = vertical acceleration of P; (6) j = Vj/2 + j2 = total acceleration of P, in the direction tan-1 (jy/jx). Note that the direction of v is along the tangent to the path of P, since Vy/xv is equal to dy/dx, or m. But the total acceleration j is not, in general, in the direction of the path of P, since jy/jx is ordinarily quite different from vy/v,. Hence j and jT are usually different.* To get jT, calculate dv/dt from (3). EXAMPLE. Let the parametric equations of the path be x = t2, y =l/t2(= t-2). To plot the path take a series of values of t: | | t = O,, 1, 3/2, 2,...; x = 0, 4, 1, 9/4, 4,...; y = o, 4, 1, 4/9, 4,... u = 2 t, ji = 2, vy = -2 t-3, jy =6 tH, v = 2/t2 + t-6. j = 2/1 + 9 t-8. Id 2i- 6 i-i ~ ~ I I 12x dv 2 t -6 t ([IV], 22.) FIG. 14. T dt t2+ t F 14. * The reason for this difference is not difficult: jT is the acceleration in the path itself; j is the total acceleration, part of its effect being precisely to make the path curved; hence, a part of j is expended not to increase the speed, but to change the direction of the speed, i.e. to bend the path. Notice that Ex. 33, p. 65, represents a straight line path; on it j, = j; this holds only on straight line paths. In uniform motion on a, circle, for example, T =0. 64 THE CALCULUS [VI, ~ 40 EXERCISES [In addition to this list, the second derivatives of some of the functions in the preceding exercises may be calculated.] Calculate the first and second derivatives in the following exercises. Interpret these exercises geometrically, and also as problems in motion, with s and t in place of y and x. 1. y=x2+5x-4. 2. y = -x2 +4x' 4. 3.y = 22-x- 15. 4.y = - 5x2- x-15. 5. y=x2 —ax-21. 6. y= — x3 + 3x2+1. 7. y 2 3 3 x2 - 36 x -20. 8. y -x4+8x2+2. 9. y = 4 - 2 x3 + 5 x2 + 2. 10. y= (1 +x) (1 -x). 11. y = V + Vx2 + 1. 12. y = (2-3 x)2 (3 + x). 13. y= (x +2)3 (x2-1). 14. y = V/1 - + +/1- x. 15. y = ax + b. 16. y = c (a constant). 17. y = ax2 + bx + c. 18. y =c (x - a). 19. y = (x- a)m (x - b). 20. y = Ax-k. 21. Show that the flexion of a straight line is everywhere zero. 22. Show that if the distance passed over by a body is proportional to the time the tangential acceleration is zero. What is the speed? 23. Show that the flexion of the curve y = ax2 + bx + c is everywhere the same, and equal to twice the coefficient of x2. 24. Show that if the space-time equation is s = at2 + bt + c, the acceleration is always the same and equal to twice the coefficient of t2. Is such a motion at all liable to occur in nature? 25. Find the flexion of the curve y = 1/x. Show that it resembles y itself in some ways. Does the slope also resemble y? 26. Can you interpret Ex. 25 as a motion problem? What is true at the beginning of the motion (t = 0)? Can a curve with a vertical asymptote represent a motion? Can a piece of such a curve? 27. Find the flexion of the curve y = (x - 2)3 (x + 3)2 (x - 4). Show that the flexion has a factor (x - 2), while the slope has a factor (x -2)2 (x + 3). VI, ~ 41] SUCCESSIVE DERIVATIVES 65 28. Show that the flexion of the curve y. (x-a)3 (x2 + 3) has a factor (x + a). 29. If the function y =f (x), where f (x) is a polynomial, has a factor (x - a)3, show that dy/dx has a factor (x - a)2, and d2y/dx2 has a factor (x - a). 30. If the equation x5 + ax4 + bx3 + CX2 + d +- e = 0 has a triple root x = a, show that the equation 20 x3 + 12 ax2 + 6 bx + 2 c =0 has a factor x - a. 31. Show how to find the double and triple roots of any algebraic equation by the Highest Common Divisor process. 32. If the equations of the curve in parameter form are x = t3, y = t2, find the slope m and the flexion b in terms of t. HINT. dy dy dy d dm dm dx -INT. m d d ' - ' dx dt dt ddx dt *dt For each of the following curves, proceed as in Ex. 32. Calculate also the values, in terms of t, of each of the six quantities mentioned in ~ 40, and the value of jT. Compare j and jT. 33. x = a bt, y = c + dt. 34. x = t2, y = t. 35. x =t, y = t-2; t = 1 and2. 36. x = 1 +t, = y t= +2. t 1 -t 3t 3 t2 37. x = y --;t = 1. 38. +t3Y;t3 = 1. 41. Concavity. Points of Inflexion. If the flexion b dm/dx is positive, the slope is increasing, and the curve turns upwards, or is concave upwards; if the flexion is negative, the slope is decreasing, and the curve is concave downwards. Thus y = x2 is concave upwards everywhere, since b = 2 is positive. For y = x3 we find b = 6 x, which is positive when x is positive, and negative when x is negative; hence y = x3 is concave upwards at the right, and concave downwards at the left of the origin. A point at which the curve changes from being concave upwards to being concave downwards, or conversely, is called a point of inflexion. The value of the flexion b changes from positive to negative, 66 THE CALCULUS [VI, ~ 41 or conversely, in passing such a point; hence, the value of b at a point of inflexion is zero, if it has any value there.* Thus the origin is a point of inflexion on the curve y = x3, for the curve is concave downwards on the left, concave upwards on the right, of the origin. 42. Second Test for Extremes. In seeking the extreme values of a function y = f(x), we find first the critical points, i.e. the points at which the tangent is horizontal. If, at a critical point, b = d2y/dx2 > 0, the curve is also concave upwards,t and the function has a minimum there; if b < 0, the curve is concave downwards, andf (x) has a maximum; that is, m dy - nd j. ^ f>d2yl> 0 \ minimum if m - = and b = l atx= a,f(a)isa ma um dx dx < o maximumj Whenever the flexion is not zero at a critical point, this method usually furnishes an easy final test for extremes. If the flexion is zero, no conclusion can be drawn directly by this method.t (See, however, ~ 34.) 43. Illustrative Examples. EXAMPLE 1. Consider the function y = x3 - 12 x + 7. See Ex. 3, p. 8, and Ex. 1, p. 54. The slope and the flexion are, respectively, dy d2y dm m = 3 X2 1'2 b = _ d = 6 x. dx -12' dx2 dx * Points where the tangent is vertical, for example, may be points of inflexion. t The curve is then also concave upwards on both sides of the point; if the curve is concave upwards on one side and downwards on the other, b must be zero if it exists at the point. t Even in this case one may decide by determining whether the curve is concave upwards or downwards on both sides of the point; but the method of ~ 34 is usually superior. VI, ~ 431 SUCCESSIVE DERIVATIVES 67 The critical points are x = ~ 2. Since 6x is positive when x is positive, b is positive for x > 0; likewise b < 0 when x < 0. Hence the curve is concave upwards when x > 0, and concave downwards when x < 0. At x = + 2, b > 0, hence by ~ 42, y has a minimum at x = + 2; at x = -2, b < 0, hence y has a maximum (compare p. 8 and p. 54). To find a point of inflexion first set b = 0; b= d - 6 x = i.e. x =0. dx dx2 Since dm/dx is negative for x < 0 and positive for x > 0, the given curve is concave downwards on the left and concave upwards on the right of this point; hence x = 0, y = 7 is a point of inflexion. (See Fig. 15, and ~ 44, p. 68.) EXAMPLE 2. Consider the function y = 3 x4 - 12 x3 - 50 (Ex. 2, p. 55). The slope and the flexion are, respectively, dy dm d2y m = 12 3-36 2; b = d = 36 X2-72x. dx dx dx2 The critical points are x = 0, x = 3. At x = 3, b = 108 > 0, hence y is a minimum there. At x = 0, b = 0, and no conclusion is reached by this method (compare, however, p. 55). To find points of inflexion, first set b = 0; b = dm dy= 36 x2- 72 x = 0, i.e. x = 0 or x = 2. dx dx2 Near x = 0, at the left, dm/dx = 36 x(x - 2) is positive; at the right, negative; the given curve is concave upwards on the left, downwards on the right, and (x = 2, y = 2) is a point of inflexion. (See Fig. 13.) EXAMPLE 3. For a body thrown vertically upwards, the distance s from the earth is: s = - gt2 + Vot, where vo is the speed with which it is thrown. The speed and the tangential acceleration are, respectively, ds d2s dv V = = -- gt + vO; jT = dt2- d= —=- dtv= d=t2 t+vo;JT= =dt= If we draw a graph of the values of s and t, the speed v (slope of the graph) is zero when v = -gt + vo = 0, i.e. t = vo/g, 68 THE CALCULUS [VI, ~ 43 that is, the point is a critical point on the graph. The tangential acceleration (flexion of the graph) is negative everywhere, hence the graph is concave downwards. In particular, at the critical point just found, b is negative; hence s has a maximum there: 1 1 V 02 = gt2 + ot = 0when t - Fig. 17 is drawn for the special values o = 64 and g = 32. Fig. 17 is drawn for the special values vo - 64 and g = 32. 44. Derived Curves. - 1 -l- l h12 7 -i 'I I -I, YI I I FIG. 15. It is very instructive to draw in the same figure graphs which give the values of the original function, its derivative, and its second derivative. These graphs of the derivatives are called the derived curves; they represent the slope (or speed in case of a motion) and the flexion (or tangential acceleration). The figures for the curves of Exs. 1 and 2 of ~ 43 are appended. The student should show that each statement made in ~ 43 and each statement made on p. 67, for each of the examples, is illustrated and verified in these figures. The similar curves for space, speed, and acceleration are drawn in Fig. 17, for the motion of a body thrown upwards: s = - i gt2 + vot for g = 32, vo = 64. Verify the statements made in Ex. 3, ~ 43. In drawing such curves, the second derivative should be drawn first of all; the information it gives should be used in drawing the graph of the first derivative, which in turn should be used in drawing the graph of the original function. VI, ~ 44] SUCCESSIVE DERIVATIVES 69 FIG. 16. I I r!i FIG. 17. 70 THE CALCULUS [VI, ~ 44 EXERCISES 1. Draw, in the order just indicated, the first and second derived curves in Ex. 1, p. 56, and show that each step of your work in that example is exhibited by these figures. 2. Draw the derived curves for Exs. 2, 4, 6, 14, p. 56; and show their connection with your previous work. 3. Draw the original and the derived curves for the function y = x3 - 6 x2 - 15 x - 6. Find the extreme values of y, and explain the figures. For what value of x is the flexion zero? Does this give a point of inflexion on the original curve? Find the extreme values of y and the points of inflexion on the following curves; in each case draw complete figures: 4. y = 2 x3- 3 x2 - 72 x. 9. y=Ax2 + Bx +C. 5. y = 4x3 6x2 -24x. 10. y = mx +n. 6. y = 3 +2. 11. y =. 7. y=x4 —6x2-40. 12. y= x3 - px q. 8. y = x(x + 2)3. 13. y = x2 - 16/x. 14. Show that the flexion of the hyperbola xy = a2 varies inversely as the cube of the abscissa x. 15. Show that the flexion of the conic Ax2 + By2 = 1 (ellipse or hyperbola) varies inversely as the cube of the ordinate y. 16. What is the effect upon the flexion of changing the sign of a in the equation y = ax2 + bx + c? 17. A beam of uniform depth is said to be of "uniform strength' (in resisting a given load) if the actual shape of its upper surface under the load is of the form y = ax2 + bx + c, where x and y represent horizontal and vertical distances measured from the middle point of the beam's surface in its original (unbent) position. Show that the flexion of such a beam is constant. 18. Show that the addition of a constant to the value of y does not affect the slope or the flexion. 19. Show that the addition of a term of the form kx + c to the value of y does not affect the flexion. What effect does it have upon the slope? VI, ~ 44] SUCCESSIVE DERIVATIVES 71 20. Show, by means of Exs. 18 and 19, that any beam in which the flexion is constant has the form specified in Ex. 17. 21. Show, by a process precisely similar to that of Ex. 20 that a motion in which the tangential acceleration is constant is defined by an equation of the form s = at2 + bt + c. 22. Find, by the methods of Exs. 18-21, what the form of y must be if the slope is: dy dy _ dyc=6 dy (a) = 0; (b) = -3; (c) 6 x; (d) =a + b. 23. What is the form of y if the flexion is 6? if the flexion is 2 x + 3? if the flexion is zero? 24. If a beam of length I is supported only at both ends, and loaded by a weight at its middle point, its deflection y at a distance x from one end is y = k (3 12x - 4 X3), provided the cross-section of the beam is constant. Find the flexion and show that there are no points of inflexion between the supports. 25. If the beam of Ex. 24 is rigidly fixed at both ends, and loaded at its middle point, the deflection of each half of the beam is y = k (3 lx2 - 4 x3), where x is measured from either end. Show that there is a point of inflexion at a distance 1/4 from the end, and that the greatest deflection is at the middle point. Find the points of inflexion and the point of maximum deflection of a uniform beam of length I whose deflection is: 26. y =k (31x2 —x3). [Beam rigidly embedded at one end, loaded at other end. Origin at fixed end.] 27. y = k (3 x212 -2 x4). [Beam freely supported at both ends, loaded uniformly. Origin at lowest point.] 28. y = k (612X2 -4 x3 + X4). [Beam embedded at one end only; loaded uniformly. Origin at fixed end.] 29. y = k (l3x-3 x3 +2 X4). [Beam embedded at one end, supported at the other end; loaded uniformly. Origin at free end.] 72 THE CALCULUS [VI, ~ 45 45. Angular Speed. If a wheel turns, the angle 8 which a given spoke makes with its original position changes with the time, i.e. 0 is a function of the time: =f (t). The time-rate of change of the angle 0 is called the angular speed; it is denoted by o: dO AO ( = angular speed = j= lim -- dt At->oat 46. Angular Acceleration. The angular speed may change; the time-rate of change of the angular speed is called the angular acceleration; it is denoted by a:. wa dwo d20 a = angular acceleration = lim = - - At-'oAt dt - dt2 EXAMPLE 1. A flywheel of an engine starts from rest, and moves for 30 seconds according to the law A4_ gele --------- - degr<ees) I- _ 30 _ w-/E - 20 FIG. 18. w (X _ - Time 10 20 3'-=t 45 FIG. 18. 1 1 0 =- t4 + t3, 1800 30 where 0 is measured in degrees, after which it rotates uniformly. Then dO 1 1t2 dt — 450 10 and do= 1 t2 + 2 t =~t = 150 - -OThis example furnishes an instance in which the derived curves, i.e. the graphs which show the values of o and of a are more important than the original curve; for the total angle described is relatively unimportant. In the figure a scale is chosen which shows particularly well the VI, ~ 461 SUCCESSIVE DERIVATIVES 73 variation of w; 0 is allowed to run off of the figure completely, since its values are uninteresting. The acceleration a is so arranged that it does not suddenly drop to zero when the flywheel is allowed to run uniformly; and the values of a are never large. Something resembling this figure is what actually occurs in starting a large flywheel. In actual practice with various machines, curves of this type are often drawn experimentally. The equations serve only as approximations to the reality, but they are often indispensable in calculating other related quantities, such as the acceleration in this example. Curves which resemble the graph of co in this example occur frequently. (See ~~ 87, 134.) EXERCISES 1. A flywheel rotates so that 0 = t3 - 1000, where 0 is the angle of rotation (in degrees) and t is the time (in seconds). Calculate the angular speed and acceleration, and draw a figure to represent each of them. 2. Suppose that a wheel rotates so that 0 = ta + 1000 where 0 is measured in radians [1 radian = 180~/7r]. Is its speed greater than or less than that of the wheel in Ex. I? What is the ratio of the speeds in the two cases? 3. If a wheel moves so that 0 = - t4/16 - t/32, where 0 is measured in radians and t in minutes, find the angular speed and acceleration in terms of radians and minutes; in terms of revolutions and minutes; in terms of radians and seconds (of time). 4. If a Ferris wheel turns so that 0 = 20 t2 while changing from rest to full speed, where 0 is in degrees and t in minutes, when will the speed reach 20 revolutions per hour? 5. If the angular speed is c = kt, as in Ex. 4, show that the acceleration a is constant. Conversely, show that if a = k, and if t is the time since starting, o = kt. 6. Express the linear speed of a point on the rim of a wheel 10 ft. in diameter when the angular speed is 4 R. P. M. 7. Find the linear speed and the tangential acceleration of a point on the rim of the wheel of Ex. 1, ~ 46, if the wheel is 5 ft. in diameter. What are they when t = 30 sec.? 74 THE CALCULUS [VI, ~ 47 47. Related Rates. If a relation between two quantities is known, the time-rate of change of one of them can be expressed in terms of the time-rate of change of the other. Thus, in a spreading circular wave caused by throwing a stone into a still pond, the circumference of the wave is (1) c= 2 7rr, where r is the radius of the circle. Hence dc dr (2) 2- = -; ~(2) ddt dt or, the time-rate at which the circumference is increasing is 2 7r times the time-rate at which the radius is increasing. Dividing both sides by dr/dt, we find dc dr dc dt. - dt= 2 Ar = d = dc - dr; dt dt dr that is, the ratio of the time-rates is the derivative of c with respect to r; or, the ratio of the time-rates is equal to the ratio of the differentials. The fact just mentioned is true in general; if y and x are any two related variables which change with the time, it is true (Rule [VIIa], p. 32) that: dy. d dy, = = dy + dx, dt dt dx that is, the ratio of the time-rates of y and x is equal to the ratio of their differentials, i.e. to the derivative dy/dx. EXAMPLE 1. Water is flowing into a cylindrical tank. Compare the rates of increase of the total volume and the increase in height of the water in the tank, if the radius of the base of the tank is 10 ft. Hence find the rate of inflow which causes a rise of 2 in. per second; and find 'the increase in height due to an inflow of 10 cu. ft. per second. Consider the same problem for a conical tank. VI, ~ 47] SUCCESSIVE DERIVATIVES 75 (A) The volume V is given in terms of the height h by the formula: V = rr2 h = 100 rh, dV dh hence - = 100 r dt dt or, the rate of increase in volume (in cubic feet per second) is 100 ir times the rate of increase in height (in feet per second). If dh/dt = 1/6 (measured in feet per second),dV/dt = 100 r/6 = (roughly) 52.3 (cubic feet per second). If dV/dt = 10, dh/dt = 10 100 r = (roughly).031 (in feet per second) = 22.3 (in inches per minute). (B) If the reservoir is conical, we have V = ~ wr2h = 7rh3tan2 a, where r is the radius of the water surface, h is the height of the water, and a is the half-angle of the cone; for r = h tan a. In this case dV dh d- =- rh2tan2 a - dt dt ' which varies with h. If a = 45~ (tana = 1), at a -. height of 10 ft., a rise 1/6 (feet per second) would mean an inflow of wrh2 X (1/6) = 100 7/6 = 52.3 h (cubic feet per second). At a height of 15 feet, a rise of 1/6 (feet per second) would mean an inflow of \ 225 7r/6 = (roughly) 117.8 (cubic feet per second). An inflow of 100 (cubic feet per second) means a rise in height of 100/rh2, which varies with the height; at a height of 5 ft., the rate of rise is 4/7 = 1.28 (feet/ FIG. 19. second). EXAMPLE 2. A body thrown upward at an angle of 45~, with an initial speed of 100 ft. per second, neglecting the air resistance, etc., travels in the parabolic path gx2 Y = 10000 + xi where x and y mean the horizontal and vertical distances from the starting point, respectively; g is the gravitational constant = 32.2 (about); and the horizontal speed has the constant value 100/V2. Find the vertical speed at any time t, and find a point where it is zero. 76 THE CALCULUS [VI, ~ 47 The horizontal speed and the vertical speed, i.e. the time-rate of change of x and y, respectively, are connected by the relation (see ~~ 8, 20). dy dx dy _ gx dt ' dt dx 5000 ' dy gx dx gx 100 hence 1 ( + ) +10 dt 5000 dt 50 /2 -V-/' This vertical speed is zero where gx 100 5000 - - + - = 0, i.e. x = 155.3 (about), 50V- -V 9 which corresponds to y = 2500/g = 77.7 (about). At this point the vertical speed is zero; just before this it is positive, just afterwards it is negative. When x = 0 the value of dy/dt is 100/V2; when x = 2500/g, dy/dt = 50/V/2; when x = 7500/g, dy/dt = - 50/v2. EXERCISES 1. Water is flowing into a tank of cylindrical shape at the rate of 100 gal. per minute. If the tank is 8 ft. in diameter, find the rate of increase in the height of the water in the tank. 2. A funnel 12 in. across the top and 9 in. deep is being emptied at the rate of 2 cu. in. per minute. How fast does the surface of the liquid fall? 3. If water flows from a hole in the bottom of a cylindrical can of radius r into another can of radius r', compare the vertical rates of rise and fall of the two water surfaces. 4. If a funnel is 12 in. wide and 9 in. deep and liquid flows from it at the rate of 5 cu. in. per minute, determine the time-rate of fall of the surface of the liquid. 6. Compare the vertical rates of the two liquid surfaces when water drains from a conical funnel into a cylindrical bottle. Compare the time-rate of flow from the funnel with the time-rate of the decrease of the wet perimeter. 6. If a wheel of radius R is turned by rolling contact with another wheel of radius R', compare their angular speeds and accelerations. VI, ~ 47] SUCCESSIVE DERIVATIVES 77 7. If the surface s of a cube increases at a given rate k (in square inches per second), what is the rate of increase of the volume? 8. If a point moves on a circle so that the arc described in time t is = t2 - 1/t2 + 1, find the angular speed and acceleration of the radius ' drawn to the moving point. 9. A point moves along the parabola y = 2 x2 -x in such a manner that the speed of the abscissa x is 4 ft./sec. Find the general expression for the speed of y; and find its value when x = 1; when x = 3. 10. In Ex. 9, find the horizontal and vertical accelerations, the total speed, the tangential acceleration, and the total acceleration. 11. A point moves on the cubical parabola y = x3 in such a way that the horizontal speed is 10 ft./sec. Find the vertical speed when x = 6. 12. In Ex. 11, find the horizontal, vertical, tangential, and total accelerations. 13. If a person walks along a sidewalk at the rate of 4 ft. per sec. toward the gate of a yard, how fast is he approaching a house in the yard which is 50 ft. from the gate in a line perpendicular to the walk, when he is 100 ft. from the gate? When 10 ft. from the gate? 14. Two ships start from the same point at the same time, one sailing due east at 10 knots an hour, the other due northwest at 12 knots an hour. How fast are they separating at any time? How fast, if the first ship starts an hour before the other? 15. If a ladder 13 ft. long rests against the side of a room, and its foot moves along the floor at a uniform rate of 2 ft./sec., how fast is the top descending when it is 5 ft. above the floor? When the top is 1 inch from the floor? 16. If the radius of a sphere increases as the square root of the time, determine the time-rate of change of the surface and that of the volume; the acceleration of the surface and that of the volume. 17. If a projectile is fired at an angle of elevation a and with muzzle velocity vo, its path (neglecting the resistance of the air) is the parabola gx2 y = x tan a - - 2 v02 Cos2 a ' x being the horizontal distance and y the vertical distance from the point of discharge. Draw the graph, taking g = 32, a = 20~, vo = 2000 ft./sec. 78 THE CALCULUS [VI, ~ 47 Calculate dy in terms of dx. In what direction is the projectile moving when x = 5000 ft., 10,000 ft., 20,000 ft.? How high will it rise? 18. If p * v = k, compare dp/dt and dv/dt in general; compare d2p/dt2 and d2v/dt2. 19. If p vn = k, compare dp/dt and dv/dt. [For air, in rapid compression, n = 1.41, nearly.] 20. If q is the quantity of one product formed in a certain chemical reaction in time t, it is known that q = ck2t/(l + ckt). The time-rate of change of q is called the speed v of the reaction. Show that ~ ck2 v (1 c2 - c(lk - q)2 (1 + ckct)2 Show also that the acceleration a of the reaction is a 2 c t = 2 C2(ka (1 + ckt3) - -c2(-q)3. CHAPTER VII REVERSAL OF RATES -INTEGRATION 48. Reversal of Rates. Up to this point, we have been engaged in finding rates of change of given functions. Often, the rate of change is known and the values of the quantity which changes are unknown; this leads to the problem of this chapter: to find the amount of a quantity whose rate of change is known. Simple instances of this occur in every one's daily experience. Thus, if the rate r (in cubic feet per second) at which water is flowing into a tank is known, the total amount A (in cubic feet) of water in the tank at any time can be computed readily,- at least if the amount originally in the tank is known: A = rt + C, where t is the time (in seconds) the water has run, and C is the amount originally in the tank, i.e. C is the value of A at the time when t = 0. If a train runs at 30 miles per hour, its total distance d, from a given point on the track, is d = 30. t + C, where t is the time (in hours) the train has run, and C is the original distance of the train from that point, i.e. C is the value of d when t = 0. (Notice that by regarding d as negative in one direction, this result is perfectly general; C may also be negative.) If a man is saving $100 a month, his total means is 100 ~ n + C, where n is the number of months counted, and C is his means at the beginning; i.e. C is his means when n = 0. If the cost for operating a printing press is 0.01 ct. per sheet the total expense of printing is T = 0.01-n + C where n is the number of copies printed, and where C is the first cost of the machine; i.e. C is the value of T when n = 0. 79 80 THE CALCULUS [VII, ~ 49 49. Principle Involved. Such simple examples require no new methods; they illustrate excellently the following fact: The total amount of a variable quantity y at any stage is determined when its rate of increase and its original value C are known. We shall see that this remains true even when the rate itself is variable. 50. Illustrative Examples. The rate R (x) at which any variable y increases with respect to an independent variable x is the derivative dy/dx; hence the general problem of ~ ~ 48 -49 may be stated as follows: given the derivative dy/dx, to find y in terms of x. In many instances our familiarity with the rules for obtaining rates of increase (differentiation) enables us to set down at once a function which has a given rate of increase. EXAMPLE 1. Thus, in each of the examples given in ~ 48, the rate is constant; using the letters of this article: dy = R (x) = k, dx where k is a known fixed number; it is obvious that a function which has this derivative is (A) y = kx + C, where C is any constant chosen at pleasure. While the examples of ~ 48 can all be solved very easily without this new method, for those which follow it is at least very convenient. The value of C in any given example is found as in ~ 48; it represents the value of y when x 0. EXAMPLE 2. Given dy/dx = x2, to find y in terms of x. Since we know that d (x3)/dx = 3 x2, and since multiplying a function by a number multiplies its derivative by the same number, we should evidently take: x3 X3 - x3 -\ y =, or else y C3 + [; check: d 3 + C = x2dxJ VII, ~ 50] REVERSAL OF RATES-INTEGRATION 81 where C is some constant. As in ~ 48, some additional information must be given to determine C. In a practical problem, such as Ex. 3, below, information of this kind is usually known. EXAMPLE 3. A body falls from a height 100 ft. above the earth's surface; given that the speed is v - gt, find its distance from the earth in terms of the time t. Let s denote the distance (in feet) of the body from the earth; we are given that (1) v= dt = - gt, or ds = vdt =- gtdt, which is negative since s is decreasing. We know that d (t2) = 2 t dt; hence it is evident that we should take: (2) s =- t2 + C; [check: ds = - gt dt]. As the body starts to fall, t = 0 and s = 100; substituting these values in (2) we find 100 = 0 + C, or C = 100. Hence we have =- t2 + 100. 2 EXAMPLE 4. Given dy/dx = xn, to find y in terms of x. Since we know that d(xn+l) = (n + 1) n dx, we should take (B) Y = x+l + C; [check: dy = x dx]. Since the rule for differentiation of a power is valid (~ 21, p. 34) for all positive and negative values of n, the formula (B) holds for all these values of n except n = - 1; when n = - 1 the formula (B) can not be used because the denominator n + 1 becomes zero. Special cases: dy x n = 1, d = x, y = 2 X2 + C; check: d ( 2) = x dx. n=-, 1, Y-xyc -c dx. n = 0, y -= 1, y = x + C; check: d (x) 1.dx. dx 1 dy 2 x2 2 dy = x1/2, - n = 1 _ l/2, y -= x3/2 + C; check: d (x3/2) = x1/2dx. n =- 2, dy -2, = 2, y= - - +C; check: d(-1 -1) = x-2 dx. Notice that these include Vx (= x1/2), 1/x2 (= x-2), etc.; other special cases are left to the student. 82 THE CALCULUS [VII, ~ 50 EXAMPLE 5. Given dy/dx = x3 + 2 x2, to find y in terms of x. Since d (x4)/dx = 4 x3 and d (x3)/dx = 3 x2, and since the derivative of a sum of two functions is equal to the sum of their derivatives, it is evident that we should write * X +2 + C. The check is dx ddx 4 + 3 +C =x3+2x2; such a check on the answer should be made in every exercise. In general, as in this example, if the given rate of increase (derivative) is the sum of two parts, the answer is found by adding the answers which would arise from the parts taken separately, since the sum of the derivatives of two variables is always the derivative of their sum. EXERCISES Determine functions whose derivatives are given below; do not forget the additive constant; check each answer. 1 = 4x. 2. =- 5x. 3. =d 3X2 4. = 2. dx dx dx dx 5. ~dy dy =dydy 5. = - 6. 6 - 105 7. =. 8. =.01 8. dx dx dx ddx In the following exercises, remember that the derivative of a sum is the sum of the derivatives of the several terms; proceed as above. 9. = 4+ 5 X2. 10. = 4X2-2 x +3. 11. ds = t3- 4 t + 7. dx dx dt 12. 3 = 3ax5 - 84. 13. = ax + b. 14. = a2 + bt + c. dx dx dt 15. Y =.006 2 -.004 x3 +.015 4. 16. =-t6 + 5t4 -6 t2+ 2. dx dt dsy t21/2 dv 17 3 18 = +/t2. 9. 3 t-2+ 4t-3. dx dt dt 20. d= X2/3. 21. d = 2X1/2 - 3X-1/2. 22. dp= kv-2.41 dx dx dv * In all the Examples of this paragraph, we have had an equation which involves dy/dx; such an equation is often called a differential equation, because it contains differentials. See also Chapter XIX. VII, ~ 51] REVERSAL OF RATES -INTEGRATION 83 51. Integral Notation. If the rate of increase dy/dx = R (x) of one variable y with respect to another variable x is given, a function y=I(x) which has precisely this given rate of increase is called an indefinite integral * of the rate R (x), and is represented by the symbol t (1) (x) = R (x) dx; that is, (2) if [ (x)] R(x), then I(x)= R (x)dx, or, what amounts to the same thing, (3) if d [I(x)] = R (x) dx, then I (x) =fR (x) dx. The results of Examples 1, 2, 3, ~ 50, written with the new symbol, are, respectively, [A] k dx=kx + C. fx dx = x3/3 + C. s =fv dt + C =f- gt dt + C = - gt2/2 + C. The first equation of Example 3 holds in general: [I] s =f v dt + C, since ds = * The common English meaning of the word integrate is "to make whole again," "to restore to its entirety," "to give the sum or total." See any dictionary, and compare ~~ 48-49. To integrate a rate R(x) is to find its integral; the process is called integration. Often the rate function R(x) which is integrated is called the integrand; thus the first part of equation (2) may be read: "the derivative of the integral is the integrand." This is the property used in checking answers. The first equation in (2) and the first in (3) are differential equations. t Note that dx is part of the symbol. As a blank symbol, it is f (blank) dx; the function R(x) to be integrated (i.e. the integrand) is inserted in place of the blank. The origin of this symbol is explained in ~ 120. 84 THE CALCULUS [VII, ~ 51 The result obtained in (B), Example 4, ~ 50, gives n7+1 [B1 Sxndx= x+ 1 + C, n7Z -1, for all positive and negative integral and fractional values of n except n = - 1, for which see ~ 65. As examples of the many special cases, we write: x2 n=1, fxdx = +C. n =0, fx~O =fldx =f dx =x+C. I n=-2, fx/2 dx =f dx= — l+C = - +C. 3-2 dx-2d =f-2 -d = -x 2 C. n =-, fx-/3x = dx = x2/3 - + C = 32 + C. 4 x f(x3 + 2 x2) dx =fx3dx+f2x2dx =4 +2 + C. The general principle used in this example is that the integral of a sum of two functions is the sum of their integrals: [C] f[R (x) + S (x)] dx =fR (x) dx +fs (x) dx, which is true because the derivative of the sum R (x) + S (x) is the sum of the derivatives: d [R S/dx = dRdx + = dS/dx. The rules (A), (B), (C) are sufficient to integrate a large number of functions, including certainly all polynomials in x. VII, ~ 511 REVERSAL OF RATES -INTEGRATION 8 85 EXERCISES 1. Express the value of y if dy/dx = 4 X2 + 3 x by means of the new signf() dx. Then find y. Check the answer by differentiation. Proceed as in Ex. 1 if y dy/dx has one of the following values: 2. x3 4x4. 6. x32. 8. x+x4 0 xn 3. X-3. 5. 4 x +5. 7. 9. 9. X3 -X4. 11. X +V In the following exercises express the given function as a sum of powers of x; then proceed as above. 12. X2 (1 ~ X). 16. (1 ~ X3) (1 - X3). 20. (1 - X2) (1 + X-2). 13. (X3 ~ 5 X2) +X2. 17 3-x 52x.21. X3/2(X - X2). 14. 3 (x -2)2. 18.X 1'2 (2 -x). 22. (X3 -2X2 + X)1/2. 15. 3 x3(4 — 31x). 19. (3 - 2x)V 23. (1 +{j) 2. Evaluate the following integrals: 24. fx-5 dx. 29, ~ dv. 34. f 12 t-7/2 dt 25. f t314 dt 30. f (6 u-3 - 7 U-4) du. 35. f 3 4y~dy. 26. f 5 s-5ds. 31. f (Z3/5 + Z-3/5) dz. 36. f (5/V\/u) du. 27. f (2 ~fu- U-1/2) du. 32. f (, - 4 y-5) dy. 37. f ( N3fu) 5du. 2b. f3 r-2/3 dr. 33. f (y1O/3 + 2 y4/3) dy. Integrate the following expressions, making use of the principle of Exs. 12- 23. 38. f (1 t)2dtA 44. f Vx(a +bx) dx. 39. f x (1 ~ V'sx) dx. 45. fxn (a ~ bx) dx. 40. fs (1-V) ds. 46. f (a + bx)2 dX. 41. Jt2 (1 - t2) dt 47. J(t2_5 - 2) t-5 dt 42. fx4 (1 ~ X + X2) dx. 48. fXl.4 (1 ~ X2)2 dx. 4.fx~~x x 9 f~t( 222t 43. fx (a + bx) dx. 49. f NIt (1.+ 2 t2)2 A 86 THE CALCULUS [VII, ~ 52 52. Fundamental Theorem. If dy/dx = 2 x, the answers y = x2, y = x2 + 5, y = x2 + C are all correct. To decide which one is wanted, additional information is needed. However, except for the additive constant C, all answers coincide. For practical purposes, there is but one answer. Stated precisely, this is the fundamental theorem of integral calculus: If the rate of increase ~(1) - _Rdy d7 = R(x) of a variable quantity y which depends on x is given, then y is determined as a function of x, I (x), except for a constant term: (2) y =fR (x) dx + C = I (x) + C. 53. Calculation by Integrals. In applications, we often care little about the actual total; it is rather the difference between two values which is important. Thus, in a motion, we care little about the real total distance a body has traveled; it is rather the distance it has traveled between two given instants. If a body falls from any height, the distance it falls is (Ex. 3, p. 81) gt2 s = vdt+C =gtdt+ C = + C, where s is counted downwards. The value of s when t = 0 is s]t=o = C; the value of s when t = 1 is st = 1 = g/2 + C. The distance traversed in the first second is found by subtracting these values: s]t=o -s] = (+C)-C= 9 = =16.1 ft., t=l. where s]t= means the space passed over between the times t = 0 and t = 1. In this calculation, we care little about where s is counted from; or its total value. The result is the same for all bodies dropped from any height. VII, ~ 54] REVERSAL OF RATES -INTEGRATION 87 Likewise, the space passed over between the times t = 2 and t = 5 is lt=5 g (25 C (4 =] -8 -( f+c +c> t]=2 t=5 t= 2 2 g.25 g.4 21 -2- -- = -9. = 338 (ft.). 2 2 2 In general the distance traversed between the times t = a and t = b is S =s- s - g-+C - - = -ga —g-=(b2 - a2). -a] b-s] -(2 C)-(a 2 C)t2 2 2( 54. Definite Integrals. The advantage realized in the example of ~ 53 in eliminating C can be gained in all problems. The numerical value of the total change in a quantity between two values of x, x = a and x = b, can be found if the rate of change dy/dx = R (x) is given. For, if y = I (x) =fR (x) dx + C, the value of y for x = a is y]xa= I (a) = [fR (x) d]x=a+ C, and the value of y for x = b is y]:,b= I (b) = [fR (x) dx] b+ C. The total change in y between the values x = a and x = b is Y]xa y]x-=- Y]x- a I (b) - I (a) =[fR (x) dx] [fR (x) dx]= This difference, found by subtracting the values of the indefinite integral at x = a from its value at x = b, is called the definite integral of R (x) between x = a and x = b; and is denoted by the symbol: R (x) dx (x) dx] - [f (x) dx]= x=a x =b x=a 88 THE CALCULUS [VII, ~ 54 It should be noticed that, in subtracting, the unknown constant C has disappeared completely; this is the reason for calling this form definite. EXAMPLE 1. Given dy/dx = x3, find the total change in y from x = tox =3. Since y =fx3 dx x 4/4 + C, it follows that J]-=l =]x=3 =]xl 4J ]=3 4J= 1 Interpreted as a problem in motion, where x means time and y means distance, this would mean: the total distance traveled by a body between the end of the first second and the end of the third second, if its speed is the cube of the time, is twenty units. Interpreted graphically, a curve whose slope m is given by the equation m = x3, rises 20 units between x = 1 and x = 3. The equation of the curve is y = X4/4 + C. EXERCISES 1. If water pours into a tank at the rate of 300 gal. per minute, how much enters in the first ten minutes? how much from the beginning of the fifth minute to the beginning of the tenth minute? 2. If a train is moving at a speed of 30 mi. per hour, how far does it go in two hours? Does this necessarily mean the distance from its last stpp? 3. If a train leaves a station with a variable speed v = t/2 (ft./sec.), find s in terms of t. How far does the train go in the first ten seconds? How far from the beginning of the fifth to the beginning of the tenth second? 4. A falling body has a speed v = gt, where t is measured from the instant it starts. How far does it go in the first four seconds? How far between the times t = 3 and t = 9? 5. A wheel rotates with a variable speed (radians/sec.) co = t2/100. How many revolutions does it make in the first fifteen seconds? How many between the times t = 1 and t = 10? VII, ~ 54] REVERSAL OF RATES —INTEGRATION 89 From the following rates of change determine the total change in the functions between the limits indicated for the independent variable. Interpret each result geometrically and as a problem in motion, and write your work in the notation used in the text. dy ds t4- 3 6. = x, x = 2 to x = 4. 11 t = 1 to t 3. dx dt t2 -7. d- =x2, x —2tox=2. 12. dv - (1 t)2 t=4tot=9. dy 5 dt t3/2 8 -—,x=-4tox =4. 13.- =, t=0.1 to. y 2-2, =Otox=10. d14. = tvt; t =1 to t i16. dx dt ds 1, d a2 a3 1. t = tot0. 15 =- t = 1 to t = 2 a. dt t2- dt t2 t3a Determine the values of the following definite integrals. [In cases where no misunderstanding could possibly arise, only the numerical values of the limits are given. In every such case, the numbers stated as limits are values of the variable whose differential appears in the integral.] z=2 27 -8 16. x dx. 21. 2. 2/3 dx. 26. sf 2/3 (2 - 2 s) ds. x-= +2 ft=7 0=.02 do 17. 3 3x dx. 22. / (1 + t) dt. 27. - x=-2 = t== 0=0o 100 =2 3 o = 100 18. 3x2 dx. 23./ 3(t2 —)dt. 28.1 (.01 +.02 0) d. J x=l Jt= -2 0=10 x=a -3 f27 1 19. 5 x2 dx. 24. f (1+s+s 2)ds. 29. J + do. J=-a -J -2 9 l3 1 s o4 7.4 20. f +/x dx. 25. d1. 30. V-1-41 dv. Jo S 2.3 31. A stone falls with a speed v = gt + 10. Find s in terms of t and find the distance passed over between the times t = 2 and t = 7. 32. A bullet is fired vertically with a speed v = - gt + 1500. How far does it go in ten seconds? How high does it rise? How long is it in the air? Make rough estimates of the answers in advance. 90 THE CALCULUS [VII, ~ 55 33. For any falling body, j = acceleration = g = const. Find the increase in speed in ten seconds. Does it matter what particular ten seconds are chosen? 34. If, in Ex. 33, the speed is 100 ft./sec. when t = 5, what is the speed when t = 15? When will the speed be 250 ft./sec.? Express v in terms of t. 55. Area under a Curve. We saw in ~ 54 that the value of any quantity could be computed if its rate of change could be found. We shall proceed to illustrate this principle by showing how to find the area bounded by any given curve, the x-axis, and any two ordinates of the curve. Let the equation of the given curve be (1) y = f(x), and let A denote the area FMQP between this curve, the Y^ x-axis, a fixed ordinate FP Tr i (Fig. 20), and a variable Q —. Ay ordinate MQ. Since A will P h^^ AXS vary as the value of x at M changes, A is a function of x. y If x changes by an amount F M N Ax = MN, the area A will x=k x=x x=x+Ax changebyanamountMNRQ, FIG. 20. which we shall call AA. Then it is evident from the figure that if the curve rises from Q to R we shall have (2) rectangle MNSQ <AA < rectangle MNRT. If the curve falls from Q to R, the inequalities would be reversed. From (2) we have (3) y Ax <AA < (y + Ay) Ax. Dividing by Ax, we find AA (4) y< <y +Ay. Axr VII, ~ 55] REVERSAL OF RATES-INTEGRATION 91 If the curve falls from Q to R, these inequalities would simply be reversed. If we now let Ax approach zero, Ay will also approach zero, and we shall have, in either case, (5) dA = lim Y= f (x). dx ax-,o Ax It follows, by ~ 54, that the area under the curve (1), between any two fixed ordinates x = a and x = b, is given by the formula x=b fx=b fx=b (6) A] = y dx= f(x) dx. x=a x=a =a EXAMPLE 1. To find the area under the curve* y = x2 between the points where x = 0 andx = 2. 4 We have, by (2) A = fy dx +C = x2dx + C = X + CY f 3 22 t=1 where A is counted from any fixed back K t=1 boundary x = k we please to assume, up to a movable boundary x = x. The area between x =0 and x==2 is O 3 given by subtracting the value of A for FIG. 21. x = 0 from the value for A for x = 2: ]X= A-=2 ] - A =J dx= X3 8 A =A - A X2- XJ=O x=2 X=O x=O 3 X=2 3 x=0 3 Likewise the area under the curve between x = 1 and x = 3 is ]x=3 rX==3 3 3 = A = 2dx = - - = - = X=l =1 3J X=3 3- X=I and the area under the curve between any two vertical lines x a and x =b is ax=b ax;=b x31 3 x31 - 3a3 A = x2dx = = — x=a x=a 3 x=b 3 Jx=a 3 * The phrase "the area under the curve" is understood in the sense used in ~ 55. When the curve is below the x-axis, this area is counted as negative, 92 THE CALCULUS [VII, ~ 55 If the equation of the curve is given in parameter form (7) X f(t), y 0= (t), the equation (5) may be replaced by the equation (8) dA dA dx dx w) ~tdat dx 'dt =Y'dt or dA df (t) (9) cp(t) dt dt and the formula (6) takes the form ]t =t2 X t2 dx (10) A = y dt, t Jt t=tl dt or (11) = t t2 r t= t2 df(t) d (I Ar(r ^ -dt, t=ti J t=ti dt which gives the area above the x-axis, below the curve (7), and between the ordinates of the points at which t has the values tl and t2, respectively. EXAMPLE 2. To find the area under the curve whose equations are X = t2 l+t between the ordinates of the points where t = 2 and t = 3. By (11), the required area is ]t=3 rt=3 1 +t A ~ 2tdt t=2 ft=2 t = (2 + 2 t) dt = 7. [CAUTION. By calculating in a similar manner the area under the curve from t = 0 to t = 1 we would find A]tt== 3. But this result would require justification by the considerations of ~ 115, p. 188.] VII, ~ 551 REVERSAL OF RATES - INTEGRATION 93 EXERCISES Find the area under each of the following curves between the ordinates x =0 and x = 1; between x = 1 and x = 4. Draw the graph and estimate the answer in advance. 1. y = X2. 4. y = x2/3. 7. y = I-. 2. y = 5. y = 1 - x2. 8. y = X2(1- X). 3. y = x3/2. 6. y = (1 - X)2. 9. y = X(1- X2). Find the area under each of the following curves and check graphically when possible. 10. y = X3 +6X2 + 15 X, ( = O to 2; x = -2to + 2). 11. y = x2/3, (X = - to+ 1; x =-a to + a). 12. y = X2 + 1/X2, (x = l to 3; x = 2 to 5; x = a to b). Find the area under each of the following curves between the ordinates determined by the indicated values of t, and check graphically. 13. x = t + 1, y = t - 1; t = O to 5. 14. X = (t - 1)/t, y = t3/8; t = 2 to 4. 15. x = 2 t, y = 3V\t; t = O to 4. 16. X = i + Vit, y =2t2; t = O to 9. 17. x,=(1 + t)2, y = (1 - t)2; t = 1to 2. 18. x = 1 - t, y = /I+t; t = - 1 to 3. 1 9. X = N/t~ jf y = -Vj2 - 1; t = 1 to 5. 20. Show the area A bounded by a curve x = 44y), the y-axis, and the two lines y = a and y = b is A S (y) dy. 21. Calculate the area between the y-axis, the curve x = y2, and the lines y = 0 and y = 1. Compare this answer with that of Ex. 14. 22. Find the area between the curve y = x3 and each of the axes separately, from the origin to a point (k, k3). Show that their sum is k4. 94 THE CALCULUS [VII, ~ 56 56. Volume of a Solid of Revolution. Let us next consider the volume of the solid of revolution which is described when the area MNLK (Fig. 22) under the curve y = f(x) from x = a to x = b is revolved about the x-axis. Any section of this solid perpendicular to the x-axis will be a circle. Let us denote by V the volume from x = a to x = x, and by AV the increase in this volume as x increases to x + Ax. The radius of the circular section at any point is the ordinate /fr^R ' \ Dividing by Ax, we have FIG. 22. y of the curve. Hence the area of the section is ~ry2. If the curve is rising steadily from P to O, it is evident that (1) ~ry2 a Ax < AV < n (y + Ay)2. Ax. Dividing by ax, we have (2) T xy2 < < T(y + Ay)2 If the curve is falling, these inequalities are reversed. If we now let Ax approach zero, Ay will approach zero, and we shall have, in either case, dV AV 2 (3) -y-=lid m - = 7ry. dx Ax-so Ax VII, ~ 57] REVERSAL OF RATES —INTEGRATION 95 It follows, by ~ 54, that the volume of the solid of revolution from x = a to x = b, is given by the formula x=b fx=b rx=b (4) V]=f= ydxd = X= f(x)}2dx. x=a x=a x= a Similarly, the volume V of a solid of revolution formed by revolving a curve x= (y) around the y-axis satisfies the relations (5) -,y=d y==d (5) d V V = x2 dy. dy ' yc y=c EXAMPLE. Find the volume generated when the area under the curve y = 1 - x2 from x = - 1 to x = + 1 revolves about the x-axis. From the symmetry of the figure, we see that the total volume required is twice the volume generated by the area from x = 0 to x = 1. Hence s=l 1 V] =2fJ,wy2ddx L-l ]a =-l1 0 /. = 2 r (1 - x2)2dx =2 Lrfo (1 - 2x2 + X4) dx 2 x- ) -1 1 =7x 3- + --- + ( 2 1\ 16 wr -2r 1-35 15 FIG. 23. 57. Volume of a Frustum of a Solid. A frustum of a solid is the portion of that solid contained between two parallel planes. The solid itself, between the limiting parallel planes, may be thought of as generated by the motion of the cross-section parallel to these planes from one extremity to the other, if the shape of the cross-section is supposed to vary in the correct manner during the motion. Thus, a frustum of a circular cone may be generated by the motion of a circle which remains parallel to its original position, if 96 THE CALCULUS [VII, ~ 57 the radius of the circle steadily increases (or diminishes) during the motion. In Fig. 24, let s denote the distance along a line AB perpendicular to the variable cross-section, measured from some fixed point A. L M-A ^M Let V denote the volume of the solid,/.-.... _.-. \ from s= a to s = s, ~Y/~-f~~-ZZ~ — Ps=s+ as and let AV denote -/ ~-. f^ ^ i j, the increase in vol/ -......... ----... jI ume as s changes to - j..-.....-ls=a s + As. Let A, de—:L A — a/& " s=0 note the area of the cross-section at the FIG. 24. position s = s, and let AA denote the increase in As as s changes from s to s + As. Then, if A, increases as s increases, we shall have (1) As * As < AV < (As + AA) ~ As. This inequality will be reversed if A, decreases. Dividing by As, and then allowing As to approach zero, AA will also approach zero, and we shall have (2) = lim =A,, ds s-oo As whence, by ~ 54, the volume of the entire frustum from x = a to x = b is given by the formula ]s=b s=b (3) V = As ds s=a s=a The formulas (2) and (3) show that the rate of change of the volume with respect to the distance s is equal to the area of the variable cross-section, and that the volume of VII, ~ 57] REVERSAL OF RATES —INTEGRATION 97 the entire frustum is obtained by integrating the area of the cross-section with respect to s. The formulas of ~ 56 are special cases of the formula (3). EXAMPLE. A circle moves with its center on a given straight line, and its plane perpendicular to that line. Its radius is proportional to the square of the distances of its center from a fixed point O of the line. Find the volume of the frustum of the solid generated as the circle moves from s = a to s = b. If s denotes the distance from the fixed point O to the center of the circle, the radius of the circle, which is to vary as the square of s, must be r = ks2, where k is a constant. Hence the area of the circle is As = rr2 = 7rk2 s4. Then the volume of the frustum from s = a to s = b is V = ' 7rk2s4 ds= rk2 f S4ds _8s=a a ga = ark2 () ] rk2 (b5 - 5). EXERCISES Find the volumes formed by revolving each of the following curves about the x-axis,. between x = 0 to x = 2; between x = - 1 to x = + 1. 1. y = X3. 3. y = 3 - x. 5. y2 =x+ 2 y. 2. y = x2 -1. 4. y = (1 + )2 6. Vtxq-+ + V+ ==4. Proceed similarly for each of the following curves, between x = 1 and x = 3; between x = a and x = b. 7. y X2. 8. xy = 1 + x2. 9. 4 - x2y2 = 1. x Find the volumes formed by revolving each of the following curves about the y-axis, between y = 0 and y = 2. 10. x = y3. 12. x = 4 y2 -y3. 14. x = y2 y. 11. x2 = y3. 13. x2 + y4 = 81. 1. x = y1/2 + yl/4. 16. Find by integration the volume of a frustum of a cone of height h, if the radii of the two bases are, respectively, r and R. 98 THE CALCULUS [VII, ~ 57 17. Find the volume of the paraboloid of revolution formed by revolving y2 = 4 x about the x-axis, between x = 0 and x = 4; between x = 1 andx = 5; between x = a and x = b. 18. Find the volume of a sphere by the formula of ~ 56. 19. Find the volume of the ellipsoid of revolution formed by revolving an ellipse (1) about its major axis; (2) about its minor axis. 20. Find the volume of the portion of the hyperboloid of revolution formed by revolving about the y-axis the portion of the hyperbola x2 _ y2 = 1 between y = 0 and y = 2. 21. Find the volume of the portion of the hyperboloid of revolution formed by revolving x2 — y2 = 1 about the x-axis, between x = 1 and x = 3. 22. Find the volume generated by a square of variable size perpendicular to the x-axis, which moves from x = 0 to x = 5, if the length of the side of the square is (1) proportional to x; (2) equal to x2. 23. Find the volume generated by a variable equilateral triangle perpendicular to the x-axis, which moves from x = 0 to x = 2, if a side of the triangle is (1) equal to x2; (2) proportional to 2 - x. 24. Find the volume generated by a variable circle which moves in a direction perpendicular to its own plane through a distance 10, if the radius varies as the cube of the distance from the original position. 25. Find the mass of a right circular cylinder of variable density, if the density varies (1) directly as the distance from the base; (2) inversely as the square root of the distance from the base. CHAPTER VIII LOGARITHMS - EXPONENTIAL FUNCTIONS 58. Necessity of Operations on Logarithms. The necessity for the introduction of logarithms in the Calculus depends not only on their own general importance, but also upon the fact that integrals of algebraic functions may involve logarithms. Thus, in ~ 51, in the case n = - 1 the integral jfx dx could not be found, although the integrand 1/x is comparatively simple. We shall see that this integral, fx-1 dx, results in a logarithm. We shall see also in ~ 68 that numerous cases arise in science in which the rate of variation of a function f (x) is precisely 1Ix. 59. Properties of Logarithms. The logarithm L of a number N to any base B is defined by the fact that the two equations (1) N = BL, logBN = L are equivalent. Thus if L =logBN and 1 = logBn, the identity BL- B' = BL+ is equivalent to the rule (2) logB (N n) = logB N + logB n, where n and N are any two numbers. Likewise BL + B = BL- gives (3) logB (N - n) = logBN - logBn; and (BL)" = BLn becomes (4) logB N = n logB N, where n may have any value whatever. The relations (1), (2), (3), (4), are the fundamental relations for logarithms. 99 100 THE CALCULUS [VIII, ~ 60 60. Computations. Graphs. To draw the graph of the equation (1) Y = logB X, for any fixed value of B, we may write the equation in the form (2) x = By. To compute the value of x when y is given, we take the common logarithm of both sides of (2): (3) loglo x = loglo Bv = y loglo B, by (4) ~ 59. But since y = logs x, we have (4) log10 x = logB x loglo B, or (5) logB x = logo x + loglo B. 5 y" ~ -7 /1 5 10 5 x FIG. 25. The relations (4) and (5) enable us to compute logarithms to any base quickly by means of a table of common logarithms. The graphs of (1) for several values of B are shown in Fig. 25. VIII, ~ 60] LOGARITHMS 101 The relation (3) enables us to compute fractional powers of any base. For, if B and y are given, as in (2), x may be found from (3) by means of a table of common logarithms. Similarly, if we take the'logarithms of both sides of (2) with respect to any other base b, we find the corresponding relations (4)' log0 x = logB X' logb B, (5)' logB = logb x + logbB. If we set x = b in (4)' and (5)', since logb b = 1, we find (6) logB b logb B = 1 and logB b = 1 logb B. EXERCISES 1. Find the value of 10x when x = 3; 0; 1.6; 2.7; - 1; - 1.9; 0.43. 2. Plot the curve y = 0l carefully, using several fractional values of x. 3. Plot the curve y = logio x by direct comparison with the figure of Ex. 2. Plot it again by use of a table of logarithms. Plot the graph of each of the following functions. 4. logo x3. 5. loglo0 x. 6. logo (1/x2). 7. logo x4/8 Do any relations exist between these graphs? Plot the graph of each of the following functions and explain its relation to graphs already drawn above. 8. logo1 (1 + x)2. 9. logo0 41 + x. 10. loglo (x 41 +lx). Plot the graphs of each of the following functions and show the relations between them. 11. logs x. 12. loggx. 13. logx2. 14. 3z. Show how to calculate most readily the values of the following expressions, and find the numerical value of each one. 15. logn,7. 17. /(5.4)6-2. 19. 10~.5 + 10-0-5. 21. logs 100. 16. 2453 8. log6S 8. 20. 2 log3 5. 22. 10 log1o 9. 102 THE CALCULUS [VIII, ~ 61 61. Napierian Logarithms. Base e. A careful examination of Fig. 25 will convince anyone that there must be a value of B for which the curve y = logBx has a slope equal to 1 at x = 1. Indeed, the equation (5), ~ 60, shows that the slope of the curve y = logBx can be found by dividing the slope of the curve y = logo x by loglo B. Hence if loglo0B is equal to the slope of the curve y = logiox at the point (1, 0), the slope of the curve y = logBx will be 1 at (1, 0). Let this value of B be denoted by the letter e. Logarithms to the base e are called Napierian logarithms,* or natural logarithms, or hyperbolic logarithms. (See Table V, C.) 62. Differentiation of loge x. To find the derivative of loge x, let us write y=l og p (1) =loge (1+h) (2) y + Ay = loge (X + Ax). 0 XA g (1+ Hence /=1 l=l+h Ay = log, (x + Ax) - log, FIG. 26. = log (1 + x) and Ay 1 log, I +g 1 log, 1+ ~ x and = x lo l )= x xl Now let u = Ax/x, so that we may write Ay 1 log (1 + u) Ax x u But the fraction {loge (1 + u)}/u is simply the slope of the secant AP of the curve logeu, and as Ax->0, so also * Named for Lord Napier, the inventor of logarithms. The value of e is stated below. No assumption is made at this point except that the logarithm curve has a tangent at (1, 0); VIIi, ~ 63] LOGARITHMS 103 u ->0; hence the secant AP becomes the tangent at A and its slope has the limit 1, by the definition of e. Hence lim loge (1 ) = 1 u —O U Ay 1 loge (1 + ) 1 Therefore m lim = m. li l ( + 1 Ax-R0 Ax x u->o u x or [VIII] dlogx 1 dx x On account of the simplicity of this formula the base e will be used henceforth in this book for all logarithms and exponentials unless the contrary is explicitly stated. 63. Differentiation of logBx. Since we have, by (4)', ~ 60, (1) y = logB x = loge, x logBe, the derivative of logBx is found by multiplying the derivative of loge x by logB e: [VIIIa]od log. log e. dx x In particular, for common logarithms, since B = 10, we have d logo x 1 (2) d logox = logo e. dx x The constant factor logio e is the value of the slope of y = logiox at (1, 0). It is called the modulus of the system of common logarithms, and is denoted by the letter M, that is logio e = M. Hence the preceding equation becomes [VIIIb] d og0 ox M dx x By means of formula VII, ~ 22, for change of variable, the formula VIII becomes d logeu _ I du dx u dx The formulas VIIIa and VIIIb may be rewritten in a similar manner. 104 THE CALCULUS [VIII, ~ 64 64. Values of M and of e. To compute approximately the value of M, that is the slope of the curve y = logio x at (1, 0), let us draw the secant connecting the points P (1, 0) and Q (1 + Ax, 0 + Ay) on that curve. Let us denote the slope of this secant PQ by mpQ. Then Ay logio (1 + Ax) mpQ= —Ax- Ax' If we choose for Ax a succession of smaller and smaller values, Ax = 0.1, 0.01, 0.001,..., we find a corresponding succession of values of mpQ: mpQ = 0.414, 0.432, 0.434,.... For the last of these values, a six or seven place logarithm table is required, while still higher place tables would be required to get a more accurate answer. Since the slope at (1, 0) is the limit of mpQ as Ax approaches zero, we have M = lim mpQ = 0.434... (approximately).* From this value of M, we can compute e, since logioe = M = 0.434.... Hence, from a table of common logarithms, e = 2.72... (approximately). 65. Illustrative Examples. EXAMPLE 1. Given y = log1o (2 x2 + 3), to find dy/dx. Method 1. Derivative notation. Set u = 2 x2 + 3, then dy dy du d logo u d (2 x2 +3) =M 4 4Mx dx du dx du dx u 2 2 3 *An independent method of calculating the values of M and of e will.be given in ~~ 147, 153. Logically, we might have waited until that time to state the value of M, but it is much more convenient, practically, to have an approximate value at once. To ten decimal places, the values are M=0.4342944819, 1/M= 2.3025850930, e=2.7182818285. - - VIII, ~ 65] LOGARITHMS 105 Method 2. Differential notation. M 4 Mx dy = d logi, (2 2 +3) = d(2x2 +3) = d 2X2 +3 a.2 +3 EXAMPLE 2. Find the area under the curve y = 1/x from x = 1 to x = 10, using formula [VIII] inversely: Ax=10 x= 10 1 -x= =10 1 1 A -X - JX dx = l ogloge 10= = 2.3026.* a=l fX j_ - =l log1o e M EXAMPLE 3. If the rate of increase dy/dx of a quantity y with respect to x is 1/x, find y in terms of x. Since dy/dx = 1/x, y =f -dx = loge x + c, where c is a constant,-the value of y when x = 1. It should be noted that logarithms to the base e occur here in a perfectly natural manner; the same remark applies in Example 2. Note that loge x = log10 x M. This case arises constantly in science. Thus, if a volume v of gas expands by an amount Av, and if the work done in the expansion is AW, the ratio AW/Av is approximately the pressure of the gas; and dW/dv = p exactly. If the temperature remains constant pv = a constant; hence dW/dv = k/v. The general expression for W is therefore W /k d = k loge v + c, and the work done in expanding from one volume vl to another volume V2 is 1V=V2 V=V2k 12 f!2 k 2J WI = = k l dv v= k loge v = loge 2 = v=v J v=1 v vl V M 1I * The number loge 10=1 M=2.302585 is important because common logarithms (base 10) are reduced to natural logarithms (base e) by multiplying by this number, since loge N=logio NXloge 10. Similarly, natural logarithms are reduced to common logarithms by multiplying by M= logio e; since logo N = M' loge N. It is easy to remember which of these two multipliers should be used in transferring from one of these bases to the other by remembering that logarithms of numbers above 1 are surely greater when e is used as base than when 10 is used. 106 THE CALCULUS [VIII, ~ 65 EXERCISES Calculate the derivative of each of the following functions; when possible, simplify the given expression first. 1. log1 x3. 2. log,,0 V 3. log10 (1 + 2 x). 4. log10 (1 + X3). 5. loge (1 + X)3. 6. loge V\3 ~ 5x. 7. loge (1/X). 8. log10 (X-3). 9. x log x2 1 / X t \ 10. loge ~ix* 11. log10 ~2- 1, 12. loge t,,,1 +X 1 t>,.,,1/ t2 loget 14. loge floge]. 1t)4. Evaluate each of the following integrals. 025 4 1 6r62 - 3X2 +X42 16.f -dx. 17.fl d. lfd 19 f20 1 + 2 dx. 20. 1100 (2 - t)3 dt. 1. e 6 t5- 2 t2 - I 1 10 X f i 1 3 t4 1 V aP~,2 X1/2 - 4 S3/2- 2 10 22. f, 3,2 dx. 23.f 5/2 ds. 24.j (1-u)(1+u2)du. 25. Calculate the area between the hyperbola xy = 1 and the x-axis, from x = 1 to 10, 10 to 100, 100 to 1000; from x = 1 to x = k. 26. Show that the slope of the curve y = logio x is a. constant times the slope of the curve y = loge x. Determine this constant factor. 27. Find the flexion of the curve y = loge x, and show that there are no points of infiexion on the curve. 28. Find the maxima and minima of the curve y = loge (x2- 2x + 3). Find the maxima and minima and the points of inflexion (if any exist), on each of the following curves: 29. y = x - loge x. 30. Y = x - loge (1 + x2). 31. y = x2 - 4 logeX2. 32. y = (2 x + logX)2. Find the areas under each of the following curves between x = 2 and x = 5: 33. y = x + 1/x. 34. y = (X2 ~ 1)/x3. 35. y = (X1/2 - x)/x2. VIII, ~ 66] LOGARITHMS 107 36. Find the volume of the solid of revolution formed by revolving that portion of the curve xy2 = 1 between x = 1 and x = 3 about the x-axis. 37. If a body moves so that its speed v = t + 1/t, calculate the distance passed over between the times t = 2 and t = 4. 38. Find the work done in compressing 10 cu. ft. of a gas to 5 cu. ft., if pv =.004. 39. Find the areas under the hyperbola xy = k2 between x = 1 and x = C, c and c2, c2 and c3, c3 and c4. 66. Differentiation of Exponential Functions. Let us consider first the function (1) y=B. Taking the logarithms of both sides of this equation with respect to the base e, (2) logey = loge B. Differentiating both sides with respect to x, we have, by VIII, I dy l dy (3) = loge, or y logeB. y dx dx Hence we have the formula [IXa] d Bx -log B. dx For the special cases B = e and B = 10, we have [IX] dex ex dx [IXb] d-x = 10 log, 10 10x(2.302585...). If u denotes a function of x, we may combine any of these formulas with formula VII, ~ 22; thus the formula IX, which we shall use most often, becomes deu eudu (4) = e dx d' 108 THE CALCULUS [VIII, ~ 67 67. Illustrative Examples. EXAMPLE 1. Given y = eX2, to find dy/dx. Method 1. Set x2 = u; then dy _dy du deu d(x2) =2x.eu =2xe:2. dx du dx du dx Method 2. dy = dex = ex2 d(x2) = 2 xe2 dx. EXAMPLE 2. Find the slope of the curve ex -t e-x (1) y = 2 and determine its extreme values. Since de-x/dx = - ex, we have dy ex - e-z (2) dx 2 To determine the extreme values, first set dy/dx = 0: ex - e-x 1 2 = 0, or e= e- e 2 ex~ Clearing of fractions, e2 = 1, whence x = 0. To determine whether y is really a maximum or a minimum at x = 0, we find d2y _ ex+ e-x (3,J) -dx2 - 2 hence d2y/dx2 = 1 when x = 0. Consequently y is a minimum (~ 42, p. 66) at x = 0. The curve (1) is called a catenary. This curve is very important because it is the form taken by a perfect inelastic cord hung between two points. The given function is often called the hyperbolic cosine of x, and is denoted by cosh x. The expression (ex - e-)/2 in (2) is called the hyperbolic sine of x, and is denoted by the symbol sinh x: ex - e-x exe+ex (4) sinhx= e, cosh x e2T 2 The equations (2) and (3) show that d coshx x, (5) dx sinh dx d sinh x d sinhx = cosh x. dx VIII, ~ 67] LOGARITHMS 109 EXAMPLE 3. If a quantity y has a rate of change dy/dx with respect to x proportional to y itself, to find y in terms of x. Given dy dy ky, dx we may write dx 1 k- = -, dy y hence kx = f dy = loge y +e, by ~ 65, Ex. 3. Transposing c, we have loge y = kx- c, or y = e- = e-cekx = Cekz, where C(= e-c)is again an arbitrary constant. The only quantity y whose rate of change is proportional to itself is Cekx where C and k are arbitrary, and k is the factor of proportionality. This principle is of the greatest importance in science; a detailed discussion of concrete cases is taken up in ~ 68. EXERCISES Find the derivative of each of the following functions: 1. e3x. 2. e2x+z2. 3. e +X. 4. elgs. 6. x'2ex. 6. (1 —x)3eX.. 7. 103+4. 8. al+x)3. 9. log ex. 10. log (1 + ex). 11. log e-2. 12. (log e2X).2 e~t-e-i2 6eq e/zx- ex - "-e-X 13. (ex+ 1)2. 14. e+ e 1. e 16. 2 2 ex +-e-x 17. Show that the slope of the curve y = ex is equal to its ordinate. 18. Show that the area under the curve y = ex between the y-axis and any value of x is y - 1. 19. Find the area under the catenary from x = 0 to x = 3; from x = - to x = +1; fromx = 0tox =a. [See Tables, V, C.] 20. Find the area under the curve y = sinh x from x = 0 to x = 3; from x = 0 to x = a. Find the maxima and minima and the points of inflexion (if any exist) on each of the following curves: 21. y = xex. 22. y = x2e. 23. y = sinh x. 24. y = e-x2. 25. y = xe-. 26. y = sech x =1 + cosh x. 110 THE CALCULUS [VIII, ~ 67 27. Show that the pair of parameter equations x = cosh t, y = sinh t represent the rectangular hyperbola x2 - y2 = 1. Hence show that the area under the hyperbola x2 - y2 = 1 from x = 1 to x = a is represented (see (9) ~ 55) by the integral t=c rt-c fJ sinh2 tdt = / [(cosh 2 t - 1)/2] dt, Jt-O vt=Q where cosh c = a. Hence show that this area is (sinh 2 c)/4 - c/2. 2 ~1 10 28. ex dx. 31. J sinh 2 x dx. 34. f (e + 1)2dx. By 2 2 29. fe-x dx. 32. fcosh 3 x dx. 3. f(ex + 3) e- dx. 30. e2x dx. 33. fsinh2 dx. 36. f(e2Z+3+1)dx. 68. Compound Interest Law. The fact proved in the Ex. 3 of ~ 67 is of great importance in science: If a variable quantity y has a rate of increase (1) ky dx with respect to an independent variable x proportional to y itself, then (2) y = Cekx, where C is an arbitrary constant. The equation (2) between two variables x and y was called by Lord Kelvin the "Compound Interest Law," on account of its crude analogy to compound interest on money. For the larger the amount y (of principal and interest) grows the faster the interest accumulates. In science instances of a rate of growth which grows as the total grows are frequent. EXAMPLE 1. Work in Expanding Gas. The example used to illustrate Ex. 3, ~ 67, can be put in this form. Since, in the work W done in the expansion at constant temperature of a gas of volume v, we found dW/dv = k/v, it follows that dv/dW = v/k; hence v = AeW/k, which agrees with the result of ~ 67. VIII, ~ 68] LOGARITHMS 111 EXAMPLE 2. Cooling in a Moving Fluid. If a heated object is cooled in running water or moving air, and if 0 is the varying difference in temperature between the heated object and the fluid, the rate of change of 0 (per second) is assumed to be proportional to 0: dO dt - k0, where t is the time and where the negative sign indicates that 0 is decreasing. It follows that 0 = C.e-kt. [Newton's Law of Cooling.] Such an equation may also be thrown in the form of ~ 67; in this example, dt/dOt= - 1/(k0), whence t = - (1/k) -loge0 + c, and the time taken to cool from one temperature 01 to another temperature 02 is 0=02 r02 dO 1 02 1 02 ] t -k = Y e loge e =-k loge J0 =01 - I k-0 k j1 ' 1 where 0 is the temperature of the body above the temperature of the surrounding fluid. < The law for the dying out of an electric current in a conductor when the power is cut off is very similar to the law for cooling in this example. See Ex. 19, p. 114. EXAMPLE 3. Bacterial Growth. If bacteria grow freely in the presence of unlimited food, the increase per second in the number in a cubic inch of culture is proportional to the number present. Hence dN =Ck t=1 d = kN, N Cekt, t = loge N + c, where N is the number of thousands per cubic inch, t is the time, and k is the rate of increase shown by a colony of one thousand per cubic inch. The time consumed in increase from one number N1 to another number N2 is AN2 fN21 dN 1 1N2 1 N2 iN, = x1 d N 1 logeN t l= 1 loge. AT J N~/k N -k &e k N, If N2 = 10 Ni, the time consumed is (1/k) loge 10 = 1/(kM). This fact is used to determine k, since the time consumed in increasing N tenfold can be measured (approximately). If this time is T, then T 1/(kM), whence k = 1/(TM), where T is known and M = 0.43 (nearly). Numerous instances similar to this occur in vegetable growth and in organic chemistry. For this reason the equation (2) on p. 110 is often called the "law of organic growth. " — (See Exs. 20, 21, p. 114.) 112 THE CALCULUS [VIII, ~ 68 EXAMPLE 4. Atmospheric Pressure. The air pressure near the surface of the earth is due to the weight of the air above. The pressure at the bottom of 1 cu. ft. of air exceeds that at the top by the weight of that cubic foot of air. If we assume the temperature constant, the volume of a given amount is inversely proportional to the pressure, hence the amount of air in 1 cu. ft. is directly proportional to the pressure, and therefore the weight of 1 cu. ft. is proportional to the pressure. It follows that the rate of decrease of the pressure as we leave the earth's surface is proportional to the pressure itself. dp 1 dp = - kp, p = Ce-h, h = loge p + c, where h is the height above the earth, and, as in Exs. 2 and 3, the difference in the height which would change the pressure from pi to P2 is ha = l- 1/kc.loge ). Jpi \Pi/ Since h]2, and p2 and pi can be found by experiment, k is determined by the last equation. 69. Percentage Rate of Increase. The principle stated in ~ 68 may be restated as follows: In the case of bacterial growth, for example, while the total rate of increase is clearly proportional to the total number in thousands to the cubic inch of bacteria, the percentage rate of increase is constant. In any case the percentage rate of increase, rp, is obtained by dividing 100 times the total rate of increase by the total amount of the quantity, 100 * (dy/dx) - y; and since the equation dy/dx = ky gives (dy/dx) - y = k, it is clear that the percentage rate of increase in any of these problems is a constant. The quotient (dy/dx) y, that is, 1/100 of the percentage rate of increase, will be called the relative rate of increase, and will be denoted by rr. In some of the exercises which follow, the statements are phrased in terms of percentage rate of increase, rp, or the relative rate of increase, rr = rp - 100. VIII, ~ 69] LOGARITHMS 113 EXERCISES Find dy/dx and (dy/dx) - y for each of the following functions: 1. 7 ex. 4. ex2. 7. (ax + b)etx. 2. 4 e-2.5x. x+. e45. 8. (x2 + pX + q)ex. 3. xes. 6. (2 + 2)ez. 9. (3 x + 2)e-x. 10. If a body cools in moving air, according to Newton's law, dO/dt = - k, where t is the time (in seconds) and 0 is the difference in temperature between the body and the air, find k if 0 falls from 40~ C. to 30~ C. in 200 seconds. 11. How soon will the difference in temperature 0 in Ex. 10 fall to 10~ C.? 12. In measuring atmospheric pressure, it is usual to express the pressure in millimeters (or in inches) of mercury in a barometer. Find C in the formula of Ex. 4, ~ 68, if p = 762 mm. when h = 0 (sea level). Find C if p = 30 in. when h = 0. 13. Using the value of C found in Ex. 12, find k in the formula for atmospheric pressure if p = 24 in. when h = 5830 ft.; if p = 600 mm. when h = 1909 m. Hence find the barometric reading at a height of 3000 ft.; 1000 m. Find the height if the barometer reads 28 in.; 650 mm. [NOTE. Pressure in pounds per square inch = 0.4908 X barometer reading in inches.] 14. If a rotating wheel is stopped by water friction, the rate of decrease of angular speed, dw/dt, is proportional to the speed. Find w in terms of the time, and find the factor of proportionality if the speed of the wheel diminishes 50% in one minute. 15. If a wheel stopped by water friction has its speed reduced at a constant rate of 2% (in revolutions per second and seconds), how long will it take to lose 50% of the speed? 16. The length I of a rod when heated expands at a constant rate per cent (= 100 k). Show that dl/dO = kl, where 0 is the temperature; if the percentage rate of increase is.001% (in feet and degrees C.), how much longer will it be when heated 200~ C.? At what temperature will the rod be 1% longer than it was originally? [NOTE. This value of k is about correct for cast iron.] 114 THE CALCULUS [VIII, ~ 69 17. The coefficient of expansion of a metal rod is the increase in length per degree rise in temperature of a rod of unit length. Show that the coefficient of expansion of any rod is the relative rate of increase in length with respect to the temperature. 18. When a belt passes around a pulley, if T is the tension (in pounds) at a distance s (in feet) from the point where the belt leaves the pulley, r the radius of the pulley, and ju the coefficient of friction, then dt/ds = 1uT/r. Express T in terms of s. If T = 30 lb. when s = 0, what is T when s = 5 ft., if r = 7 ft., and ja = 0.3? 19. When an electric circuit is cut off, the rate of decrease of the current is proportional to the current C. Show that C = Coe-kt, where Co is the value of C when t = 0. [NOTE. The assumption made is that the electric pressure, or electromotive force, suddenly becomes zero, the circuit remaining unbroken. This is approximately realized in one portion of a circuit which is short-circuited. The effect is due to self-induction: k = R/L, where R is the resistance and L the self-induction of the circuit.] 20. Radium automatically decomposes at a constant (relative) rate. Show that the quantity remaining after a time t is q = qoe-kt, where qo is the original quantity. Find k from the fact that half the original quantity disappears in 1800 yrs. How much disappears in 100 yrs.? in one year? 21. Many other chemical reactions - for example, the formation of invert sugar from sugar - proceed approximately in a manner similar to that described in Ex. 20. Show that the quantity which remains is q = qoe-kt and that the amount transformed is A = qo - q = qo( - e-kt). Show that the quantities which remain after a series of equal intervals of time are in geometric progression. 22. The amount of light which passes through a given thickness of glass, or other absorbing material, is found from the fact that a fixed per cent of the total is absorbed by any absorbing material. Express the amount which will pass through a given thickness of glass. 70. Logarithmic Differentiation. Relative Increase. In ~ 69 we defined the relative rate of increase rr of a quantity y with respect to x as the total rate of increase (dy/dx) divided by y. If y is given as a function of x, (1) y = f (x), VIII, ~ 70] LOGARITHMS 115 the relative rate of increase dy can be obtained by taking the logarithms of both sides can be obtained by taking the logarithms of both sides of (1),* (2) loge y = logef (x) and then differentiating both sides with respect to x: f(3) 1r =1 dy _dlogey _dlogef(x) y dx dx dx' This process is often called logarithmic differentiation: the logarithmic derivative of a function is its relative rate of increase, rr, or 1/100 of its percentage rate of increase. EXAMPLE 1. Given y = Cext, to find rr = (dy/dx) - y. Taking logarithms on both sides: loge y = loge C + kx; differentiating both sides with respect to x, r dy, d loge Y k. The result of Ex. 3, ~ 68, may be restated as follows: the only function of x whose relative rate of change (logarithmic derivative) is constant is Ce'. EXAMPLE 2. Given y = x2 + 3 x + 2, to find rr. dy _ dy 2x +- 3 Method 1. = 2 x + 3, hencerr- = y = 2 3 Method 2. r, dy d log yt dlog (x2 + 3 x + 2) 2 x + 3 Method 2. rr y - --- d dx dx X2 - +3x + 2 * Since log N is defined only for positive values of N, all that follows holds only for positive values of the quantities whose logarithms are used. t Here and hereafter the symbol log will be used to mean a logarithm to the base e. 116 THE CALCULUS [VIII, ~ 71 71. Logarithmic Methods. The process of logarithmic differentiation is often used apart from its meaning as a relative rate, simply as a device for obtaining the usual derivative. We shall first apply this method to prove the rule for differentiating any constant power of a variable. The equation y = x gives log y = n log x. Differentiating with respect to x, we have ldy 1 ydx x or dy Y n n- n -- dx x x = nx-1. In this proof, n may be any constant whatever. (See ~~ 17, 21.) The logarithmic method is useful also in such examples as those that follow. EXAMPLE 1. Given y = (2 x2 + 3)104X-1. Method 1. Ordinary Differentiation. dy ( d d dy = (2 2 + 3) (104X-1) + 104-1 d (2 2 + 3) dx dx dx = (2 x2 + 3) * 4. - 104-1 + 104x-1. 4 x = 4. 104x-1 (2 x2 + 3)/M + x], where M = logioe = 0.434. Method 2. Logarithmic Method. Since log y = log (2 x2 + 3) + (4 x - 1) log 10, we have 1 dy 4 x y'dx2x~q- 3+4 log!0, y dx 2 X2 + 3 VIII, ~71] LOGARITHMS 117 or d [2 2 + 4 log 10 = 4.104z-1 [x + (2 2 + 3)log 10], which agrees with the preceding result, since loge 10 = 1/logioe = 1/M. EXAMPLE 2. Given y = (3 x2 + 1)2z+4, to find dy/dx. Since no rule has been given for a variable to a variable power, ordinary differentiation cannot be used advantageously. Taking logarithms, however, we find log y = (2 x + 4) log (3 x2 + 1), whence dy = 2 log (3 x2 + 1) + 32+ (2 + 4), y dx 32 + 1 or =dy_ (3 X2 + 1)2x+4 2 log(3 x2 +1) 3 x) + I(2 +4) or (2 + The use of the logarithmic method is the only expeditious way to find the derivative in this example. EXERCISES Find the logarithmic derivatives (relative rates of increase) of each of the following functions, by each of the two methods of ~ 71. 1. e-2. 5. 0.1 elot-5. 9. (r2 + 1) e-r2. 2. 4 e4t. 6. 102x+3. 10. (2 - 3 t2) e2t-1. 3. e3t+2. 7. e-x2+kx. 11. (1 - t2 + t4) 10t3+3. 4. e-Z'. 8. 2 t2e-7t. 12. eer. Find the derivative of each of the following functions by the logarithmic method. 13. (1 + x)1+. 15. x2. 17. (1 + x) (1 + 2 x) (1 + 3 x). 14. (s2 + 1)2s3. 16. tet. 18. 1 + s2 - V -2. 19. If y = uv, show that dy y = du - u + dv - v. In general show that the relative rate of increase of a product is the sum of the relative rates of increase of the factors. 20. If a rectangular sheet of metal is heated, show that the relative rate of increase in its area is twice the coefficient of expansion of the material [see Ex. 17, List XXXI]. 21. Extend the rule of Ex. 19 to the case of any number of factors. Apply this to the expansion of a heated block of metal. 22. Show directly, and also by use of Ex. 21, that the relative rate of increase of xn with respect to x, where n is an integer, is n/x. 118 THE CALCULUS [VIII, ~ 71 23. Compare the functions e2x and e2X+3; compare their relative rates of increase; compare their derivatives; compare their second derivatives. Compare the following pairs of functions, their logarithmic derivatives, their ordinary derivatives, and their second derivatives. 24. ex and 10l. 27. e-ax and e+ax. 25. eaz and eas+b. 28. e-x2 and sech x. 26. eax and 10bx. 29. e-x2 and 1 + (a + bx2). 30. Can k be found so that keax and lOb coincide? Prove this by comparing their logarithmic derivatives, and find b in terms of a. 31. If the logarithmic derivative (dy/dx) - y is equal to 3 + 4 x, show that log y = 3 x + 2 x2 + const., or y = ke3z+2x. 32. If (dy/dx) y = f (x) show that y = keff/(x) dz Find y if the logarithmic derivative has any one of the following values: 33. 1 - x. 35. n/x. 37. ex. 34. ax + bx2. 36. a + n/x. 38. ex + n/x. CHAPTER IX TRIGONOMETRIC FUNCTIONS 72. Limit of (sin 0)/0 as 0 approaches Zero. To find the derivatives of sin x and cos x, we shall make use of the limit sin 0 lim 0-0o 0 Let 0 be the angle AOB, Fig. 27, and let us draw a circle about 0 as center with a radius r = OA, cutting OB at P. Draw PP' and BB' / B perpendicular to OA and draw OP'B'. Then 0 A (1) PP' < arc PAP' < BB', or (2) 2r sin 0 < 2r 0 < 2rtan 0, FIG. 27. since arc PAP' = 2 r * 0 if 0 is measured in circular measure. Dividing by 2 r sin 0, we have 0 1 (3) 1 < a < ' sin 0 cos 0 But cos 0 approaches 1 as 0 approaches zero. Hence 0/sin 0 must also approach 1. It follows that sin 0 (4) lim =1, 0-40 0 provided, as above, that 0 is measured in circular measure.* * On account of the simplicity of this formula and those that result from it, we shall assume henceforth that all angles are measured mn circular measure. 119 120 THE CALCULUS [IX, ~ 73 73. Derivatives of sin x and cos x. Given the equation (1) y = sin x, we proceed to find dy/dx by the fundamental process of ~ 17. We have, using the notation of ~ 17, (A) y + Ay = sin (x + Ax) (B) Ay = sin (x + Ax) - sin x (,Ax'\. Ax = 2 cos ( + sin 2 by formula 13, Tables, II, G. Dividing both sides by Ax, Ax ~Ay_ (~ sin - (C) Ay= ( + Ax)sn 2 (C) -=cos x+ ) ax 2c zx 'A 2 Ax or, if we put 0 = 2 Ay sin 0 A -cos (x + 0) Hence, by ~ 72, (D) dy = limy = cos, dx w- o Ax and we have the formula [X] dsinx=.d sin x x [X] = Cos X. dx Similarly, starting with y = cos x, we obtain the formula [XI] dc — = - sin x. dx By means of formula VII, ~ 22, these formulas may be rewritten in the form d sin u d sinu du du (2) = *-= cos u dx du dx dx' d cos u dcosu du du (3)- - = * sin udu dx du dx dX IX, ~ 74] TRIGONOMETRIC FUNCTIONS 121 EXAMPLE 1. Differentiate y = sin V1 + x2. dy d sin V/1 + x2 s d x/1 + x2 cosx x2 d x dx dx 1 +X2 EXAMPLE 2. Differentiate y = cos ex2. dy - d cos ex2 dex2 dy = d o sin e = - 2 xex sin eX2. dx dx dx 74. Derivatives of tan x, ctn x, sec x, csc x. Given y = tan x, we may write sin x d sin x d cos x d cos x - sin x dy d tan x cos X dx dx 1 _ =tan _ cosx _ ____d___________ -— 1 dx dx dx cos2 x cos2 X d tan x 1 [XII] sec X. dx cos2 x Similarly, cos x d ctn x sinx 1 C [XIII] — c ---- - sc dx dx sin2 x d I rd-ryi sec x cos x sin x [XIV] d x osn = sec x tan x. dx dx cos2 x 1 d csc x sin x - cos x [XV] = - csc x ctn x. dx dx sin2 x These may be combined with formula VII, ~ 22, as in ~ 73. EXAMPLE 1. Differentiate y = cos3 x. Setting u = cos x, we have y = u3, and dy du3 du d cos x d = = 3 U2 = 3 cos2 3 Cos2 xsin x. dx dx dx dx EXAMPLE 2. Differentiate y = cos3 (2 x2 + 1). Setting u = cos (2 x2 + 1), we have y = u3, and dy du3 = du - du ^dx S^= 3 U2 3 s2 (2 2 + 1) Wx- ctd dx dx 122 THE CALCULUS JXx, ~ 74 But du d cos (2X2 +1) =- sin(2X2 + 1) d (2X2 + 1) dx dx dx 4 x sin(2X2 +-1). Hence dy a [3 cos2 (2 X2 ~ 1)] [- 4 x sin (2 x2 + 1)] 12xcos2 (2X2 + 1) sin (2x2 + 1). EXERCISES Find the derivatives of: 1. sin 4 x. 5. sin X4. 9. X cos X. 2. cos (0/3). 6. tan (3 - 2t). 10. e6 ctn 0. 3. tan (- 2 0). 7. cos (- 3 0). 11. log sec x. 4. sin2 x. 8. se (x/2). 12. cos ez. 13. sin x - 4 cos 2 x. 17. el Cos2 (3 t - 1). 14. eldsin (7r/1O - 2 t). 18. e1+2t sin (3 t - ir/4). 15. (1 + x2) sin (3 - 2 x). 19. et/1O (cos t - 4 sin 3 t). 16. log sec esOnz. 20. cos log tan ex. 21. Find the area under the curve y = sin x from x = 0 to x = r/2; test the correctness of your result by rough comparison with the circumscribed rectangle. 22. Find the area bounded by the two axes and the curve y = cos x, in the first quadrant. Find the maxima and minima, and the points of inflexion (if any exist) on each of the following curves. 23. y = sin x. 26. y = x cos x. 29. y = e-x sin x. 24. y = cos x. 27. y = 1 - sin 2 x. 30. y = e-2 xsin x. 25. y = tan x. 28. y = sin x + cos x. 31. y = cos (2x + r/6). Find the derivative of each of the following pairs of functions, and draw conclusions concerning the functions. 32. cos x and sin (7r/2 - x). 35. sin 2 x and 2 sin x cos x. 33. cos2 x and 1 - sin2X. 36. cos 2 x and - 2 sin2 X. 34. sec x and sec (- x). 37. tan2 x and sec2 x. IX, ~ 74] TRIGONOMETRIC FUNCTIONS 123 Integrate the following expressions; in case the limits are stated, evaluate the integrals, and represent them graphically as areas. 38. Cf sin x dx. 40. f sec2 x dx. 42. cos (3 t + 7r/6) dt. JQ Jo ~/d +r"/2 39.f cos x dx. 41. fsin 2 x 7r/2 44. fo (1 + sin x) dx. 45. f(cos x + 3 sin 2 x) dx. 47. f(cos 2x- 1) dx. dx. 43. ftan t sec t dt. 46. fcos2x dx. HINT. 2 cOS2 = 1 + COS 2 X. 4 7r/2 48. f sin2xdx. JO 49. Find the derivative of sin x by showing that sin (x + Ax) - sin x = sin x (cos ax - 1) + cos x. sin Ax and remarking that, as Ax-*0, lim [(cos Ax - 1) + Ax] = 0 and lim[(sin Ax) - Ax] = 1. 50. Find the derivative of cos x as in Ex. 49. 51. Find the derivatives of the two functions (a) vers x = 1 - cos x. (b) exsec x= sec x - 1. 52. Differentiate some of the answers in the list of formulas, Tables, IV, Ea, Eb. What should the result of your differentiation be? [The teacher will indicate which formulas should be thus tested.] Find the speed of a moving particle whose motion is given in terms of the time t by one of the pairs of parameter equations which follow; and find the path in each case. 3.x = 3 cos 2 t. y = 3 sin 2 t. x = 2 cos 4 t. y = 3 sin 4 t. 55. x = sin t + cos t. y = sin t. 6. f x = sec t. 56. = tan t. 57. A flywheel 5 ft. in diameter makes 1 revolution per second. Find the horizontal and the vertical speed of a point on its rim 1 ft. above the center. 58. A point on the rim of a flywheel of radius 5 ft. which is 3 ft. above the center has a horizontal speed of 20 ft. per second. Find the angular speed, and the total linear speed of a point on the rim. 124 THE CALCULUS [IX, ~ 75 75. Simple Harmonic Motion. If a point M moves with constant speed in a circular path, the projection P of that point on any straight line is said to be Y/"s ~ in simple harmonic motion. M Let the circle have a radius a; let the constant speed be v; and let the B\ 0 PA straight line be taken as the x-axis. We may suppose the center of the circle lies on the straight line, since FI. 2. the projection of the moving point on FIG. 28. either of two parallel straight lines has the same motion. Let the center 0 of the circle be the origin. Then we have (1) x = OP = a cos 0, or x = a cos (s/a), where s = arc AM, since 0 = s/a. Moreover, since the speed v is constant, v =s/T, if T is the time since M was at A; or v = s/(t - to) if t is measured from any instant whatever, and to is the value of t when M is at A. We have therefore (2) x =a cos = a cos (t-to) =acos[kt + ]; where k = v/a, and = - kto = - vto/a. From (2), the speed dx/dt of P along BA is dx d [a cos(kt + e)] a s ( (3) = = = - adt sin (kt + E), and the acceleration of P is d2x (4) jT = d - ak2 cos(kt + e) — k2. x, or, d2x (5) ji + x = +x = - k2; IX, ~ 76] TRIGONOMETRIC FUNCTIONS 125 that is, the acceleration of x divided by x, is a negative constant, - k2. We shall see that much of the importance of simple harmonic motion arises from this fact. It is important to notice that (2) may be written in the form x = a cos (kt + e) = a [cos e cos kt - sin E sin kt], or (6) x = A sin kt + B cos kt, where A = - a sin e and B= + a cos E are both constants. The form (6) may be used to derive (5) directly. The simplest forms of the equation (6) result when k = 1 and either A = 0 and B = 1, or A = 1 and B = 0: (7) x= sin t; if k = 1, A = 1, B = 0, i.e. a =, = 3 wr/2. x = cost; if k = 1, A = 0,B = 1, i.e. a = 1, e =0. The formulas (2) and (6) are general formulas for simple harmonic motion; (7) represent two specially simple cases. 76. Vibration. The importance of simple harmonic motion, based on its property (5) of ~ 75, is evident in vibrating bodies, such as vibrating cords or wires, the prongs of a tuning fork, the atoms of water in a wave, a weight suspended by a spring. In all such cases, it is natural to suppose that the force which tends to restore the vibrating particle to its central position increases with the distance from that central position, and is proportional to that distance. (Compare Hooke's law in Physics.) It is a standard law of physics, equivalent to Newton's second law of motion, that the acceleration of any particle is proportional to the force acting upon it. 126 THE CALCULUS [IX, ~ 76 In the case of vibration, therefore, the acceleration, being proportional to the force, is proportional to the distance, x, from the central position; it follows that, in ordinary vibrations, the relative acceleration is a negative constant, - negative, because the acceleration is opposite to the positive direction of motion. For this reason, each particle of a vibrating body is supposed to have a simple harmonic motion, unless disturbing causes, such as air friction, enter to change the result. Neglecting such frictional effects temporarily, the distance x from the central position is, as in ~ 75, x = a cos (kt + E) = A sin kt + B cos kt, where t denotes the time measured from a starting time to seconds before the particle is at x = a, and where e = - tok. Moreover, from ~ 75 and also from what precedes, dx2 dt2 k2x. dt2 The quantity a is called the amplitude, 27r/k is called the period, and to = - /k is called the phase, of the vibration. EXERCISES Find the speed and the acceleration of a particle whose displacement x has one of the following values: compare the acceleration with the original expression for the displacement. 1. x =sin2t. 5. x=sin2t+0.15sin6t. 2. x = sin (t/2- 7r/4). 6. x = sin t - sin 3 t + - sin 5 t. 3. x = sin t - sin 2 t. 7. x = a sin (kt + e). 4. x = cos t + q cos 3 t. 8. x = A cos kt +B sin kt. 9. Determine the angular acceleration of a hair spring if it vibrates according to the law 0 =.2 sin 10 7rt; what is the amplitude of one vibration, the period and the extreme value of the acceleration? IX, ~ 76] TRIGONOMETRIC FUNCTIONS 127 Show that each of the following functions satisfies an equation of the form d2u/dt2 + k2u = 0 or d2u/dt2 - k2u = 0; in each case determine the value of k. 10. u = 10 sin 2 t. 15. u = 5 cos (t/3- -r/12). 11. u = 0.7 cos 15 t. 16. u = 12 cos 3 t-5 sin 3 t. 12. u = 3 e. 17. u = 3 sin 6 t +4 cos 6 t. 13. u = 20e-4t. 18. u = C1 sin 3 t + C2 cos 3t. 14. u = sin (3 t + 7r/3). 19. u = Cie7t + C2e-7t. 20. Show that the function u = A sin kt + B cos kt always satisfies the equation d2u/dt2 + k2u = 0 for any values of A and B. Check by substituting various positive and negative values for k, A, B. 21. Show that u = Aekt + Be-ke always satisfies the equation d2u d U _k2u = 0. dt2 22. When an electrical condenser discharges through a negligible resistance the current C follows the law 2C/dt2 = - a2C, where a is a constant. Express the current in terms of the time. When a = 1000, what is the frequency (number of alternations) per second? 23. Any ordinary alternating electric current varies in intensity according to the law C = a sin kt; find the maximum current and the time-rate of change of the current. 24. When a pendulum of length I swings through a small angle 0, its motion is represented by the equation d20/dt2 = - g0/, very nearly, I being in feet, 0 in radians, t in seconds. Show that 0 = C, sin kt + C2 cos kt, where k = /Vg/l. Find C1 and C2 if 0 = a and the angular speed co = 0 when t = 0; and find the time required for one full swing. 25. A needle is suspended in-a horizontal position by a torsion filament. When the needle is turned through a small angle from its position of equilibrium, the torsional restoring force produces'an angular acceleration nearly proportional to the angular displacement. Neglecting resistances, what will be the nature of the motion? 128 THE CALCULUS [IX, ~ 77 77. Inverse Trigonometric Functions. The equation (1) y = sin-l x. is equivalent to the equation '(2) sin y = x. Differentiating each side with respect to x, we find dy dy I 1 (3) cos y = 1, or - = c = 1 dx dx cos y _V - x2 Hence we have the formula d sin- 1x [XVI] dx 1-x2 dx X-2 It is evident that the radical in these expressions should have the same sign as cos y, i.e. plus when y is in the first or in the fourth quadrant, minus when y is in the second or in the third quadrant. Combining XVI with VII, ~ 22, we may write d sin- u _ 1 du (14) dx -/1 - u2 dx In a similar manner, we find from formulas XI, XII, XIII, XIV, XV, the formulas d cos-Ix - l [XVII] dcosx dx - x2 (radical + in 1st and 2d quadrants). d tan-x 1 [XVIII] 1d x =+ x2 (all quadrants). d ctnx 1 [XIX] d + (all quadrants). dx 1 + X2 d sec-x 1 ] X X \ _ _ (rdx x x2 3 (radical + in 1st and 3d quadrants). IX, ~ 79] TRIGONOMETRIC FUNCTIONS 129 [XXI] dcsc-lx -1 dX X -\/X2 (radical + in 1st and 3d quadrants). d vers-1X [XXII] dx V\2x -X2 (radical + in 1st or 2d quadrants). Each of these formulas may be combined with VII, ~ 22, as in equation (4) above. 78. Illustrative Examples. d sinlX2 _ 1 d (X2) 2x EXAMPLE 1. dx VI - (X2)2 dx -Vi - d tan-1 ex 1 dex ex EXAMPLE 2. EAP I + (ex)2 dx 1 + e2z Ed (sec'1 X)3 d sec' x Ex;ZmPLE 3. 3 (seC-1 X)2 dx dx -3 (sec-l X)2 d log (cos-l x) I d cos-1 x EXAMPLE 4 dx Cos- x dx 1 - 1 cosl x Vi - 79. Integrals of Irrational Functions. By reversal of the formulas for the derivatives of the inverse trigonometric functions, XVI-XXII, we obtain the integrals of certain important irrational functions. c dx d sin-'x 1 [XVI]1 Jvi- ~2 = sin-1 x + C, since-V1x ' X dx d tan-' x 1 fii =+tan- x+C, since d1 = 130 THE CALCULUS [IX, ~ 79 130~ TH Ecl CALCULUS dIXee-'79 Vii ~~~~dx X~x 1 [ CIIi dx =vers'1 X + C, since d vers-'x [XI] V"2 x- dx N2 x ~X2' where, C in each case denotes an arbitrary constant. Since sin-1 x + cos-'x = </2, the student may show that [XVII] leads to the same result as [XVI]. EXAMPLE. To find the area under the curve y= 1/(1 + X2) from the point where x = 0 to the point where x = 1. Since A = f y dx, we have A] f 4dx = tan-1x] = /4 -O0= 7r/4. The fact that we are using radian measure for angles appears very prominently here. Draw the curve (by first drawing y = 1 + X2) on a large scale on millimeter paper and acetually count the small squares as a check on this result. EXERCISES Differentiate each of the following functions. 1. sin-'X4. 2. cos-1 (1 - x)..1.3.. sin-1 (1/x). 4. tan-1 (3 x)..5. sin-1 VI - X2. 6. x sin-1 x. 7. tan-' (1/IX2). 8. ex cosl1 x. 9. log tan-' x. 12. sec'1(X2 + 1). 21. (log tan1 X)3. 1 22. tnx 23. sin-1 V/~ 1- x 13. csc-1 Vi 2 17. cos-1 Vi/ - X2. 1.cn (1 +e 18se1 (log tan x). 15. cos-1 (esin x). 19. e sin-. 16. tan-' (log ex). 20. 10 tanu'z. IX, ~ 79] TRIGONOMETRIC FUNCTIONS 131 Integrate the following functions; in ease limits are stated, evaluate the integral. 24. f0 1d+2 27. f2 do 1 dt dx__ 25. ____ VI- t2 28. f I + 4X2 [Set u = 2 x.] + 1 dx dx 26. 29. f d [e[Setu= 2x.] Integrate after making the change of letters u = 1 - x. dx dx dx 30. JS,-(1- x)2 31. J1+(1 )2 32. JV2xf x2 Find the areas between the x-axis and each of the following curves, between the limits stated. 33. y2 = 1 +X2y2; X = 0 to X= 1/2; x = - 1/2 to x = + 1/2. 34. y + x2y = 1; X = 0 to X = 1; x = Otox =a. 35. y2 = 1 + 4X2y2; X = 0 to X= 1/4;x - 1/4 to x = + 1/4. 36. 4X2y + y + 1 = 0; x = 1 tox =2; = - ito x = + 1. 37. Show that the derivative of tan' [(ex - e —x)/2] is 2/(&x + e-x). [NOTE. The function tan1 [(ex - ex)/2], or tan-' (sinh x), is called the Gudermannian of x and is denoted by gd x: gd x = tan-1 (sinh x). It follows from this exercise that d gd xldx = sech x.] 38. From the fact that d (sinh x) = cosh x dx, show that the derivative of the inverse hyperbolic sine (x = sinh'1 u if u = sinh x) is given by the equation d (sinh-1 u) = + du/ VI + u2. [See foot of p. 108.1 39. Show that d cosh'1 u = ~ du/ Vu2 - 1 40. Show that d tanh-1 u = du/(1 - u2). 132 THE CALCULUS [IX, ~ 80 80. Collection of Formulas for Differentiation. For convenience in reference we shall restate all of the formulas for differentiation, combining each of them with [VII] when it is desirable to do so. dc * dx d(u + v) du dv dx= dx dx' (N\ dN dD d ( D dx N dx dx D2 dy dy du dx du dx VIII. d logB u logBe 1 du "" rd =-logBe dx u ' x dBu du IX. dx = Bu log B.. dcx dx dx dxu d tan u du XII. = sec 2 dx dx d sec u du XIV. = sec u tan u - dx dx dcu du II. = c- dx dx dun du IV. -= nun- 1 _. dx dx duv dv du VI. = ud- + v-xdx dr dx dy dy dx VIa. dx dt dt VIIIa dlogu = du dx u dx deu du IXa. = eu-u Ia dx dx XI. -sin u du dx dx d csc u du XV. dx - csc u ctnud dx dx XVI XVII. id sin- u - +~1 du XVII cos- u /l- u 2dxXVIII, XIX dx ctn- u = 1 +- u2 dx' xvmxix. d* tn iui 1du2 xx xxi d sec-lul +1 du. dx csc-1 u U /U _- dx d 1 du XXII. - vers-l u = _ du dx /2 u - u2dx CHAPTER X APPLICATIONS TO CURVES LENGTH - CURVATURE 81. Introduction. The formulas obtained in Chapters VIII and IX make possible many new applications to curves. We shall treat some of these in this Chapter. 82. Length of an Arc of a Curve. Let s (x) denote the length of the arc of a given curve y = f(x), from a fixed point F to a variable point P. When x increases by t NS Q an amount Ax, let As = arc PQ be the corresponding increase in s, and let Ac = chord PQ. Then (1) AC =Ax2 + A2, whence P A / ax A B Ay R (2) lrnd Ac l + A(Y 2 Ax k x/' 0 x==K X=X FIG. 29. X=x C.x:+AaX / As _ 4I +. Ay)2 AS (3) I +U ac We now require the following fundamental axiom, which forms the basis of the mensuration of curved lines. As the chord and its arc approach zero, their ratio approaches 1, i.e. As (4) lim s = 1. Ac —o AC 133 134 THE CALCULUS [X, ~ 82 Combining (3) and (4), and passing to the limit as Ax approaches zero, we have (5) dx d + - V 1 + where m = dy/dx is the slope of the curve. It follows that the total change in s between any two fixed points x = a and x = b, is (6) Total length = s: = XS = /1 + m2 dx. X-a - =a 83. Parameter Forms. When the equation of a curve is given in parameter form (1) x=f(t), y =(t), we may square both sides of (5), ~ 82, and multiply by dx2. This gives the formula (2) ds2 = dx2 + dy2, which is called the Pythagorean differential formula. It is readily remembered by reference to the triangle PQR, Fig. 29. If we divide both sides of (2) by dt2, we find * (3) (ds)2 2 ) Y+ 2 From (3) we have ds (dx22 (dy2 (4) + dW whence (5) ]; - d) ddt) dt= whicn gives the length of the curve (1) between any two of its points. * This expresses the fact that the square of the total speed ds/dt is the sum of the squares of the horizontal speed dx/dt and the vertical speed dy/dt. This fact, proved in ~ 40, might have been used as the point of departure, and all of the formulas of ~~ 82-83 might have been deduced from it. X, ~ 84] APPLICATIONS TO CURVES 135 84. Illustrative Examples. While the square root which occurs in the formulas of ~~ 82-83 renders the integrations rather difficult in general, the work is quite easy in some examples, as illustrated below. EXAMPLE 1. Find the length of the curve y2 = X from the origin to the point where x = 5. From y2 = X3 we find y = X3/2, whence dy 3 dy 2 l 9x d8- X112 dsNI k dx 1+ dx, dx 2 dxt~ 4 and Xs=5 9X 8 1+X3/2 X==5 335 1 +- dx2 9; ]= f 4 d 7 ) x=O 27 EXAMPLE 2. Find the length of the catenary (~ 67) ex + e-x 2 from the origin to the point where x = 1. We find immediately dy ex - e e ds dxe 2 dx which reduces algebraically to the form (e2 x 2 + e-2x)1/2 ex + e —x dx- 2 hence s dx~d = - e I 1\e/ 2 X=0 2 2 0 2 — l 2 ~ (2.718 - 0.368) 2 I= 1. 175 (nearly). Compare ~ 67, and Tables III, E, and V, C. EXAMPLE 3. Find the length of one arch of the cycloid (Tables x = a (t - sin t), y = a (1 - cos t). We find dx = (a - a cos t) dt, dy ='a sin t dt, dS = VdX2 + dy2 = a V2 - 2 cos t t= 2 a sin dtA 2' whence s 1=1 2 asin- dt= -4 a cos - 4 a [cs?r - cos 0] JtoJ 102 2Jo 4 a[-1 - 1]= 8 a. 136 THE CALCULUS [X, ~ 84 EXERCISES Determine by integration the lengths of the following curves, each between the limits x = 1to x = 2, x = 2 to x = 4, x = a to x = b. Check the first three geometrically. 1. y =3x-1. 3. y =mx +c. 5. y = (2x- 1)3/2. 2. y = 3 + 2 x. 4. y= - (x - 1)3/2. 6. y= (4 x-1)3/2. Find ds, the speed v, and the length s of the path of each of the following motions, between the given limits. 7. x = + t,y = 1 - t; t = to t = 2. 8. x = (1 + t)3/2, y = (1 - t)3/2; t = 0 to t = 1. 9. x = (1-t), y = 8 t3/2/3; t = O to t = 9. 10. x = 1 + t2, y = t- t3/3; t = to t =5. 11. x = 2/t, y = t + 1/(3 t3); t =a to t = b. 12. Find the length of the cycloid (Ex. 3, ~ 84) for half of one arch, i.e., from t = 0 to t = ir; for the portion from t = 0 to t = 7r/2; from t = O to t = 7r/3. 13. Show that the element of length for the cardioid (Tables III, G4) x = 2 a cos 6 - a cos 2 0, y =2 a sin 0 - a sin 2 8, is ds = 2 a [2 -2 (cos 2 0 cos 0 + sin 20 sin 0)]1/2 d 0 = 2 a (2-2 cos 0) 1/2do. Hence show that the entire length of the cardioid, from 0 = 0 to 0 = 2 r is 16 a. Show that the length of the part of the cardioid from 0 = O to 6 = 7r/2 is 4 a (2 - 2). 14. The equation of a circle about the origin may be replaced by the parameter equations x = a cos 0, y = a sin 0, where a is the radius. Hence find by integration the length of the entire circumference. 15. Show that the element of length of the four-cusped hypocycloid (Tables III, G6) x = a cos3, y = a sin a, is ds = 3 a sin 0 cos 0 dO = a sin 2 0 d. Hence show that the length of one quarter of this curve is 3 a/2. X, ~ 85] APPLICATIONS TO CURVES 137 16. Show that the length of the general catenary eax + e-ax Y 2a 2 a from the origin to any point x = x is (ear - e-ax)/2 a. 17. Writing the equation of the simple catenary in the form used in ~ 67, y= cosh x, show that its length between the origin and any point x = x, is sinh x. 85. Areas of Surfaces of Revolution. Consider the surface generated when the arc KL of the curve (1) y =f (x) (Fig. 30) revolves about the x-axis. Let us denote by S(x) the area of the surface generated by the arc KP, and by AS the area IX Q L y as t y=fw P Y xy +A ) a==L_.?='r=Aa._ a_=h x C I N -Y FIG. 30. FIG. 30. of the surface generated by the arc PQ = As. If the curve rises from P to Q we may write, with entire accuracy, (2) 27r y As AS - 2 7 (y + Ay) As. Hence, dividing by Ax and passing to the limit, we may write dS ds (3) d 2ry 27ry + dx= 2 dx whence the area of the surface of revolution from x = a to x = b is (4) 2b ry dsdx 2 ry d x=a a 2 1 dx Similarly, if the curve is revolved about the y-axis, the area between y = c and y = d is (5 S]=d r xd d d f/)dy.\ (5) S = 2 dydy = J2-'y I + dy y —~d 138 THE CALCULUS [X, ~ 85 Finally, if the equation of the curve is written in parameter form x=f(t), y = (t), we divide (2) by At, and let At approach zero, obtaining dS ^ ds = \(dX~i 2 y M.y (3') dS 2ryd y = ( 2 dt + - (~ 83.) Hence the area of the surface of revolution formed by revolving the curve about the x-axis, between points at which t = tl, and t = t2, is t = t 2 t2 d 2 dy2 (6) S] f2 ryds =42 /()2 () dl. A similar formula can be written if the revolution is about the y-axis. EXAMPLE 1. Find the surface of the cone generated when the segment of the straight line y = x - 2 from Y, i/ x = 2 to x = 5 revolves about the x-axis. 5'~x=5 5 dy\2 I=J 2 + d 0.5 O /= J 2 7r (x - 2) A/1 + 1 dx ",,'! =2 j2 7 V2 (x- 2) dx ~ \ / JJ 2 FIG. 31. =2 7r/2 - 2x = 9r =9 /2. Check this solution by finding the area of this cone by elementary geometry. EXAMPLE 2. Find the area of the surface generated when the segment of the curve whose parameter equations are x = (4 t + 1)3/2, y = t2 + 5 between t = 1 and t = 2 is rotated about the x-axis. X, ~ 86] APPLICATIONS TO CURVES 139 ds2 = dx2 + dy2 = (4 t + 1 + 4 t2) dt2 = (2 t + 1)2 dt2 t =2 2 S = fl2 r yds = 2 T7 (t2 +5) (2t + 1) dt = 2 +3 + 5 t2 + 5 t= 2 7r [40 - 10l] = 592 7r. EXERCISES Find the area of the surface generated by each of the following lines when revolved about the x-axis, from x = 1 to x = 2; from x = 2 to x = 4; from x = a to x = b. 1. y =2 x-1. 2. y =3+4x. 3. y =3x +2. Find the area of the surface generated by each of the following curves when revolved about the x-axis, between the limits indicated. 4. y= Vl- -x2; = 0tox = 1; x = tox = 1. 5. y = /4 x-x2; x = O to x = 4; x = 1 to x = 3. 6. y = /7 + 6x - 2; = - 1 to x = 7; x = 2 tox =5. 7. Find the area of the surface generated by rotating the arc of the catenary ex + e-x Y 2 about the x-axis, between the points where x = 0 and x = 1. 8. Find the area of the surface generated by rotating the arc of the curve whose parameter equations are x = 8 t3/2/3, y = (1 - t)2 about the x-axis, between the points where t = 0 and t = 1. 9. Find the area of the surface generated by the arc of the curve 2 whose parameter equations are x = -, y = t + 1/(3 t3) about the x-axis between the points where t = 1 and t = 2. 86. Curvature. A very important concept for any plane curve is its rate of bending, or curvature. The fexion (~ 39, p. 61), dm d2y (1) b- - -dx dx dX2' 140 THE CALCULUS [X, ~ 86 is not a satisfactory measure of the bending; since it evidently depends upon the choice of axes, and changes when the axes are rotated, for example. If we consider the rate of -[TI change of the inclination of the ^ T tangent, a = tan-m, with respect PYZ to the length of arc s, that is, /Aa da (2) lim - /4A a as —os ds FIG. 32. it is evident that we have a measure of bending which does not depend on the choice of axes, since Aa and As are the same, even though the axes are moved about arbitrarily, or, indeed, before any axes are drawn. The quantity da/ds is called the curvature of the curve at the point P, and is denoted by the letter K: the curvature is the instantaneous rate of change of a per unit length of arc. Since a = tan-1 m, and since ds2 = dx2 + dy2 (~ 83, p. 134), we have, da =d tan-1 m= -- m dm, ds = /l + m2 dx, 1 + m2 where m = dy/dx; hence the curvature K is 1 dm dm (3) K _cda 1 + m2 dx _ b ds / 1 + m2 d (1 + m2)3/2 (1 + m2)3/2' where b = d2y/dx2 (= flexion), and m = dy/dx (= slope). It appears therefore that the flexion b when multiplied by the corrective factor 1/(1 + m2)3/2 gives a satisfactory measure of the bending, since K is independent of the choice of axes. X, ~ 86] APPLICATIONS TO CURVES 141 The reciprocal of K grows larger as the curve becomes flatter; it is called the radius of curvature, and is denoted by the letter R: }(4) R _1 ds (1 + m2)3/2 (4) R ds K da- b It should be noticed that this concept agrees with the elementary concept of radius in the case of a circle, T since As = rAa in any circle of radius r. Hence / ds/da = r. / Substituting the values of b and m, formulas (3) oT\ x and (4) may be written in the forms d2y dx2 ((d5Y) ] FIG. 33. [1 +~Z]L/ (5) K =1d [I + (dy)2]3/2 (6) R= d= y d2y dx2 It is preferable, however, to calculate m and b first, and then substitute these values in (3) and (4). Since V/ +- m2 = sec a the formulas may also be written in the form K = 1/R = b cos3 a. It is usual to consider only the numerical values of K, that is I K I, without regard to sign. Since K and b have the same sign, the value of K given by (3) will be negative when b is negative, i.e. when the curve is concave downwards (~ 41, p. 65). The same remarks apply to R, since R = 1/K. 142 THE CALCULUS [X, ~ 87 87. Center of Curvature. Evolute of a Curve. The center of curvature Q of a curve, corresponding to a point P on that curve, is obtained by drawing the normal to the curve at P and laying off the distance R (the radius of curvature) along this normal from P toward the concave side of the curve. Thus, in Fig. 34, denoting the coordinates of P by (x, y), and those of Q by (a, fi), we have (1) a = OB = OA - BA = x - R sin ~, (2) = BQ = AP + CQ = y + R cos0, where f is the angle which the tangent PT makes with the x-axis, (3) tan m = dy whence m 1 (4) sin = V1+m2 cos- = +M Vl + m2 1V1+ m2 It follows that we may write ~~~M(5) l+ M2) + m 2 (5) a= x- m(t +m2) where m = dy/dx, b = d2y/dx2. As the point P moves along the given curve, the point Q also describes a curve, which is called the evolute of the given curve. The parameter equations of the evolute are precisely the equations (5), in which (a, f) are the variable coordinates of the point on the evolute, x is a parameter, and y is to be replaced by its value in terms of x from the equation of the given curve. X, ~ 87] APPLICATIONS TO CURVES 143 In particular examples, it is possible to eliminate the parameter x between the y two equations (5) after having substituted for y, Q m, and b their values found from the equation of the given curve. This gives the equation of the evolute as a single equation between the variables ao and 3 FIG. 35. EXAMPLE 1. Find the values of K, R, a, y, and find the equation of the evolute for the curve y = x2/4. Here m = x/2, b = 1/2. Hence 1/2 _ 4 (1 + x2/4)3/2 (4 + x2)3/2' 1 (4 + x2)3/2 R K 4 x 1 + x2/4 x3 a X — 1 - 2/4 x2 X2 3 x2 S= + = -j+2+1 =-4 2. - 1/2 2 Eliminating x between the last two equations, we find the equation of the evolute in the variables a and 3, 27 a2 = 4 (- 2)3. EXAMPLE 2. Find the values of K, R, a, t, and the parameter equations of the evolute for the cycloid x = a (0 - sin 0), y = a (1 - cos 0). dy dy/d a sin 0 - ctndx dx/d 0 a (1 - cos 0) 2' dm dm/dO - -1 dx dx/d 0 a (1 - cos 0)2 4 40 4 a sin2 1K_1 I. (i+ctn 2 0 |=,1 4 a sin 0 4 a sin42 4a sin 144 THE CALCULUS [X, ~ 87 R 4 a sin 1.01 1 ~ ctn2 2 a a (0 - sin0) + ctn 1 a(+asin0), 2 1 ( i 4 a sin4 1 + ctn2 2 3=a (1-cos)- = - a (1-cos0). 4 a sin4 Y) ~~~CYCIoid FIG. 36. These equations for a and f are the parameter equations of the evolute; they represent a new cycloid, similar to the given one, but situated as shown in Fig. 36. EXERCISES Calculate K, R, a, 3 for each of the following curves; sketch the curve and its evolute: 1. y ='X2. 2. y = x3. 3. y2 = 4 ax. 4. xy = a2. 5. y = sin x. 6. y = Cx. 7. y = (ex + e-x)/2 = cosh x. 8. y = (ex - e-x)/2 = sinh x. 9. X2/a2 + y2/b2 = 1. 10. X2/a2 - y2/b2 = 1. 11. + -\/-Y = V\/a-. 12. X2/3 + y2/3 = a2/3 13. = a cos 0, 1 = sin 0, 1 =a cos3 0, 1 y=a sin 0. 1 y=2cos0. y =asin3 0. 16. X = t2, 17;x = /It, 18. x = sec t, = t-t3/3. Y = t. y = tan t.. x = 2 3 t, 20. x= cos t ~ t sin t, y= t2 4.. y =sin t - t gos t. X, ~ 89] APPLICATIONS TO CURVES 145 88. Properties of the Evolute. If the point P (x, y) lies on a curve y = f (x), and the point Q (a, j) is the corresponding point of the evolute, we have m +- m3 1 + m2 a= — -b ' y = + — b Let m' be the slope of the evolute at Q; then m = dfB df=/dx da da/dx But we have dp + b 2 mb- (1 + m2) db/dx dx b2 3 mb2 - (1 + m2) db/dx b2 da _ b (1 + 3 m2) b - (m + m3) db/dx dx b2 - 3 m2b2 + (m + m3) db/dx b2 It follows that,dO 1 m' - d _ da m Hence the normal to the curve at P is the tangent to its evolute at Q. Hence the radius of curvature of the curve at C is tangent to the evolute at Q. (See Fig. 35.) 89. Length of the Evolute. As Q moves along on the evolute, the rate of change of R is the rate of change of the arc of the evolute. For, since we have, in Fig. 34, (1) R2 = PQ2 = (x- a)2 + (y - )2, it follows that (2) R dR= (x - a) (dx - da) + (y - ) (dy - d,). But we have m dy = x r x a m dx Y- or (x - a)dx+(y -13) dy = O. dx Y-f3 146 THE CALCULUS IX, ~ 89 Hence (2) may be written in the form (3) dR -= (x - a) d + (y ~- ) dp (3) dR X ~ V(x- a)2 + (y - )2 By ~ 88, we have d: 1 _ y- da m x -a Substituting this in (3), we find da + y df da+ d d x - a da (4) d R = +- = =da2~ d12.: 1- -Y 2 +l(do)2 This result, however, is precisely ds, where s is the length of the arc of the evolute. Hence we have R= R2 = s = s2 (5) dR = ds, or R dR = j ds, or R2 - R1 = 2 - Sl; R= R=Ri Js = s that is: the rate of growth of the radius of curvature is equal to the rate of growth of the arc of the evolute; and the difference between two radii of curvature is the same as the length of the arc of the evolute which separates them. This fact gives rise to an interesting method of drawing the original curve (the involute) from the evolute: Imagine a string wound along the convex portion of the evolute, fastened at some point (say Q, Fig. 35, p. 143) and then stretched taut. If a pencil is inserted at any point (say P, Fig. 35) in the string, the pencil will traverse the involute as the string, still held taut, is unwound from the evolute. As exercises, the lengths of the portions of the evolutes of curves given in the preceding list of exercises may be found. CHAPTER XI POLAR COORDINATES 90. Introduction. Since the equations of curves in polar coordinates often involve trigonometric functions, the equations of curves have been written in rectangular coordinates throughout the earlier part of this book. We shall now show how to extend many of the results already found for curves whose equations are written in rectangular coordinates to curves whose equations are written in polar coordinates. 91. Angle between Radius Vector and Tangent. Let C (Fig. 37) be a curve whose equation in polar coordinates is (1) =/f (0). Consider the angle, S between the radius XD vector OPR and the tangent PT to C at T P. Let Q be a second point on C, with coordinates p + Ap, 0 + AO, and 0 FIG. 37. let o be the angle between the radius vector OQD and the secant PQS. As AO approaches zero, i.e. as Q approaches P along C, we shall have (2) = lim X, and tan I = lim tan 0. 8A0-0 A-014 147 148 THE CALCULUS [XI, ~ 91 Draw AP perpendicular to OQ; then we shall have AP AP p'sin AO (3) tan 0 tn = Q=QQ A. ~ p+Ap- pcosAG and p sin AO p sin AO (4) tan t=lim =lim a-oP +A pp cosA Ao-o P (1 - cos AG ) +AP To evaluate this limit, divide both numerator and denominator by AO and note that sin AO (5) lim- =1, A0-0 AG and AO 2 sin2 (6) limlCosAO=lim 2 z~-O AO AO K sin=lim nAO 2 0I 0. AO-*>O 2 A 0 It follows that sin AG ne (7) tan t =lim 1 - cos + AP dp AG AG dG and therefore 14 p dG tar d7 P 4p pd (8) dpd dO The angle a between the x-axis and the tangent PT can be found, after 'fr has been found, by means of the relation (9) a =G-Hir. XI, ~ 91] POLAR COORDINATES 149 EXAMPLE 1. Given the curve p = e6, to find tan 4, and 4 itself. Since p = e0, dp/d 0 = e0, and tan V = p - (dp/d ) = 1. Hence V = tan- 1 = 7r/4 = 45~. It follows that this curve cuts every radius vector at the fixed angle of 45~. EXAMPLE 2. Given the curve p = sin 2 0, find 4' at the point where o = 7/8. tan + = Pp + (dp/d 0) = sin 2 0 2 cos 2 0 = (1/2) tan 2 0. When 0 = ir/8, tan 4 = 1/2, and 4 = 26~ 34', approximately. EXERCISES Plot each of the following curves in polar coordinates; find the value of tan 4 in general, and the value of 4' in degrees when 0 = 0, 7r/6, 7r/4, ir/2, 7r. 1. p = 4 sin 0. 6. p = 0. 11. p = sin 3 0. 2. p = 6 cos - 5. 7. p = 02. 12. p = 2 cos 3 0. 3. p =3+4cos0. 8. p = l/. 13. p=3sin(30+2 r/3). 4. p = tan0. 9. p =e2. 4. p = 3cos'0 + 4 sin0. 6. p =2 + tan2 0. 10. p =e-40. 15. p = 2/(1- cos 0). 16. Show that tan a is constant for the curve p = keao. Find tan A for each of the following curves: 17. p = p/(l - e cos 0) (conic). 19. p = a (1 + cos 0) (cardioid). 18. p = a sec 0 ~ b (conchoid). 20. p2 = 2 a2 cos 2 0 (lemniscate). 92. Areas in Polar Coordinates. Let KL be an arc of a curve whose equation in polar coordinates is (1) p=f (0) Consider first the area of the sector OKP bounded by the radius vector OK, for which 0 = 01, any other radius vector OP for which 0 = 0, and the intercepted arc KP of the curve. The area is a function of the angle 0; let us denote it by A(0). 150 THE CALCULUS [XI, ~ 92 Now let 0 increase by an amount A = ZPOQ, and let PS and TQ be circular arcs whose radii are p (= OP) and p + Ap (= OQ), respectively. Then we shall have (2) sector POS < sector POQ < sector TOQ, if p increases with 0. If p L decreases when 0 increases, Q (O= O+ AO) the inequality signs must / x\,(p+A-p)dA be reversed in (2) and in /what follows. P/ / (=) The area of any circular P~/ /a~ y ^ sector is equal to half the // g-g Ko(- k) product of the angle (in A$ f (ek) circular measure) and the / /.^ A;=(k square of the radius. The 0 sector POQ is the amount FIG. 38. of increase in the area A; let us call it A A (0). Then (2) becomes p2 0 (p +Ap)2 A0 (3) < AA (0)< 2 2 2 Dividing through by A0, we have (4) p2 AA (0) (p+Ap)2 (4) < o< 2 whence (5) lim AA (0) p2 dA p2 (5) lim =, or dO = - AO-0O AO 2' du 2 It follows that the area of the sector bounded by the curve and the radii vectores for which 0 = 01, and 0 = 02, respectively, is given by the formula 02 1 '02 (6) A = = p2 da. a,2. XI, ~ 92] POLAR COORDINATES 151 EXAMPLE 1. Find the area of the sector bounded by the curve p = 1/0 and by the radii corresponding to 0 = 7r/3 and 0 = 7r/2. The curve may be plotted readily by taking corresponding values of p and 0, as in the following table. 0 0 7r/6 7/3 ir/2 2 7/3 5 7w/6 Ir etc. p.o 1.91 0.95 0.64 0.47 0.38 0.32 etc. The required area is / 2 ~rr/2 6n/ ~ R7/ A0= ~/21 '/d= 2V J d 3 0=7r/3 2 -r/3 02 2 r //3 1T/^Y1 /^Vl / I l v \7 1 2 3 = 0.16+. EXAMPLE 2. Find the area of the sector bounded by the curve p = tan 0, and by the radii corresponding to 0 = 45~ FIG. 39. and 0 = 60~. 600~ 1 -/3/3 1 fr/3 1 fr/3 d A = p2 2deO =- tan2 0dO (sec2 0 - 1) d 450 2-1/4 2 r/4 2aJ/4 o1 I r f,/3 1 (tan 0 - 0) /3 1 (3- wr/3) - (1 - 7r/4)] (a 00 7/4 2 = ( /3 - 1 - 7/12) =.2351 + EXERCISES Calculate the area formed by each of the following curves and the indicated radii, and check graphically. 1. p = O; 0 = 0 to V. 2. p = 6/02; 0 = 7r/3 to 7r/2. 3. p = V/0; 0 = r to 2 w. 4. p = 0; 0 = 0 to 2. 5. p = 4// 0; 0 = 7r/8 to 7r. 6. p = 1 + V/O; 0 = 7r/4 to 7r. 7. p = /Vl +; 0 = 1 to 3. 8. p = V + 02; 0 = 0 to 3. 152 THE CALCULUS [Xi, ~ 93 9. p = (0- 1)2; 0 = 1 to 6. 10. p = (1 ~ 0)/02; 0 = 1800 to 360' 11. p=sino0; 0 = O tow7r/2. 12. p=coso0; 0 = wr to 2 7r. 13. p =sec 0; 0= 7r/4 tow7r/3. 14. p = 1~sino0; 0 = O tow7r/2. Find the area bounded by each of the following curves: 15. p 4 cos 0. 16. p = 4 cos 2 O. 17. p2 =4 cos 2. 18. p = 1 - sinG0. 93. Lengths of Curves in Polar Coordinates. Let the equation of a curve in polar coordinates be (1) P=f (), and let s denote the length of arc from a fixed point K to a variable point P. Then s is a function of 6. We shall show Q (0==0+A) first how to obtain ds/dG, AS whence we may- proceed to P (0=0) find s itself by an integration. Let the coordinates of P be (p, 6), and those of a second K(o==k) point Q be (p + Ap, 6 + AO). 0 Denote the arc PQ by As, and the chord PQ by Ac. Then FIG. 40. from the identity As As AC (2) A = 2 we find (3) ds= limAs = lim limAc See (4),~82. dO A-*oA AO 4-ooAc AO Ao,o AO From the law of cosines, Ac = VpP + (p + Ap)2 - 2p (p + Ap) cosAO - V2(p + pAp) (- COSAO) +Ap2. It follows that Ac _ 1 2(1 - cos AO) 2'A (4) A V P + pAP). + AO A02 AO ) I XI, ~ 931 9POLAR COORDINATES153 153 But we have * (5) 1*m2(1-cos AO) plandlim dp Ao —tO A62 A-o0 AO dO Hence, by (3) and (4), (6) ds= 2 dp do P do ' Integrating both sides of equation (6) with respect to 0, we find the important formula (7) S= J ab \1p2 + dp 2do a ed-ode Equation (6) may be supplied by squaring both sides and then multiplying both sides by (dO)2. The resulting formula (8) ds2 = p2 dO + dp2 is the Pythagorean differential formula in polar coordinates. (See ~ 83.) 90g 1800 0 ) ) 00 3600 2700 FIG. 41 EXAMPLE. Find the length of the curve p = 1 - cos 0. dp = sin 0d0. d s/d 0 = /(d p/d O)2 ~ p2. -Vsin2 0 + (1 - cosj)2. - V2 (1 - cos 0) = 2 sin (0/2). The complete curve is traced when 0 varies from 0 to 2 r. Hence. 21 sin (0/2) d 4 cos]- = 8. * See and compare (6), p. 148. In this case we have 2 (1 - cos AO) 4 sin, AO sin AO 22 2 2 ~ 2 Y 2 A2 A 154 THE CALCULUS [XI, ~ 94 EXERCISES Find the length of each of the following curves, or of the portion specified. 1. p = 5 sin 0. 6. p = a csc 0; 0 = 45~ to 90~. 2. p = 3 cos 0. 7. p = sin 0 + cos 0; 0 = O to?r. 3. p = e; 0 = 0to 7r/2. 8. log p = 0; = 2 to3. 4. p = ea8; 0 = 0 to 7r. 9. log p = 3 + 2 0; 0 = 0 to 1. 6. p =sec0; 0 = to 1. 10. p = + sin. 94. Curvature in Polar Coordinates. By the definition of the curvature K of any curve (~ 86), we have da a da/dO (1) dK = ds ds/dO Since a = 0 + (~ 91), we have da/dO = 1 + d4/dO. Denoting dp/dO by p' and d2p/dO2 by p", we have (2) = tan-l -- d4 d 11 d (3) tan-1 dO d O tan p 1 + 1 p,2 - PP pp p2 _ p pPt + (p/p')2 p'2 p2 + p'2 It follows that d(A da p2 _ p p P2 + 2 p,2 - p pI (4) 1 + 1 + d '2pp" p, 2 + pp d/ d 1- P- $ p2 + p, 2 Also, from (6), ~ 93, ds (5) = Vp2 + p'2. Hence we have (6) K- dal/dO p2 +2 2 p_- pp" ds/dOe [p2 + p,23/2 EXAMPLE 1. Find the curvature for any point on the curve p = e20. Since p' = dp/dO = 2 e2, and p" = d2p/dO2 = 4 e20, we have 640 + 8 e40 - 4 e40 1 (40 4+ 4 e40)3/2 /5-20 XI, ~ 94] POLAR COORDINATES 155 EXAMPLE 2. Find the curvature at any point of the circle p= a sin 6 Here we find p' a cos 6, and p"=- a sin 6. Therefore K a2 sin2 6 + 2 a2 cos2 69 ~ a2 sin2 6q K= (a2 sin2 6 + a2 cos2 6)3/2 2 2a2 (sin2 6 ~ cos2 6) 2 a3 (sin2 6 + cos2 6)3/2 a This is the reciprocal of the radius. (~ 86.) 1EXERCISES Find the radius of curvature of each of the following curves: 1. pe=0. 4. p =cos60. 7. p=a (1+ cos60). 2. p =ao. 5. p =sin 36. 8. p =2/(1~+cos 0). 3. pO=a. 6. p=a sec 2. 9. p~a+bcoso. CHAPTER XII TECHNIQUE OF INTEGRATION 95. Question of Technique. Collection of Formulas. The discovery of indefinite integrals as reversed differentials was treated briefly, for certain algebraic functions, in Chapter VII. We proceed to show how to integrate a variety of functions, but the majority are referred to tables of integrals, since no list can be exhaustive. See Tables, IV, A-H. To every differential formula (pp. 44, 132) there corresponds a formula of integration: if d f (x) = f (x) dx, then Jf (x) dx = ) (x) + C. The numbers assigned to the following formulas correspond to the number of the differential formula from which they come. Certain omitted numbers correspond to relatively unimportant formulas. FUNDAMENTAL INTEGRALS dy [I]i If dr = 0, then y =constant. [See ~ 52, p. 86.] [The arbitrary constant C in each of the other rules results from this rule.] [II], fkf (x) dx = kf (x) dx. [III] f{f (x) + 4 (x)} dx-f f(x) dx +~fx (x) dx. [IV]i xn dx= xn+ + C, when n i - 1. (See VIII.) 156 XII, ~ 95] TECHNIQUE OF INTEGRATION 157 [VI]i UV =f (UV) fu dv +f v du. ["Parts"] [The corresponding formula [Vh for quotients is seldom used. See ~103.] [Substitution] =Cf[(x)]I4_ d~ x) Jdx -x [VIII]i = log x + C. [IX]i fex dx' = ex + C. [X] fcos xdx = sin x +C. [XI]i fin xdx = cos x +C. [XII]i fsec2 x dx =tan x ~ C. [XIII]j fSC2 xdx = - ctn x ~C. [XIV]i fsec xtan xdx =sec x +C. [XV]i fcscxctnxdx= -cscx+C. Cdx [XVIII]i +X2 =tan-' x+C -ctn1+C.XX] cdx [XX]i J='~ =sec -1 x+ C= -Csc 1 X+ C'. [XXI]i dx [XXII~i )V -x2= vers -Ix + C The remaining differential formulas referred to on p. 132 give rise to other integral formulas; these will be found in the short table of integrals, Tables, IV, A-H. 158 THE CALCULUS [XII, ~ 96 96. Polynomials. Other Simple Forms. The rules [II], [III], [IV] are evidently sufficient without further explanation to integrate any polynomials and indeed many simple radical expressions. This work has been practiced in Chapter VII extensively. Attention is called especially to the fact that the rules [II] and [III] show that integration of a sum is in general simpler than integration of a product or a quotient. If it is possible, a product or a quotient should be replaced by a sum unless the integration can be performed easily otherwise. Thus the integrand (1 + x2)/x should be written 1/x + x; (1 + x2)2 should be written 1 + 2 x2 + x4; and so on. This principle appears frequently in what follows. 97. Substitution. Use of [VII]. As we have already done in simple cases in Chapters VII and VIII, substitution of a new letter may be used extensively, based on Rule [VII]. dx EXAMPLE 1. To find r 2 Va2 _ X Set u = x/a, then du = dx/a, or dx = a du, and dx =f a du;= du 1 > dx = a du =fd du = sin-l u + C = sin-1 + C. /a2 - x2 J Va2- a2u2 V- u2 a. x 1 x\ dx/a dx CHECK. d sin1 = -) d / = = a X X2 a a x2 a2 - x2 V a2 -a-2 EXAMPLE 2. To find f sin 2 x dx. METHOD 1. Direct Substitution. fsin 2x dx = fsin (2 x) d (2 x) = -- [cos 2 x + Cl = - ~ cos 2 x + C'. Check. d (- cos 2 x) =-d cos (2 x) = sin (2 x) d (2 x) =sin 2x dx. METHOD 2. Trigonometric Transformation and Substitution. fsin 2x d =2 sin x cos x dx = - 2os d (cos x) =- (cos x)2 + K = - cos2 x + K. XII, ~ 98] TECHNIQUE OF INTEGRATION 159 Notice that cos2 x + K = (1/2) cos 2 x + C'since cos 2 x = 2 cos2 x - 1. Do not be discouraged if an answer obtained seems different from an answer given in some table or book; two apparently quite different answers both may be correct, as in this example, for they may differ only by some constant.* Whenever a prominent part of an integral is accompanied by its derivative as a coefficient of dx, there is a strong indication of a desirable substitution; thus if sin x occurs prominently and is accompanied by cos x dx, substitute ui = sin x; if log x is prominent and is accompanied by (1/x)dx, set u = log x; if any function f (x) occurs prominently and is accompanied by df (x), set u = f (x). This is further illustrated in exercises below. 98. Substitutions in Definite Integrals. In evaluating definite integrals, the new letter introduced by a substitution may either be replaced by the original one after integration, or the values of the new letter which correspond to the given limits of integration may be substituted directly without returning to the original letter. x=-r/2 EXAMPLE 1. Computefd sin x cos x dx. METHOD 1. Let u = sin x. sX= v/2 z =?r/2 r sin x cos dx d = / u du aX=0,x=0 U2 x= r/2 sin2 X = 7r/2 1 2j_ o X= 2 jX=o 2 METHOD 2. X=X/2.=1 U2u=l 1 fr/ sin x cos dx = d u du- JX=0 =0 -02 =o - 2' since u (= sin x) = 0 when x = 0, and u = 1 when x = r/2. Care must be exercised to avoid errors when double-valued functions occur. The best precaution is to sketch a figure showing the relation between the old letter and the new one. In case there seems to be any doubt, it is safer to return to the original letter. * Occasionally it is really difficult to show that two answers do actually differ by a constant in any other way than to show that the work in each case is correct and then appeal to the fundamental theorem (~ 52). 160 THE CALCULUS [XII, ~ 98 EXERCISES Integrate each of the following expressions: 1. f(1 -X)(1+ X2) dx. 5. f(&~ex -e-)2 dx. 2. dx. 6. dx. 3. fPa +bx)2cix. 7. f(l1-2 X)2V.X-dX. 4. j(x - d)x 8. f(x3 -2) (X1'/2~+X2 /3) dX. In the following integrals, carry out the. indicated substitution; in answers, the arbitrary constant is here omitted for convenience in printing. 9. f\/3 x+~2dx; setu= 3 x +2. Ans. 2 U3/2= (3 x+ 2)3/2. 1o.f d2= log (3 x + 2)=log~3 x+ 2. 1.fl+(3x+2)2 3t (x+) 12. fxV4 + x2dx; set u =4+ X2. AnS. U3/2/3 =(4 ~X2)3/2/3. 13.f4X dx =log (4~+X2) =logV4\ +X~2. 14. fsin x v'C~sxdx; set u =cos x. Ans. -2 U3/2/3 = -2(COS3/2X)/3. 15. fcos x Vsin x dx = 2 (sin3/2X) /3. 16. fcosx(1+4sinx+9sin2x)dx =sinx+2sin2X+3sin3x. 17. fsin3 x dx Pfin x (1 - COS2 x)dx COS csX + (cos3 x) /3. 18. fOs (3 x- 2) sin (3 x- 2)dx =sin2 (3 x- 2). 19. fin (1 - 3 x) cos5/2 (1 - 3 x) dx =ACOS7/2 (1 - 3 x). 20. X set u = X.Ans. -tan1 u =tan-'. fa2 + X a a a a XII, ~ 98] TECHNIQUE OF INTEGRATION 161 In the. following integrals, find a substitution by inspection and complete the integration. 21. f =~ = log(4 x +3) = log (4 x+ 3)1/4. 22. fV3-4xdx = -(3-4X)312/6. 23. fsi'n (4 x- 3) dx = —cos (4x - 3). 24. f2X2~~ log V2X2 -5 95. fcos3 x dx= sin x - (sin3 x)/3. (See Ex. 17.) 26. fCos x sin4 X dx = (sin5 x)5. 27. f2X COS(1+ x2) dx =sin (1+ X2). 28. ftan3 X sec2 x dx = (tan4 x)/4. (u = tan x.) 29. f ctn3 x csc2 x dx = - (ctn4 x)/4. (u = ctn x,.) 30. fcos2x dx = f[(1 + cos 2 x)/2] dx = x/2 + (sin 2 x)/4. 31. fsin2 x dx =frKi - cos 2 x)21 dx = x/2 - (sin 2 x)/4. 32. fcos5 x dx = sin x - 2 (sin3 x)/3 + (sin5 x)/5. 33. fctn x dx =f(cos x/sin x) dx = log sin x. (Put u =sin x). 34. ftan xdx = log COS x=log sec x. "' dx~~~ 35. f dx sin-1 36. f6 + 3 f 2 + X2 tan-(1 ) 162 THE CALCULUS' [XII, ~ 98 Compute the values of each of the following definite integrals. 38 -f 1 dx 1a-I Xx=11 I 1 _ __ 38.f =-~~~~tan1 I ~tan-1 3+X2V3 _V/1x x1/3 V/ 6V x r3 dx =sn x= sin3'1 1x=0 V3-X2 kV x=o2 40. f X dX = oe(+X2)]x= (loge 12 -loge, 3) =loge 2. 41. flogx dx __2 2-0 +x2 0 42. f sin3 x=[ o cs x)/3]'=2/3. x=1 X=1 43 f X eX2 dx = ex'2/2 =e/2 - 1/2 =.8591 + 44. f xv / + x dx. 48. 6+ dx2. f.x=2 e-2x dX. 49. fx=2 I dx. 4. sin4 xxcodxx 47. f sin 3 xdx. 51.f cos3 xdx. 52. Find the area under the witch y = 1 /(a + bX2) for a =9, b = 1, from x = 0 to x = 1; for a = 8, b = 2, from x = 1 to x =10. See Tables, iIIl J. 53. Find the volume of the solid of revolution formed by revolving one arch of the curve y = sin x about the x-axis. 54. Find the area under the general catenary y = a cosh (x/a) = a(ex/a+,e-x/a)/2 from x = 0 to x = a. 55. Find the area of one arch of the cycloid x =a (0 - sin 0), y = a (1 - cos 0). 56. Find the volume of the solid of revolution formed by revolving one arch of the cycloid about the x-axis. 57. Compare the area of one arch of the curve y = sin x with that of one arch of the curve y = sin 2 x; with that of one arch of y = sin X. XII, ~ 99] TECHNIQUE OF INTEGRATION 163 58. Show how any odd power of sin x or of cos x can be integrated by the device used in Ex. 17. 59. Show how any power of sin x multiplied by an odd power of cos x can be integrated. 99. Integration by Parts. Use of Rule [VII. -One of the most useful formulas in the reduction of an integral to a known form is [VI], which we here rewrite in the form [VI'] fu dv = uv -fv du called the formula for integration by parts. Its use is illustrated sufficiently by the following examples: EXAMPLE 1. fx sin xdx. Put u = x, dv = sin x dx; then du = dx and v =f sin x dx = - cos x; hence, fx sin x dx = -x cos x +fcos x d = -x cos x + sin x; (check). EXAMPLE 2. flog x dx. Put log x = u, dx = dv; then du = (1/x) dx, v = x, and flogd x =xlog xS - f x log x-fdx = x log x - x + C; (check). EXAMPLE 3. f/a2 - x2 dx. Put u = Va2-x2, dv=dx; then v = x, du = x dx, and a2 - x2 dx = x a2 - x 2 d -Vda2 - x2 J V\a2 - x2 but, by Algebra, r 2 d = =- \a2-2 d2 dx f XVa22 -d+ Va2 -X2 hence 2 a2 -.x2 dx = xVa2 - x2 + a2 -x2 x Va22 - 2 + a2 sin- ( + C. a This important integral gives, for example, the area of the circle x2 + y2 = a2, since one fourth of that area is Va- x_2 dx =2 [x Va2 -x2 + a2 sin-l (-) fx =02 L \/ J=o a2.] =7ra2. 2i" 2J 4 164 THE CALCULUS [XII, ~ 99 EXERCISES Carry out each of the following integrations: 1. fx cos x dx = x sin x + cos x + C. 2.fPOeTdx =ez(x - 1) +C. [HINT. U = X, dv=eZ dx.] 3. fx log x dx - X2/4 + (X2 log x)/2 + C. 4. fx3 log x dx - X4/16 + (X4 log x)/4 + C. 5. fX22e- dX = ex (X2 - 2 x ~ 2) ~ C. [HINT. Use [VI] twice.] 6. fsin-' x dx = sin1 x + V1 - -x2 + C. [HINT. u = sin-' x.] 7. ftan-1 x dx = x tan-1 x - log (1 + X2)'/2 ~ C. 8. fx2 tan-1 x dx (=X tan-' x)/3 - X2/6 + log (1 + X2)1/6 + C, 9. fx (ex - ex)/2 dx =fx sinh x dx = x cosh x - sinh x + C. 10. fX2 e3x dX = elz (9 x2 - 6 x + 2)/27 + C. 11. fez sin x dx = ex (sin x - cos x)/2. [Set u = ex; use [VI] twice.] 12. fe2x cos 3 x dx = e6x (3 sin 3 x + 2 cos 3 x)/13. 13. fe- sin 3 x dx = - e- (3 cos 3x + sin 3 x)/10. 14. fea — cos nx dx = ea (n sin nx + a cos nx)/(a2 ~ n2). 15. Show that fP (x) tan-' x dx, where P (x) is any polynomial, reduces to an algebraic integral by means of [VI]. Show how to integrate the remaining integral. 16. Show that fP (x) log x dx, where P (x) is any polynomial, reduces to an algebraic integral by means of [VI]. Show how to integrate the remaining integral. 17. Express Jxn eax dx in terms of fxn-' eax dx. Hence show that fP (x)eax dx can be integrated, where P (x) is any polynomial. XII, ~ 100] TECHNIQUE OF INTEGRATION 165 Carry out each of the following integrations: 18. (2 - 4 x + 3 x2) log x dx. 20. f(x2 - x)e-2L dx. 19. f(3 x2 +4 x + 1) tan-1 x dx. 21. f(x3 - 5)e4z dx. 22. From the rule for the derivative of a quotient, derive the formula f(l/v)du = u/v +f(u/v2) dv. Show that this rule is equivalent to [VI] if u and v in [VI] are replaced by 1/v and u, respectively. 23. Integrate f6e sin x dx by applying [VI] once with u = ex, then with u = sin x, and adding. 24. Integrate fea sin nx dx by the scheme of Ex. 23. Find the values of each of the following definite integrals: X=e x=l 25. X log x dx. 28. ] (1 + 3 x2) tan-l x dx. Jx=l x=O 26. f xe-x dx. 29. f e-2 cos 3 x dx. Jx=- - Jx=o x= 1/2 x=2 27. x sin-1 x dx. 30. f/ (e - e-) dx. x=0 Jx=l 31. Find the area under the curve y = e-' sin x from x = 0 to x = r. Find the area under each of the following curves, from x = 0 to x = 1. 32. y = xex. 33. y = x2e-Z. 34. y = x3e-. 35. Compare the area beneath the curve y = log x from x = 1 to x = e with the area between this curve and the y-axis, and the lines y = 0 and y = 1. 36. Show that the sum of the area beneath the curve y = sin x from x = 0 to x = k and that beneath the curve y = sin-1 x from x = 0 to x = sin k is the area of a rectangle whose diagonal joins (0,0) and (k, sin k). 100. Rational Fractions. Method of Partial Fractions. A fraction N/D, whose numerator and denominator are polynomials, is called a rational fraction. If such a rational fraction is to be integrated, we first note whether or not the degree of N is less than the degree of D. If not, we divide 166 THE CALCULUS [XII, ~ 100 N by D, to obtain a quotient Q and a remainder R whose degree is less than that of D. N 2x4 +x2+-2x- 2x- 1 —x2 EXAMPLE. - = - - = 2 x + D X3 + x X3 + x The integration of Q can be carried out at once. It remains only to consider the integration of rational fractions of the form R/D, where the degree of R is less than that of D. We shall proceed to discuss the integration of such forms, and we shall divide the discussion into several cases. 101. Case I. Denominator Linear or a Power of a Linear Form. If the denominator of the fraction RID (~ 100) is linear, or is a power of a linear form, i.e. if D = ax + b, or D= (ax + b)", the substitution u = ax + b transforms the integral of R/D into a new form which is readily integrated. For this reason, we shall reduce other cases to this case whenever possible. 102. Case II. Denominator the Product of Several Linear Factors. If the denominator of the fraction RID (~ 100) is the product of several linear factors, the fractions R/D may be replaced by a sum of simpler fractions. The simpler fractions which compose this sum are called partial fractions. Each linear factor ax + b of D gives rise to one such partial fraction, whose denominator is ax + b and whose numerator is a constant. Each of these partial fractions can be integrated as in ~ 101. The process is illustrated by the following example....... 6 X2 + 3x - 15 EXAMPLE. Evaluate the integral J (- 1)(x + 1)(x + 2)dx The numerator of the integrand is already of less degree than the denominator. We first set down the partial fraction sum just described: 6 x2 + 3 x - 15 A B C (x- 1)(x + 1-)(x +2) - x-1 +x + 1 x + 2' XII, ~ 103] TECHNIQUE OF INTEGRATION 167 where the constants A; B, C are as yet unknown. To find them, we clear of fractions, obtaining the equation 6x2 +3x- 15 = A(x + 1)(x +2) + B(x-1)(x+2) + C(x-l)( +1), which must hold for every value of x. If any three values of x are substituted for x in turn, we obtain three equations that may be solved for A, B, and C. Values that are particularly simple are x = 1, x = -1, and x = - 2. Substituting each of these in turn, we get A = - 1, B = 6, C = 1. It follows that we may write 6x2+3 x-15 _ 6 1 (x- 1)(x + 1)(x + 2) x- 1 x + x +2 ' Each of the fractions on the right may be integrated readily as in ~ 101. Hence 62 6 3 23x —15 d -- 1dx+f 6 d f' dx (x- l)(x + )(x + 2) xd + I + x =-log(x - 1) -+6log (x+ 1)+log(x+2) log + 1)6 (x + 2) = log-x 103. Case III. Repeated Linear Factors. If one of the linear factors of the denominator D (~ 100) is repeated, i.e., if there is a factor of D of the form (ax + b)', that factor gives rise to several partial fractions of the form A B C L ax + b (ax + b)2 (ax + b)3 (ax + b) Otherwise the process remains as in ~ 102. This case is illustrated by the following example. EXAMPLE. Evaluate the integral f 3 2)x -- )3 dx. f (x +2)(x - 2)3 According to the processes described in ~~ 102-103, we first write 3x2 -x + 1 A B C D (x + 2)(x - 3)3 x + 2 x - 3 (x - 3)2 (x - 3)3 To find A, B, C, and D, we clear of fractions 3 x2 - x + 1 = A(x - 3)3 + B(x + 2)(x - 3)2 + C(x + 2)(x - 3) + D(x + 2) 168 THE CALCULUS [XII, ~ 103 and then substitute for x, in turn, any four values of x. If we take x = - 2, 0, 1, and 3, in turn, we find A = - 3/25, B = 3/25, C = 12/5, D = 5. Hence r 3x2-x ld J (x + 2)(x -3)3 = - 3 dx 3 dx 12 dx 5dx 25(x + 2) + 25(x - 3) + 5( - 3)2 + (x - 3)3 ~3 3 12 5 - - 35log (x + 2) + log (x - 3) - 12- - -3) 25 - 25 5((x -3) 2(x - 3)2 3 x -3 24 x -47 25 x + 2 10( x-3)2 104. Case IV. Quadratic Factors. If the denominator D (~ 100) contains a quadratic factor ax2 + bx + c which cannot be factored into real linear factors, we insert in the sum of partial fractions one fraction whose denominator is that quadratic factor and whose numerator is a linear expression Ax + B. The resulting fraction Ax+B ax2 + bx + c can always be integrated. The process is illustrated in the example which follows, and in exercises in the following list. The general case is treated in the Tables, IV, B, No. 21, p. 38. EXAMPLE. Evaluate the integral f i 2 4-d f (X - 1)(x2 +-4) We first write 10 x - 5 A Bx + C (x- 1)(x2 + 4) x- 1 x2 + 4 Clearing of fractions, we have 10'x - 5 = A(x2 + 4) + (x - 1) (Bx + C). Setting x = 1, 0,- 1, in turn, we find A = 1, B = - 1, C = 9. Hence f1 x f- 5 r dd 4 r - 9 x —9 d f (x —l)(x2+4) x~ - - 1 J- 2+4 XII, ~ 105] TECHNIQUE OF INTEGRATION 169 The first integral on the right is easily evaluated by ~ 101. The second integral on the right may be broken up into two parts which can be integrated separately: f- ~ dx =f Ix dx- 9 f 2dx J^x2 + 9 do J x2 4+ 4 2 + 4 J-2+4 d = 2 log (2 + 4) (See Ex. 13, p. 160.) X dx = 1tan-1 (See Ex. 20, p. 160.) X2 + 4 2 2 Collecting all these partial results, we have f (X -)d = log (x -1 ) - log (x2 + 4)- 9 tan-l J(X- 1)(X2+4) 2 2 _x 1 I 9 x = log /x + - tan-1. VX2 + 4 J2 2 105. Case V. Repeated Quadratic Factors. If the denominator D (~ 100) contains a quadratic factor to a power, i.e., if D has a factor of the form (ax2 + bx + c)", we insert among the partial fractions a set of fractions analogous to those of ~ 103, but with numerators that are linear, and with denominators that are successive powers of the quadratic: Ax + B Bx + D Lx + M ax2 + bx + c (a +bx (ax bx c)2 x2 + bx +c) The determination of the unknown constants A, B, C, etc., is performed as before. In general, the resulting partial fractions can always be integrated, but the problems become more and more difficult as the exponent n increases. The method of integration essentially depends on the process of ~ 99 (integration by parts). Difficult examples should not be attempted without the assistance of the Tables, where the results of integration by parts are given in the general formulas 23 and 25, p. 38.* These, together with formulas * Even when it is desired to proceed without the use of tables, it is best, for difficult examples, to derive such formulas as 23 and 25, p. 38 (Tables), once for all, by integration by parts, and then to use these formulas instead of repeating the work in each example. 170 THE CALCULUS [XCII, ~ 105 21, 22, and 24, P. 38, are sufficient to integrate any such expression. The process is illustrated by the following example. EXAMPLE. Evaluate the integral f 10 - ~) ( X + 3)dx '(x - 2) (x2 + 3)2 As indicated above, we first set 10 x3 + 7 x + 4 A Bx+C Dx+E (x - 2)(X2 + 3)2 - 2 2x2 + 3 ~(2 + 3)2 Clearing of fractions and determining the constants as before, we find A = 2, B = - 2, C = 6, D = 6, and E = - 11. It follows that fl X3~+7 x~ + 4 dx =f 2 dx +f -2x+6 d fO 6x- Itd f (x- 2)X2 +3)2 - 2 X2+$ 3 (X2 + 3)2 The first integral on the right can be evaluated by ~ 101. The second one can'be evaluated as shown in the example of ~ 104. The last integral may be evaluated either by an integration by parts, or by use of the formulas 22 and 23 on p. 38 of the Tables. EXERCISES Carry out each of the following integrations. 1. f3d25=log V3 x5. dx 1 1 1 1 X-1 2. - dx =~~~~lo -Ion 3. S dx11 x +a f a2 - X2 2a ogC xfdx+2 l (x+l)2+ 4. tan (x +n3). 6. f4 X2 +6 + 20 = [tan-' (2x + 1)1/. dxn- (2 + $ 1)]/2. dx p dx X x-1 7. f A~ Iog - X2+2x-3 f (X + 1)2 - 4 - x + 3 dx d 1 2 xt - 2 X 1 8. fdx-og log = - l log ~4 X2 + 4 x - 8 12 2x+4 12 z x + 2 XII, ~ 105] TECHNIQUE OF INTEGRATION 171 In each of the following integrals, first prepare the integrand for integration as in ~~ 101-105; then complete the integrations. 9. fX2-5 x + 6 dx =log (x-2) +21og (x-3) =log [(x-2) (x-3)2]. _. __2 _ _ 5 10. J5 +3- dx = 3 log x -- log (3 x + 5). 11. f x +t + 32 l 12. f x - -1 dx =log x2+2x+2 - 2 tan-((1 + x). fx2- 2x +2 f dx 1 i 1 x 1 13. I = -tan1 - tan-1. X4 + 7 X2 + 12 V 3 -/3 2 2 14. _2dx. 15. f x3 d+2 X22- x X2+ 3 x x 2 x dx x2dx 16. x 17.. JX2- 5 + 6 j X2-4 18. dx 19. dx 18. x3 +X Z —' fX3- 7 x + 6 20. f-21. 2 f1 dx X3+2X2 2x3 +7x2 + 6 x 22. f(x - 1)2 (x- 2) 23. f(x 1) (2+1) A24. x dx 25. 2x2 + 1 24 (x + 2) ( + 3)2' 2 x + 3 26. (x+ 1) ( + 3)2 dx. 27. x2 28. x2dx 29.f 2- x +2 28. J x_-29. _x dx. f2. j7(x2 - 9)2. jx4 - 5 x2.4- 4 Derive each of the following formulas. r30. dx -1 mx +n (ax + b) (mx + n) an-bm ax + b x dx _ 1 log ( a) (x + a) (x + b) a - 2 (x + ) /;x dx b x-/x2 + Va x 32. =fa) () -az - b log a ta fX2 + a) (x + b) a + b2 x + b a +\aa 172 THE CALCULUS [XII, ~ 105 33. Derive each of the formulas Nos. 18-24, Tables, IV, A. Evaluate each of the following definite integrals. x2 = 2 8 - 3 x - x2 = 5- x+ X-X 34. dxd 35. dx=15 7X+3 x23x3 J x=l X(x+2)2 X4 +5X3 36. fX12 X (x - 4 dx. X=3 (x - 2)3 [NOTE. Further practice in definite integration may be had by inserting various limits in the previous exercises.] Carry out each of the following integrations after reducing them to algebraic form by a proper substitution. sinx os zdx 37. f 1 +cos2x dx. 38. fexX. 239. e2x 1 + Cos2 Z4 sin ~ 2 Cos X 1 + sin X o a r5 40. fsec x dx f 2 dx = -log =s nxzlog tan fs f I - sin ~~~~~~~~~~~~~ ~~~~~~2 1 - sin x 4 41. fcsc x dx. 42. fsech x dx. 43. fcsch x dx. 44. x -eC2 dx. 45. ez- Z dx. tan x - t'an2 X Ax + $ e-Z 106. Rationalization of Linear Radicals. If the integrand is rational except for a radical of the form Va6x +b, the substitution of a new letter for the radical r = Vax + b renders the new integrand rational. EXAMPLE. Find x1 ~ x + Setting r = Vx +2, we have x = r2 - 2 and dx 2 r dr; hence idx' d + -2r 22 ir dr = 2 r dr J1+ X r2-1 J r+I =2f(r~1- - 1-)dr=r2~2r-2log(r+1)+C =x +2+2 Vx + 2 - 2 log (VxT + 2 + 1) + C. XII, ~ 107] TECHNIQUE OF INTEGRATION 173 The same plan - substitution of a new letter for the essential radical - is successful in a large number of cases, including all those in which the radical is of one of the forms: /nax - b\l/n x, (ax + b) x + d) where n is an integer. Integral powers of the essential radical may also occur in the integrand. 107. Quadratic Irrationals: V/a + bx = x2. If the integral involves a quadratic irrational, either of several methods may be successful, and at least one of the following always succeeds: (A) If the quadratic Q = a + bx = x2 can be factored into real factors, we have VQ =V V(a +x) ( x) = )(a + ); \ a - ] and the method of ~ 106 can be used. The resulting expressions are sometimes not so simple, however, as those found by one of the following processes. (B) If the term in x2 is positive, either of the substitutions VQ= t+x, Q = t- x, will be found advantageous. One of these substitutions may lead to simpler forms than the other in a given example. (C) Completing the square under the radical sign throws the radical in the form VQ = V/ik i (xi=c)2; the substitution x =- c = y certainly simplifies the integral, and may throw it in a form which can be recognized instantly. THE CALCULUS[XI~10 [XIII ~ 107 (D) After completing the square under the radical sign, the radical will take one of the forms V\k2 A \2, k2 +7,2 VX2 ~-k~2. Then a trigonometric substitution often leads to a simple form. Thus: if x = k sin 6, Vk2 - x~2becomes k cos 0; if x = k tan 6, Vk2~+ x~2becomes k sec 0; if x = k sec 6. Vix2- k2 becomes k tan 0. ExAmPLE. Let V'Q = 2+ a~2; show the effect of substituting.\Q=t- X. If V\x2 + a2 = t - X, we find t2:F a2 t2~ a2 t2~~ a2 X= 2tdx = 2 2dt)VQ = t - X =-2 and the transformed integra'nd is surely rational. Carrying out these transformations in the simple examples which follow, we find C dx _(t2~+a2 t2+~a9\ dt (1) d)t2 t - = t -log t+ C VIX2 _ 2 2t 2 t =log (x +VQ/~) +C, where Q =X2 ~a2. _____ t ~ 2 t2~+a2 Ct a2 a (ii) jVx2 ~ ~a2dx =j~ _-dt + + d j2t 2t2 J4 -2 t 4tQ4 t4-a4 a2 X -IQ a2 These integrals are important and are repeated in the Table of Integrals, Tables, IV, C, 33, 45a. Many other integrals can- be reduced to these two or to that of Ex. 3, p. 163, or to Rules [XVII or [XXI by pr6cess (C) above. EXERCISES 1. fxv- dx = (1 / 15) (6 x+ 4) (x - 1) 3/2. 2. fN/ dx=2K + 1 Vx +1 -w X -\/X + 1+1 x1 2 2tan'V-1 4f 1 X dx = 1-~X~2 + sin'1X. XII, ~ 107] TECHNIQUE OF INTEGRATION 175 6,J dx = - tan-1 ) + b (ax + 2 b) Vax + b a /b \ b 6., =-_ V x -+ log /x + 1 -- 1 7. f(a + bx)3/2 dx (a + bx)5/2. 8.r dx 2 (a + bx3/2 b 'a bx Carry out each of the following integrations. 9. x/1 \+xdx. 10. f d (x- 1) 2-x 11. fr- 12. f _ x. 13. fr+Ix dx. 14. f V dx. ^ \ x J\x - 1 d 16. f -X d 21. 16. x2 dx. 17. fx /1 + xdx. 18. fx 3 x + 7 dx. 21. f X14dX 22. f dx ' 1 + x1/2 + xr/6 23.f< X dx. 24.f /4 \ dx. ^ (1 + )3/2 J l 25.f 1~ dx. 26. 3 xdx 4 1- Xo. + /1 +X x Carry out each of the following integrations by first making an appropriate substitution. 27.f oS dx. 30. fsin xd 3 -- sin x 1 - 2 v/cos x 28.f1 e —/3. 31. f/-L- I cos x dx. J 1 + e23 J 1 -f ~vnx 29. r sec2 ax dx /* sin x dx /2 + 3 tan x (2 + 3 cos x)3/2 176 THE CALCULUS [XII, ~ 107 Substitution of a new letter for the essential radical is immediately successful in the following integrals. 33. fxVT+xdx. 36. 3f. +2 2 39. fx2a+bx3dx. J J V~~~_\2 + 2 x + x J 34. /f V1 + X2 3r x dx J (a + bx2)3/2 40. f V dx. J a V+ 4 x n-l dx 41. f x V/'a + bx6. 35. fx(1 +X)3/2dx. 38. + x~ dx) Carry out each of the following integrations. 42. xx _2i = log (x + 43.2-4 43. J =d 1 tan1 -- 2. x /X2-a2 a a 44. S = - a 44. f _ dx 1 a-x J (x+ a) V/a2 - x2a a+x 45. f dx = tan-1 x (1 +2 x2) 1Vl x2 Vx2+ 1 46. f + 1 -2 dx = sin-1 x — V 21 - 2 47. 2 x2 - a2 d -2 dx = x 48. r dx x4 2 + 1 61. f x dx. x 49. r dx x V4 2 + 1 52.f /4 -X2 d. x. 2 d. 50. f dx x2 V4xx2 +1 53. 2 x2 - 1dx - Vl ----~x2 The following integrations may be performed by the methods of ~ 107; note especially method (C), which consists in completing the square under the radical. 54. dx + = log (2 x -1 + 2 Vx2- -). 55. dx = sin-12 x +1 6 n. - -- -= sm-x-_ 56. / 4 dx - = vers-1 [+ const.]. /4 ax - X2 2 2 XII, ~ 108] TECHNIQUE OF INTEGRATION 177 r x. x-2 57. f = sin — 68. J dx2 =sin-1 3x - 1 x V7x22 + 6 x-1 4x 59. dx.60. f.dx 61. -J dx 9 2x2. Vl+x-2x2 2 / - 2x -x2 62. f / x -- 65. f / 5+f x V + x2 dx. /6 x-X2-5 J 63. f - x. 66. f /3x2 + 10 x +9 dx. xV \+24-2 x + 3 dx ~ 64. (x d 4 — 67. fxV 1 +x+x2 dx. (x + 4) /X2 + 3 x - 4 Integrate by parts, [VI], each of the following integrals. 68. fx sin-l x dx. 71. (3 x - 2) sin-ix dx. ~ /*sin-1 x 2 - + r2\-x2 C O_, 69. sin- dx. 72. 2 + x- 2 d J f X2 X2 70. fx cos-1 x dx. 73. f (sin-l x + 2 x cos-1 x) dx. 108. Trigonometric Integrals. A number of trigonometric integrals have been evaluated in the preceding lists of exercises. The processes explained in Exs. 14, 15, 17, 25, 26, 28, 29, 30, 31, pp. 160-61, may be generalized and stated in the form of standard processes. They depend chiefly upon the use of well known relations between the trigonometric functions. (See Tables, pp. 12-13.) Since it is desirable to avoid the introduction of radicals that were not present in the original example, we do not ordinarily use the trigonometric formulas that involve a square root. But it is desirable to notice that (a) Any even power of sin x can be changed into a sum of powers of cos x by the relation sin2x + cos2x = 1. (b) Any even power of cos x can be changed into a sum of powers of sin x by the same relation. 178 THE CALCULUS [XII, ~ 108 (c) Similarly, sums of even powers of tan x may be changed into sums of even powers of sec x, and conversely, by the relation sec2x = 1 + tan2x. (d) All the trigonometric functions may be changed into forms in sin x and cos x without introducing any new radicals. (e) The square of sin x and the square of cos x may be expressed in terms of the first power of cos 2 x by means of the formula cos 2 x = cos2 x - sin2 x. (f) All the trigonometric functions can be expressed rationally in terms of tan (x/2). 109. Integration of Odd Powers of sin x or of cos x. Any odd power of sin x may be integrated, as in Ex. 17, p. 160, by the substitution u = cos x. Likewise, any odd power of cos x may be integrated by means of the substitution u = sin x. (See Ex. 25, p. 161.) More generally any integral of the form sin" x cosm x dx can be integrated if either m or n is an odd integer. If n is odd, set u = cos x; if m is odd, set u = sin x. This process is illustrated by Exs. 14, 15, 16, 18, 19, p. 160; and by the following example. EXAMPLE. Evaluate the integral f sin3/2 x cos3 x dx. Set u = sin x; then * sin3/2x cos3 x dx = f sin3/2x(1 - sin2 x)cos x dx f3/2 (1 - u2) du 2 1,5/2 -2 9/ 2 sin5/2 -2 sin9/2 x. 5 9 5 9 XII, ~ 110] TECHNIQUE OF INTEG'RATION 179 110. Reduction Formulas. If the integrand is an even power of sin x (or of cos x) we may use ~ 108 (e), as in Exs. 30, 31, p. 161, or we may proceed by integration by parts, as follows. Let us first write fsin x dx= sin x sin x dx and then integrate by parts (~ 99) by taking u = in, sin x, d = sin x, du = (n - 1) sinn-2 x cos x dx, v = - cos x. Then we obtain f sinn x dx = - cos x sin"n- x + (n - 1) fsinn-2x cos2 x dx. Replacing cos2x by 1 - sin2x in the last integral and breaking it up into two integrals, we have, fsin x dx = - cos x sin"- x + (n - 1) fsinn-2 x dx - (n -1) f sin x dx. Transposing the last integral and dividing by n, we find the formula (Tables, IV, E, 57) (1) fsin x dx = -cos x sin- lx + n-l sinn-2 x dx. n n Repeated applications of this formula reduce the left-hand side to integrals that involve sinn-4 x, sin-6 x, ~.., down to sin x if n is odd, or to sin~x (= 1) if n is even. In either case, the integration can be completed by the use of a standard formula. In a similar manner, we obtain the formula (Tables, IV, E, 60) 2 Scon x d sin x cosn 1 x n-1 Co n-2 d (2) fcos x dx= + cos x dx. n nfJ '180 THE CALCULUS [Xiij ~ 110 By solving these formulas for the integral on the right-hand side, we obtain (3) sin'"xdx = 2x-cisx +in n-fsinnx dx;, and n-r sin cos n-iX n ncosdx. (4) JCosn xdx= sinxcos 'xd~ These formulas raise the exponent in the integrand. Hence they are useful in integrating negative powers of sin x and cos x, i.e. positive powers of csc x and of sec x. An analogous integration by parts leads to the formula (Tables, IV, E, 64) (5) fsinnX cosmxdx * sinn+lxcosmx m- sinux cosm - x dx m+n m +n sin' 1 X cosm+l X n - n-2 in' -- ~~ ~~+ ~ sin' X Cos" x dx. m + n m n n The proof of this formula is left as an exercise for the student. If either m or n in (5) is an odd integer, the process of ~ 109 is usually quicker. But if both m and n are even integers, the formula (5) is very useful. EXAMPLE 1. Evaluate f cos4 x dx. Scos4z d~e Cos3 sin x 3,amJ f cos~x dx= 4O ~ +4f COS2xdx COs3 X sin x 3Fcosxsinx 1 1 4 4 2L 2 2J COs3 Xsin x 3 3 + Cos x sin X + - X. 4 8 8. XII, ~ 111] TECHNIQUE OF INTEGRATION 181 If, in this example, we should follow the method suggested under (e) of ~ 108, we would write I + cos2 \2 cos4 = (cos2 )2 = ( -- cs2 x 2 - (1+2 cos 2 x + cos2 2 x) 4(1 +2cos2x 1 + coss4x) 4 \I TL ~rs L~T 2 Then fCos4x dx 3 x 2 sin sin 4x jcos4X = 8 4 + 4 32 EXAMPLE 2. Evaluate f sec3 x dx. By (4), with n = - 1, we find fsecaxdx =, -/* sinX Cos-2 Jsec3 Xdx = J cos-3 x dx = _ s cos- I +-1 2 + 2j cos-1 x dx sin x sec2 x 1,d 2 21jsecxx. The integration may now be completed by means of Ex. 40, p. 172, which is essentially an application of ~ 109. EXAMPLE 3. Evaluate fsin2 x cos4 x dx. By the second part of (5), we have fsin2 xcos4 x dx = sin x cos5 dx. - x dx 6 +6f cxd. The example may now be completed by following Example 1. 111. Powers of tan x and of sec x. Any even power of tan x may be reduced to a sum of even powers of sec x (~ 108 (c) ). This may be integrated by the method mentioned in ~ 110. Another method of integrating any even power of sec x consists in making the substitution u = tan x. Since du = sec2 x dx, and since even powers of sec x may be reduced to a sum of even powers of tan x (~ 108 (c) ), the integration becomes very simple, as is illustrated by Example 1 below. 182 THE CALCULUS [XII, ~ 111 Odd powers of sec x can be most readily integrated by the method of ~ 110, as shown in Example 2, ~ 110. Any positive integral power of tan x may be integrated by reducing it to one of the two integrals: (1)ftan x dx= -log cos x= log sec x. (See Exs. 33, 34, p. 161.) (2)ftan2x dx =f(sec2 x- 1) dx = tan x - x, by means of the reduction formula (see Tables IV, E, 70): (3) ftann x dx = tan - ftan"-2 x dx. n - 1 This reduction formula is obtained as follows: ftan't x dx = ftann-2 x tan2 x dx = Stan' 2 x (sec2 x - 1) dx = ftan- 2~x sec2x dx - tan" - 2 x dx. If we set u = tan x in the first integral on the right-hand side, we obtain the formula (3). EXAMPLE 1. Evaluate the integral fsec4 x dx. Put u = tan x; then du = sec2 x dx, and we may write fsec4 x dx = fsec2 x sec2 x dx =f (1 + tan2 x) sec2 x dx u,.,,,i3. tan3 x =f(1 +2) du = u + =tan x + - EXAMPLE 2. Evaluate ftan4 x dx. As indicated above, we may write ftan4 x dx =f(sec2 x - 1)2 dx =f(sec4 x - 2 sec2 x + 1) dx = sec4 x dx - 2 tan x + x. The remaining integral may be integrated as in Example 1. The student should also apply the reduction formula (3) to this integral. EXAMPLE 3. Evaluate f tan3 x dx. Applying (3) we have n, t~anz2 x r_, tan2 x f tan3 x dx = t - tan x dx =-a 1 +log cos x. n - 1 n - 1 MLI ~ 112] TECHNIQUE OF INTEGRATION EXERCISES 183 Evaluate the following integrals. 1. fsin4 X dX. 2. fsin2 X cos2 x dx. 4. f tan5 X dX. 5. fsec2 x tan4 x dx. 7. ftan x sec4 x dx. 8. fcesc4x dx. 10. f cot4 x dx. 11. f cot5 x dx. 13. feeX o2 x dx. 14. dx x J~~~~sin x cos x 16. dx sin (x + a)dx Jsin2 x cos2 x f1 sin x 19. fsin2 x cos3 x dx. 20. fsin3 X cos2 x dx. 22. fCS dx. 23. fsin3 X cos3 x dx. 2 sin~xdx 26 in3 xd 2,f cos x J6 csx. 28. fsin4 x cos3 x dx. 29. fsin 4x cos 3xdx. 3. ftan6 x dx. 6. fcsc3 x dx. 9. fcsc6 x dx. 12. f tan2 X csc2 x dx. J5 dx 1. sin X cos3 X 18. dx,\/sin x cos3 x J cos3 x dx. Jdxl 24. Si 7n-2 x cos4 X Jcos2 x dx. 27. s~in3 X 30. fPin 3x cos 4 xdx. 31. Verify formula 111 of Table IV, E. 32. Verify formula 112 of Table IV, E. 112. Elliptic and Other Integrals. If the only irrationality is VQ, where Q is a polynomial of the third or fourth de-~ gree, the integral is called an elliptic integral. While no treatment of these integrals is given here, they are treated briefly in tables of integrals, and their values have been computed in the form of -tables.* See Tables, V, D, B. *Some idea of these quantities may be obtained by imaging soe person ignorant of logarithms. Then f (1/Ix) dx would be beyond his powers, and we, should tell him " values of the integral f (1/Ix) dk can be founid tabulated," which is precisely what is done in tables of Napierian logarithms. Of aourse as little as possible is tabulated; other allied forms are reduced to those tabulated by means of special formulas, given in the tables. 184 THE CALCULUS [XII, ~ 113 The discussion of such integrals, as well as of those in which Q is of degree higher than four, is beyond the scope of this book. 113. Binomial Differentials. Among the forms which are shown in tables of integrals to be reducible to simpler ones are the so-called binomial differentials: f (axn + b)Pxm dx. It is shown by integration by parts that such forms can be replaced by any one of the following combinations, where u stands for (ax" + b): (1) fuPxm dx = (A1) uPxr+l + (B1) fup-xm dx, (2) fuvxm dx = (A2) uP+lxm+l + (B2) f uP+lxm dx, (3) fu xm dx = (A3) uP+lxm + + (B3) fu m+n dx, (4) fuPxm dx = (A4) uP+'x-n+l + (B4) fUxmn dx, where Al, A2, A3, A4, B1, B2, B3, B4, are certain constants. These rules may be used either by direct substitution from a table of integrals in which the values of the constants are given in general * (see Tables, IV, D, 51-54), or we may denote the unknown constants by letters and find their values by differentiating both sides and comparing coefficients. * Such formulas are called reduction formulas; many other such formulas - notably for trigonometric functions - are given in tables of integrals. (See Tables, IV, Ea, 57, 60, 64, etc.) It is strongly advised that no effort be made to memorize any of these forms, - not even the skeleton forms given above. A far more profitable effort is to grasp the essential notion of the types of changes which can be made in these and other integrals, so that good judgment is formed concerning the possibility of integrating given expressions. Then the actual integration is usually performed by means of a table. See also Tables, IV, Ea, 78, 82 (b); Eb, 85, 86; Ec, 92-94; Ed, 98, 106; B, 17 (b), 25; etc. XII, ~ 114] TECHNIQUE OF INTEGRATION 185 dx x B dx b (2). EXAMPLE l.- = A __ +Bby (2).. (ax2 + b)3/2 A(aX2 +b) 1/2 (aX2+b)1/2) Differentiating and comparing coefficients of x2 and x~, we find B = 0 and A = 1/b; hence r dx x -d; (check). f (ax2 + b)3/2 b (ax2 + b) EXAMPLE 2. b xdx Ax2 B xdx EXAMPLE 2. f (ax2 + b)3/2 = (a2 + b)1/2 + B(a2 + b)3/2 Here A = 1/a, B = - 2 b/a, a x3dx ax2 + 2b and J (ax2 + b)3/2 a2 /(ax2 ' b) 114. General Remarks. The student will see that integration is largely a trial process, the success of which is dependent upon a ready knowledge of algebraic and trigonometric transformations. Skill will come only from constant practice. A very considerable help in this practice is a table of integrals (see Tables, IV, A-H). The student should apply his intelligence in the use of such tables, testing the results there given, endeavoring to see how they are obtained, studying the classification of the table; in brief, mastering the table, not becoming a slave to it. In the list which follows, many examples can be done by the processes mentioned above. The exercises 43-97 may be reserved for practice in using a table of integrals. REVIEW EXERCISES x2 + 4 x dx.2 6x15 1 (x + 2)2 (2 x + 3)2 /* dx ~ rx2q_ +1 3.f S x 4-3. x 3 dx. 5. x2 +1 6. X2+x2 —1 dx.. ( - 2)3dx. 6. (X + 1) (- 1)2 X3 + 1 d 8 3 + 2 X2 - d 7. x2 — 3 dx. 8. f dx.fX2-3x+2 x+ x2- 1 186 THE CALCULUS [XII, ~ 114 dx dx d 9. f I 10. f. 11. f dx 9J (1 - 2)2. 4 x4 + 5 x2 + 1 1- 16- x4 12.f 3 - 3 x dx. 13. 14. 6 xdx 1 2 -+ 3 x- + 4 X1 - 2 52 ( 3)3 1 x dxf x2dx -2 dx 16. x4f-4 16. X d+ 2 J -5~ 24 J; - 16.4'. a2 X+ 2 x + 5* 17 2 + 4 x + 2 18. f x +2dx. 19. dx 20. f-V dx x -ix- 1 ax + b 21. f dx 22. f x dx 23. f (x - 2) dx - (a + bx)3 (a + -x) ( - 1)3/2 f _(2_+____dx2 dx (1 d -+V x)2 24. (2 +). 26. (a + bx3)2 26. dx. dx x2 dx. 27. ( + 28. f + 2 29. f x 3 x 7 dx. x dx +2 X dx 30. f (9 2- 3)3/4' 31. f3 d. 32 (21)3/2 - 3.f x (x( - 1)/2 3 dx dx 3 (1 - x2)3/2 34. +X~ + X p/x dx JX2dxr xdx 35. Jf - <- 36. f 37. d x96S _G + S.S 37. S ~a + bx 38. f 2+ dx. 39. f x 3 x + 7 dx. 40. fx -a + b2 dx. 41. fx3 (a + x2)'/3 dx 42. f x (1 + x3)1/3 dx. In the following integrations, use Table IV freely. dx x+2 43. (x+-)2 + 45)34. 45. f23x dx 46. f (2 5x-3- dx1)2. 47. dx 48 xdx w' J (2x2 1)2dx - J(x2 + 2x + 5)2* ' 4 (+-1)2(X+2)2 91 dxaJ x2dx 51 5dx 493. 0.4 xN/ -- (7 + 4 3)2/3 52.f /(a2 - 2)3/2 dx. 53. (x2 - a2)3/2 dx. 54. x7 dx dx 56. X2dx 7 2dx f (a + bX2)3/2' (1 - x2)3/2 (a + bx2)5/2 XII, ~ 114] TECHNIQUE OF INTEGRATION 187 68., CO dx. 59. 69 3 dx. 60. cos4 x dx. 58. f sin x cos2 x. rd. O do f do 61. 2 +sin 0 2 — 3 cos 0 63 2+ 5 sin 0 64. f cos6 a da. 66. fctn2 3x dx. 66. fsin3/2 0 cos30 dO. /*l. LI2 X.. 75.2 '3x —2 67. 2 x +dx 68. ' x+ 2 dx. 69.- - - d. Jo (2 +l 2 - 2)2 j (2x-1)3 JV2x2+ 3 8. 1.. 843 x -+2 dx = 3 cosh x - x. X=2.5,=2.33 79. f e2 dx. x 872. sinhx d x. J x=1.2 x=0 x0,- 3 x1/2 x + 3 2 /x2 - 3 2 (1 +X2) dx 2 dx 2 dx 73. 74. 5 d = 7.Sh-f x Jx=0 x= Vx2 -1 1x=l21 fo -4 x2 81. 2 dx 8 7. 1 dxsinh76. (2 x + 1) (X2 + 2) 2 ' 12- X 78. ex dx. 84. 2 dx = cosh- dx. x=3.5 n-2 x=2 vx2-4 2 aX=2,s^x=3 1 1.4 ~ x e - J=2.1 X =O VX2 + ax=0 9. x=2.5 x. fx=2 h 2.3= 2 90. Jo~ 1-(1/4) s-in2- = F (2 30~), Tables, V, D. 91. 0 J 5 V - (1/4) sin2 8 do = E (1, 45~), Tables, V, E. 8 0=45~ 0 d0= 90~ 92. f _ d. 93. f85 -.25 sin2 0 d.x 80 e- dx. 86. f = cosh- 1 Jx=o =/ 1-X2 V/ -.25X2 J=o0~ v 1-.25 sin2 81-.hzi3if x8 = sin 0. 188 THE CALCULUS [XII, ~ 114 x —l.1 -1.36 x2, = 90o 95. x2 1 -.36x2 dx = i= 1 -.36 sin2 0 dO, if x = sin 0. Jx=12 12 - X2 0=300 e. rd 97. X. dox? J= 1/2 /1 - X2 N/1 -.49 X2 J= 2/2 1- X2 [NOTE. Many of the exercises in preceding lists may be used for additional practice in use of the tables.] 115. Limits Infinite. Horizontal Asymptote. If a curve approaches the x-axis as an asymptote, it is conceivable that the total area between the x-axis, the curve, and a left-hand vertical boundary may exist; by this total area we mean the limit of the area from the left-hand \iN_ _ _qe____boundary out to any vertical line k -\ - ------ x= m, as m becomes infinite. — _ --- --- _ EXAMPLE 1. The area under the -- curve y = e- from the y-axis to the _ _ - B - - _ _ _ _ ordinate x = m is _- ~Ix = m x=m M _- _ _ __ _:__ A = e- dx = 1- e-. ___ __ __ _ _ _ _ As m becomes infinite e-m approaches FIG. 42. zero; hence _a=oo a=oo ^x=m A ] e-=dx = lim ee- dx = lim (1 - e-) = 1, _x=O J x=0 - noov x= O m —oo and we say that the total area under the curve y = e-$ from x = 0 tox =+oo is. EXAMPLE 2. The area under the hyperbola y = 1/x from x = 1 to x = m is _x=m x=m dx x=m A = I -= l log x = log m. Jx =l Jx=l X _x=l As m becomes infinite, log m becomes infinite, and lim A = lim log m m —o Jx= 1 ) m —oo does not exist; hence we say that the total area between the x-axis and the hyperbola from x = 1 to x = oo does not exist.* *This is the standard short expression to denote what is quite obvious,that the area up to x =m becomes infinite as m becomes infinite. This result makes any consideration of the area up to x = oo perfectly useless; hence the expression "fails to exist," which is slightly more general. XII, ~ 116] TECHNIQUE OF INTEGRATION 189 116. Integrand Infinite. Vertical Asymptotes. If the function to be integrated becomes infinite, the situation is precisely similar to that of ~ 115; graphically, the curve whose area is represented by the integral has in this case a vertical asymptote. If f (ir) becomes infinite at one of the limits of integration, x = b, we define the integral, as in ~ 115, by a limit process: fx=~b b-c f() dx =limj f(x) dx. =a5 C ---O a A similar definition applies if f (x) becomes infinite at the lower limit, as in the following example. EXAMPLE 1. The area between the curve y = 1/V and the two axes, from x = 0 to x = 1, is A] d=1 xzdx= lim i-~dxi X=0 fz=0 -Vx c-+O fz~eV J= = lim F2\/xl li= urn2 F2-2v1=2. c —L I x=C C-O L EXAMPLE 2. The area between the hyperbola y = 1/x, the vertical line x = 1, and the two axes, does not exist. For, f dx = log x x=1 log c, but lim (- log c) as c - 0 does not exist, for - log c becomes infinite as c -0. EXAMPLE 3. The area between the curve y = 1/ 4X - 1, its asymptote x = 1, and the line x = 2 is I I I IY l Ule _ _ _ __I ~~A' I~~~ Fic. 43. 2 dX = lim f2 dx 3 3 = lim f2 dx - 3- im (1 - c 2/3) = 3 I. ~z- I c-~O e = j 2 c-:). 2 190 THE CALCULUS [XII, ~ 116 EXAMPLE 4. Show thatf~ 17w2 dx does not exist. The ordinate y = 17w2 becomes infinite as x approaches zero, i.e. the y-axis is a vertical asymptote. Hence to find the given integral we must proceed as -ai-a hC-vP rnIQ1;-;TIr +bn -r;rr;,nn1;-nlnrT-ra1 I - I Y I I I II -1 0 1 - I- - J -I-I l j j j Fme. 44. into two parts: r5=1 1 1 Z=1 ~_ - dx fx= C, 2 X 2=C C X=- - 1 1X =-C fx /J= _ 1 X2 X j 2= —1 ] i f 1 d x = li-Ii. The limit of neither exists since 1/c becomes infinite as c = 0; hence the given integral does not exist. Carelessness in such cases results in absurdly false answers; thus if no attention were paid to the nature of the curve, some person might write: A] =+1 fX+l2 1 w c i r i d i u l o e Fi g. 4 = X= -1 S5- I: which is ridiculous (see Fig. 44). - l] x=+l -= - 1- 1 = - 2, — X X=-1=+ The only general rule is to follow the principles of ~~ 115-116 in all cases of infinite limits or discontinuous integrands. Such integrals are called improper integrals. EXERCISES Verify each of the following results. +1 dx 62.-l dx if 1x2/3= 2.f i. ~1 dx is determinate if n < 1, Jo Xn non-existent if n> 1. 4.fdx =2.6 2 5. f x a 2 Va.x 7. f.2 i 8. 2 dx determinate if n < 1, 8.f.5 (2 x - 3)n1 ~non-existent if >n1. is non-existent. dx2 dx = 1. ~x —3 is non-existent _~-3 i XII, ~ 116] TECHNIQUE OF INTEGRATION 191 b dx State a similar rule forf (hx + k) fa (hx + k)' Q dx 7r ra X2 dx 3 r a2 9. - 11 - Jo Va2 - x2 2 Jo V ax -x2 8 1 x dx ' ~1 dx is non-existent. 10. = 1. 12. J +4is non-existent. fo -\lJ - x2 f- 1 X2 + 4 7r Ir 7r 13. Show that the integrals f tan x dx, f ctn x dx, f2sec x dx, I log x dx og x d ar all non-existent. o x Verify each of the following results: iA. Sr00 d^-z 1 r -dx 14. fX3 2 16. (1 + x)3/2 15. dxf dxx 15. is non-existent. 17. is non-existent. i o ( + )2/3non-existent 18 dx i.determinate if n > 1, Jo (1 + x) non-existent if n = 1. 19f: X dx =r 22. fco '- x dx is non-existent. dx w 20. a da 22 r 4 23. J V e-2 dx =. 21. -x = 1. 24. f e2 dx is non-existent. A 1 lx2v/2 - 1 2o Determine the area between each of the following curves, the xaxis, and the ordinates at the values of x indicated. 25. y3 (x- 1)2 = 1; x = to 9. Ans. 9. 26. xy2 (1+x)2 = 4; x = 0 to 4. Ans. 4 tan-1 2. 27. y2x4 (1+x) = 1; x = to 3. Ans. oo. 28. x2 y2(2 - 1) = 9; x = 1 to 2. Ans. 2 r. 29. 3 (x - 1)2 = 8 x3; z = 0 to 3. Ans. 9 42 + 9/2. 30. x2 y2 (x2 + 9) = 1; x = 4 to co. Ans. 3 log 2. 31. y2 (1 + x)4 = x; x = O to o. Ans. 7-. 32. y3 (x + 1)2 = 1: X = Oto o. Ans. oo, CHAPTER XIII INTEGRALS AS LIMITS OF SUMS 117. Step-by-step Process. The total amount of a variable quantity whose rate of change (derivative) is given [i.e. the integral of the rate] can be obtained in another way. The method about to be explained has many theoretical advantages and one decidedly practical advantage, namely in its application to the approximate evaluation of integrals when the indicated integration cannot be carried out. For example, imagine a train whose speed is increasing. The distance it travels cannot be found by multiplying the speed by the time; but we can get the total distance approximately by steps, computing (approximately) the distance traveled in each second as if the train were actually going at a constant speed during that second, and adding all these results to form a total distance traveled. If the speed increases steadily from zero to 30 mi. per hour, in 44 sec., that is, from zero to 44 ft. per second in 44 sec., the increase in speed each second (acceleration) is 1 ft. per second. Hence the speeds at the beginnings of each of the seconds are 0, 1, 2, 3, *.., etc. If we use as the speed during each second the speed at the beginning of that second, we should find the total distance (approximately) 43.44 s=0+-1+2+3+ 3 +-42+43= =946, 2 which is evidently a little too low. 192 XIII, ~ 117] INTEGRALS AS LIMITS OF SUMS 193 If we use as the speed during each second the speed at the end of that second, we should get (approximately) 44.45 s= 1 + 2 + 3 + 4 + +43 + 44= - =990 which is evidently too high. But these values differ only by 44 ft.; and we are sure that the desired distance is between 946 and 990 ft. If we reduce the length of the intervals, the result will be still more accurate; thus if, in the preceding example, the distances be computed by half seconds, it is easily shown that the distance is between 957 ft. and 979 ft.; if the steps are taken 1/10 second each, the distance is found to be between 965.8 ft. and 970.2 ft. Evidently, the exact distance is the limit approached by this step-by-step summation as the steps At approach zero: ]t=44 rt=44 d t=44 2 tdt=44 s v dt = tdt = = 968. t=o t=O t=o 2 t=o We note particularly that the two results for s are surely equal; hence we obtain the important result: C vdt liml] v At+v.v Atvl *+*At+*. t=O At=o t=0 t=At t=2At 118. Approximate Summation. This step-by-step process of summation to find a given total is of such general application, and is so valuable even in cases where no limit is taken, that we shall stop to consider a few examples, in which the methods employed are either obvious or are indicated in the discussion of the example. Thus, areas are often computed approximately by dividing them into convenient strips. We have seen in ~ 55 that if A denotes the area under a curve between x= a and 194 THE CALCULUS [XIII, ~ 118 x= b, then the rate of increase of A is the height h of the.1, _B-~~ Bcurve: y Y=R(x) dA d h = R(x), dx I/ /where R (x) is the rate of increase of A, and is 0 x=a x=b also the height of the FIG. 45. curve. For a parabola, h =x2, we may find the area A approximately between x = - 1 and x = 2 by dividing that interval into smaller pieces and computing (approximately) the areas which stand on those pieces as if the height h were constant throughout each piece. If, for example, we divide the area A into six strips of equal width, each 1/2 unit wide, and if we take the height throughout each one to be the height at the left-hand corner, the total area is (approximately) _ ---v -— I xI - 0 - FIG. 46. ( 1 I + ( )2. I + 0. I + (1)2 + ( + 1)2 + (2)2 1 = 19/8, whereas, if we take the height equal to the height at the right-hand corner we get 31/8. The area is really 3, as we find by ~ 55. Taking still smaller pieces, the result is of course better; thus with 30 pieces each -1 unit wide, the left-hand heights give 2.855, the right-hand heights 3.155. With still more numerous (smaller) pieces these approximate results approach the true value of the area. (See ~ 119, p. 196.) XIII, ~ 1181 INTEGRALS AS LIMITS OF SUMS 195 EXERCISES Approximate to about 1% the areas under the following curves, between the limits indicated. Estimate the answers roughly in advance. Use judgment with regard to scales to gain accuracy by having the figure as large as convenient. Express each area as a definite integral and check by integration when possible. 1. y=X2-4x; x = 2 to 6. 2. y = 1/x; x = 10 to 20. 3. y = X-2; 1-xz 5. y= x x = Ilto 5. 4. y = X-3/2; x = 2 to 4. x = 2 to 4. 6. Y1 1+. x=Oto2. 1 +X21 7. y = 1/1V12 - x; x = 3 to 8. 8. y = V9-x; X = 0 to 5. 9. y = 1/9\ - X2; X = 0 to 1.5. 10. y = V9 + X2; X = O to 4. cos X 11. Y= I sin X; = 0 to 7r/2. 12. Y = + e x = 0 to 1. ex +f e-x~ 1 1 13. Y + x=O to 2. 14. y = l- =Oto.5. (4 + X2)3/2; V f - X3 15. y = /9 + X4; x = 0 to 2. 16. y = (1 - CosX)3/2;X=Oto2w. 1 17. y -i = xOto 7/6. 18. y=tan-1vx; x= O tol1. '\/4 - sin2 X Cos X ___ 19. ~y 1 + sin x:' x=Oto 7r/2. 20. 1 =O0 to.5. 21. y = e-"; x=0 tol1. 22. y=x e_2'; x=O to 1. Approximate to about 1% the distance passed over between the indicated time limits, when the speed is as below; express each distance as a definite integral, and check by integration when possible. 23. v = 4 t +tV; 1 25. 2 t + t2 I t = Ilto 3. 24. v- t = 0 to 50. 1I + t' 1 t -- to 4. 26. v =; t = 0 to 100. I + t~ t 27. v 3 +t2 t = 1 to 3. 28. v = ~1 tP; t = 10 to 20. 196 THE CALCULUS [XIII, ~ 118. 29. Show how to find the volume of a cone approximately, by adding together layers perpendicular to its axis. 30. Find the volume of a sphere by imagining it divided into small pyramids with their vertices at the center and their bases in the surface, as in elementary geometry. 31. The volume of a ship is computed by means of the areas of cross sections at small distances from each other; show how the result is calculated. Show how to make a more accurate computation by the same method. 119. Exact Results. Summation Formula. Any definite integral Y t w=b y f () (1) xf(x)dx =a TMhe t(remay be thought of as the P -<^^^^'^^^^%^^/?^%?^%^% area under a curve (2) =f (x) between the ordinates x = a and x = b. O A B If the interval AB from x=a D1DDs x=b x = a to x =b be divided FIG. 47. into n parts, each of width Ax, the whole area is approximately as in ~ 118, (3) S = Ax-f (a) +Axf (a+Ax) +. x +x.f (a+ (n - 1) Ax). The term Ax f (a) is the area of the rectangle AD1N1P, since f (a) =AP. Likewise Ax f (a - Ax) is the area of DID2N2Ml, etc. Hence the sum S represents the shaded area in Fig. 47. If the curve is rising from x = a to x = b, as in Fig. 47, S is smaller than the area under the curve. On the other hand the similar sum (4) R = Ax.f(a + Ax) + Ax.f (a + 2 Ax) +-. + Ax.f (a + (n - 1) Ax) + Ax -f(b) XIII, ~ 120] INTEGRALS AS LIMITS OF SUMS 197 is represented by the shaded area in Fig. 48, and R is too large if the curve is rising. Hence Y =f() Q R> f (x) dx > S if the curve is rising. But, P subtracting (3) from (4), we have (5) R-S= Ax [f(b)-f (a)], 0 A B and this approaches zero as x=aD1D2D3 x =b Ax approaches zero. It FIG. 48. follows that the true value of the integral is the limit of either R or S as Ax approaches zero, i.e., as n becomes infinite, and we may write * (6) f (x) dx = li xf(a) +Axf(a + Ax) + * *y a ny-oo + Axf(a + (n-l )Ax) at least if the curve rises from x = a to x = b. Similarly, the formula (6) is true also if the curve falls from x = a to x = b. Finally, if the curve alternately rises and falls, we may prove (6) by separating the interval into several parts, in some of which it rises and in some of which it falls. The formula (6) is called the summation formula of the integral calculus. 120. Integrals as Limits of Sums. By far the greater number of integrations appear more naturally as limits of sums than as reversed rates. * We have written (6) as the limit of S. It would be equally correct to use R, since R-S approaches zero. THE CALCULUS [XIII, ~ 120 Thus, as a matter of fact, even the area A under a curve, treated in ~ 55 as a reversed rate, probably appears more naturally as the limit of a sum, as in (6), ~ 119. Of course the two are equivalent, since (6), ~ 119, is true; in any case the results are calculated always either approximately, as in the exercises under ~ 118, or else precisely by the methods of ~~ 52-54. Hence the method of ~ 54 was given first, because it is used for each calculation even when the problem arises by a summation process. On account of the frequent occurrence of the summation process, we may say that an integral really means* a limit of a sum, but when absolutely precise results are wanted it is calculated as a reversed differentiation. The symbol f is really a large S somewhat conventionalized, while the dx of the symbol is to remind us of the Ax which occurs in the step-by-step summation. EXERCISES Express each of the following integrals as the limit of a sum, as in (6), ~ 119. Find its approximate value to about 1%, and check by integration. 1. f210 x -+ 2 x d*4 4 1. f X~-dx. 2. 1 j-dx. 3. f V9+x2dx..d7 35. sinx dx. 6.f7ri2co 4. +xdx.. sin xdx. 6. o cos Xdx. Jo Jo o 450 300- 50 7. / tan x dx. 8. sec x dx. 9. logio xdx. * It is really a waste of time to discuss at great length here which fact about integrals is used as a definition, and which one is proved; to satisfy the demand for formal definition, the integral may be defined in either way, - as a limit of a sum, or as a reversed differentiation. The important fact is that the two ideas coincide, which is the fact stated in the Summation Formula. XIII, ~ 121] INTEGRALS AS LIMITS OF SUMS 199 Express each of the following quantities as the limit of a sum; find its approximate value to about 1%; check by integration. 10. The area under the curve y = x2 from x = 1 to x = 3; from x = - 3 to x = 3. 11. The area under the curve y = x3 from x = 0 to x = 2; -from x=-ltox = +1. 12. The area under the curve x2 y - 1 from x = 2 to x = 5. 13. The distance passed over by a body whose speed is v = 4 t + 10 from t = 0 to t = 3. 14. The distance passed over by a falling body (v = gt) from t = 2 to t = 6. 15. The increase in speed of a falling body from the fact that the acceleration is g = 32.2, from t = 0 to t = 5. 16. The increase in the speed of a train which moves so that its acceleration is j = t/100, between the times t = 0 and t = 6. The distance passed over by the same train, starting from rest, during the same interval of time. 17. The number of revolutions made in 5 min. by a wheel which moves with an angular speed co = t2/1000 (radians per second). 18. The time required by the wheel of Ex. 17 to make the first ten revolutions. 19. Repeat Ex. 18 for a wheel for which w = 100 - 10 t (degrees per second). Find the time required for the first revolution after t = 0; note that the speed is decreasing. 20. The weight of a vertical column of air 1 sq. ft. in cross section and 1 mi. high, given that the weight of air per cubic foot at a height of h feet is.0805 -.00000268 h pounds. 121. Volume of any Frustum. To illustrate the ease of application of this process, consider again the volume of a frustum of a solid. (See ~ 57.) If such a frustum be divided up into layers of thickness As, by planes parallel to the base, and if A8 represents the area of any section at a distance s from the lower bounding 200 THE CALCULUS [XIII, ~ 121 plane, the volume of each layer is, approximately, the product of its thickness As times the area As of the bottom of the layer: (1) Volume of one layer = AsAs, approximately. Now if we replace x in (3), ~ 119, by s, and if As = f(s), the sum of all such layers would be, approximately, As.f(a) + Ax xf(a + As) +...... + As f(a + (n- 1)As). Hence, by (6), ~ 119, the exact total volume is s=b s=b V= Ads = f (s) ds. s=a s=a This formula is the same as that derived in ~ 57. It may B be used to find the L= _M —4 _ ~~s — b volume of any solid, / i \ -'~ if we know how to /-~~:/-2 --- s = As find the areas of A/-I,-,s=s any such complete / -. \.,'l set of parallel cross 1-o,-'-' ---- ~~-j ---'\ i'l sections. Ht --- —— __ -sa A/ - In particular, if o the solid is a solid FIG. 49. of revolution, the preceding formula reduces to formula (4) of ~ 56. 122. Surface of a Solid of Revolution. Similarly any such formula is readily derived, and easily remembered by means of this new process. Thus the formula for the surface of a solid of revolution was derived in ~ 85. To obtain that formula, or to remember it, we may remark that the curved area of any short section is approximately AA = 2 iry As, since the curved area is approximately the area of a section XIII, ~ 123] INTEGRALS AS LIMITS OF SUMS 201 of a cone. It follows readily that the total area of such a surface is A == 2wryds_______ Y f27'y l +\ x d ). 123. Water Pressure. As another _ B x typical instance, consider the water pres-/ sure on a dam or on any container. The // pressure in water in- FIG. 50. creases directly with the depth h, and is equal in all directions at any point. The pressure p on unit area is (1) p=k h where h is the depth and k is the weight per cubic unit (about 6.24 lb. per cubic foot). Suppose water flowing in a parabolic channel, Fig. 51, whose vertical sec200Aiv / tion is the parabola 200__ \h x2 = 225y. \150-, / h\ / Let a be the depth \ 100oo / Q of the water in the \ 50- channel. If a cut-, off gate be placed -200 -150 - 100 d 50 oo 150 200 x across the channel, FIG. 51. it be required let it be required to calculate the total pressure on the gate. 202 THE CALCULUS [XIII, ~ 123 Consider a horizontal strip of height Ah, whose upper edge is h below the surface. The area of such a strip is its width, w, times its height, Ah. Hence the pressure on the strip is, approximately, pressure on horizontal strip = (k h) w Ah In this example, w = 2 x = 2 V225 y = 30 yl/2, h = a - y, and Ay = Ah. Hence pressure on horizontal strip = 30 k (a - y)yl/2 Ay, and the sum of the pressures on all such strips is the sum of such terms as this one. Hence, by (6), ~ 119, the correct total pressure is P = 30k (a - y) y/2 dy J y=o = 30 k [, ad a is t = 8 ka5 f/2 where k = 62.4 lb., and a is the total depth of the water. EXERCISES 1. Find the volume generated by revolving Vx + Vy = a about the x-axis, from x = 0 to x = a. 2. Find the volume of the paraboloid z = x2/a2 + y2/b, from z = 0 to h. 3. Find the volume of the cone z2 = x2/a2 + y2/b2, from z = 0 to h. 4. Find the volume of a regular pyramid of base B and height h. 5.6. On a system of parallel chords of a circle are constructed equilateral triangles whose bases are those chords and whose planes are perpendicular to the plane of the circle. Find the volume in which all these triangles are contained. 6. On the double ordinates of an ellipse are constructed triangles of fixed height h, with planes perpendicular to the plane of the ellipse. Find the volume containing all these triangles. 7. Find the volume generated by a variable square whose center moves along the x-axis from x = 0 to Tr, the plane of the square being perpendicular to the x-axis, and the side proportional to sin x. XIII, ~ 124] INTEGRALS AS LIMITS OF SUMS 203 8. Find the surface of the spheroid generated by revolving the ellipse y2 = (1 - e2) (a2 - x2) about the x-axis. 9. Find the surface generated by revolving the catenary y = (e- + e- )/2 about the x-axis, from x = 0 to a. 10. As in Ex. 9 for the hypocycloid x2/3 + y2/3 = a2/3, from x = -a to a. 11. As in Ex. 9 for one arch of the cycloid. 12. Find the surface generated by revolving p = a cos 0 about the initial line. 13. As in Ex. 12 for the cardioid p = a(1 + cos 0). Calculate the following pressures: 14. On one side of the gate of a dry dock, the wet area of the gate being a rectangle 80 ft. long and 30 ft. deep. 15. One side of a board 10 ft. long and 2 ft. wide, which is submerged vertically in water with the upper end 10 ft. below the surface. 16. On an equilateral triangle 20 ft. on a side, submerged in water with its plane vertical and one side in the surface. 17. On one side of a square tank 10 ft. high and 5 ft. on a side, the tank being filled with a liquid of specific gravity.8. 18. On one face of a square 10 ft. on a side, submerged so that one diagonal is vertical and one corner in the surface. 19. On one end of a parabolic trough filled with water, the depth being 3 ft. and the width across the top 4 ft. 20. On one side of an isosceles trapezoid whose upper base is 10 ft. long, parallel to the surface and 10 ft. below it, whose lower base is 20 ft. and altitude 12 ft. 21. What is the effect on the pressure if all dimensions given in Ex. 20 are multiplied by a constant c? 124. Cavalieri's Theorem. The Prismoid Formula. If two solids contained between the same two parallel planes have all their corresponding sections parallel to these planes equal, i.e. if the area A', of such a section for the first solid is the same as the area A", of the second, it follows from ~ 57 that their total volumes are equal, since the two volumes are given by the same integral. 204 THE CALCULUS [XIII, ~ 124 This fact, known as Cavalieri's Theorem, is often useful in finding the volumes of solids. If the area A8 of any section of a frustum is a quadratic function of s: * (1) As = as2 + bs + c, where, as in ~ 57, s represents the distance of the section A, from one of the two parallel truncating planes, the volume is ]s=h rs=h 3 2 s=h (2) V = s (as2 + bs + c) ds = a + b + cs s= 0 3 2 Js=o ah3 bh2 = + + ch, where h is the total height of the frustum. The area B of the base of the frustum, the area T of the top, and the area M of a section midway between the top and bottom are B =AS] = [as -+ b8 + c] = c; s=0 s=0 T =As] = as2 + bs + c] = ah2 + bh + c; s=h s=h M=As = [as2 + bs + c ah + b + c. s=2h/l2s=h/2 4 2 If we take the average of B, T, and 4 times M: B+T+4M ah2 bh 6 = -3+2+c, 6 3++C 2 this average section multiplied by the total height h turns out to be exactly the entire volume: B + T + 4 M ah3 bh2 s=h (3) B+T+4MXh= + - +ch = V 6 3 2 Js=O * It is shown in Ex. 3, p. 206, that the results of this section hold also when As is any cubic function of s: A8 = as3 + bs2 + cs + d. Notice also that any linear function bs + c is a special case of (1), for a = 0. XIII, ~ 124] INTEGRALS AS LIMITS OF SUMS 205~ This fact is known as the prismoid formula. It is easy to see by actually checking through the various formulas, that this formula holds for every solid whose volume is given in elementary geometry; the same formula holds for a great variety of other solids.* But the chief use to which the formula is put is for practical approximate computation of volumes of objects in nature: it is reasonably certain that any hill, for example, can be approximated to rather closely either by a frustum of a cone, or of a sphere, or of a cylinder, or of a pyramid, or of a paraboloid; since the prismoid formula holds for all these frusta, it is quite safe to use the formula without even troubling to see which of these solids actually approximates to the hill. Similar remarks apply to many other solids, such as metal castings, though it may be necessary to use the formula several times on separate portions of such a complicated object as the pedestal of a statue, or a large bell with attached support and tongue. EXAMPLE. The prismoid formula Y applies to any frustum of an ellipsoid of revolution cut off by planes perpendicular to the axis_; of revolution. Let the origin be situated on one of the truncating planes of the frustum, and let the axis of x be the axis of revolution. Then the equa- FIG. 52. * The formula holds also, for example, for any prismoid, i.e. for a solid with any base and top sections whatever, with sides formed by straight lines joining points of the base to points of the top section. For example, any wedge, even if the base be a polygon or a curve, is a prismoid. The solids defined by (1) include all these and many others; for example, spheres and paraboloids, which are not prismoids. The formula holds for all these solids and even (see Ex. 3, p. 206) for all cases where As is any cubic function of s. One advantage of the formula is that it is easy to remember: even the formula for the volume of a sphere is most readily remembered by remembering that the prismoid formula holds. 206 THE CALCULUS [XIII, ~ 124 tion of the generating ellipse is of the form Ax2 + By2 + Dx + F = 0. The area As of a section parallel to the bases is Try2, since the section is a circle whose radius is y. Hence A = ry2= AB - F which is a quadratic function of the distance x from one of the truncating planes of the frustum. Therefore the prismoid formula holds. Beware of applying the prismoid formula, as anything but an approximation formula, without knowing that the area of a section is a quadratic function of s, or (Ex. 3, p. 206) a cubic function of s. EXERCISES [This list includes a number of exercises which are intended for reviews.] 1. Show that the prismoid formula holds for each of the following elementary solids; hence calculate the volume of each of them by that formula: (a) sphere; (b) cone; (c) cylinder; (d) pyramid; (e) frustum of a sphere; (f) frustum of a cone. See Tables, II, F. 2. Calculate the volunme of the solid formed by revolving the area between the curve y = x2 and the x-axis about the x-axis, between x = 1 and x = 3. Find the same volume (approximately) by the prismoid formula, and show that the error is about 0.6%. 3. Calculate the volume of a frustum of a solid bounded by planes h = 0 and h = H, if the area As of a parallel cross section is a cubic function ah3 + bh2 + ch + d of the distance h from one base, first by direct integration, then by the prismoid formula. Hence prove the statement of the footnote, p. 205. 4. In which of the exercises relating to volumes on p. 97 does the prismoid formula give a precise answer? 5. How much is the percentage error made in computing the volume in Ex. 7, p. 97, from x = 1 to x = 3, by use of the prismoid formula? 6. Show, by analogy to ~ 64, that the area under any curve whose ordinate y is any quadratic function (or any cubic function) of x, between x = a and x = b, is (b - a) - [YA + 4YM + YB], where YA, YB, YM represent the values of y at x = a, x = b, x = (a + b) /2, respectively. XIII, ~ 124] INTEGRALS AS LIMITS OF SUMS 207 Calculate, first by direct integration, and then by the rule of Ex. 6, the areas under each of the following curves: 7. y =x2+2x+3 betweenx = 1 andx =5. 8. y=x2-5x+4 betweenx = andx =5. 9. y =3 + 5 x between x = 2 and x = 4. 10. Calculate approximately the area under the curve y = x4 between x = 1 and x = 3 by the rule of Ex. 6. Show that the error is about.55%. 11. Show that any integral whose integrandf (x) is a quadratic (or a cubic) function of x, can be evaluated by a process analogous to the prismoid rule: f f [(x) dx =f(a) 4f ( )+ f(b)] -a 12. Evaluate the integralf(l/x2) dx between x = 1 and x = 5 approximately, first by the rule of Ex. 11; then by applying the same rule twice in intervals half as wide; then by applying the rule to intervals of unit width. 13. Show that any integralff (x) dx can be computed approximately by using Ex. 11 with an even number of intervals of small width Ax: x=b / x d f (x= ) = 4f (a + Ax) + 2 f (a+ 2 Ax) + 4f (a+ 3 Ax) x=a +- -+f(b)] [This rule is called Simpson's Rule.] Calculate the following integrals approximately by Simpson's Rule. Notice that some of them cannot be evaluated otherwise at present. 3921 ________ 14. f x5dx. 16. Idx.. f V. + x2 dx. J o lo o 3 5 r/4 15. f(1/x) dx. 17. fV1 + x dx. 9. 9. sin x dx. 20. Find approximately the length of the arc of the curve y = x2 from x = 0 to x = 1; from x = to x = 1. 21. Find approximately the area of the convex surface of that portion of the paraboloid formed by revolving the curve y = X\x about the x-axis which is cut off by the planes x = 0 and x = ~; by x = ~ and x = 1. CHAPTER XIV MULTIPLE INTEGRALS -APPLICATIONS 125. Repeated Integration. Repeated integrations may be performed with no new principles. Thus 1 l; n l j- dx = -- + c; and j) -- +c dx =- log x + cx+ c'. The final answer might be called the second integral of 1/x2. Thus, in the case of a falling body, the tangential acceleration is constant: dv jT=- = - where g is the constant; hence v = fjTdt + const. = -gt + c; but since v = ds/dt, gt2 s = fv dt + const. = (- gt + c) dt + const. = - + ct + c'. If the body falls from a height of 100 ft., with an initial speed zero, s = 100 and v = 0 when t = 0; hence c = 0 and c' = 100, whence we find s = - gt2/2 + 100. The equations s = fv dt + const., v = fji dt + const., just obtained, apply in any motion problem, where jT is the tangential acceleration, v is the speed, and s is the distance passed over. 208 XIV, ~ 126] MULTIPLE INTEGRALS-APPLICATIONS 209 126. Successive Integration in Two Letters. A distinctly different case of successive integration which can be performed without further rules is that in which the second integration is performed with respect to a different letter. Thus, the volume of any solid is (~ 57), rh=b (1) v= As dh, h=a where As is the area of a section perpendicular to the direction in which h is measured, and where h = a and h = b denote planes which bound the solid. In many cases it is convenient first to find A, by a first integration, by the methods of ~ 55, and then integrate As to find V by (1), this second z integration being with respect 1 to the height h. EXAMPLE 1. Find the volume of the parabolic wedge y2 = X (1- z)2 between the planes z = 0 and z = 1 / and between the planes x = 0 and x = 1. x x. FIG. 53. The area A, of a section by any / plane z = h parallel to the xy-plane is twice the area between the curve y = (1 - h) V/x and the x-axis: ]x= 1 x=1 x=-1 4 - x=1 As =2 y dx = 2f (1-h) Vx dx = (1-h) X3/2 -_x=o a=0 J x=o x=o 4 (1 - ), hence this volume, by (1), is ]h=l nh=l 4 h=i 4/ 72\ 7=1 29 V ]= As dh = (1 -h)dh = h —. =. th=O= h=O 3, d 3h=O 3 2 i h=o 3 Notice that h, during the first integration, was essentially constant. 210 THE CALCULUS [XIV, ~ 126 Combining the formulas used in this example, the volume V may be written J h=\ 2 V = 2 (I- h)Vxdx dh - h=O h=O L = 3 This result is usually written without the brackets on the right: -h= 1 rh=l = 1 V = 2 (1 - h) dxdh. Jh=O Jh=O x=0 Such double integrals in two letters are very common in applications. Examples of triple integrals will be found further on in this chapter. 127. Volumes. Double Integrals. Analogous to the problem of finding the area under a given portion of a curve, (~~ 55, 119), there is the problem of finding the volume under a given portion of a surface.* This leads to double integrals. Let A'B'C'D' be a portion of a curved surface whose equation is z = F(x, y). Let ABCD be the projection of A'B'C'D' on the xy plane, curve CD being the projection of curve C'D'. The problem is to express the volume between the portion A'B'C'D' of the surface and its projection on the xy-plane. Divide this volume into slices of thickness PQ = Ax, by planes parallel to the yz-plane, and further subdivide the slices into prismatic columns by planes parallel to the xy-plane spaced at intervals Ay. The area of the section PSS'P' depends on the position * See Tables, I, b, for formulas from Solid Analytic Geometry. XIV, ~127] MULTIPLE INTEGRALS -APPLICATIONS 211 of P, hence on x, and is therefore a function of x. Therefore by the frustum formula (~ 57), z A 'P( B the required volume is V =fb Ax dx. D X=a a=a 1 But the area Ax is the area l_ under the curve.P'S' whose ordinates (heights above xy-_ a Q =b plane) are z = F (x, y), where / x has the fixed value OP and / /, y alone varies from y = 0 to S Uf(x) y = PS = f(x), this being FIG. 54. the equation of curve DC, supposed given. Hence A =f (x) A= f f F(x, y) dy. v=O Then _b x-=b ]V _= Ax dx, a ta==a or (1) i: =a j=b f(J F(x, y) dydx. a x=a -=0 This is essentially the same sort of problem as that discussed in ~ 126. This double integral may be written as the limit of a double sum. For if we consider the prismatic column on Ax Ay as base, its volume approximately is z Ax Ay or F (x, y) Ay Ax. This suggests the double sum Ax — [lim (;: )F (x, y) Ay] Ax or, as it is usually written, lim Zx=b lim Vf () F (x, y) Ay Ax. Ax —> 0ta Ax —>0 "Y=0 212 THE CAL;CULUS MM ~, 127 Here the limit of the inner summation is merely A, and then the limit of the outer summation is the integral of A or the volume. Hence we write the formula (2) lim L=a2ry=O F (x, y) AyAx Ay ---O x~b y=f'(x)F xy)ddx fxiafy_ F(xydydx This is the fundamental summation formula for double integrals. EXAMPLE. Determine the volume under the surface z = X2 + y2 between the xz-plane, the planes x = 0, x = 1, and y = 0, and the cylinder y = /X-. FIG. 55. V sf =1 1/=v= ff=ol~ ~ (X2+y2)dydx:= fi [ y + y3/3] dx f lX51 + (1/3) X3/2] dx = 44/105. XIV, ~ 127] MULTIPLE INTEGRALS -APPLICATIONS 213 EXERCISES 1. Determine a function y = f (x) whose second derivative d2yldX2 is 6 x. Ans. y = x3 + CiX + C2. 2. Determine the speed v and the distance s passed over by a particle whose tangential acceleration d2s/dt2 is 12 t. Find the values of the arbitrary constants if v = 0 and s = 0 when t = 0; if v = 100 and s = 0 when t = 0. Find the general expressions for functions whose derivatives have the following values: 3. d2yldX2 = 6 x2. 6. d2r/d02 1/1 - 0. 9. d2yldX2 = ex. 4. d2S/dt2 = 3 + 2 t. 7. d3r/d03 = 02 - 2 0. 10. d2s/dt2 = seC2 t. 5. d2s/dt2 = 1 -t. 8. d3V/dU3 = I - U2. 11. d3V/du3 = 1/u2. Determine the speed v and distance s passed over in time t, when the tangential acceleration jT and initial conditions are as below: 12. jT = sin t; v = 0 and s = 0 when t = 0. 13. jT = t + COs t; v = 0 and s = 0 when t = 0. 14. jT = V1 + t; v = 3 and s = 0 when t = 0. 15. jT = t/l'1+ tV; v = I and s = O when t = 0. Evaluate each of the following integrals, taking the inner integral sign with the inner differential: X=~1,v=1 f1=3 8=0 r-r3 2,22 16. xy dy dx. 21. f f J -dr ds dt fx2 f" z~1 fs+= 2 17. f f 6X2(2-y)dy dx. 22. (x + y) dy dx. Y I JYf J =-1 f 1=-X2 x-2 y-3 x-3 y=2x+3 18. f f(X2 + l)(4 - y2) dy dx. 23.f (X + y)2 dy dX. fv-4 u=2 _ 1 +X x+ 19. f ~ + V du dv. 24. f f f (x+y z)dddx V9~=2 fu=1 0 f x Jo (x1S_ ~7 + y + z) dz dy dx. f, e ""sin 20. f e S dy dx. 25. f f fr2 sin 0 do do dr. 26. Find the volume of the part of the elliptic paraboloid 4 x2 + 9 y2 = 36 z between the planes z = 0 and z = 1; between the planes z = a and z = b. 214 THE CALCULUS [XIV, ~ 127 27. Find the volume of the part of the cone 3 x2 + 9 y2 = 27 a2 between the planes z = 0 and z = 2; between z = a and z = b. 28. Find the volume of the part of the cylinder x2 + y2 = 25 between the planes z = 0 and z = x; between the planes z = x/2 and z = 2 x. 29. A parabola, in a plane perpendicular to the x-axis and with its axis parallel to the z-axis, moves with its vertex along the x-axis. Its latus rectum is always equal to the x-coordinate of the vertex. Find the volume inclosed by the surface so generated, from z = 0 to z = 1 and from x = 0 to x = 1. 30. Find the volume of the part of the cylinder x2 + y2 = 9 lying within the sphere x2 + y2 + z2 = 16. 31. For a beam of constant strength the deflection y is given by the fact that the flexion is constant: b = d2y/dx2 = const. if the beam is of uniform thickness. Find y in terms of x and determine the arbitrary constants if y = 0 when x = ~ 1/2. [This will occur if the beam is of length 1, and is supported freely at both ends.] 32. Determine the arbitrary constants in the case of the beam of Ex. 31, if y = 0 and dy/dx = 0 when x = 0. [This will occur if the beam is rigidly embedded at one end.] 33. For a beam of uniform cross section loaded at one end and rigidly embedded at the other, b = d2y/dx2 = k(l - x) where I is the length of the beam, x is the distance from one end, and k is a known constant which is determined by the load and the cross section of the beam. Find y in terms of x, and determine the arbitrary constants. Find y in terms of x in each of the following cases: 34. d2y/dx2 = k(12 - 2 Ix + x2); y = 0, dy/dx = 0 when x = 0. [Beam rigidly embedded at one end, loaded uniformly.] 35. d2y/dx2 = a + bx; y = 0, dy/dx = 0 when x = 0. [Beam of uniform strength of thickness proportional to (a + bx)-1, embedded at one end.] 36. d2y/dx2 = k(12/8 - x2/2); y = 0 when x = + 1/2. [Beam supported at both ends, loaded uniformly.] 37. d2y/dx2 = k/x2; y = 0, dy/dx = 0 at x = I. [Beam of uniform strength of thickness proportional to x2, embedded at x = 1.] 38. Find the angular speed w and the total angle 0 through which a wheel turns in time t, if the angular acceleration is a = d20/dt2 = 2 t, and if 0 =co = 0 when t = 0. XIV, ~ 129] MULTIPLE INTEGRALS-APPLICATIONS 215 128. Triple and Multiple Integrals. There is no difficulty in extending the ideas of ~~ 126-7 to threefold integrations or to integrations of any order. Following the same reasoning, it is possible to show that, if w = F (x, y, z) lifm l = VY=d w A Ay A im Lz =e LV = c z=a W / / A Ax-+o Ay->0 Az —}O rz=f (y=d rx=b = j F (x, y, z) dx dy dz, J z=e J y=c x=a where the three integrations are to be carried out in succession, where the limits for x may depend on y and z, and where the limits for y may depend on z: but the limits for z are, of course, constants. Thus it is readily seen that the volume mentioned in ~ 127 may be computed by dividing up the entire volume by three sets of equally spaced planes parallel to the three coordinate planes. Then the total volume is, approximately, the sum of a large number of rectangular blocks, the volume of each of which is Ax Ay Az; and its exact value is lim 2 P =f(x) Z=F(2, ) VA ---O =a y=o z= z Ay Ax Ay —0 Az ---O x=b ry=f((x) rz=F(x,y) V) = ~:.f: ()J dz dy dx, s=a y=0 J2=0 which reduces to the result of ~ 127, if we note that z=F (x, ) z = F (x, y ) dz = z = F(x, y). =0 z=0 129. Plane Areas by Double Integration. To find the area bounded by a closed curve C, we divide it into small elements of area, either by lines drawn parallel to the coordinate axes if the equation of C is given in rectangular coordinates, 216 THE CALCULUS [XIV, ~ 129 or by a system of radial lines and concentric circles if the equation of C is given in polar coordinates. (Figures 56, 57.) I Ax - - a x b 0 FIG. 56. FIG. 57. In Fig. 56 let any ordinate whose abscissa is x meet the boundary of the area in P and Q, the corresponding values of y being yi and y2. Let the extreme values of x between which the oval lies be x = a and x = b. Then the area will be (1) A =lim x=. - Ay Ax = X= _dy dx. Ax —0 O=a xda Y=YV Ay —0 This is merely a special case of (2), ~ 127, when F(x, y) = 1, because the volume of a right cylinder of height 1 equals the area of the cross section. In Fig. 57 let any radius vector whose angle is 0 meet the boundary of the area in P and Q, the corresponding values of p being OP = pi, OQ = P2. Let the extreme values of 0 be 0 = a and 0 = 3. Then the element of area is approximately pA0 * Ap and the total area is (2) A =lim I:= p -— P fPfa P dp d. Ap- *Op p AO —,O XIV, ~129] MULTIPLE INTEGRALS-APPLICATIONS 217 In formulas (1) and (2) the first integration can be carried out at once, giving (1') A = (y - yi) dx (2') A = 2 (P22- p2) dO. The last results may also be derived from the formula of ~ 55 and ~ 92, so formulas (1) and (2) may-be dispensed with so far as the mere calculation of plane areas is concerned. We shall, however, find the idea of a plane area as the limit of a double sum very useful in the following sections. EXAMPLE 1. Find the area between the parabola y2 = 4 (x - 1) and the line y = x - 1. (Fig. 58.) The extreme values of x are found to be x = 1 and x = 5. Hence x=5,,y=2 5 V _)]A = = d dx f= 2x- 1 - (x -1) dx = [4/3 (x- 1)3/2 - (x- 1)21= 8/3. 4-~_ ---- / — — _1 1j- — I o/ I;5 FIG. 58. FIG. 59. EXAMPLE 2. Find the area between the circle p = cos 0 and the cardioid p = 1 + cos 0, from 0 = 0 to 45~. (Fig. 59.) A f -,=r/4,fp2 = 1 +cos 0 f=0 J p pd co 0,r/4 = 4 [(1 + cos 0)2 - COS2 0] do -= o/" (l1 + 2 Cos 0) do = 2 (T/4 +V2). 218 THE CALCULUS [XIV, ~ 129 EXERCISES - DOUBLE INTEGRALS 1. Find the volume under the surface z = x2 + y2 between the xzplane, the planes x = 0 and x = 1, and the cylinder whose base is the curve y = x2. Find the volume between the xy-plane and each of the following surfaces cut off by the planes and surfaces mentioned in each case: 2. z=x+y cut off byy =0, x = 0, x = 1, y = /x. 3. z=x2 +y cut off byy = 0, x = 1, x =3, y =x2. 4. z = xy cut off by y = 0, x =2, x =4, y= x2 + 1. 5. '= xy + y2 cut off by y = 0, x = 1, x = 5, y = x3. 6. z = y +/x cut off by y =, x =0, x =, y =x5. 7. z =x2 +y3 cut off by x = 0, y = 1, y = 4, y2=x. 8. z = /x+ cut off byx=0, y=2, y=5, y = x. 9. z = x2 + 4 y2 cut off by y = 0 and y = 1-x2. 10. z =xy cut off by y = x2 and y =1. 11. = x2- y2 cut off by y = x2 and y = x. 12. Find the volume of the portion of the paraboloid z = 1 - x2 -4 y2 which lies in the first octant. 13. If two plane cuts are made to the same point in the center of a circular cylindrical log, one perpendicular to the axis and the other making an angle of 45~ with it, what is the volume of the wedge cut out? 14. Show that the volume common to two equal cylinders of radius a which intersect centrally at right angles is 16 a3/3. 15. Show that the volume of the ellipsoid x2/16 + y2/9 + z2/4 = 1 is 32 7r. 16. How much of the ellipsoid in Ex. 15 lies within a cube whose center is at the origin and whose edges are 6 units long and parallel to the coordinate axes? 17. Where should a plane perpendicular to the x-axis be drawn so as to divide the volume of the ellipsoid in Ex. 15 in the ratio 2:1? Calculate by double integration the areas bounded by the following curves: 18. y =x2 and y= /x. 22. x = 0, y = sin x, and y = cos x. 19. y =x2 and y =x3. 23. y =, y2 = x, and x2 - y2 = 2. 20. y = x2 and - 2 y2 = 2. 24. y = 2 x, y = 0, and y = 1-x. 21. x2 -+ 2 = 12 and y = x2. 25. y2 = x, and y = 1 - x. XIV, ~ 130] MULTIPLE INTEGRALS-APPLICATION 219 Determine the entire area, or the specified portion of the area, bounded by each of the following curves, whose equations are given in polar coordinates: 26. p =2 cos 0. 27. One loop of p = sin 2 0. 28. One loop of p = sin 3 0. 29. The cardioid p = 1 - cos 0. 30. The lemniscate p2 = cos 2 0. 31. The spiral p = 8, from 8 = 0 to ir. 32. The spiral pO = 1, from 0 = 7r/4 to r/2. 33. p =1+2 cos 0, from 0 = O to 7r. 34. p = tan 0, from = 0 to 45~. 35. p =a02, one turn. 36. p = 3 cos 0 + 2. 37. p = a sin 0 cos 0/(sin3 0 + cos3 0); folium: the loop. 130. Moment of Inertia. When a particle of volves in a plane about a point 0, with given angular speed w, its speed is v = rw, and its kinetic energy is E = 1/2 mv2 = 1/2 mr2w2. The moment of inertia of m about 0 is defined as the product of the mass m by the square of its distance from O, Fl I = r2m. Thus E is easily found where I is knc mass m rem [G. 60. )wn. y A — / K I Z I4I. '" / '. 0 - I I Given now a thin plate of metal of uniform density and thickness, whose boundary C is a given curve, let us divide the plate into small squares by lines equally spaced parallel to two rectangular axes through 0. Let P be a point in any one of these squares and let OP = r = /x2+y2. Then the mass of FIG. 61. 220 THE CALCULUS [XIV, ~ 130 the square is k Ay Ax where k denotes the constant surface density (i.e. the mass per square unit); and the moment of inertia of this square about 0 is, approximately, k r2 Ay Ax. Hence the moment of inertia I of the entire plate about 0 is: Io = li kr2 Ax Ay = k(x2 + 2) dy d. Ax —0 The limits of integration are to be taken so as to cover the area, as in ~ 129. Thus the moment of inertia of a plate bounded by the two curves y = (1 - x2) and y = (x2 - 1), about the origin (draw the figure) is: - x= +1 ~y=l-x' I k x=, (x2 + y2) dy dx ~2x=-1- J y=x2 —1 = -1 [ Y + YI3 ]=-=2 -2k 1r=-i 2k X 3 -1 =2 k f1 (1( -x6) dx = x -7 j 1 3 j7x)x - - j = -1 7 where k is the surface density. The moment of inertia about an axis is defined similarly, the distance r being replaced by the distance of P from the given axis. Thus the moments of inertia of the plate in the figure about the x- and y-axes are, respectively, I= kf 2 dy dx; I. = k 2 dy dx. The radius of gyration of an area, whether about a point or about an axis, is defined by radius of gyration = K = V/I/M. If the boundary of the plate is given in polar coordinates, the moment of inertia about 0 is calculated by dividing the area into elements r AO Ar. (See ~ 129.) Then Io = k o rrr2 rAOAr = k J = = drd A9-*0 XIV, ~131] MULTIPLE INTEGRALS -APPLICATIONS 221 Thus for a circle whose center is 0, r = f () = a, the radius. Hence the moment of inertia of a circular disk about its center is: rLC= lTr f =pa ^4 4 7 Ma2 Io = ek'. 4 -P dO= kc- -r dO =rk = Ma-2 0J 0=o L4 Jp=o =0o 4 2 2 where k is the surface density, and M = k7ra2 is the mass of disk. 131. Moments of Inertia in General. The moment of inertia of any mass about a given point may be defined as follows. Divide the mass M into elements of mass Am; multiply each element Am by the square of its distance, r2, from the given point; the limit of the sum of these products is the required moment of inertia: (1) I = lim l r2Am = fr2dm. Am-o0 When M is a plate we take Am = kAA, where k is the mass per unit area, and AA = Ay Ax or rA 6 Ar, as in ~ 129. When M is a thin wire bent into the form of a given curve, Am = kAs, where k is the mass per unit length and As the element of length. Then (2) I = kr2 s = kfr2 ds, the limits being taken to cover the given curve. For ds we use (1) of ~ 82 or (1) of ~ 93. When M is a solid body, we take Am = kAV, where k is the mass per unit volume and AV is the element of volume, Ax Ay Az. Then (3) I = lim k ZZ r2 Ax y Az = kfffr2 dxdy dz. Ax-*O Ay ---)O Az ---o '222 2THE CALCULUS [XIV, ~ 131 The limits of integration are to be taken to cover the given volume. For the moment of inertia about an axis we let r in forms (1), (2), (3), be the distance of Am from -that axis. EXAMPLE 1. Find I for a wire bent into the form of a circular quadrant, about the center. I =fkr2ds. Using polar coordinates, ds = rd 0, and I "r/2 ir/2 i=k f 3 dO = kr3 f de = kr3 r/2 = Mr2, since M = kr r/2. (x,y,z) H I J '~~~~~~~~~~~I 0/ X y FIG. 62. FIG. 63..EXAMPLE 2. Find I for a circular cylinder of height H and radius 1 about the center of one base. The equation of the surface is x2 + y2 = R2, if the z-axis is the axis of the cylinder. Then, by symmetry, I =R =4kff 2-X2 z-H fx, =0 __~ (x2+y2 + z2)dzdydx, X=R I/= VRz2-X2 = 4kfx=o Rfy0 (X2 H ( +y2 H 1+ H3/3) dy dx X=R ~ z2 =4kHJ'x y + y/3 +H2y/3) dx x2+ kHf 2 (~ 1 Xz~- 2 + (R2 -- X2)3/2/3 + II2 \/2 - X213) dx. XIV, ~ 131] MULTIPLE INTEGRALS-APPLICATIONS 223 Let x = R sin 0. Then Ikr/2 I = 4 kHR2f (R2 sin2 0 cos2 0 + R2 cos4 0/3 + H2 cos2 0/3) dO = 4 kHR2 (R2 7r/16 + R2 r/16 + 16 + H2 7r/12) = kHR2 (3 R2 +2 H2) (wr/6). EXERCISES Calculate I about the origin for the areas bounded by the following curves (Density = k = 1): 1. y =x2and y =. 6. x = 0, y =sin x, and y = cos x 2. y =x2andy =x3. 6. y =0, y2 =x, andx2 - y2 =2. 3. y=x2and -x2 + y2=2. 7. y=2x,y=0, andy=1-x. 4. x2+y2= 12andy=x2. 8. y2=x, andy = 1 - x. Find the moment of inertia and radius of gyration of each of the following shapes of thin plate: 9. A square about a diagonal. About a corner. 10. A right triangle about a side. About the vertex of the right angle. 11. A circle about its center. 12. An ellipse about either axis. About the center. 13. A circle about a diameter. 14. A triangle of given base and height, about the base. 15. A trapezoid about one of its parallel sides. 16. A thin circular plate, about a point on the circumference. 17. A thin plate bounded by two concentric circles, about the center. 18. An equilateral triangle, about its center. 19. An equilateral triangle, about one vertex. 20. p = 2 cos 0. 21. One loop of p = sin 2 0. 22. One loop of p = sin 3 0. 23. The cardioid p = 1 - cos 0. 24. The lemniscate p2 = cos 2 0. 224 THE CALCULUS [XIV, ~ 131 25. The spiral p = 0, from 0 = 0 to 7r. 26. The spiral po = 1, frohm 0 = 7r/4 to 7r/2. 27. p=1 +2 cos 0, from 0 = 0 to 7r. 28. p =tan 0, from 0=0 to45~. 29. Calculate the moment of inertia of a cube about a corner. 30. Calculate the moment of inertia of a rectangular parallelopiped about a corner. Calculate the moment of inertia about the origin of each of the solids bounded by the following surfaces and lying above the xy-plane. (See Exs. 2-11, p. 218.) 31. z=x+ y; y =0; x=0; x=1; y = -x. 32. = xy; y = x2; y =1. 33. z=x2 +y; y=0; x=1; x=3; y=x2. 34. z=xy; y=O; x=1; x=2; y=x2+1. 35. z =2-y2; y=x2; y=x. 132. Average Value. Centroid. y A< V/ IEIF V. Let there be given a function f (x, y) of two independent variables. At any point of the xy-plane, that is, for a given pair of values of x and y, this function will have a definite value. What will be the average value of f (x, y) over a given region R? Divide the region into small squares, note some point of each square / 0' FIG. 64. the value of the function at (for convenience at the corner nearest the origin). Let there be n complete squares in the region R, the fractional parts around the border being disregarded. XIV, ~133] MULTIPLE INTEGRALS -APPLICATIONS 225 Then the average value of f (x, y) at the corners of the squares will be Sum of values of f (x, y) 2; f (x, y) Number of these values n Multiplying both sides by the element of area AA, and putting nA A + A A + AA -- A...... = 2 AA, we have 2 f (x, y) AA l2AA and the limit of this, as n increases and Ax and Ay decrease, is called the average value of f(x, y) over the region R. Thus f f (x, y) dA f f f (x, y) dx dy (1) Av. Val. off (x, y) over R = f dA ffdxdy The denominator is simply the area of R; the limits are to be taken to cover the region R. In polar coordinates, replace dA by rdOdr, and f(x, y) by F(r, 0). 133. Centroid or Center of Gravity of an Area. The centroid, (x, y), of a plane area, as R in the above figure, is the point whose coordinates are the average values of the coordinates of the points of R, that is ( f xdA fydA (1) X- df A andy dA For a material plate of uniform thickness and of mass k per unit area, the element of mass is dM = kdA and its centroid is defined by f)- xdM fydM (2) x fdM; y= fdM Here the density factor k may be a given function O, x and y; if k is constant it cancels out and we get the same result as for a plane area 226 THE CALCULUS ' [XIV, ~ 134 134. Centroid of a Curved Arc. Similarly, the centroid of a curved arc is defined as the Y B point whose coordinates are the average values of the co- _ — As ordinates of the points of the,i ~ arc. Thus if AB is the arc, Ai 0 Yj we divide it into n equal _____A I___ _ segments As, and note the OF x values of x and y for some FIG. 65. point of each As. The averages of these n values of x and y are, respectively, xl + X2 + X3 + '''+ Xn Y1 +Y Y2 + Y3 + ' * + Yn n n Multiplying numerator and denominator by As and allowing n to increase, we have XAS f xds - yds (3) X = lim =s= Y ds n ---oo As fds ' f ds 135. Centroid of a Volume. Similarly, by dividing a volume into elements, and noting the coordinates (x, y, z) of a point of each element of volume, we obtain \ -(4) _ x 'dV - 'd - fzdV n g wfdV w y e fdV ' Z fdV In general we may take dV = dxdydz, and so express (4) in terms of triple integrals. In numerous applications it is simpler, however, to take for dV a slice of the volume; thus in finding x cut the volume by a plane perpendicular to the x-axis, forming a section of area Ax; then take dV = Ax dx. 136. Definition of Centroid in Mechanics. From the standpoint of mechanics the centroid is defined in a different XIV, ~136] MULTIPLE INTEGRALS -APPLICATIONS 227 manner that turns out to be equivalent to what we have done. Suppose the xyplane to be horizontal and Y a wire bent into the form of any curve as AB to lie in this plane and to be balanced on a knife-edge MN, falling on the ordi- nate whose abscissa is 0 x = x. Divide the wire IN As -r-' --- B I I I ~~~I I~ Al m I x=a x=x FIG. 66. x= b into elements of mass Am = k As, multiply each Am by its distance from the knife-edge and form the sum of the moments 2 k(x - x)As. The limit of this sum must be zero if the wire is balanced, so we have k f (x - x) ds = 0. Hence f x ds = f x ds, or, since x is a fixed value, x J ds = J x ds. Hence x = f x ds/fds. Likewise we get y by supposing the wire to be balanced on a knife-edge parallel to the x-axis. The same considerations apply to a thin plate, except that As is replaced by AA. For a solid we pass planes parallel to the coordinate planes such that the sum of the moments of the elements of mass, i.e., the products (x- x)Am, (y- y)Am, ( - z) Am, shall have the limit zero. EXERCISES Find the average value of each of the following functions, over the area under the curve y = 1 - x2, from x = 0 to 1. 1. x. 2. y. 3. x2. 4. y2 5. 1/(1 +x). 6. ex. Find the average value of each of the following functions in the volume bounded by the coordinate planes and the plane x + y + z = 1. 7. f (x, y, z) = x, or y, or z. 8. f (x, y, z) = xyz. 9. f (x, y, z) = xy. 10. f (x, y, 2) ==x2 + y2 + z2. 228 ' THE CALCULUS [XIV, ~ 136 11. Find the average ordinate of y = cos x, from x = 0 to 7r/2. 12. Find the average ordinate of a semicircle. 13. Find the average distance of the points of the area of a circular quadrant from the center. (Use polar coordinates.) 14. Find the average density of a rod in which the density varies as the distance from one end. Find the centroids of the following figures: 15. Of the segment of the parabola y2 = 4 ax, from x = 0 to h. 16. Of a semicircle. 17. Of the first quadrant of an ellipse. 18. Of the area under y = cos x, from x = 0 to ir/2. 19. Of a right circular cone. 20. Of a right pyramid. (Use sections parallel to the base.) 21. Of a semiellipsoid of revolution. 22. Of a circular quadrantal arc. 23. Of a circular arc of angle 2 a. 24. Of the arc of x2/3 + y2/3 = a2/3, in the first quadrant. 25. Of the arc of y = (ex + e-)/2, from x = 0 to 1. Find the centroids (x, y) for the areas bounded by the following curves: 26. y=x2 and y = x. 27. y=x2 and y= 3. 28. x2 + y2 = 12and y = x2. 29. y2 = xandy= 1-x. 30. y = 2 x, y = 0, and y = - x. 31. p = 2 cos 0. 32. One loop of p = sin 2 0. 33. The cardioid, p = 1 - cos 0. 34. The lemniscate, p2 = cos 2 0. 35. The spiral, p = 0, from 0 = 0 to ar. 36. p = 1 + 2 cos, from 0 = 0to r. XIV, ~ 136] MULTIPLE INTEGRALS-APPLICATIONS 229 GENERAL REVIEW EXERCISES Find the areas bounded by each of the following curves, or the part specified: 1. p = a cos + b. 2. p = a cos 3 0. 3. y2 (a - x) = x3 [cissoid]; to its asymptote x = a. 4. y2 = x2 (4 - x): the loop. Find the volume generated by revolving each of the following curves about the line specified: 5. y = 5 x/(2 + 3 x); about y = 0; x = 0 tox = 1. 6. 2 x2 + 5 y2 = 8; about y = 0; total solid. 7. y = b sin (x/a); about y = 0; x = O tox = r. 8. y = a cosh (x/a); about y = 0; x = 0 to x = a. 9. (x - a)2 + y2 = 2r2; about x = 0; total solid. 10. The cycloid; about base; one arch. 11. The cycloid; about tangent at maximum; one arch. 12. The tractrix; about asymptote; total. 13. x = a cos3 t, y = a sins t; about y = 0; total solid. Find the area, its centroid, and its moment of inertia about the origin, for each of the following curves, between the limits indicated: 14. y = a (1-x2/b2); 1st quadrant. 15. y = x/(l + x2); x = 0 to x = 1. 16. The sine curve; one arch. 17. The cycloid; one arch. 18. x2/3 + y2/3 = a2/3 [or x = a cos3 t, y = a sin3 t]; first quadrant. 19. Between the two circles p = a cos 0 and p = b cos 0; b > a. 20. x = 2 a sin2 4, y = 2 a sin2 4 tan 0; between the curve and its asymptote. Find the centroid of each of the following frusta: 21. Of the paraboloid x2 + y2 = 4 az by the plane z = c. 22. Of a hemisphere. 23. Of the upper half of the ellipsoid of revolution 42 + 4 y2+ 9 z2 =36. 230 THE CALCULUS [XIV, ~ 136 24. Of the upper half of the ellipsoid x2 + 4 y2 + 9 z2 = 36. 25. Of the solid of revolution formed by revolving half of one arch of a cycloid about its base. 26. Obtain a formula for the volume of a spherical segfnent of height h. 27. Show that the volume of an ellipsoid of three unequal semiaxes, a, b, c, is 4 wrabc/3. 28. Show that the volume bounded by the cylinder x2 + y2 = ax. the paraboloid x2 + y2 = bz, and the xy-plane is (3/32) (ra4/b). 29. Find the volume common to a sphere and a cone whose vertex lies on the surface and whose axis coincides with a diameter of the sphere. Find the lengths of the arcs of each of the following curves, between the points specified: 30. y = log x; x = a to x = b. 31. ey cos x = 1; x = 0 to x = x. 32. x = t2,y =2 at (ory2= 4 a2x); t = tl to t=t2. 33. One arch of a cycloid. 34. p = a (1 + cos 0) [cardioid]; total length. 35. Calculate the moment of inertia I for a right circular cone about its axis. Ans. (3/10) mass * square of radius. 36. Calculate the moment of inertia and the radius of gyration for the rim of a flywheel about its axis, the inner and outer radii being R1, Rz. 37. The moment of inertia of an ellipsoid about any one of its axes is (1/5) (mass) (sum of the squares of the other two semi-axes). 38. Calculate the moment of inertia for a spherical segment about the axis of the segment. 39. Show that, for any body, 2 Io = Ix + I, + I,, where Io, I,, I,, Iz denote respectively its moments of inertia about a point and three rectangular axes through that point. 40. Show that for any figure in the xy-plane, I, = I + Iy, where Ix, Iy, lz denote its moments of inertia about the three coordinate axes respectively. XIV, ~ 136] MULTIPLE INTEGRALS-APPLICATIONS 231 41. Show that the total pressure on a rectangle of height h feet and width b feet immersed vertically in water so that its upper edge is a feet below the surface and parallel to it, is 62.4 bh (a + h/2). Show that the depth of the center of pressure is at (6 a2 + 6 ah + 2 h2)/(6 a + 3 h). 42. Show that the total pressure on a circle of radius r, immersed vertically in water so that its center is at a depth a + r, is 62.4 irr2 (a + r). Show that the depth of the center of pressure is a + r + r2/(4 r + 4 a). 43. Show that the total pressure on a semicircle, immersed vertically in water with its bounding diameter in the surface, is 41.6 r3. Show that the depth of the center of pressure is 3 7rr/16. 44. Calculate the water pressure to within 1 % on a circular disk 10 ft. in diameter, if its plane is vertical and center 10 feet below the surface. 46. Show that if a triangle is immersed in a liquid with its plane vertical and one side in the surface, the center of pressure is at the middle of the median drawn to the lowest vertex. 46. Show that if a triangle is immersed in a liquid with its plane vertical and one vertex in the surface, the opposite side being parallel to the surface, the center of pressure divides the median drawn from the highest vertex in the ratio 3:1. 47. Calculate the mean ordinate of one arch of a sine-curve. The mean square ordinate. [Effective E. M. F. in an alternating electric current.] 48. Calculate the average distance of the points of a square from one corner. 49. What is the average distance of the points of a semicircular arc from the bounding diameter? 50. When a liquid flows through a pipe of radius R, the speed of flow at a distance r from the center is proportional to R2 - r2. What is the average speed over a cross section? What is the quantity of flow per unit time across any section? 51. The kinetic energy E of a moving mass is lim Am * v2/2, where Am is the element of mass moving with speed v. Show that for a disk rotating with angular speed o, E = o2/2. Calculate E for a solid car wheel of steel, 30 in. in diameter and 4 in. thick when the car is going 20 m./hr. 232 THE CALCULUS [XIV, ~ 136 52. Show that the kinetic energy E of a sphere rotating about a diameter with angular speed o is (1/5) (mass) r2W2. 53. Calculate the kinetic energy in foot-pounds of the rim of a flywheel whose inner diameter is 3 ft., cross section a square 6 in. on a side, if its angular speed is 100 R. P. M. and its density is 7. 54. The x-component of the attraction between two particles m and m', separated by a distance r, is (k ~ m ~ m'/r2) cos (r, x) where cos (r, x) denotes the cosine of the angle between r and the x-axis. Hence the x-component of the attraction between two elementary parts of two solids M and M' is (k * AM ~ AM'/r2) cos(r, x). Show that the total attraction between the two solids is expressible by a six-fold integral. 55. A uniform rod attracts an external particle m. Calculate the components of the attraction parallel and perpendicular to the rod; the resultant attraction and its direction. [HINT. Let AM be an element of the rod; then AF = k AM ~ m/r2 is the force due to AM acting on m, r being the distance from AM to m; then the components of AF are AX = AF cos a and AY = AF sin a, where a is the angle between r and the rod. Hence X kmdM nd Y kmdMsin a. x = -- r2 osa, and Y = r in.] 56. Show that in spherical coordinates, (r, 0, 4), the volume of a z 'solid is given by an integral of the form 0 L B\CO' S f f r2 cos C dC do dr. lr~ 01 [HINT. Let P be a point on AO\' ' \ a sphere of radius r, the longi\^// \ \ \ Itude of P being 0 and its latitude O. It is usual to let the xy-section of the sphere be 0 t 0 \~ \~ ~the equator and the xzsection 6/ x / x the prime meridian. Then P i/ /^^< ^ is the point (r, 0, )). PS and QR are two adjacent meridians Y and PQ, SR are two adjacent FIG. 67. parallels. Then Q =(r, 0 + AO, c), R = (r, 0 + AO, 4 + A4), S = (r, 0, + + A4). Also PS = r AO and PQ = O'P ~ AO = r cos 4 AO. XIV, ~ 136] MULTIPLE INTEGRALS -APPLICATIONS 233 Suppose r to increase to r + Ar so that P, Q, R, S move out to P', Q', R', S' respectively. In this way is formed the element of volume in spherical coordinates. Its approximate volume is PS * PQ. PP' = r AO * r cos 4 AO ' Ar = r2 cos a AOA Ar.] 57. Calculate the volume of a sphere, using spherical coordinates. 68. Calculate the volume cut from a cone of angle 2 a by twd concentric spheres with centers at the vertex of the cone. 59. Show that, in cylindrical coordinates (r, 0, z), the volume of a solid is given by an integral of the z form f fffrdOdrdz. S S Here r denotes distance from the t A RL z-axis. Q [HINT. The coordinates of P: (r, 0, z) are OM =r, xOM =, MP =.] Q = (r, 0 + A, z); R = (r, 0+A 0, z +Az); o _ S = (r, 0, z-+A z). M Increase r to r + Ar, so that we A have an element of volume whose FIG 68. FIG. 68. approximate volume is PQ PS. PP' = r A * Az. Ar. 60. Calculate the volume of a sphere using cylindrical coordinates. 61. Determine the part of the cylinder r = sin 2 0 which lies between the planes z = 0 and z = y. 62. Determine the part of the cylinder r = sin 2 0 which lies between the planes z = 0 and x + y + z = /2. CHAPTER XV EMPIRICAL CURVES -INCREMENTS INTEGRATING DEVICES 137. Empirical Curves. Some of the methods used in science to draw the curves which represent simultaneous values of two related quantities and to obtain an equation which represents that relation approximately are given in Analytic Geometry. Usually the pairs of corresponding values are plotted on squared paper first; in all that follows it is assumed that this has been done in each case. 138. Polynomial Approximations. It is advantageous to have equations which are as simple as possible. From experimental results, it is not to be expected that absolutely precise equations can be found, and the attempt is made to get an equation of simple form which approximately represents the facts, in so far as the facts themselves are known. One simple kind of function which often does approximately express the facts is a polynomial: (1) y = a + bx + cx2 + dx3 + * + kxn. 139. Logarithmic Plotting. The preceding forms of equations may not represent the facts very well unless a large number of terms (1), ~ 138, are used. If the first graph resembles one of the curves y = x2, y = x3, y = x4, etc., or y = x1/2, y = xl/3, etc., or y = 1/x, y = I/x2, etc., it is advantageous to plot the common logarithms of the quantities measured instead of the actual values of those quantities. 234 XV, ~ 139] EMPIRICAL CURVES If x and y represent the quantities measured, and u = loglox, v = logio y are their common logarithms, the values of u and v may lie very nearly on a straight line, (1) v = a +bu, where a and b are found by drawing the straight line which on the whole seems to approximate best to the points (u, v) and measuring its slope, b, and the v-intercept, a. Then from (1), since u = logo x, v = logo y, (2) logo1 y = a + b logo1 x = loglo k + loglo x = loglo (kcx), where log1o k - a; hence (3) y = ckx. This form of equation is very convenient for computation and is used in practice very extensively wherever the logarithmic graph is approximately a straight line.* This work applies equally well for negative and fractional values of b. In many cases where the process just described fails, it is sometimes advantageous to assume that the equation has the form (y - B) = k(x - A) which evidently has a horizontal tangent at the point (A, B) if n > 1, or a vertical tangent if n < 1. If the first graph (in x and y) shows such a vertical or horizontal tangent, that point (A, B) may be * To avoid the trouble of looking up the logarithms, a special paper usually described in Analytic Geometry may be purchased which is ruled with logarithmic intervals. No particular explanation of this paper is necessary except to say that it is so made that if the values of x and y are plotted directly, the graph is identical with that described above. To secure this result the successive rulings are drawn at distances proportional to log 1 (=0), log 2, log 3, * from one corner, both horizontally and vertically. Explanations and numerous figures are to be found in many books; see, e.g., Kent, "Mechanical Engineers' Pocket Book" (Wiley, 1910), p. 85; Trautwine, "Civil Engineers' Pocket Book" (Wiley), (Chapter on Hydraulics). 236 THE CALCULUS [XV, ~ 139 selected as a new origin, and the values x'= x - A and y'= x - B should be used; thus we would plot the values of u = logio x' = logio (x- A), v = logio y' = logio (y-B), in the manner described above. The values of A and B are found from the first graph (in x and y); the values of k and n are found from the logarithmic graph as above. 140. Semi-logarithmic Plotting. Variations of this process of ~ 139 are illustrated in the exercises below. In particular, if the quantities are supposed to follow a compound interest law, y = kebx, it is advantageous to take logarithms of both sides: log1o y = logio k + bx logio e, and then plot u = x, v = logio y; if the facts are approximately represented by any compound interest law, the experimental graph (in u and v) should coincide (approximately) with the straight line v = A +Bu, where A = logio k and B = b log1o e. After A and B have been measured, k and b [= B loge 10 = 2.303 B] can be found. EXERCISES 1. Find the equation of a straight line through the points (- 1, 3) and (2, 5); through (2, - 3) and (4, 5). 2. Plot the data of Exercises 37-42, page 58; draw a straight line as closely as possible through all the points without giving preference to any of them; determine the equation from this graph; compare with former results. Plot each of the following curves logarithmically, -either by plotting logio x and logio y, or else by using logarithmic paper: 3. y=2x3. 5. y=.x42. 7. y = 5.7x6. 4. y = 3 1/2. 6. y = 3-2. 8. y= - 1.4x2'4. XV, ~ 140] EMPIRICAL CURVES 237 In each of the following tables, the quantities are the results of actual experiments; the two variables are supposed theoretically to be connected by an equation of the form y = kxn. Draw a logarithmic graph and determine k and n, approximately: 9. [Steam pressure; v = volume, p = pressure.] [Saxelby.] v 2 4 6 8 10 p 68.7 31.3 19.8 14.3 11.3 10. [Gas engine mixture; notation as above.] [Gibson.] v 3.54 4.13 4.73 5.35 5.94 6.55 7.14 7.73 8.04 p 141.3 115 95 81.4 71.2 63.5 54.6 50.7 45 11. [Head of water h, and time t of discharge of a given amount.] [Gibson.] h 0.043 0.057 0.077 0.095 0.100 t 1260 540 275 170 138 12. [Heat conduction, asbestos; 0 = temperature (F.), C = coefficient of conductivity.] [Kent.] 0 32~ 212~ 392~ 572~ 752~ 1112~ C 1.048 1.346 1.451 1.499 1.548 1.644 13. [Track records: d = distance, t = record time (intercollegiate).] d 100 yd. 220 yd. 440 yd. 880 yd. 1 mi. 2 mi. t 0:094 0:211 0:484 1:56 4:174 9:273 [NOTE. See Kennelly, Fatigue, etc., Proc. Amer. Acad. Sc. XLII, No. 15, Dec. 1906; and Popular Science Monthly, Nov. 1908.] 238 THE CALCULUS [XV, ~ 140 Plot the following curves, using logarithmic values of one quantity and natural values of the other: 14. y = eC. 15. y = 10 e3. 16. y = 4 e-. 17. y=.l e-/3. Discover a formula of the type y = kea for each of the following sets of data: 18. x:.2.4.6.8 1.0 y: 4.5 6.6 9.9 15.0 22.2 19. x:.6 1.2 1.8 2.4 3.0 y: 1.5 2.2 3.3 5.0 7.4 20. x:.31.63.94 1.26 1.57 y: 2.44 2.98 3.64 4.46 5.44 21. x:.2.8 2.0 4.0 y: 8.2 4.5 1.3 0.2 22. x:.63 1.26 2.51 3.77 5.03 y: 4.02 2.70 1.20 0.54 0.24 23. x: 1 2 3 4 5 y: 3.26 2.68 2.16 1.80 1.46 24. A is the amplitude of vibration of a long pendulum, t is the time since it was set swinging. Show that they are connected by a law of the form A = ke-nt. A. in. = 10 4.97 2.47 1.22.61.30.14 tmin.= 0 1 2 3 4 5 6 141. Method of Increments. A method adapted to the case where (1) of ~ 138 has the form (1) y = a + bx + cx2, is as follows. From two pairs of values of x and y, say (x, y) and (x +-Ax, y + Ay) given by experiment, we should have (2) y = a + bx + cx2, y + Ay = a + b (x + x) + c (x + Ax)2, whence (3) Ay = bAx + 2 cxAx+c 2. XV, ~ 141] EMPIRICAL CURVES 239 If Ax is constant, i.e. if points are selected at equal x-intervals on the crudely sketched curve drawn through the experimental points, we might write (4) Y= Ay = (bh+ ch2) +2 ch x =A +Bx where h = Ax. If we should actually plot this equation, Y = A +Bx, we would get (approximately) a straight line. Now Ay = Y is the difference of two values of y; it can be found for each of the values of x selected above, and the (approximate) straight line can be drawn, so that A and B can be measured. We may repeat the preceding process; from (4) we obtain, as above, (5) AY = BAx = 2 ch2, (h = Ax), whence AY is constant if h was taken constant. Now AY is the difference between two values of Y; that is, AY is the difference between two values of Ay: AY = A (Ay) = A2y, and for that reason is called a second difference, or a second increment. If the second differences are reasonably constant, we conclude that an equation of the form (1) will reasonably represent the facts and we find c directly by solving equation (5). EXAMPLE 1. With a certain crane it is found that the forces f measured in pounds which will just overcome a weight w are f 8.5 12.8 17.0 21.4 25.6 29.9 34.2 38.5 w 100 200 300 400 500 600 700 800 What is the law connecting force with the weight that it just overcomes? [PERRY.] Plotting the values of f and w, it appears that the points are very 240 THE CALCULUS [XV, ~ 141 nearly on a straight line f = a + bw. If they were on a straight line, Af/Aw would be constant and equal to df/dw = b. As a matter of fact, for each increase of weight, Af/Aw varies only from.042 to.044, its average value being 30/700 =.0429 Taking this value for b, one gets for the equation of the line, and hence for the relation between force and weight: f = 4.21 +.0429 w, 4.21 = 8.5- 100 X.0429 Here 4.21 appears to be the force needed to start the crane if no load were to be lifted. EXAMPLE 2. If 0 is the melting point (Centigrade) of an alloy of lead and zinc containing x% of lead, it is found that x = % lead 40 50 60 70 80 90 0 = melting point 186 205 226 250 276 304 Plotting the points (x, 0) will show them not to lie in a straight line as is also shown by the difference AO. But A(AO) or A20 does run uniformly. Therefore one tries a quadratic function of x for 0, that is 0 = a + bx + cx2. It is evident that A0 = 10 b + c (20 x + 100), and A2 0 = 200 c. The average value of A2 0 is 2.25 Hence c =.01125 If we subtract cx2 from 0, we find 0 - cx2 = a + bx. These values can be calculated from the data and from c =.01125; they will be found to lie on a straight line; hence a and b can be found by any one of several preceding methods. The student will readily obtain, approximately, 0 = 133 +.875 x +.01125 x2, a formula which represents reasonably the melting point of any zinclead alloy. [SAXELBY.] EXERCISES 1. Express f (x) as a quadratic function of x, when x: 0 0.5 1.0 1.5 2.0 2.5 3.0 f(x): 2.5 1.9 1.6 1.5 1.7 2.1 2.8 2. Express f (x) as a cubic function of x, when x: 0.02.04.06.08.10.12.14 f (x): 0.020.042.064.087.111.136.163 XV, ~ 141] EMPIRICAL CURVES 241 3. Express 0 (m) as a cubic in m, when m:.01.02.03.04.05.06.07.08 X((m):.00010.00041.00093.00166.00260.00385.00530.00690 4. The specific heat S of water, at 0~ C., is 0: 0 5 10 15 20 25 30 S: 1.0066 1.0038 1.0015 1.0000 0.9995 1.0000 1.002 Express S in terms of 0. 6. Determine a relation between the vapor pressure P of mercury, and the temperature 0 C., from the data below: 0: 60 90 120 150 180 210 240 P:.03.16.78 2.93 9.23 25.12 58.8 6. The resistance R, in ohms per 1000 feet, of copper wire of diameter D mils, is D: 289 182 102 57 32 18 10 R:.126.317 1.010 3.234 10.26 32.8 105.1 Find a relation between R and D. 7. The Brown and Sharpe gauge numbers N of wire of diameter D mils, are N: 1 5 10 15 20 25 30 D: 289 182 102 57 32 18 10. Express D in terms of N. 8. Find a relation between the speed S of a train in kilometers per hour, and the horse-power (H. P.) of the engine from the data below: H. P.: 550 650 750 850 S: 26 35 52 70. 9. The energy consumed in overcoming molecular friction when iron is magnetized and demagnetized (hysteresis, H, - measured in watts per cycle per liter of iron) is given below in terms of the strength of the magnetic field (B, - measured in lines per square centimeter). What is the relation between them? B: 2000 4000 6000 8000 10000 14000 16000 18000 H:.022.048.085.138.185.320.400.475 10. Proceed as in Ex. 15, for cobalt, the hysteresis loss H being now measured in ergs per cycle per second: B: 900 2350 3100 4100 4600 5200 5850 6500 H: 450 2450 3950 6300 7400 8950 10950 13250. 242 THE CALCULUS [XV, ~ 141 The table below contains some data on the comparison of a tungsten lamp with a tantalum lamp. The voltage or electrical pressure V, is in volts, the resistance R, in ohms, the current consumed in watts per candle power; C denotes candle power, and W watts per candle power. 11. TUNGSTEN 12. TANTALUM Voltage C. P. Watts Resistance C. P. Watts Resistance per C. P. per C. P. V C W R C W R 80 14 2.51 166 5 3.80 260 90 24 1.83 173 10 2.85 265 100 36 1.49 182 18 2.05 275 110 52 1.23 190 25 1.65 283 120 71 1.10 197 38 1.35 290 130 95 0.96 202 50 1.15 300 140 128 0.83 210 62 0.95 308 150 160 0.76 216 78 0.85 315 160 196 0.58 222 100 0.75 323 170 230 0.52 227 122 0.70 327 180 270 0.50 232 156 0.70 332 190 312 0.48 238 190 0.60 340 200 340 0.47 242 235 0.55 345 For each lamp, express each of the quantities C, W, R, in terms of V. 142. Integrating Devices. It is important in many practical cases to know approximately the areas of given closed curves. Thus the volume of a ship is found by finding the areas of cross sections at small intervals. Besides the methods described above, the following devices are employed: A. Counting squares on cross-section paper. B. Weighing the figures cut from a heavy cardboard of uniform known weight per square inch. C. Integraphs. These are machines which draw the integral curve mechanically; from it values of the area may be read off as heights. XV, ~ 142] EMPIRICAL CURVES - 243 The simplest such machine is that invented by ABDANK-ABAKANOwicz. A heavy carriage CDEF on large rough rollers, R, R' is placed on the paper so that CE is parallel to the y-axis. Two sliders S and S' C 1lll D yi=x) move on the parallel sides DF and CE; to S B7 is attached a pointer P =(x P which follows the curve Y y = f(x). A grooved rod AB slides over a pivot _ at A, which lies on the X x-axis, and is fastened by pivot B to the slider S. - - A parallelogram mechanism forces a sharp wheel W attached to the slider S' to remain parallel to E X' ill F AB. A marker Q draws FIG. 69. a new curve i = (x), which obviously has a tangent parallel to W, that is, to AB. If AB makes an angle a with Ox, tan a is the slope of the new curve; but tan a is the height of S divided by the fixed horizontal distance h between A and B: d _ (x) tan height of S f (x) x ~tan a = -h-, ~ dx h h whence i-o = J f (x) dx; i x=a where a is the value of x at P when the machine starts, and io denotes the vertical height of the new curve at the corresponding point. D. Polar Planimeters. - There are machines which read off the area directly (for any smooth closed curve of simple shape) on a dial attached to a rolling wheel. The simplest such machine is that invented by AMSLER. Let us first suppose that a moving rod ab of length I always remains perpendicular to the path described by its center C. The path of C may be regarded as the limit of an inscribed polygon, and the area swept over by the.rod may be thought of as the limit of the sum of small 244 THE CALCULUS [XV, ~ 142 quadrileterals, the area AA of each of which is lAp, approximately, where Ap is the length of ' D o ---->^~ "^'^^ ^the corresponding side of l\ A / / Ad the polygon inscribed in <lw, /L the path of C. Hence the -\ / Ad ^^S8total area A swept over Ad\ I -r by the rod is evidently ip, where p is the total length FIG. 70. of the path of C. But if the rod does not remain perpendicular to the path of C during the motion, and if V is the angle between the rod and that path, the area AA becomes 1 sin A ~ Ap, approximately. The expression sin.'* Ap may be thought of as the component of Ap in a direction perpendicular to the rod. Calling this component As, we have AA = 1As, ap- o\ bo proximately; and the total area A swept over by the rod is precisely lim S AA = lim; l As =fl ds = I fds = Is, where s = fds is the total motion of C in a direction perpendicular to the rod. The quantity s = Cds can be measured mechanically by means of a wheel of which the rod is the axle, attached to the rod at C; for if 0 is the total angle through which the wheel / / turns during the motion, s = rO, where r is the radius of the wheel, and 0 is measured in T radians. Hence A = Is = lrO; the value of 0 al is read off from a dial attached to the wheel; FIG 71. I and r are known lengths. In Amsler's polar planimeter, one end b of the rod ab is forced to trace once around a given closed curve whose area is desired; the other end a is mechanically forced to move back and forth along a circular arc by being hinged at a to another rod Oa, which in its turn is hinged to a heavy metal block at 0. As b describes that part of the given curve which lies farthest from O, the rod ab sweeps over an area between the circular arc traced by a and the outer part of the given curve; as b describes the part of the curve nearest to 0, ab sweeps back over a portion of the area covered before, between the circle and the inner part of the given curve. This latter area does not count in the final total, since it has been swept over twice in opposite directions. Hence XV, ~ 142] EMPIRICAL CURVES 245 the quantity A = lrO, given by the reading of the dial on the machine, is precisely the desired area of the given closed curve, which has been swept over just once by b% the moving rod ab. b In practicing with such a machine, begin with curves of known area. The machine is b useful not only in finding areas of irregular curves whose equations are not known, but 02 b also in checking integrations performed by the standard methods, and in giving at least / / approximate values for integrals whose evaluation is difficult or impossible. For further information on integrating 0 0 devices, see: Abdank-Abakanowicz, Les in- FIG. 72 tegraphes (Paris, Gauthier-Villars); Henrici, Report on Planimeters (British Assoc. 1894, pp. 496-523); Shaw, Mechanical Integrators (Proc. Inst. Civ. Engs. 1885, pp. 75-143); Encyklopadie der Math. Wiss., Vol. II. Catalogues of dealers in instruments also contain much really valuable information. EXERCISES 1. Construct a figure of each of the types mentioned below, with dimensions selected at random, and find their areas approximately by counting squares; by Simpson's rule; by the planimeter, if one is available. (1) A right triangle; (2) An equilateral triangle; (3) A circle; (4) An ellipse. (Draw it with a thread and two pins.) (5) An arch of a sine curve; (6) An arch of a cycloid. FIG. 73 (a). 2. The figures below are reproductions of indicator cards, taken from three different types of engines. The dotted curves are entirely separate from the full lines. The average pressure on the piston is the area of one of these curves divided by the length of stroke. Find this value in each case, where the stroke is 12 in. in the first figure, and 8 in. in each of the others. (Unit of area =- 1 large square.) 246 THE CALCULUS [XV, ~ 142 [NOTE. The work done is precisely the area in question, on a proper scale, since the work is the average pressure times the length of stroke.] FIG. 73 (b). FIG. 73 (c). CHAPTER XVI LAW OF THE MEAN - TAYLOR'S FORMULA - SERIES 143. Rolle's Theorem. Let us consider a curve y =f(x), where f (x) is single-valued and continuous, and where the curve has at every point a tangent that is not vertical. If such a curve cuts the x-axis twice, at x = a and x = b, it surely Ula c Xb either has a maximum or IG74. a minimum at at least one point x = c between a and b. It was shown in ~ 33, p. 54, that the derivative at c is zero: [A] If f(a)=f(b)=0, then [d-f ()] =0, (a<c<b); k dx jx=c this fact is known as Rolle's Theorem. 144. The Law of the Mean. Rolle's Theorem is quite evident geometrically in the form: An arc of a simple smooth curve cut off by the x-axis has at least one horizontal tangent. T S The precise nature of the y C~~~? y =f(x) necessary restrictions is I /'/ {f(b)-f(a) given in ~ 129. -/ -----— _ Another similar stateS /\fI^) | [ib ment, which is true under o /.a b the same restrictions and FIG. 75. is equally obvious geometrically, is: An arc of a simple smooth curve cut off by any secant has at least one tangent parallel to that secant. 247 248 THE CALCULUS [XVI, ~ 144 If the curve is y = f (x), and if the secant S cuts it at points P:[a, f (a)] and Q:[b, f (b)], the slope of S is Ay - Ax =[f(b) -f (a) - (b -a). The slope of the tangent CT at x = c is equal to this: [B] d _ df (x) f(b) -f(a) _ (a b). dxx=c dx ]x=c b -a Ax (a b). This statement is called the law of the mean or the theorem of finite differences. It is easy to prove this statement algebraically from Rolle's Theorem. For if we subtract the height of the secant S from the height of the curve, we get a new curve whose height is: D(x) =f(x) - f (b)-f(a) (- ) +f ()]. Now D(x) is zero when x = a and when x = b. It follows by ~143 that d D (x)/dx = 0 at x = c, (a < c < b): dD (x) (df(x) f (b)-f(a) =0, (a<c < b) dx =c dx - a x=c which is nothing but a restatement of [B]. 145. Increments. The law of the mean is used to determine increments approximately, and to evaluate small errors. If y = f (x) is a given function, we have, by ~ 144, ~~dy[B] x dx]=c In practice this law is used to estimate the extreme limit of errors, that is, the extreme limit of the numerical value of Ay. It is evident that [B*] IAy |I M1 |Ax, where M, is the maximum of the numerical value of dy/dx between a and a + Ax. When Ax is very small, the slope dy/dx XVI, ~ 145] LAW OF THE MEAN 249 is practically constant from a to a + Ax in most instances, and M1 is practically the same as the value of dy/dx at any point between a and a + Ax. EXAMPLE 1. To find the correct increments in a five-place table of logarithms. The usual logarithm table contains values of L = logo1 N at intervals of size AN =.001. Hence AL [d logio N (01). [.001 log ] 00043 dN N=e(00 = -N- 10 N =c' where N < c < N +.001. Logarithms are ordinarily given from N = 1 to N = 10. Hence AL will vary from.00043 at the beginning of the table to.00004 at the end of the table. This agrees with the " differences" column in an ordinary logarithm table. EXAMPLE 2. The reading of a certain galvanometer is proportional to the tangent of the angle through which the magnetic needle swings. Find the effect of an error in reading the angle on the computed value of the electric current measured. We have C = k tan 0, where C is the current and 0 the angle reading. Hence the error EC in the computed current is EC = AC= kd tan 0] a k M kA sec2a, ( < a < ~ A), dO O=a > where Ec is the error in the computed value of the current, and A0 is the error made in reading the angle 0. Since A0 is very small, Ec = k sec2 0 A, approximately. The error Ec is extremely large if 0 is near 90~, even if AC is small; hence this form of galvanometer is not used in accurate work. EXERCISES 1. At what point on the parabola y = x2 is the tangent parallel to the secant drawn through the points where x = 0 and x = 1? 2. Proceed as in Ex. 1 for the curve y = sin x, and the points where x = 60~ and x = 75~. 3. Proceed as in Ex. 1 for the curve y = log (1 + x), for x = 1 and x = 3. 250 THE CALCULUS [XVI, ~ 145 4. Discuss the differences in a four-place table of natural sines, the argument interval being 10'. 6. Proceed as in Ex. 4 for a similar table of natural cosines; of natural tangents. 6. Discuss the differences in a four-place table of logarithmic sines, the entries being given for intervals of 10'. 7. Proceed as in Ex. 6 for a table of logarithmic tangents. 8. Calculate the difference in a seven-place table of logio sin x at the place where x = 30~; where x = 60~; where x = 85~. 9. Discuss the effect of a small change in x on the function y = log (1 + 1/x). 10. If logio N = 1.2070 ~.0002, what is the uncertainty in N? [The term +.0002 indicates the uncertainty in the value 1.2070.] 11. If the angle of elevation of a mountain peak, as measured from a point in the plain 5 mi. distant from it, is 5~ 20' + 5', what is the uncertainty in the computed height of the peak? 12. The horizontal range of a gun is R = (V2/g) sin 2 a, where V is the muzzle speed and a the angle of elevation of the gun. If V = 1200 ft./sec., discuss the effect upon R of an error of 5' in the angle of elevation. 13. The distance to the sea horizon from a point h ft. above sea level is D = V2 Rh + h2, where R is the radius of the earth. Discuss the change in D due to a change of one foot in h. (D, R, and h are all to be taken in the same units.) If D is tabulated for values of h at intervals of one foot, what is the tabular difference at the place where h = 60? 14. If the boiling point of water at height H ft. above sea level is T, H = 517 (212 ~- T) - (212~ - T)2, T being the boiling temperature in degrees F. Discuss the uncertainty in H, if T can be measured to 1~. If H be tabulated with argument T at intervals of 1~, what is the tabular entry and the tabular difference when T = 200~? 15. When a pendulum of length 1 (feet) swings through a small angle a (radians), the time (seconds) of one swing is T = 7rV/l/g (1 + a2/16). What is the effect on T of a change in a, say from 5~ to 6~? Of a change in 1 from 36 in. to 37 in.? Of a change in g from 32.16 to 32.2? XVI, ~ 146] LAW OF THE MEAN 251 16. The viscosity of water at 0~ C. is P = 1/(1 +.0337 0 +.00022 02). Discuss the change in P due to a small change in 0. What is the average value of P from 0 = 200 to 0 = 30~? 17. The quantity of heat (measured in calories) required to raise one kgm. of water from 0~ C. to 0~ C. is H = 94.21 (365 - 0)0'3125 + k. How much heat is required to raise the temperature of one kgm. of water 1~ C. when 0 = 10~? 20~? 30~? 70~? To find k, observe that H = 0 when 0 = 0. 18. The coefficient of friction of water flowing through a pipe of diameter D (inches) with a speed V (ft./sec.) is f =.0126 + (.0315 -.06 D)/VV. What is the effect on f of a small change in V? in D? 146. Limit of Error. In using the formula [B] the uncertainty in the value of c is troublesome. If the value of dy/dx at x = a is used in place of its value at x = c, the error made in finding Ay by [B] can be expressed in terms of the second derivative d2y/dx2. We shall use the convenient notation f' (x), f" (x), etc., for the derivatives of f (x): df(x) dy f'(x) -( ) = (the slope of y = f(x)). dx dx d2f(X) <dy df (X) f "(x) = x dx2 - df (x) (the flexion). dx2 - dx2 - dx Let M2 denote the maximum of the numerical value off" (x) between two points YM2 x = a and x = b, so \2i that (1) If"I(x) I. ~=f31 The area under the curve y =f"(x) be-a X- x=-a X -X x -b tween x = a and any FIG. 76 (a). point x = x between a and b is evidently not greater than the area under the horizontal line y = M2; that is, if a < x < b, (2) J f"(x)dx M2dx, or f'(x)] <M2x]1 x=a xxa xa =a 252 THE CALCULUS [XVI, ~ 146 since df'(x)/dx = f"(x), and M2 is a constant; whence, substituting the limits of integration in the usual manner, (3) 1f'(x)-f' (a) I < M (x - a), which is geometrically shown in Fig. 76 (b). It follows that the area under the curve I y =f'(x)-f'(a) is not, f/ greater than that under <i. a / the line y = M2(x - a): < f= M,(x - a) dx' 61 xx xx=a x ax=b = =x - d FIG. 76 (b). or since f' (a) and M2 are constants and df(x)/dx = f' (x), [f(x)-f'(a) -x] X < M2 a)2] x=a = 2 J;= whence, substituting the limits in the usual manner, [C] f(x) -f(a)-f'(a)(x- a) M2 (x- a)2 [C]-f 2 which holds for all values of x between x = a and x = b. This formula may be written even if x<a: [C*] f(x) =f(a) +f'(a)(x - a)+E2,where E2 <M2 (x - a) and E2 is the error made in using f'(a) in place of f' (c) in formula [C]; for (x - a)2 = x - a 2. It should be noticed that E2 is exactly the error made in substituting the tangent at x = a for the curve, i.e. it is the difference between Ay[ = f(x) - f(a)] and dy[ = f' (a) (x - a)] mentioned in ~ 26, p. 43, and shown in Fig. 8. The formula [B *] is exactly analogous to [C *]; since Ay = f(x) -f (a) if Ax = x - a, [B *] may be written [B *] f(x) =f(a) + E1, EI E < M1* Ix - al. xvIj ~ 1471 TAYLOR'S FORMULA 253 EXAMPLE 1. In Ex. 1, ~ 145, we found for L = logio N,.00043 AL = (nearly). Applying [0*], with f (N) = logio N, a = N, x = N + AN, x - a = AN.001, we find.00043 000001 AL = f (N + AN) - f (N) = + E2, I E21 < 2' 2.M2 N 2 where M2 is the maximum value of f" (N) I = (logio e)/N2 between N = 1 and N = 10. Hence E2 <.00000022 The value of AL found before was therefore quite accurate, - absolutely accurate as far as a five-place table is concerned. EXAMPLE 2. Apply [0*1 to the function f (x) = sin x, with a = 0, and show how nearly correct the values are for x <ir/90 = 20. Since f(x) = sin x, and a = 0, [0*] becomes sin x = sin (0) + cos (0).- (x - 0) + E2 =x + E2, I E21 < M2. where M2 is the maximum of f" (x) I = - sin x I between 0 and w/90. that is M2 = sin (7ir90) = sin 20 =.0349 Hence E2 <.0175 x2. Since X < 7r/90, x2 < 7r2/8100 <.0013; hence E2 <.000023, and sin x x is correct up to x $ w7/90 within.000023 Similarly, for a = 7/4, we have, by [0*], 1 >1~ ~ <M(x -7rr/4)2 sin x = $[~x-) +E2, rE V24 2 where t2 < 1. If (x - 7r/4) < 7r/90, 1 E1 < (7w/90)2 2 =.0007 147. Extended Law of the Mean. Taylor's Theorem. The formula [0*] can be extend'd very readily. Let f' (x), f" (x), f"' (x),. f " (x) denote the first n successive derivatives of f (x): f IW dt f (x) df 't-'(x) dx - dx and let the maximum of the numerical value of f"') (x) from x = a to x = b be denoted by M,. Then Ifln) (x I <~ M., and ()x~ f: d P) W dx = Mn dzrXs i =*a ~ =a 254 THE CALCULUS [XVI, ~ 147 or If(n"- (x) - fn-1) (a) ] | Mn (x - a) I for all values of x between a and b. Integrating again, we obtain, as in ~ 146: If(-2" (x) _ f(2) (a) _f(n-l)(- a) I (x - a) 2 and, continuing this process by integrations until we reach f(x), we find: [D] f (x) -f(a) -f (a) (x- a) - 2! (x-)2.. f(n-) (a) (xa)n-M n x- a n (n - 1)! a =Mn n or, [D*] f (x) =f (a) +f' (a) (x- a) +f 2( (x - a)2 +... f(n-l) (a) + (a) x- a)n- + En, where EI l < Mn I x- and where Mn is the maximum of I f(n) (x) I between x = a and x = b. This formula is known as the extended law of the mean, or Taylor's Theorem, after Taylor, who first gave such approximations as it expresses. It is one of the more important formulas of the Calculus. In particular, if a = 0, the formula becomes f"(o) [D*] f(x)=f(0)+f'(0 )x+f!2+... f(n-l) (0) + (n )xn-1 - + En, where I IEn M I x" l/n! This special case of Taylor's Theorem is often called Maclaurin's Theorem. XVI, ~ 147] TAYLOR'S FORMULA 255 The formula [D*] replaces f (x) by a polynomial of the nth degree, with an error En. These polynomials are represented graphically by curves, which are usually close to the curve which represents f (x) near x = a. See Tables, III, K. Since the expression for En above contains n! in the denominator, and since n! grows astoundingly large as n grows larger, there is every prospect that En will become smaller for larger n; hence, usually, the polynomial curves come closer and closer to f (x) as n increases, and the approximations are reasonably good farther and farther away from x = a. But it is never safe to trust to chance in this matter, and it is usually possible to see what does happen to En as n grows, without excessive work. EXAMPLE 1. Find an approximating polynomial of the third degree to replace sin x near x = 0, and determine the error in using it up to x = 7r/18 = 10~. Since f (x) = sin x and a = 0, we have f' (x) = cos x, f" (x) = - sin x, f"' (x) = - cos x, fiv (x) = + sin x, whencef (0) = 0, f' (0) = 1, f" (0) = 0,f"' (0) = - 1; and [Max. If i (x) ] = [Max. sin xl] = sin 10~ =.1736 between x = 0 and x = ir/18 = 10~. Hence (X - 0)3 +, sin x = 0 + 1 * (x- 0) + 0 + (- 1) * ( + E4 = x - + E4 where I E4 1 <(.1736) * x4/4! ~ (.1736) (7r/18)4 + 4! <.000007, when x lies between 0 and ir/18. In general, the approximation grows better as n grows larger, for If(n) (x) I is always either I sin x I or I cos xl; hence Mn ~ 1, and I En I < xn/n! which diminishes very rapidly as n increases, especially if x < 1 = 57~.3 For n = 7, the formula gives, for x > 0, sin x = x - - +E7, I E7<x7/7!. EXAMPLE 2. Express V/ + x as a quadratic in x and estimate the error if x lies between 0 and.2 Here f (0) = 1;f'(0) = 1/2;f" (0) = - 1/4; M3 If"' (0) = 3/8. Hence 1 +x = 1 + x/2- x2/8 + E3; I E < [(3/8)/(3!)] (.2)3 =.0005 Thus: 1.2 = 1 +.1 -.005 + E3 = 1.0950 + E3; I E3 1 <.0005 256 THE CALCULUS [XVI, ~ 147 EXERCISES 1. Apply the formula (C*) to obtain an approximating polynomial of the first degree for tan x, with a = 0. Show that the error, when x l < 7r/90, is less than.00003. Draw a figure to show the comparison between tan x and the approximating linear function. 2. Apply [D*] to obtain an approximating quadratic for cos x, with a = 0. Show that the error, when Ix <7r/10 is less than (7r/10)3.3! Draw a figure. 3. Apply [D*]' to obtain an approximating cubic for cos x, near x = 0. Hence show that the formula found in Ex. 2 is really correct, when l xl < r/10, to within (,t/10)4 4 4! Draw a figure. 4. Obtain an approximation of the third degree for sin x near x = 7r/3. Show that it is correct to within (7r/10)4 + 4! for angles which differ from 7r/3 by less than 7r/10. Draw a figure. Obtain an approximation of the first degree, one of the second degree, one of the third degree, for each of the following functions near the value of x mentioned; find an upper limit of the error in each case for values of x which differ from the value of a by the amount specified; draw a figure showing the three approximations in each case: 5. e, a = 0, x - a <.1 9. e-, a =2, l x - a l <.5 6. tanx,a = 0,Ix - a[ < r/90. 10. sinx, a =r/2, I x-a I < r/45. 7. log(l+x), a=0, Ix-a l<.2 11. tan x, a=7r/4, [x-a I < r/90. 8. cos x, a = r/4, I x-a | < 7/18. 12. x2+x+1,a=1,, x-a I < 1/5. 13. 2x2-x- 1, a = 1/2, x- a l < 1. 14. x3-2x2-x +, a = -2, Ix- a <.5 15. Find a polynomial which represents sin x to seven decimal places (inclusive), forlxl <10~. 16. Proceed as in Ex. 15, for cos x; for e-, when 0<x<1. 17. Show that x differs from sin x by less than.0001 for values of x less than a certain amount; and estimate this amount as well as possible. 18. The quantity of current C (in watts) consumed per candle power by a certain electric lamp in terms of voltage v is C = 2.7 + 108'007-0767v. Express C by a polynomial in v - 115 correct from v = 110 up to v = 120 to within.025 watt. XVI, ~ 148] TAYLOR'S FORMULA 257 148. Application of Taylor's Theorem to Extremes. If a function y = f (x) is given whose maxima and minima are to be found, we may find the critical points where f' (x) = 0. Let a be one solution of f'(x) = 0, that is, a critical value. Then, since f' (a) = 0, we have, by [D*], f" (a) (a - a)3 Ay =f(x) -f(a) = 0+ 2( (x-a)2+E, I E3 M3( a) =! 3~.3! where M3 If "' (x) 1. Hence the sign of Ay is determined by the sign off" (a) when (x - a) is sufficiently small. If f" (a) >0, Ay>0, and f(x) is a minimum at x = a. If f" (a) <0, Ay <0, and f (x) is a maximum at x = a. (See ~ 42, p. 66.) If f"(a) = 0, the question is not decided.* But in that case, by [D*]: Ay =f (x) f (a) = O + (a)(x - a)3+f ) (x - a)4 + E, 3! 4! wherel E < M5 x-a5/5!, M5 1 fv (x)l. From this we see that iff"' (a) 7 0 there is neither a maximum nor a minimum, for (x - a)3 changes sign near x = a. But if f"' (a) = 0, then fiv (a) determines the sign of Ay, as in the case of f" (a) above. In general, if f(k) (a) is the first one of the successive derivatives, f' (a),f"(a),, which is not zero at x = a, then there is: no extreme if k is odd; a maximum if k is even and f(k) (a) <0; a minimum if k is even and f(k) (a) >0. * The methods which follow are logically sound and can always be carried out when the derivatives can be found. But if several derivatives vanish (or, what is worse, fail to exist), the method of ~ 34, p. 54, is better in practice. 258 THE CALCULUS [XVI, ~ 148 EXAMPLE 1. Find the extremes for y = x4. Since f (x) = X4, fl (x) = 4 x3; hence the critical values are solutions of the equation 4 3 =0, and therefore x = 0 is the only such critical value. Since f" (x) = 12 x2, f"' (x) = 24 x, f iv (x) = 24, the first derivative which does not vanish at x = 0 is fiv (x), and it is positive (= 24). It follows that f (x) is a minimum when x = 0; this is borne out by the familiar graph of the given curve. EXERCISES Study the extremes in the following functions: 1. x6. 5. (x +3)5. 9. x2sinx. 2. (x-2)3. 6. X4 (2 x-1)3. 10. X4 cosX. 3. 4 x3 - 3 x4. 7. sin 3. 11. x3 tan x. 4. x3 (1 +x)3. 8. x —sin x. 12. e-1/2. 13. Discuss the extremes of the curves y = xn, for all positive integral values of n. 14. An open tank is to be constructed with square base and vertical sides so as to contain 10 cu. ft. of water. Find the dimensions so that the least possible quantity of material will be needed. 15. Show that the greatest rectangle that can be inscribed in a given circle is a square. [See Ex. 44, p. 59. Other exercises from ~ 35 may be resolved by the process of ~ 148.] 16. What is the maximum contents of a cone that can be folded from a filter paper of 8 in. diameter? 17. A gutter whose cross section is an arc of a circle is to be made by bending into shape a strip of copper. If the width of the strip is a, show that the radius of the cross section when the carrying capacity is a maximum is a/ir. [OSGOOD.] 18. A battery of internal resistance r and E. M. F. e sends a current through an external resistance R. The power given to the external circuit is Re2 (R + 7-)2 If e = 3.3 and r = 1.5, with what value of R will the greatest power be given to the external circuit? [SAXELBY.] XVI, ~ 149] TAYLOR'S FORMULA 259 149. Indeterminate Forms. The quotient of two functions is not defined at a point where the divisor is zero. Such quotients f (x).- ) (x) at x = a, where f (a) = 0 (a) = 0, are called indeterminate forms.* We may note that the graph of (1) = (x)' (f (a) = (a) = 0), - (x)' may be quite regular near x = a; hence it is natural to make the definition: ql) -f (x)~ = lim f (x) (2) ql x"a limp( x=a (X) xa x=-a q (X) If we apply [D *], we obtain, f (x) 0 +f'(a) (x - a) +E2' q ~(x) 0+' (a)(x -a)+E2"' where I E21 < M2' (x - a)2/2!, I E2"I < M2" (x - a)2/2!, and M2/ > if" (x), M2" > | " (x) |, near x = a. Hence f' (a) q- p' M ' (x - a) f (x) + M 2 q (x) (x -a)' ) () + I, M2 2 2 where p' and p" are numbers between -1 and + 1. It follows that f (x) f' (a) (3) lim q = lim - (a) zx ---a x-a-) a (x) (a) unless O' (a) = 0. But if O' (a) = 0, q becomes infinite, and the graph of (1) has a vertical asymptote at x = a unless * If, <(a) =0 but f(a) 0 the quotient q evidently becomes infinite; in that case the graph of (1) shows a vertical asymptote. 260 THE CAL~CULUS [xvII ~ 149 f' (a) = 0 also. If both f' (a) and O' (a) are zero, it follows in precisely the same manner as above, that q f) f () (a) + p'M4 (x a) __ = ( + 1)! O (x 4)() (a) + p"fMf~ '(x -a) 0 't +1 + f (k + 1)! where either f k) (a) or 04k) (a) is not zero, but all preceding derivatives of both f (x) and O (x) are zero at x = a; and where Mf+1 ~ / 1 > W~1) (X) I near x= a and where p' and p" are numbers between - 1 and + 1. It follows that f (x) f(k) (a) lim q =lim ~ x —a x —+ao) (x) P) (a) provided all previous derivatives of both f(x) and qp (x) are zero at x = a, and provided 04")' (a) 0. If 04') (a) = 0, f(k) (a) # 0, then q becomes infinite and the graph of (1) has a vertical asymptote at x = a. It should be noted that (3) is oniy a repetition of Rule [VIII, p. 31. For if u = f (x) andy = (x), since f (a). = O (a) = 0, f (x f(x) -f.(a) Au _ Au Av q (x) (x)- o(a) Av Ax Ax' where Ax = x - a; and therefore Au av Fdu dvl Ff' (x) f'(a) lim q= lim im - K d L' ' X —a AX —OAX Ax-*OAX dx (x)Ja 4'(a) provided 0' (a) is not zero (see Theorem D, p. 15). EXAMPLE 1. To find lim [(tan x) +. x]. x-*O Here f (x) = tan x, k(x) =x; f (0) = 4d0) = 0; hence tan x f' (0) [sec2 xIs=o lim = 1. z-+O x 4'(0) 1 Draw the graph q = (tan x) + x and notice that this value q 1 fits exactly where x = 0. This limit can be found directly as follows: tan h tan (O + h) - tan (0) d tanl = sec2 xl = 1. lim h- = lim +)x. X h,-O h h —O (O + hi) - (0) dx = = xvil ~ 1501 TAYLOR'S FORMULA 261 EXAMPLE 2. To find lim (1 - cos x)/x2. Heref (x) = 1 - cos x, q (X) = X2; f (0) P (0) = 0; f (0) = sin (0) = 0 and 4/ (0) = 0; f" (x) = cos x, 4/ (x) = 2; hence 1 - COS X Cos 1 lim 2 X —O X2 2 j=o 2 Draw the graph of q = (1 - cos X)/X2, and note that (x = 0, q = 1/2) fits it well. 150. Infinitesimals of Higher Order. When the quotient f (X) (1) q = approaches a finite number not zero when x is infinitesimal: f (x) (2) lim q = lim - k! X->0 X —0 Zn then f (x) is said to be an infinitesimal of order n with respect to x. An infinitesimal whose order is greater than 1 is called an infinitesimal of higher order. The equation (2) may be reduced to the form (3) lim [f (x) - kXn] = 0, or (4) f (x) = (k + E) xn, where lim E = 0. The quantity kx" is called the principal part of the infinitesimalf (x). The differencef (x) - kx' = Ex` is evidently an infinitesimal whose order is greater than n, for lim (Ex" + Xn) = lim E = 0. Thus by Example 2, ~ 149, 1 - cos x is an infinitesimal of the second order with respect to x; its principal part is X2/2. Note that 1 - COS X = X/2~2pX/33!, by [D*l, where - 1 p $ + 1; the principal part is the first term of Taylor's Theorem that does not vanish. In general, if we have f (0) =f' (0) =f" (0). =fk-1 (0) = 0, but f(k) (0) 5A 0, the formula [D*] gives, for a = 0, f (X) =f(Ik) (0). Xklk! + p Mk+l xk+l/(k + 1)! where Mk+~I1 I f (k+l) (x) I near x = 0, and - 1 ~ p < + 1. Hence f (x) is an infinitesimal of order k, and its principal part is f(k) (0) xk/k!. 262 THE CALCULUS EXERCISES [xvII ~ 150 Evaluate the indeterminate forms below, in which the notation 0 (x) a. means to determine the limit of 0 (x) when x = a. The vertical bar applies to all that precedes it. Draw the graphs as in Exs. 1, 2, above. 1. sin X/X lo. 2 - 1 1i 7. tan 2 x tan 3 x o 10. log (1 - x) sinx o 13. x -sin x x -tan o 0 16. x cos x - sin x X3 0 alogZ - x 19. alog X 1o zI 2. sin 2 x/sin 3 x lo. 5. - e-x 8. 1 - cos 2 x X2 o 14 + Cos irX 1.1-x-logx1 1.sin-1 x tan-1 o 20 log (X3 - 7) 2.X2-5x+62 sin-' -/a2 - X2~2- 2 I 22. v\'a2 -X2a 3. tan3x/xjlo. aX -1 ax — AX -t ei zI 6. x 0 12. log X2 15. log (I ~x) v\/ 1- ~ 18. - Z x - sin x o 2.sin-1 (x - 2) 21 X~2 + -62 Determine the order of each of the quantities below when the variable x is the standard infinitesimal: 23. x -sin x. 30. ~ x cos x - sin x. 24. ex - e-. 25. X2 sin X2. 26. log(1 + x) - x. 27. ex - esin x 31. sin 2 x - 2 sin x. 32. log cos X. 33. log (1 + e-1/x). 34. tan-1 x - sin-1 x. 28. ax - 1. 35. log cos x - sin 29. log [(a + )/l(a - x)]. 36. 2x-ez+e-h 37. cos1 (1- x) - V/2x - X2. 2 X. 151. Other Indeterminate Forms. The numerator and denominator can be replaced by their derivatives not only when the fraction takes tbe form 0/0, but also when it takes XVI, ~ 151] TAYLOR'S FORMULA 263 the form oo /oo (see Pierpont, Functions of a Real Variable, p. 305). Since f (x)/+ (x) = [1/ (x)] - [1/f (x)], any fraction that takes one of the two forms 0/0, oo - bo, can also be put into the other form. Thus, as x -* 7r/2, tan x and sec x both become infinite, while ctn x and cos x approach zero; hence tan x cos x lim --- = lim = 1. x-s-r/2 sec x x —r/2 ctn x Likewise, if f (x) - 0 as 4(x) becomes infinite, their product is of the form 0 X oo, and it can be put into either of the preceding forms. Thus, as x -* 0, log x becomes - oo; so that log a 1/ T lim (xlog x) ==lim gx = lim 1 = lim (- x) = 0. z-*0 x10 I/x- -- O 1/Xx2 x2-o Other indeterminate forms are oo - oo, 1X, 00, oo. All these can be made to depend on the forms already considered. For let a, A, y, 6, e, be variables simultaneously approaching, respectively, oo, oo, 1, 0, 0. Then a — A/, ya, 56, ae take, respectively, the preceding four indeterminate forms. But lim (a — ) -lim 1/ which is of the form 0/0; while the logarithms of the others, log 7y = a log y, log SE = log S, log a' = elog a, are each of the form 0 X oo. EXAMPLE 1. Thus, when x - 7r/2, (sin x)tan takes the form 10. But (sin X)tn x = e[log sin x]/ctn, which approaches the same limit as e-ctnx/csc x, as x -- 7r/2, and this limit is evidently e0 = 1. EXAMPLE 2. Similarly, when x becomes infinite, (1/x)1/(2x+l) takes the form 0~. It may be written in the form, e- log x]/(2x+1) which approaches the same limit as e-1/2x, that is, the limit is e0 = 1, as X - 0oo. EXAMPLE 3. As an example of the last form, 00o, take (l/x)$ as x -> 0. This becomes e- log x and approaches e0 = 1, as x - 0. Indeterminate forms in two variables cannot be evaluated, unless one knows a law connecting the variables as they approach their limits, which practically reduces the problem to a problem in one letter. 264 THE CALCULUS [XVI, ~ 151 EXERCISES Evaluate each of the following indeterminate forms, where (x) Ia means the limit of f (x) as x approaches a. Draw a graph in each case. 1.-* ex[x' 7. - ~ 1 x 0. 7 x ' log ctn x 8. log x log cos x x/2 2 sec log log (r/2-x) 7,/2 ' V aoo 4. x. 10. x2 ctn x lo. elog cos00 5. log 11. x.log sin2 x o 13. xsin 2. 14. (1 + x)1/x lo. 15. (1 + n/x)x Io. 16. (tan x)C~08x [1/2. 17. (sin ax)sin bx I. log sin2X- 1 6. logtan3x o 12. (tanx-secx) |/2. 18. tan x —/2-. Find the value of each of the following improper integrals, using Table V, F, when necessary after integrating by parts: 19. f x e- dx. 20. x2 e-x dx. 21. f x12 e- dx. Jo -o o 152. Infinite Series. An infinite series is an indicated sum of an unending sequence of terms: (1) ao + al + a2 + * * * + an + *; this has no meaning whatever until we make a definition, for it is impossible to add all these terms. Let us take the sum of the first n terms: Sn = ao + a + a2 + + * a-l, which is perfectly finite; if the limit of s, exists as n becomes infinite, that limit is called the sum of the series (1): (2) S = lim Sn =ao + al + -.. + an +. n->oo If lim s, = S exists, the series is called convergent; if S n —oo does not exist, the series is called divergent; if the series xvI, ~ 152] SERIES 265 formed by taking the numerical (or absolute) values of the terms of (1) converges, then (1) is called absolutely convergent. Infinite series which converge absolutely are most convenient in actual practice, for extreme precaution is necessary in dealing with other series. (See ~ 154. See also Goursat-Hedrick, Mathematical Analysis, Vol. I, Chap. VIII.) EXAMPLE 1. The series 1 + r + r2 + -* * + r + * * * is called a geometric series; the number r is called the common ratio. A geometric series converges absolutely for any value of r numerically less than 1; for 1 rn Si = 1 + r + r2 +. + r-l = - l —r 1-r' hence lim 1- S-Sn 1 = 0im r, if Irl < 1, n->oo 1- r n>oo| - r since rn decreases below any number we might name as n becomes infinite. It follows that the sum S of the infinite series is S = lim n = --- if rl < 1; n-oo - and it is easy to see that the series still converges if r is negative, when it is replaced by its numerical value r. EXAMPLE 2. Any series ao + a, + a2 + * * * + an + * * * of positive numbers can be compared with the geometric series of Ex. 1. Let on =ao+ a + a2l - +an -1; then it is evident that an increases with n. Comparing with the geometric series ao (1 + r + r2 + * * * + rn + *..), it is clear that if an < aorn, an < aosn, where sn = 1 + r +. + rn-l. Hence on approaches a limit if sn does, i.e. if 0 < r < 1. It follows that the given series converges if a value of r < 1 can be found for which an < aorn, that is, for which an - an-I < r < 1. There are, however, some convergent series for which this test cannot be applied satisfactorily. It may be applied in testing any series for absolute convergence; or in testing any series of positive terms. For example, consider the series 1 +1 1 1++ + 1! 2! WI n 266 THE CALCULUS [XVI, ~ 152 here a, = l/n!, an-i = 1/(n - 1)!, and therefore an/an-i = (n-l)!/n! = 1/n. Hence an/an-1< 1/2 when n > 2, ~1 1 1 22+ 2 l). nl+ 1 + 2 + - + ( l> 1 +(l +2 ++- + ) 2 - (n 2- 22 2> ( ~ = 1 + Sn-1, where sn- = 1 + r + * * + rn-2, r = 1/2. It follows that the given series converges and that its sum is less than 1 + 2 = 3. [Compare ~ 154, p. 269; it results that s < 3. Compare Ex. 2, p. 268.] 153. Taylor Series. General Convergence Test. Series which resemble the geometric series except for the insertion of constant coefficients of the powers of r, (1) A + Br + Cr2 +Dr3 + *, arise through application of Taylor's Theorem [D *], ~ 147, p. 254; such series are called Taylor series or power series. The properties of a Taylor series are, like those of a geometric series, comparatively simple. Comparing (1) with [D *], we. see that r takes the place of (x-a), while A, B, C, D, * * have the values: Af' (a) f' (a) Of" (a) 2!' 3! If we consider the sum of n such terms: f'(a) f"(a) )2+ Sn = f(a) + f (x - a) (x + ** +f(n- l(a) (x-a)(n-,l)( { we see by [D *], that f(x) = sn + En, where IEn| < Mn x -! Mn f(|n)(x)l; or~~~~~~~~~~- S/, =fx ' —E or sn =f (x) - En. XVI, ~ 153] SERIES 267 It follows that if En approaches zero as n becomes infinite, the infinite Taylor Series [D**] f (x) = f(a) + f! (x -a) + f-'2( (x- a)2 +... + (x - a)n +... converges, and its sum is S = im Sn = f (x).* This is certainly true, for example, whenever If( )(x) I remains, for all values of n, less than some constant C, however large, for all values of x between x = a and x = b. For in that case Ix -aln \ Ix- a[" lim IE, < lim C C lim = n — oo n —oo n i 7-o n for all values of (x - a).t When I f(n)(x) I grows larger and larger without a bound as n becomes infinite, we may still often make I En I approach zero by making (x - a) numerically small. EXAMPLE 1. Derive an infinite Taylor series in powers of x for the function f (x) = sin x. Since f (x) = sin x, we have f' (x) = cos x, f" (x) = - sin x, and, in general, f ()(x) = ~ sin x, or + cos x; hence Xn Ifn (x) < 1, lim I E < lim =r O; n->0 - n-a n therefore the infinite series [D**] for a = 0 is sin x = O+ x - 3 0- + 5 +; 3 5~ this series certainly conyerges and its sum is sin x for all values of x, since lim I En = 0. * This result is forecasted in ~ 147. t This results from the fact that n eventually exceeds (x-a) numerically; afterwards an increase in n diminishes the value of En more and more rapidly as n grows. 268 THE CALCULUS [XV12 ~ 153 EXAMP'LE 2. Derive an infinite series for Cx in powers of (x - 2). Since f (x) = ex, we have f' (x) = ex *, f (W' (x) = ex; hence f (2) = el, f' (2) = e2, l, f( ) (2) = e2, and If(n)(x) I < eb where b is the largest value of x we shall consider. Then the series e = el + 62(X - 2) + (x - 2) 2 +-.. + -e5 - 2)n + ex 2! n! = e2[1~ + (x:- 2)-+ 1 ( - 2)2 + + i - (x —2)n $ ~~ 2! I n converges and its sum is Cx, for all values of x less than b; for li n im eb Ix - 2 1n lim IEl, I~ 0 n —+oo n- n! Since b is any number we please, the series is convergent and its sum is ex for all values of x. EXERCISES Derive the following series, and show, when possible, that they converge for the indicated values of x. 1. cos x = 1 - X2 /2! + X4/4! - x6/6!~ +;allx). 2. ex = 1 + ~ + X2/2!+ X3 /3! +...; (all x). 3. e-x - 1 - X + X212! - X313 +.. ( all x). 4. tan x = X + X3/3 + 2 x5/15 + 17 X7/315 + (I;( I < 7r/4). 5. log (I + X) = X - X212 + X313 - X414 +.'.; X1 I < I). 6. sinhx = -e-)/2 = x + x3/3! + x5/5! +; (all x). 7. cosh x = (ex + eCx)/2 = 1 + X2/2!+ X4/4!; (all x). 8. tanh x = sinh x/cosh x = X - X3/3 + 2 x5/15 - 17 X7/315 +; (all x). 9. Show that the series of Ex. 6 can be obtained from those of Exs. 2 and 3 if the terms are combined separately. 10. Show that the series of Ex. 3 results from the series of Ex. 2 if x is replaced by - x. 11. Obtain the series for sin x in powers of (x - ir/4). 12. Obtain the series for ex in terms of powers of (x - 1). 13. Obtain the series for log x in powers of (x - 1). Compare it with the series of Ex. 5. XVI, ~ 154] SERIES 269 14. Obtain the series for log (1 - x) in powers of x, directly; also by replacing x by - x in Ex. 5. 15. Using the fact that log [ (1 + x)/(1 - x)] =log (1 + x) - log (1 - x), obtain the series for log [(1 + x)/(1 - x)] by combining the separate terms of the two series of Ex. 14 and of Ex. 5. This series is actually used for computing logarithms. 16. Show that the terms of the Maclaurin series for (a + x)n in powers of x are precisely those of the usual binomial theorem. 17. Show that the series for ea+x in powers of x is the same as the series for ex all multiplied by ea. 18. Show that the series for 10x is the same as the series for eZ with x replaced by x/M, where M = 2.30 - -. 154. Precautions about Infinite Series. There are several popular misconceptions concerning infinite series which yield to very commonplace arguments. (a) Infinite series are never used in computation. Contrary to a popular belief infinite series are never used in computation. What is actually done is to use a few terms (that is, a polynomial) for actual computation; one may or may not consider how much error is made in doing this, with an obvious effect on the trustworthiness of the result. Thus we may write X3 X5 x2k+1 sin x = x - +5 - - +- (2 k + 1)! (forever); but in practical computation, we decide to use a few terms, say sin x = x - x3/3! + x5/5!. The error in doing this can be estimated by ~ 147, p. 253. It is IE71 <Ix7/7!j. For reasonably small values of x [say I xl <14~ <1/4 (radians)], I E7 1 is exceedingly small. Many of the more useful series are so rapid in their convergence that it is really quite safe to use them without estimating the error made; but if one proceeds without any idea of how much the error amounts to, one usually computes more terms than necessary. Thus if it were required to calculate sin 14~ to eight decimal places, most persons would suppose it necessary to use quite a few terms of the preceding series, if they had not estimated E7. 270 THE CALCULUS [XVI, ~ 154 (b) No faith can be placed in the fact that the terms are becoming smaller. The instinctive feeling that if the terms become quite. small, one can reasonably stop and suppose the error small, is unfortunately not justified.* Thus the series 1 + +0 + + + 1J + 10 20 30 40 10 n has terms which become small rather rapidly; one instinctively feels that if about one hundred terms were computed, the rest would not affect the result very much, because the next term is.001 and the succeeding ones are still smaller. This expectation is violently wrong. As a matter of fact this series diverges; we can pass any conceivable amount by continuing the term-adding process. For + 1 > 2. 1 = 1 A+?>+2.- A>4., +I + le + ** + W- > 8 * i =, and so on; groups of terms which total more than 1/20 continue to appear forever; twenty such groups would total over 1; 200 such groups would total over 10; and so on. The preceding series is therefore very deceptive; practically it is useless for computation, though it might appear quite promising to one who still trusted the instinctive feeling mentioned above. (c) If the terms are alternately positive and negative, and if the terms are numerically decreasing with zero as their limit, the instinctive feeling just mentioned in (b) is actually correct: the series ao - a1 + a2 - a3 +.* * converges if an approaches zero; the error made in stopping with an is less than an+l.t For, the sum Sn = ao - al + -* * * an1 evidently alternates between an increase and a decrease as n increases, and this * This fallacious instinctive feeling is doubtless actually used, and it is responsible for more errors than any other single fallacy. The example here mentioned is certainly neither an unusual nor an artificial example. t One must, however, make quite sure that the terms actually approach zero, not merely that they become rather small; the addition of.0000001 to each term would often have no appreciable effect on the appearance of the first few terms, but it would make any convergent series diverge. xvI, ~ 154] SERIES 271 alternate swinging forward and then backward dies out as n increases, since an is precisely the amount of the nth swing. On each swing Sn passes a point S which it again repasses on the return swing; and its distance from that point is never more than the next swing, -never more than an+l. Since a, approaches zero, sn approaches S, as n becomes infinite. Thus the series for sin x is particularly easy to use in calculation: the error made in using x - x3/3! in place of sin x is certainly less than x5/5!. The test of ~ 147 shows, in fact, that the error IE51 <M5sx5/5!, where M5 = 1. The similar series for ex: e x =1x+ 2+ + xn + 1! 2! + + -.+n!+ is not quite so convenient, since the swings are all in one direction for positive values of x; certainly the error in stopping with any term is greater than the first term omitted. The error can be estimated by ~ 147, p. 253; thus Es (for x >0) is less than M5x5/5!, where M5 is the maximum of f (x) = ex between x = 0 and x = x, i.e. ex; hence Es<e2x5/5!. Note that el > 1 for x > 0. Another means of convincing oneself that the preceding series converges for x < 1 is by comparison with a geometric series with a ratio x/2, as in Example 2, p. 265. But this method would require the computation of a vast number of terms, to make sure that the error is small. (d) A consistently small error in the values of a function may make an enormous error in the values of its derivative. Thus the function y = x -.00001 sin (100000 x) is very well approximated by the single term y = x, -in fact the graphs drawn accurately on any ordinary scale will not show the slightest trace of difference between the two curves. Yet the slope of y = x is always 1, while the slope of y = x-.00001 sin (100000 x) oscillates between 0 and 2 with extreme rapidity. Draw the curves, and find dy/dx for the given function. One advantage in Taylor series and Taylor approximating polynomials is the known fact - proved in advanced texts 272 THE CALCULUS [XVI, ~ 154 that differentiation as well as integration is quite reliable on any valid Taylor approximation.* Thus an attempt to expand the function y = x -.00001 sin (100000x) in Taylor form gives r 1000002 1000004 y-103! x3 +- -5! x5... which would never be mistaken for y = x by any one; the series indeed converges and represents y for every value of x, but a very hasty examination is sufficient to show that an enormous number of terms would have to be taken to get a reasonable approximation, and no one would try to get the derivative by differentiating a single term. If the relation expressed by the given equation was obtained by experiment, however, no reliance can be placed in a formal differentiation, even, though Taylor approximations are used, for minute experimental errors may cause large errors in the derivative. Attention is called to the fact that the preceding example is not an unnatural one, - precisely such rapid minute variations as it contains occur very frequently in nature. EXERCISES 1. Show that the series obtained by long division for 1 + (1 + x) is the same as that given by Taylor's Series. 2. Obtain the series for log (1 + x) (see Ex. 5, ~ 153), by integrating the terms of the series found in Ex. 1 separately. 3. Find the first four terms of the series for sin- x in powers of x directly; then also by integration of the separate terms of the series for 1/vi1 - x2. 4. Proceed as in Ex. 3 for the functions tan-1 x and 1/(1 + x2). 5. Show that the series for cos x in powers of x is obtained by differentiating separately the terms of the series for sin x. 6. Show that repeated differentiation or integration of the separate terms of the series for ex always results in the same series as the original one. 7. From the series for tan-1 x compute 7r by using the identity 7r/4 = 4 tan-1 (1/5) - tan-1 (1/239). * See, e.g., Goursat-Hedrick, Mathematical Analysis, Vol. I, p. 380. X-VI, ~ 154] SERIES 273 8. The Gudermannian of x is gd (x) = 2 tan-' ex- - wr/2; expand in powers of x, calculate gd(.1) = 50 43', and gd (.7) = 370 11' 9. The "error integral" is P (x) =:= e-x' dx. Express P (x) as a series in powers of x; calculate P (.1) =.1125, P(1) =.8427, P (2)=.9953+. 1/2 10. Show that f Vi- e-t dt =.8862+. 13. Show thatfdt/V1 1- t3 =508+. 00 11. Show thatf to3 e'1 dt =.8975+. 14. Show thatfdt/x/P74=1.311+. 12. Show thatf 72Sin 5/4 x dx =.9309+. CHAPTER XVII PARTIAL DERIVATIVES -APPLICATIONS 155. Partial Derivatives. If one quantity depends upon two or more other quantities, its rate of change with respect to one of them, while all the rest remain fixed, is called a partial derivative.* If z = f (x, y) is a function of x and y, then, for a constant value of y, y = k, z is a function of x alone: z = f (x, k); the derivative of this function of x alone is called the partial derivative of z with respect to x, and is denoted by any one of the symbols az - af(x, y) = f (x, ) = df (x, k) ax ax dx = limf ( + Ax, k) -f (x, k) Ax-lO Ax A precisely similar formula defines the partial derivative of z with respect to y which is denoted by az/Oy. In general, if u is a function of any number of variables x, y, z,. *, and if one calculates the first derivative of u with respect to each of these variables, supposing all the others to be fixed, the results are called the first partial derivatives of u with respect to x, y, z,. *, respectively, and are denoted by the symbols u/ax, au/oy, au/az,. * This notion is perhaps more prevalent in the world at large than the notion of a derivative of a function of one variable, because quantities in nature usually depend upon a great many influences. The notion of partial derivative is what is expressed in the ordinary phrases "the rate at which a quantity changes, everything else being supposed equal," or "... other things being the same." 274 XVII, ~ 156] PARTIAL DERIVATIVES-APPLICATIONS 275 EXAMPLE 1. Given z = x2 + y2, to find az/Ox and az/dy. To find az/ax, think of y as constant: y = k; then az 2 (x2 + y2) d (x2 +k2) =2 az ax ax dx a = y. EXAMPLE 2. Given z = x2 sin(x + y2), to find az/ax and az/ay. az a {x2 sin (x + y2)} d {x2 sin (x +k2)}] a dx yax L d v= = 2x sin (x + y2) + x2 cos (x + y2). az a {x2 sin (x + y2)} d {k2 sin (k + y2)} ay ay dy Jx= =2 x2 y cos (x + y2). 156. Higher Partial Derivatives. Successive differentiation is carried out as in the case of ordinary differentiation. There are evidently four ways of getting a second partial derivative: differentiating twice with respect to x; once with respect to x, and then once with respect to y; once with respect to y, and then once with respect to x; twice with respect to y. These four second derivatives are denoted, respectively, by the symbols a dOza z a (dOz\ _ 2z Ox x) - - = fx(XY) Y; Y - -yf ) x = f (x, Y); da az\ d2z a (z\ d2z 02 =f (x, y); - _ x__ y^2 fY(Z'); x \ay) Oxo y =f (xy). There is no new difficulty in carrying out these operations; in fact the situation is simpler than one might suppose, for it turns out that the two cross derivatives fx, and fyx are always equal; the order of differentiation is immaterial.* A similar notation is used for still higher derivatives: f 3 _a (a/2\ a3Z_ ( 92 z\ xx 3 = ~a OX2; fyxx - ay 2 - ay \aX2)' etc., and the order of differentiation is immaterial. *At least if the derivatives are themselves continuous. See GoursatHedrick, Mathematical Analysis, Vol. I, p. 13. 276 THE CALCULUS [XVII, ~ 156 The order of a partial derivative is the total number of successive differentiations performed to obtain it. The partial derivatives of the first and second orders are very frequently represented by the letters p, q, r, s, t: az az a2 Z a2 z o2 Z 2 z p - q~ r = -~X 2 s = x 9y ~y =x t = y EXAMPLE 1. Given z = x2 sin (x + y2), show that fx = fyx. Continuing Example 2, ~ 155, we find: ayx- ) = a [2 xsin (x + 2) +x2cos(X+y2)] = 4 xy cos (x + y2) - 2 x2y sin (x + y2). a y = (a )= 2 x2y cos (x + y2) = 4 xy cos (x + y2) - 2 x2y sin (x + y2). EXERCISES Find the first and second partial derivatives, az/dx, az/ay, a02 /ax2, a2 zs/x ay, 02 z/ay ax, and a2 /lay2 for each of the following functions. In each case verify the fact that a2 z/ax ay = a2 sz/y ax. 1. z = 2 - y2. 4. z = eax+by. 7. z = (x + y) ex2+v2. 2. z = x2 y + xy2. 5. z = tan-l (y/x). 8. z = (xy - 2 y2)5/3. 3. z = sin (x2 + y2). 6. z = ex sin y. 9. z = log (x2 + y2)1/2. 10. The volume of a right circular cylinder is v = nrr2 h. Find the rate of change of the volume with respect to r when h is constant, and express it as a partial derivative. Find av/ah, and express its meaning. 11. The pressure p, the volume v, and temperature 0 of a gas are connected by the relation pv = ko, where 6 is measured from the absolute zero, - 273~ C. Assuming 0 constant, find ap/av and express its meaning. If the volume is constant, express the rate of change of pressure with respect to the temperature as a derivative, and find its value. 12. Find the rate of change of the volume of a cone with respect to its height, if the radius is constant; and the rate of change of the volume with respect to the radius, if the height is a constant. 13. Show that the functions in Exercises 1, 5, 6, and 9 satisfy the equation a2 z/ax2 + a2 z/ay2 = 0. XVII, ~ 157] PARTIAL DERIVATIVES-APPLICATIONS 277 14. A point moves parallel to the x-axis. What are the rates of change of its polar coordinates with respect to x? 15. Show that the rate of change of the total surface of a right circular cylinder with respect to its altitude is aA/Oh = 2 7r r; and that its rate of change with respect to its radius is dA/Ir = 2 r h + 4 r r. 16. Calculate the rate of change of the hypotenuse of a right triangle relative to a side, the other side being fixed; relative to an angle, the opposite side being fixed. 17. Two sides and the included angle of a parallelogram are a, b, C, respectively. Find the rate of change of the area with respect to each of them, the other two being fixed; the same for the diagonal opposite to C. 18. In a steady electric current C = V - R, where C, V, R, denote the current, the voltage (electric pressure), and the resistance, respectively. Find a C/a V and a C/a R, and express the meaning of each of them. 157. Geometric Interpretation. The first partial derivatives of a function of two independent variables z =f(x, y) can be interpreted geometrically in a simple manner. This equation represents a surface, which may J be plotted by erecting at each point of H -=(hk,) IC the xy-plane a per- A 2T pendicular of length / K f(x, y); the upper I I ends * of these per- /c pendiculars are the h/ /k points of the surface. /p(hko) k Let ABCD be a portion of this sur- FIG. 77. * If z is negative, of course the lower end is the one to take. 278 THE CALCULUS [XVII, ~ 157 face lying above an area abcd of the xy-plane. If x varies while y remains fixed, say equal to k, there is traced on the surface the curve HK, the section of the surface by the plane y = k. The slope of this curve is dz/Ox. Similarly, Oz/dy is the slope of the curve cut from the surface by a plane x = h. 158. Total Derivative. If in addition to the function z = f (x, y), a relation between x and y, say y = >(x), is given, z reduces by simple substitution to a function of one variable: z =f (x, y), ] = ~0 (x) gives z = f (x, (x)). Now any change Ax in x forces a change Ay in y; hence y cannot remain constant (unless, indeed, (x) = const.). Hence the change Az in the value of z is due both to the direct change Ax in x and also to the forced change Ay in y. We shall call Az = the total change in z =f (x + Ax, y + Ay) -f (x, y), Axz = the partial change due to Ax directly = f (x +Ax, y)-f (x, y), Az = the partial change forced by the forced change Ay = Az- AXz, = f (x +Ax, y + Ay) -f (x + Ax, y). It follows that (1) dz Az lim (f(x+Ax, Y+Ay)-f(x, )) = lim lix o Ax dx A x Ax-O \ AAx A AX-0Z +AyZ A r lim (x +Ax,y) -f(x,y) - lim A —~A z = Ax->*O Ax Axd->O Ax Ay-0 AO + f ( +Ay) f +Ax, y +A) ( ) Ay + -------- Ay ) '& i XVII, ~ 158] PARTIAL DERIVATIVES -APPLICATIONS 279 whence, if the partial derivatives exist and are continuous,* (2) dz = lim FAxz 4AZ Ay x z z dy dx o A yA dx ' or, multiplying both sides by dx(=Ax), (z Oz dy (3) dz = dx + dy, since dy = dx, where dy = -k (x) dx. Since O(x) is any function whatever, dy is really perfectly arbitrary. Hence (3) holds for any arbitrary values of dx and dy \zf(Xy) whatever, where dz = (dz/dx) dx is defined by (2); dz is called the A A B total differential of z. c: 0 These quantities are all repre- Y Q O=(Z) sented in the figure geometrically: FIG. 78. PS = AB -- zx thus Az = Axz + Az- is represented ST'=CD=Ay by the geometric equation TQ -= SR-AZ=TM=RoR-PoP MQ = A5z]x = t+ = QoQ -RoR SR + MQ. It should be noticed z=QoQ -PoP= TQ=SR+MQ that dz is the height of the plane =-xZ+^Az]x=X+ drawn tangent to the surface at P, since Oz/Ox and Oz/dy are the slopes of the sections of the surface by y = yp and x = xp, respectively. [See also ~ 163.] If the curve PoQo in the xy-plane is given in parameter form, x = (t), y = a(t), we may divide both sides of (3) by dt and write dz Oz dx Oz dy (4) = - t dt - oxdt a9ydt since dx + dt = dx/dt, dy - dt = dy/dt. *For a more detailed proof using the law of the mean, see Goursat-Hedrick, Mathematical Analysis, I, pp. 38-42. 280 THE CALCULUS [XVII, ~ 159 159. Elementary Use. In elementary cases, many of which have been dealt with successfully before ~ 158, the use of the formulas (2), (3), and (4) of ~ 158 is quite self-evident. EXAMPLE 1. The area of a cylindrical cup with no top is (1) A = 2 7rrh + 7rr2, where h is the height, and r is the radius of the base. If the volume of the cup, 7rr2 h, is known in advance, say wrr2 h = 10 (cubic inches), we actually do know a relation between h and r: 10 (2) h= 7r 7' whence 10 20 (3) A= 2 r r2 + 7rr2 = 20 + r2 'r r2 T from which dA/dr can be found. We did precisely the same work in Ex. 26, p. 57. In fact even then we might have used (1) instead of (3), and we might have written dA dh (4) - = 2 7rd +2 7 + 2 7rr, ordA = 2 w rr dh + (2 rh +2 irr) dr, dr dr where dh/dr is to be found from (2). This is precisely what formula (2), ~ 158, does for us; for OA aA (5) = 2 rh + 2 Err, =2 rr, dA dh d = (2 Arh+2 rr) + (2 7rr), or dA = (2 7h+ 2 7rr) dr+2 rr dh. dr dr We used just such equations as (4) to get the critical values in finding extremes for dA/dr = 0 at a critical point. We may now use (2), ~ 158, to find dA/dr; and the work is considerably shortened in some cases. EXAMPLE 2. The derivative dy/dx can be found from (2), ~ 158, if we know that z is constant. Thus in ~ 24, p. 39, we had the equation (1) 2 + y = 1, and we wrote: d(x2+y2) ^ dy d(1)_ (2) dx2 = 2 x d+ 2 y0 d, dx dx dx whence we found dy Xdy (3) x +y - =0, or dy = -. dx dx Y XVII, ~ 160] PARTIAL DERIVATIVES-APPLICATIONS 281 This work may be thought of as follows: Let z = x2 + y2; then dz d(x2 + y2) = z + 2zdy x dy. dx dx ax ay dx dx but z = 1 by (1) above; hence dz/dx = 0, and 2x +2y = 0, ordY = -_ dx dx y Thus the use of the formulas of ~ 158 is essentially not at all new; the preceding exercises and the work we have done in ~~ 24, 29, etc., really employ the same principle. But the same facts appear in a new light by means of ~ 158; and the new formulas are a real assistance in many examples. 160. Small Errors. Partial Differentials. Another application closely allied to the work of ~ 145, p. 248, is found in the estimation of small errors. EXAMPLE 1. The angle A of a right triangle ABC (C = 90~) may be computed by the formula a a tan A = or A = tan-l - b'~ b' where a, b, c are the sides opposite A, B, C. If an error is made in measuring a or b, the computed value of A is of course false. We may estimate the error in A caused by an error in measuring a, supposing temporarily that b is correct, by ~ 145; this gives approximately 1 aA b b AaA = - Aa = a2 a = a a, 1 +2 where 0 is used in place of d of ~ 145, since A really depends on b also, and we have simply supposed b constant temporarily. Likewise the error in A caused by an error in b is approximately, a OA b2 -a a AbA = - Abb = 1,Ab = -a -bfb. +2 a2~b2 ab +a:! CL $Lb' 282 THE CALCULUS [XVII, ~ 160 If errors are possible in both measurements, the total error in A is, approximately, the sum of these two partial errors: AA I< l AaAI +lAbAI =bal + alAb5 a2 + b2 The methods of ~ 146, p. 251, give a means of finding how nearly correct these estimates of AaA, AbA, and AA are; in practice, such values as those just found serve as a guide, since it is usually desired only to give a general idea of the amounts of such errors. This method is perfectly general. The differences in the value of a function z = (x, y) of two variables, x and y, which are caused by differences in the value of x alone, or of y alone, are denoted by Axz, Az, respectively. The total difference in z caused by a change in both x and y is az = f (x + ax, y + ay)- f (x, y) = f (x + ax, y + y) - f (x + x, y)] + [f (x + x, y) -f (x, y)] = Ayz]=x.+Ax + Az, as in ~ 158. The differences Axz and Ayz are, approximately,* az az Axz = qxAx, Az = = yAy; whence, approximately, Az = A z + AXZ = Ax + az Ay. * More precisely, these errors are Az= -.Ax + E'2, Az = y + * A + E"2, where I E'2 F and [ E"2 I are less than the maximum M2 of the values of all of the second derivatives of z near (x, y) multiplied by Ax2, or Ay2, respectively (see ~146). And since az/Oy is itself supposed to be continuous, we may write Az = ax+ t ay + E2, where l E2 1 is less than M2( Ax | + I Ay 1)2. [Law of the Mean. Compare ~146.] XVII, ~ 160] PARTIAL DERIVATIVES-APPLICATIONS 283 The products (az/ax) dx and (az/Oy) dy are often called the partial differentials of z, and are denoted by az az az = - dx, aZ =- dy, whence dz = xz + - az. ax ay We have therefore, approximately, Az = axz + aZ within an amount which can be estimated as in ~ 146 and in the preceding footnote. Similar formulas give an estimate of the values of the changes in a function u = f (x, y, z) of the variables x, y, z; we have, approximately, au au au axu = Ax, au = -ay, Azu = az, ax ay az = u A u au Au = AXu + Au + AU = u Ax + - Ay + d Az ax ay az within an amount which can be estimated as in the preceding footnote. The generalization to the case of more than three variables is obvious. EXERCISES Find dy/dx in each of the following implicit equations by method of Ex. 2, ~ 159: 1. x2+4y2 =. 3. x3+y3-3xy=0. 2. 7x2- 9 y2 = 36. 4. y2 (2 a -x ) = 3. 5. If A, B, C denote the angles, and a, b, c the sides opposite them, respectively, in a plane triangle, and if a, A, B are known by measurements, b = a sin B/sin A. Show that the error in the computed value of b due to an error da in measuring a is, approximately, dab = sin B csc A da. Likewise show that OAb = - asin B csc A ctn A dA, and asb = a cos B csc A d B; and the maximum total error is, approximately, Idbl A I abl - I4 bl + I aBb 1. Note that dA and dB are to be expressed in radian measure. 284 THE CALCULUS [XVII, ~ 160 6. The measured parts of a triangle and their probable errors are a = 100 +.01 ft., A = 100~ ~ 1', B = 40~ + 1'. Show that the partial errors in the side b are Oab = ~.007 ft., 9Ab = ~.003 ft., Oab = ~.023 ft. If these should all combine with like signs, the maximum total error would be db = ~.033 ft. 7. If a = 100 ft., B = 30~, A = 110~, and each is subject to an error of 1%, find the per cent of error in b. 8. Find the partial and total errors in angle B, when a = 100.01 ft., b = 159 +.01 ft., A = 30~ + 1'. 9. The radius of the base and the altitude of a right circular cone being measured to 1%, what is the possible per cent of error in the volume? Ans. 3%. 10. The formula for index of refraction is m = sin i/sin r, i being the angle of incidence and r the angle of refraction. If i = 50~ and r = 40~, each subject to an error of 1%, what is m, and what its actual and its percentage error? 11. Water is flowing through a pipe of length L ft., and diameter D ft., under a head of H ft. The flow, in cubic feet per minute, is HD5 Q = 2356 L + 30D If L = 1000, D = 2, and H = 100, determine the change in Q due to an increase of 1% in H; in L; in D. Compare the partial differentials with the partial increments. 161. Envelopes. The straight line (1) y = kx - k2, where k is a constant to which various values may be assigned, has a different position for each value of k. All the straight lines which (1) represents may be tangents to some one curve. If they are, the point Pt, (x, y) at which (1) is tangent to the curve, evidently depends on the value of k: (2) x = f(kc), y = (k); XVII, ~ 161] PARTIAL DERIVATIVES-APPLICATIONS 285 these equations may be considered to be the parameter equations of the required curve. The motive is to Les- I_ Lies: yl=kj - [c ifind the functions +(k) and E En epel: yL - ~ I(k) if possible. \ — -_ -/ Since Pt lies on (1) and \ - _ -= on (2), we may substitute k ____ from (2) in (1) to obtain: k,/_ (3) (k) = kk) - k- _ _, 2 which must hold for all -- - - values of k. Moreover, --- - since (1) is tangent to (2) at _ //i Pk, the values of dy/dx I _ _ found from (1) and from (2) - must coincide: FIG. 79. (4) k = =] dy or _) ' (J) from (1) &Jfrom (2) '() or (= ( k) To find k(k) and '(k) from the two equations (3) and (4), it is evident that it is expedient to differentiate both sides of (3) with respect to k: (3 *) ' '(k) = k'(k) + 0(k) - 2 k; this equation reduces by means of (4) to the form (5) 0 = 0 + c(k)-2k, or f(k) = 2 k, and then (3) gives (6) A(k) = k(2 k) -k2= k2 Hence the parameter equations (2) of the desired curve are (7) x=2k, y=k2, and the equation in usual form results by elimination of k: ~(8) Y7~ AW~x2 (8) y = x' 286 THE CALCULUS [XVII, ~ 161 It is easy to show that the tangents to (8) are precisely the straight lines (1). The preceding method is perfectly general. Given any set of curves (1)' F (x, y, k) = 0, where k may have various values, a curve to which they are all tangent is called their envelope; its equations may be written (2)' x = (k), y = (k) whence by substitution in (1)', (3)' F [ (k), ~ (k), k] = 0, for all values of k. Differentiating (3)' with respect to k, (3*) dF(x, y, k) _ +Fdx F+ dy+ OF 0 dk Ox dk Oy dk 0 Moreover, since (1)' is tangent to (2)', ( AF F _dy dy1 dy dx. Ax O' y dx from (1)' dx from (2)' dk dk' whence (3*)' reduces to the form OF (5)' Ak =0; and then (3)' and (5)' may be solved as simultaneous equations to find 0 (k) and ~ (k) as in the preceding example. The envelope may be found speedily by simply writing down the equations (1)' and (5)', and then eliminating k between them It is recommended very strongly that this should not be done until the student is familiar with the direct solution as shown in the preceding example. 162. Envelope of Normals. Evolute. If y = f(x) is a given curve and if yt and mk respectively denote the ordinate and slope when x = k, the equation of any normal may be written 1 ' (1) Y - Yk= -- (x - ), mkf or F(x, y, k) = ymk - Ykmk + x - k = 0 XVII, ~ 162] PARTIAL DERIVATIVES-APPLICATIONS 287 Hence by (5'), ~ 1161, we have, for the envelope of the system of lines (1) when k is regarded as a variable parameter, oF (2) _- = y ' bk - yk' bk - mk mk- 1 =0. o9k (Remember that in forming aF/ak, x and y are regarded as constant, and only k,,y mk, yk, are regarded as variable. We have used bk to stand for dmk/ak.) Solving (2) for y we have (3) y=y -t m2 This value of y in (1) gives FIG. 80. 1-+-m2k (4) x = k - mk Equations (3) and (4) are the parametric equations of the envelope of the system of normals (1), and are precisely equations (1) of ~ 98, with only a change of notation. Hence the envelope of the systems of normals to a given curve is the evolute of that curve. EXERCISES Find the envelopes of each of the following families of curves: 1. y = 3 kx-k3. Ans. y2 = 4 x3. 2. y = 4kx -k4. Ans. y3 = 27x4. 3. y2 = kx - k2. Ans. y = ~+ x. 4. y = kx + Vl k2. Ans. x2 y2 = 1. 5. y2 =k2x -2 k. Ans. xy2 =-1. 288 THE CALCULUS [XVII, ~ 162 6. (x — k)2 + y2 = 2 k. Ans. y2 = 2 x + 1. 7. 4 x2+ (y - C)2 =1-k2. Ans.y2 + 8 X2 = 2. 8. x cosO0 + y sinO0 = 10. Ans. x2 + y2 = 100. 9. Show that the envelope of a family of circles through the origin with their centers on the parabola y2 = 2 x is y2 (x + 1) + x3 = 0. 10. Show that the envelope of the family of straight lines ax ~ by = 1 where a ~ b = ab, is the parabola X12 + y1/2 = 1. 11. Show that the envelope of the family of parabolas represented by the equation y = x tan a - mx2 sec2 a is y = 1/(4 m) - mx2. [NOTE. If m = g/(2 v02), the given equation represents the path of a projectile fired from the origin with initial speed vo at an angle of elevation a.] 12. The lemniscate (X2 ~ y2)2 = a2(X2 -_y2) may be written in the form x = a cos t Vcos 2 t; y = a sin t V\cos 2 t. Show that the evolute is (X2/3 ~ y2/3)2 (x2/3 - y2/3) = 4 a2/9. CHAPTER XVIII CURVED SURFACES -CURVES IN SPACE 163. Tangent Plane to a Surface. (xo Yo, zo) on the surface z = f (x, y). line at Po to the curve cut from the surface by the plane y = yo and PoT2 the tangent line to the curve cut from the surface by the plane x = xo. The plane containing these two lines is the tangent plane to the surface at Po. Since this plane goes through Po, its equation can be thrown into the form Let Po be the point Let PoT, be the tangent z A FIG. 81. (1) z - o = A(x - xo) +B (y - yo). If we set y = yo we find the equation of PoT1 in the form: (2) z - Zo = A(x - Xo). But, from ~ 28, p. 49, the equation of PoT1 may be written in the form: (3) Hence (4) z - = of (x- xo). Z -- ZO - O A = 1 ar; likewise B = - * 8x o o 289 290 THE CALCULUS [XVIII, ~ 163 Thus the equation of the tangent plane is (5) Z-Zo ](x] - xo) + (t - yo); or, what is the same thing, az -xo (6) Z-ZO= x-xo) (y - yo). It is important to notice the great similarity between this equation and the equation (7) dz = d d+ - dy, ox 0 ayj o of ~ 158. Indeed (7) expresses the fact that if dx, dy are measured parallel to the x and y axes from the point of tangency (Xo, yo, zo), dz represents the height of the tangent plane above (xo, yo, zo). Equation (7) furnishes a good means of remembering (6). 164. Extremes on a Surface. If a function z = f (x, y) is represented geometrically by a surface, it is evident that the extreme values of z are represented by the points on the surface which are the highest, or the lowest, points in their neighborhood: (1) f (xo, yo) > f (xo + h, yo + k), if f (xo, yo) is a maximum, (2) f (xo, yo) < f (xo + h, yo + k), if f (xo, yo) is a minimum, for all values of h and k for which h2 + k2 is not zero and is not too large. It is evident directly from the geometry of the figure that the tangent plane at such a point is horizontal. This results also, however, from the fact that the section of the surface by the plane x = Xo must have an extreme at (xo, yo); hence [Of/Oy]o, which is the slope of this section at (xo yo), must be zero; likewise [Of/Ox]o, the slope of the XVIII, ~ 1641 CURVED SURFACES 291 section through (xo, yo) by the plane y = yo, must be zero. Hence equation (5), ~ 163, reduces to z - zo = 0 which is a horizontal plane. A point at which the tangent plane is horizontal is called a critical point on the surface. The following cases may present themselves. (1) The surface may cut through its tangent plane; then there is no extreme at (xo, yo). This is what happens at a point on a surface of the saddleback type shown by a hyperbolic paraboloid at the origin; a homelier example is the depression between the knuckles of a clenched fist. (2) The surface may just touch its tangent plane along a whole line, but not pierce through; then there is what is often called a weak extreme at (xo, yo); that is, z = f(x, y) has the same value along a whole line that it has at (xo, yo), but otherwise f (x, y) is less than [or greater than] f (xo, yo). This is what happens on the top of a surface which has a rim, such as the upper edge of a water glass, or the highest point of an anchor ring lying on its side. Most objects intended to stand on a table are provided with a rim on which to sit; they touch the table all along this rim, but do not pierce through the table. (3) The surface may touch its tangent plane only at the point (xo, yo); then z = f(x, y) is an extreme at (xo, yo): a minimum, if the surface is wholly above the tangent plane near (xo, yo); a maximum, if the surface is wholly below. The shape of the clenched fist gives many good illustrations of this type also. Examples of formal algebraic character occur below. EXAMPLE 1. For the elliptic paraboloid z = x2 + y2 the tangent plane at (xo, Yo, z2) is z - zo = 2 xO (x - xo).+ 2 Yo (Y - yo), which is horizontal if 2 x0 = 2 yo =0; this gives Xo = yo = = 0, hence (x = 0, y = 0) is the only critical point. 292 THE CALCULUS [XVIII, ~ 164 At (x = 0, y = 0), a has the value 0; for any other values of x and y, z (= x2 + y2) is surely positive. It follows that z is a minimum at x = 0, y = 0. EXAMPLE 2. In experiments with a pulley block the weight w to be lifted and the pull p necessary to lift it were found in three trials to be (in pounds) (Pi = 5, wl = 20), (P2 = 9, W2 = 50), (p3 = 15, W3 = 90). Assuming that p = aw + 3, find the values of a and t3 which make the sum S of the squares of the errors least. (Compare Ex. 37, p. 58.) Computing p by the formula aw + /, the three values are p'l 20 a + f, p'2 = 50 a + /, p'3 = 90 a + /. Hence the sum of the squares of the errors is S = (p'l - pl)2 + (P'2- P2)2 + (' 3- P3)2 = (20 a + - 5)2 + (50 a + - 9)2 + (90 a + /- 15)2. In order that S be a minimum, we must have OS = 2 [20 (20a - 5) + 50 (50a + /- 9) +90 (90a + - 15)] =0. aS = 2 [(20 a + - 5) + (50 a - + - 9) + (90 a + - 15)] =0. that is, after reduction, 1100 + 16 - 190 = 0, w = hn6 =.143, 160 a + 3- 29 = 0, ce 400 = 2.03. If the usual graph of the values of p and w is drawn, it will be seen that p = aw + /f represents these values very well for a =.143, p = 2.03 and it is evident from the geometry of the figure that these values render S a minimum, S =.0545; for any considerable increase in either a or /3 very evidently makes S increase. Since this is the only critical point, it surely corresponds to a minimum, for the function S has no singularities. This conclusion can also be reached by thinking of S as represented by the heights of a surface over an a,3 plane, and considering the section of that surface by the tangent plane at the point just found as in Ex. 3 below; but in this problem the preceding argument is simpler. It is customary to assume that the values of a and / which make S a minimum are the best compromise, or the "most probable values"; hence the most probable formula for p is p =.143 w + 2.03. The work based on more than three trials is quite similar; the only change being that S has n terms instead of 3 if n trials are made. XVIII, ~ 164] CURVED SURFACES 293 EXAMPLE 3. Find the most economical dimensions for a rectangular bin with an open top which is to hold 500 cu. ft. of grain. Let x, y, h represent the width, length, and height of the bin, respectively. Then the volume is xyh; hence xyh = 500; and the total area z of the sides and bottom is 1000 1000 (a) z = xy + 2 hy + 2 hx = xy + 10 + — 10 x y If this area (which represents the amount of material used) is to be a minimum, we must have az 1000 az 1000 (b) -= y - 2 = o, y = - = 0. ax x2 9y y2 Substituting from the first of these the value y = 1000/x2 in the second, we find (c) x-1 = 0, whence x = 0, or = 10. The lane 1000 Z= 300 10 The value x = 0 is obviously not worthy of any consideration; but the value x = 10 gives y = 1000/x2 = 10 and h = 500/(xy) =5. ( 10 X The value of z when x = 10, y = 10 is 300. If IG. the equation (a) is represented graphically by a surface, the values of z being drawn vertical, the section of the surface by the plane z = 300 is represented by the equation 1000 1000 (d) xy + 1000 + = 300, or x2y2- 300 xy + 1000 (x + y) = 0. x; y This equation is of course satisfied by x = 10, y = 10. If we attempt to plot the curve near (10, 10), - for example, if we set y = 10 + k and try to solve for x in the resulting equation: (10 + k)2 x2 - (300 k + 2000) x + 1000 (10 + k) = 0, the usual rule for imaginary roots of any quadratic ax2 + bx + c = 0 shows that b2- 4 ac = - 1000 k2 [4 k + 30] <0 for all values of k greater than - 7.5. Hence it is impossible to find any other point on the curve near (10, 10). It follows that the horizontal tangent plane z = 300 cuts the surface in a single point; hence the surface lies entirely on one side of that tangent plane. Trial of any one convenient pair of values of x and y near (10, 10) shows that z is greater near (10, 10) than at (10, 10); hence the area z is a minimum when x = 10, y = 10, which gives h = 5. 294 THE CALCULUS [XVIII, ~ 165 165. Final Tests. Final tests to determine whether a function f(x, y) has a maximum or a minimum or neither, are somewhat difficult to obtain in reliable form. Comparatively simple and natural examples are known which escape all set rules of an elementary nature.* (See Example 1 below.) One elementary fact is often useful: if the surface has a maximum at (xo, yo), every vertical section through (xo, yo) has a maximum there. Thus any critical point (x0, yo) may be discarded if the section by the plane x = x0 has no extreme at that point, or if it has the opposite sort of extreme to the section made by y = yo. EXAMPLE 1. The surface z = (y - x2) (y - 2 x2) has critical points where O=-6 y + 8 3=, =2y-3x2 =0; that is, the only critical point is (x = 0, y = 0). The tangent plane at that point is z = 0. This tangent plane cuts the surface where (y -2)(y-2x2) =0;: that is, along the two parabolas y = x2, y = 2x2. At x = 0, y = 1, the value of z is + 1; hence z is positive for points (x, y) inside a- 2X^2 2 the parabola y = 2 x2. At x = 1, y = 0, the value of z is + 2; hence z is positive for all points -- + l (x, y) outside the parabola y = + l 1 % -1- + x2. At the point x = 1, y = 1.5, the value of z is -.25; hence z is negative between the two paoI i rabolas. It is evident, therefore, 4- + that z has no extreme at x = 0, FIG. 83. Y = 0. A qualitative model of this extremely interesting surface can be made quickly by molding putty or plaster of paris in elevations in the * For a detailed discussion, see Goursat-Hedrick, Mathematical Analysis, Vol. I, p. 118. XVIII, _~ 165] CURVED SURFACES 295 unshaded regions indicated above, with a depression in the shaded portion. Another interesting fact is that every vertical section of this surface through (0, 0) has a minimum at (0, 0); this fact shows that the rule about vertical sections stated above cannot be reversed. Moreover, this surface eludes every other known elementary test except that used above. EXERCISES Find the equation of the tangent plane to each of the following surfaces at the point specified: 1. = x2 + 9 y2, (2, 1, 13). Ans. z 4 x + 18y- 13. 2. z = 2 x2-4y2, (3, 2, 2). Ans. z 12 x - 16 y - 2. 3. z = xy, (2, -3, -6). Ans. 3 x-2 y + = 6. 4. z= (x y)2, (1, 1, 4). Ans. 4x + 4y - z =4. 5. z = 2 xy2 + y3, (2, 0, 0) Ans. z = 0. 6-10. The straight line perpendicular to the tangent plane at its point of tangency is called the normal to the surface. Find the normal to each of the surfaces in Exs. 1-5, at the point specified. 11. At what angle does the plane x + 2y - z 3 =0 cut the paraboloid x2 + y2 = 4 z at the point (6, 8, 25)? 12. Find the angle between the surfaces of Exs. 1 and 2 at the point (-13, 1, 22). Find the angle between each pair of surfaces in Exs. 1- 5, at some one of their points of intersection, if they intersect. 13. Find the tangent plane to the sphere x2 + y2 + z2 = 25 at the point (3, 4, 0); at (2, 4, V5). 14. At what angles does the line x = 2 y = 3 z cut the paraboloid y =x2 + 2? 15. Find a point at which the tangent plane to the surface 1 is horizontal. Draw the contour lines of the surface near that point and show whether the point is a minimum or a maximum or neither. 296 THE CALCULUS [XVIII, ~ 165 16. Proceed as in Ex. 15 for each of the surfaces of Exs. 2-5, and verify the following facts: (2.) Horizontal tangent plane at (0, 0); no extreme. (3.) Horizontal tangent plane at (0, 0); no extreme. (4.) Horizontal tangent plane at every point on the line x + y = 0; weak minimum at each point. (5.) Horizontal tangent plane at every point where y = 0; no extreme at any point. Find the extremes, if any, on each of the following surfaces: 17. z = x2+ 4y2- 4 x. (Minimum at (2, 0, - 4).) 18. z=x3 - 3 x- y2. (See Tables, Fig. I1.) 19. z = 3- 3 x + y2 (x - 4). (See Tables, Fig. I2.) 20. z = [(x - a)2 + y2] [(x + a)2 + y2]. (Similar to Tables, Fig. I7.) 21. Z=x3 - 6 x - y2. (Draw auxiliary curve as for Fig. Ii.) 22. z = X - 4 y2 + xy2. (Draw auxiliary curve as for Fig. I2.) 23. z = x3 + y3- 3 xy. (Draw by rotating xy-plane through 7r/4.) 24. Redetermine the values of a and (3 in Example 2, ~ 164, if the additional information (p = 23, w = 135) is given. 25. Find the values of u and v for which the expression (alu + b1v - cl)2 + (a2u + b2 v - c2)2 + (a3u + bv - C3)2 becomes a minimum. (Compare Ex. 24.) 26. Show that the most economical rectangular covered box is cubical. 27. Show that the rectangular parallelopiped of greatest volume that can be inscribed in a sphere is a cube. [HINT. The equation of the sphere is x2 + y2 + z2 = 1; one corner of the parallelopiped is at (x, y, z); then V= 8 xyz, where z = i/ - x2 - y2.] 28. Show that the greatest rectangular parallelopiped which can be inscribed in an ellipsoid x2/a2 + y2/b2 + Z2/c2 = 1 has a volume V = 8 abc/(3 /V3). 29. The points (2, 4), (6, 7), (10, 9) do not lie on a straight line. Under the assumptions of ~ 164, show that the best compromise for a straight line which is experimentally determined by these values is 24y = 15x + 70. XVIII, ~ 166] CURVED SURFACES 297 30. The linear extension E (in inches) of a copper wire stretched by a load W (in pounds) was found by experiment (Gibson) to be (W = 10, E =.06), (W = 30, E =.17),. (W =60, E =.32). Find values of a and / in the formula E = a W + /3 under the assumptions of ~ 164. 31. The readings of a standard gas meter S and that of a meter T being tested were found to be (T = 4300, S = 500), (T = 4390, S = 600), (T = 4475, S = 700). Find the most probable values in the equation T = aS + 3 and explain the meaning of a and of /. 32. The temperatures 0~ C. at a depth d in feet below the surface of the ground in a mine were found to be d = 100 ft., 0 = 15~.7, d = 200 ft., 0 = 16~.5, d = 300 ft., 0 = 17~.4. Find an expression for the temperature at any depth. 33. The points (10, 3.1), (3.3, 1.6), (1.25,.7) lie very nearly on a curve of the form a/x + -/y = 1. Use the reciprocals of the given values to find the most probable values of a and /3. 34. The sizes of boiler flues and pressures under which they collapsed were found by Clark to be (d = 30, p = 76), (d = 40, p = 45), (d = 50, p = 30). These values satisfy very nearly an equation of the form p = k. ds or log p = n log d + log k, where d is the diameter in inches, and p is the pressure in pounds per square inch. Using the logarithms of the given numbers, find the most probable values for n and log k. 166. Tangent Planes. Implicit Forms. If the equation of a surface is given in implicit form, F(x, y, z) = 0, taking the total differential we find: OF dx d+ F dz0 (1) -d + adz + d = o. ax y az But, by virtue of F (x, y, z) = 0, any one of the variables, say z, is a function of the other two; hence (2) dz = adx+ -- dy. Putting this in the total differential above and rearranging: a/F aF + dz\ + z dy =., (3) + Fz + ) ( + OFyF = ~ ax z Vx ay aOz y 298 THE CALCULUS [XVIII, ~ 166 But dx and dy are independent arbitrary increments of x and of y; and since the equation is to hold for all their possible pairs of values, the coefficients of dx and dy must vanish separately. This gives (4) az O F/ax az aF/dy (4) Ax aF/Oz' ay F/Oz Substituting these values in the equation of the tangent plane, and clearing of fractions, we obtain (5) F (x- x) )+ ] ( - Y) + -]z - ZO) )=0, the equation of the tangent plane at (xo,,, z) to the surface F (x, y, z) =0. 167. Line Normal to a Surface. The direction cosines of the tangent plane to a surface whose equation is given in the explicit form z = f (x, y) are proportional (~ 163) to (1) Oz/Ox]o, az/Oy]o, and - 1. Hence the equations of the normal at (xo, yo, z0) are x-x0 y-yo z-Z0 (2) (2) Oaz/x]o oz/Oy]o - 1 The direction cosines of a surface whose equation is given in the implicit form F (x, y, z) = 0 are proportional to (3) aF/Ox]o, aF/Oy]o, OF/Oz]o, so that the equations of the normal to this surface are x-x0 y-yo _ z-zo (4) (4) d F/Ox]o F/dy]o 0 F/Oz]o 168. Parametric Forms of Equations. A surface S may also be represented by expressing the coordinates of any point on it in terms of two auxiliary variables or parameters: [S] x =f (u, v), y = (u, v), z = t (u, v). XVIII, ~ 168] CURVED SURFACES 299 If we eliminate u and v between these equations, we obtain the equation of the surface in the form F (x, y, z) = 0. Similarly a curve C may be represented by giving x, y, z in terms of a single auxiliary variable or parameter t: [C] x=f (t), y= (t), z = (t). The elimination of t from each of two pairs of these equations gives the equations of two surfaces on each of which the curve lies. In particular, taking t = x gives the curve as the intersection of the projecting cylinders: [P] y = (x), z = (x). If, in the parametric equations of a surface, one parameter (say u) is kept fixed while the other varies, a space-curve is described which lies on the surface. Now if u varies, this curve varies as a whole and describes the surface. The curve on which u keeps the value k is called the curve ut = k. Similarly, keeping v fixed while u varies gives a curve v = k'. The intersection of a curve u = k with a curve v =k' gives one or more points (k, k') on the surface. The numbers k, k' are called the curvilinear coordinates of points on the surface. Simple examples of such coordinates are the ordinary rectangular coordinate system and the polar coordinate system in a plane. Thus (2, 3) means the point at the intersection of the lines x = 2, y = 3 of the plane; in polar coordinates, (5, 30~) means the point at the intersection of the circle r = 5 with the line 0 = 300. EXAMPLE 1. The equations of the plane x + y + z = 1 may be written, in the parametric form: x =, y =v, z — u- v. Let the student draw a figure from these equations by inserting arbitrary values of u and v and finding associated values of x, y, z. Another set of parameter equations which represent the same plane is x = u +v, y = u-v, z= - -2 u+-1. Thus several different sets of parameter equations may represent the same surface. In the first form, put u = k. Then, as v varies, we obtain the straight line x =k, y =v, z= 1- k - v, 300 THE CALCULUS [XVIII, ~ 168 which lies in the given plane. As k varies this line varies; its different positions map out the entire plane. Likewise, v = k' is a line varying with k' and describing the plane. The intersection of two of these lines, one from each system, is point (k, k') of the plane. EXAMPLE 2. The sphere x2 + y2 + z2 = a2 may be represented by the equations: x =a sin 4 cos 0, y=a sin 4 sin 0, z=a cos 4. Here the parameters 4 and 0 are respectively the co-latitude and the longitude. Thus 4 = k is a parallel of latitude; 6 = k' is a meridian; and their intersection (k, k') is a point of co-latitude k and longitude k'. [If a is allowed to vary, the equations of this example define polar coordinates in space; but 90~ - 0 is often used in place of 0.] EXAMPLE 3. The equations x = a cos t, y =a sin t, z = bt, represent a space curve, namely a helix drawn on a cylinder of radius a with its axis along the z-axis. The total rise of the curve during each revolution is 2 rb. If a is replaced by a variable parameter u, the helix varies with u, and describes the surface x = u cos t, y = u sin t, z = bt, which is called a helicoid. The blade of a propeller screw is a piece of such a surface. 169. Tangent Planes and Normals. Parameter Forms. When a surface is given by means of parametric equations, (1) x =f (u, v), y = 0 (u, v), z = 4 (u, v), the equation of the tangent plane is found as follows. Elimination of u and v would give the equation in the implicit form F (x, y, z) = 0. If the parametric values of x, y, z are substituted in this equation, the resulting equation is identically true, since it must hold for all values of the independent parameters u, v; hence OF OF (2) - = 0, and =0, au -- that is OF Ox OF Oy OF Oz OF x OF Oy F z O (3) +- +- -O +- - = Oz a-x u ay au az au ax dv + y d v + d v = XVIII, ~ 1691 CURVED SURFACES 301 Solving these, we find: ay az az ax Odx Oy (aFOF OF a au u au au au (4) ax ay az ay az az ax aOxay av av av Ov Ov Ov hence the equation of the tangent plane is ay az azx a xax ay aua~u aucau auadu (x- x0) u+ (y - Yo) + (z - zo) a a 0; ay az __9X ax ay av a v aOv av0 while the equations of the normal are x-x0 Y - Yo z - Z ay az az ax ax ay au au au au au au ay az az ax ax ay av av 0 av av.0 av v I0 EXERCISES 1. Determine the tangent plane and the normal to the ellipsoid x2 + 4 y2 + z2 = 36 at the point (4, 2, 2), first by solving for z, by the methods of ~ 163; then, without solving for z, by the methods of ~~ 166- 167. Determine the tangent planes and the normals to each of the following surfaces, at the points specified: 2. X2 + y2 + Z2 = a2 at (xo, o, yz). 3. X2- 4 y + 2 = 36 at (6, 1, 2). 4. X2 - 4 y2 - 9 Z2 = 36 at (7, 1, 1). 5. X2 + y2 -Z2 = -aOt (3, 4, 5). 6. X3 + X2y - 2 Z2 = O at (1, 1, - i). 7. z2 = ex+y at (0, 2, c). 8. Find the angle between the tangent planes to the ellipsoid 4 x2 + 9 y2 + 36 z2 = 36 at the points (2, 1, zo) and (, - 1, 21). 9. At what angle does the z-axis cut the surface Z2 = e —Y? 302 THE CALCULUS [XVIII, ~ 169 10. Obtain the equation of the tangent plane to the helicoid x = ucos v, y = u sin v, z =v, at the point u = 1, v = 7r/4. 11. Taking the equations of a sphere in terms of the latitude and longitude (Example 2, ~ 168), find the equation of its tangent plane and the equations of the normal at a point where 0 = 4 = 45~; at a point where 0 = 60~, A = 30~. 12. Eliminate u and v from the equations x = u + v, y = u- v, z = uv, to obtain an equation in x, y, and z. Find the equation of the tangent plane at a point where u = 3, v = 2, by the methods of ~ 166; then by the methods of ~ 169 directly from the given equations. 13. Write the equation of the tangent plane to the surface used in Ex. 7 at any point (xo, y0, zo). At what point is the tangent plane horizontal? Is z an extreme at that, point? Proceed as in Ex. 7 for each of the following surfaces: 14. x= r cos, y=rsin 0, z =r, at r 2, 0 =/4. uv +1 u- v uv- 1 15. =, y =, at u = 2, v = - 1. u+4- U- v ' u +-v 16. x =-3u+2v, y = 2u- v, z = eU+, at (uo, ov). 17. x= 2 cos 0 cos ), y = 3 cos 8 sin 4, = sin 0, at0= =/4. 18. The surfaces z = x2 - 4 y2 and z = 6 x intersect in a curve, whose equations are the two given equations. Find the tangent line to this curve at the point (8, 2, 48) by first finding the tangent planes to each of the surfaces at that point; the line of intersection of these planes is the required line. 19. Find the tangent line to the curve defined by the two equations 16 x2- 3 y2 = 4 z and 9 x2 + 3 y2 - 2 = 20 at (1, 2, 1). 170. Area of a Curved Surface. Let S be a portion of a curved surface and R its projection on the xy-plane. In R take an element AxAy and on it erect a prism cutting an element AS out of S. At any point of AS, draw a tangent plane. The prism cuts from this an element 'AA. The smaller Ax Ay (and therefore AS) becomes, the more nearly will the ratio AA/AS approach unity, since we assume that the limit of this ratio is 1. XVIII, ~ 170] CURVED SURFACES 303 Suppose now that the area R is all divided up into elements AxAy and that on each a prism is erected. The area S will thus be divided up into elements AS and there will be cut from the tangent plane at a As point of each an element AA. One,, I thus gets (1) S = lim aA. Axy-O /But if y is the acute angle that / Y the normal to any AA makes with AtG the s-axis, we have FIG. 84. (2) AA = sec 7 Ax Ay; hence (3) S li = limA = lim (sec AxA) =ffsec d dy. Ax —m0 Ax-l X0" R Ay ---O Ay —O Of course sec 7 is a variable to be expressed in terms of x and y from the equation of the surface. The limits of integration to be inserted are the same as if the area of R were to be found by means of the integral f fdx dy. If the surface doubles back on itself, so that the projecting prisms cut it more than once, it will usually be best to calculate each piece separately. When the equation of the surface is given in the form z = f (x, y), the direction cosines of the normal are given by Oz Os cos a: cos: cos y: =:- - 1. 9x dy Taking cos r positive, that is y acute, we may write (4) see y = )2+ 1, and (I (z\2 (z\2 and S = wffJ a + 1 dx dy. The determination of sec y, when the surface is given in the form F(x, y, z) = O, is performed by straightforward transformations similar to those used in ~~ 167-169; they are left to the student. 304 THE CALCULUS [XVIII, ~ 170 EXERCISES 1. Calculate the area of a sphere by the preceding method. 2. A square hole is cut centrally through a sphere. How much of the spherical surface is removed? 3. A cylinder intersects a sphere so that an element of the cylinder coincides with a diameter of the sphere. If the diameter of the cylinder equals the radius of the sphere, what part of the spherical surface lies within the cylinder? 4. How much of the surface z = xy lies within the cylinder x2 + y2 = 1? 5. How much of the conical surface z2 = x2 + y2 lies above a square in the xy-plane whose center is the origin? 6. Show that if the region R of ~ 170 be referred to ordinary polar coordinates, AA = r sec - Ar A0, approximately. (See ~ 92, p. 149.) 7. Using the result of Ex. 6, show that S = ffr sec y dr do. 8. Show that, for a surface of revolution formed by revolving a curve whose equation is z = f (x) about the z-axis, sec Y = /1 + [df (r)/dr]2, where r = Vx2 + y2. 9. By means of Exs. 7, 8, show that the area of the surface of revolution mentioned in Ex. 8 is s= f X f r I + d[-df (r) ]2d d = 2 1 + [df (r)] dr a i t vl o r t t ed of t arc of t dra where a is the value of r at the end of the are of the generating curve. 10. Compute the area of a sphere by the method of Ex. 9. 11. Find the area of the portion of the paraboloid of revolution formed by revolving the curve z2 = 2 mx about the x axis, from x = 0 to x = k. 12. Show that the area of the surface of an ellipsoid of revolution is 2 irb [b + (a/e) sin-le], where a and b are the semiaxes and e the eccentricity, of the generating ellipse, 13. Show that the area generated by revolving one arch of a cycloid about its base is 64 7ra2/3. 14. Show that the area of the surface generated by revolving the curve x2/3 + y2/3 = a2/3 about one of the axes is 12 ira2/5. XVIII, ~ 172] CURVED SURFACES 305 171. Tangent to a Space Curve. Let the equation of the curve be given in parametric form x = f (t), y = 0 (t), z = p (t). Let Po = (xo, yo, zo) be the point on the curve where t = to. Let Q be a neighboring point on the curve where t = to + At. The direction cosines of the secant PoQ are proportional to Ax/At, Ay/At, Az/At; hence its equations are (1) xZ = Y- Yo _ - Z- z Ax/At Ay/At Az/At As At - 0, these become Q ) x-xo y-yo z- z_ Az dx/dt]o dy/dt]o dz/dt]o' O Ax y the equations of the tangent at the point Po. If the curve is given as the intersection of two projecting cylinders y =f(x), z = ((x), we may join to these the third equation x = x, thus conceiving of x, y, and z as all expressed in terms of x. The equations of the tangent then become (3) x- Xo _ Y-yo z - z0 1 dy/dx]o dz/dx]o If the curve is given as the intersection of two surfaces, f (x, y, z) = 0, F (x, y, z) =i0, and if we think of x, y, z as depending upon a parameter t, we find dfafdx- af dy f dz dt ax dt ay dt az dt dF _aFdx Fdy a F d and- + = 0. dt ax dt ay dt az dt From these equations we obtain dx/dt: dy/dt: dz/dt, and we may write the equations of the tangent at Po in the form: x- X Y — -o Z - S af f af af f of ay daz z ax ax ay OaFF aFaF aFaF |y dz o 2 da x o 0 x ay 0 172. Length of a Space Curve. The length of the chord joining two points t and t +At of the curve (1). x=f(t), y = (t), = (t), 306 THE CALCULUS [XVIII, ~ 172 is Ac = VAx2 + Ay2 + A2, or, ax2 Ay2 As22 (2) Ac = +A2 + aA<t. Defining the length of a curve between two points as the limit of the sum of the inscribed chords, we find for that length: (3) s =lim I Ac = =1 J - + ( - (ddt. AtO- dt EXERCI EXERCISES 1. At what angle does a straight line joining the earth's South pole with a point in 40~ North latitude cut the 40th parallel? 2. At what angle does the helix x = 2 cos 0, y = 2 sin 0, z = 0, cut the sphere x2 + y2 + z2 = 9? 3. Find the angle of intersection of the ellipse and parabola that are cut from the cone z2 = x2 + y2 by the planes 2 z = 1 - x and z = 1 + x respectively. 4. Show that the curves of intersection of the three surfaces z = y, z2 = y2 + z2, x2 + y 2 = 1, cut each other mutually at right angles. 5. Show the same for the curves of intersection of the surfaces 4 x2 + 9y2 + 36 z2 = 36, 3 2 + 6 y2 - 6 2 = 6, 10x2 - 15y2- 6 2 = 30. 6. Calculate the length of the curve x = t, y = t2, z = 2 t3/2, from t = 0 tot = 1. 7. Find the length of the helix x = a cos 0, y = a sin 0, z = b6, from 0 = 00 to 0 = 01. What is the length of one turn? 8. Find the length of the curve x = sin z, y = cos z, from (1, 0, 7r/2) to (0, - 1, 7r). GENERAL REVIEW EXERCISES [The exercises marked with an asterisk are of more than usual difficulty. Some of them contain new concepts of value for which it is hoped that time may be found. Those of the greatest theoretical value are marked t. Attention is called to the reviews of double and 'riple integration.] 1. Given u = xy, x = r cos 0, y = r sin 0, find ou/ar and au/8o, first by actually expressing u in terms of r and 0; then directly from the given equations. XVIII, ~ 172] CURVED SURFACES 307 2. Proceed as in Ex. 1 for the function u = tan-1 (y/x). 3. Given u = r2e20, x = r cos 0, y = r sin 0, find au/dx and au/ly, first by expressing u in terms of x and y; then directly from the given equations. [HINT. In the second part, it is convenient here to solve the last two equations for r and 0 in terms of x and y. But see Ex. 4.] 4.* If x = r cos 0 and y = r sin 0, show by differentiation that ax a = coso - r sin 0 and ay = 0 = r sin 0 + r cos 0 ax a1 cx I ax ax ax Solve these equations for ar/ax and ao0/x, and show that au/ax may be found in Ex. 3 by means of the equation au au ar au 0~ Ox Or ax a6 ax 5.* If, in general, u is a function of the two variables (r, 0), show that the last equation in Ex. 4 holds true. Find a similar equation for Ou/Oy, and evaluate au/dy in Ex. 3 by means of it. 6.*t If u is a function of any two variables p and q, and if p and q are given in terms of x and y by two equations x = f(p, q), y = 0(p, q), obtain au/ax and Ou/ay by a process analogous to that of Exs. 4, 5. Proceed as in Ex. 3, by the methods of Exs. 4, 5, in each of the following cases: 7. u = r2 - cos2. 8. u = re02. 9. u = 0 log r. 10. Find the volume of that portion of a sphere of radius 4 ft. which is bounded by two parallel planes at distances 2 ft. and 3 ft., respectively, from the center, on the same side of the center. 11. Determine the position of the center of mass of the solid described in Ex. 10. 12. What is the nature of the field of integration in the integral a/V2 a.a2-x2 / 2 f (x, y)dy dx? f0 f:x Show that the same integral may be written in the form a/Vi2 yx x a -fa2-y2 Jf (x f d(x, y) dy dx f f (x, y) dx dy. Jf 0 13. Find the volume cut from the sphere x2 + y2 + Z2 = a2 by the cylinder x2 + y2 - ax = 0. 308 THE CALCULUS [XVIII, ~ 172 14. Find the volume cut from the sphere x2 + y2 + z2 = a2 by the cone (x - a)2 + y2 - 2 = 0. 15. Show that the surface of a zone of a sphere depends only upon the radius of the sphere and the height b - a of the zone, where the bounding planes are z = a and z = b. 16. Find the area of that part of the surface k2z = xy within the cylinder x2 + y2 = k2. 17. Find the center of gravity of the portion of the surface described in Ex. 16, when k = 1. 18. Find the moment of inertia about its edge, of a wedge whose cross section, perpendicular to the edge, is a sector of a circle of radius 1 and angle 30~, if the length of the edge is 1, and the density is 1. 19. The thrust due to water flowing against an element of a surface is proportional to the area of the element and to the square of the component of the speed perpendicular to the element. Show that the total thrust on a cone whose axis lies in the direction of the flow is krr3v2/(r2 + h2)2. 20. Calculate the total thrust due to water flowing against a segment of a paraboloid of revolution whose axis lies in the direction of the flow. (See Ex. 19.) 21. Show that the thrust due to water flowing against a sphere is 2 krr2v2/3. Compare with the thrust due to the flow normally against a diametral plane of this sphere. 22. Find the critical points, if any exist, for the surface z = x2 + 2 y2 - 4 x - 4 y + 10. Is the value of z an extreme at that point? Draw the contour lines near the point. 23. Determine the greatest rectangular parallelopiped which can be inscribed in a sphere of radius a. 24. The volume of C02 dissolved in a given amount of water at temperature 0 is 0 5 10 15, v 1.80 1.45 1.18 1.00. Determine the most probable relation of the form v = a + b0. 25. Determine the most probable relation of the form S = a + b P2 from the data: P 550 650 750 850, S 26 35 52 70. XVIII, ~ 172] CURVED SURFACES 309 26. Determine the most probable relation of the form y = aebx from the data: x 1 2 3 4, y.74.27.16.04. 27. The barometric pressure P (inches) at height H (thousands feet) P 30 28 26 24 22 20 18 16, H 0 1.8 3.8 5.9 8.1 10.5 13.2 16.0. Determine the most probable values of the constants in each of the assumed relations: (a) H = a + b P; (b) H = a + bP + cP2; (c) H a + b log P or P = AeBH. Which is the best approximation? 28. t If the observed values of one quantity y are m,, m, ms, corresponding to values 11, 12, 13 of a quantity x on which y depends, and if y = ax + b, show that the sum S = (all + b - ml)2 + (a2 + b - m2)2 + (aa3 + b - m3)2 is least when 11 (all + b - mi) + 12 (al2 + b - m2) + 13 (al3 + b- m3) = 0, (all + b - ml ) + (al + b -m2) (a3 b - m) = 0; that is, when a * Z 12 + b 1 - l- mmll = 0 and a. l +3b — ml = 0, or 3 Z mnll - mml, 11 12 * M mi - 2' mill' * i 3a= Zb-[]2 l, b= ]2 where Z indicates the sum of such terms as that which follows it. [THEORY OF LEAST SQUARES.] 29. Show that the equation of the tangent plane to 2 z = x2 + y2 at (xo, yo) is z + so = XXo + YYo. 30. Determine the tangent plane and normal line to the hyperboloid x2 - 4 y2 + 9 S2 = 36 at the point (2, 1, 2). 31. Study the surface xyz = 1. Show that the volume included between any tangent plane and the coordinate planes is constant. 32. Study the surface z = (x2 + y2) (x2 + y2 - 1). Determine the extremes. 33. At what angle does a line through the origin and equally inclined to the positive axes cut the surface 2 z = x2 + y2? 310 THE CALCULUS [XVIII, ~ 172 34. Determine the tangent line and the normal plane at the point (1, 3/8, 5/8) on the curve of intersection of the surfaces x + y + z = 2 and 2 + 4 y2 - 4 z2 = 0. 35. Determine the tangent line and the normal plane to the curve x = 2 cos t, y = 2 sin t, z = t2 at t = 7r/2 and at t = r. 36. Find the length of one turn of the conical spiral x = t cos (a log t), y = t sin (a log t), z = bt, starting from t = t. 37. Determine the length of the curve x = a cos 0 cos k, y = a cos 0 sin 4, z = a sin 4, from 4 = 01 to 0 = 42, where 0 is given in terms of 4 by the equation 0 = k log ctn (7r/4 - 4/2). (Loxodrome on the sphere.) 38.*t Show that the surfaces f(x, y, z) = 0 and 4 (x, y, z) = 0 cut each other at right angles if fxOx + f, 4) +- f2zz, = 0. 39.* Show tha't the surfaces x2/(a2 + X) + y2/(b2 + X) + z2/(c2 + )) = 1, a > b > c > 0, are always (i) ellipsoids if X > - c2, (ii) hyperboloids of one sheet if - b2 < X < - c2, (iii) hyperboloids of two sheets if - a2 < X < - b2. (CONFOCAL QUADRICS.) Show also that these surfaces cut each other mutually at right angles. If x = r cos 0 cos 4, y = r cos 0 sin 4, z = r sin 0 (polar coordinates), find au/ar, au/ao, and au/af for each of the following functions: 40. u = x2 + y2 + z2. 41. = x2 + y2 - 2. 42. u = zez+v. 43. Compute au/ax, au/ay, and au/az if u = r2 (sin2 0 + sin2 4), where r, 0, 4 are defined as in Exs. 40-42. 44.t Show that the centroid (x, y) of a plane area in polar coordinates (p, 0) is ff p2 cos dpd f p2 sin Odp d ff pdp do ff dp d where the integrals are extended over the given area. CHAPTER XIX DIFFERENTIAL EQUATIONS PART I. ORDINARY DIFFERENTIAL EQUATIONS OF THE FIRST ORDER 173. Definitions. An equation involving derivatives or differentials is called a differential equation. An ordinary differential equation is one involving only total derivatives. A partial differential equation is one involving partial derivatives. The order of a differential equation is the order of the highest derivative present in it. The degree of a differential equation is the exponent of the highest power of the highest derivative, the equation having been made rational and integral in the derivatives which occur in it. EXAMPLES. (1) dg kt (First order, first degree.) dt ds d2s (2) d + d t2= ks (Second order, first degree.) (3) [1 + (dy) J = a(d 2) (Second order, second degree.) 82u 032U (4) a + a2U= 1 (Second order, first degree.) v/ X2 -y2 Such equations constantly arise in the applications of mathematics to the physical sciences. Many simple examples have already been treated in the text. 311 312 THE CALCULUS [XIX, ~ 174 174. Elimination of Constants. Differential equations also arise in the elimination of arbitrary constants from an equation. EXAMPLE 1. Thus, if A and B are arbitrary constants, then equation y = Ax + B represents a straight line in the plane, and by a proper choice of A and B represents any line one pleases in the plane except a vertical line. One differentiation gives m = dy/dx = A, which represents all lines of slope A. A second differentiation gives (1) flexion = b = d2y/dx2 = 0, which represents all non-vertical lines in the plane, since all these and on other curves have a flexion identically zero. EXAMPLE 2. Any circle whose radius is a given constant r is represented by the equation (2) (x - A)2 + (y - B)2 = r2, from which A and B may be eliminated as in the preceding example. Differentiating once, (3) x-A + (y-B)y' =0, where y' = dy/dx. Differentiating again, (4) 1 + y/2 + (y -B)y" = 0, where y" = d2y/dx2. Solving (3) and- (4) for x - A and y - B and substituting these values into (2), so as to eliminate A and B, we find (5) (1 + y'2)3 = r2y,2. This says that every one of these circles, regardless of the position of its center, has the curvature 1/r, - a statement which absolutely characterizes these circles. In general, if (6) f(x, y, c1, C2, **, cn) = 0 is an equation involving x, y, and n independent arbitrary constants cl, c2, *., Cn, n differentiations in succession with regard to x give df df - 0 (7) dx dx2- dx these equations, together with (6), form a system of n + 1 XIX, ~ 175] DIFFERENTIAL EQUATIONS 313 equations from which the constants C1, c2, c, may be eliminated. The result is a differential equation of the nth order free from arbitrary constants, and of the form (8) O(x, y,y', y",..., y()) = 0. Equation (6) is called the primitive or the general solution of (8). The term general solution is used because it can be shown that all possible solutions of an ordinary differential equation of the nth order can be produced from any solution that involves n independent arbitrary constants, with the exception of certain so-called "singular solutions" not derivable from the one general solution (6). Thus, to solve an ordinary differential equation of the nth order is understood to mean to find a relation between the variables and n arbitrary constants. These latter are called the constants of integration. If, in the general solution, particular values are assigned to the constants of integration, a particular solution of the differential equation is obtained. 175. Integral Curves. An ordinary differential equation of the first order, (1) +(x, y, y') = 0, or y' =f (x, y), where y' = dy/dx, has a general solution involving one arbitrary constant c: (2) F (x, y, c) = 0. This represents a singly infinite c set or family of curves, there being C2 c in general one curve for each value of c. Any curve of the family can be singled out by assigning to c the 0 proper value. FIG. 86. 314 THE CALCULUS [XIX, ~ 175 The differential equation determines these curves by assigning, for each pair of values of x and y, that is, at each point of the plane, a value of the slope y' [= f (x, y)] of the particular curve going through that point. Thus the curves are outlined by the directions of their tangents in much the way that iron filings sprinkled over a glass plate arrange themselves in what seem to the eye to be curves when a magnet is placed beneath the glass. Straws on water in motion create the same optical illusion. A differential equation of the second order: (x, y, y', y") = O, or y" = f(x, y, y'), has a general solution involving two arbitrary constants, F(x, y, cl, c2) = 0. This represents a doubly infinite or two-parameter family of curves; for each constant, independently of the other, can have any value whatever. The extension of these concepts to equations of higher order is obvious. The curves which constitute the solutions are called the integral curves of the differential equation. EXERCISES Find the differential equations whose general solutions are the following, the c's denoting arbitrary constants: 1. 2 + y2 = c2. Ans. x + yy' =O. 2. x2 - y2 = CX. Ans. x2 +y2 =2 xyy'. 3. y = cex - I (sin x + cos x). Ans. y' = y + sin x. 4. y = x + c2. Ans. y = y'x + y2. 6. y = cx +f(c). Ans. y = y'x +f ('). 6. y = cle2x + c2e3x. Ans. y" - 5 y' + 6 y = O. 7. y = Cleax + c2eb Ans. y" - (a + b)y'+ aby = 0. 8. xy = c + c2x. Ans. x4y'2 = y'x + y. 9. y = (cl + x)e3x + c2ex. Ans. y" -4 y' + 3 y = 2 e3. XIX, ~ 1751 DIFFERENTIAL EQUATIONS 315 10. y = le6 + c22X + c3e3X. Ans. y"' -6 y" + 11 y' -6 y = O. 11. r = csin 0. Ans. r cos 0 = r' sin 0. 12. r = ecO. Ans. r log r = r'O. 13. Assuming the differential equation found in Ex. 1, indicate the values of y'(= - x/y) at a large number of points (x, y) by short straight-line segments through each point in the correct direction. Continue doing this at points distributed over the plane until a set of curves is outlined. Are these curves given in Ex. 1? 14. Proceed as in Ex. 13 for the equation y' = y/x. Do you recognize the set of curves? Can you prove that your guess is correct? 15. Draw a figure to illustrate the meaning of y' = x2. Find y. Generalize the problem to the case y' = f(x). 16. Find that curve of the set given in Ex. 1 which passes through (1, 2). Find its slope (value of y') at that point. Do these three values of (x, y, y') satisfy the differential equation given as the answer in No. 1? 17. Proceed as in Ex. 16 for the equation of Ex. 2. 18. Proceed as in Ex. 16 for the first equation of Ex. 15. 19. Find the differential equation of all circles having their centers at the origin. 20. Find the differential equation of all parabolas with given latus rectum and axes coincident with the x-axis. 21. Find the differential equation of all parabolas with axes falling in the x-axis. 22. Find the differential equation of a system of confocal ellipses. 23. Find the differential equation of a system of confocal hyperbolas. 24. Find the differential equation of the curves in which the subtangent equals the abscissa of the point of contact of the tangent. 25. A point is moving at each instant in a direction whose slope equals the abscissa of the point. Find the differential equation of all the possible paths. 26. Write the differential equation of linear motion with constant acceleration; of linear motion whose acceleration varies as the square of the displacement. The same for angular motion of rotation. 27. A bullet is fired from a gun. Write the differential equations which govern its motion, air resistance being neglected. How must these equations be modified, if air resistance is assumed proportional to velocity? 316 THE CALCULUS [XIX, ~ 176 176. General Statement. We shall now consider methods for solving differential equations. Since the most common properties of curves involve slope and curvature, and since in the theory of motion we deal constantly with speed and acceleration, the differential equations of The first and second orders are of prime importance. Ordinary differential equations of the first order and first degree have the form (1) M + N dy = O, or M dx + N dy = 0, where M and N are functions of x and y. No general method is known for solving all such differential equations in terms of elementary functions. We proceed to give some standard methods of solution in special cases. 177. Type I. Separation of Variables. It may happen that M involves x only, and N involves y only. The variables are then said to be separated and the primitive is found by direct integration: fMdx+fNdy=C, C being an arbitrary constant. EXAMPLE 1. A particle is falling through air such that the resistance is proportional to the speed. If the particle starts from rest, what is its speed at any time? Since acceleration is dv/dt, and since this is equal to g diminished by a term proportional to v, we have dvg - av. dv dt= g-av. Separating variables: dv - =dt. g - av Integrating: log (g - av) = t + k, aor g = e-a or - av = e-a(t+-). XIX, ~ 178] DIFFERENTIAL EQUATIONS 317 Since the particle starts from rest we have v = 0 when t = 0. Substituting these values in the last equation we have g = e-a; hence g - av = e-at-a = ge-at, or v = (g/a)(1 - e-at). EXAMPLE 2. Given (x2 + 1) (y + 1) dy + y2 dx = 0. Determine the relation between x and y. Separating: Y + 1dy + dx = 0. y2 X2 + d Integrating: log y - 1/y + log Vx2 + 1 = k. Let k = - log c, rearrange and combine terms; the result is log(cy Vx21) = /y or c y /x2 + 1 = el/. 178. Type II. Homogeneous Equations. When M and N are homogeneous * in x and y and of the same degree, the equation is said to be homogeneous. If we write the equation in the form dy M dx N' and make the substitution dy xdv Y =vx, dx=v+ dx we obtain a new equation in which the variables can be separated. EXAMPLE 1. (1) (y + y2) dx + (xy - x2) dy = 0, or (2) d=dy xy - y2 (2) d- - x - xy' Substituting as above: * Polynomials are homogeneous in x and y when each term is of the same degree. In general, f (x, y) is homogeneous if f (kx, ky) = knf(x, y) for some one value of n and for all values of k. 318 ~~~~~~THE CALCULUS [XIX, 178 (3) ~~~~~dv vx2 + v2x2 _ v+v2 (3) ~ v + xdjX X2-vx2 1- VI dv 2v2 ordx 1v -v dx separating variables, 27 dv d Integrating: - - -log V = log x + C. 2 v 2 Replacing v by y/x, ~- - 1log- = log x+C, Y 2 x~~ or logxy=- — 2 c y hence (4) xy= =-ly or xy = ke-x/y, where k =e2c CHECK: Differentiating both sides of (4) with respect to x, we find (5) ydx + xdy =ke/y[ y dx-x dy] dividing the two sides of (5) by the corresponding sides of (4) respectively (6) [y dx +xdy] -'xy ydx - xdy. Y2' show that (6) agrees with (1). EXERCISES Solve the following exercises by separating the variables;: 1. xdy+ydx=O. Ans. xy =c. 2. xV1 +y2 dx -yV/1 +X2 dy =O. A ns. /1 + X2 =V/1 + y2+ C. 3. sin 0dr + rcos 0dO =O. Ans. r sin6= c. 4. x-Vl +ydx = y-\l +xdy. Solve the following homogeneous equations 5. (x+y) dx +(x -y) dy =O. Ans.X2 + 2xy -y2 =c. 6. (X2~+y2) dx =2 xydy. Ans. x2 -y2=CX. 7. (3X2 - y2) dy =2 xydx. Ans. X2-y2=Cy3. 8. (X2 +2 xy -y2) dx = (X2 - 2xy -y2) dy. Ans. X2 +y2 C (X+ y). XIX, ~ 178] DIFFERENTIAL EQUATIONS 319 The following Exs. 9-18 are intended partially for practice in recognizing types: 9. V/1 - y2 dx + /1 - 2 dy = 0. Ans. sin-1 x + sin-1 y = c. 10. x3 dx + (3 xy + 2 y3) dy = 0. Ans. x2 + 2 y2 = c/2 + y2. 11. dy + y sin x dx = sin x dx. 12. r dO = tan 0 dr. 13. (y - 1) dx = (x + 1) dy. 14. y dx + (x-y) dy = 0. 15. x(1 + y2) dx = y (1 + x2) dy. 16. (9 X2 + y2) dx = 2 xy dy. 17. x = c. 18. d- y =x. dx dx 19. In Ex. 1 above, draw a figure to represent the direction of the integral curves at various points. Hence solve the equation geometrically. 20. A point moves so that the angle between the x-axis and the direction of the motion is always double the vectorial angle. Determine the possible paths. An. xy; C >. Ans. x2 + y2c;>O 21. Proceed as in Ex. 20 for a point moving so that its radius vector always makes equal angles with the direction of the motion and the x-axis. Ans. r = c sin 0. 22. The speed of a moving point varies jointly as the displacement and the sine of the time. Determine the displacement in terms of the time. Ans. s = ce-kcos t 23. Find the value of y if its logarithmic derivative with respect to x is x2. 24. Determine the curve whose subnormal is constant and which passes through the point (2, 5). 25. Determine the curve whose subtangent at any point (x, y) is (1 + x), and which passes through (0, 3). 26. Determine the curve passing through (5, 4) such that the length of the normal at any point (~ 30) equals the distance of the point from the origin. 320 THE CALCULUS [XIX, ~ 178 27. When a wheel is driven by a belt the tension at P and the angle 0 are connected by the equation dT/dO = kT, T AB d k being a known constant. ^pR \ \ If T1 = 200 lb., = 0.1, ^t` ) and ACB = 600, what is T2? If T1 = 500 lb., A \^ -. T2 = 550 lb., FIG. 87. and ACB = 90~, FIG. 87. what is k? 28. In a chemical reaction A is the quantity of active matter originally present, q the quantity of product at time t; these are related through the equation dq/dt = k (A - q). Express q as a function of it. 29. In a bimolecular chemical reaction the original amounts of active substances are A and B; the product q formed in time t is to be determined from the equation dq/dt = k(A -q) (B-q). Express q in terms of t. Consider the special case A = B 30. The differential equation of the adiabatic expansion of a gas is kpdv + v dp = 0. Show that p = c-k. Find c if k =.001, and v = 100 when p = 10. 31. The rectilinear motion of a particle under the action of a central force which varies as the inverse square of the distance from a fixed point is v dv/dt = k2/t2, where v is the speed and t the time. Express v in terms of t. 32. Solve Helmholtz's equation for the strength of an electric current, C = E/R- (L/R) (d C/dt), E, L and R being constants. If C = 0 when t = 0, show that C = (E/R) (1 - e-Rt/L). 179. Type III. Linear Equations. This name is applied to equations of the form (1) ^ p=dy dx XIX, ~ 179] DIFFERENTIAL EQUATIONS 321 where P and Q do not involve y, but may contain x. Its solution can be obtained by first finding a particular solution of the reduced equation, dx dx + Py = 0, where y is a new quantity introduced for convenience in what follows; and where Q is replaced by zero. In (1') the variables can be separated (see ~ 179), and we get - fP d y = e-fPX as a particular solution, the constant C of integration being given the particular value 0. If we make the substitution (2) y = v I y, where v is a function of x to be determined, the equation (1) becomes d6 dv v - + Y + P vy = Q, or v - + PY + Y-d = Q. The first term vanishes by (1') leaving y = Q, or dv = dx = [Qef dx. dx d y Hence v= fdx+c= f[QefPd dx+c and (3) y = vy = e-f x [QeJfP ] dx + c This equation expresses the solution of any linear equation. It should not be used as a formula; rather, the substitution (2) should be made in each example. 322 THE CALCULUS [XIX, ~ 179 EXAMPLE. Given -(1) q+- 3 x2y -= 5, the reduced equation in the new letter y = y/v is (1') dy + 3 x2y = 0, whence y = e-x. dx Hence the substitution y = v. y becomes -), ~dy dv ) y ' v dx dx and (1) takes the form [e- d _ 3 vx2e-3] + 3 x2 [ve-13] = x. This reduces, as we foresaw in general above, to the form dv dv x e- - = X5, or = X5 X dx ' dx ' whence v = -fx5 ex dx + c = I [X3 e3 _- e'] + c, or, returning by (2) to y: (3) y v = v e- 3 - - 1] + ce-x. CHECK. Differentiating both sides, (4) d x2- 3 x2 ce-~3; eliminating c by multiplying (3) by 3 x2 and adding to (4), dy + 3 2y = x5. The result (3) may also be obtained by direct substitution from (1). Sufficient practice in the direct solution, as in the preceding example, is strongly advised. 180. Equations Reducible to Linear Equations. Certain forms of equations may be reduced to linear equations by a proper change of variable. No general rule can be given, and the proper substitution is usually to be formed by trial. dy EXAMPLE. sec2 y - + 3 x2 tan y = X5. dx Letting tan y = z, we have dz d+ 3 22z = x, which is linear in z and has the same form as the example solved in ~ 179. XIX, ~ 180] DIFFERENTIAL EQUATIONS 323 The equation dy d + Py = Qyn, (n a constant.) called the extended linear equation is always reducible to the linear type by putting y'-~ = z. EXAMPLE. Given dx+ = xy3. Put = y2. dz dy dy dz Then 2 -3 or (1/2)y3 x-x ordx dx Thus - (1/2)y3 + x 3 dz z and - 2 - 2 x. dx x Here P = - fPdx =-2 log x, efPd = x-2; so that = = x (- d + c = -2x2 log + x2 = y2, and finally x2y2 (c - 2 log x) = 1. Check this result. EXERCISES Solve the following equations and check each answer. dy dy 1. Y- xy = e2/2. 3. - +y cos x = sin 2 x. dx- dx dy dy 2. + 3 x2 y =3 x5. 4. x + y = log x. dy dr. Y+ = y3. 7. - 2 rO = r223. 6. +Y = XY3. 8. xy2 - y3X2 9. cos2x -Y + y = tan x. 10. rd-= (1 +r2) sin. ds dy 11. -d s +t. 12. + y= e-. 13. dy - y dx = sin x dx. 14. sec 0 dr + (r - 1) d = 0. 324 THE CALCULUS [XIX, ~ 180 15. (x2 + 1) dy = (xy + k) dx. 16. x dy + y dx = xy2 log x dx. 17..: The equation of a variable electric current is L d- Ri = e, where L and R are constants of the circuit, i is the current, and e the electromotive force of the circuit. Calculate i in terms of t, 1~, if e is constant; 2~, if e = eo sin ot. Ans. 2~ i= eo sin (wt - O) + ce-R/L, p = arc tan (o LIR). -\/R2 q W-o2 L2 181. Other Methods. Non-linear Equations. A variety of other methods are given in treatises on Differential Equations; some of these are indicated among the exercises which follow. Noteworthy among these are the possibility of making advantageous substitutions; and- what amounts to a special type of substitution- the possibility of writing the given equation in the form of a total differential, dz = 0, where z is a known function of x and y which leads to the general solution z = constant (see Exs. 5-12, below). Equations not linear in y' may often be solved. If the given equation can be solved for y', several values of y' may be found, each of which constitutes a differential equation: the general solution of the given equation means the totality of all of the solutions of all of these new equations. EXERCISES Solve the following equations, using the indicated substitutions: 1. y2 dy + (y3 + x) dx = O. (Put v = y3.) 2. s dt -t ds = 2 s (t- s) dt. (Puts =tv.) 3. x dy- y dx = (x2 - y2) dy. (Put y = vx.) 4. u2 v2 (u dv + v du) = (v + v2) dv. (Put uv = x, v = y.) 6. Solve the equation (3 x2 + y) dx + (x + 3 y2) dy = 0. [HINT. If we put z = x3 + xy + y3, this equation reduces to dz = 0; for d-z = (Oz/ax)dx + (Oz/dy) dy. But dz = 0 gives z = const., hence XIX, ~ 181] DIFFERENTIAL EQUATIONS 325 x3 + xy + y3 = c is the general solution. Such an equation as that given in this example is called an exact differential equation.] 6. Solve the equation x dy - y dx = 0. [HINT. This equation can be solved by previous methods; but it is easier to divide both sides by x2 and notice that the resulting equation is d (y/x) = 0; hence the general solution is y/x = c. A factor which renders an equation exact (1/x2 in this example) is called an integrating factor.] 7. Solve the. equation (x2 + 2 xy2) dx + (2 x2 y + y2) dy = 0. [HINT. Put S = X3/3 + x2y2 + Y3/3.] 8. Solve the equation (s + t sin s) ds + (t - cos s) dt = 0. [HINT. Arrange: s ds + [t sin s ds - cos s dt] + t dt = 0; integrate this, knowing that the bracketed term is - d (t cos s).] 9. Solve the equation x dy - (y - x) dx = 0. [HINT. Arrange: [x dy - ydx] + x dx = 0; divide by x2, and compare Ex. 6.] 10. Show that [f (x) + 2 xy2] dx + [2 x2 y + 4 (y)] dy = 0 can always be solved by analogy to Ex. 7. 11. Show that [f (x) + y] dx - x dy can always be solved by analogy to Ex. 9. Solve (x2 + y) dx - x dy = O. Ans. x - y/x = c. 12. Solve the equation (r - tan 0) dO + (r sec 0 + tan 0) dr = 0. [HINT. Multiply both sides by the integrating factor cos 0; - sin 0 dO +- r dr + d (r sin 0) = 0; integrate term by term.] 13. When a family of curves crosses those of another family everywhere at right angles, the curves of either family are called the orthogonal trajectories of those of the other family. Find the orthogonal trajectories of the family of circles x2 + y2 = r2 [HINT. If the differential equation of the first family be dy/dx = f (x, y), then the differential equation of the orthogonal trajectories is dx/dy = -f (x, y), for any point of intersection (x, y) the slope of the curve of one system is the negative reciprocal of the slope of the curve of the other. In this example the differential equation of the given family is x dx + y dy = 0. It is evident that the differential equation of the orthogonal 326 THE CALCULUS [XIX, ~ 181 family is obtained by replacing dy and dx by - dx and dy, respectively; hence the desired equation is x dy - y dx = 0, whence the curves are y = cx, i.e. the family of all straight lines through the origin.] 14. Find the orthogonal trajectories of the exponential curves. y = ex +k. [HINT. The differential equation is dy/dx = e". The orthogonal family is defined by the equation dy/dx = - e, whence the trajectories are y = e-$ + c. Draw the figure.] Determine the orthogonal trajectories of the following families, and draw diagrams in illustration of each: 15. x+y = k. 18. x2 + y2 = 2 logx + c. 16. xy = k. 19. 2 x2 + y2 = c2. 17. y2 = 4 k (x + k). 20. x2 +y2= kx. PART, II. ORDINARY DIFFERENTIAL EQUATIONS OF THE SECOND ORDER 182. Special Types. We first consider some very special forms of equations of the second order that are most frequently used in the application of mathematics to physics, namely: [I] d- = ~ k2y [k = constant.] [II] A d2 + B Y + Cy = 0 [A, B, C, constants.] d2Y dY [III] A d + BdY + Cy = F (x). [A, B, C, constants.] These are all special forms of the general equation of the second order 0 (x, y, dy/dx, d2 y/dx2) = 0. [IV] We shall consider other special forms also, some of which include the above; namely, the cases that arise when one or more of the quantities x, y, dy/dx, are absent from the equation. (See ~ 186, p. 334.) XIX, ~ 1831 DIFFERENTIAL EQUATIONS 327 183. Type I. This type of equation arises in problems on motion in which the tangential acceleration d2 s/dt2 is proportional to the distance passed over: dt2 (1) d -s i a form which is equivalent to [I], written in the letters s and t. If we multiply both sides of this equation by the speed v = ds/dt and then integrate with respect to t, we obtain Cdsd2s C ds (2) J ~dt dt2 = 2dt but we know that fdsd2s dv 1 C 1, lds\2 ds d -2 dt= v- dt= v dv = 2 2 d + Jdtdt2 =J dt 2 '2 =dt + C and 4- k2s ds dt = J sds = i 2 + c'; hence (2) becomes* v2 1 (ds)2 C2 C (3) -- d - - 2(S2 + C1). Case 1. If the sign before k2 is +, (3) becomes (4) v = = k/2 + C1, whence d C1 = k dt+ C2, -\/S2 + C1 (5) log (s + V/s2 + C1) = kt C2, or, solving for s, (6) s = Aekt + Be-kt * This is often called the energy integral, for if we multiply through by the mass m, the expression mv2/2 on the left is precisely the kinetic energy of the body. 328 THE CALCULUS [XIX, ~ 183 where 2 A = eC2 and 2 B = - C1 e-C2 are two new arbitrary constants. By means of the hyperbolic functions sinh u = (eu - e-)/2 and cosh u (e + e-")/2 this result may also be written in the form (7) s = a sinh (kt) + b cosh (kt), where b + a = 2 A and b- a = 2B. Case 2. If the sign before k2 is -, C must be negative also, or else v is imaginary; hence we set C1 = - a2 and write (42) v = = a2 - s2, or /a 2ds =- kdt+C2, whence (52) sin-1 ) = kt + C2; or solving for s: (62) s = a sin (kt + C2) = A sin kt + B cos kt, where A = a cos C2 and B = a sin C2 are two new arbitrary constants. Equation (62) is the characteristic equation of simple harmonic motion; the amplitude of the motion is a, the period is 2 -r/k, and the phase is - C2/k. The differential equation (1) was first found in ~ 88, p. 155. We now see that the general simple harmonic motion (62) is the only possible motion in which the tangential acceleration is a negative constant times the distance from a fixed point; i.e. it is the only possible type of natural vibration under the assumptions of ~ 76, p. 125. XIX, ~ 184] DIFFERENTIAL EQUATIONS 329 EXERCISES Solve each of the following equations: 1. d s 3 d Ss d2s d2s 1. d —~=s. 3. ~-= —s. 2. =4 s. 4. d=-t 9 s. 5. Find the curves for which the flexion (d2y/dx2) is proportional to the ordinate (y). 6. Determine the motion described by the equation of Ex. 1 if the speed v (= ds/dt) and the distance traversed s are both zero when t = 0. 7. Proceed as in Ex. 6 for Ex. 3, and explain your result. 8. Write the solution of Ex. 1 in terms of sinh t and cosh t. Determine the arbitrary constants by the conditions of Ex. 6, and show that the final answer agrees precisely with that of Ex. 6. 9. Determine the motion described by the equation of Ex. 3 if v = 2 and s = 10 when t = 0; if v = 0 and s = 5 when t = 0. 184. Type II. Homogeneous Linear Equations of the Second Order with Constant Coefficients. The form of this equation is (1) A dX + B -+ Cy =, where A, B, C are constants. The type just considered is a special case of this one. Following the indications of the results we obtained in ~ 183, it is natural to ask whether there are solutions of any one of the types we found in the special case. Trial of ekx. If we substitute y = etx in (1) we obtain the equation: (2) [Ak2 + Bk + C] ex = O. The factor ekx is never zero; hence k must satisfy the quadratic equation (1*) Ak2 + Bk + C =, 330 THE CALCULUS [XIX, ~ 184 which is called the auxiliary equation to (1). If the roots of (1*) are real and distinct, i.e. if (3) D —B2- 4 AC>O, then these roots ki and k2 are possible values for k, and the general solution of (1) is (4) y = Ceklx + C 2ek2x, since a trial is sufficient to convince one that the sum of two solutions of (1) is also a solution of (1); and that a constant times a solution is also a solution. Trial of y = eKX. v. If (3) is not satisfied, the substitution (5) y = e.v changes (1) to the form d2v dv (6) A + [2KA- B] - +[AK2+ BK C]v=0, which becomes quite simple if we determine K so that the term in dv/dx is zero: (7) 2 KA + B = 0, whence K =-B/2 A; then (6) takes the form d2v B2 -4AC (8) V K2 V dx2 4A2 v -Kv, where K = 4 AC-B2/(2 A) = -D/2 A is real if (9) D —B2 - 4 AC< O, which is the case we could not solve before. If D < 0, the solutions of (8) are (10) v = C1 sin (Kx) + C2 cos (Kx), by (62), ~ 183, p. 328; hence the solutions of (1) are (11) y = eKX. v = eKX [C1 sin (Kx) + C2 cos (Kx)], where K = -B/(2 A) and K = /-D/(2 A); these values of XIX, ~ 184] DIFFERENTIAL EQUATIONS 331 K and K are most readily found by solving (1 *) for k, since the solutions of (1*) are k=(-B+- Vb)/(2 A)= K K-1. If D = 0, K = 0, and the solutions of (8) are (12) V = CIx~+ C2; hence the solutions of (1) are (13) y = eKX.v = eKX [Cix + C2], where K = - B/(2 A) is the solution of (1 ); since when D = O, (1*) has only one root= - B/(2A). It follows that the solutions of (1) are surely of one of the three forms (4), (11), (13), according as D = B2 - 4 AC is -, or 0; that is, according as the roots of the auxiliary equation (1*) are real and distinct, imaginary, or equal; in resume: D =B'- 4 AC CHARACTER OF VALUES OF ROOTS SOLUTION OF (1) ROOTS OF (1*) OF (1*) + Real, unequal ki, k2 (4) - Imaginary K ~K K-1 (11) o Equal K (13) We illustrate by some examples put, for convenience of comparison, in tabular form. EXAMPLES 2 3 Equation (1) 3y"-4y'+y=O 3y"/-4y'+4 y-o 3y/-4y+2y =0 Auxiliary equa- 3 k2J-4k+1=0 3k2-4 k+ =0 3 k2-4k+2=0 tion (1*) _ Roots of (1*) 1,1/3 2/3, 2/3 1 (2 ~ 2j y = 62./3 (Cl COS Solution of (1) y =C1ez6'Cez/3 C y=e2x2/a3(C++C2X) V/2 -3-x+c2sin x) 3 332 THE CALCULUS [XIX, ~ 184 EXERCISES 1. y" - 4y' +3 y = 0. 9. y" - 9 y' + 14 y = 0. 2. y"+ 3 y'+ 2 y =O. 10. 2y"- 3y'+y=0. 3. 5y" -4y'+y =0. 11. 6 y" -13y'+6 y =0. 4. 9y" + 12y'+ 4 y =0. 12. y" - 3 y' = 0. 5. y"- 2 y' y =0. 13. y"- 4 y =0. 6. y"+ y'+y = O. 14.. y"+ 9 y = 0. 7. y"-2y'+3y=O. 15. y"+ky'=O. 8. 3y" +5y'+2y =. 16. y" + ky = 0. 17. If a particle is acted on by a force that varies as the distance and by a resistance proportional to its speed, the differential equation of its motion is d2x/dt2 + b dx/dt + cx = 0, where c > 0 if the force attracts, and c < 0 if the force repels. Solve the equation in each case. 18. If in Ex. 17, b = c = 1, and the particle starts from rest at a distance 1, determine its distance and speed at any time t. Is the motion oscillatory? If so, what is the period? Solve when the initial speed is vo. 19. If in Ex. 17, b = 1 and c = - 1, discuss the motion as in Ex. 18. 185. Type III. Non-homogeneous Equations. This type is of the form: (1) A 2dY+ B dY + Cy = F (x), where A B C, are constants, and F (x) is a function of x only. where A, B, C, are constants, and F (x) is a function of x only. We proceed to show that this form can be solved in a manner exactly analogous to ~ 179, p. 320. First write down the reduced equation in the new letter v: (1) A dx2 + B dv Cv = 0, dX2 d XIX, ~ 185] DIFFERENTIAL EQUATIONS 333 and solve (1*) by the method of ~ 184. Let v = < (x) be any one particular solution of (1*) (the simpler, the better, except that v = 0 is excluded). Then the substitution (2) y = u transforms (1) into (3) (Av" + Bv' + Cv) u + (2 Av' + Bv) d + Ad2 = F x); but, since v satisfies (1*), the first term of (3) is zero; and if we now set du/dx = w temporarily, this equation can be written as the linear equation: dw i 2 Av'+ Bv F (x) (4) dx - Av - A ' which is precisely of the form solved in ~ 179. Comparing (4) with (1), ~ 179, we have (5) P= Av A Av ' Q= Av Having found w by ~ 179, we have u = wdx + C2, y u = v[ wdx +c ], which is the required solution of (1). EXAMPLE 1. Given the equation (1) dgY2 + 3 dd + 2 y = sin x, we write the reduced equation d2v dv (1*) dx2 + 3 d- + 2 v =0; this is easily solved by the method of ~ 184; the simplest particular solution is v = e-. Substituting v = e- in the general work above, we find 2 A' + Bv F (x) P = - 1 and Q = = ex sin x; Av Av 334 THE CALCULUS [XIX, ~ 186 hence e-fPd- = ex, and w= e-x [fe2x sin x dx ~ Ci] = ex (2 sin x - cos x) + Cie-x, f ~~~10 u U.f dx~C =me (sin x - 33 cos x) - Cie2 x + C2,-z 10 EXERCISES 1. y"'- 3y'+ 2 y=cos x. Ans. y =I (cos x- 3sin x) + clx + c2e~ 2. y"/-4y'+2y=x. Ans. y = 1 (x + 2) + cie(2+VS~)x + c2e(2420)x. 3. Y" 3'2~x Ans. y =ex/6 -clex+ C2e-x. 4. y" -2y' + y= X. Ans. y = x+2 +ex (el +c2x). 5.y+ =ix Ans. y= - I x cosx + cisin X+ C2 COS X6. y"-y'- 2y = sin x. Ans. y (cos x - 3 sin x) + Cle-x + Cie2z. 7. y"+ 4 y=X2 +COS X. Ans. y = 18(22 1) +cosx +cel cos2 x+ c2 sin 2x. 8.y~ y' = e2x +1. Ans. y= 1x (e2x 1+el + C2enx. 9. Y1-y+ y 23T Ans. y=xe3 + Cle + C2e3x. 10. If a particle moves under the action of a periodic force through a medium resisting as the speed, the equation of motion is d2s/dt2 + Ads/dt =B sin C t. Express s and the speed in terms of t. If A = B= C = 1, what is the distance passed over and the speed after 5 seconds, the particle starting from rest? 186. Type IV. One of the quantities x, y, y' absent. Type IVa': q (Y") = 0. Solve for y", to obtain -a solution, say y" = a. Then integrate twice. The general solution for each value of y" is of the form y - ' ax2 + ClX + C2. XIX, ~ 186] DIFFERENTIAL EQUATIONS 335 In problems of motion, this type is equivalent to the statement that (jT) = 0, where jT = d2s/dt2 = dv/dt. Hence jT may have any one of the several constant values which satisfy ( (jT) = 0; but if jT = k, s = kt2/2 + cit + c2 (see Ex. 24, p. 64). Type IVb: y missing. 4(x, y', y") = 0. The substitution m = y' = dy/dx, dm/dx = d2y/dx2 = y", reduces the given equation to an equation of the first order in m, x, dm/dx. Solving, if possible, one gets a relation of the form f (m, x, c) = 0. This is again an equation of the first order in x and y, and may be integrated by methods given in Part I, ~~ 177-181. The interpretation in motion problems is particularly vivid and beautiful. Thus v = ds/dt and T = dv/dt = d2s/dt2; hence any equation in jT, v, t, with s absent, is a differential equation of the first order in v. Solving this, we get an equation in v and t; since v = ds/dt, this new equation is of the first order in s and t. EXAMPLE 1. 1 + x + x2 dy= 0 dx2 Setting dy/dx = m, 1 + x + x2 d= dx Separating variables, -dm -X dx Integrating, - m = - - + log x + cl. Integrating again, y = log x - x log x + (1 - cl) x - 2. Interpret this as a problem in motion, with s and t in place of y and x, and jr = dv/dt = d2s/dt2. EXAMPLE 2. In a certain motion the space passed over s, the speed v, and the acceleration jT are connected with the time by the relation 1 + v2 - jT = 0; find s in terms of t. Placing jT = dv/dt, the equation 1 + V2-d- = O is of the first order. The variables can be separated, and the integral is tan-1 v = t + ci or v = tan (t + cl), 336 THE CALCULUS [XIX; ~ 186 which is itself a differential equation of the first order if we replace v by ds/dt. Integrating this new equation: Jds =ftan (t + el) dt + c2, ors - log cos (t + Cl) + C2. In such a motion problem we usually know the- values of v and s for some value of t. If v = 0 and s = 10 when t = 0, for example, cl must be zero (or else a multiple of 7r) and c2 must be 10; hence s = - log cos t + 10. EXAMPLE 3y d2y dy EXAMPLE 3. 1 +- X +x x2 = 0 = 1 + xm + x2 dx dx2 dx This can be written dm/dx + mix = - 1/2, which is linear in m and x, the solution being 1 ci m = —logx +-. x X The second integration gives y = - 2 [log x]2 +ce log x + C2. Interpret this as a motion problem, and determine cl and C2 to make y = 10 and m = 3 when x = 1. Type IVc: x missing. C(y, y', y") = O. The substitution m = y' gives I,, dy' dy' dy dm y =m, Y l = - '. -d=- m; dx dy dx dym; and the transformed equation is an equation of the first. order in y and m. We solve this and then restore y' in place of m, whereupon we have left to solve another equation (in x and y) of the first order. This is precisely the way in which we solved Type I, ~ 183, Type I being only an important special case of Type IV. EXAMPLE 1. If the acceleration jT is given in terms of the distance passed over (compare ~ 188), we have d2s dv JT = (= s (), or - = (s). This is transformed by the relation dv dv ds dv JT dt = dsdt dsv XIX, ~ 186] DIFFERENTIAL EQUATIONS 337 (which is itself a most valuable formula) into dv v = v (S) in which the variables can be separated; integration gives 1V2 =f (s)ds+c, which is called the energy integral (see footnote, p. 327). The work cannot be carried further than this without knowing an exact expression for ~ (s). When p (s) is given, we proceed as in ~ 183, replacing v by ds/dt and integrating the new equation: ds =t +k..\/2Jf,(s)ds 2 c Unfortunately the indicated integrations are difficult in many cases; often they can be performed by means of a table of integrals. One case in which the integrations are comparatively easy is that already done in ~ 183. EXERCISES 1. y"2_- 4 2 = 0. Ans. y = + 1/3 x3 + cl + 2. 2. y" = 1 +y2 Ans. 2 y =cle + e-/cl + c2. 3. xy" + y' = x2. Ans. y = x3/9 + C1 log x + c2. 4. s d2s/dt2 + ds/dt = 1. Ans. s2 t2 + Ct + C2. _ d2S 1. 6. ds 6. = +k2y 7 d = e2 8.d =e 1 8.2- d2~ 9. d= 2 cos. 10. x + 3 sin x. 9. dy co x, = 11. d4 = e-cs2x 12. d2Y [ + (dY) 2]3 1.= dx ' 13. Show that Ex. 12 is equivalent to the problem, to find a curve whose radius of curvature is unity. 14. The flexion (d2y/dx2) of a beam rigidly embedded at one end, and loaded.at the other end, which is unsupported, is k (I - x), where k is a constant and I is the length of the beam. Find y, and determine the constants of integration from the fact that y = 0 and dy/dx = 0 at the embedded end, where x = 0. 338 THE CALCULUS [XIX, ~ 186 15. Find the form of a uniformly loaded beam of length 1, embedded at one end only, if the flexion is proportional to 12 - 2 Ix + x2, where x = 0 at the embedded end. 16. Find the form of a uniformly loaded beam of length 1, freely supported at both ends, if the flexion is proportional to 12 - 4 x2 in each half, where x is measured horizontally from the center of the beam. PART III. GENERALIZATIONS 187. Ordinary Equations of Higher Order. An equation whose order is greater than two is called an equation of higher order; the reason for this is the comparative rarity in applications of equations above the second order. We shall state briefly the generalizations to equations of higher order, however, since they do occur in a few problems, and since it is interesting to know that pratically the same rules apply in certain types for higher orders as those we found for order two. 188. Linear Homogeneous Type. The work of ~ 184 can be generalized to any linear homogeneous equation with constant coefficients: dy \dn-ly dy (1) d ald + * * + an-l d + any = O. dx' dx'' dx Thus if we set y = etx, as in ~ 184, we find (1*) kn + alkn-1 + * + an- k + an = 0, again called the auxiliary equation. Corresponding to any real root k1 there is therefore a solution ek'x; if all the roots are real and distinct, the general solution of (1) is (2) y = Cleklx + C2 ek2 +... + Cn eknx, where k1, k2, * *, kn are the roots of (1). Curiously enough, the chief difficulty is not in any operation of the Calculus; rather it is in solving the algebraic equation (1*). XIX, ~ 188] DIFFERENTIAL EQUATIONS 339 It is easy to show by extensions of the methods of ~ 184 that any pair of imaginary roots of (1*), k = K dt K /- 1 corresponds to a solution of the formt (3) y = eKx [C' sin (Kx) + C" cos (Kx)], which then takes the place of two of the terms of (2). Finally, if a root k = K of (1*) occurs more than once, i.e. if the left-hand side of (1*) has a factor (k - K), the corresponding solution obtained as above should be multiplied by the polynomial (4) Bo + Blx + B2x2 + * * + BpxPwhere p is the order of multiplicity of the root (i.e. the exponent of (k - K)P), and where the B's are arbitrary constants which replace those lost from (2) by the condensation of several terms into one. The proof is most easily effected by making the substitution y = e" * u, whereupon the transformed differential equation contains no derivative below dPu/dxP; hence u = the polynomial (4) is a solution of the new equation, and y = eKz times the polynomial (4) is a solution of (1). This work may be carried out by the student in any example below in which (1*) has multiple roots.t t This fact is often made plausible by the use of the equations eU'q-1 = cos u + v/- 1 sin u, e-U~-1F = cos u - V/- 1 sin u; these equations can be derived formally by using the Taylor series for e", cos u, sin u, with v = u /- 1, but they remain only plausible until after a study of the theory of imaginary numbers. The solutions eK K-l are indicated formally by (2); hence it is plausible that (3) is correct. A more direct process which avoids any uncertainty concerning imaginaries is almost as easy. For the substitution y = eK. u (see ~184) gives a new equation in u and x which, together with its auxiliary, has coefficients of the form (dnA(k)/dkn) - n!, where A (k) represents the left-hand side of (1*). Now KV/ — 1 is a solution of the new auxiliary by development of A(k) in powers of (k —K); hence u = sin (Kx) and u =cos (Kx) are solutions of the new differential equation, as a comparison of coefficients demonstrates. This process constitutes a rigorous proof of (3).: To avoid using imaginary powers of e, if that is desired, substitute y = eKx [cos (Kx) + V - 1 sin (Kx)lu, when the multiple root is imaginary, k= K+K V-1. 340 THE CALCULUS [XIX, ~ 188 These extensions of ~ 184 should be verified by the student by a direct check in each exercise. EXAMPLE 12 8 (1) y"' - y' - O yiv+6y'+12y"+8y'=O y"' +8 y =O (1*) k3-k=O k4+6k3+12k2+8k=0 k3+8 =0 k = 0,1,- 0,-2, —2,-2 -2, 1_ + /3 /-1 y cl+C2ez+c3e-x cl +e-2z (C2 + 3 +C4X2) cle-2x+e(c, cos V3x + C3 sin V3x) 189. Non-homogeneous Type. The non-homogeneous type (1) dnY d-l Y+ dy (1) -+ al d 1 + * ** + an-l + an = F (x) dxn dx - dx cannot be solved in general by an extension of ~ 185. But in the majority of cases which actually arise in practice,* a sufficient method consists in differentiating both sides of (1) repeatedly until an elimination of the right-hand sides becomes possible. The new equation will be of higher order still: drly dYm~ dy (2) dxm- + Al dxm- + ~ ~ + Aml-1 d — + Am = O, (2) dx ldxmld mdldx but its right-hand side is zero. Solve this equation by ~ 188 and then substitute the result in (1) for trial; of course there will be too many arbitrary constants; the superfluous ones are determined by comparison of coefficients, as in the examples below. EXAMPLE 1. y"' + y' = sin x. Differentiating both sides twice and adding the result to the. given equation: yV + 2 y"' + y' = 0. * For more general methods, see any work on Differential Equations; e.g. Forsyth, Differential Equations. XIX, ~ 189] DIFFERENTIAL EQUATIONS34 341 The auxiliary equation k5 + 2 k3 + k = 0 has the roots Ik = 0, kc = + -\ _1 (twice). Hence we first write as a trial solution yt the solution of the new equation: yt = C1 + (C2 + C3X) COS X + (C4 + c5x) sin x; substituting this in the given equation, we find - 2 C3 COS X - 2 C5 sin x= sin x, whence C3 = 0 and C5 =-1/2; substituting these values. in the trial solution yt gives the general solution of the given equation: Y = C1 + C2 COS X ~ (C4 -x/2) sin x. EXERCISES 1. y"'3 y" = 0. Ans. Y = C1+ C2X + C36Z. 2. y" -y"4 y' +4 y = 0. Ans. Y = c1iz ~ C2elx+csWz 3. yiv - 16 y = 0. Ans. Y = Cielx + Cle-2x + C3 c~s 2 X+C4 sin 2 x. 4yiV -6 y" +9 =0. Ans. y e'~Cl+cl)+ x1l(3 + C4X). 5. yv+6y/"~9y =O A ns. Y= Cl + (C2 ~ C3X) COS \/3 x + (C4 + C5X) sin V\3 x. 6. yvi- l6 y"' +64y =0, Ic = 2,2, - 1 ~V, -i~j Ans. y = OIx (Cl + C2X) + e-x [ (C3 ~ C4X) COS V'3fX + (C5 + C6X) sin V/~x. 7. y"f - 5 y' + 4 y = elx. Ans. y = ciex - (1/2)e2x + C2e4x. 8. 3'y" +4y' + y=sin x. 10. y"' -Y f " ~4y' +4 y ex. 9. y"'3y"+2y' =x. 11. yiv-5y" + 4 y = el 12. Solve the'equation y"' + y' = 0 by first setting y' = p. Solve the following equations by setting y' = p or else y" = q. 13. 3y"fit- 4 yif+ y=0. 16. y "t+ 3 yff+2 yi= ex. 14. y"/if+y" + y'=O. 17. yiV - y" = 0. 15. y"' + y' = sin x. 18. yiv + y" = ex. The following equations, though not linear, may be solved by first setting y' =p or y"= q or y//= r. 19. y' =y"I + 1/J +y"2. 21. 1 +x +x2 yfi = 0. 20. y" +t _y"'Ix= (y")2 X4. 22. xyiv + y"' = 2 23. Solve the equation x2 y" + xy' - y= log X. [HINT. Put x = ez; then dy = dy dz l dy d~y d (l1du\ ds 1 (dly dy\ dx dz dx x dz' dx = dz- kx d zl dx-x\ dc,/ so that the transformed equation is d~y __-y = Z, whence y = Clez + Cle'-z z = C1X + C2X_1 - log XJ] 342 THE CALCULUS [XIX, ~ 189 Solve the equations, 24. x2y" - xy'- 3 y = 0. 25. xy" - y' = log x. 26. (x + 1)2 y" - 4(x + 1)y' + 6 y = x, (x + 1 = ez). 27. (a + bx)2y" + (a + bx)y' - y = log (a + bx), (a + bx = ez). 28. x3y"' - 6 y = 1 +x. 190. Systems of Differential Equations. Let us finally consider systems of two equations, and let us suppose the equations to be linear in the derivatives, that is, to involve only the first powers of these derivatives. 191. Linear System of the First Order. Let the equations be (1) y' = ax +by + cz +d, (2) ' = alx + bly + cz + dl, where the coefficients are constant. We wish to determine y and z as functions of x. Differentiating (1) with respect to x gives (3) y" = a +by'+cz'; then the elimination of z and z' between the three equations (1), (2), (3), gives a differential equation of the second order in y, which should be solved for y. 192. dx/P = dy/Q = dz/R. Here P, Q, and R are functions of x, y, z. Let X, Mu, v be any multipliers, either constants or functions of x, y, z. Then, by the laws of algebra, (dx dy dz Xdx +- tdy ~+ dz P (Q R XP + uQ+vR Suppose that we can select from these ratios (or from these together with others obtainable from them by giving suitable values to X, pA, v) two equal ratios free from z, i.e. containing only x and y. Such an equation is an ordinary differential equation of the first order in x and y. Solving it, we obtain (2) f(x, y, ci) - 0. XIX, ~ 192] DIFFERENTIAL EQUATIONS 343 Suppose that a second pair of ratios can be found, free from another of the variables, say y. The result is an equation of the first order in x and z. Let its solution be (3) F (x, z, C2) = 0. Then (2) and (3) form the complete solution of the system. Conversely, differentiating (2) and (3) with respect to x, eliminating c1 and c2, and solving for dx: dy: dz, we find a system like (1). In selecting the second pair of ratios, the result (2) of the first integration may be utilized to eliminate the variable whose absence is desired. EXAMPLE 1. dx/x2 = dy/xy = dz/Z2. The first two ratios give dx/x = dy/y, whence y = clx. Putting this value of y in dy/xy = dz/z2 gives dy/(cly2) = dz/z2, so that 1 1 - = - + C2, cly z or, z = cy + Clc2yz = x + c2xz. Hence the solutions are given by the two equations y = clx, z = x + c2xz. Interpreted geometrically, the solu- tions represent a family of planes and Y x y a family of hyperboloids. These are the integral surfaces of the differ- 0 /ential equation. Each plane cuts each hyperboloid in a space curve, forming a doubly infinite system of FIG. 88. curves, the integral curves of the differential equation. The system may be written dx: dy: dz = x2: xy: y2. But the direction cosines of the tangent to a space curve are proportional to dx, dy, dz. Thus the given equations define at each point a direction whose cosines are proportional to x2, xy, y2. Our solution is a system of curves having at each point the proper direction. What curve of the above system goes through (4, 2, 3)? What are the angles which the tangent to the curve at this point makes with the coordinate axes? 344.THE CALCULUS [XIXI ~ 192 EXAMPLE 2. ~dx _dy _dz y -z z - x x-y Let X= = = 1. Then each of the above fractions equal dx + dy ~ dz 0 But since the given ratios are in general finite, this gives dx +dy +dz =O, whence x +y+z = cl. Again, let X = x, It = y, v = z. This gives x dx ~ ydy + zdz = 0, whence x2~+y2 + Z2 = C2. Thus the integral surfaces are planes. and spheres, and the integral curves are the circles in which they intersect. In this example the multipliers X, 1A, v have been chosen so as to get exact differentials. EXAMPLE, 3 ~~dx = dy -dz x -y X +y z The first two ratios are free from z and give arc tan (,y/X) log [cIX1NX~2/V + y2]. Using the multipliers X, = x, 1.= y, v = 0, and equating the ratio thus obtained to the last of the given ratios, we find x dx + y dy - dz wec 2+~ ~2 -, hecex2+ 2 c22 EXERCISES 1. x dx /y2 = yx zz n.x4 -y4 Cj c;z2 =~(X2.+ y) 2. dxix = dy/y =-dz/z. Ans. Vz =Ci; Y =C2X. 3. dx/yz ==dy/xz = dz/(x + y). Ans. z2 =2 (x +y) + Cj;x2- y2= C2. 4. dx/(y +z) =dy/(x +z)=dz (x +y). Ans. (x -y) = ci(x -z) (X -y)2 (x +y+ Z) =C2. 5. dx/(x2 + y2) = dy/(2 xy) =dz/(xz + yz). Ans. 2 y C (X2 -y2);X+ y = C2Z. 6.dy 2 xy dz 2 xz dx xt2- y2 - 22 dx x2 -y2 -— 2 Ans. y = C = C2 (X2+2+ Z2). XIX, ~ 192] DIFFERENTIAL EQUATIONS 345 dy z-3 x. dz 2 x-y dx 3y-2z' dx 3y-2z Ans. x 2 y +3 = c; 2 + 2 +2 C 2. 8. dx = -ky dt; dy = kx dt. Ans. x =A coskt+B sinkt; y = Asinkt- B coskt. 9. dx/dt= 3 x - y; dy/dt = x + y. Ans. x = (A + Bt) e2t; y = (A - B + Bt) e2. 10. Determine the curves in which the direction cosines of the tangent are respectively proportional to the coordinates of the point of contact; to the squares of those coordinates. 11. A particle moves in a plane so that the sum of the exial components of the speed always equals the sum of the coordinates of the particle, while the difference of the components is a constant k. Determine the possible paths. Ans. x + y = clet; x - y = kt + c2. 12. If the particle in Exercise 11 is at 1, 1) when t = 0, where is it vhen t = 5? Approximately how far has it traveled? INDEX TO TABLES References to pages of the Tables in Italic numerals. References to pages of the body of the book in Roman numerals. PAGES TABLE I. SIGNS AND ABBREVIATIONS.. 1-3 TABLE II. STANDARD FORMULAS.... 3 -TABLE III. STANDARD CURVES....... 19-34 TABLE IV. STANDARD INTEGRALS..... 35-50 TABLE V. NUMERICAL TABLES..,. 51-60 Greek Alphabet LETTERS NAMES A a Alpha B 3 Beta r 7 Gamma A a Delta E e Epsilon Z - Zeta LETTERS NAMES H V Eta 0 0 Theta I Iota K K appa A X Lambda M, Mu LETTERS NAMES N v Nu o Oxi 0 o Omicron II 7r Pi Pp Rho Z s - Sigma LETTERS NAMES T r Tau T v Upsilon, 0Phi X Chi *4 Psi 2 w Omega TABLES [Roman page numbers refer to the body of the text; italic page numbers refer to these Tables.] TABLE I SIGNS AND ABBREVIATIONS 1. Elementary signs assumed known without explanation: a. +; ~; -;;= axb a.b ab; a — bz a/b=a:b-b5 a2; a3; an; a a-=1/aa a1/n=V/a; apq VP O-1 (a F= a', a"t,..., a(n) (accents); a,, a2,..., an (subscripts). 2. Other elementary signs: ~#, not equal to. >, greater than or equal to. >, greater than. < less than or equal to. <, less than. n! (orLn),factorial n = n(n - 1)... 3. 2 1. q. p., approximately. I a, absolute or numerical value of a. 3. Signs peculiar to The Calculus and its Applications: (a) Given a plane curve y =f(x) in rectangular coiordinates (x, y); m?, = slope = dy/dx = j (x) = y' = first derivative; see p. 19. [Also occasionally Day, fx, ~, p, by some writers.] a = angle between positive x-axis and curve = tan-' m. Ay, A2y,..., A"y, first, second,_..., nth differences (or increments) of y. dy =f'(X). Ax, d2y =f"I(x) d AX,., d'cy =f(n) (). Ax", first, second, nth differentials of y. r,= relative rate of increase, or logarithmic derivative; see p. 114; '(x) -~-f(x) = (dy/dx) -~ y = d (log y)/1dx = r5 100. p= percentage rate of increase = 100 rrb = flexion = d2y/dX2 =fff(X) = y" = second derivative; see p. 62. d"y/dXn = f(n) (X) = y() nth derivative. K= curvature = 1 -B; B = radius of curvature = 1 - K; p. 140. 1 2 SIGNS AND ABBREVIATIONS [I, 3 f (x) dx = indefinite integral of f(x); see p. 83. (f x) dx = = f(x) dx = definite integral of f(x); see p. 87. s = length of arc; s = arc between x = a and x = b. x=a -b - x=b A = A = area between y = 0, y =f(x), x = a, x= b; see p. 90. a J=a (b) Given a curve p = f(6) in polar coordinates (p, 0) = (radius vector and curve) = ctn-1 [(dp/dO) - p] = ctn-1 [d (log p)/dO]. = Z (circle about 0 and curve) = tan-1 [(dp/dO) - p] = tan-1 [d (log p)/dO]. A == A = area between p =f(0), 0 = a, 0 = t; see p. 150. a JO=a (c) For problems in plane motion: s = distance. v, = horizontal speed = projection of v on Ox. t = time. vy = vertical speed = projection of v on Oy. m = mass. jx = horizontal acceleration = proj. of j on Ox. v = speed. jy = vertical acceleration = proj. of j on Oy. v = velocity (vector). jN = normal acc. = proj. ofj on the normal. j = acc. (vector). jT = tangential acc. = proj. of j on the tangent. 0 = angle (of rotation). a = angular acceleration. o = angular speed. g = acceleration due to gravity. (d) Problems in space; functions z = f(x, y,...) of several variables Previous notations are generalized when possible without ambiguity, exceptions are p = x/lx =f,; q = zy = r =2z/ax2 = f; s = s 2z/x ay = f=y = y; t = a2z/y2 =fy. [The notation (dz/dx), used by some writers for Oz/3x is ambiguous.] 4. Other letters commonly used with special meanings: 7r = ratio of circumference to diameter of circle = 3.14159-... e = base of Napierian (or hyperbolic) logarithms = 2.71828... M= loglo e = modulus of Napierian to common logarithms = 0.434... = "sum of such term as"; thus: i 2ai " = a,2 + a2 + * a2. (a, A, y), - direction angles of a line in space. (1, m, n),- direction cosines; I = cos a, etc. S. H. M. -simple harmonic motion. e or e, -eccentricity of a conic; also phase angle of a S. H. M. III Al II, A] EXPONENTS AND LOGARITHMS3 3 a, - amplitude of a S. H. M. (a., b), - semiaxes, of a conic; (a, b, c), semiaxes of a conicoid. A =difference (of two values of a quantity). p density; also radius vector, radius of curvature, radius of gyration. 5. Trigonometric, logarithmic, hyperbolic, and other transcendental functions: See Tables, II, A; II, F, 03; I1, G; II, H;and consult hidex. 6. Inverse function notations: If 'y =f (x), then f'-1(y) = x; f 1 denotes an inverse function. [This notation is ambiguous; confusion with {ff(x)}-1 = -~-f (x).] sin-' x, or arc sin x, - inverse of sin x, or anti-sine of x, or are sine x, or angle whose sine is xc. [Other inverse, trigonometric functions, and hyperbolic functions, follow the same notations. See Tables, II, G, 18; H, 7.] TABLE II STANDARD FORMULAS A. Exponents and Logarithms. (The letters B, b, etc. indicate base; L, 1,... indicate logarithm; N, n, indicate number; base arbitrary when not stated. See ~ 59, P. 99.) LAWS OF EXPONENTS RULES -OF LOGARITHMIS (1) N = BL; in particular (1)' L =logBNV, i. e. N = B10gB N. and I= BO; B= B'; 1/B= B-1. 10g1=0; 10gBB=I; lOgB(1/B)=-l. (2) BL. B' = BL~Z. (2)' log (N.- n) = logNX+ log n. (3) BL - BI = BL-1 (3)' log (N -- n) - log NV- logni. (4) (BL) = BnL. (4'log (Nn) n 1 log _N. (5) N = BL, B = b5-, NV= bkL. (5)'f log-bNX= logb B log N.9X B=e,, b=ie gives k=0.4342945=.Mflog15e; log15N=M loge _Y B=10, b=e gives k=2.302585=1 ~.Af~-Iog5 le; logeN(1~) logloN. b=N gives L==1/k-,1=1OgbB. lOgBb; e.g., 1og,1e==1-~1ogoe. L==s, gives 10X==ez-AF; ez==1OMX. NY= gives log,5 a,==2lJ. loge x; loge a'= (1~21!M) log,0sX. (6) y e x" gives v = nu + k, u = logeox, v = loglo y, k = logio c. (7) y Ceaz gives v -mx +k, v =logi10y, m = alogioee=aM, k = logio c. 4 STANDARD FORMULAS [II, B B. FPactors. (1) a2 - b2 = (a - b)(a + b)4 (2)' (a 4+ b)2 = a2 ~- 2 ab + b2. (3) an- bn = (a - b) (an-1 ~ an-2 b + an-3b2 +. + bns-1). (4) a2n~l ~ b2n+1 = (a ~ b) (a2n - a2n-lb ~ + b2n). See also Tables, IV, iNos. 16, 20, 21, 49, 50. (5) Polynomials: if f (a)= 0, f (x) has a factor x - a; in general: f(x) - (x - a) gives remainder f (a). (6) (a. 4 b)n = an~ + n an-lb ~ n(n -12an-2b2 +. + (~)nb~b. SeeJ1, E, 1,P. 7. 1 1.2 C. Solution of Equations. (1) ax2 +bx + c=0, roots: bxb2-a- S x/Ti 2a 2a 2 a 2a' where real> D = b2 -4 ac roots of (1) are coincident when D = 0. imaginaryJ <0 J (2) Xn ~ plXn-1 +I P2Xn-1 + + p,-ix ~ pn = 0. Roots: Xl, X2,.,Xn; then X =-Pi, f~XiXj= P2, ~XiXiXk -P3, etc. (3) f(x) -,0(x) = 0: roots given by intersections of y =f(x), y= (x). (Logarithmic chart often usef ul. ) Find roots approximately; redraw figure on larger scale near intersection. (Generalized Horner Process.) (4) Simultaneous Equations: f (x, y)= 0, 0 (x, y) 0: roots (x, y) are points of intersection; redraw on larger scale as in (3). (5) Linear Equations: (a)' 2 equations in 2 unknowns: ax~by c a2X ~ b2y = C2f ac1b alb I = ~~~(alC2 - a2Cl) (a15b2 - a bi). a2C2 la2b2 II) DI FORMULAS OF ALGEBRA 5 (b) n equations in n unknowns: aix, + bjX2 + + kixn =pi; t = 1, 2,...I, n. a, bi ~p... pi k a2 b2 "' P2 ~~ k2 Solutions: Xi= I:::: liD Columnofo p's replaces column ofi Solutions: ~~~~~~~~~~coefficients of xi where al bi. k1 b2... k2 b1..ki bi k a2 b2. k2 b3.. k3 b3... k3 b2 7C2 D a~~ I ~ai~ ~ -nzl I I+"a2 + -l)' k-n,~~ e b, ~ kn bn~~ k bn1 k.[Coefficient of ai skips ith row of D. The last formula is, a general definition of a determinant.] D. Applications of Algebra. 1. Interest. (P = principal; p = rate per cent; r =p +i 100; n = number of years; A, = amount after n years.) (a) Simple interest: A, = P(1 + nr). (b) Yearly compound interest: A, = P (1 + r)n. (c) Semiannually compounded: A,= P (1 + r/2)2z. (.d) Compounded once each mnth part of year: A. = P(1 + r/nm)-. (e) Continuously compounded: A5 P lim (1 + r/rn)mn Penr 003 2. Annuities. Depreciation. (I = yearly income (or depreciation or payment or charge); n ='number of years annuity, or depreciation, runs.) (a) Present worth P of yearly annuity I: P= I[(1 + r)n - 1]~ — [r(l + r)n]. (b) Annuity I purchasable by present amount P; or,. yearly deprecia. tion I of plant of value P: 1= P[r(1 + r)]~[(I + 2,)n -1]. (c) Final value An of n yearly payments: An = 1(1 + r)[(1 + r)n - 1] -rr. 6 STANDARD FORMULAS [II, D 3. Permutations P., and Combinations C.,,, of it things r at a time, without repetitions: (a) Pa,, r(n - 1)... (n - r + 1)n= n! (n - r)! (b) C., = P,,,,, - r! En(n - 1)... (n - r + 1)] -s r! 4. Chance and Probability. (a) Chance of an event = (number of favorable cases) (total number of trials)~i. Chance of successive (independent) events = product of separate chances<1. Chance of at least one of several (independent) events = sum of separate chances. (b) Probable value v of an observed quantity: v = ( mi i) n = arithmetic mean of n measurements mi, m2, *.., m; probable error in v ~.6745 (v - m)) (If the observations are unequally reliable, count each one a number of times, pi, which represents its estimated reliability; pi = " weight " of in). (c) Probable value of k, in formula v = kx: k= x2, from n measurements (Xl, vl), (x2, V2), "-I, (Xn, V1); probable error in k = ~.6745 i (kx v (n - 1) 2. See Exs. 37, p. 58; 28, P. 309. (d) Probable values of k, 1, m,., -in formula v = kx + ly ~ mz + are solutions of the equations: + 1 CXiY +m +z Xjzj =+ivi k J~xii + I 1~ Yi 1 I yizi + ~~ YIY k Yxizi + I JZ yizi + m Jzi2 +.. zsvi See also Exs. 37-42, p. 58, Example 2, p. 292, and Exs. 24-31, p. 308. (Rules for Least Squares. See also Observational Errors, No. III, J.) II, E] SERIES 7 E. Series. 1. Binomial Theorem: Expansion of (a + b)". (a) n a positive integer: (a + b)n = an+ =l Cnal rbr; [Cn,r: see No. II, D, 8, p. 6, and also II, B, 6, p. /.] (b) n fractional or negative, I a > I b I: (a +b)n=an + n an-lb + (n- 1) an-2b2+ *.* + Cn, ran-rbr+ *. (forever) 1! 2! (e) Special cases: (1f ^ a)n = ^ fm l r,( - 1) a,2+ ( - 1)( ~ - 2}a-3 +...; (I m | <1). (1 a )n = 1 q-i 1 I + I2 < 1! 2! 83! 1 = (I )-1 = 1:T + X2 TF Z3+ a4 F * —; (.. I < 1). (Geometric progression.) x/1 r =(1 O ~v)/2 = - 1 -2 - 3 (... (i1 <1). 2 22 ~ 2! 23. 3! =(1 l -)-l/2 = F + 32 1 *:F3 +..; ( <l). N//lf ~ ~ a 2 22.-! 23.-3! 2. Arithmetic series: a +(a + d)+(a + 2d)+ * — +(a+(n- 1)d); last term = I = a + (n -1)d; sum = s = n(a + )/2. 3. Geometric series: a + ar + ar2 + ar3 + ~*. rl - a r - 1 (a) n terms: I = arn-l; s: = = a — (b) infinite series, I r < 1: s = a/(1 - r). 4. 1+2+3+4+.. +(n —1)+n=n(n+ 1)/2. 5. 2 +4 +6 + 8 +.. +(2n- 2)+2n =n(n + 1). 6. 1 + 3 +5+7 +.. +(2n —3)+(2n-1) = n2 7. 12+22+32+... +(n-1)2+n2=n(n + 1)(2 n+ l) 3! 8. 13 + 23 +33 +... +(n- 1)3 + n = [n( + 1)/2]2. 9. 1+ 1/1!+ 1/2! + 1/3! + *=lim (1+ 1 e =2.71828.... 10. e = 1 + x/1 + X2/2! + X/3!...; (all x); ax = e1oga. 11. log,(1 + x)= x - x2/2 /3-X4/4X/5-...; (-1 <x<+). 12. log, [(1 + x)/(l -x)]=2 [ + x/3 + x/5+.]; (-1 < x < + 1) [Computation of log N: set NY=(1 + )/(1 - a); then a =(NA- 1)/(N+ 1); use II, A, 5'.] 8 STANDARD FORMULAS [II, E 13. sin x = x/1 - x3/3! + x5/5 - x7/7! + *..; (all x). 14. cos x = 1 - x2/2! + x4/4! - _6/6 1 +...; (all x). 15. tan x = x + x3/3 + 2 xi/15 + 17 x7/315 +..; ( x I < 7r/2). General term: 22n (22n - 1) B2n-1 - (2 n)!; see Bn, Tables, V, N, p. 60. 16. ctn= 1/X — /3- - x3/45 — 2B2n_1(2 x)2n-l- (2n)!; (0<lxI < r). 17. secx= 1 + 2/2! + 5 x4/4! + [B2 2n/(2 n)!]; ( x l < r/2). 18. cscx= l/x+x/3!+ [2(22n+1-l)B2an+lX2n+l/(2 n+2)1]; (0<l I<7r) 19. sin-lx=7r/2-cos-lx=x+x3/(2 3)+l.3x5/(2.4.5)+...;(I x 1<l) 20. tan- x =r/2- ctn- x =x-x3/3 +x5/5-x7/7 +..; ( I < 1). 21. (e2 + e-x)/2 = cosh x = 1 + x2/2! + x4/4! + x6/6! +..; (all x). 22. (ex - e-)/2 = sinh x = x + x/3! + x5/5! + 7/7! +.*; (all x). 23. e-Z2 = 1- x2 + x4/2! - x6/3! + x8/4!-..; (all x). 24. f(x) =f(a) +f(a) (x-a) +f'(a) (x-a)2/2!+ + + f(n-1a)( x- a)n-l/(n - 1)! +E5. Taylor's Theorem; Remainder En: I En <_ [Max. jf(n)(x) ] (Iw - a)n1 j -n!; En=.fn [a+p(-a)](a-a)n —i!; En =(1 -p)n-lf(n)[a.a p( - a)] (a -a)/n!; 1P|<1. Ipl<l. Set a 0: f(x) =f(O) +f'(O)ax+f"(O)xa2/2 +... +f(n-l)(O)O+n-l/(n-1)! +En; [MalclauZrin]. Set = r + h, a = r: f(r + h) =f(r) + hf'(r) + h2f"(r)/2! +... + E. 25.,f ( + h, y + k) =f/(, y) + [h.f:(x, y) + kfy(x, y)] + [hA2fx + 2 hk.fxy + k2fyy] - 2! +... + En; I En I < M( l h I + I k I)"!, M = maximum of absolute values of all 1th derivatives. 26. If f () = ao/2 + a, cos a? + a2 cos 2 x + a3 cos 3 x + ** +b1 sin a+ b2 sin 2 x+b3sin 3+...; (-7r<x<+r ). an ==_ () cos xa dx; bn = - ef(x) sin na dc. Fourier Theorem, Jx=- n X =_-r P. Geometric Magnitudes. Mensuration. I = length (or perimeter); A = area; V= volume. II, F] MENSURATION DIMENSIONS OR EQUATIONS 9 FORMULAS 1. Triangle. D cA r 2. Trapezoid. 3. Circle. Sides: a, b, c. Angles: A, B, C. I= + b + c = 2 s; Altitude from A on a = ha. A = aha/2 = bhb/2 = ch,/2 s = (a + b + c)/2; =A + B + C= 180~; r ==V/(-a)(a,-b)(.s-c) s. = (s - a) tan (A/2); c = b cos A + a cos B; c2 = a2 + b2 - 2 ab cos C. = (1/2) ab sip C, etc. = rs =x/s (s-a) (s-b) (s-c); sin A sinB sin a b c B-C b-c A tan- b +ctn2 b+e 2 h = height. bl, b2 = bases. A = h (b + b2) /2. r = radius; cl = diameter; a = COB at center arc CB = - (radians) r 1=2 A = Ir arc CB = r = 7rJ OEB = a/2; Z OBT= 90~; Z FBT=a; FBO = 90~ - a = ZFTB; 180 arc CB Chord DB= = — (degrees); 7r = a/2 = CEB, 4= 2 a = D)OB; Sector OD( sill a =FB r = 1 - csc a; cos a=OF r = 1 - r sec a; Triangle. tan a = TB - r = 1 - ctn a; vers a = FC - r = 1 - cos a; Segment D, exca =CT sec = = seca-1. tan (a/2) BF -- PFF= sin a/(l + cos a); sin (a/2) = BF- EB =/(1 - cos a)/2; cos (a/2) = EF - EB =/(1 + cos a)/2. rrr = ird = 2 A/r; r2 = rd2/4 = I r/2; a, (a in radians) ra/180, (ain degrees); = 2 r sin a = 2r sin (4/2); 7B = rr2, (a in degrees); )OB = r2 sin a cos a = (1/2)r2 sin 2 a; FBC= r,2 [rra/180 - (sin 2 a)/21. 4. Ellipse. e<1. a, b, semiaxes; r, r', radii. c =\/a2 - 02; e a = ca a2 - -2/a, (eccentricity); p = b2/e = a(1 - e2)/e; a = tan-1 (a-) = eccentric angle; 2S 2, - a cos a-, y=bsin a; = a sin ), y = b cos; 0 = 7r/2 - a. +r' = const. = 2a. A= 7rab; ab ab Wa S = OAP = a = - cos -- _; 2 2 a Arc AP= da a2 -X2 [Tables.] =ao a Vl -e2 cosz2ad a: where cos p = a/a. Arc BP =z ja2 _- e2 do - esin24d4 4 = rr/2- a; sin A = /(a. W2 Y2 a+ - =1, (origin at 0). _a2 02 or P -1 - ecos 0' (pole at F); Foci, F, F'; Center; 0. 10 STANDARD FORMULAS [II, ) DIMENsioNs op. EQUATIONS FORmULAS 5. Hyperbola. e>1; a, b: semiaxes; r, i'f: radii; c = V'aI0 + b2; e = c/a =\Val + b2/a; p = b2/c-= a (e2 -1)/e. gC2?/21 o a2 b2 1- e ns 0 (origin at 0) (pole at F). Foci: F, F' v'- r const. 2 a; S= Sector OVP ablog +W O=b cosh.I(X')=ah sinh-'Q(sP 2 _ 'a1.2 x=a cosh-M, y= blinh-;S or if tan qsinh ab, x =a seco, y =b tanr4. 6. Parabola. e=lI. U N L p =LN/2; L N =latu s rectum. OF= p/2 = LNI/4. p2 = 2px, (origin at 0); =,(pole at F). Focus: F Area ONVPMJ V'\2NI32p1J2; Arc OP =f' 1'- +(y/p)2dy. 50O (See Tables, p. A0, No. 46 (a). 7. Prism. 8. Prismoid (~ 124, p. 202). B area of base; h4 height. B - lower base (area); M f= middle section; T= upper base; h = height. Y= B.h. v~ k(B +4 Jr+ T). 6 (See also Tables, IV, G, p. 47.) [The volume of each of the solids mentioned below, except (16), follows this formulla, though not all are prismoids.] 9. Pyramid (any A = area of base' V == A. h/j3. sort). A = height. 10. Riglit Circular r == radius of base; A (curved) ==2 iak; Cylinder. 4 = height; B-base (area). A (total) = 2 7rrh, + 2 n-r2; V=,ryr2ih= Bh. I, )F] MENSURATION 1.1 DImENSIONs OR EQUATIONS FoRMULAS 11. Right Circular r radius of base; as v'\s2 + hi2; ___ Cone. See Fig. 19, p. Th. It = height; B = base; A (curved)= irs V/ri + h2 =rs's; tan a=.r/h; s =slant height; A (total)=. irs (s + r); cos a=h/s; sin a=r/s. a =half vertex angle. V= rrih1/3 = Bh/3. 12. Frustum of r radius lower base; S=(jB,) 2 +h12 Cone. R radius upper base; A (curved) = irs (B + r); B=lower base (area); h =height; s=slant height. _V = Irh (R2 + Br + r2)/ 3. T==upper base. 13. Sphere. (a) En~tire, Sphere. (b) Spherical Segmient. Other notations as above. (e) Spherical Zon~e. (x - r)2 + (y - y) + (Z - ZO)2 = ri. r radius; d = diameter; C, great circle (area). a= radius of base of segment; h =height of segment. h = height of zone; a, b = radii of bases. a = angle of lone (degrees). E= A + B +C - 1800; S=(A + B + C)/2; 3)sin (s - -y)]/sin a; A) cos (S -B) cos (S - C)] A = 4 irr2 = rd2 = 4 C; V = 4i7rri/3 = 7rdi/6 = A -r/3= 4 Cr13. (d) Spherical Lun~e. (e) Spherical Trianigle. Sides a, /3, y.~# Angles A, B, C. k = NV~sin (a3 - a) -sin (s 3 -K=='\I-cos S/[cos (S - a2= A (2 r - h) A = 2 irrh, = ir (ai + hi.); V = irh (3 a2 + h)/ = irhi (3 r -)/3. A 2irriA/9 A == 7rr2 El180;I sin A slo B sinC0. sin a in3 Sill y' cos a =cos /3 cos y + sin P sin y cos A cos A =-cos B cos C + sin B sin C cos a; tan (A/2) = k/sin (s - a); tan (a/ 2) =Kcos (8 - A). V = 4 7rabc/3. 14. Ellipsoid. Semiaxes, a, b, a. 15. Paraboloid of Revolution. 2+y2==2pz. 16. Anchor Ring. /Vxi4y \Vr2 - Z2 = B. ri2/a + y2/b2 + Z2/c2 = 1. r = radius of base; k height. V = ir7r2h/ 2 = ir ph2. r == radius, generating circle; B = mean radius of ring. A = 4iri2Rr; V = 2 urRn2. [See also Standard Applications of Integration, Tables, IV, H, p. 48.) 12 STANDARD FORMULAS [II,0 G. Trigonometric Relations. For Trigonometric Mensuration Formulas, see II, F, 1, 3, 13 e, p. 9. 1. Definitions. See also II, F, 3, p. 9. sin A= y/r; cos A =x/r; tan A= y/x; csc A =r/y; seeA= r/x; ctiiA =rr/y; vers A = 1 - cos A; exsec A =sec A - 1.' 2. Special Values, Signs, etc, for sine, cosine, and tan gent. Angle 00 300 450 600 900 1800 2700 36 00 ~ A 900 ~ A 1800 ~ A 2700~ A sin ~ 0 1/2 -V2/2 -/3-/2 1 ~0 -1 ~sinA +ceosA TsInA -cos A ens 1 V/8/2 V,2-/2 1/2 ~0 -1 ~0 +cosA TFsinA - cosA ~sin A tan ~ 0 x/313 1 x\/- ~o ~0 ~o ~tan A:ctn A ~ tanA T ~ctnA [~0and ~ oo indicate that the function changes sign.] 3. csc A 1 /sin A; sec A =1/cos A; tanA = 1/ctn A. 4. X2 +y2 - 22: cos2 A +sin2 A =1; 1~+tan2 A-=sec2A; c-un2 A + 1csc2A. 5. sin (A~B) sinA cos B +cos Asin B. 6. cos (A B) =cosA cos B:Fsin Asin B. 7. tan (A~B) =[tan A~ tan B][l:F tan Atan B]'. 8. sin 2 A=2sinA cosA; sin a-2sin (a/2) cos (a/2). 9. cos2A-cos2A-sin2A-1-2sin2A=2cos2A-1; cos jX = cos2 (a/2) - sin2 (ax/2); see also II, F, 3, p. 9. 10. sin3A=3sinA-.4 sin3 A. 11. cos3A=4COS3A-3cosA. 12. tan 2A =2 tan A~-[1 -tan.2A]. [See also1, F, 3,p. 9]. 13. 2 sinA cos B =sin (A +B) +sin (A -B); sin a~ sini = 2 sin [(a + 3)/2] cos [(aT3/] 14. 2 cos AcosB cos (A -B) ~ cos (A~+B) cos a + cosj3 2 2cos [ (a + 0) /2 ] cos [(a - f3/2]. 15. 12sin Asin B cos(A -B)-cos (A +B); cos a- cosl=2 sin [(a +f3/2] sin [(a - )/2]. II, H] TRIGONOMETRY 13 16. sin2 A - sin2 B = cos2 B- cos2 A sin (A + B)-sin (A- B). 17. cos2 A - sin2 B = cos2 B - sin2 A = cos (A + B) cos (A - B). 18o Definitions of Inverse Trigonometric Functions: (a) y = sin-i x = arc sin x - angle whose sine is x, if x = sin y; usually y is selected in 1st or 4th quadrant]. (b) y = cos-1 x = arc cos x, if x = cos y; [take y in 1st or 2d quadrant]. (c) y = tan-1 x = arc tan x, if x = tan y; [take y in 1st or 4th quadrant] 19. sin-1 x = 7r/2 - cos-1 x = cos-1 Vi - x2 = tan-1 [x/ /1 - x2] = csc-1 (l/x) = sec-'[/V1 -- x2] - ctn-1 1 i- x2/x]. 20. cos-1 x = 7r/ - sin-1 x = sin-1 /1 - x2 = tan-1 [ V 2/x] = sec-1 (1/x) = csc-1 [1/1 -- 2] = ctn-1 [x/x/1 - x2]. 21. tan-1 x = /2- ctn-lx = ctn- (1/x) = sin-l [x//1 + x2] = cos-l [1/x1 x2] = sec-ll/ + x2 = csc-1 ['/ + x2/x]. 2. Special values, correct quadrants, etc., for inverse functions. VALUE' + _- 0 1 2 1/ 1/ 2 2/2 /2/3/3 >1 -c sin-la, 1st Q 4th Q 0 r/2 -7r/2 r/6 7r/4 7T/3 0.62 - sin-1 (+ k) cos-la, 1st Q 2d Q 7r/2 0 7r T7/3 r/4 7r/6 0.96 7r-cos-l(+^) tan-la 1stQ 4thQ 0 7r/4 -7r/4 0:46 0.62 0.71 7r/6 >r/4 -tan-l(+ k) H. Hyperbolic Functions. 1. Definitions. (See figures III, E, J2, pp. 22, 30; and V, C, p. 54. sinh x = (ex - e-)/2; cosh x =(ex + e-")/2; tanh x = sinh x/cosh x = (ex -- e-x)/(ex + e-z); ctnh x = 1/tanh x; sech =- 1/cosh x; csch x = 1/sinh x. q = Gudermannian of x = gd x = tan-1 (sinh x); tan - = sinh x. = tan-1 [(ex - e-z)/2] = 2 tan-lez - 7r/2 2. cosh2 x - sinh2 x = 1. 3. 1 — tanh2 x = sech2 x, 4. 1 - ctnh2 x = csch2 x. 5. sinh (x: y) = sinhx cosh y; cosh x sinh y. 14 STANDARD FORMU)LAS [II, II 6. cosh (x + y) = cosh x cosh y ~ sinh x sinh y. 7. y = sinh-1 x = arg sinh x = inverse hyperbolic sine, if x = sinh y [Similar inverse forms corresponding to cosh x, tanh x, etc.] 8. sin-1 x =cosh-1 X-2 + 1 = csch-1 (1/x) =log (x ~ vx~ ~1) 9. cosh'1x =sinh'1 v'X2 - 1 = sech-1 (1/x) = log (x + x2 -1) 10. tanh'lx =ctih-' (l/x)=(1/2) log [(1+ x)/(1- x)]. 11. Ifq5gd x, sinh x= tan q, cosh x= ctn q5, tanh x= sinp. I. a. Plane Analytic Geometry I(x, y) or (a, b) denote a point; (xe, ys) and (X2, Y2) two points; etc.) 1. Distance 1= INP2 =v'(X2 -I +(2-Yl x+A 2. Projection of P1P2 on OX AX =X2 - XI 1Cos at, where a =Z(OX, P1P2). 3. Projection of P1P2 On OY AY= Y2 - Y = l Sin a. 4. Slope of PIP2= tan ac = (Y2 - y,)/(X2 - XI) = AY/AX. 5. Division point of P1]'2 in ratio r: (x + r Ax, yl -f r Ay). 6. Equation Ax ~ By ~ C - 0: straigrht line. (a.) y =,nix + b: slope, m; y-intercept, b. (b) y - 'yo = m (x, - xo): slope, mn; passes through (xo, yo). (C) (Y-Yl)/(Y2-Yl)=(X-Xl)/(X2-Xl): passes through (x,, yr), (X2,Y2). (d) x cos a +ycosg = p: distance to origin, p; a = Z(Ox, n); =Z/ (Oy, n); n = normal through origin. [General equation Ax + By + C = 0 redncess to this on division by V/A2 + B2. 7. Angle between lines of slopesmin,Mi2=tan1 [(Ml1-Mi2) /(l1~Ml'm2)]. [Parallel, if ns1=m2; perpendicular, if 1~mAim2=0, i.e. if Ml = -1/os2.]J 8. Transformiationx= x + h, y =y' +k. -[Translation to (h, k).] 9. Transformationx= cI, y =ky'. [Increase of scale in ratio con x-axis; in ratio k on y-axis. ] 10. Transformation, x = XI cos 6 - y/ sin 6, y = x' sin 6 + y' cos 6. [Rotation of axes through angle 6.] 11. Transformation to polar coiordinates (p, 6): x = p cos 6, y = p sin 6. Reverse transformation: p = -VIX + y2, 0 = tan-' (Y/X). ii, J] ANALYTIC GEOMETRY 15 12. Circle: (x - a)2 + (y - b)2 = r2 center, (a, b); radius, r; or (x - a) = r cos 6, (y - b) = r sin 6. (6 variable.) 13. Parabola: y2 = 2px: vertex at origin; latus rectum 2p. 14. Ellipse: X2/a2 + y2/b2 = 1: center at origin; semiaxes, a, b. (See II, F, 4, p. 9.) 15. Hyperbola: x2/a2 - y2/b2 =1; center at origin; semiaxes, a, b; asymptotes, x/a ~ y/b = 0. See II, F, 5, p. 10. (a) If a = b, X2 - y2 - a2; retangular hyperbola. (b) xy I k-, rectangular hyperbola; asymptotes: the axes. (c) y= (ax + b)/(cx c d), rectangular hyperbola; asymptotes: x d c, y = a/c. 16. Parabolic Curves: y = ao ~ aix + a2x2 ~... + a xn. [Graph of polynomial; see also Figs. A, B, pp. 19, 20.] 17. Lagrange Interpolation Formula. Given y = f(x),, the polynomial approximation of degree n - 1 [parabolic curve through n points, (X2, Yl), (X2, Y2), *., (X., YJ)I is Y = P(X)= Y1p(x) + Y2 P2(X) +... + YnPn(X), where the polynomials pi(x), p2(x),., p,,(x) are (Xi - X2)(X - X2)... (Xi - X.i-)(X - X~if)... (X - X.) [Numerator skips (x - xi); denominator skips (xi - xi). Proof by direct check.] [For a variety of other curves, see Tables, III, pp. 19-34.] I. b. Solid Analytic Geometry. [(x, y, z) denotes a point; (x1, Yi, z1) and (X2, Y2, z2) two points, etc. 0 denotes the origin, (0, 0, 0).] 1. Distance: P1P2 = V(X2 - X1)2 + (Y2 - yl)2 + (z2 - zl)2. 2. Distance from the origin: OP = vx2 ~ y2 + Z2. 3. Direction cosines of a line L: cos a, cos 3, cosy, if a, fi, -y denote the angles L makes with the x, y, z axes, respectively; and we have cos2a + cos2fl + cos2-y = 1. 4. Direction cosines proportional to given numbers: If a: b:c = cos a: cosj3: cos y, and R2 = a2 + b2 +C2, then cos a = a/R, cos3= b/R, cos y = c/l. W 16 ~STANDARD FORMULAS [I [II, I 5. Angle 0 between lines L and L' with direction cosines (1, m, n) and (iml, in') Cos 0 = 11' + mm' + nnl'. Lines parallel if Ill ~ mm' +nn' -1 or if 1-= 1' m =ml, n =nW..Lines perpendicular if 11' + mm' nn' = 0. 6. Direction cosines of a plane P =- direction cosines of any line perpendicular to P. 7. Equation of a plane P: lx +my +nz =p, or x cos a+y cos3 + z cosy =p, where (1, m, n) are the direction cosines of P, and p is the length of the perpendicular from 0 to P. 8. General equation of a plane: Ax + By ~ Cz ~ D = 0. If R2 =A2 +B2 +G -1 =cosa =A/R, m =B/R, n=U/Rl,p=-D/B. 9. Plane with intercepts a, b, c, on the axes: x/a + y/b ~ z/c =- 1. 10. Plane determined by (xi, yi, zi), (X2, Y2, Z2), (X3, Y3, Z3): x y z1 Xi Yi Zi 1 X2 Y2 Z21 x3 Y/3 Z31 11. Angle 0 between two planes (1, m, ni; p), (1, in, n1; p): cos 6 = 11' + mm' + nn'. 12. Angle 0 between planes Ax -F By + Cz ~ D =0 and A~x + B'y ~ C'z + D' =0: co =AA' ~ BB' ~ CC'Jl=2B+2B2-2B2C Planes parallel if 11' + mm' ~ nin' =1, or if A = A', B =B', (3 = C'. Planes perpendicular if AA' + BB' + CC' =0. 13. Distance d from point (xi, Yi, z1) to plane (1, m, n; p): d = lxj + my, + nz1 -p. 14. Distance d from (xi, yi, zi) to Ax ~ By + Cz ~ D = 0: d — Axi+Byi+Czi + D1?= A2 +B2~ 02. 15. Direction cosines (1, m, n) of a line determined by two planes Ax+By~ Cz+D=O, A'x+B'y+C'z+D'-0; 1: m:n- B C C A A B *see 4 BC' C ' lAl' lB II, 51 ANALYTf C GEOMETRY 17 16. Line through (x1, Yi, zj) and (X2, Y2, Z2)' _-Xi_ Y-Y - Z-Zi X2-x1 Y2 - Y1 Z2 -- Z1 17. Line through (xo, Yo, zo) in direction (1, m, n): X - Xe - Y - Ye _ Z - ze 1 r'3n n 18. Line through (xo, Ye, zo) perpendicular to plane Ax + By + z =C0: X - XO - Y - YO Z - Xg A B C 19. Plane through (xi, Yi, z1) perpendicular to line of formula 17: A (x - xi) + B (y - Yj) + C'(z - z1). = 0. 20. Sphere of center (a, b, c) and radius r: (x -- a)2 + (y - b)2 + (Z 7 C)2 = r2 21. Cones with vertex at 0: X2 y2 Z2 a2 b2 C2 Imaginary, if all signs are alike; otherwise real, and sections parallel to one of the reference planes are ellipses. 22. Ellipsoids and hyperboloids with centers at 0 (see Tables III N1, 2, 3) Signs on left: X y2 _2 -1 All +: ellipsoid ~ ~ ~ f = One -: hyperboloid of one sheet a2 b2 c2 Two -: hyperboloid of two sheets All -: surface imaginary 23. Paraboloids on z-axis with vertices at 0 (see Tables III N4, 5): ~2 cz Signs on left: a2 = Alike: elliptic paraboloid a2 b2_~ XjSin nlf I Different: hyperbolic paraboloid 24. Contour lines on curved surface F(x, y, z) = 0: Sections by z = a are F(x, y, a) = 0. 25. Curves in space: (a) Intersection of two surfaces: Fi (x, y, z)= 0, F2(x, y, z)= 0. (b) Solve for y and z: y =f(x), z = 4p(x). (c) Parameter forms: x =f(t), y = z(t), z = ~tt). 26. General cylinder with elements parallel to z-axis: f(x, y) = 0. 18 STANDARD FORMULAS [II, J J. Differential Formulas. 1. y = f(c): dy =f'(x) dx, f (z) = dy - dx = dy/dx. 2. F(x, y) = 0: Fx dx + Fydy = 0, or dy =-[F -- Fy]dx. 3. x =f(t), y=o(t): (a) dx -ft(t) dt, dy= f(t) dt, dy/dx = Qf (t) - f (t). (b) d2y/dx2 = d[dy/dx]/dx = d [b' -f f']/dx = ["lf' - f"l] * (f )3. (c) d3y/dx3 = d[d2y/dx]dx = dC[( f' -f"lf) - (f')3]/clt -f'. 4. Transformation x = f(t): y = (x) becomes y = (f(t))= t (). (a) dy/dx becomes dy/dt - f (t); [see 3 (a)]. (b) d2y/dx2 becomes [(dy/dt2) ) f'()- (dy/dt)f"1(t)] - [l(t)]3; [see 3 (b)]. 5. Transformation x = f(t, u), y = 5 (t, U): y = F(x) becomes u = (t. / c /, dy ya +, a, c0 ^n r^/ f d u ] (a) dyl/dx becomes or dx + 3qf + -f. I dt di L t au dt i at au dt J (b) d2y/dx2 becomes d [dy/dx]/dt - dx/dt; [compute as in 5 (a)]. 6. Polar Transformation x = p cos 0, y = p sin 0 dx = cos 0 dp -p sin 0 d; dy = sin 0 dp + p cos 0 dO, d2x = cos 0 d2p - 2 sin 0 dp d - p cos 0 d02, d2y = sin 0 d2p + 2 cos 0 dp dO - p sin 0 d92. 7. z F(x, y): dz=F dx + Fy dyp dx +q dy; [see I, 3 (d), p. 2]. 8. Transformation x =f(u, v), y = ~p(u, v): z F= (x, y) = (u, v), ( a a z- af aa ax axz z f az a_ (a) - + au ex au y ay ' eav ax v ay av (b) a=A + B = Cz + Dz. ax aOt av ay Oa av [A, B, C, D found by solving 8 (a) for az/ax and az/ay.] ( X2 Ox\x/ Ox au Av ax~z axax-l axa au a a u au gj Sv au Ve [Similar expressions for a2z//y2 and higher derivatives.] TABLE III STANDARD CURVES A. Curves y = xn, all pass through (1, 1); positive powers also through (0, 0); negative powers asymptotic to the y-axis. Special cases: n = 0, 1 Chart of y. = x for positive, negative, fractional values of n I fll/ i/ I /I / l Hi'i i[ I I I \'1 i"'il 111\1V I,1\~ 4 T II I A x \\I N L ri.I;-'z 'N M1 11\i\ 1i i I / / -i. \,2_,,3%\It\!'li _ /1 / ~/t: ~".r I I Eve, I,, uXF,,,4 nilI1 I I Iw.:;Z; t 3;\% N I. I I Iln~n zN IT — -nr I ' / I/ 1 N!w if I I v I 1 I 7r lII 1 ( I Y/1 X /I l 1~ 2 r<T> r'M^._-F Odd! P6wedr, an.d Rolots:-> i 74 ua iai I I I! I I I '!- _ =,I'',!.L I\ I I,, I/ IA/ k,JI'AW I'N~1 I I I I n - ' I ~ I I ~ 1 1/.1 I~~~rr~ ~~ Likg~~~~~~ WIC,~j~ l -71 \ ' - l I r i 1 I P, l I I I I I I t\ l I i I _ _\ I lI g/l /1 / I I I I tt I I I T ' i r i \ i/i /1 l i I i I FIG. A 19 20 STANDARD CURVES [III, A are straight lines; n = 2, 1/2 are ordinary parabolas; n = - 1 is an ordinary hyperbola; n = 3/2, 2/3 are semi-cubical parabolas. The curves prm- c occur in the theory of gas expansion, where p = pressure; v = volume; c and m constants. In isothermal expansion (1). 105) = 1, whence pv = c or p = cv-1; ( = - 1 in Fig. A). Choose scales so that y = 2/c and v = x. In adiabatic v2 expansion of air, m = 1.41 (nearly). The area p)dv = work done in compressing the gas from any given volumae '2 to a volume r'. vB. Logarithmic Paper; Curves y = xn, y = kXn. Logarithmic paper is used chiefly in experimental determination of the constants k and n; and for graphical tables. In Fig. B, k = 1 except where given. V rT II 4.5 I 4 3.5 I '-I I *} ^_\. \ \ I / / / '^>^- ea-L io aithm ic Chart 2.5 kLX.n 1 5- ' 13s 2.f Value s-of chfef I;=il 2.1 5 01.15 5 2.ss.3 1 7 2 21 315 4 4 5 6 7 89.95.45 TV'h.35 45 Note: ^Ttl3 7^ k^Tll^^ j~~~~~sg~~~~~in^ -^/-z-ZK~~~~~~~~~~~~ralet >vl tF t t4 f -9 l I N V.I;. thsou'h.1.15.2,25 4 3,36.4.45.-.6.7.8.9 1 FIG. B 1.5 22. 25 3 3.5 144.5 6 5 7 8 9 10 [See ~189,p.284, above; also Williams-Hazeu, Hydraulic Tables; Trautwine, En. gineers' Handbook; D'Ocagne, Nomogragphie.] The line y = a0-1 gives the reciprocals of numbers by direct readings. III, D] ELEMENTARY FUNCTIONS 21 C. Trigonometric Functions. The inverse trigonometric functions are given by reading y first. FIG. C D. Logarithms and Exponentials: y = logio x and y = log x. Note loge x = loog x loge 10 = 2.303 loglo x. The values of the exponential functions x = 10y and x = ey are given by reading y first. See E. loge x= 2.302585 X 10glog 0i1iiii110 logo x~ 0.43429448 x log. - e= 2.718281828 _ lm ] i [i ii Ei E oglI e= 0.43429448 " " """ 11gD ii FIG. D 22 STANDARD CURVES [III, E E. Exponential and Hyperbolic Functions. The catenary (hyperbolic cosine) [y = cosh x = (ex + e-x)/2] and the hyperbolic sine [y = sinh x = (ex-e-x)/2] are shown in their relation to the exponential curves y =ex, y = e-x. Notice that both hyperbolic curves are asymptotic to y = ex/2. FIG. E The curve y = e-x is the standard damping curve; see Fig. F2, and ~67, p. 108. The general catenary is y = ((/2)(eX/a + e-xla)= C cosh (w/la); it is the curve in which a flexible inelastic cord will hang. (Change the scale from 1 to a on both axes.) Ifi, Fl HARMONIC CURVES 23 F. Harmonic Curves. The general type of simple harmonic curve is y = a sin (kx + E): CURVE Ia sin (kx + e) sin x cos x sin 2 (1/2) sin (6 x - 12) amnplitudle a 1 1 1 1/2 wave-length 2 v/k/ 2 7 27r vr rr/3 phase-angle - Elk' 0 vr/2 0.2 IA compound harmonic curve is formed by superposing simple har. monies: in Fig. F1, y = sin 2 x + (1/2) sin (6 x -- 1.2) is drawn. VI = sin I2X -'I sin (O x -l1.2) = sin x+-I s n (6x 1.2) 'Jz~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 02 V1 i /F FIG. F1 Such curves occur in theories of vihrations, souud, electricity. 24 STANDARD CURVES [III, F The simplest type of damped vibrations is y = e-cx sin kx: Fig. F2 shows y =e-/2 sin 3 x. The general form is y = ae-" sin (kx + e). Such damped simple vibrations may be superposed on other damped or undamped vibrations. FIG. F2 G. The Roulettes. A roulette is the path of any point rigidly connected with a moving curve which rolls without slipping on another (fixed) curve. The Cycloid FIc. G1 Figure G1 shows the ordinary cycloid, a roulette formed by a point P on the rim of a wheel of radius a, which rolls on a straight line OX See also Fig. 36, p. 144. The equations are fx = ON- M-N — a - a sin 0, Y = N - KC = a - a cos 0. wvhere 0 = NCP. Ill, G] ROULETTES 25 Figure G2 shows the curves traced by a point on a spoke of the wheel of Fig. H, or the spoke produced. These are called trochoids; their equations are x = aO - b sinO, y = a- bcosO, b>,a b< c where b is the distance PC. If b > a, the curve is called an epit ochoid, if b < a, a hypotrochoid. Figure G3 shows the epicycloid; [ = (a+b) cos- b cos [a- b0], y (a + b) sin - b sin [ — b o], It b FIG. G3 formed by a point on the circumference of a circle of radius b rolling on the exterior of a circle of radius a. 26 STANDARD CURVES [III, G Figure G4 shows the special epicycloid, a = b, x = 2 a cos 0 - a cos 2 0, y = 2 a sin 0 - a sin 2 0, which is called the cardioid; its equation in polar cohrdinates (p, p) with poie at 0' is p = 2 a (1 - cosP). FIG. G4 Figure G5 Sh)OWS the hypocycloid: I (a-b) cosO + b cs [- b b y (a-b) sin 0 - b sin 8 formed by a point on the circumference of a circle of radius b rolling on the interior of a circle of radius a. Hypocycloid FIG. G5 Figure G6 shows the special hypocycloid, a = 4 b, (x = a cos8 6 I x = a Csin 0, or X213 + y2/3 = a2/8, i Y= asin3 0, T which is called the four-cusped hypocycloid, or Fri. G6 astroid. Yt * H. The Tractrix. - This curve is the path of a particle P drawn by a cord PQ of fixed length a attached to a point Q which moves along the x-axis from 0 to -+ o. Its equation is A p Tbe Tractrix. a ( 4I I Q FIG. II X =a log a + x/a2 - -2 x/a2 - Y III, Ii CUBICS - CONTOUR LINES 27 I. Cubic and Quartic Curves. Figure I, shows the contour lines of the surface z = x3 - 3 x - y2 cut out by the planes z = -k for k=-6, -4, - 2, 0, 2, 4; that is, the cubic curves x3 - 3 x - y2 = k. The surface has a maximum at wc - 1, p = 0; the point x = 1, y = 0 is also a critical point, hut the surface cuts through its tangent plane there, along the curve k =-2; y2 =WS -Crw+2. These curves are drawn by means of the auxiliary curve q =w2 - 3 X, itself a type of cubic curve; then y ='/q - Ic is readily computed. Contour Lines - for I 2 2iX I 0 _ _ kl=-4_ _ I c I=1 - - FI. 1 x kf-4 ~kCur ~es Au uiliaryi Curve FIc. hl FIG. 12 Figure 12 shows the contour lines of the surface Z=-3- 3 x y2 (x- 4) for z=lk=-6, -4, -2, 0, 2, 4; that is, the cubic curves Y2 = (x -3 x- k)/(4 - x). The surface has a maximum at (- I, C). At (1, C) the horizontal tangent plane z=-2 cuts the surface in the strophoid 2 - (xa 3 - Cex + 2) / (4 - c) whose equation with the Pew origin 0' is 2 cX'2 (C + ') /(3 - a'). The line xc = 4 is an asymptote for each of the curves. 28 STANDARD CURVES [ITT, 1 Figure 13 shows anotber cubic: the cissoid, famous for its use in the ancient problem of the " duplication of the cube." Its equation is Y2 -X -,or p = 2 a-tan a sinG6. 2 a - x: It can be drawn by using. an auxiliary curve as above; or by means of its geometric definition: OP = QB, when Qy and AB~are vertical tangents to the circle OQA. X3 N 2 a-F. FIG. 13 Figure 14 shows the conchoid of Nicomedes, used by the ancients in the problem of trisection of an angle. Its equation is y2 ~(_ b )+, or p=asecO~b. y2 =, X 1/ I ~onehotd II I I \ a Foilum of Descartes y 0 2 FIG. I1 Figure 15 shows the cubic x3 + y3 - 3.of Descartes; see Example, p. 45. FIG. IL axy = 0, called the Folium The W itch 8a3 = x2 +4a2 0 2a FIG. Is Figure I6 shows the witch of Agnesi: y = 8 a3/(X2 + 4 a2); see Exs. 34, 36, p. 131; 52, p. 162; and see III, J, below. III, I] QUARTICS- CONTOUR LINES 29 Figure 17 shows the Cassinian ovals, defined geometrically by the equation PP. -PFP = k-2; or by the quartic equation, [(x - a)2 + y27J[(x + a)2~+ y2] = k4, where a = OF (= 1 in Fig. 17). The special oval k2 = a2 is called the lemmiscate, (X2 + y2)2 = 2 a2 (x2 - y2) or p2 = 2 a2 cos 2 6. FIG. 17 The ovals are also the contour lines of the surface z4 - [ ( - a)2 + y21 [(x~+ a)2 ~ y2], which has minima at (x ~- a, y - 0), and a critical point with no extreme at origin. 30 STANDARD CURVES [III; J 3T. Error or Probability Curves. Figure J1. is the so-called curve of error, or probability curve -7r where h is-the Measure Zi of precision. See Tables, IV, II, 148, p. 50; and V, it G p. 56. __ __ __ __ __ __ __ __FT__ __Figure J2 shows the very 1 XX x similar curve F~~~o. ~~~1 =sech x = 2/(ex + ex-). In some instances this curve, or the witch (Fig. Is), may be used in place of Fig. J1. Any of these curves, on a proper scale,7 give I Hyperbolic good approximations Cosh x to the probable distribution of any accidentaldatawhich _ 2 11 2 3 tend to group them selves about a mean. FIG. J2 K. Polynomial Approximations. Figure K, shows the first Taylor polynomial approximations to the function y = sin x. (See ~ 147, p. 253.) raylor Approxiniationrs for Y=sinx.Successive Polynomials x - x 3/3! FIG. K, IIl, L] APPROXIMATION CURVES 31 Figure K2 shows the Simpson-Lagrange approximations: (1) by a broken line; (2) by an ordinary parabola; (3) by a cubic, which nowever degenerates into a parabola in this I I I i I i I I I i H ---- into a parabola in this ___ Simpson- Lagrange Polynomial Approximations _ example. (Lagrange I_ I- _ _ I I I _for i=j sizn xL I ___ _Interpolation Formula, Tables, p. 15.) ___ —.,._2 i_ The fourth approximation is so close that it cannot be drawn in the figure. In practice, the division points are taken closer together than is feasible in a figure. No. 1. y FIG. K2 x< O<x<n Y= y= (a)' r. n 2 0a 2 - No. 2. y = 4 - 42 = 1.273 -.405 a2, O a < 7r. Wr fr2 (Division points: a5 = 0, ar = 7r/2, ag = 7r.) No. 3. y= 9-03 _ 9'2_ - 0. a3 = 1.245 - 0.395 2. 4n 4 7T2 (Division points: ao, = r/, ax = rr/8, X3 2r/, = '.) No. 4. y = 15.38 a - 7.642 x2 - 2.36 a3 + 0.376 a4. (Agrees too closely to show in the figure.) L. Trigonometric Approximations. for ____ J 1 iu Fou rier Ap roximations (2)I s y -f(- rx),/2 ( 0xr y mX e o s - 2ionothefrms+ a, sx RA] W lh ^ a:) 2 =2 sin < I I I1 Iuccessiv Harmoic I- A \ - ( ) y =2 8in X +-S - 3 X --- FIG. L Figure L shows the approximation to the two detached line-segments y =- 7r/2, (- r < x< 0), y = 7r/2, (O < x < 7r) by means of an expression of the form a0 + a1 sin x + a2 sin 2 +x.. + a. sin nx. See II, E, 26. 32 STANDARD CURVES [III, L. FIG. Mi FIG. M2 Inr, N] SPIRALS - QUADRIC SURFACES 33 M. Spirals. Figure M1 shows the logarithmic [or equiangular spirals p = keeO] tor several values of k and a. Note that k e1 and k =- 1 give the came curve. FIG. M3 Figures M2 and Mg represent the Archimedean Spiral p = a@, and the Hyperbolic Spiral p0 = a, respectively. N. Quadric Surfaces. These are standard figures of the usual equations. Hyperboloid of one Sheet I2 Y2 Z2 -V C 2 FIa. N1 FIG. N2 .34 STANDARD INTEGRALS [III, IN Hyperbolold of two Sheets x 2-P- = 1 a b2 e FIG. Na FIG. Ns Elliptic Ppraboloid Hyperbolic Paraboloid X2 Y2=k FIG. Xi FIG. N4 TABLE IV STANDARD INTEGRALS Index; A. Fundamental General Formulas, p. 35. B. Integrand - Rational Algebraic, p. 36 C. Integrand Irrational, p. 39. (a) Linear radical r = vax.+ 6, p. 39. (b) Quadratic radical \/+ x2 ~ a2, p. 39. D. Binomial Differentials - Reduction Formulas, p. 4! E. Integrand Transcendental, p. 41. (a) Trigonometric, p. 41. (b) Trigonometric -- Algebraic, p. 44. (c) Inverse Trigonometric, p. 45. (d) Exponential and Logarithmic, p. 45. F. Important Definite Integrals, p. 46. G. Approximation Formulas, p. 47. H.' Standard Applications, p. 48. A. Fundamental General Formulas. i du dv, then u = V + constant. [Fundamnnra I Theorenm., dx dx 2. If ju dx = I, then d U M. [General Check.] 3. cu dx=f cudx. 4. Sulf [ ju dx=Sudx~Svdx. 5. fudv = uv - f v du. [Parts.l 6. LS f (u) duj =5fF,(X)J!!(Xix 'S ubstitdtion.). f d 36. STANDARD INTEGRALS [IV, B B. Integrand -Rational Algebraic. 7. f xn dx xn, n#t- 1, see 8. n + 1 NOTES. (a) f (Any Polynomial) dx, - use 3, 4, T. (b) f (Product of Two Polynomials) -dx, - expand, then use 3, 4, T. (c) fJcdx = ex, by 3, 7. 8. loge x =(login x) (log0 10) =(2.302586) log10 x. NOTES. (a) f (l/xm) dx, -use 7 with n== ssn if nt1 use 3iY s= 1. (b) f [(Any Polynomial)/xmn] dx, - use short division, then 7 and S. C dx i - J1a2-fX2 a a a a a x a a 0.fx 2-2 o alog ay-x[+const.]. NOTE. Allrational functionsaore iutegrated by reductions to 7,8, 9. The rednc.tions are performed by 3, 4, 6. No. lie and all that follow are results of this process. 11. S(ax~3 )n dx 1(ax +ib 2zl,)1 - 1. (See No. 12.) [From 7. 12. S da4x.. log (ax + b). [From 8.] NOTES. (a) Ax + Bdx, -use long division, then 7and 12. T+ b (b) An Polynomial dx, - use long division, then 7 and 12. ax +b 13C1 d =1 -1 ( )- Erm11 f (a + b)m a (m- 1) (ax+ b)m-1' m~1. [rm1. 14. 5(d 3 = -[ 3~ log (ax ~ 3)] [From 1 1, 12.] fa Ax+ B NOTES. (a) b2dx, - combine A times No. 14 and B times No. 13, m = 2. f (Any Polynomial) do, - use long division, then 7 and 14 (a); or use1. (aw + 3) 2 IV, B] RATIONAL ALGEBRAIC 37 15. [ F(x,ax + b)dx] - F b u du. [From 6.] Lui=ax+b a \VI I NOTES. (a) Restatement: put u for ax + b, ~- fer x, -d_ for ds. a a (5) J /(ax + b)3] dc, - use 15. Ans. 12 [ - + b]u= is 2U2 u2a-+bxZ () (An Polynomial) dx, - use 15; then 8 (b). (axe + b)3 (d) (Any Polynonlial d, - use 15 unless m <8,; but see 12 (b), 14 (b). (ax + b)m (e) Jfn (ax + b)m dx, - use 15 if mn > n; use 7 (b) if m < n; see also 51-54. 16. 1 - 1 [ _r a c l (ax + b)(cx + d) ad-bcLa b c + d NOTES. (a) a 7 -' -use 16, then 12. Special cases, - see 10 and 16 (b> ( (ax + b)(c? + d)) () ( b)r x:= fl --,]- dx. (Special case of 16 (a).) Jf(ax(+ 4) b ( x Sw + b] (C) r Ab +B dx, - use 16, then long division, 12. J (ax + b) (cx - +d) (d) (An y Polynomial) dx, -use 16, then long division, 7, 12. (aCx + b) (cx + d) (e) If ad - be = 0, 18 can be used. 17. [f (x, ax + b) dxl.=ax+b = F b a u - ua) Uk j]u=-J^~ \uu-a u-a) (u-a)2 NOTES, (a) Restatenmet: Putu; for a-+1. b forx; - fora + b; — bd for d? 1 - aL mt- a (n - a)2 (b) | - 1 (u - da)e+n;-2dg; then use 8b. (b) n(axe + b)n b-n+n-1 - 7m () a- = - - a lo. (d)f dx _ u2 - 4 a7- + 2 a2 logU. fx2(cax 4+ b) b 2 x3((X + b) 2 b3 18. f d b - b tan l x if a > O b >0. [See 9.] = 1 log V/a x-x/ b ifa> O b <0. [See 10.] 2 /-ab a Vax+ - b NOTES. (a) J -,'- use 8 (2nd part); b=-c. (b)J dc r d NOTES. ~LjXU~e2 - ea- Ja52 e 38 STANDARD INTEGRALS [IV, B 19. C xdx = 1 log (a2 b). J ax2 + b 2 a NOTES. (a) Jf w+ - -use long division, then 18. J A0 + B (b) dSm+, - use 18, 19. f ax2 + b *(Any Polynomial) (e) Polyno mial) d,- use long division, then 18, 19. _2 m mx - n 20. a - (mx + n)(ax2 + b) an2 + bm2 Lmx + a2 + NOTES. (a) f (mxa + n) (a,2 + b) dx, -use 20, then 12, 18, 19. AX2+- Bax + (7 A 1 B 1 +( Ah Bn\ 1 (mx + n)( (a'2 + b) a mw +n + zm a2 + b a m (ma+wln)(aw2.+b) Any Polynomial (c) j ( )(2 + ) d, - use long division, then 20 b, 12, 18, 20a. 21. a2+bx+c=a x+ b 2 - a4a. L 2 a J 4 a L - dx 1 du then 18. NOTES. (a) f ax2+ bx+ + = 2 -2 4ac'then 18. 2a 4a a - 4ac (b) [jF(a, a + bx c)d a du. [(b) I F / - aux + 2a (c) A Polynomial dx, - long division, then 7, 21, 21 b, 18, and 19. () Any Polynomial bx ( Any PlnCubic ald,-long division, then find one real factor of cubic, then use 21, 21 b. [If the cubic has a double factor, set u = that factor, then use 17 c.] 22 r xdx = 1 1 J (ax2 + b)2 2 a ax2 + b 23. x -x + 1 dx then 18. (ax2 + b)2 2 b(ax2 + b) 2 b J aX2 + b 24. ( xdx = [bt -(u= 1; then 7 or 8. 2. (aX2+b)m L2a1 Urn,ax+ 25. f dx x_ 1 2m-3 dx J (ax2+b) 2 b(rm-) (ax2 + b)m-l 2(m - 1)b (ax2+ b)mNOTES. (a) Use 25 repeatedly to reach 23 and thence 18. (b) Final forms in partial fraction reduction are of types 12, 24, 25 (by use of 21) IV, C] IRRATIONAL ALGEBRAIC 39 C. (a) Integrand Irrational: involving r = Vax + b. - b ) - 2 b r' 26. [ F(x, Vax + b) dx1 =$/( = fr2a r) dr. 27. V/ax + b dx = r 2 rax + b. 0.f a 3 a 28. 5 xVax +bddx = 2 f(r - r2) dr= 2r3 r2 b 29. dx =2 dr = 2 r. /ax + b a a x30. xva b Sr-L- b; use 9 or 10. Sx Vax + b 2 b 31. ^dx -=22a dr;use 23. x2 Vax +x b JS 2 - b) 2 NOTE. V/aa + b = (aw + b)/v/aa + b; (V/aw + b)3 = (aX + b7) V((a + b. (b) Integrand Irrational: involving x/~ x2 ~ a2. 32. = arc sin =sin-1 =- cos-l- + [const.]. - Va2-x2 a a a x33. = log (x + Vx2 a'2) = sinh-l [+ const.] for +, / VX2 ~ a2 a or cosh- x [ + const.] for - a 34. dx = sin- ) a = cos- [ + const.] f /2 ax - x \ a - a = vers-l (x/a) + const. ds x [ a 35. -dx Isec- _ = 1 cos_1a =- csc-l1[+ const.]. x/2- a a2 a x a a gg xdx = 02-2. 38. r-\/a2 X2)3. 36. v a2 - 2 38. x 2 —x2 dx= 37. ( x dx - /x2 ~ a2. 39. x Vx/2 + a2 dx= (Vx2 + a2). /x2 f a2 NOTES. (a) 32 and 33 furnish the basis for all which follow. (b) 36, 37, 38, 39 follow from xdx = d (x + const.)/2. 40 STANDARD INTEGRALS [IV, C 40. f x2dx _ X /a22 +X2 + sin-X. /a2 - X2 2 2 a 41. dx __ =-loga + a2x21 v'a2_ X2 a L X. 42. f dx =- /aqx2. x2 /a,2 _ X2 a2 43. (a) J a2d d= a2 sin-a. (b) f^ a dx-Va2 - X2-a log +-/ x x (c) - -X2 =_ 2 _X 2 _sf x2 x a ()/a2 -2dx = /a2x2 alog(x + o/a2 -- XJ Vx2 a2 2 2 45. V dx: a2 "a2 vIX2~ a2 a2x a a2 46. (a) f vx2 ~a2dx - a2 a2 log (+/ -2 a2). X2(b) /x: a'dx = /2 + a2 ~ a2 f- dx then 35 or 41. (b)J, x/x2 + a2' 5~X2 aS V2~a2 y a. (c). adx= + 2a2f d, then 32 or33. J aX2 x / X ~2 ~ a2 47. f dx = -—. 48. 5- dx (/a2 __ x2)a a2v /a2- x2 (x/x22 ~ a2)3 a2v\-2 a2 NOTES. Trigonometric Substitutions. If the desired form is not found in 32-48, try 79. Then use Nos. 55-79, see 79. (b) See also D 51-54, below. 49. /V~(ax2 + bx + c) =v'a /Vu2 k2, where u = +b and k2 b-4ac 2a 4 a2 IV, E] REDUCT1ON FORMULAS 41 50. 1ax+ b ax+b,v'(ax~~b)(cxA-d) 'cx + d \/(ax + b)(cx +d) ex ~d NOTES. (ac) Integrals containing r(~ax + ~b)/(ex + d~): use 50, then 49, then 32-48. (b) Substitution of cc =\ ~(ax + b)/I (cx + d) is successful without 50. D. Integrals of Binomial Differentials - Reduction Formulas. Symbols: u =aXn ~ b; a, b, p, rn, n, any numbers for which no de. nominator in the formula vanishes. 51. fxml(axn +b)P dx= [xnslu-P+ npb xuP1dx]. f m + ~~~~np + 1f 52. f xm"(axn + b)Pd - 1 [ xrnlup~i ~ (Di ~ a ~ lip ~ 1) SxmuP~1 dx]. 53 m a" + b) Pdx 1 [xla+luP+l - a (m ~ n + nP ~ 1) X~U x -M (r+1)bf 54 $ xmi axn ~ b)P dx. -a(nm ~1n + 1) f sn-~+1b$maPd] NOT-ES. (a) These reduction formulas useful when p, rn, or 2t. are fractional; hence applications to Irrational Integrands. (b) Repeated application may reduce to one of 32-48. (c) Do not apply if p, m, n, are all integral, unless it ~~2 and p large. Note 11, 15, 17-25. Za. Integrand Transcendental: Trigonometric Functions. 55. fsin xdx cos x. 56. $sin2 xdx~ -cos xsinux +x -sin 2 x +x. NOTE. f sin lcx do, - set kr. == u, and use 56. Likewise in 55-78. 57. Ssinn xdx= Sifla 5SCOSX2 xdx.. NOTE. If cc is odd, put sin2X cc1-coS2c and use 62. A A Nm. - -,., - ___i_ _._ 4Z 8TANDARD INTEGRALS [IV, E 58. fcosxd = sin. 59. fcos2xdx =- sin cos x + x x- = sin 2 x + x. 60. Scosn d COSx X Sin x + - 1 Scos x x dx. NOTE. If m is odd, put cos2Xa 1 - sin2x and use 63. 61. Ssin x cos xdx =- ~cos2 x = sin2x [ + const.]. n+ 1 63. fsin x cos xdx = sin +l x n 1. sn+1 ' 64. fsiln x cosm x dx - in+l C + -1 inn cosm-2 d = — sin-l x cos- + -- 1 sirn-2 x cosm xdx. m + tn m + n J NOTE. If n is an odd integer, set sin2 x = 1 - cos2 x and use 62. If m is odd, use 63. 65. fsin (x) cos (nx) dx = - cos[( n)x] _ cos(m - n)x] 2(m + i2) 2(m -n) m 4= & ne 66. fs(mx)sin(x))dx sin ([(m - n) ] sin [(m +n)x] 6.- 2(m - n) 2(m + n) 67. cos(mxcos (nx)dx= in [(m- n snin [(m + n) x] m n. 2(im - n) 2(m + n) 68. (tan xdx -log cosx. 69. ftan2 xdx = tanx-x. 70. Ctann xdx = - tann-2xdx. n n-1 d 71. 5ctnxdx =log sinx. 72. {ctn2xdx =- ctn x-. 73. cctnnxdx- Ctn1 ctn_-2 dx. n- 1 74 sec x dx = log tan2 + -) = log (sec x + tan x) [+ const.]. IV, E] TRIGONOMETRIC 43 75. fcsc x dx = log tan - log (csc x + ctn x) [ + const.. 76. sec2 dx = tan. 77. Icsc2 xdx =-ctn x. 78. Ssecmx scnx dx = - dx (See also 64.) sinn x cosm x 1 secm-1 x s x m + n - 2 secm-2 cscn x dx m- 1 m- 1 -_ _ _1_ secm-l x cscn-1 x + n - 2 sec7 x csc,-2 x dx. n- 1 n- 1 NOTES. (a) In 64 and 78 and many others, m and n may have negative values. (b) To reduce J [sinn/cosmax] d take m negative in 64. (c) To reduce [cosmw/sinn l] dx take n negative in 64. 79. Substitutions: _ u=. du sin x cos x tanx x dx (1) sin a cos w d u /1 u2 sin- tM d V"/ - i.2 /VI - '2 (2) cos - sinL d V1 - _/2 -- cos-' i - - N'a/2 - 1 1 du (4) sec x sec a tan x d - 2 - I sec-1' - ax 1 s 2u 1 -, 2 2,/ 2 d- 17 (5) ta, t an- -se d 2 tan-' + 2 2 2 1+s,22 1+- tt2 I-9s 1 + 112 Replace ctn x, sec x, csc x by 1/tan x, 1/cos x, 1/sin x, respectively. NOTES. (a) J F(sin a) cos z d, - use 79, (1). (b) J F(cos a) sin z dx,-use 79, (2). (c) J F(tan x) sec2a dc,-use 79, (3). (d) Inspection of this table shows desirable subsW.titions from trigonometric to algebraic, and conversely. Thus, if only tan, sin2x, cosa2 appear, use 79, (8). 44 STANDARD INTEGRALS [IV, E 0 dx I b1 s in-Ib asn-xif a2> b2 80. f dx 8 a - b sin ax - - b2 a + b sinx 1 b - /b- - a2 + a tan (/2) if a2< b2 V/b2 a2 b + x/b2 - a + a tan (x/2) 81. x = 2 tan-i la tan- a2 >b2 a + b cos x Ca2 - b2 La + b 2 a2> = 1 -log Vb + a + Vb/h-a tan (x/2) a2< b. V/b2 - a2 /b + a- /b- a tan (x/2) 82. 1 log tan + ~h, = sin-i1 - -a sin x + b cos x -/'a2 + b2 2 /a2 + b2 cos dx 1 NOTES. (a) J a+ b sinc - use 79, (1), = log (a + b sin a). )J f A+Bsina+ C cos cos i 7. +B 14 - c7, a+ sin x sin a+ sin b a+b sin ' then use 82 a, 80. (e) Many others similar to (a) and (b); e.g. J[sin x/(a + b cos a)] do, -use 79, (2). (d) f sin 2 cos-2 and like forms, -use 79, (3); see 79, note d. a2 sin2 a + &2 cos2 a (e) As last resort,,nse 79, (5), for any rational trigonometric integral. Eb. Integrand Transcendental: Trigonometric-Algebraic. 83. xn sin x dx =- x cos x + nm f m-l cos x dx. 84. xm cos x dx = xm sin x - m f -1 sin x dx. NOTES. (a) JI sin w dz - a cos a + - cos? dZ, - use 58. (b) Jf a^ sin a dza, - repeat 83 to reach 58. (c) f (Any Polynomial) sin a do, - split up and use 88. (d) For cos a repeat (a), (b), (e)., sin x d_ -sin x 1 rcos x 85.. ---- S + $ dx, m 1. J x (m - 1) X-1 m 1 J xm1-1 86. {cos x dx = _ cos x _1 n x 8 J ax_ ( -- 1):oL- m - d 1 J x-1 xM 1)-(in - 1) m —l in 1 ixm-i IV, E] TRANSCENDENTAL 45 87. sn -dx = [1 - +....- dx; see II, E, 13, p. 8, x 3! 6!. 88. JCO dx [L -! +!-.] dx; see 1I, E, 14, p. 8. NOTE. Other trigonometric-algebraic combinations, use 5; or 79 followed by 89-94. Ec. Integrand Transcendental: Inverse Trigonometric. 89. fsin- x dx = xsin-l x + v/1 - x2. [From 5.] 90. cos1 x dx = x cos-' x - vi - x2. 91. ftan- dx =xtan-x - log(1 + 2). 92. fxn sin-l x dx= Xn+1 sin-1 - 1 f xn+l dx then 53 or 54, 32,36. n + 1 n+l V/1-x2 93. Sx cos-1 x dx = xn+l cos-1 + 1 +l dx then 53 or 54, 32, 36. J n -+ It- 1 + /+1 -/l-x2 94. f xx tan- x dx = xll tan x 1r f x'+ldx then 19 (c). n + 1 n+l 1 + X2 NOTES. (a) Replace ctn-'1 by - tan-'1; or by tan' (l/x) and substitute Il/ =- u. (b) Replace sec-1 by cos-l (1/a), csc-1 by sin-' (l/a) and substitute l/m = u. (c) f (Any Polynomial) sin-laxZx, split up and use 92. (Similarly for cos-la, etc.) (d) f/'(ax) sin-la d, - use (5) with u = sin-la. (Similarly for cos-la and tan-'lM.) (e) Other Inverse Trigonometric Integrands, use 79 or 5. Ed. Integrand Transcendental: Exponential and LogarithAic 95. axdx = -a a = ax log e= c 0.4343. loge a log1 a logl a 96. ex dx = ex. NOTES. (a) J ekxd - ekx - k. (b) Notice ax = e(l~gea)z ekz, k = loge a. 97. Sx"ek dx = xne - n- xlek dx. J k kJ NOTES. (a) xekxdx = -xek-/k - ekx/k2. (b) J xnekXdx, - repeat 97 to reach 97 (a) (c) J (Any Polynomial) e,'dx, split up and use 97. 46 STANDARD INTEGRALS [IV, E 98. fekx- -- ekx + k ekx dx (repeat to reach 99). xm (m - 1) xxmm- m-1 J m99. dx- J ~ = +1 + 1 +. +- du, du =d kx; see Tables, V, H. X + L 2! 3! 100. f e n nx dx= ekx sin nx - n cos nx J in nx d k2 + n2 101. Sek cos mx dx - e k cos mx + m sin mx f Ak2 4- m2 102. flog xdx= x logx- x. 103. (log x) dx _ (10g x)n+. x n +1 104. $ dx =e du, u = log x; see 99 and Tables, V, H. 1lo)g x u 105. Sxa log x dx = x+rl ["io' _1 Ln +1 (n + 1)' 106. fe logd ek log - l e -k dx, see 99. k kf X F. Some Important Definite Integrals. 107. 5dx 1= if 1 > 1 (otherwise non-existent). - m alm m '1 108. odx =7r a2 + b2x2 2 ab 109. xne-x dx = r (n + 1) = n! if n is integral. See V, F, p. 56. NOTES. (a) In general, r (n + 1) = t * r (1n), as for n ', if >O0. (b) r (2) = r (i) = i, r(i /2) = x/. r (n + 1) == nII (t). 110. f lx(1 - x)dx = r (m + 1) r(n + 1). o^0 ~ r (m + n + 2) 111. Jsinnx sinmx dx= cos nx cosmxdx =, if m-i n, if m and n are integral. 112. sin2nx dx = cos2nx dx = 7r/2; n integral, see 56, 59. IV, G] DEFINITE INTEGRALS 47 113. S e-k dx = 1/k. 114. (f[(sin nx)/x] dx = r/2. 115. e-kx sin nx dx =n/(k2 + n2), if k > 0. 116. e -kx cos max dx = k/(k2 + m2), if k > 0. 117. i e-kxxdx = -n + 1) = knl, if n is integral. See 109. 118. e-k2X2dx = x/7r/(2 k). 0o vx/w e-m2/4k2 119. e-k2x2 cos mx dx = -- /4 if k > 0. fjo 2k ' 120. — 7 121. fo(logx)'dx=(-1)n! ( ekr+ee-k - o cosh kx 2- - o 122. log sin xLdx = log cos xdx =- log 2. do 2 Jo 2 123. f S112n+ d = r/2 COS2 x dx 2.4 6 n2 (n, positive 124. /2sin22+lxdx 3 5 s (2 1) (n, pitive 124. sin2'xd = oco = 1. 3.5... (2 -), positive J, - 2 * 4 * 6... 2 it 2 integer.) G. Approximation Formulas. 125. f (x) dx =f (c) (b-a), a<c< b. [Law of the Mean.] fa 126. f(x)fdx = f(b)2 (a) (- a). [Trapezoid Rule-precise a ^~2 ^yfor a straight line.] 127. f f(x)dx. [Extended Trapezoid Rule.] =[f (a)/2+ f (a+Ax) +f (a+ 2 Ax) + - + f [a+ (n-1)Ax]+f (b)/2]Ax. 128. f(x)dx f (a) + 4 f[(a + 1)/2] +f (b)(ba) 128. f f ()x) dx 6^ ^^ (b - ) Ja 6 [Prismoid Rule; or second Simpson-Lagrange approximation; precise if f (x) is any quadratic or cubic; see ~ 124, p. 202.] 129. 52f(x)dx = 3 [f(a) 4f(a + Ax) + 2f(a + 2 Ax) + 4f(a + 3 Ax) + 2f(a + 4 Ax) +... f(b)]. [Simpson's Rule; or extended prismoid rule. Ex. 13, p. 207.] 48 STANDARD INTEGRALS [IV, G 130. Sf (x)dx f(a)+ 3f [a ~ Ax] + 3ff[a + 2 Axl] f(b) (b - a); Ax (b - a)/3. 8 [A third Simpson-Lagrange Approximation. Extend as in 129.] 131. ~f(x)d, - TP(a))+ 32f [a + Ax] 12.f (a + 2 Ax]+ 32f[a + 3 Ax] + If(b) (b - a); Ax==(b - a)/4. [A fourth Simpson-Lagrange Approximation; see Lagrange interpolation formula, II, I, 17, p. 15.] H. Standard Applications of Integration. 132. Areas of Plane Figures: SdA (a) Strips AA parallel to y-axis: dA = y dx. (b) Strips AA parallel to x-axis: dA = x dy. (c) Rectangles AA = Ax Ay: dA = dx dy, A = ff dx dy. (d) Parameter form of equation: A = (1/2) f (x dy - y dx). (e) Polar sectors bounded by radii: dA (p2/2) dO. (f) Polar rectangles AA =p Ap AO: dA= p dpdO; A = f p dpdO. 133. Lengths of Plane Curves: f ds. (a) Equation in form y =f(x): ds = V/1 +[f'(x)]2 dx. (b) Equation in form x = 4(y): ds =/1 V [4f(y)f2dy. (c) Parameter equations: ds =Vd-2 + - dy2 (d) Polar equation: ds =Vdp2 + pzdO2. 134. Volumes of Solids: fdV. (a) Frsttumn (area of cross section A): d V7= A dh; V7= fA dh where', is tbe varial~e beigbt perpendicular to the cross section A. (b) Solid of revolution about x-axis: d V=' =ry2 dx. (c) Solid of revolution about y-axis: d V = 7rx2 dy. (d) Rectangular coordinate divisions: d V= dx dy dz; V = fff dz dy dx = Jf z dy dx = fJf z dy } dx=fA do. (e) Polar coordinate divisions; d V - p2 sin 0 dp do dO. IV, H] 'APPLICATIONS 49 135. Area of a Surface: 5 Jsee J dx dy., where ~1 is the angle between the element ds of the surface and its pro. jection dx dy. (a) Surface of Revolution about x-axis: A =Jf 2 7ry ds. (b) Surface of Revolution about y-axis: A = 2 7nc ds. 136. Length of twisted arcs f ds. (a) Rectangular Cobrdinates: cid =\/drI~(-JX2 + Cdy2 7+ dZ. (b) Explicit Equations y =.f (x), z= 4 (x): d= V1 + [f' (W)]2 + [4)' (X)]2. (c) n=f(t), p = (t), c= q(t) cis = /[f'(t)]2 + [0('J)]2~ [p'(t)]2. (d) Polar Co~rdinates: dIs = Vdp2 + p2d4)2 + p2 os2 4) dO2.O 137. Massofa body: M=5dM=fpdV, where p is the density (mass per uuit volume). (a) If p is constant: M= p f d V; see 134. (b) On any curve: d V= ds, if p = mass per unit length. (c) On any surface (or plane): d F= dA, ifp = mass per unit area. 138. Average value of a variable quantity q: A. F. of q.: (a) throughout a solid: q =f(n, y, z); A. V. of q. = fq dJ'- J I. (b) on an area A: A. V of q. ==f q dA f clA. (c) on an are s: A. TTJ of q= J'q ds -J'fds. 139. Center of iass, X, y, i): ~ =fxdM fdII, with similar formulas for y and Z. See dM, 137. (a) for a volume: dlJ= pdFV. (b) for an area: diff= p dA. (c) for an arc: dMf = p ds. 139.* Theoremis of Pappus or Guldin: (a) Surface generated by an arc of a plane curve revolved about an axis in its plane = length of arc x length of path of center of mass of arc. (b) Volume generated by revolving a closed plane contour about an axis in its plane = area of contour x length of path of its center of mass. 140. Moment of Inertia: I 5- 2 dl. (See 137, 139.) (a) For plane figures,, + 4 = I0, where 4x, ly, 10 are taken about the n-axis, the y-axis, the origin, respectively. (b) For space figures, Ix + ly ~ 4'= 1s* (a) I = I- + (w, - 1))1M, where I4 is taken about a line to the x-axis. 50 STANDARD INTEGRALS [IV, H 141. Iadius of Gyration: k2 = I~ 11 5 r2 dl 5-dM (In 140 and 141, r may be the distance from some fixed point, or line, or plane.] 142. Liquid pressure: P = 5ph dA, where P is the total pressure, dA is the elementary strip parallel to the surface'; h is the depth below the surface; and p is the weight per unit volume of the liquid. 143. Center of liqu.id pressure: h = 5h2 A* 5MA. 144. Work of a variableforce: W 5ff cos '1dsd, where f is the numerical magnitude of the force, ds is the element of the arc of the path, and r' is the angle between f and ds.. 145. Attraction exerted by a solid: F - F frn dfM where k is the attraction between two unit masses at unit distance, mn is the attracted particle, dM is an element of the attracting body r is the distance from m to dlLl. Components Fx, F5. F2 of F along Ox., Oy, Oz are: krcos a c7M krCos f dif cos y dJe Fx = kmj c2' Fy5 j r2' JZ= kin 2 where a. 6, v are the direction angles of a line joining ixi to ciM. 146. Work in an expanding gas: W =5p dv. 147. Distance s, speed v, tangential acceleration jT: jT =invdtj { 5S~t}t. [Similar forms for angular speed and acceleration.] 148. Errors of observation: (a) Probability of an error between w = a and x == b: P y dx, where y ie the probable number of errors of magnitude x. (b) The usual formula y = (hl/Vs) eh2xS gives: P (h2/p) J eux dx, where hs is the so-called measure of precision. (c) Probabillty of an error between x = - a and x = + a: P (a) X. (d) Probable error = (0.477)/ht = value of a for which P(a) = 1/2. (e) Mean error 5 y dx Y == I/(h/) V. NUMERICAL TABLES A. TRIGONOMETRIC FUNCTIONS [Characteristics of Logarithms omitted -determine by the usual rule from the value] De- SINE TANGENT COTANGENT COSINE Radians grees Va g Value log1o Value loglo Value loge l 61. 0000 00.0000 -0.0000 -0o.o oo 1.0000 0000 90~ 1.5708.0175 10.0175 2419.0175 2419 57.290 7581.9998 9999 89~ 1.5533.0349 2~.0349 5428.0349 5431 28.636 4569.9994 9997 88~ 1.5359.0524 30.0523 7188.0524 7194 19.081 2806.9986 9994 87~ 1.5184.0698 4~.0698 8436.0699 8446 14.301 1554.9976 9989 860 1.5010.0873 50.0872 9403.0875 9420 11.430 0580.9962 9983 85~ 1.4835.1047 6~.1045 0192.1051 0216 9.5144 9784.9945 9976 84~ 1.4661.1222 70.1219 0859.1228 0891 8.1443 9109.9925 9968 83~ 1.4486.1396 80.1392 1436.1405 1478 7.1154 8522.9903 9958 820 1.4312.1571 90.1564 1943.1584 1997 6.3138 8003.9877 9946 81~ 1.4137.1745 100.1736 2397.1763 2463 5.6713 7537.9848 9934 80~ 1.3963.1920 11~.1908 2806.1944 2887 5.1446 7113.9816 9919 790 1.3788.2094 12~.2079 3179.2126 3275 4.7046 6725.9781 9904 78~ 1.3614.2269 130.2250 3521.2309 3634 4.3315 6366.9744 9887 770 1.3439.2443 14~.2419 3837,2493 3968 4.0108 6032.9703 9869 760 1.3265.2618 15~.2588 4130.2679 4281 3.7321 5719.9659 9849 750 1.3090.2793 16~.2756 4403.2867 4575 3.4874 5425.9613 9828 74~ 1.2915.2967 170.2924 4659.3057 4853 3.2709 5147.9563 9806 730 1.2741.3142 18~.3090 4900.3249 5118 3.0777 4882.9511 9782 72~ 1.2566.3316 19~.3256 5126.3443 5370 2.9042 4630.9455 9757 71~ 1.2392.3491 20~.3420 5341.3640 5611 2.7475 4389.9397 9730 703 1.2217.3665 21~.3584 5543.3839 5842 2.6051 4158.9336 9702 69~ 1.2043.3840 220.3746 5736.4040 6064 2.4751 3936.9272 9672 680 1.1868.4014 23~.3907 5919.4245 6279 2.3559 3721.9205 9640 67~ 1.1694.4189 240.4067 6093.4452 6486 2.2460 3514.9135 9607 66~ 1.1519.4363 250.4226 6259.4663 6687 2.1445 3313.9063 9573 65~ 1.1345.4538 260.4384 6418.4877 6882 2.0503 3118.8988 9537 640 1.1170.4712 27~.4540 6570.5095 7072 1.9626 2928.8910 9499 63~ 1.0996.4887 28~.4695 6716.5317 7257 1.8807 2743.8829 9459 62~ 1,0821.5061 29~.4848 6856.5543 7438 1.8040 2562.8746 9418 61~ 1.0647.5236 300.5000 6990.5774 7614 1.7321 2386.8660 9375 60~ 1.0472.5411 31~.5150 7118.6009 7788 1.6643 2212.8572 9331 590 1.0297.5585 32~.5299 7242.6249 7958 1.6003 2042.8480 9284 58~ 1.0123.5760 330.5446 7361.6494 8125 1.5399 1875.8387 9236 570.9948.5934 340.5592 7476.6745 8290 1.4826 1710.8290 9186 56~.9774.6109 350.5736 7586.7002 8452 1.4281 1548.8192 9134 550.9599.6283 36~.5878 7692.7265 8613 1.3764 1387.8090,9080 540.9425.6458 370.6018 7795.7536 8771 1.3270 1229.7986 9023 530.9250.6632 380.6157 7893.7813 8928 1.2799 1072.7880 8965 52~.9076.6807 390.6293 7989.8098 9084 1.2349 0916.7771 8905 51~.8901.6981 400.6428 8081.8391 9238 1.1918 0762.7660 8843 500.8727.7156 41~.6561 8169.8693 9392 1.1504 0608.7547 8778 490.8552.7330 42~.6691 8255.9004 9544 1.1106 0456.7431 8711 48~.8378.7505 430.6820 8338.9325 9697 1.0724 0303.7314 8641 47~0.8203.7679 440.6947 8418.9657 9848 1.0355 0152.7193 8569 46.8029.7854 450.7071 8495 1.0000 0000 1.0000 0000.7071 8495 450.7854 Value log1o Value log1, Value log10 Value log1o De-...... m-....... P. - ___ I Q Ra... d...adians | t r - LJ'A;' I I AN(;rEN i I1NE;,. g-l 51 52 NUMERICAL TABLES [V, B B. COMMON LOGARITHMS N 0 1 2 3 4 5 6 7 8 9 D 10 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 42 11 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 38 12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 35 13 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 32 14 1461 1492 1523 1553 1584 1614 1644 1673 1703 1732 30 15 1761 1790 1818 1847 1875 1903 1931 1959 1987 2014 28 16 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 26 17 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 25 18 2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 24 19 2788 2810.,2833 2856 2878 2900 2923 2945 2967 2989 22 20 3010 3032 3054 3075 3096 3118 3139. 3160 3181 3201 21 21 3222 3243 3263 3284 3304 3324 3345 3365 3385 3404 20 22 3424 3444 3464 3483 3502 3522 3541 3560 3579 3598 19 23 3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 18 24 3802 3820 3838 3856 3874 3892 3909 3927 3945 3962 18 25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 17 26 4150 4166 4183 4200 4216 4232 4249 4265 4281 4298 16 27 4314 4330 4346 4362 4378 4393 4409 4425 4440 4456 16 28 4472 4487 4502 4518 4533 4548 4564 4579 4594 4609 15 29 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 15 30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 14 31 4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 14 32 5051 5065 5079 5092 5105 5119 5132 5145 5159 5172 13 33 5185 5198 5211 5224 5237 5250 5263 5276 5289 5302 13 34 5315 5328 5340 5353 5366 5378 5391 5403 5416 5428 13 35 5441 5453 5465 5478 5490 5502 5514 5527 5539 5551 12 36 5563 5575 5587 5599 5611 5623 5635 5647 5658 5670 12 37 5682 5694 5705 5717 5729 5740 5752 5763 5775 5786 12 38 5798 5809 5821 5832 5843 5855 5866 5877 5888 5899 11 39 5911 5922 5933 5944 5955 5966 5977 5988 5999 6010 11 40 6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 11 41 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 10 42 6232 6243 6253 6263 6274 6284 6294 6304 6314 6325 10 43 6335 6345.6355 6365 6375 6385 6395 6405 6415 6425 10 44 6435 6444 6454 6464 6474 6484 6493 6503 6513 6522 10 45 6532 6542 6551 6561 6571 6580 6590 6599 6609 6618 10 46 6628 6637 6646 6656 6665 6675 6684 6693 6702 6712 '9 47 6721 6730 6739 6749 6758 6767 6776 6785 6794 6803 9 48 6812 6821 6830 6839 6848 6857 6866 6875 6884 6893 9 49 6902 6911 6920 6928 6937 6946 6955 6964 6972 6981 9 50 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 9 51 7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 8 52 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 8 53 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 8 54 7324 7332 7340 7348 d 7356 7364 7372 7380 7388 7396 8 V, B] COMMON LOGARITHMS 53 - N 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 0 1 2 3 4 7404 7412 7419 7427 7435 7482 7490 7497 7505 7513 7559 7566 7574 7582 7589 7634 7642 7649 7657 7664 7709 7716 7723 7731 7738 7782 7789 7796 7803 7810 7853 7860 7868 7875 7882 7924 7931 7938 7945 7952 7993 8000 8007 8014 8021 8062 8069 8075 8082 8089 8129 8136 8142 8149 8156 8195 8202 8209 8215 8222 8261 8267 8274 8280 8287 8325 8331 8338 8344 8351 8388 8395 8401 8407 8414 8451 8457 8463 8470 8476 8513 8519 8525 8531 8537 8573 8579 8585 8591 8597 8633 8639 8645 8651 8657 8692 8698 8704 8710 8716 8751 8756 8762 8768 8774 8808 8814 8820 8825 8831 8865 8871 8876 8882 8887 8921 8927 8932 8938 8943 8976 8982 8987 8993 8998 9031 9036 9042 9047 9053 9085 9090 9096 9101 9106 9138 9143 9149 9154 9159 9191 9196 9201 9206 9212 9243 9248 9253 9258 9263 9294 9299 9304 9309 9315 9345 9350 9355 9360 9365 9395 9400 9405 9410 9415 9445 9450 9455 9460 9465 9494 9499 9504 9509 9513 9542 9547 9552 9557 9562 9590 9595 9600 9605 9609 9638 9643 9647 9652 9657 9685 9689 9694 9699 9703 9731 9736 974 9745 9750 9777 9782 9786 9791 9795 9823 9827 9832 9836 9841 9868 9872 9877 9881 9886 9912 9917 9921 9926 9930 9956 9961 9965 9969 9974 5 6 7 8 9 7443 7451 7459 7466 7474 7520 7528 7536 7543 7551 7597 7604 7612 7619 7627 7672 7679 7686 7694 7701 7745 7752 7760 7767 7774 7818 7825 7832 7839 7846 7889 7896 7903 7910 7917 7959 7966 7973 7980 7987 8028 8035 8041 8048 8055 8096 8102 8109 8116 8122 8162 8169 8176 8182 8189 8228 8235 8241 8248 8254 8293 8299 8306 8312 8319 8357 8363 8370 8376 8382 8420 8426 8432 8439 8445 8482 8488 8494 8500 8506 8543 8549 8555 8561 8567 8603 8609 8615 8621 8627 8663 8669 8675 8681 8686 8722 8727 8733 8739 8745 8779 8785 8791 8797 8802 8837 8842 8848 8854 8859 8893 8899 8904 8910 8915 8949 8954 8960 8965 8971 9004 9009 9015 9020 9025 9058 9063 9069 9074 9079 9112 9117 9122 9128 9133 9165 9170 9175 9180 9186 9217 9222 9227 9232 9238 9269 9274 9279 9284 9289 9320 9325 9330 9335 9340 9370 9375 9380 9385 9390 9420 9425 9430 9435 9440 9469 9474 9479 9484 9489 9518 9523 9528 9533 9538 9566 9571 9576 9581 9586 9614 9619 9624 9628 9633 9661 9666 9671 9675 9680 9708 9713 9717 9722 9727 9754 9759 9763 9768 9773 9800 9805 9809 9814 9818 9845 9850 9854 9859 9863 9890 9894,9899 9903 9908 9934 9939 9943 9948 9952 9978 9983 9987 9991 9996 D 8 8 8 7 7 7 7 7 7 7 7 7 6 6 6 6 6 6 6 6 6 6 6 6 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 4 4 4 4 54 1. NUMERICAL TABLES -. [V, C C. EXPONENTIAL AND HYPERBOLIC FUNCTIONS ex log, W Value 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 3.5 4.0 4.5 5.0 6.0 7.0 8.0 9.0 10.0 - 00 -2.303 -1.610 -1.204 -0.916 -0.693 -0.511 — 0.357 -0.223 -0.105 0.000 0.095 0.182 0.262 0.336 0.405 0.470 0.531 0.588 0.642 0.693 0.742 0.788 0.833 0.875 0.916 0.956 1.030 1.065 1.099 1.253 1.386 1.504 1.609 1.792 1.946 2.079 2.197 2.303 1.000 1.105 1.221 1.350 1.492 1.649 1.822 2.014 2.226 2.460 2.718 3.004 3.320 3.669 4.055 4.482 4.953 5.474 6.050 6.686 7.389 8.166 9.025 9.974 11.02 12.18 13.46 14.88 16.44 18.17 20.09 33.12 54.60 90.02 148.4 403.4 1096.6 2981.0 8103.1 22026. logiO Value 0.000 1.000 0.043 0.905 0.087 0.819 0.130 0.741 0.174 0.670 0.217 0.607 0.261 0.549 0 304 0.497 0.347 0.449 0.391 0.407 0.434 0.368 0.478 0.333 0.521 0.301 0.565 0.273 0.608 0.247 0.651 0.223 0.695 0.202 0.738 0.183 0.782 0.165 0.825 0.150 0.869 0.135 0.912 0.122 0.955 0.111 0.999 0.100 1.023 0.091 1.086 0.082 1.129 0.074 1.173 0.067 1.216 0.061 1.259 0.055 1.303 0.050 1.520 0.030 1.737 0.018 1.954 0.011 2.171 0.007 2.606 0.002 3.040 0.001 3.474 0.000 3.909 0.000 4.343 0.000 -x sin] log10 Value 0.090 0.000 9.957 0.100 9.913 0.201 9.870 0.305 9.826 0.411 9.783 0.521 9.739 0.637 9.696 0.759 9.653 0.888 9.609 1.027 9.566 1.175 9.522 1.336 9.479 1.509 9.435 1.698 9.392 1..904 9.349 2.129 9.305 2.376 9.262 2.646 9.218 2.942 9.175 3.268 9.131 3.627 9.088 4.022 9.045 4.457 9.001 4.937 8.958 5.466 8.914 6.050 8.871 6.695 8.827 7.406 8.784 8.192 8.741 9.060 8.697 10.018 8.480 16.543 8.263 27.290 8.046 45.003 7.829 74.203 7.394 201.7 6.960 548.3 6.526 1490.5 6.091 4051.5 5.657 11013. - cc 9.001 9.304 9.484 9.614 9.717 9.804 9.880 9.948 0.011 0.070 0.126 0.179 0.230 0.280 0.328 0.3'76 0.423 0.469 0.514 0.560 0.604 0.649 0.690 0.738 0.782 0.826 0.870 0.913 0.957 1.001 1.219 1.436 1.653 1.870 2.305 2.739 3.173 3.608 4.041 1.900 0 1.905 0.002 1.020 0.009 1.045 0.019 1.081 0.034 1.128 0.052 1.185 0.074 1.255 0.099 1.337 0.126 1.433 0.156 1.543 0.188 1.669 0.222 1.811 0.258 1.971 0.295 2.151 0.333 2.352 0.372 2.577 0.411 2.828 0.452 3.107 0.492 3.418 0.534 3.762 0.575 4.144 0.617 4.568 0.660 5.037 0.702 5.557 0.745 6.132 0.788 6.769 0.831 7.473 0.874 8.253 0.917 9.115 0.960 10.068 1.003 16.573 1.219 27.308 1.436 45.014 1.653 74.210 1.870 201.7 2.305 548.3 2.739 1490.5 3.173 4051.5 3.608 11013. 4.041 hX cosh x log10 Value log10 log, x = (loglo x)~M; M=.4342944819. logl0 ex+y = loglo ex ~ log10 ey. Sini x and cosh x approach ex/2 as x increases (see Fig. E, p. 22). The formula loglo (ex/2) - M x - log10 2 represents loglo sinh x and, loglo cosh x to three decimal places when x >3.5; four places when x > 5; to five places when x > 6; to eight places when x > 10. V, E] ELLIPTIC INTEGRALS 5 D. VALUES OF F~k, do 1 rx =sinel Vl~ k~2S1 n2 0./(I - x2) (I- k2x2)' lu = sinj [Elliptic Integral of the First Kind.] K k- "o P=10- p==15' p==30- 0b=45- p==60- 0k~75-' =-r1/36 =7ir/8.==r/12 =ir/6 ==iT/4 =ir/3 ==57T/12 O-90 -0.0 0.087 0.175 0.262 0.524 0.785 1.047 1.309 1.571 0.1 0.087 0.175 0.262 0.524 0.786 1.049 1.312 1.575 0.2 0.087 0.175 0 262 0.525 0.789 1.054 1.321 1.588 0.3 0.087 0.175 0.262 0.526 0.792 1.062 1.336 1.610 0.4 0.087 0.175 0.262 0.527 0.798 1.074 1.358 1.643 0.5 0.087 0.175 0.263 0.529 0.804 1.090 1.385 1.686 0.6 0.087 0.175 0.263 0.532 0.814 1.112 1.426 1.752 0.7 0.087 0.175 0.263 0.536 0.826 1.142 1.488 1.854 0.8 0.087 0.175 0.264 0.539 0.839 1.178 1.566 1.993 0.9 0.087 0.175 0.264 0.544 0.858 1.233 1.703 2.275 1.0 0.087 0.175 0.265 0.549 0.881 1.317, 2.028 00 E. VALUES OF E(k,c))=S'~'x/1-k2Sin20d0=5~ lk~2X2(d,x fX=Sine x/m 1- X2 u U sinlc [Elliptic Integral of the Second Kind.] -E k= ~ 5 sb=100 4150 4=300 4.=450 p==60- 4=750 =7r/36 ==7r/18.=.r/12 =7r/6 ~=aI/4 ==ir/3.=57r/12 0-90o - r/2 0.0 0 087 0.175 0.262 0.524 0.785 1.047 1.309 1.571 0.1 0.087 0.175 0.262 0.523 0.785 1.046 1.306 1.566 0.2 0.087 0.174 0.262 0. 523' 0.782 1.041 1.297 1.554 0.3 0.087 0.174 0.262 0.521 0.779 1.033 1.283 1.533 0.4 0.087 0.174 0.261 0.520 0.773 1.026 1.264 1.504 0.5 0.087 0.174 0.261 0.518 0.767 1.008 1.240 1.467 0.6 0.087 0.174 0.261 0.515 0.759 0.989 1.207 1.417 0.7 0.087 0.174 0.260 0.512 0.748 0.965 1.163 1.351 0.8 0.087 0.174 0.260 0.509 0.737 0.940 1.117 1.278 0.9 0.087 0.174 0.259 0.505 0.723 0.9.07 1.053 1.173 1.0 0.087 0.174 0.259 0.500 0.707 0.866 0.966 1.00 56 56NUMERICAL TABILES F. VALUES OF 11(p) =T(p +1) =Joe- 'xPdx p A PROPER FRACTION [II (n) = P (n + 1) = n!, if n is a positive integer.] p=0.0 p=0.1 p=0.2fp=0.3 p=0.4 p=0.5 p=0.6 p=0.7 P=0.8 p=0.9 r(p+l)= 1.000 0.951 0.918 0.897 0.887 0.886=V'/-7/2 0.894 0.909 0.931 0.962 r (k + 1) = kz (k), if k > 0; hence r(kA + 1) can be calculated at intervals of 0.1. Minimum value of P (p + 1) is.88560 atp = P.46163. G. VALUES OF THE PROBABILITY INTEGRAL: -fe- 2 dx. x.0.1.2.3.4.5.6.7.8.9 0..0000.1125.2227.3286.4284.5205.6039.6778.7421.7969 1..8427.8802.9103.9340.9523.9661.9763.9838.9891.9928 2..9953.9970.9981.9989.9993.9996.9998.9999.9999 1.0000 H. VALUES OF THE INTEGRAL [Note break at x = 0.1 n=1 n=2 n=3 n=4 n=5 n=6 n=7 n=8 n=9 x = n* -.2194 -.0489 -.0130-.0038 -.0012 -.0004-.0001 -.0000 -.0000 x=- n/10 - 1.823 - 1.223 -.9057 -.7024 -.5598 -.4544 -.3738 -.3106 -.2602 x =+ n/10 - 1.623 -.8218 -.3027 ~.1048.4542.7699 1.065 1.347 1.623 x=+n 1.895 4.954 9.934 19.63 40.18 ' 85.99 191.5 440.4 1038 * NOTE. os. Values on each side of x = 0 can be used safely. __- and e" dx reduce to the integral here tabulated; see IV, 99, 104, p. 46. o0 log x f V, I] RECIPROCALS SQUARES CUBES 57 Ii. RECIPROCALS OF NUMBERS FROM 1 TO 9.9.0.1.2.3.4.5.6.7.8.9 1 1.000 0.909 0.833 0.769 0.714 0.667 0.625 0.588 0.556 0.526 2 0.500 0.476 0.455 0.435 0.417 0.400 0.385 0.370 0.357 0.345 3 0.333 0.323 0.313 0.303 0.294 0.286 0.278 0.270 0.263 0.256 4 0.250 0.244 0.238 0.233 0.227 0.222 0.217 0.213 0.208 0.204 5 0.200 0.196 0.192 0.189 0.185 0.182 0.179 0.175 0.172 0.169 6 0.167 0.164 0.161 0.159 0.156 0.154 0.152 0.149 0.147 0.145 7 0.143 0.141 0.139 0.137 0.135 0.133 0.132 0.130 0.128 0.127 8 0.125 0.123 0.122 0.120 0.119 0.118 0.116 0.115 0.114 0.112 9 0.111 0.110 0.109 0.108 0.106 0.105 0.104 0.103 0.102 0.101 I2. SQUARES OF NUMBERS FROM 10 TO 99 0 1 2 3 4 5 6 7 8 9 1 100 121 144 169 196 225 256 289 324 361 2 400 441 484 529 576 625 676 729 784 841 3 900 961 1024 1089 1156 1225 1296 1369 1444 1521 4 1600 1681 1764 1849 1936 2025 2116 2209 2304- 2401 5 2500 2601 2704 2809 2916 3025 3136 3249 3364 3481 6 3600 3721 3844 3969 4096 4225 4356 4489 4624 4761 7 4900 5041 5184 5329 5476 5625 5776 5929 6084 6241 8 6400 6561 6724 6889 7056 7225 7396 7569 7744 7921 9 8100 8281 8464 8649 8836 9025 9216 9409 9604 9801 I3. CUBES OF NUMBERS FROM 1 TO 9.9..1.2.3.4.5.6.7.8.9 1 1.00 1.33 1.73 2.20 2.74 3.37 4.10 4.91 5.83 6.86 2 8.00 9.26 10.65 12.17 13.82 15.62 17.58 19.68 21.95 24.39 3 27.00 29.79 32.77 35.94 39.30 42.87 46.66 50.65 54.87 56.32 4 64.0 68.9 74.1 79.5 85.2 91.1 97.3. 103.8 110.6 117.6 5 125.0 132.7 140.6 148.9 157.5 166.4 175.6 185.2 195.1 205.4 6 216.0 227.0 238.3 250.0 262.1 274.6 287.5 300.8 314.4 328.5 7 343.0 357.9 373.2 389.0 405.2 421.9 439.0 456.5 474.6 493.0 8 512.0 531.4 551.4 571.8 592.7 614.1 636.1 658.5 681.5 705.0 t9 729.0 753.6 778.7 804.4 830.6.857.4 884.7 912.7 941.2 970.3 58 NUMERICAL TFABLES IV, j J1. SQUARE ROOTS OF NUMBERS FROM 1 TO 9.9.0.1.2.3.4.5.6.8.9 0 0.000 0.316 0.447 0.548 0.632 0.707 0.775 0.837 0.894 0.949 1 1.000 1.049 1.095 1.140 1.183 1.225 1.265 1.304 1.342 1.378 2 1.414 1.449 1.483 1.517 1.549 1.581 1.612 1.643 1.673 1.703 3 1.732 1.761 1.789 1.817 1.844 1.871 1.897 1.924 1.949 1.975 4 2.000 2.025 2.049 2.074 2.098 2.121 2.145 2.168 2.191 2.214 5 2.236 2.258 2.280 2.302 2.324 2.345 2,366 2.387 2.408 2.429 6 2.449 2.470 2.490 2.510 2.530 2.550 2.569 2.588 2.608 2.627 7 2.646 2.665 2.683 2.702 2.720 2.739 2.757 2.775 2.793 2.811 8 2.828 2.846 2.864 2.881 2.898 2.915 2.933 2.950 2.966 2.983 9 3.000 3.017 3.033 3.050 `3.066 3.082 3.098 3.114 3.130 3.146 J2. SQUARE ROOTS OF NUMBERS FROM 10 TO 99 0 1 2 3 4 5 6 7 8 9 1 3.162 3.317 3.464 3.606 3.742 3.873 4.000 4.123 4.243 4.359 2 4.472 4.583 4.690 4.796 4.899 5.000 5.099 5.196 5.292 5.385 3 5.477 5.568 5.657 5.745 5.831 5.916 6.000 6.083 6.164 6.245 4 6.325 6.403 6.481 6.557 6.633 6.708 6.782 6.856 6.928 7.000 5 7.071 7.141 7.211 7.280 7.348 7.416 7.483 7.550 7.616 7.681 6 7.746 7.810 7.874 7.937 8.000 8.062 8.124 8.185 8.246 8.307 7 8.367 8.426 8.485 8.544 8.602 8.660 8.718 8.775 8.832 8.888 8 8.944 9.000 9.055 9.110 9.165 9.220 9.274 9.327 9.381 9.434 9 9.487 9.539 9.592 9.644 9.695 9.747 9.798 9.849 9.899 9.950 K. RADIANS TO DEGREES RADIANS TENTHS HUNDREDTHS THOUSANDTHS T N-THlOUSAUDThlS 1 57017'44".8 5D43'46'h.5 0034'22".6 00 3'26".3 00 0'20".6 2 114035'29"'.6 11027'33".0 10 8'45".3 00 6'52".5 00 0'41''.3 3 171053'14" '4 17011'19".4 1043'07".9 0010'18".8 00 1'01".9 4 229010'59".2 22055'05".9 2017'30".6 0013'45"'.1 00 1'22".5 5 286028'44".0 28038'52".4 2051'53".2 0017'11".3 00 1'43".1 6 343046'28".8 34022'38".9 3026'15".9 0020'37".6 00 2'03".8 7 4010 4'13".6 400 6'25".4 40 0'38".5 0024'03".9 00 2'24".4 8 458021'58.4 4550'11".8 4035'01".2 0027'30"1.1 00 2'45".0 9 515039'43".3 51033'58".3 50 9'23".8 0030'56".4 00 3'05".6 V, MI CONSTANTS 59 L. IMPORTANT CONSTANTS AND THEIR COMMON LOGARITHMS N==' NUMBER VALITE OF N LOG10 IN 7r2 c = Napierian Base M= log10 e 1 M = log, 10 180 r = degrees in 1 radian 7r -- 180 = radians in 10 1r -- 10800 = radians in 1' 7r ~' (148000 = radians in 1" sin 1" tan 1" centimeters in 1 ft. feet in 1 cm. inches in 1 m. ponnds in 1 kg. kilograms in 1 lb. g (average value) weight of 1. cu. ft. of water weight of 1 cu. ft. of air cu. in. in 1 (U. S.) gallon ft. lb. per sec. in 1 H. P. kg. m. per sec. in 1 H. P. watts in 1 H. P. 3.14159265 0.31830989 9.86960440 1.77245385 2.71828183 0.43429448 2.30258509 57.2957795 0.01745329 0.0002908882 0.000004848136811095 0.000004848136811076 0.000004848136811133 30.480 0.032808 39.37 2.20462 0.453593 32.16 ft./sec./sec. 981 cm./sec./sec. 62.425 lb. (max. density) 0.0807 lb. (at 32' F.) 231 550. 76.0404 745.957 0.49714987 9.50285013 0.99429975 0.24857494 0.43429448 9.63778431 0.36221569 1.75812263 8.24187737 6.46372612 4 68557487 4.68557487 4.68557487 1.4840158 8.5159842 1.5951654 0.3433340 9.6566660 1.5073 2.9916690 1.7953586 8.907 2,3636120 2.7403627 1.8810445 2.8727135 M. DEGREES TO RADIANS 10.01745 100.17453 1000 1.74533 6'.00175 6".00003 20.03491 200.34907 1100 1.91986 7'.00204 7".00003 30.05236 300.52360 1200 2.09440 8'.00233 8".00004 40.06981 400.69813 1300 2.26893 9'.00262 9".00004 50.08727 500.87266 1400 2.44346 10'.00291 10".00005 60.10472 600 1.04720 l500 2.61799 20'.00582 20".00010 70.12217 700 1.22173 1600 2.79253 30'.00873 30".00015 80.13963 800 1.39626 1700 2.90706 40'.01164 40".00019 90.15708 900 1.57080 1800 3.14159 50'.01454 50".00024 60 NUMERICAL TABLES [V, N N., SHORT CONVERSION TABLES AND OTHER DATA: MULTIPLES, POWERS, ETC., FOR VARIOUS NUMBERS n=t n=2 n=3 n=4 n=5 n==6 n=7 n=8 n-9 7r n 3.1416 6.2832 9.4248 12.566 15.708 18.850 21.991 25.133 28.274 * n2/4.78540 3.1416 7.0686 12.566 19.635 28.274 38.485 50.265 63.617 7r n/6.52360 4.1888 14.137 33.510 65.450 113.10 179.59 268.08 381.70 +r n 3.141.6 1.5708 1.0472.78540.62382.52360.44880.39270.34907 n 7w.31831.63662.95493 1.2732 1.5915 1.9099 2.2282 2.5465 2.8618 (7r/180).n.01745.03491.05236.06981.08727.10472.12217.13963.15708 (180/ir). n 57.296 114.59 171.89 229.18 286.48 343.77 401.07 458.37 515.66 e n it 2.7183 5.4366 8.1548 10.873 13.591 16.310 19.028 21.746 24.465 M.n.43429.86859 1.3028 1.7371 2.1714 2.6057 3.0400 3.4744 3.9087 (1+ A!).n 2.3026 4.6052 6.9078 9.2103 11.513 13.816 16.118 18.421 20.723 1 -+I n 1.0000.50000.33333.25000.20000.16667.14286.12500.11111 n2 1 4. 9. 16. 25. 36. 49. 64. 81. n3 1. 8. 27. 84. 125. 216. 343. 512. 729. n 1. 16. 81. 256. 625. 1296. 2401. 4090. 6561. n5 1 32. 243. 1024. 3125. 7776. 16807. 32768..59049. 25. 2n 64. 128. 256. 512. 1024. 2048. 4096. 8192. 16384. 3fl 3. 9. 27. 81. 243. 729. 2187. 6561. 19683. vh 1. 1.4142 1.7321. 2. 2.2361 2.4495 2.6458 2.8284 3. ~i 1. 1.2599 1.4422 1.5874 1.7100 1.8171 1.9129 2. 2.0801! 1. 2. 6. 24. 120. 720. 5040. 40320. 362880. 1 + n! 1. 0.5.16667.04167.00833.00139.00020.00002.000003 Bn* 1+'.6 1. 1+30 5. 1 42 61. 1 -+-30 1385. 5+'.66 cm. innitin. 2.5400 5.0800 -7.6200 10.160 12.700 15.240 17.780 20.320 22.860 in. in 'n cm..39370.78740 1.1811 1.5748 1.9685 2.3622 2.7559 3.1496 3.5438t m. in n ft..30480.60960.91440 1.2192 1.5240 1.8288 2.1336 2.4384 2.7432 ft. in n m. 3.2808 6.5617 9.8425 13.123 16.404 19.685 22.966 26.247 29.527 km. in nmi. 1.6093 3.2187 4.8280 6.4374 8.0467 9.6561 11.265 12.875 14.484 mi. in nkm. 0.6214 1.2427 1.8641 2.4855 3.1069 3.7282 4.3496 4.9710 5.5923 kg. in nilb..45359.90719 1.3608 1.8144 2.2680 2.7216 3.1751 3.6287 4.0823 lb. in n kg. 2.2046 4.4092 6.6139 8.8185 11.023 13.228 15.432 17.637 19.842 1. inn qt..94636 1.8927, 2.8391 3.7854 4.7318 5.6782 6.6245 7.5709 8. 5 172) qt. in nl. 1.0567 2.1134 3.1700 4.2267 5.2834 6.3401 7.3968 8.4534 9.5101 *Bn =nth Bernoulli number; see II, E, 15-18, p. 8. Exact legal values in U. 8. INDEX [Numbers in roman type refer to pages of the body of the book; those in italics refer to pages of the Tables.] Absolute value, 14. Acceleration, 60, 62; angular, 72; component, 63; of a reaction, 78; tangential, 60; total, 63. Algebraic functions, 24, 41. Amplitude of S. H. M., 126. Analytic geometry, formulas, 16. See also Curves. Anchor ring, 11. Annuity, 5. Approximate integration, 193, 47. Approximation. See also Error, Lagrange, Prismoid, Simpson, Taylor. Approximations, formulas for, 47: polynomial, 234, 255, 30; SimpsonLaGrange, 31; Taylor, 30; trigonometric, 8, 31. Area, polar coordinates, 149; of a surface, 302; surface of revolution, 137, 200. Areas, 90, 215, 48. Astroid, 26. Asymptotes, 188, 189. Atmospheric pressure, 111. Attraction, 232, 50. Average value, 224, 231, 49. Bacterial growth, 111. Beams, 71, 213. Bernouilli numbers, 60. Binomial differentials, 184, 41. Binomial theorem, 269, 7. Cardioid, 136, 26. Cassinian ovals, 29. Catenary, 108, 137, 22. Cavalieri's Theorem, 202. Center of gravity, 224, 225, 226, 49. Center of mass, 49. See also Center of gravity. Center of pressure, 231. Centroid. See Center of gravity. Chance, 6. Circle, 9, 15. Circular measure, 119. See also Radian. Cissoid, 229, 28' Coefficient of expansion, 114. Combinations, 6. Compound interest law, 110. Concavity, 65. Conchoid, 28. Cone, i1. Confocal quadrics, 310. Constants, 1; of integration, 313; 60. Continuity. See Functions, continuous. Contour lines, 27. Conversion tables, 60. Cooling, in fluid, 111. Critical point, on a surface, 291. Critical points, for extremes, 53. Cubes, table of, 57. Curvature, 139, 154; center of, 142; radius of, 141. Curves, 19, see also Functions; cubic, 27; parabolic, 15, 19, see also Polynomials; quartic, 27. Curvilinear coordinates, 299. Cycloid, 136, 143, 24. Cylinder, 10; projecting, 299. Cylindrical coordinates, 233. Damping, of vibrations, 24. Definite integrals, 46. 61 62 INDEX Depreciation, 5. Derivative, 19; of a constant, 25; of a function of a function, 31; of a power, 25, 27, 34; of a product, 30; of a quotient, 28; of a sum, 25; logarithmic, see Logarithmic; partial, 274, 18; total, 278. Derivatives, notation for, 19; second, 61; of inverse trigonometric functions, 128; of exponentials, 107; of logarithms, 103; of trigonometric functions, 120, 121. Derived curves, 68. Determinant, 5. Difference quotient, 6. Differential, partial, 283; total, 279. Differential coefficient, 19. Differential equations, 82, 127, 311; exact, 325; extended linear, 323; higher order, 338; homogeneous, 317, 329, 338; linear, 320; linear, constant coefficients, 338; nonhomogeneous, 332, 340; ordinary, 311; partial, 311; second order, 326; separable, 316; special types, 334; systems of, 342. Differential formulas, 44, 132, 15. See also Derivatives. Differentials, 43; exact, 325; notation for, 43; transformation of, 18. Differentiation, 20; formulas for, 35, 44, 132, 15. Direction cosines, 16. Distribution of data, 30. Electric current, 114, 277. Elimination of constants, 312. Ellipse, 9, 15. Ellipsoid, 11, 14. Elliptic functions, 55. Elliptic intervals, 183. Empirical curves, 234. Energy integral, 337. Envelopes, 284. Epicycloid, 25. Epitrochoid, 24. Equations, differential, see Differential; in parameter form, 32, see also Parameter; solution of, 4. Error curve, 30. Errors, of observation, 50. Evolute, 142, 145, 286. Explicit functions, 40. Exponentials, 107, 22, see also Logarithms; differentiations of, 107; table of, 54. Exponents, S. Extremes, 52, 257, 290; final tests for, 54, 66, 294; weak, 291. Factors, 4. Falling bodies, 87, 208. Family, of curves, 21. Finite differences, 248. See also Increments. Flexion, 61. Flow of water, 308. Fluid pressure. See Water pressure, Atmospheric pressure, etc. Folium, 41, 28. Force, work done by, 50. Fourier's theorem, 8, 31. Frustum, of a cone, 11; formula, 96; of a solid, 95. Functions, 1; continuous, 11, 15; derived; 19; notation for, 1; implicit, etc., see Implicit, etc.; of functions, 31; algebraic, rational, etc., see Algebraic functions, etc.; classification of, 24. Gamma function, 56. Gases, expansion of, 38, 48, 78, 105, 110, 50. Geometry, of space, 16. Graphs, 2. Gudermannian, 131, 13. Guldin and Pappus, Theorem, 49. Gyration, radius of. See Radius. Harmonic functions, 23. See also Trigometric. Helicoid, 300. Helix, 300. Hooke's Law, 125. Hyperbola, 10, 15. Hyperbolic functions, 108, 13, 22, 54, inverse, see Inverse. Hyperbolic logarithm. See Logarithms. INDEX 63 Hyperboloid, 33. Hypocycloid, 26. Hypotrochoid, 25. Implicit functions, 40. Improper integrals, 190. Increments, 4, 248; method of, 238; second, 239, see also Finite differences. Indeterminate forms, 259, 262. Inertia, moment of. See Moment. Infinite series. See Series. Infinitesimal, 14; principal part, 261. Infinitesimals, higher order, 261. Infinity, 16. Inflexion, point of, 65. Integral, as limit of sun, 192, 197; fundamental theorem, 86; indefinite, 83; notation for, 83. Integral curves, 313, 343. Integrals, definite, 87, 46; double, 210; elliptic, 183, 9, 55; improper, 190; infinite limits, 188; infinite integrand, 189; multiple, 208, 215; table of, 35; triple, 208. Integral surfaces, of a differential equation, 343. Integrand, 83. Integraph, 243. Integrating factor, 325. Integration, 83; approximate, 193, see also Approximation; by parts, 163, 35: by substitution, 158, 35, 43; formulas for, 156, 35; of a sum, 84; of binomial differentials, 184, 41; of irrational functions, 129, 39; of linear radicals, 172, 39; of polynomials, 84, 158; of quadratic radicals, 173, 39; of rational functions, 165, 36; of trigonometric functions, 157, 172, 177-182, 41; reduction formulas, 184, 41, 42; repeated, 208; successive, 209. Interpolation, Lagrange's formula, 15. See also Lagrange. Inverse functions, 3. Inverse hyperbolic functions, 131, 14,54. Inverse trigonometric functions, 128. Involute, 146. Irrational functions, 25; differentiation of, 34; integration of, 164, 172. Isothermal expansion, 105. Kinetic energy, 231. Lagrange interpolation formula, 15,. 47. Law of the mean, 247, 47; extended, 253, see also Taylor's theorem. Least squares, 58, 296, 309, 6. Lemniscate, 29. Length, 133, 48; polar coordinates, 152; of a space curve, 305. Limits, 14; arc to chord, 133; properties of, 15; sin 0 to 0, 119. Liquid pressure, 50. Logarithmic derivative, 115. See also Rates, relative. Logarithmic plotting, 234, 20. Logarithms, computation of, 7; graph of, 21; hyperbolic, 102; Napierian, 102, 54; natural, 102; rules of operation, 99, 3; table of, 52. Maclaurin's Theorem, 258. See also Taylor's Theorem. Mass, 49. Mathematical symbols, 1-3. Maximum, 6. See also Extremes. Mean square ordinate, 231. Mensuration, 9. Minimum, 6. See also Extremes. Modulus, of logarithms, 103. Moment of inertia, 219, 221, 49; polar coordinates, 220. Motion. See Speed, Acceleration, etc. Napierian base e, i02. See also Logarithms. Natural logarithms. See Logarithms. Normal, 5, 49; length of, 50; to a surface, 298, 300. Notation, 1. Numbers, e, M. See Logarithms. Organic growth, law of, 111. Orthogonal trajectories, 325. 64 INDEX Pappus Theorem, 49. Parabola, 10. See also Curves, parabolic. Paraboloid, 11, 34. Parameter forms, 32, 50, 134. Partial derivative, 274, see also Derivative; order of, 275. Partial derivatives, geometric interpretation, 277; transformation, 18. Partial differential. See Differential. Partial fractions, 165. Pendulum, 127, 238, 250. Percentage rate of increase, 112. See also Rates. Period, of S. H. M., 126. Permutations, 6. Phase, of S. H. M., 126. Plane, equation of, 16. Planimeter, 243. Point of infiexion, 65. Polar coordinates, 5, 147; plane area, 216; moment of inertia, 220; space, 300. Polynomial, approximations. See Approximations. Polynomials, 24, see also Curves, parabolic; differentiation of, 25; roots of, 65; integration of, 84, 158. Power curves, 19. Power series. See Taylor series. Primitive, of a differential equation, 313. Prism, 10. Prismoid, defined, 205. Prismoid rule, 202, 10, 47. Probability, 6, see also Least Squares, Error curve, 30; integral, 50, 56. Pyramid, 10. Pythagorean formula, 134. Quadric surfaces, 33; confocal, 310. Quartic curves, 27. Radian measure, table of, 51, 58. Radium, dissipation of, 114. Radius of curvature. See Curvature. Radius of gyration, 220, 50. Rates, average, 18; instantaneous, 19; percentage, 112; related, 74; relative, 112, 114; reversal of, 79, see also Integrals; time, 10, 60. Rational functions, 24; differentiation of, 28; integration of, 165. Reactions, rates of, 78, 114. Reciprocals, table of, 57. Reduction formulas, 179, 185, 41. Relative rate of increase, 112, 114. See also Rates and Logarithmic derivative. Rolle's Theorem, 247. Roulettes, 24. Semi-logarithmic plotting, 236. Series, alternating, 270; convergence tests, 266; differentiation of, 272; geometric, 265, 7; infinite, 7; integration of, 272; precautions, 269; Taylor, 266. Simple harmonic motion, 124, 328, 23. Simpson-Lagrange approximations, 31. Simpson's rule, 207, 47. Singular solution of a differential equation, 313. Slope, 4. Solution of equations, 4. Speed, 9, 60, 62, see also Motion; component, 10; angular, 72; total, 42, 134; of a reaction, 78. Sphere, 11. Spherical coordinates, 232, 300. Spirals, 32. Square roots, table of, 58. Squares, table of, 57. Strophoid, 27. Subnormal, 50. Subtangent, 50. Summation, approximate, 193; exact, 196. Summation formula, 197. Surfaces, quadric, 33. Table of integrals, 156, 35. Tables. See special titles. Tangent, equation of, 5, 49; length of, 50; to a space curve, 305. Tangent plane, to the surface, 289, 297, 300. Taylor series, 266, 30. INDEX 65 Taylor's Theorem, 253, 8, see also Law of the Mean. Time rates. See Rates. Total derivative, 278. Total differentials, 279. Tractrix, 26. Trajectories, orthogonal, 325. Transcendental functions, 25. Trapezoid rule, 47. Trigonometric functions, table of, 12, 21, 51. Trigonometry, 9. Trochoid, 24. Variable, 1; dependent, 1; independent, 1. Velocity, 60. See also Speed. Vibration, 125, 23; electric, 125. Volume, of frustum, 95, 199; of solid of revolution, 94. Volumes, 94, 95, 210, 48. Water pressure, 201, 231, 50. Witch, 28. Work, of a force, 60; on a gas, 105, 110. ANSWERS TO EXERCISES ~6. Page 9 1. y=2x-3. 3. 8x-y-11. 5. 3x-y=2. 7. 3x+y=4. ~9. Page 12 3. 82; 114; 178; 50 +32 T. 5. 2; 6; 40; 2 T. 7. 80; 16; -80. 9, t =5/2; s= 100; v = 0. 11. 1; 2 t; VI1~4 t2. ~ 11. Pages 17-18 5. 9. 7.- 5/12. 9. - 1/2. 11. (a + b)/(c + d). 13. 1. 15. 0. 17. 3/5. 19. - 1/2. 23. 5/3. 25. 2. 27.Vap ~314. Pages 22-23 1. 2 x- 4. 3. 3 - 3 X2 5. 8-4Xr3. 7. -1I/ (X- 1)2. 9. - 2/X3. 11. 8/(x +2)2. 13. 4x+gy=24. 15. Rises whenI xI > -\v'5; falls whenIxI <x./~-; slope zero when x = ~ -,'5. 17. Rises when x > 2; f alls when x < 2; slope zero when x = 2. 19. Hor. speed, ix/ldt 15; vert. speed, dy/cdt =-32 t + 15. 21. 8irr; 3 a2; 2 7rrh/3. ~16. Pages 27-28 1. 12 X3. 3. 20 X4. 5. 50 (t4 + 1). 7. 12 t(t2 - 1). 17. 13 ay'2 - 9 aby8. 19. mkxm-l + nhXn-1. 21. 3 nt3n-1 - (n + 2)tPi+l. 23. (- 2, 0). 25. (1, 6). 27. Slopes at x = 0, 2,- 2, 4, - 4 flesp. are 0, 12, 36, 72, 120; slope is 9/2 at x =3/2 and - 1/2;slope is - 3/2 at x =1/2. ~317. Pages 29-30 1. -1/x2. 3. 6/ (x +4)2. 5-/x. 7. -28 x-8 -4x-3. 9. - (t4 + 2 t)/'(t3 - 1)2. 11. 4 U/(U2 - 1)2. 13. 16/3 + 1/u2. 15. 0. 17. - 6/Z4 - 6 Z/(Z2 + 1)2. 19. 2(1 - r2)/( r2 - r + 1)2. 25. - (v6 + 3,V2)/(VA - 1) 2. 27. -.x/(ax + b)3. 29. (y2- 2)/(y -1)4. 31. tainl(-3). 1 2 ANSWERS ~ 20. Pages 33-34 1. 6x2-2. 3. 4ax+3. 5. 12x2-5x4 -2x. 7. 4x(xz2- 1). 9. 6X2(x2 —3). 11. 24x(3x2 + 5)3. 13. 4(1 -3t)(1 + 2t-3t2). 15. 3(3 t - l)(t3 - t - 4)2. 17. 8(2 V3 + 3 v) (V4 + 3 2-2). 19. -6 t2/(t3 + 1)3. 21. (6x2- 3 4 - 18x)/(X2 + 2)4. 23. -8 s(2s2 -3)-3. 25. 6(2 + 4 2 - 5v)(3 -5)(v2 + 2)2. 27. 6ts+12 t3+12t2+-12. 29. 156 - 108 x. 31. (2/5)[3(1-2 )-2+8]. 33. 3/(2 t). 35. Horiz. tangents at x = 1/2, - 4/3, - 5/12. ~ 23. Pages 37-38 1. (4/3)x. 9. (22/3)t3 + 7 t. 13. 3 2 /2 + 3 x 19. 3(1 + x2)s 25. 4 x+x vX 1 1 3. (4/3)X. 5. (3/4)xaT. 7. -5X-2 — 5a-' 11. - 6B-4-(1/2)x4 + (2/3)ax-. 4+9u 17 2t-3 2/2 -+ 3 u 2t2- t 15x2- 16x 22x-1 21. 23. 2\/3 x- 4 2/1 -- + x2 2 u-2 2 2 t- 3t3 27. - V 29. -2 2 u2\/a2 -u / - t2 31. 2(x + x2// + X2. Tangents: 35. 2x/2y-x=2. 37. 2V/2y = 5x- 1. 41. dp/dv =- 1.41 kv-24L. ~ 25. Pages 41-42 33. 3x+4y=5. 39. tan-1(5/12). 1. - 2y/x or - 2/x3. 3. (2x - y)/x or 1 + 5/x2. 2 5. x/4y or x/(2 x2 - 36). 7. - 2/y2 or - x2/(a3 - x3). 9. 3(1 - x2)/(2 y) or 3(1 - x2)/2V/3x - a3. 11. (2 - y2)/(2 y + 2 xy) or (2 - x)/2(1 + ). 13. - y/x. 15. 1/(2 y). 17. — Vy/x. 19. 4t; 2 y=z2. 21. t; 27y2 = 4(x- 1)3. 23. (t2- 1)/(2t); x2 +2= 1. 27. dy/dx > 0 when x < 0 and v. v. 29. v = (1/2) /2 + 2 t2. ~ 27. Pages 46-48 1. (2ax+ b)dx. 3. 3 (2 ax+ b) (ax2-bx+c)2dx. 5. -a(ax;+b)-2dCx. 7. -(12t+ l)dt. 9. t(a-t)2(2a-5t)dt. 11. (3-2 t)dt/2 /3t-t2. 13. (9/2)x/t3-3t(t2 — )dt. 15. -(l+v)dv/(2v + v2). 17. -(l+v)(2 + 3V2 + v3)dv/(v3 _ 1)2. 19. 4y2(2- y3)-cy. ANSWERS 3 21. (1/2)(2 +y) (1 + y) 2dy. 25. -bmpSn-lcls/(a + bsn)p+l. 29. 31. (2X3-Xy2-x)dx 1x.( bX2 + 2 ax +by2) dx 37. y + 2 y(a - bx) 45. -(2 +O)(1 +O)2/63. 47. 9/(2 r). - 4t/( + t2)2; 2/(1 +t2). 51. dv =(b - 23. bnps-1(a + bSn)P-Ids. 33 a (X2 +y2) dX 2 y(l - ax) 49. 2(1 - t2)/(l + t2)2; - v) dp/(p - a/V2 + 2 ab/V3). ~ 31. Page 52 Tang. Norm. Subt. Subn. Tang. Norm. 1. 9x+y=5. x -9y =37. 4/9. 36. 4 V -2/ 9. 4 -82 3. 9x+4y=25. 9y-4x=32. 169. 9 4/7/9 / 5. x+3y=6. 3x-y=8. 3. 1/3. v'EY. 1- V~/ 3. Tangents: 9. y ~ yo = 2 krxx. 11. xxO0 ~ yyO = a2. 13. b2XX0 + a2yyO =a2b2. 15. axxO + b (xyo + xoy) + cyyO, + d (x + xO) + e (y + yo) + f =0. ~ 35. Pages 56-59 1. 3. 5. 7. 9. Max. x=-0. q =-1. y =O. n =1 t8/3. Min x =4. q =. y =+xV2-. n =3. t=4. 11. 13. 15. 17. 19. Max. s =0. None. x =4. (1+x\/6)/2. r =2. Min. = ~ v5/2. x=-2. x =0. (1v)/2. r =- 2. 21. A square; area 400. 23. (1 7,' ~ 7). 25. Ht. = diam. 27.. Width= 2 x depth. 29. Depth =V x breadth. 31. x/6~y/8=1. 33. Max. 3; Min. 1: the variable x does not increase steadily when the function D2 goes through its minimum or maximum, as the general theory requires. 39. Compromise: 420~ a foot; average: 42.560~ a f oot. 45. Width = 1/2 x base of A. 47. Height = (2 x/3/3) x radius of sphere. 49. Had. Nr(~5+ V5)~/10 x radius of sphere. 51. 2 ab. ~ 40. Pages, 64-65 1. 2x+5,2. 3. 4x-1,4. 5. 2 x- 5/2, 2. 7. 6(X2+x-6),6(2x~1). 9. 4X3-,6X2+lOX,12X2-12x+10. 11. 1/2-x.x+ x/ V'2 + 1, - 1/ (4 VX3) + 1/ V (2 + 1). 13. (x +2)2(5X2 +4x -3), 2(x +2)(1X2 + 16x+l1). 15. a, 0. ANSWERS 17. 2 ax + b, 2 a. 19. [mn(x - b) +n~x - a)](x - a)m-l(x - bn1 [m(m - 1)(x - b)2 + 2 mrn(x - a)(x - b)+ n(n - 1)(x - a)2] x (x - a)m-2(X - b)n-2. 25. M = -1/IX2, b = 2/X3. 27. m=2(3x2-7x-13)(x+3)(x-2)2, b=2(x-2)(l5x3-l0x2 - 110 X -10). 33. d/b. b:0. VX: b. Va:. V v'b2 +d2. jX: 0. jV: 0. j 0. jt: 0. 35. - 2 t-3. 6 t-4. 1. - 2t-3. V/I + 4 t-6. 0. 6 t -4. 6 t-4. - 12 t-4 v'/t2 + 4. 37. - (1 + t) 2/t2. 2(1 + t)3/t3. (1 + ty-2. - t-2. -x (I ~ t) 4 + t-4. - 2(1 + t)-3. 2 t3. - 2(1 + t)-5 -j2 t-54 ~ 44. Page 70 3. 5. 7. Max. x =- 1. - 11/6. 0. Mil. X =5. 5i/6. 4~V 3. Infl. x =2. - 1/2. 1 9.11 Max. - B/(2 A), A< 0; none, A> 0. None. Min. None, A< <0; - B/ (2 A), A >0. None. Infl. None. None. 23. y= 3X2 +ax +b; X3/3+ 3X2/2 +ax +b; ax +b. 27. 29. 13. None. - 2. 2 / Infl. None. Max. Defl.x= X~l1/2. x = 31/4. x = 1(1 + V%33)/16. ~ 46. Page 73 1l. co = 3 t2/1000 degr./sec.; a = 6 t/1000 degr./sec.2. 3. Co -(- t3/4 - 1/32) rad./min. -[- t3/ (S7) -1I/(64 7r) ] rev. /min. = ( - t3/240 - 11/1920) rad. /see. a = - 3 t2/4 ral. /min.2 - - 3 t2/8 ir rev./mmn.2 = - t2/4800 rad./se.2. 7. (r/720) (t'2 - t3/45) ft. /sec.; (7r/360) (t - t2/30) ft./seC.2. ANSWERS 5 ~ 47. Page 76 1. 1925/(2304 7r) ft./min. 3. Inversely as the cross-sections: dh/dt: dh'/dt = r'2: r2. 5. Inversely as areas of liquid surfaces; their ratio varies as the area of remaining liquid; d- - -P r2 cot a; dt dt 2 a = half-angle of cone, r = radius of liquid surface in the funnel. 7. (k/4) VS/6 cu. in./sec. 9. dy/dt = 4(4x -1) ft./sec.; 12 ft./sec.; 44 ft./sec. 11. Vy = 3 x2v = 30 x2 ft./sec.; 1080 ft./sec. 13. 8/V5 ft./sec.; 4/ /26 ft./sec. 15. 4 ft./sec.; Nearly 312 ft./sec. 17. 0 = 17~ 41', 15~ 18', 10~ 22'; max. h. = 62500 sin2 20~ ft. ~ 50. Page 82 1. 2 a2+c. 3. x3 + c. 5. -2 x3+c. 7. - x5/5 + c. 9. 5x2/2 +4x+c. 11. s-t4/4-2t2+7t+c. 13. y = ax2/2 + bx+c. 15. y.002 x3-.001 x4 +.003 5 + c. 17. y =-x-2/2 + c. 19. v =- 3/t- 2/t2 + c. 21. y = (4/3)x 2 - 6xZ + c. ~ 51. Page 85 1. y =f(4 2 + 3x)dx = 4 3/3 + 3 x2/2 + c. 3. y = x-3dx=-x-2/2 +c. 5. y = (4x +5)dx = 22 + 5X + c. 7. y=f 9dx=9x- c. 9. 11. y = (x + -/) dx = X2/2 2 xz-/3 15. 3X4 - 9 5/5 + c. 19. 2 x — 4 /5 + c. 4 5 23. x + 3 x3/2 + 3 x/5 + c. 25. 4 29. - 1/v- 1/(2 2) + c. 33. 3 y /13 - 6 y3/7 + c. 35. 5 39. x2/2 + (2/5)x2+ c. 41. t3/3- t 45. ax"+l1(n 1l)+ bxn+2/(n + 2) - c. 49. (2/3) (4/7). 49. (2/3)t2 +(4/7)tE + c. 4 J/7 + c. 27. 3 U3/2 - 2vii + C. J/8 + 5Z2 31. '5 z -5/2 + c. 12 V + c. 37. 3 08/8 + C,. 5/5 + C_ 43. aX2/2 + bx313 + c. = f(x3 - X4)dx = x4/4 - x5/5 + c. + c. 13. x2/2 + 5 x + c. 17. 15 x + x2/2 - 2 3/3 + c. '7 9 21. 2 x2/7 - 2 x2/9 + c. t/7+c. 27. 3u/2-22vi +c. 31. 5 z/8 5 z~/2 + c. 12 {Vy + c. 37. 3 uV/8 + c. 5/5 + c. 43. ax2/2 + bX3/3 + c. 47. - t-1-5/1.5 + t-4/2 + c. ~ 54. Page 88 1. 3000 gal.; 1500 gal. 3. s = t2/4 + c; 4; 75/4. 5. 45/(8 7r). 333/(2007r). 7. 16/5. 9. -940/3. 11. 22/3. 13. -1899. 15. 7 a/8. 17. 0. 19. 10a3/3. 21. 729/5. 23. -2. 25. 28/3; 27..0002 29. 225/4. 31. s = gt2/2 + 10 t; 770. 33. lOg; no. 6 6 ~~~ANSWERS ~ 55. Page 93 1. 1/3; 85. 3. 2/5; 62/5. 5. 2/3; -18. 7. 2(2-V2- 1)/3; 5 2 (5x52 v') /3. 9. 1/4; - 223/4. 11. 96/5; 6/5; (6/5)as. 13. 15/2. 16. 32. 17. 11/2. 19. 8/3. 21. 1/3. ~ 57. Page 97 1. 128 r/7; 2 7r/7. 3. 8567r/lOS; 16w/105. 5. For upper half of curve: For lower half of curve: O to 2 (14/3 +4V3A) r. (22/3 -4 V') 7. - 1 to 1; 47r(1 + 2v'2/3). 4 7r(1 - 2 V2/3). 7. 296 7r/81; 7r (3 X4 - 6 X2 - 1)/3 X3]b. 9. 8 mr; 7r(X~+ 3)/3x]b. 11. 4 7r. 13. 778 7/S. 15. (2w7/21) (21~+14V2/~+ 24 '8-). 17. 32 r; 48wr; 2w7r(b2 -a2). 19. 4w7rab2/ 3; 4w7ra2b/3. 21. 20wr/3. 23. 8 V'3/; 2 k/ v'. 25. wlckr2h2/2; 2ir kr2xh ~ 60. Page 101 15. logl107/loglo1ol=O0.812 17. 186.4~ 19. 3.479 21. 2.862 ~ 65. Page 106 __ 23f s.. -1. 9 2 +2 log x. x 1 +2 l I ~ Al. 13. 1 5lg. 4lg) 17. -.713 2-56t +3t2 t23 19. 150.693 21. 2(e3 -2)/3 +(e-2 -1)/6. 23. 0.219 25. 2.302; log k. 27. - X-2. 29. Max. none; Min., x = 1; luff., none. 31. Max., none; Min., x =~2; luff., none. 33. 11.416 35. - 0.396 37. 6.693 39. k2 log C. ~67. Page 109 1. 3 e3.. 3. e~l/+r-/(2v'.1 + x). 5. ex (2 x + X2). 7. (3 log 10) 103x+4. 9. 1. 11. - 2 x. 13. 2 ex(ex + 1). 15. (eY'\x + e&V\x)/4v\'x. 19. 10.02; 2.35; sinh a. 21. 23. 25. Max. None. None. X = 1. Min. X -1. None. None. Iuff. x=-2. X =0. x =2. 29. 0.632 31. 1.381 33. (elO - e-'O)/8 - 5/2. 35. 25.762 ANSWERS 7 ~69.- Page 113 7. (kax + kb + a)/lekx; (kaz+ kb + a) /(ax +b). 9. (3 -4x -6 X2)e-Z2; (3 -4 x-6 X2)/(3 X+ 2). 11. About 963 sec. after 0 =400. 13S. k = (1 /5830) log (5/4);(1 /1909) log 1. 27; 26.7 in.;67 2 minl.; 1806 ft.; 1266 m. 15. 501log 2. ~ 71. Page 117 1. - 2. 3. 3. 5. 10. 7. -2x~+3kx2. 9. - 2r3/ (r2~+1). 11. [3(1.- t2 +t4)(t2 +,1)logl10-2 t +4t3]/(1- t2 ~t4). 13. [ 1 + log(' + X) I(1 + X)l'+X. 15. X5\/xo-1/2(1 + log -V/). 17. 6 + 22 x + 18X2. 33. kex-z2 /2. 35. kXn~. a7. kee". ~74. Page 122 1. 4 cos 4. 3. - 2 sec2 2. 5. 4.x3 COS X4. 7. - 3 sin 3. 9. cosx- xsinx. 11. tan x. 13. cos x + 8 sin 2x.c 15. 2 xsin (3 -2 x) -2(1 +X2) Cos (3 -2x). 19. et /1O[(1/10) (cog t -4sin 3t) -(sin t+ 12 cos 3t)] 21. 1. Maximum Minimum Points of Inflexion 23. x= 2 nx-H+r/2. 2 n7r - 7. nl7r. 25. none. none. n7r. 27. nr - 7r/4. xx7 4~- ir/4. nir/2. 29. 2 nir+ r/4. (2 n+ 1)r + r/4. flnr. 31. nr - r/12. (2 n + 1)ir/2 - r/12. (2 n + 1)r/4 - r/12. 33, 35, 37. The functions differ at most by a constant. 39. 2. 41. - (1/2) cos 2 x+ k. 43. sec t+ kc. 45. sin x- (3/2) cos 2 x+k. 47. (1/2) sin 2 x- z k. Va, VY v Path. 53. - 6sinu2t. 6 cos 2t. 6 X 2=9 55. cos t- sin t. cost. /1 -2 sint cos t+ cos2t. X2 -2 xy+ y2=1 57. - 2 ~r ft./sec.; ~r xV~2 ft./sec. ~76. Page 126 1. 2 cos2 t, - 4sin 2t. 3. cos t- cos 2t, - sin t+ 2 sin 2t. 5. 2cos2 t +0.9 cos 6t, - 4sin 2 t-5.4 Sin 4t. 7. akc cos (ct + e), - ak2 sin (kt +e). 9. d2o/dt2=-20ir2 sinl10irt; 0. 2; 1 /5; 20 r2. 23. a; a/c cos ict. 25. S. H. M. because C120/Wt - k20. 8 ANSWERS ~ 79. Page 130 4x3 - 1 1. 3. /1 - as X/x2 - 1 -1 - 2x 5. 72 +x47. vn~a i + x4 (1 + 2) t (1 + x2) tan-] -- cos Zesi"l 15. - /1 - e2 sin x 21. 3(log tan-1 (1 4- x2) tan27. - ir/6. 33. 7r/6; 7r/3.. 11. 2 +2 x tan-12v/x x1 1+4x 17. V - x2 23. l- /2X\/ + sin-1 x/ -1 X (1 - )2 29. (1/2) sin-'2 x. 31. - 35. 7r/12; 7r/6. 3-1 13. 1 + x2 esiu-l1 19. V1- X2 25. 7r/3. - tan-1 (1 - x). ~ 84. Page 136 1. ViO; 2V10; (b-a)V\/i0. 3. /1 J+ n2; 2 1 + m2; (b - a)Vl + m2 5. (2/3)(4 - v2); (8/3)(2/2- 1); (2/3)(b6 - a )V2. 7. 9. 11. ds: /2 dt. s: 2 /2. v: 2. 2(1 + t)dt. 99. 2(1 + t). (1 + t-4)dt. (b - a) + (b — a-3)/3. 1 + t-4. ~ 85. Page 139 1. 47rv6; 207rV5; 27rx/5(b-a)(b+ a-1). 3. 13 7r V; 447rx/10; 7r/lO(b-a)(3 b + 3 a +4). 5. 87r; 4. 7. (r/2)(e2+ e-2+6). 9. 263r/64. ~ 87. Page 144 R 1. 1 ( +4 2)2 2 3 (2 + 4 a2) 1 4 a2, (1 + cos2X)" sin x 7. cosh2 x; -4 X3; 3 x + 2 a; x +2a; 1 + cos2 x. ctn z x - sinh x cosh; '3 6 x2 1 2 2 x//a(2 a-x). 1 4 cos2 X csc x 2 y. ANSWERS 9 I (b4x2 + a4J2) x(b4x2 + a4y2) a4b4 a4b2, 11.| (X+!) |; z + 2(x + y) y; 2Va a 13. Ia[; 0; 15.3 a sin2 8; a cos 0(1 + 2 sin2 0); 2t 2t 9. (1/6)(9 + 4 ); - 4t3/3; 19. |(1/6) (9+4t2)S 2-4 t3; y(b4x2 + ay2) a2b4 y+ 2(x + y) \a 0. a sin 0(1 + 2 cos2 0). 1 + 3 t4 2 t 3 t2- 1/2. ~ 91. Page 149 1. tan. 3. - (3/4) csc - ctn. 5. (1 + cos2 )/(2tan ). 7. 0/2. 9. 1/2. 11. (1/3) tan 3. 13. (1/3) tan (3 o + 2 r/3). 15. -tan (/2). 17. (e cos -1)/(e sin 0). 19. -ctn(0/2). ~ 92. Page 151 1. 73/6. 3. 3 r2/4. 5. 12/7r. 7. 3. 9. 625/2. 11. ir/8. 13. ( V3-1)/2. 15. 4 7r. 17. 4. ~ 93. Page 154 1. 57r. 3. /2(e/2 — 1). 5. tan - tan (1/2)= 1.012+. 7. rx/2. 9. (e5 - e3) 5. ~ 94. Page 155 1. p/2. 3. (a/04)(1 + 2). 5. (1+ 8cos~38))/(10 + 8cos930). 7. (2/3) V2'ap. 9. (b2-a2 + 2 ap)/(2 b2 - 2 a2 + 3 ap). ~ 98. Page 160 1. x- x2/2 + x3/3 -- 4/4 + c. 3. a2xa + abx2 + b2X3/3 + c; or, (a + bx)3/(3 b) + c'. 5. (e2_ - e-2x)/2 - 2 x + c. 3 5 7. - z- + 7 X- + c. 45..0585 47. 1/3. 49. log 3. 51. 5x/2/12. 53. ir2/2. 55. 3ra2. 57. Areas: 2, 1, 7r/2. ~ 99. Page 164 19. (2 + x + 2x2 + x3) tan-1 x- 2 - 2/2 + c. 21. e4x(32 x3- 24x2+12 x-163)/128+c. 25. 1. 27. r/12+V3/2-1. 29. (2-3 e-T)/13. 31. (1 + e-r)/2. 33. 2 - 5/e. ANSWERS ~ 105. Page 170 i6. log(X+2)2.2 —2 1 (5- 2)4(5 + g) 15. log. 17. x log 2. 19. log. X + 1 x + 2 20.(x — 1)3 21. 1logx (x + 2)3 23. 1log X I-1 tanlx. 25. logx/x2+1. 6 (2+ +3)4 2 /x2+1 2 27. log 2 + tan- 29. g 8 x+2 4 2 3 (ax - 1)((x+ 2)2 +ex 35. log2-4 log -3 37. 3- tan-1 (cosx). 39. log 1 2 1 -- eZ 1 - ex 41. log tan (x/2). 43. log -. 45. log (ex+ e-). 1 + ex ~ 107. Page 174 9. (6x -4)(1 +x)/15. 11. 2tan-l/x —2. 13. V/ -x2 + sin-l/x. 1.\ + 4 +11o /+2 - 2-' 15. _ + 4+log, 2-2. 17. 23(4x —3)(1 +X)x. X 4 x + 2 + 2 19. - (3/10) (2 x + 3) (1 - T) 21. (4/3) x -4 xT + 4 tan-1 x. 23. cos-lx-2V/(1-x)/(1+x). 25. -9-41og2. 27. 6 log (3 - /sin z) - 2 Vsin x. 29. (2/3) /2 + 3 tan x. 31. - sinx- sin x-4 log (1 -/sin x). 33. (1 + x2)/3. 35. (1 + x2)/5. 37. -(a+ bz2) /b. 39. 2(a + bx3)/9 b. 41. 2a + bx 49. log x/42 +1-1 51. x/2-1-tan-1 2-1. nb x 53. (a- V/1-X2)2/2. 59. (V2/2) log [4 x + 1 + 2/4 2 + 2 x + 2]. 61. - cos-1 x 63. - log V- + 2x 3 V*N/2 V3 x -/3- /22+2x 3 2 a4- 1 _ 65. 4 Vl + x + 2 + log (2x+1 + 2v1 + +2). 4 0 67.1 2x1 3 67.3 8 2 log (1 + 2 2 2R); R = /l+x+a2. ___ sin-i x 69. logl+ - s — 71. 6x2- x-s - 3 sin +3X-81 - 2. 4 4 73. /1 - 2 + 2 sin-1 + x cos-1 X. x 2 ANSWERS 11 ~ 111. Page 183 cos x sin3 x 3 1. — co + - ( - cos x sin x). 3. (1/5) tan5x — (1/3) tan3x + tanx - x. 5. (1/5) tan5x. tan2 x tan4 -ctn 2 ctn3 x ctn5 x 7. - + 9. - ctnx -- -. 2 4 3 5 11. - (1/2) cot2 x -log sinx. 13. - sinx - cscx. 15. (1/2) sec2 x + log tan x. 17. x cos a + sin a log sin x. 19. (1/3)sin3x-(1/5)sin5x. 21. -sinx —cscx. 23. sin4x(l+cos2x)/12. 25. (1/2)cos2 x - log cos x. 27. - (1/2) (cos x csc2 x + log tan x). 29. - (1/14)cos 7 x- (1/2) cos x. ~ 114. Page 185 a2 1 x +3 1 11-8a 1. X-2- 3. 9 log + -3 5. + log (x - 2) 9 X ' 31 2(x - 2)2 a12 - ~(x - 2)9 X 1 1 + X 7. -+3x4+log (X- 9. - ) log + - 2 > 1) 2(1-x 41-xX 11. -log + tan-l. 13. aX- +- log 32 z- 2 16 2 2(X2$-1) 4 1+l 1 z2 - 2 2 ~2-v'2 15. -log 2 - 2 17. - log +. 19. 2tan-l1x -1. 8 x2 + 2 4 x + 2 + \/2 -2 x 1 21. 23. b-bx 23. 25. - 3F. n + 27. --- n. 29. (4x- 7)(3x 7). b(n -p) ~/(a + bx)n-p 28 2 3X2 — 3 1 x2 1 31. -3(12 + ) V3 _. 33. + log - 3 1 2 c2 /1 - x2 2 X 935. 12 ( — +5 — 3 + tan- y), where y2 = x. 9 7 5 3 37. 4(3bx -4a)(a+ bx)T/(21 b2). 39. (4/405)(15x+28)(3x+7)4. ~4 aX 1 41. (3/56) (4 2- 3 a) (a + 2). 43. ( + tan-i x. 2(X2 + 1) 2 3 + 5X —9 v/3 x X +_ 1 t _X+l 45. - tan-1. 47. +1 + tan-1. 12(x2+3)2 36 \/3 8(X2 + 2 x + 5) 16 2 49 + 16 seX-l* 51 1 (4X3 21)(7 + 4 x3). 8 x2 1 -2 64 53. 2 - 5 a 2 - a2 + 3 log ( + 2 - a2). 8 0 12 12 ~~~ANSWERS 55. X.a a + bx2 X3 59. 4 shill- 3 log tanl ~r+ 57. ~~3 i 9 4 23 3 a(a~+ bX2)r -IT~ 1 2 tali (0/2) ____________ 61. tani 6 - see 6. 63. ~ log a 62)~5-v2 V21 2 tan (0/2) + 5+ V'2-1 65. - x -(1/3) etn 3x. 67. 7/2 -21log 2. 69. (3/4) log (5/3) - (2/x%/6) tan-I -%/2/3. 71.1ioCY (8 - V9) (5 + V21). 73. 7 l 71. v'~ ~ ~ ~ ~~~~3. 10lo 77. - log (3 - /) 79. 68.7 81. 0.833 83. 85. 4.037 87. 0.88 89. 2.274 91. 0.767 95. 0.902 97. 0.254 (2~V./3). 0.184 93. 0.949.1. 16/3. 1 3. x/2/8. 23. 74/3. S. 4/5. 15. 6.89 25. 0.346 ~ 118. Page 1 5. - 1.9307 17. 0.263 27. 0.550 7. 2. 9. 7r/6. 19. 0.693 11. ir/4. 21. 0.746 ~ 120. 1. 12.4 3. 17.2 11. 4; 1/2; (abs. val.). 13 19. 3500/3 rev., 4.7 see., nearly. Page 191 5. 2 ~.48. B 7. 0.35 9. 57.6 15. 161. 17. 4500/7r. 1. ira3/15. 9. 7r (e2 - e-2a + 17. 12,480 lb. ~ 123. Page 202 3. irabh3/3M. 5. 2 4 a) /4. 11. 64 7a2/3. 13. 19. 699 lb. nearly. 2: ~ 1.24. Page 206 7. 232/3. 9. 90. v'3 a3/3. 32 7ra2/5. L. Pt'= C3P. 7. ki7r/2. 15. 18,720 lb. 5. 1/2%Ol. 19. 0.293 15. 1.099 17. 7.912 ~ 127. Page 211 7. 5/60 - 64/12 + C62 + c'6 + C'1. 9. ex + cx + cf. 11. c + c'u- u log U. 13. t2/2 ~ sin t; t3/6 - cos t + 1. 17. 8. 19. (4/15) (6-1- 51~3~-2 -32). 21. 32/3. 23. 1105/2. ANSWERS 13 25. 7rr3,/3. 27. 32w 29. 8/9. 31. b( 33. y- k(31X2 - x3)/6. 35. y =(3 aX2+ bx) /6. 37. y k k[log (l/x) + x/I - 1]. ~ 129. Page 218 1. 26/105. 3. 363/'5 5. (58 -1)716 ~(510-~1)/30. 9. 26/35. 11. 1/70. 13. 2 a3/3. 17. at x= 0.904 21. V'3 ~ 27r. 23. %/-2/3 + log (1 + '/2). 25. 5x/5/6. 29. 37r/2. 31. w3/6. 33. 3wr/2. 35. 16 a2r5/5. 37. 4 X2 -12 8 7. 20477/4. 19. 1/12. 27. wr/8. a2/6. ~ 131. Page 223 1. 3/20. 3. 104',/2-/35. 5. '/~(72/16 - 31/18) + 16/9. 7. 17/162. 9. ka4/3; 2 ka4/3. 11. kwra4/2. 13. kwra4/4. 15. kh3(b + 3 b')/12. 17. kwr(a24 - a14) /2. 19. 5 ka4/xN/3; side = 2 a. 21. 3 kwI/64. 23. 35 kwr/16. 25. kw5/20. 27. 19 kw/4. 29. ka5. 31. 641 k/756. '33. 846290 k/189. 35. 21026 k/10395. ~ 136. Page 227 1. 3/8. 3. 1/5. 13. 2 a/3. 15. 19. 3 h/4 from vertex. 23. iDist. from center = 27. (3/5, 12/35). 21 35. [ (24 - 6 7w2) /w3, (2 5. 3/4. 7. 1/4. 9. 1/20. 11. 2/w. 3 h/5 17. (4 a/3r, 4 b/37r). 21. 3 a/8 from center, on axis of revolution a.: 2 asin a/(3 a). 25.(2 _____ ke~1' 4 e- 4e-1I D. (1, - 1/2). 31. (1, 0). 33. (- 5/6, 0). ~ 136. Page 229 1.r (a2/2 + b2); a> b. 3. 3w7a2/4. 5. 1. 558~ 7. wab2[w/ (2 a) - (1/4) sin (2w7/a). 9. 4wrr3/3. 11. wr2a3. 13. 32wra3/105. 15. A =log%/2- X-0.6192;y 0. 2059; IT= k(w/4 -23/48). 17. A=-3w7a mx =wa; y= 5a/6;II==w 3/3) (8wr2+ 5). 19. A-w(b2- a2)/4; x =_(a2+ ab~ 62)/2(b + a); 13 3w(b4 - a4)/32. 21. z=2c/3. 23. z=3/4. 25. x =w7a/2~+64 a/ (45w7) 29. 4 wrr3(1 - eos4 IX)/3; 2 a= angle of cone. 31. log tan (x/2 + w/4). 33. 8 a. 35. (3/10) mass times square of radius. 37. (1/5) mass times sum of squares of ot~her two semi-axes. 47. 2/w; 1/2. 49. 2 r/w. 51. 3025 wk/27. 53. 382812.5 w3 ft. lb./mmn. 61. w. 14 14 ~~~ANSWERS ~140. Page 236 1. 2 x =3 y-11; 4 x= y~ 11. 9. v = 85.2 P-876. 11. h =.66 t- 33. 13. d = 14.8 t-63 19. y = elx/l.3 1 1 ~ 3. y= 2e x. ~ 141. Page 240 1. f(x) =0. 5X2 -1.4 x +2.5 3. p (M) m2~+M3. 5. P =6~.lO-2.55 7. D = 10(50-N)/20..9. 0.012 x 10-7B1.5M. 11. Tungsten: C.00000272 V3.55; W= 10300 V,7-..96; R = 31.6 V-80 ~ 1.45. Page 249 1. x= 1/2. 3.x 1.88 11. About j 40 ft. 13S. About 420 ft. 15. T changes by about 1. 1 To, 1. 4 O/o,.06 Ol respectively. 17. 0.520, 0.530, 0.541, 0.590 calories ~ 147. Page 256 1.. tan x x. 3. Cos X= 1-X2/2. The cubics are: 5. 1 ~ X + X2/2 + X316; IEI <.000004 7. X - X2/2 + X313 I EI<.0004 9. e-2[1-(x- 2) +(x -2)2/2 -(x -2)3 /61; 1 EjI <.0004 15. X -X316 + x5/120. 17. x < 505Q0, ~ 148. Page 258 1. 3. 5. 7. 9. Max. none. x= 1. none. (2 nlw + 7r/2) 3. 131'.2 etc. none. Min. x= 0. none. none. (2 nmw -7/2)1. 291'.4 etc. x = 0. 3.~ 10. Page 262 1. 1. 3. 3. 5. - 1. 7. 2/3. 9. log (a/b). 11. 1. 13. - 1/2. 15. 0. 17. 1. 19. log a- 1. 21. 0. 23. 3. 25. 4. 27. 3. 29. 1. 31. 3. 33. oo. 35. 2. 37. 3/2. ~ 1.51. Page 264 1. 0. 3. -. 5., - 1/2. 7. 0. 9. 0. 11. 0. 13. 1. 15. en. 17. 1. 19. 1. 21. 1.102 ~ 153. Page 268 1 1. (x/2/2) [1 + (x - r/'4) -(x- -7r/4)2/2! —a- 7r/4) 3/3~ ANSWERS 15 ~ 156. Page 276 The answers for Ex's. 1-9 are in the order z,,, ZXY Z,,~, 1. 2,0, -72. 3. 2 cos(X2 +y2) -4X2 Sill(X2~+y2),- 4 xy sin (X2+ y2), 2 cos (X2 +y2) -4 y2 sin (X2~+y2). 5. Ix 2x - 2 (X2 + 'y) (X + Y2)2 (X2+ y2) 2 7. (6x~2y+4X3+4x2y)eu, 2(x+y)(.1+2xy)eu, (2x +6y~+4xy2 +4y3)eu; U u=X2 +y2. 9. /2 -X - 2xy x2 -y2 p k (X2 + y2 (X2 + y2) 2 (X2 ~ y2) 2 r v 17. Area = K Ka = bsin C, Kb =, sinGC K_-=ab cos C iDiacr D; Da =2(a -b cos C), DA=2(b -a cos C), De= 2 ab siu C. ~ 160. Page 283 1. - x1(4 y). 3. (y/ - X2)/(y2 -X). 7. Errors due to Aa, AD, AA: 1 O;.30%Oo;.84%Ol. 9.. 3%.l 11. 20.5, 19.3, 101.5 ~ 166. Page 295 7. (x-3)/12=(2-y)/16=3-z. 9. x=y, z=4. 1 1. cos-1 (2/ V39). 13. 3x-f-4y=25; 2x+4y+Nv'-z-25. 15. (0, 0, 0). 25. u and v are solutions of the "Inormal equations" ulaa + vlab = Mae; ulab + v~bb = bc; where laa = a12 + a22 + a, 2, Mab =albi + a2b2 + a3b3, etc. 31. a = 7/8, f3=11590/3. 33. a=10.30,f3 6.48 ~ 169. Page 301 1. 2x +4y +z =18; (x -4)/2 =(y -2)/4=z -2. 3. 3x-2y+ z=18; (x-6)/3=(1l-y)/2=z-2. 5. 3 x +4 y - 5z=0; (x -3)/3=(y - 4)/4=(5- z)/5. 7. ex+ ey +2z =4e; x/e =(y-2)/e =(z- e)/2. 9. cosl'(~ 2/V6),. 11. ~+ y + V2 rz -2a, xy -z/VN2. V9x/3x~y+2 z=4 a, x/ V3=y/3 =z/2. 13. xxo -yyO -2(z +zo)-0 (0, 0,0); no extreme. 15. X2 +y2l= + Z2; x +3 y +3z 1. 17. 9X2~+4 y2 +36Z2= 36; 9 + 6 y +18Y\2 z =36. 19. 8x-3y-z=1; 9x+6y-z= 20. ~ 170. Page 304 3. 2 a2(3 ir/4 - 2). 5. V'2 a a -side ofsquare. 11. (2 ir/3)(x\/m(2kIcM+3rn)3 - 16 ANSWERS ~ 172. Page 306 1. 90~. 3. cos-1 (1/5). 7. (01 - 02) Va2 + b2, 2 ir a2 + b2. ~ 172. Page 306 (Second List) 1. r sin 2 0; r2 cos 2 0. 3. 2 re20(cos 8 - sin 0); 2 re2O(cos 0 + sin 0). 7. 2 cos 0(r2 - sin2 0)/r; 2 sin 0(r2 + cos2 0)/r. 9. (O cos 0 - sin 8 log r)/r, (O sin 0 + cos 0 log r)/r. 11. x = 285/116. 13. 4 a3(3 7r - 4)/9. 17. (0, 0). 23. The inscribed cube. 25. — 7.97 +.000107 p2. 27. (a) H = 33.58 - 1.138 P. (b) H = 31.64 - 0.965 P - 0.0037 p2. (c) H = 40.85 - 10.73 log P. 33. sin-1 (1/V/3) and sin- (1/3). 35. -ax/2 =(y- 2)/0 =(4 z - 7r2)/(4 7-); (x i 2)/0 =- y/2 = ( - 7r2)/(2 7r); 8 a - 4 r + 7r3 = 0; y - rz + 7r3 = 0. 37. a/1 + k2(02-01). Value of y should be a cos 0 sin 0. ~ 175. Page 314 17. x2 - y2 =- 3x; 5/4. 19. xdx + ydy = 0. 21. yy" + y2 = 0. 23. x2 - y2- xy(yt- l/y/) = a2 - b2 25. dy = xdx. 27. d2 d=y _ 2; d2 d y. d2X__ dx dt dt2 —; - - dt dt2 dt ~ 178. Page 318 11. y = 1 + cec0.9. 13. y - 1 = c(x + 1). 15. 1 + y2 = c(l + X2). 17. 2y =cx2 + c. 23. y3.ce-3. 25. y- 3(1 +). 27. T = 200 e/30; k = (2/1r) log 1.1 29. (B- q)/(A - q) = e(B-A)(kt+c). 31 v2 = c - 2 k2/t. ~ 180. Page 323 1. y = (c + x)e 2/2. 3. y = 2 sin - 2 + ce-sinx. 5. Xy2(2 + cx) = 1. 7. 2/r = 1 - 2 + ce-O. 9. y = tanx - 1 + ce-tanx. 11. s = ce-t 1 + t. 13. y = cex -(sinx + cosx)/2. 15. y = kx + cl1 + X2. ~ 181. Page 324 1. 3 y3 1 - 3 x + ce-3z. 3. (x + y)/(x - y) = ce2y. 7. x3 + 3 32y2 + y3 = C. 9. xey/z = c. 15. (a)x- y = c. 17. y2 =4 c(x + c). 19. y2 = kx. ANSWERS 17 ~ 183. Page 329 1. s = Aet + Be-t. 3. s = a sin (t + b). 5. y = Aekt+ Be-kt or y = a sin (kt + b). 7. S. H. M. with zero amplitude; no motion. 9. a = /104, b = tan-5; a = 5, b = r/2. ~ 184. Page 332 1 = cle' + c2e3a. 3. y = clex + c2e-x/5. 5. y = (cl + c2)ez. 7. y = (cl sin V2 X + c2 cos V2 x)ex. 9. y = Cle2x + c2e7Z. 11. y = c1e2x/3 + c2e3x/2. 13. y = cle2x + c2e-2x. 15. y = c1 + c2e-kx. 17. x = e-bt/2[cleVbV24t/ 2 - + ce-/-2-4ct/2], C < b2/4; x = e-bt/2[cl sin v4 c - b2t/2 + c2cos /4 c - b2t/2], c > b2/4. 19. x = e-t/2j5 + 5 1 e-stt/2]; No; 10 10 _ e-2 -[1 + 5/5 + 2 ve-l/2 + /5 — 1- -2 vrN-/ /2 2V 5 2V5 ~186. Page 337 5. 3 t = 2(4/-2c) Vs c + c'. 7. y= e2/4 + clx + 2. 9. y =6cosx + 4xsinx - 2 cosx- cx + c. 11. y = el -(1/16) cos2 x + c1l3 + C2x2 + C3x + C4. 15. y = kx2(x2 -4 Ix + 612)/12. ~ 189. Page 341 9. y = 3 x/4 + 24/4 + cle + c2e2Z + C3. 11. y = clez + ce-x + C3e- 2z + (C4 + x/12)e2x. 13. y = clex + c2e/3 + C3. 15. y= —2 sinx + cCosx + c2sinx + c3. 17. y = clex+c2e-~-c3x+c4. 19. 2/1 + ceZ + log[(/1 + ce" - 1)/(/1 + ceX + 1)] + c'. 21. y = (X - x2/2) log X + c1x2 + C2x + C3. 25. y = cx2 - x log x + c2. 27. y = (1 - b)/b2 - log (a + bx)+ cl(a + bx)mI + ca (a + bx)m2, where m1 and m2 are the roots of m2 + (b - 1)m - b2 = 0.