-!?~1 THE CALCl ■ UN nmWnijlnffl™ .jLuJ&'M. BOUGHT WITH THE INCOME OF THE SAGE ENDOWMENT FUND THE GIFT OF 31cm n W. Sage 1891 fl,3^f^& ! ".. '■■■■ x3l.VlU Problems in the calculus 3 1924 031 220 21 olin.anx The original of this book is in the Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031220217 MATHEMATICAL TEXTS FOR COLLEGES EDITED BY PERCEY F. SMITH, Ph.D. PROFESSOR OF MATHEMATICS IN THE SHEFFIELD SCIENTIFIC SCHOOL OF YALE UNIVERSITY PROBLEMS IN THE CALCULUS WITH FORMULAS AND SUGGESTIONS BY DAVID D. LEIB, Ph.D. INSTRUCTOR IN MATHEMATICS IN THE SHEFFIELD SCIENTIFIC SCHOOL OF YALE UNIVERSITY GINN AND COMPANY BOSTON • ' NEW YORK • CHICAGO • LONDON ATLANTA • DALLAS • COLUMBUS ■ SAN FRANCISCO 5 copyright, 1915, by davjd'd. leib ALL RIGHTS RESERVED 215.10 Cfte atftetiieum Bvtse G1NN AND COMPANY • PRO- PRIETORS • BOSTON -U.S.A. PREFACE The present volume is the outgrowth of lists of problems prepared by the author from year to year to supplement the textbook used in the Sheffield Scientific School. Many of the problems have been furnished by his colleagues through the media of test and examination papers, or by direct contri- bution to this collection at some stage in its development. Since many of these in turn were doubtless adapted from other sources, no attempt has been made to assign a problem to an original source. The principles embodied in the prob- lems are surely common property. The book does not aim to be a textbook on the calculus, nor simply a collection of applied problems in science and engineering. It is believed that a teacher can find here a supplementary list of workable problems on any topic ordinarily included in a general course in the calculus. No attempt is made to explain the theory of any science involved in a problem except in so far as it is necessary to an intelligent understanding of the problem and its purpose. The text introducing the exercises aims to explain the technique of the subject and to point out some common pitfalls to the student. The answers to a large number of the problems have been purposely omitted. The general object has been to give the answer to one or more examples of each type so that the student may attack further examples of a similar nature with increased confidence. At the same time, other answers have been omitted so that the book may be used in tests and in work where it is not desirable for the student to have the answers. Comparatively few illustrative examples have been worked out in detail, and vi PROBLEMS IN THE CALCULUS practically no figures have been inserted, as these can be found in sufficient quantity in the textbook used. The drawing of suitable figures where needed is a valuable part of an example, and should be insisted upon. The book will be found to differ materially from the problem books, mostly in German, now available. A good bibliography of these is to be found in the Bulletin of the American Mathematical Society, June, 1914. The author desires to thank in particular his colleagues, Drs. Harris I\ MacNeish and George R Gundelfinger, for extensive aid — the former in the differential calculus, the latter in the integral. His thanks are also due to Captain James Gordon Steese, U.S.A., for several valuable suggestions. As no one has formally verified the answers nor read the proof, the author must assume sole responsibility for errors and omissions, which he hopes will not be sufficiently numer- ous to affect the value of the book. The author will be pleased to have his attention called to errors either in the text or in the answers. DAVID D. LED3 Sheffield Scientific School Yale University CONTENTS CHAPTER I FUNCTIONS EXERCISE I. Evaluation of functions . . II. Indeterminate forms . . PAGE 1 . 2 CHAPTER II FORMAL DIFFERENTIATION III. Differentiation by the general method . . . . . IV. Simple polynomials for oral differentiation V. The power formula. Fractional and negative exponents . VI. Products and quotients of algebraic forms VII. Review of algebraic forms ... . . VIII. Logarithms of algebraic forms, base e . IX. Logarithms of algebraic functions X. Powers of logarithms . ... . . . XI. Simple exponential functions . . . . XII. Exponential functions. Base constant . . XIII. Logarithmic differentiation . XIV. General exponential functions . . XV. Simple trigonometric functions . . XVI. Powers of trigonometric functions XVII. Forms involving trigonometric functions XVIII. Simple inverse trigonometric functions XIX. Forms involving inverse trigonometric functions . . XX. Function of a function XXI. Implicit functions . . XXII. General review of formal differentiation ... 10 11 11 12 12 13 14 15 15 16 17 17 18 CHAPTER III SIMPLE APPLICATIONS. SUCCESSIVE DIFFERENTIATION XXIII. Direction of curves . . . . ... ... 20 XXIV. Algebraic curves. Tangents, normals, and angles of inter- section 22 vii Vlll PROBLEMS IN THE CALCULUS EXEKCISE PAGE XXV. Tangents and normals. Parametric equations . . 23 XXVI. Polar curves .24 XXVII. Velocity and acceleration. Rectilinear motion ... 25 XXVIII. Multiple roots of equations .... . 27 XXIX. Successive differentiation . . . . 28 XXX. Successive differentiation. Implicit functions . 28 XXXI. Successive differentiation of parametric forms 29 CHAPTER IV MAXIMA AND MINIMA XXXII. Maximum and minimum points of algebraic curves 30 XXXIII. Maxima and minima. Transcendental functions . 31 XXXIV. Maxima and minima. Problems . . 32 XXXV. Maxima and minima. Problems (continued) . . 37 CHAPTER V DIFFERENTIALS AND RATES XXXVI. Differential of arc and approximate length of arc. Rec- tangular coordinates XXXVII. Differential of arc. Polar coordinates XXXVIII. Differentials . . XXXIX. Change of variable . . XL. Rate problems . 45 47 47 48 50 CHAPTER VI INDETERMINATE FORMS XLI. Indeterminate forms. -f- XLII. Indeterminate forms, oo -*■ ao XLIII. Indeterminate forms. • oo, oo — oo . XLIV. Other indeterminate forms .... 57 58 59 59 CHAPTER VII CURVATURE. EVOLUTES XLV. Curvature and radius of curvature . . . XLVI. Center of curvature and evolutes (Cartesian) XLVII. Evolutes. Parametric equations . . 61 62 63 CONTENTS IX CHAPTER VIII PARTIAL DERIVATIVES. APPLICATIONS EXERCISE PAGE XLVIII. Partial derivatives . .... .65 XLIX. Differentiation of implicit functions by means of partial derivatives .... .66 L. Successive partial derivatives . . 67 LI. Total differential and total derivative 68 LII. Total derivatives and differentials. Application to rates and small errors . . ... 69 LIII. Exact differentials ... . . 74 CHAPTER IX SERIES L1V. Convergence of number series . LV. Interval or region of convergence. Power series . LVI. Expansion in powers of (x — o). Taylor's theorem LVII. Expansion in a power series in x LVIII. Calculations by series . . . . LIX. Indeterminate forms. Evaluation by series . LX. Taylor's theorem — several variables LXI. Maxima and minima — two variables 76 80 81 84 85 CHAPTER X FURTHER APPLICATIONS TO GEOMETRY LXII. Envelopes . . . LXIII. Further applications to plane curves LXIV. Space geometry. Curves LXV. Space geometry. Surfaces 93 CHAPTER XI INTEGRAL CALCULUS. SIMPLE FORMAL INTEGRATION LXVI. The power form . 96 LXVII. The logarithm form . . .98 LXVIII. Exponential functions . . .99 LXIX. Integration of the fundamental trigonometric functions 100 LXX. Constant over quadratic . . 102 LXXI. Constant over the square root of a quadratic, or /(fc/-\/Q)dx 103 LXXII. Linear over quadratic . . 104 x PROBLEMS IN THE CALCULUS EXERCISE PAGE LXXIII. Linear over the square root of a quadratic 106 LXXIV. Integration of trigonometric products . . 107 LXXV. Integration of trigonometric products (continued) . 109 LXXVI. Integration by trigonometric substitution . . . Ill CHAPTER XII ELEMENTARY APPLICATIONS. DEFINITE INTEGRALS LXXVTI. The constant of integration . . 113 LXXVIII. Orthogonal trajectories . . . 115 LXXIX. The definite integral. Evaluati9Ji by direct inte- gration . *% ■ . 115 LXXX. Simple plane areas . . . !;. . . . 117 1 CHAPTER XIII | SPECIAL METHODS OF INTEGRATION. APPLICATIONS LXXXI. Integration of rational fractions . 119 LXXXII. Integration by rationalization . 122 LXXXIII. Integration of rational trigonometric functions . 124 LXXXIV. Integration by the reciprocal substitution 125 LXXXV. The definite integral. Change in limits 126 LXXXVI. Integration by various devices . . 127 LXXXVII. Integration by parts . . . . . . 129 LXXXVIII. Miscellaneous examples. Review . 131 LXXXIX. Integration by the use of series 133 XC. Examples for comparison . . 133 XCI. Integration by reduction formulas . . . . 135 XCII. Integration by the aid of tables 136 XCIII. The definite integral. Evaluation by the use of tables 138 XCIV. Applied problems involving definite integrals . 139 CHAPTER XIV INTEGRATION A SUMMATION PROCESS. GEOMETRICAL APPLICATIONS XCV. Plane area. Rectangular coordinates ... . 144 XCVI. Length of a curve. Rectangular coordinates .... 145 XCVII. Volume of a solid of revolution 146 XCVIII. Area of a surface of revolution . . . 148 XCIX. Plane area and length of arc. Polar coordinates 148 C, Volumes of miscellaneous solids 150 CONTENTS XI CHAPTER XV APPROXIMATE INTEGRATION — VARIOUS METHODS EXERCISE PAGE CI. The definite integral. Approximate evaluation by the use of series ... .... 152 CII. The definite integral. Simpson's rule 152 CIII. Geometrical integrals. Review 154 CHAPTER XVI MULTIPLE INTEGRATION CIV. Definite double and triple integrals 156 CV. Plane area by double integration. Rectangular coordinates 157 CVI. Plane area by double integration. Polar coordinates 158 CVII. Volumes by triple integration . 160 CHAPTER XVII GENERAL APPLICATIONS CVIII. Fluid pressure .... 162 CIX. Integration as a process of summation. Applied problems 164 CX. Center of gravity of an area ... 170 CXI. Average value of a function throughout a region 172 CXII. Moment of inertia. Plane areas . . 175 CXIII. Polar moment of inertia . . . .176 CHAPTER XVIII DIFFERENTIAL EQUATIONS CXIV. Differential equations of the first order . CXV. The linear differential equation . . . CXVI. The linear differential equation : second order CXVII. Miscellaneous differential equations CXVIII. Application of differential equations. Evaluating constants .... .... Answers . . . . . . Index the 178 178 180 183 185 189 221 PROBLEMS m THE CALCULUS CHAPTER I FUNCTIONS EXERCISE I Evaluation of Functions. The idea of a function is funda- mental. Remember that if f(x) is any expression involving x, whether algebraic or transcendental, f(a) is the sanie expres- sion where each x has been replaced by a. The following examples should fix this idea : 1. Given fix) = x 3 - 2 a; 2 - 13 x + 10, show that/(l) =/(2). 2. Given fix) = x 3 - 5x 2 + 8x - 4, show that/(2) =/(l). 3. Given fix) = x s - 3x 2 + 4x - 1, show that/(2) = 3/(1). 4. Given fix) = x 3 + 4x 2 - 5x + 7, show that/(2) = 3/(1). 5. Given /(x) = 4x 3 - 3x 2 - 5x- 6, show that/(3) + 6/(1) = 0. 6. Given /(x) = 2x 3 - 2x 2 - 3x - 3, show that/(- 3) = 3/(- 1). 7. Given f(x) = x 3 - 6x 2 + 11 x - 10, show that/(l) =/(2) =/(3). 8. Given $ (x) = x 8 - 9 x 2 + 26 x - 14, show that (2) = (3) = (4) = i*(6). 9. Given fix) = x + 1/x, show that /(a) =/(l/a). 10. Given fix) = sinx + cosx, show that/(w/8)/(2ir/8) = 1/2. 11 . Given (x) = tan x + tan 3 x, show that

(y - 2) = (y) . 34. Given (x) = log 1/x, show (y - 1) - (y 2 - 1) = 4> [1/(2/ + 1)] . 35. Given 0(x) = e a; - 1 , show0(x 2 ) = 0(2x 2 — x)0(l + x — x 2 ). 36. Given

(4 — y) =

X X = 0. 5. * + 3 >a 2x 2 -5 2. X • , x = 0. 3x-x 2 4. x+ 3 2x- 5' x = „ x 2 + 3 oo. 6. , x 2x- 5 7. 3x 8 -x 2 -10 , X = 00 5 x s + x + 4 13. , I = CO. 2t 4 + « 8 + i 2 8. 3x 8 -2x 2 + 5x i ,x = X 2 — X 0. 14. x 8 — 6x — » x = I x 8 -2x 2 + 3x 9. x 2 + 2 x + 1 , X = 00. x 8 — 5 x 2 + 7 15. sinx , x = 0. tanx 0, oo. ,. 3x 2 — 2x+5 ,„ tanx + secx — 1 10. ^— ,x = oo. 16. ,x = 0. x 2 — 4 x + 7 tan x — see x + 1 tn + l)<»+8)(n+8) ^ l7.i!^,* = o. (m - 1) (n - 2) (n - 3) sin 2 ,„ i 8 -3f 2 + « . A ,„ tanx tt 12.- — - — ,i = 0,oo. 18. — , x = -. t3 + 5 t 2 + 2f tan3x 2 CHAPTER II FORMAL DIFFERENTIATION EXERCISE III Differentiation by the General Method. In the following, x is regarded as the independent variable, and y is a function of x. Differentiation is then accomplished as follows : Eeplace x on the right side by x + Ax (or give x an increment Ax) and call the resulting value of the function y + Ay. Then, by subtrac- tion, we have the value of Ay (the increment of y) expressed in terms of x and Ax. This expression should be simplified as much as possible. Next divide both sides by Ax, getting a value for Ay /Ax. Then let Ax approach zero, and the left-hand member becomes the derivative, dy/dx, while the right-hand member reduces to a function of x, which may be a mere constant. Differentiate the following by this method. 1. jr = 3x + 7. 11. y = 2P-t s >. 21. y = 4/(3- 2x). 2.y = 2x-3. 12.y = t s -5t. 22. y = 3/(5- 4 x). 3. x = P + 2. 13. y = 1/x. 23. y = 2/(1 - t 2 ). 4. y = 2 1 2 - 3. 14. s = 1/3 1. 24. s = 3/(1 + t 2 ). 5. y = 2l» + l. 15. y = 1/x 2 . 25. y = (5 - 2x)/(3 - x). 6. y = 2P-t. 16. J/ = 3/(x + l). 26. y = (2x + l)/x 2 . 7. s = 3« 2 + 2i. 17.^ = 2/(1-*). 27. 2/ = x 2 /(2x + l). 8. y = x 2 + 7x. 18. 2/ = 3/(2x- 1). 28. y = x 2 /(l- 3x). 9. 2/ = 2x s -4x. 19. y = 3/(x 2 - 1). 29. s = (5 J 2 + l)/2 1. 10.3/ = 2«-«s. 20. x = 5/(2« + 3). 30. x = 2 2/ 2 /(l + 2 y) . The value of dy/dx at a point on a curve gives the slope of the tangent to the curve at that point. This enables us to find the angle between two curves at the point of intersection, i FORMAL DIFFERENTIATION by the formula tan 6 =(m 1 — m 2 )/(l+ mjnj, where the ra's are the slopes of the curves at the point. Find the angle of intersection of the following pairs of curves : 31. y = 2 - X 2 , y = x 2 . 32. y = 5 - x 2 , y = 5 + 3x. 33. y = 3x- 5, y = x 2 — 5. 34. «/ = x 2 - 2 x + 1, y = 7+ 2x — x 2 . 35. 2/ = x 2 - 4, 2/ = 4 — x 2 . 36. y = X s - 2 x 2 , y = 2x — x 2 . 37. y = x 2 + 2x+l, 2/ =9+ 2x-x 2 . 38. 2/ = 1/(1 + x), y = l/(l-x). 39. y = 2/(x - 1), 2/ = X 3 /4, at (2, 2). 40. y = x/(x — 1), y = x 2 /2, at (2, 2). 41. J/ = x 3 /4, y = 6-x 2 ,at(2,2). 42. 2/= l(x + l), 2/=l(x 2 + 2x + l). EXERCISE IV Simple Polynomials for Oral Differentiation. Differentiate the expressions below by means of the formulas : d d — x n = nx n ~ x ; — v" = nv n dx dx 1. y = x 2 -2x + 3. 13. 2/ = Jx 3 + 2x 2 -5. 25. y = (4 - x 2 ) 8 2. 2/ = 2x 2 + x + 1. 14. 2/ = 3x 2 — £x 4 . 26. y = (X s + l) 2 . 3. y = 3x — X 2 . 15. 2/ = x 3 + x 2 — x. 27. y = (l-2x) 4 . 4. y = 2x 3 — 7x + 4. 16. 2/ = ^x 3 - £x 2 . 28. y = (l-3x 2 ) 2 . 5. y = 7 - x - 2 x 2 . 17. y = x — J x 3 + x 2 . 29. y = (7-3x) 4 . 6. 2/ = 2x 3 + 5x 2 -2. 18. 2/= Jx 5 + 3x 3 . 30. y = (6 - x 2 ) 2 . 7. 2/ = 3x 2 -2x + l. 19. ?/ = (x + l) 8 . 31. y = (x 2 + 2x-l) 2 8. ^ = ^x 3 + 3x. 20, 2/ = (x-l) 2 . 32. y = (x 3 -2x) 2 . 9. 2/ = \x 2 + 2x-4. 21. y = (1 - x) 2 . 33. y = (x 3 - 1) 4 . 10. y = x 4 - 2 x 2 - 3. 22. 2/ = (l + 2x) 3 . 34. y = (4-2x 3 ) 3 . 11. 2/ = x 3 + ^x 2 — 4. 23. 2/ = (x 2 - l) 2 . 35. y = (2-x 4 ) 2 . 12. y = \ x 4 + x - 2. 24. y = (x + 6)1. EXERCISE V 36. y = (2x-x 3 ) 4 . The Power Formula. Fractional and Negative Exponents. The formula dv H /dx = nv" -1 dv/dx holds for fractional and negative exponents as well as for positive ones. In differentiating a radical expression, always rewrite with fractional exponents. Similarly, a constant over a power, for example, i/(x + 1) 8 , is 6 PROBLEMS IN THE CALCULUS most convenieatly differentiated by rewriting in the form with a negative exponent; that is, i/(x +1) 3 = 4 (a- +1)~ 8 . Differentiate the following : 1. y = 2x*-3x#. 3. y = l/x + x*. 5. r = £ 2 + 2 V« + 3ti. 2.y = 2x* + V2x. 4. s = 3ti-t§, 6. y = X s + 1/3 x*. 7. 2/ = 2x 8 + 3Vx'-3/2x 2 . 11. y = VsTe + Vi/3. 8. y = 3x$- xi 12. 2/ = 7/x 2 - 3x* - 1/Vx. 9. y = \/2x + 2 \/x. 13. B = 9 + 1/0 - 2 V#. 10. y = 3/Vx + 3Vx. 14. s = 4/ VI + Z/y/t. 15. s = 4Vx 8 +2/3x s . 21. 2/ = 3/Vx 2 - 2. 27. x = 5/ v / 3« 2 -64. 16. y = 4/(x - l) 2 . 22. y = 3/(2 x - 5) 2 . 28. x = 4/ V6«-i 2 . 17. 2/=Vx 2 -4. 23. 2/ = 4(x 2 - 2x)*. 29. x = 3 V5 J 2 - * 5 . 18. y=Vl-x s . 24. s=Vl-2J 8 . 30. j/=V4x 3 -3x 2 + 6x. 19. y = 3/(1 - x) 8 . 25. fi = 2 -^(l + i 2 ) 2 . 31. y = 2 V(l + 4x 2 ) 8 . 20. y = V4-3x 2 . 26. x = 5/(3 - 2 <) 2 . 32. y = 3 V(5-6x 2 ) 8 . EXERCISE VI Products and Quotients of Algebraic Forms. The formulas are d , . du dv d u I du dv\ „ — (uv) = v \-u— , =[v a— )/i> 2 . dx K ' dx dx dxv \ dx dx)' It should be borne in mind that u and v are both functions of x. If either of them is a mere constant, it is obviously not necessary to use these formulas. If the product is merely the product of two simple polynomials, it is preferable to multiply out into a simple polynomial before differentiating. If, in the case of a quotient, both u and v are polynomials, and u is of higher degree than v, it is possible to divide out before differen- tiating; for example, (x 8 + l)/(x — l)=x i + x+l + 2/(x — 1). Apply the above formulas in differentiating the following : 1. y = (x + 2)Vx 2 + 4x. 5. 2/ = (x-4)Vl-2x+ 4x 2 . 2. y = (2x+5)Vx 2 -3. 6. j/=(2x + 3)Vx 2 + 4x. 3. y = (2— 3t)Vl-2i 2 . 7. 2/ = (2x-l)V2x-x 2 . 4. x = (3< - 2)V2i 8 - 4t. 8. y = (2 - t)V3fi + 2t + 1. FORMAL DIFFERENTIATION 9. V- 10. y- 11. y 12. y. 13. y 14. 2/ 25. y 26. y 27. 2/ = (x 2 + 2x)V27+3. „ 2£ 2 - 15, s = 2a; + 3 3-2J 2 2 « 2 — 4 i 1-x 2 2x 2 + 3x 1 + 2x « 2 -3i-l 3t-l < 8 + 4 ,. x 8 -l 16. y = x 2 -3x 17. y = 0. y = 21. y = _ V3-t 3-x Vx 2 - 6x " l + 2«" 3-x 8 2x 2 + 5' 18. j/ = 19. y = Vl+ 2t 2 22. 1/ = 2« + 3 Vl-3x 2 y = x 2 = Vl-2x V2x 2 -3. =\ / r+Tx vi+2x. = \ / T+7 8 Vi-x 2 . Vx 2 -5 Vx-1 X 2 x 2 -x Vl-x 2 28. 2/ = Vl + x 2 Vl - x 3 . 29. 2/ = Vx 2 - 2x Vx 8 - 3x. 30. y = V3x + x 8 Vx 2 - 4. V34 2 +5 "••^-rrr- EXERCISE VII Review of Algebraic Forms. Differentiate the following : 1. 2/ = x 8 — 2x 2 — 2/Vx. 16. j/ = xV2x + 4 — £(2x + 4)1. 2. 2/ = (a* - xi + 2)/x*. 17. y = 2tVl-4t + | (1 - 4t)i 3. 2/ = 3/ Vx + x/V3 + Vx/Vs. 18. 2/ = (2 t/VT+t) - 4 VT+t 4. 2/ = 3/Vx + 2/x 8 . 5. y = V2x 8 + 4x— 5. 19. 2/ =(x 2 /Vl+ 2x)+ xVl+ 2x. 20. 2/ = (Vl + 2x/2x)-l/Vl+2x. 6. 2/ = V3x + 1/x. 7. 2/=V2«-7 + V7-3«. 8. 2/ = (x - l)Vx 2 -2x+ 2. 9. y = (3x + 2)Vl-2x-x 2 . 21. 2/ = (2x/Vx + 7)+ xVx + 7. 22. 2/ = x 2 V3x 2 -6. 23. j/ = [3/(x-7)] + [(x + 7)/3x]. 24. 2/ = [(2x + 3)/(x 2 + 4)] 2 . 25. 2/ = [(x 2 + 5)/(2x+3)f. 26. j/=V(8t + l)/(3t-l). 10. y = (3 - 2x)Vx 8 -3x. 11. 2/ = (x + x 2 )Vl-x 8 . 12. j/ = (3-x 8 )/(2x 2 +5). 13. 2/ = (3-x)/Vx 2 -6x. 14. y = Vl + 2£ 2 /(2£ + 3). 15. y = 6tVt-2-4(t-2)i. 30. 2/ = (7 + «)V3 - 2t +. 27. 2/ = (Vx+5/2x)+ 2x/Vx + 5. 28. y = (2 Vx + 3/x)/(2 Vx - 3/x). 29. 2/ = 4/V3x 2 -6x+ 5. 8 PROBLEMS IN THE CALCULUS EXERCISE VIII Logarithms of Algebraic Forms, Base e. In differentiating logarithmic functions, the forms should always be rewritten when it is possible to simplify the differentiation by so doing. For this purpose it is well to recall the following formulas used in logarithmic work generally : (a) log uv = log a + log v, (c) log v n = n log v, (b) log u/v = log u — log v, (d) log -Vv = 1/n • log v. Thus log -Vx s + 3x* = i log (x 3 + 3 x 2 ), and log (x s + 3 x^)/-y/x 2 + 6x = log (x a + 3 x\) - i log (x* + 6 x). This simplification is important. The formula for differentiat- ing the logarithm of any function is d , 1 do — log v = dx v dx This is easily remembered by the fact that the derivative is clearly a fraction with v as its denominator and dv/dx as its numerator. The first list is for oral differentiation. 1. y = log (x 2 + 1). 8. y = logV6-x 3 . 15. y = log 1/(2 - 3x). 2. y = log (3 + x 2 ). 9. y = log 1/x. 16. y = logVx. 3. y = log (x s + 3 x) . 10. y = log (x + 5) 4 . 17. y = log Vx 8 . 4. 2/ = log(3x 2 + "). 11. ^ = 31og\/l-2x 2 . 18. y = logV5 + 6x. 5. 2/ = logV2x 2 + 5. 12. 2/ = log (4-3 x) 3 . 19. y = log V3 x 2 - 6 x. 6. y = log (x 2 + 2) 2 . 13. ?/ = log (2 x + 3)*. 20. y = log V(x 2 - 4) 8 . 7. 2/ = 21og(x 2 -4x). 14. 2/ = logl/(x-3). 21. 2/ = logx 2 Vl- 2x. Differentiate the following : 1. y = log (x 2 + 4x + 6) 2 . 6. 2/ = log (2 x 2 - 4x + 3)*. 2. y = log (6x 2 + 2x + 5) 8 . 7. y = log [(x s - 3x)/(x 2 - 4)]. 3. y = log (x 8 - 3x 2 + 6x)*. 8. y = log3 (t 2 + 5t - 6) 2 . 4. 2/ = log 3 Vl— x 2 . 9. 2/ = log(2i 3 - 5« 2 — 4« + 1)*. 5. 2/ = log [(x 2 + 2)/(l - x)] . 10. y = log (x 2 + 5x + 3) (x 2 + 4 x - 1). FORMAL DIFFERENTIATION 9 11. y = log (J 3 + 3 t)i (« 2 + 1)*. 12. y = 3 log [(a; 2 + 7x - 3)/(x 2 - 2x + 5)]. 13. 2/ = log [(3x 2 - 5x)/(3x + 1)]. 14. y = log 4 (£ 3 + 4 1)». 15. ?/ = log[(3t 2 -6i + 4)/(« 2 -2t+l)]. 16. v = log (Vl + x 3 /Vl + x 2 ). 17. ?/ = log[V(3 + x 2 ) s /^(3-x 2 ) 2 ]. 22. j/ = log(2x 2 +V4x 4 + 9). 18. y = log (a; + Vl + x 2 ). 23. y = log(x 3 + Vx 6 + 4). 19. j/ = log(2x+V4x 2 + 9). 24. y = log Vx 2 + 4 Vx 3 + « .c 20. y = log (3 x + V9x 2 -l)*. 25. j/ = log [(1 + x)/x 2 Vl-x]. 21. j/ = 21og(3x +V9x 3 + 16). 26. y = 21ogx 3 V2x 2 + 6x. 27. j/ = log (V2 x 2 - 1/x Vl + x 2 ). xi. y = iugv v ix- — i/x vit x-y. 28. y = log (Vx — 1/x). 29. j/ = log [(Vx 2 + 1 - x)/(Vx 2 + 1 + 30. y = log[(x 2 Vx 2 - l)/(x 2 + 1)]. :)]• EXERCISE IX Logarithms of Algebraic Functions. When the base is changed from e to a, the formula of the preceding exercise is simply multiplied by log a e, or d , , ldv — log w = log C-— • dx ° " v dx As before, it is essential that the function be first rewritten in the best form for differentiation. For example, log 4 V 1 + x*/x 2 should be rewritten as \ log 4 (1 + x 2 ) — 2 log 4 a;. Differentiate the following : \.y = log, Vl-2x 3 . 7. /(() = log 5 [(i 2 + l)/(f 2 - 1) ] . 2. y = log 10 V3x + 7. 8.i/ = log 2 V(l + x 2 )/ (l - x 2 ) . 3. y = log 4 (2x + V6x)- 9. j/ = log 2 x 2 Vl - x 2 . 4. y = log 4 [(2 x - 6)/(5 - 6 x)] . 10. y = log 6 (1 - 2t)Vl -t 8 . 5. y = log 3 (x + 1/x). 11. y = log 4 (x/Vl.- x 2 ). 6. /(x) = log a xVT+x" 2 . 12. s = log 8 V4x-x 2 . 10 PROBLEMS IN THE CALCULUS 16. y = log s (Vl+xV^l-x 8 ). 17. y = log, [(x 2 - 2)_/(6 - 2 as")!]. 18. y = log 8 [(1 + Vx)/(l - Vx)]. 13. y = log 6 (x + Vl + x 2 ). 14. * = log, p/(l + *»)]. IB. » = log 10 [(l+0/(tVl=P)]. 19. j/ = log, [(2 + 3x)/ (x 2 Vl- x 2 )]. 20. v = log s (2 + x + V4x + x 2 ). 21. j/ = log 2 (2x + 3 + V4x 2 + 12x + 5). 22. j/ = log 3 (2 x — 1 + V4x 2 — 4x + 5). 23. y = log 4 (V2 x + V5 + 2x 2 ). 24. 2/ = log s (x + 1 + Vx 2 + 2x). 25. 2/ = log 4 [(6x + l)/2+V9x 2 ~+3x]. 26. y = log 4 (l/x 2 -V x). _ 27. 2/ = log 4 [( Vl + x - Vl - x)/(Vl + x + Vl - x)]. 28. y = log^C^x 8 - 5) 2 /V(5 - x 2 ) 8 ]. 29. 2/ = log 8 [x»(x-l)/(x + 2) 2 ]. 30. y = log 3 [(3x - l) 8 /x(3x + 1)]. EXERCISE X Powers of Logarithms. This list is designed principally to review the previous formulas and not to introduce any new ones. The student should note carefully the difference between log 2 x * and log a; 2 . The latter equals 2 log x, but not the former. To differentiate log"w = (log v) n , we must use first of all the power formula. Differentiate the following : 1. 2/ = log 2 (2x 8 -4x). 2. y =Vlog(x 2 + 6x). 3. y = log 8 (x 2 + 7x). 4. y = log 2 Vx* + 3x 2 . 5. 2/ = log 6 Vl-3x 2 . 6. j/ = 3Vlog(l-2x 2 ). 7. y = log 8 V4 — x 3 . 8. y = 2Vlog(2x 8 -5). 9. j/ = log*(2x 2 - 5x*). 10. y = log| Vl-5x 2 . 11. y = log 2 (x 4 — 3x 2 +5) 8 . 12. y = 21og 2 [(x -7)/(x-2)]. 13. y = log|(x 2 - 2/x). 14. y = log 4 (2 x 8 -1/x 2 ). 15. y = log 8 [(3- 2x)/(4 + 3x)]. 16. y = log 8 (x 8 — 2x 2 + 5x) 2 . 17. v = log»[(l- x)/(l + x)]. 18. y = log 2 [(J 2 - ■ 4)/(t 2 + 4)]*. 19. y = y = log 7 x 2 Vl log*(l + l-x. 20. c 2 ) Vl-2x. log 2 x as log (logs:) i \'\ FORMAL DIFFERENTIATION 11 EXERCISE XI Simple Exponential Functions. In the following formula v may be any function of x, the independent variable, d dv — e? = ef — ■ dx dx Differentiate the following : 1. y = e^ + 2. 11. y = e i/« 2 +4). 21. y = e* 2 '°si'*. 2. y = e l <>s s =r. 12. y = e'log*. 22. y = &+i/^x. 3. y = e~J*. 13. y = e < 2 /< 2 < + 8). 2 3. ?/ = <*^ . 4. 2/ = e"^* 2 + i. 14. i/ = e (vT+l. 24. y = e ^^ + 2x + 3 . 5. y = e iog(x z + x). is. ^ = e iog 1 (x 2 + 4>. 25. y = e'°s 2 *. 6. 2/ = e^i-x 2 . 16. y = e< ;c + 1 J /v ^. 26. 2/ = gVx+V/x. 1. y = e - ^ 8 - 2 *. 17. y = e 1 '^ 1 -". 27. j/ = e(' v ^-')Ai + '). 8. y = e 2 «-i. 18. y = e^ 1 -*'*. 28. y = e&i'+iyfeVi+i^. 9. j/ = e 8 - 1 ' 1 . 19. y = e 1 " 3 * 2 - 6 *). 29. j/ = e 1/V ^-^. 10. j/ = eO-- 0/(i+ o. 20. j/ = e (2l °B 2( . 80. 2/ = e v a- , V(i+< 1 ). EXERCISE XII Exponential Functions. Base Constant. When the base of the exponential function is a constant, a, other than e, the deriva- tive of the preceding exercise is simply multiplied by the log e a ; that is, da v /dx = a" log g adv/dx. Differentiate : 1. y = a* 3 . 9. y = 4?*'&-V. 17. y = ai^ + s. 2. y = 2<* 2 + 28 >. 10. y = 2(x+D/Wx + s>. 18. 2/ = a 3 '"^* 2 * 6 . 3. 2/ = 5 (1 °s'. 11. f(x) - &m-*>. 19. 2/ = &fc"*+i>. 4. 2/ = 2^*. 12. /(«) = ft^ 3 - 8 '. 20. 2/ = (a + b)^-^. 5. J/ = 7-1^. 13. /(x) = a 1 ' ^ "'->?. 21. y = a'V+i. 6. 2/ = 5« 2 + 8 o 2 . 14. y = S ,o e( 1 °s ar ). 22. 2/ = 3"*^ 2i /a; . 7. y = 7i Io S"- 2 . 15. ^ = 2i/ 1 "S«. 23. /(») = 2"^+ W*. $. 2/ = S^ 1 - 2 ^ 16. v = c'os 2 '/ 2 . 24. /(f) = a' + WW+D. 12 PROBLEMS IN THE CALCULUS EXERCISE XIII Logarithmic Differentiation. It is desirable to take the loga- rithm of both sides to the base e before differentiating in the following cases : (a) where there is a " variable to a variable power " involved, and (b) in the case of products and quotients involving three or more terms. In every case be sure that the right-hand member has been written in the best form for differ- entiation before proceeding. (See Exercise VIII.) Apply this process in differentiating the following : x 2 Vx 2 + 7 ,, 2 * 2 (1 — 3 i> l.y = — -=• 14. x = -vV + 1 (1-t- MV • 1 + x 2 ,. 2y s V6y + l y = .. 15. X — X s Vl - x 2 V3 y 2 + 4y + 4 3. y = x 3 (x 2 + 4)l(x 3 + 3x)i. 16. x = 2l 2 -\/3t + 6 Vf 2 + 2 1. " (2x- 5) 3 ,„ ~ 2 ' 4. ,y = - — i ; 17. y = - (l-5x 2 )Vx 2 -l 5V6x + 5- v / 3x 2 + 4 3x 2 V2x + 7 ,„ 2xV3x 2 + 2x + 1 5. ,y = 18. y = = ^ = ^ = ' ■ Vx 2 + 5 Vx*-3x 6. y=(x + l)x*. _ (3t 2 -2)V< 2 + 3 7. y = (x + l)xU*. • ' V ~ 2 1 2 — 5 1 + 6 , ..8. 2/ = x 5 .3^, 20. y = 7x V3 - 4 x V4 x 8 - 3 x 2 . 9. y = x 1 '* logx. 21. y = 4(1 - x 2 )Vl-x ^Thc. 10.;^ 1 ±lg_ - 22. y »l(*- 2 H*-3)■ Z 2 V(l + « 2 ) 3 \ (x - 4) (x - 5) 11. x = i 2 (1 - t 2 )i (1 - *»)*. 23 ^ = (2-3t) 3 VT+l «./(*) = 2X * (3- 2t) 2 Vl- t' (x-l) 2 V2x + 4 _ x 3 Vl + Iogx 13. /(x) = 4x s (l-2x)7(x+5)4. ,' ?/j (1 + x) 4 EXERCISE XIV General Exponential Functions. These functions are all vari- ables to a variable power, or of the type u v . They are to be differentiated by logarithmic differentiation. The student is cautioned to note the difference between v n , a variable to a constant power; a", a constant to a variable power; and the FORMAL DIFFERENTIATION 13 present case, u v , a variable to a variable power. In the last case only need we resort to logarithmic differentiation. Differentiate : l.y = x 1 *'. 7. y = (l/x)^*. 13. y = (2t + 5)' 2 + s. 2. y = (l-x)^. 8. y = 3 (x + l)*+i. 14. y = x^ 1 -*. 3. ?/ = (1 + x)* 2 . 9. y = (x 2 - 4)1/2*. 15. y = fe (Z . 4. y = 2x v ^. 10. y = 2x**-»*. 16. 2/ = (1- Sa:)^ 1 . 5. x = 3(t 2 - 6t)«. 11. y = 4(3x 2 -4x)3*. 17. 2/ = x«W + *>. 6. j/ = 3x :c2 + s . 12. j/ = (1+ log a;)**. 18. y = 3«' ! + s« + i. 19. y = (xlogx)*. 25. j/ = (2x+7) 1 A 2:c + 1 ). 20. y = (x + 1)**. 26. J/ = (1 - x 2 ) ^i 3 *. 21. y = &' + \ 27. y = [(1 + x)/(l - x)y. 22. y = (Vx 2 + 4)*^. 28. y = jVii + d/ii-i). 23. y = x 2 (x 2 + 4)2*. .. 29. ?y = 2(x 2 + 4x+ 3)^+1. 24. y = -v/3x 2 -6x + 7. 30. « = [log(2t+ 1)]«. EXERCISE XV Simple Trigonometric Functions. The derivatives of the six trigonometric functions are d . tfo d „ dv — sin v = cosy — ; — cot v — — csc z v — ; dx dx dx dx d . dv d dv — cos v = — sin v — ; — sec v = sec v tan v — ; dx dx dx dx d dv d dv tanv = sec 3 i> — ; — esc v = — esc v cot v — - dx dx dx dx As a mnemonic fact it should be noted that the derivatives of the co-named functions have the minus sign. Also notice that the argument, that is, the angle, does not change in the process of differentiation. Do not forget the dv/dx. Differentiate the following : 1. y = sin x 3 . 4. y = sin Vx. 7. y = cos(e 2 '). 2. y = sin(2x+ 5). 5. y = sin e 1 *. 8. y =. cos Vl- «. 3. y = sin(2/x). 6. y = cos(3 - 2x). 9. y = sin (x 2 + 1/x). 14 PROBLEMS IN THE CALCULUS 10. y = cos (log Vx). 12. y = sece*. 14. y = csc(x 2 + 1). 11. y = tan(2x 2 + 3). 13. y = cot(l/x 2 ). 15. y = tane 1 ". 16. y = cos(e 1 '* + e- 1 ' x ). 26. y = oso[(l — x)/(l + x)]. 17. y = sin[(2x + 7)/(l - 2x)]. 27. y = tan 1/Vl - x 2 . 18. y = cos (log Vx 2 — 2x). 28. j/ = sec** 2 . 19. y = cos(sinx 2 ). 29. y = csc(l/x + l) 2 . 20. y = cos(2e* + 2e-*). 30. y = tan log [(1- 3x)/(2x- 4)]. 21. 2/ = 2cos[(l-2x 2 )/(3x' 2 -2)]. 31. y = csc[(3x- 2)/( x 3 - 3)] . 22. y = sec [l/(x 2 + 2)] . 32. s = sin log (2 1 + V4J 2 -l) . 23. y = cos(« 2 — 1/t). 33. x = cote^'+D. 24. y = cot(7- 2 2 ). 34. = seclog(« 3 + 3). 25. y = cos 1/Vx. 35. 6 = cos(2/x 2 + 7). EXERCISE XVI Powers of Trigonometric Functions. No new formulas are re- quired, but care must be observed to distinguish between sin 3 a; and sin x 8 , the former being of the type v n , while the second is simply sin v. Notice that the " argument " (the angle) never changes during the process of differentiation, and the differen- tiation of the argument is the final step. That is, in differen- tiating sin 8 2/x, 2/x is the argument, and the differentiation of d I 2\ 2 2 this will be the last step. Thus — I sin 8 - J = 3 sin 2 - • cos — • 2 fi 2 2 ^ ' "" "" „ = -„ sin 2 - cos - • Differentiate : x' xxx 1. y = sin 8 x 2 . 5. y = cot 3 ^'. 9. y = 4 Vsin 2 x. 2. y = cos 2 Vx. 6. y = 2 sin 4 er 3( . 10. y = 6 Vcosx 3 . 3.2/ = tan* (1/x) . 7. s = 3 tan 2 (x + 3) . 11. y = esc 2 (3/x) . 4. 2/ = sec 6 log x. 8. 6 = 2 cos 8 (1/3 x). 12. j/ = cot 2 (1 - 2 x). 13. y = sin 2 [(x - 4)/(x + 2)] . 19. y = csc 3 Vl-2x 2 . 14. y =Vsin3x 2 . 20. y = sin 4 [x/(l - x)]. 15. y = tan 8 (e« 2 - er ««) . 21. x = 2 sec 2 [l/(£ 2 - 1)] . 16. y = 2 sec 2 (log 1/x) . 22. f(t) = sin 4 (S 2 - 1/J) . 17. y = 3 cos 4 (x logx). 23. s = cot 2 [(2 1 2 - 5)/(3 - «)]. 18. y = cot 8 a* 2 . 24. x = sin 2 (2/e»). FORMAL DIFFERENTIATION 15 25. ?/ = sin 2 (sm2x + 1). 31. y = cos i (t+l)'. 26. /(s) = esc 4 (3 - 2/s) 2 . 32. y = csc 2 ^ 1 - 2 *. 27. f(x) = cot 2 eV(^-i). 33i y _ co t 3 (tei/<). 28. /(x) = 4 tan 3 vW + e" 2 *. 34. y = sec 2 [3 eW + 2)] . 29. ?/ = tan 8 (log sin x 2 ) . 35. y = 1/tan 2 Vl - t 2 . 30. 2/ = sin 3 x*. 36. y = 4/sin (£ 2 log t). EXERCISE XVII Forms Involving Trigonometric Functions. In the following products, quotients, exponentials, etc., no additional formulas are required. 1. y = sin3£cos2£. 5. y = x emx . 9. y = e^ 1129 cos 2 0. 2. y = e 2< sin4/2. 6. 2/ = 2a 8 « c *. 10. x = log sin 8 20. 3. y = e* in2 ". 7. j/ = extant 2 . 11. s =logsec 2 t/2- 4. y = sinf/2tant/2. 8. y = 2"»" 8 *. 12. y = (sin*)'. 13. y = log (sec 3 x + tan 3 x) . 22. /(«) = &™ 2! log sin 2 «. 14. s = (Z/3) 8eo2! . 23. y = log (sin 6 sin 2 sin 3 0). 15. y = sin 2 x tan 1/x. 24. y = log xo (sin0cos30). 16. ?/ = sin 2 2/x tan 2 x/2. 25. y = (cosx)i+°<»*. 17. y = e tana: cosx' 2 . 26. ?/ = Vx cos(l— x 2 ). 18. y = 7-**cos(8- 2z). 27. 2/ = 2(sin3*)"+ 1 . 19. y = log tan 1/2 x. 28. f(x) = (sec 1/x) 21 . 20. y = (sin t) f- ' . 29. /(«) = e*»» 8 * sin 3 1. 21. /(*) = e BM 2< /sin 2 1. 30. /(<) = 3» ta = — -/(l + v 2 ), dx dx' dx dx' v — arc cos v = /Vl — f 2 , — arc cot i> = — — ■ /(l + v*), dx dx' dx dx' K d do n — - d do /— — - — arc sec v = — /»V» ! -1, — arc csct> = — — /pV» ! - 1. dx dx' dx dx' 16 PROBLEMS IN THE CALCULUS The student should notice that the derivatives of the " co-named " functions differ from the others only by the minus sign. Differentiate : 1. y — arc sin 2x. 5. y = arc sin 1/x. 9. y = arc cos l/2x. 2. y = arc cos x 2 . 6. 2/ = arc cot e*. 10. y = arc sin Vx. 3. y = arctan3x. 7. 2/ = arc tan 2/x. 11. y = arcsec(l— x). 4. y — arc secx 3 . 8. ?/ = arc esc 2x 2 . 12. 2/ = arc sin e 1 *. 13. 2/ = arc tan(logx). 21. /(x) = arc sec[(x 2 + l)/2x]. 14. y = arctan ^(e 2( - e- 2 <). 22. /(x) = arc cot[(x 2 - l)/2x]. 15. ?/ = arc esc ^ (e 2 ' ( + e~ 2 /'). 23. /(x) = arc tan(sinx). 16. y = arccos[(2x + 3)/(3x + 2)]. 24. /(x) = arc cos log (x 2 - 1) . 17. ?/ = arc sin [(1 - f)/(l + «)]. 25. y = arc sec J (a'" 2 + a-* 2 ). 18. /(t) = arc tan[(f — 1)/Vt]. 26. y = arc sin tan(2x + 5).- 19. /(x) = arccos(l/Vl+ x). 27. y = arc sin [^/(e 3 * 2 — e~ 3 * 2 )]. 20. = arc tan log Vt . 28. y = arc sec 2"*/*. EXERCISE XIX Forms involving Inverse Trigonometric Functions. In the following no new formulas are involved. Simplify algebraic results as much as possible. 1. y = x arc cos 2 x. 4. y = 3 log (t/ Vt 2 + 4) + § arc tan f/2. 2. y = e 1 arc sin e*. 5. 2/ = x arc sin x/2 + V4 — x 2 . 3. y = log arc tan x/2. 6. y = 8 x/(x 2 + 4) — 4 arc tan x/2 + x. 7. /(f) = log [(t 2 - 6 1 + 13)/t 2 ] + 10 arc tan J (t - 3). 8. x = 2 Vt — 4 + 2 arc tan £ Vt — 4. 9. j/ = ^ x 3 arc tan x — ^- x 2 + £ log(x 2 + 1). ■2 10. f(x) = | arc sin x/3 - \ x V9 - x 11. /(x) = V2 - x - x 2 + 3 arc sinV(x + 2)/3. 12. 2/ = (l— x 2 )2arcsinx — x + x 3 /3. 15. y = x arc sin (cos 2 x). 13. s = 2 Vt— 9 + 3 arc tan ^ Vt — 9. 16. y = e 2 * (arc sec 2/x) 8 . 14. y - Vt 2 - 1/2 1 2 - £ arc sec t: 17. y = Vx 2 - 1 arc esc x. 18. 2/ = (1 - x 2 )/x 3 ■ arc sin 1/x - (x 2 - 1)1/3 x 3 . FORMAL DIFFERENTIATION 17 EXERCISE XX Function of a Function. In many problems a certain variable y is a function of v, while v in turn is a known function of a third variable, x or t. Obviously one could first eliminate v, and express y as a function of x ; then differentiate as before. But it is frequently much easier to make use of the theorem (A) dy/dx = dy/dv • dv/dx. (The student is reminded that this theorem is not proved by merely canceling out the dv on the right-hand side.) Find dy/dx (or dy/di) in each of the following, using theorem (A). 1. y = v% + 4, v = 1 + logs. 5. y = z 2 + 2 z, z = arc sin (x/a). 2. y = (1 - t 2 )§, t = 1/x. 6. y = 2z/(z + 2), z = (2x + 5)/(x - 1). 3. ?y = e», « = 3*". 7. ?/ = u 2 /(u - 1), v = e* + s . 4. y = sin Vu, v = cosx 2 . 8. jr = logt 8 , t = x/(x + 4). 9. y = log [(3 v 2 + 5)/u], v = e 2 *. 10. 2/ = z log z, z = sec 1/x. ,(2 + z) 2 . 11. y = log- 1-a 12. ?/ = 2 a — 4 arc tan v/2, v = Ve* — 1. 13. y = log(l + u 2 ) + Vl + v 2 , v = tanx. 14. y = log (3 v + V9 u 2 + 4), ti = § tani. 15. 2/ = Vu + arc tan VI), v = Vl + x 2 . 16. y = 2 arc tanv (1 — x)/(l + x), x = cost. 17. y = arc tan (Vl + v 2 — v), v = cotx. 18. y = log[(3v + 2)/(2i> + 3)], d = secx + tanx. 19. y = e' + e 2! , u = log (x — x 2 ). 20. y = Vu — log ( Vu + l), d = e 2 *. EXERCISE XXI Implicit Functions. In an implicit function, the function is not explicitly expressed in terms of the independent variable, but the relation between x and y is given by an equation in- volving both variables. In differentiating products, quotients, etc., involving two variables in this way, the ordinary formulas 18 PROBLEMS IN THE CALCULUS for products, etc., are applicable, giving directly an equation involving x, y, and dy/dx, which can be solved for dy/dx. The result in general will involve both variables. Illustration : xy + x*y + y 2 = 0. Differentiating : xdy/dx + y + x 2 dy/dx -f- 2 .;•// + 2 y dy/dx = 0. Solving : dy/dx = — (y + 2 xy)/(x -\- x 2 + 2 y). Find dy/dx in each of the following : 1. xy + x 2 + y 2 = 0. 16. sin 2 x cos 2 y = 0. 2. x 2 y — 4y 2 = 0. 17. x 2 = y 2 logxy. 3. 3x 2 y + 4x-2y = 0. 18. e 2 * - log (y 2 + 4) + 7=0. 4. y 8 — 3xy + x 2 = 0. 19. &* + « + sin(x + 2y) = 0. 5. x 2 y + 3 y + x 2 = 16. 20. 3*2* = arc tan y 2 . 6. x 3 + 2 xy — y 2 = 4. 21. arc sin xy — arc tan log y = 0. 7. x 2 + 2 xy - 3 y 2 = 0. 22. (sin y)* 2 - (cos x)v* = 0. 8. x% + yb = 6. 23. (sin a;)* = y" i /Bin a: 9. x* + y*x* = 0. 24. Vx 2 + y 2 = log Vx 2 — y 2 . 10. 3x 8 — xy 2 — 2y = 0. 25. logsinxy — c + y = 0. 11. x 2 y + 2xy + y 2 = 0. 26. x 2 siny — y 2 sinx = 16. 12. y 2 — 2 xy 2 + x 2 y 2 + x 2 = 0. 27. e 2 '* sin y/2 — 0" 2 sin 2/y = 0. 13. xy 2 + 2 x 2 y + 3 xy = 5. • 28. y/Vx 2 + y 2 + arc tan y/x = 0. 14. y s + 3x 2 y + xy 2 - 3x + 5 = 0. 29. 2 log(x + y) + logx/y = 0. 15. ef c + " + €F-v = 0. 30. x l °si> + yi°e* = x + y. EXERCISE XXII General Review of Formal Differentiation. Find the derivative in each of the following : 1. y = 2** + V3x + 1/2 x 2 + e~J*. 2. /(x) = 2x/(4 + x 2 ) + arc tanx/2. 3. /(x)=^xVx 2 -l+ ilog(x + Vx 2 -l). 4. x = (2i 2 -l)/3« 8 - V« 2 + l. 5. y = log(Vx + Vx + 4). 6. y = 3x 8 + l/3x 8 +\/3x+ 3*" + e*. 7. y = x 2 - l/2x 2 + V2x + 4*« + e 2 >*. FORMAL DIFFERENTIATION 19 8. x 2 y + xy 2 - 4x + 4y - 5 = 0. 9. y = (l + e#*)/(l-e#*). 10. y = \(x- 4) V8x- x 2 + 8 arc sin ^(x- 4). 11. 2/ = xlog(4x 2 -l)-2x- ^log[(2x-l)/(2x + l)]. 12. y = xlog(4x 2 + 1)— 2x + arc tan 2 x. 13. y = (x + 1) are tan Vx — Vx + 4. 14. y = 2 arc sin 4/2 - 1 4 V4 - 4 2 . 15. x = Vi 2 - 4/8 4 2 + T ^ arc sec 4/2. 16. ?/ = Vi — x arc sin Vx — Vx. 17. 2/ = logcot(l-x 2 ). 18. y = sec x tan x + log (sec x + tan x) . 19. y = (x 8 + 2 x) Vx 2 + 4 — 81og(x + Vx 2 + 4). 20. y = log 6 [V3x 2 + 7/(x 2 V2x + 6)]. 21. y = (sin 1/x + cos l/x)/e 1/x . 24. y = arc sin [4/(0** + 4 e-**)]. 22. y = (sinx 2 ) 2 *-!. 25. y = log[(e 2 * + e- 2 *)/(l + «*)]■ 23. y = arc tanx*. 26. j/ = (1 — e- ! *)/(l + e 31 ). 27. ?/ = (2x 3 + 5x) Vx 2 + 1 + 3 log(x + Vx 2 + l). 28. 2/ = log*(x + 2). 29. y = arc cot £ (e 1 '* 2 — e- 1 '* 2 ). 30. j/ = [(1 - x)/x]a-*v*. 31. y = log(x + l) 3 (x — 1) (x 2 + 1) + 5 arc tanx. 32. y = J (1 + x 2 ) log (1 + x 2 ) - x 2 /2. 33. 2/ = (648 - 108 Vx - 18 x - 5 x Vx)/6 y/l - Vx. 34. ?/ = log 4 (V2x 2 - 3/Vl - x 3 ). 42. y = (e 2 * + 3)*. 35. 2/ = arcsin2/x + arc cotx/2. 43. y = cos 3 3 x tan 3/x. 36. 2/ = [log(«* + l)]/(e* + 1). 44. y = sin 3 3/x secx/3. 37. 2/ = e- " 3 (sin 3 4 - cos 3 4) . 45. y = e- «/* tan 2 4 4. 38. 2/ = log(e 2 *+ a 8 *) 2 . 46. y = e- (/4 (2 sin2 4— 3 cos 2 4). 39. 2/ = e 1 * sec 2 2 a. 47. xainy + y cosx + j/ 2 — X2/ = 0. 40. y = 1/x (log 1/x -1). 48. X2/— xsec2/-2/tanx-16 = 0. 41. ?/ = (l+ 2x)log(l+ 2x)-2x. 49. e*v+logxy+ sinx2/-x 2 = 0. 50. sin(x + 22/)+cos(2x + 2/)-16 = 0. 51. If a, b, and c are the roots of /(x) = 0, prove that f'(x)/f(x) = l/(x - a) + l/(x - 6) + l/(x - c). CHAPTER III SIMPLE APPLICATIONS. SUCCESSIVE DIFFERENTIATION EXERCISE XXIII Direction of Curves. In the following draw the curve and the tangents at the points considered. 1. Find the slope of the curve y = sin s for the points where x = 0, tt/6, tt/4, ir/2, rr. 2. Find the points at which the curve x 2 + y 2 — 6x — 4?/ + 12 = is (a) parallel to the x-axis ; (b) parallel to the ?/-axis. 3. At what points on the circle x- + y 2 = 25 is a, particle moving parallel to the straight line 3x— 4j/ + 7 = 0? 4. At what points on the curve y = X s — 12 x + 4 is the slope of the tangent equal to 15 ? 5. Find the points at which the curve 6y = 2 X s — 15 x 2 + 42 x— 30 has (a) inclination = 45° ; (b) slope = 3. 6. Find the points where the tangent to the curve y = ^ x 8 — x 2 + \ (a) has an inclination of 3 ir/4, and (b) is parallel to the x-axis. 7. Two particles always remaining in the same vertical line follow paths given by the curves y = 2x 2 — 8x+l and 2y = x 2 + Sx — 5. When will the two particles be moving parallel ? 8. Find the coordinates of the points on the curve s\i = 16 at which the normals pass through the origin. 9. What is the direction of the curve y = x 3 — 6x 2 + llx— 6 at the points where it croses the x-axis ? 10. Find the points at which the curve 6 y = 2 X s — 3 x 2 — 42 x + 60 has (a) inclination = 135° ; (b) slope = 5. 11. At what points on the curve y = x/(l — x 2 ) has the tangent an inclination of 45° ? 12. At what points is the tangent to the curve y = x 8 + 1 (a) parallel to the line 3 x — y = 1? (b) perpendicular to the line x + 12y — 5 = 0? 20 SUCCESSIVE DIFFERENTIATION 21 13. Where is the tangent to y = 6 (x — 6)/(x 2 — 2 x — 8) parallel to the x-axis ? 14. Find the points at which the curve 2 y = 2 x 3 — x 2 + 1 is (a) parallel to the line 10 x— y — 5 = 0; (b) perpendicular to the line x + 2y + 3 = 0. 15. Find the points at which the circle x 2 + y 2 + 4tx — Gy = 3 is (a) parallel to the x-axis ; (b) parallel to the y-axis. 16. Where is the tangent to the ellipse 4y 2 — 4xy + 2x 2 — 2x = S parallel to the line x— 2y + 4 = ? 17. Where on the ellipse y' 1 — 2 xy + 2 x 2 — 2 y — 7 = has the tangent an inclination of 45° ? 18. Find the points for which the tangent to the curve y = sin x + cos x is (a) parallel to the x-axis ; (b) has an inclination of 45°. 19. Given the curve y = | x a — J x 2 + 1. Find (a) the inclination at the points where x = 0, 1, 2 ; (b) the points for which the slope is 2. 20. At what points on the curve x 4 — 8 xy 2 + y* = is the tangent parallel to the ?/-axis ? 21. Two points always remaining in the same vertical line follow paths given by the curves y = x 2 + x — 6 and 2y = x 2 — 2x — 8 respectively. When will the points be moving (a) in parallel directions ? (b) in perpen- dicular directions ? 22. What is the inclination of the curve y = 4 x/(x 2 + 4) at the origin ? at the point (2, 1) ? 23. Find the points at which the curve y = 2 sin x + 3 cos 2 x has slope = 0. 24. Find the points at which the curve y = 5sin 2 x + 2 cos 4 x has slope = 0. 25. Find the coordinates of a point on the curve y = 2 sin 2 x where the inclination is ir/3. 26. At what points of the curve y = sin 3 x — 3 sin x is the slope = ? 27. Find the points for which the curve y = 2 sin 2 x + 3 cos 2 x has slope = 0. 28. At what point on the curve x 3 + xy 2 + 4 y 2 — 4 x 2 = is the tan- gent parallel to the y-axis ? 29. Find the angles at which the curve y s = x 2 — 7x cuts the line y = 2. 30. At what angles does the curve 4 y = x 3 — 4 x cut the x-axis ? 31. Where on the cissoid y 2 = x 3 /(4 — x) is the slope of. the tangent equal to 2 ? 22 PROBLEMS IN THE CALCULUS 32. Where on the curve y 2 = x 2 (4 — x) is the tangent parallel to x+ 2y + 5 = 0? 33. At what point on the curve xy 2 = 16 does the normal go through the origin ? 34. Given the equation of the family of curves 4 y = 4 ce~ 2 x + 2 x 2 — 2x + 1, where c is the variable parameter. Find the locus of points on the curves of this family at which the tangents are parallel to the x-axis. 35. Given the curve family 2y = 2 ke x + sinx + cosx — 2, where k is the parameter. Find the locus of points on the curves of this family where the tangents have an inclination of 45°. EXERCISE XXIV Algebraic Curves. Tangents, Normals, and Angles of Intersection. Since dy/dx is equal geometrically to the slope of the tangent to the algebraic curve represented by y =/(a:)[or by F(x, y) = 0], we can at once apply our knowledge of calculus to the follow- ing problems concerning algebraic curves. Evaluating the de- ■ rivative for the values (x v 7/ x ) of any point gives us the slope m of the tangent line. Then y — y 1 = m(x — x x ~) is the equation of this tangent line, and y — y 1 = — 1/m • (x — x x ) is the equa- tion of the normal to the curve at the point (x v y^. Now if two curves are given, we can find the value of dy/dx, or m, for each curve at this point. Call them m, and ra 2 , and then tan 8 = (m 1 — m^)/(l + m, x m^) will give us the angle be- tween the two curves at their point of intersection. In the following examples find (a) the equation of the tangent and normal to each curve at one point of intersection, (b) the angle between the two curves. Draw figures. If the origin is a point of intersection, take another point. Calculate angles in degrees. 0. [2y: Uv- = x 2 , 2, \x 2 = = X 2 + 4, 3. [2y {x 2 - = *», = x 2 + 4. :8- -2y. - 4x + 2y fx 2 - \x 2 - ■2x = 2y -1. 6. 102/ x 2 -: ix, ■6x + 2y + 5 = 0. = 2x 2 - 4x + 2C (y = x 2 -2x-3, fiy = 2x 2 - 3x, '\y = 3x-x 2 . j_4^=x 2 + 4. SUCCESSIVE DIFFERENTIATION 23 = x 3 , J 2/2 = 8— 4x s , = 32x. \2/ 2 = 4x 3 . = 16x, 13 _ J82/ = x 3 , x 2 — 2x. ' l42/ = 4x— x 2 . fx 2 + 2/2 + 2x-24 = 0, t,4x 2 + i/ 3 + 8x- 48 = 0. = 12, 18 _ [2/2 = x3 8. (x? = ±y, \2 2/ 2 = x. 10. (2 2/ L2/ 2: 9. fx 2 = 82/ 3 , \4j/ = 9x — x 2 . 11. L2/ = 14. f J/- 2 = x 3 — 4, \x 2 + 2/ 2 — 6x+ 4 = 16. fx 2 = 16 2/, \j/ 2 — 62/ = x. 17. ix 2 - 19. ri0x 2 +2/ 2 -34 = |32/- 2x 2 = 0. 0, (y 2 = \7x- + )/ 2 = 25. ' \7x-3i/-4 = 0. 2Q ; ?/ 4, x o, |^x + 2/ 2 — 6 = 0. ,,. '2x 3 -2/ 2 + 42/-4 = 0, ^ 2/ 2 - 4?/- 8x + 4 = 0. 2/ 2 = x s /(4 - x), 2/ 2 = 4x— 4. For the remaining curves calculate the length of the sub- tangent, s t , and of the subnormal, &■„, as well as the equations of the tangent and normal at the points indicated. The student should not try to remember any formulas for lengths, as they are readily calculated from the figure. The observing pupil will soon note that s t = yjm and s n = yjn, where y x is the given ordinate. 23. X2/ 2 = 16,at(2,2Vl). Show 27. x 2 + 22/ 2 -2x2/-x = 0,at(l,0). that the normal passes through the 28. y 3 = x 2 — 6x, at x = 2. origin- 29. y = (x - 3) 2 , at the points 24. 2/ 2 — 2 x 2 — 3 xy — 4 x = at where the absolute length of the the points where x = 2. subtangent is 2 units. 30. Given the three curves y = 25. x 2 + 2/ 2 -6x-16 = 0,atthe . 2 „» , , * w ' ix', 2/ = 2x 3 , and y = x i . Show point where it crosses the posiUve ^ for ^ ^^ rf x the , engthg end of the y-axis. of the SUDtangents form a har . 26. y 2 = x 8 /(4 — x), at (2, 2). monic progression. EXERCISE XXV Tangents and Normals. Parametric Equations. When x and y are both given as functions of a parameter t (or 6), dy dy dx Tx~~dt~~dt Then the slope at a point determined by any value of t, say t = a, is found by evaluating dy/dx for £ = a. This is m, and we continue as in the preceding exercise. 24 PROBLEMS IN THE CALCULUS In each of the following find the equation of the tangent and of the normal for the indicated values of the parameter, and also calculate the length of the subtangent and of the subnormal. 1. x = i 2 - 1, y = i 3 - 4, at t = 2. 5. x = P + 2, y = t s + 1, at t = 1. 2. x = t 2 + 1, y = t + 3, at t = 2. 6. a = 4i + J 2 , y = 2t 2 , at t = 1. 3. x = 3« 2 -l, y = 1-24, at 4 = 1. 7. x = 2i, y = 3/4, at 4 = 3. 4. x = 3 — 24, y = 4 3 — 3, at 4 = 1. 8. x = sintf, y = cos20, atfl = ir/6. 9. x = 4 sec 6, y = 2 tan #, at # = 7r/6. 10. x = tan 4, y = sin 2 4, at 4 = 0, tt/6, tt/4. 11. x = sec 6, y = tan 8, at = 7r/4. 12. x = J 3 + 2 4 2 + 3, y = 4 s + 3 4 + 5, at 4 = - 2. 13. x = 1/(4 + 1), v = V5-4 2 , at * = 0, 1, 2. 14. x = 1/(1- 24), ?/ = V24 2 + 1, at 4 = 0, 2. 15. x = 1/4 + 1, y = t 2 + 1, at 4 = 1. 16. x = 34/(1 + 4 s ), y = 34 2 /(l + 4 s ), at 4 = 2. 17. x = sin 8 6,y = cos 3 (9, at 6 = w/3, w/4. 18. x = 4 cos 6, y = 2 sin 2 0, at = w/3, ir/4. EXERCISE XXVI Polar Curves. The student should avoid the common error of saying that the slope of the tangent in polar coordinates is dp/dB, as this is not true. The two important angles in polar curves are (a) the angle at a point P on the curve, between the radius vector and the tangent to the curve at P, which we will call tp; and (b) the angle t between the tangent line at a point and the polar axis. These angles are given by the formulas (as proved in the text) tant^r = p -=- dp/dB = k, and r — x/f + O, where 6 is the coordinate angle. Obviously if two intersecting curves are given, the angle of intersection, <£, is ^r — f 2 , where ^ and \ji 2 are the values of ^» for the two curves at the point of intersection ; or tan^ = (ft 1 -* 2 )/( 1 + ft^)- SUCCESSIVE DIFFERENTIATION 25 Show that the following curves intersect at right angles : 1. p = a(l + sin0), p = a (1 — sin 0). 2. p = 2sin0, p = 2cos0. 3. p = 4 see 2 0/2, p sin 2 0/2 = 8. Find the angle of intersection between the following pairs of curves : 4. p = 4cos0, p = 4(1— cos0). 9. p = cos20 + 1, p = sin20. 5. p = 2(l + cos0), p = 8cos0. 10. p 2 sin2 = 8, p = 2sec0. 6. p = 4(1- sin 0), p = 4sin0. 11. p 2 sin2 = 4, p 2 = 16sin2 0. 1. p = sin 6, p = sin 2 0. 12. p = 4 sin 0, p = 4 sin tan 0. 8. p = sin 0, p = cos 2 0. 13. p = sin + cos 0, p = 2 sin 0. It is sometimes desirable to know the polar coordinates of the extreme points on a curve. As is clear from a figure, at the extreme high and low points t = 0, or tan t = ; at the extreme right and left points r = ir/2. In each of the following examples calculate the extreme points : 14. p = 2sin 8 0/3. Also find r, the inclination of the tangent for = ir/4. 13. p = sin + cos 0. 16. p = 4(1— cos0). 17. p = sin20. 18. Given the cardioid p = 5 (1 + cos 0) and the point P where tan 9 = \. Find the angle which the tangent to the curve at P makes with the polar axis. Sketch your figure and mark angles clearly. EXERCISE XXVII Velocity and Acceleration. Rectilinear Motion. When the dis- tance s of a particle from an arbitrarily fixed origin is given as a function of the time t, the velocity v, at any time t, is given by the value of the derivative ds/dt at that time. The acceleration is dv/dt, obtained by differentiating a second time. It is clear then that the. particle comes to rest and usually changes direction whenever ds/dt = 0, and that its speed is either maximum or minimum whenever dv/dt = 0. In the first 26 PROBLEMS IN THE CALCULUS 12 problems calculate the position, velocity, and acceleration for the given value of t ; calculate also when the particle first comes to rest. 1. s=248-154 2 + 364 + 10,4 = l. 7. s = 4 8 + 1/4, 4 = 2. 2. s=-24 3 + 154 2 + 36 1, * = 2. 8. s = 4sin 2 4, 4 = 1, 4 = tt/4. 3. s = 244 + 3 J 2 - 4 s , 4 = 3. 9. s = e' 3 - 1 , 4 = 1. 4. s = 24 s -27 4 2 + 1204, 4 = 3. 10. s = log(l + 4 2 ), 4 = 3. 6. s = 1204-34 2 -2 4 8 , 4 = 2. 11. s = 5< 2 , 4 = 2. 6. s = 604 + 34 2 -24 3 , 4 = 3. 12. s = sin4 + cos2«, 4 = 1. In the remaining problems calculate by the aid of a table (a) the position, velocity, and acceleration at the given time t, and (b) the two smallest values of t for which the velocity will be zero. 13. x = e-« 10 sin2(, 4 = 5. 19. x = 4e-" 2 oos34, t = 1. 14. z = 5e- ( cos4, 4 = 1. 20. x = be-v* coairt/2, 4 = 2. 15. x = 2e" 2 sin2 4, 4 = 1. 21. x = 6 er " lr > cos tt4/4, 4 = 5. 16. a: = 4 e-" 3 sin 4, 4 = 2. 22. x = 2 e- "* simr/2 (i + 1), 4 = 2. 17. X = 4e-« 4 sin24,4 = l. 23. x = 10e-" 2 coSTr/3(4 + 2), 4 = 3. 18. x = 2e-^ 5 sin4, 4 = 4. 24. x = 6e-" 5 sin7r/4(4 - 1), 4 = 3. 25. Assuming the formula s = i> 4 — 16 4 2 , calculate the time of flight of a particle projected vertically upward with a velocity of 288ft./sec. What is the velocity at the end of the time ? 26. Two particles are moving in different media, projected however from a common level. If the equations of motion are s = v t — 16 4 2 and s = b 4 — 12 1 2 , and each is projected vertically upward with initial velocity v = 96 ft./sec, which will go the higher and how much ? When and with what velocity will each return to the level from which projected ? 27. The equation of motion for a falling body may be written s = v t + 16 4 2 . If one particle is dropped from rest and 1 second later a second is projected directly downward with v = 64 ft./sec, will it overtake the first ? If so, when and with what velocity will it pass the first ? If started two seconds later than the first, show that they will remain a constant distance apart. 28. A man in a balloon drops a bomb from a height of 2400 ft. and at the same time a shot is fired directly upward with u = 400 ft./sec. How far from the earth will the shot meet the bomb, assuming the aim to be true ? SUCCESSIVE DIFFERENTIATION 27 EXERCISE XXVIII Multiple Roots of Equations. If the equation f(x) = has a as a double root, then x — a is a common factor of both f(x) and /'(a;). Hence the roots of the H.C.F of f(x) and/'(cc) are multiple roots of the original equation, and if the H.C.F is simply a constant, the equation has no multiple roots. Use this knowledge to calculate the roots of the first 16 equations. 1. x 4 -2x 3 -7x 2 + 20x- 12 = 0. 8. 9x 4 + 6x 3 -llx 2 -4x+4 = 0. 2. 4x 4 -20x 3 +37x 2 -30x + 9 = 0. 9. 4x 4 - 99x 2 - 245x- 150 = 0. 3. 8x 3 + 28 x 2 + 30x+ 9 = 0. 10. x 4 + 4x 3 + 2x 2 - 4x- 3 = 0. 4. 18x s -3x 2 -4x + 1 =0. 11. 9x 4 -6x 3 -8k 2 +6x-1 = 0. 5. 8x 4 -4x 3 - 6x 2 + 5x- 1 =0. 12. x 4 + 6x 3 + 5x 2 -24x-36 = 0. 6. 4x 4 -4x 3 -15x 2 + 16x-4=0. 13. 4x 4 + 12x 3 +x 2 - 12x + 4 = 0. 7. 2x 4 - x 3 — 9x 2 + 13x- 5 = 0. 14. x 4 -4x 3 -6 x 2 + 36x-27 = 0. 15. 4x 4 -27x 2 -25x-6 = 0. 16. x 5 + 5x 4 - 5x 3 -25x 2 + 40x-16 = 0. 17. In the following equations determine without solving which have multiple roots and which have not : (a) x 4 - 5x 3 + 2x 2 + 20x-24 = 0. (c) x 3 - x 2 - x - 2 = 0. (b) x 3 -6x 2 + llx-6 = 0. (d) 4x 4 - 19x 2 - 3x + 18 = 0. (e) x 4 -19x 2 + 6x + 72 = 0. (f) 81x 4 - 135x 3 + 54x 2 -12x-8 = 0. (g) x 4 + 2x 2 -x + 2 = 0. (h) x 4 + 4x 2 -4x + 16 = 0. (i) 8x 4 -28x 3 +18x 2 + 27x-27 = 0. (}) 8x 5 - 4x 4 -10x 3 -x 2 + 4x- 1=0. (k) 4x 4 -8x 3 + llx 2 -16x + 6 = 0. (1) 36x 4 + 36x 3 - 31x 2 -4x + 3 = 0. 18. Use the method of this exercise to find the condition that ax 2 + bx + c = shall have equal roots. 19. What is the condition on a and b that the equation x 3 + x 2 + ax + 6 = shall have a double root ? 20. What is the condition that x 3 + ax 2 + x + b = shall have a double root? 28 PROBLEMS IN THE CALCULUS EXERCISE XXIX Successive Differentiation. These exercises embody no new principles, as the idea of a second derivative was used in accel- eration. The process may be repeated indefinitely, although in all polynomials the (n + l)th and all succeeding derivatives will be zero, n being the degree of the polynomial. The student should note in each case whether there is any law governing the successive derivatives which would enable him to write them down* without' actually differentiating repeatedly. Find the second and third derivatives in all cases. Also the wth derivative for the first six. 1. 2/ = sinx+cosx. 6. y = e Sx + 1 . 11. j/=l/Vx 2 . 2.y = l/x. 7. ?/ = x 4 -12x s +48x 2 -50. 12. y = \/x*- 3. y = e 2 *. 8. y = X s - 2 sinx. 13. y = e* + '\ 4. ?/ = cos2x. 9. 2/=xarctanx. 14. y = x 8 ^. 5. y = 5/x 2 . 10. y = er* sini. 15. y = log [(2 + 3 x)/(2 - 3 x)] . 23. x = 3 e~ " s cos t/2. 16. y = xe~°?. 24. y = 8 e~ "* (sin t/2 + 2 cos t/2) . 17. y = (logx)/x. 25. y = ^tan 3 x — tanx + x. 18. y = x 2 logx. 26. y = log[(l + sinx)/(l - sinx)]. 19. !/ = arc cos x. 27. y = x arc sin x. 20. y = t sin £ + cos t. 28. j/ = arc sin (1 — x 2 ) . 21. y = 4e-" 2 sin2S. 29. y = log(x + Vl + x 2 ). 22. y = 6 e" 2 (sin 2 1 + cos 2 1). 30. 2/ = 1/x • sin 2/x. EXERCISE XXX Successive Differentiation. Implicit Functions. In finding the second and higher derivatives when the function is an implicit one, it is generally desirable to find dy/dx first and then differ- entiate both sides with respect to x. Of course the right-hand member in this second differentiation will contain dy/dx, the value of which was found in the first operation. A second method SUCCESSIVE DIFFERENTIATION 29 is to perform the second differentiation implicitly before solving for dy/dx explicitly. Find d^y/dx 2 unless otherwise stated. 1. X s — xy + y s = 0. 5. x 2 — 2xy + 4y = 0. 2. y 2 — 3 xy — 2 = 0. 6. y 2 + 2 x 2 - 4 y + 2 x - 2 = 0. 3. x 2 + 2y 2 - 2xy - x = 0. 7. x 4 + 2xV - 48 = 0. 4. x 8 + y 3 — 3x = 0. 8. sinx + cosy = 1. 9. e* + e'J = 4. 13. e 3 *- 2 " — 4x = 0. 17. e 1 + e« = e* + ». 10. x + cos(x + y) = 0. 14. (y — sinx) 2 = 4x. 18. x 2 logy = 4. 11. x + sin (x + y) = 0. IS. Vx + y = e*. 19. y 2 logx = 4. 12. &=y — x = 0. 16. e Bin * cosy = 0. 20. yeF = x + y. Find d 8 y/ a minimum point. (If the sign does not change, it gives a point of inflection with horizontal tangent.) (2) If a; is a critical value which makes dy/dx = 0, we can proceed as follows : Take the second derivative and evaluate for the critical value of x. If this value is negative, the corresponding point is a maximum; if it is positive, the corresponding point is a minimum. If it is zero, it is usually a point of inflection. At every point of inflection d^y/dx 2 equals zero, and this knowledge enables us to find the points of inflection. In the above discussion of maxima and minima the word " curve " may be replaced by "function," and "point" by "value of the function." 30 MAXIMA AND MINIMA 31 Find the coordinates of the maximum, minimum, and inflec- tion points of the following curves. Use this knowledge to sketch the graphs. 1. y = 2x 3 - 9a: 3 - 24x - 12. 5. y^-Zx* - 9x 2 - 27x + 30. 2. y = 2x 3 -3x 2 -36x + 25. 6. y = 15 + 9x-3x 2 -x 8 . 3. 32/ = x 8 -6x 2 -36x + 30. 7. 3y = x 3 -3x 2 - 9x + 3. 4. 2/ = x 3 + 12x 2 + 36x-50. 8. 2y = 2 + 3x-4x 2 -x 3 . ! + 12x 2 + 3 G) 2/ = x 4 • ■8x 3 + 22x 2 -24x + 12. 10. 61/ =x 4 '-16x 3 + 88x 2 -192x + 192. 11. 5y = 2 + 24x 2 -x 4 . 12. 2/ = x 4 - 4x 3 -2x 2 + 12x+ 6. 13. y = x 6 -7x 4 + 19x 3 -25x 2 + 16x-4. 14. ?/ =x 6 - 5x 4 + 5x 3 + 1. 15. %y = 4x 6 + 5x 4 - 60x 3 + 110x 2 - 80x + 1. 16. y = x*—6x i + 4x 3 + 9x 2 - 12x + 4. /iyj/ = 6x/(x 2 + l). 20. 2/ = (x-4) 2 (x-3) 2 . 23. y = x - 108/x 2 . /f§T^ = x + 4/x. 21. ?/ = x 2 + 128/x. 24. y = 6x/(x 2 + 1). 19. y = (x-2) 2 (x-3) 2 . 22. y = (x 2 '+ 9)/x. 25. y =x s + 48/x. 26. 2/ = (x 3 - 16)/x. 34. 2/ = (x 2 -3x + 2)/(x 2 +3x + 2). 27. X2/ = x 2 + x+l. 35. 2/ = 8x/(x 2 + 2x + 4). 28. xy + x 2 + 2 x + 4 = 0. X 36. 2/ = 4(2 - x)/(x 2 + 4). 29. xy-x* + y = 0. 37. y = x/(x 2 + 6x + 9). 30. 2/ = x (x - 2) 2 (x + 4) 2 .' ^tSj)y = (x 2 + x - l)/(x 2 - x + 1) . 31. y = (2x + l) 8 (x + 2) 2 . 39. y = (x + l)/(x 2 - 4x + 1). 32. 2/ = (x 2 — x + 9)/(x 2 + x + 9). 40. 2/ = x 8 /(2 x + 4), the trident of 33. 2/ = (x 2 -3x-19)/(x + 4). Newton. EXERCISE XXXIII Maxima and Minima. Transcendental Functions. In finding the relative maximum and minimum points on a transcendental curve, it is generally the simpler method to determine the nature of the points given by the" critical values, by means of the sign of the second derivative. If both the first and second derivatives vanish, the higher derivatives determine the nature 32 PROBLEMS IN THE CALCULUS of the point. Determine the coordinates of the maximum and minimum points on the folio-wing curves. Solve graphically or by the aid of tables where necessary. 1, y = sin 8 x. 3. y = 5 arc tan x — x. 5. y = \ cos 2 x" + sin 2 x. 2. 2/ = e*cosx. 4. y = sin 2 x + sin2x. 6. y = e 1 cos2x. 7. # = sinx + cos2x. 16. 2/ = 5e- (/2 cos2t. 8. y = sin 2 t + cost. ' 17. x = 2 e~ " 8 cos t/2. 9. y = 2sint + sin2t. 18. x = 5e- ( sin(t + 7r/3). 10. y = 2 sin t + cos 2 t. 19. x = 4 er "> sin (i + 1). 11. y = cost— ^cos2t. 20. x = 4e-" 2 cos3t. 12. y = 2 cost — sin2i. 21. x = 4e 2( (sint + cost). 13. x = e- < sin 2 1. 22. x = 5 e" 2 (sin2 1 + 2 cos 2 1). 14. y = 2 e- (/4 sin 2 1. 23. x = 4e~"5 s inir/2 • (t - 1). 15. x = 8 e" 2 sin f . 24. x = 6 e~ " 10 cos tt/2 ■ (t + 1) . EXERCISE XXXIV Maxima and Minima. Problems. In all practical problems in- volving maxima and minima, it is necessary first of all to note carefully the function or quantity, F, which is to be a maximum or minimum. This quantity must then be expressed in terms of some one independent variable, say x. Then the maximum or minimum values of the function will be for those values of x which make dF/dx = 0. The nature of the problem usually enables one to tell in advance whether the result is a maximum or minimum, without applying any formal test. The following examples give rise to algebraic functions only. 1. A rectangular flower bed is to contain 432 sq. ft. It is surrounded by a walk which is 4 ft. wide along the sides and 3 ft. wide across the ends. If the total area of bed and walk together is a minimum, what are the dimensions of the flower bed ? 2. A sheet of paper for a poster contains 12 sq. ft. The margins at top and bottom are 4 in. and on the sides 3 in. What are the dimensions if the printed area is a maximum ? 3. A letter Y of total height 16 in. and width across the top of 12 in. is to be made. Find the length of the vertical stem AB, if the length of the stem plus that of the two equal branches is a minimum. MAXIMA AND MINIMA 33 4. Divide the number 8 into two such parts that (a) the sum of the squares may be a minimum, and (b) the product of the two parts may be a maximum. 5. A rectangular box is to be made from a sheet of tin 15 x 24 in. by cutting a square from each corner and turning up the sides. Find the side of the squares cut out if the volume of the box is a maximum. 6. A rectangular garden is to be laid out along a neighbor's lot and is to contain 432 sq. rd. If the neighbor pays for half the dividing fence, what should be the dimensions of the garden so that the cost to the owner of inclosing it may be a minimum ? 7. A box with a square base and lid is to contain 360 cu. ft. If the bottom costs 4$, the lid 60, and the sides 3 per square foot, what are the dimensions for minimum cost ? 8. A right prism, the base of which is an equilateral triangle, has a volume of 2 cu. ft. Find the edge of the base for minimum total surface. 9. A vertical circular cylindrical water tank is to be built to contain 3000 V cu. ft. If the cost of the bottom, including foundations, is $4 per square foot, that of the sides $2 per square foot, and it is roofed with a hemispherical dome costing $1 per square foot, find the dimensions for minimum cost. 10. The cross section of an open irrigation canal is to be of the form of an isosceles trapezoid with the slope of the sides equal to |. The area of the cross section is to be 448 sq. ft. What are the dimensions if the retaining surface (bottom plus sloping sides) is to he a minimum ? 11. A farmer estimates that if he digs his potatoes now he will have 120 bu. worth $1 per bushel, but if he waits, the crop will grow 20 hu. per week, while the price will drop 100 per bushel per week. When should he dig them to get the largest cash returns ? 12. A telephone company agrees to put in a, new exchange for 100 subscribers or less at a uniform charge of $20 each. To encourage a larger list of subscribers they agree to deduct 100 from this uniform charge for each subscriber in excess of the 100 (i.e. if 110 subscribe, the flat rate would be $19). What number of subscribers would give the telephone company the maximum gross income ? 13. A spring is located 40 rd. at right angles from a point B on a straight road. A traveler at A, 100 rd. down the road from B, wishes to reach the spring in the shortest time. If he can walk 5 mi. per hour on the road and 4 mi. per hour across the fields, where should he leave the road ? 14. The perimeter of a sector of a circle is 16 ft. What is the radius of the circle if the area of the sector is a maximum ? (A = ^ R x arc.) 34 PROBLEMS IN THE CALCULUS 15. The shore of a lake is a straight line, and two towns A and B are located 4 and 8 mi., respectively, from D and E, the nearest points on the lake shore, DE being 9 mi. One pumping station on the lake shore is to supply both towns with water. Where must it be located so that the length of the mains to the towns may be a minimum ? 16. A farmer is 12 mi. from A, the nearest point on a straight railway. The railroad company agrees to put in a siding anywhere he designates, and to haul his produce to B, 80 mi. along the track from A, for 5£ per ton per mile. If he can haul by wagon for 130 per ton per mile, where should the siding be located that the cost of transportation of his crops to B may be a minimum ? (Assume a straight road from farm to new siding.) 17. The area of a sector of a circle is 16 sq. ft. What is the radius of the circle if the perimeter of the sector is a minimum ? (A = \ R x arc.) 18. A telephone company finds there is a net profit of f 15 per instru- ment if an exchange has 1000 subscribers or less. If there are over 1000 subscribers, the profits per instrument decrease lif for each subscriber above that number. How many subscribers would give the maximum net profit ? 19. A window composed of a rectangle surmounted by an equilateral triangle is 15 ft. in perimeter. Find its dimensions if it admits the maximum amount of light. Answer to two decimal places. 20. A window is in the form of a rectangle surmounted by an isosceles triangle, the altitude of which is •§ of its base. If the perimeter is 30 ft., find the dimensions for admitting the maximum amount of light. 21. A model is to be constructed composed of a rectangular parallelo- piped with square base topped by a pyramid of altitude § of the side of the base. Its total volume is to be 3 cu. ft. All the surface except the base is to be covered with gold leaf. Find its dimensions for minimum cost of gold leaf. 22. A model of the same character as in example 21 is to be con- structed with a volume of 7 cu. ft. Find its dimensions for minimum total surface. 23. Given a sphere of radius 6 in. Calculate the quantities desired for the inscritfed solids specified : (aj) Altitude of right circular cylinder of maximum volume. (b) Altitude of right circular cylinder of maximum total surface. (c) Altitude of right cone of maximum volume. (d) Altitude of circumscribed cone of minimum volume. MAXIMA AND MINIMA 35 24. A printing company agrees to print 50,000 posters (or less) at the rate of $10 per 1000. If the number needed exceeds 50,000, they agree to deduct 10 £ per 1000 on the whole contract, for each 1000 in excess of 50,000. What number of posters would give the printing company the maximum cash receipts, and what average price would they receive ? 25. If the radius of a cylinder is 2 ft. and increasing 1 ft. per hour while the altitude is 4 ft. and decreasing 1 ft. per hour, when will the volume be a maximum ? How much will it have increased ? 26. A grain dealer has 10,000 bu. of new wheat worth $1 per bushel in an elevator. There is a shrinkage of 200 bu. per month, a fixed storage and interest charge of. $120 per month, and a promised advance of 5^ per bushel per month. If these conditions prevail indefinitely, how long should he hold the wheat ? 27. A ship sailing due north 12 mi. per hour sights another ship dead ahead at a distance of 10 mi . and sailing east at the rate of 9 mi . per hour. If each ship keeps its course, what will be the least distance between them at any time, and how soon will this position be reached ? 28. Two boats are at A and B, A being 65 mi. due north of B. The first boat sails southeast 10 v2 mi. per hour, and the other due east 25 mi. per hour. When will they be nearest each other, and how far apart will they be? 29. The sum of the perimeters of a circle and of a square is constant. Show that for minimum total area the diameter of the circle equals the side of the square. 30. The sum of the areas of a sphere and of a cube is a constant. What is the relation between the diameter of the sphere and the edge of the cube if the total volume is a minimum ? 31. A right circular cylindrical tank is to hold 5000 cu. ft. What are its dimensions for minimum cost if the bottom costs $1 per square foot, and the sides $1.50 per square foot? (Two decimals.) 32. The perimeter of a triangle is 16 in. If one side is 6 in., what are the other two for maximum area ? 33. A triangle ABC has base AB = 6 in., and altitude 8 in. Take D any point in AB. Draw a line parallel to AB cutting the other sides in E and F respectively. What is the altitude of DEF if its area is a maximum ? 34. A given isosceles triangle has base 20 ft. and altitude 8 ft. What are the dimensions of the maximum inscribed parallelogram, one side coincid- ing with the base of the triangle, if the acute angle of the parallelogram is arc tan J ? 36 PROBLEMS IN THE CALCULUS 35. Given OX and OT, two perpendicular lines (which may be regarded as coordinate axes), and a point P (a, b). Draw a line through P cutting the positive ends of the axes at A and B. Calculate the intercepts of this line on OX and OY (a) when the area OAB is a minimum ; (b) when the length AB is a minimum ; (c) when the sum of the intercepts is a minimum ; (d) when the perpendicular distance from to AB is a maximum. Show that these results are all the same if a = b. 36. Given the parabola y 2 = 8x, and the point P (6, 0) on the axis. Find the coordinates of the points on the parabola nearest to P. 37. On a line 6 in. long construct a triangle of altitude 4 in. such that the perimeter may be a minimum. 38. Given a triangle ABC, with D and E on sides BA and BC respec- tively. If BA = a, BC = b, and 2)2? is the shortest line dividing the tri- angle into two equivalent parts, calculate BE and BD. 39. A steel girder 32 ft. in length is to be moved on rollers along a passageway and into a corridor 4 ft. in width at right angles with the passageway. How wide must the passageway be in order that the girder may go around the corner ? In the solution the horizontal width of the girder may be neglected. 40. In using the formula S = P/W to find the specific gravity of a substance, the following pairs of values for P and W were obtained from different specimens: (27,5), (31.2, 6), (35.7, 7), and (42.4, 8). If the best value for S is the one which makes {P x -S W^-f- (P 2 — S W 2 f + (P 3 — SW S ) 2 + (P 4 — SW 4 ) 2 a minimum, calculate the best value and see how much it differs from the arithmetical mean of the four values of S obtained from the separate experiments. 41. Assume Hooke's law that the tension in an elastic cord or spring is proportional to the amount it has been stretched beyond its normal length ; that is, T = kx. In an experiment with a spring, 6 lb. stretched it 2.1 in. ; 10 lb., 3.4 in. ; and 12 lb., 4.4 in. If the best value of k is the one which makes the sum of the squares of the differences T — kx a minimum, calculate the best value of k and compare it with the arith- metical mean of the values given by the individual measurements. (See preceding example.) *». 42. Assume that under similar conditions a gasoline launch uses a constant amount of gasoline per mile, or that y = kx, where y is the number of miles and x the amount of gasoline in gallons. The observed MAXIMA AND MINIMA 37 values on certain trips under similar conditions were 6 gal. in 20 mi., 8.5 gal. in 27 mi., 3.5 gal. in 12 mi., and 11 gal. in 35 mi. What is the best value for k, using the same definition as in Problem 41 ? How much does it differ from the arithmetical mean ? 43. Assume the law PV=k. A series of experimental values of P and V are (2,4), (3,2.8), and (4, 1.9). If the best value of k is the one which makes the sum of "the squares of the differences (P t — k/V x ) 2 + (P 2 — k/V 2 ) 2 + (P 3 — k/V s )' z a minimum, calculate k. Compare it with the arithmetical mean of the three separate values. EXERCISE XXXV Maxima and Minima. Problems (continued). In the following practical problems the function which is to be a maximum or minimum may be either algebraic or transcendental. The stu- dent should note that our methods require that the quantity which is to be a maximum or minimum in these problems should be expressed as a function of one independent variable or parameter. A wise choice of parameter is frequently of the utmost importance. 1. The turning effect of a ship's rudder is shown theoretically to be k cos# sin 2 #, where 6 is the angle the rudder makes with the keel, and k ' is a constant. Eor what value of 8 is the rudder most effective ? 2. Given a triangle ABC, with points D and E on BA and BC respec- tively. BA is 8 units long and BC is 10. If BE divides the triangle ABC into 2 equivalent parts, find the distance BD and BE if BE is a minimum. 3. In a triangle the base a and the angle A opposite are fixed. Find the other two sides (a) when the area is a maximum ; (b) when the perimeter is a maximum. (Use x, a variable angle, as parameter.) 4. Given the ellipse x 2 + 3y 2 = 28, and the two points (5, 1) and (— 4, 2) on it. Find the coordinates of a third point on the ellipse so that the triangle having these three points as vertices may be a maximum. 5. A regular right hexagonal prism has a volume of 36 cu. in. What are its dimensions if the total surface is a minimum ? 6. ABCB is a rectangle, and a straight line APQ cuts BC in P and BC extended in Q. Locate the point P so that the sum of the areas of the two triangles ABP and CPQ may be a minimum. 38 PROBLEMS IN THE CALCULUS 7. Two lines AB and A C of fixed length meet in A, forming an angle of fixed magnitude. Draw a line through A (dividing the angle BAC) such that the sum of the projections of AB and AC on this line may- be a minimum. 8. Find the dimensions of the cylinder of maximum volume which can be inscribed in a sphere of radius 6 in. (Use the angle 6 subtended by the radius of the base of the inscribed cylinder as a parameter. Then r = 6 sin 0, h = 12 cos 6.) 9. Solve example 8 if the convex surface of the cylinder is to be a maximum, using the same parameter. 10. A hexagonal tower is surmounted by a pyramidal roof with a 30° inclination. Find the relative dimensions (a) for minimum surface when the volume is given ; (b) for maximum volume when the surface is given. 11. An angle a is given and remains constant. On one leg of the angle two points A and B are taken, CA being 4 in. and GB being 9 in. How far from C on the other leg must a point D be taken so that the angle ADB is a maximum ? 12. A tank is formed by a vertical cylinder with a conical lid. The altitude of the lid is three fourths the radius of the base. The volume of the cylindrical portion is 104 tt cu. ft. The material for bottom and sides costs $2 per square foot and that for the lid $1 per square foot. What are the most economical dimensions for the tank ? 13. A cylinder is surmounted by a cone the altitude of which is equal to its base diameter. Find the relative dimensions of the solid (a) for maximum volume when the surface is given ; (b) for minimum surface when the volume is given. 14. A rectangular parallelopiped with a square base is surmounted by a square pyramid, the altitude of which equals § of a side of its base. Find the relative dimensions (a) for maximum volume when the total surface is given ; (b) for maximum volume when the surface not including base = 36 sq. ft. ; (c) for minimum total surface when the volume is given. 15. Two right triangles ABC and A'B'C are such that legs AB and A'B' are 3 and 4 in. respectively, legs BC + B'C = 9 J in. What are the lengths of BC and B'C if the sum of the two hypotenuses is a minimum ? 16. The volume of a right cone is given. What is the relation between its altitude and radius when (a) the lateral surface is a minimum ? (b) the total surface is a minimum ? MAXIMA AND MINIMA 39 17. Two points A and B are 100 ft. apart and there are street lights at these points of relative intensity 8 : 27. Assuming the law that the amount of light at a point is inversely proportional to the square of the distance from the source of light, how far from A on the path AB will the total amount of light from the two sources be a minimum ? (See example 60.) 18. A produce house has 10,000 lb. of beef worth 16$ in cold storage. If the beef loses 100 lb. per week in weight, the fixed charges are $60 per week, and the price advances 1$ per pound per week, how long should the meat be held, and what is the selling price ? 19. There are 20,000 bu. of wheat worth 85$ per bushel in a certain elevator. If the shrinkage from all causes is 80 bu. per week, the fixed costs are $100 per week, and the price is advancing 1$ per bushel per week, how. long should the wheat be held, and what is the selling price per bushel ? (Do not count interest on the money.) 20. The distance between the centers of two spheres of radius a and 6 respectively is c. Find from what point P on the line of centers AB the greatest amount of spherical surface is visible. (The superficial area of a zone of height A is 2 irak, where a is the radius of the sphere.) 21. The cost of fuel for running a train is proportional to the square of the speed generated in miles per hour and costs $16 per hour at 16 mi. per hour. What is the most economical speed, fixed charges $100 per hour ? 22. If the cost of running a steamboat is proportional to the cube of the speed generated, what is the most economical speed to run it against a 4 mi. per hour current ? 23. If the cost of fuel for running a steamboat is proportional to the square of the speed generated, and is $4 per hour when going 6 mi. per hour, and the fixed charges are $16 per hour, what is the most economical speed against a 3^ mi. per hour current ? 24. What is the most economical speed when the boat in example 23 is going with the current ? Compare the net progress per hour in the two cases. 25. A railway is to be built from A to B. A straight line L divides the plain region in which A is located from the hill region in which B is located. If A and B are each 12 mi. from Hand E, the nearest points on L, and the distance BE is 25 mi. (along X), locate a point Q where the tracks should cross L, assuming the relative cost per mile on the plain and through the hills to be in the ratio 3 : 4. What is the length of the track and the change in direction at Q ? 40 PROBLEMS IN THE CALCULUS 26. Given a sphere of radius 6 in., calculate the following (assume inscribed solids unless otherwise stated) : (a) Dimensions of maximum parallelopiped with square base. (b) If one edge of the parallelopiped is 4 in., what are the other two for maximum volume ? (c) Dimensions of parallelopiped with square base and maximum surface. (d) Altitude of maximum right cone. (e) Altitude of right cone of maximum convex surface. ( f ) Altitude of right cone of maximum entire surface. (g) Altitude of maximum right cone with vertex at center of sphere. (h) Altitude of right cone with vertex at center and maximum total surface. ( i ) Altitude of circumscribed right cone of minimum convex surface. ( j ) Altitude of circumscribed right cone of minimum total surface. (k) Altitude and edge of base of right pyramid with square base and maximum volume. ( 1 ) Altitude of square-based pyramid of maximum lateral surface. (m) Altitude of square-based pyramid of maximum total surface. (n) Altitude of maximum tetrahedon with equilateral base. (o) Altitude of circumscribed pyramid with square base and minimum volume. (p) Altitude of circumscribed tetrahedron with equilateral base and minimum volume. 27. A wall, with its top 8 ft. above the level of the water, surrounds an ice pond, and the ice house stands back 27 ft. horizontally from the wall. A regular incline of minimum length runs from the water to an opening in the side of the ice house, just resting on the top of the wall. "What is the length of the incline, and how high is the opening above the level of the water ? (Use the inclination as a parameter.) 28. A beam' is to be carried into a mine where it is necessary to turn from a corridor of uniform width 8 ft. into one of width 6 ft. at right angles to it. If the beam is kept horizontal, what is its maximum length, disregarding thickness ? 29. A mural painting 24 ft. high has its lower edge 8 ft. above the level of the observer's eye. How far away from the wall should he stand so as to get the most favorable view of the entire picture ? 30. In a park there is a circle 40 ft. in radius, surrounded by a flat border of flowers, and an electric light is immediately above the center. How high above the ground should this light be placed so that there may be a maximum illumination of the inner flowers ? Assume the MAXIMA AND MINIMA 41 ordinary laws that the intensity of light is proportional to the sine of the angle of incidence and inversely proportional to the square of the distance from the source of light. 31. Given a constant angle a, with its vertex at A. A fixed point M lies within the angle a, and a line drawn through this point cuts the sides of the angle at B and G respectively. Show that the area of the triangle is a minimum when M bisects BC. 82. A circular sector has a given perimeter. Prove that its area is a maximum when the angle of the sector is 2 radians, and that the maxi- mum area equals the square of the radius. 83. The range, height, and time of flight of a projectile are given by R = (v 2 sin 2 ff)/g, h = (u 2 sin 2 8)/2 g, and t = (2 v sin 8)/g, where v is a constant, the initial velocity, and g is the gravitational con- stant. Find the angle of projection 8 which will make each of these in turn a maximum. 34. If a projectile is fired from so as to strike an inclined plane which makes a constant angle a with the horizontal at 0, the range is given by the formula B = [2u 2 cos 9 sin (9 — a)]/g cos 2 a, where v and g are constants as in example 33 and 6 is the angle of elevation. Calculate the value of 8, giving the maximum range up the plane. 35. When a load is being pulled up an inclined plane of constant inclination a, by a force making an angle 8 with the horizontal, the efficiency of this device as a machine is given by the formula E = [cos(a + p— 0)sina]/[sin(ar + /3)cos(a — /3)], where /3, the angle of friction, is also a constant. At what angle must the force be applied for maximum efficiency ? 36. For a square-threaded screw with pitch 8 and angle of friction the efficiency is given by the formula E = tan(?/[tan(0 + ) +/], where / is a constant. Find the value of 8 for maximum efficiency when is a known constant angle. 37. A given weight W is to be raised by the aid of a lever weighing n pounds per linear foot. The fulcrum is at one end of the lever, and the weight W'ls suspended at a fixed distance a from that end. Find the length of the lever so that the power required to raise the weight may be a minimum. 38. The formula for the efficiency of a screw in mechanics is E=h(l- h/i)/(h + m), where p is the constant .coefficient of friction and h is the tangent of the angle of pitch of the screw. For what value of h is the efficiency a maximum ? 42 PROBLEMS IN THE CALCULUS 89. The velocity of a wave of length X in deep water is given by the formula p r = V\/a+ a/X, where a is a known constant. What is the length of the wave for minimum velocity? 40. Assuming that the formula for y, the total waste due to heat, depre- ciation, etc., which occurs in an electric conductor, is y = C 2 R + 17 2 /R, where R is the resistance in ohms per mile, and the current in amperes, find the relation between C and R for minimum waste if the current is kept constant. (17 is an arbitrary value for k.) 41. Experiments on the explosion of mixtures at the pressure of one atmosphere lead to the empirical formula p = 83 — 3.2 x, where p is the highest pressure produced in the explosion and x is the volume of air, together with products of previous explosions, added to 1 cu. ft. of coal gas before explosion. Now if the work done by a gas engine is roughly proportional to px, what value of x will make this a maximum ? 42. The velocity equations for chemical reactions in which the normal course is disturbed by autocatalysis are, for reactions of the first order, dx/dt = kx(a — x) and dx/dt = k (6 + x) (a — x), where k, a, and 6 are constants. Find in each case the values of x which give maximum velocities. 43. An electric current flows about a coil of radius r and exerts a force F on a small magnet the axis of which is on a line drawn through the center of the coil and perpendicular to its plane. This force F = x/{r 2 + x 2 )?, where x is the distance to the magnet from the center of the coil. Show that F is a maximum for x = r/2. 44. If VC denotes the input of a continuous current dynamo, S the constant losses due to friction, iron, etc., C the current, V the voltage, and rC 2 the variable losses, the efficiency formula is E = 1 - d/(VC) - (tC 2 )/(FC). Show that if the voltage is constant, the efficiency is a maximum when 5 = tC 2 . 45. The power P transmitted by a belt is given by the formula P = V(T — WV 2 /g), where T is the linear velocity of the belt in meters per second, W is the weight per linear meter of the belt in kilograms, g = 9.81 being in meters per second 2 , and T is the original tension in the belt while at rest. For what value of V is P a maximum ? 46. Given an inclined plane of variable inclination a and angle of friction 0. If a load is to be elevated by means of this plane and a con- stant horizontal force, find a for maximum efficiency E, assuming the formula E = (tan a) /tan (a + ). MAXIMA AND MINIMA 43 47. Given the law for the deflection a in a tangent galvanometer, namely, tan a is proportional to the current, or tan cr,/tan a 2 = Ci/C 2 . Show that a r — a 2 is a maximum when a\ and a s are complementary. 48. The intensity J of an alternating current is given by the formula J = J sin (kt— is the phase angle. In a coil with a coefficient of self-induction L, the induced counter-electromotive force equals L dJ/dt. At what times is this induced E.M.I?, a maximum? (An alternative form is / = J sin (2 irnt — ), where n is the frequency.) 49. Given a square A BCD of tin with each side equal to 6V2ft. A pyramid with square base is to be made as follows : Mark out a square EFGH concentric with the given square, but with EF parallel to the diagonal DB. Then cut out triangles AFB, BGC, CHD, and DEA, and form a pyramid by turning up the 4 equal triangles which are left at- tached to the square EFGH. What is the side of the. base for maximum volume of the pyramid ? 60. A square sheet of cardboard of side 8 in. is to be used in making a tray, with bottom 4 in. square and uniformly sloping sides, by cutting wedges from the corners of the cardboard and turning up the trapezoidal pieces attached to the 4-inch square center. Determine the side of the top of the tray if its volume is a maximum. 51. Given an isosceles triangle. Inscribe in it the segment of a parab- ola with its axis bisecting the vertex angle of the triangle such that the area cut from the parabola by the base of the triangle may be a maximum. (Hint. Take triangle (0, 0), (a, 6), (a, - b).) 52. A rope with a ring in one end is looped over a beam and pulled taut by a heavy weight fastened to the other end and hanging ver- tically. If the rope slips freely everywhere, and the lower side of the beam is flat, what angle will the rope make with the beam when stretched ? 53. A quadrilateral has sides of given length. Prove that for maximum area the quadrilateral can be inscribed in a circle. (Hint. Use two opposite angles as parameters ; (1) express the area in terms of them, and (2) get the added relation connecting the parameters from the length of the opposite diagonal.) 54. Given a fixed angle BCA. On CA two points D and E are taken, CD = a, and GE = b. A point F is taken on CB such that the angle EFD is a maximum. How long is CF?.(Hini. Call EFD = 6. Then = CFE- CFD, and tan CFD = a sin C/(x - a cos C), tan CFE = 6 sin C /(x - b cos O).) 44 PROBLEMS IN THE CALCULUS 55. Given a point A at a horizontal distance I from another point B. A is to he illuminated by a light L of intensity I, which can move in a groove BC. The angle ABC, or a:, is obtuse and lies in a vertical plane. How far up from B must L be located for maximum illumination ? Assume the laws that the illumination at a point is directly proportional to the intensity of the source, and to the sine of the angle of incidence, and inversely proportional to the square of the distance from the source. (Hint. Call AL = r, and angle LAB = 8, and use them as parameters, from which the relation r= %lcosd follows for maximum.) (a) Evaluate the result for I = 80 ft., a = 135°. 56. The cost of running a boat so as to drive it x miles per hour relative to the water is given by the formula G = a + bx s . It is observed that C is $44 when x = 4 mi. per hour, and $53 \ when x = 6 mi. per hour. Calculate the proper speed for most economical operation against a 2-mile current. What is the proper speed when moving with the same current? 57. A battery is to be composed of n cells arranged x in series and y in parallel. The E.M.F. of each cell is e and its resistance r, while the resistance of the external circuit is B. Assuming Ohm's law and the law of divided circuits, show that the current is a maximum when the battery can be so arranged that its internal resistance equals R. (Ohm's law is, current = electromotive force -=- resistance. For a divided circuit \/r = 1/r-! + l/r a + • ■ ■ l/r„, where rv is the resistance in the separate branches.) 58. Find the path of a ray of light from a point A in one medium to a point B in another medium, such that a minimum time will be required for light to pass from A to B, the velocity. of light in the two media being V 1 and F 2 . It is assumed that the required path is in a plane through A and B perpendicular to the plane separating the media. 59. Through a point on the axis of a parabola a chord is drawn which has minimum length. Show that it is perpendicular to the axis. 60. Two sources of heat 8 t and S 2 are at a distance a from each other. The intensities are I 1 and 7 2 . If the heat at a point is directly proportional to the intensity of the source and inversely proportional to the square of the distance from the source, locate the point on the line S^S^ at which the heat is a minimum. CHAPTER V DIFFERENTIALS AND RATES EXERCISE XXXVI Differential of Arc and Approximate Length of Arc. Rectangular Coordinates. The differential of arc ds is given by either (1) ds = -\/l + (dy/dxy ■ dx or (2) ds = Vl + (d*/'(z)dz/dx, d*y/dx 2 = " (z)(dz/dxf + ' (z)d 2 z/dx*. Formulas for higher derivatives are easily calculated. , In the- following examples change the dependent variable : 2dx 2 4y 2 \dxJ 2 dx 10. ^ + _?_ (^ + 2^ + 2^-2 = 0^ = ^8. dx 2 1 — y 2 \dx/ dx 11. (1 + yfp- _ 2y (^Y- 2(1 + y 2 )^- =0, y = tanz. da; 2 \dx/ dx y *!(1 /d2/\ 1 - 2/ 2 ' da; 2 y 2 \ 13. y8^_ 6y S^ + lly^_6 = 0,V = ««. dx 8 dx 2 dx U.p* s + yp> 2 -y 2 f-10ys = 0,y = l/z. dx 3 dx i dx (C) Change of the Independent Variable. Here we introduce a new independent variable, the defining relation being given. The procedure here is obvious, as dy/dx = dy/dt ■ dt/dx. „ „ d 2 u du . . 15. x 2 — - + x m = 0, y = logx. dx 2 dx 16. y + if + *V = , * 2 = 4t. xdx dx 2 d 2 w du 17. /j_y2\ y 1- 4m = 0, x = arooosy. dy 2 dy 18. (x - z») ^ + (1 - 3 x 2 ) ^ - x?/ = 0, x = Vl - ( 2 . dx 2 dx 19 *^ + _^L_^ + _»_ = o,z = ti U i S . dx 2 1 + x 2 dx (1 + x 2 ) 2 20 d 2 y + l_dy + y = 0x2 = it dx 2 x dx 21. (l + x 2 )^-x-^-i/ = 0, x = sinz. dx 2 dx 50 PROBLEMS IN THE CALCULUS 22. (1 - x 2 ) ^ - x ^ - 2x = 0, x = sine. dx 2 dx 23. (1 + x 2 ) ^ + 2a; — + 4 = 0, x = ta.nl. dx 2 dx 24 .x^ + xa^ + 3^ = 0,x = i. dx 2 dx t 1-x 2 d 2 y 1 dy S6 . (1 _,,(g)-_ 2l(1 _ I , ) g.| + g )+1= „, x = sin i. dx 2 1 + x 2 dx 1 + x 2 28. Vr^^-!/^ = 0, x = sinz. dx 3 dx EXERCISE XL Rate Problems. In most of these problems time, denoted by t, is the independent variable. Hence if x is any quantity which changes with the time, dx/dt is the time rate of change of x. In general, dy/dx may be defined as a rate — the relative rate of change of y with respect to x. If several dependent quanti- ties are changing simultaneously, it is necessary first to get an equation (or equations) connecting them. Then differentiating with respect to t gives a relation (or relations) connecting the several time rates. Substitute in this (or these) the particular values known at the time under consideration. Solve the result- ing equation (or equations) for the unknown rate or rates. Do not substitute a constant value for any quantity which varies with the time until after differentiation. 1. A stone is dropped into a calm lake, sending out a series of concen- Mic ripples. If the radius of the outer ripple increases uniformly 6 ft./sec, how rapidly is the disturbed area increasing at the end of 2 sec. ? 2. A rope ABC, 28 ft. long, runs over a pulley B, which is 12 ft. above the level track on which the ends A and C are moving. If the rope is taut, and the end G moves to the right 13 ft./sec, how fast will A move when C is 5 ft. from E, a point on A directly below B ? DIFFERENTIALS AND RATES 51 v- 3. A right circular cone keeps its volume constant. If the radius of the base is increasing at the rate of 2 in. /sec, how fast is the altitude changing when the altitude is 6 in. and the radius 4 in. ? | I ^ U*** 4. A right prism has constant altitude, 4 ft. If its base is an equi- lateral triangle of side 2 ft., and the sides are increasing 1 ft. per hour, how fast is the volume increasing ? How fast the surface ? 5. A boat is fastened to a rope which is wound about a windlass 24 ft. above the level at which the rope is attached to the boat. The boat is drifting away at the rate of 10 f t./sec. How fast is it unwinding the rope when 32 ft. from the point directly under the windlass ? What is the acceleration of a point on the rope ? 6. A bundle is being elevated by a rope passing over a pulley 25 ft. above the ground. A man takes hold of a ring in the extreme end of the rope, which is 50 ft. long, and walks away 10 f t./sec. How rapidly will the bundle ascend at the start ? The man's hands are assumed to be 5 ft. above the ground, w 7. A cistern is in the shape of an inverted circular cone, with its diameter equal to its height, each being 10 ft. How fast is the water pouring in when it is 5 ft. deep and rising 4 in. /min. ?/ \s 8. There is a cistern in the shape of a cone 12 ft. deep and 6 ft. in diameter at the top, into which 10 cu. ft. of water are pouring each minute. If at the time the water is 8 ft. deep it is observed to be rising only 6 in./min., how much water is leaking away ? V 9. Sand is being poured on the ground, forming a conical pile with its altitude equal to § of the radius of its base. If the sand is falling at the rate of 12 cu. ft./sec, how fast is the altitude increasing at the time * it is 5 ft. ? 10. A balloon is in the shape of an ellipsoidof revolution, its longer , axis being twice the two shorter ones. If gas is being poured in at the rate of 100 cu. ft./min., when the balloon is 10 ft. long and expanding uniformly how fast is the length increasing ? v 11. A right prism has an equilateral triangle for a base. How rapidly are its volume and surface changing when the side of the base is 3 in. and increasing 2 in./min., while the altitude is 5 in. and decreasing 3 in./min. ? V 12. A right circular cylinder is capped by a hemispherical top. If the radius of the common base increases at the rate of 2 in. /sec. and the altitude of the cylinder 3 in. /sec, at what rate is the volume increasing when the radius is 5 and the altitude 10 ? 52 PROBLEMS IN THE CALCULUS J" 13. A right prism has square bases. If the sides of the bases are 2 in. and increasing 1 in. /sec, while the altitude is 4 in. and decreasing 2 in./sec, (a) how fast is the volume increasing ? (b) When will the volume cease increasing? 14. Two ships are at the same point. The first leaves at 10 a.m., .sailing east at the rate of 9 mi./hr. The second starts at 11a.m. and sails south 12 mi./hr. How fast are they separating at noon ? v 15. A man is walking 6 ft./sec. across a bridge 60 ft. above the water level below. A man in a rowboat is rowing at right angles to the bridge 4 ft./sec. How rapidly are the men separating at the end of 5 sec. ? 16. A rod 10 ft. long moves so that its ends A and B remain constantly in the as- and the y- axis respectively. If A is 8 ft. from the origin and is moving away at the rate of 2 ft./sec, (a) at what rate is the end B coming down ? (b) at what rate is the area formed by AB and the axes changing? (c) if P is the middle point of AB, at what rate is OP changing? 17. The side in each of the following regular polygons is 12 in. and expanding 4 in. per hour. How fast is the area increasing for (a) a square ? (b) a triangle ? (c) a hexagon ? (d) an n-sided polygon ? •/18. The hour hand of a clock is 3 ft. long and the minute hand is 4 ft. long. At what rate are the ends of the hands approaching each other at (a) 3 o'clock ? (b) 4 o'clock ? * 19. An angle 8 is increasing uniformly. Show that (a) when 8 = the sine and tangent are increasing at the same rate ; (b) when 8 = ?r/6 the cosine is decreasing § as fast as the tangent is increasing ; (c) when 8 = 7r/4 the tangent is increasing V2 times as fast as the secant, while the sine is increasing at the same rate the cosine is decreasing ; (d) when 8 = ir/3 the tangent is increasing 8 times as fast as the sine, while the secant is increasing 4 times as fast as the cosine i£ decreasing. V 20. A man is crossing at right angles a street/60 ft. wide. There is a ' lamp on the opposite side, which is 10 ft. upstreet from the point which he is approaching. The lamp is on a level with the man's head, and casts his shadow against the buildings on the side he is leaving. If the man walks 10 ft./sec, what is the rate of his shadow on the buildings when he is | the way across ? 21. The adiabatic law for the expansion of air is PV 1A = C. If at a given time the volume V is observed to be 10 cu. ft. and the pressure is 50 lb. per square inch, at what rate is the pressure changing if the volume is decreasing 1 cu. ft. per second ? DIFFERENTIALS AND RATES 53 22. A kite is at an altitude of 160 ft. and there are 280 ft. of cord out. If the kite is moving horizontally 4 mi./hr. directly away from the person flying it, how fast is the cord being paid out ? \y 23. Two sides of a triangle increase at the rate of 2 in. per second and the included angle at the rate of J radian per second. At what rate is the area changing at the instant when each of the three sides is 10 in. long ? 24. A revolving light is ^mi. from a straight beach. A ray from the light illumines a spot on the beach. If the light makes a complete revo- lution per minute, how fast is the spot moving along the beach when the ray makes a 45° angle with the shore line ? How fast at the end of 10 sec. from the time it is directed toward the nearest point on the shore ? 25. I have a variable right triangle ABC, with B, the vertex of the right angle, moving along the axis of the parabola j/ 2 = 4x— 4, being fixed at the vertex of the parabola, and A tracing out the upper half of the parabola. If B starts at the vertex and recedes uniformly 2 in. /sec, how fast is A receding from the origin of coordinates at the end of 2 sec. ? How fast is the area of the triangle changing ? 26. The strength of a beam is given by the formula S = kx 3 y, where k is a constant of the material, y is the width and x the thickness of the beam. If at a given time x = 12in. and decreases 1 in./min., (a) how fast should y, which is 4 in., be increasing if the strength is to remain constant ? (b) How fast is the area of the cross section changing ? 27. A wheel of radius 5 ft. is turned by a connecting shaft 13 ft. long, fastened to a piston. If the wheel turns uniformly at the rate of 100 rev./min., (a) at what rate is the piston moving when the connection on the rim of the wheel is at its highest and lowest points ? (b) At what position has the piston maximum velocity ? (c) What is the angular velocity of the wheel at each of these times ? 28. If in Ex. 27 the radius of the wheel is 4 ft. and the connecting rod is 6 ft., what is the rate of the piston when the line joining the connection to the center O of the wheel makes a 60° angle with OP, P being the center of the piston head ? At what points does the velocity of the piston become zero ? 29. A grade crossing has a double gate, the two arms of which rotate upward about the same axis. The arm over the driveway is 12 ft. while that over the sidewalk is but 5 ft. Each arm rotates at the rate of 4 radians per minute. At what rate is the distance between their extrem- ities changing (a) when each makes a 45° angle with the horizontal ? (b) when each makes a 60° angle ? 54 PROBLEMS IN THE CALCULUS 30. A man is walking over a bridge at the rate of 3 mi. per hour, and a hoat passes under the bridge at right angles, going 7 mi. per hour. The bridge is 18 ft. above the boat. How fast are the boat and the man separating (a) 1 min. later ? (b) 4 min. later ? 31. If the x intercept of the tangent to the positive branch of the hy- perbola xy = 4 is increasing 3 units per second, find the rate and accel- eration with which the y intercept is changing at the end of 5 sec, the x intercept starting from the origin. 32. If the x intercept of the tangent to the curve y = e~ x is increasing constantly 5 units per second, find the velocity and acceleration of the y intercept at the instant when the x intercept is 12 units. 33. Given an inclined plane BC, with a 30° angle of elevation. At B, the top of the plane, is a perpendicular post BA, 10 ft. long, with a pulley at the top A . A mass M v attached to a rope, starts at B and is moved down the plane with a uniform velocity of 10 f t./sec. The rope passes over the pulley A and supports another mass M 2 . How fast is M 2 going up at the end of 1 sec. ? at the end of 2 sec. ? 34. A reservoir is in the shape of a frustum of a right circular cone with lower base of radius 10 ft., upper base 15 ft., and depth 10 ft. How fast are the volume of water and the area of the surface of the water increasing when the water is 5 ft. deep and rising 1 ft. per hour ? 35. A right triangular lot ABG has AB = 5 rd. and BC = 12 rd. There is a street lamp at A . A man walks at the uniform rate of 13 ft. per second from B to E, the mid-point of the hypotenuse A G. (a) How fast is his shadow moving when it is halfway from B to C ? (b) How fast when the man is halfway from B to E? (c) when he reaches E ? (d) What is the acceleration of the shadow when it is halfway from B to C? 36. A boy is whirling a stone on the end of a string with an angular velocity of 4 radians per second and is allowing the string to slip through his hand at the rate of 5 f t./sec. At the time the stone is 3 ft. from his hand, at what rate is it moving ? What is the component of its velocity at right angles to the string ? 37. A moving point traces out each of the following curves. Find the rate at which the arc is increasing in each case under the given conditions : (a) y 2 = 2x + 5, x = 2, dx/dt = 3. (b) 3x 2 + 4y 2 - 12 = 0, y = §, dy/dt = 1. (c) 2 s 2 - 3^/2 - 2 = 0, y = 4, dx/dt = 4. (d) y = x 2 - 2x + 3, x = 2, dx/dt = 4. (e) y = x 8 - x 2 , x = 2, dy/dt = 3. DIFFERENTIALS AND RATES 55 38. A point P moves on the curve y = 2 x s at the uniform rate of 4 ft. per second. At what rate is the inclination of the tangent to the curve at P changing as P passes through the point (1, 2) ? 39. A military observer in an aeroplane is ascending at the rate of 12 mi. per hour. How fast is the visible area of the earth's surface increasing in square miles per minute at the end of 5 min. ? at the end of 10 min. ? 40. The ends of a taut cord are A and B. The extremity A moves along the line y = 3, with the velocity of 2 units per second. The extremity B is attached to a bead which is free to move on the negative x-axis. A portion of the cord between A and B passes over the circle x 2 + y % = 9. rind the velocity of B when the bead passes through (— 5, 0) . 41. A candle is moving directly away from the center of a sphere of radius 5 ft. at the rate of 2 ft. per second. At what rate is the surface illuminated by the candle increasing when the candle is 10 ft. from the center of the sphere ? 42. A cylinder with an open top, a height of 4ft., and a base of radius 1 ft. is filled with water. It is then tipped around a tangent to the base at a uniform rate of 10° per second. At what rate is the water overflowing at the end of 3 sec, assuming that the surface of the water is always horizontal ? 43. One end A of a rod AB slides on the x-axis while the rod passes over and rests on the rim of a wheel of radius 5 in., with center at the origin O. If AB is 26 in. long and A moves 6 in. per second, at what rate is B approaching O when OA is 13 in. ? What is the velocity of B at this time ? 44. If the radius of a sphere is 12 in. and is increasing at the rate of T V in. per second, at what rate is the volume of the maximum inscribed right cylinder increasing ? 45. A horizontal cylindrical tank turns on its axis and empties through a slit cut along an element. The level of the fluid in the cylinder remains the same as the level of the slit. If the slit revolves uniformly 10° per second, at what rate is the tank emptying when the slit has turned through 30° from its extreme vertical position ? In many problems the relative rate of increase or the per- centage rate of increase is the important idea. The relative and percentage rates should be calculated for a number of the preceding examples. If dy/dt is the time rate of increase with respect to t, dy/dt -=- y is the relative rate of increase of y, and 56 PROBLEMS IN THE CALCULUS 100 times this is the percentage rate of increase. Note that the relative rate is therefore simply the logarithmic derivative of y with respect to t. Note also generally that dy/dx is an absolute rate of change in y with respect to x, while dy/dx -=- y is the relative rate. 46. If the length and breadth of a rectangle each increase -J v % per degree as a metal plate is heated, show that the area increases \% per degree. 47. A circular hill has a central vertical section of the form x 2 + 160 y — 1600 where the unit is 1 yd. If the top is being cut down in horizontal layers, at what relative rate per vertical yard is the area of the section increasing after 4 yd. have been removed ? 48. Assume the law pv 1A = k for air under pressure. If the volume is changing, what is the relative change in pressure per unit increase in volume when the volume is 28 cubic units? What is the percentage change ? 49. The deflection of a uniformly loaded beam supported at one end is given by y = h{\ Px 2 — J lx s + j^e 4 ), where I is the total length of beam and x the distance from the point of support. "What is the relative change of the deflection at the middle of the beam per unit increase in x ? at the free end ? 50. Van der Waals' equation for a gas is (p + a/v 2 ) (v — b) = k, where p is the pressure, v is the volume, and a, b, and Tc are constants. What is the change in volume per unit increase in pressure ? What is the relative change ? CHAPTER VI INDETERMINATE FORMS EXERCISE XLI Indeterminate Forms. -=- 0. If a fraction f(x)/F(x), where numerator and denominator are both functions of x, assumes the indeterminate form 0/0 for a particular value of x, its limiting value may be found by differentiating f(x) for a new numerator, and F(x) for a new denominator, giving f'(x)/F'(x). If this form is still indeterminate, repeat the process. In the case of trigonometric functions it is frequently desirable or necessary to change the form of the functions before dif- ferentiating. Note that you do not differentiate the fraction. Evaluate each of the following for the value of x given. (Assure yourself first that the forms are indeterminate.) „ xe 8 * — x „ „e a: — 1— x . „ 6x 2 — 5x + l 1 -,x = 0. 2. , x = 0. 3. — ,x = - ■ oos 2 x oos 2 x — 1 8 x 2 — 2 x — 1 2 - 2 x 2 — x + 2 a; = 1,2. 10. ,x = 0. X s — 7 x + 6 e x + e- x — x 2 — 2 s 3 + 8 n lQgx-siD(x-l) x = L 2 _Vx^3 12 sin4x+ e - 2 *-2x 2 -2x-l X 2_ 49 ' x = 7 - " x4 + 12x3 . „ , „ (2 — x 2 ) sin x — 2 x cos x . e* — sm2x— e-* „ 13. , x = 0. x = 0. x — an x ,. V3x-Vl2-x 8 - ^-g ,x = 0. 2x-3V19-5x cosx — xsinx jg V16x — x — 2V4x ^ _ p_ 2 — 2 oos x — sin 2 x 2 — \/2 x 8 57 58 PROBLEMS IN THE CALCULUS Find the slopes of the folio-wing curves at the points indicated : 16. a 2 y 2 - a 2 x 2 - x* = 0, at (0, 0). 18. x 3 - 3 axy + y 2 = 0, at (0, 0). 17. y 3 = ax 2 - x 8 , at (0, 0). 19. x 4 - a 2 xy + b 2 y"- = 0, at (0, 0). 20. (y 2 + x 2 ) 2 - 6 axy 2 - 2 ax 3 + a 2 x 2 = 0, at (0, 0) and (a, 0). 21. Given a circle with center Oand horizontal radius OA. AP is the tangent at A. A variable point Jf starts at .4 and moves about on the circle ; AP is laid off on the tangent = AM, the arc ; and PM, a straight line, is drawn cutting AO (extended) at B. If we call the variable angle A OM = 6, what is the limiting value of OB as 6 approaches zero ? EXERCISE XLII Indeterminate Forms, oo ~- oo. In ease a certain value of x makes both the numerator and the denominator infinite, the value of the fraction becomes indeterminate also. The limiting value is obtained in precisely the same manner as in the preced- ing exercise. The student should recall that it is sometimes necessary to change the form of the function before evaluating, as otherwise the value continues indeterminate indefinitely. . log sin 2 x . . cot2x . „ ir/x 1. — 2 ,x = 0. 5. ,x = 0. 9. ,x = 0. log sin x cot x cot ro/2 , x = 0. 6. — , x = oo. 10. j x = 0. cot 3 x x 3 cot x . x 6 logx log(l — x) 3. — ,x = oo. 7. — ^,x = oo. 11. — 2-i -,x = . e x of cot 7rx tanx log tan x secwx 1 4. , x = 7r/2. 8. — 2 ,x=7r/2. 12. ,x = -- tan 3 x log tan 2 x tan 3 7rx 2 , „ x 4 + x 3 . . log 1/x 13. — - — , x = oo. 16. 6 ' , x = 0. e* + 1 cotx , . log(a + x) x + logx 14. —2-^ J , x = oo. 17. — , x = oo. x xlogx seom;/2 log (x -l) + tan to/2 10. — > X — 1. IB. , x = 1. log sec to/2 cot to 15. — -,x-0. sin 8 x x" INDETERMINATE FORMS 59 EXERCISE XLIII Indeterminate Forms. • oo, oo — oo. If a product contains one factor equal to and another equal to infinity for a certain value of the independent variable, the value is again indeter- minate. But in this case the form can be rewritten so as to assume the form -^ or oo -s- oo, since • oo = -s- l/oo = -5- 0. Similarly, an indeterminate form of the type oo — oo can be changed into the form -s- or oo -s- oo and then evaluated. Evaluate : 1. xlogsinx, x = 0. ,„ x 2 — 4. ttx & ' 12. tan — ,x = 2. 2. sinx(logx), x = 0. x 4 x — 1 1 3. esc x sin (tan x), x = 0. 13. , x = 0. v " 2x 2 x(e 2 *-l) 4. (2 — x) tan 7rx/4, x = 2. i i 14. , x = 0. 5. (1 — x) tan irx/2, x = 1. log(l + x) x 6. (1 — sinx) tan x, x = tt/2. jq _^ 1 7. 2/(x 2 -l)-l/(x-l),x=l. ^ 8. x/(x - 1) - 1/logx, x = 1. 16 - 13^ - 5-x 2 -4x 8 '' X = L 9. 1/x 2 - l/(x tan x), x = 0. 17. 2* tan w/2* x = co. 10. xtanx— |7rsecx, x = ?r/2. 18. xlog(l + 1/x), x = oo. 1 1 19. secx— 1/(1— sinx), x = 7r/2. 11. -135=0. sm 2 x x 2 20. sinx (log x) 2 , x = 0. EXERCISE XLIV Other Indeterminate Forms. Certain values of the independent variable may cause a given function to assume one of the follow- ing indeterminate forms, 0°, 1", co°. These are all evaluated by taking the logarithm of both sides and then evaluating the right-hand member, which will be of the form • 00 for the value of x which made the original function indeterminate. Evaluating this gives the value of the logarithm of the function. (l-f-l/x) x , as x = 00, is a well-known illustration. Evaluate : 1. y = (sinx)™*, x = ir/2. 3. f(x) = (1 + 2 x)i/*, x = 0. 2. y = (sinx) 1 * 11 *, x = ir/2. 4. f(x) = (l/x) sin * x = 0. 60 PROBLEMS IN THE CALCULUS 5. /(x) = (1 - SB)*""™, X = 1. 12. /(x) = (X + 1)001* x = 0. 6. /(X) = (2 - Xjitanira, X = 1. 13. /(x) = (1 + 3x)< 1 - 2a: )/ a: , X = 0. 7. /(x) = [(tanx)/x]i^, x = 0. 14. /(x) = (1 - x/2)h^ »■•>', x = 2. 8. /(x) = (l/x)<»n* x = 0. 15. 2/ = (1 + x)l°e*, x = 0. 9. /(x) = (cos2/x)* x = oo. 16. y = xW-*>, x = 1. 10. /(x) = (cos2/x)^, x = oo. 17. y = {&* + 2x) 1 ^% x = 0. 11. /(x) = (cos2/x)* s , x = oo. 18. y = (logxpa-iog*^ x _ e . CHAPTER VII CURVATURE. EVOLUTES EXERCISE XLV Curvature and Radius of Curvature. If II = radius of curvature and K = curvature, the formulas to be used are (1) R = [1 + (dy/dxy\™/(d*y/dx>), (2) R = [1 + {dx/dyfY>*/{ b. 33. y = ax 2 + 6x + c. 34. x 1 ' 2 + yW = 1. Also find R at (1, 0). 35. y = x 2 — 4 x + 2. What is the nature of the point where the curvature is a maximum 1 36. Find the locus of points at which the curvature of the family of curves y = x n , x > 0, n > is a maximum. EXERCISE XLVI Center of Curvature and Evolutes (Cartesian). In the preceding exercise we constructed the circle of curvature at a given point on a curve. We now have general expressions for the coordi- nates of the center of this circle, which is called the center of curvature. The locus of centers of curvature is called the evolute. The formulas for the coordinates, calling them a and b (instead of ,x and y), are (A) a = x - dy/dx[l + {dy/dxy\/{cPy/dx% b = y+[l+ (dy/dxf]/((Py/dx*), where x and y are the coordinates of the point on the curve. The student is reminded that he should calculate the numerical CURVATURE. EVOLUTES 63 values of dy/dx and cPy/dx 2 for the point, before substituting in (A). Observation of the rule to evaluate numerically as soon as possible will save much labor in many places. In cer- tain cases, (a) where dy/dx is infinite and (b) where x is given explicitly in terms of y, it is more convenient to use the alternate forms (B) a = x + [1 + (dx/dyf] /{d*x/dy% b = y- dx/dy[l + (dx/dyf]/((Px/dy*). Calculate a and b for the following at the points indicated, and check results by showing that (a, b) is a point on the normal to the curve at the given point (x, y). Also R must equal the distance from (a, b) to (x, y). 1. j/=x 2 -6x + 10, at (3, 1). 10. j/ 2 -4x-42/ + 4 = 0,at(0,2). 2. 3y = x 8 -3x 2 -9x, at (3, -9). 11. y 2 -. x - 2y = 0, at (0, 2). 3. ?/ = x s -6x 2 + llx-6, at (1,0). 12. x 2 + 2y 2 - 4x = 2, at (0, 1). 4. xy = 30, at (3, 10). 13. 2x 2 - y 2 + 3x+ 1 = 0, at (0, 1). 5. y = &, at (0, 1). 14. y* = 4(x 2 - 8 x + 17), at (4, 2). 6. 3/ = sinx, at (w/2, 1). 15. ?/ 3 = 8x 2 - x 3 , at (4, 4). 7. y = X s - 6x + 4, at (2, 0). 16. x 2 + 2xy + 2y 2 = 10, at (2, 1). 8. y = (x 2 + 4)/x, at (2, 4). 17. x 3 + xy 2 - 6j/ 2 = 0, at (3, 3). 9. y = x 2 + 8/x, at (2, 8). 18. x 3 + y s - 4 xy = 0, at (2, 2). EXERCISE XLVII Evolutes. Parametric Equations. When the equation of a curve is given parametrically, the formulas for a and b give expressions in t. These equations, a =f(t) and b = <£ (*), are the parametric equations of the evolute of the original curve. If a value t 1 is substituted in x and y, and the same value t x in a and J, it gives us a point (x, y) on the curve, and the cor- responding point (a, b) on the evolute ; that is, (a, b) is the center of curvature. It should be noted that the normal to the curve at (x, y) passes through (a, b), and the distance from (x, y) 64 PROBLEMS IN THE CALCULUS to (a, b) is R. Recall the caution concerning the finding of cPy/dx 1 for parametric equations. Find the parametric equation of the evolute in each of the following, and sketch the curve and at least part of the evolute : 1. x = 2t, y = 2t 2 -l. 13. x = sin t, y = t. 2. x = 2t, y = t s /S. 14. x = 13 sin t, y = 5cosi. 3. x = 3- 2t, y = I 3 — 3. 15. x = 4(t - sin/), i.x = 2t + l,y = t*/S. y = 4(1 -cost). 5. x = it, y = 2/1. 16. x = 2eos*+cos2«, y = 2 sin t + sin 2 £. 17. x = 3 cos £ + cos 3 i, y = 3 sin £ + sin 3 t. 18. x = sec 3 £, 2/ = tan 3 t. 9. x = 4t+t\y = 2t*. 19 , a = 3csct,y = 4coU. 6. x = 2t, y = 4-4£ 2 . 7. x = t i -2t,y = P. 8. x = 3< 2 -l, y = l-2t. 10. x = 2 - £, « = 9 - 4 2 . 20. x = cos 4 £, ?/ = sin 4 £. 11. x = l- t 2 , y = 5-t 21. x = 2- cost, y = t-sint. 12. x = l + 2,y = P-2l-15. 22. x = t - cos t, y = t - sin i. CHAPTER VIII PARTIAL DERIVATIVES. APPLICATIONS EXERCISE XLVIII Partial Derivatives. The essential thing here is that although the function v is a function of a number of variables x, y, t, etc., nevertheless, while taking the partial derivative with re- spect to x, du/dx, all the variables except x are treated as con- stants, — and the differentiation is then the same as usual. In the following find the partial derivative of u with respect to each of the variables on the right-hand side. In an implicit function involving three or more variables, x, y, z, etc., we can find a partial derivative, say dz/dx, by regarding variables except x and z as constants during the process of differentiation. 1. u = x 3 — 3x 2 y + 3xy 2 — y s . 7. u = xe« + ye?. 2. u = x s + y 3 + z 3 — 3xyz. 8. u = sin 2 i + sinx cosy + cos 2 !/. 3. z = 2x 3 — 5x 2 y + 3xy 2 — 8y 2 9. a = sinx logy + logx cosy. — 7xy+6x. jo. u = xy sinz + xzlogy + &=y. 4. u = {ax 2 + bxy + cy 2 ) 3 . n. u = 2 xy — ye~^ 3 - x . 5. u = logx!'. 12. u = 2 e*"y + 3 a^ z + 2x 3 y 2 6. u = ycosx + xcosy. 13. u = x 2 y 3 + x 2 sinxy + y logx. 14. e 2 cosx — cosy = 0, find dz/dx and dz/dy. 15. y*/a — z 2 /b = 2 x, find dz/dx and Sz/Sy. 16. xy + yz + zx — 1G, find gz/Sx, dy/dx, dz/dy. 17. w = x 3 — x*fy — 5 xy 2 + y 3 , prove that x gu/gx + y du/By = 3 w. 18. w = x 2 y + y 2 z + z 2 x, prove that du/dx + Bu/dy + du/dz = (x + y + z) 2 . 19. u = ew + sin (x + y), prove that xdu/dx — ydu/dy — x — y cos (x + y). 20. u = e(» 2 + *J'~ :l:!! v(!< 2 + ^) ; prove that xdu/Bx + yBu/Bx = 0. 65 66 PROBLEMS IN THE CALCULUS 21. u = cos[(x+ y)/2z], prove that xdu/dx + ydu/dy + zdu/dz = 0. 22. u = (x s — y s )/xy, show that xdu/dx + ydu/dy = u. 23. it = x 3 /y — xy + x 2 y 2 /(x 2 + y 2 ), show that xdu/dx + ydu/dy = 2 m. 24. A circular cylinder of radius x and altitude j/ is surmounted by a hemisphere of the same radius. Show that (a) if the base remains con- stant, the volume increases numerically ttx 2 times as fast as the altitude, and (b) if the altitude remains constant, the volume increases 2 irx (x + y) times as fast as the radius of the base. 25. Given the frustum of a cone with B and r as the radii of the lower and upper bases respectively, and with altitude h. Show that if h and R remain constant, the volume increases numerically ^ irh (2 B + r) times as fast as the radius of the lower base. EXERCISE XLIX Differentiation of Implicit Functions by Means of Partial De- rivatives. In Exercise XXI dy/dx for implicit functions was found. In general, however, it is easier to write down dy/dx at once from the fact that if F(x, y) = is the equation, then dy/dx = — (dF/dx)-i-(dF/dy). Apply this theorem in finding dy/dx for each of the following (clear of fractions before differentiating) . 1. 9x 2 + iy- = 36. 13. & cos x- cosy = 0. 2. X s — x 2 y + xj/ 2 + ^ = 0. 14. x coty + y sinx = 0. 3. 3x 2 + 2j/ 2 -4x-4 = 0. 15. ye* + 2xy 3 = 0. 4. x 8 + y s - 3 axy = 0. 16. x 2 - (x 2 + j/ 2 )/(x 2 - y 2 ) = 0. 5. y* - (x + y)/(x -y) = 0. 17. & - x x + v = 0. 6. j/ 2 (x-2)-(x + 2) 8 =0. 18. aw+xsin?/ = 0. 7. xi + yi = ai. 19. xe» - y + 1 = 0. 8. xf + j/f = af. 20. x log y + Vx 2 + y 2 = 0. 9. 2sin 2 x+ 3sin 2 j/ = 0. 21. tan (x 2 + y 2 ) + &? + y 2 = 0. 10. j/ sin x — x cos y = 0. 22. x" — y* = 0. 11. cos (x + y) - xy = 0. 23. y/x - arc tan x/y = 0. 12. 2x + Zy + 4e*» = 0. 24. of- v _ x* = 0. PARTIAL DERIVATIVES. APPLICATIONS 67 EXERCISE L Successive Partial Derivatives. This process requires no expla- nation except the meaning of the symbol. Thus, d~u/(dxdy) means that u is first differentiated regarding x as the only variable on the right-hand side, and then this result is differen- tiated regarding y as the only variable. The student should be convinced that the order of differentiation is immaterial. In each example find all second partial derivatives, as d 2 u/8x 2 , dhi/dy 2 , ohi/dz 2 , d 2 u/(dxdy), d 2 u/(dxdz), etc., as well as any indicated relation. 1. u = 2 x s + Zx^y — Zy 2 . Show that d 2 u/(dx&y) = d 2 u/(dydx). 2. u = eFcosy — & sin x. Show that d 2 u/(dxdy) = d 2 u/(dydx). 3. u = (ax 2 + by 2 + cz 2 ) s . Show that d 3 u/(dx 2 dy) = dH/ipxdydx) = d s u/(dydx 2 ). 4. u = sin x log y + log x cos y. 5. u = xv. Show that S 2 u/(dxdy) = d 2 u/(dydx). 6. u = a x + b'J + x 2 y 2 . Eind also d s u/(dx 2 dy). 7. u = xhj 2 — 3xy i + ix 2 y 3 . Eind also d 3 u/(dxdy 2 ). 8. u = &v + ye x + xe». 9. u — y 2 — 2 ye 1 + 2 x log y. 10. u = axSfz + bxyW + cy s + dx z z 3 . Find d 4 u/(dx 2 dydz). 11. u = 2x 3 -5xhj.+ 3xy 2 -8y 2 -7xy + 6x. Fmdd s u/dx a ,d s u/(Sx 2 dy), d s y/(dxcy 2 ). 12. u = _^L. . Show that x 2 d 2 u/dx 2 + 2xyd 2 u/(dxdy) + y 2 d 2 u/dy 2 = 0. x + y 13. u = logVx 2 + 2/ 2 . Show that 3 2 u/ax 2 + d 2 u/dy 2 = 0. 14. u = 1/Vx 2 + z/ 2 + z 2 . Show that a 2 u/Sx 2 + S 2 u/S2/ 2 + d 2 u/dz 2 = 0. 15. If y = e ax + b ' + c , show that if d^y/dx 2 = A d 2 y/dt 2 + B dy/St, then a 2 = Ab 2 + Bb. 16. If u = logr, where r 2 = (x — a) 2 + (y — 6) 2 , and x - a and y - b are not zero simultaneously, show that d 2 u/dx 2 + d 2 u/dy 2 = 0. 17. If w = 1/r, where r 2 = (x - a) 2 + (y - 6) 2 + (z - c) 2 , and (x - a), (y — 6), and (z — c) do not vanish simultaneously, prove the relation d 2 u/dx 2 + d 2 u/dy 2 + d 2 u/dz 2 = 0. 68 PROBLEMS IN THE CALCULUS 18. Euler's theorem for homogeneous functions in two variables states that "Mm =/(x, y) is homogeneous, xdu/dx + y Su/dy = nu, where n is the total degree of u." Use this criterion to test (a) (x 2 + y 2 )/(y — x), (b) ewA^ + n*), (c) sin(xy), for homogeneity. 19. Show that y =

+ •••. 14. 1 + x/(2 • 1) + x 2 /(2 2 ■ 2) + x 3 /(2 8 • 3) + x 4 /(2 4 ■ 4) + - • • . 15. 1/3 + 2 x/(2 • 3 2 ) + 3 x 2 /(2 2 • 3 8 ) + 4 x 8 /(2 3 • 3 1 ) + • • • . 16. 2x 2 /3 + 4x 3 /8 + 8x715 + 16*6/24 + • • • 2»-ix»/ra 2 - 1 ■ • ■. 17. x/3* + x/3* + x 2 /3* + x 3 /3^ + • ■ • . 18. x/(l-2) + 2x 2 /(2-2-3) + 3x 3 /(2 2 • 3 • 4) + 4xV(2 8 .4-5)+ •••. The binomial theorem gives rise to an infinite power series in x when the exponent n is not an integer. We will use this theorem in the form ,as ,. x„ - n(n — 1) „ n(n — l)(n — 2) , (1) (1 + x) n = 1 + nx + -i ix 2 + -i ii '- x 3 n(n-l)(n-2)(n-3) ri 4! Clearly any binomial expression of the form (a + bx)" can be expanded by (1), by rewriting it in the form a"(l+ bx/a)' 1 - This leads to a method of extracting higher roots without recourse to a table of logarithms. Expand each of the following, and determine the interval of convergence. 19. (8 + x)*. 20. (4 + as)*. 22. (9 + 4x)i 24. (8 + x)i. This equals 2(1 + x/8)*. 21. (9 + x)i 23. (16 - x)*. 25. (16 + s)*. EXERCISE LVI Expansion in Powers of (x — a). Taylor's Theorem. It is some- times very desirable to have a function expanded in powers of (x — a), where a is some constant. This is accomplished by Taylor's theorem, F"(d] F'"(a) F(x)=F(a)+F>(a)(x- a )+ -±l(x-af+ —U.(x-af+. . . . That is, the successive coefficients are simply the successive derivatives of F(x), evaluated for x = a. The series may be 80 PROBLEMS IN THE CALCULUS finite or infinite. Observe that a may be a positive or negative quantity. Expand the following in terms of the quantities indicated. 1. tana; in powers of (x — ■jt/4) ; of (x + 7r/4) ; of (x + 2). ( 2 ) — = 0. ( 3 ) I < — ;■— r- v ' dx K ' dy v ' \dydx) dx 2 dy 2 At maximum points, (PF/dx % < 0, or d 2 F/dy 2 < 0. At minimum points, d' 2 F/dx 2 > 0, or (PF/dy 2 > 0. Find the maximum and minimum values of the following functions : 1. z = x 2 + y 2 + xy — 6x— 4y+5. 5. z = sinx + sin?/ + sin(x + y). 2. z = x 2 -6x2/ + 2/ 8 + 3x + 6j/. 6. z = xy 2 (6 - x - yf. . 3. z = 4x2/ + 1/z + 1/2/- 7. z = sin x sin j/ sin (tt — x — y). 4. z = x 2 + y* + xj/ — x — 5 j/. 8. z = x s — 6 xy + y s . 9. z = 7x 2 - 6x?/ + 32/ 2 -4x+ 7J/-12. 10. z = x 3 + j/ 2 - 6x2/ - 39a; + 18 y + 20. 11. z =x 2 + 2XJ/ + 2 2/ 2 + 4x— 4 2/ + 6. 12. z = 2x2/ — x 2 — 42/ 2 + 2x + 62/ + 12. 13. z = x 2 + X2/ + 2/ 2 — 5 J' — 4 2/ + 1. 14. Show that u = x 3 2/ 2 (1 — x — y) is a relative maximum when 15. Show that the general quadratic ax 2 + by 2 +2 kxy+ 2gx + 2fy + c a h g has the maximum value k b f 9 f e 16. The volume of a right parallelopiped is 64 cu. ft. What are its dimensions for minimum total surface ? 17. Divide the number 18 into three parts so that the continued product is a maximum. la h\ \h 6 SERIES 87 18. Prove that the sum of three positive quantities whose product is a constant is a maximum when the three quantities are all equal. 19. The sum of the sides of a triangle is 12. Show that the area is a maximum when the triangle is equilateral. 20. Divide a straight line of fixed length into three segments so that the sum of the squares of these may be a minimum. 21. Find the dimensions of the rectangular parallelopiped of maximum surface which can be inscribed in a given sphere. 22. By the former regulations of the parcel post the sum of length and girth of any parcel must not exceed 6 ft. Show that the most voluminous cylindrical package which could be sent was a cylinder 2 ft. long, with girth 4 ft., and hence of volume 2.546 cu. ft. Under the new regulations this sum has been increased to 7 ft. What is the increase allowed in the maximum cylindrical package ? 23. If a/x + b/y + c/z = 1, where a, b, and c are all positive, what are the values of x, y, and % which make the sum x + y + z a minimum. 24. If x, y, z are the lengths of the perpendiculars dropped from any point P to the three sides a, 6, c, respectively, of a triangle of area A, show that the minimum value of x 2 + y 2 + z 2 is equal to 4 A 2 /(o 2 + 6 2 + c 2 ). 25. Show that sinx + siny + cos(x + y) has a minimum when x = y = 3 7r/2 and a maximum when x = y = tt/6. 26. An open box has the form of a rectangular parallelopiped. If the volume is fixed, what are the dimensions for minimum material ? 27. Find the minimum value of x 2 y s z* if 2x + 3y + 4z = a. 28. Show that the maximum inscribed triangle and the minimum circumscribed triangle for a fixed circle are both equilateral. 29. One angle of an inscribed quadrilateral is 120°. Show that for maximum area the remaining pair of opposite angles are both 90°. 30. Given two mirrors OJOf, and OXM 2 meeting in a right angle along the edge OX. P 1 is a luminous point, and its image is reflected first from the point N r in the first mirror, then from a point iV 2 in the second mirror, and is finally caught on a screen at P 2 . Assuming P 1 A/'jiV 2 P 2 to lie in a plane perpendicular to OX, what are the positions of .ZVj and N 2 if the path P 1 N X NJP 2 is a minimum ? 31. Given a triangle ABO. Find the position of a point P so that the sum of the squares of the perpendiculars drawn to the three sides may be a minimum. 32. Given a triangle ABC, find the position of a fourth point D so that the sum AD + BD + CD may be a minimum. (Cf . Sohncke, "Sammlung," p. 148, for this and similar problems.) CHAPTER X FURTHER APPLICATIONS TO GEOMETRY EXERCISE LXII Envelopes. If an equation contains a parameter (that is, a general constant) in addition to the variables x and y, we get a system of lines or curves by assigning various numerical values to this parameter. The locus of the limiting positions of the points of intersection of "consecutive" members of this system, or family, is called the envelope of the system. It is tangent to every line or curve of the system. To get the equa- tion of the envelope, we differentiate the given equation with respect to the parameter, then eliminate the parameter between the original equation and this first derivative. Find the enve- lopes of the following systems jjf straight lines. Draw figures. 1. y = mx + m 2 . 3. y = ro 2 x — 2 m s . 5. x£ 2 + yt — 1 = 0. 2. y — x/m + m 2 . 4. y = m 2 x + 1/ro 2 . 6. xt 2 — 2yt + 4 = 0. 7. x/h + y/k = 1, where A/3 + fc/2 = 1. Find the envelopes of the following systems of circles : 8. (x-a) 2 + y 2 -4a = 0. 10. (x - «) 2 + (y - t) 2 = 4. 9. x 2 + (y- t) 2 = t. 11. (x- ty + (y- t) 2 = 2i. Find the envelopes of the following families of curves : 12. tx 2 + t 2 y = 1. 14. y - tx 2 + 1/t. 13. y = t 2 (x-t). 15. y 2 = tx* + t. 16. Find the envelope of the family of circles whose diameters are the double ordinates of the curve y 2 = x 3 . 17. A variable circle passing through the origin moves with its center on the curve y = x 3 . Find the equation of the envelope of the system of circles. 88 FURTHER APPLICATIONS TO GEOMETRY 89 18. The ellipses of a system have the same area, the same center and axes always lying in the coordinate axes. Show that the envelope consists of two rectangular hyperbolas. 19. A right triangle moves so that one leg always passes through the point (4, 0), while the vertex of the right angle, C, moves along the 2/-axis. Find the envelope of the other leg as G moves. Hint. Take the distance OG as the parameter. 20. Given the ellipse x 2 + 4 y 2 = 16. There is a family of ellipses, equal in size to the given ellipse, with their centers on the perimeter of this ellipse, and with the respective axes parallel to the coordinate axes. Find the envelope of the system. 21. Given the two concentric ellipses x" + 4?/ 2 = 64 and x 2 + 4y 2 = 16. Find the envelope of the polar lines of the points of the first ellipse with respect to the second. Note. The polar line of (a, 6) with respect to x 2 + 4y 2 = 16 is ax +iby = 16. 22. Find the envelope of the chords joining the points of contact of all pairs of mutually perpendicular tangents to the ellipse 16x 2 + 9y 2 = 144. Hint. Show first that the locus of the intersections of these pairs of tangents is the director circle x 2 + y 2 = 25. 23. Find the envelope of all straight lines such that the sum of their intercepts is constant and equal to 6. 24. Find the envelope of the system of parabolas (y — 4) 2 = k(x — k). 25. Find the envelope of the normals to the parabola (y — 2) 2 = 8 x, and show that it is the evolute, the locus of the centers of curvature. 26. Find the envelope of the normals to the ellipse 9 x 2 + 25 y 2 = 225. 27. Given the family of ellipses b 2 x 2 + a 2 y 2 = a 2 6 2 , with the restriction a + 6 = 8. Find the envelope of the family. 28. Given the parabola y 2 = 4x. Find the envelope of the family of parabolas of the same size having their foci on the given parabola. EXERCISE LXIII Further Applications to Plane Curves. (A) Asymptotes to Algebraic Curves. There are two standard methods of calculat- ing' the equations of the asymptotes to algebraic curves, (1) the method of limiting intercepts, which is strictly speaking the method of the calculus, and (2) the algebraic method of analytic geometry. The former method is given in detail in almost any 90 PROBLEMS IN THE CALCULUS textbook on the calculus. By method (2), which is in general much simpler, we arrange the equation in powers of y. If the coefficient of the highest power of y is a function of x, the real root or roots of this function equated to zero give the vertical asymptotes. To find the oblique asymptotes, assume them to be of the form y = mx + k. Substitute mx + k for y in the original equation. Equate to zero the coefficients of the two highest powers of x. Solve these two equations for the pair or pairs of values of to and k, say m^ and k v Then y = m^x + k t is an asymptote. Find the equations of the real asymptotes to the following curves by either method : 1. y 3 = x 2 (x — a). 2. 2x 2 — &xy + 3y 2 — 7x + iy — 12 = 0. 3. x s - 3xy 2 - 2y s — 4x + 8y- 9 = 0. 4. y 2 (x -4) -x s - 4x 2 = 0. 11. x 4 - y* - ixy = 0. 5. X s - x 2 y + y = 0. 12. x 3 = 8 y 2 — y s . 6. xy 2 — x + 2 y — 1 = 0. 13. x 3 — xy 2 + 4 y 2 = 0. 7. xy 2 + yx 2 -8 = 0. 14. x 2 y 2 = (x + 2) 2 (1 + x 2 ). 8. x 2 -y 2 - 2x-2y- 3 = 0. 15. 2/ 2 (x- o) = x 3 + ax 2 . 9. x 3 — xj/ 2 — 4x 2 — 4y 2 = 0. 16. x 2 (y— x) + xy + x 2 + x + 2 = 0. 10. xy 2 — 2x 2 y + x 3 — 4y = 0. 17. (x + a)2/ 2 = (y + 6)x 2 . 18. x 3 + y* - x 2 - y 2 = 0. 19. x- ixy 2 - 3x 2 + I2xy — 12y 2 + 8x + 2y + 4 = 0. 20. xV -^ x 3 j/ + x 2 - 4y 2 = 0. 21. (y— 2x)(y 2 —x 2 )- a(y-x) 2 + 4a 2 (x + y) = 0. 22. x 2 y 2 + ax{x + y) 2 - 2 a 2 ?/ 2 - a 4 = 0. 23. x 7 — xV + 0- 92 PROBLEMS m THE CALCULUS (b)Foracusp:(^j- — • ^ = 0. J / d 2 f V 8 2 f d 2 f . (c) For an isolated or conjugate point : \^~ y J ~^ 2 "df For a discussion of more complex singularities the student should consult a book on plane curves. Investigate the following curves for singular points : 1. x s + yS _ x z _ ^ = o. 5. x(x - a) 2 + 2/ 2 (x - 2 a) = 0. 2. x a - y s + (2 y - x) 2 = 0. 6. ay 2 - x 3 + bx 2 = 0. 3. x 4 -2x 2 2/-2x2/ 2 + 42/ 2 = 0. 7. x 3 + 9x 2 -2/ 2 + 24x+ 6j/ = 0. 4. x s + ya _ 3 axy - o. 8. x 3 - 4x 2 - 3z/ 2 + 12 2/ - 12 = 0. 9. x 3 + xy 2 - 2 x 2 - 2 2/ + 2 = 0. 10. x 4 + 12x 3 - 62/ 3 + 36 x 2 + 272/ 2 - 81 = 0. 11. x 4 + 2Z 4 — 2 x 2 — 2 2/ 2 + 1 = 0. Note the number of singular points. 12. x 4 + 4x 2 - 4 2/ 2 = 0. 18. x 2 - 2x?/ + 2/ 2 -x 3 = 0. 13. 2/ 2 - x 8 - x 2 = 0. 19. 42/ 2 - x 8 + 2x 2 = 0. 14. x 4 + J/ 4 - 22/ 2 - 8x 2 + 16 = 0. 20. y 2 = (2 - x) 2 (l - x). 15. x 4 - 4?/ -122/ 2 - 8x 2 + 16 = 0. 21. 3x 2 + 4j/ 2 - (x 2 + y 2 ) 2 . 16. (?/ - 2) 2 = (x - 3) 5 . 22. 4x 2 - 2/ 2 - x 3 = 0. 17. 5y 2 + 102/ - x 3 + 2x 2 + 5 = 0. 23. y = 2x 2 + x 5/2 . 24. x 4 - ax 2 ?/ + axy 2 + \ a?y 2 - 0. 25. x 5 — 4 a;/ 4 + 2 ax 3 ?/ + a 2 x?/ 2 = 0. 26. (x - yf + a(x 2 - 2/ 2 ) + a 2 2/ = 0. 27. (x 2 + ?/ 2 ) 2 + a 2 (2/ 2 - x 2 ) = 0. 28. x 2 ' 3 + 2/ z/3 = a 2/3 - 29. x 4 - 4ax 3 + 4a 2 x 3 - b 2 y 2 + ibhj - a 4 - ¥ = 0. 30. x 5 + 6x 4 — a 3 ?/ 2 = 0. 31. x 4 + j/ 4 — ay 3 — 2 ox 2 2/ + ax?/ 2 + a 2 y 2 = 0. 32. 2/ 2 (a 2 + x 2 ) = x 2 (a 2 — x 2 ) . Show that the origin is a node and that the nodal tangents bisect the angles made by the axes. 33. x = t 2 , y = t— \ I s . Find (a) radius of curvature at origin, (b) coordinates of maximum and minimum points, (c) coordinates of double point. FURTHER APPLICATIONS TO GEOMETRY 93 EXERCISE LXIV Space Geometry. Curves. When a curve is given parametri- cally, the direction cosines of the tangent line to the curve at any point are proportional,to the values of dx/dt, dy/dt, dz/dt, at the point. Call these values a, b, c. Then the tangent line is (x - x ± )/a = {y- yj/b = (z- zj/c, and the normal plane is a(x- xj + b(y - yj + c{z - z ± ) = 0. In case, however, the curve is given as the intersection of two surfaces, the direction numbers a, b, c of the tangent line are readily found. Since the parameter t may be x, obviously a:b:c = dx/dx : dy/dx : dz/dx = 1 : dy/dx : dz/dx. So it merely remains to find dy/dx and dz/dx at the required point. To do this differentiate both equations separately with respect to. x, and then in these two equations replace x, y, z by the coordinates of the point in question. This gives two simple equations to solve for dy/dx and dz/dx. Call these values A and B. Then the equation of the tangent line is (x - Xl )/1 ={y- y x )/A = (z- zJ/B and of the normal plane is (jr _ xj + A (y - y x ) + B(z-z,) = 0. In the second case, the two-plane form of the equations of the tangent line is found by getting the equation of the tangent plane to each surface at the given point. The intersection of these two planes is obviously the required tangent line. The first method seems to the author desirable, as it gives at once the direction numbers of the desired tangent line. See next exercise for the method of finding the equation of the tangent plane to a surface. Find the equations of the tangent line and the equation of the normal plane to each of the following curves at the point indicated : 1. x = t - 3, y = t 2 + 1, z = i 2 , at t = 2. 2. x = t 2 — 1, y = t + t s , z = 1 - 2t, at t = 1. 94 PROBLEMS IN THE CALCULUS 3. x = 4 2 , y = 4 3 , z = 4 2 + 4, at 4 = 1. 4. x = 2 4, y = 2/4, z = 4 2 , at 4 = 2. 5. x = 3 - 4 2 , y = 4 8 - 1, z = 4/4, at 4 = 1. 6. x = 2 4 2 - 3, y = 4/4, z = 3 4, at 4 = 2. 7. x = 4 2 , y = 1/4, z = simrt/2, * 4 = 1. 8. x = sin 4, ?/ = cos 4, z = sec 4, at t = 0. 9. x = 2 sin 4, y = 3 cos 4, z = 2 4/7r, at 4 = 7r/2. 10. x = 4/(1 + 4), j/ = (1 + 4)/4, z = 4 2 , at t = 1. 11. x 2 + j/ 2 + z 2 = 49, x 2 + j/ 2 = 13, at (3, 2, - 6). 12. z = x 2 + y 2 - 1, 3x 2 + 1y 2 + z 2 = 30, at (2, 1, 4). 13. x 2 - y 2 - z 2 = 1, x 2 - ?/ 2 + z 2 = 9, at (3, 2, 2). 14. x 2 + j/ 2 - z 2 = 16, x 2 + 42/ 2 + 4z 2 = 84, at (2, 4, 2). 15. x 2 + y l + 3z 2 = 32, 2x 2 + j/ 2 - z 2 = 0, at (2, 1, 3). 16. x 2 + 4j/ 2 - 4z 2 = 0, 2x + y + z - 24 = 0, at (8, 3, 5). 17. 2x+3y+ 2z = 9, x 2 + 2j/ 2 + 2z 2 = 8, at '(2, 1, 1). EXERCISE LXV Space Geometry. Surfaces. Since the direction cosines of the normal to a surface F(x, y, «)= at any point are proportional to dF/dx, dF/dy, and dFj'bz at that point, it is very easy to write down the equation of the tangent plane to the surface as well as the equations of the normal line. For if the values of the above three partial derivatives are A, B, and C, the equations are tangent plane, A (x — ;q) + B(j/ — j/ x ) + C(z — z x ) = ; normal line, (x — x ± )/A — (y— !/i)/B =(z— z x )/C. If a surface and a curve intersect, the angle between them is defined as the angle between the normal line to the surface and the tangent line to the curve at their common point, and can now be readily found. Find the equation of the tangent plane and the equations of the normal line to the following surfaces : 1. x 2 + y* + z 2 = 49, at (6, 2, 3). 4. x 2 + y 2 - z 2 = 25, at (5, 5, 5). 2. z = x 2 + y*-l, at (2, 1, 4). 5. 2x 2 + 3 if + 4z 2 =«, at (1, 1, 1/2). 3. x 2 - y 2 - z 2 = 1, at (3, 2, 2). 6. x + y - z 2 = 3, at (3, 4, 2). FURTHER APPLICATIONS TO GEOMETRY 95 7. x 2 + xy 2 + j/ 8 . + z + 1 = 0, at (2, - 3, 4). 8. x 2 + 2 xy + y 2 + z - 7 = 0, at (1, - 2, 6). 9. x 2 y 2 + xz - 2 J/8 _ io = 0, at (2, 1, 4) . 10. x 2 + 4y 2 - 4z = 0, at (2, 2, 5). 11. The surface x 2 — 4 j/ 2 — 4 z = is cut by the curve x = J 2 /2, y = 4/t, z = (t — 2 i 2 )/2, at the point (2,2,-3). What is the angle of intersection ? 12. The surface x 2 + y 2 + 3 z 2 = 25 and the curve x = 2 S, ?/ = 3/t, z =— 2 J 2 intersect at the point given by t = 1 on the curve. What is the angle of intersection ? 13. The ellipsoid x 2 + 2 y' 2 + 3 z 2 = 20 and the skew curve x = 3(i 2 + l)/2, y = t* + 1, z = i s meet in the point (3, 2, 1). Find the equa- tion of the tangent plane to the surface and also of the normal plane to the curve at this common point. What is the angle between the planes ? What is your conclusion ? 14. The surfaces x 2 y 2 + 2 x + z 3 = 16 and 3x 2 + i/ 2 - 2z = 9 have the point (2, 1, 2) lying on their curve of intersection. What are the equations of the respective tangent planes to the two surfaces at this point ? What is the angle between the two normal lines ? 15. Given the surfaces x 2 + 3 y 2 + 2 z 2 = 9 and x 2 + y 2 + z 2 — 8 x — 8y — 6z + 24 = 0, show that they are tangent to each other at the point (2, 1, 1). 16. Given the paraboloid 3x 2 + 2j/ 2 — 2z = l and the sphere x 2 + y 2 4-z 2 — 4y — 2z + 2 = 0, show that they cut orthogonally at the point (1, 1, 2) by finding the angle between the respective tangent planes at this point. CHAPTER XI INTEGRAL CALCULUS. SIMPLE FORMAL INTEGRATION EXERCISE LXVI The Power Form. This most important formula is / v n + l v n dv = - \-C. n + 1 It holds for n any real value except n = — 1. If the denomi- nator consists of a quantity raised to a power, it should be brought up with a negative exponent. The student should observe carefully what part of the integrand is represented by v in the formula, and then see that dv,* the differential of v, is present, and not merely dx. In fact, it is not amiss to carry out the actual substitution, even in the simpler cases, until perfectly familiar with the process. It is best to begin by putting any constant factor of the integrand outside the integral sign. Then observe whether the differential of v is present. If a constant factor is lacking, it may be supplied, but we must at the same time compensate for this by dividing outside the integral sign by the same constant. Note carefully that we cannot supply any lacking term if it contains a variable and compensate for it outside the integral sign. These principles are general and do not apply to this exercise alone. They are shown step by step in these illustrative examples. The actual integration comes last always. All answers should be verified by differentiating them, thus obtaining the integrand when correct. * The differential of one variable (usually dx in these examples) is always present. It is, in fact, a part of the integral sign designating the variable with respect to which we are integrating. FORMAL INTEGRATION 97 I. \3^(x 2 + §)xdx = 2,\(x 2 + sftxdx = %l(x 2 +rf -2xdx = |( a ; 2 + 6)Vl = (x 2 +6)Uc. II. I 3 (a; 2 + iydx = 3 I (x 2 + l) 2 cfa: (here we cannot " com- pensate," so must try some other device, namely, expansion by the binomial theorem) = 3 / (x 1 + 2 x 2 + 1) dx = $( jx*dx + 2Jx 2 dx + Jdxj = 3 (x 5 /5 + 2 x s /3 + x) + C. Be careful of signs in compensating. 1. Cx^dx. 15. f (x - 1/Vx) dx. 29. f (l- Vz) 2 /2*-dv = a'>/loga + C, f e v dv = e"+C. The student is again reminded of the importance of dv, its presence being necessary before the formulas can be applied. The above integral should not be confused with the power form I v a dv. 1. f2*dx. 8. CeU-z^dx. 15. f(secx • e tana; ) 2 ]dx. 16. f (e«"Vsin 2 x)c?x. 3. f(l/3*)dx. 10. fxe^dx. 17. C (e^e^ x /x)dx. 4. C(\/0c)dx. 11. J(e^x/Vx)dx. 18. /"[e^s < 1 -^)/(l-4x)]dx. 5. f(l/2 6 ^)dx. 12. f(e*/x 2 )dx. 19. f (tan a/eke™" *)<&;. 6. fn ar dx. 13. fe sin2a: cos2xdx. 20. f cotx ■ e lo s»™ 2 *dx. 7. fe< ix + »dx. 14. fsmx/ef> osx -dx. 21. f(e?+e ax ) 2 dx. 100 PROBLEMS IN THE CALCULUS 22. f (e 2 * + e-zxfdx. 25. f(e*'i + e) s dx. 28. f (e^ + e W)*dx. 23. f (e 2 * + e*' 2 ) 2 iix. 26. f (e* - Ve) 2 (fo. 29. f[(e 2 ^ + Vx)/x 2 ](fe. 24. f(e* + dv = sin v + C, / tan v do = — log cos v + C, or log sec v + C, I cot vdv = log sin y + C, / sec »du = log (esc v — cot v) + C, and those which follow from the formulas for differentiation, I sec 2 i>dv= tan v + C, / seci>tanu + C, I esc v cot vdv = — esc v + C; the student should have at his command the various simple relations existing among the six trigonometric functions. By- means of these, the following expressions, if not already stand- , ard integrals, are easily reduced to such. 1. CsmZxdx. 3. fsec5xdx. 5. fcsc(x/2)dx. 2. jtan3xdx. 4. fcos(x/3)dx. 6. fcot (x/7) dx. * Reduce examples 31-34 to mixed numbers before integrating. FORMAL INTEGRATION 101 1. ("tan (x/V2) dx. 12. f sec (3 x + 2) dx. 17. f x sin (1 - x 2 ) dx. 8. jsinradx. 13. fsec 2 5xdx. 18. C(cosVx/Vx)dx. 9. J csc2 7rxdx. 14. J sec 2 s tan 2 xdx. 19. J (xsec"x 3 ) 2 dx. 10. fcos(l + 2s)dx. 15. fcsc 2 (s/2)ds. 20. f [csc(l/s)/s] 2 dx. 11. fcot(l — s)dx. 16. Cxoosx 2 dx. 21. /"(sec Vx/-Vx) 2 dx. --,. f_*5_. 24. f_^_. 25. ftan 2 x/2dx. •/ eos2x J cos 2 3x J 2 tanxdx 22 /; sec 3x 26. fcot 2 3s 2 + 1)] dv = arc tan v + C = arc tan (tan x) + C. dx r dx r dx ^Tl' ' J x 2 + 4 J x 2 "^ dx . r dx „ r &% ■ Sdz 6+5z 2 • d6 70 2 + 9 • 2 dm FORMAL INTEGRATION 103 f ** ■ 25. f *? 37. f— ?*L Jx 2 + 2x -/4x 2 + 4x + 5 Jx 4 + 2x 2 + 2 r_dx_ g6 r dx 3g r x*dx . J4x-x 2 ' " ' ■' 0x 2 -6x-8' '-/x Vll-6x + x 2 ' ^ Vsin 2 x/2 + 5 5 7 Ve 2sill: > : + 4 EXEECISE LXXII Linear over Quadratic. The forms here are of the type j (L/Q)dx. Here, as previously, Q represents the general quadratic function and L the general linear function mx + n. In caseZc&c = dQ the integral reduces to that of Exercise LXVII. Otherwise the integral must be broken up into two simpler in- tegrals of the type j (l/v)dvand j (1/Q)dx. In the case where Q has no middle term the integral f — 5 dx = m / -^r^— dx J ax'+c J ax* + c + n I — 2 . ' an d both these integrals have been treated in the previous exercises. Hence the only novelty arises in the FORMAL INTEGRATION 105 case b =£ 0. In this case the integrand can be broken up by inspection, but the use of undetermined coefficients has many advantages. Thus /mx + n , C 2ax A-b ,, C ~Vadx 5——; ; ClX = A I 5 ; dx + B I ; ; ax' + Ox + c J ax 1 + bx +c J ax 2 + bx + = A log (ax" + bx+c)+B f g a f* • 7 ^y ax" 1 + foe + c This last integral is integrated as in Exercise LXX. A and B are then determined from the identity mx + n = 2Aax -\-Ab + B V a. Equating coefficients of like powers of x, we have m = 2A and n =Ab +B Va, which we can solve for A and £\ Illustration : r (2x + 3)dx ^ r 18a- + 6 r J 9a; 2 + 6;c + 4 J 9 x 2 + 6 * + 4 + J ( 3 c?x (3 x + l) 2 + 3 = 4 los (9 a 2 + 6 x + 4) + -^= arc tan i^±l . ' V3 V3 To determine 4 and 2?, we have the identity 2x + 3s 184a; ■+- 6 A + 3B, whence 184 = 2, 64 + 3.B = 3. Therefore 4 = £, .8 = f Therefore t /- (x + l)dx /- (5x + 2)fe n Wj2x+j5)dx 'J x 2 + l '■/ 5i 2 + 2 'J i'+3i ' 3 Wx-2)fe /- (2 + x)rix l2 W6-x)dx 'J x 3 - 1 ' ' J 11 + x 2 ' J 5 x - x 2 ' W2x + 3)dx g /- (7-3x)(fo 13 /- (2x+5)cZa './>4x 2 + l '«/ 5-2x 2 "Jx 2 + 2x + 5" /- (6x-l)dx g Wx+2)dx 14 /" (x-6)dx ' -1 l-9x 2 ' ' -/ x 2 + 2x' 'Jx 2 -4x-5' /- (2x-3)(fa; Wx + 3)dx w /• (3x-5)dx "J 3x 2 -2 'J 6x-x 2 ' ' 'J x 2 + 8x+ 20' / 106 PROBLEMS IN THE CALCULUS H5x-3)dx r {3x-2)dx ("J!^^: lg 'Ji!_6i-7 Jl-6x-9x 2 J5x 2 -2x + 3 /• xdx /• (2x + 3)ds 31 f (5x + 6)ds 'Jx 2 -8x + 18' '•/ 9i 2 + 6x+4' "J7x 2 -8x + 9 - 18 r Xtfa 25 f ( 3x + 2 ) dx 32 f (6s-5)ifo ■J2-6X-X 2 ' ''J9x 2 + 3x-2' " '^9x 2 -6x+7 r xdx 36 r (10x-8)dx 33 r (Ux + 2)dx "Jx 2 + x + l" '■/ 25x 2 -5x + 1 ' ' ' J 8-3x + 2x 2 ' 20 f xda: 27 f (a+2)dx 34 /• (3x-7)dx 'Jx 2 -x-l' ' J 3x 2 + 2x-l' '-/4-5x-3x 2 21 C (l-x)dx oa r (x-2)dx „ f(2x-5)dx C Q--x)dx 2g r (x-2)rix gg /■_< './4x 2 -4x-3' ' ' J 5x 2 + 4x + 2' 'Jl 5x 2 + 4x + 2 •/ l-2x-6x 2 22 /- (l-6x)cZx 29 /■ (l-x)dx 36 r (x+8)dx J "J9x 2 + 6x + 'J3x 2 -4x+3' "' ' ■/ 3x 2 + 5x - 12 EXERCISE LXXIII Linear over the Square Root of a Quadratic. The integrands here are all of the form L/y/Q. The method used in Exer- cise LXXII is also applicable to the integration of this form, the two resultant integrals being J (1/Vq)<2Q * and I (l/y/Q)dx, which are treated in Exercises LXVI and LXXI respectively. Additional examples can be had by placing radicals over the denominators in Exercise LXXII. The breaking up of the integral into two simpler integrals is accomplished precisely as in Exercise LXXII, as shown in the following illustrations. Undetermined coefficients can be used as before. r (2x + 3)dx = 2 r xdx + s r dx K) J V4x 2 + 1 J V4x 2 + 1 J V4a; 2 + 1 = jV4x 2 + l + |log(2a; + V4x 2 +l) + C. * Owing to the frequency with which this particular case of the power form (n = — 1/2) occurs, it is well that the student should include it in his list of formulas, i (l/Vw)cfo = 2 Vv + C. FORMAL INTEGRATION (2x + 3)dx 107 J V9x 2 + 6cc + 4 f (18a: + 6)/4d); (0, 1). 20. (cotx)/y ; (tt/2, 2). 14. Vl - 2 y/x* ; (1, -4). 21. [(cos?/)/x] 2 ; (1, 0). 15. V(l+x)/(l + !/); (3,8). 22. [(cosx)/?/] 2 ; (tt/2, 0). 16. V(l+j/)/(l + x); (8, 3). 23. -Vl-iyt/x; (0, J). 17. x cos 2 y ; (2, 0). 24. (1 - x 2 )/(l + y 2 ); (1, 1). 18. x*secy; (0, tt/2). 25. (1 + y)/{\ -x 2 ); (0, 0). 19. ?/ 2 sinx; (tt/2, 2). 26. xj/VVl + 2 x 2 ; (0, 0). 27. The velocity of a falling body is ds/dt = u + 32 4. Find the general expression for s if v = 16 and s = 64 when t = 1. 28. The resistance of the air to an automobile, within certain limits of speed, is proportional to the speed. Hence if F is the net force gener- ated by the motor, we have M dv/dt = F—kv. Express the velocity in terms of t, knowing that v = when t = 0. 113 114 PROBLEMS IN THE CALCULUS 29. Suppose in example 28 that the power is out off and the speed is reduced by air resistance alone. Express the subsequent speed in terms of t, assuming a velocity b when the power is cut off. 30. Assume in example 28 that the air resistance is proportional to the square of the speed, giving M dv/dt = F— kv 2 , and show that v = VF/k tanh VkFt/M. [Assume v = when t = 0. Tanh x = (e* — e~ x )/{e c + e-*)]. What is the maximum speed attainable ? 31. The resistance of air to a parachute may be regarded as propor- tional to the square of the velocity. If a man weighs 1351b. and the parachute 25 lb., and the surface of the parachute is 200 sq. ft., with what maximum velocity will he strike the earth jumping from any height ? Assume the air resistance to be 2 lb. per square foot when the velocity is 30 ft. per second. If he reaches the ground in 30 sec, with what velocity will he land ? (In this case the force downward is Mg, where M is the combined weight in pounds of man and parachute. The resultant force, as in example 28, is Mdv/dt. 32. By Hooke's law, the tension in a stretched elastic band or spring is proportional to the amount it is stretched beyond its normal length, or F = — kx. We can write F, the tension, not only Mdv/dt but also Mvdv/dx. Assuming that the band is stretched a feet when the mass M is attached (that is, x = a when v = 0), show that the velocity for any future position of the mass while this law holds is v = Vka 2 — kx 2 /VM. 33. Take the result of example 32, dx/dt = vfc(a 2 — x 2 )/M, and show that under the same conditions the position of the particle at any time is x = asm{Vk/Mt + Tr/ii). Evaluate your results of examples 32 and 33 'for k = 16, M = 41b., a = 2 ft., and check by finding the value of v when x = 0, in both cases. It is obvious that the value of v in the two cases should be the same. 34. If the band or spring is hanging vertically and therefore the gravi- tational force is also acting, the force equation is Mvdv/dx =— kx + Mg. Assuming that k = 64, M = 4 lb., g = 32 f t./sec. 2 , and that the spring is stretched 6 ft. when released, show that v = 4 V— x 2 + 4 a; + 12. 35. Taking the result of example 34, dx/dt = 4 V— x 2 + 4x + 12, and the same conditions (x = 6 when t = 0), show that the position, x, at any time, t, is x = 4 cos 4 1 + 2. 36. Any vertical section of the surface of a liquid contained in a rapidly rotating vessel is defined by the law xdx/dy = P 2 5t/4tt 2 , where P is the period of rotation in seconds, and the ?/-axis is vertical with the origin at the lowest point. Show that the surface is a paraboloid of revolution. DEFINITE INTEGRALS 115 Derive its equation if the vessel is making 120 revolutions per minute, taking g = 32. [The defining equation may also be written as dy/dx = u 2 x/<7, where « is the angular velocity.] 37. A belt passes over a pulley of radius a. Assume the law that the pull, T, in the belt at a distance s from the point where the belt leaves the pulley is given by d T/ds = ^ T/a . Calculate the general expression for T in the case where the pull at the point where the belt leaves the pulley is' 50 lb. Take /x, the coefficient of friction, as 1/4 and the radius of the pulley as 2 ft. EXERCISE LXXVIII Orthogonal Trajectories. An orthogonal trajectory of a given system of curves is a curve which cuts each curve of the given system at right angles. Therefore, to find the equation of the orthogonal trajectories we must first find dy/dx from the equation of the given system. Then set dy/dx equal to the negative reciprocal of the derivative just found and integrate. Find the equation of the system of orthogonal trajectories of each of the following systems : 1. y 2 = 2 x + G. 3. xy = :G. 5. y 2 + x 2 = C. 7. y 2 + 4x 2 = C, 2. y=$x s +C. 4. y*x - = C. 6. y 2 - x 2 = G. 8. 2/ 2 -4x 2 = C, 9. y= ilog(l-x 2 )+C. 13. y = tan x + C. 17. x 3 + y s = G. 10. y = log(7(l + x 3 ). 14. y = sec x + C. 18. x 2 + y 3 = C. 11. y = 1/(1 + x) + C. 15. x i + yi = G. 19. y 2 = ACx. 12. y = x i +G. 16. x% + y% = C. 20. y 2 + 2 x 2 = C 2 . EXERCISE LXXIX The Definite Integral. Evaluation by Direct Integration. The process of evaluating a definite integral consists of first finding the indefinite integral, dropping the constant, as it disappears any way. Evaluate this integral for the upper limit b. Call this value B. Evaluate for the lower limit a, calling this value A. Then if the function is continuous between the limits, B — A is the value of the definite integral. Before evaluating a definite integral the student should test the integrand to see whether 116 PROBLEMS IN THE CALCULUS it becomes infinite (or discontinuous) for any value of x between the limits. If the integrand becomes infinite for a limit, say b, then Jr» b r*b — h f(x) dx = limit I f(x) dx, *=» Ja which may be finite or infinite. Or if it becomes infinite for x = x v a value between a and b, the integral J f*b siTi—li fb f{x) dx = limit j f(x) dx + limit I f(x) dx, *=» J. *=» Jx 1+ h and if the limit of either of the integrals on the right is oo, the definite integral in question has no finite values. Examples 10, 16, and 24 are among those requiring examination. Why is example 24 meaningless ? 1. fV(l-V^x. 12. f'—^. 21. f"-^-. •'o ^5 Vx 2 + 144 ■'o x 2 + 4 2.^°(l + ^) 2 dx. ig ; 7cos2gdg „. j^ ^ Z" 6 _ "IT -v/Q PL oi'n 9^ 3. r 6 (l-^/2) 3 r 2 2 33. f 4 sm 3 2x = — 3. The student should be clear as to the process here. We are trying to find A,B,C, and D such that the integrand 9x + 5 = A B Cx + D and ^ x(x + l)(x 2 +2x + 5)~ x + x + 1 x 2 +2x + 5' &U is accomplished by clearing of fractions and equating coeffi- cients. By this method we get the above values ofA,B, C, and D. (20 x + 17) dx <•>/ (2 x + 1) 2 (3 a: + 5) Cdx r Adx _ r Bdx r 'J 2x + l + J {2x + lf + Jl Zx + 5 = fl og (2* + l)-^^ + flo g (3* + 5). Next we find A = 2, B = 2, C = - 3. v. /• x(8x + 2)(& , s ) " (4x 2 + 8x + l)dx W8x + 2)rfx lg /> J X — X 3 */ 14 x — x s t/ 4 x 3 — x (x + l)dx W9x 2 + 12x + l)dx |" (x + l)dx le r J x 2 + 5x + 6 # J x 2 + 5x + 6 J 9x 8 - SPECIAL METHODS OF INTEGRATION 121 17> /- (2x 2 + 5x+6)dx ^ r (x" + 6x-l)dx 'J x 3 +5x 2 + 6x ' "J (x-3) 2 (x-l)' lg _ /- (2x 2 + 7x-6)dx 36 /- (17+3x-2xVx 19 r (7x 2 -x-6)dx Wx 2 + 4x+6)dx J 6x 3 + 13x 2 +6x' J (x + 1) 8 20 /- (lQ-9x-19x 2 )dx /- (x 2 + x + l)(2x 'J 10x 8 + 29x 2 + 10x' J ( X - 1) 3 21 r 4xdx W2x 2 + 9 )dx ' ■> (x + 1) (x + 2) (x + 3) J (x - l) 8 t 2 + 9) d (x + 1) (x + 2) (x + 3) " ""' J x 8 (x-3) 22 f (z 2 +s-3)dx 4Q r (x + l)dx 'J (x-l)(x-2)(x-3)' "J x(x-l) 3 ' 2g |" (x 2 -x + 6)c?x /- (2x 3 -2x 2 -2x + l)cZx ''■/ x 4 -10x 2 + 9' "J x 4 -2x 3 + x 2 Wx 2 -3x-2)dx /• (3x 8 +12x 2 + 16x+4)cix '-/ x 4 -5x 2 + 4 ' 'J x 4 + 4 X 3 + 4 S 2 35 |" (2x 2 -l)dx W2x 2 -x + l)dx ' J (x 2 -2x)(x 2 -l)' 'J (x 2 -x) 2 26 /■ (7x 2 -4)rix W4x*-5x-l)cfc ' J (x 2 + x)(x 2 -4) 'J 4x* + 4x 3 + x 2 /■ (l + x-3x 2 )dx /• (2x 3 -x 2 + l)dx " J (x*-2x 3 -x 2 + 2x)' "J" (x-2) 4 f (2x 3 +3x 2 -4)dx W2x 8 -3x 2 + 4x-5)dx 'J x 4 + x 3 -4x 2 -4x' 'J (x+3) 5 •/ X s + X 2 J X 8 + X go r (x s -2)dx 48 _ /" (l-x)dx ' J X 3 — X 2 ' J X 3 + X si. r ta . 49. r (x + 1)2dj: . J 2 x 3 — x 2 J x 3 + x 32 /- (2-7x)tZx /■ (x 4 -2)dx J 2 x 2 - 3 x 8 ' ' j"V 5 2dx 74> /"§ 2dx 7g _ ^ 4 (3-x)dz J2 x — 1 Jo 1 — x 4 Ji x — x 8 J2 x s + x 2 EXERCISE LXXXII Integration by Rationalization. The examples in this section are largely restricted to those involving only the simplest form of an irrationality, that is, the integration of expressions con- taining fractional powers of the general linear function ax + b, or (ax + b) mM . Such expressions are readily reduced to rational fractions, which are the subject matter of the previous exercise. The rationalization is accomplished by the substitution of a new variable z, where z n = ax + b, but all answers should be expressed in terms of the original variable. Do not forget to transform dx before integrating. dx r dx r dx Vx J (1 — x) Vx ^ x — 4 Vx dx _ c dx . r dx r do. J 1 + . , f ^ _. 6 . f & 8 . r. x + 2Vx + 5 -^ x+ 6Vx x r (to /■ (to . /■ dx •^ (l + x)Vx 'x-2Vx-3 ^x(l-xi) 10 SPECIAL METHODS OF INTEGRATION 123 , r__^ i6. f xdx ■ 22. f— 1=- J x(l + xi) J Vl + x J xV3x-4 I, f fc 17. f 6 ^— • a, f V2+^dx /" dx is. f — — „, ,-V5-xcZx ,„ f 1 dx 19. f 2 **" />V2x+3dx J °Vx(9 + xi)' ^(x + ^CVl + x) «-J 3x + 2 • r dx 20. f dX 2B f V3x-2dx j Vx"(4-xV J (x-3)Vl + x "J 2x-3 dx »/C-^> "X* (2 + x)(x + l)t r_ *» 34. r ^ -; (1 + 6x)* + (1 + 6x)* J (x + 8) (x - l)i f ^ 35 /• 2dx •' a-i2xif-a-i2xit J 'J/ 29 (l-12x)«-(l-12x)7 J (x2 + 2x)VT+x 31. f (3x-10)dx ^ /■ 2dx ■^ /->• J. 9\ f 9. _ T.\i ' •/ / 32 (x + 2)(2-x)s ■> (i2 + 4x)Vx+l /• (14 — 5x)dx r dx (x+8)(l-x)S ■' (x 2 -l)Vl-x In the remaining examples, rationalize by the substitution suggested. 4dx xVx 2 -2x + 3 4xdx 38, f — Let ^/x 2 -2x+3:=Z■ 39. f i^ LetVx 2 -2x + 3 = z-x. J (x 2 -2x + 3)$ 40. f 2(fa — Let V5x-6-x 2 = (x-2)z. J V5x-6-x 2 41. f 2xdx _ . Let V5x-6-x 2 = (x-2)z. ^ V5X-6-X 2 124 PROBLEMS IN THE CALCULUS 42. f x 8 (4 + x 2 )kdx. Let 4 + a; 2 = z*. . r (4 + g2) dg. Let 4 + x 2 = z*. J x 43. 44. f (9 + 8a;8)T (fa. Let9+8x 8 = x 8 z s . J x 2 45. fli±_?!Ld x . Let 4 + x* = x 4 z 2 . J x 8 The student will find it possible to rationalize a great many forms by a properly chosen substitution. The list above contains a few of the many which might be classified. The principle of rationalizing the integrand is an important one in direct integration. EXERCISE LXXXIII Integration of Rational Trigonometric Functions. Such fractions can be reduced to rational algebraic fractions, the integration of which is practiced in Exercise LXXXI. The transforma- tion is tan x/2 = z, which gives sin x = 2 z/(l + * 2 ) ; cos x = (1 - s 2 )/(l + « 2 ) ; dx = 2 dz/(l + as 2 ). The student can derive, the formulas for the remaining four trigonometric functions as needed. All answers should be expressed in terms of the original variable by means of the relation tan x/2 = z. This method should not be used unless the expression cannot be integrated by any of the previous simpler methods, as it is often laborious. 1. r *? 5. r__*! 9 . r *> . J sinx + cosx + 1 J 5+4 oosx •/ 3 + cosx 2. f ^ 6. f— -^ 10. f *" J sinx— cosx + 1 •/4+5oosx J sinx— cosx + 1 •/4+5cosx J 2 cosx + 1 r dx 7> r dx n r / cot x + csex J 13 cosx — 5 J 1 + 2 sinx SPECIAL METHODS OF INTEGRATION 125 dx r cosxdx 2sinx— cosx +3 ' 'J 5-3 cosx dx r sinxdx 18 f ™ 16. f- J 3 sm x — 4 cos x + 5 J 5 + 4 sin x f shixdx . 19. f ^_. 21. f { - J 2 — sin 2 x J 5 sec x — 4 J < /sinxdx „„ /" dx „„ /» ' — • 20. I 22. / 4— sm 2 x •/4secx+5 ./ 4secx+5 J cosx + sinx EXERCISE LXXXIV Integration by the Reciprocal Substitution. By means of this substitution, x = 1/z, dx=— dz/z 2 , various expressions in- volving Vq can be reduced to forms similar to those of Exer- cises LXXI and LXXIII. The student should, however, express all answers in terms of the original variable. In par- ticular this substitution should be associated with the forms given in examples 9-20, that' is, (1) or , , x V Ax 2 + Bx + C (2) " • The substitution x = 1/z reduces x 2 -y/Ax 2 + Bx + C dz , ,„. , — ?-dz (1) to -=== and (2) to —=====, both of v / Vvl + Bz + Cz 2 s/a+Bz + Cz 2 which are readily integrable or can be found in the tables. Vl^tfdx , C dx « C <*" J x 4 '^xV9x 2 -4 J x* " xV9x a — 4 " x Vx 2 + 2 x dx r dx r dx , C-J?L 4. f dI ■ 6. f- J ~2 ,A,2 j. fi J r. -\A>S ?% — 1 J ct x 2 Vx 2 + 5 J xV25x 2 -l J xV3x 2 -2x r Vx 2 -3xdx 11. f **" J x^ xV4x 2 + 5x + 1 < " V2x + 5x 2 dx 12. f ^ ■ J x 5 ^ xV8x 2 +6x + l 9 . r ** is. f / *» J xV3x 2 - 2x-l -'xVx 2 + 4x-4 io. r *» i4. r / *»■ . J xVl + 4x + 5x 2 l/ xV2x 2 -6x-9 126 PROBLEMS IN" THE CALCULUS dx ,„ r dx 15 . f «* 1 8; f 21 xVl6 + 24a dx :+3x 2 xV9- 24x -5x 2 x 2 Vl + 2x- dx f 3x 2 (10x+.l)Va dx : 2 +x x 2 V3x 2 + 2x-l 16 . f <<* 19. f ■' xV9-24x-5x 2 ^ 17. f dX 20. f x 2 V27x 2 +6x-l dx x- r ^ /- Vx 2 + gdx 2S r xdx J (10x+.l)Vx 2 +x 'J x 4 J (Vx 2 + 3x) 3 r ** _ . 24. f i==. 26. K^-^) 8 ^ . •^ (3 x — 1) Vx 2 + x ^x 2 Vx 2 + x " J x 5 EXERCISE LXXXV The Definite Integral. Change in Limits. In each of the fol- lowing examples the indefinite integral is found by one of the substitutions of earlier exercises. In such problems, when the substitution is made, the limits of the new variable a (or t), corresponding to the limits originally given for x, should be calculated and used in the evaluation, thus eliminating the labor of expressing the indefinite integral in terms of x. Observe the process in these examples : (a) J a;Vl+ xdx, where the substitution is l+x = z 2 . Hence, when x = 8, s = 3, and when x = 3, z = 2. Therefore | x Wl+xdx = 2 J « 2 (s 2 — l)d» = 2 - — - =— — • In the next example the substitution is x = 2 sin t, so when x = 2, t = 7J-/2, and when x = 0, t = 0. J ,-.2 /<»/! a; 8 V4 - x 2 ,a . f- 1)S lg f* Vx^ite ^xVWVxax. I ^ ».J X — ir — 11. I .. »8 x s C?X , p 8 X(te i 2 . r 1 ^ a,, r -^ . 5- I . ■ J l xV5x 2 +4x+l Jo Vl6 — x 2 ^3 -y/1 _1_ /g 2 _i n a Va 2 X 2 6 xdx ,„ /- 2 dx 21. / dx. 13. / J a X 2 X 3 4. r b xdx f ' "*<> Vl+Tx '"'- 1 xV3x 2 -2x-l 2 ,, ,.5 3x— 1 22. J V 2^ "^-xVa^i^T" M .r" v „ :_--,,,. ,1 o-dx -- ^ 22. I , -4z s + 6a: + 3 ., 30. I — — dx. 22. / ^ rix. • ; V2x 2 + 6x + 5 ■' V5-4x-x 2 Integrals of the form J — 7 ^. — ^^- can be reduced + A„ a + b sin a; + c cos a; r dx to the form I ; ■ b y means of the substitution b = m sin a, J a + b cos x c = m cos a, where m = Vi 3 + c 2 . The actual form is then rfa; / Apply this to the next two examples. a + m cos (a; — a) 23. f d ± 27. f dX J 6 + 3 sin x + 4 cosx (1 — x 2 ) Vl + x 2 24. f ^ 28- f (2x'+3)dx J 10 + 12 sinx + 5 cosx "' * J (x 2 + (x 2 + 2x+ 5) 2 25, ' ^ J (1 + 4x)6/ 2 • 29. J o arc sin yj—ta: dx (1 + x 2 ) Vl-x 2 + dx 26 SPECIAL METHODS OF INTEGRATION 129 EXERCISE LXXXVII Integration by Parts. This device is applied to the integration of functions which may be considered as the products of two simpler functions one of which must always admit of inte- gration. The one which admits of integration is combined with the differential of the variable (dx) and denoted by dv. The other is denoted by 11. Then I vdv — uv — I vdu. Example: | x logxdx. Here there is no doubt that u = logse, do = xdx. . " . j x log x dx = \ x 2 log x — j \xdx = \x^ log x — \ x 2 . In case both factors of the product are integrable, then we select as part of dv that one which becomes simpler, or at least no more complicated, by integration. If the product consists of an algebraic factor and a trigonometric or exponential one, u is always the algebraic factor. Sometimes it may be necessary to repeat the process of integration by parts in the second integral until the new integral assumes a standard form, as in examples 7, 8, 33, 34. Certain types of products are also integrated by assuming first one factor and then the other as the integrable part, thus giving rise to two equations from which we eliminate the second integral, thereby solving for the original integral. This method is regularly employed if one factor is exponential and the other trigonometric, as in examples 37-44; also in certain trigonometric products of functions of different angles, as in examples 9-12. 1. fxsin2xdx. 5. fxsin 2 x/2dx. 9. j sin x cos 3 xdx. 2. fx cos 3 xdx. 6. jxcos 2 x/2dx. 10. f sin3xcosa;da;. 3. fxseo 2 2xdx. 7. ( x 2 sin 2 xdx. 11. f sin x sin 3 xdx. 4. fxcac 2 3xdx. 8. f x 2 cosx/2dx. VZ. f coszcos3xdx. 130 PROBLEMS IN THE CALCULUS 13. fare tan (x/3)dx. 17. Tare tan (1 — i) da;. 21. f 14. fare sin 2xdx. 18. /x arc sin (1/x) < 15. J arctan(2/x)dx. 19. f x 3 arc tanxdx. arc tanxdx 14. farcsin2xdx. 18. / x arc sin (1/x) dx. /-xarcsinxdx &&. J - — ■ J Vl - x 2 16. fare sin Vx/idx. 20. j x 8 are sinxdx. 'arcsinVxdx 23 / arctanVxdx 24. •/- Vl-x 25. f log(l — Vx)dx. 26. flog(- + lW. log (2 + x) dx 27 28 log(l + x 2 )dx 34. fx 8 e 2 *dx. 35. f(e*+x) 2 dx. 36. f(2*+x 2 ) 2 dx. 37. Ce 2x cosSxdx. 38. fe 3a: sin2xdx. 49. fixe?/ (I + x) 2 ]dx. 50. f(x 2 /e*)dx. 51. f log(l + 4x 2 )dx Jl/2 52. flog(a + hx 2 )dx. 39. /V' 2 sin2xdx. 40. fe 2 *cos(x/2)dx. 41. I cos^x ■e~ x dx. 42. J e-*sin3xdx. 43. I (F' a cosxdx. 29. | sin x log (1 + sinx)dx. 30. J sec 2 x log (sec x — 1) dx. 31. f(l-x)e«+ 2 *)dx. 32. f(l + 2x)eO-*)dx. 33. fx*e?> x dx. 44. J e 2 sinxdx. 45. |see 3 xdx. 46. j csc s xdx. 47. | tan 2 xsecxdx. 48. |see 6 xdx. 53. f log(4x 2 -l) 4 2x + sin2x dx. cos 2 x 54. f Jo 55. f xlogV9 + x 2 dx. Jo /: 66. I e 2 sin2xdx. SPECIAL METHODS OF INTEGRATION 131 EXERCISE LXXXVIII Miscellaneous Examples. Review. The following list of examples is inserted here for the purpose of reviewing the various methods of integration which precede, the determination of the method in each example being left to the student. When an example admits of more than one method of solution, the student should in each case use the shortest and least cumber- some one. The integrals may be evaluated between limits if desired. A few limits are suggested. r arc sin x dx r lldx r dx '* Vl-x 2 'J V3x 2 + 3x + l ' •> (* 2 + 4 ) 2 = • 9. |x 2 sin-dx. 16. | sin 6 -dx. x V2 x 2 + 3 x + 1 J 3 J 4 3 r (5x + 2)dx iQ /.s xdx ^ ,o dx 'J4x 2 + 4x + 5 ' Jb -^~l' ' J-3 x 2 + 4x+ 8' (5x + 2)dx 1 1 ^ (x — 1) dx i a T *" e* — 16 er x x 2 arc tan xdx. (2x + 5)dx af) f 3dx 3ec 8 x(Jx tan 4 x dx 4 r (6x + 2)ax u /- (x-l)cte lg /> 'J V4x 2 +4x+5 - Jx 2 (x 2 -2) "J g ^sin^xdx l2 r 13dx 19 f ^ COS 5 X J 1 + X + X 2 t/ 6. f (e *-e- 2 *) 2 dx. 13. f( 2 * + 5 )^ , 20. f J J Vx 2 +5x * / xV4-log 2 x „ /* . .x .X , /. ,- „, /-sec 8 xdx 7. J sm 4 -cos 2 -dx. 14. JVx(l + x 2 ) 2 dx. 21. J „ r 1 dx „. r dx 2. I 24. I == ^_ Jo9x 2 -6x + l J xV6x 2 -5x + l „ C 81 dx ' J x*(x 2 -9)§' M-JO- + x)cosvxdx. 26 ' f +xS(ix - 30. fx 4 (logx) 2 dx. 34. rsin2x.e™^dx. f (x 2 -x)dx ,- Varctanxdx 35 _ f x VT^ J x 2 + 4 ' dl, J TT^ J 2g Wx-sin2x)dx 3g /- (7x-2)dx _ 36. J J x 2 + cos 2 x •/ V7— 2x 2 29. C—to-L. 33 f (x 2 + 14)dx 37. J J xVl-x 2 'J x 4 -17x 2 +16 x 2 sin -1 xdx. dx 132 PROBLEMS IN THE CALCULUS r n : , M „ r (1 + sinx)dx p - x 38. I cosxVl + smxdx. 46. | — 54, I e a cos~dx J J sin x (2 + cos x) J 2 39. fxlog(x 2 + l)dx. ,„ C (x + l)dx 56i r^. w -fM=> 48 ./fc& *jt'-^. '-s J Vx , x 2 + 4dx K „ /"xlogxdx /"Vx ;! + 4dx ._ /-xlog J--*r- 49.r< a + 1 )f. S7 'ivr- 17 (x - 2)* 41 ^ x 42, ' r "* ' ' 58 r (^ + i)dx 7 x(Vl + x-2) 50.j"sin^dx. '^Vx(x+1)' 43 '/xSfiS%' si. JV/V*TT)(fc. 59 -/ (2 2 ~ + Z x^ - 44 r / arc sin xdx . r (x + l)dx 6Q /•» xdx J \ 1-x 2 ' J z 2 (x 2 + 2) ' "Jo V4^^' ^ r rC" + sin3x)dx /-i , /• „x 45 '/ 3x 2 -2cos3x - ".j^VT^. 61,/cos^dx. (x 3 + x 2 + x-l)dx 7n ^ 4dx ]/ 3X 2 63 /■ (x" + x-= + x-l)dx /• 4dx J x 2 (x 2 + l) '~'x 2 Vl + 2x- 71. / 2x arc tan xdx. x 2 -5x + 6 J 64. f x 2 arc sinxdx. 72 - f» x b 2 ^dx. „ /"(2x 8 -7x 2 +l)dx 6g W6 + 2x 2 -3x3) d x 73. J^ 28 _ i . X4 + 3X2 74 r (2x 3 + 6x-3)fc 66 f < a: + 1 ) fe J l-2x J (x + 2) 2 (x + 3) _. /\ ,„ . v ' v T ' 75. I tan 5 2xdx. „„ rx 2 arc tan xdx ^ 67 -J— rr^i — 76. r^ x 2 + 2 x — 3 /x-= arc tan xdx ^ 1 + X 2 76 f (x 3 -3x 2 -5x + 5)dx J x 2 + 2x-3 68. f ?-= __ />/sin^ cos0\,„ . J X 4 + X 2_ 2 77. /(__ + _^W -' \cos 2 sin 2 0/ „- < " (l+sin2x)dx f „ J C os 2 2x J arc cosa; anp r , , x . ' -H / x m (a+bx n y- 1 dx. np + m + lj (np+m + 1) (3) j x m (a+bx n ydx x m + 1 (a + bx n Y+ 1 (np+n + m + l)b r ^ , (m + l)a (m + l)a J V T ' (4) fx"{(a + bx n ydx — x m+1 (a + bx rt ) p+1 np+n + m + ir , , = L_I 1 \-JU—-L — Z_ / x m (a+bx n y +1 dx. n(p+l)a n(p + l)a J The student should observe just how each of these formulas simplifies the integral, as well as the restriction placed upon the original integrand before one of these formulas can be applied. The collected results follow : (1) diminishes m by n if np + m + 1 =£ ; (2) diminishes p by 1 if np + m + 1 =h ; (3) increases mbymifm + l^O; (4) increases p by 1 if p + 1 =£ 0. In the cases excepted above, the quantities can be integrated at once by more elementary methods. 136 PROBLEMS IN THE CALCULUS In each of the following examples the student should note carefully the numerical equivalent Of m, n, and p and should ascertain which formula or formulas to use so as to reduce the same to a standard integral. This exercise may be made an oral one throughout. If the reduction formulas fail, try some other method. . C n r- : ■, .„ r x s dx „„ n x 3 dx 1. / i 5 Vl + x*dx. 10. I 19. | •' ■' Vl+xe J (l + x 8 )f 2. / — r x n )fdx. Wa*-x 3 ^ J Vx^R ^ 22. f ^ x Jl dz . . A r x»dx J x rVl+x i dx li - f 77-, — ius" J 1 J{1+X) 23. f—^ r x*dx 1 S .J(1±|)!^. J *V-*)i r ' 16 r (i + x 8 )fdx -/ x* *"• 7. |x 5 Vl-x'dx. lb, J ~ J 25. f ta S.JxVTT^dx. n.Jx^l + x^Idx. ' J x8(4-x*)l /• (1 + xrfdx 18i f Vl + x 8 tZx . 26 rjA^_ ' J x 6 ' J x 18 " ^ •> V4 + x s EXERCISE XCII Integration by the Aid of Tables. Most of the examples of the following list are integrable by one or more of the methods already studied. They are designed, however, to be integrated here by the aid of an abbreviated table of inte- grals, such as is contained in many textbooks. They should be sufficient to acquaint the student with the range and content of his table, and to aid him in becoming accurate in its use. SPECIAL METHODS OF INTEGRATION 137 ■/: dx xVl-3x 2. J sin 2 - dx. 3. | e 2 sin 3 xdx. 4. | e 2 * cos 3 xdx. dx Vx 2 + 2x-3 dx Vx 2 + 2 x + 5 dx V3 - 2 x - x 2 dx • V3 - 4x - 4x 2 xdx */ '■/ «•/ 9. f J (4-5x) 2 10. |*V9-4x 2 dx. xdx "■/ V3 + 6x-9a 12. r- , ___ 17 x V a 2 + x 2 13. J x 4 logxdx. 14. J x 2 cos 2 xdx. 15. c-?* 16. f-^_L : - ''.f ax 4 dx (6 - ex 5 ) 2 ' /-- 2 x + 3 Vx 2 "^ 49. r./i±* ' J \2 - si '4 + x 2 3 + x 2 dx xVx + 1 18. J x sin 2 xdx Vl-x 33 21 dx. dx (x - 1) Vx + 2 /dx 3+^ 22./ 23./ cosx dx 2 + sinx (x — 1) dx Vx + 1 24. fVl-2x-3x 2 dx. 25./ 26./ 27/ 28. f (x + 2) dx Vx 2 + 9 dx (x + 2) Vx + 3 dx x 2 V9x 2 -4 dx V4-9x 2 cos 4 xdx „„ /"COS 1 !* 29 - / — ^-r- ./ Sltf X dx 30./ 31. 32 xV9 + x 2 I x 2 log 2 xdx. /cosxdx sin x log sin x • xdx. s1 ' J /; dx x 2 Vl + 3x 34. fx s (l + x 2 )tdx. *x 2 dx 35. f- 36. f- ./ 3 + cos 2 x (4j/-5)dy dx 37 38. Vy 2 + 2 y + 5 (2 x + 5) dx Vx 2 + 4x + 3 39. j sin 2 xcos 3 xdx. „„ /-sin 5 x , 40. I dx- J cos 2 X 41. f dx . " V4 x — x 2 ,„ /■ e=dx 42. I — • "^ V3 + e x — e 2a! 43. |x 2 sin-dx. J 2 • • cosxdx. „„ r^/2ax 2 — x i , 50. I dx. J x 5 52./ 53 14 r x 2 fjv. 1. I (x 2 + 16«dx. -2 dx 14. / — Jo ; 8. f — Jix(4 + 3x 2 ) 3 Jo 2 + cosx 2. f xVl + xdx. , 1K r 2 Vx 2 + 5 2 . 3. The formula for the velocity, ds/dt, of a freely falling body is ds/dt = i) + gt. Taking g = 32 ft./sec. 2 , and v = 16 ft./sec, (a) calculate the distance s traversed in the fifth second (Hint. The limits of t are 4 and 5) ; (b) calculate the total distance and the average velocity during the next 5 seconds. 4. The angular velocity, d0/dt, of a wheel is a + kt. Assume w = iir radians per second, and k = w/6 rads./sec. z Calculate the number of revolutions the wheel will make during the second minute. Note. In the formula, t is the time in seconds, 6 is the angular distance in radians, and u the initial angular velocity in radians per second. 5. A particle moves on a path along which the slope, dy/dz, is always equal to x 2 . How much does y change when x changes from 3 to 6 ? 6. The formula for the specific heat of iron for ordinary temperatures up to 200° C. is s = .1053 + .000142*, where t is temperature. Calculate the amount of heat, Q, required to raise 100 kg. of iron from 50° to 100° C, given dQ/dt = Ms, where M is the amount of iron in kilograms, and Q is in large calories. 7. A boy starts a slide on ice with a speed of 10 mi. per hour. He stops in 3§ sec, having covered 26| ft. If his speed at any time is given by ds/dt = — 32 /it + v , where v is his initial speed in feet per second and p is the coefficient of friction between the boy's shoes and the ice, calculate /*. 8. By Joule's law the quantity of heat, Q, generated in a conductor of constant resistance It is given by the equation dQ = kRPdt, where i is the current. If i is periodically variable of period P, according to the law i = i sin 2 irt/P, where i is the maximum value of i, calculate the heat generated in a period (k has a value of nearly I if Q is in calories). 140 PROBLEMS IN THE CALCULUS 9. The work done by an electric current is given by W = Eit, where E = electromotive force, i = current, t = time. In an alternating current E = E sin (kt + and 8 are the phase angles (constant), (a) Calculate the work done in a period, (b) Show it is a maximum when = 8. Hint,. dW=Eidt. 10. The theorem of Torricelli states that if the water in a vessel is flowing out through a hole in the bottom of the vessel, the velocity of the water as it leaves the orifice is exactly the same as that of a body which has fallen freely under gravity through a distance equal to the vertical distance from the surface of the water to the opening (at any instant v for a falling body is v = ^/2gh). He nce, if the original depth of the water is a, the velocity of the water is V2 g (a — x) . Therefore, if / is the area of the cross section of the opening, and F, of the surfac e of the wa ter, the rate at which the surface falls is. given by dx/dt=fV2 g(a — x)/F. (a) Calculate the time required to empty a vertical cylindrical tank of height 10 ft. and radius 2 ft. through a hole of radius 1 in. in the bottom. (b) How long does it take to lower the water 1 ft. when full ? (c) How long to empty one half of the water ? (Take g = 32 ; t is in seconds.) 11. Using the formula of example 10, calculate the time required to empty a hemispherical bowl of radius 4 ft. if the hole in the bottom is of radius 1 in. 12. For accuracy in actual practice the formula of example 10 is modified by a factor ^, called the coefficient of flow, which depends for its value, on the shape of the vessel, the shape of the opening, and other factors. The value of /* ranges from .6 to nearly unity. This corrected 1 c Fdx formula may be written dt = = | — = . (a) Calculate the gen- fifV2g J Va-x eral expression for the time required to empty a hemispherical bowl of radius B through an opening in the bottom of cross section /. (b) Evalu- ' ate this for the case B = 4 ft. and the opening is a 2-inch circular hole for which /j. = .8. (Take g = 32.) (c) Calculate the time required in (b) to empty one half of the water. 13. Torricelli's law is also applied to finding the rate at which water will escape through a vertical floodgate or spillway in the face of a dam. (a) If the floodgate is rectangular, a feet long and b feet high, and the top is h feet below the surface of the water, calculate the flow, Q, per second. The factor n must be determined by experiment {Hint. Q = /j,aV2gj x$dx\. (b) Evaluate (a) for the case where the gate is 8 ft. long and 5ft. deep and the upper edge is 4ft. below the surface. SPECIAL METHODS OF INTEGRATION 141 14. If in example 13 the. escape is over an open weir a feet wide and 6 feet deep, how fast is the water flowing out when the depth of the over- flow is h feet ? Cheek your answer by the formula Q = § pA V2gh, where A is the cross section of water in the weir and h the depth of the overflow. 15. Calculate the width of » weir required to allow 1000 cu. ft. per second to pass when the depth of the overflow is 10 ft. (Take p = .6.) 16. In organic chemistry there are various applications of the calculus. For example, in the breaking up of cane sugar into dextrose and levulose, in a weak solution, the rate of inversion at any moment is proportional to the amount of unconverted sugar. If a is the original amount of sugar, the inversion formula is dx/dt = h(a — x), where 1c is a constant, and x is the amount already inverted at the time t. Assume t in minutes, as the process is slow, (a) Calculate the time to convert an amount y calculate the velocity with which the particle returns 2 c 2 g — c 2 i) 2 to the earth and the time required in descent. Compare these results with v and the time of ascent found in example 22, (b). 25. Assume the law that the rate at which water runs out of a hole in the bottom of a tank is proportional to the square root of the height of the surface of the water above the hole. If the tank is of uniform horizontal cross section (as a cylinder, etc.), show that if half the water runs out in 30 min., it will all run out in about I hr. 42 min. 26. A vertical cylindrical tank of height /( feet and cross section F is filled with water, which is escaping through a hole of area / in the bottom. If a pipe is emptying Q t cubic feet per second into the tank, how long will it require for the surface of the water to drop k feet ? (See example 10.) CHAPTER XIV INTEGRATION A SUMMATION PROCESS. APPLICATIONS GEOMETRICAL EXERCISE XCV Plane Area. Rectangular Coordinates. In each of the following examples the student should sketch the curve and rather than substitute in the formula he should draw one of the charac- teristic rectangles and write down its area. Adding up these rectangles and calculating the limit of the sum as the number becomes infinite, he obtains the required area. This adding is ac- complished by integration. The characteristic rectangle may be selected in a vertical or in a horizontal position. That position which leads to the simplest inte- gral should be chosen. The notion of the " summation of elements " is always preferable to the use of formulas, owing to its wide application in geometrical and mechanical problems. Draw a rough figure with a typical element for each problem. (See the adjoining figure.) 1. For the following curves calculate the area in the first quadrant lying under the arc which extends from the !/-axis to the first intercept on the x-axis. (h) y = x s -8x 2 +15x. (a) y = 4 — x 2 . (b) y = 4 + 3x-x 2 . (c) y = 5x — x 2 . (d) y 2 -= 9 - x. (e) » 2 + y = 2 - x. (f) y*-2y = 3-x. (g) y = 9x — x a . (i) 2/ = (3-2x) 2 x. (j) y = x i -20x 2 + 64. (k) y = x 2 (4-x 2 ). (1) x = (2-j,)( 2/ + l) 2 . (m) ?/ 2 = (1 - x) 8 . (n) y- = x 2 — x 4 . 144 (o) 2/ 2 = x 2 -x 3 . (p) 2/ 2 = x(l-x) 2 . (q) y = e*/ 3 sin 2 x. (r) y = e x/2 cos3x. (s) y = 4x 2 — x 8 . (t) x(j/ + l) 2 = 4. (u) y — e- x/i o,os%x. •INTEGRATION A SUMMATION PROCESS 145 2. Calculate the area bounded by the x-axis, each of the following trigonometric curves and two ordinates separated by a distance equal to a period of the curve. The student should observe that areas below the x-axis obtained by integration are negative, and hence if we calculate the area under one period of the curve, using the entire period for the interval of integration, the result will be the algebraic sum of the areas above and below the axis. To obtain the actual area we must decompose the interval. Thus, if in the interval AB the curve crosses the axis at Xj, we must evaluate the two integrals I 'ydx and f ydx, then add them numerically, independent of sign. The symmetry of the curve should also be referred to in selecting the limits. (a) 2/ = sin2x. (e) y = sin 2 x. (i) y = sin 2 x — cosx. (b) y = cosx/2. (f) y = cos 2 3x. '(j) y = cos2x + 2cosx. (c) 2/ = costtx. (g) y = sin 2 x + sinx. (k) y = 3 cosx — 4sinx. (d) y = sin 2 7rx/3. .(h) y = 2sinx + sin2x. (1) y = 4 cosx + 3sinx. 3. A square is formed by the coordinate axes and the point (1, 1). Calculate the ratio of the larger to the smaller of the two areas into which it is divided by each of the following curves : (a)?/ = x 2 . (e)x 2 + ?/ 2 = l. (i) yx + y + x - 1= 0. (b)y = x s . (f)x 3 + 2/ 8 = l. h)v = e-*. (c) y = x 4 . (g) x^ + yi = 1. (k) y = sin ttx/2. (d) y 2 = x s . (h) xi + y§ = 1. (1) y = tan7rx/4. 4. Calculate the area which the given line cuts from the given parab- ola by subtracting two areas. (a) y = x; y = ix-x 2 . (d) 2y = x + 1 ; y 2 = x + 1. (b) y = x+ 7; ?/ = 9-x 2 . (e) iy = 4- x ; y 2 = 4 - x. (c) y = x + 2 ; y = 4 - x 2 . (f ) x + y = 1 ; x* + y* = 1. EXERCISE XCVI Length of a Curve. Rectangular Coordinates. By integrating the expressions for the differential of are in rectangular coordi- nates we obtain s = J ^\+{dyjdxfdx, or it is sometimes simpler to integrate the second form, s= J ^Jl-\-(dx/dy) 2 dy. 1. In the following curves calculate the length of the arc joining the two given points. (a) y 2 = X* 5 ; (0, 0), (f , 5 VfT/27). (c) y 2 = 4x - x 2 ; (0, 0), (l, VI). (b) 2 /=|(4-x)f;(0,J^),(4,0). (d) 3j/ 2 =8xS; (0, 0), (l, 2 V2/3)- ' 146 PROBLEMS IN THE CALCULUS * (e)2/2 = 2 0x; _ (f)6y = x 2 ; (g)3y = x*; (0, 0), (4, 4 V5). (0, 0), (4, 8/3). _ (0, 0), (1, 3). (h) y =V81ogx; (1, 0), (V8, V21og8). (i) y = logseex; (0, 0), (ir/3,log2). 2. (a) 3x 2 = y s ; (l/VS, l), (40 V15/3, 20). (b) 4x = 2/2-4; (0, -2), (0,2). (c) x = (e»' 2 + e-v' 2 ) from y = to y = 2. EXERCISE XCVII Volume of a Solid of Revolution. Consider the section of a solid of revolution made by a plane perpendicular to its axis. Denote the area of this section by A. If we multiply this by da, a differential distance along the axis perpendicular to the plane, we obtain the volume A da, of a characteristic cylindrical element. Using the notion of summation gives the integral / Ada, which is the required volume when evaluated for the proper limits. This method of " setting up the element " should be applied to each of the following examples. Draw a rough figure showing the cylindrical element in each case. The student should observe that A in general has the form 77-r 2 , where r is the distance to the axis from a point in the generating curve. In case the area of the section is ring- shaped (bounded by two concentric circles) A takes the form ir{r\ — rf). When the x-axis is the axis of revolution, the radius reduces to the ordinate of a point on the curve, giving the more familiar formula tt I y 2 dx, or using the y-axis as the axis of revolution, we obtain tt I x "V- J c 1. Calculate the volumes obtained by revolving the areas indicated about the x-axis. The boundary curves and lines are given. (a) y = X s , y = 0, x = 1. (d) y = sin 2x, y = 0. (b) y 2 = x 3 , y = 0, x — 5. (e) y = x 2 — 4x, y = 0. (c) y = sinx, y = 0. (i ) y = e ax , y = 0,x = 0,x = b. INTEGRATION A SUMMATION PROCESS 147 2. In these examples revolve the areas about the y-axis and compute the volumes. (a) x 2 = 4y - 2y 2 , x = 0. (c) x 2 - y 2 + 4 = 0, x = 0, x= 2. (b) 4y-y 2 + 4x = 0,x = 0. (d) x 2 = y 3 - 4 y. 3. In these examples the area between the parametric curve and the x-axis is to be revolved about the x-axis. (a) x = 2 - t, y = t 2 - 4. (c) x = t 2 , y = 4t - S 3 . (b) x = £ + 1, y = t 2 — it. (d) x = t 3 /3, y = 3t-t 2 . 4. Calculate the volume of the solid generated by revolving about each of the following lines the area which they cut from the corre- sponding curves : (g) V = x ; y = x 2 - (h) y = x ; y = 4x — x 2 . *Si) y = x + 7; y = 9-x 2 . (j) 2y-x-l = 0; y 2 = x + l. (k) X + y = 1 ; xi + y* = 1 . (1) x + y = 5; xy = 4. 5. By revolving it about the axis indicated, find the volume generated by the finite area bounded by the following pairs of curves : (a) y = 3; y = 3 + 2x-x 2 . (b) x = 3 ; x = 4 y — y 2 . (c) x = 3; y 2 — 2y + x — 3 = 0. (d) yx = 4 ; y 2 = x 3 . (e) x = 5 ; x 2 - y 2 = 16. (f ) y = 1 ; x 2 + y 2 = 4. 2. (a) y-axis, xy = 4. x + y = (b) y-axis, y 2 = x, x + y = (c) x-axis, y = 2 + x y = 2 — x. -axis, x 2 — 4 x + y = o. (e) /-axis, y 2 = X s , y 2 = 2 — x. (j (d) ^o. (f) y-axis, y 8 = 27 x, 3x = 4y — y 2 . (g) x-axis, y = (4 + x 2 )/x. y = 5. (h) x-axis, y =x 2 , y = 4x — x 2 . 6. (a) The area bounded by the curve y = sinx and the lines x : and y = is revolved about the x-axis. Find the volume. (b) The area bounded by the same arc of y = sin x, the y-axis, and the line y = 1 is revolved about the y-axis. Find the volume, (b) is a good illustration of a case where it is best to replace the differential, that is, to change the independent variable to x. 7. Given the curve x = i 2 , y = 4 1 — i s . Find (a) the area of the loop and (b) the volume generated by the area inside the loop when revolved about the x-axis. (i) x-axis, y = x 3 , y 2 = x. y-axis, V = x 3 , y 2 — x. (k) x-axis, y = x 2 , y 2 = x. (1) x-axis, y = sin ttx/2, V = x 2 . tt/2 148 PROBLEMS m THE CALCULUS 8. Eind the volume of the solid generated by revolving about the x-axis the area bounded by y = e ax and the coordinate axes. 9. Revolve the area between the two parabolas y 2 = 4x and j/ 2 = 5 — x about each axis and calculate the respective volumes. 10. Eevolve about the polar axis the part of the cardioid p = 4 + 4 cos 6 between the lines 6 = and 6 = tt/2 and compute the volume. EXERCISE XCVIII Area of a Surface of Revolution. Regarding the.axis of x as the axis of revolution, the characteristic element becomes the lateral surface of a frustum of a cone whose slant height is the differential of arc and whose mean radius is the ordinate to the curve ; hence the element of surface = 2 iryds. Using the «/-axis as the axis of revolution, the formula becomes 2 irxds, or, generally, 2 irrds. In any case, ds may be replaced by either of the forms Vl + (dy/dxf'dx, or Vl + (dx/dyfdy, according as the one or the other leads to the simpler form to integrate. The fact that the, y-axis is the axis of revolution does not demand the second form, as many students suppose. 1. In each of the following curves calculate the area of the surface generated by revolving about the x-axis the arc connecting the given points : (a) 92/ = ^; (0, 0), (2, f). _ (d) y* = 24- 4x; (3, 2V2), (6, 0). (b) 2/2 = 4x . (o, 0), (3, 2 V2). (e) x2 + 2/2 = 4 ; (l, V3), (2, 0). (c) 2/ 2 = 9x ; (0, 0), (4, 6). (f) 6y = x* ■ (0, 0), (4, §). 2. Calculate by integration the surface generated by revolving about the x-axis one arch of the curve y = sin x. 3. Calculate the entire surface of a solid generated by revolving about the x-axis the area bounded by the two parabolas y 2 = 4x ; y 2 = 3 + x. EXERCISE XCIX Plane Area and Length of Arc. Polar Coordinates. The student should plot the polar curve for each example to assist him in/ determining the proper limits for the variable angle 6. ATchar- acteristic element of area, the area between two radii vectores with an angle dO between them, — hence a circular sector of area INTEGRATION A SUMMATION PROCESS 149 \p i dQ, — should be marked on each figure, and the notion of summation should be applied as before. The element of arc in polar coordinates is Vp 2 + (dp/dOydO, therefore !=( 2 ^/p 2 +(dp/d8fd8. (a) p 2 = 4sin2 0. ( e ) p = 1 + cos #. (b) p 2 = 9cos2<2. (f ) p*=sinfl + l. (c) p = cos 2 0. (g) p = 3 — sind. (d) p = sin 3(9. (h) p = 2-cos#. (m) p = sin 3 + 2. <°) (n) p = 2 + cos 3 6». (P) 1. Calculate the total area inclosed by each of the following curves : ( i ) p = sin 2 0/2. (j) p = cos 2 0/2. (k) p = J + cos 20. (\) p = sin20- },. ( o ) p = cos 3 — cos 0. (p) p— cos 3 — 2 cos 0. 2. Calculate the ratio of the area of the larger to the area of the smaller loop of the following curves : (a) p = I + cos 0. (c) p = i + cos 2 #. (b) p = sins 0/3. (d) p = I + sin 3 0. 3. Calculate the area which the curves in each of the following pairs have in common. Each example will involve the evaluation of two definite integrals having one limit in common determined by the points of intersection of the two curves. The sum of these two integrals will give the required answer. p = 3cos#, (i) p = V2sin#, p = 1 + cos 0. p 2 = cos 2 0. p=Vh~sin0, (j) p =V2cos0, p = l+cos#. p 2 =V3sin20. p=l, (k) p = (cos0)/V§, p 2 = 2_cos 2 0. p = cos 2 0. (h)p=V6cos#, (1) p=V273sin20, p 2 = 9 cos 2 0. p 2 = cos 2 0. 4. Find the total length of each of the following polar curves : (a) p = cos 2 0/2. (b) p = sin 8 0/3. (c) p = sin 4 0/i. (d) p = 1 + cos 0. 5. Given the parabola p = 2/(1 + cos#). (a) Find the length of arc to the right of the line = 7r/2. (b) Find the area cut from the parab- ola by a line through the pole perpendicular to the polar axis. 6. (a) Find the length of arc of the entire cardioid p = 2(1— cos#). (b) Rotate the area bounded by this cardioid about the polar axis and calculate the surface generated. (a) p = cos 0, (e) p = sin 0. (b) p 2 = cos 2 0, (f) p 2 = sin 2 0. (c)p = l, (g) p = 1 + eos#. (d) p = sin 0, (h) p = 1 — cos 0. 150 PROBLEMS IN THE CALCULUS EXERCISE C. Volumes of Miscellaneous Solids. Just as we obtained the circu- lar sections of the solids of revolution in a preceding exercise, so we may obtain various sections — elliptical, triangular, rectan- gular, etc. ■ — of miscellaneous solids, and multiplying the area of such a section by the differential of that variable, say x, to whose axis the section is perpendicular, we obtain the volume of a characteristic element. It *s essential that the area of this element be a function of x. Applying the notion of summa- tion we arrive at the volume integral for any particular solid. For example, consider the volume cut from the paraboloid 4 y 2 -f- 9 z 2 = 4 x by the plane x = 9. Any element perpendicular to the x-axis is an elliptic cylinder of semiaxes Vx and § Vx, respectively (and therefore of area §■ 7rx), and of thickness dx. Hence the volume = I -| irx dx = 27 ir. 1. Calculate the volumes bounded by the following quadric surfaces and the given planes : (a) z = x 2 + iy 2 ; 2 = 1. (d) 25y 2 + 9z 2 = 1 + x 2 ; x-2 = 0. (b) 4x 2 + 9z 2 + ?/ = 0; 3/ + 1 = 0. (e) x 2 + 4y 2 + 9z 2 = l. (c) x 2 + 4?/ 2 = l + z 2 ; z±l=0. (f) z 2 = x 2 + 9 j/ 2 ; «.±1 = 0. 2. The double ordinate of the ellipse x 2 + iy 2 = 4 serves respectively as (a) the side of a square, (b) the diagonal of a square, (c) the base of an equilateral triangle, (d) the altitude of an equilateral triangle, (e) the diagonal of a regular hexagon. Calculate the volume of the solid generated by the area of each of these figures as it moves, the plane always remaining perpendicular to the x-axis. 3. Given the parabola z = 4 — x 2 in the XZ-plane and the circle x 2 + y 2 = 4 in the .XT-plane. From each point on the parabola lying above the circle two lines are drawn parallel to the FZ-plane to meet the circle. Calculate the volume of the wedge-shaped solid thus formed. INTEGRATION A SUMMATION PROCESS 151 4. The double ordinates of the circle x 2 + y 2 — 25 serve as the bases of isosceles triangles with a constant altitude of 10. Calculate the volume of the solid made up of all such triangles standing at right angles to the plane of the circle. 3. The plane sections of a solid by planes perpendicular to the x-axis are ellipses whose major and minor axes are the double ordinates of the curves j/ 2 = x and z 2 = x s respectively. Calculate the volume cut from the solid by the plane x = 2. 6. A football is 16 in. long and a plane section containing a seam is an ellipse the shorter diameter of which is 8 in. Find the volume, (a) if the leather is so stiff that every cross section is a square ; (b) if the cross section is a circle. 7. An isosceles right triangle moves with its plane perpendicular to the XT-plane and with its hypotenuse always parallel to the y-axis. One end of the hypotenuse moves on the line y = 3x and the other end on the curve x 2 + y = 0. Find the volume generated when the triangle moves from x = to x = 2. 8. The cap of a square pillar is of the form of the volume common to two equal semicircular cylinders of radius a, the axes of the cylinders intersecting at right angles. Find the volume of the cap. Hint. The horizontal cross sections are all squares. CHAPTER XV APPROXIMATE INTEGRATION — VARIOUS METHODS EXERCISE CI The Definite Integral. Approximate Evaluation by the Use of Series. Integrate the following functions by the method of Exercise LXXXIX. The interval of integration is in each case so selected that the resultant series will converge rapidly for both limits. The student should use a sufficient number of terms to have his answer correct to at least two decimal places. dx Vl-x 8 dx x* 1 sinxdx I. f Vl + x s dx. S. J" J. f v'l + ^dx. 6. f Jo J.i yY^T J. f 2 — y 1. I I dydx. 5. I I xdydx. 9. I- I y 2 dxdy. J° J ^/Ix J-lJix 2 -! Jo Vy i V2 /iSa:-! 3 3. /iV2 /iSI- X" _ Jo J*-, dydX - 6. r C +X ydydx. 10. f f ~ X xUydx. J-l J2x 2 -2 Jo Jx 2 8./ fj dydx. 7 . r f^ + 1 \ dydx . 11. rr^dydx. 4. I I dx + 2y* Jo Jo V2 T J. ^-^ dx = V 2 7r. Hence V = 4 v2 -rr. The limits of x and y were obtained from the curve of intersection of the bounding surface and plane. 1. Find by triple integration the volume of the tetrahedron formed by the plane x/a + y/b + z/c = 1 and the three coordinate planes. 2. Find the volume bounded by the planes z = 0, y = 0, z = x and the cylinder x 2 + y 2 = 9. 3. Find the volume in the first octant common to the two cylinders x 2 + y 2 = 4 and x 2 + z 2 = 4. MULTIPLE INTEGRATION 161 In the remaining examples the solids are completely bounded by the given surfaces and planes. Find the volume by triple integration. 4. y 2 = x, x + z = 1, y = 0, z = 0. 5. y 2 = x, x + y + z = 2, y = 0, z = 0. 6. y 2 + z = 1, x + y = 1, x = 0, z = 0. 7. x 2 + z = 1, 2/ 2 + z = 1, x = 0, !/ = 0, z = 0. 8. x 2 + j/ 2 - 2x = 0, 2x + z - 2 = 0, j/ = 0, z = 0. 9. y 2 - 2 x = 4, x + z = 1, z = 0. 10. x 2 + ?/ 2 = 4, z + y = 3, z = 0. 11. In cylindrical "coordinates the volume of a solid is given by the triple integrajT/l pdpdddz. Use this to calculate the volume of a cone of radius 2 and altitude 3. 12. Use cylindrical coordinates to calculate the volume of a paraboloid of revolution, the radius of the base and the altitude each being 2 units. 13. In spherical coordinates the volume of a solid is given the triple integral J ( I p 2 sin d8d ds ; but obviously the resultant pressure at right angles to the wind's direction is zero ; hence we need only consider the component of N parallel to the wind's direction, which is iVcos0. 7. Using the data of example 6, calculate the total pressure exerted by a 15-mi.-per-hour wind against a circular tower 20 ft. in diameter and 100 ft. high. (Assume k = .0804 and g = 32.) 8. The calculation of work done which is defined as Force times Dis- tance furnishes an example for summation. The work done in raising M pounds a vertical distance of h feet against gravity is defined as Mh foot-pounds. Calculate (a) the work done in pumping out a cylin- drical well of radius a and depth h feet ; (b) the same for a conical well 50 ft. deep, 4 ft. in diameter at the top, assuming that water weighs 62^ lb. per cubic foot. 9. A horizontal cylindrical oil tank is to be emptied by a pump the outlet of which is 2 ft. above the top of the tank. Calculate the work required. The tank is 40 ft. long and its diameter is 6 ft. (Assume the oil to weigh 60 lb. per cubic foot.) 10. The part of a cistern filled with water is of the form of the solid obtained by revolving x 2 = 7 — y about the j/-axis from y = 3 to y — 0, the positive end of the y-axis being downward. The unit is 1 yd., and the top of the cistern is 9 ft. above the surface of the water. Calculate the work done in pumping it out if water weighs 62 lb. per cubic foot. 11. When the limiting curve of an area has a simple polar equation, this fact can often be used in obtaining an integral over this area. For example, find the pressure against a vertical circular water gate of diameter a feet, the water being level with the top of the gate. (Draw a figure. Take the pole at the highest point of the gate and show that 166 PROBLEMS IN THE CALCULUS the element of pressure is dP = k ■ p sin 8 ■ pdpdd. Here our integral is a double integral, but simple.) 12. A horn-shaped solid is formed by rotating a variable circle C about a line in its plane. The nearer extremity of the diameter of 0, which is perpendicular to the axis of rotation, traces out a quadrant AB of a circle of radius a, and the radius of C is always k8, where 8 is the angular measure of the variable arc on AB. Find the volume. 13. The area of any curved surface may be found as a double integral. The element of surface is taken as that portion of the surface which projects an element of area on a coordinate plane, say dA = dzdy on the XF-plane. This gives dS = dx dy /cos 7, or, since cos 7 is a direction cosine, S — I I » II" ' i i- I \ 4. 1 . dxdy. The limits are determined by the pro- ■/.Mer*©' jection of the surface on the -XT-plane. Calculate the area of a hemi- sphere by this method. 14. Using the discussion of example 16 as a guide, calculate the area cut from the sphere x 2 + y 2 + z 2 = 16 by the cylinder x 2 + y 2 — 4x = 0. Calculate also the area cut from the cylindrical surface by the sphere. Hint. In the second case project on another plane, say the -5TZ-plane. 15. Find the area of the cylindrical surface x 2 + y 2 = r 2 between the planes y = mx and z = 0. 16. Calculate the area of the surface of the paraboloid y 2 + z 2 = 2 ox, cut out by the cylinder y 2 = ax and the plane x = 4 a. 17. In problems concerning the torsion of a circular pulley shaft, it is necessary to calculate the moment of the shearing force for a cross section of the shaft. Here the element is the moment of the force on a ring of radius & and infinitesimal width dx. Its area is dA = 2-irxdx. Call the shearing force F kilograms per square centimeter. Then the moment of force of the shear on this ring is FdA ■ x. Calculate the moment, of the shearing force if the radius of the shaft is a centimeters and F = F^x/a. F is the value of F at the periphery of the shaft and is to be regarded as constant. 18. Calculate the moment of the shearing force for a hollow shaft of inner radius b and outer radius a. 19. The results found in examples 17 and 18 measure the permissible turning moment which may be applied to a shaft by a belt or crank, pro- vided merely that F be the maximum allowable value of the unit shear- ing force in the material of the shaft. The value of F is determined by the elastic limit of the material and is an experimental constant. Which GENERAL APPLICATIONS 167 is the stronger, a solid shaft of radius 3 a or a hollow one of inner radius 3 a and outer radius 4a? Note that the solid shaft has the larger area of cross section . What is the relative strength ? (The two shafts are assumed to he made of the same material.) 20. The attraction between two particles is proportional to the product of their masses and inversely proportional to the square of the distance between them. Calculate the attraction of -a thin bar of length I and mass M for a small particle of mass m which is In the line of the axis of the bar and a units from one end of it. 21. Calculate the attraction of a wire bent in the form of a semicircle AB upon a particle of mass m at 0, the center of the circle. Hint. We need concern ourselves only with the attraction along OP, the radius of the circle which goes through the mid-point of AB, as the resultant at right angles to this is zero. 22. Calculate the attraction of a disk of radius a for a particle of mass m which is located h units vertically above 0, the center of the disk. Hint. Take as the element of the disk a ring of infinitesimal width with as its center. As in example 21, we consider only the component along the perpendicular. 23. A water reservoir is in the form of a hemisphere 20 ft. in diameter surmounted by a frustum of a cone 8 ft. high and 32 ft. in diameter at the top. Calculate in short tons the total pressure on the walls when the reservoir is filled with water weighing 62^ lb. per cubic foot. 24. The illumination at a point is inversely proportional to the square of the distance from the source of light, and directly proportional to the intensity of the source and the cosine of the angle between the ray of light and the surface illuminated. Find the total illumination on a cir- cular plane area of radius a due to a light of intensity I, fixed at a dis- tance h directly above the center of the disk. (Compare with example 22 on the attraction of a disk.) 25. A circle of radius 30 ft. is surrounded by a walk of width 10 ft. (a) Calculate the relative amount of light on the circle and on the walk when a light is suspended 20 ft. vertically above the center of the circle. (b) What is the ratio of the average intensity over the area to the maximum intensity ? 26. In determining the friction between a bearing and a journal in the case of a horizontal shaft, we take as the element of area of the bearing surface, ciA, an infinitesimal strip Ids or la dB. The element of friction, dF, is then the force of friction on dA. Hence, if I is the length of the 168 PROBLEMS IN THE CALCULUS journal, a its radius, and ix the coefficient of friction, dF = pndA or Pldad6, where p is the normal pressure per square unit of surface and is in general a function of 0. Here ^ may also be a function of 8, 8 being the central angle subtended by the arc of contact, measured from the vertical. In integrating, therefore, the limits of 6 are from to w/2 for one half the total friction. Assuming /t = .004 and pasa constant everywhere equal to 10 kg. per square centimeter, calculate the total force of friction F in the case of a journal 20 cm. in diameter and 25 cm. long. 27. Calculate F for the bearing in example 26, assuming p = /M —k cosd, where /* = .004 and k = .001. (It should be noted that the work done per revolution to overcome friction is 2ira ■ F.) 28. Assume the same dimensions of bearing as in example 26, but with p = p cos 8, where p = 10 kg. per square centimeter. 29. Solve example 27 for the case where p = p cos 8, and p = 10 kg. per square centimeter. 30. Solve example 26, assuming p = p cos 3/2 0, when p — 10 kg. per square centimeter. Note. Use approximate methods of integration, if necessary, in these problems. 31. The load resting on the bearing in problems of the type explained in example 26 is important and is to be found by integration. The element of load (that is, the load resting on the infinitesimal strip lad8) is the vertical component of the pressure on this strip, or lad8 -p cos0. Calculate the load resting on the bearing in example 28. 32. Calculate the load resting on the bearing in example 30. 33. Solve example 27, assuming that p = 10(1 — fc#)kg. per square centimeter if k = .2. Calculate also the load in this case. (In each of the examples of this type the student will find it desirable to express the relation between the frictional force F and the load P.) 34. When a vertical shaft revolves with a smooth, flat end turning in a shallow socket, we have a flat step bearing. Here the bearing surface is a circular area, with the axis of the shaft passing vertically through its center O. The element of area dA can be taken as a ring of infini- tesimal width with O as a center (see example 17). The force of friction dFon this element is pp dA, where p is the pressure per unit area on the step. Finally, the work done per revolution against this friction is dW = %-KxdF. Calculate the total work per revolution against friction if the radius of the shaft is a, /j. is a constant, and the pressure of the shaft is evenly distributed over the whole bearing surface. GENERAL APPLICATIONS 169 35. Suppose the vertical shaft in example 34 is hollow with inner radius 10 cm. and outer radius 20 cm. (a) Calculate W if /i = .04 and p is 2 kg. per square centimeter, (b) Calculate W if /t = .04 and p = k/x, where k is a constant. Evaluate the result for k = 30. 36. If the bearing surface of the vertical shaft of radius a in example 34 is any surface of revolution, the element of area is no longer 2 7rxdxbut 2 ttx ds, a zone of the bearing surface, where ds must be determined from the generating curve of this surface. The element of work of friction, dW, is then pdA ■ fi ■ 2ttx = 4:Tr'fi.px 2 ds. Verify this formula by the aid of a figure. Calculate the general expression for W when the bearing surface is a spherical zone of one base of generating arc a. Note. It is best to express x and ds in terms of K, the radius of the sphere, and 8, the central angle. The limits of 9 are then and a. Here p is the normal pressure per square centimeter and is to be regarded as a constant. 37. Calculate the load on the bearing in example 36. (Note that, as in example 31, the load is the sum of the vertical components of the normal pressure. Do not confuse B and a in these examples.) 38. Calculate the work of friction in a conical step bearing, the radius of the shaft being a. (Assume p constant.) 39. Calculate the work of friction when the bearing surface of the step bearing is a paraboloid of revolution with the radius of the shaft 8 cm. and the altitude of the paraboloidal pivot 4 cm. (Assume p = 30 kg. per square centimeter and /i = .05.) Calculate also the load on the bearing. 40. In general the question asked in several of the preceding prob- lems, namely, to find the load, is the converse of the problem in practice. We ordinarily know the load P and, assuming a law of distribution, cal- culate p . (a) Assume that a load of 1000 kg. rests on the flat step bearing of a hollow vertical shaft of inner radius 5 cm. and outer radius 8 cm. Assume further as a law of distribution p =p (l— T V X )- Find Po- (b) Assume that a load of 1000 kg. is supported by a journal,' as in example 26. If the bearing is 10 cm. long and the radius of the journal is 8 cm., calculate p if the law of distribution is p = p cos 6. (c) Solve (b), assuming the law p = p cos 3/2 6. 41. If a bearing of the type explained in example 36 is worn, the pressure is no longer evenly distributed, but tends to obey the law (1) px = k cos 6, where k is a constant and equal to p a a/cos a. Here a is the inclination of the profile to the horizontal at the periphery and a is the radius of the shaft. In example 36, where the bearing is a spherical one, a is equal to the central angle of the generating arc. 170 PROBLEMS IN THE CALCULUS (a) Calculate the work of friction for a spherical bearing on a shaft of radius a if the generating arc is a. (b) Calculate the value of p a if the load on the bearing is 500 kg., the radius of the shaft is 5 cm., and a = 45°, and evaluate the work of friction for this case if /j. = .04. 43. In applied mechanics cases arise where the total stress on a support can be found as an integral. A single example will illustrate the class. A lock wall 50 ft. wide contains a rectangular culvert 20 ft. wide whose axis is 20 ft. from the face of the wall. The resultant of all the forces act- ing above the plane of the culvert is vertical, its magnitude is 400,000 lb., and its action line lies 20 ft. from the face of the wall. Assume that the unit stress developed in the two columns into which the culvert divides the wall increases at a uniform rate from a minimum at the back of the wall to a maximum at the face. Find (a) the total load carried by each column, (b) the unit stresses at the two faces of each column, and (c) the rate of increase r of the unit stress per foot of width of wall. Note. The total stress developed in the two columns must equal the applied force, or 400,000 lb. The sum of the moments of the elemen- tary stresses about any point, say the back or face of the wall, must be equal to the moment of the applied force. The moment of a force about an axis is defined as the force multiplied by the distance from the axis to the point where the force is applied. Hence, if S is the unit stress at the back, the element of stress at a distance x from the back of the wall is dS = (S + rx)dx. The moment of the stress with respect to the back of the wall is dM = xdS. 43. Show that the total attraction exerted by a solid sphere of radius B and density k upon a particle of mass m located at a distance a from the center of the solid sphere is Mm/a 2 , where M is the mass of the sphere and a > fi. Note. This is added as an example where spherical coordinates (see Exercise CVII, 13) may be used to advantage. Observe that the total attraction is along the line of centers. EXERCISE CX Center of Gravity of an Area. The standard formulas are _ Jj xd y dx ^ ^ Jjy d y dx ^ M ^ I J dy d x I I d ydx GENERAL APPLICATIONS 171 Note that we are finding the average value of .»• and of y over the area ; that is, (x, y) is any point within the area bounded by the given curves. This idea of the " average value " of a function over an area is important and must not be confused in the present case with the average value of the ordinates or abscissas of the bounding curve. The numerators in the above formulas are called first moments, or moments of area, and are designated by M v and M x respectively. In the following examples plot the bounding curves and calculate the coordinates of the center of gravity (x, y). Mark that point on the figure. It is suggested that the figure in one or two examples be cut from cardboard. If the work has been done accurately, the piece of cardboard will remain balanced when supported on a pin point at its calculated center of gravity or area. If the center of gravity is not within the bounded area, this check is not feasible. 1. y 2 = x + l, 8. 42/ = 2x — x 2 , 15. 2/ = x 2 — 4x, y + x = l. 2y = 3x-x 2 . y 2 = 2ix. 2. # 2 = 8x, 9. 3y = 3 — x 2 , 16. x 2 = Wy, x + y = 6. y = 2x-x 2 + l. x = y 2 -2y. 3. y 2 = 4- x, 10. 6y = x 2 -x-2, 17. y 2 = 2x, j/2 = 4—4.c. 4y = x 2 — 2x — :J. y = x — x 2 . i. 2/ 2 = 4x, 11. x = J/ 2 - 4j/. 18. 2/ = x 8 -x, 2/ 2 = 5-x. 3x + 2/ 2 = 0. 2/ = 3x — x 3 . 5. y = x 2 + x, 12. 2/ 2 = 4 - x, 19. y = x 2 + x, 2/ = 2x 2 -2. y 2 + 2y = x. 2/ 2 = 12(l + x). 6. 2/ = x 2 -4x, 13. 42/ = 3x 2 , 20. 42/ = X s , 2/ = 6-x 2 . 22/ 2 = 9x. x = 2/-2/ 2 +L 7. 2/ = 2x - x 2 , 14. x 2 = 2 v 8 , 21. y = (x- 3) 2 , 2/ = 3x 2 -6.r. x 2 =82/. 2/ = z(a;-3) 2 . 22. Given the curves x 2 + y 2 — 4x — 4j/ = and 4 j/ = 16 - x 2 , calculate the coordinates of the center of gravity or area of (a) that part of the circle lying outside the parabola ; (b) that part of the circle lying inside the parabola. 172 PROBLEMS IN THE CALCULUS 23. Given the circle x 2 + y 2 = 10 X and the parabola 9 y = x 2 — 12 k, calculate the coordinates of the center of gravity of (a) that part of the circle lying outside the parabola ; (b) that part of the circle lying inside the parabola. •24. Calculate the coordinates of the center of gravity of the area bounded by x = 0, y = 4, xy = 4, 8y = 14 a; — Zx"-. 25. Calculate the coordinates of the center of gravity of the area bounded by y = 0, xy — 4, &y = 14x— 3x 2 . 26. Given x 2 + y 2 = 10 and y = 3 sin ttx/2. Find the coordinates of the centers of gravity (a) of the area in the first quadrant bounded by these curves and the y-axis ; (b) of the area in the second and third quadrants bounded by these curves and the y-axis. EXERCISE CXI Average Value of a Function throughout a Region. The process of finding average values may be extended to other problems and is not limited to finding average values of x and y over an area as in the preceding exercise. The average value of f(x) between the values x = a and x = S is J f(x)dx-i-(b — a). t/ a The average value oif(xy) over an area is I I f(xy)dxdy -=- A, the limits on the double integral being the same as in finding the area. The average value may equally well be along a curve, over a curved surface, throughout a volume, or may be an integral without any physical or geometrical interpretation. 1. Find the average value of x 2 from x = to x = 10. 2. Find the average value of the ordinates of y 2 = 4x from (0, 0) to (4, 4) taken uniformly along the x-axis. 3. What is the average value of xy over a quadrant of the circle x 2 + y 2 = o 2 ? for all points on the arc of the circumference ? 4. What is the average value of the abscissas of y 2 = 4 x, (a) when uniformly distributed along the i/-axis ; (b) when uniformly distributed along the curve ? (The latter is given by J xds -r- I ds. Note the difference in the results, and be able to explain this difference.) GENERAL APPLICATIONS 173 5. "What is the average value of the z coordinate of the part of the paraboloidal surface z = 4 — x 2 — y 1 which lies above the XT-plane, when the ordinates are uniformly dense over the circular base ? (This is not the distance to the center of gravity of the solid bounded by the surface and the .XT-plane.) 6. "What is the average value of sin t from t = to t = ir ; that is, throughout a half period ? 7. What is the average value of sin 2 1 throughout a half period ? 8. Find the average value of p 2 over the circle p — sin 8. This gives the square of the radius of gyration. 9. Find the average value of p 2 over the area bounded by the curve p = sin 2 8. 10. Find the average value of the length of the radii vectores of the curve p = sin 2 8. Compare with the result in example 9. 11. Given v = v + 32 i. Find the average value of v (a) during the first five seconds, starting from rest ; (b) during the first five seconds, starting with an initial velocity of 36ft./sec. ; (c) during the first two and one-half seconds, starting from rest. 12. Given v = Vu 2 + 2 gh. Find the average value of v during the first one hundred feet, (a) starting from rest ; (b) starting with an initial velocity of 60 f t./sec. (In the given formula h is the distance.) Note. The student should observe that in both examples 11, (c), and 12, (a), the distance covered is 100 ft. in 2 J- sec, but the average values calculated are different. He should go no further until he can explain clearly this difference. 13. The speed of a sled on an ice pond may be taken to equal v — iigt. If a boy strikes the ice with an initial velocity of 60 ft. per second, what is his average speed during the first ten seconds ? (/j. = -J, g = 32.) 14. A crank OA and a connecting rod AB move a slider (attached by a pivot at B) in a straight path in the direction of O, the center of the crank shaft, (a) If the crank turns about O with constant angular velocity a, what is the average value of the velocity of the slider at B ? Does it depend on the length of the connecting rod ? (b) Evaluate the result for a crank 2 ft. long making 120 revolutions per minute. 15. In examples 13, 14, and 15, Exercise XCIV, we considered the problem of water flowing through a weir or a water gate. Referring to them, calculate the average flow per second over an open weir, or spill- way, 10 ft. wide, while the water is falling regularly from a 4-f t. crest to zero overflow. (Take ^ = 1.) 174 PROBLEMS IK THE CALCULUS 16. Water is escaping through a water gate 8 ft. long and 5 ft. high. "What is the average rate of flow from the time the water is 4 ft. above the level of the top of the gate until it drops to that level ? How many cubic feet escaped if the time occupied was 15 min. ? (Take n = 3/4.) 17. Finding the center of gravity of a homogeneous solid or of a solid the density of which varies according to some fixed law is an application of average value — the average value of a coordinate throughout a mass, that is, fxdm -f- f dm gives the x coordinate. (The element dm must of course be chosen so that all parts of it are equidistant from the TZ-plane. The limits of the two integrals are the same whether single or triple. Find the center of gravity of a solid homogeneous hemisphere.) 18. Calculate the center of gravity of a solid hemisphere when the density is (a) proportional to the distance from the base ; (b) proportional to the square of the distance from the base. 19. Calculate the center of gravity of a homogeneous solid in the form of a paraboloid of revolution, with the radius of its base (5 in. and its altitude 9 in. 20. A solid is in the form of a semiellipsoid. The base has semiaxes 2 ft. and 3 ft. respectively, and its altitude is 4 ft. Find the center of gravity. 21. A rectangle moves from a fixed point 0, one side varying as the distance from the point, the other as the square of this distance. The center of the rectangle traces out a straight line OP, perpendicular to the moving plane. When OP is 2 ft., the rectangle is a square of side 3 ft. Find the volume and center of mass of the solid generated. 22. Finding the center of pressure against a dam or water gate is another example of average value. As the water pushes against the gate, the turning effect, or torque, about an arbitrary axis, of the water pressure acting on a horizontal strip is proportional to the pressure on the strip and its distance x from the arbitrary axis. The center of pressure is a point such that if P, the total pressure, were exerted at this point on the gate, the torque action would be unchanged. If the arbitrary axis is taken in the surface of the water, we have, therefore, as the formula for x, the distance of the center of pressure below the surface, x = I xdP-i-jdP, where I dP is the pressure integral found in a previous exercise and x is the distance of the typical element below the surface of the water. Calcu- late the center of pressure on a rectangular gate 8 ft. long and 4 ft. high when the water is level with the top of the gate. GENERAL APPLICATIONS 175 23. Calculate the center of pressure on the gate in example 22 when the level of the water is 6 ft. above the top of the gate. 24. Find the center of pressure on an equilateral triangular gate in » vertical dam, each side of which is 10 ft., one side of the triangle being in the surface of the water. 25. Find the center of pressure in example 24 if one vertex of the gate is in the surface and the opposite side parallel to the surface. 26. The end of a tank filled with water is parabolic in shape. It is 4 ft. across the top and 4 ft. deep. Locate the center of pressure. 27. A gate in the face of a dam is a square with a diagonal parallel to the surface of the water and 6 ft. below it. If each diagonal is 8 ft., calcu- late the total pressure and the center of pressure. (Assume W = 62.) 28. The end of a trench is in the form of a semicircle of radius 2 ft. surmounted by a 4-ft. square. A gate fits in this and is braced along a horizontal line. How far up should the braces be put, so that the gate will not tend to turn about the line of braces when the trench is full of water ? How high up should the braces be if there is only 2 ft. of water in the trench ? 29. Find the center of pressure on a, circular gate of radius 3 ft. when the top of the gate is 4 ft. below the surface of the water. 30. Calculate by triple integration the center of gravity of one eighth of a sphere. 31. Calculate by triple integration the center of gravity of the solids in Exercise CVII, examples 4, 6, and 7. 32. Find the center of gravity of a thin wire in the form of an arc of the parabola y 2 = 4 x from the origin to the positive end of the latus rectum. Find the same for the general case y 2 = 4 ax. EXERCISE CXII Moment of Inertia. Plane Areas. The general formula for the moment of inertia of any object with respect to an axis is J j^dm, where r is the common distance of all points of the element dm from the axis. From this follow the formulas for the moment of inertia of a plane area with respect to the a-axis, the y-axis, and the origin. They are I x =CCy*dxdy, I v =ffx 2 dxdy, /„ = CC(x 2 +y*)dxdy = I x +I y . 176 PROBLEMS IN THE CALCULUS The limits are precisely the same as if we were finding the area by double integration. If we wish to lay stress on the physical significance of moment of inertia, we must speak of the moment of inertia of thin plates, or laminae, of the shape defined by the area. In that case each formula must be multi- plied by 8, the mass of the plate per square unit of area. Find the moment of inertia for the areas below. The curves bounding the areas are given. 1. y 2 = 8x, y = 2x. 4. 2y = x 3 , 2x — y = 0. 2. y 2 = x, x + y = 2, y = 0. 5. \y = x 2 , y 2 — 4x. 3. y = 4 - x 2 , y = 0. 6. iy = x 8 , 2y = 4x— x 2 . 7. Given the area in the first quadrant bounded by the two curves y 2 = 2 X s and y 2 = 8 x, find (a) the area, (b) the coordinates of the cen- ter of gravity, (c) the moment of inertia with respect to each axis and the origin. 8. Given the area bounded by the parabola y 2 = x + 1 and the straight line x + y = 1. Solve as in (7). 9. If the moment of inertia is divided by fdxdy (or generally • ir 2 dm-±- C dm), we have the average of the squares of the distance of the "particles" from the axis. The square root of this average value is called the radius of gyration. This is an important quantity, and the student should find the radius of gyration for each of the preceding examples and for those in the next exercise. Further examples will be met in the study of mechanics, where the importance of both moment of inertia and radius of gyration will be realized. EXERCISE CXIII Polar Moment of Inertia. In this case the general integral I r^dm gives 1=1 jp 3 dpd0. The areas for which the student is asked to calculate the moment of inertia in this exercise include only such as are bounded by curves whose equations are given in polar coordinates, which system leads to the formula above. The examples are introduced at this point rather as an application of double integration than for their GENERAL APPLICATIONS 177 mechanical importance. The more important moments of inertia are included under Routh's rules in the study of mechanics. Calculate I for that part of the area indicated in each of the following examples : 1. The circle p = 2, which lies to the right of the line p = sec 0. 2. The circle p = cos 0, which lies outside the circle p = 1/V2. 3. The circle p = cos 0, which lies in the first quadrant outside the circle p = sin 6. 4. The circle p = sin 0, which lies outside the cardioid p = 1 — cos 6. 5. The circle p = cos 8, which lies outside the cardioid p = 1 — cos 0. 6. The lemniscate p 2 = 2 cos 2 0, which lies outside the circle p = 1. 7. Calculate I for that smaller part of the lemniscate p 2 = 9 cos 2 0, which lies outside the circle p =V6 cos#. 8. Calculate I for the area between the two circles p = 3 sin and p = 2 sin CHAPTER XVIII DIFFERENTIAL EQUATIONS EXERCISE CXIV Differential Equations of the First Order. This list includes only the simplest type of differential equations of the first order, where the variables are separable. 1. xydx + ^/l-x^dy = 0. 10. y (4 + 9 x 2 ) dy — dx - 0. 2. (x + xy*)dy-Zdx = 0. 11. 2/(4x 2 - 9)dy - 6dx = 0. 3. Vl-2/ 2 dx=Vl + x 2 cfy. 18. (x 2 + l)dy + x(y-l)dx = 0. 4. VTTy^dx- (1 - x 2 )(fy = 0. 13. (1 + y 2 )xdx + (1 + x 2 )cfy = 0. 5. sinydx + e x dy = 0. 14. (2x + l)dy + y 2 dx = 0. 6. (1 + x 2 ) dy - Vl - y 2 dx = 0. 15. (1 + 2?/)x(Jx+ (1 + x 2 )dy =0. 7. (x 2 -?/xV2/+(2/ 2 + X2/ 2 ) . The substitutions are given, but should be verified. „„ ds s , _ ^ 37. — \-- = t. Let s = v/t. dt t 38. ds \- s tan t = tan t. Let s = v cos t, dt 39. ds 2t 1 T ,,,„ , — -1 s = - • Let s = v/(t 2 + ] dt t 2 + l t /v 40. — + 2is = t 8 . Lets = ue-' 2 . dt Many differential equations of the first order can be solved by a properly chosen transformation not included in previous DIFFERENTIAL EQUATIONS 185 types. For a complete discussion, see a treatise on differential equations. „ds H 41. t 2 2st-s 8 = 0. Lets = S 2 /s. 42. (t 2 + t)ds = (t 2 + 2 si + s)dt. Let s = ui. 43. (3 + 2 si) sdi + (3 - 2 st) ids. Let s« = v. 44. (x + yf — - 2 x + 2 ^ + 5. Let x + 2/ = v. ax EXERCISE CXVIII Application of Differential Equations. Evaluating the Constants. We have already evaluated the constant of integration in Exer- cise LXXVII, where the examples were, in fact, linear differ- ential equations of the first order. In the case of differential equations of the second order there are two general constants to be evaluated. The simplest case is just the reverse of find- ing the second derivative. In the problems below, the student should know that if s represents distance, cPs/dt 2 is/, the accel- eration. From mechanics he should know that force = mass X acceleration, orP= mf. If g appears in a problem without explanation, it is to be taken as 32 ft./sec. 2 d 2 s ds 1. Given h 2 1- 5 s = 0, w ith the initial conditions s = 4, v = 0, dt 2 dt when t = 0. Express s in terms of i. d 2 s 2. Given - — h 4 s = 8 i 2 with the initial conditions s = 0, v = 0, when dt 2 t = 0. Express s in terms of t. 3. Given [- 4 s = 3 sin i, with the initial conditions s = 0, v = 5, dt 2 when t = 0. Express s in terms of i. 4. Given — 4- 4 s = 4 cos2i, with the initial conditions s = 2, v = 2, dt 2 when 4 = 0. Express s in terms of i. 5. The acceleration of a falling body is 32 ft./sec. 2 (a) If a body is projected downward with an initial velocity of 24 ft./sec, find the rela- tion between s and t. (This relation is called the equation of motion.) (b) If the body is 72 ft. above the ground when projected, how long will it take it to reach the earth's surface ? Hint. In (a) s - 0, v = 24, when t = 0. 186 PROBLEMS IN" THE CALCULUS 6. The acceleration of a body projected vertically upward is — 32 ft./sec. 2 If projected with a velocity of 96 ft./sec, find the equation of motion and determine how high the body will rise. 7. The motion of a simple pendulum is defined by the law d 2 8/dt 2 = —gem8/l. Tor small displacements we take sin8 = 8. Assuming this, find the equation of motion if 9 = a, and dff/dt, or a, the angular velocity, is zero at the start. Take the pendulum 2 ft. long. Find also the time required for a complete swing of the pendulum. 8. The force acting on a unit particle is proportional to the distance of the particle from the starting point and is directed towards the starting point (that is, force = — ks). The magnitude of the force is 16 poundals (32 poundals = 1 lb.) when s = 1ft. What is the equation of motion of a particle projected away from the origin with a velocity of 16 ft./sec. ? This is called an undamped, or harmonic, vibration. 9. In the case of a damped vibration there is another force (friction or air resistance) which is proportional to the speed and opposing the d 2 s ds motion. Hence the general equation is m — - =— k. k„s. Calculate dt 2 dt the equations of motion for a unit mass starting away from the origin with a velocity of 16 ft./sec, assuming k ± — 4 and fe 2 = 8. 10. The discharge of an electric current from a condenser of low resist- ance is defined by the law d 2 C/dt 2 =— k?C, where k is a constant. If k = 500, find the current in terms of t, the time. What is the frequency ? Note. When any motion or change of state is periodic, the frequency is the reciprocal of the period. 11. A certain flywheel being stopped by friction obeys the law d 2 8/dt 2 = — 4. Derive a general expression for 8, the angular distance in radians, in terms of t. Assume that to, the angular velocity, is 240 revo- lutions per minute at the start. When will the wheel come to rest? How many revolutions will it have made ? Hint, w = dd/dt and should be expressed in radians per second to determine the constants of integration. 8 will be in radians. 12. A constant force acts on a flywheel so that it obeys the law d 2 8/dt 2 = 2. Express 8 in terms of t, assuming the flywheel to start from rest. How long will it take for it to acquire a speed of 1000 revolutions per minute ? How many revolutions will it make in the first minute ? 13. If a particle of mass m is vibrating under the force of a spring and loses energy, owing to friction, at a rate proportional to the square of its d 2 s k ds a velocity, it obeys the law 1 1 — s = 0, where a is a constant dt 2 m dt m DIFFERENTIAL EQUATIONS 187 and k is the factor of proportionality, depending on the spring. Find the equation of motion for a unit mass. Evaluate this for the case k = J, a = If , assuming the spring was displaced (stretched) 4 units at the start. (This important law holds for fluid and electromagnetic friction, and is equivalent to saying the frictional force is proportional to the velocity.) 14. A condenser of capacity K, discharging through a circuit of resist- ance E, obeys the law 1 1 V = 0. In this equation L is ' dt 2 L dt LK the self-induction and V the potential. Express V in terms of t, assuming V= V , dV/dt = 0, when t = 0. 15. An equation similar to the one in example 14 arises in the theory ,32/3 flA of the ballistic galvanometer. The equation here is I^-y + k, 1- k,,B = 6, dt 2 dt where I is the moment of inertia of the needle, k t is a constant determined by the oil bath or other device used to retard the motion, fc 2 is a constant depending on the torsion of the fiber and the magnetic field, and 9 is the angular displacement, (a) Derive the general expression for 8, assum- ing 4 Ik 2 > fc*. (b) Evaluate the constants of integration, assuming a dis- placement a from its equilibrium position when the needle is released. 16. Suppose k?>4:Ik 2 in example 15. What is peculiar about the motion in this case ? 17. Given, as in example 14, a condenser of capacity K, self-induction L, and resistance R in the circuit. Apply to this an electromotive force of V volts and the law giving Q, the charge, is L —^ + E -^ + -^ = V. at-* dt K (a) Solve this equation for Q, assuming V to be constant, (b) What form does this assume if Q = 0, dQ/dt = 0, when t = ? 18. (a) Solve the equation of example 17, assuming that V varies with the time, that is, V=kt. (b) Assume V periodic in example 17, or V = k sin t,- and solve. 19. The problem of a beam supported in different ways and with a variously distributed load furnishes a series of very simple differential equations, a few of which follow. Calculate in the following cases the curve of mean fiber (that is, the relation between y and x) under the given conditions. In cases (a) and (b) find the deflection y of the middle of the beam. In all cases locate the points of inflection. (a) A uniform beam of length I loaded at one end and rigidly fixed at the other. The differential equation is d 2 y/dx 2 = k(l — x), with y = 0, dy/dx — 0, when x = 0. (b) A uniform beam as in (a) with load uniformly distributed along the beam. The differential equation is d 2 y/dx 2 = k(l 2 -2lx + x 2 ), with y = 0, dy/dx = 0, when x— 0. 188 PROBLEMS IN THE CALCULUS (o) Uniform beam with load uniformly distributed and supported at both ends. The differential equation is d 2 y/dx 2 = k(l 2 — 4i 2 ), with y = 0, dy/dx = 0, when x = 0. (Here the origin is at the middle of the beam and y is the vertical height o£ any point above the middle of the beam.) (d) If the beam in (c) is rigidly embedded at both ends, the equation is d 2 y/dx 2 = k(^\ P - \ x 2 ), with y = 0, dy/dx = 0, when x = 0. Show that the points of inflection in this case are nearer the center than in (c). (e) A beam of uniform strength with thickness equal to l/(a + bx), rigidly fixed at one end. The differential equation is d 2 y/dz 2 = k(a + bx), with y = 0, dy/dx = 0, when x = 0. 20. The general differential equation of the curve of mean fiber of a loaded beam (cf . example 19) is M = EI d 2 y/dx 2 , where If = the bending moment of a section x units from the end, E = the coefficient of longitudi- nal elasticity, and I = the moment of inertia of the section x. For a beam fixed at both ends and uniformly loaded with W pounds per linear inch, the bending moment at any section is given by the equation M = Wlx/2 — Wx 2 /2 — Wl 2 /V2. (a) Deduce the general equation for the curve of mean fiber, and calculate its maximum and minimum points and its points of inflection, (b) Evaluate this result for a rectangular beam of wood 8x12 in. and 24 ft. long, carrying a load of 24 lb. per linear foot. I for a rectangular cross section is bd s /12, where d is the dimension perpen- dicular to the neutral axis of the cross section. E for wood is 1,00(^.000. Calculate the amount of the maximum deflection, (c) Consider a steel beam 2 x 3 in. E for steel is 30,000,000. 21. The discharge in cubic feet per second over an open weir is Q = cbH 3 ' 2 , where o = 3.5, 6 = width of the weir, H = depth of overflow, (a) A lake covering 164 sq. mi. is controlled by an overflow weir, or spill- way, 2000 ft. wide. Assume a flood in the basin drained by the lake which is discharging 12,500 cu. ft. per second into the lake at the time water begins to overflow. The flood discharge increases to 182,800 cu. ft. per second in 29 hr. How long must this maximum discharge be con- tinued to cause the overflow to rise 5 ft. on the weir crest ? (b) Suppose the water did not begin to flow over the crest until the discharge reached 73,500 cu. ft. per second, how long would it take the water to rise 5 ft. on the crest ? (c) Suppose the flood should come at a time when the river had been at the freshet stage long enough to fill the lake and give a free flow of 12,500 cu. ft. per second (corresponding to a depth of about 1.5 ft. over the crest), how long would it require the maximum flow to be continued after it was reached to give a depth of 5 ft. on the crest ? Note. H may be assumed to increase uniformly until the flood discharge reaches its maximum. ANSWERS (In the answers the constant of integration in simple indefinite integrals is omitted.) EXERCISE I (Page 1) 13. /(I) = l,/(2) = 1.301,/(100)= 3. 16. /(0) = 2,/(l) = 9.5234. 14. /(I) = 64, /(4) = 2V2, 17. /(2) = 2.198, /(3) = 4.240. /(ll) = 1.4595. 18. /(I) = 2.624, /(2) =- 1.838. 15. /(1/2) =11.28, /(2) = 37.62. 19. /(I) = 13.868, /(4) =- 55.935. EXERCISE II (Page 2) 1. 2/3. 3. -1/2. 5. 0. 7. 3/5. 15. 1. 2. 1/3. 4. \. 6. 00. 8. - 5. 16. 1. EXERCISE III (Page 4) 13. -1/x 2 . 17. 2/(1 -t) 2 - 31. tan-14/3. 14. —1/3 i 2 . 25. — 1/(3 — x) 2 . 32, 33. tan-^andtan-^/lQ. 15. -2/x 8 . 26. - 2 (x + 1) A 3 - 34. tan- 18/15. 16. -3/(x + l) 2 . 27. 2x(x + l)/(2x + l) 2 . 38. ir/2. EXERCISE (V (Page 5) 1. X-4-2Z-*. 6. 3x 2 -^x-l ■ 16. - 8/ (x - l ) 3 . 2. 3 Vx + 1/VIx. 11. V3/2 Vx + 1/6 Vx. 17. x/V x 2 - 4. 18. — x 2 /(l— x 3 ) 2 ' 3 . 20. — 3x/V 4— 3x 2 . 19. 9/(1 -x)K 21. -3x/Vx 2 -2. EXERCISE VI (Page 6) 1. 2 (x 2 +4x+ 2)/Vi»+II. "■ (4x 2 +4x + 3)/(l + 2x) 2 . 2. (4x 2 + 5x-6)/V x 2 -3. 12. (3f 2 - 2i)/(3t- l) 2 . 3. (12 i 2 -4t- 8)/Vl3aJI. 13- (4* 3 + 3t2 - »)/(! + 2< ) 2 - 4. a5i°-6t 2 -18t + 4W2t 8 -4 t. 17. - 5/(x 2 - 5)V 2 5. (8x 2 -19x+5)/Vl-2x + 4x 2 . 18. (4 - 3x)/2x 8 Vx- 1. 10. ; 2 x 2 + 6x + 2)/(l - x 2 ) 2 . 19.-(x 3 -2x + D/d " ^ 2 - 25. (3 + 2x-6x 2 )/V(2x 2 -3)(l-2x). 189 190 PROBLEMS IN THE CALCULUS EXERCISE VII (Page 7) 1. 3 x 2 - 4 x + 1/x 3 / 2 . 8. (2 x 2 -4x + 3)/Vx 2 -2x+2. 2. 1-1/x 3 / 2 . 15. 3t/Vt-2. 3. 3/2 x 3 / 2 + 1/ V3 + 1/2 V3 x. 16. x/ V2 x + 4 . 5. (3 x 2 + 2)/V2x 3 +4x- 6. 17. (1 - 8 1)/ Vl - 4 1. 6. (3x 2 -l)/2x 2 V3x + l/x. 18. - £/(l + «) 3/2 - EXERCISE VIII (Page 8) 1. 4 (x + 2)/(x 2 + 4 x + 6) . 20. 1/V9x 2 -1. 2. 6(6x + l)/(6x 2 + 2x + 5). 28. (x 3 / 2 + 2)/2x (x 3 / 2 - 1). 4. - x /(l-x 2 ). 29. - 2/Vx 2 + 1. 18. 1/Vl + x 2 . 30. (x 4 + 3x 2 - 2)/x(x 4 - 1). 19. 2/V4x 2 + 9. EXERCISE IX (Page 9) 1. - log, e ■ 3 x 2 /(l - 2 x 3 ) . 5. logs e ■ (x 2 - l)/(x 3 + x) . 2. log 10 e/(8 x + 7) . _ 6. log a e • (2 x 2 + l)/(x 3 + x) 8. log 4 e • (2 V6x + 3)/(2 V6 x 3 ' 2 + 6 x). 7. - log 5 e ■ 4 t/(t 4 - 1) . 4. - 13 log 4 e/(x - 3) (5 - 6x). 8. log 2 e • 2x/(l - x 4 ). 9. log 2 e-(2-3x 2 )/(x-x s ). 10. log 6 e.(4i 3 - « 2 -2)/(l-2S)(l-4 3 ). EXERCISE X (Page 10) 1. 4 (3x 2 - 2)/2 x 3 - 4 x ■ log(2x 3 - 4a). 2. (x + 3)/(x 2 + 6x) Vlog(x 2 + 6x). 3. 3 (2 x + 7)/(x 2 + 7 x) • log 2 (x 2 + 7x). 4. 2 (2 x 3 + 3 x)/(x 4 + 3 x 2 ) - log Vx 4 + 3 x 2 . 5. - 15x/(l - 3x 2 ) . log 4 Vl-3x 2 . 6.-6 x/(l - 2 x 2 ) Vlog (1-2 x 2 ). EXERCISE XI (Page 11) 1. 2xe^+ 2 . 14. (3 1 + 2) e< vT+1/2 Vl + t. 2. e'o^-logge/x. 15. 2xlog 4 e ■ e lo M* ! + «/(x 2 + 4). 3. e V V2 V x, 16. (x-l)e^ + i)/^/2x 3 / 2 . 4. xe^^vVx^+l. 17. ei'^^/2 (1 - x) 3 / 2 . 5. (2x+ l)e 1 o«(^ + a: Vx 2 + x. 21. (- x + 2xlogl/x)e* 2 _ I °ei/*. 6. - xe^^^/Vl^x 2 . 22- (l - 1/2 Vx)e^ + i/^. ANSWERS 191 EXERCISE XII (Page 11) 1. 2 xoF* ■ log a. 5. —7-i'x. log 7/x 2 . 2. (2 1 + 2) 2' 2 + 2< . log 2. 16. 2/4 . logi/2 • c 1 "*; 2 " 2 \ og c . 3. (l + log<)5. EXERCISE XIX (Page 16) 1.-2 s /Vl-4 x 2 + arc cos 2 x. 6. ((x 2 - 4)/(x 2 + 4)) 2 . 2. e^/Vl - e 2 * + e x arc sine*. 7. 26(x - l)/(x 3 - 6x 2 + 13x). 3. 2/(4 + x 2 ) arc tans/2. 8. (t + 2)/t Vt - 4. 4. (5 1 + 12)/(t 3 + 4 1) . 9 . x 2 arc tan x. 5. arc sinx/2. 18. (x 2 — 3)/x 4 • arc sin 1/x. EXERCISE XX (Page 17) 1. 3 Vl + lo gx/2x. 8. 12/(x 2 + 4x). 2. 3 Vx 2 -l/x 4 . 9. 2 (3 e 4 *- 5 )/(3 e 4 * + 5). 4. — x sin x 2 cos Vcos x 2 / Vcos x 2 . 12. e* Ve* — l/(e a: + 3). 5. 2(arc sin- + 1)/ Va 2 - x 2 . 13 ' tana; ( 2 + secx )- a 16. 1. 6. - 28/(4 x + 3) 2 . ANSWERS 193 EXERCISE XXI (Page 17) 1. - (2 x + y)/(x + 2?/). 6. (3 x 2 + 2 y)/(2 y - 2 x). 2. 2xy/(8y — x 2 ). 15. (e 31 -" + e*+2/)/(e*-" — e* + ?/). 3. (6x2/ + 4)/(2-3x 2 ). 16. cot x cot?/. 4. (2 x - 3 ?/)/(3 x - 3 j/ 2 ) . 17. (2 x 2 - ?/ 2 )/xj/ (2 log x?/ + 1) . 5. - 2x(y + l)/(x 2 + 3). 18. e 2 *(2/ 2 + 4)/?,. EXERCISE XXII (Page 18) 1. 2 xe 3 ^ log e + V3/2 Vx - 1/x 3 + e^/2Vx. 2. 16/( 4 + x 2 ) 2 . 3. l/2Vx 2 + 4x. 3. x 2 /V x 2 - 1. 8. - (2 xy + y 2 - 4)/(x 2 + 2 xy + 4) . 4. l /fVi 2 + 1. 9. e*/ 2 /(l- &'*)*. 10. V8x-x 2 . 14. J2/V4 - « 2 . 18. sec 8 x. 11. log(4x 2 -l). 15. 1/i 8 Vi 2 - 4. 19. 4x 2 Vx 2 + 4. 12. log(4x 2 +l). 16. - arc sin x/2 Vl - x. 21. 2 sin l/x/xV/*. 13. arc tan Vx. 17. 2 x esc (1 - x 2 ) sec (1 - x 2 ). 33. 10x/9 (2- x 1 ' 2 )^. EXERCISE XXIII (Page 20) 1. 1, V3/2, V2/2, 0, - 1. 4. (3, - 5) ; (- 3, 13). 2. (a) (3, 3), (3, 1) ; (b) (4, 2), (2, 2). 7. When x = 4. 3. (3,-4); (-3,4). 8. (4,4). 12. (a) (1, 2), (- 1, 0) ; (b) (2, 9), (- 2, - 7). 18. (a) ((8k+ l)ir/4, V2), ((8n + 5)tt/4, - V2) ■ (b)(2^,l), (|(4n+ 3K-1). 21. (a) (-2,- 4), (-2,0); (b) (0,-6), (0, -4) and (£, - V), (*,-¥)• 22. (a) ir/4; (b) 0. 31.(2,2). 34. 2y = x 2 . 35. j = sinz. EXERCISE XXIV (Page 22) 1. 18°.4. 8. At (2, 2), 31°. 22. At (2, 2), 18°.4. 2. 71°.6. 9. At (0, 0), 24° ; 23. x + V2y = 6, 7. At (4, 5), 9°.5 ; at (8, 2), 69°.7. V2 x - y = 0, at (-1, 5/4), 33°.7. 10. 26°.6. s t = 4, s„ = 2. ■24. At (2, -2), 3x + 5y + 4 = 0, 5x- 3y- 16 = 0, s, = Jg -. *»= f; at (2, 8), 18x - 5 y + 4 = 0, 5 x + 18 y - 154 = 0, s, = - 2 F -, s„ = if ±. B. x + 6y + 10 = 0, 6x-j>- 14 = 0, s, = 12, s„ _ i — 3' 194 PROBLEMS IN THE CALCULUS EXERCISE XXV (Page 23) 1. 3x-y-5 = 0, x + 3y-15 = 0. 4. 3x + 2y+l = 0, 2x-3y-8 = 0. 2. x-4y + 15 = 0,4x + y-25 = 0. 7. x+6y-12 = 0, 6x-y-35 = 0. 3. x + 3y + l = 0, 3x-y-7 = 0. 8. 4x+ 2y-3 = 0, 2x- 4y+ 1 = 0. 9. x-y = 2V3,x + y = 10 V3/8. 11. V2x-y-l = 0, x + V2y-2\/2 = 0. 12. 15x-4y- 81 = 0, 4x + 15 y + 123 = 0. EXERCISE XXVI (Page 24) 4. ir/8. 6. 7T/3. 8. Oandtan- 1 ^ V3 = 46°.l. 5. tt/6. 7. 0andtan- 1 3V3 = 79°.l. 11. w/3. 14. High points, = 3 7r/4 and 9 tt/4 ; low points, 8 = and 3 tt/2 ; extreme left (vertical tangents), 6 = 3tt/8 and 15ir/8 ; extreme right, 6 = 9 ?r/8 and 21 ir/8. 15. High, 9 = 3 ir/8 ; low, = 7 tt/8 ; left, = 5 ir/8 ; right, = ir/8. 18. tan-!(- 9/13), or about 145°. EXERCISE XXVII (Page 25) 1. s = 33; v = 12;/ = - 18; t) = 0,i = 2. 2. s = 116 ; v = 72 ; /= 6 ; D = 0, < = 6. 3. 8 = 72; « = 15;/ = -12; v = 0, t = 4. 9. s = l; u = 3 ; / = 15 ; never at rest. 13. x =- .330 ; v =- .926 ; /= 1.522 ; t = .760 and 2.331. 14. x = .994 ; v =- 2.543 ; /= 3.097 ; t = 2.356 and 5.698. 20. x =- 3.90 ; v = .975 ; / = - 9.379 ; t = 1.899 and 3.899. 21. x =- 2.5749 ; v = 2.277 ; /= 1.158 ; t = 3.84 and 7.84. 25. 36 sec. 288 ft./sec. 27. 1/2 sec, 80ft./sec. 26. The second goes 48 ft. the higher. 28. 1824 ft. EXERCISE XXVIII (Page 27) 1. 2, 2, 1, - 3. 3. - 3/2, - 3/2, - 1/2. 17. (a), (d), (e), (f), (i), (j), 2. 1, 1, 2/3, 2/3. 4. 1/3, 1/3, - 1/2. have double roots. EXERCISE XXIX (Page 28) 1. y" = — sin x — cos x, y'" = — cos x + sin x, yin) = s i n ( x _|_ nTr/2) + cos (x + rar/2). 2. y" = 2/x 3 , y'" = — 6/x\ 2/« = (- 1)»- 1 • [n • l/x»+ 1. 8. y<»> = 2"e 2 *. 9. 2/" = 2/(1 + x 2 ) 2 , y'" = - 8 x/(l + x 2 ) 8 . 10. y" = — 2e-> cost, y'" = 2e-'(sin«+ cosi). 20. y" = cos t - t sin t, y'" = 2 sin t - « cos t. 27. y" = (2 - x 2 )/(l - x 2 ) 8 ' 2 . ANSWERS 195 EXERCISE XXX (Page 28) 1. (2xy + 3y*)/(x - 3i/ 2 ) 3 . 9. - 4e*/&v. 2. d ty/dx* = 36/(2 y - 3 x) 3 , 10. - cos (x + j/)/sin 3 (x + y). d s y/dx s = - 972 x/(2 2/ - 3 x)K 11. sin (x + ?/)/cos 3 (x + y). 3. - 2/(4 2/ - 2 x) s . 12. (1 - log x)/x 2 , d»y/dx* 8. — (sinxsinj/ + cos 2 x coty). = (21ogx — 3)/x 3 . EXERCISE XXXI (Page 29) 1. t/2. 4. 6(3t 3 + 2£)/(6S + 2) 3 . 7. -&/a 2 sin 3 t. 2. (3S 2 + 2)/4i 3 . 5. 1/t. 17. - esc 2 t/(t sin t). 3. 2/3J 3 . 6. 1/2 cos 3 2 1. 19. -(1 + 2 cost) 8 /2(2 + cosi) 3 . EXERCISE XXXII (Page 30) 1. Max., (- 1, 1) ; min., (4, - 124) ; inf., (3/2, - 61 J). 2. Max., (- 2, 69) ; min., (3, - 56) ; inf., (J, 6 J). 3. Max., (- 2, 23J) ; min., (6, -62); inf. ,"(2, - 19£). 4. Max., (- 6, - 50) ; min., (- 3, - 82) ; inf., (- 4, - 66). 5. Max., (- 1, 45) ; min., (3, - 51) ; inf., (1,-3). 9. Max., (2, 4) ; min., (1, 3) and (3, 3). 17. Max., (1, 3) ; min/, (- 1, - 3) ; inf., (0, 0). 18. Max., (-2, -4); min., (2, 4). 38. Max., (2, 5/3) ; min., (0, - 1). 39. Min., (- 3, 13J); inf., (0, 0). EXERCISE XXXIII (Page 31) . 1. Max., (jr/2, 1) ; min., (3 tt/2, - 1) ; inf., (0, 0), (tt, 0). 2. Max., (tt/4, 1.553) ; min., (5 tt/4, - 40^44) ; inf., (0, 1). 3. Max., (2, 3.515) ; min., (- 2, - 3.515). 7. Max., (.253, 1.125); min., (1.571, 0) ; inf., (.883, .580). 8. Max., (.635, 1.76); min., (2.446, - 1.751) ; inf., (1.571, 0). 13. Max., (.551, .517); min., (2.122, - .107); inf., (1.107, .265). 14. Max., (.766, 1.639); min., (2.286, - 1.123) ; inf., (1.447, .343). 16. Min., (1.448, -2.351); max., (3.02, 1.071). EXERCISE XXXIV (Page 32) 1. 18 x 24 ft. 8. 2 ft. 13. 46| rd. from^t. 2. 3 x 4ft. 9. K = 10ft.;A = 30ft. 14. B = 4. 3. 12.54 ft. 10. 40 ft. wide at top, IS. 3 mi. from D. 4. 4 and 4. 16 ft. at the bottom, 16. 5 mi. from A. 5. 3 in. 16 ft. deep. 17. R = 4. 6. 18 x 24 rd. 11. 2 wk. hence. 18. 1250. 7. 6 x 6 x 10 ft. 12. 150. 19. e = 3.51, h = 2.23. 196 PROBLEMS IN THE CALCULUS 20. Rectangle, 8 x 6 ft. 21. Parallelopiped, 2 x 2 x J f t. 22. Parallelopiped, 2 x 2 x 1J ft. 23. (a) 4 V3 ; (b) 6.31 ; (c) 8; (d) 24. 24. 75,000 @ $7.50 per M. 25. 2 hr. later; 167rou. ft. increase. 32. Each 5 in. 33. 35. (a) x = 2a, y = 26; (b) x = a+\ / o* 2 , ^ft+^o 2 *); (c) x = a + Va&, y = & + Va& ; (d) x = a + b 2 /a f y = b + a 2 /&. 36. (2, ± 4). 39. 12 V^ ft. 42.3.215. 37. a = 6 = 5 in. 40. S = 5.237, S A = 5.250. 43. 7.871. 26. 9 mo. 27. 6 mi. ; 32min. later. 28. 2 hr. later ; 54.08 mi. 30. Edge of cube = diameter oi sphere. 31. B = 13.37, h =8.91. 4 in. 34. 10x5 ft. Each 41. 2.813. EXERCISE XXXV (Page 37) 1. About 55°. 17. 40 ft. 28. 19.81ft. 2. Each is 2 VlO. 19. 20 wk. 30. 20V2ft. 4. (-1,-3). 20. a s / 2 c/(a 8/2 + & 8/2 ) 34. 9 = ir/i + a/2 5. e = 2in. 7i = 2/3V3. units from .4. 35. 6 - a + p. 6. BP = DA/V2. 21. 40mi./hr. 37. s/2aW/n. 11. 6 in. 22. 6mi./hr. 41. x = 12.97. 15. 4 in. and 5 J in. 25. 35 mi., 16 c .2. 46. a = 1 (90 - -*) 16. (a) r : h = 1 : V2. 27. Z = 13Vl3, 49. 4f ft. (b) r: ft = l:2V2. A = 26. 50. 6.152 in. , 51. Parabola : 4 ay 2 = b 2 (4x-a). 54. CF=Vab. 55. (l + I Vl + 8 sec 2 ^)^ sec 2 or ; (a) 51.23 ft. 60. a^V(^Ij + v^) from S v EXERCISE XXXVI (Page 45) 1. ds = rdx/Vr 2 — x 2 = rdy/Vr^ — y^, s = 1.014. 2. ds =V (a + x)/xcix = d?//(2a V4a 2 + y 2 ), s = 1.783. 3. ds =V4a 2 x 2 + 4a&x + & 2 + ldx, s = 2.429. 4. ds = (x 4 + 1) dx/2 x 2 , s = 3.879. 17. ds=v T+ 4t 2 d t. 19. ds = 4d«. 18. ds = tV4 + 9t 2 d<. 20. ds = 2cosiVl + 4 tan 2 idi. 25. (17) sinT = 2i/ Vl + 4i 2 , cost = 1/V l + 4t a ; (18) sin t = 3 1/ V4 + 9 i 2 , cos t = 2/ V4 + 9 i 2 . 26. (a) ^ = .97, !)„ = - 3.88 ; (b) v x = 1.789, s„ = - 3.578 ; (c) Uz = 4, Dj, = 0. 32. (a) v x = 32 V2, w„ = 16 V2 ; (b) «„, = 32 V2, »„ = ; (c) v x = 32 V2, B„ = ± 16 V§. ANSWERS 197 EXERCISE XXXVII (Page 47) 1. ds = add, sini/< = costf, cosi/- =— sinfl. 2. ds = bdd, sini/' = sinfl, cosf = cosfl. 3. ds = 5 06, sin f = J (4 sin (9 + 3 cosfl), cos^ = J (4 cos - 3 sin 6). 5. ds =V2 + 2 cos fl ag, sin./' = V(1 + cos#)/2, cos ^ = — sin fl/V2 + 2 cos 0. 12. ds = 18/ Vl + cos (9 • d0. 17. ds = sec 8 6/2d9. 13. ds = 3/(2 - cos0)V5- 4cos0d0. 18. V5 - 4 sin 3 + 8 cos 2 3 6d6. EXERCISE XXXVIII (Page 47) 1. dy = ^(cosx + cosx/2)dx. 2. (- 2/3xV3 + 2/1V8 + 5/x 2 )dx. 3. (4 x - 5 x 2 )/2 V2-x • dx. 4. 5(x 2 -Vx)V(2x-l/2Vx) -dx. 5. (2 + x)/(4x + x 2 )*-dx. 6. -3x 2 /2Vl-x 8 .dx. 12. a sin (2 a/x)/x 2 ■ dx. 13. -l/(xVl-x 2 )-dx. 19. e*(x- 3)/x 4 -dx. 23. - 1/(4 Vx - 2 x) ■ dx. 26. — 6 x cos 2 x 2 sin x 2 dx. EXERCISE XXXIX (Page 48) 1. xd 2 x/dy 2 -y(dx/dy) 2 = 0. 2. 1 + (dx/dyf -(y- 2)d 2 x/dy 2 . 5. (d 2 x/d2/ 2 ) + (dx/d2/) 2 + ?/-4=0. 9. xzd 2 z/dx 2 + zdz/dx + z 2 = 0. 10. d 2 z/dx 2 + 2dz/dx - 2 cosz = 0. 11. d 2 z/dx 2 -2dz/dx = 0. 12, 15 16. 17 19 tan zd 2 z/dx 2 + (dz/dx) 2 = 0. dhx/dy 2 + u = 0. td 2 y/dt 2 + dy/dt + y = 0. d 2 u/dx 2 + 4it = 0. dVdz 2 + y ='0. 25. dVdt 2 + j/ = 0. 26. (d 2 y/dt 2 ) 2 + (dy/di) 2 + 1 = 0. 1. 144 tt f t. 2 /sec. 2. 8£ ft./sec. 3. 6ft./sec. 4. dF/dt = 4 VI ft. 3 /hr., dS/di = 15.46 ft. 2 /hr. 18. (a) 13.823 ft./hr. ; (b) 9.840 ft./hr. 21. 7 lb. per sq. in. per sec. 22. 3.27mi./hr. 25. 14/ViT in./sec. 26. (a) 1 in./min. ; (b) increasing 8 in. 2 /min. 40. 1\ units per sec. 42. .731 ft. 3 /sec. EXERCISE XL (Page 50) 5. v = 8 ft./sec, 9. 16/75 it ft./sec. /= 9/10ft./sec. 2 10. 4/3 7rft./min. 6. 6 ft./sec. 13. (a) 8 ft. 3 /sec. ; 7. 257r/12ft. 3 /min. (b) in 2/3 sec. 8. 3.717 ft. 3 /min. 15. 3f ft./sec. 27. (a) 50 tt/3 ft./sec. 29. (a) 8/1.3 ft./sec. ; (b) 4 V3/V109 = .621 ft./sec. 31. — 16/75 units per sec. /= 32/375 units per sec. 2 33. 8.66 ft./sec. and 9.45 ft./sec. 37. 3.1134 ft./sec. 47. 1/4. 41. 5?rft. 2 /sec. 43. 6.12 in./sec. ; 6.16 in./sec. 48. Decreasing ^. 198 PROBLEMS IN THE CALCULUS ^ EXERCISE XLI (Page 57) 1.3/2. 3.1/6. 5.3/20. 14.8/69. 16. ± 1. 21. 2 r. 2.-1/4. 4.1/2,3/5. 13.-1/3. 15.296/9. 17. 3 a, 0. EXERCISE XLII (Page 58) 1. 4. 3. 0. 5. 1/2. 9. w 2 /2. 17. 0. 2. 3. 4. 3. 6. oo. 16. 0. 18. - 2. EXERCISE XLIII (Page 59) 1. 0. 3. 1. 5. 2/tt. 11. 1/3. 18. 1. 20. 2. 0. 4. 4/ir. 6. 0. 17. it. 19. 1. EXERCISE XLIV (Page 59) 1. 1. 3. e 2 . 5. 1., 7. 1. 17. e. 2. 1. 4. 1. 6. e 2 ^. 16. 1/e. 18. 1/e. EXERCISE XLV (Page 61) 1. i? = 2V2, Nis2x+2y = 3. 13. 125/12, 3x + 4y-l3 = 0. 2. 5VlO/3, x + 3y = l. 14. 5V5/2, 2x-y~G = 0. 3. 1/2, x = 2. 19. 39/8, VIx + y - 3 V3 = 0. 4. 8V2/3, x-y = 3. 21. B = 2V2. 5. 29 V29/25, 2 x + 5 j/ = 12. 22. B = 1/2. 6. 4 V2, x - 2/ + 1 = 0. 23. H = 1/36. 7. 73V73/100, 8x + 3j/-22 = 0. 24. E = 5V5. 12. 5 V5/4, 2 x + y + 2 = 0. 28. K = (x 2 + y*)»'*/cfl. 30. Max., x = nir ; min., x = (2 n + 1) tt/2 . 33. Max. at vertex. 34. Max. at (1/4, 1/4), B = 2 V2 ; B = 2. EXERCISE XL VI (Page 62) 1. (3, 3/2). 3. (8/3, - 5/6). 5. (- 2, 3). 7. (- 17J, 3 J). 2- (3, - 8f). 4. (21f 16^). 6. (»/2, 0). 8. (2, 5). 17. (- 7, 8). 18. (7/4, 7/4). / // ANSWERS 199 EXERCISE XLVII (Page 63) 1. a = -84 3 ,6 = 64 2 . 2. a = 4- |4 5 , b = £ 4 s + 2/4. 3. a = 3-4+ f 4 5 , 6 = 2/34+ 3 4= + |4 6 . 4. a =1+4- £4^, 6 = 2/4 + £4 8 - 5.0 = (12 4 4 + l)/2 4 s , 6 = (4 4* + 3)/4. 6. a = - 32 4 3 , 6 = (7 - 24 t 2 )/2. 13. a =— 2 cost cot 4, b = t + cost + cos 8 4. 14. a = 144 sin 3 4/13, 6 = 144 cos 3 4/5. 15. a = 4(4 + sini), 6 =-4(1— cos4), 16. a = ^(2 cos 4 — cos 2 4), 6= J (2 sin 4 - sin 2 4). EXERCISE XL VIII (Page 65) 1. du/dx = 3x 2 — 6xy + V 2 , du/dy = — 3x 2 + 6xy + y 2 . 2. du/dx = 3 x* - 3 yz, du/dy = 3 y 2 - 3 xz, du/dz = 3 z 2 - 3 xy. 5. Szt/8x = y/x, Stt/gy = logx. 6. du/dx =— y sinx + cosy, du/dy = cosx — x siny. 7. du/dx = & + ye* du/dy = X0i + tF. 14. Sz/3x = tan x, dz/S?/ = — tan y. EXERCISE XLIX (Page 66) 1. — 9x/4y. 10. (cosy — y cos x)/(sinx+x siny). 2. (3x 2 -2xy + y 2 )/(x 2 — 2xy — 3y 2 ). 12. - (2 + 4 ye^'/3 + 4 xe 3 *) . 3. (2—3 x)/2 y . 17. (x + x log x + y)/(x — x log x) . 4. (ay - x 2 )/(y 2 - ax). 19. 01/(2 - y). 5. (y 2 - l)/(3 y 2 - 2 xy + 1). 23. y/x. 9. — 2sin2x/3sin 2y. , 24. (x log a — y)/(xlogax). EXERCISE LI (Page 68) 1. du = (2xy — 2y — l)dx + (x 2 — 2x- 2y)dy. 2. du = [cos (x + y) - sin (x - y)] dx + [cos (x + y) + sin (x - y)] ay . 3. du = [y cosx— sin(x — y)~\dx + [sinx+ sin(x— y)]dy. i. du = yzewdx + xz&v z dy + xyewdz. 10. [2/(1 + x 2 )]dx + [1/(1 + S/ 2 )]ay- 11. (aFfe^loga + y cos xy)dx + (a 3 *" log 6 + x cos xy)dy. 12. [1/(1 + x 2 )]dx + [1/(1 + y 2 )]ay. 15. aw/a"* = (2 x - 2 y) (dx/*) + (2 y - 2 x) (dy/d4) ; dM/dx = (2 x - 2 y) + (2 y - 2 x) (dy/dx). 16. du/d4 = (9 x 2 - y 2 + 2 xz) (dx/dt) + (2 yz - 2 xy) (dy/d4) + (x 2 + y 2 ) (dz/dt) . 22. du/dt = 2 tan 4 sec 2 4-24. 23. du/dt = 2' i°s ' • log 2 (log 4 + 1) . 200 PROBLEMS IN" THE CALCULUS EXERCISE LII (Page 69) 1. (a) - 5/3 units per sec. ; 11. (a) .505 in. ; (b) 101/132%. (b) 5VlO/3unitspersec, 12. +1/20. <&/<&=- 1/3. 14. (a) .3; (b) .5; (o) 6J%. 2. (a) 8; (b) 4VB. 15. (a) .0144; 4. dy/dt=2, dz/dt=24/7 ; u=4.44. (b) 23/3600 = 23/36%. 5. dx/dt = - 1/4, dy/dt = - 1/2. 16. .0522 ohms, 47/66%. 6. Vol., ±2.16cu.in. ; 17. .0018. surface, ± 1.44 sq. in. 18. .534. 7. (a) S, .24irsq.in. ; 19. .0204 sec, 37/64%. V, .36 tt cu. in. ; 21. (a) .908 in./sec. ; (b) S, 2/3%; V, 1%. (b) 8.88 in. 2 /se c. 23. (a) v = aw [sin 6 + (c 4- a sin 9) cos 0/V6 2 — (c+ a sin 0) 2 ], where 6 is , the angle th rough w hich the crank ha s turne d ; (b) v = cau/Vb 2 — c 2 and aw ; (c) aw/V6 2 — c 2 and 0. 24. Volume, 3.0159 cu. in. ; surface, 2.818 sq. in. 27. 144tt/5cu. in.; (b) 3/10 or 30%. 29. (a) .0628%; (b) .1776%, 5.327 ft. 30. (a) dj = c/cos 2 a -da; (b) a = 0° or 180° ; (c) or = 45°. 31. (a) dl 2 = 2aa 2 I l /(a — a 2 ) 3 ■ da 2 ; (b) and (c) screen midway between the lights ; (d) ± 1.4568 candle power ; (f ) ± 3.6134 candle power, 2.62%. EXERCISE LIII (Page 74) Of the first fifteen, all are exact except examples 6, 11 and 15. EXERCISE LIV (Page 76) The divergent series are examples 2, 7, 9, 14, 15, 20. EXERCISE LV (Page 78) 1. -1)/(& + e^) + && (h — fc) 2 /2 (e* + ef) 2 + • • •. EXERCISE LXI (Page 86) 1. Min., x = 8/3, y = 2/3. 3. Min., x = y=\ V2. 2. Min., x = " 2 7 -, y = 5. 4. Min., x = — 1, 2/ = 3. 5. Max., x = y = -ir/3. 7. Max., x = y = tt/3. 16. 4 x 4 x 4 ft. 6. Max., x = 1, 2/ = 2. 8. Max., x = y = 2. 17. Equal parts. EXERCISE LXII (Page 88) 1. x 2 + 4j/ = 0. 5. j/ 2 +4x = 0. 13. 4 x 3 = 27^. 2. 27x 2 = 42/ 8 . 8. y 2 - 4x-4 = 0. 14. 2/=±2x. 3. 27 j/ = x 8 . 10. x 2 - 2xj/ + i/ 2 = 8. 19. y 2 = 16x. 4. x2/ 2 -x 2 -2x-l=0. 12. x 4 + 42/ = 0. 20. x 2 + 4 y 2 = 64. 21. x 2 + 4 y 2 = 4. 27. x 2 ' 8 + t/ 2 / 8 = 4. 22. 25x 2 /81 + 25j/ 2 /256 = 1. 28. y 2 = 8x + 8. EXERCISE LXIII (Page 89) (A) 1. 3x-3?/-4 = 0. 5. x-y = 0, x=±l. 2. x-2/ + 3 = 0, 2x-3j/-13 = 0. 6. x = 0, y = ± 1. 3. x - 2 j/ = 0. 7. x = 0, y = 0, x + 2/ = 0. 4. x = 4, x — 2/ + 4 = 0, x+2/+4=0. 8. x-y-2 = 0,x + y = 0. 9. x — 2/— 4 = 0, x — 2/ + 4 = 0, x + 4 = 0. 10. x — y — 2 = 0, x — j/ + 2 = 0. 11. x + 2/ = 0, x — j/ = 0. 12. 3x + 3?/- 8 = 0. 28. 4x-42/ + 3 = 0, 84x + 28^ + 9 = 0, 7x-Uy + l-0. 29. x-2/ = 0, x + 2/ + 2 = 0, 2x-2/ + l = 0. 30. x = 2, 2x+ 8y + 1 = 0, 6x-40y + 9 =0. 31. a; + 1 = 0, 6x- 9j/ + 28 = 0. ANSWERS 203 (B) l, 2. 5. 6. 9. 3 psin (tt/3 + 6) + 4 Vs = 0, 3 p sin (tt/3 - 0) + 4 V3 = 0. V3p sin(7r/6 + 0) = 2, V3p sin(5n-/6 + 0) + 2 = 0. (C) = 0. 14. p cos = 8. 15. psin(7r/4-0) = 4, 16. psm(57r/4-0) = 4. 17. pcos0 = 4. 18. (0, 0) conjugate point. Asl. Cusp at (0, 0). Double point at (0, 0) . Double point at (—2, 3). Conjugate point at (0, 2). Cusp at (1, 1). (2, 0), (- 2, 0), and (0, - psm0 = 1. p sin 8 = it. p sin = ± 2, p cos 9 = ± 2. p cos = — 4. 19, V2psin(0-7r/4) = l. 20, p sin = 2 tt, etc. 30. p cos = — 4. p sin = 2. 10. (0, 3), (- 3, 0), and (- 6, 3) are double points. 11. (0,1), (0,-1), (1,0), and (-1,0) are all double points, hence the curve degenerates. 14. (2, 0) and (- 2, 0) are both double points. 2) are double points. Cusp at (3, 2). 17. Conjugate point at (0, — 1). 18. Cusp at (0, 0). EXERCISE LXIV (Page 93) 1. (x + 1)/1 = (y - 5)/4 = (z - 4)/4 ; x + iy + iz = 40. 2. a;/l = ( 2 /-2)/2 = (« + l)/-l; x+2y-z = b. 3. (x - l)/2 = (y - l)/3 = (z - 5)/2 ; 2x + 3y + 2z = 15. 4. ( x -4)/4 = (2/-l)/-l=(z-4)/8; 4x-?/ + 8z = 47. 7. (x - l)/2 = (y - 1)/- 1, z = 1 ; 2ae - y - 1 = 0. 8. z-l = 0, ?/-l = 0; x = 0. 9. x=2,y/—S = ir(z-l)/2; 3iry - 2z + 2 = 0. 11. (x - 3)/2 = (?y - 2)/- 3, z + 6 = ; 2x-3y = 0. 12. (x-2)/5 = (y-l)/-ll = (z-4)/- 2; 5x- ll2/-2z+ 9 = 0. 14. (x - 2)/16 = (y - 4)/- 5 = (z - 2)/6 ; 16x - by + 6z = 24. 15. (x - 2)/6 = (2/ - 1)/- 21 = (z - 3)/l ; 6x - 21 y + z + 6 = 0. EXERCISE LXV (Page 94) 1. 6x + 2y + 3z = 49 ; (x - 6)/6 = (y - 2)/2 = (z - 3)/3. 2. 4x + 2y - z = 6 ; (x - 2)/4 = (y - l)/2 = (z - 4)/- 1. 7. 13x + 152/ + z + 15 = ; (* - 2)/13 = (y + 3)/15 = (z - 4)/l. 8. 2x + 2y-z + S = 0; (x- l)/2 = (y + 2)/2 = (z - 6)/- 1. 9. 4x + j/ + z-13 = 0; (x - 2)/4 = (y- 1)/1 = (z - 4)/l. 11. cos- 1 19/3 Vl38. 12. cos- 1 19/7 V29. 13. Each is 3x + 4?/ + 3z — 20, . . curve cuts surface orthogo nally. 14. 3x + iy + 6z = 22, 6x + y - z = 11 ; angle is cos- 1 16/V2318. 15. Tangent plane to each surface is2x + 3j/+2z = 9. 204 PROBLEMS W THE CALCULUS EXERCISE LXVI (Page 96) 1. \x e . 2. fxi 3. f xl 4. -l/4x 4 . 8. -l/2Vx. 9. 2 V2 xM/3. 13. x+ |a;8/2. 29. fz^ 3 — L2-z™ + §z 6/s . 51. Vx 2 +2x. _ 14. fx£+^x 8 . 34. ^(371+ 2) 4 . 55. - |(l-Vx]^. 15. ^x 2 -2Vx. 35. jV( 3 - 5x ) -6 - 56- -4Vl-Vx. 16. £x$ + |xt. 36. - 7 /4(1 + 2 x) 2 . 57. ^ (1 + e 8 *) 8 . 17. - i (1 - x) 4 . 39, 2 VT+ltf. 58. - % (1 - e*) 8 / 2 . 18. x + fx£ + i-x 2 . 41. ^(s 2 + 1)8. 59. l (2 + logx) 2 . 19. x- 3x2^ + 3 x 1 * 5 . 43. -1/6(3 + 2 x s ). 61. §sin 8 £x. 20. 1 + 2/x-l/3x 3 . 45. - f(l-2 x 2 )3 /2 . 62. j;tan 2 2x. 25. Jx 3 ' 2 - fx" 2 . 47. l(2x 3 ' 2 +1)5/3. 63. - §Vcos3x + 5. 26. |x s / 2 -*x5' 2 +fxi 49. ^(2x5/3 + 5) 2 «. 64. - 1/(1 + tan x). EXERCISE LXVII (Page 98) 1. log(x + l). 2. ^log(l + 3x) = log\ / l+ 3x. 3. -logVl-2x. 4. _ ^log(2-7x 2 ). 10. |log(l + X 3 / 2 ). 18. -log(l-3 8 *)/31og3. 5. §log(x 3 +l). 11. log(2 +Vx) 2 . 21. - £ log (1- log 2 x). 6. - fo log (2-5 x 4 ). 16. logV2 + e 2 *. 23. log(l + sin 2 x). 7. log (x§ + 2) 2 . 17. -log (3+ e-x). 24. - ^ log (1-2 sin 5 x). 29. J log (log tan x). 30. log (arc tan x). EXERCISE LXVIII (Page 99) 1. 2*/log2. 10. \&\ 21. ^e 2a; + \&* + \&*. 2. 5 2 */21og5. 12. -e 1 '*. 22. ^e 4 * + 2x- \e-±*. 3. — 1/3*, log 3. 13. Je sin2a: . 26. £e 2a; — 2e r + i + ex. 4. — e- x . 14. l/e 008 *. 29. — J-e 2 '* — 2x~^. 7. ±e@ x + 1 >. 15. j^tana, 31, e*- log(e* + 1). 9. ^e 2 *- 1 . 18. — Je!og(i-4*). 32. - ^e»=— e»— log(l -e*). EXERCISE LXIX (Page 100) 1. — £cos2x. 16. ^sinx 2 . 35. 2x— tanx. 2. \ log sec 3 x. 18. 2sinVx. 37. log sec e*. 3. |log(sec5x+tan5x). 19. ^tanx 8 . 38. cose- 1 . 4. 3sin(x/3). 20. cotl/x. 42. x + log(sec2x + tan2x) 7. V21ogsec(x/V2). 22. ^ sin 3 x. +^tan2x. 10. £sin(l + 2x). 25. 2tanx/2-x. 44. x-£cos4x. 11. — log sin (1 — x) . 28. — ^cos3x. 50. £tanx. 13. £tan5x. 29. £tan2x. 51. |(tan 3x + sec3x). 14. ^sec2x. 30. Jsee5x. 52. x— -^(csc3x— cot3x). 15. 21og(cscx/2— cotx/2). 34. sinx. 53. |-(sec2x+tan2x)— x. ANSWERS 205 EXERCISE LXX (Page 102) 1. arc tan x. 22. 1/V3 • arc tan (x — 3)/V§. 2. £log [(x - l)/(x + 1)]. 24. 2/V§ • arc tan (2x + 3)/ VI. 3. £log[(x + l)/(x-l)]\ 25. larctan(2x + l)/2. 5. |log[(3x-l)/(3x+l)]. 26 ' Jlog[(3x-4)/3x + 2]. 6. 1 Jjlog_[(3x + 2)/(3x-2)]. 30. V2/2 • arc tan (4 i + l)/2 V2. 7. 1/2 V5 • log [(x- V5)/(x+ VB)]. 37. £ arc tan (x 2 + 1). 8. l/V2 JL arctanxV2 1 38. ft log[(x 8 - 5)/(x 8 - 1)]. 11. 1/8 V7- arc tan (V7 6/3). 39. log (Vie- l)/(Vx + l). 13. 1 log [x/(x + 2)] . 40. 3 arc tan x*. 14. 1 log [x/(x — 4)] . 41. 1 arc tan e*. 17. ^ log [(x + l)/(x + 3)]. 43. arc tan (log x). 18. --Jtan-i [(x-l)/3]. 45. 1 arc tan (sin 3 x). 19. arc tan [(as - 4)/3] . 48. x. EXERCISE LXXI (Page 103) 3. larcsin3x/2. 11. arc sin (x — l)/4. 4. |log(2x+V4x 2 + 3). 14. log (x - 3 + Vll - 6x + x 2 ). 5. l/V7-arcsin V7x/3. 17. 2/V3 • arc sin (3x + 2)/Vl9. 6. 2/V5 • log (V5x+ V5x 2 + 2). 18. 6/V7 ■ arc sin (7x + 4 )/V79. 7. arc sin (x— 1). 19. \\og(f x + Ve 4a; _ 4). 9. J log (x + | + Vx 2 + 7x/4). 21. arcsin(logx/V3). 22. 2 log (sin x/2 + Vsin 2 (x/2) + 5 ). EXERCISE LXXII (Page 104) 3. 1 log (4 x 2 + 1) + | arc tan 2 x. 4. llog[(3x-l)/(3x + l)]-llog(9x 2 -_l). 5. 1 log (3 x 2 - 2) - V6/4 • log [(Vtx - V2)/(V3 x + V2)]. 9. 1 log (x 2 + 2 x) + \ log [x/(x + 2)]. 10. - \ log (6 x - x 2 ) - log (x - 6)/x. 13. log (x 2 + 2 x + 5) + | arc tan (x + l)/2. 18. - 1 log (2 - 6x - x 2 ) + 3/2 Vll • log [(x + 3 - Vll)/(x + 3 + Vll)]. 21. -|log(4x 2 -4x-3) + ^log^x-Sy^x + l)]. 23. - X i og (l _ 6x - 9x 2 ) +1/2 V2 • log [(3x + 1 - V2)/(3x + 1+ V2)]. 27. 1 log (3 x 2 + 2 x - 1) + ft log [(3 x - l)/(x + 1)] . 29. - 1 log (3 x 2 - 4 x + 3) + V5/15 • arc tan (3 x - 2)/ VS. 34. -llog(4-5x-3x 2 ) + ^ = log[(6x+5-V73)/(6x+5 + V73)]. 2 V73 206 PROBLEMS IN THE CALCULUS EXERCISE LXXIII (Page 106) 1. 2 V x 2 - 1 + log(x + Vx 2 - l). 2. — Vl — x 2 — arc sin x. 3. | V9x 2 + 1 + jf l og (3x + V0 x 2 + l). 4. ^ (arosin_2 x + Vl - 4 x 2 ) . 7. Vx 2 + 2x + 2 log (x + 1 + Vx 2 + 2x). 8. — V4x — x 2 + 4 a rc sin (x — 2)/2. 11. - V27 + 6x-x 2 + 3 arc sin (x - 3)/6. 14. 3 V l9 - 5 x ^-'x 2 + -V 1 log (x - 5/2 + Vl9 - 5x + x 2 ) . 17. §V4 x 2 -4x + 5- j;log(2x-l+V4x 2 -4x+ 5). 20. - 2 Vl2x-4x 2 -5 + | arc sin(2 x - 3)/2. 21. i- V3x 2 + 6x-2 - V3 log (x + 1 + Vx 2 + 2x- §) . EXERCISE LXXIV (Page 107) 1. J sin 2 2 x. 3. -cos 3 x/3. 6. § sin 2 x/3 - A sin 3 2 x/3. 2. £sin 3 3x. 5. cos s x/3— 3cosx/3. 7. \ sec 2 2 x. 9. | cos 5 x/2 — § cos 8 x/2'. 19. J- cos 2 x — log cos x. 11. 2/7 7r ■ cos% 7r x — 2/3tt • cost 7tx. 21. x/2 — | sin 2 x/3. 13. — §Vcos3x. 23. x/8— ^sinSx. 15. Jj 1 co s s x/5 — 5 cos x/5 — cos 5 x/5 . 25. 3 x/8 — ^ sin x + -j 1 ^ sin 2 x. 17. — 2 Vcos x. 27. -jig-sin 2 !— Jcosx. 29. f^-x-^in^- 2x)- £\sin(4- 4x)+ ^ sin 8 (2 — 2 x) . 33. 3 x/2— cos 2 x— ^ sin 4 x. 37. lsin 4 x. 35. 5 x/8 — § cos 3 x + 3^ sin 4 x. 39. | sin 8 x. EXERCISE LXXV (Page 109) l.,^tan 8 x. 3. J tan 2 2 x. 5. ^ sec 3 3 x — ^ sec 3 x. 2. ftan^x. 4. J tan 2 3x— £logsec3x. 6. 3sec 8 x/3. 7. T > 5 -tan 3 4x — ^-tan4x+x. 15. — fcot 4 x/5 + | cot 2 x/5 — 5 log esc x/5. 9. ftan^x. 21. \ tan 7 x + \ tan 6 x. 11. — £cot 2 3x. 23. tanx — 2cotx— ^cot 2 x. 24. — cotx + 2 tanx + ^tan s x. EXERCISE LXXVI (Page 111) 2. \ arc sec x/2. 4. — V5 — x a /5 x. 7. — V4 — x 2 /x 2 — arc sin x/2. 3. Vx 2 -3/3x. 5. x/ 2Vx 2 + 2. 8. Vx 2 — 16 — 4 arc sec x/4. 10. £log(x/(3+V9-x 2 )). 16. _(2-x 2 ) 3 / 2 (3x 2 + 4)/1 5. 12. log (x + Vx 2 - 36) - Vx 2 - 36/x. 22. T y arc sec x/2 + Vx 2 - 4/8 x 2 ■ 13. (x 2 -8) 8 ' 2 /24x 3 . 23. (x a - 2) VxMJT/3. 15. Vx 2 -' 3 (2 x 2 + 3)/27 x 3 . 24. (x 2 + 2) Vx 2 - 1/3. ANSWERS 207 EXERCISE LXXVII (Page 113) 1.2/ = x 2 + 4. . 19. 2/(1+ 2cosx)-2 = 0. 2. 2x = log2/+ 1. 21. xtany = x— 1. 3. x 2 = log2/ 2 + 4, ory = ei^- i >. 23. y = £ sin (4 Vx + ir/6) . 5. x 2 -y 2 - 4x + 4 = 0. 26. 15 2/ 2 - 2x 2 2/ 2 - 16 2/ + 4 = 0. 7. x 2 + 2/ 2 -4x + 6j/-12 = 0. 2i. v = F/k-{l-e k '/ M ). 8. y = 1/(1 — logx). 29. « = » e- i,/Jf . 9. 16 y = (x 2 + 4) 2 . 30. ?w = V¥/k. 12. 2/(1 + arc sin x) +1 = 0. 31. 18.97 ft./sec, 18.78 ft./sec. 13. x 2 - v 2 + 1 = 0. 36. 2/ = 7T 2 (x 2 + z 2 )/4. 17. 2tan2/ = x 2 -4. 37. T = 50 &&. EXERCISE LXXVIII (Page 115) 1. x + log2/=C. 11. 32/ = (l+x) 3 + C. 2. xj/ + Cx — 1 = 0. 13. 2/ + x/2 + ^ sin 2 x + C = 0. 3. x 2 - 2/ 2 = G. 15. x 3 ' 2 - 2/ 3/2 = O. 9. 2/ = logx — \ x 2 + C. 17. x — y + Cxj/ = 0. EXERCISE LXXIX (Page 115) 1. 1/21. 9. 24. 15. logVI. 25. ir/4. 35. 3/32. 3.-8. 10. 9. 17. logV2. 27. 2tt/3V3. 37.it/12. 5.2. 11. V8(V5-l). 19.^/18. 31. -ir/2. 39.4/3. 7. 1. 13. 7/5. 20. 2tt/3V3. 33. 1/3. 40. 2 7ra 2 . 41. No finite limit. 42. 12. EXERCISE LXXX (Page 117) 1. 10§. 4. 21£. 9. 3. 15. 26§ acres. 2.1i. 5. 1. ' 11. log 16. 16. |12,800. 3. 8. 7. 10|, 13. 6.524. EXERCISE LXXXI (Page 119) 1. x + log (x - 1)_. 5. x 2 /2 + x + log(x - l) 2 . 3. — 2x — logV3 — 2x. 9. x 2 — log Vx 2 + 1 + arc tanx. 11. x 2 /2-x + 2/V3-arctan[(2x + l)/V3]. 12. x + 21og(x 2 - 2x + 1) + 4/(x - 1). 13. logx 2 (l+ x)V(l- *) 5 23; log[(x+l)i/ 2 (x-3)i'V(x 2 +2x-3)^]. 15. log(2x-l) 8 (2x + l)/x. 29. - 3/x + 41og(x + l)/x. 19. log(3x+2) 2 «(2x+3)3/ 2 /x. 30. x - 2/x + logx 2 /(x - 1). 208 PROBLEMS m THE CALCULUS 33. log(x 2 + x) - l/(x + 1). 41. log(x-l ) 2 - (x -2)/(x 2 -x). 35. log[(x-l) 3 / 2 /(x-3)£]-13/(x-3). 47. log(x/Vx 2 + l). 37. log(x+l)-(4x+7)/2(x + l) 2 - ' _ 49. logx + 2 arc tanx. 51. log (2 x 2 + 3)/ x 3 + 1/ V6 ■ arc tan V2 x/ VI. 52. log [(x — 2)/Vx 2 + lj + arc tan x. 62. 2 arc tan x/2 — arc tan x. 64. 1/V3 • arc tan x/ VI + £ log (3 x 2 + 1) . 66. J- log (x 2 + 2) (x 2 + 1) + V2 arc tan x V2 — arc tan x. 72. | + log 4. 74. log 5. 73. x/6 + log V(i + VJ$)/(l - V3). 75. f + 4 log f . EXERCISE LXXXII (Page 122) 1. 2Vx — 21og(l + Vx). 2. — Vx— £log(l — 2Vx). 3. 2arctanVx. 5. log(x + 2 Vx + 5) — arc tan |-(Vx + l). 7. 2 log (Vx - 4) . 16. §<1 + x)3/2 - 2 (1 + x)V2. 9. 31og(Vx7(l-Vx)). 17. log 8/2. 13. 6-18tan-il/3. 18. log[(Vl- x-l)/(Vl- x +l)]. 15. 116 /15. 19. t t/9. 25. §V2x + 3 + Vl5/9 • log [( V6 x - 9 - Vo)/(V6 x - 9 + VI)]. 32. 2/Vl - x + 2 log [( Vl - x - 3)/(Vl - x + 3)]. 33. — 2 (x + 1) — 2 arc tan Vx + 1 ; f — ir/6. . 36. ^log( Vx + l - l)/(Vx + 1 + l) - 1/V3 • arc tan V(x + l)/3. 45. — J (V4 + x 4 /x 2 ) - \ log [(V4 + x 4 /x 2 - l)/(V4 + x 4 /x 2 + l)]. EXERCISE LXXXIII (Page 124) 1. log(l + tana/2). 6. §logJ(tanx/2 + 3)/(tanx/2 - 3)]. 3. \ log tan (x/2)— \ tan 2 (x/2). 9. 1/V2 • arc tan(l/V2 • tan \x). 5. fare tan (\ tan x/2). 13. arc tan (2 tan x/2 + 1). 15. — x/3 + £ arc tan(2 tanx/2 + 1). 17. arc tan (tan 2 x/2). 20. f arc tan (tan x/2) + T 4 B log [(tanx/2 - 3)/(tanx/2 + 3)]. EXERCISE LXXXIV (Page 125) * (4 - x 2 ) 3 ^/12 x 3 . 9. - arc sin [(1 + x)/2 x] . 2. -Vx 2 + 5/5 x. 10. log[(l + 2x+Vl+4x+ 5x 2 )/x], 3. — ^ arc sin 2/3 x. 13. - \ arc sin [(2 - x)/x V2]. 5. -Vl + 2/x. 14. - I arc sin [(3 + x)/x -Vz\. 7. f(l -3/x)8' 2 . 15. - A log [(4 + 3x + Vl6 + 24x + 3x 2 )/x]. 17. - Vl + 2x+ 3x 2 /x + log [(J. + x+Vl+ 2x + 3x 2 )/x]. 19. V27x 2 + 6x - 1/x - 3 arc sinf(l - 3x)/6xl. 21. - f ar c tan ^ V(l + x)/ 3 x. 25. - |Vx/(x + 3). 22. log[(Vx + 1 - Vx)/(Vx + 1 + Vx)]. 26. -} [(x - 2)/x]6' 2 . ANSWERS 209 EXERCISE LXXXV (Page 126) 1.2 + log 3/2. 2.2 log 4/3. 3. 46§. 5. lOfj. 10.3. 11. 1/V2 ■ log (5 + 2 V7)/(7 + 2 Vl3) = - .226. 13. -7tt/6. 16. IV3 + tt/3. 19. y-V5. 22. 18. 14. -4ir/3. _ 17. 2/3. 20. 4w._ 23. 7ra 2 /4. 15. U-/12-V3/8. 18. V3-7T/3. 21. 2V3-TT/3. 24.5/27. EXERCISE LXXXVI (Page 127) 1. 2(sin Vx — VxcosVx). 3. arc sinx +Vl — x 2 . 5. x — log^ + l). 2. sinx— cosx. 4. Jarcsinx 3/2 . 6. arc tan e x . 7. 2/3[(x + 3) 3 ' 2 +Jx + 2)3/2]. 8. - I [x 2 + xV x 2 + 1 + log(x +.Vx 2 + l)]. 9. log[(V^ + l-l)/(V ^ + l+l)]. 11. f( l+logx) 3 ' 2 . 14. \{x + l)V x 2 + 2x + 2 + |log(x + 1 + Vx 2 + 2x + 2). 15. i(x 2 -2)vT+x 2 . 16. log(Vl+x-Vl-x)/( Vl+x + V l-x)+2arctan\ / ( r-x)/(l + x ). 21. (x 3 + 4 x 2 + 24 x + 180) Vx 2 - 4x- 5 + 4501og(x - 2 + Vx 2 - 4 x- 5). 22. (6 -_2_x)V5-4x-x 2 + 25 arc sin ^(x + 2). 23. 2/Vll-arctan[(4tanx-3)/Vll(4 + 3tanx)]. 24. ^(4 x s - 6x 2 - 6x - 1)/(1 + 4x)s/ 2 . 27. r( 7r-2)/2. 30. T ! B [3 a 2 arc sin (sV8/ a i/s) - (3 a^x 1 ' 8 - 6 a 2 /Sx)Va 2 ^ - x 2 ' 8 + 8x(a 2 / 3 -x 2 ' 3 )t]. EXERCISE LXXXVII (Page 129) 1. — ^xcos2x + lsin2x. 5. lx 2 - £xsihx — £cosx. 2. ^xsin3x+ ^cos3x. 7. - ^-x 2 cos2x + J xsin2x + ^cos2x. 3. £xtan2x+ ^logcos2x. 9. ^(3 sinx sin 3x + cosx cos 3 x). 13. x arc tan x/3 — § log (x 2 + 9) . 17. I x 2 arc tan (1- x) + x/2 + log(x 2 - 2x - 2). 19. \ (x 4 - 1 ) arc ta n x + ^(Sx-x 8 ). 21. log (x/ Vl + x 2 ) — arc tan x/x. 25. (x - 1) log (l - Vx) - i (x + 2 Vx). 29. x + cosx[l- log(l + sinx)]. 31. £(3-2x)e 2 * + x . 42. £e-*(- sin3x - 3 cos3x). 33. ^^(x 2 - §x+ §). 45. |[secxtanx + log(secx+ tanx)]. 37. •ji 5 e 2 ^(3sin3x+'2cos3x). 49. e*/(l + x). 50. -e-*(x 2 + 2x + 2). 51. .584. 54. ir/2. 55.11.18. 56. .429. 13. 2Vx 2 + 5x. 17. ir/4. 20. 3 arc sin \ log a;. 22. l (7- 210 PROBLEMS IN THE CALCULUS EXEBCISE LXXXVIII (Pagk 131) 1. ^ (arc sinx) 2 . 5. 1/4 cos 4 x- 1/2 cos 2 x. 9. — 3 x 2 cos x/3 + 18 x sin x/3 + 54 cos x/3. 10. 38/3. 26. f {Vx^+l + h log[t = l/fc.log[a/(a-2/)]; (b) oo ; (c) about 55^%; (d) 106, 151, 256 min. respectively. 1. fclog(Vi)- 3. (a) 160 ft.; (b) 1280 ft.; 256 ft./sec, average. 4. 570. 5. 63 units. 6. 579.75 large calories. 7. 1/8. _ IS. (a) 14 7rr5/2/(l5ft/V2(/); (b) 11.2 min. ; (c) 4.44 min. 13. (a) %naV2g[(h + b)M-hV^;- (b) 608 cu. ft./sec. 17. (a) t = 1/fc (6 - a) log [a (6 - y)/b (a - y)] ; (b) y = o6(e a *'i — e ti! >)/( ae " Wl _ bef> kt t) ; (c) t = y/ak (a—y); (d) 1/aft. 18. (a) t = 4/a • log [2y>*(l + 6T 2 i)/T 2 V4(l + ftT^)] ; (b) 332 sec. 19. (a) v = V2gR 2 s/h(h- s) ; v = Vy + 2 gRH/h {h ^1) ; (b) Vg~R ; (c) Vy + gfl ; (d) v = Vy + 2 gfi (ft - ii)//i. ; !=Vv + 2gB. 20. (a) i = 1/fl ■ V7i/2gr[VA(7i— s) + h arc sin Vs/A].; (b) t = Vy(2 gfi 2 - v *h) ■ [(/i - s) (fc + s) + (ft + fc) arc sinV(/i-s)/(fc + s)], where fc = v Vi 2 /(2gR !i - yft). 21. (a) .55 + sec. ; (b) 9.42 +_ft./sec. ; (c) t = 1/2 c V? • log [(V? + ct)/( Vg - cu)] ; (d) a = y/g/c ■ tanh c Vj/f (tanh = hyperbolic tangent) ; (e) s = 1/2 e 2 - log [g/(g -cV)]; (f)Vff/c. 22. (a) 1/c Vg-arctanCDd/Vfif ; (b) 2.398 sec, or .727 sec. sooner than when there is no air resistance. 24. (a) 104.17 ft.; (b) v = 68.5 ft./sec, t = 2.715 sec. EXERCISE XCV (Page 144) 1. (a) 5J; (b) 13?,; (c) 20£; (d) 18; (e) 7/6; (f) 9; (g) 20J; (h) 15| ; (i) 27/16; (j) 81^; (k) 64/15; (1) 6 ; (m) 2/5; (n) 1/3; (o) 4/15; (p) 4/15; (q) 18(e^ +1 ) /37 . ( r ) 2(e^-l)/37 ; (s) 21J; (t) 4V2-2. 2. (a) 2 ; (b) 8 ; (c) 4/tt ; (d) 6/tt ; (e) tt/2 ; (f) tt/6 ; (g) 4 ; (h) 8. 3. (a) 2; (b) 3; (c) 4 ; (d) 3/2; (e) tt/{4 - tt) ■ (f) 8.4+ ; (g) 5; (h) (32 - 3 W )/3tt; (i) 1.84; (j) 1.71 ; (k) 2/(^-2) ; (1) (ir-21og2)/21og2. 4. (a), (b), (c)4i; (d)lj; (e) 10§ ; (f) 1/3. ANSWERS 213 EXERCISE XCVI (Page 145) 1. (a) 19/27 ; (b) § (5 VB - l) = 6.78 ; (c) 2 tt/3 ; (d) £ (7V7 - l) ; (e) 6 + f log 5 ; (f) ^ + 3 log 3 . (g ) j.081 ; (h) 1 + V8 log [V8 (3 + V8)/(4 + V8)] ; (i) log (2 + V3~). 2. (a) ^(512- 7 V7) ; (b) 4.56 ; (c) (e- e-*) = 2.35. EXERCISE XCVII (Page 146) 1. (a) tt/7; (b) 156^ tt; (c) W 2 ; (d) **/* ; (e) 34^ tt ; (f) (e 2 »»- 1) tt/2 a. 2. (a) 8 tt/3 ; (b) 32 tt/15 ; (c) (64 V2 - 32)tt/3 ; (d) 4 tt. 3. 8. 5290 J lb.; 10,581 J lb. 2. i|£-TT. 9. 7440 7T lb. 3. (a) 8 IF; (b) 12 IF 10. 14,8801b. 4. (a) 4 IF; (b) ->/ IF; (c) If £ W; 11. 5952 tt lb. (d) -syyi TT. 12. (a) 4036 lb. ; (b) 8711 lb. EXERCISE CIX (Page 164) 1. %kab(2I+ ftsintf). 4. 3607rfc. 2. 78851b. 5. 20581b.; 19201b.; 6961b. 3. (a) *7rte 2 (2c± Icosff); 6. Trkv 2 ah/2g. (b) 372 7rlb. and 1612 7rlb. 7. 19,1001b. 216 PROBLEMS IN THE CALCULUS 8. (a) JWOTfcft.-lb. ; 12. it 4 *; 2 (8 a + 3 w*)/192. 17. J.F>aS. (b) 156,250 jr/Sft.-lb. 14. (a) 32 (it -2); 18. irF^a* - 6*)/2 a. 9. 108,000 7rft.-lb. (b) 64. 19.41:15. 10. 248.5 ft.-tons. 15. 4otj- 2 . 20. kM m/a (a + t) . 11. 37ro 3 fc/16. 16. -y -Tra 2 . 21. 2kmM/ira 2 . 22. 2fcJfm/a 2 • (l - 7i/Vft 2 + a 2 ). 25. (a) Light on path is about 24% as 23. 349.5 tons. much as that on the circle ; 24. 2 irfcl(l - A/V/i* + a 2 ). (b) about 22%. 26. 10 ir kilograms. 28. 20 kg. 30. 17.5 kg. 27. 26.4+ kilograms. 29.16.07kg. 31. 1250irkg., or 3927kg. 32. 3590 kg. 34. %inptr i a i . 33. F = 22.05kg., 35. (a) 7369kg.; (b) 7106kg. P = 3919kg., 36. 2fipa x r s (a— sinaeosa)/sin 8 or. F/P = .0056. 37. irpa 2 kg. 38. | Tr 2 npa 2 Va 2 + h 2 or \ ir 2 npa s /cos a, where a is the half angle at the vertex. 39. 12,800 kg.-cm.; 6032 kg. 40. (a) 24.1; (b) nearly 8 kg. per sq. cm.; (c) 8.63 kg. per sq. cm. 41. (a) 2 iflykr 2 (1 - cos 2 a) ; (b) p a = 2.48, W = 346.7 kg.-cm. 42. (a) Front, 220,560.81b.; back, 179,439.21b. (b) S = 5233.64 lb. per foot of width of wall ; Si = 12,710.28 lb. ; S 2 = 20,186.92 lb. ; S s = 23,925.24. 2 (c) r = 373.832 lb./ft. EXERCISE CX (Page 170) 1- (I, -i). 3. (2, 0). 5. (J, ^). 7. (1, f). 9. (J, T V). 2. (6|, - 4). 4. (2fc 0). 6. (1, 1). 8. (2, J). 10. (2, 0). 11. (- 2^, f). 17. (if, - \i). 23. (a) (3.42, - 3.26); 12. (f, -J). 18. (8V2/15, 8V2/15). (b) (6.904, .594). "■ (A. 2 A)- 19 - (H- 2 §§)- »*■ (- 695 ' 2 - 478 >- 14. (1£, f). 20.(11,-1). 25.(2.20,-71). 15. (3|, 3^). 22. (a) (3.08, 2.95); 26. (a) (.27, 2.18); 16.(atf,lH). (b) (1.37, 1.30). (b) (-1.16, .107). EXERCISE CXI (Page 172) 11. (a) 80 ; (b) 116 ; (c) 40. 12. (a) 53J ; (b) 81f . 13. 40ft./sec. 14. (a) 2 aa/Tr ;.(b) 16 ft./sec. 15. 170§ft. 8 /sec. 1. 33£. 6. 2/ir. 2. 2|. 7. 1/2. 3. a 2 /2 7r, 2a 2 /37r. 8. 3/8. 4. (a) 1.33 ; (b) 1.64. 9. 3/8. 5. 2. 10. 1/2. ANSWERS 217 16. 675.2 ft. 3 /sec. ; 607,680 cu.ft. 22. 1£ ft. from the bottom. 17. 3 a/8. 23. 8 ft. 2 in. below the surface. 18. (a) 8 a/15 ; (b) 5 a/8. 24. 4.33 ft. from the base. 19. 3 in. from the base. 25. Three fourths of the way down. 20. \\ ft. from the base. 26. 2$ ft. . 21. i) "= 4£ cu. ft., a = If ft. 27. 11,9041b. ; 6f ft. below the surface. 28. (a) 3.76 ft. below the surface ; (b) 1.18 ft. below the surface. 29. About 7 ft. 4 in. below the surface. 30. (| a, § a, § a). 31. Example 4, (f , T %, £§); example 6, (,&, ^, JJ); example 7, (£, f , §). 32. £ = .42,^ = 1.06; x = [3/V2-lQg(3+_2V2)] [a/V2 + log(l + V2)] ; y = 4 (2V2 - 1) a/3 [ V2 + log(l + V2)]. EXERCISE CXII (Page 175) 1. Z x = 6J,I 1 ,=l|,I = 7it. 2. I x = 13/60, /„ = 1£|, Z = 1 T V 5 . 3. I x = 4096/105, l v = 128/15. 7. (a) 2^ ; (b) (f, 1|) ; (c) Z„ = 128/63, I x = 512/55. 8. (a) 4£; (b) (|, - i) ; I, = 3 A, A = 5 T ¥ + c 2 te-* + c s t 2 e-' + c 4 . 30. c x + Cje^' + c a e~^' + c 4 cos V2 J + c 5 sin V2«. EXERCISE CXVI (Page 180) 1. e S( (c 1 eos2t + c 2 sin2«) + 3. 3. c^' + e 2 e- 2! - ft + £. 2. e 2< (c 1 cosV34 + c 2 sinV3«) + 2 - 4 - e x cos3i+ c 2 sin3£ + ££ H--^. 5. Cj cos 3 1 + c L sin 3 1 + i 2 /9 - 2/81 . 6. e- ( (c 1 cosV3t + c 2 sin V3t) + t 2 /i+ t+1. 7. e- se (c 1 cosV3i + c 2 sinV3«)+ £« 2 + $t+ $. 8. c x e« + cjefi* + t* + t + l. 9. e- 2 '(c 1 cos3* + c 2 sin3*) + i 2 + « + l. 10. e> (c x cost + c 2 sini) + S 2 + 3. 15. e ( (c 1 cosV7«+ c 2 sinV7*)+ 4i 2 -22t + 17. 16. c x cos2£ + c 2 sin2t + t s + 2« 2 + 3t + 4. 17. c x e 2( + c 2 e- ( + «e 2 '. 28. c x cos 3 1 + c 2 sin 3 1 + cos 2 1. 18. c x e« + c 2 e- ( — J te-«. 29. c x cos 3 1 + c 2 sin 3 1 + £ t sin 3 J. 19. c^e 3 ' + c 2 e- ! - 1 e». 30. e^e 3 ' + c 2 e- 3 <- J cos3i. 23. c x cos3i + c 2 sin3t + ^e 3 '. 31, c x cos2t + c 2 sin2i — 2 sin 3 1. 27. c a e« + c a fe' + 8iV. 32. c 1 cos2£+ c 2 sin2« — 2tcos2«. 33. c x cos t/2 + c 2 sin t/2 — £ cos ) Q =^(1-00.^-^1). 18. (a) Q = e 2i '(Ci cos M + c 2 sin fc i*) + *^' — B ^ 2 , where fl 2 J.O, 4Z 2 ' 19. (a) 2/ = ^i:ir 2 -JfcK 3 ; (b) y = T Vfc(6Px 2 - 4fa; 3 + x 4 ) ; (c) j/ = ^ fc(3Px* -2x i ); (d) 2/ = £J;(3 ax 2 + to 8 ). INDEX (The numbers refer to the pages) Acceleration, 25, 178, 185 Air resistance, 113, 114, 142 Angle of intersection, 5, 22 of polar curves, 25 Approximate integration, 152 by Simpson's rule, 153 by the trapezoidal rule, 153 by use of series, 152 Arc, differential of, 45 length of, polar, 148 length of, rectangular, 145 Area, by double integration, 157,158 moment of, 171 polar coordinates, 148 rectangular coordinates, 117, 144 of any surface, 166 of surface of revolution, 148 Asymptotes, 89 to polar curves, 91 Attraction, 167, 170 Auxiliary equation, 179 Average value of a function, 172 Band, elastic, 114 Beam, curve of mean fiber of, 187 deflection of, 56 strength of, 53 Bearings, friction in, 167, 168, 169 load on, 168, 169 Belt, power transmitted by, 42 pull in, 115 Bimolecular reaction, law of, 141 Binomial theorem, 79 Bunsen photometer, 73 Cauchy's ratio test, 76 Center of area, 171 Center of gravity, 171 of solids of revolution, 174 by triple integration, 175 of various solids, 174 Center of pressure, 174, 175 Change of variables, 48 Circle of curvature, 62 Columns, stress in, 170 Comparison, examples for, 133 test in series, 76 Complementary function, 180 Condenser, 186, 187 Conjugate points, 92 Constant of integration, 113 Convergence, Cauchy's test for, 76 comparison test for, 76 interval of, 78 of number series, 76 Critical values, 30 Curvature, 61 center of, 62 circle of, 62 maximum, 62 radius of, 61 Curve of mean fiber, 187 Curves, angles of intersection of, 22,. 24 asymptotes to, 89, 91 direction of, 20 envelope of a system of, 88 length of, 145 maximum and minimum points of, 30 singular points on, 91 in space, 93 Cusp, 92 Cylindrical coordinates, 161 Damped harmonic motion, 186 Definite integrals, 113 applied problems in, 139 change in limits of, 126 evaluation of, 115 evaluation of, by tables, 138 Dependent variable, change of, 49 Derivatives, partial, 65 successive partial, 67 total, 68 221 222 PROBLEMS IN THE CALCULUS Differential, of arc, 45 total, 68 Differential equations, 178 applications of, 185 of the first order, 178 homogeneous, of the first order, 184 homogeneous linear, 178 linear, of the second order, 180 miscellaneous, 181 Differentials, 47 application of, to small errors, 69 exact, 74 Differentiation, of algebraic forms, 7 of exponential functions, 11 of a function of a function, 17 by the general method, 4 of implicit functions, 17, 66 of inverse trigonometric func- tions, 13 of logarithms, 8 partial, 65 of powers, 5 of products and quotients, 6 review of , 18 successive, 28 of trigonometric functions, 13 Double integration, 156 Double points, 91 Efficiency, of dynamo, 42 of inclined plane, 41 of screw, 41 Elastic band, 114 Element, of area, 144 of area of surface, 148 of length of arc, 145 of volume, 146 Envelopes, 88 Equation of motion, 185 Error, relative, 70 percentage, 70 Errors, determination of, 69 Evolutes, 62 parametric equations of, 63 Exact differentials, 74 Expansion of functions, 79 Falling bodies, 26, 113, 139, 173, 185 average velocity of, 173 with air resistance, 142 Floodgate, flow through, 140 pressure against, 163 Fluid pressure, formula for, 164 by integration, 162 Friction, flywheel stopped by, 186 for horizontal shaft, 167 for step bearing, 168, 169 Function, average value of, 172 complementary, 180 evaluation of, 1 implicit, 17, 66 maximum and minimum values of, 30 transcendental, 31 Galvanometer, ballistic, 187 error in, 73 ' tangent, 43 Gas engine, work of, 42 Geometric integrals, review of, 154 Gyration, radius of, 173, 176 Harmonic motion, 186 Hooke's law, 36, 114 Illumination, 40, 167 Independent variable, change of, 49 Indeterminate forms, 2, 57 evaluation by series, 84 Inflection, points of, 31 Inscribed solids, 34, 38, 40 Integration, 96 by aid of tables, 136 approximate, 152 constant of, 113 of constant over quadratic, 102 of constant over the square root of a quadratic, 103 of exponential functions, 99 of linear over quadratic, 104 of linear over the square root of a quadratic, 106 of the logarithm form, 98 multiple, 156 by parts, 129 of the power form, 96 of rational fractions, 119 by rationalization, 122 by reciprocal substitution, 125 by reduction formulas, 135 INDEX 223 Integration, review of, 131 a summation, applied prob- lems, 164 of trigonometric functions, 100, 107, 109, 124 by trigonometric substitution, 111 by use of series, 133 by various devices, 127 Intensity, of alternating current, 43 of heat, 44 of light, 73 Intersection, angle of, 5 of curve and surface, 95 of two curves, 22 Interval of convergence, 78 Inversion of cane sugar, 141 Joule's law, 139 Journals, 167 Least squares, 36 Length of curve, 145 Linear differential equation, 178 Load on a bearing, 168, 169 Logarithm form, integration of, 98 Logarithmic differentiation, 12 Logarithms, calculation of, 82 Lorenz's law of cooling, 141 Maclaurin's theorem, 80 Maxima and minima, 30 general problems, 37 of functions of two variables, 86 problems leading to algebraic forms, 32 Moment of area, 171 Moment of inertia, 175 polar, 176 Motion, equation of, 185 Multiple integration, 156 Multiple roots, 27 Normal lines, 22, 23 to a surface, 94 Normal plane, 93 Ohm's law, 44, 71 Orthogonal trajectory, 115 Parametric equations, 23, 29, 63, 93 Partial derivatives, 65 application to rates, 69 Partial differentiation, 65 Particular solution, 180 Parts, integration by, 129 Pendulum, 72, 186 Photometer, Bunsen, 73 Points, double, 92 extreme, 25 of inflection, 31 maximum and minimum, 30 singular, 91 Polar curves, 24 angles of intersection of, 25 area bounded by, 148, 158 asymptotes to, 91 extreme points on, 25 length of arc of, 148 Polar moment of inertia, 176 Power series, convergence of, 78 derived by binomial theorem, 79 derived by Maclaurin's theo- rem, 80 derived by Taylor's theorem, 79 Pressureacenter of, 174, 175 fluid7162, 164 wind, 165 Projectile, 41, 46 Radius, of curvature, 61 of gyration, 173, 176 Rates, percentage, 56 problems in, 50, 69 relative, 55 Rational fractions, 119 Rationalization, integration by, 122 Rectilinear motion, 25 Reduction formulas, 135 Relative errors, 70 Relative rates, 55 Revolution, solids of, 146, 148 Series, alternating, 76 calculations by, 81 convergence of, 76 indeterminate forms by, 84 in integration, 133, 152 Shaft, friction of, 167 shearing force, 166 224 PROBLEMS IN THE CALCULUS Shear in pulley shaft, 166 Simpson's rule, 153 Singular points, 91 Sohncke, 87 Solids of revolution, 146 area of surface of, 148 center of gravity of, 174 Space geometry, 93 Spherical coordinates, 161 Spillway, average flow over, 173 flow over, 140, 188 Step bearings, 169 Stress in columns, 170 Substitutions, for rationalizing, 123 for rational trigonometric functions, 124 reciprocal, 125 trigonometric, 111 Subtangent and subnormal, 23 Successive differentiation, 28 of implicit functions, 28 of parametric forms, 29 Successive partial derivatives, 67 Summation, integration as a, 144, 164 Surface, area of any curved, 166 of revolution, 148 of rotating liquid, 114 Tables, integration by, 136 evaluation of definite integrals by, 138 Tangent lines, 22, 23 to space curve, 93 Tangent planes, 94 Taylor's theorem, 79 for several variables, 85 Test, Cauchy's, 76 comparison, 76 Torricelli, theorem of, 140 Total derivative, 68 application to rates, 69 Total differential, 68 application to small errors, 69 Trajectories, orthogonal, 115 Trapezoidal rule, 153 Trigonometric functions, calcula- tion of, 82 differentiation of, 13 integration of, 100, 107, 109 Trigonometric substitution, 111 Triple integration, 156 volumes by, 160 Undetermined coefficients, in inte- gration, 105, 106, 128 in partial fractions, 120 in a particular solution, 180 in series, 81 Van der Waals' equation 56, 139 Velocity, 25 average, 173 of chemical reactions, 42 of piston, 53 of wave, 42 of wheel, 53, 139 Vibration, damped harmonic, 186 harmonic, 186 Volume, of miscellaneous solids, 150 of solid of revolution, 146 by triple integration, 160 "Waste in an electric conductor, 42 Work, by an electric current, 140 of friction in bearings, 168, 169 of a gas engine, 42 in raising fluids, 165 Worn bearings, 169 ANNOUNCEMENTS RECENT BOOKS IN HIGHER MATHEMATICS HIGHER ALGEBRA By Herbert E. 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Zobel. $1.60 A text in Fourier's Series and Heat Conduction, presenting along with the theory a large number of practical applications of special value to geologists and engineers. These include problems in the tempering of steels, freezing of con- crete, electric and thermit welding, and similar questions. The book presents an excellent first course in mathematical physics. GINN AND COMPANY Publishers ADVANCED CALCULUS By Edwin Bidweix Wilson, Professor of Mathematics in the Massachusetts Institute of Technology 8vo, cloth, ix + 566 pages, $5.00 CHAPTERS I-X. 8vo, cloth, ix + 280 pages, $2.75 TJROFESSOR WILSON'S " Advanced Calculus '' is an up- -*- to-date, practical textbook and laboratory manual of higher calculus, providing in a single volume a second course in calculus broad enough to meet the varying needs and "attainments of students and the divergent viewpoints of teachers. The material is arranged to give the maximum flexibility and adaptability. Throughout the work attention has been paid to the needs of students of engineering and mathematical physics ; and the exercises, of which a large number have been provided, aim to cultivate in the student a large reliance on himself. Two chapters in review supply the connection with elemen- tary texts. Then follow four chapters on differential calculus (Taylor's Formula, partial differentiation of explicit and of im- plicit functions, complex numbers and vectors), four chapters on the integration of differential equations, five chapters on integral calculus, and five on the theory of functions. The first ten chapters are published also in separate form. GINN AND COMPANY Publishers